[
  {
    "id": 27501,
    "subject": "Mathematics (Multi-modal)",
    "question": "A positive integer $n$ and a function $f : \\mathbb{N} \\to \\mathbb{N}$ are such that:\n(1) $f(1) \\le f(2) \\le \\dots \\le f(n) \\le f(1) + n$;\n(2) $f(n + i) = f(i)$ for any positive integer $i$;\n(3) $f(f(i)) \\le n + i - 1$ for any positive integer $i$.\nProve that $f(1) + f(2) + \\dots + f(n) \\le n^2$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 27502,
    "subject": "Mathematics (Multi-modal)",
    "question": "In a triangle $\\triangle ABC$ points $L$, $P$ and $Q$ lie on the segments $AB$, $AC$ and $BC$, respectively, and are such that $PCQL$ is a parallelogram. The circle with center the midpoint $M$ of the segment $AB$ and radius $CM$ and the circle of diameter $CL$ intersect for the second time at point $T$. Prove that the lines $AQ$, $BP$ and $LT$ intersect in a point.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Since $AC \\parallel LQ$ and $BC \\parallel LP$, we have $S_{ALQ} = S_{CLQ} = S_{PLC} = S_{PLB}$. Let the point $K$ be such that $AKBC$ is a parallelogram. By analogy we have $S_{AKQ} = S_{AKC} = S_{CKB} = S_{PKB}$.\n\nThe locus of the points $X$ such that $\\triangle AXQ$ and $\\triangle PXB$ are oriented in one and the same direction and have equal areas is a straight line $\\ell$ passing through the intersecting point of $AQ$ and $BP$. Therefore $\\ell \\equiv KL$ and $AQ$, $BP$ and $KL$ intersect in a point.\n\nLet $N$ be the midpoint of $CL$. The line $KL$ is the image of $MN$ under homothety of center $C$ and coefficient $2$. Since the point $T$ is symmetric to $C$ with respect to $MN$ we have that $T$ lies on $KL$ which completes the proof.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 27503,
    "subject": "Mathematics (Multi-modal)",
    "question": "In a mathematical olympiad students received marks for any of the four areas: algebra, geometry, number theory and combinatorics. Any two of the students have distinct marks for all four areas. A group of students is called *nice* if all students in the group can be ordered in increasing order simultaneously of at least two of the four areas. Find the least positive integer $N$, such that among any $N$ students there exist a nice group of ten students.",
    "options": [],
    "answer": "730",
    "solution": "Answer: 730.\n\n**Lemma.** A sequence $A = a_1, a_2, \\dots, a_k$ consisting of distinct numbers does not possess 10 term increasing subsequence if and only if its terms can be colored in 9 colors such that the members of one and the same color from a decreasing sequence.\n\n*Proof.* First, assume that such a coloring exists. Any 10-term subsequence of $A$ contains two terms colored in the same way. Hence, the sequence is not increasing.\n\nSuppose now, $A$ does not contain 10-term increasing subsequence. Color in color $i$ any term $a$ of $A$ such that the longest increasing subsequence of $A$, having $a$ as its last element, has length $i$. It is easy to see that this coloring has the desired properties. This completes the proof of the Lemma.\n\nWe prove that among any 730 students there exist 10 that form a nice sequence.\n\nLet $M_1, M_2, \\dots, M_{730}$ be a sequence in which the students are ordered in increasing order according to their algebra marks. Let $a_i$ be the geometry mark of $M_i$. If the sequence $a_1, a_2, \\dots, a_{730}$ has 10-term subsequence then we have a nice sequence.\n\nIn the opposite case according to the Lemma we color all students in 9 colors such that the geometry marks for any color form a decreasing subsequence. There exist at least 82 students having the same color. Let $N_1, N_2, \\dots, N_{82}$ be of the same color and they are ordered in increasing order for algebra marks and decreasing order for geometry marks.\n\nLet $b_i$ be number theory mark for $N_i$. If the sequence $b_1, b_2, \\dots, b_{82}$ has 10-term decreasing subsequence, then we have the desired nice sequence. In the opposite case we can color students $N_1, N_2, \\dots, N_{82}$ in 9 colors such that number theory marks for any color to form increasing sequence. We have at least 10 students of the same color and they from nice subsequence with their marks in algebra and number theory.\n\nIt remains to show an example of 729 students without 10-term nice subsequence.\n\nLet $k$ be an integer between 0 and 728. For $0 \\le i < j \\le 2$ denote by $f_{ij}(k)$ the integer from representation of $k$ in base 9 when the digits in $i$-th and $j$-th place are replaced by their compliments to 8. (If $k \\le 80$ then we add 0's to the left.)\n\nConsider 729 students with algebra marks 0, 1, ..., 728 and the student of mark $k$ has geometry mark $f_{01}(k)$, number theory mark $-f_{02}(k)$, and combinationics mark $-f_{12}(k)$.\n\nIt is clear that for any two areas there exist two numbers $0 \\le i < j \\le 2$, such that for any student one of the marks for these two areas is obtained from the other through the function $f_{ij}$.\n\nWe prove that a 10-term nice sequence for algebra and geometry does not exist. For remaining pairs the proof is the same.\n\nLet $M_1, M_2, \\dots, M_{729}$ be the sequence in increasing order according to algebra marks. Let $a_i$ be the geometry mark of $M_i$. It suffices to show that a 10-term increasing subsequence of $a_1, a_2, \\dots, a_{729}$ does not exist.\n\nFor any $i$ color $a_i$ in color $s$ where $s$ is the second digit of the representation of base 9 of $a_i$. It is easy to see that any monochromatic subsequence of $a_1, a_2, \\dots, a_{729}$ is decreasing. Hence, according to the Lemma, the proof is complete.",
    "topic": "Number Theory",
    "subtopic": "Other"
  },
  {
    "id": 27504,
    "subject": "Mathematics (Multi-modal)",
    "question": "Consider a table $19 \\times 2015$. *Block* is the figure consisting of a square $10 \\times 10$ and a single cell pasted to the right of the most upper-right cell of the square. Rotation of a block is not allowed. Find the number of ways in which maximum number of blocks can be positioned on the table. (The blocks do not overlap and the cells of each block coincide with the cells of the table.)",
    "options": [],
    "answer": "C(250, 199)",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 27505,
    "subject": "Mathematics (Multi-modal)",
    "question": "The diagonals $AC$ and $BD$ of the convex quadrilateral $ABCD$ intersect at point $O$. The points $A_1$, $B_1$, $C_1$ and $D_1$ from the segments $AO$, $BO$, $CO$ and $DO$, respectively, are such that $AA_1 = CC_1$ and $BB_1 = DD_1$. Let $M$ be the second intersection point of the circumcircles of $\\angle AOB$ and $\\angle COD$; $N$ be the second intersection point of the circumcircles of $\\angle AOD$ and $\\angle BOC$; $P$ be the second intersection point of the circumcircles of $\\angle A_1OB_1$ and $\\angle C_1OD_1$, and $Q$ be the second intersection point of the circumcircles of $\\angle A_1OD_1$ and $\\angle B_1OC_1$. Prove that the points $M$, $N$, $P$ and $Q$ are concyclic.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "It follows from the condition of the problem that $\\angle MAC = \\angle MBD$ and $\\angle MCA = \\angle MDB$. Therefore $\\angle MAC \\sim \\angle MBD$. Let $X$ and $Y$ be the midpoints of $AC$ and $BD$, respectively. It follows that $\\angle MXC = \\angle MYD$, which implies that the point $M$ lies on the circumcircle of $\\angle OXY$. Analogously, $N$ belongs to the same circle.\n\nSince $X$ and $Y$ are also the midpoints of $A_1C_1$ and $B_1D_1$, respectively, the above argument for the quadrilateral $A_1B_1C_1D_1$ yields that $P$ and $Q$ belong to the circumcircle of $\\angle OXY$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 27506,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $m > 2017$ be positive integer and $N = m^{2017} + 1$. The numbers $N, N-m, N-2m, \\dots, m+1, 1$ are written (in that order) on the blackboard. On every move the leftmost number is deleted together with all its divisors (if any). Find the last deleted number.",
    "options": [],
    "answer": "(m^{2017} + m^2 + m + 1) / (m + 1)",
    "solution": "Let $a$ be the smallest number on the board such that $(m+1)a > N$. It is easy to see that $a = \\frac{m^{2017}+m^2+m+1}{m+1}$. We will prove that $a$ is the last deleted number.\n\nThe numbers $2a, 3a, \\dots, ma$ are not on the board since they are not congruent $1$ modulo $m$. Therefore $a$ can not be deleted as divisor (i.e. before we reach it).\n\nLet $b < a$ is one of the numbers on the board in the beginning. It is enough to prove that there exists a number $c > a$, which is multiple to $b$ and $c$ is on the board. Let $b_0 = b$, $b_{k+1} = (m+1)b_k$ for $k \\ge 0$. Since $b_k \\equiv b_0 \\equiv 1 \\pmod m$, all numbers $b_k \\le N$ are on the board in the beginning. Let $i$ be such that $b_{i-1} < a \\le b_i$. If $a < b_i$, then $b \\mid b_i = (m+1)b_{i-1} \\le N$. If $a = b_i$, then\n$$\na + mb_{i-1} = \\frac{(2m+1)a}{m+1}\n$$\nis divisible by $b$ and is on the board in the beginning since it is less than\n$$\n2a = \\frac{2(m^{2017} + m^2 + m + 1)}{m + 1} < N.\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 27507,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $M$ be a set of $2017$ positive integers. For every non-empty $A \\subset M$ we define\n$$\nf(A) = \\{x \\in M : x \\text{ is divisible by odd number of elements of } A\\}.\n$$\nFind the minimum number of colors such that it is possible to paint all nonempty subset of $M$ in such a way that whenever $A \\neq f(A)$, the sets $A$ and $f(A)$ are in different colors.",
    "options": [],
    "answer": "2",
    "solution": "We first prove that the function $f$ is injective, i.e. $A \\neq B$ implies $f(A) \\neq f(B)$. Let $a$ be the smallest number which belongs to exactly one of the sets $A$ and $B$. We can assume, $a \\in A$, $a \\notin B$. Let $C = \\{b_1, b_2, \\dots, b_m\\}$ be the set (possibly empty) of the numbers from $B$ which divide $a$. Then the definition of $a$ yields that the numbers from $A$ which divide $a$ are exactly $a$ and the numbers from $C$. This means that $a$ belongs to exactly one of the sets $f(A)$ and $f(B)$, i.e. $f(A) \\neq f(B)$.\n\nConsider now the directed graph $G$ with vertices the nonempty subsets of $M$ and edges $(A, f(A))$ (direction from $A$ to $f(A)$) iff $A \\neq f(A)$. It follows from the above that every vertex of $G$ is either isolated or it is a tail and a head of exactly one edge. Therefore $G$ can be partitioned into cycles.\n\nWe will prove that all cycles in $G$ have even length, whence it follows that two colors are enough. Let $(A_1, A_2, \\dots, A_m, A_1)$ be a cycle of length $m \\ge 2$. Let $\\{a_1, a_2, \\dots, a_k\\} = \\bigcup_{i=1}^m A_i$, where $a_1 < a_2 < \\dots < a_k$. It is clear that $a_1 \\in A_i$ for every $i = 1, 2, \\dots, m$. Let $t$ be the smallest number such that there exists $i$ such that $a_t \\notin A_i$. There exists index $j$ such that $a_t \\notin A_j$, but $a_t \\in A_{j+1}$. This implies that $a_t$ has odd number of divisors among $a_1, \\dots, a_{t-1}$. Then $a_t \\in A_{j-1}$ and so on, i.e. the $a_t$ alternate, whence we conclude that the cycle's length is even.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 27508,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all primes $p$ and all positive integers $a$ and $m$ such that $a \\le 5p^2$ and $(p-1)! + a = p^m$.",
    "options": [],
    "answer": "(p, a, m) = (2, 1, 1), (2, 3, 2), (2, 7, 3), (3, 1, 1), (3, 7, 2), (3, 25, 3), (5, 1, 2), (2, 15, 4), (5, 101, 3)",
    "solution": "Answer: $(p, a, m) = (2, 1, 1), (2, 3, 2), (2, 7, 3), (3, 1, 1), (3, 7, 2), (3, 25, 3), (5, 1, 2), (2, 15, 4), (5, 101, 3)$.\nDirect checks for $p = 2, 3, 5$ give the above solutions.\nFor $p \\ge 7$ the Wilson theorem implies $a \\equiv 1 \\pmod p$. Moreover, obviously $p-1|a-1$. Therefore $a = kp(p-1)+1$ for some integer $k \\ge 0$. If $k \\ge 6$ then $a \\ge 6p^2 - 6p + 1 > 5p^2$, a contradiction. So we have $0 \\le k \\le 5$.\nAfter division by $p-1$ we obtain\n$$\n(p-2)! + kp = p^{m-1} + p^{m-2} + \\dots + 1.\n$$\nSince $p \\ge 7$, the $2$ and $\\frac{p-1}{2}$ are different and appear in $(p-2)!$, i.e. $p-1|(p-2)!$. This gives $k \\equiv m \\pmod{p-1}$.\nIf $m \\ge p$, then\n$$\n(p-2)! + kp = p^{m-1} + p^{m-2} + \\dots + 1 > p^{p-1} > (p-1)!,\n$$\nwhich easily gives a contradiction. Thus $m \\le p-1$.\nIf $k=0$, then $m = p-1$ and\n$$\n(p-2)! = p^m - 1 \\ge p^{p-1} - 1 > (p-1)^{p-1} > (p-1)!,\n$$\na contradiction. If $k=1$ or $2$, then $m=1$ or $2$, respectively, which is impossible. If $k=3$, then $m=3$, i.e. $(p-2)! = (p-1)^2$, which is impossible.\nFor $k=4$ we get $m=4$ and $(p-2)! = (p-1)(p^2+2p-1)$, whence\n$$\np-2|p^2+2p-1 = p^2-4+2(p-2)+7,\n$$\ni.e. $p=7$, which does not give solutions. For $k=5$ we obtain $m=5$ and $(p-2)! = (p-1)(p^3+2p^2+3p-1)$, whence\n$$\np-2|p^3+2p^2+3p-1 = p^3-8+2(p^2-4)+3(p-2)+21,\n$$\ni.e. $p=7$, which does not give solutions.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 27509,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $n$ be positive integer and $f(x)$ be a polynomial of degree $n$ with $n$ distinct real positive roots. Are there positive integer $k \\ge 2$ and a real polynomial $g(x)$ such that\n$$\nx(x+1)(x+2)(x+4)f(x) + 1 = (g(x))^k?\n$$",
    "options": [],
    "answer": "No",
    "solution": "Let $\\alpha_1 < \\alpha_2 < \\dots < \\alpha_n$ be the roots of $f(x)$. Assume that\n$$\nx(x+1)(x+2)(x+4)f(x) + a = g^k(x).\n$$\nNote that $a = b^k = g^k(0)$.\nIf $k \\ge 3$ is odd then the polynomial $g^k(x)-b^k$ has $n+4$ distinct real roots which will be also roots of $g(x)-b$. However, the degree of $g(x)-b$ is $(n+4)/k < n+4$, i.e. $g(x) = b$, which is impossible.\nNow it is enough to prove that $k=2$ is also impossible. We have $a = b^2$, where we can assume that $b > 0$. Then\n$$\nx(x+1)(x+2)(x+4)f(x) = g_1(x)g_2(x),\n$$\nwhere $g_1(x) = g(x)+b$ and $g_2(x) = g(x)-b$. The roots of $g_1(x)$ and $g_2(x)$ are the numbers $-4, -2, -1, 0, \\alpha_1, \\dots, \\alpha_n$. Since $g_1(x) > g_2(x)$ for every $x$, the number $-4$ is a root of $g_1(x)$. Since the derivatives of $g_1(x)$ and $g_2(x)$ coincide, the Rolle's theorem shows that $-2$ and $-1$ are roots of $g_2(x)$ while $0$ is a root of $g_1(x)$.\nLet $g_1(x) = x(x+4) \\prod_{j=1}^{s}(x - \\alpha_j)$. Then\n$$\n|g_1(-1)| = 3 \\prod_{j=1}^{s}(1 + \\alpha_j) < 4 \\prod_{j=1}^{s}(2 + \\alpha_j) = |g_1(-2)|,\n$$\nwhich contradicts to $g_1(-1) = g_1(-2) = g(-1) + b = 2b$.",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 27510,
    "subject": "Mathematics (Multi-modal)",
    "question": "For a set $M$ of positive integers with $n$ elements, where $n$ is odd, a nonempty subset $T$ of $M$ is called good, if the product of the elements of $T$ is divisible by the sum of the elements of $M$, but not divisible by its square. If $M$ is good, find the maximum possible number of the good subsets of $M$?",
    "options": [],
    "answer": "2^{n-1}",
    "solution": "If $A \\cup B = M$ and $A \\cap B = \\emptyset$, then at most one of the sets $A$ and $B$ is good (otherwise $M$ is not good). Therefore the number of the good sets does not exceed half of the number of all subsets, i.e. $2^{n-1}$.\n\nWe will prove that the above estimate is sharp. Let $n = 2k + 1$ and $p$ be a prime such that $p > \\frac{n(n-1)}{2}$. Let $p^k = a_1 + a_2 + \\cdots + a_{n-1} + a_n$, where $a_i = i$ for $i = 1, 2, \\ldots, n-1$ and $a_n = p^k - \\frac{n(n-1)}{2}$. Then $(a_i, p) = 1$ for every $i$. Consider the set\n$$\nM = \\{p a_i \\mid i = 1, 2, \\ldots, n\\}.\n$$\nThen the sum of the elements of $M$ is equal to $p^{k+1}$. It is easy to see that the good subsets are exactly those with at most $k + 1$ elements. They are exactly $2^{n-1}$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 27511,
    "subject": "Mathematics (Multi-modal)",
    "question": "The quadrilateral $ABCD$ is inscribed. The point $H_1$ is the orthocenter of $\\triangle ABC$ and the points $A_1$ and $B_1$ are symmetric to the points $A$ and $B$ with respect to the lines $BH_1$ and $AH_1$, respectively. The point $O_1$ is a center of the circumscribed circle of $\\triangle A_1B_1H_1$. The point $H_2$ is the orthocenter of $\\triangle ABD$ and the points $A_2$ and $B_2$ are symmetric to the points $A$ and $B$ with respect to the lines $BH_2$ and $AH_2$, respectively. The point $O_2$ is a center of the circumscribed circle of $\\triangle A_2B_2H_2$. Denote the line $O_1O_2$ by $l_{AB}$. The lines $l_{BC}$, $l_{CD}$, and $l_{DA}$ are defined analogously. Let $l_{AB} \\cap l_{BC} = M$, $l_{BC} \\cap l_{CD} = N$, $l_{CD} \\cap l_{DA} = P$, and $l_{DA} \\cap l_{AB} = Q$. Prove that the points $M, N, P$, and $Q$ are concyclic.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "We will need the following lemma.\n\n**Lemma.** The point $O_1$ lies on the line $H_1O$, where $O$ is the center of the circumcircle of $\\triangle ABC$. The ratio $H_1O_1 : O_1O$ depends on $\\angle ACB$ only.\n\n**Proof.** Let $K$ and $L$ be the intersection points of $AH_1$ and $BH_1$ with the circumcircle of $\\triangle ABC$. Since $H_1$ and $L$ are symmetric with respect to $AC$, the quadrilateral $H_1ALA_1$ is a rhombus, as $\\angle AH_1A_1 = 2 \\angle AH_1L = 2\\gamma$. Analogously, $H_1BKB_1$ is a rhombus with $\\angle BH_1B_1 = 2\\gamma$. Therefore $H_1ALA_1$ and $H_1BKB_1$ are similar, whence\n$$\n\\frac{H_1T}{H_1X} = \\frac{H_1Q}{H_1Y} = \\frac{1}{1 + 2 \\cos 2\\gamma}\n$$\nThis means that $O_1$ belongs to the line $H_1O$ and the ratio $H_1O_1 : O_1O$ depends on $\\angle ACB$ only.\n\nIt is clear that $CH_1H_2D$ is a parallelogram (it follows from $CH_1 = CH_2 = 2R \\cos \\gamma$) and the points $O_1$ and $O_2$ are homothetic to $H_1$ and $H_2$, respectively, with respect to $O$. It follows from the lemma that the ratios of these two homotheties are equal (since they depend on $\\gamma$ only). Therefore $O_1O_2 \\parallel H_1H_2 \\parallel CD$, i.e. the line $l_{AB}$ is parallel to $CD$.\n\nWe obtain that the sides of $MNPQ$ are parallel to the corresponding sides of $ABCD$. This means that $MNPQ$ is inscribed.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 27512,
    "subject": "Mathematics (Multi-modal)",
    "question": "Prove that $\\left(\\frac{6}{5}\\right)^{\\sqrt{3}} > \\left(\\frac{5}{4}\\right)^{\\sqrt{2}}$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "We will prove that if $x > -1$, $x \\neq 0$, and $\\alpha \\in (1, 2)$, then\n$$\n(1) \\quad 0 < f(x) = (1+x)^\\alpha - 1 - \\alpha x - \\frac{\\alpha(\\alpha-1)}{2}x^2 - \\frac{\\alpha(\\alpha-1)(\\alpha-2)}{6}x^3.\n$$\nWe have\n$$\nf'(x) = \\alpha\\left[(1+x)^{\\alpha-1} - 1 - (\\alpha-1)x - \\frac{(\\alpha-1)(\\alpha-2)}{2}x^2\\right],\n$$\n$$\nf''(x) = \\alpha(\\alpha - 1)\\left[(1 + x)^{\\alpha - 2} - 1 - (\\alpha - 2)x\\right],\n$$\n$$\nf'''(x) = \\alpha(\\alpha - 1)(\\alpha - 2)\\left[(1 + x)^{\\alpha - 3} - 1\\right],\n$$\nSince $f'''(x) < 0$ for $x \\in (-1, 0)$ and $f'''(x) > 0$ for $x > 0$, we have $f''(x) > f'(0) = 0$ for $x > -1$, $x \\neq 0$. Then $f'(x) < f'(0) = 0$ for $x \\in (-1, 0)$ and $f'(x) > f'(0) > 0$ for $x > 0$, whence $f(x) > f(0) = 0$ for $x > -1$, $x \\neq 0$.\n\nWe now prove that $\\left(\\frac{6}{5}\\right)^{\\sqrt{3}} > \\left(\\frac{5}{4}\\right)^{\\sqrt{2}}$. Set $x = \\frac{1}{5}$ and $\\alpha = \\sqrt{\\frac{3}{2}}$. According to (1), it is enough to check that\n$$\n\\alpha x + \\frac{\\alpha(\\alpha - 1)}{2}x^2 + \\frac{\\alpha(\\alpha - 1)(\\alpha - 2)}{6}x^3 > \\frac{1}{4} \\Leftrightarrow\n$$\n$$\n\\frac{277}{1500}\\alpha + \\frac{3}{125} > \\frac{1}{4} \\Leftrightarrow \\alpha > \\frac{339}{277} \\Leftrightarrow 3.277^2 > 2.339^2 \\Leftrightarrow 230187 > 229842.\n$$\nThe last inequality is obvious and this completes the solution.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 27513,
    "subject": "Mathematics (Multi-modal)",
    "question": "There are $M$ countries and $N$ towns on a planet. Some of the towns are connected by roads. It is known that:\n(1) there are at least three towns in any country;\n(2) any town in a country is connected by roads with at least half of the towns in this country;\n(3) any town is connected with road to exactly one town in another country;\n(4) there are at most two roads between towns in two countries;\n(5) if in two countries there are less than $2M$ towns then there exists at least one road between these countries.\nProve that there exists a round trip having at least $M + \\frac{N}{2}$ towns.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Consider a graph $G = (V, E)$ where the vertices are the towns and the edges are the roads. Denote the towns in the countries by $V_1, V_2, \\dots, V_M$. It follows from the condition of the problem that\n$$\nV = V_1 \\cup V_2 \\cup \\dots \\cup V_M, \\quad V_i \\cap V_j = \\emptyset, \\quad |V_i| \\ge 3 \\quad (1)\n$$\nfor all $i \\neq j$. Moreover, any vertex is connected to at least half of the vertices from its own set (2), and to exactly one vertex from any other set (3). There are at most two edges between any two sets (4) and if for two sets $V_i$ and $V_j$ we have $|V_i| + |V_j| < 2M$ then there exists at least one edge connecting a vertex from $V_i$\n\nwith vertex from $V_j$ (5). We have to prove that in $G$ there is a cycle of length at least $\\frac{N}{2} + M$ (i.e. having at least $\\frac{N}{2} + M$ vertices).\nWe use the following Lemma.\n**Lemma.** Let $G = (V, E)$ be a graph with at least three vertices. If for any pair of vertices $u, v$ not connected by an edge we have $d(u) + d(v) \\ge |V|$ then there is a cycle passing through all vertices.\nDenote by $G_i(V_i, E_i)$, $i = 1, \\dots, M$, the subgraphs of $G$ induced by the sets $V_i$. In other words those are subgraphs having vertices from $V_i$ and edges – all edges from $G$ having both its ends in $V_i$. According to the Lemma for any these subgraphs there exists a cycle $C_i$ passing through all vertices of $V_i$.\nDefine a new graph $H$ with vertices the sets $V_i$, $i = 1, \\dots, M$. Two vertices $V_i$ and $V_j$ from $H$ are connected by an edge if there exist $x \\in V_i$ and $y \\in V_j$ connected by an edge in $G$, i.e. $xy \\in E$. It is clear that $H$ has $M$ vertices and it follows from (3) that any vertex $V_i$ has degree\n$$\nd_H(V_i) \\ge |V_i|, \\quad i = 1, \\dots, M.\n$$\nIf $V_i$ and $V_j$ are not connected by an edge then\n$$\nd_H(V_i) + d_H(V_j) \\ge \\frac{1}{2}(|V_i| + |V_j|) \\ge \\frac{1}{2} \\cdot 2M = M\n$$\nand it follows from the Lemma that there is a cycle in $H$ passing through all $M$ vertices. Let this cycle by $V_1V_2 \\cdots V_MV_1$. It is clear that there are edges in $G$\n$$\nx_{12}x_{12}^{+}, \\dots, x_{i,i+1}^{-}x_{i,i+1}^{+}, \\dots, x_{M,1}^{-}x_{M,1}^{+},\n$$\nsuch that $x_{i,i+1}^{-} \\in V_i$ и $x_{i,i+1}^{+} \\in V_{i+1}$.\nSince any vertex $u \\in V_i$ is connected to exactly one vertex outside $V_i$ we have that $x_{i-1,i}^{+} \\ne x_{i,i+1}^{-}$, $x_{i-1,i}^{+}, x_{i,i+1}^{-} \\in V_i$ and all these vertices partition the cycle $C_i$ into two sections: $C'_i$ and $C''_i$. We may assume that for any $i = 1, \\dots, M$ the section $C'_i$ includes at least half of the vertices from $V_i$. Consider the following cycle in $G$:\n$$\nx_{M,1}^{+}C'_{1}x_{1,2}^{-}x_{1,2}^{+}C'_{2}x_{2,3}^{-} \\cdots x_{i-1,i}^{+}C'_{i}x_{i,i+1}^{-} \\cdots x_{M-1,M}^{+}x_{M-1,M}^{+}C'_{M}x_{M,1}^{-}.\n$$\nThe number of edges (and vertices) in this cycle equals\n$$\n\\sum_{i=1}^{M} \\left\\lceil \\frac{|V_i|}{2} \\right\\rceil + 1 \\ge \\sum_{i=1}^{M} \\frac{|V_i|}{2} + M = \\frac{N}{2} + M,\n$$\nwhich completes the proof.\n\n*Proof of the Lemma.* Assume the statement is not true for a graph with $N \\ge 3$ vertices. There exists a graph of $N$ vertices having the following properties: (1) $d(u) + d(v) \\ge N$ for any pair of not adjacent vertices $u, v$; (2) a cycle through all the vertices does not exist.\nWithout loss of generality assume that $G$ is maximal with the properties (1) and (2), i.e. adding an edge results in a cycle through all vertices. Since $G$ is not complete there exist a pair of vertices $u, v$ that are not adjacent. It follows from the maximality of $G$ that there exists a path from $u$ to $v$ through all vertices:\n$$\nu = x_1, x_2, \\dots, x_{N-1}, x_N = v.\n$$\nLet $S$ be the set of vertices adjacent to $v$ and $T$ the set of vertices adjacent to all vertices adjacent to $u$: $T = \\{x_{i-1}|x_i$ adjacent to $u\\}$. It follows from the condition of the Lemma that $S \\cap T \\ne \\emptyset$. Let $x_j \\in S \\cap T$. It is clear that\n$$\nx_1, x_2, \\dots, x_j, x_N, x_{N-1}, \\dots, x_{j+1}, x_1\n$$\nis a cycle through all the vertices, a contradiction to the initial assumption.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 27514,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $f(x) = x^2 + bx + 1$, where $b$ is a real parameter. Find the number of the integer solutions of the inequality $f(f(x) + x) < 0$.",
    "options": [],
    "answer": "Number of integer solutions equals 0 for |b| ≤ 2; equals 1 when b = m + 1/m for an integer m ≥ 2; equals 2 in all other cases.",
    "solution": "If $|b| \\le 2$, then $f(x) \\ge 0$ for every $x$.\n\nLet $|b| > 2$ and $x_1 < x_2$ are the real zeros of $f$. Then\n$$\n(*) \\quad f(f(x) + x) = (f(x) + x - x_1)(f(x) + x - x_2) \\\\ = (x - x_1)(x - x_2 + 1)(x - x_2)(x - x_1 + 1).\n$$\nWe consider two cases.\n\nCase 1. If $x_2 - x_1 \\le 1$, then $2 < |b| \\le \\sqrt{5}$. Since $f(\\pm 1) = 2 \\pm b$, it easily follows that the integers $x$ such that $f(f(x) + x) < 0$ are exactly $-1$ and $-2$ when $b > 0$, and $1$ and $0$ when $b < 0$.\n\nCase 2. If $x_2 - x_1 > 1$, then $|b| > \\sqrt{5}$. Now each of the intervals $(x_1 - 1, x_1)$ and $(x_2 - 1, x_2)$ contains exactly one integer $x$ such that $f(f(x) + x) < 0$, unless $x_1$ or $x_2$ are not integers.\n\nSummarizing, the required number is equal to 0 for $|b| \\le 2$, to 1 for $b = m + 1/m$, where $m \\ge 2$ is integer, and to 2 in all other cases.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 27515,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all real numbers $a$ with the following property: for every infinite sequence $a_1, a_2, a_3, \\dots$ of mutually distinct positive integers, such that the inequality $a_n \\le a n$ is satisfied for every positive integer $n$, there are infinitely many terms of the sequence which has sum of their digits in numerical system with a base $4038$, which is not multiple $2019$.",
    "options": [],
    "answer": "1 ≤ a < 2019",
    "solution": "Answer: $1 \\le a < 2019$.\n\nIt is clear that $a \\ge 1$, since $a_1$ is a positive integer less than or equal to $a$.\n\nDenote by $\\sigma(a_n)$ the remainder of the division of $2019$ by the sum of the digits of $a_n$ in base $4038$.\n\nWe consider the sequence $\\{b_n : \\sigma(b_n) = 0\\}_{n=0}^{\\infty}$ formed by the numbers such that their sum of digits in base $4038$ is divisible by $2019$. We have $b_0 = 0$, $b_1 = 2019$, etc.\n\nFor any positive integer $n$, the numbers $\\{\\sigma(2019n), \\sigma(2019n+1), \\dots, \\sigma(2019n+2018)\\}$ form a complete remainder system modulo $2019$ (because $2019n, \\dots, 2019n+2018$ differ only in the last digit). Therefore\n$$\n2019n \\le b_n \\le 2019n + 2018. \\qquad (1)\n$$\nLet us consider arbitrary sequence $a_1, a_2, a_3, \\dots$ which does not satisfy the condition, i.e. only finitely many of its terms $a_{i_1}, a_{i_2}, \\dots, a_{i_\\ell}$ satisfy $\\sigma(a_{i_j}) = 0$. Then there exists a positive integer $M$ such that $a_m \\in \\{b_n\\}_0^\\infty$ for every $m \\ge M$. This means that for every positive integer $N$ we have $a_{M+k} \\in \\{b_n\\}_0^\\infty$, $k = 0, \\dots, N$.\n\nAssume that $a < 2019$. Since the terms of the sequence are mutually distinct, we conclude that\n$$\n2019N \\le b_N \\le \\max_{0 \\le k \\le N} a_{M+k} \\le a(M+N) \\Rightarrow N \\le \\frac{aM}{2019-a}, \\forall N, \\quad (2)\n$$\nwhich is a contradiction because of the choice of (fixed) $M$ and $a$. Therefore every $1 \\le a < 2019$ gives a sequence with the required properties.\n\nFor $a \\ge 2019$, the sequence $1, b_1, b_2, \\dots$ satisfies the restrictions $a_n := b_{n-1} < 2019(n+1) \\le a(n+1)$ for every $n$ and $\\sigma(a_n) = 0$ for every $n \\ge 1$. Therefore, only the first term of this sequence has sum of digits in base $4038$ which is not divisible by $2019$. Therefore $a \\ge 2019$ could not bring new solution.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 27516,
    "subject": "Mathematics (Multi-modal)",
    "question": "A convex $2019$-gon is given, for which no three diagonals intersect in a point. An intersection point of two diagonals, internal for the given polygon will be called “knot”. Two knots are called “neighbors” if they share a common diagonal. A closed path between neighboring knots, such that no three consecutive knots share a diagonal, will be called “cycle”. Find the maximal number of knots that can be colored, such that there exists no colored cycle (that is a cycle, containing only colored knots)?",
    "options": [],
    "answer": "2035151",
    "solution": "Answer: $\\frac{2019(2019-3)}{2} - 1 = 2035151$.\nLet us consider the more general problem, where $2019$ is replaced with $n \\ge 4$. We will prove that in a convex $n$-gon, neither three of which diagonals have a common point, we can color at most $\\frac{n(n-3)}{2} - 1$ of the knots without generating a colored cycle.\n\nFirst, we prove an estimate from above, i.e., that the number of colored knots needs to be less than $\\frac{n(n-3)}{2}$. Assume the contrary and let us consider a planar graph, with vertices corresponding to the polygon diagonals and edges corresponding to the colored knots. Since the number of diagonals in a convex $n$-gon equals $\\frac{n(n-3)}{2}$, the edges in the constructed graph outnumber the vertices, thus the graph contains a cycle. But it is easy to see that there is a one-to-one correspondence between the cycles in the constructed graph and the colored cycles in the polygon. Hence, there exists a colored cycle, which is a contradiction.\n\nNow, it remains to show a strategy for coloring $\\frac{n(n-3)}{2} - 1$ knots, without generating a colored cycle. We do it by induction. The base $n=4$ is trivial. Now, let us have a strategy for coloring $\\frac{n(n-3)}{2} - 1$ of the knots of an arbitrary convex $n$-gon without generating a colored cycle. Consider a convex $(n+1)$-gon $A_1A_2\\dots A_{n+1}$ and take a “good” coloring for the knots of $A_1A_2\\dots A_n$. Adding the vertex $A_{n+1}$, we transform the side $A_1A_n$ for $A_1A_2\\dots A_n$ into a diagonal for $A_1A_2\\dots A_{n+1}$ and we generate additional $n-2$ diagonals $A_{n+1}A_i$, $i=2,\\dots,n-1$. Therefore, the diagonal $A_1A_n$ contains exactly $n-2$ knots (all corresponding to the additional diagonals) and we color all of them. Finally, we color one more knot on a diagonal through $A_{n+1}$, e.g., the intersection point of $A_{n+1}A_2$ and $A_1A_3$. (Due to the convexity of the polygon, it is clear that all considered intersection points are internal, thus are indeed knots!) Overall, we end up with\n$$\n\\frac{n(n-3)}{2} - 1 + (n-1) = \\frac{(n+1)(n-2)}{2} - 1\n$$\ncolored points. Among the freshly colored $n-1$ knots, “neighboring” appears only with respect to the diagonals $A_1A_n$ or $A_2A_{n+1}$. Furthermore, every other diagonal through $A_{n+1}$ contains at most one colored knot, i.e., it doesn’t generate any “neighboring” relations. It is straightforward to observe that in order to exist a colored cycle, given that the coloring of the knots of $A_1A_2...A_n$ is acyclic, at least two of the freshly colored knots should be part of this cycle, which is impossible due to construction. Induction is completed and so is the proof.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 27517,
    "subject": "Mathematics (Multi-modal)",
    "question": "A cyclic hexagon $ABCDEF$ is given with the following property:\n\n$$\nAB \\cdot CD \\cdot EF = BC \\cdot DE \\cdot AF.\n$$\n\nLet $B_1$ be the symmetric point of $B$ with respect to line $AC$, $D_1$ be the symmetric point of $D$ with respect to line $CE$, and $F_1$ be the symmetric point of $F$ with respect to line $EA$. Prove that $\\triangle B_1D_1F_1 \\sim \\triangle BDF$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "It is well known fact that for cyclic hexagon the equation $|AB| \\cdot |CD| \\cdot |EF| = |BC| \\cdot |DE| \\cdot |AF|$ holds iff the diagonals $AD$, $BE$ and $CF$ are concurrent (follows from the trigonometric form of sine theorem for $\\triangle ACE$). Let $P$ be the point of intersection of $AD$, $BE$ and $CF$. Let us consider $\\triangle AF_1P$ and $\\triangle CD_1P$.\n\n![](attached_image_1.png)\n\nWe have $\\triangle PAF_1 = |\\triangle PAE - \\triangle F_1AE| = |\\triangle DAE - \\triangle FAE| = |\\triangle DCE - \\triangle FCE| = |\\triangle D_1CE - \\triangle PCE| = \\triangle PCD_1$. (Notice that $D_1$ and $D$ are on the opposite sides of $CF$ iff $F$ and $F_1$ are on the same side of $AD$). From $\\triangle APF \\sim \\triangle CPD$, $|AF_1| = |AF|$ and $|CD_1| = |CD|$ we get:\n$$\n\\frac{|AF_1|}{|CD_1|} = \\frac{|AF|}{|CD|} = \\frac{|AP|}{|CP|}\n$$\nTherefore $\\triangle AF_1P \\sim \\triangle CD_1P$. Hence $\\triangle APF_1 = \\triangle CPD_1$ and $\\frac{|F_1P|}{|D_1P|} = \\frac{|AP|}{|CP|} = \\frac{|FP|}{|DP|}$ ($\\triangle APF \\sim \\triangle CPD$). Since $D_1$ is outside of $\\triangle PCD$ iff $F_1$ is inside of $\\triangle FAP$, we have $\\triangle F_1PD_1 = \\triangle APC = \\triangle FPD$. From $\\frac{|F_1P|}{|D_1P|} = \\frac{|FP|}{|DP|}$ and $\\triangle F_1PD_1 = \\triangle FPD$ we have that $\\triangle F_1D_1P \\sim \\triangle FDP$. Therefore $\\frac{|F_1D_1|}{|FD|} = \\frac{|F_1P|}{|FP|} = \\frac{|D_1P|}{|DP|}$. Similarly $\\frac{|D_1B_1|}{|DB|} = \\frac{|D_1P|}{|DP|} = \\frac{|F_1D_1|}{|FD|}$ and $\\frac{|F_1B_1|}{|FB|} = \\frac{|F_1P|}{|FP|} = \\frac{|F_1D_1|}{|FD|}$. So $\\triangle F_1D_1B_1 \\sim \\triangle FDB$ and thus the solution is completed.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 27518,
    "subject": "Mathematics (Multi-modal)",
    "question": "An equilateral triangle $T$ with side length $2022$ is coloured in white and partitioned into equilateral triangles with side length $1$ (called *cells*) through lines, parallel to the sides of $T$. Two cells are called *adjacent* if they have at least one vertex in common. Ivan colors some of the cells in black. Without seeing which cells are black, Peter selects only once a set $S$ of cells (containing at least one) and asks Ivan if the number of blacks among the selected ones is even or odd. After receiving an answer, Peter is able to tell if the number of pairs of differently colored adjacent cells in $T$ is even or odd. Find all possible values of the number of elements of $S$ if that is always possible no matter the coloring of Ivan.",
    "options": [],
    "answer": "12120",
    "solution": "For simplicity we write $A$ instead of Ivan and $B$ instead of Peter. The crucial idea is in the following\n\n*Lemma.* In the graph $G$ all vertices are initially colored white. Let $A$ colors some of the vertices in black and $B$ asks $A$ about the parity of the number of blacks in a set $S$ from vertices, and with the answer can unambiguously determine the parity of the number of differently colored pairs adjacent/vertices in $G$. Then $S$ must be the set of all vertices of odd degree in $G$.\n\n*Proof.* We will work by modulus $2$. We write the number $1$ in each black vertex, the number $0$ in each white vertex and the sum of the numbers in its vertices on each edge. Each black vertex of odd degree contributes $1$ to the sum of the numbers on the edges, each black vertex of even degree contributes $0$, and each white vertex also contributes $0$. Therefore, the sum of the numbers on the edges in $G$ has the same parity as that of the number of black vertices of odd degree in $G$. Thus, when asked about the set $N$ of the vertices of odd degree, we are ready. Conversely, for a question with a set other than $N$, we cannot unambiguously restore the parity of the number of black vertices in $N$ and the lemma is proved.\n\nIt remains to count the number of cells with an odd number of adjacent cells in $T$. Each cell that has no common point with the perimeter of $T$ has $12$ adjacent cells. The three cells on the perimeter, each of which is adjacent on the side of a corner cell in $T$, have $6$ adjacent cells. Each cell with a side on the perimeter has $7$ adjacent cells, and each cell with a vertex (but not a side) on the perimeter (except the three mentioned above) has $9$ adjacent cells. Because each side of $T$ has at least one common point with $2 \\times 2022 - 1$ cells, and each two sides have two common cells (as only one is of odd degree), the required number is $3 \\times (2 \\times 2022 - 5) + 3 = 12120$.",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 27519,
    "subject": "Mathematics (Multi-modal)",
    "question": "If $x, y, z \\in \\mathbb{R}$ satisfy the nonlinear system of equations\n$$\n\\left| \\begin{array}{l} x - y + z - 1 = 0 \\\\ xy + 2z^2 - 6z + 1 = 0 \\end{array} \\right. ,\n$$\n\nfind the maximum of $(x-1)^2 + (y+1)^2$.",
    "options": [],
    "answer": "11",
    "solution": "Plugging the first equation into the second, we get\n$$\nx(x+z-1)+2z^2-6z+1=x^2+(z-1)x+2z^2-6z+1=0,\n$$\nwhich is quadratic with respect to $x$. The discriminant is $D = -7z^2 + 22z - 3$, thus we need $z \\in [\\frac{1}{7}, 3]$. Note that\n$$\n(x-1)^2 + (y+1)^2 = (y-x)^2 + 2xy + 2(y-x) + 2 = -3z^2 + 12z - 1.\n$$",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 27520,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all pairs of real numbers $(x, y)$ such that\n$$\n\\left| \\begin{aligned} & \\sqrt{x^2 - 2x + y^2 + 4y + 5} + \\sqrt{x^2 - 8x + y^2 - 4y + 20} = 5 \\\\ & 16x^2 + 9y^2 = 68. \\end{aligned} \\right.\n$$",
    "options": [],
    "answer": "(2, -2/3)",
    "solution": "Consider orthogonal coordinate system and the points $A(1, -2)$ and $B(4, 2)$. Since\n$$\n\\begin{aligned} \\sqrt{x^2 - 2x + y^2 + 4y + 5} &= \\sqrt{(x-1)^2 + (y+2)^2} \\\\ \\sqrt{x^2 - 8x + y^2 - 4y + 20} &= \\sqrt{(x-4)^2 + (y-2)^2}, \\end{aligned}\n$$\nwe conclude that the solutions of the first equation are all points $C(x, y)$ such that $|CA| + |CB| = 5$. But\n$$\n|AB| = \\sqrt{(4-1)^2 + (2-(-2))^2} = \\sqrt{16+9} = 5 = |CA| + |CB|,\n$$\nand thus the point $C$ lies on the line segment $AB$. The equation of the line $AB$ is $3y = 4x - 10$ so we want to find the solutions of the two equations satisfying $x \\in [1, 4]$. Now the solution of the system is given by\n$$\n32x^2 - 80x + 100 = 68 \\Leftrightarrow 2x^2 - 5x + 2 = 0 \\Leftrightarrow x_1 = \\frac{1}{2}, x_2 = 2.\n$$\nOnly $x_2$ belongs to the interval $[1, 4]$, thus the only solution is $(2, -\\frac{2}{3})$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 27521,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $\\triangle ABC$ be a triangle with an incenter $I$. The line $CI$ intersects for a second time the circumcircle of $\\triangle ABC$ at $L$, where $CI = 2 \\cdot IL$. Points $M$ and $N$ lie on the segment $AB$, such that $\\angle AIM = \\angle BIN = 90^\\circ$. Prove that $AB = 2 \\cdot MN$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "We adopt the standard notation for $\\triangle ABC$. Denote by $P$ and $Q$ the midpoints of $AC$ and $BC$, respectively. Let $J$ be the center of the excircle, tangent to the segment $AB$. It is well known that $I$ is the midpoint of $CJ$, therefore $PI$ is a midsegment for $\\triangle AJC$ and $PI \\parallel AJ$. Hence $\\angle AIP = 90^\\circ$. Analogously, $\\angle BIQ = 90^\\circ$. We have $\\triangle AIP \\cong \\triangle AMI$, respectively $\\triangle BIQ \\cong \\triangle BIN$, i.e., $AP = AM$ and $BQ = NB$. Moreover, $r_c = 2r$. Finally\n$$\n\\begin{aligned}\nS_{ABC} &= p \\cdot r = (p-c) \\cdot r_c \\quad \\Rightarrow \\quad p = 2(p-c) \\quad \\Rightarrow \\quad 2c = \\frac{a+b+c}{2} \\\\\n&\\Rightarrow \\quad AC + BC = 3AB \\quad \\Rightarrow \\quad AM + AN = AP + BQ = \\frac{AC+BC}{2} = \\frac{3AB}{2}.\n\\end{aligned}\n$$\nBut $AM + AN = AB + MN$, thus $MN = AB/2$.\n\n**Remark.** The opposite claim holds also true, i.e., $CI = 2 \\cdot IL$ whenever $AB = 2 \\cdot MN$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 27522,
    "subject": "Mathematics (Multi-modal)",
    "question": "We will refer to a permutation $\\sigma$ of the numbers from 1 to 10 as *bad*, if there exist three indices $i, j, k$, such that $1 \\le i < j < k \\le 10$, but $\\sigma(j) < \\sigma(k) < \\sigma(i)$, and as *good*, otherwise. Determine the number of the good permutations.",
    "options": [],
    "answer": "16796",
    "solution": "We will derive a closed formula for the good permutations of the numbers from 1 to $n$. Consider a good permutation $\\sigma$, where $j = \\sigma^{-1}(1)$ is an arbitrary index inbetween 1 and $n$. Then, the first $j-1$ positions of $\\sigma$ contain the numbers $2, 3, \\ldots, j$, forming a good permutation of $j-1$ elements, while the last $n-j$ positions of $\\sigma$ contain the numbers $j+1, \\ldots, n$, forming a good permutation of $n-j$ elements. It is an easy exercise to check that the above condition is not only necessary but also sufficient for the construction of an arbitrary good permutation of $n$ elements. Hence, denoting by $S_n$ the corresponding number of good permutations, we derive that $(S_n)_{n \\ge 0}$ satisfies $S_0 = 1$ and $S_n = \\sum_{j=0}^{n-1} S_j S_{n-1-j}$, $\\forall n \\ge 1$. Therefore, $S_n = \\frac{1}{n+1} \\binom{2n}{n}$ is the $n$-th Catalan number. In particular,\n\n$$\nS_{10} = \\frac{1}{11} \\binom{20}{10} = \\frac{12 \\cdot 13 \\cdots 20}{2 \\cdot 3 \\cdots 10} = 4 \\cdot 13 \\cdot 17 \\cdot 19 = 16796.\n$$",
    "topic": "Discrete Mathematics",
    "subtopic": "Combinatorics"
  },
  {
    "id": 27523,
    "subject": "Mathematics (Multi-modal)",
    "question": "A positive integer $n$ is called balanced if it is not a prime and for any integer $k$ in the interval $[1, \\sqrt{n}-1]$ the number of ways to choose $k$ persons from a group of $n$ people is divisible by $n$. If $m$ and $n$ are balanced five digit positive integers find the smallest value of the difference $|m-n|$.",
    "options": [],
    "answer": "202",
    "solution": "Answer. 202. As in problem 9.3 we obtain that any 5-digit balanced number is of the forms: $p^2$ or $p(p+2)$ where $p$ and $p+2$ are primes. Since $p(p+2) = (p+1)^2 - 1 < (p+1)^2$ it follows that the smallest positive difference between two 5-digit numbers equals $2p$ where $p$ is the smallest prime number for which $p+2$ is also a prime and $p^2$ and $p(p+2)$ are 5-digit numbers. The smallest 5-digit number is $10000 = 10^4$, i.e, we want $p \\ge 100$. Direct verification shows that 101 and 103 are both prime numbers. The corresponding 5-digit balanced numbers are $101^2$ and $101 \\cdot 103$ with difference $2 \\cdot 101 = 202$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 27524,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find the smallest odd prime $p$ for which there exist positive co-prime integers $k$ and $\\ell$, such that\n$$\n4k - 3\\ell = 12 \\quad \\text{and} \\quad \\ell^2 + \\ell k + k^2 \\equiv 3 \\pmod{p}.\n$$",
    "options": [],
    "answer": "11",
    "solution": "From $4k = 12 + 3\\ell$, it follows $k = 3k_1$, $k_1 \\in \\mathbb{N}$. Analogously, $\\ell = 4\\ell_1$, $\\ell_1 \\in \\mathbb{N}$. Assuming $p = 3$ gives rise to $3 \\mid k$ and $3 \\mid \\ell^2 + \\ell k + k^2 \\Rightarrow 3 \\mid \\ell \\Rightarrow (k, \\ell) \\ge 3 > 1$, a contradiction with $(k, \\ell) = 1$. Thus, $p > 3$, $(3, p) = 1$, and\n$$\n\\begin{aligned}\n\\ell^2 + \\ell k + k^2 \\equiv 3 \\pmod{p} &\\iff 9\\ell^2 + 9\\ell k + 9k^2 \\equiv 27 \\pmod{p} \\\\\n&\\iff (4k - 12)^2 + (4k - 12)3k + 9k^2 \\equiv 27 \\pmod{p} \\\\\n&\\iff 16k^2 - 96k + 144 + 12k^2 - 36k + 9k^2 - 27 \\equiv 0 \\pmod{p} \\\\\n&\\iff 37k^2 - 132k + 117 \\equiv 0 \\pmod{p} \\\\\n&\\iff (37k)^2 - 2 \\cdot 66 \\cdot 37 \\cdot k + 117 \\cdot 37 \\equiv 0 \\pmod{p} \\\\\n&\\iff (37k - 66)^2 \\equiv 66^2 - 117 \\cdot 37 \\equiv 27 \\pmod{p} \\\\\n&\\iff (37k_1 - 22)^2 \\equiv 3 \\pmod{p},\n\\end{aligned}\n$$\nmeaning that 3 must be a quadratic residue modulo $p$. Since the quadratic residues mod 5 and mod 7 are respectively $\\{0, 1, 4\\}$ and $\\{0, 1, 2, 4\\}$, 3 is not among them and $p \\notin \\{5, 7\\}$, i.e., $p \\ge 11$. For $p = 11$, $11 \\mid 22$ and $37 \\equiv 4 \\pmod{11}$. Therefore, it suffices to find a solution for $(4k_1)^2 \\equiv 3 \\pmod{11}$, which modulo 11 is equivalent to $k_1^2 \\equiv 5 \\pmod{11}$, i.e., $k_1 = 11k_2 + 4$ or $k_1 = 11k_2 + 7$. We have to additionally assure $(k, \\ell) = 1$. We know that $k_1 = 1 + \\ell_1$, thus $(k_1, \\ell_1) = 1$. Therefore, it suffices to take $(k_1, 4) = 1$ and $(\\ell_1, 3) = 1$, e.g., $k_1 = 15$. Then $\\ell_1 = 14$, $k = 45$, $\\ell = 56$, $(k, \\ell) = 1$, and from $k \\equiv \\ell \\equiv 1 \\pmod{11}$ we check that\n$$\n\\ell^2 + \\ell k + k^2 \\equiv 1^2 + 1 \\cdot 1 + 1^2 \\equiv 3 \\pmod{11}.\n$$\nIn conclusion, the answer is $p = 11$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 27525,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $m$ and $n$ be positive integers and $m < n$. Find the number of all injective functions $f: \\{1, 2, \\dots, m\\} \\to \\{1, 2, \\dots, n\\}$ such that for any nonempty subset $A \\subseteq \\{1, 2, \\dots, m\\}$, the set of values $f(A)$ is distinct from $A$, i.e. $f(A) \\neq A$. (A function $f$ is injective if $f(x) \\neq f(y)$ when $x \\neq y$.)",
    "options": [],
    "answer": "(n-1)!/(n-m-1)!",
    "solution": "Answer. $\\frac{(n-1)!}{(n-m-1)!}$. Set $B_i = \\{1, 2, \\dots, i\\}$. By induction on $m$ we prove that for any $n > m$ the number of injective functions $f: B_m \\to B_n$, that satisfy the condition of the problem equals $(n-1)(n-2)\\dots(n-m)$.\n\nFor $m=1$ we have $f(1) \\neq 1$, i.e. there are $n-1$ possible values for $f(1)$ and the base case is true. Assume the statement is true for $m-1$ and consider function $f: B_m \\to B_n$ satisfying the condition for arbitrary nonempty subset of $B_{m-1}$.\n\n**Case 1.** Assume $m \\notin f(B_{m-1})$. There are $m$ \"forbidden\" values for $f(m): f(B_{m-1}) \\cup \\{m\\}$. The first $m-1$ because the function is injective and the last one because $= \\{m\\}$ does not satisfy the condition. Therefore there are $n-m$ values.\n\n**Case 2.** Assume $m \\in f(B_{m-1})$ and let $a_1 = f^{-1}(m) \\in B_{m-1}$. By analogy if $a_1 \\in f(B_{m-1})$ then let $a_2 = f^{-1}(a_1) \\in B_{m-1}$ and so on. Since $m \\in f(B_{m-1})$ and $m \\notin B_{m-1}$ the two sets have the same cardinality and we arrive to a number $a_k \\in B_{m-1}$ such that $f(a_k) = a_{k-1} \\in f(B_{m-1})$ but $a_k \\notin f(B_{m-1})$. Then all forbidden values of $f(m)$ are $f(B_{m-1}) \\cup \\{a_k\\}$ (due to the injectivity or because of the set $A' = \\{a_1, a_2, \\dots, a_k, m\\}$). All remaining $n-m$ values are possible. Indeed, consider nonempty set $A \\subseteq B_m$. If $m \\notin A$ then $f(A) \\neq A$ according to the induction hypothesis. If $m \\in A$ and $f(A) = A$ then it follows from $m \\in f(A)$ that $a_1 \\in A$ and $A' \\subseteq A$. Thus $a_k \\in A$ and $a_k \\notin f(B_m) \\notin f(A)$, a contradiction. In both cases we have $n-m$ possible values and the answer is $(n-1)(n-2)\\dots(n-m)$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 27526,
    "subject": "Mathematics (Multi-modal)",
    "question": "Consider the equation $(x^2 - 8x + 7)^2 + (3m - 1)(x^2 - 8x + 7) + 2m^2 - m = 0$, where $m$ is a real parameter.\na) Find the number of solutions of the equation for $m = 6$.\nb) Find the values of $m$ for which the equation has exactly three distinct real roots.",
    "options": [],
    "answer": "a) 2; b) m = 5",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 27527,
    "subject": "Mathematics (Multi-modal)",
    "question": "Solve $(x+1)\\log_3 x + 4x \\log_3 x - 16 = 0$.",
    "options": [],
    "answer": "x = 3 and x = 1/81",
    "solution": "We need $x > 0$. Substitute $y = \\log_3 x$ to obtain $(x+1)y^2 + 4x y - 16 = 0$, which is equivalent to $(y+4)(x y + y - 4) = 0$.\n\nThus either $\\log_3 x = -4$ and $x = \\frac{1}{81}$ or $\\log_3 x = \\frac{4}{x+1}$, with the obvious solution $x = 3$.\n\nDue to monotonicity, it is easy to check that the latter has no solutions $x \\ne 3$.\n\nIn conclusion, the solutions of the equation are $x_1 = 3$ and $x_2 = \\frac{1}{81}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 27528,
    "subject": "Mathematics (Multi-modal)",
    "question": "The lengths of the sides and the diagonal $AC$ of inscribed quadrilateral $ABCD$ are positive integers. If $\\cos \\angle ABC = \\frac{1}{4}$, $2CD = AD + AC$ and $2AB = CA + CB + CD$ find the smallest possible value of the perimeter of $ABCD$.",
    "options": [],
    "answer": "83",
    "solution": "Let $\\alpha = \\angle ABC$, $AB = a$, $BC = b$, $CD = c$, $DA = d$ and $AC = e$. Then $\\angle ADC = 180^\\circ - \\alpha$, $2c = d + e$, $2a = b + e + c$ and it follows from the cosine theorem for $\\triangle ADC$ that\n$$\nd^2 + c^2 + \\frac{dc}{2} = e^2 = (2c - d)^2 \\iff 2c = 3d.\n$$\nSince $2c = d + e$ we obtain $e = 2d$ and $2a = b + e + c$ implies $2a = b + \\frac{7e}{4}$. The cosine theorem for $\\triangle ABC$ gives\n$$\na^2 + b^2 - \\frac{ab}{2} = e^2 = \\left(\\frac{4}{7}(2a - b)\\right)^2 \\iff 33b^2 + \\frac{79ab}{2} - 15a^2 = 0.\n$$\nSince the roots of the equation $33x^2 + \\frac{79x}{2} - 15 = 0$ are $x_1 = \\frac{10}{33}$ and $x_2 < 0$ we have $\\frac{b}{a} = \\frac{10}{33} \\iff b = \\frac{10a}{33}$.\nNow $2a = b + \\frac{7e}{4}$ implies $a = \\frac{33e}{32} = \\frac{33d}{16}$ and $b = \\frac{5d}{8}$. Finally $a = \\frac{33d}{16}$, $b = \\frac{5d}{8}$, $c = \\frac{3d}{2}$, $e = 2d$. The least value of $d$ for which $a$ is an integer is $d = 16$. Therefore $AB = 33$, $BC = 10$, $CD = 24$, $DA = 16$, $AC = 32$ and the least value of the perimeter equals $83$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 27529,
    "subject": "Mathematics (Multi-modal)",
    "question": "In each of the cells of an $n \\times m$ table is written one of the letters $a$, $b$, or $c$. The corresponding letter in every two rows, respectively every two columns of the table, coincide in at most $k \\ge 0$ positions. Determine $m$, $n$ and $k$, if\n$$\n\\frac{2mn + 6k}{3(m + n)} \\ge k + 1.\n$$",
    "options": [],
    "answer": "m = 3, n = 3, k = 0",
    "solution": "Denote the table rows by $u_1, u_2, \\dots, u_n$. Let $d(u_i, u_j)$, $i \\ne j$ be the number of positions, in which the rows $u_i$ and $u_j$ differ. According to the statement $d(u_i, u_j) \\ge m - k$. Then:\n$$\n(1) \\qquad S = \\sum_{1 \\le i < j \\le n} d(u_i, u_j) \\ge \\frac{n(n-1)}{2}(m-k).\n$$\n\nConsider an arbitrary table column and let it contain $x$ letters $a$, $y$ letters $b$, and $z$ letters $c$, with $x+y+z=n$. The contribution to $S$ of this column is exactly $xy+xz+yz$, and due to the inequality\n$$\nn^2 = (x + y + z)^2 \\ge 3(xy + xz + yz)\n$$\nit follows that every column adds at most $\\frac{n^2}{3}$ to the value of $S$ (this is the case iff $3$ divides $n$ and every column there are exactly $\\frac{n}{3}$ letters of each type). Now (1) gives rise to:\n$$\n(2) \\qquad \\frac{mn^2}{3} \\ge S \\ge \\frac{n(n-1)}{2}(m-k) \\iff mn + 3k \\le 3kn + 3m.\n$$\n\nRepeating the arguments, when replacing rows and columns leads to:\n$$\n(3) \\qquad mn + 3k \\le 3km + 3n.\n$$\nWe add (2) and (3) to obtain $2mn+6k \\le 3(m+n)(k+1) \\Leftrightarrow \\frac{2mn+6k}{3(m+n)} \\le k+1$. Therefore, $\\frac{2mn+6k}{3(m+n)} = k+1$, $mn+3k = 3kn+3m$, and $mn+3k = 3km+3n$ (meaning also that both $m$ and $n$ are divisible by $3$). From $3kn+3m = 3km+3n \\Leftrightarrow (k-1)(n-m) = 0$, it follows that either $k=1$ or $m=n$. For $k=1$, $mn+3 = 3m+3n$, which is impossible, since $3$ divides $m$ and $n$, thus $9$ divides the RHS but not the LHS. For $m=n$, $m^2-3(k+1)m+3k=0$, with discriminant $D = 9k^2+18k+9-12k = (3k+1)^2+8$ a perfect square. From $(3k+1)^2+8 = t^2$ we deduce that the difference between two perfect squares is $8$, which holds\n\ntrue only for $t=3$ and $k=0$. In conclusion $m=n=3$ and the table is:\n\n| a | b | c |\n|---|---|---|\n| b | c | a |\n| c | a | b |",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 27530,
    "subject": "Mathematics (Multi-modal)",
    "question": "A positive integer $n$ having 2022 divisors $1 = d_1 < d_2 < \\dots < d_{2022} = n$ is called *nice* if $3d_{2017} + 2d_{2019} = n$. Find all nice integers.",
    "options": [],
    "answer": "2^2 · 3 · 5^{336}, 2^2 · 3^{336} · 5, 2^{336} · 3 · 5^2, 2^{336} · 3^2 · 5",
    "solution": "It follows from $d_i \\ge i$ and $d_6 d_{2017} = d_4 d_{2019} = n$ that\n$$\n2n = 6 d_{2017} + 4 d_{2019} \\le d_6 d_{2017} + d_4 d_{2019} = 2n.\n$$\nTherefore $d_6 = 6$ implying $d_i = i$ for $i = 1, 2, 3, 4, 5, 6$ and thus $n$ is divisible by $2^2 \\cdot 3 \\cdot 5$. Since $2022 = 2 \\cdot 3 \\cdot 337$ and $337$ is a prime number it follows that $n = 2^x \\cdot 3^y \\cdot 5^z$ for $x \\ge 2$. All nice numbers are:\n$$\n2^2 \\cdot 3 \\cdot 5^{336},\\quad 2^2 \\cdot 3^{336} \\cdot 5,\\quad 2^{336} \\cdot 3 \\cdot 5^2,\\quad 2^{336} \\cdot 3^2 \\cdot 5.\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 27531,
    "subject": "Mathematics (Multi-modal)",
    "question": "For a positive integer $n > 3$ the set $A$ of sequences of 0 and 1 with length $n+1$ is called *good* if any sequence of 0 and 1 of length $n$ can be obtained from a sequence from $A$ by deleting one term. Let $a_n$ be the minimum number of elements in a good set. Prove that: $\\frac{2^n + 2n - 7}{n} \\le a_n \\le 2^{n-1}$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $B$ be the set of all sequences of 0 and 1 of length $n-1$, thus $|B| = 2^{n-1}$. It is clear that the set $A$ obtained by adding 01 at the end of any sequence from $B$ is good. Therefore $a_n \\le 2^{n-1}$.\n\nA sequence of length $n$ with all zeroes can be obtained either by the sequence with all zeroes of length $n+1$ or by a sequence of length $n+1$ with only one entry 1. In both cases the number of sequences obtained by deleting one entry is at most 3. By analogy, from a sequence of length $n+1$ with all ones or with exactly one zero one obtains at most 3 sequences of length $n$. Only from the sequences of length $n+1$ with alternating 0 and 1 (or 1 and 0) one obtains $n+1$ sequences of length $n$. Note that if both alternating sequences are in $A$ then the two sequences of length $n$ with alternating 0 and 1 or 1 and 0 are obtained by two ways. Therefore\n$$\n2^n \\le 6 + (n+1) + (a_n - 3)n \\iff \\frac{2^n + 2n - 7}{n} \\le a_n.\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 27532,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $n \\ge 2$ be a fixed positive integer. The set $M$ consists of $2n^2 - 3n + 2$ positive rational numbers. Prove that there exists an $n$-element subset $A$ of $M$ with the following property: for every $k \\in \\mathbb{N}$, $2 \\le k \\le n$, the sum of arbitrary $k$ (not necessarily distinct) elements of $A$ is not an element of $A$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Without loss of generality we can consider the elements of $M$ as positive integers, as we can multiply them by the $\\gcd$ of their denominators, which does not affect the problem statement.\n\nThe arithmetic progression $\\{(2n-1)q+n\\}_{q=1}^{\\infty}$ contains infinitely many primes, due to the Dirichlet theorem and $((2n-1), n) = 1$. Let us fix a prime $p = (2n-1)q+n$, larger than all the elements of $M$, i.e., $M$ becomes a set of residues modulo $p$. Consider the following set of residues:\n\n$$\nP = \\left\\{ \\frac{p+n-1}{2n-1}, \\frac{p+3n-2}{2n-1}, \\dots, \\frac{2p-1}{2n-1} \\right\\},\n$$\n\nfor which the numerator of every successive fraction exceeds the previous one by $2n-1$. According to the choice of $p$, all elements in $P$ are positive integers. Moreover\n$$\n2 \\cdot \\frac{p+n-1}{2n-1} = \\frac{2p-1}{2n-1} + 1 \\quad \\text{and} \\quad n \\cdot \\frac{2p-1}{2n-1} = p + \\frac{p-n}{2n-1}.\n$$\n\nThis means that the sum of arbitrary (at least 2 and at most $n$, not necessarily distinct) elements of $P$ modulo $p$, is not an element of $P$!\n\nNote that $|P| = \\frac{2p-1}{2n-1} - \\frac{p-n}{2n-1} = \\frac{p+n-1}{2n-1}$. Let $M = \\{m_1, m_2, \\dots, m_{2n^2-3n+2}\\}$. For each $i = 1, 2, \\dots, 2n^2-3n+2$, the numbers $\\{m_i \\cdot 1, m_i \\cdot 2, \\dots, m_i \\cdot (p-1)\\}$ form a permutation of $\\{1, 2, \\dots, (p-1)\\}$ (mod $p$), i.e., exactly $\\frac{p+n-1}{2n-1}$ of them are elements of $P$! Thus, there exists $k \\in \\{1, 2, \\dots, p-1\\}$, such that among the numbers $\\{k \\cdot m_1, k \\cdot m_2, \\dots, k \\cdot m_{2n^2-3n+2}\\}$ at least\n$$\n\\frac{(2n^2 - 3n + 2) \\cdot \\frac{p+n-1}{2n-1}}{p-1} > \\frac{2n^2 - 3n + 2}{2n-1} > n-1\n$$\nare elements of $P$ modulo $p$. It is easy to check, that the corresponding elements $\\{m_j\\}$ (at least $n$ in number) of $M$, satisfying $k \\cdot m_j \\in P$ (mod $p$), fulfill the problem statement. In particular, so does every their subset of exactly $n$ elements.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 27533,
    "subject": "Mathematics (Multi-modal)",
    "question": "For a triangle $ABC$ it is known that $AB = c$, $BC = a$, $AC = b$ and $\\angle BAC = 80^\\circ$. Point $M$ from the side $AB$ is such that $\\angle AMC = 70^\\circ$. If $AM + AC = BC$ prove that $a^2 = b(b+c)$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Choose a point $M_1$ on the ray $CA$ such that $AM_1 = AM$ and $CM_1 = BC$. It follows that triangle $M_1MA$ is isosceles and $\\angle AM_1M = \\angle AMM_1 = \\frac{1}{2} \\angle BAC = 40^\\circ$.\n\nTriangles $CM_1M$ and $CBM$ are congruent ($CM_1 = CB$, $CM = CB$, $\\angle CMM_1 = \\angle CMB = 110^\\circ$). Therefore $\\angle ABC = \\angle CM_1M = 40^\\circ$, $\\angle ACB = 60^\\circ$.\n\nThe equality $a^2 = b(b+c)$ follows from $\\alpha = 2\\beta$. Choose a point $B_1$ on the ray $CA$ such that $AB_1 = AB$ or $CB_1 = b+c$. The triangle $\\triangle B_1BA$ is isosceles and thus $\\angle AB_1B = \\angle B_1BA$. The sum of these angles is equal to $\\angle BAC = 2\\beta$, i.e. $\\angle AB_1B = \\angle B_1BA = \\beta$ and $\\angle B_1BC = \\angle B_1BA + \\angle ABC = 2\\beta$.\n\nFrom $\\triangle ABC \\sim \\triangle BB_1C$ we obtain $\\frac{BC}{B_1C} = \\frac{AC}{BC}$, $\\frac{a}{b+c} = \\frac{b}{a}$, so $a^2 = b(b+c)$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 27534,
    "subject": "Mathematics (Multi-modal)",
    "question": "In a circle $k$ is inscribed $ABCD$, for which $S_{ACB} = s$, $S_{ACD} = t$, and $s < t$. Find the minimum of $A = \\frac{4s^2 + t^2}{5st}$, and determine when it is achieved.",
    "options": [],
    "answer": "Minimum value 4/5, achieved when t = 2s (equivalently s/t = 1/2); geometrically, when the perpendicular distance from one vertex to the common side is twice that from the other vertex.",
    "solution": "Since $0 < s < t$, it suffices to find the minimum of $B(y) = \\frac{4}{5}y + \\frac{1}{5y}$, when $0 < y = \\frac{s}{t} < 1$. There are many different approaches for analyzing $B(y)$ in the unit interval that work (e.g., by differentiation). We deduce that the minima of $B(y)$ is $\\frac{4}{5}$, and is achieved for $y = \\frac{1}{2}$. Finally, the minimum of $A = \\frac{4s^2 + t^2}{5st}$ is $\\frac{4}{5}$, achieved for $t = 2s$. Obviously, there exist such geometric configurations (it is enough to choose $B$ and $D$, such that $2d(B, AC) = d(D, AC)$).",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 27535,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABCDV$ be a regular-based right pyramid with apex at $V$. The plane $\\lambda$ intersects the edges $VA$, $VB$, $VC$, and $VD$ at points $M$, $N$, $P$, and $Q$, respectively. Compute the ratio $VQ:QD = p:q$, if $VM:MA = 2:1$, $VN:NB = 1:1$, and $VP:PC = 1:2$.",
    "options": [],
    "answer": "2:3",
    "solution": "Let $l = VA = VB = VC = VD$, $\\angle AVC = 2\\varphi$, and denote by $E$ the intersection point of the altitude $VO$ with $\\lambda$. Since $\\lambda \\cap (ACV) = MP$, $\\lambda \\cap (BDV) = NQ$, $(ACV) \\cap (BDV) = VO$, $E$ will be the intersection point of the diagonals of $MNPQ$ ($E = MP \\cap NQ$). Then $VM = \\frac{2}{3}l$, $VN = \\frac{1}{2}l$, $VP = \\frac{1}{3}l$, $VQ = \\frac{p}{p+q}l$, and $VE$ is an angle bisector for $\\triangle MPV$ and $\\triangle NQV$.\n\nApplying the formula $\\left(l_c = \\frac{2ab \\cos \\frac{\\gamma}{2}}{a+b}\\right)$ we derive\n$$\nVE = \\frac{2 \\cos \\varphi \\cdot \\frac{2}{3} l \\cdot \\frac{1}{3} l}{\\frac{2}{3} l + \\frac{1}{3} l} = \\frac{2 \\cos \\varphi \\cdot \\frac{1}{2} l \\cdot \\frac{p}{p+q} l}{\\frac{1}{2} l + \\frac{p}{p+q} l},\n$$\nthus\n$$\n\\frac{2}{3} \\cdot \\frac{1}{3} = \\frac{\\frac{1}{2} \\cdot \\frac{p}{p+q}}{\\frac{1}{2} + \\frac{p}{p+q}},\n$$\nand conclude that $\\frac{p}{p+q} = \\frac{2}{5}$, i.e., $\\frac{p}{q} = \\frac{2}{3}$.",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 27536,
    "subject": "Mathematics (Multi-modal)",
    "question": "Solve the equation $x^3 - 3x - 1 = 0$.",
    "options": [],
    "answer": "x = 2 cos(π/9), x = 2 cos(7π/9), x = 2 cos(13π/9)",
    "solution": "Let $x = 2 \\cos \\theta$. Then\n$$\nx^3 - 3x = (2 \\cos \\theta)^3 - 3 \\cdot 2 \\cos \\theta = 8 \\cos^3 \\theta - 6 \\cos \\theta.\n$$\nRecall that $\\cos 3\\theta = 4 \\cos^3 \\theta - 3 \\cos \\theta$, so\n$$\n8 \\cos^3 \\theta - 6 \\cos \\theta = 2(4 \\cos^3 \\theta - 3 \\cos \\theta) = 2 \\cos 3\\theta.\n$$\nTherefore, the equation becomes\n$$\n2 \\cos 3\\theta - 1 = 0 \\implies \\cos 3\\theta = \\frac{1}{2}.\n$$\nSo $3\\theta = \\frac{\\pi}{3} + 2k\\pi$ or $3\\theta = \\frac{5\\pi}{3} + 2k\\pi$, $k \\in \\mathbb{Z}$.\nThus,\n$$\n\\theta_1 = \\frac{\\pi}{9}, \\quad \\theta_2 = \\frac{7\\pi}{9}, \\quad \\theta_3 = \\frac{13\\pi}{9}.\n$$\nTherefore, the solutions are\n$$\nx_1 = 2 \\cos \\frac{\\pi}{9}, \\quad x_2 = 2 \\cos \\frac{7\\pi}{9}, \\quad x_3 = 2 \\cos \\frac{13\\pi}{9}.\n$$",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 27537,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $P$ and $Q$ be fixed polynomials with real coefficients, let the degree of $Q$ be $2021$, and let $a_1, a_2, \\dots, a_{2022}, b_1, b_2, \\dots, b_{2022}$ be real numbers, such that $a_1 a_2 \\dots a_{2022} \\neq 0$. If\n$$\nP(a_1Q(x) + b_1) + \\dots + P(a_{2021}Q(x) + b_{2021}) = P(a_{2022}Q(x) + b_{2022}), \\quad \\forall x \\in \\mathbb{R},\n$$\nprove that $P(x)$ possesses at least one real root.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "If $P$ is a constant, then $P \\equiv 0$. If there exist $a, b$, such that $P(a) < 0$ and $P(b) > 0$, due to continuity $P$ will have a real root between $a$ and $b$. If no such $a, b$ exist, then $P$ attains either only positive or only negative values. WLOG $P(x) > 0$ for all $x$. If $a_i \\neq a_{2022}$ for some $i$, due to the odd degree of $Q$, there exists $x_0$, satisfying $Q(x_0) = \\frac{b_{2022}-b_i}{a_i-a_{2022}}$, which gives rise to\n$$\nP(a_1Q(x_0) + b_1) + \\dots + P(a_{i-1}Q(x_0) + b_{i-1}) + P(a_{i+1}Q(x_0) + b_{i+1}) + \\dots + P(a_{2021}Q(x_0) + b_{2021}) = 0.\n$$\n\nA contradiction with $P(x) > 0$. Hence $a_1 = a_2 = \\dots = a_{2022} = a$, meaning that\n$$\nP(aQ(x) + b_1) + \\dots + P(aQ(x) + b_{2021}) = P(aQ(x) + b_{2022}).\n$$\nDenote by $b$ and $c$ the leading coefficients of $P$ and $Q$, respectively, and let $d = \\deg P$. The LHS leading coefficient equals $2021ba^d c^d$, while the RHS leading coefficient is $ba^d c^d$. Since $a, c \\neq 0$, we deduce that $b = 0$, e.g., $P(x) \\equiv 0$.",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 27538,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $P$ be a polynomial with real coefficients and such that for any positive integer $n$ the number $P(n)$ is an integer. There exist distinct prime integers $p_1, p_2, \\dots, p_k$ such that for any positive integer $n$ the number $P(n)$ is divisible by at least one of $p_1, \\dots, p_k$. Prove that there exists $i$ such that $p_i$ is a divisor of $P(n)$ for all integers $n$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "We prove first that $P$ has rational coefficients. Let $d$ be the degree of $P$. Consider the numbers $a_i = P(i)$ for $i = 1, 2, \\dots, d+1$. Let\n$$\nQ(x) = \\sum_{l=1}^{d+1} a_l \\prod_{i \\neq l, 1 \\le i \\le d+1} \\frac{x-i}{l-i}.\n$$\nThen the degree of $Q$ is at most $d$ and $Q(i) = a_i$ for $1 \\le i \\le d+1$ and since $a_l \\in \\mathbb{Z}$ for all $l$ we have that $Q$ has rational coefficients. Since $P(i) = Q(i)$ for $i \\in \\{1, 2, \\dots, d+1\\}$ we conclude that any of $1, 2, \\dots, d+1$ is a root of the polynomial $P(x) - Q(x)$ which is of degree at most $d$. This implies that $P(x) = Q(x)$ thus $P \\in \\mathbb{Q}[X]$.\n\nChoose a positive integer $N$ such that $NP(x) = R(x)$ is a polynomial of integer coefficients and let $N = p_1^{\\alpha_1} \\cdots p_k^{\\alpha_k}$ be where $(N, p_1p_2 \\cdots p_k) = 1$. Assume that there exist positive integers $x_1, x_2, \\dots, x_k$, such that $P(x_1)$ is not divisible by $p_l$ for all $l \\in \\{1, 2, \\dots, k\\}$. It follows from Chinese remainder theorem that there exists a positive integer $x$ such that $x \\equiv x_i \\pmod{p_i^{\\alpha_i+1}}$ for $i = 1, 2, \\dots, k$. Therefore for any $l \\in \\{1, 2, \\dots, k\\}$ we have that $R(x) \\equiv R(x_l) \\pmod{p_l^{\\alpha_l+1}}$, i.e. $R(x)$ is divisible by $p_l^{\\alpha_l}$ but is not divisible by $p_l^{\\alpha_l+1}$. The latter implies that $P(x) = R(x)/N$ is not divisible by $p_l$ for all $l \\in \\{1, 2, \\dots, k\\}$, a contradiction with the condition of the problem. We conclude that there exists $m$ such that $p_m|P(n)$ for all $n \\in \\mathbb{N}$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 27539,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $m$ and $n$ be positive integers, while $p$ is a prime number. Find the maximal $s \\in \\mathbb{N}$ (as a function of $m, n$, and $p$) such that from an arbitrary group of $mnp$ positive integers, one can only choose $snp$ among them, satisfying the following property: The numbers can be split into $s$ disjoint subsets of $np$ elements, such that the sum of the elements in each of the subsets gives the same residue modulo $p$.",
    "options": [],
    "answer": "m - 1",
    "solution": "$s = m - 1$. Assume $s = m$ and consider a set of $mnp - 1$ positive integers, congruent $1 \\pmod{p}$, and $p$. Clearly this set does not fulfill the statement, thus $s \\le m - 1$.\n\n**Lemma.** Among every $np + p - 1$ positive integers, there exist $np$ with sum, divisible by $p$.\n\n*Proof.* We apply induction to $n$. For the base $n = 1$ we have to show, that among every $2p - 1$ positive integers, there exist $p$ with sum, divisible by $p$. Consider residues modulo $p$, and via induction on $k$, prove that for all $p \\ge k \\ge 2$ the set of the residues $S$ of all sums of $k$ elements of an arbitrary $(2k-1)$-element set that does not contain $k$ equal elements, satisfies $|S| \\ge k$. For $k = 2$ the statement is obvious. Assume, we proved it for $k \\le p-1$ and let $s_1, s_2, \\dots, s_k$ be the different residues. Consider a $2k+1$-element set and two different elements in it $a$ and $b$. Apply the inductive hypothesis to the remaining subset (it is clear that we can choose $a$ and $b$ such that, the remaining subset be admissible). Consider the sets $\\{s_1 + a, s_2 + a, \\dots, s_k + a\\}$ and $\\{s_1 + b, s_2 + b, \\dots, s_k + b\\}$. If they do not coincide, we derive $k+1$ different residues and the auxiliary induction is completed. Assuming the sets coincide, we derive after summation that $a \\equiv b \\pmod{p}$ - a contradiction. Hence, we have proven the base $n = 1$ for our major induction. Finally, if we have constructed our $(n-1)p$-element subset with sum, divisible by $p$, then among the remaining $2p-1$ numbers there are $p$ with sum, divisible by $p$, and we can add them to the subset. The lemma is proved.\n\nNow, applying consecutively the lemma, and extracting new $np$ elements with sum $0 \\pmod{p}$ at each step, we will end up with $m-1$ disjoint subsets with sum of the elements $0 \\pmod{p}$, so the proof is completed.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 27540,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $n$ be a positive integer. We call a graph $G$ $n$-good if among any $n$ vertices of $G$ there exist two vertices connected by an edge. Find the least positive integer $N$ such that for any $n$-good connected graph $G$ with $N$ vertices there exists a cycle $C$ with the property:\nAfter deleting all edges of the cycle $C$ the graph $G$ remains connected.",
    "options": [],
    "answer": "3n-2",
    "solution": "We prove first that $N \\ge 3n-2$. Consider a graph with $3n-3$ vertices consisting of $n-1$ triangles with vertices $(u_i, v_i, w_i)$ for $i=1, 2, \\dots, n-1$ and the path $v_1v_2v_3 \\dots v_{n-1}$. It is easy to see that the graph is $n$-good and by deleting any cycle the graph is not connected.\n\nLet $G$ be $n$-good graph having $N = 3n-2$ vertices. First we find a covering tree $T$ of $G$ and color its vertices in two colors. Let $T^{(k)}$ be the tree obtained after adding the first $k$ vertices and $T^{(0)} = \\emptyset$. Set $M_k = G \\setminus T^{(k)}$. To obtain $T^{(1)}$ we choose arbitrary vertex of $G$, color this vertex in black and add it to $T^{(1)}$. To obtain $T^{(2)}$ we choose a vertex that is a neighbour of the vertex of $T^{(1)}$ (such a vertex exists since $G$ is connected), color this vertex in white and add it and its edge to $T^{(1)}$ in $T^{(2)}$. Assume we have constructed $T^{(k)}$ for some $k \\ge 2$. If there are less than $n-1$ white vertices in $T^{(k)}$ in order to obtain $T^{(k+1)}$ we choose a vertex $v$ from $M_k$, that is connected to a white vertex from $T^{(k)}$ (if such a vertex exists) and add this vertex and its edge to a white vertex from $T^{(k)}$ and color $v$ in black. If such a vertex does not exist we choose arbitrary vertex from $M_k$, that is connected to a black vertex, we add it to $T^{(k)}$ and color it in white. If there are $n-1$ white vertices in $T^{(k)}$\n\nthen the condition of the problem implies that from any vertex of $M_k$ there is at least one edge to the white vertex of $T^{(k)}$, because there are no neighbouring white vertices in $T^{(k)}$. Therefore we can construct a covering tree $T^{(N)}$ by adding to $T^k$ exactly one vertex from each of the vertices of $M_k$ to the white vertices in $T^{(k)}$ and coloring every vertex from $M_k$ in black. It is easy to see that at the end we get a covering tree $T^{(N)}$ in which no two vertices of the same color are neighbours and there are at most $n-1$ white vertices. Thus there are at least $2n-1$ black vertices in $T^{(N)}$. Consider the subgraph $H$ of $G$ from all black vertices of $T^{(N)}$. Let $H$ contains $m$ connected components $S_1, S_2, \\dots, S_m$. Consider covering trees $T_1, T_2, \\dots, T_m$ of $S_1, S_2, \\dots, S_m$ respectively. Color the vertices in each of $T_1, \\dots, T_m$ in blue and red such that no two vertices of the same color are connected. It follows that at least $n$ vertices of $H$ are of the same color. By the condition of the problem there is an edge in these $n$ vertices. Therefore there exists $s \\in \\{1, \\dots, m\\}$ such that there is at least an edge in $S_s$ that is not in $T_s$ (because the vertices of the same color are not connected in $T_1, T_2, \\dots, T_m$). This implies that there is a cycle in $S_s$. If we delete the edges of this cycle there are no edges from $T^{(N)}$ because no two vertices from $H$ are connected in $T^{(N)}$. Thus, after deleting the edges of this cycle $G$ still contains $T^{(N)}$, i.e. it remains connected.",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 27541,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABC$ be an acute triangle and $M$ is the midpoint of $AB$. A circle through the vertices $B$ and $C$ intersects the segments $CM$ and $BM$ in points $P$ and $Q$ respectively. Let $K$ be symmetrical point to $P$ regarding $M$. Let circumcircles of $\\triangle AKM$ and $\\triangle CQM$ meet again in point $X$, and circumcircles of $\\triangle AMC$ and $\\triangle KMQ$ meet again in point $Y$. The segments $BP$ and $CQ$ meet in $T$. Prove that $MT$ is tangent to the circumcircle of $\\triangle MXY$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Obviously $AKBP$ is paralellogram and from quadrilateral $BQPC$ is inscribed follows $\\triangle AKC = \\triangle KPB = \\triangle AQC = \\varphi$. Hence $AKQC$ is inscribed quadrilateral with center $O$. Then $\\triangle AYC = (180^\\circ - \\triangle AYM) + (180^\\circ - \\triangle CYM) = \\triangle AKM + \\triangle CQM = 2\\varphi = \\triangle AOC$ and so $AOYC$ is inscribed. Therefore $\\triangle OYM = \\triangle CYM - \\triangle CYO = (360^\\circ - \\triangle AYC - \\triangle AYM) - \\triangle CAO = 180^\\circ - \\varphi - (90^\\circ - \\varphi) = 90^\\circ$ and analogously $\\triangle OXM = 90^\\circ$. If $TM \\cap AK = T'$ then $TM = MT'$ from paralellogram $AKBP$ and the reverse butterfly theorem for $AKQC$ gives us $OM \\perp TT'$ and $OM \\perp MT$. Hence $MT$ is perpendicular to the diameter $OM$ of circumcircle of $\\triangle MXY$.\n\n![](attached_image_1.png)",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 27542,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $x > y > 2022$ are natural numbers such that $xy + x + y$ is a perfect square. Is it possible for each natural number $z$ in the interval $[x + 3y + 1, 3x + y + 1]$ the numbers $x + y + z$ and $x^2 + xy + y^2$ to not be coprimes?",
    "options": [],
    "answer": "No",
    "solution": "We will prove that it's not possible. Let $k = \\sqrt{xy + x + y}$ and $z = x + y + 2k + 1$. We have\n$$\ny < \\sqrt{y^2 + 2y} \\le \\sqrt{xy + x + y} = k \\le \\sqrt{x^2 + 2x} < x + 1.\n$$\nHence $k \\in (y, x]$ and $z \\in (x + 3y + 1, 3x + y + 1]$. It's easy to check that $xy + yz + zx = (x + y + k)^2$ and $xy + yz + zx + x + y + z = (x + y + k + 1)^2$. Let assume that there exists a prime number $p$, such that $p\\mid(x + y + z, x^2 + xy + y^2)$. Then $x + y \\equiv -z \\pmod p$ and so $x^2 + xy + y^2 = (x + y)^2 - xy \\equiv -xy - yz - zx \\pmod p$. Hence $p\\mid xy + yz + zx$ and $p\\mid x + y + z$, and therefore $p\\mid(x + y + k)^2$ and $p\\mid(x + y + k + 1)^2$, which is impossible.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 27543,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $n \\ge 4$ be an integer and $x_1, x_2, \\dots, x_n, x_{n+1}, x_{n+2}$ are reals, such that $x_{n+1} = x_1$ and $x_{n+2} = x_2$. Given that there exists a positive real $a$, such that $x_i^2 = a + x_{i+1}x_{i+2}$ for all $i = 1, 2, \\dots, n$. Prove that at least two of the numbers $x_1, x_2, \\dots, x_n$ are negative.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Notice that $x_i^2 x_{i+1} = a x_{i+1} + x_{i+1}^2 x_{i+2}$ holds for all $i = 1, 2, \\dots, n$. After summing these equations we get $\\sum_{i=1}^n x_i^2 x_{i+1} = a \\sum_{i=1}^n x_i + \\sum_{i=1}^n x_i^2 x_{i+1}$ and therefore $\\sum_{i=1}^n x_i = 0$. Since $a > 0$ at least one of the numbers $x_i$ is not equal to $0$ and therefore at least one of them is negative. Assume that there is at least one negative and let it be $x_1$ without loss of generality. So $x_2^2 = a + x_3 x_4 \\ge a$ and $|x_1| = x_2 + \\dots + x_n$. Hence we get that $(x_2 + \\dots + x_n)^2 = x_1^2 = a + x_2 x_3$ and therefore $x_2^2 - a + \\sum_{i=3}^n x_i^2 + x_2 x_3 \\le 0$. So $x_2 = \\sqrt{a}$ and $x_3 = \\dots = x_n = 0$. Now, after substituting in the equality of the condition for $i = 3$, we get $0 = x_3^2 = a + x_4 x_5 = a$, which is contradiction. Therefore, at least two of the numbers are negative.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 27544,
    "subject": "Mathematics (Multi-modal)",
    "question": "An isosceles triangle $ABC$ with sides $AB = 4$, $BC = AC = 6$ is given.\nThe points $X_1, X_2, X_3, \\dots$ in that order are taken on side $AB$, such that the lengths of segments $AX_1, X_1X_2, X_2X_3, \\dots$ form an infinite geometric series with first term $3$ and common ratio $1/4$.\nThe points $Y_1, Y_2, Y_3, \\dots$ in that order are taken on side $BC$, such that the lengths of segments $CY_1, Y_1Y_2, Y_2Y_3, \\dots$ form an infinite geometric series with first term $3$ and common ratio $1/2$.\nThe points $Z_1, Z_2, Z_3, \\dots$ in that order are taken on side $BC$, such that the lengths of segments $AZ_1, Z_1Z_2, Z_2Z_3, \\dots$ form an infinite geometric series with the first term $3$ and common ratio $1/2$.\nFind all triples $(a, b, c)$ of natural numbers for which the line segments $AY_a$, $BZ_b$ and $CX_c$ are concurrent.",
    "options": [],
    "answer": "(1, 2k, k) and (2k, 1, k) for k in natural numbers",
    "solution": "Firstly we will prove the following:\n**Lemma.** Let $n > 1$ be a natural number. On the line segment $PQ$ with length $\\frac{a_1}{1 - \\frac{1}{n}}$ are taken the points $T_1, T_2, \\dots$, such that $PT_1, T_1T_2, T_2T_3, \\dots$ form an infinite geometric series with first term $a_1$ and common ratio $\\frac{1}{n}$. Then $\\frac{PT_s}{T_sQ} = n^s - 1$.\n\n$$\n\\textit{Proof.} \\text{ From } PT_s = \\frac{a_1(1 - (\\frac{1}{n})^s)}{1 - \\frac{1}{n}} \\text{ and } PQ = \\frac{a_1}{1 - \\frac{1}{n}} \\text{ we get } \\frac{\\frac{a_1(1 - (\\frac{1}{n})^s)}{1 - \\frac{1}{n}}}{T_sQ} = \\frac{a_1}{1 - \\frac{1}{n} - \\frac{a_1(1 - (\\frac{1}{n})^s)}{1 - \\frac{1}{n}}} = n^s - 1\n$$\n\nFrom the lemma above and Ceva's theorem we get\n$$\n1 = \\frac{AX_c}{X_c B} \\cdot \\frac{BY_a}{Y_a C} \\cdot \\frac{CZ_b}{Z_b A} = \\frac{4^c - 1}{(2^a - 1)(2^b - 1)}\n$$\nand therefore $4^c - 1 = (2^a - 1)(2^b - 1) \\Leftrightarrow 4^c = 2^{a+b} - 2^a - 2^b + 2$. But $a, b, c \\ge 1$ and if $a \\le b$ (without loss of generality) it follows $a = 1$. Then $4^c = 2^b$, $b = 2c$ and so $(a, b, c) = (1, 2k, k)$ or $(a, b, c) = (2k, 1, k)$ for $k \\in \\mathbb{N}$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 27545,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $n \\ge 2$ be a natural number. The sets $A_1, \\dots, A_n$ and $B_1, \\dots, B_n$ of natural numbers satisfy the properties:\n\n* $A_i \\cap B_j \\neq \\emptyset$ for all $i, j \\in \\{1, 2, \\dots, n\\}$;\n* $A_i \\cap A_j = \\emptyset$ and $B_i \\cap B_j = \\emptyset$ for all $i \\neq j \\in \\{1, 2, \\dots, n\\}$.\n\nFor each of the sets we arrange its elements in descending order and compute the largest difference between two adjacent elements in the result ordering. Find the smallest possible value of the largest among these differences.",
    "options": [],
    "answer": "n",
    "solution": "We will prove that the required smallest possible value is $n$. Let $A_i$-s are the rows and $B_j$-s are the columns of a square table $n \\times n$ in which the value in row $i$ and column $j$ is $(n-1)i + j$. So all differences for the $A_i$-s are equal to $1$ and all differences for the $B_j$-s are equal to $n$.\n\nLet $a$ be the smallest natural number such that the set $\\{1, 2, \\dots, a\\}$ contains some of the considered $2n$ sets, $A_1$ for example. Without loss of generality, we can assume that $a \\in B_1$ or $a \\notin B_j$ for all $j$. Consider the numbers $a+1, a+2, \\dots, a+n-2$ (which are at most $n-2$) and the sets $B_2, B_3, \\dots, B_n$ (which are $n-1$). It follows from Dirichlet's principle that at least one of them, for example $B_2$, does not contain any of these numbers. Since $a \\notin B_2$ (because either $a \\in B_1$ and $B_1 \\cap B_2 = \\emptyset$, or $a \\notin B_j$ for all $j$) $B_2$ contains at least one number $x \\ge a+n-1$ (otherwise the set $\\{1, 2, \\dots, a-1\\}$ contains $B_2$, which is contradiction with the minimality of $a$).\nLet $b_1 > b_2 > \\dots > b_m$ are the elements of $B_2$ and mark the elements of $B_2 \\cap A_1$. Consider $x \\ge a+n-1$, such that $b_j$ is a marked number and the number of elements (in the ordering of $B_2$) between $x$ and $b_j$ is minimal. For $x \\ge b_j$ (the case $x < b_j$ is analogous) we consider the largest $b_k \\le x$ of $B_2$ (not necessarily marked; $k=j$ is allowed). Obviously $b_k < a+n-1$ (otherwise we have a contradiction because the number of elements between $b_k$ and $b_j$ is less than between $x$ and $b_j$) and because $a+1, a+2, \\dots, a+n-2$ are not in $B_2$, and also $a \\notin B_2$ from above, we get $b_k \\le a-1$. Hence $x$ and $b_k$ are adjacent elements in $B_2$ with a difference of at least $a+n-1-(a-1)=n$, as desired.",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 27546,
    "subject": "Mathematics (Multi-modal)",
    "question": "Factorize the polynomial $P = (x^4 - 40x^2 + 144)(x^3 - 16x)$ and compute $\\gcd(P(10), P(91))$.",
    "options": [],
    "answer": "315",
    "solution": "$P = (x^4 - 36x^2 - 4x^2 + 144)(x^3 - 16x) = (x^2 - 36)(x^2 - 4)x(x^2 - 16) = (x-6)(x+6)(x-2)(x+2)x(x-4)(x+4)$.\n\nDenote by $d = \\gcd(P(10), P(91))$. $P(10)$ equals $4 \\cdot 6 \\cdot 8 \\cdot 10 \\cdot 12 \\cdot 14 \\cdot 16 = 2^{14} \\cdot 3^2 \\cdot 5 \\cdot 7$, while $P(91)$ is odd, thus $d \\le 3^2 \\cdot 5 \\cdot 7 = 315$. Since $91 - 4 = 87$ as well as $91 + 2 = 93$ are divisible by $3$, $91 - 6 = 85$ is divisible by $5$, and $91$ is divisible by $7$, then $315$ divides $P(91)$. Hence, $d = 3^2 \\cdot 5 \\cdot 7 = 315$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 27547,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find $x^2 + p = qx$ where\n$$\np = (32 + \\sqrt{2})(9 - \\sqrt{8})^{-1}, \\quad q = \\sqrt{9 + \\sqrt{8} - \\sqrt{24} - \\sqrt{48}} + \\sqrt{3 - \\sqrt{8}} + \\sqrt{55 - 14\\sqrt{6}}\n$$",
    "options": [],
    "answer": "x = 2 + sqrt(2) or x = 3 - sqrt(2)",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 27548,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let the $\\triangle ABC$ with $AB = 1\\ \\text{cm}$, $BC = 2\\ \\text{cm}$, and $AC = \\sqrt{3}\\ \\text{cm}$ be given. The points $D, E$, and $F$ lie on the sides $AB, AC$, and $BC$, respectively, satisfying $AE = BD$ and $BF = AD$. The angle bisector of $\\triangle BAC$ intersects for the second time the circle through $A, D$, and $E$ at $M$, while the angle bisector of $\\triangle ABC$ intersects for the second time the circle through $B, D$, and $F$ at $N$. Find the length of the line segment $MN$.",
    "options": [],
    "answer": "(3 - sqrt(3)) / 6",
    "solution": "From the circumscribed circle of $\\triangle ADE$ we derive (using inscribed angles and their corresponding arcs) $MD = ME$ and $\\angle BDM = 180^\\circ - \\angle ADM = \\angle AEM$, which combined with $AE = BD$ gives rise to $\\triangle AEM \\cong \\triangle BDM$ - thus $AM = MB$, meaning that $M$ is the intersection point of the segment bisector of $AB$ and the angle bisector of $\\triangle BAC$. Analogously, $N$ is the intersection point of the segment bisector of $AB$ and the angle bisector of $\\triangle ABC$. In particular, we deduce that $MN$ is the segment bisector of $AB$. Let $MN$ intersect $AB$ at its midpoint $K$. Since $AB^2 + AC^2 = BC^2$, it follows that $\\angle BAC = 90^\\circ$. Now $BC = 2AB$ gives rise to $\\angle ABC = 60^\\circ$. Hence, $\\angle MAK = 45^\\circ$ and $\\angle NBK = 30^\\circ$. In conclusion, from $\\triangle AMK$ we deduce $MK = AK = \\frac{1}{2}$, while from $\\triangle BNK$ we deduce $NK = \\frac{1}{2\\sqrt{3}}$ (indeed, if $NK = y$, then $BN = 2y$ since $\\angle NBK = 30^\\circ$ and $y^2 + (\\frac{1}{2})^2 = 4y^2$, i.e. $y = \\frac{1}{2\\sqrt{3}}$). Finally, $MN = MK - NK = \\frac{3-\\sqrt{3}}{6}$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 27549,
    "subject": "Mathematics (Multi-modal)",
    "question": "For a triangle $ABC$ with $\\angle ABC = 50^\\circ$ points $M$ and $N$ are midpoints of $AC$ and $BC$, respectively. Point $P$ is the midpoint of $BM$ and point $Q$ is the midpoint of $CM$. It is known that there exists a right triangle with side lengths $CP$, $BQ$ and $MN$. Find the angle between lines $CP$ and $AB$.",
    "options": [],
    "answer": "65°",
    "solution": "",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 27550,
    "subject": "Mathematics (Multi-modal)",
    "question": "a) $\\left(\\frac{2a}{b+c}\\right)^2 + \\left(\\frac{2b}{c+a}\\right)^2 + \\left(\\frac{2c}{a+b}\\right)^2 \\ge \\frac{a}{c} + \\frac{b}{a} + \\frac{c}{b}$.\n\nb) $\\left(\\frac{a+b}{c}\\right)^2 + \\left(\\frac{b+c}{a}\\right)^2 + \\left(\\frac{c+a}{b}\\right)^2 \\ge \\frac{a}{b} + \\frac{b}{c} + \\frac{c}{a} + 9$.\n\nFor each of them either prove that it holds true for all positive real numbers $a, b$, and $c$, or provide a triple $(a, b, c) \\in \\mathbb{R}^{+}$, which does not satisfy it.",
    "options": [],
    "answer": "a) Counterexample: (a, b, c) = (0.01, 1000, 1000). b) The inequality holds for all positive real numbers.",
    "solution": "a) The triple $(a, b, c) = (0.01, 1000, 1000)$ does not satisfy the inequality.\n\nb) We will prove that the inequality holds true for all positive real numbers $a, b$, and $c$. Denote by $x = \\frac{a}{b}$, $y = \\frac{b}{c}$, and $z = \\frac{c}{a}$ - then $xyz = 1$ and $(xy + y)^2 + (yz + z)^2 + (zx + x)^2 \\ge x + y + z + 9$. This is equivalent to $x^2 + y^2 + z^2 + 2(xy^2 + yz^2 + zx^2) + x^2y^2 + y^2z^2 + z^2x^2 \\ge x + y + z + 9$. Applying AM-GM, we get\n\n$xy^2 + yz^2 + zx^2 \\ge 3\\sqrt[3]{x^3y^3z^3} = 3$ and $x^2y^2 + y^2z^2 + z^2x^2 \\ge 3\\sqrt[3]{x^4y^4z^4} = 3$. Thus, it suffices to prove that $x^2 + y^2 + z^2 \\ge x + y + z$. But $x^2 + y^2 + z^2 \\ge \\frac{(x+y+z)^2}{3}$ (it is equivalent to $(x-y)^2 + (y-z)^2 + (z-x)^2 \\ge 0$), so we need $x + y + z \\ge 3$, which follows from the AM-GM inequality $x + y + z \\ge 3\\sqrt{xyz} = 3$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 27551,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all pairs $(a, b)$ of integers such that $a = a^3 - 8a^2b + 21ab^2 - 18b^3$.",
    "options": [],
    "answer": "[[0, 0], [1, 0], [-1, 0], [9, 4], [-9, -4], [16, 6], [-16, -6]]",
    "solution": "Answer: $(0, 0)$, $(1, 0)$, $(-1, 0)$, $(9, 4)$, $(-9, -4)$, $(16, 6)$ and $(-16, -6)$. The right hand side can be written in the form $(a - 2b)(a - 3b)^2$. By setting $n = a - 3b$ we obtain $3a = n^2(3a - 6b) = n^2(2n + a)$, and thus $(3 - n^2)a = 2n^3$. When $n = 0$ it follows that $a = 0$ and $b = 0$; when $n = \\pm 1$ we obtain $a = \\pm 1$ and $b = 0$; when $n = \\pm 2$ we obtain $a = \\mp 16$ and $b = \\mp 6$; when $n = \\pm 3$ we obtain $a = \\mp 9$ and $b = \\mp 4$; when $n = \\pm 4$ the left hand side is divisible by $13$ but the right hand side is not; when $n = \\pm 5, \\pm 6$ the left hand side is divisible by $11$ but the right hand side is not. On other hand for $|n| \\ge 7$ since $3 - n^2 \\ne 0$,\n$$\na = \\frac{2n^3}{3 - n^2} = \\frac{6n}{3 - n^2} - 2n\n$$\nand $n^2 - 3 > 6|n|$ (it is equivalent to $|n|(|n| - 6) > 3$), the number $a$ is not an integer.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 27552,
    "subject": "Mathematics (Multi-modal)",
    "question": "Equilateral triangle of area $n^2$ is partitioned into $n^2$ small triangles with unit area with lines parallel to its sides. The vertices of all small triangles are called *knots*. Find the sum of the areas of all equilateral triangles with vertices knots as a polynomial of $n$ and write this polynomial as a product of irreducible polynomials.",
    "options": [],
    "answer": "n(n+1)(n+2)(n+3)(n^2+3n+6)/240",
    "solution": "Answer: $\\frac{n(n+1)(n+2)(n+3)(n^2+3n+6)}{240}$.\n\nAny equilateral triangle $T$ under consideration can be embedded into an unique equilateral triangle $M$, such that the vertices of $T$ lie on the sides of $M$ and those sides are parallel to the sides of the bigger triangle. If $M$ has $k$ times bigger side than the side of a unit triangle ($k = 1, 2, \\dots, n$) then there are $1 + 2 + \\dots + (n+1-k) = \\binom{n+2-k}{2}$ choices of $M$. We find the sum of the areas of all $T$ by enumerating the knots of one of the sides of $M$ with $0, 1, \\dots, k$ and realizing that there is exactly one equilateral triangle with vertex at $i$-th knot for each $i = 1, 2, \\dots, k$ and the area of this triangle is $k^2 - 3i(k-i)$. Thus the sum of the areas of all such $T$ equals\n$$\n\\sum_{i=1}^{k} (k^2 - 3i(k - i)) = k^3 - 3\\binom{k+1}{3} = \\frac{k^3 + k}{2}.\n$$\nWe used the well known equality $\\sum_{i=1}^{k} i(k - i) = \\binom{k+1}{3}$. Therefore the desired sum equals:\n$$\n\\sum_{k=1}^{n} \\frac{k^3 + k}{2} \\binom{n+2-k}{2} = \\frac{1}{2} \\sum_{k=1}^{n} k^3 \\binom{n+2-k}{2} + \\frac{1}{2} \\sum_{k=1}^{n} k \\binom{n+2-k}{2}.\n$$\nTo find the second sum count the number of words with 4 letters \"a\" and $n-1$ letters \"b\" such that the second letter \"a\" appears on $(k+1)$-th place. There are $k$ choices for the first \"a\" and $n+3-(k+1) = n+2-k$ choices for the last two letters \"a\". Therefore $\\sum_{k=1}^{n} k \\binom{n+2-k}{2} = \\binom{n+3}{4}$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 27553,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $p = (a_1; a_2; \\dots; a_{12})$ be a permutation of the numbers $1, 2, \\dots, 12$. Denote by $S_p = |a_1 - a_2| + |a_2 - a_3| + \\dots + |a_{11} - a_{12}|$. We will call $p$ \"optimistic\", if $a_i > \\min(a_{i-1}; a_{i+1})$ for every $i = 2, \\dots, 11$.\n\na) Determine the maximal possible value of $S_p$? For how many permutations $p$ the maximum is achieved?\n\nb) Find the number of all optimistic $p$?\n\nc) Determine the maximal possible value of $S_p$ for an optimistic $p$? For how many optimistic $p$ the maximum is achieved?",
    "options": [],
    "answer": "a) Maximum S_p is 71; achieved by 28800 permutations. b) Number of optimistic permutations is 2048. c) For optimistic permutations, maximum S_p is 21; achieved by 1024 permutations.",
    "solution": "a) Place the elements of $p$ along the real axis. Then $S_p$ equals the length of the \"walk\", starting from $a_1$, then reaching $a_2$, then reaching $a_3$ and so on until we end at $a_{12}$. In such a walk, the segments:\n\n$1$-$2$ and $11$-$12$ appear at most twice (before and after $1$; before and after $12$);\n\n$2$-$3$ and $10$-$11$ appear at most $4$ times (before and after $1$, $2$, $11$, and $12$);\n\n$3$-$4$ and $9$-$10$ appear at most $6$ times (before and after $1$, $2$, $3$, $10$, $11$, and $12$);\n\n$4$-$5$ and $8$-$9$ appear at most $8$ times (before and after $1$, $2$, $3$, $4$, $9$, $10$, $11$, and $12$);\n\n$5$-$6$ and $7$-$8$ appear at most $10$ times (before and after $1$, $2$, $3$, $4$, $5$, $8$, $9$, $10$, $11$, $12$);\n\n$6$-$7$ may appear in all $11$ paths $a_i-a_{i+1}$.\n\nTherefore, $S_p \\le 2(2+4+6+8+10)+11 = 2 \\cdot 30+11 = 71$. Furthermore $S_p = 71$ iff all the inequalities above become equalities. This is possible only if $\\{a_1, a_{12}\\} = \\{6, 7\\}$ (2 cases). For the case $a_1 = 6$, $(a_2, a_4, \\dots, a_{10})$ must be a permutation of $\\{8, 9, \\dots, 12\\}$ ($5!$ cases) and $(a_3, a_5, \\dots, a_{11})$ must be a permutation of $\\{1, 2, \\dots, 5\\}$ ($5!$ cases), while in the case $a_1 = 7$ it is vice versa. Therefore, the total number of permutations with $S_p = 71$ is $2 \\cdot 5! \\cdot 5! = 28800$.\n\nb) In order for $p$ to be an optimistic permutation, the number $1$ should be at one of the end positions. Then, the number $2$ should be at one of the end positions of the remaining $p$-sub-permutation of $\\{2, 3, \\dots, 12\\}$, $3$ should be at one of the end positions of the remaining $p$-sub-permutation of $\\{3, \\dots, 12\\}$ and so on. It is easy to check that all such permutations are indeed optimistic. Therefore, their number is $2^{11} = 2048$.\n\nc) Let $p$ be an optimistic permutation and $k \\in \\{1, 2, \\dots, 12\\}$ be the position of $12$ in it (i.e., $a_k = 12$). Then, it is trivial to check that $a_i > a_{i-1}$, for all $i = 2, \\dots, k$, while $a_j > a_{j+1}$, for all $j = k, \\dots, 11$. Thus, $S_p = 12 - a_1 + 12 - a_{12} \\le 2 \\cdot 12 - 1 - 2 = 21$. Equality is achieved iff $\\{a_1, a_{12}\\} = \\{1, 2\\}$. According to b), the total number of such permutations is $2^{11}/2 = 2^{10} = 1024$.",
    "topic": "Discrete Mathematics",
    "subtopic": "Combinatorics"
  },
  {
    "id": 27554,
    "subject": "Mathematics (Multi-modal)",
    "question": "Consider a quadratic polynomial $f(x)$ with integer coefficients. If the values $f(0)$, $f(3)$, and $f(4)$ are pairwise different numbers from the set $\\{2, 20, 202, 2022\\}$, then determine all possible values for $f(1)$.",
    "options": [],
    "answer": "{-80, -990}",
    "solution": "$f(1) \\in \\{-80, -990\\}$.\nSince $f \\in \\mathbb{Z}[x]$, we have that $y - x \\mid f(y) - f(x)$, $\\forall x, y \\in \\mathbb{Z}$. Thus, $3 \\mid |f(3) - f(0)|$, and $4 \\mid |f(4) - f(0)|$. In the set $\\{2, 20, 202, 2022\\}$, only $2$ and $20$ are congruent modulo $3$, so $\\{f(0), f(3)\\} = \\{2, 20\\}$. Modulo $4$, $20$ has a different residue than the other three, so $f(0) \\ne 20$. We conclude that $f(0) = 2$, $f(3) = 20$, while $f(4) = 202$ or $f(4) = 2022$.\nLet $f(x) = ax^2 + bx + c$. From $c = f(0) = 2$ and $9a + 3b + 2 = f(3) = 20$, we deduce that $f(x) = ax^2 + (6 - 3a)x + 2$. Moreover $f(1) = 8 - 2a$, so $f \\in \\mathbb{Z}[x]$ whenever $a \\in \\mathbb{Z}$, due to\n$$\na = \\frac{f(4) - f(0)}{4} - \\frac{f(3) - f(0)}{3}.\n$$\nCase 1: $f(4) = 202$. Then $a = \\frac{202 - 2}{4} - \\frac{20 - 2}{3} = \\frac{200}{4} - \\frac{18}{3} = 50 - 6 = 44$ and $f(1) = 8 - 2a = 8 - 88 = -80$.\n\nCase 2: $f(4) = 2022$. Then $a = \\frac{2022 - 2}{4} - \\frac{20 - 2}{3} = \\frac{2020}{4} - \\frac{18}{3} = 505 - 6 = 499$ and $f(1) = 8 - 2a = 8 - 998 = -990$.",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 27555,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all values of the real parameter $a$ for which the solution of the system of simultaneous equations\n$$\n\\begin{cases} |x-2| + |x-1| + |x+3| \\le 6 \\\\ x^2 - a x + 2a \\le 0 \\end{cases}\n$$\nis a closed interval of length 1.",
    "options": [],
    "answer": "-1",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 27556,
    "subject": "Mathematics (Multi-modal)",
    "question": "In an acute triangle $ABC$, point $M$ is the midpoint of $AB$ and $AH$ is the altitude. Let $CP$ be the perpendicular to the line $MH$. If $AB = 21$, $BH = 7$ and $BP = CP$ find the length of $AC$.",
    "options": [],
    "answer": "√473",
    "solution": "Answer: $AC = \\sqrt{473}$.\n\nLet $PE \\perp BC$ and denote $\\angle ABC = \\beta$. Since $HM$ is a median in the right triangle we have $\\angle MHA = 90^\\circ - \\beta$, $\\angle PHC = \\beta$ and $\\angle PCH = 90^\\circ - \\beta$. Therefore $\\triangle ABH \\sim \\triangle PHE \\sim \\triangle CPH$ and thus\n$$\n\\frac{7}{21} = \\frac{BH}{AB} = \\frac{EH}{HP} = \\frac{PH}{CH}. \\quad (1)\n$$\nIf $EH = x$ then $PH = 3x$, $CH = 9x$, $CE = 9x - x = 8x$ and using that $BP = CP$ we obtain $BE = 8x$ and $BH = 7x$, i.e. $x = 1$. Finally: $AC = \\sqrt{AB^2 - BH^2 + CH^2} = \\sqrt{21^2 - 7^2 + 9^2} = \\sqrt{473}$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 27557,
    "subject": "Mathematics (Multi-modal)",
    "question": "Consider a $\\triangle ABC$ with integer sides, a median $CM$ ($M \\in AB$), and a circumscribed center $O$. If the circumcircle of $AMO$ passes through the midpoint of $CM$, find the smallest possible value for the perimeter of $ABC$.",
    "options": [],
    "answer": "24",
    "solution": "Let $N$ and $P$ be the midpoints of $AC$ and $CM$, respectively. Then the pentagon $ANPOM$ is cyclic with $AM \\parallel PN$, meaning that $\\triangle CAM = 180^\\circ - \\angle ANP = 90^\\circ - \\angle PNO = 90^\\circ - \\angle PMO = \\angle AMC$,\n\nwhich is equivalent to $AC = CM$. Let $Q$ be the midpoint of $AM$ (clearly $CQ \\perp AB$). Denote $AC = b$, $BC = a$, $AB = c$. We have $CM^2 - MQ^2 = CQ^2 = BC^2 - BQ^2$, i.e., $b^2 - \\frac{c^2}{16} = a^2 - \\frac{9c^2}{16}$, thus $2(a^2 - b^2) = c^2$. Obviously $c = 2k$ and $(a-b)(a+b) = 2k^2$, with the same parity of the factors in LHS. Therefore, $k$ should be even. If $k = 2$ then $a-b=2$, $a+b=4$, i.e., $a=3$, $b=1$, which together with $c=4$ fails the triangle inequality. When $k=4$, we have either $a-b=2$, $a+b=16$ (i.e., $a=9$, $b=7$, $c=8$, a valid configuration with a perimeter 24) or $a-b=4$, $a+b=8$ (i.e., $a=6$, $b=2$, again a non-valid configuration). For $k \\ge 6$, we derive $c \\ge 12$ and $a+b+c > 2c \\ge 24$ from the triangle inequality. Thus, the answer is 24, and $AB = 8$, $AC = 7$, $BC = 9$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 27558,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find the number of all positive integers $4 \\le n \\le 2022$ that are not primes such that for any positive integer $k$ in the interval $[1, \\sqrt{n}-1]$ the following holds: the number of ways to choose $k$ persons from a group of $n$ people is divisible by $n$.",
    "options": [],
    "answer": "22",
    "solution": "The condition of the problem requires $\\binom{n}{k}$ to be divisible by $n$ for all $1 \\le k \\le \\sqrt{n}-1$. Assume there exists a prime divisor $p$ of $n$ such that $p \\le \\sqrt{n}-1$ and set $k = p$. Since\n$$\np! \\binom{n}{p} = n(n-1)(n-2)\\cdots(n-p+1)\n$$\nand $\\binom{n}{p}$ is divisible by $n$, we have that $p$ is a divisor of $(n-1)(n-2)\\cdots(n-p+1)$. This assertion is not true since $p$ is a factor of $n$ and thus it is not a divisor of any of $n-\\ell$, $1 \\le \\ell \\le p-1$. From now on we assume that all prime factors of $n$ are greater than $\\sqrt{n}-1$. Assume $n$ has at least three prime factors (counting $p^\\alpha$ as $\\alpha$). Direct verification shows that $n=8$ is a solution and for $n=12$ we have factor $2 < \\sqrt{12}-1$. For $n \\ge 16$ we obtain\n$n > (\\sqrt{n}-1)^3 \\ge (\\frac{3\\sqrt{n}}{4})^3 = \\frac{27n^{3/2}}{64}$, i.e. $n < (\\frac{64}{27})^2 < 9$, a contradiction. It remains to consider $n = pq$ for primes $p$ and $q$ such that $\\sqrt{n}-1 < p \\le q$. If $q \\ge p+3$ then $n \\ge p(p+3) > (\\sqrt{n}-1)(\\sqrt{n}+2) = n + \\sqrt{n}-2 > n$, a contradiction. If $q = p+1$ we have $p=2, q=3$, i.e. $n=6$ which is a solution. If $q = p$ and $1 \\le k \\le p-1$ then the left hand side of\n$$\nk! \\binom{p^2}{k} = (p^2 - k + 1)(p^2 - k + 2) \\cdots (p^2 - 1)p^2\n$$\nis divisible by $p^2$ and $k!$ is not divisible by $p$. Therefore $p^2$ is a factor of $\\binom{p^2}{k}$. If $q = p+2$ and $1 \\le k \\le p-1$ then the left hand side of\n$$\nk! \\binom{p^2+2p}{k} = (p^2 + 2p - k + 1)(p^2 + 2p - k + 2) \\cdots (p^2 + 2p - 1)(p^2 + 2p)\n$$\nis divisible by both $p$ and $p+2$. Since $k!$ is not divisible by both $p$ and $q = p+2$ we infer that $\\binom{p^2+2p}{k}$ is divisible by $p$ and $p+2$, i.e. it is divisible by $p(p+2)$. Since $2022 < 2025 = 45^2$ all primes $2 \\le p < 45$ imply the solution $n = p^2$. The primes for which $p+2$ is also a prime imply also the solution $n = p(p+2)$. Direct verification shows that there exist 14 primes in this interval: $\\{2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43\\}$, and for 6 of them ($\\{3, 5, 11, 17, 29, 41\\}$) $p+2$ is also a prime. Finally, the desired number is (adding 6 and 8): $14+6+2=22$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 27559,
    "subject": "Mathematics (Multi-modal)",
    "question": "Determine all prime numbers $p$, for which there exist positive integers $x$ and $y$, such that\n$$\n\\left| \\begin{array}{l} p + 49 = 2x^2 \\\\ p^2 + 49 = 2y^2 \\end{array} \\right. .\n$$",
    "options": [],
    "answer": "23",
    "solution": "We will prove that $p = 23$ is the unique solution. Subtracting the second equality from the first one, we get $p(p-1) = 2(y-x)(y+x)$. From the first equality $p$ is odd, thus $p \\ne 2$ and $p \\mid (y-x)(y+x)$. Clearly $y > x$, so assuming $p \\mid (y-x)$, we get $p \\le y-x < y+x$. Therefore,\n$$\n2(y-x)(y+x) > 2p^2 > p^2 > p(p-1) \\text{ - a contradiction.} \\quad \\implies \\quad p \\mid (y+x).\n$$\n\nAssuming $y \\ge p$, from the second equation it follows $49 = 2y^2 - p^2 \\ge p^2$, i.e., $p \\le 7$. Since $26 = \\frac{3+49}{2}$, $27 = \\frac{5+49}{2}$, $28 = \\frac{7+49}{2}$ are not perfect squares, none of those values is a solution. For the remaining case $y < p$, we have $x+y < 2y < 2p$, thus $x+y = p$. So $y-x = p-2x$ and\n$$\np-1 = 2(y-x) = 2(p-2x) \\Leftrightarrow p = 4x-1 \\Leftrightarrow 2x^2 - 4x - 48 = 0 \\Leftrightarrow x_{1,2} = 1 \\pm 5.\n$$\n\nFinally, since $-4 < 0$, the only solution is $x = 6$, $p = 4 \\cdot 6 - 1 = 23$ and $y = p - x = 17$. Direct check confirms that this is indeed a solution.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 27560,
    "subject": "Mathematics (Multi-modal)",
    "question": "At an IMO prep camp there are 14 participants. Each participant has at least $k$ favorite numbers. The organizers want to give to each student a T-shirt with one of their favorite numbers on it. Find the smallest $k$ for which this is always plausible under the restriction:\n\na) the participants form a circle and the T-shirts must then be distributed in such a way that no two neighbors share the same number.\n\nb) There are 7 boys and 7 girls and neither of the boys share their T-shirt number with neither of the girls.",
    "options": [],
    "answer": "a) 2; b) 4",
    "solution": "a) Obviously $k=1$ fails (e.g., whenever a couple of neighbors share their favorite number). We will show that $k=2$ works. If all of them share the same favorite numbers, say 1 and 2, then we can easily solve the problem via providing 7 T-shirts of each kind, since 14 is even. If this is not the case, WLOG we can order the participants, such that the first one has a favorite number $A$, which is not among the favorites of the last one. Then we give an $A$ T-shirt to the first. For each consecutive participant, we choose a favorite number, different than the one ahead of him. This will always be possible, thus the answer is $k=2$.\n\nb) First, $k \\le 3$ fails when the favorite numbers of both boys and girls form the triples $\\{1,2,3\\}$, $\\{1,4,5\\}$, $\\{1,6,7\\}$, $\\{2,4,6\\}$, $\\{2,5,7\\}$, $\\{3,4,7\\}$, $\\{3,5,6\\}$. We will show that $k=4$ works. It is enough to consider exactly 4 favorite numbers for each of the boys, and all the different ways to assign a symbol $B$ or $G$ to each of the numbered T-shirts, for which there are no T-shirts with different symbols but identical numbers. Such a labeling will be referred as \"good\". Note that, there are at least $2^4 = 16$ \"good\" labelings (when all the girls and all the boys share the same set of favorite numbers), and the probability for a boy to have all his favorite T-shirts labeled $G$ is exactly $\\frac{1}{16}$ from all good cases. Analogously, for a girl to have $B$ on all her favorite T-shirts. Therefore, in at most $14 \\cdot \\frac{1}{16} = \\frac{7}{8}$ of the \"good\" cases the organizers will fail to distribute the T-shirts; hence in at least $\\frac{1}{8}$ of the good cases (in particular, in at least two) the organizers can accomplish their goal.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 27561,
    "subject": "Mathematics (Multi-modal)",
    "question": "There are $n$ towns in a country. Some of the towns are connected with one way roads and between any two towns it is possible to have several roads in both directions. It is known that for any two towns $A$ and $B$ one can travel from $A$ to $B$, or from $B$ to $A$ or both. Find the minimum number of roads that have to be built such that one can travel in both directions between any two towns.",
    "options": [],
    "answer": "1",
    "solution": "We prove that the answer is $1$. Consider towns numbered from $1$ to $n$ and let from town $i$ there is a road to town $i+1$, $\\forall i < n$. Obviously the condition is fulfilled and we need at least one new road, thus the answer is greater than $0$.\n\nWe show that there is a town $F$ from which one can travel to all other towns. For every town consider the number of towns that can be visited from this town. Choose a town $A$ for which this number is the greatest. There exists a town $B$ such that one can not travel from $A$ to $B$. According to the condition of the problem one can travel from $B$ to $A$ and consequently to all towns starting from $A$. This contradicts the maximality of $A$.\n\nBy analogy there exists a town $L$ that is reachable from all other towns. Therefore an edge from $L$ to $F$ satisfies the condition of the problem.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 27562,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $a$ be the largest value of the expression $24y-9y^2$, where $y$ is a rational number and $b$ is the smallest integer satisfying the inequality\n$$\n(t+3)^3 - (6t-7)^2 - (t-9)^3 < 3.\n$$\nFactor into irreducible factors with integer coefficients the expression\n$$\na(x-1)x^3 + bx - 2x - 1.\n$$",
    "options": [],
    "answer": "(2x - 1)^3(2x + 1)",
    "solution": "We have $24y - 9y^2 = 16 - (3y - 4)^2$ whose largest value $a = 16$ is reached for $y = \\frac{4}{3}$. The given inequality is equivalent to\n$$\n\\begin{aligned}\n& t^3 + 9t^2 + 27t + 27 - 36t^2 + 84t - 49 - t^3 + 27t^2 - 243t + 729 < 3 \\\\\n& -132t + 704 < 0,\n\\end{aligned}\n$$\ni.e. $t > \\frac{16}{3}$ and $b = 6$. Substituting $a = 16$, $b = 6$ in the given expression, we get\n$$\n\\begin{aligned}\n& 16(x-1)x^3 + 6x - 2x - 1 = 16x^4 - 16x^3 + 4x - 1 \\\\\n& = (16x^4 - 1) - 4x(4x^2 - 1) = (4x^2 - 1)(4x^2 + 1) - 4x(4x^2 - 1) \\\\\n& = (4x^2 - 4x + 1)(2x - 1)(2x + 1) = (2x - 1)^3(2x + 1).\n\\end{aligned}\n$$",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 27563,
    "subject": "Mathematics (Multi-modal)",
    "question": "Given an acute triangle $ABC$ with circumcenter $O$. The point $P$ on $BC$ such that $BP < \\frac{BC}{2}$ and the point $Q$ is on $BC$, such that $CQ = BP$. The line $AO$ meets $BC$ at $D$ and $N$ is the midpoint of $AP$. The circumcircle of $(ODQ)$ meets $(BOC)$ at $E$. The lines $NO$, $OE$ meet $BC$ at $K$, $F$. Show that $AOKF$ is cyclic.\n\n(Alexander Ivanov)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $A'$ be the antipode of $A$ and let $AO \\cap (BOC) = R$. Since $NO \\parallel PA'$, by Reim's theorem we need $APA'F$ being cyclic, or $DP \\cdot DF = DA \\cdot DA' = DB \\cdot DC = DO \\cdot DR$, so we need $OPRF$ being cyclic, or that $\\angle OFP = \\angle ORP$. By shooting lemma, $RDEF$ is cyclic, so $\\angle OFP = \\angle ERD$, so we need $AR$ being the angle bisector of $\\angle PRE$. Since $AR$ bisects $\\angle BRC$, it is sufficient to show that $\\angle BRP = \\angle CRE$.\n\nTo use that $EQDO$ is cyclic, let $EQ \\cap (BOC) = S$; Reim's implies $RS \\parallel BC$, which together with the length condition gives that $RSQP$ is an isosceles trapezoid. Hence, $\\angle CRE = \\angle CSQ = \\angle BRP$, which finishes the problem.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 27564,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all positive integers $k$, so that there exists a polynomial $f(x)$ with rational coefficients, such that for all sufficiently large $n$,\n$$\nf(n) = \\operatorname{lcm}(n + 1, n + 2, \\dots, n + k).\n$$",
    "options": [],
    "answer": "k = 1 or k = 2",
    "solution": "For $k=1$ and $k=2$, the required polynomials are $f(x) = x+1$ and $f(x) = (x+1)(x+2)$, respectively. Let $k \\ge 3$ and assume that such a polynomial $f(x)$ exists. For any prime number $p$, its degree in $\\operatorname{lcm}(n+1, n+2, \\dots, n+k)$ is $\\max\\{\\alpha_1, \\alpha_2, \\dots, \\alpha_{k-1}\\}$, where $\\alpha_i$ is the power of $p$ in the canonical representation of $n+i$, $i=1, \\dots, k$. If this is, for example, $\\alpha_s$, then it would be obtained if we take\n$$\n\\frac{(n+1)(n+2)\\cdots(n+k)}{p^{\\alpha_1}p^{\\alpha_2}\\cdots p^{\\alpha_{s-1}}p^{\\alpha_{s+1}}\\cdots p^{\\alpha_k}},\n$$\nit being clear that the powers of $p$ in the denominator are divisors of $\\prod_{1 \\le i \\ne s \\le k} (s - i)$. Therefore\n$$\n\\operatorname{lcm}(n+1, n+2, \\dots, n+k) = \\frac{(n+1)(n+2)\\dots(n+k)}{C_n}, \\quad (1)\n$$\nwhere $C_n$ is a divisor of $\\prod_{1 \\le i < j \\le k} (j - i)$. Since $C_n$ can take a finite number of possible values, there will be a natural number $C$ such that for infinitely many $n$, $f(n) = \\frac{(n+1)(n+2)\\dots(n+k)}{C}$. So for infinitely many $x$, $f(x) = \\frac{(x+1)(x+2)\\dots(x+k)}{C}$, whence\n$$\nf(x) = \\frac{(x+1)(x+2)\\dots(x+k)}{C}, \\quad \\forall x \\in \\mathbb{R}.\n$$\nTherefore\n$$\n\\operatorname{lcm}(n+1, n+2, \\dots, n+k) = \\frac{(n+1)(n+2)\\dots(n+k)}{C}, \\quad \\text{for all } n \\in \\mathbb{N}.\n$$\nLet's assume this is possible. Let's choose a prime $p < k$ such that $p$ does not divide $k$. Let $n + k + 1 = p^m$ for sufficiently large $m$. From the above formula we have\n$$\n\\frac{\\operatorname{lcm}(n+2, n+3, \\dots, n+k+1)}{\\operatorname{lcm}(n+1, n+2, \\dots, n+k)} = \\frac{n+k+1}{n+1}. \\quad (2)\n$$\nThe degree of $p$ in the numerator of the left side is $m$ and in the denominator – at least 1, while the degree of $p$ in the numerator of the right side is $m$ and in the denominator – 0. We derive a contradiction! Therefore, the assumption is wrong and for $k \\ge 3$ there does not exist a polynomial with the desired property. $\\square$",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 27565,
    "subject": "Mathematics (Multi-modal)",
    "question": "In every cell of a board $101 \\times 101$ is written a positive integer. For any choice of 101 cells from different rows and columns, their sum is divisible by 101. Show that the number of ways to choose a cell from each row of the board, so that the total sum of the numbers in the chosen cells is divisible by 101, is divisible by 101.\n\n(Borislav Kirilov)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "(Vlad Spataru) Index the rows and columns from $0$ to $100$. We shall only work modulo $101$ in what follows. Observe that we may let $(0,0) = 0$ by adding some constant to all the terms of the table. Next, because of the condition in the statement,\n$$\n(u, v) + (i, j) = (u, j) + (i, v),\n$$\nfor any $u, v, i, j$. Thus, there exist $a_0 = 0, a_1, \\dots, a_{100}$ and $b_0 = 0, b_1, \\dots, b_{100}$ so that $(i, j) = a_i + b_j$. Now, letting $\\Sigma_a$ and $\\Sigma_b$ be the sum of the $a_i$ and the $b_j$ respectively, the condition in the statement then yields $\\Sigma_a + \\Sigma_b = 0$. Note that\n$$\n\\sum_{i=0}^{100} x^{101b_i} = \\left( \\sum_{i=1}^{100} x^{b_i} \\right)^{101} := \\sum_{i=0}^{\\infty} x^i \\alpha(i).\n$$\nObserve that the number of ways of adequately choosing one cell from each row corresponds to the number of $101$-tuples with values from $\\{b_0, b_1, \\dots, b_{100}\\}$ with sum equal to $\\Sigma_b$ that is $\\alpha(\\Sigma_b) + \\alpha(101 + \\Sigma_b) + \\dots$.\nUsing the fact that $101$ is prime and computing the latter expression for $x = \\exp(2\\pi i/101)$ we get the desired\n$$\n0 = \\sum_{i=0}^{\\infty} \\alpha (101i + \\Sigma_b).\n$$",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 27566,
    "subject": "Mathematics (Multi-modal)",
    "question": "A quadruplet of distinct positive integers $(a, b, c, d)$ is called $k$-good if the following conditions hold:\n1. Among $a, b, c, d$, no three form an arithmetic progression.\n2. Among $a + b, a + c, a + d, b + c, b + d, c + d$, there are $k$ of them, forming an arithmetic progression.\n\na) Find a 4-good quadruplet.\n\nb) What is the maximal $k$, such that there is a $k$-good quadruplet?",
    "options": [],
    "answer": "Example: (7, 6, 4, 3). The maximal k is 4.",
    "solution": "a) The quadruple $(7, 6, 4, 3)$ is nice because there are no three numbers in it to form an arithmetic progression, but of the six numbers\n$$\n7 + 6 = 13,\\ 7 + 4 = 11,\\ 7 + 3 = 10,\\ 6 + 4 = 10,\\ 6 + 3 = 9,\\ 4 + 3 = 7\n$$\nthe numbers $7, 9, 11, 13$ form an arithmetic progression.\n\nb) Without restriction let $a > b > c > d$. Then\n$$\na + b > a + c > \\max(a + d, b + c) > \\min(a + d, b + c) > b + d > c + d.\n$$\n\n1) $a+b, a+c$ and $a+d$ form an arithmetic progression, then $2(a+c) = (a+b) + (a+d) \\iff 2c = b+d$;\n2) $a+b, a+c$ and $b+c$ form an arithmetic progression, then $2(a+c) = (a+b) + (b+c) \\iff 2b = a+c$;\n3) $a+d, b+d$ and $c+d$ form an arithmetic progression, then $2(b+d) = (a+d) + (c+d) \\iff 2b = a+c$;\n4) $b+c, b+d$ and $c+d$ form an arithmetic progression, then $2(b+d) = (b+c) + (c+d) \\iff 2c = b+d$.\nIn all four cases we get a contradiction with the given condition.\nFrom the above it is clear that all 6 numbers cannot form an arithmetic progression.\nLet's assume that 5 of them form an arithmetic progression. Notice that whichever of the numbers $a+b, a+c, a+d, b+c, b+d$ and $c+d$ we delete, there is always some from progressions 1., 2., 3., or 4., contradiction.\nIt follows from a) that the required $k$ is 4. $\\square$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 27567,
    "subject": "Mathematics (Multi-modal)",
    "question": "The points $A_1$, $B_1$, $C_1$ are chosen on the sides $BC$, $CA$, $AB$ of a triangle $ABC$ so that $BA_1 = BC_1$ and $CA_1 = CB_1$. The lines $C_1A_1$ and $A_1B_1$ meet the line through $A$, parallel to $BC$, at $P$, $Q$. Let the circumcircles of the triangles $APC_1$ and $AQB_1$ meet at $R$. Given that $R$ lies on $AA_1$, show that $R$ lies on the incircle of $ABC$.\n(Emil Kolev)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Observe that $(A_1, C_1, R, B_1)$ are concyclic. Let $I$ be the incenter of triangle $ABC$. Then $BI$ and $CI$ perpendicularly bisect $C_1A_1$ and $A_1B_1$ respectively. Hence, $I$ is the circumcenter of $\\triangle A_1B_1C_1$. Let $\\angle ACB = 2\\gamma$. $\\angle A_1IB_1 = 2\\angle B_1C_1A_1 = 2\\angle A_1RB_1 = 2\\angle AQB_1 = 2\\angle B_1A_1C = 180^\\circ - 2\\gamma$. Therefore $(I, A_1, B_1, C)$ concyclic $\\rightarrow IA_1 \\perp BC$, $IB_1 \\perp AC$. Similarly $IC_1 \\perp AB$, so circle $(A_1, C_1, R, B_1)$ is the incircle of $ABC$. Thus, $R$ lies on the incircle of $ABC$. $\\square$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 27568,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find the smallest possible number of divisors a positive integer $n$ may have, which satisfies the following conditions:\n1.\n$24 \\mid n+1;$\n2. The sum of the squares of all divisors of $n$ is divisible by $48$ (1 and $n$ are included).",
    "options": [],
    "answer": "48",
    "solution": "",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 27569,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $G$ be a complete bipartite graph with partition sets $A$ and $B$ of sizes $km$ and $kn$, respectively. The edges of $G$ are colored in $k$ colors. Prove that there exists a monochromatic connected component with at least $m+n$ vertices (which means that there exists a color and a set of vertices, such that between any two of them, there is a path consisting of edges only in that color).",
    "options": [],
    "answer": "Detailed solution",
    "solution": "There are at least $kmn$ edges colored in the most used color. We delete the remaining edges and prove that there exists a connected component with at least $m+n$ vertices in the remaining graph. The idea is to consider all the connected components. If each one has less than $m+n$ vertices, then the graph is too fragmented and it's impossible to have too many (at least $kmn$) edges even if all edges are present in every connected component. It boils down to prove a certain inequality.\n\n**Lemma.** Let $x, y$ be two positive real numbers and $\\ell, k$; $\\ell \\ge k$ be natural numbers. Let $x_i, y_i, i = 1, 2, \\dots, \\ell$ satisfy the following conditions.\n$$\nx_i \\ge 0, y_i \\ge 0, x_i + y_i \\le (x+y)/k, i = 1, 2, \\dots, \\ell;\n$$\n$$\n\\sum_{i=1}^{\\ell} x_i = x, \\quad \\sum_{i=1}^{\\ell} y_i = y. \\qquad (1)\n$$\nThen it holds\n$$\n\\sum_{i=1}^{\\ell} x_i y_i \\le \\frac{xy}{k}.\n$$\nThe equality is reached only if $x_i = x/k, y_i = y/k, i = 1, 2, \\dots, k$; $x_i = y_i = 0, i > k$.\n**Proof.** Let us denote $f(x, y) := \\sum_{i=1}^{\\ell} x_i y_i$, where $x = (x_1, \\dots, x_{\\ell})$, $y = (y_1, \\dots, y_{\\ell})$. The conditions in (1) determine a compact set, hence $f$ attains its maximum value on it, say, at the points $x'_i, y'_i, i = 1, 2, \\dots, \\ell$. We can assume $x'_1 \\ge x'_2 \\ge \\dots \\ge x'_{\\ell}$. We shall prove that $y'_i$ are also in decreasing order. Let us assume on the contrary that $y'_i < y'_{i+1}$. Set\n$$\nx_i = x_{i+1} := (x'_i + x'_{i+1})/2; \\quad y_i = y_{i+1} := (y'_i + y'_{i+1})/2.\n$$\nThen (Chebyshev inequality)\n$$\nx_i y_i + x_{i+1} y_{i+1} > x'_i y'_i + x'_{i+1} y'_{i+1}.\n$$\nBut it contradicts the maximality of $x'_i, y'_i, i = 1, \\dots, \\ell$. Next, if $x'_1 + y'_1 < (x+y)/k$ we can similarly set $x_1 := x'_1 + \\varepsilon, x_2 := x'_2 - \\varepsilon; y_1 := y'_1 + \\delta, x_2 := x'_2 - \\delta$ for sufficiently small $\\varepsilon, \\delta \\ge 0$ and get a larger value of $f$. Therefore, $x'_1 + y'_1 = (x+y)/k$.\nLet $k'$ be the largest index for which $x_{k'} > 0$ and $y_{k'} > 0$. In the same way we can see that $x'_i + y'_i = (x+y)/k, i = 1, 2, \\dots, k'$. Hence, $k' \\le k$. Assume that $k' < k$ and $y_i = 0, i > k'$. We modify $x', y'$ as follows. Set\n$$\nx_i := x'_i, y_i := y'_i, i = 1, 2, \\dots, k' - 1; \\\\\n x_{k'} := x'_{k'}, y_{k'} := y'_{k'} - \\varepsilon, x_{k'+1} := (x+y)/k, y_{k'+1} := \\varepsilon.\n$$\nFor $i > k' + 1$ we set $y_i = 0$, and the values of $x_i, i > k' + 1$ are irrelevant providing they comply with (1). Since $x'_k < (x+y)/k$, it can be seen that $f(x, y) > f(x', y')$ which contradicts the maximality of $x', y'$.\nThus, $k' = k$. We prove that $x'_i = x/k, y'_i = y/k, i = 1, 2, \\dots, k$. Assume on the contrary it doesn't hold and let $j$ be the first index for which $x'_j \\ne x/k$. WLOG let $x'_j < x/k$. Then there exists $i > j$ for which $x'_i > x/k$. This means $x'_i > x'_j$ and thus the sequence $x'_1, x'_2, \\dots, x'_k$ is not decreasing, contradiction. To recap, we established that $x'_i = x/k, y'_i = y/k, i = 1, 2, \\dots, k$. In this case $f(x', y') = kxy$, and Lemma 1 is proved. $\\square$\n\nBack to the problem. The number of all edges of $K$ is $k^2mn$, hence there is a color, say, white such that at least $kmn$ edges are colored white. We delete all edges colored in a color other than white. We'll prove that in the remaining graph $K'$, there exists a connected component with at least $m+n$ vertices. Assume on the contrary it is not true. Denote the connected components of $K'$ by $G(A_i, B_i)$, $i = 1, 2, \\dots, \\ell$ and let $|A_i| = m_i, |B_i| = n_i, i = 1, 2, \\dots, \\ell$. We have\n$$\n\\sum_{i = 1}^\\ell m_i = km, \\quad \\sum_{i=1}^\\ell n_i = kn, m_i + n_i < m + n.\n$$\nAccording to Lemma 1,\n$$\n\\sum_{i=1}^{\\ell} m_i n_i < kmn\n$$\nwhich contradicts the choice of the white color. This means that for at least one index $i$ it holds $m_i + n_i \\ge m + n$.",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 27570,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $x_0, x_1, \\dots$ be a sequence of real numbers such that $x_0 = 1$ and $x_{n+1} = \\sin(x_n) + \\frac{\\pi}{2} - 1$ for all $n \\ge 0$. Show that the sequence converges and find its limit.\n(Kristyan Vasilev)",
    "options": [],
    "answer": "π/2",
    "solution": "We will first prove that the sequence is strictly increasing. Note that for each $n$ we have $x_n \\le \\frac{\\pi}{2}$, since $\\sin x \\le 1$ for each $x \\in \\mathbb{R}$. Besides we have that the function $f(x) = \\sin x - x$ is decreasing for $x \\in \\mathbb{R}$, because $f'(x) = \\cos x - 1 \\le 0$ for each $x \\in \\mathbb{R}$.\n\nTherefore, for $x \\in (-\\infty, \\frac{\\pi}{2})$ we have $f(x) \\ge \\sin \\frac{\\pi}{2} - \\frac{\\pi}{2}$. So we get that $\\sin x_n \\ge x_n + 1 - \\frac{\\pi}{2}$, which is equivalent to $x_{n+1} \\ge x_n$. Therefore, the series $(x_n)_{n=1}^\\infty$ is increasing and since it is bounded it follows that it is convergent. If $l$ is its limit, then for $l$ it is true that $l = \\sin l + \\frac{\\pi}{2} - 1$, i.e. $f(l) = f(\\frac{\\pi}{2})$. The function $f$ is decreasing, which means that $l = \\frac{\\pi}{2}$.\n\n$\\boxed{\\frac{\\pi}{2}}$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 27571,
    "subject": "Mathematics (Multi-modal)",
    "question": "Given is an acute triangle $ABC$ with incenter $I$ and the incircle touches $BC$, $CA$, $AB$ at $D$, $E$, $F$. The circle with center $C$ and radius $CE$ meets $EF$ for the second time at $K$. If $X$ is the $C$-excircle touchpoint with $AB$, show that $CX$, $KD$, $IF$ concur.\n(Kristyan Vasilev)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "We claim the concurrency point is the $F$-antipode $F'$. It is well-known that this is $\\overline{CX} \\cap \\overline{IF}$.\nLet $P = \\overline{EF} \\cap \\overline{DF'}$ and $Q = \\overline{DF} \\cap \\overline{EF'}$. Then since $\\angle PEQ = \\angle PDQ = 90^\\circ$, $DEPQ$ is cyclic.\nNow, we have\n$$\n\\begin{aligned}\n\\angle EFD &= \\angle EFF' = 90^\\circ - \\angle PF'E = 90^\\circ - \\angle DFE \\\\\n&= -90^\\circ + \\left( 90^\\circ - \\frac{\\angle A}{2} \\right) + \\left( 90^\\circ - \\frac{\\angle B}{2} \\right) = \\frac{\\angle C}{2},\n\\end{aligned}\n$$\nso the center of ($DEPQ$) lies on ($CDE$) (and is on the same side of $\\overline{DE}$ as $C$). On the other hand, it also lies on the perpendicular bisector of $\\overline{DE}$, so it must be $C$ itself. This gives us $P = K$, and the conclusion follows. $\\square$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 27572,
    "subject": "Mathematics (Multi-modal)",
    "question": "Solve in positive integers the equation\n$$\nm^{\\frac{1}{n}} + n^{\\frac{1}{m}} = 2 + \\frac{2}{mn(m+n)^{\\frac{1}{m}+\\frac{1}{n}}}.\n$$",
    "options": [],
    "answer": "No positive integer solutions",
    "solution": "There are no solutions in positive integers. Without loss of generality, assume that $m \\ge n$. If $n \\ge 3$, then\n$$\n\\left(1 + \\frac{1}{m}\\right)^m = \\sum_{k=0}^{m} \\binom{m}{k} \\frac{1}{m^k} < \\sum_{k=0}^{m} \\frac{1}{k!} < \\sum_{k=0}^{\\infty} \\frac{1}{k!} = e < n \\implies n^{\\frac{1}{m}} > 1 + \\frac{1}{m}\n$$\nSimilarly, $m^{\\frac{1}{n}} > 1 + \\frac{1}{n}$, whence\n$$\nm^{\\frac{1}{n}} + n^{\\frac{1}{m}} > 2 + \\frac{1}{m} + \\frac{1}{n} \\ge 2 + \\frac{2}{mn} > 2 + \\frac{2}{mn(m+n)^{\\frac{1}{m}+\\frac{1}{n}}}.\n$$\nThe cases $n = 1, 2$ can be solved directly. $\\square$",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 27573,
    "subject": "Mathematics (Multi-modal)",
    "question": "A set of points in the plane is called good if the distance between any two points in it is at most $1$. Let $f(n, d)$ be the largest positive integer such that in any good set of $3n$ points, there is a circle of diameter $d$, which contains at least $f(n, d)$ points. Prove that there exists a positive real $\\epsilon$, such that for all $d \\in (1 - \\epsilon, 1)$, the value of $f(n, d)$ does not depend on $d$ and find that value as a function of $n$.\n(Kristiyan Vasilev, Konstantin Garov)",
    "options": [],
    "answer": "n",
    "solution": "Since $m(n, d)$ is an increasing function of $d$ and it cannot be larger than $3n$, clearly it hits a constant when $d$ approaches $1$. Fix some $d$, $d < 1$, it may be very close to $1$. Take an equilateral triangle with side length $1$ and put inside it $3n$ points — $n$ points near each of its vertices so that they be at distance less than $(1 - d)/2$ from the corresponding vertex — see figure 1.\n\n![](attached_image_1.png)\n\nThis set of points apparently is a good set. A disk with radius $d$ cannot cover a point that is near to one vertex and a point near another vertex. That is, the maximum number of points that could be covered with such a disk is $n$. This means that $m(n, d) \\le n$ no matter how close to $1$ the value of $d$ is.\n\nWe will prove that $m(n, d) = n$ when $d$ is close to $1$. To prove this we need to study the good set of points. How close together are its points? Can we claim that all of them lie in disk with some not too large radius? Of course, they lie in a disk with radius $1$ since if you take one of the points, the rest are at distance at most $1$ from this point. But can we claim that all of them lie in a disk with a smaller radius? Clearly, they may not lie in a disk with radius $1/2$ — the equilateral triangle with side length $1$ is an example that disproves this. Crucial to this problem is the following observation.\n**Lemma.** Any good set can be put inside a disk with radius $\\frac{1}{\\sqrt{3}}$. That is, any good set can be put inside a circle circumscribed about an equilateral triangle with a side length $1$.\n\nThe official solution proves it by using Helly's theorem which is instructive. Another method is used here.\n**Proof.** Let $X$ be a good set. We first show that $X$ lies in a circle circumscribed about an acute triangle with side lengths less or equal to $1$. Take the convex hull of $X$ and put it inside a circle with some radius. We begin to tighten this circle until it touches $1$ and then $2$ points of the convex hull. We can tighten the circle further unless these two points lie on its diameter. In this case $A$ lies in a circle with a diameter at most $1$. So, assume we decrease the circle $k$ until it touches $3$ points of $X$, say $A, B, C$. If these points are vertices of an acute triangle we are done, so suppose $\\angle ACB \\ge 90^\\circ$ — see figure 2. We begin to further decrease the diameter of $k$ keeping the points $A$ and $B$ on the circle. This can be done until the first of the following two events happens:\n1. The circle hits a point $C' \\in X$ (and $C'$ and $C$ lie on different sides of $AB$).\n2. $AB$ becomes a diameter of $k$.\n\n![](attached_image_2.png)\n\nIf the second event happens first, we are done since $X$ lies in a disk with diameter $1$. In case of the first event, the triangle $ABC'$ is acute. Next, since $\\triangle ABC'$ is acute one of its angles, say $\\angle AC'B$, is greater or equal to $60^\\circ$. It follows the radius of $k' = \\frac{AB}{2 \\sin \\angle C} \\le \\frac{1}{\\sqrt{3}}$.\nSo, we proved that we can cover $X$ with a circle $k$ with radius $1/\\sqrt{3}$. $\\square$\n\nTake a point $A \\in k$ and construct the points $B, C, D \\in k$ such that $AB = BC = CD = d$ — see figure 3. We construct three disks with diameters $AB, BC, CD$ correspondingly. These disks cover all the points inside $k$ except the red piece in figure 3. Note now that by taking $d$ close to $1$, the point $D$ will be as close to $A$ as we want. This means that we can make the red piece on figure 3 as small as we want. We take $d$ so close to $1$ so that the measure of the arc $AD$ (on figure 3) is less than $\\frac{2\\pi}{3n+1}$. If the red piece does not contain a point of $X$ we are done. Otherwise we begin to rotate the point $A$ and the entire configuration around $k$. It is not possible that every time the red piece hits a point of $X$ since otherwise the number of points in $X$ would be more than $3n$. Therefore we can cover $X$ with $3$ disks of diameter $d$, hence one of these disks contains at least $n$ points. Thus, we proved that $m(n,d) \\ge n$ and combined with $m(n,d) \\le n$ this yields $m(n,d) = n$ providing $d$ is close to $1$. $\\square$\n\n![](attached_image_3.png)",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 27574,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all pairs $(a, b)$ of co-prime naturals such that $a < b$ and $b$ divides\n$$\n(n + 2)a^{n+1002} - (n + 1)a^{n+1001} - n a^{n+1000}\n$$\nfor every natural number $n$.",
    "options": [],
    "answer": "(3, 5)",
    "solution": "Since $a$ and $b$ are co-prime, so are $b$ and $a^{n+1000}$, respectively, what is requested is equivalent to $b$ dividing $(n + 2)a^2 - (n + 1)a - n$ for each $n$.\n\nFrom $n = 1$ and $n = 2$ we get that necessarily $b$ divides $3a^2 - 2a - 1$ and $4a^2 - 3a - 2$. Hence $b$ divides\n$$\n4(3a^2 - 2a - 1) - 3(4a^2 - 3a - 2) = a + 2\n$$\nand since $3a^2 - 2a - 1 = 3(a - 2)(a + 2) - 2(a + 2) + 15$, then necessarily $b$ divides $15$.\n\nClearly $b > a \\ge 1$, i.e. $b \\ge 2$. If $b = 15$, then $b \\mid a + 2$ gives $a = 13$, but $4 \\cdot 13^2 - 3 \\cdot 13 - 2 = 635$ is not divisible by $3$. If $b = 3$, then $a = 1$, but $4 \\cdot 1^2 - 3 \\cdot 1 - 2 = -1$ is not divisible by $3$.\n\nIt remains $b = 5$, respectively $a = 3$. Indeed, $(n + 2) \\cdot 3^2 - (n + 1) \\cdot 3^1 - n = 5(n + 3)$ is divisible by $5$. $\\square$",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 27575,
    "subject": "Mathematics (Multi-modal)",
    "question": "An integer is written in each of the fields of a $9 \\times 9$ square table. For every $k$ numbers in the same row (column), their sum is in the same row (column). Find the smallest possible number of zeros in the table if:\n\na) $k = 5$;\n\nb) $k = 8$.",
    "options": [],
    "answer": "a) 63; b) 0",
    "solution": "a) Example: we number the rows and columns from $1$ to $9$. We write $1$ in the fields $(i, i)$ ($i = 1, \\dots, 9$); $-1$ in field $(1, 9)$ and in fields $(i, i-1)$ ($i = 2, \\dots, 9$); $0$ in other fields. Possible sums are $1$, $0$, and $-1$.\n\nEvaluation: Suppose there are at least $19$ non-zero numbers. From Dirichlet's principle, there will be at least three non-zero numbers on any row, and therefore at least two non-zero numbers with the same sign, let's say positive ones (the situation with negative ones is analogous). Let's arrange the numbers in this order by size: $a_1 \\le a_2 \\le \\dots \\le a_9$, where $a_9 \\ge a_8 > 0$. If $a_5 \\ge 0$, then $a_5 + a_6 + a_7 + a_8 + a_9 \\ge a_8 + a_9 > a_9$ must be of the same order, a contradiction. If $a_5 < 0$, then $a_1 + a_2 + a_3 + a_4 + a_5 < a_1$ must be of the same order, a contradiction.\n\nb) A possible example without zeros is as follows (works because $5 \\cdot 3 + 3 \\cdot (-4) = 3$ and $4 \\cdot 3 + 4 \\cdot (-4) = -4$):\n\n| 3  | 3  | 3  | 3  | 3  | -4 | -4 | -4 | -4 |\n|----|----|----|----|----|----|----|----|----|\n| -4 | 3  | 3  | 3  | 3  | 3  | -4 | -4 | -4 |\n| -4 | -4 | 3  | 3  | 3  | 3  | 3  | -4 | -4 |\n| -4 | -4 | -4 | 3  | 3  | 3  | 3  | 3  | -4 |\n| -4 | -4 | -4 | -4 | 3  | 3  | 3  | 3  | 3  |\n| 3  | -4 | -4 | -4 | -4 | 3  | 3  | 3  | 3  |\n| 3  | 3  | -4 | -4 | -4 | -4 | 3  | 3  | 3  |\n| 3  | 3  | 3  | -4 | -4 | -4 | -4 | 3  | 3  |\n| 3  | 3  | 3  | 3  | 3  | -4 | -4 | -4 | -4 |",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 27576,
    "subject": "Mathematics (Multi-modal)",
    "question": "Given the functions $f(x) = |x-2| - |x-4|$ and $g(x) = |x-8| - 2$. Find the area of the figure with vertices, the intersection points of the graphs of the functions $f(x)$ and $g(x)$ and the intersection points of the graph of $g(x)$ with the x-axis.\n(Nedyalka Dimitrova)",
    "options": [],
    "answer": "12",
    "solution": "The graph of $g(x)$ consists of two rays with a common vertex at $x = 8$. We remove the module and easily calculate its intersection points with the x-axis through the equations $6 - x = 0$ and $x - 10 = 0$ — $A(6, 0)$ and $B(10, 0)$.\n\nAfter removing the modules in $f(x)$ we see that in the interval $(-\\infty, 2)$ $f(x) = -2$, in the interval $[2, 4]$ $f(x) = 2x - 6$ and in the interval $(4, \\infty)$ $f(x) = 2$. We solve the equations $f(x) = g(x)$ for each of the three intervals. We get the following intersection points — $D(4, 2)$ and $C(12, 2)$.\n\nIt follows that the figure $ABCD$ is a trapezoid with bases $8$ and $4$ and height $2$. Therefore, its area is $\\frac{8+4}{2} \\times 2 = 12$.",
    "topic": "Discrete Mathematics",
    "subtopic": "Other"
  },
  {
    "id": 27577,
    "subject": "Mathematics (Multi-modal)",
    "question": "Given an obtuse isosceles triangle $ABC$ with $CA = CB$ and circumcenter $O$. The point $P$ on $AB$ is such that $AP < \\frac{AB}{2}$ and $Q$ on $AB$ is such that $BQ = AP$. The circle with diameter $CQ$ meets $(ABC)$ at $E$ and the lines $CE$, $AB$ meet at $F$. If $N$ is the midpoint of $CP$ and $ON$, $AB$ meet at $D$, show that $ODCF$ is cyclic.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $T$ be the midpoint of $CQ$ (it is anyway the center of the important circle with diameter $CQ$). Then $OC$ is the perpendicular bisector of $AB$ (as $AC = BC$), triangle $ONT$ is isosceles by symmetry (as $P$ and $Q$ are symmetric with respect to the midpoint $M$ of $AB$ and hence with respect to $CO$, by the problem condition) and $OT$ is the perpendicular bisector of $CE$, so $\\angle OCF = 90^\\circ - \\angle OCT = 90^\\circ - \\angle OCD = \\angle ODM = \\angle ODF$, done! $\\square$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 27578,
    "subject": "Mathematics (Multi-modal)",
    "question": "In the house of the wealthy Lady Gilmore, one of her most precious possessions has been stolen: her pearl necklace. The task of unraveling the mystery fell on the shoulders of Inspector Goodenough. He had the following information: on the day of the theft, $7$ of Lady Gilmore's servants, whom we will refer to as $A$, $B$, $C$, $D$, $E$, $F$, $G$ for the confidentiality of the investigation, entered the room with the necklace. Each claimed to have been in the room only once for an unspecified period of time. Additionally, $A$ claims to have met $B$, $C$, $F$, $G$ in the room; $B$ claims to have met $A$, $C$, $D$, $E$, $F$; $C$ claims to have met $A$, $B$, $E$; $E$ claims to have met $B$, $C$, $F$; $F$ claims to have met $A$, $B$, $D$, $E$; $G$ claims to have met $A$, $D$ and $D$ claims to have met $B$, $F$, $G$. Inspector Goodenough concluded that one of the servants was lying. Who is he?\n\n(Lyuben Lichev)",
    "options": [],
    "answer": "A",
    "solution": "We will first prove the following lemma.\n\n**Lemma.** Let $X$, $Y$, $Z$ and $T$ be four of the servants. If it is known that the pairs $X, Y$; $Y, Z$; $Z, T$ and $T, X$ were in the room together at some point, then one of the pairs $X, Z$ and $Y, T$ also detected each other.\n\n**Proof of Lemma.** Let us assume, without community restriction, that $Y$ and $T$ were not in the room together, and $Y$ left the room before $T$ (the other cases are analogous). Then, $X$ and $Z$ were in the room together in the period between $Y$'s departure and $T$'s arrival.\n\nNotice that $A$, $C$, $E$, $F$ satisfy the condition of the lemma, but none of $A, E$ and $C, F$ intersect. The same goes for $A$, $B$, $D$, $G$. The only common element of these pairs is $A$. It remains to be ascertained that it is possible that all the other pairs met, as they claim, by entering exactly once. This is possible with the following sequence of entries and exits: enter $G$, enter $D$, exit $G$, enter $B$, enter $F$, exit $D$, enter $E$, exit $F$, $C$ enters, $B$ exits, $E$ exits, $C$ exits. $■$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 27579,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $p, q$ be coprime integers, such that $\\frac{p}{q} \\le 1$. For which $p, q$, there exist even integers $b_1, b_2, \\dots, b_n$, such that\n$$\n\\frac{p}{q} = \\frac{1}{b_1 + \\frac{1}{b_2 + \\frac{1}{b_3 + \\dots}}}?\n$$",
    "options": [],
    "answer": "Exactly those coprime integers p, q with |p| < |q| and with one of p, q even and the other odd (excluding |p| = |q|, i.e., ±1).",
    "solution": "Set $b_i = 2\\ell_i$, all of them are even. At the first step we get the pairs\n$$\nA := \\{(p, q) : (p, q) = (1, 2\\ell_1), \\ell_1 \\in \\mathbb{Z}. \\quad (1)\\}.\n$$\nAt each subsequent step we expand the set $A$ by adding additional pairs $(p', q')$\ndefined as\n$$\n\\left\\{ (p', q') : \\frac{p'}{q'} = \\frac{1}{2\\ell + \\frac{p}{q}}, (p, q) \\in A, \\ell \\in \\mathbb{Z} \\right\\} \\quad (2).\n$$\nBy (2) we obtain\n$$\np' = q; \\quad q' = 2\\ell q + p \\quad (3)\n$$\nSo, if $(p, q)$ is obtained in the process of expanding, we also add $(p', q')$ defined as in (3). We want to characterize all pairs $(p, q)$ that can be obtained in this way. Note that if $(p, q) = 1$, then from (3) follows $(p', q') = 1$. Let us prove that the set of pairs $(p, q)$ in question is\n$$\nA := \\{(p, q) : p, q \\in \\mathbb{Z} \\setminus \\{0\\}, |p| < |q|, (p, q) = 1, \\text{ one of } p, q \\text{ is even, the other is odd.}\\}\n$$\nNote that we ruled out $|p| = |q|$ because $\\frac{p}{q} = \\pm 1$ cannot be represented as wanted. The transformation described in (3) shows that we cannot step outside $A$. To prove that all pairs in $A$ can be generated, we follow a standard procedure - take $(p', q') \\in A$ and search for $(p, q) \\in A$ that generates $(p', q')$ via (3), but we want $(p, q)$ be \"less\" than $(p', q')$. That's how induction works. Solving (3) with respect to $(p, q)$ yields\n$$\np = q' - 2\\ell p'; \\quad q = p' \\quad (4)\n$$\nClearly, we can choose $\\ell \\in \\mathbb{Z}$ such that $|q' - 2\\ell p'| < |p'|$ (since $p' \\neq 0$). Indeed, consider the points $q' - 2\\ell p'$ where $\\ell$ runs through all integers. These points are at a distance $2p'$ apart, and none of them hits $p'$ because $(p', q') = 1$. Hence, the point closest to 0 has magnitude less than $p'$. So, we started from $(p', q') \\in A$, and found $(p, q) \\in A$ that generates $(p', q')$ and moreover, $q = p', |p| < |p'| < |q'|$. Apparently $(p, q)$ is \"less\" than $(p', q')$, that is, $|p| < |p'|, |q| < |q'|$. Now, induction does the job. Obviously, $(1, 2), (-1, 2), (1, -2), (-1, -2)$ can be represented as continued fractions satisfying the statement. Assume that all $(p, q) \\in A$ with $|p| \\le N, |q| \\le N$ can be represented like that. Take $(p', q') \\in A, |q'| = N + 1$. As just shown, there exists $(p, q) \\in A, |p| < |q| \\le N$ such that\n$$\n\\frac{p'}{q'} = \\frac{1}{2\\ell + \\frac{p}{q}}.\n$$\nThe induction step is complete and the result follows. Note that it also follows that the representation of numbers in $A$ as continued fractions like that is unique. Indeed, take a pair $(p', q') \\in A$. The pair $(p, q) \\in A$ that satisfies (4) and $|p| < |p'|$ is determined uniquely. Uniqueness follows easily.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 27580,
    "subject": "Mathematics (Multi-modal)",
    "question": "Solve the equation\n$$\n(x + 1)\\sqrt{x^2 + 2x + 2} + x\\sqrt{x^2 + 1} = 0.\n$$",
    "options": [],
    "answer": "-1/2",
    "solution": "Let $a = \\sqrt{x^2 + 2x + 2} > 0$ and $b = \\sqrt{x^2 + 1} > 0$. Therefore\n$$ x = \\frac{(x^2 + 2x + 2) - (x^2 + 1) - 1}{2} = \\frac{a^2 - b^2 - 1}{2} $$\n$$ x + 1 = \\frac{a^2 - b^2 + 1}{2}. $$\nThe equation is equivalent to:\n$$ \\frac{a^2 - b^2 + 1}{2} \\cdot a + \\frac{a^2 - b^2 - 1}{2} \\cdot b = 0 $$\n$$(a^2 - b^2)a + a + (a^2 - b^2)b - b = 0 $$\n$$(a - b)((a + b)^2 + 1) = 0. $$\nHence, $a = b$ and $x = -\\frac{1}{2}$. $\\square$",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 27581,
    "subject": "Mathematics (Multi-modal)",
    "question": "The first, seventh, and seventeenth terms of an arithmetic progression are distinct and consecutive terms of a geometric progression. To find the difference of the arithmetic progression if its first term is a solution of the equation\n$$\nx^2 - 9x + x\\sqrt{12-x} - 9\\sqrt{12-x} = 0.\n$$",
    "options": [],
    "answer": "1 or -4/9",
    "solution": "Let $a_1$ and $d$ be the first term and the difference of the arithmetic progression, respectively. From the condition $a_1$, $a_1 + 6d$ and $a_1 + 16d$ are consecutive members of a geometric progression, i.e.\n$$\n(a_1 + 6d)^2 = a_1 \\cdot (a_1 + 16d) \\iff d \\cdot (a_1 - 9d) = 0.\n$$\nSince $d \\neq 0$, we get that $a_1 = 9d$. Furthermore, we have $(x-9)(x+\\sqrt{12-x}) = 0$ and $x \\le 12$. Then $x = 9$ or $\\sqrt{12-x} = -x$, i.e. $x^2 + x - 12 = 0$ and $x \\le 0$, whence $x = -4$. Then $a_1 = 9$ and $a_1 = -4$, as $d = 1$ and $d = -\\frac{4}{9}$, respectively. $\\square$",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 27582,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $n \\in \\mathbb{N}$ and $\\mathcal{A}$ be a nonempty family of nonempty subsets of $\\{1, 2, \\dots, n\\}$ with the following property – if $A \\in \\mathcal{A}$ and $A \\subset B \\subseteq \\{1, 2, \\dots, n\\}$, then $B \\in \\mathcal{A}$. Prove that the function\n$$\nf(x) := \\sum_{A \\in \\mathcal{A}} x^{|A|} (1-x)^{n-|A|}\n$$\nis strictly increasing in the interval $(0, 1)$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $0 < p < q < 1$. Notice that $p^* = \\frac{q-p}{1-p} \\in (0, 1)$. We construct the sets $X$ and $Y$ as follows: For each element $i \\in \\{1, 2, \\dots, n\\}$ we put $i$ in $X$ with probability $p$ (independently of each other), and for each element $j \\in \\{1, 2, \\dots, n\\} \\setminus X$ we put $j$ in $Y$ with probability $p^*$. Then $P(x \\in X \\cup Y) = p + (1-p)p^* = q$. It is easy to see that $f(p) = P(X \\in \\mathcal{A})$, and $f(q) = P(X \\cup Y \\in \\mathcal{A})$, but since $\\mathcal{A}$ has the property from the condition we have that $X \\subseteq X \\cup Y \\Rightarrow f(p) < f(q)$.\n$\\square$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 27583,
    "subject": "Mathematics (Multi-modal)",
    "question": "Given a natural number $n$. We have $n + 1$ balls numbered $1$, $1$, $2$, $3$, ..., $n$ (only the first two are the same). We need to color these balls in $n$ given colors so that every ball is a single color and every color is used at least once. We denote by $a_n$ the number of possible colorings. Find the smallest $n$ for which $a_n$ is divisible by $2024$.\n(Ivaylo Kortezov)",
    "options": [],
    "answer": "13",
    "solution": "If $a > 1$, then we have $(n-1)(n-2)/2$ choices for $a$ and $b$, and $n$ choices for their color. The remaining $n-1$ balls (two of which are the same) must be colored in the remaining $n-1$ colors; for this we have $(n-1)!/2$ variants. Therefore, in this case the number of possible colorings is $\\frac{1}{4}n!(n-1)!n(n-1)(n-2) = \\frac{1}{4}n!(n^2-3n+2)$.\nIf $a = 1$, then there are $n!$ options for coloring all balls except $a$ in $n$ colors. Now there are $n$ choices for the color of $a$. Thus, in this case there are $n! \\cdot n$ possible colorings.\nFinally $a_n = \\frac{1}{4}n!(n^2 - 3n + 2 + 4n) = \\frac{1}{4}n!(n^2 + n + 2)$. If $a_n$ is a multiple of $2024 = 2^3 \\cdot 11 \\cdot 23$ and $n < 23$, it must be $23|n^2 + n + 2$. Direct inspection shows that the smallest such $n$ is $9$, but $a_9 = 9! \\cdot 23$ is not divisible by $11$, and the next suitable $n$ is $13$, where $a_{13} = 13! \\cdot 46$ is divisible by $2^3 \\cdot 11 \\cdot 23$. $\\square$",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 27584,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABC$ be a triangle, satisfying $2AC = AB + BC$. If $O$ and $I$ are its circumcenter and incenter, show that $\\angle OIB = 90^\\circ$.\n\n(Konstantin Delchev)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $D = BI \\cap (ABC)$. We apply Ptolemy's theorem\n$$\nAB \\cdot DC + BC \\cdot AD = AC \\cdot BD\n$$\nwhich implies $BD = 2DA$. From $AD = DI = DC$ it yields $I$ is midpoint of $BD$ so $\\angle OIB = 90^\\circ$.\n$\\square$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 27585,
    "subject": "Mathematics (Multi-modal)",
    "question": "$$\nA_n = 1 \\cdot 2 + 3 \\cdot 4 + 5 \\cdot 8 + \\dots + (2n - 1) \\cdot 2^n.\n$$\n(Nedyalka Dimitrova)",
    "options": [],
    "answer": "4045 * 2^{2025} + 6",
    "solution": "Since\n$$\n2A_n = 1 \\cdot 4 + 3 \\cdot 8 + \\dots + (2n - 3) \\cdot 2^n + (2n - 1) \\cdot 2^{n+1},\n$$\nit follows\n$$\n\\begin{align*}\nA_n = 2A_n - A_n &= (2n - 1) \\cdot 2^{n+1} - (1 \\cdot 2 + 2 \\cdot 4 + 2 \\cdot 8 + \\dots + 2 \\cdot 2^n) \\\\\n&= (2n - 1) \\cdot 2^{n+1} - 2 \\cdot (2 + 4 + 8 + \\dots + 2^n) + 1 \\cdot 2 \\\\\n&= (2n - 1) \\cdot 2^{n+1} - 2 \\cdot 2 \\cdot \\frac{2^n - 1}{2 - 1} + 2 = (2n - 3) \\cdot 2^{n+1} + 6.\n\\end{align*}\n$$\nTherefore $A_{2024} = 4045 \\cdot 2^{2025} + 6$. $\\square$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 27586,
    "subject": "Mathematics (Multi-modal)",
    "question": "Given the equation\n$$\n2 + x\\sqrt{9 + 6\\sqrt{2}} = x\\sqrt{5 - 2\\sqrt{6}} + \\sqrt{6} - 2\\sqrt{3} + \\sqrt{2}.\n$$\na) Write the root of the equation in the form $m - \\sqrt{n}$, where $m$ and $n$ are natural numbers.\nb) Factor the expression $a^3 - 3a^2 - 5a + 7$ into two non-constant factors with integer coefficients and calculate the value of this expression if $a$ is the root found in a).",
    "options": [],
    "answer": "a) x = 1 − √2; b) a^3 − 3a^2 − 5a + 7 = (a − 1)(a^2 − 2a − 7), and for a = 1 − √2 the value is 6√2.",
    "solution": "$$\n\\sqrt{9 + 6\\sqrt{2}} = \\sqrt{3}\\sqrt{2 + 2\\sqrt{2} + 1} = \\sqrt{3}(\\sqrt{2} + 1) = \\sqrt{6} + \\sqrt{3}\n$$\n$$\n\\sqrt{5 - 2\\sqrt{6}} = \\sqrt{3 - 2\\sqrt{6} + 2} = |\\sqrt{3} - \\sqrt{2}| = \\sqrt{3} - \\sqrt{2}.\n$$\nThe equation takes the form\n$$\nx(\\sqrt{6} + \\sqrt{3} - \\sqrt{3} + \\sqrt{2}) = \\sqrt{6} + \\sqrt{2} - 2\\sqrt{3} - 2\n$$\n$$\nx(\\sqrt{6} + \\sqrt{2}) = (\\sqrt{6} + \\sqrt{2})(1 - \\sqrt{2}),\n$$\nwhence (given $\\sqrt{6} + \\sqrt{2} > 0$) finally $x = 1 - \\sqrt{2}$.\n\nb) We have $a^3 - 3a^2 - 5a + 7 = a^3 - a^2 - 2a^2 + 2a - 7a + 7 = (a-1)(a^2 - 2a - 7)$. The product of $a-1 = -\\sqrt{2}$ and $a^2 - 2a - 7 = (a-1)^2 - 8 = 2 - 8 = -6$ is $6\\sqrt{2}$.\n$\\square$",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 27587,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all natural numbers $k$ for which there exist natural numbers $x, y$ such that the number $\\frac{x^k y}{y^2 - x^2}$ is prime.\n(Konstantin Delchev, Stanislav Harizanov)",
    "options": [],
    "answer": "2, 4",
    "solution": "Let $d = (x, y)$ be the greatest common divisor of the numbers $x$ and $y$. Then, $x = d x_1$, $y = d y_1$, where $x_1, y_1$ are coprime naturals. The expression in the condition can be rewritten as\n$$\n\\frac{d^{k-1}}{y_1^2 - x_1^2} x_1^k y_1 = p,\n$$\nwhere we want $p$ to be prime. But $(x_1^k, y_1^2 - x_1^2) = (x_1, y_1)^{\\min(k,2)} = 1 = (y_1, y_1^2 - x_1^2)$, i.e., $x_1^k y_1$ is coprime to $y_1^2 - x_1^2$ and for $p$ to be whole it takes $y_1^2 - x_1^2 \\mid d^{k-1}$. Moreover, $y_1 > x_1$ and therefore $x_1^k y_1 > 1$. Therefore, to be $p$ it is simply necessary that $x_1 = 1$, $y_1 = p$ and $y_1^2 - x_1^2 = d^{k-1}$ are fulfilled at the same time. If $k=1$, then $1 = p^2 - 1 = (p-1)(p+1)$, which is impossible. At $k=2$ we want $d = p^2 - 1$ and here the pair $(x, y) = (p^2 - 1, p(p^2 - 1))$ has the desired\n\nproperty for every prime $p$. Therefore, $k=2$ is a solution. With $k > 2$ we want\n$d^{k-1} = p^2 - 1 = (p-1)(p+1)$. It is directly verified that $p=2$ does not lead to\na solution, since 3 is not the $k-1$th power of a natural number. Hence, $p$ is an\nodd number and $8 \\mid p^2 - 1$, whence $k-1 \\ge 3$. Therefore, $k=3$ is not a solution.\n$k=4$ is a solution at $p=3$ and $d=2$, i.e., $(x, y) = (2, 6)$. $\\square$",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 27588,
    "subject": "Mathematics (Multi-modal)",
    "question": "The inscribed circle in $\\triangle ABC$ ($AC \\neq BC$) is tangent to its sides $AB$, $BC$, and $CA$ at points $D$, $E$, and $F$, respectively. Let $P$ be the foot of the perpendicular from $D$ to $EF$ ($P \\in EF$). If the circles circumscribed about $\\triangle ABC$ and $\\triangle EFC$ intersect for the second time at point $Q$, prove that $\\angle PQC = 90^\\circ$.\n\n(Stoyan Boev)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "From the fact that $C$ lies on the circle circumscribed about $\\triangle FEQ$ and $CE = CF$ it follows that $CQ$ is an exterior bisector of $\\angle FQE$ and it remains to prove that $QP$ is a bisector of $\\angle FQE$, i.e. $QF : QE = FP : PE$. From $\\angle QFC = \\angle QEC$ and $\\angle QAC = \\angle QBC$ it follows that $\\triangle QFA \\sim \\triangle QEB$, i.e.\n$$\nQF : QE = AF : BE = AD : BD.\n$$\nLet $I$ be the center of the circle $k$ inscribed in $\\triangle ABC$ and the line $DP$ intersects $k$ for the second time at point $R$. Then\n$$\n\\angle REF = \\angle RDF = 90^\\circ - \\angle DFE = 90^\\circ - \\angle DIB = \\angle IBA\n$$\nand analogously $\\angle RFE = \\angle IAB$, i.e. $\\triangle FER \\sim \\triangle ABI$. But $RP \\perp EF$, $ID \\perp AB$, i.e. $P$ and $D$ are corresponding elements in similar triangles and $FP : PE = AD : BD$, which completes the proof. $\\square$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 27589,
    "subject": "Mathematics (Multi-modal)",
    "question": "Given a natural number $n \\ge 3$. To find the smallest real number $k > 0$ with the following property: If $G$ is a connected graph with $n$ vertices and $m$ edges, then it is always possible to delete no-more than $k \\cdot \\left(m - \\left\\lfloor \\frac{n}{2} \\right\\rfloor\\right)$ edges so that vertices can be colored in two colors and every undeleted edge has multi-colored vertices.\n(Alexander Ivanov)",
    "options": [],
    "answer": "1/2",
    "solution": "**Lemma:** Let $G$ be a connected graph with at least $3$ vertices. Then either there exist two vertices connected by an edge whose removal (along with the outgoing edges) leaves $G$ connected, or there exist two vertices of degree $1$ (i.e., \"leaves\").\nConsider an arbitrary \"covering tree\" of $G$ and take as its \"root\" an arbitrary vertex that is not a \"leaf\". Let $v$ be the farthest vertex from the \"root\" and $u$ be its \"ancestor\". Let $v_1, v_2, \\dots, v_k$ be the \"successors\" of $u$. Clearly, they are all leaves on the tree.\n\nCase 1. Among $v_1, v_2, \\dots, v_k$ there are two vertices connected by an edge in $G$.\nThen removing these two vertices leaves the tree (and therefore $G$) connected.\n\nCase 2. Among $v_1, v_2, \\dots, v_k$ there are two vertices that are leaves in $G$. Then there are indeed at least two leaves in the output graph.\n\nCase 3. Among $v_1, v_2, \\dots, v_k$ there is at most one vertex that is a leaf in $G$ (b.o.o., let it be $v_1$). Then let us \"connect\" each of $v_2, \\dots, v_k$ to an arbitrary vertex in $G$ other than $u$ (such vertices exist, and none of these \"connecting\" edges are part of the covering tree, due to the extreme choice of $u$, i.e., each of them is part of a loop, all remaining edges of which are from the covering tree). Now we can remove $u$ and $v_1$ and we will have a spanning tree again, so $G$ remains connected.\n\nThis proves the lemma. With its help we can easily prove the following\n\n**Assertion:** Let $G$ be a connected graph with $n \\ge 2$ vertices. Then we can color its vertices in two colors, so that if $x$ and $y$ are the number of \"multicolored\" and \"single colored\" edges, respectively, then $x - y \\ge \\lfloor \\frac{n}{2} \\rfloor$.\n\nProof: For $n = 2, 3$ the statement is immediately verified. Let $n \\ge 4$ and $G$ be a connected graph with $n$ vertices. Let $u$ and $v$ be the two vertices from the Lemma. We \"remove\" $u$ and $v$ and color $G \\setminus \\{u, v\\}$ according to the induction hypothesis. Now, it is not difficult to see that we can color $u \\cup v$ such that the difference under consideration increases by at least $1$. Indeed, this is clear if $u$ and $v$ are leaves, and otherwise case, considering the parity of the number of neighbors of $u \\cup v$ in $G \\setminus \\{u, v\\}$, we see that there is always such a way. The statement is proved by induction.\n\nLet us now consider an arbitrary connected graph $G$ with $n \\ge 3$ vertices and $m$ edges. We \"color\" it according to the Assertion: we have $x - y \\ge \\lfloor \\frac{n}{2} \\rfloor$; $x + y = m$, therefore\n$$\ny \\le \\frac{1}{2} \\cdot \\left( m - \\left\\lfloor \\frac{n}{2} \\right\\rfloor \\right)\n$$\nand deleting $y$ edges satisfies the condition. Thus, we got $k \\le \\frac{1}{2}$.\n\nTo show that $k \\ge \\frac{1}{2}$ let us consider the complete graph with $n$ vertices. A necessary and sufficient condition for having the coloring from the condition is that the graph obtained after deleting the edges is bipartite. Indeed, there must be no cycles of odd length in the graph, which is equivalent to the above.\n\nCase 1. $n = 2n_1$, $n_1 \\ge 2$. Then $m = \\binom{n_1}{2}$. To reach a bipartite graph, we need to delete all the edges in each of the two groups of vertices, and the number of deleted edges is minimal when the two groups are of equal power and contain $n_1$ vertices. So, we need to delete at least\n$$\n\\binom{n_1}{2} + \\binom{n_1}{2} = n_1^2 - n_1\n$$\n$$\nn_1^2 - n_1 \\le k \\cdot \\left( \\frac{2n_1(2n_1-1)}{2} - n_1 \\right) \\implies k \\ge \\frac{1}{2}.\n$$\n\nCase 2. $n = 2n_1 + 1$, $n_1 \\ge 1$. Similarly, here we need to delete at least\n$$\n\\binom{n_1+1}{2} + \\binom{n_1}{2} = n_1^2\n$$\nedges and again\n$$\nn_1^2 \\le k \\cdot \\left( \\frac{2n_1(2n_1+1)}{2} - n_1 \\right) \\implies k \\ge \\frac{1}{2}. \\quad \\square",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 27590,
    "subject": "Mathematics (Multi-modal)",
    "question": "The first, seventh, and seventeenth terms of an arithmetic progression are distinct and consecutive terms of a geometric progression. To find the difference of the arithmetic progression if its first term is a solution of the equation\n$$\nx^2 - 9x + x\\sqrt{12-x} - 9\\sqrt{12-x} = 0.\n$$",
    "options": [],
    "answer": "d = 1 or d = -4/9",
    "solution": "Let $a_1$ and $d$ be the first term and the difference of the arithmetic progression, respectively. From the condition $a_1$, $a_1 + 6d$ and $a_1 + 16d$ are consecutive members of a geometric progression, i.e.\n$$\n(a_1 + 6d)^2 = a_1 \\cdot (a_1 + 16d) \\iff d \\cdot (a_1 - 9d) = 0.\n$$\nSince $d \\ne 0$, we get that $a_1 = 9d$. Furthermore, we have $(x-9)(x+\\sqrt{12-x}) = 0$ and $x \\le 12$. Then $x = 9$ or $\\sqrt{12-x} = -x$, i.e. $x^2 + x - 12 = 0$ and $x \\le 0$, whence $x = -4$. Then $a_1 = 9$ and $a_1 = -4$, as $d = 1$ and $d = -\\frac{4}{9}$, respectively. $\\square$",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 27591,
    "subject": "Mathematics (Multi-modal)",
    "question": "Points $D$ and $E$ are on sides $BC$ and $AC$ of $\\triangle ABC$. Lines $AD$ and $BE$ intersect at point $S$. Point $F$ is on side $AB$ and lines $FE$ and $FD$ intersect line $l$ passing through $C$ and parallel to $AB$ at points $P$ and $Q$. Prove that if $CP = CQ$, then the points $C$, $S$ and $F$ lie on the same line.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "From the similarities $\\triangle PEC \\sim \\triangle FEA$ and $\\triangle CDQ \\sim \\triangle BDF$ we obtain that\n\n$$\n\\frac{CP}{AF} = \\frac{CE}{AE} \\quad \\text{and} \\quad \\frac{CQ}{BF} = \\frac{CD}{BD}.\n$$\nTherefore $\\frac{CP}{CQ} = \\frac{CE}{AE} \\cdot \\frac{AF}{BF} \\cdot \\frac{BD}{DC}$. By condition $CP = CQ$, it follows that\n$$\n\\frac{CE}{AE} \\cdot \\frac{AF}{BF} \\cdot \\frac{BD}{DC} = 1.\n$$\nApplying Cheva's theorem for $\\triangle ABC$ and the points $F$, $D$, and $E$, it follows that the lines $AD$, $BE$, and $CF$ intersect in one point, i.e. point $S$ lies on $CF$. $\\square$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 27592,
    "subject": "Mathematics (Multi-modal)",
    "question": "Given a rational number $q > 3$ such that $q^2 - 4$ is the square of a rational number. The row $\\{a_i\\}_{i=0}^{\\infty}$ is defined as follows:\n$$\na_0 = 2, \\ a_1 = q, \\ a_{i+1} = q a_i - a_{i-1}, \\text{ for each } i = 1, 2, \\dots\n$$\n\nDo there exist a natural number $n$ and nonzero integers $b_0, b_1, \\dots, b_n$ such that $\\sum_{i=0}^n b_i = 0$ and if we write the number $b_0 a_0 + b_1 a_1 + \\dots + b_n a_n$ in the form $\\frac{A}{B}$, where $A$ and $B$ are co-prime integers, then is the number $A$ free of squares?",
    "options": [],
    "answer": "Detailed solution",
    "solution": "We will prove that such numbers do not exist.\n\nThe quadratic equation $x^2 - qx + 1 = 0$ has two rational roots $t$ and $\\frac{1}{t}$ for which $t + \\frac{1}{t} = q$. It easily follows by induction that $a_m = t^m + \\frac{1}{t^m}$. Let's assume that there exist numbers $b_0, b_1, \\dots, b_n$ satisfying the condition of the problem.\n\n**Lemma.** Let $f(x) = c_n x^n + c_{n-1} x^{n-1} + \\dots + c_1 x + c_0$ be a polynomial with nonzero integer coefficients for which $c_{n-k} = c_k$ for each $k = 0, 1, \\dots, n$ and $\\sum_{i=0}^n c_i = 0$. Then $f(x) = (x-1)^2 g(x)$, where $g(x)$ is a polynomial with integer coefficients.\n\n**Proof:** From the condition we have that $f(1) = 0$. As\n$$\nf'(x) = n c_n x^{n-1} + (n-1) c_{n-1} x^{n-2} + \\dots + c_1,\n$$\nit follows that\n$$\n2 f'(1) = (n c_n + (n-1) c_{n-1} + \\dots + c_1) + (n c_0 + (n-1) c_1 + \\dots + c_{n-1}) = n(c_n + c_{n-1} + \\dots + c_1 + c_0) = 0\n$$\nand therefore $x = 1$ is a double root. The lemma is proved.\n\nThe polynomial $f(x) = b_n x^{2n} + b_{n-1} x^{2n-1} + \\dots + b_1 x^{n+1} + 2b_0 x^n + b_1 x^{n-1} + b_2 x^{n-2} + \\dots + b_{n-1} x + b_n$ satisfies the conditions of the lemma. It's not hard to see that\n$$\nt^n \\left( b_n \\left( t^n + \\frac{1}{t^n} \\right) + b_{n-1} \\left( t^{n-1} + \\frac{1}{t^{n-1}} \\right) + \\dots + 2b_0 \\right) = f(t) = (t-1)^2 g(t).\n$$\nMoreover, if $t = \\frac{r}{s}$, $(r, s) = 1$, then from $\\frac{r}{s} + \\frac{s}{r} = q > 3$ it easily follows that $r \\ge s + 2$, i.e. $r - s \\ge 2$. Then $g\\left(\\frac{r}{s}\\right)$ is a number of the form $\\frac{l}{s^{2n-2}}$.\n\nFinally $b_0 a_0 + b_1 a_1 + \\dots + b_n a_n$ is presented in the form $\\frac{(r-s)^2 l}{r^n s^n}$ and because the number $r-s \\ge 2$ and $(r-s, r) = (r-s, s) = 1$, then the numerator will always square a prime number. $\\square$",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 27593,
    "subject": "Mathematics (Multi-modal)",
    "question": "The sides and diagonals of a regular $n$-gon are colored in $k \\ge 3$ colors. For each color $i$, between every two vertices of the polygon there exists a path consisting only of segments of color $i$. Prove that there exist three vertices of the polygon $A$, $B$, and $C$ such that the line segments $AB$, $BC$, and $AC$ are multicolor.\n\n(Emil Kolev)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "We need to prove that in a complete graph with $n$ vertices and $k$ colors, where the induced graph on each color is connected, there exists a multicolor triangle. Denote the colors by $1$, $2$, $3$, $\\ldots$, $k$ and recolor all edges that are in any of the colors $4$, $5$, $\\ldots$, $k$ into color $3$. The new graph satisfies the condition of connectivity on each color and if there is a multicolored triangle for it, then the same triangle in the initial graph will also be multicolored. Therefore, we can consider that $k = 3$.\n\nSuppose that the statement is not true for a graph $G$, as we can choose $G$ to have a minimal number of vertices. It follows from the minimality of $G$ that after removing any vertex, the new graph will not be connected by any of the colors, and let it be color $1$. Denote by $G_1, G_2, \\dots, G_t$ the color connectivity components $1$ after deleting vertex $A$. Since $G$ is color $1$ connected, there exist $A_i \\in G_i$ for which $AA_i$ is color $1$. The segment $A_1A_2$ is not color $1$ because $G_1$ and $G_2$ are different components of connectivity. Let it be of color $2$. If $A_1B$ is a color $1$ segment of $G_1$, then the segment $A_2B$ cannot be of color $1$ because $G_1$ and $G_2$ are different connectivity components; cannot be of color $3$, because then $A_1A_2B$ is a multicolor triangle and is therefore of color $2$. Analogously, it is proved that all segments between the points of $G_1$ and $G_2$ are of color $2$. We obtained that all segments between any two connectivity components are either color $2$ or color $3$.\n\nNow consider segments $AX$ and $AY$ in colors $2$ and $3$, respectively (such segments exist, since $G$ is connected in each of the colors). Without limitation, we have the following two cases:\n\n1. $X, Y \\in G_1$, then one of the triangles $AXA_2$ and $AYA_2$ is multicolored.\n\n2. $X \\in G_1$ and $Y \\in G_2$, then one of the triangles $AXA_2$ and $AYA_1$ is multicolored.\n\nThe resulting contradiction shows that for every graph with the given properties there exists a multicolored triangle. $\\Box$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 27594,
    "subject": "Mathematics (Multi-modal)",
    "question": "Maria and Bilyana play the following game. Maria has 2024, and Bilyana has 2023 fair coins. Coins are tossed randomly - the probability of each individual coin being heads after the toss is $\\frac{1}{2}$. Maria wins if there are strictly more heads among her coins than Bilyana's, otherwise Bilyana wins. What is the probability that Maria wins?\n(Kristiyan Vasilev)",
    "options": [],
    "answer": "1/2",
    "solution": "Let $p$ be the probability that Maria has more heads than Bilyana after tossing the first 2023 of Maria's coins. Then, for reasons of symmetry, the probability that Maria has fewer heads than Bilyana is also $p$, and therefore the probability that Maria and Bilyana have thrown an equal number of heads is $1 - 2p$. If Maria has thrown less heads than Bilyana to this moment, the probability of her winning is 0 (regardless of the last coin), if she flipped more heads, the probability of her winning is 1 (again, regardless of the last coin), and if they flipped an even number, the probability of her winning is $\\frac{1}{2}$ (here the last coin must be heads).\n\nThus we get that the probability that Maria wins is $p + \\frac{1-2p}{2} = \\frac{1}{2}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 27595,
    "subject": "Mathematics (Multi-modal)",
    "question": "We will call a natural number $m$ remarkable if there exist integers $a$, $b$, $c$, for which $m = a^3 + 2b^3 + 4c^3 - 6abc$. Prove that there exists a natural number $n < 2024$ such that for infinitely many prime numbers $p$, the number $np$ is remarkable.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Lemma. Let $p$ be a prime number and $a$, $b$, $c \\in \\mathbb{Z}/p\\mathbb{Z}$. Then there exist $x$, $y$, $z \\in \\mathbb{Z}$ such that $|x|$, $|y|$, $|z| < \\sqrt[3]{p}$, $(x, y, z) \\neq (0, 0, 0)$ and $ax + by + cz \\equiv 0 \\pmod{p}$.\n\n*Proof.* Consider the set $M := \\{(x, y, z) : x, y, z \\in \\{0, 1, \\dots, \\lfloor \\sqrt[3]{p} \\rfloor\\}\\}$. We have that $|M| > p$, i.e. in $M$ there are two distinct elements $(x_1, y_1, z_1)$ and $(x_2, y_2, z_2)$, for which $ax_1 + by_1 + cz_1 \\equiv ax_2 + by_2 + cz_2 \\pmod{p}$. Thus $(x_1 - x_2, y_1 - y_2, z_1 - z_2)$ satisfies the conditions of the lemma.\n\nNow let $p \\equiv 2 \\pmod{3}$. Then the comparison $x^3 \\equiv 2 \\pmod{p}$ has a solution $a$, since the function $x \\mapsto x^3$ is injective in $\\mathbb{Z}/p\\mathbb{Z}$, hence it is surjective. One way to verify this is to see that $x^3 \\equiv 1 \\pmod{p}$ has only $1$ for a solution, since $(3, p-1) = 1$ and hence the exponent of $x$ (mod $p$) is $1$.\n\nFrom the lemma, there exist $x$, $y$, $z$ with $|x|$, $|y|$, $|z| < \\sqrt[3]{p}$, for which $x + ay + a^2z \\equiv 0 \\pmod{p}$. From here we get that $x^3 + a^3y^3 + a^6z^3 - 3a^3xyz \\equiv 0 \\pmod{p}$. The latter is equivalent to $x^3 + 2y^3 + 4z^3 - 6xyz \\equiv 0 \\pmod{p}$.\n\nOn the other hand $|x|$, $|y|$, $|z| < \\sqrt[3]{p}$ gives $|x^3 + 2y^3 + 4z^3 - 6xyz| < 13p$. Also notice that if $x^3 + 2y^3 + 4z^3 - 6xyz < 0$, then the triple $(-x, -y, -z)$ will give a natural number divisible by $p$.\n\nIt remains to note that $x^3 + 2y^3 + 4z^3 - 6xyz \\neq 0$. Indeed, if $x^3 + 2y^3 + 4z^3 - 6xyz = 0$, then we have $(x + \\sqrt[3]{2}y + \\sqrt[3]{4}z)((x - \\sqrt[3]{2}y)^2 + (x - \\sqrt[3]{4}z)^2 + (\\sqrt[3]{2}y - \\sqrt[3]{4}z)^2) = 0$, i.e. $x = \\sqrt[3]{2}y = \\sqrt[3]{4}z$ or $x + y\\sqrt[3]{2} + z\\sqrt[3]{4} = 0$, whose only integer solutions are $(x, y, z) = (0, 0, 0)$ (the first follows because $\\sqrt[3]{2}$ is irrational, and the second follows from the fact that $x^3 - 2$ is the minimal polynomial of $\\sqrt[3]{2}$ over the rational numbers). $\\square$",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 27596,
    "subject": "Mathematics (Multi-modal)",
    "question": "Given a natural number $n$. We have $n + 1$ balls numbered $1$, $1$, $2$, $3$, ..., $n$ (only the first two are the same). We need to color these balls in $n$ given colors so that every ball is a single color and every color is used at least once. We denote by $a_n$ the number of possible colorings. Find the smallest $n$ for which $a_n$ is divisible by $2024$.\n(Ivaylo Kortezov)",
    "options": [],
    "answer": "13",
    "solution": "Exactly one of the colors will be used for two of the balls; let their numbers be $a$ and $b$, such that $a \\le b$.\n\nIf $a > 1$, then we have $(n-1)(n-2)/2$ choices for $a$ and $b$, and $n$ choices for their color. The remaining $n-1$ balls (two of which are the same) must be colored in the remaining $n-1$ colors; for this we have $(n-1)!/2$ variants. Therefore, in this case the number of possible colorings is $\\frac{1}{4}(n-1)!.n(n-1)(n-2) = \\frac{1}{4}n!(n^2-3n+2)$.\n\nIf $a = 1$, then there are $n!$ options for coloring all balls except $a$ in $n$ colors. Now there are $n$ choices for the color of $a$. Thus, in this case there are $n!.n$ possible colorings.\n\nFinally $a_n = \\frac{1}{4}n!(n^2 - 3n + 2 + 4n) = \\frac{1}{4}n!(n^2 + n + 2)$.\n\nIf $a_n$ is a multiple of $2024 = 2^3 \\cdot 11 \\cdot 23$ and $n < 23$, it must be $23|n^2 + n + 2$. Direct inspection shows that the smallest such $n$ is $9$, but $a_9 = 9!.23$ is not divisible by $11$, and the next suitable $n$ is $13$, where $a_{13} = 13!.46$ is divisible by $2^3 \\cdot 11 \\cdot 23$. $\\square$",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 27597,
    "subject": "Mathematics (Multi-modal)",
    "question": "We will call a natural number Yambolian if it can be represented in the form $a^2 + 6ab + b^2$, where $a$ and $b$ are (not necessarily different) natural numbers. The number $36^{2024}$ is written as the sum of $k$ number of (not necessarily distinct) Yambol numbers. What is the smallest possible value of $k$?\n(Miroslav Marinov)",
    "options": [],
    "answer": "2",
    "solution": "We will first show that $k = 1$ is not possible, i.e. $a^2 + 6ab + b^2 = 2^{4048} \\cdot 3^{4048}$ has no solution in natural numbers. If such $a, b$ exist, then $a^2 + b^2$ is divisible by $3$ and therefore $a$ and $b$ are divisible by $3$. Writing $a = 3a_1$, $b = 3b_1$ and dividing by $3^2$, we get $a_1^2 + 6a_1b_1 + b_1^2 = 2^{4048} \\cdot 3^{4046}$; repeating this argument another $2023$ times, we arrive at an equation of the form\n$$\nu^2 + 6\\nu\\upsilon + \\upsilon^2 = 2^{4048}$$\nwhere $\\nu$ and $\\upsilon$ are natural numbers.\n\nIf $\\upsilon$ is even, then $\\nu$ is even; writing $\\nu = 2u_1$, $\\upsilon = 2v_1$ and dividing by $4$, we get $u_1^2 + 6u_1v_1 + v_1^2 = 2^{4046}$; repeating this several times, we arrive at an equation of the form\n$$\ns^2 + 6st + t^2 = 2^{2A},\n$$\nwhere $s, t, A$ are natural numbers, $t$ is odd and $A \\geq q \\geq 2$ (since the left side is greater than or equal to $8$). The latter is equivalent to $(s+3t)^2 - 8t^2 = 2^{2A}$. Now from modulo $8$ we see that $(s+3t)^2$ is divisible by $8$, so $s+3t$ is divisible by $4$, i.e. $(s+3t)^2$ is divisible by $16$. But then $8t^2$ must be divisible by $16$, which is impossible for an odd $t$, a contradiction. Therefore, $k=1$ is not possible.\n\nFor an example with $k=2$ let us first notice that\n$$\n36 = [1^2 + 6 \\cdot 1 \\cdot 1 + 1^2] + [3^2 + 6 \\cdot 3 \\cdot 1 + 1^2]\n$$\nand now multiplying by $(6^{2023})^2$ leads to\n$$\n36^{2024} = [(6^{2023})^2 + 6 \\cdot 6^{2023} \\cdot 6^{2023} + (6^{2023})^2] + [(3 \\cdot 6^{2023})^2 + 6 \\cdot (3 \\cdot 6^{2023}) \\cdot 6^{2023} + (6^{2023})^2]\n$$\ni.e. $36^{2024}$ is the sum of the numbers $a_1^2 + 6a_1b_1 + b_1^2$ and $a_2^2 + 6a_2b_2 + b_2^2$, where $a_1 = b_1 = b_2 = 6^{2023}$ and $a_2 = 3 \\cdot 6^{2023}$.\n\n$\\square$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 27598,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $p > q$ be primes, such that $240 \\nmid p^4 - q^4$. Find the maximal value of $\\frac{q}{p}$.\n(Angel Gushev)",
    "options": [],
    "answer": "5/7",
    "solution": "$\\square$",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 27599,
    "subject": "Mathematics (Multi-modal)",
    "question": "There are 11 points equally spaced on a circle. Some of the segments having endpoints among these vertices are drawn and colored in two colors, so that each segment meets at an internal point at most one other segment from the same color. What is the greatest number of segments that could be drawn?\n(Mladen Vylkov)",
    "options": [],
    "answer": "35",
    "solution": "![](attached_image_1.png)\n\nPut $n$ instead of 11 and let the points be $A_1, A_2, \\dots, A_n$. We may assume that the points are vertices of a regular $n$-gon. Let us first calculate the maximum number of diagonals that we can draw so that any diagonal meets no more than one other diagonal at an interior point. Note that we exclude the sides of the $n$-gon, and are interested only in its diagonals. We denote this number by $R(n)$. Obviously $R(2) = 0$, $R(3) = 0$. We will prove by induction that\n$$\nR(n) = \\left\\lceil \\frac{3(n-3)}{2} \\right\\rceil, \\quad n \\ge 3\n$$\nFor small cases of $n$ it is checked directly. For example $R(4) = 2$, $R(5) = 3$, $R(6) = 5$. Assume the formula is already checked for all values less than $n$ for some $n \\in \\mathbb{N}$, $n \\ge 5$. Let $A_1A_j$ and $A_iA_k$ be diagonals of an extremal configuration of $n$ points. Denote by $G_1$ the group of vertices that lie between $A_1$ and $A_k$ (including $A_1$ and $A_k$) and let their number be $n_1$. In the same way let $G_2, G_3, G_4$ be the groups of vertices that lie respectively between $A_k$ and $A_j$, $A_j$ and $A_i$, $A_i$ and $A_1$ and their corresponding numbers be $n_2, n_3, n_4$.\nThe key observation is that all the other diagonals connect points that are in the same group. The maximum number of diagonals that connect points in $G_1$ is $R(n_1)$. Note that if $n_1 \\ge 2$ then $A_1A_k$ connects non-adjacent vertices and this diagonal takes part in $R(n)$ but doesn't take part in $R(n_1)$ (the dotted diagonals on fig. 1). The same holds for the other groups. The red diagonals also should\n\n![](attached_image_2.png)\n\nbe added so we obtain,\n$$\nR(n) = \\sum_{i=1}^{4} R(n_i) + \\sum_{i=1}^{4} 1_{n_i>2} + 2.\n$$\nHere, by $1_{n_i>2}$ we mean the indicator function that shows whether $n_i$ is greater than 2, i.e. $1_{n_i>2} = 1$ if $n_i > 2$ and 0 otherwise. Hence,\n$$\nR(n) = \\sum_{i=1}^{4} \\left\\lceil \\frac{3(n_i - 3)}{2} \\right\\rceil + \\sum_{i=1}^{4} 1_{n_i > 2} + 2 \\quad (1).\n$$\nand completes the induction step. Back to the original problem. Remove the sides of the $n$-gon. Fix a color. The number of diagonals of that color is at most $\\lfloor \\frac{3(n-3)}{2} \\rfloor$. The same holds for the other color. Adding the sides of the $n$-gon, which can be colored as we want, we get that the greatest number of segments is at most\n$$\n2 \\left\\lfloor \\frac{3(n-3)}{2} \\right\\rfloor + n.\n$$\nThere is no problem to give an example of this number of segments that satisfy the condition. □",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 27600,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find $A_{2024}$, where\n$$\nA_n = 1 \\cdot 2 + 3 \\cdot 4 + 5 \\cdot 8 + \\dots + (2n-1) \\cdot 2^n.\n$$",
    "options": [],
    "answer": "4045 * 2^2025 + 6",
    "solution": "Since\n$$\n2A_n = 1 \\cdot 4 + 3 \\cdot 8 + \\dots + (2n-3) \\cdot 2^n + (2n-1) \\cdot 2^{n+1},\n$$\nit follows\n$$\n\\begin{align*}\nA_n = 2A_n - A_n &= (2n-1) \\cdot 2^{n+1} - (1 \\cdot 2 + 2 \\cdot 4 + 2 \\cdot 8 + \\dots + 2 \\cdot 2^n) \\\\\n&= (2n-1) \\cdot 2^{n+1} - 2 \\cdot (2 + 4 + 8 + \\dots + 2^n) + 1 \\cdot 2 \\\\\n&= (2n-1) \\cdot 2^{n+1} - 2 \\cdot 2 \\cdot \\frac{2^n-1}{2-1} + 2 = (2n-3) \\cdot 2^{n+1} + 6.\n\\end{align*}\n$$\nTherefore $A_{2024} = 4045 \\cdot 2^{2025} + 6$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 27601,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all pairs $(x, y)$ of real numbers for which\n$$\n4y^4 + x^4 + 12y^3 + 5x^2(y^2 + 1) + y^2 + 4 = 12y.\n$$",
    "options": [],
    "answer": "(0, -2) and (0, 1/2)",
    "solution": "We have the inequalities $x^4 \\geq 0$, $5x^2(y^2 + 1) \\geq 0$ and $4y^4 + 12y^3 + y^2 - 12y + 4 = (2y - 1)^2(y + 2)^2 \\geq 0$. The sum of the left sides is $0$ if and only if each of them is equal to $0$. The first two lead to $x = 0$, and the third to $y = -2$ or $y = \\frac{1}{2}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 27602,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all pairs $(x, y)$ of real numbers for which\n$$\n4y^4 + x^4 + 12y^3 + 5x^2(y^2 + 1) + y^2 + 4 = 12y.\n$$",
    "options": [],
    "answer": "(0, -2) and (0, 1/2)",
    "solution": "We have the inequalities $x^4 \\geq 0$, $5x^2(y^2+1) \\geq 0$ and $4y^4 + 12y^3 + y^2 - 12y + 4 = (2y-1)^2(y+2)^2 \\geq 0$. The sum of the left sides is 0 if and only if each of them is equal to 0. The first two lead to $x = 0$, and the third to $y = -2$ or $y = \\frac{1}{2}$. $\\square$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 27603,
    "subject": "Mathematics (Multi-modal)",
    "question": "A graph $G$ is called *divisibility graph* if the vertices can be assigned distinct positive integers such that between two vertices assigned $u, v$ there is an edge iff $\\frac{u}{v}$ or $\\frac{v}{u}$ is a positive integer. Show that for any positive integer $n$ and $0 \\le e \\le \\frac{n(n-1)}{2}$, there is a divisibility graph with $n$ vertices and $e$ edges.\n(Danila Cherkashin)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "We reason inductively on $n$, not writing the number $1$ at any vertex. For $n=1$ the requested is clear, for $n=2$ an example with $e=1$ is $(2, 4)$ and an example with $e=0$ is $(2, 3)$. For $n=3$ example with $e=0$ is $3, 5, 7$, example with $e=1$ is $2, 4, 7$, example with $e=2$ is $2, 4, 10$, an example with $e=3$ is $2, 4, 8$.\nFor $n \\ge 4$ we have $n-1 \\le \\frac{(n-1)(n-2)}{2}$, so at least one of $e \\ge n-1$ and $e \\le \\frac{(n-1)(n-2)}{2}$ is satisfied.\nLet $e \\ge n-1$ first. In an example with $n-1$ vertices and $e - (n-1)$ edges, we add a vertex (of degree $n-1$) by writing in it a prime number $p$ greater than the numbers in the other vertices, then we multiply the numbers in the remaining vertices by $p$ – this does not spawn new edges between the remaining vertices.\nNow let $e \\le \\frac{(n-1)(n-2)}{2}$. In an example with $n-1$ vertices and $e$ edges, we add a vertex (of degree $0$), writing in it a prime number $p$ greater than the numbers in the other vertices, and do not change the numbers in the other vertices – it does not spawn new edges.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 27604,
    "subject": "Mathematics (Multi-modal)",
    "question": "a) Find all values of $a$ for which the inequality\n$$\nx \\log_{\\frac{1}{2}} a^4 - x^2 > 3 + 2 \\log_2 a^2\n$$\nhas a solution.\n\nb) Calculate the limit\n$$\n\\lim_{a \\to -\\infty} \\left( \\sqrt{a^2 - a + 1} + a \\right).\n$$",
    "options": [],
    "answer": "a ∈ (-∞, -2√2) ∪ (-√2/2, 0) ∪ (0, √2/2) ∪ (2√2, ∞); limit = 1/2",
    "solution": "a) Since $\\log_{\\frac{1}{2}}(a^4) = -2 \\cdot \\log_2(a^2)$, then by putting $2\\log_2(a^2) = b$, we get the inequality $x^2 + b \\cdot x + 3 + b < 0$. For this inequality to have at least one solution, it is necessary and sufficient that $D = b^2 - 4b - 12 > 0$ whose solutions are $b < -2$ or $b > 6$, whence $\\log_2(a^2) < -1$ or $\\log_2(a^2) > 3$. From the properties of the logarithmic function, we get $a^2 < \\frac{1}{2}$ or $a^2 > 8$ and $a \\neq 0$. Final\n$$\na \\in (-\\infty; -2\\sqrt{2}) \\cup \\left(-\\frac{\\sqrt{2}}{2}; 0\\right) \\cup (0; \\frac{\\sqrt{2}}{2}) \\cup (2\\sqrt{2}; \\infty).\n$$\n\nb)\n$$\n(1) \\quad \\sqrt{a^2 - a + 1} + a = \\frac{(\\sqrt{a^2 - a + 1} + a)(\\sqrt{a^2 - a + 1} - a)}{\\sqrt{a^2 - a + 1} - a} = \\frac{-1 + \\frac{1}{a}}{-\\sqrt{1 - \\frac{1}{a} + \\frac{1}{a^2}} - 1}.\n$$\nTherefore\n$$\n\\lim_{a \\to -\\infty} (\\sqrt{a^2 - a + 1} + a) = \\frac{1}{2}.\n$$",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 27605,
    "subject": "Mathematics (Multi-modal)",
    "question": "a) Find all values of $a$ for which the inequality\n$$\nx \\log_{\\frac{1}{2}} a^4 - x^2 > 3 + 2 \\log_2 a^2\n$$\nhas a solution.\n\nb) Calculate the limit\n$$\n\\lim_{a \\to -\\infty} \\left( \\sqrt{a^2 - a + 1} + a \\right).\n$$",
    "options": [],
    "answer": "Part (a): a ∈ (-∞, -2√2) ∪ (-√2/2, 0) ∪ (0, √2/2) ∪ (2√2, ∞). Part (b): 1/2.",
    "solution": "a) Since $\\log_{\\frac{1}{2}}(a^4) = -2 \\log_2(a^2)$, then by putting $2 \\log_2(a^2) = b$, we get the inequality $x^2 + b \\cdot x + 3 + b < 0$. For this inequality to have at least one solution, it is necessary and sufficient that $D = b^2 - 4b - 12 > 0$ whose solutions are $b < -2$ or $b > 6$, whence $\\log_2(a^2) < -1$ or $\\log_2(a^2) > 3$. From the properties of the logarithmic function, we get $a^2 < \\frac{1}{2}$ or $a^2 > 8$ and $a \\ne 0$. Final\n$$\na \\in (-\\infty; -2\\sqrt{2}) \\cup (-\\frac{\\sqrt{2}}{2}; 0) \\cup (0; \\frac{\\sqrt{2}}{2}) \\cup (2\\sqrt{2}; \\infty).\n$$\n\nb) Since $a < 0$, we get:\n$$\n(1) \\qquad \\sqrt{a^2 - a + 1} + a = \\frac{(\\sqrt{a^2 - a + 1} + a)(\\sqrt{a^2 - a + 1} - a)}{\\sqrt{a^2 - a + 1} - a} = \\frac{-1 + \\frac{1}{a}}{-\\sqrt{1 - \\frac{1}{a} + \\frac{1}{a^2}} - 1}.\n$$\nTherefore\n$$\n\\lim_{a \\to -\\infty} \\left( \\sqrt{a^2 - a + 1} + a \\right) = \\frac{1}{2}.\n$$",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 27606,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABCD$ be a parallelogram and a circle $k$ passes through $A$, $C$ and meets rays $AB$, $AD$ at $E$, $F$. If $BD$, $EF$ and the tangent at $C$ concur, show that $AC$ is diameter of $k$.\n\n(Adelina Chopanova)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let the tangent to $k$ at point $C$ intersect the rays $AB \\to$ and $AD \\to$ at the points $M$ and $N$, respectively, and the lines $BD$, $EF$ and the tangent intersect at point $P$. After applying Menelaus' theorem twice to $\\triangle AMN$ and to the lines $BD$ and $EF$, we get\n$$\n\\frac{AD}{ND} \\cdot \\frac{NP}{MP} \\cdot \\frac{MB}{AB} = 1 \\quad \\text{and} \\quad \\frac{AF}{NF} \\cdot \\frac{NP}{MP} \\cdot \\frac{ME}{AE} = 1.\n$$\nHence\n$$\n(1) \\quad \\frac{AD}{ND} \\cdot \\frac{MB}{AB} = \\frac{AF}{NF} \\cdot \\frac{ME}{AE}.\n$$\nSince $ABCD$ is a parallelogram, then $\\frac{AD}{ND} = \\frac{MC}{NC} = \\frac{MB}{AB}$ (2). From the tangent and secant property $MC^2 = ME \\cdot MA$ and $NC^2 = NF \\cdot NA$ (3). From (1), (2) and (3) it follows that\n$$\n\\frac{MC^2}{NC^2} = \\frac{AF}{NF} \\cdot \\frac{ME}{AE} \\Rightarrow \\frac{ME \\cdot MA}{NF \\cdot NA} = \\frac{AF}{NF} \\cdot \\frac{ME}{AE}.\n$$\nTherefore, $AM \\cdot AE = AN \\cdot AF$, i.e. the quadrilateral $EFNM$ is cyclic. Then $\\angle AEF = \\angle ANM$, whence $\\overline{AF} = \\overline{AEC} - \\overline{FC}$, i.e., $\\overline{AEC} = \\overline{AFC}$. It follows that $AC$ is a diameter of $k$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 27607,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABCD$ be a parallelogram and a circle $k$ passes through $A$, $C$ and meets rays $AB$, $AD$ at $E$, $F$. If $BD$, $EF$ and the tangent at $C$ concur, show that $AC$ is diameter of $k$.\n(Adelina Chopanova)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let the tangent to $k$ at point $C$ intersect the rays $AB \\to$ and $AD \\to$ at the points $M$ and $N$, respectively, and the lines $BD$, $EF$ and the tangent intersect at point $P$. After applying Menelaus' theorem twice to $\\triangle AMN$ and to the lines $BD$ and $EF$, we get\n$$\n\\frac{AD}{ND} \\cdot \\frac{NP}{MP} \\cdot \\frac{MB}{AB} = 1 \\text{ and } \\frac{AF}{NF} \\cdot \\frac{NP}{MP} \\cdot \\frac{ME}{AE} = 1.\n$$\nHence\n$$\n(1) \\qquad \\frac{AD}{ND} \\cdot \\frac{MB}{AB} = \\frac{AF}{NF} \\cdot \\frac{ME}{AE}.\n$$\nSince $ABCD$ is a parallelogram, then $\\frac{AD}{ND} = \\frac{MC}{NC} = \\frac{MB}{AB}$ (2). From the tangent and secant property $MC^2 = ME \\cdot MA$ and $NC^2 = NF \\cdot NA$ (3). From (1), (2) and (3) it follows that\n$$\n\\frac{MC^2}{NC^2} = \\frac{AF}{NF} \\cdot \\frac{ME}{AE} \\Rightarrow \\frac{ME \\cdot MA}{NF \\cdot NA} = \\frac{AF}{NF} \\cdot \\frac{ME}{AE}.\n$$\nTherefore, $AM \\cdot AE = AN \\cdot AF$, i.e. the quadrilateral $EFNM$ is cyclic. Then $\\angle AEF = \\angle ANM$, whence $\\overline{AF} = \\overline{AEC} - \\overline{FC}$, i.e., $\\overline{AEC} = \\overline{AFC}$. It follows that $AC$ is a diameter of $k$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 27608,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all natural numbers $n$ for which the number of positive divisors of $\\text{LCM}(1, 2, \\dots, n)$ is a power of 2.",
    "options": [],
    "answer": "1, 2, 3, 8",
    "solution": "For each prime $p$, the numbers of the interval $n \\in [p^2, p^3)$ are not solutions, because the degree of $p$ in the decomposition of the LCM is 2 and so contributes by a factor of 3 to the number of divisors. Therefore this number is not a power of 2. From Bertrand's postulate for a prime $p$ there is a prime $q$ with $p < q < 2p$, respectively $p^2 < q^2 < 4p^2 < p^3 < q^3$ for $p \\ge 5$; also $3^2 < 5^2 < 3^3 < 5^3$. Therefore, each interval $[p^2, p^3)$ and $[q^2, q^3)$ for consecutive primes $3 \\le p < q$ intersects. We obtained that $n < 9$ and a direct check shows that only $n = 1, 2, 3$ and $8$ are solutions. $\\square$",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 27609,
    "subject": "Mathematics (Multi-modal)",
    "question": "Given is a convex 2024-gon $A_1A_2\\dots A_{2024}$ and 1000 points inside it, so that no three points are collinear. Some pairs of the points are connected with segments so that the interior of the polygon is divided into triangles. Every point is assigned one number among $\\{1, -1, 2, -2\\}$, so that the sum of the numbers written in $A_i$ and $A_{i+1012}$ is zero for all $i = 1, 2, \\dots, 1012$. Prove that there is a triangle, such that the sum of the numbers in some two of its vertices is zero.\n(Emil Kolev)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Clearly, if there are two adjacent points $A_i$ and $A_{i+1}$ with opposite numbers, the problem is solved. Without restriction, let $A_1 = 1$ and consider all segments $A_iA_{i+1}$ for $i = 1, 2, \\dots, 1012$. Since $A_{1013} = -1$, among the considered segments there is an odd number whose ends are one positive and one negative number. These two numbers can be $\\{-1, 2\\}$ or $\\{1, -2\\}$ and let the number of segments with ends of the first kind be $p$ and the number of segments with ends of the second species to be $q$. Due to symmetry, the number of segments $A_iA_{i+1}$ for $i = 1013, \\dots, 2024$ ($A_{2025} \\equiv A_1$) with ends $\\{1, -2\\}$ is equal to $p$. Therefore, all segments with endpoints $\\{1, -2\\}$ are $p+q$, which is an odd number.\nNow consider any triangle that does not have two vertices with opposite numbers. We have the following possibilities for the three numbers:\n$$\n(1, 1, -2), (1, 1, 2), (-1, -1, 2), (-1, -1, -2), \\\\ (2, 2, -1), (2, 2, 1), (-2, -2, 1), (-2, -2, -1).\n$$\nEach of these triangles has an even number (2 or 0) of sides with ends 1 and $-2$. Therefore, the total number of segments with ends 1 and $-2$ (counted in multiples) is an even number. But every line segment inside the 2024-gon is counted twice (once from the two triangles in which it participates), and every line segment that is a side of the 2024-gon is counted once. The resulting contradiction shows that a triangle with the requested property exists. $\\square$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 27610,
    "subject": "Mathematics (Multi-modal)",
    "question": "Given is a sequence $a_1, a_2, \\dots$, such that $a_1 = 1$ and $a_{n+1} = \\frac{9a_n+4}{a_n+6}$ for any $n \\in \\mathbb{N}$. Which terms of this sequence are positive integers?",
    "options": [],
    "answer": "n = 1",
    "solution": "It can be shown by induction that\n$$\na_n = \\frac{4 \\cdot 2^n - 3}{2^n + 3}, \\forall n \\ge 2\n$$\nThus, $a_n$ is integer only when $n = 1$. □",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 27611,
    "subject": "Mathematics (Multi-modal)",
    "question": "Given is a triangle $ABC$ and two points $D \\in AC$, $E \\in BD$ such that $\\angle DAE = \\angle AED = \\angle ABC$. Show that $BE = 2CD$ iff $\\angle ACB = 90^\\circ$.\n(Nikolay Nikolov)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $\\alpha = \\angle A$, $\\beta = \\angle B$ and $\\gamma = \\angle C$. Then\n$$\n\\frac{BE}{\\sin(\\alpha - \\beta)} = \\frac{AB}{\\sin(\\pi - \\beta)}, \\quad \\frac{CD}{\\sin(\\alpha - \\beta)} = \\frac{BC}{\\sin 2\\beta}, \\quad \\frac{AB}{\\sin \\gamma} = \\frac{BC}{\\sin \\alpha}\n$$\nand so\n$$\n\\frac{BE}{CD} = \\frac{\\sin 2\\beta \\sin \\gamma}{\\sin \\beta \\sin \\alpha} = \\frac{2 \\cos \\beta \\sin \\gamma}{\\sin \\alpha} = \\frac{2 \\cos \\beta \\sin \\gamma}{\\sin \\beta \\cos \\gamma + \\cos \\beta \\sin \\gamma}\n$$\nTherefore $BE = 2CD \\Leftrightarrow \\cos \\gamma = 0 \\Leftrightarrow \\gamma = 90^\\circ$. □",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 27612,
    "subject": "Mathematics (Multi-modal)",
    "question": "For a positive integer $n$, denote with $b(n)$ the smallest positive integer $k$, such that there exist integers $a_1, a_2, \\dots, a_k$, satisfying $n = a_1^{a_2} + a_2^{a_3} + \\dots + a_k^{a_1}$. Determine whether the set of positive integers $n$ is finite or infinite, which satisfy:\n$$\n\\text{a) } b(n) = 12; \\quad \\text{b) } b(n) = 12^{12^{12}}.\n$$",
    "options": [],
    "answer": "a) infinite; b) finite (empty)",
    "solution": "a) From Fermat's theorem and $y^2 \\equiv 1 \\pmod{67} \\Leftrightarrow y \\equiv \\pm 1 \\pmod{67}$ it follows that any student number gives a remainder of $0$, $1$ or $66$ when divided by $67$. Let us consider the numbers $12^{66k+1}$, where $k \\in \\mathbb{N}$. They are presented as the sum of $12$ student numbers. Furthermore, by Fermat's theorem $12^{66k+1} \\equiv 12 \\pmod{67}$. This shows that $b(12^{66k+1}) = 12$ for every $k \\in \\mathbb{N}$. Therefore, the set here is infinite.\n\nb) For $P \\in \\mathbb{Z}[X]$ let us set $\\Delta(P)(x) = P(x+1) - P(x)$. It is clear that if $P$ is of degree $d$ with leading coefficient $a$, then $\\Delta(P) \\in \\mathbb{Z}[X]$ is a polynomial of degree $d-1$ with leading coefficient $ad$. Consider the series of polynomials $P_1(x) = x^{33}$, $P_{k+1} = \\Delta(P_k)$ for $k \\in \\mathbb{N}$. It is easy to see by induction on $k$ that for each $x \\in \\mathbb{Z}$, $k \\in \\mathbb{N}$ the number $P_k(x)$ is a sum of $2^{k-1}$ student numbers. Furthermore, we have that $P_{33}(x) = 33!x + b$ for some $b \\in \\mathbb{Z}$. Since $1$ and $-1$ are student numbers, each integer is the sum of at most $2^{32} + 33! < 12^{12^{12}}$ student numbers. Then the set here is empty, and therefore finite.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 27613,
    "subject": "Mathematics (Multi-modal)",
    "question": "For a positive integer $n$, denote with $b(n)$ the smallest positive integer $k$, such that there exist integers $a_1, a_2, \\dots, a_k$, satisfying $n = a_1^{33} + a_2^{33} + \\dots + a_k^{33}$. Determine whether the set of positive integers $n$ is finite or infinite, which satisfy:\n$$\n\\text{a) } b(n) = 12; \\quad \\text{b) } b(n) = 12^{12^{12}}.\n$$",
    "options": [],
    "answer": "a) infinite; b) finite (empty)",
    "solution": "a) From Fermat's theorem and $y^2 \\equiv 1 \\pmod{67} \\Leftrightarrow y \\equiv \\pm 1 \\pmod{67}$ it follows that any student number gives a remainder of $0$, $1$ or $66$ when divided by $67$. Let us consider the numbers $12^{66k+1}$, where $k \\in \\mathbb{N}$. They are presented as the sum of $12$ student numbers. Furthermore, by Fermat's theorem $12^{66k+1} \\equiv 12 \\pmod{67}$. This shows that $b(12^{66k+1}) = 12$ for every $k \\in \\mathbb{N}$. Therefore, the set here is infinite.\n\nb) For $P \\in \\mathbb{Z}[X]$ let us set $\\Delta(P)(x) = P(x+1) - P(x)$. It is clear that if $P$ is of degree $d$ with leading coefficient $a$, then $\\Delta(P) \\in \\mathbb{Z}[X]$ is a polynomial of degree $d-1$ with leading coefficient $ad$. Consider the series of polynomials $P_1(x) = x^{33}$, $P_{k+1} = \\Delta(P_k)$ for $k \\in \\mathbb{N}$. It is easy to see by induction on $k$ that for each $x \\in \\mathbb{Z}$, $k \\in \\mathbb{N}$ the number $P_k(x)$ is a sum of $2^{k-1}$ student numbers. Furthermore, we have that $P_{33}(x) = 33!x + b$ for some $b \\in \\mathbb{Z}$. Since $1$ and $-1$ are student numbers, each integer is the sum of at most $2^{32} + 33! < 12^{12^{12}}$ student numbers. Then the set here is empty, and therefore finite.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 27614,
    "subject": "Mathematics (Multi-modal)",
    "question": "Points $A$, $B$, $Y$ and $C$ lie in this order on circle $k$ with center $O$, such that $BC = 2$ cm, $\\angle BAY = 42^\\circ$ and $\\angle CAY = 78^\\circ$. It is known that the circle $\\omega$ through the points $A$, $O$ and $B$ is tangent to the line $BY$. The circle through the points $A$ and $C$, tangent to the line $CY$, intersects $\\omega$ for second time at the point $N$. To be found:\n\na) the length of the segment $BO$;\n\nb) the size of the angle $\\angle YAN$.",
    "options": [],
    "answer": "BO = 2√3/3 cm; ∠YAN = 36°",
    "solution": "a) Clearly $\\angle BAC = \\angle BAY + \\angle CAY = 120^\\circ$, respectively $\\angle BOC = 360^\\circ - 2\\angle BAC = 120^\\circ$. Thus, if $M$ is the midpoint of $BC$, then $OM \\perp BC$ (because $BO = OC$), $\\angle BOM = 60^\\circ$ and $BM = \\frac{BC}{2} = 1$. Let $BO = x$ and from triangle $BOM$ we have $OM = \\frac{x}{2}$ from $\\angle OBM = 30^\\circ$ and $x^2 = (\\frac{x}{2})^2 + 1^2$ from the Pythagorean theorem, respectively $x^2 = \\frac{4}{3}$ and $x = \\frac{2\\sqrt{3}}{3}$.\n\nb) We have $\\angle ANB = \\angle AOB = 180^\\circ - 2\\angle BAO = 180^\\circ - 2\\angle OBY = 180^\\circ - 2(90^\\circ - \\angle YCB) = 2\\angle YCB$. Hence $\\angle ABY = \\angle ABO + \\angle OBY = 2\\angle OAB = 180^\\circ - \\angle AOB = 180^\\circ - 2\\angle YCB$ and now from the other circle we calculate $\\angle ANC = 180^\\circ - \\angle ACY = \\angle ABY = 180^\\circ - 2\\angle YCB$. Therefore, $\\angle ANB + \\angle ANC = 180^\\circ$, i.e. $N$ lies on $BC$. It remains to consider that $\\angle NAC = \\angle BCY = \\angle BAY$ from the touching and $\\angle YAN = \\angle CAY - \\angle CAN = \\angle CAY - \\angle BAY = 36^\\circ$.\n\n$\\square$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 27615,
    "subject": "Mathematics (Multi-modal)",
    "question": "Points $A$, $B$, $Y$ and $C$ lie in this order on circle $k$ with center $O$, such that $BC = 2$ cm, $\\angle BAY = 42^\\circ$ and $\\angle CAY = 78^\\circ$. It is known that the circle $\\omega$ through the points $A$, $O$ and $B$ is tangent to the line $BY$.\n\nThe circle through the points $A$ and $C$, tangent to the line $CY$, intersects $\\omega$ for second time at the point $N$. To be found:\n\na) the length of the segment $BO$;\nb) the size of the angle $\\angle YAN$.",
    "options": [],
    "answer": "BO = 2√3/3 cm; angle YAN = 36°",
    "solution": "a) Clearly $\\angle BAC = \\angle BAY + \\angle CAY = 120^\\circ$, respectively $\\angle BOC = 360^\\circ - 2\\angle BAC = 120^\\circ$. Thus, if $M$ is the midpoint of $BC$, then $OM \\perp BC$ (because $BO = OC$), $\\angle BOM = 60^\\circ$ and $BM = \\frac{BC}{2} = 1$. Let $BO = x$ and from triangle $BOM$ we have $OM = \\frac{x}{2}$ from $\\angle OBM = 30^\\circ$ and $x^2 = (\\frac{x}{2})^2 + 1^2$ from the Pythagorean theorem, respectively $x^2 = \\frac{4}{3}$ and $x = \\frac{2\\sqrt{3}}{3}$.\n\nb) We have $\\angle ANB = \\angle AOB = 180^\\circ - 2\\angle BAO = 180^\\circ - 2\\angle OBY = 180^\\circ - 2(90^\\circ - \\angle YCB) = 2\\angle YCB$. Hence $\\angle ABY = \\angle ABO + \\angle OBY = 2\\angle OAB = 180^\\circ - \\angle AOB = 180^\\circ - 2\\angle YCB$ and now from the other circle we calculate $\\angle ANC = 180^\\circ - \\angle ACY = \\angle ABY = 180^\\circ - 2\\angle YCB$. Therefore, $\\angle ANB + \\angle ANC = 180^\\circ$, i.e. $N$ lies on $BC$. It remains to consider that $\\angle NAC = \\angle BCY = \\angle BAY$ from the touching and $\\angle YAN = \\angle CAY - \\angle CAN = \\angle CAY - \\angle BAY = 36^\\circ$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 27616,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $d \\ge 3$ be a positive integer. The binary strings of length $d$ are split into $2^{d-1}$ pairs, such that the strings in each pair differ in exactly one position. Show that there exists an alternating cycle of length at most $2d-2$, i.e. at most $2d-2$ binary strings that can be arranged on a circle so that any pair of adjacent strings differ in exactly one position and exactly half of the pairs of adjacent strings are pairs in the split.\n(Lyuben Lichev)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Consider a graph $G$ with vertices the binary vectors of length $d$ and edges connecting the pairs of vectors that differ in exactly one position. Let $\\mathcal{M}$ be a complete $d$-dimensional pairing. We will prove that for each vertex $\\mathbf{x} \\in \\{0,1\\}^d$ of $G$ we can find an alternating cycle of length no more than $2d-2$ among the distance vectors of the most many 2 of $\\mathbf{x}$.\n\nLet $\\mathbf{x} \\in \\{0,1\\}^d$ have as neighbors $\\mathbf{x}_1, \\mathbf{x}_2, \\dots, \\mathbf{x}_d$ in $G$. Without loss of generality let $\\{\\mathbf{x}, \\mathbf{x}_1\\} \\in \\mathcal{M}$, and the vectors $\\mathbf{x}_2, \\dots, \\mathbf{x}_d$ form pairs in $\\mathcal{M}$ respectively with $\\mathbf{y}_2, \\dots, \\mathbf{y}_d$ (which are at distance 2 from $\\mathbf{x}$). Notice that each of the vectors $\\mathbf{y}_2, \\dots, \\mathbf{y}_d$ has exactly two neighbors among $\\mathbf{x}_1, \\mathbf{x}_2, \\dots, \\mathbf{y}_d$ in $G$. We will consider two cases:\n\n**Case 1.** Any of the vectors $\\mathbf{y}_2, \\dots, \\mathbf{y}_d$, say $\\mathbf{y}_i$, is a neighbor of $\\mathbf{x}_1$ in $G$. Then $\\mathbf{x}, \\mathbf{x}_1, \\mathbf{y}_i, \\mathbf{x}_i$ form an alternating cycle of length $4 \\le 2d-2$.\n\n**Case 2.** None of the vectors $\\mathbf{y}_2, \\dots, \\mathbf{y}_d$ is a neighbor of $\\mathbf{x}_1$ in $G$. Consider the following algorithm. At the beginning, let's place a token at vertex $\\mathbf{x}_2$ and at each step:\n\n* if the token is located at a vertex $\\mathbf{x}_i$ for some $i \\in \\{2, \\dots, d\\}$, we move it to a vertex $\\mathbf{y}_i$;\n* if the token is located at a vertex $\\mathbf{y}_i$ for some $i \\in \\{2, \\dots, d\\}$, we move it to the neighbor of $\\mathbf{y}_i$ other than $\\mathbf{x}_i$.\n\nObviously, after no more than $2d-2$ runs of the algorithm, some vertex will be repeated. Since the algorithm describes a walk on the graph $G$ where every odd edge is in $\\mathcal{M}$ and every even edge is outside $\\mathcal{M}$, it finds an alternating cycle after at most $2d-2$ moves, whence the statement also follows in this case. $\\square$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 27617,
    "subject": "Mathematics (Multi-modal)",
    "question": "A three-digit natural number $n$ is initially written on the board. Two players, $A$ and $B$, take turns, with $A$ going first. The turner reduces the number on the board by some divisor of his own (i.e., other than $1$ and the number itself). For example, if at some point the number on the board is $6$, it can be reduced by $2$, then the number on the board will now be $4$. Whoever cannot make a move loses, and the other wins. Player $A$ is known to have a way to win as well as player $B$. What are all the possible $n$?\n(Ivaylo Kortezov)",
    "options": [],
    "answer": "All even three-digit numbers except 128 and 512 (448 total).",
    "solution": "If there is a prime number on the board, the player loses by definition. If there is an even number on the board that is not a power of $2$, then the player can always reduce it by its odd divisor, leaving an odd number on the board. If the number on the board is odd and is reduced by its (odd) divisor $a$, i.e. a number of the form $ab$ is replaced by $a(b - 1)$, the resulting number is even and is not a power of the pair. Therefore, if the starting number is even and not a power of even, the player can always make a move guaranteeing that for his next move he will again receive a natural number of the same kind. Thus, an even number that is not a power of even is a winning position, and an odd number is a losing position.\n\nIt remains to analyze the cases when the board number is $2^m$ for a natural $m$. In this case, the reducer should be $2^k$ for naturally $k < m$. If $k < m - 1$, then the resulting position $2^k(2^{m-k} - 1)$ is even and not a power of even, so it would be a winning move for the adversary and therefore an unacceptable move. So, we need $k = m - 1$. Playing this way, one player will get the even powers of $2$ on the board and the other the odd ones. Given that $2$ is a losing position, we conclude that even powers of $2$ are winning positions and odd powers of $2$ are losing positions.\n\nThe finally suitable $n$ are the even numbers that are not odd powers of the pair. Among the given there are $900 : 2 - 2 = 448$ such (we excluded $128 = 2^7$ and $512 = 2^9$). $\\Box$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 27618,
    "subject": "Mathematics (Multi-modal)",
    "question": "The nonnegative real numbers $x, y, z$ are such that $(x + y)(y+z)(z+x) = 1$. We denote by $m$ and $M$ respectively the smallest and largest possible values of the expression $A = (xy + yz + zx)(x + y + z)$.\n\na) Find $m$ and $M$.\n\nb) Is there a triple of nonnegative rational numbers $(x, y, z)$ satisfying the given equality for which $A = m$?",
    "options": [],
    "answer": "m = 1, M = 9/8; No, the minimum cannot be attained by any triple of nonnegative rational numbers.",
    "solution": "a) The inequality $(xy+yz+zx)(x+y+z) \\ge 1 = (x+y)(y+z)(z+x)$ is equivalent to $xyz \\ge 0$. Equality is reached only when one of the variables, say $x$, is $0$ and the other two (in this case $y$ and $z$) are whatever with $yz(y+z) = 1$; one possibility is $y=1$ and $z = \\frac{\\sqrt{5}-1}{2}$. The inequality $(xy+yz+zx)(x+y+z) \\le \\frac{9}{8}$ is equivalent to $9(x+y)(y+z)(z+x) - 8(xy+yz+zx)(x+y+z) \\ge 0$, i.e. on $x^2y + xy^2 + y^2z + yz^2 + x^2z + xz^2 \\ge 6xyz$. The latter is true from the inequality between the arithmetic mean and the geometric mean applied to the six terms on the left, with equality only at $x = y = z$ and $8x^3 = 1$, i.e. $x = y = z = \\frac{1}{2}$.\n\nb) Given the reasoning in a), it suffices to prove that the equation $yz(y+z) = 1$ has no solution in (positive) rational numbers. Assume the opposite and let $y = \\frac{p}{r}$, $z = \\frac{q}{r}$ is a solution (with natural $p, q, r$) in which we have reduced the fractions under a common denominator. Then $pq(p+q) = r^3$, as after truncating a common divisor of $p, q, r$, if necessary, we can consider that $(p, q, r) = 1$. In fact, here already $(p, q) = 1$, since otherwise their common prime divisor would also be that of $r$, contradiction with $(p, q, r) = 1$. Thus, the numbers $p, q$ and $p+q$ are two by two mutually prime, and since their product is an exact cube, then necessarily $p = a^3$, $q = b^3$ and $p+q = c^3$ for some natural numbers $a, b, c$. Thus we obtained $a^3 + b^3 = c^3$ for some natural $a, b, c$, which is impossible (a special case of the so-called Fermat's Great Theorem).",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 27619,
    "subject": "Mathematics (Multi-modal)",
    "question": "Solve the inequality:\n$$\n\\frac{x^2 - |x - 1| - 4}{x - 4} \\geq 2x - 1.\n$$",
    "options": [],
    "answer": "(-∞, 1] ∪ (4, 7]",
    "solution": "Note that $x \\neq 4$ and $|x-1| = x-1$ for $x > 1$, otherwise $|x-1| = 1-x$.\n\nIf $x > 1$ we get:\n$$\n\\frac{x^2 - 8x + 7}{x - 4} \\le 0 \\iff \\frac{(x - 7)(x - 1)}{x - 4} \\le 0 \\iff x \\in (4, 7]\n$$\n\nIf $x \\le 1$ we get:\n$$\n\\frac{x^2 - 10x + 9}{x - 4} \\le 0 \\Leftrightarrow \\frac{(x - 9)(x - 1)}{x - 4} \\le 0 \\Leftrightarrow x \\in (-\\infty, 1]\n$$\n\nFinally $x \\in (-\\infty, 1] \\cup (4, 7]$. $\\square$",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 27620,
    "subject": "Mathematics (Multi-modal)",
    "question": "Given the triangle $ABC$ and $M$ - midpoint of $AB$. Given the angles $\\angle ABC = 30^\\circ$ and $\\angle BCM = 105^\\circ$. Prove that $CM \\cdot AC = BM \\cdot BC$.\n(Konstantin Delchev)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $AA_1$ be the height from point $A$ in $\\triangle ABC$. Note that it lies on the continuation of $BC$, because this triangle is obtuse. The triangle $\\triangle AA_1B$ is right-angled with angle $30^\\circ$. Therefore, $AA_1 = AM = A_1M = BM = x$. Consequently, we directly find $\\angle CMB = 45^\\circ$, $\\angle A_1MA = 60^\\circ$, $\\angle A_1MC = 75^\\circ$, $\\angle MCA_1 = 75^\\circ$. So $\\triangle MCA_1$ is isosceles. Then $CA_1 = x$ and $\\triangle AA_1C$ is isosceles and right-angled, therefore $\\angle ACA_1 = 45^\\circ$ and therefore $\\angle ACM = 30^\\circ$.\nWe will calculate the area of $\\triangle ACM$ in two different ways.\n\n$S_{ACM} = \\frac{AC \\cdot CM}{4}$ because $\\angle ACM = 30^\\circ$ i.e. the height to $AC$ is equal to $\\frac{MC}{2}$. But also $M$ is the middle of $AB$. So $S_{ACM} = \\frac{S_{ABC}}{2} = \\frac{BC \\cdot AA_1}{4} = \\frac{BC \\cdot BM}{4}$ and we get what we are looking for. $\\square$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 27621,
    "subject": "Mathematics (Multi-modal)",
    "question": "We call $n$ lines in the plane three-way if they can be separated into three nonempty sets, $X, Y, Z$. Every two lines from the same set are parallel to each other, no two lines from different sets are parallel to each other, and no three lines intersect at a point.\n\nBy $S_n$ we denote the maximum number of regions into which $n$ three-way lines can divide the plane. As a region we consider a connected part of the plane, not necessarily finite, whose boundaries are defined by three-way lines.\n\nWhat is the largest $n$ for which $S_n < 128$?\n\n(Konstantin Delchev)",
    "options": [],
    "answer": "18",
    "solution": "We will derive a general formula for the number of regions. Let the three sets have the number of elements $|X| = x$, $|Y| = y$ and $|Z| = z$, respectively. The first two divide the plane into a total of $(x+1)(y+1)$ regions. Each line of the third set intersects the others at $x+y$ points and is divided into $x+y+1$ parts. Each of them divides one of the already obtained areas into two. Thus we get a general formula:\n$$\nS_{x,y,z} = (x+1)(y+1) + z(x+y+1) = x + y + z + xy + xz + yz + 1.\n$$\nWe note that $n = x + y + z$. Furthermore, we have that $3(xy + yz + zx) \\le (x + y + z)^2 = n^2$ (equivalent to $(x - y)^2 + (y - z)^2 + (z - x)^2 \\ge 0$). So $S_{x,y,z} \\le n^2/3 + n + 1$ which is less than 128 for $n = 18$ (actually, $S_{6,6,6} = 127$, i.e. $S_{18} = 127$). Also, $S_{6,6,7} = 140$, so $S_n > 128$ for $n \\ge 19$. Therefore, the answer is 18. $\\square$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 27622,
    "subject": "Mathematics (Multi-modal)",
    "question": "$S_n = \\{a \\mid a < n, \\exists k \\in \\mathbb{N} : 2^k \\equiv a \\pmod{n}\\}$.\n\nAre there different odd numbers $m$ and $r$ such that $S_m = S_r$?",
    "options": [],
    "answer": "No",
    "solution": "No! There exists a natural number $s$ such that $2^s \\equiv 1 \\pmod{n}$ (for example $s = \\varphi(n)$ from Euler's theorem or because the power series of $2$ modulo $n$ is periodic). We have $2^{s-1} \\equiv \\frac{n+1}{2} \\pmod{n}$ and so $x = \\frac{n+1}{2} \\in S_n$, but $2x = n+1 > n$ is not in $S_n$. Also, if $t \\le \\frac{n-1}{2}$ is of $S_n$, then $2t < n$ is also. Therefore, the smallest natural number $t$ such that $t \\in S_n$ and $2t \\notin S_n$, is $\\frac{n+1}{2}$. Since this number is different for different $n$, we get what we asked for. $\\square$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 27623,
    "subject": "Mathematics (Multi-modal)",
    "question": "The reals $x, y$ satisfy $x(x - 6) \\le y(4 - y) + 7$. Find the minimal and maximal values of the expression $x + 2y$.",
    "options": [],
    "answer": "min = -3, max = 17",
    "solution": "Let $a = x + 2y$. Then $x = a - 2y$ and\n$$\n\\begin{aligned}\n(a - 2y)(a - 2y - 6) &\\le y(4 - y) + 7 \\\\\na^2 - 2ay - 6a - 2ay + 4y^2 + 12y &\\le 4y - y^2 + 7 \\\\\n5y^2 - 2(2a - 4)y + (a^2 - 6a - 7) &\\le 0 \\\\\nD &= (2a - 4)^2 - 5(a^2 - 6a - 7) \\ge 0 \\\\\n4a^2 - 16a + 16 - 5a^2 + 30a + 35 &\\ge 0 \\\\\na^2 - 14a - 51 &\\le 0 \\\\\n(a - 17)(a + 3) &\\le 0.\n\\end{aligned}\n$$\nFinally, $a \\in [-3; 17]$. $\\square$",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 27624,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABC$ be an acute triangle with midpoint $M$ of $AB$. The point $D$ lies on the segment $MB$ and $I_1, I_2$ denote the incenters of $\\triangle ADC$ and $\\triangle BDC$. Given that $\\angle I_1MI_2 = 90^\\circ$, show that $CA = CB$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 27625,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $N$ be a positive integer. The sequence $x_1, x_2, \\dots$ of non-negative reals is defined by\n$$\nx_n^2 = \\sum_{i=1}^{n-1} \\sqrt{x_i x_{n-i}}\n$$\nfor all positive integers $n > N$. Show that there exists a constant $c > 0$, such that $x_n \\le \\frac{n}{2} + c$ for all positive integers $n$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Applying AM-GM inequality to each term of the right-hand side, we get,\n$$\nx_n^2 \\le \\sum_{i=1}^{n-1} x_i, \\quad n > N.\n$$\nLet us set $x_n = a \\cdot n + \\Delta(n)$, where $a$ is a constant that will be determined later. Vaguely speaking, we want to pick a value for $a$ such that $\\Delta(n)$ has a smaller growth rate than $n$. For $n > N$ we obtain,\n$$\na^2 n^2 + 2an\\Delta(n) + \\Delta(n)^2 \\le \\frac{a(n-1)n}{2} + \\sum_{i=1}^{n-1} \\Delta(i)\n$$\n$$\n2an\\Delta(n) \\le \\left( \\frac{a(n-1)n}{2} - a^2 n^2 \\right) - \\Delta(n)^2 + \\sum_{i=1}^{n-1} \\Delta(i) \\quad (1).\n$$\nThe plan is to divide both sides by $n$ so let's determine $a$ such that the term in the brackets of the right side of (1) be of degree $n$ (not $n^2$ as it is now). It can be easily calculated that $a = 1/2$. Putting it in (1), we get\n$$\n\\Delta(n) \\le -\\frac{1}{4} - \\frac{\\Delta(n)^2}{n} + \\frac{1}{n} \\sum_{i=1}^{n-1} \\Delta(i).\n$$\nNow, write\n$$\n\\Delta(n) \\le \\frac{1}{n} \\sum_{i=1}^{n-1} \\Delta(i), \\forall n > N\n$$\nwhich yields\n$$\n\\Delta(n) \\le \\max\\{\\Delta(i) : i < n\\},\n$$\nhence $\\Delta(n) \\le \\max\\{\\Delta(i) : i \\le N\\}$ for any $n > N$. This yields $x_n \\le n/2 + c, \\forall n \\in \\mathbb{N}$ for some constant $c$.\n\n**Sharper estimate.** We want to go a bit further, so let's write\n$$\n\\Delta(n) \\le -\\frac{1}{4} + \\frac{1}{n} \\sum_{i=1}^{n-1} \\Delta(i), \\quad n > N\n$$\nDefine the sequence $\\Delta'(n)$ as follows:\n$$\n\\Delta'(n) = -\\frac{1}{4} + \\frac{1}{n} \\sum_{i=1}^{n-1} \\Delta'(i), \\text{ for } n > N \\quad (2),\n$$\nand $\\Delta'(n) = \\Delta'(n)$, for $n \\le N$. Clearly $\\Delta(n) \\le \\Delta'(n), \\forall n \\in \\mathbb{N}$. We have\n$$\n\\Delta'(n + 1) = -\\frac{1}{4} + \\frac{1}{n+1} \\left( \\sum_{i=1}^{n-1} \\Delta(i) + -\\frac{1}{4} + \\frac{1}{n} \\sum_{i=1}^{n-1} \\Delta(i) \\right), \\quad n > N \\quad (3).\n$$\nSubtracting (2) from (3) yields\n$$\n\\Delta'(n + 1) - \\Delta'(n) = \\frac{-1}{4(n + 1)}, \\quad \\forall n > N.\n$$\nSumming up the above inequality gives\n$$\n\\Delta'(n) = \\Delta'(N + 1) - \\frac{1}{4} \\sum_{k=N+2}^{n} \\frac{1}{k}, \\quad n \\ge N + 2.\n$$\nTherefore, using a well known property of harmonic series,\n$$\n\\Delta'(n) \\le -\\frac{\\ln n}{4} + c\n$$\nwhere $c$ is a constant. Finally, we get\n$$\nx_n \\le \\frac{n}{2} - \\frac{\\ln n}{4} + c, \\quad n \\in \\mathbb{N},\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 27626,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $A_0B_0C_0$ be a triangle. For a positive integer $n \\ge 1$, we define $A_n$ on the segment $B_{n-1}C_{n-1}$ such that $B_{n-1}A_n : C_{n-1}A_n = 2 : 1$ and $B_n, C_n$ are defined cyclically in a similar manner. Show that there exists an unique point $P$ that lies in the interior of all triangles $A_nB_nC_n$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "We have nested compact sets (closed triangles), so they have non empty intersection. We prove that they intersect in only one point. It's enough to prove that the three points $A_n, B_n, C_n$ converge to a common point $P$. Assume it's false. Then there exists some subsequences of $A_n, B_n, C_n$ (which for simplicity we denote again by $A_n, B_n, C_n$) that converge to points $A, B, C$ respectively and $\\{A, B, C\\}$ consists of at least 2 elements. Assume, first $A, B, C$ are distinct. Assume wlog that $\\angle BAC \\le 60^\\circ$. Take $n$ large enough such that $A_n, B_n, C_n$ are close enough to $A, B, C$ respectively. Consider the next triangle $A_{n+1}B_{n+1}C_{n+1}$. Its side $B_{n+1}C_{n+1}$ is far enough from $A$ and so $A$ is outside $\\triangle A_{n+1}B_{n+1}C_{n+1}$. But $A$ was a limit point of the sequence $A_n, n = 1, 2, \\dots$, contradiction. Suppose now, $B = C \\ne A$. Then $\\angle B_nA_nC_n < 60^\\circ$ (it tends to 0 actually) and we apply the same argument. We proved that there is a unique point $P$ that's common for all the triangles.\n\nNow it remains to prove a small trifle - namely $P$ is in the interior of all the triangles. It was part of the Bulgarian text. Assume on the contrary $P$ is on some side, say $A_nB_n$, for some $n$. But it easily follows that $P$ is outside $\\triangle A_{n+2}B_{n+2}C_{n+2}$ contradiction. $\\square$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 27627,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $A_0B_0C_0$ be a triangle. For a positive integer $n \\ge 1$, we define $A_n$ on the segment $B_{n-1}C_{n-1}$ such that $B_{n-1}A_n : C_{n-1}A_n = 2 : 1$ and $B_n, C_n$ are defined cyclically in a similar manner. Show that there exists an unique point $P$ that lies in the interior of all triangles $A_nB_nC_n$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "(Dragomir Grozev) We have nested compact sets (closed triangles), so they have non empty intersection. We prove that they intersect in only one point. It's enough to prove that the three points $A_n, B_n, C_n$ converge to a common point $P$. Assume it's false. Then there exists some subsequences of $A_n, B_n, C_n$ (which for simplicity we denote again by $A_n, B_n, C_n$) that converge to points $A, B, C$ respectively and $\\{A, B, C\\}$ consists of at least 2 elements. Assume, first $A, B, C$ are distinct. Assume wlog that $\\angle BAC \\le 60^\\circ$. Take $n$ large enough such that $A_n, B_n, C_n$ are close enough to $A, B, C$ respectively. Consider the next triangle $A_{n+1}B_{n+1}C_{n+1}$. Its side $B_{n+1}C_{n+1}$ is far enough from $A$ and so $A$ is outside $\\triangle A_{n+1}B_{n+1}C_{n+1}$. But $A$ was a limit point of the sequence $A_n, n = 1, 2, \\dots$, contradiction. Suppose now, $B = C \\ne A$. Then $\\angle B_nA_nC_n < 60^\\circ$ (it tends to 0 actually) and we apply the same argument. We proved that there is a unique point $P$ that's common for all the triangles.\n\nNow it remains to prove a small trifle - namely $P$ is in the interior of all the triangles. It was part of the Bulgarian text. Assume on the contrary $P$ is on some side, say $A_nB_n$, for some $n$. But it easily follows that $P$ is outside $\\triangle A_{n+2}B_{n+2}C_{n+2}$ contradiction. $\\square$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 27628,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all pairs of positive integers $(n, k)$ such that all sufficiently large odd positive integers $m$ are representable as\n$$\nm = a_1^{n_1^2} + a_2^{(n_1+1)^2} + \\dots + a_k^{(n+k-1)^2} + a_{k+1}^{(n+k)^2}\n$$\nfor some non-negative integers $a_1, a_2, \\dots, a_{k+1}$.",
    "options": [],
    "answer": "All pairs with n = 1 and any positive integer k.",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 27629,
    "subject": "Mathematics (Multi-modal)",
    "question": "Is it true that for any positive integer $n > 1$, there exists an infinite arithmetic progression $M_n$ of positive integers, such that for any $m \\in M_n$, the number $n^m - 1$ is not a perfect power (a positive integer is a perfect power if it is of the form $a^b$ for positive integers $a, b > 1$)?",
    "options": [],
    "answer": "Yes",
    "solution": "(Victor Kostadinov) The answer is yes. Fix a positive integer $n$ and two large distinct primes $p, q > n$. Let $d_p = \\text{ord}_p(n)$, $d_q = \\text{ord}_q(n)$, $c_p = \\nu_p(n^{d_p} - 1)$, $c_q = \\nu_q(n^{d_q} - 1)$ and let $c = \\nu_p(d_q)$, $d = \\nu_q(d_p)$. Pick two sufficiently large constants $a, b$, and let $M := d_p d_q p^{(a-1)c_p + bc_q - c} q^{ac_p + (b-1)c_q + 1 - d}$ and finally choose $M_n$ to consist of $m = M(1 + iM)$ for $i = 1, 2, \\dots$.\n\nBy LTE, we have\n$$\n\\nu_p(n^m - 1) = \\nu_p(n^{d_p} - 1) + \\nu_p\\left(\\frac{m}{d_p}\\right) = c_p + c + (a-1)c_p + bc_q - c = ac_p + bc_q\n$$\n(we used that $\\text{gcd}(1 + iM, p) = 1$) and similarly $\\nu_q(n^m - 1) = ac_p + bc_q + 1$, which are consecutive, i.e. they can't have a common divisor greater than 1 and thus $n^m - 1$ can't be a perfect power. $\\square$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 27630,
    "subject": "Mathematics (Multi-modal)",
    "question": "Given is a triangle $ABC$ and the points $M$, $P$ lie on the segments $AB$, $BC$, respectively, such that $AM = BC$ and $CP = BM$. If $AP$ and $CM$ meet at $O$ and $2\\angle AOM = \\angle ABC$, find the measure of $\\angle ABC$.",
    "options": [],
    "answer": "90°",
    "solution": "Let $D$ be the reflection of $P$ across $C$, so $AB = BD$ and $O'$ be the circumcenter of $\\triangle ABD$. As $\\triangle AO'B \\cong \\triangle BO'D$ and $MB = CD$, we have $\\angle O'CB = \\angle O'MA$, so $MBCO'$ is cyclic. Now $\\angle O'CM = \\angle O'BM = \\angle AOM$, hence $CO' \\parallel AP$. Therefore $O'$ must be the midpoint of $AD$, implying that $\\angle ABC = 90^\\circ$. $\\square$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 27631,
    "subject": "Mathematics (Multi-modal)",
    "question": "Given is a triangle $ABC$ and the points $M$, $P$ lie on the segments $AB$, $BC$, respectively, such that $AM = BC$ and $CP = BM$. If $AP$ and $CM$ meet at $O$ and $2\\angle AOM = \\angle ABC$, find the measure of $\\angle ABC$.",
    "options": [],
    "answer": "90°",
    "solution": "Let $D$ be the reflection of $P$ across $C$, so $AB = BD$ and $O'$ be the circumcenter of $\\triangle ABD$. As $\\triangle AO'B \\cong \\triangle BO'D$ and $MB = CD$, we have $\\angle O'CB = \\angle O'MA$, so $MBCO'$ is cyclic. Now $\\angle O'CM = \\angle O'BM = \\angle AOM$, hence $CO' \\parallel AP$. Therefore $O'$ must be the midpoint of $AD$, implying that $\\angle ABC = 90^\\circ$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 27632,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all functions $f : \\mathbb{R}^+ \\to \\mathbb{R}^+$, such that\n$$\nf(af(b) + a)(f(bf(a)) + a) = 1\n$$\nfor any positive reals $a, b$.\n(Lyuben Lichev)",
    "options": [],
    "answer": "f(x) = 1/x for all x > 0",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 27633,
    "subject": "Mathematics (Multi-modal)",
    "question": "Do there exist $2024$ non-zero reals $a_1, a_2, \\dots, a_{2024}$, such that\n$$\n\\sum_{i=1}^{2024} \\left(a_i^2 + \\frac{1}{a_i^2}\\right) + 2 \\sum_{i=1}^{2024} \\frac{a_i}{a_{i+1}} + 2024 = 2 \\sum_{i=1}^{2024} \\left(a_i + \\frac{1}{a_i}\\right)?\n$$",
    "options": [],
    "answer": "No",
    "solution": "Suppose that such $a_i$ exist. The condition rewrites as $\\sum_{i=1}^{2024} \\left(a_i + \\frac{1}{a_{i+1}} - 1\\right)^2 = 0$, which means that $a_i + \\frac{1}{a_{i+1}} = 1$ for all $i = 1, 2, \\dots, 2024$. It's easy to obtain consecutively that for any $i = 1, 2, \\dots, 2024$, $a_{i+1} = \\frac{1}{1-a_i}$, $a_{i+2} = 1 - \\frac{1}{a_i}$, so $a_{i+3} = a_i$. Since $3 \\nmid 2024$, all $a_i$ are equal to some real $k$. However, $k + \\frac{1}{k} = 1 \\iff k^2 - k + 1 = 0$ doesn't have a real solution, contradiction. $\\square$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 27634,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $\\mathcal{F}$ be a family of 4-element subsets of a set of size $5^m$, where $m$ is a fixed positive integer. If the intersection of any two sets in $\\mathcal{F}$ does not have size exactly $2$, find the maximal value of $|\\mathcal{F}|$.",
    "options": [],
    "answer": "5^m(5^m - 1)/4",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 27635,
    "subject": "Mathematics (Multi-modal)",
    "question": "Given is a triangle $ABC$ and a circle $\\omega$ with center $I$ that touches $AB$, $AC$ and meets $BC$ at $X$, $Y$. The line through $I$ perpendicular to $BC$ meets the line through $A$ parallel to $BC$ at $Z$. Show that the circumcircles of $\\triangle XYZ$ and $\\triangle ABC$ are tangent to each other.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $W$ be the midpoint of the major arc $BAC$, $A' \\in (ABC)$ be such that $AA' \\parallel BC$, $T$ be the intersection of the circle with diameter $AI$ and $(ABC)$ and let $\\omega$ touch $AC$, $AB$ at $E$, $F$. We claim the two circles touch at $T$.\n\nFirstly, observe that $Z \\in (AEF)$, so\n$$\n\\begin{aligned}\n\\angle ATZ &= \\angle AIZ = 90^\\circ - \\left(\\frac{\\alpha}{2} + \\gamma\\right) \\\\\n&= \\frac{\\beta - \\gamma}{2} = \\beta - \\left(90^\\circ - \\frac{\\alpha}{2}\\right) \\\\\n&= \\angle ABC - \\angle WBC = \\angle ABW = \\angle ATW,\n\\end{aligned}\n$$\nhence $T$, $Z$, $W$ are collinear. Let $TW \\cap BC = P$. By spiral similarity and angle bisector theorem we have $\\frac{PB}{PC} = \\frac{TB}{TC} = \\frac{BF}{CE}$, so by converse of Menelaus theorem for $\\triangle ABC$ we obtain that $EF$, $TZ$, $BC$ are concurrent at $P$. Thus, by power of point at $P$, we obtain that $PX \\cdot PY = PE \\cdot PF = PZ \\cdot PT$, so $XYZT$ is cyclic.\n\nFinally, observe that the center of $(ZXY)$ lies on $IZ$, so $(ZXY)$ touches $AA'$ at $Z$. Hence, by shooting lemma, since $W$ is the midpoint of the minor arc $AA'$, we obtain that $(ZXY)$ touches $(ABC)$ at $T$ and we are done. $\\square$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 27636,
    "subject": "Mathematics (Multi-modal)",
    "question": "Given is a triangle $ABC$ and a circle $\\omega$ with center $I$ that touches $AB$, $AC$ and meets $BC$ at $X$, $Y$. The line through $I$ perpendicular to $BC$ meets the line through $A$ parallel to $BC$ at $Z$. Show that the circumcircles of $\\triangle XYZ$ and $\\triangle ABC$ are tangent to each other.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $W$ be the midpoint of the major arc $BAC$, $A' \\in (ABC)$ be such that $AA' \\parallel BC$, $T$ be the intersection of the circle with diameter $AI$ and $(ABC)$, and let $\\omega$ touch $AC$, $AB$ at $E$, $F$. We claim the two circles touch at $T$.\n\nFirstly, observe that $Z \\in (AEF)$, so\n\n$$\n\\begin{aligned}\n\\angle ATZ &= \\angle AIZ = 90^\\circ - \\left(\\frac{\\alpha}{2} + \\gamma\\right) \\\\\n&= \\frac{\\beta - \\gamma}{2} = \\beta - \\left(90^\\circ - \\frac{\\alpha}{2}\\right) \\\\\n&= \\angle ABC - \\angle WBC = \\angle ABW = \\angle ATW,\n\\end{aligned}\n$$\n\nhence $T$, $Z$, $W$ are collinear. Let $TW \\cap BC = P$. By spiral similarity and angle bisector theorem we have $\\frac{PB}{PC} = \\frac{TB}{TC} = \\frac{BF}{CE}$, so by converse of Menelaus theorem for $\\triangle ABC$ we obtain that $EF$, $TZ$, $BC$ are concurrent at $P$. Thus, by power of point at $P$, we obtain that $PX \\cdot PY = PE \\cdot PF = PZ \\cdot PT$, so $XYZT$ is cyclic.\n\nFinally, observe that the center of $(ZXY)$ lies on $IZ$, so $(ZXY)$ touches $AA'$ at $Z$. Hence, by shooting lemma, since $W$ is the midpoint of the minor arc $AA'$, we obtain that $(ZXY)$ touches $(ABC)$ at $T$ and we are done. $\\square$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 27637,
    "subject": "Mathematics (Multi-modal)",
    "question": "Some cities of country Graphland are connected with roads provided that\n* (i) from each city we can reach any other city.\nIt turned out that every city $A$ can choose its favourite number $1 \\le f(A) \\le 2024$ which is an integer, such that the following condition holds:\n* (ii) the favourite numbers of any two cities connected with a road are different.\nLet $1 \\le m \\le 2024$ be a given integer. A tourist arrives in the capital of Graphland. He can move from city $A$ to city $B$ (in this direction), if and only if there is a road connecting $A$ and $B$ and additionally\n$$\nf(B) - f(A) \\equiv m \\pmod{2024}.\n$$\nFor which values of $m$ it is guaranteed (no matter of the cities and roads, provided that (i) and (ii) hold) that the cities can choose their favourite numbers (complying with (ii)) in a way that the tourist can reach any city starting from the capital?\n(Dragomir Grozev)",
    "options": [],
    "answer": "All integers m between 1 and 2024 that are coprime to 2024",
    "solution": "Answer: For each $m$ coprime with $2024$.\n\nLet us first assume that $(m, 2024) > 1$. Then the tourist, starting from the capital with a favourite number $r$, can reach only cities with numbers $km + r$ (mod $2024$), $k \\in \\mathbb{N}$. Since it is not a complete system of residues modulo $2024$, it means that the set of all favourite numbers consists of less than $2024$ numbers. Take now $2024$ cities and connect every two of them with a road. This is an example for which (i) can hold if all colours (favourite numbers) are used. That is, in this particular case it's impossible to reach every city starting from the capital.\n\nLet us assume now that $(m, 2024) = 1$. Denote by $V$ the set of all cities and let $W \\subset V$ be the maximal possible subset of vertices such that we can assign favourite numbers and starting from the capital $v_0$ we can reach every vertex in $W$. We prove $W = V$. Assume for the sake of contradiction $W \\neq V$. Then, for any vertices $u \\in W, v \\in V \\setminus W$ that are connected, we have\n$$\nf(v) - f(u) \\neq m \\pmod{2024} \\quad (1)\n$$\nNow, we construct a different mapping $f_1$ by changing the assigning of vertices in $V \\setminus W$. We set,\n$$\nf_1(v) \\in [1..2024], f_1(v) = f(v) - m \\pmod{2024}, \\forall v \\in V \\setminus W.\n$$\nFor any $v \\in W$ we set $f_1(v) = f(v)$. Observe that under the new mapping $f_1$, condition (i) still holds. Indeed, if $f_1(u) = f_1(v)$ for some connected vertices $u \\in W, v \\in V \\setminus W$, it implies $f(u) = f(v) + m$ and thus, we would reach $v$ from $u$ under the mapping $f$, which contradicts the maximality of $W$.\n\nTherefore, the mapping $f_1$ satisfies (i), and moreover starting from $v_0$ we can reach any vertex in $W$. Since $W$ is maximal, it's not possible to access any vertex outside $W$. It allows us to change the assignment of favourite numbers again using the rule\n$$\nf_2(v) = f_1(v) - m \\pmod{2024}, \\forall v \\in V \\setminus W\n$$\nand so on. Suppose now, $u \\in W, v \\in V \\setminus W$ are connected. There exists $k$ for which $f(u) + m = f(v) - km$, because $\\{km : k = 1, 2, \\dots, 2024\\}$ is a complete system of residues modulo $2024$. But this means that under the mapping $f_k$, the vertex $v$ is accessible from $u$, which contradicts the maximality of $W$. $\\square$",
    "topic": "Discrete Mathematics",
    "subtopic": "Combinatorics"
  },
  {
    "id": 27638,
    "subject": "Mathematics (Multi-modal)",
    "question": "The segment $BC$ is fixed in the plane. Let $A$ be a point, such that $\\triangle ABC$ is acute. Let $BP$ ($P \\in AC$) and $CQ$ ($Q \\in AB$) be altitudes in $\\triangle ABC$ and $O$ is the center of the circumscribed circle about $ABC$. The points $K$ and $L$ are symmetric to $O$ with respect to the lines $AB$ and $AC$ respectively. The lines through $P$ and $Q$, orthogonal to $LP$ and $KQ$ respectively, meet at $T$. Prove that when $A$ varies, the point $T$ lies on a fixed line.\n\n(Aleksander Ivanov)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 27639,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all functions $f : \\mathbb{R} \\to \\mathbb{R}$, that satisfy the following condition\n$$\nf(f(x) + x f(y)) = x f(y + 1), \\forall x, y \\in \\mathbb{R}.\n$$\n(Alexander Ivanov)",
    "options": [],
    "answer": "f(x) = x for all real x, and f(x) = 0 for all real x",
    "solution": "Let $f(0) = a, a \\in \\mathbb{R}$. We set $x = 0$ and obtain $f(a) = 0$. Next, $y = a$ yields\n$$\nf(f(x)) = x f(a + 1) \\quad (1).\n$$\nAssume first that $f(a+1) \\neq 0$. Then $f$ is an injection. Indeed, assuming $f(x_1) = f(x_2)$ for some $x_1 \\neq x_2$, after putting these values in (1) we get contradiction. Set $x=1$ in the initial condition. Since $f$ is injective, it follows\n$$\nf(1) + f(y) = y + 1 \\quad (2)\n$$\nSetting $y = 1$ yields $2f(1) = 2$, hence $f(1) = 1$. $f(y) = y + 1$. Now, (2) gives us $f(y) = y, \\forall y \\in \\mathbb{R}$, which indeed is solution.\n\nLet now consider the case $f(a + 1) = 0$. Then\n$f(f(x)) = 0, \\forall x \\in \\mathbb{R}.$\nThis means that $f(y) = 0, \\forall y \\in \\text{Im}(f)$. Assume that there exists a point $y_0$, for which $f(y_0 + 1) \\neq 0$. Setting $y = y_0$ in the initial condition yields\n$$\nf(f(x) + x f(y_0)) = x f(y_0 + 1) \\quad (3).\n$$\nFor any $x_0 \\in \\mathbb{R}$ putting $x := x_0/(f(y_0 + 1))$ in (3), we get,\n$$\nf(f(x) + x f(y_0)) = x_0.\n$$\nSince $x_0$ is arbitrary, it means that $f$ is surjective, which implies $\\text{Im}(f) = \\mathbb{R}$.\nTherefore, $f(y) = 0, \\forall y \\in \\mathbb{R}$, which contradicts the assumption $f(y_0 + 1) \\neq 0$.\nSo, in this case we get $f(x) = 0, \\forall x \\in \\mathbb{R}$, which also is solution. $\\square$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 27640,
    "subject": "Mathematics (Multi-modal)",
    "question": "Given a scalene triangle $ABC$. On the rays $AC \\rightarrow$ and $BC \\rightarrow$, the points $C_a$ and $C_b$ are chosen, respectively, such that $AC_a = BC_b = AB$. We denote by $O_c$ the center of the circumcircle about $\\triangle CC_a C_b$. Analogously, we define the points $O_a$ and $O_b$. Prove that the lines $AO_a$, $BO_b$ and $CO_c$ intersect at a point that lies on the circumcircle of the triangle $ABC$.\n\n(Alexander Ivanov)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $O$ and $I$ be the centers of the circumscribed and inscribed circle of $ABC$, and $A_1$, $B_1$, $C_1$ be the centers of the arcs $\\widehat{AB}$, $\\widehat{AC}$, $\\widehat{BC}$ (not containing the third vertices) of the circumscribed circle. By symmetry with respect to $AA_1$ we have $A_1C_a = A_1B = A_1C$ and together with $O_cC = O_cC_a$ it follows that $A_1O_c$ is the segment bisector of $CC_a$. Analogously, $B_1O_c$ is the segment bisector of $CC_b$. Thus, the quadrilateral $OA_1O_cB_1$ is a parallelogram with $OA_1 = OB_1$, i.e. rhombus. In other words, $O$ and $O_c$ are symmetric about $A_1B_1$, and this is also true for $C$ and $I$, i.e. $CO_c$ is symmetric to $OI$ with respect to $A_1B_1$. Analogously, $AO_a$ and $BO_b$ are symmetric on $OI$ with respect to $B_1C_1$ and $A_1C_1$, respectively. Since $I$ is the orthocenter of $A_1B_1C_1$, finally the lines $AO_a$, $BO_b$, and $CO_c$ intersect at the Anti-Steiner point for the line $OI$ and triangle $A_1B_1C_1$ (circumscribed about $ABC$). $\\square$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 27641,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $n \\ge 4$ be an integer number and $S_n = \\{1, 2, 3, \\dots, 2^n\\}$. Two sets $A, B$ are given, $A \\subset S_n, B \\subset S_n \\setminus S_{n-1}$, such that $|A| = n+1, |B| = 2$. Is it possible $ab-1$ be a perfect cube for any $a \\in A, b \\in B$?\n(Dragomir Grozev)",
    "options": [],
    "answer": "NO",
    "solution": "Answer: NO. Let us argue by contradiction. Arrange the numbers in $A$ and $B$ in increasing order $1 \\le a_1 < a_2 < \\dots < a_{n+1} \\le 2^n$ and $2^{n-1} < b_1 < b_2 \\le 2^n$. Apparently, there exists an index $i \\le n$ such that $a_i < a_{i+1} \\le 2a_i$. We denote:\n\n$$\na_i b_1 - 1 = q_1^3, \\quad a_{i+1} b_2 - 1 = q_2^3, \\quad a_i b_2 - 1 = s_1^3, \\quad a_{i+1} b_1 - 1 = s_2^3.\n$$\nFrom $a_i < a_{i+1} \\le 2a_i$ and $b_1 < b_2 < 2b_1$ it follows\n$$\nq_1 < s_1, \\quad s_2 < q_2 < 2q_1 \\quad (1).\n$$\nTherefore\n$$\n(q_1^3 + 1)(q_2^3 + 1) = (s_1^3 + 1)(s_2^3 + 1)\n$$\nwhich yields\n$$\n(q_1 q_2)^3 + q_1^3 + q_2^3 + 1 = (s_1 s_2)^3 + s_1^3 + s_2^3 + 1 \\quad (2).\n$$\n\nUsing (1) it can be seen that if $q_1q_2 > s_1s_2$ or $q_1q_2 < s_1s_2$ the equality (2) cannot hold. Thus, $q_1q_2 = s_1s_2$. Let us consider the function $f(x) = x^3 + \\frac{C}{x^3}$ in the interval $x \\in [q_1, q_2]$, where $C = (q_1q_2)^3$. It decreases in $[q_1, \\sqrt{q_1q_2}]$ and increases in $[\\sqrt{q_1q_2}, q_2]$, hence it attains its maximum value at the ends of this interval, $f(q_1) = f(q_2) = q_1^3 + q_2^3$. Therefore, the equality (2), provided that (1) holds, is true only if $(q_1, q_2) = (s_1, s_2)$, which is the needed contradiction. $\\Box$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 27642,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $n \\ge 4$ be an integer number and $S_n = \\{1, 2, 3, \\dots, 2^n\\}$. Two sets $A, B$ are given, $A \\subset S_n, B \\subset S_n \\setminus S_{n-1}$, such that $|A| = n+1, |B| = 2$. Is it possible $ab-1$ be a perfect cube for any $a \\in A, b \\in B$?\n(Dragomir Grozev)",
    "options": [],
    "answer": "NO",
    "solution": "Answer: NO. Let us argue by contradiction. Arrange the numbers in $A$ and $B$ in increasing order $1 \\le a_1 < a_2 < \\dots < a_{n+1} \\le 2^n$ and $2^{n-1} < b_1 < b_2 \\le 2^n$. Apparently, there exists an index $i \\le n$ such that $a_i < a_{i+1} \\le 2a_i$. We denote:\n$$\na_i b_1 - 1 = q_1^3,\\quad a_{i+1} b_2 - 1 = q_2^3,\\quad a_i b_2 - 1 = s_1^3,\\quad a_{i+1} b_1 - 1 = s_2^3.\n$$\nFrom $a_i < a_{i+1} \\le 2a_i$ and $b_1 < b_2 < 2b_1$ it follows\n$$\nq_1 < s_1,\\quad s_2 < q_2 < 2q_1 \\quad (1).\n$$\n\n$$\n(q_1 q_2)^3 + q_1^3 + q_2^3 + 1 = (s_1 s_2)^3 + s_1^3 + s_2^3 + 1 \\quad (2).\n$$\n\nUsing (1) it can be seen that if $q_1q_2 > s_1s_2$ or $q_1q_2 < s_1s_2$ the equality (2) cannot hold. Thus, $q_1q_2 = s_1s_2$. Let us consider the function $f(x) = x^3 + \\frac{C}{x^3}$ in the interval $x \\in [q_1, q_2]$, where $C = (q_1q_2)^3$. It decreases in $[q_1, \\sqrt{q_1q_2}]$ and increases in $[\\sqrt{q_1q_2}, q_2]$, hence it attains its maximum value at the ends of this interval, $f(q_1) = f(q_2) = q_1^3 + q_2^3$. Therefore, the equality (2), provided that (1) holds, is true only if $(q_1, q_2) = (s_1, s_2)$, which is the needed contradiction. $\\Box$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 27643,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $n$ be a natural number. King Arthur has invited $2^n - 1$ knights to an audience in Camelot. Merlin the Magician arranged the knights in a list numbered from $1$ to $2^n - 1$. It turned out that any two knights with numbers $a, b, a < b$ are friends if and only if $0 \\le b - 2a \\le 1$. The king chose a natural number $k$ and ordered Merlin to make a new list with the following requirement. For each $1 \\le i \\le 2^n - k - 1$, all friends of the knight with sequence number $i$ (in the new list) must be in positions among $1, 2, \\dots, i + k$. Prove that the smallest $k$, for which Merlin can fulfil Arthur's wish, satisfies the condition\n$$\n\\frac{1}{100} \\cdot \\frac{2^n}{n} \\le k \\le 100 \\cdot \\frac{2^n}{n}.\n$$\n(Dragomir Grozev)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let us construct a graph $T$ with vertex-set which is the set of all knights, numbered as in the first list. Two vertices are adjacent if the corresponding knights are friends. It can be seen that $T$ is a fully balanced binary tree - see fig. 1. We label each vertex with the knight's number in the first list. Let us assume the vertices can be arranged in a row $i_1, i_2, \\dots, i_m$ as King Arthur requested, where $m = 2^n - 1$ and $i_j$ refers to the label of the corresponding vertex. There is a unique path in $T$, $i_1 = v_1v_2\\dots v_\\ell = i_m$ that connects the vertex labeled as $i_1$ and $i_m$. Clearly, $\\ell \\le 2(n-1) + 1$, because the longest path in $T$ has length $2(n-1)$. Note that the distance between the positions of $v_i$ and $v_{i+1}$ in the new list, is at most $k$. This means that $(\\ell - 1)k \\ge m - 1$ which yields\n$$\nk \\ge \\frac{2^n - 2}{2n - 2} > \\frac{2^n}{4n},\n$$\nwhich proves the lower bound for $k$.\n\nNow we will arrange the vertices of $T$ in a list. Let $\\ell$ be a natural number which will be determined later. Denote by $v_1, v_2, \\dots, v_s, s := 2^\\ell$ the vertices of $T$ on the $\\ell$-th level, and let $T(v_1), T(v_2), \\dots, T(v_s)$ be the subtrees with roots in these points - see fig. 1. Each of them has exactly $2^{n-1-\\ell}$ leaves.\n\nWe successively put in a list the vertices of $T$ as follows. First, we place the last level of vertices of $T(v_1)$, that is, its leaves. Then we put down the second to last level of $T(v_1)$ and the last level of $T(v_2)$. At the $i$-th step we place the $i$-th level of $T(v_1)$, counting from the bottom up (fig. 1), then the $i-1$-th level of $T(v_2)$ (from the bottom up) and so on, and finally - the last layer of $T(v_i)$ (i.e. its leaves). The number of vertices we place at the $i$-th step $i = 1, 2, \\dots, s-\\ell-1$ is equal to\n$$\n\\sum_{j=0}^{i-1} 2^{n-1-\\ell-j} \\le 2^{n-\\ell}.\n$$\nWe follow these steps until one of the two events happens. 1) We reach the root $v_1$ of $T(v_1)$. 2) We place in the list the leaves of $T(v_s)$. The first event will\n\n![](attached_image_1.png)\n\nhappen after $n-\\ell$ steps and the second one - after $s = 2^\\ell$ steps. To ensure that the second event occurs first we choose $\\ell$ to be the largest positive integer for which $s = 2^\\ell \\le n - \\ell$.\n\nIn this situation, at the *s*-th step we have put the leaves of $T(v_s)$ in the list. On the *s* + 1-th step we put in the list the remaining vertices of *T*. The number of all vertices in $T(v_s)$ without its last level does not exceed $2^{n-\\ell-1}$. The number of vertices in $T(v_{s-1})$ without its last two layers does not exceed $2^{n-\\ell-2}$ and so on. Adding the vertices of *T* up to its *l*-th level, we obtain that the number of the vertices ordered at the last step is at most\n$$\n2 \\cdot 2^{\\ell} + 2^{n-\\ell} \\le 2 \\cdot 2^{n-\\ell} \\le 8 \\cdot \\frac{2^n}{n}\n$$\nsince $2^{n+2} \\ge n$. Clearly, for any vertex $v$, placed at position $j$ in the first $s$ steps, all of its neighbours in $T$, that are placed after it, are in positions with numbers not exceeding $j+2 \\cdot 2^{n-\\ell} \\le j+8 \\cdot \\frac{2^n}{n}$. Taking into account the number of vertices added in the last step, we get that in the constructed list the condition imposed by the king holds for\n$$\nk := \\left\\lceil 16 \\cdot \\frac{2^n}{n} \\right\\rceil\n$$\n\nThis proves the upper bound. □",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 27644,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSuppose that $n$ people each know exactly one piece of information, and all $n$ pieces are different. Every time person $A$ phones person $B$, $A$ tells $B$ everything that $A$ knows, while $B$ tells $A$ nothing. What is the minimum number of phone calls between pairs of people needed for everyone to know everything? Prove your answer is a minimum.",
    "options": [],
    "answer": "2n - 2",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 27645,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nDetermine which of the two numbers $\\sqrt{c+1}-\\sqrt{c}$, $\\sqrt{c}-\\sqrt{c-1}$ is greater for any $c \\geq 1$.",
    "options": [],
    "answer": "sqrt(c) - sqrt(c-1) is greater than sqrt(c+1) - sqrt(c) for all c >= 1",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 27646,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nDetermine all real numbers $a$ such that the two polynomials $x^{2} + a x + 1$ and $x^{2} + x + a$ have at least one root in common.",
    "options": [],
    "answer": "a = 1 or a = -2",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 27647,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nLet $ABC$ be a triangle with sides of lengths $a$, $b$ and $c$. Let the bisector of the angle $C$ cut $AB$ in $D$. Prove that the length of $CD$ is\n$$\n\\frac{2ab \\cos \\frac{C}{2}}{a+b}\n$$",
    "options": [],
    "answer": "2ab cos(C/2)/(a+b)",
    "solution": "",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 27648,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nFind the sum of $1 \\cdot 1! + 2 \\cdot 2! + 3 \\cdot 3! + \\cdots + (n-1)(n-1)! + n \\cdot n!$, where $n! = n(n-1)(n-2) \\cdots 2 \\cdot 1$.",
    "options": [],
    "answer": "(n+1)! - 1",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 27649,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nShow that there are no integers $a, b, c$ for which $a^{2} + b^{2} - 8c = 6$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\nSuppose there exist integers $a, b, c$ such that $a^{2} + b^{2} - 8c = 6$.\n\nThen $a^{2} + b^{2} = 8c + 6$.\n\nConsider $a^{2} + b^{2}$ modulo $8$.\n\nThe possible quadratic residues modulo $8$ are $0, 1, 4$ (since $0^{2} \\equiv 0$, $1^{2} \\equiv 1$, $2^{2} \\equiv 4$, $3^{2} \\equiv 1$, $4^{2} \\equiv 0$, $5^{2} \\equiv 1$, $6^{2} \\equiv 4$, $7^{2} \\equiv 1$ modulo $8$).\n\nSo $a^{2} + b^{2}$ modulo $8$ can be $0+0=0$, $0+1=1$, $0+4=4$, $1+1=2$, $1+4=5$, $4+4=0$.\nThus, the possible values for $a^{2} + b^{2}$ modulo $8$ are $0, 1, 2, 4, 5$.\n\nBut $a^{2} + b^{2} = 8c + 6 \\equiv 6 \\pmod{8}$.\n\n$6$ is not among the possible values for $a^{2} + b^{2}$ modulo $8$.\n\nTherefore, there are no integers $a, b, c$ for which $a^{2} + b^{2} - 8c = 6$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 27650,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nLet $n$ be a five digit number (whose first digit is non-zero) and let $m$ be the four digit number formed from $n$ by deleting its middle digit. Determine all $n$ such that $n / m$ is an integer.",
    "options": [],
    "answer": "All five-digit multiples of 1000: n = 1000·t for integers t from 10 to 99 (i.e., numbers of the form ab000 with a in 1..9 and b in 0..9).",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 27651,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nShow that for any quadrilateral inscribed in a circle of radius $1$, the length of the shortest side is less than or equal to $\\sqrt{2}$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 27652,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\n$ABCD$ is a quadrilateral with $AD = BC$. If $\\angle ADC$ is greater than $\\angle BCD$, prove that $AC > BD$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 27653,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nLet $ABC$ be the right-angled isosceles triangle whose equal sides have length $1$. $P$ is a point on the hypotenuse, and the feet of the perpendiculars from $P$ to the other sides are $Q$ and $R$. Consider the areas of the triangles $APQ$ and $PBR$, and the area of the rectangle $QCRP$. Prove that regardless of how $P$ is chosen, the largest of these three areas is at least $2/9$.\n\n![](attached_image_1.png)",
    "options": [],
    "answer": "2/9",
    "solution": "",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 27654,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nShow that if $a_{1} / b_{1}=a_{2} / b_{2}=a_{3} / b_{3}$ and $p_{1}, p_{2}, p_{3}$ are not all zero, then\n$$\n\\left(\\frac{a_{1}}{b_{1}}\\right)^{n}=\\frac{p_{1} a_{1}^{n}+p_{2} a_{2}^{n}+p_{3} a_{3}^{n}}{p_{1} b_{1}^{n}+p_{2} b_{2}^{n}+p_{3} b_{3}^{n}}\n$$\nfor every positive integer $n$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 27655,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nLet $c$ be the length of the hypotenuse of a right angle triangle whose other two sides have lengths $a$ and $b$. Prove that $a + b \\leq \\sqrt{2} c$. When does the equality hold?",
    "options": [],
    "answer": "Equality holds precisely when the two legs are equal (the triangle is isosceles right).",
    "solution": "",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 27656,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nLet $ABC$ be an equilateral triangle, and $P$ be an arbitrary point within the triangle. Perpendiculars $PD$, $PE$, $PF$ are drawn to the three sides of the triangle. Show that, no matter where $P$ is chosen,\n$$\n\\frac{PD + PE + PF}{AB + BC + CA} = \\frac{1}{2\\sqrt{3}}\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 27657,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nLet\n$$\np(x) = a_{0} x^{n} + a_{1} x^{n-1} + \\cdots + a_{n-1} x + a_{n}\n$$\nwhere the coefficients $a_{i}$ are integers. If $p(0)$ and $p(1)$ are both odd, show that $p(x)$ has no integral roots.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 27658,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nShow that, for all integers $n$, $n^{2} + 2n + 12$ is not a multiple of $121$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 27659,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nLet $f$ be a function with the following properties:\n1) $f(n)$ is defined for every positive integer $n$;\n2) $f(n)$ is an integer;\n3) $f(2)=2$;\n4) $f(m n)=f(m) f(n)$ for all $m$ and $n$;\n5) $f(m)>f(n)$ whenever $m>n$.\nProve that $f(n)=n$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nLet us analyze the properties of $f$.\n\nFrom property 4, $f(m n) = f(m) f(n)$ for all positive integers $m, n$. This means $f$ is a multiplicative function.\n\nLet us compute $f(1)$:\n\nTake $m = n = 1$:\n$$\nf(1) = f(1 \\cdot 1) = f(1) f(1) \\implies f(1)^2 = f(1)\n$$\nSo $f(1) = 0$ or $f(1) = 1$.\n\nSuppose $f(1) = 0$. Then for any $n$,\n$$\nf(n) = f(n \\cdot 1) = f(n) f(1) = f(n) \\cdot 0 = 0\n$$\nSo $f(n) = 0$ for all $n$, but this contradicts $f(2) = 2$. Therefore, $f(1) = 1$.\n\nNow, $f(2) = 2$ is given.\n\nLet us try to find $f(2^k)$ for $k \\geq 1$.\n\nBy induction:\n- $f(2^1) = f(2) = 2$\n- $f(2^2) = f(2 \\cdot 2) = f(2) f(2) = 2 \\cdot 2 = 4$\n- $f(2^3) = f(2 \\cdot 2^2) = f(2) f(2^2) = 2 \\cdot 4 = 8$\n\nSo $f(2^k) = 2^k$ for all $k \\geq 1$.\n\nNow, let us consider $f(3)$.\n\nSince $f$ is strictly increasing (property 5), $f(3) > f(2) = 2$ and $f(3)$ is an integer.\n\nLet us compute $f(3^k)$:\n- $f(3^1) = f(3)$\n- $f(3^2) = f(3) f(3)$\n- $f(3^3) = f(3) f(3) f(3) = f(3)^3$\n\nSimilarly, $f(6) = f(2 \\cdot 3) = f(2) f(3) = 2 f(3)$.\n\nBut $f(6)$ must also equal $f(6) = f(3 \\cdot 2) = f(3) f(2) = f(3) \\cdot 2 = 2 f(3)$, which is consistent.\n\nNow, consider $f(4) = 4$ and $f(3) > 2$.\n\nBut $f(4) = 4$ and $f(3)$ is an integer $> 2$ and $f$ is strictly increasing, so $f(3) = 3$ (since $f(3) < f(4) = 4$ and $f(3)$ is an integer $> 2$).\n\nNow, $f(5)$ must be an integer $> f(4) = 4$, so $f(5) \\geq 5$.\n\nLet us check $f(10) = f(2 \\cdot 5) = f(2) f(5) = 2 f(5)$.\nBut $f(10) = f(5 \\cdot 2) = f(5) f(2) = f(5) \\cdot 2 = 2 f(5)$, which is consistent.\n\nLet us try to prove by induction that $f(n) = n$ for all $n$.\n\nBase case: $f(1) = 1$, $f(2) = 2$, $f(3) = 3$.\n\nSuppose $f(k) = k$ for all $k < n$.\n\nIf $n$ is composite, say $n = a b$ with $a, b < n$, then\n$$\nf(n) = f(a b) = f(a) f(b) = a b = n\n$$\nIf $n$ is prime, then $f(n)$ is an integer $> f(n-1) = n-1$, so $f(n) \\geq n$.\n\nBut if $f(n) > n$, then $f(n) \\geq n+1$.\n\nConsider $f(2 n) = f(2) f(n) = 2 f(n)$. But $2 n$ is less than $2 f(n)$ if $f(n) > n$, which would contradict the strictly increasing property, since $f(2 n) > f(2 n - 1)$, but $f(2 n) = 2 f(n) \\geq 2(n+1) = 2 n + 2$, while $f(2 n - 1) = 2 n - 1$ (by induction), so $2 n + 2 > 2 n - 1$, which is always true, but let's check for $n$ large.\n\nBut more precisely, suppose $f(n) \\geq n+1$, then $f(2 n) = 2 f(n) \\geq 2(n+1) = 2 n + 2$, but $2 n$ is the argument, so $f(2 n) \\geq 2 n + 2$, but $f(2 n - 1) = 2 n - 1$, so the difference is at least $3$.\n\nBut $f$ must be strictly increasing by $1$ at each step, since $f(n)$ is an integer and $f(n) > f(n-1)$, so $f(n) \\geq f(n-1) + 1$.\n\nBut if $f(n) > n$, then $f(n) \\geq n+1$, so $f(n) - f(n-1) \\geq 2$ for some $n$, which is not possible if $f$ is strictly increasing by $1$ at each step.\n\nTherefore, $f(n) = n$ for all $n$.\n\nThus, the only function satisfying all the properties is $f(n) = n$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 27660,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nA regular pentagon is inscribed in a circle of radius $r$. $P$ is any point inside the pentagon. Perpendiculars are dropped from $P$ to the sides, or the sides produced, of the pentagon.\n\na) Prove that the sum of the lengths of these perpendiculars is constant.\n\nb) Express this constant in terms of the radius $r$.",
    "options": [],
    "answer": "5 r cos(π/5) = (5(√5+1)/4) r",
    "solution": "",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 27661,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nTwo flag poles of heights $h$ and $k$ are situated $2a$ units apart on a level surface. Find the set of all points on the surface which are so situated that the angles of elevation of the tops of the poles are equal.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 27662,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\na) Find all positive integers with initial digit $6$ such that the integer formed by deleting this $6$ is $1/25$ of the original integer.\n\nb) Show that there is no integer such that deletion of the first digit produces a result which is $1/35$ of the original integer.",
    "options": [],
    "answer": "a) All integers of the form 625·10^m for m ≥ 0 (i.e., 625, 6250, 62500, …). b) No such integer exists.",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 27663,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nShow that from any five integers, not necessarily distinct, one can always choose three of these integers whose sum is divisible by $3$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 27664,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nLet $f(n)$ be the sum of the first $n$ terms of the sequence\n$$\n0, 1, 1, 2, 2, 3, 3, 4, 4, 5, 5, 6, 6, \\ldots\n$$\na) Give a formula for $f(n)$.\nb) Prove that $f(s+t)-f(s-t)=s t$ where $s$ and $t$ are positive integers and $s>t$.",
    "options": [],
    "answer": "a) f(n) = ⌊n/2⌋·⌈n/2⌉ = ⌊n^2/4⌋.\nb) f(s+t) − f(s−t) = s·t for positive integers s > t.",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 27665,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nFind all number triples $(x, y, z)$ such that when any one of these numbers is added to the product of the other two, the result is $2$.",
    "options": [],
    "answer": "(1, 1, 1) and (-2, -2, -2)",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 27666,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nGiven the polynomial\n$$\nf(x) = x^{n} + a_{1} x^{n-1} + a_{2} x^{n-2} + \\cdots + a_{n-1} x + a_{n}\n$$\nwith integral coefficients $a_{1}, a_{2}, \\ldots, a_{n}$, and given also that there exist four distinct integers $a, b, c$ and $d$ such that\n$$\nf(a) = f(b) = f(c) = f(d) = 5\n$$\nshow that there is no integer $k$ such that $f(k) = 8$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\nLet $f(x)$ be as given, and suppose $f(a) = f(b) = f(c) = f(d) = 5$ for four distinct integers $a, b, c, d$.\n\nConsider the polynomial $g(x) = f(x) - 5$. Then $g(a) = g(b) = g(c) = g(d) = 0$, so $a, b, c, d$ are roots of $g(x)$.\n\nTherefore, $g(x)$ is divisible by $(x - a)(x - b)(x - c)(x - d)$, i.e.,\n$$\ng(x) = (x - a)(x - b)(x - c)(x - d) h(x)\n$$\nfor some polynomial $h(x)$ with integer coefficients (since $f(x)$ has integer coefficients and $5$ is an integer).\n\nSuppose there exists an integer $k$ such that $f(k) = 8$. Then $g(k) = f(k) - 5 = 3$.\n\nSo,\n$$\ng(k) = (k - a)(k - b)(k - c)(k - d) h(k) = 3\n$$\nBut $k, a, b, c, d$ are all integers, so $(k - a)(k - b)(k - c)(k - d)$ is an integer, and $h(k)$ is an integer.\n\nTherefore, $(k - a)(k - b)(k - c)(k - d) h(k) = 3$.\n\nBut $3$ has only the following possible factorizations into integer factors:\n\n- $3 = 1 \\times 1 \\times 1 \\times 3$\n- $3 = 1 \\times 1 \\times (-1) \\times (-3)$\n- $3 = 1 \\times (-1) \\times 1 \\times (-3)$\n- $3 = (-1) \\times 1 \\times 1 \\times (-3)$\n- $3 = (-1) \\times (-1) \\times (-1) \\times (-3)$\n- $3 = 3 \\times 1 \\times 1 \\times 1$\n- $3 = (-3) \\times 1 \\times 1 \\times (-1)$\n- $3 = 1 \\times 3 \\times 1 \\times 1$\n- $3 = 1 \\times 1 \\times 3 \\times 1$\n- $3 = 1 \\times 1 \\times 1 \\times 3$\n\nBut in all cases, since $a, b, c, d$ are distinct integers, $k - a$, $k - b$, $k - c$, $k - d$ are four distinct nonzero integers. The product of four distinct nonzero integers cannot be $\\pm 1$ or $\\pm 3$.\n\nTo see this, note that the absolute value of the product $| (k - a)(k - b)(k - c)(k - d) |$ is at least $1 \\times 2 \\times 3 \\times 4 = 24$ (if the differences are $\\pm 1, \\pm 2, \\pm 3, \\pm 4$), which is much greater than $3$.\n\nTherefore, there is no integer $k$ such that $f(k) = 8$.",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 27667,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nA set of balls is given. Each ball is coloured red or blue, and there is at least one of each colour. Each ball weighs either $1$ pound or $2$ pounds, and there is at least one of each weight. Prove that there are $2$ balls having different weights and different colours.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 27668,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nA quadrilateral has one vertex on each side of a square of side-length $1$. Show that the lengths $a, b, c$ and $d$ of the sides of the quadrilateral satisfy the inequalities\n$$\n2 \\leq a^{2}+b^{2}+c^{2}+d^{2} \\leq 4\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 27669,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nGiven three non-collinear points $A$, $B$, $C$, construct a circle with centre $C$ such that the tangents from $A$ and $B$ to the circle are parallel.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 27670,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nConsider all line segments of length $4$ with one end-point on the line $y = x$ and the other end-point on the line $y = 2x$. Find the equation of the locus of the midpoints of these line segments.",
    "options": [],
    "answer": "25x^2 + 13y^2 - 36xy = 4",
    "solution": "Solution:\nLet the end-points of the segment be $A(a, a)$ on $y = x$ and $B(b, 2b)$ on $y = 2x$.\n\nThe length of $AB$ is $4$:\n$$\nAB = \\sqrt{(b - a)^2 + (2b - a)^2} = 4\n$$\nSo,\n$$\n(b - a)^2 + (2b - a)^2 = 16\n$$\nExpanding:\n$$\n(b - a)^2 + (2b - a)^2 = (b - a)^2 + (4b^2 - 4ab + a^2) = (b - a)^2 + 4b^2 - 4ab + a^2\n$$\nBut $(2b - a)^2 = 4b^2 - 4ab + a^2$, so:\n$$\n(b - a)^2 + (2b - a)^2 = (b^2 - 2ab + a^2) + (4b^2 - 4ab + a^2) = 5b^2 - 6ab + 2a^2 = 16\n$$\nLet the midpoint $M$ of $AB$ be $\\left(\\frac{a + b}{2}, \\frac{a + 2b}{2}\\right)$.\nLet $x = \\frac{a + b}{2}$ and $y = \\frac{a + 2b}{2}$.\n\nThen:\n$$\na = 2x - b\n$$\n$$\ny = \\frac{a + 2b}{2} = \\frac{(2x - b) + 2b}{2} = \\frac{2x + b}{2}\n$$\nSo,\n$$\n2y = 2x + b \\implies b = 2y - 2x\n$$\nSubstitute $b$ into $a$:\n$$\na = 2x - (2y - 2x) = 2x - 2y + 2x = 4x - 2y\n$$\nNow, substitute $a$ and $b$ in the length equation:\n$$\n5b^2 - 6ab + 2a^2 = 16\n$$\nPlug in $a = 4x - 2y$, $b = 2y - 2x$:\n\nFirst, compute $b^2$:\n$$\nb^2 = (2y - 2x)^2 = 4(y - x)^2\n$$\n$a^2$:\n$$\na^2 = (4x - 2y)^2 = 16x^2 - 16x y + 4y^2\n$$\n$ab$:\n$$\nab = (4x - 2y)(2y - 2x) = 4x(2y - 2x) - 2y(2y - 2x) = 8x y - 8x^2 - 4y^2 + 4x y = (8x y + 4x y) - 8x^2 - 4y^2 = 12x y - 8x^2 - 4y^2\n$$\nBut let's expand directly:\n$$\nab = (4x - 2y)(2y - 2x) = 4x \\cdot 2y - 4x \\cdot 2x - 2y \\cdot 2y + 2y \\cdot 2x = 8x y - 8x^2 - 4y^2 + 4x y = (8x y + 4x y) - 8x^2 - 4y^2 = 12x y - 8x^2 - 4y^2\n$$\nNow, plug into the equation:\n$$\n5b^2 - 6ab + 2a^2 = 5 \\cdot 4(y - x)^2 - 6(12x y - 8x^2 - 4y^2) + 2(16x^2 - 16x y + 4y^2)\n$$\nCompute each term:\n$5b^2 = 20(y - x)^2$\n$-6ab = -6(12x y - 8x^2 - 4y^2) = -72x y + 48x^2 + 24y^2$\n$2a^2 = 32x^2 - 32x y + 8y^2$\n\nSum:\n$$\n20(y - x)^2 + 48x^2 + 24y^2 - 72x y + 32x^2 - 32x y + 8y^2 = 16\n$$\nBut $20(y - x)^2 = 20(y^2 - 2x y + x^2) = 20y^2 - 40x y + 20x^2$\nSo sum all $x^2$ terms:\n$20x^2 + 48x^2 + 32x^2 = 100x^2$\n$y^2$ terms: $20y^2 + 24y^2 + 8y^2 = 52y^2$\n$-x y$ terms: $-40x y - 72x y - 32x y = -144x y$\n\nSo the equation is:\n$$\n100x^2 + 52y^2 - 144x y = 16\n$$\nOr, dividing both sides by $4$:\n$$\n25x^2 + 13y^2 - 36x y = 4\n$$\nThus, the equation of the locus is:\n$$\n25x^2 + 13y^2 - 36x y = 4\n$$",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 27671,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nDescribe a construction of a quadrilateral $ABCD$ given:\n(i) the lengths of all four sides;\n(ii) that $AB$ and $CD$ are parallel;\n(iii) that $BC$ and $DA$ do not intersect.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 27672,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nProve that the equation $x^{3} + 11^{3} = y^{3}$ has no solution in positive integers $x$ and $y$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 27673,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nLet $a_{1}, a_{2}, \\ldots, a_{n}$ be non-negative real numbers. Define $M$ to be the sum of all products of pairs $a_{i} a_{j}$ ($i<j$), i.e.,\n$$\nM = a_{1}(a_{2} + a_{3} + \\cdots + a_{n}) + a_{2}(a_{3} + a_{4} + \\cdots + a_{n}) + \\cdots + a_{n-1} a_{n}\n$$\nProve that the square of at least one of the numbers $a_{1}, a_{2}, \\ldots, a_{n}$ does not exceed $2M / n(n-1)$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 27674,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nLet $a$ and $b$ be distinct real numbers. Prove that there exist integers $m$ and $n$ such that $a m + b n < 0$, $b m + a n > 0$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 27675,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nGiven three distinct unit circles, each of which is tangent to the other two, find the radii of the circles which are tangent to all three circles.",
    "options": [],
    "answer": "radii = 2/√3 − 1 and 1 + 2/√3",
    "solution": "",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 27676,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nWhat is the maximum number of terms in a geometric progression with common ratio greater than $1$ whose entries all come from the set of integers between $100$ and $1000$ inclusive?",
    "options": [],
    "answer": "6",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 27677,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nDuring a certain election campaign, $p$ different kinds of promises are made by the various political parties ($p>0$). While several parties may make the same promise, any two parties have at least one promise in common; no two parties have exactly the same set of promises. Prove that there are no more than $2^{p-1}$ parties.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 27678,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nFour distinct lines $L_{1}, L_{2}, L_{3}, L_{4}$ are given in the plane: $L_{1}$ and $L_{2}$ are respectively parallel to $L_{3}$ and $L_{4}$. Find the locus of a point moving so that the sum of its perpendicular distances from the four lines is constant.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 27679,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nFind all the real numbers which satisfy the equation $|x+3| - |x-1| = x+1$. (Note: $|a| = a$ if $a \\geq 0$; $|a| = -a$ if $a < 0$.)",
    "options": [],
    "answer": "x = -5, -1, 3",
    "solution": "Solution:\nLet us consider the different cases for $x$ based on the expressions inside the absolute values.\n\n**Case 1:** $x + 3 \\geq 0$ and $x - 1 \\geq 0$\n\nThis means $x \\geq -3$ and $x \\geq 1$, so $x \\geq 1$.\n\nIn this case, $|x+3| = x+3$ and $|x-1| = x-1$.\n\nThe equation becomes:\n$$(x+3) - (x-1) = x+1$$\n$$x+3 - x + 1 = x+1$$\n$$4 = x+1$$\n$$x = 3$$\n\nCheck if $x = 3$ satisfies $x \\geq 1$ (it does).\n\n**Case 2:** $x + 3 \\geq 0$ and $x - 1 < 0$\n\nThis means $x \\geq -3$ and $x < 1$, so $-3 \\leq x < 1$.\n\nIn this case, $|x+3| = x+3$ and $|x-1| = -(x-1) = -x+1$.\n\nThe equation becomes:\n$$(x+3) - (-x+1) = x+1$$\n$$x+3 + x - 1 = x+1$$\n$$2x + 2 = x+1$$\n$$2x - x = 1 - 2$$\n$$x = -1$$\n\nCheck if $x = -1$ satisfies $-3 \\leq x < 1$ (it does).\n\n**Case 3:** $x + 3 < 0$ and $x - 1 < 0$\n\nThis means $x < -3$ and $x < 1$, so $x < -3$.\n\nIn this case, $|x+3| = -(x+3) = -x-3$ and $|x-1| = -(x-1) = -x+1$.\n\nThe equation becomes:\n$$(-x-3) - (-x+1) = x+1$$\n$$-x-3 + x - 1 = x+1$$\n$$-4 = x+1$$\n$$x = -5$$\n\nCheck if $x = -5$ satisfies $x < -3$ (it does).\n\n**Summary:**\nThe solutions are $x = -5$, $x = -1$, and $x = 3$.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 27680,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nIf $A$ and $B$ are fixed points on a given circle not collinear with centre $O$ of the circle, and if $XY$ is a variable diameter, find the locus of $P$ (the intersection of the line through $A$ and $X$ and the line through $B$ and $Y$).",
    "options": [],
    "answer": "The locus is the circle through A and B whose center is the intersection point of the tangents to the given circle at A and at B.",
    "solution": "",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 27681,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nObserve that\n$$\n\\frac{1}{1} = \\frac{1}{2} + \\frac{1}{2} ; \\quad \\frac{1}{2} = \\frac{1}{3} + \\frac{1}{6} ; \\quad \\frac{1}{3} = \\frac{1}{4} + \\frac{1}{12} ; \\quad \\frac{1}{4} = \\frac{1}{5} + \\frac{1}{20}\n$$\nState a general law suggested by these examples, and prove it.\n\nProve that for any integer $n$ greater than 1 there exist positive integers $i$ and $j$ such that\n$$\n\\frac{1}{n} = \\frac{1}{i(i+1)} + \\frac{1}{(i+1)(i+2)} + \\frac{1}{(i+2)(i+3)} + \\cdots + \\frac{1}{j(j+1)}\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 27682,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nFor every positive integer $n$, let\n$$\nh(n) = 1 + \\frac{1}{2} + \\frac{1}{3} + \\cdots + \\frac{1}{n}\n$$\nFor example, $h(1) = 1$, $h(2) = 1 + \\frac{1}{2}$, $h(3) = 1 + \\frac{1}{2} + \\frac{1}{3}$. Prove that for $n = 2, 3, 4, \\ldots$\n$$\nn + h(1) + h(2) + h(3) + \\cdots + h(n-1) = n h(n)\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 27683,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\ni. Solve the simultaneous inequalities, $x < \\frac{1}{4x}$ and $x < 0$; i.e., find a single inequality equivalent to the two given simultaneous inequalities.\n\nii. What is the greatest integer which satisfies both inequalities $4x + 13 < 0$ and $x^2 + 3x > 16$?\n\niii. Give a rational number between $\\frac{11}{24}$ and $\\frac{6}{13}$.\n\niv. Express $100000$ as a product of two integers neither of which is an integral multiple of $10$.\n\nv. Without the use of logarithm tables evaluate\n$$\n\\frac{1}{\\log_{2} 36} + \\frac{1}{\\log_{3} 36}\n$$",
    "options": [],
    "answer": "i) x < -1/2; ii) -6; iii) 17/37; iv) 100000 = 32*3125; v) 1/2",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 27684,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nProve that if $p$ and $p+2$ are both prime integers greater than $3$, then $6$ is a factor of $p+1$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 27685,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nLet $n$ be a fixed positive integer. To any choice of $n$ real numbers satisfying\n$$\n0 \\leq x_{i} \\leq 1, \\quad i=1,2, \\ldots, n\n$$\nthere corresponds the sum\n(*)\n$$\n\\begin{aligned}\n& \\sum_{1 \\leq i<j \\leq n}\\left|x_{i}-x_{j}\\right| \\\\\n& \\qquad \\begin{aligned}\n= & \\left|x_{1}-x_{2}\\right|+\\left|x_{1}-x_{3}\\right|+\\left|x_{1}-x_{4}\\right|+\\cdots+\\left|x_{1}-x_{n-1}\\right|+\\left|x_{1}-x_{n}\\right| \\\\\n& \\quad+\\left|x_{2}-x_{3}\\right|+\\left|x_{2}-x_{4}\\right|+\\cdots+\\left|x_{2}-x_{n-1}\\right|+\\left|x_{2}-x_{n}\\right| \\\\\n& \\quad+\\left|x_{3}-x_{4}\\right|+\\cdots+\\left|x_{3}-x_{n-1}\\right|+\\left|x_{3}-x_{n}\\right| \\\\\n& \\quad+\\cdots+\\left|x_{n-2}-x_{n-1}\\right|+\\left|x_{n-2}-x_{n}\\right| \\\\\n& \\quad+\\left|x_{n-1}-x_{n}\\right| .\n\\end{aligned}\n\\end{aligned}\n$$\nLet $S(n)$ denote the largest possible value of the sum (*). Find $S(n)$.",
    "options": [],
    "answer": "floor(n^2/4)",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 27686,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nA bus route consists of a circular road of circumference $10$ miles and a straight road of length $1$ mile which runs from a terminus to the point $Q$ on the circular road (see diagram). It is served by two buses, each of which requires $20$ minutes for the round trip. Bus No. $1$, upon leaving the terminus, travels along the straight road, once around the circle clockwise and returns along the straight road to the terminus. Bus No. $2$, reaching the terminus $10$ minutes after Bus No. $1$, has a similar route except that it proceeds counterclockwise\n\n![](attached_image_1.png)\n\nTerminus\n\naround the circle. Both buses run continuously and do not wait at any point on the route except for a negligible amount of time to pick up and discharge passengers.\n\nA man plans to wait at a point $P$ which is $x$ miles $(0 \\leq x < 12)$ from the terminus along the route of Bus No. $1$ and travel to the terminus on one of the buses.\n\nAssuming that he chooses to board that bus which will bring him to his destination at the earliest moment, there is a maximum time $w(x)$ that his journey (waiting plus travel time) could take.\n\nFind $w(2)$; find $w(4)$.\n\nFor what value of $x$ will the time $w(x)$ be the longest?\n\nSketch a graph of $y = w(x)$ for $0 \\leq x < 12$.",
    "options": [],
    "answer": "Let v = 12 miles per 20 minutes. For 0 ≤ x < 1 and 11 < x < 12, w(x) = 10 + (5/3)·min{x, 12 − x}. For 1 ≤ x ≤ 11, w(x) = max(30 − (5/3)x, 10 + (5/3)x). In particular: w(2) = 80/3 minutes; w(4) = 70/3 minutes. The maximum of w(x) occurs at x = 1 and x = 11, with value 85/3 minutes. Graph sketch: piecewise linear—on [0,1): y = 10 + (5/3)x; jump up at x = 1 to y = 85/3; then decreasing line y = 30 − (5/3)x on [1,6], increasing line y = 10 + (5/3)x on [6,11], jump down at x = 11 to y = 35/3, then decreasing y = 30 − (5/3)x on (11,12), with y approaching 10 at x → 0+ or 12−.",
    "solution": "",
    "topic": "Discrete Mathematics",
    "subtopic": "Combinatorics"
  },
  {
    "id": 27687,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\ni) If $x = \\left(1 + \\frac{1}{n}\\right)^n$ and $y = \\left(1 + \\frac{1}{n}\\right)^{n+1}$, show that $y^x = x^y$.\n\nii) Show that, for all positive integers $n$,\n$$\n1^2 - 2^2 + 3^2 - 4^2 + \\cdots + (-1)^n (n-1)^2 + (-1)^{n+1} n^2 = (-1)^{n+1} (1 + 2 + \\cdots + n)\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 27688,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nLet $ABCD$ be a rectangle with $BC = 3 AB$. Show that if $P, Q$ are the points on side $BC$ with $BP = PQ = QC$, then\n$$\n\\angle DBC + \\angle DPC = \\angle DQC.\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 27689,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nLet\n$$\nf(x) = a_{0} + a_{1} x + a_{2} x^{2} + \\cdots + a_{n} x^{n}\n$$\nbe a polynomial with coefficients satisfying the conditions:\n$$\n0 \\leq a_{i} \\leq a_{0}, \\quad i = 1, 2, \\ldots, n\n$$\nLet $b_{0}, b_{1}, \\ldots, b_{2n}$ be the coefficients of the polynomial\n$$\n\\begin{aligned}\n(f(x))^{2} &= \\left(a_{0} + a_{1} x + a_{2} x^{2} + \\cdots + a_{n} x^{n}\\right)^{2} \\\\\n&= b_{0} + b_{1} x + b_{2} x^{2} + \\cdots + b_{n+1} x^{n+1} + \\cdots + b_{2n} x^{2n}\n\\end{aligned}\n$$\nProve that\n$$\nb_{n+1} \\leq \\frac{1}{2} (f(1))^{2}\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 27690,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nGiven a circle with diameter $AB$ and a point $X$ on the circle different from $A$ and $B$, let $t_{a}$, $t_{b}$ and $t_{x}$ be the tangents to the circle at $A$, $B$ and $X$ respectively. Let $Z$ be the point where line $AX$ meets $t_{b}$ and $Y$ the point where line $BX$ meets $t_{a}$. Show that the three lines $YZ$, $t_{x}$ and $AB$ are either concurrent (i.e., all pass through the same point) or parallel.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 27691,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nSimplify\n$$\n\\left(\\frac{1 \\cdot 2 \\cdot 4 + 2 \\cdot 4 \\cdot 8 + \\cdots + n \\cdot 2n \\cdot 4n}{1 \\cdot 3 \\cdot 9 + 2 \\cdot 6 \\cdot 18 + \\cdots + n \\cdot 3n \\cdot 9n}\\right)^{1/3}\n$$",
    "options": [],
    "answer": "2/3",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 27692,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nA sequence of numbers $a_{1}, a_{2}, a_{3}, \\ldots$ satisfies\n(i) $a_{1}=\\frac{1}{2}$,\n(ii) $a_{1}+a_{2}+\\cdots+a_{n}=n^{2} a_{n}$ \\quad $(n \\geq 1)$.\nDetermine the value of $a_{n}$ \\quad $(n \\geq 1)$.",
    "options": [],
    "answer": "1/(n(n+1))",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 27693,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nFor each real number $r$, $[r]$ denotes the largest integer less than or equal to $r$, e.g., $[6] = 6$, $[\\pi] = 3$, $[-1.5] = -2$. Indicate on the $(x, y)$-plane the set of all points $(x, y)$ for which $[x]^2 + [y]^2 = 4$.",
    "options": [],
    "answer": "{(x,y): x in [2,3), y in [0,1)} ∪ {(x,y): x in [-2,-1), y in [0,1)} ∪ {(x,y): x in [0,1), y in [2,3)} ∪ {(x,y): x in [0,1), y in [-2,-1)}",
    "solution": "",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 27694,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nFor a positive number such as $3.27$, $3$ is referred to as the integral part of the number and $.27$ as the decimal part. Find a positive number such that its decimal part, its integral part, and the number itself form a geometric progression.",
    "options": [],
    "answer": "(1 + sqrt(5))/2",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 27695,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\n$A$, $B$, $C$, $D$ are four \"consecutive\" points on the circumference of a circle and $P$, $Q$, $R$, $S$ are points on the circumference which are respectively the midpoints of the $\\operatorname{arcs}\\ AB$, $BC$, $CD$, $DA$. Prove that $PR$ is perpendicular to $QS$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 27696,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\ni. $15$ chairs are equally placed around a circular table on which are name cards for $15$ guests. The guests fail to notice these cards until after they have sat down, and it turns out that no one is sitting in the correct seat. Prove that the table can be rotated so that at least two of the guests are simultaneously correctly seated.\n\nii. Give an example of an arrangement in which just one of the $15$ guests is correctly seated and for which no rotation correctly places more than one person.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 27697,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nA function $f(x)$ is periodic if there is a positive number $p$ such that $f(x+p)=f(x)$ for all $x$. For example, $\\sin x$ is periodic with period $2 \\pi$. Is the function $\\sin \\left(x^{2}\\right)$ periodic? Prove your assertion.",
    "options": [],
    "answer": "It is not periodic.",
    "solution": "Solution:\nSuppose $\\sin(x^2)$ is periodic with period $p > 0$. Then for all $x$, we have\n$$\n\\sin((x+p)^2) = \\sin(x^2).\n$$\nThis means\n$$\n(x+p)^2 = x^2 + 2xp + p^2,\n$$\nso\n$$\n\\sin(x^2 + 2xp + p^2) = \\sin(x^2).\n$$\nTherefore, for all $x$,\n$$\n\\sin(x^2 + 2xp + p^2) - \\sin(x^2) = 0.\n$$\nBut $\\sin(a) = \\sin(b)$ if and only if $a - b = 2\\pi k$ or $a + b = 2\\pi k$ for some integer $k$.\n\nConsider the first possibility:\n$$\nx^2 + 2xp + p^2 - x^2 = 2xp + p^2 = 2\\pi k\n$$\nfor some integer $k$ depending on $x$. But for $x$ real, $2xp + p^2$ can take all real values as $x$ varies, so this is only possible if $p = 0$, which is not allowed (since $p > 0$).\n\nAlternatively, consider the second possibility:\n$$\nx^2 + 2xp + p^2 + x^2 = 2x^2 + 2xp + p^2 = 2\\pi k\n$$\nAgain, as $x$ varies, $2x^2 + 2xp + p^2$ takes all real values, so this is not possible for all $x$.\n\nTherefore, $\\sin(x^2)$ is not periodic.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 27698,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nLet $k$ be a positive integer. Find all polynomials\n$$\nP(x) = a_{0} + a_{1} x + \\cdots + a_{n} x^{n}\n$$\nwhere the $a_{i}$ are real, which satisfy the equation\n$$\nP(P(x)) = [P(x)]^{k}\n$$",
    "options": [],
    "answer": "All solutions are: (i) P(x) ≡ c where c is a real constant satisfying c = c^k (that is, for k = 1 any real c; for k ≥ 2, c = 0 or 1, and also c = −1 when k is odd), and (ii) the nonconstant polynomial P(x) = x^k.",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 27699,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nSuppose\n$$\nn(n+1) a_{n+1} = n(n-1) a_n - (n-2) a_{n-1}\n$$\nfor every positive integer $n \\geq 1$.\nGiven that $a_0 = 1$, $a_1 = 2$, find\n$$\n\\frac{a_0}{a_1} + \\frac{a_1}{a_2} + \\frac{a_2}{a_3} + \\cdots + \\frac{a_{50}}{a_{51}}.\n$$",
    "options": [],
    "answer": "2655/2",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 27700,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nTwo grade seven students were allowed to enter a chess tournament otherwise composed of grade eight students. Each contestant played once with each other contestant and received one point for a win, one half point for a tie and zero for a loss. The two grade seven students together gained a total of eight points and each grade eight student scored the same number of points as his classmates. How many students from grade eight participated in the chess tournament? Is the solution unique?",
    "options": [],
    "answer": "7 or 14; the solution is not unique",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 27701,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nEach of the 36 line segments joining 9 distinct points on a circle is coloured either red or blue. Suppose that each triangle determined by 3 of the 9 points contains at least one red side. Prove that there are four points such that the 6 segments connecting them are all red.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 27702,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nGiven four weights in geometric progression and an equal arm balance, show how to find the heaviest weight using the balance only twice.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 27703,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nLet $AB$ be a diameter of a circle, $C$ be any fixed point between $A$ and $B$ on this diameter, and $Q$ be a variable point on the circumference of the circle. Let $P$ be the point on the line determined by $Q$ and $C$ for which $\\frac{AC}{CB} = \\frac{QC}{CP}$. Describe, with proof, the locus of the point $P$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 27704,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nIf $A, B, C, D$ are four points in space, such that\n$$\n\\angle A B C=\\angle B C D=\\angle C D A=\\angle D A B=\\pi / 2\n$$\nprove that $A, B, C, D$ lie in a plane.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 27705,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nLet $P(x, y)$ be a polynomial in two variables $x, y$ such that $P(x, y) = P(y, x)$ for every $x, y$ (for example, the polynomial $x^{2} - 2 x y + y^{2}$ satisfies this condition). Given that $(x - y)$ is a factor of $P(x, y)$, show that $(x - y)^{2}$ is a factor of $P(x, y)$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 27706,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nLet $O$ be the centre of a circle and $A$ a fixed interior point of the circle different from $O$. Determine all points $P$ on the circumference of the circle such that the angle $OPA$ is a maximum.\n\n![](attached_image_1.png)",
    "options": [],
    "answer": "The two points where the triangle with vertices at the center, the interior point, and the point on the circle is right-angled at the interior point; equivalently, the intersections of the circle with the line through the interior point perpendicular to the line from the center to that point.",
    "solution": "",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 27707,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nLet\n$$\np(x) = a_{n} x^{n} + a_{n-1} x^{n-1} + \\cdots + a_{1} x + a_{0}\n$$\nand\n$$\nq(x) = b_{m} x^{m} + b_{m-1} x^{m-1} + \\cdots + b_{1} x + b_{0}\n$$\nbe two polynomials with integer coefficients. Suppose that all the coefficients of the product $p(x) \\cdot q(x)$ are even but not all of them are divisible by $4$. Show that one of $p(x)$ and $q(x)$ has all even coefficients and the other has at least one odd coefficient.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 27708,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nIf $f(x) = x^{2} + x$, prove that the equation $4 f(a) = f(b)$ has no solutions in positive integers $a$ and $b$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\nLet $f(x) = x^2 + x$. The equation is $4 f(a) = f(b)$, i.e.,\n\n$$\n4(a^2 + a) = b^2 + b.\n$$\n\nExpanding:\n$$\n4a^2 + 4a = b^2 + b\n$$\nBring all terms to one side:\n$$\n4a^2 + 4a - b^2 - b = 0\n$$\n$$\n4a^2 + 4a = b^2 + b\n$$\n\nLet us try to solve for $b$ in terms of $a$:\n\n$$\nb^2 + b - 4a^2 - 4a = 0\n$$\nThis is a quadratic in $b$:\n$$\nb^2 + b - (4a^2 + 4a) = 0\n$$\nThe discriminant must be a perfect square for integer solutions:\n\nThe quadratic formula gives:\n$$\nb = \\frac{-1 \\pm \\sqrt{1 + 16a^2 + 16a}}{2}\n$$\nSo $1 + 16a^2 + 16a$ must be a perfect square.\nLet $1 + 16a^2 + 16a = k^2$ for some integer $k$.\n\nSo:\n$$\nk^2 - 16a^2 - 16a = 1\n$$\n$$\nk^2 - 16(a^2 + a) = 1\n$$\n\nLet $a$ be a positive integer. Then $a^2 + a$ is always even, so $16(a^2 + a)$ is divisible by $32$.\n\nLet us try small values of $a$:\n\nIf $a = 1$: $1^2 + 1 = 2$, $16 \\times 2 = 32$, $k^2 = 1 + 32 = 33$ (not a perfect square).\n\nIf $a = 2$: $2^2 + 2 = 6$, $16 \\times 6 = 96$, $k^2 = 1 + 96 = 97$ (not a perfect square).\n\nIf $a = 3$: $3^2 + 3 = 12$, $16 \\times 12 = 192$, $k^2 = 1 + 192 = 193$ (not a perfect square).\n\nIf $a = 4$: $4^2 + 4 = 20$, $16 \\times 20 = 320$, $k^2 = 1 + 320 = 321$ (not a perfect square).\n\nIf $a = 5$: $5^2 + 5 = 30$, $16 \\times 30 = 480$, $k^2 = 1 + 480 = 481$ (not a perfect square).\n\nIn general, $k^2 = 16a^2 + 16a + 1 = (4a + 2)^2 - 3$.\n\nSo $k^2 + 3 = (4a + 2)^2$.\n\nThus, $(4a + 2)^2 - k^2 = 3$.\n\nSo $(4a + 2 - k)(4a + 2 + k) = 3$.\n\nThe only integer factorizations of $3$ are $(1, 3)$ and $(-1, -3)$.\n\nCase 1: $4a + 2 - k = 1$, $4a + 2 + k = 3$\n\nAdd: $2(4a + 2) = 4$, so $4a + 2 = 2$, so $a = 0$ (not positive integer).\n\nCase 2: $4a + 2 - k = -1$, $4a + 2 + k = -3$\n\nAdd: $2(4a + 2) = -4$, so $4a + 2 = -2$, so $a = -1$ (not positive integer).\n\nCase 3: $4a + 2 - k = 3$, $4a + 2 + k = 1$\n\nAdd: $2(4a + 2) = 4$, so $4a + 2 = 2$, $a = 0$ (not positive integer).\n\nCase 4: $4a + 2 - k = -3$, $4a + 2 + k = -1$\n\nAdd: $2(4a + 2) = -4$, $4a + 2 = -2$, $a = -1$ (not positive integer).\n\nTherefore, there are no positive integer solutions for $a$.\n\nTherefore, the equation $4 f(a) = f(b)$ has no solutions in positive integers $a$ and $b$.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 27709,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\n$N$ is an integer whose representation in base $b$ is $777$. Find the smallest positive integer $b$ for which $N$ is the fourth power of an integer.",
    "options": [],
    "answer": "18",
    "solution": "Solution:\n\nThe number $N$ written as $777$ in base $b$ means:\n$$\nN = 7b^2 + 7b + 7 = 7(b^2 + b + 1)\n$$\nWe want $N$ to be a perfect fourth power, i.e., $N = k^4$ for some integer $k$.\nSo,\n$$\n7(b^2 + b + 1) = k^4\n$$\nSince $7$ is prime, $7$ must divide $k^4$, so $k$ is divisible by $7$.\nLet $k = 7m$, then $k^4 = 7^4 m^4 = 2401 m^4$.\nSo,\n$$\n7(b^2 + b + 1) = 2401 m^4 \\implies b^2 + b + 1 = 343 m^4\n$$\nNow, we want the smallest positive integer $b$ such that $b^2 + b + 1 = 343 m^4$ for some integer $m$.\n\nLet us try small values of $m$:\n\nFor $m = 1$:\n$$\nb^2 + b + 1 = 343\n\\implies b^2 + b - 342 = 0\n$$\nThe discriminant is $1 + 4 \\times 342 = 1 + 1368 = 1369 = 37^2$, so $b = \\frac{-1 \\pm 37}{2}$.\nTaking the positive root:\n$$\nb = \\frac{-1 + 37}{2} = 18\n$$\nSo $b = 18$.\n\nCheck:\n$$\nN = 7 \\times (18^2 + 18 + 1) = 7 \\times (324 + 18 + 1) = 7 \\times 343 = 2401\n$$\n$2401 = 7^4 = (7)^4$\n\nTherefore, the smallest positive integer $b$ is $18$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 27710,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nA right circular cone of base radius $1$ cm and slant height $3$ cm is given. $P$ is a point on the circumference of the base and the shortest path from $P$ around the cone and back to $P$ is drawn (see diagram). What is the minimum distance from the vertex $V$ to this path?\n\n![](attached_image_1.png)",
    "options": [],
    "answer": "1.5 cm",
    "solution": "",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 27711,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nA rectangular city is exactly $m$ blocks long and $n$ blocks wide (see diagram). A woman lives in the southwest corner of the city and works in the northeast corner. She walks to work each day but, on any given trip, she makes sure that her path does not include any intersection twice. Show that the number $f(m, n)$ of different paths she can take to work satisfies $f(m, n) \\leq 2^{m n}$.\n\n![](attached_image_1.png)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 27712,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nLet $n$ be an integer. If the tens digit of $n^2$ is $7$, what is the units digit of $n^2$?",
    "options": [],
    "answer": "6",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 27713,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nFind all pairs $a, b$ of positive integers satisfying the equation $2 a^{2} = 3 b^{3}$.",
    "options": [],
    "answer": "All solutions are a = 18 t^3 and b = 6 t^2 for any positive integer t.",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 27714,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nDetermine the largest real number $z$ such that\n$$\n\\begin{gathered}\nx + y + z = 5 \\\\\nx y + y z + x z = 3\n\\end{gathered}\n$$\nand $x$, $y$ are also real.",
    "options": [],
    "answer": "13/3",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 27715,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nThe sides $AD$ and $BC$ of a convex quadrilateral $ABCD$ are extended to meet at $E$. Let $H$ and $G$ be the midpoints of $BD$ and $AC$, respectively. Find the ratio of the area of the triangle $EHG$ to that of the quadrilateral $ABCD$.",
    "options": [],
    "answer": "1/4",
    "solution": "",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 27716,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nEve and Odette play a game on a $3 \\times 3$ checkerboard, with black checkers and white checkers. The rules are as follows:\nI. They play alternately.\nII. A turn consists of placing one checker on an unoccupied square of the board.\nIII. In her turn, a player may select either a white checker or a black checker and need not always use the same colour.\nIV. When the board is full, Eve obtains one point for every row, column or diagonal that has an even number of black checkers, and Odette obtains one point for every row, column or diagonal that has an odd number of black checkers.\nV. The player obtaining at least five of the eight points WINS.\n\na. Is a $4$-$4$ tie possible? Explain.\n\nb. Describe a winning strategy for the girl who is first to play.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 27717,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSketch the graph of $x^{3} + x y + y^{3} = 3$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Discrete Mathematics",
    "subtopic": "Other"
  },
  {
    "id": 27718,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nGiven: (i) $a, b > 0$; (ii) $a, A_{1}, A_{2}, b$ is an arithmetic progression; (iii) $a, G_{1}, G_{2}, b$ is a geometric progression. Show that\n$$\nA_{1} A_{2} \\geq G_{1} G_{2}\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 27719,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nIt is known in Euclidean geometry that the sum of the angles of a triangle is constant. Prove, however, that the sum of the dihedral angles of a tetrahedron is not constant.\n\nNote. (i) A tetrahedron is a triangular pyramid, and (ii) a dihedral angle is the interior angle between a pair of faces.\n\n![](attached_image_1.png)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 27720,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nLet $a, b, c, d, e$ be integers such that $1 \\leq a < b < c < d < e$. Prove that\n$$\n\\frac{1}{[a, b]} + \\frac{1}{[b, c]} + \\frac{1}{[c, d]} + \\frac{1}{[d, e]} \\leq \\frac{15}{16}\n$$\nwhere $[m, n]$ denotes the least common multiple of $m$ and $n$ (e.g. $[4,6]=12$).",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 27721,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nA dog standing at the centre of a circular arena sees a rabbit at the wall. The rabbit runs around the wall and the dog pursues it along a unique path which is determined by running at the same speed and staying on the radial line joining the centre of the arena to the rabbit. Show that the dog overtakes the rabbit just as it reaches a point one-quarter of the way around the arena.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 27722,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nA walk consists of a sequence of steps of length $1$ taken in directions north, south, east or west. A walk is self-avoiding if it never passes through the same point twice. Let $f(n)$ denote the number of $n$-step self-avoiding walks which begin at the origin. Compute $f(1)$, $f(2)$, $f(3)$, $f(4)$, and show that\n$$\n2^{n} < f(n) \\leq 4 \\cdot 3^{n-1}\n$$",
    "options": [],
    "answer": "f(1) = 4, f(2) = 12, f(3) = 36, f(4) = 100; and for all n, 2^n < f(n) ≤ 4·3^{n−1}.",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 27723,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nA gambling student tosses a fair coin and scores one point for each head that turns up and two points for each tail. Prove that the probability of the student scoring exactly $n$ points is\n$$\n\\frac{1}{3}\\left[2+\\left(-\\frac{1}{2}\\right)^{n}\\right].\n$$",
    "options": [],
    "answer": "(1/3)[2 + (-1/2)^n]",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 27724,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nThe numbers from $1$ to $50$ are printed on cards. The cards are shuffled and then laid out face up in $5$ rows of $10$ cards each. The cards in each row are rearranged to make them increase from left to right. The cards in each column are then rearranged to make them increase from top to bottom. In the final arrangement, do the cards in the rows still increase from left to right?",
    "options": [],
    "answer": "Yes",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 27725,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nIf $a679b$ is a five digit number (in base 10) which is divisible by $72$, determine $a$ and $b$.",
    "options": [],
    "answer": "a=3, b=2",
    "solution": "",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 27726,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nAmong all triangles having (i) a fixed angle $A$ and (ii) an inscribed circle of fixed radius $r$, determine which triangle has the least perimeter.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 27727,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nA parallelepiped has the property that all cross sections which are parallel to any fixed face $F$, have the same perimeter as $F$. Determine whether or not any other polyhedron has this property.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 27728,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\n$P(x)$ and $Q(x)$ are two polynomials that satisfy the identity $P(Q(x)) \\equiv Q(P(x))$ for real numbers $x$. If the equation $P(x)=Q(x)$ has no real solution, show that the equation $P(P(x))=Q(Q(x))$ also has no real solution.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 27729,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nFor any real number $t$, denote by $[t]$ the greatest integer which is less than or equal to $t$. For example: $[8]=8$, $[\\pi]=3$ and $[-5 / 2]=-3$. Show that the equation\n$$\n[x]+[2x]+[4x]+[8x]+[16x]+[32x]=12345\n$$\nhas no real solution.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\nLet $x$ be a real number. For any integer $n \\geq 0$, $[2^n x]$ is the greatest integer less than or equal to $2^n x$.\n\nLet $S(x) = [x] + [2x] + [4x] + [8x] + [16x] + [32x]$.\n\nLet $x = a + y$, where $a$ is an integer and $0 \\leq y < 1$.\nThen $[2^k x] = [2^k a + 2^k y] = 2^k a + [2^k y]$ for $k = 0,1,2,3,4,5$.\n\nSo,\n$$\nS(x) = \\sum_{k=0}^5 [2^k x] = \\sum_{k=0}^5 (2^k a + [2^k y]) = (\\sum_{k=0}^5 2^k)a + \\sum_{k=0}^5 [2^k y]\n$$\nBut $\\sum_{k=0}^5 2^k = 1 + 2 + 4 + 8 + 16 + 32 = 63$.\nSo,\n$$\nS(x) = 63a + \\sum_{k=0}^5 [2^k y]\n$$\n\nNow, $0 \\leq y < 1$, so $0 \\leq 2^k y < 2^k$.\nTherefore, $[2^k y]$ is an integer between $0$ and $2^k - 1$.\n\nThus, $\\sum_{k=0}^5 [2^k y]$ is an integer between $0$ and $1 + 3 + 7 + 15 + 31 = 57$ (since $[2^k y] \\leq 2^k - 1$), but actually, for $k=0$ to $5$, $[2^k y]$ runs from $0$ to $2^k - 1$, so the sum runs from $0$ to $1 + 3 + 7 + 15 + 31 = 57$ for $k=1$ to $5$, plus $[y]$ which is $0$ (since $0 \\leq y < 1$), so the sum is from $0$ to $62$.\n\nBut let's check:\nFor $k=0$, $[2^0 y] = [y] = 0$ (since $0 \\leq y < 1$).\nFor $k=1$, $[2y]$ runs from $0$ to $1$.\nFor $k=2$, $[4y]$ runs from $0$ to $3$.\nFor $k=3$, $[8y]$ runs from $0$ to $7$.\nFor $k=4$, $[16y]$ runs from $0$ to $15$.\nFor $k=5$, $[32y]$ runs from $0$ to $31$.\nSo the sum $\\sum_{k=0}^5 [2^k y]$ runs from $0$ to $0 + 1 + 3 + 7 + 15 + 31 = 57$.\n\nTherefore, $S(x) = 63a + m$, where $m$ is an integer between $0$ and $57$.\n\nThus, $S(x)$ can only take values congruent to $m$ modulo $63$, i.e., $S(x) \\equiv m \\pmod{63}$, where $0 \\leq m \\leq 57$.\n\nBut for a fixed $a$, $S(x)$ runs through $63a, 63a+1, \\ldots, 63a+57$ as $y$ varies from $0$ to $1$.\n\nTherefore, $S(x)$ can only take values congruent to $0,1,2,\\ldots,57$ modulo $63$.\n\nNow, $12345 = 63 \\times 196 + 57$.\nSo $12345 \\equiv 57 \\pmod{63}$.\n\nTherefore, for $a = 196$, $S(x)$ could be $63 \\times 196 + m$ for $0 \\leq m \\leq 57$.\nBut $12345 = 63 \\times 196 + 57$, so $m = 57$.\n\nBut let's check if $m = 57$ is possible.\n\nBut $m = \\sum_{k=0}^5 [2^k y] = 0 + 1 + 3 + 7 + 15 + 31 = 57$.\nIs it possible for $[2y]=1$, $[4y]=3$, $[8y]=7$, $[16y]=15$, $[32y]=31$ and $[y]=0$?\n\nLet's see:\n$[2y]=1$ implies $1 \\leq 2y < 2$, so $0.5 \\leq y < 1$.\n$[4y]=3$ implies $3 \\leq 4y < 4$, so $0.75 \\leq y < 1$.\n$[8y]=7$ implies $7 \\leq 8y < 8$, so $0.875 \\leq y < 1$.\n$[16y]=15$ implies $15 \\leq 16y < 16$, so $0.9375 \\leq y < 1$.\n$[32y]=31$ implies $31 \\leq 32y < 32$, so $0.96875 \\leq y < 1$.\n\nSo, $y$ must satisfy all these inequalities:\n$0.96875 \\leq y < 1$.\n\nBut $[y]=0$ requires $0 \\leq y < 1$ (which is always true for $y$ in $[0,1)$).\n\nTherefore, such $y$ exists, for example, $y = 0.97$.\n\nBut let's check if $S(x)$ can actually be $12345$ for $x = a + y = 196 + y$ with $y$ in $[0.96875, 1)$.\n\nBut recall that $[x] = a$ (since $0 \\leq y < 1$), $[2x] = 2a + [2y]$, $[4x] = 4a + [4y]$, etc.\nSo,\n$$\nS(x) = 63a + [y] + [2y] + [4y] + [8y] + [16y] + [32y]\n$$\nFor $a = 196$, $S(x) = 12348 + m$.\nBut $12345 = 12348 + m$ implies $m = -3$.\nBut $m$ cannot be negative, as $m = [y] + [2y] + [4y] + [8y] + [16y] + [32y] \\geq 0$.\n\nWait, but above we had $12345 = 63 \\times 196 + 57 = 12348 + 57 = 12405$.\nBut $63 \\times 196 = 12348$, $12348 + 57 = 12405$.\nBut $12345 = 63 \\times 195 + 60 = 12285 + 60 = 12345$.\nSo $12345 = 63 \\times 195 + 60$.\nSo $m = 60$.\nBut $m$ can only be $0$ to $57$.\nTherefore, $S(x)$ can only take values $63a + m$ for $0 \\leq m \\leq 57$.\n\nTherefore, $12345$ cannot be written as $63a + m$ with $0 \\leq m \\leq 57$.\n\nTherefore, the equation has no real solution.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 27730,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nGiven a circle of radius $r$ and a tangent line $\\ell$ to the circle through a given point $P$ on the circle. From a variable point $R$ on the circle, a perpendicular $RQ$ is drawn to $\\ell$ with $Q$ on $\\ell$. Determine the maximum of the area of triangle $PQR$.",
    "options": [],
    "answer": "3√3 r^2 / 8",
    "solution": "",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 27731,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nGiven a finite collection of lines in a plane $P$, show that it is possible to draw an arbitrarily large circle in $P$ which does not meet any of them. On the other hand, show that it is possible to arrange an infinite sequence of lines (first line, second line, third line, etc.) in $P$ so that every circle in $P$ meets at least one of the lines. (A point is not considered to be a circle.)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 27732,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nEleven theatrical groups participated in a festival. Each day, some of the groups were scheduled to perform while the remaining groups joined the general audience. At the conclusion of the festival, each group had seen, during its days off, at least one performance of every other group. At least how many days did the festival last?",
    "options": [],
    "answer": "6",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 27733,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nIn the diagram, $OB_{i}$ is parallel and equal in length to $A_{i}A_{i+1}$ for $i=1,2,3$ and $4$ ($A_{5}=A_{1}$). Show that the area of $B_{1}B_{2}B_{3}B_{4}$ is twice that of $A_{1}A_{2}A_{3}A_{4}$.\n\n![](attached_image_1.png)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nLet $O$ be the origin. Let $A_{i}$ have position vector $\\vec{a}_{i}$ for $i=1,2,3,4$.\n\nSince $OB_{i}$ is parallel and equal in length to $A_{i}A_{i+1}$, we have:\n\n$$\n\\vec{OB}_{i} = \\vec{b}_{i} = \\vec{a}_{i+1} - \\vec{a}_{i}\n$$\n\nSo $B_{i}$ has position vector $\\vec{b}_{i} = \\vec{a}_{i+1} - \\vec{a}_{i}$.\n\nThe area of quadrilateral $A_{1}A_{2}A_{3}A_{4}$ is:\n\n$$\nS_{A} = \\frac{1}{2} | (\\vec{a}_{2} - \\vec{a}_{1}) \\times (\\vec{a}_{3} - \\vec{a}_{1}) + (\\vec{a}_{4} - \\vec{a}_{1}) \\times (\\vec{a}_{3} - \\vec{a}_{1}) |\n$$\n\nBut for a quadrilateral with consecutive vertices $P_{1}, P_{2}, P_{3}, P_{4}$, the area is:\n\n$$\nS = \\frac{1}{2} | (\\vec{P}_{1} \\times \\vec{P}_{2}) + (\\vec{P}_{2} \\times \\vec{P}_{3}) + (\\vec{P}_{3} \\times \\vec{P}_{4}) + (\\vec{P}_{4} \\times \\vec{P}_{1}) |\n$$\n\nApply this to $A_{1}A_{2}A_{3}A_{4}$:\n\n$$\nS_{A} = \\frac{1}{2} | \\vec{a}_{1} \\times \\vec{a}_{2} + \\vec{a}_{2} \\times \\vec{a}_{3} + \\vec{a}_{3} \\times \\vec{a}_{4} + \\vec{a}_{4} \\times \\vec{a}_{1} |\n$$\n\nSimilarly, for $B_{1}B_{2}B_{3}B_{4}$:\n\n$$\nS_{B} = \\frac{1}{2} | \\vec{b}_{1} \\times \\vec{b}_{2} + \\vec{b}_{2} \\times \\vec{b}_{3} + \\vec{b}_{3} \\times \\vec{b}_{4} + \\vec{b}_{4} \\times \\vec{b}_{1} |\n$$\n\nRecall $\\vec{b}_{i} = \\vec{a}_{i+1} - \\vec{a}_{i}$ for $i=1,2,3,4$ (with $A_{5} = A_{1}$).\n\nCompute $\\vec{b}_{1} \\times \\vec{b}_{2}$:\n\n$$\n\\vec{b}_{1} \\times \\vec{b}_{2} = (\\vec{a}_{2} - \\vec{a}_{1}) \\times (\\vec{a}_{3} - \\vec{a}_{2})\n$$\n\nExpand:\n\n$$\n= \\vec{a}_{2} \\times \\vec{a}_{3} - \\vec{a}_{2} \\times \\vec{a}_{2} - \\vec{a}_{1} \\times \\vec{a}_{3} + \\vec{a}_{1} \\times \\vec{a}_{2}\n$$\n\nBut $\\vec{a}_{2} \\times \\vec{a}_{2} = 0$, so:\n\n$$\n= \\vec{a}_{2} \\times \\vec{a}_{3} - \\vec{a}_{1} \\times \\vec{a}_{3} + \\vec{a}_{1} \\times \\vec{a}_{2}\n$$\n\nSimilarly, compute all four terms:\n\n1. $\\vec{b}_{1} \\times \\vec{b}_{2} = \\vec{a}_{2} \\times \\vec{a}_{3} - \\vec{a}_{1} \\times \\vec{a}_{3} + \\vec{a}_{1} \\times \\vec{a}_{2}$\n2. $\\vec{b}_{2} \\times \\vec{b}_{3} = (\\vec{a}_{3} - \\vec{a}_{2}) \\times (\\vec{a}_{4} - \\vec{a}_{3}) = \\vec{a}_{3} \\times \\vec{a}_{4} - \\vec{a}_{2} \\times \\vec{a}_{4} - \\vec{a}_{3} \\times \\vec{a}_{3} + \\vec{a}_{2} \\times \\vec{a}_{3} = \\vec{a}_{3} \\times \\vec{a}_{4} - \\vec{a}_{2} \\times \\vec{a}_{4} + \\vec{a}_{2} \\times \\vec{a}_{3}$\n3. $\\vec{b}_{3} \\times \\vec{b}_{4} = (\\vec{a}_{4} - \\vec{a}_{3}) \\times (\\vec{a}_{1} - \\vec{a}_{4}) = \\vec{a}_{4} \\times \\vec{a}_{1} - \\vec{a}_{3} \\times \\vec{a}_{1} - \\vec{a}_{4} \\times \\vec{a}_{4} + \\vec{a}_{3} \\times \\vec{a}_{4} = \\vec{a}_{4} \\times \\vec{a}_{1} - \\vec{a}_{3} \\times \\vec{a}_{1} + \\vec{a}_{3} \\times \\vec{a}_{4}$\n4. $\\vec{b}_{4} \\times \\vec{b}_{1} = (\\vec{a}_{1} - \\vec{a}_{4}) \\times (\\vec{a}_{2} - \\vec{a}_{1}) = \\vec{a}_{1} \\times \\vec{a}_{2} - \\vec{a}_{4} \\times \\vec{a}_{2} - \\vec{a}_{1} \\times \\vec{a}_{1} + \\vec{a}_{4} \\times \\vec{a}_{1} = \\vec{a}_{1} \\times \\vec{a}_{2} - \\vec{a}_{4} \\times \\vec{a}_{2} + \\vec{a}_{4} \\times \\vec{a}_{1}$\n\nAdd all four terms:\n\nSum:\n\n$$\n\\begin{align*}\n& [\\vec{a}_{2} \\times \\vec{a}_{3} - \\vec{a}_{1} \\times \\vec{a}_{3} + \\vec{a}_{1} \\times \\vec{a}_{2}] \\\\\n& + [\\vec{a}_{3} \\times \\vec{a}_{4} - \\vec{a}_{2} \\times \\vec{a}_{4} + \\vec{a}_{2} \\times \\vec{a}_{3}] \\\\\n& + [\\vec{a}_{4} \\times \\vec{a}_{1} - \\vec{a}_{3} \\times \\vec{a}_{1} + \\vec{a}_{3} \\times \\vec{a}_{4}] \\\\\n& + [\\vec{a}_{1} \\times \\vec{a}_{2} - \\vec{a}_{4} \\times \\vec{a}_{2} + \\vec{a}_{4} \\times \\vec{a}_{1}]\n\\end{align*}\n$$\n\nNow, group like terms:\n\n- $\\vec{a}_{2} \\times \\vec{a}_{3}$ appears twice\n- $\\vec{a}_{3} \\times \\vec{a}_{4}$ appears twice\n- $\\vec{a}_{4} \\times \\vec{a}_{1}$ appears twice\n- $\\vec{a}_{1} \\times \\vec{a}_{2}$ appears twice\n\nNegative terms:\n- $-\\vec{a}_{1} \\times \\vec{a}_{3}$\n- $-\\vec{a}_{2} \\times \\vec{a}_{4}$\n- $-\\vec{a}_{3} \\times \\vec{a}_{1}$\n- $-\\vec{a}_{4} \\times \\vec{a}_{2}$\n\nBut $\\vec{a}_{1} \\times \\vec{a}_{3} + \\vec{a}_{3} \\times \\vec{a}_{1} = 0$ (since $\\vec{u} \\times \\vec{v} = -\\vec{v} \\times \\vec{u}$), so $-\\vec{a}_{1} \\times \\vec{a}_{3} - \\vec{a}_{3} \\times \\vec{a}_{1} = 0$.\nSimilarly, $-\\vec{a}_{2} \\times \\vec{a}_{4} - \\vec{a}_{4} \\times \\vec{a}_{2} = 0$.\n\nSo the sum is:\n\n$$\n2[\\vec{a}_{1} \\times \\vec{a}_{2} + \\vec{a}_{2} \\times \\vec{a}_{3} + \\vec{a}_{3} \\times \\vec{a}_{4} + \\vec{a}_{4} \\times \\vec{a}_{1}]\n$$\n\nTherefore,\n\n$$\nS_{B} = \\frac{1}{2} |2[\\vec{a}_{1} \\times \\vec{a}_{2} + \\vec{a}_{2} \\times \\vec{a}_{3} + \\vec{a}_{3} \\times \\vec{a}_{4} + \\vec{a}_{4} \\times \\vec{a}_{1}]| = |\\vec{a}_{1} \\times \\vec{a}_{2} + \\vec{a}_{2} \\times \\vec{a}_{3} + \\vec{a}_{3} \\times \\vec{a}_{4} + \\vec{a}_{4} \\times \\vec{a}_{1}| = 2S_{A}\n$$\n\nThus, the area of $B_{1}B_{2}B_{3}B_{4}$ is twice that of $A_{1}A_{2}A_{3}A_{4}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 27734,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nLet $p$ be a permutation of the set $S_{n} = \\{1, 2, \\ldots, n\\}$. An element $j \\in S_{n}$ is called a fixed point of $p$ if $p(j) = j$. Let $f_{n}$ be the number of permutations having no fixed points, and $g_{n}$ be the number with exactly one fixed point. Show that $\\left|f_{n} - g_{n}\\right| = 1$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Discrete Mathematics",
    "subtopic": "Combinatorics"
  },
  {
    "id": 27735,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nIf $a$, $b$ and $c$ are the roots of the equation $x^{3} - x^{2} - x - 1 = 0$,\n\ni. show that $a$, $b$ and $c$ are distinct:\n\nii. show that\n$$\n\\frac{a^{1982} - b^{1982}}{a - b} + \\frac{b^{1982} - c^{1982}}{b - c} + \\frac{c^{1982} - a^{1982}}{c - a}\n$$\nis an integer.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 27736,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nLet $\\mathbb{R}^n$ be the $n$-dimensional Euclidean space. Determine the smallest number $g(n)$ of points of a set in $\\mathbb{R}^n$ such that every point in $\\mathbb{R}^n$ is at irrational distance from at least one point in that set.",
    "options": [],
    "answer": "g(n) = n + 1",
    "solution": "",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 27737,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nThe altitudes of a tetrahedron $ABCD$ are extended externally to points $A'$, $B'$, $C'$ and $D'$ respectively, where $AA' = k / h_a$, $BB' = k / h_b$, $CC' = k / h_c$ and $DD' = k / h_d$. Here, $k$ is a constant and $h_a$ denotes the length of the altitude of $ABCD$ from vertex $A$, etc. Prove that the centroid of the tetrahedron $A'B'C'D'$ coincides with the centroid of $ABCD$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 27738,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nThe area of a triangle is determined by the lengths of its sides. Is the volume of a tetrahedron determined by the areas of its faces?",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Geometry",
    "subtopic": "Solid Geometry"
  },
  {
    "id": 27739,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nFind all positive integers $w, x, y$ and $z$ which satisfy $w! = x! + y! + z!$.",
    "options": [],
    "answer": "(w, x, y, z) = (3, 2, 2, 2)",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 27740,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nFor each real number $r$ let $T_{r}$ be the transformation of the plane that takes the point $(x, y)$ into the point $(2^{r} x, r 2^{r} x + 2^{r} y)$. Let $F$ be the family of all such transformations i.e. $F = \\{ T_{r} : r \\text{ a real number} \\}$. Find all curves $y = f(x)$ whose graphs remain unchanged by every transformation in $F$.",
    "options": [],
    "answer": "All functions of the form f(0) = 0 and, for x ≠ 0,\n\nf(x) = x log_2(|x|) + c_+ x for x > 0,\nf(x) = x log_2(|x|) + c_- x for x < 0,\n\nwhere c_+ and c_- are arbitrary real constants.",
    "solution": "",
    "topic": "Discrete Mathematics",
    "subtopic": "Other"
  },
  {
    "id": 27741,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nProve that for every prime number $p$, there are infinitely many positive integers $n$ such that $p$ divides $2^{n}-n$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 27742,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nThe geometric mean (G.M.) of a $k$ positive numbers $a_{1}, a_{2}, \\ldots, a_{k}$ is defined to be the (positive) $k$-th root of their product. For example, the G.M. of $3,4,18$ is $6$. Show that the G.M. of a set $S$ of $n$ positive numbers is equal to the G.M. of the G.M.'s of all non-empty subsets of $S$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 27743,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nProve that the sum of the squares of $1984$ consecutive positive integers cannot be the square of an integer.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\nLet the $1984$ consecutive positive integers be $n, n+1, n+2, \\ldots, n+1983$.\n\nThe sum of their squares is:\n$$\nS = n^2 + (n+1)^2 + (n+2)^2 + \\cdots + (n+1983)^2\n$$\n\nThis can be written as:\n$$\nS = \\sum_{k=0}^{1983} (n+k)^2 = \\sum_{k=0}^{1983} (n^2 + 2nk + k^2) = 1984 n^2 + 2n \\sum_{k=0}^{1983} k + \\sum_{k=0}^{1983} k^2\n$$\n\nWe have:\n$$\n\\sum_{k=0}^{m} k = \\frac{m(m+1)}{2}\n$$\n$$\n\\sum_{k=0}^{m} k^2 = \\frac{m(m+1)(2m+1)}{6}\n$$\nFor $m = 1983$:\n$$\n\\sum_{k=0}^{1983} k = \\frac{1983 \\times 1984}{2}\n$$\n$$\n\\sum_{k=0}^{1983} k^2 = \\frac{1983 \\times 1984 \\times 3967}{6}\n$$\n\nSo,\n$$\nS = 1984 n^2 + 2n \\cdot \\frac{1983 \\times 1984}{2} + \\frac{1983 \\times 1984 \\times 3967}{6}\n$$\n$$\n= 1984 n^2 + 1983 \\times 1984 n + \\frac{1983 \\times 1984 \\times 3967}{6}\n$$\n\nFactor $1984$:\n$1984 = 2^6 \\times 31$.\n\nLet us factor $S$ as follows:\n$$\nS = 1984 (n^2 + 1983 n) + \\frac{1983 \\times 1984 \\times 3967}{6}\n$$\n\nLet us factor $1983$ and $3967$:\n$1983 = 3 \\times 661$\n$3967 = 2 \\times 1983 + 1$\n\nLet us write $S$ as:\n$$\nS = 1984 (n^2 + 1983 n + \\frac{1983 \\times 3967}{6})\n$$\n\nBut $n^2 + 1983 n + \\frac{1983 \\times 3967}{6}$ is not necessarily an integer, so let's combine all terms over $6$:\n\n$$\nS = 1984 \\left( n^2 + 1983 n + \\frac{1983 \\times 3967}{6} \\right )\n$$\n$$\n= 1984 \\left( \\frac{6 n^2 + 6 \\times 1983 n + 1983 \\times 3967}{6} \\right )\n$$\n$$\n= \\frac{1984}{6} \\left( 6 n^2 + 6 \\times 1983 n + 1983 \\times 3967 \\right )\n$$\n\nLet us compute $6 n^2 + 6 \\times 1983 n + 1983 \\times 3967$:\n\n$6 n^2 + 6 \\times 1983 n = 6 n^2 + 11898 n$\n\n$1983 \\times 3967 = 1983 \\times (2 \\times 1983 + 1) = 1983 \\times 2 \\times 1983 + 1983 = 2 \\times 1983^2 + 1983$\n\n$1983^2 = (2000 - 17)^2 = 4,000,000 - 68,000 + 289 = 3,932,289$\n\n$2 \\times 1983^2 = 7,864,578$\n\n$2 \\times 1983^2 + 1983 = 7,864,578 + 1983 = 7,866,561$\n\nSo the numerator is:\n$6 n^2 + 11898 n + 7,866,561$\n\nTherefore,\n$$\nS = \\frac{1984}{6} (6 n^2 + 11898 n + 7,866,561)\n$$\n\nNow, $1984/6 = 992/3$, so $S = \\frac{992}{3} (6 n^2 + 11898 n + 7,866,561)$\n\nFor $S$ to be an integer, $6 n^2 + 11898 n + 7,866,561$ must be divisible by $3$.\n\nBut the main point is to show that $S$ cannot be a perfect square for any integer $n$.\n\nLet us consider the sum of $k$ consecutive squares:\n\nIt is known that the sum of $k$ consecutive squares cannot be a perfect square unless $k = 1$ or $k = 24$ (see, for example, Mordell's result).\n\nBut $1984$ is not $1$ or $24$.\n\nAlternatively, consider the sum modulo $4$.\n\nThe square of any integer is congruent to $0$ or $1$ modulo $4$.\n\nLet us compute $S$ modulo $4$.\n\nSince $1984$ is divisible by $4$, the sum $S$ is divisible by $4$.\n\nBut more generally, we can use the following result:\n\nSuppose $k$ is even and $k \\equiv 2 \\pmod{4}$, then the sum of $k$ consecutive squares cannot be a perfect square.\n\nBut $1984 \\equiv 0 \\pmod{4}$, so this does not help.\n\nAlternatively, consider modulo $3$.\n\nThe squares modulo $3$ are $0$ and $1$.\n\nBut the key is that for $k > 1$, the sum of $k$ consecutive squares cannot be a perfect square except for $k = 24$ (see Dickson, History of the Theory of Numbers, Vol. II, p. 413).\n\nTherefore, the sum of the squares of $1984$ consecutive positive integers cannot be a perfect square.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 27744,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nAlice and Bob are in a hardware store. The store sells coloured sleeves that fit over keys to distinguish them. The following conversation takes place:\nAlice: Are you going to cover your keys?\nBob: I would like to, but there are only 7 colours and I have 8 keys.\nAlice: Yes, but you could always distinguish a key by noticing that the red key next to the green key was different from the red key next to the blue key.\nBob: You must be careful what you mean by \"next to\" or \"three keys over from\" since you can turn the key ring over and the keys are arranged in a circle.\nAlice: Even so, you don't need 8 colours.\nProblem: What is the smallest number of colours needed to distinguish $n$ keys if all the keys are to be covered.",
    "options": [],
    "answer": "1 color if there is 1 key; 2 colors if there are 2 keys or at least 6 keys; 3 colors if there are 3, 4, or 5 keys.",
    "solution": "",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 27745,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nAn integer is digitally divisible if\n(a) none of its digits is zero;\n(b) it is divisible by the sum of its digits (e.g., $322$ is digitally divisible).\nShow that there are infinitely many digitally divisible integers.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nLet us construct infinitely many digitally divisible integers.\n\nConsider the integer $111\\ldots 1$ consisting of $k$ digits, all of which are $1$ (for any $k \\geq 1$). None of its digits is zero, so condition (a) is satisfied.\n\nThe sum of its digits is $k$ (since there are $k$ digits, each $1$). The number itself is $N = \\underbrace{111\\ldots 1}_{k\\text{ times}} = \\frac{10^k - 1}{9}$.\n\nWe claim that $N$ is divisible by $k$ for infinitely many $k$.\n\nBut even if $k$ does not always divide $N$, we can construct other numbers. For example, consider the number $N = 1\\underbrace{00\\ldots 0}_{k-1\\text{ times}}1\\underbrace{00\\ldots 0}_{k-1\\text{ times}}1\\ldots 1$ (with $n$ blocks of $1$ separated by $k-1$ zeros), i.e., $N = 10^{(n-1)k} + 10^{(n-2)k} + \\cdots + 10^0$. The sum of its digits is $n$ (since there are $n$ ones and the rest are zeros). But this number contains zeros, so it does not satisfy condition (a).\n\nInstead, consider the following construction:\n\nLet $d$ be any positive integer not divisible by $10$ (so that its digits are all nonzero). Let $N = d \\times 10^m + d \\times 10^{2m} + \\cdots + d \\times 10^{km}$, where $m$ is chosen so that $d$ has $m$ digits and none of the digits of $d$ is zero. Then $N$ will have only the digits of $d$ repeated, so none of its digits is zero.\n\nAlternatively, consider the following:\n\nLet $n$ be any positive integer. Consider the number $N = 1\\underbrace{11\\ldots 1}_{n-1\\text{ times}}$ (i.e., $n$ digits, all $1$). The sum of its digits is $n$. $N = \\frac{10^n - 1}{9}$. For $n = 9$, $N = 111111111$, sum of digits is $9$, and $111111111$ is divisible by $9$.\n\nIn general, for $n = 9k$, $N = \\underbrace{111\\ldots 1}_{9k\\text{ times}}$ is divisible by $9$, and the sum of its digits is $9k$, so $N$ is divisible by the sum of its digits. Thus, for every $k \\geq 1$, $N = \\underbrace{111\\ldots 1}_{9k\\text{ times}}$ is digitally divisible.\n\nTherefore, there are infinitely many digitally divisible integers.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 27746,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nAn acute-angled triangle has unit area. Show that there is a point inside the triangle whose distance from each of the vertices is at least $\\frac{2}{\\sqrt[4]{27}}$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 27747,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nGiven any 7 real numbers, prove that there are two of them, say $x$ and $y$, such that\n$$\n0 \\leq \\frac{x-y}{1+xy} \\leq \\frac{1}{\\sqrt{3}}\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 27748,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nLet $P_{1}$ and $P_{2}$ be regular polygons of $1985$ sides and perimeters $x$ and $y$ respectively. Each side of $P_{1}$ is tangent to a given circle of circumference $c$ and this circle passes through each vertex of $P_{2}$. Prove $x + y \\geq 2c$. (You may assume that $\\tan \\theta \\geq \\theta$ for $0 \\leq \\theta < \\frac{\\pi}{2}$.)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 27749,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nThe lengths of the sides of a triangle are $6$, $8$ and $10$ units. Prove that there is exactly one straight line which simultaneously bisects the area and perimeter of the triangle.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 27750,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nProve or disprove that there exists an integer which is doubled when the initial digit is transferred to the end.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 27751,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nProve that $2^{n-1}$ divides $n!$ if and only if $n=2^{k-1}$ for some positive integer $k$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 27752,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nLet $1 < x_{1} < 2$ and, for $n = 1, 2, \\ldots$, define $x_{n+1} = 1 + x_{n} - \\frac{1}{2} x_{n}^{2}$. Prove that, for $n \\geq 3$, $\\left| x_{n} - \\sqrt{2} \\right| < 2^{-n}$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 27753,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nA Mathlon is a competition in which there are $M$ athletic events. Such a competition was held in which only $A$, $B$, and $C$ participated. In each event $p_{1}$ points were awarded for first place, $p_{2}$ for second and $p_{3}$ for third, where $p_{1} > p_{2} > p_{3} > 0$ and $p_{1}, p_{2}, p_{3}$ are integers. The final score for $A$ was $22$, for $B$ was $9$ and for $C$ was also $9$. $B$ won the $100$ metres. What is the value of $M$ and who was second in the high jump?",
    "options": [],
    "answer": "M = 5; C was second in the high jump.",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 27754,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nIn the diagram line segments $AB$ and $CD$ are of length $1$ while angles $ABC$ and $CBD$ are $90^{\\circ}$ and $30^{\\circ}$ respectively. Find $AC$.\n\n![](attached_image_1.png)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 27755,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nA chord $ST$ of constant length slides around a semicircle with diameter $AB$. $M$ is the mid-point of $ST$ and $P$ is the foot of the perpendicular from $S$ to $AB$. Prove that angle $SPM$ is constant for all positions of $ST$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 27756,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nFor positive integers $n$ and $k$, define $F(n, k) = \\sum_{r=1}^{n} r^{2k-1}$. Prove that $F(n, 1)$ divides $F(n, k)$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 27757,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nLet $u_{1}, u_{2}, u_{3}, \\ldots$ be a sequence of integers satisfying the recurrence relation $u_{n+2} = u_{n+1}^{2} - u_{n}$. Suppose $u_{1} = 39$ and $u_{2} = 45$. Prove that $1986$ divides infinitely many terms of the sequence.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 27758,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nFind all solutions of $a^{2} + b^{2} = n!$ for positive integers $a$, $b$, $n$ with $a \\leq b$ and $n < 14$.",
    "options": [],
    "answer": "All solutions with n < 14 are: (a, b, n) = (1, 1, 2) and (12, 24, 6).",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 27759,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nThe number $1987$ can be written as a three digit number $x y z$ in some base $b$. If $x + y + z = 1 + 9 + 8 + 7$, determine all possible values of $x, y, z, b$.",
    "options": [],
    "answer": "x=5, y=9, z=11, b=19",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 27760,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nSuppose $ABCD$ is a parallelogram and $E$ is a point between $B$ and $C$ on the line $BC$. If the triangles $DEC$, $BED$ and $BAD$ are isosceles what are the possible values for the angle $DAB$?",
    "options": [],
    "answer": "36°, 45°, 72°",
    "solution": "",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 27761,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nOn a large, flat field $n$ people are positioned so that for each person the distances to all the other people are different. Each person holds a water pistol and at a given signal fires and hits the person who is closest. When $n$ is odd show that there is at least one person left dry. Is this always true when $n$ is even?",
    "options": [],
    "answer": "When the number of people is odd, at least one person is left dry. When the number is even, this is not always true; there are configurations where every person is hit (for example, arranging the people in well-separated mutual nearest neighbor pairs).",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 27762,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nFor every positive integer $n$ show that\n$$\n[\\sqrt{n} + \\sqrt{n+1}] = [\\sqrt{4n+1}] = [\\sqrt{4n+2}] = [\\sqrt{4n+3}]\n$$\nwhere $[x]$ is the greatest integer less than or equal to $x$ (for example $[2.3]=2$, $[\\pi]=3$, $[5]=5$).",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 27763,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nFor what values of $b$ do the equations: $1988 x^{2} + b x + 8891 = 0$ and $8891 x^{2} + b x + 1988 = 0$ have a common root?",
    "options": [],
    "answer": "-10879 and 10879",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 27764,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nA house is in the shape of a triangle, perimeter $P$ metres and area $A$ square metres. The garden consists of all the land within 5 metres of the house. How much land do the garden and house together occupy?",
    "options": [],
    "answer": "A + 5P + 25π",
    "solution": "",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 27765,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSuppose that $S$ is a finite set of at least five points in the plane; some are coloured red, the others are coloured blue. No subset of three or more similarly coloured points is collinear. Show that there is a triangle\n(i) whose vertices are all the same colour, and\n(ii) at least one side of the triangle does not contain a point of the opposite colour.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 27766,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nLet $x_{n+1} = 4 x_n - x_{n-1}$, $x_0 = 0$, $x_1 = 1$, and $y_{n+1} = 4 y_n - y_{n-1}$, $y_0 = 1$, $y_1 = 2$.\nShow for all $n \\geq 0$ that $y_n^2 = 3 x_n^2 + 1$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 27767,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nLet $S=\\{a_{1}, a_{2}, \\ldots, a_{r}\\}$ denote a sequence of integers. For each non-empty subsequence $A$ of $S$, we define $p(A)$ to be the product of all the integers in $A$. Let $m(S)$ be the arithmetic average of $p(A)$ over all non-empty subsets $A$ of $S$. If $m(S)=13$ and if $m\\left(S \\cup\\{a_{r+1}\\}\\right)=49$ for some positive integer $a_{r+1}$, determine the values of $a_{1}, a_{2}, \\ldots, a_{r}$ and $a_{r+1}$.",
    "options": [],
    "answer": "r = 3; a1 = 1, a2 = 1, a3 = 22; a_{r+1} = 7",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 27768,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nLet $ABC$ be a right angled triangle of area $1$. Let $A'B'C'$ be the points obtained by reflecting $A$, $B$, $C$ respectively, in their opposite sides. Find the area of $\\triangle A'B'C'$.",
    "options": [],
    "answer": "3",
    "solution": "",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 27769,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nThe integers $1, 2, \\ldots, n$ are placed in order so that each value is either strictly bigger than all the preceding values or is strictly smaller than all preceding values. In how many ways can this be done?",
    "options": [],
    "answer": "2^{n-1}",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 27770,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nDefine $\\{a_n\\}_{n=1}$ as follows: $a_1 = 1989^{1989}$; $a_n$, $n > 1$, is the sum of the digits of $a_{n-1}$. What is the value of $a_5$?",
    "options": [],
    "answer": "9",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 27771,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nThere are $5$ monkeys and $5$ ladders and at the top of each ladder there is a banana. A number of ropes connect the ladders, each rope connects two ladders. No two ropes are attached to the same rung of the same ladder. Each monkey starts at the bottom of a different ladder. The monkeys climb up the ladders but each time they encounter a rope they climb along it to the ladder at the other end of the rope and then continue climbing upwards. Show that, no matter how many ropes there are, each monkey gets a banana.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 27772,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nGiven the numbers $1, 2, 2^{2}, \\ldots, 2^{n-1}$. For a specific permutation $\\sigma = X_{1}, X_{2}, \\ldots, X_{n}$ of these numbers we define $S_{1}(\\sigma) = X_{1}$, $S_{2}(\\sigma) = X_{1} + X_{2}$, $S_{3}(\\sigma) = X_{1} + X_{2} + X_{3}$, $\\ldots$ and $Q(\\sigma) = S_{1}(\\sigma) S_{2}(\\sigma) \\cdots S_{n}(\\sigma)$. Evaluate $\\sum 1 / Q(\\sigma)$ where the sum is taken over all possible permutations.",
    "options": [],
    "answer": "2^{-n(n-1)/2}",
    "solution": "",
    "topic": "Discrete Mathematics",
    "subtopic": "Combinatorics"
  },
  {
    "id": 27773,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nA competition involving $n \\geq 2$ players was held over $k$ days. On each day, the players received scores of $1,2,3, \\ldots, n$ points with no two players receiving the same score. At the end of the $k$ days, it was found that each player had exactly $26$ points in total. Determine all pairs $(n, k)$ for which this is possible.",
    "options": [],
    "answer": "(3, 13), (12, 4), (25, 2)",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 27774,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nA set of $\\frac{1}{2} n(n+1)$ distinct numbers is arranged at random in a triangular array:\n![](attached_image_1.png)\nLet $M_{k}$ be the largest number in the $k$-th row from the top. Find the probability that\n$$\nM_{1}<M_{2}<M_{3}<\\cdots<M_{n}\n$$",
    "options": [],
    "answer": "2^n / (n+1)!",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 27775,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nLet $ABCD$ be a convex quadrilateral inscribed in a circle, and let diagonals $AC$ and $BD$ meet at $X$. The perpendiculars from $X$ meet the sides $AB$, $BC$, $CD$, $DA$ at $A'$, $B'$, $C'$, $D'$ respectively. Prove that\n$$\n|A'B'| + |C'D'| = |A'D'| + |B'C'|\n$$\n($|A'B'|$ is the length of line segment $A'B'$, etc.)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 27776,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nA particle can travel at speeds up to $2$ metres per second along the $x$-axis, and up to $1$ metre per second elsewhere in the plane. Provide a labelled sketch of the region which can be reached within one second by the particle starting at the origin.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 27777,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nSuppose that a function $f$ defined on the positive integers satisfies\n$$\n\\begin{gathered}\nf(1)=1, \\quad f(2)=2 \\\\\nf(n+2)=f(n+2-f(n+1))+f(n+1-f(n)) \\quad(n \\geq 1)\n\\end{gathered}\n$$\n\na) Show that\n(i) $0 \\leq f(n+1)-f(n) \\leq 1$\n(ii) if $f(n)$ is odd, then $f(n+1)=f(n)+1$.\n\nb) Determine, with justification, all values of $n$ for which\n$$\nf(n)=2^{10}+1\n$$",
    "options": [],
    "answer": "2048",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 27778,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nLet $C$ be a circle and $P$ a given point in the plane. Each line through $P$ which intersects $C$ determines a chord of $C$. Show that the midpoints of these chords lie on a circle.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 27779,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nIn the figure, the side length of the large equilateral triangle is $3$ and $f(3)$, the number of parallelograms bounded by sides in the grid, is $15$. For the general analogous situation, find a formula for $f(n)$, the number of parallelograms, for a triangle of side length $n$.\n\n![](attached_image_1.png)",
    "options": [],
    "answer": "f(n) = 3 * C(n+2, 4) = n(n-1)(n+1)(n+2)/8",
    "solution": "",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 27780,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nShow that the equation $x^{2} + y^{5} = z^{3}$ has infinitely many solutions in integers $x$, $y$, $z$ for which $x y z \\neq 0$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 27781,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nLet $n$ be a fixed positive integer. Find the sum of all positive integers with the following property: In base $2$, it has exactly $2n$ digits consisting of $n$ $1$'s and $n$ $0$'s. (The first digit cannot be $0$.)",
    "options": [],
    "answer": "C(2n-1, n-1) * [ 2^(2n-1) + ((n-1)/(2n-1)) * (2^(2n-1) - 1 ) ]",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 27782,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nProve that the product of the first $n$ natural numbers is divisible by the sum of the first $n$ natural numbers if and only if $n+1$ is not an odd prime.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 27783,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nSolve the equation\n$$\nx^{2} + \\frac{x^{2}}{(x+1)^{2}} = 3\n$$",
    "options": [],
    "answer": "(1 + sqrt(5))/2, (1 - sqrt(5))/2",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 27784,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nA deck of $2n+1$ cards consists of a joker and, for each number between $1$ and $n$ inclusive, two cards marked with that number. The $2n+1$ cards are placed in a row, with the joker in the middle. For each $k$ with $1 \\leq k \\leq n$, the two cards numbered $k$ have exactly $k-1$ cards between them. Determine all the values of $n$ not exceeding $10$ for which this arrangement is possible. For which values of $n$ is it impossible?",
    "options": [],
    "answer": "Possible for n = 3, 4, 7, 8. Impossible for n = 1, 2, 5, 6, 9, 10. In general, such an arrangement exists iff n ≡ 0 or 3 (mod 4).",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 27785,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nFor $x, y, z \\geq 0$, establish the inequality\n$$\nx(x-z)^2 + y(y-z)^2 \\geq (x-z)(y-z)(x+y-z)\n$$\nand determine when equality holds.",
    "options": [],
    "answer": "Equality holds if and only if either all three variables are equal, or exactly one of the variables is zero and the other two are equal.",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 27786,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nShow that the number $x$ is rational if and only if three distinct terms that form a geometric progression can be chosen from the sequence\n$$\nx, x+1, x+2, x+3, \\ldots\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 27787,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nIn triangle $ABC$, the medians to the sides $AB$ and $AC$ are perpendicular. Prove that $\\cot B + \\cot C \\geq \\frac{2}{3}$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 27788,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nA number of schools took part in a tennis tournament. No two players from the same school played against each other. Every two players from different schools played exactly one match against each other. A match between two boys or between two girls was called a single and that between a boy and a girl was called a mixed single. The total number of boys differed from the total number of girls by at most $1$. The total number of singles differed from the total number of mixed singles by at most $1$. At most how many schools were represented by an odd number of players?",
    "options": [],
    "answer": "3",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 27789,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nDetermine a triangle for which the three sides and an altitude are four consecutive integers and for which this altitude partitions the triangle into two right triangles with integer sides. Show that there is only one such triangle.",
    "options": [],
    "answer": "The unique triangle has side lengths 13, 14, and 15, with the altitude of length 12 dropped to the side of length 14.",
    "solution": "",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 27790,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nLet $y_{1}, y_{2}, y_{3}, \\ldots$ be a sequence such that $y_{1}=1$ and, for $k>0$, is defined by the relationship:\n$$\n\\begin{gathered}\ny_{2k}= \\begin{cases}2 y_{k} & \\text{ if } k \\text{ is even } \\\\\n2 y_{k}+1 & \\text{ if } k \\text{ is odd }\\end{cases} \\\\\ny_{2k+1}= \\begin{cases}2 y_{k} & \\text{ if } k \\text{ is odd } \\\\\n2 y_{k}+1 & \\text{ if } k \\text{ is even }\\end{cases}\n\\end{gathered}\n$$\nShow that the sequence $y_{1}, y_{2}, y_{3}, \\ldots$ takes on every positive integer value exactly once.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 27791,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nTwenty-five men sit around a circular table. Every hour there is a vote, and each must respond yes or no. Each man behaves as follows: on the $n^{\\text{th}}$ vote, if his response is the same as the response of at least one of the two people he sits between, then he will respond the same way on the $(n+1)^{th}$ vote as on the $n^{th}$ vote; but if his response is different from that of both his neighbours on the $n$-th vote, then his response on the $(n+1)$-th vote will be different from his response on the $n^{th}$ vote. Prove that, however everybody responded on the first vote, there will be a time after which nobody's response will ever change.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 27792,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nShow that every positive integral power of $\\sqrt{2}-1$ is of the form $\\sqrt{m}-\\sqrt{m-1}$ for some positive integer $m$. (e.g. $\\left.(\\sqrt{2}-1)^{2}=3-2 \\sqrt{2}=\\sqrt{9}-\\sqrt{8}\\right)$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 27793,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nEvaluate the sum\n$$\n\\sum_{n=1}^{1994} (-1)^n \\frac{n^2 + n + 1}{n!}\n$$",
    "options": [],
    "answer": "1/1993! + 1/1994! - 1",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 27794,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nLet $AB$ be a diameter of a circle $\\Omega$ and $P$ be any point not on the line through $A$ and $B$. Suppose the line through $P$ and $A$ cuts $\\Omega$ again in $U$, and the line through $P$ and $B$ cuts $\\Omega$ again in $V$. (Note that in case of tangency $U$ may coincide with $A$ or $V$ may coincide with $B$. Also, if $P$ is on $\\Omega$ then $P=U=V$.) Suppose that $|PU|=s|PA|$ and $|PV|=t|PB|$ for some nonnegative real numbers $s$ and $t$. Determine the cosine of the angle $APB$ in terms of $s$ and $t$.",
    "options": [],
    "answer": "cos(APB) = sqrt(s t) if P is outside the circle; cos(APB) = -sqrt(s t) if P is inside the circle; and cos(APB) = 0 if P lies on the circle",
    "solution": "",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 27795,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nLet $ABC$ be an acute angled triangle. Let $AD$ be the altitude on $BC$, and let $H$ be any interior point on $AD$. Lines $BH$ and $CH$, when extended, intersect $AC$ and $AB$ at $E$ and $F$, respectively. Prove that $\\angle EDH = \\angle FDH$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 27796,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nLet $a$, $b$, and $c$ be positive real numbers. Prove that\n$$\na^{a} b^{b} c^{c} \\geq (a b c)^{\\frac{a+b+c}{3}}\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 27797,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nLet $f(x) = \\frac{9^{x}}{9^{x} + 3}$. Evaluate the sum\n$$\nf\\left(\\frac{1}{1996}\\right) + f\\left(\\frac{2}{1996}\\right) + f\\left(\\frac{3}{1996}\\right) + \\cdots + f\\left(\\frac{1995}{1996}\\right)\n$$",
    "options": [],
    "answer": "1995/2",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 27798,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nLet $n$ be a fixed positive integer. Show that for only nonnegative integers $k$, the diophantine equation\n$$\nx_{1}^{3} + x_{2}^{3} + \\cdots + x_{n}^{3} = y^{3k+2}\n$$\nhas infinitely many solutions in positive integers $x_{i}$ and $y$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Number Theory",
    "subtopic": "Diophantine Equations"
  },
  {
    "id": 27799,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSuppose that $u$ is a real parameter with $0 < u < 1$. Define\n$$\nf(x) = \\begin{cases}\n0 & \\text{ if } 0 \\leq x \\leq u \\\\\n1 - (\\sqrt{u x} + \\sqrt{(1-u)(1-x)})^{2} & \\text{ if } u < x \\leq 1\n\\end{cases}\n$$\nand define the sequence $\\{u_n\\}$ recursively as follows:\n$$\nu_1 = f(1), \\text{ and } u_n = f(u_{n-1}) \\text{ for all } n > 1\n$$\nShow that there exists a positive integer $k$ for which $u_k = 0$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 27800,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nFind all real solutions to the following system of equations. Carefully justify your answer.\n$$\n\\left\\{\\begin{array}{l}\n\\frac{4 x^{2}}{1+4 x^{2}}=y \\\\\n\\frac{4 y^{2}}{1+4 y^{2}}=z \\\\\n\\frac{4 z^{2}}{1+4 z^{2}}=x\n\\end{array}\\right.\n$$",
    "options": [],
    "answer": "(0, 0, 0) and (1/2, 1/2, 1/2)",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 27801,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nWe denote an arbitrary permutation of the integers $1, \\ldots, n$ by $a_{1}, \\ldots, a_{n}$. Let $f(n)$ be the number of these permutations such that\n\n(i) $a_{1}=1$;\n\n(ii) $|a_{i}-a_{i+1}| \\leq 2, \\quad i=1, \\ldots, n-1$.\n\nDetermine whether $f(1996)$ is divisible by $3$.",
    "options": [],
    "answer": "Yes",
    "solution": "",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 27802,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nLet $\\triangle ABC$ be an isosceles triangle with $AB = AC$. Suppose that the angle bisector of $\\angle B$ meets $AC$ at $D$ and that $BC = BD + AD$. Determine $\\angle A$.",
    "options": [],
    "answer": "100°",
    "solution": "",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 27803,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nIf $\\alpha, \\beta, \\gamma$ are the roots of $x^{3}-x-1=0$, compute\n$$\n\\frac{1+\\alpha}{1-\\alpha}+\\frac{1+\\beta}{1-\\beta}+\\frac{1+\\gamma}{1-\\gamma}\n$$",
    "options": [],
    "answer": "-7",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 27804,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nLet $r_{1}, r_{2}, \\ldots, r_{m}$ be a given set of $m$ positive rational numbers such that $\\sum_{k=1}^{m} r_{k}=1$. Define the function $f$ by $f(n)=n-\\sum_{k=1}^{m}\\left[r_{k} n\\right]$ for each positive integer $n$. Determine the minimum and maximum values of $f(n)$. Here $[x]$ denotes the greatest integer less than or equal to $x$.",
    "options": [],
    "answer": "minimum = 0, maximum = m - 1",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 27805,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nWrite the sum\n$$\n\\sum_{k=0}^{n} \\frac{(-1)^{k}\\binom{n}{k}}{k^{3}+9 k^{2}+26 k+24}\n$$\nin the form $\\frac{p(n)}{q(n)}$, where $p$ and $q$ are polynomials with integer coefficients.",
    "options": [],
    "answer": "1/(2(n+3)(n+4))",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 27806,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nHow many pairs of positive integers $x, y$ are there, with $x \\leq y$, and such that $\\operatorname{gcd}(x, y)=5!$ and $\\operatorname{lcm}(x, y)=50!$.\n\nNote. $\\operatorname{gcd}(x, y)$ denotes the greatest common divisor of $x$ and $y$, $\\operatorname{lcm}(x, y)$ denotes the least common multiple of $x$ and $y$, and $n!=n \\times(n-1) \\times \\cdots \\times 2 \\times 1$.",
    "options": [],
    "answer": "16384",
    "solution": "",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 27807,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nThe closed interval $A = [0, 50]$ is the union of a finite number of closed intervals, each of length $1$. Prove that some of the intervals can be removed so that those remaining are mutually disjoint and have total length $\\geq 25$.\n\nNote. For $a \\leq b$, the closed interval $[a, b] := \\{ x \\in \\mathbb{R} : a \\leq x \\leq b \\}$ has length $b - a$; disjoint intervals have empty intersection.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 27808,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nProve that\n$$\n\\frac{1}{1999} < \\frac{1}{2} \\cdot \\frac{3}{4} \\cdot \\frac{5}{6} \\cdots \\frac{1997}{1998} < \\frac{1}{44}\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\nLet $P = \\frac{1}{2} \\cdot \\frac{3}{4} \\cdot \\frac{5}{6} \\cdots \\frac{1997}{1998}$.\n\nThere are $999$ terms in the product, since the numerators run through the odd numbers from $1$ to $1997$ and the denominators through the even numbers from $2$ to $1998$.\n\nWe can write:\n$$\nP = \\prod_{k=1}^{999} \\frac{2k-1}{2k}\n$$\n\nNote that:\n$$\nP = \\frac{1 \\cdot 3 \\cdot 5 \\cdots 1997}{2 \\cdot 4 \\cdot 6 \\cdots 1998}\n$$\n\nLet us estimate $P$.\n\nFirst, for the lower bound:\n\nWe have $\\frac{2k-1}{2k} > 1 - \\frac{1}{2k}$, so\n$$\nP > \\prod_{k=1}^{999} \\left(1 - \\frac{1}{2k}\\right)\n$$\n\nBut $\\prod_{k=1}^{n} (1 - \\frac{1}{k}) = \\frac{1}{n+1}$, but here the terms are $1 - \\frac{1}{2k}$.\n\nAlternatively, note that:\n$$\nP = \\frac{(2n-1)!!}{(2n)!!}\n$$\nwhere $n=999$.\n\nBut $(2n)!! = 2^n n!$, and $(2n-1)!! = \\frac{(2n)!}{2^n n!}$, so\n$$\nP = \\frac{(2n)!}{2^n n!} \\cdot \\frac{1}{2^n n!} = \\frac{(2n)!}{2^{2n} (n!)^2}\n$$\n\nSo\n$$\nP = \\frac{(2n)!}{4^n (n!)^2}\n$$\n\nBut $\\frac{(2n)!}{(n!)^2} = \\binom{2n}{n}$, so\n$$\nP = \\frac{1}{4^n} \\binom{2n}{n}\n$$\n\nFor $n=999$:\n$$\nP = \\frac{1}{4^{999}} \\binom{1998}{999}\n$$\n\nNow, using the estimate $\\binom{2n}{n} < \\frac{4^n}{\\sqrt{\\pi n}}$ and $\\binom{2n}{n} > \\frac{4^n}{2\\sqrt{n}}$ for large $n$,\n\nSo\n$$\nP < \\frac{1}{\\sqrt{\\pi n}}\n$$\n$$\nP > \\frac{1}{2\\sqrt{n}}\n$$\n\nFor $n=999$, $\\sqrt{999} \\approx 31.6$.\n\nSo\n$$\nP < \\frac{1}{\\sqrt{\\pi \\cdot 999}} \\approx \\frac{1}{56}\n$$\n$$\nP > \\frac{1}{2 \\cdot 31.6} \\approx \\frac{1}{63}\n$$\n\nBut the problem asks to show $\\frac{1}{1999} < P < \\frac{1}{44}$.\n\nSince $\\frac{1}{63} > \\frac{1}{1999}$ and $\\frac{1}{56} < \\frac{1}{44}$, but the upper bound is not tight enough.\n\nAlternatively, we can use the following:\n\nNote that $\\prod_{k=1}^{n} \\frac{2k-1}{2k} = \\frac{1}{2n}$.\n\nProof:\n\nLet us compute for small $n$:\n\nFor $n=1$: $\\frac{1}{2} = \\frac{1}{2}$\n\nFor $n=2$: $\\frac{1}{2} \\cdot \\frac{3}{4} = \\frac{3}{8}$, $\\frac{1}{4} = 0.25$, $\\frac{3}{8} = 0.375$\n\nBut $\\prod_{k=1}^{n} \\frac{2k-1}{2k} = \\frac{1 \\cdot 3 \\cdot 5 \\cdots (2n-1)}{2 \\cdot 4 \\cdot 6 \\cdots 2n} = \\frac{(2n)!}{2^n n!} \\cdot \\frac{1}{2^n n!} = \\frac{(2n)!}{2^{2n} (n!)^2}$\n\nAs above, $P = \\frac{1}{4^n} \\binom{2n}{n}$.\n\nNow, using Stirling's approximation:\n$$\nn! \\sim \\sqrt{2\\pi n} \\left(\\frac{n}{e}\\right)^n\n$$\nSo\n$$\n\\binom{2n}{n} \\sim \\frac{4^n}{\\sqrt{\\pi n}}\n$$\nSo\n$$\nP \\sim \\frac{1}{\\sqrt{\\pi n}}\n$$\nFor $n=999$, $\\sqrt{\\pi n} \\approx \\sqrt{3.1416 \\times 999} \\approx \\sqrt{3140} \\approx 56$.\nSo $P \\approx \\frac{1}{56}$.\n\nTherefore,\n$$\n\\frac{1}{1999} < P < \\frac{1}{44}\n$$\n\nsince $\\frac{1}{56}$ is between $\\frac{1}{1999}$ and $\\frac{1}{44}$.\n\nThus, the required inequalities hold.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 27809,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nThe point $O$ is situated inside the parallelogram $ABCD$ so that\n$$\n\\angle AOB + \\angle COD = 180^\\circ\n$$\nProve that $\\angle OBC = \\angle ODC$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 27810,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nDetermine the number of real solutions $a$ to the equation\n$$\n\\left[\\frac{1}{2} a\\right]+\\left[\\frac{1}{3} a\\right]+\\left[\\frac{1}{5} a\\right]=a\n$$\nHere, if $x$ is a real number, then $[x]$ denotes the greatest integer that is less than or equal to $x$.",
    "options": [],
    "answer": "30",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 27811,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nFind all real numbers $x$ such that\n$$\nx = \\left(x - \\frac{1}{x}\\right)^{1/2} + \\left(1 - \\frac{1}{x}\\right)^{1/2}\n$$",
    "options": [],
    "answer": "(1 + sqrt(5)) / 2",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 27812,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nLet $n$ be a natural number such that $n \\geq 2$. Show that\n$$\n\\frac{1}{n+1}\\left(1+\\frac{1}{3}+\\cdots+\\frac{1}{2 n-1}\\right)>\\frac{1}{n}\\left(\\frac{1}{2}+\\frac{1}{4}+\\cdots+\\frac{1}{2 n}\\right) .\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 27813,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nLet $ABC$ be a triangle with $\\angle BAC = 40^\\circ$ and $\\angle ABC = 60^\\circ$. Let $D$ and $E$ be the points lying on the sides $AC$ and $AB$, respectively, such that $\\angle CBD = 40^\\circ$ and $\\angle BCE = 70^\\circ$. Let $F$ be the point of intersection of the lines $BD$ and $CE$. Show that the line $AF$ is perpendicular to the line $BC$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 27814,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nLet $m$ be a positive integer. Define the sequence $a_{0}, a_{1}, a_{2}, \\ldots$ by $a_{0}=0$, $a_{1}=m$, and $a_{n+1}=m^{2} a_{n}-a_{n-1}$ for $n=1,2,3, \\ldots$. Prove that an ordered pair $(a, b)$ of non-negative integers, with $a \\leq b$, gives a solution to the equation\n$$\n\\frac{a^{2}+b^{2}}{a b+1}=m^{2}\n$$\nif and only if $(a, b)$ is of the form $\\left(a_{n}, a_{n+1}\\right)$ for some $n \\geq 0$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 27815,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nFind all real solutions to the equation $4x^{2} - 40[x] + 51 = 0$.\nHere, if $x$ is a real number, then $[x]$ denotes the greatest integer that is less than or equal to $x$.",
    "options": [],
    "answer": "x = sqrt(29)/2, sqrt(189)/2, sqrt(229)/2, sqrt(269)/2",
    "solution": "Let $n = [x]$, where $n$ is an integer and $n \\leq x < n+1$.\n\nThen the equation becomes:\n$$\n4x^2 - 40n + 51 = 0\n$$\nwhich gives\n$$\n4x^2 = 40n - 51\n$$\n$$\nx^2 = 10n - 12.75\n$$\n$$\nx = \\pm \\sqrt{10n - 12.75}\n$$\nBut $n \\leq x < n+1$, so $x$ must be real, i.e., $10n - 12.75 \\geq 0$, or $n \\geq 1.275$. Since $n$ is integer, $n \\geq 2$.\n\nAlso, $x$ must satisfy $n \\leq x < n+1$.\n\nLet us consider $x = \\sqrt{10n - 12.75}$ first.\n\nWe require:\n$$\nn \\leq \\sqrt{10n - 12.75} < n+1\n$$\n\nLet us solve $n \\leq \\sqrt{10n - 12.75}$:\n$$\nn^2 \\leq 10n - 12.75\n$$\n$$\nn^2 - 10n + 12.75 \\leq 0\n$$\nThe roots of $n^2 - 10n + 12.75 = 0$ are:\n$$\nn = \\frac{10 \\pm \\sqrt{100 - 4 \\times 12.75}}{2} = \\frac{10 \\pm \\sqrt{100 - 51}}{2} = \\frac{10 \\pm 7}{2}\n$$\nSo $n = \\frac{17}{2} = 8.5$ or $n = \\frac{3}{2} = 1.5$.\n\nSo $1.5 \\leq n \\leq 8.5$, i.e., $n = 2,3,4,5,6,7,8$.\n\nNow, for each $n$ in $2$ to $8$, check if $\\sqrt{10n - 12.75} < n+1$:\n\n$\\sqrt{10n - 12.75} < n+1$\n\nSquare both sides:\n$$\n10n - 12.75 < (n+1)^2 = n^2 + 2n + 1\n$$\n$$\n0 < n^2 - 8n + 13.75\n$$\nThe roots of $n^2 - 8n + 13.75 = 0$ are:\n$$\nn = \\frac{8 \\pm \\sqrt{64 - 55}}{2} = \\frac{8 \\pm 3}{2} = 5.5, 2.5\n$$\nSo $n < 2.5$ or $n > 5.5$\n\nBut $n \\geq 2$, so $n = 2$ or $n > 5.5$, i.e., $n = 6,7,8$.\n\nSo possible $n$ are $2,6,7,8$.\n\nNow, for each $n$ in $2,6,7,8$, $x = \\sqrt{10n - 12.75}$, and $n \\leq x < n+1$.\n\nCheck for $n = 2$:\n$10 \\times 2 - 12.75 = 20 - 12.75 = 7.25$\n$\\sqrt{7.25} \\approx 2.6926$\n$2 \\leq 2.6926 < 3$ is true.\n\nFor $n = 6$:\n$10 \\times 6 - 12.75 = 60 - 12.75 = 47.25$\n$\\sqrt{47.25} \\approx 6.8777$\n$6 \\leq 6.8777 < 7$ is true.\n\nFor $n = 7$:\n$10 \\times 7 - 12.75 = 70 - 12.75 = 57.25$\n$\\sqrt{57.25} \\approx 7.5697$\n$7 \\leq 7.5697 < 8$ is true.\n\nFor $n = 8$:\n$10 \\times 8 - 12.75 = 80 - 12.75 = 67.25$\n$\\sqrt{67.25} \\approx 8.2006$\n$8 \\leq 8.2006 < 9$ is true.\n\nSo for $n = 2,6,7,8$, $x = \\sqrt{10n - 12.75}$ is a solution.\n\nNow, consider $x = -\\sqrt{10n - 12.75}$.\n\nWe require $n \\leq x < n+1$, but $-\\sqrt{10n - 12.75} < 0$, so $n < 0$.\n\nBut $n \\geq 2$, so there are no solutions for $x = -\\sqrt{10n - 12.75}$.\n\nTherefore, the real solutions are:\n$$\nx = \\sqrt{10n - 12.75}, \\quad n = 2,6,7,8\n$$\nThat is,\n$$\nx_1 = \\sqrt{7.25} \\approx 2.6926\n$$\n$$\nx_2 = \\sqrt{47.25} \\approx 6.8777\n$$\n$$\nx_3 = \\sqrt{57.25} \\approx 7.5697\n$$\n$$\nx_4 = \\sqrt{67.25} \\approx 8.2006\n$$\n\nSo all real solutions are $x = \\sqrt{7.25}$, $x = \\sqrt{47.25}$, $x = \\sqrt{57.25}$, $x = \\sqrt{67.25}$.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 27816,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nLet $ABC$ be an equilateral triangle of altitude $1$. A circle with radius $1$ and center on the same side of $AB$ as $C$ rolls along the segment $AB$. Prove that the arc of the circle that is inside the triangle always has the same length.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 27817,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSuppose $a_{1}, a_{2}, \\ldots, a_{8}$ are eight distinct integers from $\\{1,2, \\ldots, 16,17\\}$. Show that there is an integer $k>0$ such that the equation $a_{i}-a_{j}=k$ has at least three different solutions. Also, find a specific set of 7 distinct integers from $\\{1,2, \\ldots, 16,17\\}$ such that the equation $a_{i}-a_{j}=k$ does not have three distinct solutions for any $k>0$.",
    "options": [],
    "answer": "{1, 2, 5, 9, 14, 16, 17}",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 27818,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nAt 12:00 noon, Anne, Beth and Carmen begin running laps around a circular track of length three hundred meters, all starting from the same point on the track. Each jogger maintains a constant speed in one of the two possible directions for an indefinite period of time. Show that if Anne's speed is different from the other two speeds, then at some later time Anne will be at least one hundred meters from each of the other runners. (Here, distance is measured along the shorter of the two arcs separating two runners.)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nBy rotating the frame of reference we may assume that Anne has speed zero, that Beth runs at least as fast as Carmen, and that Carmen's speed is positive. If Beth is no more than twice as fast as Carmen, then both are at least $100$ meters from Anne when Carmen has run $100$ meters. If Beth runs more that twice as fast as Carmen, then Beth runs a stretch of more than $200$ meters during the time Carmen runs between $100$ and $200$ meters. Some part of this stretch lies more than $100$ meters from Anne, at which time both Beth and Carmen are at least (in fact, more than) $100$ meters away from Anne.\nSolution:\n\nBy rotating the frame of reference we may assume Anne's speed to equal zero, and that the other two runners have non-zero speed. We may assume that Beth is running at least as fast as Carmen. Suppose that it takes $t$ seconds for Beth to run $200$ meters. Then at all times in the infinite set $T=\\{t, 2 t, 4 t, 8 t, \\ldots\\}$, Beth is exactly $100$ meters from Anne. At time $t$, Carmen has traveled exactly $d$ meters where $0<d \\leq 200$. Let $k$ be the least integer such that $2^{k} d \\geq 100$. Then $k \\geq 0$ and $100 \\leq 2^{k} d \\leq 200$, so at time $2^{k} t \\in T$ both Beth and Carmen are at least $100$ meters from Anne.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 27819,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nA permutation of the integers $1901, 1902, \\ldots, 2000$ is a sequence $a_{1}, a_{2}, \\ldots, a_{100}$ in which each of those integers appears exactly once. Given such a permutation, we form the sequence of partial sums\n$$\ns_{1} = a_{1}, \\quad s_{2} = a_{1} + a_{2}, s_{3} = a_{1} + a_{2} + a_{3}, \\ldots, s_{100} = a_{1} + a_{2} + \\cdots + a_{100}.\n$$\nHow many of these permutations will have no terms of the sequence $s_{1}, \\ldots, s_{100}$ divisible by three?",
    "options": [],
    "answer": "99! * 33! * 34! / 66!",
    "solution": "Solution:\n\nLet $\\{1901, 1902, \\ldots, 2000\\} = R_{0} \\cup R_{1} \\cup R_{2}$ where each integer in $R_{i}$ is congruent to $i$ modulo $3$. We note that $|R_{0}| = |R_{1}| = 33$ and $|R_{2}| = 34$. Each permutation $S = (a_{1}, a_{2}, \\ldots, a_{100})$ can be uniquely specified by describing a sequence $S' = (a_{1}', a_{2}', \\ldots, a_{100}')$ of residues modulo $3$ (containing exactly $33$ zeros, $33$ ones and $34$ twos), and three permutations (one each of $R_{0}, R_{1}$, and $R_{2}$). Note that the number of permutations of $R_{i}$ is exactly $|R_{i}|! = 1 \\cdot 2 \\cdots |R_{i}|$.\n\nThe condition on the partial sums of $S$ depends only on the sequence of residues $S'$. In order to avoid a partial sum divisible by three, the subsequence formed by the $67$ ones and twos in $S'$ must equal either $1, 1, 2, 1, 2, \\ldots, 1, 2$ or $2, 2, 1, 2, 1, \\ldots, 2, 1$. Since $|R_{2}| = |R_{1}| + 1$, only the second pattern is possible. The $33$ zero entries in $S'$ may appear anywhere among $a_{1}', a_{2}', \\ldots, a_{100}'$ provided that $a_{1}' \\neq 0$. There are $\\binom{99}{33} = \\frac{99!}{33!\\,66!}$ ways to choose which entries in $S'$ equal zero. Thus there are exactly $\\binom{99}{33}$ sequences $S'$ whose partial sums are not divisible by three. Therefore the total number of permutations $S$ satisfying this requirement is exactly\n$$\n\\binom{99}{33} \\cdot 33! \\cdot 33! \\cdot 34! = \\frac{99! \\cdot 33! \\cdot 34!}{66!}.\n$$\nIncidentally, this number equals approximately $4.4 \\times 10^{138}$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 27820,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nLet $A = (a_{1}, a_{2}, \\ldots, a_{2000})$ be a sequence of integers each lying in the interval $[-1000, 1000]$. Suppose that the entries in $A$ sum to $1$. Show that some nonempty subsequence of $A$ sums to zero.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nWe may assume no entry of $A$ is zero, for otherwise we are done. We sort $A$ into a new list $B = (b_{1}, \\ldots, b_{2000})$ by selecting elements from $A$ one at a time in such a way that $b_{1} > 0$, $b_{2} < 0$ and, for each $i = 2, 3, \\ldots, 2000$, the sign of $b_{i}$ is opposite to that of the partial sum\n$$\ns_{i-1} = b_{1} + b_{2} + \\cdots + b_{i-1}.\n$$\n(We can assume that each $s_{i-1} \\neq 0$ for otherwise we are done.) At each step of the selection process a candidate for $b_{i}$ is guaranteed to exist, since the condition $a_{1} + a_{2} + \\cdots + a_{2000} = 1$ implies that the sum of unselected entries in $A$ is either zero or has sign opposite to $s_{i-1}$.\nFrom the way they were defined, each of $s_{1}, s_{2}, \\ldots, s_{2000}$ is one of the 1999 nonzero integers in the interval $[-999, 1000]$. By the Pigeon Hole Principle, $s_{j} = s_{k}$ for some $j, k$ satisfying $1 \\leq j < k \\leq 2000$. Thus $b_{j+1} + b_{j+2} + \\cdots + b_{k} = 0$ and we are done.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 27821,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nLet $ABCD$ be a convex quadrilateral with\n$$\n\\begin{aligned}\n\\angle CBD &= 2 \\angle ADB, \\\\\n\\angle ABD &= 2 \\angle CDB \\\\\n\\text{and}\\quad AB &= CB.\n\\end{aligned}\n$$\nProve that $AD = CD$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\nExtend $DB$ to a point $P$ on the circle through $A$ and $C$ centered at $B$. Then $\\angle CPD = \\frac{1}{2} \\angle CBD = \\angle ADB$ and $\\angle APD = \\frac{1}{2} \\angle ABD = \\angle CDB$,\nso $APCD$ is a parallelogram. Now $PD$ bisects $AC$ so $BD$ is an angle bisector of isosceles triangle $ABC$. We have\n$$\n\\angle ADB = \\frac{1}{2} \\angle CBD = \\frac{1}{2} \\angle ABD = \\angle CDB\n$$\nso $DB$ is the angle bisector of $\\angle ADC$. As $DB$ bisects the base of triangle $ADC$, this triangle must be isosceles and $AD = CD$.\n\n![](attached_image_1.png)\n\nDiagram for Solution 1\nSolution:\nLet the bisector of $\\angle ABD$ meet $AD$ at $E$. Let the bisector of $\\angle CBD$ meet $CD$ at $F$. Then $\\angle FBD = \\angle BDE$ and $\\angle EBD = \\angle BDF$, which imply $BE \\parallel FD$ and $BF \\parallel ED$. Thus $BEDF$ is a parallelogram whence\n$$\nBD \\text{ intersects } EF \\text{ at its midpoint } M.\n$$\nOn the other hand since $BE$ is an angle bisector, we have $\\frac{AB}{BD} = \\frac{AE}{ED}$. Similarly we have $\\frac{CB}{BD} = \\frac{CF}{FD}$. By assumption $AB = CB$ so $\\frac{AE}{ED} = \\frac{CF}{FD}$ which implies $EF \\parallel AC$. Thus $\\triangle DEF$ and $\\triangle DAC$ are similar, which implies by (1) that $BD$ intersects $AC$ at its midpoint $N$. Since $\\triangle ABC$ is isosceles, this implies $AC \\perp BD$. Thus $\\triangle NAD$ and $\\triangle NCD$ are right triangles with equal legs and hence are congruent. Thus $AD = CD$.\n\n![](attached_image_2.png)\n\nDiagram for Solution 2",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 27822,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nSuppose that the real numbers $a_{1}, a_{2}, \\ldots, a_{100}$ satisfy\n$$\n\\begin{gathered}\na_{1} \\geq a_{2} \\geq \\cdots \\geq a_{100} \\geq 0 \\\\\na_{1}+a_{2} \\leq 100 \\\\\na_{3}+a_{4}+\\cdots+a_{100} \\leq 100\n\\end{gathered}\n$$\nDetermine the maximum possible value of $a_{1}^{2}+a_{2}^{2}+\\cdots+a_{100}^{2}$, and find all possible sequences $a_{1}, a_{2}, \\ldots, a_{100}$ which achieve this maximum.",
    "options": [],
    "answer": "Maximum value: 10000. Achieved by the sequences (100, 0, 0, ..., 0) and (50, 50, 50, 50, 0, 0, ..., 0).",
    "solution": "Solution:\nWe have $a_{1}+a_{2}+\\cdots+a_{100} \\leq 200$, so\n$$\n\\begin{aligned}\na_{1}^{2}+a_{2}^{2}+\\cdots+a_{100}^{2} & \\leq (100-a_{2})^{2}+a_{2}^{2}+a_{3}^{2}+\\cdots+a_{100}^{2} \\\\\n& = 100^{2}-200 a_{2}+2 a_{2}^{2}+a_{3}^{2}+\\cdots+a_{100}^{2} \\\\\n& \\leq 100^{2}-(a_{1}+a_{2}+\\cdots+a_{100}) a_{2}+2 a_{2}^{2}+a_{3}^{2}+\\cdots+a_{100}^{2} \\\\\n& = 100^{2}+(a_{2}^{2}-a_{1} a_{2})+(a_{3}^{2}-a_{3} a_{2})+(a_{4}^{2}-a_{4} a_{2})+\\cdots+(a_{100}^{2}-a_{100} a_{2}) \\\\\n& = 100^{2}+(a_{2}-a_{1}) a_{2}+(a_{3}-a_{2}) a_{3}+(a_{4}-a_{2}) a_{4}+\\cdots+(a_{100}-a_{2}) a_{100}\n\\end{aligned}\n$$\nSince $a_{1} \\geq a_{2} \\geq \\cdots \\geq a_{100} \\geq 0$, none of the terms $(a_{i}-a_{j}) a_{i}$ is positive. Thus $a_{1}^{2}+a_{2}^{2}+\\cdots+a_{100}^{2} \\leq 10,000$ with equality holding if and only if\n$$\na_{1}=100-a_{2} \\quad \\text{ and } \\quad a_{1}+a_{2}+\\cdots+a_{100}=200\n$$\nand each of the products\n$$\n(a_{2}-a_{1}) a_{2}, \\quad (a_{3}-a_{2}) a_{3}, \\quad (a_{4}-a_{2}) a_{4}, \\cdots, (a_{100}-a_{2}) a_{100}\n$$\nequals zero. Since $a_{1} \\geq a_{2} \\geq a_{3} \\geq \\cdots \\geq a_{100} \\geq 0$, the last condition holds if and only if for some $i \\geq 1$ we have $a_{1}=a_{2}=\\cdots=a_{i}$ and $a_{i+1}=\\cdots=a_{100}=0$. If $i=1$, then we get the solution $100,0,0,\\ldots,0$. If $i \\geq 2$, then from $a_{1}+a_{2}=100$, we get that $i=4$ and the second optimal solution $50,50,50,50,0,0, \\ldots, 0$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 27823,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nCall a positive integer $n$ practical if every positive integer less than or equal to $n$ can be written as the sum of distinct divisors of $n$.\n\nFor example, the divisors of $6$ are $\\{1, 2\\}$, $\\{3\\}$, and $\\{6\\}$. Since\n$$\n1=\\{1\\}, \\quad 2=\\{2\\}, \\quad 3=\\{3\\}, \\quad 4=\\{1\\}+\\{3\\}, \\quad 5=\\{2\\}+\\{3\\}, \\quad 6=\\{6\\},\n$$\nwe see that $6$ is practical.\n\nProve that the product of two practical numbers is also practical.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nLet $p$ and $q$ be practical. For any $k \\leq p q$, we can write\n$$\nk = a q + b \\text{ with } 0 \\leq a \\leq p,\\ 0 \\leq b < q.\n$$\nSince $p$ and $q$ are practical, we can write\n$$\na = c_1 + \\ldots + c_m, \\quad b = d_1 + \\ldots + d_n\n$$\nwhere the $c_i$'s are distinct divisors of $p$ and the $d_j$'s are distinct divisors of $q$. Now\n$$\n\\begin{aligned}\nk &= (c_1 + \\ldots + c_m) q + (d_1 + \\ldots + d_n) \\\\\n  &= c_1 q + \\ldots + c_m q + d_1 + \\ldots + d_n.\n\\end{aligned}\n$$\nEach of $c_i q$ and $d_j$ divides $p q$. Since $d_j < q \\leq c_i q$ for any $i, j$, the $c_i q$'s and $d_j$'s are all distinct, and we conclude that $p q$ is practical.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 27824,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nLet $\\Gamma$ be a circle with radius $r$. Let $A$ and $B$ be distinct points on $\\Gamma$ such that $AB < \\sqrt{3} r$. Let the circle with centre $B$ and radius $AB$ meet $\\Gamma$ again at $C$. Let $P$ be the point inside $\\Gamma$ such that triangle $ABP$ is equilateral. Finally, let $CP$ meet $\\Gamma$ again at $Q$. Prove that $PQ = r$.\n\n![](attached_image_1.png)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\nLet the center of $\\Gamma$ be $O$, the radius $r$. Since $BP = BC$, let $\\theta = \\measuredangle BPC = \\measuredangle BCP$.\n\nQuadrilateral $QABC$ is cyclic, so $\\measuredangle BAQ = 180^\\circ - \\theta$ and hence $\\measuredangle PAQ = 120^\\circ - \\theta$.\n\nAlso $\\measuredangle APQ = 180^\\circ - \\measuredangle APB - \\measuredangle BPC = 120^\\circ - \\theta$, so $PQ = AQ$ and $\\measuredangle AQP = 2\\theta - 60^\\circ$.\n\nAgain because quadrilateral $QABC$ is cyclic, $\\measuredangle ABC = 180^\\circ - \\measuredangle AQC = 240^\\circ - 2\\theta$.\n\nTriangles $OAB$ and $OCB$ are congruent, since $OA = OB = OC = r$ and $AB = BC$.\n\nThus $\\measuredangle ABO = \\measuredangle CBO = \\frac{1}{2} \\measuredangle ABC = 120^\\circ - \\theta$.\n\nWe have now shown that in triangles $AQP$ and $AOB$, $\\measuredangle PAQ = \\measuredangle BAO = \\measuredangle APQ = \\measuredangle ABO$.\n\nAlso $AP = AB$, so $\\triangle AQP \\cong \\triangle AOB$. Hence $QP = OB = r$.\n\n\nLet the center of $\\Gamma$ be $O$, the radius $r$. Since $A, P$ and $C$ lie on a circle centered at $B$, $60^\\circ = \\measuredangle ABP = 2 \\measuredangle ACP$, so $\\measuredangle ACP = \\measuredangle ACQ = 30^\\circ$.\n\nSince $Q, A$, and $C$ lie on $\\Gamma$, $\\measuredangle QOA = 2 \\measuredangle QCA = 60^\\circ$.\n\nSo $QA = r$ since if a chord of a circle subtends an angle of $60^\\circ$ at the center, its length is the radius of the circle.\n\nNow $BP = BC$, so $\\measuredangle BPC = \\measuredangle BCP = \\measuredangle ACB + 30^\\circ$.\n\nThus $\\measuredangle APQ = 180^\\circ - \\measuredangle APB - \\measuredangle BPC = 90^\\circ - \\measuredangle ACB$.\n\nSince $Q, A, B$ and $C$ lie on $\\Gamma$ and $AB = BC$, $\\measuredangle AQP = \\measuredangle AQC = \\measuredangle AQB + \\measuredangle BQC = 2 \\measuredangle ACB$.\n\nFinally, $\\measuredangle QAP = 180 - \\measuredangle AQP - \\measuredangle APQ = 90 - \\measuredangle ACB$.\n\nSo $\\measuredangle PAQ = \\measuredangle APQ$ hence $PQ = AQ = r$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 27825,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nLet $\\mathbb{N} = \\{0, 1, 2, \\ldots\\}$. Determine all functions $f: \\mathbb{N} \\rightarrow \\mathbb{N}$ such that\n$$\nx f(y) + y f(x) = (x + y) f\\left(x^{2} + y^{2}\\right)\n$$\nfor all $x$ and $y$ in $\\mathbb{N}$.",
    "options": [],
    "answer": "All constant functions: f(x) = c for any c in N.",
    "solution": "Solution:\nWe claim that $f$ is a constant function. Suppose, for a contradiction, that there exist $x$ and $y$ with $f(x) < f(y)$; choose $x, y$ such that $f(y) - f(x) > 0$ is minimal. Then\n$$\nf(x) = \\frac{x f(x) + y f(x)}{x + y} < \\frac{x f(y) + y f(x)}{x + y} < \\frac{x f(y) + y f(y)}{x + y} = f(y)\n$$\nso $f(x) < f\\left(x^{2} + y^{2}\\right) < f(y)$ and $0 < f\\left(x^{2} + y^{2}\\right) - f(x) < f(y) - f(x)$, contradicting the choice of $x$ and $y$. Thus, $f$ is a constant function. Since $f(0)$ is in $\\mathbb{N}$, the constant must be from $\\mathbb{N}$.\nAlso, for any $c$ in $\\mathbb{N}$, $x c + y c = (x + y) c$ for all $x$ and $y$, so $f(x) = c$, $c \\in \\mathbb{N}$ are the solutions to the equation.\n\n\nWe claim $f$ is a constant function. Define $g(x) = f(x) - f(0)$. Then $g(0) = 0$, $g(x) \\geq -f(0)$ and\n$$\nx g(y) + y g(x) = (x + y) g\\left(x^{2} + y^{2}\\right)\n$$\nfor all $x, y$ in $\\mathbb{N}$.\nLetting $y = 0$ shows $g\\left(x^{2}\\right) = 0$ (in particular, $g(1) = g(4) = 0$), and letting $x = y = 1$ shows $g(2) = 0$. Also, if $x, y$ and $z$ in $\\mathbb{N}$ satisfy $x^{2} + y^{2} = z^{2}$, then\n$$\ng(y) = -\\frac{y}{x} g(x) .\n$$\nLetting $x = 4$ and $y = 3$, $(*)$ shows that $g(3) = 0$.\nFor any even number $x = 2n > 4$, let $y = n^{2} - 1$. Then $y > x$ and $x^{2} + y^{2} = \\left(n^{2} + 1\\right)^{2}$. For any odd number $x = 2n + 1 > 3$, let $y = 2(n + 1)n$. Then $y > x$ and $x^{2} + y^{2} = \\left((n + 1)^{2} + n^{2}\\right)^{2}$. Thus for every $x > 4$ there is $y > x$ such that $(*)$ is satisfied.\nSuppose for a contradiction, that there is $x > 4$ with $g(x) > 0$. Then we can construct a\nsequence $x = x_{0} < x_{1} < x_{2} < \\ldots$ where $g\\left(x_{i+1}\\right) = -\\frac{x_{i+1}}{x_{i}} g\\left(x_{i}\\right)$. It follows that $\\left|g\\left(x_{i+1}\\right)\\right| > \\left|g\\left(x_{i}\\right)\\right|$ and the signs of $g\\left(x_{i}\\right)$ alternate. Since $g(x)$ is always an integer, $\\left|g\\left(x_{i+1}\\right)\\right| \\geq \\left|g\\left(x_{i}\\right)\\right| + 1$. Thus for some sufficiently large value of $i$, $g\\left(x_{i}\\right) < -f(0)$, a contradiction.\nAs for Proof 1, we now conclude that the functions that satisfy the given functional equation are $f(x) = c$, $c \\in \\mathbb{N}$.\n\n\nSuppose that $W$ is the set of nonnegative integers and that $f: W \\rightarrow W$ satisfies:\n$$\nx f(y) + y f(x) = (x + y) f\\left(x^{2} + y^{2}\\right) .\n$$\nWe will show that $f$ is a constant function.\nLet $f(0) = k$, and set $S = \\{x \\mid f(x) = k\\}$.\nLetting $y = 0$ in $(*)$ shows that $f\\left(x^{2}\\right) = k \\quad \\forall x > 0$, and so\n$$\nx^{2} \\in S \\quad \\forall x > 0\n$$\nIn particular, $1 \\in S$.\nSuppose $x^{2} + y^{2} = z^{2}$. Then $y f(x) + x f(y) = (x + y) f\\left(z^{2}\\right) = (x + y) k$. Thus,\n$$\nx \\in S \\quad \\text{iff} \\quad y \\in S\n$$\nwhenever $x^{2} + y^{2}$ is a perfect square.\nFor a contradiction, let $n$ be the smallest non-negative integer such that $f\\left(2^{n}\\right) \\neq k$. By (l) $n$ must be odd, so $\\frac{n-1}{2}$ is an integer. Now $\\frac{n-1}{2} < n$ so $f\\left(2^{\\frac{n-1}{2}}\\right) = k$. Letting $x = y = 2^{\\frac{n-1}{2}}$ in $(*)$ shows $f\\left(2^{n}\\right) = k$, a contradiction. Thus every power of 2 is an element of $S$.\nFor each integer $n \\geq 2$ define $p(n)$ to be the largest prime such that $p(n) \\mid n$.\nClaim: For any integer $n > 1$ that is not a power of 2, there exists a sequence of integers $x_{1}, x_{2}, \\ldots, x_{r}$ such that the following conditions hold:\na) $x_{1} = n$.\nb) $x_{i}^{2} + x_{i+1}^{2}$ is a perfect square for each $i = 1, 2, 3, \\ldots, r-1$.\nc) $p\\left(x_{1}\\right) \\geq p\\left(x_{2}\\right) \\geq \\ldots \\geq p\\left(x_{r}\\right) = 2$.\nProof: Since $n$ is not a power of $2$, $p(n) = p\\left(x_{1}\\right) \\geq 3$. Let $p\\left(x_{1}\\right) = 2m + 1$, so $n = x_{1} = b(2m + 1)^{a}$, for some $a$ and $b$, where $p(b) < 2m + 1$.\nCase 1: $a = 1$. Since $\\left(2m + 1, 2m^{2} + 2m, 2m^{2} + 2m + 1\\right)$ is a Pythagorean Triple, if $x_{2} = b\\left(2m^{2} + 2m\\right)$, then $x_{1}^{2} + x_{2}^{2} = b^{2}\\left(2m^{2} + 2m + 1\\right)^{2}$ is a perfect square. Furthermore, $x_{2} = 2b m(m + 1)$, and so $p\\left(x_{2}\\right) < 2m + 1 = p\\left(x_{1}\\right)$.\nCase 2: $a > 1$. If $n = x_{1} = (2m + 1)^{a} \\cdot b$, let $x_{2} = (2m + 1)^{a-1} \\cdot b \\cdot \\left(2m^{2} + 2m\\right)$, $x_{3} = (2m + 1)^{a-2} \\cdot b \\cdot \\left(2m^{2} + 2m\\right)^{2}$, $\\ldots$, $x_{a+1} = (2m + 1)^{0} \\cdot b \\cdot \\left(2m^{2} + 2m\\right)^{a} = b \\cdot 2^{a} m^{a}(m + 1)^{a}$. Note that for $1 \\leq i \\leq a$, $x_{i}^{2} + x_{i+1}^{2}$ is a perfect square and also note that $p\\left(x_{a+1}\\right) < 2m + 1 = p\\left(x_{1}\\right)$.\nIf $x_{a+1}$ is not a power of 2, we extend the sequence $x_{i}$ using the same procedure described above. We keep doing this until $p\\left(x_{r}\\right) = 2$, for some integer $r$.\nBy (2), $x_{i} \\in S$ iff $x_{i+1} \\in S$ for $i = 1, 2, 3, \\ldots, r-1$. Thus, $n = x_{1} \\in S$ iff $x_{r} \\in S$. But $x_{r}$ is a power of 2 because $p\\left(x_{r}\\right) = 2$, and we earlier proved that powers of 2 are in $S$. Therefore, $n \\in S$, proving the claim.\nWe have proven that every integer $n \\geq 1$ is an element of $S$, and so we have proven that $f(n) = k = f(0)$, for each $n \\geq 1$. Therefore, $f$ is constant, Q.E.D.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 27826,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nConsider a standard twelve-hour clock whose hour and minute hands move continuously. Let $m$ be an integer, with $1 \\leq m \\leq 720$. At precisely $m$ minutes after 12:00, the angle made by the hour hand and minute hand is exactly $1^{\\circ}$. Determine all possible values of $m$.",
    "options": [],
    "answer": "262, 458",
    "solution": "Solution:\n\nThe minute hand makes a full revolution of $360^{\\circ}$ every 60 minutes, so after $m$ minutes it has swept through $\\frac{360}{60} m = 6m$ degrees. The hour hand makes a full revolution every 12 hours (720 minutes), so after $m$ minutes it has swept through $\\frac{360}{720} m = m/2$ degrees. Since both hands started in the same position at 12:00, the angle between the two hands will be $1^{\\circ}$ if $6m - m/2 = \\pm 1 + 360k$ for some integer $k$. Solving this equation we get\n$$\nm = \\frac{720k \\pm 2}{11} = 65k + \\frac{5k \\pm 2}{11}\n$$\nSince $1 \\leq m \\leq 720$, we have $1 \\leq k \\leq 11$. Since $m$ is an integer, $5k \\pm 2$ must be divisible by 11, say $5k \\pm 2 = 11q$. Then\n$$\n5k = 11q \\pm 2 \\Rightarrow k = 2q + \\frac{q \\pm 2}{5}\n$$\nIt is now clear that only $q = 2$ and $q = 3$ satisfy all the conditions. Thus $k = 4$ or $k = 7$ and substituting these values into the expression for $m$ we find that the only possible values of $m$ are 262 and 458.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 27827,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nFind the last three digits of the number $2003^{2002^{2001}}$.",
    "options": [],
    "answer": "241",
    "solution": "Solution:\nWe must find the remainder when $2003^{2002^{2001}}$ is divided by $1000$, which will be the same as the remainder when $3^{2002^{2001}}$ is divided by $1000$, since $2003 \\equiv 3 \\pmod{1000}$. To do this we will first find a positive integer $n$ such that $3^{n} \\equiv 1 \\pmod{1000}$ and then try to express $2002^{2001}$ in the form $n k + r$, so that\n$$\n2003^{2002^{2001}} \\equiv 3^{n k + r} \\equiv (3^{n})^{k} \\cdot 3^{r} \\equiv 1^{k} \\cdot 3^{r} \\equiv 3^{r} \\pmod{1000}\n$$\nSince $3^{2} = 10 - 1$, we can evaluate $3^{2m}$ using the binomial theorem:\n$$\n3^{2m} = (10 - 1)^{m} = (-1)^{m} + 10 m (-1)^{m-1} + 100 \\frac{m(m-1)}{2} (-1)^{m-2} + \\cdots + 10^{m}.\n$$\nAfter the first 3 terms of this expansion, all remaining terms are divisible by $1000$, so letting $m = 2q$, we have that\n$$\n3^{4q} \\equiv 1 - 20q + 100q(2q-1) \\pmod{1000}\n$$\nUsing this, we can check that $3^{100} \\equiv 1 \\pmod{1000}$ and now we wish to find the remainder when $2002^{2001}$ is divided by $100$.\nNow $2002^{2001} \\equiv 2^{2001} \\pmod{100} \\equiv 4 \\cdot 2^{1999} \\pmod{4 \\cdot 25}$, so we'll investigate powers of $2$ modulo $25$. Noting that $2^{10} = 1024 \\equiv -1 \\pmod{25}$, we have\n$$\n2^{1999} = (2^{10})^{199} \\cdot 2^{9} \\equiv (-1)^{199} \\cdot 512 \\equiv -12 \\equiv 13 \\pmod{25}.\n$$\nThus $2^{2001} \\equiv 4 \\cdot 13 = 52 \\pmod{100}$. Therefore $2002^{2001}$ can be written in the form $100k + 52$ for some integer $k$, so\n$$\n2003^{2002^{2001}} \\equiv 3^{52} \\pmod{1000} \\equiv 1 - 20 \\cdot 13 + 1300 \\cdot 25 \\equiv 241 \\pmod{1000}\n$$\nusing equation (1). So the last 3 digits of $2003^{2002^{2001}}$ are $241$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 27828,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nFind all real positive solutions (if any) to\n$$\n\\begin{gathered}\nx^{3}+y^{3}+z^{3}=x+y+z, \\text{ and } \\\\\nx^{2}+y^{2}+z^{2}=x y z .\n\\end{gathered}\n$$",
    "options": [],
    "answer": "No positive real solutions exist.",
    "solution": "Solution:\nLet $f(x, y, z)=\\left(x^{3}-x\\right)+\\left(y^{3}-y\\right)+\\left(z^{3}-z\\right)$. The first equation above is equivalent to $f(x, y, z)=0$. If $x, y, z \\geq 1$, then $f(x, y, z) \\geq 0$ with equality only if $x=y=z=1$. But if $x=y=z=1$, then the second equation is not satisfied. So in any solution to the system of equations, at least one of the variables is less than 1. Without loss of generality, suppose that $x<1$. Then\n$$\nx^{2}+y^{2}+z^{2}>y^{2}+z^{2} \\geq 2 y z>y z>x y z .\n$$\nTherefore the system has no real positive solutions.\nSolution:\nWe will show that the system has no real positive solution. Assume otherwise.\nThe second equation can be written $x^{2}-(y z) x+\\left(y^{2}+z^{2}\\right)$. Since this quadratic in $x$ has a real solution by hypothesis, its discriminant is nonnegative. Hence\n$$\ny^{2} z^{2}-4 y^{2}-4 z^{2} \\geq 0\n$$\nDividing through by $4 y^{2} z^{2}$ yields\n$$\n\\frac{1}{4} \\geq \\frac{1}{y^{2}}+\\frac{1}{z^{2}} \\geq \\frac{1}{y^{2}}\n$$\nHence $y^{2} \\geq 4$ and so $y \\geq 2$, $y$ being positive. A similar argument yields $x, y, z \\geq 2$. But the first equation can be written as\n$$\nx\\left(x^{2}-1\\right)+y\\left(y^{2}-1\\right)+z\\left(z^{2}-1\\right)=0\n$$\ncontradicting $x, y, z \\geq 2$. Hence, a real positive solution cannot exist.\nSolution:\nApplying the arithmetic-geometric mean inequality and the Power Mean Inequalities to $x, y, z$ we have\n$$\n\\sqrt[3]{x y z} \\leq \\frac{x+y+z}{3} \\leq \\sqrt{\\frac{x^{2}+y^{2}+z^{2}}{3}} \\leq \\sqrt[3]{\\frac{x^{3}+y^{3}+z^{3}}{3}}\n$$\nLetting $S=x+y+z=x^{3}+y^{3}+z^{3}$ and $P=x y z=x^{2}+y^{2}+z^{2}$, this inequality can be written\n$$\n\\sqrt[3]{P} \\leq \\frac{S}{3} \\leq \\sqrt{\\frac{P}{3}} \\leq \\sqrt[3]{\\frac{S}{3}}\n$$\nNow $\\sqrt[3]{P} \\leq \\sqrt{\\frac{P}{3}}$ implies $P^{2} \\leq P^{3} / 27$, so $P \\geq 27$. Also $\\frac{S}{3} \\leq \\sqrt[3]{\\frac{S}{3}}$ implies $S^{3} / 27 \\leq S / 3$,\nso $S \\leq 3$. But then $\\sqrt[3]{P} \\geq 3$ and $\\sqrt[3]{\\frac{S}{3}} \\leq 1$ which is inconsistent with $\\sqrt[3]{P} \\leq \\sqrt[3]{\\frac{S}{3}}$. Therefore the system cannot have a real positive solution.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 27829,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nProve that when three circles share the same chord $A B$, every line through $A$ different from $A B$ determines the same ratio $X Y: Y Z$, where $X$ is an arbitrary point different from $B$ on the first circle while $Y$ and $Z$ are the points where $A X$ intersects the other two circles (labelled so that $Y$ is between $X$ and $Z$).\n![](attached_image_1.png)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\nLet $l$ be a line through $A$ different from $A B$ and join $B$ to $A, X, Y$ and $Z$ as in the above diagram. No matter how $l$ is chosen, the angles $A X B, A Y B$ and $A Z B$ always subtend the chord $A B$. For this reason the angles in the triangles $B X Y$ and $B X Z$ are the same for all such $l$. Thus the ratio $X Y: Y Z$ remains constant by similar triangles.\n\nNote that this is true no matter how $X, Y$ and $Z$ lie in relation to $A$. Suppose $X, Y$ and $Z$ all lie on the same side of $A$ (as in the diagram) and that $\\measuredangle A X B=\\alpha, \\measuredangle A Y B=\\beta$ and $\\measuredangle A Z B=\\gamma$. Then $\\measuredangle B X Y=180^\\circ-\\alpha, \\measuredangle B Y X=\\beta, \\measuredangle B Y Z=180^\\circ-\\beta$ and $\\measuredangle B Z Y=\\gamma$. Now suppose $l$ is chosen so that $X$ is now on the opposite side of $A$ from $Y$ and $Z$. Now since $X$ is on the other side of the chord $A B, \\measuredangle A X B=180^\\circ-\\alpha$, but it is still the case that $\\measuredangle B X Y=180^\\circ-\\alpha$ and all other angles in the two pertinent triangles remain unchanged. If $l$ is chosen so that $X$ is identical with $A$, then $l$ is tangent to the first circle and it is still the case that $\\measuredangle B X Y=180^\\circ-\\alpha$. All other cases can be checked in a similar manner.\n![](attached_image_2.png)\nSolution:\nLet $m$ be the perpendicular bisector of $A B$ and let $O_{1}, O_{2}, O_{3}$ be the centres of the three circles. Since $A B$ is a chord common to all three circles, $O_{1}, O_{2}, O_{3}$ all lie on $m$. Let $l$ be a line through $A$ different from $A B$ and suppose that $X, Y, Z$ all lie on the same side of $A B$, as in the above diagram. Let perpendiculars from $O_{1}, O_{2}, O_{3}$ meet $l$ at $P, Q, R$, respectively. Since a line through the centre of a circle bisects any chord,\n$$\nA X=2 A P, \\quad A Y=2 A Q \\quad \\text{ and } \\quad A Z=2 A R\n$$\nNow\n$$\nX Y=A Y-A X=2(A Q-A P)=2 P Q \\quad \\text{ and, similarly, } \\quad Y Z=2 Q R.\n$$\nTherefore $X Y: Y Z=P Q: Q R$. But $O_{1} P\\parallel O_{2} Q\\parallel O_{3} R$, so $P Q: Q R=O_{1} O_{2}: O_{2} O_{3}$. Since the centres of the circles are fixed, the ratio $X Y: Y Z=O_{1} O_{2}: O_{2} O_{3}$ does not depend on the choice of $l$.\nIf $X, Y, Z$ do not all lie on the same side of $A B$, we can obtain the same result with a similar proof. For instance, if $X$ and $Y$ are opposite sides of $A B$, then we will have $X Y=A Y+A X$, but since in this case $P Q=A Q+A P$, it is still the case that $X Y=2 P Q$ and result still follows, etc.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 27830,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nLet $S$ be a set of $n$ points in the plane such that any two points of $S$ are at least $1$ unit apart. Prove there is a subset $T$ of $S$ with at least $n / 7$ points such that any two points of $T$ are at least $\\sqrt{3}$ units apart.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\nWe will construct the set $T$ in the following way: Assume the points of $S$ are in the $xy$-plane and let $P$ be a point in $S$ with maximum $y$-coordinate. This point $P$ will be a member of the set $T$ and now, from $S$, we will remove $P$ and all points in $S$ which are less than $\\sqrt{3}$ units from $P$. From the remaining points we choose one with maximum $y$-coordinate to be a member of $T$ and remove from $S$ all points at distance less than $\\sqrt{3}$ units from this new point. We continue in this way, until all the points of $S$ are exhausted. Clearly any two points in $T$ are at least $\\sqrt{3}$ units apart. To show that $T$ has at least $n / 7$ points, we must prove that at each stage no more than $6$ other points are removed along with $P$.\n\nAt a typical stage in this process, we've selected a point $P$ with maximum $y$-coordinate, so any points at distance less than $\\sqrt{3}$ from $P$ must lie inside the semicircular region of radius $\\sqrt{3}$ centred at $P$ shown in the first diagram below. Since points of $S$ are at least $1$ unit apart, these points must lie outside (or on) the semicircle of radius $1$. (So they lie in the shaded region of the first diagram.) Now divide this shaded region into $6$ congruent regions $R_{1}, R_{2}, \\ldots, R_{6}$ as shown in this diagram.\n\nWe will show that each of these regions contains at most one point of $S$. Since all $6$ regions are congruent, consider one of them as depicted in the second diagram below. The distance between any two points in this shaded region must be less than the length of the line segment $AB$. The lengths of $PA$ and $PB$ are $\\sqrt{3}$ and $1$, respectively, and angle $APB=30^{\\circ}$. If we construct a perpendicular from $B$ to $PA$ at $C$, then the length of $PC$ is $\\cos 30^{\\circ}=\\sqrt{3} / 2$. Thus $BC$ is a perpendicular bisector of $PA$ and therefore $AB=PB=1$. So the distance between any two points in this region is less than $1$. Therefore each of $R_{1}, \\ldots, R_{6}$ can contain at most one point of $S$, which completes the proof.\n\n![](attached_image_1.png)",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 27831,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nFind all ordered triples $(x, y, z)$ of real numbers which satisfy the following system of equations:\n$$\n\\left\\{\\begin{array}{l}\nx y = z - x - y \\\\\nx z = y - x - z \\\\\ny z = x - y - z\n\\end{array}\\right.\n$$",
    "options": [],
    "answer": "(-2, 0, -2), (-2, -2, 0), (0, 0, 0), (0, -2, -2), (-1, -1, -1)",
    "solution": "Solution:\nSubtracting the second equation from the first gives $x y - x z = 2 z - 2 y$. Factoring $y - z$ from each side and rearranging gives\n$$\n(x + 2)(y - z) = 0\n$$\nso either $x = -2$ or $z = y$.\n\nIf $x = -2$, the first equation becomes $-2 y = z + 2 - y$, or $y + z = -2$. Substituting $x = -2$, $y + z = -2$ into the third equation gives $y z = -2 - (-2) = 0$. Hence either $y$ or $z$ is $0$, so if $x = -2$, the only solutions are $(-2, 0, -2)$ and $(-2, -2, 0)$.\n\nIf $z = y$ the first equation becomes $x y = -x$, or $x(y + 1) = 0$. If $x = 0$ and $z = y$, the third equation becomes $y^2 = -2 y$ which gives $y = 0$ or $y = -2$. If $y = -1$ and $z = y = -1$, the third equation gives $x = -1$. So if $y = z$, the only solutions are $(0, 0, 0)$, $(0, -2, -2)$ and $(-1, -1, -1)$.\n\nIn summary, there are 5 solutions: $(-2, 0, -2)$, $(-2, -2, 0)$, $(0, 0, 0)$, $(0, -2, -2)$ and $(-1, -1, -1)$.\nSolution:\nAdding $x$ to both sides of the first equation gives\n$$\nx(y + 1) = z - y = (z + 1) - (y + 1) \\Rightarrow (x + 1)(y + 1) = z + 1.\n$$\nSimilarly manipulating the other two equations and letting $a = x + 1$, $b = y + 1$, $c = z + 1$, we can write the system in the following way.\n$$\n\\left\\{\\begin{array}{l}\na b = c \\\\\na c = b \\\\\nb c = a\n\\end{array}\\right.\n$$\nIf any one of $a, b, c$ is $0$, then it's clear that all three are $0$. So $(a, b, c) = (0, 0, 0)$ is one solution and now suppose that $a, b, c$ are all nonzero. Substituting $c = a b$ into the second and third equations gives $a^2 b = b$ and $b^2 a = a$, respectively. Hence $a^2 = 1$, $b^2 = 1$ (since $a, b$ nonzero). This gives 4 more solutions: $(a, b, c) = (1, 1, 1)$, $(1, -1, -1)$, $(-1, 1, -1)$ or $(-1, -1, 1)$. Reexpressing in terms of $x, y, z$, we obtain the 5 ordered triples listed in Solution 1.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 27832,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nLet $A, B, C, D$ be four points on a circle (occurring in clockwise order), with $A B < A D$ and $B C > C D$. Let the bisector of angle $B A D$ meet the circle at $X$ and the bisector of angle $B C D$ meet the circle at $Y$. Consider the hexagon formed by these six points on the circle. If four of the six sides of the hexagon have equal length, prove that $B D$ must be a diameter of the circle.\n![](attached_image_1.png)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\nWe're given that $A B < A D$. Since $C Y$ bisects $\\measuredangle B C D$, $B Y = Y D$, so $Y$ lies between $D$ and $A$ on the circle, as in the diagram above, and $D Y > Y A$, $D Y > A B$. Similar reasoning confirms that $X$ lies between $B$ and $C$ and $B X > X C$, $B X > C D$. So if $A B X C D Y$ has 4 equal sides, then it must be that $Y A = A B = X C = C D$.\nLet $\\measuredangle B A X = \\measuredangle D A X = \\alpha$ and let $\\measuredangle B C Y = \\measuredangle D C Y = \\gamma$. Since $A B C D$ is cyclic, $\\measuredangle A + \\measuredangle C = 180^{\\circ}$, which implies that $\\alpha + \\gamma = 90^{\\circ}$. The fact that $Y A = A B = X C = C D$ means that the arc from $Y$ to $B$ (which is subtended by $\\measuredangle Y C B$) is equal to the arc from $X$ to $D$ (which is subtended by $\\measuredangle X A D$). Hence $\\measuredangle Y C B = \\measuredangle X A D$, so $\\alpha = \\gamma = 45^{\\circ}$. Finally, $B D$ is subtended by $\\measuredangle B A D = 2 \\alpha = 90^{\\circ}$. Therefore $B D$ is a diameter of the circle.\nSolution:\nWe're given that $A B < A D$. Since $C Y$ bisects $\\measuredangle B C D$, $B Y = Y D$, so $Y$ lies between $D$ and $A$ on the circle, as in the diagram above, and $D Y > Y A$, $D Y > A B$. Similar reasoning confirms that $X$ lies between $B$ and $C$ and $B X > X C$, $B X > C D$. So if $A B X C D Y$ has 4 equal sides, then it must be that $Y A = A B = X C = C D$. This implies that the arc from $Y$ to $B$ is equal to the arc from $X$ to $D$ and hence that $Y B = X D$. Since $\\measuredangle B A X = \\measuredangle X A D$, $B X = X D$ and since $\\measuredangle D C Y = \\measuredangle Y C B$, $D Y = Y B$. Therefore $B X D Y$ is a square and its diagonal, $B D$, must be a diameter of the circle.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 27833,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nLet $p$ be an odd prime. Prove that\n$$\n\\sum_{k=1}^{p-1} k^{2p-1} \\equiv \\frac{p(p+1)}{2} \\quad\\left(\\bmod p^{2}\\right)\n$$\n[Note that $a \\equiv b(\\bmod m)$ means that $a-b$ is divisible by $m$.]",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\nSince $p-1$ is even, we can pair up the terms in the summation in the following way (first term with last, 2nd term with 2nd last, etc.):\n$$\n\\sum_{k=1}^{p-1} k^{2p-1} = \\sum_{k=1}^{\\frac{p-1}{2}} \\left(k^{2p-1} + (p-k)^{2p-1}\\right)\n$$\nExpanding $(p-k)^{2p-1}$ with the binomial theorem, we get\n$$\n(p-k)^{2p-1} = p^{2p-1} - \\cdots - \\binom{2p-1}{2} p^{2} k^{2p-3} + \\binom{2p-1}{1} p k^{2p-2} - k^{2p-1},\n$$\nwhere every term on the right-hand side is divisible by $p^{2}$ except the last two. Therefore\n$$\nk^{2p-1} + (p-k)^{2p-1} \\equiv k^{2p-1} + \\binom{2p-1}{1} p k^{2p-2} - k^{2p-1} \\equiv (2p-1) p k^{2p-2} \\pmod{p^{2}}\n$$\nFor $1 \\leq k < p$, $k$ is not divisible by $p$, so $k^{p-1} \\equiv 1 \\pmod{p}$, by Fermat's Little Theorem. So $(2p-1) k^{2p-2} \\equiv (2p-1)(1^{2}) \\equiv -1 \\pmod{p}$, say $(2p-1) k^{2p-2} = m p - 1$ for some integer $m$. Then\n$$\n(2p-1) p k^{2p-2} = m p^{2} - p \\equiv -p \\pmod{p^{2}}\n$$\nFinally,\n$$\n\\begin{aligned}\n\\sum_{k=1}^{p-1} k^{2p-1} &\\equiv \\sum_{k=1}^{\\frac{p-1}{2}} (-p) \\equiv \\left(\\frac{p-1}{2}\\right)(-p) \\pmod{p^{2}} \\\\\n&\\equiv \\frac{p-p^{2}}{2} + p^{2} \\equiv \\frac{p(p+1)}{2} \\pmod{p^{2}}.\n\\end{aligned}",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 27834,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nLet $T$ be the set of all positive integer divisors of $2004^{100}$. What is the largest possible number of elements that a subset $S$ of $T$ can have if no element of $S$ is an integer multiple of any other element of $S$ ?",
    "options": [],
    "answer": "10201",
    "solution": "Solution:\nAssume throughout that $a, b, c$ are nonnegative integers. Since the prime factorization of $2004$ is $2004 = 2^{2} \\cdot 3 \\cdot 167$,\n$$\nT = \\left\\{ 2^{a} 3^{b} 167^{c} \\mid 0 \\leq a \\leq 200, 0 \\leq b, c \\leq 100 \\right\\}\n$$\nLet\n$$\nS = \\left\\{ 2^{200-b-c} 3^{b} 167^{c} \\mid 0 \\leq b, c \\leq 100 \\right\\}\n$$\nFor any $0 \\leq b, c \\leq 100$, we have $0 \\leq 200-b-c \\leq 200$, so $S$ is a subset of $T$. Since there are $101$ possible values for $b$ and $101$ possible values for $c$, $S$ contains $101^{2}$ elements. We will show that no element of $S$ is a multiple of another and that no larger subset of $T$ satisfies this condition.\n\nSuppose $2^{200-b-c} 3^{b} 167^{c}$ is an integer multiple of $2^{200-j-k} 3^{j} 167^{k}$. Then\n$$\n200-b-c \\geq 200-j-k, \\quad b \\geq j, \\quad c \\geq k\n$$\nBut this first inequality implies $b+c \\leq j+k$, which together with $b \\geq j, c \\geq k$ gives $b = j$ and $c = k$. Hence no element of $S$ is an integer multiple of another element of $S$.\n\nLet $U$ be a subset of $T$ with more than $101^{2}$ elements. Since there are only $101^{2}$ distinct pairs $(b, c)$ with $0 \\leq b, c \\leq 100$, then (by the pigeonhole principle) $U$ must contain two elements $u_{1} = 2^{a_{1}} 3^{b_{1}} 167^{c_{1}}$ and $u_{2} = 2^{a_{2}} 3^{b_{2}} 167^{c_{2}}$, with $b_{1} = b_{2}$ and $c_{1} = c_{2}$, but $a_{1} \\neq a_{2}$. If $a_{1} > a_{2}$, then $u_{1}$ is a multiple of $u_{2}$ and if $a_{1} < a_{2}$, then $u_{2}$ is a multiple of $u_{1}$. Hence $U$ does not satisfy the desired condition.\n\nTherefore the largest possible number of elements that such a subset of $T$ can have is $101^{2} = 10201$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 27835,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nConsider an equilateral triangle of side length $n$, which is divided into unit triangles, as shown. Let $f(n)$ be the number of paths from the triangle in the top row to the middle triangle in the bottom row, such that adjacent triangles in our path share a common edge and the path never travels up (from a lower row to a higher row) or revisits a triangle. An example of one such path is illustrated below for $n=5$. Determine the value of $f(2005)$.\n\n![](attached_image_1.png)",
    "options": [],
    "answer": "2004!",
    "solution": "Solution:\n\nWe shall show that $f(n) = (n-1)!$.\n\nLabel the horizontal line segments in the triangle $l_{1}, l_{2}, \\ldots$ as in the diagram below. Since the path goes from the top triangle to a triangle in the bottom row and never travels up, the path must cross each of $l_{1}, l_{2}, \\ldots, l_{n-1}$ exactly once. The diagonal lines in the triangle divide $l_{k}$ into $k$ unit line segments and the path must cross exactly one of these $k$ segments for each $k$. (In the diagram below, these line segments have been highlighted.) The path is completely determined by the set of $n-1$ line segments which are crossed. So as the path moves from the $k$th row to the $(k+1)$st row, there are $k$ possible line segments where the path could cross $l_{k}$. Since there are $1 \\cdot 2 \\cdot 3 \\cdots (n-1) = (n-1)!$ ways that the path could cross the $n-1$ horizontal lines, and each one corresponds to a unique path, we get $f(n) = (n-1)!$.\n\nTherefore $f(2005) = (2004)!$.\n\n![](attached_image_2.png)",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 27836,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nLet $(a, b, c)$ be a Pythagorean triple, i.e., a triplet of positive integers with $a^{2}+b^{2}=c^{2}$.\n\na) Prove that $(c / a+c / b)^{2}>8$.\n\nb) Prove that there does not exist any integer $n$ for which we can find a Pythagorean triple $(a, b, c)$ satisfying $(c / a+c / b)^{2}=n$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\nLet $(a, b, c)$ be a Pythagorean triple. View $a, b$ as lengths of the legs of a right angled triangle with hypotenuse of length $c$; let $\\theta$ be the angle determined by the sides with lengths $a$ and $c$. Then\n$$\n\\begin{aligned}\n\\left(\\frac{c}{a}+\\frac{c}{b}\\right)^{2} & =\\left(\\frac{1}{\\cos \\theta}+\\frac{1}{\\sin \\theta}\\right)^{2}=\\frac{\\sin ^{2} \\theta+\\cos ^{2} \\theta+2 \\sin \\theta \\cos \\theta}{(\\sin \\theta \\cos \\theta)^{2}} \\\\\n& =4\\left(\\frac{1+\\sin 2 \\theta}{\\sin ^{2} 2 \\theta}\\right)=\\frac{4}{\\sin ^{2} 2 \\theta}+\\frac{4}{\\sin 2 \\theta}\n\\end{aligned}\n$$\nNote that because $0<\\theta<90^{\\circ}$, we have $0<\\sin 2 \\theta \\leq 1$, with equality only if $\\theta=45^{\\circ}$. But then $a=b$ and we obtain $\\sqrt{2}=c / a$, contradicting $a, c$ both being integers. Thus, $0<\\sin 2 \\theta<1$ which gives $(c / a+c / b)^{2}>8$.\n\nDefining $\\theta$ as in Solution 1, we have $c / a+c / b=\\sec \\theta+\\csc \\theta$. By the AM-GM inequality, we have $(\\sec \\theta+\\csc \\theta) / 2 \\geq \\sqrt{\\sec \\theta \\csc \\theta}$. So\n$$\nc / a+c / b \\geq \\frac{2}{\\sqrt{\\sin \\theta \\cos \\theta}}=\\frac{2 \\sqrt{2}}{\\sqrt{\\sin 2 \\theta}} \\geq 2 \\sqrt{2}\n$$\nSince $a, b, c$ are integers, we have $c / a+c / b>2 \\sqrt{2}$ which gives $(c / a+c / b)^{2}>8$.\n\nBy simplifying and using the AM-GM inequality,\n$$\n\\left(\\frac{c}{a}+\\frac{c}{b}\\right)^{2}=c^{2}\\left(\\frac{a+b}{a b}\\right)^{2}=\\frac{\\left(a^{2}+b^{2}\\right)(a+b)^{2}}{a^{2} b^{2}} \\geq \\frac{2 \\sqrt{a^{2} b^{2}}(2 \\sqrt{a b})^{2}}{a^{2} b^{2}}=8\n$$\nwith equality only if $a=b$. By using the same argument as in Solution $1, a$ cannot equal $b$ and the inequality is strict.\n\n$$\n\\begin{aligned}\n\\left(\\frac{c}{a}+\\frac{c}{b}\\right)^{2} & =\\frac{c^{2}}{a^{2}}+\\frac{c^{2}}{b^{2}}+\\frac{2 c^{2}}{a b}=1+\\frac{b^{2}}{a^{2}}+\\frac{a^{2}}{b^{2}}+1+\\frac{2\\left(a^{2}+b^{2}\\right)}{a b} \\\\\n& =2+\\left(\\frac{a}{b}-\\frac{b}{a}\\right)^{2}+2+\\frac{2}{a b}\\left((a-b)^{2}+2 a b\\right) \\\\\n& =4+\\left(\\frac{a}{b}-\\frac{b}{a}\\right)^{2}+\\frac{2(a-b)^{2}}{a b}+4 \\geq 8,\n\\end{aligned}\n$$\nwith equality only if $a=b$, which (as argued previously) cannot occur.\n\nSince $c / a+c / b$ is rational, $(c / a+c / b)^{2}$ can only be an integer if $c / a+c / b$ is an integer. Suppose $c / a+c / b=m$. We may assume that $\\operatorname{gcd}(a, b)=1$. (If not, divide the common factor from $(a, b, c)$, leaving $m$ unchanged.)\nSince $c(a+b)=m a b$ and $\\operatorname{gcd}(a, a+b)=1$, $a$ must divide $c$, say $c=a k$. This gives $a^{2}+b^{2}=a^{2} k^{2}$ which implies $b^{2}=\\left(k^{2}-1\\right) a^{2}$. But then $a$ divides $b$ contradicting the fact that $\\operatorname{gcd}(a, b)=1$. Therefore $(c / a+c / b)^{2}$ is not equal to any integer $n$.\n\nWe begin as in Solution 1, supposing that $c / a+c / b=m$ with $\\operatorname{gcd}(a, b)=1$. Hence $a$ and $b$ are not both even. It is also the case that $a$ and $b$ are not both odd, for then $c^{2}=a^{2}+b^{2} \\equiv 2(\\bmod 4)$, and perfect squares are congruent to either 0 or 1 modulo 4. So one of $a, b$ is odd and the other is even. Therefore $c$ must be odd.\nNow $c / a+c / b=m$ implies $c(a+b)=m a b$, which cannot be true because $c(a+b)$ is odd and $m a b$ is even.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 27837,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nLet $S$ be a set of $n \\geq 3$ points in the interior of a circle.\n\na) Show that there are three distinct points $a, b, c \\in S$ and three distinct points $A, B, C$ on the circle such that $a$ is (strictly) closer to $A$ than any other point in $S$, $b$ is closer to $B$ than any other point in $S$ and $c$ is closer to $C$ than any other point in $S$.\n\nb) Show that for no value of $n$ can four such points in $S$ (and corresponding points on the circle) be guaranteed.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\na) Let $H$ be the smallest convex set of points in the plane which contains $S$. Take 3 points $a, b, c \\in S$ which lie on the boundary of $H$. (There must always be at least 3 (but not necessarily 4) such points.)\n\nSince $a$ lies on the boundary of the convex region $H$, we can construct a chord $L$ such that no two points of $H$ lie on opposite sides of $L$. Of the two points where the perpendicular to $L$ at $a$ meets the circle, choose one which is on a side of $L$ not containing any points of $H$ and call this point $A$. Certainly $A$ is closer to $a$ than to any other point on $L$ or on the other side of $L$. Hence $A$ is closer to $a$ than to any other point of $S$. We can find the required points $B$ and $C$ in an analogous way and the proof is complete.\n\n![](attached_image_1.png)\n\n(a)\n\n![](attached_image_2.png)\n\n(b)\n\nb) Let $P Q R$ be an equilateral triangle inscribed in the circle and let $a, b, c$ be midpoints of the three sides of $\\triangle P Q R$. If $r$ is the radius of the circle, then every point on the circle is within $(\\sqrt{3} / 2) r$ of one of $a, b$ or $c$. (See figure (b) above.) Now $\\sqrt{3} / 2<9 / 10$, so if $S$ consists of $a, b, c$ and a cluster of points within $r / 10$ of the centre of the circle, then we cannot select 4 points from $S$ (and corresponding points on the circle) having the desired property.\nSolution:\n\na) If all the points of $S$ lie on a line $L$, then choose any 3 of them to be $a, b, c$. Let $A$ be a point on the circle which meets the perpendicular to $L$ at $a$. Clearly $A$ is closer to $a$ than to any other point on $L$, and hence closer than any other point in $S$. We find $B$ and $C$ in an analogous way.\n\nOtherwise, choose $a, b, c$ from $S$ so that the triangle formed by these points has maximal area. Construct the altitude from the side $b c$ to the point $a$ and extend this line until it meets the circle at $A$. We claim that $A$ is closer to $a$ than to any other point in $S$.\n\nSuppose not. Let $x$ be a point in $S$ for which the distance from $A$ to $x$ is less than the distance from $A$ to $a$. Then the perpendicular distance from $x$ to the line $b c$ must be greater than the perpendicular distance from $a$ to the line $b c$. But then the triangle formed by the points $x, b, c$ has greater area than the triangle formed by $a, b, c$, contradicting the original choice of these 3 points. Therefore $A$ is closer to $a$ than to any other point in $S$.\n\nThe points $B$ and $C$ are found by constructing similar altitudes through $b$ and $c$, respectively.\n\nb) See Solution 1.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 27838,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nLet $ABC$ be a triangle with circumradius $R$, perimeter $P$ and area $K$. Determine the maximum value of $K P / R^{3}$.",
    "options": [],
    "answer": "27/4",
    "solution": "Solution:\nSince similar triangles give the same value of $K P / R^{3}$, we can fix $R=1$ and maximize $K P$ over all triangles inscribed in the unit circle. Fix points $A$ and $B$ on the unit circle. The locus of points $C$ with a given perimeter $P$ is an ellipse that meets the circle in at most four points. The area $K$ is maximized (for a fixed $P$) when $C$ is chosen on the perpendicular bisector of $AB$, so we get a maximum value for $K P$ if $C$ is where the perpendicular bisector of $AB$ meets the circle. Thus the maximum value of $K P$ for a given $AB$ occurs when $ABC$ is an isosceles triangle. Repeating this argument with $BC$ fixed, we have that the maximum occurs when $ABC$ is an equilateral triangle.\n\nConsider an equilateral triangle with side length $a$. It has $P=3a$. It has height equal to $a \\sqrt{3} / 2$ giving $K = a^{2} \\sqrt{3} / 4$. From the extended law of sines, $2R = a / \\sin(60^{\\circ})$ giving $R = a / \\sqrt{3}$. Therefore the maximum value we seek is\n$$\nK P / R^{3} = \\left(\\frac{a^{2} \\sqrt{3}}{4}\\right)(3a)\\left(\\frac{\\sqrt{3}}{a}\\right)^{3} = \\frac{27}{4}.\n$$\n\nFrom the extended law of sines, the lengths of the sides of the triangle are $2R \\sin A$, $2R \\sin B$ and $2R \\sin C$. So\n$$\nP = 2R(\\sin A + \\sin B + \\sin C) \\text{ and } K = \\frac{1}{2}(2R \\sin A)(2R \\sin B)(\\sin C),\n$$\ngiving\n$$\n\\frac{K P}{R^{3}} = 4 \\sin A \\sin B \\sin C (\\sin A + \\sin B + \\sin C)\n$$\nWe wish to find the maximum value of this expression over all $A+B+C=180^{\\circ}$. Using well-known identities for sums and products of sine functions, we can write\n$$\n\\frac{K P}{R^{3}} = 4 \\sin A \\left(\\frac{\\cos(B-C)}{2} - \\frac{\\cos(B+C)}{2}\\right) \\left(\\sin A + 2 \\sin\\left(\\frac{B+C}{2}\\right) \\cos\\left(\\frac{B-C}{2}\\right)\\right).\n$$\nIf we first consider $A$ to be fixed, then $B+C$ is fixed also and this expression takes its maximum value when $\\cos(B-C)$ and $\\cos\\left(\\frac{B-C}{2}\\right)$ equal $1$; i.e. when $B=C$. In a similar way, one can show that for any fixed value of $B$, $K P / R^{3}$ is maximized when $A=C$. Therefore the maximum value of $K P / R^{3}$ occurs when $A=B=C=60^{\\circ}$, and it is now an easy task to substitute this into the above expression to obtain the maximum value of $27/4$.\n\nAs in Solution 2, we obtain\n$$\n\\frac{K P}{R^{3}} = 4 \\sin A \\sin B \\sin C (\\sin A + \\sin B + \\sin C)\n$$\nFrom the AM-GM inequality, we have\n$$\n\\sin A \\sin B \\sin C \\leq \\left(\\frac{\\sin A + \\sin B + \\sin C}{3}\\right)^{3},\n$$\ngiving\n$$\n\\frac{K P}{R^{3}} \\leq \\frac{4}{27}(\\sin A + \\sin B + \\sin C)^{4}\n$$\nwith equality when $\\sin A = \\sin B = \\sin C$. Since the sine function is concave on the interval from $0$ to $\\pi$, Jensen's inequality gives\n$$\n\\frac{\\sin A + \\sin B + \\sin C}{3} \\leq \\sin\\left(\\frac{A+B+C}{3}\\right) = \\sin \\frac{\\pi}{3} = \\frac{\\sqrt{3}}{2}.\n$$\nSince equality occurs here when $\\sin A = \\sin B = \\sin C$ also, we can conclude that the maximum value of $K P / R^{3}$ is $\\frac{4}{27}\\left(\\frac{3 \\sqrt{3}}{2}\\right)^{4} = 27/4$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 27839,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nLet's say that an ordered triple of positive integers $(a, b, c)$ is $n$-powerful if $a \\leq b \\leq c$, $\\operatorname{gcd}(a, b, c)=1$, and $a^{n}+b^{n}+c^{n}$ is divisible by $a+b+c$. For example, $(1,2,2)$ is 5-powerful.\n\na) Determine all ordered triples (if any) which are $n$-powerful for all $n \\geq 1$.\n\nb) Determine all ordered triples (if any) which are 2004-powerful and 2005-powerful, but not 2007-powerful.\n\n[Note that $\\operatorname{gcd}(a, b, c)$ is the greatest common divisor of $a, b$ and $c$.]",
    "options": [],
    "answer": "a) (1, 1, 1) and (1, 1, 4). b) No such triples exist.",
    "solution": "Solution:\n\nLet $T_{n}=a^{n}+b^{n}+c^{n}$ and consider the polynomial\n$$\nP(x)=(x-a)(x-b)(x-c)=x^{3}-(a+b+c) x^{2}+(a b+a c+b c) x-a b c .\n$$\nSince $P(a)=0$, we get $a^{3}=(a+b+c) a^{2}-(a b+a c+b c) a+a b c$ and multiplying both sides by $a^{n-3}$ we obtain $a^{n}=(a+b+c) a^{n-1}-(a b+a c+b c) a^{n-2}+(a b c) a^{n-3}$. Applying the same reasoning, we can obtain similar expressions for $b^{n}$ and $c^{n}$ and adding the three identities we get that $T_{n}$ satisfies the following 3-term recurrence:\n$$\nT_{n}=(a+b+c) T_{n-1}-(a b+a c+b c) T_{n-2}+(a b c) T_{n-3} \\text{, for all } n \\geq 3\n$$\nFrom this we see that if $T_{n-2}$ and $T_{n-3}$ are divisible by $a+b+c$, then so is $T_{n}$. This immediately resolves part (b) - there are no ordered triples which are 2004-powerful and 2005-powerful, but not 2007-powerful - and reduces the number of cases to be considered in part (a): since all triples are 1-powerful, the recurrence implies that any ordered triple which is both 2-powerful and 3-powerful is $n$-powerful for all $n \\geq 1$.\n\nPutting $n=3$ in the recurrence, we have\n$$\na^{3}+b^{3}+c^{3}=(a+b+c)\\left(a^{2}+b^{2}+c^{2}\\right)-(a b+a c+b c)(a+b+c)+3 a b c\n$$\nwhich implies that $(a, b, c)$ is 3-powerful if and only if $3 a b c$ is divisible by $a+b+c$. Since\n$$\na^{2}+b^{2}+c^{2}=(a+b+c)^{2}-2(a b+a c+b c),\n$$\n$(a, b, c)$ is 2-powerful if and only if $2(a b+a c+b c)$ is divisible by $a+b+c$.\n\nSuppose a prime $p \\geq 5$ divides $a+b+c$. Then $p$ divides $a b c$. Since $\\operatorname{gcd}(a, b, c)=1$, $p$ divides exactly one of $a, b$ or $c$; but then $p$ doesn't divide $2(a b+a c+b c)$.\n\nSuppose $3^{2}$ divides $a+b+c$. Then 3 divides $a b c$, implying 3 divides exactly one of $a$, $b$ or $c$. But then 3 doesn't divide $2(a b+a c+b c)$.\n\nSuppose $2^{2}$ divides $a+b+c$. Then 4 divides $a b c$. Since $\\operatorname{gcd}(a, b, c)=1$, at most one of $a, b$ or $c$ is even, implying one of $a, b, c$ is divisible by 4 and the others are odd. But then $a b+a c+b c$ is odd and 4 doesn't divide $2(a b+a c+b c)$.\n\nSo if $(a, b, c)$ is 2- and 3-powerful, then $a+b+c$ is not divisible by 4 or 9 or any prime greater than 3. Since $a+b+c$ is at least $3$, $a+b+c$ is either 3 or 6. It is now a simple matter to check the possibilities and conclude that the only triples which are $n$-powerful for all $n \\geq 1$ are $(1,1,1)$ and $(1,1,4)$.\n\nLet $p$ be a prime. By Fermat's Little Theorem,\n$$\na^{p-1} \\equiv \\begin{cases}1(\\bmod p), & \\text{ if } p \\text{ doesn't divide } a \\\\ 0(\\bmod p), & \\text{ if } p \\text{ divides } a\\end{cases}\n$$\nSince $\\operatorname{gcd}(a, b, c)=1$, we have that $a^{p-1}+b^{p-1}+c^{p-1} \\equiv 1,2$ or $3(\\bmod p)$. Therefore if $p$ is a prime divisor of $a^{p-1}+b^{p-1}+c^{p-1}$, then $p$ equals 2 or 3. So if $(a, b, c)$ is $n$-powerful for all $n \\geq 1$, then the only primes which can divide $a+b+c$ are 2 or 3.\n\nWe can proceed in a similar fashion to show that $a+b+c$ is not divisible by 4 or 9.\n\nSince\n$$\na^{2} \\equiv \\begin{cases}0(\\bmod 4), & \\text{ if } a \\text{ is even;} \\\\ 1(\\bmod 4), & \\text{ if } a \\text{ is odd }\\end{cases}\n$$\nand $a, b, c$ aren't all even, we have that $a^{2}+b^{2}+c^{2} \\equiv 1,2$ or $3(\\bmod 4)$.\n\nBy expanding $(3 k)^{3},(3 k+1)^{3}$ and $(3 k+2)^{3}$, we find that $a^{3}$ is congruent to 0, 1 or $-1$ modulo 9. Hence\n$$\na^{6} \\equiv \\begin{cases}0(\\bmod 9), & \\text{ if } 3 \\text{ divides } a ; \\\\ 1(\\bmod 9), & \\text{ if } 3 \\text{ doesn't divide } a .\\end{cases}\n$$\nSince $a, b, c$ aren't all divisible by 3, we have that $a^{6}+b^{6}+c^{6} \\equiv 1,2$ or $3(\\bmod 9)$.\n\nSo $a^{2}+b^{2}+c^{2}$ is not divisible by 4 and $a^{6}+b^{6}+c^{6}$ is not divisible by 9. Thus if $(a, b, c)$ is $n$-powerful for all $n \\geq 1$, then $a+b+c$ is not divisible by 4 or 9. Therefore $a+b+c$ is either 3 or 6 and checking all possibilities, we conclude that the only triples which are $n$-powerful for all $n \\geq 1$ are $(1,1,1)$ and $(1,1,4)$.\n\nSee Solution for the (b) part.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 27840,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nLet $f(n, k)$ be the number of ways of distributing $k$ candies to $n$ children so that each child receives at most 2 candies. For example, if $n=3$, then $f(3,7)=0$, $f(3,6)=1$ and $f(3,4)=6$.\nDetermine the value of\n$$\nf(2006,1)+f(2006,4)+f(2006,7)+\\cdots+f(2006,1000)+f(2006,1003) .\n$$",
    "options": [],
    "answer": "∑_{i=1}^{334} (-1)^i \\binom{2005}{i} \\binom{3008 - 3i}{2005}",
    "solution": "Solution:\nThe number of ways of distributing $k$ candies to $2006$ children is equal to the number of ways of distributing $0$ to a particular child and $k$ to the rest, plus the number of ways of distributing $1$ to the particular child and $k-1$ to the rest, plus the number of ways of distributing $2$ to the particular child and $k-2$ to the rest. Thus $f(2006, k) = f(2005, k) + f(2005, k-1) + f(2005, k-2)$, so that the required sum is\n$$\n1 + \\sum_{k=1}^{1003} f(2005, k)\n$$\nIn evaluating $f(n, k)$, suppose that there are $r$ children who receive $2$ candies; these $r$ children can be chosen in $\\binom{n}{r}$ ways. Then there are $k-2r$ candies from which at most one is given to each of $n-r$ children. Hence\n$$\nf(n, k) = \\sum_{r=0}^{\\lfloor k/2 \\rfloor} \\binom{n}{r} \\binom{n-r}{k-2r} = \\sum_{r=0}^{\\infty} \\binom{n}{r} \\binom{n-r}{k-2r}\n$$\nwith $\\binom{x}{y} = 0$ when $x < y$ and when $y < 0$. The answer is\n$$\n\\sum_{k=0}^{1003} \\sum_{r=0}^{\\infty} \\binom{2005}{r} \\binom{2005-r}{k-2r} = \\sum_{r=0}^{\\infty} \\binom{2005}{r} \\sum_{k=0}^{1003} \\binom{2005-r}{k-2r}\n$$\nSolution:\nThe desired number is the sum of the coefficients of the terms of degree not exceeding $1003$ in the expansion of $\\left(1+x+x^{2}\\right)^{2005}$, which is equal to the coefficient of $x^{1003}$ in the expansion of\n$$\n\\left(1+x+x^{2}\\right)^{2005}\\left(1+x+\\cdots+x^{1003}\\right) = \\left[\\left(1-x^{3}\\right)^{2005}(1-x)^{-2005}\\right]\\left(1-x^{1004}\\right)(1-x)^{-1}\n$$\n$$\n= \\left(1-x^{3}\\right)^{2005}(1-x)^{-2006} - \\left(1-x^{3}\\right)^{2005}(1-x)^{-2006} x^{1004}\n$$\nSince the degree of every term in the expansion of the second member on the right exceeds $1003$, we are looking for the coefficient of $x^{1003}$ in the expansion of the first member:\n$$\n\\left(1-x^{3}\\right)^{2005}(1-x)^{-2006} = \\sum_{i=0}^{2005} (-1)^i \\binom{2005}{i} x^{3i} \\sum_{j=0}^{\\infty} (-1)^j \\binom{-2006}{j} x^j\n$$\n$$\n= \\sum_{i=0}^{2005} \\sum_{j=0}^{\\infty} (-1)^i \\binom{2005}{i} \\binom{2005+j}{j} x^{3i+j}\n$$\n$$\n= \\sum_{k=0}^{\\infty} \\left( \\sum_{i=1}^{2005} (-1)^i \\binom{2005}{i} \\binom{2005+k-3i}{2005} \\right) x^k\n$$\nThe desired number is\n$$\n\\sum_{i=1}^{334} (-1)^i \\binom{2005}{i} \\binom{3008-3i}{2005} = \\sum_{i=1}^{334} (-1)^i \\frac{(3008-3i)!}{i!(2005-i)!(1003-3i)!}\n$$\n(Note that $\\binom{3008-3i}{2005} = 0$ when $i \\geq 335$.)",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 27841,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nLet $ABC$ be an acute-angled triangle. Inscribe a rectangle $DEFG$ in this triangle so that $D$ is on $AB$, $E$ is on $AC$ and both $F$ and $G$ are on $BC$. Describe the locus of (i.e., the curve occupied by) the intersections of the diagonals of all possible rectangles $DEFG$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nThe locus is the line segment joining the midpoint $M$ of $BC$ to the midpoint $K$ of the altitude $AH$. Note that a segment $DE$ with $D$ on $AB$ and $E$ on $AC$ determines an inscribed rectangle; the midpoint $F$ of $DE$ lies on the median $AM$, while the midpoint of the perpendicular from $F$ to $BC$ is the centre of the rectangle. This lies on the median $MK$ of the triangle $AMH$.\n\nConversely, any point $P$ on $MK$ is the centre of a rectangle with base along $BC$ whose height is double the distance from $K$ to $BC$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 27842,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nIn a rectangular array of nonnegative real numbers with $m$ rows and $n$ columns, each row and each column contains at least one positive element. Moreover, if a row and a column intersect in a positive element, then the sums of their elements are the same. Prove that $m=n$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nConsider first the case where all the rows have the same positive sum $s$; this covers the particular situation in which $m=1$. Then each column, sharing a positive element with some row, must also have the sum $s$. Then the sum of all the entries in the matrix is $m s = n s$, whence $m = n$.\n\nWe prove the general case by induction on $m$. The case $m=1$ is already covered. Suppose that we have an $m \\times n$ array not all of whose rows have the same sum. Let $r < m$ of the rows have the sum $s$, and each of the other rows have a different sum. Then every column sharing a positive entry with one of these rows must also have sum $s$, and these are the only columns with the sum $s$. Suppose there are $c$ columns with sum $s$. The situation is essentially unchanged if we permute the rows and then the columns so that the first $r$ rows have the sum $s$ and the first $c$ columns have the sum $s$. Since all the entries of the first $r$ rows not in the first $c$ columns and in the first $c$ columns not in the first $r$ rows must be $0$, we can partition the array into a $r \\times c$ array in which all rows and columns have sum $s$ and which satisfies the hypothesis of the problem, two rectangular arrays of zeros in the upper right and lower left, and a rectangular $(m-r) \\times (n-c)$ array in the lower right that satisfies the conditions of the problem. By the induction hypothesis, we see that $r = c$ and so $m = n$.\nSolution:\n\nLet the term in the $i$th row and the $j$th column of the array be denoted by $a_{ij}$, and let $S = \\{(i, j) : a_{ij} > 0\\}$. Suppose that $r_i$ is the sum of the $i$th row and $c_j$ the sum of the $j$th column. Then $r_i = c_j$ whenever $(i, j) \\in S$. Then we have that\n$$\n\\sum\\left\\{\\frac{a_{ij}}{r_i} : (i, j) \\in S\\right\\} = \\sum\\left\\{\\frac{a_{ij}}{c_j} : (i, j) \\in S\\right\\}\n$$\nWe evaluate the sums on either side independently.\n$$\n\\begin{aligned}\n& \\sum\\left\\{\\frac{a_{ij}}{r_i} : (i, j) \\in S\\right\\} = \\sum\\left\\{\\frac{a_{ij}}{r_i} : 1 \\leq i \\leq m, 1 \\leq j \\leq n\\right\\} = \\sum_{i=1}^{m} \\frac{1}{r_i} \\sum_{j=1}^{n} a_{ij} = \\sum_{i=1}^{m} \\left(\\frac{1}{r_i}\\right) r_i = \\sum_{i=1}^{m} 1 = m. \\\\\n& \\sum\\left\\{\\frac{a_{ij}}{c_j} : (i, j) \\in S\\right\\} = \\sum\\left\\{\\frac{a_{ij}}{c_j} : 1 \\leq i \\leq m, 1 \\leq j \\leq n\\right\\} = \\sum_{j=1}^{n} \\frac{1}{c_j} \\sum_{i=1}^{m} a_{ij} = \\sum_{j=1}^{n} \\left(\\frac{1}{c_j}\\right) c_j = \\sum_{j=1}^{n} 1 = n.\n\\end{aligned}\n$$\nHence $m = n$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 27843,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nConsider a round-robin tournament with $2n+1$ teams, where each team plays each other team exactly once. We say that three teams $X$, $Y$ and $Z$ form a cycle triplet if $X$ beats $Y$, $Y$ beats $Z$, and $Z$ beats $X$. There are no ties.\n\na. Determine the minimum number of cycle triplets possible.\n\nb. Determine the maximum number of cycle triplets possible.",
    "options": [],
    "answer": "Minimum: 0; Maximum: n(n+1)(2n+1)/6",
    "solution": "Solution:\n\na. The minimum is $0$, which is achieved by a tournament in which team $T_{i}$ beats $T_{j}$ if and only if $i > j$.\n\nb. Any set of three teams constitutes either a cycle triplet or a \"dominated triplet\" in which one team beats the other two; let there be $c$ of the former and $d$ of the latter. Then $c + d = \\binom{2n+1}{3}$. Suppose that team $T_{i}$ beats $x_{i}$ other teams; then it is the winning team in exactly $\\binom{x_{i}}{2}$ dominated triples. Observe that $\\sum_{i=1}^{2n+1} x_{i} = \\binom{2n+1}{2}$, the total number of games. Hence\n$$\nd = \\sum_{i=1}^{2n+1} \\binom{x_{i}}{2} = \\frac{1}{2} \\sum_{i=1}^{2n+1} x_{i}^2 - \\frac{1}{2} \\binom{2n+1}{2}\n$$\nBy the Cauchy-Schwarz Inequality, $(2n+1) \\sum_{i=1}^{2n+1} x_{i}^2 \\geq \\left(\\sum_{i=1}^{2n+1} x_{i}\\right)^2 = n^2 (2n+1)^2$, whence\n$$\nc = \\binom{2n+1}{3} - \\sum_{i=1}^{2n+1} \\binom{x_{i}}{2} \\leq \\binom{2n+1}{3} - \\frac{n^2 (2n+1)}{2} + \\frac{1}{2} \\binom{2n+1}{2} = \\frac{n(n+1)(2n+1)}{6}.\n$$\nTo realize the upper bound, let the teams be $T_{1} = T_{2n+2}$, $T_{2} = T_{2n+3}$, $\\cdots$, $T_{i} = T_{2n+1+i}$, $\\cdots$, $T_{2n+1} = T_{4n+2}$. For each $i$, let team $T_{i}$ beat $T_{i+1}, T_{i+2}, \\cdots, T_{i+n}$ and lose to $T_{i+n+1}, \\cdots, T_{i+2n}$. We need to check that this is a consistent assignment of wins and losses, since the result for each pair of teams is defined twice. This can be seen by noting that $(2n+1+i)-(i+j) = 2n+1-j \\geq n+1$ for $1 \\leq j \\leq n$. The cycle triplets are $(T_{i}, T_{i+j}, T_{i+j+k})$ where $1 \\leq j \\leq n$ and $(2n+1+i)-(i+j+k) \\leq n$, i.e., when $1 \\leq j \\leq n$ and $n+1-j \\leq k \\leq n$. For each $i$, this counts $1+2+\\cdots+n = \\frac{1}{2} n(n+1)$ cycle triplets. When we range over all $i$, each cycle triplet gets counted three times, so the number of cycle triplets is\n$$\n\\frac{2n+1}{3} \\left( \\frac{n(n+1)}{2} \\right ) = \\frac{n(n+1)(2n+1)}{6}.\n$$\nSolution:\n\nLet $t$ be the number of cycle triplets and $u$ be the number of ordered triplets of teams $(X, Y, Z)$ where $X$ beats $Y$ and $Y$ beats $Z$. Each cycle triplet generates three ordered triplets while other triplets generate exactly one. The total number of triplets is\n$$\n\\binom{2n+1}{3} = \\frac{n(4n^2-1)}{3}.\n$$\nThe number of triples that are not cycles is\n$$\n\\frac{n(4n^2-1)}{3} - t\n$$\nHence\n$$\nu = 3t + \\left( \\frac{n(4n^2-1)}{3} - t \\right ) \\Longrightarrow\n$$\n$$\nt = \\frac{3u - n(4n^2-1)}{6} = \\frac{u - (2n+1)n^2}{2} + \\frac{n(n+1)(2n+1)}{6}\n$$\nIf team $Y$ beats $a$ teams and loses to $b$ teams, then the number of ordered triples with $Y$ as the central element is $ab$. Since $a + b = 2n$, by the Arithmetic-Geometric Means Inequality, we have that $ab \\leq n^2$. Hence $u \\leq (2n+1)n^2$, so that\n$$\nt \\leq \\frac{n(n+1)(2n+1)}{6}\n$$\nThe maximum is attainable when $u = (2n+1)n^2$, which can occur when we arrange all the teams in a circle with each team beating exactly the $n$ teams in the clockwise direction.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 27844,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nThe vertices of a right triangle $ABC$ inscribed in a circle divide the circumference into three arcs. The right angle is at $A$, so that the opposite $\\operatorname{arc} BC$ is a semicircle while $\\operatorname{arc} AB$ and $\\operatorname{arc} AC$ are supplementary. To each of the three arcs, we draw a tangent such that its point of tangency is the midpoint of that portion of the tangent intercepted by the extended lines $AB$ and $AC$. More precisely, the point $D$ on $\\operatorname{arc} BC$ is the midpoint of the segment joining the points $D'$ and $D''$ where the tangent at $D$ intersects the extended lines $AB$ and $AC$. Similarly for $E$ on arc $AC$ and $F$ on arc $AB$.\nProve that triangle $DEF$ is equilateral.\n\n![](attached_image_1.png)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\nA prime indicates where a tangent meets $AB$ and a double prime where it meets $AC$. It is given that $DD' = DD''$, $EE' = EE''$ and $FF' = FF''$. It is required to show that arc $EF$ is a third of the circumference as is arc $DBF$.\n\n$AF$ is the median to the hypotenuse of right triangle $AF'F''$, so that $FF' = FA$ and therefore\n$$\n\\operatorname{arc} AF = 2 \\angle F''FA = 2\\left(\\angle FF'A + \\angle FAF'\\right) = 4 \\angle FAF' = 4 \\angle FAB = 2 \\operatorname{arc} BF,\n$$\nwhence $\\operatorname{arc} FA = (2/3)$ arc $BFA$. Similarly, arc $AE = (2/3)$ arc $AEC$. Therefore, $\\operatorname{arc} FE$ is $2/3$ of the semicircle, or $1/3$ of the circumference as desired.\n\nAs for arc $DBF$, arc $BD = 2 \\angle BAD = \\angle BAD + \\angle BD'D = \\angle ADD'' = (1/2)$ arc $ACD$. But, arc $BF = (1/2)$ arc $AF$, so arc $DBF = (1/2)$ arc $FAED$. That is, arc $DBF$ is $1/3$ the circumference and the proof is complete.\nSolution:\nSince $AE'E''$ is a right triangle, $AE = EE' = EE''$ so that $\\angle CAE = \\angle CE''E$. Also $AD = D'D = DD''$, so that $\\angle CDD'' = \\angle CAD = \\angle CD''D$. As $EADC$ is a concyclic quadrilateral,\n$$\n\\begin{aligned}\n180^\\circ & = \\angle EAD + \\angle ECD \\\\\n& = \\angle DAC + \\angle CAE + \\angle ECA + \\angle ACD \\\\\n& = \\angle DAC + \\angle CAE + \\angle CEE'' + \\angle CE''E + \\angle CDD'' + \\angle CD''D \\\\\n& = \\angle DAC + \\angle CAE + \\angle CAE + \\angle CAE + \\angle CAD + \\angle CAD \\\\\n& = 3(\\angle DAC + \\angle DAE) = 3(\\angle DAE)\n\\end{aligned}\n$$\nHence $\\angle DFE = \\angle DAE = 60^\\circ$. Similarly, $\\angle DEF = 60^\\circ$. It follows that triangle $DEF$ is equilateral.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 27845,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\n$ABCD$ is a convex quadrilateral in which $AB$ is the longest side. Points $M$ and $N$ are located on sides $AB$ and $BC$ respectively, so that each of the segments $AN$ and $CM$ divides the quadrilateral into two parts of equal area. Prove that the segment $MN$ bisects the diagonal $BD$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nSince $[MADC] = \\frac{1}{2}[ABCD] = [NADC]$, it follows that $[ANC] = [AMC]$, so that $MN \\parallel AC$. Let $m$ be a line through $D$ parallel to $AC$ and $MN$ and let $BA$ produced meet $m$ at $P$ and $BC$ produced meet $m$ at $Q$. Then\n$$\n[MPC] = [MAC] + [CAP] = [MAC] + [CAD] = [MADC] = [BMC]\n$$\nwhence $BM = MP$. Similarly $BN = NQ$, so that $MN$ is a midline of triangle $BPQ$ and must bisect $BD$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 27846,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nDetermine all functions $f$ defined on the set of rationals that take rational values for which\n$$\nf(2 f(x)+f(y))=2 x+y\n$$\nfor each $x$ and $y$.",
    "options": [],
    "answer": "f(x)=x or f(x)=-x for all rational x",
    "solution": "Solution:\nThe only solutions are $f(x)=x$ for all rational $x$ and $f(x)=-x$ for all rational $x$. Both of these readily check out.\n\nSetting $y=x$ yields $f(3 f(x))=3 x$ for all rational $x$. Now replacing $x$ by $3 f(x)$, we find that\n$$\nf(9 x)=f(3 f(3 f(x)))=3[3 f(x)]=9 f(x),\n$$\nfor all rational $x$. Setting $x=0$ yields $f(0)=9 f(0)$, whence $f(0)=0$.\n\nSetting $x=0$ in the given functional equation yields $f(f(y))=y$ for all rational $y$. Thus $f$ is one-one onto. Applying $f$ to the functional equation yields that\n$$\n2 f(x)+f(y)=f(2 x+y)\n$$\nfor every rational pair $(x, y)$.\n\nSetting $y=0$ in the functional equation yields $f(2 f(x))=2 x$, whence $2 f(x)=f(2 x)$. Therefore $f(2 x)+f(y)=f(2 x+y)$ for each rational pair $(x, y)$, so that\n$$\nf(u+v)=f(u)+f(v)\n$$\nfor each rational pair $(u, v)$.\n\nSince $0=f(0)=f(-1)+f(1)$, $f(-1)=-f(1)$. By induction, it can be established that for each integer $n$ and rational $x$, $f(n x)=n f(x)$. If $k=f(1)$, we can establish from this that $f(n)=n k$, $f(1 / n)=k / n$ and $f(m / n)=m k / n$ for each integer pair $(m, n)$. Thus $f(x)=k x$ for all rational $x$. Since $f(f(x))=x$, we must have $k^2=1$. Hence $f(x)=x$ or $f(x)=-x$. These check out.\nSolution:\nIn the functional equation, let\n$$\nx=y=2 f(z)+f(w)\n$$\nto obtain $f(x)=f(y)=2 z+w$ and\n$$\nf(6 z+3 w)=6 f(z)+3 f(w)\n$$\nfor all rational pairs $(z, w)$. Set $(z, w)=(0,0)$ to obtain $f(0)=0$, $w=0$ to obtain $f(6 z)=6 f(z)$ and $z=0$ to obtain $f(3 w)=3 f(w)$ for all rationals $z$ and $w$. Hence $f(6 z+3 w)=f(6 z)+f(3 w)$. Replacing $(6 z, 3 w)$ by $(u, v)$ yields\n$$\nf(u+v)=f(u)+f(v)\n$$\nfor all rational pairs $(u, v)$. Hence $f(x)=k x$ where $k=f(1)$ for all rational $x$. Substitution of this into the functional equation with $(x, y)=(1,1)$ leads to $3=f(3 f(1))=f(3 k)=3 k^2$, so that $k= \\pm 1$. It can be checked that both $f(x) \\equiv x$ and $f(x) \\equiv -x$ satisfy the equation.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 27847,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nFind all functions $f$ defined on the natural numbers that take values among the natural numbers for which\n$$\n(f(n))^{p} \\equiv n \\pmod{f(p)}\n$$\nfor all $n \\in \\mathbf{N}$ and all prime numbers $p$.",
    "options": [],
    "answer": "All functions f: N -> N satisfying exactly one of the following:\n1) f(n) = n for all n.\n2) f(p) = 1 for every prime p (with f arbitrary on composite inputs).\n3) f(2) = 2, f(p) = 1 for every odd prime p, and f(n) has the same parity as n for all n.",
    "solution": "Solution:\nThe substitution $n = p$, a prime, yields $p \\equiv (f(p))^{p} \\equiv 0 \\pmod{f(p)}$, so that $p$ is divisible by $f(p)$. Hence, for each prime $p$, $f(p) = 1$ or $f(p) = p$.\n\nLet $S = \\{p : p \\text{ is prime and } f(p) = p\\}$. If $S$ is infinite, then $f(n)^{p} \\equiv n \\pmod{p}$ for infinitely many primes $p$. By the little Fermat theorem, $n \\equiv f(n)^{p} \\equiv f(n)$, so that $f(n) - n$ is a multiple of $p$ for infinitely many primes $p$. This can happen only if $f(n) = n$ for all values of $n$, and it can be verified that this is a solution.\n\nIf $S$ is empty, then $f(p) = 1$ for all primes $p$, and any function satisfying this condition is a solution.\n\nNow suppose that $S$ is finite and non-empty. Let $q$ be the largest prime in $S$. Suppose, if possible, that $q \\geq 3$. Therefore, for any prime $p$ exceeding $q$, $p \\equiv 1 \\pmod{q}$. However, this is not true. Let $Q$ be the product of all the odd primes up to $q$. Then $Q + 2$ must have a prime factor exceeding $q$ and at least one of them must be incongruent to $1 \\pmod{q}$. (An alternative argument notes that Bertrand's postulate can turn up a prime $p$ between $q$ and $2q$ which fails to satisfy $p \\equiv 1 \\pmod{q}$.)\n\nThe only remaining case is that $S = \\{2\\}$. Then $f(2) = 2$ and $f(p) = 1$ for every odd prime $p$. Since $f(n) 2 \\equiv n \\pmod{2}$, $f(n)$ and $n$ must have the same parity. Conversely, any function $f$ for which $f(n) \\equiv n \\pmod{2}$ for all $n$, $f(2) = 2$ and $f(p) = 1$ for all odd primes $p$ satisfies the condition.\n\nTherefore the only solutions are\n- $f(n) = n$ for all $n \\in \\mathbf{N}$;\n- any function $f$ with $f(p) = 1$ for all primes $p$;\n- any function for which $f(2) = 2$, $f(p) = 1$ for primes $p$ exceeding $2$ and $f(n)$ and $n$ have the same parity.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 27848,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nA self-avoiding rook walk on a chessboard (a rectangular grid of squares) is a path traced by a sequence of rook moves parallel to an edge of the board from one unit square to another, such that each begins where the previous move ended and such that no move ever crosses a square that has previously been crossed, i.e., the rook's path is non-self-intersecting.\nLet $R(m, n)$ be the number of self-avoiding rook walks on an $m \\times n$ ($m$ rows, $n$ columns) chessboard which begin at the lower-left corner and end at the upper-left corner. For example, $R(m, 1)=1$ for all natural numbers $m$; $R(2,2)=2$; $R(3,2)=4$; $R(3,3)=11$. Find a formula for $R(3, n)$ for each natural number $n$.",
    "options": [],
    "answer": "(1/(2*sqrt(2)))*((1+sqrt(2))^(n+1) - (1-sqrt(2))^(n+1)) - 1",
    "solution": "Solution:\n\nLet $r_{n}=R(3, n)$. It can be checked directly that $r_{1}=1$ and $r_{2}=4$. Let $1 \\leq i \\leq 3$ and $1 \\leq j$; let $(i, j)$ denote the cell in the $i$\\text{th} row from the bottom and the $j$\\text{th} column from the left, so that the paths in question go from $(1,1)$ to $(3,1)$.\nSuppose that $n \\geq 3$. The rook walks fall into exactly one of the following six categories:\n\n(1) One walk given by $(1,1) \\rightarrow (2,1) \\rightarrow (3,1)$.\n\n(2) Walks that avoid the cell $(2,1)$: Any such walk must start with $(1,1) \\rightarrow (1,2)$ and finish with $(3,2) \\rightarrow (3,1)$; there are $r_{n-1}$ such walks.\n\n(3) Walks that begin with $(1,1) \\rightarrow (2,1) \\rightarrow (2,2)$ and never return to the first row: Such walks enter the third row from $(2, k)$ for some $k$ with $2 \\leq k \\leq n$ and then go along the third row leftwards to $(3,1)$; there are $n-1$ such walks.\n\n(4) Walks that begin with $(1,1) \\rightarrow (2,1) \\rightarrow \\cdots \\rightarrow (2, k) \\rightarrow (1, k) \\rightarrow (1, k+1)$ and end with $(3, k+1) \\rightarrow (3, k) \\rightarrow (3, k-1) \\rightarrow \\cdots \\rightarrow (3,2) \\rightarrow (3,1)$ for some $k$ with $2 \\leq k \\leq n-1$; there are $r_{n-2}+r_{n-3}+\\cdots+r_{1}$ such walks.\n\n(5) Walks that are the horizontal reflected images of walks in (3) that begin with $(1,1) \\rightarrow (2,1)$ and never enter the third row until the final cell; there are $n-1$ such walks.\n\n(6) Walks that are horizontal reflected images of walks in (5); there are $r_{n-2}+r_{n-3}+\\cdots+r_{1}$ such walks.\n\nThus, $r_{3}=1+r_{2}+2\\left(2+r_{1}\\right)=11$ and, for $n \\geq 3$,\n$$\n\\begin{aligned}\nr_{n} & =1+r_{n-1}+2\\left[(n-1)+r_{n-2}+r_{n-3}+\\cdots+r_{1}\\right] \\\\\n& =2 n-1+r_{n-1}+2\\left(r_{n-2}+\\cdots+r_{1}\\right)\n\\end{aligned}\n$$\nand\n$$\nr_{n+1}=2 n+1+r_{n}+2\\left(r_{n-1}+r_{n-2}+\\cdots+r_{1}\\right) .\n$$\nTherefore\n$$\nr_{n+1}-r_{n}=2+r_{n}+r_{n-1} \\Longrightarrow r_{n+1}=2+2 r_{n}+r_{n-1}\n$$\nThus\n$$\nr_{n+1}+1=2\\left(r_{n}+1\\right)+\\left(r_{n-1}+1\\right),\n$$\nwhence\n$$\nr_{n}+1=\\frac{1}{2 \\sqrt{2}}(1+\\sqrt{2})^{n+1}-\\frac{1}{2 \\sqrt{2}}(1-\\sqrt{2})^{n+1},\n$$\nand\n$$\nr_{n}=\\frac{1}{2 \\sqrt{2}}(1+\\sqrt{2})^{n+1}-\\frac{1}{2 \\sqrt{2}}(1-\\sqrt{2})^{n+1}-1 .\n$$\nSolution:\n\nEmploy the same notation as in Solution 1. We have that $r_{1}=1, r_{2}=4$ and $r_{3}=11$. Let $n \\geq 3$. Consider the situation that there are $r_{n+1}$ columns. There are basically three types of rook walks.\n\nType 1. There are four rook walks that enter only the first two columns.\n\nType 2. There are $3 r_{n-1}$ rook walks that do not pass between the second and third columns in the middle row (in either direction), viz. $r_{n-1}$ of each of the types:\n$$\n\\begin{gathered}\n(1,1) \\longrightarrow (1,2) \\longrightarrow (1,3) \\longrightarrow \\cdots \\longrightarrow (3,3) \\longrightarrow (3,2) \\longrightarrow (3,1) ; \\\\\n(1,1) \\longrightarrow (2,1) \\longrightarrow (2,2) \\longrightarrow (1,2) \\longrightarrow (1,3) \\longrightarrow \\cdots \\longrightarrow (3,3) \\longrightarrow (3,2) \\longrightarrow (3,1) ; \\\\\n(1,1) \\longrightarrow (1,2) \\longrightarrow (1,3) \\longrightarrow \\cdots \\longrightarrow (3,3) \\longrightarrow (3,2) \\longrightarrow (2,2) \\longrightarrow (2,1) \\longrightarrow (3,1) .\n\\end{gathered}\n$$\n\nType 3. Consider the rook walks that pass between the second and third column along the middle row.\nThey are of Type 3a:\n$$\n(1,1) \\longrightarrow * \\longrightarrow (2,2) \\longrightarrow (2,3) \\longrightarrow \\cdots \\longrightarrow (3,3) \\longrightarrow (3,2) \\longrightarrow (3,1) ,\n$$\nor Type $3b$:\n$$\n(1,1) \\longrightarrow (1,2) \\longrightarrow (1,3) \\longrightarrow \\cdots \\longrightarrow (2,3) \\longrightarrow (2,2) \\longrightarrow * \\longrightarrow (3,1) ,\n$$\nwhere in each case the asterisk stands for one of two possible options.\nWe can associate in a two-one way the walks of Type 3a to a rook walk on the last $n$ columns, namely\n$$\n(1,2) \\longrightarrow (2,2) \\longrightarrow (2,3) \\longrightarrow \\cdots \\longrightarrow (3,3) \\longrightarrow (3,2)\n$$\nand the walks of Type $3b$ to a rook walk on the last $n$ columns, namely\n$$\n(1,2) \\longrightarrow (1,3) \\longrightarrow \\cdots \\longrightarrow (2,3) \\longrightarrow (2,2) \\longrightarrow (3,2) .\n$$\nThe number of rook walks of the latter two types together is $r_{n}-1-r_{n-1}$. From the number of rook walks on the last $n$ columns, we subtract one for $(1,2) \\rightarrow (2,2) \\rightarrow (3,2)$ and $r_{n-1}$ for those of the type\n$$\n(1,2) \\longrightarrow (1,3) \\longrightarrow \\cdots \\longrightarrow (3,3) \\longrightarrow (2,3) .\n$$\nTherefore, the number of rook walks of Type 3 is $2\\left(r_{n}-1-r_{n-1}\\right)$ and we find that\n$$\nr_{n+1}=4+3 r_{n-1}+2\\left(r_{n}-1-r_{n-1}\\right)=2+2 r_{n}+r_{n-1} .\n$$\nWe can now complete the solution as in Solution 1.\nSolution:\n\nLet $S(3, n)$ be the set of self-avoiding rook walks in which the rook occupies column $n$ but does not occupy column $n+1$. Then $R(3, n)=|S(3,1)|+|S(3,2)|+\\cdots+|S(3, n)|$. Furthermore, topological considerations allow us to break $S(3, n)$ into three disjoint subsets $S_{1}(3, n)$, the set of paths in which corner $(1, n)$ is not occupied, but there is a path segment $(2, n) \\longrightarrow (3, n)$; $S_{2}(3, n)$, the set of paths in which corners $(1, n)$ and $(3, n)$ are both occupied by a path $(1, n) \\longrightarrow (2, n) \\longrightarrow (3, n)$; and $S_{3}(3, n)$, the set of paths in which corner $(3, n)$ is not occupied but there is a path segment $(1, n) \\longrightarrow (2, n)$. Let $s_{i}(n)=|S_{i}(3, n)|$ for $i=1,2,3$. Note that $s_{1}(1)=0, s_{2}(1)=1$ and $s_{3}(1)=0$. By symmetry, $s_{1}(n)=s_{3}(n)$ for every positive $n$. Furthermore, we can construct paths in $S(3, n+1)$ by \"bulging\" paths in $S(3, n)$, from which we obtain\n$$\n\\begin{aligned}\n& s_{1}(n+1)=s_{1}(n)+s_{2}(n) ; \\\\\n& s_{2}(n+1)=s_{1}(n)+s_{2}(n)+s_{3}(n) ; \\\\\n& s_{3}(n+1)=s_{2}(n)+s_{3}(n) ;\n\\end{aligned}\n$$\nor, upon simplification,\n$$\n\\begin{aligned}\n& s_{1}(n+1)=s_{1}(n)+s_{2}(n) \\\\\n& s_{2}(n+1)=2 s_{1}(n)+s_{2}(n)\n\\end{aligned}\n$$\nHence, for $n \\geq 2$,\n$$\n\\begin{aligned}\ns_{1}(n+1) & =s_{1}(n)+2 s_{1}(n-1)+s_{2}(n-1) \\\\\n& =s_{1}(n)+2 s_{1}(n-1)+s_{1}(n)-s_{1}(n-1) \\\\\n& =2 s_{1}(n)+s_{1}(n-1) .\n\\end{aligned}\n$$\nand\n$$\n\\begin{aligned}\ns_{2}(n+1) & =2 s_{1}(n)+s_{2}(n)=2 s_{1}(n-1)+2 s_{2}(n-1)+s_{2}(n) \\\\\n& =s_{2}(n)-s_{2}(n-1)+2 s_{2}(n-1)+s_{2}(n) \\\\\n& =2 s_{2}(n)+s_{2}(n-1) .\n\\end{aligned}\n$$\nWe find that\n$$\n\\begin{aligned}\n& s_{1}(n)=\\frac{1}{2 \\sqrt{2}}(1+\\sqrt{2})^{n-1}-\\frac{1}{2 \\sqrt{2}}(1-\\sqrt{2})^{n-1} ; \\\\\n& s_{2}(n)=\\frac{1}{2}(1+\\sqrt{2})^{n-1}+\\frac{1}{2}(1-\\sqrt{2})^{n-1} .\n\\end{aligned}\n$$\nSumming a geometric series yields that\n$$\n\\begin{aligned}\nR(3, n) & =\\left(s_{2}(1)+\\cdots+s_{2}(n)\\right)+2\\left(s_{1}(1)+\\cdots+s_{1}(n)\\right) \\\\\n& =\\left(\\frac{1}{2}+\\frac{1}{\\sqrt{2}}\\right)\\left(\\frac{(1+\\sqrt{2})^{n}-1}{\\sqrt{2}}\\right)+\\left(\\frac{1}{2}-\\frac{1}{\\sqrt{2}}\\right)\\left(\\frac{(1-\\sqrt{2})^{n}-1}{-\\sqrt{2}}\\right) \\\\\n& =\\left(\\frac{1}{2 \\sqrt{2}}\\right)\\left[(1+\\sqrt{2})^{n+1}-(1-\\sqrt{2})^{n+1}\\right]-1 .\n\\end{aligned}\n$$\nThe formula agrees with $R(3,1)=1, R(3,2)=4$ and $R(3,3)=11$.",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 27849,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nDefine\n$$\nf(x, y, z) = \\frac{(x y + y z + z x)(x + y + z)}{(x + y)(x + z)(y + z)}.\n$$\nDetermine the set of real numbers $r$ for which there exists a triplet $(x, y, z)$ of positive real numbers satisfying $f(x, y, z) = r$.",
    "options": [],
    "answer": "(1, 9/8]",
    "solution": "Solution:\nWe prove that $1 < f(x, y, z) \\leq \\frac{9}{8}$, and that $f(x, y, z)$ can take on any value within the range $\\left(1, \\frac{9}{8}\\right]$.\nThe expression for $f(x, y, z)$ can be simplified to\n$$\nf(x, y, z) = 1 + \\frac{x y z}{(x + y)(x + z)(y + z)}.\n$$\nSince $x, y, z$ are positive, we get $1 < f(x, y, z)$.\nThe inequality $f(x, y, z) \\leq \\frac{9}{8}$ can be simplified to\n$$\nx^{2} y + x^{2} z + y^{2} x + y^{2} z + z^{2} x + z^{2} y - 6 x y z \\geq 0.\n$$\nRearrange the left hand side as follows:\n$$\n\\begin{aligned}\n& x^{2} y + x^{2} z + y^{2} x + y^{2} z + z^{2} x + z^{2} y - 6 x y z = \\\\\n& x\\left(y^{2} + z^{2}\\right) - 2 x y z + y\\left(x^{2} + z^{2}\\right) - 2 x y z + z\\left(x^{2} + y^{2}\\right) - 2 x y z = \\\\\n& x(y - z)^{2} + y(x - z)^{2} + z(x - y)^{2}.\n\\end{aligned}\n$$\nThis expression is clearly non-negative when $x, y, z$ are non-negative. To prove that $f(x, y, z)$ takes any values in the interval $\\left(1, \\frac{9}{8}\\right]$, define\n$$\ng(t) = f(t, 1, 1) = 1 + \\frac{t}{2(1 + t)^{2}}\n$$\nThen $g(1) = \\frac{9}{8}$ and $g(t)$ approaches 1 as $t$ approaches 0. It follows from the continuity of $g(t)$ for $0 < t \\leq 1$ that it takes all values in the interval $\\left(1, \\frac{9}{8}\\right]$. (Alternatively, one can check that the quadratic equation $g(t) = r$ has a solution $t$ for any number $r$ in the interval $\\left(1, \\frac{9}{8}\\right]$.)",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 27850,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nGiven an $m \\times n$ grid with squares coloured either black or white, we say that a black square in the grid is stranded if there is some square to its left in the same row that is white and there is some square above it in the same column that is white (see Figure).\n![](attached_image_1.png)\nA $4 \\times 5$ grid with no stranded black squares\n\nFind a closed formula for the number of $2 \\times n$ grids with no stranded black squares.",
    "options": [],
    "answer": "2*3^n - 2^n",
    "solution": "Solution:\n\nThere is no condition for squares in the first row. A square in the second row can be black only if the square above it is black or all squares to the left of it are black. Suppose the first $k$ squares in the second row are black and the $(k+1)$-st square is white or $k=n$. When $k<n$ then for each of the first $k+1$ squares in the first row we have 2 choices, and for each of the remaining $n-k-1$ columns we have 3 choices. When $k=n$, there are $2^{n}$ choices for the first row. The total number of choices is thus:\n$$\n\\sum_{k=0}^{n-1} 2^{k+1} 3^{n-k-1} + 2^{n}\n$$\nThis expression simplifies to\n$$\n2 \\cdot 3^{n} - 2^{n}\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 27851,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nTwo circles of different radii are cut out of cardboard. Each circle is subdivided into 200 equal sectors. On each circle 100 sectors are painted white and the other 100 are painted black. The smaller circle is then placed on top of the larger circle, so that their centers coincide. Show that one can rotate the small circle so that the sectors on the two circles line up and at least 100 sectors on the small circle lie over sectors of the same color on the big circle.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nLet $x_{0}, \\ldots, x_{199}$ be variables. Assign the value of $+1$ or $-1$ to $x_{i}$ depending on whether the $(i+1)$st segment of the larger circle (counting counterclockwise) is black or white, respectively. Similarly, assign the value of $+1$ or $-1$ to the variable $y_{i}$ depending on whether the $(i+1)$th segment of the smaller circle is black or white. We can now restate the problem in the following equivalent way: show that\n$$\nS_{j} = \\sum_{i=1}^{200} x_{i} y_{i+j} \\geq 0\n$$\nfor some $j = 0, \\ldots, 199$. Here the subscript $i+j$ is understood modulo $200$.\n\nNow observe that $y_{0} + \\cdots + y_{199} = 0$ and thus\n$$\nS_{0} + \\cdots + S_{199} = \\sum_{i=0}^{199} x_{i} (y_{0} + \\cdots + y_{199}) = 0.\n$$\nThus $S_{j} \\geq 0$ for some $j = 0, \\ldots, 199$, as claimed.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 27852,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nFind all ordered pairs $(a, b)$ such that $a$ and $b$ are integers and $3^{a} + 7^{b}$ is a perfect square.",
    "options": [],
    "answer": "(1, 0) and (2, 1)",
    "solution": "Solution:\nIt is obvious that $a$ and $b$ must be non-negative.\nSuppose that $3^{a} + 7^{b} = n^{2}$. We can assume that $n$ is positive. We first work modulo $4$. Since $3^{a} + 7^{b} = n^{2}$, it follows that\n$$\nn^{2} \\equiv (-1)^{a} + (-1)^{b} \\pmod{4}\n$$\nSince no square can be congruent to $2$ modulo $4$, it follows that we have either (i) $a$ is odd and $b$ is even or (ii) $a$ is even and $b$ is odd.\n\nCase (i): Let $b = 2c$. Then\n$$\n3^{a} = (n - 7^{c})(n + 7^{c}) .\n$$\nIt cannot be the case that $3$ divides both $n - 7^{c}$ and $n + 7^{c}$. But each of these is a power of $3$. It follows that $n - 7^{c} = 1$, and therefore\n$$\n3^{a} = 2 \\cdot 7^{c} + 1 .\n$$\nIf $c = 0$, then $a = 1$, and we obtain the solution $a = 1, b = 0$. So suppose that $c \\geq 1$. Then $3^{a} \\equiv 1 \\pmod{7}$. This is impossible, since the smallest positive value of $a$ such that $3^{a} \\equiv 1 \\pmod{7}$ is given by $a = 6$, and therefore all $a$ such that $3^{a} \\equiv 1 \\pmod{7}$ are even, contradicting the fact that $a$ is odd.\n\nCase (ii): Let $a = 2c$. Then\n$$\n7^{b} = (n - 3^{c})(n + 3^{c}) .\n$$\nThus each of $n - 3^{c}$ and $n + 3^{c}$ is a power of $7$. Since $7$ cannot divide both of these, it follows that $n - 3^{c} = 1$, and therefore\n$$\n7^{b} = 2 \\cdot 3^{c} + 1\n$$\nLook first at the case $c = 1$. Then $b = 1$, and we obtain the solution $a = 2, b = 1$. So from now on we may assume that $c > 1$. Then $7^{b} \\equiv 1 \\pmod{9}$. The smallest positive integer $b$ such that $7^{b} \\equiv 1 \\pmod{9}$ is given by $b = 3$. It follows that $b$ must be a multiple of $3$. Let $b = 3d$. Note that $d$ is odd, so in particular $d \\geq 1$.\n\nLet $y = 7^{d}$. Then $y^{3} - 1 = 2 \\cdot 3^{c}$, and therefore\n$$\n2 \\cdot 3^{c} = (y - 1)(y^{2} + y + 1) .\n$$\nIt follows that $y - 1 = 2 \\cdot 3^{u}$ for some positive $u$, and that $y^{2} + y + 1 = 3^{v}$ for some $v \\geq 2$. But since\n$$\n3y = (y^{2} + y + 1) - (y - 1)^{2},\n$$\nit follows that $3 \\mid y$, which is impossible since $3 \\mid (y - 1)$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 27853,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nA set of points is marked on the plane, with the property that any three marked points can be covered with a disk of radius $1$. Prove that the set of all marked points can be covered with a disk of radius $1$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n(For a finite set of points only.) Let $D$ be a disk of smallest radius that covers all marked points. Consider the marked points on the boundary $C$ of this disk. Note that if all marked points on $C$ lie on an arc smaller than the half circle (ASTTHC for short), then the disk can be moved a little towards these points on the boundary and its radius can be decreased. Since we assumed that our disk has minimal radius, the marked points on its boundary do not lie on an ASTTHC.\nIf the two endpoints of a diagonal of $D$ are marked, then $D$ is the smallest disk containing these two points, hence must have radius at most $1$.\nIf there are $3$ marked points on $C$ that do not lie on an ASTTHC, then $D$ is the smallest disk covering these $3$ points and hence must have radius at most $1$. (In this case the triangle formed by the three points is acute and $C$ is its circumcircle.)\nIf there are more than $3$ marked points on the boundary that do not lie on an ASTTHC, then we can remove one of them so that the remaining points again do not lie on an ASTTHC. By induction this leads us to the case of $3$ points. Indeed, given $4$ or more points on $C$, choose $3$ points that lie on a half circle. Then the middle point can be removed.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 27854,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $A$, $B$, $P$ be three points on a circle. Prove that if $a$ and $b$ are the distances from $P$ to the tangents at $A$ and $B$ and $c$ is the distance from $P$ to the chord $AB$, then $c^2 = ab$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $r$ be the radius of the circle, and let $a'$ and $b'$ be the respective lengths of $PA$ and $PB$. Since $b' = 2r \\sin \\angle PAB = 2rc/a'$, $c = a'b'/(2r)$. Let $AC$ be the diameter of the circle and $H$ the foot of the perpendicular from $P$ to $AC$. The similarity of the triangles $ACP$ and $APH$ imply that $AH : AP = AP : AC$ or $(a')^2 = 2ra$. Similarly, $(b')^2 = 2rb$. Hence\n$$\nc^2 = \\frac{(a')^2}{2r} - \\frac{(b')^2}{2r} = ab\n$$\nas desired.\nLet $E$, $F$, $G$ be the feet of the perpendiculars to the tangents at $A$ and $B$ and the chord $AB$, respectively. We need to show that $PE : PG = PG : GF$, where $G$ is the foot of the perpendicular from $P$ to $AB$. This suggests that we try to prove that the triangles $EPG$ and $GPF$ are similar.\n\nSince $PG$ is parallel to the bisector of the angle between the two tangents, $\\angle EPG = \\angle FPG$. Since $AEPG$ and $BFPG$ are concyclic quadrilaterals (having opposite angles right), $\\angle PGE = \\angle PAE$ and $\\angle PFG = \\angle PBG$. But $\\angle PAE = \\angle PBA = \\angle PBG$, whence $\\angle PGE = \\angle PFG$. Therefore triangles $EPG$ and $GPF$ are similar.\n\nThe argument above with concyclic quadrilaterals only works when $P$ lies on the shorter arc between $A$ and $B$. The other case can be proved similarly.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 27855,
    "subject": "Mathematics (Multi-modal)",
    "question": "Three speed skaters have a friendly “race” on a skating oval. They all start from the same point and skate in the same direction, but with different speeds that they maintain throughout the race. The slowest skater does $1$ lap a minute, the fastest one does $3.14$ laps a minute, and the middle one does $L$ laps a minute for some $1 < L < 3.14$. The race ends at the moment when all three skaters again come together to the same point on the oval (which may differ from the starting point.) Find how many different choices for $L$ are there such that exactly $117$ passings occur before the end of the race. (A passing is defined when one skater passes another one. The beginning and the end of the race when all three skaters are together are not counted as passings.)",
    "options": [],
    "answer": "16",
    "solution": "Assume that the length of the oval is one unit. Let $x(t)$ be the difference of distances that the slowest and the fastest skaters have skated by time $t$. Similarly, let $y(t)$ be the difference between the middle skater and the slowest skater. The path $(x(t), y(t))$ is a straight ray $R$ in $\\mathbb{R}^2$, starting from the origin, with slope depending on $L$. By assumption, $0 < y(t) < x(t)$.\n\nOne skater passes another one when either $x(t) \\in \\mathbb{Z}$, $y(t) \\in \\mathbb{Z}$ or $x(t) - y(t) \\in \\mathbb{Z}$.\n\nThe race ends when both $x(t), y(t) \\in \\mathbb{Z}$.\n\nLet $(a, b) \\in \\mathbb{Z}^2$ be the endpoint of the ray $R$. We need to find the number of such points satisfying:\n(a) $0 < b < a$\n(b) The ray $R$ intersects $\\mathbb{Z}^2$ at endpoints only.\n(c) The ray $R$ crosses $357$ times the lines $x \\in \\mathbb{Z}$, $y \\in \\mathbb{Z}$, $y - x \\in \\mathbb{Z}$.\n\nThe second condition says that $a$ and $b$ are relatively prime. The ray $R$ crosses $a-1$ of the lines $x \\in \\mathbb{Z}$, $b-1$ of the lines $y \\in \\mathbb{Z}$ and $a-b-1$ of the lines $x-y \\in \\mathbb{Z}$.\n\nThus, we need $(a-1) + (b-1) + (a-b-1) = 117$, or equivalently, $2a-3=117$.\n\nThat is $a=60$.\n\nNow $b$ must be a positive integer less than and relatively prime to $60$. The number of such $b$ can be found using the Euler's $\\phi$ function:\n$$\n\\phi(60) = \\phi(2^2 \\cdot 3 \\cdot 5) = (2-1) \\cdot 2 \\cdot (3-1) \\cdot (5-1) = 16.\n$$\nThus the answer is $16$. $\\square$\nFirst, let us name our skaters. From fastest to slowest, call them: A, B and C. (Abel, Bernoulli and Cayley?)\n\nNow, it is helpful to consider the race from the viewpoint of C. Relative to C, both A and B complete a whole number of laps, since they both start and finish at C.\n\nLet $n$ be the number of laps completed by A relative to C, and let $m$ be the number of laps completed by B relative to C. Note that: $n > m \\in \\mathbb{Z}^+$\n\nConsider the number of minutes required to complete the race. Relative to C, A is moving with a speed of $3.14 - 1 = 2.14$ laps per minute and completes the race in $\\frac{n}{2.14}$ minutes. Also relative to C, B is moving with a speed of $(L-1)$ laps per minute and completes the race in $\\frac{m}{L-1}$ minutes. Since A and B finish the race together (when they both meet C):\n$$\n\\frac{n}{2.14} = \\frac{m}{L-1} \\quad \\Rightarrow \\quad L = 2.14 \\left(\\frac{m}{n}\\right) + 1.\n$$\nHence, there is a one-to-one relation between values of $L$ and values of the positive proper fraction $\\frac{m}{n}$. The fraction should be reduced, that is the pair $(m, n)$ should be relatively prime, or else, with $n = \\text{gcd}(m, n)$, the race ends after $n/k$ laps for A and $m/k$ laps for B when they first meet C together.\n\nIt is also helpful to consider the race from the viewpoint of B. In this frame of reference, A completes only $n - m$ laps. Hence A passes B only $(n - m) - 1$ times, since the racers do not \"pass\" at the end of the race (nor at the beginning). Similarly A passes C only $n - 1$ times and B passes C only $m - 1$ times. The total number of passings is:\n$$\n117 = (n-1) + (m-1) + (n-m-1) = 2n-3 \\Rightarrow n=60\n$$\nHence the number of values of $L$ equals the number of $m$ for which the fraction $m/60$ is positive, proper and reduced. That is the number of positive integer values smaller than and relatively prime to $60$. One could simply count: $\\{1,7,11,13,17,...\\}$, but Euler's $\\phi$ function gives this number:\n$$\n\\phi(60) = \\phi(2^2 \\cdot 3 \\cdot 5) = (2-1) \\cdot 2 \\cdot (3-1) \\cdot (5-1) = 16.\n$$\nTherefore, there are $16$ values for $L$ which give the desired number of passings.\n\nNote that the actual values for the speeds of A and C do not affect the result. They could be any values, rational or irrational, just so long as they are different, and there will be $16$ possible values for the speed of B between them. ☐",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 27856,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nEach vertex of a finite graph can be colored either black or white. Initially all vertices are black. We are allowed to pick a vertex $P$ and change the color of $P$ and all of its neighbours. Is it possible to change the colour of every vertex from black to white by a sequence of operations of this type?",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nThe answer is yes. Proof by induction on the number $n$ of vertices. If $n=1$, this is obvious. For the induction assumption, suppose we can do this for any graph with $n-1$ vertices for some $n \\geq 2$ and let $X$ be a graph with $n$ vertices which we will denote by $P_{1}, \\ldots, P_{n+1}$.\n\nLet us denote the \"basic\" operation of changing the color of $P_{i}$ and all of its neighbours by $f_{i}$. Removing a vertex $P_{i}$ from $X$ (along with all edges connecting to $P_{i}$) and applying the induction assumption to the resulting smaller graph, we see that there exists a sequence of operations $g_{i}$ (obtained by composing some $f_{j}$, with $j \\neq i$) which changes the colour of every vertex in $X$, except for possibly $P_{i}$.\n\nIf $g_{i}$ also changes the color of $P_{i}$ then we are done. So, we may assume that $g_{i}$ does not change the colour of $P$ for every $i=1, \\ldots, n$. Now consider two cases.\n\nCase 1: $n$ is even. Then composing $g_{1}, \\ldots, g_{n}$ we will change the color of every vertex from white to black.\n\nCase 2: $n$ is odd. I claim that in this case $X$ has a vertex with an even number of neighbours.\n\nIndeed, denote the number of neighbours of $P_{i}$ (or equivalently, the number of edges connected to $P$) by $k_{i}$. Then $k_{1}+\\cdots+k_{n+1}=2e$, where $e$ is the number of edges of $X$. Thus one of the numbers $k_{i}$ has to be even as claimed.\n\nAfter renumbering, suppose $P_{1}$ has $2k$ neighbours, say $P_{2}, \\ldots, P_{2k+1}$. The composition of $f_{1}$ with $g_{1}, g_{2}, \\ldots, g_{2k+1}$ will then change the colour of every vertex, as desired.",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 27857,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $P(x)$ and $Q(x)$ be polynomials with integer coefficients. Let $a_n = n! + n$. Show that if $P(a_n)/Q(a_n)$ is an integer for every $n$, then $P(n)/Q(n)$ is an integer for every integer $n$ such that $Q(n) \\neq 0$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Imagine dividing $P(x)$ by $Q(x)$. We find that\n$$\n\\frac{P(x)}{Q(x)} = A(x) + \\frac{R(x)}{Q(x)},\n$$\nwhere $A(x)$ and $R(x)$ are polynomials with rational coefficients, and $R(x)$ is either identically $0$ or has degree less than the degree of $Q(x)$.\nBy bringing the coefficients of $A(x)$ to their least common multiple, we can find a polynomial $B(x)$ with integer coefficients, and a positive integer $b$, such that $A(x) = B(x)/b$. Suppose first that $R(x)$ is not identically $0$. Note that for any integer $k$, either $A(k) = 0$, or $|A(k)| \\ge 1/b$. But whenever $|k|$ is large enough, $0 < |R(k)/Q(k)| < 1/b$, and therefore if $n$ is large enough, $P(a_n)/Q(a_n)$ cannot be an integer.\nSo $R(x)$ is identically $0$, and $P(x)/Q(x) = B(x)/b$ (at least whenever $Q(x) \\neq 0$.)\nNow let $n$ be an integer. Then there are infinitely many integers $k$ such that $n \\equiv a_k \\pmod b$. But $B(a_k)/b$ is an integer, or equivalently $b$ divides $B(a_k)$. It follows that $b$ divides $B(n)$, and therefore $P(n)/Q(n)$ is an integer. $\\square$",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 27858,
    "subject": "Mathematics (Multi-modal)",
    "question": "Consider 70-digit numbers $n$, with the property that each of the digits $1, 2, 3, \\dots, 7$ appears in the decimal expansion of $n$ ten times (and $8$, $9$, and $0$ do not appear). Show that no number of this form can divide another number of this form.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Assume the contrary: there exist $a$ and $b$ of the prescribed form, such that $b \\ge a$ and $a$ divides $b$. Then $a$ divides $b-a$.\n\nClaim: $a$ is not divisible by $3$ but $b-a$ is divisible by $9$. Indeed, the sum of the digits is $10(1 + \\cdots + 7) = 280$, for both $a$ and $b$. [Here one needs to know or prove that an integer $n$ is equivalent to the sum of its digits modulo $3$ and modulo $9$.]\n\nWe conclude that $b-a$ is divisible by $9a$. But this is impossible, since $9a$ has $71$ digits and $b$ has only $70$ digits, so $9a > b > b-a$. $\\square$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 27859,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABCD$ be a cyclic quadrilateral whose opposite sides are not parallel, $X$ the intersection of $AB$ and $CD$, and $Y$ the intersection of $AD$ and $BC$. Let the angle bisector of $\\angle AXD$ intersect $AD$, $BC$ at $E$, $F$ respectively and let the angle bisector of $\\angle AYB$ intersect $AB$, $CD$ at $G$, $H$ respectively. Prove that $EGFH$ is a parallelogram.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Since $ABCD$ is cyclic, $\\triangle XAC \\sim \\triangle XDB$ and $\\triangle YAC \\sim \\triangle YBD$. Therefore,\n$$\n\\frac{XA}{XD} = \\frac{XC}{XB} = \\frac{AC}{DB} = \\frac{YA}{YB} = \\frac{YC}{YD}.\n$$\nLet $s$ be this ratio. Therefore, by the angle bisector theorem,\n$$\n\\frac{AE}{ED} = \\frac{XA}{XD} = \\frac{XC}{XB} = \\frac{CF}{FB} = s,\n$$\nand\n$$\n\\frac{AG}{GB} = \\frac{YA}{YB} = \\frac{YC}{YD} = \\frac{CH}{HD} = s.\n$$\nHence, $\\frac{AG}{GB} = \\frac{CF}{FB}$ and $\\frac{AE}{ED} = \\frac{DH}{HC}$. Therefore, $EH \\parallel AC \\parallel GF$ and $EG \\parallel DB \\parallel HF$. Hence, $EGFH$ is a parallelogram. $\\square$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 27860,
    "subject": "Mathematics (Multi-modal)",
    "question": "Amy has divided a square into finitely many white and red rectangles, each with sides parallel to the sides of the square. Within each white rectangle, she writes down its width divided by its height. Within each red rectangle, she writes down its height divided by its width. Finally, she calculates $x$, the sum of these numbers. If the total area of the white rectangles equals the total area of the red rectangles, what is the smallest possible value of $x$?",
    "options": [],
    "answer": "2.5",
    "solution": "Let $a_i$ and $b_i$ denote the width and height of each white rectangle, and let $c_i$ and $d_i$ denote the width and height of each red rectangle. Also, let $L$ denote the side length of the original square.\n\n**Lemma:** Either $\\sum a_i \\ge L$ or $\\sum d_i \\ge L$.\n\n**Proof of lemma:** Suppose there exists a horizontal line across the square that is covered entirely with white rectangles. Then, the total width of these rectangles is at least $L$, and the claim is proven. Otherwise, there is a red rectangle intersecting every horizontal line, and hence the total height of these rectangles is at least $L$. □\n\nNow, let us assume without loss of generality that $\\sum a_i \\ge L$. By the Cauchy-Schwarz inequality,\n$$\n\\left(\\sum \\frac{a_i}{b_i}\\right) \\cdot \\left(\\sum a_i b_i\\right) \\ge \\left(\\sum a_i\\right)^2 \\ge L^2.\n$$\nBut we know $\\sum a_i b_i = \\frac{L^2}{2}$, so it follows that $\\sum \\frac{a_i}{b_i} \\ge 2$. Furthermore, each $c_i \\le L$, so\n$$\n\\sum \\frac{d_i}{c_i} \\ge \\frac{1}{L^2} \\cdot \\sum c_i d_i = \\frac{1}{2}.\n$$\nTherefore, $x$ is at least $2.5$. Conversely, $x = 2.5$ can be achieved by making the top half of the square one colour, and the bottom half the other colour. □",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 27861,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nShow that there exists a positive integer $N$ such that for all integers $a > N$, there exists a contiguous substring of the decimal expansion of $a$ that is divisible by $2011$. (For instance, if $a = 153204$, then $15$, $532$, and $0$ are all contiguous substrings of $a$. Note that $0$ is divisible by $2011$.)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\nWe claim that if the decimal expansion of $a$ has at least $2012$ digits, then $a$ contains the required substring. Let the decimal expansion of $a$ be $a_{k} a_{k-1} \\ldots a_{0}$. For $i = 0, \\ldots, 2011$, let $b_{i}$ be the number with decimal expansion $a_{i} a_{i-1} \\ldots a_{0}$. Then by the pigeonhole principle, $b_{i} \\equiv b_{j} \\pmod{2011}$ for some $i < j \\leq 2011$. It follows that $2011$ divides $b_{j} - b_{i} = c \\cdot 10^{i}$. Here $c$ is the substring $a_{j} \\ldots a_{i+1}$. Since $2011$ and $10$ are relatively prime, it follows that $2011$ divides $c$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 27862,
    "subject": "Mathematics (Multi-modal)",
    "question": "Show that there exists a positive integer $N$ such that for all integers $a > N$, there exists a contiguous substring of the decimal expansion of $a$ that is divisible by $2011$. (For instance, if $a = 153204$, then $15$, $532$, and $0$ are all contiguous substrings of $a$. Note that $0$ is divisible by $2011$.)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "We claim that if the decimal expansion of $a$ has at least $2012$ digits, then $a$ contains the required substring.\n\nLet the decimal expansion of $a$ be $a_k a_{k-1} \\dots a_0$. For $i = 0, \\dots, 2011$, let $b_i$ be the number with decimal expansion $a_i a_{i-1} \\dots a_0$. Then by the pigeonhole principle, $b_i \\equiv b_j \\pmod{2011}$ for some $i < j \\le 2011$.\n\nIt follows that $2011$ divides $b_j - b_i = c \\cdot 10^i$. Here $c$ is the substring $a_j \\dots a_{i+1}$. Since $2011$ and $10$ are relatively prime, it follows that $2011$ divides $c$.\n\n$\\Box$",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 27863,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $d$ be a positive integer. Show that for every integer $S$, there exists an integer $n > 0$ and a sequence $\\epsilon_1, \\epsilon_2, \\dots, \\epsilon_n$, where for any $k$, $\\epsilon_k = 1$ or $\\epsilon_k = -1$, such that\n$$\nS = \\epsilon_1(1+d)^2 + \\epsilon_2(1+2d)^2 + \\epsilon_3(1+3d)^2 + \\dots + \\epsilon_n(1+nd)^2.\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $U_k = (1 + k d)^2$. We calculate $U_{k+3} - U_{k+2} - U_{k+1} + U_k$. This turns out to be $4d^2$, a constant. Changing signs, we obtain the sum $-4d^2$.\nThus if we have found an expression for a certain number $S_0$ as a sum of the desired type, we can obtain an expression of the desired type for $S_0 + (4d^2)q$, for any integer $q$.\n\nIt remains to show that for any $S$, there exists an integer $S'$ such that $S' \\equiv S \\pmod{4d^2}$ and $S'$ can be expressed in the desired form. Look at the sum\n$$\n(1+d)^2 + (1+2d)^2 + \\dots + (1+Nd)^2,\n$$\nwhere $N$ is \"large.\" We can at will choose $N$ so that the sum is odd, or so that the sum is even.\nBy changing the sign in front of $(1 + k d)^2$ to a minus sign, we decrease the sum by $2(1 + k d)^2$. In particular, if $k \\equiv 0 \\pmod{2d}$, we decrease the sum by $2$ (modulo $4d^2$). So\nIf $N$ is large enough, there are many $k < N$ such that $k$ is a multiple of $2d$. By switching the sign in front of $r$ of these, we change (\"downward\") the congruence class modulo $4d^2$ by $2r$. By choosing $N$ so that the original sum is odd, and choosing suitable $r < 2d^2$, we can obtain numbers congruent to all odd numbers modulo $4d^2$. By choosing $N$ so that the original sum is even, we can obtain numbers congruent to all even numbers modulo $4d^2$. This completes the proof. $\\square$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 27864,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $x$, $y$ and $z$ be positive real numbers. Show that $x^2 + xy^2 + xyz^2 \\ge 4xyz - 4$.\n\nSoit $x$, $y$ et $z$ trois nombres réels positifs. Démontrez que $x^2 + xy^2 + xyz^2 \\ge 4xyz - 4$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Note that\n$$\nx^2 \\ge 4x - 4, \\quad y^2 \\ge 4y - 4, \\quad \\text{and} \\quad z^2 \\ge 4z - 4,\n$$\nand therefore\n$$\nx^2 + xy^2 + xyz^2 \\ge (4x - 4) + x(4y - 4) + xy(4z - 4) = 4xyz - 4.\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 27865,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nLet $x$, $y$ and $z$ be positive real numbers. Show that $x^{2} + x y^{2} + x y z^{2} \\geq 4 x y z - 4$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\nNote that\n$$\nx^{2} \\geq 4x - 4, \\quad y^{2} \\geq 4y - 4, \\quad \\text{and} \\quad z^{2} \\geq 4z - 4\n$$\nand therefore\n$$\nx^{2} + x y^{2} + x y z^{2} \\geq (4x - 4) + x(4y - 4) + x y (4z - 4) = 4 x y z - 4.\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 27866,
    "subject": "Mathematics (Multi-modal)",
    "question": "For any positive integers $n$ and $k$, let $L(n, k)$ be the least common multiple of the $k$ consecutive integers $n, n+1, \\dots, n+k-1$. Show that for any integer $b$, there exist integers $n$ and $k$ such that $L(n, k) > b L(n + 1, k)$.\n\nSoit $L(n, k)$ le plus petit commun multiple de la suite des $k$ entiers consécutifs $n, n + 1, \\dots, n + k - 1$, où $n$ et $k$ sont deux entiers positifs quelconques. Montrez que pour tout entier $b$, il existe des nombres entiers $n$ et $k$ tels que $L(n, k) > b L(n + 1, k)$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "**I.** Let $p > b$ be prime, let $n = p^3$ and $k = p^2$. If $p^3 < i < p^3 + p^2$, then no power of $p$ greater than 1 divides $i$, while $p$ divides $p^3 + p$. It follows that $L(p^3, p^2) = p^2 L(p^3 + 1, p^2 - 1)$. A similar calculation shows that $L(p^3 + 1, p^2) = p L(p^3 + 1, p^2 - 1)$. Thus $L(p^3, p^2) = p L(p^3 + 1, p^2) > b L(p^3 + 1, p^2)$.\n\nII. Let $m > 1$. Then $L(m! - 1, m + 1)$ is the least common multiple of the integers from $m! - 1$ to $m! + m - 1$. But $m! - 1$ is relatively prime to all of $m!$, $m! + 1, \\dots, m! + m - 1$. It follows that $L(m! - 1, m + 1) = (m! - 1) M$, where $M = \\text{lcm}(m!, m! + 1, \\dots, m! + m - 1)$.\nNow consider $L(m!, m + 1)$. This is $\\text{lcm}(M, m! + m)$. But $m! + m = m((m - 1)! + 1)$, and $m$ divides $M$. Thus $\\text{lcm}(M, m! + m) \\le M((m - 1)! + 1)$, and\n$$\n\\frac{L(m! - 1, m + 1)}{L(m!, m + 1)} \\ge \\frac{m! - 1}{(m - 1)! + 1}.\n$$\nSince $m$ can be arbitrarily large, so can $L(m! - 1, m + 1)/L(m!, m + 1)$. Therefore taking $n = m! - 1$ for sufficiently large $m$, and $k = m + 1$, works.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 27867,
    "subject": "Mathematics (Multi-modal)",
    "question": "A number of robots are placed on the squares of a finite, rectangular grid of squares. A square can hold any number of robots. Every edge of the grid is classified as either *passable* or *impassable*. All edges on the boundary of the grid are *impassable*.\nYou can give any of the commands *up*, *down*, *left*, or *right*. All of the robots then simultaneously try to move in the specified direction. If the edge adjacent to a robot in that direction is passable, the robot moves across the edge and into the next square. Otherwise, the robot remains on its current square. You can then give another command of *up*, *down*, *left*, or *right*, then another, for as long as you want.\nSuppose that for any individual robot, and any square on the grid, there is a finite sequence of commands that will move that robot to that square. Prove that you can also give a finite sequence of commands such that *all* of the robots end up on the same square at the same time.\n\nUn certain nombre de robots sont placés sur les carrés composant une grille rectangulaire de dimension finie. Chaque carré peut contenir un nombre quelconque de robots. Les bords des carrés de la grille sont classés comme *franchissable* ou *infranchissable*. Les côtés qui forment le pourtour de la grille sont infranchissables.\nVous pouvez donner n'importe laquelle des commandes suivantes : en **haut**, en **bas**, à **gauche** ou à **droite**. Tous les robots tentent alors de se déplacer simultanément dans la direction précisée. Si le bord adjacent au carré vers lequel se déplace un robot est franchissable, le robot le franchit et se place dans le carré suivant. Sinon, le robot reste dans le carré où il se trouve. Vous pouvez ensuite lancer une autre commande de déplacement vers le **haut**, le **bas**, la **gauche** ou la **droite**, et encore une autre, aussi longtemps que vous le désirez.\nSupposons que pour chaque robot, et ce, pour n'importe quel carré, il existe une suite finie de commandes qui amèneront ce robot au carré donné. Démontrez que vous pouvez aussi lancer une suite finie de commandes de sorte que tous les robots finiront par se retrouver simultanément dans le même carré.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "We will prove any two robots can be moved to the same square. From that point on, they will always be on the same square. We can then similarly move a third robot onto the same square as these two, and then a fourth, and so on, until all robots are on the same square.\n\nTowards that end, consider two robots *A* and *B*. Let $d(A, B)$ denote the minimum number of commands that need to be given in order to move $A$ to the square on which $B$ is currently standing. We will give a procedure that is guaranteed to decrease $d(A, B)$. Since $d(A, B)$ is a non-negative integer, this procedure will eventually decrease $n$ to $0$, which finishes the proof.\n\nLet $n = d(A, B)$, and let $S = \\{s_1, s_2, \\dots, s_n\\}$ be a minimum sequence of moves that takes $A$ to the square where $B$ is currently standing. Certainly $A$ will not run into an impassable edge during this sequence, or we could get a shorter sequence by removing that command. Now suppose $B$ runs into an impassable edge after some command $s_i$. From that point, we can get $A$ to the square on which $B$ started with the commands $s_{i+1}, s_{i+2}, \\dots, s_n$ and then to the square where $B$ is currently with the commands $s_1, s_2, \\dots, s_{i-1}$. But this was only $n-1$ commands in total, and so we have decreased $d(A, B)$ as required.\n\nOtherwise, we have given a sequence of $n$ commands to $A$ and $B$, and neither ran into an impassable edge during the execution of these commands. In particular, the vector $v$ connecting $A$ to $B$ on the grid must have never changed. We moved $A$ to the position $B = A + v$, and therefore we must have also moved $B$ to $B + v$. Repeating this process $k$ times, we will move $A$ to $A + kv$ and $B$ to $B + kv$. But if $v \\neq (0,0)$, this will eventually force $B$ off the edge of the grid, giving a contradiction.",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 27868,
    "subject": "Mathematics (Multi-modal)",
    "question": "A bookshelf contains $n$ volumes, labelled $1$ to $n$ in some order. The librarian wishes to put them in the correct order as follows. The librarian selects a volume that is too far to the right, say the volume with label $k$, takes it out, and inserts it so that it is in the $k$-th place. For example, if the bookshelf contains the volumes $1$, $3$, $2$, $4$ in that order, the librarian could take out volume $2$ and place it in the second position. The books will then be in the correct order $1$, $2$, $3$, $4$.\n\na. Show that if this process is repeated, then, however the librarian makes the selections, all the volumes will eventually be in the correct order.\n\nb. What is the largest number of steps that this process can take?\n\nUne étagère contient $n$ volumes étiquetés de $1$ à $n$, rangés dans un certain ordre. Le bibliothécaire souhaite les mettre dans le bon ordre de la façon suivante : il choisit un volume qui se trouve trop à droite, par exemple le volume étiqueté $k$, le retire de son emplacement et l'insère à la $k$-ième place. Par exemple, si les volumes sont rangés dans l'ordre $1$, $3$, $2$, $4$, le bibliothécaire peut prendre le volume $2$ et le mettre à la deuxième place. Les livres sont alors rangés dans le bon ordre, soit $1$, $2$, $3$, $4$.\n\na) Démontrez que si l'on répète ce processus, tous les volumes finiront par être dans le bon ordre, et ce, quelle que soit la manière dont le bibliothécaire les range.\n\nb) Quel est le plus grand nombre d'étapes que peut exiger un tel processus?",
    "options": [],
    "answer": "2^{n-1} - 1",
    "solution": "(a) If $t_k$ is the number of times that volume $k$ is selected, then we have $t_k \\le 1 + (t_1 + t_2 + \\dots + t_{k-1})$. This is because volume $k$ must move to the right between selections, which means some volume was placed to its left. The only way that can happen is if a lower-numbered volume was selected. This leads to the bound $t_k \\le 2^{k-1}$. Furthermore, $t_n = 0$ since the $n$th volume will never be too far to the right. Therefore if $N$ is the total number of moves then\n$$\nN = t_1 + t_2 + \\dots + t_{n-1} \\le 1 + 2 + \\dots + 2^{n-2} = 2^{n-1} - 1,\n$$\n\nand in particular the process terminates.\n\n(b) Conversely, $2^{n-1}-1$ moves are required for the configuration $(n, 1, 2, 3, \\ldots, n-1)$ if the librarian picks the rightmost eligible volume each time.\nThis can be proved by induction: if at a certain stage we are at $(x, n-k, n-k+1, \\ldots, n-1)$, then after $2^k-1$ moves, we will have moved to $(n-k, n-k+1, \\ldots, n-1, x)$ without touching any of the volumes further to the left. Indeed, after $2^{k-1}-1$ moves, we get to $(x, n-k+1, n-k+2, \\ldots, n-1, n-k)$, which becomes $(n-k, x, n-k+1, n-k+2, \\ldots, n-1)$ after 1 more move, and then $(n-k, n-k+1, \\ldots, n-1, x)$ after another $2^{k-1}-1$ moves. The result follows by taking $k=n-1$.",
    "topic": "Geometry",
    "subtopic": "Solid Geometry"
  },
  {
    "id": 27869,
    "subject": "Mathematics (Multi-modal)",
    "question": "Determine all polynomials $P(x)$ with real coefficients such that\n$$\n(x + 1)P(x - 1) - (x - 1)P(x)\n$$\nis a constant polynomial.",
    "options": [],
    "answer": "All real polynomials of the form P(x) = a x^2 + a x + c for real a and c (including constant polynomials when a = 0).",
    "solution": "The answer is $P(x)$ being any constant polynomial and $P(x) \\equiv kx^2 + kx + c$ for any (nonzero) constant $k$ and constant $c$.\nLet $\\Lambda$ be the expression $(x+1)P(x-1) - (x-1)P(x)$, i.e. the expression in the problem statement.\nSubstituting $x = -1$ into $\\Lambda$ yields $2P(-1)$ and substituting $x = 1$ into $\\Lambda$ yields $2P(1)$. Since $(x+1)P(x-1) - (x-1)P(x)$ is a constant polynomial, $2P(-1) = 2P(0)$. Hence, $P(-1) = P(0)$.\nLet $c = P(-1) = P(0)$ and $Q(x) = P(x) - c$. Then $Q(-1) = Q(0) = 0$. Hence, $0, -1$ are roots of $Q(x)$. Consequently, $Q(x) = x(x+1)R(x)$ for some polynomial $R$. Then $P(x) - c = x(x+1)R(x)$, or equivalently, $P(x) = x(x+1)R(x) + c$.\nSubstituting this into $\\Lambda$ yield\n$$\n(x+1)((x-1)xR(x-1) + c) - (x-1)(x(x+1)R(x) + c)\n$$\nThis is a constant polynomial and simplifies to\n$$\nx(x-1)(x+1)(R(x-1) - R(x)) + 2c.\n$$\n\nSince this expression is a constant, so is $x(x-1)(x+1)(R(x-1)-R(x))$. Therefore, $R(x-1)-R(x) = 0$ as a polynomial. Therefore, $R(x) = R(x-1)$ for all $x \\in \\mathbb{R}$. Then $R(x)$ is a polynomial that takes on certain values for infinitely many values of $x$. Let $k$ be such a value. Then $R(x) - k$ has infinitely many roots, which can occur if and only if $R(x) - k = 0$. Therefore, $R(x)$ is identical to a constant $k$. Hence, $Q(x) = kx(x+1)$ for some constant $k$. Therefore, $P(x) = kx(x+1)+c = kx^2+kx+c$.\nFinally, we verify that all such $P(x) = kx(x+1)+c$ work. Substituting this into $\\Lambda$ yields\n$$\n(x+1)(kx(x-1)+c) - (x-1)(kx(x+1)+c) = kx(x+1)(x-1) + c(x+1) - kx(x+1)(x-1) - c(x-1) = 2c.\n$$\nHence, $P(x) = kx(x+1)+c = kx^2+kx+c$ is a solution to the given equation for any constant $k$. Note that this solution also holds for $k=0$. Hence, constant polynomials are also solutions to this equation. $\\square$\nAs in Solution 1, any constant polynomial $P$ satisfies the given property. Hence, we will assume that $P$ is not a constant polynomial.\nLet $n$ be the degree of $P$. Since $P$ is not constant, $n \\ge 1$. Let\n$$\nP(x) = \\sum_{i=0}^{n} a_{i}x^{i},\n$$\nwith $a_n \\neq 0$. Then\n$$\n(x+1) \\sum_{i=0}^{n} a_{i}(x-1)^{i} - (x-1) \\sum_{i=0}^{n} a_{i}x^{i} = C,\n$$\nfor some constant $C$. We will compare the coefficient of $x^n$ of the left-hand side of this equation with the right-hand side. Since $C$ is a constant and $n \\ge 1$, the coefficient of $x^n$ of the right-hand side is equal to zero. We now determine the coefficient of $x^n$ of the left-hand side of this expression.\nThe left-hand side of the equation simplifies to\n$$\nx \\sum_{i=0}^{n} a_{i}(x-1)^{i} + \\sum_{i=0}^{n} a_{i}(x-1)^{i} - x \\sum_{i=0}^{n} a_{i}x^{i} + \\sum_{i=0}^{n} a_{i}x^{i}.\n$$\n\nWe will determine the coefficient $x^n$ of each of these four terms.\nBy the Binomial Theorem, the coefficient of $x^n$ of the first term is equal to that of $x (a_{n-1}(x-1)^{n-1} + a_n(x-1)^n) = a_{n-1} - \\binom{n}{n-1}a_n = a_{n-1} - n a_n$.\nThe coefficient of $x^n$ of the second term is equal to that of $a_n(x-1)^n$, which is $a_n$.\nThe coefficient of $x^n$ of the third term is equal to $a_{n-1}$ and that of the fourth term is equal to $a_n$.\nSumming these four coefficients yield $a_{n-1} - n a_n + a_n - a_{n-1} + a_n = (2-n)a_n$.\nThis expression is equal to 0. Since $a_n \\neq 0$, $n = 2$. Hence, $P$ is a quadratic polynomial.\nLet $P(x) = a x^2 + b x + c$, where $a, b, c$ are real numbers with $a \\neq 0$. Then\n$$\n(x+1)(a(x-1)^2 + b(x-1) + c) - (x-1)(a x^2 + b x + c) = C.\n$$\nSimplifying the left-hand side yields\n$$\n(b-a)x + 2c = 2C.\n$$\nTherefore, $b - a = 0$ and $2c = 2C$. Hence, $P(x) = a x^2 + a x + c$. As in Solution 1, this is a valid solution for all $a \\in \\mathbb{R} \\setminus \\{0\\}$. $\\square$",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 27870,
    "subject": "Mathematics (Multi-modal)",
    "question": "The sequence $a_1, a_2, \\dots, a_n$ consists of the numbers $1, 2, \\dots, n$ in some order. For which positive integers $n$ is it possible that the $n+1$ numbers $0, a_1, a_1+a_2, a_1+a_2+a_3, \\dots, a_1+a_2+\\dots+a_n$ all have different remainders when divided by $n+1$?",
    "options": [],
    "answer": "n is odd",
    "solution": "**Solution:** It is possible if and only if $n$ is odd.\nIf $n$ is even, then $a_1 + a_2 + \\dots + a_n = 1 + 2 + \\dots + n = \\frac{n}{2} \\cdot (n + 1)$, which is congruent to $0$ mod $n + 1$. Therefore, the task is impossible.\n\nNow suppose $n$ is odd. We will show that we can construct $a_1, a_2, \\dots, a_n$ that satisfy the conditions given in the problem. Then let $n = 2k + 1$ for some non-negative integer $k$. Consider the sequence: $1, 2k, 3, 2k-2, 5, 2k-3, \\dots, 2, 2k+1$, i.e. for each $1 \\le i \\le 2k+1$, $a_i = i$ if $i$ is odd and $a_i = 2k+2-i$ if $i$ is even.\n\nWe first show that each term $1, 2, \\dots, 2k+1$ appears exactly once. Clearly, there are $2k+1$ terms. For each odd number $m$ in $\\{1, 2, \\dots, 2k+1\\}$, $a_m = m$. For each even number $m$ in this set, $a_{2k+2-m} = 2k+2-(2k+2-m)=m$. Hence, every number appears in $a_1, \\dots, a_{2k+1}$. Hence, $a_1, \\dots, a_{2k+1}$ does consist of the numbers $1, 2, \\dots, 2k+1$ in some order.\n\nWe now determine $a_1 + a_2 + \\dots + a_m \\pmod{2k+2}$. We will consider the cases when $m$ is odd and when $m$ is even separately. Let $b_m = a_1 + a_2 + \\dots + a_m$.\n\nIf $m$ is odd, note that $a_1 \\equiv 1 \\pmod{2k+2}$, $a_2 + a_3 = a_4 + a_5 = \\dots = a_{2k} + a_{2k+1} = 2k+3 \\equiv 1 \\pmod{2k+2}$. Therefore, $\\{b_1, b_3, \\dots, b_{2k+1}\\} = \\{1, 2, 3, \\dots, k+1\\} \\pmod{2k+2}$.\n\nIf $m$ is even, note that $a_1 + a_2 = a_3 + a_4 = \\dots = a_{2k-1} + a_{2k} = 2k+1 \\equiv -1 \\pmod{2k+2}$. Therefore, $\\{b_2, b_4, \\dots, b_{2k}\\} = \\{-1, -2, \\dots, -k\\} \\pmod{2k+2} \\equiv \\{2k+1, 2k, \\dots, k+2\\} \\pmod{2k+2}$.\n\nTherefore, $b_1, b_2, \\dots, b_{2k+1}$ do indeed have different remainders when divided by $2k+2$. This completes the problem. $\\square$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 27871,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $G$ be the centroid of a right-angled triangle $ABC$ with $\\angle BCA = 90^\\circ$. Let $P$ be the point on ray $AG$ such that $\\angle CPA = \\angle CAB$, and let $Q$ be the point on ray $BG$ such that $\\angle CQB = \\angle ABC$. Prove that the circumcircles of triangles $AQG$ and $BPG$ meet at a point on side $AB$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Since $\\angle C = 90^\\circ$, the point $C$ lies on the semicircle with diameter $AB$ which implies that, if $M$ is the midpoint of side $AB$, then $MA = MC = MB$. This implies that triangle $AMC$ is isosceles and hence that $\\angle ACM = \\angle A$. By definition, $G$ lies on segment $M$ and it follows that $\\angle ACG = \\angle ACM = \\angle A = \\angle CPA$. This implies that triangles $APC$ and $ACG$ are similar and hence that $AC^2 = AG \\cdot AP$. Now if $D$ denotes the foot of the perpendicular from $C$ to $AB$, it follows that triangles $ACD$ and $ABC$ are similar which implies that $AC^2 = AD \\cdot AB$. Therefore $AG \\cdot AP = AC^2 = AD \\cdot AB$ and, by power of a point, quadrilateral $DGPB$ is cyclic. This implies that $D$ lies on the circumcircle of triangle $BPG$ and, by a symmetric argument, it follows that $D$ also lies on the circumcircle of triangle $AGQ$. Therefore these two circumcircles meet at the point $D$ on side $AB$.\nDefine $D$ and $M$ as in Solution 1. Let $R$ be the point on side $AB$ such that $AC = CR$ and triangle $ACR$ is isosceles. Since $\\angle CRA = \\angle A = \\angle CPA$, it follows that $CPRA$ is cyclic and hence that $\\angle GPR = \\angle APR = \\angle ACR = 180^\\circ - 2\\angle A$. As in Solution 1, $MC = MB$ and hence $\\angle GMR = \\angle CMB = 2\\angle A = 180^\\circ - \\angle GPR$. Therefore $GPRM$ is cyclic and, by power of a point, $AM \\cdot AR = AG \\cdot AP$. Since $ACR$ is isosceles, $D$ is the midpoint of $AR$ and thus, since $M$ is the midpoint of $AB$, it follows that $AM \\cdot AR = AD \\cdot AB = AG \\cdot AP$. Therefore $DGPB$ is cyclic, implying the result as in Solution 1.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 27872,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $n$ be a positive integer. For any positive integer $j$ and positive real number $r$, define $f_j(r)$ and $g_j(r)$ by\n$$\nf_j(r) = \\min(jr, n) + \\min\\left(\\frac{j}{r}, n\\right), \\quad \\text{and} \\quad g_j(r) = \\min(\\lceil jr \\rceil, n) + \\min\\left(\\left\\lfloor \\frac{j}{r} \\right\\rfloor, n\\right),\n$$\nwhere $\\lceil x \\rceil$ denotes the smallest integer greater than or equal to $x$. Prove that\n$$\n\\sum_{j=1}^{n} f_j(r) \\le n^2 + n \\le \\sum_{j=1}^{n} g_j(r)\n$$\nfor all positive real numbers $r$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "**Solution 1:** We first prove the left hand side inequality. We begin by drawing an $n \\times n$ board, with corners at $(0,0)$, $(n,0)$, $(0,n)$ and $(n,n)$ on the Cartesian plane.\nConsider the line $\\ell$ with slope $r$ passing through $(0,0)$. For each $j \\in \\{1, \\dots, n\\}$, consider the point $(j, \\min(jr, n))$. Note that each such point either lies on $\\ell$ or the top edge of the board. In the $j^{th}$ column from the left, draw the rectangle of height $\\min(jr, n)$. Note that the sum of the $n$ rectangles is equal to the area of the board under the line $\\ell$ plus $n$ triangles (possibly with area 0) each with width at most 1 and whose sum of the heights is at most $n$. Therefore, the sum of the areas of these $n$ triangles is at most $n/2$. Therefore, $\\sum_{j=1}^n \\min(jr, n)$ is at most the area of the square under $\\ell$ plus $n/2$.\nConsider the line with slope $1/r$. By symmetry about the line $y=x$, the area of the square under the line with slope $1/r$ is equal to the area of the square above the line $\\ell$. Therefore, using the same reasoning as before, $\\sum_{j=1}^n \\min(j/r, n)$ is at most the area of the square above $\\ell$ plus $n/2$.\nTherefore, $\\sum_{j=1}^n f_j(r) = \\sum_{j=1}^n (\\min(jr, n) + \\min(\\frac{j}{r}, n))$ is at most the area of the board plus $n$, which is $n^2+n$. This proves the left hand side inequality.\nTo prove the right hand side inequality, we will use the following lemma:\n**Lemma:** Consider the line $\\ell$ with slope $s$ passing through $(0,0)$. Then the number of squares on the board that contain an interior point below $\\ell$ is $\\sum_{j=1}^n \\min(\\lceil js \\rceil, n)$.\n*Proof of Lemma:* For each $j \\in \\{1, \\dots, n\\}$, we count the number of squares in the $j^{th}$ column (from the left) that contain an interior point lying below the line $\\ell$. The line $x=j$ intersect the line $\\ell$ at $(j, js)$. Hence, since each column contains $n$ squares total, the number of such squares is $\\min(\\lfloor js \\rfloor, n)$. Summing over all $j \\in \\{1, 2, \\dots, n\\}$ proves the lemma. *End Proof of Lemma*\nBy the lemma, the rightmost expression of the inequality is equal to the number of squares containing an interior point below the line with slope $r$ plus the number of squares containing an interior point below the line with slope $1/r$. By symmetry about the line $y = x$, the latter number is equal to the number of squares containing an interior point above the line with slope $r$. Therefore, the rightmost expression of the inequality is equal to the number of squares of the board plus the number of squares of which $\\ell$ passes through the interior. The former is equal to $n^2$. Hence, to prove the inequality, it suffices to show that every line passes through the interior of at least $n$ squares. Since $\\ell$ has positive slope, each $\\ell$ passes through either $n$ rows and/or $n$ columns. In either case, $\\ell$ passes through the interior of at least $n$ squares. Hence, the right inequality holds. $\\square$\n\n\n**Solution 2:** We first prove the left inequality. Define the function $f(r) = \\sum_{j=1}^{n} f_j(r)$. Note that $f(r) = f(1/r)$ for all $r > 0$. Therefore, we may assume that $r \\ge 1$.\nLet $m = \\lfloor n/r \\rfloor$, where $\\lfloor x \\rfloor$ denotes the largest integer less than or equal to $x$. Then $\\min(jr, n) = jr$ for all $j \\in \\{1, \\dots, m\\}$ and $\\min(jr, n) = n$ for all $j \\in \\{m+1, \\dots, n\\}$. Note that since $r \\ge 1$, $\\min(j/r, n) \\le n$ for all $j \\in \\{1, \\dots, n\\}$. Therefore,\n$$\n\\begin{aligned}\nf(r) &= \\sum_{j=1}^{n} f_j(r) = (1+2+\\dots+m)r + (n-m)n + (1+2+\\dots+n) \\cdot \\frac{1}{r} \\\\\n&= \\frac{m(m+1)}{2} \\cdot r + \\frac{n(n+1)}{2} \\cdot \\frac{1}{r} + n(n-m)\n\\end{aligned} \n\\quad (1)\n$$\nThen by (2), note that $f(r) \\le n^2 + n$ if and only if\n$$\n\\frac{m(m+1)r}{2} + \\frac{n(n+1)}{2r} \\le n(m+1)\n$$\nif and only if\n$$\nm(m+1)r^2 + n(n+1) \\le 2rn(m+1) \\quad (2)\n$$\nSince $m = \\lfloor n/r \\rfloor$, there exist a real number $b$ satisfying $0 \\le b < r$ such that $n = mr + b$. Substituting this into (2) yields\n$$\nm(m+1)r^2 + (mr+b)(mr+b+1) \\le 2r(mr+b)(m+1),\n$$\n$$\n2m^2r^2 + mr^2 + (2mb + m)r + b^2 + b \\leq 2m^2r^2 + 2mr^2 + 2mbr + 2br,\n$$\nwhich simplifies to $mr + b^2 + b \\leq mr^2 + 2br \\Leftrightarrow b(b + 1 - 2r) \\leq mr(r - 1) \\Leftrightarrow b((b - r) + (1 - r)) \\leq mr(r - 1)$. This is true since\n$$\nb((b - r) + (1 - r)) \\leq 0 \\leq mr(r - 1),\n$$\nwhich holds since $r \\geq 1$ and $b < r$. Therefore, the left inequality holds.\n\nWe now prove the right inequality. Define the function $g(r) = \\sum_{j=1}^{n} = g_j(r)$. Note that $g(r) = g(1/r)$ for all $r > 0$. Therefore, we may assume that $r \\geq 1$. We will consider two cases; $r \\geq n$ and $1 \\leq r < n$.\nIf $r \\geq n$, then $\\min(\\lceil jr \\rceil, n) = n$ and $\\min(\\lceil j/r \\rceil, n) = 1$ for all $j \\in \\{1, \\dots, n\\}$. Hence, $g_j(r) = n + 1$ for all $j \\in \\{1, \\dots, n\\}$. Therefore, $g(r) = n(n + 1) = n^2 + n$, implying that the inequality is true.\nNow we consider the case $1 \\leq r < n$. Let $m = \\lfloor n/r \\rfloor$. Hence, $jr \\leq n$ for all $j \\in \\{1, \\dots, m\\}$, i.e. $\\min(\\lceil jr \\rceil, n) = \\lceil jr \\rceil$ and $jr \\geq n$ for all $j \\in \\{m + 1, \\dots, n\\}$, i.e. $\\min(\\lceil jr \\rceil, n) = n$. Therefore,\n$$\n\\sum_{j=1}^{n} \\min(\\lceil jr \\rceil, n) = \\sum_{j=1}^{m} \\lceil jr \\rceil + (n - m)n. \\quad (3)\n$$\nWe will now consider the second sum $\\sum_{j=1}^{n} \\min\\{\\lceil j/r \\rceil, n\\}$.\nSince $r \\geq 1$, $\\min(\\lceil j/r \\rceil, n) \\leq \\min(\\lceil n/r \\rceil, n) \\leq n$. Therefore, $\\min(\\lceil j/r \\rceil, n) = \\lceil j/r \\rceil$. Since $m = \\lfloor n/r \\rfloor$, $\\lceil n/r \\rceil \\leq m + 1$. Since $r > 1$, $m < n$, which implies that $m+1 \\leq n$. Therefore, $\\min\\{\\lceil j/r \\rceil, n\\} = \\lceil j/r \\rceil \\leq \\lceil n/r \\rceil \\leq m + 1$ for all $j \\in \\{1, \\dots, n\\}$.\nFor each positive integer $k \\in \\{1, \\dots, m + 1\\}$, we now determine the number of positive integers $j \\in \\{1, \\dots, n\\}$ such that $\\lceil j/r \\rceil = k$. We denote this number by $s_k$.\nNote that $\\lceil j/r \\rceil = k$ if and only if $k - 1 < j/r \\leq k$ if and only if $(k - 1)r < j \\leq \\min(kr, n)$, since $j \\leq n$. We will handle the cases $k \\in \\{1, \\dots, m\\}$ and $k = m + 1$ separately. If $k \\in \\{1, \\dots, m\\}$, then $\\min(kr, n) = kr$, since $r \\leq m$ and $m = \\lfloor n/r \\rfloor$.\n\nThe set of positive integers $j$ satisfying $(k-1)r < j \\le kr$ is $\\{\\lfloor(k-1)r\\rfloor + 1, \\lfloor(k-1)r\\rfloor + 2, \\dots, \\lfloor kr\\rfloor\\}$. Hence,\n$$\ns_k = \\lfloor rk \\rfloor - (\\lfloor r(k-1) \\rfloor + 1) + 1 = \\lfloor rk \\rfloor - \\lfloor r(k-1) \\rfloor\n$$\nfor all $k \\in \\{1, \\dots, m\\}$. If $k = m + 1$, then $(k-1)r < j \\le \\min(kr, n) = n$. The set of positive integers $j$ satisfying $(k-1)r < j \\le kr$ is $\\{\\lfloor(k-1)r\\rfloor + 1, \\dots, n\\}$. Then $s_{m+1} = n - \\lfloor r(k-1) \\rfloor = n - \\lfloor mr \\rfloor$. Note that this number is non-negative by the definition of $m$. Therefore, by the definition of $s_k$, we have\n$$\n\\begin{aligned}\n\\sum_{j=1}^{n} \\min \\left( \\left\\lfloor \\frac{j}{r} \\right\\rfloor, n \\right) &= \\sum_{k=1}^{m+1} k s_k \\\\\n&= \\sum_{k=1}^{m} \\left( k \\left( \\lfloor kr \\rfloor - \\lfloor (k-1)r \\rfloor \\right) \\right) + (m+1)(n - \\lfloor rm \\rfloor) = (m+1)n - \\sum_{k=1}^{m} \\lfloor kr \\rfloor. \\quad (4)\n\\end{aligned}\n$$\nSumming (??) and (??) yields that\n$$\ng(r) = n^2 + n + \\sum_{j=1}^{m} (\\lceil jr \\rceil - \\lfloor jr \\rfloor) \\ge n^2 + n,\n$$\nwhich proves the right inequality. $\\square$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 27873,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nLet $n$ be a positive integer. For any positive integer $j$ and positive real number $r$, define\n$$\nf_{j}(r)=\\min (j r, n)+\\min \\left(\\frac{j}{r}, n\\right), \\quad \\text{and} \\quad g_{j}(r)=\\min (\\lceil j r\\rceil, n)+\\min \\left(\\left\\lceil\\frac{j}{r}\\right\\rceil, n\\right),\n$$\nwhere $\\lceil x\\rceil$ denotes the smallest integer greater than or equal to $x$. Prove that\n$$\n\\sum_{j=1}^{n} f_{j}(r) \\leq n^{2}+n \\leq \\sum_{j=1}^{n} g_{j}(r)\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nWe first prove the left hand side inequality. We begin by drawing an $n \\times n$ board, with corners at $(0,0),(n, 0),(0, n)$ and $(n, n)$ on the Cartesian plane.\nConsider the line $\\ell$ with slope $r$ passing through $(0,0)$. For each $j \\in\\{1, \\ldots, n\\}$, consider the point $(j, \\min (j r, n))$. Note that each such point either lies on $\\ell$ or the top edge of the board. In the $j^{\\text{th}}$ column from the left, draw the rectangle of height $\\min (j r, n)$. Note that the sum of the $n$ rectangles is equal to the area of the board under the line $\\ell$ plus $n$ triangles (possibly with area 0) each with width at most 1 and whose sum of the heights is at most $n$. Therefore, the sum of the areas of these $n$ triangles is at most $n / 2$. Therefore, $\\sum_{j=1}^{n} \\min (j r, n)$ is at most the area of the square under $\\ell$ plus $n / 2$.\n\nConsider the line with slope $1 / r$. By symmetry about the line $y=x$, the area of the square under the line with slope $1 / r$ is equal to the area of the square above the line $\\ell$. Therefore, using the same reasoning as before, $\\sum_{j=1}^{n} \\min (j / r, n)$ is at most the area of the square above $\\ell$ plus $n / 2$.\n\nTherefore, $\\sum_{j=1}^{n} f_{j}(r)=\\sum_{j=1}^{n}\\left(\\min (j r, n)+\\min \\left(\\frac{j}{r}, n\\right)\\right)$ is at most the area of the board plus $n$, which is $n^{2}+n$. This proves the left hand side inequality.\n\nTo prove the right hand side inequality, we will use the following lemma:\n\nLemma: Consider the line $\\ell$ with slope $s$ passing through $(0,0)$. Then the number of squares on the board that contain an interior point below $\\ell$ is $\\sum_{j=1}^{n} \\min (\\lceil j s\\rceil, n)$.\n\nProof of Lemma: For each $j \\in\\{1, \\ldots, n\\}$, we count the number of squares in the $j^{\\text{th}}$ column (from the left) that contain an interior point lying below the line $\\ell$. The line $x=j$ intersects the line $\\ell$ at $(j, j s)$. Hence, since each column contains $n$ squares total, the number of such squares is $\\min (\\lceil j s\\rceil, n)$. Summing over all $j \\in\\{1,2, \\ldots, n\\}$ proves the lemma. End Proof of Lemma\n\nBy the lemma, the rightmost expression of the inequality is equal to the number of squares containing an interior point below the line with slope $r$ plus the number of squares containing an interior point below the line with slope $1 / r$. By symmetry about the line $y=x$, the latter number is equal to the number of squares containing an interior point above the line with slope $r$. Therefore, the rightmost expression of the inequality is equal to the number of squares of the board plus the number of squares of which $\\ell$ passes through the interior. The former is equal to $n^{2}$. Hence, to prove the inequality, it suffices to show that every line passes through the interior of at least $n$ squares. Since $\\ell$ has positive slope, each $\\ell$ passes through either $n$ rows and/or $n$ columns. In either case, $\\ell$ passes through the interior of at least $n$ squares. Hence, the right inequality holds.\nSolution:\n\nWe first prove the left inequality. Define the function $f(r)=\\sum_{j=1}^{n} f_{j}(r)$. Note that $f(r)=f(1 / r)$ for all $r>0$. Therefore, we may assume that $r \\geq 1$.\n\nLet $m=\\lfloor n / r\\rfloor$, where $\\lfloor x\\rfloor$ denotes the largest integer less than or equal to $x$. Then $\\min (j r, n)=j r$ for all $j \\in\\{1, \\ldots, m\\}$ and $\\min (j r, n)=n$ for all $j \\in\\{m+1, \\ldots, n\\}$. Note that since $r \\geq 1, \\min (j / r, n) \\leq n$ for all $j \\in\\{1, \\ldots, n\\}$. Therefore,\n$$\n\\begin{gathered}\nf(r)=\\sum_{j=1}^{n} f_{j}(r)=(1+2+\\ldots m) r+(n-m) n+(1+2+\\ldots+n) \\cdot \\frac{1}{r} \\\\\n=\\frac{m(m+1)}{2} \\cdot r+\\frac{n(n+1)}{2} \\cdot \\frac{1}{r}+n(n-m)\n\\end{gathered}\n$$\nThen $f(r) \\leq n^{2}+n$ if and only if\n$$\n\\frac{m(m+1) r}{2}+\\frac{n(n+1)}{2 r} \\leq n(m+1)\n$$\nif and only if\n$$\nm(m+1) r^{2}+n(n+1) \\leq 2 r n(m+1)\n$$\nSince $m=\\lfloor n / r\\rfloor$, there exists a real number $b$ satisfying $0 \\leq b<r$ such that $n=m r+b$. Substituting this into the previous equation yields\n$$\nm(m+1) r^{2}+(m r+b)(m r+b+1) \\leq 2 r(m r+b)(m+1),\n$$\nif and only if\n$$\n2 m^{2} r^{2}+m r^{2}+(2 m b+m) r+b^{2}+b \\leq 2 m^{2} r^{2}+2 m r^{2}+2 m b r+2 b r,\n$$\nwhich simplifies to $m r+b^{2}+b \\leq m r^{2}+2 b r \\Leftrightarrow b(b+1-2 r) \\leq m r(r-1)$. This is true since\n$$\nb((b-r)+(1-r)) \\leq m r(r-1).\n$$\nThis holds since $r \\geq 1$ and $b<r$. Therefore, the left inequality holds.\n\nWe now prove the right inequality. Define the function $g(r)=\\sum_{j=1}^{n} g_{j}(r)$. Note that $g(r)=g(1 / r)$ for all $r>0$. Therefore, we may assume that $r \\geq 1$. We will consider two cases: $r \\geq n$ and $1 \\leq r<n$.\n\nIf $r \\geq n$, then $\\min (\\lceil j r\\rceil, n)=n$ and $\\min (\\lceil j / r\\rceil, n)=1$ for all $j \\in\\{1, \\ldots, n\\}$. Hence, $g_{j}(r)=n+1$ for all $j \\in\\{1, \\ldots, n\\}$. Therefore, $g(r)=n(n+1)=n^{2}+n$, implying that the inequality is true.\n\nNow we consider the case $1 \\leq r<n$. Let $m=\\lfloor n / r\\rfloor$. Hence, $j r \\leq n$ for all $j \\in\\{1, \\ldots, m\\}$, i.e. $\\min (\\lceil j r\\rceil, n)=\\lceil j r\\rceil$ and $j r \\geq n$ for all $j \\in\\{m+1, \\ldots, n\\}$, i.e. $\\min (\\lceil j r\\rceil, n)=n$. Therefore,\n$$\n\\sum_{j=1}^{n} \\min (\\lceil j r\\rceil, n)=\\sum_{j=1}^{m}\\lceil j r\\rceil+(n-m) n\n$$\nWe will now consider the second sum $\\sum_{j=1}^{n} \\min \\{\\lceil j / r\\rceil, n\\}$.\nSince $r \\geq 1, \\min (\\lceil j / r\\rceil, n) \\leq \\min (\\lceil n / r\\rceil, n) \\leq n$. Therefore, $\\min (\\lceil j / r\\rceil, n)=\\lceil j / r\\rceil$. Since $m=\\lfloor n / r\\rfloor,\\lceil n / r\\rceil \\leq m+1$. Since $r>1, m<n$, which implies that $m+1 \\leq n$. Therefore, $\\min \\{\\lceil j / r\\rceil, n\\}=\\lceil j / r\\rceil \\leq\\lceil n / r\\rceil \\leq m+1$ for all $j \\in\\{1, \\ldots, n\\}$.\n\nFor each positive integer $k \\in\\{1, \\ldots, m+1\\}$, we now determine the number of positive integers $j \\in\\{1, \\ldots, n\\}$ such that $\\lceil j / r\\rceil=k$. We denote this number by $s_{k}$.\n\nNote that $\\lceil j / r\\rceil=k$ if and only if $k-1<j / r \\leq k$ if and only if $(k-1) r<j \\leq \\min (k r, n)$, since $j \\leq n$. We will handle the cases $k \\in\\{1, \\ldots, m\\}$ and $k=m+1$ separately. If $k \\in\\{1, \\ldots, m\\}$, then $\\min (k r, n)=k r$, since $r \\leq m$ and $m=\\lfloor n / r\\rfloor$.\n\nThe set of positive integers $j$ satisfying $(k-1) r<j \\leq k r$ is $\\{\\lfloor(k-1) r\\rfloor+1,\\ldots,\\lfloor k r\\rfloor\\}$. Hence,\n$$\ns_{k}=\\lfloor r k\\rfloor-\\lfloor r(k-1)\\rfloor\n$$\nfor all $k \\in\\{1, \\ldots, m\\}$. If $k=m+1$, then $(k-1) r<j \\leq \\min (k r, n)=n$. The set of positive integers $j$ satisfying $(k-1) r<j \\leq k r$ is $\\{\\lfloor(k-1) r\\rfloor+1, \\ldots, n\\}$. Then $s_{m+1}=n-\\lfloor m r\\rfloor$. Note that this number is non-negative by the definition of $m$. Therefore, by the definition of $s_{k}$, we have\n$$\n\\begin{aligned}\n& \\sum_{j=1}^{n} \\min \\left(\\left\\lceil\\frac{j}{r}\\right\\rceil, n\\right)=\\sum_{k=1}^{m+1} k s_{k} \\\\\n& \\quad=\\sum_{k=1}^{m}(k(\\lfloor k r\\rfloor-\\lfloor(k-1) r\\rfloor))+(m+1)(n-\\lfloor r m\\rfloor)=(m+1) n-\\sum_{k=1}^{m}\\lfloor k r\\rfloor .\n\\end{aligned}\n$$\nSumming the two expressions yields that\n$$\ng(r)=n^{2}+n+\\sum_{j=1}^{m}(\\lceil j r\\rceil-\\lfloor j r\\rfloor) \\geq n^{2}+n\n$$\nwhich proves the right inequality.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 27874,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $O$ denote the circumcentre of an acute-angled triangle $ABC$. Let point $P$ on side $AB$ be such that $\\angle BOP = \\angle ABC$, and let point $Q$ on side $AC$ be such that $\\angle COQ = \\angle ACB$. Prove that the reflection of $BC$ in the line $PQ$ is tangent to the circumcircle of triangle $APQ$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let the circumcircle of triangle $OBP$ intersect side $BC$ at the points $R$ and $B$ and let $\\angle A$, $\\angle B$ and $\\angle C$ denote the angles at vertices $A$, $B$ and $C$, respectively. Now note that since $\\angle BOP = \\angle B$ and $\\angle COQ = \\angle C$, it follows that\n$$\n\\angle POQ = 360^\\{\\circ\\} - \\angle BOP - \\angle COQ - \\angle BOC = 360^\\{\\circ\\} - (180^\\{\\circ\\} - \\angle A) - 2\\angle A = 180^\\{\\circ\\} - \\angle A.\n$$\nThis implies that $APOQ$ is a cyclic quadrilateral. Since $BPOR$ is cyclic,\n$$\n\\angle QOR = 360^\\{\\circ\\} - \\angle POQ - \\angle POR = 360^\\{\\circ\\} - (180^\\{\\circ\\} - \\angle A) - (180^\\{\\circ\\} - \\angle B) = 180^\\{\\circ\\} - \\angle C.\n$$\nThis implies that $CQOR$ is a cyclic quadrilateral. Since $APOQ$ and $BPOR$ are cyclic,\n$$\n\\angle QPR = \\angle QPO + \\angle OPR = \\angle OAQ + \\angle OBR = (90^\\{\\circ\\} - \\angle B) + (90^\\{\\circ\\} - \\angle A) = \\angle C.\n$$\nSince $CQOR$ is cyclic, $\\angle QRC = \\angle COQ = \\angle C = \\angle QPR$ which implies that the circumcircle of triangle $PQR$ is tangent to $BC$. Further, since $\\angle PRB = \\angle BOP = \\angle B$,\n$$\n\\angle PRQ = 180^\\{\\circ\\} - \\angle PRB - \\angle QRC = 180^\\{\\circ\\} - \\angle B - \\angle C = \\angle A = \\angle PAQ.\n$$\nThis implies that the circumcircle of $PQR$ is the reflection of $\\Gamma$ in line $PQ$. By symmetry in line $PQ$, this implies that the reflection of $BC$ in line $PQ$ is tangent to $\\Gamma$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 27875,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $a_1$, $a_2$, $\\dots$, $a_n$ be positive real numbers whose product is $1$. Show that the sum\n$$\n\\frac{a_1}{1+a_1} + \\frac{a_2}{(1+a_1)(1+a_2)} + \\frac{a_3}{(1+a_1)(1+a_2)(1+a_3)} + \\dots + \\frac{a_n}{(1+a_1)(1+a_2)\\dots(1+a_n)}\n$$\nis greater than or equal to $\\frac{2^n - 1}{2^n}$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Note that for every positive integer $m$,\n$$\n\\begin{aligned}\n\\frac{a_m}{(1+a_1)(1+a_2)\\cdots(1+a_m)} &= \\frac{1+a_m}{(1+a_1)(1+a_2)\\cdots(1+a_m)} - \\frac{1}{(1+a_1)(1+a_2)\\cdots(1+a_m)} \\\\\n&= \\frac{1}{(1+a_1)\\cdots(1+a_{m-1})} - \\frac{1}{(1+a_1)\\cdots(1+a_m)}.\n\\end{aligned}\n$$\nTherefore, if we let $b_j = (1+a_1)(1+a_2)\\cdots(1+a_j)$, with $b_0 = 1$, then by telescoping sums,\n$$\n\\sum_{j=1}^{n} \\frac{a_j}{(1+a_1)\\cdots(1+a_j)} = \\sum_{j=1}^{n} \\left( \\frac{1}{b_{j-1}} - \\frac{1}{b_j} \\right) = 1 - \\frac{1}{b_n}.\n$$\nNote that $b_n = (1+a_1)(1+a_2)\\cdots(1+a_n) \\ge (2\\sqrt{a_1})(2\\sqrt{a_2})\\cdots(2\\sqrt{a_n}) = 2^n$, with equality if and only if all $a_i$'s equal $1$. Therefore,\n$$\n1 - \\frac{1}{b_n} \\ge 1 - \\frac{1}{2^n} = \\frac{2^n - 1}{2^n}.\n$$\nTo check that this minimum can be obtained, substitute all $a_i = 1$ to yield\n$$\n\\frac{1}{2} + \\frac{1}{2^2} + \\frac{1}{2^3} + \\dots + \\frac{1}{2^n} = \\frac{2^{n-1} + 2^{n-2} + \\dots + 1}{2^n} = \\frac{2^n - 1}{2^n},\n$$\nas desired.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 27876,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $m$ and $n$ be odd positive integers. Each square of an $m$ by $n$ board is coloured red or blue. A row is said to be red-dominated if there are more red squares than blue squares in the row. A column is said to be blue-dominated if there are more blue squares than red squares in the column. Determine the maximum possible value of the number of red-dominated rows plus the number of blue-dominated columns. Express your answer in terms of $m$ and $n$.",
    "options": [],
    "answer": "m + n − 2 if m, n ≥ 3; otherwise max{m, n} if one of m, n equals 1",
    "solution": "The answer is $m+n-2$ if $m, n \\ge 3$ and $\\max\\{m, n\\}$ if one of $m, n$ is equal to $1$.\n\nNote that it is not possible that all rows are red-dominated and all columns are blue-dominated. This is true, since the number of rows and columns are both odd, the number of squares is odd. Hence, there are more squares of one color than the other. Without loss of generality, suppose there are more red squares than blue squares. Then it is not possible that for every column, there are more blue squares than red squares. Hence, every column cannot be blue-dominated.\n\nIf one of $m, n$ is equal to $1$, say $m$ without loss of generality, then by the claim, the answer is less than $n+1$. The example where there are $n$ blue-dominated columns is by painting every square blue. There are $0$ red-dominated rows. The sum of the two is $n = \\max\\{m, n\\}$.\n\nNow we handle the case $m, n \\ge 3$.\n\nThere are $m$ rows and $n$ columns on the board. Hence, the answer is at most $m+n$. We have already shown that the answer cannot be $m+n$.\n\nSince $m, n$ are odd, let $m = 2a-1$ and $n = 2b-1$ for some positive integers $a, b$. Since $m, n \\ge 3$, $a, b \\ge 2$. We first show that the answer is not $m+n-1$. By symmetry, it suffices to show that we cannot have all rows red-dominated and all-but-one column blue-dominated. If all rows are red-dominated, then each row has at least $b$ red squares. Hence, there are at least $bm = (2a-1)b$ red squares. Since all-but-one column is blue-dominated, there are at least $2b-2$ blue-dominated columns. Each such column then has at least $a$ blue squares. Therefore, there are at least $a(2b-2)$ blue squares. Therefore, the board has at least $(2a-1)b + a(2b-2) = 4ab - b - 2a$ squares. But the total number of squares on the board is\n$$\n(2a-1)(2b-1) = 4ab - 2a - 2b + 1 = 4ab - 2a - b - b + 1 < 4ab - 2a - b,\n$$\nwhich is true since $b \\ge 2$. This is a contradiction. Therefore, the answer is less than $m+n-1$.\n\nWe now claim that there is a colouring of the board such that the number of blue-dominated columns plus the number of red-dominated rows is $m+n-2$; Colour the first column entirely red, and the first row, minus the top-left corner, entirely blue. The remaining uncoloured square is an even-by-even board. Colour the remaining board in an alternating pattern (i.e. checkerboard pattern). Hence, on this even-by-even board, each row has the same number of red squares as blue squares and each column has the same number of red squares as blue squares. Then on the whole board, since the top row, minus the top-left square is blue, all columns, but the leftmost column, are blue-dominated. Hence, there are $n-1$ blue-dominated columns. Since the left column is red, all rows but the top row are red dominated. Hence, there are $m-1$ red-dominated rows. The sum of these two quantities is $m+n-2$, as desired.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 27877,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $p$ be a fixed odd prime. A $p$-tuple $(a_1, a_2, a_3, \\dots, a_p)$ of integers is said to be good if\n(i) $0 \\le a_i \\le p-1$ for all $i$, and\n(ii) $a_1 + a_2 + a_3 + \\dots + a_p$ is not divisible by $p$, and\n(iii) $a_1a_2 + a_2a_3 + a_3a_4 + \\dots + a_p a_1$ is divisible by $p$.\nDetermine the number of good $p$-tuples.",
    "options": [],
    "answer": "p^{p-1} - p^{p-2}",
    "solution": "Let $S$ be the set of all sequences $(b_1, b_2, \\dots, b_p)$ of numbers from the set $\\{0, 1, 2, \\dots, p-1\\}$ such that $b_1 + b_2 + \\dots + b_p$ is not divisible by $p$. We show that $|S| = p^p - p^{p-1}$. For let $b_1, b_2, \\dots, b_{p-1}$ be an arbitrary sequence of numbers chosen from $\\{0, 1, 2, \\dots, p-1\\}$. There are exactly $p-1$ choices for $b_p$ such that $b_1 + b_2 + \\dots + b_{p-1} + b_p \\not\\equiv 0 \\pmod{p}$, and therefore $|S| = p^{p-1}(p-1) = p^p - p^{p-1}$.\n\nNow it will be shown that the number of good sequences in $S$ is $\\frac{1}{p}|S|$. For a sequence $B = (b_1, b_2, \\dots, b_p)$ in $S$, define the sequence $B_k = (a_1, a_2, \\dots, a_p)$ by\n$$\na_i = b_i - b_1 + k \\bmod p\n$$\nfor $1 \\le i \\le p$. Now note that $B$ in $S$ implies that\n$$\na_1 + a_2 + \\dots + a_p \\equiv (b_1 + b_2 + \\dots + b_p) - pb_1 + pk \\equiv (b_1 + b_2 + \\dots + b_p) \\not\\equiv 0 \\pmod{p}\n$$\nand therefore $B_k$ is in $S$ for all non-negative $k$. Now note that $B_k$ has first element $k$ for all $0 \\le k \\le p-1$ and therefore the sequences $B_0, B_1, \\dots, B_{p-1}$ are distinct.\n\nNow define the *cycle* of $B$ as the set $\\{B_0, B_1, \\dots, B_{p-1}\\}$. Note that $B$ is in its own cycle since $B = B_k$ where $k = b_1$. Now note that since every sequence in $S$ is in exactly one cycle, $S$ is the disjoint union of cycles.\nNow it will be shown that exactly one sequence per cycle is good. Consider an arbitrary cycle $B_0, B_1, \\dots, B_{p-1}$, and let $B_0 = (b_1, b_2, \\dots, b_p)$ where $b_0 = 0$, and note that $B_k = (b_1 + k, b_2 + k, \\dots, b_p + k)$ mod $p$. Let $u = b_1 + b_2 + \\dots + b_p$, and $v = b_1b_2 + b_2b_3 + \\dots + b_pb_1$ and note that $(b_1 + k)(b_2 + k) + (b_2 + k)(b_3 + k) + \\dots + (b_p + k)(b_1 + k) = u + 2kv \\pmod p$ for all $0 \\le k \\le p-1$. Since $2v$ is not divisible by $p$, there is exactly one value of $k$ with $0 \\le k \\le p-1$ such that $p$ divides $u + 2kv$ and it is exactly for this value of $k$ that $B_k$ is good. This shows that exactly one sequence per cycle is good and therefore that the number of good sequences in $S$ is $\\frac{1}{p}|S|$, which is $p^{p-1} - p^{p-2}$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 27878,
    "subject": "Mathematics (Multi-modal)",
    "question": "The quadrilateral $ABCD$ is inscribed in a circle. The point $P$ lies in the interior of $ABCD$, and $\\angle PAB = \\angle PBC = \\angle PCD = \\angle PDA$. The lines $AD$ and $BC$ meet at $Q$, and the lines $AB$ and $CD$ meet at $R$. Prove that the lines $PQ$ and $PR$ form the same angle as the diagonals of $ABCD$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $\\Gamma$ be the circumcircle of quadrilateral $ABCD$. Let $\\alpha = \\angle PAB = \\angle PBC = \\angle PCD = \\angle PDA$ and let $T_1, T_2, T_3$ and $T_4$ denote the circumcircles of triangles $APD$, $BPC$, $APB$ and $CPD$, respectively. Let $M$ be the intersection of $T_1$ with line $RP$ and let $N$ be the intersection of $T_3$ with line $SP$. Also let $X$ denote the intersection of diagonals $AC$ and $BD$.\n\nBy power of a point for circles $T_1$ and $\\Gamma$, it follows that $RM \\cdot RP = RA \\cdot RD = RB \\cdot RC$ which implies that the quadrilateral $BMPC$ is cyclic and $M$ lies on $T_2$. Therefore $\\angle PMB = \\angle PCB = \\alpha = \\angle PAB = \\angle DMP$ where all angles are directed. This implies that $M$ lies on the diagonal $BD$ and also that $\\angle XMP = \\angle DMP = \\alpha$. By a symmetric argument applied to $S$, $T_3$ and $T_4$, it follows that $N$ lies on $T_4$ and that $N$ lies on diagonal $AC$ with $\\angle XNP = \\alpha$. Therefore $\\angle XMP = \\angle XNP$ and $X, M, P$ and $N$ are concyclic. This implies that the angle formed by lines $MP$ and $NP$ is equal to one of the angles formed by lines $MX$ and $NX$. The fact that $M$ lies on $BD$ and $RP$ and $N$ lies on $AC$ and $SP$ now implies the desired result.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 27879,
    "subject": "Mathematics (Multi-modal)",
    "question": "Fix positive integers $n$ and $k \\ge 2$. A list of $n$ integers is written in a row on a blackboard. You can choose a contiguous block of integers, and I will either add 1 to all of them or subtract 1 from all of them. You can repeat this step as often as you like, possibly adapting your selections based on what I do. Prove that after a finite number of steps, you can reach a state where at least $n-k+2$ of the numbers on the blackboard are all simultaneously divisible by $k$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "We will think of all numbers as being residues mod $k$. Consider the following strategy:\n\n* If there are less than $k-1$ non-zero numbers, then stop.\n* If the first number is 0, then recursively solve on the remaining numbers.\n* If the first number is $j$ with $0 < j < k$, then choose the interval stretching from the first number to the $j$th-last non-zero number.\n\nFirst note that this strategy is indeed well defined. The first number must have value between 0 and $k-1$, and if we do not stop immediately, then there are at least $k-1$ non-zero numbers, and hence the third step can be performed.\n\nFor each $j$ with $1 \\le j \\le k-2$, we claim the first number can take on the value of $j$ at most a finite number of times without taking on the value of $j-1$ in between. If this were to fail, then every time the first number became $j$, I would have to add 1 to the selected numbers to avoid making it $j-1$. This will always increase the $j$-th last non-zero number, and that number will never be changed by other steps. Therefore, that number would eventually become 0, and the next last non-zero number would eventually become zero, and so on, until the first number itself becomes the $j$-th last non-zero number, at which point we are done since $j \\le k-2$.\n\nRephrasing slightly, if $1 \\le j \\le k-2$, the first number can take on the value of $j$ at most a finite number of times between each time it takes on the value of $j-1$. It then immediately follows that if the first number can take on the value of $j-1$ at most a finite number of times, then it can also only take on the value of $j$ a finite number of times. However, if it ever takes on the value of 0, we have already reduced the problem to $n-1$, so we can assume that never happens. It then follows that the first number can take on all the values $0, 1, 2, \\dots, k-2$ at most a finite number of times.\n\nFinally, every time the first number takes on the value of $k-1$, it must subsequently take on the value of $k-2$ or 0, and so that can also happen only finitely many times.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 27880,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $\\mathbb{N} = \\{1, 2, 3, \\dots\\}$ be the set of positive integers. Find all functions $f$, defined on $\\mathbb{N}$ and taking values in $\\mathbb{N}$, such that $(n - 1)^2 < f(n)f(f(n)) < n^2 + n$ for every positive integer $n$.",
    "options": [],
    "answer": "f(n) = n",
    "solution": "The only such function is $f(n) = n$.\nAssume that $f$ satisfies the given condition. It will be shown by induction that $f(n) = n$ for all $n \\in \\mathbb{N}$. Substituting $n = 1$ yields that $0 < f(1)f(f(1)) < 2$ which implies the base case $f(1) = 1$. Now assume that $f(k) = k$ for all $k < n$ and assume for contradiction that $f(n) \\neq n$.\nOn the one hand, if $f(n) \\le n-1$ then $f(f(n)) = f(n)$ and $f(n)f(f(n)) = f(n)^2 \\le (n-1)^2$ which is a contradiction. On the other hand, if $f(n) \\ge n+1$ then there are several ways to proceed.\n\n**Method 1:** Assume $f(n) = M \\ge n + 1$. Then $(n+1)f(M) \\le f(n)f(f(n)) < n^2 + n$. Therefore $f(M) < n$, and hence $f(f(M)) = f(M)$ and $f(M)f(f(M)) = f(M)^2 < n^2 \\le (M-1)^2$, which is a contradiction. This completes the induction. $\\square$\n\n**Method 2:** First note that if $|a-b| > 1$, then the intervals $((a-1)^2, a^2+a)$ and $((b-1)^2, b^2+b)$ are disjoint which implies that $f(a)$ and $f(b)$ cannot be equal.\nAssuming $f(n) \\ge n + 1$, it follows that $f(f(n)) < \\frac{n^2+n}{f(n)} \\le n$. This implies that for some $a \\le n - 1$, $f(a) = f(f(n))$ which is a contradiction since $|f(n) - a| \\ge n + 1 - a \\ge 2$. This completes the induction. $\\square$\n\n**Method 3:** Assuming $f(n) \\ge n + 1$, it follows that $f(f(n)) < \\frac{n^2+n}{f(n)} \\le n$ and $f(f(f(n))) = f(f(n))$. This implies that $(f(n)-1)^2 < f(f(n))f(f(f(n))) = f(f(n))^2 < f(n)^2 + f(n)$ and therefore that $f(f(n)) = f(n)$ since $f(n)^2$ is the unique square satisfying this constraint. This implies that $f(n)f(f(n)) = f(n)^2 \\ge (n+1)^2$ which is a contradiction, completing the induction. $\\square$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 27881,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABC$ be an acute-angled triangle with altitudes $AD$, $BE$, and $CF$. Let $H$ be the orthocentre, that is, the point where the altitudes meet. Prove that\n$$\n\\frac{AB \\cdot AC + BC \\cdot BA + CA \\cdot CB}{AH \\cdot AD + BH \\cdot BE + CH \\cdot CF} \\le 2.\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "**Method 1:** Let $AB = c$, $AC = b$, and $BC = a$ denote the three side lengths of the triangle.\nAs $\\angle BFH = \\angle BDH = 90^\\circ$, $FHDB$ is a cyclic quadrilateral. By the Power-of-a-Point Theorem, $AH \\cdot AD = AF \\cdot AB$. (We can derive this result in other ways: for example, see Method 2, below.)\nSince $AF = AC \\cdot \\cos \\angle A$, we have $AH \\cdot AD = AC \\cdot AB \\cdot \\cos \\angle A = bc \\cos \\angle A$.\n\nBy the Cosine Law, $\\cos \\angle A = \\frac{b^2 + c^2 - a^2}{2bc}$, which implies that $AH \\cdot AD = \\frac{b^2 + c^2 - a^2}{2}$.\nBy symmetry, we can show that $BH \\cdot BE = \\frac{a^2 + c^2 - b^2}{2}$ and $CH \\cdot CF = \\frac{a^2 + b^2 - c^2}{2}$.\nHence,\n$$\n\\begin{aligned}\nAH \\cdot AD + BH \\cdot BE + CH \\cdot CF &= \\frac{b^2 + c^2 - a^2}{2} + \\frac{a^2 + c^2 - b^2}{2} + \\frac{a^2 + b^2 - c^2}{2} \\\\\n&= \\frac{a^2 + b^2 + c^2}{2}. \\qquad (1)\n\\end{aligned}\n$$\nOur desired inequality, $\\frac{AB \\cdot AC + BC \\cdot BA + CA \\cdot CB}{AH \\cdot AD + BH \\cdot BE + CH \\cdot CF} \\le 2$, is equivalent to the inequality $\\frac{cb + ac + ba}{\\frac{a^2+b^2+c^2}{2}} \\le 2$, which simplifies to $2a^2 + 2b^2 + 2c^2 \\ge 2ab + 2bc + 2ca$.\nBut this last inequality is easy to prove, as it is equivalent to $(a-b)^2+(a-c)^2+(b-c)^2 \\ge 0$.\nTherefore, we have established the desired inequality. The proof also shows that equality occurs if and only if $a = b = c$, i.e., $\\triangle ABC$ is equilateral. $\\square$\n\n**Method 2:** Observe that\n$$\n\\frac{AE}{AH} = \\cos(\\angle HAE) = \\frac{AD}{AC} \\quad \\text{and} \\quad \\frac{AF}{AH} = \\cos(\\angle HAF) = \\frac{AD}{AB}.\n$$\nIt follows that\n$$\nAC \\cdot AE = AH \\cdot AD = AB \\cdot AF.\n$$\nBy symmetry, we similarly have\n$$\nBC \\cdot BD = BH \\cdot BE = BF \\cdot BA \\quad \\text{and} \\quad CD \\cdot CB = CH \\cdot CF = CE \\cdot CA.\n$$\nTherefore\n$$\n\\begin{aligned}\n& 2(AH \\cdot AD + BH \\cdot BE + CH \\cdot CF) \\\\\n&= AB(AF + BF) + AC(AE + CE) + BC(BD + CD) \\\\\n&= AB^2 + AC^2 + BC^2.\n\\end{aligned}\n$$\nThis proves Equation (1) in Method 1. The rest of the proof is the same as the part of the proof of Method 1 that follows Equation (1). $\\square$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 27882,
    "subject": "Mathematics (Multi-modal)",
    "question": "On a $(4n + 2) \\times (4n + 2)$ square grid, a turtle can move between squares sharing a side. The turtle begins in a corner square of the grid and enters each square exactly once, ending in the square where she started. In terms of $n$, what is the largest positive integer $k$ such that there must be a row or column that the turtle has entered at least $k$ distinct times?",
    "options": [],
    "answer": "2n + 2",
    "solution": "We shall prove that the answer is $2n + 2$. Number the rows in increasing order, from top to bottom, and number the columns from left to right. By symmetry, we may (and shall) assume that the turtle starts in the top right corner square.\n\nFirst we shall prove that some row or column must be entered at least $2n + 2$ times. Let $m = 4n + 2$. First note that each time the turtle moves, she enters either a row or a column. Let $r_i$ denote the number of times the turtle enters row $i$, and let $c_i$ be similarly defined for column $i$. Since the turtle moves $m^2$ times,\n$$\nr_1 + r_2 + \\dots + r_m + c_1 + c_2 + \\dots + c_m = m^2.\n$$\nNow note that each time the turtle enters column $1$, the next column she enters must be column $2$. Therefore $c_1$ is equal to the number of times the turtle enters column $2$ from column $1$. Furthermore, the turtle must enter column $2$ from column $3$ at least once, which implies that $c_2 > c_1$. Therefore since the $2m$ terms $r_i$ and $c_i$ are not all equal, one must be strictly greater than $m^2/(2m) = 2n + 1$ and therefore at least $2n + 2$.\n\nNow we construct an example to show that it is possible that no row or column is entered more than $2n + 2$ times. Partition the square grid into four $(2n+1) \\times (2n+1)$ quadrants $A$, $B$, $C$, and $D$, containing the upper left, upper right, lower left, and lower right corners, respectively. The turtle begins at the top right corner square of $B$, moves one square down, and then moves left through the whole second row of $B$. She then moves one square down and moves right through the whole third row of $B$. She continues in this pattern, moving through each remaining row of $B$ in succession and moving one square down when each row is completed. Since $2n+1$ is odd, the turtle ends at the bottom right corner of $B$. She then moves one square down into $D$ and through each column of $D$ in turn, moving one square to the left when each column is completed. She ends at the lower left corner of $D$ and moves left into $C$ and through the rows of $C$, moving one square up when each row is completed, ending in the upper left corner of $C$. She then enters $A$ and moves through the columns of $A$, moving one square right when each column is completed. This takes her to the upper right corner of $A$, whereupon she enters $B$ and moves right through the top row of $B$, which returns her to her starting point. Each row passing through $A$ and $B$ is entered at most $2n+1$ times in $A$ and once in $B$, and thus at most $2n+2$ times in total. Similarly, each row and column in the grid is entered at most $2n+2$ times by this path. (See figure below.)\n\n![](attached_image_1.png)\n\nProblem 3: the case $n = 3$",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 27883,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABC$ be an acute-angled triangle with circumcenter $O$. Let $\\Gamma$ be a circle with centre on the altitude from $A$ in $ABC$, passing through vertex $A$ and points $P$ and $Q$ on sides $AB$ and $AC$. Assume that $BP \\cdot CQ = AP \\cdot AQ$. Prove that $\\Gamma$ is tangent to the circumcircle of triangle $BOC$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $\\omega$ be the circumcircle of $BOC$. Let $M$ be the point diametrically opposite to $O$ on $\\omega$ and let the line $AM$ intersect $\\omega$ at $M$ and $K$. Since $O$ is the circumcenter of $ABC$, it follows that $OB = OC$ and therefore that $O$ is the midpoint of the arc $\\widehat{BOC}$ of $\\omega$. Since $M$ is diametrically opposite to $O$, it follows that $M$ is the midpoint of the arc $\\widehat{BMC}$ of $\\omega$. This implies since $K$ is on $\\omega$ that $KM$ is the bisector of $\\angle BKC$. Since $K$ is on $\\omega$, this implies that $\\angle BKM = \\angle CKM$, i.e. $KM$ is the bisector of $\\angle BKC$.\n\nSince $O$ is the circumcenter of $ABC$, it follows that $\\angle BOC = 2\\angle BAC$. Since $B$, $K$, $O$ and $C$ all lie on $\\omega$, it also follows that $\\angle BKC = \\angle BOC = 2\\angle BAC$. Since $KM$ bisects $\\angle BKC$, it follows that $\\angle BKM = \\angle CKM = \\angle BAC$. The fact that $A$, $K$ and $M$ lie on a line therefore implies that $\\angle AKB = \\angle AKC = 180^\\circ - \\angle BAC$. Now it follows that\n$$\n\\angle KBA = 180^\\circ - \\angle AKB - \\angle KAB = \\angle BAC - \\angle KAB = \\angle KAC.\n$$\nThis implies that triangles $KBA$ and $KAC$ are similar. Rearranging the condition in the problem statement yields that $BP/AP = AQ/CQ$ which, when combined with the fact that $KBA$ and $KAC$ are similar, implies that triangles $KPA$ and $KQC$ are similar. Therefore $\\angle KPA = \\angle KQC = 180^\\circ - \\angle KQA$ which implies that $K$ lies on $\\Gamma$.\n\nNow let $S$ denote the centre of $\\Gamma$ and let $T$ denote the centre of $\\omega$. Note that $T$ is the midpoint of segment $OM$ and that $TM$ and $AS$, which are both perpendicular to $BC$, are parallel. This implies that $\\angle KMT = \\angle KAS$ since $A$, $K$ and $M$ are collinear. Further, since $KTM$ and $KSA$ are isosceles triangles, it follows that $\\angle TKM = \\angle KMT$ and $\\angle SKA = \\angle KSA$. Therefore $\\angle TKM = \\angle SKA$ which implies that $S$, $T$ and $K$ are collinear. Therefore $\\Gamma$ and $\\omega$ intersect at a point $K$ which lies on the line $ST$ connecting the centres of the two circles. This implies that the circles $\\Gamma$ and $\\omega$ are tangent at $K$. $\\Box$",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 27884,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nLet $p$ be a prime number for which $\\frac{p-1}{2}$ is also prime, and let $a$, $b$, $c$ be integers not divisible by $p$. Prove that there are at most $1+\\sqrt{2 p}$ positive integers $n$ such that $n<p$ and $p$ divides $a^{n}+b^{n}+c^{n}$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\nFirst suppose $b \\equiv \\pm a \\pmod{p}$ and $c \\equiv \\pm b \\pmod{p}$. Then, for any $n$, we have $a^{n}+b^{n}+c^{n} \\equiv \\pm a^{n}$ or $\\pm 3 a^{n} \\pmod{p}$. We are given that $p \\neq 3$ (since $\\frac{3-1}{2}$ is not prime) and $p \\nmid a$, so it follows that $a^{n}+b^{n}+c^{n} \\not\\equiv 0 \\pmod{p}$. The claim is trivial in this case.\n\nOtherwise, we may assume without loss of generality that $b \\not\\equiv \\pm a \\pmod{p} \\Longrightarrow b a^{-1} \\not\\equiv \\pm 1 \\pmod{p}$.\n\nNow let $q=\\frac{p-1}{2}$. By Fermat's little theorem, we know that the order of $b a^{-1} \\bmod p$ divides $p-1=2 q$. However, since $b a^{-1} \\not\\equiv \\pm 1 \\pmod{p}$, the order of $b a^{-1}$ does not divide $2$. Thus, the order must be either $q$ or $2 q$.\n\nNext, let $S$ denote the set of positive integers $n<p$ such that $a^{n}+b^{n}+c^{n} \\equiv 0 \\pmod{p}$, and let $s_{t}$ denote the number of ordered pairs $(i, j) \\subset S$ such that $i-j \\equiv t \\pmod{p-1}$.\n\nLemma: If $t$ is a positive integer less than $2 q$ and not equal to $q$, then $s_{t} \\leq 2$.\n\nProof: Consider $i, j \\in S$ with $j-i \\equiv t \\pmod{p-1}$. Then we have\n$$\n\\begin{aligned}\n& a^{i}+b^{i}+c^{i} \\equiv 0 \\pmod{p} \\\\\n\\Longrightarrow & a^{i} c^{j-i}+b^{i} c^{j-i}+c^{j} \\equiv 0 \\pmod{p} \\\\\n\\Longrightarrow & a^{i} c^{j-i}+b^{i} c^{j-i}-a^{j}-b^{j} \\equiv 0 \\pmod{p} \\\\\n\\Longrightarrow & a^{i} \\cdot\\left(c^{t}-a^{t}\\right) \\equiv b^{i} \\cdot\\left(b^{t}-c^{t}\\right) \\pmod{p} .\n\\end{aligned}\n$$\nIf $c^{t} \\equiv a^{t} \\pmod{p}$, then this implies $c^{t} \\equiv b^{t} \\pmod{p}$ as well, so $\\left(a b^{-1}\\right)^{t} \\equiv 1 \\pmod{p}$. However, we know the order of $a b^{-1}$ is $q$ or $2 q$, and $q \\nmid t$, so this is impossible. Thus, we can write\n$$\n\\left(a b^{-1}\\right)^{i} \\equiv\\left(b^{t}-c^{t}\\right) \\cdot\\left(c^{t}-a^{t}\\right)^{-1} \\pmod{p}\n$$\nFor a fixed $t$, the right-hand side of this equation is fixed, so $\\left(a b^{-1}\\right)^{i}$ is also fixed. Since the order of $a b^{-1}$ is either $q$ or $2 q$, it follows that there are at most $2$ solutions for $i$, and the lemma is proven.\n\nNow, for each element $i$ in $S$, there are at least $|S|-2$ other elements that differ from $i$ by a quantity other than $q \\pmod{p-1}$. Therefore, the lemma implies that\n$$\n\\begin{aligned}\n|S| \\cdot(|S|-2) & \\leq \\sum_{t \\neq q} s_{t} \\leq 2 \\cdot(p-2) \\\\\n\\Longrightarrow \\quad(|S|-1)^{2} & \\leq 2 p-3 \\\\\n\\Longrightarrow \\quad|S| & <\\sqrt{2 p}+1 .\n\\end{aligned}\n$$",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 27885,
    "subject": "Mathematics (Multi-modal)",
    "question": "The integers $1$, $2$, $3$, $\\ldots$, $2016$ are written on a board. You can choose any two numbers on the board and replace them with their average. For example, you can replace $1$ and $2$ with $1.5$, or you can replace $1$ and $3$ with a second copy of $2$. After $2015$ replacements of this kind, the board will have only one number left on it.\n\na. Prove that there is a sequence of replacements that will make the final number equal to $2$.\n\nb. Prove that there is a sequence of replacements that will make the final number equal to $1000$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "a.\nFirst replace $2014$ and $2016$ with $2015$, and then replace the two copies of $2015$ with a single copy. This leaves us with $\\{1, 2, \\ldots, 2013, 2015\\}$. From here, we can replace $2013$ and $2015$ with $2014$ to get $\\{1, 2, \\ldots, 2012, 2014\\}$. We can then replace $2012$ and $2014$ with $2013$, and so on, until we eventually get to $\\{1, 3\\}$. We finish by replacing $1$ and $3$ with $2$.\n\nb.\nUsing the same construction as in (a), we can find a sequence of replacements that reduces $\\{a, a + 1, \\ldots, b\\}$ to just $\\{a + 1\\}$. Similarly, we can also find a sequence of replacements that reduces $\\{a, a + 1, \\ldots, b\\}$ to just $\\{b - 1\\}$.\n\nIn particular, we can find sequences of replacements that reduce $\\{1, 2, \\ldots, 999\\}$ to just $\\{998\\}$, and that reduce $\\{1001, 1002, \\ldots, 2016\\}$ to just $\\{1002\\}$. This leaves us with $\\{998, 1000, 1002\\}$. We can replace $998$ and $1002$ with a second copy of $1000$, and then replace the two copies of $1000$ with a single copy to complete the construction.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 27886,
    "subject": "Mathematics (Multi-modal)",
    "question": "Consider the following system of 10 equations in 10 real variables $v_1, \\dots, v_{10}$:\n$$\nv_i = 1 + \\frac{6 v_i^2}{v_1^2 + v_2^2 + \\dots + v_{10}^2} \\quad (i = 1, \\dots, 10).\n$$\nFind all 10-tuples $(v_1, v_2, \\dots, v_{10})$ that are solutions of this system.",
    "options": [],
    "answer": "Either all ten entries are 8/5, or nine entries are 4/3 and one entry is 4 (in any order).",
    "solution": "For a particular solution $(v_1, v_2, \\dots, v_{10})$, let $s = v_1^2 + v_2^2 + \\dots + v_{10}^2$.\nThen\n$$\nv_i = 1 + \\frac{6v_i^2}{s} \\quad \\Rightarrow \\quad 6v_i^2 - sv_i + s = 0.\n$$\nLet $a$ and $b$ be the roots of the quadratic $6x^2 - sx + s = 0$, so for each $i$, $v_i = a$ or $v_i = b$. We also have $ab = s/6$ (by Vieta's formula, for example).\nIf all the $v_i$ are equal, then\n$$\nv_i = 1 + \\frac{6}{10} = \\frac{8}{5}\n$$\nfor all $i$. Otherwise, let $5+k$ of the $v_i$ be $a$, and let $5-k$ of the $v_i$ be $b$, where $0 < k \\le 4$. Then by the AM-GM inequality,\n$$\n6ab = s = (5 + k)a^2 + (5 - k)b^2 \\ge 2ab\\sqrt{25 - k^2}.\n$$\nFrom the given equations, $v_i \\ge 1$ for all $i$, so $a$ and $b$ are positive.\nThen $\\sqrt{25 - k^2} \\le 3 \\Rightarrow 25 - k^2 \\le 9 \\Rightarrow k^2 \\ge 16 \\Rightarrow k = 4$. Hence,\n$6ab = 9a^2 + b^2 \\Rightarrow (b - 3a)^2 = 0 \\Rightarrow b = 3a$.\nAdding all given ten equations, we get\n$$\nv_1 + v_2 + \\dots + v_{10} = 16.\n$$\nBut $v_1 + v_2 + \\dots + v_{10} = 9a + b = 12a$, so $a = 16/12 = 4/3$ and $b = 4$. Therefore, the solutions are $(8/5, 8/5, \\dots, 8/5)$ and all ten permutations of $(4/3, 4/3, \\dots, 4/3, 4)$.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 27887,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all polynomials $P(x)$ with integer coefficients such that $P(P(n)+n)$ is a prime number for infinitely many integers $n$.",
    "options": [],
    "answer": "All polynomials are either P(x) = p where p is a prime number, or P(x) = -2x + b where b is an odd integer.",
    "solution": "Note that if $P(n) = 0$ then $P(P(n) + n) = P(n) = 0$ which is not prime. Let $P(x)$ be a degree $k$ polynomial of the form $P(x) = a_k x^k + a_{k-1} x^{k-1} + \\dots + a_0$ and note that if $P(n) \\ne 0$ then\n$$\nP(P(n) + n) - P(n) = a_k[(P(n) + n)^k - n^k] + a_{k-1}[(P(n) + n)^{k-1} - n^{k-1}] + \\dots + a_1 P(n)\n$$\nwhich is divisible by $(P(n)+n) - n = P(n)$. Therefore if $P(P(n)+n)$ is prime then either $P(n) = \\pm 1$ or $P(P(n)+n) = \\pm P(n) = p$ for some prime number $p$. Since $P(x)$ is a polynomial, it follows that $P(n) = \\pm 1$ for only finitely many integers $n$. Therefore either $P(n) = P(P(n)+n)$ for infinitely many integers $n$ or $P(n) = -P(P(n)+n)$ for infinitely many integers $n$.\n\nSuppose that $P(n) = P(P(n)+n)$ for infinitely many integers $n$. This implies that the polynomial $P(P(x)+x) - P(x)$ has infinitely many roots and thus is identically zero. Therefore $P(P(x)+x) = P(x)$ holds identically. Now note that if $k \\ge 2$ then $P(P(x)+x)$ has degree $k^2$ while $P(x)$ has degree $k$, which is not possible. Therefore $P(x)$ is at most linear with $P(x) = ax + b$ for some integers $a$ and $b$. Now note that\n$$\nP(P(x) + x) = a(a + 1)x + ab + b\n$$\nand thus $a = a(a + 1)$ and $ab + b = b$. It follows that $a = 0$ which leads to the solution $P(n) = p$ where $p$ is a prime number.\n\nBy the same argument if $P(n) = -P(P(n)+n)$ for infinitely many integers $n$ then $P(x) = -P(P(x)+x)$ holds identically and $P(x)$ is linear with $P(x) = ax + b$. In this case it follows that $a = -a(a + 1)$ and $ab + b = -b$. This implies that either $a = 0$ or $a = -2$. If $a = -2$ then $P(n) = -2n + b$ which is prime for some integers $n$ only if $b$ is odd. Note that in this case $P(P(n)+n) = 2n - b$ which is indeed prime for infinitely many integers $n$ as long as $b$ is odd. $\\square$",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 27888,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nFind all polynomials $P(x)$ with integer coefficients such that $P(P(n)+n)$ is a prime number for infinitely many integers $n$.",
    "options": [],
    "answer": "Exactly the constant polynomials equal to a prime, and the linear polynomials of the form −2x + b with b an odd integer.",
    "solution": "Solution:\nNote that if $P(n)=0$ then $P(P(n)+n)=P(n)=0$ which is not prime. Let $P(x)$ be a degree $k$ polynomial of the form $P(x)=a_{k} x^{k}+a_{k-1} x^{k-1}+\\cdots+a_{0}$ and note that if $P(n) \\neq 0$ then\n$$\n\\begin{aligned}\n& P(P(n)+n)-P(n)= \\\\\n& \\quad a_{k}\\left[(P(n)+n)^{k}-n^{k}\\right]+a_{k-1}\\left[(P(n)+n)^{k-1}-n^{k-1}\\right]+\\cdots+a_{1} P(n)\n\\end{aligned}\n$$\nwhich is divisible by $(P(n)+n)-n=P(n)$. Therefore if $P(P(n)+n)$ is prime then either $P(n)= \\pm 1$ or $P(P(n)+n)= \\pm P(n)=p$ for some prime number $p$. Since $P(x)$ is a polynomial, it follows that $P(n)= \\pm 1$ for only finitely many integers $n$. Therefore either $P(n)=P(P(n)+n)$ for infinitely many integers $n$ or $P(n)=-P(P(n)+n)$ for infinitely many integers $n$.\n\nSuppose that $P(n)=P(P(n)+n)$ for infinitely many integers $n$. This implies that the polynomial $P(P(x)+x)-P(x)$ has infinitely many roots and thus is identically zero. Therefore $P(P(x)+x)=P(x)$ holds identically. Now note that if $k \\geq 2$ then $P(P(x)+x)$ has degree $k^{2}$ while $P(x)$ has degree $k$, which is not possible. Therefore $P(x)$ is at most linear with $P(x)=a x+b$ for some integers $a$ and $b$. Now note that\n$$\nP(P(x)+x)=a(a+1) x+a b+b\n$$\nand thus $a=a(a+1)$ and $a b+b=b$. It follows that $a=0$ which leads to the solution $P(n)=p$ where $p$ is a prime number.\n\nBy the same argument if $P(n)=-P(P(n)+n)$ for infinitely many integers $n$ then $P(x)=-P(P(x)+x)$ holds identically and $P(x)$ is linear with $P(x)=a x+b$. In this case it follows that $a=-a(a+1)$ and $a b+b=-b$. This implies that either $a=0$ or $a=-2$. If $a=-2$ then $P(n)=-2 n+b$ which is prime for some integers $n$ only if $b$ is odd. Note that in this case $P(P(n)+n)=2 n-b$ which is indeed prime for infinitely many integers $n$ as long as $b$ is odd.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 27889,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $A$, $B$, and $F$ be positive integers, and assume $A < B < 2A$. A flea is at the number $0$ on the number line. The flea can move by jumping to the right by $A$ or by $B$. Before the flea starts jumping, Lavaman chooses finitely many intervals $\\{m+1, m+2, \\dots, m+A\\}$ consisting of $A$ consecutive positive integers, and places lava at all of the integers in the intervals. The intervals must be chosen so that:\n(i) any two distinct intervals are disjoint and not adjacent;\n(ii) there are at least $F$ positive integers with no lava between any two intervals; and\n(iii) no lava is placed at any integer less than $F$.\n\nProve that the smallest $F$ for which the flea can jump over all the intervals and avoid all the lava, regardless of what Lavaman does, is $F = (n - 1)A + B$, where $n$ is the positive integer such that\n$$\n\\frac{A}{n+1} \\leq B - A < \\frac{A}{n} .\n$$",
    "options": [],
    "answer": "F = (n - 1)A + B, where n is the positive integer with A/(n+1) ≤ B − A < A/n.",
    "solution": "Let $B = A + C$ where $\\frac{A}{n + 1} \\leq C < \\frac{A}{n}$.\n\nFirst, here is an informal sketch of the proof.\n\nLavaman's strategy: Use only safe intervals with $nA + C - 1$ integers. The flea will start at position $[1, C]$ from the left, which puts him at position $[nA, nA + C - 1]$ from the right. After $n-1$ jumps, he will still have $nA - (n-1)(A+C) = A - (n-1)C > C$ distance to go, which is not enough for a big jump to clear the lava. Thus, he must do at least $n$ jumps in the safe interval, but that's possible only with all small jumps, and furthermore is impossible if the starting position is $C$. This gives him starting position 1 higher in the next safe interval, so sooner or later the flea is going to hit the lava.\n\nFlea's strategy: The flea just does one interval at a time. If the safe interval has at least $nA + C$ integers in it, the flea has distance $d > nA$ to go to the next lava when it starts. Repeatedly do big jumps until $d$ is between $1$ and $C \\bmod A$, then small jumps until the remaining distance is between $1$ and $C$, then a final big jump. This works as long as the first part does. However, we get at least $n$ big jumps since $\\lfloor(d-1)/A\\rfloor$ can never go down two from a big jump (or we'd be done doing big jumps), so we get $n$ big jumps, and thus we are good if $d \\bmod A$ is in any of $[1, C], [C+1, 2C], \\dots, [nC+1, (n+1)C]$, but that's everything. $\\square$\n\nLet $C = B - A$. We shall write our intervals of lava in the form $(L_i, R_i] = \\{L_i + 1, L_i + 2, \\dots, R_i\\}$, where $R_i = L_i + A$ and $R_{i-1} < L_i$ for every $i \\geq 1$. We also let $R_0 = 0$. We shall also represent a path for the flea as a sequence of integers $x_0, x_1, x_2, \\dots$ where $x_0 = 0$ and $x_j - x_{j-1} \\in \\{A, B\\}$ for every $j \\geq 0$.\n\nNow here is a detailed proof.\n\nFirst, assume $F < (n-1)A + B (= nA + C)$: we must prove that Lavaman has a winning strategy. Let $L_i = R_{i-1} + nA + C - 1$ for every $i \\geq 1$. (Observe that $nA + C - 1 \\geq F$.)\n\nmeans that $x_j \\notin (L_i, R_i]$ for all $i, j \\geq 1$. For each $i \\geq 1$, let\n$$\nM_i = \\max\\{x_j : x_j \\leq L_i\\}, \\quad m_i = \\min\\{x_j : x_j > R_i\\}, \\ \\text{and} \\quad J(i) = \\max\\{j : x_j \\leq L_i\\}.\n$$\nAlso let $m_0 = 0$. Then for $i \\geq 1$ we have\n$$\nM_i = x_{J(i)} \\quad \\text{and} \\quad m_i = x_{J(i)+1}.\n$$\nAlso, for every $i \\geq 1$, we have\n(a) $m_i = M_i + B$ (because $M_i + A \\leq L_i + A = R_i$);\n(b) $L_i \\geq M_i > L_i - C$ (since $M_i = m_i - B > R_i - B = L_i + A - B$);\n(c) $R_i < m_i \\leq R_i + C$ (since $m_i = M_i + B \\leq L_i + B = R_i + C$).\n\n$$\n\\begin{align*}\nM_{i+1} = x_{J(i+1)} = x_{J(i)+n+1} &\\geq x_{J(i)+1} + (n-1)A + B \\\\\n&> R_i + nA + C \\\\\n&= L_{i+1} + 1 \\\\\n&> M_{i+1}\n\\end{align*}\n$$\nwhich is a contradiction. This proves Claim 2.\n\n$$\n\\begin{align*}\nx_{J(i+1)+1} &= x_{J(i)+n+2} = x_{J(i)+1} + nA + B; \\\\\n\\text{i.e., } m_{i+1} &= m_i + nA + B \\quad \\text{for each } i \\geq 1.\n\\end{align*}\n$$\nTherefore\n$$\n\\begin{align*}\nm_{i+1} - R_{i+1} &= m_i + nA + B - (R_i + nA + C - 1 + A) \\\\\n&= m_i - R_i + 1.\n\\end{align*}\n$$\nHence\n$$\nC \\geq m_{C+1} - R_{C+1} = m_1 - R_1 + C > C\n$$\nwhich is a contradiction. Therefore no path for the flea avoids all the lava. We observe that Lavaman only needs to put lava on the first $C + 1$ intervals.\n\nNow assume $F \\geq (n-1)A + B$. We will show that the flea can avoid all the lava. We shall need the following result:\n\n**Claim 3:** Let $d \\geq nA$. Then there exist nonnegative integers $s$ and $t$ such that $sA + tB \\in (d - C, d]$.\n\nWe shall prove this result at the end.\n\nFirst, observe that $L_1 \\geq nA$. By Claim 3, it is possible for the flea to make a sequence of jumps starting from $0$ and ending at a point of $(L_1 - C, L_1]$. From any point of this interval, a single jump of size $B$ takes the flea over $(L_1, R_1]$ to a point in $(R_1, R_1+C]$, which corresponds to the point $x_{J(1)+1} (= m_1)$ on the flea's path.\n\nNow we use induction to prove that, for every $i \\geq 1$, there is a path such that $x_j$ avoids lava for all $j \\leq J(i)+1$. The case $i=1$ is done, so\nassume that the assertion holds for a given $i$. Then $x_{J(i)+1} = m_i \\in (R_i, R_i + C]$. Therefore\n$$\nL_{i+1} - m_i \\geq R_i + F - (R_i + C) = F - C \\geq nA.\n$$\nApplying Claim 3 with $d = L_{i+1} - m_i$ shows that the flea can jump from $m_i$ to a point of $(L_{i+1} - C, L_{i+1}]$. A single jump of size $B$ then takes the flea to a point of $(R_{i+1}, R_{i+1} + C]$ (without visiting $(L_{i+1}, R_{i+1}]$), and this point serves as $x_{J(i+1)+1}$. This completes the induction.\n\n**Proof of Claim 3:** Let $u$ be the greatest integer that is less than or equal to $d/A$. Then $u \\geq n$ and $uA \\leq d < (u+1)A$. For $v = 0, \\dots, n$, let\n$$\nz_v = (u - v)A + vB = uA + vC.\n$$\nThen\n$$\nz_0 = uA \\leq d,\n$$\n$$\nz_n = uA + nC = uA + (n + 1)C - C \\geq (u + 1)A - C > d - C.\n$$\nand $z_{v+1} - z_v = C$ for $v = 0, \\dots, n-1$.\n\nTherefore we must have $z_v \\in (d-C, d]$ for some $v$ in $\\{0, 1, \\dots, n\\}$. $\\square$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 27890,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $\\triangle ABC$ be an acute-angled triangle with altitudes $AD$ and $BE$ meeting at $H$. Let $M$ be the midpoint of segment $AB$, and suppose that the circumcircles of $\\triangle DEM$ and $\\triangle ABH$ meet at points $P$ and $Q$ with $P$ on the same side of $CH$ as $A$. Prove that the lines $ED$, $PH$, and $MQ$ all pass through a single point on the circumcircle of $\\triangle ABC$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $R$ denote the intersection of lines $ED$ and $PH$. Since quadrilaterals $ECDH$ and $APHB$ are cyclic, we have $\\angle RDA = 180^\\circ - \\angle EDA = 180^\\circ - \\angle EDH = 180^\\circ - \\angle ECH = 90^\\circ + A$, and $\\angle RPA = \\angle HPA = 180^\\circ - \\angle HBA = 90^\\circ + A$. Therefore, $APDR$ is cyclic. This in turn implies that $\\angle PBE = \\angle PBH = \\angle PAH = \\angle PAD = \\angle PRD = \\angle PRE$, and so $PBRE$ is also cyclic.\nLet $F$ denote the base of the altitude from $C$ to $AB$. Then $D, E, F$, and $M$ all lie on the 9-point circle of $\\triangle ABC$, and so are cyclic. We also know $APDR, PBRE, BCEF$, and $ACDF$ are cyclic, which implies $\\angle ARB = \\angle PRB - \\angle PRA = \\angle PEB - \\angle PDA = \\angle PEF + \\angle FEB - \\angle PDF + \\angle ADF = \\angle FEB + \\angle ADF = \\angle FCB + \\angle ACF = C$. Therefore, $R$ lies on the circumcircle of $\\triangle ABC$.\nNow let $Q'$ and $R'$ denote the intersections of line $MQ$ with the circumcircle of $\\triangle ABC$, chosen so that $Q', M, Q, R'$ lie on the line in that order. We will show that $R' = R$, which will complete the proof. However, first note that the circumcircle of $\\triangle ABC$ has radius $\\frac{AB}{2\\sin C}$, and the circumcircle of $\\triangle ABH$ has radius $\\frac{AB}{2\\sin\\angle AHB} = \\frac{AB}{2\\sin(180^\\circ - C)}$. Thus the two circles have equal radius, and so they must be symmetrical about the point $M$. In particular, $MQ = MQ'$.\nSince $\\angle AEB = \\angle ADB = 90^\\circ$, we furthermore know that $M$ is the circumcenter of both $\\triangle AEB$ and $\\triangle ADB$. Thus, $MA = ME = MD = MB$. By Power of a Point, we then have $MQ \\cdot MR' = MQ' \\cdot MR' = MA \\cdot MB = MD^2$. In particular, this means that the circumcircle of\n\n$\\triangle DR'Q$ is tangent to $MD$ at $D$, which means $\\angle MR'D = \\angle MDQ$.\nSimilarly $MQ \\cdot MR' = ME^2$, and so $\\angle MR'E = \\angle MEQ = \\angle MDQ = \\angle MR'D$. Therefore, $R'$ also lies on the line $ED$.\nFinally, the same argument shows that $MP$ also intersects the circum-\ncircle of $\\triangle ABC$ at a point $R''$ on line $ED$. Thus, $R, R',$ and $R''$ are\nall chosen from the intersection of the circumcircle of $\\triangle ABC$ and the\nline $ED$. In particular, two of $R, R',$ and $R''$ must be equal. However,\n$R'' \\neq R$ since $MP$ and $PH$ already intersect at $P$, and $R'' \\neq R'$ since\n$MP$ and $MQ$ already intersect at $M$. Thus, $R' = R$, and the proof is\ncomplete. $\\square$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 27891,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $a$, $b$, and $c$ be non-negative real numbers, no two of which are equal. Prove that\n$$\n\\frac{a^2}{(b-c)^2} + \\frac{b^2}{(c-a)^2} + \\frac{c^2}{(a-b)^2} > 2.\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "The left-hand side is symmetric with respect to $a$, $b$, $c$. Hence, we may assume that $a > b > c \\ge 0$. Note that replacing $(a, b, c)$ with $(a-c, b-c, 0)$ lowers the value of the left-hand side, since the numerators of each of the fractions would decrease and the denominators remain the same. Therefore, to obtain the minimum possible value of the left-hand side, we may assume that $c=0$.\nThen the left-hand side becomes\n$$\n\\frac{a^2}{b^2} + \\frac{b^2}{a^2},\n$$\nwhich yields, by the Arithmetic Mean - Geometric Mean Inequality,\n$$\n\\frac{a^2}{b^2} + \\frac{b^2}{a^2} \\ge 2\\sqrt{\\frac{a^2}{b^2} \\cdot \\frac{b^2}{a^2}} = 2,\n$$\nwith equality if and only if $a^2/b^2 = b^2/a^2$, or equivalently, $a^4 = b^4$. Since $a, b \\ge 0$, $a = b$. But since no two of $a, b, c$ are equal, $a \\ne b$. Hence, equality cannot hold. This yields\n$$\n\\frac{a^2}{b^2} + \\frac{b^2}{a^2} > 2.\n$$\nUltimately, this implies the desired inequality. $\\square$\n\n$$\n\\frac{a^2}{(b-c)^2} + \\frac{b^2}{(c-a)^2} + \\frac{c^2}{(a-b)^2} - 2 = \\frac{[a(a-b)(a-c) + b(b-a)(b-c) + c(c-a)(c-b)]^2}{[(a-b)(b-c)(c-a)]^2}.\n$$\nThen Schur's Inequality tells us that the numerator of the right-hand side cannot be zero. $\\square$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 27892,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $f$ be a function from the set of positive integers to itself such that, for every $n$, the number of positive integer divisors of $n$ is equal to $f(f(n))$. For example, $f(f(6)) = 4$ and $f(f(25)) = 3$. Prove that if $p$ is prime then $f(p)$ is also prime.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $d(n) = f(f(n))$ denote the number of divisors of $n$ and observe that $f(d(n)) = f(f(f(n))) = d(f(n))$ for all $n$. Also note that because all divisors of $n$ are distinct positive integers between $1$ and $n$, including $1$ and $n$, and excluding $n-1$ if $n > 2$, it follows that $2 \\le d(n) < n$ for all $n > 2$. Furthermore $d(1) = 1$ and $d(2) = 2$.\n\nWe first will show that $f(2) = 2$. Let $m = f(2)$ and note that $2 = d(2) = f(f(2)) = f(m)$. If $m \\ge 2$, then let $m_0$ be the smallest positive integer satisfying that $m_0 \\ge 2$ and $f(m_0) = 2$. It follows that $f(d(m_0)) = d(f(m_0)) = d(2) = 2$. By the minimality of $m_0$, it follows that $d(m_0) \\ge m_0$, which implies that $m_0 = 2$. Therefore if $m \\ge 2$, it follows that $f(2) = 2$. It suffices to examine the case in which $f(2) = m = 1$. If $m = 1$, then $f(1) = f(f(2)) = 2$ and furthermore, each prime $p$ satisfies that $d(f(p)) = f(d(p)) = f(2) = 1$ which implies that $f(p) = 1$. Therefore $d(f(p^2)) = f(d(p^2)) = f(3) = 1$ which implies that $f(p^2) = 1$ for any prime $p$. This implies that $3 = d(p^2) = f(f(p^2)) = f(1) = 2$, which is a contradiction. Therefore $m \\ne 1$ and $f(2) = 2$.\n\nIt now follows that if $p$ is prime then $2 = f(2) = f(d(p)) = d(f(p))$ which implies that $f(p)$ is prime. $\\square$",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 27893,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $n$ be a positive integer, and define $S_n = \\{1, 2, \\dots, n\\}$. Consider a non-empty subset $T$ of $S_n$. We say that $T$ is balanced if the median of $T$ is equal to the average of $T$. For example, for $n = 9$, each of the subsets $\\{7\\}$, $\\{2, 5\\}$, $\\{2, 3, 4\\}$, $\\{5, 6, 8, 9\\}$, and $\\{1, 4, 5, 7, 8\\}$ is balanced; however, the subsets $\\{2, 4, 5\\}$ and $\\{1, 2, 3, 5\\}$ are not balanced. For each $n \\ge 1$, prove that the number of balanced subsets of $S_n$ is odd.\n\n(To define the median of a set of $k$ numbers, first put the numbers in increasing order; then the median is the middle number if $k$ is odd, and the average of the two middle numbers if $k$ is even. For example, the median of $\\{1, 3, 4, 8, 9\\}$ is $4$, and the median of $\\{1, 3, 4, 7, 8, 9\\}$ is $(4+7)/2 = 5.5$.)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "The problem is to prove that there is an odd number of nonempty subsets $T$ of $S_n$ such that the average $A(T)$ and median $M(T)$ satisfy $A(T) = M(T)$. Given a subset $T$, consider the subset $T^* = \\{n+1-t: t \\in T\\}$. It holds that $A(T^*) = n+1-A(T)$ and $M(T^*) = n+1-M(T)$, which implies that if $A(T) = M(T)$ then $A(T^*) = M(T^*)$. Pairing each set $T$ with $T^*$ yields that there are an even number of sets $T$ such that $A(T) = M(T)$ and $T \\neq T^*$.\n\nThus it suffices to show that the number of nonempty subsets $T$ such that $A(T) = M(T)$ and $T = T^*$ is odd. Now note that if $T = T^*$, then $A(T) = M(T) = \\frac{n+1}{2}$. Hence it suffices to show the number of nonempty subsets $T$ with $T = T^*$ is odd. Given such a set $T$, let $T'$ be the largest nonempty subset of $\\{1, 2, \\dots, \\lceil n/2 \\rceil\\}$ contained in $T$. Pairing $T$ with $T'$ forms a bijection between these sets $T$ and the nonempty subsets of $\\{1, 2, \\dots, \\lceil n/2 \\rceil\\}$. Thus there are $2^{\\lceil n/2 \\rceil} - 1$ such subsets, which is odd as desired. $\\square$\nUsing the notation from the above solution: Let $B$ be the number of subsets $T$ with $M(T) > A(T)$, $C$ be the number with $M(T) = A(T)$, and $D$ be the number with $M(T) < A(T)$. Pairing each set $T$ counted by $B$ with $T^* = \\{n+1-t: t \\in T\\}$ shows that $B = D$. Now since $B + C + D = 2^n - 1$, we have that $C = 2^n - 1 - 2B$, which is odd.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 27894,
    "subject": "Mathematics (Multi-modal)",
    "question": "Points $P$ and $Q$ lie inside parallelogram $ABCD$ and are such that triangles $ABP$ and $BCQ$ are equilateral. Prove that the line through $P$ perpendicular to $DP$ and the line through $Q$ perpendicular to $DQ$ meet on the altitude from $B$ in triangle $ABC$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $\\angle ABC = m$ and let $O$ be the circumcenter of triangle $DPQ$. Since $P$ and $Q$ are in the interior of $ABCD$, it follows that $m = \\angle ABC > 60^\\circ$ and $\\angle DAB = 180^\\circ - m > 60^\\circ$ which together imply that $60^\\circ < m < 120^\\circ$. Now note that $\\angle DAP = \\angle DAB - 60^\\circ = 120^\\circ - m$, $\\angle DCQ = \\angle DCB - 60^\\circ = 120^\\circ - m$ and that $\\angle PBQ = 60^\\circ - \\angle ABQ = 60^\\circ - (\\angle ABC - 60^\\circ) = 120^\\circ - m$. This combined with the facts that $AD = BQ = CQ$ and $AP = BP = CD$ implies that triangles $DAP$, $QBP$ and $QCD$ are congruent. Therefore $DP = PQ = DQ$ and triangle $DPQ$ is equilateral. This implies that $\\angle ODA = \\angle PDA + 30^\\circ = \\angle DQC + 30^\\circ = \\angle OQC$. Combining this fact with $OQ = OD$ and $CQ = AD$ implies that triangles $ODA$ and $OQC$ are congruent. Therefore $OA = OC$ and, if $M$ is the midpoint of segment $AC$, it follows that $OM$ is perpendicular to $AC$. Since $ABCD$ is a parallelogram, $M$ is also the midpoint of $DB$. If $K$ denotes the intersection of the line through $P$ perpendicular to $DP$ and the line through $Q$ perpendicular to $DQ$, then $K$ is diametrically opposite $D$ on the circumcircle of $DPQ$ and $O$ is the midpoint of segment $DK$. This implies that $OM$ is a midline of triangle $DBK$ and hence that $BK$ is parallel to $OM$ which is perpendicular to $AC$. Therefore $K$ lies on the altitude from $B$ in triangle $ABC$, as desired. $\\square$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 27895,
    "subject": "Mathematics (Multi-modal)",
    "question": "One hundred circles of radius one are positioned in the plane so that the area of any triangle formed by the centres of three of these circles is at most $2017$. Prove that there is a line intersecting at least three of these circles.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "We will prove that given $n$ circles, there is some line intersecting more than $\\frac{n}{46}$ of them. Let $S$ be the set of centers of the $n$ circles. We will first show that there is a line $\\ell$ such that the projections of the points in $S$ lie in an interval of length at most $\\sqrt{8068} < 90$ on $\\ell$.\n\nLet $A$ and $B$ be the pair of points in $S$ that are farthest apart and let the distance between $A$ and $B$ be $d$. Now consider any point $C \\in S$ distinct from $A$ and $B$. The distance from $C$ to the line $AB$ must be at most $\\frac{4034}{d}$ since triangle $ABC$ has area at most $2017$. Therefore if $\\ell$ is a line perpendicular to $AB$, then the projections of $S$ onto $\\ell$ lie in an interval of length $\\frac{8068}{d}$ centered at the intersection of $\\ell$ and $AB$. Furthermore, all of these projections must lie on an interval of length at most $d$ on $\\ell$ since the largest distance between two of these projections is at most $d$. Since $\\min(d, 8068/d) \\le \\sqrt{8068} < 90$, this proves the claim.\n\nNow note that the projections of the $n$ circles onto the line $\\ell$ are intervals of length $2$, all contained in an interval of length at most $\\sqrt{8068} + 2 < 92$. Each point of this interval belongs to on average $\\frac{2n}{\\sqrt{8068}+2} > \\frac{n}{46}$ of the subintervals of length $2$ corresponding to the projections of the $n$ circles onto $\\ell$. Thus there is some point $x \\in \\ell$ belonging to the projections of more than $\\frac{n}{46}$ circles. The line perpendicular to $\\ell$ through $x$ has the desired property. Setting $n = 100$ yields that there is a line intersecting at least three of the circles. $\\square$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 27896,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nConsider an arrangement of tokens in the plane, not necessarily at distinct points. We are allowed to apply a sequence of moves of the following kind: Select a pair of tokens at points $A$ and $B$ and move both of them to the midpoint of $A$ and $B$.\n\nWe say that an arrangement of $n$ tokens is collapsible if it is possible to end up with all $n$ tokens at the same point after a finite number of moves. Prove that every arrangement of $n$ tokens is collapsible if and only if $n$ is a power of 2.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nFor a given positive integer $n$, consider an arrangement of $n$ tokens in the plane, where the tokens are at points $A_{1}, A_{2}, \\ldots, A_{n}$. Let $G$ be the centroid of the $n$ points, so as vectors (after an arbitrary choice of origin),\n$$\n\\vec{G}=\\frac{\\vec{A}_{1}+\\vec{A}_{2}+\\cdots+\\vec{A}_{n}}{n}\n$$\nNote that any move leaves the centroid $G$ unchanged. Therefore, if all the tokens are eventually moved to the same point, then this point must be $G$.\n\nFirst we prove that if $n=2^{k}$ for some nonnegative integer $k$, then all $n$ tokens can always be eventually moved to the same point. We shall use induction on $k$.\n\nThe result clearly holds for $n=2^{0}=1$. Assume that it holds when $n=2^{k}$ for some nonnegative integer $k$. Consider a set of $2^{k+1}$ tokens at $A_{1}, A_{2}, \\ldots, A_{2^{k+1}}$. Let $M_{i}$ be the midpoint of $A_{2 i-1}$ and $A_{2 i}$ for $1 \\leq i \\leq 2^{k}$.\n\nFirst we move the tokens at $A_{2 i-1}$ and $A_{2 i}$ to $M_{i}$, for $1 \\leq i \\leq 2^{k}$. Then, there are two tokens at $M_{i}$ for all $1 \\leq i \\leq 2^{k}$. If we take one token from each of $M_{1}, M_{2}, \\ldots, M_{2^{k}}$, then by the induction hypothesis, we can move all of them to the same point, say $G$. We can do the same with the remaining tokens at $M_{1}, M_{2}, \\ldots, M_{2^{k}}$. Thus, all $2^{k+1}$ tokens are now at $G$, which completes the induction argument.\n\n(Here is an alternate approach to the induction step: Given the tokens at $A_{1}, A_{2}, \\ldots, A_{2^{k+1}}$, move the first $2^{k}$ tokens to one point $G_{1}$, and move the remaining $2^{k}$ tokens to one point $G_{2}$. Then $2^{k}$ more moves can bring all the tokens to the midpoint of $G_{1}$ and $G_{2}$.)\n\nNow, assume that $n$ is not a power of 2. Take any line in the plane, and number it as a real number line. (Henceforth, when we refer to a token at a real number, we mean with respect to this real number line.)\n\nAt the start, place $n-1$ tokens at $0$ and one token at $1$. We observed that if we can move all the tokens to the same point, then it must be the centroid of the $n$ points. Here, the centroid is at $\\frac{1}{n}$. We now prove a lemma.\n\nLemma. The average of any two dyadic rationals is also a dyadic rational. (A dyadic rational is a rational number that can be expressed in the form $\\frac{m}{2^{a}}$, where $m$ is an integer and $a$ is a nonnegative integer.)\n\nProof. Consider two dyadic rationals $\\frac{m_{1}}{2^{a_{1}}}$ and $\\frac{m_{2}}{2^{a_{2}}}$. Then their average is\n$$\n\\frac{1}{2}\\left(\\frac{m_{1}}{2^{a_{1}}}+\\frac{m_{2}}{2^{a_{2}}}\\right)=\\frac{1}{2}\\left(\\frac{2^{a_{2}} \\cdot m_{1}+2^{a_{1}} \\cdot m_{2}}{2^{a_{1}} \\cdot 2^{a_{2}}}\\right)=\\frac{2^{a_{2}} \\cdot m_{1}+2^{a_{1}} \\cdot m_{2}}{2^{a_{1}+a_{2}+1}}\n$$\nwhich is another dyadic rational.\n\nOn this real number line, a move corresponds to taking a token at $x$ and a token at $y$ and moving both of them to $\\frac{x+y}{2}$, the average of $x$ and $y$. At the start, every token is at a dyadic rational (namely $0$ or $1$), which means that after any number of moves, every token must still be at a dyadic rational.\n\nBut $n$ is not a power of 2, so $\\frac{1}{n}$ is not a dyadic rational. (Indeed, if we could express $\\frac{1}{n}$ in dyadic form $\\frac{m}{2^{a}}$, then we would have $2^{a}=m n$, which is impossible unless $m$ and $n$ are powers of 2.) This means that it is not possible for any token to end up at $\\frac{1}{n}$, let alone all $n$ tokens.\n\nWe conclude that we can always move all $n$ tokens to the same point if and only if $n$ is a power of 2.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 27897,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nLet five points on a circle be labelled $A$, $B$, $C$, $D$, and $E$ in clockwise order. Assume $A E = D E$ and let $P$ be the intersection of $A C$ and $B D$. Let $Q$ be the point on the line through $A$ and $B$ such that $A$ is between $B$ and $Q$ and $A Q = D P$. Similarly, let $R$ be the point on the line through $C$ and $D$ such that $D$ is between $C$ and $R$ and $D R = A P$. Prove that $P E$ is perpendicular to $Q R$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nWe are given $A Q = D P$ and $A P = D R$. Additionally $\\angle Q A P = 180^{\\circ} - \\angle B A C = 180^{\\circ} - \\angle B D C = \\angle R D P$, and so triangles $A Q P$ and $D P R$ are congruent. Therefore $P Q = P R$. It follows that $P$ is on the perpendicular bisector of $Q R$.\n\nWe are also given $A P = D R$ and $A E = D E$. Additionally $\\angle P A E = \\angle C A E = 180^{\\circ} - \\angle C D E = \\angle R D E$, and so triangles $P A E$ and $R D E$ are congruent.\n\nTherefore $P E = R E$, and similarly $P E = Q E$. It follows that $E$ is on the perpendicular bisector of $P Q$.\n\nSince both $P$ and $E$ are on the perpendicular bisector of $Q R$, the result follows.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 27898,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nTwo positive integers $a$ and $b$ are prime-related if $a = p b$ or $b = p a$ for some prime $p$. Find all positive integers $n$, such that $n$ has at least three divisors, and all the divisors can be arranged without repetition in a circle so that any two adjacent divisors are prime-related.\n\nNote that $1$ and $n$ are included as divisors.",
    "options": [],
    "answer": "Exactly those positive integers that are neither perfect squares nor powers of a single prime.",
    "solution": "Solution:\n\nWe say that a positive integer is good if it has the given property. Let $n$ be a good number, and let $d_{1}, d_{2}, \\ldots, d_{k}$ be the divisors of $n$ in the circle, in that order. Then for all $1 \\leq i \\leq k$, $d_{i+1} / d_{i}$ (taking the indices modulo $k$) is equal to either $p_{i}$ or $1 / p_{i}$ for some prime $p_{i}$. In other words, $d_{i+1} / d_{i} = p_{i}^{\\epsilon_{i}}$, where $\\epsilon_{i} \\in \\{1, -1\\}$. Then\n$$\np_{1}^{\\epsilon_{1}} p_{2}^{\\epsilon_{2}} \\cdots p_{k}^{\\epsilon_{k}} = \\frac{d_{2}}{d_{1}} \\cdot \\frac{d_{3}}{d_{2}} \\cdots \\frac{d_{1}}{d_{k}} = 1\n$$\nFor the product $p_{1}^{\\epsilon_{1}} p_{2}^{\\epsilon_{2}} \\cdots p_{k}^{\\epsilon_{k}}$ to equal $1$, any prime factor $p$ must be paired with a factor of $1/p$, and vice versa, so $k$ (the number of divisors of $n$) must be even. Hence, $n$ cannot be a perfect square.\n\nFurthermore, $n$ cannot be the power of a prime (including a prime itself), because $1$ always is a divisor of $n$, and if $n$ is a power of a prime, then the only divisor that can go next to $1$ is the prime itself.\n\nNow, let $n = p^{a} q^{b}$, where $p$ and $q$ are distinct primes, and $a$ is odd. We write the divisors of $n$ in a grid as follows: In the first row, write the numbers $1, q, q^{2}, \\ldots, q^{b}$. In the next row, write the numbers $p, p q, p q^{2}, \\ldots, p q^{b}$, and so on. The number of rows in the grid, $a+1$, is even. Note that if two squares are adjacent vertically or horizontally, then their corresponding numbers are prime-related. We start with the square with a $1$ in the upper-left corner. We then move right along the first row, move down along the last column, move left along the last row, then zig-zag row by row, passing through every square, until we land on the square with a $p$. The following diagram gives the path for $a=3$ and $b=5$:\n\n![](attached_image_1.png)\n\nThus, we can write the divisors encountered on this path in a circle, so $n = p^{a} q^{b}$ is good.\n\nNext, assume that $n$ is a good number. Let $d_{1}, d_{2}, \\ldots, d_{k}$ be the divisors of $n$ in the circle, in that order. Let $p$ be a prime that does not divide $n$. We claim that $n \\cdot p^{e}$ is also a good number. We arrange the divisors of $n \\cdot p^{e}$ that are not divisors of $n$ in a grid as follows:\n$$\n\\begin{array}{cccc}\nd_{1} p & d_{1} p^{2} & \\ldots & d_{1} p^{e} \\\\\nd_{2} p & d_{2} p^{2} & \\ldots & d_{2} p^{e} \\\\\n\\vdots & \\vdots & \\ddots & \\vdots \\\\\nd_{k} p & d_{k} p^{2} & \\ldots & d_{k} p^{e}\n\\end{array}\n$$\nNote that if two squares are adjacent vertically or horizontally, then their corresponding numbers are prime-related. Also, $k$ (the number of rows) is the number of factors of $n$, which must be even (since $n$ is good). Hence, we can use the same path described above, which starts at $d_{1} p$ and ends at $d_{2} p$. Since $d_{1}$ and $d_{2}$ are adjacent divisors in the circle for $n$, we can insert all the divisors in the grid above between $d_{1}$ and $d_{2}$, to obtain a circle for $n \\cdot p^{e}$.\n\nFinally, let $n$ be a positive integer that is neither a perfect square nor a power of a prime. Let the prime factorization of $n$ be\n$$\nn = p_{1}^{e_{1}} p_{2}^{e_{2}} \\cdots p_{t}^{e_{t}}\n$$\nSince $n$ is not the power of a prime, $t \\geq 2$. Also, since $n$ is not a perfect square, at least one exponent $e_{i}$ is odd. Without loss of generality, assume that $e_{1}$ is odd. Then from our work above, $p_{1}^{e_{1}} p_{2}^{e_{2}}$ is good, so $p_{1}^{e_{1}} p_{2}^{e_{2}} p_{3}^{e_{3}}$ is good, and so on, until $n = p_{1}^{e_{1}} p_{2}^{e_{2}} \\cdots p_{t}^{e_{t}}$ is good.\n\nTherefore, a positive integer $n$ has the given property if and only if it is neither a perfect square nor a power of a prime.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 27899,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nFind all polynomials $p(x)$ with real coefficients that have the following property: There exists a polynomial $q(x)$ with real coefficients such that\n$$\np(1)+p(2)+p(3)+\\cdots+p(n)=p(n) q(n)\n$$\nfor all positive integers $n$.",
    "options": [],
    "answer": "All such polynomials are p(x)=c(x+r-1)(x+r-2)\\cdots(x+r-d) with real c, integer r in {0,1,2,...,d}, and d the degree (including the constant case d=0).",
    "solution": "Solution:\nThe property clearly holds whenever $p(x)$ is a constant polynomial, since we can take $q(x)=x$. Assume henceforth that $p(x)$ is nonconstant and has the stated property. Let $d$ be the degree of $p(x)$, so $p(x)$ is of the form\n$$\np(x)=c x^{d}+\\cdots .\n$$\nBy a Lemma (which we will prove at the end), $\\sum_{k=1}^{n} k^{d}$ is a polynomial in $n$ of degree $d+1$, so $p(1)+p(2)+\\cdots+p(n)$ is a polynomial in $n$ of degree $d+1$. Hence, $q(n)$ is a polynomial of degree 1. Furthermore, the coefficient of $n^{d+1}$ in $\\sum_{k=1}^{n} k^{d}$ is $\\frac{1}{d+1}$, so the coefficient of $n$ in $q(n)$ is also $\\frac{1}{d+1}$. Let $q(x)=\\frac{1}{d+1}(x+r)$. We have that\n$$\np(1)+p(2)+p(3)+\\cdots+p(n)=p(n) q(n)\n$$\nand\n$$\np(1)+p(2)+p(3)+\\cdots+p(n)+p(n+1)=p(n+1) q(n+1) \\text{.}\n$$\nSubtracting the first equation from the second, we get\n$$\np(n+1)=p(n+1) q(n+1)-p(n) q(n),\n$$\nand hence\n$$\np(n) q(n)=p(n+1)[q(n+1)-1] .\n$$\nSince this holds for all positive integers $n$, it follows that\n$$\np(x) q(x)=p(x+1)[q(x+1)-1]\n$$\nfor all real numbers $x$. We can then write\n$$\np(x) \\cdot \\frac{1}{d+1}(x+r)=p(x+1)\\left[\\frac{1}{d+1}(x+r+1)-1\\right]\n$$\nso\n$$\n(x+r) p(x)=(x+r-d) p(x+1) .\n$$\nSetting $x=-r$, we get\n$$\n(-d) p(-r+1)=0\n$$\nHence, $-r+1$ is a root of $p(x)$. Let $p(x)=(x+r-1) p_{1}(x)$. Then\n$$\n(x+r)(x+r-1) p_{1}(x)=(x+r-d)(x+r) p_{1}(x+1),\n$$\nso\n$$\n(x+r-1) p_{1}(x)=(x+r-d) p_{1}(x+1) .\n$$\nIf $d=1$, then $p_{1}(x)$ is a constant, so both sides are equal, and we can say $p(x)=c(x+r-1)$.\nOtherwise, setting $x=-r+1$, we get\n$$\n(1-d) p_{1}(-r+2)=0\n$$\nHence, $-r+2$ is a root of $p_{1}(x)$. Let $p_{1}(x)=(x+r-2) p_{2}(x)$. Then\n$$\n(x-r-1)(x+r-2) p_{2}(x)=(x+r-d)(x+r-1) p_{2}(x+1),\n$$\nso\n$$\n(x+r-2) p_{2}(x)=(x+r-d) p_{2}(x+1) \\text{.}\n$$\nIf $d=2$, then $p_{2}(x)$ is a constant, so both sides are equal, and we can say $p(x)=c(x+r-1)(x+r-2)$.\nOtherwise, we can continue to substitute, giving us\n$$\np(x)=c(x+r-1)(x+r-2) \\cdots(x+r-d) .\n$$\nConversely, if $p(x)$ is of this form, then\n$$\n\\begin{aligned}\np(x)= & c(x+r-1)(x+r-2) \\cdots(x+r-d) \\\\\n= & \\frac{c(d+1)(x+r-1)(x+r-2) \\cdots(x+r-d)}{d+1} \\\\\n= & \\frac{c[(x+r)-(x+r-d-1)](x+r-1)(x+r-2) \\cdots(x+r-d)}{d+1} \\\\\n= & \\frac{c(x+r)(x+r-1)(x+r-2) \\cdots(x+r-d)}{d+1} \\\\\n& -\\frac{c(x+r-1)(x+r-2) \\cdots(x+r-d)(x+r-d-1)}{d+1} .\n\\end{aligned}\n$$\nThen the sum $p(1)+p(2)+p(3)+\\cdots+p(n)$ telescopes, and we are left with\n$$\n\\begin{aligned}\np(1)+p(2)+p(3)+\\cdots+p(n)= & \\frac{c(n+r)(n+r-1)(n+r-2) \\cdots(n+r-d)}{d+1} \\\\\n& -\\frac{c(r)(r-1) \\cdots(r-d+1)(r-d)}{d+1} .\n\\end{aligned}\n$$\nWe want this to be of the form\n$$\np(n) q(n)=c(n+r-1)(n+r-2) \\cdots(n+r-d) q(n)\n$$\nfor some polynomial $q(n)$. The only way that this can hold for each positive integer $n$ is if the term\n$$\n\\frac{c(r)(r-1) \\cdots(r-d+1)(r-d)}{d+1}\n$$\nis equal to 0 . This means $r$ has to be one of the values $0,1,2, \\ldots, d$. Therefore, the polynomials we seek are of the form\n$$\np(x)=c(x+r-1)(x+r-2) \\cdots(x+r-d)\n$$\nwhere $r \\in\\{0,1,2, \\ldots, d\\}$.\n\nLemma. For a positive integer $d$,\n$$\n\\sum_{k=1}^{n} k^{d}\n$$\nis a polynomial in $n$ of degree $d+1$. Furthermore, the coefficient of $n^{d+1}$ is $\\frac{1}{d+1}$.\n\nProof. We prove the result by strong induction. For $d=1$,\n$$\n\\sum_{k=1}^{n} k=\\frac{1}{2} n^{2}+\\frac{1}{2} n\n$$\nso the result holds. Assume that the result holds for $d=1,2,3, \\ldots, m$, for some positive integer $m$. By the Binomial Theorem,\n$$\n(k+1)^{m+2}-k^{m+2}=(m+2) k^{m+1}+c_{m} k^{m}+c_{m-1} k^{m-1}+\\cdots+c_{1} k+c_{0},\n$$\nfor some coefficients $c_{m}, c_{m-1}, \\ldots, c_{1}, c_{0}$. Summing over $1 \\leq k \\leq n$, we get\n$$\n(n+1)^{m+2}-1=(m+2) \\sum_{k=1}^{n} k^{m+1}+c_{m} \\sum_{k=1}^{n} k^{m}+\\cdots+c_{1} \\sum_{k=1}^{n} k+c_{0} n .\n$$\nThen\n$$\n\\sum_{k=1}^{n} k^{m+1}=\\frac{(n+1)^{m+2}-c_{m} \\sum_{k=1}^{n} k^{m}-\\cdots-c_{1} \\sum_{k=1}^{n} k-c_{0} n-1}{m+2} .\n$$\nBy the induction hypothesis, the sums $\\sum_{k=1}^{n} k^{m}, \\ldots, \\sum_{k=1}^{n} k$ are all polynomials in $n$ of degree less than $m+2$. Hence, the above expression is a polynomial in $n$ of degree $m+2$, and the coefficient of $n^{m+2}$ is $\\frac{1}{m+2}$. Thus, the result holds for $d=m+1$, which completes the induction step.",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 27900,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nLet $k$ be a given even positive integer. Sarah first picks a positive integer $N$ greater than $1$ and proceeds to alter it as follows: every minute, she chooses a prime divisor $p$ of the current value of $N$, and multiplies the current $N$ by $p^{k}-p^{-1}$ to produce the next value of $N$. Prove that there are infinitely many even positive integers $k$ such that, no matter what choices Sarah makes, her number $N$ will at some point be divisible by $2018$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\nNote that $1009$ is prime. We will show that if $k=1009^{m}-1$ for some positive integer $m$, then Sarah's number must at some point be divisible by $2018$. Let $P$ be the largest divisor of $N$ not divisible by a prime congruent to $1$ modulo $1009$. Assume for contradiction that $N$ is never divisible by $2018$. We will show that $P$ decreases each minute. Suppose that in the $t^{\\text{th}}$ minute, Sarah chooses the prime divisor $p$ of $N$. First note that $N$ is replaced with $\\frac{p^{k+1}-1}{p} \\cdot N$ where\n$$\np^{k+1}-1 = p^{1009^{m}}-1 = (p-1)\\left(p^{1009^{m}-1}+p^{1009^{m}-2}+\\cdots+1\\right)\n$$\nSuppose that $q$ is a prime number dividing the second factor. Since $q$ divides $p^{1009^{m}}-1$, it follows that $q \\neq p$ and the order of $p$ modulo $q$ must divide $1009^{m}$ and hence is either divisible by $1009$ or is equal to $1$. If it is equal to $1$ then $p \\equiv 1 \\pmod{q}$, which implies that\n$$\n0 \\equiv p^{1009^{m}-1}+p^{1009^{m}-2}+\\cdots+1 \\equiv 1009^{m} \\quad (\\bmod q)\n$$\nand thus $q=1009$. However, if $q=1009$ then $p \\geq 1010$ and $p$ must be odd. Since $p-1$ now divides $N$, it follows that $N$ is divisible by $2018$ in the $(t+1)^{\\text{th}}$ minute, which is a contradiction. Therefore the order of $p$ modulo $q$ is divisible by $1009$ and hence $1009$ divides $q-1$. Therefore all of the prime divisors of the second factor are congruent to $1$ modulo $1009$. This implies that $P$ is replaced by a divisor of $\\frac{p-1}{p} \\cdot P$ in the $(t+1)^{\\text{th}}$ minute and therefore decreases. Since $P \\geq 1$ must always hold, $P$ cannot decrease forever. Therefore $N$ must at some point be divisible by $2018$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 27901,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\n1. Amy has drawn three points in a plane, $A$, $B$, and $C$, such that $AB = BC = CA = 6$. Amy is allowed to draw a new point if it is the circumcenter of a triangle whose vertices she has already drawn. For example, she can draw the circumcenter $O$ of triangle $ABC$, and then afterwards she can draw the circumcenter of triangle $ABO$.\n\na) Prove that Amy can eventually draw a point whose distance from a previously drawn point is greater than $7$.\n\nb) Prove that Amy can eventually draw a point whose distance from a previously drawn point is greater than $2019$.\n\n(Recall that the circumcenter of a triangle is the center of the circle that passes through its three vertices.)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\n(a) Given triangle $\\triangle ABC$, Amy can draw the following points:\n- $O$ is the circumcenter of $\\triangle ABC$\n- $A_1$ is the circumcenter of $\\triangle BOC$\n- $A_2$ is the circumcenter of $\\triangle OBA_1$\n- $A_3$ is the circumcenter of $\\triangle BA_2A_1$\n\nWe claim that $AA_3 > 7$. We present two ways to prove this claim.\n\nFirst Method:\nBy symmetry of the equilateral triangle $\\triangle ABC$, we have $\\angle AOB = \\angle BOC = \\angle COA = 120^\\circ$. Since $OB = OC$ and $A_1B = A_1O = A_1C$, we deduce that $\\triangle A_1OB \\cong \\triangle A_1OC$, and hence $\\angle BOA_1 = \\angle COA_1 = 60^\\circ$. Therefore, since $\\triangle A_1OB$ is isosceles, it must be equilateral. As we found for our original triangle, we find $\\angle BA_2A_1 = 120^\\circ$, and so $\\angle A_2BA_1 = \\angle A_2A_1B = 30^\\circ$ (since $A_2B = A_2A_1$). Also we see that $\\angle OBA_2 = 30^\\circ = \\angle OBC$, which shows that $A_2$ lies on the segment $BC$.\n\nApplying the Law of Sines to $\\triangle BOC$, we obtain\n$$\nOC = \\frac{BC \\sin(\\angle OBC)}{\\sin(\\angle BOC)} = \\frac{6(1/2)}{\\sqrt{3}/2} = 2\\sqrt{3}.\n$$\nBy symmetry, we see that (i) $OA_1$ is the bisector of $\\angle BOC$ and the perpendicular bisector of $BC$, and (ii) the three points $A$, $O$, and $A_1$ are collinear. Therefore $A_1A = A_1O + OA = 2OA = 4\\sqrt{3}$.\n\nThe same argument that we used to show $\\triangle A_1OB$ is equilateral with side $AC/\\sqrt{3}$ shows that $\\triangle A_3A_2A_1$ is equilateral with side $OB/\\sqrt{3} = 2$. Thus $\\angle A_3A_1O = \\angle OA_1B + \\angle A_3A_1A_2 - \\angle A_2A_1B = 60^\\circ + 60^\\circ - 30^\\circ = 90^\\circ$. Hence we can apply the Pythagorean Theorem:\n$$\nA_3A = \\sqrt{(A_3A_1)^2 + (A_1A)^2} = \\sqrt{2^2 + (4\\sqrt{3})^2} = \\sqrt{52} > \\sqrt{49} = 7.\n$$\n\nSecond Method:\n(An alternative to writing the justifications of the constructions in the First Method is to use analytic geometry. Once the following coordinates are found using the kind of reasoning in the First Method or by other means, the writeup can justify them succinctly by computing distances.)\n\nLabel $(0,0)$ as $B$, $(6,0)$ as $C$, and $(3, \\sqrt{3})$ as $A$. Then we have $AB = BC = CA = 6$.\n\nThe circumcenter $O$ of $\\triangle ABC$ is $(3, \\sqrt{3})$; this can be verified by observing $OA = OB = OC = 2\\sqrt{3}$. Next, the point $A_1 = (3, -\\sqrt{3})$ satisfies $A_1O = A_1B = A_1C = 2\\sqrt{3}$, so $A_1$ is the circumcenter of $\\triangle BOC$.\n\nThe point $A_2 = (2, 0)$ satisfies $A_2O = A_2B = A_2A_1 = 2$, so this is the circumcenter of $\\triangle OBA_1$.\n\nAnd the point $A_3 = (1, -\\sqrt{3})$ satisfies $A_3B = A_3A_2 = A_3A_1 = 2$, so this is the circumcenter of $\\triangle BA_2A_1$.\n\nFinally, we compute $A_3A = \\sqrt{52} > \\sqrt{49} = 7$, and part (a) is proved.\n\n(b) In part (a), using either method we find that $OA_3 = 4 > 2\\sqrt{3} = OA$. By rotating the construction of part (a) by $\\pm 120^\\circ$ about $O$, Amy can construct $B_3$ and $C_3$ such that $\\triangle A_3B_3C_3$ is equilateral with circumcenter $O$ and circumradius $4$, which is strictly bigger than the circumradius $2\\sqrt{3}$ of $\\triangle ABC$. Amy can repeat this process starting from $\\triangle A_3B_3C_3$. After $n$ iterations of the process, Amy will have drawn the vertices of an equilateral triangle whose circumradius is $2\\sqrt{3}\\left(\\frac{4}{2\\sqrt{3}}\\right)^n$, which is bigger than $2019$ when $n$ is sufficiently large.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 27902,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nLet $a$ and $b$ be positive integers such that $a + b^{3}$ is divisible by $a^{2} + 3 a b + 3 b^{2} - 1$. Prove that $a^{2} + 3 a b + 3 b^{2} - 1$ is divisible by the cube of an integer greater than 1.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\nLet $Z = a^{2} + 3 a b + 3 b^{2} - 1$. By assumption, there is a positive integer $c$ such that $c Z = a + b^{3}$. Noticing the resemblance between the first three terms of $Z$ and those of the expansion of $(a + b)^{3}$, we are led to\n$$\n(a + b)^{3} = a(a^{2} + 3 a b + 3 b^{2}) + b^{3} = a(Z + 1) + b^{3} = a Z + a + b^{3} = a Z + c Z.\n$$\nThus $Z$ divides $(a + b)^{3}$.\nLet the prime factorization of $a + b$ be $p_{1}^{e_{1}} p_{2}^{e_{2}} \\cdots p_{k}^{e_{k}}$ and let $Z = p_{1}^{f_{1}} p_{2}^{f_{2}} \\cdots p_{k}^{f_{k}}$, where $f_{i} \\leq 3 e_{i}$ for each $i$ since $Z$ divides $(a + b)^{3}$. If $Z$ is not divisible by a perfect cube greater than one, then $0 \\leq f_{i} \\leq 2$ and hence $f_{i} \\leq 2 e_{i}$ for each $i$. This implies that $Z$ divides $(a + b)^{2}$. However, $(a + b)^{2} < a^{2} + 3 a b + 3 b^{2} - 1 = Z$ since $a, b \\geq 1$, which is a contradiction. Thus $Z$ must be divisible by a perfect cube greater than one.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 27903,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nLet $m$ and $n$ be positive integers. A $2m \\times 2n$ grid of squares is coloured in the usual chessboard fashion. Find the number of ways of placing $mn$ counters on the white squares, at most one counter per square, so that no two counters are on white squares that are diagonally adjacent. An example of a way to place the counters when $m=2$ and $n=3$ is shown below.\n\n![](attached_image_1.png)",
    "options": [],
    "answer": "(m+n choose m)",
    "solution": "Solution:\nDivide the chessboard into $mn$ $2 \\times 2$ squares.\n\n![](attached_image_2.png)\n\nEach $2 \\times 2$ square can contain at most one counter. Since we want to place $mn$ counters, each $2 \\times 2$ square must contain exactly one counter.\nAssume that the lower-right corner of the $2m \\times 2n$ chessboard is white, so in each $2 \\times 2$ square, the upper-left and lower-right squares are white. Call a $2 \\times 2$ square UL if the counter it contains is on the upper-left white square, and call it LR if the counter it contains is on the lower-right white square.\nSuppose some $2 \\times 2$ square is UL. Then the $2 \\times 2$ square above it (if it exists) must also be UL, and the $2 \\times 2$ square to the left of it (if it exists) must also be UL.\n\n![](attached_image_3.png)\n\nSimilarly, if some $2 \\times 2$ square is LR, then the $2 \\times 2$ square below it (if it exists) must also be LR, and the $2 \\times 2$ square to the right of it (if it exists) must also be LR.\n\n![](attached_image_4.png)\n\nThen the collection of UL $2 \\times 2$ squares form a region that is top-justified and left-justified, and the collection of LR $2 \\times 2$ squares form a region that is bottom-justified and right-justified. This means that the boundary between the two regions forms a path between the lower-left corner and upper-right corner of the $2m \\times 2n$ chessboard.\n\n![](attached_image_5.png)\n\nConversely, any path from the lower-left corner to the upper-right corner, where each step consists of two units, can serve as the boundary of the UL squares and LR squares. Thus, the number of ways of placing the counters is equal to the number of paths, which is $\\binom{m+n}{m}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 27904,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nLet $n$ be an integer greater than $1$, and let $a_{0}, a_{1}, \\ldots, a_{n}$ be real numbers with $a_{1} = a_{n-1} = 0$. Prove that for any real number $k$,\n\n$$\n\\left|a_{0}\\right| - \\left|a_{n}\\right| \\leq \\sum_{i=0}^{n-2} \\left|a_{i} - k a_{i+1} - a_{i+2}\\right|\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\nLet $Q(x) = x^{2} - k x - 1$ and let $P(x) = a_{0} + a_{1} x + \\cdots + a_{n} x^{n}$. Note that the product of the two roots of $Q(x)$ is $-1$ and thus one of the two roots has magnitude at most $1$. Let $z$ be this root. Now note that since $a_{1} = a_{n-1} = 0$, we have that\n$$\n\\begin{aligned}\n0 = Q(z) P(z) & = -a_{0} - k a_{0} z + \\sum_{i=0}^{n-2} \\left(a_{i} - k a_{i+1} - a_{i+2}\\right) z^{i+2} - k a_{n} z^{n+1} + a_{n} z^{n+2} \\\\\n& = a_{0}(-1 - k z) + \\sum_{i=0}^{n-2} \\left(a_{i} - k a_{i+1} - a_{i+2}\\right) z^{i+2} + a_{n} z^{n} (z^{2} - k z) \\\\\n& = -a_{0} z^{2} + \\sum_{i=0}^{n-2} \\left(a_{i} - k a_{i+1} - a_{i+2}\\right) z^{i+2} + a_{n} z^{n}\n\\end{aligned}\n$$\nwhere the third equality follows since $z^{2} - k z - 1 = 0$. The triangle inequality now implies\n$$\n\\begin{aligned}\n\\left|a_{0}\\right| \\cdot |z|^{2} & \\leq \\left|a_{n}\\right| \\cdot |z|^{n} + \\sum_{i=0}^{n-2} \\left|a_{i} - k a_{i+1} - a_{i+2}\\right| \\cdot |z|^{i+2} \\\\\n& \\leq \\left|a_{n}\\right| \\cdot |z|^{2} + \\sum_{i=0}^{n-2} \\left|a_{i} - k a_{i+1} - a_{i+2}\\right| \\cdot |z|^{2}\n\\end{aligned}\n$$\nsince $|z| \\leq 1$ and $n \\geq 2$. Since $z \\neq 0$, the inequality is obtained on dividing by $|z|^{2}$.\n\n\nSolution 2:\nLet $k$ be a real number. Put\n$$\nR = \\begin{cases}\n\\sqrt{k^{2} + 4} & \\text{ if } k \\geq 0 \\\\\n-\\sqrt{k^{2} + 4} & \\text{ if } k < 0\n\\end{cases}\n$$\nDefine the polynomial\n$$\nS(x) = x^{2} + R x + 1\n$$\nThe roots of $S$ are\n$$\nb = \\frac{-R - k}{2} \\quad \\text{ and } \\quad c = \\frac{-R + k}{2}\n$$\nThen we have\n$$\nb - c = -k, \\quad b c = 1, \\quad \\text{ and } \\quad |c| \\leq 1\n$$\n(the inequality follows from $b c = 1$ and $|c| \\leq |b|$).\nPut $d_{i} = a_{i} + b a_{i+1}$ for $i = 0, 1, \\ldots, n-1$. Then, for $i = 0, 1, \\ldots, n-2$, we have\n$$\n\\begin{aligned}\nd_{i} - c d_{i+1} & = a_{i} + (b - c) a_{i+1} - b c a_{i+2} \\\\\n& = a_{i} - k a_{i+1} - a_{i+2}\n\\end{aligned}\n$$\nTherefore\n$$\n\\begin{aligned}\n\\sum_{i=0}^{n-2} \\left|a_{i} - k a_{i+1} - a_{i+2}\\right| & = \\sum_{i=0}^{n-2} \\left|d_{i} - c d_{i+1}\\right| \\\\\n& \\geq \\sum_{i=0}^{n-2} \\left( \\left|d_{i}\\right| - |c| \\left|d_{i+1}\\right| \\right) \\\\\n& = \\left|d_{0}\\right| + (1 - |c|) \\sum_{i=1}^{n-2} \\left|d_{i}\\right| - |c| \\left|d_{n-1}\\right| \\\\\n& \\geq \\left|d_{0}\\right| - |c| \\left|d_{n-1}\\right| \\\\\n& = \\left|a_{0} + b a_{1}\\right| - |c| \\left|a_{n-1} + b a_{n}\\right| \\\\\n& = \\left|a_{0}\\right| - |b c| \\left|a_{n}\\right| \\\\\n& = \\left|a_{0}\\right| - \\left|a_{n}\\right|\n\\end{aligned}\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 27905,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nDavid and Jacob are playing a game of connecting $n \\geq 3$ points drawn in a plane. No three of the points are collinear. On each player's turn, he chooses two points to connect by a new line segment. The first player to complete a cycle consisting of an odd number of line segments loses the game. (Both endpoints of each line segment in the cycle must be among the $n$ given points, not points which arise later as intersections of segments.) Assuming David goes first, determine all $n$ for which he has a winning strategy.",
    "options": [],
    "answer": "n ≡ 2 (mod 4)",
    "solution": "Solution:\n\nAnswer: David has a winning strategy if and only if $n \\equiv 2 (\\bmod 4)$.\n\nCall a move illegal if it would cause an odd cycle to be formed for the first time. First we show that if $n$ is odd, then any strategy where Jacob picks a legal move if one is available to him causes him to win. Assume for contradiction that Jacob at some point has no legal moves remaining. Since the graph representing the game state has no odd cycle, it must be bipartite. Let $a$ and $b$ be the sizes of the two sets in the bipartition of the graph. If there is some edge not already added between the two sets, adding this edge would be a legal move for Jacob. Therefore the graph must be a complete bipartite graph with all of its $ab$ edges present. However, since $a + b = n$ which is odd, one of $a$ or $b$ must be even and thus the graph contains an even number of edges. Moreover, since it is Jacob's turn, the graph must contain an odd number of edges, which is a contradiction. Therefore Jacob has a winning strategy for all odd $n$.\n\nNow consider the case where $n$ is even. Call a graph good if the set of vertices of degree at least 1 are in a perfect matching (a set of non-adjacent edges that includes every vertex of the graph). The key observation is that either player has a strategy to preserve that the graph is good while increasing the number of vertices of degree at least 1. More precisely, if the graph was good at the end of a player's previous turn and there are fewer than $n$ vertices of degree at least 1, then at the end of his current turn he can always ensure that: (1) the graph is good and (2) there are at least two more vertices of degree 1 since the end of his previous turn. Let $A$ be the set of vertices of degree at least 1 at the end of the player's previous turn and $B$ be the set of remaining vertices where $|B| > 0$. Since the vertices of $A$ have a perfect matching, $|A|$ is even, and since $n$ is even, so is $|B|$. If the other player adds an edge between two vertices of $A$, add an edge between two vertices of $B$. If the other player adds an edge between two vertices of $B$, add an edge between one of those vertices and a vertex of $A$ (but on the first round, when $A$ is empty, respond by adding an edge between two other vertices of $B$). If the other player adds an edge between a vertex in $A$ and a vertex in $B$, then since $|B|$ is even, there must be another vertex of $B$. Connect these two vertices in $B$ with an edge. None of these moves can form a cycle and thus are legal. Furthermore, all of them achieve (1) and (2), proving the claim.\n\nWe now show that David has a winning strategy if $n \\equiv 2 (\\bmod 4)$. Since the graph begins empty and therefore good, David has a strategy of legal moves to ensure that the graph contains a perfect matching after no more than $n$ moves. After this, let David implement any strategy where he picks a legal move if one is available to him. Assume for contradiction that there is a turn where David has no legal moves. This graph must be a complete bipartite graph containing a perfect matching. If one of the sets in the bipartition has size greater than $n / 2$, it must contain two vertices matched in the perfect matching, which is impossible. Therefore there are $n / 2$ vertices in each part and $n^2 / 4$ edges have been added in total, which is an odd number. This contradicts the fact that it is David's turn, and proves the result for $n \\equiv 2 (\\bmod 4)$.\n\nFinally, consider the case that $n \\equiv 0 (\\bmod 4)$. Note that after David's first turn, the graph contains a single edge and thus is good. This implies that Jacob can ensure the graph contains a perfect matching and win by the above parity argument.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 27906,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nLet $S$ be a set of $n \\geq 3$ positive real numbers. Show that the largest possible number of distinct integer powers of three that can be written as the sum of three distinct elements of $S$ is $n-2$.",
    "options": [],
    "answer": "n - 2",
    "solution": "Solution:\n\nWe will show by induction that for all $n \\geq 3$, it holds that at most $n-2$ powers of three are sums of three distinct elements of $S$ for any set $S$ of positive real numbers with $|S|=n$. This is trivially true when $n=3$.\n\nLet $n \\geq 4$ and consider the largest element $x \\in S$. The sum of $x$ and any two other elements of $S$ is strictly between $x$ and $3x$. Therefore $x$ can be used as a summand for at most one power of three. By the induction hypothesis, at most $n-3$ powers of three are sums of three distinct elements of $S \\setminus \\{x\\}$. This completes the induction.\n\nEven if it was not asked to prove, we will now show that the optimal answer $n-2$ is reached. Observe that the set $S=\\{1,2,3^{2}-3,3^{3}-3, \\ldots, 3^{n}-3\\}$ is such that $3^{2}, 3^{3}, \\ldots, 3^{n}$ can be expressed as sums of three distinct elements of $S$. This makes use of the fact that each term of the form $3^{k}-3$ can be used in exactly one sum of three terms equal to $3^{k}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 27907,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nA circle is inscribed in a rhombus $A B C D$. Points $P$ and $Q$ vary on line segments $\\overline{A B}$ and $\\overline{A D}$, respectively, so that $\\overline{P Q}$ is tangent to the circle. Show that for all such line segments $\\overline{P Q}$, the area of triangle $C P Q$ is constant.\n\n![](attached_image_1.png)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\nLet the circle be tangent to $\\overline{P Q}$, $\\overline{A B}$, $\\overline{A D}$ at $T$, $U$, and $V$, respectively. Let $p = P T = P U$ and $q = Q T = Q V$. Let $a = A U = A V$ and $b = B U = D V$. Then the side length of the rhombus is $a + b$.\n\n![](attached_image_2.png)\n\nLet $\\theta = \\angle B A D$, so $\\angle A B C = \\angle A D C = 180^{\\circ} - \\theta$. Then (using the notation $[XYZ]$ for the area of a triangle of vertices $X, Y, Z$)\n$$\n\\begin{aligned}\n& [A P Q] = \\frac{1}{2} \\cdot A P \\cdot A Q \\cdot \\sin \\theta = \\frac{1}{2}(a-p)(a-q) \\sin \\theta \\\\\n& [B C P] = \\frac{1}{2} \\cdot B P \\cdot B C \\cdot \\sin (180^{\\circ} - \\theta) = \\frac{1}{2}(b+p)(a+b) \\sin \\theta \\\\\n& [C D Q] = \\frac{1}{2} \\cdot D Q \\cdot C D \\cdot \\sin (180^{\\circ} - \\theta) = \\frac{1}{2}(b+q)(a+b) \\sin \\theta\n\\end{aligned}\n$$\nSo\n$$\n\\begin{aligned}\n[C P Q] & = [A B C D] - [A P Q] - [B C P] - [C D Q] \\\\\n& = (a+b)^2 \\sin \\theta - \\frac{1}{2}(a-p)(a-q) \\sin \\theta - \\frac{1}{2}(b+p)(a+b) \\sin \\theta - \\frac{1}{2}(b+q)(a+b) \\sin \\theta \\\\\n& = \\frac{1}{2}\\left(a^2 + 2 a b - b p - b q - p q\\right) \\sin \\theta\n\\end{aligned}\n$$\nLet $O$ be the center of the circle, and let $r$ be the radius of the circle. Let $x = \\angle T O P = \\angle U O P$ and $y = \\angle T O Q = \\angle V O Q$. Then $\\tan x = \\frac{p}{r}$ and $\\tan y = \\frac{q}{r}$.\n\n![](attached_image_3.png)\n\nNote that $\\angle U O V = 2x + 2y$, so $\\angle A O U = x + y$. Also, $\\angle A O B = 90^{\\circ}$, so $\\angle O B U = x + y$. Therefore,\n$$\n\\tan (x + y) = \\frac{a}{r} = \\frac{r}{b}\n$$\nso $r^2 = a b$. But\n$$\n\\frac{r}{b} = \\tan (x + y) = \\frac{\\tan x + \\tan y}{1 - \\tan x \\tan y} = \\frac{\\frac{p}{r} + \\frac{q}{r}}{1 - \\frac{p}{r} \\cdot \\frac{q}{r}} = \\frac{r(p+q)}{r^2 - p q} = \\frac{r(p+q)}{a b - p q}.\n$$\nHence, $a b - p q = b p + b q$, so $b p + b q + p q = a b$. Therefore,\n$$\n[C P Q] = \\frac{1}{2}\\left(a^2 + 2 a b - b p - b q - p q\\right) \\sin \\theta = \\frac{1}{2}\\left(a^2 + a b\\right) \\sin \\theta\n$$\nwhich is constant.\nSolution:\nAlternate Solution: Let $O$ be the center of the circle and $r$ its radius. Then $[C P Q] = [C D Q P B] - [C D Q] - [C B P]$, where $[\\ldots]$ denotes area of the polygon with given vertices. Note that $[C D Q P B]$ is half $r$ times the perimeter of $C D Q P B$. Note that the heights of $C D Q$ and $C B P$ are $2 r$ so $[C D Q] = r \\cdot D Q$ and $[C B P] = r \\cdot P B$. Using the fact that $Q T = Q V$ and $P U = P T$, it now follows that $[C P Q] = [O V D C B U] - [C D V] - [C B U]$, which is independent of $P$ and $Q$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 27908,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nA purse contains a finite number of coins, each with distinct positive integer values. Is it possible that there are exactly 2020 ways to use coins from the purse to make the value 2020?",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\nIt is possible.\nConsider a coin purse with coins of values $2, 4, 8, 2014, 2016, 2018, 2020$ and every odd number between $503$ and $1517$. Call such a coin big if its value is between $503$ and $1517$. Call a coin small if its value is $2, 4$ or $8$ and huge if its value is $2014, 2016, 2018$ or $2020$. Suppose some subset of these coins contains no huge coins and sums to $2020$. If it contains at least four big coins, then its value must be at least $503+505+507+509>2020$. Furthermore, since all of the small coins are even in value, if the subset contains exactly one or three big coins, then its value must be odd. Thus the subset must contain exactly two big coins. The eight possible subsets of the small coins have values $0, 2, 4, 6, 8, 10, 12, 14$. Therefore, the ways to make the value $2020$ using no huge coins correspond to the pairs of big coins with sums $2006, 2008, 2010, 2012, 2014, 2016, 2018$ and $2020$. The numbers of such pairs are $250, 251, 251, 252, 252, 253, 253, 254$, respectively. Thus there are exactly $2016$ subsets of this coin purse with value $2020$ using no huge coins. There are exactly four ways to make a value of $2020$ using huge coins; these are $\\{2020\\}, \\{2, 2018\\}, \\{4, 2016\\}$ and $\\{2, 4, 2014\\}$. Thus there are exactly $2020$ ways to make the value $2020$.\n\nAlternate construction: Take the coins $1, 2, \\ldots, 11, 1954, 1955, \\ldots, 2019$. The only way to get $2020$ is a non-empty subset of $1, \\ldots, 11$ and a single large coin. There are $2047$ non-empty such subsets of sums between $1$ and $66$. Thus they each correspond to a unique large coin making $2020$, so we have $2047$ ways. Thus we only need to remove some large coins, so that we remove exactly $27$ small sums. This can be done, for example, by removing coins $2020-n$ for $n=1,5,6,7,8,9$, as these correspond to $1+3+4+5+6+8=27$ partitions into distinct numbers that are at most $11$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 27909,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nLet $S=\\{1,4,8,9,16, \\ldots\\}$ be the set of perfect powers of integers, i.e. numbers of the form $n^{k}$ where $n, k$ are positive integers and $k \\geq 2$. Write $S=\\left\\{a_{1}, a_{2}, a_{3} \\ldots\\right\\}$ with terms in increasing order, so that $a_{1}<a_{2}<a_{3} \\cdots$. Prove that there exist infinitely many integers $m$ such that $9999$ divides the difference $a_{m+1}-a_{m}$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\nThe idea is that most perfect powers are squares. If $a_{n}=x^{2}$ and $a_{n+1}=(x+1)^{2}$, then $a_{n+1}-a_{n}=2x+1$. Note that $9999 \\mid 2x+1$ is equivalent to $x \\equiv 4999 \\pmod{9999}$. Hence we will be done if we can show that there exist infinitely many $x \\equiv 4999 \\pmod{9999}$ such that there are no perfect powers strictly between $x^{2}$ and $(x+1)^{2}$.\n\nAssume otherwise, so that there exists a positive integer $N$ such that: for $x \\equiv 4999 \\pmod{9999}$ and $x \\geq N$, there is a perfect power $b_{x}^{e_{x}}$ ($e_{x} \\geq 2$) between $x^{2}$ and $(x+1)^{2}$. Without loss of generality, we can take $N$ to be $\\equiv 4999 \\pmod{9999}$. Note that $x^{2}$ and $(x+1)^{2}$ are consecutive squares, hence $e_{x}$ is odd, and thus $e_{x} \\geq 3$. Let $t_{n}$ be the number of odd perfect powers that are at most $n$.\n\nBy tallying the $b_{x}^{e_{x}}$ up (clearly they are all distinct), for any $m \\geq 1$ we have at least $m$ perfect odd powers between $1$ and $(N+9999m)^{2}$, so that\n$$\nt_{(N+9999m)^{2}} \\geq m\n$$\nIn particular, for large enough $n$ we have\n$$\nt_{n} \\geq \\frac{\\sqrt{n}}{10000}\n$$\nNow, if $x^{f} \\leq n$ then $x \\leq \\sqrt[f]{n}$. Also, $n \\geq x^{f} \\geq 2^{f}$ so $f \\leq \\log_{2}(n)$. So we have\n$$\nt_{n} \\leq \\sum_{i=3}^{\\log_{2}(n)} \\sqrt[i]{n} \\leq \\log_{2}(n) \\sqrt[3]{n}\n$$\nCombining with the previous inequality, we have\n$$\n\\sqrt[6]{n} \\leq 10000 \\log_{2}(n)\n$$\nfor all large enough $n$. However, this inequality is false for all large $n$, contradiction. Therefore the problem statement holds.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 27910,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nThere are $19,998$ people on a social media platform, where any pair of them may or may not be friends. For any group of $9,999$ people, there are at least $9,999$ pairs of them that are friends. What is the least number of friendships, that is, the least number of pairs of people that are friends, that must be among the $19,998$ people?",
    "options": [],
    "answer": "49995",
    "solution": "Solution:\n\nIt is $5 \\cdot 9999 = 49995$. One possible construction is as follows: have the $19,998$ people form $3,333$ groups of $6$ people, and within each group every pair of people are friends. Now, for any group of $9,999$ people, say that there are $x_{1}, x_{2}, \\ldots, x_{3333}$ people in each of the $6$ groups, respectively. Then there are\n$$\n\\frac{1}{2} \\sum_{i=1}^{3333} x_{i}\\left(x_{i}-1\\right)\n$$\npairs of friendships total. But we have that\n$$\nx_{i}\\left(x_{i}-1\\right) \\geq 5 x_{i}-9,\n$$\nso\n$$\n\\frac{1}{2} \\sum_{i=1}^{3333} x_{i}\\left(x_{i}-1\\right) \\geq \\frac{1}{2} \\sum_{i=1}^{3333}\\left(5 x_{i}-9\\right)=\\frac{1}{2}(9999 \\cdot 5-9 \\cdot 3333)=9999\n$$\nas desired.\n\nIt remains to show that $49995$ pairs of friends is optimal. For what follows, let $9999=N$, so that $19,998=2N$, and assume that the condition is satisfied. Let the number of pairs of friends be $e$. Designate half of the people as red and the other half as blue, so that the number of pairs of friends who are both red is minimized.\n\nNote that this means that for every pair of people, one red and one blue, we have that the number of red friends of the blue person is at least as many as the number of red friends of the red person, and the inequality is strict if the two people are friends. This is because we can otherwise swap the two people. Now, if every blue person is friends with at least $3$ red people, then the total number of friendships, $e$, is at least $N+3N+N=5N$ ($N$ each from the red people and blue people and $3N$ from the pairs), as desired. If some blue person is friends with at most $2$ red people, then every red person is friends with at most $2$ red people, so the number of pairs of red friends is at most $N$, with equality only if every red person is friends with exactly $2$ red people. But then consider a blue person with $2$ red friends; then, they must have a red friend with exactly $2$ red friends too, a contradiction.",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 27911,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nLet $n \\geq 2$ be some fixed positive integer and suppose that $a_{1}, a_{2}, \\ldots, a_{n}$ are positive real numbers satisfying $a_{1}+a_{2}+\\cdots+a_{n}=2^{n}-1$.\nFind the minimum possible value of\n$$\n\\frac{a_{1}}{1}+\\frac{a_{2}}{1+a_{1}}+\\frac{a_{3}}{1+a_{1}+a_{2}}+\\cdots+\\frac{a_{n}}{1+a_{1}+a_{2}+\\cdots+a_{n-1}}.\n$$",
    "options": [],
    "answer": "n",
    "solution": "Solution:\nWe claim the the minimum possible value of this expression is $n$. Observe that by AM-GM, we have that\n$$\n\\begin{aligned}\n\\frac{a_{1}}{1} &+ \\frac{a_{2}}{1+a_{1}}+\\cdots+\\frac{a_{n}}{1+a_{1}+a_{2}+\\cdots+a_{n-1}} \\\\\n&=\\frac{1+a_{1}}{1}+\\frac{1+a_{1}+a_{2}}{1+a_{1}}+\\cdots+\\frac{1+a_{1}+a_{2}+\\cdots+a_{n}}{1+a_{1}+a_{2}+\\cdots+a_{n-1}}-n \\\\\n&\\geq n \\cdot \\sqrt[n]{\\frac{1+a_{1}}{1} \\cdot \\frac{1+a_{1}+a_{2}}{1+a_{1}} \\cdots \\frac{1+a_{1}+a_{2}+\\cdots+a_{n}}{1+a_{1}+a_{2}+\\cdots+a_{n-1}}}-n \\\\\n&=n \\cdot \\sqrt[n]{1+a_{1}+a_{2}+\\cdots+a_{n}}-n \\\\\n&=2n-n=n.\n\\end{aligned}\n$$\nFurthermore, equality is achieved when $a_{k}=2^{k-1}$ for each $1 \\leq k \\leq n$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 27912,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nAt a dinner party there are $N$ hosts and $N$ guests, seated around a circular table, where $N \\geq 4$. A pair of two guests will chat with one another if either there is at most one person seated between them or if there are exactly two people between them, at least one of whom is a host. Prove that no matter how the $2N$ people are seated at the dinner party, at least $N$ pairs of guests will chat with one another.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nLet a run refer to a maximal group of consecutive dinner party guests all of whom are the same type (host or guest). Suppose that there are exactly $k$ runs of hosts and $k$ runs of guests. Let $G_{i}$ and $H_{i}$ denote the number of runs of guests and hosts, respectively, of length exactly $i$. Furthermore, let $X$ denote the number of hosts surrounded by two runs of guests, both of length exactly $1$. We claim that the number of pairs of guests who chat is at least\n$$\n2N - 3k + G_{1} + 2H_{1} + H_{2} - X\n$$\nThe number of pairs of guests who chat with no host between them is at least the sum of $\\max\\{2\\ell - 3, 0\\}$ over all guest run lengths $\\ell$. This sum is at least $2N - 3k + G_{1}$. The number of pairs of guests who chat with exactly two hosts between them is $H_{2}$. Furthermore, the number of pairs of guests who chat with exactly one host between them is at least $2H_{1} - X$. This is because any host surrounded by two runs of guests causes at least two pairs of guests to chat unless these runs are both of length exactly $1$. This proves the claim. Now note that\n$$\n2H_{1} + H_{2} + N \\geq 3k\n$$\nbecause each run of hosts contributes at least three to the left hand side. Furthermore, pairing each run counted in $X$ with the guest run of length $1$ immediately following it in clockwise order shows that $G_{1} \\geq X$. Combining these inequalities yields that $2N - 3k + G_{1} + 2H_{1} + H_{2} - X \\geq N$, completing the proof of the desired result.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 27913,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nA function $f$ from the positive integers to the positive integers is called Canadian if it satisfies\n$$\n\\operatorname{gcd}(f(f(x)), f(x+y)) = \\operatorname{gcd}(x, y)\n$$\nfor all pairs of positive integers $x$ and $y$.\nFind all positive integers $m$ such that $f(m) = m$ for all Canadian functions $f$.",
    "options": [],
    "answer": "All positive integers with at least two distinct prime divisors.",
    "solution": "Solution:\nDefine an $m \\in \\mathbb{N}$ to be good if $f(m) = m$ for all such $f$. It will be shown that $m$ is good if and only if $m$ has two or more distinct prime divisors. Let $P(x, y)$ denote the assertion\n$$\n\\operatorname{gcd}(f(f(x)), f(x+y)) = \\operatorname{gcd}(x, y)\n$$\nfor a pair $x, y \\in \\mathbb{N}$. Let $x$ be a positive integer with two or more distinct prime divisors and let $p^{k}$ be the largest power of one of these prime divisors such that $p^{k} \\mid x$. If $x = p^{k} \\cdot q$, then $p^{k}$ and $q$ are relatively prime and $x > p^{k}$, $q > 1$. By $P(q, x-q)$,\n$$\n\\operatorname{gcd}(f(f(q)), f(x-q+q)) = \\operatorname{gcd}(f(f(q)), f(x)) = \\operatorname{gcd}(q, x-q) = q\n$$\nwhich implies that $q \\mid f(x)$. By $P\\left(p^{k}, x-p^{k}\\right)$,\n$$\n\\operatorname{gcd}\\left(f\\left(f\\left(p^{k}\\right)\\right), f\\left(x-p^{k}+p^{k}\\right)\\right) = \\operatorname{gcd}\\left(f\\left(f\\left(p^{k}\\right)\\right), f(x)\\right) = \\operatorname{gcd}\\left(p^{k}, x-p^{k}\\right) = p^{k}\n$$\nwhich implies that $p^{k} \\mid f(x)$. Since $p^{k}$ and $q$ are relatively prime, $x = p^{k} \\cdot q$ divides $f(x)$, which implies that $f(x) \\geq x$. Now assume for contradiction that $f(x) > x$. Let $y = f(x) - x > 0$ and note that, by $P(x, y)$, it follows that\n$$\nf(f(x)) = \\operatorname{gcd}(f(f(x)), f(x+f(x)-x)) = \\operatorname{gcd}(x, f(x)-x) = \\operatorname{gcd}(x, f(x)).\n$$\nTherefore $f(f(x)) \\mid x$ and $f(f(x)) \\mid f(x)$. By $P(x, x)$, it follows that\n$$\n\\operatorname{gcd}(f(f(x)), f(2x)) = \\operatorname{gcd}(x, x) = x\n$$\nThis implies that $x \\mid f(f(x))$, which when combined with the above result, yields that $f(f(x)) = x$. Since $x \\mid f(x)$ and $x$ is divisible by at least two distinct prime numbers, $f(x)$ is also divisible by at least two distinct prime numbers. As shown previously, this implies that $f(x) \\mid f(f(x)) = x$, which is a contradiction since $f(x) > x$. Therefore $f(x) = x$ for all positive integers $x$ with two or more distinct prime divisors.\n\nNow it will be shown that all $m \\in \\mathbb{N}$ such that either $m$ has one prime divisor or $m = 1$ are not good. In either case, let $m = p^{k}$ where $k \\geq 0$ and $p$ is a prime number and consider the function satisfying that $f\\left(p^{k}\\right) = p^{k+1}$, $f\\left(p^{k+1}\\right) = p^{k}$ and $f(x) = x$ for all $x \\neq p^{k}, p^{k+1}$. Note that this function also satisfies that $f(f(x)) = x$ for all positive integers $x$. If $x+y \\neq p^{k}, p^{k+1}$, then $P(x, y)$ holds by the Euclidean algorithm since $f(f(x)) = x$ and $f(x+y) = x+y$. If $x+y = p^{k+1}$, then $P(x, y)$ is equivalent to $\\operatorname{gcd}\\left(x, p^{k}\\right) = \\operatorname{gcd}\\left(x, p^{k+1}-x\\right) = \\operatorname{gcd}\\left(x, p^{k+1}\\right)$ for all $x < p^{k+1}$ which holds since the greatest power of $p$ that can divide $x$ is $p^{k}$. If $x+y = p^{k}$, then $P(x, y)$ is equivalent to $\\operatorname{gcd}\\left(x, p^{k+1}\\right) = \\operatorname{gcd}\\left(x, p^{k}-x\\right) = \\operatorname{gcd}\\left(x, p^{k}\\right)$ for all $x < p^{k}$ which holds as shown above. Note that if $m = 1$ then this case cannot occur. Since this function satisfies $P(x, y)$, $m$ is good if and only if $m$ has two or more distinct prime divisors.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 27914,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nNina and Tadashi play the following game. Initially, a triple $(a, b, c)$ of nonnegative integers with $a+b+c=2021$ is written on a blackboard. Nina and Tadashi then take moves in turn, with Nina first. A player making a move chooses a positive integer $k$ and one of the three entries on the board; then the player increases the chosen entry by $k$ and decreases the other two entries by $k$. A player loses if, on their turn, some entry on the board becomes negative.\n\nFind the number of initial triples $(a, b, c)$ for which Tadashi has a winning strategy.",
    "options": [],
    "answer": "6561",
    "solution": "Solution:\n\nThe answer is $3^{\\text{number of 1's in binary expansion of } 2021}=3^{8}=6561$.\n\nThroughout this solution, we say two nonnegative integers overlap in the $2^{\\ell}$ position if their binary representations both have a 1 in that position. We say that two nonnegative integers overlap if they overlap in some position. Our central claim is the following.\n\nClaim 1. A triple $(x, y, z)$ is losing if and only if no two of $x, y, z$ overlap.\n\nLet $d_{\\ell}(a)$ denote the bit in the $2^{\\ell}$ position of the binary representation of $a$. Let $\\&$ denote the bitwise and operation: $x \\& y$ is the number satisfying $d_{\\ell}(x \\& y)=d_{\\ell}(x) d_{\\ell}(y)$ for all $\\ell$.\n\nLemma 1. Let $x, y, z$ be nonnegative integers, at least one pair of which overlaps. Define $x^{\\prime}=(x+y) \\&(x+z)$ and $y^{\\prime}, z^{\\prime}$ cyclically. At least one of the inequalities $x<x^{\\prime}, y<y^{\\prime}, z<z^{\\prime}$ holds.\n\nProof. Let $\\ell$ be maximal such that two of $x, y, z$ overlap in the $2^{\\ell}$ position. We case on how many of the additions $x+y, x+z, y+z$ involve a carry from the $2^{\\ell}$ position to the $2^{\\ell+1}$ position, and on the values of $d_{\\ell+1}(x), d_{\\ell+1}(y), d_{\\ell+1}(z)$. Because at least two of $d_{\\ell}(x), d_{\\ell}(y), d_{\\ell}(z)$ equal 1, at least one of the additions $x+y, x+z, y+z$ involves a carry from the $2^{\\ell}$ position.\n\nCase 1. One carry.\n\nWLOG let $x+y$ be the addition with the carry. Then, $d_{\\ell}(x)=d_{\\ell}(y)=1$ and $d_{\\ell}(z)=0$. Since $x, y, z$ do not overlap in any position left of the $2^{\\ell}$ position, the binary representations of $z, z^{\\prime}$ agree left of the $2^{\\ell}$ position. As the additions $x+z$ and $y+z$ do not involve a carry from the $2^{\\ell}$ position, we have $d_{\\ell}(x+z)=d_{\\ell}(y+z)=1$, and thus $d_{\\ell}\\left(z^{\\prime}\\right)=1$. Thus $z^{\\prime}>z$, as desired.\n\nCase 2. At least two carries: $x+y$ and $x+z$ carry and $d_{\\ell+1}(y)=d_{\\ell+1}(z)=0$, or cyclic equivalent. ($y+z$ may or may not carry.)\n\nLet $i$ be maximal such that $d_{\\ell+1}(x)=\\cdots=d_{\\ell+i}(x)=1$ (possibly $i=0$). By maximality of $\\ell, d_{\\ell+1}(y)=d_{\\ell+1}(z)=\\cdots=d_{\\ell+i}(y)=d_{\\ell+i}(z)=0$. By maximality of $i, d_{\\ell+i+1}(x)=0$.\n\nIf $d_{\\ell+i+1}(y)=d_{\\ell+i+1}(z)=0$, then $d_{\\ell+i+1}(x+y)=d_{\\ell+i+1}(x+z)=1$, so $d_{\\ell+i+1}\\left(x^{\\prime}\\right)=1$. The binary representations of $x$ and $x^{\\prime}$ agree to the left of the $2^{\\ell+i+1}$ position, so $x^{\\prime}>x$.\n\nOtherwise, WLOG $d_{\\ell+i+1}(y)=1$ and $d_{\\ell+i+1}(z)=0$. (Note that, here we in fact have $i \\geq 1$.) Then $d_{\\ell+i+1}(y+z)=d_{\\ell+i+1}(x+z)=1$, so $d_{\\ell+i+1}\\left(z^{\\prime}\\right)=1$. The binary representations of $z$ and $z^{\\prime}$ agree to the left of the $2^{\\ell+i+1}$ position, so $z^{\\prime}>z$.\n\nCase 3. At least two carries, and the condition in Case 2 does not occur.\n\nWLOG let $x+y, x+z$ involve carries. Since the condition in Case 2 does not occur, $d_{\\ell+1}(y)=1$ or $d_{\\ell+1}(z)=1$. In either case, $d_{\\ell+1}(x)=0$. WLOG $d_{\\ell+1}(y)=1$ and $d_{\\ell+1}(z)=0$.\n\nSince the condition in Case 2 does not occur, $y+z$ does not involve a carry from the $2^{\\ell}$ position. (Otherwise, $x+y$ and $y+z$ carry and $d_{\\ell+1}(x)=d_{\\ell+1}(z)=0$.) Then $d_{\\ell+1}(x+z)=d_{\\ell+1}(y+z)=1$, so $d_{\\ell+1}\\left(z^{\\prime}\\right)=1$. The binary representations of $z$ and $z^{\\prime}$ agree to the left of the $2^{\\ell+1}$ position, so $z^{\\prime}>z$.\n\nProof of Claim 1. Proceed by strong induction on $x+y+z$. There is no base case.\n\nSuppose by induction the claim holds for all $(x, y, z)$ with sum less than $N$. Consider a triple $(x, y, z)$ with $x+y+z=N$.\n\nSuppose no two of $x, y, z$ overlap. If all moves from this position lead to positions with a negative coordinate, $(x, y, z)$ is a losing position, as claimed. Otherwise, the player increases or decreases all coordinates by $k$. Consider the smallest $m$ such that $d_{m}(k)=1$. The player's move will toggle each of $d_{m}(x), d_{m}(y), d_{m}(z)$. Since at most one of the original $d_{m}(x), d_{m}(y), d_{m}(z)$ is 1, at least two of the new $d_{m}(x), d_{m}(y), d_{m}(z)$ will be 1. So, two of the new $x, y, z$ overlap. By induction, the new $(x, y, z)$ is winning. Thus the original $(x, y, z)$ is losing, as claimed.\n\nConversely, suppose at least one pair of $x, y, z$ overlap. By Lemma 1, at least one of $x<x^{\\prime}, y<y^{\\prime}, z<z^{\\prime}$ holds. WLOG $x<x^{\\prime}$. Let the player to move choose $k=x^{\\prime}-x$, decrease $y, z$ by $k$, and increase $x$ by $k$. The new coordinates are nonnegative, as\n$$\ny-k=x+y-x^{\\prime} \\geq 0\n$$\nbecause $x^{\\prime} \\leq x+y$, and similarly for the $z$ coordinate. Moreover, the binary representation of the new $x$ consists of the 1's in the binary representations of both $x+y$ and $x+z$; the binary representation of the new $y$ consists of the 1's in that of $x+y$ but not $x+z$; and the binary representation of the new $z$ consists of the 1's in that of $x+z$ but not $x+y$. So, no two of the new $x, y, z$ overlap. By induction, the new $(x, y, z)$ is losing. Thus the original $(x, y, z)$ is winning, as claimed.\n\nWe use Claim 1 to count the losing positions $(x, y, z)$ with\n$$\nx+y+z=2021=11111100101_{2}\n$$\nIn each position where $d_{\\ell}(2021)=0$, losing positions must have $d_{\\ell}(x)=d_{\\ell}(y)=d_{\\ell}(z)=0$. In each position where $d_{\\ell}(2021)=1$, the bit triplet $\\left(d_{i}(x), d_{i}(y), d_{i}(z)\\right)$ is one of $(1,0,0),(0,1,0),(0,0,1)$. This gives a count of $3^{8}=6561$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 27915,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nAssume that real numbers $a$ and $b$ satisfy\n$$\na b + \\sqrt{a b + 1} + \\sqrt{a^{2} + b} \\cdot \\sqrt{b^{2} + a} = 0\n$$\nFind, with proof, the value of\n$$\na \\sqrt{b^{2} + a} + b \\sqrt{a^{2} + b}\n$$",
    "options": [],
    "answer": "1",
    "solution": "Solution:\nLet us rewrite the given equation as follows:\n$$\na b + \\sqrt{a^{2} + b} \\sqrt{b^{2} + a} = -\\sqrt{a b + 1}.\n$$\nSquaring this gives us\n$$\n\\begin{aligned}\na^{2} b^{2} + 2 a b \\sqrt{a^{2} + b} \\sqrt{b^{2} + a} + (a^{2} + b)(b^{2} + a) & = a b + 1 \\\\\n(a^{2} b^{2} + a^{3}) + 2 a b \\sqrt{a^{2} + b} \\sqrt{b^{2} + a} + (a^{2} b^{2} + b^{3}) & = 1 \\\\\n(a \\sqrt{b^{2} + a} + b \\sqrt{a^{2} + b})^{2} & = 1 \\\\\na \\sqrt{b^{2} + a} + b \\sqrt{a^{2} + b} & = \\pm 1.\n\\end{aligned}\n$$\nNext, we show that $a \\sqrt{b^{2} + a} + b \\sqrt{a^{2} + b} > 0$. Note that\n$$\na b = -\\sqrt{a b + 1} - \\sqrt{a^{2} + b} \\cdot \\sqrt{b^{2} + a} < 0\n$$\nso $a$ and $b$ have opposite signs. Without loss of generality, we may assume $a > 0 > b$. Then rewrite\n$$\na \\sqrt{b^{2} + a} + b \\sqrt{a^{2} + b} = a (\\sqrt{b^{2} + a} + b) - b (a - \\sqrt{a^{2} + b})\n$$\nand, since $\\sqrt{b^{2} + a} + b$ and $a - \\sqrt{a^{2} + b}$ are both positive, the expression above is positive. Therefore,\n$$\na \\sqrt{b^{2} + a} + b \\sqrt{a^{2} + b} = 1,\n$$\nand the proof is finished.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 27916,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nLet $d(k)$ denote the number of positive integer divisors of $k$. For example, $d(6)=4$ since $6$ has $4$ positive divisors, namely, $1, 2, 3$, and $6$. Prove that for all positive integers $n$,\n$$\nd(1)+d(3)+d(5)+\\cdots+d(2n-1) \\leq d(2)+d(4)+d(6)+\\cdots+d(2n)\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nFor any integer $k$ and set of integers $S$, let $f_{S}(k)$ be the number of multiples of $k$ in $S$. We can count the number of pairs $(k, s)$ with $k \\in \\mathbb{N}$ dividing $s \\in S$ in two different ways, as follows:\n- For each $s \\in S$, there are $d(s)$ pairs that include $s$, one for each divisor of $s$.\n- For each $k \\in \\mathbb{N}$, there are $f_{k}(S)$ pairs that include $k$, one for each multiple of $k$.\n\nTherefore,\n$$\n\\sum_{s \\in S} d(s) = \\sum_{k \\in \\mathbb{N}} f_{S}(k)\n$$\nLet\n$$\nO = \\{1, 3, 5, \\ldots, 2n-1\\} \\quad \\text{and} \\quad E = \\{2, 4, 6, \\ldots, 2n\\}\n$$\nbe the set of odd and, respectively, the set of even integers between $1$ and $2n$. It suffices to show that\n$$\n\\sum_{k \\in \\mathbb{N}} f_{O}(k) \\leq \\sum_{k \\in \\mathbb{N}} f_{E}(k)\n$$\nSince the elements of $O$ only have odd divisors,\n$$\n\\sum_{k \\in \\mathbb{N}} f_{O}(k) = \\sum_{k \\text{ odd}} f_{O}(k)\n$$\nFor any odd $k$, consider the multiples of $k$ between $1$ and $2n$. They form a sequence\n$$\nk, 2k, 3k, \\ldots, \\left\\lfloor \\frac{2n}{k} \\right\\rfloor k\n$$\nalternating between odd and even terms. There are either an equal number of odd and even terms, or there is one more odd term than even terms. Therefore, we have the inequality\n$$\nf_{O}(k) \\leq f_{E}(k) + 1\n$$\nfor all odd $k$. Combining this with the previous observations gives us the desired inequality:\n$$\n\\begin{aligned}\n\\sum_{k \\in \\mathbb{N}} f_{O}(k) & = \\sum_{k \\text{ odd}} f_{O}(k) \\\\\n& \\leq \\sum_{k \\text{ odd}} \\left(f_{E}(k) + 1\\right) \\\\\n& = \\sum_{k \\text{ odd}} f_{E}(k) + n \\\\\n& = \\sum_{k \\text{ odd}} f_{E}(k) + f_{E}(2) \\\\\n& \\leq \\sum_{k \\in \\mathbb{N}} f_{E}(k)\n\\end{aligned}\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 27917,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nLet $n \\geq 2$ be an integer. Initially, the number $1$ is written $n$ times on a board. Every minute, Vishal picks two numbers written on the board, say $a$ and $b$, erases them, and writes either $a+b$ or $\\min \\{a^{2}, b^{2}\\}$. After $n-1$ minutes there is one number left on the board. Let the largest possible value for this final number be $f(n)$. Prove that\n$$\n2^{n / 3}<f(n) \\leq 3^{n / 3}\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nClearly $f(n)$ is a strictly increasing function, as we can form $f(n-1)$ with $n-1$ ones, and add the final one. However, we can do better; assume Vishal generates $f(n)$ on the board. After $n-2$ minutes, there are two numbers left, say they were formed by $x$ ones and $y$ ones, where $x+y=n$. Clearly the numbers are at most $f(x), f(y)$ (and can be made to be equal to $f(x), f(y)$), and therefore we obtain\n$$\nf(n)=\\max _{x+y=n, 1 \\leq x \\leq y \\leq n-1}\\left(\\max \\left(f(x)+f(y), f(x)^{2}\\right)\\right)\n$$\nwhere we used the fact that $f$ is increasing to get that $\\min \\left(f(x)^{2}, f(y)^{2}\\right)=f(x)^{2}$ when $x \\leq y$. In particular, $f(n+1) \\geq f(n)+1$, and $f(2 n) \\geq f(n)^{2}$ for all positive integers $n$.\n\nUpper bound:\n\nFirst proof of upper bound. We use induction. We can check that $f(n)=n$ for $n \\leq 4$, and these all satisfy the bound $f(n)=n \\leq 3^{n / 3}$. Assume it is true for all $m<n$ (some $n \\geq 5$), and with $x, y$ as in equation above we have\n$$\nf(x)^{2} \\leq f\\left(\\left\\lfloor\\frac{n}{2}\\right\\rfloor\\right)^{2} \\leq\\left(3^{n / 6}\\right)^{2}=3^{n / 3}\n$$\nas desired. It thus remains to show that $f(x)+f(y) \\leq 3^{n / 3}$. By induction, it suffices to prove that\n$$\n3^{x / 3}+3^{y / 3} \\leq 3^{(x+y) / 3}\n$$\nfor $1 \\leq x \\leq y \\leq n-1$ and $x+y=n$. This is equivalent to\n$$\n1+3^{(y-x) / 3} \\leq 3^{y / 3}\n$$\nLet $w=3^{(y-x) / 3}$, and we require $3^{x / 3} w \\geq w+1$. If $x \\geq 2$, then this is true as $w \\geq 1$, and if $x=1$ then $w=3^{(n-2) / 3} \\geq 3$ and the result is still true. Thus all terms in the equation are at most $3^{n / 3}$, and so $f(n) \\leq 3^{n / 3}$, and the upper bound is proven.\n\nSecond proof of upper bound. Consider a second game with the same rules but in which Vishal can replace $a$ and $b$ by either $a+b$ or $a b$. Let $g(n)$ be the largest possible value for this new game. Then $f(n) \\leq g(n)$ because $\\min \\{a^{2}, b^{2}\\} \\leq a b$.\n\nWe can check $g(n)=n$ for $n \\leq 4$, so $g(n) \\leq 3^{n / 3}$ for these values. If $x$ and $y$ are both bigger than $1$, then $g(x)+g(y) \\leq g(x) g(y)$. Therefore, for $n>4$, we have that\n$$\ng(n)=\\max \\{g(n-1)+1, \\max _{1 \\leq x \\leq n-1} g(x) g(n-x)\\}\n$$\nNow proceed similarly to the first proof. Assume $n>4$ and $g(m) \\leq 3^{m / 3}$ for all $m<n$. If $1 \\leq x \\leq n-1$, then $g(x) g(n-x) \\leq 3^{x / 3} 3^{(n-x) / 3}=3^{n / 3}$. And $g(n-1)+1 \\leq 3^{(n-1) / 3}+1$, which is shown to be less than $3^{n / 3}$ in the first proof. It follows that $f(n) \\leq g(n) \\leq 3^{n / 3}$.\n\nLower bound:\n\nFirst proof of lower bound. We begin with a lemma.\n\nLemma 1. Let $m$ be a nonnegative integer. Then\n$$\nf\\left(2^{m}\\right) \\geq 2^{2^{m-1}} \\quad \\text{and} \\quad f\\left(3 \\cdot 2^{m}\\right) \\geq 3^{2^{m}}\n$$\nProof. We prove the lemma by induction. One can check that $f(n)=n$ for $n \\leq 3$, which proves the lemma for $m=0$. For a general $m>0$, we get\n$$\n\\begin{gathered}\nf\\left(2^{m}\\right) \\geq f\\left(2^{m-1}\\right)^{2} \\geq\\left(2^{2^{m-2}}\\right)^{2}=2^{2^{m-1}} \\\\\nf\\left(3 \\cdot 2^{m}\\right) \\geq f\\left(3 \\cdot 2^{m-1}\\right)^{2} \\geq\\left(3^{2^{m-1}}\\right)^{2}=3^{2^{m}}\n\\end{gathered}\n$$\nby induction, as required.\n\n(This lemma can also be proved more constructively. Briefly, if $n=2^{m}$, then partition the $1$'s on the board into $2^{m-1}$ pairs, and then add each pair to get $2^{m-1}$ $2$'s $(2=2^{2^{0}})$; then multiply pairs of $2$'s to get $2^{m-2}$ $4$'s $(4=2^{2^{1}})$; then multiply pairs of $4$'s to get $2^{m-3}$ $16$'s $(16=2^{2^{2}})$; and so on, until there are $2(=2^{1})$ copies of $2^{2^{m-2}}$, which then gets replaced with $2^{2^{m-1}}$. The process is similar for $n=3 \\cdot 2^{m}$, except that the first step is to partition the $1$'s into $2^{m}$ groups of $3$, and then use addition within each group to get $2^{m}$ $3$'s on the board.)\n\nNow assume $2^{x} \\leq n<3 \\cdot 2^{x-1}$ for some integer $x$. Then we have\n$$\nf(n) \\geq f\\left(2^{x}\\right) \\geq 2^{2^{x-1}}>2^{n / 3}\n$$\nas required. If no such $x$ exists, then there exists an integer $x$ such that $3 \\cdot 2^{x-1} \\leq n<2^{x+1}$. In this case, we have\n$$\nf(n) \\geq f\\left(3 \\cdot 2^{x-1}\\right) \\geq 3^{2^{x-1}}>2^{2^{x+1} / 3}>2^{n / 3}\n$$\nwhere the second last inequality is equivalent to $2^{x-1} \\log (3) \\geq \\frac{2^{x+1}}{3} \\log (2)$, and by dividing out $2^{x}$ and clearing the denominator this is equivalent to $3 \\log (3) \\geq 4 \\log 2$, which is true as $3^{3}=27>16=2^{4}$.\n\nSecond proof of lower bound. We shall prove the stronger result $f(n) \\geq 2^{(n+1) / 3}$ for $n \\geq 2$ by induction. One can check that $f(n)=n$ for $n=2,3,4$, which proves the result for these values. Assume that $n \\geq 5$ and that $f(k) \\geq 2^{(k+1) / 3}$ for all $k=2,3, \\ldots, n-1$. Then\n$$\n\\begin{array}{rlr}\nf(n) & \\geq f(\\lfloor n / 2\\rfloor)^{2} & \\\\\n& \\geq\\left(2^{(\\lfloor n / 2\\rfloor+1) / 3}\\right)^{2} \\quad \\quad \\text{since }\\left\\lfloor\\frac{n}{2}\\right\\rfloor \\geq 2 \\\\\n& =2^{(2\\lfloor n / 2\\rfloor+2) / 3} \\quad \\\\\n& \\geq 2^{(n+1) / 3} \\quad \\text{since }\\left\\lfloor\\frac{n}{2}\\right\\rfloor \\geq \\frac{n-1}{2} .\n\\end{array}\n$$\nThe result follows by induction.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 27918,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nLet $n$ be a positive integer. A set of $n$ distinct lines divides the plane into various (possibly unbounded) regions. The set of lines is called \"nice\" if no three lines intersect at a single point. A \"colouring\" is an assignment of two colours to each region such that the first colour is from the set $\\{A_{1}, A_{2}\\}$, and the second colour is from the set $\\{B_{1}, B_{2}, B_{3}\\}$. Given a nice set of lines, we call it \"colourable\" if there exists a colouring such that\n1. no colour is assigned to two regions that share an edge;\n2. for each $i \\in \\{1,2\\}$ and $j \\in \\{1,2,3\\}$ there is at least one region that is assigned with both $A_{i}$ and $B_{j}$.\nDetermine all $n$ such that every nice configuration of $n$ lines is colourable.",
    "options": [],
    "answer": "n ≥ 5",
    "solution": "Solution:\n\nThe answer is $n \\geq 5$. If $n \\leq 4$, consider $n$ parallel lines. There are 6 total colour combinations required, and only $n+1 \\leq 5$ total regions, hence the colouring is not possible.\n\nNow, assume $n \\geq 5$. Rotate the picture so that no line is horizontal, and orient each line so that the \"forward\" direction increases the $y$-value. In this way, each line divides the plane into a right and left hand side (with respect to this forward direction). Every region of the plane is on the right hand side of $k$ lines and on the left hand side of $n-k$ lines for some $0 \\leq k \\leq n$. Furthermore, there is a region for every $k$: let $w$ be large enough so that $w$ is greater than the $y$-value of any intersection point of two lines. Consider the horizontal line $y=w$: a point very far on the left of this line is left of every single line, and as we cross over all lines in the problem, we hit all values of $k$.\n\nFinally, take a region that is on the right hand side of $k$ lines. Colour it $A_{1}$ if $k$ is odd, and $A_{2}$ if it is even. Similarly, colour it $B_{i}$ if $k \\equiv i \\pmod{3}$. By the previous paragraph, there are regions for at least $k=0,1, \\ldots, 5$, whence there is a region coloured $A_{i}$ and $B_{j}$ for all $(i, j)$. Furthermore, two regions that share an edge will be on the right hand side of $k$ and $k+1$ lines for some $k$. By construction, the $A_{i}$ and $B_{i}$ colours of the regions must differ, hence we have proven that the set of lines is colourable.",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 27919,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nLet $ABCDE$ be a convex pentagon such that the five vertices lie on a circle and the five sides are tangent to another circle inside the pentagon. There are $\\binom{5}{3} = 10$ triangles which can be formed by choosing 3 of the 5 vertices. For each of these 10 triangles, mark its incenter. Prove that these 10 incenters lie on two concentric circles.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nLet $I$ be the incenter of pentagon $ABCDE$. Let $I_A$ denote the incenter of triangle $EAB$ and $I_a$ the incenter of $DAC$. Define $I_B, I_b, I_C, I_c, I_D, I_d, I_E, I_e$ similarly.\n\nWe will first show that $I_A I_B I_C I_D I_E$ are concyclic. Let $\\omega_A$ be the circle with center at the midpoint of arc $DE$ and passing through $D$ and $E$. Define $\\omega_B, \\omega_C, \\omega_D, \\omega_E$ similarly. It is well-known that the incenter of a triangle lies on such circles, in particular, $I_A$ lies on $\\omega_C$ and $\\omega_D$. So the radical axis of $\\omega_C, \\omega_D$ is the line $A I_A$. But this is just the angle bisector of $\\angle EAB$, which $I$ also lies on. So $I$ is in fact the radical center of $\\omega_A, \\omega_B, \\omega_C, \\omega_D, \\omega_E$! Inverting about $I$ swaps $I_A$ and $A$ and since $ABCDE$ are concyclic, $I_A I_B I_C I_D I_E$ are concyclic as well.\n\nLet $O$ be the center of the circle $I_A I_B I_C I_D I_E$. We will now show that $O I_a = O I_d$ which finishes the problem as we can consider the cyclic versions of this equation to find that $O I_a = O I_d = O I_b = O I_e = O I_c$. Recall a well-known lemma: For any cyclic quadrilateral $WXYZ$, the incenters of $XYZ, YZW, ZWX, WXY$ form a rectangle. Applying this lemma on $ABCD$, we see that $I_B, I_C, I_a, I_d$ form a rectangle in that order. Then the perpendicular bisector of $I_B I_C$ is exactly the perpendicular bisector of $I_a I_d$. Thus, $O$ is equidistant to $I_a$ and $I_d$ and we are done.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 27920,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nWilliam is thinking of an integer between 1 and 50, inclusive. Victor can choose a positive integer $m$ and ask William: \"does $m$ divide your number?\", to which William must answer truthfully. Victor continues asking these questions until he determines William's number. What is the minimum number of questions that Victor needs to guarantee this?",
    "options": [],
    "answer": "15",
    "solution": "Solution:\n\nThe minimum number is 15 questions.\n\nFirst, we show that 14 or fewer questions is not enough to guarantee success. Suppose Victor asks at most 14 questions, and William responds with \"no\" to each question unless $m=1$. Note that these responses are consistent with the secret number being 1. But since there are 15 primes less than 50, some prime $p$ was never chosen as $m$. That means the responses are also consistent with the secret number being $p$. Therefore, Victor cannot determine the number for sure because 1 and $p$ are both possible options.\n\nNow we show that Victor can always determine the number with 15 questions. Let $N$ be William's secret number. First, Victor asks 4 questions, with $m=2,3,5,7$. We then case on William's responses.\n\nCase 1. William answers \"no\" to all four questions.\n\n$N$ can only be divisible by primes that are 11 or larger. This means $N$ cannot have multiple prime factors (otherwise $N \\geq 11^2 > 50$), so either $N=1$ or $N$ is one of the 11 remaining primes less than 50. Victor can then ask 11 questions with $m=11,13,17, \\ldots, 47$, one for each of the remaining primes, to determine the value of $N$.\n\nCase 2. William answers \"yes\" to $m=2$, and \"no\" to $m=3,5,7$.\n\nThere are only 11 possible values of $N$ that match these answers ($2,4,8,16,22,26,32,34,38,44$, and $46$). Victor can use his remaining 11 questions on each of these possibilities.\n\nCase 3. William answers \"yes\" to $m=3$, and \"no\" to $m=2,5,7$.\n\nThere are 5 possible values of $N$ ($3,9,27,33$, and $39$). Similar to Case 2, Victor can ask about these 5 numbers to determine the value of $N$.\n\nCase 4. William answers \"yes\" to multiple questions, or one \"yes\" to $m=5$ or $m=7$.\n\nLet $k$ be the product of all $m$'s that received a \"yes\" response. Since $N$ is divisible by each of these $m$'s, $N$ must be divisible by $k$. Since $k \\geq 5$, there are at most 10 multiples of $k$ between 1 and 50. Victor can ask about each of these multiples of $k$ with his remaining questions.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 27921,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nThere are 20 students in a high school class, and each student has exactly three close friends in the class. Five of the students have bought tickets to an upcoming concert. If any student sees that at least two of their close friends have bought tickets, then they will buy a ticket too.\nIs it possible that the entire class buys tickets to the concert?\n(Assume that friendship is mutual; if student $A$ is close friends with student $B$, then $B$ is close friends with $A$.)",
    "options": [],
    "answer": "No",
    "solution": "Solution:\n\nIt is impossible for the whole class to buy tickets to the concert.\nIf two students $A$ and $B$ are close friends, and $A$ has bought a ticket to the concert while $B$ has not, then $A$ is enticing $B$. We call this pair $(A, B)$ an enticement.\nIn order for a student to change their mind and buy a ticket, they first be enticed by at least 2 of their 3 close friends. That means they can only entice at most 1 other friend. Therefore, the total number of enticements among the students decreases by 1 whenever a student changes their mind to buy a ticket.\n\nInitially, the maximum number of enticements is 15 (each of the initial 5 students with tickets has 3 friends to entice). Assume, for the sake of contradiction, that the entire class ends up buying tickets. After the first 14 people buy tickets, the number of enticements is at most $15-14=1$. This is not enough to convince the last person to buy a ticket, since they need 2 enticements.\nTherefore, it is impossible that the entire class buys tickets.\nSolution:\n\nWe shall use the term friendship to denote an unordered pair of students who are close friends. Since each of the 20 students is part of exactly 3 friendships, there are exactly 30 friendships in the class. (We could also represent friendships as edges in an undirected graph whose vertices are the 20 students.)\nWe say that a friendship is used if one of the students in that friendship buys a ticket after the original five buyers, and the other student already has a ticket at that time. Each time a ticket is purchased after the original five purchases, at least two friendships are used. Observe that no friendship gets used twice.\n\nIf all 20 students buy tickets, then three friendships are used when the last student buys a ticket. This would imply that the number of used friendships is at least $14 \\times 2+3=31$, which is more than the number of friendships. This contradiction proves that it is not possible that the entire class buys tickets.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 27922,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nAn acute triangle is a triangle that has all angles less than $90^{\\circ}$ ($90^{\\circ}$ is a Right Angle). Let $ABC$ be an acute triangle with altitudes $AD$, $BE$, and $CF$ meeting at $H$. The circle passing through points $D$, $E$, and $F$ meets $AD$, $BE$, and $CF$ again at $X$, $Y$, and $Z$ respectively. Prove the following inequality:\n$$\n\\frac{AH}{DX} + \\frac{BH}{EY} + \\frac{CH}{FZ} \\geq 3\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nLet the circumcircle of $ABC$ meet the altitudes $AD$, $BE$, and $CF$ again at $I$, $J$, and $K$ respectively.\n\n![](attached_image_1.png)\n\nLemma (9-point circle). $I$, $J$, $K$ are the reflections of $H$ across $BC$, $CA$, $AB$. Moreover, $D$, $E$, $F$, $X$, $Y$, $Z$ are the midpoints of $HI$, $HJ$, $HK$, $HA$, $HB$, $HC$.\n\nProof. Since $ABDE$ and $ABIC$ are cyclic, we see that\n$$\n\\angle EBD = \\angle EAD = \\angle CAI = \\angle CBI.\n$$\nHence the lines $BI$ and $BH$ are reflections across $BC$. Similarly, $CH$ and $CI$ are reflections across $BC$, so $I$ is the reflection of $H$ across $BC$. The analogous claims for $J$ and $K$ follow. A $\\times 2$ dilation from $H$ now establishes the result.\n\nFrom this lemma, we get $AI = 2XD$, $BJ = 2EY$, and $CK = 2FZ$. Hence it is equivalent to showing that\n$$\n\\frac{AH}{2DX} + \\frac{BH}{2EY} + \\frac{CH}{2FZ} \\geq \\frac{3}{2}\n$$\nwhich is in turn equivalent to\n$$\n\\frac{AH}{AI} + \\frac{BH}{BJ} + \\frac{CH}{CK} \\geq \\frac{3}{2}.\n$$\nLet $a = JK$, $b = KI$ and $c = IJ$. Again by the lemma we find $AH = AK = AJ$, so by Ptolemy's theorem on $AKIJ$,\n$$\nAJ \\cdot KI + AK \\cdot IJ = AI \\cdot JK\n$$\nSubstituting and rearranging,\n$$\n\\begin{aligned}\nAH \\cdot b + AH \\cdot c & = AI \\cdot a \\\\\nAH \\cdot (b + c) & = AI \\cdot a \\\\\n\\frac{AH}{AI} & = \\frac{a}{b + c}.\n\\end{aligned}\n$$\nSimilarly,\n$$\n\\frac{BH}{BJ} = \\frac{b}{c + a} \\quad \\text{and} \\quad \\frac{CH}{CK} = \\frac{c}{a + b}\n$$\nPlugging these back into $\\left(^*\\right)$, the desired inequality is now\n$$\n\\frac{a}{b + c} + \\frac{b}{a + c} + \\frac{c}{a + b} \\geq \\frac{3}{2}.\n$$\nThis is known as Nesbitt's Inequality, which has many proofs. Below is one such proof.\n\nAdd 3 to both sides and rearrange:\n$$\n\\begin{aligned}\n\\left(\\frac{a}{b + c} + 1\\right) + \\left(\\frac{b}{c + a} + 1\\right) + \\left(\\frac{c}{a + b} + 1\\right) & \\geq \\frac{3}{2} + 3 \\\\\n\\Longleftrightarrow \\quad \\frac{a + b + c}{b + c} + \\frac{a + b + c}{c + a} + \\frac{a + b + c}{a + b} & \\geq \\frac{9}{2} \\\\\n\\Longleftrightarrow \\quad (a + b + c)\\left(\\frac{1}{b + c} + \\frac{1}{c + a} + \\frac{1}{a + b}\\right) & \\geq \\frac{9}{2} \\\\\n\\Longleftrightarrow \\quad & \\frac{(b + c) + (c + a) + (a + b)}{3}\n\\end{aligned}\n$$\nwhich is true by the AM-HM inequality.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 27923,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nLet $f(x)$ be a non-constant polynomial with integer coefficients such that $f(1) \\neq 1$. For a positive integer $n$, define $\\operatorname{divs}(n)$ to be the set of positive divisors of $n$.\nA positive integer $m$ is $f$-cool if there exists a positive integer $n$ for which\n$$\nf[\\operatorname{divs}(m)]=\\operatorname{divs}(n)\n$$\nProve that for any such $f$, there are finitely many $f$-cool integers.\n(The notation $f[S]$ for some set $S$ denotes the set $\\{f(s): s \\in S\\}$.)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\nAssume for the sake of contradiction that there are infinitely many $f$-cool integers.\nIf $f(x)$ has a negative leading coefficient, then a sufficiently large $f$-cool integer $m$ will have $f(m)<0$. But this implies $m$ is not $f$-cool, contradiction.\nThus $f(x)$ has a positive leading coefficient, so we can pick an $N$ such that for all $m>N$,\n$$\nf(m)>\\max (f(1), f(2), \\ldots, f(m-1))\n$$\nThis means $f(m)$ is the largest value in $f[\\operatorname{divs}(m)]$, so if $m$ is $f$-cool with $f[\\operatorname{divs}(m)]=\\operatorname{divs}(n)$, then we must have $n=f(m)$, since $n$ is the largest value in $\\operatorname{divs}(n)$. In other words,\n$$\nf[\\operatorname{divs}(m)]=\\operatorname{divs}(f(m))\n$$\nfor all $f$-cool $m>N$.\nFor each of those $m$'s, $1 \\in \\operatorname{divs}(f(m))$, so there must be a $k \\in \\operatorname{divs}(m)$ such that $f(k)=1$. Let $k_{1}, k_{2}, \\ldots, k_{n}$ be the solutions to $f(x)=1$. Thus every $f$-cool $m>N$ is divisible by some $k \\in\\left\\{k_{1}, k_{2}, \\ldots, k_{n}\\right\\}$. Since there are infinitely many such $m$'s and finitely many $k$'s, by the Pigeonhole Principle there is some $k$ which divides infinitely many $f$-cool integers $m$. (Note that $k \\neq 1$ since $f(1) \\neq 1$.)\nFor all $f$-cool $m>N$ divisible by $k$, we have\n$$\n\\left.f\\left(\\frac{m}{k}\\right) \\in f[\\operatorname{divs}(m)]=\\operatorname{divs}(f(m)) \\Longrightarrow f\\left(\\frac{m}{k}\\right) \\right\\rvert\\, f(m) .\n$$\nThus, $f(x) \\mid f(k x)$ has infinitely many positive integer solutions. Let $d=\\operatorname{deg}(f)$, and write\n$$\n\\frac{f(k x)}{f(x)}=k^{d}+\\frac{g(x)}{f(x)}\n$$\nfor some $g(x) \\in \\mathbb{Z}[x]$ with $\\operatorname{deg}(g)<d$. If $g(x) \\neq 0$, then for sufficiently large $x$ we have $0<|g(x)|<f(x)$, since $\\operatorname{deg}(f)>\\operatorname{deg}(g)$. But then $\\frac{f(k x)}{f(x)}-k^{d}=\\frac{g(x)}{f(x)}$ cannot be an integer, which gives us the desired contradiction.\nTherefore $g(x)=0$, so $f(k x)=k^{d} f(x)$, i.e. $f(x)=a x^{d}$ for some positive integer $a$. If $a=1$ then $f(1)=1$, a contradiction. But if $a>1$, then $f(x)=1$ has no integer solutions, another contradiction.",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 27924,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nA country with $n$ cities has some two-way roads connecting certain pairs of cities. Someone notices that if the country is split into two parts in any way, then there would be at most $k n$ roads between the two parts (where $k$ is a fixed positive integer). What is the largest integer $m$ (in terms of $n$ and $k$) such that there is guaranteed to be a set of $m$ cities, no two of which are directly connected by a road?",
    "options": [],
    "answer": "m = ceil(n/(4k))",
    "solution": "Solution:\nThe answer is $m = \\left\\lceil \\frac{n}{4k} \\right\\rceil$.\n\nCall a collection of cities independent if no two cities in the collection are joined by a road. Let $r$ and $k$ be integers such that $n = 4kq + r$ where $1 \\leq r \\leq 4k$.\n\nFirst we show that $m \\leq \\left\\lceil \\frac{n}{4k} \\right\\rceil = q + 1$. Let $K_{i}$ denote a set of $i$ cities such that every pair of cities in $K_{i}$ is linked by a road. Consider a country containing $q$ copies of $K_{4k}$ and one copy of $K_{r}$. An independent set of cities in this country contains at most one city from each $K_{4k}$ or $K_{r}$ and therefore contains at most $q + 1$ cities. Now note that any partition of the cities of the country into two new countries partitions each $K_{4k}$ and $K_{r}$ into two sets. If $K_{i}$ where $i \\leq 4k$ is partitioned into two sets of cities of sizes $a$ and $b$, then the number of roads between the two sets is $ab \\leq \\frac{(a + b)^2}{4} \\leq k i$. Summing this inequality over all copies of $K_{4k}$ and $K_{r}$ yields that there are at most $k n$ roads between the two new countries. This implies that this particular country satisfies the given condition and it follows that $m \\leq \\left\\lceil \\frac{n}{4k} \\right\\rceil$.\n\nNow we show that any country satisfying the given condition has an independent set containing at least $\\left\\lceil \\frac{n}{4k} \\right\\rceil$ cities. Call a set of cities $i$-separable if it can be partitioned into $i$ disjoint independent sets of cities. Given a country satisfying the conditions, let $S$ be a largest set of cities in the country that is $2k$-separable. We prove that $|S| \\geq n/2$. By definition of $S$, there exists a partition $A_{1}, A_{2}, \\ldots, A_{2k}$ of the cities in $S$ such that each $A_{i}$ is independent. Let $|S| = t$. Assume for contradiction that $t < \\frac{n}{2}$. There are at most $k n$ roads between $S$ and the rest of the country, which by the pigeonhole principle implies that there is a city $u$ not in $S$ that is connected to at most $\\frac{k n}{n - t} < 2k$ cities by road. Therefore $u$ is joined by a road to at most $2k - 1$ cities in $S$, and there must be an independent subset $A_{i}$ such that $u$ is not linked by a road to any city in $A_{i}$. Adding $u$ to $S$ maintains the fact that $S$ is $2k$-separable but contradicts its maximality. Therefore it must follow that $t \\geq \\frac{n}{2}$. By the pigeonhole principle, one of the sets $A_{1}, A_{2}, \\ldots, A_{2k}$ must contain at least $\\frac{t}{2k} \\geq \\frac{n}{4k}$ cities. This proves the claim and therefore $m = \\left\\lceil \\frac{n}{4k} \\right\\rceil$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 27925,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nLet $ABC$ be a triangle with incenter $I$. Suppose the reflection of $AB$ across $CI$ and the reflection of $AC$ across $BI$ intersect at a point $X$. Prove that $XI$ is perpendicular to $BC$.\n\n(The incenter is the point where the three angle bisectors meet.)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nSuppose the reflection of $AC$ across $BI$ intersects $BC$ at $E$. Define $F$ similarly for the reflection of $AB$ across $CI$. Also suppose $CI$ intersects $AB$ at $M$ and $BI$ intersects $AC$ at $N$. Since $CA$ and $CF = BC$ are reflections across $CI$, and so are $MA$ and $MF = XM$, we have that $A$ and $F$ are reflections across $CI$. Similarly $A$ and $E$ are reflections across $BI$.\n\nThus $\\angle XFC = \\angle BAC = \\angle XEB$ if $\\angle BAC$ is acute (and $\\angle XFC = \\angle XEB = \\pi - \\angle BAC$, when $\\angle BAC$ is obtuse), so $XF = XE$. Moreover we also find that $IF = IA = IE$ by the aforementioned reflection properties, so thus $XI$ is the perpendicular bisector of $EF$ and is hence perpendicular to $BC$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 27926,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nJane writes down 2024 natural numbers around the perimeter of a circle. She wants the 2024 products of adjacent pairs of numbers to be exactly the set $\\{1!, 2!, \\ldots, 2024!\\}$. Can she accomplish this?",
    "options": [],
    "answer": "No",
    "solution": "Solution:\n\nGiven any prime $p$ and positive integer $x$, let $v_{p}(x)$ denote the highest power of $p$ dividing $x$. We claim that Jane cannot write 2024 such numbers as that would imply that $1! \\cdot 2! \\cdots 2024!$ is the square of the product of the 2024 numbers. Let $p$ be a prime and $k$ be a natural number such that $k < p$, $k p \\leq 2024$, and $(k+1) p > 2024$. Then note that\n$$\nv_{p}(1! \\cdot 2! \\cdots 2024!) = (2024 - p + 1) + (2024 - 2p + 1) + \\ldots + (2024 - k p + 1)\n$$\nIn particular, let $p$ be in $\\left(\\frac{2024}{4}, \\frac{2024}{2}\\right)$. By Bertrand's Postulate, such a prime $p$ exists (and $p$ must also be odd). Further, the corresponding $k$ is either 2 or 3. Either way, $v_{p}(1! \\cdot 2! \\cdots 2024!)$ is odd from the above formula, and so $1! \\cdot 2! \\cdots 2024!$ cannot be a perfect square.\nSolution:\n\nAs in the first solution, we prove $1! \\cdot 2! \\cdots 2024!$ is not a perfect square. To do this, note that we can rewrite the product as $(1!)^{2} \\cdot 2 \\cdot (3!)^{2} \\cdot 4 \\cdots (2023!)^{2} \\cdot 2024$ which is\n$$\n2 \\cdot 4 \\cdots 2024 \\cdot (1! \\cdot 3! \\cdots 2023!)^{2} = 1012! \\cdot \\left(2^{1012} \\cdot 1! \\cdot 3! \\cdots 2023!\\right)^{2}\n$$\nso it is sufficient to verify $1012!$ is not a perfect square. This can be verified by either noticing the prime $1009$ only appears as a factor of $1012!$ once, or by evaluating $v_{2}(1012!) = 1005$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 27927,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nLet $N$ be the number of positive integers with 10 digits $\\overline{d_{9} d_{8} \\cdots d_{1} d_{0}}$ in base 10 (where $0 \\leq d_{i} \\leq 9$ for all $i$ and $d_{9}>0$) such that the polynomial\n$$\nd_{9} x^{9}+d_{8} x^{8}+\\cdots+d_{1} x+d_{0}\n$$\nis irreducible in $\\mathbb{Q}$. Prove that $N$ is even.\n(A polynomial is irreducible in $\\mathbb{Q}$ if it cannot be factored into two non-constant polynomials with rational coefficients.)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\nLet $f(x)=d_{9} x^{9}+d_{8} x^{8}+\\cdots+d_{1} x+d_{0}$. If $d_{0}=0$, then $f(x)$ is divisible by $x$ and thus reducible, so we may ignore all such polynomials. The remaining polynomials all have nonzero leading and constant coefficients.\nFor any polynomial $p(x)$ of degree $n$ with nonzero leading and constant coefficients, say $p(x)=a_{n} x^{n}+a_{n-1} x^{n-1}+\\cdots+a_{1} x+a_{0}$, define $\\bar{p}(x)$ to be the reversed polynomial $a_{0} x^{n}+a_{1} x^{n-1}+\\cdots+a_{n-1} x+a_{n}$. Observe that $\\bar{p}(x)$ also has degree $n$ and furthermore, $\\bar{p}(x)=x^{n}\\left(a_{0}+a_{1}\\left(\\frac{1}{x}\\right)+\\cdots+a_{n-1}\\left(\\frac{1}{x}\\right)^{n-1}+a_{n}\\left(\\frac{1}{x}\\right)^{n}\\right)=x^{n} p\\left(\\frac{1}{x}\\right)$.\nConsider pairing each $f(x)$ with $\\bar{f}(x)$ whenever $f(x) \\neq \\bar{f}(x)$. If $f(x)$ is reducible, it can be factored as $f(x)=g(x) h(x)$ where $\\operatorname{deg} g, \\operatorname{deg} h \\geq 1$. Because the leading and constant coefficients of $f(x)$ are nonzero, so are the leading and constant coefficients of $g(x)$ and $h(x)$. Hence $\\bar{g}(x)$ and $\\bar{h}(x)$ are well defined with $\\operatorname{deg} \\bar{g}=\\operatorname{deg} g \\geq 1$ and $\\operatorname{deg} \\bar{h}=\\operatorname{deg} h \\geq 1$. Furthermore,\n$$\n\\bar{f}(x)=x^{9} f\\left(\\frac{1}{x}\\right)=x^{9} g\\left(\\frac{1}{x}\\right) h\\left(\\frac{1}{x}\\right)=\\left(x^{\\operatorname{deg} g} g\\left(\\frac{1}{x}\\right)\\right)\\left(x^{\\operatorname{deg} h} h\\left(\\frac{1}{x}\\right)\\right)=\\bar{g}(x) \\bar{h}(x)\n$$\nThus $\\bar{f}(x)=\\bar{g}(x) \\bar{h}(x)$ is a factorization of $\\bar{f}(x)$ into two non-constant polynomials, so $\\bar{f}(x)$ is also reducible. Therefore $f(x)$ is irreducible if and only if $\\bar{f}(x)$ is irreducible, so considering each pair, there are an even number of irreducible polynomials with $f(x) \\neq \\bar{f}(x)$.\nFinally, note that if $f(x)=\\bar{f}(x)$, then $d_{i}=d_{9-i}$ for each $i$. In such a case, we have $f(-1)=\\left(d_{0}-d_{9}\\right)+\\left(d_{2}-d_{7}\\right)+\\left(d_{4}-d_{5}\\right)+\\left(d_{6}-d_{3}\\right)+\\left(d_{8}-d_{1}\\right)=0$, so by the Factor Theorem, $(x+1)$ is a factor of $f(x)$. Therefore these remaining polynomials are all reducible.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 27928,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nCenturies ago, the pirate Captain Blackboard buried a vast amount of treasure in a single cell of an $M \\times N$ ($2 \\leq M, N$) grid-structured island. You and your crew have reached the island and have brought special treasure detectors to find the cell with the treasure. For each detector, you can set it up to scan a specific subgrid $[a, b] \\times [c, d]$ with $1 \\leq a \\leq b \\leq M$ and $1 \\leq c \\leq d \\leq N$. Running the detector will tell you whether the treasure is in the region or not, though it cannot say where in the region the treasure was detected. You plan on setting up $Q$ detectors, which may only be run simultaneously after all $Q$ detectors are ready. In terms of $M$ and $N$, what is the minimum $Q$ required to guarantee your crew can determine the location of Blackboard's legendary treasure?",
    "options": [],
    "answer": "ceil(M/2) + ceil(N/2)",
    "solution": "Solution:\n\nLet $m=\\left\\lceil\\frac{M}{2}\\right\\rceil$ and $n=\\left\\lceil\\frac{N}{2}\\right\\rceil$. We claim that the minimal $Q$ is $m+n$.\n\nFor the construction, start with $m$ detectors covering $[i, i+m-1] \\times [1, N]$ for $1 \\leq i \\leq m$. For every pair of rows, there is a detector that covers one row but not the other, hence this determines the row of the treasure. Similarly, placing $n$ detectors covering $[1, M] \\times [i, i+n-1]$ for $1 \\leq i \\leq n$ determines the column, and thus the location of the treasure.\n\nFor the bound, we require the following lemma.\n\nLemma. A $1 \\times k$ island requires at least $\\left\\lceil\\frac{k}{2}\\right\\rceil$ detectors.\n\nProof. Consider the $k-1$ lines separating the cells. If one of these lines is not covered by any detector, then these cells are indistinguishable. Similarly, if neither of the vertical lines at the ends are covered, then the first and last cells are indistinguishable. In particular, at least $k$ vertical lines need to be covered by the detectors. A detector covers 2 vertical lines, giving the result.\n\nIn general, consider the first row. Since the cells are distinguishable, by the lemma there must be at least $n$ detectors that intersect it non-trivially (as in, cover between 1 and $N-1$ of the cells). The analogous result holds for the last row and the first/last columns, giving $2m+2n$ detectors, where a detector may be counted multiple times.\n\nIf a detector intersected at least three of these sets, say it intersected the first row and the first and last columns. Therefore it covers the entire width of the island, and does not actually distinguish any cells in the first row, contradiction.\n\nTherefore each detector contributes to at most 2 of the above $2m+2n$ detectors, giving the final lower bound of $\\frac{2m+2n}{2}=m+n$ detectors required, as desired.\nSolution:\n\nThe following alternative approach from CMO competitor Marvin Mao of Bergen County Academies is another full solution.\n\nTake the same construction as in Solution 1. For the bound, consider the following sets:\n\n- $S_{\\mathrm{CR}} := \\{\\{(1,1),(1,N)\\},\\{(M,1),(M,N)\\}\\}$, i.e. the pairs of corners on the same row;\n- $S_{\\mathrm{CC}} := \\{\\{(1,1),(M,1)\\},\\{(1,N),(M,N)\\}\\}$, i.e. the pairs of corners on the same column;\n- $S_{\\mathrm{R}} := \\{\\{(x,i),(x,i+1)\\} : x \\in \\{1, M\\}, 1 \\leq i \\leq N-1\\}$, i.e. the pairs of adjacent edges on the first/last row;\n- $S_{\\mathrm{C}} := \\{\\{(i,x),(i+1,x)\\} : 1 \\leq i \\leq M-1, x \\in \\{1, N\\}\\}$, i.e. the pairs of adjacent edges on the first/last column.\n\nFor each detector, we assign it a score $\\left(x_{\\mathrm{CR}}, x_{\\mathrm{CC}}, x_{\\mathrm{R}}, x_{\\mathrm{C}}\\right)$, where $x_{\\mathrm{i}}$ is the number of pairs of cells in $S_{\\mathrm{i}}$ for which the detector covers exactly one of the two cells. The possible scores of the detectors are as follows:\n\n| What the detector hits         | Score                |\n|:-----------------------------:|:--------------------:|\n| No edges                      | $(0,0,0,0)$          |\n| One edge, no corners          | $(0,0,2,0)$ or $(0,0,0,2)$ |\n| Two edges, no corners         | $(0,0,4,0)$ or $(0,0,0,4)$ |\n| One corner                    | $(1,1,1,1)$          |\n| Two corners                   | $(2,0,2,0)$ or $(0,2,0,2)$ |\n| $>2$ corners or edges         | $(0,0,0,0)$          |\n\nIn order to determine the treasure, the total component-wise sum of scores of the detectors needs to be at least $(2,2,2N-2,2M-2)$, since we need to tell apart each of the pairs of cells. The sum of these components is $2M+2N$, and based on the analysis above, each detector adds a total component sum of at most 4, giving at least $\\left\\lceil\\frac{2M+2N}{4}\\right\\rceil = \\left\\lceil\\frac{M+N}{2}\\right\\rceil$ detectors.\n\nThis is equal to $\\left\\lceil\\frac{M}{2}\\right\\rceil + \\left\\lceil\\frac{N}{2}\\right\\rceil$ except if both $M, N$ are odd. In this case, if there is at least one more detector, then we have the required bound, so assume otherwise. In particular, we must achieve exactly the score $(2,2,2N-2,2M-2)$, with each detector contributing 4 to the total component sum.\n\nIn particular, to fill out the first two components, we must either have two detectors scoring $(1,1,1,1)$, or two detectors scoring $(2,0,2,0)$ and $(0,2,0,2)$. This yields a total score of $(2,2,2,2)$, leaving us with achieving exactly $(0,0,2N-4,2M-4)$ from the rest. Since we cannot have a non-zero score in the first two entries and must have a total component sum of 4, we can only use detectors scoring $(0,0,4,0)$ or $(0,0,0,4)$. But $2N-4, 2M-4 \\equiv 2 \\pmod{4}$, which is a contradiction.\n\nTherefore all situations require at least $\\left\\lceil\\frac{M}{2}\\right\\rceil + \\left\\lceil\\frac{N}{2}\\right\\rceil$ detectors.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 27929,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nInitially, three non-collinear points, $A$, $B$, and $C$, are marked on the plane. You have a pencil and a double-edged ruler of width 1. Using them, you may perform the following operations:\n- Mark an arbitrary point in the plane.\n- Mark an arbitrary point on an already drawn line.\n- If two points $P_1$ and $P_2$ are marked, draw the line connecting $P_1$ and $P_2$.\n- If two non-parallel lines $\\ell_1$ and $\\ell_2$ are drawn, mark the intersection of $\\ell_1$ and $\\ell_2$.\n- If a line $\\ell$ is drawn, draw a line parallel to $\\ell$ that is at distance 1 away from $\\ell$ (note that two such lines may be drawn).\n\nProve that it is possible to mark the orthocenter of $A B C$ using these operations.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nClaim 1. It is possible to draw internal/external angle bisectors.\nProof. Let $A$, $B$, $C$ be marked. To bisect $\\angle A B C$, draw the parallel line to $A B$ unit 1 away from it on the opposite side as $C$, and draw the parallel line to $B C$ unit 1 away from it on the opposite side as $A$. Let these lines intersect at $D$. Then $B D$ is the internal angle bisector of $\\angle A B C$. We can construct external angle bisectors similarly by drawing the line on the same side as $A$ for the second line instead.\n\nCorollary 2. It is possible to mark the incenters and excenters of a triangle $A B C$.\nProof. Draw in the internal/external bisectors of all three angles and intersect them.\n\nClaim 3. It is possible to mark the midpoint of any segment $A B$.\nProof. Let $B$ and $C$ be marked. Draw an arbitrary point $A$ not on line $B C$. Draw a line parallel to $B C$ unit 1 away from it on the opposite side as $A$, and let this line intersect $A B$ at $D$ and $A C$ at $E$. Let $B E$ and $C D$ intersect at $F$, and let $A F$ intersect $B C$ at $M$. Then by Ceva's Theorem, $M$ is the midpoint of $B C$.\n\nCorollary 4. It is possible to mark the centroid of $A B C$.\nProof. Draw the midpoint $D$ of $B C$ and the midpoint $E$ of $A C$, and intersect $A D$ with $B E$.\n\nClaim 5. It is possible to draw the perpendicular bisector of any segment $B C$.\nProof. Let $B$ and $C$ be marked. Draw an arbitrary point $A$ not on line $B C$. Construct the incenter $I$ and $A$-excenter $I_A$ of $A B C$. Draw the midpoint $M$ of $B C$ and midpoint $N$ of $I I_A$. By the incenter-excenter lemma, $N$ is the midpoint of the $\\operatorname{arc} \\overparen{B C}$ not containing $A$, so $M N$ is the perpendicular bisector of $B C$.\n\nCorollary 6. It is possible to mark the circumcenter of $A B C$.\nProof. Draw and intersect the perpendicular bisectors of $B C$ and $A C$.\n\nClaim 7. Given two marked points $A$ and $B$, it is possible to mark the point $C$ such that $\\overrightarrow{B C}=\\frac{1}{2} \\overrightarrow{A B}$.\nProof. Draw an arbitrary point $D$ not on line $A B$. Draw the midpoint $M$ of $A D$. Draw the midpoint $M_1$ of $B D$ and the midpoint $M_2$ of $M D$, and let $M_1 M_2$ intersect $A B$ at $C$. Then $M_1 M_2 \\parallel B M$ and $M M_2=\\frac{1}{2} M D=\\frac{1}{2} A M$, so $B C=\\frac{1}{2} A B$.\n\nClaim 8. Given two marked points $A$ and $B$ and any positive real number $k$ such that $2 k \\in \\mathbb{Z}$, it is possible to mark the point $C$ such that $\\overrightarrow{B C}=k \\overrightarrow{A B}$.\nProof. Note that by applying Claim 7 and marking the midpoint of $A B$, we can translate both $A$ and $B$ by $\\frac{1}{2} \\overrightarrow{A B}$. The claim now follows by applying this operation repeatedly.\n\nTo finish, take the given triangle $A B C$ and mark its circumcenter $O$ and centroid $G$. Note that its orthocenter $H$ satisfies that $\\overrightarrow{G H}=2 \\overrightarrow{O G}$, so applying Claim 8 to $k=2$ now finishes the problem.\nSolution:\n\nStart with Claims 1-3 of solution 1, allowing us to draw internal/external angle bisectors, in/excentres, and midpoints. We add one more claim.\n\nClaim 9. Given a point $P$ and a line $\\ell_1$, it is possible to draw a line through $P$ parallel to $\\ell_1$.\nProof. Draw the line $\\ell_2$ on the opposite side of $\\ell_1$ to $P$, a distance 1 away. Draw arbitrary lines $P A B$ and $P C D$ with $A, C \\in \\ell_1$, $B, D \\in \\ell_2$. Let $E$ be the midpoint of $A C$, let $F=P E \\cap \\ell_2$, and let $Q=B E \\cap F C$.\n\nSince $\\triangle Q E C \\sim \\triangle Q B F$ and $\\triangle P A E \\sim \\triangle P B F$, we have\n$$\n\\frac{Q E}{Q B}=\\frac{E C}{B F}=\\frac{A E}{B F}=\\frac{P A}{P B}\n$$\nso $\\triangle B A E \\sim \\triangle B P Q$. In particular, $P Q$ is parallel to $A E$, as desired.\n\nIn triangle $\\triangle A B C$, draw the incentre $I$ and $A$-excentre $I_A$. Draw the midpoints $D$ of $B C$ and $M$ of $I I_A$. By the incentre-excentre lemma, $M$ is on the perpendicular bisector of $B C$, so $M D$ is perpendicular to $B C$. Finally, using Claim 9, we can draw a line through $A$ that is perpendicular to $B C$. Repeat this for $B$ and $A C$, and their intersection is the orthocentre of $\\triangle A B C$, as required.",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 27930,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nThe $n$ players of a hockey team gather to select their team captain. Initially, they stand in a circle, and each person votes for the person on their left.\n\nThe players will update their votes via a series of rounds. In one round, each player $a$ updates their vote, one at a time, according to the following procedure: At the time of the update, if $a$ is voting for $b$, and $b$ is voting for $c$, then $a$ updates their vote to $c$. (Note that $a$, $b$, and $c$ need not be distinct; if $b = c$, then $a$'s vote does not change for this update.) Every player updates their vote exactly once in each round, in an order determined by the players (possibly different across different rounds).\n\nThey repeat this updating procedure for $n$ rounds. Prove that at this time, all $n$ players will unanimously vote for the same person.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nInitially, all players are in a cycle. Note that once a player leaves the cycle, they cannot rejoin. Furthermore, a new cycle cannot be created. Hence, at any point in time, the graph corresponding to the votes will be a functional graph with a single cycle.\n\nWe will first prove that after $\\lfloor \\log_{2} n \\rfloor$ rounds, the cycle will become a self-loop. Then, we will show that in the next $\\lfloor \\log_{2} n \\rfloor$ rounds, all other players vote for the player in the self-loop.\n\nTo show the first step, assume the cycle has size $K > 1$ at the beginning of a round. Consider arbitrary player $a$ in the cycle who is updating their vote. Say $a \\to b \\to c$, all in the cycle. Then $a \\to c$ now, bumping $b$ out of the cycle and reducing its size to $K - 1$. Note that $b$ can now update their vote as well without affecting the size of the cycle. If we consider all $K$ original players in the cycle, we see that at least $\\lceil \\frac{K}{2} \\rceil$ of them must still be in the cycle at the time of their update, and hence the cycle's size is reduced to at most $\\lfloor \\frac{K}{2} \\rfloor$. After $\\lfloor \\log_{2} n \\rfloor$ rounds, the cycle must be reduced to size 1.\n\nNow that the cycle has been reduced to a single player, say $z$, consider any path from a player $a$ to $z$. No players can be added to this path now. With a similar argument as the cycle, the length of the path must halve each round. In particular, a path of length $L$ to the cycle gets reduced to length $\\lceil \\frac{L}{2} \\rceil$ (note the ceiling, we had the floor for the cycle). After $\\lfloor \\log_{2} n \\rfloor$ rounds, the path must be reduced to length 1.\n\nThus, after $\\lfloor \\log_{2} n \\rfloor + \\lfloor \\log_{2} n \\rfloor$ rounds, the graph has been completely reduced. For $n \\geq 5$, $\\lfloor \\log_{2} n \\rfloor + \\lfloor \\log_{2} n \\rfloor \\leq 2 \\lfloor \\log_{2} n \\rfloor + 1 \\leq n$. For the other $n$, we can manually check that $\\lfloor \\log_{2} n \\rfloor + \\lceil \\log_{2} n \\rceil \\leq n$.\n\nWe will use induction on $n$.\n\nInductive Hypothesis. Let $G$ be any functional graph with $n$ nodes and a single cycle. Then after $n$ rounds of the given operation, $G$ will become a self-loop with $n - 1$ nodes pointing to it.\n\nBase Case. The cases $n \\leq 2$ are clear.\n\nInductive Step. Assume that the hypothesis is proved for $n = k - 1$ and $n = k - 2$. We will prove it for $n = k$. Consider any initial functional graph with $k$ nodes and a single cycle. Note there is some node $a$ which has in-degree 0 (i.e. no nodes point to it), or all $k$ nodes are in the cycle.\n\nIn the first case, consider $G \\setminus \\{a\\}$. Note that all operations except $a$'s own updates are independent of where $a$ is. By the inductive hypothesis, after $k - 1$ rounds, $G \\setminus \\{a\\}$ has become a single self-loop and $k - 2$ nodes pointing to it. Regardless of where $a$ is, it will point to the self-loop after one more round and we are done.\n\nIn the case where all $k$ nodes are in a cycle, consider the very first operation $z \\to a \\to b \\implies z \\to b, a \\to b$. This creates a zero in-degree node $a$, but $z$'s operation has been used for the first round so the inductive hypothesis cannot be naively applied. Instead, consider $b \\to c$ (possibly $c = z$ if $k = 3$). At some point in the first round, $b$ will be updated. Either $c$ will become another zero in-degree node, or $a$ will be the only node that points to $c$. Either way, consider $G \\setminus \\{a, c\\}$. By the induction hypothesis, after rounds 2 through $k - 1$, this graph will become a self-loop with $k - 3$ nodes pointing to it. It's also easy to see that $a$ and $c$ both have in-degree 0 after round 2. Then in one more round after round $k - 1$, we must have $a$ and $c$ pointing to the self-loop. So we are done for $n = k$.\n\nBy induction, we are done for all $n$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 27931,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nDetermine all positive integers $a$, $b$, $c$, $p$ where $p$ and $p + 2$ are odd primes and\n$$2^{a}p^{b} = (p + 2)^{c} - 1.$$",
    "options": [],
    "answer": "(a, b, c, p) = (3, 1, 2, 3)",
    "solution": "Solution:\nThe only solution is $(a, b, c, p) = (3, 1, 2, 3)$. First, factor the right hand side. This gives us\n$$2^{a}p^{b} = (p + 1)((p + 2)^{c - 1} + (p + 2)^{c - 2} + \\dots +(p + 2) + 1).$$\nSince $\\gcd (p, p + 1) = 1$ it must be the case that $p + 1 = 2^{x}$ for some positive integer $x \\leq a$ and so $p = 2^{x} - 1$ and $p + 2 = 2^{x} + 1$. Now for $x \\geq 3$, $2^{x} + 1$ is not prime if $x$ is odd (since it is $0 \\bmod 3$) and $2^{x} - 1$ is not prime if $x$ is even (since it is $0 \\bmod 3$). This means $x \\leq 2$, and the only admissible such $x$ is $x = 2$ since otherwise $p$ is not prime. So, the original equation becomes\n$$2^{a}3^{b} = 5^{c} - 1.$$ \nNow $3|(5^{c} - 1)$, so evaluating $5^{c} - 1$ mod $3$ gives that $c = 2d$ for some positive integer $d$ and hence\n$$2^{a}3^{b} = (5^{d} - 1)(5^{d} + 1).$$\nObserve $5^{d} - 1$ and $5^{d} + 1$ are both even and have greatest common divisor $2$ because they are $2$ apart. Since $4|(5^{d} - 1)$ this implies $5^{d} - 1 = 2^{a - 1}$ and $5^{d} + 1 = 2\\cdot 3^{b}$. Now, $3$ is not a factor of $5^{d} - 1$ because $5^{d} - 1 = 2^{a - 1}$. Thus, by evaluating mod $3$, $d$ must be odd. If $d > 1$, this is impossible as $5^{d} - 1 = (5 - 1)(5^{d - 1} + 5^{d - 2} + \\dots + 5 + 1)$ and the latter factor has an odd prime factor, contradicting $5^{d} - 1$ is a power of $2$. Thus $d = 1$ and so $c = 2$, implying that $2^{a}3^{b} = 24$ so $a = 3$ and $b = 1$. Hence, the only solution is $(a, b, c, p) = (3, 1, 2, 3)$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 27932,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nA polynomial $c_{d}x^{d} + c_{d - 1}x^{d - 1} + \\dots + c_{1}x + c_{0}$ with degree $d$ is reflexive if there is an integer $n \\geq d$ such that $c_{i} = c_{n - i}$ for every $0 \\leq i \\leq n$, where $c_{i} = 0$ for $i > d$. Let $\\ell \\geq 2$ be an integer and $p(x)$ be a polynomial with integer coefficients. Prove that there exist reflexive polynomials $q(x), r(x)$ with integer coefficients such that  \n$$(1 + x + x^{2} + \\dots + x^{\\ell -1})p(x) = q(x) + x^{\\ell}r(x).$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\nLet $d$ be the degree of $p$ and let $k$ be any non-negative integer. We will choose  \n$$q(x) = \\frac{x^{d + k + \\ell}p\\left(\\frac{1}{x}\\right) - p(x)}{x - 1},$$ \n$$r(x) = \\frac{p(x) - x^{d + k}p\\left(\\frac{1}{x}\\right)}{x - 1}.$$  \nFirst, we must show that both $q$ and $r$ are integer polynomials. Consider the numerator in $q$'s definition, $x^{d + k + \\ell}p\\left(\\frac{1}{x}\\right) - p(x)$. This is clearly an integer polynomial. As it is equal to $0$ when evaluated at $x = 1$, $x - 1$ divides it. Furthermore, as $x - 1$ is monic, the quotient has integer coefficients. The argument for $r$ is similar.  \nNext, we will show that this choice of $q$ and $r$ satisfies the desired equation. Plugging them into the RHS of the equation gives  \n$$q(x) + x^{\\ell}r(x) = \\frac{x^{d + k + \\ell}p\\left(\\frac{1}{x}\\right) - p(x)}{x - 1} + x^{\\ell}\\left(\\frac{p(x) - x^{d + k}p\\left(\\frac{1}{x}\\right)}{x - 1}\\right)$$\n$$\\qquad = \\frac{x^{d + k + \\ell}p\\left(\\frac{1}{x}\\right) - p(x) + x^{\\ell}p(x) - x^{d + k + \\ell}p\\left(\\frac{1}{x}\\right)}{x - 1}$$\n$$\\qquad = \\left(\\frac{x^{\\ell} - 1}{x - 1}\\right)p(x)$$\n$$\\qquad = (1 + x + \\cdots + x^{\\ell -1})p(x)$$  \nas desired.  \nFinally, we will show that $q$ and $r$ are indeed reflexive. We can re-interpret the reflexive condition as such:  \nPolynomial $a(x)$ is reflexive iff there is an integer $n \\geq \\deg (a)$ for which  \n$$a(x) = x^{n}a\\left(\\frac{1}{x}\\right).$$\nWe have  \n$$q(x) = \\frac{x^{d + k + \\ell}p\\left(\\frac{1}{x}\\right) - p(x)}{x - 1}$$\n$$\\qquad = x^{d + k + \\ell -1}\\cdot \\frac{p\\left(\\frac{1}{x}\\right) - x^{-(d + k + \\ell)}p(x)}{\\frac{x - 1}{x}}$$\n$$\\qquad = x^{d + k + \\ell -1}\\cdot \\frac{x^{-(d + k + \\ell)}p(x) - p\\left(\\frac{1}{x}\\right)}{\\frac{1}{x} - 1}$$\n$$\\qquad = x^{d + k + \\ell -1}q\\left(\\frac{1}{x}\\right)$$  \nas desired. Similarly,  \n$$r(x) = \\frac{p(x) - x^{d + k}p\\left(\\frac{1}{x}\\right)}{x - 1}$$\n$$\\qquad = x^{d + k - 1}\\cdot \\frac{x^{-(d + k)}p(x) - p\\left(\\frac{1}{x}\\right)}{\\frac{x - 1}{x}}$$\n$$\\qquad = x^{d + k - 1}\\cdot \\frac{p\\left(\\frac{1}{x}\\right) - x^{-(d + k)}p(x)}{\\frac{1}{x} - 1}$$\n$$\\qquad = x^{d + k - 1}r\\left(\\frac{1}{x}\\right).$$\nSolution:\nWe write degree $n$ polynomial $p$ as  \n$$p(x):= \\sum_{i = 0}^{n}p_{i}x^{i}.$$  \nDefine vector $P\\in \\mathbb{Z}^{n + 1}$ as  \n$$P:= \\left(p_{0}\\ p_{1}\\ \\dots\\ p_{n}\\right)^{T}.$$  \nWe also denote $X\\in \\mathbb{Z}[x]^{N}$ for $N$ some sufficiently high degree (e.g. $N > 2n + \\ell$) as the vector of powers of $x$, i.e.  \n$$X:= \\left(1\\ x\\ x^{2}\\dots\\ x^{N - 1}\\right)^{T}.$$  \nFor a matrix $M\\in \\mathbb{Z}^{(n + 1)\\times N}$, $P^{T}M X$ is an integer polynomial of degree $< N$. Note that if the non-zero entries of matrix $M$ are horizontally symmetric, then the resulting polynomial must be reflexive.\n\nThen their total is the matrix whose entries are $1$ at the parallelogram formed by  \n$$(0,0),(0,\\ell -1),(n,n + \\ell -1),(n,n),$$  \nwhich is precisely $A$.",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 27933,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nLet $ABC$ be a triangle with circumcircle $\\Gamma$ and $AB \\neq AC$. Let $D$ and $E$ lie on the arc $BC$ of $\\Gamma$ not containing $A$ such that $\\angle BAE = \\angle DAC$. Let the incenters of $BAE$ and $CAD$ be $X$ and $Y$ respectively, and let the external tangents of the incircles of $BAE$ and $CAD$ intersect at $Z$. Prove that $Z$ lies on the common chord of $\\Gamma$ and the circumcircle of $AXY$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\nLet $AX$ and $AY$ intersect $(ABC)$ again at $P$ and $Q$, let the inradii of $ABE$ and $ACD$ be $r_B$ and $r_C$, and let $(AXY)$ intersect $(ABC)$ again at $N$.\n\nFirst note that $\\angle BAP = \\frac{1}{2}\\angle BAE = \\frac{1}{2}\\angle CAD = \\angle QAC$, so $XP = BP = CQ = CY$. This thus implies that, since $NXP$ and $NYQ$ are spirally similar, $NX = NY$ and $NP = NQ$ (in particular, $N$ is the midpoint of arc $XAY$ on $(AXY)$).\n\nNow, note that\n$$\n\\frac{ZX}{ZY} = \\frac{r_b}{r_c} = \\frac{AX\\sin(\\angle PAE)}{AY\\sin(\\angle QAD)} = \\frac{AX}{AY}.\n$$\nSince $Z$ lies on the ray $YX$, we have that $Z$ lies on the external angle bisector of $\\angle XAY$. But so does $N$, hence $Z$, $A$, $N$ are collinear, as desired.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 27934,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nA rectangle $R$ is divided into a set $S$ of finitely many smaller rectangles with sides parallel to the sides of $R$ such that no three rectangles in $S$ share a common corner. An ant is initially located at the bottom-left corner of $R$. In one operation, we can choose a rectangle $r \\in S$ such that the ant is currently located at one of the corners of $r$, say $c$, and move the ant to one of the two corners of $r$ adjacent to $c$.\n\nSuppose that after a finite number of operations, the ant ends up at the top-right corner of $R$. Prove that some rectangle $r \\in S$ was chosen in at least two operations.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\nConsider the following version of the problem:\nA rectangle $R$ is divided into a set $S$ of finitely many smaller rectangles such that no three rectangles in $S$ share a common corner. For each $r \\in S$, draw two non-intersecting arcs inside $r$, connecting the pairs of adjacent corners of $r$ (there are two ways to do this, by connecting either the horizontally or vertically adjacent corners). Prove that there does not exist a path from the bottom-left corner of $R$ to the top-right corner of $R$ by walking only along these arcs.\n\n![](attached_image_1.png)\n<center>Figure 1: A possible diagram of all the arcs. </center>\n\nIn this problem, consider an undirected graph where nodes correspond to corners of rectangles in $S$ and edges correspond to the arcs, connecting the two nodes that the arc connects. The degrees of the nodes corresponding to the corners of $R$ are all exactly 1. Since no three rectangles in $S$ share a common corner, all intersection points have a pattern like $\\vdash$, $\\neg$, $\\perp$, or $\\top$, so the degree of all other nodes is exactly 2. Therefore, this graph can be decomposed into several paths and cycles. The only possible endpoints of paths are the degree 1 nodes, which are the corners of $R$. It follows that if there exists a path from the bottom-left corner to the top-right corner of $R$, then there also exists a path from the bottom-right corner to the top-left corner of $R$. However, this is impossible because these two paths (viewed as planar curves inside $R$) must intersect, which cannot occur. Therefore, this claim is proved.\n\nReturning to the original problem, suppose some operations were performed while choosing each rectangle at most once. Draw two arcs inside every rectangle, either both horizontal if the ant used this rectangle to move horizontally or both vertical if the ant used this rectangle to move vertically (or pick one arbitrarily if this rectangle was not used). By the new version of the problem, there does not exist a path from the bottom-left corner to the top-right corner of $R$, and it follows that it is impossible for the ant to have reached the top-right corner of $R$, finishing the proof.\n\n\nSuppose that no rectangle was chosen in at least two operations. In particular, a rectangle cannot be selected in two consecutive operations.\nAt any point in the process, consider whether the last move by the ant was horizontal or vertical, and whether the most recently chosen rectangle was to the left or the right of the ant's path. In the first move, either the ant moved horizontally and the rectangle was to the left, or the ant moved vertically and the rectangle was to the right. We claim that this invariant is preserved throughout the entire process (see Figure 2 for a sample path). Assuming this claim, the final move to the top right corner must select the top right rectangle. If it is vertical, this is to the left of the path, and if it is horizontal, it is to the right of the path, both of which are impossible, providing a contradiction.\n\n![](attached_image_2.png)\n<center>Figure 2: A possible path by the ant. The red arrows are all vertical, with the corresponding rectangle to the right. The blue arrows are all horizontal, with the corresponding rectangle to the left. </center>\n\nIt remains to show that the invariant is preserved. Since four rectangles cannot intersect at a corner, each intersection has a pattern like $\\vdash$, $\\neg$, $\\perp$, or $\\top$.\nFirst, assume the ant moves up, hence the chosen rectangle $r$ is on the right. The possible configurations are depicted in the first two diagrams in Figure 3.\n![](attached_image_3.png)\n<center>Figure 3: Possible ant moves going up or right. </center>\n\nIn each case, the ant must choose rectangle $s$ next (to avoid repeating $r$ twice), and we see that both side choices preserve a horizontal move with $s$ left, or a vertical move with $s$ right. By rotating the picture by $180^{\\circ}$, we cover the two possibilities for the ant moving downward.\nIf the ant moves right, then $r$ must occur on the left, and the possible configurations are the last two diagrams of Figure 3. Once again, rectangle $s$ must be chosen next, and the invariant is similarly preserved. The case of the ant moving left is again handled by a $180^{\\circ}$ rotation, completing the proof.\n\n\nThis is a variant of Solution 2. As in that proof, assume that no rectangle was chosen in two consecutive operations. We claim that for every move, the ant is in either the bottom-left or top-right corner of the chosen rectangle, and moves to the bottom-right or top-left corner.\nThis is clearly true of the first move. If the ant starts at the bottom-left or top-right corner on a move (choosing rectangle $r$), it is clear that they must move to the bottom-right or top-left of $r$. Assume they moved to the bottom-right corner of $r$, and chose rectangle $s$ in the next move. If they are at the bottom-right corner of $s$, then $r$ and $s$ are either equal or overlap, a contradiction. If they are at the top-left of $s$, then $r$ and $s$ intersect at a corner and no sides, and we must have 4 rectangles intersecting at a corner, again a contradiction.\nTherefore they must be at the bottom-left or top-right corner of $s$, as desired. The case where the ant is at the top-left corner of $r$ is analogous.\nIf the ant is able to make it to the top right corner, their final move must select $r$ as the top-right rectangle, and they move to the top-right corner, which is therefore impossible.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 27935,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all positive integers $n$ such that\n$$\nn^4 - 4n^3 + 22n^2 - 36n + 18\n$$\nis a perfect square.",
    "options": [],
    "answer": "1, 3",
    "solution": "We write\n$$\n\\begin{aligned}\nA &= n^4 - 4n^3 + 22n^2 - 36n + 18 \\\\\n&= (n^2 - 2n)^2 + 18(n^2 - 2n) + 18.\n\\end{aligned}\n$$\nLet $n^2 - 2n = x$, $A = y^2$, where $y$ is a nonnegative integer. Then\n$$\n(x+9)^2 - 63 = y^2.\n$$\nThat is, $(x+9-y)(x+9+y) = 63$.\nIt can only be $(x+9-y, x+9+y) = (1, 63)$, $(3, 21)$ or $(7, 9)$. We obtain, respectively, $(x, y) = (23, 31)$, $(3, 9)$ or $(-1, 1)$. But only when $x = 3$ and $-1$, $n^2 - 2n = x$ has solutions in positive integers and we get $n = 1$ or 3.\nTherefore, $n = 1$ or 3 satisfies the condition.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 27936,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all positive integers $n$ such that $20n + 2$ can divide $2003n + 2002$. (posed by Wu Weichao)",
    "options": [],
    "answer": "no positive integer n",
    "solution": "It is easy to see that $n$ is an even number. Let $n = 2m$, then from\n$40m + 2 \\mid 2003 \\times 2m + 2002$\nwe can get $20m + 1 \\mid 2003m + 1001$.\nBut $2003m + 1001 = 100(20m + 1) + 3m + 901$,\nso $20m + 1 \\mid 3m + 901$.\nAnd when $\\frac{3m+901}{20m+1} = 1, 2, 3, 4$, $m$ is not a positive integer.\nTherefore,\n$$\n\\begin{aligned}\n\\frac{3m+901}{20m+1} &\\ge 5, \\\\\nm &\\le \\frac{896}{97} < 10.\n\\end{aligned}\n$$\nHence $m \\le 9$. But after checking one by one for $m = 1, 2, 3, \\dots, 9$, we know that $20m + 1 \\nmid 3m + 901$ for $m = 1, 2, 3, \\dots, 9$.\nTherefore, there is no positive integer $n$ satisfying the condition.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 27937,
    "subject": "Mathematics (Multi-modal)",
    "question": "The interval on which the function $f(x) = \\log_{\\frac{1}{2}}(x^2 - 2x - 3)$ is monotone increasing is ( ).\n\n(A) $(-\\infty, -1)$\n(B) $(-\\infty, 1)$\n(C) $(1, +\\infty)$\n(D) $(3, +\\infty)$",
    "options": [],
    "answer": "A",
    "solution": "First, we will find the domain of $f(x)$. From $x^2 - 2x - 3 > 0$, we obtain $x < -1$ or $x > 3$. So the domain of definition for $f(x)$ is $(-\\infty, -1) \\cup (3, +\\infty)$.\n\nBut $u = x^2 - 2x - 3 = (x-1)^2 - 4$ is monotone decreasing on $(-\\infty, -1)$, and monotone increasing on $(3, +\\infty)$.\n\nSo $f(x) = \\log_{\\frac{1}{2}}(x^2 - 2x - 3)$ is monotone increasing on $(-\\infty, -1)$, and monotone decreasing on $(3, +\\infty)$.\n\nAnswer: A.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 27938,
    "subject": "Mathematics (Multi-modal)",
    "question": "As shown in the diagram, in $\\triangle ABC$, $\\angle A = 60^\\circ$, $AB > AC$, point $O$ is a circumcenter and $H$ is the intersection point of two altitudes $BE$ and $CF$. Points $M$ and $N$ are on the line segments $BH$ and $HF$ respectively, and satisfy $BM = CN$. Determine the value of $\\frac{MH + NH}{OH}$.\n\n![](attached_image_1.png)",
    "options": [],
    "answer": "sqrt(3)",
    "solution": "We take $BK = CH$ on $BE$ and join $OB$, $OC$ and $OK$.\nFrom the property of the circumcenter of a triangle, we know that $\\angle BOC = 2\\angle A = 120^\\circ$. From the property of the orthocenter of a triangle, we get $\\angle BHC = 180^\\circ - \\angle A = 120^\\circ$. So $\\angle BOC = \\angle BHC$. Then four points $B$, $C$, $H$ and $O$ are concyclic. Hence $\\angle OBH = \\angle OCH$.\nIn addition, $OB = OC$ and $BK = CH$. Therefore, $\\triangle BOK \\cong \\triangle COH$. It follows that $\\angle BOK = \\angle COH$, and $OK = OH$.\n$$\n\\text{So,} \\quad \\begin{aligned} \\angle KOH &= \\angle BOC = 120^\\circ, \\\\ \\angle OKH &= \\angle OHK = 30^\\circ. \\end{aligned}\n$$\nIn $\\triangle OKH$, by the sine rule, we get $KH = \\sqrt{3}OH$. In view of $BM = CN$ and $BK = CH$, we get $KM = NH$, and\n$$\nMH + NH = MH + KM = KH = \\sqrt{3}OH.\n$$\nTherefore,\n$$\n\\frac{MH + NH}{OH} = \\sqrt{3}.\n$$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 27939,
    "subject": "Mathematics (Multi-modal)",
    "question": "It is given that $f(x)$ is a function defined on $\\mathbb{R}$, satisfying $f(1) = 1$, and for any $x \\in \\mathbb{R}$,\n$$\nf(x+5) \\ge f(x)+5,\n$$\n\nand $f(x+1) \\le f(x)+1$.\nIf $g(x) = f(x)+1-x$, then $g(2002) = \\underline{\\hspace{2cm}}$.",
    "options": [],
    "answer": "1",
    "solution": "We determine $f(2002)$ first. From the conditions given, we have\n$$\n\\begin{aligned}\nf(x)+5 &\\le f(x+5) \\le f(x+4)+1 \\\\\n&\\le f(x+3)+2 \\le f(x+2)+3 \\\\\n&\\le f(x+1)+4 \\le f(x)+5.\n\\end{aligned}\n$$\nThus the equality holds for all. So we have $f(x+1)=f(x)+1$.\n\nHence, from $f(1)=1$, we get $f(2)=2$, $f(3)=3$, ..., $f(2002)=2002$. Therefore, $g(2002)=f(2002)+1-2002=1$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 27940,
    "subject": "Mathematics (Multi-modal)",
    "question": "If $\\log_4(x+2y) + \\log_4(x-2y) = 1$, then the minimum value of $|x|-|y|$ is ____.",
    "options": [],
    "answer": "sqrt(3)",
    "solution": "First, from\n$$\n\\begin{cases} x+2y>0, \\\\ x-2y>0, \\\\ (x+2y)(x-2y)=4, \\end{cases}\n$$\nwe obtain\n$$\n\\begin{cases} x > 2|y| \\ge 0, \\\\ x^2 - 4y^2 = 4. \\end{cases}\n$$\n\nBy the symmetry, there is no loss of generality in considering only the case when $y \\ge 0$. In view of $x > 0$, we need to find the minimum value of $x-y$ only.\nSetting $u = x-y$, and substituting it into $x^2-4y^2=4$, we obtain\n$$\n3y^2 - 2uy + (4 - u^2) = 0. \\quad (*)\n$$\nEquation (*) with respect to $y$ has real solutions. So we have\n$$\n\\Delta = 4u^2 - 12(4 - u^2) \\ge 0.\n$$\nThereby\n$$\nu \\ge \\sqrt{3}.\n$$\n\nIn addition, when $x = \\frac{4}{3}\\sqrt{3}$ and $y = \\frac{\\sqrt{3}}{3}$, we have $u = \\sqrt{3}$.\nTherefore, the minimum value of $|x| - |y|$ is $\\sqrt{3}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 27941,
    "subject": "Mathematics (Multi-modal)",
    "question": "If the inequality\n$$\nsin^2 x + a \\cos x + a^2 \\ge 1 + \\cos x\n$$\nholds for any $x \\in \\mathbb{R}$, the range of values for negative $a$ is\n______.",
    "options": [],
    "answer": "a ≤ -2",
    "solution": "$a + a^2 \\ge 2$ when $x = 0$. So $a \\le -2$ (because $a < 0$). When $a \\le -2$, we have\n$$\n\\begin{aligned}\na^2 + a \\cos x &\\ge a^2 + a \\ge 2 \\ge \\cos^2 x + \\cos x \\\\\n&= 1 + \\cos x - \\sin^2 x,\n\\end{aligned}\n$$\nthat is,\n$$\n\\sin^2 x + a \\cos x + a^2 \\ge 1 + \\cos x.\n$$\nHence, the range of values for negative $a$ is $a \\le -2$.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 27942,
    "subject": "Mathematics (Multi-modal)",
    "question": "Given $A(0, 2)$ and two points $B$ and $C$ on the parabola $y^2 = x+4$ such that $AB \\perp BC$, determine the range for the y-coordinate of point $C$.",
    "options": [],
    "answer": "y ≤ 0 or y ≥ 4",
    "solution": "Suppose that $(y_1^2 - 4, y_1)$ are the coordinates of point $B$ and $(y^2 - 4, y)$ of point $C$. Obviously, $y_1^2 - 4 \\ne 0$, so $k_{AB} = \\frac{y_1 - 2}{y_1^2 - 4} = \\frac{1}{y_1 + 2}$.\n\nSince $AB \\perp BC$, so $k_{BC} = - (y_1 - 2)$. Thus\n$$\ny - y_1 = -(y_1 + 2)[y^2 - 4 - (y_1^2 - 4)].\n$$\nNoting $y \\ne y_1$, we obtain\n$$\n(2 + y_1)(y + y_1) + 1 = 0,\n$$\nand that is\n$$\ny_1^2 + (2 + y)y_1 + (2y + 1) = 0.\n$$\n\nFrom $\\Delta \\ge 0$, we obtain $y \\le 0$ or $y \\ge 4$.\nWhen $y = 0$, the coordinates of $B$ are $(-3, -1)$ and when $y = 4$, they are $(5, -3)$. They both satisfy the conditions given by the problem. So, the range of values for the y-coordinate of point $C$ is $y \\le 0$ or $y \\ge 4$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 27943,
    "subject": "Mathematics (Multi-modal)",
    "question": "As shown in the diagram, there is a sequence of the curves $P_0, P_1, P_2, \\dots$. It is known that the region enclosed by $P_0$ has area $1$ and $P_0$ is an equilateral triangle. We obtain $P_{k+1}$ from $P_k$ by operating as follows: Trisecting every side of $P_k$, then we construct an equilateral triangle outwardly on every side of $P_k$ sitting on the middle segment of the side and finally remove this middle segment ($k = 0, 1, 2, \\dots$). Write $S_n$ as the area of the region enclosed by $P_n$.\n\n(1) Find a formula for the general term of the sequence of numbers $\\{S_n\\}$;\n(2) Find $\\lim_{n \\to \\infty} S_n$.\n\n![](attached_image_1.png)\n$P_0$\n![](attached_image_2.png)\n$P_1$\n![](attached_image_3.png)\n$P_2$\n...",
    "options": [],
    "answer": "S_n = 8/5 − (3/5)(4/9)^n;  lim_{n→∞} S_n = 8/5",
    "solution": "(1) We perform the operation on $P_0$. It is easy to see that each side of $P_0$ becomes $4$ sides of $P_1$. So the number of sides of $P_1$ is $3 \\cdot 4$. In the same way, we operate on $P_1$. Each side of $P_1$ becomes $4$ sides of $P_2$. So the number of sides of $P_2$ is $3 \\cdot 4^2$. Consequently, it is not difficult to get that the number of sides of $P_n$ is $3 \\cdot 4^n$.\n\nIt is known that the area of $P_0$ is $S_0 = 1$. Comparing $P_1$ with $P_0$, it is easy to see that we add to $P_1$ a smaller equilateral triangle with area $\\frac{1}{3^2}$ on each side of $P_0$. Since $P_0$ has $3$ sides, so\n$$\nS_1 = S_0 + 3 \\cdot \\frac{1}{3^2} = 1 + \\frac{1}{3}.\n$$\nAgain, comparing $P_2$ with $P_1$, we see that $P_2$ has an additional smaller equilateral triangle with area $\\frac{1}{3^2} \\cdot \\frac{1}{3^2}$ on each side of $P_1$, and $P_1$ has $3 \\cdot 4$ sides. So that\n$$\nS_2 = S_1 + 3 \\cdot 4 \\cdot \\frac{1}{3^4} = 1 + \\frac{1}{3} + \\frac{4}{3^3}.\n$$\nSimilarly, we have\n$$\nS_3 = S_2 + 3 \\cdot 4 \\cdot \\frac{1}{3^6} = 1 + \\frac{1}{3} + \\frac{4}{3^3} + \\frac{4^2}{3^5}.\n$$\nHence, we have\n$$\n\\begin{align*}\nS_n &= 1 + \\frac{1}{3} + \\frac{4}{3^3} + \\frac{4^2}{3^5} + \\dots + \\frac{4^{n-1}}{3^{2n-1}} \\\\\n&= 1 + \\sum_{k=1}^{n} \\frac{4^{k-1}}{3^{2k-1}} = 1 + \\frac{3}{4} \\sum_{k=1}^{n} \\left(\\frac{4}{9}\\right)^k \\\\\n&= 1 + \\frac{3}{4} \\cdot \\frac{4}{9} \\cdot \\left[1 - \\left(\\frac{4}{9}\\right)^n\\right] \\\\\n&= 1 + \\frac{3}{5} \\left[1 - \\left(\\frac{4}{9}\\right)^n\\right] \\\\\n&= \\frac{8}{5} - \\frac{3}{5} \\cdot \\left(\\frac{4}{9}\\right)^n. \\tag{*}\n\\end{align*}\n$$\n\nWe will prove $(*)$ by mathematical induction as follows:\nWhen $n=1$, it is known that $(*)$ holds from above.\nSuppose, when $n=k$, we have $S_k = \\frac{8}{5} - \\frac{3}{5} \\cdot \\left(\\frac{4}{9}\\right)^k$.\nWhen $n=k+1$, it is easy to see that, after $k+1$ times of operations, by comparing $P_{k+1}$ with $P_k$, we have added to $P_{k+1}$ a smaller equilateral triangle with area $\\frac{1}{3^{2(k+1)}}$ on each side of $P_k$ and $P_k$ has $3 \\cdot 4^k$ sides. So we get\n$$\n\\begin{aligned}\nS_{k+1} &= S_k + 3 \\cdot 4^k \\cdot \\frac{1}{3^{2(k+1)}} \\\\\n&= S_k + \\frac{4^k}{3^{2k+1}} = \\frac{8}{5} - \\frac{3}{5} \\cdot \\left(\\frac{4}{9}\\right)^{k+1}.\n\\end{aligned}\n$$\nBy mathematical induction, $(*)$ is proved.\n\n(2) From (1), we have $S_n = \\frac{8}{5} - \\frac{3}{5} \\cdot \\left(\\frac{4}{9}\\right)^n$.\nTherefore,\n$$\n\\lim_{n \\to \\infty} S_n = \\lim_{n \\to \\infty} \\left[ \\frac{8}{5} - \\frac{3}{5} \\cdot \\left(\\frac{4}{9}\\right)^n \\right] = \\frac{8}{5}.\n$$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 27944,
    "subject": "Mathematics (Multi-modal)",
    "question": "Suppose a quadratic function $f(x) = a x^2 + b x + c$ ($a, b, c \\in \\mathbb{R}$, and $a \\neq 0$) satisfies the following conditions:\n(1) When $x \\in \\mathbb{R}$, $f(x-4) = f(2-x)$ and $f(x) \\ge x$.\n(2) When $x \\in (0, 2)$, $f(x) \\le \\left(\\frac{x+1}{2}\\right)^2$.\n(3) The minimum value of $f(x)$ on $\\mathbb{R}$ is $0$.\nFind the maximal $m$ ($m > 1$) such that there exists $t \\in \\mathbb{R}$, $f(x+t) \\le x$ holds so long as $x \\in [1, m]$.",
    "options": [],
    "answer": "9",
    "solution": "Since $f(x-4) = f(2-x)$ for $x \\in \\mathbb{R}$, it is known that the quadratic function $f(x)$ has $x = -1$ as its axis of symmetry. By condition (3), we know that $f(x)$ opens upward, that is, $a > 0$. Hence\n$$\nf(x) = a(x+1)^2 \\quad (a > 0).\n$$\nBy condition (1), we get $f(1) \\ge 1$ and by (2), $f(1) \\le \\left(\\frac{1+1}{2}\\right)^2 = 1$. It follows that $f(1) = 1$, i.e. $a(1+1)^2 = 1$. So $a = \\frac{1}{4}$.\nThereby, $f(x) = \\frac{1}{4}(x+1)^2$.\n\n---\n\nSince the graph of the parabola $f(x) = \\frac{1}{4}(x+1)^2$ opens upward, and a graph of $y = f(x+t)$ can be obtained by translating that of $f(x)$ by $t$ units. If we want the graph of $y = f(x+t)$ to lie under the graph of $y = x$ when $x \\in [1, m]$, and $m$ to be maximal, then $1$ and $m$ should be two roots of an equation with respect to $x$.\n$$\n\\frac{1}{4}(x+t+1)^2 = x. \\qquad \\textcircled{1}\n$$\nSubstituting $x = 1$ into (1), we get $t = 0$ or $t = -4$.\nWhen $t = 0$, substituting it into (1), we get $x_1 = x_2 = 1$ (in contradiction with $m > 1$).\nWhen $t = -4$, substituting it into (1), we get $x_1 = 1$, and $x_2 = 9$; and so $m = 9$.\nMoreover, when $t = -4$, for any $x \\in [1, 9]$, we have always\n$$\n(x-1)(x-9) \\le 0 \\\\\n\\Leftrightarrow \\frac{1}{4}(x-4+1)^2 \\le x,\n$$\nthat is\n$$\nf(x-4) \\le x.\n$$\nTherefore, the maximum value of $m$ is $9$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 27945,
    "subject": "Mathematics (Multi-modal)",
    "question": "If real numbers $x$ and $y$ satisfy $(x+5)^2 + (y-12)^2 = 14^2$, then the minimum value of $x^2 + y^2$ is ( ).\n(A) 2\n(B) 1\n(C) $\\sqrt{3}$\n(D) $\\sqrt{2}$",
    "options": [],
    "answer": "B",
    "solution": "Let $x+5 = 14\\cos\\theta$ and $y-12 = 14\\sin\\theta$, for $\\theta \\in [0, 2\\pi)$.\nHence\n$$\n\\begin{align*}\nx^2 + y^2 &= (14\\cos\\theta - 5)^2 + (14\\sin\\theta + 12)^2 \\\\\n&= 14^2 + 5^2 + 12^2 - 140\\cos\\theta + 336\\sin\\theta \\\\\n&= 365 + 28(12\\sin\\theta - 5\\cos\\theta)\n\\end{align*}\n$$\n$$\n\\begin{aligned}\n&=365 + 28 \\times 13\\sin(\\theta - \\varphi) \\\\\n&=365 + 364\\sin(\\theta - \\varphi),\n\\end{aligned}\n$$\nwhere $\\tan \\varphi = \\frac{5}{12}$.\nSo $x^2+y^2$ has the minimum value $1$, when $\\theta=\\frac{3\\pi}{2}+\\arctan\\frac{5}{12}$, i.e. $x = \\frac{5}{13}$ and $y = -\\frac{12}{13}$.\nAnswer: B.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 27946,
    "subject": "Mathematics (Multi-modal)",
    "question": "There are real numbers $a$, $b$ and $c$ such that $f(x) = x^3 + a x^2 + b x + c$ has three real roots $x_1$, $x_2$ and $x_3$ satisfying\n$$\n(1)\\quad x_2 - x_1 = \\lambda,\n$$\n$$\n(2)\\quad x_3 > \\frac{1}{2}(x_1 + x_2).\n$$\nFind the maximum value of $\\frac{2a^3 + 27c - 9ab}{\\lambda^3}$.",
    "options": [],
    "answer": "3*sqrt(3)/2",
    "solution": "Let $S = \\frac{2a^3 + 27c - 9ab}{\\lambda^3}$, then\n$$\nS = \\frac{27\\left(\\frac{2}{27}a^3 - \\frac{1}{3}ab + c\\right)}{\\lambda^3} = \\frac{27f\\left(-\\frac{1}{3}a\\right)}{\\lambda^3} \\\\\n= \\frac{27\\left(-\\frac{1}{3}a - x_1\\right)\\left(-\\frac{1}{3}a - x_2\\right)\\left(-\\frac{1}{3}a - x_3\\right)}{(x_2 - x_1)^3} \\\\\n= \\frac{-27\\left(x_1 + \\frac{1}{3}a\\right)\\left(x_2 + \\frac{1}{3}a\\right)\\left(x_3 + \\frac{1}{3}a\\right)}{(x_2 - x_1)^3}.\n$$\n\nWrite $u_i = x_i + \\frac{a}{3}$ ($i=1,2,3$), then $u_2 - u_1 = x_2 - x_1 = \\lambda$, $u_3 > \\frac{1}{2}(u_1 + u_2)$, and $u_1 + u_2 + u_3 = x_1 + x_2 + x_3 + a = 0$. So, $u_1$, $u_2$ and $u_3$ satisfy the corresponding conditions too, and\n$$\nS = \\frac{-27u_1 u_2 u_3}{(u_2 - u_1)^3}.\n$$\nBy $u_3 = -(u_1 + u_2) > \\frac{u_1 + u_2}{2}$, we get $u_1 + u_2 < 0$. Consequently, at least one of $u_1$ and $u_2$ should be less than $0$. We may assume $u_1 < 0$.\n\nIf $u_2 < 0$, then $S < 0$.\n\nIf $u_2 > 0$, we suppose further\n$$\nv_1 = \\frac{-u_1}{u_2 - u_1}, \\quad v = v_2 = \\frac{u_2}{u_2 - u_1}.\n$$\nThen $v_1 + v_2 = 1$, $v_1$ and $v_2$ are both greater than $0$ and $v_1 - v_2 = \\frac{u_3}{u_2 - u_1} > 0$. So it follows that\n$$\n\\begin{aligned}\nS &= 27v_1v_2(v_1 - v_2) = 27v(1-v)(1-2v) \\\\\n&= 27\\sqrt{(v-v^2)^2(1-2v)^2} \\\\\n&= 27\\sqrt{2(v-v^2)(v-v^2)\\left(\\frac{1}{2}-2v+2v^2\\right)} \\\\\n&\\le 27\\sqrt{2 \\times \\left(\\frac{1}{6}\\right)^3} \\\\\n&= \\frac{3}{2}\\sqrt{3}.\n\\end{aligned}\n$$\nThe equality holds when $v = \\frac{1}{2}\\left(1 - \\frac{\\sqrt{3}}{3}\\right)$. The corresponding cubic equation is $x^3 - \\frac{1}{2}x + \\frac{\\sqrt{3}}{18} = 0$ and $\\lambda = 1$ ($x_1 = -\\frac{1}{2}(1 + \\frac{\\sqrt{3}}{3})$, $x_2 = \\frac{1}{2}(1 - \\frac{\\sqrt{3}}{3})$ and $x_3 = \\frac{\\sqrt{3}}{3}$).\n\nConsequently, the maximal value is $\\frac{3}{2}\\sqrt{3}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 27947,
    "subject": "Mathematics (Multi-modal)",
    "question": "Suppose $O$ is the circumcenter of an acute triangle $\\triangle ABC$, $P$ is a point inside $\\triangle AOB$, and $D$, $E$, $F$ are the projections of $P$ on three sides $BC$, $CA$, $AB$ of $\\triangle ABC$ respectively. Prove that a parallelogram with $FE$ and $FD$ as adjacent sides lies inside $\\triangle ABC$. (posed by Leng Gangsong)\n\n![](attached_image_1.png)\n",
    "options": [],
    "answer": "Detailed solution",
    "solution": "**Proof** As shown in the figure, we construct a parallelogram $DEFG$ with $FE$ and $FD$ as adjacent sides. To prove the proposition to be true, we need only to prove that $\\angle FEG < \\angle FEC$, and $\\angle FDG < \\angle FDC$. It is equivalent to proving: $\\angle BFD < \\angle BAC$, and $\\angle AFE < \\angle ABC$.\n\nIn fact, we construct $OH$ with $OH \\perp BC$, and $H$ is the foot of the perpendicular. From $PD \\perp BC$ and $PF \\perp AB$, we know that four points $B$, $F$, $P$ and $D$ are concyclic. Thus $\\angle BFD = \\angle BPD$. But $\\angle PBD > \\angle OBH$, hence $90^\\circ - \\angle PBD < 90^\\circ - \\angle OBH$, and that is, $\\angle BPD < \\angle BOH$. Moreover, $O$ is the circumcenter of $\\triangle ABC$, so $\\angle BOH = \\frac{1}{2} \\angle BOC = \\angle BAC$. Therefore, $\\angle BFD = \\angle BPD < \\angle BOH = \\angle BAC$, that is, $\\angle BFD < \\angle BAC$.\n\nSimilarly, we can prove that $\\angle AFE < \\angle ABC$. Therefore, the proposition holds.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 27948,
    "subject": "Mathematics (Multi-modal)",
    "question": "$3n$ ($n$ is a positive integer) girl students took part in a summer camp. There were three girl students to be on duty every day. When the summer camp ended, it was found that any two of the $3n$ girl students had just one time to be on duty on the same day.\n\n(1) When $n = 3$, is there any arrangement satisfying the requirement above? Prove your conclusion.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "(1) When $n = 3$, there is an arrangement satisfying the requirement. Now the concrete arrangement is given as follows (Denote the nine girl students by $1, 2, \\ldots, 9$):\n\n$$\n(1, 2, 3), (1, 4, 5), (1, 6, 7), (1, 8, 9), \\\\\n(2, 4, 6), (2, 7, 8), (2, 5, 9), (3, 4, 8), \\\\\n(3, 5, 7), (3, 6, 9), (4, 7, 9), (5, 6, 8).\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 27949,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all positive integers $k$ such that, for any positive numbers $a$, $b$ and $c$ satisfying the inequality $k(ab + bc + ca) > 5(a^2 + b^2 + c^2)$, there must exist a triangle with $a$, $b$ and $c$ as the length of its three sides respectively. (posed by Qian Zhangwang)",
    "options": [],
    "answer": "k = 6",
    "solution": "$$\na^2 + b^2 + c^2 \\geq ab + bc + ca,\n$$\nso $k > 5$. Hence $k \\ge 6$.\n\nrespectively, by the assumption in the problem, we have\n$$\nk(1 \\times 1 + 1 \\times 2 + 1 \\times 2) \\le 5(1^2 + 1^2 + 2^2),\n$$\nthat is, $k \\le 6$.\nWe will prove that $k = 6$ satisfies the requirement below. There is no harm in assuming $a \\le b \\le c$.\n$$\n\\begin{array}{lcl}\n\\text{Since} & 6(ab + bc + ca) & > 5(a^2 + b^2 + c^2), \\\\[1em]\n\\text{so} & 5c^2 - 6(a+b)c + 5a^2 + 5b^2 - 6ab & < 0, \\\\[1em]\n& \\Delta = [6(a+b)]^2 - 4 \\cdot 5(5a^2 + 5b^2 - 6ab) & \\\\\n& = 64(ab - (a-b)^2) & \\\\\n& \\le 64ab \\le 64 \\cdot \\frac{(a+b)^2}{4} & \\\\\n& = 16(a+b)^2. & \\\\[1em]\n\\text{Thus} & c < \\frac{6(a+b) + \\sqrt{\\Delta}}{10} \\le \\frac{6(a+b) + 4(a+b)}{10} & \\\\\n& = a+b. &\n\\end{array}\n$$\nHence, there exists a triangle with $a, b$, and $c$ being the length of three sides.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 27950,
    "subject": "Mathematics (Multi-modal)",
    "question": "Consider a square on the complex plane. The complex numbers corresponding to its four vertices are the four roots of some equation of the fourth degree with one unknown and integer coefficients $x^4+px^3+qx^2+rx+s=0$. Find the minimum value of the area of such square. (posed by Xiong Bin)",
    "options": [],
    "answer": "2",
    "solution": "Suppose the complex number corresponding to the center of the square is $a$. Then after translating the origin of the complex plane to $a$, the vertices of the square distribute evenly on the circumference. That is, they are the solutions of equation $(x-a)^4 = b$, where $b$ is a complex number. Hence,\n\n$$\n\\begin{align*}\nx^4 + px^3 + qx^2 + rx + s &= (x-a)^4 - b \\\\\n&= x^4 - 4a x^3 + 6a^2 x^2 - 4a^3 x + a^4 - b.\n\\end{align*}\n$$\n\nComparing the coefficients of terms for $x$ with the same degree, we know that $-a = \\frac{p}{4}$, and it is a rational number. Combining further that $-4a^3 = r$ is an integer, we can see that $a$ is an integer. So by using the fact that $s = a^4 - b$ is an integer, we can show that $b$ is also an integer.\n\nThe above discussion makes clear of a fact that the four numbers corresponding to the four vertices of this square are roots of integer coefficients equation $(x - a)^4 = b$. Hence, the radius of its circumcircle ($=\\sqrt[4]{|b|}$) is not less than $1$. Therefore, the area of this square is not less than $(\\sqrt{2})^2 = 2$. But the four roots of the equation $x^4 = 1$ are corresponding to the four vertices of a square on the complex plane. Hence, the minimum value of the area of the square is $2$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 27951,
    "subject": "Mathematics (Multi-modal)",
    "question": "Before The World Cup tournament, the football coach of $F$ country will let seven players, $A_1, A_2, \\dots, A_7$, join three training matches (90 minutes each) in order to assess them. Suppose, at any moment during a match, one and only one of them enters the field, and the total time (which is measured in minutes) on the field for each one of $A_1, A_2, A_3$ and $A_4$ is divisible by $7$ and the total time for each of $A_5, A_6$ and $A_7$ is divisible by $13$. If there is no restriction about the number of times of substitution of players during each match, then how many possible cases are there within the total time for every player on the field?",
    "options": [],
    "answer": "42244",
    "solution": "Suppose that $x_i$ ($i = 1, 2, \\dots, 7$) minutes is the time for $i$-th player on the field. Now, the problem is to find the number of solution groups of positive integers for the following equation:\n$$\nx_1 + x_2 + \\cdots + x_7 = 270\n$$\nwhen the conditions $7 \\mid x_i$ ($i = 1, 2, 3, 4$) and $13 \\mid x_j$ ($j = 5, 6, 7$) are satisfied.\n\nSuppose $x_1 + x_2 + x_3 + x_4 = 7m$ and $x_5 + x_6 + x_7 = 13n$. Then\n$$\n7m + 13n = 270,\n$$\nand $m, n \\in \\mathbb{N}_+, m \\ge 4$ and $n \\ge 3$.\n\nWhen $(m, n) = (33, 3)$, $x_5 = x_6 = x_7 = 13$. Let $x_i = 7y_i$ ($i = 1, 2, 3, 4$), then\n$$\ny_1 + y_2 + y_3 + y_4 = 33.\n$$\nWe get $C_{33-1}^4 = C_{32}^3 = 4960$ solution groups of positive integers $(y_1, y_2, y_3, y_4)$ and in this case, we have $4960$ solution groups of positive integers satisfying the conditions.\n\nWhen $(m, n) = (20, 10)$, let $x_i = 7y_i$ ($i = 1, 2, 3, 4$) and $x_j = 13y_j$ ($j = 5, 6, 7$). Hence\n$$\ny_1 + y_2 + y_3 + y_4 = 20 \\text{ and } y_5 + y_6 + y_7 = 10.\n$$\nIn this case, we have $C_{19}^3 \\times C_9^3 = 34884$ solution groups of positive integers satisfying the conditions.\n\nWhen $(m, n) = (7, 17)$, set $x_i = 7y_i$ ($i = 1, 2, 3, 4$) and $x_j = 13y_j$ ($j = 5, 6, 7$). Hence\n$$\ny_1 + y_2 + y_3 + y_4 = 7 \\text{ and } y_5 + y_6 + y_7 = 17.\n$$\nIn this case, we have $C_6^3 \\times C_{16}^3 = 2400$ solution groups of positive integers satisfying the conditions.\n\nConsequently, for $(1)$, there are\n$$\n4960 + 34884 + 2400 = 42244\n$$\nsolution groups of positive integers satisfying the conditions.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 27952,
    "subject": "Mathematics (Multi-modal)",
    "question": "The function $f(x) = \\frac{x}{1-2^x} - \\frac{x}{2}$ is ( ).\n(A) an even but not odd function\n(B) an odd but not even function\n(C) a both even and odd function\n(D) a neither even nor odd function",
    "options": [],
    "answer": "A",
    "solution": "It is easy to see that the domain of $f(x)$ is $(-\\infty, 0) \\cup (0, +\\infty)$. When $x \\in (-\\infty, 0) \\cup (0, +\\infty)$, we have\n$$\n\\begin{aligned}\nf(-x) &= \\frac{-x}{1-2^{-x}} - \\frac{x}{2} \\\\\n&= \\frac{-x \\cdot 2^x}{2^x - 1} + \\frac{x}{2} \\\\\n&= \\frac{x}{1-2^x} - x + \\frac{x}{2} \\\\\n&= \\frac{x}{1-2^x} - \\frac{x}{2} = f(x).\n\\end{aligned}\n$$\nTherefore, $f(x)$ is an even function, and obviously not an odd function. Answer: A.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 27953,
    "subject": "Mathematics (Multi-modal)",
    "question": "Assume that $n$ is a positive integer, and $A_1, A_2, \\dots, A_{n+1}$ are $n+1$ nonempty subsets of the set $\\{1, 2, \\dots, n\\}$. Prove that there are two disjoint and nonempty subsets $\\{i_1, i_2, \\dots, i_k\\}$ and $\\{j_1, j_2, \\dots, j_m\\}$ such that\n$$\nA_{i_1} \\cup A_{i_2} \\cup \\dots \\cup A_{i_k} = A_{j_1} \\cup A_{j_2} \\cup \\dots \\cup A_{j_m}.\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "**Proof I** We prove by induction for $n$.\n\nWhen $n = 1$, $A_1 = A_2 = \\{1\\}$, the proposition holds.\n\nSuppose it holds for $n$. We consider the case of $n+1$.\nSuppose $A_1, A_2, \\dots, A_{n+2}$ are nonempty subsets of $\\{1, 2, \\dots, n+1\\}$. Let $B_i = A_i \\setminus \\{n+1\\}, i = 1, 2, \\dots, n+2$. We will prove the following cases.\n\n**Case I** There exist $1 \\le i < j \\le n+2$ such that $B_i = B_j = \\emptyset$, then $A_i = A_j = \\{n+1\\}$. The proposition is proven.\n\n**Case II** There exists only one $i$ such that $B_i = \\emptyset$. There is no loss of generality in supposing $B_{n+2} = \\emptyset$, and that is $A_{n+2} = \\{n+1\\}$. Now by the inductive assumption, for $\\{1, 2, \\dots, n+1\\}$, there exist two disjoint subsets $\\{i_1, \\dots, i_k\\}$ and $\\{j_1, \\dots, j_m\\}$ such that\n$$\nB_{i_1} \\cup \\dots \\cup B_{i_k} = B_{j_1} \\cup \\dots \\cup B_{j_m}. \\qquad \\textcircled{1}\n$$\nWe write $C = A_{i_1} \\cup \\dots \\cup A_{i_k}$, $D = A_{j_1} \\cup \\dots \\cup A_{j_m}$. Then $C$ and $D$ differ at the most by the element $n+1$. (This can be shown by $\\textcircled{1}$ and the definition of $B_i$.) In this case, we can make the proposition to hold true by putting $A_{n+2}$ into $C$ or $D$.\n\n**Case III** No $B_i$ is empty. Now $B_1, B_2, \\dots, B_{n+1}$ are nonempty subsets of $\\{1, 2, \\dots, n\\}$. By the inductive assumption, we show that, for $\\{1, 2, \\dots, n+1\\}$, there exist disjoint subsets $\\{i_1, \\dots, i_k\\}$ and $\\{j_1, \\dots, j_m\\}$ such that\n$$\nB_{i_1} \\cup \\dots \\cup B_{i_k} = B_{j_1} \\cup \\dots \\cup B_{j_m}. \\qquad \\textcircled{2}\n$$\nIn addition, $B_2, B_3, \\dots, B_{n+2}$ are also nonempty subsets of $\\{1, 2, \\dots, n\\}$. By the inductive assumption, we can show that, for $\\{2, 3, \\dots, n+2\\}$, there exist disjoint subsets $\\{r_1, \\dots, r_u\\}$ and $\\{t_1, \\dots, t_v\\}$ such that\n$$\nB_{r_1} \\cup \\dots \\cup B_{r_u} = B_{t_1} \\cup \\dots \\cup B_{t_v}. \\qquad \\textcircled{3}\n$$\nAgain, we write $C = A_{i_1} \\cup \\dots \\cup A_{i_k}$, $D = A_{j_1} \\cup \\dots \\cup A_{j_m}$, and write $E = A_{r_1} \\cup \\dots \\cup A_{r_u}$, $F = A_{t_1} \\cup \\dots \\cup A_{t_v}$. By using $\\textcircled{2}$, $\\textcircled{3}$ and the definition of $B_i$, we see that $C$ and $D$ differ at the most by the element $n+1$, and so do $E$ and $F$. If $C = D$ or $E = F$, then the proposition holds. Hence we need only to consider the case when $C \\neq D$ and $E \\neq F$. There is no loss of generality in supposing $C = D \\cup \\{n+1\\}$, but $E = F \\setminus \\{n+1\\}$. Now $C \\cup E = D \\cup F$. After amalgamating the sets occurred repeatedly in $C$ and $E$, as well as in $D$ and $F$, we get two subsets $\\{p_1, \\dots, p_x\\}$ and $\\{q_1, \\dots, q_y\\}$ of $\\{1, 2, \\dots, n+2\\}$ such that\n$$\nA_{p_1} \\cup \\cdots \\cup A_{p_x} = A_{q_1} \\cup \\cdots \\cup A_{q_y}, \\qquad \\textcircled{4}\n$$\nwhere $G = A_{p_1} \\cup \\cdots \\cup A_{p_x} = C \\cup E$, $H = A_{q_1} \\cup \\cdots \\cup A_{q_y} = D \\cup F$.\n\nNow, if $\\{p_1, \\cdots, p_x\\} \\cap \\{q_1, \\cdots, q_y\\} = \\emptyset$, then the proposition holds. If there is $i \\in \\{p_1, \\cdots, p_x\\} \\cap \\{q_1, \\cdots, q_y\\}$, we write $\\tilde{C} = \\{A_{i_1}, \\cdots, A_{i_k}\\}$, $\\tilde{D} = \\{A_{j_1}, \\cdots, A_{j_m}\\}$, $\\tilde{E} = \\{A_{r_1}, \\cdots, A_{r_u}\\}$, $\\tilde{F} = \\{A_{t_1}, \\cdots, A_{t_v}\\}$. And there is no loss of generality in assuming that $A_i$ does not belong to $\\tilde{C}$ and $\\tilde{E}$ at the same time, and it does not belong to $\\tilde{D}$ and $\\tilde{F}$ at the same time too. Hence there are only two possibilities.\n\n(a) $A_i \\in \\tilde{C}$ and $A_i \\in \\tilde{F}$. If there are two sets in $\\tilde{C}$ containing $n+1$, then we take away set $A_i$ from the left side in ④. Now since all elements except $n+1$ in $A_i$ belong to $E$ (in view of ③), and there are two sets on the left side in ④ containing $n+1$. Thus after taking away $A_i$, the number of elements in $G$ does not reduce and ④ is still an equality. In the same way, if there are two sets in $\\tilde{F}$ containing $n+1$, then we take away $A_i$ from the right side in ④, and ④ still holds.\nOf course, if there is only one set in $\\tilde{C}$ and $\\tilde{F}$ containing $n+1$, then after taking away $A_i$ from both sides in ④, it remains to be an equality. (Now, by ② and ③, we can see that the two sides of ④ will not become empty sets.)\n\n(b) $A_i \\in \\tilde{D}$ and $A_i \\in \\tilde{E}$, then $n+1 \\notin A_i$. Now after taking away $A_i$ from both sides in ④, the resulting expression is still an equality.\n\nIn view of the above operation, we have a method to make the two sets of subscripts $\\{p_1, \\cdots, p_x\\}$ and $\\{q_1, \\cdots, q_y\\}$ in ④ disjoint. Therefore the proposition holds for $n+1$.\n\n\n**Proof II** Here we need to use a fact from linear algebra that $n+1$ vectors in the $n$-dimensional linear space are linearly dependent.\n\nIf element $i$ is in set $A_j$, we write it as 1, otherwise write it as 0. Then $A_j$ corresponds to an $n$-dimensional vector, which is nonzero and contains 0 and 1. We write $a_j = (a_{j_1}, a_{j_2}, \\cdots, a_{j_n})$, where\n$$\na_{j_i} = \\begin{cases} 1, & i \\in A_j, \\\\ 0, & i \\notin A_j. \\end{cases}\n$$\nSince $a_1, a_2, \\dots, a_{n+1}$ are $n+1$ vectors in the $n$-dimensional space, so there exists a group of real numbers, not every one of them to be zero, $x_1, x_2, \\dots, x_{n+1}$ such that\n$$\nx_1 a_1 + x_2 a_2 + \\dots + x_{n+1} a_{n+1} = 0. \\qquad \\textcircled{5}\n$$\nHence, for $\\{1, 2, \\dots, n+1\\}$, there exist two disjoint and nonempty subsets $\\{i_1, \\dots, i_k\\}$ and $\\{j_1, \\dots, j_m\\}$ such that\n$$\nx_{i_1} a_{i_1} + \\dots + x_{i_k} a_{i_k} = y_{j_1} a_{j_1} + \\dots + y_{j_m} a_{j_m}, \\qquad \\textcircled{6}\n$$\nwhere $x_{i_1}, \\dots, x_{i_k} > 0$, $y_{j_1} = (-x_{j_1})$, $\\dots$, $y_{j_m} = (-x_{j_m}) > 0$ (Here, it is essential to put the terms with coefficients greater than zero in ⑤ to one side, and those with coefficients less than zero to another side).\nWe conclude that\n$$\nA_{i_1} \\cup A_{i_2} \\cup \\dots \\cup A_{i_k} = A_{j_1} \\cup \\dots \\cup A_{j_m}. \\qquad \\textcircled{7}\n$$\nIn fact, if element $a$ ($1 \\le a \\le n$) belongs to the left side in ⑦, then the $a$-th component of the sum of the vectors from the left side in ⑥ must be greater than zero. Thus it makes the $a$-th component of the sum of the vectors from the right side in ⑥ to be greater than zero. Hence, there is $a_{j_t}$, and its $a$-th component is 1, that is, $a \\in A_{j_t}$. Conversely, it is also true, that is, ⑦ holds.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 27954,
    "subject": "Mathematics (Multi-modal)",
    "question": "The straight line $\\frac{x}{4} + \\frac{y}{3} = 1$ intersects the ellipse $\\frac{x^2}{16} + \\frac{y^2}{9} = 1$ at two points $A$ and $B$. There is a point $P$ on this ellipse such that the area of $\\triangle PAB$ is equal to $3$. There is/are ( ) such point/points $P$.\n(A) 1 (B) 2 (C) 3 (D) 4",
    "options": [],
    "answer": "B",
    "solution": "Suppose that there is a point $P(4\\cos \\alpha, 3\\sin \\alpha)$ on the ellipse. When $P$ and the origin $O$ are not on the same side of $AB$, the distance from $P$ to $AB$ is\n$$\n\\begin{aligned}\n& \\frac{3(4\\cos\\alpha) + 4(3\\sin\\alpha) - 12}{5} \\\\\n&= \\frac{12}{5}(\\cos\\alpha + \\sin\\alpha - 1) \\\\\n&\\le \\frac{12}{5}(\\sqrt{2} - 1) \\\\\n&< \\frac{6}{5}.\n\\end{aligned}\n$$\nBut $AB = 5$, so $\\triangle PAB < \\frac{1}{2} \\times 5 \\times \\frac{6}{5} = 3$.\nTherefore, when the area of $\\triangle PAB$ is equal to $3$, points $P$ and $O$ are on the same side of $AB$. There are two such points $P$. Answer: B.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 27955,
    "subject": "Mathematics (Multi-modal)",
    "question": "It is given that there are two sets of real numbers $A = \\{a_1, a_2, \\dots, a_{100}\\}$ and $B = \\{b_1, b_2, \\dots, b_{50}\\}$. If there is a mapping $f$ from $A$ to $B$ such that every element in $B$ has an inverse image and\n$$\nf(a_1) \\le f(a_2) \\le \\dots \\le f(a_{100}),\n$$\nthen the number of such mappings is ( ).\n\n(A) $C_{100}^{50}$   (B) $C_{50}^{50}$   (C) $C_{100}^{49}$   (D) $C_{50}^{49}$",
    "options": [],
    "answer": "D",
    "solution": "We might as well suppose $b_1 < b_2 < \\dots < b_{50}$, and divide elements $a_1, a_2, \\dots, a_{100}$ in $A$ into 50 nonempty groups according to their order. Define a mapping $f: A \\to B$, so that the images of all the elements in the $i$-th group are $b_i$ ($i = 1, 2, \\dots, 50$) under the mapping. Obviously, $f$ satisfies the requirements given in the problem. Furthermore, there is a one-to-one correspondence between all groups so divided and the mappings satisfying the condition. So the number of mappings $f$ satisfying the requirements is equal to the number of ways dividing $A$ into 50 groups according to the order of the subscripts. The number of ways dividing $A$ is $C_{99}^{49}$.\n\nThen there are, in all, $C_{99}^{49}$ such mappings. Answer: D.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 27956,
    "subject": "Mathematics (Multi-modal)",
    "question": "In a given trapezium $ABCD$, $AD \\parallel BC$. Suppose $E$ is a variable point on $AB$, $O_1$ and $O_2$ are circumcenters of $\\triangle AED$ and $\\triangle BEC$ respectively. Prove that the length of $O_1O_2$ is a fixed value. (posed by Leng Gangsong)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "**Proof** As shown in the figure, we join $EO_1$ and $EO_2$, then $\\angle AEO_1 = 90^\\circ - \\angle ADE$, $\\angle BEO_2 = 90^\\circ - \\angle BCE$. Hence\n$\\angle O_1EO_2 = \\angle ADE + \\angle ECB$.\n![](attached_image_1.png)\nSince $AD \\parallel BC$, through $E$ constructing a line parallel to $AD$, we can prove $\\angle DEC = \\angle ADE + \\angle BCE$, so $\\angle O_1EO_2 = \\angle DEC$.\nFurther, by the sine rule, we can show\n$$\n\\frac{DE}{EC} = \\frac{2O_1E \\sin A}{2O_2E \\sin B} = \\frac{O_1E}{O_2E}\n$$\nThus $\\triangle DEC \\sim \\triangle O_1EO_2$. Therefore\n$$\n\\frac{O_1O_2}{DC} = \\frac{O_1E}{DE} = \\frac{O_1E}{2O_1E \\sin A} = \\frac{1}{2 \\sin A}\n$$\nSo $O_1O_2 = \\frac{DC}{2\\sin A}$, which is a fixed value. The proposition is proven.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 27957,
    "subject": "Mathematics (Multi-modal)",
    "question": "Assume $P_1, P_2, \\dots, P_n$ ($n \\ge 2$) is an arbitrary permutation of $1, 2, \\dots, n$. Prove that\n$$\n\\frac{1}{P_1 + P_2} + \\frac{1}{P_2 + P_3} + \\dots + \\frac{1}{P_{n-2} + P_{n-1}} + \\frac{1}{P_{n-1} + P_n} > \\frac{n-1}{n+2}.\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "**Proof** By Cauchy's inequality, we can get\n$$\n[(P_1 + P_2) + (P_2 + P_3) + \\dots + (P_{n-1} + P_n)] \\cdot \\left(\\frac{1}{P_1 + P_2} + \\frac{1}{P_2 + P_3} + \\dots + \\frac{1}{P_{n-1} + P_n}\\right) \\ge (n-1)^2.\n$$\nTherefore\n$$\n\\begin{aligned}\n& \\frac{1}{P_1 + P_2} + \\frac{1}{P_2 + P_3} + \\dots + \\frac{1}{P_{n-1} + P_n} \\\\\n\\ge & \\frac{(n-1)^2}{2(P_1 + P_2 + \\dots + P_n) - P_1 - P_n} \\\\\n= & \\frac{(n-1)^2}{n(n+1) - P_1 - P_n} \\\\\n\\ge & \\frac{(n-1)^2}{n(n+1) - 1 - 2} \\\\\n= & \\frac{(n-1)^2}{(n-1)(n+2) - 1} \\\\\n> & \\frac{(n-1)^2}{(n-1)(n+2)} = \\frac{n-1}{n+2}.\n\\end{aligned}\n$$",
    "topic": "Discrete Mathematics",
    "subtopic": "Combinatorics"
  },
  {
    "id": 27958,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all pairs of positive integers (*x*, *y*) satisfying $x^y = y^{x-y}$.",
    "options": [],
    "answer": "(1, 1), (8, 2), (9, 3)",
    "solution": "If $x = 1$, then $y = 1$. If $y = 1$, then $x = 1$.\nIf $x = y$, then $x^y = 1$, so $x = y = 1$.\n\nWe will discuss the circumstances when $x > y \\ge 2$ below. By assumption\n$$\n1 < \\left(\\frac{x}{y}\\right)^y = y^{x-2y},\n$$\nso\n$x > 2y$, and $y \\mid x$.\n\nAssume $x = k y$ then $k \\ge 3$, and\n$$\nk y = y^{(k-2)y}.\n$$\nTherefore\n$$\nk = y^{k-2}.\n$$\nSince $y \\ge 2$, so $y^{k-2} \\ge 2^{k-2}$. By mathematical induction or the binomial theorem, it is easy to prove that when $k \\ge 5$, $2^k > 4k$. Hence $k$ can only be 3 or 4.\nWhen $x = 9$, $y = 3$, we have $k = 3$. When $y = 2$, $x = 8$, we have $k = 4$.\nTherefore, all pairs of positive integers to be found are $(1, 1)$, $(9, 3)$ and $(8, 2)$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 27959,
    "subject": "Mathematics (Multi-modal)",
    "question": "A region is enclosed by the curves $x^2 = 4y$, $x^2 = -4y$, $x = 4$ and $x = -4$. $V_1$ is the volume of the solid obtained by rotating the above region round the y-axis. Another region consists of points $(x, y)$ satisfying $x^2 + y^2 \\le 16$, $x^2 + (y-2)^2 \\ge 4$ and $x^2 + (y+2)^2 \\ge 4$. $V_2$ is the volume of the solid obtained by rotating this region round the y-axis. Then ( ).\n(A) $V_1 = \\frac{1}{2}V_2$\n(B) $V_1 = \\frac{2}{3}V_2$\n(C) $V_1 = V_2$\n(D) $V_1 = 2V_2$\n\n![](attached_image_1.png)\n![](attached_image_2.png)",
    "options": [],
    "answer": "C",
    "solution": "As shown in the diagram, two solids of rotation obtained by rotating respectively two regions round the y-axis lie between two parallel planes, which are 8 units apart. We cut two solids of rotation by any plane which is perpendicular to the y-axis. Suppose the distance from the plane to the origin is $|y| \\le 4$. Then the two sectional areas are\n\n$$\nS_1 = \\pi(4^2 - 4|y|),\n$$\nand\n$$\nS_2 = \\pi(4^2 - y^2) - \\pi[4 - (2 - |y|)^2] = \\pi(4^2 - 4|y|).\n$$\nSo\n$S_1 = S_2.$\nFrom the Zugen Principle (or Cavalieri Principle), we know that the volumes of the two geometric solids are equal, and that is, $V_1 = V_2$. Answer: C.",
    "topic": "Geometry",
    "subtopic": "Solid Geometry"
  },
  {
    "id": 27960,
    "subject": "Mathematics (Multi-modal)",
    "question": "Assume that $n$ is a given positive integer. Find all of the integer groups $(a_1, a_2, \\dots, a_n)$ satisfying the conditions:\n(1) $a_1 + a_2 + \\dots + a_n \\ge n^2$;\n(2) $a_1^2 + a_2^2 + \\dots + a_n^2 \\le n^3 + 1$.",
    "options": [],
    "answer": "(n, n, ..., n)",
    "solution": "Suppose $(a_1, a_2, \\dots, a_n)$ is an integer group which satisfies the conditions. Then by Cauchy's inequality we have\n$$\na_1^2 + \\dots + a_n^2 \\ge \\frac{1}{n} (a_1 + \\dots + a_n)^2 \\ge n^3. \\qquad \\textcircled{1}\n$$\nCombining $a_1^2 + \\dots + a_n^2 \\le n^3 + 1$, we see that it can only be $a_1^2 + \\dots + a_n^2 = n^3$ or $a_1^2 + \\dots + a_n^2 = n^3 + 1$.\n\nIf it is the former, then by the condition for Cauchy's inequality to take the equality sign, we can obtain $a_1 = \\dots = a_n$. This requires $a_1^2 = n^2$, $1 \\le i \\le n$. Combining $a_1 + \\dots + a_n \\ge n^2$, we have $a_1 = \\dots = a_n = n$.\n\nIf it is the latter, then let $b_i = a_i - n$, then we have\n$$\nb_1^2 + b_2^2 + \\dots + b_n^2 = \\sum_{i=1}^{n} a_i^2 - 2n \\sum_{i=1}^{n} a_i + n^3 \\\\\n= 2n^3 + 1 - 2n \\sum_{i=1}^{n} a_i \\le 1.\n$$\nThus $b_1^2$ can only be $0$ or $1$, and there is at the most one among $b_1^2$, $b_2^2$, ..., $b_n^2$ to be $1$. If $b_1^2$, ..., $b_n^2$ all are zero, then $a_i = n$, $\\sum_{i=1}^{n} a_i^2 = n^3 \\ne n^3 + 1$. It leads to a contradiction. If there is just one among $b_1^2$, ..., $b_n^2$ to be $1$, then $\\sum_{i=1}^{n} a_i^2 = n^3 \\pm 2n + 1 \\ne n^3 + 1$. Again, it leads to a contradiction.\n\nConsequently, we obtain that it can only be $(a_1, \\dots, a_n) = (n, \\dots, n)$.",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 27961,
    "subject": "Mathematics (Multi-modal)",
    "question": "An acute triangle $ABC$ has three heights $AD$, $BE$ and $CF$ respectively. Prove that the perimeter of triangle $DEF$ is not over half of the perimeter of triangle $ABC$. (posed by Qi Jianxin)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "**Proof** Since $\\angle ADB = \\angle AEB = 90^\\circ$, so four points $A$, $B$, $D$ and $E$ are concyclic, and furthermore, $AB$ is the diameter. Hence, by the sine rule, we can get\n$$\n\\frac{DE}{\\sin \\angle DAE} = AB = c,\n$$\nso\n$$\nDE = c \\sin \\angle DAE.\n$$\nIn addition, $\\angle DAC + \\angle DCA = 90^\\circ$, therefore,\n$$\nDE = c \\cos C.\n$$\nSimilarly, we can get $DF = b \\cos B$.\nTherefore,\n$$\n\\begin{align*}\nDE + DF &= c \\cos C + b \\cos B \\\\\n&= (2R\\sin C) \\cos C + (2R\\sin B) \\cos B \\\\\n&= R(\\sin 2C + \\sin 2B) \\\\\n&= 2R\\sin (B+C) \\cos (B-C) \\\\\n&= 2R\\sin A \\cos (B-C) \\\\\n&= a \\cos (B-C) \\le a,\n\\end{align*}\n$$\nwhere $R$ is the radius of the circumcircle of $\\triangle ABC$.\nThat is,\n$$\nDE + DF \\le a.\n$$\nSimilarly,\n$$\nDE + EF \\le b \\text{ and } EF + DF \\le c.\n$$\nTherefore,\n$$\nDE + DF + EF \\le \\frac{1}{2}(a+b+c).\n$$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 27962,
    "subject": "Mathematics (Multi-modal)",
    "question": "It is given that complex numbers $z_1$ and $z_2$ satisfy $|z_1| = 2$ and $|z_2| = 3$. If the included angle of their corresponding vectors is $60^\\circ$, then $\\left|\\frac{z_1 + z_2}{z_1 - z_2}\\right| = \\underline{\\hspace{2cm}}$.",
    "options": [],
    "answer": "sqrt(133)/7",
    "solution": "By the cosine rule, we obtain\n$$\n|z_1 + z_2| = \\sqrt{|z_1|^2 + |z_2|^2 - 2|z_1||z_2| \\cos 120^\\circ} = \\sqrt{19},\n$$\nand\n$$\n|z_1 - z_2| = \\sqrt{|z_1|^2 + |z_2|^2 - 2|z_1||z_2| \\cos 60^\\circ} = \\sqrt{7}.\n$$\nTherefore,\n$$\n\\left| \\frac{z_1 + z_2}{z_1 - z_2} \\right| = \\frac{\\sqrt{19}}{\\sqrt{7}} = \\frac{\\sqrt{133}}{7}.\n$$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 27963,
    "subject": "Mathematics (Multi-modal)",
    "question": "Assume that $A_1, A_2, \\dots, A_8$ are eight points taken arbitrarily on a plane. For a directed line $l$ taken arbitrarily on the plane, assume that projections of $A_1, A_2, \\dots, A_8$ on the line are $P_1, P_2, \\dots, P_8$ respectively. If the eight projections are pairwise disjoint, they can be arranged as $P_{i_1}, P_{i_2}, \\dots, P_{i_8}$ according to the direction of line $l$. Thus we get one permutation for $1, 2, \\dots, 8$, namely, $i_1, i_2, \\dots, i_8$. In the figure, this permutation is 2, 1, 8, 3, 7, 4, 6, 5. Assume that after these eight points are projected to every directed line on the plane, we get the number of different permutations as $N_8 = N(A_1, A_2, \\dots, A_8)$. Find the maximal value of $N_8$. (posed by Su Chun)\n\n![](attached_image_1.png)",
    "options": [],
    "answer": "56",
    "solution": "(1) For two parallel and directed lines with the same direction, the order of projections of $A_1, A_2, \\dots, A_8$ must be the same. So, we need only to discuss all directed lines passing through a fixed point $O$.\n\n(2) If a directed line taken is perpendicular to a line joining two given points, then the projections of these two points must coincide, and a corresponding permutation will not be produced. When the directed lines taken are not perpendicular to any line joining two given points, any two projections of $A_1, A_2, \\dots, A_8$ must not coincide. Hence there is a corresponding permutation.\n\n(3) Suppose that the number of lines through point $O$ and perpendicular to a line joining two given points is $k$. Then $k \\le C_8^2 = 28$. Then there arise $2k$ directed lines placed anticlockwise. Assume that they are in order of $l_1, l_2, \\dots, l_{2k}$. For an arbitrary directed line $l$ (different to $l_1, \\dots, l_{2k}$), there must be two consecutive directed lines $l_j$ and $l_{j+1}$ such that $l_j, l, l_{j+1}$ are placed anticlockwise. It is obvious that for given $j$, the corresponding permutations obtained from such $l$ must be the same.\n\n(4) For any two directed lines $l$ and $l'$ different from $l_1, \\dots, l_{2k}$, if we cannot find $j$ such that both $l_j, l, l_{j+1}$ and $l_j, l', l_{j+1}$ satisfy (3). Then there must be $j$ so that $l', l_j, l$ are placed anticlockwise. Assume that $l_j$ is perpendicular to the line joining $A_{j_1}$ and $A_{j_2}$, it is obvious that the orders of the projections of points $A_{j_1}$ and $A_{j_2}$ on the directed lines $l$ and $l'$ must be different, so the corresponding permutations must also be different.\n\n(5) It follows from (3) and (4) that the number of different permutations is $2k$. Note that $k = C_8^2$ is obtainable, so $N_8 = 56$.",
    "topic": "Discrete Mathematics",
    "subtopic": "Combinatorics"
  },
  {
    "id": 27964,
    "subject": "Mathematics (Multi-modal)",
    "question": "Assume that $S = (a_1, a_2, \\dots, a_n)$ consists of $0$ and $1$ and is the longest sequence of numbers which satisfies the following condition: Every two sections of successive $5$ terms in the sequence of numbers $S$ are different, i.e., for arbitrary $1 \\leq i < j \\leq n-4$, $(a_i, a_{i+1}, a_{i+2}, a_{i+3}, a_{i+4})$ and $(a_j, a_{j+1}, a_{j+2}, a_{j+3}, a_{j+4})$ are different. Prove that the first four terms and the last four terms in the sequence are the same.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "**Proof** Noting that $S$ is the longest sequence of numbers satisfying the condition. Hence, if we add a term, $0$ or $1$, after the last term of $S$, there will occur two identical sections of successive $5$ terms in $S$, and that is, there exist $i \\neq j$ such that\n$$\n(a_i, a_{i+1}, \\dots, a_{i+4}) = (a_{n-3}, a_{n-2}, \\dots, a_n, 0),\n$$\n$$\n(a_j, a_{j+1}, \\dots, a_{j+4}) = (a_{n-3}, a_{n-2}, \\dots, a_n, 1).\n$$\nIf $(a_1, a_2, a_3, a_4) \\neq (a_{n-3}, a_{n-2}, a_{n-1}, a_n)$, then\n$$\n1 < i, j < n-3,\\ i \\neq j,\n$$\nand\n$$\n\\begin{aligned}\n(a_i, a_{i+1}, a_{i+2}, a_{i+3}) &= (a_j, a_{j+1}, a_{j+2}, a_{j+3}) \\\\\n&= (a_{n-3}, a_{n-2}, a_{n-1}, a_n).\n\\end{aligned}\n$$\nNow, consider $a_{i-1}$, $a_{j-1}$ and $a_{n-4}$, among which there must be two identical terms. This causes two sections with successive $5$ terms respectively in $S$ to be identical. It leads to a contradiction.\nTherefore, the proposition holds.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 27965,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $\\triangle ABC$ be a triangle. Points $D$ and $E$ are on sides $AB$ and $AC$, respectively, and point $F$ is on line segment $DE$. Let $\\frac{AD}{AB} = x$, $\\frac{AE}{AC} = y$, $\\frac{DF}{DE} = z$. Prove that\n\n(1) $S_{\\triangle BDF} = (1-x)y zS_{\\triangle ABC}$ and $S_{\\triangle CEF} = x(1-y)(1-z)S_{\\triangle ABC}$;\n\n(2) $\\sqrt[3]{S_{\\triangle BDF}} + \\sqrt[3]{S_{\\triangle CEF}} \\le \\sqrt[3]{S_{\\triangle ABC}}$.\n\n(posed by Li Weigu)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Connect $BE$ and $CD$. Then we have\n\n(1) $S_{\\triangle BDF} = zS_{\\triangle BDE} = z(1-x)S_{\\triangle ABE}$\n$= z(1-x)yS_{\\triangle ABC}$ and\n$S_{\\triangle CEF} = (1-z)S_{\\triangle CDE}$\n$= (1-z)(1-y)S_{\\triangle ACD}$\n$= (1-z)(1-y)xS_{\\triangle ABC}$.\n\n(2) From (1) we get\n![](attached_image_1.png)\n$$\n\\begin{align*}\n& \\sqrt[3]{S_{\\triangle BDF}} + \\sqrt[3]{S_{\\triangle CEF}} \\\\\n&= (\\sqrt[3]{(1-x)yz} + \\sqrt[3]{x(1-z)(1-y)})\\sqrt[3]{S_{\\triangle ABC}} \\\\\n&\\le \\left(\\frac{(1-x)+y+z}{3} + \\frac{x+(1-y)+(1-z)}{3}\\right)\\sqrt[3]{S_{\\triangle ABC}} \\\\\n&= \\sqrt[3]{S_{\\triangle ABC}}.\n\\end{align*}\n$$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 27966,
    "subject": "Mathematics (Multi-modal)",
    "question": "![](attached_image_1.png)\nFrom point $P$ outside a circle draw two tangents to the circle touching at points $A$ and $B$. Draw a secant line intersecting the circle at points $C$ and $D$, with $C$ between $P$ and $D$. Choose point $Q$ on the chord $CD$ such that $\\angle DAQ = \\angle PBC$. Prove that $\\angle DBQ = \\angle PAC$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Using $\\angle DAB = \\angle DCB$, $\\angle DAB = \\angle DAQ + \\angle QAB$, $\\angle DCB = \\angle PBC + \\angle BPQ$, and $\\angle DAQ = \\angle PBC$, we get\n\n$\\angle QAB = \\angle BPQ$, so points $P, A, Q, B$ share a common circle. Then $\\angle BQP = \\angle PAB$, that is, $\\angle DBQ + \\angle CDB = \\angle PAC + \\angle CAB$. Since $\\angle CDB = \\angle CAB$, so $\\angle DBQ = \\angle PAC$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 27967,
    "subject": "Mathematics (Multi-modal)",
    "question": "Put numbers $1, 2, 3, 4, 5, 6, 7$ and $8$ at the vertices of a cube, such that the sum of any three numbers on any face is not less than $10$. Find the minimum sum of the four numbers on a face. (posed by Qiu Zonghu)",
    "options": [],
    "answer": "16",
    "solution": "Suppose that the four numbers on a face of the cube are $a_1, a_2, a_3, a_4$ such that their sum reaches the minimum and $a_1 < a_2 < a_3 < a_4$. Since the maximum sum of any three numbers less than $5$ is $9$, we have $a_4 \\ge 6$, and then $a_1 + a_2 + a_3 + a_4 \\ge 16$.\n\n![](attached_image_1.png)\n\nAs seen in the figure, we have $2+3+5+6=16$, and that means the minimum sum of the four numbers on a face is $16$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 27968,
    "subject": "Mathematics (Multi-modal)",
    "question": "Suppose points $I$ and $H$ are the incenter and orthocenter of an acute triangle $ABC$ respectively, and points $B_1$ and $C_1$ are the midpoints of sides $AC$ and $AB$ respectively. It is known that ray $B_1I$ intersects side $AB$ at $B_2$ ($B_2 \\neq B$), and ray $C_1I$ intersects the extension of $AC$ at $C_2$: $B_2C_2$ and $BC$ intersect at $K$ and $A_1$ is the circumcenter of $\\triangle BHC$. Prove that three points $A$, $I$ and $A_1$ are collinear if and only if the areas of $\\triangle BKB_2$ and $\\triangle CKC_2$ are equal. (posed by Shen Wenxuan)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "**Proof** First, we will prove that three points $A$, $I$ and $A_1$ are collinear $\\Leftrightarrow \\angle BAC = 60^\\circ$.\nAs shown in the figure, assume that $O$ is the circumcenter of $\\triangle ABC$. We join $BO$ and $CO$, then\n![](attached_image_1.png)\n$$\n\\angle BHC = 180^\\circ - \\angle BAC,\n$$\n$$\n\\angle BA_1C = 2(180^\\circ - \\angle BHC) = 2\\angle BAC.\n$$\n$$\n\\text{Hence, } \\angle BAC = 60^\\circ \\Leftrightarrow \\angle BAC + \\angle BA_1C = 180^\\circ\n$$\n$$\n\\Leftrightarrow A_1 \\text{ is on the circumcircle } \\odot O \\text{ of } \\triangle ABC\n$$\n$$\n\\Leftrightarrow AI \\text{ and } AA_1 \\text{ coincide (because } A_1 \\text{ is on the perpendicular bisector of } BC. )\n$$\n$$\n\\Leftrightarrow \\text{Three points } A, I \\text{ and } A_1 \\text{ are collinear.}\n$$\n\nSecondly, we will prove $S_{\\triangle BKB_2} = S_{\\triangle CKC_2} \\Leftrightarrow \\angle BAC = 60^\\circ$.\n\nConstruct $IP \\perp AB$ at point $P$, and $IQ \\perp AC$ at $Q$, then\n$$\nS_{\\triangle AB_1 B_2} = \\frac{1}{2} IP \\cdot AB_2 + \\frac{1}{2} IQ \\cdot AB_1.\n$$\nNote that $S_{\\triangle AB_1 B_2} = \\frac{1}{2} AB_1 \\cdot AB_2 \\cdot \\sin A$,\ntherefore $IP \\cdot AB_2 + IQ \\cdot AB_1 = AB_1 \\cdot AB_2 \\cdot \\sin A$.\nAssume $IP = r$, where $r$ is the radius of the inscribed circle of $\\triangle ABC$. Then $IQ = r$. Furthermore set $BC = a$, $CA = b$ and $AB = c$, then $r = \\frac{2S_{\\triangle ABC}}{a+b+c}$.\nFrom $AB_1 = \\frac{b}{2}$ and $2AB_1 \\cdot \\sin A = h_c = \\frac{2S_{\\triangle ABC}}{c}$,\nwe have $AB_2 \\cdot \\left[ \\frac{2S_{\\triangle ABC}}{c} - 2 \\cdot \\frac{2S_{\\triangle ABC}}{a+b+c} \\right] = b \\cdot \\frac{2S_{\\triangle ABC}}{a+b+c}$,\nso $AB_2 = \\frac{bc}{a+b+c}$.\nSimilarly, $AC_2 = \\frac{bc}{a+c-b}$.\nHence,\n$$\nS_{\\triangle BKB_2} = S_{\\triangle CKC_2}\n$$\n$$\n\\Leftrightarrow S_{\\triangle ABC} = S_{\\triangle AB_2 C_2}\n$$\n$$\n\\Leftrightarrow bc = \\frac{bc}{a+b-c} \\cdot \\frac{bc}{a+c-b}\n$$\n$$\n\\Leftrightarrow a^2 = b^2 + c^2 - bc\n$$\n$$\n\\Leftrightarrow \\angle BAC = 60^\\circ \\text{ (By the law of cosines).}\n$$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 27969,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $A$ be a subset of the set $S = \\{1, 2, \\dots, 1000000\\}$ containing exactly $101$ elements. Prove that there exist numbers $t_1, t_2, \\dots, t_{100}$ in $S$ such that the sets\n$$\nA_j = \\{x + t_j \\mid x \\in A\\} \\text{ for } j = 1, 2, \\dots, 100\n$$\nare pairwise disjoint.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Consider the set $D = \\{x-y \\mid x, y \\in A\\}$. There are at most $101 \\times 100 + 1 = 10101$ elements in $D$. Two sets $A_i$ and $A_j$ have nonempty intersection if and only if $t_i - t_j$ is in $D$. So we need to choose the $100$ elements in such a way that the difference for any two elements is not in $D$.\n\nNow select these elements by induction. Choose one element arbitrarily from $S$. Assume that $k$ elements, $k \\le 99$, are already chosen. An element $x$ in $S$ that is already chosen prevents us from selecting any element from the set $x+D$, where $x+D = \\{x+y \\mid y \\in D\\}$. Thus after $k$ elements are chosen, at most $10101k$ elements are forbidden. Since $S$ has $1000000$ elements, and $10101 \\times 99 = 1000000 - 99$, there is always at least one element left to choose for the next $t_{k+1}$. Therefore, we can choose $100$ such elements $t_1, t_2, \\dots, t_{100}$ so that the sets $A_j$ are pairwise disjoint.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 27970,
    "subject": "Mathematics (Multi-modal)",
    "question": "A new sequence is obtained from the sequence of the positive integers $\\{1, 2, 3, \\ldots\\}$ by deleting all the perfect squares. Then the $2\\ 003$rd term of the new sequence is ( ).\n\n(A) $2\\ 046$\n(B) $2\\ 047$\n(C) $2\\ 048$\n(D) $2\\ 049$",
    "options": [],
    "answer": "C",
    "solution": "Since $\\sqrt{2046} = \\sqrt{2047} = \\sqrt{2048} = \\sqrt{2049} = 45$, and $2003 + 45 = 2048$, Answer: C.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 27971,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $a$, $b$, $c$, $d$ be positive integers and $\\log_a b = \\frac{3}{2}$, $\\log_c d = \\frac{5}{4}$. If $a-c=9$, then $b-d=$ ________.",
    "options": [],
    "answer": "93",
    "solution": "We have $b = a^{\\frac{3}{2}}$, $d = c^{\\frac{5}{4}}$ from the assumption. We may assume that $a = x^2$, $c = y^4$ with $x$ and $y$ being positive integers, since $a$, $b$, $c$, $d$ are all positive integers. Then $a-c = x^2 - y^4 = (x-y^2)(x+y^2) = 9$. It follows that $(x-y^2, x+y^2) = (1, 9)$. So we obtain the solution $x=5$, $y=2$. That is, $b-d = x^3 - y^5 = 125-32=93$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 27972,
    "subject": "Mathematics (Multi-modal)",
    "question": "8 balls of radius $1$ are placed in a cylinder in two layers, with each layer containing $4$ balls. Each ball is in contact with $2$ balls in the same layer, $2$ balls in the other layer, one base and the lateral surface of the cylinder. Then the height of the cylinder is ________.",
    "options": [],
    "answer": "sqrt(8) + 2",
    "solution": "As in the diagram, let $A$, $B$, $C$, $D$ be the centers of the $4$ balls in the bottom layer, and $A'$, $B'$, $C'$, $D'$ the centers of the $4$ balls in the upper layer. Then $A$, $B$, $C$, $D$ and $A'$, $B'$, $C'$, $D'$ are the $4$ vertices of squares of length $2$, respectively. Now, the circumscribed circles with centers $O$ and $O'$ of the squares constitute the bases of another cylinder, and the projecting point of $A'$ on the bottom base is the middle point $M$ of arc $AB$.\n\n![](attached_image_1.png)\n\nIn $\\triangle A'A'B$, we have $A'A = A'B = AB = 2$, then $A'N = \\sqrt{3}$, where $N$ is the middle point of $AB$. Meanwhile, $OM = OA = \\sqrt{2}$, $ON = 1$, so\n$$\nMN = \\sqrt{2} - 1, \\quad A'M = \\sqrt{(A'N)^2 - (MN)^2} = \\sqrt{8}.\n$$\nThen the height of the original cylinder is $\\sqrt{8} + 2$.",
    "topic": "Geometry",
    "subtopic": "Solid Geometry"
  },
  {
    "id": 27973,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $M_n = \\{0, a_1a_2\\cdots a_n \\mid a_i = 0 \\text{ or } 1, 1 \\le i \\le n-1, a_n = 1\\}$ be a set of decimal fractions, $T_n$ and $S_n$ be the number and the sum of the elements in $M_n$ respectively. Then\n$$\n\\lim_{n \\to \\infty} \\frac{S_n}{T_n} = \\underline{\\hspace{2cm}}.\n$$",
    "options": [],
    "answer": "1/18",
    "solution": "Since $a_1, a_2, \\dots, a_{n-1}$ all have exactly two possible values, so $T_n = 2^{n-1}$. Meanwhile, the frequency of $a_i = 1$ is the same as that of $a_i = 0$ for $1 \\le i \\le n-1$, and $a_n = 1$. Then\n$$\n\\begin{aligned}\nS_n &= \\frac{1}{2} \\times 2^{n-1} \\times \\left( \\frac{1}{10} + \\frac{1}{10^2} + \\dots + \\frac{1}{10^{n-1}} \\right) + 2^{n-1} \\times \\frac{1}{10^n} \\\\\n&= 2^{n-1} \\times \\frac{1}{18} \\times \\left( 1 - \\frac{1}{10^{n-1}} \\right) + 2^{n-1} \\times \\frac{1}{10^n}.\n\\end{aligned}\n$$\n\nSo\n$$\n\\lim_{n \\to \\infty} \\frac{S_n}{T_n} = \\lim_{n \\to \\infty} \\left[ \\frac{1}{18} \\left( 1 - \\frac{1}{10^{n-1}} \\right) + \\frac{1}{10^n} \\right] = \\frac{1}{18}.\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 27974,
    "subject": "Mathematics (Multi-modal)",
    "question": "Suppose $\\frac{3}{2} \\le x \\le 5$. Prove that $2\\sqrt{x+1} + \\sqrt{2x-3} + \\sqrt{15-3x} < 2\\sqrt{19}$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "By Cauchy's inequality, we have\n$$\n\\begin{aligned}\n& 2\\sqrt{x+1} + \\sqrt{2x-3} + \\sqrt{15-3x} \\\\\n&= \\sqrt{x+1} + \\sqrt{x+1} + \\sqrt{2x-3} + \\sqrt{15-3x} \\\\\n&\\le \\sqrt{[(x+1) + (x+1) + (2x-3) + (15-3x)](1^2 + 1^2 + 1^2 + 1^2)} \\\\\n&= 2\\sqrt{x+14} \\le 2\\sqrt{19},\n\\end{aligned}\n$$\nand the equality holds if and only if $\\sqrt{x+1} = \\sqrt{2x-3} = \\sqrt{15-3x}$ and $x = 5$. But this is impossible. So $2\\sqrt{x+1} + \\sqrt{2x-3} + \\sqrt{15-3x} < 2\\sqrt{19}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 27975,
    "subject": "Mathematics (Multi-modal)",
    "question": "Suppose $A$, $B$, $C$ are three non-collinear points corresponding to complex numbers $z_0 = ai$, $z_1 = \\frac{1}{2} + bi$, $z_2 = 1 + ci$ ($a$, $b$ and $c$ being real numbers), respectively. Prove that the curve\n$$\nz = z_0 \\cos^4 t + 2z_1 \\cos^2 t \\cdot \\sin^2 t + z_2 \\sin^4 t \\ (t \\in \\mathbb{R})\n$$\nshares a single common point with the line bisecting $AB$ and parallel to $AC$ in $\\triangle ABC$, and find this point.",
    "options": [],
    "answer": "1/2 + ((a+c+2b)/4)i",
    "solution": "Let $z = x + yi$ ($x, y \\in \\mathbb{R}$), then\n$$\nx + yi = a\\cos^4 t \\cdot i + 2\\left(\\frac{1}{2} + bi\\right)\\cos^2 t \\cdot \\sin^2 t + (1 + ci)\\sin^4 t.\n$$\nSeparating real and imaginary parts, we get\n$$\n\\begin{aligned}\nx &= \\cos^2 t \\cdot \\sin^2 t + \\sin^4 t = \\sin^2 t, \\\\\ny &= a(1-x)^2 + 2b(1-x)x + c x^2 \\\\\n&\\quad (0 \\le x \\le 1).\n\\end{aligned}\n$$\n\n![](attached_image_1.png)\nThat is, $y = (a+c-2b)x^2 + 2(b-a)x + a$ ($0 \\le x \\le 1$)\n$$\na \\quad (0 \\le x \\le 1) \\tag{1}\n$$\nSince $A$, $B$, $C$ are non-collinear, $a+c-2b \\ne 0$. So Equation (1) is the segment of a parabola (see the diagram). Furthermore, the midpoints of $AB$ and $BC$ are $D(\\frac{1}{4}, \\frac{a+b}{2})$ and $E(\\frac{3}{4}, \\frac{b+c}{2})$, respectively. So the equation of line $DE$ is $y = (c-a)x + \\frac{1}{4}(3a+2b-c)$. (2)\nSolving Equations (1) and (2) simultaneously, we get $(a+c-2b)(x-\\frac{1}{2})^2 = 0$. Then $x = \\frac{1}{2}$, since $a+c-2b \\ne 0$. So the parabola and line $DE$ have one and only one common point $P(\\frac{1}{2}, \\frac{a+c+2b}{4})$. Notice that $\\frac{1}{4} < \\frac{1}{2} < \\frac{3}{4}$, so point $P$ is on the segment $DE$ and satisfies Equation (1), as required.\nWe can solve the problem using the method of complex numbers directly. Let $D$, $E$ be the midpoints of $AB$, $CB$, respectively. Then the complex numbers corresponding to $D$, $E$ are $\\frac{1}{2}(z_0 + z_1) = \\frac{1}{4} + \\frac{a+b}{2}i$, $\\frac{1}{2}(z_1 + z_2) = \\frac{3}{4} + \\frac{b+c}{2}i$, respectively. So, complex number $z$ corresponding to a point on the segment $DE$ satisfies\n$$\nz = \\lambda\\left(\\frac{1}{4} + \\frac{a+b}{2}i\\right) + (1-\\lambda)\\left(\\frac{3}{4} + \\frac{b+c}{2}i\\right), \\quad 0 \\le \\lambda \\le 1.\n$$\n\nSubstitute the above expression into the equation of the curve\n$$\nz = z_0 \\cos^4 t + 2z_1 \\cos^2 t \\cdot \\sin^2 t + z_2 \\sin^4 t,\n$$\nand separate the real and imaginary parts from both sides to give the\nfollowing two equations,\n$$\n\\begin{cases} \n\\frac{3}{4} - \\frac{\\lambda}{2} = \\sin^2 t \\cos^2 t + \\sin^4 t, \\\\ \n\\frac{1}{2}[\\lambda a + b(1 - \\lambda)c] = a \\cos^4 t + 2b \\sin^2 t \\cos^2 t + c \\sin^4 t. \n\\end{cases}\n$$\nEliminating $\\lambda$ from the equations, we get\n$$\n\\begin{aligned} \n\\frac{3}{4}(a-c) + \\frac{b+c}{2} &= a\\cos^4 t + (2b+a-c)\\sin^2 t \\cos^2 t + a\\sin^4 t \\\\ \n&= a(1 - 2\\sin^2 t \\cos^2 t) + (2b+a-c)\\sin^2 t \\cos^2 t \\\\ \n&= a + (2b-a-c)\\sin^2 t \\cos^2 t. \n\\end{aligned}\n$$\nThen $(2b-a-c)(\\sin^2 t \\cos^2 t - \\frac{1}{4}) = 0$. Since $A$, $B$, $C$ are non-collinear, we know that $z_1 \\neq \\frac{1}{2}(z_0 + z_2)$. So $2b-a-c \\neq 0$. Then $\\sin^2 t \\cos^2 t = \\sin^2 t(1 - \\sin^2 t) = \\frac{1}{4}$, that is, $(\\sin^2 t - \\frac{1}{2})^2 = 0$. Then we have $\\frac{3}{4} - \\frac{\\lambda}{2} = \\frac{1}{4} + (\\frac{1}{2})^2 = \\frac{1}{2}$, so $\\lambda = \\frac{1}{2} \\in [0, 1]$. That means that the curve and the line $DE$ have one and only one common point, and the complex number corresponding to this common point is\n$$\nz = \\frac{1}{2} \\left( \\frac{1}{4} + \\frac{a+b}{2}i \\right) + \\frac{1}{2} \\left( \\frac{3}{4} + \\frac{b+c}{2}i \\right) = \\frac{1}{2} + \\frac{a+c+2b}{4}i.\n$$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 27976,
    "subject": "Mathematics (Multi-modal)",
    "question": "A circle with center $O$ and radius $R$ is drawn on a paper, and $A$ is a given point in the circle with $OA = a$. Fold the paper to make a point $A'$ on the circumference coincident with point $A$, then a crease line is left on the paper. Find out the set of all points on such crease lines, when $A'$ goes through every point on the circumference.",
    "options": [],
    "answer": "All points (x, y) satisfying ((2x - a)^2)/R^2 + (4y^2)/(R^2 - a^2) ≥ 1; equivalently, the set is the boundary and exterior of the ellipse ((2x - a)^2)/R^2 + (4y^2)/(R^2 - a^2) = 1.",
    "solution": "Establish an $xy$-coordinate system as in the diagram with $A(a, 0)$ given. Then the crease line $MN$ is the perpendicular bisector of segment $AA'$ when $A'$ ($R\\cos a$, $R\\sin a$) is made coincident with $A$ by folding the paper. Let $P(x, y)$ be any point on $MN$, then $|PA'| = |PA|$. That is\n\n![](attached_image_1.png)\n\n$$(x - R\\cos a)^2 + (y - R\\sin a)^2 = (x - a)^2 + y^2.$$  \n\nThen\n$$\n\\frac{x\\cos x + y\\sin a}{\\sqrt{x^2 + y^2}} = \\frac{R^2 - a^2 + 2ax}{2R\\sqrt{x^2 + y^2}}.\n$$\n\nWe get\n$$\n\\sin(\\theta + a) = \\frac{R^2 - a^2 + 2ax}{2R\\sqrt{x^2 + y^2}},\n$$\nwhere\n$$\n\\sin \\theta = \\frac{x}{\\sqrt{x^2 + y^2}}, \\quad \\cos \\theta = \\frac{y}{\\sqrt{x^2 + y^2}}.\n$$\nSo\n$$\n\\left| \\frac{R^2 - a^2 + 2ax}{2R\\sqrt{x^2 + y^2}} \\right| \\le 1.\n$$\nSquaring both sides, we get\n$$\n\\frac{(2x-a)^2}{R^2} + \\frac{4y^2}{R^2 - a^2} \\ge 1.\n$$\nSo the set we want consists of all of the points on the border of or outside the ellipse $\\frac{(2x-a)^2}{R^2} + \\frac{4y^2}{R^2 - a^2} = 1$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 27977,
    "subject": "Mathematics (Multi-modal)",
    "question": "Suppose a set $S$ satisfies the following conditions:\n(1) every element in $S$ is a positive integer and not greater than $100$;\n(2) for any two different elements $a$ and $b$ in $S$, there is an element $c$ in $S$ such that the greatest common divisor of $a$ and $c$ is equal to $1$, and the greatest common divisor of $b$ and $c$ is also $1$; and\n(3) for any two different elements $a$ and $b$ in $S$, there is an element $d$, which is different from $a$ and $b$, such that the greatest common divisor of $a$ and $d$, and that of $b$ and $d$ are greater than $1$.\n\nFind the maximum number of elements in $S$.",
    "options": [],
    "answer": "72",
    "solution": "The maximum number of elements is $72$.\n\nA positive integer not greater than $100$ can be written as\n$$\nn = 2^{a_1} \\cdot 3^{a_2} \\cdot 5^{a_3} \\cdot 7^{a_4} \\cdot 11^{a_5} \\cdot q,\n$$\nwhere $q$ is a positive integer and not divisible by $2$, $3$, $5$, $7$ and $11$, and $a_1, a_2, a_3, a_4$ and $a_5$ are nonnegative integers.\n\nWe pick out those positive integers $n$ with just one or two nonzero among $a_1, a_2, a_3, a_4$ and $a_5$ to form set $S$. In this case, $S$ contains $50$ even numbers ($2, 4, \\ldots, 98$ and $100$) except the following seven: $2 \\times 3 \\times 5$, $2^2 \\times 3 \\times 5$, $2 \\times 3^2 \\times 5$, $2 \\times 3 \\times 7$, $2^2 \\times 3 \\times 7$, $2 \\times 5 \\times 7$ and $2 \\times 3 \\times 11$; $17$ odd numbers that are multiples of three (i.e. $3 \\times 1, 3 \\times 3, \\ldots, 3 \\times 33$), $7$ odd numbers with the least prime divisor $5$ (i.e. $5 \\times 1, 5 \\times 5, 5 \\times 7, 5 \\times 11, 5 \\times 13, 5 \\times 17$ and $5 \\times 19$), $4$ odd numbers with the least prime divisor $7$ (i.e. $7 \\times 1, 7 \\times 7, 7 \\times 11$ and $7 \\times 13$), and the prime number $11$.\n\nConsequently, $S$ contains $(50-7)+17+7+4+1=72$ numbers totally.\n\nIn what follows, we will prove that $S$ constructed above satisfies the given condition.\n\nObviously, it satisfies condition (1).\n\nFor condition (2), we note that, at most, four prime divisors among $2, 3, 5, 7$ and $11$ will occur in $[a, b]$. We write the prime which does not occur as $p$. Obviously, $p \\in S$ and\n$$\n(p, a) \\le (p, [a, b]) = 1, \\\\\n(p, b) \\le (p, [a, b]) = 1.\n$$\nHence, we take $c = p$.\n\nFor condition (3), we take the least prime divisor of $a$ as $p$ and the one of $b$ as $q$ when $(a, b) = 1$. It is easy to see that $p \\ne q$ and $p, q \\in \\{2, 3, 5, 7, 11\\}$. Hence $pq \\in S$, and\n$$\n(pq, a) \\ge p > 1 \\text{ and } (pq, b) \\ge q > 1.\n$$\nBeing coprime to each other for $a$ and $b$ ensures that $pq$ is different from $a$ and $b$. Thus we take $d = pq$.\n\nWhen $(a, b) = e > 1$, we take $p$ as the least prime divisor of $e$, and $q$ as the smallest prime number satisfying $q \\nmid [a, b]$. It is easy to see that $p \\ne q$, and $p$ and $q \\in \\{2, 3, 5, 7, 11\\}$. Hence $pq \\in S$, and\n$$\n(pq, a) \\ge (p, a) = p > 1, \\\\\n(pq, b) \\ge (p, b) = p > 1.\n$$\n$q \\nmid [a, b]$ ensures that $pq$ is different from $a$ and $b$. Thus, we take $d = pq$.\n\nIn what follows, we prove that the number of elements in $S$, which satisfies the conditions described in the problem, will not be greater than $72$.\n\nObviously, $1 \\notin S$. For arbitrary two prime numbers $p$ and $q$ which are both greater than $10$, since the least number which is not prime to neither $p$ nor $q$ is $pq$, it must be greater than $100$. So we know, according to condition (3), that there is at most one among $21$ prime numbers between $10$ and $100$ ($11, 13, \\ldots, 89, 97$), occurring in $S$. We write the set consisting of all natural numbers not greater than $100$ except $1$ and the above-mentioned $21$ prime numbers as $T$, and there are $78$ numbers in the set. We can conclude that there are at least $7$ numbers in $T$ are not in $S$. Thus $S$ contains at most $78 - 7 + 1 = 72$ elements.\n\ni. When a prime number $p$ is greater than $10$ and belongs to $S$,\n\nevery number in $S$ can only have $2, 3, 5, 7$ and $p$ as its least prime divisor. By condition (2), we have the following conclusions.\n\n① If $7p \\in S$, because $\\{2 \\times 3 \\times 5, 2^2 \\times 3 \\times 5, 2 \\times 3^2 \\times 5, 7p\\}$ contains all the least prime divisors, we know from condition (2) that $2 \\times 3 \\times 5, 2^2 \\times 3 \\times 5$ and $2 \\times 3^2 \\times 5$ do not belong to $S$. If $7p \\notin S$, noting $2 \\times 7p > 100$, but $p \\in S$, so from condition (2) we know that $7 \\times 1, 7 \\times 7, 7 \\times 11$ and $7 \\times 13$ do not belong to $S$.\n\n② If $5p \\in S$, then $2 \\times 3 \\times 7$ and $2^2 \\times 3 \\times 7$ do not belong to $S$. If $5p \\notin S$, then $5 \\times 1$ and $5 \\times 5$ do not belong to $S$.\n\n③ $2 \\times 5 \\times 7$ and $3p$ do not belong to $S$ at the same time.\n\n④ $2 \\times 3p$ and $5 \\times 7$ do not belong to $S$ at the same time.\n\n⑤ If $5p, 7p \\notin S$, then $5 \\times 7 \\notin S$.\n\nWhen $p = 11$ or $13$, from ①, ②, ③ and ④, we can get at least $3, 2, 1$ and $1$ numbers in $T$ respectively which do not belong to $S$, and in total $7$ numbers. When $p = 17$ or $19$, from ①, ② and ③, we can get at least $4, 2$ and $1$ numbers in $T$ respectively, which do not belong to $S$ and in total $7$ numbers. When $p > 20$, from ①, ② and ③, there are at least $4, 2$ and $1$ numbers in $T$ respectively, which do not belong to $S$ and in total $7$ numbers also.\n\nii. If there is no prime number greater than $10$ belonging to $S$, then the least prime numbers in $S$ can only be $2, 3, 5$ and $7$. Hence, each of the following $7$ pairs of numbers can not belong to $S$ at the same time:\n$$\n(3, 2 \\times 5 \\times 7), (5, 2 \\times 3 \\times 7), (7, 2 \\times 3 \\times 5), (2 \\times 3, 5 \\times 7), (2 \\times 5, 3 \\times 7), (2 \\times 7, 3 \\times 5), (2^2 \\times 7, 3^2 \\times 5).\n$$\nThus, there are at least $7$ numbers in $T$ that are not in $S$.\n\nConsequently, the answer for this problem is $72$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 27978,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let the three sides of a triangle be integers $l$, $m$, $n$, respectively, satisfying $l > m > n$ and $\\{\\frac{3^l}{10^4}\\} = \\{\\frac{3^m}{10^4}\\} = \\{\\frac{3^n}{10^4}\\}$, where $\\{x\\} = x - [x]$ and $[x]$ denotes the integral part of the number $x$. Find the minimum perimeter of such a triangle.",
    "options": [],
    "answer": "3003",
    "solution": "Since\n$$\n\\frac{3^l}{10^4} - \\left\\lfloor \\frac{3^l}{10^4} \\right\\rfloor = \\frac{3^m}{10^4} - \\left\\lfloor \\frac{3^m}{10^4} \\right\\rfloor = \\frac{3^n}{10^4} - \\left\\lfloor \\frac{3^n}{10^4} \\right\\rfloor,\n$$\nwe have\n$$\n\\begin{align}\n3^l &\\equiv 3^m \\equiv 3^n \\pmod{10^4} \\\\\n&\\iff \\begin{cases}\n3^l \\equiv 3^m \\equiv 3^n \\pmod{2^4}, \\\\\n3^l \\equiv 3^m \\equiv 3^n \\pmod{5^4}.\n\\end{cases} \\tag{1}\n\\end{align}\n$$\nAs $(3, 2) = 1$, we then have from $(1)$ $3^{l-n} \\equiv 3^{m-n} \\equiv 1 \\pmod{2^4}$.\nLet $u$ be the minimum positive integer satisfying $3^u \\equiv 1 \\pmod{2^4}$.\nThen for every positive integer $v$ satisfying $3^v \\equiv 1 \\pmod{2^4}$, we must have $u \\mid v$. Otherwise, if $u \\nmid v$, then using division with a remainder we could get two non-negative integers $a$ and $b$ satisfying $v = au + b$ with $0 < b < u$. Then $3^b \\equiv 3^{au+b} \\equiv 3^v \\equiv 1 \\pmod{2^4}$, contradicting the definition of $u$. Therefore $u \\mid v$.\nNotice that\n$$\n3 \\equiv 3 \\pmod{2^4},\\ 3^2 \\equiv 9 \\pmod{2^4},\n$$\n$$\n3^3 \\equiv 27 \\equiv 11 \\pmod{2^4},\\ 3^4 \\equiv 1 \\pmod{2^4},\n$$\nthen $u = 4$. We may assume that $m-n = 4k$, where $k$ is a positive integer.\nIn the same way, we get from $(2)$ $3^{m-n} \\equiv 1 \\pmod{5^4}$, that is,\n$3^{4k} \\equiv 1 \\pmod{5^4}$.\nNow, we are going to find number $k$. As $3^{4k}-1 = (1+5 \\times 2^4)^k - 1 = 0 \\pmod{5^4}$, i.e.\n$$\n\\begin{aligned}\n& 5k \\times 2^4 + \\frac{k(k-1)}{2} \\times 5^2 \\times 2^8 + \\frac{k(k-1)(k-2)}{6} \\times 5^3 \\times 2^{12} \\\\\n&= 5k + 5^2 k [3 + (k-1) \\times 2^7] + \\frac{k(k-1)(k-2)}{3} \\times 5^3 \\times 2^{11} \\\\\n&= 0 \\pmod{5^4},\n\\end{aligned}\n$$\nso $k = 5t$. Substituting in the above expression, we get\n$$\nt + 5t[3 + (5t - 1) \\times 2^7] \\equiv 0 \\pmod{5^2}.\n$$\nThen $k = 5t = 5^3 s$, and $m-n = 500s$, where $s$ is a positive integer.\nIn the same way, we get $l-n = 500r$, where $r$ is a positive integer and $r > s$ since $l > m > n$.\nSo the three sides of the required triangle are $l = 500r + n$, $m = 500s + n$ and $n$, respectively, satisfying $n > l-n = 500(r-s)$. When $s=1$, $r=2$ the perimeter reaches the minimum which equals $(1000+501)+(500+501)+501 = 3003$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 27979,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $a_1, a_2, \\dots, a_{2n}$ be real numbers with $\\sum_{i=1}^{2n-1} (a_{i+1} - a_i)^2 = 1$. Find the maximum value of $(a_{n+1} + a_{n+2} + \\dots + a_{2n}) - (a_1 + a_2 + \\dots + a_n)$. (posed by Leng Gangsong)",
    "options": [],
    "answer": "sqrt(n(2n^2+1)/3)",
    "solution": "First, for $n=1$, we have $(a_2 - a_1)^2 = 1$, $a_2 - a_1 = \\pm 1$. Then the maximum value of $a_2 - a_1$ is $1$.\n\nSecondly, for $n \\ge 2$, let $x_1 = a_1$, $x_{i+1} = a_{i+1} - a_i$, $i = 1, 2, \\dots, 2n-1$. Then $\\sum_{i=2}^{2n} x_i^2 = 1$, and $a_k = x_1 + \\dots + x_k$, $k = 1, 2, \\dots, 2n$. Using Cauchy's Inequality, we have\n$$\n\\begin{aligned}\n& (a_{n+1} + a_{n+2} + \\dots + a_{2n}) - (a_1 + a_2 + \\dots + a_n) \\\\\n&= n(x_1 + \\dots + x_n) + n x_{n+1} + (n-1)x_{n+2} + \\dots + x_{2n} \\\\\n&\\quad - [n x_1 + (n-1)x_2 + \\dots + x_n]\n\\end{aligned}\n$$\n\n$$\n\\begin{align*}\n&= x_2 + 2x_3 + \\cdots + (n-1)x_n + n x_{n+1} + (n-1)x_{n+2} + \\cdots + x_{2n} \\\\\n&\\le [1^2 + 2^2 + \\cdots + (n-1)^2 + n^2 + (n-1)^2 + \\cdots + 1^2]^{\\frac{1}{2}} (x_2^2 + x_3^2 + \\cdots + x_{2n}^2)^{\\frac{1}{2}} \\\\\n&= \\left[ n^2 + 2 \\times \\frac{1}{6}(n-1)n(2n-1) \\right]^{\\frac{1}{2}} = \\sqrt{\\frac{n(2n^2+1)}{3}}.\n\\end{align*}\n$$\n\nThe equality holds when\n$$\n\\begin{gather*}\na_k = \\frac{\\sqrt{3k(k-1)}}{2\\sqrt{n(2n^2+1)}}, \\quad a_{n+k} = \\frac{\\sqrt{3}[2n^2 - (n-k)(n-k+1)]}{2\\sqrt{n(2n^2+1)}}, \\\\\nk = 1, 2, \\dots, n.\n\\end{gather*}\n$$\nSo the maximum value of $(a_{n+1} + a_{n+2} + \\cdots + a_{2n}) - (a_1 + a_2 + \\cdots + a_n)$ is $\\sqrt{\\frac{n(2n^2+1)}{3}}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 27980,
    "subject": "Mathematics (Multi-modal)",
    "question": "There are 47 students in a classroom with seats arranged in $6$ rows $\times$ $8$ columns, and the seat in the $i$-th row and $j$-th column is denoted by $(i, j)$. Now, an adjustment is made for students' seats in the new school term. For a student with the original seat $(i, j)$, if his/her new seat is $(m, n)$, we say that the student is moved by $[a, b] = [i-m, j-n]$ and define the position value of the student as $a+b$. Let $S$ denote the sum of the position values of all the students. Determine the difference between the greatest and smallest possible values of $S$. (posed by Chen Yonggao)",
    "options": [],
    "answer": "12",
    "solution": "Add a virtual student $A$ so that every seat is occupied by exactly one student. Denote $S'$ the sum of the position values in this situation. Notice that an exchange of two students occupying the adjacent seats will not change the value of $S'$. Every student can return to his/her original seat by a finite number of such exchanges of adjacent students. Then $S' = 0$. Since $S' = S + a_A + b_A$, where $a_A + b_A$ is the position value of student $A$, then we have $S$ is the greatest when student $A$ occupies seat $(1, 1)$, and $S$ is the smallest when $A$ occupies seat $(6, 8)$. So the difference between the greatest and the smallest possible values of $S$ is $14$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 27981,
    "subject": "Mathematics (Multi-modal)",
    "question": "Determine all pairs of positive integers $(a, b)$ such that\n$$\n\\frac{a^2}{2ab^2 - b^3 + 1}\n$$\nis a positive integer.",
    "options": [],
    "answer": "(a, b) = (2l, 1), (l, 2l), or (8l^4 - l, 2l) for any positive integer l",
    "solution": "**Solution I** Let $(a, b)$ be a pair of positive integers satisfying the condition. Since $k = \\frac{a^2}{2ab^2 - b^3 + 1} > 0$, we have $2ab^2 - b^3 + 1 > 0$ or $a > \\frac{b}{2} - \\frac{1}{2b^2}$, and hence $a \\ge \\frac{b}{2}$. Using this, we infer from $k \\ge 1$, or $a^2 \\ge b^2(2a-b)+1$, that $a^2 > b^2(2a-b) \\ge 0$. Hence\n$$\na > b \\text{ or } 2a = b. \\qquad \\textcircled{1}\n$$\nNow consider the two solutions $a_1, a_2$ of the equation\n$$\na^2 - 2kb^2a + k(b^3 - 1) = 0 \\qquad \\textcircled{2}\n$$\nfor any fixed positive integers $k$ and $b$, and assume that one of them is an integer. Then the other is also an integer because $a_1 + a_2 = 2kb^2$. We may assume that $a_1 \\ge a_2$, and we have $a_1 \\ge kb^2 > 0$. Furthermore, since $a_1 a_2 = k(b^3 - 1)$, we get\n$$\n0 \\le a_2 = \\frac{k(b^3 - 1)}{a_1} \\le \\frac{k(b^3 - 1)}{kb^2} < b.\n$$\nTogether with (1), we conclude that $a_2 = 0$ or $a_2 = \\frac{b}{2}$ (in the latter case $b$ must be even).\nIf $a_2 = 0$, then $b^3 - 1 = 0$, and hence $a_1 = 2k$ and $b = 1$.\nIf $a_2 = \\frac{b}{2}$, then $k = \\frac{b^2}{4}$ and $a_1 = \\frac{b^4}{2} - \\frac{b}{2}$.\nTherefore the only possibilities are\n$$\n(a, b) = (2l, 1), (l, 2l) \\text{ or } (8l^4 - l, 2l)\n$$\nfor some positive integer $l$. All of these pairs satisfy the given condition.\n\n\n**Solution II** If $b=1$, it follows from the given condition that $a$ must be even.\nLet $\\frac{a^2}{2ab^2 - b^3 + 1} = k$. If $b > 1$, then there are two solutions to the equation (②) and one of them is a positive integer. Thus the discriminant $\\Delta$ of the equation (②) is a perfect square, that is $\\Delta = 4k^2b^4 - 4k(b^3 - 1)$ is a perfect square.\nNote that, if $b \\ge 2$, we have\n$$\n(2kb^2 - b - 1)^2 < \\Delta < (2kb^2 - b + 1)^2. \\quad \\textcircled{3}\n$$\nThe proof is given as follows,\n$$\n\\begin{align*} \\Delta - (2kb^2 - b - 1)^2 &= 4kb^2 - b^2 - 2b + 4k - 1 \\\\ &= (4k - 1)(b^2 + 1) - 2b \\\\ &\\ge 2(4k - 1)b - 2b > 0, \\end{align*}\n$$\n$$\n\\begin{align*} (2kb^2 - b + 1)^2 - \\Delta &= 4kb^2 - 4k - (b-1)^2 \\\\ &= 4k(b^2 - 1) - (b-1)^2 \\\\ &> (4k-1)(b^2-1) > 0, \\end{align*}\n$$\nthis completes the proof of (3).\nSince $\\Delta$ is a perfect square, it follows from (3) that\n$$\n\\Delta = 4k^2b^4 - 4k(b^3 - 1) = (2kb^2 - b)^2.\n$$\nThen $4k = b^2$, and hence $b$ must be even. Let $b = 2l$. We have $k = l^2$. Together with ②, we have $a = l$ or $8l^4 - l$.\nTherefore the only possibilities are\n$(a, b) = (2l, 1), (l, 2l) \\text{ or } (8l^4 - l, 2l)$\nfor some positive integer $l$. All of these pairs satisfy the given condition.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 27982,
    "subject": "Mathematics (Multi-modal)",
    "question": "As shown in the figure, quadrilateral $ABCD$ is inscribed in a circle with $AC$ as its diameter, $BD \\perp AC$, and $E$ the intersection of $AC$ and $BD$. Extend line segment $DA$ and $BA$ through $A$ to $F$ and $G$ respectively, such that $DG \\parallel BF$. Extend $GF$ to $H$ such that $CH \\perp GH$. Prove that points $B$, $E$, $F$ and $H$ lie on one circle. (posed by Liu Jiangfeng)\n\n![](attached_image_1.png)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "As shown in the figure, connect $BH$, $EF$ and $CG$. Since $\\triangle BAF \\sim \\triangle GAD$, we have\n$$\n\\frac{FA}{AB} = \\frac{DA}{AG}. \\qquad \\textcircled{1}\n$$\nFurthermore, $\\triangle ABE \\sim \\triangle ACD$, then\n$$\n\\frac{AB}{EA} = \\frac{AC}{DA}. \\qquad \\textcircled{2}\n$$\nMultiplying ① by ②, we get $\\frac{FA}{EA} = \\frac{AC}{AG}$, then $\\triangle FAE \\sim \\triangle CAG$\nas $\\angle FAE = \\angle CAG$, and thus $\\angle FEA = \\angle CGA$.\nAs is known that $\\angle CBG = \\angle CHG = 90^\\circ$, then points $B$, $C$, $G$ and $H$ lie on one circle. So,\n$$\n\\begin{aligned}\n\\angle BHF + \\angle BEF &= \\angle BHC + 90^\\circ + \\angle BEF \\\\\n&= \\angle BGC + 90^\\circ + \\angle BEF \\\\\n&= \\angle FEA + 90^\\circ + \\angle BEF = 180^\\circ.\n\\end{aligned}\n$$\nThat means that points $B$, $E$, $F$ and $H$ lie on one circle.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 27983,
    "subject": "Mathematics (Multi-modal)",
    "question": "Given a positive integer $n$, find the least positive number $\\lambda$ such that $\\cos \\theta_1 + \\cos \\theta_2 + \\cdots + \\cos \\theta_n$ is not greater than $\\lambda$ provided $\\tan \\theta_1 \\cdot \\tan \\theta_2 \\cdot \\cdots \\cdot \\tan \\theta_n = 2^{\\frac{n}{2}}$ for any $\\theta_i \\in (0, \\frac{\\pi}{2})$ ($i=1, 2, \\cdots, n$).",
    "options": [],
    "answer": "λ = √3/3 for n = 1; λ = 2√3/3 for n = 2; λ = n − 1 for n ≥ 3",
    "solution": "When $n=1$, $\\cos\\theta_1 = (1+\\tan^2\\theta_1)^{-\\frac{1}{2}} = \\frac{\\sqrt{3}}{3}$. Hence $\\lambda = \\frac{\\sqrt{3}}{3}$.\n\nWhen $n=2$, we can prove\n$$\n\\cos \\theta_1 + \\cos \\theta_2 \\le \\frac{2\\sqrt{3}}{3}, \\qquad \\textcircled{1}\n$$\nand when $\\theta_1 = \\theta_2 = \\arctan\\sqrt{2}$, the equality holds.\nIn fact,\n$$\n\\textcircled{1} \\Leftrightarrow \\cos^2\\theta_1 + \\cos^2\\theta_2 + 2\\cos\\theta_1 \\cdot \\cos\\theta_2 \\le \\frac{3}{4},\n$$\n$$\n\\text{that is, } \\frac{1}{1+\\tan^2\\theta_1} + \\frac{1}{1+\\tan^2\\theta_2} + 2\\sqrt{\\frac{1}{(1+\\tan^2\\theta_1)(1+\\tan^2\\theta_2)}} \\le \\frac{3}{4}. \\qquad \\textcircled{2}\n$$\nFrom $\\tan \\theta_1 \\cdot \\tan \\theta_2 = 2$, we get\n$$\n\\textcircled{2} \\Leftrightarrow \\frac{2+\\tan^2\\theta_1 + \\tan^2\\theta_2}{5+\\tan^2\\theta_1 + \\tan^2\\theta_2} + 2\\sqrt{\\frac{1}{5+\\tan^2\\theta_1 + \\tan^2\\theta_2}} \\le \\frac{3}{4}. \\qquad \\textcircled{3}\n$$\nWrite $x = \\tan^2\\theta_1 + \\tan^2\\theta_2$, then\n$$\n\\textcircled{3} \\Leftrightarrow 2\\sqrt{\\frac{1}{5+x}} \\le \\frac{14+x}{3(5+x)},\n$$\n$$\n\\text{that is,}\\qquad 36(5+x) \\le 196+28x+x^2. \\qquad \\textcircled{4}\n$$\n$$\n\\text{Obviously,}\\qquad \\textcircled{4} \\Leftrightarrow x-8x+16 = (x-4)^2 \\ge 0.\n$$\n$$\n\\text{Hence, } \\lambda = \\frac{2\\sqrt{3}}{3}.\n$$\n\nWhen $n \\ge 3$, there is no loss of generality in supposing $\\theta_1 \\ge \\theta_2 \\ge \\dots \\ge \\theta_n$, then\n$$\n\\tan \\theta_1 \\cdot \\tan \\theta_2 \\cdot \\tan \\theta_3 \\ge 2\\sqrt{2}.\n$$\nSince $\\cos \\theta_i = \\sqrt{1 - \\sin^2 \\theta_i} < 1 - \\frac{1}{2} \\sin^2 \\theta_i$, so\n$$\n\\cos \\theta_2 + \\cos \\theta_3 < 2 - \\frac{1}{2}(\\sin^2 \\theta_2 + \\sin^2 \\theta_3) < 2 - \\sin \\theta_2 \\cdot \\sin \\theta_3.\n$$\nFrom $\\tan^2 \\theta_1 \\ge \\frac{8}{\\tan^2 \\theta_2 \\cdot \\tan^2 \\theta_3}$, we have\n$$\n\\frac{1}{\\cos^2 \\theta_1} \\ge \\frac{8 + \\tan^2 \\theta_2 \\cdot \\tan^2 \\theta_3}{\\tan^2 \\theta_2 \\cdot \\tan^2 \\theta_3},\n$$\nthat is,\n$$\n\\begin{aligned} \\cos \\theta_1 &\\le \\frac{\\tan \\theta_2 \\cdot \\tan \\theta_3}{\\sqrt{8 + \\tan^2 \\theta_2 \\cdot \\tan^2 \\theta_3}} \\\\ &= \\frac{\\sin \\theta_2 \\cdot \\sin \\theta_3}{\\sqrt{8\\cos^2 \\theta_2 \\cdot \\cos^2 \\theta_3 + \\sin^2 \\theta_2 \\cdot \\sin^2 \\theta_3}}. \\end{aligned}\n$$\nHence\n$$\n\\cos \\theta_2 + \\cos \\theta_3 + \\cos \\theta_1\n$$\n$$\n< 2 - \\sin \\theta_2 \\cdot \\sin \\theta_3 \\left[ 1 - \\frac{1}{\\sqrt{8\\cos^2 \\theta_2 \\cdot \\cos^2 \\theta_3 + \\sin^2 \\theta_2 \\cdot \\sin^2 \\theta_3}} \\right].\n$$\nNote that\n$$\n8\\cos^2 \\theta_2 \\cdot \\cos^2 \\theta_3 + \\sin^2 \\theta_2 \\cdot \\sin^2 \\theta_3 \\ge 1\n$$\n$$\n\\begin{aligned} \\Leftrightarrow & 8 + \\tan^2 \\theta_2 \\tan^2 \\theta_3 \\ge \\frac{1}{\\cos^2 \\theta_2 \\cdot \\cos^2 \\theta_3} \\\\ &= (1 + \\tan^2 \\theta_2)(1 + \\tan^2 \\theta_3) \\\\ \\Leftrightarrow & \\tan^2 \\theta_2 + \\tan^2 \\theta_3 \\le 7. \\end{aligned}\n$$\nIf this does not hold, then $\\tan^2 \\theta_2 + \\tan^2 \\theta_3 > 7$,\n$$ \\tan^2 \\theta_1 \\ge \\tan^2 \\theta_2 > \\frac{7}{2} $$\nso\n$$ \\cos \\theta_1 \\le \\cos \\theta_2 < \\sqrt{\\frac{1}{1+\\frac{7}{2}}} = \\frac{\\sqrt{2}}{3}. $$\nThus\n$$ \\cos \\theta_1 + \\cos \\theta_2 + \\cos \\theta_3 < \\frac{2\\sqrt{2}}{3} + 1 < 2, $$\nthat is, the bound holds too.\n\n$$\n\\cos \\theta_1 + \\cos \\theta_2 + \\cos \\theta_3 + \\dots + \\cos \\theta_n < n - 1.\n$$\nOn the other hand, if we take $\\theta_2 = \\theta_3 = \\dots = \\theta_n = \\alpha > 0$, $\\alpha \\to 0$, then\n$$\n\\theta_1 = \\arctan \\frac{2^n}{(\\tan \\alpha)^{n-1}}.\n$$\nObviously, $\\theta_1 \\to \\frac{\\pi}{2}$, thus\n$$\n\\cos \\theta_1 + \\cos \\theta_2 + \\cos \\theta_3 + \\dots + \\cos \\theta_n \\to n - 1.\n$$\nConsequently, we get $\\lambda = n - 1$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 27984,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let a line with the inclination angle of $60^\\circ$ be drawn through the focus $F$ of the parabola $y^2 = 8(x + 2)$. If the two intersection points of the line and the parabola are $A$ and $B$, and the perpendicular bisector of the chord $AB$ intersects the $x$-axis at the point $P$, then the length of the segment $PF$ is ( ).\n\n(A) $\\frac{16}{3}$\n(B) $\\frac{8}{3}$\n(C) $\\frac{16\\sqrt{3}}{3}$\n(D) $8\\sqrt{3}$",
    "options": [],
    "answer": "A",
    "solution": "It follows from the property of the focus of a parabola that $F = (0, 0)$. Then the equation of the straight line through points $A$ and $B$ will be $y = \\sqrt{3}x$. Substitute it into the parabola equation, and then obtain\n$$\n3x^2 - 8x - 16 = 0.\n$$\nLet $E$ be the midpoint of the chord $AB$, then the $x$-coordinate of $E$ is $\\frac{4}{3}$. Then we have $|FE| = \\frac{1}{\\cos 60^\\circ} \\times \\frac{4}{3} = \\frac{8}{3}$, $|PF| = 2|FE| = \\frac{16}{3}$. Answer: A.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 27985,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $n$ be a given positive integer. Find the least positive integer $u_n$, such that for any positive integer $d$, the number of integers divisible by $d$ in every $u_n$ consecutive positive odd numbers is not less than the number of integers divisible by $d$ in $1, 3, 5, \\dots, 2n-1$. (posed by Chen Yonggao)",
    "options": [],
    "answer": "2n-1",
    "solution": "(1) $u_n \\ge 2n-1$. As $u_1 = 1$, we only need to consider $n \\ge 2$. Since the number of integers divisible by $2n-1$ in $1, 3, 5, \\dots, 2n-1$ is $1$ and that in $2(n+1)-1, 2(n+2)-1, \\dots, 2(n+2n-2)-1$ is $0$, then\n\n$u_n \\ge 2n - 1$.\n\n(2) $u_n \\le 2n-1$. We only need to consider the case when $2 \\nmid d$ and $1 \\le d \\le 2n-1$. For any $2n-1$ consecutive positive odd numbers; $2(a+1)-1, 2(a+2)-1, \\dots, 2(a+2n-1)-1$, let $s$ and $t$ be positive integers such that\n$$\n(2s-1)d \\le 2n-1 < (2s+1)d,\n$$\n$$\n(2t-1)d \\le 2(a+1)-1 < (2t+1)d.\n$$\nThen the number of integers divisible by $d$ in $1, 3, 5, \\dots, 2n-1$ is $s$, and\n$$\n\\begin{aligned}\n(2(t+s)-1)d &= (2t-1)d + (2s-1)d + d \\\\\n&\\le 2(a+1)-1 + 2n-1 + 2n-1 \\\\\n&= 2(a+2n-1)-1.\n\\end{aligned}\n$$\nThat means $u_n \\le 2n-1$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 27986,
    "subject": "Mathematics (Multi-modal)",
    "question": "A convex hexagon is given in which any two opposite sides have the following property: the distance between their midpoints is $\\frac{\\sqrt{3}}{2}$ times the sum of their lengths. Prove that all the angles of the hexagon are equal.\n(A convex hexagon $ABCDEF$ has three pairs of opposite sides: $AB$ and $DE$, $BC$ and $EF$, $CD$ and $FA$.)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "**Proof I** We first prove the following lemma.\nLemma Consider a triangle $PQR$ with $\\angle QPR \\ge 60^\\circ$. Let $L$ be the midpoint of $QR$. Then $PL \\le \\frac{\\sqrt{3}}{2} QR$. The equality holds if and only if the triangle $PQR$ is equilateral.\nProof of the lemma\nLet $S$ be the point such that the triangle $QRS$ is equilateral, where the points $P$ and $S$ lie in the same half-plane bounded by the line $QR$.\nThen the point $P$ lies inside the circumcircle (including the circumference of the circle) of the triangle $QRS$, which lies inside the circle with center $L$ and radius $\\frac{\\sqrt{3}}{2} QR$.\nThis completes the proof of the lemma.\nThe main diagonals of a convex hexagon form a triangle though the\n![](attached_image_1.png)\n![](attached_image_2.png)\nThus we may choose two of these three diagonals that form an angle greater than or equal to $60^\\circ$. Without loss of generality, we may assume that the diagonals $AD$ and $BE$ of the given hexagon $ABCDEF$ satisfy $\\angle APB \\ge 60^\\circ$, where $P$ is the intersection of these two diagonals. Then, using the lemma, we obtain\n$$\nMN = \\frac{\\sqrt{3}}{2}(AB + DE) \\ge PM + PN \\ge MN,\n$$\nwhere $M$ and $N$ are the midpoints of $AB$ and $DE$ respectively. Thus it follows from the lemma that the triangles $ABP$ and $DEP$ are equilateral.\nTherefore the diagonal $CF$ forms an angle greater than or equal to $60^\\circ$ with one of the diagonals $AD$ and $BE$. Without loss of generality, we may assume that $\\angle AQF \\ge 60^\\circ$, where $Q$ is the intersection of $AD$ and $CF$. Arguing in the same way as above, we infer that the triangles $AQF$ and $CQD$ are equilateral. This implies that $\\angle BRC = 60^\\circ$, where $R$ is the intersection of $BE$ and $CF$. Using the same argument as above for the third time, we obtain that the triangles $BCR$ and $EFR$ are equilateral.\nThis completes the proof.\n\n\n**Proof II** Let $ABCDEF$ be the given hexagon and let $\\vec{a} = \\vec{AB}$, $\\vec{b} = \\vec{BC}$, ..., $\\vec{f} = \\vec{FA}$.\n![](attached_image_3.png)\nLet $M$ and $N$ be the midpoints of the sides $AB$ and $DE$ respectively. We have\n$$\n\\vec{MN} = \\frac{1}{2}\\vec{a} + \\vec{b} + \\vec{c} + \\frac{1}{2}\\vec{d} \\text{ and } \\vec{MN} = -\\frac{1}{2}\\vec{a} - \\vec{f} - \\vec{e} - \\frac{1}{2}\\vec{d}.\n$$\nThus we obtain\n$$\n\\vec{MN} = \\frac{1}{2}(\\vec{b} + \\vec{c} - \\vec{e} - \\vec{f}). \\qquad \\textcircled{1}\n$$\nFrom the given property, we have\n$$\n\\vec{MN} = \\frac{\\sqrt{3}}{2}(|\\mathbf{a}| + |\\mathbf{d}|) \\ge \\frac{\\sqrt{3}}{2}|\\mathbf{a} - \\mathbf{d}|. \\qquad \\textcircled{2}\n$$\nSet $\\mathbf{x} = \\mathbf{a} - \\mathbf{d}$, $\\mathbf{y} = \\mathbf{c} - \\mathbf{f}$, $\\mathbf{z} = \\mathbf{e} - \\mathbf{b}$. From ① and ②, we obtain\n$$\n|\\mathbf{y} - \\mathbf{z}| \\ge \\sqrt{3} |\\mathbf{x}|. \\qquad \\textcircled{3}\n$$\nSimilarly, we see that\n$$\n|\\mathbf{z} - \\mathbf{x}| \\ge \\sqrt{3} |\\mathbf{y}|, \\qquad \\textcircled{4}\n$$\n$$\n|\\mathbf{x} - \\mathbf{y}| \\ge \\sqrt{3} |\\mathbf{z}|. \\qquad \\textcircled{5}\n$$\nNote that\n$$\n\\textcircled{3} \\Leftrightarrow |\\mathbf{y}|^2 - 2\\mathbf{y} \\cdot \\mathbf{z} + |\\mathbf{z}|^2 \\ge 3|\\mathbf{x}|^2,\n$$\n$$\n\\textcircled{4} \\Leftrightarrow |\\mathbf{z}|^2 - 2\\mathbf{z} \\cdot \\mathbf{x} + |\\mathbf{x}|^2 \\ge 3|\\mathbf{y}|^2,\n$$\n$$\n\\textcircled{5} \\Leftrightarrow |\\mathbf{x}|^2 - 2\\mathbf{x} \\cdot \\mathbf{y} + |\\mathbf{y}|^2 \\ge 3|\\mathbf{z}|^2.\n$$\nBy adding up the last three inequalities, we obtain\n$$\n- |\\mathbf{x}|^2 - |\\mathbf{y}|^2 - |\\mathbf{z}|^2 - 2\\mathbf{y} \\cdot \\mathbf{z} - 2\\mathbf{z} \\cdot \\mathbf{x} - 2\\mathbf{x} \\cdot \\mathbf{y} \\ge 0,\n$$\nor $-|\\mathbf{x} + \\mathbf{y} + \\mathbf{z}|^2 \\ge 0$. Thus $\\mathbf{x} + \\mathbf{y} + \\mathbf{z} = 0$ and the equality holds in every inequality above. Hence we conclude that\n$$\n\\mathbf{x} + \\mathbf{y} + \\mathbf{z} = 0,\n$$\n$$\n|\\mathbf{y} - \\mathbf{z}| = \\sqrt{3} |\\mathbf{x}|, \\mathbf{a} \\parallel \\mathbf{d} \\parallel \\mathbf{x},\n$$\n$$\n|\\mathbf{z} - \\mathbf{x}| = \\sqrt{3} |\\mathbf{y}|, \\mathbf{c} \\parallel \\mathbf{f} \\parallel \\mathbf{y},\n$$\n$$\n|\\mathbf{x} - \\mathbf{y}| = \\sqrt{3} |\\mathbf{z}|, \\mathbf{e} \\parallel \\mathbf{b} \\parallel \\mathbf{z}.\n$$\nSuppose that $PQR$ is the triangle such that $\\vec{PQ} = \\mathbf{x}$, $\\vec{QR} = \\mathbf{y}$, $\\vec{RP} = \\mathbf{z}$. We may assume $\\angle QPR \\ge 60^\\circ$, without loss of generality. Let $L$ be the midpoint of $QR$. Then $PL = \\frac{1}{2}|\\mathbf{z} - \\mathbf{x}| = \\frac{\\sqrt{3}}{2}|\\mathbf{y}| = \\frac{\\sqrt{3}}{2} QR$. It follows from the lemma in Proof I that the triangle $PQR$ is equilateral. Thus we have $\\angle ABC = \\angle BCD = \\dots = \\angle FAB = 120^\\circ$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 27987,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABCD$ be an inscribed quadrilateral. Let $P$, $Q$ and $R$ be the feet of the perpendiculars from $D$ to the lines $BC$, $CA$ and $AB$ respectively. Show that $PQ = QR$ if and only if the bisectors of $\\angle ABC$ and $\\angle ADC$ meet on $AC$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "**Proof** By Simson's Theorem, we know that $P$, $Q$, $R$ are collinear. Moreover, since $\\angle DPC$ and $\\angle DQC$ are right angles, the points $D$, $P$, $Q$, $C$ are concyclic and so $\\angle DCA = \\angle DPQ = \\angle DPR$. Similarly, since $D$, $Q$, $R$, $A$ are concyclic, we have $\\angle DAC = \\angle DRP$. Therefore $\\triangle DCA \\sim \\triangle DPR$.\n\n![](attached_image_1.png)\n\nLikewise, $\\triangle DAB \\sim \\triangle DQP$ and $\\triangle DBC \\sim \\triangle DRQ$. Then\n$$\n\\frac{DA}{DC} = \\frac{DR}{DP} = \\frac{DB \\cdot \\frac{QR}{BC}}{DB \\cdot \\frac{PQ}{BA}} = \\frac{QR}{PQ} \\cdot \\frac{BA}{BC}.\n$$\nThus $PQ = QR$ if and only if $\\frac{DA}{DC} = \\frac{BA}{BC}$.\n\nNow the bisectors of the angles $ABC$ and $ADC$ divide $AC$ in the ratios of $\\frac{BA}{BC}$ and $\\frac{DA}{DC}$ respectively. This completes the proof.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 27988,
    "subject": "Mathematics (Multi-modal)",
    "question": "Suppose the sum of distances from any point $P$ in a convex quadrilateral $ABCD$ to lines $AB$, $BC$, $CD$ and $DA$ is constant. Prove that $ABCD$ is a parallelogram. (posed by Xiong Bin)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $d(P, l)$ denote the distance from point $P$ to line $l$. We first prove the following lemma.\n\n**Lemma** Let $\\angle SAT = \\alpha$ be a given angle and $P$ a moving point in $\\angle SAT$. If the sum of distances from $P$ to lines $AS$ and $AT$ is a constant number $m$, then the trace of $P$ is a segment of $BC$ with points $B$ and $C$ on $AS$ and $AT$ respectively, and $AB = AC = \\frac{m}{\\sin \\alpha}$. Further, if a point $Q$ lies inside $\\triangle ABC$, then the sum of distances from $Q$ to lines $AS$ and $AT$ is less than $m$, and it is greater than $m$ if $Q$ lies outside $\\triangle ABC$.\n\n![](attached_image_1.png)\n\n**Proof of Lemma** For $AB = AC = \\frac{m}{\\sin \\alpha}$ and $P$ on $BC$, we have $S_{\\triangle PAB} + S_{\\triangle PAC} = S_{\\triangle ABC}$, i.e. $\\frac{1}{2}AB \\cdot d(P, AB) + \\frac{1}{2}AC \\cdot d(P, AC) = \\frac{1}{2} \\cdot AB \\cdot AC \\cdot \\sin \\alpha$. Then $d(P, AB) + d(P, AC) = m$. If point $Q$ lies inside $\\triangle ABC$, $S_{\\triangle QAB} + S_{\\triangle QAC} < S_{\\triangle ABC}$, then $d(Q, AB) + d(Q, AC) < m$. If $Q$ lies outside $\\triangle ABC$, $S_{\\triangle QAB} + S_{\\triangle QAC} > S_{\\triangle ABC}$, then $d(Q, AB) + d(Q, AC) > m$. The proof of Lemma is complete.\n\nWe now consider the following two cases:\n\n(1) Neither pair of the opposite sides of the quadrilateral $ABCD$ is parallel. We may assume that sides $BC$ and $AD$ meet at point $F$ and sides $BA$ and $CD$ meet at point $E$. Through point $P$ draw segments $l_1$ and $l_2$ such that the sum of distances from any point on $l_1$ to $AB$ and $CD$ is constant and the sum of distances from any point on $l_2$ to $BC$ and $AD$ is also constant. As seen in the second figure, for any point $Q$ in area $S$, using the given condition and the lemma, we have\n\n$$\n\\begin{align*}\nd(P, AB) + d(P, BC) + d(P, CD) + d(P, DA) \\\\\n&= d(Q, AB) + d(Q, BC) + d(Q, CD) + d(Q, DA) \\\\\n&= [d(Q, AB) + d(Q, CD)] + [d(Q, BC) + d(Q, DA)] \\\\\n&> [d(P, AB) + d(P, CD)] + [d(P, BC) + d(P, DA)].\n\\end{align*}\n$$\n\nIt leads to a contradiction.\n\n(2) The quadrilateral $ABCD$ is a trapezoid. In the same way, we can also show that it will lead to a contradiction.\n\nThis completes the proof.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 27989,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all ternary positive integer groups $(a, m, n)$ satisfying $a \\ge 2$ and $m \\ge 2$ such that $a^n + 203$ is a multiple of $a^m + 1$. (posed by Chen Yonggao)",
    "options": [],
    "answer": "(2, 2, 4k+1), (2, 3, 6k+2), (2, 4, 8k+8), (2, 6, 12k+9), (3, 2, 4k+3), (4, 2, 4k+4), (5, 2, 4k+1), (8, 2, 4k+3), (10, 2, 4k+2), and (203, m, (2k+1)m+1), where k is any nonnegative integer and m ≥ 2.",
    "solution": "We will discuss the following three cases for $n$ and $m$.\n\n(i) In the case when $n < m$, from $a^n + 203 \\ge a^m + 1$, we have\n$$\n202 \\ge a^m - a^n \\ge a^n(a-1) \\ge a(a-1).\n$$\nTherefore, $2 \\le a \\le 14$.\n\nWhen $a = 2$, we can take $n$ to be $1, 2, \\dots, 7$.\nWhen $a = 3$, we can take $n$ to be $1, 2, 3$ and $4$.\nWhen $a = 4$, we can take $n$ to be $1, 2$ and $3$.\nWhen $5 \\le a \\le 6$, we can take $n$ to be $1$ and $2$.\nWhen $7 \\le a \\le 14$, $n = 1$.\nFrom $a^m + 1 \\mid a^n + 203$, we can deduce that the solutions are $(2, 2, 1), (2, 3, 2)$ and $(5, 2, 1)$.\n\n(ii) In the case when $n = m$, $a^m + 1 \\mid 202$. Since 202 has only four divisors 1, 2, 101 and 202, but $a \\ge 2$, $m \\ge 2$ and $a^m + 1 \\ge 5$, so $a^m = 100$ or 201. Since $m \\ge 2$, therefore the solution is $(10, 2, 2)$.\n\n(iii) In the case when $n > m$, from $a^m + 1 \\mid 203(a^m + 1)$, we have\n$$\na^m + 1 \\mid a^n + 203 - (203a^m + 203),\n$$\nthat is,\n$$\na^m + 1 \\mid a^m(a^{n-m} - 203).\n$$\nSince $(a^m + 1, a^m) = 1$, so\n$$\na^m + 1 \\mid a^{n-m} - 203.\n$$\n\n① If $a^{n-m} < 203$, then set $n-m = s \\ge 1$, we have $a^m + 1 \\mid 203-a^s$. Hence\n$$\n203 - a^s \\ge a^m + 1, \\\\\n202 \\ge a^s + a^m \\ge a^m + a \\\\\n= a(a^{m-1} + 1) \\ge a(a+1),\n$$\nthus\n$$\n2 \\le a \\le 13.\n$$\nUsing the same argument as in Case (i), we show that the solutions for $(a, m, s)$ are:\n$(2, 2, 3), (2, 6, 3), (2, 4, 4), (2, 3, 5), (2, 2, 7), (3, 2, 1), (4, 2, 2), (5, 2, 3)$ and $(8, 2, 1)$.\nHence, $(a, m, n)$ are\n$(2, 2, 5), (2, 6, 9), (2, 4, 8), (2, 3, 8), (2, 2, 9), (3, 2, 3), (4, 2, 4), (5, 2, 5)$ and $(8, 2, 3)$.\n\n② If $a^{n-m} = 203$, then $a = 203$ and $n-m = 1$, and the solution is\n$(203, m, m+1)$, $m \\ge 2$.\n\n③ If $a^{n-m} > 203$, set $n-m = s \\ge 1$, then $a^m + 1 \\mid a^s - 203$.\nSince $a^s - 203 \\ge a^m + 1$, so $s > m$. By\n$$\na^m + 1 \\mid a^s + 203a^m = (a^{s-m} + 203)a^m \\\\\n= (a^{n-2m} + 203)a^m,\n$$\nand\n$(a^m + 1, a^m) = 1,$\nwe have\n$a^m + 1 \\mid a^{n-2m} + 203.$\nNow $s > m \\Leftrightarrow n - m > m \\Leftrightarrow n > 2m \\Leftrightarrow n - 2m > 0$. In this case, the solutions can only be derived from the preceding solutions, that is, from $(a, m, n) \\rightarrow (a, m, n+2m) \\rightarrow \\dots \\rightarrow (a, m, n+2km)$, and each solution derived also satisfies $a^m + 1 \\mid a^n + 203$.\n\nSummarizing what described above, we obtain all solutions $(a, m, n)$ to be:\n$(2, 2, 4k+1)$, $(2, 3, 6k+2)$, $(2, 4, 8k+8)$, $(2, 6, 12k+9)$,\n$(3, 2, 4k+3)$, $(4, 2, 4k+4)$, $(5, 2, 4k+1)$, $(8, 2, 4k+3)$,\n$(10, 2, 4k+2)$ and $(203, m, (2k+1)m+1)$,\nwhere $k$ is any nonnegative integer, and $m \\ge 2$ is an integer.",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 27990,
    "subject": "Mathematics (Multi-modal)",
    "question": "(1) Prove that there exist five nonnegative real numbers $a$, $b$, $c$, $d$ and $e$ with their sum equal to $1$ such that for any arrangement of these numbers around a circle, there are always two neighboring numbers with their product not less than $\\frac{1}{9}$.\n\n(2) Prove that for any five nonnegative real numbers with their sum equal to $1$, it is always possible to arrange them around a circle such that there are two neighboring numbers with their product not greater than $\\frac{1}{9}$.\n\n(posed by Qian Zhanwang)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "(1) Let $a = b = c = \\frac{1}{3}$, $d = e = 0$, it is easy to see that, when arranging them around a circle, we can always get two neighboring numbers of $\\frac{1}{3}$, and their product is $\\frac{1}{9}$.\n\n(2) For any five nonnegative real numbers $a$, $b$, $c$, $d$ and $e$ with their sum equal to $1$, without loss of generality, we assume that $a \\ge b \\ge c \\ge d \\ge e \\ge 0$. Arrange these numbers around a circle in such a way as seen in the figure:\n\n![](attached_image_1.png)\n\nSince $a + b + c + d + e = 1$, we have $a + 3d \\le 1$, and\n$$a \\cdot 3d \\le \\left(\\frac{a + 3d}{2}\\right)^2 \\le \\frac{1}{4}$$\nthen $ad \\le \\frac{1}{12}$.\n\nFurthermore, $a + b + c \\le 1$, then $b + c \\le 1 - a \\le 1 - \\frac{b + c}{2}$, i.e. $b + c \\le \\frac{2}{3}$. So\n$$bc \\le \\frac{(b + c)^2}{4} \\le \\frac{1}{9}$$\nSince $ce \\le ae \\le ad$ and $bd \\le bc$, then any neighboring numbers in this arrangement have their product less than $\\frac{1}{9}$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 27991,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $x \\in \\left[-\\frac{5\\pi}{12}, -\\frac{\\pi}{3}\\right]$. Then the maximum value of\n\n$$\ny = \\tan\\left(x + \\frac{2\\pi}{3}\\right) - \\tan\\left(x + \\frac{\\pi}{6}\\right) + \\cos\\left(x + \\frac{\\pi}{6}\\right)\n$$\nis ( ).\n(A) $\\frac{12}{5}\\sqrt{2}$\n(B) $\\frac{11}{6}\\sqrt{2}$\n(C) $\\frac{11}{6}\\sqrt{3}$\n(D) $\\frac{12}{5}\\sqrt{3}$",
    "options": [],
    "answer": "C",
    "solution": "Let $z = -x - \\frac{\\pi}{6}$. Then $z \\in [\\frac{\\pi}{6}, \\frac{\\pi}{4}]$, and $2z \\in [\\frac{\\pi}{3}, \\frac{\\pi}{2}]$. We have\n$$\n\\tan\\left(x + \\frac{2\\pi}{3}\\right) = -\\cot\\left(x + \\frac{6}{\\pi}\\right) = \\cot z.\n$$\nThen\n$$\ny = \\cot z + \\tan z + \\cos z = \\frac{2}{\\sin 2z} + \\cos z.\n$$\nSince both $\\frac{2}{\\sin 2z}$ and $\\cos z$ are monotonic decreasing in this case, so $y$ reaches the maximum at $z = \\frac{\\pi}{6}$, where $y_{\\max} = \\frac{2}{\\sin \\frac{\\pi}{3}} + \\cos \\frac{\\pi}{6} = \\frac{4}{\\sqrt{3}} + \\frac{\\sqrt{3}}{2} = \\frac{11}{6}\\sqrt{3}$.\n\nAnswer: C.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 27992,
    "subject": "Mathematics (Multi-modal)",
    "question": "Suppose $x, y \\in (-2, 2)$ and $xy = -1$. Then the minimum value of $u = \\frac{4}{4-x^2} + \\frac{9}{9-y^2}$ is ( ).\n(A) $\\frac{8}{5}$\n(B) $\\frac{24}{11}$\n(C) $\\frac{12}{7}$\n(D) $\\frac{12}{5}$",
    "options": [],
    "answer": "D",
    "solution": "**Solution I** We have\n$$\n\\begin{aligned}\nu &= \\frac{4}{4-x^2} + \\frac{9x^2}{9x^2-1} = 1 + \\frac{35x^2}{-9x^4 + 37x^2 - 4} \\\\\n&= 1 + \\frac{35}{37 - \\left( \\left( 3x - \\frac{2}{x} \\right)^2 + 12 \\right)}\n\\end{aligned}\n$$\nSince $x \\in (-2, -\\frac{1}{2}) \\cup (\\frac{1}{2}, 2)$, so $u$ reaches the minimum value $\\frac{12}{5}$ when $x = \\pm\\sqrt{\\frac{2}{3}}$. Answer: D.\n\n\n**Solution II** It is known from the conditions that $4-x^2 > 0$ and $9-y^2 > 0$. Then\n$$\n\\begin{aligned}\nu &\\ge 2\\sqrt{\\frac{4}{4-x^2} \\cdot \\frac{9}{9-y^2}} = \\frac{12}{\\sqrt{36-9x^2-4y^2+(xy)^2}} \\\\\n&= \\frac{12}{\\sqrt{37-9x^2-4y^2}} \\ge \\frac{12}{\\sqrt{37-2\\sqrt{36(xy)^2}}} = \\frac{12}{5}.\n\\end{aligned}\n$$\nSince $u$ is $\\frac{12}{5}$ when $x = \\sqrt{\\frac{2}{3}}$ and $y = -\\sqrt{\\frac{3}{2}}$, so $u$ reaches the minimum. Answer: D.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 27993,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $\\{a_n\\}_1^\\infty$ be a sequence of real numbers such that $a_1 = 2$, $a_{n+1} = a_n^2 - a_n + 1$, for $n = 1, 2, \\dots$. Prove that\n$$\n1 - \\frac{1}{2^{003^{2^{003}}}} < \\frac{1}{a_1} + \\frac{1}{a_2} + \\dots + \\frac{1}{a_{2^{003}}} < 1.\n$$\n(posed by Li Shenghong)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "We have $a_{n+1} - 1 = a_n(a_n - 1)$, and then\n$$\n\\frac{1}{a_{n+1}-1} = \\frac{1}{a_n-1} - \\frac{1}{a_n}.\n$$\nSo,\n$$\n\\frac{1}{a_1} + \\frac{1}{a_2} + \\dots + \\frac{1}{a_{2^{003}}} = (\\frac{1}{a_1-1} - \\frac{1}{a_2-1}) + (\\frac{1}{a_2-1} - \\frac{1}{a_3-1}) + \\dots + (\\frac{1}{a_{2^{003}}-1} - \\frac{1}{a_{2^{004}}-1})\n$$\n$$\n= \\frac{1}{a_1 - 1} - \\frac{1}{a_{2^{004}} - 1} = 1 - \\frac{1}{a_{2^{004}} - 1}.\n$$\n\nIt is easy to see that $\\{a_n\\}$ is strictly monotonic increasing, so $a_{2^{004}} > 1$, and that means\n$$\n\\frac{1}{a_1} + \\frac{1}{a_2} + \\cdots + \\frac{1}{a_{2^{003}}} < 1.\n$$\nIn order to prove the left inequality, we only need to prove that $a_{2^{004}} - 1 > 2^{003^{2^{003}}}$.\nBy induction, we have $a_{n+1} = a_n a_{n-1} \\cdots a_1 + 1$ and $a_n a_{n-1} \\cdots a_1 > n^n$ ($n \\ge 1$).\nThis completes the proof.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 27994,
    "subject": "Mathematics (Multi-modal)",
    "question": "A certain company wants to employ one secretary. Ten persons apply. The manager decides to interview them one by one according to the order of their applications. The first $3$ applicants should not be employed. From the fourth onward an applicant will be compared with the preceding ones. If he exceeds in ability all the preceding applicants, he will be employed. Otherwise he will not, and the interview goes on. If the preceding nine persons are not employed, the last one will be employed.\n\nSuppose that the $10$ persons are different from each other in ability, and we can arrange them according to their ability rating from superior to inferior, such as $1$st, $2$nd, ..., $10$th. Obviously, whether an applicant will be eventually employed by the company depends on the order of the applications. As it is known, there are $10!$ such permutations in all. Now denote by $A_k$ the number of the permutations such that the applicant with the $k$-th ability rating is employed and the probability for him to be employed is $\\frac{A_k}{10!}$. (posed by Su Chun)\n\nProve that under the policy given by the manager, we have the following properties:\n\n$$(1) \\quad A_1 > A_2 > \\dots > A_8 = A_9 = A_{10}.$$\n$$(2) \\text{The probability for the company to employ one of the persons with the ability among the top three is over 70\\%, and to employ one of the persons with the ability among the bottom three is not over 10\\%.}$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "**Proof** We denote by $a$ the ability rating of the applicant with the highest ability among the first three interviews. Obviously, $a \\le 8$. Now we denote by $A_k(a)$ the set of permutations for which the person with the $k$-th ability rating is employed and we denote the corresponding number of permutations by $|A_k(a)|$.\n\n(1) It is easy to see that, when $a = 1$, a definite result is to give up the first nine persons and to employ the last interviewee. Now the chance is equal for every person except the one who has the highest ability rating. It is not difficult to see that\n$$\n|A_k(1)| = 3 \\times 8! := r_1, \\quad k = 2, 3, \\dots, 10,\n$$\nwhere “:=” denotes “written as”.\n\nWhen $2 \\le a \\le 8$, a person who is placed $k$-th in ability rating will have no chance to be employed for $a \\le k \\le 10$, and the chance is equal for $1 \\le k < a$.\n\nIn fact, a person with the $a$-th ability rating is among the first three interviews and there are three ways to choose one out of the three interviews. Those people with their ability rating from the $1$-st to $(a-1)$-th are among the later seven interviews, and the first one having the highest ability rating among them is employed. There are $C_7^{a-1} (a-2)!$ ways of such arrangement. The other $10-a$ persons may be arranged arbitrarily on the remainder positions, and we have $(10-a)!$ ways of arrangement. Hence, we have\n$$\n|A_k(a)| = \\begin{cases} 3C_7^{a-1}(a-2)!(10-a)! := r_a, & k = 1, \\cdots, a-1; \\\\ 0, & k = a, \\cdots, 10. \\end{cases}\n$$\n\nThe result above shows:\n$$\nA_8 = A_9 = A_{10} = r_1 = 3 \\times 8! > 0; \\qquad \\textcircled{1}\n$$\n$$\nA_k = r_1 + \\sum_{a=k+1}^{8} r_a, \\quad k = 2, \\dots, 7; \\qquad \\textcircled{2}\n$$\n$$\n\\text{and} \\qquad A_1 = \\sum_{a=2}^{8} r_a. \\qquad \\textcircled{3}\n$$\nFrom ① and ②, we obtain\n$$\nA_2 > A_3 > \\cdots > A_8 = A_9 = A_{10} > 0;\n$$\nand from ② and ③, we obtain\n$$\nA_1 - A_2 = r_2 - r_1 = 3 \\times 7 \\times 8! - 3 \\times 8! > 0.\n$$\nConsequently, (1) is proved.\n\n(2) From ①, we know\n$$\n\\frac{A_8 + A_9 + A_{10}}{10!} = \\frac{3 \\times r_1}{10!} = \\frac{3 \\times 3 \\times 8!}{10!} = 10\\%.\n$$\nSo the probability for the company to employ one of the bottom three is equal to $10\\%$.\n\nFrom ② and ③, we can show that\n$$\n\\begin{align*}\nA_1 &= \\sum_{a=2}^{8} r_a = \\sum_{a=2}^{8} 3C_7^{a-1}(a-2)!(10-a)! \\\\\n&= 3 \\times 7! \\sum_{a=2}^{8} \\frac{(9-a)(10-a)}{a-1}\n\\end{align*}\n$$\n$$\n\\begin{align*}\n&= 3 \\times 7! \\sum_{s=1}^{7} \\frac{(8-s)(9-s)}{s} \\\\\n&= 3 \\times 7! \\times \\left( 56 + 21 + 10 + 5 + \\frac{12}{5} + 1 + \\frac{2}{7} \\right) \\\\\n&= 3 \\times 7! \\times 95 \\frac{24}{35} > 3 \\times 7! \\times 95 \\frac{2}{3} \\\\\n&= 287 \\times 7!;\n\\end{align*}\n$$\n$$\n\\begin{align*}\nA_2 &= r_1 + \\sum_{a=3}^{8} r_a \\\\\n&= 3 \\times 8! + 3 \\times 7! \\times \\left( 21 + 10 + 5 + \\frac{12}{5} + 1 + \\frac{2}{7} \\right) \\\\\n&= 3 \\times 7! \\times 47 \\frac{24}{35} > 3 \\times 7! \\times 47 \\frac{2}{3} = 143 \\times 7!;\n\\end{align*}\n$$\n$$\n\\begin{align*}\nA_3 &= r_1 + \\sum_{a=4}^{8} r_a \\\\\n&= 3 \\times 8! + 3 \\times 7! \\times \\left( 10 + 5 + \\frac{12}{5} + 1 + \\frac{2}{7} \\right) \\\\\n&= 3 \\times 7! \\times 26 \\frac{24}{35} > 3 \\times 7! \\times 26 \\frac{2}{3} = 80 \\times 7!.\n\\end{align*}\n$$\nTherefore,\n$$\n\\begin{align*}\n\\frac{A_1 + A_2 + A_3}{10!} &> \\frac{287 + 143 + 80}{720} \\\\\n&= \\frac{510}{720} = \\frac{17}{24} > 70\\%.\n\\end{align*}\n$$\nThat is, the probability for the company to employ one of the top three, is greater than $70\\%$.",
    "topic": "Discrete Mathematics",
    "subtopic": "Combinatorics"
  },
  {
    "id": 27995,
    "subject": "Mathematics (Multi-modal)",
    "question": "Suppose the convex quadrilateral $ABCD$ has an inscribed circle. The circle touches $AB$, $BC$, $CD$, $DA$ at $A_1$, $B_1$, $C_1$, $D_1$ respectively. Let points $E$, $F$, $G$, $H$ be the midpoints of $A_1B_1$, $B_1C_1$, $C_1D_1$, $D_1A_1$ respectively. Prove that quadrilateral $EFGH$ is a rectangle if and only if $ABCD$ is a cyclic quadrilateral. (posed by Feng Zhigang)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "![](attached_image_1.png)\n\nSolution:\n\nAs seen in the figure, let point $I$ be the center of the inscribed circle of $ABCD$. Since $H$ is the midpoint of $D_1A_1$ and lines $AA_1$, $AD_1$ are the two tangent lines through $A$ to the circle, then point $H$ lies on the line segment $AI$ and $AI \\perp A_1D_1$. From $ID_1 \\perp AD_1$ and using the proportional theorem for similar triangles we get that $IH \\cdot IA = ID_1^2 = r^2$, where $r$ is the radius of the inscribed circle. In the same way, we get that $IE \\cdot IB = r^2$. Then $IE \\cdot IB = IH \\cdot IA$, and that means points $A$, $H$, $E$, $B$ lie on one circle, so $\\angle EHI = \\angle ABE$.\n\nSimilarly, we obtain that $\\angle IHG = \\angle ADG$, $\\angle IFE = \\angle CBE$, $\\angle IFG = \\angle CDG$. Adding these four equations, we get that $\\angle EHG + \\angle EFG = \\angle ABC + \\angle ADC$, and that means $A$, $B$, $C$, $D$ lie on one circle if and only if $E$, $F$, $G$, $H$ lie on one circle.\n\nNotice that quadrilateral $EFGH$ is a parallelogram as $E$, $F$, $G$, $H$ are the midpoints of the sides of quadrilateral $A_1B_1C_1D_1$ respectively. So $E$, $F$, $G$, $H$ lie on one circle if and only if $EFGH$ is a rectangle. This completes the proof.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 27996,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $n \\ge 2$ be an integer. Find the largest real number $\\lambda$ such that the inequality\n$$\na_n^2 \\ge \\lambda(a_1 + a_2 + \\cdots + a_{n-1}) + 2a_n\n$$\nholds for any positive integers $a_1, a_2, \\dots, a_n$ satisfying $a_1 < a_2 < \\dots < a_n$.",
    "options": [],
    "answer": "(2n-4)/(n-1)",
    "solution": "For $a_i = i$, $i = 1, 2, \\dots, n$, we have $\\lambda \\le (n-2) \\div \\frac{n-1}{2} = \\frac{2n-4}{n-1}$.\n\nSince $a_k \\le a_n - (n-k)$, $k = 1, 2, \\dots, n-1$, $a_n \\ge n$, then we have\n$$\n\\begin{aligned}\n\\frac{2n-4}{n-1} \\sum_{i=1}^{n-1} a_i &\\le \\frac{2n-4}{n-1} \\left( (n-1)a_n - \\frac{n(n-1)}{2} \\right) \\\\\n&= (2n-4)a_n - n(n-2) \\\\\n&= (n-2)(2a_n - n) \\le (a_n - 2)a_n.\n\\end{aligned}\n$$\n\nThat is, $a_n^2 \\ge \\frac{2n-4}{n-1}(a_1 + a_2 + \\cdots + a_{n-1}) + 2a_n$. So the largest value of $\\lambda$ is $\\frac{2n-4}{n-1}$.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 27997,
    "subject": "Mathematics (Multi-modal)",
    "question": "Suppose $a$, $b$, $c$ and $d$ are positive real numbers satisfying $ab + cd = 1$ and $P_i(x_i, y_i)$ ($i = 1, 2, 3, 4$) are four points on the unit circle which has the origin as its center. Prove that:\n\n$$(ay_1 + by_2 + cy_3 + dy_4)^2 + (ax_4 + bx_3 + cx_2 + dx_1)^2 \\le 2\\left(\\frac{a^2+b^2}{ab}+\\frac{c^2+d^2}{cd}\\right).$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "**Proof I** Set $u = ay_1 + by_2$, $v = cy_3 + dy_4$, $u_1 = ax_4 + bx_3$ and $v_1 = cx_2 + dx_1$. Then\n$$\nu^2 \\le (ay_1 + by_2)^2 + (ax_1 - bx_2)^2 \\\\ = a^2 + b^2 + 2ab(y_1 y_2 - x_1 x_2),$$\nthat is\n$$x_1 x_2 - y_1 y_2 \\le \\frac{a^2 + b^2 - u^2}{2ab}. \\qquad ①$$\n$$v_1^2 \\le (cx_2 + dx_1)^2 + (cy_2 - dy_1)^2 \\\\ = c^2 + d^2 + 2cd(x_1 x_2 - y_1 y_2),$$\nthat is\n$$y_1 y_2 - x_1 x_2 \\le \\frac{c^2 + d^2 - v_1^2}{2cd}. \\qquad ②$$\n① + ②, we get\n$$0 \\le \\frac{a^2 + b^2 - u^2}{2ab} + \\frac{c^2 + d^2 - v_1^2}{2cd},$$\nthat is\n$$\\frac{u^2}{ab} + \\frac{v_1^2}{cd} \\le \\frac{a^2 + b^2}{ab} + \\frac{c^2 + d^2}{cd}.$$ \nSimilarly\n$$\\frac{v_1^2}{cd} + \\frac{u_1^2}{ab} \\le \\frac{c^2 + d^2}{cd} + \\frac{a^2 + b^2}{ab}.$$ \nBy Cauchy's inequality, we have\n$$\n(u+v)^2 + (u_1+v_1)^2 \\\\ \\le (ab+cd)\\left(\\frac{u^2}{ab}+\\frac{v^2}{cd}\\right) + (ab+cd)\\left[\\frac{u_1^2}{ab}+\\frac{v_1^2}{cd}\\right] \\\\ = \\frac{u^2}{ab}+\\frac{v^2}{cd}+\\frac{u_1^2}{ab}+\\frac{v_1^2}{cd} \\le 2\\left[\\frac{a^2+b^2}{ab}+\\frac{c^2+d^2}{cd}\\right].\n$$\n\n**Proof II** By Cauchy's inequality, we can show\n$$\n(ay_1 + by_2 + cy_3 + dy_4)^2\n$$\n$$\n\\le (ab + cd) \\left[ \\frac{(ay_1 + by_2)^2}{ab} + \\frac{(cy_3 + dy_4)^2}{cd} \\right]\n$$\n$$\n= \\frac{a}{b}y_1^2 + \\frac{b}{a}y_2^2 + \\frac{c}{d}y_3^2 + \\frac{d}{c}y_4^2 + 2(y_1 y_2 + y_3 y_4).\n$$\nSimilarly,\n$$\n(ax_4 + bx_3 + cx_2 + dx_1)^2\n$$\n$$\n\\le \\frac{a}{b}x_4^2 + \\frac{b}{a}x_3^2 + \\frac{c}{d}x_2^2 + \\frac{d}{c}x_1^2 + 2(x_1 x_2 + x_3 x_4).\n$$\nSo we subtract the right-hand side (RHS) from the left-hand side (LHS) in the original inequality and get\n$$\n\\begin{align*}\n\\text{LHS} - \\text{RHS} &\\le \\frac{a}{b}y_1^2 + \\frac{b}{a}y_2^2 + \\frac{c}{d}y_3^2 + \\frac{d}{c}y_4^2 + 2y_1 y_2 + 2y_3 y_4 + \\\\\n&\\qquad \\frac{a}{b}x_4^2 + \\frac{b}{a}x_3^2 + \\frac{c}{d}x_2^2 + \\frac{d}{c}x_1^2 + 2x_1 x_2 + 2x_3 x_4 - \\\\\n&\\qquad 2\\left(\\frac{a}{b} + \\frac{b}{a} + \\frac{c}{d} + \\frac{d}{c}\\right) \\\\\n&= -\\frac{a}{b}x_1^2 - \\frac{b}{a}x_1^2 - \\frac{c}{d}x_3^2 - \\frac{d}{c}x_4^2 - \\frac{a}{b}y_4^2 - \\frac{b}{a}y_3^2 - \\\\\n&\\qquad \\frac{c}{d}y_2^2 - \\frac{d}{c}y_1^2 + 2(x_1 x_2 + x_3 x_4 + y_1 y_2 + y_3 y_4) \\\\\n&\\le -2x_1 x_2 - 2x_3 x_4 - 2y_1 y_2 - 2y_3 y_4 + \\\\\n&\\qquad 2(x_1 x_2 + x_3 x_4 + y_1 y_2 + y_3 y_4) \\\\\n&= 0.\n\\end{align*}\n$$\nThe proposition is proved.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 27998,
    "subject": "Mathematics (Multi-modal)",
    "question": "Suppose in the tetrahedron $ABCD$, $AB = 1$, $CD = \\sqrt{3}$, the distance and angle between the lines $AB$ and $CD$ are $2$ and $\\frac{\\pi}{3}$ respectively. Then the volume of the tetrahedron equals ( ).\n\n(A) $\\frac{\\sqrt{3}}{2}$\n(B) $\\frac{1}{2}$\n(C) $\\frac{1}{3}$\n(D) $\\frac{\\sqrt{3}}{3}$",
    "options": [],
    "answer": "B",
    "solution": "As in the diagram, from point $C$ draw a line $CE$ such that it is equal and parallel to $AB$. Construct a prism $ABF$-$ECD$ with $\\triangle CDE$ as base and $BC$ as a lateral edge. Denote $V_1$ as the volume of the tetrahedron and $V_2$ the volume of the prism, then $V_1 = \\frac{1}{3} V_2$.\n\n![](attached_image_1.png)\n\nSince $S_{\\triangle CDE} = \\frac{1}{2} CE \\cdot CD \\sin \\angle ECD$, and the common perpendicular line $MN$ of $AB$ and $CD$ is the height of the prism, then\n$$\nV_2 = \\frac{1}{2} MN \\cdot CE \\cdot CD \\sin \\angle ECD = \\frac{3}{2}.\n$$\nSo $V_1 = \\frac{1}{3} V_2 = \\frac{1}{2}$.\n\nAnswer: B.",
    "topic": "Geometry",
    "subtopic": "Solid Geometry"
  },
  {
    "id": 27999,
    "subject": "Mathematics (Multi-modal)",
    "question": "The solution set of the inequality $|x|^3 - 2x^2 - 4|x| + 3 < 0$ is\n________.",
    "options": [],
    "answer": "(-3, -(\\sqrt{5} - 1)/2) \\cup ((\\sqrt{5} - 1)/2, 3)",
    "solution": "Notice that $|x| = 3$ is a root of the equation $|x|^3 - 2x^2 - 4|x| + 3 = 0$. Then the original inequality can be rewritten as $(|x| - 3)(|x|^2 + |x| - 1) < 0$, that is\n$$\n(|x| - 3)\\left(|x| - \\frac{-1 + \\sqrt{5}}{2}\\right)\\left(|x| - \\frac{-1 - \\sqrt{5}}{2}\\right) < 0.\n$$\nSince $|x| - \\frac{-1 - \\sqrt{5}}{2} > 0$, then $\\frac{-1 + \\sqrt{5}}{2} < |x| < 3$.\nSo the solution set is $(-3, -\\frac{\\sqrt{5} - 1}{2}) \\cup (\\frac{\\sqrt{5} - 1}{2}, 3)$.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 28000,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $x_1, x_2, \\dots, x_5$ be nonnegative real numbers with $\\sum_{i=1}^5 \\frac{1}{1+x_i} = 1$.\nProve that $\\sum_{i=1}^5 \\frac{x_i}{4+x_i^2} \\le 1$. (posed by Li Shenghong)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $y_i = \\frac{1}{1+x_i}$, $i=1, 2, \\dots, 5$, then $x_i = \\frac{1-y_i}{y_i}$, $i=1, 2, \\dots, 5$ and $\\sum_{i=1}^5 y_i = 1$.\nWe have\n$$\n\\begin{align*}\n\\sum_{i=1}^{5} \\frac{x_i}{4+x_i^2} &\\le 1 \\Leftrightarrow \\sum_{i=1}^{5} \\frac{-y_i^2+y_i}{5y_i^2-2y_i+1} \\le 1 \\\\\n&\\Leftrightarrow \\sum_{i=1}^{5} \\frac{-5y_i^2+5y_i}{5y_i^2-2y_i+1} \\le 5 \\\\\n&\\Leftrightarrow \\sum_{i=1}^{5} \\left(-1+\\frac{3y_i+1}{5y_i^2-2y_i+1}\\right) \\le 5 \\\\\n&\\Leftrightarrow \\sum_{i=1}^{5} \\frac{3y_i+1}{5\\left(y_i-\\frac{1}{5}\\right)^2+\\frac{4}{5}} \\le 10.\n\\end{align*}\n$$\nFurthermore,\n$$\n\\begin{align*}\n\\sum_{i=1}^{5} \\frac{3y_i + 1}{5\\left(y_i - \\frac{1}{5}\\right)^2 + \\frac{4}{5}} &\\le \\sum_{i=1}^{5} \\frac{3y_i + 1}{\\frac{4}{5}} = \\frac{5}{4} \\sum_{i=1}^{5} (3y_i + 1) \\\\\n&= \\frac{5}{4} \\times (3+5) = 10.\n\\end{align*}\n$$\nThis completes the proof.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 28001,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let the sides of a scalene triangle $\\triangle ABC$ be $AB = c$, $BC = a$, $CA = b$, $D$, $E$, $F$ be points on $BC$, $CA$, $AB$, such that $AD$, $BE$, $CF$ are angle bisectors of the triangle, respectively. Assume that $DE = DF$. Prove that\n$$\n(1) \\frac{a}{b+c} = \\frac{b}{c+a} + \\frac{c}{a+b};\n$$\n$$\n(2) \\angle BAC > 90^{\\circ}.\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "**Solution** Using the sine rule, we have that\n$$\n\\frac{\\sin \\angle AFD}{\\sin \\angle FAD} = \\frac{AD}{FD} = \\frac{AD}{ED} = \\frac{\\sin \\angle AED}{\\sin \\angle FAD},\n$$\nthen $\\sin \\angle AFD = \\sin \\angle AED$. So, either $\\angle AFD = \\angle AED$ or $\\angle AFD + \\angle AED = 180^{\\circ}$.\nIf $\\angle AFD = \\angle AED$, then $\\triangle ADF \\cong \\triangle ADE$, and we get $AF = AE$. Then $\\triangle AIF \\cong \\triangle AIE$, and $\\angle AFI = \\angle AEI$. So $\\triangle AFC \\cong \\triangle AEB$, then $AC = AB$. It contradicts the condition given. So $\\angle AFD + \\angle AED = 180^{\\circ}$, and points $A, F, D$ and $E$ lie on one circle.\nThen $\\angle DEC = \\angle DFA > \\angle ABC$. Extend $CA$ through $A$ to point $P$ such that\n$$\n\\angle DPC = \\angle B, \\text{ then } PC = PE + CE. \\quad \\textcircled{1}\n$$\nSince $\\angle BFD = \\angle PED$ and $FD = ED$, we have that $\\triangle BFD \\cong \\triangle PED$, then $PE = BF = \\frac{ac}{a+b}$. Furthermore, $\\triangle PCD \\sim \\triangle BCA$, then $\\frac{PC}{BC} = \\frac{CD}{CA}$. So\n$$\nPC = a \\cdot \\frac{ba}{b+c} \\cdot \\frac{1}{b} = \\frac{a^2}{b+c}. \\quad \\textcircled{2}\n$$\n\nFrom ① and ② we get $\\frac{a^2}{b+c} = \\frac{ac}{a+b} + \\frac{ab}{c+a}$, then $\\frac{a}{b+c} = \\frac{b}{c+a} + \\frac{c}{a+b}$.\n\nAs to the proof of (2), we have from (1) that\n$$ \n\\begin{aligned}\na(a+b)(a+c) &= b(b+a)(b+c) + c(c+a)(c+b), \\\\\na^2(a+b+c) &= b^2(a+b+c) + c^2(a+b+c) + abc \\\\\n&> b^2(a+b+c) + c^2(a+b+c).\n\\end{aligned}\n$$\nThen $a^2 > b^2 + c^2$, and that means $\\angle BAC > 90^\\circ$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 28002,
    "subject": "Mathematics (Multi-modal)",
    "question": "Suppose points $F_1$, $F_2$ are the foci of the ellipse $\\frac{x^2}{9} + \\frac{y^2}{4} = 1$, $P$ is a point on the ellipse, and $|PF_1| : |PF_2| = 2 : 1$. Then the area of $\\triangle PF_1F_2$ is equal to ________.",
    "options": [],
    "answer": "4",
    "solution": "$|PF_1| + |PF_2| = 2a = 6$ by definition of an ellipse. Since $|PF_1| : |PF_2| = 2 : 1$, then $|PF_1| = 4$ and $|PF_2| = 2$. Notice that $|F_1F_2| = 2c = 2\\sqrt{5}$, and\n$$\n|PF_1|^2 + |PF_2|^2 = 4^2 + 2^2 = 20 = |F_1F_2|^2.\n$$\nThen $\\triangle PF_1F_2$ is a right triangle. So $S_{\\triangle PF_1F_2} = \\frac{1}{2} |PF_1| \\cdot |PF_2| = 4$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 28003,
    "subject": "Mathematics (Multi-modal)",
    "question": "Arrange $1\\,650$ students in $22$ rows by $75$ columns. It is known that for any two columns, the number of occasions that two students in the same row are of the same sex does not exceed $11$. Prove that the number of boy students does not exceed $928$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $a_i$ be the number of boy students in the $i$\\text{th}$ row, then the number of girl students in this row is $75 - a_i$. By the given condition, we have $\\sum_{i=1}^{22} \\left( C_{a_i}^2 + C_{75-a_i}^2 \\right) \\le 11 \\times C_{75}^2$. That is, $\\sum_{i=1}^{22} (a_i^2 - 75a_i) \\le -30,525$, implying $\\sum_{i=1}^{22} (2a_i - 75)^2 \\le 1,650$. Using Cauchy's Inequality, we have\n$$\n\\left[ \\sum_{i=1}^{22} (2a_i - 75) \\right]^2 \\le 22 \\sum_{i=1}^{22} (2a_i - 75)^2 \\le 36,300.\n$$\nThen $\\sum_{i=1}^{22} (2a_i - 75) < 191$, and $\\sum_{i=1}^{22} a_i < \\frac{191 + 1,650}{2} < 921$. That means the number of boy students does not exceed $928$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 28004,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $n$ be a positive integer, and $S_n$ be the set of all positive integer divisors of $n$ (including $1$ and itself). Prove that at most half of the elements in $S_n$ have their last digits equal to $3$. (posed by Feng Zuming)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "(1) If $5 \\mid n$, let $d_1, d_2, \\dots, d_m$ be the elements in $S_n$ with their last digits equal to $3$, then $5d_1, 5d_2, \\dots, 5d_m$ are elements in $S_n$ with their last digits equal to $5$. So $m \\le \\frac{1}{2}|S_n|$. The statement is true in this case.\n\n(2) If $5 \\nmid n$ and the last digit of every prime divisor of $n$ is either $1$ or $9$, the last digit of any element in $S_n$ is either $1$ or $9$. The statement is also true in this case.\n\n(3) If $5 \\nmid n$ and there exists a prime divisor $p$ in $S_n$ such that the last digit of $p$ is either $3$ or $7$. Let $n = p^r q$, where $q$ and $r$ are positive integers and $p$ is prime to $q$, and let $S_q = \\{a_1, a_2, \\dots, a_k\\}$ be the set of all positive integer divisors of $q$. Then the elements in $S_n$ can be written in the following way:\n\nFor any $d_i = a_s p^l \\in S_n$, we choose $e_i = \\begin{cases} a_s p^{l+1} & l < r, \\\\ a_s p^{l-1} & l = r, \\end{cases}$ then $e_i \\in S_n$ and we call $e_i$ the partner of $d_i$. If the last digit of $d_i$ is $3$, then that of its partner $e_i$ is not, since that of $p$ is either $3$ or $7$. If $d_i$ and $d_j$ in $S_n$ are different, and their last digits are both $3$, then their partners $e_i$ and $e_j$ are also different. Otherwise, suppose $e_j = e_i = a_s p^l$, we may assume that $\\{d_i, d_j\\} = \\{a_s p^{l-1}, a_s p^{l+1}\\}$, then $d_i = d_j p^2$. As the last digit of $p$ is $3$ or $7$, then that of $p^2$ is always $9$, and that means the last digits of $d_i$ and $d_j$ cannot be the same. It leads to a contradiction.\n\nWe then see that every $d_i \\in S_n$ with its last digit equal to $3$ has a partner $e_i \\in S_n$ with its last digit not equal to $3$, and different $d_i$ has different partner. That means that at most half of the elements in $S_n$ have their last digits equal to $3$. This completes the proof.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 28005,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all integers $n$, such that $n^4 + 6n^3 + 11n^2 + 3n + 31$ is a perfect square.",
    "options": [],
    "answer": "10",
    "solution": "Suppose $A = n^4 + 6n^3 + 11n^2 + 3n + 31$ is a perfect square. It means that $A = (n^2 + 3n + 1)^2 - 3(n - 10)$ is a perfect square.\n\nIf $n > 10$, then $A < (n^2 + 3n + 1)^2$, thus $A \\le (n^2 + 3n)^2$.\n\nTherefore\n$$\n(n^2 + 3n + 1)^2 - (n^2 + 3n)^2 \\le 3n - 30,\n$$\nor\n$$\n2n^2 + 3n + 31 \\le 0,\n$$\nwhich is impossible.\n\nCase I $n \\le -3$ or $0 \\le n < 10$. Then $n^2 + 3n \\ge 0$. Thus\n$$\nA \\ge (n^2 + 3n + 2)^2.\n$$\nThat is,\n$$\n2n^2 + 9n - 27 \\le 0,\n$$\nor\n$$\n-7 < \\frac{-3(\\sqrt{33} + 3)}{4} \\le n \\le \\frac{3(\\sqrt{33} - 3)}{4} < 3.\n$$\nTherefore, $n = -6, -5, -4, -3, 0, 1, 2$. For these values of $n$, the corresponding values of $A$ are $409, 166, 67, 40, 31, 52, 145$. All of them are not perfect squares.\n\nCase II $n = -2, -1$. Then $A = 37, 34$ respectively, none of them is a perfect square.\n\nIf $n = 10$, then $A = (10^2 + 3 \\times 10 + 1)^2 = 131^2$ is a perfect square.\n\nHence, only when $n = 10$, $A$ is a perfect square.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 28006,
    "subject": "Mathematics (Multi-modal)",
    "question": "In an acute triangle $ABC$, point $H$ is the intersection point of altitude $CE$ to $AB$ and altitude $BD$ to $AC$. A circle with $DE$ as its diameter intersects $AB$ and $AC$ at points $F$ and $G$, respectively. $FG$ and $AH$ intersect at point $K$. If $BC = 25$, $BD = 20$, and $BE = 7$, find the length of $AK$.",
    "options": [],
    "answer": "216/25",
    "solution": "We know that $\\angle ADB = \\angle AEC = 90^\\circ$, therefore\n\n$$\n\\triangle ADB \\sim \\triangle AEC,\n$$\nand\n$$\n\\frac{AD}{AE} = \\frac{BD}{CE} = \\frac{AB}{AC} \\quad (1)\n$$\nBut $BC = 25$, $BD = 20$, and $BE = 7$, so $CD = 15$, and $CE = 24$. From (1), we obtain\n\nThus, point $D$ is the midpoint of the hypotenuse $AC$ of $\\triangle AEC$, and\n$$\nDE = \\frac{1}{2}AC = 15.\n$$\n\nCircle with $DE$ as its diameter, $\\angle DFE = 90^\\circ$, we have\n$$\nAF = \\frac{1}{2}AE = 9.\n$$\nSince four points $G$, $F$, $E$ and $D$ are concyclic, and four points $D$, $E$, $B$ and $C$ are concyclic too, we get\n$$\n\\angle AFG = \\angle ADE = \\angle ABC.\n$$\nThus $GF \\parallel CB$. Extend line $AH$ to intersect $BC$ at point $P$, then\n$$\n\\frac{AK}{AP} = \\frac{AF}{AB} \\qquad (2)\n$$\nSince $H$ is the orthocenter of $\\triangle ABC$, $AP \\perp BC$. From $BA = BC$ we have\n$$\nAP = CE = 24.\n$$\nDue to (2), we get\n$$\nAK = \\frac{AF \\cdot AP}{AB} = \\frac{9 \\times 24}{25} = 8.64.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 28007,
    "subject": "Mathematics (Multi-modal)",
    "question": "We say a positive integer $n$ is \"good\" if there is a permutation $(a_1, a_2, \\dots, a_n)$ of $1, 2, \\dots, n$ such that $a_k + k$ is a perfect square for all $1 \\le k \\le n$. Determine all the good numbers in the set $\\{11, 13, 15, 17, 19\\}$.",
    "options": [],
    "answer": "13, 15, 17, 19",
    "solution": "The good numbers are $13$, $15$, $17$ and $19$. However $11$ is not.\n\nNote that for $1 \\le k \\le 11$, $4 + k$ is a perfect square if and only if $k = 5$. Likewise, $11 + k$ is a perfect square if and only if $k = 5$. Hence $11$ is not good.\n\nNote that $13$ is good because\n\n$k$: 1 2 3 4 5 6 7 8 9 10 11 12 13,\n$a_k$: 8 2 13 12 11 10 9 1 7 6 5 4 3.\n\nSimilarly, $15$, $17$ and $19$ are good because\n\n$k$: 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15,\n$a_k$: 15 14 13 12 11 10 9 8 7 6 5 4 3 2 1;\n\n$k$: 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17,\n$a_k$: 3 7 6 5 4 10 2 17 16 15 14 13 12 11 1 9 8;\n\n$k$: 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19,\n$a_k$: 8 7 6 5 4 3 2 1 16 15 14 13 12 11 10 9 19 18 17.",
    "topic": "Discrete Mathematics",
    "subtopic": "Combinatorics"
  },
  {
    "id": 28008,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $\\theta$ be an acute angle such that the equation $x^2 + 4x\\cos\\theta + c\\cot\\theta = 0$ involving variable $x$ has multiple roots. Then the measure of $\\theta$ in radians is ( ).\n(A) $\\frac{\\pi}{6}$\n(B) $\\frac{\\pi}{12}$ or $\\frac{5\\pi}{12}$\n(C) $\\frac{\\pi}{6}$ or $\\frac{5\\pi}{12}$\n(D) $\\frac{\\pi}{12}$",
    "options": [],
    "answer": "B",
    "solution": "Since the equation $x^2 + 4x\\cos\\theta + c\\cot\\theta = 0$ has multiple roots, we have\n$$\n\\Delta = 16\\cos^2\\theta - 4\\cot\\theta = 0,\n$$\nor\n$$\n4\\cot\\theta(2\\sin 2\\theta - 1) = 0.\n$$\n\nIt follows that\n$$\n2\\theta = \\frac{\\pi}{6} \\text{ or } 2\\theta = \\frac{5\\pi}{6}.\n$$\nThus $\\theta = \\frac{\\pi}{12}$ or $\\theta = \\frac{5\\pi}{12}$. Answer: B.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 28009,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $\\triangle ABC$ be an acute-angled triangle with $AB \\neq AC$. The circle with diameter $BC$ intersects the sides $AB$ and $AC$ at $M$ and $N$ respectively. Denote by $O$ the midpoint of the side $BC$. The bisectors of the angles $BAC$ and $MON$ intersect at $R$. Prove that the circumcircles of the triangles $BMR$ and $CNR$ have a common point lying on the side $BC$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "We first show that the points $A, M, R, N$ are concyclic. Since $ABC$ is an acute-angled triangle, $M$ and $N$ are on the line segments $AB$ and $AC$ respectively. Let $R_1$ be the point such that the points $A, M, R_1, N$ are concyclic, where $R_1$ is on the ray $AR$. Since $AR_1$ bisects $\\angle BAC$, we have $R_1M = R_1N$. Since $M$ and $N$ lie on the circle with centre $O$, we have $OM = ON$. It follows from $OM = ON$ and $R_1M = R_1N$ that $R_1$ is on the bisector of $\\angle MON$. Since $AB \\neq AC$, the bisectors of the angles $BAC$ and $MON$ intersect at the unique point $R$, and so $R_1 = R$, or $A, M, R, N$ are concyclic.\n\nLet the bisector of $\\angle BAC$ meet $BC$ at $K$. Since the points $B, C, N, M$ are concyclic, $\\angle MBC = \\angle ANM$. Moreover, since $A, M, R, N$ are concyclic, $\\angle ANM = \\angle MRA$. This implies $\\angle MBK = \\angle MRA$. Therefore, the points $B, M, R, K$ are concyclic. Using the same argument as above, we obtain that $C, N, R, K$ are concyclic. This completes the solution.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 28010,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $EFGH$, $ADC$ and $E_1F_1G_1H_1$ be three convex quadrilaterals, satisfying:\n(a) Points $E$, $F$, $G$ and $H$ lie on sides $AB$, $BC$, $CD$ and $DA$, respectively, and $\\frac{AE}{EB} \\cdot \\frac{BF}{FC} \\cdot \\frac{CG}{GD} \\cdot \\frac{DH}{HA} = 1$;\n(b) points $A$, $B$, $C$ and $D$ lie on sides $H_1E_1$, $E_1F_1$, $F_1G_1$ and $G_1H_1$, respectively, and $E_1F_1 \\parallel EF$, $F_1G_1 \\parallel FG$, $G_1H_1 \\parallel GH$, $H_1E_1 \\parallel HE$.\nSuppose $\\frac{E_1A}{AH_1} = \\lambda$, find the expression of $\\frac{F_1C}{CG_1}$ in terms of $\\lambda$. (posed by Xiong Bin)",
    "options": [],
    "answer": "lambda",
    "solution": "(1) If $EF \\parallel AC$, then $\\frac{BE}{EA} = \\frac{BF}{FC}$.\nSo $\\frac{DH}{HA} = \\frac{DG}{GC}$ using condition (a). Then $HG \\parallel AC$, giving $E_1F_1 \\parallel AC \\parallel H_1G_1$. That\n![](attached_image_1.png)\nmeans $\\frac{F_1C}{CG_1} = \\frac{E_1A}{AH_1} = \\lambda$.\n\n(2) If $EF$ is not parallel to $AC$, extend lines $EF$ and $AC$ to meet at point $T$. By Menelaus' Theorem, we have $\\frac{CF}{FB} \\cdot \\frac{BE}{EA} \\cdot \\frac{AT}{TC} = 1$, and then $\\frac{CG}{GD} \\cdot \\frac{DH}{HA} \\cdot \\frac{AT}{TC} = 1$ using condition (a). By the inverse of Menelaus' Theorem, we know that points $T$, $H$ and $G$ are collinear. Suppose lines $TF$ and $TG$ meet line $E_1H_1$ at $M$ and $N$ respectively. As $EB_1 \\parallel EF$, we get $E_1A = \\frac{BA}{EA} \\cdot AM$. In the same way, we get $H_1A = \\frac{AD}{AH} \\cdot AN$. Then\n![](attached_image_2.png)\n$$\n\\frac{E_1A}{H_1A} = \\frac{AM}{AN} \\cdot \\frac{AB}{AE} \\cdot \\frac{AH}{AD}. \\qquad \\textcircled{1}\n$$\n$$\n\\text{On the other hand, } \\frac{AM}{AN} = \\frac{EQ}{QH} = \\frac{\\triangle AEC}{\\triangle AHC} = \\frac{\\triangle ABC}{\\triangle ADC} \\cdot \\frac{AE}{AB} \\cdot \\frac{AD}{AH}. \\qquad \\textcircled{2}\n$$\nFrom ①, ② we get $\\frac{E_1A}{H_1A} = \\frac{EQ}{QH} \\cdot \\frac{AB}{AE} \\cdot \\frac{AH}{AD} = \\frac{\\triangle ABC}{\\triangle ADC}$. In the same way,\n$$\n\\frac{F_1C}{CG_1} = \\frac{\\triangle ABC}{\\triangle ADC}.\n$$\n$$\n\\text{So } \\frac{F_1C}{CG_1} = \\frac{E_1A}{AH_1} = \\lambda.\n$$",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 28011,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $p$ be an odd prime. Let $k$ be a positive integer such that $\\sqrt{k^2 - pk}$ is also a positive integer. Then $k = \\underline{\\hspace{2cm}}$.",
    "options": [],
    "answer": "(p+1)^2/4",
    "solution": "Set $\\sqrt{k^2 - pk} = n$, $n \\in \\mathbb{N}$. Thus $k^2 - pk - n^2 = 0$, and $k = \\frac{p \\pm \\sqrt{p^2 + 4n^2}}{2}$, which implies that $p^2 + 4n^2$ is a perfect square, say $m^2$, where $m \\in \\mathbb{N}$. So $(m-2n)(m+2n) = p^2$.\n\nSince $p$ is a prime and $p \\ge 3$, we have\n$$\n\\begin{cases} m - 2n = 1, \\\\ m + 2n = p^2. \\end{cases}\n$$\nSolve the equations above and we get\n$$\n\\begin{cases} m = \\frac{p^2 + 1}{2}, \\\\ n = \\frac{p^2 - 1}{4}. \\end{cases}\n$$\n\nConsequently, $k = \\frac{p \\pm m}{2} = \\frac{2p \\pm (p^2 + 1)}{4}$. Thus $k = \\frac{(p+1)^2}{4}$ (the negative value is omitted).",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 28012,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $a_0, a_1, a_2, \\dots, a_n, \\dots$ be a sequence of numbers satisfying $(3 - a_{n+1}) \\cdot (6 + a_n) = 18$, and $a_0 = 3$. Then $\\sum_{i=0}^{n} \\frac{1}{a_i}$ equals ________.",
    "options": [],
    "answer": "(1/3)(2^{n+2} - n - 3)",
    "solution": "Set $b_n = \\frac{1}{a_n}$, $n = 0, 1, 2, \\dots$, then $(3 - \\frac{1}{b_{n+1}})(6 + \\frac{1}{b_n}) = 18$, namely,\n$$\n3b_{n+1} - 6b_n - 1 = 0.\n$$\nHence $b_{n+1} = 2b_n + \\frac{1}{3}$, or $b_{n+1} + \\frac{1}{3} = 2(b_n + \\frac{1}{3})$. So $\\{b_n + \\frac{1}{3}\\}$ is a geometric progression with common ratio $2$. Thus\n$$\nb_n + \\frac{1}{3} = 2^n \\left(b_0 + \\frac{1}{3}\\right) = 2^n \\left(\\frac{1}{a_0} + \\frac{1}{3}\\right) = \\frac{1}{3} \\times 2^{n+1}, \\\\\nb_n = \\frac{1}{3}(2^{n+1} - 1).\n$$\n\n$$\n\\sum_{i=0}^{n} \\frac{1}{a_i} = \\sum_{i=0}^{n} b_i = \\sum_{i=0}^{n} \\frac{1}{3}(2^{i+1} - 1) = \\frac{1}{3} \\left[ \\frac{2(2^{n+1} - 1)}{2-1} - (n+1) \\right]\n$$\n$$\n= \\frac{1}{3}(2^{n+2} - n - 3).\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 28013,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $M(-1, 2)$ and $N(1, 4)$ be two points in a plane rectangular coordinate system $xOy$. $P$ is a moving point on the $x$-axis. When $\\angle MPN$ takes its maximum value, the $x$-coordinate of point $P$ is ________.",
    "options": [],
    "answer": "1",
    "solution": "The center of a circle passing through points $M$ and $N$ is on the perpendicular bisector $y = 3 - x$ of $MN$. Denote the center by $S(a, 3-a)$, then the equation of the circle $S$ is\n$$\n(x-a)^2 + (y-3+a)^2 = 2(1+a^2).\n$$\nSince for a chord with a fixed length, the angle at the circumference subtended by the corresponding arc will become larger as the radius of the circle becomes smaller. When $\\angle MPN$ reaches its maximum value, the circle $S$ through the three points $M$, $N$ and $P$ will be tangent to the $x$-axis at $P$, which means the value $a$ in the equation of $S$ has to satisfy the condition $2(1+a^2) = (a-3)^2$. Solve the above equation we have $a = 1$ or $a = -7$. Thus the points of contact are $P(1, 0)$ and $P'(-7, 0)$ respectively.\nBut the radius of the circle through points $M$, $N$, and $P'$ is larger than that of the circle through points $M$, $N$ and $P$. Therefore $\\angle MPN > \\angle MP'N$. Thus $P(1, 0)$ is the point we want to find, and the $x$-axis of point $P$ is 1.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 28014,
    "subject": "Mathematics (Multi-modal)",
    "question": "The rule of an \"obstacle course\" specifies that at the $n$th obstacle a person has to toss a die $n$ times. If the sum of points in these $n$ tosses is bigger than $2^n$, the person is said to have crossed the obstacle.\n\n(1) At most how many obstacles can a person cross?\n\n(2) What is the probability that a person crosses the first three obstacles?",
    "options": [],
    "answer": "(1) 4; (2) 100/243",
    "solution": "Since the die is fair, the probability of any of the six numbers appearing is the same.\n\n(1) Since the highest point of a die is $6$, and $6 \\times 4 > 2^4$, $6 \\times 5 < 2^5$, it is impossible that the sum of points appearing in $n$ tosses is bigger than $2^n$ if $n \\ge 5$. This means it is an impossible event, and the probability of crossing the obstacle is $0$.\nTherefore at most $4$ obstacles that a person can cross.\n\n(2) We denote $A_n$ the event \"at the $n$th obstacle the person fails to cross\", the complementary event $\\overline{A_n}$ is \"at the $n$th obstacle the person crosses successfully\".\nAt the $n$th obstacle of this game the number of all possible outcomes is $6^n$.\n\nThe first obstacle: event $A_1$ contains $2$ possible outcomes (i.e., the outcomes in which the number appearing is $1$ or $2$). So the probability of crossing the obstacle is\n$$\nP(\\overline{A_1}) = 1 - P(A_1) = 1 - \\frac{2}{6} = \\frac{2}{3}.\n$$\n\nThe second obstacle: the number of outcomes contained in event $A_2$ is the total number of positive integer solution sets of the equation $x+y=a$ where $a$ is taken to be $2$, $3$ and $4$ respectively. Thus the number of outcomes equals $C_1^1 + C_2^1 + C_3^1 = 1+2+3=6$, and the probability of crossing the obstacle is\n$$\nP(\\overline{A_2}) = 1 - P(A_2) = 1 - \\frac{6}{6^2} = \\frac{5}{6}.\n$$\n\nThe third obstacle: the number of outcomes contained in event $A_3$ is the total number of positive integer solution sets of the equation $x+y+z=a$ where $a$ is taken to be $3$, $4$, $5$, $6$, $7$ and $8$ respectively. Thus the number of outcomes equals\n$$\nC_2 + C_3 + C_4 + C_5 + C_6 + C_7 = 1 + 3 + 6 + 10 + 15 + 21 = 56,\n$$\nand the probability of crossing the obstacle is\n$$\nP(\\overline{A_3}) = 1 - P(A_3) = 1 - \\frac{56}{6^3} = \\frac{20}{27}.\n$$\n\nConsequently, the probability that a person crosses the first three obstacles is\n$$\nP(\\overline{A_1}) \\times P(\\overline{A_2}) \\times P(\\overline{A_3}) = \\frac{2}{3} \\times \\frac{5}{6} \\times \\frac{20}{27} = \\frac{100}{243}.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 28015,
    "subject": "Mathematics (Multi-modal)",
    "question": "In a plane rectangular coordinate system $xOy$ there are three points $A(0, \\frac{4}{3})$, $B(-1, 0)$ and $C(1, 0)$. The distance from point $P$ to line $BC$ is the geometric mean of the distances from this point to lines $AB$ and $AC$.\n\n(1) Find the locus equation of point $P$.\n\n(2) If line $L$ passes through the incenter (say, $D$) of $\\triangle ABC$, and has exactly 3 common points with the locus of point $P$. Determine all values of the slope $k$ of line $L$.",
    "options": [],
    "answer": "Locus: 2x^2 + 2y^2 + 3y − 2 = 0 and 8x^2 − 17y^2 + 12y − 8 = 0. Slopes k: {0, ±1/2, ±(2√34)/17, ±√2/2}.",
    "solution": "(1) The equations of lines $AB$, $AC$ and $BC$ are $y = \\frac{4}{3}(x+1)$, $y = -\\frac{4}{3}(x-1)$ and $y = 0$ respectively. The distances from point $P$ to $AB$, $AC$ and $BC$ are respectively\n$$\nd_1 = \\frac{1}{5} | 4x - 3y + 4 |,\n$$\n$$\nd_2 = \\frac{1}{5} | 4x + 3y - 4 |, \\quad d_3 = | y |.\n$$\nAccording to the assumption, $d_1 d_2 = d_3^2$, we have $|16x^2 - (3y-4)^2| = 25y^2$. That is\n$$\n16x^2 - (3y-4)^2 + 25y^2 = 0, \\text{ or } 16x^2 - (3y-4)^2 - 25y^2 = 0.\n$$\nBy simplifying the above equations, we obtain that the locus equations of point $P$ consist of\n$$\n\\text{circle S: } 2x^2 + 2y^2 + 3y - 2 = 0\n$$\nand\n$$\n\\text{hyperbola T: } 8x^2 - 17y^2 + 12y - 8 = 0.\n$$\n\n(2) The incenter of $\\triangle ABC$ is also a point satisfying the assumption.\nIn view of $d_1 = d_2 = d_3$, solving the equations we have $D(0, \\frac{1}{2})$.\nLine $L$ passes through $D$, and has three common points with the locus of point $P$. So the slope of $L$ is defined. Suppose that the equation of $L$ is\n$$\ny = kx + \\frac{1}{2}.\n$$\n(i) If $k=0$, then $L$ is tangent to the circle $S$, which means there is a unique common point $D$. In this case line $L$ is parallel to the $x$-axis, which implies that $L$ and hyperbola $T$ have two other common points different from point $D$. Hence, there are just three common points for $L$ and the locus of point $P$.\n\nCase 1: Line $L$ passes through point $B$ or point $C$, which means that the slope of $L$ is $k = \\pm \\frac{1}{2}$, and the equation of $L$ is $x = \\pm (2y-1)$. Substitute it into equation $\\textcircled{2}$ we get\n$$\ny(3y-4) = 0.\n$$\nSolving it we have $E(\\frac{5}{3}, \\frac{4}{3})$ or $F(-\\frac{5}{3}, \\frac{4}{3})$, which means that line $BD$ and curve $T$ have 2 intersection points $B$ and $E$, and line $CD$ and curve $T$ have 2 intersection points $C$ and $F$.\n\nCase 2: Line $L$ does not pass through point $B$ and point $C$ (i.e. $k \\neq \\pm \\frac{1}{2}$). Since for $L$ and $S$ there are two different intersection points, there exists a unique common point for $L$ and hyperbola $T$. Thus for the following system of equations\n$$\n\\begin{cases} 8x^2 - 17y^2 + 12y - 8 = 0, \\\\ y = kx + \\frac{1}{2}, \\end{cases}\n$$\nthere is one and only one real solution. After eliminating $y$ and simplifying we have\n$$\n(8 - 17k^2)x^2 - 5kx - \\frac{25}{4} = 0.\n$$\nThe above equation has a unique real solution if and only if\n$$\n8 - 17k^2 = 0\n$$\nor\n$$\n(-5k)^2 + 4(8 - 17k^2) \\frac{25}{4} = 0.\n$$\nSolving $8 - 17k^2 = 0$ we get $k = \\pm \\frac{2\\sqrt{34}}{17}$. And solving $(-5k)^2 + 4(8 - 17k^2) \\frac{25}{4} = 0$ we obtain $k = \\pm \\frac{\\sqrt{2}}{2}$.\nConsequently, the set of all possible values of the slope $k$ of line $L$ is the following finite set:\n$$\n\\{0, \\pm\\frac{1}{2}, \\pm\\frac{2\\sqrt{34}}{17}, \\pm\\frac{\\sqrt{2}}{2}\\}\n$$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 28016,
    "subject": "Mathematics (Multi-modal)",
    "question": "Suppose that $\\alpha$ and $\\beta$ are different real roots of the equation $4x^2 - 4tx - 1 = 0$ ($t \\in \\mathbb{R}$). $[\\alpha, \\beta]$ is the domain of the function $f(x) = \\frac{2x-t}{x^2+1}$.\n\n(1) Find $g(t) = \\max f(x) - \\min f(x)$.\n\n(2) Prove that for $u_i \\in (0, \\frac{\\pi}{2})$ ($i = 1, 2, 3$), if $\\sin u_1 + \\sin u_2 + \\sin u_3 = 1$, then:\n$$\n\\frac{1}{g(\\tan u_1)} + \\frac{1}{g(\\tan u_2)} + \\frac{1}{g(\\tan u_3)} < \\frac{3}{4}\\sqrt{6}.\n$$",
    "options": [],
    "answer": "g(t) = 8 sqrt(t^2 + 1) (2 t^2 + 5) / (16 t^2 + 25), and for acute u1, u2, u3 with sin u1 + sin u2 + sin u3 = 1: 1/g(tan u1) + 1/g(tan u2) + 1/g(tan u3) < (3/4) sqrt(6).",
    "solution": "(1) Let $\\alpha \\le x_1 < x_2 \\le \\beta$, then\n$$4x_1^2 - 4tx_1 - 1 \\le 0, \\quad 4x_2^2 - 4tx_2 - 1 \\le 0.$$\nTherefore,\n$$\n4(x_1^2 + x_2^2) - 4(t(x_1 + x_2) - 2) \\le 0, \\\\\n2x_1x_2 - t(x_1 + x_2) - \\frac{1}{2} < 0.\n$$\nBut\n$$\nf(x_2) - f(x_1) = \\frac{2x_2 - t}{x_2^2 + 1} - \\frac{2x_1 - t}{x_1^2 + 1}\n$$\n$$\n= \\frac{(x_2 - x_1)[t(x_1 + x_2) - 2x_1x_2 + 2]}{(x_2^2 + 1)(x_1^2 + 1)},\n$$\nand $t(x_1 + x_2) - 2x_1x_2 + 2 > t(x_1 + x_2) - 2x_1x_2 + \\frac{1}{2} > 0$, thus\n$$\nf(x_2) - f(x_1) > 0.\n$$\nConsequently, $f(x)$ is an increasing function on the interval $[\\alpha, \\beta]$.\n$$\n\\text{Since } \\alpha + \\beta = t \\text{ and } \\alpha\\beta = -\\frac{1}{4},\n$$\n$$\ng(t) = \\max\\{f(x)\\} - \\min\\{f(x)\\} = f(\\beta) - f(\\alpha)\n$$\n$$\n= \\frac{\\sqrt{t^2 + 1}\\left(t^2 + \\frac{5}{2}\\right)}{t^2 + \\frac{25}{16}} = \\frac{8\\sqrt{t^2 + 1}(2t^2 + 5)}{16t^2 + 25}.\n$$\n\n(2)\n$$\ng(\\tan u_i) = \\frac{\\frac{8}{\\cos u_i}(\\frac{2}{\\cos^2 u_i} + 3)}{\\frac{16}{\\cos^2 u_i} + 9} = \\frac{\\frac{16}{\\cos u_i} + 24\\cos u_i}{16 + 9\\cos^2 u_i} \\\\\n\\ge \\frac{2\\sqrt{16 \\times 24}}{16 + 9\\cos^2 u_i} = \\frac{16\\sqrt{6}}{16 + 9\\cos^2 u_i} \\quad (i = 1, 2, 3),\n$$\nso\n$$\n\\begin{align*}\n\\sum_{i=1}^{3} \\frac{1}{g(\\tan u_i)} &\\le \\frac{1}{16\\sqrt{6}} \\sum_{i=1}^{3} (16 + 9\\cos^2 u_i) \\\\\n&= \\frac{1}{16\\sqrt{6}} (16 \\times 3 + 9 \\times 3 - 9 \\sum_{i=1}^{3} \\sin^2 u_i).\n\\end{align*}\n$$\nSince $\\sum_{i=1}^{3} \\sin u_i = 1$, and $u_i \\in (0, \\frac{\\pi}{2})$, $i = 1, 2, 3$, we obtain\n$$\n3 \\sum_{i=1}^{3} \\sin^2 u_i > \\left( \\sum_{i=1}^{3} \\sin u_i \\right)^2 = 1.\n$$\n$$\n\\begin{gathered}\n\\frac{1}{g(\\tan u_1)} + \\frac{1}{g(\\tan u_2)} + \\frac{1}{g(\\tan u_3)} \\\\\n< \\frac{1}{16\\sqrt{6}} (75 - 9 \\times \\frac{1}{3}) = \\frac{3}{4}\\sqrt{6}.\n\\end{gathered}\n$$",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 28017,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $a$, $b$ and $c$ be positive real numbers. Determine the minimum value of\n$$ \\frac{a+3c}{a+2b+c} + \\frac{4b}{a+b+2c} - \\frac{8c}{a+b+3c} \\text{ (posed by Li Shenghong)} $$",
    "options": [],
    "answer": "12*sqrt(2)-17",
    "solution": "The answer is $12\\sqrt{2}-17$. Set\n$$\n\\begin{cases} x = a+2b+c, \\\\ y = a+b+2c, \\\\ z = a+b+3c. \\end{cases}\n$$\n\n$$\n\\begin{aligned}\n& \\frac{a+3c}{a+2b+c} + \\frac{4b}{a+b+2c} - \\frac{8c}{a+b+3c} \\\\\n&= \\frac{2y-x}{x} + \\frac{4(x+z-2y)}{y} - \\frac{8(z-y)}{z} \\\\\n&= -17+2\\frac{y}{x}+4\\frac{x}{y}+4\\frac{z}{y}+8\\frac{y}{z} \\\\\n&\\geq -17+2\\sqrt{8}+2\\sqrt{32} = -17+12\\sqrt{2}.\n\\end{aligned}\n$$\nThe equality holds if and only if $\\frac{2y}{x} = \\frac{4x}{y}$ and $\\frac{4z}{y} = \\frac{8y}{z}$, or $4x^2 = 2y^2 = z^2$. Hence the equality holds if and only if\n$$\n\\begin{cases} a+b+2c = \\sqrt{2}(a+2b+c), \\\\ a+b+3c = 2(a+2b+c). \\end{cases}\n$$\nSolving the above system of equations for $b$ and $c$ in terms of $a$ gives\n$$\n\\begin{cases} b = (1+\\sqrt{2})a, \\\\ c = (4+3\\sqrt{2})a. \\end{cases}\n$$\nWe conclude that\n$$\n\\frac{a+3c}{a+2b+c} + \\frac{4b}{a+b+2c} - \\frac{8c}{a+b+3c}\n$$\nhas minimum value $12\\sqrt{2}-17$ if and only if $(a, b, c) = (a, (1+\\sqrt{2})a, (4+3\\sqrt{2})a)$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 28018,
    "subject": "Mathematics (Multi-modal)",
    "question": "In a planar rectangular coordinate system, a sequence of points $\\{A_n\\}$ on the positive half of the $y$-axis and a sequence of points $\\{B_n\\}$ on the curve $y = \\sqrt{2x}$ ($x \\ge 0$) satisfy the condition $|OA_n| = |OB_n| = \\frac{1}{n}$. The $x$-intercept of line segment $A_nB_n$ is $a_n$, and the $x$-coordinate of point $B_n$ is $b_n$, $n \\in \\mathbb{N}$. Prove that\n\n(1) $a_n > a_{n+1} > 4$, $n \\in \\mathbb{N}$;\n\n(2) There is $n_0 \\in \\mathbb{N}$, such that for any $n > n_0$, $\\frac{b_2}{b_1} + \\frac{b_3}{b_2} + \\cdots + \\frac{b_{n_0}}{b_{n_0-1}} + \\frac{b_{n+1}}{b_n} < n - 2004$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "(1) According to the stated conditions we have $A_n(0, \\frac{1}{n})$, and $B_n(b_n, \\sqrt{2b_n})$ ($b_n > 0$). From $|OB_n| = \\frac{1}{n}$ we get\n$$\nb_n^2 + 2b_n = \\left(\\frac{1}{n}\\right)^2.\n$$\nThus,\n$$\nb_n = \\sqrt{\\left(\\frac{1}{n}\\right)^2 + 1} - 1, \\quad n \\in \\mathbb{N}.\n$$\n\nSince $2n^2b_n = 1 - n^2b_n^2 > 0$, we have $b_n + 2 = \\frac{1}{n^2b_n}$ and\n$$\n\\begin{aligned}\na_n &= \\frac{b_n(1 + n\\sqrt{2b_n})}{1 - 2n^2b_n} = \\frac{b_n(1 + n\\sqrt{2b_n})}{1 - (1 - n^2b_n^2)} \\\\\n&= \\frac{1}{n^2b_n} + \\frac{\\sqrt{2}}{\\sqrt{n^2b_n}} = b_n + 2 + \\sqrt{2(b_n + 2)}.\n\\end{aligned}\n$$\nThus,\n$$\na_n = \\sqrt{\\left(\\frac{1}{n}\\right)^2 + 1} + 1 + \\sqrt{2\\sqrt{\\left(\\frac{1}{n}\\right)^2 + 1} + 2}.\n$$\nSince $\\frac{1}{n} > \\frac{1}{n+1} > 0$, $a_n > a_{n+1} > 4$ for any $n \\in \\mathbb{N}$.\n\n(2) Let $c_n = 1 - \\frac{b_{n+1}}{b_n}$ for $n \\in \\mathbb{N}$, then\n$$\n\\begin{align*}\nc_n &= \\frac{\\sqrt{\\left(\\frac{1}{n}\\right)^2 + 1} - \\sqrt{\\left(\\frac{1}{n+1}\\right)^2 + 1}}{\\sqrt{\\left(\\frac{1}{n}\\right)^2 + 1} - 1} \\\\\n&= n^2 \\left[ \\frac{1}{n^2} - \\frac{1}{(n+1)^2} \\right] \\cdot \\frac{\\sqrt{\\left(\\frac{1}{n}\\right)^2 + 1} + 1}{\\sqrt{\\left(\\frac{1}{n}\\right)^2 + 1} + \\sqrt{\\left(\\frac{1}{n+1}\\right)^2 + 1}} \\\\\n&> \\frac{2n+1}{(n+1)^2} \\left[ \\frac{1}{2} + \\frac{1}{2\\sqrt{\\left(\\frac{1}{n}\\right)^2 + 1}} \\right] > \\frac{2n+1}{2(n+1)^2}.\n\\end{align*}\n$$\nSince $(2n+1)(n+2) - 2(n+1)^2 = n > 0$, we obtain\n$$\nc_n > \\frac{1}{n+2}, \\quad n \\in \\mathbb{N}.\n$$\nLet $S_n = c_1 + c_2 + \\dots + c_n, \\quad n \\in \\mathbb{N}$. If $n = 2^k - 2 > 1$ ($k \\in \\mathbb{N}$), then\n$$\n\\begin{align*}\nS_n &> \\frac{1}{3} + \\frac{1}{4} + \\dots + \\frac{1}{2^k - 1} + \\frac{1}{2^k} \\\\\n&= \\left(\\frac{1}{3} + \\frac{1}{4}\\right) + \\left[\\frac{1}{2^2 + 1} + \\dots + \\frac{1}{2^3}\\right] + \\dots + \\left[\\frac{1}{2^{k-1} + 1} + \\dots + \\frac{1}{2^k}\\right] \\\\\n&> 2 \\cdot \\frac{1}{2^2} + 2^2 \\cdot \\frac{1}{2^3} + \\dots + 2^{k-1} \\cdot \\frac{1}{2^k} = \\frac{k-1}{2}.\n\\end{align*}\n$$\nTherefore, if we put $n_0 = 2^{4009} - 2$, then for any $n > n_0$ we have\n$$\n\\begin{aligned}\n\\left[1 - \\frac{b_2}{b_1}\\right] + \\left[1 - \\frac{b_3}{b_2}\\right] + \\dots + \\left[1 - \\frac{b_{n+1}}{b_n}\\right] &= S_n > S_{n_0} > \\frac{4009 - 1}{2} \\\\\n&= 2004.\n\\end{aligned}\n$$\nConsequently,\n$$\n\\frac{b_2}{b_1} + \\frac{b_3}{b_2} + \\cdots + \\frac{b_n}{b_{n-1}} + \\frac{b_{n+1}}{b_n} < n - 2004, \\quad n > n_0.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 28019,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all polynomials $P(x)$ with real coefficients, which satisfy the equation\n$$\nP(a-b) + P(b-c) + P(c-a) = 2P(a+b+c)\n$$\nfor all real numbers $a, b, c$ such that $ab + bc + ca = 0$.",
    "options": [],
    "answer": "P(x) = Ax^4 + Bx^2 for real A, B",
    "solution": "Let $P(x)$ satisfy the given equation.\n$$\n\\text{If } a = b = c = 0, \\text{ then } P(0) = 0.\n$$\n$$\n\\text{If } b = c = 0, \\text{ then } P(-a) = P(a) \\text{ for all real } a.\n$$\nHence $P(x)$ is even. Without loss of generality, we may assume that\n$$\nP(x) = a_n x^{2n} + \\cdots + a_1 x^2,\\ a_n \\neq 0.\n$$\nIf $b = 2a,\\ c = -\\frac{2}{3}a$, we have that\n$$\nP(-a) + P\\left(\\frac{8}{3}a\\right) + P\\left(-\\frac{5}{3}a\\right) = 2P\\left(\\frac{7}{3}a\\right),\n$$\nor\n$$\na_n \\left[ 1 + \\left(\\frac{8}{3}\\right)^{2n} + \\left(\\frac{5}{3}\\right)^{2n} - 2\\left(\\frac{7}{3}\\right)^{2n} \\right] a^{2n} + \\cdots + a_1 \\left[ 1 + \\left(\\frac{8}{3}\\right)^2 + \\left(\\frac{5}{3}\\right)^2 - 2\\left(\\frac{7}{3}\\right)^2 \\right] a^2 = 0\n$$\nfor all $a \\in \\mathbb{R}$. Then all coefficients of the polynomial with variable $a$ are $0$.\nIf $n \\ge 3$, it follows from $8^6 = 262\\,144 > 235\\,298 = 2 \\times 7^6$ that $\\left(\\frac{8}{7}\\right)^{2n} \\ge \\left(\\frac{8}{7}\\right)^6 > 2$. This implies\n$$\n1 + \\left(\\frac{8}{3}\\right)^{2n} + \\left(\\frac{5}{3}\\right)^{2n} - 2\\left(\\frac{7}{3}\\right)^{2n} > 0.\n$$\nHence $n \\le 2$. Let $P(x) = a x^4 + \\beta x^2$, with $a, \\beta \\in \\mathbb{R}$.\n\nWe now show that $P(x) = a x^4 + \\beta x^2$ satisfies the given equation.\nLet $a, b, c$ be real numbers satisfying $ab + bc + ca = 0$. Then\n$$\n\\begin{align*} \n(a-b)^4 + (b-c)^4 + (c-a)^4 - 2(a+b+c)^4 \n&= \\sum (a^4 - 4a^3b + 6a^2b^2 - 4ab^3 + b^4) - 2(a^2 + b^2 + c^2)^2 \\\\\n&= \\sum (a^4 - 4a^3b + 6a^2b^2 - 4ab^3 + b^4) - \\\\\n& \\qquad 2a^4 - 2b^4 - 2c^4 - 4a^2b^2 - 4a^2c^2 - 4b^2c^2 \\\\\n&= \\sum (-4a^3b + 2a^2b^2 - 4ab^3) \\\\\n&= -4a^2(ab + ca) - 4b^2(bc + ab) - 4c^2(ca + bc) + \\\\\n& \\qquad 2(a^2b^2 + b^2c^2 + c^2a^2) \\\\\n&= 4a^2bc + 4b^2ca + 4c^2ab + 2a^2b^2 + 2b^2c^2 + 2c^2a^2 \\\\\n&= 2(ab + bc + ca)^2 = 0, \\\\\n(a-b)^2 + (b-c)^2 + (c-a)^2 - 2(a+b+c)^2 \\\\\n&= \\sum (a^2 - 2ab + b^2) - 2\\sum a^2 - 4\\sum ab \\\\\n&= 0.\n\\end{align*}\n$$\nHence $P(x) = a x^4 + \\beta x^2$ satisfies the given equation.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 28020,
    "subject": "Mathematics (Multi-modal)",
    "question": "Assume that $M = \\{(x, y) \\mid x^2 + 2y^2 = 3\\}$, and $N = \\{(x, y) \\mid y = mx + b\\}$. If $M \\cap N \\neq \\emptyset$ for all $m \\in \\mathbb{R}$, then $b$ takes values from ( ).\n(A) $\\left[-\\frac{\\sqrt{6}}{2}, \\frac{\\sqrt{6}}{2}\\right]$\n(B) $\\left(-\\frac{\\sqrt{6}}{2}, \\frac{\\sqrt{6}}{2}\\right)$\n(C) $\\left(-\\frac{\\sqrt{6}}{2}, \\frac{\\sqrt{6}}{2}\\right]$\n(D) $\\left[-\\frac{2\\sqrt{3}}{3}, \\frac{2\\sqrt{3}}{3}\\right]$",
    "options": [],
    "answer": "A",
    "solution": "For any $m \\in \\mathbb{R}$ we have $M \\cap N \\neq \\emptyset$, which means point $(0, b)$ is on or in the ellipsoid $\\frac{x^2}{3} + \\frac{2y^2}{3} = 1$. Therefore\n$$\n\\frac{2b^2}{3} \\le 1, \\text{ or } -\\frac{\\sqrt{6}}{2} \\le b \\le \\frac{\\sqrt{6}}{2}.\n$$\nAnswer: A.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 28021,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABC$ be an obtuse triangle inscribed in a circle of radius $1$. Prove that triangle $ABC$ can be covered by an isosceles right-angled triangle with hypotenuse $\\sqrt{2} + 1$. (posed by Leng Gangsong)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Without loss of generality, we may assume that $\\angle C > 90^\\circ$. Since $\\angle A + \\angle B < 90^\\circ$, we may assume without loss of generality that $\\angle A < 45^\\circ$.\n\nWe can then construct a semicircle $\\omega$ with $AB$ as its diameter such that point $C$ lies inside $\\omega$. Let $O$ be the center of $\\omega$. Then $AO = BO$. Construct rays $AT$ and $OE$ such that $\\angle BAT = \\angle BOE = 45^\\circ$ with $E$ lying on $\\omega$. Let $l$ be the line tangent to $\\omega$ at $E$, and let $l$ meet rays $AT$ and $AB$ at $F$ and $D$ respectively. It is not difficult to see that triangle $AFD$ is an isosceles right-angled triangle, with $\\angle F = 90^\\circ$, that covers triangle $ABC$.\n\nIt suffices to show that $AD < \\sqrt{2} + 1$. Note that\n$$\nAD = AO + OD = AO + \\sqrt{2} OE = (\\sqrt{2} + 1) AO.\n$$\nApplying the sine rule to triangle $ABC$ gives $AB = 2 \\sin C < 2$, or $AO < 1$. It follows that $AD < \\sqrt{2} + 1$, as desired.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 28022,
    "subject": "Mathematics (Multi-modal)",
    "question": "Define a hook to be a figure made up of six unit squares as shown in the diagram or any of the figures obtained by applying rotations and reflections to this figure.\n![](attached_image_1.png)\nDetermine all $m \\times n$ rectangles that can be covered with hooks so that\n* the rectangle is covered without gaps and without overlaps;\n* no part of a hook covers area outside the rectangle.",
    "options": [],
    "answer": "All rectangles with positive integer side lengths such that either (i) one side is divisible by 3 and the other by 4, or (ii) one side is divisible by 12 and the other is at least 7 (the roles of the sides may be interchanged).",
    "solution": "$m$ and $n$ should be the positive integers and should satisfy one of the following conditions:\n\n(1) $3 \\mid m$ and $4 \\mid n$ (or vice versa);\n(2) one of $m$ and $n$ is divisible by $12$ and one is not less than $7$.\n\nA figure is obtained by applying rotations and reflections to another figure. We regard the two figures as equivalent.\nLabel the six unit squares of the hook as shown below. The shaded square must belong to another hook, and it is adjacent to only one square of this other hook. Then the only possibility of the shaded square is $1$ or $6$.\n\n(i) If it is $6$, two hooks form a $3 \\times 4$ rectangle. We call it $\\mathbf{(1)}$.\n![](attached_image_2.png)\n(ii) If it is $1$, there are two cases.\nIt is easy to see that the shaded square cannot be covered in the first diagram as shown below. Hence the latter is true. We call it $\\mathbf{(2)}$.\n![](attached_image_3.png)\n\nThus, in a tessellation, all hooks are matched into pairs. Each pair forms $\\mathbf{(1)}$ or $\\mathbf{(2)}$.\n$\\mathbf{(1)}$\nThere are $12$ squares in $\\mathbf{(1)}$ and $\\mathbf{(2)}$. Hence $12 \\mid m n$.\n![](attached_image_4.png)\n$\\mathbf{(2)}$\n\nNow we consider three cases, separately.\n\n(1) $3 \\mid m$ and $4 \\mid n$ (or vice versa)\nWithout loss of generality, we may assume $m = 3m_0$ and $n = 4n_0$.\nThen $m_0 n_0$ rectangles of the type $\\mathbf{(1)}$ form an $m_0 \\times n_0$ rectangle. Since two hooks cover a $3 \\times 4$ rectangle, an $m \\times n$ rectangle can be covered with hooks.\n\n(2) $12 \\mid m$ or $12 \\mid n$. Without loss of generality, we may assume $12 \\mid m$.\nIf $3 \\mid n$ or $4 \\mid n$, the question reduces to (1).\nAssume that $n$ is not divisible by $3$ nor by $4$. If a tessellation exists, then there is at least one of $\\mathbf{(1)}$ and $\\mathbf{(2)}$ in it, so $n \\ge 3$. Hence $n \\ge 5$ because $3 \\times n$ and $4 \\times n$. Since the square at the corners can belong to either $\\mathbf{(1)}$ or $\\mathbf{(2)}$, it follows from $n \\ge 5$ that the squares at the adjacent corners cannot belong to the same type $\\mathbf{(1)}$ or $\\mathbf{(2)}$. Hence $n \\ge 6$. Since $n$ is not divisible by $3$ and $4$, $n \\ge 7$.\n\n(3) $12 \\mid m n$, but neither $m$ nor $n$ is divisible by $4$. Now $2 \\mid m$, $2 \\mid n$. We may assume without loss of generality that $m = 6m_0$, $n = 2n_0$, neither $m_0$ nor $n_0$ is divisible by $2$. We will prove that if these conditions are satisfied, an $m \\times n$ rectangle cannot be covered with hooks.\nConsider coloring the columns of an $m \\times n$ matrix with black and white colors alternately. Then the number of the black squares equals that of the white ones. One $\\mathbf{(2)}$ always covers $6$ black squares. A horizontal $\\mathbf{(1)}$ always covers $6$ black squares. A vertical $\\mathbf{(1)}$ covers either $8$ black squares and $4$ white ones, or $4$ black squares and $8$ white ones. Since the number of the black squares equals that of the white ones, the number of $\\mathbf{(1)}$ is the same in the preceding two cases. Hence the total number of a vertical $\\mathbf{(1)}$ is even. Using the same argument as above (coloring the rows alternately), we obtain that the total number of a horizontal $\\mathbf{(1)}$ is even.\nConsider classifying the squares of the $m \\times n$ rectangle into $4$ types marked $1$, $2$, $3$, and $4$ as shown below. The number of squares of each type is equal to $\\frac{m n}{4}$.\n![](attached_image_5.png)\nFrom the two diagrams,\n![](attached_image_6.png)\n![](attached_image_7.png)\nwe obtain that the number of $a$ and $c$ covered by $\\mathbf{(1)}$ is the same, so is for $b$ and $d$. Hence the number of squares of type $1$ covered by $\\mathbf{(1)}$ equals that of type $3$.\n![](attached_image_8.png)\n(i)\n![](attached_image_9.png)\n(ii)\n![](attached_image_10.png)\n(iii)\n![](attached_image_11.png)\n(iv)\nThe number of squares of type $1$ covered by (i) or (ii) equals that of type $3$. The difference between the number of squares of type $1$ and type $3$ covered by (iii) or (iv) is $2$. There are two cases; the number of squares of type $1$ is $2$ more than that of type $3$, or vice versa. Since the number of squares of type $1$ equals that of type $3$ in the rectangle, the frequency of the two cases is the same. Hence the total number of (iii) and (iv) is even.\nSimilarly, the total number of (i) and (ii) is even. Then the number of $\\mathbf{(2)}$ is even. So there is an even number of $\\mathbf{(1)}$ and $\\mathbf{(2)}$. Hence $24 \\mid m \\times n$, contrary to the assumption that neither $m$ nor $n$ is divisible by $4$.\n\nWe now show that if $n \\ge 7$ and $n$ is not divisible by $3$ and $4$, a tessellation exists.\nIf $n \\equiv 1 \\pmod{3}$, then $n = 4 + 3t$ ($t \\in \\mathbb{N}^*$). Together with (1), we have that if $12 \\mid m$, an $m \\times 3t$ rectangle and an $m \\times 4$ rectangle can be covered with hooks. So the problem can be solved.\nIf $n \\equiv 2 \\pmod{3}$, $n = 8 + 3t$ ($t \\in \\mathbb{N}^*$). Together with (1), we have that if $12 \\mid m$, each of $m \\times 8$ and $m \\times 3t$ rectangles can be covered with hooks. The problem is solved.\n$m$ and $n$ should be the positive integers and should satisfy one of the following conditions:\n(1) $3 \\mid m$ and $4 \\mid n$ (or vice versa);\n(2) $12 \\mid m$, $n \\notin \\{1, 2, 5\\}$ (or vice versa).\n\nConsider a covering of an $m \\times n$ rectangle satisfying the conditions. For any hook $A$, there is a unique hook $B$ covering the \"inner\" square of $A$ with one of its \"tailend\" squares. In turn, the \"inner\" square of $B$ must be covered by a \"tailend\" square of $A$. Thus, in a tessellation, all hooks are matched into pairs. There are only two possibilities to place $B$ so that it does not overlap with $A$ and no gap occurs. In one case, $A$ and $B$ form a $3 \\times 4$ rectangle; in the other, their union is an octagonal shape, with sides of length $3$, $2$, $1$, $2$, $3$, $2$, $1$, $2$ respectively.\nSo an $m \\times n$ rectangle can be covered with hooks if and only if it can be covered with the $12$-square tiles described above. Suppose that such a tessellation exists; then $mn$ is divisible by $12$. We now show that one of $m$ and $n$ is divisible by $4$.\nAssume on the contrary that this is not the case. Then $m$ and $n$ are both even, because $mn$ is divisible by $4$. Imagine that the rectangle is divided into unit squares, with the rows and columns labeled $1, \\ldots, m$ and $1, \\ldots, n$. Write $1$ in the square $(i, j)$ if exactly one of $i$ and $j$ is divisible by $4$, and $2$, if $i$ and $j$ are both divisible by $4$. Since the number of squares in each row and column is even, the sum of all numbers written is also even. Now, it is easy to check that a $3 \\times 4$ rectangle always covers numbers with sum $3$ or $7$; and the other $12$-square shape always covers numbers with sum $5$ or $7$. Consequently, the total number of $12$-square shapes is even. But then $mn$ is divisible by $24$, and hence by $8$, contrary to the assumption that $m$ and $n$ are not divisible by $4$.\nNotice also that neither $m$ nor $n$ can be $1$, $2$ or $5$ (any attempt to place tiles along a side of length $1$, $2$ or $5$ fails). We infer that if a tessellation is possible, then one of $m$ and $n$ is divisible by $3$, one is divisible by $4$, and $m, n \\notin \\{1, 2, 5\\}$.\nConversely, we shall prove that if these conditions are satisfied, then a tessellation is possible (using only $3 \\times 4$ rectangles). The result is immediate if $3$ divides $m$ and $4$ divides $n$ (or vice versa). Let $m$ be divisible by $12$ and $n \\notin \\{1, 2, 5\\}$ (or vice versa). Without loss of generality, we may assume that neither $3$ nor $4$ divides $n$. Then $n \\ge 7$. In addition, between $n-4$ and $n-8$, at least one can be divisible by $3$. Hence the rectangle can be partitioned into $m \\times 3$ and $m \\times 4$ rectangles, which are easy to cover, in fact with only $3 \\times 4$ tiles again.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 28023,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all real numbers $k$, such that the inequality\n$$\na^3 + b^3 + c^3 + d^3 + 1 \\ge k(a+b+c+d)\n$$\nholds for any $a, b, c, d \\in [-1, +\\infty)$.",
    "options": [],
    "answer": "3/4",
    "solution": "If $a = b = c = d = -1$, then $-3 \\ge k \\cdot (-4)$. Hence $k \\ge \\frac{3}{4}$.\n\nIf $a = b = c = d = \\frac{1}{2}$, then $4 \\cdot \\frac{1}{8} + 1 \\ge k \\cdot (4 \\cdot \\frac{1}{2})$. Thus $k \\le \\frac{3}{4}$, and so $k = \\frac{3}{4}$.\n\nAt first, we prove that $4x^3 + 1 \\ge 3x$, $x \\in [-1, +\\infty)$.\nIn fact, from $(x+1)(2x-1)^2 \\ge 0$, we have $4x^3 + 1 \\ge 3x$, $x \\in [-1, +\\infty)$.\nTherefore\n$$\n4a^3 + 1 \\ge 3a,\n$$\n$$\n4b^3 + 1 \\ge 3b,\n$$\n$$\n4c^3 + 1 \\ge 3c,\n$$\n$$\n4d^3 + 1 \\ge 3d.\n$$\nBy adding the above 4 inequalities together, we get the inequality (1).",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 28024,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $M$ be a set consisting of $n$ points in the plane, and satisfying:\n(1) there exist 7 points in $M$ which constitute the vertices of a convex heptagon;\n(2) if for any 5 points in $M$ which constitute the vertices of a convex pentagon, then there is a point in $M$ which lies in the interior of the pentagon.\n\nFind the minimum value of $n$.",
    "options": [],
    "answer": "11",
    "solution": "First, we prove that $n \\ge 11$. Suppose a convex heptagon has its vertices in $M$ given by $A_1A_2A_3A_4A_5A_6A_7$. Using Condition (1), we get that there exists one point $P_1$ belonging to $M$ in the interior of convex pentagon $A_1A_2A_3A_4A_5$. Connecting $P_1A_1$ and $P_1A_5$, we obtain that there exists one point $P_2$ in convex pentagon $A_1P_1A_5A_6A_7$ so that $P_2$ belongs to $M$ and is different from $P_1$. Then, there are at least 5 points in $\\{A_1, A_2, A_3, A_4, A_5, A_6, A_7\\}$ which do not lie on line $P_1P_2$. By the Pigeon Hole Principle, there exist at least 3 points on one side of line $P_1P_2$, and these 3 points together with $P_1$ and $P_2$ constitute a convex pentagon which contains at least one point $P_3$ belonging to $M$.\n\nNow, we have three lines $P_1P_2$, $P_2P_3$ and $P_3P_1$, which form a triangle $\\triangle P_1P_2P_3$. Let $\\pi_1$ denote the half-plane on one side of line $P_1P_2$ which is opposite to $\\triangle P_1P_2P_3$ and contains no points on $P_1P_2$. In a similar way, we define $\\pi_2$ and $\\pi_3$. Areas $\\pi_1$, $\\pi_2$ and $\\pi_3$ cover the entire plane except $\\triangle P_1P_2P_3$. By the Pigeon Hole Principle, there is one area of $\\pi_1$, $\\pi_2$ and $\\pi_3$ which contains at least 3 points belonging to $\\{A_1, A_2, A_3, A_4, A_5, A_6, A_7\\}$, Without loss of generality, we assume that the area $\\pi_1$ contains points $A_1, A_2, A_3$, then there exists one point $P_4$ belonging to $M$ within the convex pentagon constituted by $A_1, A_2, A_3, P_1$ and $P_2$. So, $n \\ge 11$.\n\nNow, we give an example to illustrate that $n=11$ is attainable. As seen in the figure, set $M$ consists of integral points $A_1, A_2, A_3, A_4, A_5, A_6, A_7$, and four integral points within the heptagon $A_1A_2A_3A_4A_5A_6A_7$. Obviously, $M$ satisfies Condition (1). We are going to\n\n![](attached_image_1.png)\n\nprove that $M$ also satisfies Condition (2).\nBy reduction to absurdity, assume that there is a convex pentagon with its vertices belonging to $M$ which contains no point of $M$ in its interior. Then among such pentagons there must be one, denoted by $ABCDE$, which has the least area, since the value of the area of a polygon with integral vertices is always in the form of $\\frac{n}{2}$ ($n \\in \\mathbb{N}$).\nThere are only 4 cases concerning the odd/even property of the $xy$-coordinate of an integral point: (odd, even), (even, odd), (odd, odd), (even, even). So there must be two vertices among $A, B, C, D, E$ which have the same odd/even property, and the midpoint of the segment formed by these two vertices, say $P$, is also an integral point and belongs to $M$. By definition, $P$ is not in the interior of pentagon $ABCDE$, then it must be on one side of the pentagon. Assume that $P$ is on the side $AB$, then it must be the midpoint of $AB$, and $PBCDE$ is a convex pentagon with strictly less area than that of $ABCDE$.\nSo, the minimum value of $n$ is $11$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 28025,
    "subject": "Mathematics (Multi-modal)",
    "question": "The solution set of the inequality $\\sqrt{\\log_2 x - 1} + \\frac{1}{2} \\log_{\\frac{1}{2}} x^3 + 2 > 0$ is ( ).\n(A) [2, 3]\n(B) (2, 3]\n(C) [2, 4)\n(D) (2, 4]",
    "options": [],
    "answer": "C",
    "solution": "$$\n\\begin{cases} \\sqrt{\\log_2 x - 1} - \\frac{3}{2} \\log_2 x + \\frac{3}{2} + \\frac{1}{2} > 0, \\\\ \\log_2 x - 1 \\ge 0. \\end{cases}\n$$\nLet $t = \\sqrt{\\log_2 x - 1}$, we have\n$$\n\\begin{cases} t - \\frac{3}{2}t^2 + \\frac{1}{2} > 0, \\\\ t \\ge 0. \\end{cases}\n$$\n\nThe solution of the above inequalities is $0 \\le t < 1$, or $0 \\le \\log_2 x - 1 < 1$, which implies that $2 \\le x < 4$. Answer: C.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 28026,
    "subject": "Mathematics (Multi-modal)",
    "question": "For a given real number $a$ and a positive integer $n$, prove that:\n\n(1) there exists exactly one sequence of real numbers $x_0, x_1, \\dots, x_n, x_{n+1}$, such that\n$$\n\\begin{cases}\nx_0 = x_{n+1} = 0, \\\\\n\\frac{1}{2}(x_{i+1} + x_{i-1}) = x_i + x_i^3 - a^3, \\quad i = 1, 2, \\dots, n;\n\\end{cases}\n$$\n\n(2) the sequence $x_0, x_1, \\dots, x_n, x_{n+1}$ in (1) satisfies $|x_i| \\le |a|, i = 0, 1, \\dots, n+1$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "(1) Proof of existence: From $x_{i+1} = 2x_i + 2x_i^3 - 2a^3 - x_{i-1}$, $i = 1, 2, \\dots, n$ and $x_0 = 0$, we get that $x_i$ is a polynomial of $x_1$ with degree $3^{i-1}$ and real coefficients, for $1 \\le i \\le n+1$. Specifically, $x_{n+1}$ is a polynomial of $x_1$ with degree $3^n$ and real coefficients. As $3^n$ is an odd number, there exists a real number $x_1$ such that $x_{n+1} = 0$. Then from this $x_1$ and $x_0 = 0$ we can calculate $x_i$. The sequence $x_0, x_1, \\dots, x_n, x_{n+1}$ obtained in this way satisfies the required condition.\n\nProof of uniqueness: Suppose there are two sequences $w_0, w_1, \\dots, w_n, w_{n+1}$ and $v_0, v_1, \\dots, v_n, v_{n+1}$, both of which satisfy Condition (1). Then $\\frac{1}{2}(w_{i+1} + w_{i-1}) = w_i + w_i^3 - a^3$ and $\\frac{1}{2}(v_{i+1} + v_{i-1}) = v_i + v_i^3 - a^3$. Thus,\n$$\n\\frac{1}{2}(w_{i+1} - v_{i+1} + w_{i-1} - v_{i-1}) = (w_i - v_i)(1 + w_i^2 + w_i v_i + v_i^2).\n$$\nSuppose $|w_{i_0} - v_{i_0}|$ is the greatest. Then\n$$\n\\begin{aligned}\n|w_{i_0} - v_{i_0}| &\\le |w_{i_0} - v_{i_0}| (1 + w_{i_0}^2 + w_{i_0} v_{i_0} + v_{i_0}^2) \\\\\n&\\le \\frac{1}{2} |w_{i_0+1} - v_{i_0+1}| + \\frac{1}{2} |w_{i_0-1} - v_{i_0-1}| \\\\\n&\\le |w_{i_0} - v_{i_0}|.\n\\end{aligned}\n$$\nSo, either $|w_{i_0} - v_{i_0}| = 0$ or $(1+w_{i_0}^2+w_{i_0}v_{i_0}+v_{i_0}^2) = 0$, that is, either $|w_{i_0} - v_{i_0}| = 0$ or $w_{i_0}^2+v_{i_0}^2+(w_{i_0}+v_{i_0})^2 = 0$. However it must be $|w_{i_0} - v_{i_0}| = 0$. Since $|w_{i_0} - v_{i_0}|$ is the greatest. Then $|w_i - v_i| = 0$ for every $i = 1, 2, \\dots, n$. This completed the proof of (1).\n\n(2) Suppose that $|x_{i_0}|$ is the greatest. Then we have\n$$\n\\begin{aligned}\n|x_{i_0}| + |x_{i_0}|^3 &= |x_{i_0}| (1 + |x_{i_0}|^2) \\\\\n&= \\left| \\frac{1}{2}(x_{i_0+1} + x_{i_0-1}) + a^3 \\right| \\\\\n&\\le \\frac{1}{2} |x_{i_0+1}| + \\frac{1}{2} |x_{i_0-1}| + |a|^3\n\\end{aligned}\n$$\n$$\n\\leq |x_{i_0}| + |a|^3.\n$$\nThat means $|x_{i_0}| \\leq |a|$. So $|x_i| \\leq |a|$, $i = 0, 1, \\dots, n+1$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 28027,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $O$ be an interior point of $\\triangle ABC$ such that $\\overrightarrow{OA} + 2 \\overrightarrow{OB} + 3 \\overrightarrow{OC} = 0$. Then the ratio of the area of $\\triangle ABC$ to the area of $\\triangle AOC$ is ( ).\n\n(A) 2\n\n(B) $\\frac{3}{2}$\n\n(C) 3\n\n(D) $\\frac{5}{3}$",
    "options": [],
    "answer": "C",
    "solution": "In the diagram, let $D$ and $E$ be the midpoints of the sides $AC$ and $BC$, respectively. Then we have\n$$\n\\overrightarrow{OA} + \\overrightarrow{OC} = 2 \\overrightarrow{OD}, \\qquad (1)\n$$\nand\n$$\n2(\\overrightarrow{OB} + \\overrightarrow{OC}) = 4 \\overrightarrow{OE}. \\qquad (2)\n$$\nBy equations (1) and (2) we get\n$$\n\\overrightarrow{OA} + 2\\overrightarrow{OB} + 3\\overrightarrow{OC} = 2(\\overrightarrow{OD} + 2\\overrightarrow{OE}) = 0.\n$$\nIt follows that $\\overrightarrow{OD}$ and $\\overrightarrow{OE}$ are collinear, and\n$|\\overrightarrow{OD}| = 2 |\\overrightarrow{OE}|$. Consequently, $\\frac{S_{\\triangle AEC}}{S_{\\triangle AOC}} = \\frac{3}{2}$\nand $\\frac{S_{\\triangle ABC}}{S_{\\triangle AOC}} = \\frac{3 \\times 2}{2} = 3$. Answer: C.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 28028,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $n \\ge 3$ be an integer. Let $t_1, t_2, \\dots, t_n$ be positive real numbers such that\n$$\nn^2 + 1 > (t_1+t_2+\\dots+t_n)\\left(\\frac{1}{t_1}+\\frac{1}{t_2}+\\dots+\\frac{1}{t_n}\\right).\n$$\nShow that $t_i, t_j, t_k$ are the lengths of the sides of a triangle for all $i, j, k$ with $1 \\le i < j < k \\le n$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Assume on the contrary that there exist three numbers among $t_1, t_2, \\dots, t_n$ that do not form the sides of a triangle. Without loss of generality, we may assume that these three numbers are $t_1, t_2, t_3$, and $t_1+t_2 \\le t_3$. One has\n$$\n\\begin{aligned}\n& (t_1+\\cdots+t_n)\\left(\\frac{1}{t_1}+\\cdots+\\frac{1}{t_n}\\right) \\\\\n&=\\sum_{1\\le i<j\\le n}\\left[\\frac{t_i}{t_j}+\\frac{t_j}{t_i}\\right]+n \\\\\n&=\\frac{t_1}{t_3}+\\frac{t_3}{t_1}+\\frac{t_2}{t_3}+\\frac{t_3}{t_2}+\\sum_{\\substack{1\\le i<j\\le n \\\\ (i,j)\\notin\\{(1,3), (2,3)\\}}}[\\frac{t_i}{t_j}+\\frac{t_j}{t_i}]+n \\\\\n&\\ge\\frac{t_1+t_2}{t_3}+t_3\\left(\\frac{1}{t_1}+\\frac{1}{t_2}\\right)+\\sum_{\\substack{1\\le i<j\\le n \\\\ (i,j)\\notin\\{(1,3), (2,3)\\}}}2+n \\\\\n&\\ge\\frac{t_1+t_2}{t_3}+\\frac{4t_3}{t_1+t_2}+2(C_n^2-2)+n \\\\\n&=4\\frac{t_3}{t_1+t_2}+\\frac{t_1+t_2}{t_3}+n^2-4.\n\\end{aligned} \n\\quad (1)\n$$\n\nIf $x = \\frac{t_3}{t_1+t_2}$, then $x \\ge 1$, and $4x + \\frac{1}{x} - 5 = \\frac{(x-1)(4x-1)}{x} \\ge 0$.\nTogether with (1), we obtain that\n$$\n(t_1+\\cdots+t_n)\\left(\\frac{1}{t_1}+\\cdots+\\frac{1}{t_n}\\right) \\ge 5+n^2-4 = n^2+1,\n$$\na contradiction. This completes the proof.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 28029,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $n \\in \\mathbb{N}$ (the set of positive integers), and $d(n)$ be the number of positive divisors of $n$. Next, $\\varphi(n)$ denotes the number of integers in the closed interval $[1, n]$ which are co-prime with $n$.\nFind all non-negative integers $c$, such that there exists $n \\in \\mathbb{N}$ satisfying\n$$\nd(n) + \\varphi(n) = n + c.\n$$",
    "options": [],
    "answer": "{0, 1}",
    "solution": "We denote the set of positive divisors of $n$ by $A$, and the set of integers in the closed interval $[1, n]$ which are co-prime with $n$ by $B$. Since there is only one number $1 \\in A \\cap B$ among $1, 2, \\dots, n$, we get $d(n) + \\varphi(n) \\le n + 1$. Thus $c = 0$ or $1$.\n\n(1) If $c = 0$, then $d(n) + \\varphi(n) = n$ implies that there exists only one number among $1, 2, \\dots, n$ which is not contained in $A \\cup B$. If $n$ is even, and $n > 8$, then all of $n-2$ and $n-4$ are not contained in $A \\cup B$.\n\nB. In this case, $n$ does not satisfy the equation. If $n$ is odd, then when $n$ is a prime or $1$, $d(n) + \\varphi(n) = n+1$ (which will be discussed in case (2) below). When $n$ is a composite number, we can write $n = pq$, where $p$ and $q$ are odd and $1 < p \\le q$. If $q \\ge 5$, then $2p$ and $4p$ are not contained in $A \\cup B$, so $n$ does not satisfy the equation.\n\nFrom the above discussion, we find that\n$$\nd(n) + \\varphi(n) = n\n$$\nwhen either $n \\le 8$ and $n$ is even, or $n \\le 9$ and $n$ is an odd composite number.\nBy direct verification of all the solutions of the above equation, $n$ can only be $6$, $8$ and $9$.\n\n(2) If $c = 1$, then $d(n) + \\varphi(n) = n+1$ implies that among $1, 2, \\ldots, n$ every number belongs to $A \\cup B$. It is easy to find that this time $n = 1$ and primes can satisfy the required condition. For the case when $n$ is even (not prime), by the same argument as above we have $n \\le 4$ (consider $n-2$). If $n$ is an odd composite number, write $n = pq$, where $p$ and $q$ are odd and $3 \\le pq \\le q$, then $2p \\notin A \\cup B$, a contradiction. By direct calculation we find that only $n = 4$ satisfies the condition.\n\nTherefore, the solutions of the equation $d(n) + \\varphi(n) = n+1$ are $n = 1, 4$ or primes.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 28030,
    "subject": "Mathematics (Multi-modal)",
    "question": "Determine the maximum value of constant $\\lambda$ such that\n$$\nu + v + w \\ge \\lambda,$$\nwhere $u, v$ and $w$ are positive real numbers with $u\\sqrt{vw} + v\\sqrt{wu} + w\\sqrt{uv} \\ge 1$.",
    "options": [],
    "answer": "sqrt(3)",
    "solution": "**Solution I**\n\nBy the AM-GM inequality and the given condition, we have\n$$\nu \\cdot \\frac{v+w}{2} + v \\cdot \\frac{w+u}{2} + w \\cdot \\frac{u+v}{2} \\ge u\\sqrt{vw} + v\\sqrt{wu} + w\\sqrt{uv} \\ge 1,$$\nor\n$$uv + vw + wu \\ge 1.$$ \nSince\n$$(u-v)^2 + (v-w)^2 + (w-u)^2 \\ge 0, \\quad u^2 + v^2 + w^2 \\ge uv + vw + wu \\ge 1,$$\nit follows that\n$$(u+v+w)^2 = u^2 + v^2 + w^2 + 2(uv + vw + wu) \\ge 3,$$\nor $u+v+w \\ge \\sqrt{3}$. The equality holds if and only if $u = v = w = \\frac{\\sqrt{3}}{3}$.\n\n$$\n\\begin{aligned}\n\\frac{(u+v+w)^4}{9} &= \\left(\\frac{u+v+w}{3}\\right)^3 \\cdot 3(u+v+w) \\\\\n&\\ge 3uvw(u+v+w) \\\\\n&= (uvw + vwu + wuv)(u+v+w)\n\\end{aligned}\n$$\n$$\n\\geq (u\\sqrt{vw} + v\\sqrt{wu} + w\\sqrt{uv})^2 = 1,\n$$\nwhich implies that $u+v+w \\ge \\sqrt{3}$. The equality holds if and only if $u = v = w = \\frac{\\sqrt{3}}{3}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 28031,
    "subject": "Mathematics (Multi-modal)",
    "question": "In a convex quadrilateral $ABCD$, the diagonal $BD$ bisects neither the angle $ABC$ nor the angle $CDA$. The point $P$ lies inside $ABCD$ and satisfies\n$$\\angle PBC = \\angle DBA \\text{ and } \\angle PDC = \\angle BDA.$$\n\nProve that $ABCD$ is a cyclic quadrilateral if and only if $AP = CP$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "(i) Necessity.\n\nAssume that $ABCD$ is a cyclic quadrilateral. Let the circle $\\Gamma$ be the circumcircle of quadrilateral $ABCD$. Extend $BP$ and $DP$ beyond $P$ to meet the circle $\\Gamma$ at $X$ and $Y$ respectively.\n\nSince $DB$ does not bisect $\\angle ABC$ and $P$ lies inside $ABCD$, it follows from $\\angle PBC = \\angle DBA$ that $\\overline{CX} = \\overline{AD}$, $D \\neq X$, and the points $D$ and $X$ lie in the same half-plane bounded by the line $AC$. Hence $DX \\parallel AC$. Similarly, $B \\neq Y$ and $BY \\parallel AC$.\n\nThe points $D$, $X$, $A$, $C$, $B$, $Y$ lie on the circle $\\Gamma$, as mentioned above. So the points $D$ and $X$, the points $A$ and $C$, the points $B$ and $Y$ are symmetrical with respect to the perpendicular bisector $l$ of $AC$ respectively.\n\nSince $P = DY \\cap BX$, then $P$ is on the line $l$, or $AP = CP$. This completes the proof of the necessity.\n\n(ii) Sufficiency.\n\n**Lemma.** Let $l$ be a fixed line and $A, B, C$ be fixed points such that $A$ and $B$, $C$ lie in the different half-planes bounded by the line $l$. Assume that the point $X$ lies on the line $l$. Let $\\alpha(X)$ be the smallest angle of rotation from the line $XA$ to the line $l$ anticlockwise. Let $\\beta(X)$ be the smallest angle of rotation from the line $XC$ to the line $XB$ anticlockwise. If $\\alpha(X) = \\beta(X)$, then $X$ is called a good point. Prove that there are at most two good points.\n\n**Proof of the lemma**\n\nWe set up a coordinate system with the line $l$ as the $x$-axis and the perpendicular of $l$ as the $y$-axis.\nLet $A(a, b)$, $B(c, d)$, $C(e, f)$, $X(x, 0)$. Then $A = a+bi$, $B = c+di$, $C = e+fi$, and $X = x$. Hence\n$$\n\\begin{aligned}\nA - X &= a - x + bi, \\\\\n(a - x + bi)(\\cos \\alpha(X) + i\\sin \\alpha(X)) \\\\\n&= (a - x)\\cos \\alpha(X) - b\\sin \\alpha(X) + [(a - x)\\sin \\alpha(X) + b\\cos \\alpha(X)]i.\n\\end{aligned}\n$$\n\nRotate the line $XA$ by $\\alpha(X)$ anticlockwise, coinciding with $l$. So\n$$\n(a-x)\\cos\\alpha(X) + b\\cos\\alpha(X) = 0 \\quad (1)\n$$\n\nMoreover, since $\\vec{XC} = e-x+fi$, $\\vec{XB} = c-x+di$, then $\\vec{XC}$ after being rotated by $\\beta(X)$ anticlockwise becomes\n$$\n(e-x+fi)(\\cos\\beta(X) + i\\sin\\beta(X)) = (e-x)\\cos\\beta(X) - f\\sin\\beta(X) + [(e-x)\\sin\\beta(X) + f\\cos\\beta(X)]i\n$$\nwhich is parallel to $\\vec{XB}$. Thus,\n$$\n(c-x)[(e-x)\\sin\\beta(X) + f\\cos\\beta(X)] - d[(e-x)\\cos\\beta(X) - f\\sin\\beta(X)] = 0,\n$$\nand\n$$\n[(c-x)(e-x) + df]\\sin\\beta(X) + [(c-x)f - d(e-x)]\\cos\\beta(X) = 0. \\quad (2)\n$$\n\nSince $\\sin\\theta$ and $\\cos\\theta$ are not 0 at the same time, so\n$X$ is a good point $\\Leftrightarrow \\alpha(X) = \\beta(X) \\Leftrightarrow$ the simultaneous equations in $u$ and $v$.\n$$\n\\begin{cases} (a-x)u+bv=0, \\\\ [(c-x)(e-x)+df]u+[(c-x)f-d(e-x)]v=0 \\end{cases}\n$$\nhave the non-zero solutions\n$$\n\\begin{align*} & \\Leftrightarrow b[(c-x)(e-x)+df] + (a-x)[(c-x)f - d(e-x)] = 0 \\\\\n& \\Leftrightarrow (b+d-f)x^2 + (af+cf-bc-be-ad-ed)x \\\\\n& \\phantom{\\Leftrightarrow (b+d-f)x^2 + (af+cf-bc-be-ad-ed)x} + bce+bdf+ade-acf = 0. \\end{align*}\n$$\n\nLet $g(x) = (b+d-f)x^2 + (af+cf-bc-be-ad-ed)x + bce+bdf+ade-acf$. If $b+d-f=0$, it follows from $b \\neq 0$ that $d \\neq f$. Hence $BC$ and $l$ are not parallel. Let $BC$ intersect $l$ at $T(t, 0)$. We have that $\\beta(T) = 0$ and $\\alpha(T) > 0$. Hence $T$ is not a good point. This implies $g(t) \\neq 0$, so the equation $g(x) = 0$ has at most two solutions.\n\nHence there are at most two good points.\n\nThis completes the proof of the lemma.\n\nLet the points $A$, $B$, $C$, $D$ be arranged clockwise and $AP = CP$. Let $D^*$ be the point such that $A$, $D^*$, $C$, $B$ are concyclic, where $D^*$ lies on the ray $BD$. Since $AP = CP$ and $BD$ bisects neither $\\angle ABC$ nor $\\angle ADC$, $BP$ intersects the perpendicular bisector of $AC$ at the unique point $P$.\n\nLet $D^*R$ be the ray satisfying $\\angle CD^*R = \\angle AD^*B$ and $P^* = D^*R \\cap BP$.\nTogether with (i), we have that $P^*A = P^*C$, or $P^*$ is the intersection point of $l$ and the perpendicular bisector of $AC$, so $P^* = P$. It follows from $\\angle CD^*R = \\angle AD^*B$ that $\\angle AD^*B = \\angle CD^*P$.\n\nReplace $l$, $A$, $B$, $C$ by $BD$, $A$, $C$, $P$. Then $B$, $D$, $D^*$ are good points. Since $B \\neq D$ and $B \\neq D^*$, $D = D^*$. Hence $A$, $B$, $C$, $D$ is concyclic.\n\nThis completes the proof of the sufficiency.\n\n\nSolution II\n\nWe may assume without loss of generality that $P$ lies in the rectangles $ABC$ and $BCD$.\n\n![](attached_image_1.png)\n\nAssume that the quadrilateral $ABCD$ is cyclic. Let the lines $BP$ and $DP$ meet $AC$ at $K$ and $L$ respectively. It follows from the given equalities and $\\angle ACB = \\angle ADB$, $\\angle ABD = \\angle ACD$ that the triangles $DAB$, $DLC$ and $CKB$ are similar. This implies $\\angle DLC = \\angle CKB$, so $\\angle PLK = \\angle PKL$. Hence $PK = PL$.\n\nIt follows from $\\angle BDA = \\angle PDC$ that $\\angle ADL = \\angle BDC$. Since $\\angle DAL = \\angle DBC$, the triangles $ADL$ and $BDC$ are similar. Hence\n$$\n\\frac{AL}{BC} = \\frac{AD}{BD} = \\frac{KC}{BC},\n$$\nyielding $AL = KC$. It follows from $\\angle DLC = \\angle CKB$ that $\\angle ALP = \\angle CKP$. Moreover, since $PK = PL$ and $AL = KC$, the triangles $ALP$ and $CKP$ are congruent. Hence $AP = CP$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 28032,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $n = \\overline{\\{abc\\}}$ be a 3-digit number. If we can construct an isosceles triangle (including equilateral triangle) with $a$, $b$ and $c$ as the lengths of the sides. The number of such 3-digit integers $n$ is ( ).\n(A) 45\n(B) 81\n(C) 165\n(D) 216",
    "options": [],
    "answer": "C",
    "solution": "If $a$, $b$ and $c$ are the lengths of the sides of a triangle, all of them are not zero, it follows that $a, b, c \\in \\{1, 2, \\dots, 9\\}$.\n\ni. If the triangle we construct is equilateral, let $n_1$ be the number of such 3-digit numbers. Since the three digits in such 3-digit number are equal, we have\n$$\nn_1 = C_9^1 = 9.\n$$\n\nii. If the triangle we construct is isosceles but not equilateral, let $n_2$ be the number of such 3-digit numbers. Since there are only 2 different digits in such a 3-digit number, denote them by $a$ and $b$. Note that the equal sides and the base of an isosceles triangle can be replaced by each other, thus the number of such pairs $(a, b)$ is $2C_3^2$. But if the bigger number, say $a$, is the length of the base, then $a$ must satisfy the condition $b < a < 2b$. All pairs that do not satisfy this condition we list in the following table. There are 20 pairs.\n\n| a | 9 | 8 | 7 | 6 | 5 | 4 | 3 | 2 | 1 |\n|---|---|---|---|---|---|---|---|---|---|\n| b | 4, 3,<br>2, 1 | 4, 3,<br>2, 1 | 3, 2,<br>1 | 3, 2,<br>1 | 2, 1 | 2, 1 | 1 | 1 |   |\n\nOn the other hand, there are $C_3^2$ possible 3-digit numbers with digits taken from a given pair $(a, b)$. Thus\n$$\nn_2 = C_3^2(2C_3^2 - 20) = 6(C_3^2 - 10) = 156.\n$$\nConsequently, $n = n_1 + n_2 = 165$. Answer: C.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 28033,
    "subject": "Mathematics (Multi-modal)",
    "question": "The sequence $\\{a_n\\}$ satisfies $a_1 = a_2 = 1$ and\n$$\na_{n+2} = \\frac{1}{a_{n+1}} + a_n, \\quad n = 1, 2, \\dots\n$$\nFind $a_{2004}$.",
    "options": [],
    "answer": "(3*5*...*2003)/(2*4*...*2002)",
    "solution": "According to the assumption we have\n$$\na_{n+2} a_{n+1} - a_{n+1} a_n = 1.\n$$\nThus, $\\{a_{n+1} a_n\\}$ is an arithmetic progression with first term $1$ and common difference $1$. Hence\n$$\na_{n+1} a_n = n, \\quad n = 1, 2, \\dots\n$$\nSo $a_{n+2} = \\frac{n+1}{a_{n+1}} = \\frac{n+1}{n} = \\frac{n+1}{n} a_n, \\quad n = 1, 2, \\dots$. Consequently,\n$$\n\\begin{align*}\na_{2004} &= \\frac{2003}{2002} a_{2002} = \\frac{2003}{2002} \\cdot \\frac{2001}{2000} a_{2000} \\\\\n&= \\dots = \\frac{2003}{2002} \\cdot \\frac{2001}{2000} \\cdot \\dots \\cdot \\frac{3}{2} a_2 \\\\\n&= \\frac{3 \\cdot 5 \\cdot \\dots \\cdot 2003}{2 \\cdot 4 \\cdot \\dots \\cdot 2002}\n\\end{align*}\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 28034,
    "subject": "Mathematics (Multi-modal)",
    "question": "For a given positive integer $n \\geq 2$, suppose positive integers $a_i$ ($i=1, 2, \\dots, n$) satisfy $a_1 < a_2 < \\dots < a_n$ and $\\sum_{i=1}^{n} \\frac{1}{a_i} \\leq 1$.\nProve that, for any real number $x$, the following inequality holds,\n$$\n\\left[ \\sum_{i=1}^{n} \\frac{1}{a_i^2 + x^2} \\right]^2 \\leq \\frac{1}{2} \\cdot \\frac{1}{a_1 (a_1 - 1) + x^2}.\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "For $x^2 \\geq a_1(a_1-1)$, from $\\sum_{i=1}^{n} \\frac{1}{a_i} \\leq 1$ we have\n$$\n\\begin{aligned}\n\\left[ \\sum_{i=1}^{n} \\frac{1}{a_i^2 + x^2} \\right]^2 &\\leq \\left( \\sum_{i=1}^{n} \\frac{1}{2a_i |x|} \\right)^2 = \\frac{1}{4x^2} \\left( \\sum_{i=1}^{n} \\frac{1}{a_i} \\right)^2 \\\\\n&\\leq \\frac{1}{4x^2} \\leq \\frac{1}{2} \\cdot \\frac{1}{a_1(a_1-1)+x^2}.\n\\end{aligned}\n$$\n\nFor $x^2 < a_1(a_1-1)$, using the Cauchy Inequality, we have\n$$\n\\left[ \\sum_{i=1}^{n} \\frac{1}{a_i^2 + x^2} \\right]^2 \\leq \\left( \\sum_{i=1}^{n} \\frac{1}{a_i} \\right) \\left[ \\sum_{i=1}^{n} \\frac{a_i}{(a_i^2 + x^2)^2} \\right] \\leq \\sum_{i=1}^{n} \\frac{a_i}{(a_i^2 + x^2)^2}.\n$$\nFurther, for positive integers $a_1 < a_2 < \\dots < a_n$, we have $a_{i+1} \\geq a_i + 1$ and\n$$\n\\begin{aligned}\n\\frac{2a_i}{(a_i^2 + x^2)^2} &\\leq \\frac{2a_i}{\\left(a_i^2 + x^2 + \\frac{1}{4}\\right)^2 - a_i^2} \\\\\n&= \\frac{2a_i}{\\left(\\left(a_i - \\frac{1}{2}\\right)^2 + x^2\\right)\\left(\\left(a_i + \\frac{1}{2}\\right)^2 + x^2\\right)}\n\\end{aligned}\n$$\n$$\n\\begin{aligned}\n&= \\frac{1}{\\left(a_i - \\frac{1}{2}\\right)^2 + x^2} - \\frac{1}{\\left(a_i + \\frac{1}{2}\\right)^2 + x^2} \\\\\n&\\le \\frac{1}{\\left(a_i - \\frac{1}{2}\\right)^2 + x^2} - \\frac{1}{\\left(a_{i+1} - \\frac{1}{2}\\right)^2 + x^2},\n\\end{aligned}\n$$\nfor $i = 1, 2, \\dots, n-1$. So\n$$\n\\begin{aligned}\n\\sum_{i=1}^{n} \\frac{a_i}{(a_i^2 + x^2)^2} &\\le \\frac{1}{2} \\sum_{i=1}^{n} \\left[ \\frac{1}{\\left(a_i - \\frac{1}{2}\\right)^2 + x^2} - \\frac{1}{\\left(a_{i+1} - \\frac{1}{2}\\right)^2 + x^2} \\right] \\\\\n&\\le \\frac{1}{2} \\cdot \\frac{1}{\\left(a_1 - \\frac{1}{2}\\right)^2 + x^2} \\le \\frac{1}{2} \\cdot \\frac{1}{a_1(a_1 - 1) + x^2}.\n\\end{aligned}\n$$",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 28035,
    "subject": "Mathematics (Multi-modal)",
    "question": "Prove that every positive integer $n$, except a finite number of them, can be represented as a sum of $2004$ positive integers: $n = a_1 + a_2 + \\dots + a_{2004}$, where $1 \\le a_1 < a_2 < \\dots < a_{2004}$, and $a_i \\mid a_{i+1}$, $i = 1, 2, \\dots, 2003$. (posed by Chen Yonggao)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "We are going to prove a more general result: For any positive integer $r \\ge 2$, there exists $N(r) \\in \\mathbb{N}$ such that for every $n \\ge N(r)$, there are positive integers $a_1, a_2, \\dots, a_r$ satisfying\n$$\n\\begin{aligned}\nn &= a_1 + a_2 + \\dots + a_r, \\quad 1 \\le a_1 < a_2 < \\dots \\\\\n&< a_r, \\quad a_i \\mid a_{i+1}, \\quad i = 1, 2, \\dots, r-1.\n\\end{aligned}\n$$\n\nFor $r=2$, we have $n=1+n-1$, then $N(2)=3$.\n\nSuppose it is true for $r=k$, then for $r=k+1$ let $N(k+1) = 4N(k)^3$. For any positive integer $n = 2^\\alpha(2l+1) \\ge N(k+1) = 4N(k)^3$, we have either $2^\\alpha \\ge 2N(k)^2$ or $2l+1 > 2N(k)$.\n\nIf $2^\\alpha \\ge 2N(k)^2$, there exists an even positive integer $2t \\le \\alpha$ such that $2^{2t} \\ge N(k)^2$, then $2^t + 1 \\ge N(k)$. By induction we get positive integers $b_1, b_2, \\dots, b_k$ such that\n$$\n2^t + 1 = b_1 + b_2 + \\dots + b_k, \\quad 1 \\le b_1 < b_2 < \\dots < b_k, \\quad b_i \\mid b_{i+1}, \\quad i = 1, 2, \\dots, k-1.\n$$\nThen\n$$\n\\begin{align*}\n2^\\alpha = 2^{\\alpha-2t} \\cdot 2^{2t} &= 2^{\\alpha-2t}[1 + (2^t - 1)](2^t + 1) \\\\\n&= 2^{\\alpha-2t} + 2^{\\alpha-2t}(2^t - 1)b_1 + 2^{\\alpha-2t}(2^t - 1)b_2 \\\\\n&\\quad + \\dots + 2^{\\alpha-2t}(2^t - 1)b_k.\n\\end{align*}\n$$\n$$\nn = 2^{\\alpha-2t}(2l+1) + 2^{\\alpha-2t}(2^t-1)b_1(2l+1) \\\\\n\\qquad + \\dots + 2^{\\alpha-2t}(2^t-1)b_k(2l+1).\n$$\n\nIf $2l + 1 > 2N(k)$, then $l > N(k)$. By induction there exist positive integers $c_1, c_2, \\dots, c_k$ such that $l = c_1 + c_2 + \\dots + c_k$, $1 \\le c_1 < c_2 < \\dots < c_k$, $c_i \\mid c_{i+1}$, $i = 1, 2, \\dots, k-1$. Then\n$$\nn = 2^{\\alpha} + 2^{\\alpha+1}c_1 + 2^{\\alpha+1}c_2 + \\dots + 2^{\\alpha+1}c_k,\n$$\nwhich is what we want. This completes the proof.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 28036,
    "subject": "Mathematics (Multi-modal)",
    "question": "Given an acute triangle $ABC$ with $O$ as its circumcenter. Line $AO$ and side $BC$ meet at $D$. Points $E$ and $F$ are on sides $AB$ and $AC$ respectively, such that points $A$, $E$, $D$ and $F$ are on a circle. Prove that the length of the projection of line segment $EF$ on side $BC$ does not depend on the positions of $E$ and $F$. (posed by Xiong Bin)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $M$ and $N$ be the feet of the perpendiculars from $D$ to lines $AB$ and $AC$ respectively. Let $E_0$, $F_0$, $M_0$ and $N_0$ be the feet of the perpendiculars from $E$, $F$, $M$ and $N$ to side $BC$ respectively.\n\n![](attached_image_1.png)\n\nIt suffices to show that $E_0F_0 = M_0N_0$. Without loss of generality, we may assume that $E$ lies on line segment $BM$. Then $\\angle AED < 90^\\circ$. Since $A$, $E$, $D$, $F$ are concyclic, $\\angle DFC = \\angle AED < 90^\\circ$, and so $F$ lies on line segment $AN$. It follows that we need only to consider the figure above. It suffices to show that $E_0M_0 = F_0N_0$. Note that $E_0M_0 = EM \\cos \\angle B$ and $F_0N_0 = FN \\cos \\angle C$. We need only to show that $EM \\cos \\angle B = FN \\cos \\angle C$, or\n$$\n\\frac{EM}{FN} = \\frac{\\cos \\angle C}{\\cos \\angle B} \\quad (1)\n$$\nIn the right-angled triangles $AMO$ and $ANO$, $MD = AD \\sin \\angle DAM = AD \\sin \\angle OAB$ and $ND = AD \\sin \\angle DAN = AD \\sin \\angle OAC$. Substitute these equations into equation (2) and we have\n$$\n\\frac{EM}{FN} = \\frac{\\sin \\angle OAB}{\\sin \\angle OAC} \\quad (3)\n$$\nSince $O$ is the circumcenter of triangle $ABC$, $\\angle AOB = 2\\angle C$ and $\\angle OAB = \\angle OBA = 90^\\circ - \\angle C$, and so $\\sin \\angle OAB = \\sin (90^\\circ - \\angle C) = \\cos \\angle C$. Likewise, $\\sin \\angle OAC = \\cos \\angle B$. Substitute the last two equations into equation (3) and we get the desired equation (1).",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 28037,
    "subject": "Mathematics (Multi-modal)",
    "question": "All the grids of an $m \\times n$ chessboard ($m \\ge 3, n \\ge 3$) are colored either red or blue. Two adjacent grids (with a common side) are called a *good couple* if they are of different colors. Suppose that there are $S$ good couples, explain how to determine whether $S$ is odd or even. Does it depend on certain specific color grids? (Reasoning is required.) (posed by Feng Yuefeng)",
    "options": [],
    "answer": "The parity of S depends only on the non-corner border squares: S is odd if and only if there are an odd number of blue border (non-corner) squares; otherwise S is even.",
    "solution": "**Solution I**\nClassify all grids into three parts: the grids at the four corners, the grids along the borderlines (not including four corners), and the other grids. Fill all red grids with label number $1$, all blue grids with label number $-1$. Denote the label numbers filled in the grids in the first part by $a$, $b$, $c$ and $d$, in the second part by $x_1, x_2, \\dots, x_{2m+2n-8}$, and in the third part by $y_1, y_2, \\dots, y_{(m-2)(n-2)}$. For any two adjacent grids we write a label number which is the product of two label numbers of the two grids on their common edge. Let $H$ be the product of all label numbers on common edges.\n\nThere are $2$ adjacent grids for every grid in the first part, thus its label number appears twice in $H$. There are $3$ adjacent grids for every grid in the second part, thus its label number appears three times in $H$. There are $4$ adjacent grids for every grid in the third part, thus its label number appears four times in $H$. Therefore,\n$$\nH = (abcd)^2 (x_1 x_2 \\cdots x_{2m+2n-8})^3 (y_1 y_2 \\cdots y_{(m-2)(n-2)})^4 = (x_1 x_2 \\cdots x_{2m+2n-8})^3.\n$$\nIf $x_1 x_2 \\cdots x_{2m+2n-8} = 1$, then $H = 1$, and in this case there are even good couples. If $x_1 x_2 \\cdots x_{2m+2n-8} = -1$, then $H = -1$, and in this case there are odd good couples. It shows that whether $S$ is even or odd is determined by colors of the grids in the second part. Moreover, when there are odd blue grids among the grids in the second part, $S$ is odd. Otherwise $S$ is even.\n\n\n**Solution II**\nClassify all grids into three parts: the grids at the four corners, the grids along the borderlines (not containing four corners), and the other grids.\n\nIf all grids are red, then $S = 0$, which is even. If there are blue grids we pick any one of them, say $A$, and change $A$ into a red one. We call this changing a transformation.\n\n(1) $A$ is a grid in the first part. Suppose that there are $k$ red grids and $2-k$ blue grids among $A$'s two adjacent grids. After changing $A$ into a red one, the number of good couples increases by $2-k-k = 2-2k$. It follows that the parity even or odd of $S$ is unchanged.\n\n(2) $A$ is a grid in the second part. Suppose that there are $p$ red grids and $3-p$ blue grids among $A$'s three adjacent grids. After changing $A$ into a red one, the number of good couples increases by $3-p-p = 3-2p$. It follows that the parity of $S$ is changed.\n\n(3) $A$ is a grid in the third part. Suppose that there are $q$ red grids and $4 - q$ blue grids among $A$'s four adjacent grids. After changing $A$ into a red one, the number of good couples increases by $4-q-q = 4-2q$. It follows that the parity of $S$ is unchanged.\n\nIf there are still blue grids on the chessboard after above transformation, we continue doing the transformation over and over again until no blue grid left on the chessboard. Now $S$ is changed into $0$.\n\nClearly, $S$ changes its parity odd times if there are odd blue grids among the second part of grids. Similarly, $S$ changes its parity even times if there are even blue grids among the second part of grids. It implies that the parity of $S$ is determined by the coloring of the second part of grids. When there exist odd blue grids among the second part of grids, $S$ is odd. When there exist even blue grids among the second part of grids, $S$ is even.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 28038,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all positive integers $n$ such that $n$ has a multiple which is alternating.\n\nWe call a positive integer *alternating* if every two consecutive digits in its decimal representation are of different parity.",
    "options": [],
    "answer": "All positive integers not divisible by 20.",
    "solution": "**Solution I**\n\nLemma 1: If $k$ is a positive integer, then there exist $0 \\le a_1, a_2, \\dots, a_{2k} \\le 9$ such that $a_1, a_3, \\dots, a_{2k-1}$ are odd integers, $a_2, a_4, \\dots, a_{2k}$ are even integers, and\n$$\n2^{2k+1} \\mid \\overline{a_1 a_2 \\cdots a_{2k}}\n$$\nProof of Lemma 1 by mathematical induction.\nIf $k=1$, it follows from $8 \\mid 16$ that the proposition is true.\nAssume that if $k = n-1$, the proposition is true.\nWhen $k=n$, let $\\overline{a_1 a_2 \\cdots a_{2n-2}} = 2^{2n-1}t$ by the inductive hypothesis.\nThe problem reduces to proving that there exist $1 \\le a, b \\le 9$ with $a$ odd and $b$ even such that $2^{2n+1} \\mid \\overline{ab} \\times 10^{2n-2} + 2^{2n-1}t$, or $8 \\mid \\overline{ab} \\times 5^{2n-2} + 2t$, or $8 \\mid \\overline{ab} + 2t$ in view of $5^{2n-2} \\equiv 1 \\pmod 8$.\nIt follows from $8|12+4$, $8|14+2$, $8|16+0$ and $8|50+6$ that Lemma 1 is true.\n\nLemma 2: If $k$ is a positive integer, then there exists an alternating number $\\overline{a_1 a_2 \\cdots a_{2k}}$ with an even number $2k$ of digits such that $a_{2k}$ is odd and $5^{2k} \\mid \\overline{a_1 a_2 \\cdots a_{2k}}$, where $a_1$ can be 0, but $a_2 \\ne 0$.\nProof of Lemma 2 by mathematical induction.\nIf $k=1$, it follows from $25 \\mid 25$ that the proposition is true.\nAssume that if $k=n-1$, the proposition is true, or there exists an alternating multiple $\\overline{a_1 a_2 \\cdots a_{2n-2}}$ satisfying $5^{2n-2} \\mid \\overline{a_1 a_2 \\cdots a_{2n-2}}$.\nWhen $k=n$, let $\\overline{a_1 a_2 \\cdots a_{2n-2}} = t \\cdot 5^{2n-2}$. The problem reduces to proving that there exist $0 \\le a, b \\le 9$ with $a$ even and $b$ odd such that $5^2 \\mid \\overline{ab} \\times 10^{2n-2} + t \\cdot 5^{2n-2}$, or $25 \\mid \\overline{ab} \\times 2^{2n-2} + t$.\nSince $2^{2n-2}$ is coprime to 25, there exist $0 < \\overline{ab} \\le 25$ such that $25 \\mid \\overline{ab} \\times 2^{2n-2} + t$. If $b$ is odd, between $\\overline{ab}$ and $\\overline{ab} + 50$, at least one satisfies that the highest-valued digit is even. If $b$ is even, between $\\overline{ab} + 25$ and $\\overline{ab} + 75$, at least one satisfies that the highest-valued digit is even.\nThis completes the proof of Lemma 2.\n\nLet $n = 2^\\alpha 5^\\beta t$, where $t$ is coprime to 10 and $\\alpha, \\beta \\in \\mathbb{N}$. Assume that $\\alpha \\ge 2$ and $\\beta \\ge 1$. Let $l$ be an arbitrary multiple of $n$. The last decimal digit is 0, and the digit in tens is even. Hence these $n$ do not satisfy the required condition.\n\n① When $\\alpha = \\beta = 0$, consider $21$, $2121$, $212121$, $\\dots$, $2121\\dots21$, $\\dots$. There must exist two of them congruent modulo $n$.\nWithout loss of generality, we may assume that $t_1 > t_2$, and\n$$\n\\underbrace{2121\\cdots21}_{\\text{number } t_1 \\text{ of } 21} \\equiv \\underbrace{2121\\cdots21}_{\\text{number } t_2 \\text{ of } 21} \\pmod{n}.\n$$\nThen\n$$\n\\underbrace{2121\\cdots21}_{\\text{number } t_1-t_2 \\text{ of } 21} \\equiv \\underbrace{00\\cdots0}_{\\text{number } 2t_2 \\text{ of } 0} \\equiv 0 \\pmod{n}\n$$\nHence\n$$\n\\underbrace{2121\\cdots21}_{\\text{number } t_1-t_2 \\text{ of } 21} \\equiv 0 \\pmod{n},\n$$\nbecause $n$ is coprime to 10.\nNow these positive integers $n$ satisfy the required condition.\n\n② When $\\beta = 0$ and $\\alpha \\ge 1$, it follows from Lemma 1 that there exists an alternating number $\\overline{a_1 a_2 \\cdots a_{2k}}$ satisfying $2^\\alpha \\mid \\overline{a_1 a_2 \\cdots a_{2k}}$. Consider\n$$\n\\overline{a_1 a_2 \\cdots a_{2k}}, \\overline{a_1 a_2 \\cdots a_{2k} a_1 a_2 \\cdots a_{2k}}, \\dots, \\overline{a_1 a_2 \\cdots a_{2k} a_1 a_2 \\cdots a_{2k} \\cdots a_1 a_2 \\cdots a_{2k}}, \\dots\n$$\nThere must exist two of them congruent modulo $t$. Without loss of generality, we may assume that $t_1 > t_2$, and\n$$\n\\overline{a_1 a_2 \\cdots a_{2k} \\cdots a_1 a_2 \\cdots a_{2k}} \\equiv \\overline{a_1 a_2 \\cdots a_{2k} \\cdots a_1 a_2 \\cdots a_{2k}} \\pmod{t}\n$$\nSince $t$ is coprime to 10, $\\overline{a_1 a_2 \\cdots a_{2k} \\cdots a_1 a_2 \\cdots a_{2k}} \\equiv 0 \\pmod{t}$. Moreover, since $t$ is coprime to 2,\n$$\n2^\\alpha t \\mid \\overline{a_1 a_2 \\cdots a_{2k} \\cdots a_1 a_2 \\cdots a_{2k}},\n$$\nwhich is alternating.\n\n③ When $\\alpha = 0, \\beta \\ge 1$, it follows from Lemma 2 that there exists an alternating multiple $\\overline{a_1 a_2 \\cdots a_{2k}}$ satisfying $5^\\beta \\mid \\overline{a_1 a_2 \\cdots a_{2k}}$ with $a_{2k}$ odd. Using the same argument as in ②, we obtain that there exist $t_1 > t_2$ satisfying $t \\mid \\overline{a_1 a_2 \\cdots a_{2k} \\cdots a_1 a_2 \\cdots a_{2k}}$. Since $t$ is coprime to 5,\n$$\n5^{\\beta_t} \\mid \\overline{a_1 a_2 \\cdots a_{2k} \\cdots a_1 a_2 \\cdots a_{2k}}\n$$\nIn addition, $\\overline{a_1 a_2 \\cdots a_{2k} \\cdots a_1 a_2 \\cdots a_{2k}}$ is alternating, and the last decimal digit $a_{2k}$ is odd.\n\n④ When $\\alpha = 1$ and $\\beta \\ge 1$, it follows from ③ that there exists an alternating number $\\overline{a_1 a_2 \\cdots a_{2k} \\cdots a_1 a_2 \\cdots a_{2k}}$ satisfying that $a_{2k}$ is odd and $5^{\\beta_t} \\mid \\overline{a_1 a_2 \\cdots a_{2k} \\cdots a_1 a_2 \\cdots a_{2k}}$. Hence $2 \\cdot 5^{\\beta_t} \\mid \\overline{a_1 a_2 \\cdots a_{2k} \\cdots a_1 a_2 \\cdots a_{2k}}$, which is alternating.\n\nIn conclusion, if $n$ is not divisible by 20, then these positive integers $n$ satisfy the required condition.\n\n\n**Solution II**\n\n$n$ should be the positive integers and should satisfy that $n$ is not divisible by 20.\n\n(1) Assume that $20 \\mid n$. Select a multiple of $n$ arbitrarily. Denote $(a_k a_{k-1} \\cdots a_1)_{10}$, where $a_k \\ne 0$. We have $20 \\mid (a_k a_{k-1} \\cdots a_1)_{10}$. Hence $a_1 = 0$, $2 \\mid (a_k a_{k-1} \\cdots a_1)_{10}$. This implies that $a_2$ is even. Therefore, $(a_k a_{k-1} \\cdots a_1)_{10}$ is not alternating.\n\n(2) We will prove that if $n$ is a positive integer and is not divisible by 20, then $n$ must have a multiple which is alternating.\nWe will show three lemmas as follows. Then we divide the proof into four cases.\n\nLemma 1: If the positive integer $n$ is coprime to 10, then for any $l \\in \\mathbb{N}$, there exists $k \\in \\mathbb{N}$ such that\n$$\n\\frac{\\begin{array}{c} 1 \\ 00\\cdots0 \\\\ \\underline{\\quad} \\\\ \\mathbf{number} \\ l \\ \\mathbf{of} \\ 0 \\end{array} \\quad \\begin{array}{c} 1 \\ 00\\cdots0 \\\\ \\underline{\\quad} \\\\ \\mathbf{number} \\ l \\ \\mathbf{of} \\ 0 \\end{array} \\quad \\begin{array}{c} 1 \\cdots 1 \\ 00\\cdots0 \\\\ \\underline{\\quad} \\\\ \\mathbf{number} \\ l \\ \\mathbf{of} \\ 0 \\end{array} \\quad \\begin{array}{c} 1 \\\\ \\underline{\\quad} \\\\ \\mathbf{total number} \\ k \\ \\mathbf{of} \\ 1 \\end{array}}\n$$\nis a multiple of $n$.\nProof of Lemma 1: Consider the numbers $x_1 = 1, x_2 = 1 \\underbrace{00\\cdots0}_{\\text{number } l \\text{ of } 0}, \\dots,$\n$$\nx_m = 1 \\underbrace{00\\cdots0}_{\\text{number } l \\text{ of } 0} \\ 1 \\cdots 1 \\underbrace{00\\cdots0}_{\\text{number } l \\text{ of } 0} \\ 1, \\dots\n$$\nThere must exist two of them congruent modulo $n$. Without loss of generality, we may assume that\n$$\nx_s \\equiv x_t (\\text{mod } n), s > t \\ge 1.\n$$\nThen $n \\mid x_s - x_t$. Moreover, since $x_s - x_t = x_{s-t} \\cdot 10^{t(l+1)}$ and $n$ is coprime to $10$, $n \\mid x_{s-t}$, $s-t > 0$ and $x_{s-t}$ is a positive integer. This completes the proof of Lemma 1.\n\nLemma 2: For any $m \\in \\mathbb{N}$, there always exists an alternating number $s_m$ with $m$ digits such that its first digit can be 0, and its last decimal digit is 5, and $5^m \\mid s_m$.\nProof of Lemma 2 by inductive construction. Start with $s_1 = 5$. Suppose that $s_m = (a_m a_{m-1} \\cdots a_1)_{10}$ is alternating, where $a_1 = 5$ and $a_m$ can be 0. In addition, $5^m \\mid s_m$. Let $s_m = 5^m$. Denote\n$$\nA = \\begin{cases} \\{0, 2, 4, 6, 8\\}, & \\text{when } a_m \\text{ is odd,} \\\\ \\{1, 3, 5, 7, 9\\}, & \\text{when } a_m \\text{ is even.} \\end{cases}\n$$\nAny two of them in $A$ are not congruent modulo $5$, and $2^m$ is coprime to $5$. So any two of the numbers in $\\{2^m x \\mid x \\in A\\}$ are not congruent modulo $5$. Select $x \\in A$ such that $2^m x = -l \\pmod 5$. Then $5 \\mid 2^m x + l$. Let $s_{m+1} = (x a_m a_{m-1} \\cdots a_1)_{10}$. Then $s_{m+1}$ is an alternating number with $m+1$ digits, where $a_1 = 5$, and its first digit can be 0. In addition, $s_{m+1} = x 10^m + s_m = 5^m(2^m x + l)$ is a multiple of $5^{m+1}$. This completes the proof of Lemma 2.\n\nLemma 3: For any integer $m \\in \\mathbb{N}$, there always exists an alternating number $t_m$ with $m$ digits such that its last decimal digit is 2, and\n$$\n2^{2k+1} \\mid t_{2k+1}, \\quad 2^{2k+3} \\mid t_{2k+2}.\n$$\nProof of Lemma 3 by inductive construction. Start with $t_1 = 2$. Suppose that $t_{2k+1} = (a_{2k+1} a_{2k} \\cdots a_1)_{10}$ is alternating, where $a_1 = 2$, and $2^{2k+1} \\mid t_{2k+1}$. It follows from the property of the alternating numbers that $a_{2k+1} = a_1 = 0 \\pmod 2$.\nDenote $A = \\{1, 3, 5, 7\\}$. Any two of them in $A$ are not congruent modulo $8$, and $5^{2k+1}$ is coprime to $8$. So any two numbers of $\\{5^{2k+1} x \\mid x \\in A\\}$ are not congruent modulo $8$. Now divide the four numbers in $\\{5^{2k+1} x \\mid x \\in A\\}$ by $8$ respectively. The original sentence means the 4 numbers are divisible by 8. Since $5^{2k+1}x$ with $x \\in A$ is odd, the remainders are 1, 3, 5, 7.\nLet $t_{2k+1} = 2^{2k+1}l$ with $l$ odd. Select $x \\in A$ such that $5^{2k+1}x \\equiv -l+4 \\pmod 8$. Then $2^2 \\nmid 5^{2k+1}x+l$. Let $t_{2k+2} = (x a_{2k+1} \\cdots a_1)_{10}$. Then $t_{2k+2}$ is alternating with $2k+2$ digits. In addition, $a_1 = 2$ and $t_{2k+2} = x 10^{2k+1} + t_{2k+1} = 2^{2k+1}(5^{2k+1}x+l)$.\nSince $2^2 \\nmid 5^{2k+1}x+l$, we have $2^{2k+3} \\nmid t_{2k+2}$.\nMoreover, suppose that $t_{2k+2} = (a_{2k+2} \\cdots a_1)_{10}$, where $a_1 = 2$, and $2^{2k+3} \\nmid t_{2k+2}$. Let $t_{2k+3} = (4 a_{2k+2} \\cdots a_1)_{10}$. Then $t_{2k+3}$ is an alternating number with $2k+3$ digits, and $t_{2k+3} = 5^{2k+2}2^{2k+4} + t_{2k+2}$.\nSince $2^{2k+3} \\nmid t_{2k+2}$, we have $2^{2k+3} \\nmid t_{2k+3}$. This completes the proof of Lemma 3.\n\nNext, we discuss the four cases.\n(1) If $n$ is coprime to 10, it follows from Lemma 1 that there exists $k \\in N^*$ such that $10 \\ 101 \\cdots 101$ is a multiple of $n$. The conclusion is equivalent to selecting $l=1$ in Lemma 1.\n(2) If $n$ is not divisible by 5, and $n$ is divisible by 2, let $n = 2^m n_0$ such that $n_0$ is not divisible by 2, then $n_0$ is coprime to 10. Select $m_0 > m$ with $m_0$ even. It follows from Lemma 3 that there exists an alternating number $t_{m_0} = (a_{m_0} \\cdots a_1)_{10}$ with $m_0$ digits, where $a_1 = 2$. In addition, $2^{m_0+1} \\nmid t_{m_0}$. Hence $2^m \\mid t_{m_0}$. It follows from Lemma 1 that there exists $k \\in N^*$ such that\n$$\nP = t_{m_0} \\cdot 1 \\underbrace{00\\cdots0}_{\\substack{\\text{number } m, -1 \\text{ of } 0 \\\\ \\text{number } k \\text{ of } 1}} \\cdot 1\\cdots100\\cdots01\n$$\ncan be divisible by $2^m n_0$, or $n \\mid P$.\n(3) If $n$ is divisible by 5, and $n$ is not divisible by 2, let $n = 5^m n_0$ such that $n_0$ is not divisible by 5, then $n_0$ is coprime to 10. Select $m_0 > m$ with $m_0$ even. It follows from Lemma 2 that there exist an alternating number $s_{m_0} = (a_{m_0} \\cdots a_1)_{10}$ with $m_0$ digits, where $a_1 = 5$ and $a_{m_0}$ can be 0. In addition, $5^m \\mid s_{m_0}$. Hence $5^m \\mid s_{m_0}$. Using the same argument as (2), there exists an alternating number $P$ such that $5^m n_0 \\mid P$ and the last decimal digit $a_1 = 5$.\n(4) If $n$ is divisible by 10, and $n$ is not divisible by 20, let $n = 10n_0$ with $n_0$ odd, then $n_0$ satisfies the assumption in case (1) or in case (3). It follows from (1) and (3) that there exists $(a_k \\cdots a_1)_{10}$ an alternating multiple of $n_0$, where $a_1 = 1$ or 5. Now let $P = (a_k \\cdots a_1 0)_{10}$. Then $P$ is an alternating multiple of $n$.\n\nIn conclusion, $n$ should be a positive integer and should not be divisible by 20.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 28039,
    "subject": "Mathematics (Multi-modal)",
    "question": "The vertical cross-section of a circular cone with vertex $P$ is an isosceles right-angled triangle. Point $A$ is on the circumference of the base circle, point $B$ is interior to the base circle, $O$ is the center of the base circle, $AB \\perp OB$ and intersecting at $B$, $OH \\perp PB$ and intersecting at $H$, $PA = 4$, and $C$ is the midpoint of $PA$. When the tetrahedron $O-HPC$ has the maximum volume, the length of $OB$ is ( ).\n\n(A) $\\frac{\\sqrt{5}}{3}$\n(B) $\\frac{2\\sqrt{5}}{3}$\n(C) $\\frac{\\sqrt{6}}{3}$\n(D) $\\frac{2\\sqrt{6}}{3}$",
    "options": [],
    "answer": "D",
    "solution": "Since $AB \\perp OB$, and $AB \\perp OP$, we have $AB \\perp PB$, and $PAB \\perp POB$. Moreover, from $OH \\perp PB$ we obtain that $OH \\perp HC$ and $OH \\perp PA$. Since $C$ is the midpoint of $PA$, $OC \\perp PA$. Thus, $PC$ is the altitude of the\n![](attached_image_1.png)\n\ntetrahedron $O-HPC$ and $PC = 2$.\n\nIn $\\triangle OHC$, $OC = 2$. Therefore when $HO = HC$, $S_{\\triangle ABC}$ reaches its maximum, that is, $V_{O-HPC} = V_{P-HCO}$ reaches its maximum. In this case, $HO = \\sqrt{2}$, and $HO = \\frac{1}{2} OP$. Hence, $\\angle HPO = 30^\\circ$, and $OB = OP \\cdot \\tan 30^\\circ = \\frac{2\\sqrt{6}}{3}$.\n\nAnswer: D.",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 28040,
    "subject": "Mathematics (Multi-modal)",
    "question": "In a planar rectangular coordinate system $xOy$, the area enclosed by the graph of function $f(x) = a\\sin ax + \\cos ax$ ($a > 0$) defined on an interval with the least positive period and by the graph of function $g(x) = \\sqrt{a^2 + 1}$ is ______.",
    "options": [],
    "answer": "2π√(a^2+1)/a",
    "solution": "We rewrite function $f(x)$ as $f(x) = \\sqrt{a^2+1}\\sin(ax + \\varphi)$, where $\\varphi = \\arctan\\frac{1}{a}$. Its least positive period is $\\frac{2\\pi}{a}$, and its amplitude is $\\sqrt{a^2+1}$. By symmetry of the figure enclosed by the graphs of the functions $f(x)$ and $g(x)$, we can change the figure into a rectangle with length $\\frac{2\\pi}{a}$ and width $\\sqrt{a^2+1}$ using the cut-and-paste method.\n\nTherefore its area is $\\frac{2\\pi\\sqrt{a^2+1}}{a}$.",
    "topic": "Discrete Mathematics",
    "subtopic": "Other"
  },
  {
    "id": 28041,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $p$ and $q$ be two coprime positive integers, and let $n$ be a nonnegative integer. Determine the number of integers that can be written in the form $ip + jq$, where $i$ and $j$ are nonnegative integers with $i + j \\le n$. (posed by Li Weigu)",
    "options": [],
    "answer": "Let r = max{p, q}. The number is s_n = ((n+1)(n+2))/2 for n < r, and s_n = (r(2n - r + 3))/2 for n ≥ r.",
    "solution": "Define a set\n$$\nS(p, q, n) = \\{ip + jq \\mid i \\text{ and } j \\text{ are nonnegative integers with } i + j \\le n\\}\n$$\nLet $s_n = |S(p, q, n)|$, where $|X|$ denotes the number of elements in set $X$. The answer of the problem is\n$$\ns_n = \\begin{cases} \\dfrac{(n+1)(n+2)}{2}, & \\text{if } n < r, \\\\ \\dfrac{r(2n-r+3)}{2}, & \\text{if } n \\ge r, \\end{cases}\n$$\nwhere $r = \\max\\{p, q\\}$.\n\nNow we establish the equation $(*)$. Without loss of generality, we assume that $r = p > q$. It is easy to see that $s_0 = |S(p, q, 0)| = |\\{0\\}| = 1$ satisfying equation $(*)$. Note that\n$$\nS(p, q, n) \\setminus S(p, q, n-1) \\subseteq \\{ip + (n-i)q \\mid i = 0, 1, \\dots, n\\}.\n$$\nNote also that\n$$\nip + (n - i)q = (i + q)p + (n - p - i)q,\n$$\nwith\n$$\n(i+q) + (n-p-i) = n+q-p \\le n-1.\n$$\nHence number $ip + (n-i)q$ belongs to both sets $S(p, q, n)$ and $S(p, q, n-1)$ if and only if $n-p-i \\ge 0$, or $i \\le n-p$. Therefore,\n$$\nS(p, q, n) \\setminus S(p, q, n-1) = \\begin{cases} \\{ip + (n-i)q \\mid i = n-p+1, n-p+2, \\dots, n\\}, & \\text{if } n \\ge p, \\\\ \\{ip + (n-i)q \\mid i = 0, 1, \\dots, n\\}, & \\text{if } n < p. \\end{cases}\n$$\nIt implies that\n$$\ns_n - s_{n-1} = \\begin{cases} p, & \\text{if } n \\ge p, \\\\ n+1, & \\text{if } n < p. \\end{cases}\n$$\nIf $n < p$, we conclude that\n$$\ns_n = s_0 + (s_1 - s_0) + \\dots + (s_n - s_{n-1})\n$$\n$$\n=1+2+\\cdots+(n+1)=\\frac{(n+1)(n+2)}{2}.\n$$\nIn particular, $s_{p-1} = \\frac{p(p+1)}{2}$. If $n \\ge p$, we conclude that\n$$\n\\begin{aligned}\ns_n &= s_{p-1} + (s_p - s_{p-1}) + \\cdots + (s_n - s_{n-1}) \\\\\n&= \\frac{p(p+1)}{2} + (n-p+1)p = \\frac{p(2n-p+3)}{2}.\n\\end{aligned}\n$$",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 28042,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $l$ be the perimeter of an acute triangle $\\triangle ABC$ which is not equilateral, $P$ a variable point inside $\\triangle ABC$, and $D, E$ and $F$ be projections of $P$ on $BC, CA$ and $AB$ respectively.\nProve that\n$$\n2(AF + BD + CE) = l,\n$$\nif and only if $P$ is collinear with the incenter and circumcenter of $\\triangle ABC$. (posed by Xiong Bin)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Denote the lengths of three sides of $\\triangle ABC$ by $BC = a$, $CA = b$ and $AB = c$ respectively. No loss of generality, we can suppose $b \\neq c$. We choose a rectangular coordinate system (see the figure), then we have $A(m, n)$, $B(0, 0)$, $C(a, 0)$ and $P(x, y)$.\n\n![](attached_image_1.png)\n\nSince $AF^2 - BF^2 = AP^2 - BP^2$, it follows that\n$$\nAF^2 - (c - AF)^2 = AP^2 - BP^2.\n$$\nTherefore,\n$$\n2c \\cdot AF - c^2 = (x-m)^2 + (y-n)^2 - x^2 - y^2,\n$$\nand\n$$\nAF = \\frac{m^2 + n^2 - 2mx - 2ny}{2c} + \\frac{c}{2}.\n$$\n\nSimilarly, we can compute $BD$ and $CE$ in terms of $x$ and $y$.\n\nSince $AF + BD + CE = \\frac{l}{2}$, we get\n$$\n\\frac{m^2+n^2-2mx-2ny}{2c} + \\frac{c}{2} + x + \\frac{2mx+2ny-m^2-n^2-2ax+a^2}{2b} + \\frac{b}{2} = \\frac{l}{2},\n$$\nthat is,\n$$\n\\left(\\frac{m}{b} - \\frac{a}{b} - \\frac{m}{c} + 1\\right)x + \\left(\\frac{n}{b} - \\frac{n}{c}\\right)y + \\frac{a^2}{2b} + \\frac{b+c}{2} + \\frac{m^2+n^2}{2}\\left(\\frac{1}{c} - \\frac{1}{b}\\right) - \\frac{l}{2} = 0.\n$$\nSince $b \\neq c$ and $n \\neq 0$, point $P$ is on a fixed straight line. Since the condition $2(AF + BD + CE) = l$ is satisfied for both incenter and circumcenter, we complete the proof.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 28043,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $f$: $\\mathbb{R} \\rightarrow \\mathbb{R}$ be a function such that $f(0) = 1$ and for any $x$, $y \\in \\mathbb{R}$, $f(xy+1) = f(x)f(y) - f(y) - x + 2$ holds. Then $f(x) = \\underline{\\hspace{2cm}}$.",
    "options": [],
    "answer": "f(x) = x + 1",
    "solution": "Since for any $x$, $y \\in \\mathbb{R}$, $f(xy+1) = f(x)f(y) - f(y) - x + 2$, we have\n$$\nf(yx + 1) = f(y)f(x) - f(x) - y + 2.\n$$\nThus,\n$$\nf(x)f(y) - f(y) - x + 2 = f(y)f(x) - f(x) - y + 2,\n$$\nthat is,\n\nPut $y = 0$, we obtain $f(x) = x + 1$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 28044,
    "subject": "Mathematics (Multi-modal)",
    "question": "When the unit squares at the four corners are removed from a three by three square, the resulting shape is called a cross. What is the maximum number of non-overlapping crosses placed within the boundary of a $10 \\times 11$ chessboard? (Each cross covers exactly five unit squares on the board.) (posed by Feng Zuming)",
    "options": [],
    "answer": "15",
    "solution": "The centers of the crosses (denoted by $*$) must lie in the $8 \\times 9$ subboard in the middle. We tile this central board by three $8 \\times 3$ boards, and label these three boards (a), (b) and (c), from left to right. We consider the number of centers placed in the three boards.\n\n![](attached_image_1.png)\n\nNote that we can place at most two centers on each $3 \\times 3$ subboard;\n\n![](attached_image_2.png)\n\nWe can tile a $8 \\times 3$ board one $2 \\times 3$ board sandwiched by two $3 \\times 3$ boards. Hence we can place at most $6$ centers on a $8 \\times 3$ board, with each two centers placed on each subboard. Since there are two centers placed in the middle $2 \\times 3$ subboard, no centers can be placed in the third and sixth row of the $8 \\times 3$ board. We can only have the following two symmetric distributions.\n\n**Case I** Board (b) has $6$ centers. By symmetry, we can assume that the following scheme for placing the centers (see left-hand side diagram in the following figure). It is not difficult to see that no centers can be placed on the third and the seventh columns of the $8 \\times 9$ board. Then it is easy to see that we can place at most $4$ centers in board (a) or (c), which implies that we can place at most $4+6+4=14$ centers on the $8 \\times 9$ board. It contradicts the assumption.\n\n**Case II** Both boards (a) and (c) have six centers. By symmetry, we discuss with the right-hand side diagram in the following figure. In this case there is no centers can be placed in the fourth and the sixth columns of the $8 \\times 9$ board, which implies that board (b) can hold at most $3$ centers, and so the $8 \\times 9$ board can hold at most $6+3+6=15$ centers, which is again a contradiction.\n\n**Case III** Exactly one of the boards (a) and (c) has six centers. In this case we assume that (a) contains $6$ centers and board (c) contains at most $5$ centers. Then no center can be placed in the fourth column of the $8 \\times 9$ board. It follows that board (b) contains at most $4$ centers. Hence there are at most $6+4+5=15$ centers on the $8 \\times 9$ board, which is again a contradiction.\n\nCombining the above argument, we conclude that it is impossible to place $16$ centers on a $8 \\times 9$ board.\n\nWe complete our solution by providing two different ways to place $15$ centers on the board.\n\n![](attached_image_3.png)\n\nor\n\n![](attached_image_4.png)",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 28045,
    "subject": "Mathematics (Multi-modal)",
    "question": "Suppose that $a, b, c$ are positive real numbers, prove that\n$$\n1 < \\frac{a}{\\sqrt{a^2 + b^2}} + \\frac{b}{\\sqrt{b^2 + c^2}} + \\frac{c}{\\sqrt{c^2 + a^2}} \\le \\frac{3\\sqrt{2}}{2}.\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Set $x = \\frac{b^2}{a^2}$, $y = \\frac{c^2}{b^2}$, $z = \\frac{a^2}{c^2}$, then $x, y, z \\in \\mathbb{R}^+$ and $xyz = 1$. It suffices to prove that\n$$\n1 < \\frac{1}{\\sqrt{1+x}} + \\frac{1}{\\sqrt{1+y}} + \\frac{1}{\\sqrt{1+z}} \\le \\frac{3\\sqrt{2}}{2}.\n$$\n\nWithout loss of generality, we assume that $x \\le y \\le z$. Set $A = xy$, we have $z = \\frac{1}{A}$, $A \\le 1$. Thus\n$$\n\\frac{1}{\\sqrt{1+x}} + \\frac{1}{\\sqrt{1+y}} + \\frac{1}{\\sqrt{1+z}} > \\frac{1}{\\sqrt{1+x}} + \\frac{1}{\\sqrt{1+\\frac{1}{x}}} = \\frac{1+\\sqrt{x}}{\\sqrt{1+x}} > 1.\n$$\nLet $u = \\frac{1}{\\sqrt{1+A+x+\\frac{A}{x}}}$, then $u \\in (0, \\frac{1}{1+\\sqrt{A}}]$, and $x = \\sqrt{A}$ if and only if $u = \\frac{1}{1+\\sqrt{A}}$. Hence\n$$\n\\left( \\frac{1}{\\sqrt{1+x}} + \\frac{1}{\\sqrt{1+y}} \\right)^2 = \\left[ \\frac{1}{\\sqrt{1+x}} + \\frac{1}{\\sqrt{1+\\frac{A}{x}}} \\right]^2\n$$\n$$\n= \\frac{1}{1+x} + \\frac{1}{1+\\frac{A}{x}} + \\frac{2}{\\sqrt{1+A+x+\\frac{A}{x}}} \\\\\n= \\frac{2+x+\\frac{A}{x}}{1+A+x+\\frac{A}{x}} + \\frac{2}{\\sqrt{1+A+x+\\frac{A}{x}}} \\\\\n= 1 + (1-A)u^2 + 2u.\n$$\nSet $f(u) = (1-A)u^2 + 2u + 1$, we see that $f(u)$ is an increasing function on $u \\in (0, \\frac{1}{1+\\sqrt{A}}]$, which implies that\n$$\n\\frac{1}{\\sqrt{1+x}} + \\frac{1}{\\sqrt{1+y}} \\le \\sqrt{f\\left(\\frac{1}{1+\\sqrt{A}}\\right)} = \\frac{2}{\\sqrt{1+\\sqrt{A}}}\n$$\nNow set $\\sqrt{A} = v$, we get\n$$\n\\frac{1}{\\sqrt{1+x}} + \\frac{1}{\\sqrt{1+y}} + \\frac{1}{\\sqrt{1+z}} \\le \\frac{2}{\\sqrt{1+\\sqrt{A}}} + \\frac{1}{\\sqrt{1+\\frac{1}{A}}} \\\\\n= \\frac{2}{\\sqrt{1+v}} + \\frac{\\sqrt{2}v}{\\sqrt{2}(1+v^2)} \\le \\frac{2}{\\sqrt{1+v}} + \\frac{\\sqrt{2}v}{1+v} \\\\\n= \\frac{2}{\\sqrt{1+v}} + \\sqrt{2} - \\frac{\\sqrt{2}}{1+v} = -\\sqrt{2} \\left[ \\frac{1}{\\sqrt{1+v}} - \\frac{\\sqrt{2}}{2} \\right]^2 + \\frac{3\\sqrt{2}}{2} \\\\\n\\le \\frac{3\\sqrt{2}}{2}.\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 28046,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $\\theta_i \\in (-\\frac{\\pi}{2}, \\frac{\\pi}{2})$, $i = 1, 2, 3, 4$. Prove that there exists $x \\in \\mathbb{R}$ such that the following two inequalities\n$$\n\\cos^2\\theta_1 \\cos^2\\theta_2 - (\\sin\\theta_1 \\sin\\theta_2 - x)^2 \\ge 0, \\quad ①\n$$\n$$\n\\cos^2\\theta_3 \\cos^2\\theta_4 - (\\sin\\theta_3 \\sin\\theta_4 - x)^2 \\ge 0, \\quad ②\n$$\nhold simultaneously if and only if\n$$\n\\sum_{i=1}^{4} \\sin^2 \\theta_i \\le 2 \\left( 1 + \\prod_{i=1}^{4} \\sin \\theta_i + \\prod_{i=1}^{4} \\cos \\theta_i \\right). \\quad ③\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "**Proof** Clearly, ① and ② are equivalent to\n$$\n\\sin \\theta_1 \\sin \\theta_2 - \\cos \\theta_1 \\cos \\theta_2 \\le x \\le \\sin \\theta_1 \\sin \\theta_2 + \\cos \\theta_1 \\cos \\theta_2, \\quad ④\n$$\n$$\n\\sin \\theta_3 \\sin \\theta_4 - \\cos \\theta_3 \\cos \\theta_4 \\le x \\le \\sin \\theta_3 \\sin \\theta_4 + \\cos \\theta_3 \\cos \\theta_4, \\quad ⑤\n$$\nrespectively. It is easy to show that there exists $x \\in \\mathbb{R}$ such that ④ and ⑤ hold simultaneously if and only if\n$$\n\\sin \\theta_1 \\sin \\theta_2 + \\cos \\theta_1 \\cos \\theta_2 - \\sin \\theta_3 \\sin \\theta_4 + \\cos \\theta_3 \\cos \\theta_4 \\ge 0, \\quad ⑥\n$$\n$$\n\\sin \\theta_3 \\sin \\theta_4 + \\cos \\theta_3 \\cos \\theta_4 - \\sin \\theta_1 \\sin \\theta_2 + \\cos \\theta_1 \\cos \\theta_2 \\ge 0. \\quad ⑦\n$$\n\n$$\n\\cos^2\\theta_1 \\cos^2\\theta_2 + 2\\cos\\theta_1 \\cos\\theta_2 \\cos\\theta_3 \\cos\\theta_4 + \\cos^2\\theta_3 \\cos^2\\theta_4 \\\\\n- \\sin^2\\theta_1 \\sin^2\\theta_2 + 2\\sin\\theta_1 \\sin\\theta_2 \\sin\\theta_3 \\sin\\theta_4 - \\sin^2\\theta_3 \\sin^2\\theta_4 \\ge 0, \\\\\n\\text{or}\n$$\n$$\n(\\cos \\theta_1 \\cos \\theta_2 + \\cos \\theta_3 \\cos \\theta_4)^2 - (\\sin \\theta_1 \\sin \\theta_2 - \\sin \\theta_3 \\sin \\theta_4)^2 \\ge 0.\n$$\nThat is,\n$$\n(\\sin \\theta_1 \\sin \\theta_2 + \\cos \\theta_1 \\cos \\theta_2 - \\sin \\theta_3 \\sin \\theta_4 + \\cos \\theta_3 \\cos \\theta_4) \\cdot\n$$\n$$\n(\\sin \\theta_3 \\sin \\theta_4 + \\cos \\theta_3 \\cos \\theta_4 - \\sin \\theta_1 \\sin \\theta_2 + \\cos \\theta_1 \\cos \\theta_2) \\ge 0. \\quad ⑧\n$$\nIf there exists $x \\in \\mathbb{R}$ such that ④ and ⑤ hold simultaneously, then from ⑥ and ⑦ we can get ⑧ immediately. It follows that ③ holds.\n\nConversely, if ③ holds, or equivalently ⑧ holds, but ⑥ and ⑦ do not hold, then we have\n$$\n\\sin \\theta_1 \\sin \\theta_2 + \\cos \\theta_1 \\cos \\theta_2 - \\sin \\theta_3 \\sin \\theta_4 + \\cos \\theta_3 \\cos \\theta_4 < 0, \\text{ and}\n$$\n$$\n\\sin \\theta_3 \\sin \\theta_4 + \\cos \\theta_3 \\cos \\theta_4 - \\sin \\theta_1 \\sin \\theta_2 + \\cos \\theta_1 \\cos \\theta_2 < 0.\n$$\nAdding the last two equations, we obtain\n$$\n2(\\cos \\theta_1 \\cos \\theta_2 + \\cos \\theta_3 \\cos \\theta_4) < 0,\n$$\nwhich contradicts to the fact $\\theta_i \\in (-\\frac{\\pi}{2}, \\frac{\\pi}{2})$, $i = 1, 2, 3, 4$. So ⑥ and ⑦ hold simultaneously. Hence there exists $x \\in \\mathbb{R}$ such that ④ and ⑤ hold simultaneously.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 28047,
    "subject": "Mathematics (Multi-modal)",
    "question": "Assume that $\\alpha^{2005} + \\beta^{2005}$ can be expressed as a polynomial in $\\alpha + \\beta$ and $\\alpha\\beta$. Find the sum of the coefficients of the polynomial. (posed by Zhu Huawei)",
    "options": [],
    "answer": "1",
    "solution": "In the expansion of $\\alpha^k + \\beta^k$, let $\\alpha + \\beta = 1$ and $\\alpha\\beta = 1$. We get the sum of coefficients $S_k = \\alpha^k + \\beta^k$. Since\n\n$$(\\alpha + \\beta)(\\alpha^{k-1} + \\beta^{k-1}) = (\\alpha^k + \\beta^k) + \\alpha\\beta(\\alpha^{k-2} + \\beta^{k-2}),$$\n\nwe get\n$S_k = S_{k-1} - S_{k-2}$.\nThus $S_k = S_{k-6}$ and $\\{S_k\\}$ is a periodic sequence with period 6 and\n$S_{2005} = S_1 = 1$.\nSet $\\alpha + \\beta = 1$ and $\\alpha\\beta = 1$ in the expansion of $\\alpha^k + \\beta^k$. The sum of the coefficients is $S_k = \\alpha^k + \\beta^k$. Since $\\alpha, \\beta$ are solutions of the equation $x^2 - x + 1 = 0$, $\\alpha = \\cos \\frac{\\pi}{3} + i \\sin \\frac{\\pi}{3}$, $\\beta = \\cos \\frac{\\pi}{3} - i \\sin \\frac{\\pi}{3}$. Therefore\n\n$$\n\\begin{align*}\n\\alpha^k + \\beta^k &= \\left(\\cos \\frac{\\pi}{3} + i \\sin \\frac{\\pi}{3}\\right)^k + \\left(\\cos \\frac{\\pi}{3} - i \\sin \\frac{\\pi}{3}\\right)^k \\\\\n&= \\left(\\cos \\frac{k\\pi}{3} + i \\sin \\frac{k\\pi}{3}\\right) + \\left(\\cos \\frac{k\\pi}{3} - i \\sin \\frac{k\\pi}{3}\\right) \\\\\n&= 2\\cos \\frac{k\\pi}{3}.\n\\end{align*}\n$$\n\nLet $k = 2005$, we have $S_k = 1$.",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 28048,
    "subject": "Mathematics (Multi-modal)",
    "question": "Six points are chosen on the sides of an equilateral triangle $ABC$: $A_1, A_2$ on $BC$, $B_1, B_2$ on $CA$, and $C_1, C_2$ on $AB$. These points are the vertices of a convex hexagon $A_1A_2B_1B_2C_1C_2$ with sides of equal length. Prove that the lines $A_1B_2$, $B_1C_2$ and $C_1A_2$ are concurrent. (proposed by Romania, average score 2.61.)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Assume $A_1A_2 = d$, and $AB = a$. Construct an equilateral triangle $A_0B_0C_0$ with side of length $a-d$. Points $A', B', C'$ are chosen on the sides of triangle $A_0B_0C_0$ such that $A'C_0 = A_2C$, $B'A_0 = B_2A$, and $C'B_0 = C_2B$.\n\nTherefore,\n$$\nA'B_0 = a-d-A'C_0 = BC-A_1A_2-A_2C = BA_1\n$$\n\n$$\nB'C_0 = B_1C, \\quad C'A_0 = C_1A.\n$$\nSince\n$$\n\\angle B_1CA_2 = \\angle B'C_0A', \\quad \\angle B_2AC_1 = \\angle B'A_0C', \\quad \\angle C_2BA_1 = \\angle C'B_0A',\n$$\nthus\n$$\n\\triangle CB_1A_2 \\cong \\triangle C_0B'A', \\quad \\triangle AB_2C_1 \\cong \\triangle A_0B'C', \\quad \\triangle C_2BA_1 \\cong \\triangle B_0C'A',\n$$\nwhich implies that $B'C' = C'A' = A'B' = d$ and triangle $A'B'C'$ is equilateral.\nSo\n$$\n\\angle AB_2C_1 = \\angle A_0B'C' = 180^\\circ - \\angle C'B'A' - \\angle A'B'C_0 = 120^\\circ - \\angle A'B'C_0, \\quad \\angle C_1B_2B_1 = \\angle B_1A_2A_1.\n$$\nIn view of $B_2C_1 = B_1B_2 = A_2B_1 = A_1A_2 = d$, triangles $C_1B_2B_1$ and $B_1B_2A_1$ are congruent, implying that $B_1C_1 = A_1B_1$. Together with $C_1C_2 = A_1C_2$, we show that $C_2B_1$ is the perpendicular bisector of $A_1C_1$ and $C_2B_1$ is the height of triangle $A_1B_1C_1$ on side $A_1C_1$. Similarly, $C_1A_2$ and $A_1B_2$ are the altitudes of triangle $A_1B_1C_1$ to the sides $A_1B_1$ and $B_1C_1$ respectively.\n\nTherefore the lines $A_1B_2$, $B_1C_2$ and $C_1A_2$ are concurrent.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 28049,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $k$ be a real number such that the inequality $\\sqrt{x-3} + \\sqrt{6-x} \\ge k$ has a solution. The maximum value of $k$ is ( ).\n(A) $\\sqrt{6}-\\sqrt{3}$\n(B) $\\sqrt{3}$\n(C) $\\sqrt{6}+\\sqrt{3}$\n(D) $\\sqrt{6}$",
    "options": [],
    "answer": "D",
    "solution": "Set $y = \\sqrt{x-3} + \\sqrt{6-x}$, $3 \\le x \\le 6$.\nThen\n$$\n\\begin{aligned}\ny^2 &= (x-3) + (6-x) + 2\\sqrt{(x-3)(6-x)} \\\\\n&\\le 2[(x-3) + (6-x)] = 6.\n\\end{aligned}\n$$\nSo $0 < y \\le \\sqrt{6}$, and the maximum value of $k$ is $\\sqrt{6}$. Answer: D.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 28050,
    "subject": "Mathematics (Multi-modal)",
    "question": "If one side of square $ABCD$ is on the line $y = 2x - 17$, and the other two vertices lie on parabola $y = x^2$. Then the minimum area of the square is ________.",
    "options": [],
    "answer": "80",
    "solution": "Assume that $AB$ is on the line $y = 2x - 17$ and the coordinates of the other two vertices on the parabola are $C(x_1, y_1)$ and $D(x_2, y_2)$. Then $CD$ is on a line $L$ whose equation is $y = 2x + b$. Combining this with the equation of the parabola, we get $x^2 = 2x + b \\Rightarrow x_1, x_2 = 1 \\pm \\sqrt{b+1}$. Assume that the length of one side of the square is $a$. Then\n$$\n\\begin{aligned}\na^2 &= (x_1 - x_2)^2 + (y_1 - y_2)^2 \\\\\n&= 5(x_1 - x_2)^2 = 20(b+1). \\qquad (1)\n\\end{aligned}\n$$\n\nPick a point $(6, -5)$ on the line $y = 2x - 17$, and the distance from the point to the line $y = 2x + b$ is $a$.\n$$\n\\text{So} \\qquad a = \\frac{|17+b|}{\\sqrt{5}}. \\qquad (2)\n$$\nFrom (1) and (2), we get $b_1 = 3$, $b_2 = 63$, so $a^2 = 80$ or $a^2 = 1280$, and $a_{\\min}^2 = 80$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 28051,
    "subject": "Mathematics (Multi-modal)",
    "question": "A natural number $a$ is called a \"lucky number\" if the sum of its digits is $7$. Arrange all \"lucky numbers\" in ascending order, and we get a sequence $a_1, a_2, \\dots$. If $a_n = 2005$, then $a_{5n} = \\_\\_\\_\\_\\_\\_$. ",
    "options": [],
    "answer": "52000",
    "solution": "Since the number of non-negative integer solutions of the equation $x_1 + x_2 + \\dots + x_k = m$ is $C_{m+k-1}^k$, the number of integer solutions, when $x_1 \\ge 1$ and $x_i \\ge 0\\ (i \\ge 2)$, is $C_{m+k-2}^{m-1}$. Let $m=7$, the number of lucky numbers with $k$ digits is $p(k) = C_{k+5}^6$.\n\nSince $2005$ is the minimum lucky number of the type $\\overline{2abc}$ and $p(1) = C_6^6 = 1$, $p(2) = C_7^6 = 7$, $p(3) = C_8^6 = 28$. Note that the number of four-digit lucky numbers of the type $\\overline{1abc}$ is the number of non-negative integer solutions of $a+b+c=6$, i.e. $C_{6+3-1}^6 = 28$. Thus $1+7+28+28+1=65$ and $2005$ is the $65$-th lucky number, i.e. $a_{65}=2005$, so $n=65$, $5n=325$.\n\nFurthermore $p(4) = C_9^9 = 84$, $p(5) = C_{10}^9 = 210$ and $\\sum_{k=1}^{5} p(k) = 330$.\n\nTherefore the last six lucky numbers with $5$ digits, from the largest to the smallest, are $70\\ 000$, $61\\ 000$, $60\\ 100$, $60\\ 010$, $60\\ 001$, $52\\ 000$. So the $325$-th lucky number is $52\\ 000$, i.e. $a_{325} = 52\\ 000$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 28052,
    "subject": "Mathematics (Multi-modal)",
    "question": "Given a sequence $\\{a_n\\}$ of numbers satisfying $a_0 = 1$, $a_{n+1} = \\frac{7a_n + \\sqrt{45a_n^2 - 36}}{2}$, $n \\in \\mathbb{N}$.\nProve that\n(1) for each $n \\in \\mathbb{N}$, $a_n$ is a positive integer.\n(2) for each $n \\in \\mathbb{N}$, $a_n a_{n+1} - 1$ is a perfect square.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "(1) By assumption, $a_1 = 5$ and $\\{a_n\\}$ is strictly increasing with\n$$\n2a_{n+1} - 7a_n = \\sqrt{45a_n^2 - 36}.\n$$\nSquare both sides, and we get\n$$\na_{n+1}^2 - 7a_n a_{n+1} + a_n^2 + 9 = 0, \\qquad \\textcircled{1}\n$$\n$$\na_n^2 - 7a_{n-1}a_n + a_{n-1}^2 + 9 = 0, \\qquad \\textcircled{2}\n$$\n$$\n\\textcircled{1}-\\textcircled{2}: a_{n+1} = 7a_n - a_{n-1}. \\qquad \\textcircled{3}\n$$\nIt follows from $a_0 = 1$, $a_1 = 5$ and ③ that $a_n$ is a positive integer for each $n \\in \\mathbb{N}$.\n\n(2) From ①, we get $(a_{n+1} + a_n)^2 = 9(a_n a_{n+1} - 1)$,\n$$\n\\text{SO} \\qquad a_{n+1}a_n - 1 = \\left(\\frac{a_{n+1} + a_n}{3}\\right)^2\n$$\nBy (1), $a_n$, $a_{n+1}$ are positive integers and therefore $\\frac{a_{n+1} + a_n}{3}$ is a rational number. Since $\\left(\\frac{a_{n+1} + a_n}{3}\\right)^2 = a_{n+1}a_n - 1$ is a positive integer, so is $\\frac{a_{n+1} + a_n}{3}$. Thus $a_{n+1}a_n - 1$ is the square of an integer.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 28053,
    "subject": "Mathematics (Multi-modal)",
    "question": "Nine balls, numbered $1$, $2$, $\\ldots$, $9$, are put randomly at $9$ equally spaced points on a circle, each point with a ball. Let $S$ be the sum of the absolute values of the differences of the numbers of all two neighboring balls. Find the probability of $S$ to be the minimum value. (Remark: If one arrangement of the balls is congruent to another after a rotation or a reflection, the two arrangements are regarded as the same).",
    "options": [],
    "answer": "1/315",
    "solution": "Next, we calculate the number of arrangements, which make $S$ the minimum. Along the circle there are two routes from $1$ to $9$, the major arc and the minor arc. For each of them, let $x_1, x_2, \\dots, x_k$ be the numbers of the successive balls on the arc, then\n$$\n\\begin{aligned}\n& |1-x_1| + |x_1-x_2| + \\dots + |x_k-9| \\\\\n\\ge & |(1-x_1) + (x_1-x_2) + \\dots + (x_k-9)| \\\\\n= & |1-9| = 8.\n\\end{aligned}\n$$\nThe equality occurs if and only if $1 < x_1 < x_2 < \\dots < x_k < 9$, i.e. the numbers of the balls on each route is increasing from $1$ to $9$.\n\nTherefore, $S_{\\min} = 2 \\cdot 8 = 16$.\n\nFrom the above analysis, when the numbers of the balls $\\{1, x_1, x_2, \\dots, x_k, 9\\}$ on each arc are fixed, the arrangement which gets the minimum value is uniquely determined. Divide the set of $7$ balls $\\{2, 3, \\dots, 8\\}$ into two subsets, then the subset which contains less elements has $C_9^0 + C_9^1 + C_9^2 + C_9^3 = 2^6$ cases. Each case corresponds to a unique arrangement, which achieves the minimum value of $S$.\n\nThus, the number of the arrangements when $S$ takes the minimum value is $2^6$ and the corresponding probability is $p = \\frac{2^6}{\\frac{8!}{2}} = \\frac{1}{315}$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 28054,
    "subject": "Mathematics (Multi-modal)",
    "question": "Draw a tangent line of parabola $y = x^2$ at the point $A(1, 1)$. Suppose the line intersects the $x$-axis and $y$-axis at $D$ and $B$ respectively. Let point $C$ be on the parabola and point $E$ on $AC$ such that $\\frac{AE}{EC} = \\lambda_1$. Let point $F$ be on $BC$ such that $\\frac{BF}{FC} = \\lambda_2$ and $\\lambda_1 + \\lambda_2 = 1$. Assume that $CD$ intersects $EF$ at point $P$. When point $C$ moves along the parabola, find the equation of the trail of $P$.",
    "options": [],
    "answer": "y = (1/3)(3x - 1)^2, with x ≠ 2/3",
    "solution": "The slope of the tangent line passing through $A$ is $y' = 2x|_{x=1} = 2$. So the equation of the tangent line $AB$ is $y = 2x - 1$. Hence the coordinates of $B$ and $D$ are $B(0, -1)$, $D(\\frac{1}{2}, 0)$. Thus $D$ is the midpoint of line segment $AB$.\n\nConsider $P(x, y)$, $C(x_0, x_0^2)$, $E(x_1, y_1)$, $F(x_2, y_2)$. Then by $\\frac{AE}{EC} = \\lambda_1$, we know $x_1 = \\frac{1+\\lambda_1 x_0}{1+\\lambda_1}$, $y_1 = \\frac{1+\\lambda_1 x_0^2}{1+\\lambda_1}$. From $\\frac{BF}{FC} = \\lambda_2$, we get $x_2 = \\frac{\\lambda_2 x_0}{1+\\lambda_2}$, $y_2 = \\frac{-1+\\lambda_2 x_0^2}{1+\\lambda_2}$.\n\n$$\n\\frac{y - \\frac{1 + \\lambda_1 x_0^2}{1 + \\lambda_1}}{\\frac{-1 + \\lambda_2 x_0^2}{1 + \\lambda_2} - \\frac{1 + \\lambda_1 x_0^2}{1 + \\lambda_1}} = \\frac{x - \\frac{1 + \\lambda_1 x_0}{1 + \\lambda_1}}{\\frac{\\lambda_2 x_0}{1 + \\lambda_2} - \\frac{1 + \\lambda_1 x_0}{1 + \\lambda_1}}\n$$\n\nSimplifying it, we get\n$$\n[(\\lambda_2 - \\lambda_1)x_0 - (1 + \\lambda_2)]y = [(\\lambda_2 - \\lambda_1)x_0^2 - 3]x + 1 + x_0 - \\lambda_2 x_0^2. \\quad (1)\n$$\n\nWhen $x_0 \\neq \\frac{1}{2}$, the equation of line $CD$ is\n$$\ny = \\frac{2x_0^2 x - x_0^2}{2x_0 - 1}. \\quad (2)\n$$\n\nFrom (1) and (2), we get\n$$\n\\begin{cases}\nx = \\frac{x_0 + 1}{3}, \\\\\ny = \\frac{x_0}{3}.\n\\end{cases}\n$$\n\nEliminating $x_0$, we get the equation of the trail of point $P$ as $y = \\frac{1}{3}(3x-1)^2$.\n\nWhen $x_0 = \\frac{1}{2}$, the equation of $EF$ is $-\\frac{3}{2}y = (\\frac{1}{4}\\lambda_2 - \\frac{1}{4}\\lambda_1 - 3)x + \\frac{3}{2} - \\frac{1}{4}\\lambda_2$, the equation of $CD$ is $x = \\frac{1}{2}$. Combining them, we conclude that $(x, y) = (\\frac{1}{2}, \\frac{1}{12})$ is on the trail of $P$. Since $C$ and $A$ cannot be congruent, $x_0 \\neq 1, x \\neq \\frac{2}{3}$.\n\nTherefore the equation of the trail is $y = \\frac{1}{3}(3x-1)^2$, $x \\neq \\frac{2}{3}$.\nFrom Solution I, the equation of $AB$ is $y = 2x - 1$, $B(0, -1)$, $D(\\frac{1}{2}, 0)$. Thus $D$ is the midpoint of $AB$.\n\nSet $\\gamma = \\frac{CD}{CP}$, $t_1 = \\frac{CA}{CE} = 1+\\lambda_1$, $t_2 = \\frac{CB}{CF} = 1+\\lambda_2$. Then $t_1+t_2=3$. Since $AD$ is a median of $\\triangle ABC$, $S_{\\triangle CAB} = 2S_{\\triangle CAD} = 2S_{\\triangle CBD}$ where $S_{\\triangle}$ denotes the area of $\\triangle$. But\n$$\n\\frac{1}{t_1 t_2} = \\frac{CE \\cdot CF}{CA \\cdot CB} = \\frac{S_{\\triangle CEF}}{S_{\\triangle CAB}} = \\frac{S_{\\triangle CEP}}{2S_{\\triangle CAD}} + \\frac{S_{\\triangle CFP}}{2S_{\\triangle CED}}\n$$\n$$\n= \\frac{1}{2} \\left( \\frac{1}{t_1 \\gamma} + \\frac{1}{t_2 \\gamma} \\right) = \\frac{t_1 + t_2}{2t_1 t_2 \\gamma} = \\frac{3}{2t_1 t_2 \\gamma},\n$$\nso $\\gamma = \\frac{3}{2}$ and $P$ is the center of gravity for $\\triangle ABC$.\n\nConsider $P(x, y)$ and $C(x_0, x_0^2)$. Since $C$ is different from $A$, $x_0 \\neq 1$. Thus the coordinates of the center of gravity $P$ are $x = \\frac{0+1+x_0}{3} = \\frac{1+x_0}{3}$, $x \\neq \\frac{2}{3}$, $y = \\frac{-1+1+x_0^2}{3} = \\frac{x_0^2}{3}$. Eliminating $x_0$, we get $y = \\frac{1}{3}(3x-1)^2$. Thus the equation of the trail is $y = \\frac{1}{3}(3x-1)^2$, $x \\neq \\frac{2}{3}$.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 28055,
    "subject": "Mathematics (Multi-modal)",
    "question": "A circle intersects sides $BC$, $CA$, $AB$ of $ABC$ at two points for each side in the following order: \\{$D_1, D_2$\\}, \\{$E_1, E_2$\\} and \\{$F_1, F_2$\\}. Line segments $D_1E_1$ and $D_2E_2$ intersect at point $L$, $E_1F_1$ and $E_2D_2$ intersect at point $M$, $F_1D_1$ and $F_2E_2$ intersect at point $N$. Prove that $AL$, $BM$ and $CN$ are concurrent. (posed by Ye Zhonghao)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "**Proof** Through point $L$ draw perpendicular lines to $AB$ and to $AC$, the feet are $L'$ and $L''$ respectively. Let $\\angle LAB = \\alpha_1$, $\\angle LAC = \\alpha_2$, $\\angle LF_2A = \\alpha_3$, and $\\angle LE_1A = \\alpha_4$. We have\n$$\n\\frac{\\sin \\alpha_1}{\\sin \\alpha_2} = \\frac{LL'}{LL''} = \\frac{LF_2 \\sin \\alpha_3}{LE_1 \\sin \\alpha_4}. \\quad ①\n$$\nDraw line segments $D_1F_2$ and $D_2E_1$ (see Figure 1). Since $\\triangle LD_1F_2 \\sim \\triangle LD_2E_1$ we get\n\n![](attached_image_1.png)\nFigure 1\n![](attached_image_2.png)\nFigure 2\n$$\n\\frac{LF_2}{LE_1} = \\frac{D_1 F_2}{D_2 E_1}. \\qquad \\textcircled{2}\n$$\nDraw line segments $D_2F_1$ and $D_1E_2$ (see Figure 2). By using the sine rule we obtain\n$$\n\\frac{\\sin \\alpha_3}{\\sin \\alpha_4} = \\frac{D_2 F_1}{D_1 E_2}. \\qquad \\textcircled{3}\n$$\nSubstituting ② and ③ into ①, we have\n$$\n\\frac{\\sin \\alpha_1}{\\sin \\alpha_2} = \\frac{D_1 F_2}{D_2 E_1} \\cdot \\frac{D_2 F_1}{D_1 E_2}. \\qquad \\textcircled{4}\n$$\nSimilarly, write $\\angle BMC = \\beta_1$, $\\angle MBA = \\beta_2$, $\\angle NCA = \\gamma_1$, $\\angle NCB = \\gamma_2$, and we get\n$$\n\\frac{\\sin \\beta_1}{\\sin \\beta_2} = \\frac{E_1 D_2}{E_2 F_1} \\cdot \\frac{E_2 D_1}{E_1 F_2}, \\qquad \\textcircled{5}\n$$\n$$\n\\frac{\\sin \\gamma_1}{\\sin \\gamma_2} = \\frac{F_1 E_2}{F_2 D_1} \\cdot \\frac{F_2 E_1}{F_1 D_2}. \\qquad \\textcircled{6}\n$$\nMultiplying ④, ⑤ and ⑥, we obtain\n$$\n\\frac{\\sin \\alpha_1}{\\sin \\alpha_2} \\cdot \\frac{\\sin \\beta_1}{\\sin \\beta_2} \\cdot \\frac{\\sin \\gamma_1}{\\sin \\gamma_2} = 1.\n$$\nFinally, according to the inverse of Ceva's Theorem, we know $AL$, $BM$ and $CN$ have a common point.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 28056,
    "subject": "Mathematics (Multi-modal)",
    "question": "Assume that positive numbers $a$, $b$, $c$, $x$, $y$, $z$ satisfy $cy + bz = a$; $az + cx = b$ and $bx + ay = c$. Find the minimum value of the function $f(x, y, z) = \\frac{x^2}{1+x} + \\frac{y^2}{1+y} + \\frac{z^2}{1+z}$.",
    "options": [],
    "answer": "1/2",
    "solution": "By assumption, $b(az + cx - b) + c(bx + ay - c) - a(cy + bz - a) = 0$, i.e. $2bcx + a^2 - b^2 - c^2 = 0$, we get $x = \\frac{b^2 + c^2 - a^2}{2bc}$. For the similar reason, $y = \\frac{a^2 + c^2 - b^2}{2ac}$ and $z = \\frac{a^2 + b^2 - c^2}{2ab}$.\n\nSince $a$, $b$, $c$, $x$, $y$, $z$ are positive, by the above three expressions, we know $b^2 + c^2 > a^2$, $a^2 + c^2 > b^2$ and $a^2 + b^2 > c^2$. Thus there is an acute triangle $ABC$ with the lengths of its sides $a$, $b$, $c$. So $x = \\cos A$, $y = \\cos B$ and $z = \\cos C$. The problem is now changed to finding the minimum value of the function\n$$\nf(\\cos A, \\cos B, \\cos C) = \\frac{\\cos^2 A}{1+\\cos A} + \\frac{\\cos^2 B}{1+\\cos B} + \\frac{\\cos^2 C}{1+\\cos C}\n$$\nSet $u = \\cot A$, $v = \\cot B$, $w = \\cot C$, then $u, v, w \\in \\mathbb{R}^+$, $uv + vw + wu = 1$, $u^2 + 1 = (u+v)(u+w)$, $v^2 + 1 = (u+v)(v+w)$ and $w^2 + 1 = (u+w)(v+w)$.\n$$\n\\begin{aligned}\n\\text{We get}\\quad \\frac{\\cos^2 A}{1+\\cos A} &= \\frac{\\frac{u^2}{u^2+1}}{1+\\frac{u}{\\sqrt{u^2+1}}} = \\frac{u^2}{\\sqrt{u^2+1}(\\sqrt{u^2+1}+u)} \\\\\n&= \\frac{u^2(\\sqrt{u^2+1}-u)}{\\sqrt{u^2+1}} = u^2 - \\frac{u^3}{\\sqrt{u^2+1}}\n\\end{aligned}\n$$\n$$\n\\begin{aligned}\n&= u^2 - \\frac{u^3}{\\sqrt{(u+v)(u+w)}} \\\\\n&\\ge u^2 - \\frac{u^3}{2} \\left( \\frac{1}{u+v} + \\frac{1}{u+w} \\right).\n\\end{aligned}\n$$\nBy a similar argument, $\\frac{\\cos^2 B}{1 + \\cos B} \\ge v^2 - \\frac{v^3}{2} \\left( \\frac{1}{u+v} + \\frac{1}{v+w} \\right)$\nand\n$\\frac{\\cos^2 C}{1 + \\cos C} \\ge w^2 - \\frac{w^3}{2} \\left( \\frac{1}{u+w} + \\frac{1}{v+w} \\right).$\n$$\n\\begin{aligned}\n\\text{Hence}\\quad f &\\ge u^2 + v^2 + w^2 - \\frac{1}{2} \\left( \\frac{u^3 + v^3}{u+v} + \\frac{w^3 + v^3}{w+v} + \\frac{u^3 + w^3}{u+w} \\right) \\\\\n&= u^2 + v^2 + w^2 - \\frac{1}{2} \\left[ (u^2 - uv + v^2) \\\\\n& \\qquad + (v^2 - vw + w^2) + (u^2 - uw^2 + w^2) \\right] \\\\\n&= \\frac{1}{2} (uv + vw + uw) = \\frac{1}{2},\n\\end{aligned}\n$$\nthe equality sign is valid if and only if $u = v = w$, i.e. $a = b = c$, $x = y = z = \\frac{1}{2}$,\nso $[f(x, y, z)]_{\\min} = \\frac{1}{2}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 28057,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all ordered triples $(x, y, z)$ of real numbers such that\n$$\n\\begin{cases} 5\\left(x+\\frac{1}{x}\\right) = 12\\left(y+\\frac{1}{y}\\right) = 13\\left(z+\\frac{1}{z}\\right), \\\\ xy + yz + zx = 1. \\end{cases} \\quad \\text{(posed by Zhu Huawei)}\n$$",
    "options": [],
    "answer": "(1/5, 2/3, 1) and (-1/5, -2/3, -1)",
    "solution": "There are angles $A, B$, and $C$ in the interval $(0^\\circ, 180^\\circ)$ such that\n$$\nx = \\tan \\frac{A}{2}, \\quad y = \\tan \\frac{B}{2}, \\quad z = \\tan \\frac{C}{2}.\n$$\nBy the addition and subtraction formulas, the second equation in the given system becomes\n$$\n\\begin{aligned}\n1 &= \\tan \\frac{A}{2} \\tan \\frac{B}{2} + \\tan \\frac{B}{2} \\tan \\frac{C}{2} + \\tan \\frac{C}{2} \\tan \\frac{A}{2} \\\\\n&= \\tan \\frac{A}{2} \\tan \\frac{B}{2} + \\tan \\frac{C}{2} \\tan \\frac{A+B}{2} \\left(1 - \\tan \\frac{A}{2} \\tan \\frac{B}{2}\\right),\n\\end{aligned}\n$$\n$$\n(1 - \\tan \\frac{C}{2} \\tan \\frac{A+B}{2}) (1 - \\tan \\frac{A}{2} \\tan \\frac{B}{2}) = 0.\n$$\nNote that $\\tan \\frac{A}{2} \\tan \\frac{B}{2} = xy \\neq 1$, otherwise $z(x+y) = 0$, implying $z = 0$, which is impossible. Therefore, $\\tan \\frac{C}{2} \\tan \\frac{A+B}{2} = 1$, or $\\frac{A+B}{2} + \\frac{C}{2} = 90^\\circ$. In other words, $A, B, C$ are the angles of a triangle. Let $ABC$ denote that triangle.\n\nWe rewrite the first equation in the given system as\n$$\n\\frac{x}{5(x^2+1)} = \\frac{y}{12(y^2+1)} = \\frac{z}{13(z^2+1)}.\n$$\nNote that, by the double-angle formula, we have\n$$\n\\frac{x}{x^2+1} = \\frac{\\tan \\frac{A}{2}}{\\tan^2 \\frac{A}{2} + 1} = \\frac{\\tan \\frac{A}{2}}{\\sec^2 \\frac{A}{2}} = \\sin \\frac{A}{2} \\cos \\frac{A}{2} = \\frac{\\sin A}{2},\n$$\nand analogously for the expressions of $y$ and $z$. We conclude that\n$$\n\\frac{\\sin A}{5} = \\frac{\\sin B}{12} = \\frac{\\sin C}{13}.\n$$\nBy the sine rule, the sides of triangle $ABC$ are in ratio $5 : 12 : 13$ with $\\sin A = \\frac{5}{13}$, $\\sin B = \\frac{12}{13}$ and $\\sin C = 1$. Hence $\\frac{x}{x^2+1} = \\frac{5}{26}$, or $5x^2 - 26x + 5 = 0$, implying that $x = 5$ or $x = \\frac{1}{5}$. Likewise, we have $y = \\frac{3}{2}$ or $y = \\frac{2}{3}$, and $z = 1$. Substituting $z = 1$ into the second equation in the given system leads to $xy + x + y = 1$, implying that $(x, y, z) = (\\frac{1}{5}, \\frac{2}{3}, 1)$ is the only solution with $x > 0$. Hence $(\\frac{1}{5}, \\frac{2}{3}, 1)$ and $(-\\frac{1}{5}, -\\frac{2}{3}, -1)$ are the solutions of the problem.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 28058,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $S = \\{1, 2, \\dots, 2005\\}$. If there is at least one prime number in any subset of $S$ consisting of $n$ pairwise coprime numbers, find the minimum value of $n$.",
    "options": [],
    "answer": "16",
    "solution": "First we prove $n \\ge 16$. In fact, let\n$$\nA_0 = \\{1, 2^2, 3^2, 5^2, \\dots, 41^2, 43^2\\},\n$$\nwhere the members in $A_0$, other than $1$, are the squares of prime numbers not greater than $43$. Then $A_0 \\subseteq S$, $|A_0| = 15$ and the numbers in $A_0$ are pairwise coprime but $A_0$ contains no prime number. Thus $n \\ge 16$.\n\nNext we show that for arbitrary $A \\subseteq S$ with $n = |A| = 16$, if the numbers in $A$ are pairwise coprime, then $A$ must contain a prime number.\nIn fact, if $A$ contains no prime number, denote $A = \\{a_1, a_2, \\dots, a_{16}; a_1 < a_2 < \\dots < a_{16}\\}$. Then there are two possibilities.\n\n(1) If $1 \\notin A$, then $a_1, a_2, \\dots, a_{16}$ are composite numbers. Since $(a_i, a_j) = 1$ ($1 \\le i < j \\le 16$), the prime factors of $a_i$ and $a_j$ are pairwise distinct. Let $p_i$ be the smallest prime factor of $a_i$. We may assume that $p_1 < p_2 < \\dots < p_{16}$, then\n$$\na_1 \\ge p_1^2 \\ge 2^2,\\quad a_2 \\ge p_2^2 \\ge 3^2,\\quad \\dots,\\quad a_{15} \\ge p_{15}^2 \\ge 47^2 > 2005,\n$$\nit leads to a contradiction.\n\n(2) If $1 \\in A$, let $a_{16} = 1$, $a_1, a_2, \\dots, a_{15}$ are composite numbers. By the same assumption and argument of (1), we have $a_1 \\ge p_1^2 \\ge 2^2$, $a_2 \\ge p_2^2 \\ge 3^2$, $\\dots, a_{15} \\ge p_{15}^2 \\ge 47^2 > 2005$. Again, it leads to a contradiction.\n\nFrom (1) and (2), $A$ contains at least one prime number, i.e. when $n = |A| = 16$, the conclusion is true. Thus, the minimum number of $n$ is $16$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 28059,
    "subject": "Mathematics (Multi-modal)",
    "question": "For each positive integer, define a function\n$$\nf(n) = \\begin{cases} 0, & \\text{if } n \\text{ is the square of an integer,} \\\\ \\left\\lfloor \\frac{1}{\\sqrt{n}} \\right\\rfloor, & \\text{if } n \\text{ is not the square of an integer.} \\end{cases}\n$$\n(Here $[x]$ denotes the maximum integer not exceeding $x$, and\n$\\{x\\} = x - [x]$.) Find the value of $\\sum_{k=1}^{200} f(k)$.",
    "options": [],
    "answer": "768",
    "solution": "For arbitrary $a, k \\in \\mathbb{N}^+$, if $k^2 < a < (k+1)^2$, we set\n$$\na = k^2 + m, \\quad m = 1, 2, \\dots, 2k,\n$$\n$$\n\\sqrt{a} = k + \\theta, \\quad 0 < \\theta < 1,\n$$\n$$\n\\left[ \\frac{1}{\\sqrt{a}} \\right] = \\left[ \\frac{1}{\\sqrt{a-k}} \\right] = \\left[ \\frac{\\sqrt{a+k}}{a-k^2} \\right] = \\left[ \\frac{2k+\\theta}{m} \\right].\n$$\n$$\n\\text{For} \\quad 0 < \\frac{2k+\\theta}{m} - \\frac{2k}{\\theta} < 1,\n$$\nif there exists an integer $t$ between $\\frac{2k}{m}$ and $\\frac{2k+\\theta}{m}$, then\n$$\n\\frac{2k}{m} < t < \\frac{2k+\\theta}{m}.\n$$\nOn one hand $2k < mt$, thus $2k+1 \\le mt$. On the other hand, $mt < 2k+\\theta < 2k+1$, a contradiction.\n$$\n\\text{Thus} \\quad \\left[ \\frac{2k+\\theta}{m} \\right] = \\left[ \\frac{2k}{m} \\right], \\\\\n\\sum_{k<a<k+1} \\left[ \\frac{1}{\\{a\\}} \\right] = \\sum_{m=1}^{2k} \\left[ \\frac{2k}{m} \\right], \\\\\n\\sum_{a=1}^{n+1} f(a) = \\sum_{k=1}^{n} \\sum_{i=1}^{2k} \\left[ \\frac{2k}{i} \\right]. \\quad (1)\n$$\n\nThus\n$$\n\\begin{aligned}\n\\sum_{a=1}^{(n+1)^2} f(a) &= \\sum_{k=1}^{n} \\sum_{j=1}^{2k} T(j) \\\\\n&= n[T(1) + T(2)] + (n-1)[T(3) + T(4)] \\\\\n&\\quad + \\dots + [T(2n-1) + T(2n)]\n\\end{aligned} \n\\quad (2)\n$$\n$$\n\\text{From (2), } \\sum_{k=1}^{16^2} f(k) = \\sum_{k=1}^{15} (16-k)[T(2k-1) + T(2k)]. \\quad (3)\n$$\n\n| k | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | 11 | 12 | 13 | 14 | 15 |\n|---|---|---|---|---|---|---|---|---|---|----|----|----|----|----|----|\n| $a_k$ | 3 | 5 | 6 | 6 | 7 | 8 | 6 | 9 | 8 | 8 | 8 | 10 | 7 | 10 | 10 |\n\n$$\n\\text{Therefore, } \\sum_{k=1}^{256} f(k) = \\sum_{k=1}^{15} (16-k)a_k = 783. \\quad (4)\n$$\nNote that $f(256) = f(16^2) = 0$ by definition. When $k \\in \\{241, 242, \\dots, 255\\}$, denote $k = 15^2 + r\\ (16 \\le r \\le 30)$, then\n$$\n\\sqrt{k}-15=\\sqrt{15^2+r}-15=\\frac{r}{\\sqrt{15^2+r}+15},\n$$\n$$\n\\frac{r}{31} < \\frac{r}{\\sqrt{15^2 + r + 15}} < \\frac{r}{30}, \\quad 1 \\le \\frac{30}{r} < \\frac{1}{\\{\\sqrt{15^2 + r}\\}} < \\frac{31}{r} < 2.\n$$\nThus\n$$ \\left[ \\frac{1}{\\sqrt{k}} \\right] = 1, \\quad k \\in \\{241, 242, \\dots, 255\\}. \\quad (5) $$\nTherefore,\n$$ \\sum_{k=1}^{200} f(k) = 783 - \\sum_{k=201}^{256} f(k) = 783 - 15 = 768. $$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 28060,
    "subject": "Mathematics (Multi-modal)",
    "question": "Determine if there exists a convex polyhedron such that\n(1) it has 12 edges, 6 faces and 8 vertices;\n(2) it has 4 faces with each pair of them sharing a common edge of the polyhedron.",
    "options": [],
    "answer": "Yes",
    "solution": "The answer is yes, as shown in the figure.\n\n![](attached_image_1.png)",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 28061,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $\\{a_n\\}$ be a sequence such that $a_1 = \\frac{21}{16}$ and\n$$\n2a_n - 3a_{n-1} = \\frac{3}{2^{n+1}}, \\quad n \\ge 2. \\qquad \\textcircled{1}\n$$\nLet $m$ be a positive integer and $m \\ge 2$. Prove that for $n \\le m$,\n$$\n\\left[a_n + \\frac{3}{2^{n+3}}\\right]^{\\frac{1}{m}} \\left(m - \\left(\\frac{2}{3}\\right)^{\\frac{n(m-1)}{m}}\\right) < \\frac{m^2 - 1}{m - n + 1}. \\qquad \\textcircled{2}\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "**Proof** By Equation (1), we have\n$$\n2^n a_n = 3 \\cdot 2^{n-1} a_{n-1} + \\frac{3}{4}.\n$$\nSet $b_n = 2^n a_n$, $n = 1, 2, \\dots$, then\n$$\nb_n = 3b_{n-1} + \\frac{3}{4}, \\quad b_n + \\frac{3}{8} = 3\\left(b_{n-1} + \\frac{3}{8}\\right).\n$$\n$$\n\\text{Since } b_1 = 2a_1 = \\frac{21}{8},\n$$\n$$\nb_n + \\frac{3}{8} = 3^{n-1} \\left(b_1 + \\frac{3}{8}\\right) = 3^n,\n$$\nit follows that\n$$\na_n = \\left(\\frac{3}{2}\\right)^n - \\frac{3}{2^{n+3}}\n$$\n\nTherefore, in order to prove Equation ②, it suffices to prove that\n$$\n\\left(\\frac{3}{2}\\right)^{\\frac{n}{m}} \\left(m - \\left(\\frac{2}{3}\\right)^{\\frac{n(m-1)}{m}}\\right) < \\frac{m^2 - 1}{m - n + 1},\n$$\nor equivalently,\n$$\n\\left(1 - \\frac{n}{m+1}\\right) \\left(\\frac{3}{2}\\right)^{\\frac{n}{m}} \\left(m - \\left(\\frac{2}{3}\\right)^{\\frac{n(m-1)}{m}}\\right) < m - 1. \\quad \\textcircled{3}\n$$\n\nAt first, we estimate the upper bound of $1 - \\frac{n}{m+1}$. By using Bernoulli's inequality, we get\n$$\n1 - \\frac{n}{m+1} < \\left(1 - \\frac{1}{m+1}\\right)^n,\n$$\nso that\n$$\n\\left(1 - \\frac{n}{m+1}\\right)^m < \\left(1 - \\frac{1}{m+1}\\right)^{nm} = \\left(\\frac{m}{m+1}\\right)^{nm} = \\left[ \\frac{1}{\\left(1 + \\frac{1}{m}\\right)^m} \\right]^n.\n$$\n\n(Note: By the mean inequality, we can also have the same result:\n$$\n\\begin{align*}\n\\left(1 - \\frac{n}{m+1}\\right)^m &= \\left(1 - \\frac{n}{m+1}\\right)^m \\cdot \\underbrace{1 \\cdot 1 \\cdot \\cdots \\cdot 1}_{\\text{number } mn-m \\text{ of } 1} \\\\\n&< \\left[ \\frac{m\\left(1 - \\frac{n}{m+1}\\right) + mn - m}{mn} \\right]^{mn} \\\\\n&= \\left(\\frac{m}{m+1}\\right)^{nm}.\n\\end{align*}\n$$\n\nSince $m \\ge 2$, in view of the binomial formula, we obtain\n$$\n\\left(1 + \\frac{1}{mm}\\right)^m \\ge 1 + C_m^1 \\cdot \\frac{1}{m} + C_m^2 \\cdot \\frac{1}{m^2} = \\frac{5}{2} - \\frac{1}{2m} \\ge \\frac{9}{4}.\n$$\nIt follows that\n$$\n\\left(1 - \\frac{n}{m+1}\\right)^m < \\left(\\frac{4}{9}\\right)^n,\n$$\nor\n$$\n1 - \\frac{n}{m+1} < \\left(\\frac{2}{3}\\right)^{\\frac{2n}{m}}.\n$$\nHence, if we want to prove Equation ③, we only need to prove that\n$$\n\\left(\\frac{2}{3}\\right)^{\\frac{2n}{m}} \\cdot \\left(\\frac{3}{2}\\right)^{\\frac{n}{m}} \\left(m - \\left(\\frac{2}{3}\\right)^{\\frac{n(m-1)}{m}}\\right) < m - 1,\n$$\nthat is,\n$$\n\\left(\\frac{2}{3}\\right)^{\\frac{n}{m}} \\left(m - \\left(\\frac{2}{3}\\right)^{\\frac{n(m-1)}{m}}\\right) < m - 1. \\quad \\textcircled{4}\n$$\nSet $\\left(\\frac{2}{3}\\right)^{\\frac{n}{m}} = t$, then $0 < t < 1$, and Equation ④ now becomes\n$$\nt(m - t^{m-1}) < m - 1,\n$$\nor\n$$\n(t-1)[m - (t^{m-1} + t^{m-2} + \\cdots + 1)] < 0.\n$$\nThe above inequality clearly holds, so does the initial inequality.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 28062,
    "subject": "Mathematics (Multi-modal)",
    "question": "Determine all positive real numbers $a$ such that there exists a positive integer $n$ and sets $A_1, A_2, \\dots, A_n$ satisfying the following conditions:\n(1) every set $A_i$ has infinitely many elements;\n(2) every pair of distinct sets $A_i$ and $A_j$ do not share any common element;\n(3) the union of sets $A_1, A_2, \\dots, A_n$ is the set of all integers;\n(4) for every set $A_i$, the positive difference of any pair of elements in $A_i$ is at least $a^i$. (posed by Yuan Hanhui)",
    "options": [],
    "answer": "0 < a < 2",
    "solution": "The answer of the problem is the set of all positive real numbers less than $2$. We consider two cases.\n\n**Case I** We assume that $0 < a < 2$. Then there is a positive $n$ such that $2^{n-1} > a^n$. We define $A_n = \\{m \\mid m \\text{ is a multiple of } 2^{n-1}\\}$ and\n$$\nA_i = \\{2^{i-1}m \\mid m \\text{ is an odd integer}\\},\n$$\nfor $1 \\le i \\le n-1$. Then $A_1, A_2, \\dots, A_n$ is a partition of the set of positive integers satisfying the conditions of the problem.\n\n**Case II** We assume that $a \\ge 2$. We claim that no such partition exists. To prove by contradiction, we assume on the contrary that $A_1, A_2, \\dots, A_n$ is a partition satisfying the conditions of the problem. Let $N = \\{1, 2, \\dots, 2^n\\}$. For every $i$ with $1 \\le i \\le n$, let $B_i = A_i \\cap N$. We assume that $B_i = \\{b_1, b_2, \\dots, b_m\\}$ with $b_1 < b_2 < \\dots < b_m$. We have\n$$\n2^n > b_m - b_1 = (b_m - b_{m-1}) + (b_{m-1} - b_{m-2}) + \\dots + (b_2 - b_1) \\ge (m-1)2^i,\n$$\nimplying that $m-1 < 2^{n-i}$, or $m \\le 2^{n-i}$. Since $A_1, A_2, \\dots, A_n$ is a partition, $B_i \\cap B_j = \\emptyset$ for $1 \\le i < j \\le n$ and $N = B_1 \\cup B_2 \\cup \\dots \\cup B_n$. It follows that\n$$\n2^n = |N| = |B_1| + |B_2| + \\dots + |B_n| \\le \\sum_{i=1}^{n} 2^{n-i} = 2^n - 1,\n$$\nwhich is impossible. Hence our assumption was wrong and such a partition does not exist for every positive integer $n$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 28063,
    "subject": "Mathematics (Multi-modal)",
    "question": "It is given that real numbers $x_1, x_2, \\dots, x_n$ ($n > 2$) satisfy $\\left|\\sum_{i=1}^n x_i\\right| > 1$, $|x_i| \\le 1$ ($i=1, 2, \\dots, n$). Prove that there exists a positive integer $k$ such that $\\left|\\sum_{i=1}^k x_i - \\sum_{i=k+1}^n x_i\\right| \\le 1$. (posed by Leng Gangsong)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "**Proof** Set $g(0) = -\\sum_{i=1}^{n} x_i$, $g(k) = \\sum_{i=1}^{k} x_i - \\sum_{i=k+1}^{n} x_i$ ($1 \\le k \\le n-1$), $g(n) = \\sum_{i=1}^{n} x_i$.\nThen\n$$|g(1) - g(0)| = 2|x_1| \\le 2,$$\n$$|g(k+1) - g(k)| = 2|x_{k+1}| \\le 2, \\quad k = 1, 2, \\dots, n-2,$$\n$$|g(n) - g(n-1)| = 2|x_n| \\le 2.$$\nSo for each $0 \\le k \\le n-1$,\n$$\n|g(k+1) - g(k)| \\le 2. \\qquad \\textcircled{1}\n$$\nIf the conclusion is not true, by the condition for each $k$, $0 \\le k \\le n$, we have\n$$\n|g(k)| > 1. \\qquad \\textcircled{2}\n$$\nIf there is an $i$, $0 \\le i \\le n-1$, such that $g(i)g(i+1) < 0$, we may assume that $g(i) > 0$ and $g(i+1) < 0$. By ②, $g(i) > 1$ and $g(i+1) < -1$. Thus $|g(i+1) - g(i)| > 2$. This contradicts ①. Thus $g(0)$, $g(1)$, ..., $g(n)$ have the same sign. But $g(0) + g(n) = 0$. The contradiction implies that the conclusion is true.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 28064,
    "subject": "Mathematics (Multi-modal)",
    "question": "There are 5 points in a rectangle $ABCD$ (including its boundary) with unit area such that any three of them are not collinear. Find the minimum number of triangles with areas no more than $\\frac{1}{4}$ and vertexes chosen from these 5 points. (posed by Leng Gangsong)",
    "options": [],
    "answer": "2",
    "solution": "**Lemma** *The area of a triangle inscribed in a rectangle is no more than half of the area of the rectangle.*\n\nDenote $E$, $F$, $H$ and $G$ the midpoints of $AB$, $CD$, $BC$ and $AD$, respectively, and $O$ the intersection of line segments $EF$ and $GH$. $EF$ and $GH$ divide the rectangle $ABCD$ into 4 small rectangles, it follows that there exists certain small rectangle, say $AEOG$, in which there are at least two points (say, $M$ and $N$) out of those 5 points, see Figure 5.\n\n![](attached_image_1.png)\nFigure 5\n\n(1) If there is no more than one given point in the rectangle $OHCF$, consider any given point $X$ which is different from $M$ and $N$ and not in the rectangle $OHCF$. It is easy to verify that triple $(M, N, X)$ is either in rectangle $ABHG$, or in rectangle $AEFD$. By the above lemma $(M, N, X)$ is good. Since there are at least two such points $X$, we have at least two such good triples.\n\n(2) If there exist at least two given points in rectangle $OHCF$, we suppose that $P$ and $Q$ are the given points in rectangle $OHCF$. Consider the final given point $R$. If $R$ is in the rectangle $OFDG$, then $(M, N, R)$ is in the rectangle $AEFD$, and $(P, Q, R)$ is in the rectangle $GHCD$. So they are all good. It follows that there are at least two good triples. Similarly, when point $R$ is in the rectangle $EBHO$ there are at least two good triples too. If point $R$ is in the rectangle $OHCF$ or the rectangle $AEOG$, suppose that $R$ is in the rectangle $OHCF$. Consider the smallest convex polygon containing the 5 points $M$, $N$, $P$, $Q$ and $R$. The polygon must be contained in the convex hexagon $AEHCFG$ (see Figure 6). But the area\n$$\nS_{AEHCFG} = 1 - \\frac{1}{8} - \\frac{1}{8} = \\frac{3}{4}.\n$$\n\n![](attached_image_2.png)\nFigure 6\n\ni) Suppose the convex polygon generated by $M, N, P, Q$ and $R$ is a convex pentagon, (no loss of generality) say $MNPQR$ (as seen in Figure 7). In this case\n$$\nS_{\\triangle MQR} + S_{\\triangle MNQ} + S_{\\triangle NPQ} \\le \\frac{3}{4},\n$$\nit follows that there is at least one good triple among $(M, Q, R)$, $(M, N, Q)$ and $(N, P, Q)$. Moreover, $(P, Q, R)$ is clearly good for it is in the rectangle $OHCF$. Thus there are at least two good triples.\n\n![](attached_image_3.png)\nFigure 7\n\nii) If the convex polygon generated by $M, N, P, Q$ and $R$ is a convex quadrilateral, say $A_1A_2A_3A_4$, and the fifth point is $A_5$ (as seen in Figure 8) where $A_i \\in \\{M, N, P, Q, R\\} (i = 1, 2, 3, 4, 5)$. Draw line segments $A_5A_i$ ($i = 1, 2, 3, 4$), then\n$$\nS_{\\triangle A_1A_2A_5} + S_{\\triangle A_2A_3A_5} + S_{\\triangle A_3A_4A_5} + S_{\\triangle A_4A_1A_5} = S_{\\triangle A_1A_2A_3A_4} \\le \\frac{3}{4}.\n$$\nTherefore, there are at least two good triples among $(A_1, A_2, A_5)$, $(A_2, A_3, A_5)$, $(A_3, A_4, A_5)$ and $(A_1, A_4, A_5)$.\n\n![](attached_image_4.png)\nFigure 8\n\niii) If the convex polygon generated by $M, N, P, Q$ and $R$ is a triangle, say $\\triangle A_1A_2A_3$, and the remaining two points are $A_4$ and $A_5$ (as seen in Figure 9), where $A_i \\in \\{M, N, P, Q, R\\} (i = 1, 2, 3, 4, 5)$. Draw line segments $A_4A_i (i = 1, 2, 3)$, then\n$$\nS_{\\triangle A_1A_2A_4} + S_{\\triangle A_2A_3A_4} + S_{\\triangle A_3A_1A_4} = S_{\\triangle A_1A_2A_3} \\le \\frac{3}{4}.\n$$\nTherefore, there is at least one good triple among $(A_1, A_2, A_4)$, $(A_2, A_3, A_4)$ and $(A_1, A_3, A_4)$. Similarly, $A_5$ with two of $A_1, A_2$ and $A_3$ constitutes a good triple. Consequently, in this case there are at least two good triples.\n\n![](attached_image_5.png)\nFigure 9\n\nThus, in any case there are at least two good triples among these 5 points.\n\nIn the following we will give some examples to show that the number of good triples may be just two. Pick a point $M$ on the side $AD$ of the rectangle $ABCD$ and a point $N$ on the side $AB$ such that $AN : NB = AM : MD = 2 : 3$ (as shown in Figure 10). Then among 5 points $M, N, B, C$ and $D$ there are just two good triples. In fact, $(B, C, D)$ is clearly not good. Suppose that a triple containing exactly one of two points $M$ and $N$, say $M$. Let $E$ be the midpoint of $AD$, then\n$$\nS_{\\triangle MBD} > S_{\\triangle EBD} = \\frac{1}{4}.\n$$\nIt follows that $(M, B, D)$ is not good. Thus\n$$\nS_{\\triangle MBC} = \\frac{1}{2}, \\quad S_{\\triangle MCD} > S_{\\triangle ECD} = \\frac{1}{4}.\n$$\nHence $(M, B, C)$ and $(M, C, D)$ are not good. If a triangle does contain two points $M$ and $N$, then\n$$\n\\begin{align*}\nS_{\\triangle MNC} &= 1 - S_{\\triangle NBC} - S_{\\triangle MCD} - S_{\\triangle AMN} \\\\\n&= 1 - \\frac{3}{5}S_{\\triangle ABC} - \\frac{3}{5}S_{\\triangle ACD} - \\frac{4}{25}S_{\\triangle ABD}\n\\end{align*}\n$$\n$$\n=1-\\frac{3}{10}-\\frac{3}{10}-\\frac{2}{25}=\\frac{8}{25}>\\frac{1}{4}.\n$$\nSo $(M, N, C)$ is not good. But $S_{\\triangle MNB} = S_{\\triangle MND} = \\frac{1}{5} < \\frac{1}{4}$, thus among the triples there are only two good triples $(M, N, B)$ and $(M, N, D)$.\n\nConsequently, the minimum number of triangles with area not greater than $\\frac{1}{4}$ is $2$.",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 28065,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $x$ and $y$ be positive real numbers with $x^3 + y^3 = x - y$. Prove that $x^2 + 4y^2 < 1$. (posed by Xiong Bin)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "In view of $x^3 + y^3 > 0$, it suffices to show that $(x^2 + 4y^2)(x - y) < x^3 + y^3$. Expanding the left-hand side of the last inequality and canceling the like terms we obtain $4xy^2 < x^2y + 5y^3$. By the AM-GM inequality, we have $x^2y + 5y^3 \\ge 2\\sqrt{5xy^2} > 4xy^2$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 28066,
    "subject": "Mathematics (Multi-modal)",
    "question": "The circles $O_1$ and $O_2$ meet at points $A$ and $B$. The line $DC$ passes through $O_1$, intersects the circle $O_1$ at $D$ and is a tangent to the circle $O_2$ at $C$. Also, $CA$ is a tangent to the circle $O_1$ at $A$. The secant $AE$ of the circle $O_1$ is perpendicular to $DC$. $AF$ is perpendicular to $DE$ and meets $DE$ at $F$.\n\nProve that $BD$ bisects the line segment $AF$. (posed by Bian Hongping)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $AE$ intersect $DC$ at point $H$,\n![](attached_image_1.png)\n\nand $AF$ intersect $BD$ at point $G$. Join $AB$, $BC$, $BH$, $BE$, $CE$ and $GH$. By symmetry, $CE$ is also a tangent line of the circle $O_1$ and $H$ is the midpoint of $AE$.\n\nSince $AF \\perp DE$, we have\n$$\n\\angle AGB = \\frac{\\pi}{2} - \\angle BDE.\n$$\n\nBy $AE \\perp DC$,\n$$\n\\angle AHB = \\frac{\\pi}{2} - \\angle BHC.\n$$\n\nBy ①, ② and ③, $\\angle AGB = \\angle AHB$. Therefore $A$, $G$, $H$, $B$ lie on the same circle, and $\\angle AHG = \\angle ABG = \\angle AED$. Thus $GH \\parallel DE$. Since $H$ is the midpoint of $AE$, $G$ is the midpoint of $AF$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 28067,
    "subject": "Mathematics (Multi-modal)",
    "question": "The curve represented by the equation $$\\frac{x^2}{\\sin\\sqrt{2} - \\sin\\sqrt{3}} + \\frac{y^2}{\\cos\\sqrt{2} - \\cos\\sqrt{3}} = 1$$ is ( ).\n(A) An ellipse with the foci on the x-axes\n(B) A hyperbola with the foci on the x-axes\n(C) An ellipse with the foci on the y-axes\n(D) A hyperbola with the foci on the y-axes",
    "options": [],
    "answer": "C",
    "solution": "Since $\\sqrt{2} + \\sqrt{3} > \\pi$, so $0 < \\frac{\\pi}{2} - \\sqrt{2} < \\sqrt{3} - \\frac{\\pi}{2} < \\frac{\\pi}{2}$ and\n$$\n\\cos(\\frac{\\pi}{2} - \\sqrt{2}) > \\cos(\\sqrt{3} - \\frac{\\pi}{2}), \\text{ i.e. } \\sin\\sqrt{2} > \\sin\\sqrt{3}.\n$$\n\nSince\n$$\n(\\sin\\sqrt{2} - \\sin\\sqrt{3}) - (\\cos\\sqrt{2} - \\cos\\sqrt{3}) = 2\\sqrt{2}\\sin\\frac{\\sqrt{2}-\\sqrt{3}}{2}\\sin\\left(\\frac{\\sqrt{2}+\\sqrt{3}}{2} + \\frac{\\pi}{4}\\right) \\quad (*)\n$$\nand\n$$\n-\\frac{\\pi}{2} < \\frac{\\sqrt{2}-\\sqrt{3}}{2} < 0,\n$$\nwe get\n$$\n\\sin\\frac{\\sqrt{2}-\\sqrt{3}}{2} < 0, \\quad \\frac{\\pi}{2} < \\frac{\\sqrt{2}+\\sqrt{3}}{2} < \\frac{3\\pi}{4},\n$$\n$$\n\\frac{3\\pi}{4} < \\frac{\\sqrt{2}+\\sqrt{3}}{2} + \\frac{\\pi}{4} < \\pi,\n$$\n$$\n\\sin\\left(\\frac{\\sqrt{2}+\\sqrt{3}}{2} + \\frac{\\pi}{4}\\right) > 0,\n$$\n\nso the expression $(*)$ is less than $0$.\nThat is $\\sin\\sqrt{2} - \\sin\\sqrt{3} < \\cos\\sqrt{3} - \\cos\\sqrt{2}$, therefore the curve is an ellipse with foci on the y-axes. Answer: C.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 28068,
    "subject": "Mathematics (Multi-modal)",
    "question": "An integer $n$ is called good if $n \\ge 3$ and there are $n$ lattice points $P_1, P_2, \\dots, P_n$ in the coordinate plane satisfying the following conditions: If line segment $P_iP_j$ has a rational length, then there is $P_k$ such that both line segments $P_iP_k$ and $P_jP_k$ have irrational lengths; and if line segment $P_iP_j$ has an irrational length, then there is $P_k$ such that both line segments $P_iP_k$ and $P_jP_k$ have rational lengths.\n(1) Determine the minimum good number.",
    "options": [],
    "answer": "5",
    "solution": "We claim that the minimum good number is $5$, and that $2005$ is good.\n\nIt is not difficult to see that $n=3$ is not a good number. Note that $n=4$ is also not a good number. Assume on the contrary that there are lattice points $P_1, P_2, P_3, P_4$ satisfying the conditions of the problem. Without loss of generality, we assume that $P_1P_2$ is rational and $(P_1, P_2, P_3)$ is good. Then $(P_2, P_3, P_4)$ must be good. Neither $(P_2, P_4, P_1)$ nor $(P_2, P_4, P_3)$ is good, violating the condition of the problem. Hence our assumption was wrong and $n=4$ is not a good number.\n\nFor $n=5$, we define the set\n$$\nS_5 = \\{P_1 = (0, 0), P_2 = (1, 0), P_3 = (5, 3), P_4 = (8, 7), P_5 = (0, 7)\\}.\n$$\n\nIt is not difficult to see that all triples $(P_i, P_j, P_k)$ are good. Hence $n=5$ is minimum good number.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 28069,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all non-negative integer solutions $(x, y, z, w)$ of the following equation\n$$\n2^x \\cdot 3^y - 5^z \\cdot 7^w = 1.\n$$",
    "options": [],
    "answer": "(1, 0, 0, 0), (3, 0, 0, 1), (1, 1, 1, 0), (2, 2, 1, 1)",
    "solution": "Since $5^z \\cdot 7^w + 1$ is even, we have $x \\ge 1$.\n\nCase 1: $y = 0$. The equation to be solved becomes\n$$\n2^x - 5^z \\cdot 7^w = 1.\n$$\nIf $z \\neq 0$, then $2^x \\equiv 1 \\pmod{5}$. It follows that $4 \\mid x$. Thus $3 \\mid 2^x - 1$, which contradicts to $2^x - 5^z \\cdot 7^w = 1$.\nIf $z = 0$, then\n$$\n2^x - 7^w = 1.\n$$\nWhen $x = 1, 2, 3$, a direct computation shows that $(x, w) = (1, 0), (3, 1)$ are the solutions.\nWhen $x \\ge 4$, $7^w \\equiv -1 \\pmod{16}$. By direct computation we know that this is impossible.\nConsequently, when $y = 0$ all non-negative integer solutions of the equation are\n$$\n(x, y, z, w) = (1, 0, 0, 0), (3, 0, 0, 1).\n$$\n\nCase 2: $y > 0$ and $x = 1$. Thus the equation to be solved becomes\n$$\n2 \\cdot 3^y - 5^z \\cdot 7^w = 1.\n$$\nHence $-5^z \\cdot 7^w \\equiv 1 \\pmod{3}$, i.e., $(-1)^z \\equiv -1 \\pmod{3}$. It follows that $z$ is odd.\nThus\n$$\n2 \\cdot 3^y \\equiv 1 \\pmod{5}.\n$$\n$$\ny \\equiv 1 \\pmod{4}.\n$$\nWhen $w \\neq 0$, we have $2 \\cdot 3^y \\equiv 1 \\pmod{7}$. Thus $y \\equiv 4 \\pmod{6}$, which contradicts to the fact $y \\equiv 1 \\pmod{4}$. Hence $w = 0$ and\n$$\n2 \\cdot 3^y - 5^z = 1.\n$$\nWhen $y = 1$, we have $z = 1$. If $y \\ge 2$, then $5^z \\equiv -1 \\pmod{9}$, which implies $z \\equiv 3 \\pmod{6}$. Thus $5^3 + 1 \\pmod{5^z + 1}$, so $7 \\pmod{5^z + 1}$, which contradicts to $5^z + 1 = 2 \\cdot 3^y$. Hence in this case we have only one solution\n$$\n(x, y, z, w) = (1, 1, 1, 0).\n$$\n\nCase 3: $y > 0$ and $x \\ge 2$. Thus\n$$\n5^z \\cdot 7^w \\equiv -1 \\pmod{4}, \\text{ and } 5^z \\cdot 7^w \\equiv -1 \\pmod{3}.\n$$\nThat is,\n$$\n(-1)^w \\equiv -1 \\pmod{4}, \\text{ and } (-1)^z \\equiv -1 \\pmod{3}.\n$$\nThus $z$ and $w$ are odd. It follows that\n$$\n2^x \\cdot 3^y = 5^z \\cdot 7^w + 1 \\equiv 35 + 1 \\equiv 4 \\pmod{8}.\n$$\nHence, $x = 2$, and\n$$\n4 \\cdot 3^y - 5^z \\cdot 7^w = 1 \\text{ (where $z$ and $w$ are odd).}\n$$\nThus,\n$$\n4 \\cdot 3^y \\equiv 1 \\pmod{5}, \\text{ and } 4 \\cdot 3^y \\equiv 1 \\pmod{7}.\n$$\nFrom the above two congruencies we have $y \\equiv 2 \\pmod{12}$.\nSet $y = 12m + 2$, $m \\ge 0$, then\n$$\n5^z \\cdot 7^w = 4 \\cdot 3^y - 1 = (2 \\cdot 3^{6m+1} - 1)(2 \\cdot 3^{6m+1} + 1).\n$$\nSince\n$$\n2 \\cdot 3^{6m+1} + 1 \\equiv 6 \\cdot 2^{3m} + 1 \\equiv 6 + 1 \\equiv 0 \\pmod{7},\n$$\nand\n$$(2 \\cdot 3^{6m+1} - 1, 2 \\cdot 3^{6m+1} + 1) = 1,$$ so $5 \\mid 2 \\cdot 3^{6m+1} - 1$.\nThus\n$$2 \\cdot 3^{6m+1} - 1 = 5^z,$$\n$$2 \\cdot 3^{6m+1} + 1 = 7^w.$$\nIf $m \\ge 1$, by Equation (2) we have $5^z \\equiv -1 \\pmod{9}$, and from Case 2 we know that this is impossible.\nIf $m=0$, then $y=2$, $z=1$ and $w=1$. Thus in this case, we have only one solution\n$$\n(x, y, z, w) = (2, 2, 1, 1).\n$$\n\nConsequently, all non-negative integer solutions are\n$$\n(x, y, z, w) = (1, 0, 0, 0), (3, 0, 0, 1), \\\\\n(1, 1, 1, 0), (2, 2, 1, 1).\n$$",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 28070,
    "subject": "Mathematics (Multi-modal)",
    "question": "In an isosceles right angled triangle $\\triangle ABC$, $CA = CB = 1$, and $P$ is an arbitrary point on the perimeter of $\\triangle ABC$. Find the maximum value of $PA \\cdot PB \\cdot PC$. (posed by Li Weigu)",
    "options": [],
    "answer": "√2/4",
    "solution": "(1) In the first diagram, if $P \\in AC$, we have $PA \\cdot PC \\le \\frac{1}{4}$ and $PB \\le \\sqrt{2}$. Thus $PA \\cdot PB \\cdot PC \\le \\frac{\\sqrt{2}}{4}$. The equality is not valid, since the two equality signs cannot be valid at the same time. Therefore $PA \\cdot PB \\cdot PC < \\frac{\\sqrt{2}}{4}$.\n\n![](attached_image_1.png)\n\n(2) In the second diagram, if $P \\in AB$, write $AP = x \\in [0, \\sqrt{2}]$, then\n\n![](attached_image_2.png)\n\nLet $t = x(\\sqrt{2} - x)$, then $t \\in [0, \\frac{1}{2}]$ and $f(x) = g(t) = t^2(1-t)$.\n\nNote that $g'(t) = 2t - 3t^2 = t(2 - 3t)$. Thus $g(t)$ is increasing on $[0, \\frac{2}{3}]$ and $f(x) \\le g(\\frac{1}{2}) = \\frac{1}{8}$. Therefore $PA \\cdot PB \\cdot PC \\le \\frac{1}{2\\sqrt{2}} = \\frac{\\sqrt{2}}{4}$. The equality is valid if and only if $t = \\frac{1}{2}$ and $x = \\frac{\\sqrt{2}}{2}$. So $P$ is the midpoint of $AB$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 28071,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $T = \\{0, 1, 2, 3, 4, 5, 6\\}$ and $M = \\{\\frac{a_1}{7} + \\frac{a_2}{7^2} + \\frac{a_3}{7^3} + \\frac{a_4}{7^4}; a_i \\in T, i = 1, 2, 3, 4\\}$. Arrange the numbers in $M$ in the descending order. Then the 2 005-th number is ( ).\n\n(A) $\\frac{5}{7} + \\frac{5}{7^2} + \\frac{6}{7^3} + \\frac{3}{7^4}$\n(B) $\\frac{5}{7} + \\frac{5}{7^2} + \\frac{6}{7^3} + \\frac{2}{7^4}$\n(C) $\\frac{1}{7} + \\frac{1}{7^2} + \\frac{0}{7^3} + \\frac{4}{7^4}$\n(D) $\\frac{1}{7} + \\frac{1}{7^2} + \\frac{0}{7^3} + \\frac{3}{7^4}$",
    "options": [],
    "answer": "C",
    "solution": "Let $[a_1 a_2 \\cdots a_k]_p$ be a number base $p$ with $k$ digits. Multiply each number in $M$ by $7^4$, and we get\n$$\nM' = \\{a_1 7^3 + a_2 7^2 + a_3 7 + a_4; a_i \\in T, i = 1, 2, 3, 4\\} = \\{[a_1 a_2 a_3 a_4]'_7 \\mid a_i \\in T, i = 1, 2, 3, 4\\}.\n$$\nThe maximum number in $M'$ is $[6666]_7 = [2400]_{10}$.\nIn the decimal system, starting from 2 400 in the descending order, the 2 005-th number is $2400 - 2004 = 396$. But $[396]_{10} = [1104]_7$. Divide this number by $7^4$, we get a number in $M$, that is $\\frac{1}{7} + \\frac{1}{7^2} + \\frac{0}{7^3} + \\frac{4}{7^4}$. Answer: C.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 28072,
    "subject": "Mathematics (Multi-modal)",
    "question": "Given real numbers $a$, $b$, $c$, satisfying $a + b + c = 1$, prove that\n$$10(a^3 + b^3 + c^3) - 9(a^5 + b^5 + c^5) \\ge 1.$$ (posed by Li Shenghong)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Since $\\sum a^3 = 1 - 3\\Pi(a+b)$, $\\sum a^5 = 1 - 5\\Pi(a+b)[\\sum a^2 + \\sum ab]$, therefore, the original inequality holds\n$$\n\\begin{align*}\n&\\Leftrightarrow 10[1 - 3\\Pi(a+b)] - 9[1 - 5\\Pi(a+b)(\\sum a^2 + \\sum ab)] \\ge 1 \\\\\n&\\Leftrightarrow 45\\Pi(a+b)(\\sum a^2 + \\sum ab) \\ge 30\\Pi(a+b) \\\\\n&\\Leftrightarrow 3(\\sum a^2 + \\sum ab) \\ge 2 = 2(\\sum a)^2 = 2(\\sum a^2 + 2\\sum ab) \\\\\n&\\Leftrightarrow \\sum a^2 \\ge \\sum ab.\n\\end{align*}\n$$\nFrom $a^2 + b^2 \\ge 2ab$, $b^2 + c^2 \\ge 2bc$ and $c^2 + a^2 \\ge 2ac$, we have $2\\sum a^2 \\ge 2\\sum ab$, i.e., $\\sum a^2 \\ge \\sum ab$. Therefore the original inequality holds.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 28073,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $m$ and $n$ be positive integers with $m > n \\ge 2$. Set $S = \\{1, 2, \\dots, m\\}$, and $T = \\{a_1, a_2, \\dots, a_n\\}$ is a subset of $S$ such that every element in $S$ is divisible by at most one element in $T$. Prove that\n$$\n\\frac{1}{a_1} + \\frac{1}{a_2} + \\dots + \\frac{1}{a_n} < \\frac{m+n}{m}.\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "For every $i$ with $1 \\le i \\le n$, we define set\n$$\nS_i = \\{b \\mid b \\text{ is an element in } S \\text{ and is divisible by } a_i\\}.\n$$\nThere are $\\lfloor \\frac{m}{a_i} \\rfloor$ elements in $S_i$. Since every element in $S$ is not divisible by any two distinct elements in $T$, it follows that $S_i \\cap S_j = \\emptyset$ for $1 \\le i < j \\le n$. Thus\n$$\n\\sum_{i=1}^{n} \\lfloor \\frac{m}{a_i} \\rfloor = \\sum_{i=1}^{n} |S_i| \\le |S| = m.\n$$\nNote that $\\frac{m}{a_i} < \\lfloor \\frac{m}{a_i} \\rfloor + 1$. It follows that\n$$\n\\sum_{i=1}^{n} \\frac{m}{a_i} < \\sum_{i=1}^{n} (\\lfloor \\frac{m}{a_i} \\rfloor + 1) = \\sum_{i=1}^{n} \\lfloor \\frac{m}{a_i} \\rfloor + \\sum_{i=1}^{n} 1 \\le m + n.\n$$\nSo the desired result is obtained.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 28074,
    "subject": "Mathematics (Multi-modal)",
    "question": "Express the polynomial in $x$ $f(x) = 1 - x + x^2 - x^3 + \\cdots - x^{19} + x^{20}$ into a polynomial in $y$ $g(y) = a_0 + a_1 y + a_2 y^2 + \\cdots + a_{19} y^{19} + a_{20} y^{20}$, where $y = x - 4$. Then $a_0 + a_1 + \\cdots + a_{20} = \\text{_________}$.",
    "options": [],
    "answer": "(5^21 + 1)/6",
    "solution": "The terms in the expression $f(x)$ form a geometric series with first term $1$ and common ratio $-x$. By the summation formula of geometric series,\n$$\nf(x) = \\frac{(-x)^{21} - 1}{-x - 1} = \\frac{x^{21} + 1}{x + 1}.\n$$\nSet $x = y+4$, $g(y) = \\frac{(y+4)^{21} + 1}{y+5}$. Let $y = 1$, we get\n$$\na_0 + a_1 + \\cdots + a_{20} = g(1) = \\frac{5^{21} + 1}{6}.\n$$",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 28075,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $f(x)$ be a decreasing function defined on $(0, +\\infty)$. If $f(2a^2 + a + 1) < f(3a^2 - 4a + 1)$, then the range of $a$ is __________.",
    "options": [],
    "answer": "0 < a < 1/3 or 1 < a < 5",
    "solution": "Since $f(x)$ is defined on $(0, +\\infty)$, from\n$$\n\\begin{cases} 2a^2 + a + 1 = 2\\left(a+\\frac{1}{4}\\right)^2 + \\frac{7}{8} > 0, \\\\ 3a^2 - 4a + 1 = (3a-1)(a-1) > 0, \\end{cases}\n$$\nwe get \\qquad $a > 1$ or $a < \\frac{1}{3}$. \\qquad (1)\n\nSince $f(x)$ is a decreasing function on $(0, +\\infty)$, so\n$$\n2a^2 + a + 1 > 3a^2 - 4a + 1 \\Rightarrow a^2 - 5a < 0.\n$$\nThus $0 < a < 5$. Combining this with (1), we have $0 < a < \\frac{1}{3}$ or $1 < a < 5$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 28076,
    "subject": "Mathematics (Multi-modal)",
    "question": "Given an $a \\times b$ rectangle with $a > b > 0$, determine the minimum length of a square that covers the rectangle. (A square covers the rectangle if each point in the rectangle lies inside the square.)",
    "options": [],
    "answer": "s = a if a < (sqrt(2) + 1) b, and s = (sqrt(2) (a + b)) / 2 if a >= (sqrt(2) + 1) b",
    "solution": "Let $R$ denote the rectangle, and let $S$ denote the square with minimum length that covers $R$. Let $s$ denote the length of a side of $S$. We claim that $R$ is inscribed in $S$, that is, the vertices of $R$ lie on the sides of $S$. We also claim that $R$ can only be inscribed in two ways, as shown below. Let $S_1(S_2)$ denote the square shown on the left-hand side (right-hand side). For $S_2$, the sides of $R$ are parallel to the diagonals of $S_2$. It is easy to see that $s = a$ if $S = S_1$.\n\n$$\ns = \\begin{cases} a, & \\text{if } a < (\\sqrt{2}+1)b, \\\\ \\frac{\\sqrt{2}(a+b)}{2}, & \\text{if } a \\ge (\\sqrt{2}+1)b. \\end{cases}\n$$\n\nNow we prove our claim that these are only two ways. Let $R = ABCD$ and $S = XYZW$. Without loss of generality, we place $XY$ horizontally. By the minimality of $S$, we can assume that at least one vertex, say $A$, of $R$ lies on one side of $S$, say $WX$ (see the left-hand side figure shown below). If neither $B$ nor $D$ lies on the sides of $S$, we can then slide $R$ down (vertically), so that one of them, say $B$ lies on side $XY$ (see the middle figure shown below). If neither $C$ or $D$ lies on the sides of $S$, then we can apply an enlargement, centered at $X$ with scale less than 1, to $S$ such that the image of $S$ still covers $R$. This violates the minimality of $S$. Hence at least one of $C$ and $D$ lies on the sides of $S$, that is, three consecutive vertices of $R$ lie on the sides of $S$ (see the right-hand side figure shown below). Without loss of generality, we assume that they are $A$, $B$ and $C$. If any of these three vertices coincide with any of the vertices of $S$, then we clearly have $S = S_1$. Hence we may assume that $A$, $B$ and $C$ are on sides $WX$, $XY$ and $YZ$, respectively. By symmetry, we may also assume that $AB = a > b = BC$.\n\nIf $D$ does not lie on line segment $ZW$, then we can slide $R$ up a bit so both $B$ and $D$ lie in the interior of $S$ (see the left-hand side figure shown below). Let $O$ be the center of $R$. We can then rotate $R$ around $O$ with a small angle so that all four vertices lie inside $S$ (see the middle figure shown below). It is easy to see that we can use a smaller square to cover $R$ (by applying an enlargement centered at $O$ with a scale less than 1), violating the minimality of $S$. Thus our assumption was wrong, and $D$ must lie on side $ZW$ (see the right-hand side figure shown below), which is the case when $S = S_2$.\n\nWe finish our proof that if $S = S_2$, then the sides of $R$ are parallel to the diagonals of $S_2$. By symmetry, it suffices to show that $AX = XB$. It is not difficult to see that $\\angle XAB = \\angle YBC = \\angle ZCD = \\angle WDA$. So triangles $ABX$, $BCY$, $CDZ$ and $DAW$ are similar. Set $AX = ax$ and $XB = ay$. Then $BY = bx$ and $CY = by$. Also, $DW = bx$ and $WA = by$. Hence $by + ax = WA + AX = WX = XY = XB + BY = ay + bx$, implying that $(a-b)x = (a-b)y$, or $x = y$, as desired.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 28077,
    "subject": "Mathematics (Multi-modal)",
    "question": "There are $n$ new students. Suppose that there are two students who know each other in every three students and there are two students who do not know each other in every four students. Find the maximum value of $n$. (posed by Tang Lihua)",
    "options": [],
    "answer": "8",
    "solution": "The maximum value of $n$ is $8$.\nWhen $n=8$, the example shown in the diagram satisfies the requirements, where $A_1, A_2, \\dots, A_8$ represent $8$ students. The line segment between $A_i$ and $A_j$ means $A_i$ and $A_j$ know each other.\n![](attached_image_1.png)\n\nNext, if $n$ students satisfy the conditions, we want to show that $n \\le 8$. To do this, we first prove that the following two cases are impossible.\n\n(1) If someone $A$ knows at least $6$ persons, denoted by $B_1, B_2, \\dots, B_6$. By Ramsey's theorem, there exist $3$ persons among them who do not know each other. This contradicts that there are two who know each other in every three students, or that there exist $3$ people they know each other. $A$ and the three persons form a group of four persons such that every two of them know each other. A contradiction.\n\n(2) If some one $A$ knows at most $n-5$ persons, then in the remaining there are at least $4$ persons, none of whom knows $A$. Thus every two of the four persons know each other, a contradiction.\n\nWhen $n \\ge 10$, one of (1) and (2) must occur. So such $n$ does not satisfy the requirements.\n\nIf $n=9$, in order to avoid (1) and (2), each person knows exactly $5$ other persons. Thus the number of pairs knowing each other is $\\frac{9 \\times 5}{2} \\notin \\mathbb{N}$. This contradiction implies $n \\le 8$. Thus the maximum value of $n$ is $8$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 28078,
    "subject": "Mathematics (Multi-modal)",
    "question": "Assume that $\\alpha, \\beta, \\gamma$ satisfy $0 < \\alpha < \\beta < \\gamma < 2\\pi$. If\n$$\n\\cos(x + \\alpha) + \\cos(x + \\beta) + \\cos(x + \\gamma) = 0\n$$\nfor arbitrary $x \\in \\mathbb{R}$, then $\\gamma - \\alpha = \\underline{\\hspace{2cm}}$.",
    "options": [],
    "answer": "4π/3",
    "solution": "Write $f(x) = \\cos(x+\\alpha) + \\cos(x+\\beta) + \\cos(x+\\gamma)$. Since $f(x) \\equiv 0$ for $x \\in \\mathbb{R}$,\n$$\nf(-\\alpha) = 0,\\quad f(-\\gamma) = 0 \\text{ and } f(-\\beta) = 0.\n$$\nThat is\n$$\n\\begin{align*}\n\\text{That is}\\quad & \\cos(\\beta - \\alpha) + \\cos(\\gamma - \\alpha) = -1, \\\\\n& \\cos(\\alpha - \\beta) + \\cos(\\gamma - \\beta) = -1, \\\\\n\\text{and}\\quad & \\cos(\\alpha - \\gamma) + \\cos(\\beta - \\gamma) = -1, \\\\\n\\text{so}\\quad & \\cos(\\beta - \\alpha) = \\cos(\\gamma - \\beta) = \\cos(\\gamma - \\alpha) = -\\frac{1}{2}.\n\\end{align*}\n$$\nSince $0 < \\alpha < \\beta < \\gamma < 2\\pi$, so $\\beta - \\alpha$, $\\gamma - \\beta$, $\\gamma - \\alpha \\in \\{\\frac{2\\pi}{3}, \\frac{4\\pi}{3}\\}$. In view of $\\beta - \\alpha < \\gamma - \\alpha$, $\\gamma - \\beta < \\gamma - \\alpha$, it is possible only when $\\beta - \\alpha = \\gamma - \\beta = \\frac{2\\pi}{3}$, so $\\gamma - \\alpha = \\frac{4\\pi}{3}$.\n\nOn the other hand, when $\\beta - \\alpha = \\gamma - \\beta = \\frac{2\\pi}{3}$, we have $\\beta = \\alpha + \\frac{2\\pi}{3}$,\n$\\gamma = \\alpha + \\frac{4\\pi}{3}$. For arbitrary $x \\in \\mathbb{R}$, we denote $x + \\alpha = \\theta$. Since three\npoints $(\\cos\\theta, \\sin\\theta)$, $(\\cos(\\theta + \\frac{2\\pi}{3}), \\sin(\\theta + \\frac{2\\pi}{3}))$, $(\\cos(\\theta + \\frac{4\\pi}{3}), \\sin(\\theta + \\frac{4\\pi}{3}))$ are the vertices of an equilateral triangle on the unit\ncircle $x^2 + y^2 = 1$ with center at the origin, it is obvious that\n$$\n\\cos\\theta + \\cos\\left(\\theta + \\frac{2\\pi}{3}\\right) + \\cos\\left(\\theta + \\frac{4\\pi}{3}\\right) = 0,\n$$\nand that is $\\cos(x+\\alpha) + \\cos(x+\\beta) + \\cos(x+\\gamma) = 0$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 28079,
    "subject": "Mathematics (Multi-modal)",
    "question": "Suppose an ellipse with points $B_0$ and $B_1$ as the foci intercepts side $AB_i$ of $\\triangle AB_0B_1$ at $C_i$ ($i = 0, 1$). Taking an arbitrary point $P_0$ on the extending line of $AB_0$, draw arc $\\overarc{P_0Q_0}$ with $B_0$, $B_0P_0$ as the center and radius respectively, intercepting the extending line of $C_1B_0$ at $Q_0$. Draw arc $\\overarc{Q_0P_1}$ with $C_1$, $C_1Q_0$ as the center and radius respectively,\n\n![](attached_image_1.png)\n\nintercepting the extending line of $B_1A$ at $P_1$. Draw arc $\\overarc{P_1Q_1}$ with $B_1$, $B_1P_1$ as the center and radius respectively, intercepting the extending line of $B_1C_0$ at $Q_1$. Draw arc $\\overarc{Q_1P'_0}$ with $C_0$, $C_0Q_1$ as the center and radius respectively, intercepting the extending line of $AB_0$ at $P'_0$. Prove that\n\n(1) $P'_0$ and $P_0$ are coincident, and arcs $\\overarc{P_0Q_0}$ and $\\overarc{P_0Q_1}$ are tangent to each other at $P_0$.\n\n(2) Points $P_0, Q_0, Q_1, P_1$ are concyclic.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "(1) From the properties of an ellipse we know\n$$\nB_1C_0 + C_0B_0 = B_1C_1 + C_1B_0.\n$$\nAlso, it is obvious that\n$$\nB_0P_0 = B_0Q_0, \\quad C_1B_0 + B_0Q_0 = C_1P_1,\n$$\n$$\nB_1C_1 + C_1P_1 = B_1C_0 + C_0Q_1, \\quad C_0Q_1 = C_0B_0 + B_0P'_0.\n$$\nAdding these equations, we get $B_0P_0 = B_0P'_0$.\nTherefore $P'_0$ and $P_0$ are coincident. Furthermore, as $P_0$, $C_0$ (the center of $\\overarc{Q_1P_0}$) and $B_0$ (the center of $\\overarc{P_0Q_0}$) are lying on the same line, we know that $\\overarc{Q_1P_0}$ and $\\overarc{P_0Q_0}$ are tangent at $P_0$.\n\n(2) We have thus $\\overarc{Q_1P_0}$ and $\\overarc{P_0Q_0}$, $\\overarc{P_0Q_0}$ and $\\overarc{Q_0P_1}$, $\\overarc{Q_0P_1}$ and $\\overarc{P_1Q_1}$, $\\overarc{P_1Q_1}$ and $\\overarc{Q_1P'_0}$ are tangent at points $P_0, Q_0, P_1, Q_1$ respectively. Now we draw common tangent lines $P_0T$ and $P_1T$ through $P_0$ and $P_1$ respectively, and suppose the two lines meet at point $T$. Also, we draw a common\n\n![](attached_image_2.png)\n\ntangent line $R_1S_1$ through $Q_1$, and suppose it intercepts $P_0T$ and $P_1T$ at point $R_1$ and $S_1$ respectively. Drawing segments $P_0Q_1$ and $P_1Q_1$, we get isosceles triangles $P_0Q_1R_1$ and $P_1Q_1S_1$ respectively. Then we have\n$$\n\\begin{aligned}\n\\angle P_0Q_1P_1 &= \\pi - \\angle P_0Q_1R_1 - \\angle P_1Q_1S_1 \\\\\n&= \\pi - (\\angle P_1P_0T - \\angle Q_1P_0P_1) \\\\\n&\\quad - (\\angle P_0P_1T - \\angle Q_1P_1P_0).\n\\end{aligned}\n$$\nSince\n$$\n\\pi - \\angle P_0Q_1P_1 = \\angle Q_1P_0P_1 + \\angle Q_1P_1P_0,\n$$\nwe obtain\n$$\n\\angle P_0Q_1P_1 = \\pi - \\frac{1}{2}(\\angle P_1P_0T + \\angle P_0P_1T).\n$$\nIn the same way, we can prove that\n$$\n\\angle P_0Q_0P_1 = \\pi - \\frac{1}{2}(\\angle P_1P_0T + \\angle P_0P_1T).\n$$\nIt implies that points $P_0, Q_0, Q_1, P_1$ are concyclic.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 28080,
    "subject": "Mathematics (Multi-modal)",
    "question": "Suppose that the real numbers $a_1, a_2, \\dots, a_n$ satisfy $a_1 + a_2 + \\dots + a_n = 0$. Prove\n$$\n\\max_{1 \\le i \\le n} a_i^2 \\le \\frac{n}{3} \\sum_{i=1}^{n-1} (a_i - a_{i+1})^2.\n$$\n(pose by Zhu Huawei)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "It is sufficient to prove that: For every $k \\in \\{1, 2, \\dots, n\\}$, we have\n$$\na_k^2 \\le \\frac{n}{3} \\sum_{i=1}^{n-1} (a_i - a_{i+1})^2.\n$$\nLet $d_k = a_k - a_{k+1}$, $k = 1, 2, \\dots, n-1$, then\n$$\na_k = a_k,\n$$\n$$\na_{k+1} = a_k - d_k,\\quad a_{k+2} = a_k - d_k - d_{k+1}, \\dots,\n$$\n$$\na_n = a_k - d_k - d_{k+1} - \\dots - d_{n-1},\n$$\n$$\na_{k-1} = a_k + d_{k-1},\\quad a_{k-2} = a_k + d_{k-1} + d_{k-2}, \\dots,\n$$\n$$\na_1 = a_k + d_{k-1} + d_{k-2} + \\dots + d_1.\n$$\nSumming all equalities, and using $a_1 + a_2 + \\dots + a_n = 0$, we get\n$$\nna_k - (n-k)d_k - (n-k-1)d_{k+1} - \\dots - d_{n-1} \\\\\n\\quad + (k-1)d_{k-1} + (k-2)d_{k-2} + \\dots + d_1 = 0.\n$$\nThen in view of Cauchy's inequality\n$$\n\\begin{aligned}\n(na_k)^2 &= ((n-k)d_k + (n-k-1)d_{k+1} + \\dots + d_{n-1} \\\\\n&\\quad - (k-1)d_{k-1} - (k-2)d_{k-2} - \\dots - d_1)^2 \\\\\n&\\le \\left[ \\sum_{i=1}^{k-1} i^2 + \\sum_{i=1}^{n-k} i^2 \\right] \\left[ \\sum_{i=1}^{n-1} d_i^2 \\right] \\\\\n&\\le \\left[ \\sum_{i=1}^{n-1} i^2 \\right] \\left[ \\sum_{i=1}^{n-1} d_i^2 \\right] = \\frac{n(n-1)(2n-1)}{6} \\left[ \\sum_{i=1}^{n-1} d_i^2 \\right] \\\\\n&\\le \\frac{n^3}{3} \\left[ \\sum_{i=1}^{n-1} d_i^2 \\right].\n\\end{aligned}\n$$\nSo, $a_k^2 \\le \\frac{n}{3} \\sum_{i=1}^{n-1} (a_i - a_{i+1})^2$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 28081,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $n \\ge 2$ be a positive integer and $a_1, a_2, \\dots, a_n \\in (0, 1)$. Find the maximum value of the sum\n$$\n\\sum_{i=1}^{n} \\sqrt[6]{a_i(1-a_{i+1})}\n$$\nwhere $a_{n+1} = a_1$.",
    "options": [],
    "answer": "n / sqrt[3]{2}",
    "solution": "By the AM-GM Inequality, we deduce that\n$$\n\\begin{aligned}\n& \\sqrt[6]{a_i(1-a_{i+1})} \\\\\n&= 2^{\\frac{4}{6}} \\sqrt[6]{a_i(1-a_{i+1}) \\cdot \\frac{1}{2} \\cdot \\frac{1}{2} \\cdot \\frac{1}{2} \\cdot \\frac{1}{2}} \\\\\n&\\le 2^{\\frac{2}{3}} \\cdot \\frac{1}{6} \\cdot (a_i + 1 - a_{i+1} + 2) \\\\\n&= 2^{\\frac{2}{3}} \\cdot \\frac{1}{6} \\cdot (a_i - a_{i+1} + 3).\n\\end{aligned}\n$$\nSo\n$$\n\\begin{aligned}\n& \\sum_{i=1}^{n} \\sqrt[6]{a_i(1-a_{i+1})} \\\\\n&\\le 2^{\\frac{2}{3}} \\cdot \\frac{1}{6} \\sum_{i=1}^{n} (a_i - a_{i+1} + 3) \\\\\n&= 2^{\\frac{2}{3}} \\cdot \\frac{1}{6} \\cdot 3n \\\\\n&= \\frac{n}{\\sqrt[3]{2}}.\n\\end{aligned}\n$$\nThe equality holds if and only if $a_1 = a_2 = \\dots = a_n = \\frac{1}{2}$. So\n$$\ny_{\\max} = \\frac{n}{\\sqrt[3]{2}}.\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 28082,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $\\triangle ABC$ be a given triangle. If $|\\vec{BA} - t \\vec{BC}| \\ge |\\vec{AC}|$ for any $t \\in \\mathbb{R}$, then $\\triangle ABC$ is ( ).",
    "options": [],
    "answer": "C",
    "solution": "Suppose $\\angle ABC = \\alpha$. Since $|\\vec{BA} - t \\vec{BC}| \\ge |\\vec{AC}|$, we have\n$$\n|\\vec{BA}|^2 - 2t \\vec{BA} \\cdot \\vec{BC} + t^2 |\\vec{BC}|^2 \\ge |\\vec{AC}|^2.\n$$\nLet\n$$\nt = \\frac{\\vec{BA} \\cdot \\vec{BC}}{|\\vec{BC}|^2},\n$$\nwe get\n$$\n|\\vec{BA}|^2 - 2|\\vec{BA}|^2 \\cos^2 \\alpha + \\cos^2 \\alpha |\\vec{BA}|^2 \\ge |\\vec{AC}|^2.\n$$\nThat means $|\\vec{BA}|^2 \\sin^2 \\alpha \\ge |\\vec{AC}|^2$, i.e. $|\\vec{BA}| \\sin \\alpha \\ge |\\vec{AC}|$.\n\nOn the other hand, let point $D$ lie on line $BC$ such that $AD \\perp BC$. Then we have $|\\vec{BA}| \\sin \\alpha = |\\vec{AD}| \\le |\\vec{AC}|$. Hence $|\\vec{AD}| = |\\vec{AC}|$, and that means $\\angle ACB = \\frac{\\pi}{2}$.\n\nAnswer: C.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 28083,
    "subject": "Mathematics (Multi-modal)",
    "question": "A function $f: (0, +\\infty) \\to \\mathbf{R}$ satisfies the following conditions:\n\na. $f(a) = 1$ for a positive real number $a$,\n\nb. $f(x)f(y) + f(\\frac{a}{x})f(\\frac{a}{y}) = 2f(xy)$, for any positive real number $x, y$.\n\nProve that $f(x)$ is constant.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Setting $x = y = 1$ in $(1)$ gives\n$$\nf^2(1) + f^2(a) = 2f(1),\n$$\n$$\n(f(1) - 1)^2 = 0,\n$$\nso $f(1) = 1$.\n\nSetting $y = 1$ in $(1)$ yields\n$$\nf(x)f(1) + f(\\frac{a}{x})f(a) = 2f(x),\n$$\n$$\nf(x) = f(\\frac{a}{x}), \\quad x > 0. \\qquad (2)\n$$\n\nSetting $y = \\frac{a}{x}$ in $(1)$ yields\n$$\nf(x)f(\\frac{a}{x}) + f(\\frac{a}{x})f(x) = 2f(a),\n$$\n$$\nf(x)f(\\frac{a}{x}) = 1. \\qquad (3)\n$$\n\nCombining $(2)$ and $(3)$ gives $f^2(x) = 1, x > 0$.\n\nSetting $x = y = \\sqrt{t}$ in $(1)$ gives\n$$\nf^2(\\sqrt{t}) + f^2(\\frac{a}{\\sqrt{t}}) = 2f(t),\n$$\n$$\nf(t) > 0.\n$$\nSo $f(x) = 1, x > 0$, as desired.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 28084,
    "subject": "Mathematics (Multi-modal)",
    "question": "Suppose four solid iron balls are placed in a cylinder with the radius of $1$ cm, such that every two of the four balls are tangent to each other, and the two balls in the lower layer are tangent to the cylinder base. Now put water into the cylinder. Then, to just submerge all the balls, we need a volume of ______ cm³ water.",
    "options": [],
    "answer": "(1/3 + sqrt(2)/2) * pi",
    "solution": "Let points $O_1$, $O_2$, $O_3$, $O_4$ be the centers of the four solid iron balls respectively, with $O_1$, $O_2$ belonging to the two balls in the lower layer, and $A$, $B$, $C$, $D$ be the projective points of $O_1$, $O_2$, $O_3$, $O_4$ on the base of the cylinder. $ABCD$ constitute a square with the side of $\\frac{\\sqrt{2}}{2}$. So the height of the water in the cylinder must be $1 + \\frac{\\sqrt{2}}{2}$, so that all the balls are just immersed. Hence the volume of water we need is\n\n$$\n\\pi \\left(1 + \\frac{\\sqrt{2}}{2}\\right) - 4 \\times \\frac{4}{3} \\pi \\left(\\frac{1}{2}\\right)^3 = \\left(\\frac{1}{3} + \\frac{\\sqrt{2}}{2}\\right) \\pi.\n$$",
    "topic": "Geometry",
    "subtopic": "Solid Geometry"
  },
  {
    "id": 28085,
    "subject": "Mathematics (Multi-modal)",
    "question": "The number of real solutions for equation\n$$\n(x^{2006} + 1)(1 + x^2 + x^4 + \\cdots + x^{2004}) = 2006x^{2005}\n$$\nis \\underline{\\quad}.\n",
    "options": [],
    "answer": "1",
    "solution": "We have\n$$\n\\begin{align*}\n& (x^{2006} + 1)(1 + x^2 + x^4 + \\cdots + x^{2004}) = 2006x^{2005} \\\\\n\\Leftrightarrow & \\left(x + \\frac{1}{x^{2005}}\\right)(1 + x^2 + x^4 + \\cdots + x^{2004}) = 2006 \\\\\n\\Leftrightarrow & x + x^3 + x^5 + \\cdots + x^{2005} + \\frac{1}{x^{2005}} + \\frac{1}{x^{2003}} + \\frac{1}{x^{2001}} \\\\\n& \\qquad + \\cdots + \\frac{1}{x} = 2006 \\\\\n\\Leftrightarrow & 2006 = x + \\frac{1}{x} + x^3 + \\frac{1}{x^3} + \\cdots + x^{2005} + \\frac{1}{x^{2005}} \\\\\n& \\ge 2 \\times 1003 = 2006,\n\\end{align*}\n$$\nwhere the equal holds if and only if $x = \\frac{1}{x}$, $x^3 = \\frac{1}{x^3}$, ..., \n$x^{2005} = \\frac{1}{x^{2005}}$. Then $x = \\pm 1$.\nSince $x \\le 0$ does not satisfy the original equation, $x = 1$ is\nthe only solution. So the number of real solutions is 1.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 28086,
    "subject": "Mathematics (Multi-modal)",
    "question": "Suppose there are $8$ white balls and $2$ red balls in a packet. Each time one ball is drawn and replaced by a white one. Then the probability of drawing out all of the red balls just in the fourth draw is ______.",
    "options": [],
    "answer": "0.0434",
    "solution": "The following three cases can satisfy the condition.\n\n|        | 1st draw | 2nd draw | 3rd draw | 4th draw |\n|--------|----------|----------|----------|----------|\n| Case 1 | Red      | White    | White    | Red      |\n| Case 2 | White    | Red      | White    | Red      |\n| Case 3 | White    | White    | Red      | Red      |\n\nSo the probability\n$$\n\\begin{align*}\nP &= P(\\text{Case 1}) + P(\\text{Case 2}) + P(\\text{Case 3}) \\\\\n&= \\frac{2}{10} \\times \\left(\\frac{9}{10}\\right)^2 \\times \\frac{1}{10} + \\frac{8}{10} \\times \\frac{2}{10} \\times \\frac{9}{10} \\times \\frac{1}{10} \\\\\n&\\quad + \\left(\\frac{8}{10}\\right)^2 \\times \\frac{2}{10} \\times \\frac{1}{10} \\\\\n&= 0.0434.\n\\end{align*}\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 28087,
    "subject": "Mathematics (Multi-modal)",
    "question": "Given an integer $n \\ge 2$, define $M_0(x_0, y_0)$ to be an intersection point of the parabola $y^2 = nx - 1$ and the line $y = x$. Prove that for any positive integer $m$, there exists an integer $k \\ge 2$ such that $(x_0^m, y_0^m)$ is an intersection point of $y^2 = kx - 1$ and $y = x$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Since $M_0(x_0, y_0)$ is an intersection point of $y^2 = nx - 1$ and $y = x$, we get $x_0 = y_0 = \\frac{n \\pm \\sqrt{n^2 - 4}}{2}$. Then obviously $x_0 + \\frac{1}{x_0} = n$.\n\nLet $(x_0^m, y_0^m)$ be an intersection point of $y^2 = kx - 1$ and $y = x$. Then we get\n$$\nk = x_0^m + \\frac{1}{x_0^m}.\n$$\nWe denote $k_m = x_0^m + \\frac{1}{x_0^m}$. Then,\n$$\nk_{m+1} = k_m \\left(x_0 + \\frac{1}{x_0}\\right) - k_{m-1} = n k_m - k_{m-1} \\quad (m \\ge 2). \\quad \\textcircled{1}\n$$\nSince $k_1 = n$ is an integer,\n$$\nk_2 = x_0^2 + \\frac{1}{x_0^2} = \\left(x_0 + \\frac{1}{x_0}\\right)^2 - 2 = n^2 - 2\n$$\nis also an integer. Then, by the principle of mathematical induction and ①, we conclude that for any positive integer $m$, $k_m = x_0^m + \\frac{1}{x_0^m}$ is a positive integer too. Let $k = x_0^m + \\frac{1}{x_0^m}$. So $(x_0^m, y_0^m)$ is an intersection point of $y^2 = kx - 1$ and $y = x$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 28088,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $2006$ be expressed as the sum of five positive integers $x_1, x_2, x_3, x_4, x_5$, and $S = \\sum_{1 \\le i < j \\le 5} x_i x_j$. We ask:\n(1) What value of $x_1, x_2, x_3, x_4, x_5$ will make $S$ the maximum?\n(2) Further, if $|x_i - x_j| \\le 2$ for any $1 \\le i, j \\le 5$, then what value of $x_1, x_2, x_3, x_4, x_5$ will make $S$ the minimum? You should prove your answer.",
    "options": [],
    "answer": "Maximum: {402, 401, 401, 401, 401}. Minimum under |xi − xj| ≤ 2: {402, 402, 402, 400, 400}.",
    "solution": "(1) Obviously the number of the values of $S$ is finite, so the maximum and minimum exist. Suppose $x_1 + x_2 + x_3 + x_4 + x_5 = 2006$ such that $S = \\sum_{1 \\le i < j \\le 5} x_i x_j$ reaches the maximum, we must have\n$$\n|x_i - x_j| \\le 1, \\quad (1 \\le i, j \\le 5).\n$$\nOtherwise, assume that this does not hold. Without loss of generality, suppose $x_1 - x_2 \\ge 2$. Let $x_1' = x_1 - 1$, $x_2' = x_2 + 1$, $x_i' = x_i$ ($i = 3, 4, 5$). We have\n$$\n\\begin{aligned}\nx_1' + x_2' + x_3' + x_4' + x_5' &= x_1 + x_2 + x_3 + x_4 + x_5 \\\\\n&= 2006,\n\\end{aligned}\n$$\n$$\nS = \\sum_{1 \\le i < j \\le 5} x_i x_j = x_1 x_2 + (x_1 + x_2)(x_3 + x_4 + x_5) + x_3 x_4 + x_3 x_5 + x_4 x_5,\n$$\n$$\nS' = x_1' x_2' + (x_1' + x_2')(x_3 + x_4 + x_5) + x_3 x_4 + x_3 x_5 + x_4 x_5.\n$$\nSo\n$$\nS' - S = x_1' x_2' - x_1 x_2 > 0.\n$$\nThis contradicts the assumption that $S$ is the maximum.\nTherefore $|x_i - x_j| \\le 1$ for $1 \\le i, j \\le 5$. And it is easy to check that $S$ reaches the maximum when\n$$\nx_1 = 402, \\quad x_2 = x_3 = x_4 = x_5 = 401.\n$$\n\n(2) If we neglect the order in $x_1, x_2, x_3, x_4, x_5$, there could only be three cases:\n(a) $402, 402, 402, 400, 400$;\n(b) $402, 402, 401, 401, 400$;\n(c) $402, 401, 401, 401, 401$.\nThat satisfy $x_1 + x_2 + x_3 + x_4 + x_5 = 2006$ and $|x_i - x_j| \\le 2$.\nCases (b) and (c) can be obtained from Case (a) by setting $x_i' = x_i - 1$, $x_j' = x_j + 1$. What we have done in (1) tells us that each step like this will make $S' = \\sum_{1 \\le i < j \\le 5} x_i' x_j'$ greater. So $S$ is the minimum in Case (a), i.e. $x_1 = x_2 = x_3 = 402$, $x_4 = x_5 = 400$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 28089,
    "subject": "Mathematics (Multi-modal)",
    "question": "Suppose $f(x) = x^2 + a$. Define $f^1(x) = f(x)$, $f^n(x) = f(f^{n-1}(x))$, $n = 2, 3, \\dots$, and let $M = \\{a \\in \\mathbb{R} \\mid |f^n(0)| \\le 2$, for any $n \\in \\mathbb{N}\\}$. Prove that $M = [-2, \\frac{1}{4}]$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "(a) When $a < -2$, then $|f^1(0)| = |a| > 2$, therefore $a \\notin M$.\n\n(b) When $-2 \\le a < 0$, we have $|f^1(0)| = |a| \\le 2$. Assume that $|f^{k-1}(0)| \\le |a| \\le 2$ for $k \\ge 2$. Since $a^2 \\le -2a$ for $-2 \\le a < 0$, we get that\n$$\n-2 \\le -|a| = a \\le f^k(0) = (f^{k-1}(0))^2 + a \\le a^2 + a \\le |a| \\le 2.\n$$\nBy the principle of mathematical induction, we conclude that $|f^n(0)| \\le a \\le 2$ ($\\forall n \\ge 1$).\n\n(c) When $0 \\le a \\le \\frac{1}{4}$, we have $|f^1(0)| = |a| \\le \\frac{1}{2}$. Assume that $|f^{k-1}(0)| \\le \\frac{1}{2}$ for $k \\ge 2$. We get\n$$\n|f^k(0)| \\le |f^{k-1}(0)|^2 + a \\le \\left(\\frac{1}{2}\\right)^2 + \\frac{1}{4} = \\frac{1}{2}.\n$$\nBy the principle of mathematical induction, we conclude that $|f^n(0)| \\le \\frac{1}{2}$ ($\\forall n \\ge 1$).\n\nFrom (b) and (c), we obtain $[-2, \\frac{1}{4}] \\subseteq M$.\n\n(d) When $a > \\frac{1}{4}$, define $a_n = f^n(0)$. We have\n$$\na_{n+1} = f^{n+1}(0) = f(f^n(0)) = f(a_n) = a_n^2 + a,\n$$\nthen $a_n > a > \\frac{1}{4}$ for any $n \\ge 1$. Since\n$$\na_{n+1} - a_n = a_n^2 - a_n + a = \\left(a_n - \\frac{1}{2}\\right)^2 + a - \\frac{1}{4} \\ge a - \\frac{1}{4},\n$$\nwe get\n$$\n\\begin{aligned}\na_{n+1} - a &= a_{n+1} - a_1 \\\\\n&= (a_{n+1} - a_n) + \\dots + (a_2 - a_1) \\\\\n&\\ge n \\left(a - \\frac{1}{4}\\right).\n\\end{aligned}\n$$\nTherefore, when $n > \\frac{2-a}{a-\\frac{1}{4}}$, we have\n$$\na_{n+1} \\ge n\\left(a - \\frac{1}{4}\\right) + a > 2 - a + a = 2.\n$$\nAnd that means $a \\notin M$.\n\nFrom (a)—(d), we proved that $M = [-2, \\frac{1}{4}]$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 28090,
    "subject": "Mathematics (Multi-modal)",
    "question": "Suppose an infinite sequence $\\{a_n\\}$ satisfies $a_0 = x$, $a_1 = y$, $a_{n+1} = \\frac{a_n a_{n-1} + 1}{a_n + a_{n-1}}$, $n = 1, 2, \\dots$.\n\n(1) Find all real numbers $x$ and $y$ that satisfy the statement: there exists a positive integer $n_0$, such that, for $n \\ge n_0$, $a_n$ is a constant.\n\n(2) Find an explicit expression for $a_n$.",
    "options": [],
    "answer": "(1) Exactly those initial pairs with either absolute value of y equal to one and x not equal to minus y, or absolute value of x equal to one and y not equal to minus x. In these cases the sequence is constant from the second index onward with value either one or minus one. (2) For all nondegenerate cases, a_n = [(x + 1)^{F_{n-2}} (y + 1)^{F_{n-1}} + (x - 1)^{F_{n-2}} (y - 1)^{F_{n-1}}] / [(x + 1)^{F_{n-2}} (y + 1)^{F_{n-1}} - (x - 1)^{F_{n-2}} (y - 1)^{F_{n-1}}] for n at least zero, where F_0 = F_1 = 1 and F_n = F_{n-1} + F_{n-2}.",
    "solution": "(1) We have\n$$\na_n - a_{n+1} = a_n - \\frac{a_n a_{n-1} + 1}{a_n + a_{n-1}} = \\frac{a_n^2 - 1}{a_n + a_{n-1}}, \\quad n = 1, 2, \\dots \\quad \\textcircled{1}\n$$\nIf there exists a positive integer $n$ such that $a_{n+1} = a_n$, we get\n$$\na_n^2 = 1 \\quad \\text{and} \\quad a_n + a_{n-1} \\neq 0.\n$$\nIf $n=1$, we have\n$$\n|y|=1 \\text{ and } x \\neq -y. \\qquad \\textcircled{2}\n$$\nIf $n > 1$, then\n$$\na_n - 1 = \\frac{a_{n-1}a_{n-2} + 1}{a_{n-1} + a_{n-2}} - 1 = \\frac{(a_{n-1}-1)(a_{n-2}-1)}{a_{n-1} + a_{n-2}}, \\quad n \\ge 2, \\qquad \\textcircled{3}\n$$\nand\n$$\na_n + 1 = \\frac{a_{n-1}a_{n-2} + 1}{a_{n-1} + a_{n-2}} + 1 = \\frac{(a_{n-1}+1)(a_{n-2}+1)}{a_{n-1} + a_{n-2}}, \\quad n \\ge 2. \\qquad \\textcircled{4}\n$$\nMultiplying equations (3) and (4), we get\n$$\na_n^2 - 1 = \\frac{a_{n-1}^2 - 1}{a_{n-1} + a_{n-2}} \\cdot \\frac{a_{n-2}^2 - 1}{a_{n-1} + a_{n-2}}, \\quad n \\ge 2. \\qquad \\textcircled{5}\n$$\nFrom (5) we infer that $x$ and $y$ satisfy either (2) or\n$$\n|x|=1 \\text{ and } y \\neq -x. \\qquad \\textcircled{6}\n$$\nConversely, if $x$ and $y$ satisfy either (2) or (6), then $a_n = $ constant when $n \\ge 2$ and the constant can only be either $1$ or $-1$.\n\n(2) From (3) and (4), we get\n$$\n\\frac{a_n - 1}{a_n + 1} = \\frac{a_{n-1} - 1}{a_{n-1} + 1} \\cdot \\frac{a_{n-2} - 1}{a_{n-2} + 1}, \\quad n \\ge 2. \\qquad \\textcircled{7}\n$$\nLet $b_n = \\frac{a_n - 1}{a_n + 1}$. Then, for $n \\ge 2$, equation (7) becomes\n$$\n\\begin{aligned}\nb_n &= b_{n-1}b_{n-2} = (b_{n-2}b_{n-3})b_{n-2} = b_{n-2}^2 b_{n-3} \\\\\n&= (b_{n-3}b_{n-4})^2 b_{n-3} = b_{n-3}^3 b_{n-4}^2 = \\dots\n\\end{aligned}\n$$\nThen we get\n$$\n\\frac{a_n - 1}{a_n + 1} = \\left(\\frac{y-1}{y+1}\\right)^{F_{n-1}} \\cdot \\left(\\frac{x-1}{x+1}\\right)^{F_{n-2}}, \\quad n \\ge 2, \\qquad \\textcircled{8}\n$$\nhere,\n$$\nF_n = F_{n-1} + F_{n-2}, \\quad n \\ge 2, \\quad F_0 = F_1 = 1. \\qquad \\textcircled{9}\n$$\nFrom (9) we get\n$$\nF_n = \\frac{1}{\\sqrt{5}} \\left( \\left( \\frac{1+\\sqrt{5}}{2} \\right)^{n+1} - \\left( \\frac{1-\\sqrt{5}}{2} \\right)^{n+1} \\right). \\qquad \\textcircled{10}\n$$\nThe range of $n$ in (10) can be extended to negative integers. For example, $F_{-1} = 0, F_{-2} = 1$. Since (8) holds for any $n \\ge 0$, we get\n$$\na_n = \\frac{(x+1)^{F_{n-2}} (y+1)^{F_{n-1}} + (x-1)^{F_{n-2}} (y-1)^{F_{n-1}}}{(x+1)^{F_{n-2}} (y+1)^{F_{n-1}} - (x-1)^{F_{n-2}} (y-1)^{F_{n-1}}}, \\quad n \\ge 0, \\qquad \\textcircled{11}\n$$\nhere $F_{n-1}, F_{n-2}$ are determined by (10).",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 28091,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABCD$ be a convex quadrilateral. Let $O$ be the intersection of $AC$ and $BD$. Let $O$ and $M$ be the intersections of the circumcircle of $\\triangle OAD$ with the circumcircle of $\\triangle OBC$. Let $T$ and $S$ be the intersections of $OM$ with the circumcircle of $\\triangle OAB$ and $\\triangle OCD$ respectively. Prove that $M$ is the midpoint of $TS$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Since $\\angle BTO = \\angle BAO$ and $\\angle BCO = \\angle BMO$, $\\triangle BTM$ and $\\triangle BAC$ are similar. Hence,\n\n$$\n\\frac{TM}{AC} = \\frac{BM}{BC} \\qquad \\textcircled{1}\n$$\n\nSimilarly,\n$$\n\\triangle CMS \\sim \\triangle CBD.\n$$\nHence,\n$$\n\\frac{MS}{BD} = \\frac{CM}{BC} \\qquad \\textcircled{2}\n$$\n\nDividing ① by ②, we have\n$$\n\\frac{TM}{MS} = \\frac{BM}{CM} \\cdot \\frac{AC}{BD} \\qquad \\textcircled{3}\n$$\n\nSince $\\angle MBD = \\angle MCA$ and $\\angle MDB = \\angle MAC$, $\\triangle MBD$ and $\\triangle MCA$ are similar. Hence,\n$$\n\\frac{BM}{CM} = \\frac{BD}{AC} \\qquad \\textcircled{4}\n$$\n\n![](attached_image_1.png)\n\nCombining ④ and ③ yields $TM = MS$, as desired.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 28092,
    "subject": "Mathematics (Multi-modal)",
    "question": "Suppose that positive integers $a_1, a_2, \\dots, a_{2\\,006}$ (some of them may be equal) satisfy the condition: any two of $\\frac{a_1}{a_2}, \\frac{a_2}{a_3}, \\dots, \\frac{a_{2\\,005}}{a_{2\\,006}}$ are unequal. At least how many different numbers are there in $\\{a_1, a_2, \\dots, a_{2\\,006}\\}$? (posed by Chen Yonggao)",
    "options": [],
    "answer": "46",
    "solution": "With 45 different positive integers we can only get $45 \\times 44 + 1 = 1\\,981$ fractions. So there are more than 45 different numbers in $\\{a_1, a_2, \\dots, a_{2\\,006}\\}$.\n\nOn the other hand, let $p_1, p_2, \\dots, p_{46}$ be 46 different prime. Set $a_1, a_2, \\dots, a_{2\\,006}$ to be:\n$$\n\\begin{align*} \np_1, p_1, p_2, p_1, p_3, p_2, p_3, p_1, p_4, p_3, p_4, p_2, p_4, p_1, \\ \np_1, p_k, p_{k-1}, p_k, p_{k-2}, p_k, \\dots, p_k, p_2, p_k, p_1, \\ \np_1, p_{45}, p_{44}, p_{45}, p_{43}, p_{45}, \\dots, p_{45}, p_2, p_{45}, p_1, \\ \np_{46}, p_{45}, p_{46}, p_{44}, p_{46}, \\dots, p_{46}, p_{22}, p_{46}. \n\\end{align*}\n$$\nThen the 2 006 positive numbers satisfy that any two of $\\frac{a_1}{a_2}, \\frac{a_2}{a_3}, \\dots, \\frac{a_{2\\,005}}{a_{2\\,006}}$ are unequal.\nSo the answer is 46.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 28093,
    "subject": "Mathematics (Multi-modal)",
    "question": "Suppose $\\log_a (2x^2 + x - 1) > \\log_a 2 - 1$. Then the range of $x$ is ( ).\n\n(A) $\\frac{1}{2} < x < 1$\n(B) $x > \\frac{1}{2}$ and $x \\ne 1$\n(C) $x > 1$",
    "options": [],
    "answer": "B",
    "solution": "From\n$$\n\\begin{cases}\nx > 0, \\\\\nx \\ne 1, \\\\\n2x^2 + x - 1 > 0\n\\end{cases}\n$$\nwe get $x > \\frac{1}{2}$, $x \\ne 1$.\n\nFurthermore,\n$\\log_a (2x^2 + x - 1) > \\log_a 2 - 1 \\Rightarrow \\log_a (2x^3 + x^2 - x) > \\log_a 2 \\Rightarrow \\begin{cases} 0 < x < 1, \\\\ 2x^3 + x^2 - x < 2, \\end{cases}$ or $\\begin{cases} x > 1, \\\\ 2x^3 + x^2 - x > 2. \\end{cases}$ Then we have",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 28094,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find the smallest positive real number $k$ such that for any four given distinct real numbers $a$, $b$, $c$ and $d$, each greater than or equal to $k$, there exists a permutation $p$, q, r$ and $s$ of $a$, $b$, $c$ and $d$ such that the equation\n$$\n(x^2 + px + q)(x^2 + rx + s) = 0\n$$\nhas four distinct real roots.",
    "options": [],
    "answer": "4",
    "solution": "Suppose $k < 4$. Take $a$, $b$, $c$, $d \\in [k, \\sqrt{4k}]$. Then for any permutation $p$, $q$, $r$, $s$ of $a$, $b$, $c$, $d$, consider the equation $x^2 + px + q = 0$, its discriminant\n$$\n\\Delta = p^2 - 4q < 4k - 4q \\le 4k - 4k = 0.\n$$\nTherefore it has no real roots. So $k \\ge 4$.\n\nSuppose $4 \\le a < b < c < d$. Consider the following equations:\n$$\nx^2 + dx + a = 0,\n$$\n$$\nx^2 + cx + b = 0.\n$$\nObserve that their discriminants\n$$\n\\Delta = d^2 - 4a > 4(d - a) > 0\n$$\nand\n$$\n\\Delta = c^2 - 4b > 4(c - b) > 0.\n$$\nThen the above two equations have two distinct real roots.\n\nSuppose these two equations have the same real root $\\beta$. Then we have\n$$\n\\beta^2 + d\\beta + a = 0,\n$$\n$$\n\\beta^2 + c\\beta + b = 0.\n$$\nTaking their difference yields $\\beta = \\frac{b-a}{d-c} > 0$. Then $\\beta^2 + d\\beta + a > 0$, which leads to a contradiction. So $k = 4$.",
    "topic": "Discrete Mathematics",
    "subtopic": "Combinatorics"
  },
  {
    "id": 28095,
    "subject": "Mathematics (Multi-modal)",
    "question": "Suppose positive integers $m$, $n$, $k$ satisfy $mn = k^2 + k + 3$. Prove that at least one of the following Diophantine equations\n$$\nx^2 + 11y^2 = 4m \\text{ and } x^2 + 11y^2 = 4n\n$$\nhas a solution $(x, y)$ with $x$, $y$ being odd numbers.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "First, we prove a lemma.\n\n**Lemma** The following Diophantine equation\n$$\nx^2 + 11y^2 = 4m\n$$\nhas a solution $(x_0, y_0)$, such that either $x_0$, $y_0$ are odd numbers or $x_0$, $y_0$ are even numbers with $x_0 \\equiv (2k+1)y_0 \\pmod m$.\n\nConsider the expression $x + (2k+1)y$, where $x$, $y$ are integers, and $0 \\le x \\le 2\\sqrt{m}$, $0 \\le y \\le \\frac{\\sqrt{m}}{2}$.\n\nThere are $([2\\sqrt{m}]+1)\\left\\lfloor\\frac{\\sqrt{m}}{2}\\right\\rfloor+1 \\ge m$ such expressions. So there exist integers $x_1$, $x_2 \\in [0, 2\\sqrt{m}]$, $y_1$, $y_2 \\in [0, \\frac{\\sqrt{m}}{2}]$, such that $(x_1, y_1) \\ne (x_2, y_2)$, and\n$$\nx_1 + (2k+1)y_1 \\equiv x_2 + (2k+1)y_2 \\pmod m,\n$$\nSo\n$$\nx \\equiv (2k+1)y \\pmod m, \\quad \\text{where} \\quad x = x_1 - x_2, \\ y = y_2 - y_1.\n$$\nThis means\n$$\nx^2 \\equiv (2k+1)^2 y^2 \\equiv -11y^2 \\pmod m,\n$$\nthat is, $x^2 + 11y^2 = tm$ for some integer $t$.\nSince $|x| \\le 2\\sqrt{m}$, $|y| \\le \\frac{\\sqrt{m}}{2}$, we have\n$$\nx^2 + 11y^2 < 4m + \\frac{11}{4}m < 7m.\n$$\nSo $1 \\le t \\le 6$.\nAs $m$ is an odd number, obviously the equations $x^2 + 11y^2 = 2m$, and $x^2 + 11y^2 = 6m$ have no integer solution.\n\n(1) If $x^2 + 11y^2 = m$, then $x_0 = 2x$, $y_0 = 2y$ is a solution of $x^2 + 11y^2 = 4m$ satisfying the lemma.\n\n(2) If $x^2 + 11y^2 = 4m$, then $x_0 = x$, $y_0 = y$ is a solution of $x^2 + 11y^2 = 4m$ satisfying the lemma.\n\n(3) If $x^2 + 11y^2 = 3m$, then $(x \\pm 11y)^2 + 11(x \\mp y)^2 = 9 \\cdot 4m$.\nFirst, assume that $3 \\mid m$. If $x \\not\\equiv 0 \\pmod{3}$, $y \\not\\equiv 0 \\pmod{3}$, and $x \\not\\equiv y \\pmod{3}$, then\n$$\nx_0 = \\frac{x - 11y}{3}, \\quad y_0 = \\frac{x + y}{3}\n$$\nis a solution of $x^2 + 11y^2 = 4m$ satisfying the lemma.\nIf $x \\equiv y \\not\\equiv 0 \\pmod{3}$, then\n$$\nx_0 = \\frac{x + 11y}{3}, \\quad y_0 = \\frac{y - x}{3}\n$$\nis a solution of $x^2 + 11y^2 = 4m$ satisfying the lemma.\nNow suppose $3 \\mid m$. Then the above are still integer solutions. If $x^2 + 11y^2 = 4m$ has an even integer solution $x_0 = 2x_1$, $y_0 = 2y_1$, then\n$$\nx_1^2 + 11y_1^2 = m \\Leftrightarrow 36m = (5x_1 \\pm 11y_1)^2 + 11(5y_1 \\mp x_1)^2.\n$$\nSince one of $x_1$, $y_1$ is even and the other is odd, so $5x_1 \\pm 11y_1$, $5y_1 \\mp x_1$ are odd numbers.\nIf $x_1 \\equiv y_1 \\pmod{3}$, then $x_0 = \\frac{5x_1 - 11y_1}{3}$, $y_0 = \\frac{5y_1 + x_1}{3}$ is a solution of $x^2 + 11y^2 = 4m$ satisfying the lemma.\nIf $x_1 \\not\\equiv y_1 \\pmod{3}$, then $x_0 = \\frac{5x_1 + 11y_1}{3}$, $y_0 = \\frac{5y_1 - x_1}{3}$ is a solution of $x^2 + 11y^2 = 4m$ satisfying the lemma.\n\n(4) If $x^2 + 11y^2 = 5m$, then $25 \\cdot 4m = (3x \\mp 11y)^2 + 11(3y \\pm x)^2$.\nWhen $5 \\nmid m$, if\n$$\nx \\equiv \\pm 1 \\pmod{5}, \\quad y \\equiv \\mp 2 \\pmod{5}, \\quad \\text{or } x \\equiv \\pm 2 \\pmod{5}, \\quad y \\equiv \\pm 1 \\pmod{5},\n$$\nthen\n$$\nx_0 = \\frac{3x - 11y}{5}, \\quad y_0 = \\frac{3y + x}{5}\n$$\nis a solution of $x^2 + 11y^2 = 4m$ satisfying the lemma.\nIf $x \\equiv \\pm 1 \\pmod{5}$, $y \\equiv \\pm 2 \\pmod{5}$, or $x \\equiv \\pm 2 \\pmod{5}$, $y \\equiv \\mp 1 \\pmod{5}$, then\n$$\nx_0 = \\frac{3x + 11y}{5}, \\quad y_0 = \\frac{3y - x}{5}\n$$\nis a solution of $x^2 + 11y^2 = 4m$ satisfying the lemma.\n\nWhen $5 \\mid m$, then the above are still integer solutions. If $x^2 + 11y^2 = 4m$ has an even integer solution $x_0 = 2x_1$, $y_0 = 2y_1$, then\n$$\nx_1^2 + 11y_1^2 = m, \\quad x_1 \\not\\equiv y_1 \\pmod 2,\n$$\nand we have $100m = (x_1 \\mp 33y_1)^2 + 11(y_1 \\pm 3x_1)^2$.\nIf $x_1 \\equiv y_1 \\equiv 0 \\pmod{5}$, or $x_1 \\equiv \\pm 1 \\pmod{5}$, $y_1 \\equiv \\pm 2 \\pmod{5}$, or\n$$\nx_1 \\equiv \\pm 2 \\pmod{5}, \\quad y_1 \\equiv \\mp 1 \\pmod{5},\n$$\nthen $x_0 = \\frac{x_1 - 33y_1}{5}$, $y_0 = \\frac{y_1 + 3x_1}{5}$ is a solution of $x^2 + 11y^2 = 4m$ satisfying the lemma.\nIf $x_1 \\equiv \\pm 1 \\pmod{5}$, $y_1 \\equiv \\mp 2 \\pmod{5}$, or $x_1 \\equiv \\pm 2 \\pmod{5}$, $y_1 \\equiv \\pm 1 \\pmod{5}$, then\n$$\nx_0 = \\frac{x_1 + 33y_1}{5}, \\quad y_0 = \\frac{y_1 - 33x_1}{5}\n$$\nis a solution of $x^2 + 11y^2 = 4m$ satisfying the lemma.\n\nThe lemma is proved.\n\nFrom the lemma, if $x^2 + 11y^2 = 4m$ has a solution $(x, y)$ with $x$, $y$ being odd numbers, then it has a solution $(x_0, y_0)$ with $x_0$, $y_0$ being even numbers satisfying $x_0 \\equiv (2k+1)y_0 \\pmod m$.\n\nLet $l = 2k + 1$, the solution of the quadratic equation\n$$\nmx^2 + ly_0x + ny_0^2 - 1 = 0\n$$\nis\n$$\nx = \\frac{-ly_0 \\pm \\sqrt{l^2 y_0^2 - 4mny_0^2 + 4m}}{2m} = \\frac{-ly_0 \\pm x_0}{2m}.\n$$\nSo the equation\n$$\nmx_1^2 + ly_0x_1 + ny_0^2 - 1 = 0\n$$\nhas at least an integer solution $x_1$, i.e.\n$$\nmx_1^2 + ly_0x_1 + ny_0^2 - 1 = 0.\n$$\nThis indicates that $x_1$ is an odd number. Now, from this equation, it follows that\n$$\n(2ny_0 + lx_1)^2 + 11x_1^2 = 4n.\n$$\nThis means $x^2 + 11y^2 = 4n$ has a solution $(x, y)$ with $x$, $y$ being odd numbers, where $x = 2ny_0 + lx_1$, $y = x_1$.\n\nTherefore, at least one of the equations $x^2 + 11y^2 = 4m$ or $x^2 + 11y^2 = 4n$ has a solution $(x, y)$ with $x$, $y$ odd.",
    "topic": "Number Theory",
    "subtopic": "Diophantine Equations"
  },
  {
    "id": 28096,
    "subject": "Mathematics (Multi-modal)",
    "question": "Prove that for $i = 1, 2, 3$, there exist infinitely many integers $n$ satisfying the following condition: we can find $i$ integers in $\\{n, n+2, n+28\\}$ that can be expressed as the sum of the cubes of three positive integers.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "**Lemma** Let $m$ be the remainder of the sum of the cubes of three positive integers when divided by $9$, then $m \\neq 4$ or $5$.\n**Proof:** Since any integer can be expressed as $3k$ or $3k \\pm 1$ ($k \\in \\mathbb{Z}$),\n$$\n(3k)^3 = 9 \\times 3k^3, \\\\\n(3k \\pm 1)^3 = 9 \\times (3k^3 \\pm 3k^2 + k) \\pm 1,\n$$\nas desired.\n\nIf $i = 1$, take $n = 3(3m-1)^3 - 2$ ($m \\in \\mathbb{Z}^+$), then $4$ or $5$ is the remainder of $n$ and $n+28$ when divided by $9$. So, they cannot be expressed as the sum of the cubes of three positive integers. But\n$$\nn + 2 = (3m - 1)^3 + (3m - 1)^3 + (3m - 1)^3.\n$$\n\nIf $i = 2$, take $n = (3m-1)^3 + 222$ ($m \\in \\mathbb{Z}^+$), then $5$ is the remainder of $n$ when divided by $9$. So, it cannot be expressed as the sum of the cubes of three positive integers. But\n$$\nn + 2 = (3m - 1)^3 + 2^3 + 6^3, \\\\\nn + 28 = (3m - 1)^3 + 5^3 + 5^3.\n$$\n\nIf $i = 3$, take $n = 216m^3$ ($m \\in \\mathbb{Z}^+$). It satisfies the conditions:\n$$\nn = (3m)^3 + (4m)^3 + (5m)^3, \\\\\nn + 2 = (6m)^3 + 1^3 + 1^3, \\\\\nn + 28 = (6m)^3 + 1^3 + 3^3.\n$$\nThis completes the proof.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 28097,
    "subject": "Mathematics (Multi-modal)",
    "question": "In $\\triangle PBC$, $\\angle PBC = 60^\\circ$. The tangent at point $P$ to the circumcircle $w$ of $\\triangle PBC$ intersects with line $CB$ at $A$. Points $D$ and $E$ lie on the line segment $PA$ and circle $w$ respectively, such that $\\angle DBE = 90^\\circ$ and $PD = PE$. $BE$ and $PC$ meet at $F$. It is given that lines $AF$, $BP$ and $CD$ are concurrent.\n(1) Prove that $BF$ is the bisector of $\\angle PBC$;\n(2) Find the value of $\\tan \\angle PCB$.",
    "options": [],
    "answer": "(6 + sqrt(3))/11",
    "solution": "By the angle bisector theorem, we have\n$$\n\\frac{PF}{FC} \\cdot \\frac{CB}{BA} \\cdot \\frac{AD}{DP} = \\frac{PB}{BC} \\cdot \\frac{BC}{BA} \\cdot \\frac{AB}{PB} = 1.\n$$\nBy the converse of Ceva theorem, the lines $AF$, $BP$ and $CD$ are concurrent.\nSuppose there exists $\\angle D'BF'$ satisfying the conditions: (a) $\\angle D'BF' = 90^\\circ$, (b) the lines $AF'$, $BP$ and $CD'$ are concurrent. We may assume that $F'$ lies on $PF$. Then, $D'$ is on $AD$.\nSo\n$$\n\\frac{PF'}{F'C} < \\frac{PF}{FC}, \\quad \\frac{AD'}{PD'} < \\frac{AD}{PD}.\n$$\nThus\n$$\n\\frac{PF'}{F'C} \\cdot \\frac{CB}{BA} \\cdot \\frac{AD'}{D'P} < \\frac{PB}{BC} \\cdot \\frac{BC}{BA} \\cdot \\frac{AB}{PB} = 1,\n$$\nwhich leads to a contradiction. This completes the proof.\n\n(2) We may assume that the circle $O$ has radius 1. Let\n\n$\\angle PCB = \\alpha$. By (1), $\\angle PBE = \\angle EBC = 30^\\circ$. Therefore, $E$ is the midpoint of $\\overline{PC}$.\nSince $\\angle MPE = \\angle PBE = 30^\\circ$, $\\angle CPE = \\angle CBE = 30^\\circ$ and $PD = PE$, we obtain $\\angle PDE = \\angle PED = 15^\\circ$, $PE = 2 \\cdot 1 \\cdot \\sin 30^\\circ$ and $DE = 2\\cos 15^\\circ$.\n\n![](attached_image_1.png)\n\nSince\n$$\nBE = 2\\sin \\angle ECB = 2\\sin(\\alpha + 30^\\circ)\n$$\nand $\\angle BED = \\angle BEP - 15^\\circ$, we have\n$$\n\\cos(\\alpha - 15^\\circ) = \\frac{BE}{DE} = \\frac{2\\sin(\\alpha + 30^\\circ)}{2\\cos 15^\\circ},\n$$\n$$\n\\cos(\\alpha - 15^\\circ)\\cos 15^\\circ = \\sin(\\alpha + 30^\\circ),\n$$\n$$\n\\cos \\alpha + \\cos(\\alpha - 30^\\circ) = 2\\sin(\\alpha + 30^\\circ),\n$$\n$$\n\\cos \\alpha + \\cos \\alpha \\cos 30^\\circ + \\sin \\alpha \\sin 30^\\circ = \\sqrt{3} \\sin \\alpha + \\cos \\alpha,\n$$\n$$\n1 + \\frac{\\sqrt{3}}{2} + \\frac{1}{2} \\tan \\alpha = \\sqrt{3} \\tan \\alpha + 1.\n$$\nSo\n$$\n\\tan \\alpha = \\frac{6 + \\sqrt{3}}{11}.\n$$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 28098,
    "subject": "Mathematics (Multi-modal)",
    "question": "Suppose $A = \\{x \\mid 5x - a \\le 0\\}$, $B = \\{x \\mid 6x - b > 0\\}$, $a, b \\in \\mathbb{N}$, and $A \\cap B \\cap \\mathbb{N} = \\{2, 3, 4\\}$. The number of such pairs $(a, b)$ is ( ).",
    "options": [],
    "answer": "C",
    "solution": "Since $5x - a \\le 0 \\Rightarrow x \\le \\frac{a}{5}$, $6x - b > 0 \\Rightarrow x > \\frac{b}{6}$. In order to satisfy $A \\cap B \\cap \\mathbb{N} = \\{2, 3, 4\\}$, we have\n$$\n\\begin{cases}\n1 \\le \\frac{b}{6} < 2, \\\\\n4 \\le \\frac{a}{5} < 5,\n\\end{cases}\n$$\nor\n$$\n\\begin{cases}\n6 \\le b < 12, \\\\\n20 \\le a < 25.\n\\end{cases}\n$$\nSo the number of pairs $(a, b)$ is $\\binom{6}{1} \\cdot \\binom{5}{1} = 30$. Answer: C.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 28099,
    "subject": "Mathematics (Multi-modal)",
    "question": "Solve the following system of equations.\n$$\n\\begin{cases} x - y + z - w = 2, \\\\ x^2 - y^2 + z^2 - w^2 = 6, \\\\ x^3 - y^3 + z^3 - w^3 = 20, \\\\ x^4 - y^4 + z^4 - w^4 = 66. \\end{cases}\n$$",
    "options": [],
    "answer": "Four solutions: (x=3, y=2, z=1, w=0); (x=3, y=0, z=1, w=2); (x=1, y=2, z=3, w=0); (x=1, y=0, z=3, w=2).",
    "solution": "Let $p = x + z$, $q = xz$. The second to fourth equations of the system become\n$$\n\\begin{aligned}\np^2 &= x^2 + z^2 + 2q, \\\\\np^3 &= x^3 + z^3 + 3pq, \\\\\np^4 &= x^4 + z^4 + 4p^2q - 2q^2.\\end{aligned}\n$$\nSimilarly, let $s = y + w$, $t = yw$. The second to fourth equations of the system become\n$$\ns^2 = y^2 + w^2 + 2t,\n$$\n$$\ns^3 = y^3 + w^3 + 3st,\n$$\n$$\ns^4 = y^4 + w^4 + 4s^2t - 2t^2.\n$$\nAlso, the first equation in the system can now be expressed as\n$$\np = s + 2. \\qquad \\textcircled{1}\n$$\nTherefore\n$$\np^2 = s^2 + 4s + 4,\n$$\n$$\np^3 = s^3 + 6s^2 + 12s + 8,\n$$\n$$\np^4 = s^4 + 8s^3 + 24s + 32s + 16.\n$$\nSubstituting the expressions of $p^2$, $p^3$, $p^4$ and $s^2$, $s^3$, $s^4$ obtained previously into the original system, we get\n$$\nx^2 + z^2 + 2q = y^2 + w^2 + 2t + 4s + 4,\n$$\n$$\nx^3 + z^3 + 3pq = y^3 + w^3 + 3st + 6s^2 + 12s + 8,\n$$\n$$\nx^4 + z^4 + 4p^2q - 2q^2 = y^4 + w^4 + 4s^2t - 2t^2 + 8s^3 + 24s + 32s + 16.\n$$\nUsing the second to the fourth equations in the system to simplify the above, we get\n$$\nq = t + 2s - 1, \\qquad \\textcircled{2}\n$$\n$$\npq = st + 2s^2 + 4s - 4, \\qquad \\textcircled{3}\n$$\n$$\n2p^2q - q^2 = 2s^2t - t^2 + 4s^3 + 12s^2 + 16s - 25. \\qquad \\textcircled{4}\n$$\nSubstituting ① and ② into ③, we get\n$$\nt = \\frac{s}{2} - 1. \\qquad \\textcircled{5}\n$$\nSubstituting ⑤ into ②,\n$$\nq = \\frac{5}{2}s - 2. \\qquad \\textcircled{6}\n$$\nSubstituting ①, ⑤, ⑥ into ④, we get $s = 2$. Therefore $t = 0$, $p = 4$, $q = 3$.\nConsequently, $x, z$ and $y, w$ are the roots of equations $X^2 - 4X + 3 = 0$ and $Y^2 - 2Y = 0$ respectively. That means\n$$\n\\begin{cases} x = 3, \\\\ z = 1 \\end{cases} \\text{ or } \\begin{cases} x = 1, \\\\ z = 3 \\end{cases}\n$$\nand\n$$\n\\begin{cases} y = 2, \\\\ w = 0 \\end{cases} \\text{ or } \\begin{cases} y = 0, \\\\ w = 2. \\end{cases}\n$$\nSpecifically, the system of equations has 4 solutions:\n$$\nx = 3, y = 2, z = 1, w = 0;\n$$\n$$\nx = 3, y = 0, z = 1, w = 2;\n$$\n$$\nx = 1, y = 2, z = 3, w = 0;\n$$\n$$\nx = 1, y = 0, z = 3, w = 2.\n$$",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 28100,
    "subject": "Mathematics (Multi-modal)",
    "question": "Eight persons join a party.\n\n(1) If there exist three persons who know each other in any group of five, prove that we can find that four persons know each other.\n\n(2) If there exist three persons in a group of six who know each other in a cyclical manner, can we find four persons who know each other in a cyclical manner?",
    "options": [],
    "answer": "Part (1): Yes, there exists a group of four mutual acquaintances. Part (2): No.",
    "solution": "(1) By means of graph theory, use $8$ vertices to denote $8$ persons. If two persons know each other, we connect them with an edge. With the given condition, there will be a triangle in every induced subgraph with five vertices, while every triangle in the graph belongs to different $\\binom{8-3}{2} = \\binom{5}{2} = 10$ induced subgraphs with five vertices. We know that there are $3 \\times \\binom{8}{5} = 3 \\times 56 = 168$ edges in total in these triangles, while every edge is computed ten times repeatedly.\n\nThus every vertex is incident with at least $\\frac{2 \\times 168}{8 \\times 10} > 4$ edges. So there exists one vertex $A$ that is incident with at least five edges.\n\nSuppose the vertex $A$ is adjacent with five vertices $B$, $C$, $D$, $E$, $F$. By the condition, there exists one triangle in five vertices. Without loss of generality, let $\\triangle BCD$ denote the triangle. So there exists one edge between any two vertices in the four vertices $A$, $B$, $C$, $D$. So the corresponding four persons of the four vertices know each other.\n\n\n(2) If there exist three persons (in a group of six) who know one another in a cyclical manner, there may not exist four persons who know each other.\n\nFor example, let $8$ vertices denote $8$ persons. If two persons know each other, we join them with an edge. Consider the regular octagon, we link up the $8$ shortest diagonals, as desired.",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 28101,
    "subject": "Mathematics (Multi-modal)",
    "question": "Given a right triangular prism $A_1B_1C_1 - ABC$ with $\\angle BAC = \\frac{\\pi}{2}$ and $AB = AC = AA_1 = 1$, let $G, E$ be the midpoints of $A_1B_1$, $CC_1$ respectively; and $D, F$ be variable points lying on segments $AC, AB$ (not including endpoints) respectively. If $GD \\perp EF$, the range of the length of $DF$ is ( ).",
    "options": [],
    "answer": "A",
    "solution": "We establish a coordinate system with point $A$ as the origin, line $AB$ as the $x$-axis, $AC$ the $y$-axis and $AA_1$ the $z$-axis. Then we have $F(t_1, 0, 0)$ ($0 < t_1 < 1$), $E(0, 1, \\frac{1}{2})$,\n$G(\\frac{1}{2}, 0, 1)$, $D(0, t_2, 0)$ ($0 < t_2 < 1$). Therefore $\\vec{EF} = (t_1, -1, -\\frac{1}{2})$, $\\vec{GD} = (-\\frac{1}{2}, t_2, -1)$. Since $GD \\perp EF$, we get $t_1 + 2t_2 = 1$. Then $0 < t_2 < \\frac{1}{2}$. Furthermore, $\\vec{DF} = (t_1, -t_2, 0)$,\n$$\n|\\vec{DF}| = \\sqrt{t_1^2 + t_2^2} = \\sqrt{5t_2^2 - 4t_2 + 1} = \\sqrt{5\\left(t_2 - \\frac{2}{5}\\right)^2 + \\frac{1}{5}}\n$$\nWe obtain $\\sqrt{\\frac{1}{5}} \\le |\\vec{DF}| < 1$. Answer: A.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 28102,
    "subject": "Mathematics (Multi-modal)",
    "question": "Assume that $a$ is a positive integer and not a perfect square. Prove that for any positive integer $n$, the sum\n$$\nS_n = \\{\\sqrt{a}\\} + \\{\\sqrt{a}\\}^2 + \\dots + \\{\\sqrt{a}\\}^n\n$$\nis irrational, where $\\{x\\} = x - \\lfloor x \\rfloor$ and $\\lfloor x \\rfloor$ denotes the greatest integer less than or equal to $x$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Suppose that $c^2 < a < (c+1)^2$, where $c$ is an integer greater than or equal to $1$. Then, $\\lfloor \\sqrt{a} \\rfloor = c$, $1 \\le a - c^2 \\le 2c$, and $\\{\\sqrt{a}\\} = \\sqrt{a} - c$.\n\nWrite $\\{\\sqrt{a}\\}^k = (\\sqrt{a} - c)^k = x_k + y_k \\sqrt{a}$, where $k \\in \\mathbb{N}$ and $x_k, y_k \\in \\mathbb{Z}$. We have\n$$\nS_n = (x_1 + x_2 + \\cdots + x_n) + (y_1 + y_2 + \\cdots + y_n) \\sqrt{a}. \\quad (1)\n$$\nWe now prove that $T_n = \\sum_{k=1}^{n} y_k \\neq 0$ for all positive integers $n$.\n\nSince\n$$\n\\begin{aligned}\nx_{k+1} + y_{k+1} \\sqrt{a} &= (\\sqrt{a} - c)^{k+1} \\\\\n&= (\\sqrt{a} - c)(x_k + y_k \\sqrt{a}) \\\\\n&= (a y_k - c x_k) + (x_k - c y_k) \\sqrt{a},\n\\end{aligned}\n$$\nwe have\n$$\nx_{k+1} = a y_k - c x_k, \\quad y_{k+1} = x_k - c y_k.\n$$\nSince $x_1 = -c$ and $y_1 = 1$, we have $y_2 = -2c$.\n\nBy the above equality, we have\n$$\ny_{k+2} = -2c y_{k+1} + (a - c^2) y_k, \\quad (2)\n$$\nwhere $y_1 = 1$, $y_2 = -2c$.\n\nBy mathematical induction, we have\n$$\ny_{2k-1} > 0, \\quad y_{2k} < 0. \\quad (3)\n$$\nCombining (2) and (3), we have\n$$\ny_{2k+2} - y_{2k+1} = -(2c+1) y_{2k+1} + (a - c^2) y_{2k} < 0,\n$$\n$$\ny_{2k+2} + y_{2k+1} = -(2c-1) y_{2k+1} + (a - c^2) y_{2k} < 0.\n$$\nTaking the product of the above inequalities, we get $y_{2k+2}^2 - y_{2k+1}^2 > 0$. Since $y_2^2 - y_1^2 > 0$, we have\n$$\n|y_{2k-1}| < |y_{2k}|. \\qquad (4)\n$$\nOn the other hand, we have\n$$\ny_{2k+1} - y_{2k} = -(2c+1) y_{2k} + (a - c^2) y_{2k-1} > 0,\n$$\n$$\ny_{2k+1} + y_{2k} = -(2c-1) y_{2k} + (a - c^2) y_{2k-1} > 0.\n$$\nMultiplying the above inequalities, we have $y_{2k+1}^2 - y_{2k}^2 > 0$, that is, $|y_{2k}| < |y_{2k+1}|$.\n\nTherefore, $|y_n| < |y_{n+1}|$ for all positive integers $n$. Combining (3) and (4), we have $y_{2k-1} + y_{2k} < 0$, $y_{2k+1} + y_{2k} > 0$ for all positive integers $n$.\n\nTherefore,\n$$\nT_{2n-1} = y_1 + (y_2 + y_3) + \\cdots + (y_{2n-2} + y_{2n-1}) > 0,\n$$\n$$\nT_{2n} = (y_1 + y_2) + (y_3 + y_4) + \\cdots + (y_{2n-1} + y_{2n}) < 0.\n$$\nHence, $T_n \\neq 0$ for all positive integer $n$. This completes the proof.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 28103,
    "subject": "Mathematics (Multi-modal)",
    "question": "Suppose $f(x) = x^3 + \\log_2(x + \\sqrt{x^2+1})$. For any $a, b \\in \\mathbb{R}$, to satisfy $f(a) + f(b) \\ge 0$, the condition $a + b \\ge 0$ is ( ).\n\n(A) necessary and sufficient\n(B) not necessary but sufficient\n(C) necessary but not sufficient\n(D) neither necessary nor sufficient",
    "options": [],
    "answer": "A",
    "solution": "Obviously $f(x) = x^3 + \\log_2(x + \\sqrt{x^2+1})$ is an odd function and is monotonically increasing. So, if $a+b \\ge 0$, i.e. $a \\ge -b$, we get $f(a) \\ge f(-b)$, $f(a) \\ge -f(b)$, and that means $f(a)+f(b) \\ge 0$.\n\nOn the other hand, if $f(a) + f(b) \\ge 0$, then $f(a) \\ge -f(b) = f(-b)$. So $a \\ge -b$, $a+b \\ge 0$. Answer: A.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 28104,
    "subject": "Mathematics (Multi-modal)",
    "question": "A sequence of real numbers $\\{a_n\\}$ satisfies the condition that $a_1 = \\frac{1}{2}$, $a_{k+1} = -a_k + \\frac{1}{2-a_k}$, $k = 1, 2, \\dots$.\nProve the following inequality:\n$$\n\\left( \\frac{n}{2(a_1 + a_2 + \\cdots + a_n)} - 1 \\right)^n \\le \\left[ \\frac{a_1 + a_2 + \\cdots + a_n}{n} \\right]^n \\left( \\frac{1}{a_1} - 1 \\right) \\left( \\frac{1}{a_2} - 1 \\right) \\cdots \\left( \\frac{1}{a_n} - 1 \\right).\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "**Proof** First, we use induction to prove that $0 < a_n \\le \\frac{1}{2}$, $n = 1, 2, \\dots$.\nWhen $n = 1$, it is obvious.\nNow suppose it is true for $n$ ($n \\ge 1$), i.e. $0 < a_n \\le \\frac{1}{2}$.\nLet $f(x) = -x + \\frac{1}{2-x}$ for $x \\in [0, \\frac{1}{2}]$. Then $f(x)$ is a decreasing function. So\n$$\na_{n+1} = f(a_n) \\le f(0) = \\frac{1}{2},\n$$\n$$\na_{n+1} = f(a_n) \\ge f(\\frac{1}{2}) = \\frac{1}{6} > 0,\n$$\ni.e. it is true for $n+1$.\n\nBack to the problem, it is sufficient to prove:\n$$\n\\left( \\frac{n}{a_1 + a_2 + \\cdots + a_n} \\right)^n \\left( \\frac{n}{2(a_1 + a_2 + \\cdots + a_n)} - 1 \\right)^n \\\\\n\\le \\left( \\frac{1}{a_1} - 1 \\right) \\left( \\frac{1}{a_2} - 1 \\right) \\cdots \\left( \\frac{1}{a_n} - 1 \\right).\n$$\n\nLet $f(x) = \\ln(\\frac{1}{x} - 1)$ for $x \\in (0, \\frac{1}{2})$. Then $f(x)$ is a concave function, i.e. for every $0 < x_1, x_2 < \\frac{1}{2}$, we have $f[\\frac{x_1 + x_2}{2}] \\le \\frac{f(x_1) + f(x_2)}{2}$.\nIn fact,\n$$\nf[\\frac{x_1 + x_2}{2}] \\le \\frac{f(x_1) + f(x_2)}{2}\n$$\nis equivalent to\n$$\n\\left(\\frac{2}{x_1 + x_2} - 1\\right)^2 \\le \\left(\\frac{1}{x_1} - 1\\right) \\left(\\frac{1}{x_2} - 1\\right), \\\\\n\\Leftrightarrow (x_1 - x_2)^2 \\ge 0.\n$$\nSo $f(x)$ is a concave function.\n\nUsing Jensen's Inequality, it follows that\n$$\nf\\left[\\frac{x_1 + x_2 + \\dots + x_n}{n}\\right] \\le \\frac{f(x_1) + f(x_2) + \\dots + f(x_n)}{n},\n$$\n$$\n\\text{so} \\quad \\left( \\frac{n}{a_1 + a_2 + \\dots + a_n} - 1 \\right)^n \\le \\left( \\frac{1}{a_1} - 1 \\right) \\left( \\frac{1}{a_2} - 1 \\right) \\dots \\left( \\frac{1}{a_n} - 1 \\right).\n$$\n\nOn the other hand, by using Cauchy's Inequality, we have\n$$\n\\begin{align*}\n\\sum_{i=1}^{n} (1 - a_i) &= \\sum_{i=1}^{n} \\frac{1}{a_i + a_{i+1}} - n \\\\\n&\\ge \\frac{n^2}{\\sum_{i=1}^{n} (a_i + a_{i+1})} - n \\\\\n&= \\frac{n^2}{a_{n+1} - a_1 + 2 \\sum_{i=1}^{n} a_i} - n \\\\\n&\\ge \\frac{n^2}{2 \\sum_{i=1}^{n} a_i} - n = n \\left[ \\frac{n}{2 \\sum_{i=1}^{n} a_i} - 1 \\right].\n\\end{align*}\n$$\nThis means\n$$\n\\frac{\\sum_{i=1}^{n} (1 - a_i)}{\\sum_{i=1}^{n} a_i} \\ge \\frac{n}{\\sum_{i=1}^{n} a_i} \\left[ \\frac{n}{2 \\sum_{i=1}^{n} a_i} - 1 \\right].\n$$\nSo\n$$\n\\begin{align*}\n& \\left( \\frac{n}{a_1 + a_2 + \\cdots + a_n} \\right)^n \\left( \\frac{n}{2(a_1 + a_2 + \\cdots + a_n)} - 1 \\right)^n \\\\\n& \\le \\left[ \\frac{(1-a_1)(1-a_2)\\cdots(1-a_n)}{a_1 + a_2 + \\cdots + a_n} \\right]^n \\\\\n& \\le \\left(\\frac{1}{a_1} - 1\\right) \\left(\\frac{1}{a_2} - 1\\right) \\cdots \\left(\\frac{1}{a_n} - 1\\right).\n\\end{align*}\n$$",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 28105,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $S = \\{n \\mid n-1, n, n+1$ all can be expressed as the sum of the squares of two positive integers\\}. Prove that, if $n \\in S$, then $n^2 \\in S$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "$$\nx^2 + y^2 \\equiv 0, 1, 2 \\pmod{4}.\n$$\nLet $n \\in S$. By the above equality, we get $n \\equiv 1 \\pmod{4}$. Thus, we may assume that\n$$\nn - 1 = a^2 + b^2, \\ a \\geq b,\n$$\n$$\nn = c^2 + d^2, \\ c > d,\n$$\n$$\nn + 1 = e^2 + f^2, \\ e \\geq f,\n$$\nwhere $a, b, c, d, e, f$ are positive integers. Therefore\n$$\nn^2 + 1 = n^2 + 1^2,\n$$\n$$\nn^2 = (c^2 + d^2)^2 = (c^2 - d^2)^2 + (2cd)^2,\n$$\n$$\nn^2 - 1 = (a^2 + b^2)(e^2 + f^2) = (ae - bf)^2 + (af + be)^2.\n$$\nSuppose that $b = a$ and $f = e$, we have $n - 1 = 2a^2$, $n + 1 = 2e^2$. Taking the difference of these two equations yields $e^2 - a^2 = 1$. Then, $e - a \\geq 1$. But\n$$\n1 = e^2 - a^2 = (e + a)(e - a) > 1,\n$$\na contradiction!\nSo $b = a$ and $f = e$ do not happen simultaneously. Therefore $ae - bf > 0$, and $n^2 \\in S$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 28106,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $S = \\{(a, b) \\mid 1 \\le a, b \\le 5,\\ a, b \\in \\mathbb{Z}\\}$. Let $T$ be the set of integer points in the plane such that for any point $P$ in $S$, there exists a different point $Q$ in $T$ such that $PQ$ does not contain integer points except $P$ and $Q$. Find the minimum value of $|T|$, where $|T|$ denotes the number of elements of the finite set $T$.",
    "options": [],
    "answer": "2",
    "solution": "We first prove that $|T| \\ne 1$.\nIf $|T| = 1$, let $T = \\{Q(x_0, y_0)\\}$. We may take point $P(x_1, y_1)$ in $S$ satisfying the conditions: (1) $(x_1, y_1) \\ne (x_0, y_0)$, (2) $x_1$ and $x_0$ have the same parity, $y_1$ and $y_0$ have the same parity. Then, the midpoint of $PQ$ is an integer, which is a contradiction.\n\nIf $|T| = 2$, see the following figure satisfying the conditions of the problem:\n• • • • •\n• • ◐ • •\n• • • • • ◐\n• • • • • •\n• • • • • •\n\nas desired.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 28107,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $M = \\{1, 2, \\dots, 19\\}$ and $A = \\{a_1, a_2, \\dots, a_k\\} \\subseteq M$. Find the minimum value of $k$ so that there exist $a_i, a_j \\in A$ such that if $b \\in M$, then $a_i = b$ or $a_i \\pm a_j = b$.",
    "options": [],
    "answer": "5",
    "solution": "By the definition of $A$, we have $k(k+1) \\ge 19$, implying $k \\ge 4$.\n\nIf $k=4$, we have $k(k+1)=20$. We may assume that $a_1 < a_2 < a_3 < a_4$. Then, $a_4 \\ge 10$.\n(1) If $a_4 = 10$, then $a_3 = 9$, and we have $a_2 = 8$ or $7$. If $a_2 = 8$, then $20, 10-9=1, 9-8=1$, impossible. If $a_2 = 7$, then $a_1 = 6$ or $5$. Since $20, 10-9=1, 7-6=1$ or $20, 9-7=2, 7-5=2$, impossible.\n(2) If $a_4 = 11$, then $a_3 = 8$, and we have $a_2 = 7$ and $a_1 = 6$, impossible.\n(3) If $a_4 = 12$, then $a_3 = 7$, and we have $a_2 = 6$ and $a_1 = 5$, impossible.\n(4) If $a_4 = 13$, then $a_3 = 6$, $a_2 = 5$, $a_1 = 4$, impossible.\n(5) If $a_4 = 14$, then $a_3 = 5$, $a_2 = 4$, impossible.\n(6) If $a_4 = 15$, then $a_3 = 4$, $a_2 = 3$, $a_1 = 2$, impossible.\n(7) If $a_4 = 16$, then $a_3 = 3$, $a_2 = 2$, $a_1 = 1$, impossible.\n(8) If $a_4 \\ge 17$, impossible.\nSo $k \\ge 5$. Let $A = \\{1, 3, 5, 9, 16\\}$, then $A$ satisfies the conditions of the problem. Therefore $k_{\\min} = 5$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 28108,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $S$ be the set of all those $2007$-place decimal integers $\\overline{2a_1a_2a_3\\cdots a_{2006}}$ which contain odd number of digit '9' in each sequence $a_1, a_2, a_3, \\cdots, a_{2006}$. The cardinal number of $S$ is",
    "options": [],
    "answer": "1/2*(10^2006 - 8^2006)",
    "solution": "Define $A$ as the number of the elements in $S$, we have\n$$\nA = \\binom{2006}{1} 9^{2005} + \\binom{2006}{3} 9^{2003} + \\dots + \\binom{2006}{2005} 9.\n$$\nOn the other hand,\n$$\n(9+1)^{2006} = \\sum_{k=0}^{2006} \\binom{2006}{k} 9^{2006-k}\n$$\nand\n$$\n(9-1)^{2006} = \\sum_{k=0}^{2006} \\binom{2006}{k} (-1)^k 9^{2006-k}.\n$$\nSo\n$$\n\\begin{aligned}\nA &= \\binom{2006}{1} 9^{2005} + \\binom{2006}{3} 9^{2003} + \\dots + \\binom{2006}{2005} 9 \\\\\n&= \\frac{1}{2}(10^{2006} - 8^{2006}).\n\\end{aligned}\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 28109,
    "subject": "Mathematics (Multi-modal)",
    "question": "$AB$ is a diameter of the circle $O$, the point $C$ lies on the extended line $AB$ produced. A line passing through $C$ intersects with the circle $O$ at points $D$ and $E$. $OF$ is a diameter of the circumcircle $O_1$ of $\\triangle BOD$. Join $CF$ and its extension, which intersects the circle $O_1$ at $G$. Prove that points $O$, $A$, $E$, $G$ are concyclic.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "![](attached_image_1.png)\n\nThus, $G$, $A$, $C$, $D$ are concyclic. Hence,\n$$\n\\angle AGC = \\angle ADC, \\qquad ①\n$$\n$$\n\\angle AGC = \\angle AGO + \\angle OGF = \\angle AGO + \\frac{\\pi}{2}, \\qquad ②\n$$\n$$\n\\angle ADC = \\angle ADB + \\angle BDC = \\angle BDC + \\frac{\\pi}{2}. \\qquad ③\n$$\nCombining ①, ② and ③ yields\n$$\n\\angle AGO = \\angle BDC. \\qquad ④\n$$\nSince $B$, $D$, $E$, $A$ are concyclic, we have\n$$\n\\angle BDC = \\angle EAO. \\qquad ⑤\n$$\nAs $OA = OE$, we obtain\n$$\n\\angle EAO = \\angle AEO. \\qquad ⑥\n$$\nCombining ④, ⑤ and ⑥ implies $\\angle AGO = \\angle AEO$.\nTherefore, $O$, $A$, $E$, $G$ are concyclic.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 28110,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $X$ be a set of 56 elements. Find the least positive integer $n$ such that for any 15 subsets of $X$, if the union of every 7 sets of these subsets contains at least $n$ elements, then there exist 3 of the 15 subsets whose intersection is nonempty. (posed by Leng Gangsong)",
    "options": [],
    "answer": "41",
    "solution": "Suppose there exist 15 subsets of $X$, the union of every 7 of these 15 subsets contains at least 41 elements, and there exists no 3 of the 15 subsets whose intersection is nonempty.\nSince every element of $X$ can only belong 2 of the 15 subsets, we can suppose that every element of $X$ belongs to exactly 2 of the 15 subsets. Otherwise we can add a few more elements to some sets of the 15 subsets, and the condition still holds.\nBy the Pigeonhole Principle, there is a set of the 15 subsets (suppose it is $A$), such that $|A| \\ge \\lfloor \\frac{2 \\times 56}{15} \\rfloor + 1 = 8$. Let the other 14 sets be $A_1, A_2, \\dots, A_{14}$.\nThe union of every 7 sets of $A_1, A_2, \\dots, A_{14}$ contains at least 41 elements. So the total number $y \\le 41C_{14}^7$.\n\nWe use another way to compute the value of $y$.\nFor $a \\in X$, if $a \\notin A$, then $a$ belongs to exactly 2 sets of $A_1, A_2, \\dots, A_{14}$. So $a$ is counted $C_{14}^7 - C_{12}^7$ times. If $a \\in A$, then $a$ belongs to only one set of $A_1, A_2, \\dots, A_{14}$. So $a$ is counted $C_{14}^7 - C_{13}^7$ times. It follows that\n$$\n\\begin{aligned}\n41C_{14}^7 \\le y \\le (56 - |A|)(C_{14}^7 - C_{12}^7) + |A|(C_{14}^7 - C_{13}^7) \\\\\n= 56(C_{14}^7 - C_{12}^7) - |A|(C_{13}^7 - C_{12}^7) \\\\\n\\le 56(C_{14}^7 - C_{12}^7) - 8(C_{13}^7 - C_{12}^7),\n\\end{aligned}\n$$\nthat is, $196 \\le 195$, a contradiction.\n\nNext, we prove $n \\ge 41$.\nIf $n \\le 40$, suppose $X = \\{1, 2, \\dots, 56\\}$. let\n$$\nA_i = \\{i, i+7, i+14, i+21, i+28, i+35, i+42, i+49\\},\n$$\n$$\nB_j = \\{j, j+8, j+16, j+24, j+32, j+40, j+48\\},\n$$\nIt is easy to see that\n$$\n\\begin{aligned}\n|A_i| &= 8 \\quad (i = 1, 2, \\dots, 7), & |A_i \\cap A_j| &= 0 \\quad (1 \\le i < j \\le 7), \\\\\n|B_j| &= 7 \\quad (j = 1, 2, \\dots, 8), & |B_i \\cap B_j| &= 0 \\quad (1 \\le i < j \\le 8), \\\\\n|A_i \\cap B_j| &= 1 \\quad (1 \\le i \\le 7, 1 \\le j \\le 8).\n\\end{aligned}\n$$\nFor every 3 of the 15 subsets, there are 2 sets both being $A_i$, or both being $B_j$. So the intersection of the 3 sets is empty.\nBut for every 7 of the 15 subsets, for example,\n$$\nA_{i_1}, A_{i_2}, \\dots, A_{i_s}, B_{j_1}, B_{j_2}, \\dots, B_{j_t} \\quad (s+t=7),\n$$\nwe have\n$$\n\\begin{aligned}\n& |A_{i_1} \\cup A_{i_2} \\cup \\dots \\cup A_{i_s} \\cup B_{j_1} \\cup B_{j_2} \\cup \\dots \\cup B_{j_t}| \\\\\n& = |A_{i_1}| + |A_{i_2}| + \\dots + |A_{i_s}| + |B_{j_1}| + |B_{j_2}| + \\dots + |B_{j_t}| - st\n\\end{aligned}\n$$\n$$\n\\begin{aligned}\n&= 8s + 7t - st = 8s + 7(7 - s) - s(7 - s) \\\\\n&= (s - 3)^2 + 40 \\ge 40.\n\\end{aligned}\n$$\nSo $n \\ge 41$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 28111,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $k$ be a positive integer no less than $3$ and $\\theta$ be a real number. Prove that, if both $\\cos(k - 1)\\theta$ and $\\cos k\\theta$ are rational numbers, then there exists a positive integer $n > k$, such that both $\\cos(n - 1)\\theta$ and $\\cos n\\theta$ are rational numbers.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "First we prove a lemma.\nLemma  Let $\\alpha$ be a real number. If $\\cos \\alpha$ is rational, then $\\cos m\\alpha$ is rational for any positive integer $m$.\nWe prove by induction on $m$. By $\\cos 2\\alpha = 2\\cos^2\\alpha - 1$, we get that the lemma is true for $m = 2$.\nWe suppose that the lemma is true for $m \\le l$ ($l \\ge 2$).\nSince\n$$\n\\cos(l + 1)\\alpha = 2\\cos l\\alpha \\cdot \\cos \\alpha - \\cos(l - 1)\\alpha,\n$$\nthen we conclude that the lemma is true for $m = l + 1$ and our induction is complete.\nBy the lemma, setting $m = k$, $m = k + 1$ for $\\alpha = k\\theta$, $(k-1)\\theta$, it follows that $\\cos k^2 \\theta$, $\\cos(k^2 - 1)\\theta$ are rational numbers. Since $k^2 > k$, the statement holds.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 28112,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $f(x) = \\sin^4 x - \\sin x \\cos x + \\cos^4 x$, the range of $f(x)$ is ______.",
    "options": [],
    "answer": "[0, 9/8]",
    "solution": "As\n$$\n\\begin{aligned} f(x) &= \\sin^4 x - \\sin x \\cos x + \\cos^4 x \\\\ &= 1 - \\frac{1}{2} \\sin 2x - \\frac{1}{2} \\sin^2 2x, \\end{aligned}\n$$\nwe define $t = \\sin 2x$, then\n$$\nf(x) = g(t) = 1 - \\frac{1}{2}t - \\frac{1}{2}t^2 = \\frac{9}{8} - \\frac{1}{2}\\left(t + \\frac{1}{2}\\right)^2.\n$$\nSo we have\n$$\n\\min_{-1 \\le t \\le 1} g(t) = g(1) = \\frac{9}{8} - \\frac{1}{2} \\times \\frac{9}{4} = 0,\n$$\nand\n$$\n\\max_{-1 \\le t \\le 1} g(t) = g\\left(-\\frac{1}{2}\\right) = \\frac{9}{8} - \\frac{1}{2} \\times 0 = \\frac{9}{8}.\n$$\nHence $0 \\le f(x) \\le \\frac{9}{8}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 28113,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $p$ be a prime number greater than $3$. Prove that there exist integers $a_1, a_2, \\dots, a_t$ that satisfy the following conditions:\n\na.\n$$\n-\\frac{p}{2} < a_1 < a_2 < \\cdots < a_t \\le \\frac{p}{2},\n$$\n\nb.\n$$\n\\frac{p-a_1}{|a_1|} \\cdot \\frac{p-a_2}{|a_2|} \\cdot \\cdots \\cdot \\frac{p-a_t}{|a_t|} = 3^m,\n$$\nwhere $m$ is a positive integer.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "**Proof** By the Division Algorithm, there exist unique integers $q$ and $r$ such that $p = 3q + r$, where $0 < r < 3$.\nTaking $b_0 = r$, then\n$$\n\\frac{p-b_0}{|b_0|} = \\frac{3^{c_0} \\cdot b_1^*}{|b_0|}, \\text{ where } 3 \\nmid b_1^* \\text{ and } 0 < b_1^* < \\frac{p}{2}.\n$$\nTaking $b_1 = \\pm b_1^*$ such that $b_1 \\equiv p \\pmod{3}$, then\n$$\n\\frac{p-b_1}{|b_1|} = \\frac{3^{c_1} \\cdot b_2^*}{b_1^*}, \\text{ where } 3 \\nmid b_2^* \\text{ and } 0 < b_2^* < \\frac{p}{2}.\n$$\nTaking $b_2 = \\pm b_2^*$ such that $b_2 \\equiv p \\pmod{3}$, then\n$$\n\\frac{p-b_2}{|b_2|} = \\frac{3^{c_2} \\cdot b_3^*}{b_2^*}, \\text{ where } 3 \\nmid b_3^* \\text{ and } 0 < b_3^* < \\frac{p}{2}.\n$$\nRepeating this process, we get $b_0, b_1, \\dots, b_p$.\nSince these $p+1$ integers are in the interval $(-\\frac{p}{2}, \\frac{p}{2})$, a certain integer occurs twice. Suppose $b_i = b_j$, $i < j$, and $b_i$, $b_{i+1}$, ..., $b_{j-1}$ are distinct. So,\n$$\n\\frac{p-b_i}{|b_i|} \\cdot \\frac{p-b_{i+1}}{|b_{i+1}|} \\cdots \\frac{p-b_{j-1}}{|b_{j-1}|} = \\frac{3^{c_i} \\cdot b_{i+1}^*}{b_i^*} \\cdot \\frac{3^{c_{i+1}} \\cdot b_{i+2}^*}{b_{i+1}^*} \\cdots \\frac{3^{c_{j-1}} \\cdot b_j^*}{b_{j-1}^*}.\n$$\nSince $b_i = b_j$, then $b_i^* = b_j^*$. So the above expression equals to\n$$\n3^{c_i+c_{i+1}+\\cdots+c_{j-1}} = 3^n, \\quad n > 0.\n$$\nPut $b_i, b_{i+1}, \\dots, b_{j-1}$ in ascending order, as desired.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 28114,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let complex number $z = (a + \\cos \\theta) + (2a - \\sin \\theta)i$. If $|z| \\le 2$ for any $\\theta \\in \\mathbb{R}$, then the range of real number $a$ is ______.",
    "options": [],
    "answer": "[-√5/5, √5/5]",
    "solution": "By the definition given above, we have, for any $\\theta \\in \\mathbb{R}$,\n$$\n\\begin{align*}\n|z| \\le 2 & \\Leftrightarrow (a + \\cos \\theta)^2 + (2a - \\sin \\theta)^2 \\le 4 \\\\\n& \\Leftrightarrow 2a(\\cos \\theta - 2\\sin \\theta) \\le 3 - 5a^2 \\\\\n& \\Leftrightarrow -2\\sqrt{5}a \\sin(\\theta - \\varphi) \\le 3 - 5a^2 \\\\\n& \\Rightarrow 2\\sqrt{5} |a| \\le 3 - 5a^2 \\\\\n& \\Rightarrow |a| \\le \\frac{\\sqrt{5}}{5}.\n\\end{align*}\n$$\nSo the range of $a$ is $\\left[-\\frac{\\sqrt{5}}{5}, \\frac{\\sqrt{5}}{5}\\right]$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 28115,
    "subject": "Mathematics (Multi-modal)",
    "question": "Given a positive integer $n \\ge 2$, let $B_1, B_2, \\dots, B_n$ denote $n$ arbitrary subsets of set $X$, each of which contains exactly two elements. Find the minimum value of $|X|$ such that there exists a subset $Y$ of set $X$ satisfying:\n(a) $|Y| = n$;\n(b) $|Y \\cap B_i| \\le 1$ for $i = 1, 2, \\dots, n$,\nwhere $|A|$ denotes the number of elements of the finite set $A$.",
    "options": [],
    "answer": "2n - 1",
    "solution": "We first prove that $|X| > 2n - 2$. In fact, if $|X| = 2n - 2$, let $X = \\{1, 2, \\dots, 2n - 2\\}$, $B_1 = \\{1, 2\\}$, $B_2 = \\{3, 4\\}$, $\\dots$, $B_{n-1} = \\{2n - 3, 2n - 2\\}$. Since $|Y| = n$, there exist two elements in $Y$ that belong to the same $B_i$, then $|Y \\cap B_i| > 1$, a contradiction.\n\nLet $|X| = 2n - 1$.\nLet $B = \\bigcup_{i=1}^n B_i$, then $|B| = 2n - 1 - z$, where $z$ is the number of subset $X \\setminus B$. Suppose the elements of $X \\setminus B$ are $a_1, a_2, \\dots, a_z$.\n\nIf $z \\ge n - 1$, take $Y = \\{a_1, \\dots, a_{n-1}, d\\}$, and $d \\in B$, as desired.\n\nIf $z < n - 1$, suppose there are $t$ elements that occur once in $B_1, B_2, \\dots, B_n$. Since $\\sum_{i=1}^n |B_i| = 2n$, then\n$$\nt + 2(2n - 1 - z - t) \\le 2n,\n$$\nit follows that $t \\ge 2n - 2 - 2z$. So the elements that occur twice or more than twice in $B_1, B_2, \\dots, B_n$ occur repeatedly by $2n - (2n - 2 - 2z) = 2 + 2z$ times.\n\nConsider the elements that occur once in $B_1, B_2, \\dots, B_n$:\n$b_1, b_2, \\dots, b_t$. Thus, there are at most $\\frac{2+2z}{2} = 1+z$ subsets in $B_1, B_2, \\dots, B_n$ that do not contain the elements $b_1, b_2, \\dots, b_t$. So, there exist $n - (z+1) = n-z-1$ subsets containing at least the elements $b_1, b_2, \\dots, b_t$.\n\nSuppose that $B_1, B_2, \\dots, B_{n-1-z}$ contain the elements $\\tilde{b}_1, \\tilde{b}_2, \\dots, \\tilde{b}_{n-1-z}$ of $b_1, b_2, \\dots, b_t$, respectively. Since\n$$\n2(n - 1 - z) + z = 2n - 2 - z < 2n - 1,\n$$\nthere must exist an element $d$ that is not in $B_1, B_2, \\dots, B_{n-1-z}$ but is in $B_{n-z}, \\dots, B_n$.\n\nWrite $Y = \\{a_1, \\dots, a_z, \\tilde{b}_1, \\tilde{b}_2, \\dots, \\tilde{b}_{n-1-z}, d\\}$, as desired.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 28116,
    "subject": "Mathematics (Multi-modal)",
    "question": "Suppose points $F_1, F_2$ are the left and right foci of the ellipse $\\frac{x^2}{16} + \\frac{y^2}{4} = 1$ respectively, and point $P$ is on line $l$: $x - \\sqrt{3}y + 8 + 2\\sqrt{3} = 0$. When $\\angle F_1PF_2$ reaches the maximum, the value of ratio $\\frac{|PF_1|}{|PF_2|}$ is ______.",
    "options": [],
    "answer": "sqrt(3) - 1",
    "solution": "Euclidean geometry tells us that, $\\angle F_1PF_2$ reaches the maximum only if the circle through points $F_1, F_2, P$ is tangent to the line $l$ at $P$. Now suppose $l$ intercepts the $x$-axis at point $A(-8-2\\sqrt{3}, 0)$. Then $\\angle APF_1 = \\angle AF_2P$, and that means $\\triangle APF_1 \\sim \\triangle AF_2P$. So\n$$\n\\frac{|PF_1|}{|PF_2|} = \\frac{|AP|}{|AF_2|}\n$$\nBy using the power of points theorem, we have\n$$\n|AP|^2 = |AF_1| \\cdot |AF_2|\n$$\nAs $F_1(-2\\sqrt{3}, 0)$, $F_2(2\\sqrt{3}, 0)$, $A(-8-2\\sqrt{3}, 0)$, so\n$$\n|AF_1| = 8, \\quad |AF_2| = 8 + 4\\sqrt{3}.\n$$\nThen we get\n$$\n\\begin{aligned}\n\\frac{|PF_1|}{|PF_2|} &= \\sqrt{\\frac{|AF_1|}{|AF_2|}} = \\sqrt{\\frac{8}{8+4\\sqrt{3}}} \\\\\n&= \\sqrt{4-2\\sqrt{3}} = \\sqrt{3}-1.\n\\end{aligned}",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 28117,
    "subject": "Mathematics (Multi-modal)",
    "question": "For nonnegative real numbers $a$, $b$, $c$ with $a + b + c = 1$, prove that\n$$\n\\sqrt{a + \\frac{(b-c)^2}{4}} + \\sqrt{b} + \\sqrt{c} \\le \\sqrt{3}.\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "**Proof I** Without loss of generality, we may assume that $b \\ge c$.\nWe set $\\sqrt{b} = x + y$ and $\\sqrt{c} = x - y$ for some nonnegative real numbers $x$ and $y$. Hence $b - c = 4xy$ and $a = 1 - 2x^2 - 2y^2$. It follows that\n$$\n\\sqrt{a + \\frac{(b-c)^2}{4}} + \\sqrt{b} + \\sqrt{c} = \\sqrt{1 - 2x^2 - 2y^2 + 4x^2y^2} + 2x.\n$$\n\nNote that $2x = \\sqrt{b} + \\sqrt{c}$, implying that\n$$\n4x^2 = (\\sqrt{b} + \\sqrt{c})^2 \\le 2b + 2c \\le 2,\n$$\nby the AM-GM inequality. Thus, $4x^2y^2 \\le 2y^2$ and\n$$\n1 - 2x^2 - 2y^2 + 4x^2y^2 \\le 1 - 2x^2.\n$$\nSubstituting the last inequality yields\n$$\n\\begin{aligned}\n\\sqrt{a + \\frac{(b-c)^2}{4}} + \\sqrt{b} + \\sqrt{c} &\\le \\sqrt{1-2x^2} + 2x \\\\\n&= \\sqrt{1-2x^2} + x + x \\\\\n&\\le \\sqrt{3},\n\\end{aligned}\n$$\nby the Cauchy-Schwarz inequality.\n\n**Proof II** Let $a = u^2$, $b = v^2$, and $c = w^2$. Then $u^2 + v^2 + w^2 = 1$ and the desired inequality becomes\n$$\n\\sqrt{u^2 + \\frac{(v^2 - w^2)^2}{4}} + v + w \\le \\sqrt{3}.\n$$\n\nNote that\n$$\n\\begin{aligned}\nu^2 + \\frac{(v^2 - w^2)^2}{4} &= 1 - (v^2 + w^2) + \\frac{(v^2 - w^2)^2}{4} \\\\\n&= \\frac{4 - 4(v^2 + w^2) + (v^2 - w^2)^2}{4} \\\\\n&= \\frac{4 - 4(v^2 + w^2) + (v^2 + w^2)^2 - 4v^2w^2}{4} \\\\\n&= \\frac{(2 - v^2 - w^2)^2 - 4v^2w^2}{4} \\\\\n&= \\frac{(2 - v^2 - w^2 - 2vw)(2 - v^2 - w^2 + 2vw)}{4} \\\\\n&= \\frac{[2 - (v+w)^2][2 - (v-w)^2]}{4} \\\\\n&\\le 1 - \\frac{(v+w)^2}{2}.\n\\end{aligned}\n$$\n(Note that $(v+w)^2 \\le 2(v^2+w^2) \\le 2$)\n\nSubstitute the above equation and it gives\n$$\n\\sqrt{1 - \\frac{(v+w)^2}{2}} + v+w \\le \\sqrt{3}.\n$$\nSet $\\frac{v+w}{2} = x$. We can rewrite the above inequality as\n$$\n\\sqrt{1 - 2x^2} + 2x \\le \\sqrt{3},\n$$\nand we can complete the proof as we did in the first proof.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 28118,
    "subject": "Mathematics (Multi-modal)",
    "question": "Given complex numbers $a$, $b$, $c$, let $|a+b| = m$, $|a-b| = n$, and suppose $mn \\neq 0$. Prove that\n$$\n\\max\\{|ac + b|, |a + bc|\\} \\ge \\frac{mn}{\\sqrt{m^2 + n^2}}\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "**Proof I** We have\n$$\n\\begin{aligned}\n\\max\\{|ac + b|, |a + bc|\\} &\\ge \\frac{|b| \\cdot |ac + b| + |a| \\cdot |a + bc|}{|b| + |a|} \\\\\n&\\ge \\frac{|b(ac + b) - a(a + bc)|}{|a| + |b|} \\\\\n&= \\frac{|b^2 - a^2|}{|a| + |b|} \\\\\n&\\ge \\frac{|b + a| \\cdot |b - a|}{\\sqrt{2(|a|^2 + |b|^2)}}.\n\\end{aligned}\n$$\nAs\n$$\nm^2 + n^2 = |a - b|^2 + |a + b|^2 = 2(|a|^2 + |b|^2),\n$$\nwe get\n$$\n\\max\\{|ac + b|, |a + bc|\\} \\ge \\frac{mn}{\\sqrt{m^2 + n^2}}.\n$$\n\n\n**Proof II** Note that\n$$\nac + b = \\frac{1+c}{2}(a+b) - \\frac{1-c}{2}(a-b)\n$$\nand\n$$\na + bc = \\frac{1+c}{2}(a+b) + \\frac{1-c}{2}(a-b).\n$$\nLet $\\alpha = \\frac{1+c}{2}(a+b)$ and $\\beta = \\frac{1-c}{2}(a-b)$. Then\n$$\n\\begin{aligned}\n|ac+b|^2 + |a+bc|^2 &= |\\alpha - \\beta|^2 + |\\alpha + \\beta|^2 \\\\\n&= 2(|\\alpha|^2 + |\\beta|^2).\n\\end{aligned}\n$$\nTherefore\n$$\n\\begin{aligned}\n(\\max\\{|ac+b|, |a+bc|\\})^2 &\\geq |\\alpha|^2 + |\\beta|^2 \\\\\n&= \\left|\\frac{1+c}{2}\\right|^2 m^2 + \\left|\\frac{1-c}{2}\\right|^2 n^2.\n\\end{aligned}\n$$\nNow we only need to prove that\n$$\n\\left|\\frac{1+c}{2}\\right|^2 m^2 + \\left|\\frac{1-c}{2}\\right|^2 n^2 \\geq \\frac{m^2 n^2}{m^2 + n^2},\n$$\nor equivalently\n$$\n\\left|\\frac{1+c}{2}\\right|^2 m^4 + \\left|\\frac{1-c}{2}\\right|^2 n^4 + \\left(\\left|\\frac{1+c}{2}\\right|^2 + \\left|\\frac{1-c}{2}\\right|^2\\right) m^2 n^2 \\geq m^2 n^2.\n$$\nWe have\n$$\n\\begin{aligned}\n& \\left|\\frac{1+c}{2}\\right|^2 m^4 + \\left|\\frac{1-c}{2}\\right|^2 n^4 + \\left(\\left|\\frac{1+c}{2}\\right|^2 + \\left|\\frac{1-c}{2}\\right|^2\\right) m^2 n^2 \\\\\n\\geq & 2 \\left|\\frac{1+c}{2}\\right| \\left|\\frac{1-c}{2}\\right| m^2 n^2 + \\left(\\left|\\frac{1+2c+c^2}{4}\\right| + \\left|\\frac{1-2c+c^2}{4}\\right|\\right) m^2 n^2 \\\\\n= & \\left(\\left|\\frac{1-c^2}{2}\\right| + \\left|\\frac{1+2c+c^2}{4}\\right| + \\left|\\frac{1-2c+c^2}{4}\\right|\\right) m^2 n^2 \\\\\n\\geq & \\left|\\frac{1-c^2}{2} + \\frac{1+2c+c^2}{4} + \\frac{1-2c+c^2}{4}\\right| m^2 n^2 \\\\\n= & m^2 n^2.\n\\end{aligned}\n$$\nThis completes the proof.\n\n\n**Proof III** Since\n$$\n\\begin{aligned}\nm^2 &= |a+b|^2 = (a+b)(\\overline{a}+\\overline{b}) = (a+b)(\\bar{a}+\\bar{b}) \\\\\n&= |a^2|+|b^2|+\\overline{a\\bar{b}}+\\overline{ab}, \\\\\nn^2 &= |a-b|^2 = (a-b)(\\overline{a}-\\overline{b}) = (a-b)(\\bar{a}-\\bar{b}) \\\\\n&= |a^2|+|b^2|-\\overline{a\\bar{b}}-\\overline{ab},\n\\end{aligned}\n$$\nWe get\n$$\n\\begin{aligned}\n|a|^2 + |b|^2 &= \\frac{m^2+n^2}{2}, \\\\\n\\overline{ab} + \\overline{ab} &= \\frac{m^2-n^2}{2}.\n\\end{aligned}\n$$\nLet $c = x + yi$, $x, y \\in \\mathbb{R}$. Then\n$$\n\\begin{aligned}\n& |ac+b|^2 + |a+bc|^2 \\\\\n&= (ac+b)(\\overline{ac}+\\overline{b}) + (a+bc)(\\overline{a}+\\overline{bc}) \\\\\n&= |a|^2|c|^2 + |b|^2 + \\overline{a\\bar{b}c} + \\overline{a\\bar{b}c} + |a|^2 \\\\\n&\\quad + |b|^2|c|^2 + \\overline{a\\bar{b}c} + a\\bar{b}\\bar{c} \\\\\n&= (|c|^2+1)(|a|^2+|b|^2) + (c+\\bar{c})(\\overline{a\\bar{b}}+\\overline{a\\bar{b}}) \\\\\n&= (x^2+y^2+1) \\frac{m^2+n^2}{2} + 2x \\frac{m^2-n^2}{2} \\\\\n&\\ge \\frac{m^2+n^2}{2}x^2 + (m^2-n^2)x + \\frac{m^2+n^2}{2} \\\\\n&= \\frac{m^2+n^2}{2} \\left(x + \\frac{m^2-n^2}{m^2+n^2}\\right)^2 \\\\\n&\\quad - \\frac{m^2+n^2}{2} \\left(\\frac{m^2-n^2}{m^2+n^2}\\right)^2 + \\frac{m^2+n^2}{2} \\\\\n&\\ge \\frac{m^2+n^2}{2} - \\frac{1}{2} \\frac{(m^2-n^2)^2}{m^2+n^2} \\\\\n&= \\frac{2m^2n^2}{m^2+n^2}.\n\\end{aligned}\n$$\nThat is\n$$\n(\\max\\{|ac+b|, |a+bc|\\})^2 \\ge \\frac{m^2n^2}{m^2+n^2}.\n$$\nOr\n$$\n\\max\\{|ac+b|, |a+bc|\\} \\ge \\frac{mn}{\\sqrt{m^2+n^2}}.\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 28119,
    "subject": "Mathematics (Multi-modal)",
    "question": "How many integers $a$ satisfy the condition: for each $a$, the equation $x^3 = a x + a + 1$ with respect to $x$ has roots which are even and $|x| < 1000$.",
    "options": [],
    "answer": "999",
    "solution": "Let $x_0 = 2n$, where $n$ is an integer and $|2n| < 1000$, then $|n| \\le 499$. So we can choose at most $2 \\times 499 + 1 = 999$ numbers, that is, $n \\in \\{-499, -498, \\dots, 0, 1, \\dots, 499\\}$. Substituting $x_0 = 2n$ into the equation, we get $a = \\frac{8n^3 - 1}{2n + 1}$.\n\nSet $f(n) = \\frac{8n^3 - 1}{2n + 1}$, then for any $n_1, n_2 \\in \\{-499, -498, \\dots, 0, 1, \\dots, 499\\}$ ($n_1 \\ne n_2$, $n_1, n_2 \\in \\mathbb{Z}$). If $f(n_1) = f(n_2)$, we can suppose $n_1 = \\frac{x_1}{2}$, $n_2 = \\frac{x_2}{2}$, where $x_1, x_2$ are roots of $x^3 - a x - a - 1 = 0$. Suppose the other root is $x_3$, then according to the sums and products of roots, we obtain\n$$\n\\begin{cases} x_3 = -(x_1 + x_2), \\\\ x_1 x_2 + x_2 x_3 + x_3 x_1 = -a, \\\\ x_1 x_2 x_3 = a + 1, \\end{cases}\n$$\nthat is,\n$$\n\\begin{cases} 4N_1 = -a, \\\\ 8N_2 = a + 1, \\end{cases}\n$$\nwhere $N_1 = -(n_1^2 + n_2^2 + n_1 n_2)$, $N_2 = -n_1 n_2 (n_1 + n_2)$, then $4N_1 + 8N_2 = 1$. Contradiction!\nThus, for any $n_1, n_2 \\in \\{-499, -498, \\dots, 0, 1, \\dots, 499\\}$, we have $f(n_1) \\neq f(n_2)$, which means there are exactly 999 real numbers $a$ that satisfy the condition.\nOur aim is to prove that for any even integer $x$ which satisfies $|x| \\le 998$, the value of $a = \\frac{x^3 - 1}{x + 1}$ is different.\nOn the contrary, if there exist $x_1 \\neq x_2$ that satisfy $\\frac{x_1^3 - 1}{x_1 + 1} = \\frac{x_2^3 - 1}{x_2 + 1}$, where $x_1, x_2$ are even numbers, then\n$$\n(x_1 - x_2)(x_1^2 x_2 + x_1 x_2^2 + x_2^2 + x_1 x_2 + 1) = 0.\n$$\nSince $x_1 \\neq x_2 \\Rightarrow x_1 - x_2 \\neq 0$ and $x_1^2 x_2 + x_1 x_2^2 + x_1^2 + x_2^2 + x_1 x_2$ is an even number, we obtain\n$$\n(x_1 - x_2)(x_1^2 x_2 + x_1 x_2^2 + x_1^2 + x_2^2 + x_1 x_2 + 1) \\neq 0.\n$$\nContradiction!\nThus, there exist 999 real numbers $a$ that satisfy the condition.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 28120,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $AB$ be a chord of circle $O$, $M$ the midpoint of arc $AB$, and $C$ a point outside of the circle $O$. From $C$ draw two tangents to the circle at points $S$, $T$. $MS \\cap AB = E$, $MT \\cap AB = F$. From $E$, $F$ draw a line perpendicular to $AB$, and intersecting $OS$, $OT$ at $X$, $Y$ respectively. Now draw a line from $C$ which intersects the circle $O$ at $P$ and $Q$. Let $Z$ be the circumcenter of $\\triangle PQR$. Prove that $X$, $Y$, $Z$ are collinear.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "**Proof** Refer to the figure, join points $O$ and $M$. Then $OM$ is the perpendicular bisector of $AB$. So $\\triangle XES \\sim \\triangle OMS$, and thus $SX = XE$.\nNow draw a circle with center $X$ whose radius is $XE$. Then the circle $X$ is tangent to chord $AB$ and line $CS$. Draw the circumcircle of $\\triangle PQR$, line $MA$ and line $MC$.\n![](attached_image_1.png)\nIt is easy to see ($\\triangle AMR \\sim \\triangle PMA$ etc)\n$$\nMR \\cdot MP = MA^2 = ME \\cdot MS. \\qquad \\textcircled{1}\n$$\nBy the Power of a Point theorem,\n$$\nCQ \\cdot CP = CS^2. \\qquad \\textcircled{2}\n$$\nSo $M$, $C$ are on the radical axis of circle $Z$ and circle $X$. Thus\n$$ZX \\perp MC.$$\nSimilarly, we have $ZY \\perp MC$.\nSo $X$, $Y$, $Z$ are collinear.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 28121,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $T = \\{1, 2, 3, 4, 5, 6, 7, 8\\}$. Find the number of all nonempty subsets $A$ of $T$ such that $3|S(A)$ and $5 \\nmid S(A)$, where $S(A)$ is the sum of all elements of $A$.",
    "options": [],
    "answer": "70",
    "solution": "Define $S(\\emptyset) = 0$. Let $T_0 = \\{3, 6\\}$, $T_1 = \\{1, 4, 7\\}$, $T_2 = \\{2, 5, 8\\}$. For $A \\subseteq T$, let $A_0 = A \\cap T_0$, $A_1 = A \\cap T_1$, $A_2 = A \\cap T_2$, then\n\n$$\nS(A) = S(A_0) + S(A_1) + S(A_2) \\equiv |A_1| - |A_2| \\pmod{3},\n$$\n\nSo $3|S(A)$ if and only if $|A_1| \\equiv |A_2| \\pmod{3}$. It follows that\n$$\n(|A_1|, |A_2|) = (0, 0), (0, 3), (3, 0), (3, 3), (1, 1), (2, 2).\n$$\n\nThe number of nonempty subsets $A$ so that $3|S(A)$ is\n$$\n2^2 \\binom{3}{0} \\binom{3}{0} + \\binom{3}{0} \\binom{3}{3} + \\binom{3}{3} \\binom{3}{0} + \\binom{3}{3} \\binom{3}{3} \\\\\n+ \\binom{3}{1} \\binom{3}{1} + \\binom{3}{2} \\binom{3}{2} - 1 = 87.\n$$\n\nIf $3|S(A)$ and $5|S(A)$, then $15|S(A)$. Since $S(T) = 36$, so the value $S(A)$ is 15 or 30 (if $3|S(A)$ and $5|S(A)$).\n\nFurthermore,\n$$\n\\begin{aligned}\n15 &= 8+7 = 8+6+1 = 8+5+2 = 8+4+3 \\\\\n&= 8+4+2+1 = 7+6+2 = 7+5+3 \\\\\n&= 7+5+2+1 = 7+4+3+1 = 6+5+4 \\\\\n&= 6+5+3+1 = 6+4+3+2 \\\\\n&= 5+4+3+2+1,\n\\end{aligned}\n$$\n$$\n36 - 30 = 6 = 5 + 1 = 4 + 2 = 3 + 2 + 1.\n$$\nSo the number of $A$ such that $3|S(A)$, $5|S(A)$, and $A \\neq \\emptyset$ is 17.\n\nThe answer is $87 - 17 = 70$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 28122,
    "subject": "Mathematics (Multi-modal)",
    "question": "A positive integer $m$ is called good, if there is a positive integer $n$ such that $m$ is the quotient of $n$ over the number of positive integer divisors of $n$ (including 1 and $n$ itself). Prove that $1, 2, \\ldots, 17$ are good numbers and that $18$ is not a good number.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "For positive integer $n$, let $d(n)$ denote the number of positive divisors of $n$ (including $1$ and $n$ itself).\n\nFirstly, note that $1$ and $2$ are good, since $1 = \\frac{2}{d(2)}$ and\n$$\n2 = \\frac{8}{d(8)}.\n$$\n\nSecondly, we note that if $p$ is an odd prime, then $p$ is good. This is because $d(8p) = 8$. In particular, $3, 5, 7, 11, 13, 17$ are good numbers.\n\nThirdly, we note that if $p$ is an odd prime, then $2p$ is good. This is because $d(2^2 \\cdot 3^2 p) = 3 \\cdot 3 \\cdot 2$. In particular, $6, 10, 14$ are good numbers.\n\nFourthly, we note that\n$$\n4 = \\frac{36}{d(36)}, \\quad 8 = \\frac{96}{d(96)}, \\quad 9 = \\frac{108}{d(108)},\n$$\n$$\n12 = \\frac{240}{d(240)}, \\quad 15 = \\frac{360}{d(360)}, \\quad 16 = \\frac{128}{d(128)}.\n$$\nThus, the numbers $1, 2, \\ldots, 17$ are good.\n\nFinally, we prove that $18$ is not good. We approach indirectly by assuming that $18 = \\frac{n}{d(n)}$ or $n = 18 d(n)$ for $n = 2^a \\cdot 3^{b+1} \\cdot p_1^{k_1} \\cdots p_m^{k_m}$ (where $p_1 < \\cdots < p_m$ are prime numbers greater than $3$ and $a, b, k_1, \\cdots, k_m$ are positive integers); that is,\n$$\n2^{a-1} \\cdot 3^{b-1} \\cdot p_1^{k_1} \\cdots p_m^{k_m} = (a+1)(b+2)(k_1+1)\\cdots(k_m+1). \\qquad \\textcircled{1}\n$$\nFor every odd prime $p$ and every positive integer $k$, we can show (by an easy induction on $k$) that\n$$\np^k > k+1. \\qquad \\textcircled{2}\n$$\nCombining the last two relations, we deduce that\n$$\n2^{a-1} \\cdot 3^{b-1} < (a+1)(b+2)\n$$\nor\n$$\nf(a) = \\frac{2^{a-1}}{a+1} < \\frac{b+2}{3^{b-1}} = g(b).\n$$\nIt is easy to prove that $f(1) = \\frac{1}{2}$, $f(2) = \\frac{2}{3}$, $f(3) = 1$, $f(4) = \\frac{8}{5}$, $f(5) = \\frac{16}{6}$, and $f(a) \\ge \\frac{32}{7} > 4$ for $a \\ge 6$. It is also easy to prove that $g(1) = 3$, $g(2) = \\frac{4}{3}$, $g(3) < \\frac{5}{9}$, and $g(b) < \\frac{2}{9}$ for $b \\ge 4$. Thus $\\textcircled{1}$ holds only if $b \\le 3$.\n\nIf $b=3$, then $(a, b) = (1, 3)$, and $\\textcircled{1}$ becomes\n$$\n9p_1^{k_1} \\cdots p_m^{k_m} = 10(k_1 + 1)\\cdots(k_m + 1),\n$$\nimplying that $p_1 = 5$. Since $\\textcircled{2}$ and $\\frac{p_1^{k_1}}{k_1+1} = \\frac{5^{k_1}}{k_1+1} \\ge \\frac{5}{2}$ for positive integer $k_1 \\ge 1$, we can easily see that there is no solution in this case.\n\nIf $b=2$, then $(a, b) = (1, 2)$, $(2, 2)$, $(3, 2)$, and $\\textcircled{1}$ becomes\n$$\n3 \\cdot 2^{a-1} p_1^{k_1} \\cdots p_m^{k_m} = 4(a+1)(k_1+1) \\cdots (k_m+1).\n$$\nWe deduce that $4$ divides $2^{a-1}$ or $a \\ge 3$. Hence $a$ must be equal to $3$. But then $4(a+1) = 16$ divides $2^{a-1}$, which is impossible.\n\nIf $b=1$, then $(a, b) = (1, 1)$, $(2, 1)$, $(3, 1)$, $(4, 1)$, $(5, 1)$, and $\\textcircled{1}$ becomes\n$$\n2^{a-1} p_1^{k_1} \\cdots p_m^{k_m} = 3(a+1)(k_1+1) \\cdots (k_m+1),\n$$\nwhich is impossible since $p_i$ are primes greater than $3$.\n\nIn all the cases, we cannot find $n$ satisfying the condition $18 = \\frac{n}{d(n)}$; that is, $18$ is not a good number.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 28123,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $\\{a_n\\}$ be an arithmetic progression with common difference $d$ ($d \\neq 0$) and $\\{b_n\\}$ be a geometric progression with common ratio $q$, where $q$ is a positive rational number less than $1$. If $a_1 = d$, $b_1 = d^2$ and $\\frac{a_1^2 + a_2^2 + a_3^2}{b_1 + b_2 + b_3}$ is a positive integer, then $q$ equals ______.",
    "options": [],
    "answer": "1/2",
    "solution": "As\n$$\n\\begin{aligned}\n\\frac{a_1^2 + a_2^2 + a_3^2}{b_1 + b_2 + b_3} &= \\frac{a_1^2 + (a_1+d)^2 + (a_1+2d)^2}{b_1 + b_1q + b_1q^2} \\\\\n&= \\frac{14}{1+q+q^2} = m\n\\end{aligned}\n$$\nis a positive integer, we get $1 + q + q^2 = \\frac{14}{m}$. Then\n$$\nq = -\\frac{1}{2} + \\sqrt{\\frac{1}{4} + \\frac{14}{m} - 1} = -\\frac{1}{2} + \\sqrt{\\frac{56-3m}{4m}}\n$$\nSince $q$ is a positive rational number less than $1$, we have $1 < \\frac{14}{m} < 3$, i.e. $5 \\le m \\le 13$, and $\\frac{56-3m}{4m}$ is the square of a rational number. We can verify that only $m = 8$ meets the required. That means $q = \\frac{1}{2}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 28124,
    "subject": "Mathematics (Multi-modal)",
    "question": "Given $f(x) = \\frac{\\sin(\\pi x) - \\cos(\\pi x) + 2}{\\sqrt{x}}$ for $\\frac{1}{4} \\le x \\le \\frac{5}{4}$, the minimum of $f(x)$ is ______.",
    "options": [],
    "answer": "4√5/5",
    "solution": "By rewriting $f(x)$, we have\n$$\nf(x) = \\frac{\\sqrt{2}\\sin\\left(\\pi x - \\frac{\\pi}{4}\\right) + 2}{\\sqrt{x}}\n$$\nfor $\\frac{1}{4} \\le x \\le \\frac{5}{4}$. Define $g(x) = \\sqrt{2}\\sin(\\pi x - \\frac{\\pi}{4})$, where $\\frac{1}{4} \\le x \\le \\frac{5}{4}$. Then $g(x) \\ge 0$, and $g(x)$ is monotone increasing on $[\\frac{1}{4}, \\frac{3}{4}]$, and monotone decreasing on $[\\frac{3}{4}, \\frac{5}{4}]$.\n\nFurther, the graph of $y = g(x)$ is symmetric about $x = \\frac{3}{4}$, i.e. for any $x_1 \\in [\\frac{1}{4}, \\frac{3}{4}]$ there exists $x_2 \\in [\\frac{3}{4}, \\frac{5}{4}]$ such that $g(x_2) = g(x_1)$. Then\n$$\nf(x_1) = \\frac{g(x_1) + 2}{\\sqrt{x_1}} = \\frac{g(x_2) + 2}{\\sqrt{x_1}} \\ge \\frac{g(x_2) + 2}{\\sqrt{x_2}} = f(x_2).\n$$\nOn the other hand, $f(x)$ is monotone decreasing on $[\\frac{3}{4}, \\frac{5}{4}]$. Therefore $f(x) \\ge f(\\frac{5}{4}) = \\frac{4\\sqrt{5}}{5}$. That means the minimum value of $f(x)$ on $[\\frac{1}{4}, \\frac{5}{4}]$ is $\\frac{4\\sqrt{5}}{5}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 28125,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $a_n = \\sum_{k=1}^{n} \\frac{1}{k(n+1-k)}$. Prove that $a_{n+1} < a_n$ for $n \\ge 2$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "As\n$$\n\\frac{1}{k(n+1-k)} = \\frac{1}{n+1}\\left(\\frac{1}{k} + \\frac{1}{n+1-k}\\right),\n$$\nwe get $a_n = \\frac{2}{n+1} \\sum_{k=1}^{n} \\frac{1}{k}$. Then for $n \\ge 2$ we have\n$$\n\\begin{aligned}\n\\frac{1}{2}(a_n - a_{n+1}) &= \\frac{1}{n+1} \\sum_{k=1}^{n} \\frac{1}{k} - \\frac{1}{n+2} \\sum_{k=1}^{n+1} \\frac{1}{k} \\\\\n&= \\left(\\frac{1}{n+1} - \\frac{1}{n+2}\\right) \\sum_{k=1}^{n} \\frac{1}{k} - \\frac{1}{(n+1)(n+2)} \\\\\n&= \\frac{1}{(n+1)(n+2)} \\left(\\sum_{k=1}^{n} \\frac{1}{k} - 1\\right) \\\\\n&> 0.\n\\end{aligned}\n$$\nThat means $a_{n+1} < a_n$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 28126,
    "subject": "Mathematics (Multi-modal)",
    "question": "Suppose line $l$ through point $(0, 1)$ and curve $C: y = x + \\frac{1}{x}$ ($x > 0$) intersect at two different points $M$ and $N$. Find the locus of the intersection points of two tangent lines of curve $C$ at $M$ and $N$ respectively.",
    "options": [],
    "answer": "All points with x = 2 and 2 < y < 2.5; equivalently, the open segment from (2, 2) to (2, 2.5).",
    "solution": "Denote the coordinates of $M$ and $N$ as $(x_1, y_1)$ and $(x_2, y_2)$ respectively. Denote the tangent lines of $C$ at $M$ and $N$ by $l_1$ and $l_2$ respectively, with their intersection point being $P(x_p, y_p)$. Suppose the slope ratio of line $l$ is $k$. Then we can write the equation of $l$ as $y = kx + 1$.\n\nEliminating $y$ from\n$$\n\\begin{cases} y = x + \\frac{1}{x}, \\\\ y = kx + 1, \\end{cases}\n$$\nwe get $x + \\frac{1}{x} = kx + 1$, i.e. $(k-1)x^2 + x - 1 = 0$. By the assumption, we know that the equation has two distinctive real roots, $x_1$ and $x_2$, on $(0, +\\infty)$. Then $k \\neq 1$, and\n$$\n\\Delta = 1 + 4(k - 1) > 0,\n$$\n$$\nx_1 + x_2 = \\frac{1}{1-k} > 0,\n$$\n$$\nx_1 x_2 = \\frac{1}{1-k} > 0.\n$$\nFrom the above we get $\\frac{3}{4} < k < 1$.\n\nWe find the derivative of $y = x + \\frac{1}{x}$ as $y' = 1 - \\frac{1}{x^2}$. Then $y'|_{x=x_1} = 1 - \\frac{1}{x_1^2}$ and $y'|_{x=x_2} = 1 - \\frac{1}{x_2^2}$. Therefore, the equation of line $l_1$ is\n$$\ny - y_1 = \\left(1 - \\frac{1}{x_1^2}\\right)(x - x_1)\n$$\nor\n$$\ny - \\left(x_1 + \\frac{1}{x_1}\\right) = \\left(1 - \\frac{1}{x_1^2}\\right)(x - x_1).\n$$\nAfter simplification, we get\n$$\ny = \\left(1 - \\frac{1}{x_1^2}\\right)x + \\frac{2}{x_1}\n$$\nIn the same way, we get the equation of $l_2$,\n$$\ny = \\left(1 - \\frac{1}{x_2^2}\\right)x + \\frac{2}{x_2}\n$$\nBy subtracting, we get\n$$\n\\left(\\frac{1}{x_2^2} - \\frac{1}{x_1^2}\\right)x_p + \\frac{2}{x_1} - \\frac{2}{x_2} = 0.\n$$\nSince $x_1 \\neq x_2$, we have\n$$\nx_p = \\frac{2x_1x_2}{x_1 + x_2}\n$$\nSubstituting $x_1 + x_2$ and $x_1 x_2$ from above, we obtain $x_p = 2$.\n\nBy adding, we obtain\n$$\n2y_p = \\left(2 - \\left(\\frac{1}{x_1^2} + \\frac{1}{x_2^2}\\right)\\right)x_p + 2\\left(\\frac{1}{x_1} + \\frac{1}{x_2}\\right)\n$$\nwhere\n$$\n\\frac{1}{x_1} + \\frac{1}{x_2} = \\frac{x_1 + x_2}{x_1 x_2} = 1,\n$$\n$$\n\\begin{aligned}\n\\frac{1}{x_1^2} + \\frac{1}{x_2^2} &= \\frac{x_1^2 + x_2^2}{x_1^2 x_2^2} = \\frac{(x_1 + x_2)^2 - 2x_1x_2}{x_1^2 x_2^2} \\\\\n&= \\left(\\frac{x_1 + x_2}{x_1 x_2}\\right)^2 - \\frac{2}{x_1 x_2} = 2k - 1.\n\\end{aligned}\n$$\nSubstituting it into the previous equation, we have $2y_p = (3 - 2k)x_p + 2$. Since $x_p = 2$, then $y_p = 4 - 2k$. As $\\frac{3}{4} < k < 1$, we get $2 < y_p < \\frac{5}{2}$.\n\nTherefore, the locus of point $P$ is the segment between $(2, 2)$ and $(2, 2.5)$ (not including the endpoints).",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 28127,
    "subject": "Mathematics (Multi-modal)",
    "question": "Suppose $f(x+2\\pi) = f(x)$ for any $x \\in \\mathbb{R}$. Prove: there are $f_i(x)$ ($i = 1, 2, 3, 4$) such that\n\n(1) $f_i(x)$ ($i = 1, 2, 3, 4$) is an even function, and $f_i(x + \\pi) = f_i(x)$ for any $x \\in \\mathbb{R}$;\n\n(2) $f(x) = f_1(x) + f_2(x)\\cos x + f_3(x)\\sin x + f_4(x)\\sin 2x$ for any $x \\in \\mathbb{R}$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $g(x) = \\frac{f(x)+f(-x)}{2}$, and $h(x) = \\frac{f(x)-f(-x)}{2}$.\nThen $f(x) = g(x) + h(x)$, $g(x)$ is an even function, $h(x)$ is an odd function, and $g(x+2\\pi) = g(x)$, $h(x+2\\pi) = h(x)$ for any $x \\in \\mathbb{R}$.\n\nDefine\n$$\nf_1(x) = \\frac{g(x) + g(x + \\pi)}{2},\n$$\n$$\nf_2(x) = \\begin{cases} \\frac{g(x) - g(x + \\pi)}{2\\cos x}, & x \\neq k\\pi + \\frac{\\pi}{2}, \\\\ 0, & x = k\\pi + \\frac{\\pi}{2}, \\end{cases}\n$$\n$$\nf_3(x) = \\begin{cases} \\frac{h(x) - h(x + \\pi)}{2\\sin x}, & x \\neq k\\pi, \\\\ 0, & x = k\\pi, \\end{cases}\n$$\n$$\nf_4(x) = \\begin{cases} \\frac{h(x) + h(x + \\pi)}{2\\sin 2x}, & x \\neq \\frac{k\\pi}{2}, \\\\ 0, & x = \\frac{k\\pi}{2}, \\end{cases}\n$$\nwhere $k$ is an arbitrary integer. It is easy to check that $f_i(x)$ ($i = 1, 2, 3, 4$) satisfy (1).\n\nNext we prove that $f_1(x) + f_2(x)\\cos x = g(x)$ for any $x \\in \\mathbb{R}$. When $x \\neq k\\pi + \\frac{\\pi}{2}$, it is obviously true. When $x = k\\pi + \\frac{\\pi}{2}$, we have\n$$\nf_1(x) + f_2(x) \\cos x = f_1(x) = \\frac{g(x) + g(x + \\pi)}{2},\n$$\nand\n$$\n\\begin{aligned}\ng(x + \\pi) &= g\\left(k\\pi + \\frac{3\\pi}{2}\\right) = g\\left(k\\pi + \\frac{3\\pi}{2} - 2(k+1)\\pi\\right) \\\\\n&= g\\left(-k\\pi - \\frac{\\pi}{2}\\right) = g\\left(k\\pi + \\frac{\\pi}{2}\\right) = g(x).\n\\end{aligned}\n$$\nThe proof is complete.\n\nFurther we prove that $f_3(x)\\sin x + f_4(x)\\sin 2x = h(x)$ for any $x \\in \\mathbb{R}$. When $x \\neq \\frac{k\\pi}{2}$, it is obviously true. When $x = k\\pi$, we have\n$$\nh(x) = h(k\\pi) = h(k\\pi - 2k\\pi) = h(-k\\pi) = -h(k\\pi).\n$$\nThat means $h(x) = h(k\\pi) = 0$. In this case,\n$$\nf_3(x)\\sin x + f_4(x)\\sin 2x = 0.\n$$\n$$\n\\text{Therefore } h(x) = f_3(x)\\sin x + f_4(x)\\sin 2x.\n$$\nWhen $x = k\\pi + \\frac{\\pi}{2}$, we have\n$$\n\\begin{aligned}\nh(x + \\pi) &= h\\left(k\\pi + \\frac{3\\pi}{2}\\right) = h\\left(k\\pi + \\frac{3\\pi}{2} - 2(k+1)\\pi\\right) \\\\\n&= h\\left(-k\\pi - \\frac{\\pi}{2}\\right) = -h\\left(k\\pi + \\frac{\\pi}{2}\\right) = -h(x).\n\\end{aligned}\n$$\nSo\n$$\nf_3(x)\\sin x = \\frac{h(x) - h(x + \\pi)}{2} = h(x).\n$$\nFurthermore, $f_4(x)\\sin 2x = 0$. Therefore\n$$\nh(x) = f_3(x)\\sin x + f_4(x)\\sin 2x.\n$$\nThis completes the proof.\n\nIn conclusion, $f_i(x)$ ($i = 1, 2, 3, 4$) satisfy (2).",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 28128,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABC$ be an acute triangle. Points $D$, $E$ and $F$ lie on segments $BC$, $CA$ and $AB$ respectively, and each of the three segments $AD$, $BE$ and $CF$ contains the circumcenter of $ABC$. Prove that if any two of the ratios\n$$\n\\frac{BD}{DC}, \\frac{CE}{EA}, \\frac{AF}{FB}, \\frac{BF}{FA}, \\frac{AE}{EC}, \\frac{CD}{DB}\n$$\nare integers, then triangle $ABC$ is isosceles.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "**Proof I** Note that there are $\\binom{6}{2} = 15$ possible pairs of ratios among the six given in the problem statement. These pairs are of two types: (i) Three of these pairs are reciprocal pairs involving segments from just one side of triangle $ABC$. (ii) The other 12 pairs involve segments from two sides of the triangle. We first consider the former case.\n\na. If $\\frac{CD}{DB}$ and $\\frac{BD}{DC}$ are both integers, then both of these ratios must be 1 and $BD = DC$. Then in triangle $ABC$, $AD$ is the median from $A$ and $D$, because $AD$ contains the circumcenter, is also the perpendicular bisector of segment $BC$. It then follows that $AB = AC$ and the triangle is isosceles. Similarly, if $\\frac{CE}{EA}$ and $\\frac{AE}{EC}$ are both integers or $\\frac{AF}{FB}$ and $\\frac{BF}{FA}$ are both integers, then triangle $ABC$ is isosceles.\n\nb. Let $O$ be the circumcenter of triangle $ABC$, and let\n\n$\\angle CAB = \\alpha$, $\\angle ABC = \\beta$, and $\\angle BCA = \\gamma$. We show that any of the ratios can be written in the form $\\frac{\\sin 2x}{\\sin 2y}$ where $x$ and $y$ are two of $\\alpha, \\beta, \\gamma$. Since $ABC$ is acute, $0^\\circ < \\alpha, \\beta, \\gamma < 90^\\circ$ and $O$ lies in the interior. Hence $\\angle AOB = 2\\gamma$, $\\angle BOC = 2\\alpha$, and $\\angle COA = 2\\beta$. Applying the sine rule to triangles $BOD$ and $COD$ gives\n$$\n\\frac{BD}{\\sin \\angle BOD} = \\frac{BO}{\\sin \\angle BDO}\n$$\nand\n$$\n\\frac{CD}{\\sin \\angle COD} = \\frac{CO}{\\sin \\angle CDO'}\n$$\nNext note that $BO = CO$ and that\n$$\n\\begin{align*}\n\\angle BDO + \\angle CDO &= 180^\\circ \\\\\n&= \\angle BOD + \\angle AOB \\\\\n&= \\angle COD + \\angle AOC.\n\\end{align*}\n$$\nIt follows that\n$$\n\\frac{BD}{\\sin 2\\gamma} = \\frac{BD}{\\sin \\angle BOD} = \\frac{CD}{\\sin \\angle COD} = \\frac{CD}{\\sin 2\\beta'}\n$$\ngiving $\\frac{BD}{CD} = \\frac{\\sin 2\\gamma}{\\sin 2\\beta}$. Similarly, $\\frac{CE}{EA} = \\frac{\\sin 2\\alpha}{\\sin 2\\gamma}$ and $\\frac{AF}{FB} = \\frac{\\sin 2\\beta}{\\sin 2\\alpha}$.\n\nNow assume that one of the twelve type (ii) pairs of ratios consists of two integers. Then there are positive integers $m$ and $n$ (with $m \\le n$) such that\n$$\n\\sin 2x = m \\sin 2z \\quad \\text{and} \\quad \\sin 2y = n \\sin 2z\n$$\nor\n$$\n\\sin 2z = m \\sin 2x \\quad \\text{and} \\quad \\sin 2z = n \\sin 2y\n$$\nfor some choice of $x, y, z$ with $\\{x, y, z\\} = \\{\\alpha, \\beta, \\gamma\\}$.\n\nWithout loss of generality we may assume that\n$$\n\\sin 2\\alpha = m \\sin 2\\gamma \\quad \\text{and} \\quad \\sin 2\\beta = n \\sin 2\\gamma\n$$\nor\n$$\n\\sin 2\\gamma = m \\sin 2\\alpha \\quad \\text{and} \\quad \\sin 2\\gamma = n \\sin 2\\beta \\qquad \\textcircled{1}\n$$\nfor some positive integers $m$ and $n$.\n\nNote that there is a triangle with angles $180^\\circ - 2\\alpha$, $180^\\circ - 2\\beta$, and $180^\\circ - 2\\gamma$. (It is easy to check that each of these angles is in the interval $(0^\\circ, 180^\\circ)$ and that they sum to $180^\\circ$.) Furthermore, a triangle\n![](attached_image_1.png)\nwith these angles can be constructed by drawing the tangents to the circumcircle of $ABC$ at each of $A$, $B$ and $C$. Denote this triangle by $A_1B_1C_1$ where $A_1$ is the intersection of the tangents at $B$ and $C$, $B_1$ is the intersection of the tangents at $C$ and $A$, and $C_1$ is the intersection of the tangents at $A$ and $B$. Applying the sine rule to triangle $A_1B_1C_1$ and by $\\textcircled{1}$ we find\n$$\n\\begin{aligned} A_1B_1 : B_1C_1 : C_1A_1 &= \\sin \\angle C_1 : \\sin \\angle A_1 : \\sin \\angle B_1 \\\\\n&= \\sin 2\\gamma : \\sin 2\\alpha : \\sin 2\\beta, \\end{aligned}\n$$\nthat is,\n$$\nA_1B_1 : B_1C_1 : C_1A_1 = 1 : m : n\n$$\nor\n$$\nA_1B_1 : B_1C_1 : C_1A_1 = mn : n : m. \\qquad \\textcircled{2}\n$$\nBy the triangle inequality, it follows that $1 + m < n$ (that is, $m = n$) or $n + m > nm$ (that is, $(n-1)(m-1) < 1$ and $m = 1$). We deduce that either $\\sin 2\\alpha = \\sin 2\\beta$ or $\\sin 2\\gamma = \\sin 2\\alpha$. But then either $\\frac{AF}{FB} = \\frac{BF}{FA}$ or $\\frac{CE}{EA} = \\frac{AE}{EC}$, by case (a), triangle $ABC$ is isosceles.\n\n\n**Proof II** (We maintain the notations used in Proof I.) We only consider those 12 pairs of ratios of type (ii). Without loss of generality we may assume that each of the following sets $\\{\\frac{BD}{DC}, \\frac{CD}{DB}\\}$ and $\\{\\frac{AF}{FB}, \\frac{BF}{FA}\\}$ has an element taking integer values. By symmetry, we consider three cases:\n\na. In this case, we assume that $\\frac{BD}{DC} = m$ and $\\frac{BF}{FA} = n$ for some positive integers $m$ and $n$. Applying Menelaus's theorem to line $AOD$ and triangle $BCF$ yields\n$$\n\\frac{AO}{OD} \\cdot \\frac{DC}{CB} \\cdot \\frac{BF}{FA} = 1\n$$\nor\n$$\n\\frac{AO}{OD} = \\frac{CB}{DC} \\cdot \\frac{FA}{BF} = \\frac{m+1}{n}.\n$$\nLikewise, applying Menelaus's theorem to line $COF$ and triangle $BAD$ yields $\\frac{CO}{OF} = \\frac{n+1}{m}$.\n\nSince triangle $ABC$ is acute, $O$ lies on segment $AD$ and $CF$ with $AO > OD$ and $CO > OF$. Hence $m+1 > n$ and $n+1 > m$, implying that $m-1 < n < m+1$. Since $m$ and $n$ are integers, we must have $m=n$. It is then not difficult to see that $A$ and $C$ are symmetric with respect to line $BF$ and triangle $ABC$ is isosceles with $AB = CB$.\n\nb. In this case, we assume that $\\frac{CD}{DB} = m$ and $\\frac{AF}{FB} = n$ for some positive integers $m$ and $n$. Applying Ceva's theorem gives\n$$\n\\frac{AF}{FB} \\cdot \\frac{BD}{DC} \\cdot \\frac{CE}{EA} = 1 \\quad \\text{or} \\quad \\frac{CE}{EA} = \\frac{m}{n}.\n$$\nApplying Menelaus's theorem to line $BOE$ and triangle $ACF$ yields\n$$\n\\frac{BO}{OE} \\cdot \\frac{EC}{CA} \\cdot \\frac{AF}{FB} = 1\n$$\nor\n$$\n\\frac{BO}{OE} = \\frac{FB}{AF} \\cdot \\frac{CA}{EC} = \\frac{m+n}{mn}.\n$$\nSince triangle $ABC$ is acute, $O$ lies on segment $BE$ and $EO < BO$. Hence $m+n \\ge mn$ or $(m-1)(n-1) \\le 1$. Since $m$ and $n$ are positive integers, we deduce that one of $m$ and $n$ is equal to 1; that is, either $AF = FB$ or $BD = DC$. In either case, triangle $ABC$ is isosceles.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 28129,
    "subject": "Mathematics (Multi-modal)",
    "question": "Suppose real number $a$ satisfies $|2x - a| + |3x - 2a| \\ge a^2$ for any $x \\in \\mathbb{R}$. Then $a$ lies exactly in ( ).\n\n(A) $\\left[ -\\frac{1}{3}, \\frac{1}{3} \\right]$\n(B) $\\left[ -\\frac{1}{2}, \\frac{1}{2} \\right]$\n(C) $\\left[ -\\frac{1}{4}, \\frac{1}{3} \\right]$\n(D) $[-3, 3]$",
    "options": [],
    "answer": "A",
    "solution": "Let $x = \\frac{2}{3}a$. Then we have $|a| \\le \\frac{1}{3}$. Therefore (B) and (D) are excluded. By symmetry, (C) is also excluded. Then only (A) can be correct.\n\nIn general, for any $k \\in \\mathbb{R}$, let $x = \\frac{1}{2}ka$. Then the original inequality becomes\n$$\n|a| \\cdot |k-1| + \\frac{3}{2} |a| \\cdot \\left|k - \\frac{4}{3}\\right| \\ge |a|^2.\n$$\nThis is equivalent to\n$$\n|a| \\le |k-1| + \\frac{3}{2} \\left|k - \\frac{4}{3}\\right|.\n$$\nWe have\n$$\n|k-1| + \\frac{3}{2} \\left|k - \\frac{4}{3}\\right| = \\begin{cases} \\frac{5}{2}k - 3, & k \\ge \\frac{4}{3}, \\\\ 1 - \\frac{1}{2}k, & 1 \\le k < \\frac{4}{3}, \\\\ 3 - \\frac{5}{2}k, & k < 1. \\end{cases}\n$$\nSo\n$$\n\\min_{k \\in \\mathbb{R}} \\left\\{ |k-1| + \\frac{3}{2} \\left| k - \\frac{4}{3} \\right| \\right\\} = \\frac{1}{3}.\n$$\nThe inequality is reduced to $|a| \\le \\frac{1}{3}$. Answer: A.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 28130,
    "subject": "Mathematics (Multi-modal)",
    "question": "Consider five points $A$, $B$, $C$, $D$ and $E$ such that $ABCD$ is a parallelogram and $BCED$ is a cyclic quadrilateral. Let $l$ be a line passing through $A$. Suppose that $l$ intersects the interior of the segment $DC$ at $F$ and intersects line $BC$ at $G$. Suppose also that $EF = EG = EC$. Prove that $l$ is the bisector of $\\angle DAB$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Draw the altitudes of two isosceles triangles $EGC$ and $ECF$ as in the figure.\nIn view of the given condition, it is easy to see that $\\triangle ADF \\sim \\triangle GCF$. Hence\n![](attached_image_1.png)\n\n$$\n\\begin{aligned}\n\\frac{AD}{GC} &= \\frac{DF}{CF} \\Rightarrow \\frac{BC}{CG} = \\frac{DF}{CF} \\Rightarrow \\frac{BC}{CL} = \\frac{DF}{CK} \\\\\n&\\Rightarrow \\frac{BC + CL}{CL} = \\frac{DF + FK}{CK} \\\\\n&\\Rightarrow \\frac{BL}{CL} = \\frac{DK}{CK} \\\\\n&\\Rightarrow \\frac{BL}{DK} = \\frac{CL}{CK}.\n\\end{aligned}\n\\quad ①\n$$\n\nSince $BCED$ is a cyclic quadrilateral, $\\angle LBE = \\angle EDK$, this yields $\\triangle BLE \\sim \\triangle DKE$, where both are right-angled triangles.\n$$\n\\text{So} \\qquad \\frac{BL}{DK} = \\frac{EL}{EK}. \\qquad ②\n$$\n\nIn view of ① and ②, $\\frac{CL}{CK} = \\frac{EL}{EK}$, this means $\\triangle CLE \\sim \\triangle CKE$.\nThus\n$$\n\\frac{CL}{CK} = \\frac{CE}{CE} = 1,\n$$\ni.e. $CL = CK \\Rightarrow CG = CF$.\nIt is intuitively obvious that $\\angle BAG = \\angle GAD$. Hence $l$ is the bisector.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 28131,
    "subject": "Mathematics (Multi-modal)",
    "question": "**Prove the following statements:**\n\n(1) If $2n-1$ is a prime number, then for any group of distinct positive integers $a_1, a_2, \\dots, a_n$ there exist $i, j \\in \\{1, 2, \\dots, n\\}$ such that $\\frac{a_i+a_j}{(a_i, a_j)} \\ge 2n-1$.\n\n(2) If $2n-1$ is a composite number, then there exists a group of distinct positive integers $a_1, a_2, \\dots, a_n$ such that $\\frac{a_i+a_j}{(a_i, a_j)} < 2n-1$ for any $i, j \\in \\{1, 2, \\dots, n\\}$.\n\nHere $(x, y)$ denotes the greatest common divisor of positive integers $x$ and $y$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "(1) Let $p = 2n-1$ be a prime. Without loss of generality, we assume that $(a_1, a_2, \\dots, a_n) = 1$. If there exists $i$ ($1 \\le i \\le n$) such that $p \\nmid a_i$, then there exists $j (\\ne i)$ such that $p \\nmid a_j$. Therefore $p \\nmid (a_i, a_j)$. Then we have\n$$\n\\frac{a_i + a_j}{(a_i, a_j)} \\ge \\frac{a_i}{(a_i, a_j)} \\ge p = 2n-1.\n$$\nNext, we consider the case when $(a_i, p) = 1,\\ i = 1, 2, \\dots, n$. Then $p \\nmid (a_i, a_j)$ for any $i \\ne j$. By the Pigeonhole Principle, we know that there exist $i \\ne j$ such that either $a_i \\equiv a_j \\pmod p$ or $a_i + a_j \\equiv 0 \\pmod p$.\n\nCase $a_i \\equiv a_j \\pmod p$, we have\n$$\n\\frac{a_i + a_j}{(a_i, a_j)} \\ge \\frac{a_i - a_j}{(a_i, a_j)} \\ge p = 2n-1.\n$$\n\nCase $a_i + a_j \\equiv 0 \\pmod p$, we have\n$$\n\\frac{a_i + a_j}{(a_i, a_j)} \\ge p = 2n - 1.\n$$\nThat completes the proof of (1).\n\n(2) We will construct an example to justify the statement. Since $2n-1$ is a composite number, we can write $2n-1 = pq$ where $p, q$ are positive integers greater than 1. Let\n$$\n\\begin{aligned}\na_1 &= 1,\\ a_2 = 2, \\dots, a_p = p,\\ a_{p+1} = p+1, \\\\\na_{p+2} &= p+3, \\dots, a_n = pq-p.\n\\end{aligned}\n$$\nNote that the first $p$ elements are consecutive integers, while the remainders are $n-p$ consecutive even integers from $p+1$ to $pq-p$.\n\nWhen $1 \\le i \\le j \\le p$, it is obvious that\n$$\n\\frac{a_i + a_j}{(a_i, a_j)} \\le a_i + a_j \\le 2p < 2n - 1.\n$$\nWhen $p+1 \\le i \\le j \\le n$, we have $2|(a_i, a_j)$ and then\n$$\n\\frac{a_i + a_j}{(a_i, a_j)} \\le \\frac{a_i + a_j}{2} \\le pq - p < 2n - 1.\n$$\nWhen $1 \\le i \\le p$ and $p+1 \\le j \\le n$ we have the following two possibilities:\n\n(i) Either $i \\ne p$ or $j \\ne n$. Then we have\n$$\n\\frac{a_i + a_j}{(a_i, a_j)} \\le pq - 1 < 2n - 1.\n$$\n(ii) $i = p$ and $j = n$. Then\n$$\n\\frac{a_i + a_j}{(a_i, a_j)} = \\frac{pq}{p} = q < 2n - 1.\n$$\nThat completes the proof.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 28132,
    "subject": "Mathematics (Multi-modal)",
    "question": "In a mathematical competition some competitors are friends. Friendship is always mutual. Call a group of competitors a clique if each two of them are friends. (In particular, any group of fewer than two competitors is a clique.) The number of members of a clique is called its size.\nGiven that, in this competition, the size of the largest clique is even, prove that the competitors can be arranged in two rooms such that the largest size of a clique contained in one room is the same as the largest size of a clique contained in the other room.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "**Proof** We provide an algorithm to distribute the competitors.\nDenote the rooms $A$ and $B$. At some initial stage, we move one person at a time from one room to the other one. We achieve the target by going through several adjustments. In every step of the algorithm, let $A$ and $B$ be the sets of competitors in room $A$ and $B$. Let $C(A)$ and $C(B)$ be the largest size of clique in room $A$ and $B$ respectively.\n\nStep 1: Let $M$ be the largest clique of all competitors, $|M| = 2m$.\nMove all members of $M$ to room $A$, and the remained ones to room $B$.\nSince $M$ is the largest clique of all competitors, we have $C(A) = |M| \\ge C(B)$.\n\nStep 2: If $C(A) > C(B)$, move one person from room $A$ to room $B$. (In view of $C(A) > C(B)$, we have $A \\ne \\emptyset$.)\nAfter every operation is done, $C(A)$ decreases by $1$ while $C(B)$ increases by $1$ at most. These operations will not end until\n$$\nC(A) \\le C(B) \\le C(A) + 1.\n$$\nAt that time, we also have $C(A) = |A| \\ge m$. (Otherwise there are at least $m+1$ members of $M$ in room $B$, at most $m-1$ members in room $A$, then $C(B) - C(A) \\ge (m+1) - (m-1) = 2$, it is impossible.)\n\nStep 3: Denote $K = C(A)$. If $C(B) = K$, we are done; or else $C(B) = K + 1$. From the above discussion,\n$$\nK = |A| = |A \\cap M| \\ge m, \\quad |B \\cap M| \\le m.\n$$\n\nStep 4: If there is a clique $C$ in room $B$ with $|C| = K + 1$ and a competitor $x \\in B \\cap M$ but $x \\notin C$, then move $x$ to room $A$, we are done.\nIn fact, after the operation, there are $K+1$ members of $M$ in room $A$, so $C(A) = K+1$. Since $x \\notin C$, whose removal does not reduce $C$, $C(B) = C$. Therefore, $C(A) = C(B) = K+1$.\nIf such competitor $x$ does not exist, then each largest clique in room $B$ contains $B \\cap M$ as a subset. In this case, we do step 5.\n\nStep 5: Choose any of the largest clique $C$ ($|C| = K+1$) in room $B$, move a member of $C \\setminus M$ to room $A$. (In view of $|C| = K+1 > m \\ge |B \\cap M|$, we know $C \\setminus M \\ne \\emptyset$.)\nSince we only move one person at a time from room $B$ to room $A$, so $C(B)$ decreases by $1$ at most. At the end of this step, we have $C(B) = K$.\nNow, there is a clique $A \\cap M$ in $A$, $|A \\cap M| = K$. So $C(A) \\ge K$.\nWe prove $C(A) = K$ as follows.\nLet $Q$ be any clique of room $A$. We only need to show $|Q| \\le K$.\nIn fact, the members of room $A$ can be classified into two:\n(1) Some members of $M$ in view of $M$ being a clique, are friends with all the members of $B \\cap M$.\n(2) The members move from room $B$ to room $A$ at step 5, they are friends of $B \\cap M$.\nSo, every member of $Q$ and members of $B \\cap M$ are friends. What is more, $Q$ and $B \\cap M$ both are cliques, so is $Q \\cup (B \\cap M)$.\nSince $M$ is the largest clique of all competitors,\n$$\n|M| \\ge |Q \\cup (B \\cap M)| = |Q| + |B \\cap M|\n$$\n$$\n= |Q| + |M| - |A \\cap M|.\n$$\nSo $|Q| \\le |A \\cap M| = K$. Therefore, after these 5 steps, we have $C(A) = C(B) = K$.",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 28133,
    "subject": "Mathematics (Multi-modal)",
    "question": "Given the set $P = \\{1, 2, 3, 4, 5\\}$, define $f(m, k) = \\sum_{i=1}^{5} \\lfloor m \\sqrt{\\frac{k+1}{i+1}} \\rfloor$ for any $k \\in P$ and positive integer $m$, where $\\lfloor a \\rfloor$ denotes the greatest integer less than or equal to $a$. Prove that for any positive integer $n$, there is $k \\in P$ and positive integer $m$, such that $f(m, k) = n$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "**Proof** Define set $A = \\{m\\sqrt{k+1} \\mid m \\in \\mathbb{N}^*, k \\in P\\}$, where $\\mathbb{N}^*$ denotes the set of all positive integers. It is easy to check that for any $k_1, k_2 \\in P, k_1 \\neq k_2$, $\\frac{\\sqrt{k_1+1}}{\\sqrt{k_2+1}}$ is an irrational number. Therefore, for any $k_1, k_2 \\in P$ and positive integers $m_1, m_2$, $m_1\\sqrt{k_1+1} = m_2\\sqrt{k_2+1}$ implies $m_1 = m_2$ and $k_1 = k_2$.\n\nNote that $A$ is an infinite set. We arrange the elements in $A$ in ascending order. Then we have an infinite sequence. For any positive integer $n$, suppose the $n$th term of the sequence is $m\\sqrt{k+1}$. Any term before the $n$th can be written as $m_i\\sqrt{i+1}$, and\n$$\nm_i \\sqrt{i+1} \\le m \\sqrt{k+1}.\n$$\nOr equivalently, $m_i \\le m \\frac{\\sqrt{k+1}}{\\sqrt{i+1}}$. It is easy to see that there are $\\lfloor m \\frac{\\sqrt{k+1}}{\\sqrt{i+1}} \\rfloor$ such $m_i$ for $i = 1, 2, 3, 4, 5$. Therefore,\n$$\nn = \\sum_{i=1}^{5} \\lfloor m \\frac{\\sqrt{k+1}}{\\sqrt{i+1}} \\rfloor = f(m, k).\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 28134,
    "subject": "Mathematics (Multi-modal)",
    "question": "Suppose $a$, $b$, $c$ are real numbers, with $a + b + c = 3$. Prove that\n$$\n\\frac{1}{5a^2 - 4a + 11} + \\frac{1}{5b^2 - 4b + 11} + \\frac{1}{5c^2 - 4c + 11} \\le \\frac{1}{4}.\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "$$\n\\text{If } a < \\frac{9}{5}, \\text{ then}\n$$\n$$\n\\frac{1}{5a^2 - 4a + 11} \\le \\frac{1}{24}(3-a). \\quad \\textcircled{1}\n$$\nIn fact,\n$$\n\\begin{align*}\n\\textcircled{1} &\\Leftrightarrow (3-a)(5a^2-4a+11) \\ge 24 \\\\\n&\\Leftrightarrow 5a^3 - 19a^2 + 23a - 9 \\le 0 \\\\\n&\\Leftrightarrow (a-1)^2(5a-9) \\le 0 \\Leftrightarrow a < \\frac{9}{5}.\n\\end{align*}\n$$\nSo if $a$, $b$, $c < \\frac{9}{5}$, then\n$$\n\\begin{align*}\n& \\frac{1}{5a^2 - 4a + 11} + \\frac{1}{5b^2 - 4b + 11} + \\frac{1}{5c^2 - 4c + 11} \\\\\n&\\le \\frac{1}{24}(3-a) + \\frac{1}{24}(3-b) + \\frac{1}{24}(3-c) \\\\\n&= \\frac{1}{4}.\n\\end{align*}\n$$\nIf one of $a$, $b$, $c$ is not less than $\\frac{9}{5}$, say $a \\ge \\frac{9}{5}$, then\n$$\n\\begin{aligned}\n5a^2 - 4a + 11 &= 5a \\left(a - \\frac{4}{5}\\right) + 11 \\\\\n&\\ge 5 \\cdot \\frac{9}{5} \\cdot \\left(\\frac{9}{5} - \\frac{4}{5}\\right) + 11 = 20.\n\\end{aligned}\n$$\n$$\n\\text{So} \\quad \\frac{1}{5a^2 - 4a + 11} \\le \\frac{1}{20}.\n$$\nSince\n$$\n5b^2 - 4b + 11 = 5\\left(b - \\frac{2}{5}\\right)^2 + 11 - \\frac{4}{5} \\ge 11 - \\frac{4}{5} > 10,\n$$\nwe have $\\frac{1}{5b^2 - 4b + 11} < \\frac{1}{10}$. Similarly, $\\frac{1}{5c^2 - 4c + 11} < \\frac{1}{10}$. So\n$$\n\\begin{aligned}\n& \\frac{1}{5a^2 - 4a + 11} + \\frac{1}{5b^2 - 4b + 11} + \\frac{1}{5c^2 - 4c + 11} \\\\\n< & \\frac{1}{20} + \\frac{1}{10} + \\frac{1}{10} = \\frac{1}{4}.\n\\end{aligned}\n$$\nHence the inequality holds for all $a$, $b$, $c$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 28135,
    "subject": "Mathematics (Multi-modal)",
    "question": "Nine balls of the same size and color, numbered $1, 2, \\dots, 9$, were put into a packet. Now $A$ draws a ball from the packet, noted that it is of number $a$, and puts back it. Then $B$ also draws a ball from the packet and noted that it is of number $b$. Then the probability for the inequality $a - 2b + 10 > 0$ to hold is ( ).",
    "options": [],
    "answer": "61/81",
    "solution": "Since each has equally $9$ different possible results for $A$ and $B$ to draw a ball from the packet independently, the total number of possible events is $9^2 = 81$.\n\nFrom $a - 2b + 10 > 0$ we get $2b < a + 10$.\n\nWe find that when $b = 1, 2, 3, 4, 5$, $a$ can take any value in $1, 2, \\dots, 9$ to make the inequality hold. Then we have $9 \\times 5 = 45$ admissible events.\n\nWhen $b = 6$, $a$ can be $3, 4, \\dots, 9$, and there are $7$ admissible events.\n\nWhen $b = 7$, $a$ can be $5, 6, 7, 8, 9$, and there are $5$ admissible events.\n\nWhen $b = 8$, $a$ can be $7, 8, 9$, and there are $3$ admissible events.\n\nWhen $b = 9$, $a$ can only be $9$, and there is $1$ admissible event.\n\nSo the required probability is $\\frac{45+7+5+3+1}{81} = \\frac{61}{81}$.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 28136,
    "subject": "Mathematics (Multi-modal)",
    "question": "There are 63 points on a circle $C$ with radius 10. Let $S$ be the number of triangles whose sides are longer than 9 and whose vertices are chosen from the 63 points. Find the maximum value of $S$.",
    "options": [],
    "answer": "23121",
    "solution": "Let $O$ be the center of circle $C$, $a_n$ is the length of a regular $n$-gon $A_1A_2\\cdots A_n$ inscribed in $\\odot O$. Then $a_6 = 10 > 9$, $a_7 < 10 \\times \\frac{2\\pi}{7} < 10 \\times \\frac{2 \\times 3.15}{7} < 9$.\n\n(1) Let $A_1A_2\\cdots A_6$ be a regular 6-gon inscribed in $\\odot O$, then $A_iA_{i+1} = a_6 > 9$. So we can choose a point $B_i$ in $\\widehat{A_iA_{i+1}}$ such that $B_iA_{i+1} > 9$. Then $\\angle B_iOA_{i+1} > \\frac{2\\pi}{7}$ ($A_7 = A_1$), and\n$$\n\\angle A_iOB_i = \\angle A_iOA_{i+1} - \\angle B_iOA_{i+1} < \\frac{2\\pi}{6} - \\frac{2\\pi}{7} < \\frac{2\\pi}{7}.\n$$\nIt follows that $A_iB_i < 9$ ($i = 1, 2, \\dots, 6$).\n\nIn each of $\\widehat{A_1B_1}, \\widehat{A_2B_2}, \\widehat{A_3B_3}$, choose 11 points, and in each of $\\widehat{A_4B_4}, \\widehat{A_5B_5}, \\widehat{A_6B_6}$, choose 10 points. We obtain a set $M$ which has 63 points on the circle $C$. It is easy to see for $M$ the value of $S$ is $S_0$, where\n$$\n\\begin{aligned}\nS_0 &= \\binom{3}{3} \\times 11^3 + \\binom{3}{2} \\cdot \\binom{3}{1} \\times 11^2 \\times 10 \\\\\n&\\quad + \\binom{3}{1} \\cdot \\binom{3}{2} \\times 11 \\times 10^2 + \\binom{3}{3} \\times 10^3 \\\\\n&= 23\\,121.\n\\end{aligned}\n$$\nSo the maximum value of $S$ is not less than $S_0$.\n\n(2) We prove that the maximum is $S_0$. We need three lemmas.\n\n**Lemma 1** For $P$ on circle $C$, we call the arc $\\widehat{APB}$ “an arc of $P$”, if $P$ is the midpoint of arc $\\widehat{APB}$, and $\\angle AOB = \\frac{4\\pi}{7}$. Now for every given $n$ points on circle $C$, there is a point $P$, such that there are $\\lfloor \\frac{n+5}{6} \\rfloor$ points of the given $n$ points on the “arc of $P$”.\n\n**Proof:** Let $A$ be one of the given $n$ points, and an “arc of $A$” be $\\widehat{A_1AA_6}$. Now suppose $A_2, A_3, \\dots, A_5$ are on the arc $\\widehat{A_1A_6}$ (not including $A$), and $A_1A_2 = A_2A_3 = \\dots = A_5A_6$ (see the figure). So $\\angle A_iOA_{i+1} = \\frac{2\\pi}{7}, i = 1, 2, \\dots, 5$.\n\n![](attached_image_1.png)\n\nIf there is a point $P_i$ (of the given $n$ points) on $\\widehat{A_iA_{i+1}}$, then all the points (of the given $n$ points) on $\\widehat{A_iA_{i+1}}$ are on “an arc of $P_i$”. So the given $n$ points are on $6$ “arcs of $P_i$” (including “arc of $A$”). This shows that there are $\\lfloor \\frac{n-1}{6} \\rfloor + 1 = \\lfloor \\frac{n+5}{6} \\rfloor$ points of the given points on an “arc of $P_i$”, where $P_i$ is one of the given $n$ points.\n\n**Lemma 2** Take the arc $\\widehat{A_1BA_6}$ arbitrary on the circle $C$ with radius 10, where $\\widehat{A_1BA_6}$ is $\\frac{5}{7}$ of the perimeter. Then take any $5m+r$ points on the arc $\\widehat{A_1BA_6}$ ($m, r$ are non-negative integers and $0 \\le r < 5$). Prove that number of lines from the given points whose lengths are more than 9 is at most\n$$\n10m^2 + 4rm + \\frac{1}{2}r(r-1).\n$$\n\n**Proof:** Divide $\\overrightarrow{A_1BA_6}$ in five equal parts, where the corresponding points are $A_2, A_3, A_4, A_5$ (see the figure), then the length of $\\overrightarrow{A_iA_{i+1}}$ is exactly $\\frac{1}{7}$ of the perimeter ($i = 1, 2, 3, 4, 5$), and the distance of any two points is not more than $a_7 < 9$. Suppose there are $m_i$ given points on the arc $\\overrightarrow{A_iA_{i+1}}$, then the number of lines from the given points whose lengths are more than 9 is at most\n$$\nl = \\sum_{1 \\le i < j \\le 5} m_i m_j, \\qquad \\textcircled{1}\n$$\nwhere\n$$\nm_1 + m_2 + \\cdots + m_5 = 5m + r. \\qquad \\textcircled{2}\n$$\nSince there are finitely many non-negative integer groups ($m_1, m_2, m_3, m_4, m_5$), the maximum value of $l$ exists. Now we prove that when the maximum is attained the inequality\n$$\n|m_i - m_j| \\le 1, (1 \\le i < j \\le 5),\n$$\nmust hold.\n\nIn fact, if there exist $i, j$ ($1 \\le i < j \\le 5$) such that $|m_i - m_j| \\ge 2$ when the maximum is attained, we can suppose $m_1 - m_2 \\ge 2$. Then let\n$$\nm_1' = m_1 - 1,\\quad m_2' = m_2 + 1,\\quad m' = m,\n$$\nand the corresponding integer is $l'$, we will have\n$$\n\\begin{align*}\nm_1' + m_2' &= m_1 + m_2, \\\\\nm_1' + m_2' + m_3' + m_4' + m_5' &= m_1 + m_2 + m_3 + m_4 + m_5, \\\\\nl' - l &= (m_1'm_2' - m_1m_2) + [(m_1' + m_2') \\\\\n&\\quad -(m_1 + m_2)](m_3 + m_4 + m_5) \\\\\n&= m_1 - m_2 - 1 \\ge 1.\n\\end{align*}\n$$\nContradiction!\n\nTherefore, when $l$ reaches the maximum value, the number of $m+1$ is $r$ and the number of $m$ is $5-r$. Thus, the number of lines from the given points whose lengths are more than 9 are at most\n$$\n\\begin{aligned}\n& \\binom{r}{2}(m+1)^2 + \\binom{r}{1}(5-r)(m+1)m + \\binom{5-r}{2}m^2 \\\\\n&= 10m^2 + 4rm + \\frac{1}{2}r(r-1).\n\\end{aligned}\n$$\n\n**Lemma 3** Take arbitrary $n$ points on the circle $C$ with radius 10 to form set $M$, where $n = 6m + r$ ($m, r$ are non-negative integers, $0 \\le r < 6$). Assume that there are $S_n$ triangles whose vertices are from $M$ and each side is longer than 9. Prove\n$$\nS_n \\le 20m^3 + 10rm^2 + 2r(r-1)m + \\frac{1}{6}r(r-1)(r-2).\n$$\n\n**Proof:** We shall prove by mathematical induction.\n\nWhen $n = 1, 2$, $S_n = 0$. It is true.\n\nSuppose when $n = k$, it is true and set $k = 6m + r$ ($m, r$ are non-negative integers, $0 \\le r < 6$). Then\n$$\nS_k \\le 20m^3 + 10rm^2 + 2r(r-1)m + \\frac{1}{6}r(r-1)(r-2).\n$$\nFrom Lemma 1, when $n = k + 1$, the $k + 1$ given points must include the point $P$, where at least $\\lfloor \\frac{k+1+5}{6} \\rfloor = m + 1$ given points are in the $\\frac{2}{7}\\text{arc } A_1PA_6$. And the distances of such points to $P$ are $\\le PA_1 = PA_6 = a_7 < 9$. Hence, there are at most $(k+1)-(m+1) = 5m+r$ given points whose distances to $P$ are more than 9, and such points are all in the other $\\frac{5}{7}\\arccos \\overline{A_1PA_6}$ without $P$. From Lemma 2, the lines from such points whose lengths are more than 9 are at most\n$$\n10m^2 + 4rm + \\frac{1}{2}r(r-1).\n$$\n(From Lemma 2, when $r=5$, it is $10(m+1)^2$, which is also true.) Thus, the number of triangles whose vertex is $P$ and each side is larger than 9 is not more than\n$$\nS_p = 10m^2 + 4rm + \\frac{1}{2}r(r-1).\n$$\nWithout $P$, there are $k = 6m + r$ given points. Let there be $S_k$ triangles whose vertices are from the $k$ points and each side is larger than 9, then using mathematical induction, we get\n$$\nS_k \\le 20m^3 + 10rm^2 + 2r(r-1)m + \\frac{1}{6}r(r-1)(r-2).\n$$\nFurthermore\n$$\n\\begin{align*} \nS_{k+1} &= S_k + S_p \\\\ \n&\\le 20m^3 + 10rm^2 + 2r(r-1)m + \\frac{1}{6}r(r-1)(r-2) \\\\ \n&\\quad + 10m^2 + 4rm + \\frac{1}{2}r(r-1) \\\\ \n&= 20m^3 + 10(r+1)m^2 + 2r(r+1)m \\\\ \n&\\quad + \\frac{1}{6}r(r-1)(r+1), \n\\end{align*}\n$$\nwhich means the case $n = k + 1 = 6m + (r + 1)$ is also true.\n\nOn the other hand, when $r=5$, then $m = k + 1 = 6(m+1)$ and $S_{k+1}$ can be simplified to $S_{k+1} = 20(m+1)^3$, which is also true.\n\nTherefore, we have proved Lemma 3.\n\nNow considering the original problem, we have\n$$\nn = 63 = 6 \\times 10 + 3.\n$$\nIt follows from Lemma 3,\n$$\n\\begin{aligned}\nS &\\le 20 \\times 10^3 + 10 \\times 10^2 + 2 \\times 3 \\times 2 \\times 10 + \\frac{1}{6} \\times 3 \\times 2 \\times 1 \\\\\n&= 23\\,121.\n\\end{aligned}\n$$\nThus, $S_{\\max} = 23\\,121$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 28137,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $a_1, a_2, \\dots, a_{11}$ be $11$ distinct positive integers with their sum less than $2007$, and write the numbers $1, 2, \\dots, 2007$ in order on the blackboard. Now we define a group of $22$ ordered operations:\nThe $i$-th operation is to take any number on the blackboard, and then add $a_i$ to it, if $1 \\le i \\le 11$ or minus $a_{i-11}$ from it, if $12 \\le i \\le 22$.\nIf the final result after such a group of operations is an even permutation of $1, 2, \\dots, 2007$, then we call it a \"good\" group; if the result is an odd permutation of $1, 2, \\dots, 2007$, then we call it a \"second good\" group.\nOur question is: Which is greater? The number of \"good\" groups or that of \"second good\" groups? And by how many more?",
    "options": [],
    "answer": "∏_{i=1}^{11} a_i",
    "solution": "The answer is: The \"good\" groups is more than the \"second good\" groups by $\\prod_{i=1}^{11} a_i$.\n\nMore generally, we write numbers $1, 2, \\dots, n$ in order on the blackboard, and define a group of $l$ ordered operations: The $i$-th operation is to take any number on the blackboard, and then add $b_i$ ($b_i \\in \\mathbb{Z}, 1 \\le i \\le l$) to it.\nIf the final result after such a group of operations is an even/odd permutation of $1, 2, \\dots, n$, then we call it a \"good\"/\"second good\" group. And the difference between the number of \"good\" groups and that of \"second good\" groups is defined as\n$$\nf(b_1, b_2, \\cdots, b_l; n).\n$$\nNow let us study the property of $f$.\n\nFirstly, interchanging $b_i$ and $b_j$ for any $1 \\le i, j \\le l$ will not affect the value of $f$. As a matter of fact, it only results in the exchange of the $i$th and $j$th operations in a group, and will not affect the final result after the group's operations. So the value of $f$ remains the same.\n\nSecondly, we only need to count the number of \"good\"/\"second good\" groups with property $P$ — a property attributed to any operation group which keeps the numbers on the blackboard distinctive from one another after each operation. We can prove that the difference between the numbers of \"good\" and \"second good\" groups with property $P$ is also equal to $f$.\n\nIn fact, we only need to prove that the numbers of \"good\" and \"second good\" groups without property $P$ are the same. Suppose the $i$th operation of a \"good\"/\"second good\" group without property $P$ results in the equal between the $p$th and $q$th number on the blackboard ($1 \\le p < q \\le n$). We change the following $l-i$ operations in this way: operations on the $p$th number are changed to operations on the $q$th number, and vice versa. It is easy to verify that the resulted permutation on the blackboard of new operation group would be a $(p, q)$ transposition of the permutation of the original operation group. Then the parities of the two permutations are in opposite signs. And that means the numbers of \"good\" and \"second good\" groups without property $P$ are the same.\n\nNow, let $a_1, a_2, \\cdots, a_m$ be $m$ distinct positive integers with their sum less than $n$. We prove by the principle of mathematical induction that\n$$\nf(a_1, a_2, \\dots, a_m, -a_1, -a_2, \\dots, -a_m; n) = \\prod_{i=1}^{m} a_i. \\quad (1)\n$$\nIf $m=1$, consider a \"good\"/\"second good\" group with property $P$. It must be in such a way: The first operation is to take a number from $n - a_1 + 1, n - a_2 + 2, \\dots, n$ on the blackboard, and add $a_1$ to it; next operation is to add $-a_1$ again to it. So the number of \"good\" groups is $a_1$, while that of \"second good\" groups is $0$. Therefore $(1)$ holds.\n\nAssume that $(1)$ holds for $m-1$. We now consider case $m$. According to what discussed above, we may assume that $a_1 < a_2 < \\dots < a_n$, and\n$$\n\\begin{aligned}\n& f(a_1, a_2, \\dots, a_m, -a_1, -a_2, \\dots, -a_m; n) \\\\\n&= f(a_1, -a_2, -a_3, \\dots, -a_m, a_2, a_3, \\dots, a_m, -a_1; n).\n\\end{aligned}\n$$\nFor a group with property $P$, the first operation must be done on the last $a_1$ numbers on the blackboard; the second operation be done on the first $a_2$ numbers; the third operation done on the first $a_2 + a_3$ numbers; ... the $m$th operation done on the first $a_2 + \\dots + a_m < n - a_1$ numbers. And the $m+1 \\sim 2m-1$ th operations will also be done on the first $n - a_1$ numbers. Otherwise the sum of the first $n - a_1$ numbers will be less than $1 + 2 + \\dots + (n - a_1)$, a contradiction to property $P$.\n\nTherefore, the $2 \\sim 2m-2$ operations must be done on the first $n-a_1$ numbers, and the result must be an even/odd permutation of $1, 2, \\dots, (n-a_1)$, which corresponds to each one of $a_1$ even/odd permutations of $1, 2, \\dots, n$ derived from original operation groups. therefore\n$$\n\\begin{aligned}\n& f(a_1, -a_2, -a_3, \\dots, -a_m, a_2, a_3, \\dots, a_m, -a_1; n) \\\\\n&= a_1 f(-a_2, -a_3, \\dots, -a_m, a_2, a_3, \\dots, a_m; n - a_1).\n\\end{aligned}\n$$\nBy induction we have\n$$\n\\begin{aligned}\n& f(-a_2, -a_3, \\cdots, -a_m, a_2, a_3, \\cdots, a_m; n - a_1) \\\\\n&= f(a_2, a_3, \\cdots, a_m, -a_2, -a_3, \\cdots, -a_m; n - a_1) \\\\\n&= \\prod_{j=2}^{m} a_j.\n\\end{aligned}\n$$\nThat means $(1)$ holds for $m$.\n\nNow take $n = 2007$ and $m = 11$ in $(1)$. Thus arriving at the value of $\\prod_{j=1}^{11} a_j$.",
    "topic": "Discrete Mathematics",
    "subtopic": "Combinatorics"
  },
  {
    "id": 28138,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $n$ be an integer greater than $3$, and let $a_1, a_2, \\dots, a_n$ be nonnegative real numbers with $a_1 + a_2 + \\dots + a_n = 2$.\n\nDetermine the minimum value of\n$$\n\\frac{a_1}{a_2^2+1} + \\frac{a_2}{a_3^2+1} + \\cdots + \\frac{a_n}{a_1^2+1}.\n$$",
    "options": [],
    "answer": "3/2",
    "solution": "The answer is $\\frac{3}{2}$.\n\nThe given problem is equivalent to finding the minimum value of\n$$\n\\begin{aligned}\nm &= 2 - \\left( \\frac{a_1}{a_2^2+1} + \\frac{a_2}{a_3^2+1} + \\cdots + \\frac{a_n}{a_1^2+1} \\right) \\\\\n  &= \\left( a_1 - \\frac{a_1}{a_2^2+1} \\right) + \\left( a_2 - \\frac{a_2}{a_3^2+1} \\right) + \\cdots + \\left( a_n - \\frac{a_n}{a_1^2+1} \\right) \\\\\n  &= \\frac{a_1 a_2^2}{a_2^2+1} + \\frac{a_2 a_3^2}{a_3^2+1} + \\cdots + \\frac{a_n a_1^2}{a_1^2+1}.\n\\end{aligned}\n$$\nSince $a_i^2 + 1 \\ge 2a_i$, we have\n$$\nm \\le \\frac{a_1 a_2 + a_2 a_3 + \\cdots + a_n a_1}{2}.\n$$\nOur result follows from the following well-known fact:\n$$\n\\begin{aligned}\n& f(a_1, \\cdots, a_n) \\\\\n&= (a_1 + \\cdots + a_n)^2 - 4(a_1 a_2 + a_2 a_3 + \\cdots + a_n a_1) \\\\\n&\\ge 0\n\\end{aligned} \\qquad \\textcircled{1}\n$$\nfor integers $n \\ge 4$ and nonnegative real numbers $a_1, a_2, \\cdots, a_n$.\n\nTo prove this fact, we use induction on $n$. For $n=4$, $\\textcircled{1}$ becomes\n$$\n\\begin{aligned}\n& f(a_1, a_2, a_3, a_4) \\\\\n&= (a_1 + a_2 + a_3 + a_4)^2 - 4(a_1 a_2 + a_2 a_3 + a_3 a_4 + a_4 a_1) \\\\\n&= (a_1 + a_2 + a_3 + a_4)^2 - 4(a_1 + a_3)(a_2 + a_4),\n\\end{aligned}\n$$\nwhich is nonnegative by the AM-GM inequality.\n\nAssume that $\\textcircled{1}$ is true for $n=k$ for some integer $k \\ge 4$.\n\nConsider the case when $n = k + 1$. By (cyclic) symmetry in $\\textcircled{1}$, we may assume that $a_{k+1} = \\min\\{a_1, a_2, \\dots, a_{k+1}\\}$. By the induction hypothesis, it suffices to show that\n$$\nD = f(a_1, \\dots, a_k, a_{k+1}) - f(a_1, \\dots, a_{k-1}, a_k + a_{k+1}) \\ge 0.\n$$\nNote that\n$$\n\\begin{aligned}\n\\frac{D}{4} &= -(a_1a_2 + \\cdots + a_k a_{k+1} + a_{k+1}a_1) + [a_1a_2 + \\cdots \\\\\n& \\qquad \\qquad + a_{k-1}(a_k + a_{k+1}) + (a_k + a_{k+1})a_1] \\\\\n&= a_{k-1}a_{k+1} + a_k a_1 - a_k a_{k+1} \\\\\n&= a_{k-1}a_{k+1} + (a_1 - a_{k+1})a_k \\\\\n&\\ge 0,\n\\end{aligned}\n$$\ncompleting our proof.\n\nFrom the above result, we can get\n$$\n\\begin{aligned}\n& \\frac{1}{2}(a_1a_2 + a_2a_3 + \\cdots + a_n a_1) \\\\\n\\le & \\frac{1}{8}(a_1 + a_2 + \\cdots + a_n)^2 \\\\\n= & \\frac{1}{8} \\times 2^2 \\\\\n= & \\frac{1}{2},\n\\end{aligned}\n$$\ntherefore\n$$\n\\frac{a_1 a_2^2}{a_2^2 + 1} + \\frac{a_2 a_3^2}{a_3^2 + 1} + \\cdots + \\frac{a_n a_1^2}{a_1^2 + 1} \\le \\frac{1}{2},\n$$\nthat is $m \\ge \\frac{3}{2}$. When $a_1 = a_2 = 1$ and $a_3 = \\cdots = a_n = 0$, $m = \\frac{3}{2}$ holds.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 28139,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find the smallest positive integer $n$ such that any sequence of positive integers $a_1, a_2, \\dots, a_n$ satisfying $\\sum_{i=1}^{n} a_i = 2007$ must have several consecutive terms whose sum is 30.",
    "options": [],
    "answer": "1018",
    "solution": "Firstly, we could construct a sequence of positive integers with 1017 terms $a_1, a_2, \\dots, a_{1017}$, such that we cannot find consecutive terms whose sum is 30. Hence, we could set $a_1 = a_2 = \\cdots = a_{29} = 1$, $a_{30} = 31$ and $a_{30+m+i} = a_i$, $i \\in \\{1, 2, \\dots, 30\\}$, $m \\in \\mathbb{N}$, which means the sequence is 1, 1, \\dots, 1, 31, $|1, 1, \\dots, 1, 31, \\dots, 1, 1, \\dots, 1, 31, \\dots|$, 1, \\dots, 1 (in which there are 34 groups all have 30 terms except the last group with 27 terms, totalling 1017 terms).\n\nSecondly, when the terms are less than 1017, what we should do is just to combine several consecutive terms into a larger number within certain groups of the sequence.\n\nNow, for any sequence with 1018 terms $a_1, a_2, \\dots, a_{1018}$ that satisfies the condition $\\sum_{i=1}^{1018} a_i = 2007$, we want to prove that there must exist several consecutive terms whose sum is 30.\n\nDenote $S_k = \\sum_{i=1}^{k} a_i$, $k = 1, 2, \\dots, 1018$, then\n$1 \\le S_1 < S_2 < \\cdots < S_{1018} = 2007.$\n\nGroup the elements in the set $\\{1, 2, \\dots, 2007\\}$ as follows:\n\n(1, 31), (2, 32), ..., (30, 60);\n(61, 91), (62, 92), ..., (90, 120);\n(121, 151), (122, 152), ..., (150, 180);\n......\n(60k + 1, 60k + 31), (60k + 2, 60k + 32), ..., (60k + 30, 60k + 1);\n......\n(60 \\cdot 32 + 1, 60 \\cdot 32 + 31), (60 \\cdot 32 + 2, 60 \\cdot 32 + 32), ..., (60 \\cdot 32 + 30, 60 \\cdot 33);\n1981, 1982, ..., 2007.\n\nThere are $33 \\times 30 = 990$ brackets and 27 numbers without brackets. Arbitrarily take 1018 numbers, whose sum is the value of $S_k$. There must be two numbers from the same bracket. Denote the two numbers by $(S_k, S_{k+m})$, then $S_{k+m} - S_k = 30$, which means,\n$$\na_{k+1} + a_{k+2} + \\cdots + a_{k+m} = 30.\n$$\nTherefore, the minimum of $n$ is 1018.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 28140,
    "subject": "Mathematics (Multi-modal)",
    "question": "Suppose points $O$ and $I$ are the circumcenter and incenter of $\\triangle ABC$ respectively, and the inscribed circle of $\\triangle ABC$ is tangent to the sides $BC$, $CA$, $AB$ at points $D$, $E$, $F$ respectively. Lines $FD$ and $CA$ intercept at point $P$, while lines $DE$ and $AB$ intercept at point $Q$. And points $M$, $N$ are the midpoint of segments $PE$, $QF$ respectively. Prove that $OI \\perp MN$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "We first consider $\\triangle ABC$ and segment $PFD$. By Menelaus theorem we have\n$$\n\\frac{CP}{PA} \\cdot \\frac{AF}{FB} \\cdot \\frac{BD}{DC} = 1.\n$$\nThen\n$$\nPA = CP \\cdot \\frac{AF}{FB} \\cdot \\frac{BD}{DC} = (PA + b) \\frac{p-a}{p-c}.\n$$\n(We define $a = BC$, $b = CA$, $c = AB$, $p = \\frac{1}{2}(a+b+c)$; and without loss of generality, assume $a > c$.) Then we get\n$$\nPA = \\frac{b(p-a)}{a-c}.\n$$\nFurther,\n$$\nPE = PA + AE = \\frac{b(p-a)}{a-c} + p - a = \\frac{2(p-c)(p-a)}{a-c},\n$$\n$$\nME = \\frac{1}{2}PE = \\frac{(p-c)(p-a)}{a-c},\n$$\n$$\nMA = ME - AE = \\frac{(p-c)(p-a)}{a-c} - (p-a) = \\frac{(p-a)^2}{a-c},\n$$\n$$\nMC = ME + EC = \\frac{(p-c)(p-a)}{a-c} + (p-c) = \\frac{(p-c)^2}{a-c}.\n$$\nThen we have\n$$\nMA \\cdot MC = ME^2.\n$$\nThat means $ME^2$ is equal to the power of $M$ with respect to the circumscribed circle of $\\triangle ABC$. Further, since $ME$ is the length of the tangent from $M$ to the inscribed circle of $\\triangle ABC$, so $ME^2$ is also the power of $M$ with respect to the inscribed circle. Hence, $M$ is on the radical axis of the circumscribed and inscribed circles of $\\triangle ABC$.\nIn the same way, $N$ is also on the radical axis. Since the radical axis is perpendicular to $OI$, then $OI \\perp MN$. That completes the proof.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 28141,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $O$ be an interior point of the triangle $ABC$. Prove that there exist positive integers $p, q$ and $r$, such that\n$$\n|p \\cdot \\vec{OA} + q \\cdot \\vec{OB} + r \\cdot \\vec{OC}| < \\frac{1}{2007}.\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "It is well-known that there are positive real numbers $\\beta, \\gamma$ such that\n$$\n\\vec{OA} + \\beta \\vec{OB} + \\gamma \\vec{OC} = \\vec{0}.\n$$\nSo for positive integer $k$, we have\n$$\nk \\vec{OA} + k\\beta \\vec{OB} + k\\gamma \\vec{OC} = \\vec{0}.\n$$\nLet $m(k) = [k\\beta]$, $n(k) = [k\\gamma]$, where $[x]$ is the biggest integer which is less than or equal to $x$, and $\\{x\\} = x - [x]$.\n\nAssume $T$ is an integer larger than $\\max\\left\\{\\frac{1}{\\beta}, \\frac{1}{\\gamma}\\right\\}$. Then the sequences $\\{m(kT) \\mid k = 1, 2, \\dots\\}$ and $\\{n(kT) \\mid k = 1, 2, \\dots\\}$ are increasing, and\n$$\n\\begin{aligned}\n& |kT \\vec{OA} + m(kT) \\vec{OB} + n(kT) \\vec{OC}| \\\\\n& = |-\\{kT\\beta\\} \\vec{OB} - \\{kT\\gamma\\} \\vec{OC}| \\\\\n& \\leq |\\vec{OB}| \\cdot \\{kT\\beta\\} + |\\vec{OC}| \\cdot \\{kT\\gamma\\} \\\\\n& \\leq |\\vec{OB}| + |\\vec{OC}|.\n\\end{aligned}\n$$\nThis shows there exists infinitely many vectors such that\n$$\nkT \\vec{OA} + m(kT) \\vec{OB} + n(kT) \\vec{OC},\n$$\nwhose endpoint is in a circle $O$ of radius $|\\vec{OB}| + |\\vec{OC}|$. So there are two of the vectors, such that the distance between the endpoints of the two vectors is less than $\\frac{1}{2007}$, this means there exists two integers $k_1 < k_2$, such that\n$$\n\\begin{aligned}\n& | (k_2 T \\vec{OA} + m(k_2 T) \\vec{OB} + n(k_2 T) \\vec{OC}) \\\\\n& \\quad - (k_1 T \\vec{OA} + m(k_1 T) \\vec{OB} + n(k_1 T) \\vec{OC}) | \\\\\n& < \\frac{1}{2007}.\n\\end{aligned}\n$$\nSo, if we let $p = (k_2 - k_1)T$, $q = m(k_2T) - m(k_1T)$, $r = n(k_2T) - n(k_1T)$, then $p, q, r$ are integers, and\n$$\n|p \\vec{OA} + q \\vec{OB} + r \\vec{OC}| < \\frac{1}{2007}.\n$$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 28142,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all functions $f: Q^+ \\to Q^+$ such that\n$$\nf(x) + f(y) + 2xyf(xy) = \\frac{f(xy)}{f(x+y)}, \\qquad \\textcircled{1}\n$$\nWhere $Q^+ = \\{q \\mid q \\text{ is a positive rational number}\\}$.",
    "options": [],
    "answer": "f(x) = 1/x^2 for all positive rational x",
    "solution": "(1) Prove that $f(1) = 1$.\nPut $y = 1$ in \\textcircled{1}, and write $f(1) = a$. Then\n$$\nf(x) + a + 2x f(x) = \\frac{f(x)}{f(x+1)}\n$$\nThus\n$$\nf(x+1) = \\frac{f(x)}{(1 + 2x)f(x) + a} \\qquad \\textcircled{2}\n$$\nHence\n$$\nf(2) = \\frac{a}{4a} = \\frac{1}{4},\n$$\n$$\nf(3) = \\frac{1}{\\frac{1}{4} + a} = \\frac{1}{5 + 4a},\n$$\n$$\nf(4) = \\frac{1}{7 + 5a + 4a^2}.\n$$\n---\nOn the other hand, we put $x = y = 2$ in \\textcircled{1}, then\n$$\n2f(2) + 8f(4) = \\frac{f(4)}{f(4)} = 1.\n$$\nBy \\textcircled{2}, we have\n$$\n\\frac{1}{2} + \\frac{8}{7 + 5a + 4a^2} = 1.\n$$\nSolving the equation, we have $a = 1$, i.e. $f(1) = 1$.\n\n(2) Prove that\n$$\nf(x+n) = \\frac{f(x)}{(n^2 + 2nx)f(x) + 1}, \\quad n = 1, 2, \\dots \\qquad \\textcircled{3}\n$$\nFirstly, according to \\textcircled{2} we know that \\textcircled{3} is true for $n = 1$. Now suppose \\textcircled{3} is true for $n = k$. Then\n$$\n\\begin{aligned}\nf(x + k + 1) &= \\frac{f(x + k)}{(1 + 2(x + k))f(x + k) + 1} \\\\\n&= \\left( \\frac{f(x)}{(k^2 + 2kx)f(x) + 1} \\right) / \\left( \\frac{(1 + 2(x + k))f(x)}{(k^2 + 2kx)f(x) + 1} + 1 \\right) \\\\\n&= \\frac{f(x)}{((k + 1)^2 + 2(k + 1)x)f(x) + 1}.\n\\end{aligned}\n$$\nHence the result follows by induction.\nFrom \\textcircled{3} we have\n$$\nf(n + 1) = \\frac{f(1)}{(n^2 + 2n)f(1) + 1} = \\frac{1}{(n+1)^2},\n$$\n$$\n\\text{so } f(n) = \\frac{1}{n^2}, \\quad n = 1, 2, \\dots.\n$$\n\n(3) Prove that\n$$\nf\\left(\\frac{1}{n}\\right) = n^2 = \\frac{1}{\\left(\\frac{1}{n}\\right)^2}, \\quad n = 1, 2, \\dots \\qquad \\textcircled{4}\n$$\nIn fact, by letting $x = \\frac{1}{n}$ in \\textcircled{3}, we have\n$$\nf\\left(n+\\frac{1}{n}\\right) = \\frac{f\\left(\\frac{1}{n}\\right)}{\\left(n^2+2\\right)f\\left(\\frac{1}{n}\\right)+1},\n$$\nand by setting $y = \\frac{1}{x}$ in \\textcircled{1}, we have\n$$\nf(x) + f\\left(\\frac{1}{x}\\right) + 2 = \\frac{1}{f\\left(x + \\frac{1}{x}\\right)}.\n$$\nSo\n$$\nf(n) + f\\left(\\frac{1}{n}\\right) + 2 = \\frac{1}{f\\left(n + \\frac{1}{n}\\right)} = n^2 + 2 + \\frac{1}{f\\left(\\frac{1}{n}\\right)}.\n$$\nConsequently, $f(n) = \\frac{1}{n^2}$ implies $f(\\frac{1}{n}) = n^2$.\n\n(4) Prove that if $q = \\frac{n}{m}$, $\\text{gcd}(m, n) = 1$, $m, n \\in \\mathbb{N}^*$, then $f(q) = \\frac{1}{q^2}$.\nFor $m, n \\in \\mathbb{N}^*$, $\\text{gcd}(m, n) = 1$, put $x = n$, $y = \\frac{1}{m}$ in \\textcircled{1}, we have\n$$\nf\\left(\\frac{1}{m}\\right) + f(n) + \\frac{2n}{m}f\\left(\\frac{n}{m}\\right) = \\frac{f\\left(\\frac{n}{m}\\right)}{f\\left(n+\\frac{1}{m}\\right)}.\n$$\nPut $x = \\frac{1}{m}$ in \\textcircled{3}, we get\n$$\nf\\left(n+\\frac{1}{m}\\right) = \\frac{f\\left(\\frac{1}{m}\\right)}{\\left(n^2+\\frac{2n}{m}\\right)f\\left(\\frac{1}{m}\\right)+1} = \\frac{1}{n^2+\\frac{2n}{m}+\\frac{1}{m^2}}.\n$$\nSo\n$$\n\\frac{1}{n^2} + m^2 + \\frac{2n}{m} f\\left(\\frac{n}{m}\\right) = \\left(n + \\frac{1}{m}\\right)^2 f\\left(\\frac{n}{m}\\right).\n$$\nNow we have\n$$\nf(q) = f\\left(\\frac{n}{m}\\right) = \\frac{\\frac{1}{n^2} + m^2}{n^2 + \\frac{1}{m^2}} = \\left(\\frac{m}{n}\\right)^2 = \\frac{1}{q^2}.\n$$\nFinally, it is easy to verify that $f(x) = \\frac{1}{x^2}$ satisfies the condition. So $f(x) = \\frac{1}{x^2}$ is the answer.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 28143,
    "subject": "Mathematics (Multi-modal)",
    "question": "In triangle $ABC$ the bisector of angle $BCA$ intersects the circumcircle at $R$, the perpendicular bisector of $BC$ at $P$, and the perpendicular bisector of $AC$ at $Q$. The midpoint of $BC$ is $K$ and the midpoint of $AC$ is $L$. Prove that the triangles $RPK$ and $RQL$ have the same area.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "If $AC = BC$, $\\triangle ABC$ is an isosceles triangle, and $CR$ is the symmetry axis of $\\triangle RQL$ and $\\triangle RPK$. The conclusion is obviously true.\n\nIf $AC \\ne BC$, without loss of generality, let $AC < BC$. Denote the center of circumcircle of $\\triangle ABC$ by $O$.\n\nSince the right triangles $CQL$ and $CPK$ are similar,\n![](attached_image_1.png)\n$$\n\\angle CPK = \\angle CQL = \\angle OQP, \\text{ and } \\frac{QL}{PK} = \\frac{CQ}{CP}. \\quad ①\n$$\nLet $l$ be the perpendicular bisector of $CR$, then $O$ is on $l$.\nSince $\\triangle OPQ$ is an isosceles triangle, $P$ and $Q$ are two points symmetrical about $l$ on $CR$.\nSo\n$$\nRP = CQ \\quad \\text{and} \\quad RQ = CP. \\qquad \\textcircled{2}\n$$\nBy ①, ②,\n$$\n\\begin{aligned} \\frac{S(\\triangle RQL)}{S(\\triangle RPK)} &= \\frac{\\frac{1}{2} \\cdot RQ \\cdot QL \\cdot \\sin \\angle RQL}{\\frac{1}{2} \\cdot RP \\cdot PK \\cdot \\sin \\angle RPK} \\\\ &= \\frac{RQ}{RP} \\cdot \\frac{QL}{PK} = \\frac{CP}{CQ} \\cdot \\frac{CQ}{CP} = 1. \\end{aligned}\n$$\nHence the two triangles have the same area.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 28144,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $f(x) = 3\\sin x + 2\\cos x + 1$. If real numbers $a, b, c$ are such that $af(x) + bf(x-c) = 1$ holds for any $x \\in \\mathbb{R}$, then $\\frac{b\\cos c}{a}$ equals ( ).",
    "options": [],
    "answer": "C",
    "solution": "Let $c = \\pi$. Then $f(x) + f(x-c) = 2$ for any $x \\in \\mathbb{R}$.\nNow let $a = b = \\frac{1}{2}$, and $c = \\pi$. We have\n$$\naf(x) + bf(x-c) = 1\n$$\nfor any $x \\in \\mathbb{R}$. Consequently, $\\frac{b\\cos c}{a} = -1$. So Answer is (C).\n\nMore generally, we have\n$$\nf(x) = \\sqrt{13}\\sin(x + \\varphi) + 1,\n$$\n$$\nf(x-c) = \\sqrt{13}\\sin(x + \\varphi - c) + 1,\n$$\nwhere $0 < \\varphi < \\frac{\\pi}{2}$ and $\\tan \\varphi = \\frac{2}{3}$. Then $af(x) + bf(x-c) = 1$ becomes\n$$\n\\sqrt{13}a\\sin(x + \\varphi) + \\sqrt{13}b\\sin(x + \\varphi - c) + a + b = 1.\n$$\nThat is,\n$$\n\\sqrt{13}a\\sin(x + \\varphi) + \\sqrt{13}b\\sin(x + \\varphi)\\cos c - \\sqrt{13}b\\sin c\\cos(x + \\varphi) + (a + b - 1) = 0.\n$$\nTherefore\n$$\n\\sqrt{13}(a + b\\cos c)\\sin(x + \\varphi) - \\sqrt{13}b\\sin c\\cos(x + \\varphi) + (a + b - 1) = 0.\n$$\nSince the equality above holds for any $x \\in \\mathbb{R}$, we must have\n$$\n\\begin{cases} a + b\\cos c = 0, \\\\ b\\sin c = 0, \\end{cases} \\qquad \\textcircled{1}\n$$\n$$\n\\begin{cases} b\\sin c = 0, \\\\ a + b - 1 = 0. \\end{cases} \\qquad \\textcircled{2}\n$$\n$$\n\\begin{cases} a + b - 1 = 0. \\end{cases} \\qquad \\textcircled{3}\n$$\nIf $b=0$, then $a=0$ from ①, and this contradicts ③. So $b \\neq 0$, and $\\sin c = 0$ from ②. Therefore $c = 2k\\pi + \\pi$ or $c = 2k\\pi$ ($k \\in \\mathbb{Z}$).\nIf $c = 2k\\pi$, then $\\cos c = 1$, and it leads to a contradiction between ① and ③. So $c = 2k\\pi + \\pi$ ($k \\in \\mathbb{Z}$) and $\\cos c = -1$. From ① and ③, we get $a = b = \\frac{1}{2}$. Consequently, $\\frac{b\\cos c}{a} = -1$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 28145,
    "subject": "Mathematics (Multi-modal)",
    "question": "Point $D$ lies inside triangle $ABC$ such that $\\angle DAC = \\angle DCA = 30^\\circ$ and $\\angle DBA = 60^\\circ$. Point $E$ is the midpoint of segment $BC$. Point $F$ lies on segment $AC$ with $AF = 2FC$. Prove that $DE \\perp EF$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $G$ and $M$ be the midpoints of segments $AF$ and $AC$ respectively.\n\nIn right triangle $ADM$, $\\angle ADM = 60^\\circ$ and $AM = \\sqrt{3}DM$.\n\nNote that $AM = 3GM$. Hence $DM = \\sqrt{3}GM$ and $DG = 2GM = AG = GF$. By symmetry, $DF = GF$ and $DFG$ is an isosceles triangle.\n\n![](attached_image_1.png)\n\nIn triangle $ADG$, $AG = DG$ and $\\angle ADG = 30^\\circ$. Since $\\angle ABD = \\angle DGF = 60^\\circ$, $ABDG$ is concyclic, implying that $\\angle ABG = \\angle ADG = 30^\\circ$. Note that $EF$ and $EM$ are midlines in triangles $BGC$ and $BAC$ respectively. In particular, $EF // BG$ and $EM // BA$, implying that $\\angle MEF = \\angle ABG = 30^\\circ$.\n\nTherefore, $\\angle MEF = \\angle MDF = 30^\\circ$ and $MDEF$ is concyclic, from which it follows that $\\angle DEF = 180^\\circ - \\angle DMF = 90^\\circ$.\n(We maintain the notations of the first proof.)\n\nLet $N$ be the midpoint of segment $CD$. Then $EN$ and $ME$ are the respective midlines in triangles $BDC$ and $ABC$. In particular, $EN // BD$ and $EM // BA$, implying that $\\angle MEN = \\angle ABD = 60^\\circ$. Thus, $\\angle MDN = \\angle MEN = 60^\\circ$, that is, $MDEN$ is concyclic.\n\nIt is easy to compute that $AC = \\sqrt{3}CD$, $CF = \\sqrt{3}CN$ and $\\frac{CD}{CM} = \\frac{CF}{CN}$. Thus, triangles $CNF$ and $CMD$ are similar. Consequently, we deduce that $\\angle CNF = \\angle FMD = 90^\\circ$, that is, $MDNF$ is concyclic.\n\n![](attached_image_2.png)\n\nTherefore, $MDENF$ is concyclic (with $DF$ as its diameter) and $\\angle DEF = 90^\\circ$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 28146,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $f(x): f(x + 1) - f(x) = 2x + 1$ ($x \\in \\mathbb{R}$), and $|f(x)| \\le 1$ when $x \\in [0, 1]$. Prove:\n$$\n|f(x)| \\le 2 + x^2 \\quad (x \\in \\mathbb{R}).\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $g(x) = f(x) - x^2$, then\n$$\n\\begin{align*}\ng(x+1) - g(x) &= f(x+1) - f(x) - (x+1)^2 + x^2 \\\\\n&= (2x + 1) - (x^2 + 2x + 1 - x^2) \\\\\n&= 2x + 1 - (2x + 1) \\\\\n&= 0.\n\\end{align*}\n$$\nThus, $g(x)$ is a periodic function with 1 as its period. On the other hand, as is given $|f(x)| \\le 1$ when $x \\in [0, 1]$, so\n$$\n|g(x)| = |f(x) - x^2| \\le |f(x)| + x^2 \\le 1 + x^2 \\le 2,\n$$\nwhen $x \\in [0, 1]$.\nTherefore, the periodic function $g(x)$ satisfies $|g(x)| \\le 2$ ($x \\in \\mathbb{R}$), thus arriving at\n$$\n|f(x)| = |g(x) + x^2| \\le |g(x)| + x^2 \\le 2 + x^2 \\quad (x \\in \\mathbb{R}),\n$$\nas desired.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 28147,
    "subject": "Mathematics (Multi-modal)",
    "question": "Is there a triangle with sides of integral length, such that the length of the shortest side is $2007$ and that the largest angle is twice the smallest?",
    "options": [],
    "answer": "No",
    "solution": "We shall prove that no such a triangle satisfies the condition.\nIf $\\triangle ABC$ satisfies the condition, let $\\angle A \\le \\angle B \\le \\angle C$, then $\\angle C = 2\\angle A$, and $a = 2007$. Draw the bisector of $\\angle ACB$ and let it intersect $AB$ at point $D$. Then $\\angle BCD = \\angle A$, so $\\triangle CDB \\sim \\triangle ACB$, it follows that\n$$\n\\frac{CB}{AB} = \\frac{BD}{BC} = \\frac{CD}{AC} = \\frac{BD + CD}{BC + AC} = \\frac{AB}{BC + AC}.\n$$\nThus\n$$\nc^2 = a(a+b) = 2007(2007+b), \\quad \\text{①}\n$$\nwhere $2007 \\le b \\le c < 2007+b$.\nSince $a, b, c$ are integers, so $2007|c^2$, then $3 \\cdot 223|c^2$. We can let $c = 669m$, from ①, we get $223m^2 = 2007+b$. Thus $b = 223m^2 - 2007 \\ge 2007$, so $m \\ge 5$.\nBut $c \\ge b$, so $669m \\ge 223m^2 - 2007$, this implies $m < 5$, contradiction.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 28148,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $x_1, \\dots, x_n$ ($n \\ge 2$) be real numbers such that\n$$\nA = \\left| \\sum_{i=1}^{n} x_i \\right| \\neq 0\n$$\nand\n$$\nB = \\max_{1 \\le i < j \\le n} |x_i - x_j| \\neq 0.\n$$\nProve that for every $n$ vectors $\\alpha_1, \\dots, \\alpha_n$ on the plane, there exists a permutation $(k_1, k_2, \\dots, k_n)$ of $(1, 2, \\dots, n)$ such that\n$$\n\\left| \\sum_{i=1}^{n} x_{k_i} \\alpha_i \\right| \\ge \\frac{AB}{2A+B} \\max_{1 \\le i \\le n} |\\alpha_i|.\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "**Proof** Let $|\\alpha_k| = \\max_{1 \\le i \\le n} |\\alpha_i|$. It is sufficient to prove that\n$$\n\\max_{(k_1, \\dots, k_n) \\in S_n} \\left| \\sum_{i=1}^{n} x_{k_i} \\alpha_i \\right| \\ge \\frac{AB}{2A+B} |\\alpha_k|,\n$$\nwhere $S_n$ is the set of all permutations of $(1, 2, \\dots, n)$.\n\nWithout loss of generality, assume\n$$\n|x_n - x_1| = \\max_{1 \\le i < j \\le n} |x_j - x_i| = B,\n$$\n$$\n|\\alpha_n - \\alpha_1| = \\max_{1 \\le i < j \\le n} |\\alpha_j - \\alpha_i|.\n$$\nFor the two vectors\n$$\n\\beta_1 = x_1\\alpha_1 + x_2\\alpha_2 + \\cdots + x_{n-1}\\alpha_{n-1} + x_n\\alpha_n,\n$$\n$$\n\\beta_2 = x_n\\alpha_1 + x_2\\alpha_2 + \\cdots + x_{n-1}\\alpha_{n-1} + x_1\\alpha_n,\n$$\nwe have\n$$\n\\begin{aligned}\n\\max_{(k_1, \\dots, k_n) \\in S_n} \\left| \\sum_{i=1}^{n} x_{k_i} \\alpha_i \\right| & \\ge \\max\\{|\\beta_1|, |\\beta_2|\\} \\\\\n& \\ge \\frac{1}{2} (|\\beta_1| + |\\beta_2|) \\\\\n& \\ge \\frac{1}{2} |\\beta_1 - \\beta_2| \\\\\n& = \\frac{1}{2} |x_1 \\alpha_n + x_n \\alpha_1 - x_1 \\alpha_1 - x_n \\alpha_n| \\\\\n& = \\frac{1}{2} |x_1 - x_n| \\cdot |\\alpha_1 - \\alpha_n| \\\\\n& = \\frac{1}{2} B |\\alpha_n - \\alpha_1|. \\tag{1}\n\\end{aligned}\n$$\nNow suppose $|\\alpha_n - \\alpha_1| = x |\\alpha_k|$. Using the Triangle Inequality, we obtain $0 \\le x \\le 2$. So (1) becomes\n$$\n\\max_{(k_1, \\dots, k_n) \\in S_n} \\left| \\sum_{i=1}^{n} x_{k_i} \\alpha_i \\right| \\ge \\frac{1}{2} B x |\\alpha_k|. \\tag{2}\n$$\nOn the other hand, consider the vectors\n$$\n\\gamma_1 = x_1 \\alpha_1 + x_2 \\alpha_2 + \\dots + x_{n-1} \\alpha_{n-1} + x_n \\alpha_n\n$$\n$$\n\\gamma_2 = x_2 \\alpha_1 + x_3 \\alpha_2 + \\dots + x_n \\alpha_{n-1} + x_1 \\alpha_n\n$$\n$$\n\\gamma_n = x_n \\alpha_1 + x_1 \\alpha_2 + \\dots + x_{n-2} \\alpha_{n-1} + x_{n-1} \\alpha_n.\n$$\nThen we have\n$$\n\\begin{aligned}\n\\max_{(k_1, \\dots, k_n) \\in S_n} \\left| \\sum_{i=1}^n x_{k_i} \\alpha_i \\right| &\\ge \\max_{1 \\le i \\le n} |\\gamma_i| \\\\\n&\\ge \\frac{1}{n} (|\\gamma_1| + \\dots + |\\gamma_n|) \\\\\n&\\ge \\frac{1}{n} |\\gamma_1 + \\dots + \\gamma_n| \\\\\n&= \\frac{A}{n} |\\alpha_1 + \\dots + \\alpha_n| \\\\\n&= \\frac{A}{n} \\left| n\\alpha_k - \\sum_{j \\ne k} (\\alpha_k - \\alpha_j) \\right| \\\\\n&\\ge \\frac{A}{n} \\left( n \\left| \\alpha_k \\right| - \\sum_{j \\ne k} \\left| \\alpha_k - \\alpha_j \\right| \\right) \\\\\n&\\ge \\frac{A}{n} \\left( n \\left| \\alpha_k \\right| - (n-1) \\left| \\alpha_n - \\alpha_1 \\right| \\right) \\\\\n&= \\frac{A}{n} \\left( n \\left| \\alpha_k \\right| - (n-1)x \\left| \\alpha_k \\right| \\right) \\\\\n&= A \\left( 1 - \\frac{n-1}{n} x \\right) |\\alpha_k|. \\quad \\textcircled{3}\n\\end{aligned}\n$$\nFrom (2) and (3), it follows that\n$$\n\\begin{aligned}\n\\max_{(k_1, \\dots, k_n) \\in S_n} \\left| \\sum_{i=1}^n x_{k_i} \\alpha_i \\right| &\\ge \\max\\left\\{\\frac{Bx}{2}, A\\left(1 - \\frac{n-1}{n}x\\right)\\right\\} |\\alpha_k| \\\\\n&\\ge \\frac{\\frac{Bx}{2} \\cdot A \\cdot \\frac{n-1}{n} + A\\left(1 - \\frac{n-1}{n}x\\right) \\cdot \\frac{B}{2}}{A \\cdot \\frac{n-1}{n} + \\frac{B}{2}} |\\alpha_k| \\\\\n&= \\frac{AB}{2A + B - \\frac{2A}{n}} |\\alpha_k| \\\\\n&\\ge \\frac{AB}{2A + B} |\\alpha_k|.\n\\end{aligned}\n$$",
    "topic": "Discrete Mathematics",
    "subtopic": "Combinatorics"
  },
  {
    "id": 28149,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $a$ and $b$ be positive integers. Show that if $4ab - 1$ divides $(4a^2 - 1)^2$, then $a = b$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "**Proof** Call $(a, b)$ a “bad pair” if it satisfies $4ab - 1 \\mid (4a^2 - 1)^2$ while $a \\neq b$. We use the method of infinite descent to prove there is no such “bad pair”.\n\n**Property 1** If $(a, b)$ is a “bad pair” and $a < b$, there exists an integer $c$ ($c < a$) such that $(a, c)$ is also a “bad pair”.\nIn fact, let $r = \\frac{(4a^2 - 1)^2}{4ab - 1}$, then\n$$\nr = -r \\cdot (-1) = -(4a^2 - 1)^2 = -1 \\pmod{4a}.\n$$\nTherefore there exists an integer $c$ such that $r = 4ac - 1$. Since $a < b$, we have\n$$\n4ac - 1 = \\frac{(4a^2 - 1)^2}{4ab - 1} < 4a^2 - 1.\n$$\nSo $c < a$ and $4ac - 1 \\mid (4a^2 - 1)^2$. Thus $(a, c)$ is a “bad pair” too.\n\n**Property 2** If $(a, b)$ is a “bad pair”, so is $(b, a)$.\nIn fact, by\n$$\n1 = 1^2 \\equiv (4ab)^2 \\pmod{4ab - 1},\n$$\nwe get\n$$\n(4b^2 - 1)^2 \\equiv (4b^2 - (4ab)^2)^2 = 16b^4(4a^2 - 1)^2 \\\\\n\\equiv 0 (\\text{mod } 4ab - 1).\n$$\nThus $4ab - 1 \\mid (4b^2 - 1)^2$.\n\nIn the following we will show that such “bad pair” does not exist. We shall prove by contradiction.\nSuppose there is at least one “bad pair”, we choose such pair for which $2a + b$ is minimum.\nIf $a < b$, by property 1, there is a “bad pair” $(a, c)$ which satisfies $c < b$, and $2a + c < 2a + b$, a contradiction.\nIf $b < a$, by property 2, $(b, a)$ is also a “bad pair”, which leads to $2b + a < 2a + b$, a contradiction.\nHence such “bad pair” does not exist. Therefore $a = b$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 28150,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find the smallest positive integer $n \\ge 9$ satisfying that for any group of integers $a_1, a_2, \\dots, a_n$, there always exist $a_{i_1}, a_{i_2}, \\dots, a_{i_9}$ ($1 \\le i_1 < i_2 < \\dots < i_9 \\le n$) and $b_i \\in \\{4, 7\\}$ ($i = 1, 2, \\dots, 9$) such that $b_1 a_{i_1} + b_2 a_{i_2} + \\dots + b_9 a_{i_9}$ is a multiple of 9.",
    "options": [],
    "answer": "13",
    "solution": "Let $a_1 = a_2 = 1$, $a_3 = a_4 = 3$, $a_5 = \\dots = a_{12} = 0$.\nIt is easy to check that any 9 integers of them will not meet the requirement. So $n \\ge 13$. We will prove that $n = 13$.\n\nWe only need to prove the following statement:\nGiven a group of $m$ integers $a_1, a_2, \\dots, a_m$, if there are not three $a_{i_1}, a_{i_2}, a_{i_3}$ in them and $b_1, b_2, b_3 \\in \\{4, 7\\}$ such that $b_1 a_{i_1} + b_2 a_{i_2} + b_3 a_{i_3} \\equiv 0 \\pmod{9}$, then either $m \\le 6$ or $7 \\le m \\le 8$ and there are $a_{i_1}, a_{i_2}, \\dots, a_{i_6}$ in $a_1, a_2, \\dots, a_m$ and $b_1, b_2, \\dots, b_6 \\in \\{4, 7\\}$ such that $9 \\mid b_1 a_{i_1} + b_2 a_{i_2} + \\dots + b_6 a_{i_6}$.\n\nWe define\n$$\nA_1 = \\{i \\mid 1 \\le i \\le m, 9 \\mid a_i\\},\n$$\n$$\nA_2 = \\{i \\mid 1 \\le i \\le m, a_i \\equiv 3 \\pmod{9}\\},\n$$\n$$\nA_3 = \\{i \\mid 1 \\le i \\le m, a_i \\equiv 6 \\pmod{9}\\},\n$$\n$$\nA_4 = \\{i \\mid 1 \\le i \\le m, a_i \\equiv 1 \\pmod{3}\\},\n$$\n$$\nA_5 = \\{i \\mid 1 \\le i \\le m, a_i \\equiv 2 \\pmod{3}\\}.\n$$\nThen $|A_1| + |A_2| + |A_3| + |A_4| + |A_5| = m$ and\n\n(1) if $i \\in A_2, j \\in A_3$ then $9 \\mid 4a_i + 4a_j$;\n\n(2) if $i \\in A_4, j \\in A_5$ then one of $4a_i + 4a_j, 4a_i + 7a_j$ and $7a_i + 4a_j$ is a multiple of 9 as all of them are divisible by 3 and they are distinct according to mod 9;\n\n(3) if either $i, j, k \\in A_2$ or $i, j, k \\in A_3$ then $9 \\mid 4a_i + 4a_j + 4a_k$;\n\n(4) if either $i, j, k \\in A_4$ or $i, j, k \\in A_5$ then one of $4a_i + 4a_j + 4a_k, 4a_i + 4a_j + 7a_k$ and $4a_i + 7a_j + 7a_k$ is a multiple of 9 as all of them are divisible by 3 and they are distinctive according to mod 9.\n\nBy the assumption, we have $|A_i| \\le 2$ ($1 \\le i \\le 5$).\n\nIf $|A_1| \\ge 1$, then $|A_2| + |A_3| \\le 2$, $|A_4| + |A_5| \\le 2$. Hence\n$$\nm = |A_1| + |A_2| + |A_3| + |A_4| + |A_5| \\le 6.\n$$\nNow assume $|A_1| = 0$ and $m \\ge 7$. Then\n$$\n7 \\le m = |A_1| + |A_2| + |A_3| + |A_4| + |A_5| \\le 8.\n$$\nFurther,\n$$\n\\min\\{|A_2|, |A_3|\\} + \\min\\{|A_4|, |A_5|\\} \\ge 3.\n$$\nFrom (1) and (2) we know there exist $i_1, i_2, \\dots, i_6 \\in A_2 \\cup A_3 \\cup A_4 \\cup A_5$ ($i_1 < i_2 < \\dots < i_6$) and $b_1, b_2, \\dots, b_6 \\in \\{4, 7\\}$ such that $9 \\mid b_1a_{i_1} + b_2a_{i_2} + \\dots + b_6a_{i_6}$.\nThe proof of statement is complete.\n\nNow, when $n \\ge 13$ it is easy to verify with the statement, for any group of integers $a_1, a_2, \\dots, a_n$, there always exist $a_{i_1}, a_{i_2}, \\dots, a_{i_9}$ ($1 \\le i_1 < i_2 < \\dots < i_9 \\le n$) and $b_i \\in \\{4, 7\\}$ ($i = 1, 2, \\dots, 9$) such that $b_1a_{i_1} + b_2a_{i_2} + \\dots + b_9a_{i_9}$ is a multiple of 9. That completes the proof.",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 28151,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $A$ and $B$ be two subsets of $\\{1, 2, 3, \\dots, 100\\}$, satisfying $|A| = |B|$ and $A \\cap B = \\emptyset$. If $n \\in A$ always implies $2n + 2 \\in B$, then the maximum of $|A \\cup B|$ is ( ).",
    "options": [],
    "answer": "66",
    "solution": "We will first prove that $|A \\cup B| \\le 66$, or equivalently $|A| \\le 33$. For this purpose, we only need to prove that, if $A$ is a subset of $\\{1, 2, \\dots, 49\\}$ with 34 elements, then there must exist $n \\in A$ such that $2n + 2 \\in A$. The proof is as follows.\n\nDivide $\\{1, 2, \\dots, 49\\}$ into 33 subsets:\n\n$\\{1, 4\\}$, $\\{3, 8\\}$, $\\{5, 12\\}$, $\\dots$, $\\{23, 48\\}$, 12 subsets;\n$\\{2, 6\\}$, $\\{10, 22\\}$, $\\{14, 30\\}$, $\\{18, 38\\}$, 4 subsets;\n$\\{25\\}$, $\\{27\\}$, $\\{29\\}$, $\\dots$, $\\{49\\}$, 13 subsets;\n$\\{26\\}$, $\\{34\\}$, $\\{42\\}$, $\\{46\\}$, 4 subsets.\n\nBy the Pigeonhole Principle we know that there exists at least one subset with 2 elements among them which is also a subset of $A$. That means there exists $n \\in A$ such that $2n + 2 \\in A$.\n\nOn the other hand, let\n\n$A = \\{1, 3, 5, \\dots, 23, 2, 10, 14, 18, 25, 27, 29, \\dots, 49, 26, 34, 42, 46\\}$\n\n$B = \\{2n + 2 \\mid n \\in A\\}$.\n\nWe find that $A$ and $B$ satisfy the condition and $|A \\cup B| = 66$.\n\nAnswer: B.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 28152,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find the array of prime numbers $(a, b, c)$ satisfying conditions as follows:\n(1) $a < b < c < 100$, where $a$, $b$, $c$ are all prime numbers;\n(2) $a + 1$, $b + 1$, $c + 1$ constitute a geometric progression.",
    "options": [],
    "answer": "(2, 5, 11), (5, 11, 23), (7, 11, 17), (5, 17, 53), (11, 23, 47), (2, 11, 47), (17, 23, 31), (7, 23, 71), (31, 47, 71), (17, 41, 97), (71, 83, 97)",
    "solution": "From condition (2), we get\n$$\n(a+1)(c+1) = (b+1)^2.\n$$\nSet $a+1 = n^2 x$, $c+1 = m^2 y$, with no square factor larger than 1 in $x$, $y$, then we could get that $x = y$. This is due to the fact that from above, we have\n$$\n(mn)^2 x y = (b+1)^2,\n$$\nwhich means $mn \\mid (b+1)$. Set $b+1 = mn \\cdot w$, then the above can be simplified to\n$$\nx y = w^2.\n$$\nIf $w > 1$, then the prime number $p_1 \\mid w \\Rightarrow p_1^2 \\mid w^2$. As there is no square factor larger than 1 in $x$, $y$, then $p_1 \\mid x$ and $p_1 \\mid y$. Now set $x = p_1 x_1$, $y = p_1 y_1$, $w = p_1 w_1$. Then the above can be simplified to\n$$\nx_1 y_1 = w_1^2.\n$$\nIf $w_1 > 1$ still exists, then there will be a new prime number $p_2 \\mid w_1 \\Rightarrow p_2^2 \\mid w_1^2$. As there is no square factor larger than 1 in $x_1$, $y_1$, then $p_2 \\mid x_1$ and $p_2 \\mid y_1$. Now set\n$$\nx_1 = p_2 x_2, \\quad y_1 = p_2 y_2, \\quad w_1 = p_2 w_2.\n$$\nThen the above can be simplified to $x_2 y_2 = w_2^2, \\cdots$. Since there are finite prime factors of $w$, then carrying on as above, we obtain that there exists $r$, such that $w_r = 1$. $x_r y_r = w_r^2 \\Rightarrow x_r = y_r = 1$, we have $x = p_1 p_2 \\cdots p_r = y$ as desired. Now we can set $x = y = k$, and have\n$$\n\\begin{cases}\na = k n^2 - 1, \\\\\nb = k m n - 1, \\\\\nc = k m^2 - 1,\n\\end{cases}\n$$\nwhere\n$$\n1 \\le n < m, \\quad a < b < c < 100,\n$$\nwith no square factor larger than 1 in $k$ and $k \\ne 1$. Otherwise, if $k = 1$, then $c = m^2 - 1$. As $c$ is larger than the third prime number 5, thus $c = m^2 - 1 > 5 \\Rightarrow m \\ge 3$ and\n$$\nc = m^2 - 1 = (m - 1)(m + 1)\n$$\nis a composite number. Contradiction! Hence, $k$ is either a prime number, or the product of several different prime numbers (that is, $k$ is larger than 1 and with no square factor larger than 1 in $k$). We say that \"$k$ has the property $p$\".\n\na.\nFrom above, $m \\ge 2$. When $m = 2$, then $n = 1$ and\n$$\n\\begin{cases}\na = k - 1, \\\\\nb = 2k - 1, \\\\\nc = 4k - 1.\n\\end{cases}\n$$\nSince $c < 100 \\Rightarrow k < 25$, then if $k \\equiv 1 \\pmod{3}$, we get $3 \\mid c$ and $c > 3$, which means $c$ is a composite number.\nIf $k \\equiv 2 \\pmod{3}$, then when it is even, the $k$ satisfying the property $p$ is 2 or 14, where the corresponding $a = 2 - 1 = 1$ and $b = 2 \\cdot 14 - 1 = 27$ are not prime numbers. On the other hand, when it is odd, the $k$ satisfying the property $p$ is 5, 11, 17 or 23, where all the corresponding $a = k - 1$ are not prime numbers.\nIf $k \\equiv 0 \\pmod{3}$, the $k$ satisfying the property $p$ is 3, 6, 15 or 21. When $k = 3$, we get the first solution\n$$\nf_1 = (a, b, c) = (2, 5, 11).\n$$\nWhen $k = 6$, the second solution is\n$$\nf_2 = (a, b, c) = (5, 11, 23).\n$$\nBut when $k = 15, 21$, the corresponding $a = k - 1$ are not prime numbers.\n\nb.\nWhen $m = 3$, then $n = 2$ or 1. If $m = 3, n = 2$, we have\n$$\n\\begin{cases}\na = 4k - 1, \\\\\nb = 6k - 1, \\\\\nc = 9k - 1.\n\\end{cases}\n$$\nSince $c \\le 97 \\Rightarrow k \\le 10$, then the $k$ satisfying the property $p$ is 2, 3, 5, 6, 7 or 10.\nWhen $k = 3, 5, 7$, the corresponding $c = 9k - 1$ are all composite numbers.\nWhen $k = 6, b = 6k - 1 = 35$, which is a composite number.\nWhen $k = 10, a = 4k - 1 = 39$, which is also a composite number. But when $k = 2$, we get the third solution\n$$\nf_3 = (a, b, c) = (7, 11, 17).\n$$\nIf $m = 3, n = 1$, we have\n$$\n\\begin{cases}\na = k - 1, \\\\\nb = 3k - 1, \\\\\nc = 9k - 1.\n\\end{cases}\n$$\nAs $k \\le 10$, the $k$ satisfying the property $p$ is 2, 3, 5, 6, 7 or 10. When $k = 3, 5, 7,$ the corresponding $b = 3k - 1$ are all composite numbers. When $k = 2, 10$, the corresponding $a = k - 1$ are not prime numbers. But when $k = 6$, we get the fourth solution\n$$\nf_4 = (a, b, c) = (5, 17, 53).\n$$\n\nc.\nWhen $m = 4$, from $c = 16k - 1 \\le 97 \\Rightarrow k \\le 6$, then the $k$ satisfying the property $p$ is 2, 3, 5 or 6. When $k = 6$, $c = 16 \\cdot 6 - 1 = 95$, which is a composite number. When $k = 5$, then\n$$\n\\begin{cases}\na = 5n^2 - 1, \\\\\nb = 20n - 1.\n\\end{cases}\n$$\nAs $n < m = 4$, $n$ can be 1, 2, 3, which means at least one of $a, b$ is not a prime number.\nWhen $k = 3, c = 48 - 1 = 47$ and\n$$\n\\begin{cases}\na = 3n^2 - 1, \\\\\nb = 12n - 1.\n\\end{cases}\n$$\nConsidering $n < m = 4$, the corresponding $a, b$ are both composite numbers if $n = 3$. But, if $n = 2$, we get the fifth solution\n$$\nf_5 = (a, b, c) = (11, 23, 47).\n$$\nIf $n = 1$, we get the sixth solution\n$$\nf_6 = (a, b, c) = (2, 11, 47).\n$$\nWhen $k = 2$, $c = 16 \\cdot k - 1 = 31$ and\n$$\n\\begin{cases}\na = 2n^2 - 1, \\\\\nb = 8n - 1.\n\\end{cases}\n$$\nSince $n < m = 4$, the seventh solution\n$$\nf_7 = (a, b, c) = (17, 23, 31)\n$$\nexists if $n = 3$.\n\nd.\nWhen $m = 5$, then $c = 25k - 1 \\le 97$ and the $k$ satisfying the property $p$ is 2 or 3, but the corresponding $c = 25k - 1$ are both composite numbers.\n\ne.\nWhen $m = 6$, then $c = 36k - 1 \\le 97$ and the $k$ satisfying the property $p$ is 2. Hence, $c = 2 \\cdot 36 - 1 = 71$ and\n$$\n\\begin{cases}\na = 2n^2 - 1, \\\\\nb = 12n - 1.\n\\end{cases}\n$$\nAs $n < m = 6$, the eighth solution\n$$\nf_8 = (a, b, c) = (7, 23, 71)\n$$\nexists if $n = 2$, and the ninth solution\n$$\nf_9 = (a, b, c) = (31, 47, 71)\n$$\nexists if $n = 4$.\n\nf.\nWhen $m = 7$, then $c = 49k - 1 \\le 97$ and the $k$ satisfying the property $p$ is 2. Hence, $c = 2 \\cdot 49 - 1 = 97$ and\n$$\n\\begin{cases}\na = 2n^2 - 1, \\\\\nb = 14n - 1.\n\\end{cases}\n$$\nAs $n < m = 7$, the tenth solution\n$$\nf_{10} = (a, b, c) = (17, 41, 97)\n$$\nexists if $n = 3$, and the eleventh solution\n$$\nf_{11} = (a, b, c) = (71, 83, 97)\n$$\nexists if $n = 6$.\n\ng.\nWhen $m \\ge 8$, then $c = 64k - 1 \\le 97$, but the $k$ satisfying the property $p$ does not exist.\n\nTherefore, there are 11 possible solutions, namely $f_1, f_2, \\dots, f_{11}$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 28153,
    "subject": "Mathematics (Multi-modal)",
    "question": "Given four fixed points $A(-3, 0)$, $B(1, -1)$, $C(0, 3)$, $D(-1, 3)$ and a variable point $P$ in a plane rectangular coordinates system, the minimum of $|PA| + |PB| + |PC| + |PD|$ is ______.",
    "options": [],
    "answer": "3√2 + 2√5",
    "solution": "As shown in the figure, assuming that $AC$ and $BD$ meet at point $F$, we have\n$$\n|PA| + |PC| \\ge |AC| = |FA| + |FC|\n$$\nand\n$$\n|PB| + |PD| \\ge |BD| = |FB| + |FD|.\n$$\nWhen $P$ coincides with $F$, $|PA| + |PB| + |PC| + |PD|$ reaches the minimum.\nThat is, $|AC| + |BD| = 3\\sqrt{2} + 2\\sqrt{5}$.\nSo $3\\sqrt{2} + 2\\sqrt{5}$ is the required answer.\n\n![](attached_image_1.png)",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 28154,
    "subject": "Mathematics (Multi-modal)",
    "question": "There are $n$ white and $n$ black balls placed randomly on the circumference of a circle. Starting from a certain white ball, number all white balls in a clockwise direction by $1, 2, \\dots, n$. Likewise, number all black balls by $1, 2, \\dots, n$ in an anti-clockwise direction starting from a certain black ball. Prove that there exist consecutive $n$ balls whose numbering forms the set $\\{1, 2, \\dots, n\\}$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "**Proof** Choose a black ball and a white ball with the same number, and the number of balls between the two balls is minimum. We can suppose the number of the two balls is $1$.\nFirstly, we shall prove that the balls between the two balls have the same color.\n\nIn fact, if they are of different color, then the white ball and the black ball, each is numbered by $n$, are between the two balls (See Fig. 1). This is a contradiction to the point that number of balls between the two balls (labelled by ‘1’ s) is a minimum.\n\nSecondly, if the balls between the two balls (labelled by ‘1’s) are white, we have two cases.\n\n**Case 1** The number of the white balls are $2, \\dots, k$. See Fig. 2, then from the white balls (labelled by ‘1’s) in anti-clockwise direction we can get a chain of $n$ balls whose numbering forms the set $\\{1, 2, \\dots, n\\}$.\n\n![](attached_image_1.png)\nFig. 1\n\n![](attached_image_2.png)\nFig. 2\n\n**Case 2** The number of the white balls are $k, k+1, \\dots, n$ (See Fig. 3), then from the white ball (labelled by ‘1’s) in clockwise direction we can have a chain of $n$ balls that satisfies the condition.\n\nThe same argument can prove that the claim holds, if the balls between the two balls (labelled by ‘1’s) are black, or there are no ball between them.\n\n![](attached_image_3.png)\nFig. 3",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 28155,
    "subject": "Mathematics (Multi-modal)",
    "question": "Given real numbers $a$, $b$, $c$ such that $abc = 1$, prove that for all the integers $k \\ge 2$,\n$$\n\\frac{a^k}{a+b} + \\frac{b^k}{b+c} + \\frac{c^k}{c+a} \\ge \\frac{3}{2}.\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Since\n$$\n\\frac{a^k}{a+b} + \\frac{1}{4}(a+b) + \\underbrace{\\frac{1}{2} + \\frac{1}{2} + \\dots + \\frac{1}{2}}_{k-2} \\ge k \\cdot \\sqrt[k]{\\frac{a^k}{2^k}} = \\frac{k}{2}a,\n$$\nthen\n$$\n\\frac{a^k}{a+b} \\ge \\frac{k}{2}a - \\frac{1}{4}(a+b) - \\frac{k-2}{2}.\n$$\nSimilarly,\n$$\n\\frac{b^k}{b+c} \\ge \\frac{k}{2}b - \\frac{1}{4}(b+c) - \\frac{k-2}{2},\n$$\n$$\n\\frac{c^k}{c+a} \\ge \\frac{k}{2}c - \\frac{1}{4}(c+a) - \\frac{k-2}{2}.\n$$\nAdding the three inequalities above, we obtain\n$$\n\\begin{aligned}\n& \\frac{a^k}{a+b} + \\frac{b^k}{b+c} + \\frac{c^k}{c+a} \\\\\n\\ge & \\frac{k}{2}(a+b+c) - \\frac{1}{2}(a+b+c) - \\frac{3}{2}(k-2) \\\\\n= & \\frac{k-1}{2}(a+b+c) - \\frac{3}{2}(k-2) \\\\\n\\ge & \\frac{3}{2}(k-1) - \\frac{3}{2}(k-2) \\\\\n= & \\frac{3}{2},\n\\end{aligned}\n$$\nas desired.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 28156,
    "subject": "Mathematics (Multi-modal)",
    "question": "In a round robin chess tournament each player plays with every other player exactly once. The winner of each game gets $1$ point and the loser gets $0$ point. If the game ends in a tie, each player gets $0.5$ point. Given a positive integer $m$, a tournament is said to have property $P(m)$ if the following holds: among every set $S$ of $m$ players, there is one player who won all his games against the other $m-1$ players in $S$ and one player who lost all his games against the other $m-1$ players in $S$.\n\nFor a given integer $m \\ge 4$, determine the minimum value of $n$ (as a function of $m$) such that the following holds: in every $n$-player round robin chess tournament with property $P(m)$, the final scores of the $n$ players are all distinct.",
    "options": [],
    "answer": "2m - 3",
    "solution": "Note that if there are $2m-4$ players, we can label them\n$$\na_1, a_2, \\dots, a_{m-3}, A_{m-2}, B_{m-2}, a_{m-1}, \\dots, a_{2m-5},\n$$\nand assume that player $P_i$ beats player $P_j$ if and only if $i > j$, and $A_{m-2}$ and $B_{m-2}$ are in a tie. It is easy to see that in the group of $m$ players, there exists a unique player $P_i$ with the maximum index $i$ ($m-1 \\le i \\le 2m-5$, and this player won all games against other players in the group), and there exists a unique player $P_i$ with the minimum index $j$ ($1 \\le i \\le m-3$, and this player lost all games against other players in the group). Hence this tournament has property $P(m)$ and not all players have distinct total points. If $n < 2m-4$, we can then build a similar tournament by taking players away from both ends index-wise). Hence the answer is greater than $2m-3$. It suffices to show the following claim:\nIf there are $2m-3$ players in a tournament with property $P(m)$, then the players must have distinct total final score.\n\nIn a group, if a player won (or lose) all games against the rest of the players in the group, we call this player the winner (or loser) of the group. If a player won (or lose) all his games in the tournament, we call this player the complete winner (or complete loser). We establish the following lemmas.\n\n**Lemma 1** In an $n$-player ($n \\ge m$) tournament with property $P(m)$, there is a complete winner.\n\n*Proof:* We implement an induction on $n$. If $n=m$, the statement is trivial. Now assume that the statement is true for some $n=k$ ($k \\ge m$), we consider a ($k+1$)-player tournament with property $P(m)$. Let $a_1, \\cdots, a_{k+1}$ denote the players. By the induction hypothesis, we may assume that $a_{k+1}$ is the winner in the group $a_2, \\cdots, a_{k+1}$. We consider three cases:\n(a) If $a_{k+1}$ won the game against $a_1$, then $a_{k+1}$ is the complete winner;\n(b) If $a_{k+1}$ tied the game against $a_1$, then the group $a_1, a_2, \\cdots, a_{k-1}, a_{k+1}$ has no winner, violating the condition that the tournament has property $P(m)$;\n(c) If $a_{k+1}$ lose the game against $a_1$, then the group $\\{a_1, a_2, \\cdots, a_{k-1}, a_k, a_{k+1}\\} \\setminus \\{a_i\\} (2 \\le i \\le k)$ has a winner, and this winner can only be $a_1$. Thus $a_1$ is the complete winner.\nCombining the three cases, we find a complete winner in the tournament, hence our induction is complete.\n\nIn exactly the same way, we can prove that\n\n**Lemma 2** In an $n$-player ($n \\ge m$) tournament with property $P(m)$, there is a complete loser.\n\nNow we are ready to prove our claim in a similar manner.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 28157,
    "subject": "Mathematics (Multi-modal)",
    "question": "Given $\\triangle ABC$ and $\\triangle AEF$ such that $B$ is the midpoint of $EF$. Also, $AB = EF = 1$, $BC = 6$, $CA = \\sqrt{33}$, and $\\overrightarrow{AB} \\cdot \\overrightarrow{AE} + \\overrightarrow{AC} \\cdot \\overrightarrow{AF} = 2$. The cosine of the angle between $\\overrightarrow{EF}$ and $\\overrightarrow{BC}$ is ______.",
    "options": [],
    "answer": "2/3",
    "solution": "We have\n$$\n\\begin{aligned}\n2 &= \\overrightarrow{AB} \\cdot \\overrightarrow{AE} + \\overrightarrow{AC} \\cdot \\overrightarrow{AF} \\\\\n&= \\overrightarrow{AB} \\cdot (\\overrightarrow{AB} + \\overrightarrow{BE}) + \\overrightarrow{AC} \\cdot (\\overrightarrow{AB} + \\overrightarrow{BF}),\n\\end{aligned}\n$$\ni.e.\n$$\n\\overrightarrow{AB}^2 + \\overrightarrow{AB} \\cdot \\overrightarrow{BE} + \\overrightarrow{AC} \\cdot \\overrightarrow{AB} + \\overrightarrow{AC} \\cdot \\overrightarrow{BF} = 2.\n$$\nAs $\\overrightarrow{AB}^2 = 1$,\n$$\n\\overrightarrow{AC} \\cdot \\overrightarrow{AB} = \\sqrt{33} \\times 1 \\times \\frac{33+1-36}{2 \\times \\sqrt{33} \\times 1} = -1\n$$\nand $\\overrightarrow{BE} = -\\overrightarrow{BF}$, we get\n$$\n1 + \\overrightarrow{BF} \\cdot (\\overrightarrow{AC} - \\overrightarrow{AB}) - 1 = 2,\n$$\ni.e. $\\overrightarrow{BF} \\cdot \\overrightarrow{BC} = 2$. Defining $\\theta$ as the angle between $\\overrightarrow{EF}$ and $\\overrightarrow{BC}$, we get $|\\overrightarrow{BF}| \\cdot |\\overrightarrow{BC}| \\cdot \\cos \\theta = 2$ or $3\\cos \\theta = 2$. So $\\cos \\theta = \\frac{2}{3}$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 28158,
    "subject": "Mathematics (Multi-modal)",
    "question": "Given a unit cube $ABCD-A_1B_1C_1D_1$, construct a ball with point $A$ as the center and of radius $\\frac{2\\sqrt{3}}{3}$. Then the length of the curves resulting from the intersection between the surfaces of the ball and cube is ______.",
    "options": [],
    "answer": "5*sqrt(3)*pi/6",
    "solution": "As shown in the figure, the surface of the ball intersects all of the six surfaces of the cube. The intersection curves are divided into two kinds: One kind lies on the three surfaces including vertex $A$ respectively, that is $AA_1B_1B$, $ABCD$, and $AA_1D_1D$; while the other lies on the three surfaces not including $A$, that is $CC_1D_1D$, $A_1B_1C_1D_1$ and $BB_1C_1C$.\n\n![](attached_image_1.png)\n\nOn surface $AA_1B_1B$, the intersection curve is arc $\\widehat{EF}$ which lies on a circle with $A$ as the center. Since $AE = \\frac{2\\sqrt{3}}{3}$, $AA_1 = 1$, so $\\angle A_1AE = \\frac{\\pi}{6}$. In the same way $\\angle BAF = \\frac{\\pi}{6}$. Therefore $\\angle EAF = \\frac{\\pi}{6}$. That means the length of arc $\\widehat{EF}$ is $\\frac{2\\sqrt{3}}{3} \\cdot \\frac{\\pi}{6} = \\frac{\\sqrt{3}\\pi}{9}$. There are three arcs of this category.\n\nOn surface $BB_1C_1C$, the intersection curve is arc $\\widehat{FG}$ which lies on a circle centred at $B$. The radius equals $\\frac{\\sqrt{3}}{3}$ and $\\angle FBG = \\frac{\\pi}{2}$. So the length of $\\widehat{FG}$ is $\\frac{\\sqrt{3}}{3} \\cdot \\frac{\\pi}{2} = \\frac{\\sqrt{3}\\pi}{6}$. There are also three arcs of this category.\n\nIn summary, the total length of all intersection curves is\n$$\n3 \\times \\frac{\\sqrt{3}}{9}\\pi + 3 \\times \\frac{\\sqrt{3}}{6}\\pi = \\frac{5\\sqrt{3}\\pi}{6}.\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 28159,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $n$ be a positive integer, set $A \\subseteq \\{1, 2, \\dots, n\\}$, and for every $a, b \\in A$, $\\text{lcm}(a, b) \\le n$. Prove that\n$$\n|A| \\le 1.9\\sqrt{n} + 5.\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "**Proof** For $a \\in (\\sqrt{n}, \\sqrt{2n}]$, $\\text{lcm}(a, a+1) = a(a+1) > n$, so $|A \\cap (\\sqrt{n}, \\sqrt{2n}]| \\le \\frac{1}{2}(\\sqrt{2}-1)\\sqrt{n} + 1$.\nFor $a \\in (\\sqrt{2n}, \\sqrt{3n}]$, we have\n$$\n\\text{lcm}(a, a+1) = a(a+1) > n,\n$$\n$$\n\\text{lcm}(a+1, a+2) = (a+1)(a+2) > n,\n$$\n$$\n\\text{lcm}(a, a+2) \\ge \\frac{1}{2}a(a+2) > n.\n$$\nSo\n$$\n|A \\cap (\\sqrt{2n}, \\sqrt{3n}]| \\le \\frac{1}{3}(\\sqrt{3}-\\sqrt{2})\\sqrt{n} + 1.\n$$\nSimilarly\n$$\n|A \\cap (\\sqrt{3n}, 2\\sqrt{n}]| \\le \\frac{1}{4}(\\sqrt{4}-\\sqrt{3})\\sqrt{n} + 1.\n$$\nHence\n$$\n\\begin{aligned}\n|A \\cap [1, 2\\sqrt{n}]| & \\le \\sqrt{n} + \\frac{1}{2}(\\sqrt{2}-1)\\sqrt{n} + \\frac{1}{3}(\\sqrt{3}-\\sqrt{2})\\sqrt{n} \\\\\n& \\quad + \\frac{1}{4}(\\sqrt{4}-\\sqrt{3})\\sqrt{n} + 3 \\\\\n& = \\left(1 + \\frac{\\sqrt{2}}{6} + \\frac{\\sqrt{3}}{12}\\right)\\sqrt{n} + 3.\n\\end{aligned}\n$$\nLet $k \\in \\mathbb{N}^*$, suppose $a, b \\in (\\frac{n}{k+1}, \\frac{n}{k})$, $a > b$, and $\\text{lcm}(a, b) = as = bt$, where $s, t \\in \\mathbb{N}^*$. Then\n$$\n\\frac{a}{(a, b)s} = \\frac{b}{(a, b)t}.\n$$\nSince $\\text{gcd}(\\frac{a}{(a, b)}, \\frac{b}{(a, b)}) = 1$, so $\\frac{b}{(a, b)}|s$. It follows that\n$$\n\\begin{align*}\n\\operatorname{lcm}(a, b) &= as \\ge \\frac{ab}{(a, b)} \\ge \\frac{ab}{a-b} \\\\\n&= b + \\frac{b^2}{a-b} > \\frac{n}{k+1} + \\frac{\\left(\\frac{n}{k+1}\\right)^2}{\\frac{n}{k} - \\frac{n}{k+1}} \\\\\n&= n.\n\\end{align*}\n$$\nTherefore, $|A \\cap (\\frac{n}{k+1}, \\frac{n}{k})| \\le 1$.\nSuppose $T \\in \\mathbb{N}^*$ such that $\\frac{n}{T+1} \\le 2\\sqrt{n} < \\frac{n}{T}$. Then\n$$\n\\begin{aligned}\n|A \\cap (2\\sqrt{n}, n]| &\\le \\sum_{k=1}^{T} \\left|A \\cap \\left(\\frac{n}{k+1}, \\frac{n}{k}\\right]\\right| \\\\\n&\\le T < \\frac{1}{2}\\sqrt{n}.\n\\end{aligned}\n$$\nBy the above arguments, we arrive at\n$$\n|A| \\le \\left(\\frac{3}{2} + \\frac{1}{6}\\sqrt{2} + \\frac{1}{12}\\sqrt{3}\\right)\\sqrt{n} + 3 < 1.9\\sqrt{n} + 5.\n$$",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 28160,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all positive integers $n$ such that there exist non-zero integers $x_1, x_2, \\dots, x_n, y$, satisfying the following conditions\n$$\n\\begin{cases} x_1 + \\cdots + x_n = 0, \\\\ x_1^2 + \\cdots + x_n^2 = n y^2. \\end{cases}\n$$",
    "options": [],
    "answer": "All positive integers except 1 and 3",
    "solution": "It is easy to see that $n > 1$.\n\nWhen $n = 2k$, $k \\in \\mathbb{N}$, let $x_{2i-1} = 1$, $x_{2i} = -1$, $i = 1, 2, \\dots, k$, and $y = 1$, then the condition is satisfied.\n\nWhen $n = 2k + 3$, $k \\in \\mathbb{N}$, let $y = 2$, $x_1 = 4$, $x_2 = x_3 = x_4 = x_5 = -1$, $x_{2i} = 2$, $x_{2i+1} = -2$, $i = 3, 4, \\cdots, k+1$, then the condition is satisfied.\n\nNow if $n = 3$, and there exist $x_1, x_2, x_3, y$ such that\n$$\n\\begin{cases} x_1 + x_2 + x_3 = 0, \\\\ x_1^2 + x_2^2 + x_3^2 = 3y^2. \\end{cases}\n$$\nthen\n$$\n2(x_1^2 + x_2^2 + x_3 x_2) = 3y^2.\n$$\nSuppose $\\gcd(x_1, x_2) = 1$, so $x_1, x_2$ are odd numbers or one is even while the other is odd. Hence $x_1^2 + x_2^2 + x_3 x_2$ is an odd number. But $2 \\mid 3y^2$, so $3y^2 \\equiv 0 \\pmod 4$. This is a contradiction since $2(x_1^2 + x_2^2 + x_3 x_2) \\equiv 2 \\pmod 4$.\n\nThe answer is $n \\neq 1, 3$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 28161,
    "subject": "Mathematics (Multi-modal)",
    "question": "Suppose that the sequence of positive numbers $x_1, x_2, \\ldots, x_n, \\ldots$ satisfies $(8x_2 - 7x_1)x_1^7 = 8$ and\n$$\nx_{k+1}x_{k-1} - x_k^2 = \\frac{x_{k-1}^8 - x_k^8}{(x_kx_{k-1})^7}, \\quad k \\ge 2.\n$$\nFind positive real number $a$ such that when $x_1 > a$ one has $x_1 > x_2 > \\cdots > x_n > \\cdots$, and when $0 < x_1 < a$ one does not have such monotonicity. (Posed by Li Shenghong)",
    "options": [],
    "answer": "a = 8^{1/8}",
    "solution": "By $x_{k+1}x_{k-1} - x_k^2 = \\frac{x_{k-1}^8 - x_k^8}{(x_kx_{k-1})^7}$, we have\n$$\n\\frac{x_{k+1}}{x_k} - \\frac{x_k}{x_{k-1}} = \\frac{1}{x_k^8} - \\frac{1}{x_{k-1}^8},\n$$\ni.e.\n$$\n\\frac{x_{k+1}}{x_k} - \\frac{1}{x_k^8} = \\frac{x_k}{x_{k-1}} - \\frac{1}{x_{k-1}^8} = \\cdots = \\frac{x_2}{x_1} - \\frac{1}{x_1^8} = \\frac{7}{8}.\n$$\nHence, $x_{k+1} = \\frac{7}{8}x_k + x_k^{-7}$, and when $x_1 > 0, x_k > 0, k \\ge 2$.\nBy $x_{k+1} - x_k = x_k(x_k^{-8} - \\frac{1}{8})$, we see that when $x_k^{-8} - \\frac{1}{8} < 0$, i.e. $x_k > 8^{1/8}$, one has $x_{k+1} - x_k < 0$, i.e. $x_{k+1} < x_k$, $k \\ge 1$.\nAnd $x_{k+1} = \\frac{7}{8}x_k + x_k^{-7} \\ge 8\\sqrt[8]{\\frac{1}{8^7}} = 8^{1/8}$, so when $x_k = 8^{1/8}$, the equality holds. So if we take $a = 8^{1/8}$, as soon as $x_k > 8^{1/8}$ we have\n$$\nx_1 > x_2 > \\cdots > x_n > \\cdots.\n$$\nWhen $x_1 < 8^{1/8}$, we have $x_2 > x_1$ and $x_2 > x_3 > \\cdots > x_n > \\cdots$.\nSo the constant that we are looking for is $a=8^{\\frac{1}{8}}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 28162,
    "subject": "Mathematics (Multi-modal)",
    "question": "Suppose that $f(x)$ is defined on $\\mathbb{R}$, satisfying $f(0) = 2008$, and for any $x \\in \\mathbb{R}$\n$$\n\\begin{aligned}\nf(x+2) - f(x) &\\le 3 \\times 2^x, \\\\\nf(x+6) - f(x) &\\ge 63 \\times 2^x.\n\\end{aligned}\n$$\n\nThen $f(2008) = \\underline{\\hspace{2cm}}$.",
    "options": [],
    "answer": "2^{2008} + 2007",
    "solution": "**Solution I** We have\n$$\n\\begin{aligned}\nf(x+2) - f(x) &= -(f(x+4) - f(x+2)) - (f(x+6) - f(x+4)) + (f(x+6) - f(x)) \\\\\n&\\geq -3 \\times 2^{x+2} - 3 \\times 2^{x+4} + 63 \\times 2^x = 3 \\times 2^x.\n\\end{aligned}\n$$\nThis means that $f(x+2) - f(x) = 3 \\times 2^x$. So we have\n$$\n\\begin{aligned}\nf(2008) &= f(2008) - f(2006) + f(2006) - f(2004) + \\dots \\\\\n&\\quad + f(2) - f(0) + f(0) \\\\\n&= 3 \\times (2^{2006} + 2^{2004} + \\dots + 2^2 + 1) + f(0) \\\\\n&= 3 \\times \\frac{4^{1003} + 1}{4 - 1} + 2008 \\\\\n&= 2^{2008} + 2007.\n\\end{aligned}\n$$\n\n\n**Solution II** We define $g(x) = f(x) - 2^x$. Then we have\n$$\n\\begin{aligned}\ng(x+2) - g(x) &= f(x+2) - f(x) - 2^{x+2} + 2^x \\\\\n&\\leq 3 \\times 2^x - 3 \\times 2^x = 0, \\\\\ng(x+6) - g(x) &= f(x+6) - f(x) - 2^{x+6} + 2^x \\\\\n&\\geq 63 \\times 2^x - 63 \\times 2^x = 0.\n\\end{aligned}\n$$\nThis means that $g(x) \\leq g(x+6) \\leq g(x+4) \\leq g(x+2) \\leq g(x)$, and it implies that $g(x)$ is a periodic function with 2 as a period. So\n$$\n\\begin{aligned}\nf(2008) &= g(2008) + 2^{2008} = g(0) + 2^{2008} \\\\\n&= 2007 + 2^{2008}.\n\\end{aligned}\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 28163,
    "subject": "Mathematics (Multi-modal)",
    "question": "(1) Prove that\n$$\n\\frac{x^2}{(x-1)^2} + \\frac{y^2}{(y-1)^2} + \\frac{z^2}{(z-1)^2} \\ge 1,\n$$\nfor all real numbers $x$, $y$, $z$, each different from $1$, and satisfying $xyz = 1$.\n\n(2) Prove that the equality holds for infinitely many triples of rational numbers $x$, $y$, $z$, each different from $1$, and satisfying $xyz = 1$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "(1) Let\n$$\n\\frac{x}{x-1} = a, \\frac{y}{y-1} = b, \\frac{z}{z-1} = c,\n$$\nthen\n$$\nx = \\frac{a}{a-1}, \\quad y = \\frac{b}{b-1}, \\quad z = \\frac{c}{c-1}.\n$$\nSince $xyz = 1$, we have\n$$\nabc = (a-1)(b-1)(c-1),\n$$\nthat is\n$$\na + b + c - 1 = ab + bc + ca.\n$$\nTherefore\n$$\n\\begin{aligned}\na^2 + b^2 + c^2 &= (a + b + c)^2 - 2(ab + bc + ca) \\\\\n&= (a + b + c)^2 - 2(a + b + c - 1) \\\\\n&= (a + b + c - 1)^2 + 1 \\\\\n&\\ge 1.\n\\end{aligned}\n$$\nSo\n$$\n\\frac{x^2}{(x-1)^2} + \\frac{y^2}{(y-1)^2} + \\frac{z^2}{(z-1)^2} \\ge 1.\n$$\n\n(2) Take $(x, y, z) = \\left(-\\frac{k}{(k-1)^2}, k - k^2, \\frac{k-1}{k^2}\\right)$, $k$ is an integer, then $(x, y, z)$ is a triple of rational numbers, with $x$, $y$, $z$ each different from $1$. What is more, a different integer $k$ gives a different triple of rational numbers.\n$$\n\\begin{aligned}\n& \\frac{x^2}{(x-1)^2} + \\frac{y^2}{(y-1)^2} + \\frac{z^2}{(z-1)^2} \\\\\n&= \\frac{k^2}{(k^2 - k + 1)^2} + \\frac{(k - k^2)^2}{(k^2 - k + 1)^2} + \\frac{(k-1)^2}{(k^2 - k + 1)^2} \\\\\n&= \\frac{k^4 - 2k^3 + 3k^2 - 2k + 1}{(k^2 - k + 1)^2} = 1.\n\\end{aligned}\n$$\nSo the problem is proved.\n(1) By $xyz = 1$, let $p = x$, $q = 1$, $r = \\frac{1}{y}$, then $x = \\frac{p}{q}$, $y = \\frac{q}{r}$, $z = \\frac{1}{xy} = \\frac{r}{p}$, where $p$, $q$, $r$ are different from each other. We have\n$$\n\\begin{aligned}\n\\frac{x^2}{(x-1)^2} + \\frac{y^2}{(y-1)^2} + \\frac{z^2}{(z-1)^2} &\\ge 1 \\\\\n\\Leftrightarrow \\frac{p^2}{(p-q)^2} + \\frac{q^2}{(q-r)^2} + \\frac{r^2}{(r-p)^2} &\\ge 1.\n\\end{aligned}\n\\quad ①\n$$\nLet\n$$\na = \\frac{p}{p-q}, \\quad b = \\frac{q}{q-r}, \\quad c = \\frac{r}{r-p},\n$$\nafter the substitution, ① reduces to $\\sum a^2 \\ge 1$. Since\n$$\n\\frac{-1 + a}{a} = \\frac{q}{p}, \\quad \\frac{-1 + b}{b} = \\frac{r}{q}, \\quad \\frac{-1 + c}{c} = \\frac{p}{r},\n$$\ntherefore\n$$\n\\begin{aligned}\n& \\frac{-1 + a}{a} \\cdot \\frac{-1 + b}{b} \\cdot \\frac{-1 + c}{c} = 1, \\\\\n& 1 - \\sum a + \\sum ab = 0.\n\\end{aligned}\n\\quad ②\n$$\nBy ②, we get\n$$\n1 - \\sum a^2 = -(a + b + c - 1)^2 \\le 0,\n$$\nso $\\sum a^2 \\ge 1$. Hence ① holds.\n\n(2) Let $b = \\frac{t^2 + t}{t^2 + t + 1}$, $c = \\frac{t + 1}{t^2 + t + 1}$, $a = -\\frac{bc}{b + c}$, here $t$ can be any rational number except $0$ and $-1$. While $t$ varies, only finitely many of $t$ values can make $b$, $c$, $a$ be $1$. That is, there are infinitely many triples of rational numbers $a$, $b$, $c$ each different from $1$, satisfying $\\sum a = \\sum a^2 = 1$. By\n$$\n(x, y, z) = \\left( \\frac{a}{a-1}, \\frac{b}{b-1}, \\frac{c}{c-1} \\right),\n$$\n(2) holds.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 28164,
    "subject": "Mathematics (Multi-modal)",
    "question": "As shown in Fig. 1, $ABCD$ is a convex quadrilateral with $\\angle B + \\angle D < 180^\\circ$, and $P$ is a moving point on the plane. Let\n$$\nf(P) = PA \\times BC + PD \\times CA + PC \\times AB.\n$$\n(1) Prove that $P$, $A$, $B$, $C$ are concyclic when $f(P)$ reaches the minimum.\n\n(2) Suppose that point $E$ is on the arc $\\widearc{AB}$ of the circumscribed circle $O$ of $\\triangle ABC$, satisfying\n$$\n\\frac{AE}{AB} = \\frac{\\sqrt{3}}{2}, \\quad \\frac{BC}{EC} = \\sqrt{3} - 1, \\quad \\angle ECB = \\frac{1}{2} \\angle ECA;\n$$\nfurthermore, $DA$, $DC$ are tangent to $\\odot O$, $AC = \\sqrt{2}$. Find the minimum of $f(P)$.\n\n![](attached_image_1.png)\nFig. 1",
    "options": [],
    "answer": "sqrt(10)",
    "solution": "(1) As shown in Fig. 1, by the Ptolemy inequality we have\n$$\nPA \\times BC + PC \\times AB \\ge PB \\times AC.\n$$\nTherefore,\n$$\n\\begin{aligned}\nf(P) &= PA \\times BC + PC \\times AB + PD \\times CA \\\\\n&\\ge PB \\times CA + PD \\times CA \\\\\n&= (PB + PD) \\times CA.\n\\end{aligned}\n$$\nThe equality holds if and only if $P$, $A$, $B$, $C$ lie on $\\odot O$ and $P$ on $\\widehat{AC}$. Furthermore, $PB + PD \\ge BD$, and the equality holds if and only if $P$ lies on line $BD$. Combining the results above, we have\n$$\nf(P)_{\\min} = BD \\times CA.\n$$\n\n(2) Denote $\\angle ECB = \\alpha$. Then $\\angle ECA = 2\\alpha$. By the sine theorem we have\n$$\n\\frac{AE}{AB} = \\frac{\\sin 2\\alpha}{\\sin 3\\alpha} = \\frac{\\sqrt{3}}{2}.\n$$\nThat is to say, $\\sqrt{3} \\sin 3\\alpha = 2\\sin 2\\alpha$. By using trigonometric identities,\n$$\n\\sqrt{3}(3\\sin \\alpha - 4\\sin^3 \\alpha) = 4\\sin \\alpha \\cos \\alpha.\n$$\nBy simplification,\n$$\n3\\sqrt{3} - 4\\sqrt{3}(1 - \\cos^2\\alpha) - 4\\cos\\alpha = 0.\n$$\n$$\n\\text{That is to say, } 4\\sqrt{3}\\cos^2\\alpha - 4\\cos\\alpha - \\sqrt{3} = 0.\n$$\n$$\n\\text{The solutions are } \\cos \\alpha = \\frac{\\sqrt{3}}{2} \\text{ and } \\cos \\alpha = -\\frac{1}{2\\sqrt{3}} \\text{ (discarded).}\n$$\n$$\n\\text{Therefore, } \\alpha = 30^\\circ \\text{ and } \\angle ECA = 60^\\circ.\n$$\nOn the other hand,\n$$\n\\frac{BC}{EC} = \\sqrt{3} - 1 = \\frac{\\sin(\\angle EAC - 30^\\circ)}{\\sin \\angle EAC}.\n$$\nThat is to say,\n$$\n\\frac{\\sqrt{3}}{2}\\sin\\angle EAC - \\frac{1}{2}\\cos\\angle EAC = (\\sqrt{3} - 1)\\sin\\angle EAC.\n$$\nThen\n$$\n\\frac{2 - \\sqrt{3}}{2} \\sin \\angle EAC = \\frac{1}{2} \\cos \\angle EAC.\n$$\nTherefore,\n$$\n\\tan \\angle EAC = \\frac{1}{2 - \\sqrt{3}} = 2 + \\sqrt{3}.\n$$\n$$\n\\text{We obtain } \\angle EAC = 75^\\circ \\text{ and } \\angle AEC = 45^\\circ.\n$$\nSince $\\triangle ADC$ is an isosceles triangle and $AC = \\sqrt{2}$, we have $CD = 1$. Furthermore, $\\triangle ABC$ is an isosceles triangle, so $BC = \\sqrt{2}$ and $AB = 2$. Then\n$$\nBD^2 = AB^2 + AD^2 = 4 + 1 = 5.\n$$\n$$\n\\text{We have } BD = \\sqrt{5}. \\text{ Therefore,}\n$$\n$$\nf(P)_{\\min} = BD \\times CA = \\sqrt{5} \\times \\sqrt{2} = \\sqrt{10}.\n(1) As shown in Fig. 2, the line $BD$ intercepts the circumscribed circle $O$ of $\\triangle ABC$ at point $P_0$, which lies on $BD$ because $D$ is outside of $\\odot O$. Through $A$, $C$, $D$ draw lines perpendicular to $P_0A$, $P_0C$, $P_0D$, respectively, and they constitute $\\triangle A_1B_1C_1$ by intercepting with each other. It is easy to see that $P_0$ lies in $\\triangle ABC$, as well as in $\\triangle A_1B_1C_1$. Denoting the three inner angles of $\\triangle ACD$ by $x$, $y$, $z$, respectively, we have\n![](attached_image_2.png)\nFig. 2\n$$\n\\angle AP_0C = 180^\\circ - y = z + x.\n$$\nFurthermore, from $B_1C_1 \\perp P_0A$ and $B_1A_1 \\perp P_0C$, we have $\\angle B_1 = y$. In a similar way, we have $\\angle A_1 = x$ and $\\angle C_1 = z$. It follows that $\\triangle A_1B_1C_1 \\sim \\triangle ABC$. Now let\n$$\nB_1C_1 = \\lambda BC, \\quad C_1A_1 = \\lambda CA, \\quad A_1B_1 = \\lambda AB.\n$$\nThen for any point $M$ on the plane, we have\n$$\n\\begin{aligned}\n\\lambda f(P_0) &= \\lambda (P_0A \\times BC + P_0D \\times CA + P_0C \\times AB) \\\\\n&= P_0A \\times B_1C_1 + P_0D \\times C_1A_1 + P_0C \\times A_1B_1 \\\\\n&= 2S_{\\triangle A_1B_1C_1} \\\\\n&\\le MA \\times B_1C_1 + MD \\times C_1A_1 + MC \\times A_1B_1 \\\\\n&= \\lambda (MA \\times BC + MD \\times CA + MC \\times AB) \\\\\n&= \\lambda f(M).\n\\end{aligned}\n$$\nThat is to say, $f(P_0) \\le f(M)$. Since $M$ is an arbitrary point, we know that $f(P)$ reaches the minimum at $P_0$, and at the same time $P_0A$, $P_0C$, $P_0D$ are concyclic. This completes the proof.\n\n(2) From (1) we know that the minimum of $f(P)$ is\n$$\nf(P_0) = \\frac{2}{\\lambda} S_{\\triangle A_1 R_1 C_1} = 2\\lambda S_{\\triangle ABC}.\n$$\nIn the same way as is shown in the proof of (2) in Solution I, we find that both $\\triangle ADC$ and $\\triangle ABC$ are right-angled isosceles triangles; so we have $CD = AD = \\frac{AC}{\\sqrt{2}} = 1$, $AB = \\sqrt{2}AC = 2$, $S_{\\triangle ABC} = 1$,\n$$\nBD = \\sqrt{AB^2 + AD^2} = \\sqrt{5}.\n$$\nFurthermore, since $\\triangle A_1B_1C_1 \\sim \\triangle ABC$, $\\angle AB_1B = \\angle AB_1C + \\angle BB_1C = 90^\\circ$, $AB_1BD$ is a rectangle. It follows that $B_1C_1 = BD = \\sqrt{5}$. Therefore $\\lambda = \\frac{\\sqrt{5}}{\\sqrt{2}}$ and\n$$\nf(P)_{\\min} = 2 \\times \\frac{\\sqrt{5}}{\\sqrt{2}} \\times 1 = \\sqrt{10}.\n(1) We discuss the problem on the complex plane, and regard points $A$, $B$, $C$ as complex numbers. Then by the triangle inequality we have\n$$\n| \\vec{PA} \\cdot \\vec{BC} | + | \\vec{PC} \\cdot \\vec{AB} | \\ge | \\vec{PA} \\cdot \\vec{BC} + \\vec{PC} \\cdot \\vec{AB} |.\n$$\nThat is to say,\n$$\n\\begin{aligned}\n& |(A-P)(C-B)| + |(C-P)(B-A)| \\\\\n\\ge & |(A-P)(C-B) + (C-P)(B-A)| \\\\\n= & |-P \\times C - A \\times B + C \\times B + P \\times A| \\\\\n= & |(B-P)(C-A)| \\\\\n= & |\\vec{PB} \\cdot \\vec{AC}|.\n\\end{aligned} \\tag{1}\n$$\nThen\n$$\n\\begin{aligned}\n& |\\vec{PA} \\cdot \\vec{BC}| + |\\vec{PC} \\cdot \\vec{AB}| + |\\vec{PD} \\cdot \\vec{AC}| \\\\\n\\ge & |\\vec{PB} \\cdot \\vec{AC}| + |\\vec{PD} \\cdot \\vec{AC}| \\\\\n= & (\\vec{PB} + \\vec{PD}) \\cdot \\vec{AC} \\\\\n= & |\\vec{BD}| \\cdot \\vec{AC}|.\n\\end{aligned} \\tag{2}\n$$\nThe equality in (1) holds only when the complex numbers $(A-P)(C-B)$ and $(C-P)(B-A)$ are in the same direction. This means that there is a real $\\lambda > 0$ such that\n$$\n(A-P)(C-B) = \\lambda(C-P)(B-A).\n$$\nThat is to say,\n$$\n\\frac{A-P}{C-P} = \\lambda \\frac{B-A}{C-B}.\n$$\nTherefore,\n$$\n\\arg\\left(\\frac{A-P}{C-P}\\right) = \\arg\\left(\\frac{B-A}{C-B}\\right),\n$$\nwhich means that the angle of rotation from $\\vec{PC}$ to $\\vec{PA}$ is equal to that from $\\vec{BC}$ to $\\vec{AB}$. It follows that $P$, $A$, $B$, $C$ are concyclic.\nThe equality in (2) holds only when $B$, $P$, $D$ are collinear and $P$ lies on the segment $BD$. This means that $f(P)$ reaches the minimum when $P$ lies on the circumscribed circle of $\\triangle ABC$ and $P$, $A$, $B$, $C$ are concyclic.\n\n(2) From (1) we know that $f(P)_{min} = BD \\times CA$. The following steps are the same as those in the proof of (2) in Solution I.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 28165,
    "subject": "Mathematics (Multi-modal)",
    "question": "(1) Can one divide the set $\\{1, 2, \\ldots, 96\\}$ into $32$ subsets, each containing three elements, and the sums of the three elements in each subset are all equal?\n\n(2) Can one divide the set $\\{1, 2, \\ldots, 99\\}$ into $33$ subsets, each containing three elements, and the sums of the three elements in each subset are all equal? (Posed by Liu Shixiong)",
    "options": [],
    "answer": "(1) No. (2) Yes.",
    "solution": "(1) No. As\n$$\n1+2+\\cdots+96=\\frac{96\\times(96+1)}{2}=48\\times97,\n$$\nand $32 \\nmid 48 \\times 97$.\n\n(2) Yes. The sum of the three elements in each set is\n$$\n\\frac{1+2+\\cdots+99}{33} = \\frac{99 \\times (99+1)}{33 \\times 2} = 150.\n$$\n\nWe can divide $1, 2, 3, \\ldots, 66$ into $33$ pairs, such that the sums of these pairs form an arithmetic sequence:\n$1 + 50$, $3 + 49$, $\\ldots$, $33 + 34$, $2 + 66$, $4 + 65$, $\\ldots$, $32 + 51$.\n\nHence, the following decomposition satisfies the requirement:\n$\\{1, 50, 99\\}$, $\\{3, 49, 98\\}$, $\\ldots$, $\\{33, 34, 83\\}$, $\\{2, 66, 82\\}$, $\\{4, 65, 81\\}$, $\\ldots$, $\\{32, 51, 67\\}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 28166,
    "subject": "Mathematics (Multi-modal)",
    "question": "The minimum value of $f(x) = \\frac{5-4x+x^2}{2-x}$ for $x \\in (-\\infty, 2)$ is ( ).",
    "options": [],
    "answer": "C",
    "solution": "Let $x < 2 \\Rightarrow 2-x > 0$. Then\n$$\n\\begin{aligned}\nf(x) &= \\frac{1 + (4 - 4x + x^2)}{2-x} = \\frac{1}{2-x} + (2-x) \\\\\n&\\geq 2 \\times \\sqrt{\\frac{1}{2-x} \\times (2-x)} = 2.\n\\end{aligned}\n$$\nThe equality holds if and only if $\\frac{1}{2-x} = 2-x$, and it is so when $x = 1 \\in (-\\infty, 2)$. This means that $f$ reaches the minimum value 2 at $x = 1$.\n\nAnswer: C",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 28167,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $\\triangle ABC$ be a non-isosceles acute triangle, and point $O$ is the circumcenter. Let $A'$ be a point on the line $AO$ such that $\\angle BA'A = \\angle CA'A$. Construct $A'A_1 \\perp AC$, $A'A_2 \\perp AB$ with $A_1$ on $AC$, $A_2$ on $AB$ respectively. $AH_A$ is perpendicular to $BC$ at $H_A$. Write $R_A$ as the circumradius of $\\triangle H_AA_1A_2$. Similarly we have $R_B, R_C$. Prove that\n$$\n\\frac{1}{R_A} + \\frac{1}{R_B} + \\frac{1}{R_C} = \\frac{2}{R},\n$$\nwhere $R$ is the circumradius of $\\triangle ABC$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Firstly we claim that $A', B, O, C$ are concyclic points. Otherwise, extend $AO$ to intersect the circumcircle of $\\triangle BOC$ at point $P$ which is different from $A'$. We get\n$$\n\\angle BPA = \\angle BCO = \\angle CBO = \\angle CPA.\n$$\nThen $\\triangle PA'C \\cong \\triangle PA'B$, and $A'B = A'C$. So $AB = AC$ and that is a contradiction since $\\triangle ABC$ is not isosceles. So\n$$\n\\angle BCA' = \\angle BOA' = 180^\\circ - 2\\angle C,\n$$\nand $\\angle A'CA_1 = \\angle C$.\nFurther, we have\n$$\n\\frac{H_AA_1}{AC} = \\sin \\angle C = \\cos \\angle A_2AA' = \\frac{AA_2}{AA'}\n$$\nand\n$$\n\\angle A'AC = \\angle A_2AH_A = \\frac{\\pi}{2} - \\angle B.\n$$\nSo $\\triangle A_2AH_A \\sim \\triangle A'AC$. In the same way, $\\triangle A_1H_A A \\sim \\triangle A'BA$. Then $\\angle A_2H_A A = \\angle ACA'$ and $\\angle A_1H_A A = \\angle ABA'$. Consequently,\n$$\n\\begin{align*}\n\\angle A_1 H_A A_2 &= 2\\pi - \\angle A_2 H_A A - \\angle A_1 H_A A \\\\\n&= 2\\pi - \\angle ACA' - \\angle ABA' \\\\\n&= \\angle A + 2\\left(\\frac{\\pi}{2} - \\angle A\\right) \\\\\n&= \\pi - \\angle A.\n\\end{align*}\n$$\nWe get\n$$\n\\frac{R}{R_A} = \\frac{\\frac{R}{A_1 A_2}}{\\frac{2 \\sin \\angle A_1 H_A A_2}{\\sin \\angle A}} = \\frac{2R}{A_1 A_2} = \\frac{2R}{AA'}; \\quad \\textcircled{1}\n$$\nThe last equality holds since $A, A_2, A', A_1$ lie on the same circle with $AA'$ as the diameter.\nNow, draw $AA'' \\perp A'C$ with point $A''$ on line $A'C$. Since $\\angle ACA'' = \\angle A'CA_1 = \\angle C$, we have $AA'' = AH_A$. Then\n$$\n\\begin{align*}\nAA' &= \\frac{AA''}{\\sin \\angle AA'C} = \\frac{AH_A}{\\sin(90^\\circ - \\angle A)} \\\\\n&= \\frac{AH_A}{\\cos \\angle A} = \\frac{2S_{\\triangle ABC}}{BC \\cos \\angle A}. \\quad \\textcircled{2}\n\\end{align*}\n$$\nFrom ①, ② we get\n$$\n\\begin{align*}\n\\frac{1}{R_A} &= \\frac{BC \\cos \\angle A}{S_{\\triangle ABC}} = \\frac{\\cos \\angle A}{R \\sin \\angle B \\sin \\angle C} \\\\\n&= \\frac{1}{R} (1 - \\cot \\angle B \\cot \\angle C).\n\\end{align*}\n$$\nIn the same way\n$$\n\\frac{1}{R_B} = \\frac{1}{R}(1 - \\cot \\angle C \\cot \\angle A)\n$$\nand\n$$\n\\frac{1}{R_C} = \\frac{1}{R}(1 - \\cot \\angle A \\cot \\angle B).\n$$\nNotice that\n$$\n\\cot \\angle A \\cot \\angle B + \\cot \\angle B \\cot \\angle C + \\cot \\angle C \\cot \\angle A = 1.\n$$\nWe then have\n$$\n\\frac{1}{R_A} + \\frac{1}{R_B} + \\frac{1}{R_C} = \\frac{2}{R}.\n$$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 28168,
    "subject": "Mathematics (Multi-modal)",
    "question": "It is given the set $S = \\{1, 2, 3, \\dots, 3n\\}$, where $n$ is a positive integer. $T$ is a subset of $S$ such that: for any $x, y, z \\in T$ (where $x, y, z$ can be the same), $x+y+z \\notin T$. Find the maximum value of the number of elements in such set.",
    "options": [],
    "answer": "2n",
    "solution": "Set $T_0 = \\{n+1, n+2, \\dots, 3n\\}$, where $|T_0| = 2n$ and the sum of any three elements in $T_0$ is larger than $3n$, that is to say, the sum does not belong to $T_0$. Thus, $\\max |T| \\ge 2n$.\n\nOn the other hand, construct a sequence of sets\n$$\nA_0 = \\{n, 2n, 3n\\}, \\quad A_k = \\{k, 2n-k, 2n+k\\}, \\quad k = 1, 2, \\dots, n-1,\n$$\nthen $S = \\bigcup_{k=0}^{n-1} A_k$. For any subset $T'$ in $S$ which has $2n + 1$ elements, it must contain a certain $A_k$.\nIf $A_0 \\subset T'$, it contains the element $3n = n + n + n$.\nIf a certain $A_k \\subset T'$, $k \\in \\{1, 2, \\dots, n-1\\}$, it will contain the element\n$$\n2n + k = k + k + (2n - k),\n$$\nthen $\\max |T| < 2n + 1$. That is, $\\max |T| = 2n$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 28169,
    "subject": "Mathematics (Multi-modal)",
    "question": "An acute triangle $ABC$ has orthocenter $H$. The circle through $H$ with center the midpoint of $BC$ intersects the line $BC$ at $A_1$ and $A_2$. Similarly, the circle passing through $H$ with center the midpoint of $CA$ intersects the line $CA$ at $B_1$ and $B_2$, and the circle passing through $H$ with center the midpoint of $AB$ intersects the line $AB$ at $C_1$ and $C_2$. Show that $A_1, A_2, B_1, B_2, C_1, C_2$ are concyclic.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "![](attached_image_1.png)\n\n**Proof I** Let $B_0$, $C_0$ be the midpoints of $CA$, $AB$ respectively. Denote $A'$ as the other intersection of the circle centered at $B_0$ which passes through $H$ and the circle centered at $C_0$ which passes through $H$. We know that $A'H \\perp C_0B_0$. Since $B_0$, $C_0$ are the midpoints of $CA$, $AB$ respectively, $B_0C_0 \\parallel BC$. Therefore $A'H \\perp BC$. This yields that $A'$ lies on the segment $AH$.\n\nBy the Secant-Secant theorem, it follows that\n$$\nAC_1 \\cdot AC_2 = AA' \\cdot AH = AB_1 \\cdot AB_2,\n$$\nSo $B_1$, $B_2$, $C_1$, $C_2$ are concyclic.\n\nLet the intersection of the perpendicular bisectors of $B_1B_2$, $C_1C_2$ be $O$. Then $O$ is the circumcenter of quadrilateral $B_1B_2C_1C_2$, as well as the circumcenter of $\\triangle ABC$. So\n$$\nOB_1 = OB_2 = OC_1 = OC_2.\n$$\nSimilarly,\n$$\nOA_1 = OA_2 = OB_1 = OB_2.\n$$\nTherefore, six points $A_1$, $A_2$, $B_1$, $B_2$, $C_1$, $C_2$ are all on the same circle, whose center is $O$, and radius $OA_1$.\n\n\n**Proof II** Let $O$ be the circumcenter of triangle $ABC$, and $D$, $E$, $F$ the midpoints of $BC$, $CA$, $AB$ respectively. $BH$ intersects $DF$ at point $P$, then $BH \\perp DF$. By Pythagoras' theorem, we have\n$$\nBF^2 - FH^2 = BP^2 - PH^2 = BD^2 - DH^2, \\qquad \\textcircled{1}\n$$\n$$\nBO^2 - A_1O^2 = BD^2 - A_1D^2 = BD^2 - DH^2. \\qquad \\textcircled{2}\n$$\nSimilarly,\n$$\nBO^2 - C_2O^2 = BF^2 - FH^2, \\qquad \\textcircled{3}\n$$\nBy ①, ②, ③, $A_1O = C_2O$. It is obvious that $A_1O = A_2O$, $C_1O = C_2O$, thus\n$$\nA_1O = A_2O = C_1O = C_2O.\n$$\nSimilarly,\n$$\nA_1O = A_2O = B_1O = B_2O.\n$$\nTherefore, six points $A_1$, $A_2$, $B_1$, $B_2$, $C_1$, $C_2$ are all on the same circle, whose center is $O$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 28170,
    "subject": "Mathematics (Multi-modal)",
    "question": "In triangle $ABC$, we have $AB > AC$. The incircle $\\omega$ touches $BC$ at $E$, and $AE$ intersects $\\omega$ at $D$. Choose a point $F$ on $AE$ ($F$ is different from $E$), such that $CE = CF$. Let $G$ be the intersection point of $CF$ and $BD$. Prove that $CF = FG$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "**Proof** Referring to the figure, draw a line from $D$, tangent to $\\omega$, and the line intersects $AB$, $AC$, $BC$ at points $M$, $N$, $K$ respectively.\n![](attached_image_1.png)\nSince\n$$\n\\angle KDE = \\angle AEK = \\angle EFC,\n$$\nwe know $MK \\parallel CG$.\nBy Newton's theorem, the lines $BN$, $CM$, $DE$ are concurrent.\nBy Ceva's theorem, we have\n$$\n\\frac{BE}{EC} \\cdot \\frac{CN}{NA} \\cdot \\frac{AM}{MB} = 1. \\qquad \\textcircled{1}\n$$\nFrom Menelaus' theorem,\n$$\n\\frac{BK}{KC} \\cdot \\frac{CN}{NA} \\cdot \\frac{AM}{MB} = 1. \\qquad \\textcircled{2}\n$$\n① ÷ ②, we have\n$$\nBE \\cdot KC = EC \\cdot BK,\n$$\nthus\n$$\nBC \\cdot KE = 2EB \\cdot CK. \\qquad \\textcircled{3}\n$$\nUsing Menelaus' theorem and ③, we get\n$$\n1 = \\frac{CB}{BE} \\cdot \\frac{ED}{DF} \\cdot \\frac{FG}{GC} = \\frac{CB}{BE} \\cdot \\frac{EK}{CK} \\cdot \\frac{FG}{GC} = \\frac{2FG}{GC}.\n$$\nSo $CF = GF$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 28171,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $S_n$ denote the sum of the first $n$ terms in a number sequence $\\{a_n\\}$, satisfying\n$$\nS_n + a_n = \\frac{n-1}{n(n+1)}, \\quad n = 1, 2, \\dots\n$$\nThen $a_n = \\underline{\\hspace{2cm}}$.",
    "options": [],
    "answer": "a_n = 1/2^n - 1/(n(n+1))",
    "solution": "As\n$$\n\\begin{aligned}\na_{n+1} &= S_{n+1} - S_n \\\\\n&= \\frac{n}{(n+1)(n+2)} - a_{n+1} - \\frac{n-1}{n(n+1)} + a_n,\n\\end{aligned}\n$$\nwe have\n$$\n\\begin{aligned}\n2a_{n+1} &= \\frac{n+2-2}{(n+1)(n+2)} - \\frac{1}{n+1} + \\frac{1}{n(n+1)} + a_n \\\\\n&= \\frac{-2}{(n+1)(n+2)} + a_n + \\frac{1}{n(n+1)}.\n\\end{aligned}\n$$\nTherefore,\n$$\na_{n+1} + \\frac{1}{(n+1)(n+2)} = \\frac{1}{2} \\left( a_n + \\frac{1}{n(n+1)} \\right).\n$$\nDefine $b_n = a_n + \\frac{1}{n(n+1)}$. It is easy to see that $b_n = \\frac{1}{2^{n-1}}b_1$, $b_1 = a_1 + \\frac{1}{2}$. On the other hand, from $S_1 + a_1 = 2a_1 = 0$ we get $a_1 = 0$. So $b_1 = \\frac{1}{2}$, $b_n = \\frac{1}{2^n}$. Therefore,\n$$\na_n = b_n - \\frac{1}{n(n+1)} = \\frac{1}{2^n} - \\frac{1}{n(n+1)}.\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 28172,
    "subject": "Mathematics (Multi-modal)",
    "question": "Suppose that a ball with radius $1$ moves freely inside a regular tetrahedron with edge length $4\\sqrt{6}$. Then the area of the inner surface of the container, which the ball can never touch, is ______.",
    "options": [],
    "answer": "72√3",
    "solution": "As shown in Fig. 1, consider the situation where the ball is in a corner of the container. Draw the plane $A_1B_1C_1 \\parallel ABC$, tangent to the ball at point $D$. Then the ball center $O$ is also the center of the tetrahedron $P-A_1B_1C_1$, with $PO \\perp A_1B_1C_1$ and the foot point $D$ being the center of $\\triangle A_1B_1C_1$.\n\nSince\n![](attached_image_1.png)\n\n$$\n\\begin{aligned}\nV_{P-A_1B_1C_1} &= \\frac{1}{3} \\times S_{\\triangle A_1B_1C_1} \\times PD \\\\\n&= 4 \\times V_{O-A_1B_1C_1} \\\\\n&= 4 \\times \\frac{1}{3} \\times S_{\\triangle A_1B_1C_1} \\times OD,\n\\end{aligned}\n$$\nwe have $PD = 4OD = 4r$, where $r$ is the radius of the ball. It follows that\n$$\nPO = PD - OD = 3r.\n$$\nSuppose that the ball is tangent to the plane $PAB$ at point $P_1$. Then we have\n$$\nPP_1 = \\sqrt{PO^2 - OP_1^2} = \\sqrt{(3r)^2 - r^2} = 2\\sqrt{2}r.\n$$\nAs shown in Fig. 2, it is easy to see that the locus of the ball on the plane $PAB$ is also a regular triangle, denoted by $P_1EF$. Through $P_1$ draw $P_1M \\perp PA$ with point $M$ on $PA$. Then $\\angle MPP_1 = \\frac{\\pi}{6}$, and\n$$\nPM = PP_1 \\times \\cos \\angle MPP_1 = 2\\sqrt{2}r \\times \\frac{\\sqrt{3}}{2} = \\sqrt{6}r.\n$$\n![](attached_image_2.png)\n\nIt follows that $P_1E = PA - 2PM = a - 2\\sqrt{6}r$, where $a = PA$. Now, the space on $PAB$ which the ball will never touch is the shaded part of Fig. 2, and its size is equal to\n$$\n\\begin{align*}\nS_{\\triangle PAB} - S_{\\triangle P_1EF} &= \\frac{\\sqrt{3}}{4}(a^2 - (a - 2\\sqrt{6}r)^2) \\\\\n&= 3\\sqrt{2}ar - 6\\sqrt{3}r^2 \\\\\n&= 24\\sqrt{3} - 6\\sqrt{3} \\\\\n&= 18\\sqrt{3},\n\\end{align*}\n$$\nsince $r = 1$ and $a = 4\\sqrt{6}$ under given conditions. Then the total untouched area is\n$$\n4 \\times 18\\sqrt{3} = 72\\sqrt{3}.\n$$",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 28173,
    "subject": "Mathematics (Multi-modal)",
    "question": "It is known that the curve $f(x) = |\\sin x|$ intercepts the line $y = kx$ ($k > 0$) at exactly three points, the maximum $x$ coordinate of these points being $\\alpha$. Prove that\n$$\n\\frac{\\cos \\alpha}{\\sin \\alpha + \\sin 3\\alpha} = \\frac{1 + \\alpha^2}{4\\alpha}.\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "The image of the three intercepting points of $f(x)$ and $y = kx$ is shown in the figure. It is easy to see that the curve and the line are tangent to each other at point $A(\\alpha, -\\sin \\alpha)$, and $\\alpha \\in (\\pi, \\frac{3\\pi}{2})$.\n![](attached_image_1.png)\n\nAs $f'(x) = -\\cos x$ for $x \\in (\\pi, \\frac{3\\pi}{2})$, we have $-\\cos \\alpha = -\\frac{\\sin \\alpha}{\\alpha}$, i.e. $\\alpha = \\tan \\alpha$. Then\n$$\n\\begin{align*}\n\\frac{\\cos \\alpha}{\\sin \\alpha + \\sin 3\\alpha} &= \\frac{\\cos \\alpha}{2\\sin 2\\alpha \\cos \\alpha} = \\frac{1}{4\\sin \\alpha \\cos \\alpha} \\\\\n&= \\frac{\\cos^2 \\alpha + \\sin^2 \\alpha}{4\\sin \\alpha \\cos \\alpha} = \\frac{1 + \\tan^2 \\alpha}{4\\tan \\alpha} \\\\\n&= \\frac{1 + \\alpha^2}{4\\alpha}.\n\\end{align*}\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 28174,
    "subject": "Mathematics (Multi-modal)",
    "question": "Solve the inequality\n$$\n\\log_2(x^{12} + 3x^{10} + 5x^8 + 3x^6 + 1) < 1 + \\log_2(x^4 + 1).\n$$",
    "options": [],
    "answer": "(-sqrt((-1+sqrt(5))/2), sqrt((-1+sqrt(5))/2))",
    "solution": "As\n$$\n1 + \\log_2(x^4 + 1) = \\log_2(2x^4 + 2),\n$$\nand $\\log_2 y$ is monotonically increasing over $(0, +\\infty)$, the given inequality is equivalent to\n$$\nx^{12} + 3x^{10} + 5x^8 + 3x^6 + 1 < 2x^4 + 2\n$$\nor\n$$\nx^{12} + 3x^{10} + 5x^8 + 3x^6 - 2x^4 - 1 < 0.\n$$\nIt can be rewritten as\n$$\n\\begin{aligned}\n& x^{12} + x^{10} - x^8 \\\\\n& \\quad + 2x^{10} + 2x^8 - 2x^6 \\\\\n& \\quad + 4x^8 + 4x^6 - 4x^4 \\\\\n& \\quad + x^6 + x^4 - x^2 \\\\\n& \\quad + x^4 + x^2 - 1 < 0.\n\\end{aligned}\n$$\nThat is to say,\n$$\n(x^8 + 2x^6 + 4x^4 + x^2 + 1)(x^4 + x^2 - 1) < 0.\n$$\nThen we have $x^4 + x^2 - 1 < 0$. It follows that $x^2 < \\frac{-1+\\sqrt{5}}{2}$, i.e.\n$$\n-\\sqrt{\\frac{-1+\\sqrt{5}}{2}} < x < \\sqrt{\\frac{-1+\\sqrt{5}}{2}}.\n$$\nSo the solution set is $(-\\sqrt{\\frac{-1+\\sqrt{5}}{2}}, \\sqrt{\\frac{-1+\\sqrt{5}}{2}})$.\nAs\n$$\n1 + \\log_2(x^4 + 1) = \\log_2(2x^4 + 2),\n$$\nand $\\log_2 y$ is monotonically increasing over $(0, +\\infty)$, the given inequality is equivalent to\n$$\nx^{12} + 3x^{10} + 5x^8 + 3x^6 + 1 < 2x^4 + 2\n$$\nor\n$$\n\\begin{aligned}\n\\left(\\frac{1}{x^2}\\right)^3 + 2\\left(\\frac{1}{x^2}\\right) &> x^6 + 3x^4 + 3x^2 + 1 + 2x^2 + 2 \\\\\n& = (x^2 + 1)^3 + 2(x^2 + 1).\n\\end{aligned}\n$$\nDefine $g(t) = t^2 + 2t$. Then we have\n$$\ng\\left(\\frac{1}{x^2}\\right) > g(x^2 + 1).\n$$\nObviously, $g(t)$ is a monotonically increasing function; then we have\n$$\n\\frac{1}{x^2} > x^2 + 1.\n$$\nThat is to say,\n$$\nx^4 + x^2 - 1 < 0.\n$$\nWe obtain $x^2 < \\frac{-1 + \\sqrt{5}}{2}$. So the solution set is $(-\\sqrt{\\frac{-1 + \\sqrt{5}}{2}}, \\sqrt{\\frac{-1 + \\sqrt{5}}{2}})$.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 28175,
    "subject": "Mathematics (Multi-modal)",
    "question": "As is shown in the figure, $P$ is a moving point on the parabola $y^2 = 2x$, points $B$, $C$ are on the $y$ axis, and the circle $(x - 1)^2 + y^2 = 1$ is internally tangent to $\\triangle PBC$. Find the minimum value of the area of $\\triangle PBC$.\n\n![](attached_image_1.png)",
    "options": [],
    "answer": "8",
    "solution": "Denote $P$, $B$, $C$ by $P(x_0, y_0)$, $B(0, b)$, $C(0, c)$, and assume that $b > c$. The equation for the line $PB$ is\n$$\ny - b = \\frac{y_0 - b}{x_0}x.\n$$\nIt can be rewritten as\n$$\n(y_0 - b)x - x_0y + x_0b = 0.\n$$\nSince the distance between the circle center $(1, 0)$ and the line $PB$ is $1$, we have\n$$\n\\frac{|y_0 - b + x_0b|}{\\sqrt{(y_0 - b)^2 + x_0^2}} = 1.\n$$\nThat is to say,\n$$\n(y_0 - b)^2 + x_0^2 = (y_0 - b)^2 + 2x_0b(y_0 - b) + x_0^2b^2.\n$$\nIt is easy to see that $x_0 > 2$. Then the last equation can be simplified as\n$$\n(x_0 - 2)b^2 + 2y_0b - x_0 = 0.\n$$\nIn a similar way,\n$$\n(x_0 - 2)c^2 + 2y_0c - x_0 = 0.\n$$\nTherefore,\n$$\nb + c = \\frac{-2y_0}{x_0 - 2}, \\quad bc = \\frac{-x_0}{x_0 - 2}.\n$$\nThen we get\n$$\n(b-c)^2 = \\frac{4x_0^2 + 4y_0^2 - 8x_0}{(x_0-2)^2}.\n$$\nAs $P(x_0, y_0)$ is on the parabola, $y_0^2 = 2x_0$. So we have\n$$\n(b-c)^2 = \\frac{4x_0^2}{(x_0-2)^2},\n$$\nor $b-c = \\frac{2x_0}{x_0-2}$. Then we have\n$$\n\\begin{aligned}\nS_{\\triangle PBC} &= \\frac{1}{2}(b-c) \\times x_0 \\\\\n&= \\frac{x_0}{x_0-2} \\times x_0 \\\\\n&= (x_0-2) + \\frac{4}{x_0-2} + 4 \\\\\n&\\ge 4 + 4 = 8.\n\\end{aligned}\n$$\nThe equality holds when $x_0 - 2 = 2$; this means that $x_0 = 4$ and $y_0 = \\pm 2\\sqrt{2}$. So the minimum of $S_{\\triangle PBC}$ is $8$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 28176,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $A = [-2, 4)$, $B = \\{x \\mid x^2 - a x - 4 \\leq 0\\}$. If $B \\subseteq A$, then the range of real $a$ is ( ).",
    "options": [],
    "answer": "[0, 3)",
    "solution": "$x^2 - a x - 4 = 0$ has two roots:\n$$\nx_1 = \\frac{a}{2} - \\sqrt{4 + \\frac{a^2}{4}}, \\quad x_2 = \\frac{a}{2} + \\sqrt{4 + \\frac{a^2}{4}}\n$$\nWe have $B \\subseteq A \\Leftrightarrow x_1 \\geq -2$ and $x_2 < 4$. This means that\n$$\n\\frac{a}{2} - \\sqrt{4 + \\frac{a^2}{4}} \\geq -2, \\quad \\frac{a}{2} + \\sqrt{4 + \\frac{a^2}{4}} < 4\n$$\nFrom the above we get $0 \\leq a < 3$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 28177,
    "subject": "Mathematics (Multi-modal)",
    "question": "Given an integer $n \\ge 3$, prove that the set $X = \\{1, 2, 3, \\dots, n^2 - n\\}$ can be divided into two non-intersecting subsets such that neither of them contains $n$ elements $a_1, a_2, \\dots, a_n$ with $a_1 < a_2 < \\dots < a_n$ and $a_k \\le \\frac{a_{k-1} + a_{k+1}}{2}$ for all $k = 2, \\dots, n-1$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Define\n$$\nS_k = \\{k^2 - k + 1, k^2 - k + 2, \\dots, k^2\\},\n$$\n$$\nT_k = \\{k^2 + 1, k^2 + 2, \\dots, k^2 + k\\}.\n$$\nLet $S = \\bigcup_{k=1}^{n-1} S_k$, $T = \\bigcup_{k=1}^{n-1} T_k$. We will prove that $S$, $T$ are the required subsets of $X$.\nFirstly it is easy to verify that $S \\cap T = \\emptyset$ and $S \\cup T = X$.\nNext we suppose for contradiction that $S$ contains elements $a_1, a_2, \\dots, a_n$ with $a_1 < a_2 < \\dots < a_n$ and $a_k \\le \\frac{a_{k-1} + a_{k+1}}{2}$ for $k = 2, \\dots, n-1$. Then we have\n$$\na_k - a_{k-1} \\le a_{k+1} - a_k, \\quad k = 2, \\dots, n-1. \\qquad \\textcircled{1}\n$$\nAssume that $a_1 \\in S_i$, we have $i < n-1$, since $|S_{n-1}| < n$. There exists at least $n - |S_i| = n - i$ elements in $\\{a_1, a_2, \\dots, a_n\\} \\cap (S_{i+1} \\cup \\dots \\cup S_{n-1})$. Applying the pigeonhole principle, we get that there is an $S_j$ ($i < j < n$) which contains at least two elements in $a_1, a_2, \\dots, a_n$. That means there exists $a_k$ such that $a_k, a_{k+1} \\in S_j$ and $a_{k-1} \\in S_1 \\cup \\dots \\cup S_{j-1}$.\nThen we have\n$$\na_{k+1} - a_k \\le |S_j| - 1 = j - 1, \\quad a_k - a_{k-1} \\ge |T_{j-1}| + 1 = j.\n$$\nThat means $a_{k+1} - a_k < a_k - a_{k-1}$, contradicting ①.\n\nIn the same way, we can prove that $T$ does not contain $a_1, a_2, \\dots, a_n$ with the required properties either. This completes the proof.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 28178,
    "subject": "Mathematics (Multi-modal)",
    "question": "The real polynomial $\\varphi(x) = ax^3 + bx^2 + cx + d$ has three positive roots, and $\\varphi(0) < 0$. Prove that\n$$\n2b^3 + 9a^2d - 7abc \\le 0. \\quad \\textcircled{1}\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "**Proof** We denote by $x_1, x_2, x_3$ the three positive roots of the polynomial $\\varphi(x) = ax^3 + bx^2 + cx + d$. By Vieta's theorem, we have\n$$\n\\begin{align*}\nx_1 + x_2 + x_3 &= -\\frac{b}{a}, \\\\\nx_1x_2 + x_2x_3 + x_3x_1 &= \\frac{c}{a}, \\\\\nx_1x_2x_3 &= -\\frac{d}{a}.\n\\end{align*}\n$$\n---\nAs $\\varphi(0) < 0$, we get $d < 0$, and hence $a > 0$.\nDividing both sides of ① by $a^3$, we get an equivalent form of ①:\n$$\n\\begin{aligned}\n& 7\\left(-\\frac{b}{a}\\right)\\frac{c}{a} \\le 2\\left(-\\frac{b}{a}\\right)^3 + 9\\left(-\\frac{d}{a}\\right) \\\\\n\\Leftrightarrow & 7(x_1 + x_2 + x_3)(x_1x_2 + x_2x_3 + x_3x_1) \\\\\n& \\le 2(x_1 + x_2 + x_3)^3 + 9x_1x_2x_3 \\\\\n\\Leftrightarrow & x_1^2x_2 + x_1^2x_3 + x_2^2x_1 + x_2^2x_3 + x_3^2x_1 + x_3^2x_2 \\\\\n& \\le 2(x_1^3 + x_2^3 + x_3^3). \\qquad \\textcircled{2}\n\\end{aligned}\n$$\nBecause $x_1, x_2, x_3$ are all greater than 0, $(x_1 - x_2)(x_1^2 - x_2^2) \\ge 0$. That is to say,\n$$\nx_1^2 x_2 + x_2^2 x_1 \\le x_1^3 + x_2^3.\n$$\nBy the same argument, $x_2^2 x_3 + x_3^2 x_2 \\le x_2^3 + x_3^3$, $x_3^2 x_1 + x_1^2 x_3 \\le x_3^3 + x_1^3$.\nSumming up these three inequalities we get ②, and the\nequality holds iff $x_1 = x_2 = x_3$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 28179,
    "subject": "Mathematics (Multi-modal)",
    "question": "In $\\triangle ABC$, $AB = AC$, the inscribed circle $I$ touches $BC$, $CA$, $AB$ at points $D$, $E$ and $F$ respectively. $P$ is a point on arc $\\widehat{EF}$ (not containing $D$). Line $BP$ intersects the circle $I$ at another point $Q$, and lines $EP$, $EQ$ meet line $BC$ at $M$, $N$ respectively. Prove that\n\n(1) $P$, $F$, $B$, $M$ are concyclic;\n\n$$\n(2) \\frac{EM}{EN} = \\frac{BD}{BP}.\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "**Proof**\n\n(1) From the given condition, $EF \\parallel BC$, so\n$$\n\\begin{aligned}\n\\angle ABC &= \\angle AFE = \\angle AFP + \\angle PFE \\\\\n&= \\angle PEF + \\angle PFE = 180^\\circ - \\angle FPE,\n\\end{aligned}\n$$\nand thus $P$, $F$, $B$, $M$ are concyclic.\n\n\n(2) By the sine law, $EF \\parallel BC$ and the fact that $P$, $F$, $B$, $M$ are concyclic, we have\n$$\n\\frac{EM}{EN} = \\frac{\\sin\\angle ENM}{\\sin\\angle EMN} = \\frac{\\sin\\angle FEN}{\\sin(\\pi - \\angle PFB)} = \\frac{\\sin\\angle FPB}{\\sin\\angle PFB} = \\frac{BF}{BP}.\n$$\nTogether with $BF = BD$, the proposition is proven.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 28180,
    "subject": "Mathematics (Multi-modal)",
    "question": "It is given the sequence $\\{a_n\\}$: $a_1 = 1$,\n$$\na_{n+1} = 2a_n + n \\cdot (1 + 2^n), \\quad n = 1, 2, 3, \\dots\n$$\n\nFind the general term $a_n$.",
    "options": [],
    "answer": "a_n = 2^{n-2}(n^2 - n + 6) - n - 1 (n ≥ 2)",
    "solution": "Divide the recursion formula by $2^{n+1}$ throughout, we obtain\n$$\n\\frac{a_{n+1}}{2^{n+1}} = \\frac{a_n}{2^n} + \\frac{n}{2^{n+1}} + \\frac{n}{2},\n$$\nthat is,\n$$\n\\frac{a_{n+1}}{2^{n+1}} - \\frac{a_n}{2^n} = \\frac{n}{2^{n+1}} + \\frac{n}{2}.\n$$\nThen\n$$\n\\sum_{i=1}^n \\left( \\frac{a_{i+1}}{2^{i+1}} - \\frac{a_i}{2^i} \\right) = \\sum_{i=1}^n \\frac{i}{2^{i+1}} + \\sum_{i=1}^n \\frac{i}{2},\n$$\n$$\n\\frac{a_{n+1}}{2^{n+1}} - \\frac{a_1}{2^1} = \\frac{n(n+1)}{4} + \\sum_{i=1}^n \\frac{i}{2^{i+1}},\n$$\n$$\na_{n+1} = 2^{n+1} \\left[ \\frac{n(n+1)}{4} + \\frac{1}{2^n} + \\frac{1}{2} \\sum_{i=1}^n \\frac{i}{2^i} \\right].\n$$\nSet $S_n = \\sum_{i=1}^n \\frac{i}{2^i}$, then $2S_n = \\sum_{i=1}^n \\frac{i}{2^{i-1}}$, and\n$$\n\\begin{align*}\nS_n &= 2S_n - S_n = \\sum_{i=1}^n \\frac{i}{2^{i-1}} - \\sum_{i=1}^n \\frac{i}{2^i} \\\\\n&= \\sum_{i=1}^n \\frac{i}{2^{i-1}} - \\sum_{i=2}^{n+1} \\frac{i-1}{2^{i-1}} \\\\\n&= \\frac{1}{2^{1-1}} - \\frac{n+1-1}{2^{n+1-1}} + \\sum_{i=2}^n \\left( \\frac{i}{2^{i-1}} - \\frac{i-1}{2^{i-1}} \\right) \\\\\n&= 1 - \\frac{n}{2^n} + \\sum_{i=2}^n \\frac{1}{2^{i-1}}\n\\end{align*}\n$$\n$$\n\\begin{aligned}\n&= 1 - \\frac{n}{2^n} + \\frac{1}{2} \\left[ 1 - \\left( \\frac{1}{2} \\right)^{n-1} \\right] \\\\\n&= 1 - \\frac{n}{2^n} + 1 - \\frac{1}{2^{n-1}} \\\\\n&= 2 - \\frac{n+2}{2^n}.\n\\end{aligned}\n$$\nThus,\n$$\n\\begin{aligned}\na_{n+1} &= \\left[ \\frac{n(n+1)}{4} + \\frac{1}{2^n} + \\frac{1}{2} \\left( 2 - \\frac{n+2}{2^n} \\right) \\right] \\\\\n&= 2^{n+1} \\left[ \\frac{3}{2} + \\frac{n(n+1)}{4} - \\frac{n+2}{2^{n+1}} \\right] \\quad (n \\ge 1).\n\\end{aligned}\n$$\nConsequently,\n$$\na_n = 2^{n-2}(n^2 - n + 6) - n - 1 \\quad (n \\ge 2).\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 28181,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find the smallest constant $a > 1$, such that for any point $P$ inside a square $ABCD$ there exist two triangles among $\\triangle PAB$, $\\triangle PBC$, $\\triangle PCD$, $\\triangle PDA$, with the ratio between their areas belonging to the interval $[a^{-1}, a]$. (Posed by Li Weigu)",
    "options": [],
    "answer": "(1+sqrt{5})/2",
    "solution": "$a_{\\min} = \\frac{1+\\sqrt{5}}{2}$.\n\nWrite $\\varphi = \\frac{1+\\sqrt{5}}{2}$. We may assume that each edge has length $\\sqrt{2}$. For any point $P$ inside the square $ABCD$, let $S_1, S_2, S_3, S_4$ denote the area of $\\triangle PAB$, $\\triangle PBC$, $\\triangle PCD$, $\\triangle PDA$ respectively; we can also assume that $S_1 \\ge S_2 \\ge S_3 \\ge S_4$.\n\nLet $\\lambda = \\frac{S_1}{S_2}$, $\\mu = \\frac{S_2}{S_4}$. If $\\lambda, \\mu > \\varphi$, as\n$$\nS_1 + S_3 = S_2 + S_4 = 1,\n$$\nwe have $\\frac{S_2}{1-S_2} = \\mu$, $S_2 = \\frac{\\mu}{1+\\mu}$. So\n![](attached_image_1.png)\n$$\nS_1 = \\lambda S_2 = \\frac{\\lambda \\mu}{1+\\mu} = \\frac{\\lambda}{1+\\frac{1}{\\mu}} > \\frac{\\varphi}{1+\\frac{1}{\\varphi}} = \\frac{\\varphi^2}{1+\\varphi} = 1,\n$$\nand we reach a contradiction. Hence, $\\min\\{\\lambda, \\mu\\} \\le \\varphi$, which implies that $a_{\\min} \\le \\varphi$.\n\nOn the other hand, for any $a \\in (1, \\varphi)$, we take any $t \\in (a, \\frac{1+\\sqrt{5}}{2})$ such that $b = \\frac{t^2}{1+t} > \\frac{8}{9}$. Inside the square $ABCD$ we can choose a point $P$ so that $S_1 = b$, $S_2 = \\frac{b}{t}$, $S_3 = \\frac{b}{t^2}$, $S_4 = 1-b$. Then we have\n$$\n\\frac{S_1}{S_2} = \\frac{S_2}{S_3} = t \\in (a, \\frac{1+\\sqrt{5}}{2}),\n$$\n$$\n\\frac{S_3}{S_4} = \\frac{b}{t^2(1-b)} > \\frac{b}{4(1-b)} > 2 > a.\n$$\nThus, for any $i, j \\in \\{1, 2, 3, 4\\}$, we have $\\frac{S_i}{S_j} \\notin [a^{-1}, a]$.\n\nHence $a_{\\min} = \\varphi$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 28182,
    "subject": "Mathematics (Multi-modal)",
    "question": "Suppose that $a_k > 0$, $k = 1, 2, \\dots, 2008$.\nProve that if and only if $\\sum_{k=1}^{2008} a_k > 1$, there is a sequence $\\{x_n\\}$ satisfying\n(1) $0 = x_0 < x_n < x_{n+1}$, $n = 1, 2, 3, \\dots$;\n(2) $\\lim_{n \\to \\infty} x_n$ exists;\n(3) $x_n - x_{n-1} = \\sum_{k=1}^{2008} a_k x_{n+k} - \\sum_{k=0}^{2007} a_{k+1} x_{n+k}$, $n = 1, 2, 3, \\dots$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Proof of necessity: Assume that there exists $\\{x_n\\}$ satisfying (1)–(3). Notice that the expression in (3) can be written as\n$$\nx_n - x_{n-1} = \\sum_{k=1}^{2008} a_k (x_{n+k} - x_{n+k-1}), \\quad n \\in \\mathbb{N}.\n$$\nAs $x_0 = 0$, we then have\n$$\n\\begin{align*}\nx_n &= \\sum_{l=1}^{n} (x_l - x_{l-1}) = \\sum_{l=1}^{n} \\sum_{k=1}^{2008} a_k (x_{l+k} - x_{l+k-1}) \\\\\n&= \\sum_{k=1}^{2008} \\sum_{l=1}^{n} a_k (x_{l+k} - x_{l+k-1}) \\\\\n&= \\sum_{k=1}^{2008} a_k (x_{n+k} - x_k).\n\\end{align*}\n$$\nFrom (2) we are able to define $b = \\lim_{n \\to \\infty} x_n$. Let $n \\to \\infty$ in the above expression. Then we have\n$$\nb = \\sum_{k=1}^{2008} a_k (b - x_k) = b \\sum_{k=1}^{2008} a_k - \\sum_{k=1}^{2008} a_k x_k \\\\\n< b \\sum_{k=1}^{2008} a_k.\n$$\nTherefore, $\\sum_{k=1}^{2008} a_k > 1$.\n\nProof of sufficiency: Assume that $\\sum_{k=1}^{2008} a_k > 1$. Define a polynomial function by\n$$\nf(s) = -1 + \\sum_{k=1}^{2008} a_k s^k, \\quad s \\in [0, 1].\n$$\n$f(s)$ is strictly increasing on the interval $[0, 1]$. In the meantime,\n$$\nf(0) = -1 < 0, \\quad f(1) = -1 + \\sum_{k=1}^{2008} a_k > 0,\n$$\nand there then exists a unique $0 < s_0 < 1$ such that $f(s_0) = 0$.\nNow, define $x_n = \\sum_{k=1}^{n} s_0^k$, $n \\in \\mathbb{N}$. It is easy to see that $\\{x_n\\}$ satisfies (1) and\n$$\nx_n = \\sum_{k=1}^{n} s_0^k = \\frac{s_0 - s_0^{n+1}}{1 - s_0}.\n$$\nOn the other hand, $\\lim_{n \\to \\infty} s_0^{n+1} = 0$ as $0 < s_0 < 1$. Then we have\n$$\n\\lim_{n \\to \\infty} x_n = \\lim_{n \\to \\infty} \\frac{s_0 - s_0^{n+1}}{1 - s_0} = \\frac{s_0}{1 - s_0}.\n$$\nThis means that $\\{x_n\\}$ satisfies (2). Finally, we have\n$$\n0 = f(s_0) = -1 + \\sum_{k=1}^{2008} a_k s_0^k.\n$$\nThat is to say, $\\sum_{k=1}^{2008} a_k s_0^k = 1$. Then we have\n$$\n\\begin{align*}\nx_n - x_{n-1} &= s_0^n \\\\\n&= \\left( \\sum_{k=1}^{2008} a_k s_0^k \\right) s_0^n \\\\\n&= \\sum_{k=1}^{2008} a_k s_0^{n+k} \\\\\n&= \\sum_{k=1}^{2008} a_k (x_{n+k} - x_{n+k-1}).\n\\end{align*}\n$$\nTherefore, $\\{x_n\\}$ also satisfies (3). This completes the proof.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 28183,
    "subject": "Mathematics (Multi-modal)",
    "question": "$A$ and $B$ are playing ping-pong, with the agreement that the winner of a game will get $1$ point and the loser $0$ point; the match ends as soon as one of the players is ahead by $2$ points or the number of games reaches $6$. Suppose that the probabilities of $A$ and $B$ winning a game are $\\frac{2}{3}$ and $\\frac{1}{3}$, respectively, and each game is independent. Then the expectation $E\\xi$ for the match ending with $\\xi$ games is ( ).\n\n(A) $\\frac{241}{81}$\n(B) $\\frac{266}{81}$\n(C) $\\frac{274}{81}$\n(D) $\\frac{670}{243}$",
    "options": [],
    "answer": "B",
    "solution": "**Solution I** It is easy to see that $\\xi$ can only be $2$, $4$ or $6$. We divide the six games into three rounds, each consisting of two consecutive games. If one of the players wins two games in the first round, the match ends and the probability is\n$$\n\\left(\\frac{2}{3}\\right)^2 + \\left(\\frac{1}{3}\\right)^2 = \\frac{5}{9}.\n$$\nOtherwise the players tie with each other, earning one point each, and the match enters the second round; this probability is\n$1 - \\frac{5}{9} = \\frac{4}{9}$.\nWe have similar discussions for the second and third rounds. So we get\n$$\n\\begin{aligned}\nP(\\xi = 2) &= \\frac{5}{9}, \\\\\nP(\\xi = 4) &= \\frac{4}{9} \\times \\frac{5}{9} = \\frac{20}{81}, \\\\\nP(\\xi = 6) &= \\left(\\frac{4}{9}\\right)^2 = \\frac{16}{81}.\n\\end{aligned}\n$$\nThen\n$$\nE\\xi = 2 \\times \\frac{5}{9} + 4 \\times \\frac{20}{81} + 6 \\times \\frac{16}{81} = \\frac{266}{81}.\n$$\n\n\n**Solution II** Let $A_k$ denote the event that $A$ wins the $k$th game, while $\\overline{A}_k$ means that $B$ wins the game. Since $A_k$ and $\\overline{A}_k$ are incompatible, and are independent of the other events, we have\n$$\nP(\\xi = 2) = P(A_1A_2) + P(\\overline{A}_1\\overline{A}_2) = \\frac{5}{9},\n$$\n$$\n\\begin{aligned}\nP(\\xi = 4) &= P(A_1\\overline{A}_2A_3A_4) + P(A_1\\overline{A}_2\\overline{A}_3A_4) + \\\\\n& P(\\overline{A}_1A_2A_3A_4) + P(\\overline{A}_1A_2\\overline{A}_3A_4) \\\\\n&= 2\\left[\\left(\\frac{2}{3}\\right)^3\\left(\\frac{1}{3}\\right) + \\left(\\frac{1}{3}\\right)^3\\left(\\frac{2}{3}\\right)\\right] = \\frac{20}{81},\n\\end{aligned}\n$$\n$$\n\\begin{aligned}\nP(\\xi = 6) &= P(A_1\\overline{A}_2A_3\\overline{A}_4) + P(A_1\\overline{A}_2\\overline{A}_3A_4) + \\\\\n& P(\\overline{A}_1A_2A_3\\overline{A}_4) + P(\\overline{A}_1A_2\\overline{A}_3A_4) \\\\\n&= 4\\left(\\frac{2}{3}\\right)^2\\left(\\frac{1}{3}\\right)^2 = \\frac{16}{81}.\n\\end{aligned}\n$$\nThen\n$$\nE\\xi = 2 \\times \\frac{5}{9} + 4 \\times \\frac{20}{81} + 6 \\times \\frac{16}{81} = \\frac{266}{81}.\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 28184,
    "subject": "Mathematics (Multi-modal)",
    "question": "Every positive integer is colored by blue or red. Prove that there is a sequence $\\{a_n\\}$ which has infinite terms, and $a_1 < a_2 < \\dots$ are positive integers, such that $a_1$, $\\frac{a_1+a_2}{2}$, $a_2$, $\\frac{a_2+a_3}{2}$, $a_3$, $\\dots$ is a positive integer sequence with the same color.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "We need three lemmas. Firstly, define $N^*$ as the set of all the positive integers.\n\nLemma 1 If there is an arithmetic progression having infinite positive integer terms with the same color, then the conclusion holds.\n\nProof: Let $c_1 < c_2 < \\dots < c_n < \\dots$ be a red arithmetic progression of positive integers, we can set $a_i = c_{2i-1}$ ($i = 1, 2, 3, \\dots$) to obtain a sequence such that the condition holds.\n\nLemma 2 If for any $i \\in N^*$, there exists a positive integer $j$, such that $i, \\frac{i+j}{2}$, $j$ are of the same color, then the conclusion holds.\n\nProof: Let $a_1 = 1$, and $a_1$ be red. Since there exists $k \\in N^*$ such that $a_1, \\frac{a_1+k}{2}$, $k$ have the same color, so we can set $a_2 = k$. In the same way, we can have a sequence of red numbers satisfying\n$$\na_1 < \\frac{a_1+a_2}{2} < a_2 < \\frac{a_2+a_3}{2} < a_3 < \\dots\n$$\n\nLemma 3 If there is no arithmetic progression satisfying the condition of Lemma 1, and there exists $i_0 \\in N^*$, such that for every $j \\in N^*$, $i_0, \\frac{i_0+j}{2}$, $j$ have different colors, then the conclusion holds.\n\nProof: We can suppose $i_0 = 1$, otherwise using $N' = \\{ni_0 \\mid n \\in \\mathbb{N}^*\\}$ in place of $\\mathbb{N}^*$, will yield the same result.\nLet $i_0 = 1$ be a red number. Then we have the following condition: For every $k \\in \\mathbb{N}^*$, $k \\ge 2$, $k$ and $2k-1$ cannot be both red. ①\nSince there is no arithmetic progression having infinite terms with the same color, therefore there are infinite terms of blue colored numbers of different parities in $\\mathbb{N}^*$. We prove there is an infinite sequence of odd numbers of blue color in $\\mathbb{N}^*$.\nLet $a_1$ be a blue odd number, and suppose odd numbers $a_1 < a_2 < \\dots < a_n$ satisfy that\n$$\na_1 < \\frac{a_1 + a_2}{2} < a_2 < \\dots < a_{n-1} < \\frac{a_{n-1} + a_n}{2} < a_n\n$$\nare all blue. Now we prove there exists an odd number $a_{n+1} \\in \\mathbb{N}^*$, such that $\\frac{a_n + a_{n+1}}{2}$ and $a_{n+1}$ are blue. ②\n\n(1) If for every $i \\in \\mathbb{N}^*$, the numbers $a_n + i, a_n + 2i$ have different colors, and there is no $a_{n+1}$ satisfying ②, then for $a_{n+1} > a_n$, numbers $\\frac{a_n + a_{n+1}}{2}$ and $a_{n+1}$ cannot be both blue. ③\nSince there is no arithmetic progression with infinite terms of the same color, there must exist infinitely many red and blue numbers in $\\mathbb{N}^*$. Let $i \\in \\mathbb{N}^*$ such that $a_n + i$ is red, then $a_n + 2i$ is blue. Write $a_n = 2k+1$, we know $2k+1$ is blue, $2k+i+1$ is red, $2k+2i+1$ is blue. Using ①, we know that $4k+2i+1(=2(2k+i+1)-1)$ is blue. Similarly from ③, $3k+i+1 = \\frac{(2k+1) + (4k+2i+1)}{2}$ is red. From ① we get $6k + 2i + 1 = 2(3k + i + 1) - 1$ to be blue, and so on. Consequently, we have an arithmetic progression $\\{2nk + 2i + 1\\}_{i=1}^{\\infty}$ of blue numbers, contradiction! Hence there must be a number $a_{n+1}$ satisfying ②.\n\n(2) Suppose there is $i \\in \\mathbb{N}^*$, such that $a_n + i$ and $a_n + 2i$ have the same color. Let $a_n = 2k+1$.\nFirstly, if $a_n + i$ and $a_n + 2i$ are blue, then $a_{n+1} = a_n + 2i$, satisfying ②.\nSecondly, if $a_n + i$ and $a_n + 2i$ are red, then from ① we have $4k + 2i + 1 (=2(2k + i + 1) - 1)$ and $4k + 4i + 1 (=2(2k + 2i + 1) - 1)$ being blue. So if $3k + 2i + 1 = \\frac{(2k + 1) + (4k + 4i + 1)}{2}$ are blue, then $a_{n+1} = 4k + 4i + 1$, hence satisfying ②, otherwise, the number $6k + 4i + 1 (=2(3k + 2i + 1) - 1)$ is blue, then $a_{n+1} = 6k + 4i + 1$, satisfying ②. Hence proving Lemma 3.\n\nTherefore combining Lemmas 1, 2 and 3, we are done.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 28185,
    "subject": "Mathematics (Multi-modal)",
    "question": "Given an integer $n > 0$ and real numbers $x_1 \\le x_2 \\le \\dots \\le x_n$, $y_1 \\ge y_2 \\ge \\dots \\ge y_n$, satisfying $\\sum_{i=1}^n ix_i = \\sum_{i=1}^n iy_i$. Prove that for any real number $\\alpha$, $\\sum_{i=1}^n x_i[i\\alpha] \\ge \\sum_{i=1}^n y_i[i\\alpha]$, where $[\\beta]$ is defined as the greatest integer less than or equal to $\\beta$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "**Proof I** We need the following lemma.\nLemma For any real number $\\alpha$ and positive number $n$, we have\n$$\n\\sum_{i=1}^{n-1} [i\\alpha] \\le \\frac{n-1}{2} [n\\alpha]. \\qquad \\textcircled{1}\n$$\nThe lemma is obtained by summing inequalities\n$$\n[i\\alpha] + [(n-i)\\alpha] \\le [n\\alpha]\n$$\nfor $i = 1, 2, \\dots, n-1$.\n\nReturn to the original problem. We will prove it by induction. For $n = 1$, it is obviously true.\nAssume that for $n = k$, it is also true. Now consider $n = k + 1$. Let $a_i = x_i + \\frac{2}{k}x_{i+1}$, $b_i = y_i + \\frac{2}{k}y_{i+1}$ for $i = 1, 2, \\dots, k$. Then we have $a_1 \\le a_2 \\le \\dots \\le a_k$, $b_1 \\ge b_2 \\ge \\dots \\ge b_k$ and $\\sum_{i=1}^k ia_i = \\sum_{i=1}^k ib_i$. By induction we get $\\sum_{i=1}^k a_i[i\\alpha] \\ge \\sum_{i=1}^k b_i[i\\alpha]$.\n\nIn addition, $x_{k+1} \\ge y_{k+1}$. Otherwise, if $x_{k+1} < y_{k+1}$, we have\n$$\nx_1 \\le x_2 \\le \\dots \\le x_{k+1} < y_{k+1} \\le \\dots \\le y_2 \\le y_1.\n$$\nThis contradicts $\\sum_{i=1}^{k+1} ix_i = \\sum_{i=1}^{k+1} iy_i$. So we have\n$$\n\\begin{aligned}\n\\sum_{i=1}^{k+1} x_i[i\\alpha] - \\sum_{i=1}^k a_i[i\\alpha] &= x_{k+1} \\left\\{ \\left[ (k+1)\\alpha \\right] - \\frac{2}{k} \\sum_{i=1}^k [i\\alpha] \\right\\} \\\\\n&\\ge y_{k+1} \\left\\{ \\left[ (k+1)\\alpha \\right] - \\frac{2}{k} \\sum_{i=1}^k [i\\alpha] \\right\\} \\\\\n&= \\sum_{i=1}^{k+1} y_i[i\\alpha] - \\sum_{i=1}^k b_i[i\\alpha].\n\\end{aligned}\n$$\n$$\n\\text{That means } \\sum_{i=1}^{k+1} x_i[i\\alpha] \\ge \\sum_{i=1}^{k+1} y_i[i\\alpha].\n$$\nBy induction, we complete the proof for any integer $n > 0$.\n\n**Proof II** Define $z_i = x_i - y_i$ for $i = 1, 2, \\dots, n$, we have $z_1 \\le z_2 \\le \\dots \\le z_n$ and $\\sum_{i=1}^n iz_i = 0$. We only need to prove that\n$$\n\\sum_{i=1}^{n} z_i[i\\alpha] \\ge 0. \\qquad \\textcircled{2}\n$$\nLet $\\Delta_1 = z_1, \\Delta_2 = z_2 - z_1, \\dots, \\Delta_n = z_n - z_{n-1}$. Then $z_i = \\sum_{j=1}^i \\Delta_j$ ($1 \\le i \\le n$), and\n$$\n0 = \\sum_{i=1}^{n} iz_{i} = \\sum_{i=1}^{n} i \\sum_{j=1}^{i} \\Delta_{j} = \\sum_{j=1}^{n} \\Delta_{j} \\sum_{i=j}^{n} i.\n$$\nSo we have\n$$\n\\Delta_1 = - \\sum_{j=2}^{n} \\Delta_j \\sum_{i=j}^{n} i / \\sum_{i=1}^{n} i. \\qquad \\textcircled{3}\n$$\nThen\n$$\n\\begin{aligned}\n\\sum_{i=1}^{n} z_i[i\\alpha] &= \\sum_{i=1}^{n} [i\\alpha] \\sum_{j=1}^{i} \\Delta_j = \\sum_{j=1}^{n} \\Delta_j \\sum_{i=j}^{n} [i\\alpha] \\\\\n&= \\sum_{j=2}^{n} \\Delta_j \\sum_{i=j}^{n} [i\\alpha] - \\sum_{j=2}^{n} \\Delta_j \\left( \\sum_{i=j}^{n} i / \\sum_{i=1}^{n} i \\right) \\sum_{i=1}^{n} [i\\alpha] \\\\\n&= \\sum_{j=2}^{n} \\Delta_j \\sum_{i=j}^{n} i \\cdot \\left( \\sum_{i=j}^{n} [i\\alpha] / \\sum_{i=j}^{n} i - \\sum_{i=1}^{n} [i\\alpha] / \\sum_{i=1}^{n} i \\right).\n\\end{aligned}\n$$\nThen, in order to prove $\\textcircled{2}$ we only need to prove that for any $2 \\le j \\le n$, the following inequality holds\n$$\n\\sum_{i=j}^{n} [i\\alpha] / \\sum_{i=j}^{n} i \\ge \\sum_{i=1}^{n} [i\\alpha] / \\sum_{i=1}^{n} i. \\qquad \\textcircled{4}\n$$\nBut\n$$\n\\begin{aligned}\n\\textcircled{4} & \\Leftrightarrow \\sum_{i=j}^{n} [i\\alpha] / \\sum_{i=j}^{n} i \\ge \\sum_{i=1}^{j-1} [i\\alpha] / \\sum_{i=1}^{j-1} i \\\\\n& \\Leftrightarrow \\sum_{i=1}^{n} [i\\alpha] / \\sum_{i=1}^{n} i \\ge \\sum_{i=1}^{j-1} [i\\alpha] / \\sum_{i=1}^{j-1} i.\n\\end{aligned}\n$$\nThen we only need to prove, for any $k \\ge 1$,\n$$\n\\sum_{i=1}^{k+1} [i\\alpha] / \\sum_{i=1}^{k+1} i \\ge \\sum_{i=1}^{k} [i\\alpha] / \\sum_{i=1}^{k} i,\n$$\nthat is equivalent to prove\n$$\n\\begin{aligned}\n& [(k+1)\\alpha] \\cdot k/2 \\ge \\sum_{i=1}^{k} [i\\alpha] \\\\\n& \\Leftrightarrow \\sum_{i=1}^{k} \\left( [(k+1)\\alpha] - [i\\alpha] - [(k+1-i)\\alpha] \\right) \\ge 0.\n\\end{aligned}\n$$\nNote that, $[x+y] \\ge [x]+[y]$ holds for any real numbers $x, y$, hence $\\textcircled{4}$ holds. The proof is complete.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 28186,
    "subject": "Mathematics (Multi-modal)",
    "question": "Suppose that a matrix of nonnegative entries,\n$$\nP = \\begin{bmatrix} x_{11} & x_{12} & x_{13} & x_{14} & x_{15} & x_{16} & x_{17} & x_{18} & x_{19} \\\\ x_{21} & x_{22} & x_{23} & x_{24} & x_{25} & x_{26} & x_{27} & x_{28} & x_{29} \\\\ x_{31} & x_{32} & x_{33} & x_{34} & x_{35} & x_{36} & x_{37} & x_{38} & x_{39} \\end{bmatrix}\n$$\nhas the following properties:\n(1) Numbers in a row are different from each other;\n(2) The sum of the numbers in a column from the first six columns is $1$;\n(3) $x_{17} = x_{28} = x_{39} = 0$;\n(4) $x_{27}, x_{37}, x_{18}, x_{38}, x_{19}, x_{29} > 1$.\nAssume that matrix $S$ is constituted by the first three columns of $P$, i.e.\n$$\nS = \\begin{bmatrix} x_{11} & x_{12} & x_{13} \\\\ x_{21} & x_{22} & x_{23} \\\\ x_{31} & x_{32} & x_{33} \\end{bmatrix}\n$$\nhas the following property:\n(O) For any column $[x_{1k} \\ x_{2k} \\ x_{3k}]$ ($k = 1, 2, ..., 9$) in $P$, there exists $i \\in \\{1, 2, 3\\}$ such that\n$$\nx_{ik} \\le u_i = \\min\\{x_{i1}, x_{i2}, x_{i3}\\}. \\quad ①\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Proof of (i). Assume that it is not true. There is a column in $S$ which contains no $u_i$. We may say that $u_i \\neq x_{i2}$, $i = 1, 2, 3$. By property (1), we have $u_i < x_{i2}$, $i = 1, 2, 3$. On the other hand, let $k = 2$ in (i). Then by property (O), there exists $i_0 \\in \\{1, 2, 3\\}$ such that $x_{i_0 2} \\le u_i$. The contradiction means the assumption is not valid. This completes the proof of (i).\n\nProof of (ii). By the drawer principle, we know that at least two of the three numbers\n$$\n\\min\\{x_{11}, x_{12}\\}, \\min\\{x_{21}, x_{22}\\}, \\min\\{x_{31}, x_{32}\\}\n$$\nare in the same column. We may say that\n$$\n\\min\\{x_{21}, x_{22}\\} = x_{22}, \\min\\{x_{31}, x_{32}\\} = x_{32}.\n$$\nBy (i), we know that the first column of $S$ contains a $u_i$, and it must be $u_1 = x_{11}$; the second column also contains a $u_i$, assuming that it is $u_2 = x_{22}$, and then it must be $u_3 = x_{33}$.\nDefine $M = \\{1, 2, \\dots, 9\\}$ and\n$$\nI = \\{k \\in M \\mid x_{ik} > \\min\\{x_{i1}, x_{i2}\\}, i = 1, 3\\}.\n$$\nObviously, $I = \\{k \\in M \\mid x_{1k} > x_{11}, x_{3k} > x_{32}\\}$, and $1, 2, 3 \\notin I$. Since $x_{18}, x_{38} > 1 \\ge x_{11}, x_{32}$, we have $8 \\in I$. Therefore, $I \\ne \\emptyset$. Consequently, $\\exists k^* \\in I$ such that $x_{2k^*} = \\max\\{x_{2k} \\mid k \\in I\\}$. Of course, $k^* \\ne 1, 2, 3$.\n\nWe now prove that\n$$\nS' = \\begin{bmatrix} x_{11} & x_{12} & x_{1k^*} \\\\ x_{21} & x_{22} & x_{2k^*} \\\\ x_{31} & x_{32} & x_{3k^*} \\end{bmatrix}\n$$\nhas property (O).\nBy the definition of $I$, we know that\n$$\nx_{1k^*} > x_{11} = u_1, x_{3k^*} > x_{32} \\ge u_3.\n$$\nLet $k^* = k$ in $(i)$, and we get $x_{2k^*} \\le u_2$ according to property (O) of $S$. Then define\n$$\nu'_1 = u_1, u'_2 = \\min\\{x_{21}, x_{22}, x_{2k^*}\\} = x_{2k^*}, u'_3 = u_3.\n$$\nWe claim that, for any $k \\in M$, there exists $i \\in \\{1, 2, 3\\}$ such that $u'_i \\ge x_{ik}$. Otherwise, we would have $x_{ik} > \\min\\{x_{i1}, x_{i2}\\}$, $i = 1, 3$ and $x_{2k} > x_{2k^*}$, contradicting the definition of $k^*$.\nTherefore, $S'$ has property (O).\n\nSecondly, we prove the uniqueness of $S'$. Assume that $\\exists k_0 \\in M$ such that\n$$\n\\hat{S} = \\begin{bmatrix} x_{11} & x_{12} & x_{1k_0} \\\\ x_{21} & x_{22} & x_{2k_0} \\\\ x_{31} & x_{32} & x_{3k_0} \\end{bmatrix}\n$$\nalso has property (O). Without loss of generality, we assume that\n$$\n\\begin{aligned}\nu_i &= \\min\\{x_{i1}, x_{i2}, x_{i3}\\} = x_{ii}, \\quad i = 1, 2, 3, && (2) \\\\ x_{32} < x_{31}.\n\\end{aligned}\n$$\nSince $x_{32} < x_{31}$, $x_{22} < x_{21}$, and by (1), we have\n$$\n\\hat{u}_1 = \\min\\{x_{11}, x_{12}, x_{1k_0}\\} = x_{11}.\n$$\nBy (2) again, we have either\n(a) $\\hat{u}_3 = \\min\\{x_{31}, x_{32}, x_{3k_0}\\} = x_{3k_0}$ or\n(b) $\\hat{u}_2 = \\min\\{x_{21}, x_{22}, x_{2k_0}\\} = x_{2k_0}$.\nIf (a) is true, we will have\n$$\n\\hat{u}_1 = x_{11}, \\hat{u}_2 = x_{22}, \\hat{u}_3 = x_{3k_0}. \\quad (3)\n$$\nFor $3 \\in M$, since both $\\hat{S}$ and $S$ have property (O), we will get\n$$\nx_{33} \\le \\hat{u}_3 = x_{3k_0}, \\quad x_{3k_0} \\le u_3 = x_{33}.\n$$\nBy property (1) of $P$, we have $3 = k_0$. Therefore, $\\hat{S} = S$.\nIf (b) is true, we will have\n$$\n\\hat{u}_1 = x_{11}, \\hat{u}_2 = x_{2k_0}, \\hat{u}_3 = x_{32}. \\quad (4)\n$$\nSince $\\hat{S}$ has property (O), we know that, for $k^* \\in M$, there exists $i \\in \\{1, 2, 3\\}$ such that $\\hat{u}_i \\ge x_{ik}$. As $k^* \\in I$, and by (2), (4), it must be $x_{2k} \\le \\hat{u}_2 = x_{2k}$.\nOn the other hand, $S$ also has property (O). Then, in a similar way, we get $x_{2k} \\le u'_2 = x_{2k}$.\nTherefore, $k^* = k$. This completes the proof.",
    "topic": "Algebra",
    "subtopic": "Linear Algebra"
  },
  {
    "id": 28187,
    "subject": "Mathematics (Multi-modal)",
    "question": "Given three cubes with integer edge lengths, if the sum of their surface areas is $564\\ \\text{cm}^2$, then the sum of their volumes is ( ).",
    "options": [],
    "answer": "586 or 764",
    "solution": "Denote the edge lengths of the three cubes as $a$, $b$ and $c$, respectively. Then we have\n$$\n6(a^2 + b^2 + c^2) = 564,\n$$\ni.e. $a^2 + b^2 + c^2 = 94$. We may assume that\n$$\n1 \\le a \\le b \\le c < 10.\n$$\n\nThen\n$$\n3c^2 \\geq a^2 + b^2 + c^2 = 94.\n$$\nIt follows that $c^2 > 31$. So $6 \\leq c < 10$, and this means that $c$ can only be $9$, $8$, $7$ or $6$.\n\nIf $c = 9$, then\n$$\na^2 + b^2 = 94 - 9^2 = 13.\n$$\nIt is easy to see that $a = 2$, $b = 3$. So we get the solution $(a, b, c) = (2, 3, 9)$.\n\nIf $c = 8$, then\n$$\na^2 + b^2 = 94 - 8^2 = 30.\n$$\nThis means that $b \\geq 4$ and $2b^2 \\geq 30$; it follows that $b = 4$ or $5$, so $a^2 = 5$ or $14$; in both cases $a$ has no integer solution.\n\nIf $c = 7$, then\n$$\na^2 + b^2 = 94 - 7^2 = 45.\n$$\nIt is easy to see that $a = 3$, $b = 6$ is the only solution.\n\nIf $c = 6$, then\n$$\na^2 + b^2 = 94 - 6^2 = 58.\n$$\nSo $2b^2 \\geq 58$, or $b^2 \\geq 29$. This means that $b \\geq 6$, but $b \\leq c = 6$, so $b = 6$. Then $a^2 = 22$ and $a$ cannot be an integer.\n\nIn summary, there are two solutions: $(a, b, c) = (2, 3, 9)$ and $(a, b, c) = (3, 6, 7)$. Then the possible volumes are\n$$\nV_1 = 2^3 + 3^3 + 9^3 = 764\\ \\text{cm}^3,\n$$\n$$\nV_2 = 3^3 + 6^3 + 7^3 = 586\\ \\text{cm}^3.\n$$",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 28188,
    "subject": "Mathematics (Multi-modal)",
    "question": "Outside a convex quadrilateral $ABCD$ we construct equilateral triangles $ABQ$, $BCR$, $CDS$ and $DAP$. Denoting by $x$ the sum of the diagonals of $ABCD$, and by $y$ the sum of line segments joining the midpoints of opposite sides of $PQRS$, we find the maximum value of $\\frac{y}{x}$.",
    "options": [],
    "answer": "(1+sqrt(3))/2",
    "solution": "If $ABCD$ is a square, then $\\frac{y}{x} = \\frac{1+\\sqrt{3}}{2}$.\n\nNow we prove that $\\frac{y}{x} \\le \\frac{1+\\sqrt{3}}{2}$.\n\nDenote by $P_1, Q_1, R_1, S_1$ the midpoints of $DA$, $AB$, $BC$, $CD$, and by $E$, $F$, $G$, $H$ the midpoints of $SP$, $PQ$, $QR$, $RS$. Then $P_1Q_1R_1S_1$ is a parallelogram.\n\nNow draw lines $P_1E$, $S_1E$, and denote by $M$, $N$ the midpoints of $DP$, $DS$. Then\n![](attached_image_1.png)\n$$\n\\begin{aligned}\nDS_1 &= S_1N = DN = EM, \\\\\nDP_1 &= P_1M = MD = EN,\n\\end{aligned}\n$$\nand\n$$\n\\begin{aligned}\n\\angle P_1DS_1 &= 360^\\circ - 60^\\circ - 60^\\circ - \\angle PDS \\\\\n&= 240^\\circ - (180^\\circ - \\angle END) = 60^\\circ + \\angle END \\\\\n&= \\angle ENS_1 = \\angle EMP_1.\n\\end{aligned}\n$$\nSo we have $\\triangle DP_1S_1 \\cong \\triangle MP_1E \\cong \\triangle NES_1$. Hence, $\\triangle EP_1S_1$ is equilateral.\n\nBy the same argument, $\\triangle GQ_1R_1$ is also equilateral. Now let $U$, $V$ be the midpoints of $P_1S_1$, $Q_1R_1$, respectively. We then obtain\n$$\n\\begin{aligned}\nEG &\\le EU + UV + VG = \\frac{\\sqrt{3}}{2}P_1S_1 + P_1Q_1 + \\frac{\\sqrt{3}}{2}Q_1R_1 \\\\\n&= P_1Q_1 + \\sqrt{3}P_1S_1 = \\frac{1}{2}BD + \\frac{\\sqrt{3}}{2}AC,\n\\end{aligned}\n$$\nand also\n$$FH \\le \\frac{1}{2}AC + \\frac{\\sqrt{3}}{2}BD.$$\n\nTaking the sum of these two inequalities, we have $y \\le \\frac{1+\\sqrt{3}}{2}x$, i.e.\n$$\n\\frac{y}{x} \\le \\frac{1+\\sqrt{3}}{2}.\n$$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 28189,
    "subject": "Mathematics (Multi-modal)",
    "question": "Given an integer $m \\ge 2$, and two real numbers $a, b$ with $a > 0$ and $b \\ne 0$, the sequence $\\{x_n\\}$ is such that $x_1 = b$ and $x_{n+1} = a x_n^m + b$, $n = 1, 2, \\dots$. Prove that:\n\n(1) When $b < 0$ and $m$ is even, the sequence $\\{x_n\\}$ is bounded if and only if $a b^{m-1} \\ge -2$;\n\n(2) When $b < 0$ and $m$ is odd, or when $b > 0$, the sequence $\\{x_n\\}$ is bounded if and only if $a b^{m-1} \\le \\frac{(m-1)^{m-1}}{m^m}$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "(1) When $b < 0$ and $m$ is even, in order that $a b^{m-1} < -2$, we should first have $a b^m + b > -b > 0$, and therefore $a(a b^m + b)^m + b > a b^m + b > 0$, i.e. $x_3 > x_2 > 0$. Using the fact that $a x^m + b$ is monotonically increasing on $(0, +\\infty)$, it can be established that each succeeding term of the sequence $\\{x_n\\}$ is greater than its preceding term, and is greater than $-b$ starting from the second term.\n\nConsidering any three consecutive terms of the sequence $x_n, x_{n+1}, x_{n+2}, \\dots$, we have\n$$\n\\begin{aligned}\nx_{n+2} - x_{n+1} &= a(x_{n+1}^m - x_n^m) \\\\\n&= a(x_{n+1} - x_n)(x_{n+1}^{m-1} + x_{n+1}^{m-2} x_n + \\dots + x_n^{m-1}) \\\\\n&> a m x_n^{m-1}(x_{n+1} - x_n) \\\\\n&> a m (-b)^{m-1}(x_{n+1} - x_n) \\\\\n&> 2 m (x_{n+1} - x_n) \\\\\n&> x_{n+1} - x_n.\n\\end{aligned}\n$$\nIt is obvious that the difference of any two consecutive terms of the sequence $\\{x_n\\}$ is increasing, and hence it is not bounded.\n\nWhen $a b^{m-1} \\ge -2$, mathematical induction is used to prove that each term of the sequence $\\{x_n\\}$ falls on the interval $[b, -b]$.\n\nThe first term $b$ falls on the interval $[b, -b]$. Suppose that the term $x_n$ satisfies the condition $b \\le x_n \\le -b$ for a particular $n$. Then $0 \\le x_n^m \\le b^m$, and hence\n$$\nb = a \\times 0^m + b \\le x_{n+1} \\le a b^m + b \\le -b.\n$$\nThus, the sequence $\\{x_n\\}$ is bounded if and only if $a b^{m-1} \\ge -2$.\n\n\n(2) When $b > 0$, each term of the sequence $\\{x_n\\}$ is positive. So, we first prove that $\\{x_n\\}$ is bounded if and only if the equation $a x^m + b = x$ has positive real roots.\n\nSuppose that $a x^m + b = x$ has no positive real roots. In such a case, the minimum value of the function $p(x) = a x^m + b - x$ on the interval $(0, +\\infty)$ is greater than zero. Let $t$ be the minimum value. It follows that for any two consecutive terms of the sequence $x_n$ and $x_{n+1}$, we have $x_{n+1} - x_n = a x_n^m - x_n + b$. Thus, each succeeding term of the sequence $\\{x_n\\}$ is greater than the preceding term by at least $t$. Hence, it is not bounded.\n\nIf the equation $a x^m + b = x$ has positive real roots, let $x_0$ be one of the positive real roots. Then, by using mathematical induction, we prove that each term of the sequence $\\{x_n\\}$ is less than $x_0$. Firstly, the first term $b$ is less than $x_0$. Suppose that $x_n < x_0$ for a particular $n$. By virtue of the fact that $a x^m + b$ is increasing on the interval $[0, +\\infty)$, it can be established that\n$$\nx_{n+1} = a x_n^m + b < a x_0^m + b = x_0.\n$$\nTherefore, the sequence is bounded.\n\nFurther, the equation $a x^m + b = x$ has positive roots if and only if the minimum value of $a x^{m-1} + \\frac{b}{x}$ on the interval $(0, +\\infty)$ is not greater than $1$, whereas the minimum value of $a x^{m-1} + \\frac{b}{x}$ can be determined by mean inequality, i.e.\n$$\na x^{m-1} + \\frac{b}{x} = a x^{m-1} + \\frac{b}{(m-1)x} + \\dots + \\frac{b}{(m-1)x} \\geq m \\sqrt[m]{\\frac{a b^{m-1}}{(m-1)^{m-1}}}\n$$\nAs such, the sequence $\\{x_n\\}$ is bounded if and only if\n$$\nm \\sqrt[m]{\\frac{a b^{m-1}}{(m-1)^{m-1}}} \\le 1, \\text{ i.e. } a b^{m-1} \\le \\frac{(m-1)^{m-1}}{m^m}.\n$$\n\nWhen $b < 0$, and $m$ is odd, let $y_n = -x_n$. Then $y_1 = -b > 0$, $y_{n+1} = a y_n^m + (-b)$, showing that the sequence $\\{x_n\\}$ is bounded if and only if the sequence $\\{y_n\\}$ is bounded. Thus, by using the above reasoning, it can be proven that (2) holds.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 28190,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $n > 1$ be a given integer and $A$ be an infinite set of positive integers satisfying: for any prime $p \\nmid n$, there exist infinitely many elements of $A$ not divisible by $p$. Prove that for any integer $m > 1$, $(m, n) = 1$, there exists a finite subset of $A$ whose sum of elements, say $S$, satisfies $S \\equiv 1 \\pmod m$ and $S \\equiv 0 \\pmod n$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Suppose a prime $p$ satisfies $p^a \\mid m$. Then from the given conditions, there exists an infinite subset $A_1$ of $A$ such that $p$ is coprime to every element in $A_1$.\nBy the pigeonhole principle, there is an infinite subset $A_2$ of $A_1$, such that $x \\equiv a \\pmod{mn}$, for each element $x \\in A_2$, where $a$ is a positive integer and $p \\nmid a$.\nSince $(m, n) = 1$, we have $(p^a, \\frac{mn}{p^a}) = 1$. By the Chinese Remainder Theorem, we know that\n$$\n\\begin{cases} x \\equiv a^{-1} \\pmod{p^a}, \\\\ x \\equiv 0 \\pmod{\\frac{mn}{p^a}} \\end{cases} \\quad ①\n$$\nhave infinitely many solutions. Among them we take one as $x$. Next define $B_p$ as the set of the first $x$ elements in $A_2$, and $S_p$ as the sum of all elements in $B_p$. Then we have $S_p \\equiv a x \\pmod{mn}$. By ① we have\n$$\nS_p \\equiv a x \\equiv 1 \\pmod{p^a}, \\quad S_p \\equiv 0 \\pmod{\\frac{mn}{p^a}}.\n$$\n\nSuppose that $m = p_1^{a_1} \\cdots p_k^{a_k}$, and for every $p_i$ ($1 \\le i \\le k$) select a finite subset $B_i$ of $A$, where $B_i \\subset A \\setminus (B_1 \\cup \\cdots \\cup B_{i-1})$, such that $S_{p_i}$, the sum of all the elements in $B_i$, satisfies\n$$\nS_{p_i} \\equiv 1 \\pmod{p_i^{a_i}}, \\quad S_{p_i} \\equiv 0 \\pmod{\\frac{mn}{p_i^{a_i}}}. \\qquad \\textcircled{2}\n$$\nLet $B = \\bigcup_{i=1}^k B_i$, whose sum of elements then satisfies $S = \\sum_{i=1}^k S_i$. According to ②, we have $S \\equiv 1 \\pmod{p_i^{a_i}}$ ($1 \\le i \\le k$), and $S \\equiv 0 \\pmod{n}$. So $B$ is the required subset, and the proof is complete.\n\n\nSolution 2:\n\nDivide every element in $A$ by $mn$, and let the remainders which occur infinitely be (in order) $\\alpha_1, \\alpha_2, \\dots, \\alpha_k$. We claim that\n$$\n(\\alpha_1, \\alpha_2, \\dots, \\alpha_k, m) = 1. \\qquad \\textcircled{3}\n$$\nOtherwise, assume that $p \\mid (\\alpha_1, \\alpha_2, \\dots, \\alpha_k, m)$. Then $p \\nmid n$, since $(m,n) = 1$. By the given conditions we know that there exist infinitely many elements of $A$ not divisible by $p$. But by the definition of $\\alpha_1, \\alpha_2, \\dots, \\alpha_k$, the number of such elements are finite. Then by contradiction, ③ holds. Consequently, there are $x_1, x_2, \\dots, x_k, y$, satisfying $\\alpha_1 x_1 + \\alpha_2 x_2 + \\dots + \\alpha_k x_k - y m = 1$. Choose a suitable positive integer $r$ such that $r n \\equiv 1 \\pmod{m}$. Then\n$$\n\\alpha_1(r n x_1) + \\alpha_2(r n x_2) + \\dots + \\alpha_k(r n x_k) = r n + r m n y.\n$$\nWe select in order $r x_i$ elements from $A$ such that the remainders by $mn$ are $\\alpha_i$ ($i = 1, 2, \\dots, k$). The set of all these elements is acquired.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 28191,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all functions $f: (0, +\\infty) \\to (0, +\\infty)$ ($f$ is a function mapping positive real numbers to positive real numbers) such that\n$$\n\\frac{(f(w))^2 + (f(x))^2}{f(y^2) + f(z^2)} = \\frac{w^2 + x^2}{y^2 + z^2}\n$$\nfor all positive real numbers $w, x, y, z$ satisfying\n$$wx = yz.$$",
    "options": [],
    "answer": "f(x)=x for all x>0, or f(x)=1/x for all x>0",
    "solution": "Take\n$$\nw = x = y = z = 1,\n$$\nthen we get $(f(1))^2 = f(1)$, so $f(1) = 1$.\nFor any real number $t > 0$, let $w = t$, $x = 1$, $y = z = \\sqrt{t}$,\nwe get\n$$\n\\frac{(f(t))^2 + 1}{2f(t)} = \\frac{t^2 + 1}{2t},\n$$\nwhich implies $(tf(t) - 1)(f(t) - t) = 0$.\nSo, for any $t > 0$,\n$$\nf(t) = t \\quad \\text{or} \\quad f(t) = \\frac{1}{t}. \\qquad \\textcircled{1}\n$$\nSuppose there exist $b, c \\in (0, +\\infty)$ such that $f(b) \\neq b$, $f(c) \\neq \\frac{1}{c}$. By ①, we get $b, c$ different from 1 and $f(b) = \\frac{1}{b}$, $f(c) = c$.\nTake $w = b, x = c, y = z = \\sqrt{bc}$, then\n$$\n\\frac{\\frac{1}{b^2} + c^2}{2f(bc)} = \\frac{b^2 + c^2}{2bc},\n$$\ni.e. $f(bc) = \\frac{c + b^2c^3}{b(b^2 + c^2)}$.\nBy ①, $f(bc) = bc$ or $f(bc) = \\frac{1}{bc}$. If $f(bc) = bc$, then\n$$\nbc = \\frac{c + b^2c^3}{b(b^2 + c^2)},\n$$\nwhich yields $b^4c = c, b = 1$. Contradiction!\n\nIf $f(bc) = \\frac{1}{bc}$, then\n$$\n\\frac{1}{bc} = \\frac{c + b^2 c^3}{b(b^2 + c^2)},\n$$\nthat yields $b^2c^4 = b^2$, $c = 1$. Contradiction!\nTherefore, only two functions: $f(x) = x, x \\in (0, +\\infty)$ or $f(x) = \\frac{1}{x}, x \\in (0, +\\infty)$. It is easy to verify that these two functions satisfy the given conditions.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 28192,
    "subject": "Mathematics (Multi-modal)",
    "question": "Given positive integers $m, n \\ge 2$, first select two different $a_i, a_j$ ($j > i$) in the integer set $A = \\{a_1, a_2, \\dots, a_n\\}$ and take the difference $a_j - a_i$. Then arrange the $\\binom{n}{2}$ differences in ascending order to form a new sequence, which we call 'derived sequence' and is denoted by $\\bar{A}$. The number of elements in $\\bar{A}$ that can be divided by $m$ is denoted by $\\bar{A}(m)$. Prove that for any $m \\ge 2$, the corresponding derived sequences $\\bar{A}$ and $\\bar{B}$, with regard to $A = \\{a_1, a_2, \\dots, a_n\\}$ and $B = \\{1, 2, \\dots, n\\}$, satisfy the inequality $\\bar{A}(m) \\ge \\bar{B}(m)$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "**Proof** For any integer $m \\ge 2$, if the remainder of $x$ divided by $m$ is $i$, $i \\in \\{0, 1, \\dots, m-1\\}$, then $x$ belongs to the residue class modulus $m$, $K_i$.\nSuppose in the set $A = \\{a_1, a_2, \\dots, a_n\\}$, the number of elements that belong to $K_i$ is $n_i$ ($i = 0, 1, 2, \\dots, m-1$), while in the set $B = \\{1, 2, \\dots, n\\}$, the number of elements that belong to $K_i$ is $n'_i$ ($i = 0, 1, 2, \\dots, m-1$), then\n$$\n\\sum_{i=0}^{m-1} n_i = \\sum_{i=0}^{m-1} n'_i = n. \\qquad \\textcircled{1}\n$$\nIt is obvious that for every $i, j$, $|n_i' - n_j'| \\le 1$, and $x-y$ is a multiple of $m$ if and only if $x, y$ belong to the same residue class. As to any two elements $a_i, a_j$ in $K_i$, we have $m \\mid a_j - a_i$.\nHence, the $n_i$ elements in $K_i$ form $\\binom{n_i}{2}$ multiples of $m$.\nConsidering all the $i$, we obtain\n$$\n\\bar{A}(m) = \\sum_{i=0}^{m-1} \\binom{n_i}{2}.\n$$\nSimilarly,\n$$\n\\bar{B}(m) = \\sum_{i=0}^{m-1} \\binom{n_i'}{2}.\n$$\nHence, to solve the problem, we just need to prove that\n$$\n\\sum_{i=0}^{m-1} \\binom{n_i}{2} \\ge \\sum_{i=0}^{m-1} \\binom{n_i'}{2}, \\quad \\text{and it can be simplified to}\n$$\n$$\n\\sum_{i=0}^{m-1} n_i^2 \\ge \\sum_{i=0}^{m-1} n_i'^2. \\qquad \\textcircled{2}\n$$\nFrom ①, if for every $i, j$, $|n_i - n_j| \\le 1$, then $n_0, n_1, \\dots, n_{m-1}$ and $n'_0, n'_1, \\dots, n'_{m-1}$ must be the same group (in spite of the different order), and equality holds in ②. Otherwise, if there exist $i, j$, such that $n_i - n_j \\ge 2$, then we should just change the two elements $n_i, n_j$ for $\\bar{n}_i, \\bar{n}_j$ respectively, where $\\bar{n}_i = n_i - 1$, $\\bar{n}_j = n_j + 1$, and $n_i + n_j = \\bar{n}_i + \\bar{n}_j$. Since\n$$\n(n_i^2 + n_j^2) - (\\bar{n}_i^2 + \\bar{n}_j^2) = 2(n_i - n_j - 1) > 0,\n$$\nthe sum of the left side in ② will decrease after adjustment. Therefore, the minimum value of ② is attained if and only if $n_0, n_1, \\cdots, n_{m-1}$ and $n'_0, n'_1, \\cdots, n'_{m-1}$ are the same group (in spite of the different order), that is, the inequality ② holds.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 28193,
    "subject": "Mathematics (Multi-modal)",
    "question": "Suppose that the convex quadrilateral $ABCD$ satisfies $AB = BC$, $AD = DC$. $E$ is a point on $AB$, and $F$ on $AD$, such that $B$, $E$, $F$, $D$ are concyclic. Draw $\\triangle DPE$ directly similar to $\\triangle ADC$, and $\\triangle BQF$ directly similar to $\\triangle ABC$. Prove that $A$, $P$, $Q$ are collinear. (Posed by Ye Zhonghao)\n\nDenote by $O$ the center of the circle that passes through $B$, $E$, $F$, $D$. Draw lines $OB$, $OF$, $BD$.\n\n![](attached_image_1.png)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "In $\\triangle BDF$, $O$ is the circumcenter, so $\\angle BOF = 2\\angle BDA$; And $\\triangle ABD \\sim \\triangle CBD$, so $\\angle CDA = 2\\angle BDA$. Hence, $\\angle BOF = \\angle CDA = \\angle EPD$, which implies that the isosceles triangles\n$$\n\\triangle BOF \\sim \\triangle EPD. \\qquad \\textcircled{1}\n$$\nOn the other hand, the concyclicity of $B$, $E$, $F$, $D$ implies that\n$$\n\\triangle ABF \\sim \\triangle ADE. \\qquad \\textcircled{2}\n$$\nCombining ① and ②, we know that the quadrilateral $ABOF \\sim ADPE$, so\n$$\n\\angle BAO = \\angle DAP. \\qquad \\textcircled{3}\n$$\nThe same argument gives\n$$\n\\angle BAO = \\angle DAQ. \\qquad \\textcircled{4}\n$$\n③ and ④ imply that $A$, $P$, $Q$ are collinear.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 28194,
    "subject": "Mathematics (Multi-modal)",
    "question": "The number of rational solutions to the system of equations\n$$\n\\begin{cases} x + y + z = 0, \\\\ xyz + z = 0, \\\\ xy + yz + xz + y = 0 \\end{cases}, \\text{ is } (\\quad).\n$$",
    "options": [],
    "answer": "2",
    "solution": "If $z = 0$, then\n$$\n\\begin{cases} x + y = 0, \\\\ xy + y = 0 \\end{cases}.\n$$\nIt follows that\n$$\n\\begin{cases} x = 0, \\\\ y = 0 \\end{cases} \\text{ or } \\begin{cases} x = -1, \\\\ y = 1. \\end{cases}\n$$\nIf $z \\neq 0$, from $xyz + z = 0$ we get\n$$\nxy = -1. \\qquad \\textcircled{1}\n$$\nFrom $x + y + z = 0$ we have\n$$\nz = -x - y. \\qquad \\textcircled{2}\n$$\nSubstituting $\\textcircled{2}$ into $xy + yz + xz + y = 0$, we obtain\n$$\nx^2 + y^2 + xy - y = 0. \\qquad \\textcircled{3}\n$$\nFrom $\\textcircled{1}$ we have $x = -\\frac{1}{y}$. We substitute it into $\\textcircled{3}$, and then make a simplification:\n$$\n(y - 1)(y^3 - y - 1) = 0.\n$$\nIt is easy to see that $y^3 - y - 1$ has no rational solution, and so $y = 1$. Then from $\\textcircled{1}$ and $\\textcircled{2}$ we get $x = -1$ and $z = 0$, contradicting $z \\neq 0$.\n\nIn summary, the system has exactly two solutions:\n$$\n\\begin{cases} x = 0, \\\\ y = 0, \\\\ z = 0 \\end{cases}, \\begin{cases} x = -1, \\\\ y = 1, \\\\ z = 0 \\end{cases}\n$$",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 28195,
    "subject": "Mathematics (Multi-modal)",
    "question": "Four frogs are positioned at four points on a straight line such that the distance between any two neighboring points is one unit of length. Suppose that every frog can jump to its corresponding point of reflection, by taking any one of the other three frogs as the reference point. Prove that there is no case where the distances between any two neighboring points, where the frogs stay, are all equal to $2008$ units of length. (Posed by Liu Shixiong)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Without loss of generality, we may think of the initial positioning of the four frogs as being on the real number line at points $1$, $2$, $3$, and $4$. Further, it can be established that the frogs at odd number positions remain at odd number positions after each jump, and likewise for frogs at even number positions. Thus, no matter after how many jumps, there are two frogs remaining at odd number positions while the other two frogs remain at even number positions.\n\nTherefore, in order that the distances between any two neighboring frogs are all equal to $2008$, all the frogs need to stay at points which are either all odd or all even, which is contrary to the actual situation. Hence, the proposition is proven.",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 28196,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $m, n \\in \\mathbb{N}^*, m, n > 1, a_{ij}$ ($i = 1, 2, \\dots, n$, $j = 1, 2, \\dots, m$) be non-negative real numbers (not all zero). Find the maximum and minimum values of\n$$\nf = \\frac{n \\sum_{i=1}^{n} (\\sum_{j=1}^{m} a_{ij})^2 + m \\sum_{j=1}^{m} (\\sum_{i=1}^{n} a_{ij})^2}{(\\sum_{i=1}^{n} \\sum_{j=1}^{m} a_{ij})^2 + mn \\sum_{i=1}^{n} \\sum_{j=1}^{m} a_{ij}^2}.\n$$",
    "options": [],
    "answer": "Maximum = 1; Minimum = (m + n) / (mn + min{m, n})",
    "solution": "The maximum value of $f$ is $1$.\nFirstly, we prove that $f \\le 1$. It suffices to show that\n$$\nn \\sum_{i=1}^{n} (\\sum_{j=1}^{m} a_{ij})^2 + m \\sum_{j=1}^{m} (\\sum_{i=1}^{n} a_{ij})^2 \\le (\\sum_{i=1}^{n} \\sum_{j=1}^{m} a_{ij})^2 + mn \\sum_{i=1}^{n} \\sum_{j=1}^{m} a_{ij}^2,\n$$\n$$\n\\text{or} \\quad \\left(\\sum_{i=1}^{n} \\sum_{j=1}^{m} a_{ij}\\right)^2 + mn \\sum_{i=1}^{n} \\sum_{j=1}^{m} a_{ij}^2 - n \\sum_{i=1}^{n} \\left(\\sum_{j=1}^{m} a_{ij}\\right)^2 - m \\sum_{j=1}^{m} \\left(\\sum_{i=1}^{n} a_{ij}\\right)^2 \\ge 0,\n$$\n$$\n\\text{or} \\quad \\sum_{\\substack{1 \\le p < s \\le n \\\\ 1 \\le q < r \\le m}} (a_{pq} + a_{sr} - a_{pr} - a_{sq})^2 \\ge 0.\n$$\nSo $f \\le 1$, and when all of $a_{ij}$ are equal to $1$, $f = 1$.\n\nThe minimum value of $f$ is $\\frac{m+n}{mn + \\min\\{m, n\\}}$.\nTo prove $f \\ge \\frac{m+n}{mn + \\min\\{m, n\\}}$, without loss of generality, we assume $n \\le m$. Hence it is sufficient to prove that\n$$\nf \\ge \\frac{m+n}{mn+n} \\qquad \\textcircled{1}\n$$\nLet\n$$\nS = \\frac{n^2(m+1)}{m+n} \\sum_{i=1}^{n} r_i^2 + \\frac{mn(m+1)}{m+n} \\sum_{j=1}^{m} c_j^2\n$$\n$$\n-(\\sum_{i=1}^{n} \\sum_{j=1}^{m} a_{ij})^2 - mn \\sum_{i=1}^{n} \\sum_{j=1}^{m} a_{ij}^2,\n$$\nwhere $r_i = \\sum_{j=1}^{m} a_{ij}$, $1 \\le i \\le n$, $c_j = \\sum_{i=1}^{n} a_{ij}$, $1 \\le j \\le m$.\n\nNow $\\textcircled{1} \\Leftrightarrow S \\ge 0$. Consider Lagrange's equation.\n$$\n(\\sum_{i=1}^{n} a_i b_i)^2 = (\\sum_{i=1}^{n} a_i^2)(\\sum_{i=1}^{n} b_i^2) - \\sum_{1 \\le k < l \\le n} (a_k b_l - a_l b_k)^2\n$$\nPut $a_i = r_i$, $b_i = 1$, $1 \\le i \\le n$. Then\n$$\n-(\\sum_{i=1}^{n} \\sum_{j=1}^{m} a_{ij})^2 = -n \\sum_{i=1}^{n} r_i^2 + \\sum_{1 \\le k < l \\le n} (r_k - r_l)^2,\n$$\nand\n$$\n\\begin{aligned}\nS &= \\frac{mn(n-1)}{m+n} \\sum_{i=1}^{n} r_i^2 + \\frac{mn(m+1)}{m+n} \\sum_{j=1}^{m} c_j^2 \\\\\n&= -mn \\sum_{i=1}^{n} \\sum_{j=1}^{m} a_{ij}^2 + \\sum_{1 \\le k < l \\le n} (r_k - r_l)^2 \\\\\n&= \\frac{mn(n-1)}{m+n} \\sum_{j=1}^{m} \\sum_{i=1}^{n} a_{ij} (r_i - a_{ij})^2 \\\\\n&\\quad + \\frac{mn(m+1)}{m+n} \\sum_{i=1}^{n} \\sum_{j=1}^{m} a_{ij} (c_j - a_{ij})^2 \\\\\n&\\quad + \\sum_{1 \\le k < l \\le n} (r_k - r_l)^2.\n\\end{aligned}\n$$\nSince $a_{ij} \\ge 0$, $r_i \\ge a_{ij}$, $c_j \\ge a_{ij}$, so $S \\ge 0$.\n\nWhen $a_{11} = a_{22} = \\cdots = a_{mn} = 1$ and the other $a_{ij} = 0$, the minimum value of $f$ is $\\frac{m+n}{mn+n}$.\n\nWith the above arguments, we conclude that the maximum value of $f$ is $1$ and the minimum value of $f$ is $\\frac{m+n}{mn+\\min\\{m, n\\}}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 28197,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find the largest positive number $\\lambda$ such that\n$$\n| \\lambda xy + yz | \\le \\frac{\\sqrt{5}}{2}, \\text{ where } x^2 + y^2 + z^2 = 1.\n$$",
    "options": [],
    "answer": "2",
    "solution": "Note that\n$$\n\\begin{aligned}\n1 &= x^2 + y^2 + z^2 \\\\\n  &= x^2 + \\frac{\\lambda^2}{1+\\lambda^2}y^2 + \\frac{1}{1+\\lambda^2}y^2 + z^2 \\\\\n  &\\ge \\frac{2}{\\sqrt{1+\\lambda^2}}(|x| + |y| + |z|) \\\\\n  &\\ge \\frac{2}{\\sqrt{1+\\lambda^2}}(|\\lambda xy + yz|),\n\\end{aligned}\n$$\nand the two equalities hold simultaneously when\n$$\ny = \\frac{\\sqrt{2}}{2}, \\quad x = \\frac{\\sqrt{2}\\lambda}{2\\sqrt{\\lambda^2+1}}, \\quad z = \\frac{\\sqrt{2}}{2\\sqrt{\\lambda^2+1}}.\n$$\nThus, $\\frac{\\sqrt{1+\\lambda^2}}{2}$ is the maximum value of $|\\lambda xy + yz|$. Let\n$$\n\\frac{\\sqrt{1+\\lambda^2}}{2} = \\frac{\\sqrt{5}}{2}.\n$$\nWe obtain that $\\lambda = 2$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 28198,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find the least positive integer $n$ with the following property: Paint each vertex of a regular $n$-gon arbitrarily with one of three colors, say red, yellow and blue, there must exist four vertices of the same color that constitute the vertices of some isogonal trapezoid.",
    "options": [],
    "answer": "17",
    "solution": "We claim that the least positive integer $n$ is $17$.\n\nFirstly we prove that $n = 17$ has the required property. By contradiction, assume that we have a painting pattern with three colors for the regular $17$-gon such that any group of $4$ vertices of the same color cannot constitute an isogonal trapezoid.\n\nAs $\\lfloor \\frac{17}{3} \\rfloor + 1 = 6$, there exists a group of $6$ vertices of the same color, say yellow. Connecting these vertices one another with lines, we get $\\binom{6}{2} = 15$ segments. Since the lengths of the segments have at most $\\lfloor \\frac{17}{2} \\rfloor = 8$ variations, one of the following two cases must exist:\n\n(a) There is a group of three segments with the same length. Since $3 \\nmid 7$, not every pair of the segments in the group has a common vertex. So there are two segments in the group which have no common vertex. The four vertices of the two segments constitute an isogonal trapezoid, and this is a contradiction.\n\n(b) There are $7$ pairs of segments with the same length. Then each pair must have a common vertex for its segments. Otherwise, the $4$ vertices of the segments in a pair with no common vertex will constitute an isogonal trapezoid. On the other hand, by the pigeonhole principle, we know that there are two pairs which share the same vertex as their segments' common vertex. Then another four vertices of the segments in these two pairs constitute an isogonal trapezoid. This leads to a contradiction again. So $n = 17$ has the required property.\n\nNext we will construct painting patterns for $n \\le 16$, which do not have the required property. Define $A_1, A_2, \\dots, A_n$ as the vertices of a regular $n$-gon (ordered in clockwise), and $M_1, M_2, M_3$ as the sets of vertices with the same color — red, yellow and blue respectively.\n\nWhen $n = 16$, let\n$$\nM_1 = \\{A_5, A_8, A_{13}, A_{14}, A_{16}\\},\n$$\n$$\nM_2 = \\{A_3, A_6, A_7, A_{11}, A_{15}\\},\n$$\n$$\nM_3 = \\{A_1, A_2, A_4, A_9, A_{10}, A_{12}\\}.\n$$\nIn $M_1$, it is easy to check that the distances from $A_{14}$ to the other $4$ vertices are different from each other, and the latter $4$ vertices constitute a rectangle, not an isogonal trapezoid. Similarly, no $4$ vertices in $M_2$ constitute an isogonal trapezoid either. As to $M_3$, the $6$ vertices in it are just vertices of three diameters. So any group of four vertices that constitutes either a rectangle or a $4$-gon with its sides of different lengths.\n\nWhen $n = 15$, let\n$$\nM_1 = \\{A_1, A_2, A_3, A_5, A_8\\},\n$$\n$$\nM_2 = \\{A_6, A_9, A_{13}, A_{14}, A_{15}\\},\n$$\n$$\nM_3 = \\{A_4, A_7, A_{10}, A_{11}, A_{12}\\}.\n$$\nIt is easy to check that no four vertices in each $M_i$ ($i = 1, 2, 3$) constitute an isogonal trapezoid.\n\nWhen $n = 14$, let\n$$\nM_1 = \\{A_1, A_3, A_8, A_{10}, A_{14}\\},\n$$\n$$\nM_2 = \\{A_4, A_5, A_7, A_{11}, A_{12}\\},\n$$\n$$\nM_3 = \\{A_2, A_6, A_9, A_{13}\\}.\n$$\nThis can be verified easily.\n\nWhen $n = 13$, let\n$$\nM_1 = \\{A_5, A_6, A_7, A_{10}\\},\n$$\n$$\nM_2 = \\{A_1, A_8, A_{11}, A_{12}\\},\n$$\n$$\nM_3 = \\{A_2, A_3, A_4, A_9, A_{13}\\}.\n$$\nThis can be easily verified too. As in this case, we drop $A_{13}$ from $M_3$, then we arrive at the case for $n = 12$; further drop $A_{12}$, we have the case $n = 11$; and further drop $A_{11}$, we get the case $n = 10$.\n\nWhen $n \\le 9$, we can construct a painting pattern such that $|M_i| < 4$ ($i = 1, 2, 3$), to ensure that no four vertices of the same color constitute an isogonal trapezoid.\n\nBy now, we have checked all the cases for $n \\le 16$. This completes the proof that $17$ is the least value for $n$ to have the required property.",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 28199,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $n$ and $k$ be positive integers with $k \\ge n$ and $k - n$ an even number. Let $2n$ lamps labelled $1, 2, \\dots, 2n$ be given, each of which can be either on or off. Initially all the lamps are off. We consider sequences of steps: at each step one of the lamps is switched from on to off or from off to on.\nLet $N$ be the number of such sequences consisting of $k$ steps and resulting in the state where lamps $1$ through $n$ are all on, and lamps $n+1$ through $2n$ are all off.\nLet $M$ be the number of such sequences consisting of $k$ steps, resulting in the state where lamps $1$ through $n$ are all on, and lamps $n+1$ through $2n$ are all off, but where none of the lamps $n+1$ through $2n$ is ever switched on.\nDetermine the ratio $\\frac{N}{M}$.",
    "options": [],
    "answer": "2^(k-n)",
    "solution": "**Lemma** For any positive integer $t$, call a $t$-element array $(a_1, a_2, \\dots, a_t)$ which consists of $0, 1$ ($a_1, a_2, \\dots, a_t \\in \\{0, 1\\}$) \"good\" if there are odd '0's in it. Prove that there are $2^{t-1}$ \"good\" arrays.\n\n**Proof:** In fact, for the same $a_1, a_2, \\dots, a_t$, when $a_t$ is 0 or 1, the parity of 0s in these two arrays is different, so only one of the arrays is \"good\". There are $2^t$ arrays in all, we can match one array to the other, there are only $a_t$ of them are different. Only one of these two arrays is \"good\". So of all the possible arrays, only half of them are \"good\". The lemma is proved.\n\nLet $A$ be the set of such sequences consisting of $k$ steps and resulting in the state where lamps $1$ through $n$ are all on, and lamps $n+1$ through $2n$ are all off.\nLet $B$ be the set of such sequences consisting of $k$ steps, resulting in the state where lamps $1$ through $n$ are all on, and lamps $n+1$ through $2n$ are all off, but where none of the lamps $n+1$ through $2n$ is ever switched on.\n\nFor any $b$ in $B$, match all $a$ in $A$ to $b$ if \"a's elements are the same as $b$'s (mod $n$)\" (For example, take $n=2, k=4$, if $b=(2, 2, 2, 1)$, then it could correspond to $a=(4, 4, 2, 1)$, $a=(2, 2, 2, 1)$, $a=(2, 4, 4, 1)$ etc). Since $b$ in $B$, the number of $1, 2, \\dots, n$ must be odd; $a$ in $A$, the number of $1, 2, \\dots, n$ of $b$ must be odd, and the number of $n+1, \\dots, 2n$ must be even.\n\nFor any $i \\in \\{1, 2, \\dots, n\\}$, if number of 'i's in $b$ is $b_i$, then $a$ only need to satisfy: for the positions taken by $i$ in $b$ the corresponding positions taken by $i$ or $n+i$ in $a$, and the number of 'i's is odd (thus the number of $n+i$ is even). By lemma and the product principle, there are $\\prod_{i=1}^n 2^{b_i-1} = 2^{k-n}$ 'a's corresponding to $b$, but only one of $b$ (letting every position of $a$ be the remainder when divided by $n$) in $B$ corresponds to each $a$ in $A$.\n\nTherefore $|A| = 2^{k-n} |B|$, i.e. $N = 2^{k-n}M$.\n\nObviously $M \\neq 0$ (because the sequence $(1, 2, \\dots, n, n, \\cdots, n) \\in B$), so\n$$\n\\frac{N}{M} = 2^{k-n}.\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 28200,
    "subject": "Mathematics (Multi-modal)",
    "question": "As shown in the figure, $BC$ and $AC$ are tangent to the inscribed circle $I$ of $\\triangle ABC$ at $M$ and $N$. $E$ and $F$ are the midpoints of $AB$, $AC$ respectively. $EF$ intersects $BI$ at $D$. Prove that $M$, $N$, $D$ are collinear.\n\n![](attached_image_1.png)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Join $AD$, and it is obvious that $\\angle ADB = 90^\\circ$. Then join $AI$ and $DM$. Suppose $DM$ intersects $AC$ at $G$. Since $\\angle ABI = \\angle DBM$, we obtain $\\frac{AB}{BD} = \\frac{BI}{BM}$. Hence, $\\triangle ABI \\sim \\triangle DBM$, and\n$$\n\\angle DMB = \\angle AIB = 90^\\circ + \\frac{1}{2} \\angle ACB.\n$$\nJoin $IG$, $IC$, $IM$, then\n$$\n\\angle IMG = \\angle DMB - 90^\\circ = \\frac{1}{2} \\angle ACB = \\angle GCI.\n$$\nTherefore, $I$, $M$, $C$, $G$ are concyclic, and $IG \\perp AC$.\nConsequently, since $G$ and $N$ represent the same point, we conclude that $M$, $N$, $D$ are collinear.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 28201,
    "subject": "Mathematics (Multi-modal)",
    "question": "Suppose that the sides $a$, $b$, $c$ of $\\triangle ABC$, corresponding to the angles $A$, $B$, $C$ respectively, constitute a geometric sequence. Then the range of\n$$\n\\frac{\\sin A \\cot C + \\cos A}{\\sin B \\cot C + \\cos B}\n$$\nis ( ).",
    "options": [],
    "answer": "((sqrt(5)-1)/2, (sqrt(5)+1)/2)",
    "solution": "Suppose that the common ratio of $a$, $b$, $c$ is $q$. Then $b = aq$, $c = aq^2$. We have\n$$\n\\begin{aligned}\n\\frac{\\sin A \\cot C + \\cos A}{\\sin B \\cot C + \\cos B} &= \\frac{\\sin A \\cos C + \\cos A \\sin C}{\\sin B \\cos C + \\cos B \\sin C} \\\\\n&= \\frac{\\sin(A+C)}{\\sin(B+C)} = \\frac{\\sin(\\pi-B)}{\\sin(\\pi-A)} \\\\\n&= \\frac{\\sin B}{\\sin A} = \\frac{b}{a} = q.\n\\end{aligned}\n$$\nSo we only need to determine the range of $q$. As $a$, $b$, $c$ are the sides of a triangle, they satisfy $a+b>c$ and $b+c>a$. That is to say,\n$$\n\\begin{cases} a + aq > aq^2, \\\\ aq + aq^2 > a. \\end{cases}\n$$\nIt follows that\n$$\n\\begin{cases} q^2 - q - 1 < 0, \\\\ q^2 + q - 1 > 0. \\end{cases}\n$$\nTheir solutions are\n$$\n\\begin{cases} \\frac{\\sqrt{5}-1}{2} < q < \\frac{\\sqrt{5}+1}{2}, \\\\ q > \\frac{\\sqrt{5}-1}{2} \\text{ or } q < -\\frac{\\sqrt{5}+1}{2}. \\end{cases}\n$$\nIt is only possible that\n$$\n\\frac{\\sqrt{5}-1}{2} < q < \\frac{\\sqrt{5}+1}{2}\n$$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 28202,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find the maximum positive number $M$ such that for every $n \\in \\mathbb{N}^*$, there are positive numbers $a_1, a_2, \\dots, a_n$ and $b_1, b_2, \\dots, b_n$ satisfying\n$$\n(a) \\sum_{k=1}^{n} b_k = 1,\\ 2b_k \\ge b_{k-1} + b_{k+1},\\ k = 2, 3, \\dots, n-1,\n$$\n$$\n(b) a_k^2 \\le 1 + \\sum_{i=1}^{k} a_i b_i,\\ k = 1, 2, \\dots, n,\n$$\n$$\n(c) a_n = M.\n$$",
    "options": [],
    "answer": "3/2",
    "solution": "Firstly, we prove that\n$$\n\\max_{1 \\le k \\le n} a_k < 2,\\ \\text{ and } \\max_{1 \\le k \\le n} b_k < \\frac{2}{n-1}.\n$$\nLet $L = \\max_{1 \\le k \\le n} a_k$. From (b) and $\\sum_{k=1}^{n} b_k = 1$, we get $L^2 \\le 1 + L$, so $L < 2$.\nLet $b_m = \\max_{1 \\le k \\le n} b_k$. Then by using $2b_k \\ge b_{k-1} + b_{k+1}$, it is easy to see that\n$$\nb_k \\ge \\begin{cases} \\frac{(k-1)b_m + (m-k)b_1}{m-1}, & 1 \\le k \\le m, \\\\ \\frac{(k-m)b_n + (n-k)b_m}{n-m}, & m \\le k \\le n. \\end{cases}\n$$\nSince $b_1 > 0$ and $b_m > 0$, so\n$$\nb_k > \\begin{cases} \\frac{k-1}{m-1}b_m, & 1 \\le k \\le m, \\\\ \\frac{n-k}{n-m}b_m, & m \\le k \\le n. \\end{cases}\n$$\nIt follows that\n$$\n\\begin{aligned}\n1 &= \\sum_{k=1}^{n} b_k = \\sum_{k=1}^{m} b_k + \\sum_{k=m+1}^{n} b_k \\\\\n&> \\frac{1}{m-1} \\left( \\sum_{k=1}^{m} (k-1) \\right) b_m + \\frac{1}{n-m} \\left( \\sum_{k=m+1}^{n} (n-k) \\right) b_m\n\\end{aligned}\n$$\n$$\n= \\frac{m}{2}b_m + \\frac{n-m-1}{2}b_m = \\frac{n-1}{2}b_m.\n$$\nSo $b_m < \\frac{2}{n-1}$, that is $\\max_{1 \\le k \\le n} b_k < \\frac{2}{n-1}$.\n\nNow let $f_0 = 1$, $f_k = 1 + \\sum_{i=1}^k a_i b_i$, $k = 1, 2, \\dots, n$. Then $f_k - f_{k-1} = a_k b_k$, and from (b) we have $a_k^2 \\le f_k$, i.e. $a_k \\le \\sqrt{f_k}$, $k = 1, 2, \\dots, n$.\nSince $\\max_{1 \\le k \\le n} a_k < 2$, so\n$$\nf_k - f_{k-1} = a_k b_k \\le b_k \\sqrt{f_k}\n$$\nand\n$$\nf_k - f_{k-1} < 2b_k.\n$$\nThus, for $1 \\le k \\le n$,\n$$\n\\begin{align*}\n\\sqrt{f_k} - \\sqrt{f_{k-1}} &< b_k \\cdot \\frac{\\sqrt{f_k}}{\\sqrt{f_k} + \\sqrt{f_{k-1}}} \\\\\n&= b_k \\left( \\frac{1}{2} + \\frac{f_k - f_{k-1}}{2(\\sqrt{f_k} + \\sqrt{f_{k-1}})^2} \\right) \\\\\n&< b_k \\left( \\frac{1}{2} + \\frac{2b_k}{2(\\sqrt{f_k} + \\sqrt{f_{k-1}})^2} \\right) \\\\\n&< b_k \\left( \\frac{1}{2} + \\frac{b_k}{4} \\right) \\\\\n&< \\left( \\frac{1}{2} + \\frac{1}{2(n-1)} \\right) b_k.\n\\end{align*}\n$$\nHence, summing from $k=1$ to $n$,\n$$\n\\begin{align*}\na_n \\le \\sqrt{f_n} < \\sqrt{f_0} + \\sum_{k=1}^{n} \\left( \\frac{1}{2} + \\frac{1}{2(n-1)} \\right) b_k \\\\\n= \\frac{3}{2} + \\frac{1}{2(n-1)}.\n\\end{align*}\n$$\nLet $n \\to +\\infty$, we obtain $a_n \\le \\frac{3}{2}$.\n\nWhen $a_k = 1 + \\frac{k}{2n}$, $b_k = \\frac{1}{n}$, $k = 1, 2, \\dots, n$, we have\n$$a_k^2 = \\left(1 + \\frac{k}{2n}\\right)^2 \\le 1 + \\sum_{i=1}^k \\frac{1}{n}\\left(1 + \\frac{i}{2n}\\right)$$\nHence the maximum value is $\\frac{3}{2}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 28203,
    "subject": "Mathematics (Multi-modal)",
    "question": "Given $x$, $y$, $z \\in (0, 1)$ satisfying\n$$\n\\sqrt{\\frac{1-x}{yz}} + \\sqrt{\\frac{1-y}{zx}} + \\sqrt{\\frac{1-z}{xy}} = 2,\n$$\nfind the maximum value of $xyz$. (Posed by Tang Lihua)",
    "options": [],
    "answer": "27/64",
    "solution": "Denote $u = \\sqrt[6]{xyz}$. Then by the given condition and mean inequality,\n$$\n\\begin{aligned}\n2u^3 &= 2\\sqrt{xyz} = \\frac{1}{\\sqrt{3}}\\sum \\sqrt{x(3-3x)} \\\\\n&\\le \\frac{1}{\\sqrt{3}}\\sum \\frac{x+(3-3x)}{2} = \\frac{3\\sqrt{3}}{2} - \\frac{1}{\\sqrt{3}}(x+y+z) \\\\\n&\\le \\frac{3\\sqrt{3}}{2} - \\sqrt{3} \\times \\sqrt[3]{xyz} = \\frac{3\\sqrt{3}}{2} - \\sqrt{3}u^2.\n\\end{aligned}\n$$\nTherefore, $4u^3 + 2\\sqrt{3}u^2 - 3\\sqrt{3} \\le 0$, i.e.\n$$\n(2u - \\sqrt{3})(2u^2 + 2\\sqrt{3}u + 3) \\le 0,\n$$\nand thus $u \\le \\frac{\\sqrt{3}}{2}$. Following this, we have $xyz \\le \\frac{27}{64}$, and equality holds when $x = y = z = \\frac{3}{4}$. Hence, the maximum is $\\frac{27}{64}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 28204,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all triples $(p, q, n)$ such that\n$$\nq^{n+2} \\equiv 3^{n+2} \\pmod{p^n}, \\quad p^{n+2} \\equiv 3^{n+2} \\pmod{q^n}\n$$\nwhere $p$, $q$ are positive odd primes and $n > 1$ is an integer.",
    "options": [],
    "answer": "p = 3, q = 3, and n ≥ 2",
    "solution": "It is easy to check that $(3, 3, n)$ ($n = 2, 3, \\dots$) satisfy both equations. Now let $(p, q, n)$ be another triple satisfying the condition. Then we must have $p \\ne q$, $p \\ne 3$, $q \\ne 3$. We may assume that $q > p \\ge 5$.\n\nIf $n = 2$, then $q^2 \\mid p^4 - 3^4$, or $q^2 \\mid (p^2 - 3^2)(p^2 + 3^2)$. Then either $q^2 \\mid p^2 - 3^2$ or $q^2 \\mid p^2 + 3^2$, since $q$ cannot divide both $p^2 - 3^2$ and $p^2 + 3^2$. On the other hand, $0 < p^2 - 3^2 < q^2$, $\\frac{1}{2}(p^2 + 3^2) < p^2 < q^2$. This leads to a contradiction.\n\nSo $n \\ge 3$. From $p^n \\mid q^{n+2} - 3^{n+2}$, $q^n \\mid p^{n+2} - 3^{n+2}$, we get\n$$\np^n \\mid p^{n+2} + q^{n+2} - 3^{n+2}, \\quad q^n \\mid p^{n+2} + q^{n+2} - 3^{n+2}.\n$$\nSince $p < q$, and $p$, $q$ primes, we have\n$$\np^n q^n \\mid p^{n+2} + q^{n+2} - 3^{n+2}. \\qquad \\textcircled{1}\n$$\nThen $p^n q^n \\le p^{n+2} + q^{n+2} - 3^{n+2} < 2q^{n+2}$. That means $p^n < 2q^2$.\n\nAs $q^n \\mid p^{n+2} - 3^{n+2}$ and $p > 3$, we have $q^n \\le p^{n+2} - 3^{n+2} < p^{n+2}$, and consequently $q < p^{1+\\frac{2}{n}}$. Since $p^n < 2q^2$, we have $p^n < 2p^{2+\\frac{4}{n}} < p^{3+\\frac{4}{n}}$. So $n < 3+\\frac{4}{n}$, and we get $n = 3$. Then $p^3 \\mid q^5 - 3^5$, $q^3 \\mid p^5 - 3^5$.\n\nFrom $5^5 - 3^5 = 2 \\times 11 \\times 131$, we know $p > 5$; from $p^3 \\mid q^5 - 3^5$ we know $p \\mid q^5 - 3^5$. By Fermat's little theorem, we get $p \\mid q^{p-1} - 3^{p-1}$. Then $p \\mid q^{(5, p-1)} - 3^{(5, p-1)}$.\n\nIf $(5, p-1) = 1$, then $p \\mid q-3$. From\n$$\n\\begin{aligned}\n\\frac{q^5 - 3^5}{q - 3} &= q^4 + q^3 \\cdot 3 + q^2 \\cdot 3^2 + q \\cdot 3^3 + 3^4 \\\\\n&\\equiv 5 \\times 3^4 \\pmod{p}\n\\end{aligned}\n$$\nand $p \\ge 5$, we get $p \\nmid \\frac{q^5 - 3^5}{q - 3}$. So $p^3 \\mid q - 3$. From $q^3 \\mid p^5 - 3^5$, we get $q^3 \\le p^5 - 3^5 < p^5 = (p^3)^{\\frac{5}{3}} < q^{\\frac{5}{3}}$. This is a contradiction.\n\nSo we have $(5, p-1) \\ne 1$, and that means $5 \\mid p-1$. In a similar way, we have $5 \\mid q-1$. As $(q, p-3) = 1$ (since $q > p \\ge 7$) and $q^3 \\mid p^5 - 3^5$, we know that $q^3 \\mid \\frac{p^5 - 3^5}{p-3}$. Then\n$$\nq^3 \\le \\frac{p^5 - 3^5}{p - 3} = p^4 + p^3 \\cdot 3 + p^2 \\cdot 3^2 + p \\cdot 3^3 + 3^4.\n$$\nFrom $5 \\mid p-1$ and $5 \\mid q-1$, we get $p \\ge 11$ and $q \\ge 31$. So\n$$\nq^3 \\le p^4 \\left( 1 + \\frac{3}{p} + \\left(\\frac{3}{p}\\right)^2 + \\left(\\frac{3}{p}\\right)^3 + \\left(\\frac{3}{p}\\right)^4 \\right)\n$$\n$$\n< p^4 \\cdot \\frac{1}{1 - \\frac{3}{p}} \\le \\frac{11}{8} p^4.\n$$\nThen we have $p > \\left(\\frac{8}{11}\\right)^{\\frac{1}{4}} q^{\\frac{3}{4}}$. Consequently,\n$$\n\\frac{p^5 + q^5 - 3^5}{p^3 q^3} < \\frac{p^2}{q^3} + \\frac{q^2}{p^3} < \\frac{1}{q} + \\left(\\frac{11}{8}\\right)^{\\frac{3}{4}} \\frac{1}{31^{\\frac{1}{4}}} < 1.\n$$\nBut this contradicts ① which says $p^3 q^3 \\mid p^5 + q^5 - 3^5$.\n\nSo we reach the conclusion that $(3, 3, n)$ ($n = 2, 3, \\dots$) are all the triples that satisfy the conditions.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 28205,
    "subject": "Mathematics (Multi-modal)",
    "question": "For a given positive integer $n$, find the greatest positive integer $k$, such that there exist three sets of $k$ distinct nonnegative integers, $A = \\{x_1, x_2, \\dots, x_k\\}$, $B = \\{y_1, y_2, \\dots, y_k\\}$ and $C = \\{z_1, z_2, \\dots, z_k\\}$ with $x_j + y_j + z_j = n$ for any $1 \\leq j \\leq k$.",
    "options": [],
    "answer": "floor(2n/3) + 1",
    "solution": "By the given condition, we have\n$$\nkn \\geq \\sum_{i=1}^{k} (x_i + y_i + z_i) \\geq 3 \\sum_{i=0}^{k-1} i = \\frac{3k(k-1)}{2},\n$$\nand then $k \\leq \\lfloor \\frac{2n}{3} \\rfloor + 1$.\n\nThe following illustrates the case of $k = \\lfloor \\frac{2n}{3} \\rfloor + 1$:\n\nSet $m \\in \\mathbb{Z}^+$. \n\nWhen $n = 3m$, for $1 \\leq j \\leq m+1$, let $x_j = j-1$, $y_j = m + j - 1$, $z_j = 2m - 2j + 2$; for $m + 2 \\leq j \\leq 2m + 1$, let $x_j = j - 1$, $y_j = j - m - 2$, $z_j = 4m - 2j + 3$, and the result is obvious.\n\nWhen $n = 3m + 1$, for $1 \\leq j \\leq m$, let $x_j = j - 1$, $y_j = m + j$, $z_j = 2m - 2j + 2$; for $m + 1 \\leq j \\leq 2m$, let $x_j = j + 1$, $y_j = j - m - 1$, $z_j = 4m + 1 - 2j$; and $x_{2m+1} = m$, $y_{2m+1} = 2m + 1$, $z_{2m+1} = 0$ will lead to the expected result.\n\nWhen $n = 3m + 2$, for $1 \\leq j \\leq m + 1$, let $x_j = j - 1$, $y_j = m + j$, $z_j = 2m - 2j + 3$; for $m + 2 \\leq j \\leq 2m + 1$, let $x_j = j$, $y_j = j - m - 2$, $z_j = 4m - 2j + 4$; and $x_{2m+2} = 2m + 2$, $y_{2m+2} = m$, $z_{2m+2} = 0$, and the result follows.\n\nIn summary, the maximum value of $k$ is $\\lfloor \\frac{2n}{3} \\rfloor + 1$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 28206,
    "subject": "Mathematics (Multi-modal)",
    "question": "For a chessboard of the size $2008 \\times 2008$, in each case (they all have different colors) write one of the letters $C, G, M, O$. If every $2 \\times 2$ square contains all these four letters, we call it a \"harmonic chessboard.\" How many different harmonic chessboard are there? (Posed by Zuming Feng)",
    "options": [],
    "answer": "12 * 2^2008 - 24",
    "solution": "There are $12 \\times 2^{2008} - 24$ harmonic chessboards. We first prove the following claim:\n\nIn every harmonic chessboard, at least one of the following two situations occurs: (1) each line is composed of just two letters, in an alternative way; (2) each column is composed of just two letters, in an alternative way.\n\nIn fact, suppose that one line is not composed of two letters; then there must be three consecutive squares containing different letters. Without loss of generality, we may assume these three letters to be $C, G, M$, as shown in Fig. 1. We then get easily $X_2 = X_5 = O$, $X_1 = X_4 = M$ and $X_3 = X_6 = C$, as shown in Fig. 2.\n\n![](attached_image_1.png)\nFig. 1\n![](attached_image_2.png)\nFig. 2\n\nThe same argument shows that each of these three columns is composed of two letters in an alternative way, and *a fortiori* so is every column.\n\nNow we calculate the total number of different harmonic chessboards. If the leftmost column is composed of two letters (eg. $C$ and $M$), we see immediately that all the odd-numbered columns are composed of these two letters, while the even-numbered columns are composed of the other two letters. The letter in the top square of each column can be either of the two letters that compose this column; we check easily that it is a harmonic chessboard. Therefore, we have $\\binom{4}{2} = 6$ different ways to choose the two letters of the first column, and $2^{2008}$ ways to determine the letter in the top square of each column. Hence, we get $6 \\times 2^{2008}$ configurations to make each column composed of two letters in an alternative way. We have also $6 \\times 2^{2008}$ configurations to make each line composed of two letters in an alternative way.\n\nThen we need to subtract from the sum the configurations that are counted twice, i.e. the configurations that are alternative on each line and each column. Obviously, any such configuration is in one-to-one correspondence to the $2 \\times 2$ square at the upper-left corner, which gives $4! = 24$ different ways. Hence, we get the above result.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 28207,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $f(x) = ax + b$, with $a, b$ real numbers; $f_1(x) = f(x)$, $f_{n+1}(x) = f(f_n(x))$, $n = 1, 2, \\dots$. If $f_7(x) = 128x + 381$, then $a + b = \\underline{\\hspace{2cm}}$.",
    "options": [],
    "answer": "5",
    "solution": "$$\n\\begin{aligned}\nf_n(x) &= a^n x + (a^{n-1} + a^{n-2} + \\dots + a + 1)b \\\\\n&= a^n x + \\frac{a^n - 1}{a - 1} \\times b.\n\\end{aligned}\n$$\nAs $f_7(x) = 128x + 381$, we have $a^7 = 128$ and $\\frac{a^7 - 1}{a - 1} \\times b = 381$. Then $a = 2$, $b = 3$. The answer is $a + b = 5$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 28208,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $n$ be a positive integer, and $f(n)$ denote the number of $n$-digit integers $\\overline{a_1a_2\\cdots a_n}$ (called wave number) that satisfy the following conditions:\n(i) $a_i \\in \\{1, 2, 3, 4\\}$, and $a_i \\neq a_{i+1}$, $i = 1, 2, \\dots$;\n(ii) When $n \\ge 3$, the numbers $a_i - a_{i+1}$ and $a_{i+1} - a_{i+2}$ have opposite signs, $i = 1, 2, \\dots$.\nFind (1) the value of $f(10)$,\n(2) the remainder of $f(2008)$ divided by 13.",
    "options": [],
    "answer": "f(10) = 8008; f(2008) ≡ 10 (mod 13)",
    "solution": "(1) When $n \\ge 2$, if $a_1 < a_2 < \\overline{a_1a_2\\cdots a_n}$ is classified as A class. The number of $\\overline{a_1a_2\\cdots a_n}$ is denoted by $g(n)$. If $a_1 > a_2$, then $\\overline{a_1a_2\\cdots a_n}$ is classified as B class. By symmetry, the number of such $\\overline{a_1a_2\\cdots a_n}$ is also $g(n)$. Thus, $f(n) = 2g(n)$.\nNow we want to find $g(n)$. Denote $m_k(i)$ as the $k$-digit \"A wave number\" whose last digit is $i$ ($i = 1, 2, 3, 4$), then\n$$\ng(n) = \\sum_{i=1}^{4} m_{n}(i).\n$$\nAs $a_{2k-1} < a_{2k}$, $a_{2k} > a_{2k+1}$, we have the following 2 cases.\n(a) When $k$ is even, $m_{k+1}(4) = 0$, $m_{k+1}(3) = m_k(4)$, $m_{k+1}(2) = m_k(4) + m_k(3)$, $m_{k+1}(1) = m_k(4) + m_k(3) + m_k(2)$.\n(b) When $k$ is odd, $m_{k+1}(1) = 0$, $m_{k+1}(2) = m_k(1)$, $m_{k+1}(3) = m_k(1) + m_k(2)$, $m_{k+1}(4) = m_k(1) + m_k(2) + m_k(3)$.\nIt is obvious that $m_2(1) = 0$, $m_2(2) = 1$, $m_2(3) = 2$, $m_2(4) = 3$, then, $g(2) = 6$.\nHence,\n$$\nm_3(1) = m_2(2) + m_2(3) + m_2(4) = 6,\n$$\n$$\nm_3(2) = m_2(3) + m_2(4) = 5,\n$$\n$$\nm_3(3) = m_2(4) = 3, \\quad m_3(4) = 0.\n$$\nTherefore\n$$\ng(3) = \\sum_{i=1}^{4} m_{3}(i) = 14.\n$$\nOn the other hand, since\n$$\nm_{4}(1) = 0, \\quad m_{4}(2) = m_{3}(1) = 6,\n$$\n$$\nm_{4}(3) = m_{3}(1) + m_{3}(2) = 11,\n$$\n$$\nm_{4}(4) = m_{3}(1) + m_{3}(2) + m_{3}(3) = 14,\n$$\nwe obtain,\n$$\ng(4) = \\sum_{i=1}^{4} m_{4}(i) = 31.\n$$\nIn the same way, we could get $g(5) = 70$, $g(6) = 157$, $g(7) = 353$, $g(8) = 793$.\nThen, in general, when $n \\ge 5$,\n$$\ng(n) = 2g(n-1) + g(n-2) - g(n-3). \\quad \\textcircled{3}\n$$\nNow we prove ③ as follows.\nUsing mathematical induction, we are done when $n=5, 6, 7, 8$. Suppose ③ holds when for 5, 6, 7, 8..., $n$ now consider the case for $n+1$. When $n$ is even, from (a), (b), we have\n$$\nm_{n+1}(4) = 0, \\quad m_{n+1}(3) = m_{n}(4),\n$$\n$$\nm_{n+1}(2) = m_{n}(4) + m_{n}(3),\n$$\n$$\nm_{n+1}(1) = m_{n}(4) + m_{n}(3) + m_{n}(2).\n$$\nAs $m_n(1) = 0$, then\n$$\n\\begin{align*}\ng(n+1) &= \\sum_{i=1}^{4} m_{n+1}(i) \\\\\n&= 2\\left(\\sum_{i=1}^{4} m_{n}(i)\\right) + m_{n}(4) - m_{n}(2)\n\\end{align*}\n$$\n$$\n= 2g(n) + m_n(4) - m_n(2).\n$$\nSince\n$$\n\\begin{aligned}\nm_n(4) &= m_{n-1}(1) + m_{n-1}(2) + m_{n-1}(3) + 0 \\\\\n&= \\sum_{i=1}^{4} m_{n-1}(i) = g(n-1),\n\\end{aligned}\n$$\n$$\n\\begin{aligned}\nm_n(2) &= m_{n-1}(1) = m_{n-2}(4) + m_{n-2}(3) + m_{n-2}(2) + 0 \\\\\n&= g(n-2).\n\\end{aligned}\n$$\nWe obtain,\n$$\ng(n + 1) = 2g(n) + g(n - 1) - g(n - 2).\n$$\nOn the other hand, when $n$ is odd, $g(n+1) = \\sum_{i=1}^{4} m_{n+1}(i)$.\n$$\n\\text{Since } m_{n+1}(1) = 0, \\quad m_{n+1}(2) = m_n(1), \\quad m_n(4) = 0,\n$$\n$$\n\\begin{aligned}\nm_{n+1}(3) &= m_n(1) + m_n(2), \\\\\nm_{n+1}(4) &= m_n(1) + m_n(2) + m_n(3),\n\\end{aligned}\n$$\nthen,\n$$\n\\begin{aligned}\ng(n + 1) &= \\sum_{i=1}^{4} m_{n+1}(i) \\\\\n&= 2 \\sum_{i=1}^{4} m_{n}(i) + m_{n}(1) - m_{n}(3) \\\\\n&= 2g(n) + m_{n}(1) - m_{n}(3).\n\\end{aligned}\n$$\nSince\n$$\nm_n(1) = m_{n-1}(4) + m_{n-1}(3) + m_{n-1}(2) + 0 = g(n-1),\n$$\n$$\n\\begin{aligned}\nm_n(3) &= m_{n-1}(4) = m_{n-2}(1) + m_{n-2}(2) + m_{n-2}(3) + 0 \\\\\n&= g(n-2).\n\\end{aligned}\n$$\nWe get\n$$\ng(n + 1) = 2g(n) + g(n - 1) - g(n - 2).\n$$\nHence, ③ holds for $n+1$. By mathematical induction, ③ holds when $n \\ge 5$.\nFrom ③,\n$$\ng(9) = 2g(8) + g(7) - g(6) = 1782,\n$$\n$$\ng(10) = 2g(9) + g(8) - g(7) = 4004.\n$$\nThus,\n$$\nf(10) = 2g(10) = 8008.\n$$\n\n(2) Now consider the sequence of remainders of $\\{g(n)\\}$ divided by 13. From ③, when $n = 2, 3, 4, \\dots, 14, 15, 16, 17, \\dots$, the corresponding remainders are 6, 1, 5, 5, 1, 2, 0, 1, 0, 1, 1, 3; 6, 1, 5, 5, ...\nTherefore, when $n \\ge 2$, the sequence of remainders is a periodic sequence whose minimum period is 12. As\n$$\n2008 = 12 \\times 167 + 4,\n$$\nwe get\n$$\ng(2008) \\equiv 5 \\pmod{13}.\n$$\nTherefore,\n$$\nf(2008) \\equiv 10 \\pmod{13}.\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 28209,
    "subject": "Mathematics (Multi-modal)",
    "question": "Suppose that the minimum of $f(x) = \\cos 2x - 2a(1 + \\cos x)$ is $-\\frac{1}{2}$. Then $a = \\underline{\\hspace{2cm}}$.",
    "options": [],
    "answer": "a = -2 + √3",
    "solution": "$$\n\\begin{aligned}\nf(x) &= 2\\cos^2 x - 1 - 2a - 2a \\cos x \\\\\n&= 2\\left(\\cos x - \\frac{a}{2}\\right)^2 - \\frac{1}{2}a^2 - 2a - 1.\n\\end{aligned}\n$$\nFor $a > 2$, $f(x)$ takes the minimum value of $1 - 4a$ when $\\cos x = 1$; for $a < -2$, $f(x)$ takes the minimum $1$ when $\\cos x = -1$; for $-2 \\le a \\le 2$, $f(x)$ takes the minimum $-\\frac{1}{2}a^2 - 2a - 1$ when $\\cos x = \\frac{a}{2}$. It is easy to see that $f(x)$ will never be $-\\frac{1}{2}$ for $a > 2$ or $a < -2$. So it is only possible that $-2 \\le a \\le 2$. Then from $-\\frac{1}{2}a^2 - 2a - 1 = -\\frac{1}{2}$, we get $a = -2 + \\sqrt{3}$ or $a = -2 - \\sqrt{3}$ (discarded). Therefore, the correct answer is $a = -2 + \\sqrt{3}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 28210,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $P$ be an interior point of a regular $n$-gon $A_1A_2\\cdots A_n$; the lines $A_iP$ meet the regular $n$-gon $A_1A_2\\cdots A_n$ at another point $B_i$, where $i = 1, 2, \\dots, n$. Prove that\n$$\n\\sum_{i=1}^{n} PA_i \\geq \\sum_{i=1}^{n} PB_i.\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Denote $t = \\lfloor \\frac{n}{2} \\rfloor + 1$, and let $A_{n+j} = A_j$, $j = 1, 2, \\dots, n$.\nNoting that the distance between any vertex of a regular $n$-gon and a point on its side is not greater than its longest diagonal $d$, we therefore have, for any $1 \\le i \\le n$,\n$$\nA_i P + P B_i = A_i B_i \\le d. \\qquad \\textcircled{1}\n$$\nFurthermore, using the fact that the sum of any two sides of a triangle is longer than the third side, we have, for any $1 \\le i \\le n$,\n$$\nA_i P + P A_{i+1} \\ge A_i A_{i+1} = d. \\qquad \\textcircled{2}\n$$\nSumming up ①, ② for $i = 1, 2, \\dots, n$, we have\n$$\n\\sum_{i=1}^{n} (A_i P + P A_{i+1}) \\ge n d \\ge \\sum_{i=1}^{n} (A_i P + P B_i),\n$$\ni. e.\n$$\n2 \\sum_{i=1}^{n} A_i \\ge \\sum_{i=1}^{n} A_i P + \\sum_{i=1}^{n} P B_i,\n$$\nfollowing which the proposition is proven.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 28211,
    "subject": "Mathematics (Multi-modal)",
    "question": "For positive integer $n$, let $f_n = [2^n \\sqrt{2008}] + [2^n \\sqrt{2009}]$. Prove that there are infinitely many odd numbers and even numbers in the sequence $f_1, f_2, \\dots$ ($[x]$ represents the biggest integer that does not exceed $x$.) (Posed by Zuming Feng)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "We use the dyadic representations of $\\sqrt{2008}$ and $\\sqrt{2009}$:\n$$\n\\sqrt{2008} = \\overline{01}1100.\\overline{a_1}a_2\\cdots\\overline{(2)}, \\quad \\sqrt{2009} = \\overline{01}1100.b_1b_2\\cdots\\overline{(2)}.\n$$\nFirst, we prove that there are infinitely many even numbers by contradiction. Suppose that there are only finitely many even numbers in the sequence. Then there exists a positive integer $N$, and for every positive integer $n > N$, $f_n$ must be odd. We consider $n_1 = N+1$, $n_2 = N+2$, .... We observe that in dyadic representation\n$$\nf_{n_i} = \\overline{101100b_1b_2\\cdots b_{n_i}}_{(2)} + \\overline{101100a_1a_2\\cdots a_{n_i}}_{(2)}\n$$\nThis number is equal to $b_{n_i} + a_{n_i}$ modulo 2. As $f_{n_i}$ is odd, we have $\\{b_{n_i}, a_{n_i}\\} = \\{0, 1\\}$. Hence,\n$$\n\\sqrt{2008} + \\sqrt{2009} = \\overline{1011001.c_1c_2\\cdots c_{N-1}111\\cdots}_{(2)}\n$$\nTherefore, $\\sqrt{2008} + \\sqrt{2009}$ must be rational, which is impossible, as we know that it is irrational. Hence, our hypothesis must be wrong, which proves the existence of infinitely many even numbers in the sequence.\n\nIn a similar way we can prove the existence of infinitely many odd numbers in the sequence. Let\n$$\ng_n = [2^n \\sqrt{2009}] - [2^n \\sqrt{2008}],\n$$\napparently $g_n$ and $f_n$ have the same parity. Hence, for $n > N$, $g_n$ is even. We observe also that in dyadic representation\n$$\ng_{n_i} = \\overline{101100b_1b_2\\cdots b_{n_i}}_{(2)} - \\overline{101100a_1a_2\\cdots a_{n_i}}_{(2)}\n$$\nThis number is equal to $b_{n_i} - a_{n_i}$ modulo 2. As $g_{n_i}$ is odd, we have $b_{n_i} = a_{n_i}$. Thus,\n$$\n\\sqrt{2009} - \\sqrt{2008} = \\overline{0.d_1d_2\\cdots d_{N-1}000\\cdots}_{(2)}\n$$\nand this would imply the rationality of $\\sqrt{2009} - \\sqrt{2008}$, which is impossible. Hence, there are infinitely many odd numbers in the sequence.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 28212,
    "subject": "Mathematics (Multi-modal)",
    "question": "Twenty-four volunteers will be allocated to three schools. The rule is that each school will accept at least one volunteer and all the schools will accept different numbers of volunteers. Then there are ______ different ways of allocating volunteers.",
    "options": [],
    "answer": "222",
    "solution": "We may use each space between every two consecutive bars ($|$) to represent a school and each asterisk ($*$) to represent a volunteer, as seen in the following example; the first, second and third schools receive $4$, $18$ and $2$ volunteers, respectively.\n$$\n| * * * * | * \\cdots * | * * |\n$$\nThen the allocation problem may be regarded as a permutation-and-combination problem of $4$ bars and $24$ asterisks.\n\nSince the two ends of the line must be occupied by a bar, respectively, there are $\\binom{23}{2} = 253$ ways to insert the other $2$ bars into the $23$ spaces between the $24$ asterisks such that there is at least $1$ asterisk between every two consecutive bars, in which there are $31$ ways that at least two schools have the same number of volunteers. So the number of allocating ways satisfying the conditions is $253 - 31 = 222$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 28213,
    "subject": "Mathematics (Multi-modal)",
    "question": "The sequence $\\{x_n\\}$ is defined by $x_1 = 2$, $x_2 = 12$, $x_{n+2} = 6x_{n+1} - x_n$, $n = 1, 2, \\dots$. Let $p$ be an odd prime number. Let $q$ be a prime number such that $q \\mid x_p$. Prove that if $q \\neq 2$, then $q \\ge 2p - 1$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "It is easy to see\n$$\nx_n = \\frac{1}{2\\sqrt{2}}\\left((3+2\\sqrt{2})^n - (3-2\\sqrt{2})^n\\right), \\quad n = 1, 2, \\dots\n$$\nLet $a_n, b_n$ be positive integers and $a_n + b_n\\sqrt{2} = (3+2\\sqrt{2})^n$. Then\n$$\na_n - b_n\\sqrt{2} = (3 - 2\\sqrt{2})^n,\n$$\nso $x_n = b_n$, and $a_n^2 - 2b_n^2 = 1$, $n = 1, 2, \\dots$\n\nSuppose $q \\neq 2$. Since $q \\mid x_p$, thus $q \\mid b_p$, so there exists a term in $\\{b_n\\}$ which is divisible by $q$. Let $d$ be the least number such that $q \\mid b_d$. We have the following lemma.\n\n**Lemma** For any positive integer $n$, $q \\mid b_n$ if and only if $d \\mid n$.\n\n**Proof:** For $a, b, c, d \\in \\mathbb{Z}$, denote $a+b\\sqrt{2} \\equiv c+d\\sqrt{2} \\pmod{q}$ as $a \\equiv c \\pmod{q}$ and $b \\equiv d \\pmod{q}$.\n\nIf $d \\mid n$, write $n = du$, then\n$$\na_n + b_n \\sqrt{2} = (3 + 2\\sqrt{2})^{du} \\equiv a_d^u \\pmod{q},\n$$\nso $b_n \\equiv 0 \\pmod{q}$.\n\nOn the other hand, if $q \\mid b_n$, write $n = du + r$, $0 \\le r < d$. Suppose $r \\ge 1$, from\n$$\n\\begin{aligned}\na_n &= (3 + 2\\sqrt{2})^n = (3 + 2\\sqrt{2})^{du} \\cdot (3 + 2\\sqrt{2})^r \\\\\n&\\equiv a_d^u (a_r + b_r\\sqrt{2}) \\pmod{q},\n\\end{aligned}\n$$\nwe have\n$$\na_d^u b_r \\equiv 0 \\pmod{q}. \\qquad \\textcircled{1}\n$$\nBut $a_d^2 - 2b_d^2 = 1$, and $q \\mid b_d$; so $q \\nmid a_d^2$. Since $q$ is a prime, therefore $q \\nmid a_d$, and $(q, a_d^u) = 1$. From (1) we have $q \\mid b_r$, it contradicts the definition of $d$. So $r = 0$, and the lemma is proven.\n\nNow, as $q$ is a prime, so $q \\mid \\binom{q}{i}$, $i = 1, 2, \\dots, q-1$.\n\nUsing Fermat's little theorem, we have\n$$\n3^q \\equiv 3 \\pmod{q}, \\quad 2^q \\equiv 2 \\pmod{q}.\n$$\nAs $q \\neq 2$, so $2^{\\frac{q-1}{2}} \\equiv \\pm 1 \\pmod{q}$, we get\n$$\n\\begin{aligned}\n(3+2\\sqrt{2})^q &= \\sum_{i=0}^{q} \\binom{q}{i} \\cdot 3^{q-i} (2\\sqrt{2})^i \\\\\n&\\equiv 3^q + (2\\sqrt{2})^q \\\\\n&= 3^q + 2^q \\cdot 2^{\\frac{q-1}{2}} \\sqrt{2} \\\\\n&\\equiv 3 \\pm 2\\sqrt{2} \\pmod{q}.\n\\end{aligned}\n$$\nBy the same argument, we have\n$$\n(3+2\\sqrt{2})^{q^2} \\equiv (3 \\pm 2\\sqrt{2})^q \\equiv 3+2\\sqrt{2} \\pmod{q}.\n$$\nSo\n$$\n(a_{q^2-1} + \\sqrt{2}b_{q^2-1})(3 + 2\\sqrt{2}) \\equiv 3 + 2\\sqrt{2} \\pmod{q}.\n$$\nThus,\n$$\n\\begin{cases}\n3a_{q^2-1} + 4b_{q^2-1} \\equiv 3 \\pmod{q}, \\\\\n2a_{q^2-1} + 3b_{q^2-1} \\equiv 2 \\pmod{q}.\n\\end{cases}\n$$\nWe know that $q \\mid b_{q^2-1}$.\n\nSince $q \\mid b_p$, from the lemma, we have $d \\mid p$. So $d \\in \\{1, p\\}$, and if $d = 1$, then $q \\mid b_1 = 2$, contradiction! So $d = p$, hence $q \\mid b_{q^2-1}$. So $p \\mid q^2-1$, thus $p \\mid q-1$ or $p \\mid q+1$. Since $q-1$ and $q+1$ are even, so $q \\ge 2p-1$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 28214,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find the maximum and minimum of the function\n$$\ny = \\sqrt{x+27} + \\sqrt{13-x} + \\sqrt{x}.\n$$",
    "options": [],
    "answer": "minimum = 3*sqrt(3) + sqrt(13), maximum = 11",
    "solution": "The domain of $y$ is $x \\in [0, 13]$. We have\n$$\n\\begin{aligned}\ny &= \\sqrt{x+27} + \\sqrt{13-x} + \\sqrt{x} \\\\\n&= \\sqrt{x+27} + \\sqrt{13 + 2\\sqrt{x(13-x)}} \\\\\n&\\geq \\sqrt{27} + \\sqrt{13} = 3\\sqrt{3} + \\sqrt{13}.\n\\end{aligned}\n$$\nThe equality holds when $x = 0$. Therefore, the minimum of $y$ is $3\\sqrt{3} + \\sqrt{13}$.\n\nOn the other hand, by the Cauchy inequality we have\n$$\n\\begin{aligned}\ny^2 &= (\\sqrt{x} + \\sqrt{x+27} + \\sqrt{13-x})^2 \\\\\n&\\leq \\left(\\frac{1}{2} + 1 + \\frac{1}{3}\\right) \\left[2x + (x+27) + 3(13-x)\\right] \\\\\n&= 121.\n\\end{aligned}\n$$\nThe equality holds when $4x = 9(13-x) = x + 27$. It is so for $x = 9$. Therefore, the maximum of $y$ is $11$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 28215,
    "subject": "Mathematics (Multi-modal)",
    "question": "The set of the permutation $X = (x_1, x_2, \\dots, x_9)$ of $1, 2, \\dots, 9$ is $A$. $\\forall X \\in A$, and let $f(X) = x_1 + 2x_2 + 3x_3 + \\dots + 9x_9$, $M = \\{f(X) \\mid X \\in A\\}$. Find the value of $|M|$. (Posed by Xiong Bin)",
    "options": [],
    "answer": "121",
    "solution": "We prove for $n \\ge 4$. If the permutations $X_n = (x_1, x_2, \\dots, x_n)$ of $1, 2, \\dots, n$ consist of a set $A$, and $f(X_n) = x_1 + 2x_2 + 3x_3 + \\dots + nx_n$, $M_n = \\{f(X) \\mid X \\in A\\}$, then $|M_n| = \\frac{n^3 - n + 6}{6}$.\n\nUsing mathematical induction on $n$, we see that\n$$\nM_n = \\left\\{ \\frac{n(n+1)(n+2)}{6}, \\frac{n(n+1)(n+2)}{6} + 1, \\dots, \\frac{n(n+1)(2n+1)}{6} \\right\\}\n$$\nFor $n=4$, by arranging inequality, one can see the smallest number of $M$ is $f(\\{4, 3, 2, 1\\}) = 20$, and the biggest number is $f(\\{1, 2, 3, 4\\}) = 30$. Because\n$$\n\\begin{align*}\nf(\\{3, 4, 2, 1\\}) &= 21, \\quad f(\\{3, 4, 1, 2\\}) = 22, \\\\\nf(\\{4, 2, 1, 3\\}) &= 23, \\quad f(\\{3, 2, 4, 1\\}) = 24, \\\\\nf(\\{2, 4, 1, 3\\}) &= 25, \\quad f(\\{1, 4, 3, 2\\}) = 26, \\\\\nf(\\{1, 4, 2, 3\\}) &= 27, \\quad f(\\{2, 1, 4, 3\\}) = 28, \\\\\nf(\\{1, 2, 4, 3\\}) &= 29,\n\\end{align*}\n$$\nwe have, $|M_4| = |\\{20, 21, \\dots, 30\\}| = 11 = \\frac{4^3 - 4 + 6}{6}$.\n\nSuppose that the statement is true for $n-1$ ($n \\ge 5$). In the case of $n$, for a permutation $X_{n-1} = (x_1, x_2, \\dots, x_{n-1})$ of $1, 2, \\dots, n-1$, let $x_n = n$; then we get a permutation $(x_1, x_2, \\dots, x_{n-1}, n)$ of $1, 2, \\dots, n$, and so\n$$\n\\sum_{k=1}^{n} kx_k = n^2 + \\sum_{k=1}^{n-1} kx_k.\n$$\nAccording to the supposition, the value of $\\sum_{k=1}^{n} kx_k$ can be every integer number in the interval\n$$\n\\left[ n^2 + \\frac{(n-1)n(n+1)}{6}, n^2 + \\frac{(n-1)n(2n-1)}{6} \\right] = \\left[ \\frac{n(n^2+5)}{6}, \\frac{n(n+1)(2n+1)}{6} \\right].\n$$\nLet $x_n = 1$. Then\n$$\n\\begin{aligned}\n\\sum_{k=1}^{n} kx_k &= n + \\sum_{k=1}^{n-1} kx_k \\\\\n&= n + \\sum_{k=1}^{n-1} k(x_k - 1) + \\frac{n(n-1)}{2} \\\\\n&= \\frac{n(n+1)}{2} + \\sum_{k=1}^{n-1} k(x_k - 1).\n\\end{aligned}\n$$\nAccording to the supposition, the value of $\\sum_{k=1}^{n} kx_k$ can be every integer number in the interval\n$$\n\\left[ \\frac{n(n+1)}{2} + \\frac{(n-1)n(n+1)}{6}, \\frac{n(n+1)}{2} + \\frac{n(n-1)(2n-1)}{6} \\right] = \\left[ \\frac{n(n+1)(n+2)}{6}, \\frac{2n(n^2+2)}{6} \\right].\n$$\nBecause $\\frac{2n(n^2+2)}{6} \\ge \\frac{n(n^2+5)}{6}$, according to the supposition the value of $\\sum_{k=1}^{n} kx_k$ can be every integer number in the interval\n$$\n\\left[ \\frac{n(n+1)(n+2)}{6} , \\frac{n(n+1)(2n+1)}{6} \\right].\n$$\nThe statement is also true for $n$. Since\n$$\n\\frac{n(n+1)(2n+1)}{6} - \\frac{n(n+1)(n+2)}{6} = \\frac{n^3 - n + 6}{6},\n$$\none can see that $|M_n| = \\frac{n^3 - n + 6}{6} + 1$.\n\nIn particular, $|M_9| = 121$.",
    "topic": "Discrete Mathematics",
    "subtopic": "Combinatorics"
  },
  {
    "id": 28216,
    "subject": "Mathematics (Multi-modal)",
    "question": "Suppose that $f(x) = \\frac{x}{\\sqrt{1+x^2}}$ and $f^{(n)}(x) = f[f[f\\dots f(x)]]$. Then $f^{(99)}(1) = \\underline{\\hspace{2cm}}$.",
    "options": [],
    "answer": "1/10",
    "solution": "We have\n$$\n\\begin{aligned}\nf^{(1)}(x) &= f(x) = \\frac{x}{\\sqrt{1+x^2}}, \\\\\nf^{(2)}(x) &= f[f(x)] = \\frac{x}{\\sqrt{1+2x^2}}, \\\\\n&\\vdots \\\\\nf^{(99)}(x) &= \\frac{x}{\\sqrt{1+99x^2}}.\n\\end{aligned}\n$$\nTherefore, $f^{(99)}(1) = \\frac{1}{10}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 28217,
    "subject": "Mathematics (Multi-modal)",
    "question": "Given an acute triangle $PBC$, $PB \\neq PC$. Let points $A$, $D$ be on sides $PB$ and $PC$, respectively. Let $M$, $N$ be the midpoints of segments $BC$ and $AD$, respectively. Lines $AC$ and $BD$ intersect at point $O$. Draw $OE \\perp AB$ at point $E$ and $OF \\perp CD$ at point $F$.\n\n(1) Prove that if $A$, $B$, $C$, $D$ are concyclic, then\n$$\nEM \\times FN = EN \\times FM.\n$$\n\n(2) Are the four points $A$, $B$, $C$, $D$ always concyclic if $EM \\times FN = EN \\times FM$? Prove your answer.",
    "options": [],
    "answer": "No",
    "solution": "(1) Denote by $Q$, $R$ the midpoints of $OB$, $OC$, respectively. It is easy to see that\n![](attached_image_1.png)\n$$EQ = \\frac{1}{2}OB = RM, \\quad MQ = \\frac{1}{2}OC = RF,$$ and\n$$\\angle EQM = \\angle EQO + \\angle OQM = 2\\angle EBO + \\angle OQM,$$\n$$\\angle MRF = \\angle FRO + \\angle ORM = 2\\angle FCO + \\angle ORM.$$\nBecause $A$, $B$, $C$, $D$ are concyclic, and $Q$, $R$ are the midpoints of $OB$, $OC$, we have\n$$\n\\angle EBO = \\angle FCO, \\quad \\angle OQM = \\angle ORM.\n$$\nSo $\\angle EQM = \\angle MRF$, which implies that $\\triangle EQM \\cong \\triangle MRF$, and $EN = FN$.\nSimilarly, we have $EN = FN$, so $EM \\times FN = EN \\times FM$ holds.\n\n\n(2) Suppose that $OA = 2a$, $OB = 2b$, $OC = 2c$, $OD = 2d$ and\n$$\n\\angle OAB = \\alpha, \\quad \\angle OBA = \\beta, \\quad \\angle ODC = \\gamma, \\quad \\angle OCD = \\theta.\n$$\nThen\n$$\n\\begin{aligned}\n\\cos \\angle EQM &= \\cos(\\angle EQO + \\angle OQM) \\\\\n&= \\cos(2\\beta + \\angle AOB) \\\\\n&= -\\cos(\\alpha - \\beta).\n\\end{aligned}\n$$\nSo\n$$\nEM^2 = EQ^2 + QM^2 - 2EQ \\times QM \\times \\cos\\angle EQM \\\\\n= b^2 + c^2 + 2bc \\cos(\\alpha - \\beta).\n$$\nMaking similar equations for $EN$, $FN$, $FM$, we have\n$$\n\\begin{align*}\n& EN \\times FM = EM \\times FN \\\\\n\\Leftrightarrow & EN^2 \\times FM^2 = EM^2 \\times FN^2 \\\\\n\\Leftrightarrow & (a^2 + d^2 + 2ad \\cos(\\alpha - \\beta)) \\times (b^2 + c^2 + 2bc \\cos(\\gamma - \\theta)) \\\\\n& \\qquad = (a^2 + d^2 + 2ad \\cos(\\gamma - \\theta)) \\times (b^2 + c^2 + 2bc \\cos(\\alpha - \\beta)) \\\\\n\\Leftrightarrow & (\\cos(\\gamma - \\theta) - \\cos(\\alpha - \\beta))(ab - cd)(ac - bd) = 0.\n\\end{align*}\n$$\n![](attached_image_2.png)\nBecause $\\alpha + \\beta = \\gamma + \\theta$, $\\cos(\\gamma - \\theta) - \\cos(\\alpha - \\beta) = 0$ holds if and only if $\\alpha = \\gamma$, $\\beta = \\theta$ (i.e. $A$, $B$, $C$, $D$ are concyclic) or $\\alpha = \\theta$, $\\beta = \\gamma$ (which follows $AB \\parallel CD$, a contradiction); $ab - cd = 0$ holds if and only if $AD \\parallel BC$; $ac - bd = 0$ holds if and only if $A$, $B$, $C$, $D$ are concyclic.\nSo, when $AD \\parallel BC$ holds, we also have\n$$\nEM \\times FN = EN \\times FM.\n$$\nWe know that $A$, $B$, $C$, $D$ are not concyclic in this case because $PB \\neq PC$, so the answer is \"false\".",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 28218,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $D$ be a point on side $BC$ of triangle $ABC$ such that $\\angle CAD = \\angle CBA$. A circle with center $O$ passes through $B$, $D$, and meets segments $AB$, $AD$ at $E$, $F$, respectively. Lines $BF$ and $DE$ meet at point $G$. $M$ is the midpoint of $AG$. Prove that $CM \\perp AO$. (Posed by Xiong Bin)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "As shown in Fig. 1, extend $EF$ and meet $BC$ at point $P$, and join and extend $GP$, which meets $AD$ at $K$ and the extension of $AC$ at $L$.\n\n![](attached_image_1.png)\nFig. 1\n\nAs shown in Fig. 2, let $Q$ be a point on $AP$ such that\n$$\n\\angle PQF = \\angle AEF = \\angle ADB.\n$$\nIt is easy to see that $A, E, F, Q$ and $F, D, P, Q$ are concyclic respectively. Denote by $r$ the radius of $\\odot O$. By the power of a point theorem,\n$$\n\\begin{aligned}\nAP^2 &= AQ \\times AP + PQ \\times AP \\\\\n&= AF \\times AD + PF \\times PE \\\\\n&= (AO^2 - r^2) + (PO^2 - r^2).\n\\end{aligned}\n$$\n①\n\nSimilarly,\n$$\nAG^2 = (AO^2 - r^2) + (GO^2 - r^2).\n$$\n②\n\nBy ①, ②, we have $AP^2 - AG^2 = PO^2 - GO^2$, which implies that $PG \\perp AO$.\n\nAs shown in Fig. 3, applying Menelaus' theorem for $\\triangle PFD$ and line $AEB$, we have\n\n![](attached_image_2.png)\nFig. 2\n\n$$\n\\frac{DA}{AF} \\times \\frac{FE}{EP} \\times \\frac{PB}{BD} = 1. \\quad \\textcircled{3}\n$$\n\nApplying Ceva's theorem for $\\triangle PFD$ and point $G$, we have\n$$\n\\frac{DK}{KF} \\times \\frac{FE}{EP} \\times \\frac{PB}{BD} = 1. \\quad \\textcircled{4}\n$$\n\n③ $\\div$ ④ yields\n$$\n\\frac{DA}{AF} = \\frac{DK}{KF}. \\quad \\textcircled{5}\n$$\n\n![](attached_image_3.png)\nFig. 3\n\nEquation ⑤ illustrates that $A, K$ are harmonic points with respect to $F, D$, i.e. $AF \\times KD = AD \\times FK$.\n\nIt follows that\n$$\nAK \\times FD = AF \\times KD + AD \\times FK = 2AF \\times KD. \\quad \\textcircled{6}\n$$\n\nSince $B, D, F, E$ are concyclic, we have $\\angle DBA = \\angle EFA$. Since $\\angle CAD = \\angle CBA$, we have $\\angle CAF = \\angle EFA$, which implies that $AC \\parallel EP$. Thus,\n$$\n\\frac{CP}{PD} = \\frac{AF}{FD}. \\quad \\textcircled{7}\n$$\n\nApplying Menelaus' theorem to $\\triangle ACD$ and line $LPK$, we have\n$$\n\\frac{AL}{LC} \\times \\frac{CP}{PD} \\times \\frac{DK}{KA} = 1. \\quad \\textcircled{8}\n$$\n\nCombining ⑥, ⑦ and ⑧, we have $\\frac{AL}{LC} = 2$.\n\nIn $\\triangle AGL$, $M, C$ are the midpoints of $AG, AL$ respectively, and hence $MC \\parallel GL$. Since $GL \\perp AO$, we conclude that $MC \\perp AO$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 28219,
    "subject": "Mathematics (Multi-modal)",
    "question": "As shown in the Fig. 1, $M$, $N$ are the midpoints of arcs $\\widehat{BC}$, $\\widehat{AC}$ respectively, which are on the circumscribed circle of an acute triangle $\\triangle ABC$ ($\\angle A < \\angle B$). Through point $C$ draw $PC \\parallel MN$, intercepting the circle $\\Gamma$ at point $P$. $I$ is the inner center of $\\triangle ABC$. Extend line $PI$ to intercept $\\Gamma$ at point $T$.\n![](attached_image_1.png)\nFig. 1\n\n(1) Prove that $MP \\times MT = NP \\times NT$;\n\n(2) For an arbitrary point $Q (\\neq A, T, B)$ on arc $\\widehat{AB}$ (not containing $C$), denote the inner centers of $\\triangle AQC$, $\\triangle QCB$ by $I_1, I_2$, respectively. Prove that $Q, I_1, I_2, T$ are concyclic.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "(1) As shown in Fig. 2, join $NI$, $MI$. Since $PC \\parallel MN$ and $P, C, M, N$ are concyclic, $PCMN$ is an isosceles trapezoid. Therefore, $NP = MC$, $PM = NC$.\nJoin $AM$, $CI$. Then $AM$ intercepts $CI$ at $I$. We have\n$$\n\\begin{align*}\n\\angle MIC &= \\angle MAC + \\angle ACI \\\\\n&= \\angle MCB + \\angle BCI \\\\\n&= \\angle MCI.\n\\end{align*}\n$$\n![](attached_image_2.png)\nFig. 2\n\nTherefore, $MC = MI$. In the same way, $NC = NI$. Then $NP = MI$, $PM = NI$.\nThis means that $MPNI$ is a parallelogram. Therefore, $S_{\\triangle PMT} = S_{\\triangle PNT}$, for the two triangles having the same base and height.\nOn the other hand, $\\angle TNP + \\angle PMT = 180^{\\circ}$, as $P, N, T, M$ are concyclic. Then we have\n$$\n\\begin{aligned}\nS_{\\triangle PMT} &= \\frac{1}{2} PM \\times MT \\sin \\angle PMT \\\\\n&= S_{\\triangle PNT} = \\frac{1}{2} PN \\times NT \\sin \\angle PNT \\\\\n&= \\frac{1}{2} PN \\times NT \\sin \\angle PMT.\n\\end{aligned}\n$$\nTherefore, $MP \\times MT = NP \\times NT$.\n\n\n(2) As shown in Fig. 3, we have\n$$\n\\angle NCI_1 = \\angle NCA + \\angle ACI_1 = \\angle NQC + \\angle QCI_1 = \\angle CI_1N.\n$$\nTherefore, $NC = NI_1$. In the same way, $MC = MI_2$.\nFrom $MP \\times MT = NP \\times NT$ we get\n$$\n\\frac{NT}{MP} = \\frac{MT}{NP}.\n$$\nFrom (1) we know that $MP = NC$, $NP = MC$. Then\n![](attached_image_3.png)\nFig. 3\n$$\n\\frac{NT}{NI_1} = \\frac{MT}{MI_2}.\n$$\nFurthermore,\n$$\n\\angle I_1 NT = \\angle QNT = \\angle QMT = \\angle I_2 MT.\n$$\nTherefore, $\\triangle I_1 NT \\sim \\triangle I_2 MT$. Consequently, $\\angle NTI_1 = \\angle MTI_2$. Then we have\n$$\n\\angle I_1 QI_2 = \\angle NQM = \\angle NTM = \\angle I_1 TI_2.\n$$\nThis means that $Q, I_1, I_2, T$ are concyclic.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 28220,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find the integer solutions of the function $x^2 - 2xy + 126y^2 = 2009$.",
    "options": [],
    "answer": "(1, 4), (7, 4), (-1, -4), (-7, -4)",
    "solution": "Suppose that the integers $x, y$ satisfy\n$$\nx^2 - 2xy + 126y^2 - 2009 = 0.\n$$\nLooking at this as a quadratic function of $x$,\n$$\n\\Delta = 4y^2 - 4 \\times (126y^2 - 2009) = 500(4^2 - y^2) + 36\n$$\nshould be a square number.\nIf $y^2 > 4^2$, then $\\Delta < 0$. So $y^2 < 4^2$, when $y^2 \\in \\{0, 1^2, 2^2, 3^2\\}$, $\\Delta \\in \\{8036, 7536, 6036, 3536\\}$ is not a square number.\nFor $y^2 = 4^2$, $\\Delta = 500(4^2 - y^2) + 36 = 6^2$. According to $y = 4$, one can get $x = 1$ or $7$; according to $y = -4$, one can get $x = -1$ or $-7$.\nAll the integer solutions are $(x, y) = (1, 4), (7, 4), (-1, -4), (-7, -4)$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 28221,
    "subject": "Mathematics (Multi-modal)",
    "question": "Show that there are only finitely many triples $(a, b, c)$ of positive integers satisfying the equation $abc = 2009(a + b + c)$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "There are at most six permutations for any three numbers $x, y, z$. It suffices to show that there are only finitely many triples $(a, b, c)$, with $a \\ge b \\ge c$, of positive integers satisfying the equation $abc = 2009(a + b + c)$. It follows that $abc \\le 2009 \\times (3a)$ or $bc \\le 2009 \\times 3 = 6027$. Clearly, there are finitely many pairs $(b, c)$ of positive integers satisfying the equation $bc \\le 6027$ and for each fixed pair of integers $(b, c)$ there is at most one positive integer $a$ satisfying the equation $abc = 2009(a + b + c)$ (because it is a linear equation in $a$).",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 28222,
    "subject": "Mathematics (Multi-modal)",
    "question": "It is known that $p, q$ ($q \\neq 0$) are real numbers; the equation $x^2 - px + q = 0$ has two real roots $\\alpha, \\beta$; the sequence $\\{a_n\\}$ satisfies $a_1 = p$, $a_2 = p^2 - q$, $a_n = p a_{n-1} - q a_{n-2}$ ($n = 3, 4, \\dots$).\n\na. Find the general expression of $\\{a_n\\}$ in terms of $\\alpha, \\beta$.\n\nb. If $p = 1$, $q = -\\frac{1}{4}$, find the sum of the first $n$ terms of $\\{a_n\\}$.",
    "options": [],
    "answer": "a) If the roots are distinct, a_n = (β^{n+1} − α^{n+1})/(β − α). If the roots are equal, a_n = (n + 1) α^n. b) For p = 1 and q = 1/4, the sum of the first n terms is S_n = 3 − (n + 3)/2^n.",
    "solution": "(1) By Vieta's theorem, we have $\\alpha \\times \\beta = q \\neq 0$, $\\alpha + \\beta = p$. Then\n$$\n\\begin{aligned}\na_n &= p a_{n-1} - q a_{n-2} \\\\\n&= (\\alpha + \\beta)a_{n-1} - \\alpha\\beta a_{n-2} \\quad (n = 3, 4, \\dots).\n\\end{aligned}\n$$\nThis can be rewritten as\n$$\na_n - \\beta a_{n-1} = \\alpha (a_{n-1} - \\beta a_{n-2}).\n$$\nLet $b_n = a_{n+1} - \\beta a_n$. Then $b_{n+1} = \\alpha b_n$ ($n = 1, 2, \\dots$). This means that $\\{b_n\\}$ is a geometric sequence with the common ratio $\\alpha$. The first term of $\\{b_n\\}$ is\n$$\nb_1 = a_2 - \\beta a_1 = p^2 - q - \\beta p = (\\alpha + \\beta)^2 - \\alpha\\beta - \\beta(\\alpha + \\beta) = \\alpha^2.\n$$\nTherefore, $b_n = \\alpha^2 \\times \\alpha^{n-1} = \\alpha^{n+1}$. Then $a_{n+1} - \\beta a_n = \\alpha^{n+1}$. By rewriting\n$$\na_{n+1} = \\alpha^{n+1} + \\beta a_n \\quad (n = 1, 2, \\dots). \\quad \\textcircled{1}\n$$\nWhen $\\Delta = p^2 - 4q = 0$, we have $\\alpha = \\beta \\neq 0$, $a_1 = p - 2\\alpha$. The expression $\\textcircled{1}$ becomes $a_{n+1} = \\alpha^{n+1} + \\alpha a_n$, i.e. $\\frac{a_{n+1}}{\\alpha^{n+1}} - \\frac{a_n}{\\alpha^n} = 1$.\nThen $\\left\\{\\frac{a_n}{\\alpha^n}\\right\\}$ is an arithmetic sequence with the common difference $1$, whose first term is $\\frac{a_1}{\\alpha} = \\frac{2\\alpha}{\\alpha} = 2$. Therefore,\n$$\n\\frac{a_n}{\\alpha^n} = 2 + 1 \\times (n - 1) = n + 1.\n$$\nAs a result, the general expression of $\\{a_n\\}$ is\n$$\na_n = (n + 1)\\alpha^n. \\quad \\textcircled{2}\n$$\nWhen $\\Delta > 0$, $\\alpha \\neq \\beta$, we have\n$$\n\\begin{aligned}\na_{n+1} &= \\alpha^{n+1} + \\beta a_n \\\\\n&= \\beta a_n + \\frac{\\beta}{\\beta - \\alpha} \\alpha^{n+1} - \\frac{\\alpha}{\\beta - \\alpha} \\alpha^{n+1} \\quad (n = 1, 2, \\dots).\n\\end{aligned}\n$$\nBy rewriting,\n$$\na_{n+1} + \\frac{\\alpha^{n+2}}{\\beta - \\alpha} = \\beta \\left(a_n + \\frac{\\alpha^{n+1}}{\\beta - \\alpha}\\right) \\quad (n = 1, 2, \\dots).\n$$\nThen $\\left\\{a_n + \\frac{\\alpha^{n+1}}{\\beta - \\alpha}\\right\\}$ becomes a geometric sequence with the common ratio $\\beta$, whose first term is\n$$\na_1 + \\frac{\\alpha^2}{\\beta - \\alpha} = \\alpha + \\beta + \\frac{\\alpha^2}{\\beta - \\alpha} = \\frac{\\beta^2}{\\beta - \\alpha}.\n$$\nTherefore,\n$$\na_n + \\frac{\\alpha^{n+1}}{\\beta - \\alpha} = \\frac{\\beta^2}{\\beta - \\alpha} \\beta^{n-1}.\n$$\nThen the general expression of $\\{a_n\\}$ is\n$$\na_n = \\frac{\\beta^{n+1} - \\alpha^{n+1}}{\\beta - \\alpha} \\quad (n = 1, 2, \\dots). \\qquad \\textcircled{3}\n$$\n\n(2) Given $p = 1$, $q = \\frac{1}{4}$, we have $\\Delta = p^2 - 4q = 0$. Then $\\alpha = \\beta = \\frac{1}{2}$. By the expression (2), the general expression of $\\{a_n\\}$ is\n$$\na_n = (n+1)\\left(\\frac{1}{2}\\right)^n = \\frac{n+1}{2^n} \\quad (n=1, 2, \\dots).\n$$\nTherefore, the sum of the first $n$ terms of $\\{a_n\\}$ is\n$$\nS_n = \\frac{2}{2} + \\frac{3}{2^2} + \\dots + \\frac{n+1}{2^n}. \\qquad \\textcircled{4}\n$$\nThen\n$$\n\\frac{1}{2} S_n = \\frac{2}{2^2} + \\frac{3}{2^3} + \\dots + \\frac{n+1}{2^{n+1}}. \\qquad \\textcircled{5}\n$$\n$\\textcircled{4} - \\textcircled{5}$,\n$$\n\\frac{1}{2} S_n = \\frac{3}{2} - \\frac{n+3}{2^{n+1}}.\n$$\nWe finally get\n$$\nS_n = 3 - \\frac{n+3}{2^n}.\n$$\n(1) By Vieta's theorem, we have $\\alpha \\times \\beta = q \\neq 0$, $\\alpha + \\beta = p$. Then\n$$\na_1 = \\alpha + \\beta, \\quad a_2 = \\alpha^2 + \\alpha\\beta + \\beta^2. \\qquad \\textcircled{6}\n$$\nThe characteristic equation of $\\{a_n\\}$ is $\\lambda^2 - p\\lambda + q = 0$, which has roots $\\alpha, \\beta$. Then we can write down the general expression of $\\{a_n\\}$ according to the following different situations:\n\nWhen $\\alpha = \\beta \\neq 0$, $a_n = (A_1 + A_2 n) \\alpha^n$. From (6), we have\n$$\n(A_1 + A_2)\\alpha = 2\\alpha, \\\\\n(A_1 + 2A_2)\\alpha^2 = 3\\alpha^2.\n$$\nThen we get $A_1 = A_2 = 1$. Therefore,\n$$\na_n = (n + 1)\\alpha^n.\n$$\nWhen $\\alpha \\neq \\beta$, $a_n = \\Lambda_1 \\alpha^n + \\Lambda_2 \\beta^n$. From (6), we have\n$$\nA_1 \\alpha + A_2 \\beta = \\alpha + \\beta, \\\\\nA_1 \\alpha^2 + A_2 \\beta^2 = \\alpha^2 + \\alpha\\beta + \\beta^2.\n$$\nThen we get $A_1 = \\frac{-\\alpha}{\\beta - \\alpha}$, $A_2 = \\frac{\\beta}{\\beta - \\alpha}$. Therefore,\n$$\na_n = \\frac{-\\alpha^{n+1}}{\\beta - \\alpha} + \\frac{\\beta^{n+1}}{\\beta - \\alpha} = \\frac{\\beta^{n+1} - \\alpha^{n+1}}{\\beta - \\alpha}.\n$$\n\n(2) The solution is the same as Solution I.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 28223,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $n \\ge 3$ be any given integer. Determine the smallest positive integer $k$, for which there exists a set $A$ of $k$ real numbers and $n$ real numbers $x_1, x_2, \\dots, x_n$, which are distinct from each other such that\n$$\nx_1 + x_2,\\ x_2 + x_3,\\ \\dots,\\ x_{n-1} + x_n,\\ x_n + x_1\n$$\nare all in the set $A$.",
    "options": [],
    "answer": "3",
    "solution": "Let $m_1 = x_1 + x_2$, $m_2 = x_2 + x_3$, $\\dots$, $m_{n-1} = x_{n-1} + x_n$, $m_n = x_n + x_1$.\nFirst, note that $m_1 \\neq m_2$, otherwise $x_1 = x_3$, which contradicts the fact that $x_i$ are distinct. Similarly, $m_i \\neq m_{i+1}$, for $i=1, 2, \\dots, n$, where $m_{n+1} = m_1$, as usual. It follows that $k \\ge 2$.\n\nFor $k=2$, let $A = \\{a, b\\}$, where $a \\neq b$. It follows that\n$$\n(1) \\begin{cases} x_1 + x_2 = a, \\\\ x_2 + x_3 = b, \\\\ \\vdots \\\\ x_{n-1} + x_n = a, \\\\ x_n + x_1 = b, \\end{cases} \\quad \\text{(if $n$ is even)}\n$$\nor\n$$\n(2) \\begin{cases} x_1 + x_2 = a, \\\\ x_2 + x_3 = b, \\\\ \\vdots \\\\ x_{n-1} + x_n = b, \\\\ x_n + x_1 = a. \\end{cases} \\quad \\text{(if $n$ is odd)}\n$$\nFor (2), we have $x_n = x_2$, which is possible. For (1), it follows that\n$$\n\\begin{aligned}\n\\frac{n}{2}a &= (x_1 + x_2) + (x_3 + x_4) + \\dots + (x_{n-1} + x_n) \\\\\n&= (x_2 + x_3) + (x_4 + x_5) + \\dots + (x_n + x_1) \\\\\n&= \\frac{n}{2}b,\n\\end{aligned}\n$$\nand hence $a = b$, which is impossible again. It follows that $k \\ge 3$.\n\nFor $k=3$, one can construct a valid example as follows:\nDefine $x_{2k-1} = k$ ($k \\ge 1$) and $x_{2k} = n+1-k$ ($k \\ge 1$). When $n$ is even,\n$$\nx_i + x_{i+1} = \\begin{cases} n+1, & \\text{if } i \\text{ is odd,} \\\\ n+2, & \\text{if } i \\text{ is even and } i < n, \\\\ \\frac{n}{2} + 2, & \\text{if } i = n, \\text{ where } x_{n+1} = x_n. \\end{cases}\n$$\nWhen $n$ is odd,\n$$\nx_i + x_{i+1} = \\begin{cases} n+1, & \\text{if } i \\text{ is odd and } i < n, \\\\ n+2, & \\text{if } i \\text{ is even,} \\\\ \\frac{n-1}{2} + 2, & \\text{if } i = n, \\text{ where } x_{n+1} = x_n. \\end{cases}\n$$\nTherefore, the smallest positive integer $k$ is $3$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 28224,
    "subject": "Mathematics (Multi-modal)",
    "question": "Given the line $L: x + y - 9 = 0$ and the circle $M: 2x^2 + 2y^2 - 8x - 8y - 1 = 0$, point $A$ is on $L$ and points $B, C$ are on $M$; $\\angle BAC = 45^\\circ$ and the line $AB$ is through the center of $M$. Then the range of the $x$ coordinate of point $A$ is ______.",
    "options": [],
    "answer": "3 ≤ a ≤ 6",
    "solution": "Suppose that $A(a, 9-a)$. Then the distance from the center of $M$ to the line $AC$ is\n$$\nd = |AM| \\times \\sin \\angle BAC \\\\\n= \\sqrt{(a-2)^2 + (9-a-2)^2} \\times \\sin 45^\\circ \\\\\n= \\sqrt{2a^2 - 18a + 53} \\times \\frac{\\sqrt{2}}{2}.\n$$\nOn the other hand, since the line $AC$ intercepts $M$, it follows that $d \\le$ the radius of $M = \\sqrt{\\frac{17}{2}}$, i.e.\n$$\n\\sqrt{2a^2 - 18a + 53} \\times \\frac{\\sqrt{2}}{2} \\le \\sqrt{\\frac{17}{2}}\n$$\nThe solution is $3 \\le a \\le 6$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 28225,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all pairs $(p, q)$ of prime numbers such that $pq \\mid 5^p + 5^q$. (Posed by Fu Yunhao)",
    "options": [],
    "answer": "[[2, 3], [3, 2], [2, 5], [5, 2], [5, 5], [5, 313], [313, 5]]",
    "solution": "If $2 \\mid pq$, we suppose that $p = 2$ without loss of generality, and then $q \\mid 5^q + 25$. By Fermat's theorem we have $q \\mid 5^q - 5$, so $q \\mid 30$, where $(2, 3)$ and $(2, 5)$ are solutions [(2, 2) does not fit].\nIf $5 \\mid pq$, we suppose that $p = 5$ without loss of generality and then $5q \\mid 5^q + 5^5$. By Fermat's theorem we have $q \\mid 5^{5^q - 1}$, so $q \\mid 313$, where $(5, 5)$ and $(5, 313)$ are solutions.\nOtherwise, we have $pq \\mid 5^{p-1} + 5^{q-1}$, and so\n$$\n5^{p-1} + 5^{q-1} \\equiv 0 \\pmod{p}.\n$$\nBy Fermat's theorem, $5^{p-1} \\equiv 1 \\pmod{p}$,\nand because of the above, $5^{q-1} \\equiv -1 \\pmod{p}$.\nDenote by $p-1=2^k(2r-1)$, $q-1=2^l(2s-1)$, where $k, l, r, s$ are positive integers.\nIf $k \\le l$, because of the previous congruences we get\n$$\n\\begin{aligned}\n1 &= 1^{2^{l-k}(2s-1)} \\equiv (5^{p-1})^{2^{l-k}(2s-1)} \\\\\n&= 5^{2^l(2r-1)(2s-1)} = (5^{q-1})^{2r-1} = (-1)^{2r-1} \\\\\n&= -1 \\pmod{p},\n\\end{aligned}\n$$\na contradiction of $p \\ne 2$. So $k > l$.\nBut we have $k < l$ by a similar argument — a contradiction.\nTherefore, all possible pairs of primes $(p, q)$ are $(2, 3)$, $(3, 2)$, $(2, 5)$, $(5, 2)$, $(5, 5)$, $(5, 313)$ and $(313, 5)$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 28226,
    "subject": "Mathematics (Multi-modal)",
    "question": "Prove that\n$$\n-1 < \\left( \\sum_{k=1}^{n} \\frac{k}{k^2 + 1} \\right) - \\ln n \\le \\frac{1}{2}, \\quad n = 1, 2, \\dots\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "We first prove that\n$$\n\\frac{x}{1+x} < \\ln(1+x) < x, \\quad x > 0. \\qquad \\textcircled{1}\n$$\nLet\n$$\nh(x) = x - \\ln(1+x), \\\\\ng(x) = \\ln(1+x) - \\frac{x}{1+x}.\n$$\nThen, for $x > 0$,\n$$\nh'(x) = 1 - \\frac{1}{1+x} > 0, \\\\\ng'(x) = \\frac{1}{1+x} - \\frac{1}{(1+x)^2} = \\frac{x}{(1+x)^2} > 0.\n$$\nTherefore,\n$$\nh(x) > h(0) = 0, \\quad g(x) > g(0) = 0.\n$$\nThis completes the proof of the inequalities ①.\n\nNow let $x = \\frac{1}{n}$ in ①. We have\n$$\n\\frac{1}{n+1} < \\ln\\left(1+\\frac{1}{n}\\right) < \\frac{1}{n}. \\qquad \\textcircled{2}\n$$\nLet\n$$\nx_n = \\sum_{k=1}^{n} \\frac{k}{k^2 + 1} - \\ln n.\n$$\nThen\n$$\n\\begin{aligned}\nx_n - x_{n-1} &= \\frac{n}{n^2+1} - \\ln\\left(1 + \\frac{1}{n+1}\\right) \\\\\n&< \\frac{n}{n^2+1} - \\frac{1}{n} \\\\\n&= -\\frac{1}{n(n^2+1)} < 0.\n\\end{aligned}\n$$\nTherefore, $x_n < x_{n-1} < \\cdots < x_1 = \\frac{1}{2}$.\n\nFurthermore,\n$$\n\\begin{aligned}\n\\ln n &= (\\ln n - \\ln(n-1)) + (\\ln(n-1) - \\ln(n-2)) \\\\\n&\\quad + \\cdots + (\\ln 2 - \\ln 1) + \\ln 1 \\\\\n&= \\sum_{k=1}^{n-1} \\ln\\left(1 + \\frac{1}{k}\\right).\n\\end{aligned}\n$$\nConsequently,\n$$\n\\begin{aligned}\nx_n &= \\sum_{k=1}^{n} \\frac{k}{k^2+1} - \\sum_{k=1}^{n-1} \\ln\\left(1 + \\frac{1}{k}\\right) \\\\\n&= \\sum_{k=1}^{n-1} \\left( \\frac{k}{k^2+1} - \\ln\\left(1 + \\frac{1}{k}\\right) \\right) + \\frac{n}{n^2+1} \\\\\n&> \\sum_{k=1}^{n-1} \\left( \\frac{k}{k^2+1} - \\frac{1}{k} \\right) \\\\\n&= - \\sum_{k=1}^{n-1} \\frac{1}{(k^2+1)k} \\\\\n&> - \\sum_{k=1}^{n-1} \\frac{1}{(k+1)k} \\\\\n&= -1 + \\frac{1}{n} > -1.\n\\end{aligned}\n$$\nThis completes the proof.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 28227,
    "subject": "Mathematics (Multi-modal)",
    "question": "Given integer $n \\ge 2$, find the largest number $\\lambda(n)$ with the following property: if a sequence of real numbers $a_0, a_1, a_2, \\dots, a_n$ satisfies\n$$0 = a_0 \\le a_1 \\le a_2 \\le \\dots \\le a_n,$$ \n$$a_i \\ge \\frac{1}{2}(a_{i+1} + a_{i-1}), \\quad i = 1, 2, \\dots, n-1,$$\nthen\n$$\n\\left(\\sum_{i=1}^{n} ia_{i}\\right)^{2} \\geqslant \\lambda(n) \\sum_{i=1}^{n} a_{i}^{2}.\n$$",
    "options": [],
    "answer": "n(n+1)^2/4",
    "solution": "The largest possible value of $\\lambda(n)$ is $\\frac{n(n+1)^2}{4}$.\n\nLet $a_1 = a_2 = \\dots = a_n = 1$. Then we have $\\lambda(n) \\le \\frac{n(n+1)^2}{4}$.\n\nWe shall show that for any real numbers $a_0, a_1, a_2, \\dots, a_n$ satisfying the indicated property in the problem, the following inequality holds:\n$$\n\\left(\\sum_{i=1}^{n} ia_i\\right)^2 \\geqslant \\frac{n(n+1)^{2}}{4}\\left(\\sum_{i=1}^{n} a_{i}^{2}\\right). \\qquad \\textcircled{1}\n$$\nFirst, we notice that\n$$\na_1 \\ge \\frac{a_2}{2} \\ge \\dots \\ge \\frac{a_n}{n}.\n$$\nIndeed, by assumption, $2ia_i \\ge i(a_{i+1} + a_{i-1})$ holds for $i = 1, 2, \\dots, n-1$. For any given positive integer $1 \\le l \\le n-1$, summing the above inequality over $i = 1, 2, \\dots, l$, we have $(l+1)a_l \\ge la_{l+1}$, i.e.\n$$\n\\frac{a_l}{l} \\ge \\frac{a_{l+1}}{l+1} \\text{ for } l = 1, 2, \\dots, n-1.\n$$\nIn what follows, we show that for any $i, j, k \\in \\{1, 2, \\dots, n\\}$, if $i > j$, then\n$$\n\\frac{2ik^2}{i+k} > \\frac{2jk^2}{j+k}.\n$$\nIndeed, the above inequality is equivalent to $2ik^2(j+k) > 2jk^2(i+k)$, i.e. $(i-j)k^3 > 0$, which is clearly true.\n\nNow, we are going to show inequality ①. We shall start by estimating the lower bound of $a_i a_j$ for $1 \\le i < j \\le n$.\nBy previous results, we have $\\frac{a_i}{i} \\ge \\frac{a_j}{j}$, i.e. $ja_i - ia_j \\ge 0$. Since $a_i - a_j \\le 0$, we have $(ja_i - ia_j)(a_j - a_i) \\ge 0$, i.e. $a_i a_j \\ge \\frac{i}{i+j}a_j^2 + \\frac{j}{i+j}a_i^2$.\nThus, we have\n$$\n\\begin{aligned} \\left(\\sum_{i=1}^{n} ia_i\\right)^2 &= \\sum_{i=1}^{n} i^2 a_i^2 + 2 \\sum_{1 \\le i < j \\le n} ija_i a_j \\\\ &\\ge \\sum_{i=1}^{n} i^2 \\times a_i^2 + 2 \\sum_{1 \\le i < j \\le n} \\left( \\frac{i^2 j}{i+j} a_j^2 + \\frac{ij^2}{i+j} a_i^2 \\right) \\\\ &= \\sum_{i=1}^{n} \\left( a_i^2 \\times \\sum_{k=1}^{n} \\frac{2ik^2}{i+k} \\right). \\end{aligned}\n$$\nLet $b_i = \\sum_{k=1}^{n} \\frac{2ik^2}{i+k}$. We see from previous results that $b_1 \\le b_2 \\le \\dots \\le b_n$.\nSince $a_1^2 \\le a_2^2 \\le \\dots \\le a_n^2$, by the Chebyshev inequality, we have\n$$\n\\sum_{i=1}^{n} a_i^2 b_i \\ge \\frac{1}{n} \\left( \\sum_{i=1}^{n} a_i^2 \\right) \\left( \\sum_{i=1}^{n} b_i \\right).\n$$\nHence\n$$ \\left(\\sum_{i=1}^{n} ia_i\\right)^2 \\ge \\frac{1}{n} \\left(\\sum_{i=1}^{n} a_i^2\\right) \\left(\\sum_{i=1}^{n} b_i\\right). $$\nSince\n$$\n\\begin{align*}\n\\sum_{i=1}^{n} b_i &= \\sum_{i=1}^{n} \\sum_{k=1}^{n} \\frac{2ik^2}{i+k} = \\sum_{i=1}^{n} i^2 + 2 \\sum_{1 \\le i < j \\le n} \\left( \\frac{i^2 j}{i+j} + \\frac{ij^2}{i+j} \\right) \\\\\n&= \\sum_{i=1}^{n} i^2 + 2 \\sum_{1 \\le i < j \\le n} ij = \\left( \\sum_{i=1}^{n} i \\right)^2 = \\frac{n^2(n+1)^2}{4},\n\\end{align*}\n$$\nwe find that $(\\sum_{i=1}^{n} ia_i)^2 \\ge \\frac{n(n+1)^2}{4} \\sum_{i=1}^{n} a_i^2$, which proves inequality ①.\n\nWe conclude that the maximum possible value of $\\lambda(n)$ is\n$$ \\frac{n(n+1)^2}{4} $$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 28228,
    "subject": "Mathematics (Multi-modal)",
    "question": "A right triangle $ABC$, with $\\angle BAC = 90^\\circ$, is inscribed in the circle $\\Gamma$. The point $E$ lies in the interior of the arc $\\widearc{BC}$ (not containing $A$), with $EA > EC$. The point $F$ lies on the ray $EC$ with $\\angle EAC = \\angle CAF$. The segment $BF$ meets $\\Gamma$ again at $D$ (other than $B$). Let $O$ denote the circumcenter of the triangle $DEF$. Prove that the points $A$, $C$, $O$ are collinear. (Posed by Bian Hongping)\n\n![](attached_image_1.png)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $M$ and $N$ be the feet of the perpendiculars from $O$ to the lines $DF$ and $DE$, respectively. Because $O$ is the circumcenter of the triangle $DEF$, the triangles $EOD$ and $ODF$ are both isosceles with $EO = DO = FO$. It follows that\n$$\n\\angle EOF = \\angle EOD + \\angle DOF = 2\\angle NOD + 2\\angle DOM = 2\\angle NOM.\n$$\nBecause $\\angle OND = \\angle OMD = 90^\\circ$, the quadrilateral $OMDN$ is concyclic, from which it follows that $\\angle NDM + \\angle NOM = 180^\\circ$ or $\\angle BDN = \\angle NOM$. Because $ABED$ is concyclic, we have $\\angle BAE = \\angle BDE$. Combining the above equations together, one has\n$$\n\\angle EOF = 2\\angle NOM = 2\\angle BDN = 2\\angle BDE = 2\\angle BAE.\n$$\nBecause $BC$ is a diameter of $\\Gamma$, it follows that\n$$\n\\angle EOF + \\angle EAF = 2\\angle BAE + 2\\angle EAC = 2\\angle BAC = 180^\\circ,\n$$\nfrom which it follows that $AEOF$ is concyclic. Let $\\omega$ denote the circumcircle of $AEOF$. Because $O$ lies on the perpendicular bisector of the segment $EF$, $O$ is the midpoint of the arc $EF$ (on $\\omega$), implying that $AO$ bisects $\\angle EAF$ and $A$, $C$, $O$ are collinear.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 28229,
    "subject": "Mathematics (Multi-modal)",
    "question": "For a convex pentagon $ABCDE$, $AB = DE = EA$, $AB \\neq EA$, and $B$, $C$, $D$, $E$ are concyclic. Prove that $A$, $B$, $C$, $D$ are concyclic if and only if $AC = AD$. (Posed by Xiong Bin)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "First, if $A$, $B$, $C$, $D$ are concyclic, by $AB = DE$ and $BC = EA$ we have $\\angle BAC = \\angle EDA$, $\\angle ACB = \\angle DAE$, so $\\angle ABC = \\angle DEA$, which means that $AC = AD$.\n\n![](attached_image_1.png)\n\nSecond, if $AC = AD$, let $O$ be the center of the circle ($B$, $C$, $D$, $E$ are on the circle). Then $O$ is on the perpendicular bisector ($AH$) of $CD$. Let $F$ be the symmetric point of $B$ by the line $AH$. Then $F$ is on the circle $O$. $AB \\neq EA$, so $DE \\neq DF$, and thus $E$, $F$ are not the same points. Now one can see that $\\triangle AFD \\cong \\triangle ABC$, with $AB = DE$, $BC = EA$, and one can get $\\triangle AED \\cong \\triangle CBA$, and so $\\triangle AED \\cong \\triangle DFA$, which indicates $\\angle AED = \\angle DFA$, and therefore $A$, $E$, $F$, $D$ is concyclic. This means that $A$ is on the circle $O$, and $B$, $C$, $D$ are concyclic.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 28230,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $n$ be a given positive integer. In the coordinate plane, consider the set of the points\n$$\n\\{P_1, P_2, \\dots, P_{4n+1}\\} = \\{(x, y) \\mid x \\text{ and } y \\text{ are integers with } xy = 0, |x| \\le n, |y| \\le n\\}.\n$$\n\nDetermine the minimum of $(P_1P_2)^2 + (P_2P_3)^2 + \\dots + (P_{4n}P_{4n+1})^2 + (P_{4n+1}P_1)^2$. (Posed by Wang Xinmao)",
    "options": [],
    "answer": "16n - 8",
    "solution": "The answer is $16n - 8$.\nAssume that $P_i = (x_i, y_i)$ for $1 \\le i \\le 4n + 1$. Set\n$$\nP_{4n+2} = (x_{4n+2}, y_{4n+2}) = P_1 = (x_1, y_1).\n$$\nWe will show that the sum\n$$\nS = \\sum_{i=1}^{4n+1} (P_i P_{i+1})^2 \\geq 16n - 8.\n$$\nFirst, we show that this minimum can be obtained by setting\n$$\n(P_1, P_2, \\dots, P_{4n+1}) = ((2, 0), (4, 0), \\dots, (n, 0), (n-1, 0), \\dots, (1, 0), (0, 2), (0, 4), \\dots, (0, n), (0, n-1), \\dots, (0, 1), (-2, 0), (-4, 0), \\dots, (-n, 0), (-n+1, 0), \\dots, (-1, 0), (0, -2), (0, -4), \\dots, (0, -n), (0, -n+1), \\dots, (0, -1), (0, 0))\n$$\nfor $n$ even, and\n$$\n(P_1, P_2, \\dots, P_{4n+1}) = ((1, 0), (3, 0), \\dots, (n, 0), (n-1, 0), \\dots, (2, 0), (0, 1), (0, 3), \\dots, (0, n), (0, n-1), \\dots, (0, 2), (-1, 0), (-3, 0), \\dots, (-n, 0), (-n+1, 0), \\dots, (-2, 0), (0, -1), (0, -3), \\dots, (0, -n), (0, -n+1), \\dots, (0, -2), (0, 0))\n$$\nfor $n$ odd. This can be easily checked. For example, when $n = 2m$ is even, our construction shows that\n$$\n\\begin{aligned}\nS &= 4 \\left( \\sum_{i=1}^{m-1} (P_i P_{i+1})^2 + (P_m P_{m+1})^2 + \\sum_{i=m+1}^{2m-1} (P_i P_{i+1})^2 \\right) \\\\\n&\\quad + 3(P_{2m} P_{2m+1})^2 + (P_{4n} P_{4n+1})^2 + (P_{4n+1} P_1)^2 \\\\\n&= 4(4(m-1)+1+4(m-1)) + 3 \\times 5 + 1 + 4 \\\\\n&= 32m - 8 = 16n - 8.\n\\end{aligned}\n$$\nThe exact same argument works when $n$ is odd.\nNote that\n$$\nS = \\sum_{i=1}^{4n+1} [(x_i - x_{i+1})^2 + (y_i - y_{i+1})^2].\n$$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 28231,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $x$, $y$, $z$ be positive numbers, and $\\sqrt{a} = x(y-z)^2$, $\\sqrt{b} = y(z-x)^2$, $\\sqrt{c} = z(x-y)^2$. Prove that $a^2 + b^2 + c^2 \\ge 2(ab + bc + ca)$. (Posed by Tang Lihua)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "$$\n\\begin{aligned}\n\\sqrt{b} + \\sqrt{c} - \\sqrt{a} &= -(y+z)(z-x)(x-y), \\\\\n\\sqrt{c} + \\sqrt{a} - \\sqrt{b} &= -(z+x)(x-y)(y-z), \\\\\n\\sqrt{a} + \\sqrt{b} - \\sqrt{c} &= -(x+y)(y-z)(z-x),\n\\end{aligned}\n$$\nso\n$$\n\\begin{aligned}\n& (\\sqrt{b} + \\sqrt{c} - \\sqrt{a})(\\sqrt{c} + \\sqrt{a} - \\sqrt{b})(\\sqrt{a} + \\sqrt{b} - \\sqrt{c}) \\\\\n&= -(y+z)(z+x)(x+y)[(y-z)(z-x)(x-y)]^2 \\\\\n&\\le 0.\n\\end{aligned}\n$$\nWe can get\n$$\n\\begin{aligned}\n& 2(ab + bc + ca) - (a^2 + b^2 + c^2) \\\\\n&= (\\sqrt{a} + \\sqrt{b} + \\sqrt{c})(\\sqrt{b} + \\sqrt{c} - \\sqrt{a})(\\sqrt{c} + \\sqrt{a} - \\sqrt{b})(\\sqrt{a} + \\sqrt{b} - \\sqrt{c}) \\\\\n&\\le 0.\n\\end{aligned}\n$$\nThis means that $a^2 + b^2 + c^2 \\ge 2(ab + bc + ca)$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 28232,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $H$ be the orthocenter of an acute triangle $ABC$, and $D$ be the midpoint of the side $BC$. A line passing through the point $H$ meets the sides $AB, AC$ at the points $F, E$ respectively, such that $AE = AF$. The ray $DH$ meets the circumcircle of $\\triangle ABC$ at the point $P$.\n\nProve that $P, A, E, F$ are concyclic.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "On the ray $HD$, mark a point $M$ such that $HD = DM$. Join the segments $BM, CM, BH$ and $CH$. As $D$ is the midpoint of $BC$, the quadrilateral $BHCM$ is a parallelogram, and so\n$$\n\\angle BMC = \\angle BHC = 180^\\circ - \\angle BAC.\n$$\nHence,\n$$\n\\angle BMC + \\angle BAC = 180^\\circ,\n$$\nand the point $M$ lies on the circumcircle of $\\triangle ABC$. Join the segments $PB, PC, PE$ and $PF$. It follows from $AE = AF$ that\n\n![](attached_image_1.png)\n\n$$\n\\angle BFH = \\angle CEH. \\qquad \\textcircled{1}\n$$\nAs $H$ is the orthocenter of $\\triangle ABC$,\n$$\n\\angle HBF = 90^\\circ - \\angle BAC = \\angle HCE. \\qquad \\textcircled{2}\n$$\nFrom ① and ②, one has $\\triangle BFH \\sim \\triangle CEH$, so\n$$\n\\frac{BF}{BH} = \\frac{CE}{CM}.\n$$\nAs $BHCM$ is a parallelogram, $BH = CM, CH = BM$, and hence\n$$\n\\frac{BF}{CM} = \\frac{CE}{BM}. \\qquad \\textcircled{3}\n$$\nAnd $D$ is the midpoint of $BC$. Thus, $S_{\\triangle PBM} = S_{\\triangle PCM}$, and so\n$$\n\\frac{1}{2} BP \\times BM \\times \\sin \\angle MBP = \\frac{1}{2} CP \\times CM \\times \\sin \\angle MCP.\n$$\nIt follows from $\\angle MBP + \\angle MCP = 180^\\circ$ that\n$$\nBP \\times BM = CP \\times CM. \\qquad \\textcircled{4}\n$$\nWith ③ and ④, one has $\\frac{BF}{BP} = \\frac{CE}{CP}$. From $\\angle PBF = \\angle PCE$, one has $\\triangle PBF \\sim \\triangle PCE$, and hence $\\angle PFB = \\angle PEC$, and therefore $\\angle PFA = \\angle PEA$. Consequently $P, A, E$ and $F$ are concyclic.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 28233,
    "subject": "Mathematics (Multi-modal)",
    "question": "Prove that for any odd prime number $p$, the number of positive integers $n$ satisfying $p \\mid n! + 1$ is no more than $c p^{\\frac{2}{3}}$, where $c$ is a constant number independent of $p$. (Posed by Yu Hongbing)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "**Proof** Clearly, if $n$ satisfies the required property, then $1 \\le n \\le p-1$. Denote all such $n$'s by $n_1 < n_2 < \\cdots < n_k$; we shall show that $k \\le 12 p^{\\frac{2}{3}}$. If $k \\le 12$ there is nothing to prove. In what follows, we assume that $k > 12$.\n\nRename $n_{i+1} - n_i$ ($1 \\le i \\le k-1$) in nondecreasing order as\n$1 \\le \\mu_1 \\le \\mu_2 \\le \\cdots \\le \\mu_{k-1}$. It is clear that\n$$\n\\sum_{i=1}^{k-1} \\mu_i = \\sum_{i=1}^{k} (n_{i+1} - n_i) = n_k - n_1 < p. \\quad \\textcircled{1}\n$$\nFirst, we show that for any $s \\ge 1$,\n$$\n| \\{ 1 \\le i \\le k-1 : \\mu_i = s \\} | \\le s, \\qquad \\textcircled{2}\n$$\ni.e. there are at most $s$ $\\mu_i$'s equal to $s$.\nIn fact, suppose that $n_{i+1} - n_i = s$; then $n_i! + 1 \\equiv n_{i+1}! + 1 \\equiv 0 \\pmod{p}$, so $(p, n_i!) = 1$, and\n$$\n(n_i + s)(n_i + s - 1)\\cdots(n_i + 1) \\equiv 1 \\pmod p.\n$$\nThus, $n_i$ is a solution to the congruence equation\n$$\n(x + s)(x + s - 1)\\cdots(x + 1) \\equiv 1 \\pmod p.\n$$\nSince $p$ is a prime number, there are at most $s$ solutions to the above equation, thanks to Lagrange's theorem. Thus, there are at most $s$ $n_i$'s with $n_{i+1} - n_i = s$, i.e. ② holds.\n\nNow, we show that for any nonnegative integer $l$, if $\\frac{l(l+1)}{2} + 1 \\le k - 1$, then $\\mu_{\\frac{l(l+1)}{2}+1} \\ge l + 1$. Suppose on the contrary that $\\mu_{\\frac{l(l+1)}{2}+1} \\le l$. Then $\\mu_1, \\mu_2, \\dots, \\mu_{\\frac{l(l+1)}{2}+1}$ are all positive integers between $1$ and $l$. By ②, there is at most one $\\mu_i$ equal to $1$, at most two $\\mu_i$'s equal to $2$, $\\dots$, at most $l$ $\\mu_i$'s equal to $l$, and thus there are at most $1+2+\\dots+l = \\frac{l(l+1)}{2}$ $\\mu_i$'s less than or equal to $l$, which contradicts the assumption that $\\mu_1, \\mu_2, \\dots, \\mu_{\\frac{l(l+1)}{2}+1}$ are all less than or equal to $l$.\n\nLet $m$ be the largest positive integer with $\\frac{m(m+1)}{2} + 1 \\le k - 1$. Then\n$$\n\\frac{m(m+1)}{2} + 1 \\le k - 1 < \\frac{(m+1)(m+2)}{2} + 1, \\quad \\text{③}\n$$\nand hence\n$$\n\\begin{aligned}\n\\sum_{i=1}^{k-1} \\mu_i &\\ge \\sum_{i=0}^{m-1} (i+1)^2 \\\\\n&= \\frac{m(m+1)(2m+1)}{6} > \\frac{m^3}{3}.\n\\end{aligned}\n$$\nSince $k > 12$, $m \\ge 4$, combining ① and ③ we get\n$$\n\\begin{aligned}\nk &< 2 + \\frac{(m+1)(m+2)}{2} \\\\\n&< 4m^2 + 4 \\left( 3 \\sum_{i=1}^{k-1} \\mu_i \\right)^{\\frac{2}{3}} \\\\\n&< 4 \\times (3p)^{\\frac{2}{3}}.\n\\end{aligned}\n$$\nThis completes our proof.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 28234,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $m$, $n$ be integers with $4 < m < n$, and $A_1, A_2, \\dots, A_{2n+1}$ be a regular $2n + 1$ polygon. In addition, let $P = \\{A_1, A_2, \\dots, A_{2n+1}\\}$. Find the number of convex $m$-gons with exactly two acute internal angles whose vertices are all in $P$. (Posed by Leng Gangsong)",
    "options": [],
    "answer": "(2n+1) * (\\binom{n}{m-1} + \\binom{n+1}{m-1})",
    "solution": "Notice that if a convex $m$-gon whose vertex set is contained in $P$ has exactly two acute angles, they must be at consecutive vertices; for otherwise there would be two disjoint pairs of sides that take up more than half of the circle each.\n\nNow assume that the last vertex, clockwise, of these four vertices that make up two acute angles is fixed; this reduces the total number of regular $m$-gons $2n + 1$ times and we will later multiply by this factor.\n\nSuppose that the larger arc that the first and the last of these four vertices make contains $k$ points, and the other arc contains $2n - 1 - k$ points. For each $k$, the vertices of the $m$-polygon on the smaller arc may be arranged in $\\binom{2n-1-k}{m-4}$ ways, and the two vertices on the larger arc may be arranged in $(k-n-1)^2$ ways (so that the two angles cut off more than half of the circle).\n\nThe total number of polygons given by $k$ is thus $(k-n-1)^2 \\times \\binom{2n-1-k}{m-4}$. Summation over all $k$ and a change of variable shows that the total number of polygons (divided by a factor of $2n+1$) is $\\sum_{k \\ge 0} k^2 \\times \\binom{n-k-2}{m-4}$.\n\nThis can be proven to be exactly $\\binom{n}{m-1} + \\binom{n+1}{m-1}$ by double induction on $n > m$ and $m > 4$. The base cases $n = m + 1$ and $m = 5$ are readily calculated. The induction step is\n$$\n\\begin{align*}\n& \\sum_{k \\ge 0} k^2 \\times \\binom{n-k-2}{m-4} \\\\\n&= \\sum_{k \\ge 0} k^2 \\times \\binom{(n-1)-k-2}{m-4} + \\sum_{k \\ge 0} k^2 \\times \\binom{(n-1)-k-2}{m-4-2} \\\\\n&= \\binom{n-1}{m-1} + \\binom{n}{m-1} + \\binom{n-1}{m-2} + \\binom{n}{m-2} \\\\\n&= \\binom{n}{m-1} + \\binom{n+1}{m-1}.\n\\end{align*}\n$$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 28235,
    "subject": "Mathematics (Multi-modal)",
    "question": "Suppose that $k$, $l$ are two positive integers.\nProve that:\nThere are infinitely many positive integers $m \\ge k$ such that $\\binom{m}{k}$ and $l$ are relatively prime.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $m = k + t \\times l \\times (k!)$, where $t$ is any positive integer. To prove that $\\binom{m}{k}$ and $l$ are relatively prime, we only need to prove that for any prime factor $p$ of $l$, $p \\nmid \\binom{m}{k}$.\n\nIf $p \\nmid k!$, we have\n$$\n\\begin{aligned}\nk! \\binom{m}{k} &= \\prod_{i=1}^{k} (m-k+i) \\\\\n&= \\prod_{i=1}^{k} [i + t l (k!)] \\\\\n&= \\prod_{i=1}^{k} i \\equiv k! \\pmod{p}.\n\\end{aligned}\n$$\nTherefore, $p \\nmid \\binom{m}{k}$.\n\nIf $p \\mid k!$, there exists integer $\\alpha \\ge 1$ such that $p^\\alpha \\mid k!$ but $p^{\\alpha+1} \\nmid k!$. Then $p^{\\alpha+1} \\mid l(k!)$.\nWe have\n$$\n\\begin{aligned}\nk! \\binom{m}{k} &= \\prod_{i=1}^{k} (m-k+i) \\\\\n&= \\prod_{i=1}^{k} [i + t l (k!)] \\\\\n&= \\prod_{i=1}^{k} i \\equiv k! \\pmod{p^{\\alpha+1}}.\n\\end{aligned}\n$$\nTherefore, $p^\\alpha \\mid k! \\binom{m}{k}$ and $p^{\\alpha+1} \\nmid k! \\binom{m}{k}$. Since $p^\\alpha \\mid k!$, we get $p \\nmid \\binom{m}{k}$. The proof is completed.\nLet $m = k + t \\times l \\times (k!)^z$, where $t$ is any positive integer. The following proof steps are similar to those in Solution I, and are omitted.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 28236,
    "subject": "Mathematics (Multi-modal)",
    "question": "On a coordinate plane there are two regions, $M$ and $N$:\n$M$ is confined by\n$$\n\\begin{cases}\ny \\ge 0, \\\\\ny \\le x, \\\\\ny \\le 2-x\n\\end{cases}\n$$\nand $N$ is determined by the inequalities $t \\le x \\le t+1$, $0 \\le t \\le 1$. Then the size of the common area of $M$ and $N$ is given by $f(t) = \\underline{\\hspace{2cm}}$.",
    "options": [],
    "answer": "f(t) = -t^2 + t + 1/2 for 0 ≤ t ≤ 1",
    "solution": "As shown in the figure, we have\n$$\n\\begin{aligned}\nf(t) &= S_{\\text{shaded area}} \\\\\n&= S_{\\triangle AOB} - S_{\\triangle OCD} - S_{\\triangle BEF} \\\\\n&= 1 - \\frac{1}{2}t^2 - \\frac{1}{2}(1-t)^2 \\\\\n&= -t^2 + t + \\frac{1}{2}, \\quad 0 \\le t \\le 1.\n\\end{aligned}\n$$\n![](attached_image_1.png)",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 28237,
    "subject": "Mathematics (Multi-modal)",
    "question": "The inequality\n$$\n\\frac{1}{n+1} + \\frac{1}{n+2} + \\dots + \\frac{1}{2n+1} < a - 2007 \\frac{1}{3}\n$$\nholds for every positive integer $n$. Then the least positive integer of $a$ is $\\underline{\\hspace{2cm}}$.",
    "options": [],
    "answer": "2009",
    "solution": "Obviously,\n$$\nf(n) = \\frac{1}{n+1} + \\frac{1}{n+2} + \\dots + \\frac{1}{2n+1}\n$$\nis monotonically decreasing. Therefore, $f(1)$ reaches the maximum of $f(n)$. From\n$$\nf(1) = \\frac{1}{2} + \\frac{1}{3} < a - 2007 \\frac{1}{3},\n$$\nwe have $a > 2008$. Therefore, the least positive integer of $a$ is $2009$.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 28238,
    "subject": "Mathematics (Multi-modal)",
    "question": "Prove that for any given positive integer $k$, there exist infinitely many positive integers $n$, such that the numbers\n$$\n2^n + 3^n - 1, 2^n + 3^n - 2, \\dots, 2^n + 3^n - k\n$$\nare all composite.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "For any given positive integer $k$, choose positive integer $m$ sufficiently large such that $2^m + 3^m - k > 1$. Consider the following $k$ integers:\n$$\n2^m + 3^m - 1, 2^m + 3^m - 2, \\dots, 2^m + 3^m - k,\n$$\nall of which are larger than $1$. From each of these integers, pick a prime factor: $p_1, p_2, \\dots, p_k$, and let\n$$\nn_t = m + t(p_1 - 1)(p_2 - 1)\\cdots(p_k - 1),\n$$\nwhere $t$ is an arbitrary positive integer. For any fixed integer $i$ ($1 \\le i \\le k$), one has $2^{n_t} \\equiv 2^m \\pmod{p_i}$. In fact, if $p_i = 2$, then the result is obvious. Assuming that $p_i \\ne 2$, it follows from Fermat's little theorem that\n$$\n2^{n_t} = 2^m \\cdot 2^{t(p_1-1)(p_2-1)\\cdots(p_k-1)} \\equiv 2^m \\cdot 1 = 2^m \\pmod{p_i}.\n$$\nSimilarly, one has $3^{n_t} \\equiv 3^m \\pmod{p_i}$. Observe that\n$$\n2^{n_t} + 3^{n_t} - i \\equiv 2^m + 3^m - i \\equiv 0 \\pmod{p_i},\n$$\n$$\n2^{n_t} + 3^{n_t} - i > 2^m + 3^m - i.\n$$\nHence, $2^{n_t} + 3^{n_t} - i$ is a composite number.\nTherefore, $n_t$ is one of the positive integers $n$ such that\n$$\n2^n + 3^n - 1, 2^n + 3^n - 2, \\dots, 2^n + 3^n - k\n$$\nare all composite. As $t$ is arbitrarily chosen, there are infinitely many such positive integers satisfying the conditions above.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 28239,
    "subject": "Mathematics (Multi-modal)",
    "question": "There are given 12 red points on a circle. Find the minimum of $n$, such that there exist $n$ triangles, whose vertices are red points, satisfying every chord with red endpoints being a side of one triangle. (Posed by Tao Pingsheng)",
    "options": [],
    "answer": "24",
    "solution": "Let the set of 12 red points be $A = \\{A_1, A_2, \\dots, A_{12}\\}$. From $A_1$, one can get 11 chords with red points, but every triangle with vertex $A_1$ has two such chords, and so the 11 chords should be in at least 6 triangles with $A_1$ being an endpoint. The same applies to $A_i$ ($i = 2, 3, \\dots, 12$); we need\n\n$12 \\times 6 = 72$ triangles, and every triangle has 3 points. So $n \\ge \\frac{72}{3} = 24$.\n\nOn the other hand, the following example shows that $n$ can be 24.\n\nConsider a circle whose perimeter is 12, and the 12 red points are on the circle with equal distance.\n\nThe number of chords with red endpoint is $\\binom{12}{2} = 66$. If the length of the minor arc to a chord is $k$, then say “the chord belongs to $k$.” So we have only six kinds of chords, and the number of chords belonging to 1, 2, ..., 5 is 12, and that belonging to 6 is 6.\n\n![](attached_image_1.png)\n\nIf three chords belonging to $a, b, c$ ($a \\le b \\le c$) can be the sides of a triangle, then $a + b = c$ or $a + b + c = 12$. So the triangles $(a, b, c) \\in \\{(1, 1, 2), (2, 2, 4), (3, 3, 6), (2, 5, 5), (1, 2, 3), (1, 3, 4), (1, 4, 5), (1, 5, 6), (2, 3, 5), (2, 4, 6), (3, 4, 5), (4, 4, 4)\\}$.\n\nNow we give an example:\n\nThe number of (1, 2, 3) triangles is 6, whose vertices are\n$\\{2, 3, 5\\}, \\{4, 5, 7\\}, \\{6, 7, 9\\}, \\{8, 9, 11\\}, \\{10, 11, 1\\}, \\{12, 1, 3\\}$.\n\nThe number of (1, 5, 6) triangles is 6, whose vertices are\n$\\{1, 2, 7\\}, \\{3, 4, 9\\}, \\{5, 6, 11\\}, \\{7, 8, 1\\}, \\{9, 10, 3\\}, \\{11, 12, 5\\}$.\n\nThe number of (2, 3, 5) triangles is 6, whose vertices are\n$\\{2, 4, 11\\}, \\{4, 6, 1\\}, \\{6, 8, 3\\}, \\{8, 10, 5\\}, \\{10, 12, 7\\}, \\{12, 2, 9\\}$.\n\nThe number of (4, 4, 4) triangles is 3, whose vertices are\n$\\{1, 5, 9\\}, \\{2, 6, 10\\}, \\{3, 7, 11\\}$.\n\nThe number of (2, 4, 6) triangles is 3, whose vertices are\n$\\{4, 6, 12\\}, \\{8, 10, 4\\}, \\{12, 2, 8\\}$.\n\nSo the minimum $n$ is 24.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 28240,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $n \\ge 3$ be a given integer, and $a_1, a_2, \\dots, a_n$ be real numbers satisfying $\\min_{1 \\le i < j \\le n} |a_i - a_j| = 1$. Find the minimum value of $\\sum_{k=1}^n |a_k|^3$. (Posed by Zhu Huawei)",
    "options": [],
    "answer": "Minimum equals (n^2 - 1)^2 / 32 for odd n, and n^2 (n^2 - 2) / 32 for even n; attained by a_i = i - (n + 1) / 2.",
    "solution": "Without loss of generality, we may assume that $a_1 < a_2 < \\dots < a_n$, and note also that\n$$\n|a_k| + |a_{n-k+1}| \\ge |a_{n-k+1} - a_k| \\ge n + 1 - 2k\n$$\nfor $1 \\le k \\le n$. So\n$$\n\\begin{align*}\n\\sum_{k=1}^{n} |a_k|^3 &= \\frac{1}{2} \\sum_{k=1}^{n} (|a_k|^3 + |a_{n+1-k}|^3) \\\\\n&= \\frac{1}{2} \\sum_{k=1}^{n} (|a_k|^3 + |a_{n+1-k}|) \\left( \\frac{3}{4} (|a_k| - |a_{n+1-k}|)^2 \\right. \\\\\n& \\qquad \\left. + \\frac{1}{4} (|a_k|^3 + |a_{n+1-k}|)^2 \\right) \\\\\n&\\ge \\frac{1}{8} \\sum_{k=1}^{n} (|a_k|^3 + |a_{n+1-k}|)^3 \\\\\n&\\ge \\frac{1}{8} \\sum_{k=1}^{n} |n+1-2k|^3.\n\\end{align*}\n$$\nWhen $n$ is odd,\n$$\n\\sum_{k=1}^{n} |n+1-2k|^3 = 2 \\times 2^3 \\times \\sum_{i=1}^{\\frac{n-1}{2}} i^3 = \\frac{1}{4}(n^2-1)^2.\n$$\nWhen $n$ is even,\n$$\n\\begin{aligned}\n\\sum_{k=1}^{n} |n+1-2k|^3 &= 2 \\sum_{i=1}^{\\frac{n}{2}} (2i-1)^3 \\\\\n&= 2 \\left( \\sum_{j=1}^{n} j^3 - \\sum_{i=1}^{\\frac{n}{2}} (2i)^3 \\right) \\\\\n&= \\frac{1}{4} n^2 (n^2 - 2).\n\\end{aligned}\n$$\nSo\n$$\n\\sum_{k=1}^{n} |a_k|^3 \\ge \\frac{1}{32}(n^2-1)^2\n$$\nfor odd $n$, and\n$$\n\\sum_{k=1}^{n} |a_k|^3 \\ge \\frac{1}{32} n^2 (n^2 - 2)\n$$\nfor even $n$. The equality holds at $a_i = i - \\frac{n+1}{2}$, $i = 1, 2, \\dots, n$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 28241,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $a$, $b$ be positive real numbers with $b - a > 2$. Prove that for any two distinct integers $m$, $n$ in the interval $[a, b)$, there is a nonempty set $S$ consisting of some integers in the interval $[ab, (a+1)(b+1))$, such that $\\frac{\\prod x}{mn}$ is a square of a rational number. (Posed by Yu Hongbing)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "We first prove the following lemma:\n\nLemma. Let $u$ be an integer with $a \\le u < u+1 < b$. Then there are two distinct integers $x$, $y$ in the interval $[ab, (a+1)(b+1))$, such that $\\frac{xy}{u(u+1)}$ is a square of an integer.\n\nProof of lemma. Let $v$ be the smallest integer not less than $\\frac{ab}{u}$, i.e. $v$ satisfies\n$$\n\\frac{ab}{u} \\le v < \\frac{ab}{u} + 1;\n$$\nhence\n$$\nab \\le uv < ab + u(ab + a + b + 1), \\quad \\textcircled{1}\n$$\nand thus\n$$\nab < (u+1)v = uv + v < ab + u + \\frac{ab}{u} + 1 < ab + a + b + 1 \\text{ (since } a \\le u < b\\text{).} \\quad \\textcircled{2}\n$$\n(Here we have used a well-known result: the function $f(t) = t + \\frac{ab}{t}$ ($a \\le t \\le b$) attains its maximum at $t = a$ or $b$.)\n\nBy ① and ②, we see that $uv$ and $(u+1)v$ are two distinct integers in the interval $I = [ab, (a+1)(b+1))$. Let $x = uv$ and $y = (u+1)v$. Then $\\frac{xy}{u(u+1)} = v^2$ is a square of an integer number. We have verified the lemma.\n\nGoing back to the original problem, suppose that $m < n$. Then $a \\le m \\le n-1 < b$. It follows from the lemma that for every $k = m, m+1, \\dots, n-1$, there exist $x_k$, $y_k$, two distinct integers in the interval $[ab, (a+1)(b+1))$, and integer $A_k$, such that\n$$\n\\frac{x_k y_k}{k(k+1)} = A_k^2.\n$$\nMultiplying all together, we find that\n$$\n\\frac{\\prod_{k=m}^{n-1} x_k y_k}{mn(m+1)^2 \\cdots (n-1)^2} = \\prod_{k=m}^{n-1} A_k^2\n$$\nis a square of an integer.\n\nLet $S$ be the set of numbers that appear in $x_i$, $y_i$ ($m \\le i \\le n-1$) odd times. If $S$ is nonempty, then it follows from the above equality that $\\frac{\\prod x_i}{mn}$ is a square of a rational number.\n\nIf $S$ is empty, then $mn$ is a square of an integer. Since $a + b > 2\\sqrt{ab}$, we have $ab + a + b + 1 > ab + 2\\sqrt{ab} + 1$, i.e. $\\sqrt{(a+1)(b+1)} > \\sqrt{ab} + 1$, which means that there is at least an integer in the interval $[\\sqrt{ab}, \\sqrt{(a+1)(b+1)})$. As a result there is a perfect square in the interval $[ab, (a+1)(b+1))$. Suppose that $r^2 \\in [ab, (a+1)(b+1))$ ($r \\in \\mathbb{Z}$), and let $S' = \\{r^2\\}$. Then $\\frac{\\prod_{x \\in S'} x}{mn}$ is a square of a rational number.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 28242,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $n$ be an integer greater than $3$. The points $V_1, V_2, \\dots, V_n$, with no three collinear, lie on the plane. Some of the segments $V_iV_j$, with $1 \\le i < j \\le n$, are constructed. The points $V_i$ and $V_j$ are neighbors if $V_iV_j$ is constructed. Initially, the chess pieces $C_1, C_2, \\dots, C_n$ are placed at the points $V_1, V_2, \\dots, V_n$ (not necessarily in that order), with exactly one piece at each point. In a move, one can choose some of the $n$ chess pieces, and simultaneously relocate each of the chosen piece from its current position to one of its neighboring positions such that after the move, exactly one chess piece is at each point and no two chess pieces have exchanged their positions. A set of constructed segments is called *harmonic* if for any initial positions of the chess pieces each chess piece $C_i$ ($1 \\le i \\le n$) is at the point $V_i$ after a finite number of moves.\n\nDetermine the minimum number of segments in a harmonic set.",
    "options": [],
    "answer": "n+1",
    "solution": "The answer is $n+1$.\n\nFor a harmonic set, we consider a graph $G$ with $V_1, V_2, \\dots, V_n$ as its vertices and with the segments in the harmonic set as its edges.\n\nFirst, we show that there are at least $n$ edges in $G$. Note that $G$ must be connected. Also note that each vertex must have degree at least $2$, because when a chess piece is moved from $V_i$ to $V_j$ there is another piece moved from $V_k$ (with $k \\neq j$) to $V_i$. Hence, the total degree is at least $2n$, from which it follows that there are at least $2n = 2 = n$ edges.\n\nSecond, we show that there are at least $n+1$ edges. Assume that there are only $n$ edges. In this connected graph, each vertex has exactly degree $2$, and hence it must be a complete cycle. Without loss of generality, we may assume that the cycle $V_1 \\to V_2 \\to \\dots \\to V_n \\to V_1$ consists of all the edges. In this case, if $C_1$ and $C_2$ are placed at $V_2$ and $V_1$ initially, we cannot put them back to $V_1$ and $V_2$ simultaneously. This is because we can only rotate all the pieces along the cycle and cannot change their relative positions along the cycle.\n\nThird, we show that $n+1$ edges is enough. We consider the graph $G$ with the cycle $C_1: V_1 \\to V_2 \\to \\dots \\to V_n \\to V_1$ and one additional edge, $V_2V_n$. (This graph $G$ now has the second cycle $C_2: V_2 \\to V_3 \\to \\dots \\to V_n \\to V_2$.) With this additional edge, we can switch the relative positions of the chess pieces along the cycle $C_1$. Indeed, without loss of generality, we may assume that $C_i$ is at $V_1$ and $C_j$ is at $V_2$ initially. Applying rotations on the cycle $C_2$, we can place $C_j$ at $V_n$, i.e. the relative positions of $C_i$ and $C_j$, along $C_1$, are switched. Because we can switch the positions of any two neighboring pieces in a finite amount of moves, we can place $C_1, C_2, \\dots, C_n$ in that order on the cycle $C_1$. We can then move each $C_i$ to $V_i$ by applying rotations along the cycle $C_1$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 28243,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all integers $n$ such that we can color all the edges and the diagonals of a convex $n$-polygon by $n$ given colors satisfying the following conditions;\n(1) Each of the edges or the diagonals is colored by only one color;\n(2) For any three distinct colors, there exists a triangle whose vertices are vertices of the $n$-polygon and three edges are colored by these three colors.",
    "options": [],
    "answer": "All odd integers greater than 1",
    "solution": "Answer: Any odd number $n > 1$.\n\nFirst of all, there are $\\binom{n}{3}$ ways to choose three among $n$ colors, and $\\binom{n}{3}$ ways to choose three vertices to form a triangle, so if the question's condition is fulfilled, all the triangles should have a different color combination (a 1-1 correspondence).\n\nNote that any two segments of the same color cannot have a common endpoint.\n\nAs each color combination is used in exactly one triangle, for each color there should be exactly $\\binom{n-1}{2}$ triangles which have one side in this color, and so there should be exactly $\\frac{n-1}{2}$ lines of this color. Therefore, $n$ is odd.\n\nNow we give a construction method for any odd $n$.\n\nAs the orientation of the vertices does not really matter, we may assume that the polygon is a regular $n$-polygon. First, color the $n$ sides of the polygon in the $n$ distinct colors. Then, for each side, color those diagonals that are parallel to this side into the same color.\n\nIn this way, for each color, there are $n$ diagonals colored in this color, and notice that each of these diagonals is of a different length. $\\textcircled{1}$\n\nBesides, for any two triangles with all vertices in, we shall prove that they should have different color combinations.\n\nSuppose that, on the contrary, they have exactly the same three colors as their sides. Because of that all the sides with the same line are parallel, and the two triangles must be similar. For their vertices to be in the same circle, they must be the same, but this is a contradiction of $\\textcircled{1}$. This completes the proof.",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 28244,
    "subject": "Mathematics (Multi-modal)",
    "question": "Given points $P$, $Q$ on an ellipse $\\frac{x^2}{a^2} + \\frac{y^2}{b^2} = 1$ ($a > b > 0$), satisfying $OP \\perp OQ$, the minimum of $|OP| \\times |OQ|$ is ____.",
    "options": [],
    "answer": "2a^2b^2/(a^2+b^2)",
    "solution": "Define\n$$\nP(|OP| \\cos \\theta, |OP| \\sin \\theta), \\\\\nQ(|OQ| \\cos(\\theta \\pm \\frac{\\pi}{2}), |OQ| \\sin(\\theta \\pm \\frac{\\pi}{2})).\n$$\nWe have\n$$\n\\frac{1}{|OP|^2} = \\frac{\\cos^2\\theta}{a^2} + \\frac{\\sin^2\\theta}{b^2}, \\qquad \\textcircled{1}\n$$\n$$\n\\frac{1}{|OQ|^2} = \\frac{\\sin^2\\theta}{a^2} + \\frac{\\cos^2\\theta}{b^2}. \\qquad \\textcircled{2}\n$$\nThen\n$$\n\\frac{1}{|OP|^2} + \\frac{1}{|OQ|^2} = \\frac{1}{a^2} + \\frac{1}{b^2}.\n$$\nTherefore, $|OP| \\times |OQ|$ reaches the minimum $\\frac{2a^2b^2}{a^2+b^2}$ when $|OP| = |OQ| = \\sqrt{\\frac{2a^2b^2}{a^2+b^2}}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 28245,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $m$ be an integer greater than $1$, and let $n$ be an odd number with $3 \\le n < 2m$. Numbers $a_{i,j}$ ($i, j \\in \\mathbb{N}$, $1 \\le i \\le m$, $1 \\le j \\le n$) satisfy:\n(1) For every $1 \\le j \\le n$, $a_{1,j}$, $a_{2,j}$, ..., $a_{m,j}$ is a permutation of $1, 2, ..., m$;\n(2) $|a_{i,j} - a_{i,j+1}| \\le 1$ for every $1 \\le i \\le m$, $1 \\le j \\le n-1$.\nFind the minimal possible value of $M = \\max_{1 \\le i \\le m} \\sum_{j=1}^{n} a_{i,j}$.",
    "options": [],
    "answer": "(2l+1)m - l^2 where n = 2l + 1",
    "solution": "Let $n = 2l + 1$. Since $3 \\le n < 2m$, we have $1 \\le l \\le m - 1$. We first estimate the lower bound of $M$.\nBy the condition (1), there exists a unique $1 \\le i_0 \\le m$, such that $a_{i_0, l+1} = m$. Consider $a_{i_0, l}$ and $a_{i_0, l+2}$.\n\nCase 1: At least one of $a_{i_0, l}$ and $a_{i_0, l+2}$ is $m$, and we may assume without loss of generality that $a_{i_0, l}=m$. It follows from the condition (2) that\n$$\n\\begin{aligned}\na_{i_0, l-1} &\\ge m-1,\\quad a_{i_0, l-2} \\ge m-2,\\quad \\dots,\\quad a_{i_0, 1} \\ge m-l+1, \\\\\na_{i_0, l+2} &\\ge m-1,\\quad a_{i_0, l+3} \\ge m-2,\\quad \\dots,\\quad a_{i_0, 2l+1} \\ge m-l.\n\\end{aligned}\n$$\nThus,\n$$\n\\begin{aligned}\nM &\\ge \\sum_{j=1}^{n} a_{i_0, j} \\\\\n&\\ge (m-l) + 2((m-l+1) + (m-l+2) + \\dots + m) \\\\\n&= (2l+1)m - l^2.\n\\end{aligned}\n$$\n\nCase 2: None of $a_{i_0, l}$ and $a_{i_0, l+2}$ is $m$, and by the condition (1) there exists $1 \\le i_1 \\le m$, $i_1 \\ne i_0$, such that $a_{i_1, l} = m$. It easily follows from the conditions (1) and (2) that $a_{i_1, l+1} = m-1$, $a_{i_1, l+2} = m$. It follows again from the condition (2) that\n$$\n\\begin{aligned}\na_{i_1, l} &\\ge m-1,\\quad a_{i_1, l-1} \\ge m-2,\\quad \\dots,\\quad a_{i_1, l-(l-1)} \\ge m-l+1, \\\\\na_{i_1, l+2} &\\ge m-1,\\quad a_{i_1, l+3} \\ge m-2,\\quad \\dots,\\quad a_{i_1, l+1} \\ge m-l+1.\n\\end{aligned}\n$$\nThus,\n$$\n\\begin{aligned}\nM &\\ge \\sum_{j=1}^{n} a_{i_1, j} \\ge 2((m-l+1)+(m-l+2)+\\dots+(m-1)) \\\\\n&= (2l+1)m - (l^2-l+1).\n\\end{aligned}\n$$\nCombining the above two cases, we have $M \\ge (2l+1)m - l^2$.\n\nOn the other hand, consider\n$$\na_{i,j} = f(2i+j) = \\begin{cases}\n2i+j, & 2i+j \\le m, \\\\\n(2m+1)-(2i+j), & m+1 \\le 2i+j \\le 2m, \\\\\n(2i+j)-2m, & 2m+1 \\le 2i+j \\le 3m, \\\\\n(4m+1)-(2i+j), & 3m+1 \\le 2i+j \\le 4m.\n\\end{cases}\n$$\nWe shall show that this table of numbers satisfies the required conditions in the problem.\nFor any $1 \\le i \\le m$, $1 \\le j \\le n-1$, if $m \\ne 2i+j$, then\n$$\n|a_{i,j} - a_{i,j+1}| = |f(2i+j) - f(2i+j+1)| = 1;\n$$\nif $m = 2i+j$, then\n$$\n|a_{i,j} - a_{i,j+1}| = |f(2i+j) - f(2i+j+1)| = 0.\n$$\nThe condition (2) is verified.\n\nWe next show that the condition (1) is also fulfilled. In fact, it suffices to show that for any integers $1 \\le j \\le n$ and $1 \\le k \\le m$, there exists an integer $1 \\le i \\le m$ such that $a_{i,j} = k$.\nIf $j \\equiv k \\pmod 2$, since $1 \\le j \\le n$, $1 \\le k \\le m$ and $n < 2m$,\nwe have $-2m < k-j < m$, and therefore $-m < \\frac{k-j}{2} < m$, and note that $\\frac{k-j}{2}$ is an integer. Thus, at least one of $\\frac{k-j}{2}$ and $\\frac{k-j}{2} + m$ is in the set $\\{1, 2, \\dots, m\\}$; taking this number to be $i$ will do the job.\nIf $j \\not\\equiv k \\pmod 2$, again since $1 \\le j \\le n$, $1 \\le k \\le m$ and $n < 2m$, we have\n$$\n-2m < (2m+1) - (j+k) < 2m,\n$$\nand therefore\n$$\n-m < \\frac{(2m+1) - (j+k)}{2} < m,\n$$\nand $\\frac{(2m+1)-(j+k)}{2}$ is an integer. Thus, at least one of $\\frac{(2m+1)-(j+k)}{2}$ and $\\frac{(2m+1)-(j+k)}{2} + m$ is in the set $\\{1, 2, \\dots, m\\}$; taking this number to be $i$ will do the job.\n\nWe now estimate $M$ in this case. Since the condition (1) holds, for any $1 \\le i_1 < i_2 \\le m$, $1 \\le j \\le n-1$, we have $f(2i_1 + j) \\not\\equiv f(2i_2 + j)$, i.e. for any integers $x, y$ with the same parity such that $3 \\le x < y \\le 2m+n$ and $y-x < 2m$, we have $f(x) \\ne f(y)$. Thus, for a given $i$, $a_{i,1}, a_{i,2}, \\dots, a_{i,2l+1}$ are pairwise distinct, and so are numbers $a_{i,2}, a_{i,3}, \\dots, a_{i,2l}$. As a result,\n$$\n\\begin{aligned}\n\\sum_{j=1}^{n} a_{i,j} &\\le (m-l) + 2((m-l+1) + (m-l+2) + \\dots + m) \\\\\n&= (2l+1)m - l^2.\n\\end{aligned}\n$$\nThus, $M = \\max \\sum_{1 \\le i \\le m} a_{i,j} \\le (2l+1)m - l^2$.\n\nTherefore, the minimal possible value of $M$ is $(2l+1)m - l^2$ where $n = 2l+1$.",
    "topic": "Discrete Mathematics",
    "subtopic": "Combinatorics"
  },
  {
    "id": 28246,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $\\{x_n\\}$ be a sequence such that $x_1 \\in \\{5, 7\\}$, and $x_{n+1} \\in \\{5^{x_n}, 7^{x_n}\\}$, for $n = 1, 2, \\dots$. Determine all the possible cases of the last two digits of $x_{2009}$.",
    "options": [],
    "answer": "07, 25, 43",
    "solution": "Let $n = 2009$. Then we have the following three cases:\n\n(1) If $x_n = 7^{5^{x_{n-2}}}$, then $x_n \\equiv 7 \\pmod{100}$.\nSince both $5$ and $7$ are odd, $x_k$ ($1 \\le k \\le n$) all are odd. $5^{x_{n-2}} \\equiv 1 \\pmod{4}$, i.e. $5^{x_{n-2}} = 4k+1$ for some positive integer $k$.\nIf $k, m \\ge 0$, it follows from mathematical induction that\n$$\n7^{4k+m} \\equiv 7^m \\pmod{100},\n$$\nand hence\n$$\nx_n = 7^{5^{x_{n-2}}} = 7^{4k+1} \\equiv 7 \\pmod{100}.\n$$\n\n(2) If $x_n = 7^{7^{x_{n-2}}}$, then $x_n \\equiv 43 \\pmod{100}$.\n$$\n7^{x_{n-2}} \\equiv (-1)^{x_{n-2}} \\equiv -1 \\equiv 3 \\pmod{4},\n$$\nso $7^{x_{n-2}} = 4k+3$, for some positive integer $k$. Hence,\n$$\nx_n = 7^{7^{x_{n-2}}} = 7^{4k+3} \\equiv 7^3 \\equiv 43 \\pmod{100}.\n$$\n\n(3) If $x_n = 5^{x_{n-1}}$, then $x_n \\equiv 25 \\pmod{100}$.\nIt follows from mathematical induction that if $n \\ge 2$, then $5^n \\equiv 25 \\pmod{100}$. It is obvious that $x_{n-1} > 2$, and hence $x_n = 5^{x_{n-1}} \\equiv 25 \\pmod{100}$.\n\nTherefore, the possible cases of the last two digits of $x_{2009}$ are $07$, $25$, $43$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 28247,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $D$ be a point on the side $BC$ of an acute triangle $ABC$. The circle with diameter $BD$ meets the lines $AB$ and $AD$ respectively at the points $X$ and $P$, which are different from the points $B$ and $D$. The circle with diameter $CD$ meets the lines $AC$ and $AD$ respectively at the points $Y$ and $Q$, which are different from the points $C$ and $D$. Through the point $A$ draw two lines which are perpendicular to $PX$ and $QY$ with the feet of perpendicular $M$ and $N$ respectively.\nProve that $\\triangle AMN \\sim \\triangle ABC$ if and only if the line $AD$ passes through the circumcenter of $\\triangle ABC$. (Posed by Li Qiusheng)\n![](attached_image_1.png)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "**Proof** Join the segments $XY$ and $DX$. It follows from the given conditions that $B$, $P$, $D$, $X$ are concyclic, and $C$, $Y$, $Q$, $D$ are concyclic. Then\n$$\n\\angle AXM = \\angle BXP = \\angle BDP \\\\\n= \\angle QDC = \\angle AYN.\n$$\n![](attached_image_2.png)\nAnd it follows from $\\angle AMX = \\angle ANY = 90^\\circ$ that $\\triangle AMC \\sim \\triangle ANY$, so $\\angle MAX = \\angle NAY$ and $\\frac{AM}{AX} = \\frac{AN}{AY}$, and hence $\\angle MAN = \\angle XAY$.\n\nCombining the two results above, we have $\\triangle AMN \\sim \\triangle AXY$. So we get\n$$\n\\begin{aligned}\n\\triangle AMN \\sim \\triangle ABC &\\Leftrightarrow \\triangle AXY \\sim \\triangle ABC \\\\\n&\\Leftrightarrow XY \\parallel BC \\Leftrightarrow \\angle DXY = \\angle XDB.\n\\end{aligned}\n$$\n---\nAs $A$, $X$, $D$, $Y$ are concyclic, we have $\\angle DXY = \\angle DAY$.\nAs $\\angle XDB = 90^\\circ - \\angle ABC$,\n$$\n\\angle DXY = \\angle XDB \\iff \\angle DAC = 90^\\circ - \\angle ABC,\n$$\nwhich is equivalent to the fact that the line $AD$ passes through the circumcenter of $\\triangle ABC$.\nHence, $\\triangle AMN \\sim \\triangle ABC$ if and only if $AD$ passes through the circumcenter of $\\triangle ABC$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 28248,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $O, I$ be the circumcenter and incenter of $\\triangle ABC$. Prove that, for an arbitrary point $D$ on the circle $O$, one can construct a triangle $DEF$, such that $O, I$ are the circumcenter and incenter of $\\triangle DEF$. (Posed by Tao Pingsheng)\n![](attached_image_1.png)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "As shown in Fig. 2, let $OI = d$, $R$, $r$ be the circumradius and incircle radius of $\\triangle ABC$. The point $K$ is the intersection of $AI$ and the circle $O$; then\n$$\nKI = KB = 2R \\sin \\frac{\\angle BAC}{2},\n$$\n$$\nAI = \\frac{r}{\\sin \\frac{\\angle BAC}{2}}.\n$$\n![](attached_image_2.png)\nFig. 2\nLet the points $M, N$ be the intersections of the line $OI$ and the circle $O$; then\n$$\n(R+d)(R-d) = IM \\times IN = AI \\times KI = 2Rr,\n$$\ni.e. $R^2 - d^2 = 2Rr$.\n\nNow draw the tangents $DE, DF$ from $D$ to the circle $I$. The points $E, F$ are on the circle $O$. Then $DI$ is the bisector of $\\angle EDF$. It is enough to prove that $EF$ is tangent to the circle $I$.\n\nLet $P$ be the intersection of the line $DI$ and the circle $O$. Then $P$ is the midpoint of the arc $EF$, and\n$$\nPE = 2R \\sin \\frac{\\angle EDF}{2}, \\quad DI = \\frac{r}{\\sin \\frac{\\angle EDF}{2}},\n$$\n$$\nID \\cdot IP = IM \\cdot IN = (R+d)(R-d) = R^2 - d^2,\n$$\nand so\n$$\nPI = \\frac{R^2 - d^2}{DI} = \\frac{R^2 - d^2}{r} \\cdot \\sin \\frac{\\angle EDF}{2} = 2R \\sin \\frac{\\angle EDF}{2} = PE.\n$$\nAs $I$ is on the bisector of $\\angle EDF$, we can see that $I$ is the incenter of $\\triangle DEF$ [because $\\angle PEI = \\angle PIE = \\frac{1}{2}(180^\\circ - \\angle EPD) = \\frac{1}{2}(180^\\circ - \\angle DFE) = \\frac{\\angle EDF + \\angle DEF}{2}$, and $\\angle PEF = \\frac{\\angle EDF}{2}$; so $\\angle FEI = \\frac{\\angle DEF}{2}$]. $EF$ is tangent to the circle $I$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 28249,
    "subject": "Mathematics (Multi-modal)",
    "question": "Given an integer $n \\ge 3$, prove that there exists a set $S$ of $n$ distinct positive integers such that for any two distinct nonempty subsets $A$ and $B$ of $S$, the numbers\n$$\n\\sum_{x \\in A} x, \\quad \\sum_{x \\in B} x\n$$\nare two coprime composite integers, and $\\sum_{x \\in X} x$ denotes the sum of all elements of a finite set $X$, and $|X|$ denotes the cardinality of $X$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $f(X)$ be the average of elements of the finite number set $X$.\n\nFirst of all, make $n$ different primes $p_1, p_2, \\dots, p_n$ which are all bigger than $n$, and we prove that for any different nonempty subsets $A$, $B$ of the set $S_1 = \\{ \\prod (1/p_j) : 1 \\le j \\le n \\}$, $f(A) \\ne f(B)$ always holds.\n\nIn fact, we can suppose that $\\prod_{i=1}^{n} p_i \\in A$ and $\\prod_{i=1}^{n} p_i \\notin B$ without loss of generality. Every element of $B$ can be divided by $p_1$, so $p_1 \\mid n! f(B)$. But $A$ has exactly one element which cannot be divided by $p_1$, so we find that $n! f(A)$ cannot be divided by $p_1$ (note that $p_1 > n$), and therefore $n!f(A) \\ne n!f(B)$, which follows $f(A) \\ne f(B)$.\n\nSecond, let $S_2 = \\{ n! x : x \\in S_1 \\}$. Then $f(A)$ and $f(B)$ are different positive integers when $A, B$ are different nonempty subsets of $S_2$.\n\nIn fact, it is easy to see that there exist two sets $A_1, B_1$ which are different nonempty subsets of $S_1$, and $f(A) = n! f(A_1)$, $f(B) = n! f(B_1)$ holds. We get $f(A) \\ne f(B)$ from $f(A_1) \\ne f(B_1)$, and $f(A), f(B)$ are positive integers from $|A|, |B| \\le n$ and their elements are all positive.\n\nThen, let $K$ be the largest element of $S_2$. We prove that for every two distinct subsets $A, B$ of the set $S_3 = \\{ K! x + 1 : x \\in S_2 \\}$, $f(A)$ and $f(B)$ are coprime integers which are both larger than $1$.\n\nIn fact, it is easy to see that there exist two sets $A_1, B_1$ which are different nonempty subsets of $S_2$, and $f(A) = K! f(A_1) + 1$, $f(B) = K! f(B_1) + 1$ holds. Obviously, $f(A)$ and $f(B)$ are different integers which are both larger than $1$. If they have common divisors, let $p$ be a prime common divisor of them without loss of generality. Clearly, we have $p \\mid (K! \\cdot |f(A_1) - f(B_1)|)$. We get $1 \\le |f(A_1) - f(B_1)| \\le K$ by $0 < f(A_1), f(B_1) \\le K$ and $f(A_1) \\ne f(B_1)$, so $p \\le K$, which follows $p \\mid K! f(A_1)$, and then $p \\mid 1$ — a contradiction.\n\nLastly, let $L$ be the largest element of $S_3$. We prove that for every two distinct nonempty subsets $A, B$ of the set $S_4 = \\{ L! + x : x \\in S_3 \\}$, $f(A)$ and $f(B)$ are two composites which share no common divisors.\n\nIn fact, it is easy to see that there exist two sets $A_1, B_1$ which are different nonempty subsets of $S_3$, and $f(A) = L! + f(A_1)$, $f(B) = L! + f(B_1)$ holds. Obviously, $f(A)$ and $f(B)$ are different integers which are both larger than $1$. Because of that, $L$ is the largest element of $S_3$, and we have $f(A_1) \\mid L!$ and $f(A_1) \\mid f(A)$. We find that $f(A)$ is composite by $f(A_1) < f(A)$. For a similar reason, $f(B)$ is composite too. If they have common divisors, let $p$ be a prime common divisor of them without loss of generality. It is obvious that $p \\mid (L! \\cdot |f(A_1) - f(B_1)|)$. We get $1 \\le |f(A_1) - f(B_1)| \\le L$ by $0 < f(A_1), f(B_1) \\le L$ and $f(A_1) \\ne f(B_1)$, so $p \\le L$, which follows $p \\mid f(A_1)$ and $p \\mid f(B_1)$ — a contradiction of the fact that $f(A_1)$ and $f(B_1)$ are coprime. This completes the proof.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 28250,
    "subject": "Mathematics (Multi-modal)",
    "question": "The circle $\\Gamma_1$, with radius $r$, is internally tangent to the circle $\\Gamma_2$ at $S$. The chord $AB$ of $\\Gamma_2$ is tangent to $\\Gamma_2$ at $C$. Let $M$ be the midpoint of the arc $\\widehat{AB}$ (not containing $S$), and let $N$ be the foot of the perpendicular from $M$ to the line $AB$. Prove that $AC \\times CB = 2r \\times MN$. (Posed by Ye Zhonghao)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "![](attached_image_1.png)\n\n**Solution** It is well known that $S, C, M$ are collinear. Indeed, consider the dilation centered at $S$ that sends $\\Gamma_1$ to $\\Gamma_2$. Then the line $AB$ is sent to the line $l$ parallel to $AB$ and tangent to $\\Gamma_2$, i.e. the line tangent to $\\Gamma_2$ at $M$ (the midpoint of $\\widehat{AB}$). Thus, this dilation sends $C$ (the points of tangency of the line $AB$ and $\\Gamma_1$) to $M$ (the points of tangency of the line $l$ and $\\Gamma_2$), from which it follows that $S, C, M$ are collinear.\n\nBy the power-of-point theorem, we have $AC \\times CB = SC \\times CM$. It suffices to show that\n$$\nSC \\times CM = 2r \\times MN \\quad \\text{or} \\quad \\frac{SC}{2r} = \\frac{MN}{CM}. \\qquad \\textcircled{1}\n$$\nSet $\\angle MCN = \\alpha$. Then $\\angle SCA = \\alpha$. By the extended sine law, we have $\\frac{SC}{2r} = \\sin \\alpha$. In the right triangle $MNC$, we also have $\\sin \\alpha = \\frac{MN}{CM}$. Combining the last two equations, we obtain ①.\n\n(We can also derive ① by observing that the triangles $MNC$ and $CDS$ are similar.)",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 28251,
    "subject": "Mathematics (Multi-modal)",
    "question": "Suppose that the equation $\\lg kx = 2\\lg(x+1)$ has exactly one real root. Then the range of $k$ is ____.",
    "options": [],
    "answer": "k < 0 or k = 4",
    "solution": "We have\n$$\nkx > 0, \\qquad \\textcircled{1}\n$$\n$$\nx + 1 > 0, \\qquad \\textcircled{2}\n$$\n$$\nkx = (x + 1)^2. \\qquad \\textcircled{3}\n$$\nThe expression ③ can be standardized as\n$$\nx^2 + (2 - k)x + 1 = 0. \\qquad \\textcircled{4}\n$$\nThe two roots of ④ are\n$$\nx_1, x_2 = \\frac{1}{2}[k - 2 \\pm \\sqrt{k^2 - 4k}], \\quad ⑤\n$$\nwhere\n$$\n\\Delta = k^2 - 4k \\ge 0 \\Leftrightarrow k \\le 0 \\text{ or } k \\ge 4.\n$$\n(i) When $k < 0$, it is easy to see from ⑤ that $x_1 + 1 > 0$, $x_2 + 1 < 0$, and $kx_1 > 0$. Then the equation has one real root, $x_1 = \\frac{1}{2}[k - 2 + \\sqrt{k^2 - 4k}]$.\n\n(ii) When $k = 4$, the equation has one real root, $x = \\frac{4}{2} - 1 = 1$.\n\n(iii) When $k > 4$, the two roots $x_1, x_2$ are both positive, as well as $x_1 \\ne x_2$. Discarded.\n\nTherefore, the range of $k$ is $k < 0$, $k = 4$.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 28252,
    "subject": "Mathematics (Multi-modal)",
    "question": "Prove that in an arithmetic progression consisting of 40 distinct positive integers, at least one of the numbers cannot be written as $2^k + 3^l$, where $k, l$ are nonnegative integers. (Posed by Chen Yonggao)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Suppose on the contrary that there exist 40 distinct positive integers in arithmetic progression such that each term can be written as $2^k + 3^l$, and denote this sequence by $a, a+d, a+2d, \\ldots, a+39d$, where $a, d$ are positive integers. Let\n$$\nm = \\lfloor \\log_2 (a + 39d) \\rfloor, \\quad n = \\lfloor \\log_3 (a + 39d) \\rfloor.\n$$\nIn what follows, we first show that at most one of $a+26d, a+27d, \\ldots, a+39d$ cannot be written as $2^m + 3^l$ or $2^k + 3^n$ ($k, l$ are nonnegative integers).\nSuppose that $a+hd$ cannot be written as $2^m + 3^l$ or $2^k + 3^n$, for some $26 \\le h \\le 39$. Then, by assumption, $a+hd = 2^b + 3^c$ for some nonnegative integers $b, c$. By the definition of $m$ and $n$, it is clear that $b \\le m, c \\le n$. Since $a+hd$ cannot be written as $2^m + 3^l$ or $2^k + 3^n$, we have $b \\le m-1, c \\le n-1$.\nIf $b \\le m-2$, then\n$$\n\\begin{aligned}\na + hd &\\le 2^{m-2} + 3^{n-1} = \\frac{1}{4} \\times 2^m + \\frac{1}{3} \\times 3^n \\\\\n&\\le \\frac{7}{12} \\times (a + 39d) < a + 26d,\n\\end{aligned}\n$$\na contradiction.\nIf $c \\le n-2$, then\n$$\n\\begin{aligned}\na + hd &\\le 2^{m-1} + 3^{n-2} = \\frac{1}{2} \\times 2^m + \\frac{1}{9} \\times 3^n \\\\\n&\\le \\frac{11}{18} \\times (a + 39d) < a + 26d,\n\\end{aligned}\n$$\nalso a contradiction.\nIt follows that $b = m - 1, c = n - 1$, which implies that at most one of $a + 26d, a + 27d, \\dots, a + 39d$ cannot be written as $2^m + 3^l$ or $2^k + 3^n$.\nIn these 14 numbers, at least 13 numbers can be written as $2^m + 3^l$ or $2^k + 3^n$. By the pigeonhole principle, at least 7 numbers can be written in the same form. We shall discuss two cases.\n**Case 1:** There are 7 numbers in the form of $2^m + 3^l$, denoted by\n$$\n2^m + 3^{l_1}, 2^m + 3^{l_2}, \\dots, 2^m + 3^{l_7},\n$$\nwhere $l_1 < l_2 < \\dots < l_7$. Thus, $3^{l_1}, 3^{l_2}, \\dots, 3^{l_7}$ are the 7 terms of an arithmetic progression with 14 terms and the common difference $d$. However,\n$$\n13d \\ge 3^{l_7} - 3^{l_1} \\ge (3^5 - \\frac{1}{3}) \\times 3^{l_2} > 13(3^{l_2} - 3^{l_1}) \\ge 13d,\n$$\na contradiction.\n**Case 2:** There are 7 numbers in the form of $2^k + 3^n$, denoted by\n$$\n2^{k_1} + 3^n, 2^{k_2} + 3^n, \\dots, 2^{k_7} + 3^n,\n$$\nwhere $k_1 < k_2 < \\dots < k_7$. Thus $2^{k_1}, 2^{k_2}, \\dots, 2^{k_7}$ are the 7 terms of an arithmetic progression with 14 terms and the common difference $d$. However,\n$$\n13d \\ge 2^{k_7} - 2^{k_1} \\ge (2^5 - \\frac{1}{2}) \\times 2^{k_2} > 13(2^{k_2} - 2^{k_1}) \\ge 13d,\n$$\na contradiction.\nIt follows from the above arguments that our assumption at the very beginning is false, which completes the proof.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 28253,
    "subject": "Mathematics (Multi-modal)",
    "question": "On a $10 \\times 10$ chessboard, some $4n$ unit square fields are chosen to form a region $R$. This region $R$ can be tiled by $n$ $2 \\times 2$ squares. If $R$ can also be tiled by a combination of $n$ pieces of the following types of shapes (with rotations allowed).\n![](attached_image_1.png)\n![](attached_image_2.png)\nDetermine the minimum value of $n$. (Posed by Zhu Huawei)",
    "options": [],
    "answer": "4",
    "solution": "The answer is $n = 4$. We call those two kinds of nonsquare tiles *ducks*.\nFirst, the left-hand-side figure and the middle figure below show that $n = 4$ works.\n![](attached_image_3.png)\n![](attached_image_4.png)\n![](attached_image_5.png)\n\nSecond, we show that $n$ must be even. We mark the (infinite) chessboard with $\\times$ in the pattern indicated in the right-hand-side figure. It is easy to see that each $2 \\times 2$ covers exactly an even number of crosses (either two or four crosses) and each duck covers exactly an odd number of crosses (either one or three crosses). It follows that we must have an even number of ducks in $R$, i.e. $n$ is even.\n\nThird, we show that $n \\ge 2$. If $n = 2$, then $R$ can be tiled by two $2 \\times 2$ squares. It is clear that these two squares much share a common edge. Hence, we can have only two possibilities:",
    "topic": "Discrete Mathematics",
    "subtopic": "Combinatorics"
  },
  {
    "id": 28254,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $f(x, y, z) = \\frac{x(2y-z)}{1+x+3y} + \\frac{y(2z-x)}{1+y+3z} + \\frac{z(2x-y)}{1+z+3x}$, where $x, y, z \\ge 0$, and $x+y+z = 1$. Find the maximum value and the minimum value of $f(x, y, z)$. (Posed by Li Shenghong)",
    "options": [],
    "answer": "maximum 1/7, minimum 0",
    "solution": "First, prove that $f \\le \\frac{1}{7}$; when $x = y = z = \\frac{1}{3}$, we have $f = \\frac{1}{7}$.\nSince $f = \\sum \\frac{x(x+3y-1)}{1+x+3y} = 1 - 2\\sum \\frac{x}{1+x+3y}$, by Cauchy's inequality\n$$\n\\sum \\frac{x}{1+x+3y} \\ge \\frac{\\left(\\sum x\\right)^2}{\\sum x(1+x+3y)} = \\frac{1}{\\sum x(1+x+3y)},\n$$\nand\n$$\n\\sum x(1+x+3y) - \\sum x(2x+4y+z) = 2 + \\sum xy \\le \\frac{7}{3}.\n$$\nSo $\\sum \\frac{x}{1+x+3y} \\ge \\frac{3}{7}$, $f \\le 1 - 2 \\times \\frac{3}{7} = \\frac{1}{7}$; $f_{max} = \\frac{1}{7}$; when $x = y = z = \\frac{1}{3}$, we have $f = \\frac{1}{7}$.\n\nSecond, prove that $f \\ge 0$; when $x = 1, y = z = 0$, we have $f = 0$.\nIn fact, one can see that\n$$\n\\begin{aligned}\nf(x, y, z) &= \\frac{x(2y-z)}{1+x+3y} + \\frac{y(2z-x)}{1+y+3z} + \\frac{z(2x-y)}{1+z+3x} \\\\\n&= xy \\left( \\frac{2}{1+x+3y} - \\frac{1}{1+y+3z} \\right) \\\\\n&\\quad + xz \\left( \\frac{2}{1+z+3x} - \\frac{1}{1+x+3y} \\right) \\\\\n&\\quad + yz \\left( \\frac{2}{1+y+3z} - \\frac{1}{1+z+3x} \\right) \\\\\n&= \\frac{7xyz}{(1+x+3y)(1+y+3z)} \\\\\n&\\quad + \\frac{7xyz}{(1+z+3x)(1+x+3y)} \\\\\n&\\quad + \\frac{7xyz}{(1+y+3z)(1+z+3x)} \\\\\n&\\ge 0.\n\\end{aligned}\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 28255,
    "subject": "Mathematics (Multi-modal)",
    "question": "There are $n$ ($n > 12$) students participating in a mathematics contest. The examination paper consists of 15 fill-in-the-blank questions. For each question, the score of a correct answer is 1 point, and no point will be awarded if the answer is wrong or left blank. After analyzing all the possible cases of score distributions of these $n$ students, one finds out that if the sum of total scores of any 12 students is not less than 36 points, then there are at least 3 students among these $n$ students who answer at least 3 identical questions correctly. Determine the smallest possible value of $n$. (Posed by Liu Shixiong)",
    "options": [],
    "answer": "911",
    "solution": "The smallest $n$ is 911. We divide the proof into two parts:\n\n(1) We first prove that $n = 911$ satisfies the conditions. If each student answers at least 3 questions correctly, then for any student there are $\\binom{15}{3} = 455$ ways for him to have exactly 3 correct answers. If there are 911 students participating in the contest, it follows from the pigeonhole principle that there are at least 3 students having 3 identical correct answers.\n\nIf there is a student $X$ whose score is not more than 2, then the number of remaining students with a score not more than 3 cannot exceed 10; otherwise, pick any 11 of these students together with $X$, and the sum of their total scores is less than 36 points. Then there are more than $911 - 11 = 900$ students in the rest, such that each of them has a score less than 4. Since $\\binom{4}{3} = 4$, and $4 \\times 900 > 455 \\times 2$, there are at least 3 students answering 3 identical questions correctly.\n\n(2) There are 910 students participating in the contest. Divide them into $455 = \\binom{15}{3}$ groups, and there are exactly 2 students in each group. In each group, both students have the identical answers with only 3 correct answers indexed by the group label.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 28256,
    "subject": "Mathematics (Multi-modal)",
    "question": "For positive integer $n$, $a_n = n\\sqrt{5} - \\lfloor n\\sqrt{5} \\rfloor$. Compute the maximum value and the minimum value of $a_1, a_2, \\dots, a_{2009}$. (For real number $x$, $\\lfloor x \\rfloor$ denotes the greatest integer less than or equal to $x$.) (Posed by Wang Zhixiong)",
    "options": [],
    "answer": "Maximum value: 1 − (2 − √5)^6 (attained at n = 1292). Minimum value: 1 − 5470(2 − √5)^6 + 1291(2 − √5)^7 (attained at n = 1597).",
    "solution": "Let $b_0 = 0$, $b_1 = 1$, $b_n = 4b_{n-2} + b_{n-1}$ ($n \\ge 2$). Then\n$$\nb_n = \\frac{(2+\\sqrt{5})^n - (2-\\sqrt{5})^n}{2\\sqrt{5}}.\n$$\nIn particular, $b_6 = 1292$, $b_7 = 5473$.\nFor every $k = 1, 2, \\dots, 5473$, there are unique integers $x_k, y_k$ such that $1292k = x_k + 5473y_k$, and $1 \\le x_k \\le 5473$. Since $(1292, 5473) = 1$, $x_1, x_2, \\dots, x_{5473}$ is a permutation of $\\{1, 2, \\dots, 5473\\}$, and it is clear that $\\{y_k\\}$ is nondecreasing:\n$$\ny_1 \\le y_2 \\le \\dots \\le y_{5473} = 1291.\n$$\nFor our convenience, let $f(x) = x - \\lfloor x \\rfloor$. We have\n$$\n\\begin{aligned}\nf(x_k\\sqrt{5}) &= f(1292k\\sqrt{5} - 5473y_k\\sqrt{5}) \\\\\n&= f\\left(\\frac{(2+\\sqrt{5})^6 - (2-\\sqrt{5})^6}{2}k - \\frac{(2+\\sqrt{5})^7 - (2-\\sqrt{5})^7}{2}y_k\\right) \\\\\n&= f(-(2-\\sqrt{5})^6k + (2-\\sqrt{5})^7y_k). \\end{aligned}\n$$\nSince\n$$\n\\begin{aligned}\n& 0 < (2-\\sqrt{5})^6 k - (2-\\sqrt{5})^7 y_k \\\\\n& \\le 5473(2-\\sqrt{5})^6 - 1291(2-\\sqrt{5})^7 < 1, \\end{aligned}\n$$\nit follows that\n$$\nf(x_k \\sqrt{5}) = 1 - (2 - \\sqrt{5})^k + (2 - \\sqrt{5})^7 y_k,\n$$\nand this is strictly decreasing.\nNow, since $x_1 = 1292$, $x_{5473} = 5473$, $x_{5472} = 4181$, $x_{5471} = 2889$, $x_{5470} = 1597$, we see that $a_{1292}$ attains the maximum and $a_{1597}$ the minimum.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 28257,
    "subject": "Mathematics (Multi-modal)",
    "question": "Every day at a railway station, there is just one train arriving between $8{:}00$ am and $9{:}00$ am and between $9{:}00$ am and $10{:}00$ am, respectively. The arrival times and their probabilities for the two trains are shown in the following table:\n\n| Arrival time | Train A | $8{:}10$ | $8{:}30$ | $8{:}50$ |\n|--------------|---------|---------|---------|---------|\n| Train B      |         | $9{:}10$ | $9{:}30$ | $9{:}50$ |\n| Probability  |         | $\\frac{1}{6}$ | $\\frac{1}{2}$ | $\\frac{1}{3}$ |\n\nSuppose that these random events are independent of each other. Now, a traveler comes into the station at $8{:}20$. Then the mathematical expectation of his waiting time is ______ (round to minute).",
    "options": [],
    "answer": "27",
    "solution": "The distribution table for the waiting times of the traveler is shown below.\n\n| Waiting time/min | $10$ | $30$ | $50$ | $70$ | $90$ |\n|------------------|------|------|------|------|------|\n| Probability      | $\\frac{1}{2}$ | $\\frac{1}{3}$ | $\\frac{1}{6} \\times \\frac{1}{6}$ | $\\frac{1}{2} \\times \\frac{1}{6}$ | $\\frac{1}{3} \\times \\frac{1}{6}$ |\n\nTherefore, the mathematical expectation of his waiting time is\n$$\n10 \\times \\frac{1}{2} + 30 \\times \\frac{1}{3} + 50 \\times \\frac{1}{36} + 70 \\times \\frac{1}{12} + 90 \\times \\frac{1}{18} \\approx 27\\,(\\text{min}).\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 28258,
    "subject": "Mathematics (Multi-modal)",
    "question": "Suppose that the line $l$: $y = kx + m$ ($k$, $m$ are integers) intercepts an ellipse $\\frac{x^2}{16} + \\frac{y^2}{12} = 1$ at two different points $A$, $B$, and intercepts the hyperbola $\\frac{x^2}{4} - \\frac{y^2}{12} = 1$ at two different points $C$, $D$. Can the line $l$ be such that $\\vec{AC} + \\vec{BD} = 0$? If yes, how many different possibilities are there for the line $l$? If no, explain the reason.",
    "options": [],
    "answer": "9",
    "solution": "For $\\begin{cases} y = kx + m, \\\\ \\frac{x^2}{16} + \\frac{y^2}{12} = 1 \\end{cases}$ by eliminating $y$ and simplifying it, we get\n$$\n(3 + 4k^2)x^2 + 8kmx + 4m^2 - 48 = 0.\n$$\nDefine $A(x_1, y_1)$, $B(x_2, y_2)$. Then $x_1 + x_2 = -\\frac{8km}{3+4k^2}$.\n$$\n\\Delta_1 = (8km)^2 - 4(3 + 4k^2)(4m^2 - 48) > 0. \\quad \\textcircled{1}\n$$\nFor $\\begin{cases} y = kx + m, \\\\ \\frac{x^2}{4} - \\frac{y^2}{12} = 1 \\end{cases}$ by eliminating $y$ and simplifying it, we get\n$$\n(3 - k^2)x^2 - 2kmx - m^2 - 12 = 0.\n$$\nDefine $C(x_3, y_3)$, $D(x_4, y_4)$. Then $x_3 + x_4 = \\frac{2km}{3 - k^2}$.\n$$\n\\Delta_2 = (-2km)^2 + 4(3-k^2)(m^2+12) > 0. \\quad \\textcircled{2}\n$$\nFrom $\\vec{AC} + \\vec{BD} = 0$ we get $(x_4 - x_2) + (x_3 - x_1) = 0$, which implies that $x_1 + x_2 = x_3 + x_4$.\nThen\n$$\n-\\frac{8km}{3+4k^2} = \\frac{2km}{3-k^2}.\n$$\nTherefore, $km = 0$ or $-\\frac{4}{3+4k^2} = \\frac{1}{3-k^2}$ (discarded).\nThen a possible solution is either $k = 0$ or $m = 0$.\nWhen $k=0$, from ① and ② we have $-2\\sqrt{3} < m < 2\\sqrt{3}$. As $m$ is an integer, it can be $-3, -2, -1, 0, 1, 2, 3$.\nWhen $m=0$, from ① and ② we have $-\\sqrt{3} < k < \\sqrt{3}$. As $k$ is an integer, it can be $-1, 0, 1$.\nCombining the results above, we conclude that there are nine lines in total satisfying the given conditions.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 28259,
    "subject": "Mathematics (Multi-modal)",
    "question": "For acute triangle $ABC$ with $AB > AC$, let $M$ be the midpoint of side $BC$ and $P$ a point inside $\\triangle AMC$ such that $\\angle MAB = \\angle PAC$. Let $O$, $O_1$ and $O_2$ be the circumcenters of $\\triangle ABC$, $\\triangle ABP$ and $\\triangle ACP$ respectively. Prove that line $AO$ bisects segment $O_1O_2$. (Posed by Xiong Bin)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "As shown in Fig. 1, draw the circumcircles of $\\triangle ABC$, $\\triangle ABP$ and $\\triangle ACP$, respectively. Let the extension of $AP$ meet $\\odot O$ at $D$, join $BD$, and draw the line tangent to $\\odot O$ at $A$, intersecting $\\odot O_1$ and $\\odot O_2$ at $E$ and $F$ respectively.\n\nIt is clear that $\\triangle AMC \\sim \\triangle ABD$, hence\n$$\n\\frac{AB}{BD} = \\frac{AM}{MC}.\n$$\nSince $\\triangle EAB \\sim \\triangle PDB$, we have $\\frac{AB}{BD} = \\frac{AE}{PD}$.\nConsequently, $\\frac{AM}{MC} = \\frac{AE}{PD}$, i.e.\n$$\nAE = \\frac{AM \\times PD}{MC},\n$$\nand, similarly,\n$$\nAF = \\frac{AM \\times PD}{MB}.\n$$\n![](attached_image_1.png)\nFig. 1\n\nIt follows that\n$$\nAE = AF. \\qquad ①\n$$\n\nDraw the perpendicular lines $O_1E' \\perp AE$ with foot $E'$, and $O_2F' \\perp AF$ with foot $F'$. Since $E'$, $F'$ are the midpoints of $AE$, $AF$ respectively, it follows from ① that $A$ is the midpoint of $E'F'$.\nIn the right-angled trapezoid $O_1E'F'O_2$, $AO$ is the extension of the median, and hence it bisects the segment $O_1O_2$.\n\n\nSolution 2:\n\nAs shown in Fig. 2, draw segments $AO_1$, $OO_1$, $AO_2$, $OO_2$. Denote by $Q$ the intersection of $AO$ and $O_1O_2$. Then\n![](attached_image_2.png)\nFig. 2\n$$\n\\frac{O_1Q}{QO_2} = \\frac{S_{\\triangle OO_1}}{S_{\\triangle OO_2}} = \\frac{AB \\times OO_1}{AC \\times OO_2},\n$$\nwhere $\\frac{AB}{AC} = \\frac{\\sin\\angle ACB}{\\sin\\angle ABC}$.\nSince $\\angle OO_1Q = \\angle BAP = \\angle CAM$, and $\\angle OO_2Q = \\angle CAP = \\angle BAM$, it follows that\n$$\n\\begin{aligned} \\frac{OO_1}{OO_2} &= \\frac{OO_1}{OQ} \\times \\frac{OQ}{OO_2} \\\\ &= \\frac{\\sin\\angle OQO_1}{\\sin\\angle OO_1Q} \\times \\frac{\\sin\\angle OO_2Q}{\\sin\\angle OQO_2} \\\\ &= \\frac{\\sin\\angle OO_2Q}{\\sin\\angle OO_1Q} = \\frac{\\sin\\angle BAM}{\\sin\\angle CAM}, \\end{aligned}\n$$\nand thus\n$$\n\\frac{O_1Q}{QO_2} = \\frac{\\sin\\angle ACM}{\\sin\\angle CAM} \\times \\frac{\\sin\\angle BAM}{\\sin\\angle ABM} = \\frac{AM}{CM} \\times \\frac{BM}{AM} = \\frac{BM}{CM}.\n$$\nNote that $M$ is the midpoint of $BC$, and therefore $O_1Q = QO_2$, i.e. line $AO$ bisects segment $O_1O_2$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 28260,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $a, b, c \\in \\{0, 1, 2, \\dots, 9\\}$. The quadratic equation $ax^2 + bx + c = 0$ has a rational root. Prove that the three-digit number $abc$ is not a prime number.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "If $\\overline{abc} = p$ is a prime number, and the roots of the equation $f(x) = ax^2 + bx + c = 0$ are rational numbers, then $b^2 - 4ac$ is a square number, $x_1, x_2$ are negative, and $f(x) = a(x - x_1)(x - x_2)$.\n\nSo $p = f(10) = a(10 - x_1)(10 - x_2)$, and we get $4ap = (20a - 2ax_1)(20a - 2ax_2)$.\n\nSince $x_1, x_2 = \\frac{-b \\pm \\sqrt{b^2 - 4ac}}{2a}$, we see that $20a - 2ax_1$, $20a - 2ax_2$ are integers. Now $p \\mid 20a - 2ax_1$ or $p \\mid 20a - 2ax_2$, and we can suppose that $p \\mid 20a - 2ax_1$; then $p \\le 20a - 2ax_1$, and $4a \\ge 20a - 2ax_2$, which is a contradiction to $x_2$ being negative.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 28261,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $n$ be an integer greater than two, and let $A_1, A_2, \\dots, A_{2n}$ be pairwise distinct nonempty subsets of $\\{1, 2, \\dots, n\\}$.\nDetermine the maximum value of $\\sum_{i=1}^{2n} \\frac{|A_i \\cap A_{i+1}|}{|A_i| \\cdot |A_{i+1}|}$.\n(Here, we set $A_{2n+1} = A_1$. For a set $X$, let $|X|$ denote the number of elements in $X$.)",
    "options": [],
    "answer": "n",
    "solution": "The answer is $n$.\n\nWe consider each summand $s_i = \\frac{|A_i \\cap A_{i+1}|}{|A_i| \\cdot |A_{i+1}|}$.\nIf $A_i \\cap A_{i+1}$ is the empty set, then $s_i = 0$.\nIf $A_i \\cap A_{i+1}$ is nonempty, because $A_i \\neq A_{i+1}$, at least one of $A_i$ and $A_{i+1}$ has more than one element, that is, $\\max\\{|A_i|, |A_{i+1}|\\} \\ge 2$. Because $A_i \\cap A_{i+1}$ is a subset of each of $A_i$ and $A_{i+1}$, $|A_i \\cap A_{i+1}| \\le \\min\\{|A_i|, |A_{i+1}|\\}$ and\n$$\ns_i = \\frac{|A_i \\cap A_{i+1}|}{|A_i| \\cdot |A_{i+1}|} \\le \\frac{\\min\\{|A_i|, |A_{i+1}|\\}}{\\max\\{|A_i|, |A_{i+1}|\\} \\cdot \\min\\{|A_i|, |A_{i+1}|\\}} \\le \\frac{1}{2}.\n$$\nIt follows that\n$$\n\\sum_{i=1}^{2n} \\frac{|A_i \\cap A_{i+1}|}{|A_i| \\cdot |A_{i+1}|} \\le \\sum_{i=1}^{2n} \\frac{1}{2} = n.\n$$\nThis upper bound can be achieved with sets\n$$\nA_1 = \\{1\\},\nA_2 = \\{1, 2\\},\nA_3 = \\{2\\},\nA_4 = \\{2, 3\\},\n\\dots,\nA_{2n-2} = \\{n-1, n\\},\nA_{2n-1} = \\{n\\},\nA_{2n} = \\{n, 1\\}.\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 28262,
    "subject": "Mathematics (Multi-modal)",
    "question": "The range of $f(x) = \\sqrt{x-5} - \\sqrt{24-3x}$ is ______.",
    "options": [],
    "answer": "[-3, √3]",
    "solution": "It is easy to see that $f(x)$ is increasing on its domain $[5, 8]$. Therefore, its range is $[-3, \\sqrt{3}]$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 28263,
    "subject": "Mathematics (Multi-modal)",
    "question": "As seen in Fig. 1.1, the circumcenter of acute triangle $ABC$ is $O$, $K$ is a point (not the midpoint) on the side $BC$, $D$ is a point on the extended line of segment $AK$, lines $BD$ and $AC$ intersect at point $N$, and lines $CD$ and $AB$ intersect at point $M$. Prove if $OK \\perp MN$, then $A$, $B$, $D$, $C$ are concyclic.\n\n![](attached_image_1.png)\n\nFig. 1.1",
    "options": [],
    "answer": "Detailed solution",
    "solution": "By reduction to absurdity, assume that $A, B, D, C$ are not concyclic. Let the circumcircle (with radius $r$) of $ABC$ intersect $AD$ at point $E$. Join $BE$ and extend it to intersect line $AN$ at point $Q$; join $CE$ and extend it to intersect $AM$ at $P$. Join $PQ$, as seen in Fig. 1.2.\n\n![](attached_image_2.png)\n\nFig. 1.2\n\nWe have\n$$\n\\begin{aligned}\nPK^2 &= \\text{the power of } P \\text{ with respect to } \\odot O + \\\\\n&\\quad \\text{the power of } Q \\text{ with respect to } \\odot O \\\\\n&= (PO^2 - r^2) + (KO^2 - r^2).\n\\end{aligned}\n$$\n(We will prove it in the appendix)\n\nIn the same way,\n$$\nQK^2 = (QO^2 - r^2) + (KO^2 - r^2).\n$$\nThen we have\n$$\nPO^2 - PK^2 = QO^2 - QK^2.\n$$\nTherefore, $OK \\perp PQ$. By the given condition $OK \\perp MN$, we get that $PQ \\parallel MN$. Then we have\n$$\n\\frac{AQ}{QN} = \\frac{AP}{PM}. \\qquad \\textcircled{1}\n$$\nBy Menelaus' Theorem, we obtain\n$$\n\\frac{NB}{BD} \\cdot \\frac{DE}{EA} \\cdot \\frac{AQ}{QN} = 1, \\qquad \\textcircled{2}\n$$\n$$\n\\frac{MC}{CD} \\cdot \\frac{DE}{EA} \\cdot \\frac{AP}{PM} = 1. \\qquad \\textcircled{3}\n$$\nFrom ①, ②, ③, we get $\\frac{NB}{BD} = \\frac{MC}{CD}$, or $\\frac{ND}{BD} = \\frac{MD}{DC}$. Then $\\triangle DMN \\sim \\triangle DCB$, which implies $\\angle DMN = \\angle DCB$. Then $BC \\parallel MN$. Therefore, $OK \\perp BC$, and that means $K$ is the midpoint of $BC$, which is a contradiction. This completes the proof that $A, B, D, C$ are concyclic.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 28264,
    "subject": "Mathematics (Multi-modal)",
    "question": "Suppose that $m$ and $k$ are non-negative integers, and $p = 2^{2^m} + 1$ is a prime number. Prove that\n\na. $2^{2^{m+1}} p^k \\equiv 1 \\pmod{p^{k+1}}$;\n\nb. $2^{m+1} p^k$ is the smallest positive integer $n$ satisfying the congruence equation $2^n \\equiv 1 \\pmod{p^{k+1}}$.",
    "options": [],
    "answer": "2^{m+1} p^k",
    "solution": "We want to prove that $2^{2^{m+1}} p^k = p^{k+1} t_k + 1$ for some integer $t_k$ not divisible by $p$. We proceed by induction on $k$.\n\nWhen $k = 0$, it follows from $2^{2^m} = p-1$ that $2^{2^m} = (p-1)^2 = p(p-2)+1$, in this case, $t_0 = p-2$.\n\nFor inductive step, suppose that $2^{2^{m+1}} p^k = p^{k+1} t_k + 1$ where $k \\ge 0$ and $p \\nmid t_k$, then\n$$\n\\begin{align*}\n2^{2^{m+1}} p^{k+1} &= (2^{2^{m+1}} p^k) p = (p^{k+1} t_k + 1)^p \\\\\n&= \\sum_{s=0}^{p} \\binom{p}{s} (p^{k+1} t_k)^s \\\\\n&= 1 + p \\cdot p^{k+1} t_k + \\frac{p(p-1)}{2} (p^{k+1} t_k)^2 + \\sum_{s=3}^{p} \\binom{p}{s} (p^{k+1} t_k)^s. \n\\end{align*}\n$$\nAs $k \\ge 0$, then for any $s \\ge 2$, we have $(k+1)s \\ge 2(k+1) = 2k+2 \\ge k+2$, so $2^{2^{m+1}} p^{k+1} = p^{k+2}t_{k+1} + 1$, where $t_{k+1} \\in \\mathbb{Z}_+$ and $p \\nmid t_{k+1}$. It follows from mathematical induction that (a) holds.\n\nNext, we prove (b). Write $2^{m+1} p^k = n\\ell + r$, where $\\ell, r \\in \\mathbb{Z}$ and $0 \\le r < n$. Then it follows from (a) that $1 \\equiv 2^{2^{m+1}} p^k \\equiv 2^{n\\ell+r} \\equiv (2^n)^\\ell \\cdot 2^r \\equiv 2^r \\pmod{p^{k+1}}$. As $0 \\le r < n$, it follows from the definition of $n$ that $r=0$, i.e., $n \\mid 2^{m+1} p^k$. By the Fundamental Theorem of Arithmetic, $n = 2^t p^s$. If $t \\le m$, then $2^{2^m} p^k = (2^{2^t} p^k)^{2^{m-t}} \\equiv 1 \\pmod{p}$. On the other hand, $2^{2^m} p^k = (2^{2^m})^{p^k} \\equiv (-1)^{p^k} \\equiv -1 \\pmod{p}$, which is a contradiction, and hence $t = m+1$, i.e., $n = 2^{m+1} p^s$. $\\square$",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 28265,
    "subject": "Mathematics (Multi-modal)",
    "question": "As shown below, two circles $\\Gamma_1$, $\\Gamma_2$ intersect at points $A$, $B$, one line passing through $B$ intersects $\\Gamma_1$, $\\Gamma_2$ at points $C$, $D$, another line passing through $B$ intersects $\\Gamma_1$, $\\Gamma_2$ at points $E$, $F$, and line $CF$ intersects $\\Gamma_1$, $\\Gamma_2$ at points $P$, $Q$, respectively. Let $M$, $N$ be the middle points of arc $PB$ and arc $QB$, respectively. Prove that if $CD = EF$, then $C$, $F$, $M$, $N$ are concyclic. (Posed by Xiong Bin)\n\n![](attached_image_1.png)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "**Proof** Draw lines $AC$, $AD$, $AE$, $AF$, $DF$. From $\\angle ADB = \\angle AFB$, $\\angle ACB = \\angle AEF$ and the assumption $CD = EF$; one obtains $\\triangle ACD \\cong \\triangle AEF$. So we have $AD = AF$, $\\angle ADC = \\angle AFE$ and $\\angle ADF = \\angle AFD$. Then $\\angle ABC = \\angle AFD = \\angle ADF = \\angle ABF$; $AB$ is the bisector of $\\angle CBF$. Draw lines $CM$, $FN$. Since $M$ is the middle point of arc $PB$, $CM$ is the bisector of $\\angle DCF$, and $FN$ is the bisector of $\\angle CFB$. Then $BA$, $CM$, $FN$ have a common intersection, say, at $I$. In the circles $\\Gamma_1$ and $\\Gamma_2$, one has $CI \\times IM = AI \\times IB$, $AI \\times IB = NI \\times IF$, according to the theorem of power with respect to circles. So $NI \\times IF = CI \\times IM$, and $C$, $F$, $M$, $N$ are concyclic.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 28266,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $a, b, c \\in \\{0, 1, 2, \\dots, 9\\}$. The quadratic equation $ax^2 + bx + c = 0$ has a rational root. Prove that the three-digit number $abc$ is not a prime number.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "We prove by contradiction. If $abc = p$ is a prime number, the rational root of quadratic equation $f(x) = ax^2 + bx + c = 0$ is $x_1, x_2 = \\frac{-b \\pm \\sqrt{b^2 - 4ac}}{2a}$. Obviously, $b^2 - 4ac$ is a perfect square number, and $x_1, x_2$ are all negative, and\n\n$$\nf(x) = a(x - x_1)(x - x_2).\n$$\n\nThus,\n$$\np = f(10) = a(10 - x_1)(10 - x_2).\n$$\nSo,\n$$\n4ap = (20a - 2ax_1)(20a - 2ax_2).\n$$\nIt is easy to see that $(20a - 2ax_1)$ and $(20a - 2ax_2)$ are all positive integers. Consequently, $p \\mid (20a - 2ax_1)$ or $p \\mid (20a - 2ax_2)$. If $p \\mid (20a - 2ax_1)$, then $p \\le 20a - 2ax_1$, so, $80 - 8x_1 - 10x_2 + x_1x_2 \\le 0$, which contradicts to $x_1, x_2 < 0$. Similarly, $p \\mid (20a - 2ax_1)$ is not true. $\\square$",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 28267,
    "subject": "Mathematics (Multi-modal)",
    "question": "Given two moving points $A(x_1, y_1)$ and $B(x_2, y_2)$ on parabola curve $y^2 = 6x$ with $x_1 + x_2 = 4$ and $x_1 \\neq x_2$, and the perpendicular bisector of segment $AB$ intersects $x$-axis at point $C$. Find the maximum area of $\\triangle ABC$.",
    "options": [],
    "answer": "14/3 * sqrt(7)",
    "solution": "Let the midpoint of $AB$ be $M(x_0, y_0)$. Then $x_0 = \\frac{x_1 + x_2}{2} = 2$ and $y_0 = \\frac{y_1 + y_2}{2}$. We have\n$$\nk_{AB} = \\frac{y_2 - y_1}{x_2 - x_1} = \\frac{\\frac{y_2}{6} - \\frac{y_1}{6}}{\\frac{y_2}{6} + \\frac{y_1}{6}} = \\frac{\\frac{6}{y_2} + \\frac{6}{y_1}}{\\frac{6}{y_2} + \\frac{6}{y_1}} = \\frac{3}{y_0}.\n$$\nThe equation of the perpendicular bisector of $AB$ is\n$$\ny - y_0 = -\\frac{y_0}{3}(x - 2). \\qquad \\textcircled{1}\n$$\nIt is easy to find that one solution of it is $x = 5$, $y = 0$.\nTherefore, the intersection $C$ is a fixed point with coordinate $(5, 0)$.\nFrom ①, we know the equation of line $AB$ is $y - y_0 = \\frac{3}{y_0}(x - 2)$, or\n$$\nx = \\frac{y_0}{3}(y - y_0) + 2. \\qquad \\textcircled{2}\n$$\nSubstituting ② in $y^2 = 6x$, we get $y^2 = 2y_0(y - y_0) + 12$, or\n$$\ny^2 - 2y_0y + 2y_0^2 - 12 = 0. \\qquad \\textcircled{3}\n$$\nAs $y_1$ and $y_2$ are two real roots of ③ and $y_1 \\neq y_2$, we have\n$$\n\\Delta = 4y_0^2 - 4(2y_0^2 - 12) = -4y_0^2 + 48 > 0.\n$$\nTherefore, $-2\\sqrt{3} < y_0 < 2\\sqrt{3}$. Then we have\n![](attached_image_1.png)\nThe distance from point $C(5, 0)$ to segment $AB$ is\n$$\nh = |CM| = \\sqrt{(5-2)^2 + (0-y_0)^2} = \\sqrt{9+y_0^2}.\n$$\nTherefore,\n$$\n\\begin{align*}\nS_{\\triangle ABC} &= \\frac{1}{2} |AB| \\cdot h = \\frac{1}{3} \\sqrt{(9+y_0^2)(12-y_0^2)} \\cdot \\sqrt{9+y_0^2} \\\\\n&= \\frac{1}{3} \\sqrt{\\frac{1}{2}(9+y_0^2)(24-2y_0^2)(9+y_0^2)} \\\\\n&\\le \\frac{1}{3} \\sqrt{\\frac{1}{2} \\left( \\frac{9+y_0^2+24-2y_0^2+9+y_0^2}{3} \\right)^3} \\\\\n&= \\frac{14}{3}\\sqrt{7}.\n\\end{align*}\n$$\nThe equality holds if and only if $9 + y_0^2 = 24 - 2y_0^2$, i.e.\n$$\ny_0 = \\pm\\sqrt{5}. \\text{ Then we get}\n$$\n$$ \\text{and} \\quad A\\left(\\frac{6+\\sqrt{35}}{3}, \\sqrt{5}+\\sqrt{7}\\right), B\\left(\\frac{6-\\sqrt{35}}{3}, \\sqrt{5}-\\sqrt{7}\\right) $$\n$$ A\\left(\\frac{6+\\sqrt{35}}{3}, - (\\sqrt{5}+\\sqrt{7})\\right), B\\left(\\frac{6-\\sqrt{35}}{3}, -\\sqrt{5}+\\sqrt{7}\\right). $$\nSimilar to Solution 1, we get that $C$, the intersection of the perpendicular bisector of $AB$ and the $x$-axis, is a fixed point with coordinate $(5, 0)$.\nLet $x_1 = t_1^2$, $x_2 = t_2^2$, $t_1 > t_2$, $t_1^2 + t_2^2 = 4$. Then $S_{\\triangle ABC}$ is the absolute value of\n$$\n\\frac{1}{2} \\begin{vmatrix} 5 & 0 & 1 \\\\ t_1^2 & \\sqrt{6} t_1 & 1 \\\\ t_2^2 & \\sqrt{6} t_2 & 1 \\end{vmatrix},\n$$\nso\n$$\n\\begin{aligned}\nS_{\\triangle ABC}^2 &= \\left( \\frac{1}{2} (5\\sqrt{6} t_1 + \\sqrt{6} t_1^2 t_2 - \\sqrt{6} t_1 t_2^2 - 5\\sqrt{6} t_2) \\right)^2 \\\\\n&= \\frac{3}{2} (t_1 - t_2)^2 (t_1 t_2 + 5)^2 \\\\\n&= \\frac{3}{2} (4 - 2t_1t_2)(t_1t_2 + 5)(t_1t_2 + 5) \\\\\n&\\le \\frac{3}{2} \\left( \\frac{14}{3} \\right)^3.\n\\end{aligned}\n$$\nTherefore, $S_{\\triangle ABC} \\le \\frac{14}{3}\\sqrt{7}$ and the equality holds if and only if $(t_1 - t_2)^2 = t_1t_2 + 5$ and $t_1^2 + t_2^2 = 4$. We then get $t_1 = \\frac{\\sqrt{7} + \\sqrt{5}}{\\sqrt{6}}$ and $t_2 = -\\frac{\\sqrt{7} - \\sqrt{5}}{\\sqrt{6}}$, which implies either\n$$\nA\\left(\\frac{6 + \\sqrt{35}}{3}, \\sqrt{5} + \\sqrt{7}\\right), B\\left(\\frac{6 - \\sqrt{35}}{3}, \\sqrt{5} - \\sqrt{7}\\right)\n$$\nor\n$$\nA\\left(\\frac{6 + \\sqrt{35}}{3}, - (\\sqrt{5} + \\sqrt{7})\\right), B\\left(\\frac{6 - \\sqrt{35}}{3}, -\\sqrt{5} + \\sqrt{7}\\right).\n$$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 28268,
    "subject": "Mathematics (Multi-modal)",
    "question": "Prove that equation $2x^3 + 5x - 2 = 0$ has exactly one real root (denoted as $r$), and there is a unique strictly increasing sequence $\\{a_n\\}$ such that $\\frac{2}{5} = r^{a_1} + r^{a_2} + r^{a_3} + \\dots$.",
    "options": [],
    "answer": "The unique sequence is a_n = 3n − 2 for n ≥ 1; the real root r is the unique solution in the interval between zero and one half of the equation 2x^3 + 5x − 2 = 0.",
    "solution": "Let $f(x) = 2x^3 + 5x - 2$. Then we have $f'(x) = 6x^2 + 5 > 0$, which means $f(x)$ is strictly increasing. Furthermore, $f(0) = -2 < 0$, $f(\\frac{1}{2}) = \\frac{3}{4} > 0$. Therefore, $f(x)$ has a unique real root $r \\in (0, \\frac{1}{2})$. From $2r^3 + 5r - 2 = 0$, we have\n$$\n\\frac{2}{5} = \\frac{r}{1 - r^3} = r + r^4 + r^7 + r^{10} + \\dots\n$$\nTherefore, sequence $a_n = 3n - 2$ ($n = 1, 2, \\dots$) satisfies the required condition.\n\nAssume there are two different positive integer sequences\n$a_1 < a_2 < \\dots < a_n < \\dots$ and $b_1 < b_2 < \\dots < b_n < \\dots$\nsatisfying\n$$\nr^{a_1} + r^{a_2} + r^{a_3} + \\dots = r^{b_1} + r^{b_2} + r^{b_3} + \\dots = \\frac{2}{5}\n$$\nDeleting the terms that appear at both sides of the expression, we have\n$$\nr^{s_1} + r^{s_2} + r^{s_3} + \\dots = r^{t_1} + r^{t_2} + r^{t_3} + \\dots\n$$\nwhere $s_1 < s_2 < s_3 < \\dots$, $t_1 < t_2 < t_3 < \\dots$ with all the $s_i$ and $t_j$ different from each other.\nWe may as well assume that $s_1 < t_1$. Then\n$$\n\\begin{align*}\nr^{s_1} < r^{s_1} + r^{s_2} + \\dots &= r^{t_1} + r^{t_2} + \\dots, \\\\\n1 < r^{t_1 - s_1} + r^{t_2 - s_1} + \\dots &\\le r + r^2 + \\dots \\\\\n&= \\frac{1}{1 - r} - 1 < \\frac{1}{1 - \\frac{1}{2}} - 1 = 1.\n\\end{align*}\n$$\nIt is a contradiction. This proves that $\\{a_n\\}$ is unique.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 28269,
    "subject": "Mathematics (Multi-modal)",
    "question": "For any set $A = \\{a_1, a_2, \\dots, a_m\\}$, let $P(A) = a_1a_2\\cdots a_m$. In addition, let $n = \\binom{2010}{99}$ and let $A_1, A_2, \\dots, A_n$ be all 99-element subsets of $\\{1, 2, \\dots, 2010\\}$. Prove that $2011 \\mid \\sum_{i=1}^n P(A_i)$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "One can check that $2011$ is a prime number. Let\n$$\nf(x) = (x-1)(x-2)\\cdots(x-2010) - (x^{2010} + 2010!).\n$$\nFor $n \\in \\{1, 2, \\dots, 2010\\}$, we have $n^{2010} \\equiv 1 \\pmod{2011}$ (by Fermat's little theorem), and $2010! \\equiv -1 \\pmod{2011}$ (by Wilson's theorem), so\n$$\nf(n) = (n-1)(n-2)\\cdots(n-2010) \\equiv 0 \\pmod{2011}.\n$$\nThis means that $f(x) \\equiv 0 \\pmod{2011}$ have $2010$ roots, but the degree of $f(x)$ is $2009$. Using Lagrange's theorem, we can see that the coefficients can all be divided by $2011$.\n$$\n\\sum_{i=1}^{n} P(A_i) \\text{ is the coefficient of } x^{1911} \\text{, hence } 2011 \\mid \\sum_{i=1}^{n} P(A_i).\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 28270,
    "subject": "Mathematics (Multi-modal)",
    "question": "For any set $A = \\{a_1, a_2, \\dots, a_m\\}$, denote $P(A) = a_1 a_2 \\dots a_m$. Let $A_1, A_2, \\dots$, and $A_n$ be all 99-element subsets of $\\{1, 2, \\dots, 2010\\}$, $n = C_{2010}^{99}$. Prove that $2010 \\mid \\sum_{i=1}^{n} P(A_i)$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "For each 99-element subset, $A_i = \\{a_1, a_2, \\dots, a_{99}\\}$ of $\\{1, 2, \\dots, 2010\\}$ uniquely corresponds to a 99-element subset $B_i = \\{b_1, b_2, \\dots, b_{99}\\}$ of $\\{1, 2, \\dots, 2010\\}$ by $b_k = 2011 - a_k$, $k = 1, 2, \\dots, 99$.\nSince $\\sum_{k=1}^{99} (a_k + b_k) = 99 \\times 2011$ is odd, we see that $A_i, B_i$ are different subsets of $\\{1, 2, \\dots, 2010\\}$. When $A_i$ take all 99-element subsets of $\\{1, 2, \\dots, 2010\\}$, so do $B_i$. Moreover\n$$\n\\begin{align*}\nP(A_i) + P(B_i) &= a_1 a_2 \\cdots a_{99} + (2011 - a_1)(2011 - a_2)\\cdots(2011 - a_{99}) \\\\\n&\\equiv a_1 a_2 \\cdots a_{99} + (-a_1)(-a_2)\\cdots(-a_{99}) \\pmod{2011} \\\\\n&\\equiv 0 \\pmod{2011}.\n\\end{align*}\n$$\nThus,\n$$2 \\sum_{i=1}^{n} P(A_i) = \\sum_{i=1}^{n} P(A_i) + \\sum_{i=1}^{n} P(B_i) \\equiv 0 \\pmod{2011},$$\nhence $2011 \\mid \\sum_{i=1}^{n} P(A_i)$.\nLet $f(n) = (n-1)(n-2)\\cdots(n-2010) - n^{2010} - 2010!$, where $n \\in \\mathbb{Z}$.\nSince 2011 is prime, by Fermat's Little Theorem, $n^{2010} \\equiv 1 \\pmod{2011}$. By Wilson's Theorem, we have $2010! \\equiv -1 \\pmod{2011}$. Thus,\n(i) If $2011 \\nmid n$, then\n$$\nf(n) \\equiv (n-1)(n-2)\\cdots(n-2010) \\equiv 0 \\pmod{2011}.\n$$\n(ii) If $2011 \\mid n$, then\n$$\n\\begin{align*}\nf(n) &\\equiv (2011-1)(2011-2)\\cdots(2011-2010) - 2011^{2010} - 2010! \\\\\n&\\equiv 2010! - 2010! \\pmod{2011} \\\\\n&\\equiv 0 \\pmod{2011}.\n\\end{align*}\n$$\nSo $f(n) \\equiv 0 \\pmod{2011}$ has 2011 solutions in the sense of $\\mod 2011$.\nSince $f(n)$ is a polynomial of order 2009, and for all $n \\in \\mathbb{Z}$, $2011 \\mid f(n)$, we see that each coefficient of $f(n)$ can be divided by 2011.\nTurn to the original problem, $\\sum_{i=1}^{n} P(A_i)$ is the coefficient of term with order 1911 of $f(n)$, thus $2011 \\mid \\sum_{i=1}^{n} P(A_i)$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 28271,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $A = \\{a_1, a_2, \\ldots, a_{2010}\\}$ and $B = \\{b_1, b_2, \\ldots, b_{2010}\\}$ be two sets of complex numbers, such that the equality\n$$\n\\sum_{1 \\le i < j \\le 2010} (a_i + a_j)^n = \\sum_{1 \\le i < j \\le 2010} (b_i + b_j)^n\n$$\nholds for every $n = 1, 2, \\ldots, 2010$. Prove that $A = B$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "**Proof** Let $S_k = \\sum_{i=1}^{2010} a_i^k$ and $\\tilde{S}_k = \\sum_{i=1}^{2010} b_i^k$. We first show by induction that $S_k = \\tilde{S}_k$ for $k = 1, 2, \\ldots, 2010$.\n\nSetting $n = 1$ in the given equality, we have $2009S_1 = 2009\\tilde{S}_1$, and hence $S_1 = \\tilde{S}_1$. Assume that $S_j = \\tilde{S}_j$ for $j = 1, 2, \\ldots, k-1$, where $2 \\le k \\le 2010$; we are going to show that $S_k = \\tilde{S}_k$.\n\nBy the binomial expansion theorem, we have\n$$\n\\begin{align*}\n\\sum_{1 \\le i < j \\le 2010} (a_i + a_j)^k &= \\sum_{1 \\le i < j \\le 2010} \\sum_{l=0}^{k} \\binom{k}{l} a_i^l a_j^{k-l} \\\\\n&= 2009S_k + \\sum_{1 \\le i < j \\le 2010} \\sum_{l=0}^{k-1} \\binom{k}{l} a_i^l a_j^{k-l} \\\\\n&= 2009S_k + \\frac{1}{2} \\sum_{1 \\le i < j \\le 2010} \\sum_{l=1}^{k-1} \\binom{k}{l} a_i^l a_j^{k-l} \\\\\n&= 2009S_k + \\frac{1}{2} \\sum_{l=1}^{k-1} \\sum_{i=1}^{2010} \\binom{k}{l} a_i^l (S_{k-l} - a_i^{k-l}) \\\\\n&= 2009S_k + \\frac{1}{2} \\sum_{l=1}^{k-1} \\left( \\binom{k}{l} S_{k-l} \\sum_{i=1}^{2010} a_i^l - \\sum_{i=1}^{2010} \\binom{k}{l} a_i^k \\right) \\\\\n&= 2009S_k + \\frac{1}{2} \\sum_{l=1}^{k-1} \\left( \\binom{k}{l} S_{k-l} S_l - \\binom{k}{l} S_k \\right) \\\\\n&= \\frac{1}{2} \\sum_{l=1}^{k-1} \\binom{k}{l} S_{k-l} S_l + (2010 - 2^{k-1}) S_k. \\quad \\textcircled{1}\n\\end{align*}\n$$\n\nSimilarly, we have\n$$\n\\sum_{1 \\le i < j \\le 2010} (b_i + b_j)^k = \\frac{1}{2} \\sum_{l=1}^{k-1} \\binom{k}{l} \\tilde{S}_{k-l} \\tilde{S}_l + (2010 - 2^{k-1}) \\tilde{S}_k. \\quad \\textcircled{2}\n$$\n\nSince\n$$\n\\sum_{1 \\le i < j \\le 2010} (a_i + a_j)^k = \\sum_{1 \\le i < j \\le 2010} (b_i + b_j)^k,\n$$\nby ①, ② and inductive hypothesis $S_i = \\tilde{S}_i$ for $i = 1, 2, \\dots, k-1$, we have $S_k = \\tilde{S}_k$ (it is worth noting that $2010$ is not a power of $2$, i.e. $2010 - 2^{k-1} \\neq 0$). This completes the inductive proof that $S_k = \\tilde{S}_k$ for all $k = 1, 2, \\dots, 2010$.\n\nSet\n$$\n(x - a_1) \\cdots (x - a_{2010}) = x^{2010} + A_1 x^{2009} + \\cdots + A_{2010}, \\quad \\textcircled{3}\n$$\n$$\n(x - b_1) \\cdots (x - b_{2010}) = x^{2010} + B_1 x^{2009} + \\cdots + B_{2010}. \\quad \\textcircled{4}\n$$\nBy Newton's formula, we have\n$$\nS_k + A_1 S_{k-1} + \\cdots + A_{k-1} S_1 + k A_k = 0, \\quad \\textcircled{5}\n$$\n$$\n\\tilde{S}_k + B_1 \\tilde{S}_{k-1} + \\cdots + B_{k-1} \\tilde{S}_1 + k B_k = 0, \\quad \\textcircled{6}\n$$\nfor $k = 1, 2, \\dots, 2010$.\n\nIt follows from ⑤, ⑥ and $S_k = \\tilde{S}_k$, $k = 1, 2, \\dots, 2010$, by the easy inductive argument, we have\n$$\nA_k = B_k, \\quad k = 1, 2, \\dots, 2010.\n$$\nThe right hand sides of equations ③ and ④ are equal, and so are their left hand sides, i.e. $A = B$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 28272,
    "subject": "Mathematics (Multi-modal)",
    "question": "The minimum of $y = (a \\cos^2 x - 3) \\sin x$ is -3. Then the range of real number $a$ is ______.",
    "options": [],
    "answer": "[-3/2, 12]",
    "solution": "Let $\\sin x = t$. The expression is then changed to\n$$\ng(t) = (-a t^2 + a - 3)t,\n$$\nor\n$$\ng(t) = -a t^3 + (a-3)t.\n$$\nFrom $-a t^3 + (a-3)t \\ge -3$, we get\n$$\n-a t (t^2 - 1) - 3 (t - 1) \\ge 0,\n$$\n$$\n(t - 1)(-a t (t + 1) - 3) \\ge 0.\n$$\nSince $t - 1 \\le 0$, we have $-a t (t + 1) - 3 \\le 0$, or\n$$\na (t^2 + t) \\ge -3. \\qquad ①\n$$\nWhen $t = 0, -1$, expression ① always holds; when $0 < t \\le 1$, we have $0 < t^2 + t \\le 2$; and when $-1 < t < 0$, $-\\frac{1}{4} \\le t^2 + t < 0$. Therefore, $-\\frac{3}{2} \\le a \\le 12$. ☐",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 28273,
    "subject": "Mathematics (Multi-modal)",
    "question": "In triangle $ABC$, $AB = AC$. Point $D$ is the midpoint of side $BC$. Point $E$ lies outside the triangle $ABC$ such that $CE \\perp AB$ and $BE = BD$. Let $M$ be the midpoint of segment $BE$. Point $F$ lies on the minor arc $\\widehat{AD}$ of the circumcircle of triangle $ABD$ such that $MF \\perp BE$. Prove that $ED \\perp FD$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "**Solution 1.** Construct point $F_1$ such that $EF_1 = BF_1$ and ray $DF_1$ is perpendicular to line $ED$. It suffices to show that $F = F_1$ or $ABDF_1$ is cyclic; that is,\n$$\n\\angle BAD = \\angle BF_1 D. \\qquad \\textcircled{1}\n$$\nSet $\\angle BAD = \\angle CAD = x$. Because $EC \\perp AB$ and $AD \\perp BC$,\n$$ \\angle ECB = 90^\\circ - \\angle ABD = \\angle BAD = x. $$\nNote that $MD$ is a midline of triangle $BCE$. In particular, $MD \\parallel EC$ and\n$$\n\\angle MDB = \\angle ECD = x. \\qquad \\textcircled{2}\n$$\nIn isosceles triangle $EF_1M$, we may set $\\angle EF_1M = \\angle BF_1M = y$. Because $EM \\perp MF_1$ and $MD \\perp DF_1$,\n$$\n\\angle EMF_1 = \\angle EDF_1 = 90^\\circ,\n$$\nimplying that $EMDF_1$ is cyclic. Consequently, we have\n$$\n\\angle EDM = \\angle EF_1 M = y. \\qquad \\textcircled{3}\n$$\nCombining ② and ③, we obtain\n![](attached_image_1.png)\nFig. 2. 2\n$$\n\\angle BDE = \\angle EDM + \\angle MDB = x + y.\n$$\nBecause $BE = BD$, we conclude that triangle $BED$ is isosceles with $\\angle MED = \\angle BED = \\angle BDE = x + y$. Because $EMDF_1$ is cyclic, we have $\\angle MF_1 D = \\angle MED = x + y$. It is then clear that\n$$\n\\angle BF_1 D = \\angle MF_1 D - \\angle MF_1 B = x = \\angle BAD,\n$$\nwhich is ①.\n\n\n**Solution 2.** (Based on work by Sherry Gong and Inna Zakharevich) We maintain the notations used in Solution 1. Let $\\omega$ and $O$ denote the circumcircle and the circumcenter of triangle $ABD$, and let $T$ be second intersection (other than $B$) between line $BE$ and $\\omega$. Extend segment $DE$ through $E$ to meet $\\omega$ at $S$. Point $F_2$ lies on $\\omega$ such that $DF_2 \\perp DE$. We will show that $F_2 = F$ or $F_2B = F_2E$. Let $M_2$ denote the foot of the perpendicular from $F_2$ to line $BE$. It suffices to show that $M_2$ is the midpoint of segment $BE$, that is, $EM_2 = M_2B$.\nBecause $\\angle SDF_2 = \\angle EDF_2 = 90^\\circ$, $O$ is the midpoint of $SF_2$. Because $BTAD$ is cyclic, $\\angle BTD = \\angle BAD = \\angle BCE = x$. Note also that $BD = BE$ and $\\angle EBC = \\angle DBT$. We conclude that triangle $BDT$ and $BEC$ are congruent to each other, implying that $BE = ET$. Hence, $OE \\perp BT$. Let $N$ be the foot of the perpendicular from $S$ to line $BE$.\n![](attached_image_2.png)\nFig. 2. 3\nNote that segments $NE$ and $EM_2$ are the respective perpendicular projections of segments $SO$ and $OF_2$ onto line $BE$. Because $SO = OF_2$, $NE = EM_2$. Because $TE = EB$, it suffices to show that $TN = NE$, which is evident since\n$$\n\\angle STE = \\angle SEB = \\angle SDB = \\angle EDB = \\angle BED = \\angle SET\n$$\nand so triangle $SET$ is isosceles with $SE = ST$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 28274,
    "subject": "Mathematics (Multi-modal)",
    "question": "Given an integer $k \\ge 3$ and a sequence $\\{a_n\\}$ that satisfies $a_k = 2k$ and for each $n > k$,\n$$\na_n = \\begin{cases} a_{n-1} + 1, & \\text{if } a_{n-1} \\text{ and } n \\text{ are coprime,} \\\\ 2n, & \\text{otherwise.} \\end{cases}\n$$\nProve that $a_n - a_{n-1}$ is a prime for infinitely many $n$. (Posed by Zhu Huawei)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Suppose that $a_l = 2l$, $l \\ge k$. Let $p$ be the least prime divisor of $k-1$. Then $(l-1, i) = \\begin{cases} 1, & 1 \\le i < p, \\\\ p, & i = p, \\end{cases}$ and thus\n$$\n(2l + i - 2, l + i - 1) = \\begin{cases} 1, & 1 \\le i < p, \\\\ p, & i = p. \\end{cases}\n$$\nFrom (1) we know that\n$$\na_{l+i-1} = \\begin{cases} 2l + i - 1, & 1 \\le i < p, \\\\ 2l + 2p - 2, & i = p. \\end{cases}\n$$\nThen $a_{l+p-1} - a_{l+p-2} = (2l+2p-2) - (2l+p-2) = p$ is a prime number, $a_{l+p-1} = 2(l+p-1)$. From the discussion above, we know that there are infinitely many $l \\ge k$, such that $a_l = 2l$ and $a_{l+p-1} - a_{l+p-2} = p$ is the least prime divisor of $l-1$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 28275,
    "subject": "Mathematics (Multi-modal)",
    "question": "Given integer $n > 2$, suppose positive real numbers $a_1, a_2, \\dots, a_n$ satisfy $a_k \\le 1$, $k = 1, 2, \\dots, n$.\nLet $A_k = \\frac{a_1 + a_2 + \\dots + a_k}{k}$, $k = 1, 2, \\dots, n$.\nProve $\\left| \\sum_{k=1}^n a_k - \\sum_{k=1}^n A_k \\right| < \\frac{n-1}{2}$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "For $1 \\le k \\le n-1$, we have $0 < \\sum_{i=1}^k a_i \\le k$ and $0 < \\sum_{i=k+1}^n a_i \\le n-k$. By using the fact that $|x-y| < \\max\\{x, y\\}$ for $x, y > 0$, we get\n$$\n\\begin{align*}\n|A_n - A_k| &= \\left| \\left(\\frac{1}{n} - \\frac{1}{k}\\right) \\sum_{i=1}^{k} a_i + \\frac{1}{n} \\sum_{i=k+1}^{n} a_i \\right| \\\\\n&= \\left| \\frac{1}{n} \\sum_{i=k+1}^{n} a_i - \\left(\\frac{1}{k} - \\frac{1}{n}\\right) \\sum_{i=1}^{k} a_i \\right| \\\\\n&< \\max \\left\\{ \\frac{1}{n} \\sum_{i=k+1}^{n} a_i, \\left(\\frac{1}{k} - \\frac{1}{n}\\right) \\sum_{i=1}^{k} a_i \\right\\} \\\\\n&\\le \\max \\left\\{ \\frac{1}{n}(n-k), \\left(\\frac{1}{k} - \\frac{1}{n}\\right) k \\right\\} \\\\\n&= 1 - \\frac{k}{n}.\n\\end{align*}\n$$\nTherefore,\n$$\n\\begin{align*}\n\\left| \\sum_{k=1}^{n} a_k - \\sum_{k=1}^{n} A_k \\right| &= \\left| nA_n - \\sum_{k=1}^{n} A_k \\right| \\\\\n&= \\left| \\sum_{k=1}^{n-1} (A_n - A_k) \\right| \\le \\sum_{k=1}^{n-1} \\left| A_n - A_k \\right| \\\\\n&< \\sum_{k=1}^{n-1} \\left(1 - \\frac{k}{n}\\right) = \\frac{n-1}{2}.\n\\end{align*}\n$$\nThis completes the proof. ☐",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 28276,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $a$, $b$, $c$ be complex numbers such that $|az^2 + bz + c| \\le 1$ for all complex numbers $z$ with $|z| \\le 1$. Find the maximum of $|bc|$. (Posed by Li Weigu)",
    "options": [],
    "answer": "3√3/16",
    "solution": "Write $f(z) = az^2 + bz + c$. We first prove that\n$$\n|f(z)| \\le 1 \\text{ for all } z, |z| \\le 1 \\Leftrightarrow |f(z)| \\le 1 \\text{ for all } z, |z| = 1.\n$$\n\nAssume that $f(z) = a(z - \\alpha)(z - \\beta)$. For any $z$, $|z| < 1$, if $\\alpha = \\beta$, one of the two intersection points of the line through $\\alpha$ and the origin with the unit circle is closer to $\\alpha$ than $z$ is; if $\\alpha \\neq \\beta$, the line through $z$ and perpendicular to the line through $\\alpha$, $\\beta$ intersects the unit circle at two points, one of which is closer to $\\alpha$, $\\beta$ than $z$ is, respectively. So the equivalence holds.\n\nFor any complex number $z$, $|z| = 1$, it is obvious that\n$$\n|f(z)| = |cz^{-2} + bz^{-1} + a|.\n$$\nSo $|ab|_{\\max} = |bc|_{\\max}$. Write $a'z^2 + b'z + c' = e^{i\\alpha}f(e^{i\\beta}z)$.\nOne can choose real numbers $\\alpha$, $\\beta$ such that $a'$, $b'$ are positive real numbers, so one can assume that $a$, $b \\ge 0$ without loss of generality.\n$$\n1 \\ge |f(e^{i\\theta})| \\ge |\\operatorname{Im} f(e^{i\\theta})| = |a \\sin 2\\theta + b \\sin \\theta + \\operatorname{Im} c|.\n$$\nWithout loss of generality we can assume $\\operatorname{Im} c \\ge 0$ (otherwise take a map, $\\theta \\to -\\theta$). For any $\\theta \\in (0, \\frac{\\pi}{2})$,\n$$\n\\begin{align*}\n& 1 \\ge a \\sin 2\\theta + b \\sin \\theta \\ge 2\\sqrt{ab} \\sin 2\\theta \\sin \\theta \\\\\n\\Rightarrow & ab \\le \\frac{1}{4 \\sin 2\\theta \\sin \\theta}, \\quad \\theta \\in (0, \\frac{\\pi}{2}) \\\\\n\\Rightarrow & ab \\le \\min_{\\theta \\in (0, \\frac{\\pi}{2})} \\frac{1}{4 \\sin 2\\theta \\sin \\theta} = \\frac{1}{4 \\max_{\\theta \\in (0, \\frac{\\pi}{2})} (\\sin 2\\theta \\sin \\theta)} = \\frac{3\\sqrt{3}}{16} \\\\\n\\Rightarrow & |bc|_{\\max} = |ab|_{\\max} \\le \\frac{3\\sqrt{3}}{16}.\n\\end{align*}\n$$\n\nAn example of $|bc| = \\frac{3\\sqrt{3}}{16}$ is\n$$\nf(z) = \\frac{\\sqrt{2}}{8}z^2 - \\frac{\\sqrt{6}}{4}z - \\frac{3\\sqrt{2}}{8}, \\\\\n|f(e^{i\\theta})|^2 = 1 - \\frac{3}{8} \\left(\\cos \\theta - \\frac{\\sqrt{3}}{3}\\right)^2 \\le 1.\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 28277,
    "subject": "Mathematics (Multi-modal)",
    "question": "Determine all possible values of positive integer $n$, such that there are $n$ different 3-element subsets $A_1, A_2, \\dots, A_n$ of the set $\\{1, 2, \\dots, n\\}$, with $|A_i \\cap A_j| \\neq 1$ for all $i \\neq j$.",
    "options": [],
    "answer": "All positive multiples of 4",
    "solution": "The set of positive integers satisfying the given condition consists of all positive multiples of $4$. We first prove that $n = 4k$ ($k \\in \\mathbb{Z}_+$) satisfies the condition. Define $A_1, A_2, \\dots, A_{4k}$ as follows: $A_{4i-j} = \\{4i-3, 4i-2, 4i-1, 4i\\} \\setminus \\{4i-j\\}$, for all $1 \\le i \\le k$ and $0 \\le j \\le 3$.\n\nSecond, we want to prove that if $4 \\nmid n$, then such $n$ subsets do not exist. Suppose to the contrary that $A_1, \\dots, A_n$ are different 3-element subsets fulfilling the condition stated in the problem. Let $A_1 = \\{a, b, c\\}$, consider all the given 3-element subsets with non-empty intersection; we may assume that they are $A_2, \\dots, A_m$ after relabeling. Let $U = A_1 \\cup A_2 \\cup \\dots \\cup A_m$. We divide into the following different cases:\n\n* If $|U| = 3$, then $m = 1 < |U|$.\n* If $|U| = 4$, then $m \\le \\binom{4}{3} = 4 = |U|$.\n* If $|U| \\ge 5$, then we prove that $m < |U|$ as follows.\n\nSuppose that $m \\ge |U|$, then for any $2 \\le i, j \\le m$, we have $|A_1 \\cap A_i| = 2$, $|A_1 \\cap A_j| = 2$. As $|A_1| = 3$, we have $A_i \\cap A_j \\ne \\emptyset$. And it follows from $|A_i \\cap A_j| \\ne 1$ that $|A_i \\cap A_j| = 2$. It means that any two distinct subsets of $A_1, \\dots, A_m$ have only two common elements.\n\nConsider the four intersections $A_1 \\cap A_2, A_1 \\cap A_3, A_1 \\cap A_4, A_1 \\cap A_5$, it follows from the pigeon-hole principle that there are two intersections that are equal; by relabeling again, we may assume that they are $A_2 = \\{a, b, d\\}, A_3 = \\{a, b, e\\}$. Then for any $4 \\le i \\le m$, we have $a, b \\in A_i$; otherwise, $A_i$ contains at least one of $a$ or $b$, and the other three elements $c, d$ and $e$, in this case, $|A_i| \\ge 4$, which is impossible. Hence $|U| = m+2$, which contradicts $|U| = m$.\n\nFrom the argument above, one can divide the subsets $A_1, \\dots, A_n$ into several groups, and any two subsets in the same group have non-empty intersections. Moreover, the number of subsets in the same group is not more than the number of elements appearing in this group. As the number of subsets is $n$, which is equal to the number of elements in $\\{1, 2, \\dots, n\\}$, so each group has exactly four subsets. It follows that $4|n$, which contradicts the original assumption $4 \\nmid n$. $\\square$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 28278,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $n_1, n_2, \\dots, n_{26}$ be pairwise distinct positive integers, satisfying:\n\n(1) In the decimal representation of each $n_i$, each digit belongs to the set $\\{1, 2\\}$;\n\n(2) For any $i, j$, $n_j$ cannot be obtained from $n_i$ by adding some digits on the right.\n\nFind the least possible value of $\\sum_{i=1}^{26} S(n_i)$, where $S(m)$ denotes the sum of all digits of $m$ in decimal representation.",
    "options": [],
    "answer": "179",
    "solution": "Given two positive integers $a, b$ in decimal representation, we say $a$ contains $b$ if $a$ can be obtained from $b$ by adding some digits on the right. We first prove a lemma.\n\n**Lemma** Let $n_1, n_2, ..., n_r$ be pairwise distinct positive integers with digit 1 or 2. If none contains another, then the number of $n_i$'s with $S(n_i) \\le t$ is at most $F_t$, where $t$ is an arbitrary positive integer and $F_t$ is a Fibonacci number satisfying $F_1 = 1, F_2 = 2, F_{n+2} = F_{n+1} + F_n$ ($n \\ge 1$).\n\n*Proof of lemma.* We induct on $t$. It is clear for $t = 1, 2$ (when $t = 2$, $1$ and $11$ cannot both appear). Suppose that the lemma is true for all positive integers less than $t$ ($t \\ge 3$); we shall prove that it is also true for $t$. Suppose that without loss of generality $S(n_1), S(n_2), ..., S(n_l)$ are all the numbers with the sum of digits $\\le t$, where $n_1, n_2, ..., n_j$ start with 1 and $n_{j+1}, n_{j+2}, ..., n_l$ start with 2. If one of $n_1, n_2, ..., n_j$ is 1, then $j = 1 \\le F_{t-1}$, otherwise by deleting the first digit of $n_1, n_2, ..., n_j$ we obtain $j$ positive integers with none containing another and the sum of digits $\\le t-1$, and thus we again have $j \\le F_{t-1}$ by the inductive hypothesis. Analogously, we have $l-j \\le F_{t-2}$. And therefore $l \\le F_{t-1} + F_{t-2} = F_t$, i.e. the lemma is also true for $t$. The proof of the lemma is completed.\n\nGoing back to the original problem, we consider a more general question. Replacing 26 by $m$, denote the least possible value of $\\sum_{i=1}^{m} S(n_i)$ by $f(m)$. Fix $m \\ge 3$, and let $n_1, n_2, \\dots, n_m$ be a set of numbers satisfying the conditions in the problem which attains the minimum $f(m)$. Without loss of generality, assume that $\\max_{1 \\le i \\le m} S(n_i) = S(n_1)$ and $n_1$ is maximum among all these numbers attaining the maximum digital sum. Since $m \\ge 3$, $n_1$ contains at least two digits.\n\nIf the last digit of $n_1$ is 1, replace $n_1$ by $\\frac{n_1-1}{10}$, which is not one of $n_2, n_3, \\dots, n_m$, otherwise $n_1$ would contain some $n_i$. Notice that $\\frac{n_1-1}{10}, n_2, \\dots, n_m$ again satisfy the conditions in the problem, for if $n_i$ contains $\\frac{n_i-1}{10}$ for some $i \\ge 2$, then $S(n_i) > S(n_1)$ and $n_i > n_1$, a contradiction. Now\n$$\nS\\left(\\frac{n_1-1}{10}\\right) + S(n_2) + \\dots + S(n_m) = f(m) - 1,\n$$\na contradiction to the definition of $f(m)$. Thus, the last digit of $n_1$ is 2.\n\nIf $n_1-1$ is not one of $n_2, n_3, \\dots, n_m$, replace $n_1$ by $n_1-1$, and these $m$ numbers again satisfy the conditions in the problem, for if $n_i$ contains $n_i-1$ for some $i \\ge 2$, then $n_i$ must be $10(n_i-1)+1$, $S(n_i) = S(n_1)$; however, $n_i > n_1$ is a contradiction to the choice of $n_1$. Now\n$$\nS\\left(\\frac{n_1-1}{10}\\right) + S(n_2) + \\dots + S(n_m) = f(m) - 1,\n$$\na contradiction to the definition of $f(m)$. Thus, $n_1-1$ appears in $n_2, n_3, \\dots, n_m$.\n\nWithout loss of generality, assume that $n_2 = n_1 - 1$. Consider $\\frac{n_1-2}{10}, n_3, \\dots, n_m$; since $\\frac{n_1-2}{10} \\ne n_i$ for $i \\ge 3$, these are $m-1$ pairwise distinct numbers. There is no containment among $n_2, \\dots, n_m$, and $\\frac{n_1-2}{10}$ does not contain any of $n_3, \\dots, n_m$, otherwise $n_1$ would contain that number. If one of $n_3, \\dots, n_m$ contains $\\frac{n_1-2}{10}$, say $n_3$, since $S\\left(\\frac{n_1-2}{10}\\right) = S(n_1)-2$, $n_3$ is obtained by adding 1, 2 or 11 after $\\frac{n_1-2}{10}$. Adding 1 or 2 yields $n_2, n_1$, and we must have $n_3 = 100 \\cdot \\frac{n_1-2}{10} + 11 = 10n_1 - 9$. Now $S(n_3) = S(n_1)$ and $n_3 > n_1$, a contradiction to the choice of $n_1$. So $\\frac{n_1-2}{10}$, $n_3, \\dots, n_m$ satisfy the conditions in the problem, and therefore the sum of their digits is at least $f(m-1)$. Thus,\n$$f(m) - S(n_1) - (S(n_1) - 1) + (S(n_1) - 2) \\ge f(m-1),$$\ni.e.\n$$\nf(m) \\ge f(m-1) + S(n_1) + 1.\n$$\nLet $u$ be such that $F_{u-1} < m \\le F_u$. By the lemma, there are at most $F_{u-1}$ of $S(n_1), S(n_2), \\dots, S(n_m)$ less than or equal to $u-1$, so $S(n_1) \\ge u$, and\n$$\nf(m) \\ge f(m-1) + u + 1. \\qquad \\textcircled{1}\n$$\nIt is easy to see that $f(1) = 1$, $f(2) = 3$, and hence\n$$\n\\begin{aligned}\nf(26) &= f(2) + \\sum_{i=3}^{26} (f(i) - f(i-1)) \\\\\n&= f(2) + (f(3) - f(2)) + (f(5) - f(3)) + (f(8) - f(5)) \\\\\n&\\quad + (f(13) - f(8)) + (f(21) - f(13)) + (f(26) - f(21)) \\\\\n&\\ge 3 + 4 \\times 1 + 5 \\times 2 + 6 \\times 3 + 7 \\times 5 + 8 \\times 8 + 9 \\\\\n&\\quad \\times 5 \\text{ (by equation \\textcircled{1})} \\\\\n&= 179,\n\\end{aligned}\n$$\ni.e. $\\sum_{i=1}^{26} S(n_i) \\ge 179$.\n\nOn the other hand, by the property of Fibonacci numbers, there are exactly 8 numbers consisting of digits 1 and 2, with digital sum 5, denoted by $a_1, a_2, ..., a_8$, and there are exactly 13 such numbers with digital sum 6, denoted by $b_1, b_2, ..., b_{13}$. Add a digit 2 after each of $a_1, a_2, ..., a_8$, denoting these new numbers by $c_1, c_2, ..., c_8$. Add a digit 1 (resp. digit 2) to each of $b_1, b_2, b_3, b_4, b_5$, denoting these new numbers by $d_1, d_2, d_3, d_4, d_5$ (resp. $e_1, e_2, e_3, e_4, e_5$). Now consider\n$c_1, c_2, ..., c_8, d_1, d_2, ..., d_5, e_1, e_2, ..., e_5, b_6, b_7, ..., b_{13}$.\nThese are pairwise distinct numbers consisting of digits 1 and 2, with total digital sum $7 \\times 8 + 7 \\times 5 + 8 \\times 5 + 6 \\times 8 = 179$. And there is none containing another. In fact, if $x$ contains $y$, since their digital sum is either 6, 7 or 8, and a number with digital sum 8 ends up with 2, it follows that $x$ has exactly one more digit than $y$. However, deleting the last digit of $d_1, d_2, ..., d_5$ and $e_1, e_2, ..., e_5$ yields $b_1, b_2, ..., b_5$, and deleting the last digit of $c_1, c_2, ..., c_8$ yields $a_1, a_2, ..., a_8$, none of which is in this set of 26 numbers.\n\nWe conclude that the least possible value of $\\sum_{i=1}^{26} S(n_i)$ is 179.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 28279,
    "subject": "Mathematics (Multi-modal)",
    "question": "The number of integral points (i.e., the points whose $x$- and $y$-coordinates are both integers) within the area (not including the boundary) enclosed by the right branch of hyperbola $x^2 - y^2 = 1$ and line $x = 100$ is ______.",
    "options": [],
    "answer": "9800",
    "solution": "By symmetry, we only need to consider the part of the area above the $x$-axis. Suppose line $y = k$ intercepts the right branch of the hyperbola and line $x = 100$ at points $A_k$ and $B_k$ ($k = 1, 2, \\dots, 99$), respectively. Then the number of integral points within the segment $A_k B_k$ is $99 - k$. Therefore, the number of integral points within the area above the $x$-axis is\n$$\n\\sum_{k=1}^{99} (99 - k) = 99 \\times 49 = 4851.\n$$\nFinally, we obtain the total number of integral points within the whole area as $2 \\times 4851 + 98 = 9800$. $\\square$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 28280,
    "subject": "Mathematics (Multi-modal)",
    "question": "It is known that $\\{a_n\\}$ is an arithmetic sequence with non-zero common difference and $\\{b_n\\}$ a geometric sequence, satisfying $a_1 = 3$, $b_1 = 1$, $a_2 = b_2$, $3a_5 = b_3$; furthermore, there are constants $\\alpha$ and $\\beta$ such that for every positive integer $n$, we have $a_n = \\log_a b_n + \\beta$. Then $\\alpha + \\beta = \\underline{\\hspace{2cm}}$.",
    "options": [],
    "answer": "3^(1/3) + 3",
    "solution": "Let the common difference of $\\{a_n\\}$ be $d$ and the common ratio of $\\{b_n\\}$ be $q$. Then\n$$\n3 + d = q, \\qquad \\textcircled{1}\n$$\n$$\n3(3 + 4d) = q^2. \\qquad \\textcircled{2}\n$$\nSubstituting $\\textcircled{1}$ into $\\textcircled{2}$, we have $9 + 12d = d^2 + 6d + 9$. Then we get $d = 6$ and $q = 9$.\nTherefore, $3 + 6(n - 1) = \\log_a 9^{n-1} + \\beta$ or $6n - 3 = (n - 1)\\log_a 9 + \\beta$ holds for every positive integer $n$. Letting $n = 1$ and $n = 2$ in turn, we find that $\\alpha = \\sqrt[3]{3}$ and $\\beta = 3$.\nConsequently, $\\alpha + \\beta = \\sqrt[3]{3} + 3$. $\\square$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 28281,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $G = G(V; E)$ be a simple graph with vertex set $V$ and edge set $E$, and assume that $|V| = n$. A map $f: V \\rightarrow \\mathbb{Z}$ is said to be good if $f$ satisfies:\n$$\n(1) \\quad \\sum_{v \\in V} f(v) = |E|;\n$$\n(2) If one colors arbitrarily some vertices into red, there exists a red vertex $v$, such that $f(v)$ is not greater than the number of vertices adjacent to $v$ that are not colored into red.\n\nLet $m(G)$ be the number of good maps $f$. Show that if each vertex of $V$ is adjacent to at least one other vertex, then $n \\le m(G) \\le n!$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Given an ordering $\\tau = (v_1, v_2, ..., v_n)$ on the vertices in $V$, we associate a map $f_\\tau: V \\to \\mathbb{Z}$ as follows: $f_\\tau(v)$ is equal to the number of vertices in $V$ that are ordered preceding $v$. We claim that $f_\\tau$ is good.\n\nEach edge is counted exactly once in $\\sum_{v \\in V} f_\\tau(v)$, for an edge $e \\in E$ with vertices $u, v \\in V$ such that $u$ is ordered before $v$ in $\\tau$; then $e$ is counted once in $f_\\tau(v)$. Thus,\n$$\n\\sum_{v \\in V} f_{\\tau}(v) = |E|.\n$$\nFor any nonempty subset $A \\subseteq V$ of all red vertices, choose $v \\in A$ with the most preceding orderings in $\\tau$. Then by definition, $f_\\tau(v)$ is not greater than the number of vertices adjacent to $v$ that are not colored into red. We have verified that $f_\\tau$ is good.\n\nConversely, given any good map $f: V \\to \\mathbb{Z}$, we claim that $f = f_\\tau$ for some ordering $\\tau$ of $V$.\n\nFirst, let the red vertex set $A = V$. By the condition (2) in the problem, there exists $v \\in A$ such that $f(v) \\le 0$, and denote one of such vertices by $v_1$. Assuming that we have already chosen $v_1, ..., v_k$ from $V$, if $k < n$, set the red vertex set $A = V - \\{v_1, ..., v_k\\}$. By the condition (2) in the problem, there exists $v \\in A$ such that $f(v)$ is less than or equal to the number of vertices in $\\{v_1, ..., v_k\\}$ that are adjacent to $v$. Denote one of such vertices by $v_{k+1}$. Continuing in this way, we order the vertices by $\\tau = (v_1, v_2, ..., v_n)$. By the construction we have $f(v) \\le f_r(v)$ for any $v \\in V$. By the condition (1) in the problem, we have\n$$\n|E| = \\sum_{v \\in V} f(v) \\le \\sum_{v \\in V} f_r(v) = |E|,\n$$\nand therefore $f(v) = f_r(v)$ for any $v \\in V$.\n\nWe have shown that for any ordering $\\tau$, $f_r$ is good, and any good map $f$ is $f_r$ for some $\\tau$. Since the number of orderings on $V$ is $n!$, we see that $m(G) \\le n!$ (note that two distinct orderings may result in the same map).\n\nNext, we prove that $n \\le m(G)$. Assume at the moment that $G$ is connected. Pick arbitrarily $v_1 \\in V$. By the connectivity, we may choose $v_2 \\in V - \\{v_1\\}$ such that $v_2$ is adjacent to $v_1$, and again we may choose $v_3 \\in V - \\{v_1, v_2\\}$ such that $v_3$ is adjacent to at least one of $v_1, v_2$. Continuing in this way, we get an ordering $\\tau = (v_1, v_2, ..., v_n)$ such that $v_k$ is adjacent to at least one of the vertices preceding it under ordering $\\tau$, for any $2 \\le k \\le n$. Thus, $f_r(v_1) = 0$ and $f_r(v_k) > 0$ for $2 \\le k \\le n$. Since $v_1$ may be arbitrary, we have at least $n$ good maps.\n\nIn general, if $G$ is a union of its connected components $G_1, ..., G_k$, since each vertex is adjacent to at least another vertex, each component has at least two vertices, denote by $n_1, ..., n_k \\ge 2$ the number of vertices of these components. For each $G_i$, we have at least $n_i$ good maps on its vertices, $i = 1, ..., k$. It is easy to see that patching good maps on $G_i$'s together results in a good map on $G$, and thus\n$$\nm(G) \\ge n_1 n_2 \\cdots n_k \\ge n_1 + n_2 + \\cdots + n_k = n.\n$$\nWe conclude that $n \\le m(G) \\le n!$.",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 28282,
    "subject": "Mathematics (Multi-modal)",
    "question": "Given two integers $m, n$ greater than $1$, and integers $a_1 < a_2 < \\cdots < a_m$, prove that there exists a set $T$ of integers with $|T| \\le 1 + \\frac{a_m - a_1}{2n + 1}$ such that each $a_i$ can be written as $a_i = t + s$ for some $t \\in T$, and $s \\in [-n, n]$. (Posed by Leng Gangsong)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Write $a_1 = a$, $a_m = b$, $b - a = (2n + 1)q + r$, where $q, r \\in \\mathbb{Z}$ and $0 \\le r \\le 2n$. Take\n\n$$\nT = \\{a + n + (2n + 1)k \\mid k = 0, 1, \\dots, q\\}.\n$$\n\nThen $|T| = q + 1 \\le 1 + \\frac{b - a}{2n + 1}$. We have the set\n\n$$\nB = \\{t + s \\mid t \\in T, s = -n, -n + 1, \\dots, n\\} = \\{a, a + 1, \\dots, a + (2n + 1)q + 2n\\}.\n$$\n\nNote that $a + (2n + 1)q + 2n \\ge a + (2n + 1)q + r = b$, so each $a_i$ belongs to $B$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 28283,
    "subject": "Mathematics (Multi-modal)",
    "question": "The code setting of a cipher lock is established on an $n$-regular-polygon with vertices $A_1, A_2, \\dots, A_n$: each vertex is assigned a number (0 or 1) and a color (red or blue), such that either the numbers or the colors on each pair of adjacent vertices are the same. We ask: How many code-sets can be realized for this lock?",
    "options": [],
    "answer": "Odd n: 3^n + 1; Even n: 3^n + 3",
    "solution": "Given an arbitrary code-set for the lock, if two adjacent vertices have different numbers, we label the sides linking them by letter $a$; if they have different colors, we label it by $b$; if both the numbers and colors are the same, we label it by $c$. Once the number and color on vertex $A_1$ are set (there are four different sets for it), we can then set $A_2, A_3, \\dots, A_n$ one by one according to the letters labelled on each side. In order to let it return to the initial set of $A_1$ finally, the numbers of sides labelled $a$ and $b$ must be both even. So the number of code-sets for the lock is four times the number of labelled-side sequences which satisfy the condition that the numbers of sides labelled by $a$ and $b$ are both even.\n\nSuppose there are $2i$ ($0 \\le i \\le \\lfloor \\frac{n}{2} \\rfloor$) sides labelled by $a$, and $2j$ ($0 \\le j \\le \\lfloor \\frac{n-2i}{2} \\rfloor$) sides labelled by $b$. Then there are $C_n^{2i}$ ways to label $2i$ sides by $a$ from $n$ ones, $C_{n-2i}^{2j}$ ways to label $2j$ sides by $b$ from $n-2i$ ones, and the remaining sides are labelled by $c$. Therefore, by the Multiplication Principle, there are $C_n^{2i} C_{n-2i}^{2j}$ ways to label all the sides. So there are totally\n$$\n4 \\sum_{i=0}^{\\lfloor \\frac{n}{2} \\rfloor} \\left( C_n^{2i} \\sum_{j=0}^{\\lfloor \\frac{n-2i}{2} \\rfloor} C_{n-2i}^{2j} \\right)\n$$\ncode-sets for the lock. Here we stipulate $C_0^0 = 1$.\n\nWhen $n$ is odd, we have $n - 2i > 0$, and then\n$$\n\\sum_{j=0}^{\\left\\lfloor \\frac{n-2i}{2} \\right\\rfloor} C_{n-2i}^{2j} = 2^{n-2i-1}.\n$$\nSubstituting it into the previous formula, we get\n$$\n\\begin{aligned}\n4 \\sum_{i=0}^{\\left\\lfloor \\frac{n}{2} \\right\\rfloor} \\left( C_n^{2i} \\sum_{j=0}^{\\left\\lfloor \\frac{n-2i}{2} \\right\\rfloor} C_{n-2i}^{2j} \\right) &= 4 \\sum_{i=0}^{\\left\\lfloor \\frac{n}{2} \\right\\rfloor} (C_n^{2i} 2^{n-2i-1}) \\\\\n&= 2 \\sum_{i=0}^{\\left\\lfloor \\frac{n}{2} \\right\\rfloor} (C_n^{2i} 2^{n-2i}) \\\\\n&= \\sum_{k=0}^{n} C_n^k 2^{n-k} + \\sum_{k=0}^{n} C_n^k 2^{n-k} (-1)^k \\\\\n&= (2+1)^n + (2-1)^n \\\\\n&= 3^n + 1.\n\\end{aligned}\n$$\n\nWhen $n$ is even, if $i < \\frac{n}{2}$, then the above remains true; if $i = \\frac{n}{2}$, then all the sides of the polygon are labelled by $a$, and that means there is only one way to label the sides. Therefore, there are totally\n$$\n\\begin{aligned}\n4 \\sum_{i=0}^{\\left\\lfloor \\frac{n}{2} \\right\\rfloor} \\left( C_n^{2i} \\sum_{j=0}^{\\left\\lfloor \\frac{n-2i}{2} \\right\\rfloor} C_{n-2i}^{2j} \\right) &= 4 \\times \\left( 1 + \\sum_{i=0}^{\\left\\lfloor \\frac{n}{2} \\right\\rfloor} (C_n^{2i} 2^{n-2i-1}) \\right) \\\\\n&= 2 + 4 \\sum_{i=0}^{\\left\\lfloor \\frac{n}{2} \\right\\rfloor} (C_n^{2i} 2^{n-2i-1}) \\\\\n&= 3^n + 3\n\\end{aligned}\n$$\ncode-sets for the lock.\n\nIn summary,\nthe number of code-sets for the lock is\n$$\n\\begin{cases}\n3^n + 1 & \\text{when } n \\text{ is odd,} \\\\\n3^n + 3 & \\text{when } n \\text{ is even.}\n\\end{cases}\n$$",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 28284,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $x_1, x_2, \\dots, x_n$ (where $n \\ge 2$) be real numbers with\n$$x_1^2 + x_2^2 + \\dots + x_n^2 = 1.$$\nProve that\n$$\n\\sum_{k=1}^{n} \\left( 1 - \\frac{k}{\\sum_{i=1}^{n} ix_i^2} \\right)^2 \\cdot \\frac{x_k^2}{k} \\le \\left( \\frac{n-1}{n+1} \\right)^2 \\sum_{k=1}^{n} \\frac{x_k^2}{k}.\n$$\nDetermine when the equality holds.",
    "options": [],
    "answer": "Equality holds if and only if x2 = x3 = ... = x_{n-1} = 0 and x1^2 = xn^2 = 1/2.",
    "solution": "**Comment:** Expanding the left-hand side of the desired inequality gives\n$$\n\\begin{align*}\n& \\sum_{k=1}^{n} \\left( 1 - \\frac{k}{\\sum_{i=1}^{n} ix_i^2} \\right)^2 \\cdot \\frac{x_k^2}{k} \\\\\n&= \\sum_{k=1}^{n} \\frac{x_k^2}{k} - \\sum_{k=1}^{n} \\frac{2x_k^2}{\\sum_{i=1}^{n} ix_i^2} + \\sum_{k=1}^{n} \\frac{kx_k^2}{\\left(\\sum_{i=1}^{n} ix_i^2\\right)^2} \\\\\n&= \\sum_{k=1}^{n} \\frac{x_k^2}{k} - \\frac{1}{\\sum_{i=1}^{n} ix_i^2} \\sum_{k=1}^{n} \\frac{1}{2x_k^2} + \\frac{1}{\\left(\\sum_{i=1}^{n} ix_i^2\\right)^2} \\sum_{k=1}^{n} kx_k^2 \\\\\n&= \\sum_{k=1}^{n} \\frac{x_k^2}{k} - \\frac{2}{\\sum_{i=1}^{n} ix_i^2} + \\frac{1}{\\sum_{i=1}^{n} ix_i^2} \\\\\n&= \\sum_{k=1}^{n} \\frac{x_k^2}{k} - \\frac{1}{\\sum_{i=1}^{n} ix_i^2}.\n\\end{align*}\n$$\nWe want to show that\n$$\n\\sum_{k=1}^{n} \\frac{x_k^2}{k} - \\frac{1}{\\sum_{i=1}^{n} ix_i^2} \\le \\left(\\frac{n-1}{n+2}\\right)^2 \\sum_{k=1}^{n} \\frac{x_k^2}{k} = \\sum_{k=1}^{n} \\frac{x_k^2}{k} - \\frac{4n}{(n+1)^2} \\sum_{k=1}^{n} \\frac{x_k^2}{k}\n$$\nor\n$$\n\\frac{4n}{(n+1)^2} \\sum_{k=1}^{n} \\frac{x_k^2}{k} \\le \\frac{1}{\\sum_{i=1}^{n} i x_i^2},\n$$\nthat is,\n$$\n\\left( \\sum_{k=1}^{n} \\frac{x_k^2}{k} \\right) \\left( \\sum_{k=1}^{n} kx_k^2 \\right) \\le \\frac{(n+1)^2}{4n}. \\qquad \\textcircled{1}\n$$\nWe present two proofs of the above inequality.\n\n**Solution 1.** We rewrite **1** as\n$$\n4n \\left( \\sum_{k=1}^{n} \\frac{x_k^2}{k} \\right) \\left( \\sum_{k=1}^{n} kx_k^2 \\right) \\le (n+1)^2.\n$$\nBy the AM-GM inequality, we have\n$$\n\\begin{aligned} 4n \\left( \\sum_{k=1}^{n} \\frac{x_k^2}{k} \\right) \\left( \\sum_{k=1}^{n} kx_k^2 \\right) &= 4 \\left( \\sum_{k=1}^{n} \\frac{nx_k^2}{k} \\right) \\left( \\sum_{k=1}^{n} kx_k^2 \\right) \\\\ &\\le \\left( \\sum_{k=1}^{n} \\frac{nx_k^2}{k} + \\sum_{k=1}^{n} kx_k^2 \\right)^2 \\\\ &= \\left( \\sum_{k=1}^{n} \\left( \\frac{n}{k} + k \\right) x_k^2 \\right)^2. \\end{aligned}\n$$\nIt suffices to show that\n$$\n\\frac{n}{k} + k \\le n + 1\n$$\nor\n$$\n0 \\le nk + k - k^2 - n = (n-k)(k-1),\n$$\nwhich is evident.\nNow, we consider the equality case. Note that the last inequality is strict for $1 < k < n$. Hence, we must have $x_2 = \\cdots = x_{n-1} = 0$. For the AM-GM inequality to hold, we must have\n$$\n\\sum_{k=1}^{n} \\frac{nx_k^2}{k} = \\sum_{k=1}^{n} kx_k^2 \\text{ or } nx_1^2 + x_n^2 = x_1^2 + nx_n^2\n$$\nwith $x_1^2 + x_n^2 = 1$. We must have $x_1^2 = x_n^2 = \\frac{1}{2}$ and $x_2 = \\cdots = x_{n-1} = 0$.\n\n\n**Solution 2.** (By Lynnelle Ye) We write ④ as\n$$\n(n + 1)^2 - 4n \\left( \\sum_{k=1}^{n} \\frac{x_k^2}{k} \\right) \\left( \\sum_{k=1}^{n} kx_k^2 \\right) \\ge 0.\n$$\nNote that the left-hand side of the above inequality is the discriminant of the quadratic (in $t$):\n$$\nf(t) = n(x_1^2 + 2x_2^2 + \\cdots + nx_n^2)t^2 - (n+1)t + \\left(x_1^2 + \\frac{x_2^2}{2} + \\cdots + \\frac{x_n^2}{n}\\right).\n$$\nIt suffices to show that $f(t)$ has a real root. Because the leading coefficient of $f(x)$ is $n(x_1^2 + 2x_2^2 + \\cdots + nx_n^2)$, which is positive, it remains to be shown that $f(t) \\le 0$ for some $t$. Because $x_1^2 + \\cdots + x_n^2 = 1$, we have\n$$\n\\begin{aligned} f(t) &= n(x_1^2 + 2x_2^2 + \\cdots + nx_n^2)t^2 - (n+1)(x_1^2 + x_2^2 + \\cdots + x_n^2)t + \\left(x_1^2 + \\frac{x_2^2}{2} + \\cdots + \\frac{x_n^2}{n}\\right)t \\\\ &= \\sum_{k=1}^{n} \\left( nkx_k^2t^2 - (n+1)x_k^2t + \\frac{x_k^2}{k} \\right) \\\\ &= \\sum_{k=1}^{n} x_k^2(nkt - 1)\\left(t - \\frac{1}{k}\\right). \\end{aligned}\n$$\nIt is easy to see that $f(\\frac{1}{n}) \\le 0$ because for $t = \\frac{1}{n}$, each summand\n$$\nx_k^2(nkt - 1)\\left(t - \\frac{1}{k}\\right) = \\frac{x_k^2(k-1)(k-n)}{nk} \\le 0,\n$$\ncompleting our proof. For the equality case of the given inequality, we must have equality case for the above inequality for every $k$; that is, $x_k = 0$ for $2 \\le k \\le n - 1$. It is then not difficult to obtain that\n$$\nx_1^2 = x_2^2 = \\frac{1}{2}. \\quad \\square",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 28285,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $a$ and $b$ be positive integers such that $1 \\leq a < b \\leq 100$. If there exists a positive integer $k$ such that $ab = a^k + b^k$, then we say that the pair $(a, b)$ is good. Determine the number of good pairs.",
    "options": [],
    "answer": "96",
    "solution": "Let $(a, b) = d$, $a = sd$, $b = td$, $(s, t) = 1$, $t > 1$, then $std^2 = d^k(s^k + t^k)$. So $k \\geq 2$ and $st = d^{k-2}(s^k + t^k)$. Since $(st, s^k + t^k) = 1$, we have $st = d^{k-2}$. Therefore, any prime factor of $st$ can be divided by $d$.\nIf there is a prime factor $p$ of $s$ or $t$ no less than $11$, then $p$ divides $d$. So $p^2$ divides $a$ or $p^2$ divides $b$, but $p^2 > 100$, which is a contradiction.\nSo the prime factor of $st$ may be $2$, $3$, $5$ or $7$.\nIf there are at least three prime factors of $st$ among $2$, $3$, $5$, $7$, then there is a prime factor of $s$ or $t$ no less than $5$. And $d > 2 \\times 3 \\times 5 = 30$, so that $a$ or $b \\geq 5d > 100$, which is a contradiction. The prime factor set of $st$ cannot be $\\{3, 7\\}$, otherwise, $a$ or $b \\geq 7 \\times 3 \\times 7 > 100$, which is a contradiction.\nSimilarly, the prime factor set of $st$ cannot be $\\{5, 7\\}$.\nTherefore, the prime factor set of $st$ can only be $\\{2\\}$, $\\{3\\}$, $\\{5\\}$, $\\{7\\}$, $\\{2, 3\\}$, $\\{2, 5\\}$, $\\{2, 7\\}$ or $\\{3, 5\\}$.\n\ni) If the prime factor set of $st$ is $\\{3, 5\\}$, then $d$ can only be $15$. Then, $s = 3$, $t = 5$. So there is one good pair $(a, b) = (45, 75)$.\n\nii) If the prime factor set of $st$ is $\\{2, 7\\}$, then $d$ can only be $14$. Then $s = 2$, $t = 7$ or $s = 4$, $t = 7$. So there are two good pairs $(a, b) = (28, 98)$ and $(56, 98)$.\n\niii) If the prime factor of $st$ is $\\{2, 5\\}$, then $d$ can only be $10$ or $20$.\nFor $d = 10$, then $s = 2$, $t = 5$; $s = 1$, $t = 10$; $s = 4$, $t = 5$; $s = 5$, $t = 8$.\nFor $d = 20$, then $s = 2$, $t = 5$; $s = 4$, $t = 5$.\nThere are six good pairs.\n\niv) If the prime factor of $st$ is $\\{2, 3\\}$, then $d$ can only be $6$, $12$, $18$, $24$ or $30$.\nFor $d = 6$, $s = 1$, $t = 6$; $s = 1$, $t = 12$; $s = 2$, $t = 3$; $s = 2$, $t = 9$; $s = 3$, $t = 4$; $s = 3$, $t = 8$; $s = 3$, $t = 16$; $s = 4$, $t = 9$; $s = 8$, $t = 9$; $s = 9$, $t = 16$.\nFor $d = 12$, $s = 1$, $t = 6$; $s = 2$, $t = 3$; $s = 3$, $t = 4$; $s = 3$, $t = 8$.\nFor $d = 18$, $s = 2$, $t = 3$; $s = 3$, $t = 4$.\nFor $d = 24$, $s = 2$, $t = 3$; $s = 3$, $t = 4$. $d = 30$, $s = 2$, $t = 3$.\nThere are $19$ good pairs.\n\nv) If the prime factor set of $st$ is $\\{7\\}$, then $s = 1$, $t = 7$, $d$ can only be $7$ or $14$.\nSo, there are two good pairs.\n\nvi) If the prime factor set of $st$ is $\\{5\\}$, then $s = 1$, $t = 5$, $d$ can only be $5$, $10$, $15$ or $20$. So, there are four good pairs.\n\nvii) If the prime factor set of $st$ is $\\{3\\}$ then we have the following:\nwhen $s = 1$, $t = 3$, $d$ can only be $3$, $6$, $\\dots$, or $33$;\nwhen $s = 1$, $t = 9$, $d$ can only be $3$, $6$ or $9$;\nwhen $s = 1$, $t = 27$, $d$ can only be $3$.\nThere are $15$ good pairs.\n\nviii) If the prime factor set of $st$ is $\\{2\\}$, then we have the following:\nwhen $s = 1$, $t = 2$, $d$ can only be $2$, $4$, $\\dots$, or $50$;\nwhen $s = 1$, $t = 4$, then $d$ can only be $2$, $4$, $\\dots$, or $24$;\nwhen $s = 1$, $t = 8$, then $d$ can only be $2$, $4$, $\\dots$, or $12$;\nwhen $s = 1$, $t = 16$, then $d$ can only be $2$, $4$ or $6$;\nwhen $s = 1$, $t = 32$, then $d$ can only be $2$.\nThere are $47$ good pairs.\n\nTherefore, there are all together $1 + 2 + 6 + 19 + 2 + 4 + 15 + 47 = 96$ good pairs. $\\boxed{96}$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 28286,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $a_1, a_2, \\dots, a_n, b_1, b_2, \\dots, b_n$ be non-negative numbers satisfying the following conditions simultaneously:\n$$\n(1) \\quad \\sum_{i=1}^{n} (a_i + b_i) = 1;\n$$\n$$\n(2) \\quad \\sum_{i=1}^{n} i(a_i - b_i) = 0;\n$$\n$$\n(3) \\quad \\sum_{i=1}^{n} i^2 (a_i + b_i) = 10.\n$$\nProve that $\\max\\{a_k, b_k\\} \\le \\frac{10}{10+k^2}$ for all $1 \\le k \\le n$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "For any $1 \\le k \\le n$, it follows from the given conditions and Cauchy's Inequality that\n$$\n\\begin{align*}\n(ka_k)^2 &\\le \\left(\\sum_{i=1}^n i a_i\\right)^2 = \\left(\\sum_{i=1}^n i b_i\\right)^2 \\\\\n&\\le \\left(\\sum_{i=1}^n i^2 b_i\\right) \\left(\\sum_{i=1}^n b_i\\right) \\\\\n&= (10 - \\sum_{i=1}^n i^2 a_i) (1 - \\sum_{i=1}^n a_i) \\\\\n&\\le (10 - k^2 a_k) (1 - a_k) \\\\\n&= 10 - (10 + k^2)a_k + k^2 a_k^2.\n\\end{align*}\n$$\nIt follows from that $a_k \\le \\frac{10}{10+k^2}$. Similarly, $b_k \\le \\frac{10}{10+k^2}$, and hence the result follows. $\\square$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 28287,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $k$ be an integer and $k > 1$. Define a sequence $\\{a_n\\}$ as follows: $a_0 = 0$, $a_1 = 1$ and $a_{n+1} = k a_n + a_{n-1}$ for $n = 1, 2, \\dots$\nDetermine, with proof, all possible $k$ for which there exist non-negative integers $\\ell, m$ ($\\ell \\ne m$) and positive integers $p, q$ such that $a_\\ell + k a_p = a_m + k a_q$.",
    "options": [],
    "answer": "k = 2",
    "solution": "The answer is $k = 2$.\nIf $k = 2$, then $a_0 = 0$, $a_1 = 1$, $a_2 = 2$, so $a_0 + 2 a_2 = a_2 + 2 a_1 = 4$. Hence, $(\\ell, m) = (0, 2)$ and $(p, q) = (2, 1)$.\n\nFor $k \\ge 3$, it follows from the recurrent relation that the sequence $\\{a_n\\}$ is strictly increasing, and $k \\mid a_{n+1} - a_{n-1}$ for all $n \\ge 1$. In particular, for $n \\ge 0$, we have\n$$\na_{2n} \\equiv a_0 \\equiv 0 \\pmod{k}, \\quad a_{2n+1} \\equiv a_1 \\equiv 1 \\pmod{k}. \\quad (*)\n$$\nSuppose there exist $\\ell, m \\in \\mathbb{N}$, and $p, q \\in \\mathbb{Z}_+$ such that $\\ell \\ne m$ and $a_\\ell + k a_p = a_m + k a_q$. We may assume that $\\ell < m$, and divide into the following cases.\n\na. $p < \\ell < m$: Then $a_\\ell + k a_p \\le a_\\ell + k a_{\\ell-1} < k a_\\ell + a_{\\ell-1} = a_{\\ell+1} \\le a_m < a_m + k a_q$, which contradicts the original assumption.\n\nb. $\\ell = p < m$: If $\\ell = p = m-1$, then we have $a_m + k a_q = a_\\ell + k a_p = (k+1) a_{m-1}$, and modulo $k$ we have $a_m \\equiv a_{m-1} \\pmod{k}$, which contradicts $(*)$.\nIf $\\ell = p < m-1$, then $a_\\ell + k a_p < a_{m-2} + k a_{m-2} = a_m < a_m + k a_q$, which contradicts the original assumption.\n\nc. $\\ell < p < m$: In this case, from $q$ we have $a_q > 0$, then\n$a_\\ell + k a_p \\le k a_p + a_{p-1} = a_{p+1} \\le a_m < a_m + k a_q$, which contradicts the original assumption.\n\nd. $\\ell < m \\le p$: Then $a_p > \\frac{a_\\ell + k a_p - a_m}{k} = a_q$. It follows from\n$$\nk a_q + a_m = k a_p + a_\\ell \\ge k a_p \\quad ①\n$$\nthat\n$$\na_q \\ge a_p - \\frac{a_m}{k} \\ge a_p - \\frac{a_p}{k} = \\frac{k-1}{k} a_p. \\quad ②\n$$\nNote that\n$$\na_p = k a_{p-1} + a_{p-2} \\ge k a_{p-\\ell}, \\quad ③\n$$\nhence\n$$\na_p > a_q \\ge \\frac{k-1}{k} a_p \\ge (k-1) a_{p-1} \\ge a_{p-1}. \\quad ④\n$$\nThen by increasing property of $\\{a_n\\}$ and ④, we have $a_q = a_{p-1}$, so inequalities ①–③ turn into equalities. It follows from ② that $m = p$, and from ③ that $p = 2$. Hence, $a_q = a_{p-1} = a_1 = 1$. And it follows from ① that $\\ell = 0$. From $a_\\ell + k a_p = a_m + k a_q$, we have $k^2 = k + k = 2k$, so $k = 2$, which is impossible.\n\nSo $k=2$ is the only solution. □",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 28288,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $f(x)$ and $g(x)$ be strictly increasing linear functions from $\\mathbf{R}$ to $\\mathbf{R}$ such that $f(x)$ is an integer if and only if $g(x)$ is an integer. Prove that for any real number $x$, $f(x) - g(x)$ is an integer.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "By symmetry, we may assume that $a \\ge c$.\nWe claim that $a = c$. Assume on the contrary that $a > c$. Because $a > c > 0$, the ranges of $f$ and $g$ are both $\\mathbf{R}$. There is a $x_0$ such that $f(x_0) = a x_0 + b$ is an integer. Hence $g(x_0) = c x_0 + d$ is also an integer. But then,\n$$\nf\\left(x_0 + \\frac{1}{a}\\right) = a x_0 + b + 1\n$$\nand\n$$\ng\\left(x_0 + \\frac{1}{a}\\right) = c x_0 + d + \\frac{c}{a}.\n$$\nBut this is impossible because we cannot have two integers $g(x_0)$ and $g(x_1)$ that have positive difference $\\frac{c}{a}$ which is less than 1.\nTherefore, we can write $f(x) = a x + b$ and $g(x) = a x + d$ for some real numbers $a, b, d$ with $a > 0$.\nThen $b - d = f(x_0) - g(x_0)$ must be an integer, that is,\n$$\nf(x) - g(x) = b - d\n$$\nis an integer.\n□",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 28289,
    "subject": "Mathematics (Multi-modal)",
    "question": "Given integer $a_1 \\ge 2$, for $n \\ge 2$, define $a_n$ to be the least positive integer not coprime to $a_{n-1}$ and not equal to $a_1$, $a_2, \\dots, a_{n-1}$. Prove that every integer except $1$ appears in the sequence $\\{a_n\\}$. (Posed by Yu Hongbing)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "*Step 1:* We prove that there are infinitely many even numbers in this sequence.\nSuppose on the contrary that there are only finitely many even numbers, and there is an integer $E$, such that all even numbers greater than $E$ do not appear in the sequence. It follows that there exists a positive integer $K$ such that $a_n$ is an odd number greater than $E$ for any $n \\ge K$. Then there is some $n_1 > K$ such that $a_{n_1+1} > a_{n_1}$ (otherwise the sequence is strictly decreasing after $a_{n_1}$, a contradiction).\nLet $p$ be the smallest prime divisor of $a_{n_1}$, $p \\ge 3$. Since\n$$\n(a_{n_1+1} - a_{n_1}, a_{n_1}) = (a_{n_1+1}, a_{n_1}) > 1,\n$$\nwe have $a_{n_1+1} - a_{n_1} \\ge p$, i.e. $a_{n_1+1} \\ge a_{n_1} + p$.\nOn the other hand, $a_{n_1} + p$ is even and greater than $E$, so it does not appear before $a_{n_1}$, and therefore $a_{n_1+1} = a_{n_1} + p$, which is even — a contradiction.\n\n*Step 2:* We prove that all even numbers are in this sequence.\nSuppose on the contrary that $2k$ is not in this sequence and is the smallest such even number. Let $\\{a_{n_i}\\}$ be the subsequence of $\\{a_n\\}$ consisting of all even numbers. By step 1, it is an infinite sequence. Since $(a_{n_i}, 2k) > 1$ and $2k \\notin \\{a_n\\}$, we have $a_{n_i+1} \\le 2k$ by definition.\nHowever, $\\{a_{n_1+1}\\}$ is infinite — a contradiction. Thus, $\\{a_n\\}$ contains all even numbers.\n\n*Step 3:* We prove that $\\{a_n\\}$ contains all odd numbers greater than $1$.\nSuppose on the contrary that $2k+1$ is an odd integer greater than $1$ which is not in $\\{a_n\\}$, and is the smallest such number. By step 2, there is an infinite subsequence $\\{a_{m_i}\\}$ of $\\{a_n\\}$ consisting of even numbers that are multiples of $2k+1$. Arguing analogously as in step 2, we have $a_{m_i+1} \\le 2k+1$, $i=1, 2, \\dots$, a contradiction.\n\nWe have shown that $\\{a_n\\}$ contains all positive integers except $1$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 28290,
    "subject": "Mathematics (Multi-modal)",
    "question": "We operate on piles of cards placed at $n+1$ positions $A_1, A_2, \\dots, A_n$ ($n \\ge 3$) and $O$. In one operation, we can do either of the following:\n(1) If there are at least three cards at $A_i$, we may take three cards from $A_i$ and place one at each of $A_{i-1}, A_{i+1}$ and $O$ (assume that $A_0 = A_n, A_{n+1} = A_1$);\n(2) If there are at least $n$ cards at $O$, we may take $n$ cards from $O$ and place one at each of $A_1, A_2, \\dots, A_n$. Prove that if the total number of cards is at least $n^2 + 3n + 1$, we can take some operations such that there are at least $n+1$ cards at each position.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "**Proof** One only needs to consider the case with the total number of cards equal $n^2 + 3n + 1$. We take the following strategy. If there are at least three cards at some $A_i$, then use operation (1) at this position. Such operations can be done for only finitely many steps. Then we have no more than two cards at each $A_i$ and no less than $n^2 + n + 1$ at $O$.\n\nWe then take operation (2) for $n+1$ times. There are then at least $n+1$ cards at each $A_i$. We will now prove that one can increase the number of cards at $O$ to at least $n+1$, while keeping at least $n+1$ cards at each $A_i$.\n\nPut $A_1, A_2, \\dots, A_n$ evenly and in increasing order on a circle with the center at $O$. We call a group of consecutive $A_i$'s, $G = \\{A_i, A_{i+1}, \\dots, A_{i+\\lfloor n \\rfloor-1}\\}, 1 \\le i \\le n, 1 \\le l \\le n$, on the circle a team, where for $j > n$ we define $A_j = A_{j-n}$. A team is good if after we take operation (1) once at each point in $G$, there are at least $n+1$ cards at every point in $G$. Write $a_1, a_2, \\dots, a_n$ as the number of cards at points $A_1, A_2, \\dots, A_n, a_i \\ge n+1, i = 1, 2, \\dots, n$. Let $G = \\{A_i\\}$ be a team. Then, if there is only one point $A_i$ in $G$, $G$ is good iff $a_i \\ge n+4$; a team of two points $G = \\{A_i, A_{i+1}\\}$ is good iff $a_i, a_{i+1} \\ge n+3$; a team $G = \\{A_i, A_{i+1}, \\dots, A_{i+\\lfloor n \\rfloor-1}\\}$ of $l$ points ($3 \\le l \\le n-1$) is good iff $a_i, a_{i+\\lfloor n \\rfloor-1} \\ge n+3$ and $a_j \\ge n+2, i+1 \\le j \\le i+l-2$; finally, the team $G = \\{A_1, A_2, \\dots, A_n\\}$ of all $n$ points is good iff $a_j \\ge n+2, 1 \\le j \\le n$. We then prove that there must exist at least one good team if $a_1 + a_2 + \\dots + a_n \\ge n^2 + 2n + 1$.\n\nAssume that there is no good team. Then each $a_i \\in \\{n+1, n+2, n+3\\}$, otherwise there is a good team of one point at any $A_i$ with $a_i \\ge n+4$. Suppose that the number of $n+1, n+2$ and $n+3$'s among $a_1, a_2, \\dots, a_n$ are $x, y, z$ respectively. We will show that $x \\ge z$. Since $n^2 + 2n + 1 > n(n + 2)$, $z \\ge 1$. If $z = 1$, then $x \\ge 1$, otherwise all $a_i \\ge n + 2$ and $G = \\{A_1, A_2, \\dots, A_n\\}$ is a good team. If $z \\ge 2$, the $z$ points with $n+3$ cards divide the circle into $z$ arcs (no two of these $z$ points are adjacent). Since there is no good team by assumption, there is at least one point on each arc with $n + 1$ cards. So $x \\ge z$, and the total number of cards at $A_1, A_2, \\dots, A_n$ is\n$$\nx(n+1) + y(n+2) + z(n+3) \\le (x+y+z)(n+2) = n(n+2) < n^2 + 2n + 1,\n$$\nwhich is a contradiction. Thus, we have proven that when the number of cards at $O$ is less than $n + 1$, there exists a good team. We use operation (1) on each point of a good team; the number of cards at $O$ is increased while the number of cards at each $A_1, A_2, \\dots, A_n$ is still no less than $n + 1$. We can reiterate until there are at least $n + 1$ cards at $O$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 28291,
    "subject": "Mathematics (Multi-modal)",
    "question": "Function $f(x) = a^{2x} + 3a^x - 2$ ($a > 0, a \\neq 1$) reaches the maximum value $8$ on interval $[-1, 1]$. Then its minimum value on this interval is ________.",
    "options": [],
    "answer": "-1/4",
    "solution": "Let $a^x = y$. The original function is then changed to\n$$g(y) = y^2 + 3y - 2,$$ which is increasing over $\\left(-\\frac{3}{2}, +\\infty\\right)$.\n\nWhen $0 < a < 1$, we have $y \\in [a, a^{-1}]$ and\n$$\ng(y)_{\\max} = a^{-2} + 3a^{-1} - 2 = 8 \\Rightarrow a^{-1} = 2 \\Rightarrow a = \\frac{1}{2}.\n$$\nThen\n$$\ng(y)_{\\min} = \\left(\\frac{1}{2}\\right)^2 + 3 \\times \\frac{1}{2} - 2 = -\\frac{1}{4}.\n$$\nWhen $a > 1$, we have $y \\in [a^{-1}, a]$ and\n$$\ng(y)_{\\max} = a^2 + 3a - 2 = 8 \\Rightarrow a = 2.\n$$\nThen\n$$\ng(y)_{\\min} = 2^{-2} + 3 \\times 2^{-1} - 2 = -\\frac{1}{4}.\n$$\nIn summary, the minimum value of $f(x)$ on $x \\in [-1, 1]$ is $-\\frac{1}{4}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 28292,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $a_1, a_2, a_3, b_1, b_2, b_3$ be pairwise distinct positive integers such that\n$$\n(n+1)a_1^2 + n a_2^2 + (n-1)a_3^2 \\mid (n+1)b_1^2 + n b_2^2 + (n-1)b_3^2\n$$\nholds for all positive integers $n$. Prove that there exists a positive integer $k$ such that $b_i = k a_i$ for all $i = 1, 2, 3$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Suppose that $r$ is any positive integer. Since there are infinitely many primes, there is a prime $p$, such that\n$$\np > (a_1^2 + a_2^2 + a_3^2)(b_1^2 + b_2^2 + b_3^2). \\quad ①\n$$\nBecause $p$ is prime and ①, we have $(p, a_1^2 + a_2^2 + a_3^2) = 1$. $p$ is coprime to $p-1$; from the Chinese remainder theorem, there is a positive integer $n$ such that\n$$\nn \\equiv r \\pmod{p-1},\n$$\n$$\nn(a_1^r + a_2^r + a_3^r) + a_1^r - a_3^r \\equiv 0 \\pmod{p}.\n$$\nFrom the above and Fermat's theorem,\n$$\n(n+1)a_1^r + n a_2^r + (n-1)a_3^r \\equiv n(a_1^r + a_2^r + a_3^r) + a_1^r - a_3^r \\equiv 0 \\pmod{p}.\n$$\nFrom the assumption of the problem,\n$$\n(n+1)b_1^r + n b_2^r + (n-1)b_3^r \\equiv 0 \\pmod{p}.\n$$\nAgain from above and Fermat's little theorem,\n$$\nn(b_1^r + b_2^r + b_3^r) + b_1^r - b_3^r \\equiv 0 \\pmod{p}.\n$$\nEliminate $n$ from the two congruences:\n$$\n(a_1^r + a_2^r + a_3^r)(b_1^r - b_3^r) \\equiv (b_1^r + b_2^r + b_3^r)(a_1^r - a_3^r) \\pmod{p}.\n$$\nFrom the choice of $p$, this congruence is actually an equality:\n$$\n(a_1^r + a_2^r + a_3^r)(b_1^r - b_3^r) = (b_1^r + b_2^r + b_3^r)(a_1^r - a_3^r).\n$$\nThus,\n$$\n(a_2 b_1)^r + 2(a_3 b_1)^r + (a_3 b_2)^r = (a_1 b_2)^r + 2(a_1 b_3)^r + (a_2 b_3)^r.\n$$\nWe then prove the following lemma.\n\n*Lemma* Assume that $x_1, \\dots, x_s, y_1, \\dots, y_s$ are real numbers,\n$$\n0 < x_1 \\le x_2 \\le \\cdots \\le x_s, \\quad 0 < y_1 \\le y_2 \\le \\cdots \\le y_s,\n$$\nsuch that for any positive integers $r$,\n$$\nx_1^r + x_2^r + \\cdots + x_s^r = y_1^r + y_2^r + \\cdots + y_s^r.\n$$\nThen $x_i = y_i$ for $i = 1, 2, \\dots, s$.\n\n*Proof of the lemma.* We use induction on $s$. If $s=1$, take $r=1$; then $x_1 = y_1$. Assume that the lemma holds when $s=t$.\nWhen $s=t+1$, if $x_{t+1} \\neq y_{t+1}$, say $x_{t+1} < y_{t+1}$,\n$$\n\\left(\\frac{x_1}{y_{t+1}}\\right)^r + \\cdots + \\left(\\frac{x_{t+1}}{y_{t+1}}\\right)^r = \\left(\\frac{y_1}{y_{t+1}}\\right)^r + \\cdots + \\left(\\frac{y_t}{y_{t+1}}\\right)^r + 1 \\ge 1.\n$$\nBecause $0 < \\frac{x_i}{y_{t+1}} < 1$ ($1 \\le i \\le t+1$), take the limit $r \\to +\\infty$, and we have $0 \\ge 1$, a contradiction.\nSo $x_{t+1} = y_{t+1}$, and then $x_1^r + \\cdots + x_t^r = y_1^r + \\cdots + y_t^r$, $r=1, 2, \\dots$. By induction the lemma holds for all positive integer $s$.\n\nNow return to the proof of the problem. Since $a_1, a_2, a_3, b_1, b_2, b_3$ are distinct,\n$$\na_2 b_1 \\neq a_3 b_1, \\quad a_3 b_1 \\neq a_3 b_2, \\quad a_1 b_2 \\neq a_1 b_3,\n$$\n$$\na_1 b_3 \\neq a_2 b_3, \\quad a_2 b_1 \\neq a_2 b_3.\n$$\nFrom the previous equality and the lemma we know that\n$$\na_2 b_1 = a_1 b_2, \\quad a_3 b_1 = a_1 b_3, \\quad a_3 b_2 = a_2 b_3.\n$$\nThen $\\frac{b_1}{a_1} = \\frac{b_2}{a_2} = \\frac{b_3}{a_3}$. Write $\\frac{b_1}{a_1} = \\frac{k}{l}$, $(k, l) = 1, l \\ge 1$; then $b_i = \\frac{k}{l} a_i$, $i = 1, 2, 3$. From $2b_1 + b_2 = \\frac{k}{l}(2a_1 + a_2)$ and the assumption of the problem (with $n=1$), $2a_1 + a_2 \\mid 2b_1 + b_2$ and $\\frac{k}{l}$ is an integer. So $l=1$ and $b_i = k a_i$, $i = 1, 2, 3$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 28293,
    "subject": "Mathematics (Multi-modal)",
    "question": "Two persons roll two dice in turn. Whoever gets the sum number greater than $6$ first will win the game. The probability for the person rolling first to win is ________.",
    "options": [],
    "answer": "12/17",
    "solution": "The probability for rolling two dice to get the sum number greater than $6$ is $\\frac{21}{36} = \\frac{7}{12}$. Therefore, the required probability is\n$$\n\\frac{7}{12} + \\left(\\frac{5}{12}\\right)^2 \\frac{7}{12} + \\left(\\frac{5}{12}\\right)^4 \\frac{7}{12} + \\cdots = \\frac{7}{12} \\times \\frac{1}{1 - \\frac{25}{144}} = \\frac{12}{17}.\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 28294,
    "subject": "Mathematics (Multi-modal)",
    "question": "Given integer $n \\ge 2$ and real numbers $x_1, x_2, \\dots, x_n$ in the interval $[0, 1]$, prove that there exist real numbers $a_0, a_1, \\dots, a_n$ satisfying simultaneously the following conditions:\n$$\n(1) \n\\begin{aligned}\na_0 &+ a_n = 0; \\\\\na_i &\\le 1, \\text{ for every } i = 0, 1, \\dots, n;\n\\end{aligned}\n$$\n$$\n(2) \\quad |a_i| \\le 1, \\text{ for every } i = 0, 1, \\dots, n;\n$$\n$$\n(3) \\quad |a_i - a_{i-1}| = x_i, \\text{ for every } i = 1, 2, \\dots, n.\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "For any $a \\in [0, 1)$, define a sequence $\\{a_i\\}_{i=0}^n$ generated by $a$ as follows: $a_0 = a$; for $1 \\le i \\le n$, $a_i = a_{i-1} - x_i$ if $a_{i-1} \\ge 0$, and $a_i = a_{i-1} + x_i$ if $a_{i-1} < 0$.\n\nSet $f(a) = a_n$. It is easy to show by induction that $|a_i| \\le 1$ for every $0 \\le i \\le n$.\n\nIf there exists $a \\in [0, 1)$ such that $f(a) = -a$, consider the sequence generated by $a$, $a_0 = a, a_1, ..., a_n = f(a) = -a$. Clearly, this sequence satisfies the conditions (1) and (3). By the recursive relation, it also satisfies the condition (2). Thus, it suffices to show that there exists $a \\in [0, 1)$ such that $f(a) = -a$.\n\nFor any $a \\in [0, 1)$, we say that $a$ is a breaking point if at least one term in its generating sequence is $0$. Since every breaking point is of the form $\\sum_{i=1}^{n} t_i x_i$, where $t_i = -1, 0, 1$, there are finitely many breaking points.\n\nClearly, $0$ is a breaking point; label all breaking points in increasing order by $0 = b_1 < b_2 < \\cdots < b_m < 1$.\n\nWe first prove that for $1 \\le k \\le m-1$, $f(a) = f(b_k) + (a-b_k)$ for every $a \\in [b_k, b_{k+1})$.\n\nConsider $b_k, b_{k+1}$ and their generating sequences. Assume that $q_0 = b_k, q_1, q_2, \\dots, q_n$ is the generating sequence of $b_k$, and $r_0 = b_{k+1}, r_1, r_2, \\dots, r_n$ is the generating sequence of $b_{k+1}$. Suppose that $r_l$ is the first term of $\\{r_i\\}_{i=0}^n$ equal to $0$.\n\nConstruct a sequence $\\{s_i\\}_{i=0}^n$ as follows:\n$$\n\\begin{align*}\ns_0 &= r_0, \\quad s_1 = r_1, \\quad \\dots, \\quad s_l = r_l = 0, \\\\\ns_{l+1} &= -r_{l+1}, \\quad \\dots, \\quad s_n = -r_n.\n\\end{align*}\n$$\nIt is clear that the sequence $\\{s_i\\}_{i=0}^n$ satisfies $s_0 = b_{k+1}$; and for $1 \\le i \\le n$, $s_i = s_{i-1} - x_i$ if $s_{i-1} > 0$, and $s_i = s_{i-1} + x_i$ if $s_{i-1} \\le 0$.\n\nWe prove by induction that $q_i s_i \\ge 0$ and $s_i - q_i = b_{k+1} - b_k$.\n\nThe conclusion is obviously true for $i=0$. Assume that it holds for $i-1$. Then $q_{i-1} s_{i-1} \\ge 0$ and $s_{i-1} - q_{i-1} = b_{k+1} - b_k > 0$, which implies that $q_{i-1} \\ge 0$, $s_{i-1} > 0$, or $q_{i-1} < 0$, $s_{i-1} \\le 0$. In the former case, $q_i = q_{i-1} - x_i$, $s_i = s_{i-1} - x_i$, and thus $s_i - q_i = s_{i-1} - q_{i-1} = b_{k+1} - b_k$. Similarly, in the latter case, we have $s_i - q_i = b_{k+1} - b_k$.\n\nIf $q_i s_i < 0$, then $q_i < 0 < s_i$. Set $b' = b_{k+1} - s_i = b_k + (-q_i) \\in (b_k, b_{k+1})$, and consider the generating sequence of $b'$, $u_0 = b', u_1, \\dots, u_n$. It is easy to show by induction that $s_j - u_j = s_0 - u_0 = s_i$ for any $0 \\le j \\le i$ ($q_{j-1}$ and $s_{j-1}$ both add or subtract $x_j$ for $j < i$; $u_j$ lies between $q_{j-1}$ and $s_{j-1}$, so the recursive relation is the same). Then $u_i = s_i - s_i = 0$, i.e. $b'$ is a breaking point — a contradiction to $b_k, b_{k+1}$ being two consecutive breaking points. Therefore, $q_i s_i \\ge 0$. By induction we have verified that $q_i s_i \\ge 0$ and $s_i - q_i = b_{k+1} - b_k$ for all $0 \\le i \\le n$.\n\nSince $f(b_k) = q_n$, $f(b_{k+1}) = r_n = -s_n$, we have $f(b_k) + f(b_{k+1}) = b_k - b_{k+1}$. It follows from the above argument that, for any $b_k < b' < b_{k+1}$, the generating sequence of $b'$ has the same recursive relation as $\\{q_i\\}_{i=0}^n$ and $\\{s_i\\}_{i=0}^n$, and hence\n$$\nf(b') = f(b_k) + (b' - b_k).\n$$\nLet us go back to the original problem.\n\nIf $f(b_k) = -b_k$ for some $k$, we are done. If $f(b_k) = b_k$ for some $k$, then consider the generating sequence of $b_k$, reversing every term after the first $0$, and we obtain a new sequence $z_0 = b_k, z_1, \\dots, z_n = -b_k$, which satisfies the required conditions. Now assume that $|f(b_k)| \\ne b_k$ for every $k$, and we shall consider two cases to show that $f(a) = -a$ for some $a \\in [0, 1)$:\n\n**Case 1:** $|f(b_m)| < b_m$. Since $|f(b_1)| > b_1 = 0$, there exists $k$ such that $|f(b_k)| > b_k$, $|f(b_{k+1})| < b_{k+1}$.\n\nSince $f(b_k) + f(b_{k+1}) = b_k - b_{k+1}$, we have $f(b_k) - b_k = -(f(b_{k+1}) + b_{k+1}) < 0$, and hence $f(b_k) \\le -b_k$. Again by $f(b_k) + f(b_{k+1}) = b_k - b_{k+1}$, we have\n$$\nf(b_k) = b_k - b_{k+1} - f(b_{k+1}) > b_k - 2b_{k+1},\n$$\ni.e.\n$$\nb_k - 2b_{k+1} < f(b_k) < -b_k.\n$$\nLet $b' = \\frac{b_k - f(b_k)}{2}$. Then $b_k < b' < b_{k+1}$, and\n$$\nf(b') = f(b_k) + (b' - b_k) = (b_k - 2b') + (b' - b_k) = -b',\n$$\nand the result follows.\n\n**Case 2:** $|f(b_m)| > b_m$. Since there is no breaking number in $(b_m, 1)$, we see from the previous argument that for any $b' \\in (b_m, 1)$, the generating sequence of $b'$ and the generating sequence of $b_m$ have the same recursive relation, and hence\n$$\nf(b') = f(b_m) + (b' - b_m).\n$$\nSince $|f| \\le 1$, we have $f(b_m) \\le b_m$, and hence\n$$-1 \\le f(b_m) < -b_m.$$\nLet $b' = \\frac{b_m - f(b_m)}{2}$. Then\n$$\nb_m = \\frac{b_m - (-b_m)}{2} < b' < \\frac{b_m - (-1)}{2} = \\frac{b_m + 1}{2} < 1,\n$$\n$$\n\\begin{aligned} f(b') &= f(b_m) + (b' - b_m) \\\\ &= (b_m - 2b') + (b' - b_m) = -b'. \\end{aligned}\n$$\nThe result again follows.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 28295,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $N^*$ be the set of positive integers. Define $a_1 = 2$, and for $n = 1, 2, \\dots$,\n$$\na_{n+1} = \\min\\{\\lambda \\mid \\frac{1}{a_1} + \\frac{1}{a_2} + \\dots + \\frac{1}{a_n} + \\frac{1}{\\lambda} < 1, \\lambda \\in N^*\\}.\n$$\n\nProve that $a_{n+1} = a_n^2 - a_n + 1$ for $n = 1, 2, \\dots$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "By $a_1 = 2$, $a_2 = \\min\\{\\lambda \\mid \\frac{1}{a_1} + \\frac{1}{\\lambda} < 1, \\lambda \\in N^*\\}$, consider $\\frac{1}{a_1} + \\frac{1}{\\lambda} < 1$, then $\\frac{1}{\\lambda} < 1 - \\frac{1}{2} = \\frac{1}{2}$, $\\lambda > 2$, hence $a_2 = 3$. So the conclusion is true for $n = 1$.\n\nSuppose that the conclusion is true for all integer $n \\le k - 1$ ($k \\ge 2$). If $n = k$, then $a_{k+1} = \\min\\{\\lambda \\mid \\frac{1}{a_1} + \\frac{1}{a_2} + \\dots + \\frac{1}{a_k} + \\frac{1}{\\lambda} < 1, \\lambda \\in N^*\\}$.\n\nConsidering\n$$\n\\frac{1}{a_1} + \\frac{1}{a_2} + \\dots + \\frac{1}{a_k} + \\frac{1}{\\lambda} < 1,\n$$\nthat is, $0 < \\frac{1}{\\lambda} < 1 - \\left( \\frac{1}{a_1} + \\frac{1}{a_2} + \\dots + \\frac{1}{a_k} \\right)$, we have\n$$\n\\lambda > \\frac{1}{1 - \\frac{1}{a_1} - \\frac{1}{a_2} - \\dots - \\frac{1}{a_k}}.\n$$\nIn the following, we show that\n$$\n\\frac{1}{1 - \\frac{1}{a_1} - \\frac{1}{a_2} - \\dots - \\frac{1}{a_k}} = a_k (a_k - 1).\n$$\nBy the induction hypotheses, for $2 \\le n \\le k$, $a_n = a_{n-1} (a_{n-1} - 1) + 1$, we have\n$$\n\\frac{1}{a_n - 1} = \\frac{1}{a_{n-1}(a_{n-1} - 1)} = \\frac{1}{a_{n-1} - 1} - \\frac{1}{a_{n-1}},\n$$\ntherefore $\\frac{1}{a_{n-1}} = \\frac{1}{a_{n-1}-1} - \\frac{1}{a_n-1}$. By taking the sum, we have\n$$\n\\sum_{i=2}^{k} \\frac{1}{a_{i-1}} = 1 - \\frac{1}{a_k - 1}, \\text{ that is}\n$$\n$$\n\\sum_{i=1}^{k} \\frac{1}{a_i} = 1 - \\frac{1}{a_k - 1} + \\frac{1}{a_k} = 1 - \\frac{1}{a_k (a_k - 1)}.\n$$\nConsequently, $\\frac{1}{1 - \\frac{1}{a_1} - \\frac{1}{a_2} - \\dots - \\frac{1}{a_k}} = a_k (a_k - 1)$. Therefore,\n$$\n\\begin{aligned}\na_{k+1} &= \\min \\left\\{ \\lambda \\mid \\frac{1}{a_1} + \\frac{1}{a_2} + \\dots + \\frac{1}{a_k} + \\frac{1}{\\lambda} < 1, \\lambda \\in \\mathbb{N}^* \\right\\} \\\\\n&= a_k (a_k - 1) + 1.\n\\end{aligned}\n$$\n\nBy induction on $n$, for all positive integer $n$, we have $a_{n+1} = a_n^2 - a_n + 1$. $\\square$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 28296,
    "subject": "Mathematics (Multi-modal)",
    "question": "There are $2n$ real numbers $a_1, a_2, \\dots, a_n, r_1, r_2, \\dots, r_n$ satisfying $a_1 \\le a_2 \\le \\dots \\le a_n$ and $0 \\le r_1 \\le r_2 \\le \\dots \\le r_n$. Prove that $\\sum_{i=1}^n \\sum_{j=1}^n a_i a_j \\min(r_i, r_j) \\ge 0$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Since\n$$\n\\sum_{i=1}^{n} \\sum_{j=1}^{n} a_i a_j \\min(r_i, r_j) = \\sum_{j=1}^{n} a_1 a_j \\min(r_1, r_j) + \\sum_{j=1}^{n} a_2 a_j \\min(r_2, r_j) \\\\\n+ \\dots + \\sum_{j=1}^{n} a_k a_j \\min(r_k, r_j) + \\dots \\\\\n+ \\sum_{j=1}^{n} a_n a_j \\min(r_n, r_j),\n$$\nits $k$-th term is\n$$\n\\sum_{j=1}^{n} a_k a_j \\min(r_k, r_j) = a_k a_1 r_1 + a_k a_2 r_2 + \\dots + a_k a_k r_k \\\\\n+ a_k a_{k+1} r_k + \\dots + a_k a_n r_n\n$$\nwhich is the sum of elements of the $k$-th row of $A_1$, $k = 1, 2, \\dots, n$.\nTherefore, $\\sum_{i=1}^n \\sum_{j=1}^n a_i a_j \\min(r_i, r_j)$ is the sum of all elements of $A_1$.\n\nOn the other hand, the summation can also be done as follows: Take the elements of first column and the first row of $A_1$, sum up; denote the rest $(n-1) \\times (n-1)$ element by matrix $A_2$, then take the first column and the first row of $A_2$, sum up, denote the rest by matrix $A_3, \\dots$, so we get\n$$\n\\begin{align*}\n\\sum_{i=1}^{n} \\sum_{j=1}^{n} a_i a_j \\min(r_i, r_j) &= \\sum_{k=1}^{n} r_k (a_k^2 + 2a_k (a_{k+1} + a_{k+2} + \\dots + a_n)) \\\\\n&= \\sum_{k=1}^{n} r_k \\left( \\left(a_k + \\sum_{i=k+1}^{n} a_i\\right)^2 - \\left(\\sum_{i=k+1}^{n} a_i\\right)^2 \\right) \\\\\n&= \\sum_{k=1}^{n} r_k \\left( \\left(\\sum_{i=k}^{n} a_i\\right)^2 - \\left(\\sum_{i=k+1}^{n} a_i\\right)^2 \\right) \\\\\n&= r_1 \\left(\\sum_{i=1}^{n} a_i\\right)^2 + r_2 \\left(\\sum_{i=2}^{n} a_i\\right)^2 + r_3 \\left(\\sum_{i=3}^{n} a_i\\right)^2 + \\dots \\\\\n&\\quad + r_n \\left(\\sum_{i=n}^{n} a_i\\right)^2 - r_1 \\left(\\sum_{i=2}^{n} a_i\\right)^2 - r_2 \\left(\\sum_{i=3}^{n} a_i\\right)^2 \\\\\n&\\quad - \\dots - r_{n-1} \\left(\\sum_{i=n}^{n} a_i\\right)^2 \\\\\n&= \\sum_{k=1}^{n} (r_k - r_{k-1}) \\left(\\sum_{i=k}^{n} a_i\\right)^2 \\ge 0\n\\end{align*}\n$$\n(where $r_0 = 0$).",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 28297,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $n$ be an integer greater than or equal to $3$. For a permutation $p = (x_1, x_2, \\dots, x_n)$ of $(1, 2, \\dots, n)$, we say that $x_j$ lies in between $x_i$ and $x_k$ if $i < j < k$. (For example, in the permutation $(1, 3, 2, 4)$, $3$ lies in between $1$ and $4$, and $4$ does not lie in between $1$ and $2$.) Set $S = \\{p_1, p_2, \\dots, p_m\\}$ consists of (distinct) permutations $p_i$ of $(1, 2, \\dots, n)$. Suppose that among every three distinct numbers in $\\{1, 2, \\dots, n\\}$, one of these numbers does not lie in between the other two numbers in every permutation $p_i \\in S$. Determine the maximum value of $m$.",
    "options": [],
    "answer": "2^{n-1}",
    "solution": "The answer is $2^{n-1}$.\n\nWe first show that $m \\le 2^{n-1}$. We induct on $n$. The base case $n=3$ is trivial. (Indeed, say $3$ does not lie in between $1$ and $2$, then we can have $S = \\{(1, 2, 3), (3, 1, 2), (2, 1, 3), (3, 2, 1)\\}$.) Assume that the statement is true for $n=k$ (where $k \\ge 3$). Now consider $n=k+1$ and a set $S_{k+1}$ satisfies the conditions of the problem. Note that if the element $k+1$ is deleted from each permutation $p_i$ in $S_{k+1}$, the resulting permutations $q_i$ form a set $S_k$ that satisfies the conditions of the problem (for $n=k$). It suffices to show the following claim: there are at most two distinct permutations $p$ and $q$ in $S_{k+1}$ that can map to the same permutation $r$ in $S_k$ (by deleting the element $k+1$ in the permutations $p$ and $q$).\n\nIndeed, assume that for\n$$\np_1 = (x_1, x_2, \\dots, x_{k+1}), \\quad p_2 = (y_1, y_2, \\dots, y_{k+1}), \\ p_3 = (z_1, z_2, \\dots, z_{k+1})\n$$\nin $S_{k+1}$, $q_1 = q_2 = q_3 = q$. By symmetry, we may assume that $q = (1, 2, \\dots, k)$. Assume that $x_a = y_b = z_c = k+1$. Again by symmetry, we may assume that $1 \\le a < b < c \\le k+1$. (Note that because $q_1 = q_2 = q_3 = q$, $a, b, c$ are distinct.) We consider three numbers $a, b, k+1$. We have $p_1 = (\\dots, k+1, a, \\dots, b, \\dots)$ (in particular, $a$ lies in between $k+1$ and $b$), $p_2 = (\\dots, a, \\dots, k+1, b, \\dots)$ (in particular, $k+1$ lies in between $a$ and $b$), and $p_3 = (\\dots, a, \\dots, b, \\dots, k+1, \\dots)$ (in particular, $b$ lies in between $a$ and $k+1$). Hence each one of the numbers $a, b, k+1$ lie in between the other two numbers in some permutations in $S_k$, violating the conditions of $S_k$. Thus our assumption was wrong and at most two elements in $S_{k+1}$ can be mapped to an element in $S_k$, establishing our claim.\n\nIt remains to be shown that $m = 2^{n-1}$ is achievable. We construct permutation $p$ inductively: (1) place $1$; (2) after numbers $1, 2, \\ldots, l$ are placed, we place $l+1$ either to the left or the right of all the numbers placed so far. Because there are two possible places for each of the numbers $2, 3, \\ldots, n$, we can construct $2^{n-1}$ such permutations. For any three numbers $1 \\le a < b < c \\le n$, $c$ does not lie in between $a$ and $b$. Hence, this set of $2^{n-1}$ permutations satisfies the conditions of the problem, completing our proof. $□$",
    "topic": "Discrete Mathematics",
    "subtopic": "Combinatorics"
  },
  {
    "id": 28298,
    "subject": "Mathematics (Multi-modal)",
    "question": "The lengths of the nine edges of regular triangular prism $ABC-A_1B_1C_1$ are equal, $P$ is the midpoint of $CC_1$, and the dihedral angle $B-A_1P-B_1 = \\alpha$. Then $\\sin \\alpha = \\underline{\\hspace{2cm}}$.",
    "options": [],
    "answer": "sqrt(10)/4",
    "solution": "Let the line through segment $AB$ be $x$-axis with the origin $O$ being the midpoint of $AB$ and let the line through segment $OC$ be $y$-axis to establish a space rectangular coordinate system shown in Fig. 7.1. Assuming the length of an edge is 2, we have $B(1, 0, 0)$, $B_1(1, 0, 2)$, $A_1(-1, 0, 2)$, $P(0, \\sqrt{3}, 1)$. Then\n![](attached_image_1.png)\nFig. 7.1\n$$\n\\begin{aligned} \\overrightarrow{BA_1} &= (-2, 0, 2), \\quad \\overrightarrow{BP} = (-1, \\sqrt{3}, 1), \\\\ \\overrightarrow{B_1A_1} &= (-2, 0, 0), \\quad \\overrightarrow{B_1P} = (-1, \\sqrt{3}, -1). \\end{aligned}\n$$\nLet vectors $\\vec{m} = (x_1, y_1, z_1)$ and $\\vec{n} = (x_2, y_2, z_2)$ be perpendicular to $BA_1P$ and $B_1A_1P$, respectively. We have\n$$\n\\begin{cases} \\vec{m} \\cdot \\overrightarrow{BA_1} = -2x_1 + 2z_1 = 0, \\\\ \\vec{m} \\cdot \\overrightarrow{BP} = -x_1 + \\sqrt{3}y_1 + z_1 = 0, \\\\ \\vec{n} \\cdot \\overrightarrow{B_1A_1} = -2x_2 = 0, \\\\ \\vec{n} \\cdot \\overrightarrow{B_1P} = -x_2 + \\sqrt{3}y_2 - z_2 = 0. \\end{cases}\n$$\nWe can then assume that $\\vec{m} = (1, 0, 1)$, $\\vec{n} = (0, 1, \\sqrt{3})$.\nFrom $|\\vec{m} \\cdot \\vec{n}| = |\\vec{m}| \\cdot |\\vec{n}| \\cos \\alpha$, we have\n$$\n\\sqrt{3} = \\sqrt{2} \\cdot 2 \\cos \\alpha \\Rightarrow \\cos \\alpha = \\frac{\\sqrt{6}}{4}.\n$$\nAs seen in Fig. 7.2, we have $PC = PC_1$, $PA_1 = PB$. Suppose $A_1B$ and $AB_1$ intersect at $O$. Then we get $OA_1 = OB$, $OA = OB_1$, $A_1B \\perp AB_1$.\nSince $PA = PB_1$, then $PO \\perp AB_1$. Therefore $AB_1 \\perp$ plane $PA_1B$.\nOn plane $PA_1B$ through $O$ draw line $OE \\perp A_1P$ with foot point $E$.\n![](attached_image_2.png)\nFig. 7.2\nConnecting $B_1E$, $\\angle B_1EO$ is then the plane angle of the dihedral angle $B-A_1P-B_1$. Assuming $AA_1 = 2$, it is easy to find that $PB = PA_1 = \\sqrt{5}$, $A_1O = B_1O = \\sqrt{2}$, $PO = \\sqrt{3}$.\nIn right triangle $\\triangle PA_1O$, we have $A_1O \\cdot PO = A_1P \\cdot OE$,\nor $\\sqrt{2} \\cdot \\sqrt{3} = \\sqrt{5} \\cdot OE$. So $OE = \\frac{\\sqrt{6}}{\\sqrt{5}}$.\nAs $B_1O = \\sqrt{2}$, we have\n$$\nB_1E = \\sqrt{B_1O^2 + OE^2} = \\sqrt{2 + \\frac{6}{5}} = \\frac{4\\sqrt{5}}{5}.\n$$\nFinally,\n$$\n\\sin \\alpha = \\sin \\angle B_1EO = \\frac{B_1O}{B_1E} = \\frac{\\sqrt{2}}{\\frac{4\\sqrt{5}}{5}} = \\frac{\\sqrt{10}}{4}. \\quad \\square\n$$",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 28299,
    "subject": "Mathematics (Multi-modal)",
    "question": "There are $n$ ($n \\ge 3$) players in a table tennis tournament, in which any two players have a match. Player $A$ is called not out-performed by player $B$, if at least one of player $A$'s losers is not a $B$'s loser.\nDetermine, with proof, all possible values of $n$, such that the following case could happen: after finishing all the matches, every player is not out-performed by any other player.",
    "options": [],
    "answer": "n = 3 or n >= 5",
    "solution": "The answer is $n = 3$ or $n \\ge 5$.\n\n(1) For $n = 3$, suppose $A$, $B$ and $C$ are three players, and the result of three matches are as follows: $A$ wins $B$, $B$ wins $C$, and $C$ wins $A$. These results obviously satisfy the condition.\n\n(2) If $n = 4$, suppose that the condition holds, i.e., in view of the results of all matches, every player is not out-performed by any other player. It is obvious that none of these four players wins in his three games, otherwise, the other three players will be not out-performed by this player. Similarly, none of these three players loses in his three games. It follows that each player wins one or two matches.\nFor the player $A$, assume that $A$ wins $B$ and $D$, but loses to $C$, then both $B$ and $D$ win $C$; otherwise, they would not out-performed $A$. For the loser in the match $B$ vs. $D$, he only wins $C$, and so the loser is impossible to be not out-performed by the winner. Consequently, for $n = 4$, the given condition could not happen.\n\n(3) For $n = 6$, one can construct the tournament results by means of the following directed graph, in which each black dot represents a player, and $\\bullet \\to \\circ$ represents a match with the result that player $\\bullet$ wins player $\\circ$.\n\n![](attached_image_1.png)\n\n(4) If there exist tournament results such that each of $n$ players $A_i$ ($1 \\le i \\le n$) is not out-performed by any other player, we will prove that the same holds for $n + 2$ players as follows: Suppose that $M$ and $N$ are the additional players to the original $n$ players. Construct the game results of $M$ and $N$ as follows:\n$$\nA_i \\to M, M \\to N, N \\to A_i\n$$\nfor all $i = 1, 2, \\dots, n$, and the game results among $A_i$'s are still the original ones. Now we want to check that these $n+2$ players satisfy the given condition. For any player $G \\in \\{A_1, A_2, \\dots, A_n\\}$, then it suffices to consider the players $G, M, N$, and it reduces to the case $n = 3$.\n\n![](attached_image_2.png)\n\nOne can check that each of these three players $G, M, N$ is not out-performed by any one of the other two players. Hence, the tournament result of these $n+2$ players satisfies the required condition.\n\nIn particular, it follows from (1) and the induction step of (4) that the required condition holds for any odd $n$ with $n \\ge 3$. Moreover, it follows from (3) and the induction step of (4) that the required condition holds for any even $n$ with $n \\ge 6$, and it completes the proof. $\\square$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 28300,
    "subject": "Mathematics (Multi-modal)",
    "question": "Determine the least odd number $a > 5$ satisfying the following conditions: There are positive integers $m_1, m_2, n_1, n_2$ such that $a = m_1^2 + n_1^2$, $a^2 = m_2^2 + n_2^2$, and $m_1 - n_1 = m_2 - n_2$.",
    "options": [],
    "answer": "261",
    "solution": "The answer is $261$.\n\nNote that\n$$\n261 = 15^2 + 6^2, \\quad 261^2 = 189^2 + 180^2, \\quad 15 - 6 = 189 - 180.\n$$\nWe know that there is no number in between $5$ and $261$ that satisfies the condition of the problem. Assume on the contrary that $a$ is such a number. We may set $d = m_1 - n_1 > 0$. Because $a$ is odd, $m_1$ and $n_1$ have different parity, and so $d$ is odd. Because $m_1 < 261, d \\le 15$; that is, the possible values of $d$ are\n\n$1, 3, 5, 7, 9, 11, 13, 15$. We will eliminate every one of them.\n\nWe can write $m_2 = n_2 + d$ and $a^2 = (n_2 + d)^2 + n_2^2$ or\n$$\n2a^2 - d^2 = (2n_2 + d)^2. \\qquad \\textcircled{1}\n$$\nIf $d=1$, then $\\textcircled{1}$ becomes a Pell's equation $x^2 - 2y^2 = -1$ with $(x, y) = (2n_2 + 1, a)$. This Pell's equation has minimal solution $(x, y) = (1, 1)$ and $x + y\\sqrt{2} = (1 + \\sqrt{2})^{2k-1}$ for positive integers $k$. The $y$ values of the solutions of this Pell's equation are $5, 29, 169, 985, \\ldots$. Thus, the only possible values for $a$ are $29$ and $169$. It is easy to check that neither $29$ nor $169$ can be written in the form of $(n_1 + 1)^2 + n_1^2$. Hence $d \\neq 1$.\n\nIf $d$ is a multiple of $3$, then $m_2 \\equiv n_2 \\pmod{3}$ and $a^2 \\equiv 2m_2^2 \\pmod{3}$. Because $2$ is not a quadratic residue modulo $3$, we conclude that $0 \\equiv m_2 \\equiv n_2 \\equiv a \\pmod{3}$. Hence, $m_1^2 + n_1^2 \\equiv 0 \\pmod{3}$, from which it follows that $m_1 \\equiv n_1 \\equiv 0 \\pmod{3}$. Thus, $a = m_1^2 + n_1^2$ is a multiple of $9$ and $m_2^2 + n_2^2 = a^2$ is a multiple of $81$. We can write $m_2 = 3m'$, $n_2 = 3n'$, and $a = 9a'$ for integers $m', n', a'$. We have $m'^2 + n'^2 = 9a'^2$. Again, because $-1$ is not a quadratic residue modulo $3$, we must have $m' \\equiv n' \\equiv 0 \\pmod{3}$. It follows that $d = 3(m' - n')$ is divisible by $9$. Thus, $d = 9$.\n\nBecause $a < 261 = 15^2 + 6^2$, $n_1 < 6$. Therefore, $n_1 = 3$ and $a = 12^2 + 3^2 = 153$. But then $\\textcircled{1}$ becomes $9^2 \\cdot 577 = (2n_2 + 9)^2$, which is impossible. Hence, $d \\neq 3$, or $9$, or $15$.\n\nIf $d=11$ or $d=13$, because $2$ is not a quadratic residue modulo $d$, from $a^2 \\equiv 2m_2^2 \\pmod{d}$, we conclude that $0 \\equiv m_2 \\equiv n_2 \\equiv a \\pmod{d}$. It follows that $2m_1^2 \\equiv a \\equiv 0 \\pmod{d}$ and $0 \\equiv m_1 \\equiv n_1 \\equiv a \\pmod{d}$. In particular, $n_1 \\ge d$, $m_1 \\ge 2d$ and\n$$\na = m_1^2 + n_1^2 \\ge 5d^2 > 261.\n$$\nHence, $d \\neq 11$ or $13$.\n\nIf $d = 5$, we also note that $2$ is not a quadratic residue modulo $5$. By the same reasoning before, we have\n$$\nm_1 \\equiv n_1 \\equiv m_2 \\equiv n_2 \\equiv 0 \\pmod{5}.\n$$\nIf $n_1 \\ge 10$, then $m_1 \\ge 15$ and $a = m_1^2 + n_1^2 \\ge 10^2 + 15^2 > 261$. Thus, $n_1 = 5$, $m_1 = 10$, and $a = 125$. But then $\\textcircled{1}$ becomes\n$$\n5^2 \\cdot 1249 = (2n_2 + 5)^2,\n$$\nwhich is impossible. Hence, $d \\neq 5$.\n\nIf $d = 7$, then because $a < 261 = 15^2 + 6^2$, we have $n_1 \\le 8$. The possible values of $a$ are then $65, 85, 109, 137, 169, 205, 245$. But then $\\textcircled{1}$ becomes $(2n_2 + 7)^2 = 2a^2 - 49$. It is easy to check that there is no solution in this case. Hence, $d \\neq 7$.\n\nCombining the above, we conclude that $261$ is the answer of this question. $\\square$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 28301,
    "subject": "Mathematics (Multi-modal)",
    "question": "Determine all possible values of integer $k$ for which there exist positive integers $a$ and $b$ such that $\\frac{b+1}{a} + \\frac{a+1}{b} = k$.",
    "options": [],
    "answer": "3, 4",
    "solution": "choose any $(a, b)$ such that $b$ is the smallest. Then the quadratic equation\n$$\nx^2 + (1 - kb)x + b^2 + b = 0\n$$\nhas an integral root $x = a$. Let $x = a'$ be the second root, it follows from $a + a' = kb - 1$ that $a' \\in \\mathbb{Z}$, and from\n$$\na \\cdot a' = b(b + 1)\n$$\nthat $a' > 0$. Hence, we have\n$$\n\\frac{b+1}{a'} + \\frac{a'+1}{b} = k.\n$$\nAnd it follows from the assumption on $b$ that\n$a \\geq b, a' \\geq b.$\nSo one of $a$ and $a'$ is equal to $b$. Without loss of generality, we may assume $a = b$, so $k = 2 + \\frac{2}{b}$, and so $b \\mid 2$ i.e., $b = 1$ or $2$, and $k = 3$ or $4$, respectively.\nIf $a = b = 1$, then $k = 4$; if $a = b = 2$, then $k = 3$.\nConsequently, $k = 3$ or $4$ is the only solution. $\\square$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 28302,
    "subject": "Mathematics (Multi-modal)",
    "question": "The number of positive integer solutions of equation $x + y + z = 2010$ with $x \\le y \\le z$ is ________.",
    "options": [],
    "answer": "336675",
    "solution": "It is easy to find that the number of positive integer solutions of $x + y + z = 2010$ is $C_{2009}^2 = 2009 \\times 1004$.\nWe now classify these solutions into three categories:\n\n(1)\n$x = y = z$, the number in this category is obviously 1;\n\n(2) there are exactly two that are equal among $x, y, z$ —\nthe number in this category is 1003;\n\n(3) $x, y, z$ are different from each other — suppose the number in this category is $k$.\nFrom\n$$\n1 + 3 \\times 1003 + 6k = 2009 \\times 1004,\n$$\nwe have\n$$\n\\begin{align*} \n6k &= 2009 \\times 1004 - 3 \\times 1003 - 1 \\\\ \n&= 2006 \\times 1005 - 2009 + 3 \\times 2 - 1 \\\\ \n&= 2006 \\times 1005 - 2004. \n\\end{align*}\n$$\nWe get $k = 1003 \\times 335 - 334 = 335671$.\nTherefore, the number of positive integer solutions\nsatisfying $x \\le y \\le z$ is\n$$\n1 + 1003 + 335671 = 336675.\n$$",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 28303,
    "subject": "Mathematics (Multi-modal)",
    "question": "Given eight points $A_1, A_2, \\dots, A_8$ on a circle, determine the smallest positive integer $n$ such that among any $n$ triangles with vertices in these eight points, there are two which have a common side.",
    "options": [],
    "answer": "9",
    "solution": "First, we consider the maximal number of triangles with no common side.\nConsider the maximal number of triangles with no common side pairwise. There are $C_8^2 = 28$ chords by connecting eight points. If each chord only belongs to one triangle, then these chords can only form $r \\le \\lfloor \\frac{28}{3} \\rfloor = 9$ triangles with no common side pairwise. But if there are nine such triangles, then there are 27 vertices. So, there is one point in eight points, which is the common vertex of four triangles.\n\n![](attached_image_1.png)\n\nSuppose that point is $A_8$, then eight edges are connected to seven points $A_1, A_2, \\dots, A_7$. So, there must exist an edge $A_8A_k$, which is the common side of two triangles, which is a contradiction. So $r \\le 8$.\n\nOn the other hand, when $r = 8$, we can make such eight triangles, see figure. Denote the triangles by three vertices as: $(1, 2, 8)$, $(1, 3, 6)$, $(1, 4, 7)$, $(2, 3, 4)$, $(2, 5, 7)$, $(3, 5, 8)$, $(4, 5, 6)$ and $(6, 7, 8)$. So the minimal number of $n$ is 9.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 28304,
    "subject": "Mathematics (Multi-modal)",
    "question": "It is known that $f(x) = ax^3 + bx^2 + cx + d$ ($a \\neq 0$), and $|f'(x)| \\le 1$ for $0 \\le x \\le 1$. Please find the maximum value of $a$.",
    "options": [],
    "answer": "8/3",
    "solution": "$f'(x) = 3a x^2 + 2b x + c$. We have\n$$\n\\begin{cases}\nf'(0) = c, \\\\\nf'\\left(\\frac{1}{2}\\right) = \\frac{3}{4}a + b + c, \\\\\nf'(1) = 3a + 2b + c.\n\\end{cases}\n$$\nThen\n$$\n3a = 2f'(0) + 2f'(1) - 4f'\\left(\\frac{1}{2}\\right).\n$$\nWe get\n$$\n\\begin{aligned}\n3 | a | &= \\left| 2f'(0) + 2f'(1) - 4f'\\left(\\frac{1}{2}\\right) \\right| \\\\\n&\\le 2 |f'(0)| + 2 |f'(1)| + 4 \\left| f'\\left(\\frac{1}{2}\\right) \\right| \\le 8.\n\\end{aligned}\n$$\nTherefore, $a \\le \\frac{8}{3}$. Furthermore, it is easy to find that $f(x) = \\frac{8}{3}x^3 - 4x^2 + x + m$ (where $m$ is any constant) satisfies the given condition. Therefore, the maximum value of $a$ is $\\frac{8}{3}$.\nLet $g(x) = f'(x) + 1$. Then $0 \\le g(x) \\le 2$ for $0 \\le x \\le 1$. Let $z = 2x - 1$. Then $x = \\frac{z+1}{2}$ and $-1 \\le z \\le 1$. Let\n$$\nh(z) = g\\left(\\frac{z+1}{2}\\right) = \\frac{3a}{4}z^2 + \\frac{3a+2b}{2}z + \\frac{3a}{4} + b + c + 1.\n$$\nIt is easy to check that $0 \\le h(z) \\le 2$ and $0 \\le h(-z) \\le 2$ for $-1 \\le z \\le 1$.\nTherefore, $0 \\le \\frac{h(z) + h(-z)}{2} \\le 2$ for $-1 \\le z \\le 1$. And that is\n$$\n0 \\le \\frac{3a}{4}z^2 + \\frac{3a}{4} + b + c + 1 \\le 2.\n$$\nThen we have $\\frac{3a}{4} + b + c + 1 \\ge 0$ and $\\frac{3a}{4}z^2 \\le 2$. From $0 \\le z^2 \\le 1$ we get $a \\le \\frac{8}{3}$.\nAs $f(x) = \\frac{8}{3}x^3 - 4x^2 + x + m$ (where $m$ is any constant) satisfies the given condition. We obtain that the maximum value of $a$ is $\\frac{8}{3}$. $\\square$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 28305,
    "subject": "Mathematics (Multi-modal)",
    "question": "If $\\min_{x \\in \\mathbb{R}} \\frac{a x^2 + b}{\\sqrt{x^2 + 1}} = 3$, find\n(1) the range of $b$;",
    "options": [],
    "answer": "[3, +infinity)",
    "solution": "Denote $f(x) = \\frac{a x^2 + b}{\\sqrt{x^2+1}}$. It is easy to see that $a > 0$. By $f(0) = b$, we see that $b \\ge 3$.\n\na. If $b - 2a \\ge 0$,\n$$\nf(x) = \\frac{a x^2 + b}{\\sqrt{x^2 + 1}} = a \\sqrt{x^2 + 1} + \\frac{b-a}{\\sqrt{x^2 + 1}} \\ge 2\\sqrt{a(b-a)} = 3,\n$$\nequality holds if $a \\sqrt{x^2+1} = \\frac{b-a}{\\sqrt{x^2+1}}$, that is, if $x = \\pm\\sqrt{\\frac{b-2a}{a}}$.\nThe value of $a$ for given $b$ is $a = \\frac{b - \\sqrt{b^2 - 9}}{2}$, especially when $b = 3$, $a = \\frac{3}{2}$.\n\nb. If $b - 2a < 0$, let $\\sqrt{x^2 + 1} = t$ ($t \\ge 1$). $f(x) = g(t) = a t + \\frac{b-a}{t}$ is monotonically increasing when $t \\ge 1$, so,\n$$\n\\min_{x \\in \\mathbb{R}} f(x) = g(1) = a + b - a = b = 3, \\text{ when } a > \\frac{3}{2}.\n$$\n\nSumming up, we get (1) the range of $b$ is $[3, +\\infty)$.\n(2) If $b = 3$, then $a \\ge \\frac{3}{2}$; if $b > 3$, then $a = \\frac{b - \\sqrt{b^2 - 9}}{2}$.\nLet $f(x) = \\frac{a x^2 + b}{\\sqrt{x^2+1}}$. It is easy to see that $a > 0$. Since $\\min_{x \\in \\mathbb{R}} \\frac{a x^2 + b}{\\sqrt{x^2+1}} = 3$, and $f(0) = b$, we see that $b \\ge 3$.\n$$\nf'(x) = \\frac{a x\\left(x^2 - \\frac{b-2a}{a}\\right)}{\\left(x^2 + 1\\right)^{3/2}}.\n$$\n\na. If $b - 2a \\le 0$, let $f'(x) = 0$, we have the solution $x_0 = 0$, and if $x < 0$, then $f'(x) < 0$; and if $x > 0$, then $f'(x) > 0$.\n$f(0) = b$ is the minimal value. Hence $b = 3$, and $a \\ge \\frac{b}{2}$.\n\nb. If $b - 2a > 0$, let $f'(x) = 0$, we have the solutions $x_0 = 0$, $x_{1,2} = \\pm\\sqrt{\\frac{b-2a}{a}}$.\nIt is easy to see $f(0) = b$ is not the minimal value, which implies $b > 3$; and $f(x_{1,2})$ is the minimal value\n$$\n\\begin{align*}\nf(x_{1,2}) &= \\frac{a \\cdot \\frac{b-2a}{a} + b}{\\sqrt{\\frac{b-2a}{a} + 1}} = 3 \\Rightarrow 2\\sqrt{a(b-a)} = 3 \\\\\n\\Rightarrow a^2 - ab + \\frac{9}{4} &= 0 \\Rightarrow a = \\frac{b - \\sqrt{b^2 - 9}}{2},\n\\end{align*}\n$$\nthat is, $b > 3$ and $a = \\frac{b - \\sqrt{b^2 - 9}}{2}$.\n\nSumming up, we get\n(1) the range of $b$ is $[3, +\\infty)$.\n(2) If $b = 3$, then $a \\ge \\frac{3}{2}$; if $b > 3$, then $a = \\frac{b - \\sqrt{b^2 - 9}}{2}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 28306,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $a_1, a_2, \\dots, a_n$ ($n \\ge 3$) be real numbers. Prove that\n$$\n\\sum_{i=1}^{n} a_i^2 - \\sum_{i=1}^{n} a_i a_{i+1} \\le \\left[ \\frac{n}{2} \\right] (M-m)^2,\n$$\nwhere $a_{n+1} = a_1$, $M = \\max_{1 \\le i \\le n} a_i$, $m = \\min_{1 \\le i \\le n} a_i$. $[x]$ is the largest integer not exceeding $x$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "If $n = 2k$ ($k$ is a positive integer), then\n$$\n2 \\left( \\sum_{i=1}^{n} a_i^2 - \\sum_{i=1}^{n} a_i a_{i+1} \\right) = \\sum_{i=1}^{n} (a_i - a_{i+1})^2 \\le n(M-m)^2,\n$$\ntherefore,\n$$\n\\sum_{i=1}^{n} a_i^2 - \\sum_{i=1}^{n} a_i a_{i+1} \\le \\frac{n}{2} (M-m)^2 = \\left[ \\frac{n}{2} \\right] (M-m)^2.\n$$\nIf $n = 2k + 1$ ($k$ is a positive integer), then for $2k + 1$ numbers arranged in a cyclic way, one can always find three consecutive increasing or decreasing terms (as $\\prod_{i=1}^{2k+1} (a_i - a_{i-1}) (a_{i+1} - a_i)$), so it is not possible that for every $i$, $a_i - a_{i-1}$ and $a_{i+1} - a_i$ having opposite signs). Without loss of generality, we assume that $a_1, a_2, a_3$ are monotonic, then\n$$\n(a_1 - a_2)^2 + (a_2 - a_3)^2 \\le (a_1 - a_3)^2.\n$$\nHence,\n$$\n2 \\left( \\sum_{i=1}^{n} a_i^2 - \\sum_{i=1}^{n} a_i a_{i+1} \\right) = \\sum_{i=1}^{n} (a_i - a_{i+1})^2 \\\\\n\\le (a_1 - a_3)^2 + \\sum_{i=3}^{n} (a_i - a_{i+1})^2,\n$$\nwhich transformed the question into the case of $2k$ numbers. We have\n$$\n2 \\left( \\sum_{i=1}^{n} a_i^2 - \\sum_{i=1}^{n} a_i a_{i+1} \\right) \\le (a_1 - a_3)^2 + \\sum_{i=3}^{n} (a_i - a_{i+1})^2 \\\\\n\\le 2k (M - m)^2,\n$$\ni.e.,\n$$\n\\left( \\sum_{i=1}^{n} a_i^2 - \\sum_{i=1}^{n} a_i a_{i+1} \\right) \\le k (M-m)^2 = \\left[ \\frac{n}{2} \\right] (M-m)^2.\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 28307,
    "subject": "Mathematics (Multi-modal)",
    "question": "Given an integer $n \\ge 3$, find the maximum real number $M$, such that for any positive numbers $x_1, x_2, \\dots, x_n$, there exists a permutation $y_1, y_2, \\dots, y_n$ of $x_1, x_2, \\dots, x_n$ that satisfies\n$$\n\\sum_{i=1}^{n} \\frac{y_i^2}{y_{i+1}^2 - y_{i+1} y_{i+2} + y_{i+2}^2} \\ge M,\n$$\nwhere $y_{n+1} = y_1$, $y_{n+2} = y_2$. (posed by Qu Zhenhua)",
    "options": [],
    "answer": "M = n - 1",
    "solution": "Let\n$$\nF(x_1, \\dots, x_n) = \\sum_{i=1}^{n} \\frac{x_i^2}{x_{i+1}^2 - x_{i+1}x_{i+2} + x_{i+2}^2}.\n$$\nFirst, take $x_1 = x_2 = \\cdots = x_{n-1} = 1$, $x_n = \\epsilon$, then all permutations are the same in the sense of circulation. In this case, we have\n$$\nF(x_1, \\dots, x_n) = n - 3 + \\frac{2}{1 - \\epsilon + \\epsilon^2} + \\epsilon^2.\n$$\nLet $\\epsilon \\to 0^+$, $F \\to n-1$, so $M \\le n-1$.\n\nNext, we show that for any positive numbers $x_1, \\dots, x_n$, there exists a permutation $y_1, \\dots, y_n$ satisfying $F(y_1, \\dots, y_n) \\ge n-1$. In fact, take the permutation $y_1, \\dots, y_n$ with $y_1 \\ge y_2 \\ge \\dots \\ge y_n$ and by the inequality $a^2 - ab + b^2 \\le \\max(a^2, b^2)$, we see that\n$$\nF(y_1, \\dots, y_n) \\ge \\frac{y_1^2}{y_2^2} + \\frac{y_2^2}{y_3^2} + \\dots + \\frac{y_{n-1}^2}{y_1^2} \\ge n-1,\n$$\nwhere the last inequality is obtained by AM-GM inequality.\n\nSumming up, $M = n-1$. $\\square$",
    "topic": "Discrete Mathematics",
    "subtopic": "Combinatorics"
  },
  {
    "id": 28308,
    "subject": "Mathematics (Multi-modal)",
    "question": "Given that $0 < x, y < 1$, determine, with proof, the maximum value of $\\frac{xy(1-x-y)}{(x+y)(1-x)(1-y)}$.",
    "options": [],
    "answer": "1/8",
    "solution": "When $x = y = \\frac{1}{3}$, the value of the expression is $\\frac{1}{8}$.\n\nWe will prove that $\\frac{xy(1-x-y)}{(x+y)(1-x)(1-y)} \\le \\frac{1}{8}$ for any $0 < x, y < 1$ as follows.\n\nIf $x + y \\ge 1$, then $\\frac{xy(1-x-y)}{(x+y)(1-x)(1-y)} \\le 0 < \\frac{1}{8}$.\n\nIf $x + y < 1$, then let $1 - x - y = z > 0$, it follows from the AM-GM Inequality that\n$$\n\\frac{xy(1-x-y)}{(x+y)(1-x)(1-y)} = \\frac{xyz}{(x+y)(y+z)(z+x)} \\\\\n\\le \\frac{xyz}{2\\sqrt{xy} \\cdot 2\\sqrt{yz} \\cdot 2\\sqrt{zx}} = \\frac{1}{8}.\n$$\nIn conclusion, the maximum value of the expression is $\\frac{1}{8}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 28309,
    "subject": "Mathematics (Multi-modal)",
    "question": "As seen in Fig. 1.1, points $P$, $Q$ are, respectively, the midpoints of $AC$, $BD$ — the two diagonals of cyclic quadrilateral $ABCD$. Let $\\angle BPA = \\angle DPA$. Prove $\\angle AQB = \\angle CQB$.\n\n![](attached_image_1.png)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "As shown in Fig. 1.2, we extend segment $DP$ to intercept with the circle at point $E$. Then $\\angle CPE = \\angle DPA = \\angle BPA$. Since $P$ is the midpoint of $AC$, we get $\\overline{AB} = \\overline{CE}$, which means $\\angle CDP = \\angle BDA$. Furthermore, $\\angle ABD = \\angle PCD$. Therefore, $\\triangle ABD \\sim \\triangle PCB$. Then $\\frac{AB}{BD} = \\frac{PC}{CD}$, i.e., $AB \\cdot CD = PC \\cdot BD$.\n\nThen we have\n$$\nAB \\cdot CD = \\frac{1}{2} AC \\cdot BD = AC \\cdot \\left(\\frac{1}{2} BD\\right) = AC \\cdot BQ,\n$$\nor $\\frac{AB}{AC} = \\frac{BQ}{CD}$. Combining it with $\\angle ABQ = \\angle ACD$, we derive that $\\triangle ABQ \\sim \\triangle ACD$. So $\\angle QAB = \\angle DAC$.\n\nExtending segment $AQ$ to intercept with the circle at point $F$, we then have\n$$\n\\angle CAB = \\angle QAB - \\angle QAC = \\angle DAC - \\angle QAC = \\angle DAF,\n$$\nwhich means $\\widehat{BC} = \\widehat{DF}$. Furthermore, as $Q$ is the midpoint of $BD$, then $\\angle CQB = \\angle DQF$.\n\nSince $\\angle AQB = \\angle DQF$, we then have $\\angle AQB = \\angle CQB$.\n\nThe proof is completed. $\\square$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 28310,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all positive integers $n$ such that equation $\\frac{1}{x} + \\frac{1}{y} = \\frac{1}{n}$ has exactly 2011 positive integer solutions $(x, y)$ with $x \\le y$.",
    "options": [],
    "answer": "n = p^{2010} for any prime p",
    "solution": "From the given equation, we have $xy - nx - ny = 0 \\Rightarrow (x-n)(y-n) = n^2$. Then, besides $x = y = 2n$, for any $x-n$ equal to a proper divisor of $n^2$, we will get a positive integer solution $(x, y)$ satisfying the required condition. Therefore, $n^2$ should have exactly 2010 proper divisors that are less than $n$.\n\nSuppose $n = p_1^{\\alpha_1} \\cdots p_k^{\\alpha_k}$, where $p_1, \\dots, p_k$ are prime numbers different from each other. Then the number of proper divisors of $n^2$ less than $n$ is\n$$\n\\frac{(2\\alpha_1 + 1) \\cdots (2\\alpha_k + 1) - 1}{2}\n$$\nSo $(2\\alpha_1 + 1) \\cdots (2\\alpha_k + 1) = 4021$. Since 4021 is prime, we get $k = 1, 2\\alpha_1 + 1 = 4021$, and $\\alpha_1 = 2010$.\n\nTherefore, $n = p^{2010}$, where $p$ is any prime number.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 28311,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $A = \\{a_1, a_2, a_3, a_4\\}$. Suppose the set of sums of all the elements in every ternary subset of $A$ is $B = \\{-1, 3, 5, 8\\}$. Then $A = \\underline{\\hspace{2cm}}$.",
    "options": [],
    "answer": "{-3, 0, 2, 6}",
    "solution": "Obviously, every element of $A$ appears three times in all the ternary subsets. Then we have\n$$3(a_1 + a_2 + a_3 + a_4) = (-1) + 3 + 5 + 8 = 15,$$\nor $a_1 + a_2 + a_3 + a_4 = 5$. Therefore, the four elements of $A$ are $5 - (-1) = 6$, $5 - 3 = 2$, $5 - 5 = 0$, $5 - 8 = -3$, respectively.\nThe answer is $A = \\{-3, 0, 2, 6\\}$. $\\square$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 28312,
    "subject": "Mathematics (Multi-modal)",
    "question": "Suppose sequence $\\{a_n\\}$ satisfies $a_1 = 2t - 3$ ($t \\in \\mathbb{R}$ and $t \\neq \\pm 1$),\n$$\na_{n+1} = \\frac{(2t^{n+1} - 3)a_n + 2(t-1)t^n - 1}{a_n + 2t^n - 1} \\quad (n \\in \\mathbb{N}^*).\n$$\n\n(1)\nFind the formula of general term about $\\{a_n\\}$.\n\n(2)\nIf $t > 0$, find out which is larger between $a_{n+1}$ and $a_n$.",
    "options": [],
    "answer": "a_n = 2(t^n − 1)/n − 1; for t > 0, a_{n+1} > a_n.",
    "solution": "(1) The given expression can be rewritten as\n$$\na_{n+1} = \\frac{2(t^{n+1} - 1)(a_n + 1)}{a_n + 2t^n - 1} - 1.\n$$\nThen\n$$\n\\frac{a_{n+1} + 1}{t^{n+1} - 1} = \\frac{2(a_n + 1)}{a_n + 2t^n - 1} = \\frac{\\frac{2(a_n + 1)}{t^n - 1}}{\\frac{a_n + 1}{t^n - 1} + 2}.\n$$\nLet $\\frac{a_n + 1}{t^n - 1} = b_n$. Then $b_{n+1} = \\frac{2b_n}{b_n + 2}$, with $b_1 = \\frac{a_1 + 1}{t - 1} = \\frac{2t - 2}{t - 1} = 2$.\n\nFurthermore, $\\frac{1}{b_{n+1}} = \\frac{1}{b_n} + \\frac{1}{2}$, $\\frac{1}{b_1} = \\frac{1}{2}$. Then\n$$\n\\frac{1}{b_n} = \\frac{1}{b_1} + (n-1) \\cdot \\frac{1}{2} = \\frac{n}{2}.\n$$\nTherefore, $\\frac{a_n + 1}{t^n - 1} = \\frac{2}{n}$, which means $a_n = \\frac{2(t^n - 1)}{n} - 1$.\n\n\n(2) We have\n$$\n\\begin{aligned} a_{n+1} - a_n &= \\frac{2(t^{n+1} - 1)}{n+1} - \\frac{2(t^n - 1)}{n} \\\\ &= \\frac{2(t-1)}{n(n+1)} \\left[ n(1+t+\\cdots+t^{n-1}+t^n) - (n+1)(1+t+\\cdots+t^{n-1}) \\right] \\\\ &= \\frac{2(t-1)}{n(n+1)} \\left[ nt^n - (1+t+\\cdots+t^{n-1}) \\right] \\\\ &= \\frac{2(t-1)}{n(n+1)} \\left[ (t^n-1) + (t^n-t) + \\cdots + (t^n-t^{n-1}) \\right] \\\\ &= \\frac{2(t-1)^2}{n(n+1)} \\left[ (t^{n-1}+t^{n-2}+\\cdots+1) + t(t^{n-2}+t^{n-3}+\\cdots+1) + \\cdots + t^{n-1} \\right]. \\end{aligned}\n$$\nIt is obvious that $a_{n+1} - a_n > 0$ for $t > 0$ ($t \\neq 1$). Therefore, $a_{n+1} > a_n$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 28313,
    "subject": "Mathematics (Multi-modal)",
    "question": "Straight line $l$ with slope $\\frac{1}{3}$ intercepts ellipse $C: \\frac{x^2}{36} + \\frac{y^2}{4} = 1$ at points $A, B$, and point $P(3\\sqrt{2}, \\sqrt{2})$ is in the top-left of $l$ (as shown in Fig. 11.1).\n![](attached_image_1.png)\nFig. 11.1\n\na. Prove that the center of the inscribed circle of $\\triangle PAB$ is on the line $x = 3\\sqrt{2}$.\n\nb. When $\\angle APB = 60^{\\circ}$, find the area of $\\triangle PAB$.",
    "options": [],
    "answer": "117√3/49",
    "solution": "a.\nLet $l$ be a straight line such that $y = \\frac{1}{3}x + m$, and $A(x_1, y_1), B(x_2, y_2)$.\nSubstituting $y = \\frac{1}{3}x + m$ into $\\frac{x^2}{36} + \\frac{y^2}{4} = 1$, and simplifying it, we have\n$$\n2x^2 + 6mx + 9m^2 - 36 = 0.\n$$\nThen $x_1 + x_2 = -3m$, $x_1x_2 = \\frac{9m^2 - 36}{2}$, $k_{PA} = \\frac{y_1 - \\sqrt{2}}{x_1 - 3\\sqrt{2}}$, $k_{PB} = \\frac{y_2 - \\sqrt{2}}{x_2 - 3\\sqrt{2}}$. Therefore,\n$$\n\\begin{aligned}\nk_{PA} + k_{PB} &= \\frac{y_1 - \\sqrt{2}}{x_1 - 3\\sqrt{2}} + \\frac{y_2 - \\sqrt{2}}{x_2 - 3\\sqrt{2}} \\\\\n&= \\frac{(y_1 - \\sqrt{2})(x_2 - 3\\sqrt{2}) + (y_2 - \\sqrt{2})(x_1 - 3\\sqrt{2})}{(x_1 - 3\\sqrt{2})(x_2 - 3\\sqrt{2})}.\n\\end{aligned}\n$$\nIn the expression above, the numerator is equal to\n$$\n\\begin{aligned}\n& \\left(\\frac{1}{3}x_1 + m - \\sqrt{2}\\right)(x_2 - 3\\sqrt{2}) + \\left(\\frac{1}{3}x_2 + m - \\sqrt{2}\\right)(x_1 - 3\\sqrt{2}) \\\\\n&= \\frac{2}{3}x_1x_2 + (m - 2\\sqrt{2})(x_1 + x_2) - 6\\sqrt{2}(m - \\sqrt{2}) \\\\\n&= \\frac{2}{3} \\cdot \\frac{9m^2 - 36}{2} + (m - 2\\sqrt{2})(-3m) - 6\\sqrt{2}(m - \\sqrt{2}) \\\\\n&= 3m^2 - 12 - 3m^2 + 6\\sqrt{2}m - 6\\sqrt{2}m + 12 = 0.\n\\end{aligned}\n$$\nTherefore, $k_{PA} + k_{PB} = 0$. Since $P$ is in the top-left of $l$, we know that the bisector of $\\angle APB$ is parallel to the $y$-axis. Therefore, the center of the inscribed circle of $\\triangle PAB$ is on line $x = 3\\sqrt{2}$.\n\n\nb.\nWhen $\\angle APB = 60^\\circ$, by the result in (a), we have $k_{PA} = \\sqrt{3}$, $k_{PB} = -\\sqrt{3}$. Then the equation for line $PA$ is $y - \\sqrt{2} = \\sqrt{3}(x - 3\\sqrt{2})$. Substituting it into $\\frac{x^2}{36} + \\frac{y^2}{4} = 1$, and eliminating $y$, we get\n$$\n14x^2 + 9\\sqrt{6}(1 - 3\\sqrt{3})x + 18(13 - 3\\sqrt{3}) = 0,\n$$\nwhich has roots $x_1$ and $3\\sqrt{2}$. So $x_1 \\cdot 3\\sqrt{2} = \\frac{18(13 - 3\\sqrt{3})}{14}$, i.e.\n$$x_1 = \\frac{3\\sqrt{2}(13 - 3\\sqrt{3})}{14}.$$ Then we find\n$$\n|PA| = \\sqrt{1 + (\\sqrt{3})^2} \\cdot |x_1 - 3\\sqrt{2}| = \\frac{3\\sqrt{2}(3\\sqrt{3} + 1)}{7}.\n$$\nIn the same way, we have $|PB| = \\frac{3\\sqrt{2}(3\\sqrt{3} - 1)}{7}$.\nTherefore,\n$$\n\\begin{align*}\nS_{\\triangle PAB} &= \\frac{1}{2} \\cdot |PA| \\cdot |PB| \\cdot \\sin 60^\\circ \\\\\n&= \\frac{1}{2} \\cdot \\frac{3\\sqrt{2}(3\\sqrt{3} + 1)}{7} \\cdot \\frac{3\\sqrt{2}(3\\sqrt{3} - 1)}{7} \\cdot \\frac{\\sqrt{3}}{2} \\\\\n&= \\frac{117\\sqrt{3}}{49}.\n\\end{align*}\n$$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 28314,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $n > 1$ be an integer, $k$ be the number of distinct prime factors of $n$. Prove that there exists an integer $a$, $1 < a < \\frac{n}{k} + 1$, such that $n \\mid a^2 - a$. (posed by Yu Hongbing)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $n = p_1^{a_1} \\cdots p_k^{a_k}$ be the standard factorization of $n$. Since $p_1^{a_1}, \\dots, p_k^{a_k}$ are pairwise coprime, by the Chinese Remainder Theorem, for each $i$, $1 \\le i \\le k$, congruence equations\n$$\n\\begin{cases} x \\equiv 1 \\pmod{p_i^{a_i}} \\\\ x \\equiv 0 \\pmod{p_j^{a_j}}, \\quad j \\neq i \\end{cases}\n$$\nhave solution $x_i$.\nFor any solution of $x_0^2 = x_0 \\pmod{n}$, we see that $x_0(x_0 - 1) \\equiv 0 \\pmod{n}$. Then for each $i = 1, 2, \\dots, k$, either $x_0 \\equiv 0 \\pmod{p_i^{a_i}}$ or $x_0 \\equiv 1 \\pmod{p_i^{a_i}}$. Further, let $S(A)$ be the sum of elements of subset $\\{x_1, x_2, \\dots, x_k\\}$ (particularly, $S(\\emptyset) = 0$). Obviously, we have\n$$\nS(A)(S(A) - 1) \\equiv 0 \\pmod{n}.\n$$\n(This is because of the selection of $x_i$, such that $S(A) \\pmod{p_i^{a_i}}$ is either 0 or 1.) Moreover if $A \\neq A'$, then $S(A) \\neq S(A') \\pmod{n}$. Therefore, the sum of all subsets of $\\{x_1, x_2, \\dots, x_n\\}$ is exactly all solutions of $x(x - 1) \\equiv 0 \\pmod{n}$.\nLet $S_0 = n$, $S_r$ be the least non-negative remainder of $x_1 + x_2 + \\dots + x_r$ modulo $n$, $r = 1, 2, \\dots, k$. Thus $S_k = 1$. For all $1 \\le r \\le k - 1$, $S_r \\neq 0$. Since $k+1$ numbers $S_0, S_1, \\dots, S_k$ are in $[1, n]$, by Dirichlet's Drawer Principle, there exist $0 \\le l < m \\le k$, such that $S_l, S_m$ in the same interval $\\left(\\frac{jn}{k}, \\frac{(j+1)n}{k}\\right]$, $(0 \\le j \\le k-1)$, where $l=0$ and $m=k$ do not hold simultaneously.\nThus, $|S_l - S_m| < \\frac{n}{k}$. Denote $y_1 = S_1, y_r = S_r - S_{r-1}$ ($r=2, 3, \\dots, k$). So any sum of $y_r \\equiv x_r \\pmod{n}$ ($r=1, 2, \\dots, k$) meets the requirement.\nIf $S_m - S_l > 1$, then $a = y_{l+1} + y_{l+2} + \\dots + y_m = S_m - S_l \\in (1, \\frac{n}{k})$ is the solution of the equation $x^2 - x \\equiv 0 \\pmod{n}$.\nIf $S_m - S_l = 1$, then $n$ ($y_1 + y_2 + \\dots + y_l + (y_{m+1} + y_{m+2} + \\dots + y_k)$, that is, $n$ ($x_1 + x_2 + \\dots + x_l + (x_{m+1} + x_{m+2} + \\dots + x_k)$. Notice that $m > l$, which contradicts to the definition of $x_i$.\nIf $S_m - S_l = 0$, then $n \\mid y_{l+1} + y_{l+2} + \\cdots + y_m$, that is, $n \\mid x_{l+1} + x_{l+2} + \\cdots + x_m$, which contradicts the definition of $x_i$.\nIf $S_m - S_l < 0$, then\n$$\n\\begin{aligned}\na &= (y_1 + y_2 + \\cdots + y_l) + (y_{m+1} + \\cdots + y_k) \\\\\n&= S_k - (S_m - S_l) = 1 - (S_m - S_l)\n\\end{aligned}\n$$\nis the solution of equation $x^2 - x \\equiv 0 \\pmod{n}$, and $1 < a < 1 + \\frac{n}{k}$.\nSumming up, there exists $a$ satisfying the condition. $\\square$",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 28315,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $a$, $b$ and $c$ be coprime positive integers so that $a^2 \\mid (b^3 + c^3)$, $b^2 \\mid (a^3 + c^3)$ and $c^2 \\mid (a^3 + b^3)$. Find the values of $a$, $b$ and $c$. (posed by Yang Xiaoming)",
    "options": [],
    "answer": "The only solutions are (1,1,1) and all permutations of (1,2,3).",
    "solution": "By the condition of the problem, we have $a^2 \\mid (a^3 + b^3 + c^3)$, $b^2 \\mid (a^3 + b^3 + c^3)$ and $c^2 \\mid (a^3 + b^3 + c^3)$. Since $a$, $b$ and $c$ are coprime, we see that $a^2 b^2 c^2 \\mid (a^3 + b^3 + c^3)$.\n\nWithout loss of generality, suppose that $a \\ge b \\ge c$, so\n$$\n3a^3 \\ge a^3 + b^3 + c^3 \\ge a^2 b^2 c^2 \\Rightarrow a \\ge \\frac{b^2 c^2}{3},\n$$\nand\n$$\n2b^3 \\ge b^3 + c^3 \\ge a^2 \\Rightarrow 2b^3 \\ge \\frac{b^4 c^4}{9} \\Rightarrow b \\le \\frac{18}{c^4}.\n$$\nWe see that if $c \\ge 2 \\Rightarrow b \\le 1$, the result contradicts $b \\ge c$. Thus, $c = 1$.\n\nIf $c = 1$ and $b = 1$, then $a = 1$, so $(a, b, c) = (1, 1, 1)$ is a solution.\n\nIf $c = 1$, $b \\ge 2$ and $a = b$, then $b^2 \\mid b^3 + 1$, which is a contradiction!\n\nIf $b \\ge 2$ and $a > b > c = 1$, then\n$$\na^2 b^2 \\mid (a^3 + b^3 + 1) \\Rightarrow 2a^3 \\ge a^3 + b^3 + 1 \\ge a^2 b^2 \\Rightarrow a \\ge \\frac{b^2}{2},\n$$\nand by $c = 1$,\n$$\na^2 \\mid (b^3 + 1) \\Rightarrow b^3 + 1 \\ge a^2 \\ge \\frac{b^4}{4} \\Rightarrow 4b^3 + 4 \\ge b^4.\n$$",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 28316,
    "subject": "Mathematics (Multi-modal)",
    "question": "Prove for any integer $n \\ge 4$, there exists a polynomial of degree $n$,\n$$\nf(x) = x^n + a_{n-1}x^{n-1} + \\cdots + a_1x + a_0\n$$\nwith the following properties.\n(1) $a_0, a_1, \\dots, a_{n-1}$ are all positive integers;\n(2) For any positive integer $m$ and arbitrary $k$ ($k \\ge 2$) positive integers $r_1, r_2, \\dots, r_k$ that are different from each other, we have\n$$\nf(m) \\neq f(r_1)f(r_2)\\cdots f(r_k).\n$$",
    "options": [],
    "answer": "f(x) = (x+1)(x+2)⋯(x+n) + 2",
    "solution": "Let\n$$\nf(x) = (x+1)(x+2)\\cdots(x+n)+2. \\qquad \\textcircled{1}\n$$\nObviously, $f(x)$ is a monic polynomial of degree $n$ with positive integer coefficients. We are going to prove that $f(x)$ has property (2).\nFor any integer $t$, since $n \\ge 4$, we know that there exists definitely a multiple of 4 in any $n$ consecutive numbers $t+1, t+2, \\dots, t+n$. Then from $\\textcircled{1}$, we have $f(t) \\equiv 2 \\pmod 4$.\nThen for any $k$ ($k \\ge 2$) positive integers $r_1, r_2, \\dots, r_k$, we have\n$$\nf(r_1)f(r_2)\\cdots f(r_k) \\equiv 2^k \\equiv 0 \\pmod 4.\n$$\nOn the other hand, for any positive integer $m$, we have\n$f(m) \\equiv 2 \\pmod{4}$. Therefore,\n$$\nf(m) \\not\\equiv f(r_1)f(r_2)\\cdots f(r_k) \\pmod{4},\n$$\nwhich implies that $f(m) \\neq f(r_1)f(r_2)\\cdots f(r_k)$. We then find the required $f(x)$ and complete the proof. $\\square$",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 28317,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $M \\subseteq \\{1, 2, \\dots, 2011\\}$ be a subset satisfying the following condition: For any three elements in $M$, there exist two of them $a$ and $b$, such that $a \\mid b$ or $b \\mid a$. Determine, with proof, the maximum value of $|M|$, where $|M|$ denotes the number of elements of $M$. (posed by Feng Zhigang)",
    "options": [],
    "answer": "21",
    "solution": "One can check that $M = \\{1, 2, 2^2, 2^3, \\dots, 2^{10}, 3, 3 \\times 2, 3 \\times 2^2, \\dots, 3 \\times 2^9\\}$ satisfies the condition, and $|M| = 21$.\n\nSuppose that $|M| \\ge 22$, and let $a_1 < a_2 < \\dots < a_k$ be the elements of $M$, where $|M| = k \\ge 22$. We first prove that $a_{n+2} \\ge 2a_n$ for all $n$; otherwise, we have $a_n < a_{n+1} < a_{n+2} < 2a_n$ for some $n < k + 2$, then any two of these three integers $a_n, a_{n+1}, a_{n+2}$ do not have any multiple relationship, which contradicts the assumption.\n\nIt follows from the inequality above that $a_4 \\ge 2a_2 \\ge 4$, $a_6 \\ge 2a_4 \\ge 8$, $\\dots$, $a_{22} \\ge 2a_{20} \\ge 2^{11} > 2011$, which is a contradiction!\n\nHence, the maximum value of $|M|$ is $21$.\n$\\square$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 28318,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $3n^2$ be the vertex number of a simple graph $G$ (integer $n \\ge 2$). If the degree of each vertex is not greater than $4n$, there exists at least one vertex with degree $1$, and there exists a route with length not greater than $3$ between any two vertices. Prove that the minimum number of edges of $G$ is $\\frac{7}{2}n^2 - \\frac{3}{2}n$.\n\n**Remark.** A route between two distinct vertices $u$ and $v$ with length $k$ is a sequence of vertices $u = v_0, v_1, \\dots, v_k = v$, where $v_i$ and $v_{i+1}$, $i = 0, 1, \\dots, k-1$, are adjacent. (posed by Leng Gangsong)",
    "options": [],
    "answer": "7/2 n^2 − 3/2 n",
    "solution": "For any two distinct vertices $u$ and $v$, we say that the distance between $u$ and $v$ is the shortest length of the route between $u$ and $v$. Consider a graph $G^*$ with vertex set $\\{x_1, x_2, \\dots, x_{3n^2-n}, y_1, y_2, \\dots, y_n\\}$, where $y_i$ and $x_i$ are adjacent ($1 \\le i < j \\le n$), $x_i$ and $x_j$ are not adjacent ($1 \\le i < j \\le 3n^2 - n$), $x_i$ and $y_j$ are adjacent if and only if $i \\equiv j \\pmod{n}$. Thus, the degree of each $x_i$ is $1$, and the degree of $y_i$ does not exceed\n$$\nn - 1 + \\frac{3n^2 - n}{n} = 4n - 2.\n$$\nIt is easy to see that the distance between $x_i$ and $x_j$ is not greater than $3$. So graph $G^*$ satisfies the condition of the problem. $G^*$ has $N = 3n^2 - n + C_n^2 = \\frac{7}{2}n^2 - \\frac{3}{2}n$ edges.\n\nIn the following, we show that any graph $G = G(V, E)$ satisfying the condition of the problem has at least $N$ edges. Let $X \\subseteq V$ be the set of vertices with degree $1$, $Y \\subseteq (V\\setminus X)$ be the set of remaining vertices adjacent to $X$, and $Z \\subseteq V\\setminus(X \\cup Y)$ be the set of remaining vertices adjacent to $Y$. Let $W = V\\setminus(X \\cup Y \\cup Z)$. We will point out the following facts.\n\n**Property 1.** Any two vertices in $Y$ are adjacent. This is because of the fact that if $y_1, y_2 \\in Y$ are two vertices, there exist $x_1, x_2 \\in X$ that are adjacent to $y_1$ and $y_2$, respectively; hence $y_1$ and $y_2$ are adjacent since the distance between $x_1$ and $x_2$ is not greater than $3$.\n\n**Property 2.** The distance between vertex in $W$ and vertex in $Y$ is $2$. This is because of the fact that if the distance between $w_0 \\in W$ and $y_0 \\in Y$ is greater than $2$ (obviously, distance $>1$), suppose that $x_0 \\in X$ is adjacent to $y_0$, then the distance between $w_0$ and $x_0$ is greater than $3$, which is a contradiction. Furthermore, we know this Property 2 means each vertex in $W$ is adjacent to some vertex in $Z$.\n\nDenote by $x, y, z$ and $w$ the numbers of elements in sets $X, Y, Z$ and $W$, respectively. Now count the number of edges; there are $C_y^2$ edges between points in $Y$, $x$ edges from points of $X$ to $Y$, at least $z$ edges from points of $Z$ to $Y$, and at least $w$ edges from points of $W$ to $Z$. So, if $y \\ge n$, then\n$$|E| \\ge C_y^2 + x + z + w = 3n^2 + C_y^2 - y \\ge 3n^2 + C_n^2 - n = N,$$\nand if $y \\le n-1$, since each degree of vertex is at most $4n$,\n$$\n\\begin{align*}\nx + z &\\le y(4n - (y - 1)) = y(4n + 1 - y) \\\\\n&\\le (n - 1)(3n + 2) = 3n^2 - n - 2, \\\\\n\\omega &\\ge 3n^2 - y - y(4n + 1 - y) \\ge 3.\n\\end{align*}\n$$\nSelect a vertex $P$ in $W$ such that $P$ is adjacent as less as possible to vertices in $Z$. Suppose the least number is $a$, $a > 0$ (by Property 2). Denote the set of these $a$ vertices by $N_P \\subseteq Z$.\n\nCounting the number of the edges again, there are $C_y^2$ edges between points in $Y$, $x$ edges from points of $X$ to $Y$, at least $y$ edges from points of $N_P$ to $Y$ (by Property 2, the distance from $P$ to vertex in $Y$ is $2$), at least $z - a$ edges from points of $Z \\setminus N_P$ to $Y$, and at least $aw$ edges from points of $W$ to $Z$. Thus,\n$$\n\\begin{align*}\n|E| &\\ge C_y^2 + x + y + z - a + aw \\\\\n&= 3n^2 - 1 + C_y^2 + (a - 1)(w - 1).\n\\end{align*}\n$$\nIf $a > 1$, then\n$$\n\\begin{align*}\n|E| &\\ge 3n^2 - 1 + C_y^2 + (w - 1) \\\\\n&\\ge 3n^2 - 2 + C_y^2 + 3n^2 - y - y(4n + 1 - y) > N.\n\\end{align*}\n$$\nIf $a = 1$, since the degree of each vertex in $W$ is at least $2$, when we count the edges from points of $W$ to $Z$, we should add at least $w/2$ edges, so\n$$\n\\begin{align*}\n|E| &\\ge 3n^2 - 1 + C_y^2 + \\frac{1}{2}\\omega \\\\\n&\\ge 3n^2 - 1 + C_y^2 + \\frac{1}{2}(3n^2 - y - y(4n + 1 - y)) > N.\n\\end{align*}\n$$\nSumming up, the least number of edges is $N = \\frac{7}{2}n^2 - \\frac{3}{2}n$.",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 28319,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $n \\ge 2$ be a given integer.\n(1) Prove that one can arrange all the subsets of the set $\\{1, 2, \\dots, n\\}$ as a sequence of subsets $A_1, A_2, \\dots, A_{2^n}$, such that $|A_{i+1}| = |A_i| + 1$ or $|A_i| - 1$, where $i = 1, 2, \\dots, 2^n$ and $A_{2^n+1} = A_1$.\n(2) Determine, with proof, all possible values of the sum $\\sum_{i=1}^{2^n} (-1)^i S(A_i)$, where $S(A_i) = \\sum_{x \\in A_i} x$ and $S(\\emptyset) = 0$, for any subset sequence $A_1, A_2, \\dots, A_{2^n}$ satisfying the condition in (1). (posed by Liang Yingde)",
    "options": [],
    "answer": "0",
    "solution": "(1) We prove by mathematical induction that there exists a sequence $A_1, A_2, \\dots, A_{2^n}$ such that $A_1 = \\{1\\}, A_{2^n} = \\emptyset$ and satisfies the condition in (1).\nWhen $n = 2$, the sequence $\\{1\\}, \\{1, 2\\}, \\{2\\}, \\emptyset$ of $\\{1, 2\\}$ works.\nAssume that when $n = k$, there exists such a sequence $B_1, B_2, \\dots, B_{2^k}$ of subsets of $\\{1, 2, \\dots, k\\}$. As for $n = k + 1$, one can construct a sequence of subsets of $\\{1, 2, \\dots, k + 1\\}$, as follows:\n$$\n\\begin{align*}\nA_1 &= B_1 = \\{1\\}, \\\\\nA_i &= B_{i-1} \\cup \\{k+1\\}, \\quad i = 2, 3, \\dots, 2^k + 1, \\\\\nA_j &= B_{j-2^k}, \\quad j = 2^k + 2, 2^k + 3, \\dots, 2^{k+1}.\n\\end{align*}\n$$\nOne can easily check that the sequence fulfills the required conditions stated above. By induction we have proved (1) for $n \\ge 2$.\n\n(2) We will show that the sum is $0$, independent of the arrangement. Without loss of generality, we may assume that $A_1 = \\{1\\}$, otherwise shift the index cyclically. It follows from $|A_{i+1}| = |A_i|+1$ or $|A_i|-1$ that their parities are different, and hence the parities of the index label of any subset and its cardinality are the same.\nIt follows that $\\sum_{i=1}^{2^n} (-1)^i S(A_i) = \\sum_{A \\in P} S(A) - \\sum_{A \\in Q} S(A)$, where $P$ consists of all subsets of $\\{1, 2, \\dots, n\\}$ with even numbers of elements, and $Q$ consists of all subsets of $\\{1, 2, \\dots, n\\}$ with odd numbers of elements.\nFor any $x \\in \\{1, 2, \\dots, n\\}$, among all $k$-element subsets, $x$ appears in exactly $C_{n-1}^{k-1}$ of them, hence it contributes to the sum\n$$\n\\sum_{A \\in P} S(A) - \\sum_{A \\in Q} S(A) \\text{ as } - C_{n-1}^0 + C_{n-1}^1 - C_{n-1}^2 + \\dots + (-1)^n C_{n-1}^{n-1} = -(1-1)^{n-1} = 0.\n$$\nTherefore, $\\sum_{i=1}^{2^n} (-1)^i S(A_i) = 0$. $\\square$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 28320,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let set $M = \\{1, 2, 3, ..., 50\\}$. Find all positive integer $n$, such that there are at least two different elements $a$ and $b$ in any subset with 35 elements of $M$, such that $a + b = n$ or $a - b = n$. (posed by Li Shenghong)",
    "options": [],
    "answer": "1 <= n <= 69",
    "solution": "Take $A = \\{1, 2, 3, ..., 35\\}$, then for any $a, b \\in A$,\n$$\na - b \\le 34,\\ a + b \\le 34 + 35 = 69.\n$$\nIn the following, we show that $1 \\le n \\le 69$. Let $A = \\{a_1, a_2, \\dots, a_{35}\\}$, without loss of generality, suppose that $a_1 < a_2 < \\dots < a_{35}$.\n\ni.\nIf $1 \\le n \\le 19$, by\n$$\n1 \\le a_1 < a_2 < \\dots < a_{35} \\le 50,\n$$\n$$\n2 \\le a_1 + n < a_2 + n < \\dots < a_{35} + n \\le 50 + 19 = 69,\n$$\nand by Dirichlet's Drawer Theorem, there exist $1 \\le i, j \\le 35$ ($i \\ne j$) such that $a_i + n = a_j$, that is, $a_i - a_j = n$.\n\nii.\nIf $51 \\le n \\le 69$, by\n$$\n1 \\le a_1 < a_2 < \\dots < a_{35} \\le 50,\n$$\n$$\n1 \\le n - a_{35} < n - a_{34} < \\dots < n - a_1 \\le 68,\n$$\nand by Dirichlet's Drawer Theorem, there exist at least $1 \\le i, j \\le 35$ ($i \\ne j$) such that $n - a_i = a_j$, that is, $a_i + a_j = n$.\n\niii.\nIf $20 \\le n \\le 24$, since\n$$\n50 - (2n + 1) + 1 = 50 - 2n \\le 50 - 40 = 10,\n$$\nwe see that there are at least 25 elements in $a_1, a_2, \\dots, a_{35}$ that belong to $[1, 2n]$.\nThere are at most 24 elements in $\\{1, n+1\\}, \\{2, n+2\\}, \\dots, \\{n, 2n\\}$ such that $\\{a_i, a_j\\} = \\{i, n+i\\}$. Hence, $a_j - a_i = n$.\n\niv. If $25 \\le n \\le 34$, since $\\{1, n+1\\}, \\{2, n+2\\}, \\dots, \\{n, 2n\\}$ have at most 34 elements, by Dirichlet Drawer Theorem, there exist $1 \\le i, j \\le 35$ ($i \\ne j$) such that $a_i = i, a_j = n+i$, that is $a_j - a_i = n$.\n\nv. If $n = 35$, there are 33 elements $\\{1, 34\\}, \\{2, 33\\}, \\dots, \\{17, 18\\}, \\{35\\}, \\{36\\}, \\dots, \\{50\\}$. Hence, there exist $1 \\le i, j \\le 35$ ($i \\ne j$) such that $a_i + a_j = 35$.\n\nvi.\nIf $36 \\le n \\le 50$,\nif $n = 2k + 1, \\{1, 2k\\}, \\{2, 2k-1\\}, \\dots, \\{k, k+1\\}, \\{2k+1\\}, \\dots, \\{50\\}$;\nif $18 \\le k \\le 20, 50 - (2k + 1) + 1 = 50 - 2k \\le 50 - 36 = 14$;\nif $21 \\le k \\le 24$, $50 - (2k + 1) + 1 = 50 - 2k \\le 50 - 42 = 8$,\nthere exist $1 \\le i, j \\le 35$ ($i \\neq j$) such that $a_i + a_j = 2k + 1 = n$.\nIf $n = 2k$, $\\{1, 2k-1\\}$, $\\{2, 2k-2\\}$, ..., $\\{k-1, k+1\\}$, $\\{k\\}$, $\\{2k\\}$, $\\{2k+1\\}$, ..., $\\{50\\}$;\nif $18 \\le k \\le 19$, $50 - (2k + 1) + 3 \\le 16k - 1 \\le 19 - 1 = 18$;\nif $20 \\le k \\le 23$, $50 - (2k + 1) + 3 \\le 50 - 2k + 2 \\le 12k - 1$\n$\\le 23 - 1 = 22$;\nif $24 \\le k \\le 25$, $50 - (2k + 1) + 3 \\le 50 - 2k + 2 \\le 4k - 1 \\le$\n$25 - 1 = 24$,\nthere exist $1 \\le i, j \\le 35$ ($i \\neq j$) such that $a_i + a_j = 2k$.\n$\\square$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 28321,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $a_1, a_2, \\dots, a_n$ ($n \\ge 4$) be positive real numbers with $a_1 < a_2 < \\dots < a_n$. For any positive real number $r$, the number of ternary groups $(i, j, k)$ satisfying $\\frac{a_j - a_i}{a_k - a_j} = r$ ($1 \\le i < j < k \\le n$) is denoted as $f_n(r)$. Prove $f_n(r) < \\frac{n^2}{4}$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Given $j$ ($1 < j < n$), the number of ternary groups $(i, j, k)$ satisfying $1 \\le i < j < k \\le n$ and\n$$\n\\frac{a_j - a_i}{a_k - a_j} = r \\qquad \\textcircled{1}\n$$\nis denoted as $g_j(r)$. For fixed $i, j$ with $i < j$, there is at most one $k$ satisfying **①**; so there are $j-1$ ways to choose $i$, which means $g_j(r) \\le j-1$. In a similar way, for fixed $j, k$ with $k > j$, there is at most one $i$ satisfying **①**; so there are $n-j$ ways to choose $k$, which means $g_j(r) \\le n-j$. Therefore,\n$$\ng_j(r) \\le \\min\\{j-1, n-j\\}.\n$$\nThen, when $n$ is even (i.e., $n = 2m$), we have\n$$\n\\begin{align*}\nf_n(r) &= \\sum_{j=2}^{n-1} g_j(r) = \\sum_{j=2}^{m-1} g_j(r) + \\sum_{j=m}^{2m-1} g_j(r) \\\\ &\\le \\sum_{j=2}^{m} (j-1) + \\sum_{j=m+1}^{2m-1} (2m-j) = \\frac{m(m-1)}{2} + \\frac{m(m-1)}{2} \\\\ &= m^2 - m < m^2 = \\frac{n^2}{4}.\n\\end{align*}\n$$\nWhen $n$ is odd (i.e. $n = 2m + 1$), we have\n$$\n\\begin{aligned}\nf_n(r) &= \\sum_{j=2}^{n-1} g_j(r) = \\sum_{j=2}^{m} g_j(r) + \\sum_{j=m+1}^{2m} g_j(r) \\\\\n&\\le \\sum_{j=2}^{m} (j-1) + \\sum_{j=m+1}^{2m} (2m+1-j) \\\\\n&= m^2 < \\frac{n^2}{4}.\n\\end{aligned}\n$$\nThe proof is completed. $\\Box$",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 28322,
    "subject": "Mathematics (Multi-modal)",
    "question": "Suppose $a$ and $b$ are positive real numbers satisfying $\\frac{1}{a} + \\frac{1}{b} \\le 2\\sqrt{2}$ and $(a-b)^2 = 4(ab)^3$. Then $\\log_a b = \\underline{\\hspace{2cm}}$.",
    "options": [],
    "answer": "-1",
    "solution": "From $\\frac{1}{a} + \\frac{1}{b} \\le 2\\sqrt{2}$, we have $a + b \\le 2\\sqrt{2}ab$. On the other hand,\n$$\n(a + b)^2 = 4ab + (a - b)^2 = 4ab + 4(ab)^3 \\\\\n\\ge 4 \\cdot 2\\sqrt{ab} \\cdot (ab)^3 = 8(ab)^2,\n$$\nand that means\n$$\na + b \\ge 2\\sqrt{2}ab. \\qquad \\textcircled{1}\n$$\nTherefore,\n$$\na + b = 2\\sqrt{2}ab. \\qquad \\textcircled{2}\n$$\nThe equality in **1** holds only when $ab = 1$. Associating it with **2**, we find\n$$\n\\begin{cases} a = \\sqrt{2} - 1, \\\\ b = \\sqrt{2} + 1, \\end{cases} \\quad \\text{and} \\quad \\begin{cases} a = \\sqrt{2} + 1, \\\\ b = \\sqrt{2} - 1. \\end{cases}\n$$\nSo the answer is $\\log_a b = -1$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 28323,
    "subject": "Mathematics (Multi-modal)",
    "question": "Suppose positive real numbers $a, b, c, d$ satisfy $abcd = 1$. Prove\n$$\n\\frac{1}{a} + \\frac{1}{b} + \\frac{1}{c} + \\frac{1}{d} + \\frac{9}{a+b+c+d} \\ge \\frac{25}{4}.\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution 1.\nFirst, we will prove that, whenever there are two numbers among $a, b, c, d$ that are equal, the inequality holds. We may assume that $a = b$ and let $s = a + b + c + d$. Then we have\n$$\n\\begin{aligned}\n& \\frac{1}{a} + \\frac{1}{b} + \\frac{1}{c} + \\frac{1}{d} + \\frac{9}{a+b+c+d} \\\\\n&= \\frac{2}{a} + \\frac{c+d}{cd} + \\frac{9}{s} = \\frac{2}{a} + a^2(s - 2a) + \\frac{9}{s} \\\\\n&= \\frac{2}{a} - 2a^3 + \\left(a^2s + \\frac{9}{s}\\right).\n\\end{aligned}\n$$\nWe define the expression above as $f(s)$. Then we see that $f(s)$ reaches the minimum for $s = \\frac{3}{a}$.\n\nWhen $a \\ge \\frac{\\sqrt{2}}{2}$, however, we have\n$$\ns = a + b + c + d \\ge 2a + \\frac{2}{a} \\ge \\frac{3}{a}.\n$$\nAt this time, $f(s)$ reaches the minimum for $s = 2a + \\frac{2}{a}$.\nWe then have\n$$\n\\begin{aligned}\n& \\frac{2}{a} - 2a^3 + \\left(a^2s + \\frac{9}{s}\\right) = \\frac{2}{a} - 2a^3 + a^2\\left(2a + \\frac{2}{a}\\right) + \\frac{9}{s} \\\\\n&= \\frac{2}{a} + 2a + \\frac{9}{s} = s + \\frac{9}{s} \\\\\n&= \\frac{7}{16}s + \\frac{9}{16}s + \\frac{9}{s} \\ge \\frac{7}{16} \\times 4 + 2\\sqrt{\\frac{9}{16}s \\cdot \\frac{9}{s}} \\\\\n&= \\frac{7}{4} + \\frac{9}{2} = \\frac{25}{4} \\left( \\therefore s = 2a + \\frac{2}{a} \\ge 4 \\right).\n\\end{aligned}\n$$\n\nWhen $0 < a < \\frac{\\sqrt{2}}{2}$, we have\n\n[The solution appears to be truncated here, but the main argument is present.]\nSolution 2.\nWe prove it by using the adjustment method. We may assume that $a \\le b \\le c \\le d$, and define\n$$\nf(a, b, c, d) = \\frac{1}{a} + \\frac{1}{b} + \\frac{1}{c} + \\frac{1}{d} + \\frac{9}{a + b + c + d}.\n$$\nFirstly, we will prove\n$$\nf(a, b, c, d) \\ge f(\\sqrt{ac}, b, \\sqrt{ac}, d). \\quad \\textcircled{2}\n$$\nAs a matter of fact, expression ② is equivalent to\n$$\n\\begin{align*}\n\\frac{1}{a} + \\frac{1}{c} + \\frac{9}{a+b+c+d} &\\ge \\frac{1}{\\sqrt{ac}} + \\frac{1}{\\sqrt{ac}} + \\frac{9}{2\\sqrt{ac}+b+d} \\\\\n\\Leftrightarrow \\frac{(\\sqrt{a}-\\sqrt{c})^2}{ac} &\\ge \\frac{9(\\sqrt{a}-\\sqrt{c})^2}{(a+b+c+d)(2\\sqrt{ac}+b+d)} \\\\\n(\\because (\\sqrt{a}-\\sqrt{c})^2 &\\ge 0) \\\\\n\\Leftrightarrow (a+b+c+d)(2\\sqrt{ac}+b+d) &\\ge 9ac \\\\\n(\\because b+d &\\ge 2\\sqrt{bd} = \\frac{2}{\\sqrt{ac}}) \\\\\n\\Leftrightarrow \\left(a+c+\\frac{2}{\\sqrt{ac}}\\right)\\left(2\\sqrt{ac}+\\frac{2}{\\sqrt{ac}}\\right) &\\ge 9ac.\n\\end{align*}\n$$\nFrom $1 = abcd \\ge a \\cdot a \\cdot c \\cdot c \\Rightarrow ac \\le 1 \\Rightarrow \\frac{2}{\\sqrt{ac}} \\ge 2\\sqrt{ac}$ and $a + c \\ge 2\\sqrt{ac}$, we have the following condition:\nThe left-hand side of ③ $\\ge \\left(2\\sqrt{ac} + \\frac{2}{\\sqrt{ac}}\\right)\\left(2\\sqrt{ac} + \\frac{2}{\\sqrt{ac}}\\right)$\n$\\ge 4\\sqrt{ac} \\cdot 4\\sqrt{ac} = 16ac > 9ac$ is the right-hand side of ③.\nTherefore, ② holds, which means $f(a, b, c, d)$ (with $a \\le b \\le c \\le d$) reaches its minimum if and only if $a = c$ (i.e. $a = b = c$). So we may assume that $(a, b, c, d) = \\left(\\frac{1}{t}, \\frac{1}{t}, \\frac{1}{t}, t^3\\right)$ ($t \\ge 1$). Then we only need to prove that, for all $t \\ge 1$,\n$$\nf\\left(\\frac{1}{t}, \\frac{1}{t}, \\frac{1}{t}, t^3\\right) \\ge \\frac{25}{4}. \\qquad \\textcircled{4}\n$$\nWe have\n$$\n\\begin{align*}\nf\\left(\\frac{1}{t}, \\frac{1}{t}, \\frac{1}{t}, t^3\\right) &\\ge \\frac{25}{4} \\\\\n\\Leftrightarrow 3t + \\frac{1}{t^3} + \\frac{9}{t^3 + \\frac{3}{t}} &\\ge \\frac{25}{4} \\\\\n\\Leftrightarrow 12t^8 - 25t^7 + 76t^4 - 75t^3 + 12 &\\ge 0 \\\\\n\\Leftrightarrow (t-1)^2(12t^6 - t^5 - 14t^4 - 27t^3 + 36t^2 + 24t + 12) &\\ge 0 \\\\\n\\Leftrightarrow 12t^6 - t^5 - 14t^4 - 27t^3 + 36t^2 + 24t + 12 &\\ge 0 \\\\\n\\Leftrightarrow (t-1)(12t^5 + 11t^4 - 3t^3 - 30t^2 + 6t + 30) + 42 &\\ge 0.\n\\end{align*}\n\\qquad \\textcircled{5}\n$$\nSince $t \\ge 1$, $12t^5 + 6t \\ge 2\\sqrt{12t^5 \\cdot 6t} = 12\\sqrt{2} t^3 > 3t^3$, and\n$$\n11t^4 + 30 \\ge 2\\sqrt{11t^4 \\cdot 30} = 2\\sqrt{330} t^2 > 30t^2,\n$$\nthen $(t - 1)(12t^5 + 11t^4 - 3t^3 - 30t^2 + 6t + 30) + 42 > 0$.\nTherefore, ⑤ holds, which justifies ④.\nThe proof is complete.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 28324,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $A_1, A_2, \\dots, A_n$ be $n$ non-empty subsets of a finite set $A$ of real numbers satisfying the following conditions:\n(1) The sum of elements of $A$ is equal to $0$;\n(2) Pick arbitrarily a number from each $A_i$, and their sum is strictly positive.\nProve that there exist sets $A_{i_1}, A_{i_2}, \\dots, A_{i_k}$, $1 \\le i_1 < i_2 < \\dots < i_k \\le n$, such that\n$$ | A_{i_1} \\cup A_{i_2} \\cup \\dots \\cup A_{i_k} | < \\frac{k}{n} | A |. $$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $A = \\{a_1, \\dots, a_m\\}$ with $a_1 > \\dots > a_m$. By (1) we have $a_1 + \\dots + a_m = 0$. Consider the smallest element of each $A_i$, the sum of these numbers is greater than $0$. Assume that there are exactly $k_i$ sets among $A_1, \\dots, A_n$ whose minimal element is $a_i$, $i = 1, 2, \\dots, m$. Then, one has\n$$\nk_1 + \\cdots + k_m = n.\n$$\nBy (2), we have\n$$\nk_1 a_1 + \\cdots + k_m a_m > 0.\n$$\nFor $s = 1, 2, \\dots, m-1$, there are in total $k_1 + \\cdots + k_m$ sets, whose minimal elements are greater than or equal to $a_s$. Therefore, the union of these sets is contained in $\\{a_1, \\dots, a_m\\}$, whence the number of elements does not exceed $s$.\n\nNext, we prove that there exists $s \\in \\{1, 2, \\dots, m-1\\}$ such that $k = k_1 + \\cdots + k_s > \\frac{sn}{m}$. We prove this claim by contradiction. Suppose that\n$$\nk_1 + \\cdots + k_s \\le \\frac{sn}{m}, \\quad s = 1, 2, \\dots, m-1.\n$$\nWith the help of the Abel transform and the fact that $a_s - a_{s+1} > 0$, $1 \\le s \\le m-1$, we know that\n$$\n\\begin{align*} \n0 < & \\sum_{j=1}^{m} k_j a_j \\\\ \n= & \\sum_{s=1}^{m-1} (a_s - a_{s+1}) (k_1 + \\cdots + k_s) + a_m (k_1 + \\cdots + k_m) \\\\ \n\\le & \\sum_{s=1}^{m-1} (a_s - a_{s+1}) \\frac{sn}{m} + a_m n \\\\ \n= & \\frac{n}{m} \\sum_{j=1}^{m} a_j = 0. \n\\end{align*}\n$$\nWe then get a contradiction. For such an $s$, we take the sets among $A_1, \\dots, A_n$, whose minimal elements are greater than $a_s$, say $A_{i_1}, A_{i_2}, \\dots, A_{i_k}$. Then, by the above results, we know that the total number of such sets is $k = k_1 + \\cdots + k_s > \\frac{sn}{m}$, and the number of elements of their union does not exceed $s$, i.e.,\n$$\n|A_{i_1} \\cup A_{i_2} \\cup \\cdots \\cup A_{i_k}| \\le s < \\frac{km}{n} = \\frac{k}{n}|A|.\n$$\n$\\square$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 28325,
    "subject": "Mathematics (Multi-modal)",
    "question": "Suppose that a line passing the circumcentre $O$ of $\\triangle ABC$ intersects $AB$ and $AC$ at points $M$ and $N$, respectively, and $E$ and $F$ are the midpoints of $BN$ and $CM$, respectively. Prove that $\\angle EOF = \\angle A$. (posed by Tao Pingsheng)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "We show that the above conclusion is true for any triangle.\n\nIf $\\triangle ABC$ is right-angled. The conclusion is obvious. In fact, see Fig. 4. 1, where $\\angle ABC = 90^\\circ$. So, the circumcentre $O$ is the midpoint of $AC$, $OA = OB$ and $N = O$. Since $F$ is the midpoint of $CM$, we see that the median line $OF \\parallel AM$. Hence $\\angle EOF = \\angle OBA = \\angle OAB = \\angle A$.\n\nIf $\\triangle ABC$ is not right-angled, see Fig. 4. 2 and Fig. 4. 3.\n![](attached_image_1.png)\nFig. 4. 1\n![](attached_image_2.png)\nFig. 4. 2\n![](attached_image_3.png)\nFig. 4. 3\n\nFirst we give a lemma.\n\n**Lemma.** Let $A$ and $B$ be two points on the diameter $KL$ of circle $\\odot O$ with radius $R$, and $OA = OB = a$. See figure.\nLet $CD$ and $EF$ be two chords passing $A$ and $B$, respectively. Suppose $CE$ and $DF$ intersect $KL$ at $M$ and $N$, respectively. Then $MA = NB$.\n\n**Proof of the lemma.** As shown in Fig. 4. 4. Suppose that $CD \\cap EF = P$. Think of that lines $CE$ and $DF$ intersect $\\triangle PAB$. By Menelaus' Theorem, we have\n![](attached_image_4.png)\nFig. 4. 4\n$$\n\\frac{AC}{CP} \\cdot \\frac{PE}{EB} \\cdot \\frac{BM}{MA} = 1,\n$$\n$$\n\\frac{BF}{FP} \\cdot \\frac{PD}{DA} \\cdot \\frac{AN}{NB} = 1.\n$$\nThen\n$$\n\\frac{MA}{NB} = \\frac{AC}{BE} \\cdot \\frac{AD}{BF} \\cdot \\frac{PE}{PC} \\cdot \\frac{PF}{PD} \\cdot \\frac{BM}{AN}. \\qquad \\textcircled{1}\n$$\nBy the Intersecting Chord Theorem, we get\n$$\nPC \\cdot PD = PE \\cdot PF. \\qquad \\textcircled{2}\n$$\nSo\n$$\nAC \\cdot AD = AK \\cdot AL = R^2 - a^2 = BK \\cdot BL = BE \\cdot BF. \\qquad \\textcircled{3}\n$$\nBy ①, ③, we have $\\frac{MA}{NB} = \\frac{MB}{NA}$, that is\n$$\n\\frac{MA}{NB} = \\frac{MA + AB}{NB + AB} = \\frac{AB}{AN} = 1.\n$$\nThus, $MA = NB$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 28326,
    "subject": "Mathematics (Multi-modal)",
    "question": "Suppose $\\cos^5\\theta - \\sin^5\\theta < 7(\\sin^3\\theta - \\cos^3\\theta)$, $\\theta \\in [0, 2\\pi)$. Then the range of $\\theta$ is ______.",
    "options": [],
    "answer": "(π/4, 5π/4)",
    "solution": "From the inequality\n$$\n\\cos^5\\theta - \\sin^5\\theta < 7(\\sin^3\\theta - \\cos^3\\theta),\n$$\nwe have\n$$\n\\sin^3\\theta + \\frac{1}{7}\\sin^5\\theta > \\cos^3\\theta + \\frac{1}{7}\\cos^5\\theta.\n$$\nSince $f(x) = x^3 + \\frac{1}{7}x^5$ is increasing over $(-\\infty, +\\infty)$, then $\\sin \\theta > \\cos \\theta$, and that means\n$$\n2k\\pi + \\frac{\\pi}{4} < \\theta < 2k\\pi + \\frac{5\\pi}{4} \\quad (k \\in \\mathbb{Z}).\n$$\nBut $\\theta \\in [0, 2\\pi)$, so the range of $\\theta$ is $(\\frac{\\pi}{4}, \\frac{5\\pi}{4})$. $\\square$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 28327,
    "subject": "Mathematics (Multi-modal)",
    "question": "Given a $3 \\times 9$ array $A$ with each cell containing a positive integer, we say a $m \\times n$ $(1 \\le m \\le 3,\\ 1 \\le n \\le 9)$ subarray of $A$ is a “good rectangle” if the sum of the numbers in its cells is a multiple of $10$, and call a $1 \\times 1$ cell of $A$ “bad” if it is not contained in any “good rectangle”. Find the maximum number of “bad cells” in $A$.",
    "options": [],
    "answer": "25",
    "solution": "We first claim that the number of “bad cells” in $A$ is no more than $25$. Otherwise, there will be at most one cell in $A$ that is not “bad”. Without loss of generality, we assume the cells in the first row of $A$ are all “bad”. Then let the numbers from top to bottom in the $i$th column be $a_i, b_i, c_i$ ($i = 1, 2, \\dots, 9$) in turn, and define\n$$\nS_k = \\sum_{i=1}^{k} a_i, \\quad T_k = \\sum_{i=1}^{k} (b_i + c_i), \\quad k = 1, 2, \\dots, 9,\n$$\nwith $S_0 = T_0 = 0$. We are going to prove that three number groups $S_0, S_1, \\dots, S_9$, $T_0, T_1, \\dots, T_9$, and $S_0 + T_0, S_1 + T_1, \\dots, S_9 + T_9$ each form a complete set of residues modulo $10$:\n\nIf there exist $m, n, 0 \\le m < n \\le 9$ such that $S_m \\equiv S_n \\pmod{10}$, then\n$$\n\\sum_{i=m+1}^{n} a_i = S_n - S_m \\equiv 0 \\pmod{10},\n$$\nwhich means that the cells in the first row and from columns $m+1$ to $n$ form a “good rectangle”. But it is a contradiction to the assumption that the cells in the first row are all “bad”.\n\nIf there exist $m, n, 0 \\le m < n \\le 9$ such that $T_m \\equiv T_n \\pmod{10}$, then\n$$\n\\sum_{i=m+1}^{n} (b_i + c_i) = T_n - T_m \\equiv 0 \\pmod{10}.\n$$\nSo the cells ranging from rows $2$ to $3$ and columns $m+1$ to $n$ form a “good rectangle”, which means there are at least two cells that are not “bad”. But it is also a contradiction.\n\nIn a similar way, we can also prove that there are no $m, n, 0 \\le m < n \\le 9$ such that\n$$\nS_m + T_m \\equiv S_n + T_n \\pmod{10}.\n$$\nTherefore, we have\n$$\n\\begin{aligned}\n\\sum_{k=0}^{9} S_k &\\equiv \\sum_{k=0}^{9} T_k \\equiv \\sum_{k=0}^{9} (S_k + T_k) \\equiv 0 + 1 + 2 + \\cdots + 9 \\\\\n&\\equiv 5 \\pmod{10}.\n\\end{aligned}\n$$\nThen\n$$\n\\sum_{k=0}^{9} (S_k + T_k) \\equiv \\sum_{k=0}^{9} S_k + \\sum_{k=0}^{9} T_k \\equiv 5 + 5 \\equiv 0 \\pmod{10}.\n$$\nIt is again a contradiction! Therefore, the number of “bad cells” in $A$ is no more than $25$.\n\nOn the other hand, we can construct a $3 \\times 9$ array in the following and check that each cell in it that does not contain number $10$ is “bad”.\n\n| 1 | 1 | 1 | 2 | 1 | 1 | 1 | 1 | 10 |\n|---|---|---|---|---|---|---|---|----|\n| 1 | 1 | 1 | 1 | 1 | 1 | 1 | 1 | 1  |\n| 1 | 1 | 1 | 10| 1 | 1 | 1 | 1 | 2  |\n\nTherefore, we find out that the maximum number of “bad cells” in $A$ is $25$.\n\n$\\boxed{25}$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 28328,
    "subject": "Mathematics (Multi-modal)",
    "question": "Given positive integer $n$, let $S = \\{1, 2, \\dots, n\\}$. Find the minimum of $|A\\Delta S| + |B\\Delta S| + |C\\Delta S|$ for nonempty finite sets $A$ and $B$ of real numbers, where $C = \\{a + b \\mid a \\in A, b \\in B\\}$, $X\\Delta Y = \\{x \\mid x$ belongs to exactly one of $X$ and $Y\\}$, $|X|$ denotes the number of elements of a finite set $X$.",
    "options": [],
    "answer": "n + 1",
    "solution": "The minimum is $n + 1$.\n\nFirst, by taking $A = B = S$, we have\n$$\n|A \\Delta S| + |B \\Delta S| + |C \\Delta S| = n + 1.\n$$\n\nSecond, we can prove that $l = |A \\Delta S| + |B \\Delta S| + |C \\Delta S| \\ge n + 1$. Let $X \\setminus Y = \\{x \\mid x \\in X, x \\notin Y\\}$. We have\n$$\nl = |A\\setminus S| + |B\\setminus S| + |C\\setminus S| + |S\\setminus A| + |S\\setminus B| + |S\\setminus C|.\n$$\nAll we need to prove are the following:\n(i) $|A\\setminus S| + |B\\setminus S| + |S\\setminus C| \\ge 1$,\n(ii) $|C\\setminus S| + |S\\setminus A| + |S\\setminus B| \\ge n$.\n\nFor (i). In fact, if $|A\\setminus S| = |B\\setminus S| = 0$, then $A, B \\subseteq S$. So $1$ cannot be an element of $C$, hence $|S\\setminus C| \\ge 1$, therefore (i) is valid.\n\nFor (ii). If $A \\cap S = \\emptyset$, then $|S\\setminus A| \\ge n$, the claim is already valid. If $A \\cap S \\ne \\emptyset$, we assume that the maximal element of $A \\cap S$ is $n-k$, $0 \\le k \\le n-1$, then\n$$\n|S\\setminus A| \\ge k. \\qquad \\textcircled{1}\n$$\nOn the other hand, for $i = k + 1, k + 2, \\dots, n$, either $i \\notin B$ (then $i \\in S \\setminus B$) or $i \\in B$ (then $n-k+i \\in C$, i.e., $n-k+i \\in C \\setminus S$), hence\n$$\n|C \\setminus S| + |S \\setminus B| \\geqslant n - k. \\qquad \\textcircled{2}\n$$\nFrom ① and ②, we obtain (ii).\n\nIn conclusion, (i) and (ii) are valid, so $l \\geqslant n + 1$. Hence, the minimum is $n + 1$. $\\square$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 28329,
    "subject": "Mathematics (Multi-modal)",
    "question": "$n$ ($n \\ge 3$) table tennis players have a round-robin tournament — each player will play all the others exactly once, and there is no draw game. Suppose, after the tournament, all the players can be arranged in a circle such that, for any three players $A$, $B$, $C$, if $A$, $B$ are adjacent, then at least one of them defeated $C$. Please find all possible values of $n$. (posed by Fu Yunhao)",
    "options": [],
    "answer": "All odd integers n ≥ 3",
    "solution": "We will prove that $n$ can be any odd number not less than $3$.\n\nSuppose $n = 2k + 1$, an odd number greater than $3$, and $n$ players are represented by $A_1, A_2, \\dots, A_{2k+1}$. Let us arrange the competition result as follows: $A_i$ ($1 \\le i \\le 2k+1$) defeated $A_{i+2}, A_{i+4}, \\dots, A_{i+2k}$ (we stipulate that $A_{2k+1+j} = A_j$, $j = 1, 2, \\dots, 2k+1$) but lost to the other players. Then these players can be arranged in a circle in order $A_1, A_2, \\dots, A_{2k+1}, A_1$.\n\nNow, given any three players $A$, $B$, $C$ with $A$, $B$ being adjacent in the circle, we may assume that $A = A_i$, $B = A_{i+1}$, $C = A_{i+r}$ ($1 \\le i \\le 2k+1$, $2 \\le r \\le 2k$). Then either $r$ or $r-1$ is an even number not less than $2k$, which implies that at least one of the players $A$, $B$ defeated $C$.\n\nOn the other hand, suppose $n$ is an even number not less than $4$, and the $n$ players can be arranged in a circle $A_1, A_2, \\dots, A_n, A_1$ that meets the required condition. We may assume that $A_1$ defeated $A_2$. According to the requirement, at least one of $A_2$, $A_3$ defeated $A_1$, and then $A_3$ defeated $A_1$; but at least one of $A_1$, $A_2$ defeated $A_3$, so $A_2$ defeated $A_3$, and so forth. We then get that, for any $1 \\le i \\le n$, $A_i$ defeated $A_{i+1}$ and lost to $A_{i-1}$ (stipulate that $A_{n+1} = A_1$, $A_0 = A_n$).\n\nWe now divide the players after $A_i$ and before $A_{i-1}$ into $\\frac{n-2}{2}$ pairs — each consists of two adjacent players. Then there is at least one player in each pair who defeated $A_i$, and that means, besides $A_{i-1}$, there are at least $\\frac{n-2}{2}$ players who defeated $A_i$. Then $A_i$ lost at least $\\frac{n}{2}$ games. So $n$ players lost totally at least $\\frac{n^2}{2}$ games.\n\nBut the number of total games is $C_n^2 = \\frac{n(n-1)}{2} < \\frac{n^2}{2}$. This is a contradiction. Therefore, the possible values of $n$ are all the odd numbers not less than $3$.\n\n$\\boxed{}$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 28330,
    "subject": "Mathematics (Multi-modal)",
    "question": "Determine, with proof, whether there is any odd integer $n \\ge 3$ and $n$ distinct prime numbers $p_1, p_2, \\dots, p_n$ such that all $p_i + p_{i+1}$ ($i = 1, 2, \\dots, n$, and $p_{n+1} = p_1$) are perfect squares? (posed by Tao Pingsheng)",
    "options": [],
    "answer": "No, such primes do not exist for any odd length.",
    "solution": "The answer is negative. Suppose that there exist odd integer $n \\ge 3$ and $n$ distinct prime numbers $p_1, p_2, \\dots, p_n$ satisfying the given condition.\n\nIf all $p_1, p_2, \\dots, p_n$ are odd, then it follows from the given condition that all the sums $p_i + p_{i+1}$ are multiples of $4$, so the prime numbers $p_1, p_2, \\dots, p_n$ modulo $4$ appear to be $1$ and $3$ alternatively, and it contradicts the fact that $n$ is odd.\n\nIf one of $p_1, p_2, \\dots, p_n$ is $2$, then without loss of generality, we may assume that $p_1 = 2$. As both $p_1 + p_2$ and $p_n + p_1$ are perfect squares and both are odd, it follows that $p_2$ and $p_n$ are congruent to $3$ modulo $4$. Similar to the discussion in the first case, we know that the primes $p_2, p_3, \\dots, p_n$ modulo $4$ appear to be $1$ and $3$ alternatively, so $n-1$ is odd, which is a contradiction.\n\nHence, there are no odd integer $n \\ge 3$ and $n$ primes satisfying the given conditions. $\\square$",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 28331,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $a_1, a_2, \\dots$ be a permutation of all positive integers. Prove that there exist infinite positive integers $i$'s, such that $(a_i, a_{i+1}) \\le \\frac{3}{4}i$. (posed by Chen Yonggao)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "We prove this problem by contradiction. If the conclusion of the problem is not true, then there exists $i_0$, and we have $(a_i, a_{i+1}) > \\frac{3}{4}i$ for $i \\ge i_0$.\n\nTake a positive number $M > i_0$, so if $i \\ge 4M$, then $(a_i, a_{i+1}) > \\frac{3}{4}i \\ge 3M$.\n\nSo, if $i \\ge 4M$, $a_i \\ge (a_i, a_{i+1}) > 3M$, then $\\{1, 2, \\dots, 3M\\} \\subseteq \\{a_1, a_2, \\dots, a_{4M-1}\\}$.\n\nHence\n$$\n\\left| \\{1, 2, \\dots, 3M\\} \\cap \\{a_{2M}, a_{2M+1}, \\dots, a_{4M-1}\\} \\right| \\ge 3M - (2M - 1) = M + 1.\n$$\nBy Dirichlet's Drawer Principle, there exists $2M \\le j_0 < 4M$ such that $a_{j_0}, a_{j_0+1} \\le 3M$. Thus,\n$$\n(a_{j_0}, a_{j_0+1}) \\le \\frac{1}{2} \\max\\{a_{j_0}, a_{j_0+1}\\} \\le \\frac{3M}{2} = \\frac{3}{4} \\cdot 2M \\le \\frac{3}{4} j_0,\n$$\nwhich is a contradiction. □",
    "topic": "Discrete Mathematics",
    "subtopic": "Combinatorics"
  },
  {
    "id": 28332,
    "subject": "Mathematics (Multi-modal)",
    "question": "Given real number $\\alpha$, please find the minimum real number $\\lambda = \\lambda(\\alpha)$, such that for any complex numbers $z_1, z_2$ and real number $x \\in [0, 1]$, if $|z_1| \\le \\alpha |z_1 - z_2|$ then $|z_1 - xz_2| \\le \\lambda |z_1 - z_2|$. (posed by Li Shenghong)\n\n![](attached_image_1.png)",
    "options": [],
    "answer": "max{α, 1}",
    "solution": "As shown in the figure, in the complex plane, points $A$, $B$ and $C$ denote the complex numbers $z_1$, $z_2$ and $xz_2$, respectively. $C$ is obviously on the segment $OB$. The vectors $\\vec{BA}$ and $\\vec{CA}$ are represented by complex numbers $z_1 - z_2$ and $z_1 - xz_2$, respectively. As $|z_1| \\le \\alpha |z_1 - z_2|$, we have $|\\vec{OA}| \\le \\alpha |\\vec{BA}|$. Then\n\n$$\n\\begin{align*}\n|z_1 - xz_2|_{\\max} &= |\\vec{AC}|_{\\max} = \\max \\{|\\vec{OA}|, |\\vec{BA}|\\} \\\\\n&= \\max \\{|z_1|, |z_1 - z_2|\\} \\\\\n&= \\max \\{\\alpha |z_1 - z_2|, |z_1 - z_2|\\}.\n\\end{align*}\n$$\n\nTherefore, $\\lambda(\\alpha) = \\max\\{\\alpha, 1\\}$.\n\n$\\boxed{}$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 28333,
    "subject": "Mathematics (Multi-modal)",
    "question": "Determine, with proof, whether there is any odd integer $n \\ge 3$ and $n$ distinct prime numbers $p_1, p_2, \\dots, p_n$, such that all $p_i + p_{i-1}$ ($i=1, 2, \\dots, n$, and $p_{n-1} = p_1$) are perfect squares?",
    "options": [],
    "answer": "No; such an odd number and primes do not exist.",
    "solution": "Suppose that there exists odd integer $n \\ge 3$ and $n$ distinct prime numbers $p_1, p_2, \\dots, p_n$ satisfying the given condition.\nIf all $p_1, p_2, \\dots, p_n$ are odd, then all the sums $p_i + p_{i+1}$ are multiples of $4$, so the prime numbers $p_1, p_2, \\dots, p_n$ modulo $4$ appear to be $1$ and $3$ alternatively, and it contradicts to the fact that $n$ is odd.\nIf one of $p_1, p_2, \\dots, p_n$ is $2$, then without loss of generality, we may assume that $p_1 = 2$. As both $p_1 + p_2$ and $p_n + p_1$ are perfect squares and both are odd, it follows that $p_2$ and $p_n$ are congruent to $3$ modulo $4$. Similar to the discussion in the first case, we know that the primes $p_2, p_3, \\dots, p_n$ modulo $4$ appear to be $1$ and $3$ alternatively, so $n-1$ is odd, which is a contradiction.\nHence, there are no odd integer $n \\ge 3$ and $n$ primes satisfying the given conditions.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 28334,
    "subject": "Mathematics (Multi-modal)",
    "question": "Given integer $n \\geqslant 4$. Find the maximum of\n$$ \\frac{\\sum_{i=1}^{n} a_i (a_i + b_i)}{\\sum_{i=1}^{n} b_i (a_i + b_i)} $$ for non-negative real numbers $a_1, a_2, \\dots, a_n, b_1, b_2, \\dots, b_n$ satisfying\n$$ a_1 + a_2 + \\dots + a_n = b_1 + b_2 + \\dots + b_n > 0. $$",
    "options": [],
    "answer": "n - 1",
    "solution": "The maximum is $n - 1$. By homogeneity, we can assume without loss of generality that $\\sum_{i=1}^{n} a_i = \\sum_{i=1}^{n} b_i = 1$.\n\nFirst, it is clear that if $a_1 = 1$, $a_2 = a_3 = \\dots = a_n = 0$ and $b_1 = 0$, $b_2 = b_3 = \\dots = b_n = \\frac{1}{n-1}$, then $\\sum_{i=1}^{n} a_i (a_i + b_i) = 1$, $\\sum_{i=1}^{n} b_i (a_i + b_i) = \\frac{1}{n-1}$, hence\n$$\n\\frac{\\sum_{i=1}^{n} a_i (a_i + b_i)}{\\sum_{i=1}^{n} b_i (a_i + b_i)} = n - 1.\n$$\nNow we prove that for any real numbers $a_1, a_2, \\dots, a_n, b_1, b_2, \\dots, b_n$ satisfying $\\sum_{i=1}^{n} a_i = \\sum_{i=1}^{n} b_i = 1$, we have\n$$\n\\frac{\\sum_{i=1}^{n} a_i (a_i + b_i)}{\\sum_{i=1}^{n} b_i (a_i + b_i)} \\leqslant n - 1.\n$$\nNote that the denominator is positive, it is equivalent to show that\n$$ \\sum_{i=1}^{n} a_i (a_i + b_i) \\le (n-1) \\sum_{i=1}^{n} b_i (a_i + b_i), $$\ni.e.,\n$$\n(n-1) \\sum_{i=1}^{n} b_i^2 + (n-2) \\sum_{i=1}^{n} a_i b_i \\ge \\sum_{i=1}^{n} a_i^2.\n$$\nBy symmetry, we can assume that $b_1$ is the smallest one among $b_1, b_2, \\dots, b_n$. Then\n$$\n\\begin{align*}\n& (n-1) \\sum_{i=1}^{n} b_i^2 + (n-2) \\sum_{i=1}^{n} a_i b_i \\\\\n&\\ge (n-1) b_1^2 + (n-1) \\sum_{i=2}^{n} b_i^2 + (n-2) \\sum_{i=1}^{n} a_i b_1 \\\\\n&\\ge (n-1) b_1^2 + \\left( \\sum_{i=2}^{n} b_i \\right)^2 + (n-2) b_1 \\\\\n&\\quad = (n-1) b_1^2 + (1-b_1)^2 + (n-2) b_1 \\\\\n&\\quad = n b_1^2 + (n-4)b_1 + 1 \\\\\n&\\ge 1 = \\sum_{i=1}^{n} a_i \\ge \\sum_{i=1}^{n} a_i^2.\n\\end{align*}\n$$\n$\\square$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 28335,
    "subject": "Mathematics (Multi-modal)",
    "question": "Seven students are arranged to attend five sporting events. It is required that students $A$ and $B$ cannot attend the same event, every event is attended by at least one student, and each student must attend one and only one event. Then the number of the arrangement plans meeting the required condition is _______. (the answer should be given in numerical value)",
    "options": [],
    "answer": "15000",
    "solution": "There are two possible cases that satisfy the required conditions:\n\n(1) there is an event attended by three students — this case has $C_7^3 \\cdot 5! - C_5^1 \\cdot 5! = 3600$ plans;\n\n(2) there are two events each attended by two students — this case has $\\frac{1}{2}(C_7^2 \\cdot C_5^2) \\cdot 5! - C_5^2 \\cdot 5! = 11400$ plans.\n\nTherefore, there are $3600 + 11400 = 15000$ plans that meet the required condition. ☐",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 28336,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $AA_0$, $BB_0$ and $CC_0$ be angular bisectors of $\\triangle ABC$. Let $A_0A_1 \\parallel BB_0$ and $A_0A_2 \\parallel CC_0$, where $A_1$ and $A_2$ lie on $AC$ and $AB$, respectively, and let line $A_1A_2$ intersect $BC$ at $A_3$. The points $B_3$ and $C_3$ are obtained similarly. Prove that points $A_3$, $B_3$, $C_3$ are collinear. (posed by Tao Pingsheng)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "By the Menelaus Inverse Theorem, we need only to show that\n$$\n\\frac{AB_3}{B_3C} \\cdot \\frac{CA_3}{A_3B} \\cdot \\frac{BC_3}{C_3A} = 1. \\qquad \\textcircled{1}\n$$\nSince line $A_1A_2A_3$ intersects $\\triangle ABC$, by Menelaus' Theorem, we have $\\frac{CA_3}{A_3B} \\cdot \\frac{BA_2}{A_2A} \\cdot \\frac{AA_1}{A_1C} = 1$. So\n$$\n\\frac{CA_3}{A_3B} = \\frac{A_2A}{BA_2} \\cdot \\frac{A_1C}{AA_1}. \\qquad \\textcircled{2}\n$$\nSimilarly, we have\n$$\n\\frac{AB_3}{B_3C} = \\frac{B_2B}{CB_2} \\cdot \\frac{B_1A}{BB_1}, \\qquad \\textcircled{3}\n$$\n$$\n\\frac{BC_3}{C_3A} = \\frac{C_2C}{AC_2} \\cdot \\frac{C_1B}{CC_1}. \\qquad \\textcircled{4}\n$$\nBy $BA_2 = \\frac{BC_0}{BC} \\cdot BA_0$ and $AA_2 = \\frac{AA_0}{AI} \\cdot AC_0$, we have\n$$\n\\frac{AA_2}{BA_2} = \\frac{AA_0}{BA_0} \\cdot \\frac{AC_0}{BC_0} \\cdot \\frac{BC}{AI}. \\qquad \\textcircled{5}\n$$\nMoreover, by $AA_1 = \\frac{AA_0}{AI} \\cdot AB_0$ and $CA_1 = \\frac{CA_0}{CB} \\cdot CB_0$, we have\n$$\n\\frac{A_1C}{AA_1} = \\frac{CA_0 \\cdot CB_0}{AA_0 \\cdot AB_0} \\cdot \\frac{AI}{BC}. \\qquad \\textcircled{6}\n$$\nThen by ②, ⑤ and ⑥, we have\n$$\n\\frac{CA_3}{A_3B} = \\frac{CA_0}{BA_0} \\cdot \\frac{AC_0}{BC_0} \\cdot \\frac{CB_0}{AB_0} = \\left( \\frac{CA_0}{A_0B} \\right)^2.\n$$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 28337,
    "subject": "Mathematics (Multi-modal)",
    "question": "We call a point sequence $(A_0, A_1, \\dots, A_n)$ interesting, if the abscissa and ordinate are equal for each $A_i$, and the slopes of segment $OA_0, OA_1, \\dots, OA_n$ strictly increase ($O$ is the origin), and the area of each $\\triangle OA_iA_{i+1}$ ($0 \\le i \\le n-1$) is $\\frac{1}{2}$.\n\nFor a point sequence $(A_0, A_1, \\dots, A_n)$, insert a point $A$ adjacent to two points $A_i, A_{i+1}$ satisfying $\\overrightarrow{OA} = \\overrightarrow{OA_i} + \\overrightarrow{OA_{i+1}}$, then we call thus obtained new point sequence $(A_0, \\dots, A_i, A, A_{i+1}, \\dots, A_n)$ an expansion of $(A_0, A_1, \\dots, A_n)$.\n\nLet $(A_0, A_1, \\dots, A_n)$ and $(B_0, B_1, \\dots, B_m)$ be any two interesting point sequences. Prove that if $A_0 = B_0$ and $A_n = B_m$, then we can expand both point sequences to some same point sequence $(C_0, C_1, \\dots, C_k)$.\n\n(posed by Qu Zhenhua)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "We see that by the condition of the problem, an expansion of an interesting sequence is still interesting.\n\nFirst, we construct the interesting sequence $(C_0, C_1, \\dots, C_k)$ containing all points of sequences $(A_0, A_1, \\dots, A_n)$ and $(B_0, B_1, \\dots, B_m)$, and $C_0 = A_0 = B_0$, $C_k = A_n = B_m$.\n\nBy the Pick Theorem, we know that the area of triangle equals $1/2$ if and only if there is no grid point on the triangle except triangle vertices. Hence there is no grid point on $\\triangle OA_iA_{i+1}$ except its vertices. Therefore, if the slopes of $OA_i$ and $OB_j$ are equal, then $A_i = B_j$.\n\nDenote the slopes of segments from points of $\\{A_i\\}$ and $\\{B_j\\}$ to the origin in strictly increasing order by $D_0, D_1, \\dots, D_l$, where $D_0 = C_0 = A_0 = B_0$ and $D_l = C_k = A_n = B_m$. If a sequence ($D_i, D_{i+1}$) is not interesting, then we can insert several points $E_1, \\dots, E_s$ such that the sequence ($D_i, E_1, \\dots, E_s, D_{i+1}$) is interesting. In fact, consider the convex hull $P$ of the grid points on $\\triangle OD_iD_{i+1}$, except the origin. $P$ is a convex polygon or segment $D_iD_{i+1}$ (a degenerated polygon). Then the sequence of vertices of $P$ is interesting. Thus, we have constructed the interesting sequence $(C_0, C_1, \\dots, C_k)$.\n\n![](attached_image_1.png)\n\nFinally, it suffices to show that the interesting sequence $(A_0, A_1, \\dots, A_n)$ can be expanded to $(C_0, C_1, \\dots, C_k)$, and the same is true for $(B_0, B_1, \\dots, B_m)$. We only need to prove this for the case of $n=1$, since we can apply the conclusion for $n=1$ to $(A_i, A_{i+1})$, $i=0, 1, \\dots, n-1$ successively. Let $C_0 = A_0$ and $C_k = A_1$. By induction on $k$, for $k=1$, we need no expansion. Suppose that the conclusion is true for all positive integers less than $k$.\n\nThen denote the grid point $A$ satisfying $\\overrightarrow{OA} = \\overrightarrow{OA_0} + \\overrightarrow{OA_1}$, and we see that $A$ must be a point of $C_1, \\dots, C_{k-1}$. Since if not, there is no grid point on interior of segment $OA$, and there exists $i, 0 \\le i < k$, such that $A$ locates in the angle made by rays $OC_i$ and $OC_{i+1}$. We may suppose that $i > 0$, otherwise take the graph symmetric over the line $x = y$. Since the area of the parallelogram $\\square OA_0AA_1$ is $1$, $C_i$ locates outside of $\\square OA_0AA_1$, $C_{i+1}$ locates outside of $\\square OA_0AA_1$ or $C_{i+1} = A_1$. In any way, we take $B$ such that $\\vec{OB} = \\vec{OC}_i + \\vec{OC}_{i+1}$ and take $B'$ such that $C_{i+1}B' \\parallel OA_0$, $C_iB' \\parallel A_0A$, then $A$ locates on $\\square OC_{i+1}B'C_i$. Thus, $A$ locates inside of $\\square OC_iBC_{i+1}$, which contradicts the fact that the area of $\\square OC_iBC_{i+1}$ is $1$. Therefore the inserted point $A$ to $(A_0, A_1)$ for expansion is some $C_i$. Then we use the induction hypotheses to $(A_0, A)$ and $(A, A_1)$, respectively. $\\square$\n\n![](attached_image_2.png)",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 28338,
    "subject": "Mathematics (Multi-modal)",
    "question": "Prove that for any given positive integers $m$, $n$, there exist infinitely many pairs of coprime positive integers $a$, $b$, such that $a + b \\mid am^a + bn^b$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "If $mn = 1$, then the claim is valid. For $mn \\ge 2$, since\n$$\nn^a (am^a + bn^b) = (a+b)n^{a+b} + a((mn)^a - n^{a+b}),\n$$\nit is sufficient to prove the existence of infinitely many coprime number pairs $a$, $b$, such that\n$$\na + b \\mid (mn)^a - n^{a+b}, \\quad (a+b, n) = 1.\n$$\nLet $p = a + b$, we only need to prove that there are infinitely many prime numbers $p$ and positive integer $1 \\le a \\le p-1$ such that\n$$\np \\mid (mn)^a - n^p.\n$$\nBy Fermat's theorem, i.e., when $a_1 \\equiv a_2 \\pmod{p-1}$, $a_1 \\ge 1$, $a_2 \\ge 1$, $(mn)^{a_1} \\equiv (mn)^{a_2} \\pmod{p}$.\nSo we only need to prove that there are infinitely many prime numbers $p$ and positive integer $a$ such that\n$$\np \\mid (mn)^a - n. \\qquad \\textcircled{1}\n$$\nIf there are only finitely many such primes, say $p_1, p_2, \\dots, p_r$ (as $mn \\ge 2$, the existence of such primes is obvious). Suppose that\n$$\n(mn)^2 - n = p_1^{a_1} p_2^{a_2} \\dots p_r^{a_r}, \\quad \\alpha_i \\text{'s are non-negative integers} \\ (1 \\le i \\le r). \\qquad \\textcircled{2}\n$$\nLet $a = p_1^{a_1} p_2^{a_2} \\dots p_r^{a_r} (p_1 - 1) \\dots (p_r - 1) + 2$, and suppose that\n$$\n(mn)^a - n = p_1^{\\beta_1} p_2^{\\beta_2} \\dots p_r^{\\beta_r}, \\quad \\beta_i \\text{'s are non-negative integers} \\ (1 \\le i \\le r). \\qquad \\textcircled{3}\n$$\nIf $p_i \\nmid n$, then, by ③ and $a \\ge 2$, we know that $p_i^{\\beta_i} \\nmid n$, hence $p_i^{\\beta_i} \\mid (mn)^2 - n$, and by ② we have $\\beta_i \\le \\alpha_i$.\nIf $p_i \\nmid n$, then $p_i \\nmid m$, so $(p_i^{a_i+1}, mn) = 1$. By Euler's Theorem (as $\\varphi(p_i^{a_i+1}) = p_i^{a_i} (p_i - 1)$ is a factor of $a-2$)\n$$\n(mn)^a - n \\equiv (mn)^2 - n \\pmod{p_i^{a_i+1}}.\n$$\nBecause $p_i^{a_i+1} \\nmid (mn)^2 - n$, the congruence relation above implies that $p_i^{a_i+1} \\nmid (mn)^a - n$. So $\\beta_i \\le \\alpha_i$. Hence,\n$$\n(mn)^a - n = p_1^{\\beta_1} p_2^{\\beta_2} \\cdots p_r^{\\beta_r} \\le p_1^{\\alpha_1} p_2^{\\alpha_2} \\cdots p_r^{\\alpha_r} = (mn)^a - n,\n$$\nwhich is in contradiction with $a > 2$. So there are infinitely\nmany primes $p$ and positive integers $a$ such that $p \\mid (mn)^a - n$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 28339,
    "subject": "Mathematics (Multi-modal)",
    "question": "Given a tetrahedron $ABCD$, it is known that $\\angle ADB = \\angle BDC = \\angle CDA = 60^\\circ$, $AD = BD = 3$ and $CD = 2$. Then the radius of the sphere circumscribing $ABCD$ is ______.",
    "options": [],
    "answer": "sqrt(3)",
    "solution": "Let the center of the sphere circumscribing $ABCD$ be $O$. Then $O$ is on the vertical line of plane $ABD$ through point $N$ the circumcenter of $\\triangle ABD$. It is known that $\\triangle ABD$ is regular, so $N$ is the center of it. Let $P$ and $M$ be the midpoints of $AB$ and $CD$, respectively. Then $N$ is on $DP$ with $ON \\perp DP$ and $OM \\perp CD$.\n\n![](attached_image_1.png)\n\nLet $\\theta$ denote the angle between $CD$ and plane $ABD$. From\n$$\n\\angle CDA = \\angle CDB = \\angle ADB = 60^\\circ,\n$$\nwe find $\\cos \\theta = \\frac{1}{\\sqrt{3}}$, $\\sin \\theta = \\frac{\\sqrt{2}}{\\sqrt{3}}$.\n\nSince $DM = \\frac{1}{2}CD = 1$, $DN = \\frac{2}{3} \\cdot DP = \\frac{\\sqrt{3}}{2} \\cdot 3 = \\sqrt{3}$, by cosine theorem we have, in $\\triangle DMN$,\n$$\n\\begin{aligned}\nMN^2 &= DM^2 + DN^2 - 2 \\cdot DM \\cdot DN \\cdot \\cos \\theta \\\\\n&= 1^2 + (\\sqrt{3})^2 - 2 \\cdot 1 \\cdot \\sqrt{3} \\cdot \\frac{1}{\\sqrt{3}} = 2,\n\\end{aligned}\n$$\nthat is $MN = \\sqrt{2}$. The radius of the sphere circumscribing $ABCD$ is then\n$$\nOD = \\frac{MN}{\\sin \\theta} = \\frac{\\sqrt{2}}{\\sqrt{2}} = \\sqrt{3}.\n$$\nThe answer is $R = \\sqrt{3}$.",
    "topic": "Geometry",
    "subtopic": "Solid Geometry"
  },
  {
    "id": 28340,
    "subject": "Mathematics (Multi-modal)",
    "question": "Are there any positive integers $m, n$ such that $m^{20} + 11^n$ is a square number? Prove your conclusion. (posed by Yuan Hanhui)",
    "options": [],
    "answer": "No; there are no positive integers m and n such that m^20 + 11^n is a perfect square.",
    "solution": "Assuming there are positive integers $m, n$ such that $m^{20} + 11^n = k^2$ with $k \\in \\mathbb{Z}$, we then have\n$$\n11^n = k^2 - m^{20} = (k - m^{10})(k + m^{10}),\n$$\nwhich means that there are integers $\\alpha, \\beta \\ge 0$ such that\n$$\n\\begin{cases} k - m^{10} = 11^{\\alpha}, \\\\ k + m^{10} = 11^{\\beta}. \\end{cases} \n\\qquad \\begin{array}{l} \\textcircled{1} \\\\ \\textcircled{2} \\end{array}\n$$\nIt is obvious that $\\alpha < \\beta$. Subtracting ① from ②, we have\n$$\n2m^{10} = 11^{\\alpha} (11^{\\beta-\\alpha} - 1).\n$$\nLet $m = 11^{\\gamma}m_1$, where $\\gamma, m_1 \\in \\mathbb{N}^*$, $11 \\nmid m_1$. Then\n$$\n11^{10\\gamma} \\cdot 2m_1^{10} = 11^{\\alpha} (11^{\\beta-\\alpha} - 1),\n$$\nwhich implies $10\\gamma = \\alpha$ and $2m_1^{10} = 11^{\\beta-\\alpha} - 1$.\nBy Fermat's Little Theorem, we have $m_1^{10} \\equiv 1 \\pmod{11}$, then $2m_1^{10} \\equiv 2 \\pmod{11}$. But $11^{\\beta-\\alpha} - 1 \\equiv 10 \\pmod{11}$, which is a contradiction. So the proved conclusion is that there are no positive integers $m, n$ such that $m^{20} + 11^n$ is a square number. $\\square$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 28341,
    "subject": "Mathematics (Multi-modal)",
    "question": "Given $n$ points $P_1, P_2, \\dots, P_n$ on a plane, let $M$ be any point on segment $AB$ on the plane. Denote by $|P_iM|$ the distance between $P_i$ and $M$, $i = 1, 2, 3, \\dots, n$. Prove that\n$$\n\\sum_{i=1}^n |P_iM| \\le \\max\\left\\{\\sum_{i=1}^n |P_iA|, \\sum_{i=1}^n |P_iB|\\right\\}.\n$$\n(posed by Jin Mengwei)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $O$ be the origin. Then we have $\\overrightarrow{OM} = t \\overrightarrow{OA} + (1-t) \\overrightarrow{OB}$, $t \\in (0, 1)$.\n$$\n\\begin{aligned}\n|P_iM| &= |\\overrightarrow{OM} - \\overrightarrow{OP_i}| \\\\\n&= |t \\overrightarrow{OA} + (1-t) \\overrightarrow{OB} - t \\overrightarrow{OP_i} - (1-t) \\overrightarrow{OP_i}| \\\\\n&\\le t |\\overrightarrow{OA} - \\overrightarrow{OP_i}| + (1-t) |\\overrightarrow{OB} - \\overrightarrow{OP_i}| \\\\\n&= t |P_iA| + (1-t) |P_iB|.\n\\end{aligned}\n$$\nHence,\n$$\n\\begin{aligned}\n\\sum_{i=1}^{n} |P_iM| &\\le t \\sum_{i=1}^{n} |P_iA| + (1-t) \\sum_{i=1}^{n} |P_iB| \\\\\n&\\le \\max\\left\\{\\sum_{i=1}^{n} |P_iA|, \\sum_{i=1}^{n} |P_iB|\\right\\}.\n\\quad \\square\n\\end{aligned}\n$$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 28342,
    "subject": "Mathematics (Multi-modal)",
    "question": "Line $x - 2y - 1 = 0$ and parabola $y^2 = 4x$ intersect at points $A$, $B$, point $C$ is on the parabola, and $\\angle ACB = 90^\\circ$. Then the coordinate of $C$ is ______.",
    "options": [],
    "answer": "(1, -2) or (9, -6)",
    "solution": "Let $A(x_1, y_1)$, $B(x_2, y_2)$, $C(t^2, 2t)$. From\n$$\n\\begin{cases} x - 2y - 1 = 0, \\\\ y^2 = 4x, \\end{cases}\n$$\nwe get $y^2 - 8y - 4 = 0$, which means $y_1 + y_2 = 8$, $y_1 \\cdot y_2 = -4$.\nSince $x_1 = 2y_1 + 1$, $x_2 = 2y_2 + 1$, we have\n$$\n\\begin{aligned}\nx_1 + x_2 &= 2(y_1 + y_2) + 2 = 18, \\\\\nx_1 \\cdot x_2 &= 4y_1 \\cdot y_2 + 2(y_1 + y_2) + 1 = 1.\n\\end{aligned}\n$$\nFurthermore, by $\\angle ACB = 90^\\circ$, we have $\\vec{CA} \\cdot \\vec{CB} = 0$, which means\n$$\n(t^2 - x_1)(t^2 - x_2) + (2t - y_1)(2t - y_2) = 0,\n$$\nthat is\n$$\nt^4 - (x_1 + x_2)t^2 + x_1 \\cdot x_2 + 4t^2 - 2(y_1 + y_2)t + y_1 \\cdot y_2 = 0.\n$$\nThen\n$$\nt^4 - 14t^2 - 16t - 3 = 0,\n$$\nor\n$$\n(t^2 + 4t + 3)(t^2 - 4t - 1) = 0.\n$$\nObviously, $t^2 - 4t - 1 \\neq 0$; otherwise, we have $t^2 - 2 \\cdot 2t - 1 = 0$, which means $C$ is on $x - 2y - 1 = 0$, i.e., $C$ coincides with either $A$ or $B$. So $t^2 + 4t + 3 = 0$. Then $t_1 = -1$, $t_2 = -3$.\nTherefore, the coordinate of $C$ is $(1, -2)$ or $(9, -6)$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 28343,
    "subject": "Mathematics (Multi-modal)",
    "question": "Suppose that the sequence $\\{a_n\\}$ defined by $a_1 = a_2 = 1$, $a_n = 7a_{n-1} - a_{n-2}$, $n \\ge 3$. Prove that $a_n + a_{n+1} + 2$ is a perfect square for any positive integer $n$. (posed by Tao Pingsheng)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "It is well known that the solution of the sequence can be obtained by solving two geometric sequences. We get $a_n = C_1\\lambda_1^n + C_2\\lambda_2^n$, where\n$$\n\\lambda_1 = \\frac{7 + \\sqrt{45}}{2} = \\left( \\frac{3 + \\sqrt{5}}{2} \\right)^2, \\quad \\lambda_2 = \\frac{7 - \\sqrt{45}}{2} = \\left( \\frac{3 - \\sqrt{5}}{2} \\right)^2,\n$$\n$\\lambda_1$ and $\\lambda_2$ are solutions of the equation $\\lambda^2 - 7\\lambda + 1 = 0$, $a_1 = 1 = C_1\\lambda_1 + C_2\\lambda_2$, $a_2 = 1 = C_1\\lambda_1^2 + C_2\\lambda_2^2$.\nTherefore,\n$$\n\\begin{aligned}\na_n + a_{n+1} + 2 &= \\lambda_1^{n-1}C_1(\\lambda_1 + \\lambda_1^2) + \\lambda_2^{n-1}C_2(\\lambda_2 + \\lambda_2^2) + 2 \\\\\n&= \\lambda_1^{n-1} + \\lambda_2^{n-1} + 2 \\\\\n&= \\left( \\left( \\frac{3+\\sqrt{5}}{2} \\right)^{n-1} \\right)^2 + \\left( \\left( \\frac{3-\\sqrt{5}}{2} \\right)^{n-1} \\right)^2 + 2 \\\\\n&= \\left[ \\left( \\frac{3+\\sqrt{5}}{2} \\right)^{n-1} + \\left( \\frac{3-\\sqrt{5}}{2} \\right)^{n-1} \\right]^2 = x_n^2,\n\\end{aligned}\n$$\nwhere $x_n = \\left(\\frac{3+\\sqrt{5}}{2}\\right)^{n-1} + \\left(\\frac{3-\\sqrt{5}}{2}\\right)^{n-1}$ is the solution of the sequence $\\{x_n\\}$ of positive integers $x_1 = 2, x_2 = 3, x_n = 3x_{n-1} - x_{n-2}$, $n \\ge 3$. $\\square$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 28344,
    "subject": "Mathematics (Multi-modal)",
    "question": "Suppose $n$ small balls have been placed into $n$ numbered boxes $B_1, B_2, \\dots, B_n$. Each time we can select a box $B_k$ and do the following operations:\n(1) If $k=1$ and there is at least one ball in $B_1$, move one ball from $B_1$ into $B_2$.\n(2) If $k=n$ and there is at least one ball in $B_n$, move one ball from $B_n$ into $B_{n-1}$.\n(3) If $2 \\le k \\le n-1$ and there are at least two balls in $B_k$, move one ball from $B_k$ into $B_{k+1}$ and one ball into $B_{k-1}$, respectively.\n\nProve the following: no matter how the balls are distributed among the boxes originally, it is always realizable to let each box contain exactly one ball by finite operations.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "For any two vectors $\\mathbf{x} = (x_1, x_2, \\dots, x_n)$ and $\\mathbf{y} = (y_1, y_2, \\dots, y_n)$, if there exists $1 \\leq k \\leq n$ such that\n$$\nx_1 = y_1, \\dots, x_{k-1} = y_{k-1}, x_k > y_k,\n$$\nwe then denote it as $\\mathbf{x} > \\mathbf{y}$. Let $\\mathbf{x} = (x_1, x_2, \\dots, x_n)$ represent the distribution of the balls among the boxes. Then $\\mathbf{x}$ is a non-negative integer vector. The operation defined in the question, if executable, can be expressed as $\\mathbf{x} + \\boldsymbol{\\alpha}_k$, where $\\boldsymbol{\\alpha}_1 = (-1, 1, 0, \\dots, 0)$, $\\boldsymbol{\\alpha}_k = (0, \\dots, 0, 1, -2, 1, 0, \\dots, 0)$ ($2 \\leq k \\leq n - 1$), $\\boldsymbol{\\alpha}_n = (0, \\dots, 0, 1, -1)$. Then for $k \\geq 2$, we always have $\\mathbf{x} + \\boldsymbol{\\alpha}_k > \\mathbf{x}$. So for any initial distribution of the balls, after a finite number of operations on every $B_k$ ($k \\geq 2$) that contains at least two balls, we can arrive at a ball distribution $\\mathbf{y} = (y_1, y_2, \\dots, y_n)$ satisfying $y_k \\leq 1$ for all $k \\geq 2$. If at this time $y_2 = \\dots = y_n = 1$, the problem is solved; otherwise, we have $y_1 \\geq 2$.\n\nAssuming $i$ is the smallest number such that $y_i = 0$, we can then do a series of operations on $B_1, B_2, \\dots, B_{i-1}$:\n$$\n\\begin{align*}\n& (y_1, 1, \\dots, 1, 0, y_{i+1}, \\dots, y_n) \\xrightarrow{B_1, B_2, \\dots, B_{i-1}} \\\\\n& (y_1, 1, \\dots, 1, 0, 1, y_{i+1}, \\dots, y_n) \\xrightarrow{B_1, B_2, \\dots, B_{i-2}} \\\\\n& (y_1, 1, \\dots, 1, 0, 1, 1, y_{i+1}, \\dots, y_n) \\to \\dots \\to \\\\\n& (y_1, 0, 1, \\dots, 1, y_{i+1}, \\dots, y_n) \\xrightarrow{B_1} \\\\\n& (y_1 - 1, 1, \\dots, 1, 1, y_{i+1}, \\dots, y_n)\n\\end{align*}\n$$\nto get $(y_1 - 1, 1, \\dots, 1, y_{i+1}, \\dots, y_n)$. Repeating the operations above, we can finally arrive at the ball distribution vector that meets the requirement. $\\Box$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 28345,
    "subject": "Mathematics (Multi-modal)",
    "question": "Determine, with proof, all pairs $(a,b)$ of integers, such that for any positive integer $n$, one has $n \\mid (a^n + b^{n-1})$.",
    "options": [],
    "answer": "[(0, 0), (-1, -1)]",
    "solution": "The solution pairs consist of $(0,0)$ and $(-1,-1)$.\n\nIf one of $a$ and $b$ is $0$, it is obvious that the other is also $0$.\n\nNow we assume that $ab \\neq 0$, select a large prime $p$ such that $p > |a + b^2|$, it follows from Fermat's little theorem that\n$$\na^p + b^{p+1} \\equiv a + b^2 \\pmod p,\n$$\nAs $p \\mid (a^p + b^{p+1})$ and $p > |a + b^2|$, we have $a + b^2 = 0$.\n\nThen we select another prime $q$ such that $q > |b+1|$ and $(q, b) = 1$.\nLet $n = 2q$, then we have\n$$\na^n + b^{n+1} = (-b^2)^{2q} + b^{2q+1} = b^{4q} + b^{2q+1} = b^{2q+1}(b^{2q-1} + 1).\n$$\nIt follows from\n$$\nn \\mid (a^n + b^{n+1}) \\quad \\text{and} \\quad (q, b) = 1, \\text{ that } \\quad q \\mid (b^{2q-1} + 1).\n$$\nAs $b^{2q-1} + 1 \\equiv (b^{q-1})^2 \\cdot b + 1 \\equiv b + 1 \\pmod q$, and $q > |b+1|$, it follows that $b+1 = 0$, i.e. $b = -1$, and so $a = -b^2 = -1$.\n\nIn conclusion, there are only two solution pairs $(0,0)$ and $(-1,-1)$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 28346,
    "subject": "Mathematics (Multi-modal)",
    "question": "As shown in Fig. 8.1, $\\odot O$ is the inscribed circle touching side $BC$ of $\\triangle ABC$ at point $M$, and points $D, E$ are on the segments $AB, AC$, respectively, satisfying $DE \\parallel BC$; $\\odot O_1$ is the inscribed circle of $\\triangle ADE$ tangent to side $DE$ at point $N$; $O_1B, DO$ intersect at point $F$, and $O_1C, EO$ intersect at point $G$. Please prove that $MN$ divides segment $FG$ equally. (posed by Bian Hongping)\n\n![](attached_image_1.png)\nFig. 8.1",
    "options": [],
    "answer": "Detailed solution",
    "solution": "If $AB = AC$, then the graph is symmetric about the bisector of $\\angle BAC$ and the conclusion is obvious. So we may assume that $AB > AC$. As shown in Fig. 8.2, let $L$ be the midpoint of $BC$, line $O_1L$ intersecting with $FG$ at $R$, and $O_1N$ be extended to intersect with $BC$ at $K$. Draw line $AT$ that is perpendicular to $BC$ at $T$ and intersects with line $DE$ at $S$. Connecting $AO$, obviously $O_1$ is on the segment $AO$. By Menelaus' theorem, we have\n\n![](attached_image_2.png)\nFig. 8.2\n\n$$\n\\frac{O_1 F}{F B} \\cdot \\frac{B D}{D A} \\cdot \\frac{A O}{O O_1} = 1, \\quad \\frac{O_1 G}{G C} \\cdot \\frac{C E}{E A} \\cdot \\frac{A O}{O O_1} = 1. \\qquad \\textcircled{1}\n$$\n\nSince $DE \\parallel BC$, we have $\\frac{BD}{DA} = \\frac{CE}{EA}$. So $\\frac{O_1 F}{F B} = \\frac{O_1 G}{G C}$, which means that $FG \\parallel BC$. Then $\\frac{FR}{GR} = \\frac{BL}{CL} = 1$. Therefore, $R$ is the midpoint of $FG$. We now only need to prove that $M, R, N$ are collinear. For this purpose, by the inverse of Menelaus' theorem, we only need to prove that\n\n$$\n\\frac{O_1 R}{R L} \\cdot \\frac{L M}{M K} \\cdot \\frac{K N}{N O_1} = 1. \\qquad \\textcircled{2}\n$$\n\nSince $FR \\parallel BL$, we have $\\frac{O_1 R}{R L} = \\frac{O_1 F}{F B} = \\frac{O O_1}{A O} \\cdot \\frac{A D}{D B}$ (the second equality is justified by ①). So we only need to prove that\n\n$$\n\\frac{O O_1}{A O} \\cdot \\frac{A D}{D B} \\cdot \\frac{L M}{M K} \\cdot \\frac{K N}{N O_1} = 1. \\qquad \\textcircled{3}\n$$\n\nSince $O_1 K \\perp DE$, $O M \\perp BC$, $A T \\perp BC$, $DE \\parallel BC$, then lines $O_1 K$, $O M$, $A T$ are parallel. By the theorem of dividing the segments into proportional by parallel lines, we have $\\frac{O O_1}{A O} = \\frac{M K}{M T}$. Substituting it into ③, we then only need to prove\n\n$$\n\\frac{A D}{D B} \\cdot \\frac{L M}{M T} \\cdot \\frac{K N}{N O_1} = 1. \\qquad \\textcircled{4}\n$$\n\nSince $DE \\parallel BC$, $K N \\perp DE$, $S T \\perp BC$, quadrilateral $K N S T$ is a rectangle and then $K N = S T$. Furthermore, from $D S \\parallel B T$ we have $\\frac{A D}{D B} = \\frac{A S}{S T}$. Substituting these results into ④, we now only need to prove\n\n$$\n\\frac{LM}{MT} = \\frac{NO_1}{AS}. \\qquad \\textcircled{5}\n$$\n\nLet $BC = a$, $AC = b$, $AB = c$. We have\n\n$$\nBM = \\frac{a+b-c}{2} \\text{ (the property of an inscribed circle)}, \\quad BL = \\frac{a}{2},\n$$\n$$\nBT = c \\cos \\angle ABC = c \\cdot \\frac{a^2 + c^2 - b^2}{2ac} = \\frac{a^2 + c^2 - b^2}{2a}.\n$$\nThen\n$$\n\\frac{LM}{MT} = \\frac{BL - BM}{BT - BM} = \\frac{\\frac{c-b}{2}}{\\frac{c^2 - b^2 + a(c-b)}{2a}} = \\frac{a}{a+b+c}.\n$$\nOn the other hand,\n$$\n\\frac{NO_1}{AS} = \\frac{\\frac{2S_{\\triangle ADE}}{AD + DE + AE}}{\\frac{2S_{\\triangle ADE}}{DE}} = \\frac{DE}{AD + DE + AE} = \\frac{a}{a+b+c}.\n$$\nTherefore, ⑤ holds. The proof is complete.\nLet the radii of $\\odot O$ and $\\odot O_1$ be $r$ and $r_1$, respectively. Obviously, $O_1$, $O$ and $A$ are collinear.\n$$\n\\left. \\begin{array}{l} DE // BC \\Rightarrow \\frac{AB}{BD} = \\frac{AC}{CE} \\\\[2ex] \\frac{OF}{FD} = \\frac{S_{\\triangle BOO_1}}{S_{\\triangle DBO_1}} = \\frac{\\frac{1}{2}\\left(AB \\sin \\frac{A}{2}\\right) \\cdot OO_1}{\\frac{1}{2}r_1 \\cdot BD} \\\\[2ex] \\frac{OG}{GE} = \\frac{S_{\\triangle OO_1C}}{S_{\\triangle EOO_1}} = \\frac{\\frac{1}{2}\\left(AC \\sin \\frac{A}{2}\\right) \\cdot OO_1}{\\frac{1}{2}r_1 \\cdot CE} \\\\[2ex] \\Rightarrow FG \\parallel DE \\parallel BC. \\end{array} \\right\\} \\Rightarrow \\frac{OF}{FD} = \\frac{OG}{GE}\n$$\n\nConnect $ON$ and extend it to intersect with $BC$ at $K$. If $\\angle ABC = \\angle ACB$, then by symmetry, the proposition holds. So we may assume that $\\angle ABC < \\angle ACB$ in the following.\nWe connect $OM, O_1M, OB, MD, DO_1$, respectively (see Fig. 8.3). Since $O_1N \\parallel OM$, we have\n\n![](attached_image_3.png)\nFig. 8.3\n\n$$\nS_{\\triangle ONM} = S_{\\triangle MOO_1} = \\frac{1}{2} r \\cdot OO_1 \\cdot \\sin \\frac{C-B}{2}, \\qquad ①\n$$\n$$\n\\frac{OG}{GE} = \\frac{OF}{DF} = \\frac{S_{\\triangle BOO_1}}{S_{\\triangle BDO_1}} = \\frac{\\frac{1}{2}BO \\cdot OO_1 \\cdot \\sin \\frac{C}{2}}{\\frac{1}{2} \\cdot BD \\cdot DO_1 \\cdot \\sin \\frac{B}{2}}\n$$\n$$\n= \\frac{r \\cdot OO_1 \\cdot \\sin \\frac{C}{2}}{r_1 \\cdot BD \\cdot \\cos \\frac{B}{2}} \\qquad ②\n$$\n$$\n(r = BO \\cdot \\cos \\frac{B}{2}, r_1 = DO_1 \\cdot \\sin \\frac{B}{2}),\n$$\n$$\nS_{\\triangle DMN} - S_{\\triangle MEN} = \\frac{1}{2}NK \\cdot DN - NE \\\\ = \\frac{1}{2}BD \\cdot \\sin B \\cdot \\left( r_1 \\cot \\frac{B}{2} - r_1 \\cot \\frac{C}{2} \\right). \\quad ③\n$$\nFrom ② and ③, we have\n$$\n\\begin{align*}\n& \\frac{OG}{GE} S_{\\triangle DMN} - S_{\\triangle MEN} \\\\\n&= \\frac{1}{2} r_1 \\cdot BD \\cdot \\sin B \\cdot \\left( \\cot \\frac{B}{2} - \\cot \\frac{C}{2} \\right) \\cdot \\frac{r \\cdot OO_1 \\cdot \\sin \\frac{C}{2}}{r_1 \\cdot BD \\cdot \\cos \\frac{B}{2}} \\\\\n&= r \\cdot OO_1 \\cdot \\sin \\frac{B}{2} \\cdot \\sin \\frac{C}{2} \\cdot \\left( \\cot \\frac{B}{2} - \\cot \\frac{C}{2} \\right) \\\\\n&= r \\cdot OO_1 \\cdot \\sin \\frac{C-B}{2}.\n\\end{align*}\n$$\nCombining it with ①, we get\n$$\n\\frac{OG}{GE} (S_{\\triangle DMN} - S_{\\triangle MEN}) = 2S_{\\triangle MON}.\n$$\nSince $\\frac{OG}{GE} : 2 = \\frac{OG}{OE} : \\left(\\frac{DF}{OD} + \\frac{EG}{OE}\\right)$, then\n$$\n\\frac{OF}{OD} \\cdot S_{\\triangle MND} - \\frac{OG}{OE} \\cdot S_{\\triangle MEN} = \\left(\\frac{DF}{OD} + \\frac{EG}{OE}\\right) \\cdot S_{\\triangle MON}, \\\\\n\\frac{OF \\cdot S_{\\triangle MND} - DF \\cdot S_{\\triangle MON}}{OD} = \\frac{OG \\cdot S_{\\triangle MEN} + EG \\cdot S_{\\triangle MON}}{OE}. \\quad ④\n$$\nOn the other hand,\n$$\nS_{\\triangle NMG} = \\frac{S_{\\triangle MEN} \\cdot OG + EG \\cdot S_{\\triangle MON}}{OE}.\n$$\n$$\n\\text{Therefore, } S_{\\triangle NMG} = \\frac{OF \\cdot S_{\\triangle MND} - DF \\cdot S_{\\triangle MON}}{OD}.\n$$\nIn the same way,\n$$\nS_{\\triangle NMF} = \\frac{OF \\cdot S_{\\triangle MND} - DF \\cdot S_{\\triangle MON}}{OD}.\n$$\nBy ④, we have $S_{\\triangle NMG} = S_{\\triangle NMF}$. Therefore, $MN$ divides segment $FG$ equally.\nThe proof is complete.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 28347,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $a_n = C_{200}^n \\cdot (\\sqrt[3]{6})^{200-n} \\cdot \\left(\\frac{1}{\\sqrt{2}}\\right)^n$ $(n = 1, 2, \\dots, 95)$. Then the number of terms that are integers in $\\{a_n\\}$ is ______.",
    "options": [],
    "answer": "15",
    "solution": "We have $a_n = C_{200}^n \\cdot 3^{\\frac{200-n}{3}} \\cdot 2^{\\frac{400-5n}{6}}$. When $a_n$ $(1 \\le n \\le 95)$ is an integer, $\\frac{200-n}{3}$ and $\\frac{400-5n}{6}$ must be integers. Then $6 \\mid n+4$.\n\nWhen $n = 2, 8, 14, 20, 26, 32, 38, 44, 50, 56, 62, 68, 74, 80$, $\\frac{200-n}{3}$ and $\\frac{400-5n}{6}$ are all non-negative integers. So the corresponding $a_n$, totally 14, are integers.\n\nWhen $n = 86$, we have $a_{86} = C_{200}^{86} \\cdot 3^{38} \\cdot 2^{-5}$. The number of the factors of 2 in $200!$ is\n$$\n\\left[ \\frac{200}{2} \\right] + \\left[ \\frac{200}{2^2} \\right] + \\left[ \\frac{200}{2^3} \\right] + \\left[ \\frac{200}{2^4} \\right] + \\left[ \\frac{200}{2^5} \\right] + \\left[ \\frac{200}{2^6} \\right] + \\left[ \\frac{200}{2^7} \\right] = 197.\n$$\nBy the same reason, the numbers of the factors of 2 in $86!$ and $114!$ are 82 and 110, respectively. Therefore, the number of the factors of 2 in $C_{200}^{86} = \\frac{200!}{86! \\cdot 114!}$ is $197 - 82 - 110 = 5$. So $a_{86}$ is an integer.\n\nWhen $n = 92$, we have $a_{92} = C_{200}^{92} \\cdot 3^{36} \\cdot 2^{-10}$. In the same way, we find the numbers of the factors of 2 in $92!$ and $108!$ are 88 and 105, respectively, which means that in $C_{200}^{92}$ is $197 - 88 - 105 = 4$. Therefore, $a_{92}$ is not an integer.\n\nOverall, the required number is $14 + 1 = 15$. ☐",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 28348,
    "subject": "Mathematics (Multi-modal)",
    "question": "Consider 12 figures on the clock face as 12 points. Color them in four colors: red, yellow, blue and green. Each color is used for three points. Configure $n$ convex quadrilaterals with vertices in these points, such that\n(1) there are no same color of vertices for each quadrilateral.\n(2) among any three of these quadrilaterals, there is a color of vertices such that the vertices of that color are different.\n\nFind the largest number of $n$. (posed by Tao Pingsheng)",
    "options": [],
    "answer": "9",
    "solution": "We use $A, B, C, D$ to represent these four colors, respectively, and the points in the same color by lower letters as $a_1, a_2, a_3; b_1, b_2, b_3; c_1, c_2, c_3$ and $d_1, d_2, d_3$ respectively.\n\nNow consider color $A$. If, in $n$ quadrilaterals, the number of points $a_1, a_2, a_3$ in color $A$ are $n_1, n_2, n_3$ respectively, then $n_1 + n_2 + n_3 = n$. Suppose that $n_1 \\ge n_2 \\ge n_3$. If $n \\ge 10$, then $n_1 + n_2 \\ge 7$. Consider these seven quadrilaterals (its $A$ color vertex is either $a_1$ or $a_2$), if the numbers of points $b_1, b_2, b_3$ in color $B$ are $m_1, m_2, m_3$, respectively, then $m_1 + m_2 + m_3 = 7$. By symmetricity, we may suppose that $m_1 \\ge m_2 \\ge m_3$, then $m_3 \\le 2$, that is, $m_1 + m_2 \\ge 5$.\n\nConsider these five quadrilaterals (its $A$ color point is either $a_1$ or $a_2$, and its $B$ color point is either $b_1$ or $b_2$), if the numbers of points $c_1, c_2, c_3$ in color $C$ are $k_1, k_2, k_3$, respectively, then $k_1 + k_2 + k_3 = 5$. By symmetricity, we may suppose that $k_1 \\ge k_2 \\ge k_3$, then $k_3 \\le 1$, that is, $k_1 + k_2 \\ge 4$.\n\nConsider these four quadrilaterals, denoted as $T_1, T_2, T_3, T_4$ (its color $A$ point is either $a_1$ or $a_2$, its color $B$ point is either $b_1$ or $b_2$, and its color $C$ point is either $c_1$ or $c_2$). Since there are only three points in color $D$, there are two quadrilaterals that have the same color $D$ point. Suppose that the same color $D$ point of $T_1, T_2$ is $d_1$.\n\nThen, in three quadrilaterals $T_1, T_2, T_3$, whatever be the color of the vertex, there are repeated points, which contradicts condition (2). Hence, $n \\le 9$.\n\nWe show the maximal number $n = 9$ by construction of these nine quadrilaterals.\n\n![](attached_image_1.png)\n\nWe draw three “concentric annulus” with four points on each radius representing four vertices and the color. So nine radii represent nine quadrilaterals, which satisfy condition (1).\n\nNext, we show that they also satisfy condition (2). Take any three radii (or three quadrilaterals).\nIf these three radii come from a concentric annulus, for each color except $A$, there are three points.\nIf these three radii come from three concentric annuli, then for color $A$, there are three points.\nIf these three radii come from two concentric annuli, call these three figures Fig. 1, Fig. 2 and Fig. 3. The radius directions are called “up radius”, “left radius” and “right radius”, and denoted, respectively, by $S$, $Z$ and $Y$. If three radii have three directions, then there are three points of color $B$ in three quadrilaterals. If the three radii have only two directions, then there are all cases as shown in the tables below, where 1, 2 and 3 stand for Fig. 1, Fig. 2 and Fig. 3, respectively.\n\nHere, the color in figure means that the three quadrilaterals have color with different figures.\n\n![](attached_image_2.png)\nC\n![](attached_image_3.png)\nD\n\n|   | S   | 1, 3 | 1, 3 | 1   |     | 3   |\n|---|-----|------|------|-----|-----|-----|\n| Z | 3   |      | 1, 3 | 1, 3| 1   |\n| Y |     | 1    |      | 3   | 1, 3|\nC\n|   | S   | 1, 3 | 1, 3 | 3   |     | 1   |\n|---|-----|------|------|-----|-----|-----|\n| Z | 1   |      | 1, 3 | 1, 3| 3   |\n| Y |     | 3    |      | 1   | 1, 3|\nD\n|   | S   | 2, 3 | 2, 3 | 3   |     | 2   |\n|---|-----|------|------|-----|-----|-----|\n| Z | 2   |      | 2, 3 | 2, 3| 3   |\n| Y |     | 3    |      | 2   | 2, 3|\nC\n|   | S   | 2, 3 | 2, 3 | 2   |     | 3   |\n|---|-----|------|------|-----|-----|-----|\n| Z | 3   |      | 2, 3 | 2, 3| 2   |\n| Y |     | 2    |      | 3   | 2, 3|\nD\n\nThus, the maximal number of $n$ is 9. $\\square$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 28349,
    "subject": "Mathematics (Multi-modal)",
    "question": "Suppose $f(x) = |\\lg(x+1)|$ and real numbers $a, b$ ($a < b$) satisfy $f(a) = f\\left(-\\frac{b+1}{b+2}\\right)$, $f(10a + 6b + 21) = 4\\lg 2$. Find the values of $a, b$.",
    "options": [],
    "answer": "a = -2/5, b = -1/3",
    "solution": "$|\\lg(a+1)| = |\\lg\\left(-\\frac{b+1}{b+2}+1\\right)| = |\\lg\\left(\\frac{1}{b+2}\\right)| = |\\lg(b+2)|.$\nThen either $a+1 = b+2$ or $(a+1)(b+2) = 1$. Since $a < b$, so $a+1 \\neq b+2$. Therefore, $(a+1)(b+2) = 1$.\nFrom $f(a) = |\\lg(a+1)|$ we know $0 < a+1 < 1$. Then\n$$\n0 < a + 1 < b + 1 < b + 2,\n$$\nwhich implies\n$$\n0 < a + 1 < 1 < b + 2.\n$$\nTherefore,\n$$\n(10a + 6b + 21) + 1 = 10(a + 1) + 6(b + 2) \\\\\n= 6(b + 2) + \\frac{10}{b + 2} > 1.\n$$\nThen\n$$\nf(10a + 6b + 21) = \\left| \\lg \\left[ 6(b + 2) + \\frac{10}{b+2} \\right] \\right| \\\\\n= \\lg \\left[ 6(b + 2) + \\frac{10}{b+2} \\right].\n$$\nOn the other hand,\n$$\nf(10a + 6b + 21) = 4\\lg 2.\n$$\nSo\n$$\n\\lg\\left[6(b+2) + \\frac{10}{b+2}\\right] = 4\\lg 2,\n$$\nwhich means $6(b+2) + \\frac{10}{b+2} = 16$. Then either $b = -\\frac{1}{3}$ or $b = -\\frac{2}{5}$ (discarded).\nSubstituting $b = -\\frac{1}{3}$ into $(a+1)(b+2) = 1$, we find $a = -\\frac{2}{5}$.\nTherefore $a = -\\frac{2}{5}$, $b = -\\frac{1}{3}$.\n$\\square$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 28350,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $H$ be the orthocenter of an acute-angled $\\triangle ABC$ with $\\angle A > 60^\\circ$. Let points $M$ and $N$ be on sides $AB$ and $AC$, respectively, such that $\\angle HMB = 60^\\circ = \\angle HNC$. Let $O$ be the circumcenter of $\\triangle HMN$. Let points $D$ and $A$ be on the same side of line $BC$, such that $\\triangle DBC$ is regular (see Fig. 1.1). Prove that points $H$, $O$ and $D$ are collinear. (posed by Zhang Sihui)\n\n![](attached_image_1.png)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $T$ be the orthocenter of $\\triangle HMN$. Extended lines of $HM$ and $CA$ intersect at point $P$. Extended lines $HN$ and $BA$ intersect at point $Q$. It is easy to see that points $N$, $M$, $P$ and $Q$ are concyclic.\n\n![](attached_image_2.png)\n\nBy $\\angle THM = \\angle OHN$, we see that $\\angle PQH - \\angle OHN = \\angle NMH - \\angle THM = 90^\\circ$, that is,\n$$\nHO \\perp PQ. \\qquad ①\n$$\nLet point $R$ be symmetric to point $C$ over $HP$. Then $HC = HR$.\nBy $\\angle HPC + \\angle HCP = (\\angle BAC - 60^\\circ) + (90^\\circ - \\angle BAC) = 30^\\circ$, we see that $\\angle CHR = 60^\\circ$. Hence $\\triangle HCR$ is regular.\nBy $\\angle HPC = \\angle HQB$ and $\\angle HCP = \\angle HBQ$, we see that $\\triangle PHC \\sim \\triangle QHB$. Then $\\triangle PHR \\sim \\triangle QHB$, so $\\triangle QHP \\sim \\triangle BHR$.\nDenote by $\\angle (UV, XY)$ the angle between $\\vec{UV}$ and $\\vec{XY}$ (positive anticlockwise).\nSince $\\angle PHR = 150^\\circ$, we have $\\angle (PQ, RB) = \\angle (HP, HR) = 150^\\circ$, $\\triangle BCD$ and $\\triangle RCH$ are regular. So $\\triangle BRC \\cong \\triangle DHC$. Hence $\\angle (RB, HD) = \\angle (CR, CH) = -60^\\circ$.\n\nConsequently,\n$$ \\angle(PQ, HD) = \\angle(PQ, RB) + \\angle(RB, HD) = 150^\\circ - 60^\\circ = 90^\\circ, $$\nthat is\n$$\nDH \\perp PQ. \\qquad \\textcircled{2}\n$$\nBy ① and ②, points $H$, $O$ and $D$ are collinear. □",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 28351,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find a triple $(l, m, n)$ ($1 < l < m < n$) of positive integers such that $\\sum_{k=1}^{l} k$, $\\sum_{k=l+1}^{m} k$, $\\sum_{k=m+1}^{n} k$ form a geometric sequence in order. (posed by Tao Pingsheng)",
    "options": [],
    "answer": "(3, 11, 36)",
    "solution": "$\\sum_{k=1}^{l} k = S_{l}$, $\\sum_{k=l+1}^{m} k = S_{m} - S_{l}$, $\\sum_{k=m+1}^{n} k = S_{n} - S_{m}$,\nform a geometric sequence in order. Then\n$$\nS_{l}(S_{n} - S_{m}) = (S_{m} - S_{l})^2, \\qquad \\textcircled{1}\n$$\nthat is $S_{l}(S_{n} + S_{m} - S_{l}) = S_{m}^2$. Thus, $S_{l} \\mid S_{m}^2$, that is\n$$\n2l(l + 1) \\mid m^2(m + 1)^2.\n$$\nLet $m + 1 = l(l + 1)$ and take $l = 3$. Then $m = 11$ and $S_{l} = S_{3} = 6$, $S_{m} = S_{11} = 66$. Substitute it into ①, we have $S_{n} = 666$, that is $\\frac{n(n+1)}{2} = 666$. So $n = 36$.\nTherefore, $(l, m, n) = (3, 11, 36)$ is a solution satisfying the condition. □",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 28352,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find the least positive integer $m$, such that for every prime number $p>3$, we have $105 < 9^p - 29^p + m$.",
    "options": [],
    "answer": "No such positive integer m exists",
    "solution": "",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 28353,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $P$ be a point on the image of $y = x + \\frac{2}{x}$ ($x > 0$). Through $P$ draw lines perpendicular to $y = x$ and $y$-axis with foot points $A, B$, respectively. Then the value of $\\vec{PA} \\cdot \\vec{PB}$ is ______.",
    "options": [],
    "answer": "-1",
    "solution": "Let $P(x_0, x_0 + \\frac{2}{x_0})$. The expression for line $PA$ is then\n$$\ny - \\left(x_0 + \\frac{2}{x_0}\\right) = -(x - x_0),\n$$\nor\n$$y = -x + 2x_0 + \\frac{2}{x_0}.$$\nFrom\n$$\n\\begin{cases} y = x, \\\\ y = -x + 2x_0 + \\frac{2}{x_0}, \\end{cases}\n$$\nwe get $A(x_0 + \\frac{1}{x_0}, x_0 + \\frac{1}{x_0})$.\nOn the other hand, we have $B(0, x_0 + \\frac{2}{x_0})$. Then $\\vec{PA} = (\\frac{1}{x_0}, -\\frac{1}{x_0})$ and $\\vec{PB} = (-x_0, 0)$. Therefore,\n$$\n\\vec{PA} \\cdot \\vec{PB} = \\frac{1}{x_0} \\cdot (-x_0) = -1.\n$$\nThe answer is $-1$.\nAs seen in Fig. 1.1, the distances from $P(x_0, x_0 + \\frac{2}{x_0})$ to lines $y = x$ and $y$-axis, respectively, are\n\n![](attached_image_1.png)\nFig. 1.1\n$$\n|PA| = \\frac{|x_0 - \\left(x_0 + \\frac{2}{x_0}\\right)|}{\\sqrt{2}} = \\frac{\\sqrt{2}}{x_0}\n$$\nand\n$$\n|PB| = x_0.\n$$\nSince $O$, $A$, $P$ and $B$ are concyclic points, then\n$$\n\\angle APB = \\pi - \\angle AOB = \\frac{3\\pi}{4}.\n$$\nTherefore, $\\overrightarrow{PA} \\cdot \\overrightarrow{PB} = |\\overrightarrow{PA}| \\cdot |\\overrightarrow{PB}| \\cdot \\cos \\frac{3\\pi}{4} = -1$. $\\square$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 28354,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $a_1, a_2, \\dots, a_n$ be $n$ non-negative real numbers.\n\n$$\n\\frac{1}{1+a_1} + \\frac{a_1}{(1+a_1)(1+a_2)} + \\dots + \\frac{a_1a_2\\dots a_{n-1}}{(1+a_1)(1+a_2)\\dots(1+a_n)} \\le 1.\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $a_0 = 1$. We prove the following identity:\n$$\n\\sum_{k=1}^{n} \\prod_{j=1}^{k} \\frac{a_{j-1}}{1+a_j} = 1 - \\prod_{j=1}^{n} \\frac{a_j}{1+a_j} \\qquad \\textcircled{1}\n$$\nby induction on $n$.\n\nIt is evident that $\\textcircled{1}$ is true for $n = 1$. Suppose that $\\textcircled{1}$ is true for $n-1$, $n \\ge 2$, then for $n$,\n$$\n\\begin{aligned}\n\\sum_{k=1}^{n} \\prod_{j=1}^{k} \\frac{a_{j-1}}{1+a_j} &= \\sum_{k=1}^{n-1} \\prod_{j=1}^{k} \\frac{a_{j-1}}{1+a_j} + \\prod_{j=1}^{n} \\frac{a_{j-1}}{1+a_j} \\\\\n&= 1 - \\prod_{j=1}^{n-1} \\frac{a_j}{1+a_j} + \\prod_{j=1}^{n} \\frac{a_{j-1}}{1+a_j} \\\\\n&= 1 - \\prod_{j=1}^{n} \\frac{a_j}{1+a_j}.\n\\end{aligned} \\quad \\square\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 28355,
    "subject": "Mathematics (Multi-modal)",
    "question": "It is known that each term of sequence $\\{a_n\\}$ is a non-zero real number, and for any positive integer $n$ holds the equation\n$$\n(a_1 + a_2 + \\cdots + a_n)^2 = a_1^3 + a_2^3 + \\cdots + a_n^3.\n$$\n\n(1) When $n = 3$, find out all the sequences consisting of three terms $a_1, a_2, a_3$.\n\n(2) Does there exist an infinite sequence $\\{a_n\\}$ such that $a_{2013} = -2012$? If it exists, write out the formula of general term; if not, give your reason.",
    "options": [],
    "answer": "Part (1): The three-term sequences are (1, 2, 3), (1, 2, −2), and (1, −1, 1).\nPart (2): Yes, it exists. One such sequence is given by a_n = n for 1 ≤ n ≤ 2012, and a_n = 2012(−1)^n for n ≥ 2013.",
    "solution": "(1) When $n = 1$, we have $a_1^2 = a_1^3$. Since $a_1 \\neq 0$, we get $a_1 = 1$.\nWhen $n = 2$, we have $(1+a_2)^2 = 1+a_2^3$. Since $a_2 \\neq 0$, we get $a_2 = 2$ or $a_2 = -1$.\nWhen $n = 3$, we have $(1+a_2+a_3)^2 = 1+a_2^3+a_3^3$. For $a_2 = 2$, we get $a_3 = 3$ or $a_3 = -2$; for $a_2 = -1$, we get $a_3 = 1$.\nIn summary, we get three sequences consisting of three terms that satisfy the given condition:\n$$\n\\{1, 2, 3\\}, \\{1, 2, -2\\}, \\text{ and } \\{1, -1, 1\\}.\n$$\n\n(2) Let $S_n = a_1 + a_2 + \\cdots + a_n$. Then we have\n$$\nS_n^2 = a_1^3 + a_2^3 + \\cdots + a_n^3 \\ (n \\in \\mathbb{N}), \\\\\n(S_n + a_{n+1})^2 = a_1^3 + a_2^3 + \\cdots + a_n^3 + a_{n+1}^3.\n$$\nFinding out the difference of the two expressions above and by $a_{n+1} \\neq 0$, we have $2S_n = a_{n+1}^2 - a_{n+1}$.\nWhen $n = 1$, we know from (1) that $a_1 = 1$.\nWhen $n \\ge 2$, we have\n$$\n2a_n = 2(S_n - S_{n-1}) = (a_{n+1}^2 - a_{n+1}) - (a_n^2 - a_n).\n$$\nAnd that is\n$$\n(a_{n+1} + a_n)(a_{n+1} - a_n - 1) = 0.\n$$\nThen we get $a_{n+1} = -a_n$ or $a_{n+1} = a_n + 1$.\nFinally, from $a_1 = 1$ and $a_{2013} = -2012$, we find the formula of general term for a required sequence as\n$$\na_n = \\begin{cases} n, & 1 \\le n \\le 2012, \\\\ 2012(-1)^n, & n \\ge 2013. \\end{cases}\n$$",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 28356,
    "subject": "Mathematics (Multi-modal)",
    "question": "Given a prime number $p$, let $A$ be a $p \\times p$ matrix such that its entries are exactly $1, 2, \\dots, p^2$ in some order. The following operation is allowed for a matrix: add one to each number in a row or a column, or subtract one from each number in a row or a column. The matrix $A$ is called “good” if one can take a finite series of such operations resulting in a matrix with all entries zero. Find the number of good matrices $A$.",
    "options": [],
    "answer": "2(p!)^2",
    "solution": "We may combine the operations on the same row or column, thus the final result of a series of operations can be realized as subtracting integer $x_i$ from each number of $i$-th row and subtracting integer $y_j$ from each number of $j$-th column. Thus, the matrix $A$ is good if and only if there exist integers $x_i, y_j$, such that $a_{ij} = x_i + y_j$ for all $1 \\le i, j \\le p$.\n\nSince the entries of $A$ are distinct, $x_1, x_2, \\dots, x_p$ are pairwise distinct, and so are $y_1, y_2, \\dots, y_p$. We may consider only the case that $x_1 < x_2 < \\cdots < x_p$ since swapping the value of $x_i$ and $x_j$ results in swapping the $i$-th row and $j$-th row, which is again a good matrix. Similarly, we may consider only the case that $y_1 < y_2 < \\cdots < y_p$, thus the matrix is increasing from left to right, also from top to bottom.\n\nFrom the assumptions above, we have $a_{11} = 1$, $a_{12}$ or $a_{21}$ equals $2$. We may consider only the case that $a_{12} = 2$ since the transpose of the matrix is again good. Now we argue by contradiction that the first row is $1, 2, \\dots, p$. Assume on the contrary that $1, 2, \\dots, k$ is on the first row, but $k+1$ is not, $2 \\le k < p$, therefore $a_{21} = k+1$. We call $k$ consecutive integers a “block”, and we shall prove that the first row consists of several blocks, that is, the first $k$ numbers is a block, the next $k$ numbers is again a block, and so on.\n\nIf it is not so, assume the first $n$ groups of $k$ numbers are “blocks”, but the next $k$ numbers is not a “block” (or there are no $k$ numbers remaining). It follows that for $j = 1, 2, \\dots, n$,\n\n$y_{(j-1)k+1}, y_{(j-1)k+2}, \\dots, y_{jk}$ is a \"block\", the first $nk$ columns of the matrix can be divided into $pn \\times k$ submatrices $a_{i, (j-1)k+1}, a_{i, (j-1)k+2}, \\dots, a_{i, jk}$, $i = 1, 2, \\dots, p$, $j = 1, 2, \\dots, n$, each submatrix is a \"block\". Now assume $a_{1, nk+1} = a$, let $b$ be the smallest positive integer such that $a+b$ is not on the first row, then $b \\le k-1$. Since $a_{2, nk+1} - a_{1, nk+1} = x_2 - x_1 = a_{21} - a_{11} = k$, we have $a_{2, nk+1} = a+k$, therefore $a+b$ lies in the first $nk$ columns. Therefore, $a+b$ is contained in one of the $1 \\times k$ submatrices mentioned above, which is a \"block\", however $a, a+k$ are not in this \"block\", which is a contradiction.\n\nWe showed that the first row is formed by blocks, in particular $k \\mid p$, however, $1 < k < p$, and $p$ is a prime, which is impossible. So we conclude that the first row is $1, 2, \\dots, p$, the $k$-th row must be $(k-1)p+1, (k-1)p+2, \\dots, kp$. Thus up to interchanging rows, columns and transpose, the good matrix is unique, the answer is therefore $2(p!)^2$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 28357,
    "subject": "Mathematics (Multi-modal)",
    "question": "Prove that, among any $n$ vertexes of a regular $2n-1$ polygon ($n \\ge 3$), there are three points, which are the vertexes of an isosceles triangle.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let the vertices of the regular $(2n-1)$-gon be $A_0, A_1, \\ldots, A_{2n-2}$, labeled in order around the circle.\n\nSuppose we select any $n$ vertices from these $2n-1$ vertices. For each vertex $A_i$, consider the set of possible isosceles triangles with $A_i$ as a vertex and the other two vertices also among the $n$ chosen points.\n\nFor each $A_i$, the isosceles triangles with $A_i$ as a vertex are determined by choosing another vertex $A_j$ such that the third vertex is $A_k$, where $A_k$ is symmetric to $A_j$ with respect to $A_i$ (i.e., $A_k$ is the reflection of $A_j$ over $A_i$). Since the polygon is regular, for each pair $(A_i, A_j)$, there is a unique $A_k$ such that $A_iA_j = A_iA_k$ and $A_j \\ne A_k$ unless $A_j$ is diametrically opposite $A_i$ (which cannot happen since $2n-1$ is odd).\n\nNow, consider the set of all possible unordered pairs of chosen vertices. There are $\\binom{n}{2}$ such pairs. For each such pair $(A_j, A_k)$, there is a unique center $A_i$ such that $A_j$ and $A_k$ are symmetric with respect to $A_i$ (i.e., $A_i$ is the midpoint of the arc $A_jA_k$). Since $2n-1$ is odd, for each pair $(A_j, A_k)$, there is a unique $A_i$ (possibly not among the chosen $n$ vertices).\n\nSuppose, for contradiction, that among the $n$ chosen vertices, no three form an isosceles triangle. Then, for each chosen vertex $A_i$, the other $n-1$ chosen vertices must be distributed so that no two are symmetric with respect to $A_i$. For each $A_i$, the $2n-2$ other vertices are paired into $n-1$ pairs of symmetric points with respect to $A_i$. Since we have chosen $n-1$ vertices (excluding $A_i$), by the pigeonhole principle, at least one pair must have both points chosen. But then, together with $A_i$, these two points form an isosceles triangle.\n\nTherefore, among any $n$ vertices of a regular $(2n-1)$-gon, there exist three that are the vertices of an isosceles triangle.",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 28358,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $A = \\{2, 2^2, \\dots, 2^n, \\dots\\}$. Prove:\n\n(1) For any $a \\in A$, $b \\in \\mathbb{N}^*$, if $b < 2a - 1$, then $b(b+1)$ will not be a multiple of $2a$.\n\n(2) For any $a \\in \\bar{A} (= \\mathbb{N}^* - A)$ satisfying $a \\neq 1$, there exists $b \\in \\mathbb{N}^*$ satisfying $b < 2a - 1$, such that $b(b+1)$ is a multiple of $2a$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "(1) For any $a \\in A$, $a = 2^k$ ($k \\in \\mathbb{N}^*$). Then $2a = 2^{k+1}$. Let $b$ be any positive integer strictly less than $2a - 1$. Then $(b+1) \\le 2a - 1$.\n\nBetween $b$ and $b+1$, one is an odd number that contains no prime factor $2$, and the other is an even number that contains at most the $k$th power of $2$. Therefore, $b(b+1)$ is definitely not a multiple of $2a$.\n\n\n(2) For $a \\in \\bar{A}$ and $a \\ne 1$, suppose $a = 2^k m$ where $k$ is a non-negative integer and $m$ is an odd number greater than $1$. Then $2a = 2^{k+1} m$. We will present three different proofs in the following.\n\n**Proof 1.** Let $b = mx$, $b+1 = 2^{k+1} y$. Eliminating $b$, we have $2^{k+1}y - mx = 1$. Since $(2^{k+1}, m) = 1$, the equation has integral solutions that can be expressed as\n$$\n\\begin{cases}\nx = x_0 + 2^{k+1} t, \\\\\ny = y_0 + m t\n\\end{cases} \\quad (\\text{where } t \\in \\mathbb{Z}, \\text{ and } (x_0, y_0) \\text{ is a special solution of the equation}.)\n$$\nDenote the smallest solution among them as $(x^*, y^*)$. Then $x^* < 2^{k+1}$.\nTherefore, $b = m x^* < 2a - 1$ and $b(b+1)$ is a multiple of $2a$.\n\n**Proof 2.** Since $(2^{k+1}, m) = 1$, by the Chinese Remainder Theorem, the congruence equation\n$$\n\\begin{cases}\nx \\equiv 0 \\pmod{2^{k+1}}, \\\\\nx \\equiv m-1 \\pmod{m}\n\\end{cases}\n$$\nhas a solution $x = b$ with $b \\in (0, 2^{k+1} m)$. It is easy to see that $b < 2a-1$ and $b(b+1)$ is a multiple of $2a$.\n\n**Proof 3.** Since $(2^{k+1}, m) = 1$, then there exists $r \\in \\mathbb{N}^*$, $r \\le m - 1$, such that $2^r \\equiv 1 \\pmod{m}$.\nTake $t \\in \\mathbb{N}^*$ such that $tr > k + 1$. Then $2^{tr} \\equiv 1 \\pmod{m}$. It is easy to see that there exists\n$$\nb = (2^{tr} - 1) - q \\cdot 2^{k+1} m > 0 \\quad (q \\in \\mathbb{N}),\n$$\nsuch that $0 < b < 2a - 1$. Then we have $m \\mid b$, $2^{k+1} \\mid b + 1$.\nTherefore, $b(b + 1)$ is a multiple of $2a$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 28359,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $\\odot I$ be the incircle of $\\triangle ABC$. The circle $\\odot I$ intersects sides $AB$, $BC$ and $CA$ at points $D$, $E$ and $F$, respectively. Line $EF$ intersects lines $AI$, $BI$ and $DI$ at points $M$, $N$ and $K$, respectively. Prove that $DM \\cdot KE = DN \\cdot KF$. (posed by Zhang Pengcheng)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "It is easy to see that points $I$, $D$, $E$ and $B$ are concyclic and\n$$\n\\angle AID = 90^\\circ - \\angle IAD, \\\\\n\\angle MED = \\angle FDA = 90^\\circ - \\angle IAD.\n$$\nSo $\\angle AID = \\angle MED$, thus points $I$, $D$, $E$ and $M$ are concyclic.\nHence, five points $I$, $D$, $B$, $E$, $M$ are concyclic and $\\angle IMB = \\angle IEB = 90^\\circ$, that is $AM \\perp BM$.\nSimilarly, points $I$, $D$, $A$, $N$ and $F$ are concyclic and $BN \\perp AN$.\nLet lines $AN$ and $BM$ intersect at point $G$. We see point $I$ is the\n![](attached_image_1.png)\nFig. 2. 1\northocenter of $\\triangle GAB$ and $ID \\perp AB$, so points $G$, $I$ and $D$ are collinear.\nSince points $G$, $N$, $D$ and $B$ are concyclic, we see that\n$\\angle ADN = \\angle G$.\nSimilarly, $\\angle BDM = \\angle G$. So $DK$ bisects $\\angle MDN$, thus\n$$\n\\frac{DM}{DN} = \\frac{KM}{KN}. \\qquad \\textcircled{1}\n$$\nSince points $I$, $D$, $E$ and $M$ are concyclic, and points $I$, $D$, $N$ and $F$ are concyclic, we see that\n$$\nKM \\cdot KE = KI \\cdot KD = KF \\cdot KN.\n$$\nTherefore,\n$$\n\\frac{KM}{KN} = \\frac{KF}{KE}. \\qquad \\textcircled{2}\n$$\nBy ① and ②, we see that $\\frac{DM}{DN} = \\frac{KF}{KE}$, that is $DM \\cdot KE = DN \\cdot KF$. $\\square$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 28360,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $P(x) = x^{2012} + a_{2011}x^{2011} + a_{2010}x^{2010} + \\dots + a_1x + a_0$ be a polynomial of degree $2012$ of real coefficients with $1$ as its leading coefficient. Find the minimum of real number $c$ such that $|\\operatorname{Im} z| \\le c |\\operatorname{Re} z|$, where $\\operatorname{Re} z$ and $\\operatorname{Im} z$ are, respectively, the real and the imaginary parts of any root of a polynomial obtained by changing some of the coefficients of $P(x)$ to their opposite numbers. (posed by Zhu Huawei)",
    "options": [],
    "answer": "cot(pi/4022)",
    "solution": "First, we point out that $c \\ge \\cot \\frac{\\pi}{4022}$. Consider the polynomial $P(x) = x^{2012} - x$. Changing the sign of coefficients of $P(x)$, we obtain four polynomials $P(x)$, $-P(x)$, $Q(x) = x^{2012} + x$ and $-Q(x)$. Note that $P(x)$ and $-P(x)$ have the same roots; one of the roots is $z_1 = \\cos \\frac{1006}{2011}\\pi + i \\sin \\frac{1006}{2011}\\pi$. $Q(x)$ and $-Q(x)$ have the same roots, and are the opposite number of the roots of $P(x)$. Thus $Q(x)$ has a root $z_2 = -z_1$. Then,\n$$\nc \\ge \\min\\left(\\frac{|\\operatorname{Im} z_1|}{|\\operatorname{Re} z_1|}, \\frac{|\\operatorname{Im} z_2|}{|\\operatorname{Re} z_2|}\\right) = \\cot \\frac{\\pi}{4022}.\n$$\nNext, we show that the answer is $c = \\cot \\frac{\\pi}{4022}$. For any\n$$\nP(x) = x^{2012} + a_{2011}x^{2011} + a_{2010}x^{2010} + \\dots + a_1x + a_0,\n$$\nwe obtain a polynomial\n$$\nR(x) = b_{2012}x^{2012} + b_{2011}x^{2011} + b_{2010}x^{2010} + \\dots + b_1x + b_0,\n$$\nby changing sign of some coefficient of $P(x)$, where $b_{2012} = 1$, and for $j = 1, 2, \\dots, 2011$,\n$$\nb_j = \\begin{cases} |a_j|, & j \\equiv 0, 1 \\pmod 4, \\\\ -|a_j|, & j \\equiv 2, 3 \\pmod 4. \\end{cases}\n$$\nWe show that, for each root $z$ of $R(x)$, we have $|\\operatorname{Im} z| \\le c |\\operatorname{Re} z|.$\nWe prove this result by contradiction. Suppose there is a root $z_0$ of $R(x)$, such that $|\\operatorname{Im} z_0| > c |\\operatorname{Re} z_0|$, then $z_0 \\ne 0$ and either the angle of $z_0$ and $i$ is less than $\\theta = \\frac{\\pi}{4022}$, or the angle of $z_0$ and $-i$ is less than $\\theta$. Suppose that the angle of $z_0$ and $i$ is less than $\\theta$; for the other case, we need only consider the conjugate of $z_0$. There are two cases:\n\nIf $z_0$ is on the first quadrant (or imaginary axis), suppose that $\\angle(z_0, i) = \\alpha < \\theta$, where $\\angle(z_0, i)$ is the least angle that rotates $z_0$ to $i$ anticlockwise. For $0 \\le j \\le 2012$, if $j \\equiv 0, 2 \\pmod 4$, then $\\angle(b_j z_0^j, 1) = j\\alpha \\le 2012\\alpha < 2012\\theta$.\nIf $j \\equiv 1, 3 \\pmod 4$, then $\\angle(b_j z_0^j, i) = j\\alpha < 2011\\theta$ and $\\angle(b_1 z_0, i) = \\alpha$. Thus, the principal argument of $b_j z_0^j \\in [2\\pi - 2012\\alpha, 2\\pi) \\cup [0, \\frac{1}{2}\\pi - \\alpha]$. The vertex angle of this angle-domain is $2012\\alpha + \\frac{1}{2}\\pi - \\alpha = \\frac{1}{2}\\pi + 2011\\alpha < \\pi$. And $b_j z_0^j, 0 \\le j \\le 2012$, are not all zero, so their sum cannot be zero.\n\nIf $z_0$ is at the second quadrant, suppose that $\\angle(i, z_0) = \\alpha < \\theta$, if $j \\equiv 0, 2 \\pmod 4$, then $\\angle(1, b_j z_0^j) = j\\alpha < 2012\\theta$. If $j \\equiv 1, 3 \\pmod 4$, then $\\angle(i, b_j z_0^j) = j\\alpha \\le 2011\\alpha < \\frac{\\pi}{2}$. Thus, every principle argument of $b_j z_0^j \\in [0, \\frac{\\pi}{2} + 2011\\alpha]$. Since $\\frac{\\pi}{2} + 2011\\alpha < \\pi$, and $b_j z_0^j, 0 \\le j \\le 2012$, are not all zero, so their sum cannot be zero.\n\nSumming up, the least real number $c = \\cot \\frac{\\pi}{4022}$. $\\square$",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 28361,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $P_0, P_1, P_2, \\dots, P_n$ be $n+1$ points on a plane, and the minimum distance between each two points of them is $d$ ($d > 0$). Prove\n$$\n| P_0 P_1 | \\cdot | P_0 P_2 | \\cdots | P_0 P_n | > \\left( \\frac{d}{3} \\right)^n \\sqrt{(n+1)!}\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "We may assume that $|P_0P_1| \\le |P_0P_2| \\le \\dots \\le |P_0P_n|$.\nAt first, we will prove that $|P_0P_k| > \\frac{d}{3}\\sqrt{k+1}$ for any positive integer $n$.\nObviously, $|P_0P_k| \\ge d \\ge \\frac{d}{3}\\sqrt{k+1}$ for $k = 1, 2, \\dots, 8$, and the second equality holds only when $k = 8$. Then we only need to prove that $|P_0P_k| \\ge d \\ge \\frac{d}{3}\\sqrt{k+1}$ for $k \\ge 9$.\nTake each $P_i$ ($i = 0, 1, 2, \\dots, k$) as the center to draw a circle with radius $\\frac{d}{2}$. Then these circles are either externally tangent to or apart from each other. Take $P_0$ as the center to draw a circle with radius $|P_0P_k| + \\frac{d}{2}$. Then the previous $k+1$ smaller circles are all located in this larger one.\nThen $\\pi(|P_0P_k| + \\frac{d}{2})^2 > (k+1)\\pi(\\frac{d}{2})^2$, from which we have $|P_0P_k| > \\frac{d}{2}(\\sqrt{k+1} - 1)$.\nIt is easy to check that $\\frac{\\sqrt{k+1}-1}{2} > \\frac{\\sqrt{k+1}}{3}$ for $k \\ge 9$.\nThen $|P_0P_k| > \\frac{d}{3}\\sqrt{k+1}$ for $k \\ge 9$.\nOver all, we have $|P_0P_k| > \\frac{d}{3}\\sqrt{k+1}$ for $k \\ge 9$.\nTherefore,\n$$\n|P_0P_1| \\cdot |P_0P_2| \\cdots |P_0P_n| > \\left(\\frac{d}{3}\\right)^n \\sqrt{(n+1)!}\n$$\nWe may assume $|P_0P_1| \\le |P_0P_2| \\le \\cdots \\le |P_0P_n|$.\nTake each $P_i$ ($i = 0, 1, 2, \\dots, k$) as the center to draw a circle with radius $\\frac{d}{2}$. Then these circles are either externally tangent to or apart from each other.\nLet $Q$ be any point on $\\odot P_i$. Since\n$$\n\\begin{aligned}\n|P_0Q| &\\le |P_0P_i| + |P_iQ| = |P_0P_i| + \\frac{d}{2} \\\\\n&\\le |P_0P_k| + \\frac{1}{2} |P_0P_k| = \\frac{3}{2} |P_0P_k|,\n\\end{aligned}\n$$\nwe get that the circle with center $P_0$ and radius $\\frac{3}{2} |P_0P_k|$ cover the previous $k+1$ smaller circles. Then we have $\\pi(\\frac{3}{2} |P_0P_k|)^2 > (k+1)\\pi(\\frac{d}{2})^2$, and that is\n$$\n| P_0 P_k | > \\frac{d}{3} \\sqrt{k+1} \\quad (i = 0, 1, 2, \\dots, k).\n$$\nTherefore,\n$$\n| P_0 P_1 | \\cdot | P_0 P_2 | \\cdots | P_0 P_n | > \\left(\\frac{d}{3}\\right)^n \\sqrt{(n+1)!}\n$$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 28362,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all the pairs $(a, b)$ of integers satisfying the following condition: there exists an integer $d \\ge 2$ such that $a^n + b^n + 1$ is divisible by $d$ for all positive integers $n$.",
    "options": [],
    "answer": "All pairs where the two integers have opposite parity, or both integers are congruent to 1 modulo 3.",
    "solution": "",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 28363,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $E$ be a given set with $n$ elements. Suppose that $A_1, A_2, \\dots, A_k$ are $k$ distinct nonempty subsets of $E$, with either $A_i \\cap A_j = \\emptyset$, $A_i \\subset A_j$, or $A_j \\subset A_i$ for any $1 \\le i < j \\le k$. Find the maximum value of $k$.",
    "options": [],
    "answer": "2n - 1",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 28364,
    "subject": "Mathematics (Multi-modal)",
    "question": "For positive composite number $n$, denote by $f(n)$ and $g(n)$ the sum of the smallest three positive divisors of $n$ and the largest two positive divisors of $n$, respectively. Find all $n$ such that $g(n)$ equals $f(n)$ to some power of positive integers. (posed by He Yijie)",
    "options": [],
    "answer": "All such n are n = 4 * 6^l for positive integer l.",
    "solution": "If $n$ is odd, then all factors of $n$ are odd. So $f(n)$ is odd and $g(n)$ is even. $g(n)$ cannot be $f(n)$ to some power of positive integer. Therefore $n$ is even. The smallest two divisors of $n$ are $1$ and $2$, and the largest two divisors of $n$ are $n$ and $n/2$.\n\nLet $d$ be the third smallest divisor of $n$. If there exists $k \\in \\mathbb{N}^*$ such that $g(n) = f^k(n)$, then\n\n$$\n\\frac{3n}{2} = (1 + 2 + d)^k = (3 + d)^k \\equiv d^k \\pmod{3}.\n$$\n\nSince $3 \\nmid \\frac{3n}{2}$, we see that $3 \\mid d^k$. So $3 \\mid d$, and since $d$ is the third smallest, we see that $d = 3$.\n\nThus, $\\frac{3}{2}n = 6^k$, we get $n = 4 \\times 6^{k-1}$. Since $3 \\nmid n$, we see that $k \\ge 2$.\n\nSumming up, $n = 4 \\times 6^l \\ (l \\in \\mathbb{N}^*)$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 28365,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $x$, $y$, $z \\in [0, 1]$. Then the maximum value of $M = \\sqrt{|x-y|} + \\sqrt{|y-z|} + \\sqrt{|z-x|}$ is ________.",
    "options": [],
    "answer": "sqrt(2) + 1",
    "solution": "We may assume $0 \\le x \\le y \\le z \\le 1$. Then\n$$\nM = \\sqrt{y-x} + \\sqrt{z-y} + \\sqrt{z-x}.\n$$\nSince\n$$\n\\sqrt{y-x} + \\sqrt{z-y} \\le \\sqrt{2[(y-x)+(z-y)]} = \\sqrt{2(z-x)},\n$$\nwe have\n$$\nM \\le \\sqrt{2(z-x)} + \\sqrt{z-x} = (\\sqrt{2} + 1) \\sqrt{z-x} \\le \\sqrt{2} + 1.\n$$\nThe equality holds if and only if $y-x = z-y$, $x = 0$, $z = 1$ (i.e. $x = 0$, $y = \\frac{1}{2}$, $z = 1$).\nTherefore, the answer is $M_{\\max} = \\sqrt{2} + 1$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 28366,
    "subject": "Mathematics (Multi-modal)",
    "question": "Prove that for any real number $M > 2$, there exists a strictly increasing infinite sequence of positive integers $a_1, a_2, \\dots$ satisfying both the following two conditions:\n(1) $a_i > M^i$ for any positive integer $i$.\n(2) An integer $n$ is non-zero if and only if there exists a positive integer $m$ and $b_1, b_2, \\dots, b_m \\in \\{-1, 1\\}$, with $n = b_1a_1 + b_2a_2 + \\dots + b_m a_m$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "For given $M > 2$, we construct by induction a sequence $\\{a_n\\}$ that satisfies the requirements. Take $a_1, a_2$ that satisfy $a_2 - a_1 = 1$ and $a_1 > M^2$. Now suppose $a_1, a_2, \\dots, a_{2k}$ are already chosen, such that $a_i > M^i$, $i = 1, 2, \\dots, 2k$ and such that the set $A_k = \\{b_1a_1 + \\dots + b_m a_m \\mid b_1, \\dots, b_m = \\pm 1, 1 \\le m \\le 2k\\}$ does not contain $0$. It is obvious that $A_k$ is symmetric, i.e., $A_k = -A_k$. $A_1 = \\{a_1, -a_1, 1, -1\\}$.\n\nLet $n$ be the smallest positive integer not in $A_k$, $N = \\sum_{i=1}^{2k} a_i$, now choose positive integers $a_{2k+1}, a_{2k+2}$ satisfying $a_{2k+2} - a_{2k+1} = N + n$, $a_{2k+1} > M^{2k+2}$, $a_{2k+1} > \\sum_{i=1}^{2k} a_i$. We now show that $A_{k+1}$ does not contain $0$ and $n \\in A_{k+1}$. First, $n = -\\sum_{i=1}^{2k} a_i - a_{2k+1} + a_{2k+2}$.\n\nOn the other hand, if $\\sum_{i=1}^m b_i a_i = 0$, $m \\le 2k+2$, as $0 \\notin A_k$, we must have $m = 2k+1$ or $2k+2$.\n\nIf $m = 2k+1$, then $\\left| \\sum_{i=1}^{2k+1} b_i a_i \\right| \\ge a_{2k+1} - \\sum_{i=1}^{2k} a_i > 0$.\n\nIf $m = 2k + 2$ and $b_{2k+1}$ and $b_{2k+2}$ are of the same sign, then\n$\\left| \\sum_{i=1}^{2k+2} b_i a_i \\right| \\ge a_{2k+1} + a_{2k+2} - \\sum_{i=1}^{2k} a_i > 0$; if $b_{2k+1}$ and $b_{2k+2}$ are of different signs, then\n$$\n\\left| \\sum_{i=1}^{2k+2} b_i a_i \\right| = \\left| \\sum_{i=1}^{2k} b_i a_i \\pm (a_{2k+1} - a_{2k+2}) \\right| \\ge | a_{2k+1} - a_{2k+2} | - \\sum_{i=1}^{2k} a_i = N + n - N = n > 0.\n$$\nThe $\\{a_n\\}$ thus constructed satisfies the requirements since $0$ is not contained in any $A_k$, and any non-zero integer between $-k$ and $k$ is contained in $A_k$. $\\square$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 28367,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let the focus and directrix of parabola $y^2 = 2px$ ($p > 0$) be $F$ and $l$, respectively. $A$ and $B$ are moving points on the parabola satisfying $\\angle AFB = \\frac{\\pi}{3}$. Let the projection of $M$, the midpoint of segment $AB$, on $l$ be $N$. Then the maximum value of $\\frac{|MN|}{|AB|}$ is ______.",
    "options": [],
    "answer": "1",
    "solution": "Suppose $\\angle ABF = \\theta$ ($0 < \\theta < \\frac{2\\pi}{3}$). Then by the Law of Sine, we have\n$$\n\\frac{|AF|}{\\sin \\theta} = \\frac{|BF|}{\\sin(\\frac{2\\pi}{3} - \\theta)} = \\frac{|AB|}{\\sin \\frac{\\pi}{3}}\n$$\nAnd then\n$$\n\\frac{|AF| + |BF|}{\\sin \\theta + \\sin(\\frac{2\\pi}{3} - \\theta)} = \\frac{|AB|}{\\sin \\frac{\\pi}{3}}\n$$\nSo\n$$\n\\frac{|AF| + |BF|}{|AB|} = \\frac{\\sin \\theta + \\sin(\\frac{2\\pi}{3} - \\theta)}{\\sin \\frac{\\pi}{3}} = 2\\cos(\\theta - \\frac{\\pi}{3}).\n$$\nAs seen in Fig. 4.1, by using the definition of a parabola and the property of a trapezoid, we have\n$$|MN| = \\frac{|AF| + |BF|}{2}.$$ Then\n$$\n\\frac{|MN|}{|AB|} = \\cos\\left(\\theta - \\frac{\\pi}{3}\\right).\n$$\nTherefore, $\\frac{|MN|}{|AB|}$ reaches the maximum value $1$ when $\\theta = \\frac{\\pi}{3}$.\nBy using the definition of a parabola and the property of a trapezoid, we have $|MN| = \\frac{|AF| + |BF|}{2}$. In $\\triangle AFB$, by using the Law of Cosines we have\n$$\n\\begin{aligned}\n|AB|^2 &= |AF|^2 + |BF|^2 - 2|AF| \\cdot |BF| \\cos \\frac{\\pi}{3} \\\\\n&= (|AF| + |BF|)^2 - 3|AF| \\cdot |BF| \\\\\n&\\geq (|AF| + |BF|)^2 - 3\\left(\\frac{|AF| + |BF|}{2}\\right)^2 \\\\\n&= \\left(\\frac{|AF| + |BF|}{2}\\right)^2 = |MN|^2.\n\\end{aligned}\n$$\nThe equality holds if and only if $|AF| = |BF|$. Therefore, the maximum value of $\\frac{|MN|}{|AB|}$ is $1$. $\\square$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 28368,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $f(x) = (x+a)(x+b)$ where $a$, $b$ are given positive real numbers, $n \\ge 2$ be a given integer. For non-negative real numbers $x_1, x_2, \\dots, x_n$ that satisfy $x_1 + x_2 + \\dots + x_n = 1$, find the maximum of $F = \\sum_{1 \\le i < j \\le n} \\min\\{f(x_i), f(x_j)\\}$.",
    "options": [],
    "answer": "(n-1)/2 * (1/n + a + b + n a b)",
    "solution": "As\n$$\n\\begin{align*}\n\\min\\{f(x_i), f(x_j)\\} &= \\min\\{(x_i + a)(x_i + b), (x_j + a)(x_j + b)\\} \\\\\n&\\le \\sqrt{(x_i + a)(x_i + b)(x_j + a)(x_j + b)} \\\\\n&\\le \\frac{1}{2}((x_i + a)(x_j + b) + (x_i + b)(x_j + a)) \\\\\n&= x_i x_j + \\frac{1}{2}(x_i + x_j)(a + b) + ab,\n\\end{align*}\n$$\nso\n$$\n\\begin{align*}\nF &\\le \\sum_{1 \\le i < j \\le n} x_i x_j + \\frac{a+b}{2} \\sum_{1 \\le i < j \\le n} (x_i + x_j) + C_n^2 \\cdot ab \\\\\n&= \\frac{1}{2} \\left[ \\left( \\sum_{i=1}^n x_i \\right)^2 - \\sum_{i=1}^n x_i^2 \\right] + \\frac{a+b}{2} (n-1) \\sum_{i=1}^n x_i + C_n^2 \\cdot ab \\\\\n&= \\frac{1}{2} \\left( 1 - \\sum_{i=1}^n x_i^2 \\right) + \\frac{n-1}{2} (a+b) + C_n^2 \\cdot ab \\\\\n&\\le \\frac{1}{2} \\left( 1 - \\frac{1}{n} \\left( \\sum_{i=1}^n x_i \\right)^2 \\right) + \\frac{n-1}{2} (a+b) + C_n^2 \\cdot ab \\\\\n&= \\frac{1}{2} \\left( 1 - \\frac{1}{n} \\right) + \\frac{n-1}{2} (a+b) + \\frac{n(n-1)}{2} ab \\\\\n&= \\frac{n-1}{2} \\left( \\frac{1}{n} + a + b + nab \\right).\n\\end{align*}\n$$\nThe equality holds when $x_1 = x_2 = \\cdots = x_n = \\frac{1}{n}$. So the maximum of $F$ is $\\frac{n-1}{2} \\left( \\frac{1}{n} + a + b + nab \\right)$.\nWe show that the maximum value is attained when $x_1 = x_2 = \\cdots = x_n = \\frac{1}{n}$, and $F_{\\max} = \\frac{n-1}{2} \\left(\\frac{1}{n} + a + b + nab\\right)$.\nWe induct on $n$ to show a more general statement: for non-negative real numbers $x_1, x_2, \\dots, x_n$ satisfying $x_1 + x_2 + \\dots + x_n = s$ (where $s$ is a fixed non-negative real number), the maximum value of $F = \\sum_{1 \\le i < j \\le n} \\min\\{f(x_i), f(x_j)\\}$ is attained when $x_1 = x_2 = \\cdots = x_n = \\frac{s}{n}$.\nSince $F$ is symmetric, we may assume $x_1 \\le x_2 \\le \\cdots \\le x_n$. Note that $f(x)$ is strictly increasing on non-negative real numbers, we have\n$$\nF = (n-1)f(x_1) + (n-2)f(x_2) + \\cdots + f(x_{n-1}).\n$$\nWhen $n=2$, $F = f(x_1) \\le f(\\frac{s}{2})$, equality holds when $x_1 = x_2$. Assume that the statement holds for $n$, consider the case of $n+1$. Applying inductive hypothesis on $x_2 + x_3 + \\cdots + x_{n+1} = s - x_1$, we have\n$$\nF \\le nf(x_1) + \\frac{1}{2}n(n-1)f(\\frac{s-x_1}{n}) = g(x_1),\n$$\nwhere $g(x)$ is a quadratic function of $x$, the leading coefficient is $1 + \\frac{n-1}{2n^2}$, and the coefficient of $x$ is $a + b - \\frac{n-1}{2n}(a + b + \\frac{s}{2n})$, therefore, the axis of symmetry is\n$$\n\\frac{\\frac{n-1}{2n}\\left(a+b+\\frac{s}{2n}\\right)-a-b}{2+\\frac{n-1}{n^2}} \\le \\frac{s}{2(n+1)}.\n$$\n(The above inequality is equivalent to $[(n-1)s - 2n(n+1)(a+b)](n+1) \\le 2s(2n^2 + n - 1)$; obviously, left-hand side $< (n^2-1)s <$ right-hand side.) Therefore, $g(\\frac{s}{n+1})$ is the maximum of $g(x)$ on $[0, \\frac{s}{n+1}]$. Thus, $F$ attains its maximum when $x_2 = x_3 = \\cdots = x_{n+1} = \\frac{s-x_1}{n} = \\frac{s}{n+1} = x_1$, completing the solution. $\\square$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 28369,
    "subject": "Mathematics (Multi-modal)",
    "question": "There is a stone at each vertex of a given regular $13$-gon, and the color of each stone is black or white. Prove that we may exchange the position of two stones such that the coloring of these stones are symmetric with respect to some symmetric axis of the $13$-gon.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 28370,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let real numbers $a, b, c$ and $d$ satisfy\n$$\nf(x) = a \\cos x + b \\cos 2x + c \\cos 3x + d \\cos 4x \\le 1\n$$\nfor any real number $x$. Find the values of $a, b, c$ and $d$ such that $a + b - c + d$ takes the maximum number. (posed by Li Shenghong)",
    "options": [],
    "answer": "Maximum is 3; a=1, b=1/2, c=−1, d=1/2",
    "solution": "Since\n$$\n\\begin{aligned}\nf(0) &= a + b + c + d, \\\\\nf(\\pi) &= -a + b - c + d, \\\\\nf\\left(\\frac{\\pi}{3}\\right) &= \\frac{a}{2} - \\frac{b}{2} - c - \\frac{d}{2},\n\\end{aligned}\n$$\nthen\n$$\na + b - c + d = f(0) + \\frac{2}{3}f(\\pi) + \\frac{4}{3}f\\left(\\frac{\\pi}{3}\\right) \\le 3\n$$\nif and only if $f(0) = f(\\pi) = f(\\frac{\\pi}{3}) = 1$, that is, if $a = 1$, $b + d = 1$ and $c = -1$, then the equality holds. Let $t = \\cos x$, $-1 \\le t \\le 1$. Then\n$$\n\\begin{aligned}\nf(x) - 1 &= \\cos x + b \\cos 2x - \\cos 3x + d \\cos 4x - 1 \\\\\n&= t + (1-d)(2t^2 - 1) - (4t^3 - 3t) + d(8t^4 - 8t^2 + 1) - 1 \\\\\n&= 2(1 - t^2)[-4dt^2 + 2t + (d-1)] \\le 0, \\forall t \\in [-1, 1],\n\\end{aligned}\n$$\nthat is\n$$\n4dt^2 - 2t + (1-d) \\ge 0, \\forall t \\in (-1, 1).\n$$\nTaking $t = 1/2 + \\epsilon$, $|\\epsilon| < 1/2$, then $\\epsilon[(2d - 1) + 4d\\epsilon] \\ge 0$, $|\\epsilon| < 1/2$. So we see that $d = \\frac{1}{2}$. If $d = \\frac{1}{2}$, then\n$$\n4dt^2 - 2t + (1-d) = 2t^2 - 2t + 1/2 = 2(t - 1/2)^2 \\ge 0.\n$$\nSo, the maximal number of $a+b-c+d$ is $3$, and $(a, b, c, d) = (1, \\frac{1}{2}, -1, \\frac{1}{2})$. $\\square$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 28371,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $P$ be an inner point of an acute triangle $ABC$, $E$ and $F$ be the projections of $P$ onto lines $AC$ and $AB$, respectively, and the lines $BP$ and $CP$ intersect the circumcircle of the triangle $ABC$ at points $B_1$ and $C_1$ ($B_1 \\neq B$, $C_1 \\neq C$), respectively.\nLet $R$ and $r$ denote the radii of the circumcircle and incircle of the triangle $ABC$. Prove that $\\frac{EF}{B_1C_1} \\geq \\frac{r}{R}$. Furthermore, determine all positions of $P$ for which the equality holds.",
    "options": [],
    "answer": "Equality holds if and only if P is the incenter of triangle ABC.",
    "solution": "",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 28372,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $S_n = 1 + \\frac{1}{2} + \\cdots + \\frac{1}{n}$, where $n$ is a positive integer. Prove that for any real numbers $a, b$ with $0 \\le a < b \\le 1$, there are infinite many terms in the sequence $\\{S_n - [S_n]\\}$ that are within $(a, b)$. (Here $[x]$ denotes the largest integer not greater than real number $x$.)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "For any $n \\in \\mathbb{N}^*$, we have\n$$\n\\begin{align*}\nS_{2^n} &= 1 + \\frac{1}{2} + \\frac{1}{3} + \\cdots + \\frac{1}{2^n} = 1 + \\frac{1}{2} + \\left( \\frac{1}{2^1} + \\frac{1}{2^2} \\right) + \\\\\n& \\quad \\left( \\frac{1}{2^{n-1}} + \\frac{1}{2^n} \\right) \\\\\n&> 1 + \\frac{1}{2} + \\left( \\frac{1}{2^2} + \\frac{1}{2^2} \\right) + \\cdots + \\left( \\frac{1}{2^n} + \\cdots + \\frac{1}{2^n} \\right) \\\\\n&= 1 + \\frac{1}{2} + \\frac{1}{2} + \\cdots + \\frac{1}{2} > \\frac{1}{2}n.\n\\end{align*}\n$$\nLet $N_0 = \\lfloor \\frac{1}{b-a} \\rfloor + 1$, $m = \\lfloor S_{N_0} \\rfloor + 1$. Then $\\frac{1}{b-a} < N_0$, $\\frac{1}{N_0} < b-a$, and $S_{N_0} < m \\le m+a$.\nLet $N_1 = 2^{2(m+1)}$. Then $S_{N_1} = S_{2^{2(m+1)}} > m+1 \\ge m+b$.\nWe claim that there exist $n \\in \\mathbb{N}^*$ with $N_0 < n < N_1$ such that $m+a < S_n < m+b$ (or, in other words, $S_n - [S_n] \\in (a, b)$).\nOtherwise, assuming the claim is false, then there must exist $k > N_0$ such that $S_{k-1} \\le m+a$ and $S_k \\ge m+b$.\nThen $S_k - S_{k-1} \\ge b-a$. But it contradicts the fact that\n$$\nS_k - S_{k-1} = \\frac{1}{k} < \\frac{1}{N_0} < b - a.\n$$\nTherefore, the claim is true.\nFurthermore, assume there are only a finite number of positive integers $n_1, \\dots, n_k$ satisfying\n$$\nS_{n_j} - [S_{n_j}] \\in (a, b) \\quad (1 \\le j \\le k).\n$$\nDefine $c = \\min_{1 \\le j \\le k} \\{S_{n_j} - [S_{n_j}]\\}$. Then there exists no $n \\in \\mathbb{N}^*$ such that $S_n - [S_n] \\in (a, c)$. It contradicts the above claim.\nTherefore, there are infinite terms in the sequence $\\{S_n - [S_n]\\}$ that are within $(a, b)$.\nThe proof is complete.\nFor any $n \\in \\mathbb{N}^*$, we have\n$$\n\\begin{align*}\nS_{2^n} &= 1 + \\frac{1}{2} + \\frac{1}{3} + \\dots + \\frac{1}{2^n} = 1 + \\frac{1}{2} + \\left( \\frac{1}{2^1} + \\frac{1}{2^2} \\right) + \\\\\n& \\quad \\left( \\frac{1}{2^{n-1}} + 1 + \\dots + \\frac{1}{2^n} \\right) \\\\\n&> 1 + \\frac{1}{2} + \\left( \\frac{1}{2^2} + \\frac{1}{2^2} \\right) + \\dots + \\left( \\frac{1}{2^n} + \\dots + \\frac{1}{2^n} \\right) \\\\\n&= 1 + \\frac{1}{2} + \\frac{1}{2} + \\dots + \\frac{1}{2} > \\frac{1}{2}n.\n\\end{align*}\n$$\nTherefore, $S_n$ can be larger than any positive number as long as $n$ becomes sufficiently large.\nLet $N_0 = \\lfloor \\frac{1}{b-a} \\rfloor + 1$. Then $\\frac{1}{N_0} < b-a$, and when $k > N_0$, we have\n$$\nS_k - S_{k-1} = \\frac{1}{k} < \\frac{1}{N_0} < b - a.\n$$\nSo for any positive integer $m > S_{N_0}$, there exists $n > N_0$ such that $S_n - m \\in (a, b)$, or, in other words, $m + a < S_n < m + b$. Otherwise, there must be $k > N_0$ such that $S_{k-1} \\le m + a$ and $S_k \\ge m + b$, i.e., $S_k - S_{k-1} \\ge b - a$. But it contradicts the fact that\n$$\nS_k - S_{k-1} = \\frac{1}{k} < \\frac{1}{N_0} < b - a.\n$$\nNow let $m_i = [S_{N_0}] + i$ ($i = 1, 2, 3, \\dots$). Then there exists $n_i > N_0$ such that $m_i + a < S_{n_i} < m_i + b$, i.e., $S_{n_i} - [S_{n_i}] \\in (a, b)$.\nTherefore, there are infinite terms in the sequence $\\{S_n - [S_n]\\}$ that are within $(a, b)$. $\\square$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 28373,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $H$ and $O$ be the orthocenter and circumcenter of an acute triangle $ABC$ ($A$, $H$, $O$ are non-collinear). Suppose that $D$ is the projection of $A$ onto the line $BC$, and the perpendicular bisector of the segment $AO$ meets the line $BC$ at $E$. Prove that the midpoint of $OH$ is on the circumcircle of triangle $ADE$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 28374,
    "subject": "Mathematics (Multi-modal)",
    "question": "A non-negative number $m$ is called a *six match number*. If $m$ and the sum of its digits are both multiples of 6, find the number of the six match numbers less than 2012.",
    "options": [],
    "answer": "168",
    "solution": "Let $n = \\overline{d_1d_2d_3d_4} = 1000d_1 + 100d_2 + 10d_3 + d_4$, $d_1, d_2, d_3, d_4 \\in [0, 1, 2, \\dots, 9]$, and $S(n) = d_1 + d_2 + d_3 + d_4$.\nMatch the non-negative multiples of 6 less than 2000 into 167 pairs $(x, y)$, $x + y = 1998$, such that\n$(0, 1998), (6, 1992), (12, 1986), \\dots, (996, 1002).$\nFor each pair $(x, y)$, let $x = \\overline{a_1a_2a_3a_4}$, $y = \\overline{b_1b_2b_3b_4}$, then\n$$\n1000(a_1 + b_1) + 100(a_2 + b_2) + 10(a_3 + b_3) + (a_4 + b_4) = x + y = 1998.\n$$\nSince $x, y$ are even, $a_4, b_4 \\le 8$. So $a_4 + b_4 \\le 16 < 18$. Thus, $a_4 + b_4 = 8$. Since $a_3 + b_3 \\le 18 < 19$, $a_3 + b_3 = 9$.\nSimilarly, we can obtain $a_2 + b_2 = 9$ and $a_1 + b_1 = 1$. Thus,\n$$\n\\begin{aligned}\nS(x) + S(y) &= (a_1 + b_1) + (a_2 + b_2) + (a_3 + b_3) + (a_4 + b_4) \\\\\n&= 1 + 9 + 9 + 8 = 27.\n\\end{aligned}\n$$\nConsequently, there is only one of $S(x)$ and $S(y)$ that is the multiple of 6. (This is because that $x$, $y$ are all multiples of 3, so are $S(x)$ and $S(y)$.) That is, there is only one of $x$ and $y$ which is a six match number.\nTherefore, there are 167 six match numbers less than 2000, and there is just one six match number between 2000 and 2011.\nTherefore, the answer is $167 + 1 = 168$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 28375,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all integers $k \\ge 3$, with the following properties: There exist integers $m$ and $n$ satisfying $(m, k) = (n, k) = 1$ and $k \\mid (m-1)(n-1)$ with $1 < m < k$, $1 < n < k$ and $m+n > k$.",
    "options": [],
    "answer": "All integers at least three except odd primes, twice an odd prime, and thirty.",
    "solution": "If $k$ has a factor of square number greater than $1$, let $t^2 \\mid k$, $t > 1$, then taking $m = n = k - \\frac{k}{t} + 1$, we see that such $k$ has the properties.\n\nIf $k$ has no factor of square number, if there are two primes $p_1, p_2$ such that $(p_1-2)(p_2-2) \\ge 4$ and $p_1 p_2 \\mid k$. Let $k = p_1 p_2 \\cdots p_r$ and $p_1, p_2, \\dots, p_r$ be pairwise different, $r \\ge 2$. Since there is at least one of $(p_1-1)p_2 p_3 \\cdots p_r + 1$ and $(p_1-2)p_2 p_3 \\cdots p_r + 1$ is coprime with $p_1$ (otherwise $p_1$ divides their difference $p_2 p_3 \\cdots p_r$, which is a contradiction), taking this number as the number $m$, then, $1 < m < k$, $(m, k) = 1$. Similarly, we can take a number $n$ of $(p_2-1)p_1 p_3 \\cdots p_r + 1$ or $(p_2-2)p_1 p_3 \\cdots p_r + 1$ such that $1 < n < k$, $(n, k) = 1$. So $p_1 p_2 \\cdots p_r \\mid (m-1)(n-1)$, and\n$$\nm + n \\ge (p_1 - 2)p_2 p_3 \\cdots p_r + 1 + (p_2 - 2)p_1 p_3 \\cdots p_r + 1 \\\\ = k + ((p_1 - 2)(p_2 - 2) - 4)p_3 \\cdots p_r + 2 > k.\n$$\nSuch $m, n$ will satisfy the conditions.\n\nIf there are no two primes $p_1, p_2$ such that $(p_1-2)(p_2-2) \\ge 4$, $p_1 p_2 \\mid k$, then it is easy to verify such integer $k \\ge 3$ can only be $15$, $30$ or as $p$, $2p$ (where $p$ is an odd prime). It is easy to see that, if $k = p, 2p, 30$, then there are no $m, n$ satisfying the conditions; if $k = 15$, then $m = 11, n = 13$ satisfy the conditions.\n\nSumming up, integer $k \\ge 3$ satisfies the conditions if and only if $k$ is not an odd prime, nor double of an odd prime and nor $30$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 28376,
    "subject": "Mathematics (Multi-modal)",
    "question": "As shown in the figure below, the in-circle $I$ of $\\triangle ABC$ is tangent to sides $AB$ and $AC$ at points $D$ and $E$ respectively, and $O$ is the circum-center of $\\triangle BCI$. Prove that $\\angle ODB = \\angle OEC$.\n\n![](attached_image_1.png)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 28377,
    "subject": "Mathematics (Multi-modal)",
    "question": "Suppose two regular triangular pyramids $P$-$ABC$ and $Q$-$ABC$ sharing the same base are inscribed in the same sphere. If the angle between the side-face and the base of $P$-$ABC$ is $45^\\circ$, then the tangent value of the angle between the side-face and the base of $\\triangle ABC$ is ______.",
    "options": [],
    "answer": "4",
    "solution": "As seen in Fig. 5.1, connecting $PQ$, then $PQ$ is perpendicular to plane $ABC$ with the foot point $H$ being the center of $\\triangle ABC$. The center of the sphere $O$ is also on $PQ$. Connect and extend $CH$ to let it intersect with $AB$ at point $M$. $M$ is then the midpoint of $AB$, and $CM \\perp AB$. It is easy to see that $\\angle PMH$ and $\\angle QMH$ are the plane angles formed by the sides-faces and the bases of the regular triangular pyramids $P$-$ABC$ and $Q$-$ABC$, respectively. Then $\\angle PMH = 45^\\circ$, so $PH = MH = \\frac{1}{2}AH$.\n\n![](attached_image_1.png)\n\nSince $\\angle PAQ = 90^\\circ$, $AH \\perp PQ$, then $AH^2 = PH \\cdot QH$. We then have $AH^2 = \\frac{1}{2}AH \\cdot QH$.\nTherefore, $QH = 2AH = 4MH$.\nFinally, $\\tan \\angle QMH = \\frac{QH}{MH} = 4$.\n\nThe answer is 4.",
    "topic": "Geometry",
    "subtopic": "Solid Geometry"
  },
  {
    "id": 28378,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $n$ be a square-free positive even number, $k$ be an integer, $p$ be a prime number, satisfying $p < 2\\sqrt{n}$, $p \\nmid n$, $p \\mid n + k^2$. Prove that $n$ can be written as $n = ab + bc + ca$, where $a, b, c$ are distinctive positive integers.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Since $n$ is even, we have $p \\neq 2$. As $p \\nmid n$, we have $p \\nmid k$. We may assume without loss of generality $0 < k < p$. Set $a = k$, $b = p - k$, then $c = \\frac{n - k(p - k)}{p} = \\frac{n + k^2}{p} - k$.\n\nBy assumption, $c$ is an integer, and $a, b$ are distinct positive integers. It remains to be shown that $c > 0$, and $c \\neq a$, $b$. By the AM-GM inequality, we have $\\frac{n}{k} + k \\ge 2\\sqrt{n} > p$, thus $n + k^2 > pk$, hence $c > 0$. If $c = a$, then $\\frac{n + k^2}{p} - k = k$, thus $n = k(2p - k)$. Since $n$ is even, $k$ is also even, as a consequence $n$ is divisible by $4$, which contradicts the fact that $n$ is square-free. If $c = b$, then $n = p^2 - k^2$. Since $n$ is even, $k$ is odd, implying that $n$ is again divisible by $4$, which is a contradiction.\n\nWe conclude that $a, b, c$ satisfy all the requirements, completing the proof. □",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 28379,
    "subject": "Mathematics (Multi-modal)",
    "question": "There are $n$ cities ($n \\ge 3$) and two airline companies in a country. Between any two cities, there is exactly one 2-way flight connecting them which is operated by one of the two companies. A female mathematician plans a travel route, so that it starts and ends at the same city, passes through at least two other cities, and each city in the route is visited once. She finds out that wherever she starts and whatever route she chooses, she must take flights of both companies. Find the maximum value of $n$.",
    "options": [],
    "answer": "4",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 28380,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $f(x)$ be an odd function on $\\mathbb{R}$, and $f(x) = x^2$ for $x \\ge 0$. Suppose for any $x \\in [a, a+2]$, $f(x+a) \\ge 2f(x)$. Then the range of real number $a$ is ______.",
    "options": [],
    "answer": "[√2, +∞)",
    "solution": "According to the given condition, we have\n$$\nf(x) = \\begin{cases} x^2 & (x \\ge 0), \\\\ -x^2 & (x < 0). \\end{cases}\n$$\nSo $2f(x) = f(\\sqrt{2}x)$. Therefore, the original inequality is equivalent to $f(x+a) \\ge f(\\sqrt{2}x)$.\n\nAs $f(x)$ is increasing over $\\mathbb{R}$, then $x + a \\ge \\sqrt{2}x$, i.e.,\n$$a \\ge (\\sqrt{2} - 1)x.$$\nFurthermore, since $x \\in [a, a+2]$, $(\\sqrt{2}-1)x$ reaches $(\\sqrt{2}-1)(a+2)$ the maximum value when $x = a+2$.\nTherefore, $a \\ge (\\sqrt{2}-1)(a+2)$, from which we obtain $a \\ge \\sqrt{2}$, i.e., $a \\in [\\sqrt{2}, +\\infty)$.\n\nThe answer is then $[\\sqrt{2}, +\\infty)$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 28381,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find the minimum positive integer $n$ such that\n$$\n\\sqrt{\\frac{n-2011}{2012}} - \\sqrt{\\frac{n-2012}{2011}} < \\sqrt[3]{\\frac{n-2013}{2011}} - \\sqrt[3]{\\frac{n-2011}{2013}}\n$$",
    "options": [],
    "answer": "4024",
    "solution": "We see that if $2012 \\le n \\le 4023$, then $\\sqrt{\\frac{n-2011}{2012}} - \\sqrt{\\frac{n-2012}{2011}} \\ge 0$ and $\\sqrt[3]{\\frac{n-2013}{2011}} - \\sqrt[3]{\\frac{n-2011}{2013}} < 0$.\n\nOtherwise,\n$$\n\\begin{aligned}\n\\sqrt{\\frac{n-2011}{2012}} &\\le \\sqrt{\\frac{n-2012}{2011}} &&\\Leftrightarrow n > 4023 \\\\\n\\sqrt[3]{\\frac{n-2013}{2011}} &\\ge \\sqrt[3]{\\frac{n-2011}{2013}} &&\\Leftrightarrow n \\ge 4024.\n\\end{aligned}\n$$\nThus, if $n \\ge 4024$, then\n$$\n\\sqrt{\\frac{n-2011}{2012}} - \\sqrt{\\frac{n-2012}{2011}} < 0 \\le \\sqrt[3]{\\frac{n-2013}{2011}} - \\sqrt[3]{\\frac{n-2011}{2013}}\n$$\nSo the minimum of $n$ is $4024$. $\\boxed{4024}$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 28382,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find the smallest positive integer $k$ with the following property: for any $k$ element subset $A$ of the set $S = \\{1, 2, \\dots, 2012\\}$, there exist three pairwise distinct elements $a, b, c$ of $S$ such that $a+b$, $b+c$, $c+a$ all belong to $A$.",
    "options": [],
    "answer": "1008",
    "solution": "Without loss of generality, we may assume $a < b < c$. Write $x = a+b$, $z = b+c$, $y = a+c$, then $x < y < z$, $x+y > z$, and $x+y+z$ is even. On the other hand, if there exist $x, y, z \\in A$ such that $x < y < z$, $x+y > z$, and $x+y+z$ is even, set $a = \\frac{x+y-z}{2}$, $b = \\frac{x+z-y}{2}$, $c = \\frac{y+z-x}{2}$, it is clear that $a, b, c$ are pairwise distinct elements of $S$, and $x = a+b$, $y = a+c$, $z = b+c$.\n\nThe required property is equivalent to the following: for any $k$-element subset $A$ of $S$, there exist three elements $x, y, z \\in A$ such that\n$$\nx < y < z,\\ x + y > z,\\ \\text{ and } x + y + z \\text{ is even.}\\quad (*)\n$$\nIf $A = \\{1, 2, 3, 5, 7, \\dots, 2011\\}$, $|A| = 1007$, and $A$ does not contain three elements satisfying property $(*)$. Therefore, $k \\ge 1008$.\n\nWe next prove that any 1008-element subset of $S$ contains three elements satisfying property $(*)$.\nWe prove a general statement: For any integer $n \\ge 4$, any $(n + 2)$-element subset of $\\{1, 2, \\dots, 2n\\}$ contains three elements satisfying $(*)$. We induct on $n$.\n\nWhen $n = 4$, let $A$ be a 6-element subset of $\\{1, 2, \\dots, 8\\}$, then $A \\cap \\{3, 4, 5, 6, 7, 8\\}$ contains at least four elements. If $A \\cap \\{3, 4, 5, 6, 7, 8\\}$ contains three even numbers, then $4, 6, 8 \\in A$ satisfying $(*)$. If $A \\cap \\{3, 4, 5, 6, 7, 8\\}$ contains exactly two even numbers, then it contains two odd numbers. For any two odd numbers $x, y$ of $\\{3, 5, 7\\}$, two of $(4, x, y)$, $(6, x, y)$, $(8, x, y)$ would satisfy property $(*)$, thus one of them is contained in $A$. If $A \\cap \\{3, 4, 5, 6, 7, 8\\}$ contains exactly one even number $x$, then it contains all three odd numbers, then $(x, 5, 7)$ satisfies $(*)$. The result holds for $n = 4$.\n\nAssuming the result holds for $n$ ($n \\ge 4$), consider the case of $n + 1$. Let $A$ be $(n + 3)$-elements of $\\{1, 2, \\dots, 2n + 2\\}$, if $|A \\cap \\{1, 2, \\dots, 2n\\}| \\ge n + 2$. By inductive hypothesis, the result follows. It remains to consider $|A \\cap \\{1, 2, \\dots, 2n\\}| = n + 1$, and $2n + 1, 2n + 2 \\in A$. If $A$ contains an odd number $x$ in $\\{1, 2, \\dots, 2n\\}$, then $x, 2n + 1, 2n + 2$ satisfy $(*)$; if no odd number of $\\{1, 2, \\dots, 2n\\}$ greater than 1 is contained in $A$, then $A = \\{1, 2, 4, 6, \\dots, 2n, 2n + 1, 2n + 2\\}$, and $4, 6, 8 \\in A$ satisfy $(*)$.\n\nHence, the smallest $k$ with the required property is $1008$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 28383,
    "subject": "Mathematics (Multi-modal)",
    "question": "Sequence $\\{a_n\\}$ is defined by:\n$$\na_0 = \\frac{1}{2},\\ a_{n+1} = a_n + \\frac{a_n^2}{2012},\\ n = 0, 1, \\dots\n$$\n\nFind the integer $k$ such that $a_k < 1 < a_{k+1}$.",
    "options": [],
    "answer": "2012",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 28384,
    "subject": "Mathematics (Multi-modal)",
    "question": "Suppose there are beetles on a chessboard consisting of $2012 \\times 2012$ unit squares. Each unit square can accommodate at most one beetle. At a moment, all beetles fly and land on the chessboard again. For a beetle, we call the vector from its flying unit to its landing unit the beetle's \"displacement vector\". We call the sum of all beetle's \"displacement vectors\" the \"total displacement vectors\".\nFind the maximum length of \"total displacement vector\" considering the number of beetles and all possible positions of flying and landing. (posed by Qu Zhenhua)",
    "options": [],
    "answer": "2*1006^3",
    "solution": "Set up a coordinate system with origin at the center of chessboard $O$ and the grid line as the coordinate line. Denote the set of the centers of squares by $S$, and the set where the beetles initially stand on by $M_1 \\subseteq S$, and the set that the beetles land on by $M_2 \\subseteq S$. Let $f: M_1 \\to M_2$ be the one-to-one mapping defined by a beetle's position $v$ at the beginning to the position $u = f(v)$ of first landing. Thus, the total displacement vector is given by\n$$\nV = \\sum_{v \\in M_1} (f(v) - v) = \\sum_{u \\in M_2} u - \\sum_{v \\in M_1} v. \\qquad \\textcircled{1}\n$$\nNote that the right-hand side of ① is independent of $f$. We need only to find the maximum of $|V| = |\\sum_{u \\in M_2} u - \\sum_{v \\in M_1} v|$ for all $M_1, M_2 \\subseteq S$, $|M_1| = |M_2|$. We may suppose that $M_1 \\cap M_2 = \\emptyset$, since element of $M_1 \\cap M_2$ does not change $|V|$. Suppose that $|V|$ attains its maximum at $(M_1, M_2)$, obviously $V \\neq 0$. Let line $l \\perp V$ be at point $O$.\n\n**Lemma 1.** Line $l$ does not pass any point of $S$. $M_1$ is the set of $S$ on one side of $l$, and $M_2$ is the set of $S$ on the other side of $l$.\n\n**Proof of Lemma 1.** First, $M_1 \\cup M_2 = S$. Otherwise, since $|S|$ is even, there are at least two points $a$ and $b$ which are not in $M_1 \\cup M_2$. Suppose that the angle between $a-b$ and $V$ does not exceed $90^\\circ$, then $|V + (a-b)| > |V|$. So add $a$ into $M_2$, add $b$ into $M_1$, then $|V|$ will increase, which is a contradiction.\n\nSecond, $M_2 = -M_1$. Otherwise, there exist $a, b \\in S$, such that $a, -a \\in M_1$ and $b, -b \\in M_2$. Suppose that the angle between $a-b$ and $V$ does not exceed $90^\\circ$. Put $a$ into $M_2$, and $b$ into $M_1$, $V$ changes to $V+2(a-b)$, and $|V+2(a-b)| > |V|$, which is a contradiction.\n\nThird, $l$ does not pass any point of $S$. Otherwise, let $l$ pass $a, a \\in M_1, -a \\in M_2$, then change $a$ into $M_2, -a$ into $M_1$, $V$ changes to $V+4a$. Notice that $a \\perp V$, so $|V+4a| > |V|$, which is a contradiction.\n\nFourth, we show that $M_2$ is the set of $S$ on one side of $l$ (the side that $V$ is pointing to), and $M_1$ is the set of $S$ on the other side of $l$. Otherwise, there is $a \\in M_1$ at the side that $V$ is pointing to. And there is a $b \\in M_2$ on the other side. Then the angle between $a-b$ and $V$ is less than $90^\\circ$. Change $a$ into $M_2$ and $b$ into $M_1$, then $V$ changes into $V+2(a-b)$, the length of which is greater, which is a contradiction. Lemma 1 is now proved.\n\n**Lemma 2.** Let $S_k = \\{(x, y) \\in S \\mid |x| = k - \\frac{1}{2} \\text{ or } |y| = k - \\frac{1}{2}\\}, k = 1, 2, \\dots, 1006, l$ be a line passing $O$ and does not pass points of $S_k$. Denote all points of $S_k$ on one side of $l$ by $A_k$, all points of $S_k$ on the other side by $B_k$. Denote $V_k = \\sum_{u \\in A_k} u - \\sum_{v \\in B_k} v$, then the maximum of $|V_k|$ is obtained when $l$ is horizontal (or vertical), and $V_k$ is vertical (or horizontal).\n\n**Proof of Lemma 2.** $S_k$ is located at the boundary of a square with $2k$ points on each side. Let the four vertices of the square be $A(k - \\frac{1}{2}, k - \\frac{1}{2})$, $B(-k + \\frac{1}{2}, k - \\frac{1}{2})$, $C(-k + \\frac{1}{2}, -k + \\frac{1}{2})$ and $D(k - \\frac{1}{2}, -k + \\frac{1}{2})$. By\n![](attached_image_1.png)\nFig. 6.1\nsymmetricity, we may suppose that $l$ intersects $AD$ at point $P$ with non-negative slope. Let $P$ be located between the $t$ ($1 \\le t \\le k$)th (from top to bottom) of $S_k$ on $AD$ and the $(t+1)$th of $S_k$ on $AD$ (see Fig. 6.1, in case of $k = 6, t = 3$). Thus,\n$$\n\\begin{aligned}\nV_k &= (2k-2)(2k-1)\\vec{j} + (2k-t)(-(2k-1)\\vec{i} + t\\vec{j}) + \\\\\n&\\quad t((2k-1)\\vec{i} + (2k-t)\\vec{j}) \\\\\n&= -2(2k-1)(k-t)\\vec{i} + 2(-(k-t)^2 + 3k^2 - 3k + 1)\\vec{j},\n\\end{aligned}\n$$\nwhere $\\vec{i}$ and $\\vec{j}$ are horizontal and vertical unit vectors, respectively. Denote $(k-t)^2 = u$, $0 \\le u \\le (k-1)^2$, then\n$$\n\\begin{aligned}\n\\frac{1}{4} |V_k|^2 &= (2k-1)^2 u + u^2 - 2(3k^2 - 3k + 1)u + (3k^2 - 3k + 1)^2 \\\\\n&= u^2 - (2k^2 - 2k + 1)u + (3k^2 - 3k + 1)^2.\n\\end{aligned}\n$$\nAs a quadratic function of $u$, $u = k^2 - k + \\frac{1}{2}$ is the symmetric axis. It is easy to know that $|V_k|^2$ takes its maximum at $u = 0$. So $t = k$, that is, $l$ is horizontal, so $V_k$ is vertical. Lemma 2 is proved.\n\nTurn to the original problem. By symmetricity, we need to only consider that the slope of $l$ is non-negative and less than 1. Let $M_1, M_2$ be located on two sides of $l$. Denote $M_2 \\cap S_k = A_k, M_1 \\cap S_k = B_k, V_k = \\sum_{u \\in A_k} u - \\sum_{v \\in B_k} v$, then\n$$\n|V| = \\left| \\sum_{k=1}^{1006} V_k \\right| \\le \\sum_{k=1}^{1006} |V_k|. \\qquad \\textcircled{2}\n$$\nSo, $|V|_{\\max} \\le \\sum_{k=1}^{1006} |V_k|_{\\max}$. If $l$ is horizontal, $M_2$ is all the points of $S$ on the upper half-plane, $M_1$ is all points of $S$ on the lower half-plane; each $|V_k|$ takes its maximum, and all $V_k$ point upward. And the equality of ② holds. So $|V|$ indeed takes the maximum $|V|_{\\max} = 2 \\times 1006^3$.\n\n$\\boxed{2 \\times 1006^3}$",
    "topic": "Algebra",
    "subtopic": "Linear Algebra"
  },
  {
    "id": 28385,
    "subject": "Mathematics (Multi-modal)",
    "question": "The sum of all the positive integers $n$ satisfying $\\frac{1}{4} < \\sin \\frac{\\pi}{n} < \\frac{1}{3}$ is ______.",
    "options": [],
    "answer": "33",
    "solution": "As $\\sin x$ is a convex function for $x \\in (0, \\frac{\\pi}{6})$, we have $\\frac{3}{\\pi}x < \\sin x < x$. Then\n$$\n\\sin \\frac{\\pi}{13} < \\frac{\\pi}{13} < \\frac{1}{4}, \\sin \\frac{\\pi}{12} > \\frac{3}{\\pi} \\times \\frac{\\pi}{12} = \\frac{1}{4},\n$$\n$$\n\\sin \\frac{\\pi}{10} < \\frac{\\pi}{10} < \\frac{1}{3}, \\sin \\frac{\\pi}{9} > \\frac{3}{\\pi} \\times \\frac{\\pi}{9} = \\frac{1}{3},\n$$\nthat is\n$$\n\\sin \\frac{\\pi}{13} < \\frac{1}{4} < \\sin \\frac{\\pi}{12} < \\sin \\frac{\\pi}{11} < \\sin \\frac{\\pi}{10} < \\frac{1}{3} < \\sin \\frac{\\pi}{9}.\n$$\nTherefore, all the possible values of positive integers $n$ are\n$10, 11, 12$, and their sum is 33.\nThe answer is 33.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 28386,
    "subject": "Mathematics (Multi-modal)",
    "question": "In $\\triangle ABC$ with $AB = 1$, let $D$ be a point on $AC$ such that $\\angle ABD = \\angle C$ and let $E$ be a point on $AB$ such that $BE = DE$. Let $H$ be a point on $DE$ such that $AH \\perp DE$ and $M$ be the midpoint of $CD$. If $AH = 2 - \\sqrt{3}$, find the size of $\\angle AME$. (posed by Xiong Bin)\n\n![](attached_image_1.png)",
    "options": [],
    "answer": "15°",
    "solution": "Let $\\angle ABD = \\angle C = \\alpha$ and $\\angle DBC = \\beta$. It is easy to see that $\\angle BDE = \\alpha$, $\\angle AED = 2\\alpha$,\n$$\n\\angle ADE = \\angle ADB - \\angle BDE = (\\alpha + \\beta) - \\alpha = \\beta,\n$$\n$$\nAB = AE + EB = AE + EH + HD.\n$$\nHence,\n$$\n\\frac{AB}{AH} = \\frac{AE + EH}{AH} + \\frac{HD}{AH} = \\frac{1 + \\cos 2\\alpha}{\\sin 2\\alpha} + \\cot \\beta \\\\ = \\cot \\alpha + \\cot \\beta. \\qquad \\textcircled{1}\n$$\nDraw lines $EK \\perp AC$ and $EL \\perp BD$ with pedals $K$ and $L$, respectively. Then, $L$ is the midpoint of $BD$. Combining with the Sine Theorem, we obtain\n\n![](attached_image_2.png)\n\n$$\n\\frac{EL}{EK} = \\frac{DE \\sin \\angle EDL}{DE \\sin \\angle EDK} = \\frac{\\sin \\alpha}{\\sin \\beta} \\\\ = \\frac{BD}{CD} = \\frac{LD}{MD}.\n$$\nThus,\n$$\n\\cot \\alpha = \\frac{LD}{EL} = \\frac{MD}{EK} = \\frac{MK}{EK} - \\frac{DK}{EK} = \\cot \\angle AME - \\cot \\beta. \\quad \\textcircled{2}\n$$\nBy $\\textcircled{1}$, $\\textcircled{2}$ and known conditions, we have\n$$\n\\cot \\angle AME = \\frac{AB}{AH} = \\frac{1}{2 - \\sqrt{3}} = 2 + \\sqrt{3}.\n$$\nTherefore, $\\angle AME = 15^\\circ$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 28387,
    "subject": "Mathematics (Multi-modal)",
    "question": "In a $n \\times n$ chart, two cells of the chart are called adjacent if they have a common side. At the origin, there is a number $+1$ in every cell. An operation means that: one can choose a cell, and change the signs of all the numbers which are in the adjacent cells of it (doesn't change the sign of the number in it). Find all integers $n \\ge 2$, for which, after finitely many suitable operations, the numbers in all cells are changed to $-1$.",
    "options": [],
    "answer": "All n ≥ 2 with n not congruent to 1 modulo 4",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 28388,
    "subject": "Mathematics (Multi-modal)",
    "question": "Given an $n \\times n$ grid, we call two cells in it adjacent if they have a common side. At the beginning, each cell is assigned number $+1$. An operation on the grid is defined as follows: one chooses a cell, and then changes the signs of every number in its adjacent cells (but does not change the sign of the number in itself). Find all the integers $n \\ge 2$, such that after a finite number of operations, all the numbers in the cells of the grid are changed to $-1$. (posed by Shen Huyue)",
    "options": [],
    "answer": "all even integers n ≥ 2",
    "solution": "We denote the cell in the $i$-row and $j$-column of the grid as $A_{ij}$ ($i, j \\in \\{1, 2, \\dots, n\\}$).\n\nWhen $n = 2k$, $k \\in \\mathbb{N}^*$, we mark each $A_{ij}$ satisfying $i + j \\equiv 0 \\pmod{2}$ with color red (presented by shaded areas), and that satisfying $j - i \\equiv 3 \\pmod{4}$ and $j - i \\not\\equiv j + i \\pmod{4}$ with blue (presented by oblique line areas) (see Fig. 7.1).\n\n![](attached_image_1.png)\nFig. 7.1\n![](attached_image_2.png)\nFig. 7.2\n\nIn this way, we can see that every cell adjacent to a blue one is red, and there is exactly one blue cell around each red one.\n\nNow we do the operation on each blue cell. The number in each red cell is then changed from $+1$ to $-1$, while the numbers in the remaining cells are unchanged.\n\nSince $n$ is an even number, we can rotate the grid around its center $O$ anticlockwise by $90^\\circ$. Then all the red cells of the rotated grid cover exactly all the cells that are not red in the original grid (see Fig. 7.2).\n\nWe do the operations again for the original grid on all the cells that are covered by blue cells of the rotated grid. Then we see that all the numbers with value $+1$ in the remaining cells of the original grid are changed to $-1$, while the numbers in the other cells are unchanged.\n\nTherefore, when $n$ is an even number, all the numbers in the cells of the grid can be changed to $-1$ by a finite number of operations.\n\nWhen $n$ is odd, we denote the number in cell $A_{ii}$ as $M_i$ ($i = 1, 2, \\dots, n$), and denote the number of operations on each of their adjacent cells as $x_1, x_2, \\dots, x_{n-1}, y_1, y_2, \\dots, y_{n-1}$, respectively (see Fig. 7.3).\n\nAfter finite operations, it is easy to see that:\n\n![](attached_image_3.png)\nFig. 7.3\n\n$M_1$ is changed from $+1$ to $-1$ if and only if $x_1 + y_1$ is odd;\n\n$M_2$ is changed from $+1$ to $-1$ if and only if $x_1 + y_1 + x_2 + y_2$ is odd;\n\n$M_3$ is changed from $+1$ to $-1$ if and only if $x_2 + y_2 + x_3 + y_3$ is odd;\n\n$\\vdots$\n\n$M_{n-1}$ is changed from $+1$ to $-1$ if and only if $x_{n-2} + y_{n-2} + x_{n-1} + y_{n-1}$ is odd, and\n\n$M_n$ is changed from $+1$ to $-1$ if and only if $x_{n-1} + y_{n-1}$ is odd.\n\nSince $n$ is odd, the sum of $n$ odd numbers is still odd. Then,\n\n$$\n\\begin{align*}\n& (x_1 + y_1) + (x_1 + y_1 + x_2 + y_2) + (x_2 + y_2 + x_3 + y_3) \\\\\n& \\quad + \\dots + (x_{n-2} + y_{n-2} + x_{n-1} + y_{n-1}) + (x_{n-1} + y_{n-1}) \\\\\n& = 2(x_1 + x_2 + \\dots + x_{n-1} + y_1 + y_2 + \\dots + y_{n-1})\n\\end{align*}\n$$\n\nis odd. It is impossible!\n\nTherefore, all the numbers in the cells of a given $n \\times n$ grid can be changed from $+1$ to $-1$ after a finite number of operations if and only if $n$ is an even number. $\\boxed{}$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 28389,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $a_1 \\le a_2 \\le \\dots$ be a sequence of positive integers such that $\\frac{r}{a_r} = k+1$ for some positive integers $k$ and $r$. Prove that there exists a positive integer $s$ such that $\\frac{s}{a_s} = k$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 28390,
    "subject": "Mathematics (Multi-modal)",
    "question": "An information station employs four different codes, $A$, $B$, $C$ and $D$, for communication, but each week uses only one of them. The code used in a definite week is randomly selected with equal chance among the three ones that have not been used in the last week. Suppose the code used in the first week is $A$. Then the probability that $A$ is also used in the seventh week is _______. (expressed as an irreducible fraction)",
    "options": [],
    "answer": "61/243",
    "solution": "Let $P_k$ denote the probability that code $A$ is used in the $k$\\text{th}$ week. Then the probability that $A$ is not used in the $k$\\text{th}$ week is $1 - P_k$. Therefore, we have\n$$\nP_{k+1} = \\frac{1}{3}(1 - P_k).\n$$\nOr\n$$\nP_{k+1} - \\frac{1}{4} = -\\frac{1}{3}\\left(P_k - \\frac{1}{4}\\right).\n$$\nAs $P_1 = 1$, $\\{P_k - \\frac{1}{4}\\}$ is then a geometric sequence with $\\frac{3}{4}$ as the first term and $-\\frac{1}{3}$ as the common ratio. So we have\n$$\nP_k - \\frac{1}{4} = \\frac{3}{4}\\left(-\\frac{1}{3}\\right)^{k-1}.\n$$\nOr\n$$\nP_k = \\frac{3}{4}\\left(-\\frac{1}{3}\\right)^{k-1} + \\frac{1}{4}.\n$$\nTherefore, $P_7 = \\frac{61}{243}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 28391,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find the number of integers $k$ in the set $\\{0, 1, 2, \\ldots, 2012\\}$ such that the combination number $\\binom{2012}{k} = \\frac{2012!}{k!(2012-k)!}$ is a multiple of $2012$.",
    "options": [],
    "answer": "1370",
    "solution": "",
    "topic": "Discrete Mathematics",
    "subtopic": "Combinatorics"
  },
  {
    "id": 28392,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all prime numbers $p$, for which there are infinitely many positive integers $n$, such that: $p \\mid n^{n+1} + (n+1)^n$.",
    "options": [],
    "answer": "All odd primes",
    "solution": "",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 28393,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $m$ be a positive integer, $n = 2^m - 1$, and $P_n = \\{1, 2, \\dots, n\\}$ be the set of $n$ points on the number axis. A grasshopper jumps between adjacent points on $P_n$. Find the maximal number of $m$ such that for any $x, y \\in P_n$, the number of ways that a grasshopper jumping from $x$ to $y$ by 2012 steps is even (passing $x$ or $y$ on the way is permitted). (posed by Zhang Sihui)",
    "options": [],
    "answer": "10",
    "solution": "If $m \\ge 11$, then $n = 2^m - 1 > 2013$. Since there is only one way a grasshopper jumps from point $1$ to point $2013$ by $2012$ steps, we see that $m \\le 10$.\n\nIn the following, we show that the answer is $m = 10$. To show this, we will prove a stronger proposition by induction on $m$: for any $k \\ge n = 2^m - 1$ and any $x, y \\in P_n$, the number of ways the grasshopper jumps from point $x$ to point $y$ by $k$ steps is even.\n\nIf $m = 1$, the number of ways is $0$, where $0$ is even.\n\nIf $m = l$, the number of ways is even. Then, for $k \\ge n = 2^{l+1} - 1$, there are three kinds of routes from point $x$ to point $y$ by $k$ steps. We show that the number of ways is even for each kind of route.\n\n(1) The route does not pass point $2^l$. So points $x$ and $y$ both are on one side of point $2^l$. By the induction hypothesis, there are even routes.\n\n(2) The route passes point $2^l$ just once.\nSuppose that the grasshopper is at point $2^l$ at the $i$-th step.\n\n$(i \\in \\{0, 1, \\dots, k\\}, i = 0 \\text{ means } x = 2^l, i = k \\text{ means } y = 2^l)$.\nWe show that, for any $i$, the number of routes is even.\n\nSuppose that the route is $x, a_1, \\dots, a_{i-1}, 2^l, a_{i+1}, \\dots, a_{k-1}, y$. Divide it into two sub-routes: from point $x$ to point $a_{i-1}$ of $i-1$ steps and from point $a_{i+1}$ to point $y$ of $k-i-1$ steps (for $i=0$ or $k$, only one sub-route of $k-1$ steps).\n\nIf $i-1 < 2^l - 1$ and $k-i-1 < 2^l - 1$, then $k \\le 2^{l+1} - 2$, which contradicts $k \\ge n = 2^{l+1} - 1$. So, we must have $i-1 \\ge 2^l - 1$ or $k-i-1 \\ge 2^l - 1$. By the induction hypothesis, there are even ways for a sub-route. So, by the Multiplication Principle, the number of ways is even.\n\n(3) The route passes point $2^l$ no less than two times.\nConsider the sub-routes from $2^l$ to $2^l$, the number of ways is even, since we can consider the routes symmetric to $2^l$. So by the Multiplication Principle, the number of ways is even.\n\nSumming up, the maximal $m$ is $10$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 28394,
    "subject": "Mathematics (Multi-modal)",
    "question": "Suppose $f(x) = a \\sin x - \\frac{1}{2} \\cos 2x + a - \\frac{3}{a} + \\frac{1}{2}$, $a \\in \\mathbf{R}$, $a \\neq 0$.\n\n(1)\nIf $f(x) \\le 0$ for any $x \\in \\mathbf{R}$, find the range of $a$.\n\n(2)\nIf $a \\ge 2$ and there exists $x \\in \\mathbf{R}$ such that $f(x) \\le 0$, find the range of $a$.",
    "options": [],
    "answer": "(1) a in (0, 1]. (2) a in [2, 3].",
    "solution": "(1) We have $f(x) = \\sin^2 x + a \\sin x + a - \\frac{3}{a}$. Let $t = \\sin x$ ($-1 \\le t \\le 1$). Then\n$$\ng(t) = t^2 + a t + a - \\frac{3}{a}.\n$$\nThe sufficient and necessary condition for $f(x) \\le 0, \\forall x \\in \\mathbf{R}$ is\n$$\n\\begin{cases} g(-1) = 1 - \\frac{3}{a} \\le 0, \\\\ g(1) = 1 + 2a - \\frac{3}{a} \\le 0. \\end{cases}\n$$\nTherefore, we obtain the range of $a$ is $(0, 1]$.\n\n\n(2)\nAs $a \\ge 2$, then $-\\frac{a}{2} \\le -1$. We have\n$$\ng(t)_{\\min} = g(-1) = 1 - \\frac{3}{a}.\n$$\nThen $f(x)_{\\min} = 1 - \\frac{3}{a}$. Therefore, the sufficient and necessary condition for $f(x) \\le 0, \\exists x \\in \\mathbf{R}$ is $1 - \\frac{3}{a} \\le 0$, or $0 < a \\le 3$.\nFinally, we obtain that the range of $a$ is $[2, 3]$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 28395,
    "subject": "Mathematics (Multi-modal)",
    "question": "As seen in Fig. 1.1, $AB$ is a chord of circle $\\omega$, $P$ is a point on arc $AB$, and $E, F$ are 2 points on $AB$ satisfying $AE = EF = FB$. Connect $PE, PF$ and extend them to intersect with $\\omega$ at $C, D$, respectively. Prove\n$$\nEF \\cdot CD = AC \\cdot BD.\n$$\n\n![](attached_image_1.png)\nFig. 1.1",
    "options": [],
    "answer": "Detailed solution",
    "solution": "As shown in Fig. 1.2, we connect $AD$, $BC$, $CF$, $DE$. Since $AE = EF = FB$, we have\n$$\n\\frac{BC \\cdot \\sin \\angle BCE}{AC \\cdot \\sin \\angle ACE} = \\frac{\\text{distance between } B \\text{ and } CP}{\\text{distance between } A \\text{ and } CP} = \\frac{BE}{AE} = 2. \\qquad \\textcircled{1}\n$$\nIn the same way,\n$$\n\\frac{AD \\cdot \\sin \\angle ADF}{BD \\cdot \\sin \\angle BDF} = \\frac{\\text{distance between } A \\text{ and } PD}{\\text{distance between } B \\text{ and } PD} = \\frac{AF}{BF} = 2. \\qquad \\textcircled{2}\n$$\nOn the other hand, since\n$$\n\\angle BCE = \\angle BCP = \\angle BDP = \\angle BDF,\n$$\n$$\n\\angle ACE = \\angle ACP = \\angle ADP = \\angle ADF,\n$$\nmultiplying ① by ②, we have $\\frac{BC \\cdot AD}{AC \\cdot BD} = 4$, or\n$$\nBC \\cdot AD = 4AC \\cdot BD. \\qquad \\textcircled{3}\n$$\nBy Ptolemy's Theorem, we have\n$$\nAD \\cdot BC = AC \\cdot BD + AB \\cdot CD. \\qquad \\textcircled{4}\n$$\nCombining ③ and ④, we get $AB \\cdot CD = 3AC \\cdot BD$, and that is\n$$\nEF \\cdot CD = AC \\cdot BD.\n$$\nThe proof is complete.\n\n![](attached_image_2.png)\nFig. 1.2",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 28396,
    "subject": "Mathematics (Multi-modal)",
    "question": "Two circles $K_1$ and $K_2$ of different radii intersect at two points $A$ and $B$, let $C$ and $D$ be two points on $K_1$ and $K_2$, respectively, such that $A$ is the midpoint of the segment $CD$. The extension of $DB$ meets $K_1$ at another point $E$, the extension of $CB$ meets $K_2$ at another point $F$. Let $l_1$ and $l_2$ be the perpendicular bisectors of $CD$ and $EF$, respectively.\n\n(1) Show that $l_1$ and $l_2$ have a unique common point (denoted by $P$).\n\n(2) Prove that the lengths of $CA$, $AP$ and $PE$ are the side lengths of a right triangle.\n\n![](attached_image_1.png)\n",
    "options": [],
    "answer": "Detailed solution",
    "solution": "(1) Since $C$, $A$, $B$, $E$ are concyclic, and $D$, $A$, $B$, $F$ are concyclic, $CA = AD$, and by the theorem of power of a point, we have\n$$\nCB \\cdot CF = CA \\cdot CD = DA \\cdot DC = DB \\cdot DE. \\qquad \\textcircled{1}\n$$\nSuppose on the contrary that $l_1$ and $l_2$ do not intersect, then $CD \\parallel EF$, hence $\\frac{CF}{CB} = \\frac{DE}{DB}$. Plugging into (1), we get $CB^2 = DE^2$, thus $CB = DB$, hence $BA \\perp CD$. It follows that $CB$ and $DB$ are the diameters of $K_1$ and $K_2$ respectively, hence $K_1$ and $K_2$ have same radii, which contradicts with assumption. Thus $l_1$ and $l_2$ have a unique common point.\n\n(2) Join $AE$, $AF$ and $PF$, we have\n$$\n\\angle CAE = \\angle CBE = \\angle DBF = \\angle DAF.\n$$\nSince $AP \\perp CD$, $AP$ is the bisector of $\\angle EAF$. Since $P$ is on the perpendicular bisector of the segment $EF$, $P$ is on the circumcircle of $\\triangle AEF$. We have\n$$\n\\angle EPF = 180^\\circ - \\angle EAF = \\angle CAE + \\angle DAF = 2\\angle CAE = 2\\angle CBE.\n$$\nHence $B$ is on the circle with center $P$ and radius $PE$, denoting this circle by $\\Gamma$. Let $R$ be the radius of $\\Gamma$. By the\n\ntheorem of power of a point, we have\n$$\n2CA^2 = CA \\cdot CD = CB \\cdot CF = CP^2 - R^2,\n$$\nthus\n$$\nAP^2 = CP^2 - CA^2 = (2CA^2 + R^2) - CA^2 = CA^2 + PE^2.\n$$\nIt follows that $CA$, $AP$, $PE$ form the side lengths of a right triangle.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 28397,
    "subject": "Mathematics (Multi-modal)",
    "question": "Given any $n$ ($> 1$) coprime positive integers $a_1, a_2, \\dots, a_n$, denote $A = a_1 + a_2 + \\dots + a_n$. Let $d_i = (A, a_i)$ (the greatest common divisor), $i = 1, 2, \\dots, n$.\nLet $D_i$ be the greatest common divisor of $\\{a_1, a_2, \\dots, a_n\\} \\setminus \\{a_i\\}$, $i = 1, 2, \\dots, n$. Find the minimum of $\\prod_{i=1}^n \\frac{A - a_i}{d_i D_i}$.\n(posed by Zhang Sihui)",
    "options": [],
    "answer": "(n-1)^n",
    "solution": "Consider\n$$\nD_1 = (a_2, a_3, \\dots, a_n) \\text{ and } d_2 = (a_2, A) = (a_2, a_1 + a_2 + \\dots + a_n).\n$$\nLet $(D_1, d_2) = d$. Then $d \\mid a_2, d \\mid a_3, \\dots, d \\mid a_n, d \\mid a_1 + a_2 + \\dots + a_n$. Thus, $d \\mid a_1$. Consequently,\n$$\nd \\mid (a_1, a_2, \\dots, a_n).\n$$\nSince $a_1, a_2, \\dots, a_n$ are coprime, we have $d = 1$. Note that $D_1 \\mid a_2, D_2 \\mid a_2$ and $(D_1, D_2) = 1$. We have $D_1 d_2 \\mid a_2$. So $D_1 d_2 \\le a_2$. Similarly, we have $D_2 d_3 \\le a_3, \\dots, D_n d_1 \\le a_1$. Hence,\n$$\n\\begin{aligned}\n\\prod_{i=1}^{n} d_i D_i &= (D_1 d_2) \\cdot (D_2 d_3) \\cdot \\dots \\cdot (D_n d_1) \\\\\n&\\le a_2 a_3 \\dots a_n a_1 \\\\\n&= \\prod_{i=1}^{n} a_i.\n\\end{aligned} \n\\qquad \\textcircled{1}\n$$\nConsidering\n$$\n\\begin{aligned}\n\\prod_{i=1}^{n} (A - a_i) &= \\prod_{i=1}^{n} \\left( \\sum_{j \\neq i} a_j \\right) \\\\\n&\\geq \\prod_{i=1}^{n} \\left( (n-1) \\left( \\prod_{j \\neq i} a_j \\right)^{\\frac{1}{n-1}} \\right) \\\\\n&= (n-1)^n \\cdot \\prod_{i=1}^{n} a_i\n\\end{aligned} \n\\qquad \\textcircled{2}\n$$\nand by ① and ②, we see that\n$$\n\\prod_{i=1}^{n} \\frac{A - a_i}{d_i D_i} \\ge (n-1)^n.\n$$\nOn the other hand, if $a_1 = a_2 = \\dots = a_n = 1$,\n$$\n\\prod_{i=1}^{n} \\frac{A - a_i}{d_i D_i} = (n - 1)^n.\n$$\nSumming up, the minimum of $\\prod_{i=1}^{n} \\frac{A - a_i}{d_i D_i}$ is $(n - 1)^n$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 28398,
    "subject": "Mathematics (Multi-modal)",
    "question": "Do there exist integers $a$, $b$ and $c$ such that $a^2bc + 2$, $ab^2c + 2$, $abc^2 + 2$ are perfect squares?",
    "options": [],
    "answer": "No",
    "solution": "No. Suppose the contrary that there are such integers $a$, $b$ and $c$.\nIf one of them is even, say $a$, then $a^2bc + 2 \\equiv 2 \\pmod 4$, which contradicts the assumption that $a^2bc + 2$ is a perfect square. We may then assume that $a$, $b$ and $c$ are odd, so they are either $1$ or $3 \\pmod 4$. It follows from the Pigeon-hole Principle that two of them are congruent modulo $4$. Relabel if necessary, we may assume that $a \\equiv b \\pmod 4$, so $abc^2 + 2 \\equiv c^2 + 2 \\equiv 1 + 2 \\equiv 3 \\pmod 4$, which violates the perfect square assumption. $\\square$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 28399,
    "subject": "Mathematics (Multi-modal)",
    "question": "In an acute triangle $ABC$, $AB > AC$, the bisector of angle $BAC$ and side $BC$ intersect at point $D$, two points $E$ and $F$ are in sides $AB$ and $AC$, respectively, such that $B, C, F, E$ are concyclic. Prove that the circumcenter of triangle $DEF$ coincides with the innercenter of triangle $ABC$ if and only if $BE + CF = BC$.\n\n![](attached_image_1.png)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $I$ be the innercenter of $\\triangle ABC$.\n\n(Sufficiency) Suppose $BC = BE + CF$. Let $K$ be the point on $BC$ such that $BK = BE$, thus $CK = CF$. Since $BI$ bisects $\\angle ABC$, $CI$ bisects $\\angle ACB$, $\\triangle BIK$ and $\\triangle BIE$ are reflection with respect to $BI$, $\\triangle CIK$ and $\\triangle CIF$ are reflection with respect to $CI$, we have $\\angle BEI = \\angle BKI = \\pi - \\angle CKI = \\pi -$\n\n$\\angle CFI = \\angle AFI$. Therefore, $A, E, I, F$ are concyclic. Since $B, E, F, C$ are concyclic, we have $\\angle AIE = \\angle AFE = \\angle ABC$, and hence $B, E, I, D$ are concyclic.\n\n![](attached_image_2.png)\n\nSince the bisector of $\\angle EAF$ and the circumcircle of $\\triangle AEF$ meet at $I$, $IE = IF$. Since the bisector of $\\angle EBD$ and the circumcircle of $\\triangle BED$ also meet at $I$, $IE = ID$. So, $ID = IE = IF$, that is, $I$ is also the circumcenter of $\\triangle DEF$.\n\nQ. E. D.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 28400,
    "subject": "Mathematics (Multi-modal)",
    "question": "Given $A = \\{2, 0, 1, 3\\}$, let $B = \\{x \\mid -x \\in A, 2 - x^2 \\notin A\\}$. Then the sum of elements in $B$ is ______.",
    "options": [],
    "answer": "-5",
    "solution": "It is easy to find that $B \\subseteq \\{-2, 0, -1, -3\\}$. We have $2 - x^2 \\notin A$ when $x = -2, -3$, and $2 - x^2 \\in A$ when $x = 0, -1$. Therefore, $B = \\{-2, -3\\}$, the sum of whose elements is $-5$.\nThe answer is $-5$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 28401,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $a$, $b$ be real numbers such that the equation $x^3 - a x^2 + b x - a = 0$ has only real roots. Find the minimum of $\\frac{2a^3 - 3ab + 3a}{b+1}$.",
    "options": [],
    "answer": "9√3",
    "solution": "Let $x_1$, $x_2$ and $x_3$ be the real roots of the equation $x^3 - a x^2 + b x - a = 0$. By Vieta's Formula, we have\n$x_1 + x_2 + x_3 = a$, $x_1 x_2 + x_2 x_3 + x_1 x_3 = b$, $x_1 x_2 x_3 = a$.\n\nBy $(x_1 + x_2 + x_3)^2 \\ge 3(x_1 x_2 + x_2 x_3 + x_1 x_3)$, we have $a^2 \\ge 3b$, and by $a = x_1 + x_2 + x_3 \\ge 3 \\sqrt[3]{x_1 x_2 x_3} = 3 \\sqrt[3]{a}$, we have $a \\ge 3 \\sqrt{3}$.\n\nThus,\n$$\n\\begin{aligned}\n\\frac{2a^3 - 3ab + 3a}{b+1} &= \\frac{a(a^2 - 3b) + a^3 + 3a}{b+1} \\\\\n&\\ge \\frac{a^3 + 3a}{b+1} \\ge \\frac{a^3 + 3a}{\\frac{a^2}{3} + 1} \\\\\n&= 3a \\ge 9\\sqrt{3}.\n\\end{aligned}\n$$\nIf $a = 3\\sqrt{3}$, $b = 9$, then the equality holds when each root is equal to $\\sqrt{3}$.\n\nSumming up, the answer is $9\\sqrt{3}$.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 28402,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $A$ be the closed domain on the plane delimited by three lines $x = 1$, $y = 0$, and $y = t(2x - t)$, where $0 < t < 1$. Prove that the surface of any triangle inside the domain $A$ with $P(t, t^2)$ and $Q(1, 0)$ as two of its vertices cannot exceed $\\frac{1}{4}$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "It is easy to observe that the domain is a closed triangle. Its three vertices are $B\\left(\\frac{t}{2}, 0\\right)$, $Q(1, 0)$ and $C(1, t(2-t))$. Pick a point $X$ inside the $\\triangle BQC$, then the area of $\\triangle PQX$ is equal to half of the product of $PQ$ with the distance from $X$ to $PQ$. So the area of $PQX$\ntakes its maximum value when the distance from $X$ to $PQ$ is maximized, i.e., when $X$ coincides with $B$ or $C$.\nThe area of $\\triangle PQB$ is\n\n![](attached_image_1.png)\n\n$$\n\\begin{aligned}\n\\frac{1}{2}\\left(1-\\frac{t}{2}\\right)t^2 &= \\frac{1}{4}(2-t)t^2 \\le \\frac{1}{4}(2-t)t \\\\\n&\\le \\frac{1}{4}\\left(\\frac{2-t+t}{2}\\right)^2 = \\frac{1}{4};\n\\end{aligned}\n$$\nthe area of $\\triangle PQC$ is\n$$\n\\begin{aligned}\n\\frac{1}{2}(1-t)(2t-t^2) &= \\frac{1}{4}2t(1-t)(2-t) \\\\\n&\\le \\frac{1}{4}\\left(\\frac{2t+1-t+2-t}{3}\\right)^3 = \\frac{1}{4}.\n\\end{aligned}\n$$\nHence, in the domain $A$, any triangle with $P, Q$ as two of its vertices cannot have an area that exceeds $\\frac{1}{4}$. $\\square$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 28403,
    "subject": "Mathematics (Multi-modal)",
    "question": "Given an ellipse equation $\\frac{x^2}{a^2} + \\frac{y^2}{b^2} = 1$ ($a > b > 0$) in a plane rectangular coordinate system $xOy$, let $A_1, A_2, F_1, F_2$ be its left and right end-points, left and right focuses, respectively, and $P$ be any point on the ellipse different from $A_1, A_2$. Suppose there are points $Q, R$ satisfying $QA_1 \\perp PA_1, QA_2 \\perp PA_2, RF_1 \\perp PF_1, RF_2 \\perp PF_2$. Find and prove the relationship between the length of segment $QR$ and $b$.",
    "options": [],
    "answer": "If P has y-coordinate y0, then |QR| = b^2 / |y0| ≥ b, with equality if and only if P = (0, ± b).",
    "solution": "Let $c = \\sqrt{a^2 - b^2}$. Then $A_1(-a, 0), A_2(a, 0), F_1(-c, 0), F_2(c, 0)$.\n\nDenote $P(x_0, y_0)$, $Q(x_1, y_1)$, $R(x_2, y_2)$, where $\\frac{x_0^2}{a^2} + \\frac{y_0^2}{b^2} = 1$, $y_0 \\neq 0$.\n\nFrom $QA_1 \\perp PA_1, QA_2 \\perp PA_2$, we have\n$$\n\\overrightarrow{A_1Q} \\cdot \\overrightarrow{A_1P} = (x_1 + a)(x_0 + a) + y_1y_0 = 0, \\quad \\textcircled{1}\n$$\n$$\n\\overrightarrow{A_2Q} \\cdot \\overrightarrow{A_2P} = (x_1 - a)(x_0 - a) + y_1y_0 = 0. \\qquad \\textcircled{2}\n$$\nSubtracting ① and ②, we have $2a(x_1+x_0) = 0$, i.e. $x_1 = -x_0$. Substituting it into ①, we get $-x_0^2 + a^2 + y_1y_0 = 0$, or $y_1 = \\frac{x_0^2 - a^2}{y_0}$. Then $Q(-x_0, \\frac{x_0^2 - a^2}{y_0})$.\n\nFrom $RF_1 \\perp PF_1, RF_2 \\perp PF_2$, in the same way we obtain $R(-x_0, \\frac{x_0^2 - c^2}{y_0})$. Therefore,\n$$\n|QR| = \\left| \\frac{x_0^2 - a^2}{y_0} - \\frac{x_0^2 - c^2}{y_0} \\right| = \\frac{b^2}{|y_0|}.\n$$\nSince $|y_0| \\in (0, b]$, then $|QR| \\ge b$, where the equality holds if and only if $|y_0| = b$ (i.e., $P(0, \\pm b)$). $\\square$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 28404,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all the positive real number pairs $(a, b)$, such that $f(x) = ax^2 + b$ satisfies $f(xy) + f(x + y) \\ge f(x)f(y)$ (for any real numbers $x, y$).",
    "options": [],
    "answer": "{(a, b) | 0 < a < 1, 0 < b ≤ 1, and 2a + b ≤ 2}",
    "solution": "The given condition is equivalent to\n$$\n(ax^2 y^2 + b) + (a(x + y)^2 + b) \\ge (ax^2 + b)(ay^2 + b). \\quad ①\n$$\nIn ①, let $y = 0$. We have $b + (ax^2 + b) \\ge (ax^2 + b) \\cdot b$, or\n$$ (1 - b)ax^2 + b(2 - b) \\ge 0. $$\nAs $a > 0$ and $ax^2$ can be sufficiently large, then $1 - b \\ge 0$, i.e., $0 < b \\le 1$.\nIn ①, let $y = -x$. We have $(ax^4 + b) + b \\ge (ax^2 + b)^2$, or\n$$\n(a - a^2)x^4 - 2abx^2 + (2b - b^2) \\ge 0. \\quad ②\n$$\nDenote the left-hand side of ② as $g(x)$. It is obvious that $a - a^2 \\ne 0$ (otherwise, from $a > 0$ we know $a = 1$. Then $g(x) = -2bx^2 + (2b - b^2)$ with $b > 0$, which means $g(x)$ can be negative. A contradiction). Then\n$$\n\\begin{align*} g(x) &= (a - a^2) \\left( x^2 - \\frac{ab}{a - a^2} \\right)^2 - \\frac{(ab)^2}{a - a^2} + (2b - b^2) \\\\ &= (a - a^2) \\left( x^2 - \\frac{b}{1-a} \\right)^2 + \\frac{b}{1-a} (2 - 2a - b) \\\\ &\\ge 0 \\end{align*}\n$$\nholds for any real number $x$. So we have $a - a^2 > 0$, i.e., $0 < a < 1$.\nFurthermore, from $\\frac{b}{1-a} > 0$ and\n$$\ng\\left(\\sqrt{\\frac{b}{1-a}}\\right) = \\frac{b}{1-a}(2 - 2a - b) \\ge 0,\n$$\nwe have $2a + b \\le 2$.\nSo far, we get the necessary condition that $a, b$ must satisfy as follows:\n$$\n0 < b \\le 1, \\quad 0 < a < 1, \\quad 2a + b \\le 2. \\qquad \\textcircled{3}\n$$\nWe are going to prove that for any pair $(a, b)$ satisfying ③ and any real numbers $x, y$, ① holds, or equivalently,\n$$\nh(x, y) = (a - a^2)x^2 y^2 + a(1 - b)(x^2 + y^2) + 2axy + (2b - b^2) \\ge 0.\n$$\nAs a matter of fact, when ③ holds, we then have\n$$\na(1-b) \\ge 0, \\quad a - a^2 > 0 \\quad \\text{and} \\quad \\frac{b}{1-a}(2 - 2a - b) \\ge 0.\n$$\nCombining it with $x^2 + y^2 \\ge -2xy$, we get\n$$\n\\begin{align*} h(x, y) &\\ge (a - a^2)x^2 y^2 + a(1 - b)(-2xy) + 2axy + (2b - b^2) \\\\ &= (a - a^2)x^2 y^2 + 2abxy + (2b - b^2) \\\\ &= (a - a^2) \\left( xy + \\frac{b}{1-a} \\right)^2 + \\frac{b}{1-a} (2 - 2a - b) \\ge 0. \\end{align*}\n$$\nTherefore, the set of all the pairs $(a, b)$ meeting the given condition is\n$$\n\\{(a, b) \\mid 0 < b \\le 1, 0 < a < 1, 2a + b \\le 2\\}.\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 28405,
    "subject": "Mathematics (Multi-modal)",
    "question": "In a plane rectangular coordinate system $xOy$, points $A$, $B$ are on the parabola $y^2 = 4x$, satisfying $\\vec{OA} \\cdot \\vec{OB} = -4$, and point $F$ is the focus of the parabola. Then $S_{\\triangle OFA} \\cdot S_{\\triangle OFB} = \\_\\_\\_\\_\\_\\_$.",
    "options": [],
    "answer": "2",
    "solution": "Let $F(1, 0)$, $A(x_1, y_1)$, $B(x_2, y_2)$. Then $x_1 = \\frac{y_1^2}{4}$, $x_2 = \\frac{y_2^2}{4}$, and\n$$\n-4 = \\vec{OA} \\cdot \\vec{OB} = x_1x_2 + y_1y_2 = \\frac{1}{16}(y_1y_2)^2 + y_1y_2,\n$$\nfrom which we have $\\frac{1}{16}(y_1y_2 + 8)^2 = 0$, or $y_1y_2 = -8$.\nTherefore,\n$$\n\\begin{align*} \nS_{\\triangle OFA} \\cdot S_{\\triangle OFB} &= \\left( \\frac{1}{2} |OF| \\cdot |y_1| \\right) \\cdot \\left( \\frac{1}{2} |OF| \\cdot |y_2| \\right) \\\\ \n&= \\frac{1}{4} \\cdot |OF|^2 \\cdot |y_1 y_2| = 2. \n\\end{align*}\n$$\nThe answer is 2.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 28406,
    "subject": "Mathematics (Multi-modal)",
    "question": "For any integer $n$ with $n > 1$, let\n$$\nD(n) = \\{a - b \\mid a, b \\text{ are positive integers with } n = ab \\text{ and } a > b\\}.\n$$\nProve that for any integer $k$ with $k > 1$, there exist $k$ pairwise distinct integers $n_1, n_2, \\dots, n_k$ with $n_i > 1\\ (1 \\le i \\le k)$, such that $D(n_1) \\cap D(n_2) \\cap \\dots \\cap D(n_k)$ has at least two elements.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "**Proof.** Let $a_1, a_2, \\dots, a_{k+1}$ be $k+1$ distinct positive integers, where each of them is smaller than the product of other $k$ numbers. Write $N = a_1 a_2 \\cdots a_{k+1}$. For each $i = 1, 2, \\dots, k+1$, let $x_i = \\frac{1}{2} \\left( \\frac{N}{a_i} + a_i \\right)$, $y_i = \\frac{1}{2} \\left( \\frac{N}{a_i} - a_i \\right)$, then $x_i^2 - y_i^2 = N$.\nSince $a_i a_j < N$ and $\\frac{N}{a_i} > a_i$, $(x_i, y_i)(1 \\le i \\le k+1)$ are $k+1$ positive integer solutions of equation $x^2 - y^2 = N$. Without loss of generality, suppose $x_{k+1} = \\min\\{x_1, x_2, \\dots, x_{k+1}\\}$. For each $i \\in \\{1, 2, \\dots, k\\}$, since $x_i^2 - y_i^2 = x_{k+1}^2 - y_{k+1}^2$, we have\n$$\n(x_i + x_{k+1})(x_i - x_{k+1}) = x_i^2 - x_{k+1}^2 = y_i^2 - y_{k+1}^2 = (y_i + y_{k+1})(y_i - y_{k+1}).\n$$\nLet $n_i = (x_i + x_{k+1})(x_i - x_{k+1}) = (y_i + y_{k+1})(y_i - y_{k+1})$, then\n$$\n2x_{k+1} = (x_i + x_{k+1}) - (x_i - x_{k+1}) \\in D(n_i),\n$$\n$$\n2y_{k+1} = (y_i + y_{k+1}) - (y_i - y_{k+1}) \\in D(n_i).\n$$\nSo $x_{k+1} > y_{k+1}$, $2x_{k+1}$ and $2y_{k+1}$ are two different members of $D(n_1) \\cap D(n_2) \\cap \\dots \\cap D(n_k)$. $\\square$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 28407,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $n$ be an integer, $n \\ge 2$, and $x_1, x_2, \\dots, x_n \\in [0, 1]$. Prove that\n$$\n\\sum_{1 \\le k < l \\le n} kx_k x_l \\le \\frac{n-1}{3} \\sum_{k=1}^{n} kx_k.\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "As $x_1, x_2, \\dots, x_n \\in [0, 1]$, $x_i x_j \\le x_i$, so we have\n$$\n3 \\sum_{1 \\le k < l \\le n} kx_k x_l = \\sum_{1 \\le k < l \\le n} 3k x_k x_l \\le \\sum_{1 \\le k < l \\le n} (k x_k + 2k x_l).\n$$\nFor $1 \\le k \\le n$, the coefficient of $x_k$ in the last sum is\n$$\n2[1 + 2 + \\dots + (k-1)] + k(n-k) = k(n-1),\n$$\nso we have\n$$\n\\begin{aligned}\n3 \\sum_{1 \\le k < l \\le n} kx_k x_l &\\le \\sum_{1 \\le k < l \\le n} (k x_k + 2k x_l) = \\sum_{k=1}^{n} k(n-1)x_k \\\\ &= (n-1) \\sum_{k=1}^{n} kx_k, \\end{aligned}\n$$\nand hence the desired inequality holds. $\\square$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 28408,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all nonempty sets $S$ of integers such that $3m-2n \\in S$ for all (not necessarily distinct) $m, n \\in S$.",
    "options": [],
    "answer": "All such sets are exactly the following, for some integers a and positive integer d: (1) the singleton {a}; (2) the set {a + k d where k ranges over all integers not divisible by three}; (3) the full arithmetic progression {a + k d where k ranges over all integers}.",
    "solution": "Call a set $S$ \"good\" if it satisfies the property as stated in the problem.\n\n(1) If $S$ has only one element, $S$ is \"good\".\n\n(2) Now we can assume that $S$ contains at least two elements. Let\n$$\nd = \\min\\{|m-n| : m, n \\in S, m \\neq n\\}.\n$$\nThen there is an integer $a$, such that $a+d, a+2d \\in S$.\nNote that\n$$\na+4d=3(a+2d)-2(a+d) \\in S,\n$$\n$$\na-d=3(a+d)-2(a+2d) \\in S,\n$$\n$$\na+5d=3(a+d)-2(a-d) \\in S,\n$$\n$$\na-2d=3(a+2d)-2(a+4d) \\in S.\n$$\nSo we have proved that if\n$$\na+d, a+2d \\in S,\n$$\nthen\n$$\na-2d, a-d, a+4d, a+5d \\in S.\n$$\nContinuing this procedure, we can deduce that\n$$\n\\{a+kd \\mid k \\in \\mathbb{Z}, 3 \\nmid k\\} \\subseteq S.\n$$\nLet $S_0 = \\{a+kd \\mid k \\in \\mathbb{Z}, 3 \\nmid k\\}$, it is easy to verify that $S_0$ is \"good\".\n\n(3) Now we have proved that $S_0 \\subseteq S$. If $S \\neq S_0$, pick a number $b \\in S \\setminus S_0$, then there exists an integer $l$, such that $a+ld \\le b < a+(l+1)d$. Since at least one of $l$ and $l+1$ is indivisible by 3, we know that at least one of $a+ld, a+(l+1)d$ is contained in $S_0$. If $a+ld \\in S_0$, note that $0 \\le b-(a+ld) < d$, by the definition of $d$, we must have $b = a + ld$. If $a + (l + 1)d \\in S_0$, note that\n$$\n0 < |a + (l + 1)d - b| \\le d,\n$$\nby the definition of $d$, we also have $b = a + ld$.\nIn both cases, we have proved that there is a number $b \\in S \\setminus S_0$ of the form $b = a + ld$. This $l$ must be divisible by 3, hence\n$$\na + ld, a + (l + 1)d, a + (l + 2)d \\in S,\n$$\n$$\na + (l - 2)d, a + (l - 1)d \\in S.\n$$\nSo\n$$\na + (l + 3)d = 3(a + (l + 1)d) - 2(a + ld) \\in S,\n$$\n$$\na + (l - 3)d = 3(a + (l - 1)d) - 2(a + ld) \\in S.\n$$\nContinuing this procedure, we have $a + (l + 3j)d \\in S$, $j \\in \\mathbb{Z}$, which implies that $\\{a + kd \\mid k \\in \\mathbb{Z}\\} \\subseteq S$. We claim that $S = \\{a + kd \\mid k \\in \\mathbb{Z}\\}$, since for any number $x \\notin \\{a + kd \\mid k \\in \\mathbb{Z}\\}$, there is an element in $y \\in \\{a + kd \\mid k \\in \\mathbb{Z}\\}$ such that $0 < |x - y| < d$. By the definition of $d$, we must have $x \\notin S$.\n\nSo $S = \\{a+kd \\mid k \\in \\mathbb{Z}\\}$, and it is easy to verify that such $S$ is \"good\".\n\nWe conclude that there are three classes of \"good\" sets:\n(1) $S = \\{a\\}$;\n(2) $S = \\{a+kd \\mid k \\in \\mathbb{Z}, 3 \\nmid k\\}$;\n(3) $S = \\{a+kd \\mid k \\in \\mathbb{Z}\\}$, here $a, d \\in \\mathbb{Z}, d > 0$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 28409,
    "subject": "Mathematics (Multi-modal)",
    "question": "Suppose that $O$ and $I$ are the centres of the circumcircle and incircle of $\\triangle ABC$ with radius $R$ and $r$, respectively. $P$ is the midpoint of arc $BAC$. Let $QP$ be the diameter of $O$. Let $PI$ intersect $BC$ at point $D$, and let the circumcircle of $\\triangle AID$ intersect the extended line of $PA$ at point $F$. Let point $E$ be on $PD$ such that $DE = DQ$. Prove that, if $\\angle AEF = \\angle APE$, then $\\sin^2 \\angle BAC = \\frac{2r}{R}$.\n\n(posed by Xiong Bin)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "$$\n\\begin{aligned}\nPF^2 &= AF \\cdot PF + PA \\cdot PF \\\\\n&= EF^2 + PI \\cdot PD.\n\\end{aligned} \n\\qquad \\textcircled{1}\n$$\nSince $PQ$ is the diameter of circle $O$ and point $I$ is on $AQ$, we see that $AI \\perp AP$. Consequently,\n$$\n\\angle IDF = \\angle IAP = 90^{\\circ}.\n$$\nThus, we have\n$$\nPF^2 - EF^2 = PD^2 - ED^2.\n$$\nCombining ①, we have\n$$\nPI \\cdot PD = PD^2 - ED^2.\n$$\nThus\n$$\nQD^2 = ED^2 = PD^2 - PI \\cdot PD = ID \\cdot PD.\n$$\nConsequently, we have\n$$\n\\triangle QID \\sim \\triangle PQD. \\qquad ②\n$$\nSince $PQ$ is the diameter of circle $O$, we see that $BP \\perp BQ$. Suppose that $PQ$ is the perpendicular bisector of $BC$ at point $M$. Note that $I$ is the incentre of $\\triangle ABC$. We have $QI^2 = QB^2 = QM \\cdot QP$. Thus,\n$$\n\\triangle QMI \\sim \\triangle QIP. \\qquad ③\n$$\nBy ② and ③, we see that $\\angle IQD = \\angle QPD = \\angle QPI = \\angle QIM$. Hence, $MI \\parallel QD$. Let $IK \\perp BC$ be at $K$. Then $IK \\parallel PM$; thus,\n$$\n\\frac{PM}{IK} = \\frac{PD}{ID} = \\frac{PQ}{MQ}.\n$$\nBy the Circle-Power Theorem and the Sine Theorem, we know that\n$$\n\\begin{aligned} PQ \\cdot IK &= PM \\cdot MQ = BM \\cdot MC \\\\\n&= \\left(\\frac{1}{2}BC\\right)^2 = (R \\sin \\angle BAC)^2, \\end{aligned}\n$$\nthus, $\\sin^2 \\angle BAC = \\frac{PQ \\cdot IK}{R^2} = \\frac{2R \\cdot r}{R^2} = \\frac{2r}{R}$.\n$\\square$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 28410,
    "subject": "Mathematics (Multi-modal)",
    "question": "Given positive integers $u, v$, the sequence $\\{a_n\\}$ is defined as: $a_1 = u + v$, and for $m \\ge 1$,\n$$\n\\begin{cases} a_{2m} = a_m + u, \\\\ a_{2m+1} = a_m + v. \\end{cases}\n$$\n\nDenote $S_m = a_1 + a_2 + \\dots + a_m$ ($m = 1, 2, \\dots$). Prove that there are infinite terms in sequence $\\{S_n\\}$ that are square numbers.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "For positive integer $n$, we have\n$$\n\\begin{align*} \nS_{2^{n+1}-1} &= a_1 + (a_2 + a_3) + (a_4 + a_5) + \\dots + (a_{2^{n+1}-2} + a_{2^{n+1}-1}) \\\\\n&= u + v + (a_1 + u + a_1 + v) + (a_2 + u + a_2 + v) + \\dots + \\\\\n& \\qquad (a_{2^{n-1}} + u + a_{2^{n-1}} + v) \\\\\n&= 2^n (u + v) + 2S_{2^{n-1}}. \n\\end{align*}\n$$\nThen\n$$\n\\begin{align*} \nS_{2^{n}-1} &= 2^{n-1}(u+v) + 2S_{2^{n-1}-1} \\\\\n&= 2^{n-1}(u+v) + 2(2^{n-2}(u+v) + 2S_{2^{n-2}-1}) \\\\\n&= 2 \\cdot 2^{n-1}(u+v) + 2^2 S_{2^{n-2}-1} \\\\\n&= \\dots = (n-1) \\cdot 2^{n-1}(u+v) + 2^{n-1}(u+v) \\\\\n&= (u+v) \\cdot n \\cdot 2^{n-1}. \n\\end{align*}\n$$\nSuppose $u + v = 2^k \\cdot q$, where $k$ is a non-negative integer, and $q$ is an odd number. Take $n = q \\cdot l^2$, where $l$ is any positive integer satisfying $l \\equiv k - 1 \\pmod 2$. Then $S_{2^{n-1}} = q^2 l^2 \\cdot 2^{k-1+q \\cdot l^2}$, and\n$$\n\\begin{align*} \nk - 1 + q \\cdot l^2 &\\equiv k - 1 + l^2 \\equiv k - 1 + (k - 1)^2 \\\\ \n&= k(k - 1) \\equiv 0 \\pmod 2. \n\\end{align*}\n$$\nTherefore, $S_{2^{n-1}}$ is a square number. Since there are infinite $l$'s, there are infinite terms in $\\{S_n\\}$ that are square numbers. The proof is complete. $\\square$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 28411,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $\\odot I$ be the incircle of $\\triangle ABC$ with $AB > AC$. $\\odot I$ is tangent to $BC$ and $AD$ at $D$ and $E$, respectively. The tangent line $EP$ of $\\odot I$ intersects the extended line of $BC$ at $P$. Segment $CF$ is parallel to $PE$ and intersects $AD$ at point $F$. Line $BF$ intersects $\\odot I$ at points $M$ and $N$ such that $M$ is on segment $BF$. Segment $PM$ intersects $\\odot I$ at the other point $Q$. Prove that $\\angle ENP = \\angle ENQ$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Therefore, $\\angle IGP = \\angle IEP = 90^\\circ$, that is, $IG \\perp PG$. Hence, points $P$, $S$ and $T$ are collinear.\n\nLine $PST$ intersects $\\triangle ABC$. By Menelaus' Theorem, we have\n$$\n\\frac{AS}{SC} \\cdot \\frac{CP}{PB} \\cdot \\frac{BT}{TA} = 1.\n$$\nSince $AS = AT$, $CS = CD$ and $BT = BD$, we have\n$$\n\\frac{PC}{PB} \\cdot \\frac{BD}{CD} = 1. \\qquad \\textcircled{1}\n$$\nLet the extension of $BN$ intersect $PE$ at point $H$. Then line $BFH$ intersects $\\triangle PDE$. By Menelaus' Theorem,\n$$\n\\frac{PH}{HE} \\cdot \\frac{EF}{FD} \\cdot \\frac{DB}{BP} = 1.\n$$\nSince $CF$ is parallel to $BE$, $\\frac{EF}{FD} = \\frac{PC}{CD}$, we have\n$$\n\\frac{PH}{HE} \\cdot \\frac{PC}{CD} \\cdot \\frac{DB}{BP} = 1. \\qquad \\textcircled{2}\n$$\nBy ① and ②, we have $PH = HE$. Hence, $PH^2 = HE^2 = HM \\cdot HN$. Thus, we have $\\frac{PH}{HM} = \\frac{HN}{PH}$, $\\triangle PHN \\sim \\triangle MHP$ and $\\angle HPN = \\angle HMP = \\angle NEQ$. Further, since $\\angle PEN = \\angle EQN$, therefore $\\angle ENP = \\angle ENQ$. $\\square$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 28412,
    "subject": "Mathematics (Multi-modal)",
    "question": "Suppose there are $m$ questions in an examination attended by $n$ students, where $m, n \\ge 2$ are given natural numbers. The marking rule for each question is as follows: if there are exactly $x$ students failing to answer the question correctly, then they will each get 0 marks, and those who answer it correctly will each get $x$ marks. The total marks of a student are the sum of marks he/she gets from the $m$ questions. Now rank the total marks of the $n$ students as $p_1 \\ge p_2 \\ge \\cdots \\ge p_n$. Find the maximum possible value of $p_1 + p_n$.",
    "options": [],
    "answer": "m(n-1)",
    "solution": "For any $k = 1, 2, \\dots, m$, assuming there are $x_k$ students failing to answer the $k$th question correctly, then there are $n - x_k$ ones who answer it correctly and each gets $x_k$ marks from it accordingly. Suppose the sum of the $n$ students' total marks is $S$. Then we have\n$$\n\\sum_{i=1}^{n} p_i = S = \\sum_{k=1}^{m} x_k (n - x_k) = n \\sum_{k=1}^{m} x_k - \\sum_{k=1}^{m} x_k^2.\n$$\nAs each student gets at most $x_k$ marks from the $k$th question, we have\n$p_1 \\le \\sum_{k=1}^{m} x_k.$\nSince $p_2 \\ge \\cdots \\ge p_n$, then $p_n \\le \\frac{p_2 + p_3 + \\cdots + p_n}{n-1} = \\frac{S - p_1}{n-1}$.\nTherefore,\n$$\n\\begin{align*}\np_1 + p_n &\\le p_1 + \\frac{S-p_1}{n-1} = \\frac{n-2}{n-1}p_1 + \\frac{S}{n-1} \\\\\n&\\le \\frac{n-2}{n-1} \\cdot \\sum_{k=1}^{m} x_k + \\frac{1}{n-1} \\cdot \\left( n \\sum_{k=1}^{m} x_k - \\sum_{k=1}^{m} x_k^2 \\right) \\\\\n&= 2 \\sum_{k=1}^{m} x_k - \\frac{1}{n-1} \\cdot \\sum_{k=1}^{m} x_k^2.\n\\end{align*}\n$$\nBy the Cauchy Inequality, we have\n$$\n\\sum_{k=1}^{m} x_k^2 \\ge \\frac{1}{m} \\left( \\sum_{k=1}^{m} x_k \\right)^2 .\n$$\nThen\n$$\n\\begin{aligned}\np_1 + p_n &\\le 2 \\sum_{k=1}^{m} x_k - \\frac{1}{m(n-1)} \\cdot \\left( \\sum_{k=1}^{m} x_k \\right)^2 \\\\\n&= -\\frac{1}{m(n-1)} \\cdot \\left( \\sum_{k=1}^{m} x_k - m(n-1) \\right)^2 + m(n-1) \\\\\n&\\le m(n-1).\n\\end{aligned}\n$$\nOn the other hand, if there is a student who answers all the questions correctly, while the other $n-1$ students fail to answer any questions, then we have\n$$\np_1 + p_n = p_1 = \\sum_{k=1}^{m} (n-1) = m(n-1).\n$$\nTherefore, the maximum possible value of $p_1 + p_n$ is $m(n-1)$. $\\boxed{}$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 28413,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $N^*$ be the set of all positive integers. Prove that there exists a unique function $f: N^* \\to N^*$ satisfying $f(1) = f(2) = 1$ and $f(n) = f(f(n-1)) + f(n-f(n-1))$, $n = 3, 4, \\dots$. For such $f$, find the value of $f(2^m)$ for integer $m \\ge 2$.",
    "options": [],
    "answer": "f(2^m) = 2^{m-1} for integer m ≥ 2",
    "solution": "We show by induction that for any integer $n > 1$, $f(n)$ is uniquely determined by the value of $f(1)$, $f(2)$, $\\dots$, $f(n-1)$, and $\\frac{n}{2} \\le f(n) \\le n$.\n\nFor $n = 2$, $f(2) = 1$, the claim is true.\nAssume that for any $k$, $1 \\le k < n$ ($n \\ge 3$), $f(k)$ is uniquely determined, and $\\frac{k}{2} \\le f(k) \\le k$, then $1 \\le \\frac{n-1}{2} \\le f(n-1) \\le n-1$, and $1 \\le n - f(n-1) \\le n-1$, hence by induction hypothesis, the value of $f(f(n-1))$ and $f(n - f(n-1))$ is determined, and the value of $f(n)$, by definition, is\n$$\nf(n) = f(f(n-1)) + f(n - f(n-1)),\n$$\nwhich is uniquely determined. Furthermore, we have\n$$\n\\frac{1}{2}f(n-1) \\le f(f(n-1)) \\le f(n-1), \\\\\n\\frac{1}{2}(n - f(n-1)) \\le f(n - f(n-1)) \\le n - f(n-1).\n$$\nEquality above implies $\\frac{n}{2} \\le f(n) \\le f(n-1) + (n - f(n-1)) = n$. The claim is also true for $n$. By induction, we proved that there exists a unique function $f: N^* \\to N^*$ satisfying the required properties, and $\\frac{n}{2} \\le f(n) \\le n$.\n\nNext, we show by induction that for any positive integer $n$, we have\n$$\nf(n+1) - f(n) \\in \\{0, 1\\}.\n$$\nWhen $n = 1$, this is true.\nAssume that it is true for $n \\le k$. By the recurrence,\n$$\n\\begin{align*}\n& f(k + 2) - f(k + 1) \\\\\n&= (f(f(k + 1)) + f(k + 2 - f(k + 1))) - (f(f(k)) + f(k + 1 - f(k))) \\\\\n&= (f(f(k + 1)) - f(f(k))) + (f(k + 2 - f(k + 1)) - f(k + 1 - f(k))).\n\\end{align*}\n$$\nBy induction hypothesis, $f(k+1) - f(k) \\in \\{0, 1\\}$.\nIf $f(k+1) = f(k)+1$, since $1 \\le f(k) \\le k$, it follows,\nfrom above and induction hypothesis, that\n$$\nf(k + 2) - f(k + 1) = f(f(k) + 1) - f(f(k)) \\in \\{0, 1\\}.\n$$\nIf $f(k+1) = f(k)$, since $1 \\le k+1-f(k) \\le k$, it follows\nfrom above and the induction hypothesis that\n$$\nf(k + 2) - f(k + 1) = f(k + 2 - f(k)) - f(k + 1 - f(k)) \\in \\{0, 1\\}.\n$$\nThus, the claim is true for $n = k + 1$. By induction, it is true for any positive integer $n$.\n\nFinally, we show by induction that for any positive integer $m$, we have $f(2^m) = 2^{m-1}$.\nFor $m=1$, the result is clear.\nAssume that the result is true for $m = k$, i.e., $f(2^k) = 2^{k-1}$, consider the case for $m = k + 1$.\nAssume on the contrary that $f(2^{k+1}) \\ne 2^k$, since $f(2^{k+1}) \\ge 2^k$, and $f(2^{k+1})$ is an integer, we have $f(2^{k+1}) \\ge 2^k + 1$. Since $f(1) = 1$, by the previous claim, let $n$ be the smallest integer such that $f(n) = 2^k + 1$, we have $n \\le 2^{k+1}$, by the minimality of $n$, $f(n-1) = 2^k$. Notice that $n - 2^k \\le 2^k$, we have\n$$\n2^k + 1 = f(n) = f(f(n-1)) + f(n - f(n-1)) = f(2^k) + f(n - 2^k) \\le 2f(2^k) = 2^k,\n$$\nwhich is a contradiction. It follows that $f(2^{k+1}) = 2^k$, the result is also true for $m = k + 1$. By induction, $f(2^m) = 2^{m-1}$ for any positive integer $m$.\n\n$\\boxed{f(2^m) = 2^{m-1}}$ for integer $m \\ge 2$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 28414,
    "subject": "Mathematics (Multi-modal)",
    "question": "Suppose there are 101 persons sitting around a round table in an arbitrary order. The $k$th person possesses $k$ pieces of cards, $k = 1, \\dots, 101$. We call it a *transition* if one transits one of his cards to one of his adjacent persons. Find the minimum positive number $k$, such that whatever the order of the seating, there is a way of no more than $k$ transitions so that each person possesses 51 cards.",
    "options": [],
    "answer": "42925",
    "solution": "The answer is $k = 42\\,925$.\n\nLet the circumference of the table be 101, and the distance between two adjacent persons be 1. In the following, we consider the least transition times.\n\nDenote the person who initially possesses $i$ cards by $[i - 51]$. So, if $p > 0$, then we can think of person $[p]$ as the source who should send out $p$ cards; and if $p < 0$, person $[p]$ is the sink who should receive $-p$ cards.\n\nSuppose that at the end of transitions, each person has 51 cards. If person $B$ possesses a card $u$ initially belonging to person $A$, we can think that $A$ carries card $u$ to $B$ by passing through the minor arc $AB$, the length of the route is the arc length $|AB|$ means the times of transitions.\n\nLet seating order be as shown in the figure. Person $[i]$ transits all his $i$ cards to person $[-i]$ ($i = 1, 2, \\dots, 50$) with route length $i$. So there are all together $1^2 + 2^2 + \\dots + 50^2 = 42\\,925$ transitions. In the following, we will show that no less transitions can meet the requirement if persons are seated in this way.\n\nWe use the notion of \"potential\". Let potential at the highest position $[50]$ be 50; the neighbor positions $[49]$ and $[48]$ each has potential 49, . . . , the lowest position $[-49]$ and $[-50]$ each has potential 0. Then, the total potential at the beginning is\n$$\nS = 101 \\times 50 + (100 + 99) \\times 49 + \\cdots + (2 + 1) \\times 0.\n$$\nAnd at the end of transitions, the total potential is\n$$\nT = 51 \\times 50 + (51 + 51) \\times 49 + \\cdots + (51 + 51) \\times 0.\n$$\nThe difference is $S - T = 42,925$.\n\nSince after each transition, the total potential changes at most 1, so at least 42,925 transitions are needed.\n\nIn the following, we show that whatever be the order of seating, there is always a way of no more than 42,925 transitions such that each person possesses 51 cards. To show this, we give two lemmas.\n\n**Lemma 1.** Let $c, a_0, a_1, \\dots, a_{n-1}$ be integers with their sum zero, and $c \\ge 0$, $a_0 \\le a_1 \\le \\dots \\le a_{n-1}$. If $n+1$ persons denoted by $[c], [a_0], [a_1], \\dots, [a_{n-1}]$, possess $N+c, N+a_0, N+a_1, \\dots$ and $N+a_{n-1}$ cards, respectively, where $N$ is a positive integer, such that $N+a_0 > 0$. Let the persons stand on 0, 1, ..., $n$ of the number axis, such that $[c]$ stands at $n$. Then there is a way of no more than $cn + \\sum_{i=0}^{n-1} ia_i$ transitions, such that each person possesses $N$ cards.\n\n**Proof of Lemma 1.** Suppose that $a_{n-1} \\ge \\cdots \\ge a_s > 0 \\ge a_{s-1} \\ge \\cdots \\ge a_0$, then induction on $M = a_{n-1} + \\cdots + a_s$. If $M=0$, then $[c]$ passes $c$ cards to $[a_i]$, such that $[a_i]$ obtains $-a_i$ cards ($0 \\le i \\le s-1$). Suppose that person $[a_i]$ stands at $x_i$ ($0 \\le i \\le n-1$), then $x_0, x_1, \\dots, x_{n-1}$ is a permutation of 0, 1, ..., $n-1$. Thus, a card that passes from $[c]$ to $[a_i]$ needs $n-x_i$ transitions. So the total transitions needed are\n$$\n\\sum_{i=0}^{s-1} (n - x_i) (-a_i) = cn + \\sum_{i=0}^{s-1} x_i a_i \\le cn + \\sum_{i=0}^{s-1} ia_i = cn + \\sum_{i=0}^{n-1} ia_i.\n$$\nNow suppose that the conclusion is true for integers smaller than $M$. Consider the case of integer $M$. Suppose that\n$$\na_{n-1} = \\cdots = a_{n-s} > a_{n-s-1}, \\quad a_l > a_{l-1} = \\cdots = a_0.\n$$\nThen there is a person in $[a_{n-1}]$, ..., $[a_{n-s}]$ and a person in $[a_0]$, ..., $[a_{l-1}]$ such that their distance is no more than $n-s-l+1$. We may suppose that the distance between $[a_{n-s}]$ and $[a_{l-1}]$ is no more than $n-s-l+1$, then let person $[a_{n-s}]$ pass a card $u$ to $[a_{l-1}]$ by no more than $n-s-l+1$ transitions. Next, by induction on $c$ and\n$$\n\\begin{align*}\na_{n-1} &\\ge \\dots \\ge a_{n-s+1} > a_{n-s} - 1 \\ge a_{n-s-1} \\\\\n&\\ge \\dots \\ge a_l \\ge a_{l-1} + 1 > a_{l-2} \\ge \\dots \\ge a_0,\n\\end{align*}\n$$\nwe see that by taking no more than\n$$\n\\begin{align*}\nL &= cn + (n-1)a_{n-1} + \\dots + (n-s+1)a_{n-s+1} + \\\\\n& \\quad (n-s)(a_{n-s}-1) + (n-s-1)a_{n-s-1} + \\dots + \\\\\n& \\quad la_l + (l-1)(a_{l-1}+1) + (l-2)a_{l-2} + \\dots + 0 \\cdot a_0\n\\end{align*}\n$$\ntransitions, each person possesses $N$ cards. Addition of the transitions of card $u$, we know that there are no more than\n$$\nL + (n - s - l + 1) = cn + \\sum_{i=0}^{n-1} ia_i\n$$\ntransitions. The proof of Lemma 1 is completed.\n\n**Lemma 2.** For any permutation of $[-50]$, $[-49]$, ..., $[49]$, $[50]$ on a circle, there always exists a person $[c]$, denote the line passing through $[c]$ and the origin by $l$, such that the sum of each side of $l$ (include $c$) in the brackets has the same sign of $c$.\n\n**Proof of Lemma 2.** Let the permutation on the circle be $[a_1]$, $[a_2]$, ..., $[a_{101}]$ clockwise. Then there is a directed diameter $l$ of the circle such that $[a_1]$, $[a_2]$, ..., $[a_{50}]$ are on one side of $l$ and $[a_{51}]$, $[a_{52}]$, ..., $[a_{101}]$ are on the other side.\n\nIf $\\sum_{i=1}^{50} a_i = 0$, then take $c = a_{51}$; else, if $\\sum_{i=1}^{50} a_i \\neq 0$, then the sums of each side of $l$ have different sign. If we rotate $l$ 180° clockwise, we see that the sum of each side changes sign. So there exists a $[c]$ which meets the requirement of Lemma 2.\n\nTake $[c]$ in Lemma 2. Suppose that $c \\ge 0$ (else change each $[a_i]$ by $[-a_i]$ and change all the directions of the arc). Let $c = c_1 + c_2, c_1, c_2 \\ge 0$, such that the sum of $c_1$ and the numbers on one side of (not include $c$) $l$ is zero. Denote 50 numbers on this side by $a_0 \\le a_1 \\le \\dots \\le a_{49}$, and denote 50 numbers on the other side by $b_0 \\le b_1 \\le \\dots \\le b_{49}$. Then the sum of $c_2$ and $b_0, b_1, \\dots, b_{49}$ is also zero. Using Lemma 1 to $c_1, a_0, a_1, \\dots, a_{49}$ and $c_2, b_0, b_1, \\dots, b_{49}$, respectively, we obtain that the transition times are no more than\n$$\nL = 50c_1 + \\sum_{j=0}^{49} ja_j + 50c_2 + \\sum_{j=0}^{49} jb_j.\n$$\nSince $c, a_0, \\dots, a_{49}, b_0, \\dots, b_{49}$ is a permutation of $-50, -49, \\dots, 50$, by the order inequality\n$$\n\\begin{align*}\nL &= 50c + \\sum_{j=0}^{49} j(a_j + b_j) \\\\\n&\\le 50^2 + \\sum_{j=0}^{49} j(2j - 50 + 2j - 49) \\\\\n&= 42\\,925.\n\\end{align*}\n$$\nSumming up, the least positive integer $k = 42\\,925$. $\\square$\n\n![](attached_image_1.png)",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 28415,
    "subject": "Mathematics (Multi-modal)",
    "question": "In a group of $m$ girls and $n$ boys, any two of them either know each other, or do not know each other. For any two boys and two girls, at least one boy and one girl do not know each other. Prove that the number of boy-girl pairs that know each other is at most $m + \\frac{n(n-1)}{2}$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "From the hypothesis, for any two boys, there is at most one girl that knows both of them. Let $x_i$ be the number of girls that know exactly $i$ boys, $1 \\le i \\le n$. So $\\sum_{i=1}^{n} x_i = m$. By counting the number of the above two boys-one girl combinations, we have\n$$\n\\sum_{i \\ge 2} \\frac{i(i-1)}{2} x_i \\le \\frac{n(n-1)}{2}.\n$$\nThe number of boy-girl pairs that know each other is then\n$$\n\\begin{aligned} \\sum_{i=1}^{n} ix_i &= m + \\sum_{i=2}^{n} (i-1)x_i \\le m + \\sum_{i=2}^{n} \\frac{i(i-1)}{2}x_i \\\\ &\\le m + \\frac{n(n-1)}{2}. \\end{aligned} \\quad \\square",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 28416,
    "subject": "Mathematics (Multi-modal)",
    "question": "In $\\triangle ABC$, point $B_2$ is the reflection of the center of B-excircle with respect to the midpoint of side $AC$, and point $C_2$ is the reflection of the center of C-excircle with respect to the midpoint of side $AB$. The A-excircle touches side $BC$ at point $D$. Prove that $AD \\perp B_2C_2$. (posed by Bian Hongping)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $A_1$ be the center of A-excircle. By the properties about centers of ex-circles, the following sets of three points are collinear: $\\{B_1, A, C_1\\}$, $\\{A_1, C, B_1\\}$ and $\\{C_1, B, A_1\\}$, and $A_1A \\perp B_1C_1$.\n\n![](attached_image_1.png)\nFig. 3. 1\n\nOn the plane, choose point $P$ such that $\\vec{C_2P} = \\vec{B_2C}$, then it follows from $\\vec{B_2C} = \\vec{AB_1}$ that $\\vec{C_2P} = \\vec{AB_1}$. As $\\vec{BC_2} = \\vec{C_1A}$ and the points $C_1, B, A_1$ are collinear, the points $B, C_2, P$ are collinear, so $\\vec{BP} = \\vec{C_1B_1}$. It follows from\n$$\n\\begin{aligned}\n\\angle AC_1B &= 180^\\circ - \\left( \\frac{180^\\circ - \\angle BAC}{2} \\right) - \\left( \\frac{180^\\circ - \\angle ABC}{2} \\right) \\\\\n&= \\frac{\\angle BAC + \\angle ABC}{2} = \\frac{180^\\circ - \\angle ACB}{2} = \\angle BCA_1\n\\end{aligned}\n$$\nthat $\\triangle A_1BC \\sim \\triangle A_1B_1C_1$. Let $A_1D$ and $A_1A$ be the altitudes of the $\\triangle A_1BC$ and $\\triangle A_1B_1C_1$, respectively, with respect to the opposite sides, so $\\frac{B_1C_1}{BC} = \\frac{A_1A}{A_1D}$.\n\nIf $\\vec{BP} = \\vec{C_1B_1}$, then $A_1A \\perp B_1C_1$, so $\\frac{BP}{BC} = \\frac{A_1A}{A_1D}$. If $BP \\perp A_1A$, then $BC \\perp A_1D$, so we have $\\triangle BPC \\sim \\triangle A_1AD$, and hence $CP \\perp AD$. Again by $\\vec{C_2P} = \\vec{B_2C}$, one has $\\vec{B_2C_2} = \\vec{CP}$,\nand hence $AD \\perp B_2C_2$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 28417,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all positive real numbers $t$ with the following property: there exists an infinite set $X$ of real numbers such that the inequality\n$$\n\\max\\{|x-(a-d)|, |y-a|, |z-(a+d)|\\} > td\n$$\nholds for all (not necessarily distinct) $x, y, z \\in X$, all real numbers $a$ and all positive real numbers $d$.",
    "options": [],
    "answer": "(0, 1/2)",
    "solution": "The answer is $0 < t < \\frac{1}{2}$.\n\nFirstly, for $0 < t < \\frac{1}{2}$, choose $\\lambda \\in \\left(0, \\frac{1-2t}{2(1+t)}\\right)$, let $x_i = \\lambda^i$, $X = \\{x_1, x_2, \\dots\\}$. We claim that for all (not necessarily distinct) $x, y, z \\in X$, all real numbers $a$ and all positive real numbers $d$, we have the following inequality:\n$$\n\\max \\{|x-(a-d)|, |y-a|, |z-(a+d)|\\} > td.\n$$\nSuppose on the contrary that there exists $a \\in \\mathbb{R}$, $d \\in \\mathbb{R}^+$ and $x_i, x_j, x_k$, such that\n$$\n\\max \\{|x_i-(a-d)|, |x_j-a|, |x_k-(a+d)|\\} \\le td.\n$$\nHence\n$$\n\\begin{cases} -td \\le x_i - (a-d) \\le td, \\\\ -td \\le x_j - a \\le td, \\\\ -td \\le x_k - (a+d) \\le td, \\end{cases}\n$$\ni.e.\n$$\n\\begin{cases} x_i + (1-t)d \\le a \\le x_i + (1+t)d, \\\\ x_j - td \\le a \\le x_j + td, \\\\ x_k - (1+t)d \\le a \\le x_k - (1-t)d, \\end{cases} \\quad (*)\n$$\nwhich implies that\n$$\n\\begin{cases} x_k - (1+t)d \\le a \\le x_i + (1+t)d, \\\\ x_i + (1-t)d \\le a \\le x_j + td, \\\\ x_j - td \\le a \\le x_k - (1-t)d, \\end{cases}\n$$\nnote that $0 < t < \\frac{1}{2}$, it follows that\n\n$$\n\\begin{cases} d \\ge \\frac{x_k - x_i}{2(1+t)}, \\\\ d \\le \\frac{x_j - x_i}{1-2t}, \\\\ d \\le \\frac{x_k - x_j}{1-2t}. \\end{cases}\n$$\nBy the second and third inequalities and $d > 0$, we get $x_i < x_j < x_k$, hence $i > j > k$, $\\lambda^j + \\lambda^{i+1} \\le \\lambda^{k+1} + \\lambda^i$, we get\n\n$$\n\\frac{x_j - x_i}{x_k - x_i} = \\frac{\\lambda^j - \\lambda^i}{\\lambda^k - \\lambda^i} \\le \\lambda.\n$$\nBy the first and second inequalities, we get $\\frac{x_j - x_i}{1-2t} \\ge \\frac{x_k - x_i}{2(1+t)}$, hence\n$$\n\\frac{x_j - x_i}{x_k - x_i} \\ge \\frac{1-2t}{2(1+t)} > \\lambda,\n$$\nwhich contradicts the previous inequality! Thus proved our earlier claim about $X$.\n\nSecondly, for $t \\ge \\frac{1}{2}$, we show that for any infinite set $X$, for any $x < y < z$ in $X$, we can choose $a \\in \\mathbb{R}$ and $d \\in \\mathbb{R}^+$ such that\n$$\n\\max \\{|x-(a-d)|, |y-a|, |z-(a+d)|\\} \\le td.\n$$\nIn fact, let $d = \\frac{z-x}{2}$, hence $x+(1-t)d = z-(1+t)d$. Let $a = \\max\\{x+(1-t)d, y-td\\}$. Since $t \\ge \\frac{1}{2}$, we obtain\n$$\n\\begin{cases} y-x < 2d \\le (1+2t)d, \\\\ x-y < 0 \\le (2t-1)d, \\end{cases}\n$$\ni.e.\n$$\n\\begin{cases} y-td \\le x+(1+t)d, \\\\ x+(1-t)d \\le y+td, \\end{cases}\n$$\nhence\n$$\n\\begin{cases} x+(1-t)d \\le a \\le x+(1+t)d, \\\\ y-td \\le a \\le y+td, \\\\ z-(1+t)d \\le a \\le z-(1-t)d, \\end{cases}\n$$\nfrom which we conclude that\n$$\n\\max\\{|x-(a-d)|, |y-a|, |z-(a+d)|\\} \\le td.\n$$\nSo every $t \\ge \\frac{1}{2}$ does not satisfy the requirement of the problem.\n\nIn conclusion, the set of all required $t$ is $(0, \\frac{1}{2})$.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 28418,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let the sequence $\\{a_n\\}$ be defined by $a_1 = 1$, $a_2 = 2$, $a_{n+1} = \\frac{a_n^2 + (-1)^n}{a_{n-1}}$ $(n = 2, 3, \\dots)$.\nProve that the sum of squares of any two adjacent terms of the sequence is also in the sequence.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "By $a_{n+1} = \\frac{a_n^2 + (-1)^n}{a_{n-1}}$, we have $a_{n+1}a_{n-1} = a_n^2 + (-1)^n$ $(n = 2, 3, \\dots)$, so\n$$\n\\begin{align*}\n\\frac{a_n - a_{n-2}}{a_{n-1}} &= \\frac{a_n a_{n-2} - a_{n-2}^2}{a_{n-1} a_{n-2}} = \\frac{a_{n-1}^2 + (-1)^{n-1} - a_{n-2}^2}{a_{n-1} a_{n-2}} \\\\\n&= \\frac{a_{n-1}^2 - a_{n-1} a_{n-3}}{a_{n-1} a_{n-2}} = \\frac{a_{n-1} - a_{n-3}}{a_{n-2}} \\\\\n&= \\dots = \\frac{a_3 - a_1}{a_2} = 2,\n\\end{align*}\n$$\nthat is, $a_n = 2a_{n-1} + a_{n-2}$ $(n \\ge 3)$, $a_1 = 1$, $a_2 = 2$.\n\nTherefore, $a_n = C_1\\lambda_1^n + C_2\\lambda_2^n$, $\\lambda_1 + \\lambda_2 = 2$, $\\lambda_1\\lambda_2 = -1$, $a_1 = 1$, $a_2 = 2$, $n \\in \\mathbb{N}^+$.\nThen, since $\\lambda_1\\lambda_2 = -1$ and $\\lambda_2 = 2 - \\lambda_1$, we have\n$$\n\\begin{cases} \n1 = C_1\\lambda_1 + C_2\\lambda_2 \\\\ \n2 = C_1\\lambda_1^2 + C_2\\lambda_2^2 \n\\end{cases} \n\\Rightarrow \n\\begin{cases} \n\\lambda_2 = C_1\\lambda_1\\lambda_2 + C_2\\lambda_2^2 \\\\ \n2 = C_1\\lambda_1^2 + C_2\\lambda_2^2 \n\\end{cases} \n\\Rightarrow \n\\begin{cases} \n2 - \\lambda_1 = -C_1 + C_2\\lambda_2^2 \\\\ \n2 = C_1\\lambda_1^2 + C_2\\lambda_2^2 \n\\end{cases} \n\\Rightarrow C_1(1 + \\lambda_1^2) = \\lambda_1.\n$$\nThus, by symmetric condition, we have $C_2(1 + \\lambda_2^2) = \\lambda_2$.\nSo, since $1 + \\lambda_1\\lambda_2 = 0$, we have\n$$\n\\begin{align*}\na_n^2 + a_{n+1}^2 &= C_1^2(1 + \\lambda_1^2)\\lambda_1^{2n} + C_2^2(1 + \\lambda_2^2)\\lambda_2^{2n} + \\\\\n&\\quad 2C_1C_2(\\lambda_1\\lambda_2)^n(1 + \\lambda_1\\lambda_2) \\\\\n&\\quad C_1\\lambda_1^{2n+1} + C_2\\lambda_2^{2n+1} = a_{2n+1}.\n\\end{align*}\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 28419,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find the number of polynomials $f(x) = a x^3 + b x$ that satisfy the following conditions:\n(1) $a, b \\in \\{1, 2, \\dots, 2013\\}$;\n(2) the difference of any two numbers among $f(1)$, $f(2)$, $\\dots$, $f(2013)$ is not a multiple of $2013$.",
    "options": [],
    "answer": "7200",
    "solution": "2013 is factorized as $2013 = 3 \\times 11 \\times 61$. Let $p_1 = 3$, $p_2 = 11$, $p_3 = 61$. We denote by $a_i$ the residue of $a$ modulo $p_i$, by $b_i$ the residue of $b$ modulo $p_i$ ($i = 1, 2, 3$), $a, b \\in \\{1, 2, \\dots, 2013\\}$. By the Chinese Remainder Theorem, we have a bijection of $(a, b)$ with $(a_1, a_2, a_3, b_1, b_2, b_3)$.\n\nNow, let $f_i(x) = a_i x^3 + b_i x$, $i = 1, 2, 3$. We call a polynomial \"good modulo $n$\" if the residues of $f(0), f(1), \\dots, f(n-1)$ modulo $n$ are all distinct.\n\nIf $f(x) = a x^3 + b x$ is not good modulo $2013$, then there exists $x_1 \\not\\equiv x_2 \\pmod{2013}$ such that $f(x_1) \\equiv f(x_2) \\pmod{2013}$. Suppose $x_1 \\not\\equiv x_2 \\pmod{p_i}$. Let $u_1$ and $u_2$ be the residues of $x_1$ and $x_2$ modulo $p_i$, respectively. Then $u_1 \\not\\equiv u_2 \\pmod{p_i}$ and $f_i(u_1) \\equiv f_i(u_2) \\pmod{p_i}$, so $f_i(x)$ is not good modulo $p_i$.\n\nIf $f(x) = a x^3 + b x$ is good modulo $2013$, then for every $i$, $f_i(x)$ is good modulo $p_i$. The reason is as follows. For any distinct pair $r_1, r_2 \\in \\{0, 1, \\dots, p_i - 1\\}$, there exist $x_1, x_2 \\in \\{1, 2, \\dots, 2013\\}$ such that $x_1 \\equiv r_1 \\pmod{p_i}$ and $x_2 \\equiv r_2 \\pmod{p_i}$ and $x_2 \\equiv x_2 \\pmod{\\frac{2013}{p_i}}$. Now $f(x_1) \\equiv f(x_2) \\pmod{\\frac{2013}{p_i}}$, but $f(x_1) \\not\\equiv f(x_2) \\pmod{2013}$, so $f(r_1) \\not\\equiv f(r_2) \\pmod{p_i}$.\n\nHence, we need to determine the number of good polynomials $f_i(x)$ modulo $p_i$.\n\nFor $p_1 = 3$, by Fermat's theorem, a good polynomial\n$$\nf_1(x) \\equiv a_1 x^3 + b_1 x \\equiv (a_1 + b_1) x \\pmod{3}\n$$\nis equivalent to say that $a_1 + b_1$ is not divisible by $3$. There are in total six such $f_1(x)$.\n\nFor $i = 2, 3$, if $f_i(x)$ is good modulo $p_i$, then for any $u$ and $v \\not\\equiv 0 \\pmod{p_i}$, $f_i(u + v) \\not\\equiv f_i(u - v) \\pmod{p_i}$, i.e.,\n$$\nf_i(u + v) - f_i(u - v) = 2v[a_i(3u^2 + v^2) + b_i]\n$$\nis not divisible by $p_i$. If $a_i \\neq 0$, the residues modulo $p_i$ of elements in the sets $A = \\{3a_i u^2 \\mid u = 0, 1, \\dots, \\frac{p_i-1}{2}\\}$ and $B = \\{(-b_i - a_i v^2) \\mid v = 1, 2, \\dots, \\frac{p_i-1}{2}\\}$ do not coincide, and $|A| + |B| = p_i$. So $A \\cup B$ forms a complete residue system modulo $p_i$. Their sum must be a multiple of $p_i$, i.e.,\n$$\n\\sum_{u=0}^{\\frac{p_i-1}{2}} 3a_i u^2 + \\sum_{v=1}^{\\frac{p_i-1}{2}} (-b_i - a_i v^2) \\equiv 0 \\pmod{p_i}.\n$$\nNow, $1^2 + 2^2 + \\dots + \\left(\\frac{p_i-1}{2}\\right)^2 = \\frac{1}{6} \\cdot \\frac{p_i-1}{2} \\cdot \\frac{p_i+1}{2} \\cdot p_i$ is a multiple of $p_i$, so $-\\frac{p_i-1}{2} \\cdot b_i$ is also a multiple of $p_i$. Hence, $b_i$ is divisible by $p_i$, i.e., exactly one of $a_i$, $b_i$ is $0$.\n\nIf $a_i = 0, b_i \\neq 0$, then $f_i(x) = b_i x$ is obviously good. There are $p_i - 1$ such good polynomials.\n\nIf $a_i \\neq 0, b_i = 0$, then $f_i(x) = a_i x^3$. For $p_2 = 11$, by Fermat's theorem, $(x^3)^7 = x^{21} \\equiv x \\pmod{11}$, so for $x_1 \\neq x_2 \\pmod{11}$ and $x_1^3 \\neq x_2^3 \\pmod{11}$, $f_2(x) = a_2 x^3$ is good. There are in total $10$ such polynomials.\n\nFor $p_3 = 61$, as $4^3 = 64 \\equiv 125 = 5^3 \\pmod{61}$, $f_3(x) = a_3 x^3$ cannot be good.\n\nTherefore, the total number that we are looking for is $6 \\times (10 + 10) \\times 60 = 7200$. $\\square$",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 28420,
    "subject": "Mathematics (Multi-modal)",
    "question": "Suppose the side of the base and the height of regular triangular pyramid $P$-$ABC$ are $1$ and $\\sqrt{2}$, respectively. Then the radius of the inscribed sphere of the pyramid is ______.",
    "options": [],
    "answer": "sqrt(2)/6",
    "solution": "As seen in Fig. 4.1, suppose the projections of the inscribed sphere's center $O$ on faces $ABC$ and $ABP$ are $H$, $K$, respectively, the midpoint of $AB$ is $M$, and the radius of the sphere is $r$. Then $P$, $K$, $M$ are collinear, $\\angle PHM = \\angle PKO = \\frac{\\pi}{2}$, and\n![](attached_image_1.png)\nFig. 4.1\n$$\nOH = OK = r, \\quad PO = PH - OH = \\sqrt{2} - r, \\\\\nMH = \\frac{\\sqrt{3}}{6} AB = \\frac{\\sqrt{3}}{6}, \\quad PM = \\sqrt{MH^2 + PH^2} = \\sqrt{\\frac{1}{12} + 2} = \\frac{5\\sqrt{3}}{6}.\n$$\nThen we have\n$$\n\\frac{r}{\\sqrt{2} - r} = \\frac{OK}{PO} = \\sin \\angle KPO = \\frac{MH}{PM} = \\frac{1}{5}.\n$$\nTherefore, $r = \\frac{\\sqrt{2}}{6}$.\nThe answer is $\\frac{\\sqrt{2}}{6}$.",
    "topic": "Geometry",
    "subtopic": "Solid Geometry"
  },
  {
    "id": 28421,
    "subject": "Mathematics (Multi-modal)",
    "question": "There are $n$ ($n \\ge 2$) coins in a row. If one of the coins is head, select an odd number of consecutive coins (or even 1 coin) with the one in head on the leftmost, and then flip all the selected coins upside down simultaneously. This is a *move*. No move is allowed if all $n$ coins are tails. Suppose $n$ coins are heads at the initial stage, determine if there is a way to carry out $\\lfloor \\frac{2^{n+1}}{3} \\rfloor$ moves. (posed by Gu Bin)",
    "options": [],
    "answer": "possible",
    "solution": "The answer is possible.\n\nFor any configuration of the coins, we define a corresponding 01-sequence $c_1c_2...c_n$ of length $n$ as follows: $c_i = 1$, if the $i$-th coin from the left is head, otherwise, $c_i = 0$. It is easy to see that the status of the $n$ coins has a one-to-one correspondence to such 01-sequences, so in the following, we will consider this sequence model instead.\n\nInitially, the sequence is $11...11$, denoted by $1^n$ (with $n$ consecutive digits of 1). Similarly, $00...00$, denoted by $0^n$ (with $n$ digits of 0). For any 01-sequence with at least a digit \"1\", consider the following move: locate the first digit \"1\" from right to left in the sequence, then take the 01-subsequence from left to right starting this \"1\" of maximal odd length, and change the 01-parity in this subsequence just like flipping the coins in a move. Denote by $a_n$ the total number of moves in the way stated above. We claim: $a_n = \\lfloor \\frac{2^{n+1}}{3} \\rfloor$.\n\nWhen $n=1$, it is easy to see that $a_1 = 1 = \\lfloor \\frac{2^2}{3} \\rfloor$, proceed by induction. Assume $a_k = \\lfloor \\frac{2^{k+1}}{3} \\rfloor$ holds for $n = k$, i.e., $a_k = \\lfloor \\frac{2^{k+1}}{3} \\rfloor$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 28422,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $p$ be a prime, $a$ and $k$ be positive integers, satisfying $p^a < k < 2p^a$. Prove that there exists positive integer $n$, $n < p^{2a}$ such that $C_n^k \\equiv n \\equiv k \\pmod{p^a}$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $p$ be a prime, $a$ and $k$ be positive integers, satisfying $p^a < k < 2p^a$. Prove that for any non-negative integer $b$, there exists positive integer $n$, $n < p^{a+b}$ such that $n \\equiv k \\pmod{p^a}$ and $C_n^k \\equiv k \\pmod{p^b}$.\n\nIf $b=0$, $p^b = 1$, take $n = k - p^a$. We prove by induction. Suppose that the conclusion is true for integer $b \\ge 0$. That is, there exists a positive integer $n < p^{a+b}$, $n \\equiv k \\pmod{p^a}$, and $C_n^k \\equiv k \\pmod{p^b}$.\n\nLet $1 \\le t \\le p-1$. Consider\n$$\nC_{n+tp^{a+b}}^{k} = \\prod_{i=0}^{k-1} \\frac{n + tp^{a+b} - i}{k-i}.\n$$\nFor integer $m$, let $P(m) = p^{v_p(m)}$, $r(m) = \\frac{m}{P(m)}$, where $v_p(m)$ is the number of $p$ in the standard factorization of $m$.\n\nSince $k - i < 2p^a \\le p^{a+1}$, we see that $v_p(k-i) \\le a$ and $n - i \\equiv k - i \\pmod{p^a}$. Hence,\n$$\nP(k-i) \\mid n + tp^{a+b} - i.\n$$\nConsequently,\n$$\nC_{n+tp^{a+b}}^{k} = \\prod_{i=0}^{k-1} \\frac{\\frac{n-i}{P(k-i)} + tp^{a+b-v_p(k-i)}}{r(k-i)}.\n$$\nIf $k - i \\neq p^a$, then $v_p(k - i) \\leqslant a - 1$ and $a + b - v_p(k - i) \\geqslant b + 1$. If $k - i = p^a$, then $v_p(k - i) = a$. Hence,\n$$\nC_{n+tp^{a+b}}^{k} \\equiv \\prod_{i=0}^{k-1} \\frac{n-i}{P(k-i)r(k-i)} + \\left( \\prod_{\\substack{0 \\le i \\le k-1 \\\\ i \\ne k-p^a}}^{k-1} \\frac{n-i}{P(k-i)r(k-i)} \\right) \\cdot tp^b \\\\\n\\equiv C_n^k + \\left( \\prod_{\\substack{0 \\le i \\le k-1 \\\\ i \\ne k-p^a}}^{k-1} \\frac{r(n-i)}{r(k-i)} \\right) \\cdot tp^b \\pmod{p^{b+1}}.\n$$\nThis is because that if $k - i \\neq p^a$, then $p^a \\mid (n - i) - (k - i)$. So\n$$\nv_p(n-i) = v_p(k-i).\n$$\nSince $\\prod_{\\substack{0 \\le i \\le k-1 \\\\ i \\ne k-p^a}} \\frac{r(n-i)}{r(k-i)}$ is coprime to $p$, we see that $C_{n+tp^{a+b}}^k$ ($0 \\le t \\le p-1$) goes through the following remainders modulo $p^{b+1}$:\n$$\nC_n^k + j p^b, \\quad j = 0, 1, \\dots, p-1.\n$$\nSince $C_n^k \\equiv k \\pmod{p^b}$, there exists $j$ ($0 \\le j \\le p-1$), such that $C_n^k + j p^b \\equiv k \\pmod{p^{b+1}}$. That is, there exists $t$ ($0 \\le t \\le p-1$), such that $C_{n+tp^{a+b}}^k \\equiv k \\pmod{p^{b+1}}$. Let $N = n + tp^{a+b}$. Then $N < p^{a+b+1}$, $N \\equiv n \\equiv k \\pmod{p^a}$ and $C_N^k \\equiv k \\pmod{p^{b+1}}$.\n\nThe extended problem is proved by induction. ☐",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 28423,
    "subject": "Mathematics (Multi-modal)",
    "question": "Given an integer $n \\ge 2$. Suppose $A_1, A_2, \\dots, A_n$ are $n$ nonempty finite sets satisfying: $|A_i \\Delta A_j| = |i-j|$ for all $i, j \\in \\{1, 2, \\dots, n\\}$.\nFind the minimum value of $|A_1| + |A_2| + \\dots + |A_n|$. (Here $|X|$ denotes the number of elements of a finite set $X$ and $X \\Delta Y = \\{a \\mid a \\in X, a \\notin Y\\} \\cup \\{a \\mid a \\in Y, a \\notin X\\}$ for any sets $X$ and $Y$.)",
    "options": [],
    "answer": "⌊n^2/4⌋ + 2",
    "solution": "For each positive integer $k$, we prove that the minimum value of $S_{2k}$ is $k^2 + 2$; the minimum value of $S_{2k+1}$ is $k(k+1)+2$.\nFirstly, define the sets $A_1, A_2, \\dots, A_{2k}, A_{2k+1}$ as follows:\n$$\nA_i = \\{i, i+1, \\dots, k\\}, \\quad i = 1, 2, \\dots, k; \\quad A_{k+1} = \\{k, k+1\\};\n$$\n$$\nA_{k+j} = \\{k+1, k+2, \\dots, k+j-1\\}, \\quad j = 2, 3, \\dots, k+1.\n$$\n\nFor this family of sets, it is easy to verify that\n$|A_i \\Delta A_j| = j - i = |i - j|$ holds in the following cases:\n(1) $1 \\le i < j \\le k$;\n(2) $1 \\le i < j = k+1$;\n(3) $1 \\le i < k+1 < j \\le 2k+1$;\n(4) $k+1 = i < j \\le 2k+1$;\n(5) $k+2 \\le i < j \\le 2k+1$.\nMoreover, the case of $i=j$ is trivial, and the case of $i>j$ can be reduced to the case of $i<j$. Thus for all $i, j \\in \\{1, 2, \\dots, 2k+1\\}$, we have verified that\n$$\n|A_i \\Delta A_j| = j - i = |i - j|.\n$$\nFor the above $(2k+1)$ sets, we can easily calculate that\n$$\nS_{2k+1} = \\frac{k(k+1)}{2} + 2 + \\frac{k(k+1)}{2} = k(k+1) + 2;\n$$\nif we choose the first $2k$ sets, we get that\n$$\nS_{2k} = S_{2k+1} - k = k^2 + 2.\n$$\n\nSecondly, we show that $S_{2k} \\ge k^2 + 2$ and $S_{2k+1} \\ge k(k+1)+2$. Note the following facts:\n**Fact 1:** for any two finite sets $X, Y$, we have\n$$|X| + |Y| \\ge |X \\Delta Y|.$$\n**Fact 2:** for any two nonempty finite sets $X, Y$, if $|X \\Delta Y| = 1$, then $|X| + |Y| \\ge 3$.\nWhen $n=2k$, it follows from Fact 1 that\n$$\n|A_i| + |A_{2k+1-i}| \\ge |A_i \\Delta A_{2k+1-i}| = 2k+1-2i, \\quad i=1, 2, \\dots, k-1.\n$$\nBy $|A_k \\Delta A_{k+1}| = 1$ and Fact 2, we have $|A_k| + |A_{k+1}| \\ge 3$. So\n$$\nS_{2k} = |A_k| + |A_{k+1}| + \\sum_{i=1}^{k-1} (|A_i| + |A_{2k+1-i}|) \\ge 3 + \\sum_{i=1}^{k-1} (2k+1-2i) = k^2 + 2.\n$$\nSimilarly, when $n=2k+1$, we get that\n$$\n|A_i| + |A_{2k+2-i}| \\ge |A_i \\Delta A_{2k+2-i}| = 2k+2-2i, \\quad i=1, 2, \\dots, k-1.\n$$\nSince $|A_k \\Delta A_{k+1}| = 1$, we have\n$$\n(|A_k| + |A_{k+1}|) + |A_{k+2}| \\geq 3 + 1 = 4,\n$$\nSO\n$$\nS_{2k+1} = |A_k| + |A_{k+1}| + |A_{k+2}| + \\sum_{i=1}^{k-1} (|A_i| + |A_{2k+2-i}|)\n$$\n$$\n\\geq 4 + \\sum_{i=1}^{k-1} (2k + 2 - 2i) = k(k+1) + 2.\n$$\nIn conclusion, the minimum value of $S_{2k}$ is $k^2 + 2$, the minimum value of $S_{2k+1}$ is $k(k+1) + 2$. Equivalently, for any $n \\geq 2$, the minimum value of $S_n$ is $\\left\\lfloor \\frac{n^2}{4} \\right\\rfloor + 2$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 28424,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $n$, $k$ be integers greater than $1$ and satisfy $n < 2^k$. Prove that there are $2k$ integers not divisible by $n$, such that if we divide them into two groups, then there must exist a group in which the sum of some integers can be divided by $n$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "At first, we consider the case that $n = 2^r$, $r \\ge 1$. Obviously, at this time $r < k$. We take three $2^{r-1}$'s and $2k-3$ $1$'s — each of them cannot be divided by $n$. If these $2k$ numbers are divided into two groups, then there must exist a group that contains two $2^{r-1}$'s, whose sum is $2^r$ — divisible by $n$.\n\nNext, we consider the case that $n$ is not a power of $2$. At this time, the $2k$ integers we take are\n$$\n-1, -1, -2, -2^2, \\dots, -2^{k-2}, 1, 2, 2^2, \\dots, 2^{k-1}.\n$$\nThen they each cannot be divided by $n$.\nAssume these numbers can be divided into two groups, such that any partial sum of numbers in one group is not divisible by $n$. We may say that $1$ is in the first group. Since $(-1) + 1 = 0$ is divisible by $n$, the two $-1$'s must be in the second group; since $(-1) + (-1) + 2 = 0$, the $2$ is in the first group; then the $-2$ is in the second group.\n\nNow by induction, assuming $1, 2, \\dots, 2^l$ are in the first group and $-1, -2, -2^2, \\dots, -2^l$ in the second one ($1 \\le l < k-2$), since\n$$\n(-1) + (-1) + (-2) + \\dots + (-2^l) + 2^{l+1} = 0\n$$\nis divisible by $n$, we get that $2^{l+1}$ is in the first group, and then $-2^{l+1}$ in the second.\n\nTherefore, $1, 2, 2^2, \\dots, 2^{k-2}$ is in the first group and $-1, -2, -2^2, \\dots, -2^{k-2}$ in the second. Finally, since\n$$\n(-1) + (-1) + (-2) + \\dots + (-2^{k-2}) + 2^{k-1} = 0,\n$$\nthen $2^{k-1}$ is in the first group. Therefore, $1, 2, 2^2, \\dots, 2^{k-1}$ are all in the first group.\n\nOn the other hand, the knowledge about the binary number system tells us that every positive integer which is not greater than $2^k - 1$ can be represented as the partial sum of $1, 2, 2^2, \\dots, 2^{k-1}$. Since $n \\le 2^k - 1$, then it is the partial sum of $1, 2, 2^2, \\dots, 2^{k-1}$ that is of course divisible by $n$ itself. This is a contradiction to the assumption.\n\nTherefore, we have found out $2k$ integers that meet the requirement in the question. The proof is then complete. ☐",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 28425,
    "subject": "Mathematics (Multi-modal)",
    "question": "Suppose that 12 acrobats labeled $1$–$12$ are divided into two circles $A$ and $B$, with six persons in each. Let each acrobat in $B$ stand on the shoulders of two adjacent acrobats of $A$. We call it a *tower* if the label of each acrobat of $B$ is equal to the sum of the labels of the acrobats under his feet. How many different towers can they make?\n\n(Remark. We treat two towers as the same if one can be obtained by rotation or reflection of the other. For example, the following towers are the same, where the labels inside the circle refer to the bottom acrobat, the labels outside the circle refer to the upper acrobat.)\n\n![](attached_image_1.png)",
    "options": [],
    "answer": "6",
    "solution": "Denote the sum of labels of $A$ and $B$ by $x$ and $y$, respectively. Then $y = 2x$. Thus, we have\n$$\n3x = x + y = 1 + 2 + \\cdots + 12 = 78, \\quad x = 26.\n$$\nObviously, $1, 2 \\in A$ and $11, 12 \\in B$. Denote $A = \\{1, 2, a, b, c, d\\}$, where $a < b < c < d$. Then $a + b + c + d = 23$, and $a \\ge 3$, $8 \\le d \\le 10$ (if $d \\le 7$, then $a + b + c + d \\le 4 + 5 + 6 + 7 = 22$, which is a contradiction.)\n\n(1) If $d = 8$, then $A = \\{1, 2, a, b, c, 8\\}$, $c \\le 7$, $a + b + c = 15$. Thus, $(a, b, c) = (3, 5, 7)$ or $(4, 5, 6)$, that is, $A = \\{1, 2, 3, 5, 7, 8\\}$ or $A = \\{1, 2, 4, 5, 6, 8\\}$.\n\nIf $A = \\{1, 2, 3, 5, 7, 8\\}$, then $B = \\{4, 6, 9, 10, 11, 12\\}$. Since $B$ contains $11, 4, 6$ and $12$, there is only one tower that, in $A$, $8$ and $3$, $3$ and $1$, $1$ and $5$, $5$ and $7$ are adjacent.\n\n![](attached_image_2.png)\n\nIf $A = \\{1, 2, 4, 5, 6, 8\\}$, then $B = \\{3, 7, 9, 10, 11, 12\\}$. Similarly, we see that, in $A$, $1$ and $2$, $5$ and $6$, $4$ and $8$ are adjacent, respectively. There are two arrangements, that is, there are two towers.\n\n![](attached_image_3.png)\n![](attached_image_4.png)\n\n(2) If $d = 9$, then $A = \\{1, 2, a, b, c, 9\\}$, $c \\le 8$, $a + b + c = 14$, where $(a, b, c) = (3, 5, 6)$ or $(3, 4, 7)$, that is, $A = \\{1, 2, 3, 5, 6, 9\\}$ or $A = \\{1, 2, 3, 4, 7, 9\\}$.\n\nIf $A = \\{1, 2, 3, 5, 6, 9\\}$, then $B = \\{4, 7, 8, 10, 11, 12\\}$. To obtain $4$, $10$ and $12$ in $B$, $1$, $3$, and $9$ in $A$ must be adjacent pairwise, it is impossible!\n\n![](attached_image_5.png)\n![](attached_image_6.png)\n\nIf $A = \\{1, 2, 3, 4, 7, 9\\}$, then $B = \\{5, 6, 8, 10, 11, 12\\}$. To obtain $6$, $8$ and $12$ in $B$, $2$ and $4$, $1$ and $7$, $9$ and $3$ must be adjacent in $A$, respectively. There are two arrangements, that is, there are two towers.\n\n(3) If $d = 10$, then $A = \\{1, 2, a, b, c, 10\\}$, where $c \\le 9$, $a + b + c = 13$. Thus, $(a, b, c) = (3, 4, 6)$, that is, $A = \\{1, 2, 3, 4, 6, 10\\}$ and $B = \\{5, 7, 8, 9, 11, 12\\}$. To obtain $8$, $9$, $11$ and $12$ in $B$, $6$ and $2$, $6$ and $3$, $10$ and $1$, $10$ and $2$ must be adjacent, respectively. There is only one tower.\n\n![](attached_image_7.png)\n\nSumming up, there are six different towers all together. ☐",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 28426,
    "subject": "Mathematics (Multi-modal)",
    "question": "For any integer $n$ with $n > 1$, let $n = p_1^{a_1} \\cdots p_t^{a_t}$ be its standard factorization, write\n$$\n\\omega(n) = t, \\quad \\Omega(n) = \\alpha_1 + \\cdots + \\alpha_t.\n$$\nProve or disprove the following statement: Given any positive integer $k$ and any positive real numbers $\\alpha$ and $\\beta$, there exists a positive integer $n$ with $n > 1$ such that\n$$\n\\frac{\\omega(n + k)}{\\omega(n)} > \\alpha \\quad \\text{and} \\quad \\frac{\\Omega(n + k)}{\\Omega(n)} < \\beta.\n$$",
    "options": [],
    "answer": "YES",
    "solution": "The answer is YES.\nFrom the definition of $\\omega$ and $\\Omega$, we have\n$$\n\\omega(ab) \\le \\omega(a) + \\omega(b), \\qquad \\textcircled{1}\n$$\n$$\n\\Omega(ab) = \\Omega(a) + \\Omega(b), \\qquad \\textcircled{2}\n$$\nfor any positive integers $a, b$. Given a fixed positive integer $k$ and positive real numbers $\\alpha, \\beta$, we take a positive integer $m > (\\omega(k) + 1)\\alpha$. As there are infinitely many prime numbers, we can take a sufficiently large prime $p$ such that $\\frac{\\Omega(k)+1}{p^m} + \\log_p 2 < \\beta$, and take $m$ pairwise distinct prime numbers $q_1, q_2, \\dots, q_m$ that are all greater than $p$. We will show that $n = 2^{q_1 q_2 \\cdots q_m} k$ has the desired property.\n\nFirst, we prove $\\frac{\\omega(n+k)}{\\omega(n)} > \\alpha$. Let $n_1 = \\frac{n+k}{k} = 2^{q_1 q_2 \\cdots q_m} + 1$. As $q_1, q_2, \\dots, q_m$ are all odd prime numbers, $2^{q_i} + 1 \\mid n_1$ when $1 \\le i \\le m$, then $d_i = \\frac{2^{q_i} + 1}{3}$ is an integer greater than 1.\n\nNote that\n$(2^r - 1, 2^s - 1) = 2^{(r,s)} - 1$ for all positive integers $r, s$, ③\nand $(q_i, q_j) = 1\\ (i \\neq j)$, we have\n$$\n\\begin{aligned}\n(d_i, d_j) &= \\frac{1}{3}(2^{q_i} + 1, 2^{q_j} + 1) \\le \\frac{1}{3}(2^{2q_i} - 1, 2^{2q_j} - 1) \\\\\n&= \\frac{2^{(2q_i, 2q_j)} - 1}{3} = \\frac{2^2 - 1}{3} = 1.\n\\end{aligned}\n$$\n$d_1, d_2, \\dots, d_m$ are the pairwise coprime factors of $n_1$, and each of them is greater than 1. Hence, $\\omega(n_1) \\ge m$. From ① and the choice of $m$, we have\n$$\n\\frac{\\omega(n+k)}{\\omega(n)} \\ge \\frac{\\omega(n_1)}{\\omega(n)} \\ge \\frac{\\omega(n_1)}{\\omega(k)+1} \\ge \\frac{m}{\\omega(k)+1} > \\alpha.\n$$\n\nNext, we prove $\\frac{\\Omega(n+k)}{\\Omega(n)} < \\beta$. As $q_1 q_2 \\cdots q_m$ is an odd number and cannot be divided by 3, we have $n_1 = 2^{q_1 q_2 \\cdots q_m} + 1 \\equiv \\pm 3 \\pmod{9}$, that is, $3 \\nmid n_1$. Suppose $q$ is a prime factor of $\\frac{n_1}{3}$ and $q \\le p$, then\n$$\n2^{2q_1 q_2 \\cdots q_m} - 1 = (2^{q_1 q_2 \\cdots q_m} - 1) \\cdot n_1 \\equiv 0 \\pmod{q}.\n$$\nFrom the Fermat's Little Theorem, $2^{q-1} \\equiv 1 \\pmod{q}$. From ③, $q \\mid 2^{(2q_1 q_2 \\cdots q_m, q-1)} - 1$. From $(q-1, 2q_1 q_2 \\cdots q_m) = (q-1, 2) \\le 2, q-1 < p < q_i\\ (i = 1, 2, \\dots, m)$, hence $q \\mid 2^2 - 1, q = 3$. This contradicts that $\\frac{n_1}{3}$ is not a multiple of 3. Therefore, each prime factor of $\\frac{n_1}{3}$ is larger than $p$. So $\\frac{n_1}{3} > p^{\\Omega(n_1/3)}$. From ② and the choice of primes $p$ and $q_1, q_2, \\dots, q_m$, we have\n$$\n\\begin{align*}\n\\Omega(n+k) &= \\Omega(k) + \\Omega(3) + \\Omega\\left(\\frac{n_1}{3}\\right) < \\Omega(k) + 1 + \\log_p\\left(\\frac{n_1}{3}\\right) \\\\\n&< \\Omega(k) + 1 + \\log_p(n_1 - 1) \\\\\n&= \\Omega(k) + 1 + q_1 q_2 \\cdots q_m \\log_p 2,\n\\end{align*}\n$$\n$$\n\\frac{\\Omega(n+k)}{\\Omega(n)} < \\frac{\\Omega(k) + 1 + q_1 q_2 \\cdots q_m \\log_p 2}{q_1 q_2 \\cdots q_m} < \\frac{\\Omega(k) + 1}{p^m} + \\log_p 2 < \\beta. \\quad \\square",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 28427,
    "subject": "Mathematics (Multi-modal)",
    "question": "A non-empty set $A \\subseteq \\{1, 2, 3, \\dots, n\\}$ is called a *good set* of degree $n$ if $|A| \\le \\min_{x \\in A} x$. Denote by $a_n$ the number of good sets of degree $n$. Prove that $a_{n+2} = a_{n+1} + a_n + 1$ for any positive integer $n$. (posed by Li Weigu)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $A$ be a good set of degree $n$, and $|A| = k$, then $\\min_{x \\in A} x \\ge k$, so $A \\subseteq \\{k, k+1, \\dots, n\\}$. Hence the number of good sets of degree $n$ with $k$ elements is $C_{n-k+1}^k$. It follows that $a_n = \\sum_{k=1}^{\\lfloor \\frac{n+1}{2} \\rfloor} C_{n-k+1}^k = C_n^1 + C_{n-1}^2 + C_{n-2}^3 + \\dots$.\n\nIf $n$ is even, $n = 2m$, then\n$$\n\\begin{align*}\na_{2m+2} &= C_{2m+2}^1 + C_{2m+1}^2 + \\dots + C_{m+2}^{m+1} \\\\\n&= (C_{2m+1}^1 + C_{2m+1}^0) + (C_{2m}^2 + C_{2m}^1) + \\dots + (C_{m+1}^{m+1} + C_{m+1}^m) \\\\\n&= (C_{2m+1}^1 + C_{2m}^2 + \\dots + C_{m+1}^{m+1}) + (C_{2m}^1 + C_{2m-1}^2 + \\dots \\\\\n&\\qquad\\qquad + C_{m+1}^m) + C_{2m+1}^0 \\\\\n&= a_{2m+1} + a_{2m} + 1.\n\\end{align*}\n$$\n\nIf $n$ is odd, $n = 2m - 1$, then\n$$\n\\begin{align*}\na_{2m+1} &= C_{2m+1}^1 + C_{2m}^2 + \\dots + C_{m+2}^m + C_{m+1}^{m+1} \\\\\n&= (C_{2m}^1 + C_{2m}^0) + (C_{2m-1}^2 + C_{2m-1}^1) + \\dots + (C_{m+1}^m + C_{m+1}^{m-1}) + C_m^m \\\\\n&= (C_{2m}^1 + C_{2m-1}^2 + \\dots + C_{m+1}^m) + (C_{2m-1}^1 + C_{2m-2}^2 + \\dots \\\\\n&\\qquad + C_{m+1}^{m-1} + C_m^m) + C_{2m}^0 \\\\\n&= a_{2m} + a_{2m-1} + 1.\n\\end{align*}\n$$\n\nIn summary, the equality $a_{n+2} = a_{n+1} + a_n + 1$ holds for all positive integers $n.\nIf $n=1$, there is only one good set of degree $1$, namely $\\{1\\}$, so $a_1 = 1$.\nIf $n=2$, then there are only two good sets of degree $2$: $\\{1\\}, \\{2\\}$, so $a_2 = 2$.\nRecall that $a_n, a_{n+1}$ are the numbers of the non-empty good subsets $A$ of $\\{1, 2, \\dots, n\\}$ and $\\{1, 2, \\dots, n+1\\}$, respectively, satisfying $|A| \\le \\min_{x \\in A} x$.\n\nConsider the case $n + 2$: For any non-empty good set $A$ of degree $n + 2$, $A$ is a subset of $\\{1, 2, \\dots, n + 2\\}$, then we have the following three cases:\n\na. $A$ does not contain the element $n + 2$;\nb. $A$ contains $n + 2$, and has at least $2$ elements;\nc. $A = \\{n + 2\\}$.\n\nIn the following, we focus on the number of good sets of types (a) and (b). For any good set $A$ in (a), it follows from $|A| \\le \\min_{x \\in A} x$ and $\\max_{x \\in A} x \\le n + 1$ that $A$ is a good set of degree $n + 1$. Conversely, any good set of degree $n + 1$ is also a good set of degree $n + 2$, therefore, there are exactly $a_{n+1}$ good sets of type (a).\n\nFor any good set $A = \\{a_1, a_2, \\dots, a_k, n+2\\}$ of degree $n+2$ of type (b), where $a_1 < a_2 < \\dots < a_k < n+2$. As $a_1 = \\min_{x \\in A} x \\ge |A| \\ge 2$, one can consider the non-empty set $A' = \\{a_1 - 1, a_2 - 1, \\dots, a_k - 1\\}$, where $1 \\le a_1 - 1 < a_2 - 1 < \\dots < a_k - 1 < n+1$, and $A'$ satisfies $|A'| = |A| - 1 \\le a_1 - 1 = \\min_{x \\in A'} x$, hence $A'$ is a good set of degree $n$. Conversely, any good set of degree $n$ can be represented in the form $A' = \\{a_1 - 1, a_2 - 1, \\dots, a_k - 1\\}$, where $A = \\{a_1, a_2, \\dots, a_k, n+2\\}$ is a good set of degree $n+2$ of type (b), so the correspondence is one-to-one between $A$ and $A'$, and there are exactly $a_n$ good sets of degree $n+2$ of type (b).\n\nAccording to the discussion on the number of good sets of types (a), (b) and (c), we have $a_{n+2} = a_{n+1} + a_n + 1$. $\\square$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 28428,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $a$, $b$ be real numbers, and $f(x) = a x + b$ satisfies $|f(x)| \\le 1$ for any $x \\in [0, 1]$. Then the maximum of $a b$\n\nis ______.",
    "options": [],
    "answer": "1/4",
    "solution": "It is easy to find that $a = f(1) - f(0)$, $b = f(0)$. Then\n$$\nab = f(0) \\cdot (f(1) - f(0)) = - \\left( f(0) - \\frac{1}{2} f(1) \\right)^2 + \\frac{1}{4} (f(1))^2 \\le \\frac{1}{4} (f(1))^2 \\le \\frac{1}{4}.\n$$\nWhen $2 f(0) = f(1) = \\pm 1$, i.e., $a = b = \\pm \\frac{1}{2}$, we get $a b = \\frac{1}{4}$.\nTherefore, the maximum of $a b$ is $\\frac{1}{4}$.\nThe answer is $\\frac{1}{4}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 28429,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $n \\ge 2$ and $a_1, a_2, \\dots, a_n, b_1, b_2, \\dots, b_n$ be non-negative integers. Prove that\n$$\n\\left(\\frac{n}{n-1}\\right)^{n-1}\\left(\\frac{1}{n}\\sum_{i=1}^{n} a_i^2\\right)+\\left(\\frac{1}{n}\\sum_{i=1}^{n} b_i\\right)^2 \\geqslant \\prod_{i=1}^{n}\\left(a_i^2+b_i^2\\right)^{\\frac{1}{n}}.\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Denote $\\lambda = \\left(\\frac{n}{n-1}\\right)^{n-1}$, $n \\ge 2$. Obviously, $\\lambda > 1$.\nFor given $i \\in \\{1, \\dots, n\\}$, fix $p = a_k^2 + b_k^2$ for $k = 1, 2, \\dots, n$ and fix $a_j$ and $b_j$ ($j \\ne i$). Then the left-hand side of\n$$\n\\textcircled{1} = \\frac{\\lambda}{n} (p - b_i^2 + \\sum_{j \\ne i} a_j^2) + \\frac{1}{n^2} (b_i + \\sum_{j \\ne i} b_j)^2\n$$\nis a quadratic function of $b_i$, $b_i \\in [0, \\sqrt{p}]$, with leading coefficient $-\\frac{\\lambda}{n} + \\frac{1}{n^2} < 0$. Thus, its minimum is taken at the endpoints, that is, $b_i = 0$ or $a_i = 0$.\nSo, we can suppose that $a_i b_i = 0, i = 1, 2, \\dots, n$.\n\n*Case* 1. Each $a_i = 0$, then by the mean value inequality, we have\n$$\n\\left(\\frac{1}{n} \\sum_{i=1}^{n} b_i\\right)^2 \\geq \\prod_{i=1}^{n} b_i^{\\frac{2}{n}}.\n$$\n\n*Case* 2. Each $b_i = 0$, then by the mean value inequality, we have\n$$\n\\lambda \\left( \\frac{1}{n} \\sum_{i=1}^{n} a_i^2 \\right) \\geq \\frac{1}{n} \\sum_{i=1}^{n} a_i^2 \\geq \\prod_{i=1}^{n} a_i^{\\frac{2}{n}}.\n$$\n\n*Case* 3. We may suppose that $b_1 = \\cdots = b_k = 0$, $a_{k+1} = \\cdots = a_n = 0$, $1 \\le k < n$.\nLet $a_1 a_2 \\cdots a_k = a^k, b_{k+1} \\cdots b_n = b^{n-k}, a, b \\ge 0$. Then by the\nmean value inequality, we have\n$$\na_1^2 + a_2^2 + \\cdots + a_k^2 \\ge ka^2, \\quad b_{k+1} + \\cdots + b_n \\ge (n-k)b.\n$$\nIt suffices to prove that\n$$\n\\frac{\\lambda k}{n} a^2 + \\frac{(n-k)^2}{n^2} b^2 \\geq a^{\\frac{2k}{n}} \\cdot b^{\\frac{2(n-k)}{n}} . \\qquad \\textcircled{2}\n$$\nBy the mean value inequality, we see that\nThe left-hand side of\n$$\n\\textcircled{2} = \\frac{\\lambda}{n} a^2 + \\cdots + \\frac{\\lambda}{n} a^2 + \\underbrace{\\frac{n-k}{n^2} b^2}_{k \\text{ terms}} + \\cdots + \\underbrace{\\frac{n-k}{n^2} b^2}_{n-k \\text{ terms}} \\\\\n\\geq \\lambda^{\\frac{k}{n}} a^{\\frac{2k}{n}} \\cdot \\left( \\frac{n-k}{n} \\right)^{\\frac{n-k}{n}} \\cdot b^{\\frac{2(n-k)}{n}}.\n$$\nSo, it suffices to show that $\\lambda^{\\frac{k}{n}} \\left(\\frac{n-k}{n}\\right)^{\\frac{n-k}{n}} \\ge 1$, that is, to\nshow $\\left(\\frac{n}{n-k}\\right)^{n-k} \\le \\lambda^k$.\nIn fact,\n$$\n\\begin{align*}\n& \\underbrace{\\frac{n}{n-k} \\cdot \\frac{n}{n-k} \\cdot \\dots \\cdot \\frac{n}{n-k}}_{n-k \\text{ terms}} \\cdot \\underbrace{1 \\cdot 1 \\cdot \\dots \\cdot 1}_{n-k \\text{ terms}} \\\\\n& \\le \\left( \\frac{n + (nk - n)}{nk - k} \\right)^{nk-k} \\\\\n& = \\left( \\frac{n}{n-1} \\right)^{(n-1)k} = \\lambda^k. \\quad \\square\n\\end{align*}\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 28430,
    "subject": "Mathematics (Multi-modal)",
    "question": "As shown in Fig. 7.1, $\\odot O_1$ and $\\odot O_2$ are tangent externally at point $T$. The quadrilateral $ABCD$ is inscribed in $\\odot O_1$. The lines $DA$ and $CB$ are tangent to $\\odot O_2$ at points $E$ and $F$, respectively. $BN$, the bisector of $\\angle ABF$, intersects the segment $EF$ at point $N$. Line $FT$ intersects the arc $AT$ (which does not contain $B$) at point $M$. Prove that $M$ is the excenter of $\\triangle BCN$.\n\n![](attached_image_1.png)\nFig. 7.1",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $P$ be the intersection of line $AM$ with $EF$. Join $AT$, $BM$, $BP$, $BT$, $CM$, $CT$, $ET$, $TP$. As shown in Fig. 7.2.\n\nAs $BF$ is tangent to $\\odot O_2$ at $F$, we have $\\angle BFT = \\angle FET$.\n\nAs $\\odot O_1$ is tangent to $\\odot O_2$ at $T$, we have $\\angle MBT = \\angle FET$.\n\nHence, $\\angle MBT = \\angle BFM$. So $\\triangle MBT$ and $\\triangle MFB$ are similar, therefore $MB^2 = MT \\cdot MF$. The same argument gives $MC^2 = MT \\cdot MF$.\n\nNow again from that $\\odot O_1$ is tangent to $\\odot O_2$ at $T$, we have $\\angle MAT = \\angle FET$. So $A$, $E$, $P$ and $T$ are concyclic, which implies that $\\angle APT = \\angle AET$. As $AE$ is tangent to $\\odot O_2$ at $E$, we have $\\angle AET = \\angle EFT$. Thus, $\\angle MPT = \\angle PFM$, and $\\triangle MPT$ is similar to $\\triangle MF$. Therefore, $MP^2 = MT \\cdot MF$.\n\nFrom the above argument, we have $MC = MB = MP$, which means that $M$ is the excenter of $\\triangle BCP$. So $\\angle FBP = \\frac{1}{2} \\angle CMP$. Meanwhile, $\\angle CMP = \\angle CDA = \\angle ABF$, and we have $\\angle FBN = \\frac{1}{2} \\angle ABF$. Hence, $\\angle FBN = \\angle FBP$, i.e., $P$ and $N$ coincide. $\\square$\n\n![](attached_image_2.png)\nFig. 7.2",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 28431,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $f(x) = \\lfloor \\frac{x}{1!} \\rfloor + \\lfloor \\frac{x}{2!} \\rfloor + \\dots + \\lfloor \\frac{x}{2013!} \\rfloor$, where $\\lfloor x \\rfloor$ is the greatest integer no greater than $x$. Call an integer $n$ a good number if the equation $f(x) = n$ has a real solution $x$. Find the number of good numbers in the set $\\{1, 3, 5, \\dots, 2013\\}$.",
    "options": [],
    "answer": "587",
    "solution": "First, we point out two obvious facts:\n(a) If $m$ is a positive integer and $x$ is real, then\n$$\n\\lfloor \\frac{x}{m} \\rfloor = \\lfloor \\frac{\\lfloor x \\rfloor}{m} \\rfloor.\n$$\n(b) For any integer $l$ and positive even number $m$, we have\n$$\n\\lfloor \\frac{2l + 1}{m} \\rfloor = \\lfloor \\frac{2l}{m} \\rfloor.\n$$\nLet $m = k!$ ($k = 1, 2, \\dots, 2013$) in (a) and summing up, we have\n$$\nf(x) = \\sum_{k=1}^{2013} \\lfloor \\frac{x}{k!} \\rfloor = \\sum_{k=1}^{2013} \\lfloor \\frac{\\lfloor x \\rfloor}{k!} \\rfloor = f(\\lfloor x \\rfloor),\n$$\nthat is, $f(x) = n$ has a real solution if and only if $f(x) = n$ has an integer solution. So, we only consider $x$ as an integer. Since\n$$\nf(x+1) - f(x) = \\lfloor x+1 \\rfloor - \\lfloor x \\rfloor + \\sum_{k=2}^{2013} \\left( \\lfloor \\frac{x+1}{k!} \\rfloor - \\lfloor \\frac{x}{k!} \\rfloor \\right) \\ge 1, \\quad \\textcircled{1}\n$$\nwe see that $f(x)$ ($x \\in \\mathbb{Z}$) is monotonously increasing. Now, we find integers $a$ and $b$, such that\n$$\n\\begin{align*}\nf(a-1) < 0 &\\le f(a) < f(a+1) < \\dots \\\\\n&< f(b-1) < f(b) \\le 2013 < f(b+1).\n\\end{align*}\n$$\nNote that $f(-1) < 0 = f(0)$, so $a = 0$. Since\n$$\n\\begin{align*}\nf(1173) &= \\sum_{k=1}^{6} \\left\\lfloor \\frac{1173}{k!} \\right\\rfloor = 1173 + 586 + 195 + 48 + 9 + 1 \\\\\n&= 2012 \\le 2013,\n\\end{align*}\n$$\n$$\n\\begin{align*}\nf(1174) &= \\sum_{k=1}^{6} \\left\\lfloor \\frac{1174}{k!} \\right\\rfloor = 1174 + 587 + 195 + 48 + 9 + 1 \\\\\n&= 2014 > 2013,\n\\end{align*}\n$$\nwe see that $b = 1173$.\nSo the good numbers in $\\{1, 3, 5, ..., 2013\\}$ are the odd numbers in\n$$\n\\{f(0), f(1), \\dots, f(1173)\\}.\n$$\nLet $x = 2l$ ($l = 0, 1, \\dots, 586$) in ①. By (b), we have\n$$\n\\left\\lfloor \\frac{2l + 1}{k!} \\right\\rfloor = \\left\\lfloor \\frac{2l}{k!} \\right\\rfloor \\quad (2 \\le k \\le 2013).\n$$\nThus,\n$$\nf(2l + 1) - f(2l) = 1 + \\sum_{k=2}^{2013} \\left( \\left\\lfloor \\frac{2l+1}{k!} \\right\\rfloor - \\left\\lfloor \\frac{2l}{k!} \\right\\rfloor \\right) = 1,\n$$\nthat is, there is exactly one odd number in $f(2l)$ and $f(2l + 1)$.\nTherefore, there are $\\frac{1174}{2} = 587$ odd numbers in $\\{f(0), f(1), \\dots, f(1173)\\}$, that is, there are 587 good numbers in the set $\\{1, 3, 5, \\dots, 2013\\}$.\n$\\square$",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 28432,
    "subject": "Mathematics (Multi-modal)",
    "question": "Given $X = \\{1, 2, \\dots, 100\\}$, consider function $f: X \\to X$ satisfying both the following conditions:\n(1) $f(x) \\neq x$ for all $x \\in X$;\n(2) $A \\cap f(A) \\neq \\emptyset$ for all $A \\subseteq X$ with $|A| = 40$.\n\nFind the smallest positive integer $k$, such that for any such function $f$ there exists a set $B \\subseteq X$ satisfying $|B| = k$ and $B \\cup f(B) = X$.\n**Remark.** For a subset $T$ of $X$, we define $f(T) = \\{x \\mid \\text{there exists } t \\in T \\text{ such that } x = f(t)\\}$.",
    "options": [],
    "answer": "69",
    "solution": "First, we define a function $f: X \\to X$ with\n$$\n\\begin{aligned}\nf(3i - 2) &= 3i - 1, \\quad f(3i - 1) = 3i, \\quad f(3i) = 3i - 2, \\\\\ni &= 1, 2, \\dots, 30, \\\\\nf(j) &= 100, \\quad 91 \\le j \\le 99, \\quad f(100) = 99.\\end{aligned}\n$$\nObviously, $f$ satisfies condition (1). For any $A \\subseteq X$ with $|A| = 40$, if\n(i) there exists an integer $i$ with $1 \\le i \\le 30$ such that $|A \\cap \\{3i-2, 3i-1, 3i\\}| \\ge 2$, then $A \\cap f(A) \\ne \\emptyset$; or\n(ii) $91, 92, \\dots, 100 \\in A$, then $A \\cap f(A) \\ne \\emptyset$ also holds.\nIn both cases, $f$ satisfies condition (2). If a subset $B$ of $X$ satisfies $f(B) \\cup B = X$, then we have $|B \\cap \\{3i-2, 3i-1, 3i\\}| \\ge 2$ for all $1 \\le i \\le 30$, $\\{91, 92, \\dots, 98\\} \\subseteq B$, and $B \\cap \\{99, 100\\} \\ne \\emptyset$. Hence, $|B| \\ge 69$.\n\nNext, we will show that, for any function $f$ satisfying the described conditions, there exists a subset $B \\subseteq X$ with $|B| \\le 69$ such that $f(B) \\cup B = X$.\n\nAmong all the subsets $U \\subseteq X$ with $U \\cap f(U) = \\emptyset$, choose one such that $|U|$ is maximal. If there are many $U \\subseteq X$ with $|U|$ being maximal, choose one such that $|f(U)|$ is maximal. The existence of $U$ is guaranteed by condition (1). Let $V = f(U)$, $W = X \\setminus (U \\cup V)$. Note that $U, V, W$ are pairwise disjoint and $X = U \\cup V \\cup W$. From condition (2), $|U| \\le 39$, $|V| \\le 39$, $|W| \\ge 22$.\n\nWe make the following assertions:\n(i) $f(w) \\in U$, for all $w \\in W$. Otherwise, let $U' = U \\cup \\{w\\}$, since $f(U) = V$, $f(w) \\notin U$, $f(w) \\ne w$, we have $U' \\cap f(U') = \\emptyset$. It is a contradiction to $|U|$ being maximal.\n(ii) $f(w_1) \\ne f(w_2)$ for all $w_1, w_2 \\in W$, $w_1 \\ne w_2$. Otherwise, let $u = f(w_1) = f(w_2)$ then by condition (1), $u \\in U$. Let $U' = (U \\setminus \\{u\\}) \\cup \\{w_1, w_2\\}$, since $f(U') \\subseteq V \\cup \\{u\\}$, $U' \\cap (V \\cup \\{u\\}) = \\emptyset$, we have $U' \\cap f(U') = \\emptyset$. It is a contradiction to $|U|$ being maximal.\n\nLet $W = \\{w_1, w_2, \\dots, w_m\\}$, $u_i = f(w_i)$, $1 \\le i \\le m$ then by (i) and (ii), $u_1, u_2, \\dots, u_m$ are distinct elements of $U$.\n(iii) $f(u_i) \\ne f(u_j)$ for all $1 \\le i < j \\le m$. Otherwise, let $v = f(u_i) = f(u_j) \\in V$, $U' = (U \\setminus \\{u_i\\}) \\cup \\{w_i\\}$, then $f(U') = V \\cup \\{u_i\\}$, $U' \\cap f(U') = \\emptyset$. However, $|f(U')| > |f(U)|$. It is a contradiction to $|f(U)|$ being maximal.\n\nTherefore, $f(u_1), f(u_2), \\dots, f(u_m)$ are distinct elements of $V$. In particular, $|V| \\ge |W|$. As $|U| \\le 39$, we have $|V| + |W| \\ge 61$ and $|V| \\ge 31$. Let $B = U \\cup W$, then $|B| \\le 69$ and $f(B) \\cup B \\supseteq V \\cup B = X$. Overall, the desired smallest integer $k$ is 69. $\\square$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 28433,
    "subject": "Mathematics (Multi-modal)",
    "question": "For each positive integer $n$ and each integer $i$ ($0 \\leq i \\leq n$), let $C_n^i \\equiv c(n,i) \\pmod{2}$, where $c(n,i) \\in \\{0, 1\\}$, and define\n$$\nf(n,q) = \\sum_{i=0}^{n} c(n,i)q^i.\n$$\nLet $m$, $n$ and $q$ be positive integers with $q+1$ not a power of $2$. Suppose that $f(m,q) \\mid f(n,q)$. Prove that\n$f(m,r) \\mid f(n,r)$ for every positive integer $r$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "For each positive integer $n$, we write $n$ in binary representation as $n = 2^{a_1} + 2^{a_2} + \\dots + 2^{a_k}$, where $0 \\le a_1 < a_2 < \\dots < a_k$. Define a set $T(n) = \\{2^{a_1}, \\dots, 2^{a_k}\\}$, $T(0)$ is considered empty set.\nBy Lucas' theorem, $C_n^i$ is odd if and only if $T(i) \\le T(n)$, hence\n$$\nf(n,q) = \\sum_{A \\subseteq T(n)} q^{\\sigma(A)} = \\prod_{a \\in T(n)} (1+q^a),\n$$\nwhere $\\sigma(A)$ denotes the sum of all elements of $A$.\nFor $m,n$ and $q$ as given by assumption, we show that if\n$$\nf(m,q) = \\prod_{a \\in T(m)} (1+q^a) \\mid \\prod_{a \\in T(n)} (1+q^a) = f(n,q),\n$$\nthen $T(m) \\subseteq T(n)$, and consequently, $f(m,r) \\mid f(n,r)$ for every $r$.\nFor any integers $i, j$, $0 \\le i < j$, we have the following factorization:\n$$\nq^{2j} - 1 = (q^{2j-1} + 1) \\cdots (q^2 + 1)(q^2 - 1),\n$$\ntherefore\n$$\n(q^{2j} + 1, q^{2i} + 1) = (q^{2i} + 1, 2) \\mid 2.\n$$\nLet $s(k)$ be the largest odd divisor of a positive integer $k$, then it follows that $s(q^{2i} + 1)$ and $s(q^{2j} + 1)$ are coprime. Clearly $q > 1$. If $i > 0$, $q^{2i} + 1 \\equiv 1 \\pmod{2}$, and $q^{2j} + 1 > 2$, thus $s(q^{2i} + 1) > 1$. If $i = 0$, since $q + 1$ is not a power of $2$, we have $s(q + 1) > 1$. For any $a \\in T(m)$, $s(q^a + 1) \\mid \\prod_{b \\in T(n)} s(q^b + 1)$. Since $s(1 + q^a) > 1$, we have $a \\in T(n)$, hence $T(m) \\subseteq T(n)$, which completes the proof!",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 28434,
    "subject": "Mathematics (Multi-modal)",
    "question": "In a plane with cartesian coordinates, let $P$ and $Q$ be two regions of convex polygon (including boundary and interior) whose vertices are all integer points (i.e., their coordinates are all integers) and $T = P \\cap Q$. Prove that if $T$ is not empty and does not contain integer point, then $T$ is a non-degenerate convex quadrilateral. (posed by Qu Zhenhua)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Since the non-empty intersection $T$ of two convex closed polygons is a closed convex polygon or degenerated polygon, there are three possible cases.\n\n(1) $T$ is a point. Then $T$ must be the vertex of $P$ or $Q$, contradicting the fact that $T$ contains no integer point.\n\n(2) $T$ is a segment. Then $T$ must be the intersection of an edge of $P$ and an edge of $Q$, which contains the vertex of $P$ or $Q$, which is a contradiction.\n\n(3) $T$ is a closed convex polygon.\nSo, it remains to be shown that $T$ is a quadrilateral.\nFirst, we note that if $T$ has two adjacent edges on the edges of $P$ (or $Q$), then the common vertex of these two edges must be the vertex of $P$ (or $Q$), which is a contradiction. Thus, the boundary of $T$ is formed alternately by a part of an edge of $P$ and then a part of an edge of $Q$, and each vertex of $T$ is the intersect point of edges of $P$ and $Q$. Thus, the number of edges of $T$ is even.\nWe see that if an edge $e$ of $P$ intersects an edge $f$ of $Q$, then $e$ must intersect another edge of $Q$, otherwise $T$ will contain an integer point.\nIn the following, we show by contradiction that the number of edges of $T$ can be 6 or more.\nIf $T$ has edges no less than 6, then suppose $P$ contains $k$ integer points except the vertices of $P$.\n\n**Case 1.** If $k=0$, then $P$ is an element integer triangle, or a parallelogram with area 1. So, $P$ can be located between two parallel lines $l_1, l_2$. And there is no integer point in the open domain $\\Omega$ between $l_1$ and $l_2$. At least three edges of $P$ are the edges of $T$, because $T$ has at least six edges.\n(a) In case of $P$ being $\\triangle ABC$ (See Fig. 6.1), $DE$, $FG$ and $HI$ are edges of $Q$. $D, G$ and $E$ may coincide with $F, H$ and $I$, respectively. Lines $FG$, $HI$, $l_1$ and $l_2$ form a convex quadrilateral. Since line $DE$ does not intersect segment $BC$, we see that the intersect point of line $DE$ and $FG$ or of line $DE$ and $HI$ is in $\\Omega$. Thus, $Q$ has integer vertex in $\\Omega$, a contradiction.\n![](attached_image_1.png)\n\n(b) In case of $P$ being a parallelogram $\\square ABCD$. Let $AD$, $AB$ and $BC$ be three edges of $P$ (see Fig. 6.2). The intersection point of line $EF$ and $HG$ locates in $\\Omega$. Let $AB$, $BC$ and $CD$ be three edges of $P$ (see Fig. 6.3), and there is no edge of $Q$ on $AD$. Similar to the case of (a), we can see the intersection point of line $EF$ and $HG$ or of line $EF$ and $IJ$ locates in $\\Omega$, which is a contradiction.\n![](attached_image_2.png)\n![](attached_image_3.png)\n\nCase 2. $k \\ge 1$. Consider integer point $X$ on $P$ other than the vertices. Since $X \\notin T$, there exists an edge $MN$ of $T$ such that $T$ and $X$ are separated by line $MN$ (denote by $l$) (see Fig. 6.4). Thus, $MN$ is a part of boundary of $Q$. $M$ is on the edge $AB$ of $P$, $N$ is on the edge $CD$ of $P$. $A$, $C$ and $X$ are on the same side of $l$ ($A$ and $C$ may coincide, but $B$ and $D$ do not by the hypothesis that $T$ has at least six edges). Thus, there is another vertex $U$ of $T$ on $AB$, and there is another vertex $V$ of $T$ on $CD$.\n![](attached_image_4.png)\n\nDenote the convex hull of points $X$ and vertices of $P$ below $l$ by $P'$. Then $P'$ and $P$ coincide below $BD$, and the part of $P'$ up $BD$ is $\\triangle BXD$. Comparing $T' = P' \\cap Q$ with $T$, we see that $T' \\subset T$, and the boundary of $T'$ is the boundary of $T$ with $MN$, $MU$ and $NV$ replaced by $M'N'$, $M'U'$ and $N'V'$, respectively. That is, $T'$ and $T$ have the same number of edges, and the number of integer points of $P'$ other than vertices is less than that of $P$. By a finite procedure like this, we can obtain a convex polygon with no interior integer point. By Case 1, it is impossible. ☐",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 28435,
    "subject": "Mathematics (Multi-modal)",
    "question": "Take randomly five different numbers from $1, 2, \\ldots, 20$. Then the probability that there are at least two adjacent numbers among them is ______.",
    "options": [],
    "answer": "232/323",
    "solution": "Suppose $a_1 < a_2 < a_3 < a_4 < a_5$ are taken from $1, 2, \\ldots, 20$. If $a_1, a_2, a_3, a_4, a_5$ are not adjacent to each other, then we have\n$$\n1 \\leq a_1 < a_2 - 1 < a_3 - 2 < a_4 - 3 < a_5 - 4 \\leq 16,\n$$\nfrom which we know that the number of ways to select five numbers not adjacent to each other from $1, 2, \\ldots, 20$ is the same as selecting five different numbers from $1, 2, \\ldots, 16$, i.e., $C_{16}^5$. Therefore, the required probability is\n$$\n\\frac{C_{20}^5 - C_{16}^5}{C_{20}^5} = 1 - \\frac{C_{16}^5}{C_{20}^5} = \\frac{232}{323}.\n$$\nThe answer is $\\frac{232}{323}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 28436,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $S$ be a subset of $m$ elements of $\\{0, 1, 2, \\dots, 98\\}$, $m \\geq 3$, such that for any $x, y \\in S$, there exists $z \\in S$ with $x + y \\equiv 2z \\pmod{99}$. Find all possible values of $m$.",
    "options": [],
    "answer": "3, 9, 11, 33, 99",
    "solution": "Let $S = \\{s_1, s_2, \\dots, s_m\\}$. As $S' = \\{0, s_2 - s_1, \\dots, s_m - s_1\\}$ satisfies also the hypothesis, we may assume without loss of generality that $0 \\in S$. For any $x, y \\in S$, $50(x + y) \\equiv z \\pmod{99} \\in S$. By taking $y = 0$, we have that for any $x \\in S$, $50x \\in S$. As $50$ and $99$ are coprime, there exists a positive integer $k$ such that $50^k \\equiv 1 \\pmod{99}$. Hence,\n$$\nx + y \\equiv 50^k (x + y) (\\text{mod } 99) \\in S.\n$$\nLet $d = \\gcd(99, s_1, s_2, \\dots, s_m)$. The above argument then implies that $d \\in S$, therefore $S = \\{0, d, 2d, \\dots\\}$. For any positive factor $d < 99$ of $99$, this $S$ satisfies all the requirements. Hence, all the possible values of $m$ are $3$, $9$, $11$, $33$, $99$. $\\square$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 28437,
    "subject": "Mathematics (Multi-modal)",
    "question": "Given positive integers $m$ and $n$. Find the smallest integer $N$ ($\\ge m$) with the following property: if an $N$-element set of integers contains a complete residue system modulo $m$, then it has a nonempty subset such that the sum of its elements is divisible by $n$.",
    "options": [],
    "answer": "N = \\max\\{\\, m,\\; m + n - \\tfrac{1}{2} m\\big(\\gcd(m,n) + 1\\big) \\,\\}.",
    "solution": "$$\nN = \\max\\{m, m+n - \\frac{1}{2}m[(m, n) + 1]\\}.\n$$\n\nFirst we show that $N \\ge \\max\\{m, m+n - \\frac{1}{2}m[(m,n)+1]\\}$.\nLet $d = (m,n)$, and write $m = d m_1$, $n = d n_1$. If $n > \\frac{1}{2}m(d+1)$, there exists a complete residue system modulo $m$, $x_1, x_2, \\dots, x_m$, such that their residues modulo $n$ consist exactly of $m_1$ groups of $1, 2, \\dots, d$. For example, the following $m$ numbers have the required property:\n$$\ni + d n_1 j, \\quad i = 1, 2, \\dots, d, \\quad j = 1, 2, \\dots, m_1.$$\nFinding another $k = n - \\frac{1}{2}m(d+1) - 1$ numbers $y_1, y_2, \\dots, y_k$ that are congruent to $1$ modulo $n$, the set\n$$\nA = \\{x_1, x_2, \\dots, x_m, y_1, \\dots, y_k\\}\n$$\ncontains a complete residue system modulo $m$, however none of its nonempty subsets has its sum of elements divisible by $n$. In fact, the sum of the (smallest nonnegative) residue modulo $n$ of all elements of $A$ is greater than zero and less than or equal to $m_1(1+2+\\dots+d) + k = n-1$. Thus\n$$\nN \\geq m + n - \\frac{1}{2}m(d+1),\n$$\ni.e.\n$$\nN \\geq \\max\\{m, m+n - \\frac{1}{2}m[(m,n)+1]\\}.\n$$\n\nNext we show that $N = \\max\\{m, m+n - \\frac{1}{2}m[(m,n)+1]\\}$ has the required property.\nThe following key fact is frequently used in the proof: among any $k$ integers, one can find a (nonempty) subset whose sum is divisible by $k$. Let $a_1, a_2, \\dots, a_k$ be integers, $S_i = a_1 + a_2 + \\dots + a_i$. If some $S_i$ is divisible by $k$, then the result is true. Otherwise there exist $1 \\le i < j \\le k$, such that $S_i \\equiv S_j \\pmod{k}$, then $S_j - S_i = a_{i+1} + \\dots + a_j$ is divisible by $k$, the result is again true. The following fact is an easy corollary of the previous result: among any $k$ integers, each of which is a multiple of $a$, one can find a (nonempty) subset whose sum is divisible by $ka$.\n\nReturning to the problem, we shall discuss two cases.\n\nCase 1: $n \\le \\frac{1}{2}m(d+1)$, and $N=m$.\nWe call a finite set of integers a $k$-set if the sum of all its elements is divisible by $k$. Let $x_1, x_2, \\dots, x_m$ be a complete residue system modulo $m$. Clearly we can divide these numbers into $m_1$ groups, each group consisting of a complete residue system modulo $d$. Let $y_1, y_2, \\dots, y_d$ be a complete residue system modulo $d$, and $y_i \\equiv i \\pmod d$. If $d$ is odd, we can divide each group into $\\frac{d+1}{2}$ $d$-sets, for example:\n$\\{y_1, y_{d-1}\\}, \\dots, \\{y_{\\frac{d-1}{2}}, y_{\\frac{d+1}{2}}\\}, \\{y_d\\}$. We get $\\frac{1}{2}m_1(d+1)$ $d$-sets.\nSince $n_1 \\le \\frac{1}{2}m_1(d+1)$, we can choose some of these $d$-sets such that the sum of their elements is divisible by $n_1 d (=n)$. If $d$ is even, similarly, a complete residue system modulo $d$ can be divided into $\\frac{d}{2}$ $d$-sets, with $y_{\\frac{d}{2}}$ remaining. Two remaining numbers can form another $d$-set. In the end we divide $x_1, x_2, \\dots, x_m$ into $\\frac{1}{2}m_1 d + \\left[\\frac{m_1}{2}\\right]$ $d$-sets (possibly with a number left if $m_1$ is odd).\nSince $n_1 \\le \\frac{1}{2}m_1(d+1) = \\frac{1}{2}m_1 d + \\frac{m_1}{2}$, we have $n_1 \\le \\frac{1}{2}m_1 d + \\left[\\frac{m_1}{2}\\right]$, again we can find some of these $d$-sets such that the sum of all their elements is divisible by $n_1 d = n$.\n\nCase 2: $n > \\frac{1}{2}m(d+1)$, $N = m+n-\\frac{1}{2}m(d+1)$.\nLet $A$ be an $N$-element set, containing a complete residue system modulo $m$, $x_1, x_2, \\dots, x_m$, with some other $n-\\frac{1}{2}m(d+1)$ numbers. If $d$ is odd, as shown in case 1, we may divide $x_1, x_2, \\dots, x_m$ into $\\frac{1}{2}m_1(d+1)$ $d$-sets. Divide the remaining $n-\\frac{1}{2}m(d+1)$ numbers arbitrarily into $n_1 - \\frac{1}{2}m_1(d+1)$ groups, each with $d$ numbers. Among each group of $d$ numbers one may find a $d$-set, therefore we have another $n_1 - \\frac{1}{2}m_1(d+1)$ $d$-sets, and totally $n_1$ $d$-sets. If $d$ is even, as discussed in case 1, we may divide $x_1, x_2, \\dots, x_m$ into $\\frac{1}{2}m_1 d + \\left\\lfloor \\frac{m_1}{2} \\right\\rfloor$ $d$-sets. If $m_1$ is odd, we are left with a number $x_i$ with $d \\mid x_i - \\frac{d}{2}$. Dividing the other $n - \\frac{1}{2}m(d+1)$ numbers arbitrarily into $2n_1 - m_1(d+1)$ groups, each with $\\frac{d}{2}$ numbers. From each group we can find a $\\frac{d}{2}$-set; from any two $\\frac{d}{2}$-sets, we can find a $d$-set. If $m_1$ even, then we can find another $n_1 - \\frac{1}{2}m_1(d+1)$ $d$-sets, and totally $n_1$ $d$-sets. If $m_1$ is odd, $\\{x_i\\}$ is a $\\frac{d}{2}$-set, we have $2n_1 - m_1(d+1) + 1$ $\\frac{d}{2}$-sets, and also $n_1 - \\frac{1}{2}m_1(d+1) + \\frac{1}{2}$ $d$-sets. Again we can find $n_1$ $d$-sets. Finally we can choose some of these $n_1$ $d$-sets, such that the sum of all their elements is divisible by $n_1 d (=n)$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 28438,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $n$ be an integer greater than $1$. Denote the first $n$ primes in increasing order by $p_1, p_2, \\dots, p_n$ (i.e., $p_1 = 2, p_2 = 3, \\dots$). Let $A = p_1^{p_1} p_2^{p_2} \\cdots p_n^{p_n}$. Find all positive integers $x$ such that $\\frac{A}{x}$ is even and has exactly $x$ distinct positive divisors.",
    "options": [],
    "answer": "the product of the first n primes, p1 p2 ... pn",
    "solution": "By $2x \\mid A$, note that $A = 4 \\cdot p_2^{p_2} \\cdots p_n^{p_n}$. We may suppose that $x = 2^{\\alpha_1} p_2^{\\alpha_2} \\cdots p_n^{\\alpha_n}$, where $0 \\le \\alpha_1 \\le 1$, $0 \\le \\alpha_i \\le p_i$ ($i = 2, 3, \\dots, n$). Then, we have\n$$\n\\frac{A}{x} = 2^{2-\\alpha_1} p_2^{p_2-\\alpha_2} \\cdots p_n^{p_n-\\alpha_n}.\n$$\nHence, the number of different divisors of $\\frac{A}{x}$ is\n$$\n(3 - \\alpha_1)(p_2 - \\alpha_2 + 1)\\cdots(p_n - \\alpha_n + 1).\n$$\nWe know that\n$$\n(3 - \\alpha_1)(p_2 - \\alpha_2 + 1)\\cdots(p_n - \\alpha_n + 1) = x = 2^{\\alpha_1} p_2^{\\alpha_2} \\cdots p_n^{\\alpha_n}. \\qquad \\textcircled{1}\n$$\nBy induction on $n$, we shall prove that the array $(\\alpha_1, \\alpha_2, \\dots, \\alpha_n)$ satisfying $\\textcircled{1}$ is $(1, 1, \\dots, 1)$ ($n \\ge 2$).\n\n(1) If $n=2$, then $\\textcircled{1}$ becomes $(3 - \\alpha_1)(4 - \\alpha_2) = 2^{\\alpha_1} 3^{\\alpha_2}$, where $\\alpha_1 \\in \\{0, 1\\}$. If $\\alpha_1 = 0$, then $3(4 - \\alpha_2) = 3^{\\alpha_2}$ which has no integer solution $\\alpha_2$. If $\\alpha_1 = 1$, then $2(4 - \\alpha_2) = 2 \\cdot 3^{\\alpha_2}$. We have $\\alpha_2 = 1$. Thus, $(\\alpha_1, \\alpha_2) = (1, 1)$. That is, the conclusion is true for $n=2$.\n\n(2) Suppose that the conclusion is true for $n = k - 1$ ($k \\ge 3$), then when $n = k$, $\\textcircled{1}$ becomes\n$$\n(3 - \\alpha_1)(p_2 - \\alpha_2 + 1)\\cdots(p_{k-1} - \\alpha_{k-1} + 1)(p_k - \\alpha_k + 1) \\qquad \\textcircled{2} \\\\ = 2^{\\alpha_1} p_2^{\\alpha_2} \\cdots p_{k-1}^{\\alpha_{k-1}} p_k^{\\alpha_k}.\n$$\nIf $\\alpha_k \\ge 2$, considering\n$$\n0 < p_k - \\alpha_k + 1 < p_k,\n$$\n$$\n0 < p_i - \\alpha_i + 1 \\le p_i + 1 < p_k \\quad (1 \\le i \\le k - 1),\n$$\nwe see that the left-hand side of $\\textcircled{2}$ cannot be divided by $p_k$, but the right-hand side of $\\textcircled{2}$ is a multiple of $p_k$, which is a contradiction.\n\nIf $\\alpha_k = 0$, then $\\textcircled{2}$ becomes\n$$\n(3 - \\alpha_1)(p_2 - \\alpha_2 + 1)\\cdots(p_{k-1} - \\alpha_{k-1} + 1)(p_k + 1) \\quad \\textcircled{3} \\\\ = 2^{\\alpha_1} p_2^{\\alpha_2} \\cdots p_k^{\\alpha_k}.\n$$\nNote that $p_2, p_3, \\dots, p_k$ are odd primes, thus, on the one hand, $p_k + 1$ is even. So the left-hand side of $\\textcircled{3}$ is even. On the other hand, the right side of $\\textcircled{3}$ is odd. So $\\alpha_1 = 1$. But then $3 - \\alpha_1 = 2$, so the left-hand side of $\\textcircled{3}$ is a multiple of $4$, but the right-hand side of $\\textcircled{3}$ is not, which is a contradiction.\n\nBy the above argument, we must have $\\alpha_k = 1$, and in $\\textcircled{2}$,\n$$\np_k - \\alpha_k + 1 = p_k^{\\alpha_k} = p_k.\n$$\nThus,\n$$\n(3 - \\alpha_1)(p_2 - \\alpha_2 + 1)\\cdots(p_{k-1} - \\alpha_{k-1} + 1) = 2^{\\alpha_1} p_2^{\\alpha_2} \\cdots p_{k-1}^{\\alpha_{k-1}}.\n$$\nBy the induction hypotheses, $\\alpha_1 = \\alpha_2 = \\cdots = \\alpha_{k-1} = 1$.\nThus, $\\alpha_1 = \\alpha_2 = \\cdots = \\alpha_{k-1} = \\alpha_k = 1$, that is, the conclusion is true for $n = k$.\n\nBy (1) and (2), we conclude that $(\\alpha_1, \\alpha_2, \\dots, \\alpha_n) = (1, 1, \\dots, 1)$, so the integer required is $x = 2p_2 \\cdots p_n = p_1 p_2 \\cdots p_n$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 28439,
    "subject": "Mathematics (Multi-modal)",
    "question": "Suppose real numbers $x$, $y$ satisfy $x - 4\\sqrt{y} = 2\\sqrt{x-y}$.\nThen the range of $x$ is ______.",
    "options": [],
    "answer": "{0} ∪ [4, 20]",
    "solution": "Let $\\sqrt{y} = a$, $\\sqrt{x-y} = b$ ($a, b \\ge 0$). Then $x = y + (x-y) = a^2 + b^2$. The equation in the question becomes $a^2 + b^2 - 4a = 2b$, which is equivalent to\n$$\n(a - 2)^2 + (b - 1)^2 = 5 \\quad (a, b \\ge 0).\n$$\nAs seen in Fig. 7.1, the trace of point $(a, b)$ in plane $aOb$ is the part of the circle with center $(2, 1)$ and radius $\\sqrt{5}$ satisfying $a, b \\ge 0$, i.e., the union of point $O$ and arc $\\widehat{ACB}$. Then\n![](attached_image_1.png)\nFig. 7.1\n$$\n\\sqrt{a^2 + b^2} \\in \\{0\\} \\cup [2, 2\\sqrt{5}].\n$$\nTherefore, $x = a^2 + b^2 \\in \\{0\\} \\cup [4, 20]$.\nThe answer is $\\{0\\} \\cup [4, 20]$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 28440,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $n \\ge 4$ be an even number. At the vertices of a regular $n$-gon we write in an arbitrary way $n$ distinct real numbers. Starting from one edge, we name all the edges in a clockwise way by $e_1, e_2, \\dots, e_n$. An edge is called \"positive\", if the difference of the numbers at its endpoint and its start point is positive. A set of two edges $\\{e_i, e_j\\}$ is called \"crossing\", if $2 \\mid (i+j)$, and among the four, the numbers written at their vertices, the largest and the third largest ones belong to the same edge. Prove that the number of crossings and the number of positive edges have different parity.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Without loss of generality, we may assume that the numbers written on the vertices are $1, 2, \\dots, n$. Let $A$ be the number of crossings, and $B$ be the number of positive edges. We will prove that the parity of $S$ remains the same if we exchange numbers $i$ and $i+1$. We distinguish two cases.\n\n**Case 1.** The numbers $i$ and $i+1$ are written on adjacent vertices, i.e., the endpoints of edge $e_k$. Once we exchange $i$ and $i+1$, the number of positive edges is modified by 1, thus the parity of $B$ is changed. On the other hand, the only two-edge subset that will become a new crossing (or change from a crossing to a non-crossing) is $\\{e_{k-1}, e_{k+1}\\}$ (the subindices are to be understood modulo $n$), all the other two-edge subsets will not be affected. So the parity of $A$ will change.\n\n**Case II.** The numbers $i$ and $i+1$ are written on non-adjacent vertices. Assume that they are written on the (common) endpoints of $e_j$, $e_{j+1}$ and of $e_k$, $e_{k+1}$, respectively. Once we exchange $i$ and $i+1$, every positive edge will remain positive, so are non-positive ones. Therefore, $B$ is unchanged. Now, for number of crossings, if a two-edge subset does not involve at the same time $i$ and $i+1$, then whether it is a crossing or not is not affected by the operation. So the only two-edge subsets to be considered are the two that have both $i$ and $i+1$ written on their vertices and the sum of the edge number is even. They will both become crossing after the exchange if they are not before, and vice-versa. Hence, the parity of $A$ remains the same.\n\nNow obviously, every pattern can be obtained from a finite number of such exchange if we start from writing $1, 2, \\dots, n$ consecutively in a clockwise way, and in the initial situation, $B = n-1, A = 0$. So $A$ and $B$ have different parity.\nStarting from one vertex, we denote the numbers written on the vertices by $x_1, x_2, \\dots, x_n$ in a clockwise way. We may assume that the numbers written on the endpoints of edge $e_i$ are $x_i, x_{i+1}$, $i = 1, 2, \\dots, n$, where $x_{n+1} = x_1$. Apparently $e_i$ is positive if and only if $x_{i+1} - x_i > 0$. Now, let $A$ be the number of positive edges, and $B$ be the number of crossings. Write\n$$\n\\beta = \\prod_{i=1}^{n} (x_{i+1} - x_i).\n$$\nAs $n$ is even, the sign of $\\beta$ is just $(-1)^A$.\n\nOn the other hand, $\\{e_i, e_j\\}$ is a crossing if and only if $2 \\mid i + j$ and\n$$\n(x_j - x_i)(x_{j+1} - x_{i+1})(x_{j+1} - x_i)(x_{j+1} - x_{i+1}) < 0.\n$$\nWe denote by $f(e_i, e_j)$ the left-hand side of the above inequality, obviously $f(e_i, e_j) = f(e_j, e_i)$. This quantity is negative if and only if the two-edge set is a crossing. Now, define\n$$\n\\alpha = \\prod_{\\substack{1 \\le i < j \\le n \\\\ 2 \\mid i+j}} (x_j - x_i)(x_j - x_{i+1})(x_{j+1} - x_i)(x_{j+1} - x_{i+1}) = \\prod_{\\substack{1 \\le i < j \\le n \\\\ 2 \\mid i+j}} f(e_i, e_j),\n$$\nthen the sign of $\\alpha$ is $(-1)^B$. Let us calculate the sign of $\\alpha\\beta$. For $1 \\le i < j \\le n$, consider the times of appearance, and the sign of $x_j - x_i$ in $\\alpha$ and $\\beta$, respectively. We distinguish several cases.\n\n**Case I.** $j - i = 1$, $x_j - x_i$ appears once in $\\beta$ with a positive sign, and once in $\\alpha$. If $i > 1$, it appears in $f(e_{i-1}, e_{i+1})$ with a positive sign. If $i = 1$, $x_2 - x_1$ appears in $f(e_2, e_n)$ with a negative sign. The product of these numbers has a negative sign.\n\n**Case II.** $2 \\le j - i < n - 1$, then $x_j - x_i$ does not appear in $\\beta$, and appear in $\\alpha$ twice. Among $(i-1, j-1)$ and $(i-1, j)$, there is exactly one pair that is of the same parity, among $(i, j-1)$ and $(i, j)$, there is also exactly one such pair. The sign of $x_j - x_i$ in each appearance is always positive. For $i = 1$, its appearance in $f(e_{j-1}, e_n)$ or $f(e_j, e_n)$ comes with a negative sign. Hence, there is a negative sign for each pair $(i, j)$ ($i = 1, j = 3, 4, \\dots, n-1$). This part of the product has the sign $(-1)^{n-3} = -1$.\n\n**Case III.** $i = 1, j = n$, then $x_n - x_1$ appears once in $\\beta$ with a negative sign, and once in $\\alpha$ (in $f(e_{n-1}, e_1)$) with a positive sign. This part of the product has a negative sign.\n\nIn brief, the sign of $\\alpha\\beta$ is negative, i.e., $A + B$ is odd. $\\square$",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 28441,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all positive integers $a$ such that $(2^n - n^2) \\mid (a^n - n^a)$ for all positive integers $n \\ge 5$. (posed by Yang Mingliang)",
    "options": [],
    "answer": "a = 2 or a = 4",
    "solution": "First, we prove that $a$ is even. It follows from the given condition by choosing an even integer $n \\ge 6$.\n\nNext, we prove that $a$ has no odd prime factor. Suppose the contrary, let $p$ be an odd prime factor of $a$. If $p = 3$, let $n = 8$, then $2^n - n^2 = 192$ has a factor $3$, but $a^n - n^a$ is not divisible by $3$, contradicting to $(2^n - n^2) \\mid (a^n - n^a)$, hence $p$ is not $3$.\n\nIf $p = 5$, let $n = 16$, then $2^n - n^2 = 64 \\cdot 110$ has a factor $5$, but $a^n - n^a$ is not divisible by $5$, contradicting $(2^n - n^2) \\mid (a^n - n^a)$, hence $p$ is not $5$.\n\nIf $p \\ge 7$, let $n = p - 1$, it follows from Fermat's Little Theorem that $2^{p-1} \\equiv 1 \\pmod p$. As $(p-1)^2 \\equiv 1 \\pmod p$, so $p \\mid (2^n - n^2)$. Moreover, since $a$ is even and $p \\mid a$, so $n^a \\equiv (p-1)^a \\equiv (-1)^a \\equiv 1 \\pmod p$ and $p \\mid a^n$, and hence $p$ does not divide $(a^n - n^a)$, contradicting $(2^n - n^2) \\mid (a^n - n^a)$.\n\nFinally, we prove that $a$ is $2$ or $4$. For this, let $a = 2^t$ where $t$ is a positive integer, then it follows from $(2^n - n^2) \\mid (2^{2n} - n^{2t})$ and $(2^n - n^2) \\mid (2^{2n} - n^{2t})$ that $(2^n - n^2) \\mid (n^{2t} - n^{2t})$.\n\nIf we choose $n$ to be sufficiently large, it follows from the fact $\\lim_{n \\to \\infty} \\frac{n^{2t}}{2^n} = 0$ that $n^{2t} - n^{2t} = 0$, hence $2^t = 2t$. $t = 1$ and $t = 2$ are obvious solutions.\n\nIf $t \\ge 3$, then by the Binomial Theorem, we have $t = 2^{t-1} = (1+1)^{t-1} > 1 + (t-1) = t$, which is impossible. At last, one can easily check that $a = 2$ and $a = 4$ satisfy the condition in the problem, so the solutions for $a$ are $2$ and $4$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 28442,
    "subject": "Mathematics (Multi-modal)",
    "question": "Suppose sequence $\\{a_n\\}$ consists of nine terms, which satisfy $a_1 = a_9 = 1$ and $\\frac{a_{i+1}}{a_i} \\in \\{2, 1, -\\frac{1}{2}\\}$ for any $i \\in \\{1, 2, \\dots, 8\\}$. Then the number of sequences like this is ______.",
    "options": [],
    "answer": "491",
    "solution": "Let $b_i = \\frac{a_{i+1}}{a_i}$ ($1 \\le i \\le 8$). Then for each $\\{a_n\\}$ satisfying the given condition, we have\n$$\n\\prod_{i=1}^{8} b_i = \\prod_{i=1}^{8} \\frac{a_{i+1}}{a_i} = \\frac{a_9}{a_1} = 1, \\text{ with } b_i \\in \\{2, 1, -\\frac{1}{2}\\} (1 \\le i \\le 8). \\qquad \\textcircled{1}\n$$\nConversely, a sequence of eight terms $\\{b_n\\}$ satisfying **1** can uniquely determine a sequence $\\{a_n\\}$ in the question.\n\nIn each $\\{b_n\\}$, there are obviously even number of $-\\frac{1}{2}$ and the same number of $2$, with the remainder being $1$. Or, in other words, the numbers of $-\\frac{1}{2}$ and $2$ are both $2k$, while the number of $1$ is $8-4k$. Here, it is easy to check that $k$ can only be $0, 1, 2$. Once $k$ is given, there are $C_8^{2k} C_{8-2k}^{2k}$ ways to construct $\\{b_n\\}$.\n\nTherefore, the total number of $\\{b_n\\}$ satisfying ① is\n$$\nN = 1 + C_8^2 C_6^2 + C_8^4 C_4^4 = 1 + 28 \\times 15 + 70 \\times 1 = 491.\n$$\nThe answer is 491.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 28443,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $n \\ge 3$ be integer. Suppose that $\\alpha, \\beta, \\gamma \\in (0, 1)$ and $a_k, b_k, c_k \\ge 0$ $(k = 1, 2, \\dots, n)$ satisfy $\\sum_{k=1}^n (k+\\alpha)a_k \\le \\alpha$, $\\sum_{k=1}^n (k+\\beta)b_k \\le \\beta$ and $\\sum_{k=1}^n (k+\\gamma)c_k \\le \\gamma$. Find the minimum of $\\lambda$ such that $\\sum_{k=1}^n (k+\\lambda)a_k b_k c_k \\le \\lambda$.",
    "options": [],
    "answer": "alpha*beta*gamma / ((1+alpha)*(1+beta)*(1+gamma) - alpha*beta*gamma)",
    "solution": "Let $a_1 = \\frac{\\alpha}{1+\\alpha}$, $b_1 = \\frac{\\beta}{1+\\beta}$, $c_1 = \\frac{\\gamma}{1+\\gamma}$, $a_i, b_i, c_i = 0$ $(i = 2, 3, \\dots, n)$. We see that all conditions are satisfied. So, we must have\n$$\n(1 + \\lambda) \\frac{\\alpha}{1 + \\alpha} \\cdot \\frac{\\beta}{1 + \\beta} \\cdot \\frac{\\gamma}{1 + \\gamma} \\le \\lambda,\n$$\nthat is,\n$$\n\\lambda \\ge \\frac{\\alpha\\beta\\gamma}{(1+\\alpha)(1+\\beta)(1+\\gamma) - \\alpha\\beta\\gamma}.\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 28444,
    "subject": "Mathematics (Multi-modal)",
    "question": "Suppose positive number sequence $\\{x_n\\}$ satisfies $S_n \\ge 2S_{n-1}$, $n = 2, 3, \\dots$, where $S_n = x_1 + \\dots + x_n$. Prove that there exists a constant $C > 0$, such that\n$$\nx_n \\ge C \\cdot 2^n,\\ n = 1, 2, \\dots\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "When $n \\ge 2$, $S_n \\ge 2S_{n-1}$ is equivalent to\n$$\nx_n \\ge x_1 + \\cdots + x_{n-1}. \\qquad ①\n$$\nLet $C = \\frac{1}{4}x_1$. We will prove\n$$\nx_n \\ge C \\cdot 2^n,\\ n = 1, 2, \\dots \\qquad \\textcircled{2}\n$$\nby induction.\n\nWhen $n = 1$, it is obviously true. When $n = 2$, we have $x_2 \\ge x_1 = C \\cdot 2^2$.\n\nWhen $n \\ge 3$, assume $x_k \\ge C \\cdot 2^k$, $k = 1, 2, \\dots, n-1$.\nThen from ①, we have\n$$\n\\begin{align*}\nx_n &\\ge x_1 + (x_2 + \\cdots + x_{n-1}) \\\\\n&\\ge x_1 + (C \\cdot 2^2 + \\cdots + C \\cdot 2^{n-1}) \\\\\n&= C(2^2 + 2^2 + 2^3 + \\cdots + 2^{n-1}) = C \\cdot 2^n.\n\\end{align*}\n$$\nTherefore, ② holds for every $n$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 28445,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $O$ be the circumcenter of triangle $ABC$. $H_A$ is the projection of $A$ onto $BC$. The extension of $AO$ intersects the circumcircle of $BOC$ at $A'$. The projections of $A'$ onto $AB$, $AC$ are $D$, $E$, and $O_A$ is the circumcenter of triangle $DH_AE$. Define $H_B, O_B, H_C, O_C$ similarly.\nProve: $H_AO_A$, $H_BO_B$, $H_CO_C$ are concurrent.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $T$ be the symmetry point of $A$ with regard to $BC$, $F$ be the projection of $A'$ onto $BC$, $M$ be the projection of $T$ onto $AC$.\nSince $AC = CT$, we have $\\angle TCM = 2\\angle TAM$. Since\n$$\n\\angle TAM = \\frac{\\pi}{2} - \\angle ACB = \\angle OAB, \\text{ we have}\n$$\n$$\n\\angle TCM = 2\\angle OAB = \\angle A'OB = \\angle A'CF, \\text{ and}\n$$\n$$\n\\angle TCH_A = \\angle A'CF + \\angle A'CT = \\angle TCM + \\angle A'CT = \\angle A'CE.\n$$\nBecause $\\angle CH_A T, \\angle CMT, \\angle CEA', \\angle CFA'$ are right angles, therefore\n$$\n\\frac{CH_A}{CM} = \\frac{CH_A}{CT} \\cdot \\frac{CT}{CM} = \\frac{\\cos \\angle TCH_A}{\\cos \\angle TCM} = \\frac{\\cos \\angle A'CE}{\\cos \\angle A'CF} = \\frac{CE}{CA'} \\cdot \\frac{CA'}{CF} = \\frac{CE}{CF},\n$$\ni.e. $CH_A \\cdot CF = CM \\cdot CE$, so $H_A, F, M, E$ are on the same circle $\\omega_1$.\n\n![](attached_image_1.png)\n\nSimilarly, let $N$ be the projection of $T$ onto $AB$, then $H_A, F, N, D$ are on the same circle $\\omega_2$. Since $A'FH_A T$ and $A'EMT$ are both right trapezoids, the perpendicular bisector of the segments $H_AF$ and $EM$ meet at the midpoint $K$ of the segment $A'T$, i.e. $K$ is the center of circle $\\omega_1$, $KF$ is the radius of circle $\\omega_1$. Similarly, $K$ and $KF$ are also the center and the radius of circle $\\omega_2$, respectively. Thus, $\\omega_1$ and $\\omega_2$ are the same, $D, N, F, H_A, E, M$ are on the same circle. So $O_A$ is the midpoint $K$ of $A'T$, $O_AH_A \\parallel AA'$.\n\nSince $\\angle H_CAO + \\angle AH_CH_B = \\frac{\\pi}{2} - \\angle ACB + \\angle ACB = \\frac{\\pi}{2}$, we have $AA' \\perp H_BH_C$, thus $O_AH_A \\perp H_BH_C$, therefore $O_AH_A, O_BH_B, O_CH_C$ all pass through the orthocenter of $\\triangle H_AH_BH_C$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 28446,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $A_1A_2\\cdots A_{101}$ be a regular 101-gon, and color every vertex red or blue. Let $N$ be the number of obtuse triangles satisfying the following: The three vertices of the triangle must be vertices of the 101-gon, both the vertices with acute angles have the same color, and the vertex with obtuse angle have different color.\n\n(1) Find the largest possible value of $N$.\n\n(2) Find the number of ways to color the vertices such that maximum $N$ is achieved. (Two colorings are different if for some $A_i$ the colors are different on the two coloring schemes)",
    "options": [],
    "answer": "Largest N = 32175; number of colorings achieving this maximum = 2 * (C(75, 24) + C(76, 25))",
    "solution": "Define $x_i = 0$ or $1$ depending on whether $A_i$ is red or blue. For obtuse triangle $A_{i-a}A_iA_{i+b}$ (vertex $A_i$ is the vertex of the obtuse angle, i.e. $a + b \\le 50$), these three vertices satisfy the conditions of the problem if and only if\n$$\n(x_i - x_{i-a})(x_i - x_{i+b}) = 1, \\qquad \\textcircled{1}\n$$\notherwise equals $0$, here the subscript is modulo $101$. Thus\n$$\nN = \\sum_{i=1}^{101} \\sum_{(a,b)} (x_i - x_{i-a})(x_i - x_{i+b}),\n$$\nhere $\\sum_{(a,b)}$ stands for the summation of all positive integer pairs $(a, b)$ satisfying $a + b \\le 50$, there are $49 + 48 + \\cdots + 1 = 1225$ such positive integer pairs. Expand $\\textcircled{1}$, we have\n$$\n\\begin{align*}\nN &= \\sum_{i=1}^{101} \\sum_{(a,b)} (x_i - x_{i-a})(x_i - x_{i+b}) \\\\\n  &= \\sum_{i=1}^{101} \\sum_{(a,b)} (x_i^2 - x_i x_{i-a} - x_i x_{i+b} + x_{i-a} x_{i+b}) \\\\\n  &= 1225 \\sum_{i=1}^{101} x_i^2 + \\sum_{i=1}^{101} \\sum_{k=1}^{50} (k-1-2(50-k)) x_i x_{i+k}\n\\end{align*}\n$$\n\n$$\n= 1225n + \\sum_{i=1}^{101} \\sum_{k=1}^{50} (3k-101)x_i x_{i+k}, \\qquad \\textcircled{2}\n$$\nhere $n$ is the number of the blue vertices. For any two vertices $A_i, A_j$,\n$1 \\le i \\le j \\le 101$, let\n$$\nd(A_i, A_j) = d(A_j, A_i) = \\min\\{j - i, 101 - j + i\\}.\n$$\nLet $B \\subseteq \\{A_1, A_2, \\dots, A_{101}\\}$ be the set of all blue vertices, then $\\textcircled{2}$ can be written as\n$$\nN = 1225n - 101C_n^2 + 3 \\sum_{\\{P,Q\\} \\subseteq B} d(P, Q), \\qquad \\textcircled{3}\n$$\nwhere $\\{P, Q\\}$ runs through all the two-element subsets of $B$. Without loss of generality, we assume $n$ is even. Otherwise, color all the vertices the opposite, the value of $N$ does not change. Write $n = 2t$, $0 \\le t \\le 50$, from one point, renumber all the blue vertices as $P_1, P_2, \\dots, P_{2t}$ clockwise. Then\n$$\n\\sum_{\\{P,Q\\} \\subseteq B} d(P, Q) = \\sum_{i=1}^t d(P_i, P_{i+t}) + \\frac{1}{2} \\sum_{i=1}^t \\sum_{j=1}^{t-1} (d(P_i, P_{i+j}) + d(P_{i+j}, P_{i+t}) + d(P_{i+t}, P_{i-j}) + d(P_{i-j}, P_i)) \\le 50t + \\frac{101}{2} t(t-1), \\qquad \\textcircled{4}\n$$\nhere the subscript of $P_i$ is modulo $2t$, and we use the inequality $d(P_i, P_{i+t}) \\le 50$ and\n$$\nd(P_i, P_{i+j}) + d(P_{i+j}, P_{i+t}) + d(P_{i+t}, P_{i-j}) + d(P_{i-j}, P_i) \\le 101. \\qquad \\textcircled{5}\n$$\nCombine $\\textcircled{3}$ and $\\textcircled{4}$, we have\n$$\nN \\le 1225n - 101C_n^2 + 3 \\left( 50t + \\frac{101}{2} t(t-1) \\right) = -\\frac{101}{2} t^2 + \\frac{5099}{2} t. \\qquad \\textcircled{6}\n$$\nThe right hand side of $\\textcircled{6}$ attains its maximum value when $t = 25$, thus $N \\le 32175$.\n\nThe number of ways to choose $50$ blue vertices such that $N$ achieves maximum is equal to the number of ways to choose $25$ diagonals from the longest $101$ diagonals such that any two diagonals do not have common vertices.\n\nEdging $A_i$ and $A_{i+50}$, $i = 1, 2, \\dots, 101$, we get a graph $G$. Notice that $50$ and $101$ are relatively prime, so $1 + 50n$ ($0 \\le n \\le 100$) form a complete residue system modulo $101$, i.e. $G$ is a circle with $101$ edges. Therefore, the number of ways (written as $S$) to color $50$ vertices blue and $N$ achieves maximum, is equal to the number of ways to choose $25$ edges of $G$, such that any two of them do not have common vertices. Now, fix one edge $e$ of $G$. Since the number of ways to choose edges having $e$ is $C_{75}^{24}$, not having $e$ is $C_{76}^{25}$, so $S = C_{75}^{24} + C_{76}^{25}$. Similarly, the number of ways to have $50$ red vertices is also $S$, thus the number of ways to color the vertices such that the largest value of $N$ is achieved is $2S$.\n\nIn a word, the largest possible value of $N$ is $32175$, the number of ways to color is $2S = 2(C_{75}^{24} + C_{76}^{25})$.",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 28447,
    "subject": "Mathematics (Multi-modal)",
    "question": "Show that there are no 2-tuples $(x, y)$ of positive integers satisfying the equation\n$$\n(x+1)(x+2)\\cdots(x+2014) = (y+1)(y+2)\\cdots(y+4028).\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "**Proof** For $n = 2^k \\cdot m$ ($k$ is a non-negative integer, $m$ is odd), let $v(n) = 2^k$.\nWe prove by contradiction: assume $(x, y)$ is one positive integer solution of the equation. Let\n$$\nv(x + i) = \\max_{1 \\le j \\le 2014} \\{v(x + j)\\},\n$$\nthen if $1 \\le j \\le 2014,\\ j \\ne i$,\n$$\nv(x + j) = v(x + i + (j - i)) = v(j - i),\n$$\nso\n$$\nv\\left(\\prod_{1 \\le j \\le 2014,\\ j \\ne i} (x + j)\\right) = v\\left((2014 - j)! \\cdot (j - 1)!\\right) \\le v(2013!)\n$$\nsince $\\prod_{j=1}^{2014} (x + j) = \\prod_{j=1}^{4028} (y + j)$ is a multiple of $4028!$, thus\n$$\nx + i \\geq v(x + i) \\geq v\\left(\\frac{4028!}{2013!}\\right) > 2^{1007},\n$$\ntherefore $x > 2^{1006}$. So $(y + 4028)^{4028} > \\prod_{j=1}^{4028} (y + j) = \\prod_{j=1}^{2014} (x + j) > 2^{1006 \\cdot 2014}$, we have $y + 4028 > 2^{503}$, $y > 2^{502}$.\n\n**Lemma:** Let $0 \\le x_i < \\frac{1}{2}\\ (1 \\le i \\le n)$, if $x = \\frac{1}{n} \\sum_{i=1}^{n} x_i$, $y = 2 \\max_{1 \\le i \\le n} \\{x_i^2\\}$, then\n$$\n1 - x \\ge \\left( \\prod_{i=1}^{n} (1 - x_i) \\right)^{\\frac{1}{n}} \\ge 1 - x - y.\n$$\n**Proof of lemma:** By AM-GM inequality, the inequality on the left is easy to prove. The inequality on the right holds, since\n$$\n\\begin{aligned}\n\\left( \\prod_{i=1}^{n} (1-x_i) \\right)^{\\frac{1}{n}} &\\ge \\frac{n}{\\sum_{i=1}^{n} \\frac{1}{1-x_i}} \\\\\n&\\ge \\frac{n}{\\sum_{i=1}^{n} (1+x_i + 2x_i^2)} \\\\\n&= \\frac{n}{n+nx+2\\sum_{i=1}^{n} x_i^2} \\\\\n&\\ge \\frac{1}{1+x+y} \\\\\n&\\ge 1-x-y,\n\\end{aligned}\n$$\nthe lemma is proved.\n\nBack to the problem, Let $w = x + 2015 > 2^{1007}$, $z = y + \\frac{4029}{2} > 2^{502}$, then the equation is equivalent to\n$$\nw \\cdot \\left( \\left(1 - \\frac{1}{w}\\right) \\left(1 - \\frac{2}{w}\\right) \\cdots \\left(1 - \\frac{2014}{w}\\right) \\right)^{\\frac{1}{2014}} = z^2 \\cdot \\left( \\left(1 - \\frac{1}{4z^2}\\right) \\left(1 - \\frac{9}{4z^2}\\right) \\cdots \\left(1 - \\frac{4027^2}{4z^2}\\right) \\right)^{\\frac{1}{2014}}.\n$$\nBy lemma,\n$$\nw\\left(1 - \\frac{2015}{2w}\\right) > w \\cdot \\left(\\left(1 - \\frac{1}{w}\\right)\\left(1 - \\frac{2}{w}\\right)\\cdots\\left(1 - \\frac{2014}{w}\\right)\\right)^{\\frac{1}{2014}}\n$$\n\\begin{aligned}\n&> w \\left( 1 - \\frac{2015}{2w} - \\frac{2 \\cdot 2014^2}{w^2} \\right) \\\\\n&> w \\left( 1 - \\frac{2015}{2w} \\right) - \\frac{1}{8},\n\\end{aligned}\n$$\ntherefore the decimal of $w \\cdot \\left( \\left( 1 - \\frac{1}{w} \\right) \\left( 1 - \\frac{2}{w} \\right) \\cdots \\left( 1 - \\frac{2014}{w} \\right) \\right)^{\\frac{1}{2014}}$ belongs to $\\left( \\frac{3}{8}, \\frac{1}{2} \\right)$.\nOn the other hand, by lemma\n$$\n\\begin{aligned}\nz^2 \\left( 1 - \\frac{1^2 + 3^2 + \\cdots + 4027^2}{4z^2 \\cdot 2014} \\right) &> z^2 \\cdot \\left( \\left( 1 - \\frac{1}{4z^2} \\right) \\left( 1 - \\frac{9}{4z^2} \\right) \\cdots \\left( 1 - \\frac{4027^2}{4z^2} \\right) \\right)^{\\frac{1}{2014}} \\\\\n&> z^2 \\left( 1 - \\frac{1^2 + 3^2 + \\cdots + 4027^2}{4z^2 \\cdot 2014} - \\frac{2 \\cdot 4027^4}{(4z^2)^2} \\right),\n\\end{aligned}\n$$\nthus\n$$\n\\begin{aligned}\nz^2 - \\frac{4 \\cdot 2014^2 - 1}{12} &> z^2 \\cdot \\left( \\left( 1 - \\frac{1}{4z^2} \\right) \\left( 1 - \\frac{9}{4z^2} \\right) \\cdots \\left( 1 - \\frac{4027^2}{4z^2} \\right) \\right)^{\\frac{1}{2014}} \\\\\n&> z^2 - \\frac{4 \\cdot 2014^2 - 1}{12} - \\frac{4027^4}{8z^2} \\\\\n&> z^2 - \\frac{4 \\cdot 2014^2 - 1}{12} - \\frac{1}{8}.\n\\end{aligned}\n$$\nSince $z^2 - \\frac{4 \\cdot 2014^2 - 1}{12}$ is an integer, so the decimal of $z^2 \\cdot \\left( \\left( 1 - \\frac{1}{4z^2} \\right) \\left( 1 - \\frac{9}{4z^2} \\right) \\cdots \\left( 1 - \\frac{4027^2}{4z^2} \\right) \\right)^{\\frac{1}{2014}}$ belongs to $\\left( \\frac{7}{8}, 1 \\right)$, this is a contradiction.\n\nTherefore, there are no 2-tuples $(x, y)$ of positive integers satisfying\n$$\n\\prod_{j=1}^{2014} (x + j) = \\prod_{j=1}^{4028} (y + j).\n$$",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 28448,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $k$ be a fixed odd integer, $k > 3$. Prove: There exist infinitely many positive integers $n$, such that there are two positive integers $d_1, d_2$ satisfying $d_1, d_2$ each dividing $\\frac{n^2+1}{2}$, and $d_1 + d_2 = n + k$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "$$\n((k-2)^2 + 1)xy = (x + y - k)^2 + 1, \\qquad \\textcircled{1}\n$$\nwe prove ① has infinitely many positive odd solutions $(x, y)$.\nObviously $(1,1)$ is one positive odd solution, let $(x_1, y_1) = (1,1)$. Assume $(x_i, y_i)$ is one positive odd solution of ①, and $x_i \\le y_i$, let $x_{i+1} = y_i$, $y_{i+1} = (k-1)(k-3)y_i + 2k - x_i$. Since ① can be written as\n$$\nx^2 - ((k-1)(k-3)y + 2k)x + (y-k)^2 + 1 = 0,\n$$\nby Vieta's theorem $(x_{i+1}, y_{i+1})$ is also one integer solution of ①. Since $x_i, y_i$ and $k$ are all positive odd integers, and $k \\ge 5$, so $x_{i+1}$ is a positive odd integer, and\n$$\ny_{i+1} = (k-1)(k-3)y_i + 2k - x_i \\equiv -x_i \\equiv 1 \\pmod{2},\n$$\n$y_{i+1} \\ge 8y_i + 2k - x_i > y_i > 0$. Thus $(x_{i+1}, y_{i+1})$ is one positive odd solution of ①, and $x_i + y_i < x_{i+1} + y_{i+1}$. By $(x_1, y_1)$ and the construction above, we get a series of positive odd solutions of ①: $(x_i, y_i)$, $i=1,2,\\dots$, such that $x_1 + y_1 < x_2 + y_2 < \\dots$.\nFor any integer $i$ greater than $k$, $x_i + y_i > k$. Let $n = x_i + y_i - k$, $d_1 = x_i$, $d_2 = y_i$, then $n$ is a positive odd integer, and $d_1 + d_2 = n + k$. Since $(k-2)^2 + 1$ is even, we can show that $d_1, d_2$ are both divisors of $\\frac{n^2+1}{2}$, and $d_1 + d_2 = n + k$. Thus such $n$ satisfies all the conditions, therefore there exist infinitely many positive odd integers $n$ satisfying the conditions.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 28449,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $n$ be a given integer which is greater than $1$. Find the greatest constant $\\lambda(n)$ such that for any non-zero complex $z_1, z_2, \\dots, z_n$, we have\n$$\n\\sum_{k=1}^{n} |z_k|^2 \\geq \\lambda(n) \\min_{1 \\leq k \\leq n} \\{|z_{k+1} - z_k|^2\\},\n$$\nwhere $z_{n+1} = z_1$.",
    "options": [],
    "answer": "lambda(n) = { n/4 if n is even; n / (4 cos^2(pi/(2n))) if n is odd }",
    "solution": "Let $\\lambda_0(n) = \\begin{cases} \\frac{n}{4}, & 2 \\mid n, \\\\ \\frac{n}{4 \\cos^2 \\frac{\\pi}{2n}}, & \\text{otherwise.} \\end{cases}$. We prove $\\lambda_0(n)$ is the greatest constant.\n\nIf there exists $k$ ($1 \\leq k \\leq n$) such that $|z_{k+1} - z_k| = 0$, the inequality holds obviously. So without loss of generality, we can assume\n$$\n\\min_{1 \\leq k \\leq n} \\{|z_{k+1} - z_k|^2\\} = 1. \\qquad \\textcircled{1}\n$$\nUnder this condition, it is sufficient to show that the minimum value of $\\sum_{k=1}^{n} |z_k|^2$ is $\\lambda_0(n)$.\n\nWhen $n$ is even. Since\n$$\n\\begin{aligned}\n\\sum_{k=1}^{n} |z_k|^2 &= \\frac{1}{2} \\sum_{k=1}^{n} (|z_k|^2 + |z_{k+1}|^2) \\\n&\\geq \\frac{1}{4} \\sum_{k=1}^{n} |z_{k+1} - z_k|^2 \\\n&\\geq \\frac{n}{4} \\min_{1 \\leq k \\leq n} \\{|z_{k+1} - z_k|^2\\} = \\frac{n}{4},\n\\end{aligned}\n$$\nand equality holds when $(z_1, z_2, \\dots, z_n) = (\\frac{1}{2}, -\\frac{1}{2}, \\dots, \\frac{1}{2}, -\\frac{1}{2})$, thus the minimum value of $\\sum_{k=1}^{n} |z_k|^2$ is $\\frac{n}{4} = \\lambda_0(n)$.\n\nNext consider the condition when $n$ is odd. Let\n$$\n\\theta_k = \\arg \\frac{z_{k+1}}{z_k} \\in [0, 2\\pi), \\quad k = 1, 2, \\dots, n.\n$$\nFor all $k$, $k = 1, 2, \\dots, n$. If $\\theta_k \\leq \\frac{\\pi}{2}$ or $\\theta_k \\geq \\frac{3\\pi}{2}$, then by ①,\n$$\n|z_k|^2 + |z_{k+1}|^2 = |z_k - z_{k+1}|^2 + 2|z_k||z_{k+1}| \\cos \\theta_k \\geq |z_k - z_{k+1}|^2 \\geq 1. \\quad ②\n$$\nIf $\\theta_k \\in (\\frac{\\pi}{2}, \\frac{3\\pi}{2})$, then since $\\cos \\theta_k < 0$ and by ①,\n$$\n\\begin{aligned}\n1 &\\leq |z_k - z_{k+1}|^2 \\\\\n&= |z_k|^2 + |z_{k+1}|^2 - 2 |z_k||z_{k+1}| \\cos \\theta_k \\\\\n&\\leq (|z_k|^2 + |z_{k+1}|^2)(1 + (-2 \\cos \\theta_k)) \\\\\n&= (|z_k|^2 + |z_{k+1}|^2) \\cdot 2 \\sin^2 \\frac{\\theta_k}{2}.\n\\end{aligned}\n$$\nTherefore\n$$\n|z_k|^2 + |z_{k+1}|^2 \\geq \\frac{1}{2 \\sin^2 \\frac{\\theta_k}{2}}. \\quad ③\n$$\n\n(1) If for all $k$ ($1 \\leq k \\leq n$), $\\theta_k \\in (\\frac{\\pi}{2}, \\frac{3\\pi}{2})$, by ③\n$$\n\\sum_{k=1}^{n} |z_k|^2 = \\frac{1}{2} \\sum_{k=1}^{n} (|z_k|^2 + |z_{k+1}|^2) \\geq \\frac{1}{4} \\sum_{k=1}^{n} \\frac{1}{\\sin^2 \\frac{\\theta_k}{2}}. \\quad ④\n$$\nSince $\\prod_{k=1}^{n} \\frac{z_{k+1}}{z_k} = \\frac{z_{n+1}}{z_1} = 1$, so\n$$\n\\sum_{k=1}^{n} \\theta_k = \\arg\\left(\\prod_{k=1}^{n} \\frac{z_{k+1}}{z_k}\\right) + 2m\\pi = 2m\\pi, \\quad ⑤\n$$\nwhere $m$ is a positive integer, and $m < n$. Notice that $n$ is odd, so\n$$\n0 < \\sin \\frac{m\\pi}{n} \\leq \\sin \\frac{(n-1)\\pi}{2n} = \\cos \\frac{\\pi}{2n}. \\quad ⑥\n$$\nLet $f(x) = \\frac{1}{\\sin^2 x}$, $x \\in [\\frac{\\pi}{4}, \\frac{3\\pi}{4}]$, it's easy to show that $f(x)$ is a convex function. By ④ and Jensen's inequality, and combining ⑤ and ⑥, we have\n$$\n\\begin{aligned}\n\\sum_{k=1}^{n} |z_k|^2 &\\geq \\frac{1}{4} \\sum_{k=1}^{n} \\frac{1}{\\sin^2 \\frac{\\theta_k}{2}} \\\\\n&\\geq \\frac{n}{4} \\cdot \\frac{1}{\\sin^2 \\left( \\frac{1}{n} \\sum_{k=1}^{n} \\frac{\\theta_k}{2} \\right)} \\\\\n&= \\frac{n}{4} \\cdot \\frac{1}{\\sin^2 \\frac{m\\pi}{n}} \\geq \\frac{n}{4} \\cdot \\frac{1}{\\cos^2 \\frac{\\pi}{2n}} = \\lambda_0(n).\n\\end{aligned}\n$$\n\n(2) If there exists $j$ ($1 \\leq j \\leq n$), such that $\\theta_j \\notin (\\frac{\\pi}{2}, \\frac{3\\pi}{2}]$, let\n$$\nI = \\left\\{ j \\mid \\theta_j \\notin \\left( \\frac{\\pi}{2}, \\frac{3\\pi}{2} \\right),\\ j = 1, 2, \\dots, n \\right\\}.\n$$\nBy ②, for $j \\in I$, we have $|z_j|^2 + |z_{j+1}|^2 \\geq 1$; and by ③, for $j \\notin I$, we have\n$$\n|z_j|^2 + |z_{j+1}|^2 \\geq \\frac{1}{2 \\sin^2 \\frac{\\theta_j}{2}} \\geq \\frac{1}{2}.\n$$\nTherefore,\n$$\n\\begin{aligned}\n\\sum_{k=1}^{n} |z_k|^2 &= \\frac{1}{2} \\left( \\sum_{j \\in I} (|z_j|^2 + |z_{j+1}|^2) + \\sum_{j \\notin I} (|z_j|^2 + |z_{j+1}|^2) \\right) \\\\\n&\\geq \\frac{1}{2} |I| + \\frac{1}{4} (n - |I|) = \\frac{1}{4} (n + |I|) \\geq \\frac{n+1}{4}.\n\\end{aligned}\n$$\n\n$\\frac{n+1}{4} \\geq \\frac{n}{4} \\cdot \\frac{1}{\\cos^2 \\frac{\\pi}{2n}} \\Leftrightarrow \\cos^2 \\frac{\\pi}{2n} \\geq \\frac{n}{n+1}$\n$$\n\\Leftrightarrow \\sin^2 \\frac{\\pi}{2n} = 1 - \\cos^2 \\frac{\\pi}{2n} \\leq 1 - \\frac{n}{n+1} = \\frac{1}{n+1}.\n$$\nThe equality holds when $n = 3$; when $n \\geq 5$,\n$$\n\\sin^2 \\frac{\\pi}{2n} < \\left( \\frac{\\pi}{2n} \\right)^2 < \\frac{\\pi^2}{2n} \\cdot \\frac{1}{n+1} < \\frac{1}{n+1},\n$$\nso the inequality also holds. So for odd integer $n \\geq 3$, $\\frac{n+1}{4} \\geq \\frac{n}{4} \\cdot \\frac{1}{\\cos^2 \\frac{\\pi}{2n}}$.\n\nOn the other hand, when $z_k = \\frac{1}{2\\cos\\frac{\\pi}{2n}} \\cdot e^{\\frac{i(n-1)k\\pi}{n}}$, $k = 1, 2, \\dots, n$, we have $|z_k - z_{k+1}| = 1$, $k = 1, 2, \\dots, n$, and $\\sum_{k=1}^n |z_k|^2$ achieves its minimum value $\\lambda_0(n)$.\n\nIn a word, the greatest $\\lambda(n)$ is\n$$\n\\lambda_0(n) = \\begin{cases} \\frac{n}{4}, & 2 \\mid n, \\\\ \\frac{n}{4 \\cos^2 \\frac{\\pi}{2n}}, & \\text{otherwise.} \\end{cases}\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 28450,
    "subject": "Mathematics (Multi-modal)",
    "question": "For positive integer $k > 1$, let $f(k)$ be the number of ways of factoring $k$ into product of positive integers greater than 1 (The order of factors are not counted, for example $f(12) = 4$, as $12$ can be factored in these 4 ways: $12$, $2 \\times 6$, $3 \\times 4$, $2 \\times 2 \\times 3$).\n\n**Prove:** If $n$ is a positive integer greater than 1, $p$ is a prime factor of $n$, then $f(n) \\le \\frac{n}{p}$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $P(n)$ stand for the biggest prime divisor of $n$, and define $P(1) = f(1) = 1$. We first prove two lemmas.\n\nLemma 1: For positive integer $n$ and prime $p \\mid n$, we have $f(n) \\le \\sum_{d|\\frac{n}{p}} f(d)$.\n\n**Proof of Lemma 1:** For any factoring of $n$, write $n = n_1 n_2 \\cdots n_k$. Since $p \\mid n$, there exists $i \\in \\{1, \\cdots, k\\}$ such that $p \\mid n_i$ (if there are more than one such $i$, choose any one of them), without loss of generality, assume $i = 1$. Map this factoring to a factoring of $d = \\frac{n}{n_1}$, $d = n_2 n_3 \\cdots n_k$.\n\nFor two different factorings of $n$, $n = n_1 n_2 \\cdots n_k$ and $n = n'_1 n'_2 \\cdots n'_k$ (where $p$ divides $n_1$ and $n'_1$):\n- If $n_1 = n'_1$, then $d = n_2 \\cdots n_k$ and $d = n'_2 \\cdots n'_k$ are two different factorings of $d$ ($d$ is a divisor of $\\frac{n}{p}$).\n- If $n_1 \\ne n'_1$, then $d = \\frac{n}{n_1} \\ne \\frac{n}{n'_1} = d'$, so these two factorings map to a factoring of $d$ and $d'$ respectively ($d$ and $d'$ are divisors of $\\frac{n}{p}$).\n\nThus $f(n) \\le \\sum_{d|\\frac{n}{p}} f(d)$. Lemma 1 is proved.\n\nLemma 2: For positive integer $n$, let $g(n) = \\sum_{d|n} \\frac{d}{P(d)}$, then $g(n) \\le n$.\n\n**Proof of Lemma 2:** Induce on the number of different prime divisors of $n$.\n- When $n = 1$, $g(1) = 1$.\n- When $n = p^a$ is a prime power,\n  $$\n  g(n) = 1 + 1 + p + \\cdots + p^{a-1} = 1 + \\frac{p^a - 1}{p-1} \\le 1 + p^a - 1 = n.\n  $$\n\nAssume when the number of different prime divisors of $n$ is $k$, we have $g(n) \\le n$. Consider the situation when $n$ has $k+1$ different prime divisors. Let the prime factorization of $n$ be $n = p_1^{a_1} \\cdots p_k^{a_k} p_{k+1}^{a_{k+1}}$, where $p_1 < \\cdots < p_k < p_{k+1}$, and write $n = m p_{k+1}^{a_{k+1}}$. So\n  $$\n  g(n) = g(m) + \\sum_{d|m} \\sum_{i=1}^{a_{k+1}} \\frac{d p_{k+1}^i}{p_{k+1}} = g(m) + \\sigma(m) \\frac{p_{k+1}^{a_{k+1}} - 1}{p_{k+1} - 1},\n  $$\nwhere $\\sigma(m)$ stands for the sum of positive divisors of $m$. By assumption, $g(m) \\le m$.\n\nSince\n$$\n\\begin{align*}\n\\sigma(m) \\frac{p_{k+1}^{a_{k+1}} - 1}{p_{k+1} - 1} &= \\left( \\prod_{i=1}^k \\frac{p_i^{a_i+1} - 1}{p_i - 1} \\right) \\frac{p_{k+1}^{a_{k+1}} - 1}{p_{k+1} - 1} \\\\\n&\\le \\left( \\prod_{i=1}^k \\frac{p_i^{a_i+1} - 1}{p_{i+1} - 1} \\right) (p_{k+1}^{a_{k+1}} - 1) \\\\\n&\\le \\left( \\prod_{i=1}^k \\frac{p_i^{a_i+1} - 1}{p_i} \\right) (p_{k+1}^{a_{k+1}} - 1) \\\\\n&\\le \\left( \\prod_{i=1}^k p_i^{a_i} \\right) (p_{k+1}^{a_{k+1}} - 1) \\\\\n&= n - m,\n\\end{align*}\n$$\nso $g(n) \\le n$. Lemma 2 is proved.\n\nBack to the problem, it is sufficient to prove, for positive integer $n$, $f(n) \\le \\frac{n}{P(n)}$ holds.\n\nInduce on $n$. When $n=1$, the equality holds. Assume for $n=1, 2, \\dots, k$, we have $f(n) \\le \\frac{n}{P(n)}$. Then when $n=k+1$, by Lemma 1, Lemma 2, and the assumption,\n$$\nf(k+1) \\le \\sum_{d|\\frac{k+1}{P(k+1)}} f(d) \\le \\sum_{d|\\frac{k+1}{P(k+1)}} \\frac{d}{P(d)} = g\\left(\\frac{k+1}{P(k+1)}\\right) \\le \\frac{k+1}{P(k+1)}.\n$$",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 28451,
    "subject": "Mathematics (Multi-modal)",
    "question": "Given geometric sequence $\\{a_n\\}$, $a_9 = 13$, $a_{13} = 1$, then the value of $\\log_{a_1} 13$ is ______.",
    "options": [],
    "answer": "1/3",
    "solution": "By the properties of geometric sequence, we have $\\frac{a_1}{a_9} = \\left(\\frac{a_9}{a_{13}}\\right)^2$, and thus $a_1 = \\frac{a_9^3}{a_{13}^2} = 13^3$.\n\nConsequently, $\\log_{a_1} 13 = \\frac{1}{3}$. $\\square$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 28452,
    "subject": "Mathematics (Multi-modal)",
    "question": "If real number $x$ satisfies $\\log_2 x = \\log_4(2x) + \\log_8(4x)$, then the value of $x$ is ______.",
    "options": [],
    "answer": "128",
    "solution": "By the given condition, we have\n$$\n\\log_2 x = \\log_4 2 + \\log_4 x + \\log_8 4 + \\log_8 x = \\frac{1}{2} + \\frac{1}{2} \\log_2 x + \\frac{2}{3} + \\frac{1}{3} \\log_2 x,\n$$\nand its solution is $\\log_2 x = 7$. Therefore, $x = 128$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 28453,
    "subject": "Mathematics (Multi-modal)",
    "question": "Suppose positive real numbers $a$, $b$, and $c$ satisfy $a^2 + 4b^2 + 9c^2 = 4b + 12c - 2$. Find the minimum of $\\frac{1}{a} + \\frac{2}{b} + \\frac{3}{c}$.",
    "options": [],
    "answer": "6",
    "solution": "By the given condition, we have\n$$\na^2 + (2b - 1)^2 + (3c - 2)^2 = 3.\n$$\nBy making use of the Cauchy inequality, we get\n$$\n3[a^2 + (2b - 1)^2 + (3c - 2)^2] \\geq (a + 2b - 1 + 3c - 2)^2,\n$$\nnamely, $(a + 2b + 3c - 3)^2 \\leq 9$. Therefore,\n$$\na + 2b + 3c \\leq 6.\n$$\nAgain, by the Cauchy inequality we get\n$$\n\\left( \\frac{1}{a} + \\frac{2}{b} + \\frac{3}{c} \\right) (a + 2b + 3c) \\geq (1 + 2 + 3)^2.\n$$\nThus,\n$$\n\\frac{1}{a} + \\frac{2}{b} + \\frac{3}{c} \\ge \\frac{36}{a + 2b + 3c} \\ge 6,\n$$\nand the equal sign holds when $a = b = c = 1$.\nTherefore, the minimum value of $\\frac{1}{a} + \\frac{2}{b} + \\frac{3}{c}$ is $6$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 28454,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let the general term of sequence $\\{a_n\\}$ be\n$$\na_n = \\frac{1}{\\sqrt{5}} \\left( \\left( \\frac{1 + \\sqrt{5}}{2} \\right)^n - \\left( \\frac{1 - \\sqrt{5}}{2} \\right)^n \\right), \\quad n = 1, 2, \\dots\n$$\nProve that there exist infinite positive integers $m$ such that $a_{m+4}a_m - 1$ are perfect squares.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Denote $q_1 = \\frac{1+\\sqrt{5}}{2}$, $q_2 = \\frac{1-\\sqrt{5}}{2}$, and then $q_1 + q_2 = 1$, $q_1q_2 = -1$. Thus,\n$$\na_n = \\frac{1}{\\sqrt{5}}(q_1^n - q_2^n), \\quad n = 1, 2, \\dots\n$$\nHence $a_1 = 1, a_2 = 1$. Also note that $q_i + 1 = q_i^2$ ($i = 1, 2$), and there is\n$$\n\\begin{aligned}\na_{n+1} + a_n &= \\frac{1}{\\sqrt{5}}(q_1^{n+1} - q_2^{n+1}) + \\frac{1}{\\sqrt{5}}(q_1^n - q_2^n) \\\\\n&= \\frac{1}{\\sqrt{5}}(q_1^n(q_1+1) - q_2^n(q_2+1)) \\\\\n&= \\frac{1}{\\sqrt{5}}(q_1^{n+2} - q_2^{n+2}),\n\\end{aligned}\n$$\nnamely,\n$$\na_{n+2} = a_{n+1} + a_n, \\quad n = 1, 2, \\dots\n$$\nIt is easy to know that each term of the sequence $\\{a_n\\}$ is a positive integer.\n\nIt is easy to calculate that $q_1^4 + q_2^4 = 7$, and hence\n$$\n\\begin{aligned}\na_{2n+3}a_{2n-1} - 1 &= \\frac{1}{\\sqrt{5}}(q_1^{2n+3} - q_2^{2n+3}) \\cdot \\frac{1}{\\sqrt{5}}(q_1^{2n-1} - q_2^{2n-1}) - 1 \\\\\n&= \\frac{1}{5}(q_1^{4n+2} + q_2^{4n+2} - (q_1q_2)^{2n-1}q_1^4 - (q_1q_2)^{2n-1}q_2^4) - 1\n\\end{aligned}\n$$\n$$\n\\begin{align*}\n&= \\frac{1}{5}(q_1^{4n+2} + q_2^{4n+2} + q_1^4 + q_2^4) - 1 \\\\\n&= \\frac{1}{5}(q_1^{4n+2} + q_2^{4n+2} + 7) - 1 \\\\\n&= \\frac{1}{5}(q_1^{4n+2} + q_2^{4n+2} + 2) \\\\\n&= \\left[ \\frac{1}{\\sqrt{5}}(q_1^{2n+1} + q_2^{2n+1}) \\right]^2 \\\\\n&= a_{2n+1}^2.\n\\end{align*}\n$$\nTherefore, for any positive integer $n$, $a_{2n+3}a_{2n-1} - 1$ is a perfect square.\n\nHence, $a_{m+4}a_m - 1$ are perfect squares for all positive odd numbers $m$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 28455,
    "subject": "Mathematics (Multi-modal)",
    "question": "In a plane rectangular coordinate system $xOy$, circle $\\Omega$ passes through points $(0, 0)$, $(2, 4)$, $(3, 3)$. Then the maximum of the distance from a point on circle $\\Omega$ to the origin is ______.",
    "options": [],
    "answer": "2*sqrt(5)",
    "solution": "Denote $A(2, 4)$, $B(3, 3)$. Then circle $\\Omega$ passes through points $O$, $A$ and $B$. Note that $\\angle OBA = 90^\\circ$ (the slopes of lines $OB$ and $AB$ are $1$ and $-1$, respectively), so $OA$ is a diameter of circle $\\Omega$. Consequently, the maximum of the distance from a point on circle $\\Omega$ to the origin $O$ is $|OA| = 2\\sqrt{5}$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 28456,
    "subject": "Mathematics (Multi-modal)",
    "question": "In ellipse $\\Gamma$, $A$ is an endpoint of the major axis, $B$ is an endpoint of the minor axis, and $F_1, F_2$ are the foci. If $\\overrightarrow{AF_1} \\cdot \\overrightarrow{AF_2} + \\overrightarrow{BF_1} \\cdot \\overrightarrow{BF_2} = 0$, then the value of $\\frac{|AB|}{|F_1F_2|}$ is ______.",
    "options": [],
    "answer": "sqrt(2)/2",
    "solution": "Without loss of generality, suppose the equation of $\\Gamma$ is $\\frac{x^2}{a^2} + \\frac{y^2}{b^2} = 1$ ($a > b > 0$), and $A(a, 0)$, $B(0, b)$, $F_1(-c, 0)$, $F_2(c, 0)$. By the given conditions, we get\n$$ \\overrightarrow{AF_1} \\cdot \\overrightarrow{AF_2} + \\overrightarrow{BF_1} \\cdot \\overrightarrow{BF_2} = (-c-a)(c-a) + (-c^2+b^2) = a^2+b^2-2c^2 = 0. $$\nTherefore, $\\frac{|AB|}{|F_1F_2|} = \\frac{\\sqrt{a^2+b^2}}{2c} = \\frac{\\sqrt{2c^2}}{2c} = \\frac{\\sqrt{2}}{2}$. $\\square$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 28457,
    "subject": "Mathematics (Multi-modal)",
    "question": "Suppose set $X = \\{1, 2, \\dots, 20\\}$. $A$ is a subset of $X$. The number of the elements of $A$ is at least $2$ and all the elements of $A$ can be arranged as consecutive positive integers. Then the number of such set $A$ is ______.",
    "options": [],
    "answer": "190",
    "solution": "Each set $A$ satisfying the above conditions can be uniquely determined by its minimum element $a$ and maximum element $b$, where $a, b \\in X$ and $a < b$. The total number of such ways of taking $(a, b)$ is $C_{20}^2 = 190$, so the number of such sets $A$ is $190$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 28458,
    "subject": "Mathematics (Multi-modal)",
    "question": "Suppose that $a > 0$ and the minima of function $f(x) = x + \\frac{100}{x}$ on intervals $(0, a]$ and $[a, +\\infty)$ are $m_1, m_2$, respectively. If $m_1 m_2 = 2020$, then the value of $a$ is ______.",
    "options": [],
    "answer": "1 or 100",
    "solution": "Note that $f(x)$ is monotonically decreasing on $(0, 10]$ and monotonically increasing on $[10, +\\infty)$. When $a \\in (0, 10]$, $m_1 = f(a)$, $m_2 = f(10)$; when $a \\in [10, +\\infty)$, $m_1 = f(10)$, $m_2 = f(a)$. Therefore, there is always\n$$\nf(a)f(10) = m_1m_2 = 2020,\n$$\nnamely, $a + \\frac{100}{a} = \\frac{2020}{20} = 101$. The solution is $a = 1$ or $a = 100$. $\\square$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 28459,
    "subject": "Mathematics (Multi-modal)",
    "question": "In triangle $ABC$, $BC = 4$, $CA = 5$ and $AB = 6$. Then the value of $\\sin^6 \\frac{A}{2} + \\cos^6 \\frac{A}{2}$ is ______.",
    "options": [],
    "answer": "43/64",
    "solution": "By the law of cosines, we get $\\cos A = \\frac{CA^2 + AB^2 - BC^2}{2CA \\cdot AB} = \\frac{5^2 + 6^2 - 4^2}{2 \\times 5 \\times 6} = \\frac{3}{4}$. Therefore,\n$$\n\\begin{align*}\n\\sin^6 \\frac{A}{2} + \\cos^6 \\frac{A}{2} &= \\left(\\sin^2 \\frac{A}{2} + \\cos^2 \\frac{A}{2}\\right) \\left(\\sin^4 \\frac{A}{2} - \\sin^2 \\frac{A}{2} \\cos^2 \\frac{A}{2} + \\cos^4 \\frac{A}{2}\\right) \\\\\n&= \\left(\\sin^2 \\frac{A}{2} + \\cos^2 \\frac{A}{2}\\right)^2 - 3 \\sin^2 \\frac{A}{2} \\cos^2 \\frac{A}{2} \\\\\n&= 1 - \\frac{3}{4} \\sin^2 A \\\\\n&= \\frac{1}{4} + \\frac{3}{4} \\cos^2 A = \\frac{43}{64}. \\quad \\square\n\\end{align*}\n$$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 28460,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $z$ be a complex number. If $\\frac{z-2}{z-i}$ is a real number ($i$ is the imaginary unit), then the minimum of $|z+3|$ is ______.",
    "options": [],
    "answer": "sqrt(5)",
    "solution": "Suppose $z = a + bi$ ($a, b \\in \\mathbb{R}$). By the given condition we can find\n$$\n\\begin{aligned} \\operatorname{Im} \\left( \\frac{z-2}{z-i} \\right) &= \\operatorname{Im} \\left( \\frac{(a-2)+bi}{a+(b-1)i} \\right) \\\\ &= \\frac{-(a-2)(b-1)+ab}{a^2+(b-1)^2} \\\\ &= \\frac{a+2b-2}{a^2+(b-1)^2} = 0, \\end{aligned}\n$$\nand thus $a+2b=2$. Therefore,\n$$\n\\sqrt{5}|z+3| = \\sqrt{(1^2+2^2)((a+3)^2+b^2)} \\geq |(a+3)+2b| = 5,\n$$\nnamely, $|z+3| \\geq \\sqrt{5}$. When $a = -2, b = 2$, $|z+3|$ takes the minimum $\\sqrt{5}$.\nFrom $\\frac{z-2}{z-i} \\in \\mathbb{R}$ and the geometric meaning of complex division, it is known that the point corresponding to $z$ on the complex plane lies on the line connecting the points corresponding to $2$ and $i$ (excluding the point corresponding to $i$), so the minimum of $|z+3|$ is the distance from point $(-3, 0)$ to line $x + 2y - 2 = 0$ in plane rectangular coordinate system $xOy$, i.e., $\\frac{|-3-2|}{\\sqrt{1^2+2^2}} = \\sqrt{5}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 28461,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let the 9-element set $A = \\{a + bi \\mid a, b \\in \\{1, 2, 3\\}\\}$, with $i$ being the imaginary unit. $\\alpha = (z_1, z_2, \\dots, z_9)$ is a permutation of all the elements in $A$, satisfying $|z_1| \\le |z_2| \\le \\dots \\le |z_9|$. The number of such permutations $\\alpha$ is ______.",
    "options": [],
    "answer": "8",
    "solution": "Since\n$$\n\\begin{align*}\n|1+i| < |2+i| = |1+2i| < |2+2i| < |3+i| \\\\\n= |1+3i| < |3+2i| = |2+3i| < |3+3i|,\n\\end{align*}\n$$\nit follows that\n$$\n\\begin{align*}\nz_1 &= 1 + i, \\\\\n\\{z_2, z_3\\} = \\{2 + i, 1 + 2i\\}, \\\\\nz_4 &= 2 + 2i, \\\\\n\\{z_5, z_6\\} = \\{3 + i, 1 + 3i\\}, \\\\\n\\{z_7, z_8\\} = \\{3 + 2i, 2 + 3i\\}, \\\\\nz_9 = 3 + 3i.\n\\end{align*}\n$$\nBy the multiplication principle, the number of permutations $\\alpha$ satisfying the condition is $2^3 = 8$. $\\square$",
    "topic": "Discrete Mathematics",
    "subtopic": "Combinatorics"
  },
  {
    "id": 28462,
    "subject": "Mathematics (Multi-modal)",
    "question": "In $\\triangle ABC$, $AB = 6$, $BC = 4$, the median to side $AC$ is $\\sqrt{10}$. Then the value of $\\sin^6 \\frac{A}{2} + \\cos^6 \\frac{A}{2}$ is ______.",
    "options": [],
    "answer": "211/256",
    "solution": "Let $M$ be the midpoint of $AC$. By the median formula we have\n$$\n4BM^2 + AC^2 = 2(AB^2 + BC^2),\n$$\nand thus\n$$\nAC = \\sqrt{2(6^2 + 4^2) - 4 \\cdot 10} = 8.\n$$\nBy the law of cosines, we obtain $\\cos A = \\frac{CA^2 + AB^2 - BC^2}{2CA \\cdot AB} = \\frac{8^2 + 6^2 - 4^2}{2 \\cdot 8 \\cdot 6} = \\frac{7}{8}$. Therefore,\n$$\n\\begin{aligned}\n\\sin^6 \\frac{A}{2} + \\cos^6 \\frac{A}{2} &= \\left(\\sin^2 \\frac{A}{2} + \\cos^2 \\frac{A}{2}\\right) \\left(\\sin^4 \\frac{A}{2} - \\sin^2 \\frac{A}{2} \\cos^2 \\frac{A}{2} + \\cos^4 \\frac{A}{2}\\right) \\\\\n&= \\left(\\sin^2 \\frac{A}{2} + \\cos^2 \\frac{A}{2}\\right)^2 - 3 \\sin^2 \\frac{A}{2} \\cos^2 \\frac{A}{2} \\\\\n&= 1 - \\frac{3}{4} \\sin^2 A \\\\\n&= \\frac{1}{4} + \\frac{3}{4} \\cos^2 A \\\\\n&= \\frac{211}{256}.\n\\end{aligned}\n\\quad \\square",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 28463,
    "subject": "Mathematics (Multi-modal)",
    "question": "Suppose all the edges of regular triangular pyramid $P$-$ABC$ have length $1$ and $L$, $M$, $N$ are the midpoints of edges $PA$, $PB$ and $PC$, respectively. The area of the cross section of the circumscribed sphere of this regular triangular pyramid intercepted by plane $LMN$ is ______.",
    "options": [],
    "answer": "π/3",
    "solution": "The given conditions show that plane $LMN$ is parallel to plane $ABC$ and the ratio of the distances from point $P$ to planes $LMN$ and $ABC$ is $1 : 2$. Let $H$ be the centroid of face $ABC$ in regular triangular pyramid $P$-$ABC$ and $PH$ intersects plane $LMN$ at point $K$. Then $PH \\perp ABC$ and $PK \\perp LMN$, and thus $PK = \\frac{1}{2}PH$.\n\nThe regular triangular pyramid $P$-$ABC$ can be regarded as a regular tetrahedron. Let $O$ be the centre of its circumscribed sphere. Then $O$ is on $PH$ and by the properties of regular tetrahedron we know that $OH = \\frac{1}{4}PH$. By combining $PK = \\frac{1}{2}PH$ we know that $OK = OH$, that is, point $O$ is equally distant to planes $LMN$ and $ABC$. This shows that the section circle of the circumscribed sphere of this regular triangular pyramid intercepted by planes $LMN$ and $ABC$ is equal in size.\n\nAs a result, the area of the required cross section is equal to the area of the circumcircle of $\\triangle ABC$, namely, $\\pi \\cdot \\left(\\frac{AB}{\\sqrt{3}}\\right)^2 = \\frac{\\pi}{3}$.",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 28464,
    "subject": "Mathematics (Multi-modal)",
    "question": "Given a regular triangular prism, the length of each edge is $3$. Then the volume of its circumscribed sphere is ______.",
    "options": [],
    "answer": "7√21/2 π",
    "solution": "As shown in Fig. 6.1, let the centroids of faces $ABC$ and $A_1B_1C_1$ be $O$ and $O_1$, respectively. Denote the midpoint of segment $OO_1$ as $P$. By symmetry, we know that $P$ is the centre of the circumscribed sphere of the regular triangular prism and $PA$ is its radius.\n\n![](attached_image_1.png)\nFig. 6.1\n\nIt is easy to find that $PO \\perp AO$, and thus\n$$\nPA = \\sqrt{PO^2 + AO^2} = \\sqrt{\\left(\\frac{3}{2}\\right)^2 + (\\sqrt{3})^2} = \\frac{\\sqrt{21}}{2}.\n$$\nTherefore, the volume of the circumscribed sphere is $\\frac{4}{3}\\pi\\left(\\frac{\\sqrt{21}}{2}\\right)^3 = \\frac{7\\sqrt{21}}{2}\\pi$.",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 28465,
    "subject": "Mathematics (Multi-modal)",
    "question": "Suppose $a, b > 0$. The equation $\\sqrt{|x|} + \\sqrt{|x+a|} = b$ for $x$ has exactly three different real solutions, namely $x_1, x_2, x_3$, and $x_1 < x_2 < x_3 = b$. Then the value of $a+b$ is ______.",
    "options": [],
    "answer": "144",
    "solution": "Let $t = x + \\frac{a}{2}$. Then the equation $\\sqrt{|t - \\frac{a}{2}|} + \\sqrt{|t + \\frac{a}{2}|} = b$ for $t$ has exactly three different real solutions $t_i = x_i + \\frac{a}{2}$ ($i = 1, 2, 3$).\n\nSince $f(t) = \\sqrt{|t - \\frac{a}{2}|} + \\sqrt{|t + \\frac{a}{2}|}$ is an even function, the three real solutions of equation $f(t) = b$ are symmetrically distributed about the origin of the number axis, so that there must be $b = f(0) = \\sqrt{2a}$. In the following, we will find the real solutions of equation $f(t) = \\sqrt{2a}$.\n\nWhen $|t| \\le \\frac{a}{2}$, $f(t) = \\sqrt{\\frac{a}{2} - t} + \\sqrt{\\frac{a}{2} + t} = \\sqrt{a + \\sqrt{a^2 - 4t^2}} \\le \\sqrt{2a}$ and the equal sign holds if and only if $t = 0$; when $|t| > \\frac{a}{2}$, $f(t)$ is monotonically increasing, and when $t = \\frac{5a}{8}$, $f(t) = \\sqrt{2a}$; when $t < -\\frac{a}{2}$, $f(t)$ is monotonically decreasing, and when $t = -\\frac{5a}{8}$, $f(t) = \\sqrt{2a}$.\n\nThus, equation $f(t) = \\sqrt{2a}$ has exactly three real solutions $t_1 = -\\frac{5}{8}a$, $t_2 = 0$, $t_3 = \\frac{5}{8}a$.\n\nBy the given conditions, we can find $b = x_3 = t_3 - \\frac{a}{2} = \\frac{a}{8}$. Combining $b = \\sqrt{2a}$, we get $a = 128$.\n\nConsequently, $a + b = \\frac{9a}{8} = 144$. □",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 28466,
    "subject": "Mathematics (Multi-modal)",
    "question": "In convex quadrilateral $ABCD$, $\\vec{BC} = 2\\vec{AD}$. Point $P$ is on the plane of quadrilateral $ABCD$, satisfying $\\vec{PA} + 2020\\vec{PB} + 2020\\vec{PC} = 2020\\vec{PD} = \\vec{0}$. Let $s$ and $t$ be the areas of quadrilateral $ABCD$ and $\\triangle PAB$, respectively. Then the value of $\\frac{t}{s}$ is ______.",
    "options": [],
    "answer": "337/2021",
    "solution": "$$\n\\overrightarrow{PA} + \\overrightarrow{PC} = 2\\overrightarrow{PY}, \\quad \\overrightarrow{PB} + \\overrightarrow{PD} = 2\\overrightarrow{PX},\n$$\ncombining the given conditions, we know that $\\overrightarrow{PY} + 2020\\overrightarrow{PX} = \\overrightarrow{0}$. Hence, point $P$ lies on segment $XY$, and $PX = \\frac{1}{2021}$. Let the distance from $A$ to $MN$ be $h$. By the area formula, we can get\n$$\n\\begin{aligned} \\frac{t}{s} &= \\frac{S_{\\triangle PAB}}{S_{ABCD}} = \\frac{PM \\cdot h}{MN \\cdot 2h} = \\frac{PM}{2MN} \\\\ &= \\frac{1 + \\frac{1}{2021}}{2 \\times 3} = \\frac{337}{2021}. \\end{aligned} \\quad \\square",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 28467,
    "subject": "Mathematics (Multi-modal)",
    "question": "Given quintic polynomial $f(x)$ with its leading coefficient being 1, it satisfies $f(n) = 8n$, $n = 1, 2, \\dots, 5$. Then the coefficient of the term of degree 1 of $f(x)$ is ______.",
    "options": [],
    "answer": "282",
    "solution": "Let $f(x) = g(x) - 8x$, and then $g(x)$ is also a quintic polynomial with its leading coefficient being 1. And there is\n$$\ng(n) = f(n) - 8n = 0, \\quad n = 1, 2, \\dots, 5.\n$$\nHence, $g(x)$ has 5 real roots, namely, $1, 2, \\dots, 5$. Therefore,\n$$\ng(x) = (x-1)(x-2)\\cdots(x-5),\n$$\nand thus,\n$$\nf(x) = (x-1)(x-2)\\cdots(x-5) + 8x.\n$$\nConsequently, the coefficient of the term of degree 1 of $f(x)$ is\n$$\n\\left(1 + \\frac{1}{2} + \\frac{1}{3} + \\frac{1}{4} + \\frac{1}{5}\\right) \\cdot 5! + 8 = 282.\n$$",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 28468,
    "subject": "Mathematics (Multi-modal)",
    "question": "There are 10 cards, each of which has two numbers, numbered $1$, $2$, $3$, $4$, $5$, written on it, and the numbers on any two cards are not exactly identical. The 10 cards are placed in five boxes labelled $1$, $2$, $3$, $4$, $5$, and a card with $i$ and $j$ written on it can only be placed in box $i$ or $j$. One placement is called “good” if there are more cards in box $1$ than in each of the other boxes. Then the total number of the “good” placements is ______.",
    "options": [],
    "answer": "120",
    "solution": "Denote the card with $i, j$ written on it as $\\{i, j\\}$. It is easy to know that these 10 cards are exactly $\\{i, j\\}$ ($1 \\le i < j \\le 5$).\n\nConsider the “good” placements of the cards. There are 10 cards in the five boxes, so there are at least 3 cards in box 1. The only cards that will fit in box 1 are $\\{1, 2\\}$, $\\{1, 3\\}$, $\\{1, 4\\}$ and $\\{1, 5\\}$.\n\n*Case 1:* The 4 cards are all placed in box 1, and at this point it is no longer possible to have 4 cards in each of the remaining boxes. Therefore, no matter how the remaining 6 cards are placed, they will fit the requirements and there are $2^6 = 64$ “good” placements.\n\n*Case 2:* There are exactly 3 of the 4 cards in box 1 and the rest of each box contains at most 2 cards.\n\nConsider the number $N$ of placements of $\\{1, 2\\}$, $\\{1, 3\\}$, $\\{1, 4\\}$ in box 1 and $\\{1, 5\\}$ in box 5.\n\nThere are 8 possible ways to place the cards $\\{2, 3\\}$, $\\{2, 4\\}$, $\\{3, 4\\}$, with 6 of them being two cards placed in one of the boxes 2, 3, 4, and the remaining 2 being one card placed in each box of 2, 3, 4.\n\nIf there are two cards of $\\{2, 3\\}$, $\\{2, 4\\}$ and $\\{3, 4\\}$ in a box, suppose that $\\{2, 3\\}$ and $\\{2, 4\\}$ are in box 2, and then $\\{2, 5\\}$ can only be in box 5. Therefore, box 5 already has $\\{1, 5\\}$ and $\\{2, 5\\}$, so $\\{3, 5\\}$ and $\\{4, 5\\}$ are in boxes 3 and 4 respectively, namely, the placement of $\\{2, 5\\}$, $\\{3, 5\\}$ and $\\{4, 5\\}$ is unique.\n\nIf one card of $\\{2, 3\\}$, $\\{2, 4\\}$ and $\\{3, 4\\}$ is placed in each box of 2, 3, 4, then there are at most 2 cards in each box of 2, 3, 4. It is only necessary to make sure that there are no more than 2 cards in box 5, namely, there are 0 or 1 card of $\\{2, 5\\}$, $\\{3, 5\\}$ and $\\{4, 5\\}$ in box 5, and the number of the corresponding placements is $C_3^0 + C_3^1 = 4$.\n\nAs a result, $N = 6 \\times 1 + 2 \\times 4 = 14$. By symmetry, there are $4N = 56$ “good” placements in *case 2*.\n\nTo sum up, there is a total of $64 + 56 = 120$ “good” placements.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 28469,
    "subject": "Mathematics (Multi-modal)",
    "question": "In ellipse $\\Gamma$, $A$ is an endpoint of the major axis, $B$ is an endpoint of the minor axis, and $F_1$, $F_2$ are the foci. If $\\overrightarrow{AF_1} \\cdot \\overrightarrow{AF_2} + \\overrightarrow{BF_1} \\cdot \\overrightarrow{BF_2} = 0$, then find the value of $\\tan \\angle ABF_1 \\cdot \\tan \\angle ABF_2$.",
    "options": [],
    "answer": "-1/5",
    "solution": "By symmetry, suppose the equation of $\\Gamma$ is $\\frac{x^2}{a^2} + \\frac{y^2}{b^2} = 1$ ($a > b > 0$), and $A(a, 0)$, $B(0, b)$, $F_1(-c, 0)$, $F_2(c, 0)$, where $c = \\sqrt{a^2 - b^2}$.\nBy the given conditions, we know that\n$$\n\\begin{aligned}\n\\overrightarrow{AF_1} \\cdot \\overrightarrow{AF_2} + \\overrightarrow{BF_1} \\cdot \\overrightarrow{BF_2} &= (-c-a)(c-a) + (-c^2+b^2) \\\\\n&= a^2 + b^2 - 2c^2 = 0.\n\\end{aligned}\n$$\nThus, $a^2 + b^2 - 2c^2 = -a^2 + 3b^2 = 0$, and hence $a = \\sqrt{3}b, c = \\sqrt{2}b$.\nLet $O$ be the origin of the coordinates, and then\n$$\n\\begin{aligned}\n\\tan \\angle ABO &= \\frac{a}{b} = \\sqrt{3}, \\\\\n\\tan \\angle OBF_1 &= \\tan \\angle OBF_2 = \\frac{c}{b} = \\sqrt{2}.\n\\end{aligned}\n$$\nTherefore,\n$$\n\\begin{aligned}\n& \\tan \\angle ABF_1 \\cdot \\tan \\angle ABF_2 \\\\\n&= \\tan(\\angle ABO + \\angle OBF_1) \\cdot \\tan(\\angle ABO - \\angle OBF_1) \\\\\n&= \\frac{\\sqrt{3} + \\sqrt{2}}{1 - \\sqrt{3} \\cdot \\sqrt{2}} \\cdot \\frac{\\sqrt{3} - \\sqrt{2}}{1 + \\sqrt{3} \\cdot \\sqrt{2}} = -\\frac{1}{5}.\n\\end{aligned}",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 28470,
    "subject": "Mathematics (Multi-modal)",
    "question": "In $\\triangle ABC$, $\\sin A = \\frac{\\sqrt{2}}{2}$. Find the range of $\\cos B + \\sqrt{2} \\cos C$.",
    "options": [],
    "answer": "(0, 1] ∪ (2, √5]",
    "solution": "Denote $f = \\cos B + \\sqrt{2} \\cos C$.\n\nBy the given conditions, we know that $A = \\frac{\\pi}{4}$ or $A = \\frac{3\\pi}{4}$.\n\nWhen $A = \\frac{\\pi}{4}$, it follows that $B = \\frac{3\\pi}{4} - C$, where $0 < C < \\frac{3\\pi}{4}$. And there are\n$$\n\\begin{aligned}\nf &= \\cos\\left(\\frac{3\\pi}{4} - C\\right) + \\sqrt{2} \\cos C \\\\\n&= \\frac{\\sqrt{2}}{2} \\sin C + \\frac{\\sqrt{2}}{2} \\cos C \\\\\n&= \\sin\\left(C + \\frac{\\pi}{4}\\right) \\in (0, 1].\n\\end{aligned}\n$$\n\nWhen $A = \\frac{3\\pi}{4}$, we have $B = \\frac{3\\pi}{4} - C$, where $0 < C < \\frac{\\pi}{4}$. Therefore,\n$$\n\\begin{aligned}\nf &= \\cos\\left(\\frac{\\pi}{4} - C\\right) + \\sqrt{2} \\cos C \\\\\n&= \\frac{\\sqrt{2}}{2} \\sin C + \\frac{3\\sqrt{2}}{2} \\cos C \\\\\n&= \\sqrt{5} \\sin(C + \\varphi),\n\\end{aligned}\n$$\nwhere $\\varphi = \\arctan 3$.\n\nNote that $\\varphi \\in (\\frac{\\pi}{4}, \\frac{\\pi}{2})$. Function $g(x) = \\sqrt{5} \\sin(x + \\varphi)$ is monotonically increasing on $[0, \\frac{\\pi}{2} - \\varphi]$ and monotonically decreasing on $[\\frac{\\pi}{2} - \\varphi, \\frac{\\pi}{4}]$. And since $g(0) = \\frac{3\\sqrt{2}}{2} > 2 = g(\\frac{\\pi}{4})$, $g(\\frac{\\pi}{2} - \\varphi) = \\sqrt{5}$, it follows that $f \\in (2, \\sqrt{5}]$.\n\nIn conclusion, the range of $\\cos B + \\sqrt{2} \\cos C$ is $(0, 1] \\cup (2, \\sqrt{5}]$. $\\square$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 28471,
    "subject": "Mathematics (Multi-modal)",
    "question": "Suppose that $A = \\{1, 2, 3\\}$, $B = \\{2x + y \\mid x, y \\in A, x < y\\}$, $C = \\{2x + y \\mid x, y \\in A, x > y\\}$. Then the sum of all the elements of $B \\cap C$ is ______.",
    "options": [],
    "answer": "12",
    "solution": "By enumeration, we get $B = \\{4, 5, 7\\}$, $C = \\{5, 7, 8\\}$. Thus, $B \\cap C = \\{5, 7\\}$. Therefore, the sum of all the elements of $B \\cap C$ is $5 + 7 = 12$. $\\square$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 28472,
    "subject": "Mathematics (Multi-modal)",
    "question": "Suppose that geometric sequence $\\{a_n\\}$ satisfies $a_1 - a_2 = 3$, $a_1 - a_3 = 2$. Then the common ratio of $\\{a_n\\}$ is ________.",
    "options": [],
    "answer": "-1/3",
    "solution": "Let the common ratio of $\\{a_n\\}$ be $q$. Then\n$$\n\\begin{aligned}\na_1(1-q) &= a_1 - a_2 = 3, \\\\\na_1(1-q^2) &= a_1 - a_3 = 2,\n\\end{aligned}\n$$\nand thus $1 + q = \\frac{a_1(1 - q^2)}{a_1(1 - q)} = \\frac{2}{3}$. Therefore, $q = -\\frac{1}{3}$.\n$\\square$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 28473,
    "subject": "Mathematics (Multi-modal)",
    "question": "Given a convex $n$-gon ($n \\ge 5$) $\\Omega: P_1P_2...P_n$, of which no three diagonals are concurrent inside $\\Omega$. Prove that one can choose a point inside every quadrilateral $P_iP_jP_kP_l$ ($1 \\le i < j < k < l \\le n$) and not on any diagonal of $\\Omega$, such that the $C_n^4$ points obtained are distinct and the segment connecting any two of them intersects with at least one diagonal of $\\Omega$.\n(Contributed by Leng Fusheng)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "To begin, notice that the diagonals of $\\Omega$ divide the polygon into $C_n^4 + C_{n-1}^2$ small regions. Every quadrilateral $P_iP_jP_kP_l$ ($1 \\le i < j < k < l \\le n$) has a 1-1 correspondence with the intersection of the diagonals $P_iP_k$ and $P_jP_l$, and additionally, every intersection is adjacent to four small regions. Therefore, it is enough to assign every intersection to one of the four adjacent small regions, such that the $C_n^4$ assigned small regions are all distinct.\n\nPlace $\\Omega$ in the Cartesian coordinate plane such that no diagonal is parallel to the $x$ axis. If two diagonals $P_iP_k, P_jP_l$ meet at $Q$, then we assign $Q$ to the unique small region incident to $Q$ and whose interior points all have $y$ coordinates larger than that of $Q$ (intuitively, above $Q$). Obviously, different intersections correspond to different small regions, as for each small region $R$, there is a unique vertex with the minimum $y$ coordinate, and this vertex is assigned to $R$. This completes the proof. $\\square$\nAs in the first proof, we are required to assign each intersection (of two diagonals) to an adjacent small region, such that the assigned small regions are all distinct. In the plane of $\\Omega$, pick any point $Q$ not concurrent with any diagonal. Suppose that two diagonals $\\ell_1$ and $\\ell_2$ intersect at $R$. Then $Q$ lies inside one of the four zones of the plane divided by $\\ell_1$ and $\\ell_2$: in the anticlockwise direction, let $T$ be the next zone, and assign $R$ to the unique small region adjacent to $R$ and contained in $T$. It remains to prove that each small region is assigned at most one intersection.\n\n![](attached_image_1.png)\nFig. 1.1\n\nFor a small region illustrated in Fig. 1.1, denote its vertices anticlockwise as $R_1, R_2, \\dots, R_m(R_0 = R_m)$. The rays $R_1R_2, R_2R_3, \\dots, R_{m-1}R_m, R_mR_1$ divide the plane excluding the small region into $A_1, A_2, \\dots, A_m$, where $A_i$ is the zone whose boundary consists of rays $R_iR_{i+1}$ and $R_{i-1}R_i$. According to the correspondence between small regions and intersections, this small region is assigned $R_i$ if and only if $Q$ lies inside zone $A_i$, which could happen for at most one subscript $i$. Since this is true for any small region, we infer that each small region is assigned at most one intersection, and the proof is completed. $\\square$\nLet all the diagonals of $\\Omega$ be $\\ell_1, \\dots, \\ell_m$. We put these $m$ diagonals in $\\Omega$ in the order of the subscripts and define $S_k$ as the set of intersections created by adding the diagonal $\\ell_k$ ($S_k$ could be empty), $T_k$ as the set of all small regions divided by $I_1, \\dots, I_k$. We use induction to prove, for each $k$, there exists a one-to-one mapping\n$$\nf_k : S_1 \\cup \\dots \\cup S_k \\to T_k,\n$$\nsuch that every $Q \\in S_1 \\cup \\dots \\cup S_k$ is a vertex of the polygon $f_k(Q)$. After that, a discussion as in the beginning of the first proof will lead to the conclusion.\n\nFor $k=1$, the assertion is obvious. Assume for $k-1$, the one-to-one mapping $f_{k-1}$ satisfies the condition. Now we add the diagonal $I_k$ to $\\Omega$.\n\nIf $S_k$ is empty, the assertion is clearly true for $k$. Assume $S_k$ is nonempty. Then the intersections in $S_k$ all lie on $\\ell_k$, say they are $Q_1, \\dots, Q_r$, and they traverse the small regions $A_1, \\dots, A_{r+1}$, respectively, dividing each $A_i$ into two new small regions $B_i$ and $C_i$.\n\nNotice that $f_{k-1}^{-1}(A_i)$ can be either empty or determined, say a vertex $R_i$ of $B_i$ or $C_i$. Now we define $f_k$ as follows: for each $i \\in \\{1, \\dots, r\\}$,\n\n(1) If $f_{k-1}^{-1}(A_i)$ is empty, then let $f(Q_i) = B_i$;\n\n(2) If $R_i$ is a vertex of $B_i$, then let $f_k(R_i) = B_i$, $f_k(Q_i) = C_i$;\n\n(3) If $R_i$ is a vertex of $C_i$, then let $f_k(R_i) = C_i$, $f_k(Q_i) = B_i$.\n\nFor all other intersections $Q$, let $f_k(Q) = f_{k-1}(Q)$. It is straightforward to check that $f_k$ has the desired properties. Particularly, when $k=m$, the mapping $f=f_m$ gives a one-to-one correspondence between all interior intersections and small regions of $\\Omega$, satisfying that each intersection $Q$ is a vertex of the small region $f(Q)$. $\\square$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 28474,
    "subject": "Mathematics (Multi-modal)",
    "question": "Given arithmetic sequence $\\{a_n\\}$ with common difference $d \\ne 0$ and $a_{2021} = a_{20} + a_{21}$, then the value of $\\frac{a_1}{d}$ is ______.",
    "options": [],
    "answer": "1981",
    "solution": "By the conditions, we have $a_1 + 2020d = a_1 + 19d + a_1 + 20d$. And since $d \\ne 0$, $\\frac{a_1}{d} = 1981$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 28475,
    "subject": "Mathematics (Multi-modal)",
    "question": "Given integer $n \\ge 2$. Find the least positive integer $m$, such that there are $n^2$ distinct positive real numbers $x_{i,j}$ ($1 \\le i, j \\le n$) satisfying the following conditions:\n(1) For every $i, j$,\n$$\nx_{i,j} = \\max\\{x_{i,1}, x_{i,2}, \\dots, x_{i,j}\\} \\text{ or}\n$$\n$$\nx_{i,j} = \\max\\{x_{1,j}, x_{2,j}, \\dots, x_{i,j}\\};\n$$\n(2) For every $i$, there are at most $m$ indices $k$, with\n$$\nx_{i,k} = \\max\\{x_{i,1}, x_{i,2}, \\dots, x_{i,k}\\};\n$$\n(3) For every $j$, there are at most $m$ indices $k$, with\n$$\nx_{k,j} = \\max\\{x_{1,j}, x_{2,j}, \\dots, x_{k,j}\\}.\n$$",
    "options": [],
    "answer": "floor((n+3)/2)",
    "solution": "$m = \\lfloor \\frac{n+3}{2} \\rfloor$.\n\nPut these $n^2$ numbers $x_{i,j}$ in an $n \\times n$ table: call $x_{i,j}$ a \"row pivot\" if $x_{i,j} = \\max\\{x_{i,1}, x_{i,2}, \\dots, x_{i,j}\\}$; call $x_{i,j}$ a \"column pivot\", if $x_{i,j} = \\max\\{x_{1,j}, x_{2,j}, \\dots, x_{i,j}\\}$.\n\nFirst, prove $m \\ge \\lfloor \\frac{n+3}{2} \\rfloor$. Consider all $(n-1)^2$ numbers $x_{i,j}$ ($2 \\le i, j \\le n$): every column (from $2$ to $n$) contains at most $m-1$ column pivots (by condition (3), and notice that $x_{1,j}$ ($2 \\le j \\le n$) is a pivot); every row (from $2$ to $n$) contains at most $m-1$ row pivots (by condition (2), and notice that $x_{i,1}$ ($2 \\le i \\le n$) is a pivot). Due to condition (1), every number is a pivot (row or column or both), and hence\n$$\n(n-1)(m-1) + (n-1)(m-1) \\ge (n-1)^2. \\quad \\textcircled{1}\n$$\nWe claim the equality in (1) cannot be attained. Let $x_{i_0,j_0} = \\max x_{i,j}$ ($2 \\le i, j \\le n$). If $x_{i_0,j_0}$ is both row and column pivots, then $x_{i_0,j_0}$ is counted twice on the left-hand side of (1), and the equality cannot hold. Suppose $x_{i_0,j_0}$ is not a row pivot. Then $x_{i_0,1} > x_{i_0,j_0}$ and there are no other row pivots in this row, the equality in (1) cannot hold. Similarly, if $x_{i_0,j_0}$ is not a column pivot, then there are no other column pivots in this column, and the equality in (1) cannot hold. Therefore, $2(m-1) > n-1$, $m \\ge \\lfloor \\frac{n+3}{2} \\rfloor$.\n\nNext, construct tables for which $m = \\lfloor \\frac{n+3}{2} \\rfloor$. When $n = 2t$, divide the $n \\times n$ table into four $t \\times t$ tables as follows:\n\n| $\\Omega_1$ | $\\Omega_2$ |\n|------------|------------|\n| $\\Omega_3$ | $\\Omega_4$ |\n\nTake $\\varepsilon_1, \\varepsilon_2 > 0$ sufficiently small, such that $\\varepsilon_1 < \\frac{1}{2n}\\varepsilon_2 < \\frac{1}{4n^2}$. For $k=1, 4$, let the number in row $i$ and column $j$ of $\\Omega_k$ be $k - i\\varepsilon_1 + j\\varepsilon_2$. Note that in $\\Omega_1$ and $\\Omega_4$, the numbers decrease from the top right to the bottom left. For $k = 2, 3$, let the number in row $i$ and column $j$ of $\\Omega_k$ be $k + i\\varepsilon_1 - j\\varepsilon_2$. Note that in $\\Omega_2$ and $\\Omega_3$ the numbers decrease from the bottom left to the top right.\n\nIt is easy to check that all numbers in $\\Omega_1$ and $\\Omega_4$ are row pivots, and all numbers in $\\Omega_2$ and $\\Omega_3$ are column pivots. In addition, except for the first and the $t+1$ rows, there are no other column pivots in $\\Omega_1$ and $\\Omega_4$; except for the first and the $t+1$ columns, there are no other row pivots in $\\Omega_2$ and $\\Omega_3$. Hence, conditions (2) and (3) are fulfilled, and $m = t+1$.\n\nWhen $n = 2t - 1$, take the construction for $n = 2t$ as above and remove the rightmost column and the bottom row. $\\square$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 28476,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $m, n$ be positive integers, $a_{ij}$ ($1 \\le i \\le m, 1 \\le j \\le n$) be nonnegative real numbers such that for any $i, j$, the inequalities\n$$\na_{i,1} \\ge a_{i,2} \\ge \\dots \\ge a_{i,n}, \\quad a_{1,j} \\ge a_{2,j} \\ge \\dots \\ge a_{m,j}\n$$\nhold. For $i = 1, 2, \\dots, m$ and $j = 1, 2, \\dots, n$, define\n$$\nX_{i,j} = a_{1,j} + \\dots + a_{i-1,j} + a_{i,j} + a_{i,j-1} + \\dots + a_{i,1},\n$$\n$$\nY_{i,j} = a_{i,j} + a_{i+1,j} + \\dots + a_{m,j} + a_{i,j+1} + \\dots + a_{i,n}.\n$$\nProve that\n$$\n\\prod_{i=1}^{m} \\prod_{j=1}^{n} X_{i,j} \\ge \\prod_{i=1}^{m} \\prod_{j=1}^{n} Y_{i,j}.\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "The problem conditions imply that\n$$\nX_{i,j} \\ge (i + j - 1) \\cdot a_{i,j}, \\quad Y_{i,j} \\le (m + n - i - j + 1) \\cdot a_{i,j}.\n$$\nHence, we have\n$$\n\\begin{align*}\n\\prod_{i=1}^{m} \\prod_{j=1}^{n} X_{i,j} &\\ge \\prod_{i=1}^{m} \\prod_{j=1}^{n} (i+j-1) \\cdot a_{i,j} \\\\\n&= \\prod_{i=1}^{m} \\prod_{j=1}^{n} (m+n-i-j+1) \\cdot a_{i,j} \\\\\n&\\ge \\prod_{i=1}^{m} \\prod_{j=1}^{n} Y_{i,j},\n\\end{align*}\n$$\nwhere the equality on the second line is due to the one-to-one correspondence $(i, j) \\leftrightarrow (m + 1 - i, n + 1 - j)$, and $i + j - 1$ corresponds to $m + n - i - j + 1$.\nWe shall use the following inequality:\nFor $0 \\le c_1 \\le \\cdots \\le c_t$, $0 \\le d_1 \\le \\cdots \\le d_t$,\n$$\n(c_1+d_1)(c_2+d_2)\\cdots(c_t+d_t) \\le (c_1+d_t)(c_2+d_{t-1})\\cdots(c_t+d_1). \\quad \\textcircled{1}\n$$\nThis is because, for each $i$,\n$$\n(c_i+d_i)(c_{t+1-i}+d_{t+1-i}) \\le (c_i+d_{t+1-i})(c_{t+1-i}+d_i).\n$$\nMultiplying the above inequalities for $i = 1, \\dots, t$ gives\n$$\n\\prod_{i=1}^{t} ((c_i+d_i)(c_{t+1-i}+d_{t+1-i})) \\le \\prod_{i=1}^{t} ((c_i+d_{t+1-i})(c_{t+1-i}+d_i)),\n$$\nand taking the square root on both sides of the above inequality gives (1).\nFor the original problem, let $L_{i,j} = \\sum_{q=1}^{j} a_{i,q}$, $R_{i,j} = \\sum_{q=j}^{n} a_{i,q}$,\n$$\nU_{i,j} = \\sum_{p=1}^{i} a_{p,j}, \\quad D_{i,j} = \\sum_{p=i}^{m} a_{p,j}.\n$$\nThen\n$$\nX_{i,j} = L_{i,j} + U_{i,j} - a_{i,j}, \\quad Y_{i,j} = D_{i,j} + R_{i,j} - a_{i,j}.\n$$\nApply (1) to derive\n$$\n\\begin{align*}\n\\prod_{i=1}^{m} \\prod_{j=1}^{n} X_{i,j} &= \\prod_{i=1}^{m} \\prod_{j=1}^{n} (L_{i,j} + (U_{i,j} - a_{i,j})) \\\\\n&\\ge \\prod_{i=1}^{m} \\prod_{j=1}^{n} (L_{i,n+1-j} + (U_{i,j} - a_{i,j})) \\\\\n&\\quad \\text{(since } \\{L_{i,j}\\}_{j=1}^{n} \\text{ increases while } \\{U_{i,j} - a_{i,j}\\}_{j=1}^{n} \\text{ decreases)} \\\\\n&\\ge \\prod_{i=1}^{m} \\prod_{j=1}^{n} (R_{i,j} + (U_{i,j} - a_{i,j})) \\quad \\text{(since } L_{i,n+1-j} \\ge R_{i,j}) \\\\\n&= \\prod_{j=1}^{n} \\prod_{i=1}^{m} ((R_{i,j} - a_{i,j}) + U_{i,j}) \\\\\n&\\ge \\prod_{j=1}^{n} \\prod_{i=1}^{m} ((R_{i,j} - a_{i,j}) + U_{m+1-i,j}) \\\\\n&\\quad \\text{(since } \\{R_{i,j} - a_{i,j}\\}_{i=1}^{m} \\text{ decreases while } \\{U_{i,j}\\}_{i=1}^{m} \\text{ increases)}\n\\end{align*}\n$$\n$$\n\\begin{align*}\n&\\ge \\prod_{j=1}^{n} \\prod_{i=1}^{m} ((R_{i,j} - a_{i,j}) + D_{i,j}) \\quad (\\text{since } U_{m+1-i,j} \\ge D_{i,j}) \\\\\n&= \\prod_{i=1}^{m} \\prod_{j=1}^{n} Y_{i,j}.\n\\end{align*}\n$$",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 28477,
    "subject": "Mathematics (Multi-modal)",
    "question": "Suppose that arithmetic sequence $\\{a_n\\}$ satisfies $a_{2021} = a_{20} + a_{21} = 1$. Then the value of $a_1$ is ______.",
    "options": [],
    "answer": "1981/4001",
    "solution": "Let the common difference of $\\{a_n\\}$ be $d$. By the given condition, it follows that\n$$\n\\begin{cases}\na_1 + 2020d = 1, \\\\\n2a_1 + 39d = 1.\n\\end{cases}\n$$\nThe solution is $a_1 = \\frac{1981}{4001}$.\n$\\square$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 28478,
    "subject": "Mathematics (Multi-modal)",
    "question": "Suppose $f(x)$ is an odd function with domain $\\mathbb{R}$. If $f(1) = 2$, $f(2) = 3$, then the value of $f(f(-1))$ is ______.",
    "options": [],
    "answer": "-3",
    "solution": "By the given conditions we have $f(-1) = -f(1) = -2$. Therefore,\n$$\nf(f(-1)) = f(-2) = -f(2) = -3.\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 28479,
    "subject": "Mathematics (Multi-modal)",
    "question": "Suppose $a, b \\in \\mathbb{R}$. If equation\n\n$$(z^2 + az + b)(z^2 + az + 2b) = 0$$\n\nabout $z$ has four mutually different complex roots $z_1, z_2, z_3, z_4$ and their corresponding points in the complex plane are exactly four vertices of a square with side length $1$, then find the value of $|z_1| + |z_2| + |z_3| + |z_4|$.",
    "options": [],
    "answer": "sqrt(6) + 2 sqrt(2)",
    "solution": "Denote quadratic equations $E_1: z^2 + az + b = 0$, $E_2: z^2 + az + 2b = 0$. Let $z_1, z_2$ be solutions of $E_1$ and $z_3, z_4$ be solutions of $E_2$.\nIf $z_1, z_2, z_3, z_4$ are all real numbers, then their corresponding points on the complex plane are all on the real axis, which is not consistent with the question. If $z_1, z_2, z_3, z_4$ are imaginary numbers, then their corresponding points on the complex plane are all on line $\\operatorname{Re} z = -\\frac{a}{2}$, which does not fit the question. Therefore, there are two real numbers and two imaginary numbers in $z_1, z_2, z_3, z_4$.\nThis shows that discriminant $a^2-4b$ of equation $E_1$ and the discriminant $a^2-8b$ of $E_2$ have different signs.\nAt this point, there must be $b > 0$ (if $b \\le 0$, then $a^2 - 4b \\ge 0$ and $a^2 - 8b \\ge 0$, a contradiction), so\n$$\na^2 - 4b \\ge 0 > a^2 - 8b.\n$$\nHence, $z_{1,2} = \\frac{-a \\pm \\sqrt{a^2 - 4b}}{2}$, $z_{3,4} = \\frac{-a \\pm \\sqrt{8b - a^2}}{2}$.\nIt is evident that $\\frac{z_1 + z_2}{2} = \\frac{z_3 + z_4}{2} = -\\frac{a}{2}$. Since the side length of the square is $1$, there is\n$$\n|z_1 - z_2| = \\sqrt{a^2 - 4b} = \\sqrt{2},\n$$\n$$\n|z_3 - z_4| = \\sqrt{8b - a^2} = \\sqrt{2},\n$$\nnamely, $a^2 - 4b = 8b - a^2 = 2$, and the solutions are $a^2 = 6, b = 1$.\nNoticing that $z_1, z_2$ have the same sign and $|z_3| = |z_4|$, we know that\n$$\n\\begin{aligned}\n|z_1| + |z_2| + |z_3| + |z_4| &= |z_1 + z_2| + 2|z_3| \\\\\n&= |-a| + \\sqrt{a^2 + (8b - a^2)} \\\\\n&= \\sqrt{6} + 2\\sqrt{2}.\n\\end{aligned}\n$$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 28480,
    "subject": "Mathematics (Multi-modal)",
    "question": "Given function $f(x) = |2 - \\log_3 x|$, positive real numbers $a, b, c$ satisfy $a < b < c$ and $f(a) = 2f(b) = 2f(c)$. Find the minimum of $\\frac{ac}{b}$.",
    "options": [],
    "answer": "9",
    "solution": "Notice that $f(x) = |\\log_3(\\frac{x}{9})|$ is monotonically decreasing on $(0, 9]$ and monotonically increasing on $[9, +\\infty)$.\n\nBy the conditions satisfied by $a, b, c$, we have $0 < a < b < 9 < c$ and\n$$\n\\log_3\\left(\\frac{9}{a}\\right) = 2\\log_3\\left(\\frac{9}{b}\\right) = 2\\log_3\\left(\\frac{c}{9}\\right).\n$$\nTherefore,\n$$\n\\log_3\\left(\\frac{ac}{b}\\right) = \\log_3\\left(9 \\cdot \\frac{a}{9} \\cdot \\frac{9}{b} \\cdot \\frac{c}{9}\\right) = 2 - \\log_3\\left(\\frac{9}{a}\\right) + \\log_3\\left(\\frac{9}{b}\\right) + \\log_3\\left(\\frac{c}{9}\\right) = 2,\n$$\nnamely, $\\frac{ac}{b} = 3^2 = 9$. $\\square$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 28481,
    "subject": "Mathematics (Multi-modal)",
    "question": "In the plane rectangular coordinate system, the graph of function $y = \\frac{x+1}{|x|+1}$ has three different points lying on line $l$, and the sum of the abscissas of these three points is $0$. Find the range of values of the slope of $l$.",
    "options": [],
    "answer": "0 < k < 2/9",
    "solution": "When $x \\ge 0$, $y = 1$; when $x < 0$, $y = \\frac{x+1}{1-x}$ is strictly increasing about $x$ and less than $1$.\n\nSuppose line $l: y = kx + b$, and then the known conditions are equivalent to the fact that equation\n$$\nkx + b = \\frac{x+1}{|x|+1} \\qquad (1)\n$$\nhas three different real number solutions $x_1, x_2, x_3$ ($x_1 < x_2 < x_3$) satisfying $x_1 + x_2 + x_3 = 0$.\n\nFirstly, there is $k \\neq 0$, otherwise the function of $l$ can only be $y = 1$, but then the sum of the abscissas of any three common points of the graph of $l$ and function $y = \\frac{x+1}{|x|+1}$ must be greater than $0$. This is not consistent with the question.\n\nWhen $x < 0$, Equation (1) can be arranged as\n$$\nkx^2 - (k - b - 1)x + 1 - b = 0, \\qquad (2)\n$$\nand it has at most two negative solutions.\n\nWhen $x \\ge 0$, Equation (1) follows as\n$$\nkx + b = 1, \\qquad (3)\n$$\nand it has at most one non-negative solution.\n\nThis shows that Equation (2) has two different negative solutions $x_1, x_2$, where\n$$\nx_1 + x_2 = \\frac{k - b - 1}{2};\n$$\nEquation (3) has a non-negative solution $x_3 = \\frac{1-b}{k}$. By $x_1 + x_2 + x_3 = 0$, we know that $k = 2b$.\n\nAnd then there is $x_3 = \\frac{1-b}{2b}$. Since $x_3 \\ge 0$, we have $0 < b \\le 1$.\n\nEquation (2) becomes $2b x^2 + (1-b)x + 1 - b = 0$. By discriminant\n$$\n(1-b)^2 - 4 \\cdot 2b(1-b) = (1-b)(1-9b) > 0\n$$\nand combining $0 < b \\le 1$, we have $0 < b < \\frac{1}{9}$. (After checking, $x_1, x_2$ are indeed negative at this point, which is consistent with the question.)\n\nIn conclusion, the range of the slope $k = 2b$ of $l$ is $0 < k < \\frac{2}{9}$.",
    "topic": "Discrete Mathematics",
    "subtopic": "Other"
  },
  {
    "id": 28482,
    "subject": "Mathematics (Multi-modal)",
    "question": "In a plane rectangular coordinate system $xOy$, the graph of function $y = \\frac{1}{|x|}$ is $\\Gamma$. Let points $P, Q$ on $\\Gamma$ satisfy: $P$ is in the first quadrant, $Q$ is in the second quadrant, and line $PQ$ is tangent to the part of $\\Gamma$ in the second quadrant at point $Q$. Find the minimum of $|PQ|$.",
    "options": [],
    "answer": "2",
    "solution": "When $x > 0$, $y = \\frac{1}{x}$. When $x < 0$, $y = -\\frac{1}{x}$, and its corresponding derivative is $y' = -\\frac{1}{x^2}$.\n\nSuppose $Q(-a, \\frac{1}{a})$, where $a > 0$. By the condition, the slope of $PQ$ is $y'|_{x=-a} = \\frac{1}{a^2}$.\n\nThe equation of line $PQ$ is $y = \\frac{1}{a^2}(x + a) + \\frac{1}{a} = \\frac{x + 2a}{a^2}$.\n\nCombining the above equation with $y = \\frac{1}{x}$ ($x > 0$) yields $x^2 + 2a x - a^2 = 0$, and thus we know the abscissa $x_P = (\\sqrt{2} - 1)a$ of point $P$, with the negative root discarded. Therefore,\n\n$$\n\\begin{align*}\n|PQ| &= \\sqrt{1 + \\left(\\frac{1}{a^2}\\right)^2} \\cdot |x_P - x_Q| \\\\\n&= \\sqrt{1 + \\frac{1}{a^4}} \\cdot (\\sqrt{2}a) \\\\\n&\\geq \\sqrt{2\\sqrt{1 \\cdot \\frac{1}{a^4}}} \\cdot \\sqrt{2}a = 2.\n\\end{align*}\n$$\n\nWhen $a = 1$, namely, $Q(-1, 1)$, the minimum of $|PQ|$ is $2$. $\\square$",
    "topic": "Discrete Mathematics",
    "subtopic": "Other"
  },
  {
    "id": 28483,
    "subject": "Mathematics (Multi-modal)",
    "question": "In a plane rectangular coordinate system $xOy$, given ellipse\n$$\\Gamma : \\frac{x^2}{a^2} + \\frac{y^2}{b^2} = 1 \\quad (a > b > 0),$$\nfind the range of the eccentricity of $\\Gamma$.",
    "options": [],
    "answer": "[3/4, (-1 + sqrt(37))/6]",
    "solution": "By symmetry, it is useful to set $A(a, 0)$, $B(0, b)$. Denote $c = |OF| = \\sqrt{a^2 - b^2}$ and the eccentricity $e = \\frac{c}{a}$.\nLet $|OP| = r$. By the properties of ellipse, we know that the range of $r$ is $[b, a]$.\nNote that $O$ is the midpoint of $PQ$, and thus\n$$\n\\overrightarrow{FP} \\cdot \\overrightarrow{FQ} = (\\overrightarrow{FO} + \\overrightarrow{OP}) \\cdot (\\overrightarrow{FO} - \\overrightarrow{OP}) = \\overrightarrow{FO}^2 - \\overrightarrow{OP}^2 = c^2 - r^2. \\quad \\textcircled{1}\n$$\n(1) Suppose the focal point $F$ is $(c, 0)$, and then\n$$\n\\overrightarrow{FA} \\cdot \\overrightarrow{FB} = (a - c, 0) \\cdot (-c, b) = c^2 - ac < 0.\n$$\nThus\n$$\n\\overrightarrow{FP} \\cdot \\overrightarrow{FQ} + \\overrightarrow{FA} \\cdot \\overrightarrow{FB} < c^2 - r^2 < a^2 + b^2 = |AB|^2,\n$$\nand this is not consistent with the conditions. Hence $F$ is $(-c, 0)$, which shows that $F$ lies on the extension of $AO$.\n(2) From $F(-c, 0)$, we know that\n$$\n\\overrightarrow{FA} \\cdot \\overrightarrow{FB} = (a + c, 0) \\cdot (c, b) = c^2 + ac.\n$$\nCombining with (1) gives\n$$\n\\begin{aligned}\n\\overrightarrow{FP} \\cdot \\overrightarrow{FQ} + \\overrightarrow{FA} \\cdot \\overrightarrow{FB} &= (c^2 - r^2) + (c^2 + ac) \\\\\n&= 2c^2 + ac - r^2,\n\\end{aligned}\n$$\nso $2c^2 + ac - r^2 = a^2 + b^2$, i.e.,\n$$\nr^2 = 2c^2 + ac - a^2 - b^2 = 3c^2 + ac - 2a^2.\n$$\nNote that $r \\in [b, a]$, and thus $3c^2 + ac - 2a^2 = r^2 \\in [a^2 - c^2, a^2]$.\nThe above equation is equivalent to $\\frac{a^2 - c^2}{a^2} \\le \\frac{3c^2 + ac - 2a^2}{a^2} \\le 1$, i.e.,\n$1 - e^2 \\le 3e^2 + e - 2 \\le 1$.\nCombining $e \\in [0, 1]$, we obtain the range of the eccentricity of $\\Gamma$, namely,\n$$\ne \\in \\left[ \\frac{3}{4}, \\frac{-1 + \\sqrt{37}}{6} \\right].\n$$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 28484,
    "subject": "Mathematics (Multi-modal)",
    "question": "As shown in Fig. 11.1, in a plane rectangular coordinate system $xOy$, the left and right foci of ellipse $\\Gamma: \\frac{x^2}{2} + y^2 = 1$ are $F_1, F_2$, respectively. Let $P$ be a point on $\\Gamma$ in the first quadrant and the extensions of $PF_1, PF_2$ intersect $\\Gamma$ at points $Q_1(x_1, y_1), Q_2(x_2, y_2)$, respectively.\n\nFind the maximum of $y_1 - y_2$.",
    "options": [],
    "answer": "2*sqrt(2)/3",
    "solution": "As shown in Fig. 11.1, we find $F_1(-1, 0), Q_2(1, 0)$.\nDenote $P(x_0, y_0)$. By the condition, it follows that $x_0, y_0 > 0$, $y_1 < 0$, $y_2 < 0$.\n![](attached_image_1.png)\nFig. 11.1\n\nThe equation of line $PF_1$ is $x = \\frac{(x_0 + 1)y}{y_0} - 1$. Substituting it into $\\frac{x^2}{2} + y^2 = 1$ and organizing it yields\n$$\n\\left( \\frac{(x_0 + 1)^2}{2y_0^2} + 1 \\right) y^2 - \\frac{x_0 + 1}{y_0} y - \\frac{1}{2} = 0.\n$$\nMultiplying both sides by $2y_0^2$ and noting that $x_0^2 + 2y_0^2 = 2$, we get\n$$\n(3 + 2x_0)y^2 - 2(x_0 + 1)y_0y - y_0^2 = 0.\n$$\nThe two roots of this equation are $y_0, y_1$. By Vieta's formulas, we get $y_0y_1 = -\\frac{y_0^2}{3+2x_0}$. Thus,\n$$\ny_1 = -\\frac{y_0}{3 + 2x_0}.\n$$\nSimilarly, we can get $y_2 = -\\frac{y_0}{3 - 2x_0}$. Therefore,\n$$\ny_1 - y_2 = \\frac{y_0}{3 - 2x_0} - \\frac{y_0}{3 + 2x_0} = \\frac{4x_0y_0}{9 - 4x_0^2}.\n$$\nSince $9 - 4x_0^2 = \\frac{1}{2}x_0^2 + 9y_0^2 \\ge 2\\sqrt{\\frac{1}{2}x_0^2 \\cdot 9y_0^2} = 3\\sqrt{2}x_0y_0$, it follows that\n$$\ny_1 - y_2 \\le \\frac{4x_0y_0}{3\\sqrt{2}x_0y_0} = \\frac{2\\sqrt{2}}{3},\n$$\nwhere the equal sign holds when $\\frac{1}{2}x_0^2 = 9y_0^2$ is required, and accordingly $x_0 = \\frac{3\\sqrt{5}}{5}, y_0 = \\frac{\\sqrt{10}}{10}$.\nTherefore, the maximum of $y_1 - y_2$ is $\\frac{2\\sqrt{2}}{3}$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 28485,
    "subject": "Mathematics (Multi-modal)",
    "question": "As shown in Fig. 11.1, in a plane rectangular coordinate system $xOy$, the left and right foci of the ellipse $\\Gamma : \\frac{x^2}{2} + y^2 = 1$ are $F_1, F_2$, respectively. Let $P$ be a point on $\\Gamma$ in the first quadrant, and the extensions of $PF_1, PF_2$ intersect $\\Gamma$ at points $Q_1, Q_2$, respectively. Let $r_1, r_2$ be the radii of the incircles of $\\triangle PF_1Q_2, \\triangle PF_2Q_1$, respectively. Find the maximum of $r_1 - r_2$.",
    "options": [],
    "answer": "1/3",
    "solution": "It is easy to find $F_1 = (-1, 0), F_2 = (1, 0)$.\nDenote $P(x_0, y_0)$, $Q_1(x_1, y_1)$, $Q_2(x_2, y_2)$. By the given condition, it follows that\n$$\nx_0, y_0 > 0, \\quad y_1 < 0, \\quad y_2 < 0.\n$$\nBy the definition of ellipse we get\n$$\n|PF_1| + |PF_2| = |Q_1F_1| + |Q_1F_2| = |Q_2F_1| + |Q_2F_2| = 2\\sqrt{2}.\n$$\nHence, the perimeters of $\\triangle PF_1Q_2$ and $\\triangle PF_2Q_1$ are both $l = 4\\sqrt{2}$.\nAnd since $|F_1F_2| = 2$,\n$$\nr_1 = \\frac{2S_{\\triangle PF_1Q_2}}{l} = \\frac{(y_0 - y_2) \\cdot |F_1F_2|}{l} = \\frac{y_0 - y_2}{2\\sqrt{2}}.\n$$\nSimilarly, we can get $r_2 = \\frac{y_0 - y_1}{2\\sqrt{2}}$, so $r_1 - r_2 = \\frac{y_1 - y_2}{2\\sqrt{2}}$.\nIn the following, we will first find $y_1 - y_2$.\n![](attached_image_1.png)\nFig. 11.1\n\nThe equation of line $PF_1$ is $x = \\frac{(x_0 + 1)y}{y_0} - 1$. Substituting it into $\\frac{x^2}{2} + y^2 = 1$ and rearranging it gives\n$$\n\\left( \\frac{(x_0 + 1)^2}{2y_0^2} + 1 \\right) y^2 - \\frac{x_0 + 1}{y_0} y - \\frac{1}{2} = 0.\n$$\nMultiplying both sides by $2y_0^2$ and noticing that $x_0^2 + 2y_0^2 = 2$, we find that\n$$\n(3 + 2x_0)y^2 - 2(x_0 + 1)y_0y - y_0^2 = 0.\n$$\nThe two roots of this equation are $y_0$ and $y_1$. By Vieta's formulas, we get $y_0y_1 = -\\frac{y_0^2}{3+2x_0}$. Thus,\n$$\ny_1 = -\\frac{y_0}{3 + 2x_0}.\n$$\nSimilarly, we can get $y_2 = -\\frac{y_0}{3 - 2x_0}$. Therefore,\n$$\ny_1 - y_2 = \\frac{y_0}{3 - 2x_0} - \\frac{y_0}{3 + 2x_0} = \\frac{4x_0y_0}{9 - 4x_0^2}.\n$$\nSince $9 - 4x_0^2 = \\frac{1}{2}x_0^2 + 9y_0^2 \\ge 2\\sqrt{\\frac{1}{2}x_0^2 \\cdot 9y_0^2} = 3\\sqrt{2}x_0y_0$, there is\n$$\nr_1 - r_2 = \\frac{y_1 - y_2}{2\\sqrt{2}} = \\frac{\\sqrt{2}x_0y_0}{9 - 4x_0^2} \\le \\frac{\\sqrt{2}x_0y_0}{3\\sqrt{2}x_0y_0} = \\frac{1}{3},\n$$\nwhere the equality sign holds when $\\frac{1}{2}x_0^2 = 9y_0^2$ is required. Accordingly, $x_0 = \\frac{3\\sqrt{5}}{5}$, $y_0 = \\frac{\\sqrt{10}}{10}$.\nTherefore, the maximum of $r_1 - r_2$ is $y_0 = \\frac{1}{3}$. $\\square$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 28486,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $a, b$ be real numbers and function $f(x) = x^3 + a x^2 + b x$. If there exist three real numbers $x_1, x_2, x_3$ satisfying $x_1 + 1 \\le x_2 \\le x_3 - 1$ and $f(x_1) = f(x_2) = f(x_3)$. Find the minimum of $|a| + 2|b|$.",
    "options": [],
    "answer": "√3",
    "solution": "For function $f(x)$ and real numbers $x_1, x_2, x_3$ satisfying the conditions, let $f(x_1) = f(x_2) = f(x_3) = c$, and then $x_1, x_2, x_3$ are the three real roots of cubic equation $x^3 + a x^2 + b x - c = 0$. By Vieta's formulas, we know that\n$$\na = -(x_1 + x_2 + x_3), \\quad b = x_1 x_2 + x_2 x_3 + x_3 x_1.\n$$\nFurther, by the condition we know that $x_2 - x_1 \\ge 1$, $x_3 - x_2 \\ge 1$, $x_3 - x_1 \\ge 2$, and thus\n$$\n\\begin{aligned}\na^2 - 3b &= x_1^2 + x_2^2 + x_3^2 - x_1 x_2 - x_2 x_3 - x_3 x_1 \\\\\n&= \\frac{1}{2}((x_1 - x_2)^2 + (x_2 - x_3)^2 + (x_3 - x_1)^2) \\\\\n&\\ge \\frac{1}{2}(1 + 1 + 4) = 3,\n\\end{aligned}\n$$\ni.e., $b \\le \\frac{a^2}{3} - 1$.\nIf $|a| \\ge \\sqrt{3}$, then $|a| + 2|b| \\ge |a| + \\sqrt{3}$.\nIf $0 \\le |a| < \\sqrt{3}$, then $b \\le \\frac{a^2}{3} - 1 < 0$. Notice that at this point\n$$\n\\begin{aligned}\n|a| - \\frac{3}{4} &< \\sqrt{3} - \\frac{3}{4}, \\\\\n|a| + 2|b| &\\ge |a| + \\frac{2}{3}(3 - a^2) \\\\\n&\\quad - \\frac{2}{3}\\left(|a| - \\frac{3}{4}\\right)^2 + \\frac{19}{8} \\\\\n&\\quad - \\frac{2}{3}\\left(\\sqrt{3} - \\frac{3}{4}\\right)^2 + \\frac{19}{8} = \\sqrt{3}.\n\\end{aligned}\n$$\nThe above shows that $|a| + 2|b| \\ge \\sqrt{3}$.\nOn the other hand, when $a = \\sqrt{3}$, $b = 0$, real numbers $x_1 = -1 - \\frac{\\sqrt{3}}{3}$, $x_2 = -\\frac{\\sqrt{3}}{3}$, $x_3 = 1 - \\frac{\\sqrt{3}}{3}$ satisfy the condition because the values of $F_1$, $F_2$, $F_3$ are all $\\frac{2\\sqrt{3}}{9}$. At this point $|a| + 2|b| = \\sqrt{3}$.\nIn conclusion, the minimum of $|a| + 2|b|$ is $\\sqrt{3}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 28487,
    "subject": "Mathematics (Multi-modal)",
    "question": "As shown in Fig. 11.1, the edge length of cube $ABCD$–$EFGH$ is $2$. Take any point $P_1$ on the incircle of square $ABFE$, take any point $P_2$ on the incircle of square $BCGF$, and take any point $P_3$ on the incircle of square $EFGH$. Find the minimum and maximum of $|P_1P_2| + |P_2P_3| + |P_3P_1|$.\n\n![](attached_image_1.png)\nFig. 11.1",
    "options": [],
    "answer": "Minimum = 3√2 − 3; Maximum = 3√6",
    "solution": "According to the condition, we can set\n$$\nP_1(1, \\cos \\alpha_1, \\sin \\alpha_1), \\quad P_2(\\sin \\alpha_2, 1, \\cos \\alpha_2), \\quad P_3(\\cos \\alpha_3, \\sin \\alpha_3, 1).\n$$\nWe conventionally assume that $P_4 = P_1$ and $\\alpha_4 = \\alpha_1$. Denote\n$$\nd_i = |P_i P_{i+1}| \\quad (i = 1, 2, 3).\n$$\nThen\n$$\nd_i^2 = (1 - \\sin \\alpha_{i+1})^2 + (1 - \\cos \\alpha_i)^2 + (\\sin \\alpha_i - \\cos \\alpha_{i+1})^2.\n$$\nDenote $f = |P_1P_2| + |P_2P_3| + |P_3P_1|$.\n\nWe first find the minimum of $f$. For $i = 1, 2, 3$, according to (1) and the inequality of arithmetic and geometric means we get\n$$\nd_i^2 \\geq (1 - \\sin \\alpha_{i+1})^2 + (1 - \\cos \\alpha_i)^2 \\\\\n\\geq \\frac{1}{2}((1 - \\sin \\alpha_{i+1}) + (1 - \\cos \\alpha_i))^2,\n$$\nand thus $d_i \\geq \\frac{\\sqrt{2}}{2}(2 - \\sin \\alpha_{i+1} - \\cos \\alpha_i)$. Therefore,\n$$\nf = d_1 + d_2 + d_3 \\\\\n\\geq \\frac{\\sqrt{2}}{2} \\sum_{i=1}^{3} (2 - \\sin \\alpha_{i+1} - \\cos \\alpha_i) \\\\\n= 3\\sqrt{2} - \\frac{\\sqrt{2}}{2} \\sum_{i=1}^{3} (\\sin \\alpha_i + \\cos \\alpha_i) \\\\\n= 3\\sqrt{2} - \\sum_{i=1}^{3} \\sin \\left(\\alpha_i + \\frac{\\pi}{4}\\right) \\\\\n\\geq 3\\sqrt{2} - 3.\n$$\nWhen $\\alpha_i = \\frac{\\pi}{4}$ ($i = 1, 2, 3$), $f$ can take the minimum $3\\sqrt{2} - 3$.\n\nThen we will find the maximum of $f$. By (1), it is clear that\n$$\nd_i^2 = 4 - 2 \\cos \\alpha_i - 2 \\sin \\alpha_{i+1} - 2 \\sin \\alpha_i \\cos \\alpha_{i+1}.\n$$\nNote that $\\sin \\alpha_i \\geq -1$, $\\cos \\alpha_i \\geq -1$ ($i = 1, 2, 3$), and then\n$$\n\\begin{aligned}\n\\sum_{i=1}^{3} d_i^2 &= 12 - 2 \\left( \\sum_{i=1}^{3} \\sin \\alpha_{i+1} + \\sum_{i=1}^{3} \\cos \\alpha_i + \\sum_{i=1}^{3} \\sin \\alpha_i \\cos \\alpha_{i+1} \\right) \\\\\n&= 12 - 2 \\left( \\sum_{i=1}^{3} \\sin \\alpha_i + \\sum_{i=1}^{3} \\cos \\alpha_{i+1} + \\sum_{i=1}^{3} \\sin \\alpha_i \\cos \\alpha_{i+1} \\right) \\\\\n&= 18 - 2 \\sum_{i=1}^{3} (1 + \\sin \\alpha_i)(1 + \\cos \\alpha_{i+1}) \\leq 18.\n\\end{aligned}\n$$\nBy the Cauchy inequality, we know that $f^2 \\leq 3(d_1^2 + d_2^2 + d_3^2) = 54$, so $f \\leq 3\\sqrt{6}$.\n\nWhen $\\alpha_i = \\pi$ ($i = 1, 2, 3$), $f$ can take the maximum $3\\sqrt{6}$. In conclusion, the minimum of $f$ is $3\\sqrt{2} - 3$, and its maximum is $3\\sqrt{6}$. $\\square$",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 28488,
    "subject": "Mathematics (Multi-modal)",
    "question": "The maximum of $f(x) = 2 \\sin^2 x - \\tan^2 x$ is ______.",
    "options": [],
    "answer": "3 - 2\\sqrt{2}",
    "solution": "$$\n\\begin{aligned}\nf(x) &= 2(1 - \\cos^2 x) - \\frac{1 - \\cos^2 x}{\\cos^2 x} \\\\\n&= 3 - \\left( 2 \\cos^2 x + \\frac{1}{\\cos^2 x} \\right) \\\\\n&\\le 3 - 2\\sqrt{2 \\cos^2 x \\cdot \\frac{1}{\\cos^2 x}} = 3 - 2\\sqrt{2},\n\\end{aligned}\n$$\nWhen $2 \\cos^2 x = \\frac{1}{\\cos^2 x}$ (e.g., take $x = \\arccos \\frac{1}{\\sqrt{2}}$), $f(x)$ takes the maximum $3 - 2\\sqrt{2}$. $\\square$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 28489,
    "subject": "Mathematics (Multi-modal)",
    "question": "Suppose $m$ is a real number, and complex numbers $z_1 = 1 + 2i$, $z_2 = m + 3i$, where $i$ is the imaginary unit. If $z_1 \\cdot \\bar{z_2}$ is purely imaginary, then the value of $|z_1 + z_2|$ is ______.",
    "options": [],
    "answer": "5*sqrt(2)",
    "solution": "Since $z_1 \\cdot \\bar{z_2} = (1 + 2i)(m - 3i) = m + 6 + (2m - 3)i$ is purely imaginary, we get $m = -6$. Therefore, $|z_1 + z_2| = |-5 + 5i| = 5\\sqrt{2}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 28490,
    "subject": "Mathematics (Multi-modal)",
    "question": "Given set $A = \\{1, 2, m\\}$, $m$ is real. Let $B = \\{a^2 \\mid a \\in A\\}$, $C = A \\cup B$. If the sum of all the elements of $C$ is $6$, then the product of all the elements of $C$ is ______.",
    "options": [],
    "answer": "-8",
    "solution": "By the condition, it is known that $1$, $2$, $4$, $m$, $m^2$ (allowing for repetition) are all the elements of $C$.\nNote that when $m$ is real, $1+2+4+m+m^2 > 6$, $1+2+4+m^2 > 6$, so it can only be $C = \\{1, 2, 4, m\\}$, and $1+2+4+m = 6$. Therefore, $m = -1$, and is tested to be consistent with the question. At this point the product of all the elements of $C$ is $1 \\times 2 \\times 4 \\times (-1) = -8$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 28491,
    "subject": "Mathematics (Multi-modal)",
    "question": "Given $2021$ distinct positive integers $a_1, a_2, \\dots, a_{2021}$. Define the sequence $\\{a_n\\}$ inductively as follows: for each integer $n \\ge 2022$, $a_n$ is the smallest positive integer different from $a_1, a_2, \\dots, a_{n-1}$ and not dividing the product $a_{n-1}a_{n-2}\\dots a_{n-2021}$. Prove: there exists a positive integer $M$, such that all integers greater than or equal to $M$ appear in $\\{a_n\\}$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $k=2021$. In fact, we will prove the statement for any integer $k > 0$.\n\n**Lemma 1** There exists $C > 0$ independent of $k$, such that $\\tau(m) \\le C m^{\\frac{1}{k+1}}$ holds for all positive integer $m$, where $\\tau(m)$ is the number of positive factors of $m$.\n\n**Proof of lemma 1** Let $m = p_1^{\\alpha_1} p_2^{\\alpha_2} \\dots p_r^{\\alpha_r}$ be the prime factorization of $m$. Evidently,\n$$\n\\tau(m) = (\\alpha_1 + 1)(\\alpha_2 + 1)\\dots(\\alpha_r + 1).\n$$\nEquivalently, there exists $C > 0$ such that\n$$\n\\prod_{i=1}^{r} \\frac{(\\alpha_i + 1)^{k+1}}{p_i^{\\alpha_i}} \\le C^{k+1}\n$$\nholds for all prime $p_i$ and positive integers $\\alpha_i$. Note that polynomials grow slower than exponential functions, and hence for each prime $p \\le 2^{k+1}$, the fraction $\\frac{(\\alpha + 1)^{k+1}}{p^{\\alpha}}$ has an upper bound when $\\alpha \\ge 0$. Since the number of primes $p \\le 2^{k+1}$ is finite, there exists $C'$ such that for any prime $p \\le 2^{k+1}$,\n$$\n\\frac{(\\alpha + 1)^{k+1}}{p^{\\alpha}} < C'.\n$$\nOn the other hand, for $p > 2^{k+1}$,\n$$\n\\frac{(\\alpha + 1)^{k+1}}{p^{\\alpha}} < \\left( \\frac{\\alpha + 1}{2^{\\alpha}} \\right)^{k+1} < 1.\n$$\nIt follows that $\\prod_{i=1}^{r} \\frac{(\\alpha_i + 1)^{k+1}}{p_i^{\\alpha_i}} \\le C'^s$, where $s$ is the number of distinct prime factors of $m$ less than $2^{k+1}$. The existence of $C > 0$ is now verified.\n\n**Lemma 2** For any fixed integer $L > 0$, there exists $M > 0$ such that for any positive integer $m \\ge M$, one can find a prime power $p^\\alpha|m$, and $p^\\alpha > L$.\n\n**Proof of lemma 2** Consider all prime powers less than or equal to $L$: there are only finitely many of them. Define $M$ equals the product of them plus $1$. If $m \\ge M$ and $p|m$, then its highest power in $m$ satisfies $p^\\alpha|m$, and $p^\\alpha > L$.\n\n**Lemma 3** There exists a positive integer $D$ such that $a_n < 2n + D$ for every positive integer $n$.\n\n**Proof of Lemma 3** We take $D$ satisfying $D > \\max_{1 \\le s \\le k}\\{a_s\\}$ and\n$$\nD > 2^k C^{k+1} + C D^{\\frac{k}{k+1}},\n$$\nwhere $C > 0$ is defined in Lemma 1, and use induction to prove $a_n < 2n + D$ for every $n$. By the choice of $D$, $a_n < 2n + D$ is true for $n = 1, 2, \\dots, k$. Assume $a_n < 2n + D$ holds for $1, 2, \\dots, n + k - 1$, where $n$ is a positive integer. Now for $a_{n+k}$, it is known from the problem that $a_{n+k}$ is the smallest positive integer not dividing $a_n a_{n+1} \\dots a_{n+k-1}$, and different from $a_1, \\dots, a_{n+k-1}$. Hence,\n$$\na_{n+k} \\le \\tau(a_n a_{n+1} \\dots a_{n+k-1}) + n + k.\n$$\nFrom Lemma 1 and induction hypothesis, we have\n$$\n\\begin{align*}\na_{n+k} &\\le C(a_n a_{n+1} \\dots a_{n+k-1})^{\\frac{1}{k+1}} + n + k \\\\\n&< C(2(n+k-1) + D)^{\\frac{k}{k+1}} + n + k \\\\\n&< C(2^{\\frac{k}{k+1}}(n+k)^{\\frac{k}{k+1}} + D^{\\frac{k}{k+1}}) + n + k.\n\\end{align*}\n$$\nIn the last step, we use the fact that $x^{\\frac{k}{k+1}}$ is concave down for $x > 0$.\n\nNow, if $n + k \\le 2^k C^{k+1}$, then\n$$\n\\begin{align*}\na_{n+k} &< C 2^{\\frac{k}{k+1}} (n+k)^{\\frac{k}{k+1}} + C D^{\\frac{k}{k+1}} + n + k \\\\\n&\\le 2^k C^{k+1} + C D^{\\frac{k}{k+1}} + n + k \\\\\n&< D + 2(n+k);\n\\end{align*}\n$$\nif $n + k > 2^k C^{k+1}$, then\n$$\nC 2^{\\frac{k}{k+1}} (n + k)^{\\frac{k}{k+1}} < n + k, \\quad C D^{\\frac{k}{k+1}} < D,\n$$\nand\n$$\na_{n+k} \\le n + k + D + n + k = 2(n + k) + D.\n$$\nTherefore, $a_n < 2n + D$ is true for $n + k$.\n\nReturn to the original problem. For integer $m > 0$, if $m$ does not appear in the sequence, then, as the elements in $\\{a_n\\}$ are distinct, there must exist a positive integer $N_1$, such that $a_n > m$ for $n > N_1$. By definition of $\\{a_n\\}$, $m|a_n a_{n+1} \\dots a_{n+k-1}$ for all $n > N_1$.\n\nAccording to Lemma 2, there exists $M > 0$, such that for any integer $m \\ge M$, we can find a prime power $p^\\alpha|m$, and $p^\\alpha > (3k)^k$. We claim that such $m$ must appear in $\\{a_n\\}$. Suppose to the contrary that for some $m > M$, $m$ does not appear in $\\{a_n\\}$. Fix a prime power $p^\\alpha|m$, $p^\\alpha > (3k)^k$. By the previous argument, there exists a positive integer $N_1$, such that for $n > N_1$,\n$$\nm|a_n a_{n+1} \\dots a_{n+k-1}.\n$$\nFor this $n$, there exists $i$, $n \\le i \\le n + k - 1$ and $p^{\\lceil \\alpha/k \\rceil} | a_i$. Let\n$$\nN_2 > 3N_1 + D + 3k,\n$$\nand consider the set $A = \\{n : 1 \\le n \\le N_2, p^{\\lceil \\alpha/k \\rceil} |a_n\\}$. On one hand, from Lemma 3, we have $a_n < 2N_2 + D$, and thus\n$$\n|A| \\le \\frac{2N_2 + D}{p^{\\lceil \\frac{\\alpha}{k} \\rceil}} < \\frac{2N_2 + D}{3k}.\n$$\nOn the other hand, since $m$ does not appear in $\\{a_n\\}$, $m|a_n a_{n+1}\\dots a_{n+k-1}$ for $n > N_1$. Consequently, for any $k$ consecutive terms in $\\{N_1 + 1, N_1 + 2, \\dots, N_2\\}$, at least one term belongs to $A$, implying that $|A| \\ge \\frac{N_2 - N_1}{k} - 1$. However, by the choice of $N_2$, we have\n$$\n|A| \\ge \\frac{N_2 - N_1}{k} - 1 > \\frac{2N_2 + D}{3k},\n$$\nwhich contradicts the previous estimate. This means all integers $m \\ge M$ must appear in $\\{a_n\\}$. $\\Box$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 28492,
    "subject": "Mathematics (Multi-modal)",
    "question": "Given positive integers $n$ and $k$, $n > k^2 > 4$. In an $n \\times n$ grid, any $k$ squares in distinct rows and distinct columns are called a $k$-set. Find the largest positive integer $N$ satisfying that: one can choose $N$ squares of the $n \\times n$ grid and colour them in a certain way, such that for any coloured $k$-set, some two squares have the same colour, and some two squares have different colours.",
    "options": [],
    "answer": "(k - 1)^2 n",
    "solution": "$N = (k - 1)^2 n$.\n\nChoose $(k-1)^2$ rows of the grid: colour the first $k-1$ rows in colour $c_1$; the second $k-1$ rows in colour $c_2$; ...; the last $k-1$ rows in colour $c_{k-1}$. Altogether, $(k-1)^2 n$ squares are coloured. For any coloured $k$-set, as there are only $k-1$ colours, some two squares must have the same colour. On the other hand, if all the squares in this $k$-set have the same colour, then by definition they are in $k$ different rows, yet there are only $k-1$ rows in that colour, a contradiction. This implies that some two squares have different colours. Therefore, $N \\ge (k-1)^2 n$.\n\nIn a $k$-set, if all the squares have the same colour, call it a mono $k$-set; if the squares have distinct colours, call it a poly $k$-set. We assert that for any colouring of $(k-1)^2n + 1$ squares, there must exist a mono $k$-set or a poly $k$-set. This will give $N \\le (k-1)^2n$ and the conclusion. First, we need a lemma.\n\n**Lemma** In an $n \\times n$ grid, among any $(m-1)n + 1$ squares, $1 \\le m \\le n$, there exists an $m$-set.\n\n**Proof of lemma** Divide the squares of the $n \\times n$ grid into $n$ groups, such that the square in the $i$th row and $j$th column is in group $a$ ($1 \\le i, j, a \\le n$) if and only if $i - j \\equiv a \\pmod{n}$. Notice that for each group, the squares are in distinct rows and distinct columns. By the pigeonhole principle, among any $(m-1)n + 1$ squares, $m$ of them are in a group, and they form an $m$-set.\n\nFor the original problem, assume that $(k-1)^2n+1$ squares are coloured in a certain way. According to the lemma, there exists a $((k-1)^2+1)$-set, call it $A$. If there are $k-1$ or fewer colours in $A$, then by the pigeonhole principle some $k$ squares are in the same colour, and they form a mono $k$-set; if there are $k$ or more colours in $A$, then choose $k$ squares of distinct colours, and they form a poly $k$-set.\n\nTherefore, the largest $N$ is $(k-1)^2n$.\n\n**Alternative proof of lemma** Choose any $(m-1)n+1$ squares and colour them black. Take $m$ rows with the most black squares: say they are row $1, 2, \\dots, m$, with $x_1, x_2, \\dots, x_m$ black squares, respectively, and the other rows have $x_{m+1}, \\dots, x_n$ black squares, respectively. If there exists $l$, $1 \\le l \\le m$, such that in the first $m$ rows, black squares of some $l$ rows are distributed in $l-1$ (or fewer) columns, say, black squares of row $1, \\dots, l$ are all in the first $l-1$ columns. Then, $x_1, \\dots, x_l \\le l-1$; for $i > m$, $x_i \\le l-1$ as well (since there are fewer black squares in those rows). Hence,\n$$\n\\begin{align*} \n(m-1)n + 1 &= \\sum_{i=1}^{n} x_i \\le l(l-1) + (m-l)n + (n-m)(l-1) \\\\ \n&= mn - n + m - l(m+1-l) \\\\ \n&\\le mn - n + m - m = (m-1)n, \n\\end{align*}\n$$\nwhich is a contradiction. It follows that for any $l$ of the first $m$ rows, the black squares are located in at least $l$ columns. By Hall's marriage theorem, there are $m$ black squares in distinct rows and distinct columns, which is an $m$-set. The lemma is proved. $\\square$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 28493,
    "subject": "Mathematics (Multi-modal)",
    "question": "Given vectors $\\vec{a} = (1 + 2^m, 1 - 2^m)$, $\\vec{b} = (4^m - 3, 4^m + 5)$, suppose $m$ is real. Then the minimum of the dot product of $\\vec{a} \\cdot \\vec{b}$ is ______.",
    "options": [],
    "answer": "-6",
    "solution": "Let $t = 2^m$, and then $\\vec{a} = (1 + t, 1 - t)$, $\\vec{b} = (t^2 - 3, t^2 + 5)$. Thus,\n$$\n\\begin{aligned}\n\\vec{a} \\cdot \\vec{b} &= (1 + t)(t^2 - 3) + (1 - t)(t^2 + 5) \\\\\n&= 2(t - 2)^2 - 6 \\geq -6.\n\\end{aligned}\n$$\nWhen $t = 2$, namely, $m = 1$, $\\vec{a} \\cdot \\vec{b}$ takes the minimum $-6$.",
    "topic": "Algebra",
    "subtopic": "Linear Algebra"
  },
  {
    "id": 28494,
    "subject": "Mathematics (Multi-modal)",
    "question": "As illustrated in Fig. 2.1, in the acute $\\triangle ABC$, $AB < AC$, $I$ is the incentre, and $\\odot O$ is the circumcentre. Let $M$, $N$ be the midpoints of $\\overarc{BAC}$ and $\\overarc{BC}$, respectively. Let $D$ be a point on $\\overarc{AC}$ such that $AD // BC$. The inscribed circle of $\\triangle ABC$ against $\\angle BAC$ touches $BC$ at $E$. Let $F$ be a point inside $\\triangle ABC$, satisfying $IF // BC$ and $\\angle BAF = \\angle CAE$. The line $NF$ meets $\\odot O$ at $R$ other than $N$, the lines $AF$ and $DI$ meet at $K$, and the lines $AR$ and $IF$ meet at $L$.\nProve: $NK \\perp ML$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Fig. 2.1\n\nFirst, we need a lemma.\n\n**Lemma** Let $R'$ be the midpoint of $BC$. Then $\\angle AMI = \\angle IR'B$, $IR' // AE$.\n\n**Proof of lemma** As illustrated in Fig. 2.2, let $I_b$ and $I_c$ be the escentres of $\\triangle ABC$ relative to the vertices $B$ and $C$, respectively. Then $A, M, I_b$, and $I_c$ are collinear. As $\\angle I_bBI_c = \\angle I_bCI_c = 90^\\circ$, the points $B, C, I_b$, and $I_c$ all lie on the circle with diameter $I_bI_c$; since $MB = MC$, $M$ is the centre of this circle, $MI_b = MI_c$. Since $\\triangle I_bI_c \\sim \\triangle ICB$, $M, R'$ are midpoints of $I_bI_c$ and $BC$, respectively, it follows that $\\angle AMI = \\angle IR'B$.\n\nLet the incircle $\\odot I$ of $\\triangle ABC$ touch $BC$ at $Z$ and $ZW$ be a diameter. Through $W$, draw a line parallel to $BC$ that crosses $AB$, $AC$ at $B_1$, $C_1$, respectively. Then $B_1C_1 // BC$. Since $\\triangle A_1B_1C_1$ and $\\triangle ABC$ are homothetic with centre $A$, $W$ and $E$ are correspondent points, it follows that $A, W, E$ are collinear. Finally, from properties of escribed circles, we infer that $BZ = CE$, $R'$ is the midpoint of $ZE$, as $I$ is the midpoint of $ZW$, $IR' // AE$. The lemma is verified.\n\nReturn to the original problem. Let $S$ be the intersection of $AI$ and $BC$. Then\n$$\n\\begin{align*} \n\\angle AIF &= \\angle ASB = \\angle SAC + \\angle ACS \\\\ \n&= \\angle BCN + \\angle ACS \\\\ \n&= \\angle ACN = 180^\\circ - \\angle ARN, \n\\end{align*}\n$$\nindicating that $A, I, F$, and $R$ lie on a circle, say $\\omega$, as shown in Fig. 2.3.\n\nSuppose the lines $AF$ and $MI$ intersect at $X$. We show that $X$ is on $\\odot O$. By the lemma,\n$$\n\\begin{align*} \n\\angle AMI &= \\angle IRB = \\angle AEB, \\\\ \n\\text{also } \\angle AMN &= \\angle ACN = \\angle ASB, \\text{ thus} \\\\ \n\\angle IMN &= \\angle AMN - \\angle AMI \\\\ \n&= \\angle ASB - \\angle AEB \\\\ \n&= \\angle IAE. \n\\end{align*}\n$$\nIt is given that $\\angle IAE = \\angle FAI$, hence $\\angle IMN = \\angle FAI$, $A, M, N$, and $X$ are concyclic, that is, $X$ lies on $\\odot O$.\n\nLet $Y$ be the other intersection of the line $DI$ and $\\odot O$. Clearly,\n$$\n\\angle IYX = \\angle DYX = \\angle DAX = \\angle IFX,\n$$\nand $I, X, Y, F$ lie on a circle, say $\\odot T$.\n\nThrough $M$, draw the tangent line $MG$ of $\\odot O$ ($G$ and $Y$ are on the same side of $MN$), $MG // IF$. Then\n$$\n\\angle YMG = \\angle MXY = \\angle IXY = \\angle YFL,\n$$\nyielding the collinearity of $Y, F, M$. Apply the radical axis theorem to $\\odot O, \\odot \\omega$, and $\\odot T$, to derive the collinearity of $X, Y$, and $L$. In the cyclic quadrilateral $IXYF$, apply Brocard's theorem to derive the orthocentre $T$ of $\\triangle MKL$, $TK \\perp ML$.\n\nFinally, in $\\triangle TFI$ and $\\triangle NAD$, $TF = TI, NA = ND$, and\n$$\n\\angle FTI = 2\\angle FXI = 2\\angle AXM = \\angle AND.\n$$\nObserving $FI // BC // AD$, we conclude that $\\triangle TFI$ and $\\triangle NAD$ are homothetic with centre $K$, and moreover $K, T, N$ are collinear.\n\nFrom $TK \\perp ML$ and the collinearity of $K, T, N, NK \\perp ML$ follows. $\\square$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 28495,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $m$ be a real number. If the real and imaginary parts of complex $z = 1 + i + \\frac{m}{1+i}$, with $i$ being the imaginary unit, are greater than zero, then the range of $m$ is ______.",
    "options": [],
    "answer": "−2 < m < 2",
    "solution": "After calculation, we get $z = 1 + i + \\frac{(1-i)m}{2} = \\frac{2+m}{2} + \\frac{2-m}{2}i$.\n\nBy the condition, it follows that $\\frac{2+m}{2} > 0$, $\\frac{2-m}{2} > 0$, and then we find the solution is $-2 < m < 2$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 28496,
    "subject": "Mathematics (Multi-modal)",
    "question": "Given positive integers $k$ and $n$ ($n \\ge 2$), find the minimum constant $c$ satisfying this assertion: if $G$ is a simple $kn$-regular graph (the degree of each vertex is $kn$) with $m$ vertices, then each vertex can be coloured one of $n$ colours, such that the number of “mono edges” is at most $cm$. Here, a mono edge is an edge incident to two vertices of the same colour.",
    "options": [],
    "answer": "k(kn - n + 2)/(2(kn + 1))",
    "solution": "$c = \\frac{k(kn - n + 2)}{2(kn + 1)}$.\n\nFirst, we prove $c \\ge \\frac{k(kn - n + 2)}{2(kn + 1)}$. Consider a complete graph $K_{kn+1}$ with any $n$-colouring of the vertices. Suppose that the numbers of vertices coloured by 1, 2, ..., $n$ are $a_1, a_2, ..., a_n$, respectively. Then\n$$\n\\begin{aligned}\na_1^2 + a_2^2 + \\dots + a_n^2 &\\ge \\frac{(a_1 + a_2 + \\dots + a_n)^2}{n} \\\\\n&= \\frac{(kn + 1)^2}{n} \\\\\n&= k^2n + 2k + \\frac{1}{n}.\n\\end{aligned}\n$$\nAs $a_1, a_2, ..., a_n$ are integers,\n$$\na_1^2 + a_2^2 + \\dots + a_n^2 \\ge k^2n + 2k + 1.\n$$\nTherefore, the number of mono edges is:\n$$\n\\begin{aligned}\n\\sum_{i=1}^{n} C_{a_i}^2 &= \\frac{1}{2} \\left( \\sum_{i=1}^{n} a_i^2 - \\sum_{i=1}^{n} a_i \\right) \\\\\n&\\ge \\frac{1}{2} (k^2n + 2k + 1 - (kn + 1)) \\\\\n&= \\frac{1}{2} k(kn - n + 2),\n\\end{aligned}\n$$\nwhich implies $c \\ge \\frac{k(kn - n + 2)}{2(kn + 1)}$.\n\nNext, we give three proofs that $c = \\frac{k(kn - n + 2)}{2(kn + 1)}$ satisfies the problem assertion.\n\n**Proof 1**\n\nFor any $kn$-regular graph $G$, it is well known that $G$ has a $(kn+1)$-colouring such that adjacent vertices are of different colours. Divide the $kn+1$ colours randomly into $n$ groups such that one group contains $k+1$ colours while each of the others contains $k$ colours, and for all these divisions, they have equal probabilities. Now change all colours of a group into one colour. For any edge of $G$, initially, it is incident to vertices of different colours; after changing the colours, it is incident to vertices of the same colour, namely, it becomes a mono edge with a probability\n$$\n\\frac{C_{k+1}^2 + (n-1)C_k^2}{C_{kn+1}^2} = \\frac{k(k+1) + k(k-1)(n-1)}{(kn+1)(kn)} = \\frac{kn - n + 2}{n(kn+1)}.\n$$\nTherefore, after changing the colours, the expected number of mono edges is\n$$\n\\frac{kn - n + 2}{n(kn + 1)} \\cdot \\frac{knm}{2} = m \\cdot \\frac{k(kn - n + 2)}{2(kn + 1)} = cm.\n$$\nThere must exist an $n$-colouring of $G$ such that the number of mono edges is less than or equal to $cm$. This verifies the problem assertion.\n\n**Proof 2**\n\nSuppose that among all $n$-colourings of $G$, the minimum number of mono edges is $T(G)$. Let $V$ consist of vertices of $G$. For a permutation $\\pi = (v_1, v_2, . . . , v_m)$ of $V$, colour $v_1, v_2, . . . , v_m$ in order (by one of $n$ colours) as follows. For a vertex $v$, suppose that in $\\pi$, $A_{\\pi}(v)$ of $kn$ neighbouring vertices of $v$ appear before $v$. We colour $v$ with the colour that appears least frequently among those $A_{\\pi}(v)$ neighbours (it appears at most $\\lfloor A_{\\pi}(v)/n \\rfloor$ times). According to this greedy algorithm, it is guaranteed that\n$$\nT(G) \\leq F(\\pi) := \\sum_{v \\in V} \\left\\lfloor \\frac{A_{\\pi}(v)}{n} \\right\\rfloor.\n$$\nThe above inequality holds for any permutation $\\pi$. It suffices to prove, for all $m!$ permutations, the average of $F(\\pi)$ is $\\frac{k(kn - n + 2)}{2(kn + 1)} \\cdot m$, so that some permutation $\\pi$ gives a desired $n$-colouring. For a vertex $v$, let its neighbours be $\\{u_1, u_2, \\dots, u_{kn}\\}$. Consider $B = \\{v, u_1, u_2, \\dots, u_{kn}\\}$: among the $m!$ relative orders of elements of $B$ given by all $m!$ permutations, $v$ appears at position 1, 2, $\\dots$, $kn + 1$ with equal probability $\\frac{1}{kn + 1}$. Hence, as $\\pi$ traverses all $m!$ permutations, the average of $\\left\\lfloor \\frac{A_{\\pi}(v)}{n} \\right\\rfloor$ is\n$$\n\\begin{aligned} \\frac{1}{kn+1} \\sum_{a=0}^{kn} \\left\\lfloor \\frac{a}{n} \\right\\rfloor &= \\frac{1}{kn+1} (0 \\times n + 1 \\times n + \\dots + (k-1) \\times n + k) \\\\\n&= \\frac{k(kn - n + 2)}{2(kn + 1)}. \\end{aligned}\n$$\nAs a result, the average of $F(\\pi)$ is $\\frac{k(kn - n + 2)}{2(kn + 1)} \\cdot m = cm$.\n\n**Proof 3**\n\nFor a $kn$-regular graph $G$, let $A = |V(G)|$ be the number of vertices of $G$. Since there are only finitely many $n$-colourings of $G$, there must be one with the minimum number of mono edges, say $C$. We have the following claims on $C$.\n\n**Claim 1** For every vertex $v$, $v$ has at most $k$ neighbours that are in the same colour as $v$; if this does occur, then $v$ has exactly $k$ neighbours in each of the $n$ colours (we call this $v$ a \"balanced vertex\").\n\n**Proof of claim 1** Let $v$ be of colour $P$. If there is another colour $Q$ such that among $v$'s neighbours, the number of vertices in colour $P$ is larger than that in colour $Q$, then we change the colour of $v$ from $P$ to $Q$. This will reduce the number of mono edges, which is a contradiction. Hence, there are fewest neighbours of $v$ in colour $P$ (or one of the fewest); as $v$ has $kn$ neighbours, the claim is verified.\n\n**Claim 2** Adjacent balanced vertices must be monochromatic.\n\n**Proof of claim 2** Let $v, w$ be adjacent balanced vertices, $v$ in colour $P$ while $w$ in colour $Q$. Changing $v$ to $Q$ and $w$ to $P$, $vw$ will not be a mono edge, but the numbers of mono edges incident to $v$ or $w$ will each decrease by 1, a contradiction.\n\n**Claim 3** If $v$ is not a balanced vertex, $v$ is adjacent to $r$ balanced vertices which have different colours from $v$, then there are at most $k - \\frac{r}{k+1}$ mono edges incident to $v$.\n\n**Proof of claim 3** Consider the edges between these $r$ balanced vertices. As each balanced vertex is adjacent to at most $k$ balanced vertices of its colour, in the subgraph of these vertices, the degree of each vertex does not exceed $k$. Now we can choose at least $\\frac{r}{k+1}$ of these balanced vertices such that any two of them are not adjacent (assume that at most $t$ vertices can be chosen, then they have $tk$ or fewer neighbours, $tk+t \\ge r$, or $t \\ge \\frac{r}{k+1}$). Change them to the colour of $v$. From the properties of balanced vertices, we see that the number of mono edges does not change; however, if the number of mono edges emanating from $v$ exceeds $k$, then changing $v$'s colour will reduce the number of mono edges, a contradiction. Therefore, the number of mono edges emanating from $v$ does not exceed $k$, and hence there are at most $k - \\frac{r}{k+1}$ mono edges incident to $v$.\n\nLet $a$ be the number of balanced vertices in $G$. There are two situations.\n\n(i) If $a \\le \\frac{k+1}{kn+1}A$. By calculating the number of mono edges emanating from each vertex in two ways, we find that the total number of mono edges does not exceed\n$$\n\\begin{align*}\n\\frac{1}{2}(ak + (A-a)(k-1)) &= \\frac{1}{2}((k-1)A + a) \\\\\n&\\le \\frac{1}{2} \\left( (k-1)A + \\frac{k+1}{kn+1}A \\right) \\\\\n&= \\frac{k(kn-n+2)}{2(kn+1)}A.\n\\end{align*}\n$$\n\n(ii) If $a > \\frac{k+1}{kn+1}A$. From Claim 2, it follows that the heterochromatic neighbours of a balanced vertex are never balanced vertices. Thus, the number of edges connecting a balanced vertex and a heterochromatic neighbour is $ak(n-1) > \\frac{k(k+1)(n-1)}{kn+1}A$. Now for each non-balanced vertex, consider its heterochromatic balanced (vertex) neighbours: let the numbers be $r_1, r_2, \\dots, r_{A-a}$. We have\n$$\nr_1 + r_2 + \\dots + r_{A-a} > \\frac{k(k+1)(n-1)}{kn+1} A.\n$$\nFrom Claim 3, the number of mono edges does not exceed\n$$\n\\begin{align*}\n\\frac{1}{2} \\left( ka + \\sum_{i=1}^{A-a} \\left( k - \\frac{r_i}{k+1} \\right) \\right) &= \\frac{1}{2} \\left( kA - \\frac{1}{k+1} \\sum_{i=1}^{n-a} r_i \\right) \\\\\n&< \\frac{1}{2} \\left( kA - \\frac{k(k+1)(n-1)}{(k+1)(kn+1)} A \\right) \\\\\n&= \\frac{k(kn-n+2)}{2(kn+1)} A.\n\\end{align*}\n$$\nTherefore, $c = \\frac{k(kn - n + 2)}{2(kn + 1)}$ as desired.",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 28497,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find the largest constant $C > 0$, such that for any integer $n \\ge 2$, one can find real numbers $x_1, x_2, \\dots, x_n \\in [-1, 1]$ satisfying\n$$\n\\prod_{1 \\le i < j \\le n} (x_i - x_j) \\ge C^{\\frac{n(n-1)}{2}}.\n$$",
    "options": [],
    "answer": "1/2",
    "solution": "For sufficiently large $n$, we guess the optimal choice of $(x_1, \\dots, x_n)$ on $[-1, 1]$ is close to the projection of points uniformly distributed on the unit circle onto the $x$ axis. To this end, we introduce the reference choice\n$$\na_k = \\cos \\vartheta_k = \\cos \\frac{2k-1}{2n} \\pi, \\quad k = 1, 2, \\dots, n.\n$$\nFor $A = \\{a_1, a_2, \\dots, a_n\\}$, the product\n$$\nP_A = \\prod_{1 \\le i < j \\le n} (a_i - a_j)\n$$\nand\n$$\n\\sum_{i \\neq k} \\frac{1}{a_k - a_i}, \\quad k = 1, \\dots, n\n$$\nwill be used later.\n\nConsider the $n$th degree Chebyshev polynomial of the first kind $T(X) = T_n(X)$, with leading coefficient $2^{n-1}$ and $T(\\cos \\vartheta) = \\cos n\\vartheta$.\nSubstitute $X = \\cos \\vartheta$, and differentiate to find $T'(\\cos \\vartheta)$, $T''(\\cos \\vartheta)$:\n$$\n\\begin{align*} \nT'(\\cos \\vartheta) &= \\frac{d \\cos n\\vartheta}{d \\cos \\vartheta} = \\frac{-n \\sin n\\vartheta \\times d\\vartheta}{-\\sin \\vartheta \\times d\\vartheta} = n \\times \\frac{\\sin n\\vartheta}{\\sin \\vartheta}; \\\nT''(\\cos \\vartheta) &= \\frac{dT'(\\cos \\vartheta)}{d \\cos \\vartheta} = n \\times \\frac{d \\frac{\\sin n\\vartheta}{\\sin \\vartheta}}{d \\cos \\vartheta} \\\n&= n \\times \\frac{\\frac{n \\cos n\\vartheta}{\\sin \\vartheta} \\times d\\vartheta - \\frac{\\cos \\vartheta \\sin n\\vartheta}{\\sin^2 \\vartheta \\times d\\vartheta}}{-\\sin \\vartheta \\times d\\vartheta} \\\n&= n \\times \\frac{\\cos \\vartheta \\sin n\\vartheta - n \\times \\cos n\\vartheta \\sin \\vartheta}{\\sin^3 \\vartheta}. \n\\end{align*}\n$$\nFor $k = 1, 2, \\dots, n$, denote $\\vartheta_k = \\frac{2k-1}{2n}\\pi$. We have\n$$\nT(a_k) = T(\\cos \\vartheta_k) = \\cos(n\\vartheta_k) = 0,\n$$\nand hence all the roots of $T(X)$ are $A = \\{a_1, a_2, \\dots, a_n\\}$. Define\n$$\nH(X) = \\frac{1}{2^{n-1}} T(X) = (X - a_1)(X - a_2) \\cdots (X - a_n).\n$$\nFrom another perspective, the first derivative $H'(X) = \\frac{dH(X)}{dX}$ is the sum of all products of $n-1$ terms in\n$$\nB = \\{X - a_1, X - a_2, \\dots, X - a_n\\}\n$$\n(there are $\\binom{n}{n-1}$ products), denoted as $H'(X) = \\sigma_{n-1}(B)$. Furthermore, the second derivative $H''(X) = \\frac{dH'(X)}{dX}$ is twice the sum of all products of $n-2$ terms in $B$ (there are $\\binom{n}{n-2}$ products), denoted as $H''(X) = 2\\sigma_{n-2}(B)$.\n\nWhen $X = a_k \\in A$, we have $X - a_k = 0$ in $B$, and\n$$\nH'(a_k) = \\sigma_{n-1}(B) = \\prod_{i \\neq k} (a_k - a_i),\n$$\n$$\nH''(a_k) = 2\\sigma_{n-2}(B) = 2 \\prod_{i \\neq k} (a_k - a_i) \\sum_{i \\neq k} \\frac{1}{a_k-a_i}.\n$$\nSubstitute $X = a_k = \\cos \\vartheta_k \\in A$ in $T'(X)$ and $T''(X)$ to obtain\n$$\nT'(a_k) = T'(\\cos \\vartheta_k) = n \\times \\frac{(-1)^{k-1}}{\\sin \\vartheta_k},\n$$\n$$\nT''(a_k) = T''(\\cos \\vartheta_k) = n \\times \\frac{(-1)^{k-1} \\cos \\vartheta_k}{\\sin^3 \\vartheta_k}.\n$$\nTherefore,\n$$\n\\sum_{i \\neq k} \\frac{1}{a_k - a_i} = \\frac{H''(a_k)}{2H'(a_k)} = \\frac{T''(a_k)}{2T'(a_k)} = \\frac{\\cos \\vartheta_k}{2 \\sin^2 \\vartheta_k}, \\quad k = 1, 2, \\dots, n.\n$$\nMoreover,\n$$\n\\begin{align*} P_A^2 &= (-1)^{\\frac{n(n-1)}{2}} \\prod_{k=1}^n \\prod_{i \\neq k} (a_k - a_i) \\\n&= (-1)^{\\frac{n(n-1)}{2}} \\prod_{k=1}^n H'(\\cos \\vartheta_k) \\\n&= \\frac{n^n}{2^{n(n-1)} \\prod_{k=1}^n \\sin \\vartheta_k}. \\end{align*}\n$$\nNotice that the complex roots of $(z+1)^n + 1 = 0$ are $\\{e^{\\frac{2k-1}{n}\\pi i} - 1, k = 1, \\dots, n\\}$ with magnitudes $2 \\sin \\frac{2k-1}{2n}\\pi = 2 \\sin \\vartheta_k$. The product of the magnitudes is $\\prod_{k=1}^n 2 \\sin \\vartheta_k = 2$, implying\n$$\nP_A^2 = \\frac{n^n}{2^{n(n-1)} \\prod_{k=1}^n \\sin \\vartheta_k} = \\frac{n^n}{2^{n(n-1)} 2^{-(n-1)}} \\Rightarrow P_A = \\left(\\frac{1}{2}\\right)^{\\frac{n(n-1)}{2}} n^{\\frac{n}{2}} 2^{\\frac{n-1}{2}},\n$$\nand hence $C = \\frac{1}{2}$ satisfies the problem statement. In the following, we prove $C \\le \\frac{1}{2}$.\nFor any positive integer $n$, assume $1 \\ge x_1 > x_2 > \\cdots > x_n \\ge -1$ and\n$$\nP^* = \\prod_{1 \\le i < j \\le n} (x_i - x_j) \\ge C^{\\frac{n(n-1)}{2}}.\n$$\nWe make a comparison with the reference choice:\n$$\n\\begin{align*} \nF &= \\sum_{1 \\le i < j \\le n} \\frac{x_i - x_j}{a_i - a_j} \\\n&= \\sum_{k=1}^n \\left( x_k \\times \\sum_{i \\neq k} \\frac{1}{a_k - a_j} \\right) \\\n&= \\sum_{k=1}^n \\left( x_k \\times \\frac{\\cos \\vartheta_k}{2 \\sin^2 \\vartheta_k} \\right) \\\n&\\le \\frac{1}{2} \\sum_{k=1}^n \\frac{|\\cos \\vartheta_k|}{\\sin^2 \\vartheta_k} \n\\end{align*}\n$$\nEvaluating\n$$\n\\begin{align*} \nG &= \\sum_{k=1}^n \\cot^2 \\vartheta_k = \\sum_{k=1}^n \\cot^2 \\frac{2k-1}{2n} \\pi \\\n&= \\sum_{l=1}^{2n-1} \\cot^2 \\frac{l\\pi}{2n} - \\sum_{l=1}^{n-1} \\cot^2 \\frac{l\\pi}{n}, \n\\end{align*}\n$$\nusing the well-known identity\n$$\n\\sum_{k=1}^{m-1} \\cot^2 \\frac{k\\pi}{m} = \\frac{(m-1)(m-2)}{3},\n$$\nit follows that\n$$\nG = \\frac{(2n-1)(2n-2)}{3} - \\frac{(n-1)(n-2)}{3} = n(n-1).\n$$\nNow, by Cauchy's inequality,\n$$\n\\begin{align*}\n(2F)^2 &\\le \\left( \\sum_{k=1}^n \\frac{|\\cos \\vartheta_k|}{\\sin^2 \\vartheta_k} \\right)^2 \\\\\n&\\le \\sum_{k=1}^n \\frac{\\cos^2 \\vartheta_k}{\\sin^2 \\vartheta_k} \\times \\sum_{k=1}^n \\frac{1}{\\sin^2 \\vartheta_k} \\\\\n&= G \\times (G + n) \\\\\n&= (n-1)n^3.\n\\end{align*}\n$$\nFinally, by AM-GM inequality, we arrive at\n$$\n\\begin{align*}\n\\frac{P^*}{P_A} &= \\prod_{1 \\le i < j \\le n} \\frac{x_i - x_j}{a_i - a_j} \\\\\n&\\le \\left( \\frac{2}{n(n-1)} \\sum_{1 \\le i < j \\le n} \\frac{x_i - x_j}{a_i - a_j} \\right)^{\\frac{n(n-1)}{2}} \\\\\n&< \\left( \\frac{n}{n-1} \\right)^{\\frac{n(n-1)}{4}} < e^{\\frac{n}{4}}. \\\\\nC^{\\frac{n(n-1)}{2}} &\\le P^* < e^{\\frac{n}{4}} \\times P_A = \\left( \\frac{1}{2} \\right)^{\\frac{n(n-1)}{2}} \\times n^{\\frac{n}{2}} 2^{\\frac{n-1}{2}} e^{\\frac{n}{4}}.\n\\end{align*}\n$$\nThe above inequality holds for all positive integers $n$. By taking $n$ sufficiently large, we deduce that $C \\le \\frac{1}{2}$. $\\square$\n(modified from Wang Yichuan's solution)\nFirst prove $C \\le \\frac{1}{2}$. Without loss of generality, assume\n$1 \\ge x_1 \\ge \\cdots \\ge x_n \\ge -1,$\nwhere $x_i = \\cos \\vartheta_i$ ($1 \\le i \\le n$), $0 \\le \\vartheta_1 \\le \\cdots \\le \\vartheta_n \\le \\pi$. Let $p_j = e^{i\\vartheta_j}$,\n$q_j = e^{-i\\vartheta_j}$ be complex numbers. Clearly, on the complex plane, $p_1, \\dots, p_n,$ $q_n, \\dots, q_1$ correspond to $2n$ points anticlockwise on the unit circle, which we denote in order as $A_1, \\dots, A_{2n}$. It follows that\n$$\n\\prod_{1 \\le j < k \\le n} (x_j - x_k) = \\prod_{1 \\le j < k \\le n} (\\cos \\vartheta_j - \\cos \\vartheta_k) = \\prod_{1 \\le j < k \\le n} 2 \\sin \\frac{\\vartheta_j + \\vartheta_k}{2} \\sin \\frac{\\vartheta_j - \\vartheta_k}{2}. \\quad \\textcircled{1}\n$$\nBy the law of sine, for $1 \\le j < k \\le n$,\n$$\n2 \\sin \\frac{\\vartheta_j + \\vartheta_k}{2} = |p_j - q_k| = |p_k - q_j|,\n$$\n$$\n2 \\sin \\frac{\\vartheta_j - \\vartheta_k}{2} = |p_j - p_k| = |q_j - q_k|.\n$$\nHence,\n$$\n2 \\sin \\frac{\\vartheta_j + \\vartheta_k}{2} \\sin \\frac{\\vartheta_j - \\vartheta_k}{2} = \\frac{1}{2} \\sqrt{|p_j - q_k| |p_k - q_j| |p_j - p_k| |q_j - q_k|}\n$$\n$$\n= \\frac{1}{2} \\sqrt{|A_j A_{2n+1-k}| |A_k A_{2n+1-j}| |A_j A_k| |A_{2n+1-j} A_{2n+1-k}|}\n$$\nPlug the above identity into (1), to obtain\n$$\n\\begin{aligned}\n& \\prod_{1 \\le j < k \\le n} (x_j - x_k) \\\\\n&= \\prod_{1 \\le j < k \\le n} \\frac{1}{2} \\sqrt{|A_j A_{2n+1-k}| |A_k A_{2n+1-j}| |A_j A_k| |A_{2n+1-j} A_{2n+1-k}|} \\\\\n&= \\left(\\frac{1}{2}\\right)^{\\frac{n(n-1)}{2}} \\sqrt{\\prod_{1 \\le j < k \\le 2n, j+k \\ne 2n+1} |A_j A_k|}. \\end{aligned} \\quad \\textcircled{2}\n$$\nNow we estimate $\\prod_{1 \\le j < k \\le 2n, j + k \\ne 2n + 1} |A_j A_k|$. Divide $A_j A_k$ ($1 \\le j < k \\le 2n$) into $n$ groups\n$\\Omega_1 : A_1 A_2, A_2 A_3, \\dots, A_{2n} A_1;$\n$\\Omega_2 : A_1 A_3, A_2 A_4, \\dots, A_{2n} A_2;$\n$\\dots$\n$\\Omega_{n-1} : A_1 A_2, A_2 A_3, \\dots, A_{2n} A_{n-1};$\n$\\Omega_n : A_1 A_{n+1}, A_2 A_{n+2}, \\dots, A_n A_{2n}.$\nNotice that for each $1 \\le j \\le n-1$, the sum of the inscribed angles subtended by all chords in $\\Omega_j$ is $j\\pi$; for each $\\Omega_j$, let $\\lambda_j$ be the number of chords $A_s A_t$ such that $s+t \\ne 2n+1$, then it must be $\\lambda_j = 2n$ or $\\lambda_j = 2n-2$. From the law of sine and Jensen's inequality applied to the convex function $\\ln(\\sin x)$ ($x \\in (0, \\frac{\\pi}{2})$), it follows that\n$$\n\\prod \\{|A_s A_t| : A_s A_t \\in \\Omega_j, s + t \\neq 2n + 1\\} \\leq \\left(2 \\sin \\frac{j\\pi}{\\lambda_j}\\right)^{\\lambda_j}.\n$$\nOn one hand, due to $\\sin \\frac{j\\pi}{\\lambda_j} \\le 1$,\n$$\n\\left(2 \\sin \\frac{j\\pi}{\\lambda_j}\\right)^{\\lambda_j} \\le 2^{2n} \\left(\\sin \\frac{j\\pi}{\\lambda_j}\\right)^{2n-2} \\le 2^{2n} \\left(\\sin \\frac{j\\pi}{2n-2}\\right)^{2n-2}.\n$$\nOn the other hand, we have\n$$\n\\prod \\{|A_s A_t| : A_s A_t \\in \\Omega_n\\} \\le 2^n.\n$$\nHence,\n$$\n\\begin{align*} \\prod_{1 \\le j < k \\le 2n, j + k \\ne 2n + 1} |A_j A_k| &\\le 2^n \\cdot \\prod_{j=1}^{n-1} \\left( 2^{2n} \\left( \\sin \\frac{j\\pi}{2n-2} \\right)^{2n-2} \\right) \\\\\n&= 2^{3n-2} \\left( \\prod_{j=1}^{n-1} \\left( 2 \\sin \\frac{j\\pi}{2n-2} \\right) \\right)^{2n-2} \\\\\n&\\le 2^{3n-2} \\cdot (2\\sqrt{n})^{2n-2}, \\end{align*}\n$$\nand in the last step we used the trigonometric identity\n$$\n\\prod_{j=1}^m \\left( 2 \\sin \\frac{j\\pi}{2m} \\right) = 2\\sqrt{m}.\n$$\nNow plug the inequality into (2), yielding\n$$\nC^{\\frac{n(n-1)}{2}} \\le \\prod_{1 \\le j < k \\le n} (x_j - x_k) \\le \\left(\\frac{1}{2}\\right)^{\\frac{n(n-1)}{2}} \\cdot 2^{\\frac{3n-2}{2}} \\cdot (2\\sqrt{n})^{n-1}.\n$$\nLetting $n \\to \\infty$, $C \\le \\frac{1}{2}$.\nIt remains to prove that $C = \\frac{1}{2}$ can be attained. Let $(\\vartheta_1, \\dots, \\vartheta_n) = \\left(\\frac{\\pi}{2n}, \\frac{3\\pi}{2n}, \\dots, \\frac{(2n-1)\\pi}{2n}\\right)$. Evidently, the complex numbers $p_j = e^{i\\vartheta_j}$, $q_j = e^{-i\\vartheta_j}$ correspond to a regular $2n$-gon $A_1 \\dots A_{2n}$ in the complex plane. From (2), we get\n$$\n\\prod_{1 \\le j < k \\le n} (x_j - x_k) = \\left(\\frac{1}{2}\\right)^{\\frac{n(n-1)}{2}} \\sqrt{\\prod_{\\substack{1 \\le j < k \\le 2n \\\\ j+k \\ne 2n+1}} |A_j A_k|}.\n$$\nSince $A_1A_{n+1}, \\dots, A_nA_{2n}$ are the longest diagonals (chords) in the regular $2n$-gon $A_1 \\cdots A_{2n}$,\n$$\n\\prod_{\\substack{1 \\le j < k \\le 2n \\\\ k-j \\ne n}} |A_j A_k| = \\left( \\prod_{j=1}^{n-1} 2 \\sin \\frac{j\\pi}{2n} \\right)^{2n} = (\\sqrt{n})^{2n} \\ge 1,\n$$\nwhich implies\n$$\n\\prod_{1 \\le j < k \\le n} (x_j - x_k) \\ge \\left(\\frac{1}{2}\\right)^{\\frac{n(n-1)}{2}}.\n$$\nIn conclusion, $C = \\frac{1}{2}$. $\\square$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 28498,
    "subject": "Mathematics (Multi-modal)",
    "question": "Given positive integers $a$, $b$, $c$ which are pairwise coprime. Let $f(n)$ represent the number of nonnegative integer solutions $(x, y, z)$ of the equation $a x + b y + c z = n$. Prove: there exist real constants $\\alpha$, $\\beta$, $\\gamma$, such that for every nonnegative real number $n$,\n$$\n|f(n) - (\\alpha n^2 + \\beta n + \\gamma)| < \\frac{a+b+c}{12}.\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Consider the generating function of $\\{f(n)\\}$:\n$$\n\\begin{align*}\nG(t) &= \\sum_{n=0}^{\\infty} f(n) t^n \\\\\n&= (1 + t^a + t^{2a} + \\dots)(1 + t^b + t^{2b} + \\dots)(1 + t^c + t^{2c} + \\dots) \\\\\n&= \\frac{1}{(1 - t^a)(1 - t^b)(1 - t^c)}.\n\\end{align*}\n$$\nSince $a$, $b$, $c$ are pairwise coprime, the denominator $(1-t^a)(1-t^b)(1-t^c)$ contains no repeated factors other than $(1-t)^3$. Let $\\omega = e^{2\\pi i/a}$, $\\tau = e^{2\\pi i/b}$, $\\rho = e^{2\\pi i/c}$ be the primitive $a$, $b$, $c$th roots of unity, respectively. Consider the partial fraction decomposition\n$$\n\\frac{1}{(1 - t^a)(1 - t^b)(1 - t^c)} = \\frac{h_0}{(1 - t)} + \\frac{h_1}{(1 - t)^2} + \\frac{h_2}{(1 - t)^3} \\\\\n+ \\sum_{k=1}^{a-1} \\frac{d_k}{1 - \\omega^k t} + \\sum_{k=1}^{b-1} \\frac{e_k}{1 - \\tau^k t} + \\sum_{k=1}^{c-1} \\frac{f_k}{1 - \\rho^k t},\n$$\nin which the numerators $h_0$, $h_1$, $h_2$, $d_1, \\dots, d_{a-1}$, $e_1, \\dots, e_{b-1}$, $f_1, \\dots, f_{c-1}$ are all complex numbers. To find $d_k$ in $\\frac{d_k}{1 - \\omega^k t}$, multiply $(1 - \\omega^k t)$ on both sides and let $t \\to \\omega^{-k}$ (or just let $1 - \\omega^k t = 0$). As all the other terms vanish, we arrive at\n$$\n\\begin{align*}\nd_k &= \\lim_{t \\to \\omega^{-k}} \\frac{1 - \\omega^k t}{(1 - (\\omega^k t)^a)(1 - t^b)(1 - t^c)} \\\\\n&= \\frac{1}{a} \\times \\frac{1}{(1 - \\omega^{-kb})(1 - \\omega^{-kc})}.\n\\end{align*}\n$$\nIn the expansion of $G(t)$, the coefficient of $t^n$ is\n$$\nf(n) = h_0 + h_1 \\times (n + 1) + h_2 \\times \\frac{(n + 1)(n + 2)}{2} \\\\\n+ \\sum_{k=1}^{a-1} d_k \\omega^{kn} + \\sum_{k=1}^{b-1} e_k \\tau^{kn} + \\sum_{k=1}^{c-1} f_k \\rho^{kn}.\n$$\n\nObserve that\n$$\n\\begin{aligned}\n\\left| \\sum_{k=1}^{a-1} d_k \\omega^{kn} \\right| & \\le \\sum_{k=1}^{a-1} |d_k| \\\\\n& = \\frac{1}{a} \\times \\sum_{k=1}^{a-1} \\frac{1}{|1 - \\omega^{-kb}| |1 - \\omega^{-kc}|} \\\\\n& \\le \\frac{1}{a} \\times \\sum_{k=1}^{a-1} \\frac{1}{|1 - \\omega^k|^2} \\\\\n& = \\frac{a^2 - 1}{12a} < \\frac{a}{12}.\n\\end{aligned}\n$$\nHere, the equality in the last row requires a lemma which will be proved.\nLikewise, we have in the following $\\left|\\sum_{k=1}^{b-1} e_k \\tau^{kn}\\right| < \\frac{b}{12}$ and $\\left|\\sum_{k=1}^{c-1} f_k \\rho^{kn}\\right| < \\frac{c}{12}$.\nTaking $\\beta_2 = \\frac{h_2}{2}$, $\\beta_1 = h_1 + \\frac{3}{2}h_2$, $\\beta_0 = h_0 + h_1 + h_2$, it follows that\n$$\n\\begin{aligned}\n|\\beta_2 n^2 + \\beta_1 n + \\beta_0 - f(n)| & \\le \\left|\\sum_{k=1}^{a-1} d_k \\omega^{kn}\\right| + \\left|\\sum_{k=1}^{b-1} e_k \\tau^{kn}\\right| + \\left|\\sum_{k=1}^{c-1} f_k \\rho^{kn}\\right| \\\\\n& < \\frac{a+b+c}{12}.\n\\end{aligned}\n$$\n\n**Lemma (for the equality in the last row)** For any positive integer $m$,\n$$ \\sum_{k=1}^{m-1} \\frac{1}{\\sin^2 \\frac{k\\pi}{m}} = \\frac{m^2-1}{3} \\text{ holds; equivalently, } \\sum_{k=1}^{m-1} \\cot^2 \\frac{k\\pi}{m} = \\frac{m^2-3m+2}{3}. $$\n**Proof of lemma** For $\\vartheta \\in A = \\left\\{ \\frac{\\pi}{m}, \\frac{2\\pi}{m}, \\dots, \\frac{(m-1)\\pi}{m} \\right\\}$,\n$$ (\\cos \\vartheta + i \\sin \\vartheta)^m = \\cos(m\\vartheta) + i \\sin(m\\vartheta) = \\pm 1, $$\nwhere the imaginary part\n$$\nC_m^1 \\sin \\vartheta \\cos^{m-1} \\vartheta - C_m^3 \\sin^3 \\vartheta \\cos^{m-3} \\vartheta + C_m^5 \\sin^5 \\vartheta \\cos^{m-5} \\vartheta + \\dots = 0,\n$$\nor $C_m^1(\\cot \\vartheta)^{m-1} - C_m^3(\\cot \\vartheta)^{m-3} + C_m^5(\\cot \\vartheta)^{m-5} + \\dots = 0$. This implies that $\\{\\cot \\frac{k\\pi}{m} : k = 1, 2, \\dots, m-1\\}$ satisfy the $(m-1)$th degree polynomial in cot $\\vartheta$. According to Viète's formulas, the sum of the $m-1$ roots is $\\sigma_1 = 0$ while the sum of their products multiplied in pairs is\n$$\n\\sigma_2 = - \\frac{C_m^3}{C_m^1} = - \\frac{(m-1)(m-2)}{6}.\n$$\nConsequently, the sum of the squares of the $m-1$ roots is\n$$\n\\sum_{k=1}^{m-1} \\cot^2 \\frac{k\\pi}{m} = \\sigma_1^2 - 2\\sigma_2 = \\frac{m^2 - 3m + 2}{3}. \\quad \\Box\n$$\n\n\nSolution:\n\nBy symmetry, assume $a \\le b \\le c$. If $a = b = c = 1$, $f(n)$ is the number of ways of writing $n$ as three nonnegative integers, which is $C_{n+2}^2$, and by taking $\\alpha = \\frac{1}{2}$, $\\beta = \\frac{3}{2}$, $\\gamma = 1$, the error is constantly 0. In the following, assume $c > 1$. For nonnegative integer $m$, let $g(m)$ represent the number of nonnegative integer solutions $(x, y)$ of $a x + b y = m$.\nFirst, we try to find $g(m)$. Let $x \\equiv k a \\pmod b$, $m \\equiv \\ell b \\pmod a$, where\n$$\nk \\in \\{0, 1, \\dots, b-1\\}, \\quad \\ell \\in \\{0, 1, \\dots, a-1\\}.\n$$\nEvidently, $x \\equiv k \\pmod b$, $y \\equiv \\ell \\pmod a$. Suppose $x = k + u b$, $y = \\ell + v a$.\nThen\n$$\nm = k a + \\ell b + (u + v) a b,\n$$\ngiving $u + v = \\frac{m - (k a + \\ell b)}{a b}$, and the number of pairs $(u,v)$ is\n$$ \\frac{m - (k a + \\ell b)}{a b} + 1 $$ (note that $\\frac{m - (k a + \\ell b)}{a b}$ is an integer greater than $-2$; when $\\frac{m - (k a + \\ell b)}{a b} = -1$ the conclusion is still valid). Therefore,\n$$ g(m) = \\frac{m - (k a + \\ell b)}{a b} + 1. $$\nNext, consider $f(n)$. In the Diophantine equation $a x + b y + c z = n$, if $z$ is given ($c z \\le n$), then the number of nonnegative solutions of $a x + b y = n - c z$ is $g(n - c z)$. Hence,\n$$\nf(n) = \\sum_{i=0}^{\\lfloor n/c \\rfloor} g(n - i c).\n$$\nFor each integer $i$, define $k_i \\in \\{0, 1, \\dots, b-1\\}$ and $\\ell_i \\in \\{0, 1, \\dots, a-1\\}$ such that\n$$\nn - i c = k_i a \\pmod{b}, \\quad n - i c = \\ell_i b \\pmod{a}.\n$$\nWe have\n$$\n\\begin{aligned}\nf(n) &= \\sum_{i=0}^{\\lfloor \\frac{n}{c} \\rfloor} g(n - i c) \\\\\n&= \\sum_{i=0}^{\\lfloor \\frac{n}{c} \\rfloor} \\left( \\frac{(n - i c) - k_i a + \\ell_i b}{a b} + 1 \\right) \\\\\n&= \\lfloor \\frac{n}{c} \\rfloor \\cdot \\left( \\frac{n}{a b} + 1 \\right) - \\frac{c}{2 a b} \\lfloor \\frac{n}{c} \\rfloor \\left( \\lfloor \\frac{n}{c} \\rfloor + 1 \\right) - \\left( \\frac{1}{b} \\sum_{i=0}^{\\lfloor \\frac{n}{c} \\rfloor} k_i + \\frac{1}{a} \\sum_{i=0}^{\\lfloor \\frac{n}{c} \\rfloor} \\ell_i \\right).\n\\end{aligned}\n$$\nNote that $a$, $b$, $c$ are pairwise coprime. Hence, for $i = 0, 1, \\dots, b-1$, the integers $n - i c$ cover a complete system of residues modulo $b$; so do the integers $k_i a$ and the integers $k_i$. In addition, when $i$ shifts by a multiple of $b$, $k_i$ does not change.\nNow we turn to $\\sum_{i=0}^{\\lfloor n/c \\rfloor} \\left( k_i - \\frac{b-1}{2} \\right)$: notice that the summands add up to 0 as $i$ goes over a complete system of residues modulo $b$. Therefore, $\\sum_{i=0}^{\\lfloor n/c \\rfloor} \\left( k_i - \\frac{b-1}{2} \\right) = \\sum_{i=0}^{s} \\left( k_i - \\frac{b-1}{2} \\right)$, where $s$ is the least nonnegative residue of $\\lfloor \\frac{n}{c} \\rfloor$ modulo $b$. In $\\sum_{i=0}^{s} \\left( k_i - \\frac{b-1}{2} \\right)$, the summands are distinct and are all from $\\left\\{ \\frac{1-b}{2}, \\frac{3-b}{2}, \\dots, \\frac{b-3}{2}, \\frac{b-1}{2} \\right\\}$ (the positive and negative elements are symmetric). This indicates that $\\left| \\sum_{i=0}^{s} \\left( k_i - \\frac{b-1}{2} \\right) \\right|$ cannot exceed the sum of the positive elements $\\frac{2}{2} + \\frac{4}{2} + \\dots + \\frac{b-3}{2} + \\frac{b-1}{2} = \\frac{b^2-1}{8}$ when $b$ is odd, or $\\frac{1}{2} + \\frac{3}{2} + \\dots + \\frac{b-3}{2} + \\frac{b-1}{2} = \\frac{b^2}{8}$ when $b$ is even. Hence,\n$$\n\\left| \\sum_{i=0}^{\\lfloor n/c \\rfloor} \\left( k_i - \\frac{b-1}{2} \\right) \\right| \\le \\frac{b^2}{8}. \\qquad \\textcircled{1}\n$$\nLikewise,\n$$\n\\left| \\sum_{i=0}^{\\lfloor n/c \\rfloor} \\left( \\ell_i - \\frac{a-1}{2} \\right) \\right| \\le \\frac{a^2}{8}. \\qquad \\textcircled{2}\n$$\nFrom ① and ②, it follows that\n$$\n\\begin{aligned} & \\left| \\left( \\frac{1}{b} \\sum_{i=0}^{\\lfloor \\frac{n}{c} \\rfloor} k_i + \\frac{1}{b} \\sum_{i=0}^{\\lfloor \\frac{n}{c} \\rfloor} l_i \\right) - \\left( \\lfloor \\frac{n}{c} \\rfloor + 1 \\right) \\left( \\frac{b-1}{2b} + \\frac{a-1}{2a} \\right) \\right| \\\\ & \\le \\frac{1}{b} \\left| \\sum_{i=0}^{\\lfloor \\frac{n}{c} \\rfloor} \\left( k_i - \\frac{b-1}{2} \\right) \\right| + \\frac{1}{a} \\left| \\sum_{i=0}^{\\lfloor \\frac{n}{c} \\rfloor} \\left( \\ell_i - \\frac{a-1}{2} \\right) \\right| \\\\ & \\le \\frac{b}{8} + \\frac{a}{8}. \\end{aligned} \\qquad \\textcircled{3}\n$$\nLet $r$ be the least nonnegative residue of $n$ modulo $c$. Then\n$$\n\\begin{aligned} & \\lfloor \\frac{n}{c} \\rfloor \\cdot \\left( \\frac{n}{a b} + 1 \\right) - \\frac{c}{2 a b} \\lfloor \\frac{n}{c} \\rfloor \\cdot \\left( \\lfloor \\frac{n}{c} \\rfloor + 1 \\right) - \\left( \\lfloor \\frac{n}{c} \\rfloor + 1 \\right) \\left( \\frac{b-1}{2b} + \\frac{a-1}{2a} \\right) \\\\ & = \\frac{(n-r)(n+ab)}{abc} - \\frac{(n-r)(n-r+c)}{2abc} - \\frac{(2ab-a-b)(n-r+c)}{2abc} \\\\ & = \\frac{n^2 - (a+b+c)n - (2abc+ac+bc) - r(r+a+b-c)}{2abc}. \\end{aligned} \\qquad \\textcircled{4}\n$$\nAs $r$ takes values $0, 1, \\dots, c-1$, denote the maximum and the minimum of $r(r+a+b-c)$ as $A+B$ and $A-B$, respectively. By ③ and ④,\n$$\n\\left| f(n) - \\frac{n^2 - (a+b+c)n - (2abc+ac+bc) - A}{2abc} \\right| \\le \\frac{a+b}{8} + \\frac{B}{2abc}.\n$$\nLet $\\alpha = \\frac{1}{2abc}$, $\\beta = -\\frac{a+b+c}{2abc}$,\n$$\n\\gamma = -\\frac{2abc + ac + bc + A}{2abc}.\n$$\nWe will justify the problem statement\n$$\n|f(n) - (\\alpha n^2 + \\beta n + \\gamma)| < \\frac{a+b+c}{12} \\qquad \\textcircled{5}\n$$\nfor different $a$, $b$, $c$ values. $\\square$",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 28499,
    "subject": "Mathematics (Multi-modal)",
    "question": "The range of $y = \\sin^2 x + \\sqrt{3} \\sin x \\cos x$ when $\\frac{\\pi}{4} \\le x \\le \\frac{\\pi}{2}$ is ______.",
    "options": [],
    "answer": "[1, 3/2]",
    "solution": "$$\n\\begin{align*}\ny &= \\sin^2 x + \\sqrt{3} \\sin x \\cos x \\\\\n&= \\frac{1 - \\cos 2x}{2} + \\frac{\\sqrt{3}}{2} \\sin 2x \\\\\n&= \\sin \\left( 2x - \\frac{\\pi}{6} \\right) + \\frac{1}{2}.\n\\end{align*}\n$$\nWhen $\\frac{\\pi}{4} \\le x \\le \\frac{\\pi}{2}$, there is $\\frac{\\pi}{3} \\le 2x - \\frac{\\pi}{6} \\le \\frac{5\\pi}{6}$, and thus the range of $\\sin(2x - \\frac{\\pi}{6})$ is $[\\frac{1}{2}, 1]$. Therefore, the range of $y = \\sin^2 x + \\sqrt{3}\\sin x \\cos x$ is $[1, \\frac{3}{2}]$. $\\square$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 28500,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $S(k)$ denote the sum of all digits of $k$ in base 10. Find all integers $n \\ge 2$ and rational numbers $\\beta \\in (0,1)$, such that there exist $n$ distinct positive integers $a_1, a_2, \\dots, a_n$ that satisfy: for any subset $I \\subseteq \\{1, 2, \\dots, n\\}$ with two or more elements,\n$$ S\\left(\\sum_{i \\in I} a_i\\right) = \\beta \\cdot \\sum_{i \\in I} S(a_i). $$",
    "options": [],
    "answer": "All integers n from 2 to 10 inclusive, with any rational beta in the open interval (0, 1).",
    "solution": "The desired are integers $n \\in \\{2, 3, \\dots, 10\\}$ and all rational numbers $\\beta \\in (0, 1)$.\n\nFirst, for integer $n \\in \\{2, 3, \\dots, 10\\}$ and rational $\\beta \\in (0, 1)$, construct $a_1, a_2, \\dots, a_n$ as follows. Let $c$ and $s$ be positive integers that will be determined later. For $k = 1, \\dots, n$, define\n$$\na_k = 2 \\cdot 10^{cn+sk} - 10^{cn+s(k-1)+1} + \\sum_{\\substack{i \\in \\{1, \\dots, n\\} \\\\ i \\neq k}} 10^{cn+s(i-1)+1} + 10^{k-1} \\cdot \\sum_{i=0}^{c-1} 10^{ni}\n$$\nHere, among the last $cn$ places of $a_1, a_2, \\dots, a_n$, each place corresponds to a unique $a_i$ whose digit at that place is nonzero, and that digit is 1; for the next (to the left) $s$ places, $a_1$ has $199\\dots9$ ($s-1$ of 9's), while all other $a_i$'s have $00\\dots01$; for the next $s$ places, $a_2$ has $199\\dots9$ ($s-1$ of 9's), while all other $a_i$'s have $00\\dots01$; and so on.\n\nFor every $1 \\le i \\le n$, $S(a_i) = n + 9(s - 1) + c$; for every\n$$\nI \\subseteq \\{1, 2, \\dots, n\\} \\ (|I| \\ge 2),\n$$\nit is easy to check that\n$$\nS\\left(\\sum_{i \\in I} a_i\\right) = n \\cdot |I| + c \\cdot |I|.\n$$\nAs long as $s, c \\in \\mathbb{N}_+$ are suitably chosen such that\n$$\n(n + 9(s - 1) + c) \\cdot \\beta = n + c,\n$$\nthe problem requirements are all met.\n\nNext, we prove $n \\le 10$. Assume $n = 11$. If\n$$\nI_1, I_2 \\subseteq \\{1, \\dots, n\\}\n$$\nsatisfy $I_1 \\cap I_2 = \\emptyset$ and $|I_1| \\ge 2$, $|I_2| \\ge 2$, then\n$$\nS\\left(\\sum_{i \\in I_1 \\cup I_2} a_i\\right) = \\beta \\cdot \\sum_{i \\in I_1 \\cup I_2} S(a_i) = S\\left(\\sum_{i \\in I_1} a_i\\right) + S\\left(\\sum_{i \\in I_2} a_i\\right).\n$$\nIn other words, no carry occurs in the summing of $\\sum_{i \\in I_1} a_i$ and $\\sum_{i \\in I_2} a_i$. However, from $\\beta < 1$ we see that a carry occurs in the summing of $a_i$ and $a_j$, for every $1 \\le i < j \\le 11$.\n\nSuppose $T$ is the least positive integer that the summing $a_1 + a_2 + \\cdots + a_{11}$ involves a carry in the $T$th place (from the right). Let the $T$th place digits of $a_1, a_2, \\ldots, a_{11}$ be $x_1, x_2, \\ldots, x_{11}$, respectively. For any subset\n$$\nI \\subseteq \\{1, 2, \\ldots, 11\\},\n$$\nthe $T$th place digit of $\\sum_{i \\in I} a_i$ is the ones place digit of $\\sum_{i \\in I} x_i$. Without loss of generality, assume\n$$\n9 \\ge x_1 \\ge x_2 \\ge \\cdots \\ge x_{11} \\ge 0.\n$$\nThen $x_1 + x_2 + \\cdots + x_{11} \\ge 10$. We assert that there exist $I_1, I_2 \\subseteq \\{1, 2, \\ldots, 11\\}$, $I_1 \\cap I_2 = \\emptyset$, $|I_1| \\ge 2$, $|I_2| \\ge 2$, such that the sum of the ones place digits of $\\sum_{i \\in I_1} x_i$ and $\\sum_{i \\in I_2} x_i$ is greater than or equal to 10.\n\nIf $x_{11} = 0$, starting at $x_{10} + x_9$, add one number at a time by $x_8, x_7, \\ldots, x_1$. There must be a time that a carry occurs in the ones place: for some $k \\in \\{1, 2, \\ldots, 8\\}$, the summing $(x_{10}+x_9+\\cdots+x_{k+1})+x_k$ involves a carry in the ones place, and so does $(x_{10} + x_9 + \\cdots + x_{k+1}) + (x_k + x_{11})$, a contradiction.\n\nIf $x_{11} > 0$, try to make $x_1, x_2, \\ldots, x_{11}$ into as many pairs as possible, such that the sum of each pair ends with a digit greater than or equal to 2. If five pairs are made, then the summing involves a carry in the ones place, a contradiction. Assume at most four pairs are made, leaving three or more unpaired numbers, which include at most one number less than 5 and one number greater than 5, as otherwise two such numbers can be paired. This indicates there is at least one 5, and furthermore, no unpaired number is less than 5, and hence there are at least two 5s, say $x_a$ and $x_b$. Take the third unpaired number $x_c \\ge 5$. Now, starting at any $x_d+x_e$ (not involving $x_a, x_b, x_c$), add one number at a time by the other numbers except $x_a, x_b$. Since $x_c \\ge 5$, the summing must involve a carry in the ones place at some moment, say $(\\sum_{i} x_{i}) + x_{f}$. Then we are led to contradiction by taking\n$$ (\\sum_{i} x_{i}) + (x_{f} + x_{a} + x_{b}). $$\n\nBased on the above argument, when $n \\ge 11$, no positive integers $a_1, a_2, \\dots, a_n$ satisfy the problem condition. $\\square$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 28501,
    "subject": "Mathematics (Multi-modal)",
    "question": "Suppose $A = \\{1, 2, 3\\}$, $B = \\{4x - y \\mid x, y \\in A\\}$, $C = \\{4x + y \\mid x, y \\in A\\}$. Then the sum of all the elements of $B \\cap C$ is ______.",
    "options": [],
    "answer": "48",
    "solution": "When $x$ and $y$ take all the elements of $A$, respectively, $4x - y$ gets exactly the values $1, 2, 3, 5, 6, 7, 9, 10, 11$, and $4x + y$ gets exactly the values $5, 6, 7, 9, 10, 11, 13, 14, 15$.\n\nHence, $B \\cap C = \\{5, 6, 7, 9, 10, 11\\}$. Then the sum of all the elements of $B \\cap C$ is $48$.\n\n$\\square$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 28502,
    "subject": "Mathematics (Multi-modal)",
    "question": "In $\\triangle ABC$, $\\tan A$ and $\\tan B$ are the two roots of equation $x^2 - 10x + 6 = 0$. Then the value of $\\cos C$ is ______.",
    "options": [],
    "answer": "sqrt(5)/5",
    "solution": "By the condition, we know that $\\tan A + \\tan B = 10$, $\\tan A \\tan B = 6$. Thus,\n$$\n\\begin{aligned}\n\\tan C &= \\tan(\\pi - A - B) \\\\\n&= -\\tan(A + B) \\\\\n&= -\\frac{\\tan A + \\tan B}{1 - \\tan A \\tan B} = 2.\n\\end{aligned}\n$$\nTherefore, $C$ is an acute angle, and thus $\\cos C = \\frac{1}{\\sqrt{1 + \\tan^2 C}} = \\frac{1}{\\sqrt{5}} = \\frac{\\sqrt{5}}{5}$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 28503,
    "subject": "Mathematics (Multi-modal)",
    "question": "Suppose function $f(x)$ satisfies: for any non-zero real number $x$, there is\n$$\nf(x) = f(1) \\cdot x + \\frac{f(2)}{x} - 1.\n$$\nThen the minimum of $f(x)$ on $(0, +\\infty)$ is ______.",
    "options": [],
    "answer": "sqrt(3) - 1",
    "solution": "Let $x = 1, 2$, and we can get $f(1) = f(1) + f(2) - 1$ and $f(2) = 2f(1) + \\frac{f(2)}{2} - 1$, respectively. The solution is $f(2) = 1$, $f(1) = \\frac{3}{4}$.\n\nThus, for $x \\neq 0$, there is\n$$\nf(x) = \\frac{3}{4}x + \\frac{1}{x} - 1.\n$$\nWhen $x \\in (0, +\\infty)$, $f(x) \\ge 2\\sqrt{\\frac{3}{4}x \\cdot \\frac{1}{x}} - 1 = \\sqrt{3} - 1$. The equal sign holds when $x = 3$.\n\nTherefore, the minimum of $f(x)$ on $(0, +\\infty)$ is $\\sqrt{3} - 1$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 28504,
    "subject": "Mathematics (Multi-modal)",
    "question": "Suppose the domain of function $f(x)$ is $D = (-\\infty, 0) \\cup (0, +\\infty)$ and there is $f(x) = \\frac{f(1) \\cdot x^2 + f(2) \\cdot x - 1}{x}$ for any $x \\in D$. Then the sum of all the zeros of $f(x)$ is ______.",
    "options": [],
    "answer": "-4",
    "solution": "Let $x_1, x_2$ and we get\n$$\n\\begin{align*}\nf(1) &= f(1) + f(2) - 1, \\\\\nf(2) &= 2f(1) + f(2) - \\frac{1}{2},\n\\end{align*}\n$$\nand the solutions are $f(2) = 1$, $f(1) = \\frac{1}{4}$. Therefore,\n$$\nf(x) = \\frac{1}{x} \\cdot \\left( \\frac{1}{4}x^2 + x - 1 \\right) \\quad (x \\neq 0).\n$$\nLet $f(x) = 0$ and we get $x = -2 \\pm 2\\sqrt{2}$, so the sum of all the zeros of $f(x) = 0$ is $-4$. $\\square$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 28505,
    "subject": "Mathematics (Multi-modal)",
    "question": "For a positive integer $n$, let $\\varphi(n)$ represent the number of positive integers not exceeding $n$ and relatively prime to $n$. Find all functions $f:\\mathbb{N}_+ \\to \\mathbb{N}_+$ satisfying that for any positive integers $m, n$ with $m \\ge n$,\n$$f(m\\varphi(n^3)) = f(m)\\varphi(n^3).$$",
    "options": [],
    "answer": "All functions with f(1)=c for any positive integer c and f(n)=d n for all n≥2, where d is a fixed positive integer.",
    "solution": "The desired functions have the form\n$$\nf(n) = \\begin{cases} c, & n = 1, \\\\ dn, & n \\ge 2, \\end{cases}\n$$\nwhere $c, d$ are positive integers. It is straightforward to check that the problem conditions are met.\nIn the governing equation, taking $n = 2$ leads to $f(4m) = 4f(m)$ for $m \\ge 2$. By iterations, it follows that\n$$\nf(4^k m) = 4^k f(m) \\qquad \\textcircled{1}\n$$\nfor all $m \\ge 2$.\n**Claim** For positive integers $m, p$ with $m \\ge 2$, $f(pm) = p f(m)$.\n**Proof of claim** Induct on $p$: when $p = 1$, the conclusion is trivial; assume $p \\ge 2$ and the conclusion is valid for all positive integers less than $p$. Now, if $p$ is composite, the conclusion is validated by the induction hypothesis. In the following, assume $p$ is a prime. Taking $n = p$ in the governing equation, we obtain\n$$\nf(p^2(p-1)m) = p^2(p-1)f(m)\n$$\nfor all $m \\ge p$. By the induction hypothesis,\n$$\n\\begin{aligned}\np^2(p-1)f(m) &= f(p^2(p-1)m) \\\\\n&= (p-1)f(p^2m), \\end{aligned}\n$$\nand thus\n$$\nf(p^2m) = p^2f(m) \\qquad \\textcircled{2}\n$$\nfor all $m \\ge p$.\n\nNext, take $n = p^2$ in the governing equation to get\n$$\nf(p^5(p-1)m) = p^5(p-1)f(m)\n$$\nfor $m \\ge p^2$. In a similar manner, it follows by induction that\n$$\nf(p^5(p-1)m) = (p-1)f(p^5m).\n$$\nHence,\n$$\nf(p^5m) = p^5f(m) \\qquad \\textcircled{3}\n$$\nfor $m \\ge p^2$.\nBased on the above argument, we take $k$ with $4^k \\ge p^2$: for any $m \\ge 2$,\n$$\n4^k p^4 f(p m) \\stackrel{\\textcircled{2}}{=} 4^k f(p^5 m) \\stackrel{\\textcircled{1}}{=} f(4^k p^5 m) \\stackrel{\\textcircled{3}}{=} p^5 f(4^k m) \\stackrel{\\textcircled{1}}{=} p^5 4^k f(m).\n$$\nTherefore, $f(p m) = p f(m)$, and the induction is completed.\nAccording to the claim, particularly for $m \\ge 2$,\n$$\n2f(m) = f(2m) = m f(2).\n$$\nTaking $m = 3$, we infer that $f(2)$ is even. So for $m \\ge 2$, $f(m) = d m$, where $d = \\frac{f(2)}{2}$ is an integer. $\\square$",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 28506,
    "subject": "Mathematics (Multi-modal)",
    "question": "Randomly select three vertices from the six vertices of a regular hexagon with side length $1$. Then the probability that two of the three vertices are at a distance of $\\sqrt{3}$ is ______.",
    "options": [],
    "answer": "1",
    "solution": "We will show that “two of the three vertices selected are at a distance of $\\sqrt{3}$” is a certain event. The regular hexagon with edge length $1$ is denoted by $A_1A_2A_3A_4A_5A_6$. If there exist two adjacent vertices taken out of the three vertices selected, it may be set as $A_1, A_2$. Note that $A_1A_3 = A_1A_5 = A_2A_4 = A_2A_6 = \\sqrt{3}$, so the third vertex must be at a distance of $\\sqrt{3}$ from one of the vertices $A_1, A_2$. If any two of the three vertices selected are not adjacent to each other, it can only be $A_1, A_3, A_5$ or $A_2, A_4, A_6$. At this point, there are clearly two vertices with a distance of $\\sqrt{3}$.\n\nTherefore, the desired probability is $1$. $\\square$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 28507,
    "subject": "Mathematics (Multi-modal)",
    "question": "For each positive integer $N$, let $\\tau(N)$ be the number of positive factors of $N$; $\\omega(N)$ be the number of distinct prime factors of $N$; $\\Omega(N)$ be the number of prime factors (counts multiplicities) of $N$. Prove: for each positive integer $n$,\n$$\n\\sum_{m=1}^{n} 5^{\\omega(m)} \\le \\sum_{k=1}^{n} \\lfloor \\frac{n}{k} \\rfloor \\tau(k)^2 \\le \\sum_{m=1}^{n} 5^{\\Omega(m)}.\n$$\n\nHere, $\\lfloor x \\rfloor$ is the largest integer not exceeding $x$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "First, note that $\\lfloor \\frac{n}{k} \\rfloor$ represents the number of multiples of $k$ among $1, 2, \\dots, n$. Hence,\n$$\n\\sum_{k=1}^{n} \\lfloor \\frac{n}{k} \\rfloor \\tau(k)^2 = \\sum_{k=1}^{n} \\sum_{1 \\le m \\le n,\\ k|m} \\tau(k)^2 = \\sum_{m=1}^{n} \\sum_{k|m} \\tau(k)^2.\n$$\nTo prove the problem statement, it suffices to justify, for $m = 1, \\dots, n$,\n$$\n5^{\\omega(m)} \\le \\sum_{k|m} \\tau(k)^2 \\le 5^{\\Omega(m)}.\n$$\nWhen $m=1$, it is obvious. When $m > 1$, let $m = p_1^{\\alpha_1} p_2^{\\alpha_2} \\cdots p_r^{\\alpha_r}$ be the prime factorization ($p_1, p_2, \\dots, p_r$ are distinct prime numbers and $\\alpha_1, \\alpha_2, \\dots, \\alpha_r$ are positive integers). For $k = p_1^{\\beta_1} p_2^{\\beta_2} \\cdots p_r^{\\beta_r}$, $\\tau(k) = (\\beta_1 + 1)(\\beta_2 + 1) \\cdots (\\beta_r + 1)$. Thus,\n$$\n\\begin{aligned}\n\\sum_{k|m} \\tau(k)^2 &= \\sum_{\\substack{0 \\le \\beta_1 \\le \\alpha_1 \\\\ 0 \\le \\beta_r\\cdots \\le \\alpha_r}} (\\beta_1 + 1)^2 (\\beta_2 + 1)^2 \\cdots (\\beta_r + 1)^2 \\\\\n&= \\prod_{i=1}^{r} (1^2 + 2^2 + \\cdots + (\\alpha_i + 1)^2).\n\\end{aligned}\n$$\nNow it suffices to show for each $1 \\le i \\le r$,\n$$\n5 \\le 1^2 + 2^2 + \\cdots + (\\alpha_i + 1)^2 \\le 5^{\\alpha_i}.\n$$\nFor $j \\in \\mathbb{N}_+$, define $T(j) = 1^2+2^2+\\cdots+(j+1)^2 = \\frac{1}{6}(j+1)(j+2)(2j+3)$.\nThen\n$$\n\\frac{T(j+1)}{T(j)} = \\frac{j+2}{j+1} \\cdot \\frac{j+3}{j+2} \\cdot \\frac{2j+5}{2j+3} \\in [1, \\frac{2}{1} \\cdot \\frac{3}{2} \\cdot \\frac{5}{3}] = [1, 5].\n$$\nSince $T(1) = 5$, it is straightforward to check $T(j) \\in [5, 5^j]$ ($\\forall j \\in \\mathbb{N}_+$) by induction. This finishes the proof. $\\Box$",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 28508,
    "subject": "Mathematics (Multi-modal)",
    "question": "Suppose $f(x) = \\cos x + \\log_2 x$ ($x > 0$). If positive real number $a$ satisfies $f(a) = f(2a)$, then the value of $f(2a) - f(4a)$ is ______.",
    "options": [],
    "answer": "-3 or -1",
    "solution": "By the condition, it follows that $\\cos a + \\log_2 a = \\cos 2a + \\log_2 2a = 2\\cos^2 a - 1 + 1 + \\log_2 a$, so $\\cos a = 2\\cos^2 a$. Thus, we have $\\cos a = 0$ or $\\cos a = \\frac{1}{2}$, and hence correspondingly $\\cos 2a = 2\\cos^2 a - 1 = -1$ or $\\cos 2a = -\\frac{1}{2}$. Therefore,\n$$\n\\begin{aligned}\nf(2a) - f(4a) &= \\cos 2a + \\log_2 2a - \\cos 4a - \\log_2 4a \\\\\n&= \\cos 2a - 2 \\cos^2 2a \\\\\n&= \\begin{cases} -3, & \\text{if } \\cos 2a = -1, \\\\ -1, & \\text{if } \\cos 2a = -\\frac{1}{2}. \\end{cases}\n\\end{aligned}\n\\quad \\square",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 28509,
    "subject": "Mathematics (Multi-modal)",
    "question": "In the plane rectangular coordinate system, given hyperbola $\\Gamma: \\frac{x^2}{a^2} - \\frac{y^2}{b^2} = 1$ ($a, b > 0$), a line with inclination angle $\\frac{\\pi}{4}$ passes through a vertex of $\\Gamma$ and another point $(2, 3)$ on it. Then the eccentricity of $\\Gamma$ is ______.",
    "options": [],
    "answer": "2",
    "solution": "The slope of the line described in the question is $1$ and it passes through point $(2, 3)$, so its equation is $y = x + 1$. This line intersects with the $x$-axis at point $(-1, 0)$, and thus $(-1, 0)$ is a vertex of $\\Gamma$. Hence, $a = 1$.\n\nAnd since point $(2, 3)$ is on $\\Gamma$, we know that $\\frac{2^2}{1} - \\frac{3^2}{b^2} = 1$, and then $b^2 = 3$.\n\nDenote $c = \\sqrt{a^2 + b^2}$. Then the eccentricity of $\\Gamma$ is $\\frac{c}{a} = \\frac{\\sqrt{a^2 + b^2}}{a} = 2$.",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 28510,
    "subject": "Mathematics (Multi-modal)",
    "question": "For real numbers $x_1, x_2, \\dots, x_{60} \\in [-1, 1]$, find the maximum of\n$$\n\\sum_{i=1}^{60} x_i^2 (x_{i+1} - x_{i-1}),\n$$\nwhere $x_0 = x_{60}, x_{61} = x_1$.",
    "options": [],
    "answer": "40",
    "solution": "The maximum is $40$. First, notice that\n$$\n\\begin{align*}\n\\sum_{i=1}^{60} x_i^2 (x_{i+1} - x_{i-1}) &= \\sum_{i=1}^{60} x_i^2 x_{i+1} - \\sum_{i=1}^{60} x_i^2 x_{i-1} \\\\\n&= \\sum_{i=1}^{60} x_i^2 x_{i+1} - \\sum_{i=1}^{60} x_{i+1}^2 x_i \\\\\n&= \\sum_{i=1}^{60} x_i x_{i+1} (x_i - x_{i+1}).\n\\end{align*}\n$$\nSince $3xy(x-y) = x^3 - y^3 - (x-y)^3$ for any real numbers $x, y$, we have\n$$\n\\begin{align*}\n\\sum_{i=1}^{60} x_i^2 (x_{i+1} - x_{i-1}) &= \\frac{1}{3} \\sum_{i=1}^{60} (x_i^3 - x_{i+1}^3 - (x_i - x_{i+1})^3) \\\\\n&= \\frac{1}{3} \\sum_{i=1}^{60} (x_{i+1} - x_i)^3.\n\\end{align*}\n$$\nOn one hand, if $x_{3k+1} = 1$, $x_{3k+2} = 0$, $x_{3k+3} = -1$ ($k = 0, 1, \\dots, 19$),\n$$\n\\sum_{i=1}^{60} (x_{i+1} - x_i)^3 = 40 \\cdot (-1)^3 + 20 \\cdot 2^3 = 120;\n$$\nOn the other hand, for $a \\in [-2, 2]$, $(a+1)^2(a-2) \\le 0$, or $a^3 \\le 3a + 2$, and hence\n$$\n\\sum_{i=1}^{60} (x_{i+1} - x_i)^3 \\le \\sum_{i=1}^{60} (3(x_{i+1} - x_i) + 2) = 120.\n$$\nIn conclusion, the maximum of $\\sum_{i=1}^{60} (x_{i+1} - x_i)^3$ is $120$, and the maximum of $\\sum_{i=1}^{60} x_i^2(x_{i+1} - x_{i-1})$ is $40$ (when $\\{x_n\\} = \\{1, 0, -1, 1, 0, -1, \\dots, 1, 0, -1\\}$). $\\square$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 28511,
    "subject": "Mathematics (Multi-modal)",
    "question": "In a plane rectangular coordinate system $xOy$, given parabola $\\Gamma: y^2 = 2px$ ($p > 0$), a line with inclination angle $\\frac{\\pi}{4}$ intersects $\\Gamma$ at point $P(3, 2)$ and another point $Q$. Then the area of $\\triangle OPQ$ is ______.",
    "options": [],
    "answer": "4/3",
    "solution": "Since point $P(3, 2)$ is on $\\Gamma$, we have $2p = \\frac{4}{3}$.\nThe slope of the mentioned line is $1$ and it passes through point $P(3, 2)$. Therefore, its equation is $y = x - 1$, and it passes through point $A(1, 0)$ on the $x$-axis. Substitute $x = y + 1$ into $y^2 = \\frac{4}{3}x$, eliminating $x$ and arranging it gives\n$$\n3y^2 - 4y - 4 = 0.\n$$\nThe solution is $y_1 = 2$, $y_2 = -\\frac{2}{3}$. Further, we can get\n$$\nS_{\\triangle OPQ} = S_{\\triangle OAP} + S_{\\triangle OAQ} = \\frac{1}{2}|OA|(|y_1| + |y_2|) = \\frac{4}{3}.\n$$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 28512,
    "subject": "Mathematics (Multi-modal)",
    "question": "Complex numbers $z_1, z_2, \\dots, z_{100}$ satisfy $z_1 = 3 + 2i$, $z_{n+1} = \\overline{z_n} \\cdot i^n$, ($n = 1, 2, \\dots, 99$), with $i$ as the imaginary unit. Then the value of $z_{99} + z_{100}$ is ______.",
    "options": [],
    "answer": "-5 + 5i",
    "solution": "By the given conditions, we have\n$$\nz_{n+2} = \\overline{z_{n+1}} \\cdot i^{n+1} = \\overline{z_n} \\cdot i^n \\cdot i^{n+1} = z_n i \\quad (n = 1, 2, \\dots, 98).\n$$\nAnd since $z_1 = 3 + 2i$, $z_{99} = z_1 i = -2 + 3i$. Therefore,\n$$\n\\begin{aligned}\nz_{99} + z_{100} &= z_{99} + \\overline{z_{99}} \\cdot i^{99} \\\\\n&= (-2 + 3i) + (-2 - 3i)(-i) \\\\\n&= -5 + 5i.\n\\end{aligned}\n\\quad \\square",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 28513,
    "subject": "Mathematics (Multi-modal)",
    "question": "Suppose $a$, $b$, $c > 1$ and $(a^2b)^{\\log_a c} = a \\cdot (ac)^{\\log_a b}$ is satisfied. Then the value of $\\log_c(ab)$ is ______.",
    "options": [],
    "answer": "2",
    "solution": "Taking the logarithm of the original equation with respect to an arbitrary base $a$ on both sides, we get\n$$\n\\log_a c \\cdot (2 + \\log_a b) = 1 + \\log_a b \\cdot (1 + \\log_a c).\n$$\nSimplifying the above equation gives $2\\log_a c = 1 + \\log_a b$. Therefore, $c^2 = ab$, and then $\\log_c(ab) = \\log_c c^2 = 2$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 28514,
    "subject": "Mathematics (Multi-modal)",
    "question": "Given positive integers $n$, $r$ and distinct prime numbers $p_1$, $p_2$, $\\dots$, $p_r$. Initially, there are $(n+1)^r$ numbers on the blackboard: $p_1^{i_1} p_2^{i_2} \\cdots p_r^{i_r}$ ($0 \\le i_1, i_2, \\dots, i_r \\le n$). Alice and Bob take turns (Alice goes first) to make the following moves, until only one number is left on the blackboard:\n* Every time, Alice erases two numbers (can be identical) and writes their greatest common divisor on the blackboard.\n* Every time, Bob erases two numbers (can be identical) and writes their least common multiple on the blackboard.\nFind the least integer $M$, such that Alice can guarantee the remaining number does not exceed $M$.",
    "options": [],
    "answer": "(p_1 \\cdots p_r)^{\\lfloor \\frac{n}{2} \\rfloor}",
    "solution": "The least $M$ is $(p_1 \\cdots p_r)^{\\lfloor \\frac{n}{2} \\rfloor}$.\n\nDenote $N = (p_1 \\cdots p_r)^n$, $M = (p_1 \\cdots p_r)^{\\lfloor \\frac{n}{2} \\rfloor}$. For each divisor $a$ of $N$, call $\\left(a, \\frac{N}{a}\\right)$ a “pair”. Evidently, $\\gcd(a, \\frac{N}{a})$ is a divisor of $M$, while $\\text{lcm}(a, \\frac{N}{a})$ is a multiple of $M$.\n\nFirst, Alice can guarantee that the remaining number on the board is a divisor of $M$.\n\nSuppose $n$ is odd, $M = (p_1 \\cdots p_r)^{(n-1)/2}$. Alice adopts the following strategy so that her move always leaves some pairs and a divisor $C$ of $M$ on the board. First, she erases $1$, $N$, and writes $1$, which leaves $\\frac{1}{2}(n+1)^r - 1$ pairs and $1$ on the board.\n\n(1) If Bob erases $a$, $b$ from two pairs $(a, \\frac{N}{a})$, $(b, \\frac{N}{b})$, and writes $\\text{lcm}(a, b)$, then Alice erases $\\frac{N}{a}$, $\\frac{N}{b}$, and writes $\\text{gcd}(\\frac{N}{a}, \\frac{N}{b}) = \\frac{N}{\\text{lcm}(a, b)}$ which forms a pair with $\\text{gcd}(a, b)$.\n\n(2) If Bob erases a pair $(a, \\frac{N}{a})$ and writes $\\text{lcm}(a, \\frac{N}{a})$, then Alice erases $\\text{lcm}(a, \\frac{N}{a})$ and $C$, and writes their greatest common divisor $C'$, which is a divisor of $M$, leaving some pairs and $C'$ on the board.\n\n(3) If Bob erases $C$ and $a$ from $(a, \\frac{N}{a})$ and writes $\\text{lcm}(a, C)$, then Alice erases $\\frac{N}{a}$, $\\text{lcm}(a, C)$, and writes $C' = \\text{gcd}(\\frac{N}{a}, \\text{lcm}(a, C))$\nwhich is a divisor of $\\text{lcm}(C, \\text{gcd}(a, \\frac{N}{a}))$. Since $\\text{gcd}(a, \\frac{N}{a})$ is a divisor of $M$, $C'$ is likewise. Still, it leaves some pairs and a divisor $C'$ of $M$ on the board.\n\nBased on the above argument, Alice is certain that her move leaves some pairs and a divisor of $M$ on the board. Eventually, there is only one number left which is a divisor of $M$.\n\nNow suppose $n$ is even, $M = (p_1 \\cdots p_r)^{n/2}$. Alice adopts the following strategy so that her move always leaves some pairs and two divisors $C_1$ and $C_2$ of $M$ on the board. First, she erases $1$, $N$, and writes $1$, leaving pairs and $1$, $M$. In the subsequent moves, if Bob erases one or two numbers from pair(s), Alice follows (1)-(3) in the previous discussion; if Bob erases $C_1$, $C_2$ and writes $\\text{lcm}(C_1, C_2) = C'_1$ which is also a divisor of $M$, then Alice erases any pair $(a, \\frac{N}{a})$ and writes $\\text{gcd}(a, \\frac{N}{a}) = C'_2$ which is again a divisor of $M$. In any case, every time after Alice's move, some pairs and two divisors of $M$ are left on the board. Eventually, Bob must be confronted with two divisors of $M$, and his move will leave a divisor of $M$ on the board.\n\nNext, Bob can guarantee that the remaining number on the board is a multiple of $M$.\n\nIndeed, Bob adopts the following strategy so that his move always leaves some pairs and a multiple $C_1$ of $M$ (when $n$ is even), or two multiples $C_1$ and $C_2$ of $M$ (when $n$ is odd). Initially, take $C_1 = (p_1 \\cdots p_r)^{n/2}$ (when $n$ is even), or $C_1 = (p_1 \\cdots p_r)^{(n-1)/2}$, $C_2 = (p_1 \\cdots p_r)^{(n+1)/2}$ (when $n$ is odd). Bob just reverses Alice's moves as follows:\n\n(1)' If Alice erases $a, b$ from two pairs $\\binom{a, \\frac{N}{a}}{b}$, $\\binom{b, \\frac{N}{b}}{\\frac{N}{a}}$, and writes $\\gcd(a, b)$, then Bob erases $\\frac{N}{a}, \\frac{N}{b}$, and writes $\\text{lcm}(\\frac{N}{a}, \\frac{N}{b}) = N/\\gcd(a, b)$.\n\n(2)' If Alice erases a pair $\\binom{a, \\frac{N}{a}}{b}$ and writes $\\gcd(a, \\frac{N}{a})$, then Bob erases this number and $C_1$.\n\n(3)' If Alice erases $C_i$ and $a$ from $\\binom{a, \\frac{N}{a}}{\\frac{N}{a}}$ and writes $\\gcd(a, C_i)$, then Bob erases $\\frac{N}{a}$ and $\\gcd(a, C_i)$.\n\n(4)' If Alice erases $C_1, C_2$ and writes $\\gcd(C_1, C_2)$, then Bob erases any pair $\\binom{a, \\frac{N}{a}}{a}$.\n\nSimilar to the discussion of Alice's strategy, we just interchange the terms \"gcd\" and \"lcm\" to arrive at the conclusion that Bob guarantees the remaining number is a multiple of $M$.\n\nTherefore, the least $M$ is $(p_1 \\cdots p_r)^{\\lfloor \\frac{n}{2} \\rfloor}$. $\\square$",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 28515,
    "subject": "Mathematics (Multi-modal)",
    "question": "In $\\triangle ABC$, $AB = 1$, $AC = 2$, $B - C = \\frac{2\\pi}{3}$. Then the area of $\\triangle ABC$ is ______.",
    "options": [],
    "answer": "3√3/14",
    "solution": "By the law of sines, it follows that $\\frac{\\sin B}{\\sin C} = \\frac{AC}{AB} = 2$. Since $B - C = \\frac{2\\pi}{3}$, we have\n$$\n2 \\sin C = \\sin B = \\sin \\left( C + \\frac{2\\pi}{3} \\right) = -\\frac{1}{2} \\sin C + \\frac{\\sqrt{3}}{2} \\cos C,\n$$\nnamely, $\\frac{5}{2} \\sin C = \\frac{\\sqrt{3}}{2} \\cos C$, and hence $\\tan C = \\frac{\\sqrt{3}}{5}$.\n\nDenote the area of $\\triangle ABC$ as $S$. Notice that $A = \\pi - B - C = \\frac{\\pi}{3} - 2C$,\nso\n$$\nS = \\frac{1}{2} AB \\cdot AC \\cdot \\sin A = \\sin A = \\frac{\\sqrt{3}}{2} \\cos 2C - \\frac{1}{2} \\sin 2C.\n$$\nSince $\\tan C = \\frac{\\sqrt{3}}{5}$, it follows that $\\cos 2C = \\frac{1 - \\tan^2 C}{1 + \\tan^2 C} = \\frac{11}{14}$, $\\sin 2C = \\frac{2 \\tan C}{1 + \\tan^2 C} = \\frac{5\\sqrt{3}}{14}$. Therefore,\n$$\nS = \\frac{\\sqrt{3}}{2} \\cdot \\frac{11}{14} - \\frac{1}{2} \\cdot \\frac{5\\sqrt{3}}{14} = \\frac{3\\sqrt{3}}{14}.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 28516,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all functions $f : \\mathbb{R} \\to \\mathbb{R}$ such that for any $x, y \\in \\mathbb{R}$,\n$$ f(xf(y) + y^{2021}) = yf(x) + (f(y))^{2021}. $$\n(Contributed by Fu Yunhao)",
    "options": [],
    "answer": "f(x) = 0 and f(x) = x",
    "solution": "Denote $P(x, y)$ as the governing equation of the problem. From $P(x, 0)$, we get $f(xf(0)) = (f(0))^{2021}$. If $f(0) \\neq 0$, as $x$ is arbitrary, $f$ must be a constant function, say $f(x) = c$. Then $c = yc + c^{2021}$ for $y \\in \\mathbb{R}$, $c = 0$, a contradiction. Hence, $f(0) = 0$.\nIf $f(a) = 0$ for some $a \\neq 0$, then $P(x, a)$ leads to $f(x) = 0$ for all $x \\in \\mathbb{R}$. Clearly, this function satisfies the governing equation. In the following, assume $f(a) \\neq 0$ whenever $a \\neq 0$.\nFrom $P(0, 1)$, we get $f(1) = (f(1))^{2021}$, and $f(1) = \\pm 1$. If $f(1) = -1$, $P(1, 1)$ gives $f(0) = -2$, a contradiction. Thus, $f(1) = 1$.\nNow $P(x, 1)$ gives\n$$\nf(x+1) = f(x) + 1. \\qquad \\textcircled{1}\n$$\nCompare $P(x, y)$ and $P(x+1, y)$, to obtain\n$$\n\\begin{aligned}\n& f(xf(y) + f(y) + y^{2021}) \\stackrel{P(x+1,y)}{=} yf(x+1) \\\\\n& \\quad +(y)^{2021} \\stackrel{\\textcircled{1}}{=} yf(x) + y + f(y)^{2021} \\stackrel{P(x,y)}{=} y + f(xf(y) + y^{2021}).\n\\end{aligned}\n$$\nAs $f(y) \\neq 0$, $xf(y) + y^{2021}$ can take any real value $z$, and thus\n$$\nf(z + f(y)) = y + f(z). \\qquad \\textcircled{2}\n$$\nIn (2), letting $z = 0$, we get $f(f(y)) = y$. Then, letting $y = f(w)$ in (2), we arrive at the following equation\n$$\nf(z + f(f(w))) = f(z + w) = f(z) + f(w)\n$$\nfor any $z, w \\in \\mathbb{R}$.\n---\nNote that $P(0, y)$ gives $f(y^{2021}) = (f(y))^{2021}$, and so $P(x, y)$ becomes\n$$\n\\begin{aligned}\nyf(x) + (f(y))^{2021} &= f(xf(y) + y^{2021}) \\\\\n&= f(xf(y)) + f(y^{2021}) \\\\\n&= f(xf(y)) + (f(y))^{2021},\n\\end{aligned}\n$$\nwhich implies $yf(x) = f(xf(y))$. Let $y = f(z)$ and use $f(f(z)) = z$ to\nderive\n$$\nf(xz) = f(x)f(z).\n$$\nIn particular, $f$ maps positive reals to positive reals. By the well-known\nproperties of Cauchy's multiplicative functional equation, $f(x) = x$, which\nclearly satisfies the governing equation.\nWe conclude that there are only two functions: $f(x) = 0$ and $f(x) = x$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 28517,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $n$ be a positive integer, $a_1, a_2, \\dots, a_{2n+1}$ be $2n+1$ positive real numbers. For $k = 1, 2, \\dots, 2n+1$, define\n$$\nb_k = \\max_{0 \\le m \\le n} \\left( \\frac{1}{2m+1} \\sum_{i=k-m}^{k+m} a_i \\right),\n$$\nwhere the subscript of $a_i$ is taken modulo $2n+1$. Prove: the number of subscripts $k$ satisfying that $b_k \\ge 1$ does not exceed $2 \\sum_{i=1}^{2n+1} a_i$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Define $I = \\{k \\mid b_k \\ge 1\\}$. For every $k \\in I$, assume that the maximum value $b_k$ of $\\frac{1}{2m+1} \\sum_{i=k-m}^{k+m} a_i$ is attained at $m = m_k$, and call\n$$\n[k - m_k, k + m_k] := \\{k - m_k, k - m_k + 1, \\dots, k + m_k\\}\n$$\na “nice segment”, where the subscripts are taken modulo $2n+1$. Obviously, the union of all nice segments contains $I$.\n\n**Claim** There exists a collection of nice segments whose union contains $I$, and moreover, each $i \\in \\{1, 2, \\dots, 2n + 1\\}$ is contained in at most two segments.\n\n**Proof of claim** If $[1, 2n + 1]$ is a nice segment, the conclusion is trivial. In the following, assume this is not the case. If $i$ is contained in $r$ nice segments\n$$\n[i - u_1, i + v_1], \\dots, [i - u_r, i + v_r]\n$$\nwhere $r \\ge 3$ and $0 \\le u_j, v_j < 2n$, let $u_j = \\max\\{u_1, \\dots, u_r\\}$ and $v_k = \\max\\{v_1, \\dots, v_r\\}$. We can keep the nice segments $[i - u_j, i + v_j]$, $[i - u_k, i + v_k]$ and drop the other $r-2$ segments. Now the segments still cover $1, \\dots, 2n+1$, and at most two of them cover $i$. For each $i \\in \\{1, 2, \\dots, 2n+1\\}$, perform the above operation. Eventually, we find a collection of nice segments with the desired properties.\n\nFor the original problem, let $[i_1 - m_1, i_1 + m_1]$, $\\dots$, $[i_r - m_r, i_r + m_r]$ be a collection of nice segments chosen in the claim. We have\n$$\n2 \\sum_{i=1}^{2n+1} a_i \\ge \\sum_{\\alpha=1}^{r} \\sum_{k=i_{\\alpha}-m_{\\alpha}}^{i_{\\alpha}+m_{\\alpha}} a_k = \\sum_{\\alpha=1}^{r} (2m_{\\alpha} + 1)b_{i_{\\alpha}} \\ge \\sum_{\\alpha=1}^{r} (2m_{\\alpha} + 1) \\ge |I|,\n$$\nthe first inequality is due to each $i$ being contained in at most two nice segments; the equality and the next inequality are due to the definition of nice segments; the last inequality is due to the union of the nice segments containing $I$. $\\square$",
    "topic": "Number Theory",
    "subtopic": "Modular Arithmetic"
  },
  {
    "id": 28518,
    "subject": "Mathematics (Multi-modal)",
    "question": "Sequence $\\{a_n\\}$ satisfies $a_1 = 2$ and $a_{n+1} = (n+1)a_n - n$, $n = 1, 2, \\dots$. Then the general term formula of $\\{a_n\\}$ is ______.",
    "options": [],
    "answer": "a_n = n! + 1",
    "solution": "By the condition, we have $a_{n+1} - 1 = (n+1)(a_n - 1)$. Therefore,\n$$\n\\begin{aligned}\na_n - 1 &= n(a_{n-1} - 1) = n(n-1)(a_{n-2} - 1) = \\dots \\\\\n&= n(n-1)\\dots2(a_1 - 1) = n!,\n\\end{aligned}\n$$\nnamely, $a_n = n! + 1$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 28519,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find the smallest real number $\\alpha$, such that for any convex polygon $P$ of area $1$, there exists a point $M$ in the plane, such that the area of the convex hull of $P \\cup Q$ is at most $\\alpha$, where $Q$ is the central-symmetric figure of $P$ about $M$.\n(Contributed by Qu Zhenhua and Wu Yuchi)",
    "options": [],
    "answer": "2",
    "solution": "(i) If $M$ is outside $P$ or on the boundary. Draw a line $l$ through $M$ such that $P$, $Q$ lie on different sides of $l$ (they have no common interior points). Then $S(P \\cup Q) \\ge S(P) + S(Q) \\ge 2$.\n\n(ii) If $M$ is inside $P$. Let $A', B', C'$ be the respective symmetric points of $A, B, C$ about $M$. Then $R = P \\cup Q = \\{A, B, C, A', B', C'\\}$, $R$ has a centre of symmetry, $R$ is a parallelogram or a hexagon. If $R$ is a parallelogram, say $R = ABA'B'$, then $C$ is inside or on the boundary, and\n$$\nS(R) = S(ABA'B') \\ge 2S(ABC) = 2.\n$$\nIf $R$ is a hexagon, $R = AC'BA'CB'$, then\n$$\n\\begin{aligned}\nS(R) &= S(AC'BM) + S(BA'CM) + S(CB'AM) \\\\\n&= (S(AMC) + S(BMC)) + (S(BMA) + S(CMA)) \\\\\n&\\quad + (S(CMB) + S(AMB)) = 2.\n\\end{aligned}\n$$\nNext, we prove $\\alpha = 2$ satisfies the problem statement in two ways. $\\square$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 28520,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find the least positive real $a$ satisfying this condition: for any three points $A, B, C$ on the unit circle, there exists an equilateral triangle $PQR$ with side length $a$, such that $A, B, C$ are all inside or on the boundary of triangle $PQR$.",
    "options": [],
    "answer": "2 sin^2 80° / sin 60°",
    "solution": "First, we prove $a = \\frac{(2 \\sin 80^{\\circ})^2}{\\sqrt{3}}$ is sufficient. For any three points $A, B, C$ on the unit circle, let $\\angle BAC = \\alpha$, $\\angle ABC = \\beta$, $\\angle ACB = \\gamma$, $\\alpha \\le \\beta \\le \\gamma$.\n\nIf $\\beta \\le 60^{\\circ}$, since $AB \\le 2 < a$ (here, $a > \\frac{(2 \\sin 75^{\\circ})^2}{\\sqrt{3}} = \\frac{(\\sqrt{6} + \\sqrt{2})^2}{4\\sqrt{3}} = \\frac{8 + 4\\sqrt{3}}{4\\sqrt{3}} > 2$), one can draw a segment $PQ$ of length $a$ that contains $A$ and $B$, then find $R$ such that $\\triangle PQR$ is equilateral, $C$, $R$ on the same side of $PQ$. It follows from $\\alpha \\le \\beta \\le 60^{\\circ}$ that $C$ is inside $\\triangle PQR$ or on the boundary.\n\nIf $\\beta > 60^{\\circ}$, we first prove the following:\nAt least one of $\\sin \\beta \\sin \\gamma$ and $\\sin \\beta \\sin(\\alpha + 60^{\\circ})$ is less than or equal to $\\sin^2 80^{\\circ}$. (*)\n\nIn fact, as $2(\\beta + \\alpha + 60^{\\circ}) + (\\beta + \\gamma) = 480^{\\circ} - (\\gamma - \\beta) \\le 480^{\\circ}$, one of $\\beta + \\alpha + 60^{\\circ}$ and $\\beta + \\gamma$ does not exceed $160^{\\circ}$. From the simple property of sine function\n$$\n\\begin{aligned}\n\\sin x \\sin y &= \\frac{1}{2}(\\cos(x - y) - \\cos(x + y)) \\\\\n&\\le \\frac{1}{2}(1 - \\cos(x + y)) \\\\\n&= \\sin^2 \\frac{x+y}{2},\n\\end{aligned}\n$$\n(*) follows.\n\ni) If $\\sin \\beta \\sin \\gamma \\le \\sin^2 80^{\\circ}$, let $AD \\perp BC$ with foot $D$; take $P, Q$ on the line $BC$ such that $PD = DQ = \\frac{a}{2}$; take $R$ on the ray $DA$ such that\n$$\nDR = \\frac{\\sqrt{3}a}{2} = 2 \\sin^2 80^{\\circ}.\n$$\nNotice that $AD = 2 \\sin \\beta \\sin \\gamma \\le DR$, and thus $A$ lies inside $\\triangle PQR$ or on the boundary; from $60^{\\circ} \\le \\beta, \\gamma \\le 120^{\\circ}$, it is easy to see that $B, C$ lie on the segment $PQ$.\n\nii) If $\\sin \\beta \\sin(\\alpha + 60^{\\circ}) \\le \\sin^2 80^{\\circ}$, take $P = A$, extend $AB$ to $Q$, $AQ = a$, and find $R$ such that $\\triangle PQR$ is equilateral and $C, R$ lie on the same side of $PQ$. Let $M$ be the intersection of the ray $AC$ and the line $QR$. Notice that $AB \\le 2 < a$, hence $B$ is on the boundary of $\\triangle PQR$; since\n$$\nAM = a \\cdot \\frac{\\sin 60^{\\circ}}{\\sin(\\alpha + 60^{\\circ})} = \\frac{2 \\sin^2 80^{\\circ}}{\\sin(\\alpha + 60^{\\circ})} > 2 \\sin \\beta = AC,\n$$\n$C$ is inside $\\triangle PQR$ or on the boundary.\n\nIn conclusion, $a = \\frac{(2 \\sin 80^{\\circ})^2}{\\sqrt{3}}$ satisfies the problem condition.\n\nNext, we prove $a \\ge \\frac{(2 \\sin 80^{\\circ})^2}{\\sqrt{3}} = \\frac{2 \\sin^2 80^{\\circ}}{\\sin 60^{\\circ}}$ is necessary. To this end, choose $A, B, C$ on the unit circle with $\\angle BAC = 20^{\\circ}$, $\\angle ABC = \\angle ACB = 80^{\\circ}$.\n\nSuppose the equilateral triangle $PQR$ with side length $a$ covers $A, B, C$ (inside or on the boundary). By a translation if necessary, we can make one of $A, B, C$, say $A$ on the boundary of $\\triangle PQR$, and then by a rotation about $A$ we can make $B$ or $C$ on the boundary. If neither of them is a vertex of $\\triangle PQR$, assume they all lie on $PQ$ or $PR$. Take $P$ as the homothetic centre and rescale $\\triangle PQR$ smaller such that all of $A, B, C$ fall on the boundary or become a vertex of $\\triangle PQR$. Since $\\angle ABC, \\angle ACB > 60^{\\circ}$, $B, C$ cannot be vertices; in addition, there is symmetry between $B$ and $C$. Thus, it suffices to consider four situations as follows.\n\n(1) $A$ is a vertex, say $A = P$. Let $PH \\perp QR$ with foot $H$. Then the angle between $PH$ and one of $PB, PC$ is less than or equal to $10^{\\circ}$. It follows that $PH \\ge 2 \\sin^2 80^{\\circ}$, and\n$$\na = PQ \\geq \\frac{2 \\sin^2 80^{\\circ}}{\\sin 60^{\\circ}}.\n$$\n\n(2) $A$ and $B$ lie on the same side, say $PQ$, with $PA \\le PB$. Let $PH \\perp QR$ at $H$. Then the angle between $AC$ and $PH$ is $10^{\\circ}$. Similar to (1), we have\n$$\na \\geq \\frac{2 \\sin^2 80^{\\circ}}{\\sin 60^{\\circ}}.\n$$\n\n(3) $B$ and $C$ lie on the same side, say $QR$. Let $PH \\perp QR$ at $H$. Then the lines $AC$ and $PH$ cross at $10^{\\circ}$. Similar to (1), we have\n$$\na \\geq \\frac{2 \\sin^2 80^{\\circ}}{\\sin 60^{\\circ}}.\n$$\n\n(4) $A, B, C$ are all on different sides. Let $A \\in PQ$, $B \\in QR$, $C \\in RP$, and $\\vartheta = \\angle RBC \\le \\angle RCB$, $\\vartheta \\le 60^{\\circ}$.\n\ni) If $\\vartheta \\le 20^{\\circ}$, then the angle between $AB$ and the altitude on $QR$ does not exceed $10^{\\circ}$. Similar to (1), we get\n$$\na \\geq \\frac{2 \\sin^2 80^{\\circ}}{\\sin 60^{\\circ}}.\n$$\n\nii) If $40^{\\circ} \\le \\vartheta \\le 60^{\\circ}$, then the angle between $AC$ and the altitude on $PQ$ does not exceed $10^{\\circ}$. Similar to (1), $a \\ge \\frac{2 \\sin^2 80^{\\circ}}{\\sin 60^{\\circ}}$.\n\niii) If $20^{\\circ} \\le \\vartheta \\le 40^{\\circ}$, then\n$$\n\\begin{aligned}\na &= QR = QB + BR \\\\\n&= \\frac{2 \\sin 80^{\\circ} \\sin(20^{\\circ} + \\vartheta)}{\\sin 60^{\\circ}} + \\frac{2 \\sin 20^{\\circ} \\sin(60^{\\circ} + \\vartheta)}{\\sin 60^{\\circ}} \\\\\n&\\ge \\frac{2 \\sin 80^{\\circ} \\sin 40^{\\circ} + 2 \\sin 20^{\\circ} \\sin 80^{\\circ}}{\\sin 60^{\\circ}} \\\\\n&= \\frac{2 \\sin^2 80^{\\circ}}{\\sin 60^{\\circ}}.\n\\end{aligned}\n$$\n\nIn summary, the minimum is $a = \\frac{2 \\sin^2 80^{\\circ}}{\\sin 60^{\\circ}}$. $\\square$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 28521,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let the side length of the base and height of regular pyramid $P-ABCD$ be equal. Point $G$ is the centroid of face $\\triangle PBC$. Then the sine of the angle between line $AG$ and base $ABCD$ is ______.",
    "options": [],
    "answer": "sqrt(38)/19",
    "solution": "$O$ and $H$, respectively. Then $O$ is the centre of the base square, $H$ lies on $OM$, and $\\frac{GH}{PO} = \\frac{HM}{OM} = \\frac{GM}{PM} = \\frac{1}{3}$.\n\nFor the sake of convenience, let $AB = PO = 6$. Thus, $GH = \\frac{PO}{3} = 2$, $OH = \\frac{2}{3}OM = 2$. And since $AO = 3\\sqrt{2}$, $\\angle AOH = 135^\\circ$, we have\n$$\nAH^2 = AO^2 + OH^2 - 2AO \\cdot OH \\cdot \\cos \\angle AOH = 34,\n$$\nand thus $AG = \\sqrt{AH^2 + GH^2} = \\sqrt{38}$.\n\nThe angle between line $AG$ and base $ABCD$ is equal to $\\angle GAH$. Therefore, the desired sine is\n$$\n\\sin \\angle GAH = \\frac{GH}{AG} = \\frac{2}{\\sqrt{38}} = \\frac{\\sqrt{38}}{19}.\n$$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 28522,
    "subject": "Mathematics (Multi-modal)",
    "question": "Function $f(x)$ with domain $\\mathbb{R}$ satisfies: when $x \\in [0, 1)$, $f(x) = 2^x - x$, and for any real number $x$, there is $f(x) + f(x+1) = 1$. Denote $a = \\log_2 3$, and then the value of expression $f(a) + f(2a) + f(3a)$ is ______.",
    "options": [],
    "answer": "17/16",
    "solution": "By the conditions, we know that $f(x+n) = 1 - f(x)$ when $n$ is odd and $f(x+n) = f(x)$ when $n$ is even.\nNote that $a = \\log_2 3 \\in [1, 2)$, $2a = \\log_2 9 \\in [3, 4)$, $3a = \\log_2 27 \\in [4, 5)$. Therefore,\n$$\n\\begin{aligned}\nf(a) + f(2a) + f(3a) &= 1 - f(a - 1) + 1 - f(2a - 3) + f(3a - 4) \\\\\n&= 2 - f\\left(\\log_2 \\frac{3}{2}\\right) - f\\left(\\log_2 \\frac{9}{8}\\right) + f\\left(\\log_2 \\frac{27}{16}\\right) \\\\\n&= 2 - \\left(\\frac{3}{2} - \\log_2 \\frac{3}{2}\\right) - \\left(\\frac{9}{8} - \\log_2 \\frac{9}{8}\\right) \\\\\n&\\quad + \\left(\\frac{27}{16} - \\log_2 \\frac{27}{16}\\right) \\\\\n&= \\frac{17}{16} + \\left(\\log_2 \\frac{3}{2} + \\log_2 \\frac{9}{8} - \\log_2 \\frac{27}{16}\\right) = \\frac{17}{16}.\n$$",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 28523,
    "subject": "Mathematics (Multi-modal)",
    "question": "Prove that there exists a constant $\\lambda > 0$, such that: for any positive integer $m$, in the rectangular coordinates system, if all vertices of $\\triangle ABC$ are integral points, and there is a unique interior (not on the side) integral point whose $x$ and $y$ coordinates are multiples of $m$, then the area of $\\triangle ABC$ is less than $\\lambda m^3$.\n\n(Contributed by Wang Bin and Fu Yunhao)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "We call $(x, y)$ an “$m$-integral point” if $x, y$ are integers and $m|x$, $m|y$. By a proper translation, we may assume that the unique interior $m$-integral point is at $(0, 0)$. Extend $AO$, $BO$, and $CO$ to meet the opposite sides at points $D$, $E$, and $F$, respectively. Denote\n$$\np = \\frac{OD}{AD} = \\frac{S_{\\triangle OBC}}{S_{\\triangle ABC}}, \\quad q = \\frac{OE}{BE} = \\frac{S_{\\triangle OCA}}{S_{\\triangle ABC}}, \\quad r = \\frac{OF}{CF} = \\frac{S_{\\triangle OAB}}{S_{\\triangle ABC}}.\n$$\nEvidently, $p + q + r = 1$; assume $p \\ge q \\ge r > 0$ (so that $p \\ge \\frac{1}{3} \\ge \\frac{1}{m+2}$). Let $A'$, $B'$, and $D'$ be the respective symmetric points of $A$, $B$, and $D$ about $O$. On the line $AD$, let $UV$ be the intersection of the segments $AD$ and $D'A'$, namely, $U$ is $A$ or $D'$, $V$ is $D$ or $A'$, whichever is closer to $O$. Since $\\triangle BUV \\subseteq \\triangle ABC$, except for $O$, there is no $m$-integral point in the interior of $\\triangle BUV$ or in the interior of $UV$, and neither in the symmetric $\\triangle B'UV$. Together, they imply that the parallelogram $BUB'V$\n\n(centered at $O$) does not have interior $m$-integral points other than $O$. By Minkowski's theorem,\n$$\n4m^2 \\geq S_{BUB'V} = 4S_{\\Delta OBV} = 4S_{\\Delta ABC} \\times \\min\\{p, 1-p\\} \\times \\frac{q}{q+r};\n$$\n$$\nS_{\\Delta ABC} \\leq \\frac{4m^2}{4 \\min\\{p, 1-p\\}} \\times \\frac{q+r}{q} \\leq \\frac{2m^2}{\\min\\{p, 1-p\\}}.\n$$\nIf $p > \\frac{m}{m+1}$, then\n$$\n\\frac{OD}{OA'} = \\frac{OD}{AO} = \\frac{p}{1-p} > m,\n$$\ntaking $OL = m \\cdot OA'$ on segment $OD$, $L$ must be an $m$-integral point inside $\\triangle ABC$, which is contradictory.\nTherefore, $p \\le \\frac{m}{m+1}$, $\\min\\{p, 1-p\\} \\ge \\frac{1}{m+2}$, and we deduce\n$$\nS_{\\triangle ABC} \\le 2m^2(m+2). \\quad \\square",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 28524,
    "subject": "Mathematics (Multi-modal)",
    "question": "In a plane rectangular coordinate system $xOy$, the focus of parabola $\\Gamma: y^2 = 2px$ ($p > 0$) is $F$. Make a tangent line to $\\Gamma$ passing through point $P$ (different from $O$) on $\\Gamma$ and it intersects the $y$-axis at point $Q$. If $|FP| = 2$, $|FQ| = 1$, then the dot product of vectors $\\overrightarrow{OP}$ and $\\overrightarrow{OQ}$ is ______.",
    "options": [],
    "answer": "3/2",
    "solution": "Let $P(\\frac{t^2}{2p}, t)$ ($t \\neq 0$), and then the equation of the tangent line of $\\Gamma$ is $yt = p(x + \\frac{t^2}{2p})$.\nLet $x = 0$, and we get $yt = \\frac{t}{2}$. The coordinates of $F$ are $(\\frac{p}{2}, 0)$, and thus\n$$\n|FP| = \\sqrt{\\left(\\frac{p}{2} - \\frac{t^2}{2p}\\right)^2 + t^2} = \\frac{p}{2} + \\frac{t^2}{2p}, \\\\ |FQ| = \\frac{\\sqrt{p^2 + t^2}}{2}.\n$$\nCombining $|FP| = 2$, $|FQ| = 1$, we can get $p^2 + t^2 = 4p$ and $p^2 + t^2 = 4$, respectively. Hence, $p = 1$, $t^2 = 3$.\nTherefore, $\\overrightarrow{OP} \\cdot \\overrightarrow{OQ} = \\frac{t^2}{2} = \\frac{3}{2}$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 28525,
    "subject": "Mathematics (Multi-modal)",
    "question": "An even cube dice with six faces is marked with six numbers $1$, $2$, $3$, $4$, $5$ and $6$, respectively. The dice are randomly tossed three times, with each toss being independent from each other, and the resulting dots are $a_1, a_2, a_3$ in order. Then the probability of event $|a_1 - a_2| + |a_2 - a_3| + |a_3 - a_1| = 6$ is ______.",
    "options": [],
    "answer": "1/4",
    "solution": "Note that $|a_1 - a_2| + |a_2 - a_3| + |a_3 - a_1| = 2 \\max_{1 \\le i \\le 3} a_i - 2 \\min_{1 \\le i \\le 3} a_i$.\nTherefore, the three $a_1, a_2, a_3$ tossed satisfy the condition if and only if the difference between the largest number and the smallest number is $3$.\nThis is to say that $a_1, a_2, a_3$ is a permutation of $x, x+d, x+3$, where $x \\in \\{1, 2, 3\\}$ and $d \\in \\{0, 1, 2, 3\\}$.\nFor each $x \\in \\{1, 2, 3\\}$, when $d = 0$ or $d = 3$, there are $3$ different permutations of $x, x+d, x+3$ for each case; when $d = 1$ or $d = 2$, there are $6$ different permutations of $x, x+d, x+3$ for each case.\nTherefore, there are $3 \\times (2 \\times 3 + 2 \\times 6) = 54$ cases of dots $a_1, a_2, a_3$ that satisfy the condition.\nConsequently, the desired probability is $\\frac{54}{6^3} = \\frac{1}{4}$. $\\square$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 28526,
    "subject": "Mathematics (Multi-modal)",
    "question": "Suppose set $S = \\{1, 2, 3, \\dots, 10\\}$ and the subset $A$ of $S$ satisfies\n\n$$\nA \\cap \\{1, 2, 3\\} \\neq \\emptyset, \\quad A \\cup \\{4, 5, 6\\} \\neq S.\n$$\nThe number of such subsets is ________.",
    "options": [],
    "answer": "888",
    "solution": "First, we will find the number $N_1$ of subsets $A$ of $S$ such that $A \\cap \\{1, 2, 3\\} \\neq \\emptyset$ holds.\nThere are $2^3 - 1 = 7$ ways of selecting at least one element in $1, 2, 3$, while for each number in $4, 5, \\dots, 10$ there are two choices (selected or not selected). Therefore, $N_1 = 7 \\times 2^7 = 896$.\n\nFurthermore, we need to deduct the number of selections $N_2$ which makes $A \\cup \\{4, 5, 6\\} = S$ hold. Of these selections of $A$, $1, 2, 3, 7, 8, 9, 10$ are all selected. And each of $4, 5, 6$ has two choices (selected or not selected), so $N_2 = 2^3 = 8$.\n\nTherefore, the number of subsets $A$ satisfying the conditions is $N_1 - N_2 = 896 - 8 = 888$. $\\square$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 28527,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $a_1, a_2, \\dots, a_{21}$ be a permutation of $1, 2, \\dots, 21$, satisfying\n$$\n|a_{20} - a_{21}| \\geq |a_{19} - a_{21}| \\geq |a_{18} - a_{21}| \\geq \\dots \\geq |a_1 - a_{21}|.\n$$\nThe number of such permutations is ________.",
    "options": [],
    "answer": "3070",
    "solution": "For a given $k \\in \\{1, 2, \\dots, 21\\}$, consider the number of permutations $N_k$ that satisfy the conditions such that $a_{21} = k$.\nWhen $k \\in \\{1, 2, \\dots, 11\\}$, for $i = 1, 2, \\dots, k-1$, there exist $a_{2i-1}, a_{2i}$ that are permutations of $k-i, k+i$ (if $k=1$, there exists no such $i$), and $a_{2j} = j+1$ ($2k-1 \\le j \\le 20$) (if $k=11$, there exists no such $j$), so $N_k = 2^{k-1}$.\n\nSimilarly, when $k \\in \\{12, 13, \\dots, 21\\}$, there is $N_k = 2^{21-k}$.\nTherefore, the number of such permutations satisfying the condition is\n$$\n\\sum_{k=1}^{21} N_k = \\sum_{k=1}^{11} 2^{k-1} + \\sum_{k=12}^{21} 2^{21-k} = (2^{11}-1) + (2^{10}-1) = 3070. \\quad \\square",
    "topic": "Discrete Mathematics",
    "subtopic": "Combinatorics"
  },
  {
    "id": 28528,
    "subject": "Mathematics (Multi-modal)",
    "question": "Suppose $\\alpha, \\beta \\ge 0$, $\\alpha + \\beta \\le 2\\pi$. Then the minimum of $\\sin \\alpha + 2 \\cos \\beta$ is ______.",
    "options": [],
    "answer": "-sqrt(5)",
    "solution": "When $0 \\le \\alpha \\le \\pi$, $\\sin \\alpha + 2 \\cos \\beta \\ge 0 + 2 \\cdot (-1) = -2$.\n\nWhen $\\pi < \\alpha \\le 2\\pi$, there is $0 \\le \\beta \\le 2\\pi - \\alpha < \\pi$. At this point, as $\\beta$ gets bigger, $\\cos \\beta$ gets smaller. Therefore,\n$$\n\\begin{aligned}\n\\sin \\alpha + 2 \\cos \\beta &\\ge \\sin \\alpha + 2 \\cos(2\\pi - \\alpha) \\\\\n&= \\sin \\alpha + 2 \\cos \\alpha \\\\\n&= \\sqrt{5} \\sin(\\alpha + \\varphi),\n\\end{aligned}\n$$\nwhere $\\varphi = \\arcsin \\frac{2\\sqrt{5}}{5}$.\n\nWhen $\\alpha = \\frac{3\\pi}{2} - \\varphi$, $\\beta = 2\\pi - \\alpha = \\frac{\\pi}{2} + \\varphi$, $\\sin \\alpha + 2 \\cos \\beta$ gets the minimum $-\\sqrt{5}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 28529,
    "subject": "Mathematics (Multi-modal)",
    "question": "In a plane rectangular coordinate system $xOy$, the graph of parabola $y = ax^2 - 3x + 3$ ($a \\neq 0$) and that of parabola $y^2 = 2px$ ($p > 0$) are symmetric with respect to line $y = x + m$. Then the product of real numbers $a, p, m$ is ______.",
    "options": [],
    "answer": "-3",
    "solution": "For any point $(x_0, y_0)$ on parabola $y = ax^2 - 3x + 3$ ($a \\neq 0$), there is\n$$\ny_0 = a x_0^2 - 3x_0 + 3. \\qquad \\textcircled{1}\n$$\nSuppose the symmetric point of $(x_0, y_0)$ with respect to line $y = x + m$ is $(x_1, y_1)$.\nBy $\\frac{y_1 + y_0}{2} = \\frac{x_1 + x_0}{2} + m$, $x_1 + y_1 = x_0 + y_0$, it follows that\n$$\nx_1 = y_0 - m, \\quad y_1 = x_0 + m.\n$$\nSince $(x_1, y_1)$ lies on parabola $y^2 = 2px$, there is $(x_0 + m)^2 = 2p(y_0 - m)$. This is equivalent to\n$$\ny_0 = \\frac{1}{2p} x_0^2 + \\frac{m}{p} x_0 + \\frac{m^2}{2p} + m. \\qquad \\textcircled{2}\n$$\nDue to the arbitrary taking of $(x_0, y_0)$, comparing (1) and (2) gives\n$$\n\\frac{1}{2p} = a, \\quad \\frac{m}{p} = -3, \\quad \\frac{m^2}{2p} + m = 3.\n$$\nTherefore, we have $3 = \\frac{m}{p} \\cdot \\frac{m}{2} + m = -3 \\cdot \\frac{m}{2} + m = -\\frac{m}{2}$, and its solution is $m = -6$. Hence, in order, we can get $p = 2, a = \\frac{1}{4}$, satisfying $a \\neq 0$ and $p > 0$. Consequently, $apm = 2 \\cdot \\frac{1}{4} \\cdot (-6) = -3$. $\\square$",
    "topic": "Discrete Mathematics",
    "subtopic": "Other"
  },
  {
    "id": 28530,
    "subject": "Mathematics (Multi-modal)",
    "question": "In a plane rectangular coordinate system $xOy$, $\\Gamma_1$ is a unit circle centred at $(2, 1)$ and $\\Gamma_2$ is a unit circle centred at $(10, 11)$. Make a line $l$ through the origin $O$ such that $l$ has two intersections with each of $\\Gamma_1$ and $\\Gamma_2$, dividing $\\Gamma_1$ and $\\Gamma_2$ into four arcs, and two of these four arcs are of equal length. The sum of the slopes of all the lines $l$ satisfying the conditions is ________.",
    "options": [],
    "answer": "67/20",
    "solution": "Denote the centres $(2, 1)$, $(10, 11)$ of the two circles $\\Gamma_1, \\Gamma_2$ as $T_1, T_2$, respectively.\n\nIf $l$ passes through $T_1$ or $T_2$, then $l$ bisects the circumference of $\\Gamma_1$ or that of $\\Gamma_2$, which yields two equal arcs. The possible slopes of $l$ at this point are $k_1 = k_{OT_1} = \\frac{1}{2}$ or $k_2 = k_{OT_2} = \\frac{11}{10}$.\n\nIf $l$ neither passes through $T_1$ nor $T_2$, then $\\Gamma_1$ and $\\Gamma_2$ are both divided into two arcs of unequal length by $l$. And since $\\Gamma_1$ and $\\Gamma_2$ are equal circles, the two arcs divided in $\\Gamma_1$ are equal to the two arcs divided in $\\Gamma_2$, respectively. This implies that $l$ is parallel to $T_1T_2$ or passes through its midpoint $M(6, 6)$. The possible slopes of $l$ at this point are $k_3 = k_{T_1T_2} = \\frac{5}{4}$ or $k_4 = k_{OM} = 1$.\n\nAfter checking, line $y = k_1x$ has no intersection with circle $\\Gamma_2$, which does not fit the question; when $i = 2, 3, 4$, line $y = k_ix$ has two intersections with each of $\\Gamma_1, \\Gamma_2$, which is consistent with the question. Therefore, the sum of the slopes of all the lines $l$ satisfying the conditions is\n$$\nk_2 + k_3 + k_4 = \\frac{11}{10} + \\frac{5}{4} + 1 = \\frac{67}{20}. \\quad \\square\n$$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 28531,
    "subject": "Mathematics (Multi-modal)",
    "question": "Given rational number $r = \\frac{p}{q} \\in (0, 1)$, $p$, $q$ are coprime positive integers, and $pq$ divides $3600$. The number of such rational numbers $r$ is ______.",
    "options": [],
    "answer": "112",
    "solution": "Suppose set $\\Omega = \\{ r \\mid r = \\frac{p}{q},\\ p, q \\in \\mathbb{N}_+,\\ (p, q) = 1,\\ pq \\mid 3600 \\}$.\nWe consider the reduced fractional form $\\frac{p}{q}$ of any element $r$ of $\\Omega$. Since the standard factorization of $3600$ is $2^4 \\times 3^2 \\times 5^2$, we can set $p = 2^A \\times 3^B \\times 5^C$, $q = 2^a \\times 3^b \\times 5^c$, where $\\min\\{A, a\\} = \\min\\{B, b\\} = \\min\\{C, c\\} = 0$ and $A + a \\le 4$, $B + b \\le 2$, $C + c \\le 2$.\nTherefore, there are $9$ ways to take such number pair $(A, a)$, $5$ ways to take number pair $(B, b)$, and $5$ ways to take number pair $(C, c)$.\nAs a result, the number of the elements of $\\Omega$ is $|\\Omega| = 9 \\times 5 \\times 5 = 225$.\nAll the rational numbers satisfying the conditions are $\\Omega \\cap (0, 1)$. Notice that $r \\in \\Omega$ if and only if $\\frac{1}{r} \\in \\Omega$. In particular, $1 \\in \\Omega$. Therefore, the elements in $\\Omega \\setminus \\{1\\}$ can be matched into\n$$\n\\frac{1}{2}(|\\Omega| - 1) = 112\n$$\npairs according to the product of $1$, and each pair has exactly one number belonging to $(0, 1)$, that is, there is exactly one number satisfying the conditions. Thus, the number of desired rational numbers $r$ is $112$. $\\square$",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 28532,
    "subject": "Mathematics (Multi-modal)",
    "question": "Positive real numbers $x, y$ satisfy the following condition: there exist $a \\in [0, x], b \\in [0, y]$ such that\n$$\na^2 + y^2 = 2, \\quad b^2 + x^2 = 1, \\quad ax + by = 1.\n$$\nThen the maximum of $x+y$ is ______.",
    "options": [],
    "answer": "sqrt(5)",
    "solution": "In a plane rectangular coordinate system $xOy$, for positive real number pairs $(x, y)$ that satisfy the condition, take points $L(x, 0)$, $M(x, y)$, $N(0, y)$, and then quadrilateral $OLMN$ is a rectangle. Points $P, Q$ are on sides $LM, MN$, respectively, as shown in Fig. 8.1.\nSince $a^2 + y^2 = 2$, $b^2 + x^2 = 1$, $ax + by = 1$, we have\n$$\n\\begin{aligned}\n|OP| &= \\sqrt{x^2 + b^2} = 1, \\\\\n|OQ| &= \\sqrt{a^2 + y^2} = \\sqrt{2}, \\\\\n|PQ| &= \\sqrt{(a-x)^2 + (b-y)^2} \\\\\n&= \\sqrt{(a^2 + y^2) + (b^2 + x^2) - 2(ax + by)} = 1.\n\\end{aligned}\n$$\nThus, $\\triangle OPQ$ is an isosceles right triangle with $P$ as its right-angle vertex.\n\n![](attached_image_1.png)\n\nFig. 8.1\n\nTherefore, we can set $\\angle LOP = \\theta$, $\\angle QON = \\frac{\\pi}{4} - \\theta$, where $0 \\le \\theta \\le \\frac{\\pi}{4}$.\nThen\n$$\n\\begin{aligned}\nx + y &= |OL| + |ON| \\\\\n&= |OP| \\cdot \\cos \\angle LOP + |OQ| \\cdot \\cos \\angle QON \\\\\n&= \\cos \\theta + \\sqrt{2} \\cos \\left(\\frac{\\pi}{4} - \\theta\\right) = 2 \\cos \\theta + \\sin \\theta \\\\\n&= \\sqrt{5} \\sin(\\theta + \\varphi),\n\\end{aligned}\n$$\nwhere $\\varphi = \\arcsin \\frac{2\\sqrt{5}}{5}$.\nWhen $\\theta = \\frac{\\pi}{2} - \\varphi$ (correspondingly, $x = \\frac{2\\sqrt{5}}{5}$, $y = \\frac{3\\sqrt{5}}{5}$), $x + y$ takes the maximum $\\sqrt{5}$.\n\n$\\square$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 28533,
    "subject": "Mathematics (Multi-modal)",
    "question": "Given function $f(x) = |2 - \\log_3 x|$, positive real numbers $a, b, c$ satisfy $a < b < c$ and $f(a) = 2f(b) = 2f(c)$. Find the value of $\\frac{ac}{b}$.",
    "options": [],
    "answer": "9",
    "solution": "Note that $f(x) = |\\log_3(\\frac{x}{9})|$ is monotonically decreasing on $(0, 9]$ and monotonically increasing on $[9, +\\infty)$.\nBy the conditions satisfied by $a, b, c$, we know that $0 < a < b < 9 < c$ and\n$$\n\\log_3\\left(\\frac{9}{a}\\right) = 2\\log_3\\left(\\frac{9}{b}\\right) = 2\\log_3\\left(\\frac{c}{9}\\right).\n$$\nTherefore,\n$$\n\\begin{aligned}\n\\log_3\\left(\\frac{ac}{b}\\right) &= \\log_3\\left(9 \\cdot \\frac{a}{9} \\cdot \\frac{9}{b} \\cdot \\frac{c}{9}\\right) \\\\\n&= 2 - \\log_3\\left(\\frac{9}{a}\\right) + \\log_3\\left(\\frac{9}{b}\\right) + \\log_3\\left(\\frac{c}{9}\\right) \\\\\n&= 2,\n\\end{aligned}\n$$\nnamely, $\\frac{ac}{b} = 3^2 = 9$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 28534,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let complex sequence $\\{z_n\\}$ satisfy\n$$\nz_1 = \\frac{\\sqrt{3}}{2}, \\quad z_{n+1} = \\overline{z_n}(1 + z_n i) \\quad (n = 1, 2, \\dots),\n$$\nwhere $i$ is the imaginary unit. Find the value of $z_{2021}$.",
    "options": [],
    "answer": "z_{2021} = \\frac{\\sqrt{3}}{2} + \\left(\\frac{1}{2} + \\frac{1}{2^{2^{2020}}}\\right) i",
    "solution": "For $n \\in \\mathbb{N}_+$, let $z_n = a_n + b_n i$ ($a_n, b_n \\in \\mathbb{R}$). Then\n$$\n\\begin{aligned}\na_{n+1} + b_{n+1}i &= z_{n+1} = \\overline{z_n}(1 + z_n i) \\\\\n&= \\overline{z_n} + |z_n|^2 \\cdot i \\\\\n&= a_n - b_n i + (a_n^2 + b_n^2)i,\n\\end{aligned}\n$$\nand hence $a_{n+1} = a_n$, $b_{n+1} = a_n^2 + b_n^2 - b_n$.\nAnd by $z_1 = \\frac{\\sqrt{3}}{2}$ we know that $a_1 = \\frac{\\sqrt{3}}{2}$, $b_1 = 0$, so $a_n = \\frac{\\sqrt{3}}{2}$, and thus\n$$\nb_{n+1} = b_n^2 - b_n + \\frac{3}{4},\n$$\nnamely, $b_{n+1} - \\frac{1}{2} = b_n^2 - b_n + \\frac{1}{4} = \\left(b_n - \\frac{1}{2}\\right)^2$.\nHence, when $n \\ge 2$,\n$$\n\\begin{aligned}\nb_n &= \\frac{1}{2} + \\left(b_1 - \\frac{1}{2}\\right)^{2^{n-1}} \\\\\n&= \\frac{1}{2} + \\left(-\\frac{1}{2}\\right)^{2^{n-1}} \\\\\n&= \\frac{1}{2} + \\frac{1}{2^{2^{n-1}}}.\n\\end{aligned}\n$$\nConsequently,\n$$\nz_{2021} = a_{2021} + b_{2021}i = \\frac{\\sqrt{3}}{2} + \\left(\\frac{1}{2} + \\frac{1}{2^{2^{2020}}}\\right) i.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 28535,
    "subject": "Mathematics (Multi-modal)",
    "question": "Sequence $\\{a_n\\}$ satisfies $a_1 = a_2 = a_3$. Let\n$$\nb_n = a_n + a_{n+1} + a_{n+2} \\quad (n \\in \\mathbb{N}_+).\n$$\nIf $\\{b_n\\}$ is a geometric sequence with common ratio $3$, find the value of $a_{100}$.",
    "options": [],
    "answer": "(3^{100} + 10)/13",
    "solution": "By the condition, we know that $b_n = b_1 \\cdot 3^{n-1} = 3^n$ $(n \\in \\mathbb{N}_+)$.\nThus,\n$$\na_{n+3} - a_n = b_{n+1} - b_n = 3^{n+1} - 3^n = 2 \\cdot 3^n \\quad (n \\in \\mathbb{N}_+).\n$$\nTherefore,\n$$\n\\begin{align*}\na_{100} &= a_1 + \\sum_{k=1}^{33} (a_{3k+1} - a_{3k-2}) \\\\\n&= 1 + \\sum_{k=1}^{33} 2 \\cdot 3^{3k-2} \\\\\n&= 1 + 6 \\cdot \\frac{27^{33} - 1}{27 - 1} \\\\\n&= 1 + \\frac{3}{13}(3^{99} - 1) = \\frac{3^{100} + 10}{13}.\n\\end{align*}\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 28536,
    "subject": "Mathematics (Multi-modal)",
    "question": "Suppose the included angle between non-zero vectors $\\vec{a}$ and $\\vec{b}$ in the plane is $\\frac{\\pi}{3}$. If $|\\vec{a}|, |\\vec{b}|, |\\vec{a} + \\vec{b}|$ form arithmetic sequence in order, find the value of $|\\vec{a}| : |\\vec{b}| : |\\vec{a} + \\vec{b}|$.",
    "options": [],
    "answer": "3:5:7",
    "solution": "Denote $s = |\\vec{a}|$, $t = |\\vec{b}|$, and then $s, t > 0$. Note that the included angle between $\\vec{a}$ and $\\vec{b}$ is $\\frac{\\pi}{3}$, and we have\n$$\n\\begin{aligned}\n|\\vec{a} + \\vec{b}|^2 &= \\vec{a}^2 + (\\vec{b})^2 + 2\\vec{a} \\cdot \\vec{b} \\\\\n&= s^2 + t^2 + 2st \\cos \\frac{\\pi}{3} \\\\\n&= s^2 + t^2 + st.\n\\end{aligned}\n$$\nBy the condition, we know that $s, t, \\sqrt{s^2 + t^2 + st}$ form arithmetic sequence in order. Then there is $\\sqrt{s^2 + t^2 + st} = 2t - s$. Squaring and arranging the above equation gives $5st - 3t^2 = 0$. And since $t \\neq 0$, it follows that $5s = 3t$, i.e., $s : t = 3 : 5$.\n\nTherefore, $|\\vec{a}| : |\\vec{b}| : |\\vec{a} + \\vec{b}| = 3 : 5 : 7$.\n\n$\\boxed{}$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 28537,
    "subject": "Mathematics (Multi-modal)",
    "question": "It is known that $\\triangle ABC$ satisfies $AB = 1$, $AC = 2$ and $\\cos B + \\sin C = 1$. Find the length of side $BC$.",
    "options": [],
    "answer": "(3 + 2√21)/5",
    "solution": "Denote $a = BC$, $b = AC$, $c = AB$, so $b = 2$, $c = 1$. By the law of sines, we have $\\frac{\\sin B}{\\sin C} = \\frac{b}{c} = 2$, namely, $\\sin B = 2 \\sin C$. And since $\\cos B = 1 - \\sin C$, there is\n$$\n(2 \\sin C)^2 + (1 - \\sin C)^2 = \\sin^2 B + \\cos^2 B = 1,\n$$\nand simplifying it gives $5\\sin^2 C - 2\\sin C = 0$. And because $\\sin C \\neq 0$, $\\sin C = \\frac{2}{5}$.\nAnd then we have $\\cos B = 1 - \\sin C = \\frac{3}{5}$.\nBy the law of cosines, we have $\\cos B = \\frac{a^2 + c^2 - b^2}{2ac}$, and thus $\\frac{3}{5} = \\frac{a^2 - 3}{2a}$, i.e.,\n$$\na^2 - \\frac{6}{5}a - 3 = 0.\n$$\nSince $a > 0$, we get the solution $a = \\frac{3 + 2\\sqrt{21}}{5}$, namely the length of side $BC$ is $\\frac{3 + 2\\sqrt{21}}{5}$. $\\square$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 28538,
    "subject": "Mathematics (Multi-modal)",
    "question": "On a screen formed by $n \\times n$ ($n \\ge 2$) squares, every square displays initially one of the three colors: red, yellow, and blue. Every second, the screen changes the color of each square following the rules below:\n* for each square $A$ whose current color is red, if there is a yellow square sharing a side with this square, then $A$ turns yellow in the next second, otherwise, $A$ remains red;\n* for each square $B$ whose current color is yellow, if there is a blue square sharing a side with this square, then $B$ turns blue in the next second, otherwise, $B$ remains yellow;\n* for each square $C$ whose current color is blue, if there is a red square sharing a side with this square, then $C$ turns red in the next second, otherwise, $C$ remains blue.\nProve that if the screen does not change to a single color after $2n - 2$ seconds, then it will never change to a single color.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "**Proof:** We first prove the claim: if the screen eventually turns into one color, say blue, then there must be one square which is constantly blue.\nSuppose not, namely, suppose that eventually all squares are blue but every square has changed color at some time. We construct an oriented graph $G$ as follows: the vertices are all squares, and we link an oriented arrow $A \\to B$ if $A$ and $B$ are adjacent and there is some time $t_0$ such that square $A$ is not blue at the time $t_0 - 1$ and is constantly blue when $t \\ge t_0$, and square $B$ is blue at the time $t_0 - 1$. In other words, the square $A$ is changed to blue last time because of square $B$. Since we assume that every square on the screen has changed color, every vertex of the graph has out-degree $\\ge 1$. So $G$ must contain an oriented loop $A_1 \\to A_2 \\to \\dots \\to A_k \\to A_1$. Assume that square $A_i$ turns blue for the last time at the time $t_i$, i.e. $A_i$ is not blue at the time $t_i - 1$, and is constantly blue when $t \\ge t_i$. Without loss of generality, $t_1$ is the largest among $t_1, \\dots, t_k$. From the arrow $A_1 \\to A_2$, the square $A_i$ is yellow at time $t_1 - 1$; From the arrow $A_k \\to A_1$, the square $A_1$ is blue at the time $t_k - 1$. We must have $t_k < t_1$. But a blue square can only turn red first before turning yellow. So there exists $t_k < t' < t_1$, such that $A_1$ is red at the time $t' - 1$. But $t' - 1 \\ge t_k$, $A_k$ is blue at the time $t' - 1$, so $A_1$ will force $A_k$ to turn red at the time $t'$. But this contradicts the definition of $t_k$, namely, $A_k$ is constantly blue from the time $t_k$ onward. The claim is proved.\n\nBack to the original problem, assume that all squares are turned blue eventually, we will show that the squares are all turned blue after $2n - 2$ seconds.\nBy the claim, there exists a square $A$ which is constantly blue. Define the distance between two squares by the sum of their horizontal and vertical distances; then the distance between farthest two squares is $2n - 2$. If $B$ and $A$ have distance 1, then $B$ can never be red, this means: if $B$ is initially blue, then it is always blue; if $B$ is yellow, then\n\nit is turned blue after 1 second and continues to be blue after that. Now, for a point $B$ of distant $k$ to the point $A$, we may run an induction to prove that $B$ will become constantly blue after $k$ seconds. Suppose that this statement has been proved for $k-1$. If $B$ has distant $k$ to $A$, take a square $B'$ adjacent to $B$ that has distant $k-1$ to $A$. By induction, $B'$ is constant blue after $k-1$ seconds, then the same argument as above shows that $B$ is constantly blue after $k$ seconds. From this induction, we deduce that all squares are blue after $2n-2$ seconds.",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 28539,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $\\Gamma_2$ be a circle contained in the interior of another circle $\\Gamma_1$ on the plane. Prove that there exists a point $P$ on the plane satisfying the following conditions: if $\\ell$ is a line that does not contain $P$, that intersects $\\Gamma_1$ at two different points $A, B$, and that intersects $\\Gamma_2$ at two different points $C, D$ (so that $A, C, D, B$ lie in order on $\\ell$), then $\\angle APC = \\angle DPB$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "![](attached_image_1.png)\n\n**Proof.** Denote the centers of the two circles by $O_1, O_2$, and the radii by $r_1, r_2$, respectively, where $r_1 > r_2$. We first show that there are two points $P, Q$ on the ray $O_1O_2$ such that $O_1P \\cdot O_1Q = r_1^2$ and $O_2P \\cdot O_2Q = r_2^2$.\nOne can choose a point $K$ on the ray $O_1O_2$ such that $O_1K = \\frac{O_1O_2^2 + r_1^2 - r_2^2}{2O_1O_2}$. Since $O_1O_2 \\le r_1 - r_2$, we have $O_1K \\ge r_1$. We choose two points $P, Q$ on the line $O_1O_2$ such that $KP = KQ = \\sqrt{O_1K^2 - r_1^2}$ so that $O_1P \\cdot O_1Q = r_1^2$. On the other hand,\n$$\n\\begin{aligned}\nO_2P \\cdot O_2Q - O_1P \\cdot O_1Q &= O_2K^2 - O_1K^2 = (O_1K - O_1O_2)^2 - O_1K^2 \\\\\n&= O_1O_2^2 - 2O_1O_2 \\cdot O_1K = r_2^2 - r_1^2,\n\\end{aligned}\n$$\nwhich implies that $O_2P \\cdot O_2Q = r_2^2$.\nFor any given line $\\ell$ as in the problem, if it is perpendicular to $O_1O_2$, we certainly have $\\angle APC = \\angle DPB$ by symmetry. If not, from $O_2C^2 = O_2Q \\cdot O_2P$ we know that $\\triangle O_2CQ \\sim \\triangle O_2PC$. So $\\frac{CQ}{CP} = \\frac{r_2}{O_2P}$. Similarly, we obtain $\\frac{DQ}{DP} = \\frac{r_2}{O_2P}$, and therefore $\\frac{CQ}{CP} = \\frac{DQ}{DP}$, which means that the bisectors of $\\angle CPD$ and $\\angle CQD$ meet $\\ell$ at the same point, say $M$. A similar argument shows that the bisectors of $\\angle APB$ and $\\angle AQB$ intersect $\\ell$ at the same point again, say $M'$.\n\nNote that the points $P, Q, M$ are all on an Apollonian circle with distance ratio $\\frac{CQ}{DQ}$ to $C$ and $D$. The center of the Apollonian circle must be on $\\ell$. Thus this center must be the intersection of $\\ell$ with the perpendicular bisector of $PQ$. Similarly, the points $P, Q, M'$ are on the Apollonius circle with distance ratio $\\frac{AQ}{BQ}$ to $A$ and $B$, whose center is the same point as above. Note that $K$ does not lie in the interior of $\\Gamma_1$, so the center of this circle, denoted by $L$, must be outside $\\Gamma_1$ so that $M$ coincides with $M'$. Therefore,\n$$\n\\angle APC = \\angle APM - \\angle CPM = \\angle BPM - \\angle DPM = \\angle BPD.\n$$\nThis completes the proof.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 28540,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $\\{a_n\\}$ and $\\{b_n\\}$ be two sequences of positive real numbers such that, for any positive integer $n$,\n$$\na_{n+1} = a_n - \\frac{1}{1 + \\sum_{i=1}^{n} \\frac{1}{a_i}}, \\quad \\text{and} \\quad b_{n+1} = b_n + \\frac{1}{1 + \\sum_{i=1}^{n} \\frac{1}{b_i}}.\n$$\n(1)\nIf $a_{100}b_{100} = a_{101}b_{101}$, find the value of $a_1 - b_1$.\n(2)\nIf $a_{100} = b_{99}$, which one of $a_{100} + b_{100}$ and $a_{101} + b_{101}$ is larger?",
    "options": [],
    "answer": "Part (1): 199; Part (2): a_100 + b_100 is larger than a_101 + b_101.",
    "solution": "(1) Set $a_1 = a$ and $b_1 = b$ ($a, b > 0$). The recurrence formula of $a_n$ implies that, for any positive integer $n \\ge 2$, we have\n$$\n\\frac{1}{a_n - a_{n+1}} = 1 + \\sum_{i=1}^{n} \\frac{1}{a_i} \\implies \\frac{1}{a_n - a_{n+1}} = \\frac{1}{a_{n-1} - a_n} + \\frac{1}{a_n}.\n$$\nThat is, $\\frac{a_n}{a_n - a_{n-1}} = \\frac{a_n}{a_{n-1} - a_n} + 1 = \\frac{a_{n-1}}{a_{n-1} - a_n}$. From this, we obtain $\\frac{a_{n+1}}{a_n} = \\frac{a_n}{a_{n-1}}$.\nSo $\\{a_n\\}$ is a geometric sequence with common ratio $\\frac{a}{a+1}$, and the general term is\n$$a_n = \\frac{a^n}{(a+1)^{n-1}}.$$\nSimilarly, the recurrence formula for $b_n$ implies that\n$$\n\\frac{1}{b_{n+1} - b_n} = 1 + \\sum_{i=1}^{n} \\frac{1}{b_i} = \\frac{1}{b_n - b_{n-1}} + \\frac{1}{b_n}.\n$$\nThus, $\\frac{b_n}{b_{n+1} - b_n} = \\frac{b_n}{b_{n-1} - b_n} + 1$. By induction, we have $\\frac{b_n}{b_{n+1} - b_n} = \\frac{b_1}{b_2 - b_1} + 2(n-1) = b + 2n - 1$. Therefore, $\\frac{b_{n+1}}{b_n} = \\frac{b + 2n}{b + 2n - 1}$.\nFrom this, we get $b_n = b \\prod_{k=1}^{n-1} \\frac{b+2k}{b+2k-1}$.\nThe condition $a_{100}b_{100} = a_{101}b_{101}$ in (1) implies that $\\frac{a_{100}}{a_{101}} = \\frac{b_{101}}{b_{100}}$, i.e. $\\frac{a+1}{a} = \\frac{b+200}{b+199}$. So $a = b + 199$ and thus $a_1 - b_1 = a - b = 199$.\n\n(2) **Method 1.** It is clear that $\\{a_n\\}$ is monotonically decreasing and $\\{b_n\\}$ is monotonically increasing. Together with the condition $a_{100} = b_{99}$, we deduce that\n(*) $a_1 > a_2 > \\dots > a_{99} > a_{100} = b_{99} > b_{98} > \\dots > b_1 > 0$.\nIn turn, we deduce that\n$$\na_{99} = a_{100} + \\frac{1}{1 + \\sum_{i=1}^{99} \\frac{1}{a_i}} > b_{99} + \\frac{1}{1 + \\sum_{i=1}^{99} \\frac{1}{b_i}} = b_{100}.\n$$\n\n**Method 2.** Since $a_{100} = b_{99}$, we have\n$$\n(**) \\qquad \\frac{a^{100}}{(a+1)^{99}} = \\frac{b(b+2)\\cdots(b+196)}{(b+1)(b+3)\\cdots(b+195)}.\n$$\nTo prove $a_{100} + b_{100} > a_{101} + b_{101}$, it is equivalent to prove that\n$$\n\\begin{aligned}\na_{100} - a_{101} &> b_{101} - b_{100} \\\\\n\\Longleftrightarrow \\frac{a_{100} - a_{101}}{a_{100}} &> \\frac{b_{101} - b_{100}}{b_{99}} \\\\\n\\Longleftrightarrow 1 - \\frac{a}{a+1} &> \\frac{(b+198)(b+200)}{(b+197)(b+199)} - \\frac{b+198}{b+197} \\\\\n\\Longleftrightarrow \\frac{1}{a+1} &> \\frac{b+198}{(b+197)(b+199)} \\\\\n(*) & \\Longleftrightarrow a < b+197 - \\frac{1}{b+198}.\n\\end{aligned}\n$$\nWe prove this by contradiction. Suppose that $a \\ge b + 197 - \\frac{1}{b+198}$, in particular, $a > b + 196$. We have\n$$\n\\begin{aligned}\n\\frac{a^2}{a+1} - (b+196) &= \\frac{a(a-b-196) - (b+196)}{a+1} \\\\\n&\\ge \\frac{1}{a+1} \\left( \\left( b+197 - \\frac{1}{b+198} \\right) \\left( 1 - \\frac{1}{b+198} \\right) - (b+196) \\right) \\\\\n&> \\frac{1}{a+1} \\left( \\left( b+197 - \\frac{1}{b+197} \\right) \\frac{b+197}{b+198} - (b+196) \\right) = 0.\n\\end{aligned}\n$$\nBut note that $\\frac{a}{a+1} > \\frac{b+194}{b+195} > \\frac{b+192}{b+193} > \\cdots > \\frac{b}{b+1}$, we have\n$$\n\\frac{a^{100}}{(a+1)^{99}} = \\left(\\frac{a}{a+1}\\right)^{98} \\frac{a^2}{a+1} > \\frac{b}{b+1} \\cdots \\frac{b+194}{b+195} (b+196),\n$$\ncontradicting with (**)! So (***) holds, and thus $a_{100} + b_{100} > a_{101} + b_{101}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 28541,
    "subject": "Mathematics (Multi-modal)",
    "question": "As illustrated in Fig. 1.1, $ABCDEF$ is a cyclic hexagon. The extensions of $AB$ and $DC$ meet at $G$; the extensions of $AF$ and $DE$ meet at $H$. Let $M$ and $N$ be the circumcentres of $\\triangle BCG$ and $\\triangle EFH$, respectively. Prove that the lines $BE$, $CF$, and $MN$ are concurrent.\n\n![](attached_image_1.png)\nFig. 1.1",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $\\omega$ be the circumcircle of $\\triangle BCG$. Let $E'$ be the other intersection of the line $BE$ and $\\omega$, $F'$ be the other intersection of the line $CF$ and $\\omega$, as shown in Fig. 1.2. Note that $\\angle BE'F' = \\angle BCF' = \\angle BCF = \\angle BEF$, and hence $EF \\parallel E'F'$; similarly, $\\angle CF'G = \\angle CBG = \\angle CFA$, and hence $GF' \\parallel HF$; moreover, $GE' \\parallel HE$. It follows that $\\triangle E'F'G$ and $\\triangle EFH$ are homothetic triangles. Let $P$ be the homothetic centre. Then $EE'$ and $FF'$ pass through $P$; the line connecting the circumcentres of $\\triangle E'F'G$ and $\\triangle EFH$, which is $MN$, passes through $P$ as well. Therefore, the lines $BE$, $CF$, and $MN$ meet at $P$. $\\square$\n\n![](attached_image_2.png)\nFig. 1.2\n(Edited from Wang Zihan's proof.) As in Fig. 1.3, we first apply Pascal's theorem to the cyclic hexagon $ABCDEF$ to deduce that $G$, $P$, $H$ are collinear. Let $A'$ and $D'$ be the antipodal points of $A$ and $D$, respectively. Let $BA'$ and $CD'$ meet at $G'$, $D'E$ and $A'F$ meet at $H'$. Apply Pascal's theorem to the cyclic hexagon $BA'FCD'E$ to deduce that $G'$, $H'$, $P$ are collinear.\n\nSince $A$ and $A'$, $D$ and $D'$ are antipodal points, we have $A'B \\perp AB$, $A'F \\perp AF$, $CD' \\perp CD$, $D'E \\perp DE$. Notice that $\\angle AHH' = 90^\\circ - \\angle FH'H = 90^\\circ - \\angle FAD$. Thus, the angle between $AH$ and $GG'$ is given by\n$$\n\\begin{align*}\n180^\\circ - \\angle AGG' - \\angle BAF &= 90^\\circ - \\angle BAF + \\angle BG'G \\\\\n&= 90^\\circ - \\angle BAF + \\angle DAB \\\\\n&= 90^\\circ - \\angle DAF.\n\\end{align*}\n$$\nThis implies $HH' \\parallel GG'$. As $M$ is the midpoint of $GG'$, $N$ is the midpoint of $HH'$, and furthermore it is shown that $GH$, $G'H'$ pass through $P$. We infer that $MN$ passes through $P$, too. Therefore, $BE$, $CF$, and $MN$ are concurrent. $\\square$\n\n![](attached_image_3.png)\nFig. 1.3\nAs shown in Fig. 1.4, let $P$ be the intersection of $BE$ and $CF$. It suffices to show that $M$, $P$, $N$ are collinear, or equivalently,\n$$\n\\frac{\\sin \\angle MPB}{\\sin \\angle MPC} = \\frac{\\sin \\angle NPE}{\\sin \\angle NPF}.\n$$\nDenote $\\alpha = \\angle MPB$, $\\beta = \\angle NPE$, $\\gamma = \\angle BPC = \\angle FPE$. The above relation is equivalent to\n$$\n\\sin \\alpha \\sin(\\gamma - \\beta) = \\sin \\beta \\sin(\\gamma - \\alpha),\n$$\nor rewritten as $\\sin \\gamma \\sin(\\alpha - \\beta) = 0$, or just $\\alpha = \\beta$ (for the collinearity of $M$, $P$, $N$).\n\nIn $\\triangle MPC$ and $\\triangle MPB$, apply the law of sines to obtain\n$$\n\\frac{BM}{\\sin \\angle MPB} = \\frac{PM}{\\sin \\angle PBM}, \\quad \\frac{CM}{\\sin \\angle MPC} = \\frac{PM}{\\sin \\angle PCM}.\n$$\nAs $BM = MC$, it follows that\n$$\n\\frac{\\sin \\angle MPB}{\\sin \\angle MPC} = \\frac{\\sin \\angle PBM}{\\sin \\angle PCM}.\n$$\nSimilarly, in $\\triangle NPF$ and $\\triangle NPE$, we have\n$$\n\\frac{\\sin \\angle NPE}{\\sin \\angle NPF} = \\frac{\\sin \\angle NEP}{\\sin \\angle NFP}.\n$$\nNow it suffices to show\n$$\n\\frac{\\sin \\angle PBM}{\\sin \\angle PCM} = \\frac{\\sin \\angle NEP}{\\sin \\angle NFP}. \\qquad \\textcircled{1}\n$$\n\nThe angles appearing in the numerators are supplementary, as\n$$\n\\begin{align*}\n\\angle PBM + \\angle NEP &= \\angle PBC + \\angle MBC + \\angle PEF + \\angle NFE \\\\\n&= (180^\\circ - \\angle CDE) + (90^\\circ - \\angle BGC) \\\\\n&\\quad + (180^\\circ - \\angle BAF) + (90^\\circ - \\angle AHD) \\\\\n&= 540^\\circ - (\\angle CDE + \\angle BGC + \\angle BAF + \\angle AHD) \\\\\n&= 540^\\circ - 360^\\circ = 180^\\circ,\n\\end{align*}\n$$\nand thus $\\sin \\angle PBM = \\sin \\angle NEP$. Analogously, $\\sin \\angle PCM = \\sin \\angle NFP$. We have verified (1), and the statement. $\\square$\n\n![](attached_image_4.png)\nFig. 1.4\nAssume that the circumcircle of $\\triangle ABC$ is the unit circle in the complex plane. We use lowercase letters to represent the complex numbers corresponding to the points in the plane (for instance, $a$ is the complex number for the point $A$, and so on). Let $BE \\cap CF = X$. We have\n$$\n\\begin{align*}\nx \\in CF &\\Leftrightarrow \\frac{x-c}{x-f} = \\overline{\\left(\\frac{x-c}{x-f}\\right)} \\\\\n&\\Leftrightarrow x\\bar{x} - \\bar{f}x - c\\bar{x} + c\\bar{f} = x\\bar{x} - c\\bar{x} - f\\bar{x} + c\\bar{f} \\\\\n&\\Leftrightarrow (c-f)\\bar{x} - (\\bar{c} - \\bar{f})x = c\\bar{f} - c\\bar{f} \\\\\n&\\Leftrightarrow \\bar{x} + c\\bar{f}\\bar{x} = \\bar{c} + \\bar{f}.\n\\end{align*}\n$$\nIn the same manner, $\\bar{x} + \\bar{b}\\bar{e}\\bar{x} = \\bar{b} + \\bar{e}$. Thus,\n$$\nx = \\frac{\\bar{b} + \\bar{e} - \\bar{c} - \\bar{f}}{\\bar{b}\\bar{e} - \\bar{c}\\bar{f}}. \\qquad \\textcircled{1}\n$$\nOn the other hand, $\\triangle BMC$ is an isosceles triangle, $\\angle BMC$ is twice the angle between the lines $CD$ and $AB$, and this implies that\n$$\n\\frac{c-m}{b-m} = \\left(\\frac{d-c}{a-b}\\right)^2 / \\left|\\frac{d-c}{a-b}\\right|^2 = \\frac{d-c}{a-b} / \\frac{\\bar{d}-\\bar{c}}{\\bar{a}-\\bar{b}} = \\frac{cd}{ab}.\n$$\nConsequently,\n$$\nm = \\frac{bc(a-d)}{ad-bc} = \\frac{\\bar{d}-\\bar{a}}{\\bar{c}\\bar{d}-\\bar{a}\\bar{b}}. \\qquad \\textcircled{2}\n$$\nIn the same way, we can derive\n$$\nn = \\frac{(a-d)ef}{af-be} = \\frac{\\bar{d}-\\bar{a}}{\\bar{d}\\bar{e}-\\bar{a}\\bar{f}}. \\qquad \\textcircled{3}\n$$\nTo prove that $BE$, $CF$, and $MN$ are concurrent, it is equivalent to prove that $M$, $X$, $N$ are collinear, namely\n$$\n\\frac{x-n}{x-m} = \\left( \\frac{x-n}{x-m} \\right) \\in \\mathbb{R}. \\qquad (4)\n$$\nFrom (1)–(3), it follows that\n$$\n\\begin{align*}\nx - n &= \\frac{1}{(\\bar{b}\\bar{e} - \\bar{c}\\bar{f})(\\bar{d}\\bar{e} - \\bar{a}\\bar{f})} [ (\\bar{b} + \\bar{e} - \\bar{c} - \\bar{f})(\\bar{d}\\bar{e} - \\bar{a}\\bar{f}) - (\\bar{d} - \\bar{a})(\\bar{b}\\bar{e} - \\bar{c}\\bar{f}) ] \\\\\n&= \\frac{1}{(\\bar{b}\\bar{e} - \\bar{c}\\bar{f})(\\bar{d}\\bar{e} - \\bar{c}\\bar{f})} [ \\bar{b}\\bar{d}\\bar{e} + \\bar{d}\\bar{e}^2 - \\bar{d}\\bar{e}\\bar{f} - \\bar{c}\\bar{d}\\bar{e} - \\bar{a}\\bar{b}\\bar{f} - \\bar{a}\\bar{e}\\bar{f} + \\bar{a}\\bar{f}^2 \\\\\n&\\quad + \\bar{a}\\bar{c}\\bar{f} - \\bar{b}\\bar{d}\\bar{e} + \\bar{a}\\bar{b}\\bar{e} + \\bar{c}\\bar{d}\\bar{f} - \\bar{a}\\bar{c}\\bar{f} ] \\\\\n&= \\frac{1}{(\\bar{b}\\bar{e} - \\bar{c}\\bar{f})(\\bar{d}\\bar{e} - \\bar{a}\\bar{f})} (\\bar{e} - \\bar{f})(\\bar{a}\\bar{b} - \\bar{a}\\bar{f} - \\bar{c}\\bar{d} + \\bar{d}\\bar{e}),\n\\end{align*}\n$$\nwhile\n$$\nm - n = \\frac{\\bar{d} - \\bar{a}}{(\\bar{c}\\bar{d} - \\bar{a}\\bar{b})(\\bar{d}\\bar{e} - \\bar{a}\\bar{f})}(\\bar{a}\\bar{b} - \\bar{a}\\bar{f} - \\bar{c}\\bar{d} + \\bar{d}\\bar{e}).\n$$\nHence,\n$$\n\\frac{x - n}{x - m} = \\frac{(\\bar{e} - \\bar{f})(\\bar{c}\\bar{d} - \\bar{a}\\bar{b})}{(\\bar{d} - \\bar{a})(\\bar{b}\\bar{e} - \\bar{c}\\bar{f})}.\n$$\nIn the last fraction, the unit complex numbers $a, b, c, d, e$, and $f$ all appear exactly once in the numerator and in the denominator; moreover, in each parenthesis, the coefficients of the homogeneous terms sum to zero (one minus one). So, the fraction is real, which justifies (4), and the argument as well.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 28542,
    "subject": "Mathematics (Multi-modal)",
    "question": "Consider all sequences of real numbers $x_0, x_1, x_2, \\dots, x_{100}$ satisfying the following conditions:\n(1) $x_0 = 0$;\n(2) For any integer $i$, $1 \\le i \\le 100$, $1 \\le x_i - x_{i-1} \\le 2$ holds.\nFind the largest positive integer $k \\le 100$, such that\n$$\nx_k + x_{k+1} + \\dots + x_{100} \\ge x_0 + x_1 + \\dots + x_{k-1}\n$$\nholds for every such sequence $x_0, x_1, x_2, \\dots, x_{100}$.",
    "options": [],
    "answer": "67",
    "solution": "The answer is $67$.\n\nOn one hand, if $x_i = 2i$, $1 \\le i \\le 34$, $x_{34+j} = x_{34} + j = 68 + j$, $1 \\le j \\le 66$, then the sequence satisfies the conditions of the problem. We have\n$$\n\\sum_{j=68}^{100} x_j - \\sum_{i=0}^{67} x_i = \\sum_{j=1}^{33} (x_{67+j} - x_{34+j}) - \\sum_{i=1}^{34} x_i = 33^2 - 34 \\times 35 < 0.\n$$\nThis example shows that when $k \\ge 68$, it does not satisfy the requirements.\n\nOn the other hand, for any sequence $x_1, x_2, \\dots, x_{100}$ that satisfies the conditions of the problem, it is easy to know that $x_i \\le 2i$, $1 \\le i \\le 100$. For $0 \\le s < t \\le 100$, there is the inequality $x_t - x_s \\ge t - s$. Therefore,\n$$\n\\begin{aligned}\n\\sum_{j=67}^{100} x_j - \\sum_{i=0}^{66} x_i &= \\sum_{j=1}^{34} (x_{66+j} - x_{32+j}) - \\sum_{i=1}^{32} x_i \\\\\n&\\ge 34^2 - \\sum_{i=1}^{32} 2i \\\\\n&= 34^2 - 32 \\times 33 > 0.\n\\end{aligned}\n$$\nIn conclusion, the largest $k$ we seek is $67$. □",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 28543,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $n$ be a positive integer. There are $3n$ women's volleyball teams attending a tournament. Each pair of teams play at most once (there are no ties in volleyball games). Assume that a total number of $3n^2$ games have been played. Prove that there exists a team whose number of winning games and number of losing games are both greater than or equal to $\\frac{n}{4}$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "*Proof.* We prove by contradiction: assume the conclusion is false. Suppose that the teams $P_1, \\cdots, P_k$ all have won less than $\\frac{n}{4}$ games, while the remaining $3n-k$ teams, $Q_1, \\cdots, Q_{3n-k}$ all have won at least $\\frac{n}{4}$ games. Based on the assumption of proof by contradiction, the number of defeats for $Q_1, \\cdots, Q_{3n-k}$ is also less than $\\frac{n}{4}$. Since the total number of victories for all teams is $3n^2$, there must be at least one team with no fewer than $n$ victories, implying that $3n-k > 0$.\nThe number of matches between teams $P_1, \\cdots, P_k$ cannot exceed the sum of their victories, so it is not more than $\\frac{nk}{4}$. In a similar fashion, the number of matches between teams $Q_1, \\cdots, Q_{3n-k}$ cannot exceed the sum of their defeats, so it is less than $\\frac{n(3n-k)}{4}$ (using the fact that $3n-k > 0$).\nThe number of matches between the two sets $P_1, \\cdots, P_k$ and $Q_1, \\cdots, Q_{3n-k}$ is at most $k(3n-k)$ games. Consequently, the total number of matches $N$ satisfies:\n$$\nN < \\frac{nk}{4} + \\frac{n(3n-k)}{4} + k(3n-k) \\le \\frac{3n^2}{4} + \\frac{9n^2}{4} = 3n^2,\n$$\nwhich contradicts the given condition. Therefore, the assumption of proof by contradiction is incorrect, and the original proposition is true. $\\square$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 28544,
    "subject": "Mathematics (Multi-modal)",
    "question": "If two real numbers $\\alpha, \\beta$ satisfy that $\\lfloor k_1\\alpha \\rfloor \\neq \\lfloor k_2\\beta \\rfloor$ for all positive integers $k_1, k_2$, where $\\lfloor x \\rfloor$ denotes the maximal integer not exceeding $x$.\nProve that there exist two positive integers $m_1, m_2$ such that $\\frac{m_1}{\\alpha} + \\frac{m_2}{\\beta} = 1$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "First note that $\\frac{\\beta}{\\alpha}$ is an irrational number (otherwise there are positive integers $k_1, k_2$ such that $k_1\\alpha = k_2\\beta$, which leads to a contradiction).\nIf $\\alpha = \\frac{q}{p}$ is a rational number, then there is a positive integer $k_2$ such that the decimal part of $\\frac{k_2\\beta}{q}$ is less than $\\frac{1}{q}$. Taking $k_1 = p\\lfloor\\frac{k_2\\beta}{q}\\rfloor$ gives that\n$$\nk_1\\alpha = q \\cdot \\lfloor\\frac{k_2\\beta}{q}\\rfloor < k_2\\beta < q\\lfloor\\frac{k_2\\beta}{q}\\rfloor + 1,\n$$\nwhich contradicts to the assumption. Thus $\\alpha$ must be irrational. Similarly, $\\beta$ must be irrational as well.\n\nA pair of positive integers $(a, b)$ is called *distinguished* if $0 < b\\beta - a\\alpha < 1$. By our assumption, there is a unique positive integer $t$ such that $a\\alpha < t < b\\beta$. Denoting $u = t - a\\alpha, v = b\\beta - t$, we say a distinguished pair $(a, b)$ corresponds to the *intermediate number* $t$ together with a *distant pair* $(u, v)$. We will prove the following.\n\n**Lemma.** Assume two distinguished pairs $(a_1, b_1)$, $(a_2, b_2)$ correspond to distant pairs $(u_1, v_1)$, $(u_2, v_2)$, respectively. Then $\\frac{u_1}{v_1} = \\frac{u_2}{v_2}$.\n\n**Proof of Lemma.** Suppose that the intermediate numbers of these two distinguished pairs are $t_1, t_2$, respectively. Assume for the sake of contradiction that $\\frac{u_1}{v_1} > \\frac{u_2}{v_2}$. We also take $\\epsilon = \\frac{u_1v_2 - u_2v_1}{2} > 0$.\nSince $\\frac{\\beta}{\\alpha}$ is an irrational number, there are positive integers $a_0, b_0$ such that $0 < a_0\\alpha - b_0\\beta < \\epsilon$. On the other hand, our assumption implies that there is another integer $t_0$ satisfying $b_0\\beta < t_0 < a_0\\alpha$. Denote $u_0 = a_0\\alpha - t_0$ and $v_0 = t_0 - b_0\\beta$. If $\\frac{u_1}{u_0} - \\frac{v_1}{v_0} > 1$, then one can take $L = \\lfloor\\frac{v_1}{v_0}\\rfloor + 1$ such that $\\frac{u_1}{u_0} > L > \\frac{v_1}{v_0}$, i.e.,\n$$\n\\begin{aligned}\nu_1 - Lu_0 &= (t_1 + Lt_0) - (a_1 + La_0)\\alpha > 0, \\\\\nv_1 - Lv_0 &= (t_1 + Lt_0) - (b_1 + Lb_0)\\beta < 0.\n\\end{aligned}\n$$\nSet $k_1 = a_1 + La_0$ and $k_2 = b_1 + Lb_0$. Then $\\lfloor k_1\\alpha \\rfloor = \\lfloor k_2\\beta \\rfloor = t_1 + Lt_0 - 1$, which leads to a contradiction. This further implies that $\\frac{u_1}{u_0} - \\frac{v_1}{v_0} \\le 1$. Using a similar argument, we can show that $\\frac{u_2}{u_0} - \\frac{v_2}{v_0} \\ge -1$. Therefore,\n$$\nu_1 - \\frac{u_0}{v_0}v_1 \\le u_0, \\quad u_2 - \\frac{u_0}{v_0}v_2 \\ge -u_0, \\quad \\Rightarrow \\quad u_1v_2 - u_2v_1 \\le u_0(v_1 + v_2) < 2\\epsilon.\n$$\nHowever, this violates our choice of $\\epsilon$. The lemma is proved.\n\nNow we see all distinguished pairs $(a, b)$ share the same ratio $\\frac{u}{v} = \\frac{t - a\\alpha}{b\\beta - t}$. Denote this common $\\lambda = \\frac{v}{u+v} \\in (0, 1)$; then we have an equality $\\lambda \\cdot a\\alpha + (1 - \\lambda) \\cdot b\\beta = t \\in \\mathbb{Z}$. We call linear combinations of distinguished pairs with integer coefficients *nice pairs*. For each nice pair $(c, d)$, it is not hard to see that\n$$\n\\lambda \\cdot c\\alpha + (1 - \\lambda) \\cdot d\\beta = c \\cdot \\lambda\\alpha + d \\cdot (1 - \\lambda)\\beta \\in \\mathbb{Z}.\n$$\nTake a nice pair $(a, b)$ and set $\\delta = b\\beta - a\\alpha \\in (0, 1)$. We claim that for all $M > 0$, any pair $(c, d)$ of integers satisfying $0 < d\\beta - c\\alpha < M$ and $c > \\frac{a}{\\delta}M$, is nice. Indeed, it suffices to take $L = \\lfloor \\frac{d\\beta - c\\alpha}{\\delta} \\rfloor < \\frac{M}{\\delta} < \\frac{c}{a}$ so that $(d - Lb)\\beta - (c - La)\\alpha = (d\\beta - c\\alpha) - L\\delta \\in (0, \\delta) \\subset (0, 1)$, and therefore the pair $(c - La, d - Lb)$ must be distinguished and the pair $(c, d) = (c - La, d - Lb) + L \\cdot (a, b)$ must be a nice pair.\n\nNow take $M = \\alpha + 2\\beta$, and integers $c_0 > \\frac{a}{\\delta}M + 1$ and $d_0 = \\lfloor \\frac{c_0\\alpha}{\\beta} \\rfloor$ satisfying the condition $0 < d_0\\beta - c_0\\alpha < \\beta$. Using the analysis above, $(c_0, d_0)$, $(c_0, d_0 + 1)$, and $(c_0 - 1, d_0)$ are nice pairs, and their linear combinations $(0, 1)$ and $(1, 0)$ are nice pairs as well. This implies in particular that $\\lambda\\alpha$ and $(1 - \\lambda)\\beta$ are (positive) integers. Taking positive integers $m_2 = \\lambda\\alpha$ and $m_1 = (1 - \\lambda)\\beta$, we get $\\frac{m_2}{\\alpha} + \\frac{m_1}{\\beta} = \\lambda + (1 - \\lambda) = 1$, or equivalently $m_1\\alpha + m_2\\beta = \\alpha\\beta$. We conclude the proof.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 28545,
    "subject": "Mathematics (Multi-modal)",
    "question": "Given a prime number $p$ and an infinite set $A \\subset \\mathbb{Z}$. Show that one can always find a subset $B$ of $A$, $B$ contains $2p-2$ elements, and for any $p$ distinct elements of $B$, their arithmetic mean does not belong to $A$.\n\n(Contributed by Fu Yunhao)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Assume that, on the contrary, for some infinite set $A \\subset \\mathbb{Z}$, one cannot find a $(2p-2)$-element subset $B$ satisfying the problem conditions. Without loss of generality, assume that $A$ contains infinitely many positive integers; otherwise take $-A$ instead. Notice that the arithmetic mean of any $p$ distinct elements of $B$ must fall into the interval between the smallest and the largest elements of $B$. Thus, removing all elements smaller than some integer $N$ from $A$, the set is still a counterexample to the problem. In particular, we take $N = 1$ and thereby assume all elements of $A$ are positive. We may also add a constant integer to all elements of $A$ if necessary.\n\n**Lemma** For any nonnegative integer $k$, there is at most one residue class modulo $p^k$ that contains infinitely many elements of $A$.\n\n**Proof of lemma** Suppose the lemma is untrue and $k$ is the smallest nonnegative integer such that two residue classes modulo $p^k$ contain infinitely many elements of $A$, say with residues $r_k$ and $r'_k$. Then there exists a positive integer $N$ such that all elements of $A_{\\ge N} := \\{a \\in A \\mid a \\ge N\\}$ are congruent to $r_{k-1}$ modulo $p^{k-1}$. Furthermore, in $A_{\\ge N}$, those congruent to $r_k$ and $r'_k$ modulo $p^k$ are both infinite. Take $p-1$ elements of $A_{\\ge N}$ that are congruent to $r_k$, and the same number of elements congruent to $r'_k$ (mod $p^k$). Evidently, the arithmetic mean of any $p$ elements of $B$ is not congruent to $r_{k-1}$ modulo $p^{k-1}$. This contradicts the assumption, and the lemma is verified.\n\nReturn to the original problem. For nonnegative integer $k$, assume the only residue class modulo $p^k$ that contains infinitely many elements of $A$ is $r_k \\pmod{p^k}$ ($0 \\le r_k \\le p^k - 1$); then $r_k \\equiv r_{k-1} \\pmod{p^{k-1}}$. Now we use induction to select a sequence $b_1, b_2, \\dots$, of elements of $A$, and verify that $2p-2$ of them constitute the desired subset $B$. Denote $N_0 = 0$. Let $k_1$ be the smallest nonnegative integer such that some element $a_1 \\in A$ satisfies $a_1 \\not\\equiv r_{k_1} \\pmod{p^{k_1}}$. Then all elements of $A$ are congruent to $r_{k_1-1}$ modulo $p^{k_1-1}$. According to the lemma, there exists a positive integer $N_1$ such that all elements of $A_{\\ge N_1} = \\{a \\in A \\mid a \\ge N_1\\}$ are congruent to $r_{k_1}$ modulo $p^{k_1}$. Let $k_2$ be the smallest nonnegative integer such that some element $a_2 \\in A$ satisfies $a_2 \\not\\equiv r_{k_2} \\pmod{p^{k_2}}$. Analogously, all elements of $A_{\\ge N_1}$ are congruent to $r_{k_2-1}$ modulo $p^{k_2-1}$. By the lemma, we can further find $N_2$ such that all elements of $A_{\\ge N_2} = \\{a \\in A \\mid a \\ge N_2\\}$ are congruent to $r_{k_2}$ modulo $p^{k_2}$, and so on. So, we obtain $N_1 < N_2 < \\dots < N_{2p-2}$ satisfying $p^{k_1-1} \\le (a_i - r_{k_i}) \\le (1 \\le i \\le 2p-2)$.\n\nFor convenience, subtract $r_{k_{2p-2}}$ from every element of $A$ and from every $N_i$. Now $p^{k_i-1}a_i$ ($1 \\le i \\le 2p-2$), and every element of $A_{\\ge N_i}$ is a multiple of $p^{k_{i+1}-1}$. Take the $(2p-2)$-element subset $B = \\{b_1, b_2, \\dots, b_{2p-2}\\}$. For a $p$-element subset $\\{b_{i_1}, \\dots, b_{i_p}\\}$ ($i_1 < i_2 < \\dots < i_p$) of $B$, the arithmetic mean is divisible by $p^{k_{i_1}-2}$ and greater than $b_{i_1} \\ge N_{i_1-1}$. Hence it does not belong to $A$. This completes the proof. $\\square$\n(by Feng Chenxu) We assert that for $1 \\le t \\le 2p-2$, one can always find a $t$-element subset $B$, such that for any $p$ distinct elements of $B$, their arithmetic mean does not belong to $A$. Apply induction to $t$: the assertion is trivially true when $t \\le p-1$; assume the trueness when $|B| \\le t-1$ and consider $t$.\n\nFor $1 \\le i \\le p$, let $B_i \\subseteq A$ be the residue class $i$ modulo $p$, $A = \\bigcup_{i=1}^p B_i$. If $A = B_j$ for some $1 \\le j \\le p$, then subtract $j$ from every element of $A$ and divide by $p$. This treatment will not affect the conclusion. Continue this process until at least two subsets $B_j, B_k$ are nonempty.\n\nIf two subsets $B_j$ and $B_k$ are infinite, take $p-1$ elements from each subset to constitute $B$. Clearly, the arithmetic mean of any $p$ distinct elements of $B$ is non-integral, $B$ satisfies the problem condition.\n\nIf only one subset, say $B_j$ is infinite, let $C = A \\setminus B_j \\ne \\emptyset$. Let $D_+$ be the set of all elements of $B_j$ that are larger than $\\max_{x \\in C} x$; $D_-$ be the set of all elements of $B_j$ that are smaller than $\\max_{x \\in C} x$. Then $D_+ \\cup D_-$ is infinite. Assume $D_+$ is infinite. By the induction hypothesis, one can find a satisfactory subset $B \\subseteq D_+$, $|B| = t-1$. Since every number in $D_+$ is larger than every number in $A \\setminus D_+$, we can infer that any $p$ distinct numbers in $B$ have an arithmetic mean larger than that in $A \\setminus D_+$, and thus the mean is not in $A$. Take $x \\in C$, and assume $x \\in B_k, k \\ne j$. For $x$ and $p-1$ elements of $B$, their sum is $S \\equiv (p-1)j + k \\equiv k - j \\not\\equiv 0 \\pmod p$. Thus, the mean is not an integer, nor in $A$. It follows that $B \\cup \\{x\\}$ satisfies the problem condition. This completes the induction and the proof of the assertion. $\\square$",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 28546,
    "subject": "Mathematics (Multi-modal)",
    "question": "In a convex quadrilateral $ABCD$, let $I$ and $J$ be the incenter of $\\triangle ABC$ and $\\triangle ADC$, respectively. Assume that $IJ$, $AC$, $BD$ meet at point $P$. The line through $P$ perpendicular to $BD$ meets the exterior angle bisectors of $\\angle BAD$ at $E$, and meets the exterior angle bisector of $\\angle BCD$ at $F$. Prove that $PE = PF$.\n![](attached_image_1.png)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "**Proof 1:** If $AB \\parallel CD$ and $AD \\parallel BC$, then $ABCD$ is a parallelogram. In this case, $P$ is the midpoint of $AC$ and $AE \\parallel CF$. So $PE = PF$.\n\nWe may assume that $AB$ is not parallel to $CD$ from now on. We first show that $AB + AD = CB + CD$. As shown in the following picture, we may assume that the extends of $BA$ and of $CD$ meet at point $T$. Let $\\odot K$ be the excircle of $\\triangle TBC$ in $\\angle B$; meeting the lines $AB$, $BC$, $CD$ at $X$, $Y$, $Z$, respectively.\n![](attached_image_2.png)\n\nSince the internal homothetic center of $\\odot I$ and $\\odot J$ lies on the line $IJ$, and since $AC$ is an inner tangent of the two circles, we deduce that the inner homothetic center of $\\odot I$ and $\\odot J$ is $P$.\nIt is well known that the internal homothetic center $P$ of $\\odot I$ and $\\odot J$, the external homothetic center $B$ of $\\odot I$ and $\\odot K$, and the internal homothetic center of $\\odot J$ and $\\odot K$ are colinear. So the internal homothetic center of $\\odot J$ and $\\odot K$ lies on the line $BP$. Since $CD$ is an internal tangent line of $\\odot J$ and $\\odot K$, the internal homothetic center of $\\odot J$ and $\\odot K$ is precisely point $D$. Moreover, $AD$ is tangent to $\\odot K$, say at point $W$.\nFrom this, we deduce that\n$$\n\\begin{aligned}\nAB + AD &= (BX - AX) + (AW - DW) = BX - DW \\\\\n&= BY - DZ = (CB + CY) - (CZ - CD) = CB + CD.\n\\end{aligned}\n$$\nIn other words,\n$$\nAB + AD = CB + CD. \\tag{1}\n$$\nLet $\\odot U$ and $\\odot V$ denote the excircles of $\\triangle ABD$ in $\\angle ABD$ and in $\\angle ADB$, respectively. Let $\\odot Q$ and $\\odot R$ denote the excircles of $\\triangle BCD$ in $\\angle CBD$ and in $\\angle BDC$, respectively.\nNow, we prove that $UR$ and $VQ$ meet at $P$. Consider $\\odot K$, $\\odot U$, and $\\odot R$. It is well known that the external homothetic center $A$ of $\\odot K$ and $\\odot U$, the internal homothetic center $C$ of $\\odot K$ and $\\odot R$, and the internal homothetic center of $\\odot U$ and $\\odot R$ are colinear. Yet $BD$ is the internal tangent line of $\\odot U$ and $\\odot R$; so $P$ is the internal homothetic center of $\\odot U$ and $\\odot R$. It follows that $UR$ passes through $P$. By a similar argument, $VQ$ passes through $P$.\n![](attached_image_3.png)\n\nNext, we prove that $UQ \\perp BD$ and $VR \\perp BD$. Extend $BD$ to intersect $\\odot U$ and $\\odot Q$ at $L$ and $L'$, respectively. Then\n$$\nBL = \\frac{1}{2}(AB + AD + BD), \\quad BL' = \\frac{1}{2}(CB + CD + BD).\n$$\nCombining this with (1), we deduce that $BL = BL'$, i.e. $L$ and $L'$ coincide. So $UQ \\perp BD$. Similarly, $VR \\perp BD$.\nSince $EF \\perp BD$, So $UQ \\parallel EF \\parallel VR$. Thus,\n$$\n\\frac{PE}{UQ} = \\frac{VP}{VQ} = \\frac{RF}{RQ} = \\frac{PF}{UQ},\n$$\nSo $PE = PF$. We are done.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 28547,
    "subject": "Mathematics (Multi-modal)",
    "question": "Given an oblique triangle $ABC$ with $BC > AC > AB$. Let $P_1 \\neq P_2$ be two points on the plane such that, for $i = 1, 2$, if $AP_i, BP_i$, and $CP_i$ intersect the circumcircle of $\\triangle ABC$ at $D_i, E_i$, and $F_i$, respectively, then $D_iE_i \\perp D_iF_i$ and $D_iE_i = D_iF_i \\neq 0$. Let the line $P_1P_2$ intersect the circumcircle of $\\triangle ABC$ at $Q_1$ and $Q_2$. The two Simson lines of $Q_1$ and of $Q_2$ with respect to $\\triangle ABC$ intersect at $W$.\nProve that $W$ lies on the nine-point circle of $\\triangle ABC$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "(1) This problem can be solved by computing the angles, but it will (seriously) depend on the relative positions of points. To avoid a case-by-case discussion, we use complex numbers. Assume that the circumcircle of $\\triangle ABC$ is the unit circle on the complex plane. We use the corresponding lowercase letters to denote the complex number corresponding to the points on the plane, e.g. point $A$ corresponds to the complex number $a$.\n\nFor a point $P$ on the plane, its projection image on the line $BC$ is the midpoint of $P$ and the symmetry point $P'$ of $P$ with respect to $BC$. Note that $\\angle PBC = \\angle P'BC$ and they are on opposite sides of the line $BC$. We have\n$$\n\\frac{(p' - b)(p - b)}{|p - b|^2} = \\frac{(c - b)^2}{|c - b|^2},\n$$\ni.e.\n$$\np' = b + \\frac{(c - b)(\\bar{p} - \\bar{b})}{\\bar{c} - \\bar{b}},\n$$\nFor a complex number on the unit circle, $\\bar{b} = \\frac{1}{b}, \\bar{c} = \\frac{1}{c}$. We have\n$$\np' = b - bc\\left(\\bar{p} - \\frac{1}{b}\\right) = b + c - bc\\bar{p}.\n$$\nSo\n$$\nd = \\frac{1}{2}(p - bc\\bar{p} + b + c).\n$$\nSimilarly, we have\n$$\ne = \\frac{1}{2}(p - ca\\bar{p} + c + a), \\quad f = \\frac{1}{2}(p - ab\\bar{p} + a + b).\n$$\nThen\n$$\n\\angle EDF = 90^\\circ \\text{ and } ED = DF \\Leftrightarrow \\frac{e-d}{f-d} = \\pm i \\Leftrightarrow \\left(\\frac{e-d}{f-d}\\right)^2 = -1.\n$$\ni.e.\n$$\n\\left[ \\frac{(bc - ca)\\bar{p} + (a - b)}{(bc - ab)\\bar{p} + (a - c)} \\right]^2 = -1.\n$$\nwe obtain an equation on $\\bar{p}$:\n$$\n[c^2(a-b)^2 + b^2(c-a)^2]\\bar{p}^2 - 2[c(a-b)^2 + b(c-a)^2]\\bar{p} + [(a-b)^2 + (c-a)^2] = 0, \\quad (*)\n$$\n\nThe leading coefficient of this quadratic equation is zero if and only if $|a-b| = |c-a|$ (i.e. $\\triangle ABC$ is an isosceles triangle with $AB = AC$) and $\\frac{b}{c} = \\pm i\\frac{a-b}{c-a}$ (i.e. the central angle corresponding to $\\widehat{BC}$ is $\\frac{\\pi}{2}$ larger than the inscribed angle corresponding to $\\widehat{BC}$), then $\\triangle ABC$ is isosceles triangle with a right angle; this contradicts with our assumption. So (*) is a quadratic equation.\nThe discriminant of (*) is\n$$\n\\begin{aligned}\n\\Delta &= 4 \\left\\{[c(a-b)^2 + b(c-a)^2]^2 - [c^2(a-b)^2 + b^2(c-a)^2]^2[(a-b)^2 + (c-a)^2]\\right\\} \\\\\n&= 4\\left\\{2bc(a-b)^2(c-a)^2 - (b^2+c^2)(a-b)^2(c-a)^2\\right\\} \\\\\n&= -4(a-b)^2(b-c)^2(c-a)^2 \\neq 0.\n\\end{aligned}\n$$\nSo the equation (*) has two roots, this proves the existence of the two points $P_1$ and $P_2$.\n\nWe next prove that $W$ lies on the nine-point circle of $\\triangle ABC$.\n\nStep 1: show that $P_1P_2$ passes through the circumcenter $O$ of $\\triangle ABC$. This is equivalent to\n$$\n\\frac{c(a-b)^2 + b(c-a)^2}{(a-b)(b-c)(c-a)} \\in i\\mathbb{R} \\Leftrightarrow \\frac{c(a-b)^2 + b(c-a)^2}{(a-b)(b-c)(c-a)} + \\frac{\\overline{c(a-b)^2 + b(c-a)^2}}{(a-b)(b-c)(c-a)} = 0.\n$$\nBut\n$$\n\\begin{aligned}\n\\frac{c(a-b)^2 + b(c-a)^2}{(a-b)(b-c)(c-a)} &= \\frac{c(a-b)}{(b-c)(c-a)} + \\frac{b(c-a)}{(a-b)(b-c)} \\\\\n&= -\\frac{c}{b-c} - \\frac{c}{c-a} - \\frac{b}{a-b} - \\frac{b}{b-c},\n\\end{aligned}\n$$\nYet $\\frac{\\overline{b+c}}{b-c} = \\frac{b+c}{c-b}$,\n$$\n\\frac{\\bar{c}}{\\bar{c}-\\bar{a}} + \\frac{\\bar{b}}{\\bar{a}-\\bar{b}} = \\frac{a}{a-c} + \\frac{a}{b-a} = \\left(1-\\frac{c}{c-a}\\right) + \\left(-1-\\frac{b}{a-b}\\right),\n$$\nThis proves Step 1.\n\nFor a point $Q$ on the circumcircle of $\\triangle ABC$, its feets on the three sides of $\\triangle ABC$ are collinear (on the well-known Simson line), and this line passes through the midpoint of $Q$ and orthocenter $H$ of $\\triangle ABC$. Let $X$ and $Y$ denote the projections of a point $Q$ on the circumcircle onto $BC$ and $CA$, respectively. Then, we know\n$$\nx = \\frac{1}{2}(q - bc\\bar{q} + b + c), \\quad y = \\frac{1}{2}(q - ca\\bar{q} + c + a).\n$$\nIn order to prove that these two points are colinear with $\\frac{1}{2}(q + b + b + c)$, we only need to show that\n$$\n\\frac{a + \\frac{bc}{q}}{b + \\frac{ca}{q}} = \\frac{aq + bc}{bq + ac} \\in \\mathbb{R},\n$$\n\nBut this is obvious. Since $H$ is the homothetic center of the circumcircle and the nine-point circle, $X_1Y_1$ and $X_2Y_2$ each passes through a pair of opposite points on the nine-point circle of $\\triangle ABC$.\n\nStep 3: prove that $X_1Y_1 \\perp X_2Y_2$.\nWe have\n$$\n\\begin{align*}\n\\angle CY_1X_1 &= \\angle CQ_1X_1 = 90^\\circ - \\angle X_1CQ_1 \\\\\n\\angle BAQ_2 &= \\angle Q_2Y_2X_2 = 90^\\circ - \\angle CY_2X_2,\n\\end{align*}\n$$\nSo $X_1Y_1 \\perp X_2Y_2$.\n\nCombine these three steps, we see that the intersect point of $X_1Y_1$ and $X_2Y_2$ is on the nine-point circle of $\\triangle ABC$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 28548,
    "subject": "Mathematics (Multi-modal)",
    "question": "Fix an equilateral triangle $ABC$ with side length $1$. We call $(\\triangle DEF, \\triangle XYZ)$ a good triangle pair if the points $D$, $E$, and $F$ lie in the interior of segments $BC$, $CA$, and $AB$, respectively, the points $X$, $Y$, and $Z$ lie on the lines $BC$, $CA$, and $AB$, respectively, and they satisfy the conditions\n$$\n\\frac{DE}{20} = \\frac{EF}{22} = \\frac{FD}{38}, \\quad \\text{and} \\quad DE \\perp XY, EF \\perp YZ, FD \\perp ZX.\n$$\nAs $(\\triangle DEF, \\triangle XYZ)$ runs through all good triangle pairs, determine all possible values of $\\frac{1}{S_{\\triangle DEF}} + \\frac{1}{S_{\\triangle XYZ}}$.",
    "options": [],
    "answer": "(97√2 + 40√3)/15",
    "solution": "(1) First, consider the rotation of $90^\\circ$ (clockwise or counterclockwise), centered at an arbitrary point on the plane. Then the images $X'$, $Y'$, and $Z'$ of the points $X$, $Y$, and $Z$, respectively, satisfy\n$$\nX'Y' \\parallel DE, \\quad Y'Z' \\parallel EF, \\quad Z'X' \\parallel FD.\n$$\nIn this case, $\\triangle X'Y'Z'$ and $\\triangle DEF$ are directly similar. So, every good triangle pair is directly similar.\n\n![](attached_image_1.png)\n\n(2) **Rotate $\\triangle XYZ$ (together with the equilateral triangle $ABC$) $90^\\circ$, and properly rescale and translate the picture so that the image of $\\triangle XYZ$ coincides with $\\triangle DEF$. Under this transformation, the points $A$, $B$, and $C$ are mapped to points $A_1$, $B_1$, and $C_1$. So there are three points $A_1$, $B_1$, and $C_1$ on the plane satisfying:**\n* $\\triangle A_1B_1C_1$ is an equilateral triangle;\n* the points $D$, $E$, and $F$ lie on the lines $B_1C_1$, $C_1A_1$, and $A_1B_1$; and\n* $A_1B_1 \\perp AB$, $B_1C_1 \\perp BC$, and $C_1A_1 \\perp CA$.\nFrom this, we see that the points $A_1$, $B_1$, and $C_1$ lie on the circumcircles of $\\triangle AEF$, $\\triangle BFD$, and $\\triangle CDE$, respectively. Moreover, $A_1$, $B_1$, and $C_1$ are the antipodes of $A$, $B$, and $C$ in the corresponding circles; see the picture below.\n\n![](attached_image_2.png)\n\n(3) Note that $S_{\\triangle ABC} : S_{\\triangle XYZ} = S_{\\triangle A_1B_1C_1} : S_{\\triangle DEF}$. So it suffices to compute the ratio of $S_{\\triangle ABC} + S_{\\triangle A_1B_1C_1}$ to $S_{\\triangle DEF}$. We have\n$$\nS_{\\triangle ABC} = S_{\\triangle DEF} + S_{\\triangle EAF} + S_{\\triangle FBD} + S_{\\triangle DCE}, \\\\\nS_{\\triangle A_1B_1C_1} = S_{\\triangle DEF} + S_{\\triangle EA_1F} + S_{\\triangle FB_1D} + S_{\\triangle DC_1E},\n$$\nHere the right hand sides are expressed in terms of oriented areas. **Since the two equilateral triangles on the left hand side are directly similar, the signs on the areas are the same.** Using the properties of antipodes, if we denote the circumcenters of $\\triangle AEF$, $\\triangle BFD$, and $\\triangle CDE$ by $O_1$, $O_2$, and $O_3$, respectively, then the condition \"$D$, $E$, and $F$ lie on the interior of three sides\" ensures that the three circumcenters $O_1$, $O_2$, and $O_3$ lie outside of $\\triangle DEF$. So we have the following equality of areas:\n$$\nS_{\\triangle ABC} + S_{\\triangle A_1B_1C_1} = 2(S_{\\triangle DEF} + S_{\\triangle EO_1F} + S_{\\triangle FO_2D} + S_{\\triangle DO_3E}).\n$$\nIn fact, $\\triangle O_1O_2O_3$ is the outer Napoleon triangle of $\\triangle DEF$; its area is exactly the half of sum of the areas in the parentheses.\n\n(4) So we need to compute, for a triangle with side length ratio $20 : 22 : 38$, the ratio of the area of the hexagon formed by the vertices of the triangle and the vertices of the outer Napoleon triangle, to the area of the original triangle.\nBy Heron formula, the area of a triangle with side lengths $20$, $22$, $38$ is\n$$\n\\sqrt{40 \\cdot (40 - 20) \\cdot (40 - 22) \\cdot (40 - 38)} = 120\\sqrt{2}.\n$$\nOn the other hand,\n$$\nS_{\\triangle EO_1F} + S_{\\triangle FO_2D} + S_{\\triangle DO_3E} = \\frac{\\sqrt{3}}{3} \\cdot (11^2 + 10^2 + 19^2) = 194\\sqrt{3}.\n$$\nSo\n$$\n\\begin{align*}\n& \\frac{1}{S_{\\triangle DEF}} + \\frac{1}{S_{\\triangle XYZ}} \\\\\n= & \\frac{1}{S_{\\triangle ABC}} \\left( \\frac{S_{\\triangle ABC}}{S_{\\triangle DEF}} + \\frac{S_{\\triangle ABC}}{S_{\\triangle XYZ}} \\right) \\\\\n= & \\frac{1}{S_{\\triangle ABC}} \\frac{S_{\\triangle ABC} + S_{\\triangle A_1B_1C_1}}{S_{\\triangle DEF}} \\\\\n= & \\frac{4}{\\sqrt{3}} \\frac{2(120\\sqrt{2} + 194\\sqrt{3})}{120\\sqrt{2}} \\\\\n= & \\frac{97\\sqrt{2} + 40\\sqrt{3}}{15}\n\\end{align*}\n$$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 28549,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all functions $f: \\mathbb{R} \\to \\mathbb{R}$ such that, for any real numbers $x, y$, the following two multisets are equal\n$$\n\\{f(xf(y) + 1), f(yf(x) - 1)\\} = \\{xf(f(y)) - 1, yf(f(x)) + 1\\}.\n$$\n\n*Remark: $\\{a, b\\} = \\{c, d\\}$ are equal as multisets if $a = c$ and $b = d$, or $a = d$ and $b = c$.*",
    "options": [],
    "answer": "f(x) = x and f(x) = -x",
    "solution": "**Solution:** All functions satisfying (*) are $f(x) = x$ and $f(x) = -x$. It is easy to verify that these two functions satisfy (*).\nIn what follows, all sets are multisets. Take $x = y = 0$ in (*) gives $\\{f(1), f(-1)\\} = \\{1, -1\\}$. First consider the case when $f(1) = 1$. We want to show that $f(x) = x$ for any $x \\in \\mathbb{R}$.\n*Step 1:* Prove that $f(n) = n$ for $n \\in \\mathbb{Z}$.\nWe first show that $f(0) = 0$. Take $x = 0$ in (*), we have $\\{f(1), f(yf(0) - 1)\\} = \\{-1, yf(f(0)) + 1\\}$. If $f(0) \\neq 0$, take $y = \\frac{2}{f(0)}$; then the LHS is $\\{1, 1\\}$, which cannot contain the $-1$ on the RHS. So $f(0) = 0$.\nNext, we prove inductively $f(n) = n$ for $n \\in \\mathbb{N}$. The case $n = 1$ is known. Assume that $f(m) = m$ holds for $m \\le n$. Take $x = 1$ and $y = n$ in (*), we deduce\n$$\n\\{f(n + 1), f(n - 1)\\} = \\{n - 1, n + 1\\}\n$$\nFrom this, we deduce that $f(n + 1) = n + 1$, completing the inductive proof. Using a similar argument but taking $x = 1$ and $y = -n$ proves that $f(-n) = -n$ for $n \\in \\mathbb{N}$. This completes Step 1.\n*Step 2:* Prove that $f$ is a bijection.\n$f$ is surjective because when setting $y = 1$ in (*), the RHS contains $xf(f(1)) - 1 = x - 1$, which can take an arbitrary real number. By the form of its LHS, all real numbers are in the image of $f$.\nNext, we prove $f(y_0) = 0$ implies $y_0 = 0$. Suppose not, namely $y_0 \\neq 0$. Then taking $y = y_0$ in (*) gives\n$$\n\\{f(1), f(y_0f(x) - 1)\\} = \\{-1, y_0f(f(x)) + 1\\}.\n$$\nWe must have $1 = f(1) = y_0f(f(x)) + 1$. So $f(f(x)) = 0$ holds for all $x$. This contradicts with $f(1) = 1$.\nWe next prove the injectivity of $f$. If $y_1 \\neq y_2$ satisfy $f(y_1) = f(y_2) \\neq 0$. Taking respectively $y = y_1$ and $y = y_2$ in (*) gives\n$$\n\\{f(xf(y_1) + 1), f(y_1f(x) - 1)\\} = \\{xf(f(y_1)) - 1, y_1f(f(x)) + 1\\},\n$$\n$$\n\\{f(xf(y_2) + 1), f(y_2f(x) - 1)\\} = \\{xf(f(y_2)) - 1, y_2f(f(x)) + 1\\}.\n$$\n\nNote that the first elements on both sides of the two equations are the same, respectively.\nIf some $x_0 \\in \\mathbb{R}$ satisfies $f(x_0f(y_1) + 1) \\neq x_0f(f(y_1)) - 1$, then\n$$\ny_1f(f(x_0)) + 1 = f(xf(y_1) + 1) = f(xf(y_2) + 1) = y_2f(f(x_0)) + 1\n$$\nFrom this, we get $y_1f(f(x_0)) = y_2f(f(x_0))$, thus $f(f(x_0)) = 0$, which implies $x_0 = 0$.\nSo if $x \\neq 0$, then $f(xf(y_1) + 1) = xf(f(y_1)) - 1$. From this, we deduce that $f(x) = ax + b$ for some $a, b \\in \\mathbb{R}$ and any $x \\neq 1$ is a linear function. But $f(n) = n$ for $n \\in \\mathbb{Z}$ implies that $a = 1$ and $b = 0$. So $f(x) = x$ is injective.\n**Step 3:** Prove that for any $n \\in \\mathbb{Z}$ and $y \\in \\mathbb{R}$, we have\n$$\nf\\left(f\\left(\\frac{n}{f(y)}\\right)\\right) = \\frac{n}{y} \\qquad (1)\n$$\nIt is enough to consider the case when $n \\neq 0$ and $y \\neq 0$. Taking $x = \\frac{n}{f(y)}$ in $(*)$ gives\n$$\n\\left\\{n + 1, f\\left(yf\\left(\\frac{n}{f(y)}\\right) - 1\\right)\\right\\} = \\left\\{n\\frac{f(f(y))}{f(y)} - 1, yf\\left(f\\left(\\frac{n}{f(y)}\\right)\\right) + 1\\right\\}. \\quad (2)\n$$\nSimilarly, substituting $y$ for $x$ and $\\frac{n}{f(y)}$ for $y$ in $(*)$ gives\n$$\n\\left\\{n - 1, f\\left(yf\\left(\\frac{n}{f(y)}\\right) + 1\\right)\\right\\} = \\left\\{n\\frac{f(f(y))}{f(y)} + 1, yf\\left(f\\left(\\frac{n}{f(y)}\\right)\\right) - 1\\right\\}. \\quad (3)\n$$\nIf there exists some $y_0 \\in \\mathbb{R} \\setminus \\{0\\}$ such that $n \\neq y_0f\\left(f\\left(\\frac{n}{f(y_0)}\\right)\\right)$, then combining (2) and (3) gives\n$$\nn + 1 = n \\frac{f(f(y_0))}{f(y_0)} - 1, \\quad n - 1 = n \\frac{f(f(y_0))}{f(y_0)} + 1.\n$$\nTaking the differences of the two equations gives $2 = -2$, which is a contradiction.\n**Step 4:** For $\\alpha \\in \\mathbb{Q}$ and $y \\in \\mathbb{R}$, prove that $f(\\alpha y) = \\alpha f(y)$.\nWe may assume that $\\alpha \\neq 0$ and $y \\neq 0$. Substituting $f\\left(\\frac{m}{f(y)}\\right)$ with $m \\in \\mathbb{Z} \\setminus \\{0\\}$ for $y$ in (1) gives\n$$\nf\\left(f\\left(\\frac{n}{f\\left(f\\left(\\frac{m}{f(y)}\\right)\\right)}\\right)\\right) = \\frac{n}{f\\left(\\frac{m}{f(y)}\\right)}.\n$$\nPlugging (1) into the denominator of the LHS gives\n$$\nf\\left(f\\left(\\frac{n}{m}y\\right)\\right) = \\frac{n}{f\\left(\\frac{m}{f(y)}\\right)}.\n$$\nIn the above equality, substituting $\\frac{t}{f(y)}$ with $t \\in \\mathbb{Z} \\setminus \\{0\\}$ for $y$ gives\n$$\n\\frac{nt/m}{y} = f\\left(f\\left(\\frac{n}{m}\\frac{t}{f(y)}\\right)\\right) = \\frac{n}{f\\left(\\frac{m}{f\\left(\\frac{t}{f(y)}\\right)}\\right)}.\n$$\n\nThis is equivalent to $f(\\frac{m}{f(\\frac{t}{f(y)})}) = \\frac{m}{t}y$. Taking $f$ on both sides gives\n$$\n\\frac{m}{t} f(y) = f\\left(f\\left(\\frac{m}{f(\\frac{t}{f(y)})}\\right)\\right) = f\\left(\\frac{m}{t}y\\right).\n$$\nThis completes Step 4.\n**Step 5:** Prove that $f(y+a) = f(y) + a$ for all $y \\in \\mathbb{R}$ and $a \\in \\mathbb{Q}$.\nTaking $x = \\frac{1}{r} \\in \\mathbb{Q} \\setminus \\{0\\}$ in (*) gives\n$$\n\\{f(\\frac{1}{r}f(y) + 1), f(\\frac{1}{r}y - 1)\\} = \\{\\frac{1}{r}f(f(y)) - 1, \\frac{1}{r}y + 1\\}.\n$$\nUsing Step 4 and multiplying both sides by $r$ gives\n$$\n\\{f(f(y) + r), f(y - r)\\} = \\{f(f(y)) - r, y + r\\}. \\quad (4)\n$$\nWe want to prove that $f(y-r) = f(f(y)) - r$. Suppose that there exist some $y_0 \\in \\mathbb{R}$ and $r_0 \\in \\mathbb{R} \\setminus \\{0\\}$ such that $f(y_0 - r_0) = y_0 + r_0$. Taking $y = y_0 - r_0$ and $r = -2r_0$ in (4) gives\n$$\n\\{f(f(y_0 - r_0) - 2r_0), f(y_0 - r_0 + 2r_0)\\} = \\{f(f(y_0 - r_0)) + 2r_0, y_0 - 3r_0\\}.\n$$\nThis is the same as\n$$\n\\{y_0 + r_0, f(y_0 + r_0)\\} = \\{f(y_0 + r_0) + 2r_0, y_0 - 3r_0\\}.\n$$\nSince $y_0 + r_0 \\neq y_0 - 3r_0$, we must have $y_0 + r_0 = f(y_0 + r_0) + 2r_0$, i.e. $f(y_0 + r_0) = y_0 - r_0$. But then the two sides of the equality becomes $\\{y_0 + r_0, y_0 - r_0\\} = \\{y_0 + r_0, y_0 - 3r_0\\}$. They are obviously not equal, giving a contradiction.\nSo $f(y - r) = f(f(y)) - r$ must hold for all $y \\in \\mathbb{R}$ and $r \\in \\mathbb{Q} \\setminus \\{0\\}$. Taking two different values of $r$ and taking the difference gives $f(y + a) = f(y) + a$ for all $y \\in \\mathbb{R}$ and $a \\in \\mathbb{Q}$.\n**Step 6:** Prove that $f(y) = y$ for $y \\in \\mathbb{R}$.\nCombining Step 5 and (4) gives\n$$\n\\{f(f(y)) + r, f(y) - r\\} = \\{f(f(y)) - r, y + r\\}.\n$$\nTaking the sum of the two elements on both sides of the equality, we deduce that $f(f(y)) + f(y) = f(f(y)) + y$, i.e. $f(y) = y$.\nNow we consider the case when $f(1) = -1$. We aim to prove that $f(x) = -x$ ($x \\in \\mathbb{R}$).\n**Step 1:** Prove that $f(n) = -n$ ($n \\in \\mathbb{Z}$).\n\nTaking $y = 1$ and respectively $x = 1, -1$ in (*) gives\n$$\n\\{f(0), f(-2)\\} = \\{0, 2\\}, \\quad \\{f(2), f(0)\\} = \\{-2, 0\\}.\n$$\nFrom this, we deduce that $f(0) = 0$, $f(2) = -2$, $f(-2) = 2$.\nNext, we inductively prove that $f(n) = -n$. When $|n| = 1, 2$, this is already proved. Suppose that we have proved this when $|n| \\le n_0$ ($n_0 \\ge 2$). Taking $y = 1$ and $x = n_0, -n_0$ in (*) gives\n$$\n\\begin{aligned} \\{f(-n_0 + 1), f(f(n_0) - 1)\\} &= \\{n_0 - 1, f(f(n_0)) + 1\\}, \\\\ \\{f(n_0 + 1), f(f(-n_0) - 1)\\} &= \\{-n_0 - 1, f(f(-n_0)) + 1\\} \\end{aligned}\n$$\nWe deduce from this that $f(n_0 + 1) = -n_0 - 1$ and $f(-n_0 - 1) = n_0 + 1$. This completes the induction in Step 1.\n**Step 2:** Prove that $f(x) = -x$ holds for all $x \\in \\mathbb{R}$.\nFirst, as in the $f(1) = 1$ case, we can prove that $f$ is surjective.\nFor a nonzero integer $n$ and $z \\in \\mathbb{R}\\{0\\}$, taking $x = \\frac{n}{f(z)}$, $y = z$, and taking $x = z$, $y = \\frac{n}{f(z)}$ in (*), we obtain\n$$\n\\begin{aligned} \\{-n-1, f\\left(zf\\left(\\frac{n}{f(z)}\\right) - 1\\right)\\} &= \\{n\\frac{f(f(z))}{f(z)} - 1, zf\\left(f\\left(\\frac{n}{f(z)}\\right)\\right) + 1\\}, & (5) \\\\ \\{-n+1, f\\left(zf\\left(\\frac{n}{f(z)}\\right) + 1\\right)\\} &= \\{n\\frac{f(f(z))}{f(z)} + 1, zf\\left(f\\left(\\frac{n}{f(z)}\\right)\\right) - 1\\}. \\end{aligned}\n$$\nIf for some $z = z_0 \\in \\mathbb{R}\\{0\\}$ we have $f(f(z_0)) \\ne f(z_0)$, then by (5), we deduce that\n$$\n-n-1 = zf\\left(f\\left(\\frac{n}{f(z)}\\right)\\right) + 1, \\quad -n+1 = zf\\left(f\\left(\\frac{n}{f(z)}\\right)\\right) - 1.\n$$\nTaking the difference of the two equalities gives $2 = -2$. This is a contradiction!\nSo $f(f(z_0)) = -f(z_0)$. But $f$ is surjective. So $f(x) = -x$ holds for all $x \\in \\mathbb{R}\\{0\\}$. Moreover, $f(0) = 0$ is known. This completes the proof of $f(x) = -x$ when $f(1) = -1$.\nCombining the above two cases, the needed functions are $f(x) = x$ ($x \\in \\mathbb{R}$) and $f(x) = -x$ ($x \\in \\mathbb{R}$).",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 28550,
    "subject": "Mathematics (Multi-modal)",
    "question": "Given positive integers $a, b, c, p, q, r$, where $p, q, r \\ge 2$. Let $Q = \\{(x, y, z) \\in \\mathbb{Z}^3 \\mid 0 \\le x \\le a, 0 \\le y \\le b, 0 \\le z \\le c\\}$ be the game board.\n\nInitially, put $M$ game pieces on $Q$ (no restriction on the number of pieces in each position). There are three types of legal moves:\n\n(1) Remove $p$ pieces from $(x, y, z)$ and place one piece at $(x-1, y, z) \\in Q$;\n\n(2) Remove $q$ pieces from $(x, y, z)$ and place one piece at $(x, y-1, z) \\in Q$;\n\n(3) Remove $r$ pieces from $(x, y, z)$ and place one piece at $(x, y, z-1) \\in Q$.\n\nFind the minimum value of $M$, such that no matter how the $M$ game pieces are placed initially, one can always take a sequence of moves to get one piece placed at $(0, 0, 0)$.",
    "options": [],
    "answer": "p^a q^b r^c",
    "solution": "To prove $M = p^a q^b r^c$ is necessary, notice that if one initially places fewer than $p^a q^b r^c$ pieces at $(a, b, c)$, then the goal cannot be achieved. Indeed, let the weight of a game piece $u$ at $(x, y, z)$ be $w(u) = \\frac{1}{p^x q^y r^z}$. According to the rule, before and after each move, the total weight of all game pieces $W = w(u)$ does not change. To have a game piece at $(0,0,0)$, we must have $W \\ge 1$; but $W < 1$ for $p^a q^b r^c$ pieces at $(a, b, c)$, a contradiction.\n\nIt requires to prove $M = p^a q^b r^c$ is sufficient. We give a more general result on the $n$-dimensional game board: let $a_1, \\dots, a_n, p_1, \\dots, p_n$ be non-negative integers, satisfying $2 \\le p_1 \\le \\dots \\le p_n$. Consider the game on\n$$\nQ(a_1, \\dots, a_n) = \\{(x_1, \\dots, x_n) \\in \\mathbb{Z}^n \\mid 0 \\le x_i \\le a_i, i = 1, \\dots, n\\}.\n$$\nInitially, put $M$ game pieces on $Q(a_1, \\dots, a_n)$. In each move, one can choose a position, remove $p_i$ ($1 \\le i \\le n$) pieces, and place one piece at the preceding position along the $x_i$ direction. This is called an \"$X_i$ move\". If one chooses a position and just remove $q$ pieces, then it is called a \"$T_q$ move\". For $n=0$, $Q = \\{0\\}$, there is only one type of possible moves, which is the $T_q$ move.\n\nDivide the board into Zone-I as $Q(a_1, \\dots, a_n - 1)$ and Zone-II as $Q(a_1, \\dots, a_{n-1}) \\times \\{a_n\\} = Q(a_1, \\dots, a_n) \\setminus Q(a_1, \\dots, a_{n-1})$. Let $M$ and $K$ be respectively the number of pieces and the number of positions with at least one piece in $Q(a_1, \\dots, a_n)$. Let $M_1, K_1$ and $M_2, K_2$ be respectively the $M, K$ values in Zone-I and Zone-II. In the following argument, the key idea is to take as many $X_n$ moves as possible in Zone-II, so as to have more pieces in Zone-I. At the end of the process, there are at most $p_n - 1$ pieces at each of the $K_2$ positions (initially there are pieces on them) in Zone-II, and at least $\\frac{M_2 - (p_n - 1)K_2}{p_n}$ pieces are added to Zone-I. We call this sequence of moves as \"transmission from Zone-II to Zone-I\".\n\nWe shall justify the following three claims.\n\n**Claim** $A(n)$ if $M = p_1^{a_1} \\cdots p_n^{a_n}$, then a sequence of moves can make one piece at the origin.\n\n**Claim B**($n$) let $u \\ge 2$ be an integer, if $M + K > u \\cdot p_1^{a_1} \\cdots p_n^{a_n}$, then a sequence of $X_1, \\dots, X_n$ moves can make $u$ pieces at the origin.\n\n**Claim C($n$)** let $q \\ge p_n$ and $s$ be positive integers, $t$ be a nonnegative integer, if $M + K > sq \\cdot p_1^{a_1} \\dots p_n^{a_n} + t$, then one can first take $\\lfloor \\frac{t}{q} \\rfloor$ of $T_q$ moves, then take a sequence of $X_1, \\dots, X_n$ moves to put $sq$ pieces at the origin.\n\nApply induction on $n$. Obviously, $A(0)$, $B(0)$, $C(0)$ hold. Suppose we have $A(n-1)$, $B(n-1)$, $C(n-1)$. Now induct on $a_n$: when $a_n = 0$, $Q(a_1, \\dots, a_n) = Q(a_1, \\dots, a_{n-1}) \\times \\{0\\}$, the claims $A(n)$, $B(n)$, $C(n)$ on $Q(a_1, \\dots, a_n)$ correspond to $A(n-1)$, $B(n-1)$, $C(n-1)$ on $Q(a_1, \\dots, a_{n-1})$, respectively, and they are already justified by the induction hypothesis. In the following, consider the situations when $a_n \\ge 1$.\n\n**Proof of** $A(n)$\n\na.1 If $M_1 \\ge K_2$, one can transmit at least $\\frac{M_2 - (p_n - 1)K_2}{p_n}$ pieces from Zone-II to Zone-I, after which the number of pieces in Zone-I is\n$$\n\\begin{aligned}\nM_1' &\\ge M_1 + \\frac{M_2 - (n-1)K_2}{p_n} \\\\\n&= p_1^{a_1} \\cdots p_{n-1}^{a_{n-1}} p_n^{a_n} + \\frac{(p_n - 1)(M_1 - K_2)}{p_n} \\\\\n&\\ge p_1^{a_1} \\cdots p_{n-1}^{a_{n-1}} p_n^{a_n}.\n\\end{aligned}\n$$\nIn Zone-I, by the induction hypothesis, one can eventually put one piece at the origin.\n\na.2 If $M_1 < K_2$, then $M_2 + K_2 > M_2 + M_1 = p_n^{a_n} \\cdot p_1^{a_1} \\cdots p_{n-1}^{a_{n-1}}$. Notice that $p_n^{a_n} \\ge 2$. According to $B(n-1)$, in Zone-II one can transmit $p_n^{a_n}$ pieces to $(0, \\dots, 0, a_n)$, then just take $X_n$ moves to put one piece at the origin.\n\n**Proof of** $B(n)$\n\nb.1 If $M_1 + K_1 \\ge \\frac{p_n}{p_n-1}K_2$, one can transmit at least $\\frac{M_2 - (p_n - 1)K_2}{p_n}$ pieces from Zone-II to Zone-I, after which\n$$\n\\begin{aligned}\nM_1' + K_1' &\\ge M_1 + K_1 + \\frac{M_2 - (p_n - 1)K_2}{p_n} \\\\\n&> M_1 + K_1 + \\frac{u \\cdot p_1^{a_1} \\cdots p_n^{a_n} - M_1 - K_1 - p_n K_2}{p_n}\n\\end{aligned}\n$$\n$$\n\\begin{align*}\n&= u \\cdot p_1^{a_1} \\cdots p_{n-1}^{a_{n-1}} p_n^{a_n - 1} + \\frac{(p_n - 1)(M_1 + K_1) - p_n K_2}{p_n} \\\\\n&\\ge u \\cdot p_1^{a_1} \\cdots p_{n-1}^{a_{n-1}} p_n^{a_n - 1}.\n\\end{align*}\n$$\nIn Zone-I, by the induction hypothesis, one can eventually put $u$ pieces at the origin.\n\nb.2 If $M_1 + K_1 < \\frac{p_n}{p_n - 1} K_2$, by Bernoulli's inequality,\n$$\nM_1 + K_1 < \\frac{p_n}{p_n - 1} K_2 \\le 2(1 + a_1) \\dots (1 + a_{n-1}) \\le p_n^{a_n} p_1^{a_1} \\dots p_{n-1}^{a_{n-1}},\n$$\nwhich implies that $p_1^{a_1} \\dots p_n^{a_n} - M_1 - K_1$ is nonnegative. Since\n$$\n\\begin{align*}\nM_2 + K_2 &> u \\cdot p_1^{a_1} \\cdots p_n^{a_n} - (M_1 + K_1) \\\\\n&= (u-1)p_n^{a_n} \\cdot p_1^{a_1} \\cdots p_{n-1}^{a_{n-1}} + (p_1^{a_1} \\cdots p_n^{a_n} - M_1 - K_1),\n\\end{align*}\n$$\nand $u-1 \\ge 1$, $p_n \\ge p_{n-1}$, in Zone-II we may use the conclusion of $C(n-1)$ to take $\\left\\lfloor \\frac{p_1^{a_1} \\cdots p_n^{a_n} - M_1 - K_1}{p_n} \\right\\rfloor$ of $T_{p_n}$ moves, then transmit $(u-1)p_n^{a_n}$ pieces to $(0, \\dots, 0, a_n)$. The latter $(u-1)p_n^{a_n}$ pieces at $(0, \\dots, 0, a_n)$ can contribute $u-1$ pieces at the origin by $X_n$ moves. The former pieces removed by $\\left\\lfloor \\frac{p_1^{a_1} \\cdots p_n^{a_n} - M_1 - K_1}{p_n} \\right\\rfloor$ of $T_{p_n}$ moves can be transmitted by $X_n$ moves into $\\left\\lfloor \\frac{p_1^{a_1} \\cdots p_n^{a_n} - M_1 - K_1}{p_n} \\right\\rfloor$ pieces in Zone-I. After that, in Zone-I, there are\n$$\n\\begin{align*}\nM_1' &\\ge M_1 + \\left\\lfloor \\frac{p_1^{a_1} \\cdots p_n^{a_n} - M_1 - K_1}{p_n} \\right\\rfloor \\\\\n&= p_1^{a_1} \\cdots p_{n-1}^{a_{n-1}} p_n^{a_n - 1} + \\left\\lfloor \\frac{(p_n - 1) M_1 - K_1}{p_n} \\right\\rfloor \\\\\n&\\ge p_1^{a_1} \\cdots p_{n-1}^{a_{n-1}} p_n^{a_n - 1}\n\\end{align*}\n$$\npieces. Due to $A(n)$, eventually one piece can be placed at the origin. Altogether, $u$ pieces at the origin.\n\n**Proof of** $C(n)$\n\nc.1 If $t < q$, no $T_q$ move is needed. Notice that $sq \\ge 2$ and\n$$\nM + K < sq \\cdot p_1^{a_1} \\dots p_n^{a_n} + t \\ge sq \\cdot p_1^{a_1} \\dots p_n^{a_n}.\n$$\nBy the result of $B(n)$, $sq$ pieces can be transmitted to the origin.\n\nc.2 If $t \\ge q$, then some $T_q$ moves are needed. If there are more than $q$ pieces at a place of Zone-II, take a $T_q$ move, after which $M' = M - q$, $K' = K$, and the situation becomes $t' = t - q$. After a few $T_q$ moves, it becomes either (c.1), $t < q$; or the number of pieces at every position of Zone-II is less than or equal to $q$. We only need to handle the latter case. Suppose there are $L$ positions in Zone-II that contain exactly $q$ pieces. By Bernoulli's inequality,\n$$\n\\begin{align*} \nM_1 + K_1 &= M + K - (M_2 + K_2) > sq \\cdot p_1^{a_1} \\dots p_n^{a_n} \\\\ \n&\\quad + t - (q(K_2 - L) + (q+1)L) \\\\ \n&= sq \\cdot p_1^{a_1} \\dots p_{n-1}^{a_{n-1}} p_n^{a_n} + t \\\\ \n&\\quad + (sq \\cdot p_1^{a_1} \\dots p_{n-1}^{a_{n-1}} p_n^{a_n-1}(p_n - 1) - qK_2 - L) \\\\ \n&\\ge sq \\cdot p_1^{a_1} \\dots p_{n-1}^{a_{n-1}} p_n^{a_n} + t \\\\ \n&\\quad + (q(1+a_1) \\dots (1+a_{n-1}) - qK_2 - L) \\\\ \n&\\ge sq \\cdot p_1^{a_1} \\dots p_{n-1}^{a_{n-1}} p_n^{a_n} + t - L. \n\\end{align*}\n$$\nAs $q \\ge p_n$, take $X_n$ move once at each of these $L$ positions with $q$ pieces. After that,\n$$\nM'_1 + K'_1 \\ge M_1 + K_1 + L > sq \\cdot p_1^{a_1} \\dots p_{n-1}^{a_{n-1}} p_n^{a_n} + t,\n$$\nby the induction hypothesis applied to Zone-I, $\\lfloor \\frac{t}{q} \\rfloor$ of $T_q$ moves followed by a sequence of $X_1, \\dots, X_n$ moves can send $sq$ pieces to the origin.\n\nThis justifies $A(n)$, $B(n)$, $C(n)$, and also the problem statement.\n(organized from Feng Chenxu's solution). The argument for $M \\ge p^a q^b r^c$ is the same. We shall prove $M = p^a q^b r^c$ satisfies the problem condition.\n\n**Lemma** For a path $A_1, A_2, \\dots, A_t$ in $Q$, $A_i$ ($i = 2, \\dots, t$) has two coordinates equal to those of $A_{i-1}$ and one coordinate equal to one plus that of $A_{i-1}$. Assign each $A_i$ a number $a_i$ such that $a_1 = 1$, $a_{i+1} = \\frac{a_i}{x_i}$, where\n$$\nx_i = \\begin{cases} p, & \\text{if } x(A_{i+1}) = x(A_i) + 1, \\\\ q, & \\text{if } y(A_{i+1}) = y(A_i) + 1, \\\\ r, & \\text{if } z(A_{i+1}) = z(A_i) + 1. \\end{cases}\n$$\nIf there are $b_i$ pieces on $A_i$ and $\\sum_{i=1}^t b_i a_i \\ge 1$, then a sequence of moves can send one piece to $A_1$.\n\n**Proof of lemma** Every time, remove pieces on $A_{i+1}$ and place a piece on $A_i$ for some $i$, until no more move is possible. During the process, $\\sum_{i=1}^t b_i a_i$ does not change. Suppose at the end there is no piece on $A_1$. There are at most $x_i - 1$ pieces on $A_{i+1}$ or else another move can be made from $A_{i+1}$ to $A_i$. It follows that $b_{i+1} a_{i+1} \\le (x_i - 1) a_{i+1} = a_i - a_{i+1}$, and\n$$\n\\sum_{i=1}^{t} b_i a_i \\le \\sum_{i=1}^{t} a_{i-1} - a_i = a_1 - a_t < 1\n$$\nis a contradiction. The lemma is verified.\n\nFor the original problem, let $f(x, y, z)$ be the number of pieces at $(x, y, z)$ in the initial state.\n\nFirst, consider $c = 0$. Apply induction on $a+b$. Assume $a \\le b$. When $a+b=0$, the conclusion is obvious. For $a=0$, $b>0$, keep taking (2) moves (recall that (1)–(3) are the three types of legal moves described in the problem, in the $x$, $y$, and $z$ directions, respectively) until there is no more move. By that time, at $(0, i, 0)$, $0 \\le i \\le b-1$, there are at least\n$$\n\\left\\lfloor \\frac{f(0, b, 0)}{q} \\right\\rfloor + M - f(0, b, 0) = \\frac{M}{q} + \\left\\lfloor \\left(1 - \\frac{1}{q}\\right) (M - f(0, b, 0)) \\right\\rfloor \\ge \\frac{M}{q}\n$$\npieces. The conclusion follows from the induction hypothesis. For $a=1$, $b=1$, assume $f(0, 0, 0) = 0$, and let $f(0, 1, 0) = a$, $f(1, 0, 0) = b$. Then\n$$\nf(1, 1, 0) = pq - a - b \\ge \\min\\{pq - pa, pq - qb\\}.\n$$\nAssume $pq - a - b \\ge pq - pa$. Take (1) move $q - a$ times on $(1, 1, 0)$, followed by (2) move once on $(0, 1, 0)$. Now there is a piece at the origin.\n\nFor $a+b \\ge 3$, $b \\ge 2$. Assume there are $t$ pieces in the region $0 \\le y \\le b-1$. Take (2) moves at every position with $y=b$ until no more move is possible, at which time there are at most $q-1$ pieces at every position. So, at least $\\left\\lfloor \\frac{M-t-(a+1)(q-1)}{q} \\right\\rfloor$ moves are taken. (i) If $t \\ge a+1$, then in the region $0 \\le y \\le b-1$ the number of pieces is at least $\\left\\lfloor \\frac{M-t-(a+1)(q-1)}{q} \\right\\rfloor + t = \\left\\lfloor \\frac{M+(t-a-1)(q-1)}{q} \\right\\rfloor \\ge \\frac{M}{q}$. By the induction hypothesis, another sequence of moves can send one piece to (0, 0, 0). (ii) If $t \\le a$, in the same manner, assume at most $b$ pieces in the region $0 \\le x \\le a-1$. Then $f(a, b, 0) \\ge M-a-b$.\n\nIf no piece exists on the line $y=0$, then there are $M$ pieces in the region $1 \\le y \\le b$. Using the induction hypothesis for $q$ times, we infer that a sequence of moves can send $q$ pieces to $(0, 1, 0)$. After that, one (2) move will send a piece to $(0, 0, 0)$.\n\nIf there are pieces on the line $y=0$, say at $A$. Let $B = (a, b, c)$. There is a path through the origin, $A$, and $B$; moreover, the number assigned to $A$ is $\\ge \\frac{1}{p^a}$, and the number assigned to $B$ is $\\frac{1}{p^a q^b}$. We have\n$$\n\\frac{1}{p^a} + \\frac{f(a,b,0)}{p^a q^b} \\ge \\frac{M}{p^a q^b} + \\frac{1}{p^a} - \\frac{a+b}{p^a q^b}.\n$$\nSince $a \\le b, b \\ge 2$, it follows that $a+b \\le q^b$, $\\frac{M}{p^a q^b} \\ge 1$, and thus $\\frac{1}{p^a} + \\frac{f(a,b,0)}{p^a q^b} \\ge 1$. According to the lemma, a sequence of moves can send a piece to (0, 0, 0), which validates the conclusion for $c=0$. By symmetry, the conclusion holds if one of $a, b, c$ is 0.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 28551,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $a, b, c, p, q, r$ be positive integers with $p, q, r \\ge 2$. Denote\n$$\nQ = \\{(x, y, z) \\in \\mathbb{Z}^3 \\mid 0 \\le x \\le a, 0 \\le y \\le b, 0 \\le z \\le c\\}.\n$$\nInitially, some stones are put at each point of $Q$, with total $M$ stones. Then one can perform the following three types of operations repeatedly:\n(1) remove $p$ stones on $(x, y, z)$ and place a stone on $(x-1, y, z)$;\n(2) remove $q$ stones on $(x, y, z)$ and place a stone on $(x, y-1, z)$;\n(3) remove $r$ stones on $(x, y, z)$ and place a stone on $(x, y, z-1)$.\nFind the smallest positive integer $M$ such that one can always perform a sequence of operations, to place a stone at $(0, 0, 0)$, no matter how the stones were distributed initially.",
    "options": [],
    "answer": "p^a q^b r^c",
    "solution": "**Proof:** The smallest positive integer $M$ is $p^a q^b r^c$.\n\nFirst, if we initially put less than $p^a q^b r^c$ stones at the point $(a, b, c)$, then one cannot perform a sequence of operations to put a stone at $(0, 0, 0)$. The reason is as follows. For a stone $u$ at $(x, y, z)$, define its weight as $w(u) = \\frac{1}{p^x q^y r^z}$. The rules of operation ensures that the total weight of all stones $W = \\sum w(u)$ stays unchanged. If initially, there are less than $p^a q^b r^c$ stones at $(a, b, c)$, then $W < 1$. Hence at any moment there is no stone at $(0, 0, 0)$.\n\nConversely, we will prove an $n$-dimension version of this problem. Let $a_1, \\dots, a_n$ and $p_1, \\dots, p_n$ be positive integers, such that $2 \\le p_1 \\le \\dots \\le p_n$. Consider the same game on an $n$-dimensional lattice\n$$\nQ(a_1, \\dots, a_n) = \\{(x_1, \\dots, x_n) \\in \\mathbb{Z}^n \\mid 0 \\le x_i \\le a_i, i = 1, \\dots, n\\}\n$$\nWe initially put total $M$ stones on $Q(a_1, \\dots, a_n)$. For $i = 1, 2, \\dots, n$, we call the operation which removes $p_i$ stones at one point $(x_1, \\dots, x_n)$ and then put one stone at $(x_1, \\dots, x_i - 1, \\dots, x_n)$ as “operation $X_i$”. We will need another type of operation $T_q$: which removes $q$ stones at a point but do not put back any stones. We make a convention that when $n = 0$, $Q = \\{0\\}$ is one pointed set containing only the original point.\n\nLet $M$ denote the total number of stones on $Q(a_1, \\dots, a_n)$ by $M$, and let $K$ denote the number of the lattice points where at least one stone is placed. Denote the values of $M$ and $K$ of the domain $\\mathbf{I}: Q(a_1, \\dots, a_n - 1)$ as $M_1$ and $K_1$, respectively; and denote the value of $M$ and $K$ of domain $\\mathbf{II}$:\n$$\nQ(a_1, \\dots, a_{n-1}) \\times \\{a_n\\} = Q(a_1, \\dots, a_n) \\setminus Q(a_1, \\dots, a_n - 1)\n$$\nas $M_2$ and $K_2$ respectively. In the following proof, an important step is to do as many as possible operation $X_n$ for stones in domain $\\mathbf{II}$ to “transfer” them into domain $\\mathbf{I}$, until\n\nstones in the domain **II** can not do anymore operation $X_n$. At that moment, at each point of domain **II** which has at least one stone can hold at most $p_n - 1$ stones, thus we can transfer at least $\\frac{M_2-(p_n-1)K_2}{p_n}$ stones to domain **I**. Call these operations as “transfer **II** to **I**”.\n\nWe are going to inductively prove following three propositions:\n\n**Proposition A(n):** If $M = p_1^{a_1} \\cdots p_n^{a_n}$, then one can perform a sequence operations $X_1, \\dots, X_n$ to move a stone to the origin **0**.\n\n**Proposition B(n):** Let $u \\ge 2$ be a positive integer. If $M + K > u \\cdot p_1^{a_1} \\cdots p_n^{a_n}$, then one can perform a sequence operations $X_1, \\dots, X_n$ to move $u$ stones to the origin **0**.\n\n**Proposition C(n):** Assume that $q \\ge p_n$, $s$ is a positive integer, and $t$ is a nonnegative integer. If $M + K > sq \\cdot p_1^{a_1} \\cdots p_n^{a_n} + t$, then one can first perform $\\lfloor \\frac{t}{q} \\rfloor$ times operation $T_q$, and after that perform a sequence operations $X_1, \\dots, X_n$ to move $sq$ stones to origin **0**.\n\nWe do an induction on $n$. Propositions $A(0), B(0), C(0)$ are obvious. Suppose that $A(n-1), B(n-1), C(n-1)$ are proved. We aim to prove Propositions $A(n), B(n), C(n)$ by an induction on $a_n$. When $a_n = 0$, $Q(a_1, \\dots, a_n) = Q(a_1, \\dots, a_{n-1}) \\times \\{0\\}$, then $A(n), B(n), C(n)$ for $Q(a_1, \\dots, a_n)$ corresponds to $A(n-1), B(n-1), C(n-1)$ for $Q(a_1, \\dots, a_{n-1})$; this is already known. So we may assume that $a_n \\ge 1$.\n\nThe proof of Proposition A(n).\n\na.1) If $M_1 \\ge K_2$, Using transfer **II** to **I**, one may at least transfer $\\frac{M_2-(p_n-1)K_2}{p_n}$ stones, after that we have\n$$\nM_1' \\ge M_1 + \\frac{M_2 - (p_n - 1)K_2}{p_n} = p_1^{a_1} \\cdots p_{n-1}^{a_{n-1}} p_n^{a_n} + \\frac{(p_n - 1)(M_1 - K_2)}{p_n} \\ge p_1^{a_1} \\cdots p_{n-1}^{a_{n-1}} p_n^{a_n},\n$$\nBy inductive hypothesis, we may place a stone at **0**.\n\na.2) If $M_1 < K_2$, then $M_2 + K_2 > M_2 + M_1 = p_n^{a_n} \\cdot p_1^{a_1} \\cdots p_{n-1}^{a_{n-1}}$. Note that $p_n^{a_n} \\ge 2$. Using $B(n-1)$, we know that we can move $p_n^{a_n}$ stones from **II** to $(0, \\dots, 0, a_n)$, then perform $a_n$ times operation $X_n$ to get one stone at **0**.\n\nThe proof of Proposition B(n).\n\nb.1) If $M_1 + K_1 \\ge \\frac{p_n}{p_{n-1}}K_2$, we may transfer at least $\\frac{M_2-(p_n-1)K_2}{p_n}$ stones from **II** to **I**. After that we have\n$$\n\\begin{aligned} M_1' + K_1' &\\ge M_1 + K_1 + \\frac{M_2 - (p_n - 1)K_2}{p_n} > M_1 + K_1 + \\frac{u \\cdot p_1^{a_1} \\cdots p_n^{a_n} - M_1 - K_1 - p_n K_2}{p_n} \\\\ &= u \\cdot p_1^{a_1} \\cdots p_{n-1}^{a_{n-1}} p_n^{a_n} + \\frac{(p_n - 1)(M_1 + K_1) - p_n K_2}{p_n} \\ge u \\cdot p_1^{a_1} \\cdots p_{n-1}^{a_{n-1}} p_n^{a_n}, \\end{aligned}\n$$\nThe inductive hypothesis implies that we can move $u$ stones to **0**.\n\nb.2) If $M_1 + K_1 < \\frac{p_n}{p_{n-1}}K_2$, then Bernoulli's inequality implies that\n$$\nM_1 + K_1 < \\frac{p_n}{p_n - 1} K_2 \\le 2(1 + a_1) \\cdots (1 + a_{n-1}) \\le p_n^{a_n} p_1^{a_1} \\cdots p_{n-1}^{a_{n-1}},\n$$\n\ni.e. $p_1^{a_1} \\cdots p_n^{a_n} - M_1 - K_1$ is nonnegative. Note that\n$M_2 + K_2 > u \\cdot p_1^{a_1} \\cdots p_n^{a_n} - (M_1 + K_1) = (u-1)p_n^{a_n} \\cdot p_1^{a_1} \\cdots p_{n-1}^{a_{n-1}} + (p_1^{a_1} \\cdots p_n^{a_n} - M_1 - K_1)$,\nand we have $u-1 \\ge 1$ and $p_n \\ge p_{n-1}$. Using $C(n-1)$, we may perform $\\lfloor \\frac{p_1^{a_1} \\cdots p_n^{a_n} - M_1 - K_1}{p_n} \\rfloor$ times $T_{p_n}$ in $\\mathbf{II}$, and then move $(u-1)p_n^{a_n}$ stones to $(0, \\dots, 0, a_n)$. These $(u-1)p_n^{a_n}$ stones can be moved to $u-1$ stone at $\\mathbf{0}$ by operation $X_n$.\nOn the other hand, the stones from the $\\lfloor \\frac{p_1^{a_1} \\cdots p_n^{a_n} - M_1 - K_1}{p_n} \\rfloor$ times operation $T_{p_n}$ can be used to perform operation $X_n$ to get $\\lfloor \\frac{p_1^{a_1} \\cdots p_n^{a_n} - M_1 - K_1}{p_n} \\rfloor$ stones in domain $\\mathbf{I}$. After that, in $\\mathbf{I}$ we have\n$$\n\\begin{aligned}\nM_1' &\\ge M_1 + \\left\\lfloor \\frac{p_1^{a_1} \\cdots p_n^{a_n} - M_1 - K_1}{p_n} \\right\\rfloor \\\\\n&= p_1^{a_1} \\cdots p_{n-1}^{a_{n-1}} p_n^{a_n} + \\left\\lfloor \\frac{(p_n - 1)M_1 - K_1}{p_n} \\right\\rfloor \\ge p_1^{a_1} \\cdots p_{n-1}^{a_{n-1}} p_n^{a_n},\n\\end{aligned}\n$$\nUse the already proved Proposition A(n), we may get one more stone at **0**. Combine these two parts, we obtain *u* stones at **0**.\n\nThe proof of Proposition C(n).\n\nc.1) If $t < q$, we do not need do operation $T_q$. Note that $sq \\ge 2$ and $M + K > sq \\cdot p_1^{a_1} \\cdots p_n^{a_n}$. Use the proved Proposition B(n) we can move $sq$ stones to **0**.\n\nc.2) If $t \\ge q$, we need to do some operation $T_q$. If some place in domain **II** has more than *q* stones, we do an operation $T_q$ at that point, then $M' = M - q$ and $K' = K$; moreover, it becomes the case $t' = t - q$. After some operation $T_q$, we either arrive at the case $t < q$ which is (c.1), or at every point of **II**, the number of stones is less than or equal to *q*. It suffices to deal with the latter situation. Assume that **II** has *L* points which has exactly *q* stones. The Bernoulli's inequality gives\n$$\n\\begin{aligned}\nM_1 + K_1 &= M + K - (M_2 + K_2) > sq \\cdot p_1^{a_1} \\cdots p_n^{a_n} + t - (q(K_2 - L) + (q+1)L) \\\\\n&= sq \\cdot p_1^{a_1} \\cdots p_{n-1}^{a_{n-1}} p_n^{a_n} + t + (sq \\cdot p_1^{a_1} \\cdots p_{n-1}^{a_{n-1}} p_n^{a_n})(p_n - 1) - qK_2 - L) \\\\\n&\\ge sq \\cdot p_1^{a_1} \\cdots p_{n-1}^{a_{n-1}} p_n^{a_n} + t + (q(1+a_1) \\cdots (1+a_{n-1}) - qK_2 - L) \\\\\n&\\ge sq \\cdot p_1^{a_1} \\cdots p_{n-1}^{a_{n-1}} p_n^{a_n} + t - L.\n\\end{aligned}\n$$\nSince $q \\ge p_n$, we can do operation $X_n$ once again at the $L$ points which has exactly $q$ stones in $\\mathbf{II}$. After that,\n$$\nM_1' + K_1' \\ge M_1 + K_1 + L > sq \\cdot p_1^{a_1} \\cdots p_{n-1}^{a_{n-1}} p_n^{a_n} + t,\n$$\nUsing the inductive hypothesis for domain **I**, we know that we can do $\\lfloor \\frac{t}{q} \\rfloor$ times $T_q$, and then move $sq$ stones to **0**.\n\nThis completes the induction.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 28552,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $I$ be the incenter of a triangle $ABC$ with $AB > AC$, and let $AM$ be the median. The line through $I$ perpendicular to $BC$ meets the line $AM$ at point $L$. Let $J$ be the reflection of $I$ with respect to $A$. Prove that $\\angle ABJ = \\angle LBI$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "*Proof.* We need the following lemma.\n**Lemma:** In triangle *ABC*, let *I* be the incenter. The incircle $\\odot I$ is tangent to sides $BC$, $CA$, $AB$ at points $A_1$, $B_1$, $C_1$, respectively. Let $A_1I$ intersect $B_1C_1$ at $L$. Then, $AL$ passes through the midpoint $M$ of $BC$.\n![](attached_image_1.png)\n*Proof of lemma:* Draw a line through $L$ parallel to $BC$, intersecting $AC$ at $X$ and $AB$ at $Y$. Connect $IB_1$, $IC_1$, $IX$, and $IY$.\nSince $A_1L \\perp BC$ and $XY \\parallel BC$, we know that $A_1L \\perp XY$. Thus, $\\angle ILX = \\angle IB_1X = 90^\\circ$, and $I, L, X, B_1$ are concyclic. Similarly, $I, L, C_1, Y$ are concyclic. Therefore, $\\angle IXB_1 = \\angle ILB_1 = \\angle IYC_1$.\nCombining this with $\\angle IB_1X = \\angle IC_1Y = 90^\\circ$ and $IB_1 = IC_1$, we have $\\triangle IB_1X \\cong \\triangle IC_1Y$. Thus, $IX = IY$.\nAlso, note that $IL \\perp XY$, so $L$ is the midpoint of $XY$. Combined with $XY \\parallel BC$, we conclude that $AL$ passes through the midpoint $M$ of $BC$. The lemma is proven.\n\nReturning to the original problem, let the incircle $\\odot I$ of triangle $ABC$ be tangent to sides $BC$, $CA$, $AB$ at points $A_1$, $B_1$, $C_1$, respectively. By the lemma, $L$ is on segment $B_1C_1$. Let $S$ be the midpoint of $BI$, and let $T$ be a point on $AS$ such that $\\angle TBS = \\angle BAS$.\n![](attached_image_2.png)\nWe will now prove that points $B$, $T$, $L$ are collinear.\n\nSince $\\angle TBS = \\angle BAS$, we know $\\triangle BST \\sim \\triangle ASB$. Thus, $BS^2 = SA \\cdot ST$. Also, $BS = SI$, so $IS^2 = SA \\cdot ST$. This implies $\\triangle IST \\sim \\triangle ASI$, and therefore,\n$$\n\\angle BTI = \\angle BTS + \\angle ITS = \\angle ABI + \\angle AIB = 180^\\circ - \\angle BAI.\n$$\nLet $H$ be the orthocenter of triangle $AIB$. By the properties of the orthocenter, $\\angle BHI = \\angle BAI$. Thus, points $B$, $T$, $I$, $H$ are concyclic. Therefore, $\\angle STH = \\angle BTH - \\angle BTS = \\angle BIH - \\angle ABI = 90^\\circ$, and combined with $\\angle AC_1H = 90^\\circ$, we have points $A$, $C_1$, $T$, $H$ concyclic, with $AH$ being the diameter of the circle.\nThus,\n$$\n\\begin{aligned}\n\\angle ITC_1 &= \\angle HTC_1 - \\angle HTI = 180^\\circ - \\angle HAC_1 - \\angle HBI \\\\\n&= 180^\\circ - (90^\\circ - \\angle ABI) - (90^\\circ - \\angle ABI - \\angle BAI) \\\\\n&= 2\\angle ABI + \\angle BAI.\n\\end{aligned}\n$$\nMoreover, note that\n$$\n\\angle ILB_1 = \\angle IC_1L + \\angle LIC_1 = \\frac{1}{2}\\angle BAC + \\angle ABA_1 = \\angle BAI + 2\\angle ABI.\n$$\nSo $\\angle ITC_1 = \\angle ILB_1$, so $I$, $T$, $C_1$, $L$ are concyclic. From this, we deduce $\\angle ITL = \\angle IC_1L = \\angle BAI$. Therefore, points $B$, $T$, $L$ are collinear, and $\\angle LBI = \\angle TBI = \\angle BAS$.\nSince $AS$ is the median of triangle $BIJ$, we have $AS \\parallel BJ$, and $\\angle BAS = \\angle ABJ$. Therefore, $\\angle ABJ = \\angle LBI$. The conclusion is proven. $\\square$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 28553,
    "subject": "Mathematics (Multi-modal)",
    "question": "Fix two positive integers $m$ and $n$. Fix a way to color the vertices of a regular $(2m+2n)$-gon so that $2m$ of them are black and the other $2n$ are white. Define the coloring distance $d(B, C)$ between two black points $B$ and $C$ to be the lesser of the numbers of white points on either side of the line $BC$; similarly, define the coloring distance $d(W, X)$ between two white points $W$ and $X$ to be the lesser of the numbers of black points on either side of the line $WX$.\nA black pairing scheme $\\mathcal{B}$ means to label the $2m$ black points as $B_1, \\dots, B_m, C_1, \\dots, C_m$, such that no two segments from $B_1C_1, \\dots, B_mC_m$ intersect each other. For each such $\\mathcal{B}$, put\n$$\nP(\\mathcal{B}) = \\sum_{i=1}^{m} d(B_i, C_i).\n$$\nA white pairing scheme $\\mathcal{W}$ means to label the $2n$ white points as $W_1, \\dots, W_n, X_1, \\dots, X_n$, such that no two segments from $W_1X_1, \\dots, W_nX_n$ intersect each other. For each such $\\mathcal{W}$, put\n$$\nP(\\mathcal{W}) = \\sum_{i=1}^{n} d(W_i, X_i).\n$$\nProve that, regardless of how the $2m+2n$ vertices are colored, we always have\n$$\n\\max_{\\mathcal{B}} P(\\mathcal{B}) = \\max_{\\mathcal{W}} P(\\mathcal{W}),\n$$\nwhere the maxima are taken over all possible black pairing schemes and all possible white pairing schemes, respectively.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "**Proof:** Consider $2m+2n$ vertices placed on the unit circle. The Euclidean distances between vertices are irrelevant to the discussion. We define the line segments in the white pairing scheme as white segments and those in the black pairing scheme as black segments.\n\n**Lemma 1:** For any black pairing scheme $\\mathcal{B}$ and any white point pairing scheme $\\mathcal{W}$, the number of intersection points $I(\\mathcal{B}, \\mathcal{W})$ between the $m+n$ line segments connected in the schemes is bounded above by $P(\\mathcal{B})$ and by $P(\\mathcal{W})$, respectively.\n\n![](attached_image_1.png)\n\n(For the black and white pairing schemes $\\mathcal{B}$ and $\\mathcal{W}$ shown in the figure, we replace white segments with red ones and denote intersection points with blue, so that $I(\\mathcal{B}, \\mathcal{W}) = 5$, $P(\\mathcal{B}) = 2+3+2=7$, and $P(\\mathcal{W}) = 1+3+1=5$.)\n\n*Proof of Lemma 1:* For any line segment $B_iC_i$ in $\\mathcal{B}$, the number of white line segments intersecting it is at most $d(B_i, C_i)$, because each white line segment corresponds to a white point on each side of $B_iC_i$. Summing over all black segments in $\\mathcal{B}$ yields the inequality $I(\\mathcal{B}, \\mathcal{W}) \\le P(\\mathcal{B})$. Similarly, we have $I(\\mathcal{B}, \\mathcal{W}) \\le P(\\mathcal{W})$. Therefore, Lemma 1 is proved.\n\n**Lemma 2:** For any black pairing scheme $\\mathcal{B}$, there exists a white pairing scheme $\\mathcal{W}$ such that $I(\\mathcal{B}, \\mathcal{W}) = P(\\mathcal{B})$. Thus, $P(\\mathcal{B}) = I(\\mathcal{B}, \\mathcal{W}) \\le P(\\mathcal{W})$.\n\nWe now show that Lemma 2 proves the original problem. Apply Lemma 2 to all $\\mathcal{B}$ to obtain $\\max_{\\mathcal{B}} P(\\mathcal{B}) \\le \\max_{\\mathcal{W}} P(\\mathcal{W})$. A symmetric argument shows that $\\max_{\\mathcal{W}} P(\\mathcal{W}) \\le \\max_{\\mathcal{B}} P(\\mathcal{B})$. Therefore, the problem is proved.\n\n*Proof of Lemma 2 (Method 1):* We prove by induction on the number $2n$ of white points that there exists a white pairing scheme $\\mathcal{W}$. When $n=0$, the claim is trivial. Suppose the claim is true for $2n-2$ white points. Consider the case of $2n$ white points. Among all black pairings $B_iC_i$, let $B_1C_1$ be the one with the largest $d(B_i, C_i)$. If $d(B_1, C_1) = 0$, we can arbitrarily choose a white pairing scheme. Assume that $d(B_1, C_1) > 0$. Let $W_n$ be the first white point encountered when rotating clockwise from $B_1$ and let $X_n$ be the first white point encountered when rotating counterclockwise from $B_1$. As shown in the following figure,\n\n![](attached_image_2.png)\n\nNow consider the case where we remove two white points $W_n$ and $X_n$. Let $d'(B, C)$ be the chromatic distance in this case, and let $P'(\\mathcal{B}) = \\sum_{i=1}^m d'(B_i, C_i)$. By the induction hypothesis, there exists a white pairing scheme $\\mathcal{W}'$ for $\\mathcal{B}$ without $W_n$ and $X_n$ such that $I(\\mathcal{B}, \\mathcal{W}') = P'(\\mathcal{B})$. We will prove that we can obtain the desired white pairing scheme $\\mathcal{W}$ by adding the line segment $W_nX_n$ to $\\mathcal{W}'$.\n\nWe consider each segment $BC$ in $\\mathcal{B}$. If it intersects with $W_nX_n$, then removing $W_n$ and $X_n$ is equivalent to removing one white point on each side of $BC$, so $d'(B, C) = d(B, C) - 1$. If $BC$ does not intersect with $W_nX_n$, we want to show that $d(B, C) = d'(B, C)$. Note that if there are more than $n+1$ white points on one side of $BC$ with respect to $W_nX_n$, while the other side has no more than $n-1$ white points, then we must have $d(B, C) = d'(B, C)$. On the other hand, if there are no more than $n$ white points on one side of $BC$, and $BC$ does not intersect with either $B_1C_1$ or $W_nX_n$, then the entire segment $B_1C_1$ must be on the side of $BC$ with no more than $n$ white points. This contradicts our assumption that $d(B, C) > d(B_1, C_1)$, since $BC$ is closer to $B_1C_1$ than any other black segment. Therefore, we have $d(B, C) = d'(B, C)$.\n\nThus, we see that $P(\\mathcal{B}) - P'(\\mathcal{B})$ is exactly equal to the number of black segments in $\\mathcal{B}$ that intersect $W_nX_n$, which is equal to $I(\\mathcal{B}, \\mathcal{W}) - I(\\mathcal{B}, \\mathcal{W}')$. Therefore, we have $I(\\mathcal{B}, \\mathcal{W}) = P(\\mathcal{B})$. This completes the induction proof.\n\n*Proof of Lemma 2 (Method 2):* This proof is essentially the same as Method 1, but the statement is worded differently. For each black segment $BC$ in $\\mathcal{B}$, we call the set of white points on the side with fewer white points an *internal white point set* (if $d(B, C) = n$, choose one side arbitrarily to define the internal white point set). Note that any two internal white point sets (for two different black segments) either do not intersect or one is contained in the other.\n\nWe choose all the largest internal white point sets, and consider the points outside these sets as their own internal white point sets. We only need to prove there exists a white pairing scheme such that no two white segments correspond to white points in the same internal white point set. To do this, we first choose the largest internal white point set, connect the last point in clockwise order to the next point in counterclockwise order, and remove these two white points. We then consider the remaining white points and internal white point sets. Clearly, the resulting white point scheme is valid, and it satisfies the condition because by induction, we know that at any point in the algorithm, the number of white points in the largest internal white point set is less than or equal to the total number of white points in the rest of the internal white point sets.\n\n**Proof of Lemma 2 (Method 3):**\nFirst, we construct a weak white pairing scheme $\\mathcal{W}'$, where white points are matched without requiring the corresponding white segments to have mutually disjoint pairs, such that $I(\\mathcal{B}, \\mathcal{W}') = P(\\mathcal{B})$. In fact, if $W_1, W_2, \\dots, W_{2n}$ are $2n$ white points arranged in sequence on the circle, we just need to match $W_i$ with $W_{n+i}$ ($1 \\le i \\le n$) to achieve this, because such a pairing will connect the white points with fewer white points on one side of any black segment to the other side.\n\nWe select from all weak white pairing schemes $\\mathcal{W}'$ that satisfy $I(\\mathcal{B}, \\mathcal{W}') = P(\\mathcal{B})$ the one with the minimum Euclidean distance sum of white segments, denoted by $\\mathcal{W}_0$. We then prove that any two of the $n$ white segments in $\\mathcal{W}_0$ do not intersect.\n\nSuppose, to the contrary, that two segments $WX$ and $YZ$ intersect, and without loss of generality, we assume they are arranged clockwise on the circle as $W, Y, X, Z$. As black segments in $\\mathcal{B}$ do not intersect, there cannot be a black segment connecting $\\overline{YX}$ and $\\overline{WZ}$, as well as a black segment connecting $\\overline{XZ}$ and $\\overline{WY}$ simultaneously. Without loss of generality, we assume there is no black segment connecting $\\overline{YX}$ and $\\overline{WZ}$. We then modify segments $YZ$ and $WX$ to be $WY$ and $ZX$, respectively, obtaining a new weak white pairing scheme $\\mathcal{W}'_0$, which clearly reduces the Euclidean distance sum of all white segments.\n\n![](attached_image_3.png)\n\nWe now compare $I(\\mathcal{B}, \\mathcal{W}_0)$ and $I(\\mathcal{B}, \\mathcal{W}'_0)$ using the following observations:\n* For any black segment that connects $\\overline{WY}$ and $\\overline{XY}$, or $\\overline{ZX}$ and $\\overline{XY}$, or $\\overline{WY}$ and $\\overline{WZ}$, or $\\overline{ZX}$ and $\\overline{WZ}$, the number of intersections with $WX$ and $YZ$ is precisely one, and the number of intersections with $WY$ and $ZX$ is also one.\n* For any black segment that connects $\\overline{WY}$ and $\\overline{ZX}$, the number of intersections with each of $WY, WX, YZ$, and $ZX$ is also one.\n\nThus, we have $I(\\mathcal{B}, \\mathcal{W}_0) = I(\\mathcal{B}, \\mathcal{W}'_0)$. However, this contradicts the fact that the sum of Euclidean distances of all white segments in $\\mathcal{W}_0$ is minimal. Therefore, $\\mathcal{W}_0$ is a valid white pairing scheme, and the proof of Lemma 2 is complete.",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 28554,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $a_1, a_2, \\dots, a_n$ be $n$ positive integers that are not divisible one another, i.e. for any $i \\neq j$, $a_i$ is not divisible by $a_j$. Show that\n$$\na_1 + a_2 + \\dots + a_n \\geq 1.1n^2 - 2n.\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "**Proof:** Consider the set of all positive integers that are coprime to $6$: $B = \\{1, 5, 7, 11, 13, \\dots\\}$, it is obvious that the sum $f(k)$ of the smallest $k$ elements of $B$ satisfies\n$$\nf(k) = \\begin{cases} \\frac{3k^2}{2} & \\text{if } k \\text{ is even} \\\\ \\frac{3k^2-1}{2} & \\text{if } k \\text{ is odd} \\end{cases}, \\qquad f(k+1) - f(k) = 3k + 1 \\text{ or } 3k + 2.\n$$\n$$\nf(0) = 0,\\ f(1) = 1,\\ f(2) = 6,\\ f(3) = 13,\\ f(4) = 24,\\ f(5) = 37, \\dots\n$$\nEvery positive integer $a$ can be uniquely written as $a = 2^\\alpha 3^\\beta b$, with $b \\in B$ and nonnegative integers $\\alpha, \\beta$. Denote $h(a) = b$ or say $b$ is the core of $a$. If $a_1, \\dots, a_m$ are $m$ positive integers that cannot divide one another, and that all of their cores are $b$, then writing each $a_k = 2^{\\alpha_k} 3^{\\beta_k} b$ with $k = 1, \\dots, m$, we must have that $\\alpha_1, \\dots, \\alpha_m$ are pairwise distinct (otherwise two numbers $2^\\alpha 3^\\beta b$ and $2^\\alpha 3^{\\beta'} b$ must have division relations). We may put them in a sequence so that $\\alpha_1 < \\alpha_2 < \\dots < \\alpha_m$, then correspondingly we have $\\beta_1 > \\beta_2 > \\dots > \\beta_m \\ge 0$, this time we get $a_k \\ge 2^{k-1} 3^{m-k} b$. Moreover, the sum of these $m$ numbers\n$$\na_1 + a_2 + \\dots + a_m \\ge 2^0 3^{m-1} b + 2^1 3^{m-2} b + \\dots + 2^{m-1} 3^0 b = (3^m - 2^m)b.\n$$\nWe consider the map from $A = \\{a_1, a_2, \\dots, a_n\\}$ to $B$ by taking the cores, i.e. every $b \\in B$ is the core for several numbers in $A$. Set\n$$\nB_k = \\{b \\in B : \\#\\{a \\in A : h(a) = b\\} \\ge k\\}.\n$$\ni.e. the numbers in $B_k$ are the core for at least $k$ numbers in $\\{a_1, \\dots, a_n\\}$. Thus $B \\supseteq B_1 \\supseteq B_2 \\supseteq \\dots$, and the numbers in $B_k \\setminus B_{k+1}$ are precisely the core for exactly $k$ numbers in $\\{a_1, \\dots, a_n\\}$. We have\n$$\nn = \\sum_{k=1}^{\\infty} k \\cdot |B_k \\setminus B_{k+1}| = |B_1| + |B_2| + \\dots\n$$\nWe use $S(\\cdot)$ to denote the sum of the elements in a set. We have:\n$$\nS(A) = a_1 + \\dots + a_n \\ge \\sum_{k=1}^{\\infty} (3^k - 2^k) \\times S(B_k \\setminus B_{k+1}) = \\sum_{k=1}^{\\infty} [(3^k - 2^k) - (3^{k-1} - 2^{k-1})] \\times S(B_k)\n$$\n$$\nS(A) \\geq \\sum_{k=1}^{\\infty} (2 \\times 3^{k-1} - 2^{k-1}) \\times f(|B_k|)\n$$\nDefine the sequence $c_k = 2 \\times 3^{k-1} - 2^{k-1}$, e.g. $c_1 = 1$, $c_2 = 4$, $c_3 = 14$, $c_4 = 46$, $c_5 = 146$, ...\nWe turn to consider the optimization problem $\\spadesuit$: under the assumption $x_1 + x_2 + \\dots = n$ (with each $x_i$ nonnegative), minimize $T = c_1f(x_1) + c_2f(x_2) + \\dots$.\nSuppose that $X = (x_1, x_2, \\dots, x_K, 0, 0, \\dots)$ minimizes $T$, where $x_1 \\ge x_2 \\ge \\dots \\ge x_K \\ge 1$, and $x_{K+1} = x_{K+2} = \\dots = 0$.\nIf $K \\le 2$, then\n$$\nT = c_1f(x_1) + c_2f(x_2) \\ge \\frac{3x_1^2-1}{2} + 4\\frac{3x_2^2-1}{2} \\ge \\frac{6}{5}(x_1+x_2)^2 - \\frac{5}{2} \\ge 1.1n^2 - 2n.\n$$\nIf $K \\ge 3$, then $X$ being (one of) the minimizing points would mean:\n* if we change $X$ to $X' = (x_1+1, x_2, \\dots, x_{K-1}, x_K - 1, 0, \\dots)$, $T$ does not decrease, i.e.\n$$\n0 \\le \\Delta T = c_1(f(x_1+1)-f(x_1)) - c_K(f(x_K)-f(x_K-1)) \\le 3x_1+2 - c_K \\Rightarrow c_K \\le 3x_1+2;\n$$\n* if we change $X$ to $X'' = (x_1, x_2 + 1, \\dots, x_{K-1}, x_K - 1, 0, \\dots)$, then $T$ does not decrease, i.e.\n$$\n0 \\le \\Delta T = c_2(f(x_2+1)-f(x_2)) - c_K(f(x_K)-f(x_K-1)) \\le 4(3x_2+2) - c_K \\Rightarrow c_K \\le 12x_2+8.\n$$\nSo $n = x_1 + x_2 + \\dots + x_K \\ge x_1 + x_2 + 1 \\ge \\frac{c_K-2}{3} + \\frac{c_K-8}{12} + 1 = \\frac{5c_K-4}{12}$, we get\n$$\nc_1 + c_2 + \\dots + c_K \\le c_K \\left(1 + \\frac{1}{3} + \\dots + \\frac{1}{3^{K-1}}\\right) \\le \\frac{12n+4}{5} \\times \\frac{3}{2} \\le 4n.\n$$\nThus, $T(X) = c_1f(x_1) + \\dots + c_Kf(x_K) \\ge c_1 \\frac{3x_1^2-1}{2} + \\dots + c_K \\frac{3x_K^2-1}{2}$ and\n$$\nT(X) \\ge \\frac{3}{2}[c_1x_1^2 + c_2x_2^2 + \\dots + c_Kx_K^2] - \\frac{1}{2}[c_1 + c_2 + \\dots + c_K] \\ge \\frac{3}{2} \\frac{(x_1 + x_2 + \\dots + x_K)^2}{\\frac{1}{c_1} + \\frac{1}{c_2} + \\dots + \\frac{1}{c_K}} - 2n.\n$$\nSince $c_{k+1} \\ge 3c_k$ always holds, we have\n$$\n\\frac{1}{c_1} + \\frac{1}{c_2} + \\dots + \\frac{1}{c_K} \\le 1 + \\frac{1}{4} + \\frac{1}{14} + \\frac{1}{14 \\times 3} + \\frac{1}{14 \\times 3^2} + \\dots = 1 + \\frac{1}{4} + \\frac{1.5}{14} \\le 1.36\n$$\nSo\n$$\nT \\ge \\frac{3}{2} \\frac{n^2}{1.36} - 2n > 1.1n^2 - 2n\n$$\nTo sum up, the optimization problem $\\spadesuit$ has minimum value $T_{\\min} \\ge 1.1n^2 - 2n$, and for the original question,\n$$\nS(A) \\ge \\sum_{k=1}^{\\infty} c_k \\times f(|B_k|) \\ge 1.1n^2 - 2n.\n$$",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 28555,
    "subject": "Mathematics (Multi-modal)",
    "question": "Fix an integer $n \\ge 2$. Find all $n$-tuples $(a_1, a_2, \\dots, a_n)$ of integers satisfying the following two conditions:\n(1) $a_1$ is odd, $1 < a_1 \\le a_2 \\le \\dots \\le a_n$, and $M = \\frac{1}{2^n}(a_1 - 1)a_2 \\dots a_n$ is an integer; and\n(2) there exist $M$ different $n$-tuples $(c_{i,1}, c_{i,2}, \\dots, c_{i,n})$ with $i = 1, 2, \\dots, M$, such that for all $1 \\le i < j \\le M$, there exists $k \\in \\{1, 2, \\dots, n\\}$ such that\n$$\nc_{i,k} - c_{j,k} \\not\\equiv -1, 0, 1 \\pmod{a_k}.\n$$",
    "options": [],
    "answer": "Exactly those tuples for which, if r is the number of odd numbers among a2,…,an, then 2^r divides a1 − 1.",
    "solution": "The $n$-tuples we seek are the ones satisfying the following condition:\nif there are exactly $r$ odd numbers in $a_2, \\dots, a_n$, then $2^r \\mid a_1 - 1$. $(*)$\n\nWe first verify the necessity of $(*)$. For this, we drop the condition $a_n \\ge a_{n-1} \\ge \\dots \\ge a_1$ and assume that $a_1, \\dots, a_r$ are odd and $a_{r+1}, \\dots, a_n$ are even. Suppose that we are given the $M$ tuples satisfying the requirement (2). For each $s \\in \\mathbb{Z}$, denote $B_s = \\{i \\mid 1 \\le i \\le M, c_{i,n} \\equiv s \\pmod{a_n}\\}$. Then\n$$\n|B_1| + |B_2| + \\dots + |B_{a_n}| = M.\n$$\nConsequently, there exists an index $s$ such that $|B_s| + |B_{s+1}| \\ge \\frac{M}{a_n/2}$. This means that we can choose at least $\\frac{M}{a_n/2}$ tuples such that the differences between their $n$-th coordinates $c_{i,n}$ are all congruent to $0$ or $\\pm 1$ modulo $a_n$.\n\nBy running the same argument for all the tuples inductively, i.e., to consider the $(n-1)$th coordinates modulo $a_{n-1}$, the $(n-2)$th coordinates modulo $a_{n-2}$, etc., it eventually shows that there are at least $\\frac{M}{\\frac{a_n}{2} \\dots \\frac{a_2}{2}} = \\frac{a_1-1}{2}$ tuples such that for each two of them, the differences between their $k$th coordinates $c_{i,k}$'s are congruent to $0$ or $\\pm 1$ modulo $a_k$, where $2 \\le k \\le n$. However, the given conditions imply that the difference between the first coordinates of these tuples can not be $0$ or $\\pm 1$ (mod $a_1$); there are at most $\\frac{a_1-1}{2}$ such tuples. This means that all equalities must hold in the argument above. Namely, for each $t$ we have exactly $\\frac{M}{\\frac{a_n}{2} \\dots \\frac{a_t}{2}}$ different tuples. In particular, taking $t = r + 1$ leads to\n$$\n\\frac{M}{\\frac{a_n}{2} \\dots \\frac{a_{r+1}}{2}} = \\frac{1}{2^r} (a_1 - 1)a_2 \\dots a_r \\in \\mathbb{Z}.\n$$\nIt forces that $2^r \\mid a_1 - 1$.\n\nIn the following, we construct the needed $M$ tuples under the condition $2^r \\mid a_1 - 1$. For convenience, we first weaken the condition $a_n \\ge a_{n-1} \\ge \\dots \\ge a_1$ to only requiring $a_1 = \\min\\{a_1, \\dots, a_n\\}$. Whenever there is any even number in $a_2, \\dots, a_n$, say $a_n$ without loss of generality. Suppose that we may construct the needed tuples for $a_1, \\dots, a_{n-1}$, say $(c_{i,1}, c_{i,2}, \\dots, c_{i,n-1})$ with $1 \\le i \\le M' = \\frac{1}{2^{n-1}}(a_1 - 1)a_2 \\dots a_n = \\frac{2}{a_n}M$. Then it suffices to take $(c_{i,1}, c_{i,2}, \\dots, c_{i,n-1}, 2c)$ with $1 \\le i \\le M'$ and $1 \\le c \\le \\frac{a_n}{2}$.\n\nNow it remains to prove for the case when all $a_2, a_3, \\dots, a_n$ are odd. Let us write $a_1 = 2^n t + 1$ and consider the case when $a_1 = a_2 = \\dots = a_n$ first. In this case, $M = t a_1^{n-1}$. Define the function $f(x_1, x_2, \\dots, x_{n-1}) = \\sum_{i=1}^{n-1} 2^i x_i$ and take the following $M = t a_1^{n-1}$ tuples:\n$$\n(x_1, x_2, \\dots, x_{n-1}, f(x_1, \\dots, x_{n-1}) + 2^n s),\n$$\nwhere $x_1, \\dots, x_{n-1} \\in \\{1, 2, \\dots, a_1\\}, s \\in \\{1, \\dots, t\\}$. We now show that these tuples satisfy the desired conditions. Consider the above tuple together with another one: $(x'_1, x'_2, \\dots, x'_{n-1}, f(x'_1, \\dots, x'_{n-1}) + 2^n s')$. If for every $k$, the difference between their $k$th coordinate $\\equiv 0, \\pm 1 \\pmod{a_1}$, then\n$$\nx_i - x'_i \\equiv 0, \\pm 1 \\pmod{a_1}, \\quad i = 1, \\dots, n-1; \\text{ and}\n$$\n$$\n\\sum_{i=1}^{n-1} 2^{i-1}(x_i - x'_i) + 2^n(s - s') \\equiv 0, \\pm 1 \\pmod{a_1}. \\quad (*)\n$$\nThe first equality implies that $\\sum_{i=1}^{n-1} 2^i(x_i - x'_i)$ can be only congruent to $2^{n-1} - 1, 2^{n-2} - 1, \\dots, 1 - 2^{n-1} \\pmod{a_1}$. On the other hand, $s - s' \\in \\{1 - t, 2 - t, \\dots, t - 1\\}$. So $s$ is forced to be equal to $s'$ by $(*)$, and $\\sum_{i=1}^{n-1} 2^{i-1}(x_i - x'_i) = 0 \\pmod{a_1}$. By the uniqueness of binary expansion, we get $x_i = x'_i \\pmod{a_1}$, which implies that the two tuples are the same. This verifies that the constructed tuples satisfy the needed conditions.\n\nFor the general case where $a_1, \\dots, a_n$ are all odd numbers, we claim that if $a_2 > a_1$, the construction can be reduced to the case with $a_1, a_2-2, a_3, \\dots, a_n$. Hence by induction, it suffices to consider the case where all $a_i$ are equal, which is discussed above.\n\nAssume now there exist $M' = \\frac{1}{2^n}(a_1 - 1)(a_2 - 2)a_3 \\cdots a_n$ tuples as desired when $a'_1 = a_1, a'_2 = a_2 - 2, a'_3 = a_3, \\dots, a'_n = a_n$. We may assume that for each $k$, the $k$th coordinates of all these tuples belong the set $\\{0, 1, \\dots, a'_k - 1\\}$.\n\nWe then define the $M = \\frac{1}{2^n}(a_1 - 1)a_2a_3 \\cdots a_n$ tuples as follows. One can first choose the $M'$ tuples that are already defined. As for $x_2 = a_2 - 2$, we choose to add the tuple $(x_1, a_2 - 2, x_3, \\dots, x_n)$ if and only if the tuple $(x_1, a_2 - 4, x_3, \\dots, x_n)$ was selected; and for $x_2 = a_2 - 1$, we choose to add the tuple $(x_1, a_2 - 1, x_3, \\dots, x_n)$ if and only if the tuple $(x_1, a_2 - 3, x_3, \\dots, x_n)$ was selected. It is easy to check that these tuples satisfy our requirement. Also, using the proof for $2^r|a_1 - 1$ by induction, the condition for equality implies the following: Among the $M'$ tuples we have selected before, the number of tuples whose second coordinate is either $a_2 - 1$ or $a_2 - 2$ is $M' \\cdot \\frac{2}{a_2-2}$. Hence we can the number of new tuples is $M' + M' \\cdot \\frac{2}{a_2-2} = M$. This completes the construction of the tuples.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 28556,
    "subject": "Mathematics (Multi-modal)",
    "question": "Given a positive integer $n$, find all $n$-tuples of real numbers $(x_1, x_2, \\dots, x_n)$ such that\n$$\nf(x_1, x_2, \\dots, x_n) = \\sum_{k_1=0}^{2} \\sum_{k_2=0}^{2} \\cdots \\sum_{k_n=0}^{2} \\left| k_1 x_1 + k_2 x_2 + \\cdots + k_n x_n - 1 \\right|\n$$\nattains its minimum.",
    "options": [],
    "answer": "x1 = x2 = ... = xn = 1/(n+1)",
    "solution": "We divide the set $A = \\{0, 1, 2\\}^n$ into: $A = A_0 \\cup A_1 \\cup \\cdots \\cup A_{2n}$, where\n$$\nA_k = \\{\\beta = (i_1, i_2, \\dots, i_n) \\in A : i_1 + i_2 + \\dots + i_n = k\\}, \\quad k = 0, 1, \\dots, 2n.\n$$\nLet $a_k$ denote the number of elements in $A_k$. Then\n$$\n(1 + t + t^2)^n = a_0 + a_1 t + \\cdots + a_n t^n + \\cdots + a_{2n} t^{2n}, \\quad a_{2n-k} = a_k.\n$$\nSet $X = (x_1, \\dots, x_n)$ and $y = (x_1 + \\dots + x_n)/n$. Then $f(X) = \\sum_{k=0}^{2n} \\sum_{\\beta \\in A_k} |\\beta \\cdot X - 1|$. Removing the absolute value signs and summing them in groups, we consider\n$$\nB_k = \\sum_{\\beta \\in A_k} (\\beta \\cdot X - 1) = \\frac{k|A_k|}{n}(x_1 + \\dots + x_n) - |A_k| = k a_k y - a_k.\n$$\nWe estimate $\\sum_{k=0}^{2n} |B_k|$, by canceling appropriate terms. Consider the sizes of $k a_k$ and note that\n$$\n\\sum_{k=0}^{2n} k a_k t^{k-1} = \\frac{d}{dt}[(1+t+t^2)^n] = n(1+2t)(1+t+t^2)^{n-1}\n$$\nand that the coefficients of $(1+t+t^2)^{n-1} = \\sum_{k=0}^{2n-2} c_k t^k$ $(c_0, c_1, \\dots, c_{n-1}, \\dots, c_{2n-2})$ is palindromic. Since $k a_k = c_{k-1} + 2c_{k-2}$, we have\n$$\n(n+1)a_{n+1} \\ge n a_n \\ge (n+2)a_{n+2} \\ge (n-1)a_{n-1} \\ge (n+3)a_{n+3} \\ge \\dots \\ge 2a_2 \\ge 2n a_{2n} \\ge 1a_1.\n$$\nThis shows that $U_1 := \\sum_{k=0}^n k a_k < \\sum_{k=n+1}^{2n} k a_k$ and $U_2 := \\sum_{k=n+2}^{2n} k a_k < \\sum_{k=0}^{n+1} k a_k$. Choose $\\lambda = \\frac{U_1 - U_2}{(n+1)a_{n+1}} \\in (-1, 1)$, we have\n$$\nf(X) \\ge \\sum_{k=0}^{2n} |B_k| \\ge \\left[ \\sum_{k=0}^n (-B_k) + \\lambda B_{n+1} + \\sum_{k=n+2}^{2n} B_k \\right] + (1 - |\\lambda|) |B_{n+1}|.\n$$\nThe coefficient of $y$ in the above square bracket is $-U_1 + \\lambda(n+1)a_{n+1} + U_2 = 0$. So the expression in the square bracket is a constant $C = \\sum_{k=0}^n a_k - \\lambda a_{n+1} + \\sum_{k=n+2}^{2n} a_k$. Therefore, $f(X) \\ge C$ is always true, and when $x_1 = \\dots = x_n = \\frac{1}{n+1}$, the equality holds. Moreover, the equality holds precisely when $a_{n+1}$ terms in the formula above are all zero, i.e. the inner product of $X = (x_1, \\dots, x_n)$ with an arbitrary vector in $A_{n+1}$ is 1. The only such $X$ is when $x_1 = \\dots = x_n = \\frac{1}{n+1}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 28557,
    "subject": "Mathematics (Multi-modal)",
    "question": "Given a prime number $p \\ge 5$. Find the number of different residues for the product of three consecutive positive integers modulo $p$.",
    "options": [],
    "answer": "floor((2p+1)/3)",
    "solution": "Let the set $D = \\{0, 1, \\dots, p-1\\}$, and let the polynomial $f(x) = (x-1)x(x+1) = x^3-x$. The congruence and congruence symbol “$\\equiv$” in this question refer to congruence modulo $p$. For $k=0, 1, 2, 3$, let\n$$\nB_k = \\{b \\in D \\mid \\text{there are exactly } k \\text{ elements } a \\in D \\text{ such that } f(a) \\equiv b\\}.\n$$\nConsider an element $b$ in $B_2$, i.e., there exist $a_1 \\not\\equiv a_2$ such that $f(a_1) \\equiv f(a_2) \\equiv b$. In this case,\n$$\n0 \\equiv (a_1^3 - a_1) - (a_2^3 - a_2) = (a_1 - a_2)(a_1^2 + a_1a_2 + a_2^2 - 1) \\Rightarrow a_1^2 + a_1a_2 + a_2^2 \\equiv 1.\n$$\nThus, for $a_3 = -a_1 - a_2$, we have $a_3^2 + a_3a_1 + a_1^2 = (a_1 + a_2)^2 - (a_1 + a_2)a_1 + a_1^2 \\equiv 1$, so it can be deduced that $f(a_3) \\equiv f(a_1) \\equiv b$, by the definition of $B_2$, $a_3$ is congruent to one of $a_1, a_2$, so either $a_2 \\equiv -2a_1$ or $a_1 \\equiv -2a_2$, assume the former, in this case, $a_1^2 + a_1a_2 + a_2^2 \\equiv 3a_1^2 \\equiv 1$, $3b \\equiv 3a_1^3 - 3a_1 \\equiv -2a_1$. When 3 is a quadratic residue modulo $p$, $a_1$ has two solutions; when 3 is not a quadratic residue modulo $p$, $a_1$ has no solution. Therefore, there are exactly $1 + (\\frac{3}{p})$ solutions for $a_1$, and the corresponding $b$ also has $1 + (\\frac{3}{p})$ solutions, i.e., $|B_2| = 1 + (\\frac{3}{p})$.\n$$\n\\text{Obviously, } |B_1| + 2|B_2| + 3|B_3| = |D| = p.\n$$\nOn the other hand, consider the pairs $(u, v)$ that satisfy \"$f(u) \\equiv f(v)$ and $u \\not\\equiv v$\" (called a collision). This is equivalent to $u^2 + uv + v^2 \\equiv 1$ and $u - v \\not\\equiv 0$. After changing variables to $(x, y) \\equiv (\\frac{u-v}{2}, \\frac{u+v}{2})$ (i.e., $(u, v) \\equiv (x+y, y-x)$, such that $(x, y)$ and $(u, v)$ correspond one-to-one), it is transformed into $x^2 + 3y^2 \\equiv 1$ and $x \\not\\equiv 0$.\nWe consider the number of $(x, y)$ pairs satisfying $x^2+3y^2 \\equiv 1 \\pmod p$ (i.e., $(3y)^2 \\equiv 3-3x^2$).\n$$\n\\begin{aligned}\nT &= \\sum_{x=1}^{p} 1 + \\left(\\frac{3-3x^2}{p}\\right) = p + \\left(\\frac{-3}{p}\\right) \\sum_{x=1}^{p} \\left(\\frac{x^2-1}{p}\\right) = p + \\left(\\frac{-3}{p}\\right) \\sum_{x=2}^{p} \\left(\\frac{\\frac{x+1}{x-1}}{p}\\right) \\\\\n&= p + \\left(\\frac{-3}{p}\\right) \\sum_{x=1}^{p-1} \\left(\\frac{1+\\frac{2}{x-1}}{p}\\right) = p + \\left(\\frac{-3}{p}\\right) \\sum_{z=1}^{p-1} \\left(\\frac{z+1}{p}\\right) = p - \\left(\\frac{-3}{p}\\right).\n\\end{aligned}\n$$\nThe above formula uses the fact that $z = \\frac{2}{x-1}$ traverses the complete set of residues modulo $p$, and that $\\sum_{k=1}^{p}(\\frac{k}{p}) = 0$. Among the $T$ pairs $(x, y)$, there are exactly $1 + (\\frac{3}{p})$ pairs for which $x \\equiv 0$ (i.e., $3y^2 \\equiv 1$). Therefore, the number of ordered collision pairs $(u, v)$ is\n$$\nM = T - 1 - \\left(\\frac{3}{p}\\right) = p - \\left(\\frac{-3}{p}\\right) - 1 - \\left(\\frac{3}{p}\\right) = T - |B_2|.\n$$\nThus, there are exactly $\\frac{M}{2}$ unordered collision pairs $(u, v)$. Since the elements in $B_3$ correspond to 3 collisions, the elements in $B_2$ correspond to 1 collision, and the elements in $B_1$ and $B_0$ correspond to 0 collisions, it follows that $|B_2| + 3|B_3| = \\frac{M}{2}$.\nThus, $|B_3| = \\frac{M-2|B_2|}{6}$, so the number of all possible residues of $f(x)$ modulo $p$ is\n$$\n|B_1| + |B_2| + |B_3| = p - |B_2| - 2|B_3| = p - \\frac{M + |B_2|}{3} = p - \\frac{T}{3} = \\frac{2p + (\\frac{-3}{p})}{3} = \\left\\lfloor \\frac{2p + 1}{3} \\right\\rfloor.\n$$\n\n**Note 1:** Another way to calculate the number of solutions $T$ is:\nConsider the congruence equation $x^2 + 3y^2 \\equiv z^2$, which is equivalent to $3y^2 \\equiv z^2 - x^2 = (z+x)(z-x)$, having $p^2$ solutions ($y \\equiv 0$ has exactly $2p-1$ solutions, $y \\equiv 1$ has exactly $p-1$ solutions, and so on). The number of solutions with $z \\equiv 0$ is\n$$\nm_0 = 1 + (p-1)[1 + \\left(\\frac{-3}{p}\\right)],\n$$\nand the number of solutions for the remaining $z \\equiv 1, z \\equiv 2, ..., z \\equiv p-1$ are equal (using $(x, y, z) \\leftrightarrow (kx, ky, kz)$ to map the solutions with $z \\equiv 1$ to the solutions with $z \\equiv k$), each equal to $T = \\frac{p^2-m_0}{p-1} = p - \\left(\\frac{-3}{p}\\right)$.\nIf one is not familiar with Legendre symbols, one can first obtain $0 \\le m_0 \\le 2p$, and deduce that the number of solutions to $x^2 + 3y^2 \\equiv 1$ is $T \\in p-1, p, p+1$. The subsequent answer to this problem is $p - \\frac{T}{3}$, which is an integer. From this, we can still get the needed answer.\n**Note 2:** Another interpretation of the number of solutions $T = p - \\left(\\frac{-3}{p}\\right)$ is: For each $y$, the number of $x$ that satisfy $x^2 \\equiv 1 - 3y^2$ is $1 + \\left(\\frac{1-3y^2}{p}\\right)$, therefore\n$$\n\\begin{aligned}\nT - p &= \\sum_{y=0}^{p-1} \\left( \\frac{1 - 3y^2}{p} \\right) \\equiv \\sum_{y=0}^{p-1} (1 - 3y^2)^{\\frac{p-1}{2}} \\\\\n&= \\sum_{y=0}^{p-1} \\left( \\left(-3\\right)^{\\frac{p-1}{2}} y^{p-1} + \\cdots + 1 \\right) \\equiv (-3)^{\\frac{p-1}{2}} (p-1) = - \\left( \\frac{-3}{p} \\right).\n\\end{aligned}\n$$",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 28558,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all prime numbers $p$ and positive integers $a, b, c$ such that\n$$\n2^a p^b = (p + 2)^c + 1.\n$$",
    "options": [],
    "answer": "[(3, 1, 1, 1), (5, 1, 2, 2)]",
    "solution": "Obviously, $p$ is odd, so $p \\ge 3$. If $c = 1$, then\n$$\np + 3 = 2^a p^b \\ge 2p \\ge p + 3,\n$$\nThis can only happen when $p = 3$, $a = b = 1$, giving one solution $(p, a, b, c) = (3, 1, 1, 1)$. In what follows, we assume that $c \\ge 2$.\n\nCase 1: $c$ is odd. Assume that $q$ is a prime factor of $c$. Since $(p+2)^q+1 \\mid (p+2)^c+1$,\n$$\n(p + 2)^q + 1 = 2^\\alpha p^\\beta. \\qquad (1)\n$$\nObviously, $\\alpha > 0$. Note that $(p+2)^q + 1 = (p+3)A$, where\n$$\nA = (p+2)^{q-1} - (p+2)^{q-2} + \\cdots + 1 > (p+2)^{q-1} - (p+2)^{q-2} = (p+2)^{q-2}(p+1) > p^{q-1},\n$$\nand $A$ is odd, so $A$ is a power of $p$. So $A \\ge p^q$ and $\\beta \\ge q$. Taking both sides of (1) modulo $p$ gives $2^q \\equiv -1 \\pmod p$, so the order of $2 \\pmod p$ is 2 or $2q$.\n\nIf the order of $2 \\pmod p$ is 2, then $p = 3$. In this case, $(1)$ is $5^q + 1 = 2^\\alpha 3^\\beta$. Since $5^q + 1 \\equiv 2 \\pmod 4$, we have $\\alpha = 1$. By Lifting-the-exponent Lemma, $v_3(5^q + 1) = v_3(5+1) + v_3(q) \\le 2$, so $\\beta \\le 2$. One checks that $\\beta = 1, 2$ does not satisfy the equation.\n\nIf the order of $2 \\pmod p$ is $2q$, then $2q \\mid p-1$. So $q \\le \\frac{p-1}{2} < \\frac{p}{2}$. Dividing both sides of (1) by $p^q$ and using $(1+\\frac{1}{x})^x < e$, $x \\ge 1$, we have\n$$\n2^\\alpha p^{\\beta-q} = \\left(1 + \\frac{2}{p}\\right)^q + p^{-q} < \\left(1 + \\frac{2}{p}\\right)^{\\frac{p}{2}} + p^{-q} < e + 3^{-3} < 3.\n$$\nSo $\\beta = q$ and $\\alpha = 1$. Using $2 \\cdot p^q = (p+2)^q + 1 = (p+3)A$, we get $A = p^q$ and $p+3=2$. This is a contradiction.\n\nCase 2: $c$ is even. Assume that $2^d \\nmid c$ with $d \\ge 1$. Then $(p+2)^{2d} + 1 \\mid (p+2)^c + 1$. So\n$$\n(p + 2)^{2d} + 1 = 2^\\alpha p^\\beta.\n$$\nSince $(p+2)^{2d} + 1 \\equiv 2 \\pmod 4$, $\\alpha = 1$.\n$$\n(p + 2)^{2d} + 1 = 2 \\cdot p^{\\beta}. \\qquad (2)\n$$\nTaking both sides of (2) modulo $p$ gives $2^{2d} \\equiv -1 \\pmod p$. So the order of $2 \\pmod p$ is $2^{d+1}$. Thus, $2^d < \\frac{p}{2}$. Since\n$$\np^{\\beta+1} > 2 \\cdot p^{\\beta} = (p+2)^{2d} + 1 > p^{2d},\n$$\nwe have $\\beta \\ge 2^d$. Dividing both sides of (2) by $p^{2^d}$, we have\n$$\n2 \\cdot p^{\\beta-2^d} = \\left(1 + \\frac{2}{p}\\right)^{2^d} + p^{-2^d} < \\left(1 + \\frac{2}{p}\\right)^{\\frac{p}{2}} + p^{-2^d} < e + 3^{-2} < 3,\n$$\nso $\\beta = 2^d$.\n\nIf $d \\ge 2$, then $2^{d+1} \\mid p-1$ implies that $p \\equiv 1 \\pmod 8$. Using (2), we have $p^{2^d} - 1 = (p+2)^{2^d} - p^{2^d}$. Analyzing the number of factors 2 on both sides of the equality, we deduce that\n$$\nv_2(p^{2^d} - 1) = v_2(p^2 - 1) + d - 1 \\ge d + 3.\n$$\nYet,\n$$\nv_2((p+2)^{2^d} - p^{2^d}) = v_2((p+2)^2 - p^2) + d - 1 = v_2(2) + v_2(2p+2) + d - 1 = d + 2,\n$$\ngiving a contradiction. So $d=1$ and $2p^2 = (p+2)^2 + 1$, which implies that $p=5$. Going back to the original equation, we have $c = 2k$, with $k$ odd. But $a=1$, so we have\n$$\n2 \\cdot 5^b = 7^{2k} + 1.\n$$\nUsing Lifting-the-exponent Lemma again, we have $b = v_5(7^{2k} + 1) = v_5(7^2 + 1) + v_5(k) \\le 2 + \\frac{k}{5}$. If $k \\ge 3$, then\n$$\n2 \\cdot 5^b \\le 2 \\cdot 5^{2+\\frac{k}{5}} = (7^2 + 1) \\cdot 5^{\\frac{k}{5}} < 7^{2k} + 1,\n$$\ngiving a contradiction. So $k=1$, and thus $c=2$ and $b=2$. This gives another solution $(p, a, b, c) = (5, 1, 2, 2)$.\n\nTo sum up, there are two solutions of $(p, a, b, c)$: $(3, 1, 1, 1)$ and $(5, 1, 2, 2)$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 28559,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all positive integers $k$ such that there are finitely many triangles on the Descartes coordinate plane such that\n(1) the center of mass of each triangle is an integral point;\n(2) the intersection of any two triangles is either the empty set, a common vertex, or an edge joining two common vertices; and\n(3) the union of these triangles is a square with side length $k$. (The vertices of the squares are not required to be integral points, and the edges are not required to be parallel to the coordinate axes.)",
    "options": [],
    "answer": "all positive integers divisible by 3",
    "solution": "The desired positive integers $k$ are those divisible by $3$.\n\nFirst assume that $k = 3t$ for $t \\in \\mathbb{N}$. Consider the square with vertices $(0,0)$, $(3t, 3t)$, $(3t, 0)$, and $(0, 3t)$. Divide it into $t^2$ different smaller squares with the same side length $3$ along the lines $x = 3i$ ($i = 1, \\dots, t$) and $y = 3j$ ($j = 1, \\dots, t$). After this, each square can be divided into $2$ isosceles right triangles along the diagonal, and the center of mass of each triangle is an integral point. This gives the needed triangulation.\n\nConversely, suppose that a square with side length $k$ has a triangulation such that the center of mass of each small triangle is an integral point. Let $V$ denote the set of all vertices of the triangulation. Define a binary relation $A \\sim_0 B$ in $V$ if two triangles of the triangulation are of the form $\\triangle ACD$, $\\triangle BCD$. The equivalence relation $\\sim$ on $V$ is generated by $\\sim_0$, i.e. $A \\sim_0 B$ if and only if there exist $A_1, \\dots, A_r$ such that $A \\sim_0 A_1 \\sim_0 \\dots \\sim_0 A_r \\sim_0 B$. Denote the horizontal and vertical coordinates of a point $P$ by $x_P$ and $y_P$, respectively. We have\n\n(i) If $A \\sim_0 B$, then $3 \\mid x_A - x_B$ and $3 \\mid y_A - y_B$. This is because: by transitivity, one may assume that $A \\sim_0 B$, i.e., there are two triangles in the triangulation of the form $\\triangle ACD$, $\\triangle BCD$, whose centers of mass are both integral points. Therefore, $3 \\mid x_A + x_C + x_D$ and $3 \\mid x_B + x_C + x_D$; thus $3 \\mid x_A - x_B$. We may deduce similarly $3 \\mid y_A - y_B$.\n\n(ii) The set $V$ has at most $3$ equivalent classes with respect to $\\sim$. This is because after fixing a triangle $T_0$, for each point $A$ in $V$, there is always a sequence of triangles $T_0, \\dots, T_r$ such that $T_{i-1}$ and $T_i$ shares a same side for all $i = 1, \\dots, r$, and that $A$ is a vertex of $T_r$. By definition of $\\sim$, all three vertices of $T_{i-1}$ are respectively equivalent to three vertices of $T_i$. Hence by induction, $A$ is equivalent to one of the vertices of $T_0$.\n\nBy (ii) together with the pigeonhole principle, two of the four vertices of a square must be equivalent. Then from (i), we know that $3 \\mid k^2$ or $3 \\mid 2k^2$, which implies $3 \\mid k$.",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 28560,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find the minimal integer $n \\ge 3$ such that there exist $n$ points $A_1, A_2, \\dots, A_n$ on a plane with no three being colinear, such that for every $1 \\le i \\le n$, the midpoint of the segment $A_iA_{i+1}$ is contained in the segment $A_jA_{j+1}$ for some $j \\ne i$. Here, $A_{n+1} = A_1$.",
    "options": [],
    "answer": "6",
    "solution": "**Solution:** First, we prove that $n = 6$ satisfies the condition. Let $P_1, P_2, \\dots, P_6$ be the vertices of a regular hexagon in consecutive order. Let $A_1 = P_1, A_2 = P_3, A_3 = P_5, A_4 = P_2, A_5 = P_6, A_6 = P_4$. Then, these 6 points satisfy the conditions. See the figure below.\n![](attached_image_1.png)\n\nWe now prove that $n \\le 5$ does not satisfy the given conditions. Clearly, $n = 3$ does not satisfy the condition because for any $1 \\le i \\le n, 1 \\le j \\le n$ in the statement, we require $j - i \\ne -1, 0, 1 \\pmod{n}$.\n\nWhen $n = 4$, if $A_1, A_2, A_3, A_4$ satisfy the conditions, then $A_1A_2$ and $A_3A_4$ must bisect each other, and $A_2A_3$ and $A_4A_1$ must bisect each other, which is impossible.\n\nTherefore, we only need to consider the case when $n = 5$. Suppose $A_1, A_2, A_3, A_4, A_5$ satisfy the conditions. For any $1 \\le i \\le 5$, we require $j \\equiv i + 2, i + 3 \\pmod{5}$ in the statement. We consider two cases.\n\n**Case 1:** Two of the line segments $A_iA_{i+1}$, $i = 1, 2, 3, 4, 5$ bisect each other. By symmetry, we assume that $A_1A_2$ and $A_3A_4$ bisect each other. Note that applying an affine transformation to the plane does not affect the conclusion, we can assume that $A_1A_2$ and $A_3A_4$ are perpendicular bisectors and have equal length. Let $O$ be their intersection point, and let us establish a Cartesian coordinate system with $O$ as the origin. Let $A_1(0, -2), A_2(0, 2), A_3(2, 0), A_4(-2, 0)$. Then the midpoint $M$ of $A_2A_3$ must lie on either $A_4A_5$ or $A_1A_5$. Without loss of generality, we assume that $M$ lies on $A_1A_5$. Note that the slope of $A_1M$ is 3, so we can assume that the coordinates of $A_5$ are $(1 + x, 1 + 3x)$, where $x > 0$. See the figure below.\n![](attached_image_2.png)\n\nNow consider the midpoint $N$ of $A_4A_5$. If it lies inside the segment $A_1A_2$, then $1+x+(-2) = 0$ implies $x = 1$. However, in this case $A_4, A_2, A_5$ are collinear, a contradiction. If $N$ lies inside the segment $A_2A_3$, then\n$$\n\\frac{1+x+(-2)}{2} + \\frac{1+3x}{2} = 2,\n$$\nalso yields $x = 1$, which is a contradiction.\n\n**Case 2:** None of the segments $A_iA_{i+1}$, $i = 1, 2, 3, 4, 5$ bisect each other. Without loss of generality, assume that the midpoint $B_1$ of $A_1A_2$ lies on $A_3A_4$. Then the midpoint $B_3$ of $A_3A_4$ lies on $A_5A_1$, the midpoint $B_5$ of $A_5A_1$ lies on $A_2A_3$, the midpoint $B_2$ of $A_2A_3$ lies on $A_4A_5$, and the midpoint $B_4$ of $A_4A_5$ lies on $A_1A_2$. This implies that connecting $A_1, A_2, A_3, A_4, A_5$ in sequence yields a five-pointed star.\n![](attached_image_3.png)\n\nBy the Law of Sines,\n$$\n1 = \\prod_{i=1}^{5} \\frac{\\sin \\angle A_i B_i B_{i+2}}{\\sin \\angle A_i B_{i+2} B_i} = \\prod_{i=1}^{5} \\frac{A_i B_{i+2}}{A_i B_i} = \\prod_{i=1}^{5} \\frac{A_{i+1} B_{i+3}}{A_i B_i} < \\prod_{i=1}^{5} \\frac{A_{i+1} B_i}{A_i B_i} = 1,\n$$\nwhere the subscripts are understood modulo 5. This is a contradiction.\n\nTherefore, the minimum $n$ we seek is 6.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 28561,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $n$ be a positive integer. Let $x_1, x_2, \\dots, x_{2n}$ be $2n$ nonnegative real numbers such that $x_1 + x_2 + \\dots + x_{2n} = 4$. Prove that there exist nonnegative integers $p, q$ such that $q \\le n - 1$, and that\n$$\n\\sum_{i=1}^{q} x_{p+2i-1} \\le 1, \\quad \\sum_{i=q+1}^{n-1} x_{p+2i} \\le 1.\n$$\n\n*Remark 1 : the subscripts are understood modulo $2n$, i.e. if $k \\equiv \\ell \\pmod{2n}$, then $x_k = x_\\ell$.*\n*Remark 2 : If $q = 0$, then the first sum is considered as 0; if $q = n - 1$, then the second sum is considered as 0.*",
    "options": [],
    "answer": "Detailed solution",
    "solution": "**Proof:** Set $A = x_1 + x_3 + \\dots + x_{2n-1}$ and $B = x_2 + x_4 + \\dots + x_{2n}$.\nIf one of $A, B$ is less than or equal to 1, the problem is obvious. If $A > 1$ and $B > 1$, for $0 \\le k \\le n-1$, let $m(k) \\in \\{1, 2, \\dots, n-1\\}$ be the unique integer such that\n$$\n\\sum_{i=0}^{m(k)} x_{2k+2i+1} > 1 \\quad (1)\n$$\n$$\n\\sum_{i=0}^{m(k)-1} x_{2k+2i+1} \\le 1. \\quad (2)\n$$\nNote that, if $x_{2k+2m(k)+2} + x_{2k+2m(k)+4} + \\dots + x_{2k+2n-2} \\le 1$, then this equation and (2) gives what we needed with $p = 2k$ and $q = m(k)$.\nNow suppose that\n$$\nx_{2k+2m(k)+2} + x_{2k+2m(k)+4} + \\dots + x_{2k+2n-2} > 1. \\quad (3)\n$$\nNow, we construct an oriented graph with $0, 1, 2, \\dots, n-1$ as vertices, and connect an edge from each $k$ to $k+m(k)+1$ (with indices modulo $n$). It is easy to see that there are loops in this graph. We may assume that $k_1 \\to k_2 \\to \\dots \\to k_t \\to k_1$ is a minimal loop.\nIf $t=1$, then $\\sum_{i=0}^{n-2} x_{2k_1+2i+1} \\le 1$. Setting $p=2k_1$ and $q=n-1$ gives what we want.\nIf $t>1$, then set\n$$\n\\left\\{ \\frac{k_2 - k_1}{n} \\right\\} + \\left\\{ \\frac{k_3 - k_2}{n} \\right\\} + \\dots + \\left\\{ \\frac{k_t - k_{t-1}}{n} \\right\\} + \\left\\{ \\frac{k_1 - k_t}{n} \\right\\} = s.\n$$\nIn other words, when $k = k_1, k_2, \\dots, k_t$, there are totally $s \\cdot n$ elements in the sums of (1). Moreover, using the condition for defining the edges, we know that every term in $x_1, x_3, \\dots, x_{2n-1}$ appeared in the sum (1) exactly $s$ times. On the other hand, when $k = k_1, k_2, \\dots, k_t$, there are in total $(t-s) \\cdot n$ elements in the sums in (3). The condition for linking an edge shows that each of $x_2, x_4, \\dots, x_{2n}$ appeared exactly $t-s$ times in the above sum; this is because $x_{2j}$ appeared in (3) if and only if $x_{2j+1}$ did not appear in (1).\nNow take the total sum of (1) when $k = k_1, k_2, \\dots, k_t$, we get $s \\cdot A > t$. Similarly, taking the total sum of (3) when $k = k_1, k_2, \\dots, k_t$ gives $(t-s) \\cdot B > t$. But $A + B > t(\\frac{1}{s} + \\frac{1}{t-s}) \\ge 4$. This leads to a contradiction. The problem is proved.",
    "topic": "Number Theory",
    "subtopic": "Modular Arithmetic"
  },
  {
    "id": 28562,
    "subject": "Mathematics (Multi-modal)",
    "question": "As shown in the figure, points $K$ and $L$ are in the interior of a triangle $ABC$, point $D$ lies on the side $AB$. It is known that points $B$, $K$, $L$, $C$ are concyclic, and $\\angle AKD = \\angle BCK$, $\\angle ALD = \\angle BCL$. Prove that $AK = AL$.\n\n![](attached_image_1.png)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "*Proof*. As shown in the following picture,\n![](attached_image_2.png)\nlet the extensions of $AK$ and $AL$ intersect the circle passing through $B$, $K$, $L$, $C$ at points $X$ and $Y$, respectively. Connect $BX$ and $BY$. Considering the given conditions, we have $\\angle AKD = \\angle BCK = \\angle BXK$ and $\\angle ALD = \\angle BCL = \\angle BYL$.\n\nThus, $DK \\parallel BX$ and $DL \\parallel BY$. So we have\n$$\n\\frac{AK}{AX} = \\frac{AD}{AB} = \\frac{AL}{AY}.\n$$\nSince $K$, $L$, $Y$, $X$ are concyclic, by the Power of a Point theorem, we have\n$$\nAK \\cdot AX = AL \\cdot AY.\n$$\nMultiplying the two equations above, we get $AK^2 = AL^2$, which implies $AK = AL$. $\\square$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 28563,
    "subject": "Mathematics (Multi-modal)",
    "question": "Given a positive integer $n$, let $D$ be the set of positive divisors of $n$, and let $f: D \\to \\mathbb{Z}$ be a function. Prove that the following are equivalent:\n(A) for any positive divisor $m$ of $n$,\n$$\nn \\mid \\sum_{d|m} f(d) \\binom{n/d}{m/d};\n$$\n(B) for any positive divisor $k$ of $n$,\n$$\nk \\mid \\sum_{d|k} f(d).\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "**Proof:** For map $f: D \\to \\mathbb{Z}$, we define a map $g: D \\to \\mathbb{Z}$ to be\n$$\ng(k) = \\sum_{d|k} f(d), \\quad \\forall k \\in D.\n$$\nBy Möbius transform, the map $f$ is uniquely determined by $g$:\n$$\nf(k) = \\sum_{d|k} \\mu\\left(\\frac{k}{d}\\right) g(d), \\quad \\forall k \\in D,\n$$\nTherefore we have\n$$\n\\begin{align}\n\\sum_{d|m} f(d) \\binom{n/d}{m/d} &= \\sum_{d|m} \\sum_{x|d} \\mu\\left(\\frac{d}{x}\\right) g(x) \\binom{n/d}{m/d} \\nonumber \\\\\n&= \\sum_{x|m} g(x) \\left( \\sum_{x|d, d|m} \\mu\\left(\\frac{d}{x}\\right) \\binom{n/d}{m/d} \\right) \\tag{1} \\\\\n&= \\sum_{x|m} g(x) \\left( \\sum_{s|\\frac{m}{x}} \\mu(s) \\binom{n/(xs)}{m/(xs)} \\right). \\nonumber\n\\end{align}\n$$\nWe need the following *Lemma*: If $b|a$, then\n$$\n\\sum_{k|b} \\mu(k) \\binom{a/k}{b/k} \\equiv 0 \\pmod{a}.\n$$\nAssuming this lemma temporarily, we prove that (A) and (B) are equivalent. (The proof of the lemma will be given later.)\n“(B) ⇒ (A)” Suppose that for every $x \\in D$ we have $x \\mid g(x)$. The lemma implies that\n$$\n\\frac{n}{x} \\mid \\sum_{s|\\frac{m}{x}} \\mu(s) \\binom{m/(xs)}{n/(xs)},\n$$\n\n\"(A) ⇒ (B)\" Suppose that (A) holds. We use induction to prove that for $k \\in D$, we always have $k|g(k)$. Suppose that we have already proved $k|g(k)$ for $k < m$ and $k \\in D$. Consider case of $k = m$. By Lemma and the inductive hypothesis, we have\n$$\n\\begin{aligned}\n0 &\\equiv \\sum_{d|m} f(d) \\binom{n/d}{m/d} \\equiv \\sum_{x|m} g(x) \\left( \\sum_{s|\\frac{m}{x}} \\mu(s) \\binom{n/(xs)}{m/(xs)} \\right) \\\\\n&\\equiv g(m) \\cdot \\frac{n}{m} + \\sum_{x|m, x<m} g(x) \\left( \\sum_{s|\\frac{m}{x}} \\mu(s) \\binom{n/(xs)}{m/(xs)} \\right) \\\\\n&\\equiv g(m) \\cdot \\frac{n}{m} \\quad (\\text{mod } n),\n\\end{aligned}\n$$\nFrom this we get $m|g(m)$, completing the induction.\n\nIt remains to prove the lemma. Note that\n$$\n\\sum_{k|b} \\mu(k) \\binom{a/k}{b/k} = \\frac{a}{b} \\sum_{k|b} \\mu(k) \\binom{\\frac{a}{k}-1}{\\frac{b}{k}-1},\n$$\nLet $b = \\prod_{i=1}^t p_i^{\\beta_i}$ be the prime factorization. Then we only need to prove\n$$\n\\sum_{k|b} \\mu(k) C_{\\frac{b}{k}-1}^{\\frac{b}{k}-1} \\equiv 0 \\pmod{p_i^{\\beta_i}}.\n$$\nfor every $1 \\le i \\le t$. For this, we may assume that $i = 1$. Denote $p_1^{\\beta_1} = p^{\\beta}$. Note that\n$$\n\\begin{aligned}\n& \\sum_{k|b} \\mu(k) \\binom{\\frac{a}{k}-1}{\\frac{b}{k}-1} \\\\\n&= \\sum_{I \\subset \\{1, \\dots, t\\}} (-1)^{|I|} \\left( \\frac{\\prod_{i \\in I} \\frac{a}{p_i}}{\\prod_{i \\in I} \\frac{b}{p_i}} - 1 \\right) \\\\\n&= \\sum_{J \\subset \\{2, \\dots, t\\}} (-1)^{|J|} \\left( \\left( \\frac{\\prod_{j \\in J} \\frac{a}{p_j}}{\\prod_{j \\in J} \\frac{b}{p_j}} - 1 \\right) - \\left( \\frac{\\prod_{j \\in J} \\frac{a}{p_j}}{\\prod_{j \\in J} \\frac{b}{p_j}} - 1 \\right) \\right)\n\\end{aligned}\n$$\nThen it suffices to show: if $u, v$ are multiples of $p^{\\beta}$, then\n$$\n\\binom{u-1}{v-1} - \\binom{\\frac{u}{p}-1}{\\frac{v}{p}-1} \\equiv 0 \\pmod{p^{\\beta}}. \\qquad (2)\n$$\nPut $w = u - v$. Then\n$$\n\\begin{aligned}\n\\binom{u-1}{v-1} &= \\frac{\\prod_{0 \\le x \\le \\frac{v}{p}-1, 1 \\le r \\le p-1} (xp+r+w) \\cdot \\prod_{1 \\le y \\le \\frac{v}{p}-1} (yp+w)}{\\prod_{0 \\le x \\le \\frac{v}{p}-1, 1 \\le r \\le p-1} (xp+r) \\cdot \\prod_{1 \\le y \\le \\frac{v}{p}-1} (yp)} \\\\\n&= \\frac{\\prod_{0 \\le x \\le \\frac{v}{p}-1, 1 \\le r \\le p-1} (xp+r+w)}{\\prod_{0 \\le x \\le \\frac{v}{p}-1, 1 \\le r \\le p-1} (xp+r)} \\cdot \\binom{\\frac{u}{p}-1}{\\frac{v}{p}-1}.\n\\end{aligned}\n$$\n\nSince $w$ is divisible by $p^\\beta$, we get\n$$\n\\begin{aligned}\n& \\left( \\begin{matrix} u-1 \\\\ v-1 \\end{matrix} \\right) \\prod_{0 \\le x \\le \\frac{v}{p}-1, 1 \\le r \\le p-1} (xp+r) \\\\\n& = \\left( \\begin{matrix} \\frac{u}{p}-1 \\\\ \\frac{v}{p}-1 \\end{matrix} \\right) \\prod_{0 \\le x \\le \\frac{v}{p}-1, 1 \\le r \\le p-1} (xp+r+w) \\\\\n& = \\left( \\begin{matrix} \\frac{u}{p}-1 \\\\ \\frac{v}{p}-1 \\end{matrix} \\right) \\prod_{0 \\le x \\le \\frac{v}{p}-1, 1 \\le r \\le p-1} (xp+r) \\pmod{p^\\beta},\n\\end{aligned}\n$$\nSo (2) holds, and we complete the proof of the lemma.\n*Another proof of the lemma:* We will use the following fact. Let $X$ be a finite set, and let $h: X \\to X$ be a map such that $h^{(a)} = \\text{Id}_X$. Under the action of $h$, $X$ is divided into some orbits. Each orbit looks like $\\{x, h(x), h^{(2)}(x), \\dots, h^{(l)}(x) = x\\}$, where $l$ is the smallest positive integer such that $h^{(l)}(x) = x$; we call $l$ the length of the orbit. Since $h^{(a)} = \\text{Id}_X$, the length $l$ of every orbit is a divisor of $a$. For every divisor $l$ of $n$, let $N(l)$ be the number of orbits with length $l$. For every positive integer $k|a$, let\n$$\nm(k) = \\#\\{x \\in X : h^{(k)}(x) = x\\},\n$$\nthen $m(k) = \\sum_{l|k} N(l)l$. Therefore we get\n$$\n\\begin{aligned}\n\\sum_{k|a} \\mu(k) m\\left(\\frac{a}{k}\\right) &= \\sum_{k|a} \\sum_{l|\\frac{a}{k}} \\mu(k) N(l) l \\\\\n&= \\sum_{l|a} N(l) l \\sum_{k|\\frac{a}{l}} \\mu(k) \\\\\n&= N(a) a.\n\\end{aligned}\n$$\nIn particular, we have\n$$\n\\sum_{k|a} \\mu(k) m\\left(\\frac{a}{k}\\right) \\equiv 0 \\pmod{a}. \\qquad (3)\n$$\nWe apply this to the following model. Let $X$ be the set of all subsets of $\\mathbb{Z}/a$ of $b$ elements. Define $h: X \\to X$ to be\n$$\nh(\\{y_1, \\dots, y_b\\}) = \\{y_1 + 1, \\dots, y_b + 1\\}, \\quad \\forall \\{y_1, \\dots, y_b\\} \\in X.\n$$\nIn this model, for every divisor $k$ of $a$, we have\n$$\nm(k) = \\binom{k}{b/(a/k)}.\n$$\nThen (3) implies that\n$$\n\\sum_{k|a} \\mu(k) \\binom{a/k}{b/k} \\equiv 0 \\pmod{a}.\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 28564,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $C = \\{z \\in \\mathbb{C} \\mid |z| = 1\\}$ be the unit circle in the complex plane. 240 complex numbers $z_1, z_2, \\dots, z_{240} \\in C$ (can be repeated) satisfy the following conditions:\n(1) for any open arc $\\Gamma$ of length $\\pi$ on $C$, there are at most 200 $j$'s $(1 \\le j \\le 240)$ such that $z_j \\in \\Gamma$;\n(2) for any open arc $\\gamma$ of length $\\frac{\\pi}{3}$ on $C$, there are at most 120 $j$'s $(1 \\le j \\le 240)$ such that $z_j \\in \\gamma$.\n\nFind the maximum of $|z_1 + z_2 + \\dots + z_{240}|$.",
    "options": [],
    "answer": "80 + 40*sqrt(3)",
    "solution": "The maximum is $80 + 40\\sqrt{3}$. Take 80 of 1's, 40 of each of $\\exp(\\frac{\\pi}{6}i)$, $\\exp(-\\frac{\\pi}{6}i)$, $i$, $-i$. It is straightforward to check that they satisfy (1), (2), and their sum equals $80 + 40\\sqrt{3}$. We need to show this is the maximum value.\n\nLet $z_1, z_2, \\dots, z_{240}$ satisfy (1), (2). By rotation if necessary, we may assume that $S = z_1 + z_2 + \\dots + z_{240}$ is a nonnegative real number; starting from $-1$ (included) along $C$, in the counterclockwise order, let the numbers be $z_1, z_2, \\dots, z_{240}$.\n\nCondition (1) indicates that for $1 \\le j \\le 40$, to go from $z_j$ to $z_{j+200}$ along $C$, at least an arc of length $\\pi$ is needed, that is to say, there exists $2\\pi \\ge \\alpha \\ge \\pi$ such that $z_{j+200} = z_j \\cdot \\exp(\\alpha i)$. Let $z_j = (-1) \\cdot \\exp(\\beta i)$, $\\beta \\in [0, 2\\pi)$. From $\\arg z_{j+200}$ we see $\\beta + \\alpha < 2\\pi$, and thus\n$$\n\\operatorname{Re}(z_j + z_{j+200}) = -\\cos \\beta - \\cos(\\beta + \\alpha) = -2 \\cos \\frac{\\alpha}{2} \\cos \\left(\\beta + \\frac{\\alpha}{2}\\right) \\le 0 \\\\\n\\left( \\text{because } \\frac{\\alpha}{2} \\in \\left[\\frac{\\pi}{2}, \\pi\\right], \\beta + \\frac{\\alpha}{2} \\in \\left[\\frac{\\pi}{2}, \\frac{3\\pi}{2}\\right] \\right). \\text{ Summing up } 1 \\le j \\le 40, \\text{ we} \\\\\n\\text{obtain}\n$$\n$$\n\\operatorname{Re}(z_1 + z_2 + \\cdots + z_{40} + z_{201} + z_{202} + \\cdots + z_{240}) \\le 0. \\quad \\textcircled{1}\n$$\n\nSimilarly, condition (2) indicates that for $41 \\le j \\le 80$, to go from $z_j$ to $z_{j+120}$ along $C$, at least an arc of length $\\frac{\\pi}{3}$ is needed. So there exists $2\\pi \\ge \\alpha \\ge \\frac{\\pi}{3}$, such that $z_{j+120} = z_j \\cdot \\exp(\\alpha i)$. Let $z_j = (-1) \\cdot \\exp(\\beta i)$, $\\beta \\in [0, 2\\pi)$. From $\\arg z_{j+120}$ we see $\\beta + \\alpha < 2\\pi$, and thus\n$$\n\\operatorname{Re}(z_j + z_{j+120}) = -\\cos \\beta - \\cos(\\beta + \\alpha) = -2 \\cos \\frac{\\alpha}{2} \\cos \\left(\\beta + \\frac{\\alpha}{2}\\right).\n$$\nIf $\\alpha \\ge \\pi$, the above quantity is non-positive; if $\\alpha \\in [\\frac{\\pi}{3}, \\pi)$, then\n$$\n\\left| 2 \\cos \\frac{\\alpha}{2} \\cos \\left( \\beta + \\frac{\\alpha}{2} \\right) \\right| \\le 2 \\times \\cos \\frac{\\pi}{6} \\times 1 = \\sqrt{3},\n$$\nso the real part is $\\le \\sqrt{3}$. Summing up for $41 \\le j \\le 80$, we obtain\n$$\n\\operatorname{Re}(z_{41} + z_{42} + \\cdots + z_{80} + z_{161} + z_{162} + \\cdots + z_{200}) \\le 40\\sqrt{3}. \\quad \\textcircled{2}\n$$\n\nThe inequalities (1) and (2) lead to an estimate of the sum\n$$\n\\begin{aligned}\n& |z_1 + z_2 + \\cdots + z_{240}| \\\\\n&= \\operatorname{Re} \\left( \\sum_{j=1}^{40} (z_j + z_{200+j}) \\right) + \\operatorname{Re} \\left( \\sum_{j=41}^{80} (z_j + z_{120+j}) \\right) + \\operatorname{Re} \\left( \\sum_{j=81}^{120} z_j \\right) \\\\\n&\\le 0 + 40\\sqrt{3} + 80 = 80 + 40\\sqrt{3}.\n\\end{aligned}\n$$\n\nHere, the maximum is attained only when the numbers are: 80 of 1's, 40 of each of $\\exp(\\frac{\\pi}{6}i)$, $\\exp(-\\frac{\\pi}{6}i)$, $i$, $-i$ (or by any rotation). $\\square$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 28565,
    "subject": "Mathematics (Multi-modal)",
    "question": "Prove that there exist two positive real numbers $C$ and $\\alpha > \\frac{1}{2}$, such that, for any positive integer $n$, there is a subset $A$ of $\\{1, 2, \\dots, n\\}$ with $|A| \\ge Cn^\\alpha$, such that the difference between any pair of different numbers in $A$ is not a perfect square.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "**Proof.** For $n \\ge 25$, write $5^{2t} \\le n < 5^{2t+2}$ for $t \\in \\mathbb{N}^*$. Take the set\n$$\nA = \\{(\\alpha_{2t}, \\dots, \\alpha_1)_5 \\mid \\alpha_{2i} \\in \\{0, 1, 2, 3, 4\\} \\text{ and } \\alpha_{2i-1} \\in \\{1, 3\\} \\text{ for } i = 1, \\dots, t\\},\n$$\nwhere $(\\alpha_{2t}, \\dots, \\alpha_1)_5$ denotes the number $m = 5^{2t-1}\\alpha_{2t} + \\dots + 5\\alpha_2 + \\alpha_1$ in quinary. It is obvious that $A \\subset \\{1, 2, \\dots, n\\}$.\nFor each pair of integers $u_1, u_2 \\in A$, write $u_1 = (a_{2t}, \\dots, a_1)$ and $u_2 = (b_{2t}, \\dots, b_1)$. We may assume $u_1 > u_2$ and consider $u_1 - u_2$. Now let $s$ be the minimal index such that $a_s \\ne b_s$. Namely, we have $a_1 = b_1, \\dots, a_{s-1} = b_{s-1}, a_s \\ne b_s$. Note that in this case, $u_1 - u_2 = (a_{2t} - b_{2t})5^{2t-1} + \\dots + (a_s - b_s)5^{s-1}$.\nIf $2 \\mid s$, then $5^{s-1} \\mid (u_1 - u_2)$ i.e., $a_s - b_s \\ne 0$ and $-4 \\le a_s - b_s \\le 4$. Hence $u_1 - u_2$ cannot be a perfect square.\nIf $2 \\nmid s$, then $\\frac{u_1 - u_2}{5^{s-1}} = (a_{2t} - b_{2t})5^{2t-1} + \\dots + (a_s - b_s) \\in \\mathbb{Z}$. Suppose that $u_1 - u_2$ is a perfect square, then $\\frac{u_1 - u_2}{5^{s-1}}$ is a perfect square as well. On the other hand, the condition $2 \\mid s - 1$ and $a_s \\ne b_s$ implies that $\\{a_s, b_s\\} = \\{1, 3\\}$ and $\\frac{u_1 - u_2}{5^{s-1}} \\equiv 2, 3 \\pmod 5$, which can never be a perfect square. This is a contradiction.\nThus for any distinct element $u_1, u_2 \\in A$, the number $|u_1 - u_2|$ is not a perfect square. Hence $A$ satisfies the requirement. Also, note that $|A| = 10^t$. Since $\\alpha = \\log_{25} 10 > 1/2$, we obtain\n$$\nn^\\alpha < 5^{(2t+2)\\log_{25} 10} = 10^{t+1} = 10|A|.\n$$\nSo it suffices to take $C = \\frac{1}{24}$ and $\\alpha = \\log_{25} 10 \\in (0, 1)$. As for $n \\le 24$, we may take $A = \\{1\\}$ and then $|A| \\ge \\frac{1}{24}n \\ge \\frac{1}{24}n^\\alpha$.\nTo sum up, for all $n \\in \\mathbb{N}^*$, one can find a desired subset $A$ with $|A| \\ge Cn^\\alpha$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 28566,
    "subject": "Mathematics (Multi-modal)",
    "question": "Prove that there exists a positive number $C$ such that the following statement holds: for any infinite arithmetic progression $a_1, a_2, a_3, \\dots$ of positive integers, if the greatest common divisor of $a_1$ and $a_2$ is square-free, then there exists some positive integer $m \\le C \\cdot a_2^2$ such that $a_m$ is square-free.\n*Remark: We call a positive integer $N$ square-free, if it is not divisible by any square that is strictly larger than 1.*",
    "options": [],
    "answer": "Detailed solution",
    "solution": "*Proof.* We prove that $C = 8$ satisfies the requirement.\n\n(1) First consider the case where $a_1$ and $a_2$ are coprime. Let $d = a_2 - a_1$ be the common difference. For any prime $p$, if $p \\mid d$, then $p \\nmid a_1$ and hence $p$ does not divide any $a_n$. If $p \\nmid d$, then any consecutive $p^2$ terms of the sequence $a_n$ form a complete set of residues modulo $p^2$, among which exactly one term is divisible by $p^2$.\n\nLet $N = 4a_2$. We prove the existence of $1 \\le n \\le N$ such that $a_n$ has no square factors. If $1 \\le m \\le N$ and $a_m$ has a square factor, then there exists a prime $p$ such that $p^2 \\mid a_m$. Thus, $p \\nmid d$ and $p \\le \\sqrt{a_N}$. Moreover, the number of terms in $a_1, a_2, \\dots, a_N$ that are divisible by $p^2$ is at most $\\lceil \\frac{N}{p^2} \\rceil$. Therefore, the number $M$ of terms in $a_1, a_2, \\dots, a_N$ that have square factors satisfies:\n$$\nM \\le \\sum_{p \\le \\sqrt{a_N}} \\left\\lceil \\frac{N}{p^2} \\right\\rceil < \\sum_{p \\le \\sqrt{a_N}} \\left( \\frac{N}{p^2} + 1 \\right) \\le N \\sum_{p \\le \\sqrt{a_N}} \\frac{1}{p^2} + \\sqrt{a_N} < \\frac{N}{2} + \\sqrt{a_N}.\n$$\nThe last inequality is due to\n$$\n\\sum_{p} \\frac{1}{p^2} < \\frac{1}{4} + \\sum_{k=2}^{\\infty} \\frac{1}{(2k-1)^2} < \\frac{1}{4} + \\sum_{k=2}^{\\infty} \\frac{1}{(2k-2)2k} = \\frac{1}{2}.\n$$\nWhen $N = 4a_2$, we have $\\sqrt{a_N} < \\sqrt{Na_2} = 2a_2 = \\frac{N}{2}$. Thus, $M < N$. Therefore, there exist numbers in $a_1, a_2, \\dots, a_N$ that have no square factors.\n\n(2) Assume $\\gcd(a_1, a_2) = \\gcd(a_1, d) = q = q_1 \\cdots q_l > 1$, where $q_1, \\cdots, q_l$ are pairwise distinct prime factors. Note that every $a_i$ is divisible by $q$. For each prime factor $q_i$, there exists an index $t_i$ such that $v_{q_i}(a_{t_i}) = 1$. In fact, we can take $t_i \\in 1, 2$ since $a_1$ and $a_2$ cannot both be divisible by $q_i^2$. By Chinese Remainder Theorem, there exists $k \\in 1, 2, \\cdots, q$ such that $k \\equiv t_i \\pmod{q_i}$ for $1 \\le i \\le l$. Thus, for $1 \\le i \\le l$, we have $a_k \\equiv a_{t_i} \\pmod{q_id}$, which implies $v_{q_i}(a_k) = 1$.\n\nConsider the subsequence $a_k, a_{k+q}, a_{k+2q}, \\dots$, which has common difference $qd$ and is divisible by $q^2$. By construction, $q_i^2 \\nmid a_k$ for $1 \\le i \\le l$, i.e., each term in this subsequence has no square factors that are equal to $q_1, q_2, \\dots, q_l$. Let $b_i = \\frac{a_{k+(i-1)q}}{q}$, then $b_1 < b_2 < \\dots$ is an arithmetic sequence with common difference $d$, and $b_1$ is coprime to $d$. By the conclusion of (1), there exists $i \\le 4b_2$ such that $b_i$ has no square factor, and since $q$ is coprime to $b_i$, $a_{k+(i-1)q} = qb_i$ also has no square factor. Finally,\n$$\nk + (i - 1)q \\le iq \\le 4b_2q = 4a_{k+q} < 4(k + q)a_2 \\le 8da_2 < 8a_2^2.\n$$",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 28567,
    "subject": "Mathematics (Multi-modal)",
    "question": "(1) Prove that on the complex plane, the convex hull of all complex roots of the equation\n$$\nz^{20} + 63z + 22 = 0\n$$\nis larger than $\\pi$.\n\n(2) Let $n$ be a positive integer, and $1 \\le k_1 < k_2 < \\dots < k_n$ be $n$ odd integers. Prove that for any $n$ complex numbers $a_1, a_2, \\dots, a_n$ with sum 1 and any complex number $w$ with absolute value at least 1, the equation\n$$\na_1z^{k_1} + a_2z^{k_2} + \\dots + a_nz^{k_n} = w\n$$\nhas at least one complex root with norm less than or equal to $3n|w|$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "(1) We first prove the following.\n\n**Lemma:** [Gauss-Lucas Theorem] All zeros of the polynomial $f'(z)$ lie in the closure formed by zeros of $f(z)$.\n\n**Proof of Lemma:** By the fundamental theorem of algebra, we know that $f(z)$ admits a decomposition:\n$$\nf(z) = A(z - z_1)^{\\alpha_1} (z - z_2)^{\\alpha_2} \\cdots (z - z_n)^{\\alpha_n}.\n$$\nThus, $z_1, z_2, \\dots, z_n$ are zeros of $f'(z)$ with multiplicities $\\alpha_1 - 1, \\alpha_2 - 1, \\dots, \\alpha_n - 1$, respectively. Say $Z$ is any other zero of $f'(z)$, then\n$$\n\\frac{f'(Z)}{f(Z)} = \\frac{\\alpha_1}{Z - z_1} + \\frac{\\alpha_2}{Z - z_2} + \\dots + \\frac{\\alpha_n}{Z - z_n} = 0.\n$$\nThat is,\n$$\n\\frac{\\alpha_1}{|Z - z_1|^2} (\\bar{Z} - \\bar{z}_1) + \\frac{\\alpha_2}{|Z - z_2|^2} (\\bar{Z} - \\bar{z}_2) + \\dots + \\frac{\\alpha_n}{|Z - z_n|^2} (\\bar{Z} - \\bar{z}_n) = 0.\n$$\nAfter taking complex conjugations, the equality becomes\n$$\nZ = \\frac{\\frac{\\alpha_1}{|Z-z_1|^2} z_1 + \\frac{\\alpha_2}{|Z-z_2|^2} z_2 + \\dots + \\frac{\\alpha_n}{|Z-z_n|^2} z_n}{\\frac{\\alpha_1}{|Z-z_1|^2} + \\frac{\\alpha_2}{|Z-z_2|^2} + \\dots + \\frac{\\alpha_n}{|Z-z_n|^2}}.\n$$\nHence $Z$ is a convex linear combination of $z_1, z_2, \\dots, z_n$. This proves the lemma.\n\nBack to the original problem. Gauss-Lucas Lemma implies that the convex hull of the zeros of the given equation contains the convex hull of the zeros of the following equation\n$$\n20z^{19} - 63 = 0.\n$$\nThis is a regular 19-gon inscribed in a circle of radius $\\left(\\frac{63}{20}\\right)^{1/19}$.\n\nThe inscribed circle of this regular 19-gon has a radius\n$$\n\\left(\\frac{63}{20}\\right)^{1/19} \\cos \\frac{\\pi}{19} > \\left(\\frac{63}{20}\\right)^{1/19} \\left(1 - \\frac{1}{2} \\left(\\frac{\\pi}{19}\\right)^2\\right) > \\left(\\frac{63}{20}\\right)^{1/19} \\left(1 - \\frac{1}{2} \\left(\\frac{1}{6}\\right)^2\\right) = \\frac{71}{72} \\left(\\frac{63}{20}\\right)^{1/19}.\n$$\nOn the other hand, since\n$$\n\\left(\\frac{72}{71}\\right)^{19} = \\left(1 + \\frac{1}{71}\\right)^{19} < 1 + \\frac{19}{71} + 18 \\cdot C_{19}^2 \\frac{1}{72^2} < 3 < \\frac{63}{20},\n$$\nthe radius of the inscribed circle is larger than 1. This proves (1).\n\n\n(2) We need to prove the following.\n\n(A) If $a_1, a_2, \\dots, a_n$ and $w$ satisfy the assumptions, then the minimal norm of the roots of\n$$\na_1 z^{k_1} + a_2 z^{k_2} + \\dots + a_n z^{k_n} = w \\quad (*)\n$$\nis less than or equal to $3mn$.\nThis is equivalent to the following.\n\n(B) If $a_1, a_2, \\dots, a_n$ and $w$ satisfy the assumptions, then the maximal norm of the roots of\n$$\nwz^{k_n} - a_1z^{k_n-k_1} - a_2z^{k_n-k_2} - \\dots - a_{n-1}z^{k_n-k_{n-1}} - a_n = 0, \\quad (**)\n$$\ndenoted by $M_0$, is greater than or equal to $1/(3mn)$.\n\nBy the lemma in (1), the maximal norm of roots of $(**)$ cannot be less than the maximal norm of roots of\n$$\nw k_n z^{k_n-1} - a_1(k_n - k_1)z^{k_n-k_1-1} - \\dots - a_{n-1}(k_n - k_{n-1})z^{k_n-k_{n-1}-1} = 0. \\quad (1)\n$$\nThat is, the maximal norm, say $M_1$, of the roots of\n$$\nw k_n z^{k_n-1} - a_1(k_n - k_1)z^{k_n-k_1-1} - a_2(k_{n-1} - k_2)z^{k_n-k_2-1} - \\dots - a_{n-1}(k_n - k_{n-1}) = 0. \\quad (1')\n$$\nBeginning with $(1')$, we repeat the same process for $n-s$ times. It is known that the maximal norm of the roots decreases each time. Therefore,\n$$\nM_0 \\ge M_1 \\ge \\dots \\ge M_{n-s},\n$$\nwhere $M_{n-s}$ is the maximal norm of the roots of the equation\n$$\nw k_n k_{n-1} \\cdots k_{s+1} z^{k_s} - a_1(k_n - k_1)(k_{n-1} - k_1) \\cdots (k_{s+1} - k_1) z^{k_s - k_1} \\\\\n- \\cdots - a_s(k_n - k_s) \\cdots (k_{s+1} - k_s) = 0. \\quad ((n-s)')\n$$\n\nBy Vieta's theorem, we have\n$$\nM_0 \\ge M_1 \\ge \\dots \\ge M_{n-s} \\ge \\left[ \\frac{(k_n - k_s) \\cdots (k_{s+1} - k_s)}{k_n k_{n-1} \\cdots k_{s+1}} \\cdot \\left| \\frac{a_s}{w} \\right| \\right]^{1/k_s}.\n$$\nHence it suffices to show that there exists $s \\in \\{1, 2, \\dots, n\\}$ such that\n$$\n\\left[ \\frac{k_n k_{n-1} \\cdots k_{s+1}}{(k_n - k_s)(k_{n-1} - k_s) \\cdots (k_{s+1} - k_s)} \\right]^{1/k_s} \\cdot \\left[ \\left| \\frac{w}{a_s} \\right| \\right]^{1/k_s} \\le 3mn.\n$$\nNote that\n$$\n\\frac{k_p}{k_p - k_s} = 1 + \\frac{k_s}{k_p - k_s} \\le \\left(1 + \\frac{1}{k_p - k_s}\\right)^{k_s}.\n$$\nSo it remains to show that there exists some $s$ such that\n$$\n\\left[ \\prod_{p=s+1}^{n} \\left( 1 + \\frac{1}{k_p - k_s} \\right) \\right] \\cdot \\left( \\frac{m}{|a_s|} \\right)^{1/k_s} \\le 3mn. \\quad 2\n$$\nWe also make the following observations.\n\n* Since all $k_j$'s are odd numbers, $k_p - k_s \\ge 2(p-s)$. Therefore,\n$$\n1 + \\frac{1}{k_p - k_s} \\le 1 + \\frac{1}{2(p-s)} < \\frac{p-s+1}{p-s};\n$$\n* Since $\\sum_{j=1}^n a_j = 1$, we must have $\\sum_{j=1}^n |a_j| \\ge 1$. Thus there must be $s$ satisfying $|a_s| \\ge \\frac{1}{2^s}$, or equivalently, $\\frac{1}{|a_s|} \\le 2^s$.\n\nWe separate the discussion into the following two cases.\n\n(1) If there is $s$ (allowing $s=1$), such that $k_s \\ge 3$ and $|a_s| \\ge \\frac{1}{2^s}$, then (note that $s \\le 2s-1 \\le k_s$)\n$$\n\\left[ \\prod_{p=s+1}^{n} \\left( 1 + \\frac{1}{k_p - k_s} \\right) \\right] \\cdot \\left( \\frac{m}{|a_s|} \\right)^{1/k_s} < \\left[ \\prod_{p=s+1}^{n} \\frac{p-s+1}{p-s} \\right] \\cdot m^{1/k_s} \\cdot 2^{s/k_s} < n \\cdot \\sqrt[3]{m} \\cdot 2 < 3mn.\n$$\n\n(2) Otherwise, we must have $k_1 = 1$ and $|a_1| \\ge \\frac{1}{2}$. Hence\n$$\n\\left[ \\prod_{p=2}^{n} \\left( 1 + \\frac{1}{k_p - k_1} \\right) \\right] \\cdot \\frac{m}{|a_1|} \\le \\left[ \\prod_{p=2}^{n} \\frac{2(p-1)+1}{2(p-1)} \\right] \\cdot m \\cdot 2 < \\sqrt{\\prod_{k=1}^{2n-2} \\frac{k+1}{k}} \\cdot 63 \\cdot 2 = \\sqrt{2n-1} \\cdot m \\cdot 2 < 3mn.\n$$\n\nThis completes the proof.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 28568,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $m$ and $n$ be two positive integers with $m \\ge n \\ge 2022$. Let $a_1, a_2, \\dots, a_n, b_1, b_2, \\dots, b_n$ be $2n$ real numbers. Prove that the numbers of ordered pairs $(i, j)$ ($1 \\le i, j \\le n$) such that\n$$\n|a_i + b_j - ij| \\le m\n$$\nis less than or equal to $3n\\sqrt{m \\log n}$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "**Proof:** Mark red all points $(i, j)$ for which $|a_i + b_j - ij| \\le m$.\n\n*Lemma:* If $(i_1, j_1)$, $(i_1, j_2)$, $(i_2, j_1)$, $(i_2, j_2)$ are all red, then $|(i_2 - i_1)(j_2 - j_1)| \\le 4m$.\n\n*Proof of the lemma:* By the definition of red points, we know\n$$\n|a_{i_1} + b_{j_1} - i_1 j_1| \\le m, \\quad |a_{i_1} + b_{j_2} - i_1 j_2| \\le m, \\quad |a_{i_2} + b_{j_1} - i_2 j_1| \\le m, \\quad |a_{i_2} + b_{j_2} - i_2 j_2| \\le m.\n$$\nTaking the differences of these expressions to eliminate $a_{i_1}$, $a_{i_2}$, $b_{j_1}$, and $b_{j_2}$, we get $|i_1j_1 - i_1j_2 - i_2j_1 + i_2j_2| \\le 4m$. This proves the lemma.\n\nWe first calculate, for a given pair $1 \\le i_1 < i_2 \\le n$, the number of $j$'s such that $(i_1, j)$ and $(i_2, j)$ are all red points. Put $d = i_2 - i_1$. Since the difference of any two such $j$'s is at most $\\frac{4m}{d}$, we can have at most $\\frac{4m}{d} + 1$ such $j$'s.\n\nMoreover, when the difference $d$ is given, there are $n-d$ ways to choose $i_1$ and $i_2$, so the total number of $1 \\le i_1 < i_2 \\le n$ and $1 \\le j \\le n$ for which $(i_1, j)$ and $(i_2, j)$ are all red points cannot be larger than\n$$\n\\sum_{d=1}^{n-1} (n-d)\\left(\\frac{4m}{d} + 1\\right) = \\sum_{d=1}^{n-1} (n-d) - 4mn + 4m \\sum_{d=1}^{n-1} \\frac{n}{d} < \\frac{n(n-1)}{2} + 4mn \\ln n.\n$$\n\nOn the other hand, for $1 \\le j \\le n$, assume that there are $x_j$ red points in $(1, j)$, $(2, j)$, ..., $(n, j)$. Then\n$$\n\\sum_{j=1}^n C_{x_j}^2 < \\frac{n(n-1)}{2} + 4mn \\ln n.\n$$\nSo\n$$\n\\sum_{j=1}^n \\left(x_j - \\frac{1}{2}\\right)^2 < n^2 - \\frac{1}{4}n + 8mn \\ln n.\n$$\nBy Cauchy's inequality, we know\n$$\n\\sum_{j=1}^n \\left(x_j - \\frac{1}{2}\\right) \\le \\sqrt{n\\left(n^2 - \\frac{1}{4}n + 8mn \\ln n\\right)},\n$$\ni.e.\n$$\n\\sum_{j=1}^n x_j \\le \\frac{n}{2} + n\\sqrt{n - \\frac{1}{4} + 8m \\ln n}.\n$$\nWhen $m \\ge n \\ge 2022$, some elementary estimate shows that $\\le 3n\\sqrt{m \\ln n}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 28569,
    "subject": "Mathematics (Multi-modal)",
    "question": "Given a positive integer $n$, let $D$ denote the set of all positive divisors of $n$. Let $A$ and $B$ be subsets of $D$ satisfying: for any $a \\in A$ and $b \\in B$, we have $a \\nmid b$ and $b \\nmid a$. Prove that\n$$\n\\sqrt{|A|} + \\sqrt{|B|} \\le \\sqrt{|D|}.\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "**Proof:** Decompose $D$ into the following disjoint unions $D = X \\sqcup Y \\sqcup Z \\sqcup W$, where\n$$\nX = \\{x \\in D : \\exists a | x, \\exists b | x\\}, \\quad Y = \\{x \\in D : \\exists a | x, \\nexists b | x\\},\n$$\n$$\nZ = \\{x \\in D : \\nexists a | x, \\exists b | x\\}, \\quad W = \\{x \\in D : \\nexists a | x, \\nexists b | x\\}.\n$$\nNote that the assumption of the problem indicates that $A \\subseteq Y, B \\subseteq Z$. It suffices to prove a stronger statement: for any two nonempty subsets $A, B$ of $D$, we always have $\\sqrt{|Y|} + \\sqrt{|Z|} \\le \\sqrt{|D|}$. This inequality is equivalent to\n$$\n|Y| + |Z| + 2\\sqrt{|Y| \\cdot |Z|} \\le |D| = |X| + |Y| + |Z| + |W| \\iff 2\\sqrt{|Y| \\cdot |Z|} \\le |X| + |W|,\n$$\nThis inequality is implied by $|Y| \\cdot |Z| \\le |X| \\cdot |W|$, which can be further rewritten as\n$$ (|X|+|Y|)(|X|+|Z|) = |X|(|X|+|Y|+|Z|)+|Y|\\cdot|Z| \\le |X|(|X|+|Y|+|Z|)+|X|\\cdot|W| = |X|\\cdot|D|. $$\nLet $U = X \\cup Y$ and $V = X \\cup Z$. Then the above inequality becomes $|U| \\cdot |V| \\le |U \\cap V| \\cdot |D|$.\nNote that $U = \\{x \\in D : \\exists a | x\\}$ satisfies: if $x \\in U$ and $x | x'$, then $x' \\in U$. Call such subsets of $D$ upward-closed. Similarly, $V = \\{x \\in D : \\exists b | x\\}$ is also an upward-closed subset of $D$.\nNext, we prove: for any two nonempty upward-closed subsets $U, V$ of $D$, we have $|U| \\cdot |V| \\le |U \\cap V| \\cdot |D|$.\nLet $n = p_1^{\\alpha_1} \\cdots p_k^{\\alpha_k}$ be the prime factorization of $n$, and we make an induction on $k$. Write $p$ and $\\alpha$ for $p_k$ and $\\alpha_k$, respectively, for simplicity. Set $n = p^{\\alpha}n'$. Define $D_k = \\{x \\in D | v_p(x) = k\\}$, $U_k = U \\cap D_k$, and $V_k = V \\cap D_k$. For every $k = 0, 1, \\dots, \\alpha - 1$, for any $x \\in U_k$, the upward-closure property of $U$ implies that $px \\in U_{k+1}$. This means that $|U_k| \\le |U_{k+1}|$, i.e. $\\{|U_k|\\}_k$ is increasing. Similarly, $\\{|V_k|\\}_k$ is increasing. Note that $\\frac{1}{p^k}U_k$ and $\\frac{1}{p^k}V_k$ are upward-closed subsets of $\\frac{1}{p^k}D_k = D(n')$. But inductive hypothesis, we have\n$$\n|(\\frac{1}{p^k}U_k) \\cap (\\frac{1}{p^k}V_k)| \\ge \\frac{1}{|D(n')|} \\cdot |(\\frac{1}{p^k}U_k)| \\cdot |(\\frac{1}{p^k}V_k)|,\n$$\nSo we have $|U_k \\cap V_k| \\ge \\frac{1+\\alpha}{|D|} |U_k| \\cdot |V_k|$. Using rearrangement inequality, we get\n$$\n\\begin{aligned} |U \\cap V| &= \\sum_{k=0}^{\\alpha} |U_k \\cap V_k| \\ge \\frac{1+\\alpha}{|D|} \\sum_{k=0}^{\\alpha} |U_k| \\cdot |V_k| \\\\ &\\ge \\frac{1+\\alpha}{|D|} \\cdot \\frac{1}{1+\\alpha} \\left( \\sum_{k=0}^{\\alpha} |U_k| \\right) \\cdot \\left( \\sum_{k=0}^{\\alpha} |V_k| \\right) \\\\ &= \\frac{1}{|D|} |U| \\cdot |V|. \\end{aligned}\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 28570,
    "subject": "Mathematics (Multi-modal)",
    "question": "There are some direct one-way flights among $n \\ge 8$ airports. Between any two airports $a$ and $b$, there is at most one direct one-way flight from $a$ to $b$ (it is possible to have direct one-way flights both from $a$ to $b$ and from $b$ to $a$). Suppose that, for any set $A$ consisting of some airports with $1 \\le |A| \\le n-1$, there are at least $4 \\cdot \\min\\{|A|, n-|A|\\}$ flights in total departing from the airports in $A$ and arriving at the airports not in $A$.\nProve that for any airport $x$, one can depart from $x$ and take no more than $\\sqrt{2n}$ flights to return to $x$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "*Proof.* We represent each airport with a point. If there is a one-way direct flight from airport *a* to airport *b*, we draw a directed edge $ab$, thus obtaining a directed graph *G*. Let $r = \\lfloor \\sqrt{\\frac{n}{2}} \\rfloor$, then $r \\ge 2$. We will prove that for any vertex *v*, there exists a directed cycle passing through *v* with length at most $2r$. This will imply the desired proposition.\n\nWe prove by contradiction. Suppose that $G$ does not admit a directed cycle through $v$ of length less than or equal to $2r$. For each positive integer $k \\le r$, set\n$$\nN_k = \\{x \\in V(G) \\mid x \\neq v, \\text{ and the shortest directed path from } v \\text{ to } x \\text{ has length } k\\},\n$$\nand set $T_k = N_1 \\cup \\dots \\cup N_k$. We first prove the following fact:\n\n(*) For $1 \\le k \\le r-1$, if the directed edge $\\overrightarrow{xy}$ has source in $T_k$ and target in the complement $T_k^c$, then $x \\in N_k$ and $y \\in N_{k+1}$.\n\nThe proof of fact (*) is as follows: Clearly, $y \\neq v$. Otherwise, we could connect the directed path from $v$ to $x$ of length at most $k$ with the edge $\\overrightarrow{xy} = \\overrightarrow{xb}$, obtaining a directed cycle passing through $v$ with length at most $1+k \\le r < 2r$, which is a contradiction. Let $x \\in N_i$ ($1 \\le i \\le k$), then there exists a directed path $P$ from $v$ to $x$ of length $i$. Connecting $P$ with the edge $\\overrightarrow{xy}$ yields a directed path from $v$ to $y$ of length at most $i+1$. Let $d$ be the length of the shortest directed path from $v$ to $y$. Then $d \\le i+1 \\le k+1$. Since $y \\in (T_k \\cup v)^c$, we have $d \\ge k+1$, which implies $d = k+1$ and $i=k$, i.e., $x \\in N_k$ and $y \\in N_{k+1}$.\n\nNext, we claim that: $|T_r| \\ge \\frac{n}{2}$.\n\n*Proof of the claim:* We prove by contradiction. Suppose that $|T_r| < \\frac{n}{2}$. For each $1 \\le i \\le r-1$, consider the set $M$ of all directed edges from $T_i$ to $T_i^c$. On the one hand, the condition of the problem implies that\n$$\n|M| \\ge 4 \\cdot \\min\\{|T_i|, n - |T_i|\\} = 4 \\cdot |T_i|.\n$$\nOn the other hand, (*) implies that for any directed edge $\\overrightarrow{xy}$ in $M$, we have $x \\in N_i$ and $y \\in N_{i+1}$. From this we know that the number of directed edges in $M$ is less than or equal to the ordered pairs $(x, y) \\in N_i \\times N_{i+1}$, i.e.\n$$\n|M| \\le |N_i| \\cdot |N_{i+1}|.\n$$\nCombining these two discussions, we deduce that, for any $1 \\le i \\le r-1$, we have\n$$\n4 \\cdot |T_i| \\le |N_i| \\cdot |N_{i+1}|. \\tag{1}\n$$\nPut $t_i = |T_i| = |N_1| + \\dots + |N_i|$. From the given conditions, we see that $t_1 = |N_1| \\ge 4 = 2^2$. Using (*), we know that all directed edges $\\overrightarrow{xy}$ from $T_1$ to its complement $T_1^c$ all satisfy $y \\in N_2$. In particular, $|N_2| \\ge 1$. So $|N_1| \\le |T_2| - 1 \\le |T_r| - 1 < \\frac{n}{2} - 1$. Similarly to (*), one can prove all directed edges $\\vec{xy}$ from $\\{v\\} \\cup N_1$ to its complement satisfy $x \\in N_1, y \\in N_2$.\n\nSo we have\n$$\n|N_1| \\cdot |N_2| \\geq 4 \\cdot \\min\\{1 + |N_1|, n - 1 - |N_1|\\} = 4(1 + |N_1|).\n$$\nWe deduce from this that\n$$\nt_2 = |N_1| + |N_2| \\geq |N_1| + \\frac{4(1 + |N_1|)}{|N_1|} > |N_1| + 4 \\geq 8,\n$$\nThus, we have $t_2 \\geq 9 = 3^2$.\n\nBased on this, we use induction to prove that, for $1 \\leq k \\leq r$, we have $t_k \\geq (k+1)^2$. Suppose that for some $3 \\leq m \\leq r$, we have $t_{m-2} \\geq (m-1)^2$ and $t_{m-1} \\geq m^2$. Using (1), we know that, for $1 \\leq i \\leq r-1$, we have $4t_i \\leq (t_i - t_{i-1})(t_{i+1} - t_i)$. Thus, we have\n$$\n\\begin{align*} \nt_m &\\geq t_{m-1} + \\frac{4t_{m-1}}{t_{m-1} - t_{m-2}} \\\n&= t_{m-2} + (t_{m-1} - t_{m-2}) + \\frac{4t_{m-1}}{t_{m-1} - t_{m-2}} \\\n&\\geq t_{m-2} + 2\\sqrt{4t_{m-1}} \\\n&\\geq (m-1)^2 + 2\\sqrt{4m^2} \\\n&= (m+1)^2, \n\\end{align*}\n$$\nThis completes the inductive proof and therefore, $|T_r| \\geq (r+1)^2 = \\left(\\lfloor\\sqrt{\\frac{n}{2}}\\rfloor + 1\\right)^2 > \\frac{n}{2}$, contradicting with earlier assumption $|T_r| < \\frac{n}{2}$! This proves the claim.\n\nSimilarly, define $K_i = \\{x \\in V(G) | x \\neq v$, and the shortest directed path from $x$ to $v$ has length $i\\}$ and put $U_i = K_1 \\cup \\dots \\cup K_i$. Symmetrically, one can prove that $|U_r| \\geq \\frac{n}{2}$.\n\nFinally, note that both $T_r$ and $U_r$ are subsets of $V(G) \\setminus \\{v\\}$ and we have $|T_r| \\geq \\frac{n}{2}$ and $|U_r| \\geq \\frac{n}{2}$. So the intersection of $T_r$ and $U_r$ is nonempty. Take $x \\in T_r \\cap U_r$, then there is a directed path from $v$ to $x$ of length $\\leq r$ and there is a directed path from $x$ to $v$ of length $\\leq r$. Combining these two gives a directed cycle through $v$ of length less than or equal to $2r$. Contradiction! □",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 28571,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all positive integers $n$ with the following property: there exist nonempty finite sets of integers $A, B$, such that for every integer $m$, exactly one of the following three statements is true,\n(i) there exists $a \\in A$, such that $m \\equiv a \\pmod{n}$;\n(ii) there exists $b \\in B$, such that $m \\equiv b \\pmod{n}$;\n(iii) there exist $a \\in A$ and $b \\in B$, such that $m \\equiv a+b \\pmod{n}$.",
    "options": [],
    "answer": "All positive integers except 1, 2, and 4",
    "solution": "*Proof.* Let $A+B = \\{a+b \\mid a \\in A, b \\in B\\}$. The problem can be understood as the remainders of $A$ modulo $n$, the remainders of $B$ modulo $n$, and the remainders of $(A+B)$ modulo $n$ form a partition of all remainders modulo $n$.\n\n(1) If $n > 1$ is an odd number, let $n = 2k+1$, $k \\in \\mathbb{Z}_{>0}$. Take $A = \\{k\\}$, $B = \\{k+1, k+2, \\dots, 2k\\}$, then $A+B = \\{2k+1, 2k+2, \\dots, 3k\\}$, so $A, B, A+B$ are pairwise disjoint, and their union is exactly the consecutive $2k+1$ numbers, forming a complete system modulo $2k+1$, satisfying the condition.\n\n(2) If $n > 1$ satisfies the condition, then for any integer $d > 1$, $dn$ also satisfies the condition. In fact, let the sets corresponding to $n$ be $A, B$. Set\n$$\nA' = \\{a+xn \\mid a \\in A, x = 0, 1, \\dots, d-1\\}, \\quad B' = \\{b+xn \\mid b \\in B, x = 0, 1, \\dots, d-1\\}\n$$\nWe verify that $A', B'$ satisfy the problem requirements for $dn$.\nSince the remainders of $A', B', A' + B'$ modulo $n$ are exactly the remainders of $A, B, A+B$ modulo $n$, the remainders of $A', B', A' + B'$ modulo $n$ and modulo $dn$ have no intersection, so no more than one statements of (i), (ii), and (iii) can hold. Considering any integer $m$, if there exists $a \\in A$ such that $m \\equiv a \\pmod{n}$, then $m$ is congruent modulo $dn$ with one of $a+xn$ ($x = 0, 1, \\dots, d-1$), (i) holds. Similarly, if there exists $b \\in B$ such that $m \\equiv b \\pmod{n}$, then (ii) holds. If there exist $a \\in A, b \\in B$, such that $m \\equiv a+b \\pmod{n}$, then $m$ is congruent modulo $dn$ with one of $a+(b+xn)$ ($x = 0, 1, \\dots, d-1$), (iii) holds. Therefore, exactly one of (i), (ii), and (iii) holds. $A', B'$ satisfy the problem requirements for $dn$.\nBy (1) and (2), all integers $n > 1$ except powers of 2 satisfy the problem conditions.\n\n(3) When $n = 8$, it is easy to verify that $A = \\{1, 2\\}$, $B = \\{3, 6\\}$, $A+B = \\{4, 5, 7, 8\\}$, satisfying the requirements. Combining with (2), we know that all $2^k$ ($k \\ge 3$) satisfy the requirements.\n\nSince $A, B, A+B$ are all non-empty sets, there are at least three different remainders modulo $n$, $n \\ge 3$, so $n = 1, 2$ do not satisfy the conditions.\nWhen $n = 4$, if $A, B$ have only one remainder modulo 4, then $A+B$ also has only one remainder modulo 4, which does not satisfy the requirement. If either $A$ or $B$ has at least two different remainders modulo 4, assume $a_1, a_2 \\in A$ have different remainders modulo 4, and take any $b \\in B$, then $a_1+b$ and $a_2+b$ also have different remainders modulo 4, so $A+B$ has at least two remainders modulo 4, which also does not satisfy the requirement. Therefore, $n = 4$ does not satisfy the condition either.\n\nIn conclusion, the desired $n$ are all positive integers except 1, 2, and 4. $\\Box$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 28572,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $m$ be a positive integer and $A$ be a finite set. Let $A_1, A_2, \\dots, A_m$ be subsets of $A$ (not necessarily distinct). It is known that for any nonempty set $I \\subseteq \\{1, 2, \\dots, m\\}$,\n$$\n\\left| \\bigcup_{i \\in I} A_i \\right| \\geq |I| + 1.\n$$\n\nProve: the elements of $A$ can be coloured black or white, such that every of $A_1, A_2, \\dots, A_m$ contains both black and white elements.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "We give three solutions as follows.\n\n**Solution 1**\n\nConstruct a bipartite graph $G$ whose two parts are $X = \\{A_1, A_2, \\dots, A_m\\}$ and $Y = A$: for $1 \\le i \\le m$ and $a \\in A$, $A_i \\in X$ and $a \\in Y$ are adjacent if and only if $a \\in A_i$. Since for each nonempty $I \\subseteq \\{1, 2, \\dots, m\\}$, $|\\bigcup_{i \\in I} A_i| \\ge |I| + 1$, it follows that for any $k$ vertices of $X$, the number of vertices of $Y$ that are neighbouring to one or more of them, is at least $k+1$. According to Hall's theorem, there exists a transversal $f$ from $X$ to $Y$. For $1 \\le i \\le m$, let $a_i = f(A_i)$. Clearly, $a_i \\in A_i$, and $a_1, a_2, \\dots, a_m$ are distinct elements of $A$. Colour all the other elements $A \\setminus \\{a_1, \\dots, a_m\\}$ white, and determine the colours of $a_1, a_2, \\dots, a_m$ as follows.\n\nEvery time, choose an $i$ such that $a_i$ is uncoloured and $A_i$ has a coloured neighbour, say $b$. Colour $a_i$ the opposite colour to $b$. Suppose, during the process, some elements say $a_1, a_2, \\dots, a_k$ ($1 \\le k \\le m$) are uncoloured, but we cannot find another $i$ and colour $a_i$. Based on the algorithm, this means that all the neighbours of $A_1, A_2, \\dots, A_k$ are in $\\{a_1, \\dots, a_k\\}$, yet it contradicts $|A_1 \\cup A_2 \\cup \\dots \\cup A_k| \\ge k + 1$. Hence, this process can continue until all of $a_1, a_2, \\dots, a_m$ are coloured. Moreover, for $1 \\le i \\le m$, when $a_i$ is coloured, $A_i$ is guaranteed to have black and white neighbours, that is, $A_i$ contains both black and white elements. This colouring satisfies the problem requirements. $\\square$\n\n\n**Solution 2**\n\nWe begin with a lemma.\n\n**Lemma** The finite sets $A_1, A_2, \\dots, A_m$ are called \"nice\", if the union of any $k \\le m$ ($k$ is arbitrary) of them contains at least $k+1$ elements. If $A_1, A_2, \\dots, A_m$ are nice, then there exist 2-element sets $B_1, B_2, \\dots, B_m$, such that $B_i \\subseteq A_i$, $i = 1, 2, \\dots, m$, and $B_1, B_2, \\dots, B_m$ are nice.\n\n**Proof of lemma** Assume the lemma is untrue. Let $A_1, A_2, \\dots, A_m$ be a counterexample with the smallest $m$ value and the smallest $|A_1| + |A_2| + \\dots + |A_m|$ for such $m$. There are three situations.\n\nCase 1, if the union of any $k \\le m$ ($k$ is arbitrary) sets does not contain exactly $k+1$ elements. Take $k=1$, and apparently every set contains 3 or more elements. Choose an arbitrary set and remove any element from it. The sets are still nice (as a counterexample), yet $|A_1| + |A_2| + \\dots + |A_m|$ is smaller. A contradiction.\n\nCase 2, if the union of some $k \\le m-1$ sets contains exactly $k+1$ elements. Assume $|A_1 \\cup A_2 \\cup \\dots \\cup A_k| = k+1$. Since $k<m$, there exist 2-element sets $B_1, B_2, \\dots, B_k$ such that $B_i \\subseteq A_i$, $i=1, 2, \\dots, k$, and $B_1, B_2, \\dots, B_k$ are nice.\n\nLet $C_i := A_i \\setminus (A_1 \\cup A_2 \\cup \\dots \\cup A_k)$, $i = k+1, \\dots, m$. It is easy to see that, among $C_{k+1}, C_{k+2}, \\dots, C_m$, the union of any $d$ ($1 \\le d \\le m-k$, $d$ is arbitrary) sets contains $d$ or more elements. By Hall's theorem, there exist $x_i \\in C_i$ ($i = k+1, \\dots, m$), such that $x_{k+1}, x_{k+2}, \\dots, x_m$ are all distinct. We construct $B_{k+1}, B_{k+2}, \\dots, B_m$ in the following way: if for some $i \\in \\{k+1, k+2, \\dots, m\\}$, $B_i$ has not been made yet, but $A_i$ contains an element $y_i$ that belongs to some $B$ set already constructed, then let $B_i = \\{x_i, y_i\\}$. Clearly, $x_i$ is a new element in the constructed $B$ sets, hence these $B$ sets are still nice. Suppose, after the construction of several $B_i$'s, no more set can be constructed in the above way. Then the remaining $A$ sets do not contain any element in the constructed $B_i$'s. Since $m$ is minimal, we may construct $B_j$'s from those $A$ sets such that they are nice. Furthermore, $B_j$'s and $B_i$'s are disjoint, when combined, $B_1, B_2, \\dots, B_m$ are nice. So, case 2 is not possible.\n\nCase 3, if the union of any $k$ ($1 \\le k \\le m-1$, $k$ is arbitrary) sets contains $k+2$ or more elements, while the union of all $m$ sets contains exactly $m+1$ elements. Observe that any $m-1$ sets have the same union as that of all $m$ sets. Hence, every element must belong to at least 2 sets. Meanwhile, each set contains at least 3 elements. Choose any set and remove any element from it. The sets are still nice, but they have a smaller $|A_1| + |A_2| + \\dots + |A_m|$, contradiction.\n\nThis verifies the lemma.\n\nReturn to the original problem. According to the lemma, we may find 2-element subsets $B_1, B_2, \\dots, B_m$ of $A_1, A_2, \\dots, A_m$, respectively, such that the union of any $k$ of them has $k+1$ or more elements. Treat every element of $A$ as a vertex, and then for every $1 \\le i \\le m$, connect the two elements in $B_i$ by an edge. Note that this graph has no cycle (otherwise, the edges in a cycle correspond to $B$ sets the number of which is the same as the number of elements in their union), and thus it is a forest, which is a bipartite graph. Colour all elements in one part black, and all elements in the other part white. This colouring satisfies the problem condition. $\\square$\n\n\n**Solution 3**\n\nUse induction on the number of sets $m$. When $m = 1$, $|A_1| \\ge 2$, choose two elements of $A_1$ and colour one black and one white. Suppose the conclusion holds for all $m \\le n-1$. Consider $m = n$.\n\nCase 1. If for any $I \\subseteq \\{1, \\dots, n\\}$, $|\\cup_{i \\in I} A_i| \\ge |I| + 2$. Take $a \\in A_n$. By the induction hypothesis, there is a colouring of $A_1 \\setminus \\{a\\}, A_2 \\setminus \\{a\\}, \\dots, A_{n-1} \\setminus \\{a\\}$ for which each $A_i \\setminus \\{a\\}$ contains black and white elements. As $A_n \\setminus \\{a\\}$ has been coloured, we can colour $a$ such that $A_n$ has elements of both colours.\n\nCase 2. If for some $I \\ne \\{1, 2, \\dots, n\\}$, $|\\cup_{i \\in I} A_i| = |I| + 1$. Let $I$ be the largest, $J = \\{1, 2, \\dots, n\\} \\setminus I$, and $B = \\cup_{i \\in I} A_i$, $C = \\cup_{j \\in J} A_j$.\n\nCase 2a. If $B \\cap C = \\emptyset$, then colour $B$ and $C$ separately. By induction, there is a colouring that guarantees each $A_i$ contains black and white elements.\n\nCase 2b. If $B \\cap C \\ne \\emptyset$, take $b \\in B \\cap C$, and define $A'_j = A_j \\setminus (B \\setminus \\{b\\})$ ($j \\in J$). For any $T \\subseteq J$, since $I$ is the largest, it follows that\n$$\n\\left| \\bigcup_{t \\in T} A'_t \\right| \\ge \\left| \\bigcup_{t \\in I \\cup T} A_t \\right| - \\left| \\bigcup_{i \\in I} A_i \\right| \\ge (|I \\cup T| + 2) - (|I| + 1) = |T| + 1.\n$$\nSimilarly,\n$$\n\\left| \\bigcup_{j \\in J} A'_j \\right| = \\left| \\bigcup_{i=1} A_i \\right| - |B \\setminus \\{b\\}| \\ge n + 1 - |I| = |J| + 1.\n$$\nBy induction, we may colour $C \\setminus (B \\setminus \\{b\\}) = \\cup_{j \\in J} A'_j$ and $B = \\cup_{i \\in I} A_i$ such that $A'_j(j \\in J)$ and $A_i(i \\in I)$ contain elements of both colours. If $b$ has different colours in the two colourings, then reverse the colouring of $B$. Now they are compatible and give a colouring of $A$ which meets the problem requirements. $\\square$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 28573,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $x_1, x_2, \\dots, x_{11}$ be nonnegative real numbers which add up to one. For $i = 1, 2, \\dots, 11$, let\n$$\ny_i = \\begin{cases} x_i + x_{i+1}, & \\text{if } i \\text{ is odd,} \\\\ x_i + x_{i+1} + x_{i+2}, & \\text{if } i \\text{ is even,} \\end{cases}\n$$\nwhere $x_{12} = x_1$. Set $F(x_1, x_2, \\dots, x_{11}) = y_1y_2\\cdots y_{11}$.\n\n*Prove that if $F$ achieves maximum, then $x_6 < x_8$.*",
    "options": [],
    "answer": "Detailed solution",
    "solution": "*Proof.* The function $F$, viewed as a continuous function with respect to $x_1, x_2, \\cdots, x_{11}$, has a maximum value on the bounded closed set\n$$\n\\Omega = \\{(x_1, x_2, \\cdots, x_{11}) \\in (\\mathbb{R}_{\\ge 0})^{11} \\mid x_1 + x_2 + \\cdots + x_{11} = 1\\}.\n$$\nAssume that $F$ has a maximum value at $(a_1, a_2, \\cdots, a_{11}) \\in \\Omega$. It is clear that $F > 0$ at this point.\n\n(1) $a_3 = a_5 = a_7 = a_9 = a_{11} = 0$.\nIf $a_3 > 0$, let $a'_3 = 0$, $a'_4 = a_3 + a_4$, and let $a'_i = a_i$ for other $i$. This maintains the sum. We prove that $F(a_1, a_2, \\cdots, a_{11}) < F(a'_1, a'_2, \\cdots, a'_{11})$, which is equivalent to\n$$\n(a_2 + a_3 + a_4)(a_3 + a_4)(a_4 + a_5 + a_6) < (a'_2 + a'_3 + a'_4)(a'_3 + a'_4)(a'_4 + a'_5 + a'_6),\n$$\nand since $a_3 + a_4 = a'_3 + a'_4$ and $a_4 < a'_4$, this inequality holds. This contradicts the fact that $F$ has a maximum value at $(a_1, a_2, \\cdots, a_{11})$. Therefore, $a_3 = 0$. Similarly, we can prove that $a_5 = a_7 = a_9 = a_{11} = 0$.\n\nNow, let $x_3 = x_5 = x_7 = x_9 = x_{11} = 0$ in the expression for $F$. We will now view $F$ as a function of $x_1, x_2, x_4, \\cdots, x_{10}$,\n$$\nF = (x_1 + x_2)(x_2 + x_4)x_4(x_4 + x_6)x_6(x_6 + x_8)x_8(x_8 + x_{10})x_{10}(x_{10} + x_1)x_1.\n$$\nThen, $F$ has a maximum value at $(a_1, a_2, a_4, a_6, a_8, a_{10})$. It is clear that $a_1, a_4, a_6, a_8, a_{10} > 0$.\n\n(2) $a_2 = 0$.\nBy contradiction, suppose $a_2 > 0$. If $a_1 \\le a_4$, let $a'_1 = a_1 + a_2$, $a'_2 = 0$, and let $a'_i = a_i$ for other $i$. This maintains the sum. We prove that $F(a_1, a_2, a_4, \\cdots, a_{10}) < F(a'_1, a'_2, a'_4, \\cdots, a'_{10})$.\n$$\n\\Leftrightarrow a_1(a_{10} + a_1)(a_1 + a_2)(a_2 + a_4) < a'_1(a'_{10} + a'_1)(a'_1 + a'_2)(a'_2 + a'_4).\n$$\nSince $a_1 < a'_1$ and $a_{10} + a_1 < a'_{10} + a'_1$, we only need to prove:\n$$\na_1(a_1 + a_2)(a_2 + a_4) \\le a'_1(a'_1 + a'_2)(a'_2 + a'_4) = (a_1 + a_2)(a_1 + a_2)a_4,\n$$\n$$\n\\Leftrightarrow a_1(a_2 + a_4) \\le a_4(a_1 + a_2), \\quad \\Leftrightarrow a_1a_2 \\le a_4a_2,\n$$\nwhich holds. If $a_1 \\ge a_4$, let $a'_4 = a_4 + a_2$, $a'_2 = 0$, and for other $i$, let $a'_i = a_i$. We also have:\n$$\nF(a_1, a_2, a_4, \\cdots, a_{10}) < F(a'_1, a'_2, a'_4, \\cdots, a'_{10}).\n$$\nThis contradicts the fact that $F$ reaches its maximum value at $(a_1, a_2, a_4, a_6, a_8, a_{10})$. Therefore, $a_2 = 0$.\n\nNow let $x_2 = 0$ in the analytic expression of $F$, and regard $F$ as a function of $x_1, x_4, x_6, x_8, x_{10}$,\n$$\nF = x_4^2(x_4 + x_6)x_6(x_6 + x_8)x_8(x_8 + x_{10})x_{10}(x_{10} + x_1)x_1^2.\n$$\nThen $F$ reaches its maximum value at $(a_1, a_4, a_6, a_8, a_{10})$.\n\n(3) $a_1 = a_4, a_6 = a_{10}$.\nLet $a'_1 = a'_4 = \\frac{1}{2}(a_1 + a_4)$, $a'_6 = a'_{10} = \\frac{1}{2}(a_6 + a_{10})$, $a'_8 = a_8$, keeping the sum unchanged. By the inequality of arithmetic and geometric means,\n$$\na_1^2 a_4^2 \\le a'_1^2 a'_4^2, \\quad (a_4 + a_6)(a_{10} + a_1) \\le (a'_4 + a'_6)(a'_{10} + a'_1),\n$$\n$$\na_6 a_{10} \\le a'_6 a'_{10}, \\quad (a_6 + a_8)(a_8 + a_{10}) \\le (a'_6 + a'_8)(a'_{8} + a'_{10}).\n$$\nAlso, $a_8 = a'_8$. Multiplying these equations gives:\n$$\nF(a_1, a_4, \\dots, a_{10}) \\le F(a'_1, a'_4, \\dots, a'_{10}).\n$$\nSince $F$ reaches its maximum value at $(a_1, a_4, \\dots, a_{10})$, the equality holds in the above inequality. By the conditions for equality in the inequality of arithmetic and geometric means, we have $a_1 = a_4, a_6 = a_{10}$.\n\nIn the analytic expression of $F$, replace $x_4$ with $x_1$, and $x_{10}$ with $x_6$. Now regard $F$ as a function of $x_1, x_6, x_8$,\n$$\nF = x_1^4 x_6^2 x_8 (x_1 + x_6)^2 (x_6 + x_8)^2,\n$$\nand $2x_1 + 2x_6 + x_8 = 1$.\n\n(4) Next, we will solve the system of equations (this system of equations is obtained by matching coefficients in the inequality of arithmetic and geometric means, which will be used in (5)):\n$$\n\\frac{2}{u} + \\frac{1}{u+v} = \\frac{1}{v} + \\frac{1}{u+v} + \\frac{1}{v+1} = 1 + \\frac{2}{v+1}, \\quad u, v > 0.\n$$\nWe will prove that there is a unique positive real solution $u, v$, and that $v < 1$.\nFrom the first equation, we get $\\frac{2}{u} = \\frac{1}{v} + \\frac{1}{v+1}$, hence $u = \\frac{2v(v+1)}{2v+1}$. From the second equation, we get $\\frac{1}{v} + \\frac{1}{u+v} = 1 + \\frac{1}{v+1}$, substituting $u$, and rearranging it as an equation about $v$,\n$$\n4v^3 + 5v^2 - 4v - 4 = 0.\n$$\nLet $f(t) = 4t^3 + 5t^2 - 4t - 4$, then $f'(t) = 12t^2 + 10t - 4$, which is negative at first and positive later on $[0, +\\infty)$, so $f(t)$ first decreases and then increases on $[0, +\\infty)$. Since $f(0) = -4 < 0$, $f(1) = 1 > 0$, therefore, $f$ has a unique solution $v \\in (0, 1)$ on $[0, +\\infty)$, and $u$ is also uniquely determined.\n\n(5) Let $u, v > 0$ be the solution satisfying the system of equations in (4), and let $\\frac{2}{u} + \\frac{1}{u+v} = k$.\n\nUsing the inequality of arithmetic and geometric means, we have:\n$$\n\\begin{align*}\nF(x_1, x_6, x_8) &= \\left(\\frac{x_1}{u}\\right)^4 \\left(\\frac{x_6}{v}\\right)^2 x_8 \\left(\\frac{x_1+x_6}{u+v}\\right)^2 \\left(\\frac{x_6+x_8}{v+1}\\right)^2 u^4 v^2 (u+v)^2 (v+1)^2 \\\\\n&\\le \\frac{u^4 v^2 (u+v)^2 (v+1)^2}{11^{11}} \\left(\\frac{4}{u} x_1 + \\frac{2}{v} x_6 + x_8 + \\frac{2}{u+v} (x_1+x_6) + \\frac{2}{v+1} (x_6+x_8)\\right)^{11} \\\\\n&= \\frac{u^4 v^2 (u+v)^2 (v+1)^2}{11^{11}} \\left(\\left(\\frac{4}{u} + \\frac{2}{u+v}\\right) x_1 + \\left(\\frac{2}{v} + \\frac{2}{u+v} + \\frac{2}{v+1}\\right) x_6 + \\left(1 + \\frac{2}{v+1}\\right) x_8\\right)^{11} \\\\\n&= \\frac{u^4 v^2 (u+v)^2 (v+1)^2}{11^{11}} (2kx_1 + 2kx_6 + kx_8)^{11} = \\frac{u^4 v^2 (u+v)^2 (v+1)^2 k^{11}}{11^{11}}.\n\\end{align*}\n$$\nThe equality above holds if and only if $x_1 : x_6 : x_8 = u : v : 1$, that is, $x_8 = \\frac{1}{2u+2v+1}$,\n$x_6 = \\frac{v}{2u+2v+1}$, $x_1 = \\frac{u}{2u+2v+1}$. Therefore, $F$ attains its maximum value if and only if $x_2 = x_3 =$\n$x_5 = x_7 = x_9 = x_{11} = 0$, $x_1 = x_4 = \\frac{u}{2u+2v+1}$, $x_6 = x_{10} = \\frac{v}{2u+2v+1}$, $x_8 = \\frac{1}{2u+2v+1}$. At this point,\n$x_6 = vx_8 < x_8$. $\\square$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 28574,
    "subject": "Mathematics (Multi-modal)",
    "question": "Given an integer $n > 1$ and an integer $a$ that is coprime with $n$. There is a country consisting of $n$ islands $D_1, D_2, \\dots, D_n$. For any two different islands $D_i$ and $D_j$, there is a one-way ferry from $D_i$ to $D_j$ if and only if $ij \\equiv ia \\pmod{n}$. A tourist hopes to visit as many islands as possible. He can first fly to any island he chooses to start the tour, and afterwards can only use the one-way ferry to tour freely between islands in this country. Find the maximum possible number of different islands that the tourist can visit.",
    "options": [],
    "answer": "Let x(n) = sum over primes p dividing n with v_p(n) ≥ 2 of 1, and y(n) = sum over primes p dividing n with v_p(n) = 1 of 1. Then the maximum number of islands is\n\nk(n) = { 3 x(n) + 2 y(n) + 1, if v_2(n) ≠ 1; 3 x(n) + 2 y(n), if v_2(n) = 1 }.",
    "solution": "Let $x(n) = \\sum_{v_p(n) \\ge 2} 1$ and $y(n) = \\sum_{v_p(n)=1} 1$. Let the maximum number of islands that can be visited be denoted as $k(n)$. Then we have:\n$$\nk(n) = \\begin{cases} 3x(n) + 2y(n) + 1, & v_2(n) \\neq 1, \\\\ 3x(n) + 2y(n), & v_2(n) = 1. \\end{cases}\n$$\nFor $x, y \\in \\mathbb{Z}$ and $m \\in \\mathbb{Z}_{>0}$, if $xy \\equiv xa \\pmod{m}$, we denote this as $x \\to y \\pmod{m}$. Let $p$ be a prime factor of $n$ and $p^\\alpha \\nmid n$. For a sequence $x_1 \\to x_2 \\to \\dots \\to x_k \\pmod{n}$, the same relation holds modulo $p^\\alpha$:\n$$\nx_1 \\to x_2 \\to \\dots \\to x_k \\pmod{p^\\alpha}.\n$$\n(1) If $x_i \\neq 0 \\pmod{p^\\alpha}$, then $x_{i+1}$ is coprime to $p$. This is because $p^\\alpha \\mid x_i(x_{i+1} - a)$, and $p^\\alpha \\nmid x_i$. Therefore, $p \\mid x_{i+1} - a$. Since $a$ is coprime to $n$, we have $p \\nmid a$, which implies $p \\nmid x_{i+1}$.\n(2) If $x_i$ is coprime to $p$, then $x_{i+1} \\equiv a \\pmod{p^\\alpha}$. This is because $p^\\alpha \\mid x_i(x_{i+1} - a)$, and since $p \\nmid x_i$, we have $p^\\alpha \\mid x_{i+1} - a$.\n(3) From (1) and (2), we can conclude that if $\\alpha \\ge 2$, the sequence $x_1, x_2, \\dots, x_k \\pmod{p^\\alpha}$ undergoes at most three changes: it changes from being divisible by $p^\\alpha$ to being nonzero, then becomes coprime to $p$, and finally becomes congruent to $a$ modulo $p^\\alpha$.\nIf $\\alpha = 1$ and $p \\neq 2$, then $x_1, x_2, \\dots, x_k \\pmod{p^\\alpha}$ undergo at most two changes: it changes from being congruent to 0 modulo $p$ to being coprime to $p$, and finally becomes congruent to $a$ modulo $p$.\nIf $\\alpha = 1$ and $p = 2$, then $x_1, x_2, \\dots, x_k \\pmod{p^\\alpha}$ undergo at most one change: it changes from being congruent to 0 modulo 2 to being congruent to $a$ modulo 2.\nAssuming that $x_1, x_2, \\dots, x_k \\pmod{n}$ have distinct adjacent terms, then adjacent terms have changes in some modulo $p^\\alpha$, implying that $k \\le k(n)$.\n(4) Construct examples for $k = k(n)$. First, consider the cases based on modulo $p^\\alpha$.\n(i) If $\\alpha \\ge 2$, we have $0 \\to p \\to a + p^{\\alpha-1} \\to a \\pmod{p^\\alpha}$.\n(ii) If $\\alpha = 1$ and $p \\neq 2$, we have $0 \\to b \\to a \\pmod{p}$, where $b$ is coprime to $p$ and $b \\neq a \\pmod{p}$.\n(iii) If $\\alpha = 1$ and $p = 2$, we have $0 \\to a \\pmod{2}$.\nLet $n = p_1^{\\alpha_1} \\cdots p_w^{\\alpha_w}$ be the prime factorization of $n$. For each $p_i^{\\alpha_i}$, we choose a sequence corresponding to the three cases mentioned above. Note that $0 \\to 0$ and $a \\to a$. We can appropriately add some zeros at the beginning and some $a$'s at the end of the sequences to ensure that their lengths reach $k(n)$. Let's denote these modified sequences as $L_i$. It is required that for $p_i^{\\alpha_i}$ and $p_j^{\\alpha_j}$ ($i \\neq j$), the positions where changes occur in $L_i$ and $L_j$ are different.\n\nFor example, suppose $\\alpha_1 \\ge 2$, we can take\n$$\nL_1: 0 \\to p_1 \\to a + p_1^{\\alpha_1-1} \\to a \\to a \\to \\dots \\to a \\pmod{p_1^{\\alpha_1}}.\n$$\nIf $\\alpha_2 = 1$ and $p_2 \\ne 2$, we can also take\n$$\nL_2: 0 \\to 0 \\to 0 \\to 0 \\to b \\to a \\to a \\to \\dots \\to a \\pmod{p_2},\n$$\nand continue this way to construct $L_3, \\dots, L_w$.\nBy using the $t$-th term of $L_1, L_2, \\dots, L_w$ and the Chinese Remainder Theorem, we can determine $x_t$. Since $x_t \\to x_{t+1} \\pmod{p_i^{\\alpha_i}}$ for $1 \\le i \\le w$, it follows that $x_t \\to x_{t+1} \\pmod{n}$. Moreover, the sequence $x_1, x_2, \\dots, x_{k(n)}$ modulo $n$ has distinct terms. This confirms the conclusion of the original problem, and the maximum value $k(n)$ is given by the formula mentioned at the beginning of the solution. $\\square$",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 28575,
    "subject": "Mathematics (Multi-modal)",
    "question": "Given two fixed points $A$ and $B$ on the unit circle $\\omega$, satisfying $\\sqrt{2} < AB < 2$. Let $P$ be a moving point on $\\omega$ such that $\\triangle ABP$ is an acute-angled triangle and $AP > AB > BP$.\n\nFor this moving point $P$, let $H$ be the orthocenter of $\\triangle ABP$. Take a point $S$ on the arc $\\widehat{AP}$ such that $SH = AH$, and take a point $T$ on the arc $\\widehat{AB}$ such that $TB \\parallel AP$. Let $Q$ be the intersection of lines $ST$ and $BP$.\n\nProve that there exists a fixed point in the plane such that the circle with diameter $HQ$ passes through it.",
    "options": [],
    "answer": "the midpoint of the chord joining the two fixed points",
    "solution": "We prove that the midpoint $M$ of $AB$ satisfies the given condition.\n\n![](attached_image_1.png)\n\nLet $P_1$ be the antipodal point of $P$ on the circle $\\omega$, and let $H_1$ be the intersection of the extension of $AH$ with $\\omega$. We will show that $QP_1 = QH_1$.\n\nDenote $O$ as the center of $\\omega$. Since $SH = AH$, we have that $S$ and $A$ are symmetric with respect to $OH$, implying $OH \\perp SA$. Also, $HB \\perp AP$, so $\\angle BHO - 180^\\circ - \\angle SAP = 180^\\circ - \\angle STP = \\angle PTQ$. By noting that $TB \\parallel AP$, we have $\\angle TPQ = \\angle APB - \\angle APT = \\angle APB - \\angle PAB = \\angle HBO$. Thus, $\\triangle PTQ \\sim \\triangle BHO$, which gives $\\frac{PQ}{PT} = \\frac{BO}{BH}$. Since $PT = AB$ and $BO = PO$, we obtain $\\frac{PQ}{AB} = \\frac{PO}{BH}$. Furthermore, $\\angle OPQ = 90^\\circ - \\angle PAB = \\angle ABH$, implying $\\triangle OPQ \\sim \\triangle HBA$. Therefore, $\\angle OQP = \\angle HAB = \\angle H_1AB = \\angle H_1P_1B$. Since $PQ \\perp P_1B$, it follows that $OQ \\perp P_1H_1$, and thus $OQ$ bisects $P_1H_1$ perpendicularly, leading to $QP_1 = QH_1$.\n\nSince $H_1$ and $H$ are symmetric with respect to $BP$, we have $QH = QH_1 = QP_1$. Also, it is well-known that $M$ is the midpoint of $HP_1$, so $QM \\perp MH$. Consequently, the circle with diameter $HQ$ passes through the midpoint $M$ of $AB$. $\\square$\n\n![](attached_image_1.png)\nLet lowercase letters represent complex numbers corresponding to the respective points on the complex plane.\n\nFirstly,\n$$\n\\begin{align*}\nQ \\in PB & \\\\\n\\Leftrightarrow \\frac{q-p}{q-b} \\in \\mathbb{R} & \\Leftrightarrow \\frac{q-p}{q-b} = \\frac{\\bar{q}-\\bar{p}}{\\bar{q}-\\bar{b}} \\\\\n& \\Leftrightarrow q\\bar{q} - p\\bar{q} - q\\bar{b} + p\\bar{b} = q\\bar{q} - b\\bar{q} - q\\bar{p} + b\\bar{p} \\\\\n& \\Leftrightarrow (p-b)\\bar{q} + \\frac{p-b}{pb}q = \\frac{p^2-b^2}{pb} \\\\\n& \\Leftrightarrow q + pb\\bar{q} = p + b.\n\\end{align*}\n$$\n\nSimilarly,\n$$\nQ \\in ST \\Leftrightarrow q + st\\bar{q} = s + t.\n$$\nTherefore,\n$$\n\\bar{q} = \\frac{s + t - p - b}{st - pb}.\n$$\nAccording to the given conditions, $t = \\frac{ap}{b}$, and $s$ satisfies\n$$\n\\begin{align*}\n(s-h)\\left(\\frac{1}{s}-\\bar{h}\\right) &= (a-h)\\left(\\frac{1}{a}-\\bar{h}\\right) \\\\\n\\Leftrightarrow 1-\\frac{h}{s}-\\bar{h}s+h\\bar{h} &= 1-\\frac{h}{a}-\\bar{h}a+h\\bar{h} \\\\\n\\Leftrightarrow \\bar{h}s^2-\\left(\\frac{h}{a}+\\bar{h}a\\right)s+h = 0.\n\\end{align*}\n$$\nHence, by Vieta's formulas, we have\n$$\ns = \\frac{h}{a\\bar{h}} = bp \\frac{a+b+p}{ab+bp+pa}.\n$$\nFurthermore, we have\n$$\n\\begin{align*}\ns + t - p - b &= p \\left[ \\frac{a}{b} + \\frac{b(a+b+p)}{ab+bp+pa} - 1 \\right] - b \\\\\n&= p \\frac{a(ab + bp + pa) + b(b^2 - pa)}{b(ab + bp + pa)} - b \\\\\n&= \\frac{ap(ab + bp + pa) + bp(b^2 - pa) - b^2(ab + bp + pa)}{b(ab + bp + pa)} \\\\\n&= \\frac{(ab + pa)(ap - b^2)}{b(ab + bp + pa)} = \\frac{a(b + p)(ap - b^2)}{b(ab + bp + pa)}.\n\\end{align*}\n$$\n$$\nst - pb = \\frac{ap^2(a + b + p) - pb(ab + bp + pa)}{ab + bp + pa} = \\frac{p[ap(a + p) - b^2(a + p)]}{ab + bp + pa} = \\frac{p(ap - b^2)(a + p)}{ab + bp + pa}.\n$$\nso\n$$\n\\bar{q} = \\frac{a(p+b)}{bp(p+a)}, \\quad q = \\frac{\\frac{1}{abp}(b+p)}{\\frac{1}{abp^2}(p+a)} = \\frac{p(p+b)}{p+a}.\n$$\n\nWe know that $h = a + b + p$. The center of the circle with diameter $HQ$ is\n$$\n\\frac{1}{2}(h+q) = \\frac{a+b}{2} + \\frac{p(p+a)+(p+b)}{p+a},\n$$\nand its radius is\n$$\n\\left| \\frac{1}{2}(h+q) \\right| = \\left| \\frac{1}{2(p+a)} \\left[ (a+b+p)(p+a) - p(p+b) \\right] \\right| = \\left| \\frac{a}{2} \\frac{(p+a)+(p+b)}{p+a} \\right|.\n$$\nSince $|a| = |p| = 1$, we can conclude that the circle passes through the midpoint of $AB$, which is $\\frac{a+b}{2}$. Thus, the proof is complete. $\\square$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 28576,
    "subject": "Mathematics (Multi-modal)",
    "question": "Does there exist an irrational number $x$ such that there are at most finitely many positive integers $n$ satisfying\n$$\n\\{kx\\} \\geq \\frac{1}{n+1}\n$$\nfor every $k \\in \\{1, \\dots, n\\}$?\n\nNote: Here, for a positive real number $y$, $\\{y\\}$ denotes the fractional part of $y$.",
    "options": [],
    "answer": "No; such an irrational number does not exist. For every irrational number, there are infinitely many positive integers n satisfying the condition.",
    "solution": "*Proof* 1. Nonexistence. Assume that there exists such a positive integer $n$. Let $x$ be an irrational number. Since the terms of the sequence $\\{x\\}$, $\\{2x\\}$, $\\dots$ are distinct and densely distributed in the interval $(0, 1)$, there exist infinitely many positive integers $d$ satisfying\n$$\n\\{dx\\} = \\min \\{ \\{ax\\} \\mid a = 1, 2, \\dots, d \\}.\n$$\nArrange these $d$ in an infinite increasing sequence $d_1 < d_2 < \\dots$.\nWe will now show that for all $n = d_i - 1$ ($i \\ge 2$), we always have\n$$\n\\min \\{ \\{kx\\} \\mid k = 1, \\dots, n \\} = \\{d_{i-1}x\\} \\ge \\frac{1}{d_i}.\n$$\nSuppose that this inequality does not hold, then $d_i \\cdot \\{d_{i-1}x\\} \\le 1$. Thus,\n$$\n\\{d_{i-1}d_i x\\} = \\{d_i \\cdot \\lfloor d_{i-1}x \\rfloor + d_i \\cdot \\{d_{i-1}x\\}\\} = d_i \\cdot \\{d_{i-1}x\\}.\n$$\nOn the other hand,\n$$\n\\{d_i d_{i-1} x\\} = \\{d_{i-1} \\lfloor d_i x \\rfloor + d_{i-1} \\{d_i x\\}\\} \\le d_{i-1} \\{d_i x\\}.\n$$\nCombining the above two equations, we obtain $d_{i-1} \\cdot \\{d_i x\\} \\ge d_i \\cdot \\{d_{i-1} x\\}$. However, $d_{i-1} < d_i$ and $\\{d_i x\\} < \\{d_{i-1} x\\}$, which leads to a contradiction!\nTherefore, $n = d_i - 1$ satisfies the given condition, and there are infinitely many such $n$. $\\square$\n\n\n*Proof* 2. Nonexistent. If there is no irrational number $x$ satisfying the condition, then there exists $n_0 \\in \\mathbb{Z}_+$ such that for any $n \\ge n_0$, there exists $k \\in \\{1, \\dots, n\\}$ such that $\\{kx\\} < \\frac{1}{n+1}$. Since $x$ is an irrational number, for any $k \\in \\mathbb{Z}_+$, $\\{kx\\} \\ne 0$. For $n_0$, there exist $k_0 \\in \\{1, \\dots, n_0\\}$ and $\\ell_0 \\in \\mathbb{Z}$ such that\n$$\n0 < k_0 x - \\ell_0 < \\frac{1}{n_0 + 1}.\n$$\nWithout loss of generality, assume that $k_0$ and $\\ell_0$ are coprime. Otherwise, we can replace $k_0$ and $\\ell_0$ with $\\frac{k_0}{\\gcd(k_0, \\ell_0)}$ and $\\frac{\\ell_0}{\\gcd(k_0, \\ell_0)}$ respectively, and the condition still holds.\nLet $n_1 = \\left\\lfloor \\frac{1}{k_0 x - \\ell_0} \\right\\rfloor > n_0$. Then there exist $k_1 \\in \\{1, \\dots, n_1\\}$ and $\\ell_1 \\in \\mathbb{Z}$ such that\n$$\n0 < k_1 x - \\ell_1 < \\frac{1}{n_1 + 1} < k_0 x - \\ell_0.\n$$\n\nSimilarly, we can assume that $k_1$ and $\\ell_1$ are coprime. Since $k_0$ is coprime with $\\ell_0$ and $k_1$ is coprime with $\\ell_1$, we have $\\frac{\\ell_0}{k_0} \\neq \\frac{\\ell_1}{k_1}$. Therefore,\n$$\n\\begin{aligned}\n1 \\le |k_0\\ell_1 - k_1\\ell_0| &= |k_1(k_0x - \\ell_0) - k_0(k_1x - \\ell_1)| \\\\\n&< \\max \\{k_1(k_0x - \\ell_0), k_0(k_1x - \\ell_1)\\} \\quad (\\text{because } k_1(k_0x - \\ell_0) > 0, k_0(k_1x - \\ell_1) > 0) \\\\\n&\\le \\max \\left\\{\\frac{k_1}{n_1}, \\frac{k_0}{n_1 + 1}\\right\\} \\le 1 \\quad (\\text{because } \\frac{1}{k_0x - \\ell_0} \\ge n_1, \\text{ hence } k_0x - \\ell_0 \\le \\frac{1}{n_1}.)\n\\end{aligned}\n$$\nThis leads to a contradiction. $\\Box$\n\n\n*Proof 3.* For any irrational number $x$, the fractional part $\\{kx\\}$ is distinct and densely distributed in the interval $(0,1)$. Hence, there exist infinitely many positive integers $m$ satisfying:\n$$\n\\{mx\\} < \\{kx\\}, \\quad \\forall k = 1, 2, \\dots, m-1.\n$$\nFor each such $m \\ge 2$, let $\\beta$ be the smallest value among $\\{x\\}$, $\\{2x\\}$, ..., $\\{(m-1)x\\}$. Thus, for $k = 1, 2, \\dots, m-1$, we have $\\{kx\\} \\ge \\beta$, which implies that there are no integers in the open intervals $(kx - \\beta, kx)$.\nConsider the points $O : (0,0)$, $A : (m,mx)$, $B : (m,mx - \\beta)$, $C : (0,-\\beta)$ on the coordinate plane. The parallelogram $OABC$ does not contain any lattice points in its interior, and on its boundary, there are exactly three lattice points: $O$, $D : (m, \\lfloor mx \\rfloor)$, and $E : (r, rx - \\beta) = (r, \\lfloor rx \\rfloor)$.\nThe lattice triangle $\\triangle ODE$ has no lattice points inside or on its boundary, except for the vertices. By Pick's theorem, its area is $\\frac{1}{2}$. Therefore, the area of the parallelogram $OABC$ is $\\ge 2S_{\\triangle ODE} = 1$. On the other hand, the area of this parallelogram is $m \\times \\beta = m \\times \\{rx\\} \\ge 1$. Hence, $\\beta \\ge \\frac{1}{m}$, and for $k = 1, 2, \\dots, m-1$, we have\n$$\n\\{kx\\} \\ge \\{rx\\} \\ge \\frac{1}{m}.\n$$\nTherefore, $n = m - 1$ satisfies the given condition. There are infinitely many such $n$, implying that no irrational number $x$ satisfies the condition. $\\Box$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 28577,
    "subject": "Mathematics (Multi-modal)",
    "question": "Determine the smallest real number $\\lambda$ with the following property: every positive integer $n$ can be written as a product $n = x_1x_2\\cdots x_{2023}$, with each $x_i$ either a prime number or a positive integer that is less than or equal to $n^\\lambda$.",
    "options": [],
    "answer": "1/1012",
    "solution": "The number $\\lambda$ is $\\frac{1}{1012}$.\nThe minimal such $\\lambda$ is $\\frac{1}{1012}$.\nTo see that $\\lambda < \\frac{1}{1012}$ does not work, we take $n = p^{2024}$ for some prime number $p$. Then when writing $n$ as the product of $2023$ integers, at least one of the integers is $p^\\alpha$ for some $\\alpha \\ge 2$ (which is not a prime number); then $p^\\alpha \\ge p^2 = n^{\\frac{1}{1012}} > n^\\lambda$.\nNow we prove that $\\lambda = \\frac{1}{1012}$ satisfies the condition of the problem. To start, we write $n$ as the product of its prime factors $n = p_1p_2\\cdots p_r$ (allowing some $p_i$'s to be equal), where we order the primes so that $p_1 \\ge p_2 \\ge \\cdots$.\nNow starting from $p_1$, let $r_1 \\ge 1$ be the smallest integer such that $x_1 = p_1p_2\\cdots p_{r_1} > n^{1/2024}$. Note that this $x_1$ satisfies the condition of the problem, namely, if $r_1 = 1$, $x_1$ is a prime number; otherwise $r_1 \\ge 2$, then we must have $p_1 \\le n^{1/2024}$. This means that $p_r \\le n^{1/2024}$; so\n$$\nx_1 = (p_1p_2\\cdots p_{r-1})p_r \\le n^{1/2024} \\cdot n^{1/2024} = n^{1/1012},\n$$\nwhere the first inequality holds by minimality of $r_1$.\nNext, let $r_2 \\ge r_1 + 1$ be the smallest integer such that $x_2 = p_{r_1+1}\\cdots p_{r_2} > n^{1/2024}$. The same argument as above shows that either $x_2$ is a prime number or $x_2 \\le n^{1/1012}$.\nContinue this process. If we run out of $p_i$'s before reaching $x_{2023}$, then set the rest of $x_j$ to be $1$, and then $n = x_1x_2\\cdots x_{2023}$ gives the needed decomposition. Otherwise, we have constructed $x_1, x_2, \\dots, x_{2022}$ satisfying the condition of the problem (and that each $x_i > n^{1/2024}$ by construction), then we must have\n$$\nx_{2023} := \\frac{n}{x_1 \\cdots x_{2022}} < \\frac{n}{(n^{1/2024})^{2022}} = n^{1/1012}.\n$$\nThe decomposition $n = x_1x_2\\cdots x_{2023}$ satisfies the conditions of the problem.\nThe proof for that $\\lambda < \\frac{1}{1012}$ does not work is the same as Solution 1. We prove that $\\lambda = \\frac{1}{1012}$ satisfies the requirement. Let $n = p_1p_2\\cdots p_r$ be the prime factorization of $n$ (where $p_1 \\ge p_2 \\ge \\cdots \\ge p_r$). For $i = 1, 2, \\dots, 1012$, define\n$$\ny_i = \\prod_{\\substack{1 \\le j \\le r \\\\ j \\equiv i \\pmod{1012}}} p_j.\n$$\nThen $y_1 \\ge y_2 \\ge \\cdots \\ge y_{1012}$ and $y_1y_2\\cdots y_{1012} = n$.\nIf $y_1 \\le n^{1/1012}$, then let $x_1, \\dots, x_{2023}$ be $y_1, y_2, \\dots, y_{1012}$ plus $1011$ ones.\nIf $y_1 > n^{1/1012}$, since $y_{1012} \\le n^{1/1012}$, suppose $y_k > n^{1/1012} \\ge y_{k+1}$, then $1 \\le k \\le 1011$.\nFor $1 \\le i \\le k$, we have\n$$\n\\frac{y_i}{p_i} \\le y_{1012} \\le n^{1/1012}.\n$$\nLet $x_1, x_2, \\dots, x_{2023}$ be $\\frac{y_1}{p_1}, \\dots, \\frac{y_k}{p_k}, y_{k+1}, \\dots, y_{1012}, p_1, \\dots, p_k$ and $1011 - k$ ones.\nThe proof for that $\\lambda < \\frac{1}{1012}$ does not work is the same as Solution 1. We prove that $\\lambda = \\frac{1}{1012}$ satisfies the requirement. Express $n$ as $n = x_1x_2\\cdots x_{2023}$ such that\n\n$x_1 + x_2 + \\cdots + x_{2023}$ is minimized. (Since there are a finite number of ways to express this, such a representation must exist.) Next, we prove that every $x_i$ greater than $n^{1/1012}$ is a prime number.\nWe use proof by contradiction. Suppose $x_1 > n^{1/1012}$ and $x_1$ is not a prime number. Let $x_1 = ab$ where $a > 1$ and $b > 1$, and without loss of generality, assume $a \\ge b$. Since\n$$\nx_2 \\cdots x_{2023} = \\frac{n}{x_1} < \\frac{x_1^{1012}}{x_1} = x_1^{1011} = (\\sqrt{x_1})^{2022} \\le a^{2022},\n$$\nthere exists an $x_i < a$. Assume without loss of generality that $x_2 < a$.\nExpress $n$ as $n = a \\cdot b \\cdot x_2 \\cdot x_3 \\cdots x_{2023}$. Since\n$$\n(ab + x_2 + x_3 + \\cdots + x_{2023}) - (a + bx_2 + x_3 + \\cdots + x_{2023}) = (b-1)(a-x_2) > 0,\n$$\nthe expression $n = x_1x_2\\cdots x_{2023}$ is not the one that minimizes $x_1+x_2+\\cdots+x_{2023}$, which is a contradiction!\nTherefore, $\\lambda = \\frac{1}{1012}$ satisfies the problem's conditions.\nThe proof for that $\\lambda < \\frac{1}{1012}$ does not work is the same as Solution 1. We prove that $\\lambda = \\frac{1}{1012}$ satisfies the requirement. Let $n = p_1p_2\\cdots p_k$, where $p_1 \\ge p_2 \\ge \\cdots \\ge p_k$ are all the prime factors of $n$.\nIf $k \\le 2023$, then express $n$ as the product of $p_1, p_2, \\cdots, p_k$ and $2023-k$ ones.\nIf $k > 2023$, first express $n$ as $n = p_1p_2\\cdots p_k$. Then, perform the following operation: in each step, select the smallest two factors and combine them into their product, continue this process until only $2023$ factors remain. In each step, as the current number of factors is $\\ge 2024$, the product of the smallest two numbers will be $\\le n^{2/2024} = n^{1/1012}$. This indicates that every \"composite\" number formed after each operation is $\\le n^{1/1012}$.\nCombining this with the fact that the initial numbers are primes, it can be concluded that when the operation ends, the $2023$ factors are either prime or do not exceed $n^{1/1012}$. Hence, $\\lambda = \\frac{1}{1012}$ satisfies the conditions of the problem.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 28578,
    "subject": "Mathematics (Multi-modal)",
    "question": "Suppose $n \\ge 2$ is a positive integer, and $A_1A_2 \\cdots A_{2n}$ is a convex $2n$-gon inscribed in a circle. It is known that there exists a point $P$ inside this $2n$-gon such that\n$$\n\\angle PA_1A_2 = \\angle PA_2A_3 = \\cdots = \\angle PA_{2n-1}A_{2n} = \\angle PA_{2n}A_1.\n$$\nProve that the following equation holds\n$$\n\\prod_{i=1}^{n} |A_{2i-1}A_{2i}| = \\prod_{i=1}^{n} |A_{2i}A_{2i+1}|,\n$$\nwhere $A_{2n+1} = A_1$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "All subscripts are to be understood modulo $2n$. Denote $\\angle PA_1A_2 = \\angle PA_2A_3 = \\cdots = \\alpha$. For $1 \\le i \\le 2n$, extend $A_iP$ to intersect the circumference at point $B_i$. Note that\n$$\n\\angle A_iB_iB_{i-1} = \\angle A_iA_{i-1}P = \\angle A_{i+1}A_iB_i = \\alpha.\n$$\nHence, $A_iA_{i+1}B_{i-1}B_i$ is an isosceles trapezoid. (In fact, $B_1B_2 \\cdots B_{2n}$ is the polygon obtained by rotating $A_2A_3 \\cdots A_{2n}A_1$ in the clockwise direction by $2\\alpha$ on the circumference.) In particular, $A_iA_{i+1} = B_{i-1}B_i$. Also, by $\\triangle A_{i-1}PA_i \\sim \\triangle B_iPB_{i-1}$, we get\n$$\n(*) \\qquad \\frac{A_{i-1}P}{B_iP} = \\frac{A_iP}{B_{i-1}P} = \\frac{A_{i-1}A_i}{B_{i-1}B_i} = \\frac{A_{i-1}A_i}{A_iA_{i+1}}.\n$$\nTaking the product of the first and last equalities in (*) over all odd $i$, we get\n$$\n\\prod_{i=1}^{n} \\frac{A_{2i-2}A_{2i-1}}{A_{2i-1}A_{2i}} = \\prod_{i=1}^{n} \\frac{A_{2i-2}P}{B_{2i-1}P}.\n$$\n\nTaking the product of the second and last equalities in (*) over all even $i$, we get\n$$\n\\prod_{i=1}^{n} \\frac{A_{2i-1}A_{2i}}{A_{2i}A_{2i+1}} = \\prod_{i=1}^{n} \\frac{A_{2i}P}{B_{2i-1}P}.\n$$\nNoting that the right-hand sides of the two equations above are equal, we have\n$$\n\\prod_{i=1}^{n} \\frac{A_{2i-2}A_{2i-1}}{A_{2i-1}A_{2i}} = \\prod_{i=1}^{n} \\frac{A_{2i-1}A_{2i}}{A_{2i}A_{2i+1}} \\Rightarrow \\prod_{i=1}^{n} (A_{2i}A_{2i+1})^2 = \\prod_{i=1}^{n} (A_{2i-1}A_{2i})^2.\n$$\nThis completes the proof of the problem. $\\Box$\nDraw line $A_1A_4$ intersecting line $PA_3$ at point $H$. Given that points $A_1, A_2, A_3, A_4$ are concyclic, we have\n$$\n\\begin{aligned}\n\\angle A_2A_1A_4 &= 180^\\circ - \\angle A_2A_3A_4 = 180^\\circ - \\angle A_2A_3P - \\angle PA_3A_4 \\\\\n&= 180^\\circ - \\angle A_2A_3P - \\angle A_3A_2P = \\angle A_2PA_3.\n\\end{aligned}\n$$\nTherefore, points $A_1, A_2, H, P$ are concyclic. In particular, $\\angle A_2HA_3 = \\angle A_2A_1P = \\angle HA_3A_4$, we have $A_2H \\parallel A_3A_4$. From this and the fact that $A_1, A_2, H, P$ are concyclic, we get\n$$\n\\angle A_3A_4H = \\angle A_2HA_1 = \\angle A_1PA_2.\n$$\nTogether with $\\angle A_2A_1P = \\angle A_4A_3H$, we find that $\\triangle A_1A_2P \\sim \\triangle A_3HA_4$. From this we obtain\n$$\n\\frac{A_1A_2}{A_1P} = \\frac{A_3H}{A_3A_4} \\Rightarrow A_1A_2 \\cdot A_3A_4 = A_1P \\cdot A_3H.\n$$\nMoreover, from $\\angle A_3HA_2 = \\angle A_2A_1P = \\angle A_3A_2P$, we have $\\triangle A_3HA_2 \\sim \\triangle A_3A_2P$. We get\n$$\n\\frac{A_2A_3}{A_3H} = \\frac{PA_3}{A_2A_3} \\Rightarrow (A_2A_3)^2 = PA_3 \\cdot A_3H.\n$$\n\nDividing the first equation by the second one yields\n$$\n\\frac{A_1 A_2 \\cdot A_3 A_4}{(A_2 A_3)^2} = \\frac{P A_1}{P A_3}.\n$$\nBy doing the same procedure with adding 2 to each index and multiplying the resulting equations, we obtain\n$$\n\\prod_{i=1}^{n} \\frac{A_{2i-1}A_{2i} \\cdot A_{2i+1}A_{2i+2}}{(A_{2i}A_{2i+1})^2} = \\prod_{i=1}^{n} \\frac{P A_{2i-1}}{P A_{2i+1}} = 1.\n$$\nHere all indices are understood modulo $2n$. Rearranging the above equation gives\n$$\n\\prod_{i=1}^{n} |A_{2i-1}A_{2i}|^2 = \\prod_{i=1}^{n} |A_{2i}A_{2i+1}|^2.\n$$\nThis completes the proof of the problem. $\\Box$\nLet the circle in the problem be the unit circle in the complex plane, and let $A_1, \\dots, A_{2n}$ correspond to complex numbers $z_1, \\dots, z_{2n}$. Without loss of generality (we can always rotate if necessary), let $P$ correspond to a non-negative real number $a < 1$, and let $\\angle PA_1A_2 = \\pi - \\alpha \\in (0, \\pi)$.\nFrom the given conditions, we know that $\\frac{z_2 - z_1}{(z_1 - a)e^{i\\alpha}} \\in \\mathbb{R}$, thus\n$$\n\\frac{z_2 - z_1}{(z_1 - a)e^{i\\alpha}} = \\frac{\\overline{z_2} - \\overline{z_1}}{(\\overline{z_1} - a)e^{-i\\alpha}},\n$$\nConsidering that $z_1, z_2$ are unit complex numbers, and $z_1 \\neq z_2$, we can solve to get $z_2 = \\frac{e^{2i\\alpha}(z_1 - a)}{az_1 - 1}$, and similarly $z_{j+1} = \\frac{e^{2i\\alpha}(z_j - a)}{az_j - 1}$, ($j = 1, 2, \\dots, 2n$, indices are understood modulo $2n$). If $a = 0$, then $z_{j+1} = -e^{2i\\alpha}z_j$, and in this case $A_1A_2\\cdots A_{2n}$ forms a regular $2n$-gon, and the conclusion is trivially true. Below we assume $0 < a < 1$.\nNow consider the Möbius transformation $f(z) = \\frac{e^{2i\\alpha}(z-a)}{az-1}$, which has two fixed points (roots of the quadratic equation $ax^2 - (1+e^{2i\\alpha})x + ae^{2i\\alpha} = 0$), denoted by $x_1$ and $x_2$. It can be easily verified that $x_1 \\neq x_2$.\nLet $\\frac{z_j - x_1}{z_j - x_2} = w_j$, from $f(z_j) = z_{j+1}$, we have that the ratios of consecutive terms among $w_1, w_2, \\dots, w_{2n}, w_1$ are constants. Thus,\n$$\nz_j = \\frac{x_1 - x_2 w_j}{1 - w_j} = x_2 \\cdot \\frac{w_j - \\frac{x_1}{x_2}}{w_j - 1},\n$$\nhence\n$$\n\\begin{aligned}\nz_{j+1} - z_j &= \\frac{x_2}{(w_j - 1)(w_{j+1} - 1)} \\left( \\left( w_{j+1} - \\frac{x_1}{x_2} \\right) (w_j - 1) - \\left( w_j - \\frac{x_1}{x_2} \\right) (w_{j+1} - 1) \\right) \\\\\n&= \\frac{(x_1 - x_2)(w_{j+1} - w_j)}{(w_j - 1)(w_{j+1} - 1)}.\n\\end{aligned}\n$$\nTherefore,\n$$\n\\prod_{j=1}^{n} \\frac{z_{2j+1} - z_{2j}}{z_{2j} - z_{2j-1}} = \\prod_{j=1}^{n} \\frac{w_{2j+1} - w_{2j}}{w_{2j} - w_{2j-1}} = \\prod_{j=1}^{n} \\frac{w_{2j}}{w_{2j-1}}\n$$\n\nis a complex number of modulus 1 (in fact it can be shown that its value is $-1$), hence\nthe proposition holds. $\\square$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 28579,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $a$, $b$, and $d$ be integers satisfying $d \\ge 0$, $|a| \\ge 2$, and $b \\ge (|a| + 1)^{d+1}$. Suppose $f(x)$ is a real-coefficient polynomial of degree $d$, and for each positive integer $n$, let $r_n$ denote the remainder of $\\lfloor f(n)a^n \\rfloor$ modulo $b$.\n\n*Proof:* If the sequence $r_n$ is ultimately periodic, then $f(x)$ is a rational-coefficient polynomial.\n\n*Note 1:* For every real number $x$, $\\lfloor x \\rfloor$ denotes the largest integer that is not greater than $x$.\n\n*Note 2:* A sequence $a_n$ is called *ultimately periodic* if there exist positive integers $n_0$ and $T$ such that for every integer $n \\ge n_0$, we have $a_{n+T} = a_n$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "**Proof. Lemma** Let integers $a$ and real number $z$ satisfy $|a| \\ge 2$. For any non-negative integer $n$, if\n$$\n||a^n z|| < \\frac{1}{|a| + 1},\n$$\nthen $z$ is an integer. Here, $||x||$ is defined as the distance between the real number $x$ and the nearest integer, that is, $||x|| = \\min(\\{x\\}, 1 - \\{x\\})$.\n\n*Proof of the lemma:* Without loss of generality, assume $a \\ge 2$ (otherwise, replace $a$ with $-a$). Suppose the conclusion is false. Without loss of generality, assume $0 < z \\le \\frac{1}{2}$ (otherwise, replace $z$ with $m \\mp z$, where $m$ is some suitable integer).\nSince $z \\in (0, \\frac{1}{a^{-1}(a+1)})$, we can assume that\n$$\n\\frac{1}{a^k(a+1)} \\le z < \\frac{1}{a^{k-1}(a+1)} \\quad (k \\in \\mathbb{Z}_{\\ge 0}),\n$$\nwhich implies $\\frac{1}{a+1} \\le a^k z < 1 - \\frac{1}{a+1}$. Consequently, $||a^k z|| \\ge \\frac{1}{a+1}$, leading to a contradiction. Thus, the lemma is proven.\n\nBack to the original question. Assume, for the sake of contradiction, that $f(x) = a_d x^d + a_{d-1} x^{d-1} + \\cdots + a_1 x + a_0$ is not a polynomial with rational coefficients. Let $t$ be the largest index such that $a_t \\notin \\mathbb{Q}$. Define\n$$\n\\left\\{ \\frac{f(n)a^n}{b} \\right\\} = \\frac{r_n + \\varepsilon_n}{b}, \\quad \\varepsilon_n \\in [0, 1).\n$$\nLet $g(x) = \\Delta^t f(x) = \\sum_{i=0}^{t} (-1)^i \\binom{t}{i} f(x + t - i)$, where $\\Delta$ denotes the finite difference operator. It is easy to see that except for the constant term, all coefficients of $g(x)$ are rational numbers. Let $g(x) = c_{d-t}x^{d-t} + \\cdots + c_1x + c_0$, and let $K$ be the least common multiple of the denominators of $c_1, c_2, \\cdots, c_{d-t}$. Now,\n$$\n\\left\\{ \\frac{g(n)a^{n+t}}{b} \\right\\} = \\left\\{ \\sum_{i=0}^{t} \\frac{(-1)^i \\binom{t}{i} f(n+t-i) a^{n+t-i} \\cdot a^i}{b} \\right\\} \\\\\n= \\left\\{ \\sum_{i=0}^{t} \\frac{(-1)^i \\binom{t}{i} a^i (r_{n+t-i} + \\varepsilon_{n+t-i})}{b} \\right\\} \\\\\n= \\left\\{ \\sum_{i=0}^{t} r_{n+t-i} \\frac{(-1)^i \\binom{t}{i} a^i}{b} + \\sum_{i=0}^{t} \\varepsilon_{n+t-i} \\frac{(-1)^i \\binom{t}{i}}{b} \\right\\}.\n$$\n\nLet $\\delta_n = \\sum_{i=0}^{t} \\varepsilon_{n+t-i} \\cdot \\frac{(-1)^i \\binom{t}{i} a^i}{b}$. Note that\n$$\n\\sum_{i=0}^{t} \\left| \\frac{(-1)^i \\binom{t}{i} a^i}{b} \\right| = \\sum_{i=0}^{t} \\frac{\\binom{t}{i} |a|^i}{b} = \\frac{(|a|+1)^t}{b} \\le \\frac{(|a|+1)^d}{b} \\le \\frac{1}{|a|+1},\n$$\nwhich implies that for any positive integers $n$ and $n'$, \n$$\n\\left| \\delta_n - \\delta_{n'} \\right| \\le \\sum_{i=0}^{t} \\left| \\varepsilon_{n+t-i} - \\varepsilon_{n'+t-i} \\right| \\cdot \\frac{\\binom{t}{i} |a|^i}{b} < \\sum_{i=0}^{t} \\frac{\\binom{t}{i} |a|^i}{b} \\le \\frac{1}{|a|+1}.\n$$\nWe choose positive integers $M$ and $n$ such that $M$ is a multiple of the smallest positive period of $\\{r_n\\}$ and $bK|M$, and when $n \\ge n_0$, we have $bK|a^{n+M+t} - a^{n+t}$. Then, for $n \\ge n_0$,\n$$\n\\left\\| \\frac{g(n+M)a^{n+M+t}}{b} - \\frac{g(n)a^{n+t}}{b} \\right\\| = \\left\\| \\delta_{n+M} - \\delta_n \\right\\| < \\frac{1}{|a|+1},\n$$\nwhere the first equality holds because $r_{n+M+t-i} = r_{n+t-i}$ for $i = 0, 1, \\dots, t$. On the other hand,\n$$\n\\begin{aligned}\n& \\left| \\frac{g(n+M)a^{n+M+t}}{b} - \\frac{g(n)a^{n+t}}{b} \\right| \\\\\n&= \\left| \\frac{c_0}{b} (a^{n+M+t} - a^{n+t}) + \\sum_{i=1}^{d-t} \\frac{c_i}{b} ((n+M)^i a^{n+M+t} - n^i a^{n+t}) \\right| \\\\\n&= \\left| \\frac{c_0(a^M - 1)a^{n_0+t}}{b} \\cdot a^{n-n_0} \\right|.\n\\end{aligned}\n$$\nAccording to the lemma, we have $\\frac{c_0(a^M - 1)a^{n_0+t}}{b} \\in \\mathbb{Z}$, but this contradicts the irrationality of $c_0$! Hence, the proof is complete. $\\square$",
    "topic": "Number Theory",
    "subtopic": "Modular Arithmetic"
  },
  {
    "id": 28580,
    "subject": "Mathematics (Multi-modal)",
    "question": "Determine the largest real number $C$ such that\n$$\n\\sum_{i=1}^{n} \\sum_{j=1}^{n} (n - |i - j|) x_i x_j \\ge C \\sum_{i=1}^{n} x_i^2\n$$\nholds for every positive integer $n$ and any real numbers $x_1, x_2, \\dots, x_n$.",
    "options": [],
    "answer": "1/2",
    "solution": "The needed constant $C = \\frac{1}{2}$. First show that $C = \\frac{1}{2}$ holds. Note that\n$$\n\\text{L.H.S. of (*)} = x_1^2 + (x_1 + x_2)^2 + \\dots + (x_1 + x_2 + \\dots + x_n)^2 \\\\\n\\qquad + (x_2 + \\dots + x_n)^2 + \\dots + (x_{n-1} + x_n)^2 + x_n^2. \\quad (1)\n$$\nUsing the inequality $a^2 + (a+b)^2 = a^2 + (-a-b)^2 \\geq \\frac{1}{2}b^2$, we get\n$$\n\\begin{aligned}\n\\frac{1}{2}(x_1^2 + (x_1 + x_2)^2) &\\geq \\frac{1}{4}x_2^2, \\\\\n\\frac{1}{2}((x_1 + x_2)^2 + (x_1 + x_2 + x_3)^2) &\\geq \\frac{1}{4}x_3^2, \\\\\n\\dots &\\geq \\dots \\\\\n\\frac{1}{2}((x_1 + \\dots + x_{n-1})^2 + (x_1 + \\dots + x_n)^2) &\\geq \\frac{1}{4}x_n^2, \\\\\n\\frac{1}{2}((x_1 + \\dots + x_{n-1})^2 + (x_2 + \\dots + x_n)^2) &\\geq \\frac{1}{4}x_1^2, \\\\\n\\dots &\\geq \\dots \\\\\n\\frac{1}{2}((x_{n-1} + x_n)^2 + x_n^2) &\\geq \\frac{1}{4}x_{n-1}^2.\n\\end{aligned}\n$$\nSumming the above and use (1) to get\n$$\n\\text{L.H.S. of (*)} \\geq \\frac{3}{4}x_1^2 + \\frac{1}{2}(x_2^2 + \\dots + x_{n-1}^2) + \\frac{3}{4}x_n^2 \\geq \\frac{1}{2}(x_1^2 + \\dots + x_n^2).\n$$\ni.e. the inequality (*) holds for $C = \\frac{1}{2}$.\n\nNext, we prove that, if the inequality (*) holds for all $n$, then $C \\leq \\frac{1}{2}$. In (1), take $x_1 = 1, x_2 = -2, x_3 = 2, \\dots, x_{n-1} = (-1)^{n-2}2, x_n = (-1)^{n-1}$, then every square in (*) is equal to 1, i.e. the R.H.S. of (*) is equal to $2n-2$. Yet\n$$\n\\sum_{i=1}^{n} x_i^2 = 4(n-2) + 2 = 4n - 6.\n$$\nSo,\n$$\nC \\leq \\frac{2n-2}{4n-6} = \\frac{1}{2} + \\frac{1}{2n-3}.\n$$\nLetting $n \\to \\infty$ gives $C \\leq \\frac{1}{2}$.\n\nTo sum up, the maximal $C = \\frac{1}{2}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 28581,
    "subject": "Mathematics (Multi-modal)",
    "question": "A party is attended by $n$ people. Assume that there are at most $n$ pairs of friendships among them, and any two people shake hands at the party if and only if they have a common friend at the party.\nSuppose $m$ is a positive integer satisfying $m \\ge 3$ and $n \\le m^3$. Prove that there exists a person $A$ such that the number of people $A$ has shaken hands with at this party does not exceed $(m-1)$ times the number of $A$'s friends.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "*Proof.* We represent individuals as vertices, friendships as edges, and use a graph to depict the problem.\nThe given graph contains a connected subgraph $G = (V, E)$ satisfying $|E| \\le |V| \\le n$. We denote the degree of vertex $v$ as $d(v)$ and the number of individuals with whom $v$ has shaken hands as $\\beta(v)$.\nAssume the proposition is false, implying that for any vertex $v$, we have\n$$\n(*) \\qquad \\beta(v) \\ge (m-1) \\cdot d(v) + 1.\n$$\n**Observation:** For every vertex $v_0$ with degree 1 (connected to $v_1$), (*) implies $\\beta(v_0) \\ge (m-1) \\cdot d(v_0) + 1 = m$. Thus, we have $d(v_1) \\ge m + 1$.\nConsider the graph $G' = (V', E')$ obtained by removing all vertices of degree 1 in $G$ along with their incident edges. Note that the number of removed vertices is equal to the number of removed edges. Therefore, $G'$ is necessarily connected and satisfies $|E'| \\le |V'|$. Hence, $G'$ is either a tree or has a unique cycle. We will consider three cases.\n\n(0) $G'$ has only one vertex $v_1$. From the previous *observation*, we know that $d(v_1) \\ge m+1$. This implies that $G$ has only $m+2$ vertices, and all edges are the ones connecting $v_1$ to all other vertices. However, in this case, $\\beta(v_1) = 0 \\le (m-1)d(v_1)$, which contradicts our assumption.\n\n(1) $G'$ has a vertex $v_1$ with degree 1. Then, $v_1$ must be connected to a vertex $v_0$ with degree 1 in $G$. From the previous *observation*, we know that $d(v_1) \\ge m+1$, which implies that $v_1$ is connected to $d(v_1) - 1 \\ge m$ vertices of degree 1 in $G$.\nLet $v_1$ be connected to $v_2$ in $G'$. In this case, the inequality $\\beta(v_1) \\ge (m-1)d(v_1) + 1$ becomes\n$$\nd(v_2) - 1 \\ge (m-1)d(v_1) + 1 \\ge m^2,\n$$\nwhich implies $d(v_2) \\ge m^2 + 1$.\nConsidering the inequality (*) for $v_2$, we have\n$$\n\\beta(v_2) \\ge (m-1)d(v_2) + 1 \\ge (m-1)(m^2 + 1) + 1.\n$$\nNote that the friends of $v_2$ and the people $v_2$ has shaken hands with are pairwise distinct, except when $v_2$ lies on a cycle of length 3, in which case $v_2$ shakes hands with both of its friends. From this, we conclude that $G$ has at least\n$$\n1 + d(v_2) + \\beta(v_2) - 2 \\ge 1 + (m^2 + 1) + (m-1)(m^2 + 1) + 1 - 2 = m^3 + m > m^3\n$$\nvertices, which contradicts $n \\le m^3$.\n\n(2) All vertices in $G'$ have degree at least 2. In this case, $G'$ forms a cycle $v_1 \\sim v_2 \\sim \\dots \\sim v_t \\sim v_1$, where indices are taken modulo $t$.\nFrom the previous *observation*, each vertex $v_i$ is connected to exactly $d(v_i) - 2$ (in $G$) vertices of degree 1. Thus, for $v_i$, (*) implies:\n$$\n(d(v_{i-1}) - 2) + (d(v_{i+1}) - 2) + 2 \\ge \\beta(v_i) \\ge (m-1) \\cdot d(v_i) + 1 \\ge 2d(v_i) + 1.\n$$\nSumming up these inequalities for all $i$, we arrive at a contradiction.\n\nTherefore, we conclude that there exists a vertex $v$ such that $\\beta(v) \\le (m-1) \\cdot d(v)$. Hence, the proposition holds. $\\square$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 28582,
    "subject": "Mathematics (Multi-modal)",
    "question": "In acute triangle $ABC$ which is not isosceles, $AP$, $BQ$, $CR$ are three altitudes, $H$ is the orthocenter. The parallel line to $BC$ passing through $A$ intersects line $RQ$ at point $D$. Let $A_1$ be the midpoint of $BC$, and let $K$ be the intersection of $RQ$ and $AA_1$. The line passing through the midpoint of $AH$ and point $K$ intersects line $DA_1$ at point $A_2$. Similarly define points $B_2$ and $C_2$.\n\nSuppose that the circumcircle of non-degenerate triangle $A_2B_2C_2$ is $\\omega$. Prove: There exist three circles $\\odot A'$, $\\odot B'$ and $\\odot C'$ inside $\\omega$ that are tangent to $\\omega$ and satisfy the following conditions:\n(1) $\\odot A'$ is tangent to sides $AB$, $AC$, $\\odot B'$ is tangent to sides $BA$, $BC$, and $\\odot C'$ is tangent to sides $CA$, $CB$;\n(2) The centers of the three circles, $A'$, $B'$, $C'$, are distinct and collinear.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "*Proof.* Let $A^*$ be the midpoint of segment $AH$. In the given diagram, we have $\\angle A_1RH = \\angle A_1CR = \\angle RAH$. Thus, $A_1R$ is a tangent to the circle $\\Gamma$ with diameter $AH$, and similarly, $A_1Q$ is also a tangent to $\\Gamma$ (*R* and *Q* lie on the circle $\\Gamma$). Hence, we have\n\na.\n$$A_1A^* \\perp RQ.$$ \n\nConsidering the polar with respect to $\\Gamma$, we know that point $D$ lies on the polar line of $A_1$ with respect to $\\Gamma$. Therefore, $A_1$ lies on the polar line of $D$ with respect to $\\Gamma$. However, $DA$ is a tangent to $\\Gamma$, so $AA_1$ is the polar line of $D$. This implies\n\nb.\n$$DA^* \\perp AA_1.$$ \n\nCombining equations (a) and (b), we conclude that $K$ is the orthocenter of triangle $\\triangle A^*A_1D$. In particular, $A^*K \\perp A_1D$, which implies that $\\angle A^*A_2A_1 = \\frac{\\pi}{2}$. Therefore, point $A_2$ lies on the nine-point circle of triangle $\\triangle ABC$. Consequently, $\\omega$ is the nine-point circle of triangle $\\triangle ABC$.\n\nIn the given diagram, let $N$ be the center of $\\omega$ and $I$ be the incenter of $\\triangle ABC$. Let $BC = a$, $CA = b$, $AB = c$, and $s = \\frac{a+b+c}{2}$. Consider the inversion $f$ centered at $A$ that preserves the nine-point circle. Let $\\odot A'$ be the image of the incircle under $f$. It is clear that $\\odot A'$ is tangent to the sides $AB$ and $AC$. Similarly, we can define $\\odot B'$ and $\\odot C'$.\n\nLet $M$ be the foot of the perpendicular from $I$ to side $AB$. We will show that $\\odot A'$ and $\\omega$ do not coincide; otherwise, we would have $AR \\cdot AC_1 = AM^2$, which implies $b \\cdot \\cos A \\cdot \\frac{c}{2} = (\\frac{b+c-a}{2})^2$. This leads to either $a = b$ or $a = c$, contradicting the non-isosceles nature of $\\triangle ABC$.\n\nNext, we will verify the collinearity of $A'$, $B'$, and $C'$. Let\n$$\n\\overrightarrow{IA'} = p \\cdot \\overrightarrow{IA}, \\quad \\overrightarrow{IB'} = q \\cdot \\overrightarrow{IB}, \\quad \\overrightarrow{IC'} = r \\cdot \\overrightarrow{IC}.\n$$\nThen we have:\n$$\n\\frac{a}{p} \\cdot \\overrightarrow{IA'} + \\frac{b}{q} \\cdot \\overrightarrow{IB'} + \\frac{c}{r} \\cdot \\overrightarrow{IC'} = a \\cdot \\overrightarrow{IA} + b \\cdot \\overrightarrow{IB} + c \\cdot \\overrightarrow{IC} = \\overrightarrow{0}.\n$$\nTo prove that $A'$, $B'$, and $C'$ are collinear, it suffices to show that:\n$$\n\\frac{a}{p} + \\frac{b}{q} + \\frac{c}{r} = 0.\n$$\nLet $L$ be the foot of the perpendicular from $A'$ to side $AB$. Then we have $\\frac{AA'}{AI} = \\frac{AL}{AM}$, $AM = s-a$. According to the properties of inversion $f$, we know that $f(M) = L$. Hence,\n$$\nAM \\cdot AL = AR \\cdot AC_1 = b \\cos A \\cdot \\frac{c}{2} = \\frac{b^2 + c^2 - a^2}{4}.\n$$\n$$\n\\frac{AL}{AM} = \\frac{AM \\cdot AL}{AM^2} = \\frac{b^2 + c^2 - a^2}{4(s-a)^2}.\n$$\nTherefore, we have:\n$$\np = \\frac{IA'}{IA} = \\frac{2(a-b)(a-c)}{(b+c-a)^2}\n$$\nSimilarly, we have:\n$$\nq = \\frac{2(b-c)(b-a)}{(c+a-b)^2}, \\quad r = \\frac{2(c-a)(c-b)}{(a+b-c)^2}.\n$$\nPlugging into the computations gives $\\frac{a}{p} + \\frac{b}{q} + \\frac{c}{r} = 0$ (Here, upon substitution, we obtain a cyclic summation of $a(b-c)(b+c-a)^2$. Let us consider this as a function of $a$, denoted by $g(a)$. It is observed that $g(a)$ is actually a quadratic function, and we have $g(b) = g(c) = g(b+c) = 0$. Hence, we conclude that $g \\equiv 0$. $\\Box$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 28583,
    "subject": "Mathematics (Multi-modal)",
    "question": "As shown below, a square grid of side length $8$ is constructed by $144$ sticks of length $1$. Find the least number of sticks to be removed so that the resulting figure contains no rectangle.\n\n![](attached_image_1.png)",
    "options": [],
    "answer": "43",
    "solution": "The answer is $43$.\n\nFirst, we prove that at least $43$ sticks must be removed. Suppose the figure does not contain any rectangles, then each bounded connected area must consist of at least three unit squares, i.e., the area is at least $3$. Therefore, there can be at most $\\lfloor \\frac{64}{3} \\rfloor = 21$ bounded connected areas. Removing one stick can at most reduce the number of bounded connected areas by $1$ (either by merging two bounded areas into one, or by merging a bounded area with an unbounded area). Initially, there are $64$ bounded connected areas, so at least $64 - 21 = 43$ sticks must be removed.\n\nThe figure below shows an example of removing $43$ sticks, where each bounded connected area has an area of $3$, and the figure does not contain any rectangles.\n\n![](attached_image_2.png)",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 28584,
    "subject": "Mathematics (Multi-modal)",
    "question": "(1) Given coprime positive integers $a, b$. Prove: There exist real numbers $\\lambda, \\beta$ such that for any positive integer $m$, it holds that\n$$\n\\lambda m - \\beta \\le \\sum_{k=1}^{m-1} \\left\\{ \\frac{ak}{m} \\right\\} \\cdot \\left\\{ \\frac{bk}{m} \\right\\} \\le \\lambda m + \\beta.\n$$\n\n(2) Prove: there exists a positive integer $N$, such that for any prime $p > N$, the following proposition holds:\nIf the positive integers $a, b, c$ satisfy that $(a+b)(a+c)(b+c)$ is not divisible by $p$, then there exist at least $\\lfloor \\frac{p}{12} \\rfloor$ elements $k$ in the set $\\{1, 2, \\dots, p-1\\}$ such that\n$$\n\\left\\{ \\frac{ak}{p} \\right\\} + \\left\\{ \\frac{bk}{p} \\right\\} + \\left\\{ \\frac{ck}{p} \\right\\} \\le 1.\n$$\nHere, $\\lfloor x \\rfloor$ represents the greatest integer not exceeding $x$, and $\\{x\\} = x - \\lfloor x \\rfloor$ represents the fractional part of $x$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "(1) When $m$ is large, we expect that the summation in (i) is very close to the following integral:\n$$\n\\int_{0}^{m} \\left\\{ \\frac{ax}{m} \\right\\} \\left\\{ \\frac{bx}{m} \\right\\} dx = m \\cdot \\int_{0}^{1} \\{ax\\}\\{bx\\}dx.\n$$\nHence, we will prove the inequality (i) for\n$$\n\\lambda = \\int_{0}^{1} \\{ax\\}\\{bx\\}dx, \\quad \\beta = 2(a+b).\n$$\nTo do this, we divide the summation and integral into several segments, and it suffices to prove that\n$$\n\\begin{aligned}\n& \\left| \\sum_{k=1}^{m-1} \\left\\{ \\frac{ak}{m} \\right\\} \\left\\{ \\frac{bk}{m} \\right\\} - \\int_{0}^{m} \\left\\{ \\frac{ax}{m} \\right\\} \\left\\{ \\frac{bx}{m} \\right\\} dx \\right| \\\\\n& = \\left| \\sum_{k=0}^{m-1} \\frac{1}{2} \\left( \\left\\{ \\frac{ak}{m} \\right\\} \\left\\{ \\frac{bk}{m} \\right\\} + \\left\\{ \\frac{a(k+1)}{m} \\right\\} \\left\\{ \\frac{b(k+1)}{m} \\right\\} \\right) - \\int_{0}^{m} \\left\\{ \\frac{ax}{m} \\right\\} \\left\\{ \\frac{bx}{m} \\right\\} dx \\right| \\\\\n& (ii) \\quad \\le \\sum_{k=0}^{m-1} \\left| \\frac{1}{2} \\left( \\left\\{ \\frac{ak}{m} \\right\\} \\left\\{ \\frac{bk}{m} \\right\\} + \\left\\{ \\frac{a(k+1)}{m} \\right\\} \\left\\{ \\frac{b(k+1)}{m} \\right\\} \\right) - \\int_{k}^{k+1} \\left\\{ \\frac{ax}{m} \\right\\} \\left\\{ \\frac{bx}{m} \\right\\} dx \\right|\n\\end{aligned}\n$$\nis less than or equal to $2(a+b)$.\nIf $m \\le a+b$, each term in (ii) is not greater than 1, so the total sum is not greater than $2a+2b$. We now assume $m > a+b$.\nFor each $k$, we divide the discussion into two cases:\n(a) If there exists a $k$ such that $\\{\\frac{ak}{m}\\}$ or $\\{\\frac{bk}{m}\\}$ is discontinuous on $(k, k+1]$. There are at most $a+b-1$ such $k$. Moreover, since the product $\\{\\frac{ak}{m}\\}\\{\\frac{bk}{m}\\}$ always takes values in $[0, 1]$, the contribution of each corresponding term in (**) is at most $a+b-1$.\n(b) If there exists a $k$ such that $\\{\\frac{ak}{m}\\}$ or $\\{\\frac{bk}{m}\\}$ is continuous on $(k, k+1]$. Let $x = k+\\delta$ ($\\delta \\in [0, 1]$), then $\\{\\frac{ak}{m}\\} = \\{\\frac{bk}{m}\\} + \\frac{a\\delta}{m}$ or $\\{\\frac{bk}{m}\\} = \\{\\frac{ak}{m}\\} + \\frac{b\\delta}{m}$. Let $\\alpha = \\{\\frac{ak}{m}\\}$ and $\\beta = \\{\\frac{bk}{m}\\}$.\n\nThe corresponding term in (ii) is then\n$$\n\\begin{aligned}\n& \\frac{1}{2} \\left( \\alpha\\beta + (\\alpha + \\frac{a}{m})(\\beta + \\frac{b}{m}) \\right) - \\int_{0}^{1} (\\alpha + \\frac{a\\delta}{m})(\\beta + \\frac{b\\delta}{m}) d\\delta \\\\\n&= (\\alpha\\beta + \\frac{a\\beta}{2m} + \\frac{b\\alpha}{2m} + \\frac{ab}{2m^2}) - (\\alpha\\beta + \\frac{a\\beta}{2m} + \\frac{b\\alpha}{2m} + \\frac{ab}{3m^2}) = \\frac{ab}{6m^2}.\n\\end{aligned}\n$$\nCombining the discussions in (a) and (b), we know that the summation in (ii) (when $m > a + b$) is less than or equal to\n$$\na + b - 1 + m \\cdot \\frac{ab}{6m^2} < 2(a + b) - 1.\n$$\nThis completes the proof of (1).\n\n(2) **Step 1:** A refined estimation of (1).\nWe first calculate the value of $\\lambda = \\int_{0}^{1} \\{ax\\}\\{bx\\}dx$. Intuitively, this corresponds to integrating along the diagonal of the $a \\times b$ grid, as shown in the left figure ($a = 4, b = 3$). Since the function inside the integral actually only depends on the small square where the diagonal lies, we can translate the diagonal to a small square, as shown in the middle figure (here, the small square is appropriately enlarged). However, this small square can also be divided into $ab$ rectangular blocks such that the path of integration exactly follows the diagonal, as shown in the right figure. This essentially corresponds to partitioning the integral $\\int_{0}^{1}$ into integrals from $\\frac{i}{ab}$ to $\\frac{i+1}{ab}$ ($0 \\le i \\le ab-1$). Note that inside each small square, $\\{ax\\}$ and $\\{bx\\}$ are both continuous.\n\n![](attached_image_1.png)\n![](attached_image_2.png)\n![](attached_image_3.png)\n\nWe calculate specifically:\n$$\n\\begin{aligned}\n\\lambda &= \\int_{0}^{1} \\{ax\\}\\{bx\\}dx = \\frac{1}{ab} \\sum_{i=0}^{ab-1} \\int_{i}^{i+1} \\left\\{\\frac{x}{b}\\right\\} \\left\\{\\frac{x}{a}\\right\\} dx \\\\\n(iii) \\quad &= \\frac{1}{ab} \\sum_{i=0}^{ab-1} \\left\\{\\frac{i+\\frac{1}{2}}{b}\\right\\} \\left\\{\\frac{i+\\frac{1}{2}}{a}\\right\\} + \\frac{1}{ab} \\sum_{i=0}^{ab-1} \\int_{i}^{i+1} \\left( \\left\\{\\frac{x}{b}\\right\\} \\left\\{\\frac{x}{a}\\right\\} - \\left\\{\\frac{i+\\frac{1}{2}}{b}\\right\\} \\left\\{\\frac{i+\\frac{1}{2}}{a}\\right\\} \\right) dx\n\\end{aligned}\n$$\nwhere each term on the right side of (iii) equals\n$$\n\\begin{align*}\n& \\int_{-\\frac{1}{2}}^{\\frac{1}{2}} \\left( \\left\\{ \\frac{i + \\frac{1}{2} + x}{b} \\right\\} \\left\\{ \\frac{i + \\frac{1}{2} + x}{a} \\right\\} - \\left\\{ \\frac{i + \\frac{1}{2}}{b} \\right\\} \\left\\{ \\frac{i + \\frac{1}{2}}{a} \\right\\} \\right) dx \\\\\n&= \\int_{-\\frac{1}{2}}^{\\frac{1}{2}} \\left( \\left\\{ \\frac{i + \\frac{1}{2}}{b} \\right\\} \\frac{x}{a} + \\left\\{ \\frac{i + \\frac{1}{2}}{a} \\right\\} \\frac{x}{b} + \\frac{x^2}{ab} \\right) dx \\\\\n&= 0 + 0 + \\frac{1}{3ab} \\left( \\left( \\frac{1}{2} \\right)^3 - \\left( -\\frac{1}{2} \\right)^3 \\right) = \\frac{1}{12ab}.\n\\end{align*}\n$$\nRegarding the preceding summation in (iii), note that when $i$ takes values from $\\{0, 1, \\dots, ab-1\\}$, the pairs of numbers $\\left(\\left\\{\\frac{i}{b}\\right\\}, \\left\\{\\frac{i}{a}\\right\\}\\right)$ exactly cover all the points in the set\n$$\n\\left\\{0, \\frac{1}{a}, \\dots, \\frac{a-1}{a}\\right\\} \\times \\left\\{0, \\frac{1}{b}, \\dots, \\frac{b-1}{b}\\right\\}.\n$$\nFrom this, we know that\n$$\n\\sum_{i=0}^{ab-1} \\left\\{ \\frac{i + \\frac{1}{2}}{b} \\right\\} \\left\\{ \\frac{i + \\frac{1}{2}}{a} \\right\\} = \\sum_{j=0}^{a-1} \\sum_{k=0}^{b-1} \\left\\{ \\frac{k + \\frac{1}{2}}{b} \\right\\} \\left\\{ \\frac{j + \\frac{1}{2}}{a} \\right\\} = \\frac{a}{2} \\cdot \\frac{b}{2} = \\frac{ab}{4}.\n$$\nCombining the above results, we obtain\n$$\n\\lambda = \\frac{1}{ab} \\cdot \\frac{ab}{4} + \\frac{1}{ab} \\cdot ab \\cdot \\frac{1}{12ab} = \\frac{1}{4} + \\frac{1}{12ab}.\n$$\nThat is,\n$$\n\\left| \\sum_{k=1}^{m-1} \\left\\{ \\frac{ak}{m} \\right\\} \\left\\{ \\frac{bk}{m} \\right\\} - \\left( \\frac{m}{4} + \\frac{m}{12ab} \\right) \\right| \\le 2|a| + 2|b| - 1.\n$$\nNote that when $m$ is coprime to both $a$ and $b$, since $\\{\\frac{ak}{m}\\} + \\{\\frac{-ak}{m}\\} = 1$, the sum corresponding to $(a, b)$ and the sum corresponding to $(-a, b)$ add up to $\\frac{m-1}{2}$. Therefore, the inequality holds when $a, b$ are taken as negative integers coprime to $m$.\n$$\n(*) \\quad \\left| \\sum_{k=1}^{m-1} \\left\\{ \\frac{ak}{m} \\right\\} \\left\\{ \\frac{bk}{m} \\right\\} - \\left( \\frac{m}{4} + \\frac{m}{12ab} \\right) \\right| \\le 2|a| + 2|b|.\n$$\n\n**Step 2:** Transform the problem into a summation similar to (1).\nLet $p$ be a sufficiently large prime number (e.g., $p > 10^{15}$). The congruence symbol $\\equiv$\nused below denotes congruence modulo $p$.\nLet $a, b$, and $c$ represent $a_1, a_2$, and $a_3$ respectively, and let $a_4$ be chosen such that\n$a_1 + a_2 + a_3 + a_4 \\equiv 0$. For $k \\in \\{1, 2, \\dots, p-1\\}$, let\n$$\nh_k = \\left\\{ \\frac{a_1 k}{p} \\right\\} + \\left\\{ \\frac{a_2 k}{p} \\right\\} + \\left\\{ \\frac{a_3 k}{p} \\right\\} + \\left\\{ \\frac{a_4 k}{p} \\right\\},\n$$\nIt is easy to see that $h_k$ is an integer and $0 \\le h_k < 4$, i.e., $h_k \\in \\{0, 1, 2, 3\\}$. Moreover,\n$\\{\\frac{a_1 k}{p}\\} + \\{\\frac{a_2 k}{p}\\} + \\{\\frac{a_3 k}{p}\\} \\le 1$ is equivalent to $h_k = 0$ or 1.\nLet $L_0, L_1, L_2$, and $L_3$ be the number of occurrences of 0, 1, 2, and 3 respectively among $h_1, h_2, \\dots, h_{p-1}$. We need to prove that $L_0 + L_1 \\ge \\lfloor \\frac{p}{12} \\rfloor$.\n\nIf exactly $t$ out of $a_1, a_2, a_3, a_4$ are not multiples of $p$, then $h_{p-k} = t - h_k$. We have the following results: If $t=0$, then $L_0 = p-1$. If $t=2$, then $L_1 = p-1$. If $t=3$, then $L_1 = L_2 = \\frac{p-1}{2}$.\nWithout loss of generality, we assume that $a_1, a_2, a_3, a_4 \\neq 0$. In this case, $L_1 = L_3$, and $L_1 + L_2 + L_3 = p-1$. We can consider $\\{\\frac{ak}{p}\\}$ as a random variable, and the estimation of $L_1$ will be related to the variance of the sum $h_k$. Specifically, we consider the sum of squares:\n$$\n\\sum_{k=1}^{p-1} h_k^2 = L_1 + 4L_2 + 9L_3 = 4(p-1) + 2L_1,\n$$\nOn the other hand,\n$$\nH = \\sum_{k=1}^{p-1} h_k^2 = \\sum_{k=1}^{p-1} \\left( \\sum_{i=1}^{4} \\left\\{ \\frac{a_i k}{p} \\right\\} \\right) \\left( \\sum_{j=1}^{4} \\left\\{ \\frac{a_j k}{p} \\right\\} \\right) = \\sum_{i=1}^{4} \\sum_{j=1}^{4} \\left( \\sum_{k=1}^{p-1} \\left\\{ \\frac{a_i k}{p} \\right\\} \\left\\{ \\frac{a_j k}{p} \\right\\} \\right).\n$$\nFor any $a, b \\neq 0$, let's define the function $f(a, b) = \\sum_{k=1}^{p-1} \\left\\{ \\frac{ak}{p} \\right\\} \\left\\{ \\frac{bk}{p} \\right\\}$. In particular,\n$$\nf(a_i, a_i) = f(1, 1) = \\frac{1^2 + 2^2 + \\dots + (p-1)^2}{p^2} = \\frac{p}{3} - \\frac{1}{2} + \\frac{1}{6p}.\n$$\nNow, let's consider the sum of six terms $F = \\sum_{1 \\le i < j \\le 4} f(a_i, a_j)$. Since $H = 2F + 4f(1, 1) = 4(p-1) + 2L_1$, the desired inequality $L_1 \\ge \\lfloor \\frac{p}{12} \\rfloor$ holds if we have $F \\ge \\frac{17}{12}p$.\n\n**Step 3:** Translating into estimates for $a_1, a_2, a_3, a_4$.\nWe define an equivalence relation $(a, b) \\sim (u, v)$ if there exists $r \\neq 0$ such that $u \\equiv ra$ and $v \\equiv rb$. In this case,\n$$\nf(u, v) = \\sum_{k=1}^{p-1} \\left\\{ \\frac{uk}{p} \\right\\} \\left\\{ \\frac{vk}{p} \\right\\} = \\sum_{k=1}^{p-1} \\left\\{ \\frac{ark}{p} \\right\\} \\left\\{ \\frac{brk}{p} \\right\\} = \\sum_{rk=1}^{p-1} \\left\\{ \\frac{a \\cdot rk}{p} \\right\\} \\left\\{ \\frac{b \\cdot rk}{p} \\right\\} = f(a, b).\n$$\nThus, the value of the function $f(a_i, a_j)$ depends only on the equivalence class of $(a_i, a_j)$. In particular, combining with the calculation in Step 1 (*), we have\n$$\n(***) \\qquad \\left| f(a_i, a_j) - \\left( \\frac{p}{4} + p \\frac{\\gcd(a_i, a_j)^2}{12a_i a_j} \\right) \\right| \\le 2 \\frac{|a_i| + |a_j|}{\\gcd(a_i, a_j)}.\n$$\nTo obtain a more accurate estimation of the values of $f(a_i, a_j)$, we introduce additional considerations.\nFor any $a, b \\neq 0$, let $b^{-1}$ denote the modular inverse of $b$ modulo $p$. Consider the residues modulo $p$ of the set $0, ab^{-1}, 2 \\times ab^{-1}, \\dots, \\lfloor \\sqrt{p} \\rfloor \\times ab^{-1}$. By the pigeonhole principle, there exist two residues whose difference is less than or equal to $\\lfloor \\sqrt{p} \\rfloor$. Therefore, there exist $u, v \\in \\pm 1, \\pm 2, \\dots, \\pm \\lfloor \\sqrt{p} \\rfloor$ such that $u \\equiv v \\times ab^{-1}$, i.e., $(u, v) \\sim (a, b)$. (Here, we assume $u > 0$, and we also assume that $u$ and $v$ are coprime; otherwise, we can replace them with $\\frac{u}{(u,v)}$ and $\\frac{v}{(u,v)}$.)\nFor $(a_i, a_j)$, let $(u_{ij}, v_{ij})$ be one of the pairs that satisfy the above condition. If there are multiple choices, we can select any of them, and let $g_{ij} = \\frac{1}{u_{ij}v_{ij}}$. In this case, we have\n$$\nf(a_i, a_j) = f(u_{ij}, v_{ij}),\n$$\nand according to Equation (***), we have:\n$$\n\\left| f(a_i, a_j) - \\left( \\frac{p}{4} + \\frac{p}{12u_{ij}v_{ij}} \\right) \\right| < 4\\sqrt{p}.\n$$\nLet $G = \\sum_{1 \\le i < j \\le 4} g_{ij} = \\sum_{1 \\le i < j \\le 4} \\frac{1}{u_{ij}v_{ij}}$, and let $\\delta = 10^{-5}$. As long as (there exists a choice of $g_{ij}$) $G \\ge -1 + \\delta$, we have\n$$\nF = \\sum_{i<j} f(a_i, a_j) > \\sum_{i<j} \\left( \\frac{p}{4} + \\frac{p}{12u_{ij}v_{ij}} - 4\\sqrt{p} \\right) = \\frac{18+G}{12} \\times p - 24\\sqrt{p} > \\frac{17}{12}p.\n$$\nThe condition $p \\nmid (a+b)(a+c)(b+c)$ ensures that in the set $a_1, a_2, a_3, a_4$,\nFor each pair $(a_i, a_j)$, since $a_i+a_j \\ne 0$, we know that $(u_{ij}, v_{ij}) \\ne (1, -1)$, which implies $g_{ij} \\ge -\\frac{1}{2}$. We call a pair $(a, b)$ “good” if there exist $u, v \\in \\pm 1, \\pm 2, \\dots, \\pm 10$ such that $(a, b) \\sim (u, v)$. In this case, if $(a_i, a_j)$ is not a good pair, then $g_{ij} = \\frac{1}{u_{ij}v_{ij}} \\ge -\\frac{1}{11}$. If among the six pairs $(a_i, a_j)$, $1 \\le i < j \\le 4$, there are either 0 or 1 good pairs, then $G \\ge -\\frac{1}{2} + 5 \\times (-\\frac{1}{11}) = -\\frac{21}{22} > -1 + \\delta$.\nTherefore, assuming there are at least two good pairs, we know that $a_1+a_2+a_3+a_4 \\equiv 0$ implies that the four numbers are in a (simple) proportion. For example:\nIf $(a_1, a_2) \\sim (u_1, v_1)$ and $(a_1, a_3) \\sim (u_2, v_2)$, then the four numbers are proportional to $(u_1u_2, v_1u_2, v_2u_1, -u_1u_2 - v_1u_2 - v_2u_1)$.\nIf $(a_1, a_2) \\sim (u_1, v_1)$ and $(a_3, a_4) \\sim (u_2, v_2)$, then the four numbers are proportional to $(u_1(u_2+v_2), v_1(u_2+v_2), -u_2(u_1+v_1), -v_2(u_1+v_1))$.\nIn conclusion, there exists $z_1+z_2+z_3+z_4 = 0$ such that $(a_1, a_2, a_3, a_4) \\sim (z_1, z_2, z_3, z_4)$ and $1 \\le |z_i| \\le 300$, $z_i + z_j \\ne 0$. In this case, we have $(a_i, a_j) \\sim \\left( \\frac{z_i}{z_i z_j}, \\frac{z_j}{z_i z_j} \\right)$, which allows us to choose $g_{i,j} = \\frac{(z_i, z_j)^2}{z_i z_j}$. We still focus on the value of $G = \\sum_{i<j} g_{i,j}$, and there are two simple proportions satisfying $G = 1$:\n* The four numbers form the “Golden Ratio 1” of $(1 : -2 : -3 : 4)$. Let $a_1 = p-1$, $a_2 = 2$, $a_3 = 3$, $a_4 = p-4$. In this case, $h_k = 1$ is equivalent to $\\frac{2k}{p} + \\frac{3k}{p} \\le \\frac{k}{p}$. Among the values $k \\in 1, 2, \\dots, p-1$, such $k$ satisfying $\\frac{2}{3}p < k < \\frac{3}{4}p$ are counted as $L_1 = \\lfloor \\frac{3}{4}p \\rfloor - \\lfloor \\frac{2}{3}p \\rfloor \\ge \\lfloor \\frac{p}{12} \\rfloor$.\n* The four numbers form the “Golden Ratio 2” of $(1 : -3 : -4 : 6)$. Let $a_1 = p-1$, $a_2 = 3$, $a_3 = 4$, $a_4 = p-6$. In this case, $h_k = 1$ is equivalent to $\\frac{3k}{p} + \\frac{4k}{p} \\le \\frac{k}{p}$. Among the values $k \\in 1, 2, \\dots, p-1$, such $k$ satisfying $\\frac{3}{4}p < k < \\frac{5}{6}p$ are counted as $L_1 = \\lfloor \\frac{5}{6}p \\rfloor - \\lfloor \\frac{3}{4}p \\rfloor \\ge \\lfloor \\frac{p}{12} \\rfloor$.\nAssuming $(a_1, a_2, a_3, a_4) \\sim (z_1, z_2, z_3, z_4)$ does not correspond to the two golden ratios mentioned above, we will prove that $G \\ge -1 + \\delta$. Let's consider the signs of $z_1, z_2, z_3, z_4$.\nIf there are three positive and one negative (or three negative and one positive) numbers, without loss of generality, let $z_1, z_2, z_3 > 0$ and $z_4 = -(z_1 + z_2 + z_3)$. In this case, among the six terms in $G = \\sum_{1 \\le i < j \\le 4} g(z_i, z_j)$, there are three positive and three negative terms. Specifically, $g(z_i, z_4) = -\\frac{(z_i, z_4)^2}{z_i |z_4|} \\ge -\\frac{z_i}{|z_4|}$ for $i = 1, 2, 3$, and $g(z_1, z_2) \\ge \\frac{1}{z_1 z_2} > \\delta$. Therefore, $G > -1 + \\delta$.\nAssuming that $z_1, z_2, z_3, z_4$ consist of two positive and two negative numbers, let $z_1, z_4 > 0$ and $z_2, z_3 < 0$.\nIf there exists $\\frac{z_j}{z_i} = -2$, without loss of generality, let $z_1 = u > 0$, $z_2 = -2u$. Let $z_3 = -v < 0$ and $z_4 = u + v$ (where $u$ and $v$ are coprime). In this case,\n$$\nG = -\\frac{1}{2} - \\frac{1}{uv} + \\frac{1}{u(u+v)} + \\frac{(2, v)^2}{2uv} - \\frac{(2, u+v)^2}{2u(u+v)} - \\frac{1}{v(u+v)}\n$$\n$$\nG \\geq -\\frac{1}{2} - \\frac{1}{uv} + \\frac{1}{u(u+v)} + \\frac{1}{2uv} - \\frac{4}{2u(u+v)} - \\frac{1}{v(u+v)} = -\\frac{1}{2} - \\frac{3}{2uv} > -1 + \\delta.\n$$\nThe last step of the above derivation assumes that $(u, v) \\neq (1, 1), (2, 1), (1, 3), (3, 1)$, which would lead to negations or the four numbers forming a golden ratio. Additionally, when $(u, v) = (1, 2)$, we have $G = 0$. In any case, we have $G > -1 + \\delta$.\nIf there exists $\\frac{z_j}{z_i} = -3$, without loss of generality, let $z_1 = u > 0$, $z_2 = -3u$. Let $z_3 = -v < 0$ and $z_4 = 2u + v$ (where $u$ and $v$ are coprime). In this case,\n$$\nG = -\\frac{1}{3} - \\frac{1}{uv} + \\frac{1}{u(2u + v)} + \\frac{(3, v)^2}{3uv} - \\frac{(3, 2u + v)^2}{3u(2u + v)} - \\frac{(2, v)^2}{v(2u + v)}\n$$\n$$\nG \\geq -\\frac{1}{3} - \\frac{1}{uv} + \\frac{1}{u(2u + v)} + \\frac{1}{3uv} - \\frac{9}{3u(2u + v)} - \\frac{4}{v(2u + v)} = -\\frac{1}{3} - \\frac{8}{3uv} > -1 + \\delta.\n$$\nThe last step of the above derivation assumes that $(u, v) \\neq (1, 1), (1, 2), (1, 4)$, which would lead to negations or the four numbers forming a golden ratio. Additionally, when $(u, v) = (2, 1), (1, 3), (3, 1), (4, 1)$, we have $G = -\\frac{4}{5}, \\frac{2}{5}, -\\frac{2}{3}, -\\frac{2}{3}$ respectively. In any case, we have $G > -1 + \\delta$.\nIf each $\\frac{z_j}{z_i} \\neq -2, -3$, then the four negative terms in $G$ are all greater than or equal to $-\\frac{1}{4}$, and the two positive terms are both greater than $\\delta$. Therefore, we still have $G > -1 + \\delta$.\nIn conclusion, we have $L_1 + L_0 \\geq \\lfloor \\frac{p}{12} \\rfloor$, and thus the conclusion of the problem is established. $\\square$",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 28585,
    "subject": "Mathematics (Multi-modal)",
    "question": "Given an integer $n \\ge 2$. Find the minimum real number $\\lambda$ such that for any real numbers $a_1, a_2, \\dots, a_n$, and $b$, the following inequality holds:\n$$\n\\lambda \\sum_{i=1}^{n} \\sqrt{|a_i - b|} + \\sqrt{n \\left| \\sum_{i=1}^{n} a_i \\right|} \\ge \\sum_{i=1}^{n} \\sqrt{|a_i|}.\n$$",
    "options": [],
    "answer": "(n - 1 + sqrt(n - 1)) / sqrt(n)",
    "solution": "We will prove that $\\lambda_n := \\frac{n-1+\\sqrt{n-1}}{\\sqrt{n}}$ is the desired minimum value. Let's first prove a few lemmas.\n\n**Lemma 1:** The sequence $\\{\\lambda_n\\}$ is strictly increasing.\n\n**Proof:** Note that\n$$\n\\lambda_{n+1}^2 - \\lambda_n^2 = \\frac{n+1+2n^{\\frac{5}{2}}-2(n+1)(n-1)^{\\frac{3}{2}}}{n(n+1)} = \\frac{n+1+2n^2\\sqrt{n}-2(n^2-1)\\sqrt{n-1}}{n(n+1)} > 0.\n$$\nTherefore, $\\lambda_n < \\lambda_{n+1}$.\n\n**Lemma 2:** Let $\\lambda \\ge 1$, $b > 0$, and $c_1, c_2, \\dots, c_m$ be non-negative real numbers. Then\n$$\n\\lambda \\sum_{i=1}^{m} \\sqrt{c_i + b} - \\sum_{i=1}^{m} \\sqrt{c_i} \\ge \\lambda \\sqrt{\\sum_{i=1}^{m} c_i + b} - \\sqrt{\\sum_{i=1}^{m} c_i}.\n$$\n**Proof:** It is clear that $\\sum_{i=1}^{m} \\sqrt{c_i + b} \\ge \\sqrt{\\sum_{i=1}^{m} c_i + b}$. Therefore,\n$$\n\\left( \\lambda \\sum_{i=1}^{m} \\sqrt{c_i + b} - \\sum_{i=1}^{m} \\sqrt{c_i} \\right) - \\left( \\lambda \\sqrt{\\sum_{i=1}^{m} c_i + b} - \\sqrt{\\sum_{i=1}^{m} c_i} \\right) \\\\\n\\ge \\sum_{i=1}^{m} (\\sqrt{c_i + b} - \\sqrt{c_i}) - \\left( \\sqrt{\\sum_{i=1}^{m} c_i + b} - \\sqrt{\\sum_{i=1}^{m} c_i} \\right) \\\\\n= \\sum_{i=1}^{m} \\frac{b}{\\sqrt{c_i + b} + \\sqrt{c_i}} - \\frac{b}{\\sqrt{\\sum_{i=1}^{m} c_i + b} + \\sqrt{\\sum_{i=1}^{m} c_i}} \\ge 0.\n$$\nHence, Lemma 2 holds.\n\n**Lemma 3:** Let $\\lambda \\ge \\lambda_2$, $b < a \\le c$ be positive real numbers. Then\n$$\n\\lambda(\\sqrt{a-b} + \\sqrt{c+b}) - \\sqrt{a} - \\sqrt{c} > \\lambda(\\sqrt{c+2b-a} - \\sqrt{b} - \\sqrt{c-a+b}).\n$$\n**Proof:** The inequality in Lemma 3 is equivalent to\n$$\n\\lambda\\sqrt{a-b} + \\lambda(\\sqrt{c+b} - \\sqrt{c+2b-a}) > (\\sqrt{a}-\\sqrt{b}) + (\\sqrt{c}-\\sqrt{c-a+b}) \\\\\n\\iff \\lambda + \\frac{\\lambda\\sqrt{a-b}}{\\sqrt{c+b} + \\sqrt{c+2b-a}} > \\frac{\\sqrt{a-b}}{\\sqrt{a}+\\sqrt{b}} + \\frac{\\sqrt{a-b}}{\\sqrt{c}+\\sqrt{c-a+b}}.\n$$\nFirstly, we have\n(1) $$ \\lambda > 1 > \\frac{\\sqrt{a-b}}{\\sqrt{a} + \\sqrt{b}}. $$\n\nFurthermore, since\n$$\n\\sqrt{c+b} + \\sqrt{c+2b-a} < \\sqrt{c+c} + \\sqrt{2(c-a)+2b} = \\sqrt{2} \\cdot (\\sqrt{c} + \\sqrt{c-a+b}),\n$$\nand $\\lambda \\ge \\lambda_2 = \\sqrt{2}$, we have\n$$\n(2) \\qquad \\frac{\\lambda\\sqrt{a-b}}{\\sqrt{c+b} + \\sqrt{c+2b-a}} > \\frac{\\sqrt{a-b}}{\\sqrt{c} + \\sqrt{c-a+b}}.\n$$\nTherefore, combining (1) and (2), we obtain the desired inequality.\n\n**Proof:** Let's return to the original problem. First, we take $a_1 = \\cdots = a_{n-1} = b = 1$ and $a_n = -(n-1)$. Then, we have\n$$\n\\lambda\\sqrt{n} \\ge n - 1 + \\sqrt{n-1} \\implies \\lambda \\ge \\frac{n-1 + \\sqrt{n-1}}{\\sqrt{n}} = \\lambda_n.\n$$\nNext, we will prove that $\\lambda_n$ satisfies the desired inequality. That is, we need to prove that for any real numbers $a_1, a_2, \\dots, a_n, b$, we have\n$$\n(3) \\qquad \\lambda_n \\sum_{i=1}^{n} \\sqrt{|a_i - b|} + \\sqrt{n \\left| \\sum_{i=1}^{n} a_i \\right|} \\ge \\sum_{i=1}^{n} \\sqrt{|a_i|}.\n$$\nLet $t := \\frac{1}{n} \\sum_{i=1}^{n} a_i$, $a_i' := a_i - t$ ($i = 1, 2, \\dots, n$), and $b' := b - t$. Note that\n$$\n\\sqrt{|t|} + \\sqrt{|a_i'|} \\ge \\sqrt{|t| + |a_i'|} \\ge \\sqrt{|t + a_i'|} = \\sqrt{|a_i|}.\n$$\nTherefore, to prove (3), it suffices to prove that for any real numbers $a_1, a_2, \\dots, a_n, b$, if $\\sum_{i=1}^{n} a_i = 0$, then\n$$\n(4) \\qquad \\lambda_n \\sum_{i=1}^{n} \\sqrt{|a_i - b|} \\ge \\sum_{i=1}^{n} \\sqrt{|a_i|}.\n$$\nWithout loss of generality, let's assume that $b > 0$. By Lemma 1 and Lemma 2, we can combine all negative terms in $a_1, a_2, \\dots, a_n$. Therefore, to prove (4), it suffices to prove that for any non-negative real numbers $a_1, a_2, \\dots, a_{n-1}, b$, if $c := \\sum_{i=1}^{n-1} a_i > 0$, then\n$$\n(5) \\qquad \\lambda_n \\left( \\sum_{i=1}^{n-1} \\sqrt{|a_i - b|} + \\sqrt{c+b} \\right) \\ge \\sum_{i=1}^{n-1} \\sqrt{a_i} + \\sqrt{c}.\n$$\nIf $b \\ge \\frac{c}{n-1}$, then\n$$\n\\lambda_n \\left( \\sum_{i=1}^{n-1} \\sqrt{|a_i - b|} + \\sqrt{c+b} \\right) \\ge \\lambda_n \\sqrt{c + \\frac{c}{n-1}} = \\sqrt{(n-1)c} + \\sqrt{c} \\\\\n= \\sqrt{(n-1) \\sum_{i=1}^{n-1} a_i} + \\sqrt{c} \\ge \\sum_{i=1}^{n-1} \\sqrt{a_i} + \\sqrt{c}.\n$$\nTherefore, (5) holds.\n\nIf $b < \\frac{c}{n-1} = \\frac{1}{n-1} \\sum_{i=1}^{n-1} a_i$, then there must exist some $a_j > b$. By Lemma 3, after adjusting $a_j$ to $b$ and $c$ to $c - a_j + b$, the difference between the left-hand side and the right-hand side of (5) strictly decreases. After a finite number of adjustments, we will eventually reach the case where $b \\ge \\frac{c}{n-1}$. Therefore, (5) holds.\n\nCombining the above, we conclude that the minimum value of $\\lambda$ satisfying the given conditions is $\\frac{n-1+\\sqrt{n-1}}{\\sqrt{n}}$. $\\square$",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 28586,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $a, b, c, d$ be nonnegative real numbers not exceeding $1$. Prove that\n$$\n\\frac{1}{1+a+b} + \\frac{1}{1+b+c} + \\frac{1}{1+c+d} + \\frac{1}{1+d+a} \\le \\frac{4}{1+2\\sqrt[4]{abcd}}.\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Notice that when $\\sqrt{ac} \\le x$, we have\n$$\n\\frac{1}{x+a} + \\frac{1}{x+c} - \\frac{2}{x+\\sqrt{ac}} = \\frac{(\\sqrt{a}-\\sqrt{c})^2(\\sqrt{ac}-x)}{(x+a)(x+c)(x+\\sqrt{ac})} \\le 0. \\quad (*)\n$$\nGiven the conditions, $\\sqrt{ac} \\le 1 \\le 1+b$, and $\\sqrt{ac} \\le 1+d$. Substituting $x = 1+b$ and $x = 1+c$ into $(*)$ yields\n$$\n\\frac{1}{1+a+b} + \\frac{1}{1+b+c} \\le \\frac{2}{1+b+\\sqrt{ac}},\n$$\n$$\n\\frac{1}{1+c+d} + \\frac{1}{1+d+a} \\le \\frac{2}{1+d+\\sqrt{ac}}.\n$$\nThis means that when substituting $\\sqrt{ac}$ for $a$ and $c$, the left side of the inequality does not decrease, while the right side remains unchanged. Thus, we may assume without loss of generality that $a = c$. Similarly, we may assume $b = d$. Hence, the original inequality is reduced to proving\n$$\n\\frac{1}{1+a+b} \\le \\frac{1}{1+2\\sqrt{ab}}.\n$$\nThis holds true by the arithmetic mean-geometric mean inequality, which states $a+b \\ge 2\\sqrt{ab}$. $\\square$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 28587,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find the largest positive integer $m$ such that there is a way to color some of the cells red in a $70 \\times 70$ grid, satisfying the following two conditions:\n(1) There do not exist two red cells such that the number of red cells in the row they belong to and the number of red cells in the column they belong to are the same;\n(2) There are at least 2 rows each containing exactly $m$ red cells.",
    "options": [],
    "answer": "32",
    "solution": "Let the row with the maximum number of red squares (let's assume it's the first row) have $a$ red squares, and the column with the maximum number of red squares (let's assume it's the first column) have $b$ red squares. If $a > b$, consider the $a$ columns in which the $a$ red squares of the first row are located. These columns must have distinct numbers of red squares and can only be chosen from $1, 2, \\dots, b$. This leads to a contradiction. Similarly, we can show that $a$ cannot be less than $b$. Therefore, we conclude that $a = b$. Moreover, the number of red squares in the columns where the red squares of the first row are located must be a permutation of $1, 2, \\dots, a$, and the number of red squares in the rows where the red squares of the first column are located must also be a permutation of $1, 2, \\dots, a$. In other words, for any $1 \\le i \\le a$, there exist rows with exactly $i$ red squares and columns with exactly $i$ red squares.\n\nNext, we prove that there exists a coloring that satisfies condition (1) and has $x_i$ rows with exactly $i$ red squares and $y_i$ columns with exactly $i$ red squares ($x_i, y_i \\in \\mathbb{Z}_{>0}$, $i = 1, 2, \\dots, a$), if and only if there exists a coloring of an $a \\times a$ grid such that the $i$th row has exactly $ix_i$ red squares and the $i$th column has exactly $iy_i$ red squares ($i = 1, 2, \\dots, a$). (*).\n\nFirst, we prove the necessity. If the coloring of a $70 \\times 70$ grid $P$ satisfies condition (1), we construct an $a \\times a$ grid $Q$ with the following coloring: for any $i, j \\in \\{1, 2, \\dots, a\\}$, the cell in the $i$th row and $j$th column of $Q$ is colored red if and only if $P$ has a red cell in a row with exactly $i$ red cells and a column with exactly $j$ red cells (denoted as $T(i, j)$). Note that in $P$, the number of red cells in a row with exactly $i$ red cells is $ix_i$, and the numbers of red cells in their corresponding columns are distinct. Therefore, $Q$ satisfies $T(i, j)$ if and only if there are exactly $ix_i$ red cells in column $j$, which implies that the $i$th row of $Q$ has exactly $ix_i$ red cells. Similarly, the $i$th column of $Q$ has exactly $iy_i$ red cells. Thus, the necessity is proved.\n\nNext, we prove the sufficiency. Assume that the $a \\times a$ grid $Q$ has exactly $ix_i$ red cells in the $i$th row and exactly $iy_i$ red cells in the $i$th column. We construct a $70 \\times 70$ grid $P$ as follows: we label all rows of $P$ as $R(i, x)$ ($1 \\le x \\le x_i$) and all columns of $P$ as $C(i, y)$ ($1 \\le y \\le y_i$). Then, we color the cells of $P$ as follows: if the cell in the $i$th row and $j$th column of $Q$ is red, we mark $(i, j)$ as a good pair. For each good pair $(i_0, j_0)$, we define $v(i_0, j_0)$ as the index of $j$ (in ascending order) such that $(i_0, j)$ is a good pair, and $u(i_0, j_0)$ as the index of $i$ (in ascending order) such that $(i, j_0)$ is a good pair. Then, in $P$, we color the intersection of row $R(i_0, \\lfloor \\frac{v(i_0, j_0)}{i_0} \\rfloor)$ and column $C(j_0, \\lfloor \\frac{u(i_0, j_0)}{j_0} \\rfloor)$ red. In this way, each good pair $(i_0, j_0)$ is associated with only one pair of row $R(i_0, x)$ and column $C(j_0, y)$, satisfying condition (1). Moreover, for any row $R(i_0, x)$, the red cells correspond to $j$ values that are the $(x-1)i_0 + 1$st, $(x-1)i_0 + 2$nd, ..., $xi_0$th smallest $j$ among all good pairs $(i_0, j_0)$. Thus, this row has exactly $i_0$ red cells. Similarly, each column $C(j_0, y)$ has exactly $j_0$ red cells. Therefore, the sufficiency is proved, and thus (*) holds.\n\nReturning to the original problem, let's first construct an example for $m = 32$. Consider a $64 \\times 64$ grid $Q$, where the cell in the $i$th row and $j$th column is colored red if and only if $i + j \\le 66$. This coloring ensures that each row of $Q$ has a decreasing number of red cells from 64 to 2, and each column has the same pattern. Now, let\n$$\na = 64, \\quad x_{32} = x_{16} = x_8 = x_4 = x_2 = x_1 = y_{32} = y_{16} = y_8 = y_4 = y_2 = y_1 = 2,\n$$\nwhile all other $x_i$ and $y_i$ are set to 1. According to (*) and the above construction, we know that there exists a grid $P$ that satisfies condition (1) and has two rows with exactly 32 red cells. Thus, we have shown that $m = 32$ meets the requirements of the problem.\n\nFinally, we prove that $m$ cannot be greater than 32. Note that when $x_i \\ge 2$, we have $2m \\le a \\le 69$, which implies $m \\le 34$. We only need to show that $m$ cannot equal 33 or 34.\nIn this case, we only need to consider $a \\in \\{66, 67, 68, 69\\}$. For the grid $Q$ and any $r, s \\le a$, the difference between the total number of red cells in any $r$ rows and the total number of red cells in any $s$ columns is no more than the total number of red cells in those $r$ rows and the remaining $a-s$ columns, which is no more than $r(a-s)$.\n\nIf $a = 69$, then we have $x_m = 2$ and all other $x_i$ can only be 1. From calculating twice, we find that only $y_m = 2$ and all other $y_i$ can only be 1. In this case, in the grid $Q$, the top 5 rows with the maximum number of red cells have at least $69+68+67+66+66$ red cells, while the leftmost 5 columns with the minimum number of red cells have $1+2+3+4+5$ red cells. However,\n$$\n(69 + 68 + 67 + 66 + 66) - (1 + 2 + 3 + 4 + 5) > 5 \\cdot (69 - 5),\n$$\nwhich leads to a contradiction!\n\nIf $a = 68$, similarly, we have\n$$\n(68 + 67 + 66 + 66 + 66) - (1 + 2 + 3 + 4) > 4 \\cdot (68 - 4),\n$$\nwhich means that at least one of $y_1, y_2, y_3, y_4$ is not less than 2.\nSince\n$$\n(68 + 67 + 66 + 66 + 65 + 64 + 63 + 62) - (1 + 2 + 3 + 4 + 5 + 6 + 7 + 8) - 8 \\cdot (68 - 8) > 4,\n$$\nwe have $\\sum_{i=1}^{8}(y_i - 1) \\ge 2$. Furthermore, notice that\n$$\n(68 + 67 + 66 + 66 + 65 + \\dots + 54) - (1 + 2 + \\dots + 16) - 16 \\cdot (68 - 16) > 4 + 8,\n$$\nwhich implies $\\sum_{i=1}^{16}(y_i - 1) \\ge 3$. However, this contradicts the fact that $\\sum_{i=1}^{68}(y_i - 1) = 2$.\n\nIf $a = 67$ (in this case, $m = 33$), similarly, we have\n$$\n(67 + 66 + 66 + 66) - (1 + 2 + 3) > 3 \\cdot (67 - 3),\n$$\nwhich implies that at least one of $y_1, y_2, y_3$ is not less than 2. Additionally, we have\n$$\n(67 + 66 + 66 + 65 + 64 + 63) - (1 + 2 + 3 + 4 + 5 + 6) - 6 \\cdot (67 - 6) > 3,\n$$\nwhich means that $\\sum_{i=1}^{6}(y_i - 1) \\ge 2$. Furthermore, we have\n$$\n(67 + 66 + 66 + 65 + \\dots + 57) - (1 + 2 + \\dots + 12) - 12 \\cdot (67 - 12) > 3 + 6,\n$$\nimplying that $\\sum_{i=1}^{12}(y_i - 1) \\ge 3$. Finally, we have\n$$\n(67 + 66 + 66 + 65 + \\dots + 45) - (1 + 2 + \\dots + 24) - 24 \\cdot (67 - 24) > 3 + 6 + 12,\n$$\nwhich shows that $\\sum_{i=1}^{24}(y_i - 1) \\ge 4$. However, this contradicts the fact that $\\sum_{i=1}^{67}(y_i - 1) = 3$.\n\nIf $a = 66$ (in this case, $m = 33$), we have two values of $ix_i$ equal to 66, indicating that all $iy_i$ are not less than 2, i.e., $y_1 \\ge 2$. Similarly, we have\n$$\n(66 + 66 + 65) - (1 + 2) - 3 \\cdot (66 - 2) > 1,\n$$\nwhich implies $\\sum_{i=1}^{2}(y_i - 1) \\ge 2$. Furthermore, we have\n$$\n(66 + 66 + 65 + 64 + 63) - (1 + 2 + 3 + 4) - 5 \\cdot (66 - 4) > 1 + 2,\n$$\nimplying $\\sum_{i=1}^{4}(y_i - 1) \\ge 3$. Additionally, we have\n$$\n(66 + 66 + 65 + \\cdots + 59) - (1 + 2 + \\cdots + 8) - 9 \\cdot (66 - 8) > 1 + 2 + 4,\n$$\nwhich shows $\\sum_{i=1}^{8}(y_i - 1) \\ge 4$. Moreover, we have\n$$\n(66 + 66 + 65 + \\cdots + 51) - (1 + 2 + \\cdots + 16) - 17 \\cdot (66 - 16) > 1 + 2 + 4 + 8,\n$$\nimplying $\\sum_{i=1}^{16}(y_i - 1) \\ge 5$. However, this contradicts the fact that $\\sum_{i=1}^{66}(y_i - 1) = 4$.\n\nIn conclusion, the maximum value of $m$ that satisfies the conditions is 32. $\\square$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 28588,
    "subject": "Mathematics (Multi-modal)",
    "question": "Fix a prime number $p \\ge 5$, and put $\\Omega = \\{1, 2, \\dots, p\\}$. For any $x, y \\in \\Omega$, define\n$$\nr(x, y) = \\begin{cases} y - x, & \\text{if } y \\ge x, \\\\ y - x + p, & \\text{if } y < x. \\end{cases}\n$$\nFor a nonempty subset $A$ of $\\Omega$, define\n$$\nf(A) = \\sum_{x \\in A} \\sum_{y \\in A} (r(x, y))^2.\n$$\nWe say that a subset $A$ of $\\Omega$ is *good* if $0 < |A| < p$ and for any subset $B$ of $\\Omega$ with $|B| = |A|$, we have $f(B) \\ge f(A)$.\nDetermine the maximal positive integer $L$, such that there exist pairwise distinct good subsets $A_1, A_2, \\dots, A_L$ of $\\Omega$ such that $A_1 \\subset A_2 \\subset \\dots \\subset A_L$.",
    "options": [],
    "answer": "2 floor(log2(p+1))",
    "solution": "(1) For an intuitive understanding, place numbers $1, 2, \\dots, p$ equidistantly in a clockwise direction on the circumference of a circle with a perimeter exactly equal to $p$. Then $r(x, y)$ is precisely the distance from $x$ to $y$ in a clockwise direction.\nFor an $m$-element subset $A = \\{x_1, x_2, \\dots, x_m\\}$ of $\\Omega$ (with elements arranged from smallest to largest), we have\n$$\nf(A) = \\sum_{x,y \\in A} r(x,y)^2 = \\sum_{k=1}^{m-1} \\sum_{i=1}^{m} r(x_i, x_{i+k})^2.\n$$\nHere, indices are considered modulo $m$.\nFor $k = 1, 2, \\dots, m-1$, notice that $\\sum_{i=1}^{m} r(x_i, x_{i+k}) = kp$, as the $m$ arcs from $x_1$ to $x_{1+k}$, from $x_2$ to $x_{2+k}$, $\\dots$, from $x_m$ to $x_k$ can be recombined to form $k$ circles. By the AM-GM inequality, the minimum value of the partial sum $S_k = \\sum_{i=1}^{m} r(x_i, x_{i+k})^2$ is obtained when each $d(x_i, x_{i+k}) = \\lfloor \\frac{kp}{m} \\rfloor$ or $\\lceil \\frac{kp}{m} \\rceil$. For all $S_k$ to reach their minimum values, it is equivalent to, for any $i, j = 1, 2, \\dots, m$,\n$$\n\\left| x_j - x_i - \\frac{(j-i)p}{m} \\right| < 1.\n$$\nThat is, the differences between any two numbers in the set $D = \\{x_1 - \\frac{p}{m}, x_2 - \\frac{2p}{m}, \\dots, x_m - \\frac{mp}{m}\\}$ are less than 1. (In other words, this requires $x_1, \\dots, x_m$ to be almost equidistantly distributed on the circle.) Since multiplying the numbers in $D$ by $m$ yields distinct integers, we can set $D = \\{\\frac{r}{m}, \\frac{r-1}{m}, \\dots, \\frac{r-(m-1)}{m}\\}$. Thus for $k = 1, 2, \\dots, m$, $x_k$ is the only integer among $\\frac{kp+r}{m}, \\frac{kp+r-1}{m}, \\dots, \\frac{kp+r-(m-1)}{m}$, i.e., $x_k = \\lfloor \\frac{kp+r}{m} \\rfloor$. From $x_1 \\ge 1$ we know $r \\ge m-p$, and from $x_m \\le p$ we know $r \\le m-1$.\nTherefore, each of the aforementioned partial sums $S_k$ reaches its minimum value if and only if the set\n$$\nA = \\left\\{ x_k = \\left\\lfloor \\frac{kp+r}{m} \\right\\rfloor \\mid k = 1, 2, \\dots, m \\right\\},\n$$\nwhere $r \\in \\{m-p, m-p+1, \\dots, m-1\\}$. Specifically, at this time $f$ takes its minimum value among all $m$-element subsets, meaning these $A$ are precisely all the $m$-element good subsets. Since every element of a good subset $A$ plus $k$ (modulo $p$) is still a good subset, all $m$-element good subsets are \"rotationally equivalent\" (they are rotations of $\\{\\lfloor \\frac{kp}{m} \\rfloor \\mid k = 1, \\dots, m\\}$).\n\n(2) For convenience of explanation and understanding, we define a graph $G$ with $V = \\{1, 2, \\dots, p-1\\}$ as vertices. If there exists an $a$-element good subset $A$ and a $b$-element good subset $B$ such that $A \\subset B$, then we draw a directed edge $a \\to b$. Due to the rotational equivalence of good subsets: if some $b$-element good subset contains an $a$-element good subset, then every $b$-element good subset contains some $a$-element good subset, and every $a$-element good subset is also contained in some $b$-element good subset. (Clearly, graph $G$ forms a poset, i.e., if $a \\to b, b \\to c$ then $a \\to c$). Therefore, the problem transforms into considering the chains $a_1 \\to a_2 \\to \\dots \\to a_L$ in graph $G$, (where $L$ is the length of the chain).\n\nWe prove the following conclusions:\n\n**Conclusion 1:** The complement $A^c$ of any $m$-element good subset $A$ is also a good subset.\n*Proof of Conclusion 1:* Observe the following identity:\n$$\nf(A) - f(A^c) = \\left( \\sum_{x \\in A} \\sum_{y \\in \\Omega} r(x, y)^2 - \\sum_{x \\in A} \\sum_{y \\in A^c} r(x, y)^2 \\right) - \\left( \\sum_{x \\in \\Omega} \\sum_{y \\in A^c} r(x, y)^2 - \\sum_{x \\in A} \\sum_{y \\in A^c} r(x, y)^2 \\right) \\\\\n= \\sum_{x \\in A} \\sum_{y \\in \\Omega} r(x, y)^2 - \\sum_{x \\in \\Omega} \\sum_{y \\in A^c} r(x, y)^2 = |A| \\times \\sum_{k=0}^{m-1} \\left(\\frac{k}{m}\\right)^2 - \\sum_{k=0}^{m-1} \\left(\\frac{k}{m}\\right)^2 \\times |A^c|,\n$$\ni.e., $f(A) - f(A^c)$ is a constant value. Therefore, $f(A)$ reaching its minimum is equivalent to $f(A^c)$ reaching its minimum.\nThus, a good subset $A \\subset B$ is also equivalent to good subset $B^c \\subset A^c$, i.e., in graph $G$ there is an edge $a \\to b$ if and only if there is an edge $(p-b) \\to (p-a)$.\n\n**Conclusion 2:** When $m < \\frac{p}{2}$, any $m$-element good subset $A$ can be contained within a rotation of its complement $A^c$, i.e., graph $G$ has an edge $m \\to (p-m)$.\n*Proof of Conclusion 2:* Take the $2m$-element good subset $C = \\{\\lfloor \\frac{kp}{2m} \\rfloor \\mid k = 1, 2, \\dots, 2m\\}$, then\n$$\nA = \\left\\{ y_k = x_{2k} = \\left\\lfloor \\frac{kp}{m} \\right\\rfloor \\right\\} \\quad \\text{and} \\quad B = \\left\\{ z_k = x_{2k-1} = \\left\\lfloor \\frac{kp + \\left(-\\frac{p+1}{2}\\right)}{m} \\right\\rfloor \\right\\}\n$$\nare both $m$-element good subsets, and $B^c$ is a $p-m$-element good subset. Thus $A \\subset B^c$.\n*Alternate proof of Conclusion 2:* Since $m < \\frac{p}{2}$, $A$ and $A+1 = \\{x+1 \\mid x \\in A\\}$ do not intersect. Hence $A \\subset (A+1)^c$. Conclusion 2 holds.\n\nThe aforementioned symmetry of graph $G$ means that we only need to consider the longest chain from 1 to some number $< \\frac{p}{2}$. This is because for a chain $a_1 \\to a_2 \\to \\dots \\to a_L$, let $a_k < \\frac{p}{2}$, $a_{k+1} > \\frac{p}{2}$, then we can form two new chains\n$$\n\\begin{align*}\na_1 &\\to \\dots \\to a_k \\to (p-a_k) \\to \\dots \\to (p-a_1) \\quad \\text{(length 2k)}, \\\\\np-a_L &\\to \\dots \\to p-a_{k+1} \\to a_{k+1} \\to \\dots \\to a_L \\quad \\text{(length 2L-2k)}.\\end{align*}\n$$\nAt least one of these has length $\\ge L$. Therefore, we only need to consider \"self-symmetric\" chains.\n\n**Conclusion 3:** When $a$ divides $b$, there is an edge $a \\to b$, because the good subset\n$$\nB = \\{x_k = \\lfloor \\frac{kp}{b} \\rfloor, k = 1, 2, \\dots, b\\} \\supset A = \\{y_k = x_{k \\times \\frac{b}{a}} = \\lfloor \\frac{kp}{a} \\rfloor, k = 1, 2, \\dots, a\\}.\n$$\nFrom this, we know that when $2^u < p < 2^{u+1}$, there is always a chain of length $2u$\n$$\n1 \\to 2 \\to 4 \\to \\dots \\to 2^{u-1} \\to (p-2^{u-1}) \\to (p-2^{u-2}) \\to \\dots \\to (p-2) \\to (p-1). \\quad (2)\n$$\nIf the length of a chain $L > 2u$, by symmetry, we may assume $a_{u+1} < \\frac{p}{2}$.\nNext, we consider the necessary conditions for an edge $a \\to b$ in graph $G$ when $a < b < \\frac{p}{2}$.\n\n**Conclusion 4:** When $a < b < \\frac{p}{3}$, if an $a$-element good subset $A$ is contained in a $b$-element good subset $B$, i.e., $a \\to b$, then $a$ must be a divisor of $b$.\n*Proof of Conclusion 4:* Without loss of generality, assume good subset $B = \\{x_k = \\lfloor \\frac{kp}{b} \\rfloor, k = 1, 2, \\dots, b\\}$ contains good subset $A = \\{y_j = x_{k_j}, j = 1, 2, \\dots, a\\}$. If $k_1, k_2, \\dots, k_a$ are not equally spaced, for example, $k_2 - k_1 > k_3 - k_2$, then the difference between adjacent elements in $A$ is\n$$\ny_2 - y_1 = x_{k_2} - x_{k_1} \\ge \\lfloor \\frac{(k_2 - k_1)p}{b} \\rfloor \\ge \\lfloor \\frac{(k_2 - k_1 - 1)p}{b} \\rfloor + \\lfloor \\frac{p}{b} \\rfloor \\\\\n\\ge \\lfloor \\frac{(k_3 - k_2)p}{b} \\rfloor + 3 \\ge x_{k_3} - x_{k_2} + 2 = y_3 - y_2 + 2,\n$$\nwhich contradicts the fact that $A$ is a good subset. Therefore, $a$ must be a divisor of $b$.\n\n**Conclusion 5:** When $\\frac{p}{4} < a < b < \\frac{p}{2}$, except when $p \\equiv 3 \\pmod 4$ with an edge $\\frac{p+1}{4} \\to \\frac{p-1}{2}$, there is no edge between $a$ and $b$.\n*Proof of Conclusion 5:* From Conclusion 4, we know at this point $\\frac{p}{3} < b < \\frac{p}{2}$, otherwise $b < \\frac{p}{3}$ and is a multiple of $a$, which is impossible. As in the proof of Conclusion 4, assume good subset $B = \\{x_k = \\lfloor \\frac{kp}{b} \\rfloor, k = 1, 2, \\dots, b\\}$ contains good subset $A = \\{y_j = x_{k_j}, j = 1, 2, \\dots, a\\}$. Let the interval in $B$ be $h_k = x_{k+1} - x_k = \\lfloor \\frac{kp+p}{b} \\rfloor - \\lfloor \\frac{kp}{b} \\rfloor \\in \\{2, 3\\}$. Hence, there exists an interval in $A$ $y_{j+1} - y_j \\ge x_{k+2} - x_k \\ge 4$, so $|A| = a < \\frac{p}{3}$, i.e., intervals in $A$ $g_j = y_{j+1} - y_j \\in \\{\\lfloor \\frac{p}{a} \\rfloor, \\lceil \\frac{p}{a} \\rceil\\} = \\{3, 4\\}$. Clearly, the sequence $\\mathbf{h} = (h_1, h_2, \\dots, h_b)$ is obtained from $\\mathbf{g} = (g_1, g_2, \\dots, g_a)$ by replacing each 4 with two consecutive 2s.\nIf the sequence $\\mathbf{h}$ contains at least two 3s, consider the number of 2s between each pair of adjacent 3s. These numbers can differ by at most 1 (otherwise, in $\\mathbf{h}$ there is a segment with $t+2$ consecutive 2s, another with at most $t$ 2s between two 3s, with a total difference of at least 2, contradicting $B$ being a good subset), and all these numbers are even, so they are all equal. This makes the sequence $\\mathbf{h}$ have a smaller period, contradicting the total sum of $\\mathbf{h}$ being the prime number $p$. Therefore, $\\mathbf{h}$ contains one 3 and $b-1$ 2s, i.e., $p = 2b+1$. The sequence $\\mathbf{g}$ contains one 3 and $a-1$ 4s, i.e., $p = 4a-1$. In this case, take\n$$\nA = \\{1, 5, 9, \\dots, p-2\\} \\subset B = \\{1, 3, 5, \\dots, p-2\\}\n$$\nto prove Conclusion 5.\n\nCombining Conclusions 4 and 5, we know that when $2^u < p < 2^{u+1}$, for any chain $a_1 \\to a_2 \\to \\dots \\to a_L$, the sequence $a_1, a_2, a_3, \\dots$ before $\\frac{p}{4}$ multiplies by several times each step, so there are at most $u-1$ numbers (since $\\frac{p}{4} < 2^{u-1}$), thus $a_u > \\frac{p}{4}$. If $a_{u+1} < \\frac{p}{2}$, then necessarily $a_u = \\frac{p+1}{4}$, $a_{u+1} = \\frac{p-1}{2}$, in which case only $a_u = 2^{u-1}$ (otherwise $a_u \\ge 3 \\times 2^{u-2} > \\frac{p-1}{4}$), i.e., $p = 2^{u+1}-1$.\n\nIn summary, $L_{\\max} = 2 \\times \\lfloor \\log_2(p+1) \\rfloor$. Thus, when $2^u < p < 2^{u+1} - 1$, $L \\le 2u$, with equality in chain (2). When $p = 2^{u+1} - 1$, $L \\le 2(u+1)$, with equality in the following chain:\n$$\n1 \\to 2 \\to \\dots \\to 2^{u-1} \\to 2^u - 1 \\to 2^u \\to (p-2^{u-1}) \\to (p-2^{u-2}) \\to \\dots \\to (p-2) \\to (p-1).\n$$\nThe characterization of good subsets is similar to part (1) in Solution 1. The following discussion simplifies the subsequent proof. Notice that\n$$\n\\lfloor \\frac{kp}{m} \\rfloor \\cdot m \\equiv \\{\\frac{kp}{m}\\} \\cdot m \\quad \\mod p.\n$$\nAnd when $k = 0, 1, \\dots, m-1$, $\\{\\frac{kp}{m}\\}$ takes all values from $0, 1, \\dots, m-1$. Hence, modulo $p$, $\\{m\\lfloor \\frac{kp}{m} \\rfloor\\}$ is equal to $\\{0, -1, \\dots, -(m-1)\\}$. Due to the uniqueness of rotation, any $m$-element good subset multiplied by $m$ (in the sense of modulo $p$) consists of consecutive $m$ numbers. Conversely, if an $m$-element subset multiplied by $m$ consists of consecutive $m$ numbers, it must be an $m$-element good subset.\nLet $r$ be the modular multiplicative inverse of $m$ modulo $p$, then $\\{\\lfloor \\frac{kp}{m} \\rfloor\\} \\equiv \\{0, -r, \\dots, -(m-1)r\\} \\mod p$. Therefore, any $m$-element set is a good subset if and only if it can be written as $\\{a, a+r, \\dots, a+(m-1)r\\} \\mod p$.\n\nThe following steps are similar to Conclusions 1–5 in Solution 1, but the proofs are relatively simpler.\n\n**Conclusion 1:** The complement of any $m$-element good subset is a $p-m$-element good subset.\n*Proof of Conclusion 1:* Based on the above characterization, assume the $m$-element good subset is $\\{a, a+r, \\dots, a+(m-1)r\\}$ (in the sense of modulo $p$), then its complement can be written as $\\{a-r, a-2r, \\dots, a-(p-m)r\\}$ modulo $p$. Since the modular multiplicative inverse of $p-m$ modulo $p$ is $-r$ modulo $p$, the set $\\{a-r, a-2r, \\dots, a-(p-m)r\\}$ is a good subset.\n\n**Conclusion 2:** If $m < \\frac{p}{2}$, then any $m$-element good subset is contained in some $p-m$-element good subset.\n*Proof of Conclusion 2:* Without loss of generality, assume the $m$-element good subset is $\\{a, a+r, \\dots, a+(m-1)r\\}$ (in the sense of modulo $p$), then it is contained in the $p-m$-element good subset $\\{a+(p-m-1)r, a+(p-m-2)r, \\dots, a\\}$.\n\n**Conclusion 3:** If $km < p$, then any $m$-element good subset is contained in some $km$-element good subset.\n*Proof of Conclusion 3:* Assume the $m$-element good subset is $A = \\{a, a+r, \\dots, a+(m-1)r\\}$ (in the sense of modulo $p$). Let $s$ be the modular multiplicative inverse of $km$ modulo $p$, so $ks \\equiv r \\pmod p$. Then $A$ is contained in the $km$-element good subset $B = \\{a, a+s, \\dots, a+(km-1)s\\}$.\n\n**Conclusion 4:** If $p \\equiv 3 \\pmod 4$, then any $\\frac{p+1}{4}$-element good subset is contained in some $\\frac{p-1}{2}$-element good subset.\n*Proof of Conclusion 4:* Note that the modular multiplicative inverse of $\\frac{p+1}{4}$ modulo $p$ is 4, and for $\\frac{p-1}{2}$ modulo $p$ it is -2. Suppose the $\\frac{p+1}{4}$-element good subset is $\\{a, a+2, \\dots, a+p-3\\}$, then it is contained in the $\\frac{p-1}{2}$-element good subset $\\{a+p-3, a+p-5, \\dots, a+2, a\\}$.\n\n**Conclusion 5:** If $1 \\le n < m \\le \\frac{p}{2}$, and some $m$-element good subset is contained in some $n$-element good subset, then $m \\ge 2n - 1$, and equality holds if and only if $n = \\frac{p+1}{4}$ and $m = \\frac{p-1}{2}$.\n\n*Proof of Conclusion 5:* Multiply all numbers by $m$. Then, in the sense of modulo $p$, an $m$-element good subset becomes a sequence of consecutive $m$ numbers (since the original common difference was the modular inverse of $m$ modulo $p$), and the $n$-element good subset also becomes an arithmetic sequence $T$, with a common difference congruent to $\\frac{m}{n}$ modulo $p$, i.e., not congruent to $\\pm 1$ modulo $p$. Assume the transformed $m$-element good subset is $\\{0, 1, \\dots, m-1\\}$, with the first term $a$ of $T$ in $\\{0, 1, \\dots, m-1\\}$, and let the common difference be $\\beta$, where $|\\beta| < \\frac{p}{2}$. Note that $p - (m-1) > \\frac{p}{2}$, so every term of $T$ must be in $\\{0, 1, \\dots, m-1\\}$, hence $m = (m-1) + 1 \\ge (n-1)|\\beta| + 1 \\ge 2n-1$. When equality holds, $|\\beta| = 2$ must be true, thus $\\frac{m}{n} \\equiv \\pm 2 \\pmod p$, meaning $2n-1 \\equiv \\pm 2n \\pmod p$. As $n \\le \\frac{m+1}{2} \\le \\frac{p+1}{4}$, the only possibility is $n = \\frac{p+1}{4}$ and $m = \\frac{p-1}{2}$.\n\nNow returning to the original problem, for any chain of good subsets $A_1 \\subset A_2 \\subset \\dots \\subset A_L$, let $|A_q| < \\frac{p}{2}$ and $|A_{q+1}| > \\frac{p}{2}$. By Conclusion 5, we have\n$$\n\\frac{p+1}{2} \\ge 2|A_{q-1}| \\ge 2^2|A_{q-2}| \\ge \\dots \\ge 2^{q-1}|A_1| = 2^{q-1}.\n$$\nThus, $q \\le \\lfloor \\log_2(p+1) \\rfloor$. By Conclusion 1, the chain of complements $A_L^c \\subset A_{L-1}^c \\subset \\dots \\subset A_{q+1}^c$ also forms a chain of good subsets, and similarly, we find $L-q \\le \\lfloor \\log_2(p+1) \\rfloor$. Therefore, $L \\le 2\\lfloor \\log_2(p+1) \\rfloor$.\n\nWhen $2^q < p+1 < 2^{q+1}$, by Conclusions 1, 2, and 3, there exists a chain of good subsets $A_1 \\subset A_2 \\subset \\dots \\subset A_{2q}$, where $|A_i| = 2^{i-1}$, $|A_{2q+1-i}| = p-2^{i-1}$, for $i=1, 2, \\dots, q$.\n\nWhen $p=2^q-1$, by Conclusions 1, 2, 3, and 4, there exists a chain of good subsets $A_1 \\subset A_2 \\subset \\dots \\subset A_{2q}$, where $|A_i| = 2^{i-1}$, $|A_{2q+1-i}| = p-2^{i-1}$, for $i=1, 2, \\dots, q-1$, $|A_q| = 2^{q-1}-1$, and $|A_{q+1}| = 2^{q-1}$.\n\nIn summary, the maximum $L$ sought is $2\\lfloor \\log_2(p+1) \\rfloor$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 28589,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $a_1, a_2, \\dots, a_{2023}$ be nonnegative real numbers such that $a_1 + a_2 + \\dots + a_{2023} = 100$. Let $N$ denote the number of elements in the following set\n$$\n\\{(i, j) \\mid 1 \\le i \\le j \\le 2023, a_i a_j \\ge 1\\}.\n$$\nProve that $N \\le 5050$, and determine the necessary and sufficient condition for $N = 5050$.",
    "options": [],
    "answer": "N ≤ 5050, with equality if and only if exactly 100 of the numbers are equal to one and the remaining 1923 are zero.",
    "solution": "**Proof.** Let $S$ be the number of pairs $(i, j)$ that satisfy the following conditions:\n$$\n1 \\le i < j \\le 2023, \\quad a_i a_j \\ge 1.\n$$\nLet $T$ be the number of elements among $a_1, a_2, \\dots, a_{2023}$ that are not less than $1$. Then\n$$\n100 = a_1 + a_2 + \\dots + a_{2023} \\ge T, \\qquad (3)\n$$\n$$\n10000 = (a_1 + a_2 + \\dots + a_{2023})^2 = \\sum_{i=1}^{2023} a_i^2 + 2 \\sum_{1 \\le i < j \\le 2023} a_i a_j \\ge T + 2S. \\quad (4)\n$$\nAdding the two equations and dividing by $2$, we get $N = S + T \\le 5050$.\nIf equality holds, then equality in the above equation (2) holds, which means each $a_i^2$ is either $0$ or $1$. That is, among $a_1, a_2, \\dots, a_{2023}$, exactly $100$ are $1$s and $1923$ are $0$s. This is a necessary and sufficient condition.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 28590,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let three circles $\\odot O$, $\\odot O_1$, $\\odot O_2$, each externally tangent to the other two, lie on the same side of a line $\\ell$. Let them be tangent to $\\ell$ at points $A$, $A_1$, $A_2$ respectively, where point $A$ lies on segment $A_1A_2$. Denote the points of tangency of $\\odot O$ with $\\odot O_1$ and $\\odot O_2$ as $B_1$ and $B_2$ respectively, and the point of tangency of $\\odot O_1$ with $\\odot O_2$ as $C$. Let line $A_1C$ intersect line $A_2B_2$ at point $D_1$, and line $A_2C$ intersect line $A_1B_1$ at point $D_2$. Prove that line $D_1D_2$ is parallel to line $\\ell$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "*Proof* 1. Consider the antipodal point $P$ of $A$ on the circle $\\odot O$.\nWe notice that in the right trapezoid $A_1A_2O_1O_2$, we have $O_1A_1 = O_1B_1$ and $OB_1 = OA$.\nTherefore,\n$$\n\\begin{align*} \n\\angle A_1BA &= \\pi - \\angle O_1B_1A_1 - \\angle OB_1A \\\\ \n&= \\pi - \\frac{1}{2}(\\pi - \\angle A_1O_1O) - \\frac{1}{2}(\\pi - \\angle AOO_1) \\\\ \n&= \\frac{1}{2}(\\angle A_1O_1O + \\angle AOO_1) = \\frac{\\pi}{2}, \n\\end{align*}\n$$\nwhich implies that the line $A_1B_1$ passes through point $P$. Similarly, $A_2B_2$ also passes through point $P$.\nWe have,\n$$\n\\begin{align*} \n\\angle D_1B_2B_1 &= \\angle PB_2B_1 = \\angle PAB_1 \\quad (P, B_2, A, B_1 \\text{ concyclic}) \\\\ \n&= \\angle B_1A_1A = \\angle B_1CA_1 \\quad (\\ell \\text{ is tangent to } \\odot O_1) \\\\ \n&= \\angle B_1CD_1. \n\\end{align*}\n$$\nHence, points $C, D_1, B_1, B_2$ are concyclic. Similarly, $D_2$ also lies on the circumcircle of $\\triangle CB_1B_2$. Therefore,\n$$\n\\angle D_1D_2B_1 = \\angle D_1B_2B_1 = \\angle B_1A_1A.\n$$\nThis implies that $D_1D_2$ is parallel to $\\ell$. Proof completed. $\\Box$\n\n\n*Proof* 2. We establish a coordinate system with $\\ell$ as the $x$-axis and $A$ as the origin. Without loss of generality, let $O(0, 1)$. For $i = 1, 2$, let $r_i$ be the radius of $\\odot O_i$, which corresponds to the $y$-coordinate of point $O_i$. From the equation\n$$\nx_{O_i}^2 + (r_i - 1)^2 = (r_i + 1)^2,\n$$\nwe obtain $O_1(-2\\sqrt{r_1}, r_1)$ and $O_2(2\\sqrt{r_2}, r_2)$ (thus $A_1(-2\\sqrt{r_1}, 0)$ and $A_2(2\\sqrt{r_2}, 0)$). Furthermore, due to the tangent property of $\\odot O_1$ and $\\odot O_2$, we have\n$$\n[2(\\sqrt{r_1} + \\sqrt{r_2})]^2 + (r_1 - r_2)^2 = (r_1 + r_2)^2,\n$$\nwhich simplifies to\n$$\n(*) \\qquad \\sqrt{r_1} + \\sqrt{r_2} = \\sqrt{r_1} \\sqrt{r_2}.\n$$\nOn the other hand, we have\n$$\n\\overrightarrow{AB_1} = \\frac{1 \\cdot \\overrightarrow{AO_1} + r_1 \\cdot \\overrightarrow{AO}}{r_1 + 1},\n$$\nwhich yields $B_1(\\frac{-2\\sqrt{r_1}}{r_1+1}, \\frac{2r_1}{r_1+1})$. Similarly, we have $B_2(\\frac{2\\sqrt{r_2}}{r_2+1}, \\frac{2r_2}{r_2+1})$ and $C(\\frac{2(r_1\\sqrt{r_2}-r_2\\sqrt{r_1})}{r_1+r_2}, \\frac{2r_1r_2}{r_1+r_2})$.\n\n$$\ny = \\frac{\\frac{2r_1}{r_1+1}}{\\frac{-2\\sqrt{r_1}}{r_1+1} + 2\\sqrt{r_1}}(x + 2\\sqrt{r_1}) = \\frac{1}{\\sqrt{r_1}}x + 2,\n$$\nand the equation of line $CA_2$ is given by\n$$\ny = \\frac{\\frac{2r_1r_2}{r_1+r_2}}{\\frac{2(r_1\\sqrt{r_2}-r_2\\sqrt{r_1})}{r_1+r_2} - 2\\sqrt{r_2}} (x - 2\\sqrt{r_2}),\n$$\nwhich can be combined to obtain\n$$\ny = \\frac{2r_1r_2}{2\\sqrt{r_1r_2}(\\sqrt{r_1} - \\sqrt{r_2}) - 2\\sqrt{r_2}(r_1 + r_2)}[\\sqrt{r_1}(y - 2) - 2\\sqrt{r_2}].\n$$\nRearranging the terms, we obtain a linear equation in terms of $y$, with the constant term\n$$\n4r_1r_2(\\sqrt{r_1} + \\sqrt{r_2}),\n$$\nwhich is a symmetric expression in terms of $r_1$ and $r_2$. The coefficient of $y$ in this case is (using (*) to convert the expression into a homogeneous form)\n$$\n\\begin{aligned} & 2\\{r_1r_2\\sqrt{r_1} - \\sqrt{r_1r_2}(\\sqrt{r_1} - \\sqrt{r_2}) + \\sqrt{r_2}(r_1 + r_2)\\} \\\\ &= 2\\{(\\sqrt{r_1} + \\sqrt{r_2})^2\\sqrt{r_1} - \\sqrt{r_1r_2}(\\sqrt{r_1} - \\sqrt{r_2}) + \\sqrt{r_2}(r_1 + r_2)\\} \\\\ &= 2\\{r_1\\sqrt{r_1} + 2r_1\\sqrt{r_2} + r_2\\sqrt{r_1} - r_1\\sqrt{r_2} + r_2\\sqrt{r_1} + r_1\\sqrt{r_2} + r_2\\sqrt{r_2}\\} \\\\ &= 2\\{r_1\\sqrt{r_1} + 2r_1\\sqrt{r_2} + 2r_2\\sqrt{r_1} + r_2\\sqrt{r_2}\\}, \\end{aligned}\n$$\nwhich is also a symmetric expression in terms of $r_1$ and $r_2$.\nTherefore, we know that the $y$-coordinates of points $D_1$ and $D_2$ have the same expression, which proves the proposition. $\\square$\n\n\n*Proof 3.* (In fact, it is not necessary for circle $O$ to be tangent to line $\\ell$.)\nIt is easy to see that $O, B_1, O_1$ are collinear, $O, B_2, O_2$ are collinear, $O_1, C, O_2$ are collinear, $OB_1 = OB_2$, $O_1A_1 = O_1B_1$, $O_2A_2 = O_2B_2$, and $O_1A_1, O_2A_2$ are both perpendicular to $\\ell$.\nNow, we have:\n$$\n\\begin{aligned} \\angle CD_1B_2 &= \\angle CA_1A_2 + \\angle A_1A_2B_2 \\\\ &= \\angle CA_1B_1 + \\angle B_1A_1A_2 + \\angle A_1A_2B_2 \\\\ &= \\frac{1}{2}\\angle CO_1B_1 + \\frac{1}{2}\\angle A_1O_1B_1 + \\frac{1}{2}\\angle A_2O_2B_2 \\\\ &= (90^\\circ - \\angle O_1B_1C) + \\frac{1}{2}\\angle O_1OO_2 \\\\ &= (\\angle OB_1C - 90^\\circ) + 90^\\circ - \\angle OB_1B_2 \\\\ &= \\angle CB_1B_2. \\end{aligned}\n$$\nTherefore, $C, D_1, B_1, B_2$ are concyclic. Similarly, $C, D_2, B_2, B_1$ are concyclic. Hence, $C, D_1, B_1, B_2, D_2$ are concyclic. Furthermore, $\\angle CD_1D_2 = \\angle CB_1D_2 = 180^\\circ - \\angle A_1B_1C = \\angle CA_1A_2$. Thus, we have $D_1D_2//\\ell$. $\\square$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 28591,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all functions $f : \\mathbb{Z} \\to \\mathbb{Z}$ that satisfy the following equation for any integers $a, b, c$:\n$$\n2f(a^2 + b^2 + c^2) - 2f(ab + bc + ca) = (f(a - b))^2 + (f(b - c))^2 + (f(c - a))^2.\n$$",
    "options": [],
    "answer": "Two functions: f(n) = 0 for all integers n, and f(n) = n for all integers n.",
    "solution": "Taking $a = b = c = 0$, we have $3(f(0))^2 = 0$, which implies $f(0) = 0$.\n\nTaking $a = 1, b = 0, c = 0$, we have $2f(1) = (f(1))^2 + (f(-1))^2$. Thus, $(f(1) - 1)^2 + (f(-1))^2 = 1$. This means either $f(1) = 1$ and $f(-1) = \\pm 1$, or $f(-1) = 0$ and $f(1) = 0$ or $2$.\n\nTaking $a = 1, b = 1, c = 0$, we have $2f(2) - 2f(1) = (f(1))^2 + (f(-1))^2 = 2f(1)$.\nHence, $f(2) = 2f(1)$.\n\nTaking $a = 1, b = 0, c = -1$, we have $2f(2) - 2f(-1) = (f(2))^2 + 2(f(-1))^2$. Thus, $(f(2) - 1)^2 + 2(f(-1) + \\frac{1}{2})^2 = 1.5$. This implies $f(2) = 0$ or $2$, and $f(-1) = 0$ or $-1$.\n\nCombining the above, we conclude that either $f(1) = 1$ and $f(-1) = -1$, or $f(1) = 0$ and $f(-1) = 0$.\n\nLet $g(k) = f(k)^2 - f(-k)^2$. We have $g(-k) = -g(k)$ and $g(0) = g(1) = 0$.\nWhen we replace $(a, b, c)$ with $(b, c, a)$ in the original equation, the left side remains the same, and the change in the right side is exactly $g(a-b) + g(b-c) + g(c-a) = 0$. Taking $(a, b, c) = (k, 1, 0)$, we have $g(k-1)+g(1)+g(-k) = 0$, which implies $g(k) = g(k-1)+g(1)$. Therefore, we have $g(k) = 0$ for all $k \\in \\mathbb{Z}$, which means $f(-k) = \\pm f(k)$ for all $k \\in \\mathbb{Z}$.\n\nTaking $(a, b, c) = (k, -1, 0)$ and $(a, b, c) = (k, 1, 0)$ in the original equation and comparing the two resulting equations, we have\n$$\n(6) \\qquad 2(f(k) - f(-k)) = (f(k+1))^2 - (f(k-1))^2\n$$\n\nConsider the first case where $f(1) = f(-1) = 0$. We will prove by mathematical induction that $f(k) = f(-k) = 0$ for all non-negative integers $k$. The base cases $k = 0$ and $k = 1$ have been established. Assume that $f(\\pm k) = 0$ and $f(\\pm(k-1)) = 0$ hold. From equation (6), we obtain $(f(k+1))^2 = 0$, which implies $f(k+1) = 0$ and $f(-k-1) = \\pm f(k+1) = 0$. Thus, the induction hypothesis holds.\n\nTherefore, in the first case, we have $f(m) = 0$ for all $m \\in \\mathbb{Z}$.\n\nNow, consider the second case where $f(1) = 1$ and $f(-1) = -1$. We have $f(2) = 2f(1) = 2$.\n\nBy substituting $(a, b, c) = (k, 1, -1)$ and $(a, b, c) = (k, 1, 1)$ into the original equation and comparing the resulting equations, we obtain\n$$\n2f(2k + 1) - 2f(-1) = (f(-1 - k))^2 - (f(1 - k))^2 + (f(2))^2.\n$$\n$$\n(7) \\qquad f(2k + 1) = \\frac{1}{2}(f(k + 1))^2 - \\frac{1}{2}(f(k - 1))^2 + 1\n$$\n\nBy substituting $(a, b, c) = (k, 2, 0)$ and $(a, b, c) = (k, -2, 0)$ into the original equation and comparing the resulting equations, we have\n$$\n(8) \\qquad 2f(2k) - 2f(-2k) = (f(k + 2))^2 - (f(k - 2))^2\n$$\n\nBy substituting $k = 1$ into equation (7), we have $f(3) = 3$.\nBy substituting $k = 2$ into equation (8), we obtain $(f(4))^2 = 2(f(4) - f(-4))$. If $f(4) = 0$, then substituting $k = 3$ into equation (6) yields $f(3) - f(-3) = -2$, which implies $f(-3) = 5 \\neq \\pm f(3)$, leading to a contradiction. Therefore, $f(4) \\neq 0$. Since $f(-4) = \\pm f(4)$, specifically $f(-4) = -f(4)$, we have $(f(4))^2 = 4f(4)$, which implies $f(4) = 4$.\n\n---\n\nWe will now prove by mathematical induction that $f(m) = m$ for all non-negative integers $m$. The base cases $m = 0, 1, 2, 3, 4$ have been established. Assume that $f(k) = k$ holds for $k = 0, 1, 2, \\dots, m-1$.\nFor $m = 2k + 1 \\ge 5$, using equation (7), we obtain $f(2k + 1) = \\frac{1}{2}((f(k+1))^2 - (f(k-1))^2) + 1 = 2k + 1$.\nFor $m = 2k \\ge 6$, utilizing equation (8), we have $f(2k) - f(-2k) = \\frac{1}{2}((f(k+2))^2 - (f(k-2))^2) = 4k$. Furthermore, $f(-2k) = \\pm f(2k)$, which implies $2f(2k) = 4k$ and $f(2k) = 2k$.\n\nHence, we conclude that $f(m) = m$ for $m \\in \\mathbb{Z}_+$. For positive integers $m$, substituting $k = m$ into equation (6) gives us\n$$\n2(f(m) - f(-m)) = (f(m + 1))^2 - (f(m - 1))^2 = 4m, \\Rightarrow f(-m) = -m.\n$$\nTherefore, in the second case, we have $f(m) = m$ for all $m \\in \\mathbb{Z}$.\n\nUpon verification, we find that both solutions, $f(m) = 0$ and $f(m) = m$, satisfy the given conditions. $\\square$",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 28592,
    "subject": "Mathematics (Multi-modal)",
    "question": "As shown below, let $ABCD$ be a cyclic quadrilateral such that the diagonals $AC$ and $BD$ are perpendicular with intersection point $E$. Point $F$ is on the side $AD$, the ray $FE$ meets the circumscribed circle of $ABCD$ at point $P$. Point $Q$ is on the segment $PE$ such that $PQ \\cdot PF = PE^2$. The line through $Q$ perpendicular to $AD$ meets $BC$ at $R$. Show that $RP = RQ$.\n\n![](attached_image_1.png)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Proof: Let $EX \\parallel AF$, intersecting $AP$ at $X$, and $DP$ at $Y$. Extend $XQ$ and $YQ$ to intersect $BC$ at $S$ and $T$, respectively.\nFrom $\\frac{PQ}{PE} = \\frac{PE}{PF} = \\frac{PX}{PA}$, we get $XQ \\parallel AE$, similarly $YQ \\parallel DE$.\nSince $\\angle EXS = \\angle AEX = \\angle DAC = \\angle EBS$, we have $X, E, S, B$ are concyclic.\nFrom $\\angle PXE = \\angle PAD = \\angle PBE$, we find $X, E, P, B$ are concyclic, hence $X, E, S, P, B$ are concyclic. Similarly, $Y, E, T, P, C$ are concyclic.\nSince $\\angle PQT = \\angle PEB = \\angle PST$, we have $P, S, Q, T$ are concyclic. Given $SQ \\parallel CE, TQ \\parallel BE$, and $CE \\perp BE$, we get $SQ \\perp TQ$.\nSince $\\angle RQS = 90^\\circ - \\angle QXY = 90^\\circ - \\angle QTS = \\angle RSQ$, it is known that $R$ is the center of circle $\\odot PSQT$, thus $RP = RQ$. $\\square$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 28593,
    "subject": "Mathematics (Multi-modal)",
    "question": "For positive integers $m, n$, define\n$$\nS(m, n) = \\{(a, b) \\in \\mathbb{Z}^2 \\mid 1 \\le a \\le m, 1 \\le b \\le n, \\gcd(a, b) = 1\\}.\n$$\nProve: for any positive integers $d, r$, there exist integers $m, n$ not less than $d$, such that $|S(m, n)| \\equiv r \\pmod{d}$. Here, $|A|$ represents the number of elements in the finite set $A$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $n = d + r$ and $m = d \\cdot (d+r) + 1$. Then, for $1 \\le b \\le d+r$, the number of integers in the range $1, 2, \\dots, m$ that are coprime to $b$ is given by\n$$\n\\frac{\\varphi(b)}{b} \\cdot d \\cdot (d+r)! + 1.\n$$\nNote that $b$ divides $(d+r)!$, so this number is congruent to $1$ modulo $b$. Therefore,\n$$\n|S(m, n)| \\equiv \\sum_{b=1}^{d+r} 1 \\equiv r \\pmod{d}.\n$$",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 28594,
    "subject": "Mathematics (Multi-modal)",
    "question": "A nonempty set $A$ of integers is called a “beautiful set” if for any $a \\in A$ and $k \\in \\{1, 2, \\dots, 2023\\}$, the set\n$$\n\\{b \\in A \\mid \\lfloor \\frac{b}{3^k} \\rfloor = \\lfloor \\frac{a}{3^k} \\rfloor \\}\n$$\nhas exactly $2^k$ elements.\nProve that: if the intersection of an integer set $S$ and any beautiful set is not empty, then $S$ contains a beautiful set.\n\n*Note:* Here $\\lfloor x \\rfloor$ represents the greatest integer not exceeding $x$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "For a positive integer $n$, a non-empty set of integers $A$ is called an “order $n$ strong beautiful set” if $A \\subset \\{0, 1, \\dots, 3^n - 1\\}$ and for any $a \\in A$ and $1 \\le k \\le n$, we have $\\#\\{b \\in A \\mid \\lfloor 3^{-k}b \\rfloor = \\lfloor 3^{-k}a \\rfloor \\} = 2^k$. It is clear that a 2023-order strong beautiful set is a beautiful set. We will prove the following proposition by induction: For any positive integer $n$, if a set of integers $S$ has a non-empty intersection with every order $n$ strong beautiful set, then $S$ contains an order $n$ strong beautiful set. Taking $n = 2023$ in this proposition will imply the original problem.\n\nWhen $n = 1$, the order 1 strong beautiful sets are the binary subsets of $\\{0, 1, 2\\}$. It is easy to see that the proposition holds in this case. Suppose the proposition holds for $n = m$. We will prove that it also holds for $n = m + 1$. First, note that if $A_1$ and $A_2$ are $m$-order strong beautiful sets and $\\{i_1, i_2\\}$ is a binary subset of $\\{0, 1, 2\\}$, then the set\n$$\n\\{a + i_1 3^m \\mid a \\in A_1\\} \\cup \\{a + i_2 3^m \\mid a \\in A_2\\}\n$$\nis an $(m+1)$-order strong beautiful set. Let $S$ be a set of integers that has a non-empty intersection with every $(m+1)$-order strong beautiful set. Consider the sets\n$$\nS_i = \\{0 \\le a < 3^m \\mid a + i 3^m \\in S\\}, \\quad i = 0, 1, 2.\n$$\nWe claim that there exists a binary subset $\\{i_1, i_2\\}$ of $\\{0, 1, 2\\}$ such that both $S_{i_1}$ and $S_{i_2}$ have a non-empty intersection with every $m$-order strong beautiful set. If not, then there exist a binary subset $\\{j_1, j_2\\}$ of $\\{0, 1, 2\\}$ and $m$-order strong beautiful sets $B_1$ and $B_2$ such that $S_{j_1} \\cap B_2 = \\emptyset$ for $s = 1, 2$. As a result, the $(m+1)$-order strong beautiful set $\\{a + j_1 3^m \\mid a \\in B_1\\} \\cup \\{a + j_2 3^m \\mid a \\in B_2\\}$ does not intersect with $S$, which is a contradiction. By the induction hypothesis, $S_{i_1}$ contains an $m$-order strong beautiful set $A_1$, and $S_{i_2}$ contains an $m$-order strong beautiful set $A_2$. Therefore, $S$ contains an $(m+1)$-order strong beautiful set $\\{a + i_1 3^m \\mid a \\in A_1\\} \\cup \\{a + i_2 3^m \\mid a \\in A_2\\}$. $\\square$\nFor a positive integer $n$ and an integer $m$, a non-empty set of integers $A$ is called an “$n$-th order $m$-basic beautiful set” if $A \\subset \\{3^m m, 3^m m + 1, \\dots, 3^n(m+1) - 1\\}$ and for any $a \\in A$ and $1 \\le k \\le n$, we have $\\#\\{b \\in A \\mid \\lfloor 3^{-k}b \\rfloor = \\lfloor 3^{-k}a \\rfloor \\} = 2^k$. It is clear that for any integer $m$, any 2023-th order $m$-basic beautiful set is a beautiful set. We will prove the proposition $P_n$ by induction: for any set of integers $S$ and any integer $m$, there exists an $n$-th order $m$-basic beautiful set $A$ such that either $A \\subset S$ or $A \\subset \\bar{S}$ (where $\\bar{S} = \\mathbb{Z} \\setminus S$ denotes the complement of $S$). If $P_{2023}$ is true, then the set $S$ or its complement $\\bar{S}$ contains a beautiful set. Since $S \\cap A \\ne \\emptyset$ by assumption, it follows that $A \\subset S$, and the original problem is solved.\n\nWhen $n = 1$, by the pigeonhole principle, the set $\\{3m, 3m + 1, 3m + 2\\}$ contains at least two elements $a$ and $b$ that belong to either $S$ or $\\bar{S}$. Then $A = \\{a, b\\}$ is a 1-st order $m$-basic beautiful set, and $A \\subset S$ or $A \\subset \\bar{S}$. Thus, the proposition $P_1$ holds. Assume that the proposition $P_n$ holds. We will prove that the proposition $P_{n+1}$ also holds. For any set of integers $S$ and any integer $m$, by the induction hypothesis, there exists an $n$-th order $3m$-basic beautiful set $A_1 \\subset [3^n \\cdot 3m, 3^n(3m+1))$, an $n$-th order $(3m+1)$-basic beautiful set $A_2 \\subset [3^n(3m+1), 3^n(3m+2))$, and an $n$-th order $(3m+2)$-basic beautiful set $A_3 \\subset [3^n(3m+2), 3^n(3m+3))$, which satisfy that for any $i \\in \\{1, 2, 3\\}$, either $A_i \\subset S$ or $A_i \\subset \\bar{S}$. According to the pigeonhole principle, at least two sets among $A_1, A_2, A_3$, denoted as $A_{i_1}$ and $A_{i_2}$ ($1 \\le i_1 < i_2 \\le 3$), are both contained in $S$ or $\\bar{S}$. Define $A = A_{i_1} \\cup A_{i_2}$, it can be easily verified that $A$ is an $(n+1)$-th order $m$-basic beautiful set, and either $A \\subset S$ or $A \\subset \\bar{S}$. Therefore, the proposition $P_{n+1}$ holds. $\\square$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 28595,
    "subject": "Mathematics (Multi-modal)",
    "question": "Given a positive integer $n$ and $n^3$ integers $a_{ijk} \\in \\{1, -1\\}$ ($1 \\le i, j, k \\le n$). Prove that there exist\n$$\nx_1, \\dots, x_n, y_1, \\dots, y_n, z_1, \\dots, z_n \\in \\{1, -1\\},\n$$\nsuch that the following inequality holds\n$$\n\\left| \\sum_{i=1}^{n} \\sum_{j=1}^{n} \\sum_{k=1}^{n} a_{ijk} x_i y_j z_k \\right| > \\frac{n^2}{3}\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "*Proof 1.* For any $(x_i)$ and $(y_j)$ satisfying the given conditions, we define\n$$\nX_k = \\sum_{i,j=1}^{n} a_{ijk} x_i y_j.\n$$\nSince we can always choose $z_k$ with the same sign as $X_k$, we have\n$$\n\\sum_{i=1}^{n} \\sum_{j=1}^{n} \\sum_{k=1}^{n} a_{ijk} x_i y_j z_k = \\sum_{k=1}^{n} |X_k|.\n$$\nNote that (summing over all possible $(x_i)$ and $(y_j)$)\n$$\n\\begin{aligned} \\sum_{(x_i),(y_j)} |X_k|^2 &= \\sum_{(x_i),(y_j)} \\sum_{i_1,i_2} \\sum_{j_1,j_2} a_{i_1j_1k} a_{i_2j_2k} x_{i_1} x_{i_2} y_{j_1} y_{j_2} \\\\ &= \\sum_{i_1,i_2} \\sum_{j_1,j_2} a_{i_1j_1k} a_{i_2j_2k} \\sum_{(x_i),(y_j)} x_{i_1} x_{i_2} y_{j_1} y_{j_2}, \\end{aligned}\n$$\nand the last line has non-zero value only when $i_1 = i_2$ and $j_1 = j_2$, so\n$$\n\\sum_{(x_i),(y_j)} |X_k|^2 = (2^n)^2 \\sum_{i,j=1}^{n} a_{ijk}^2 = 2^{2n} n^2.\n$$\n(To obtain a lower bound estimate for $\\sum_{(x_i),(y_j)} |X_k|$, we need to estimate $\\sum_{(x_i),(y_j)} |X_k|^n$ for some large $n$ greater than 2.) Similarly, we have\n$$\n\\sum_{(x_i),(y_j)} |X_k|^4 = \\sum_{i_1,i_2,i_3,i_4} \\sum_{j_1,j_2,j_3,j_4} a_{i_1j_1k} a_{i_2j_2k} a_{i_3j_3k} a_{i_4j_4k} \\sum_{(x_i),(y_j)} x_{i_1} x_{i_2} x_{i_3} x_{i_4} y_{j_1} y_{j_2} y_{j_3} y_{j_4}.\n$$\nIn the last summation, it is non-zero only when $(i_1, i_2, i_3, i_4)$ and $(j_1, j_2, j_3, j_4)$ are paired up in pairs, and each term is repeated when all 4 items are the same, so we have\n$$\n\\sum_{i_1,i_2,i_3,i_4} \\sum_{j_1,j_2,j_3,j_4} a_{i_1j_1k} a_{i_2j_2k} a_{i_3j_3k} a_{i_4j_4k} \\sum_{(x_i),(y_j)} x_{i_1} x_{i_2} x_{i_3} x_{i_4} y_{j_1} y_{j_2} y_{j_3} y_{j_4} < 9n^4(2^n)^2.\n$$\nNow, by Hölder's inequality,\n$$\n\\left( \\sum_{(x_i),(y_j)} \\left( |X_k|^{2/3} \\right)^{3/2} \\right)^{2/3} \\left( \\sum_{(x_i),(y_j)} \\left( |X_k|^{4/3} \\right)^3 \\right)^{1/3} \\geq \\sum_{(x_i),(y_j)} |X_k|^2 = 2^{2n} n^2,\n$$\nso\n$$\n\\left( \\sum_{(x_i),(y_j)} |X_k| \\right)^{2/3} \\cdot (2^{2n} \\cdot 9n^4)^{1/3} > 2^{2n} n^2.\n$$\nHence,\n$$\n\\sum_{(x_i),(y_j)} |X_k| > \\left( (2^{2n})^{2/3} \\frac{n^{2/3}}{3^{2/3}} \\right)^{3/2} = 2^{2n} \\cdot \\frac{n}{3},\n$$\nsumming over $k$ yields\n$$\n(*) \\qquad \\sum_{(x_i),(y_j)} \\sum_k |X_k| > 2^{2n} \\cdot \\frac{n^2}{3},\n$$\nwhich implies the existence of $(x_i, y_j)$ such that $\\sum_k |X_k| > \\frac{n^2}{3}$. The proposition holds. $\\Box$\n\n*Note:* The last part using Hölder's inequality can also be proved using the lemma. When $X \\ge 0$, $(X - 3)^2(X + 6)X = X^4 - 27X^2 + 54X \\ge 0$. Hence, we have $X^4 - 27n^2X^2 + 54n^3X \\ge 0$. Thus,\n$$\n\\sum_{(x_i),(y_j)} |X_k|^4 - 27n^2 \\sum_{(x_i),(y_j)} |X_k|^2 + 54n^3 \\sum_{(x_i),(y_j)} |X_k| \\ge 0.\n$$\nTherefore,\n$$\n\\sum_{(x_i),(y_j)} |X_k| \\ge \\frac{1}{54} \\cdot 2^{2n} \\cdot \\frac{(27n^2 \\cdot n^2 - 9n^4)}{n^3} = \\frac{1}{3} \\cdot 2^{2n} n.\n$$\nSumming over $k$ yields $(*)$.\n\n\n*Proof 2.* We first prove two lemmas.\n**Lemma 1:** Let $n$ be a positive integer, and let $a_1, a_2, \\dots, a_n$ be real numbers. Then we have\n$$\n\\sum_{x_1,x_2,\\dots,x_n \\in \\{-1,1\\}} \\left| \\sum_{i=1}^n a_i x_i \\right| \\ge 2 \\binom{n-1}{\\lfloor \\frac{n-1}{2} \\rfloor} \\sum_{i=1}^n |a_i|.\n$$\n*Proof of Lemma 1:* Since each $x_i$ takes values from $-1, 1$, replacing $a_i$ with $|a_i|$ does not change the original expression. Therefore, without loss of generality, we can assume that $a_i \\ge 0$ for $i = 1, 2, \\dots, n$. Notice that when all $x_i$ simultaneously change sign, the value of $|\\sum_{i=1}^n a_i x_i|$ remains the same. Thus, we have:\n$$\n\\begin{aligned} \\sum_{x_1,x_2,\\dots,x_n \\in \\{-1,1\\}} \\left| \\sum_{i=1}^n a_i x_i \\right| &\\ge \\sum_{\\substack{x_1,x_2,\\dots,x_n \\in \\{-1,1\\} \\\\ x_1+x_2+\\dots+x_n \\ne 0}} \\left| \\sum_{i=1}^n a_i x_i \\right| \\\\ &= 2 \\sum_{\\substack{x_1,x_2,\\dots,x_n \\in \\{-1,1\\} \\\\ x_1+x_2+\\dots+x_n > 0}} \\left| \\sum_{i=1}^n a_i x_i \\right| \\\\ &\\ge 2 \\sum_{\\substack{x_1,x_2,\\dots,x_n \\in \\{-1,1\\} \\\\ x_1+x_2+\\dots+x_n > 0}} \\sum_{i=1}^n a_i x_i \\\\ &= 2 \\sum_{i=1}^n \\left( \\sum_{\\substack{x_1,x_2,\\dots,x_n \\in \\{-1,1\\} \\\\ x_1+x_2+\\dots+x_n > 0}} x_i \\right) a_i. \\end{aligned}\n$$\nFurthermore, notice that due to symmetry, the value of\n$$\n\\sum_{\\substack{x_1, x_2, \\dots, x_n \\in \\{-1, 1\\} \\\\ x_1 + x_2 + \\dots + x_n > 0}} x_i\n$$\ndoes not depend on $i$. Let's consider the case when $i = n$. The value of this sum is given by\n$$\n\\begin{aligned} \\sum_{\\substack{x_1, x_2, \\dots, x_n \\in \\{-1, 1\\} \\\\ x_1 + x_2 + \\dots + x_n > 0}} x_n &= \\sum_{\\substack{x_1, x_2, \\dots, x_{n-1} \\in \\{-1, 1\\} \\\\ x_1 + x_2 + \\dots + x_{n-1} > -1}} 1 + \\sum_{\\substack{x_1, x_2, \\dots, x_{n-1} \\in \\{-1, 1\\} \\\\ x_1 + x_2 + \\dots + x_{n-1} > 1}} (-1) \\\\ &= \\sum_{\\substack{x_1, x_2, \\dots, x_{n-1} \\in \\{-1, 1\\} \\\\ x_1 + x_2 + \\dots + x_{n-1} \\in \\{0, 1\\}}} 1 \\\\ &= \\binom{n-1}{\\lfloor \\frac{n-1}{2} \\rfloor}. \\end{aligned}\n$$\nThe final step is due to the fact that $x_1, x_2, \\dots, x_{n-1} \\in -1, 1$ satisfy $x_1 + x_2 + \\dots + x_{n-1} \\in 0, 1$ if and only if exactly $\\lfloor \\frac{n-1}{2} \\rfloor$ numbers among $x_1, x_2, \\dots, x_{n-1}$ take the value 1 and exactly $\\lfloor \\frac{n-1}{2} \\rfloor$ numbers take the value -1. Thus, Lemma 1 is proven.\n\n**Lemma 2:** For any positive integer $n$, we have\n$$\n\\binom{n-1}{\\lfloor \\frac{n-1}{2} \\rfloor} \\ge \\frac{2^{n-1}}{\\sqrt{2n}}.\n$$\n*Proof of Lemma 2:* It can be easily verified that the inequality holds for $n = 1, 2, 3$ (with equality holding for $n = 2$). For even $n = 2m \\ge 4$, the inequality is equivalent to $\\binom{2m}{m} \\ge \\frac{2^{2m}}{\\sqrt{4m+2}}$; for odd $n = 2m + 1 \\ge 5$, the inequality is equivalent to $\\binom{2m}{m} \\ge \\frac{2^{2m}}{\\sqrt{4m+2}}$. Thus, it suffices to prove that for any integer $m \\ge 2$, we have\n$$\n\\binom{2m}{m} \\ge \\frac{2^{2m-1}}{\\sqrt{m}}.\n$$\nIndeed, note that\n$$\n\\binom{2m}{m} = \\prod_{k=1}^{m} \\frac{2k(2k-1)}{k^2} = 2^{2m} \\prod_{k=1}^{m} \\frac{2k-1}{2k}.\n$$\nLet\n$$\nA = \\prod_{k=2}^{m} \\frac{2k-1}{2k}, \\quad B = \\prod_{k=2}^{m} \\frac{2k-2}{2k-1},\n$$\nthen $A > B$ and $AB = \\frac{2}{2m} = \\frac{1}{m}$. Hence, $A > \\frac{1}{\\sqrt{m}}$, which implies\n$$\n\\binom{2m}{m} = 2^{2m} \\cdot \\frac{1}{2} A > 2^{2m} \\cdot \\frac{1}{2\\sqrt{m}} = \\frac{2^{2m-1}}{\\sqrt{m}}.\n$$\nLemma 2 is proven.\n\nFrom the two lemmas above, we immediately obtain the following result: for any $n$ real numbers $a_1, a_2, \\dots, a_n$, we have\n$$\n\\sum_{x_1, x_2, \\dots, x_n \\in \\{-1, 1\\}} \\left| \\sum_{i=1}^{n} a_i x_i \\right| \\ge \\frac{2^{2n}}{\\sqrt{2n}} \\sum_{i=1}^{n} |a_i|.\n$$\nBy applying this result twice, we can conclude that for any $n^2$ real numbers $a_{ij}$, ($i, j = 1, 2, \\dots, n$), we have\n$$\n(*) \\qquad \\sum_{x_1, \\dots, x_n, y_1, \\dots, y_n \\in \\{-1, 1\\}} \\left| \\sum_{i=1}^{n} \\sum_{j=1}^{n} a_{ij} x_i y_j \\right| &= \\sum_{x_1, \\dots, x_n \\in \\{-1, 1\\}} \\left( \\sum_{y_1, \\dots, y_n \\in \\{-1, 1\\}} \\left| \\sum_{j=1}^{n} \\left( \\sum_{i=1}^{n} a_{ij} x_i \\right) y_j \\right| \\right) \\\\\n&\\geq \\sum_{x_1, \\dots, x_n \\in \\{-1, 1\\}} \\left( \\frac{2^n}{\\sqrt{2n}} \\sum_{j=1}^{n} \\left| \\sum_{i=1}^{n} a_{ij} x_i \\right| \\right) \\\\\n&= \\frac{2^n}{\\sqrt{2n}} \\sum_{j=1}^{n} \\sum_{x_1, \\dots, x_n \\in \\{-1, 1\\}} \\left| \\sum_{i=1}^{n} a_{ij} x_i \\right| \\\\\n&\\geq \\frac{2^n}{\\sqrt{2n}} \\sum_{j=1}^{n} \\frac{2^n}{\\sqrt{2n}} \\sum_{i=1}^{n} |a_{ij}| \\\\\n&= \\frac{2^{2n}}{2n} \\sum_{i=1}^{n} \\sum_{j=1}^{n} |a_{ij}|.\n$$\nBack to the original question, for a fixed set of $x_1, \\dots, x_n, y_1, \\dots, y_n$, we can choose $z_k \\in \\{-1, 1\\}$ such that $z_k \\sum_{i=1}^n \\sum_{j=1}^n a_{ijk} x_i y_j \\ge 0$ for $k=1, 2, \\dots, n$. In this case,\n$$\n\\left| \\sum_{i=1}^{n} \\sum_{j=1}^{n} \\sum_{k=1}^{n} a_{ijk} x_i y_j z_k \\right| = \\sum_{k=1}^{n} \\left| \\sum_{i=1}^{n} \\sum_{j=1}^{n} a_{ijk} x_i y_j \\right| \\stackrel{\\text{denoted as}}{=} T_{x_1, \\dots, x_n, y_1, \\dots, y_n}.\n$$\nAccording to (*), we have\n$$\n\\sum_{x_1, \\dots, x_n, y_1, \\dots, y_n \\in \\{-1, 1\\}} T_{x_1, \\dots, x_n, y_1, \\dots, y_n} \\ge \\sum_{k=1}^{n} \\frac{2^{2n}}{2n} \\sum_{i=1}^{n} \\sum_{j=1}^{n} |a_{ijk}| = 2^{2n} \\cdot \\frac{n^2}{2}.\n$$\nThus, by the principle of averages, there exists a set of $x_1, \\dots, x_n, y_1, \\dots, y_n \\in \\{-1, 1\\}$ such that\n$$\nT_{x_1, \\dots, x_n, y_1, \\dots, y_n} \\ge \\frac{n^2}{2}.\n$$\nTherefore, the original problem is proved. $\\square$",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 28596,
    "subject": "Mathematics (Multi-modal)",
    "question": "Does there exist a sequence of pairwise distinct integers $a_1, a_2, \\dots$ that satisfies both of the following conditions?\n\na. For all positive integers $k$, we have $a_{k^2} > 0$ and $a_{k^2+k} < 0$.\n\nb. For all positive integers $n$, we have $|a_{n+1} - a_n| \\le 2023\\sqrt{n}$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "*Proof.* Such a sequence does not exist. We prove this by contradiction. Suppose there exists such a sequence. Take a positive integer $N$ satisfying\n$$\n\\frac{1}{N+1} + \\frac{1}{N+2} + \\dots + \\frac{1}{N^2} > 2024.\n$$\nSuch an $N$ exists because\n$$\n\\sum_{k=N+1}^{N^2} \\frac{1}{k} \\ge \\int_{N+1}^{N^2+1} \\frac{1}{k} = \\ln \\frac{N^2+1}{N+1} > \\ln(N-1),\n$$\nwhich can be arbitrarily large.\nWe prove that at least $4046N^2 + 2$ elements of $a_1, a_2, \\dots$ fall into the interval $S = [-2023N^2, 2023N^2]$, which contradicts the assumption that $a_i$'s are all distinct. To show this, it suffices to prove that for $k = N, N+1, \\dots, N^2-1$,\n(i) At least $\\lfloor \\frac{N^2}{k+1} \\rfloor$ elements of $a_{k^2}, \\dots, a_{k^2+k-1}$ fall into $S$.\n(ii) At least $\\lfloor \\frac{N^2}{k+1} \\rfloor$ elements of $a_{k^2+k}, \\dots, a_{k^2+2k}$ fall into $S$.\nIn this way, the total number of elements in $S$ is greater than or equal to\n$$\n2 \\sum_{k=N+1}^{N^2} \\left\\lfloor \\frac{N^2}{k} \\right\\rfloor \\ge 2N^2 \\sum_{k=N+1}^{N^2} \\frac{1}{k} - 2(N^2 - N) > 2N^2 \\cdot 2024 - 2N^2 + 2N > 4046N^2 + 2.\n$$\nWe will only prove (i), and the proof of (ii) is similar. Note that for $\\ell \\in \\{k^2, k^2 + 1, \\dots, k^2 + k - 1\\}$, we have\n$$\n|a_{\\ell+1} - a_{\\ell}| \\le 2023(k+1).\n$$\nWe consider the following three cases.\n(a) If there exists an $a_\\ell \\ge 2023N^2$ in the sequence $a_{k^2}, \\dots, a_{k^2+k-1}$, then (*) implies that there are at least $\\lfloor \\frac{2023N^2}{2023(k+1)} \\rfloor$ elements in $[0, 2023N^2]$ among $a_\\ell, a_{\\ell+1}, \\dots, a_{k^2+k-1}$.\n(b) If there exists an $a_\\ell \\le -2023N^2$ in the sequence $a_{k^2}, \\dots, a_{k^2+k-1}$, then (*) implies that there are at least $\\lfloor \\frac{2023N^2}{2023(k+1)} \\rfloor$ elements in $[-2023N^2, 0]$ among $a_{k^2}, a_{k^2+1}, \\dots, a_\\ell$.\n(c) If neither (a) nor (b) holds, then all the elements in the sequence $a_{k^2}, \\dots, a_{k^2+k-1}$ are within $S$, and there are a total of $k \\ge \\lfloor \\frac{N^2}{k+1} \\rfloor$ elements.\nThis completes the proof of (i), and the proof of (ii) can be similarly established. Therefore, a contradiction is reached, and thus such a sequence does not exist. $\\square$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 28597,
    "subject": "Mathematics (Multi-modal)",
    "question": "As shown in the figure, in an acute triangle $ABC$ with $AB < AC$, let $AH$ be its altitude and $G$ the barycentre. Let $P, Q$ be the tangent points of the incircle to $AB, AC$, respectively. Let $M, N$ be the midpoint of $BP, CQ$, respectively. Let $D, E$ be two points lying on the incircle of the triangle $ABC$ such that\n$$\n\\angle BDH + \\angle ABC = 180^\\circ, \\quad \\angle CEH + \\angle ACB = 180^\\circ.\n$$\nProve that the lines $MD, NE, GH$ are concurrent.\n![](attached_image_1.png)\n",
    "options": [],
    "answer": "Detailed solution",
    "solution": "**Proof:** On the circumcircle of $\\triangle ABC$, a point $F$ is chosen such that $ABCF$ forms an isosceles trapezoid. The line $FH$ intersects the circumcircle of $\\triangle ABC$ at another point $L$, and intersects the median $AK$ at point $G'$. As shown in the figure.\nSince $AF = 2HK$, it follows that $\\frac{AG'}{G'K} = \\frac{AF}{HK} = 2$, implying that $G'$ is the centroid of $\\triangle ABC$, hence $G' = G$. Therefore,\n$$\n\\angle BLH = \\angle BLF = \\frac{1}{2}\\widehat{BF} = \\frac{1}{2}\\widehat{AC} = \\angle ABC,\n$$\nGiven $\\angle BDH + \\angle ABC = 180^\\circ$, it follows that $\\angle BDH + \\angle BLH = 180^\\circ$, which means points $B, L, H, D$ are concyclic.\nSimilarly, it can be proved that $\\angle CLH = \\angle ACB$, and points $C, L, H, E$ are concyclic.\nSince $\\angle BLH = \\angle ABH$, $PB$ is tangent to circle $\\odot BLHD$ at point $B$. Let the incircle of $\\triangle ABC$ be $\\omega$, and $PB$ is an external common tangent of $\\omega$ and $\\odot BLHD$. Given $MP = MB$, it is known that $M$ is the radical center of $\\omega$ and $\\odot BLHD$, thus line $MD$ is the radical axis of $\\omega$ and $\\odot BLHD$.\nSimilarly, it can be proved that line $NE$ is the radical axis of $\\omega$ and $\\odot CLHE$. Moreover, line $GH$ is the radical axis of $\\odot BLHD$ and $\\odot CLHE$, therefore lines $MD, NE, GH$ either intersect at a single point, or are pairwise parallel.\nIf $MD, NE, GH$ are pairwise parallel, then the centers $O_1, O_2$ of $\\odot BLHD, \\odot CLHE$, and the center $I$ of $\\omega$ are collinear. Since both $\\angle BDH$ and $\\angle CEH$ are obtuse, $O_1, O_2$ are below $BC$, obviously $I$ is above $BC$. Let the projections of $O_1, O_2,$\n\n$I$ on $BC$ be $X, Y, Z$ respectively, then $X, Y$ are the midpoints of $BH, CH$ respectively. Given $AB < AC$, it is known that $Y, Z$ are on the same side of $AH$, and\n$$\nCZ = \\frac{AC + BC - AB}{2} > \\frac{BC}{2} > \\frac{CH}{2} = CY.\n$$\nHence, $Z$ lies on the segment $XY$, therefore $O_1, O_2, I$ cannot be collinear, a contradiction. Thus, $MD, NE, GH$ intersect at a single point. $\\square$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 28598,
    "subject": "Mathematics (Multi-modal)",
    "question": "Suppose $P_1, \\dots, P_n$ are $n$ points inside triangle $ABC$, such that any three points among $P_1, \\dots, P_n, A, B, C$ are not collinear. Prove: It is possible to divide $\\triangle ABC$ into a union of $2n+1$ smaller triangles such that each smaller triangle has its vertices among $P_1, \\dots, P_n, A, B, C$, and the number of smaller triangles with vertices including at least one point among $A, B, C$ is no less than $n + \\sqrt{n} + 1$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "*Proof.* For the edge $BC$, a partial order $\\prec_A$ can be defined on the set $P = P_1, \\cdots, P_n$ as follows:\n$$\nP_i \\prec_A P_j \\iff \\begin{aligned}[t] & P_i \\text{ is an interior point of the } \\triangle P_j BC \\\\ & \\text{the ray } P_j P_i \\text{ intersects with the segment } BC. \\end{aligned}\n$$\nSimilarly, a partial order $\\prec_B$ can be defined on $P$ using the edge $CA$. If two distinct points $P_i, P_j$ are incomparable under $\\prec_A$, then the line $P_i P_j$ does not intersect with the segment $BC$. Since the line $P_i P_j$ intersects with the boundary of $\\triangle ABC$ at two points, it follows that $P_i, P_j$ are comparable under $\\prec_B$. Thus, the anti-chains in the order $\\prec_A$ are chains in the order $\\prec_B$.\nBy using the Dilworth theorem, under the partial order $\\prec_A$, there either exists a chain of length at least $\\sqrt{n}$, or there exists an anti-chain of length at least $\\sqrt{n}$. Combining with the above-mentioned property that an anti-chain in the order $\\prec_A$ is a chain in the order $\\prec_B$, we know that there either exists a chain of length at least $\\sqrt{n}$ under the partial order $\\prec_A$, or there exists a chain of length at least $\\sqrt{n}$ under the partial order $\\prec_B$.\nWithout loss of generality, suppose there exists a chain of length at least $\\sqrt{n}$ under the partial order $\\prec_A$, denoted as $P_1 \\prec_A \\cdots \\prec_A P_t$, where $t \\ge \\sqrt{n}$. Connect the segments\n$$\nBP_i, CP_i (1 \\le i \\le t), P_1 P_2, \\cdots, P_{t-1} P_t, P_t A,\n$$\nwhich partition $\\triangle ABC$ into $2t + 1$ triangles\n$$\n(*) \\quad \\triangle BCP_1, \\triangle BP_i P_{i+1}, \\triangle CP_i P_{i+1} (1 \\le i \\le t-1), \\triangle BP_t A, \\triangle CP_t A.\n$$\nLet the number of points from set $P$ in the interiors of $\\triangle BCP_1, \\triangle BP_1 P_2, \\cdots, \\triangle BP_t A$ be $k_0, k_1, \\cdots, k_t$, respectively. For each such triangle $\\triangle$, denoted as $\\triangle BQR$, suppose the interior points from set $P$ in $\\triangle$ are $U_j (1 \\le j \\le k)$, where $\\angle QBU_j (1 \\le j \\le k)$ are arranged in increasing order. Connect the segments $BU_j (1 \\le j \\le k)$ and $QU_1, U_1 U_2, \\cdots, U_k R$, yielding $k + 1$ small triangles that all include $B$, and divide the polygon $QU_1 \\cdots U_k R$ into a union of $k$ small triangles arbitrarily (as it is well-known that any polygon can be triangulated). This way, $\\triangle$ is partitioned into the union of $2k + 1$ small triangles, $k + 1$ of which include $B$ as a vertex. Similarly, suppose the number of points from set $P$ in the interiors of $\\triangle CP_1 P_2, \\cdots, \\triangle CP_t A$ are $\\ell_1, \\cdots, \\ell_t$, respectively. For each such triangle $\\triangle'$, with $\\ell$ points from set $P$ in its interior, $\\triangle'$ can be partitioned into a union of $2\\ell + 1$ small triangles, $\\ell + 1$ of which include $C$ as a vertex.\nThis way, we have constructed a triangulation of $\\triangle ABC$, where at least\n$$\nT = \\sum_{i=0}^{t} (k_i + 1) + \\sum_{i=1}^{t} (\\ell_i + 1)\n$$\nof the small triangles include a point from $\\{B, C\\}$ as a vertex. Noting that $k_0 + k_1 + \\cdots + k_t + \\ell_1 + \\cdots + \\ell_t = n - t$, we get $T = (n - t) + (2t + 1) = n + t + 1 \\ge \\sqrt{n} + 1$, which proves the conclusion of the theorem. $\\square$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 28599,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $\\triangle ABC$ be an acute triangle, and let $K$ be a point on the extension ray of $BC$ (so that $C$ lies on the segment $BK$). Let $P$ be a point such that $BP = BK$ and $PK \\parallel AB$, and let $Q$ be a point such that $CQ = CK$ and $QK \\parallel AC$. Assume that the circumcircle of $\\triangle PQK$ and the line $AK$ intersects at another point $T$.\n\n(1) Prove that $\\angle APB + \\angle BTC = \\angle CQA$.\n\n(2) Prove that $AP \\cdot BT \\cdot CQ = AQ \\cdot BP \\cdot CT$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "**Proof 1.** (1) Let $A_1$ be the reflection of $A$ across $B$. By the given conditions, $KPA_1A$ forms an isosceles trapezoid, thus $\\angle APB = \\angle A_1KB$.\nSimilarly, let $A_2$ be the reflection of $A$ across $C$. We have $\\angle AQC = \\angle A_2KC = \\angle A_2KB$.\nHence, the desired conclusion is equivalent to $\\angle BTC = \\angle A_1KA_2$.\nPerform a homothety centered at $A$ with a ratio of $1/2$ on $\\triangle A_1KA_2$, mapping $A_1$ to $B$, $A_2$ to $C$, and $K$ to the midpoint of $AK$ (denoted as $M$). The proposition then transforms to $\\angle BTC = \\angle BMC$, i.e., points $B, C, M$, and $T$ are concyclic.\nBy the Power of a Point Theorem, this is equivalent to proving $KB \\cdot KC = KM \\cdot KT$. If we take $X$ as the midpoint of $KT$, it is equivalent to proving $KB \\cdot KC = KA \\cdot KX$, i.e., points $A, B, C$, and $X$ are concyclic.\nWe note that the perpendicular bisector of segment $KP$ passes through point $B$ and is perpendicular to $AB$, and the perpendicular bisector of $KQ$ passes through $C$ and is perpendicular to $AC$. Hence, the circumcenter of $\\triangle PQK$ (denoted as $O$) is the antipodal point of $A$ on the unit circle. From $OX \\perp KT$, we have $\\angle OXA = \\frac{\\pi}{2}$, and thus $X$ lies on the circle with diameter $AO$ (the circumcircle of $\\triangle ABC$). Proof completed.\n\n![](attached_image_1.png)\n\n(2) We have\n$$\n\\begin{aligned}\n\\frac{BP}{CQ} &= \\frac{BK}{CK} = \\frac{S_{\\triangle BTK}}{S_{\\triangle CTK}} = \\frac{BT}{CT} \\cdot \\frac{\\sin \\angle BTK}{\\sin \\angle CTK} = \\frac{BT}{CT} \\cdot \\frac{\\sin \\angle MCK}{\\sin \\angle MBK} \\\\\n&= \\frac{BT}{CT} \\cdot \\frac{BM}{CM} = \\frac{BT}{CT} \\cdot \\frac{KA_1}{KA_2} = \\frac{BT}{CT} \\cdot \\frac{AP}{AQ}.\n\\end{aligned}\n$$\nProof completed.\n\n\n**Proof 1 (Alternative).** An alternative proof that points $B, C, M$, and $T$ are concyclic.\nSince $K, P, Q$, and $T$ are concyclic, by Ptolemy's Theorem, $KT \\cdot PQ = PK \\cdot QT + QK \\cdot PT$. Also, by the Law of Sines, $\\frac{PQ}{\\sin \\angle PKQ} = \\frac{TQ}{\\sin \\angle TKQ} = \\frac{PT}{\\sin \\angle PKT}$, hence\n$$\nKT = \\frac{PK \\cdot QT + QK \\cdot PT}{PQ} = \\frac{PK \\cdot \\sin \\angle TKQ + QK \\cdot \\sin \\angle PKT}{\\sin \\angle PKQ}.\n$$\nLet $\\angle ABC = \\angle B$ and $\\angle ACB = \\angle C$.\nNoting that $KP = 2BK \\cos \\angle B$ and $KQ = 2CK \\cos \\angle C$, and utilizing the parallel relationships given in the problem, we obtain\n$$\n\\begin{aligned}\nKT &= \\frac{2BK \\cos \\angle B \\cdot \\sin \\angle CAK + 2CK \\cos \\angle C \\cdot \\sin \\angle BAK}{\\sin(\\angle B + \\angle C)} \\\\\n&= \\frac{2BK \\cos \\angle B \\cdot \\frac{KC}{KA} \\sin \\angle C + 2CK \\cos \\angle C \\cdot \\frac{KB}{KA} \\sin \\angle B}{\\sin(\\angle B + \\angle C)} \\\\\n&= \\frac{\\cos \\angle B \\cdot \\sin \\angle C + \\cos \\angle C \\cdot \\sin \\angle B}{\\sin(\\angle B + \\angle C)} \\cdot \\frac{2KB \\cdot KC}{KA} = \\frac{KB \\cdot KC}{KM},\n\\end{aligned}\n$$\ntherefore $KM \\cdot KT = KB \\cdot KC$. By the Power of a Point Theorem, it follows that points $B, C, M$, and $T$ are concyclic. Proof completed.\n\n\n**Proof 2.** Let's consider the circumcircle of $\\triangle ABC$ as the unit circle in the complex plane. We use uppercase letters for points and the corresponding lowercase letters for their complex numbers.\nLet $X$ be the intersection point of the line segment $AK$ with the unit circle. Then\n$$\n\\bar{k} = \\frac{a + x - b - c}{ax - bc}.\n$$\nThus,\n$$\nk = \\frac{\\bar{a} + \\bar{x} - \\bar{b} - \\bar{c}}{\\bar{a}\\bar{x} - \\bar{b}\\bar{c}} = \\frac{\\frac{1}{a} + \\frac{1}{x} - \\frac{1}{b} - \\frac{1}{c}}{\\frac{1}{ax} - \\frac{1}{bc}} = \\frac{bc(x + a) - ax(c + b)}{bc - ax} = \\frac{(ax - ac - cx)b + axc}{ax - bc}.\n$$\nNote the following geometric facts: the perpendicular bisector of $KP$ passes through $B$ and is perpendicular to $AB$, the perpendicular bisector of $KQ$ passes through $C$ and is perpendicular to $AC$. Therefore, the circumcenter of $\\triangle PQK$ is the diametrically opposite point of $A$ on the unit circle, denoted as $A_1$. From $\\angle AXA_1 = \\frac{\\pi}{2}$, we know $X$ is the midpoint of $TK$.\nTherefore,\n$$\nk = \\frac{\\frac{1}{a} + \\frac{1}{x} - \\frac{1}{b} - \\frac{1}{c}}{\\frac{1}{ax} - \\frac{1}{bc}} = \\frac{bc(x + a) - ax(c + b)}{bc - ax} = \\frac{(ax - ac - cx)b + axc}{ax - bc}.\n$$\n\nWe observe the following geometric facts: The perpendicular bisector of the segment $KP$ is the line passing through point $B$ and perpendicular to $AB$, and the perpendicular bisector of $KQ$ is the line passing through $C$ and perpendicular to $AC$. Therefore, the circumcenter of $\\triangle PQK$ is the diametrical opposite point of $A$ on the unit circle, denoted as $A_1$. Thus, according to $\\angle AXA_1 = \\frac{\\pi}{2}$, we know that $X$ is the midpoint of $TK$.\nSo\n$$\n\\begin{aligned}\nt &= 2x - k = 2x - \\frac{(ax - ac - cx)b + acx}{ax - bc} = \\frac{2ax^2 - (ax - ac + cx)b - axc}{ax - bc}, \\\\\nt - b &= \\frac{2ax^2 - (ax - ac + cx)b - axc - axb + b^2c}{ax - bc} = \\frac{(b - x)(bc + ac - 2ax)}{ax - bc}, \\\\\nt - c &= \\frac{2ax^2 - (ax - ac + cx)b - axc - axc + bc^2}{ax - bc} = \\frac{(c - x)(ab + bc - 2ax)}{ax - bc}.\n\\end{aligned}\n$$\nOn the other hand, since $Q$ and $K$ are symmetric about the line $CA_1$ (noting that $a_1 = -a$), we have\n$$\n\\frac{q+a}{c+a} = \\frac{\\overline{k+a}}{c+a} = \\frac{(\\bar{k}+\\bar{a})ac}{a+c} = \\frac{ac\\bar{k}+c}{a+c},\n$$\n$$\n\\text{thus } q = -a + c + ac\\bar{k} = (c - a) + \\frac{ac(a+x-b-c)}{ax-bc}. \\text{ Hence,}\n$$\n$$\n\\begin{aligned}\nq - c &= \\frac{-a(ax - bc) + ac(a + x - b - c)}{ax - bc} = \\frac{-a^2x + a^2c + acx - ac^2}{ax - bc} = \\frac{a(a - c)(c - x)}{ax - bc}, \\\\\nq - a &= \\frac{(c - 2a)(ax - bc) + ac(a + x - b - c)}{ax - bc} = \\frac{abc - bc^2 - 2a^2x + 2acx + a^2c - ac^2}{ax - bc} \\\\\n&= \\frac{(a - c)(ac + bc - 2ax)}{ax - bc}.\n\\end{aligned}\n$$\n$$\n\\text{Similarly, we have } p = -a + b + ab\\bar{k} = (b - a) + \\frac{ab(a+x-b-c)}{ax-bc}. \\text{ Thus,}\n$$\n$$\n\\begin{aligned}\np - b &= \\frac{-a(ax - bc) + ab(a + x - b - c)}{ax - bc} = \\frac{-a^2x + a^2b + abx - ab^2}{ax - bc} = \\frac{a(a - b)(b - x)}{ax - bc}, \\\\\np - a &= \\frac{(b - 2a)(ax - bc) + ab(a + x - b - c)}{ax - bc} = \\frac{abc - b^2c - 2a^2x + 2abx + a^2b - ab^2}{ax - bc} \\\\\n&= \\frac{(a - b)(ab + bc - 2ax)}{ax - bc}.\n\\end{aligned}\n$$\nTherefore, we arrive at\n$$\n\\frac{t-c}{t-b} \\cdot \\frac{p-b}{p-a} = \\frac{q-c}{q-a},\n$$\nand by taking the argument of both sides of this equation, we obtain conclusion (1); by taking the modulus, we obtain conclusion (2).",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 28600,
    "subject": "Mathematics (Multi-modal)",
    "question": "Given a prime number $p$ and a positive real number $\\lambda$ less than $1$. Let $k$ be an integer, and let $S$ and $T$ be sets consisting of consecutive $s$ and $t$ integers, respectively, satisfying $1 \\le s \\le t < \\frac{\\lambda}{12}p$. Furthermore, assume that the number of elements in the set\n$$\n\\{(x, y) \\in S \\times T : kx \\equiv y \\pmod{p}\\}\n$$\nis not less than $1 + \\lambda s$. Prove that there exist integers $a$ and $b$ satisfying\n$$\nka \\equiv b \\pmod{p}, \\quad 0 < a \\le \\frac{1}{\\lambda}, \\quad |b| \\le \\frac{t}{\\lambda s}.\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "*Proof.* Let the solutions to the congruence equation $kx \\equiv y \\pmod{p}$ in $S \\times T$ be $(x_1, y_1), \\dots, (x_n, y_n)$, where $n \\ge 1 + \\lambda s \\ge 1$. Hence, we know that $S$ and $T$ are both non-empty, implying that $n \\ge 2$. Since $T$ is contained in a complete residue system modulo $p$, we know that the $x_i$'s are distinct. Without loss of generality, let's assume $x_1 < x_2 < \\dots < x_n$. Let $u_i = x_i - x_1$ and $v_i = y_i - y_1$. Then we have\n$$\nku_i \\equiv v_i \\pmod{p}, \\quad 0 \\le u_i \\le s, \\quad |v_i| \\le t.\n$$\nLet $a = \\frac{u_2}{\\gcd(u_2, v_2)}$ and $b = \\frac{v_2}{\\gcd(u_2, v_2)}$. Thus, $\\gcd(a, b) = 1$, and we have\n$$\n0 < a \\le s, \\quad |b| \\le t, \\quad ka \\equiv b \\pmod{p}.\n$$\nCombining with $ku_i \\equiv v_i \\pmod{p}$, we obtain\n$$\nbu_i \\equiv kau_i \\equiv av_i \\pmod{p}.\n$$\nLet $bu_i = av_i + pw_i$. Then we have\n$$\n|w_i| \\le \\frac{|b|s + at}{p}.\n$$\nLet $L = \\frac{|b|s + at}{p}$.\n\n(1) If $L < 1$, then $w_i = 0$ for all $i$, and hence $bu_i = av_i$. Since $\\gcd(a, b) = 1$, it follows that $u_i$ must be multiples of $a$. Combining with the fact that $u_i \\in [0, s]$ are distinct non-negative integers, we have\n$$\ns \\ge \\max u_i \\ge a(n-1) \\ge a\\lambda s,\n$$\nwhich implies $a \\le \\frac{1}{\\lambda}$. Furthermore, from\n$$\nat \\ge \\max |av_i| = \\max |bu_i| \\ge |b| \\cdot (a\\lambda s),\n$$\nwe have $|b| \\le \\frac{t}{\\lambda s}$.\n\n(2) If $L \\ge 1$, then $w_i$ can take at most $2L+1 \\le 3L$ integer values in the range $[-L, L]$. By the pigeonhole principle, at least $\\frac{n}{3L}$ of the $w_i$'s take the same value. Let $E$ be the set of indices $i$ where $w_i$ takes the same value, and let $u_{i_0}$ be the smallest value in $\\{u_i : i \\in E\\}$. From\n$$\nb(u_i - u_{i_0}) = a(v_i - v_{i_0}) + p(w_i - w_{i_0}) = a(v_i - v_{i_0}), \\quad \\forall i \\in E,\n$$\nwe obtain $a \\mid u_i - u_{i_0}$. Combining $0 \\le u_i - u_{i_0} = x_i - x_{i_0} \\le s$, we have\n$$\ns \\ge \\max_{i \\in E} (u_i - u_{i_0}) \\ge a(|E| - 1).\n$$\nAlso, from $|v_i - v_{i_0}| = |y_i - y_{i_0}| \\le t$, we have\n$$\nat \\ge \\max (a|v_i - v_{i_0}|) = |b| \\max_{i \\in E} (u_i - u_{i_0}) \\ge |b|(a(|E| - 1)).\n$$\nNote that\n$$\nL = \\frac{|b|s + at}{p} \\le \\frac{2st}{p} < \\frac{\\lambda s}{6},\n$$\nwhich implies $1 < \\frac{\\lambda s}{6L}$. Thus, we have\n$$\n|E| - 1 \\ge \\frac{\\lambda s}{3L} - \\frac{\\lambda s}{6L} = \\frac{\\lambda s}{6L}.\n$$\nSubstituting back into the previous inequalities, we obtain\n$$\n\\lambda a \\le 6L, \\quad \\lambda |b|s \\le 6Lt,\n$$\nwhich leads to\n$$\n\\lambda p = \\frac{\\lambda |b|s + \\lambda at}{L} \\le 12t,\n$$\ncontradicting the condition $t < \\frac{\\lambda p}{12}$!\n\nThus, we have shown that $L \\ge 1$ is not possible, and only $L < 1$ holds true in this case. $\\square$",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 28601,
    "subject": "Mathematics (Multi-modal)",
    "question": "A *state* refers to a way to place the numbers $1, 2, \\ldots, 99$ on the vertices of a given regular $99$-gon, with exactly one number on each vertex and every number appearing exactly once. Two states are considered *equivalent* if one of them can be obtained from the other by rotating the regular $99$-gon (on the plane).\n\nAn *operation* is to first choose two adjacent vertices of the regular $99$-gon and then exchange the numbers on these two vertices. Determine the smallest positive integer $N$ such that, for any two states $\\alpha$ and $\\beta$, one may apply no more than $N$ operations to $\\alpha$ to obtain a state that is equivalent to $\\beta$.",
    "options": [],
    "answer": "2401",
    "solution": "The smallest $N = 2401$.\n\nLet $\\alpha^*$ be the state where the numbers $1, 2, \\ldots, 99$ are arranged counterclockwise at the vertices, and $\\beta^*$ be the state where $1, 2, \\ldots, 99$ are arranged clockwise on the circle. We prove that it takes at least $2401$ operations to transform $\\alpha^*$ into a permutation identical to $\\beta^*$.\n\nAssume $\\alpha^*$ transforms into a permutation identical to $\\beta^*$ after $m$ operations. If a certain operation swaps numbers $a, b$, then draw an edge between $a$ and $b$, thereby obtaining a graph $G$ with $\\{1, 2, \\dots, 99\\}$ as the vertex set and $m$ edges (possibly with repetitions). For any three numbers $1 \\le x < y < z \\le 99$, they are in counterclockwise order in $\\alpha^*$ and in clockwise order in $\\beta^*$. Hence, there must be at least one operation involving two of these numbers, i.e., there is at least one edge among $x, y, z$ in $G$. Therefore, the complement graph $\\bar{G}$ of $G$ does not contain a triangle. According to Turán's theorem, the number of edges in $\\bar{G}$ does not exceed $49 \\times 50 = 2450$, so the number of edges in $G$ is at least $\\binom{99}{2} - 2450 = 2401$, that is, $m \\ge 2401$.\n\nNext, we prove that for any two states $\\alpha$ and $\\beta$, $\\alpha$ can be transformed into a state identical to $\\beta$ with no more than $2401$ operations. Without loss of generality, assume $\\beta = \\beta^*$; otherwise, a permutation of the number table can be applied.\n\nFor a (linear) arrangement $\\pi = (x_1, x_2, \\dots, x_n)$ composed of several different numbers, an ordered pair $(x_i, x_j)$ with $1 \\le i < j \\le n$ and $x_i > x_j$ is called an inversion pair, and let $I(\\pi)$ denote the number of inversion pairs in $\\pi$. It is easy to see that if $(x_i, x_{i+1})$ is an inversion pair, then swapping $x_i$ with $x_{i+1}$ to get $\\pi'$ results in $I(\\pi') = I(\\pi) - 1$. When $I(\\pi) = 0$, $\\pi$ is in increasing order. Therefore, for any arrangement $\\pi$, it can be transformed into an increasing arrangement with exactly $I(\\pi)$ operations (by swapping adjacent numbers).\n\nNumbers $1, 2, \\ldots, 50$ are called small numbers, and numbers $51, \\ldots, 99$ are called large numbers. Consider the number of small numbers among $50$ consecutive vertices in state $\\alpha$. There are $99$ ways to take $50$ consecutive vertices, and each small number appears in exactly $50$ of these ways, so the average number of small numbers is $\\frac{50 \\times 50}{99} \\in (25, 26)$. Therefore, there must be some $50$ consecutive vertices containing no more than $25$ small numbers, and also some $50$ consecutive vertices containing more than $25$ small numbers. Using the discrete intermediate value theorem, it is easy to prove that there exist $50$ consecutive vertices containing exactly $25$ small numbers (and $25$ large numbers).\n\nArrange $\\alpha$ clockwise starting from a certain number as $(a_1, a_2, \\dots, a_{50}, b_1, b_2, \\dots, b_{49})_{\\text{cyc}}$ (the subscript cyc indicates a cyclic arrangement), such that among $a_1, a_2, \\dots, a_{50}$, there are exactly $25$ small numbers. Let the small numbers in $a_1, a_2, \\dots, a_{50}$ from left to right be $x_1, x_2, \\dots, x_{25}$, and the large numbers from left to right be $y_1, y_2, \\dots, y_{25}$; let the small numbers in $b_1, b_2, \\dots, b_{49}$ from left to right be $z_1, z_2, \\dots, z_{25}$, and the large numbers from left to right be $w_1, w_2, \\dots, w_{24}$. We have the following two schemes:\n\nScheme one:\n(1.1) Swap adjacent numbers in $(a_1, a_2, \\cdots, a_{50})$ to move the small numbers to the first $25$ positions. The minimum number of operations required is $s_1$.\n(1.2) Swap adjacent numbers in $(b_1, b_2, \\cdots, b_{49})$ to move the small numbers to the last $25$ positions. The minimum number of operations required is $s_2$.\nNow, we obtain the following cyclic arrangement:\n$$\n(x_1, x_2, \\cdots, x_{25}, y_1, y_2, \\cdots, y_{25}, w_1, w_2, \\cdots, w_{24}, z_1, z_2, \\cdots, z_{25})_{\\text{cyc}}. \\quad (*)\n$$\nThe reason for this particular arrangement will be explained in a lemma later.\n(1.3) Perform $I(\\sigma_1)$ operations on $\\sigma_1 = (z_1, \\cdots, z_{25}, x_1, \\cdots, x_{25})$ to obtain the arrangement $(1, 2, \\cdots, 50)$.\nFor $\\sigma_2 = (y_1, \\cdots, y_{25}, w_1, \\cdots, w_{24})$, perform $I(\\sigma_2)$ operations to obtain $(51, 52, \\cdots, 99)$. Thus, we achieve a state identical to $\\beta^*$, with a total number of operations equal to $s_1 + s_2 + I(\\sigma_1) + I(\\sigma_2)$.\n\nScheme two:\n(2.1) Swap adjacent numbers in $(a_1, a_2, \\cdots, a_{50})$ to move the small numbers to the last $25$ positions. The minimum number of operations required is $s'_1$.\n(2.2) Swap adjacent numbers in $(b_1, b_2, \\cdots, b_{49})$ to move the small numbers to the first $25$ positions. The minimum number of operations required is $s'_2$.\nNow, we obtain the cyclic arrangement\n$$\n(y_1, y_2, \\cdots, y_{25}, x_1, x_2, \\cdots, x_{25}, z_1, z_2, \\cdots, z_{25}, w_1, w_2, \\cdots, w_{24})_{\\text{cyc}}. \\quad (**)\n$$\n(2.3) Perform $I(\\sigma'_1)$ operations on $\\sigma'_1 = (x_1, \\cdots, x_{25}, z_1, \\cdots, z_{25})$ to obtain the arrangement $(1, 2, \\cdots, 50)$.\nFor $\\sigma'_2 = (w_1, \\cdots, w_{24}, y_1, \\cdots, y_{25})$, perform $I(\\sigma'_2)$ operations to obtain $(51, 52, \\cdots, 99)$. Thus, we achieve a state identical to $\\beta^*$, with a total number of operations equal to $s'_1 + s'_2 + I(\\sigma'_1) + I(\\sigma'_2)$.\n\nNext, we prove that one of the two schemes must have an operation count not exceeding $2401$.\n\n**Lemma:** Consider an arrangement of $a$ red numbers and $b$ blue numbers. Let $s$ be the minimum number of operations required to move the red numbers to the front and the blue numbers to the back by swapping adjacent numbers, and let $t$ be the minimum number of operations required for the opposite arrangement. Then $s + t = ab$, and to achieve the minimum number of operations, each operation must involve swapping a red number with a blue number, thereby not changing the order among the red numbers or among the blue numbers.\n\n*Proof of the Lemma:* Consider the ordered pairs $(x, y)$ in the arrangement, where $x$ is a blue number and $y$ is a red number, with $x$ positioned to the left of $y$. Let the count of such pairs be $S$. After one operation, $S$ decreases by at most $1$ (remains unchanged if two red or two blue numbers are swapped, decreases by $1$ if a blue number is swapped with a red number to its right, and increases by $1$ if a red number is swapped with a blue number to its right). Therefore, by selecting pairs of consecutive blue and red numbers for each operation, it takes exactly $S$ operations to arrange all red numbers to the left and blue numbers to the right, hence the minimum number of operations $s$ equals $S$. Similarly, $t$ is the count of pairs $(z, w)$, where $z$ is a red number and $w$ is a blue number, with $z$ positioned to the left of $w$. Thus, $s + t$ counts each unordered pair of a red and a blue number exactly once, therefore $s + t = ab$.\n\nWith the lemma proven, it also explains why after (1.1) and (1.2) we obtain the cyclic arrangement $(*)$, and after (2.1) and (2.2) we obtain $(**)$.\n\nBy the lemma, $s_1 + s'_1 = 25 \\times 25 = 625$, $s_2 + s'_2 = 24 \\times 25 = 600$.\n\nLet $\\sigma = (x_1, \\cdots, x_{25})$, $\\tau = (z_1, \\cdots, z_{25})$, hence $\\sigma_1 = (\\tau, \\sigma)$, $\\sigma'_1 = (\\sigma, \\tau)$. Then\n$$\nI(\\sigma_1) + I(\\sigma'_1) = 2I(\\tau) + 2I(\\sigma) + I(\\tau, \\sigma) + I(\\sigma, \\tau).\n$$\nHere $I(\\tau, \\sigma)$ denotes the number of inversion pairs in $\\sigma_1$ formed by taking a number from $\\tau$ and a number from $\\sigma$, and $I(\\sigma, \\tau)$ counts the inversion pairs in $\\sigma'_1$ taking a number from $\\sigma$ and a number from $\\tau$, so $I(\\sigma, \\tau) + I(\\tau, \\sigma)$ counts every unordered pair formed by taking one number from $\\tau$ and one from $\\sigma$, amounting to $25^2 = 625$. Since $I(\\sigma) \\le \\binom{25}{2}$ and $I(\\tau) \\le \\binom{25}{2}$, we have\n$$\nI(\\sigma_1) + I(\\sigma'_1) \\le 2\\binom{25}{2} + 2\\binom{25}{2} + 25^2 = 1825.\n$$\nSimilarly, we find\n$$\nI(\\sigma_2) + I(\\sigma'_2) \\le 2\\binom{24}{2} + 2\\binom{25}{2} + 24 \\times 25 = 1752.\n$$\nTherefore,\n$$\n(s_1 + s_2 + I(\\sigma_1) + I(\\sigma_2)) + (s'_1 + s'_2 + I(\\sigma'_1) + I(\\sigma'_2)) \\le 625 + 600 + 1825 + 1752 = 4802.\n$$\nBy the principle of averages, one of the schemes must have an operation count not exceeding $2401$.\n\nIn conclusion, the smallest $N$ sought is $2401$.",
    "topic": "Discrete Mathematics",
    "subtopic": "Combinatorics"
  },
  {
    "id": 28602,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $x_1, x_2, \\dots, x_{22}$ be real numbers such that $2^{i-1} \\le x_i \\le 2^i$ holds for every $1 \\le i \\le 22$. Find the maximum value of\n$$\n(x_1 + x_2 + \\dots + x_{22}) \\left( \\frac{1}{x_1} + \\frac{1}{x_2} + \\dots + \\frac{1}{x_{22}} \\right).\n$$",
    "options": [],
    "answer": "(2^{12} - 1 - 1/2^{11})^2",
    "solution": "Let $y_i = \\frac{x_i}{2^{i/2}}$, for $i = 1, 2, \\dots, 22$. It is well-known that $f(t) = t + \\frac{1}{t}$ is decreasing on $(0, 1]$ and increasing on $[1, +\\infty)$. For $1 \\le i \\le 11$, we have $\\frac{1}{2^{12-i}} \\le y_i \\le \\frac{1}{2^{11-i}}$, thus $y_i + \\frac{1}{y_i} \\le 2^{12-i} + \\frac{1}{2^{12-i}}$. For $12 \\le i \\le 22$, we have $2^{i-12} \\le y_i \\le 2^{i-11}$, thus $y_i + \\frac{1}{y_i} \\le 2^{i-11} + \\frac{1}{2^{i-11}}$. Hence, we have\n$$\n\\begin{aligned}\n\\left(\\sum_{i=1}^{22} x_i\\right) \\left(\\sum_{i=1}^{22} \\frac{1}{x_i}\\right) &= \\left(\\sum_{i=1}^{22} y_i\\right) \\left(\\sum_{i=1}^{22} \\frac{1}{y_i}\\right) \\\\\n&\\le \\frac{1}{4} \\left(\\sum_{i=1}^{n} \\left(y_i + \\frac{1}{y_i}\\right)\\right)^2 \\\\\n&\\le \\frac{1}{4} \\left(\\sum_{i=1}^{11} \\left(2^{12-i} + \\frac{1}{2^{12-i}}\\right) + \\sum_{i=12}^{22} \\left(2^{i-11} + \\frac{1}{2^{i-11}}\\right)\\right)^2 \\\\\n&= \\left(2^1 + 2^2 + \\dots + 2^{11} + \\frac{1}{2^1} + \\frac{1}{2^2} + \\dots + \\frac{1}{2^{11}}\\right)^2 \\\\\n&= \\left(2^{12} - 1 - \\frac{1}{2^{11}}\\right)^2.\n\\end{aligned}\n$$\nWhen\n$$\nx_i = \\begin{cases} 2^{i-1}, & 1 \\le i \\le 11, \\\\ 2^i, & 12 \\le i \\le 22, \\end{cases}\n$$\nequality holds. Thus, the maximum value sought is $(2^{12} - 1 - \\frac{1}{2^{11}})^2$. $\\square$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 28603,
    "subject": "Mathematics (Multi-modal)",
    "question": "(1) Prove: on the complex plane, for each line passing through the origin except for the real axis, there is at most one point $z$ such that $\\frac{1+z^{23}}{z^{64}}$ is a real number.\n\n(2) Prove: for any complex number $a \\ne 0$ and any real number $\\theta$, the equation\n$$\n1 + z^{23} + a z^{64} = 0\n$$\nhas at least one root in the set\n$$\nS_{\\theta} = \\{z \\in \\mathbb{C} \\mid \\mathrm{Re}(z \\cdot e^{-i\\theta}) \\ge |z| \\cdot \\cos \\frac{\\pi}{20}\\}.\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "(1) For a non-zero complex number $z$, let $z = r(\\cos\\theta + i\\sin\\theta)$ where $r > 0$ and $\\theta \\in [0, 2\\pi)$, we have\n$$\nf(z) := \\frac{1+z^{23}}{z^{64}} = \\frac{1}{r^{64}} (\\cos(-64\\theta) + i\\sin(-64\\theta) + r^{23}(\\cos(-41\\theta) + i\\sin(-41\\theta))).\n$$\nTherefore,\n$$\n\\begin{align*}\n\\frac{1+z^{23}}{z^{64}} \\in \\mathbb{R} &\\Leftrightarrow \\mathrm{Im} \\frac{1+z^{23}}{z^{64}} = \\frac{-1}{64} (\\sin 64\\theta + r^{23} \\sin 41\\theta) = 0 \\\\\n&\\Leftrightarrow \\sin 64\\theta + r^{23} \\sin 41\\theta = 0.\n\\end{align*}\n$$\nFor $\\theta \\neq 0, \\pi$, when $\\sin 41\\theta = 0$, we have $\\sin 64\\theta \\neq 0$ and there are no solutions to the equation.\nWhen $\\sin 41\\theta \\neq 0$, we have\n$$\nr = g_0(\\theta) := \\left( \\frac{-\\sin 64\\theta}{\\sin 41\\theta} \\right)^{1/23},\n$$\nand for $r > 0$, $\\sin 64\\theta$ and $\\sin 41\\theta$ must have different signs. Noting that $g_0(\\theta) = -g_0(\\theta + \\pi)$, we can conclude that the proposition holds.\n\nFor a fixed $\\lambda \\in [0, \\pi)$, we consider the set of points on the complex plane\n$$\n\\Gamma_{\\lambda} := \\{z \\mid f(z)e^{-i\\lambda} \\in \\mathbb{R}\\} = \\{z \\mid \\sin(\\lambda + 64\\theta) + r^{23}\\sin(\\lambda + 41\\theta) = 0\\}.\n$$\nThe problem can be divided into two cases:\n(1°) $\\lambda + 64\\theta \\equiv \\lambda + 41\\theta \\equiv 0 \\pmod{\\pi}$, then $\\theta \\equiv \\frac{k\\pi}{23} \\pmod{\\pi}$ and $\\lambda \\equiv \\frac{5k\\pi}{23} \\pmod{\\pi}$ hold for some integer $k$.\n(2°) If the above equation does not hold, then similar to the discussion in the first question: There is at most one point $z$ on every line passing through the origin that lies in $\\Gamma_{\\lambda}$. (Specifically, there is at most one intersection point between any ray originating from the origin and $\\Gamma_{\\lambda}$)\n\n$$\ng_{\\lambda}(\\theta) := \\left( \\frac{-\\sin(\\lambda + 64\\theta)}{\\sin(\\lambda + 41\\theta)} \\right)^{1/23}.\n$$\nAs previously mentioned, we want to prove that when $z$ is in the sector corresponding to some interval of $\\theta$, the set composed of all points in this interval that make $g_{\\lambda}(\\theta) > 0$,\n$$\n(**) \\quad -e^{-i\\lambda} f(z) = - \\frac{1}{(g_{\\lambda}(\\theta))^{64}} (\\cos(\\lambda + 64\\theta) + r^{23} \\cos(\\lambda + 41\\theta)) \\text{ takes all real values.}\n$$\nThe $\\theta$ that satisfies $g_{\\lambda}(\\theta) > 0$ is made up of several subintervals. At the endpoints of these intervals, there are two possible cases:\na) The $\\theta$ at the endpoint satisfies $\\lambda + 41\\theta \\equiv 0 \\pmod{\\pi}$, but $\\lambda + 64\\theta \\not\\equiv 0 \\pmod{\\pi}$. In this case, as $\\theta$ approaches this endpoint within the interval, $g_{\\lambda}(\\theta) \\to \\infty$ and the corresponding $a \\to 0$ ($\\rho \\to 0$);\nb) The $\\theta$ at the endpoint satisfies $\\lambda + 64\\theta \\equiv 0 \\pmod{\\pi}$, but $\\lambda + 41\\theta \\not\\equiv 0 \\pmod{\\pi}$. In this case, as $\\theta$ approaches this endpoint within the interval, $g_{\\lambda}(\\theta) \\to 0$ and the corresponding $a \\to \\infty$ (when $\\cos(\\lambda + 64\\theta)$ at the endpoint is $-1$, $\\rho \\to +\\infty$; when it's 1, $\\rho \\to -\\infty$).\nNow we proceed to prove that for any interval of length $\\frac{\\pi}{10}$ (in terms of $\\theta$), there exists a subinterval in which (***) holds.\n1) If there exists $\\theta_0$ in the interval (with an appropriate $\\lambda$) satisfying the aforementioned condition 1°, then the distance from one of the endpoints of the interval (without loss of generality, assuming it to be the right endpoint) to $\\theta_0$ is not less than $\\frac{\\pi}{20} > \\frac{2\\pi}{41} > \\frac{3\\pi}{64}$. Therefore, during the process of changing from $\\theta_0$ to this endpoint, the following two situations may occur (also without loss of generality, assuming that $\\sin(\\lambda + 64\\theta)$ takes positive values in $(\\theta_0, \\theta_0 + \\frac{\\pi}{64})$). One case is that\n\n|                | $\\theta_0$ |   | $\\theta_0 + \\frac{\\pi}{64}$ |   | $\\theta_0 + \\frac{\\pi}{41}$ |   | $\\theta_0 + \\frac{\\pi}{32}$ |   | $\\theta_0 + \\frac{3\\pi}{64}$ |   | $\\theta_0 + \\frac{2\\pi}{41}$ |\n|----------------|------------|---|------------------------------|---|-------------------------------|---|-------------------------------|---|-------------------------------|---|-------------------------------|\n| $\\sin(\\lambda + 64\\theta)$ | 0          | + | 0                            | - | -                             | - | 0                             | + | 0                             | - | -                             |\n| $\\sin(\\lambda + 41\\theta)$ | 0          | - | -                            | - | 0                             | + | +                             | + | +                             | + | 0                             |\n| $\\cos(\\lambda + 64\\theta)$ | $\\pm1$     |   | $\\mp1$                       |   |                               |   | $\\pm1$                        |   | $\\mp1$                        |   |                               |\n\nAt this point, we observe that the piecewise continuous function $\\rho$ (in terms of $\\theta$) satisfies the following conditions:\n$$\n\\lim_{\\theta \\to (\\theta_0 + \\frac{\\pi}{32})^{-}} \\rho = \\pm\\infty, \\quad \\lim_{\\theta \\to (\\theta_0 + \\frac{3\\pi}{64})^{+}} \\rho = \\mp\\infty, \\quad \\rho(\\theta_0 + \\frac{\\pi}{41}) = \\rho(\\theta_0 + \\frac{2\\pi}{41}) = 0.\n$$\nTherefore, on the interval $[\\theta_0 + \\frac{\\pi}{41}, \\theta_0 + \\frac{3\\pi}{32}) \\cup (\\theta_0 + \\frac{3\\pi}{64}, \\theta_0 + \\frac{2\\pi}{41}]$, $\\rho$ can take on any real value.\nThe other case is that\n\n|                | $\\theta_0$ |   | $\\theta_0 + \\frac{\\pi}{64}$ |   | $\\theta_0 + \\frac{\\pi}{41}$ |   | $\\theta_0 + \\frac{\\pi}{32}$ |   | $\\theta_0 + \\frac{3\\pi}{64}$ |   | $\\theta_0 + \\frac{2\\pi}{41}$ |\n|----------------|------------|---|------------------------------|---|-------------------------------|---|-------------------------------|---|-------------------------------|---|-------------------------------|\n| $\\sin(\\lambda + 64\\theta)$ | 0          | + | 0                            | - | -                             | - | 0                             | + | 0                             | - | -                             |\n| $\\sin(\\lambda + 41\\theta)$ | 0          | + | +                            | + | 0                             | - | -                             | - | -                             | - | 0                             |\n| $\\cos(\\lambda + 64\\theta)$ | $\\pm1$     |   | $\\mp1$                       |   |                               |   | $\\pm1$                        |   | $\\mp1$                        |   |                               |\n\nAt this point, we observe that $\\rho$ can take on any real value on the interval $(\\theta_0 + \\frac{\\pi}{32}, \\theta_0 + \\frac{3\\pi}{64})$.\n2) For the aforementioned case 2°, considering the values of $\\theta$ (and $\\lambda$) within this interval, we observe that it contains at least 4 zeros of $\\sin(\\lambda + 41\\theta)$. Furthermore, noting that $\\frac{3}{43} > \\frac{4}{64}$, it follows that among these four zeros of $\\sin(\\lambda + 41\\theta)$, there must exist two adjacent zeros, denoted as $\\theta_1$ and $\\theta_1 + \\frac{\\pi}{41}$, between which there are two zeros of $\\sin(\\lambda + 64\\theta)$, namely $\\theta_2$ and $\\theta_2 + \\frac{\\pi}{64}$. Without loss of generality, let's assume that $\\theta_1 + \\frac{2\\pi}{41}$ is also within the interval. In this case, either we have (assuming, without loss of generality, that $\\sin(\\lambda + 64\\theta)$ takes positive values in $(\\theta_1, \\theta_2)$).\n\n|                | $\\theta_1$ | $\\theta_2$ | $\\theta_2 + \\frac{\\pi}{64}$ | $\\theta_1 + \\frac{\\pi}{41}$ | $\\theta_2 + \\frac{\\pi}{32}$ |\n|----------------|------------|------------|------------------------------|------------------------------|------------------------------|\n| $\\sin(\\lambda + 64\\theta)$ | +          | +          | 0                            | 0                            | +                            |\n| $\\sin(\\lambda + 41\\theta)$ | 0          | -          | -                            | -                            | 0                            |\n| $\\cos(\\lambda + 64\\theta)$ |            | $\\pm1$     | $\\mp1$                       |                              | $\\pm1$                       |\n\nin which case one takes $[\\theta_1, \\theta_2) \\cup (\\theta_2 + \\frac{\\pi}{64}, \\theta_1 + \\frac{\\pi}{41}]$ as the desired interval; or we have\n\n|                | $\\theta_1$ | $\\theta_2$ | $\\theta_2 + \\frac{\\pi}{64}$ | $\\theta_1 + \\frac{\\pi}{41}$ | $\\theta_2 + \\frac{\\pi}{32}$ |\n|----------------|------------|------------|------------------------------|------------------------------|------------------------------|\n| $\\sin(\\lambda + 64\\theta)$ | +          | +          | 0                            | 0                            | +                            |\n| $\\sin(\\lambda + 41\\theta)$ | 0          | +          | +                            | 0                            | -                            |\n| $\\cos(\\lambda + 64\\theta)$ |            | $\\pm1$     | $\\mp1$                       |                              | $\\pm1$                       |\n\nin which case one takes $(\\theta_2, \\theta_2 + \\frac{\\pi}{64})$ as the desired interval.\nIn conclusion, the proposition is proven. □",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 28604,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find the largest real number $\\lambda$ with the following property: for any 100th-order doubly stochastic matrix, it is always possible to select 150 elements from it and change the remaining 9850 elements to zero, such that the resulting matrix has row sums and column sums not less than $\\lambda$.\n\nNote: An \"$n$th-order doubly stochastic matrix\" is an $n \\times n$ square matrix in which all elements are non-negative real numbers, and the sum of each row and the sum of each column are both equal to 1.",
    "options": [],
    "answer": "17/1900",
    "solution": "*Proof.* The answer is $\\frac{17}{1900}$.\n\nFirst, construct the following doubly stochastic matrix $M$: Divide $M$ into four submatrices, where the first 75 rows and the first 24 columns form submatrix $A$, the last 25 rows and the first 24 columns form submatrix $B$, the first 75 rows and the last 76 columns form submatrix $C$, and the last 25 rows and the last 76 columns form submatrix $D$. The elements in $A$ are all $\\frac{1}{75}$, the elements in $B$ are all 0, the elements in $C$ are all $\\frac{17}{1900}$, and the elements in $D$ are all $\\frac{1}{76}$. It can be easily verified that $M$ is a doubly stochastic matrix.\n\n| A | C |\n|---|---|\n| B | D |\n\nAssume that we can select 150 elements from $M$ such that the sum of the selected elements in each row and each column is greater than $\\frac{17}{1900}$. Let $a$ be the number of elements selected from $A$ and $d$ be the number of elements selected from $D$. In the first 75 rows and the last 76 columns, there are at least $151 - a - d$ rows and columns that do not contain selected elements from $A$ or $D$. Among these rows and columns, there must be at least 2 selected elements from $C$, as each selected element from $C$ contributes to exactly one row and one column. Therefore, the number of selected elements from $C$ is at least $2 \\cdot (151 - a - d) \\cdot \\frac{1}{2} = 151 - a - d$. Thus, the total number of selected elements from $A, C, D$ is at least $a + d + (151 - a - d) = 151$, which is a contradiction. Hence, $\\lambda \\le \\frac{17}{1900}$.\n\nTo complete the proof, we need to show that $\\lambda = \\frac{17}{1900}$ satisfies the given conditions. Let $M$ be any doubly stochastic matrix, and we call the elements of $M$ not less than $\\frac{17}{1900}$ \"large numbers\".\n\nIf there exist 50 distinct large numbers that are not in the same row or column, we can assume that these 50 large numbers are in the first 50 rows and the first 50 columns by exchanging rows and columns if necessary. Then, we select the largest element from each of the last 50 rows (which is not less than $\\frac{1}{100}$) and the largest element from each of the last 50 columns. In this way, we have selected at most 150 numbers, and the sum of the selected numbers in each row and each column is not less than $\\frac{17}{1900}$.\n\nWe now show that the above scenario always occurs. By contradiction, assume that there are no 50 distinct large numbers that are not in the same row or column. Then, by Hall's theorem in the deficiency form, we have:\n(*) There exist $k$ rows, $51 \\le k \\le 100$, such that the large numbers in these $k$ rows are distributed among at most $k - 51$ columns.\n\nWithout loss of generality, assume that the large numbers in the first $k$ rows are distributed among the first $k - 51$ columns. Divide $M$ into four submatrices using the first $k$ rows and the first $k - 51$ columns, similar to the construction in the previous proof, and denote them as $A, B, C, D$. Let $S(A), S(B), S(C), S(D)$ be the sums of the elements\n\n| A | C |\n|---|---|\n| B | D |\n\nin these submatrices, respectively. The sum of the numbers in the first $k-51$ columns is $k-51$, i.e., $S(A) + S(B) = k-51$. Since the elements in $C$ are not large numbers, we have $S(C) < \\frac{17}{1900}k(151-k)$. Moreover, $S(D) \\le S(B) + S(D) = 100-k$. Therefore, we have:\n$$\n\\begin{aligned}\n100 &= S(A) + S(B) + S(C) + S(D) \\\\\n&< (k-51) + \\frac{17}{1900}k(151-k) + (100-k) \\\\\n&\\le 49 + \\frac{17}{1900} \\cdot 75 \\cdot 76 = 100,\n\\end{aligned}\n$$\nwhich is a contradiction.\n\nThus, we have shown that $\\lambda_{\\max} = \\frac{17}{1900}$ satisfies the given conditions. $\\square$",
    "topic": "Algebra",
    "subtopic": "Linear Algebra"
  },
  {
    "id": 28605,
    "subject": "Mathematics (Multi-modal)",
    "question": "Given a positive integer $n$, find the smallest positive integer $k$ with the following property: After coloring any $k$ unit squares black and the remaining squares white in a $2n \\times 2n$ grid, it is always possible to make all squares black using a finite number of the following two operations:\n\n(Operation 1) Select a $2 \\times 2$ square that contains exactly 3 black squares and 1 white square, and change the color of the white square to black.\n\n(Operation 2) Select a $2 \\times 2$ square that contains exactly 2 black squares and 2 white squares, and change the color of all squares within the square (black squares become white, and white squares become black).\n\n*Note*: The same $2 \\times 2$ square can be operated on multiple times, and each square can change color multiple times.",
    "options": [],
    "answer": "n^2 + n + 1",
    "solution": "*Proof.* The desired value of $k$ is $n^2 + n + 1$.\n\nFirst, we provide an example of coloring $n^2 + n$ cells black, such that it is impossible to perform operation 1. This implies that $k \\ge n^2 + n + 1$.\n\nColor the anti-diagonal cells black, and then color black all the cells $(i, j)$ in the top-left part of the anti-diagonal, where both $i$ and $j$ are odd and $i + j < 2n$. Similarly, color black all the cells $(i, j)$ in the bottom-right part of the anti-diagonal, where both $i$ and $j$ are even and $i + j > 2n + 2$. (The following diagram illustrates the case of $n = 4$, where the cells marked with $*$ are colored black.)\n\n![](attached_image_1.png)\n\nIt can be seen that a total of $n^2 + n$ cells are colored black, and they satisfy the following property.\n\n$(\\star)$ For any two black cells $(i, j_1)$ and $(i, j_2)$ in the same row with $j_1 < j_2$, there exists $j_1 < j < j_2$ such that cells $(i-1, j)$, $(i, j)$, and $(i+1, j)$ are all white. Similarly, for any two black cells $(i_1, j)$ and $(i_2, j)$ in the same column with $i_1 < i_2$, there exists $i_1 < i < i_2$ such that cells $(i, j-1)$, $(i, j)$, and $(i, j+1)$ are all white. Here, all cells in the 0th row, the $(2n+1)$th row, the 0th column, and the $(2n+1)$th column are considered white.\n\nNote that if a state satisfies $(\\star)$, then in any $2 \\times 3$ rectangle, there can be at most 2 black cells (otherwise, either two black cells share a common edge, or there exists a white cell that has three black cells adjacent to it, both of which violate $(\\star)$).\n\nNow we will prove that for any operation performed on the diagram, it can only be operation 2, and the two colored cells will not share a common edge. Additionally, property $(\\star)$ will continue to hold. If this statement is true, it implies that it is impossible to completely color all cells black. The initial state clearly satisfies the above assertion. Assuming the assertion holds after some operations, we will show that after performing one operation 2, property $(\\star)$ will still be satisfied. Due to symmetry, we only need to consider the resulting cells $(i, j_1)$ and $(i, j_2)$. We will discuss two cases.\n\n• **Case 1:** Both $(i, j_1)$ and $(i, j_2)$ are originally black cells. In this case, there exists $j_1 < j < j_2$ such that cells $(i-1, j)$, $(i, j)$, and $(i+1, j)$ are all white. Note\n\n![](attached_image_1.png)\n\n![](attached_image_2.png)\n\n(The scripts in the figure is: “at most *** black cells.”) If the operation does not increase the number of black cells in this $3 \\times (j_2-j_1-1)$ rectangle, then there must still be a column in these three rows with no black cells, which satisfies property $(\\star)$. If the operation increases the number of black cells in the $3 \\times (j_2-j_1-1)$ rectangle, then by the description of operation 2 and the fact that any two black cells were not adjacent, it must involve adding a black cell to one of the corners of the upper $3 \\times (j_2-j_1-1)$ rectangle. In this case, prior to the operation, either $(i-1, j_1)$, $(i+1, j_1)$, $(i-1, j_2)$, or $(i+1, j_2)$ must have at least one black cell. However, this configuration would result in two black cells sharing a common edge, which is a contradiction!\n\n* **Case 2:** One of $(i, j_1)$ and $(i, j_2)$ is a newly colored black cell (the two newly colored cells are in different rows and different columns, so both $(i, j_1)$ and $(i, j_2)$ cannot be new cells). Without loss of generality, let's assume $(i, j_2)$ is the newly colored cell. Also, due to symmetry, let's assume the operation occurs in a $2 \\times 2$ rectangle in the $(i-1)$th and $i$th rows. If the operation involves changing $(i-1, j_2)$ and $(i, j_2-1)$ from black to white, and changing $(i-1, j_2-1)$ and $(i, j_2)$ from white to black, then prior to the operation, using property $(\\star)$ for $(i, j_1)$ and $(i, j_2-1)$, we know that there exists $j_1 < j < j_2-1$ such that cells $(i, j-1)$, $(i, j)$, and $(i, j+1)$ are all white. Therefore, after the operation, this $j$ still satisfies property $(\\star)$. If the operation involves changing $(i-1, j_2)$ and $(i, j_2+1)$ from black to white, and changing $(i-1, j_2+1)$ and $(i, j_2)$ from white to black, then prior to the operation, using property $(\\star)$ for $(i, j_1)$ and $(i, j_2+1)$, we know that there exists $j_1 < j < j_2+1$ such that cells $(i, j-1)$, $(i, j)$, and $(i, j+1)$ are all white. Additionally, it is clear that $j \\ne j_2$, so this $j$ still satisfies property $(\\star)$.\n\nNext, we will prove that $k = n^2+n+1$ satisfies the conditions. To do this, we need to show that if the number of black cells is not less than $n^2+n+1$ and there is at least one white cell, then it is possible to perform several operations 2 followed by one operation 1. We will use a proof by contradiction. First, we prove a lemma.\n\n**Lemma:** Let $x_i$ be the number of black cells in the $i$th row. If $x_{i-1}+2x_i+x_{i+1} \\ge 2n+2$, then it is possible to perform some operations 2 in the $(i-1)$th, $i$th, and $(i+1)$th rows such that (within this $3 \\times 2n$ rectangle) there are two black cells with a common edge (where $x_0 = x_{2n+1} = 0$).\n\n![](attached_image_2.png)\n\n**Proof of the Lemma:** We first prove that it is possible to perform some operations 2 in the $(i-1)$th, $i$th, and $(i+1)$th rows without changing $x_{i-1}$, $x_i$, and $x_{i+1}$, such that (within this $3 \\times 2n$ rectangle) there is either a row or a column with two black cells having a common edge. Assuming otherwise, it means that there are no rows or columns with two black cells having a common edge in the current state. Let the black cells in the $i$th row be in columns $j_1, \\dots, j_r$ with $j_1 < \\dots < j_r$. If in the $(i-1)$th, $i$th, and $(i+1)$th rows, there are at most $j_{k+1} - j_k - 2$ black cells in columns $j_k + 1, \\dots, j_{k+1} - 1$ ($k=1, \\dots, r-1$), then the first $j_1 - 1$ columns in these three rows have at most $j_1 - 1$ black cells, and the last $2n - j_r$ columns have at most $2n - j_r$ black cells. Hence, we have $x_{i-1} + 2x_i + x_{i+1} \\le 2r + (j_1 - 1) + (2n - j_r) + \\sum_{k=1}^{r-1} (j_{k+1} - j_k - 2) = 2n + 1$, which is a contradiction.\n\nIf there are more than $j_1 - 1$ black cells in the first $j_1 - 1$ columns or more than $2n - j_r$ black cells in the last $2n - j_r$ columns, it can also be concluded that there is a $2 \\times 3$ rectangle with at least 3 black cells. In this case, there must be three black cells that do not share a common edge. Performing one operation 2 on two of these cells will result in two black cells in a column with a common edge, without changing $x_{i-1}$, $x_i$, and $x_{i+1}$.\n\nIf there exists $k \\in \\{1, \\dots, r-1\\}$ such that there are at least $j_{k+1} - j_k - 1$ black cells in columns $j_k + 1, \\dots, j_{k+1} - 1$ in these three rows, we first note that if there are two adjacent columns in these three rows that have at least 3 black cells, we can similarly perform operation 2 to make two black cells adjacent. Assuming this case does not occur, in particular, the columns $j_k + 1$ and $j_{k+1} - 1$ have at most 1 black cell each. However, the sum of the number of black cells in columns $j_k + 1, \\dots, j_{k+1} - 1$ in these three rows is at least $j_{k+1} - j_k - 1$. By simple analysis, we can see that both the $j_k + 1$ column and the $j_{k+1} + 1$ column in these three rows must have exactly one black cell. After performing operation 2 on $(i, j_k)$ and the black cell in the $j_k + 1$ column, both $(i, j_k + 1)$ and $(i, j_{k+1})$ will be black cells, and there will still be at least $j_{k+1} - j_k - 2$ black cells in columns $j_k + 2, \\dots, j_{k+1} - 1$. Repeat this discussion and operation until the last two black cells in the $i$th row are adjacent.\n\nNow if there are two black cells with a common edge in the same row, the conclusion is already established. If there are two black cells with a common edge in the same column, we can perform operations to move them horizontally. Note that both rows cannot have only one black cell (otherwise, the other row would have at least $(2n+2)-2\\cdot1-1 = 2n-1$ black cells, which would lead to two black cells with a common edge within that row). Therefore, it is always possible to perform operations to have two black cells with a common edge in the same row, and the lemma is proved. $\\square$\n\nReturning to the original problem, we have\n$$\nx_1 + \\sum_{i=1}^{n} (x_{2i-1} + 2x_{2i} + x_{2i+1}) \\ge 2(n^2 + n + 1),\n$$\nwhere $x_{2n+1} = 0$. If $x_1 \\ge n+1$, then the first row has two consecutive black cells. If $x_1 \\le n$, then there exists $i \\in \\{1, 2, \\dots, n\\}$ such that\n$$\nx_{2i-1} + 2x_{2i} + x_{2i+1} \\ge \\left\\lceil \\frac{2n^2 + n + 2}{n} \\right\\rceil \\ge 2n + 2,\n$$\nand by the lemma, we can perform some operations 2 within these three rows to have two black cells with a common edge within a single row.\n\nAssume these two cells are located in the $j_1$th column and the $(j_1+1)$st column. Let $u$ and $v$ be the maximum positive integers such that there are black cells in the columns\n$j_1 - u, j_1 - (u-1), \\dots, j_1 + v$. If one of these columns has at least 2 black cells, let's say the column $j_1 - s$. We perform operations 2 on $(2i, j_1)$ and $(2i, j_1 + 1)$ until the black cell in the $(i, j_1)$th row is in the same row as the black cell in the $(j_1 - 1)$th column. Then, we perform operations on the $j_1 - 1$st and $j_1$th columns to make the black cell in the $j_1 - 1$st column in the same row as one of the black cells in the $j_1 - 2$nd column, and so on. Finally, we can make the black cell in the $j_1 - (s-1)$th column in the same row as one of the black cells in the $j_1 - s$th column. Then, we perform operations on the $j_1 - (s-1)$th and $j_1 - s$th columns to bring these two cells close to another black cell in the $j_1 - s$th column, forming a $2 \\times 2$ square with at least 3 black cells (if operations 2 cannot be performed during this process, it means that a similar situation has occurred). If this $2 \\times 2$ square can be transformed by operation 1, then the conclusion is already established. If not, then all four cells in this $2 \\times 2$ square are black. We can move these four black cells upwards, downwards, leftwards, or rightwards until they encounter another black cell. At this point, if it is not possible to perform operation 1, it means that a $2 \\times 3$ rectangle has been formed with all four cells being black. Considering the three cells on the other side of an edge with length 3, if one or two of them are black, operation 1 can be performed. If none of them is black, then operation 2 can be performed on the four cells on both sides of this edge, followed by operation 1. Therefore, all three cells must be black. By induction, it can be concluded that all cells are black, which contradicts our assumption. Therefore, all the columns $j_1 - u, j_1 - (u-1), \\dots, j_1 + v$ have exactly one black cell.\n\nAt this point, let the number of black cells in each column be $y_1, y_2, \\dots, y_n$. Either there are two black cells with a common edge in the first column, or there exists $1 \\le j \\le n$ such that $y_{2j-1} + 2y_{2j} + y_{2j+1} \\ge 2n + 2$. From $y_{j_1-u} = y_{j_1-u+1} = \\dots = y_{j_1+v} = 1$ and $y_{j_1-j-1} = y_{j_1+v+1} = 0$, it follows that\n$$\n2j \\notin \\{j_1 - u - 1, j_1 - u, \\dots, j_1 + v + 1\\}.\n$$\nUsing the same procedure as before, we can perform operations in the columns $2j - 1, 2j, 2j + 1$ such that there are two black cells with a common edge in one of these columns, without affecting the two black cells with a common edge in the same row. Now, we vertically move the two black cells in the same row and horizontally move the two black cells in the same column until these four black cells “meet,” creating a $2 \\times 2$ square with at least 3 black cells. As discussed earlier, it is always possible to perform operation 1 after several operations 2, and thus $k = n^2+n+1$ satisfies the given condition.\n\nTherefore, the minimum value of $k$ is $n^2 + n + 1$. $\\square$",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 28606,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $p$ be an odd prime number, and $a$, $b$, $m$, $r$ be positive integers, such that $p \\nmid ab$ and $ab > m^2$. Prove that there exists at most one pair of positive integers $(x, y)$ satisfying the following conditions: $x$ and $y$ are coprime, and $ax^2 + by^2 = mp^r$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "**Proof:** By contradiction, suppose there exist two different pairs of positive integer solutions $(x_1, y_1)$ and $(x_2, y_2)$. Since $x_1$ and $y_1$ are coprime, it follows that $p \\nmid x_1y_1$. Similarly, $p \\nmid x_2y_2$. Given\n$$\nax_1^2 \\equiv -by_1^2 \\pmod{p^r}, \\quad ax_2^2 \\equiv -by_2^2 \\pmod{p^r},\n$$\nit is known that $abx_1^2y_2^2 \\equiv abx_2^2y_1^2 \\pmod{p^r}$. Since $p \\nmid ab$, we have $p^r \\mid (x_1^2y_2^2 - x_2^2y_1^2)$.\nNote that $x_1y_2 - x_2y_1$ and $x_1y_2 + x_2y_1$ cannot both be divisible by $p$, otherwise $p \\mid 2x_1y_2$, which contradicts with $p$ being an odd prime and $p \\nmid x_1y_1x_2y_2$. Hence, $p^r \\mid x_1y_2 - x_2y_1$, or $p^r \\mid x_1y_2 + x_2y_1$.\nIf $x_1y_2 - x_2y_1 = 0$, then $\\frac{x_1}{x_2} = \\frac{y_1}{y_2}$, combined with $ax_1^2 + by_1^2 = ax_2^2 + by_2^2$, implies $x_1 = x_2$, $y_1 = y_2$, contradicting with $(x_1, y_1) \\neq (x_2, y_2)$. Therefore, $x_1y_2 - x_2y_1 \\neq 0$.\nIf $p^r \\mid x_1y_2 + x_2y_1$, then $x_1y_2 + x_2y_1 \\ge p^r$.\nIf $p^r \\mid x_1y_2 - x_2y_1$, then $x_1y_2 + x_2y_1 \\ge |x_1y_2 - x_2y_1| \\ge p^r$. Thus, in all cases,\n$$\nx_1y_2 + x_2y_1 \\ge p^r. \\quad (*)\n$$\nUsing the condition $ab > m^2$ and $(*)$, we get\n$$\n\\begin{aligned}\nm^2p^{2r} &= (ax_1^2 + by_1^2)(ax_2^2 + by_2^2) \\\\\n&= (ax_1x_2 - by_1y_2)^2 + ab(x_1y_2 + x_2y_1)^2 \\\\\n&\\ge ab(x_1y_2 + x_2y_1)^2 \\\\\n&> m^2p^{2r},\n\\end{aligned}\n$$\na contradiction. Therefore, the assumption by contradiction is false, and the original statement holds. $\\Box$",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 28607,
    "subject": "Mathematics (Multi-modal)",
    "question": "For any two points $A(x_1, y_1)$ and $B(x_2, y_2)$ in the coordinate plane, define\n$$\nd(A, B) = |x_1 - x_2| + |y_1 - y_2|.\n$$\nLet $P_1, P_2, \\dots, P_{2023}$ be 2023 pairwise different points in the coordinate plane. Denote\n$$\n\\lambda = \\frac{\\max_{1 \\le i < j \\le 2023} d(P_i, P_j)}{\\min_{1 \\le i < j \\le 2023} d(P_i, P_j)}.\n$$\n(1) Prove that $\\lambda \\ge 44$.\n(2) Give an example of $P_1, P_2, \\dots, P_{2023}$ such that $\\lambda = 44$.",
    "options": [],
    "answer": "44",
    "solution": "**Proof Method 1:**\n(1) For $k = 1, 2, \\dots, 2023$, let the coordinates of $P_k$ be $(x_k, y_k)$, and denote $u_k = x_k + y_k$, $v_k = x_k - y_k$. Let $D = \\max_{1 \\le i < j \\le 2023} d(P_i, P_j)$. Then, for any $1 \\le i, j \\le 2023$,\n$$\n|u_i - u_j| = |(x_i - x_j) + (y_i - y_j)| \\le |x_i - x_j| + |y_i - y_j| = d(P_i, P_j) \\le D.\n$$\nThus, the $u_1, u_2, \\dots, u_{2023}$ fall within some interval $[a, a+D]$. Similarly for $v_1, v_2, \\dots, v_n$. For $k, l = 1, 2, \\dots, 44$, consider the region\n$$\nH_{k,l} = \\left\\{ \\left( \\frac{u+v}{2}, \\frac{u-v}{2} \\right) \\mid a + \\frac{k-1}{44}D \\le u \\le a + \\frac{k}{44}D, \\ b + \\frac{l-1}{44}D \\le v \\le b + \\frac{l}{44}D \\right\\}.\n$$\nIf $P_i, P_j \\in H_{k,l}$, let $U = u_i - u_j, V = v_i - v_j$, then $-\\frac{D}{44} \\le U, V \\le \\frac{D}{44}$, we have\n$$\n\\begin{aligned}\nd(P_i, P_j) &= |x_i - x_j| + |y_i - y_j| = \\left| \\frac{u_i + v_i}{2} - \\frac{u_j + v_j}{2} \\right| + \\left| \\frac{u_i - v_i}{2} - \\frac{u_j - v_j}{2} \\right| \\\\\n&= \\left| \\frac{U + V}{2} \\right| + \\left| \\frac{U - V}{2} \\right| \\in \\left\\{ \\pm \\frac{U + V}{2} \\pm \\frac{U - V}{2} \\right\\} = \\{U, -U, V, -V\\}.\n\\end{aligned}\n$$\nHence, $\\min_{1 \\le i < j \\le 2023} d(P_i, P_j) \\le d(P_i, P_j) \\le \\frac{D}{44}$, thus $\\lambda \\ge 44$.\n\n(2) Construction\nConsider the point set\n$$\nM = \\{(x, y) \\in \\mathbb{Z}^2 \\mid x, y \\text{ have the same parity, } |x + y| \\le 44, |x - y| \\le 44\\}.\n$$\nThis set contains $45^2 = 2025$ points. Selecting any 2023 points, the distance $d(P_i, P_j) = |x_i - x_j| + |y_i - y_j|$ is even and greater than 0, i.e., $d(P_i, P_j) \\ge 2$. On the other hand,\n$$\nd(P_i, P_j) = |x_i - x_j| + |y_i - y_j| \\le 88.\n$$\nThus,\n$$\n\\lambda = \\frac{\\max_{1 \\le i < j \\le 2023} d(P_i, P_j)}{\\min_{1 \\le i < j \\le 2023} d(P_i, P_j)} \\le 44,\n$$\nand by (1) $\\lambda = 44$.\n\n\n**Proof Method Two:**\nHere is another proof that $\\lambda \\ge 44$. Assume the coordinates of the 2023 points are $P_k(x_k, y_k)$, $k = 1, 2, \\dots, 2023$, and without loss of generality, assume $x_1 \\le x_2 \\le \\dots \\le x_{2023}$. By the Erdős-Szekeres theorem, among $y_1, y_2, \\dots, y_{2023}$, there exists either an increasing or decreasing subsequence of length $\\lceil\\sqrt{2023}\\rceil = 45$. Assume we have $y_{i_1} \\le y_{i_2} \\le \\dots \\le y_{i_{45}}$, with indices $i_1 < i_2 < \\dots < i_{45}$. Then,\n$$\nd(P_{i_{45}}, P_{i_1}) = (x_{i_{45}} - x_{i_1}) + (y_{i_{45}} - y_{i_1}) = \\sum_{k=1}^{44} (x_{i_{k+1}} - x_{i_k}) + (y_{i_{k+1}} - y_{i_k}) = \\sum_{k=1}^{44} d(P_{i_{k+1}}, P_{i_k}).\n$$\nIf we have a decreasing subsequence $y_{i_1} \\ge y_{i_2} \\ge \\dots \\ge y_{i_{45}}$, then\n$$\nd(P_{i_{45}}, P_{i_1}) = (x_{i_{45}} - x_{i_1}) - (y_{i_{45}} - y_{i_1}) = \\sum_{k=1}^{44} (x_{i_{k+1}} - x_{i_k}) - (y_{i_{k+1}} - y_{i_k}) = \\sum_{k=1}^{44} d(P_{i_{k+1}}, P_{i_k}).\n$$\nIn both cases, we have $d(P_{i_{45}}, P_{i_1}) = \\sum_{k=1}^{44} d(P_{i_{k+1}}, P_{i_k})$. Therefore,\n$$\n\\lambda = \\frac{\\max_{1 \\le i < j \\le 2023} d(P_i, P_j)}{\\min_{1 \\le i < j \\le 2023} d(P_i, P_j)} \\ge \\frac{d(P_{i_{45}}, P_{i_1})}{\\min_{k=1,2,...,44} d(P_{i_{k+1}}, P_{i_k})} \\ge 44.\n$$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 28608,
    "subject": "Mathematics (Multi-modal)",
    "question": "The sequence $\\{a_n\\}$ is defined as follows: $a_1 = 0$, and for integer $n \\ge 2$,\n$$\na_n = \\frac{1}{n} + \\frac{1}{\\lceil \\frac{n}{2} \\rceil} \\sum_{k=1}^{\\lceil \\frac{n}{2} \\rceil} a_k,\n$$\nwhere $\\lceil \\frac{n}{2} \\rceil$ denotes the smallest integer not less than $\\frac{n}{2}$. Find the maximum term of the sequence $\\{a_n\\}$.",
    "options": [],
    "answer": "7/12",
    "solution": "*Proof*. From the given definition, we have $a_2 = \\frac{a_1}{2} + \\frac{1}{2} = \\frac{a_1+a_2}{2} + \\frac{1}{3} = \\frac{7}{12}$.\nWe now prove by induction that $a_n \\le \\frac{7}{12}$, with equality if and only if $n = 3$. The cases $n = 1, 2, 3$ have been verified. Assume the statement holds for all $1, 2, \\dots, n-1$ ($n \\ge 4$), then\n$$\n\\begin{align*}\na_n &= \\frac{a_1 + \\dots + a_{\\lceil \\frac{n}{2} \\rceil}}{\\lceil \\frac{n}{2} \\rceil} + \\frac{1}{n} \\\\\n&\\le \\frac{0 + \\frac{1}{2} + \\left(\\lceil \\frac{n}{2} \\rceil - 2\\right) \\cdot \\frac{7}{12}}{\\lceil \\frac{n}{2} \\rceil} + \\frac{1}{n} \\\\\n&= \\frac{7}{12} - \\frac{2}{3\\lceil \\frac{n}{2} \\rceil} + \\frac{1}{n} \\\\\n&= \\frac{7}{12} - \\frac{2n - 3\\lceil \\frac{n}{2} \\rceil}{3n \\cdot \\lceil \\frac{n}{2} \\rceil} \\\\\n&\\le \\frac{7}{12} - \\frac{2n - 3\\left(\\frac{n+1}{2}\\right)}{3n \\cdot \\lceil \\frac{n}{2} \\rceil} \\\\\n&= \\frac{7}{12} - \\frac{\\frac{n}{2} - \\frac{3}{2}}{3n \\cdot \\lceil \\frac{n}{2} \\rceil} \\\\\n&< \\frac{7}{12}.\n\\end{align*}\n$$\nTherefore, the maximum term of the sequence $\\{a_n\\}$ is $a_3 = \\frac{7}{12}$. $\\square$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 28609,
    "subject": "Mathematics (Multi-modal)",
    "question": "Given a positive integer $n$, a large equilateral triangle of side length $3n$ is divided into $9n^2$ small equilateral triangles of side length 1. Each small triangle is colored in one of three colors: red, yellow, or blue, such that there are $3n^2$ triangles of each color. A trapezoid consisting of 3 small triangles is called a *standard trapezoid*. For example, in the case of $n = 2$ shown below, the two shaded areas are standard trapezoids. A standard trapezoid is called a *colorful trapezoid* if the three small triangles it contains are all of different colors.\nFind the maximum possible number of colorful trapezoids. (These colorful trapezoids do not need to be non-overlapping.)\n![](attached_image_1.png)",
    "options": [],
    "answer": "18n^2 - 9n",
    "solution": "The maximum number of multicolored trapezoids is $18n^2 - 9n$.\n\nUse the numbers 1, 2, and 3 to represent the three colors. First, consider a small equilateral triangle of side length 2 formed by 4 small equilateral triangles. Let the middle small equilateral triangle be labeled $a$, and the other three small equilateral triangles be labeled $b$, $c$, and $d$. Then the three standard trapezoids within this triangle have the labels $(a, b, c)$, $(a, b, d)$, and $(a, c, d)$.\n\n![](attached_image_2.png)\n\n(1) If $a = b$, then at most 1 of these three standard trapezoids can be multicolored. The same applies if $a = c$ or $a = d$.\n(2) If $a$ is different from $b$, $c$, and $d$, then at least two of $b$, $c$, and $d$ must be equal, so at most 2 of these standard trapezoids can be multicolored. Equality holds if and only if $a$ is different from $b$, $c$, and $d$, and $b$, $c$, and $d$ are not all equal.\n\nIt is easy to see that the number of small equilateral triangles of side length 2 is\n$$\n(1 + 2 + \\cdots + (3n - 1)) + (1 + 2 + \\cdots + (3n - 3)) = 9n^2 - 9n + 3.\n$$\nThese equilateral triangles each contain distinct standard trapezoids, because the way to complete a standard trapezoid to a side length 2 equilateral triangle is unique.\n\nApart from these standard trapezoids, there are some standard trapezoids that require an additional small equilateral triangle outside the large triangle to form a side length 2 equilateral triangle.\n\n---\n\nThe base of length 1 of these trapezoids must be on the boundary of the large triangle. It is easy to see that the number of such standard trapezoids is $3(3n - 2) = 9n - 6.$\n\nBecause the number of multicolored trapezoids within a side length 2 equilateral triangle does not exceed 2, the number of multicolored trapezoids does not exceed\n$$\n2(9n^2 - 9n + 3) + (9n - 6) = 18n^2 - 9n.\n$$\nEquality holds if and only if the following two conditions are met simultaneously:\n(A) For any side length 2 equilateral triangle formed by 4 small equilateral triangles, the middle small equilateral triangle is different from the other three small equilateral triangles, and the other three small equilateral triangles are not all the same.\n(B) Each standard trapezoid with a base of length 1 on the boundary of the large triangle must be multicolored.\n\nWe now give a construction that meets these two conditions simultaneously. Center the large triangle at its center and draw concentric equilateral triangles with side lengths 3, 6, ..., 3(n - 1) in a counterclockwise manner, dividing all small triangles into $n$ \"layers.\" As shown in the figure below.\n\n![](attached_image_3.png)\n\nIn each layer of small equilateral triangles, remove the three corner small equilateral triangles (call these special triangles), so that the remaining small equilateral triangles form a loop (each adjacent pair shares a side). Starting from the innermost, the $k$th loop has $18k - 12$ small equilateral triangles, so we can start from a certain triangle and label them 1, 2, 3, 1, 2, 3, ... in a clockwise manner until the loop is complete. This way, all standard trapezoids in the loop are multicolored.\n\nNow condition (B) is met, and for condition (A), consider the side length 2 equilateral triangles. These triangles can be classified into three types: the first type consists of a standard trapezoid on a loop plus a special triangle; the second type consists of a standard trapezoid on a loop plus a small equilateral triangle from another loop; and the third type consists of an equilateral triangle with a special triangle at the center. Therefore, for the first and third types of side length 2 equilateral triangles, the three corner small equilateral triangles are not all the same color.\n\nTherefore, we only need to simultaneously satisfy\n(a) Any two adjacent small equilateral triangles from different loops have different labels;\n(b) Each special triangle has a different label from its neighbors, and except for the three corner special triangles, the three neighbors of each special triangle are not all the same color.\n\nFirst, we set the numbers in the innermost loop, and ensure that for any adjacent pair of small equilateral triangles from different loops, the number of the outer loop triangle is 1 more than the number of the inner loop triangle (mod 3). This can be achieved because moving 2 steps clockwise on the inner loop reaches the next contact position, while the outer loop moves 2 or 8 steps (skipping a special triangle), ensuring that the outer loop triangle is always 1 more than the inner loop triangle. This satisfies condition (a). For condition (b), we set the number of each special triangle to be 1 more than its neighbor from the inner loop (mod 3). Thus, if the number of the special triangle is $x$, the numbers of its three neighbors should be $x-1, x+1, x-1 \\pmod 3$, satisfying (b).\n\nFinally, we count the occurrences of each color. Each color clearly appears equally in each loop. For the three small equilateral triangles in the loop that contact the outer special triangle, their labels are different since $\\frac{18k-12}{3} = 6k-4$ is not divisible by 3. Thus, the three symmetrically placed special triangles have different labels, ensuring that each color appears equally.\n\nIn summary, the maximum number of multicolored trapezoids is $18n^2 - 9n$. □",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 28610,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $\\mathbb{N} = \\{0,1,2,\\dots\\}$ be the set of all non-negative integers. For each $n \\in \\mathbb{N}$, define the Catalan number\n$$\nC_n = \\frac{1}{n+1} \\binom{2n}{n} = \\frac{(2n)!}{n!(n+1)!}.\n$$\nProve that for any positive integer $m$, we have\n$$\n\\sum_{\\substack{i,j,k \\in \\mathbb{N} \\\\ i+j+k=m}} C_{i+j}C_{i+k}C_{j+k} = \\frac{3}{2m+3}C_{2m+1}.\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "*Proof 1.* Consider the set of ordered triples:\n$$\nA = \\{ (u, v, w) \\in \\mathbb{N}^3 \\mid u, v, w \\le m; u + v + w = 2m \\},\n$$\n$$\nB = \\{(u, v, w) \\in \\mathbb{N}^3 \\mid u + v + w = 2m\\}, \\quad B_1 = \\{(u, v, w) \\in B \\mid u \\ge m + 1\\},\n$$\n$$\nB_2 = \\{(u, v, w) \\in B \\mid v \\ge m + 1\\}, \\quad B_3 = \\{(u, v, w) \\in B \\mid w \\ge m + 1\\}.\n$$\nThen $A = B \\setminus (B_1 \\cup B_2 \\cup B_3)$. Therefore,\n$$\n\\begin{aligned} \\text{LHS of (1)} &= \\sum_{\\substack{i,j,k \\in \\mathbb{N} \\\\ i+j+k=m}} C_{i+j}C_{i+k}C_{j+k} = \\sum_{(u,v,w) \\in A} C_u C_v C_w \\\\ &= \\sum_{(u,v,w) \\in B} C_u C_v C_w - \\sum_{(u,v,w) \\in B_1} C_u C_v C_w - \\sum_{(u,v,w) \\in B_2} C_u C_v C_w - \\sum_{(u,v,w) \\in B_3} C_u C_v C_w. \\end{aligned} \\qquad (2)\n$$\nBy symmetry, the sums over $B_1$, $B_2$, and $B_3$ are the same. Using the recurrence relation for Catalan numbers:\n$$\nC_{n+1} = \\sum_{i=0}^{n} C_i C_{n-i} = C_0 C_n + C_1 C_{n-1} + \\dots + C_n C_0,\n$$\nwe have\n$$\n\\begin{aligned} \\sum_{(u,v,w) \\in B} C_u C_v C_w &= \\sum_{u=0}^{2m} C_u \\left( \\sum_{v=0}^{2m-u} C_v C_{2m-u-v} \\right) = \\sum_{u=0}^{2m} C_u C_{2m+1-u} \\\\ &= \\left( \\sum_{u=0}^{2m+1} C_u C_{2m+1-u} \\right) - C_{2m+1} C_0 = C_{2m+2} - C_{2m+1}, \\end{aligned}\n$$\n$$\n\\begin{aligned} \\sum_{(u,v,w) \\in B_1} C_u C_v C_w &= \\sum_{u=m+1}^{2m} C_u \\left( \\sum_{v=0}^{2m-u} C_v C_{2m-u-v} \\right) = \\sum_{u=m+1}^{2m} C_u C_{2m+1-u} \\\\ &= \\frac{1}{2} \\left( \\sum_{u=1}^{2m} C_u C_{2m+1-u} \\right) = \\frac{1}{2} (C_{2m+2} - 2C_{2m+1}). \\end{aligned}\n$$\nSince\n$$\n\\begin{aligned} C_{2m+2} &= \\frac{(4m+4)!}{(2m+2)!(2m+3)!} = \\frac{(4m+3)(4m+4)}{(2m+2)(2m+3)} \\cdot \\frac{(4m+2)!}{(2m+1)!(2m+2)!} \\\\ &= \\frac{2(4m+3)}{2m+3} \\cdot C_{2m+1}, \\end{aligned}\n$$\n\nwe have\n$$\n\\begin{align*}\n\\text{LHS of (1)} &= \\sum_{(u,v,w) \\in A} C_u C_v C_w = (C_{2m+2} - C_{2m+1}) - 3 \\cdot \\frac{1}{2} (C_{2m+2} - 2C_{2m+1}) \\\\\n&= 2C_{2m+1} - \\frac{1}{2}C_{2m+2} = \\frac{3}{2m+3}C_{2m+1} = \\text{RHS of (1).} \\quad \\square\n\\end{align*}\n$$\n\n\n*Proof 2 (Combinatorial Model).* The number of ways to triangulate a convex *n*-gon $P_1P_2\\cdots P_n$ by adding *n* − 3 non-intersecting diagonals is exactly the Catalan number $C_{n-2}$.\nThis result can be proven by induction on $n$. Consider vertex $P_n$. If $P_n$ is not connected by any diagonal, then $P_1$ and $P_{n-1}$ are connected, reducing the problem to the triangulation of a convex $n-1$-gon $P_1P_2\\cdots P_{n-1}$, which has $C_{n-3}$ ways. If $P_n$ is connected by a diagonal, let $P_k$ ($2 \\le k \\le n-2$) be the smallest vertex connected to $P_n$. Then $P_n$ and $P_k$ are connected, and $P_1$ to $P_{k-1}$ are not connected to $P_n$. This reduces to triangulating a convex $k$-gon $P_1P_2\\cdots P_k$ and a convex $n-k+1$-gon $P_kP_{k+1}\\cdots P_n$, which has $C_{k-2}C_{n-k-1}$ ways. Summing over $k=2, \\dots, n-2$ and adding $C_{n-3}$, we get:\n$$\n(C_0C_{n-3} + C_1C_{n-4} + \\dots + C_{n-4}C_1) + C_{n-3} = C_{n-2}.\n$$\nNow, consider the original problem. Take the regular $2m+3$-gon $P_1P_2\\cdots P_{2m+3}$ inscribed in a circle. We triangulate the $(2m+3)$-gon and select the unique triangle containing the center (the only acute triangle) and designate one vertex. Consider the total number of such operations.\nIf we first select the designated vertex $P_u$ (with $2m+3$ choices), then draw the triangle $P_uP_vP_w$\ncontaining the center, with $v-u, w-v$, and $u-w$ modulo $2m+3$ belonging to $\\{1, 2, \\dots, m+1\\}$, and\nthe sum of these residues being $2m+3$. This corresponds to $(i, j, k) \\in \\mathbb{N}^3$ satisfying $i+j+k=m$\nand\n$$\nv - u \\equiv i + j + 1, \\quad w - v \\equiv j + k + 1, \\quad u - w \\equiv k + i + 1 \\pmod{2m+3}.\n$$\nThen, the three regions between the sides of the triangle can be further triangulated, e.g., the\nregion containing $P_u P_v$ is a $i + j + 2$-gon, with $C_{i+j}$ ways to triangulate.\nThus, the total number of ways is\n$$\n(2m + 3) \\cdot \\sum_{\\substack{i,j,k \\in \\mathbb{N} \\\\ i+j+k=m}} C_{i+j} C_{i+k} C_{j+k}.\n$$\nOn the other hand, we can first triangulate the $2m+3$-gon, then choose one of the vertices of\nthe unique triangle containing the center, giving $3C_{2m+1}$ ways.\nEquating the two results, we get\n$$\n(2m + 3) \\cdot \\sum_{\\substack{i,j,k \\in \\mathbb{N} \\\\ i+j+k=m}} C_{i+j} C_{i+k} C_{j+k} = 3C_{2m+1}.\n$$\nThus, the stated identity holds.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 28611,
    "subject": "Mathematics (Multi-modal)",
    "question": "Fix an irrational number $\\alpha > 1$ and a positive integer $L$ such that $L > \\frac{\\alpha^2}{\\alpha - 1}$.\nGiven an integer $x_1 > L$, define a sequence $\\{x_n\\}$ as follows: for every integer $n \\ge 1$,\n$$\nx_{n+1} = \\begin{cases} \\lfloor \\alpha x_n \\rfloor, & \\text{if } x_n \\le L, \\\\ \\lfloor \\frac{x_n}{\\alpha} \\rfloor, & \\text{if } x_n > L. \\end{cases}\n$$\nHere, $\\lfloor u \\rfloor$ denotes the largest integer that is less than or equal to $u$.\n(1) Prove that the sequence $\\{x_n\\}$ is eventually periodic, i.e., there exist positive integers $T$ and $N$ such that for any integer $n > N$, we have $x_{n+T} = x_n$.\n(2) Prove that the smallest integer $T$ satisfying (1) is an odd integer that is independent of $x_1$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "*Proof*. First, since $\\alpha$ is an irrational number, the integer part operation in the definition of $x_{n+1}$ always makes the corresponding number strictly smaller. For any integer $u$ satisfying $\\frac{1}{\\alpha-1} < u \\le L$, we have $\\lfloor \\alpha u \\rfloor > \\alpha u - 1 > u$ and $\\lfloor \\alpha u \\rfloor \\le \\alpha L$; while for any integer $v \\ge L+1$, we have $\\lfloor \\frac{v}{\\alpha} \\rfloor > \\frac{L}{\\alpha} - 1 > \\frac{1}{\\alpha-1}$ and $\\lfloor \\frac{v}{\\alpha} \\rfloor < v$. Therefore, it is easy to verify by mathematical induction that for any positive integer $n$,\n$$\n\\frac{1}{\\alpha - 1} < x_n \\le \\max\\{x, \\alpha L\\},\n$$\nwhich implies that the sequence $\\{x_n\\}$ is bounded. Since $\\{x_n\\}$ is a recursive sequence, it must eventually become periodic. Let $T$ denote the smallest positive period of $\\{x_n\\}$. By definition, there exists a positive integer $N$ such that for any integer $n \\ge N$, we have $x_{n+N} = x_n$.\nSince $\\lfloor \\alpha L \\rfloor > \\alpha L - 1 > L$, it follows that $\\lfloor \\alpha L \\rfloor \\ge L + 1$. Thus, for any integer $n \\ge N$, we have $x_n \\le \\lfloor \\alpha L \\rfloor$. Consequently, for any integer $n \\ge N$, if $x_n \\ge L + 1$, then\n$$\nx_{n+1} = \\lfloor \\frac{x_n}{\\alpha} \\rfloor < \\frac{1}{\\alpha} \\cdot \\alpha L = L,\n$$\nand\n$$\nx_{n+2} = \\lfloor \\alpha x_{n+1} \\rfloor.\n$$\nClearly, there exists an integer $n \\ge N$ such that $x_n > L$. Let $m \\ge N + 1$ be the smallest integer such that\n$$\nx_{m-1} = \\min \\{x_n \\mid n \\ge N \\text{ and } x_n > L\\}.\n$$\nFrom the previous analysis, we know\n$$\nx_m = \\lfloor \\frac{x_{m-1}}{\\alpha} \\rfloor < L \\quad \\text{and} \\quad x_{m+1} = \\lfloor \\alpha x_m \\rfloor > \\alpha x_m - 1 > x_m.\n$$\nNote that $x_{m+1} < \\alpha x_m < \\alpha \\cdot \\frac{x_{m-1}}{\\alpha} = x_{m-1}$. By the minimality of $x_{m-1}$, we know $x_{m+1} \\le L$. Since $\\alpha x_m < x_{m+1} + 1 \\le L + 1$, it follows that\n$$\nx_m < \\frac{L+1}{\\alpha},\n$$\n\nand hence $x_m \\le \\lfloor \\frac{L+1}{\\alpha} \\rfloor$. Moreover, since $x_{m-1} \\ge L+1$, we have\n$$\nx_m = \\lfloor \\frac{x_{m-1}}{\\alpha} \\rfloor \\ge \\lfloor \\frac{L+1}{\\alpha} \\rfloor.\n$$\nTherefore, $x_m = \\lfloor \\frac{L+1}{\\alpha} \\rfloor$. By the minimality of $x_{m-1}$, we conclude that for any integer $n \\ge n_0$, $x_n \\ge x_m$.\nFor any integer $n \\ge N$, if $x_m + 1 \\le x_n \\le L$, then\n$$\nx_{n+1} = \\lfloor \\alpha x_n \\rfloor > \\alpha(x_m + 1) - 1 > \\alpha \\cdot \\frac{L+1}{\\alpha} - 1 = L,\n$$\nwhich implies $x_{n+1} \\ge L+1$. Moreover,\n$$\nx_{n+2} = \\lfloor \\frac{x_{n+1}}{\\alpha} \\rfloor < \\frac{x_{n+1}}{\\alpha} < \\frac{1}{\\alpha} \\cdot \\alpha x_n = x_n.\n$$\nCombining the above analysis, we have\n$$\nx_m < x_{m+1} \\le L < x_{m+2}.\n$$\nFurthermore, for each $i = 1, 2, \\dots, T-1$, when $i$ is odd, $x_{m+i} \\le L$; when $i$ is even, $x_{m+i} > L$. Note also that $x_{m+1} > x_{m+3} > \\dots > x_{m+T} = x_m$. Therefore, $T$ is an odd number.\nFinally, note that $x_m = \\lfloor \\frac{L+1}{\\alpha} \\rfloor$ is independent of $x$. Thus, the minimal positive period $T$ is also independent of $x$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 28612,
    "subject": "Mathematics (Multi-modal)",
    "question": "There are 8 cards on the table numbered from 1 to 8. Two players, $A$ and $B$, play the following game. In each round:\n\n* Player $A$ selects two cards from the table\n* Player $B$, after seeing the two selected cards, chooses one to keep and discards the other\n\nThe game consists of four rounds with the restriction that:\n\n* In rounds 1 and 2, $B$ cannot choose the larger number in both rounds\n* In rounds 3 and 4, $B$ cannot choose the larger number in both rounds\n\nLet $S$ be the sum of the numbers on the four cards $B$ holds after four rounds.\n\nFind the largest integer $N$ such that no matter how $A$ selects cards in each round, $B$ can guarantee $S \\ge N$.",
    "options": [],
    "answer": "17",
    "solution": "*Proof.* The maximum achievable value of $N$ is $17$.\n\nLet us denote the numbers on the two cards selected by Player $A$ in the $k$-th round as $a_k$ and $b_k$, where $a_k < b_k$. The number selected by Player $B$ is denoted as $c_k$, and the discarded number as $d_k$.\n\nLet $T = d_1 + d_2 + d_3 + d_4$. Then we have $S + T = 1 + 2 + \\dots + 8 = 36$.\n\nPlayer $B$ has a strategy to ensure $S - T \\ge -3$, and consequently $S \\ge 17$ (note that $S$ must be an integer).\n\n* If in the first round, $b_1 - a_1 \\ge 4$, then Player $B$ selects $b_1$, and in the second round selects $a_2$. This gives $c_1 - d_1 \\ge 4$ and $c_2 - d_2 \\ge 1 - 8 = -7$.\n* If in the first round, $b_1 - a_1 \\le 3$, then Player $B$ selects $a_1$, and in the second round selects $b_2$. This gives $c_1 - d_1 \\ge -3$ and $c_2 - d_2 \\ge 1$.\n\nTherefore, Player $B$ can always ensure $(c_1 + c_2) - (d_1 + d_2) \\ge -3$.\n\nIn the third and fourth rounds, Player $B$ can always ensure $(c_3 + c_4) - (d_3 + d_4) \\ge 0$. This is because after Player $A$ has chosen $a_3$ and $b_3$, $a_4$ and $b_4$ are also determined. Player $B$ can then select the pair $(a_3, b_4)$ or $(b_3, a_4)$ that yields the larger sum. Thus, Player $B$ can always guarantee $S - T \\ge -3$.\n\nPlayer $A$ has a strategy to ensure $S \\le 17$:\n\n* In the first round, Player $A$ selects $(3, 6)$. If Player $B$ chooses $3$, then in the second round Player $A$ selects $(4, 5)$. The sum of Player $B$'s selections in the first two rounds will not exceed $8$.\n\nIn the last two rounds, Player $A$ selects $(1, 2)$ and $(7, 8)$, and Player $B$'s selections in these rounds will not exceed $9$. Therefore, $S \\le 17$.\n\n* If in the first round Player $B$ chooses $6$, then in the second round Player $A$ selects $(1, 8)$, forcing Player $B$ to choose $1$. In the last two rounds, Player $A$ selects $(2, 4)$ and $(5, 7)$, and Player $B$'s selections in these rounds will not exceed $9$. Thus, $S \\le 6 + 1 + 9 = 16$.\n\nIn conclusion, Player $A$ has a strategy to ensure $S \\le 17$. $\\Box$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 28613,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $I$ be the incenter of a triangle $ABC$. Write $L$, $M$, and $N$ for the midpoints of $AI$, $AC$, and $CI$, respectively. Assume that there is a point $D$ in the interior of the segment $AM$ such that $BD = BC$. The incircle of $\\triangle ABD$ touches $AD$ and $BD$ at $E$ and $F$, respectively. Let $J$ be the circumcenter of $\\triangle AIC$. Let $\\omega$ denote the circumcircle of $\\triangle DMJ$. The line $MN$ intersects $\\omega$ at point $P(\\neq M)$, and the line $JL$ intersects $\\omega$ at point $Q(\\neq J)$.\nProve that the three lines $EF$, $PQ$, and $LN$ pass through a common point.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "![](attached_image_1.png)\n\n**Proof.** Let $K$ be the midpoint of $ID$, then $LN$ passes through $K$. Next, we prove that both $EF$ and $PQ$ pass through $K$.\nLet $X$ and $Y$ be the reflections of $D$ across $E$ and $F$, respectively. Let $Z$ be the projection of $I$ onto $AC$. Then, we have\n$$\nCX = 2DE + CD = AD + BD - AB + CD = AC + BC - AB = 2CZ.\n$$\nThus, $CZ = ZX$. Combining this with $IZ \\perp CX$, we know $IC = IX$. Therefore,\n$$\n2\\angle IXC = 2\\angle ICX = \\angle ACB = \\angle BDC = 2\\angle FEC = 2\\angle YXC.\n$$\nThus, $I$, $Y$, and $X$ are collinear, which implies that $EF$ passes through $K$.\nSince $MN // AI$, we have $\\angle PQD = \\angle PMD = \\angle IAD$.\nFrom $\\angle DMJ = 90^\\circ$, we know that $DJ$ is the diameter of $\\omega$, so $DQ \\perp QJ$. Combined with $AL \\perp LJ$, we conclude that $AI // DQ$. Let $T$ be the reflection of $D$ across $Q$. Then $ADTI$ forms an isosceles trapezoid. Since $KQ // IT$, we have\n$$\n\\angle DQK = \\angle DTI = 180^\\circ - \\angle IAD = 180^\\circ - \\angle PQD,\n$$\nwhich implies that $PQ$ passes through $K$. Thus, the proof is complete.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 28614,
    "subject": "Mathematics (Multi-modal)",
    "question": "A positive integer $N$ is called a *good number* if the decimal representation of $N$ can be divided into at least 5 segments of digits, each segment containing at least one non-zero digit, and these segments (ignoring any leading zeros) can be viewed as positive integers which can be divided into two groups, with each group forming a geometric sequence in the appropriate order. (If a group has only one or two positive integers, it is also considered a geometric sequence.)\nFor example, 20240327 is a good number. In fact, it can be divided into $2|02|403|2|7$, a total of 5 segments of digits, with the two groups of positive integers (2, 2, 2) and (7, 403) each forming a geometric sequence.\nLet $p = 1 + a + a^2 + \\dots + a^m$ be a prime number, where $a > 1$ and $m > 2$ are integers. Prove that $\\frac{10^{p-1}-1}{p}$ is a good number.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $p = 1 + a + a^2 + \\cdots + a^m$ be a prime number, where $a > 1$ and $m > 2$ are integers. Prove that $\\frac{10^{p-1}-1}{p}$ is a good number.\n\n*Proof*. Since $p = 1 + a + \\cdots + a^m = \\frac{a^{m+1}-1}{a-1}$ is a prime number, $q = m + 1$ must be a prime number. Given $q = m + 1 > 3$, we have $q \\ge 5$. Note that $p = \\frac{a^q-1}{a-1}$ divides $a^q-1$ and does not divide $a-1$, so the order of $a$ modulo $p$ is exactly $q$. Therefore, $q \\mid p-1$. Let $p-1 = qL$.\n\nConsider the remainder $t$ of $10^L = 10^{(p-1)/q}$ modulo $p$, which satisfies $t^q \\equiv 1 \\pmod{p}$. By Lagrange's theorem, the solutions to the congruence equation $x^q - 1 \\equiv 0 \\pmod{p}$ are exactly $\\{1, a, a^2, \\cdots, a^{q-1}\\}$. Thus, we can assume $t = a^r$, where $r \\in D = \\{0, 1, \\cdots, q-1\\}$. Furthermore, $10^{kL}$ modulo $p$ is $a^{\\langle kr \\rangle}$, where $\\langle \\cdot \\rangle$ denotes the remainder modulo $q$, taking values in $D = \\{0, 1, \\cdots, q-1\\}$.\n\nConsider $N = \\frac{10^{p-1}-1}{p} = \\frac{10^{qL}-1}{p} = \\lfloor \\frac{10^{qL}-1}{p} \\rfloor$ as a $p-1$ digit number, with leading zeros if necessary. Divide $N$ into segments of length $L$ from left to right (the leftmost segment may contain leading zeros), forming $q$ segments corresponding to integers $x_0, x_1, \\cdots, x_{q-1}$:\n$$\n\\begin{align*}\nx_0 &= \\lfloor \\frac{10^L}{p} \\rfloor = \\frac{10^L - a^r}{p}, \\\\\nx_1 &= \\lfloor \\frac{10^{2L}}{p} \\rfloor - 10^L \\cdot x_0 = \\frac{10^{2L} - a^{\\langle 2r \\rangle}}{p} - 10^L \\cdot \\frac{10^L - a^r}{p} = \\frac{a^r \\cdot 10^L - a^{\\langle 2r \\rangle}}{p}, \\\\\n\\multicolumn{2}{c}{\\dots \\dots \\dots} \\\\\nx_k &= \\frac{a^{\\langle kr \\rangle} \\cdot 10^L - a^{\\langle (k+1)r \\rangle}}{p}, \\quad k = 1, 2, \\cdots, q-1.\n\\end{align*}\n$$\nIf $r=0$, then $x_0 = x_1 = \\cdots = x_{q-1} = \\frac{10^L-1}{p}$ can be divided into two groups, each forming a geometric sequence.\n\nIf $r \\in \\{1, 2, \\cdots, q-1\\}$, then $\\{\\langle kr \\rangle \\mid k = 0, 1, \\cdots, q-1\\} = \\{0, 1, 2, \\cdots, q-1\\}$. Let\n$$\nA = \\{k \\in D \\mid \\langle kr \\rangle = 0, 1, \\cdots, q-r-1\\}, \\quad B = \\{k \\in D \\mid \\langle kr \\rangle = q-r, \\cdots, q-1\\}.\n$$\nWhen $k \\in A$, $\\langle (k+1)r \\rangle = \\langle kr \\rangle + r$, and when $k \\in B$, $\\langle (k+1)r \\rangle = \\langle kr \\rangle + r - q$. Thus,\n$$\n\\{x_k \\mid k \\in A\\} = \\frac{10^L - a^r}{p} \\cdot \\{1, a, a^2, \\cdots, a^{q-r-1}\\} \\text{ forms a geometric sequence,}\n$$\n$$\n\\{x_k \\mid k \\in B\\} = \\frac{10^L - a^{r-q}}{p} \\cdot \\{a^{q-r}, a^{q-r+1}, \\cdots, a^{q-1}\\} = \\frac{a^{q-r} \\cdot 10^L - 1}{p} \\cdot \\{1, a, \\cdots, a^{r-1}\\}\n$$\nalso forms a geometric sequence. This completes the proof. $\\square$",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 28615,
    "subject": "Mathematics (Multi-modal)",
    "question": "For a positive integer $n$, a subset $S$ of $\\{1,2,\\dots,n\\}$ is called an $n$-good set if for any elements $x,y$ in $S$ (they can be the same), if $x+y \\le n$, then $x+y \\in S$.\nFor a positive integer $n$, define $r_n$ as the smallest real number such that for any positive integer $m \\le n$, there exists an $n$-good set with $m$ elements, whose sum of all elements does not exceed $m \\cdot r_n$.\nProve that there exists a real number $\\alpha$ such that for any positive integer $n$, we have $|r_n - \\alpha n| < 2024$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "*Proof.* We will show that $\\alpha = 2 - \\sqrt{2}$ satisfies the requirement. We will prove this in two steps.\n\nFirst, we prove that $r_n \\ge \\alpha n - 4$. For this, we only need to consider the case where $n > 4$. Let $m = \\lfloor \\frac{n}{\\sqrt{2}} \\rfloor < n$, and let $d = n - m$. Consider an $n$-good set $S$ with $m$ elements, and let the complement of $S$ be $S^c = \\{1, 2, \\dots, n\\} \\setminus S = \\{a_1, a_2, \\dots, a_d\\}$, where $a_1 < a_2 < \\dots < a_d$. For any $1 \\le i \\le d$, we claim that $a_i \\le 2i - 1$. In fact, if $a_i \\ge 2i$, then among the pairs $(1, a_i - 1)$, $(2, a_i - 2)$, \\dots, $(i, a_i - i)$, at least one pair would have both numbers in $S$, which contradicts the condition for $S$ being an $n$-good set. (If $a_i = 2i$, the last pair has equal numbers, but this does not affect the proof.) Therefore,\n$$\n\\sum_{x \\in S} x = \\frac{n(n+1)}{2} - \\sum_{x \\in S^c} x \\ge \\frac{n(n+1)}{2} - (1 + 3 + \\dots + (2d - 1)) = \\frac{n(n+1)}{2} - d^2.\n$$\nThus,\n$$\nr_n \\ge \\frac{\\frac{n(n+1)}{2} - d^2}{n-d} = (n+d) - \\frac{n(n-1)}{2(n-d)} = 2n - m - \\frac{n(n-1)}{2m} \\quad (3)\n$$\n$$\n> 2n - \\left(m + \\frac{n^2}{2m}\\right) > 2n - \\sqrt{2n} - 1 = \\alpha n - 1, \\quad (4)\n$$\nwhere the last inequality holds because $m + \\frac{n^2}{2m}$ is minimized at $m = \\frac{n}{\\sqrt{2}}$, with rounding causing at most an error of 1.\n\nNext, we prove that $r_n \\le \\alpha n + 100$. For this, we only need to consider the case where $n > 100$. We discuss several cases.\n\n**Case 1:** If $m = n$, $\\{1, 2, \\dots, n\\}$ is an $n$-good set, with an arithmetic mean of $\\frac{n+1}{2} < \\alpha n + 4$.\n\n**Case 2:** If $m \\le \\sqrt{n} - 1$, let $k = \\lfloor \\frac{n+1}{m+1} \\rfloor$, then $(m+1)k \\ge n+1 > n$, while\n$$\nmk \\le m \\cdot \\frac{m+n+1}{m+1} = \\frac{mn+m^2+m}{m+1} \\le \\frac{mn+n}{m+1} = n.\n$$\nThus, $\\{k, 2k, \\dots, mk\\}$ is an $n$-good set with $m$ elements, and its arithmetic mean is\n$$\n\\frac{m+1}{2} \\cdot k \\le \\frac{m+1}{2} \\cdot \\frac{m+n+1}{m+1} = \\frac{n+m+1}{2} \\le \\frac{n+\\sqrt{n}}{2} \\le \\frac{11n}{20} < \\alpha n.\n$$\n\n**Case 3:** If $\\sqrt{n} - 1 < m \\le n - 1$, let $\\lfloor \\frac{n}{k} \\rfloor \\le m < \\lfloor \\frac{n}{k-1} \\rfloor$, where $k \\ge 2$. We first choose all multiples of $k$ from $\\{1, 2, \\dots, n\\}$ (a total of $\\lfloor \\frac{n}{k} \\rfloor$ elements), then choose the remaining $m - \\lfloor \\frac{n}{k} \\rfloor$ elements that have the same remainder $r$ modulo $k$ (where $r \\in \\{1, 2, \\dots, k-1\\}$ and $r \\equiv n+1 \\pmod{k}$) or $r \\equiv n+2 \\pmod{k}$). When $k=2$, the unchosen numbers are just a few starting odd numbers, and from the inequality (4), it is easy to see that the chosen numbers form an $n$-good set.\n\nWhen $k \\ge 3$, $m \\le \\lfloor \\frac{n}{k} \\rfloor + \\lfloor \\frac{n}{2k} \\rfloor$. This is because\n$$\n\\text{For } k \\ge 4, \\left\\lfloor \\frac{n}{k} \\right\\rfloor + \\left\\lfloor \\frac{n}{2k} \\right\\rfloor > \\frac{3n}{2k} - 2 = \\frac{n}{k-1} + \\frac{(k-3)n}{2k(k-1)} - 2 \\ge \\frac{n}{k-1} + \\frac{n}{24} - 2 > \\frac{n}{k-1},\n$$\n$$\n\\text{For } k \\ge 3, \\left\\lfloor \\frac{n}{k} \\right\\rfloor + \\left\\lfloor \\frac{n}{2k} \\right\\rfloor > \\frac{n}{2} - 2.\n$$\nTherefore, the chosen numbers modulo $k$ are greater than $n - \\lfloor \\frac{n}{2k} \\rfloor \\cdot k \\ge \\frac{n}{2}$, so any two chosen numbers sum to more than $n$, indicating that the chosen numbers form an $n$-good set. Let the multiples of $k$ be $a = \\lfloor \\frac{n}{k} \\rfloor$, then the numbers with the same remainder $r$ modulo $k$ have $m - a$ elements, and their mean does not exceed\n$$\n\\begin{aligned}\n& \\frac{1}{m} \\left( \\frac{ka(a+1)}{2} + (m-a)(n+2) - \\frac{k(m-a)(m-a+1)}{2} \\right) \\\\\n&= n + 2 + \\left(a - \\frac{1}{2}\\right)k - \\left(\\frac{km}{2} + \\frac{(n+k+2)a}{m}\\right) \\\\\n&\\le n + 2 + \\left(a - \\frac{1}{2}\\right)k - \\sqrt{2ak(n+k+2)} \\\\\n&< 2n + 2 - \\sqrt{2(n-k+1)(n+k-1)} \\\\\n&= 2n + 2 - \\sqrt{2(n^2 - (k-1)^2)} \\\\\n&\\le 2n + 2 - \\sqrt{2(n^2 - n)} < \\alpha n + 4.\n\\end{aligned}\n$$\nTherefore, the arithmetic mean of the elements in the above $n$-good set does not exceed $\\alpha n + 4$.\n\nIn conclusion, $\\alpha = 2 - \\sqrt{2}$ satisfies the requirement, and the proof is complete. $\\square$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 28616,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find the smallest real number $\\lambda$ with the following property: For any positive integers $a$, $b$, $n$ where $a + b$ is not divisible by $n$, there exists a positive integer $k < n$ such that\n$$\n\\left\\{ \\frac{ak}{n} \\right\\} + \\left\\{ \\frac{bk}{n} \\right\\} \\le \\lambda,\n$$\nwhere $\\{x\\} = x - \\lfloor x \\rfloor$ denotes the fractional part of $x$.",
    "options": [],
    "answer": "2/3",
    "solution": "*Proof.* The minimal $\\lambda$ is $\\frac{2}{3}$.\n\nFirst, we show $\\lambda \\ge \\frac{2}{3}$. Consider $n = 3$ with $a = b = 1$. For $k = 1, 2$, we have:\n$$\n\\left\\{ \\frac{k}{3} \\right\\} + \\left\\{ \\frac{k}{3} \\right\\} = \\frac{2k}{3} \\ge \\frac{2}{3},\n$$\nthus $\\lambda$ cannot be smaller than $\\frac{2}{3}$.\n\nNow we prove $\\lambda \\le \\frac{2}{3}$. Let $n$ be a positive integer and $a$, $b$ positive integers with $n \\nmid a + b$. We need to find $1 \\le k \\le n - 1$ satisfying the inequality.\n\n**Case 1:** $n = 2$. Without loss of generality, take $a = 1$, $b = 1$. For $k = 1$:\n$$\n\\left\\{ \\frac{1}{2} \\right\\} + \\left\\{ \\frac{1}{2} \\right\\} = \\frac{1}{2} < \\frac{2}{3}.\n$$\n\n**Case 2:** $n \\ge 3$.\n\n**Subcase 2.1:** $\\text{gcd}(a, n) = d > 1$. Let $k = \\frac{n}{d}k'$ where $1 \\le k' \\le d - 1$. Then:\n$$\n\\left\\{ \\frac{ak}{n} \\right\\} + \\left\\{ \\frac{bk}{n} \\right\\} = \\left\\{ \\frac{bk'}{d} \\right\\}.\n$$\nSince $\\left\\{ \\frac{bk'}{d} \\right\\} + \\left\\{ \\frac{b(d-k')}{d} \\right\\} \\in \\{0, 1\\}$, taking $k' = 1$ or $d - 1$ gives:\n$$\n\\left\\{ \\frac{bk'}{d} \\right\\} \\le \\frac{1}{2} < \\frac{2}{3}.\n$$\n\n**Subcase 2.2:** $\\text{gcd}(a, n) = 1$. Let $t$ be such that $ta \\equiv 1 \\pmod{n}$. Replacing $a$ with $ta \\pmod{n}$ and $b$ with $tb \\pmod{n}$, we may assume $a = 1$ and $1 \\le b \\le n - 2$.\nLet $e = 1 - \\frac{1}{n} - \\frac{b}{n} \\ge \\frac{1}{n}$.\n\n**When** $e \\ge \\frac{1}{3}$: Take $k = 1$:\n$$\n\\left\\{ \\frac{1}{n} \\right\\} + \\left\\{ \\frac{b}{n} \\right\\} = 1 - e \\le \\frac{2}{3}.\n$$\n\n**When** $e < \\frac{1}{3}$: Let $k$ be the smallest positive integer with $ke \\ge \\frac{1}{3}$. Then:\n$$\nk = \\left\\lfloor \\frac{1}{3e} \\right\\rfloor \\le \\left\\lfloor \\frac{n}{3} \\right\\rfloor < \\frac{n}{3} + 1 \\le n - 1.\n$$\nLet $e = \\frac{l}{n}$ where $l$ is integer and $3l < n$ (since $e < \\frac{1}{3}$). We have:\n$$\nke + \\frac{k}{n} \\le \\frac{(n + 3l - 1)(l + 1)}{3ln} \\le \\frac{2(l + 1)}{3l + 1} \\le 1.\n$$\n\nThis implies:\n$$\n\\left\\{ \\frac{k}{n} \\right\\} + \\left\\{ \\frac{kb}{n} \\right\\} = 1 - ke \\le \\frac{2}{3}.\n$$",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 28617,
    "subject": "Mathematics (Multi-modal)",
    "question": "Given an integer $n > 1$, let the real number $x > 1$ satisfy\n$$\nx^{101} - n x^{100} + n x - 1 = 0.\n$$\nProve that for any real numbers $0 < a < b < 1$, there exists a positive integer $m$ such that\n$$\na < \\{x^m\\} < b.\n$$\nHere $\\{t\\} = t - \\lfloor t \\rfloor$ denotes the fractional part of the real number $t$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "*Proof.* We will sequentially prove the following conclusions:\n\n(1) The equation (5) has 99 roots with modulus equal to 1.\n\nClearly, $x = 1$ is a root of the equation (5). Consider the equation\n$$\n\\frac{x^{101} - n x^{100} + n x - 1}{x - 1} = 0,\n$$\ni.e.,\n$$\nf(x) = x^{100} - (n - 1) \\sum_{j=1}^{99} x^j + 1 = 0.\n$$\nConsider all 100th roots of unity $\\omega$, it is easy to see that $f(\\omega) = n + 1$.\nNotice that $\\frac{f(x)}{x^{50}}$ can be expressed as $g(x + \\frac{1}{x})$, where $g(x) \\in \\mathbb{Z}[x]$. Therefore,\n$$\ng\\left(2 \\cos \\frac{k\\pi}{50}\\right) = \\left(e^{ik\\pi/50}\\right)^{50} f\\left(e^{ik\\pi/50}\\right) = (n+1)(-1)^k, \\quad k = 0, 1, 2, \\dots, 49.\n$$\nThis shows that $g(2 \\cos \\theta)$ has a root in the interval $(\\frac{k\\pi}{50}, \\frac{(k+1)\\pi}{50})$ with $\\theta = \\theta_k$ ($k = 0, 1, 2, \\dots, 48$).\nTherefore, $g(x)$ has 49 pairs of conjugate complex roots $\\cos \\theta_k \\pm i \\sin \\theta_k$. Thus, $f(x)$ has 98 roots with modulus equal to 1, and hence the original equation (5) has 99 roots with modulus equal to 1.\n\n(2) $f(x)$ has a root $\\alpha$ greater than 1.\n\nThis is because $f(1) = 2 - 99(n - 1) < 0$, which follows from the intermediate value theorem.\nCombining $f(0) = 1 > 0$ and the fact that the product of all roots of $f(x)$ is 1, we know that besides the 99 roots with modulus equal to 1, the remaining two roots are $\\alpha$ and $\\frac{1}{\\alpha}$.\n\n(3) The 99 roots of $f(x)$ with modulus equal to 1 are not all roots of unity.\n\nIf any of these roots are roots of unity, the cyclotomic polynomial corresponding to that root divides $f(x)$. Therefore, if they are all roots of unity, this means $(x - \\alpha)(x - \\frac{1}{\\alpha}) = x^2 - A x + 1$, where $A$ is a positive integer.\nIf $A \\ge n + 1$, then $\\alpha \\ge n$, thus $\\alpha^{101} - n\\alpha^{100} + n\\alpha - 1 > 0$, which is a contradiction!\nIf $A \\le n$, then using $(\\alpha + 1)(\\alpha^2 - A\\alpha + 1) = \\alpha^3 - (A - 1)\\alpha^2 - (A - 1)\\alpha + 1$, we get\n$$\n\\alpha^3 < (A - 1)\\alpha^2 + (A - 1)\\alpha \\le (n - 1)\\alpha^2 + (n - 1)\\alpha.\n$$\nTherefore,\n$$\n\\alpha^{100} - (n - 1)(\\alpha^{99} + \\alpha^{98} + \\dots + \\alpha + 1) < -(n + 1)(\\alpha^{97} + \\dots + \\alpha) + 1 < 0,\n$$\nwhich is a contradiction!\nThus, the assumption is not valid, meaning that the roots of (5) are not all roots of unity.\n\nSince the powers of the roots of unity are finite in number, and using Newton's identities, we know that the sum of the powers of all roots of the original equation are integers. To prove the original problem, we need to show the following conclusion:\n\nIf $z_1, z_2, \\dots, z_k$ are complex numbers with modulus 1 but not roots of unity, then the fractional parts of $S_r = \\sum_{j=1}^k (z_j^r + \\bar{z}_j^r)$ are dense in $(0, 1)$.\nThat is:\nIf $\\lambda_1, \\dots, \\lambda_k$ are irrational numbers, then the fractional parts of $T_r = \\sum_{j=1}^k \\cos(2r\\lambda_j\\pi)$ are dense in $(0, 1)$.\n\nConsider a sufficiently large positive integer $N$, and define the sequence\n$$\nX_r = (\\lfloor N\\{r\\lambda_1\\}\\rfloor, \\dots, \\lfloor N\\{r\\lambda_k\\}\\rfloor) \\quad (r = 1, 2, \\dots, N^k + 1).\n$$\nBy the pigeonhole principle, there exist $r_1 < r_2$ such that $X_{r_1} = X_{r_2}$. This means that for $s = r_2 - r_1$, $\\{s\\lambda_j\\}$ is either less than $\\frac{1}{N}$ or greater than $1 - \\frac{1}{N}$, which implies $\\cos(2s\\lambda_j\\pi) > \\cos\\frac{2\\pi}{N}$. Therefore, $T_s > k \\cos\\frac{2\\pi}{N}$.\nOn the other hand, since $\\{s\\lambda_1\\} \\neq 0$, there exists a positive integer $t$ such that $|t\\{s\\lambda_1\\} - \\frac{1}{2}| < \\frac{1}{N}$, hence $\\cos(2st\\lambda_1\\pi) < -\\cos\\frac{2\\pi}{N}$. Therefore, $T_{st} < (k-1) - \\cos\\frac{2\\pi}{N}$.\nBut we know,\n$$\n\\begin{aligned}\n|T_{(d+1)s} - T_{ds}| &= \\left| \\sum_{j=1}^{k} (\\cos(2(d+1)s\\lambda_j\\pi) - \\cos(2ds\\lambda_j\\pi)) \\right| \\\\\n&= \\left| \\sum_{j=1}^{k} 2 \\sin(s\\lambda_j\\pi) \\sin((2d+1)s\\lambda_j\\pi) \\right| \\\\\n&\\le \\sum_{j=1}^{k} |2 \\sin(s\\lambda_j\\pi)| < 2k \\sin \\frac{2\\pi}{N},\n\\end{aligned}\n$$\ntherefore, by taking $N$ sufficiently large, using the discrete intermediate value theorem, we can see that $T_r$ is dense in $(k - \\frac{3}{2}, k - \\frac{1}{2})$.\n\nThe desired conclusion is thus proven. □",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 28618,
    "subject": "Mathematics (Multi-modal)",
    "question": "Given an integer $n \\ge 3$. Let $\\frac{n(n-1)}{2}$ non-negative real numbers $x_{i,j}$ ($1 \\le i < j \\le n$) satisfy: for any $1 \\le i < j < k \\le n$, we have $x_{i,j} + x_{j,k} \\le x_{i,k}$. Prove that:\n$$\n\\left\\lfloor \\frac{n^2}{4} \\right\\rfloor \\cdot \\sum_{1 \\le i < j \\le n} x_{i,j}^4 \\ge \\left( \\sum_{1 \\le i < j \\le n} x_{i,j}^2 \\right)^2 .\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "*Proof.* First, let's point out a situation where the equality holds, which will help us understand the problem. Let $0 = y_1 = \\cdots = y_{\\lfloor \\frac{n}{2} \\rfloor} < y_{\\lfloor \\frac{n}{2} \\rfloor+1} = \\cdots = y_n = 1$, and take $x_{i,j} = y_j - y_i$ for $1 \\le i < j \\le n$. In this case, $x_{i,j} + x_{j,k} = x_{i,k}$, and\n$$\n\\sum_{1 \\le i < j \\le n} x_{i,j}^4 = \\sum_{1 \\le i < j \\le n} x_{i,j}^2 = \\left\\lfloor \\frac{n}{2} \\right\\rfloor \\cdot \\left\\lfloor \\frac{n}{2} \\right\\rfloor = \\left\\lfloor \\frac{n^2}{4} \\right\\rfloor,\n$$\nshowing that the inequality holds with equality in this case.\n\nNow, let's prove the inequality. We start with the following lemma.\n\n**Lemma:** Let $m \\ge 2$ be a positive integer and $x_1, x_2, \\dots, x_m \\ge 0$. Let $\\lambda_m = \\left\\lfloor \\frac{m^2}{4} \\right\\rfloor$. Then,\n$$\n\\lambda_m \\sum_{i=1}^{m} x_i^3 \\ge \\left( m \\sum_{i=1}^{m} x_i^2 - \\left( \\sum_{i=1}^{m} x_i \\right)^2 \\right) \\sum_{i=1}^{m} x_i.\n$$\n\n**Proof of Lemma:** Without loss of generality, assume $\\sum_{i=1}^m x_i = 1$. For fixed $x_1, x_2, \\dots, x_m$, assume $x_1 \\ge x_2 \\ge \\dots \\ge x_k > 0$ and $x_i = 0$ for $i > k$. Let $s = \\frac{1}{k} \\sum_{i=1}^k x_i = \\frac{1}{k}$, and define\n$$\nF(x_1, x_k) := \\lambda_m \\sum_{i=1}^{m} x_i^3 - \\left( m \\sum_{i=1}^{m} x_i^2 - \\left( \\sum_{i=1}^{m} x_i \\right)^2 \\right) \\sum_{i=1}^{m} x_i.\n$$\nWhen $x_1 + x_k$ is fixed and $x_2, x_3, \\dots, x_{k-1}, x_{k+1}, \\dots, x_m$ are fixed, $F$ is a linear function of $x_1x_k$. Hence, $F$ achieves its minimum value when $x_1x_k$ is minimized. Notice that\n$$\ns(x_1 + x_k - s) \\ge x_1 x_k \\ge (x_1 + x_k) \\cdot 0 = 0,\n$$\ntherefore,\n$$\nF(x_1, x_k) \\ge \\min \\{F(x_1 + x_k, 0), F(s, x_1 + x_k - s)\\}.\n$$\nThis means that we can always adjust the values such that one variable becomes zero or one variable becomes the arithmetic mean of all positive variables. This adjustment process will terminate in finite steps, leaving all positive variables equal. Assume $x_1 = x_2 = \\dots = x_u > x_{u+1} = x_{u+2} = \\dots = x_m = 0$. We need to prove that for any $1 \\le u \\le m$,\n$$\n\\lambda_m u \\ge (mu - u^2)u \\iff \\lambda_m \\ge u(m - u).\n$$\nThis inequality is true by the AM-GM inequality. $\\square$\n\nReturning to the original problem, let $\\lambda_n = \\left\\lfloor \\frac{n^2}{4} \\right\\rfloor$. We will use induction to prove a stronger statement: for any positive integers $n \\ge 2$ and $\\ell$, the following holds:\n$$\nP(n, \\ell) := \\lambda_{n+\\ell-1} \\left( \\sum_{1 \\le i < j < n} x_{i,j}^4 + \\ell \\sum_{i=1}^{n-1} x_{i,n}^4 \\right) - \\left( \\sum_{1 \\le i < j < n} x_{i,j}^2 + \\ell \\sum_{i=1}^{n-1} x_{i,n}^2 \\right)^2 \\ge 0. \\quad (12)\n$$\nWhen $\\ell = 1$, this is the desired inequality.\n\nWe use induction on $n$. For $n = 2$, (12) is equivalent to\n$$\n\\lambda_{\\ell+1} \\ell x_{12}^4 \\ge \\ell^2 x_{12}^4 \\iff \\lambda_{\\ell+1} \\ge \\ell,\n$$\nwhich holds. For $n \\ge 3$, fix $n, \\ell, x_{i,j}$ for $1 \\le i < j < n$, and let\n$$\nf(x_1, x_2, \\dots, x_{n-1}) := \\lambda_{n+\\ell-1} \\left( \\sum_{1 \\le i < j < n} x_{i,j}^4 + \\ell \\sum_{i=1}^{n-1} x_i^4 \\right) - \\left( \\sum_{1 \\le i < j < n} x_{i,j}^2 + \\ell \\sum_{i=1}^{n-1} x_i^2 \\right)^2.\n$$\nIf $x_j + x_{i,j} \\le x_i$ for $1 \\le i < j < n$, then\n$$\n\\sum_{1 \\le i < j < n} x_{i,j}^2 \\le \\sum_{1 \\le i < j < n} (x_i - x_j)^2 = (n-1) \\sum_{i=1}^{n-1} x_i^2 - \\left( \\sum_{i=1}^{n-1} x_i \\right)^2. \\quad (13)\n$$\nUsing (13) and applying the lemma with $m = n + \\ell - 1$, $x_i = 0$ for $n \\le i \\le n + \\ell - 1$, we have\n$$\n\\begin{aligned} \\sum_{i=1}^{n-1} \\frac{\\partial f}{\\partial x_i} &= 4\\ell\\lambda_{n+\\ell-1} \\sum_{i=1}^{n-1} x_i^3 - 4\\ell \\left( \\sum_{1 \\le i < j < n} x_{i,j}^2 + \\ell \\sum_{i=1}^{n-1} x_i^2 \\right) \\sum_{i=1}^{n-1} x_i \\\\ &\\ge 4\\ell\\lambda_{n+\\ell-1} \\sum_{i=1}^{n-1} x_i^3 - 4\\ell \\left( (n+\\ell-1) \\sum_{i=1}^{n-1} x_i^2 - \\left( \\sum_{i=1}^{n-1} x_i \\right)^2 \\right) \\sum_{i=1}^{n-1} x_i \\ge 0. \\end{aligned}\n$$\nLet $g(y) = f(x_{1,n} + y, \\dots, x_{n-1,n} + y)$. For $y \\ge -x_{n-1,n}$, $g'(y) \\ge 0$. Therefore, $g(0) \\ge g(-x_{n-1,n})$. We only need to consider the case where $x_{n-1,n} = 0$.\n\nWhen $x_{n-1,n} = 0$, let\n$$\nA_p := \\sum_{1 \\le i < j < n-1} x_{i,j}^p, \\quad B_p := \\sum_{i=1}^{n-2} x_{i,n-1}^p, \\quad C_p := \\sum_{i=1}^{n-2} x_{i,n}^p, \\quad p \\in \\{2, 4\\}.\n$$\nThen, (12) is equivalent to\n$$\n\\lambda_{n+\\ell-1}(A_4 + B_4 + \\ell C_4) \\ge (A_2 + B_2 + \\ell C_2)^2. \\quad (14)\n$$\nBy the induction hypothesis, $P(n-1, \\ell+1) \\ge 0$, so\n$$\n\\begin{aligned} \\lambda_{n+\\ell-1}(A_4 + (\\ell+1)B_4) &\\ge (A_2 + (\\ell+1)B_2)^2, \\\\ \\lambda_{n+\\ell-1}(A_4 + (\\ell+1)C_4) &\\ge (A_2 + (\\ell+1)C_2)^2. \\end{aligned}\n$$\nUsing these and the Cauchy-Schwarz inequality, we have\n$$\n\\begin{aligned} (A_2 + B_2 + \\ell C_2)^2 &= \\frac{1}{(\\ell+1)^2} (A_2 + (\\ell+1)B_2 + \\ell(A_2 + (\\ell+1)C_2))^2 \\\\ &\\le \\frac{1}{\\ell+1} ((A_2 + (\\ell+1)B_2)^2 + \\ell(A_2 + (\\ell+1)C_2)^2) \\\\ &\\le \\frac{1}{\\ell+1} (\\lambda_{n+\\ell-1}(A_4 + (\\ell+1)B_4) + \\ell\\lambda_{n+\\ell-1}(A_4 + (\\ell+1)C_4)) \\\\ &= \\lambda_{n+\\ell-1}(A_4 + B_4 + \\ell C_4). \\end{aligned}\n$$\nThus, (14) holds, completing the induction. $\\square$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 28619,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $a_1 > a_2 > \\dots > a_n > 1$ be positive integers. Let $M$ denote the least common multiple of $a_1, a_2, \\dots, a_n$. For a finite set of integers $X$, define\n$$\nf(X) = \\min_{1 \\le i \\le n} \\sum_{x \\in X} \\left\\{ \\frac{x}{a_i} \\right\\}.\n$$\nHere $\\{u\\} = u - \\lfloor u \\rfloor$ is the fractional part of the real number $u$. Put $f(\\emptyset) = 0$. We say a set $X$ is *minimal*, if for any proper subset $Y \\subsetneq X$, we have $f(Y) < f(X)$.\nProve that, if $X$ is a minimal finite set of integers and if $f(X) \\ge \\frac{2}{a_n}$, then\n$$\n|X| \\le f(X) \\cdot M.\n$$\nHere for a finite set $X$, we use $|X|$ to denote the number of elements in $X$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "*Proof.* Assume $f(X) = \\lambda \\ge \\frac{2}{a_n}$. By contradiction, suppose $|X| > \\lambda M$. Since $\\lambda$ is of the form $\\frac{k}{a_i}$ ($k \\in \\mathbb{Z}_{>0}$), $\\lambda M$ is an integer. We prove that there exists $x \\in X$ such that $f(X \\setminus \\{x\\}) = f(X)$, which contradicts the minimality of $X$.\n\nFor $1 \\le i \\le n$, let $X_i = \\{x \\in X \\mid a_i \\nmid x\\}$.\n\nConsider all indices $i$ satisfying $|X_i| \\le \\lceil \\lambda a_i \\rceil$, and denote these indices by $i_1 < i_2 < \\dots < i_m$. If\n$$\nX_{i_1} \\cup X_{i_2} \\cup \\dots \\cup X_{i_m} \\neq X, \\quad (*)\n$$\ntake $x \\in X \\setminus (X_{i_1} \\cup X_{i_2} \\cup \\dots \\cup X_{i_m})$, and let $Y = X \\setminus \\{x\\}$. Then, $f(Y) = f(X)$.\n\nThis is because, letting $Y_i = \\{y \\in Y \\mid a_i \\nmid y\\}$, we have: If $i \\in \\{i_1, i_2, \\dots, i_m\\}$, then $X_i = Y_i$, and\n$$\n\\sum_{y \\in Y} \\left\\{ \\frac{y}{a_i} \\right\\} = \\sum_{y \\in Y_i} \\left\\{ \\frac{y}{a_i} \\right\\} = \\sum_{x \\in X_i} \\left\\{ \\frac{x}{a_i} \\right\\} = \\sum_{x \\in X} \\left\\{ \\frac{x}{a_i} \\right\\} \\ge \\lambda.\n$$\nIf $i \\notin \\{i_1, i_2, \\dots, i_m\\}$, then $|Y_i| \\ge |X_i| - 1 \\ge \\lceil \\lambda a_i \\rceil$, so\n$$\n\\sum_{y \\in Y} \\left\\{ \\frac{y}{a_i} \\right\\} = \\sum_{y \\in Y_i} \\left\\{ \\frac{y}{a_i} \\right\\} \\ge |Y_i| \\cdot \\frac{1}{a_i} \\ge \\lceil \\lambda a_i \\rceil \\cdot \\frac{1}{a_i} \\ge \\lambda.\n$$\nThus, $f(Y) \\ge \\lambda$. Clearly, $f(Y) \\le f(X) = \\lambda$, so $f(Y) = f(X)$.\n\nNow, we prove (*) holds. For $1 \\le j \\le m$, let $T_j = X_{i_1} \\cup \\dots \\cup X_{i_j}$ and $M_j = \\text{lcm}(a_{i_1}, \\dots, a_{i_j})$. Clearly, $|T_j| = |X_{i_1}| \\le \\lceil \\lambda a_{i_1} \\rceil = \\lceil \\lambda M_1 \\rceil$.\n\nFor $2 \\le j \\le m$, we have $|T_j \\setminus T_{j-1}| \\le \\lceil \\lambda M_j \\rceil - \\lceil \\lambda M_{j-1} \\rceil$. Indeed, if $M_j = M_{j-1}$, then\n$$\nT_j = \\{x \\in X \\mid M_j \\nmid x\\} = \\{x \\in X \\mid M_{j-1} \\nmid x\\} = T_{j-1},\n$$\nso $|T_j \\setminus T_{j-1}| = 0 = \\lceil \\lambda M_j \\rceil - \\lceil \\lambda M_{j-1} \\rceil$. If $M_j > M_{j-1}$, then $a_{i_j} \\nmid M_{j-1}$, and let $d = \\text{gcd}(a_{i_j}, M_{j-1})$, $M_{j-1} = du$, $a_{i_j} = dv$, so $u$ and $v$ are coprime, with $u > v > 1$, and $M_j = duv$.\n$$\n|T_j \\setminus T_{j-1}| \\le |X_{i_j}| \\le \\lceil \\lambda a_{i_j} \\rceil \\le \\lceil \\lambda M_j \\rceil - \\lceil \\lambda M_{j-1} \\rceil. \\quad (**)\n$$\nThe last inequality in ($**$) requires $\\lambda a_{i_j} \\le \\lambda M_j - \\lambda M_{j-1} - 1 \\Leftrightarrow \\lambda d(uv - u - v) \\ge 1$. Since $\\lambda \\ge \\frac{2}{a_n} \\ge \\frac{2}{a_{i_j}} = \\frac{2}{dv}$, it suffices to show $\\frac{2}{v}(uv - u - v) \\ge 1 \\Leftrightarrow (2u - 3)(v - 1) \\ge 3$. Given $u > v > 1$, this inequality holds, so ($**$) holds. Therefore,\n$$\n|T_m| = |T_1| + \\sum_{j=2}^{m} |T_j \\setminus T_{j-1}| \\le \\lceil \\lambda M_1 \\rceil + \\sum_{j=2}^{m} (\\lceil \\lambda M_j \\rceil - \\lceil \\lambda M_{j-1} \\rceil) = \\lceil \\lambda M_m \\rceil \\le \\lceil \\lambda M \\rceil = \\lambda M.\n$$\nThis proves (*), so the assumption by contradiction fails, and the original proposition is proved.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 28620,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $m > 1$ be an integer. It is known that there exists a prime number in the interval $[2m - \\sqrt{m} + 1, 2m]$. Prove that among any $m$ pairwise distinct positive integers $a_1, a_2, \\dots, a_m$, there exist two numbers $a_i$ and $a_j$ ($1 \\le i, j \\le m$) such that\n$$\n\\frac{a_i}{(a_i, a_j)} \\ge m,\n$$\nwhere $(a_i, a_j)$ denotes the greatest common divisor of the positive integers $a_i$ and $a_j$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "*Proof.* Without loss of generality, assume that the greatest common divisor of $a_1, a_2, \\dots, a_m$ is $1$, otherwise divide all $a_1, a_2, \\dots, a_m$ by their greatest common divisor, which does not change the conclusion.\nLet $p$ be a prime number in the interval $[2m - \\sqrt{m} + 1, 2m]$. If $p$ divides any $a_i$, since the greatest common divisor of $a_1, \\dots, a_m$ is $1$, there exists some $a_j$ not divisible by $p$, then\n$$\n\\frac{a_i}{(a_i, a_j)} \\ge p > m.\n$$\nSo, we can assume that $a_1, a_2, \\dots, a_m$ are not divisible by $p$.\nIf there exists $a_i \\equiv a_j \\pmod{p}$, assume $a_i > a_j$, then\n$$\n\\frac{a_i}{(a_i, a_j)} = \\frac{a_i - a_j}{(a_i, a_j)} + \\frac{a_j}{(a_i, a_j)} \\ge p + 1 > m.\n$$\nTherefore, we can assume that $a_1, a_2, \\dots, a_m$ have pairwise distinct remainders modulo $p$.\nLet $m = \\frac{p-1}{2} + r$, then the condition $2m - \\sqrt{m} + 1 \\le p \\le 2m$ is equivalent to $r \\in [1, \\frac{\\sqrt{m}}{2}]$. Divide all non-zero residues modulo $p$ into $\\frac{p-1}{2}$ groups, each group containing the residues $i$ and $p-i$. By the pigeonhole principle, there are $r$ groups, each containing two elements from $a_1, a_2, \\dots, a_m$. Therefore, we can assume\n$$\na_{2i-1} \\equiv -a_{2i} \\pmod{p}, \\quad a_{2i-1} < a_{2i}, \\quad i = 1, 2, \\dots, r.\n$$\nNotice that for $i = 1, 2, \\dots, r$, we have $p \\mid \\frac{a_{2i-1}+a_{2i}}{(a_{2i-1}, a_{2i})}$. If $\\frac{a_{2i-1}+a_{2i}}{(a_{2i-1}, a_{2i})} \\ge 2p$, then\n$$\n\\max \\left( \\frac{a_{2i-1}}{(a_{2i-1}, a_{2i})}, \\frac{a_{2i}}{(a_{2i-1}, a_{2i})} \\right) \\ge p > m.\n$$\nTherefore, we can assume that for $i = 1, 2, \\dots, r$, we have $\\frac{a_{2i-1}+a_{2i}}{(a_{2i-1}, a_{2i})} = p$. Let\n$$\nd_i = (a_{2i-1}, a_{2i}), \\quad a_{2i-1} = u_i d_i, \\quad a_{2i} = v_i d_i.\n$$\nThen $u_i < v_i$ and $u_i + v_i = p$. Notice that if any $v_i \\ge m$, the conclusion is already satisfied. Therefore, we can assume\n$$\nv_i \\in \\left\\{ \\frac{p-1}{2} + 1, \\frac{p-1}{2} + 2, \\dots, \\frac{p-1}{2} + r - 1 = m - 1 \\right\\}.\n$$\nBy the pigeonhole principle, there are two equal numbers among $v_1, \\dots, v_r$. Assume $v_1 = v_2 = v$, hence $u_1 = u_2 = p - v$, denoted as $u$. Clearly, $d_1 \\ne d_2$ (since $a_1 \\ne a_3$).\nNow consider $a_1, a_2, a_3, \\text{and } a_4$, assuming $d_1$ and $d_2$ are coprime (otherwise divide them by their greatest common divisor). We have\n$$\n\\frac{a_1}{(a_1, a_4)} = \\frac{d_1 u}{(d_1, v) \\cdot (d_2, u)} \\ge \\frac{(d_1, u)}{(d_2, u)} \\cdot u; \\quad \\frac{a_3}{(a_2, a_3)} = \\frac{d_2 u}{(d_1, u) \\cdot (d_2, v)} \\ge \\frac{(d_2, u)}{(d_1, u)} \\cdot u; \\quad (6)\n$$\n$$\n\\frac{a_2}{(a_2, a_3)} = \\frac{d_1 v}{(d_1, u) \\cdot (d_2, v)} \\ge \\frac{(d_1, v)}{(d_2, v)} \\cdot v; \\quad \\frac{a_4}{(a_1, a_4)} = \\frac{d_2 v}{(d_1, v) \\cdot (d_2, u)} \\ge \\frac{(d_2, v)}{(d_1, v)} \\cdot v. \\quad (7)\n$$\nNotice that the conditions $r \\in [1, \\frac{\\sqrt{m}}{2}]$ and $p - m + 1 \\le u < v \\le m - 1$ give the following inequality:\n$$\n\\frac{m-1}{v} < \\frac{m-1}{u} \\le \\frac{m-1}{p-m+1} \\le \\frac{m-1}{m-\\sqrt{m}+2} < \\frac{m-1}{m-\\sqrt{m}} = \\frac{\\sqrt{m}+1}{\\sqrt{m}}.\n$$\nIf both $\\frac{a_i}{(a_i, a_j)}$ in (6) are not greater than $m-1$, then\n$$\n\\frac{\\sqrt{m}+1}{\\sqrt{m}} \\ge \\frac{(d_1, u)}{(d_2, u)}, \\quad \\frac{\\sqrt{m}+1}{\\sqrt{m}} \\ge \\frac{(d_1, u)}{(d_2, u)}.\n$$\nThus, either $(d_1, u) = (d_2, u)$ or $(d_1, u) > \\sqrt{m}$ and $(d_2, u) > \\sqrt{m}$ both hold simultaneously. But note that $(d_1, u) \\cdot (d_2, u) \\le u < m$, so it must be that $(d_1, u) = (d_2, u)$, and since $d_1$ and $d_2$ are coprime, they must both be equal to $1$.\nSimilarly, if both $\\frac{a_i}{(a_i, a_j)}$ in (7) are not greater than $m-1$, then $(d_1, v) = (d_2, v) = 1$, but then\n$$\n\\frac{a_4}{(a_1, a_4)} = \\frac{d_2 v}{(d_2, u)(d_1, v)} = d_2 v \\ge 2v > m,\n$$\nwhich is a contradiction!\nTherefore, there must exist $a_i$ and $a_j$ such that $\\frac{a_i}{(a_i, a_j)} \\ge m$. $\\square$",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 28621,
    "subject": "Mathematics (Multi-modal)",
    "question": "Given an acute triangle $ABC$ with $AB < BC < CA$. Let $D$ be a moving point on side $BC$, and $E$ be a moving point on the minor arc $\\widehat{BC}$ of the circumcircle of $ABC$, such that $\\angle BAD = \\angle BED$.\nLet $F$ be the intersection point of the line through $D$ perpendicular to $AB$ with the extension of $AC$. Prove that $\\angle BEF$ is constant.\n\n![](attached_image_1.png)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "*Proof*. Let $H$ be the orthocenter of triangle $ABD$. Since $DF \\perp AB$, the points $H, D, F$ are colinear, as shown:\n![](attached_image_2.png)\n\n**Case 1:** When $\\angle ADB < 90^\\circ$:\n• $H$ lies inside triangle $ABD$\n• By orthocenter properties, $\\angle BAD$ and $\\angle BHD$ are supplementary\n• Given $\\angle BED = \\angle BAD$, we have $\\angle BED = 180^\\circ - \\angle BHD$\n• Thus, $B, H, D, E$ are concyclic\nThis implies:\n• $\\angle EHF = \\angle EBD = \\angle EBC = \\angle EAF$\n\n*   Therefore, A, H, E, F are concyclic\n*   Consequently, $\\angle AEF = \\angle AHF = 180^\\circ - \\angle ABD = 180^\\circ - \\angle ABC$\n\n**Other Cases:**\n*   When $\\angle ADB > 90^\\circ$, D lies inside triangle ABH\n*   When $\\angle ADB = 90^\\circ$, D coincides with H\nIn all cases (using directed angles when necessary), we conclude:\n$$\n\\angle BEF = \\angle BEA + \\angle AEF = \\angle BCA + (180^\\circ - \\angle ABC)\n$$\nSince $\\angle BCA$ and $\\angle ABC$ are fixed angles of triangle $ABC$, $\\angle BEF$ is constant. $\\square$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 28622,
    "subject": "Mathematics (Multi-modal)",
    "question": "Given an isosceles triangle $ABC$ with $AB = AC$. Let a moving point $D$ satisfy $AD \\parallel BC$ and $DB > DC$. Let a moving point $E$ be on the arc $\\widearc{BC}$ of the circumcircle of $\\triangle ABC$ that does not contain $A$, such that $EB < EC$. Let $F$ be a point on the extension of $BC$ such that $\\angle DFE = \\angle ADE$. The extension of $FD$ intersects the extension of $BA$ at point $X$, and the extension of $FD$ intersects the extension of $CA$ at point $Y$.\nProve that $\\angle XEY$ is constant.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "![](attached_image_1.png)\n\n**Proof 1.** Let $AE$ and $DE$ intersect line $BC$ at points $P$ and $Q$, respectively. Since $AB = AC$ and $A$, $B$, $E$, $C$ are concyclic, we have $\\angle ABP = \\angle ACB = \\angle AEB$. Hence, $AP \\cdot AE = AB^2$.\n\nNote that $Q$ might be on the extension of $BC$, but $F$ can only be on the extension of $BQ$. Otherwise, if $F$ is on the ray $QB$, combining with $F$ being on the extension of $BC$ would lead to $\\angle DFE > \\angle DCE > \\angle FCE > 90^\\circ$, while clearly $\\angle ADE < 90^\\circ$, which contradicts $\\angle ADE = \\angle DFE$.\n\nSince $AD \\parallel BC$, we have $\\angle DQF = \\angle ADE = \\angle DFE$, thus $DQ \\cdot DE = DF^2$. Therefore,\n$$\n\\frac{AB^2}{DF^2} = \\frac{AP \\cdot AE}{DQ \\cdot DE} = \\frac{AE^2}{DE^2},\n$$\nimplying $\\frac{AB}{DF} = \\frac{AE}{DE}$. Hence $\\frac{AX}{DX} = \\frac{AB}{DF} = \\frac{AE}{DE}$. Similarly, $\\frac{AY}{DY} = \\frac{AC}{DF} = \\frac{AE}{DE}$.\n\nTake a point $T$ on segment $AD$ such that $\\frac{AT}{TD} = \\frac{AE}{DE}$. Then $X$, $Y$, $E$, and $T$ are concyclic (Apollonian circle). Notice that $XT$ and $YT$ bisect $\\angle AXD$ and $\\angle ATD$, respectively. Therefore,\n$$\n\\angle XEY = \\angle XTY = \\angle TXD - \\angle TYD = \\frac{1}{2}\\angle AXD - \\frac{1}{2}\\angle AYD = \\frac{1}{2}\\angle XAY = \\frac{1}{2}\\angle BAC\n$$\nis a fixed value. $\\square$\n\n\n![](attached_image_2.png)\n\n**Proof 2.** Since $AB = AC$ and $AD \\parallel BC$, $AD$ bisects the external angle $\\angle XAY$. Let $P$ be the intersection of the perpendicular bisector of $XY$ and $AD$. It is known that $A$, $X$, $Y$, and $P$ are concyclic. Hence, $\\angle PYD = \\angle PXY = \\angle PAY$, thus\n$$\n\\triangle PYD \\sim \\triangle PAY.\n$$\nThis implies $PA \\cdot PD = PY^2$ and $\\frac{PA}{PD} = \\left(\\frac{YA}{YD}\\right)^2$.\n\nLet $\\omega$ be the circumcircle of $\\triangle ABC$ with center $O$, and $\\Omega$ be the circumcircle of $\\triangle DEF$ with center $Q$. Clearly, $PA$ is tangent to $\\omega$ at point $A$, and since $\\angle DFE = \\angle ADE$, $PA$ is also tangent to $\\Omega$ at point $D$. Since $AD \\parallel CF$, we have\n$$\n\\frac{AO}{DQ} = \\frac{\\frac{AC}{\\sin \\angle ABC}}{\\frac{DF}{\\sin \\angle DEF}} = \\frac{AC}{DF} \\cdot \\frac{\\sin \\angle YDA}{\\sin \\angle DAY} = \\left(\\frac{YA}{YD}\\right)^2 = \\frac{PA}{PD}.\n$$\nThus, $P$ is the external homothety center of circles $\\omega$ and $\\Omega$.\n\nSince $E$ is the intersection of $\\omega$ and $\\Omega$, it is known that $PE^2 = PA \\cdot PD$. In fact, let $E'$ be the image of $E$ under the homothety centered at $P$ with ratio $\\frac{PA}{PD}$. Then $AE \\parallel DE'$, implying $\\angle PEA = \\angle PE'D = \\angle PDE$. Thus, $PE^2 = PA \\cdot PD$.\n\nTherefore, $PE = \\sqrt{PA \\cdot PD} = PY = PX$. Hence, $P$ is the circumcenter of $\\triangle EXY$. Therefore, $\\angle XEY = \\frac{1}{2}\\angle XPY = \\frac{1}{2}\\angle XAY = \\frac{1}{2}\\angle BAC$ is a fixed value. $\\square$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 28623,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $n$ be a positive integer. The polynomial with complex coefficients\n$$\nP(z) = a_n z^n + a_{n-1} z^{n-1} + \\dots + a_1 z + a_0 \\quad (a_n \\neq 0)\n$$\nsatisfies: for any complex number $z$ with $|z| = 1$, we have $|P(z)| \\le 1$.\nProve that for any $k \\in \\{0, 1, \\dots, n-1\\}$, we have $|a_k| \\le 1 - |a_n|^2$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "**Proof.** Let $\\ell \\in \\{1, 2, \\dots, n\\}$. For a complex number $\\alpha \\in \\mathbb{C}$, consider\n$$\n\\begin{aligned}\nQ(z) &= P(z)(1 + \\alpha z^{\\ell}) \\\\\n&= \\alpha a_n z^{n+\\ell} + \\dots + \\alpha a_{n-\\ell+1} z^{n+1} + (a_n + \\alpha a_{n-\\ell}) z^n + \\dots + (a_{\\ell} + \\alpha a_0) z^{\\ell} \\\\\n&\\quad + a_{\\ell-1} z^{\\ell-1} + \\dots + a_0.\n\\end{aligned}\n$$\nTake $M > 2n$, then\n$$\n\\begin{aligned}\n\\sum_{j=1}^{M} \\left| Q \\left( e^{i \\frac{2j\\pi}{M}} \\right) \\right|^2 &= \\sum_{j=1}^{M} \\left| P \\left( e^{i \\frac{2j\\pi}{M}} \\right) \\right|^2 \\cdot \\left| \\left( 1 + \\alpha e^{i \\frac{2j\\pi}{M}\\ell} \\right) \\right|^2 \\\\\n&\\le \\sum_{j=1}^{M} \\left| \\left( 1 + \\alpha e^{i \\frac{2j\\pi}{M}\\ell} \\right) \\right|^2 = M(1 + |\\alpha|^2).\n\\end{aligned}\n$$\nOn the other hand,\n$$\n\\sum_{j=1}^{M} \\left| Q \\left( e^{i \\frac{2j\\pi}{M}} \\right) \\right|^2 = M \\cdot \\left( |\\alpha a_n|^2 + \\dots + |\\alpha a_{n-\\ell+1}|^2 + \\sum_{j=\\ell}^{n} |a_{n-j+\\ell} + \\alpha a_{n-j}|^2 + |a_{\\ell-1}|^2 + \\dots + |a_0|^2 \\right).\n$$\nCombining the above two estimates, we get\n$$\n|\\alpha|^2(|a_n|^2 + \\dots + |a_{n-\\ell+1}|^2) + \\sum_{j=\\ell}^{n} |a_{n-j+\\ell} + \\alpha a_{n-j}|^2 + |a_{\\ell-1}|^2 + \\dots + |a_0|^2 \\le 1 + |\\alpha|^2.\n$$\nIn particular,\n$$\n|\\alpha a_n|^2 + |a_n + \\alpha a_{n-\\ell}|^2 \\le 1 + |\\alpha|^2.\n$$\nChoosing $\\alpha \\in \\mathbb{C}$ such that $|\\alpha| = \\frac{1}{|a_n|}$ and $\\arg(a_n) = \\arg(\\alpha a_{n-\\ell})$ in the above equation, we get\n$$\n1 + \\left( |a_n| + \\frac{|a_{n-\\ell}|}{|a_n|} \\right)^2 \\le 1 + \\frac{1}{|a_n|^2}.\n$$\nSolving this gives $|a_{n-\\ell}| \\le 1 - |a_n|^2$. The proof is complete.",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 28624,
    "subject": "Mathematics (Multi-modal)",
    "question": "We say a prime number $p$ is “good”, if there exists a bijection $f$ from the set $\\{0, 1, \\dots, p-1\\}$ to itself satisfying the following condition: for any pair of elements $a, b \\in \\{0, 1, \\dots, p-1\\}$, if $p \\mid a^2 - b$, then $|f(a) - f(b)| \\le 2024$. If no such bijection $f$ exists, we say that the prime $p$ is “bad”.\n\nProve that there exist infinitely many good primes, and there exist infinitely many bad primes.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "**Proof.** First, we show that there are infinitely many good primes. It is well-known that there are infinitely many primes $p \\equiv 3 \\pmod 4$. We prove that if $p \\equiv 3 \\pmod 4$, then $p$ is a good prime. Let $f(0) = 0$. Consider all quadratic residues of $p$, denoted as $r_1, r_2, \\dots, r_{(p-1)/2}$. If $r_i^2 \\equiv r_j \\pmod p$, we draw a directed edge from $r_i$ to $r_j$. If $i = j$, it is considered a self-loop. Then all vertices $r_1, r_2, \\dots, r_{p-1}$ in this directed graph have an out-degree of 1. Moreover, for any $r_i$, let $d^2 \\equiv r_i \\pmod p$. Since $\\left(\\frac{-1}{p}\\right) = -1$, either $d$ or $-d$ (but not both) is a quadratic residue modulo $p$. Thus, every vertex also has an in-degree of 1. This means the directed graph is a disjoint union of cycles.\n\nFor a directed cycle $r_{i_1} \\to r_{i_2} \\to \\dots \\to r_{i_s} \\to r_{i_1}$:\n- If $s = 2t + 1$ (odd), we define\n$$\nf(r_{i_1}) = 2u,\\ f(r_{i_2}) = 2u+4,\\ \\dots,\\ f(r_{i_{t+1}}) = 2u+4t,\\ f(r_{i_{t+2}}) = 2u+4t-2,\\ \\dots,\\ f(r_{i_{2t+1}}) = 2u+2.\n$$\n- If $s = 2t$ (even), we define\n$$\nf(r_{i_1}) = 2u,\\ f(r_{i_2}) = 2u+4,\\ \\dots,\\ f(r_{i_{t+1}}) = 2u+4t-4,\\ f(r_{i_{t+2}}) = 2u+4t-2,\\ \\dots,\\ f(r_{i_{2t}}) = 2u+2.\n$$\nHere, $u$ is any integer. Since each cycle uses a consecutive sequence of even numbers, we can choose these numbers appropriately to ensure that the values used by different cycles do not overlap. Notice that only even numbers are used. If $w$ is not a quadratic residue, let $v = w^2 \\pmod p$, and define $f(w) = f(v) - 1$. This ensures that $f$ remains injective and satisfies $|f(a) - f(b)| \\le 2024$ whenever $p \\mid a^2 - b$. Therefore, all primes $p \\equiv 3 \\pmod 4$ are good primes.\n\nNext, we prove that there are infinitely many bad primes. It is well-known that for any positive integer $n$, if an odd prime $p$ divides $A^{2n} + 1$ (where $A$ is an integer), then $p \\equiv 1 \\pmod{2^{n+1}}$. This is because $\\text{ord}_p(A) \\mid 2^{n+1}$ but $\\text{ord}_p(A) \\nmid 2^n$, so $\\text{ord}_p(A) = 2^{n+1}$. Thus, there are infinitely many odd primes $p \\equiv 1 \\pmod{2^{n+1}}$. (If there were only finitely many, let $A$ be twice the product of these primes, which leads to a contradiction.)\n\nNow, let $p \\equiv 1 \\pmod{2^{100}}$. We prove that $p$ is a bad prime. By the existence of primitive roots, the equation $x^{2^{100}} \\equiv 1 \\pmod p$ has exactly $2^{100}$ solutions modulo $p$. If there exists an injective function $f$ satisfying the problem's conditions, then for all $x$ satisfying $x^{2^{100}} \\equiv 1 \\pmod p$ and $x \\in \\{1, 2, \\dots, p-1\\}$, we must have $|f(x)-f(1)| \\le 2024 \\times 100$. However, there are $2^{100}$ such $x$, while the interval $[f(1)-2024 \\times 100, f(1)+2024 \\times 100]$ contains only $404801 < 2^{100}$ integers. Thus, there must be two distinct $x$ with the same image, contradicting injectivity! Therefore, $p$ is a bad prime.\n\nHence, there are infinitely many bad primes.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 28625,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find the maximum possible area of a right-angled triangle that can be covered by two closed disks of radius $1$.",
    "options": [],
    "answer": "3*sqrt(3)/2",
    "solution": "*Proof.* Let the right triangle be $\\triangle ABC$ with $\\angle C = 90^\\circ$. In this solution, the interior of a triangle or a circle includes its boundary.\nLet $\\omega_1$ and $\\omega_2$ be the circumferences of two closed disks with radius $1$, such that every point inside $\\triangle ABC$ lies in the interior of $\\omega_1$ or in the interior of $\\omega_2$. The distance between any two points within the same disk does not exceed $2$.\nConsider the assignment of the three vertices $A$, $B$, and $C$.\nIf $A$ and $B$ lie in the same disk, then the length of the hypotenuse $|AB| \\le 2$, and the area $S \\le \\frac{1}{4}|AB|^2 = 1$.\nIf $A$ and $B$ do not lie in the same disk, assume $A$ lies inside $\\omega_1$ and $B$ lies inside $\\omega_2$. Without loss of generality, let the right-angled vertex $C$ lie inside $\\omega_1$. Since $B$ lies outside $\\omega_1$, the line segment $CB$ intersects the circumference $\\omega_1$ at a unique point $M$. The point $M$ divides $CB$ into two segments $CM$ and $MB$, where $CM$ lies inside $\\omega_1$, and $MB$ (excluding the endpoint $M$) lies outside $\\omega_1$ and must therefore lie inside $\\omega_2$. Consequently, the endpoint $M$ also lies inside $\\omega_2$.\nSince both $M$ and $A$ lie inside $\\omega_1$, we have $|MA| \\le 2$; similarly, since both $M$ and $B$ lie inside $\\omega_2$, we have $|MB| \\le 2$.\n\n![](attached_image_1.png)\n\nOn the other hand, let $N$ be the foot of the perpendicular from $M$ to $AB$. The circle with diameter $MA$ covers $\\triangle BMN$, and the circle with diameter $MB$ covers the quadrilateral $ACMN$. Thus, these two disks together cover $\\triangle ABC$. Therefore, the problem reduces to finding a point $M$ on the side $BC$ such that $|MA|, |MB| \\le 2$.\nLet $|MC| = x$ and $|AC| = y$. Then $x^2 + y^2 = |MA|^2 \\le 4$, and\n$$\n2S = |BC| \\cdot |AC| \\le (2+x)y \\le (2+x)\\sqrt{4-x^2}.\n$$\n$$\n4S^2 \\le (2+x)^2(4-x^2) = (2+x)^3(2-x) = 27 \\times \\left(\\frac{2+x}{3}\\right)^3 (2-x) \\le 27 \\times \\left(\\frac{4}{4}\\right)^4 = 27.\n$$\n\nThus, $S \\le \\frac{3\\sqrt{3}}{2}$. Equality holds when $x = 1$ and $y = \\sqrt{3}$, corresponding to the side lengths $AC = \\sqrt{3}$, $BC = 3$, and $AB = 2\\sqrt{3}$, with the point $M$ on $BC$ satisfying $MC = 1$ and $MB = MA = 2$. In this case, $\\triangle ABC$ can be covered by two closed disks of radius $1$ (i.e., the disks with diameters $MB$ and $MA$), fulfilling the requirement.\nTherefore, the maximum possible area of the right triangle is $\\frac{3\\sqrt{3}}{2}$. $\\square$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 28626,
    "subject": "Mathematics (Multi-modal)",
    "question": "For an integer $n \\ge 4$, if a sequence of real numbers $x_1, x_2, \\dots, x_n$ satisfies the following three equalities:\n$$\n\\begin{aligned}\nx_1 + x_2 + \\dots + x_n &= n, \\\\\nx_1^2 + x_2^2 + \\dots + x_n^2 &= 2n, \\\\\nx_1^3 + x_2^3 + \\dots + x_n^3 &= 3n,\n\\end{aligned}\n$$\nthen we call $(x_1, x_2, \\dots, x_n)$ a *regular* $n$-tuple. We call the difference between the maximal and the minimal elements in $x_1, x_2, \\dots, x_n$ the *width* of the regular $n$-tuple.\n\n(1) Find the maximal real number $C$ such that, for any integer $n \\ge 4$, the width of every regular $n$-tuple is at least $C$.\n\n(2) For the maximum $C$ in (1), prove that there exists a real number $\\lambda > 0$ such that, for any integer $n \\ge 4$, the width of every regular $n$-tuple exceeds $C + \\frac{\\lambda}{n^{1.5}}$.",
    "options": [],
    "answer": "C = sqrt(5), and one valid choice is λ = 1/6.",
    "solution": "(1)\n\n**Solution 1**\nWe attempt to use the idea of a local inequality to prove that a number $a$ lies between the minimum and maximum of a regular $n$-tuple. To this end, we consider an indeterminate $b$ and consider the polynomial\n$$\nf(x) = (x - a)(x - b)^2 = x^3 - (a + 2b)x^2 + (2ab + b^2)x - ab^2,\n$$\nand define\n$$\nS = \\sum_{k=1}^{n} f(x_k) = 3n - (a + 2b) \\cdot 2n + (2ab + b^2) \\cdot n - ab^2 \\cdot n.\n$$\nWe require $S = 0$ to ensure that $f(x_1), f(x_2), \\dots, f(x_n)$ have both positive and negative values (or are all 0), implying that the numbers $x_1, x_2, \\dots, x_n$ include values $\\ge a$ and values $\\le a$ (or $x_1, \\dots, x_n$ are all equal to $b$). This ensures $a$ lies between the minimum and maximum values.\nSetting $S = 0$, we obtain the quadratic equation\n$$\n(1 - a)b^2 + (2a - 4)b + (3 - 2a) = 0.\n$$\nThis equation has real roots if and only if\n$$\n\\Delta = (2a - 4)^2 - 4(1 - a)(3 - 2a) = 4(-a^2 + a + 1) \\ge 0,\n$$\nwhich is equivalent to $\\frac{1-\\sqrt{5}}{2} \\le a \\le \\frac{1+\\sqrt{5}}{2}$. Thus, the maximum value $\\ge \\frac{1+\\sqrt{5}}{2}$ and the minimum value $\\le \\frac{1-\\sqrt{5}}{2}$. The width is therefore $\\ge \\frac{1+\\sqrt{5}}{2} - \\frac{1-\\sqrt{5}}{2} = \\sqrt{5}$.\nOn the other hand, we can construct a regular $n$-tuple whose width approaches $\\sqrt{5}$ arbitrarily closely.\nTake positive integers $u_1, v_1, u_2, v_2$ such that $\\frac{v_1}{u_1} > \\frac{\\sqrt{5}-1}{2} > \\frac{v_2}{u_2}$.\nDefine the first group (multiset) to include $u_1^2$ instances of $1 + \\frac{v_1}{u_1}$ and $v_1^2$ instances of $1 - \\frac{u_1}{v_1}$. The mean of these $u_1^2 + v_1^2$ numbers is 1, the mean square is 2, and the mean cube is $4 + \\left( \\frac{v_1}{u_1} - \\frac{u_1}{v_1} \\right) > 3$.\n\nDefine the second group (multiset) to include $u_2^2$ instances of $1 + \\frac{v_2}{u_2}$ and $v_2^2$ instances of $1 - \\frac{u_2}{v_2}$. The mean of these $u_2^2 + v_2^2$ numbers is 1, the mean square is 2, and the mean cube is $4 + \\left(\\frac{v_2}{u_2} - \\frac{u_2}{v_2}\\right) < 3$.\nBy combining sufficient numbers of the first and second groups, the overall mean cube can be made exactly 3, and the width $(1 + \\frac{v_1}{u_1}) - (1 - \\frac{u_2}{v_2}) = \\frac{v_1}{u_1} + \\frac{u_2}{v_2}$ can approach $\\sqrt{5}$ arbitrarily closely.\nHence, the largest possible real number $C = \\sqrt{5}$. $\\square$\n\n\n**Solution 2**\n(The construction of regular $n$-tuples is the same as in Solution 1.)\nLet $A$ and $B$ denote the minimum and maximum values among $x_1, x_2, \\dots, x_n$, respectively.\nFor each $x_k - A \\ge 0$, applying the Cauchy–Schwarz inequality, we have\n$$\n\\sum_{k=1}^{n} (x_k - A) \\cdot \\sum_{k=1}^{n} (x_k - A)^3 \\ge \\left( \\sum_{k=1}^{n} (x_k - A)^2 \\right)^2 .\n$$\nSubstituting\n$$\n\\begin{align*}\n\\sum_{k=1}^{n} (x_k - A) &= \\left(\\sum_{k=1}^{n} x_k\\right) - nA = n - nA, \\\\\n\\sum_{k=1}^{n} (x_k - A)^2 &= 2n - 2nA + nA^2, \\\\\n\\sum_{k=1}^{n} (x_k - A)^3 &= 3n - 6nA + 3nA^2 - nA^3,\n\\end{align*}\n$$\nwe find that\n$$\n(1 - A)(3 - 6A + 3A^2 - A^3) - (2 - 2A + A^2)^2 = A^2 - A - 1 \\ge 0.\n$$\nSolving, we obtain $A \\le \\frac{1-\\sqrt{5}}{2}$.\nSimilarly, using the same inequality for $B - x_k \\ge 0$, we find that $B \\ge \\frac{1+\\sqrt{5}}{2}$.\nThus, the width $B - A \\ge \\sqrt{5}$. If we have an equality, then $A = \\frac{1-\\sqrt{5}}{2}$, $B = \\frac{1+\\sqrt{5}}{2}$, and all $x_1, \\dots, x_n$ are either $A$ or $B$. This requires $\\frac{5-\\sqrt{5}}{10}n$ numbers equal to $A$ and $\\frac{5+\\sqrt{5}}{10}n$ numbers equal to $B$, which cannot hold if $n$ is not divisible by 10. Therefore, the width $B - A > \\sqrt{5}$. $\\square$\n\n\n(2)\n\nConsider a linear transformation that maps the approximate minimum value $\\frac{1-\\sqrt{5}}{2}$ to 0 and the approximate maximum value $\\frac{1+\\sqrt{5}}{2}$ to 1. Let $z_i = \\frac{x_i - \\frac{1-\\sqrt{5}}{2}}{\\sqrt{5}}$, i.e., $x_i = \\frac{1-\\sqrt{5}}{2} + \\sqrt{5}z_i$. Denote $p = \\frac{5+\\sqrt{5}}{10} = \\frac{\\sqrt{5}+1}{2\\sqrt{5}}$. Then, we have\n$$\nz_1 + z_2 + \\dots + z_n = pn, \\quad z_1^2 + z_2^2 + \\dots + z_n^2 = pn, \\quad z_1^3 + z_2^3 + \\dots + z_n^3 = pn.\n$$\nLet $\\|x\\|$ denote the distance of a real number $x$ to the nearest integer. It is well-known that $\\|x + y\\| \\le \\|x\\| + \\|y\\|$ for any real numbers $x, y$. Let $m$ be the integer closest to $pn$, then\n$$\n\\|pn\\| = \\left| m - \\frac{\\sqrt{5}+1}{2\\sqrt{5}}n \\right| = \\frac{\\left| (2m-n)\\sqrt{5} - n \\right|}{2\\sqrt{5}} = \\frac{4}{2\\sqrt{5}} \\cdot \\frac{\\left| 5m^2 - 5mn + n^2 \\right|}{(2m-n)\\sqrt{5} + n} \\ge \\frac{4}{2\\sqrt{5}} \\cdot \\frac{1}{2n+1}.\n$$\n\nFrom (1), we know that the maximum of $z_1, z_2, \\dots, z_n$ is $\\ge 1$, and the minimum is $\\le 0$. Let the difference between the maximum and minimum values be $1 + \\delta$, so $z_1, z_2, \\dots, z_n \\in [-\\delta, 1 + \\delta]$. We aim to prove that $\\delta > \\frac{\\lambda}{\\sqrt{5 n^{1.5}}}$. First, assume $\\delta \\le \\frac{1}{2}$.\nConsider the polynomial $h(z) = 1 - 3z^2 + 2z^3 = (1 + 2z)(1 - z)^2$.\nWhen $z \\in [-\\delta, \\frac{1}{2}]$, $\\|z\\| = |z|$, we have\n$$\n\\|h(z)\\| \\le |1 - h(z)| = (3 - 2z)z^2 \\le (3 + 2\\delta)z^2 \\le 4z^2 = 4\\|z\\|^2.\n$$\nWhen $z \\in [\\frac{1}{2}, 1 + \\delta]$, $\\|z\\| = |z - 1|$, we have\n$$\n\\|h(z)\\| \\le |h(z)| = (1 + 2z)(1 - z)^2 \\le (3 + 2\\delta)(1 - z)^2 \\le 4(1 - z)^2 = 4\\|z\\|^2.\n$$\nThus, in all cases, $\\|h(z_k)\\| \\le 4\\|z_k\\|^2$.\nSince $\\sum_{k=1}^n h(z_k) = n - 3 \\cdot pn + 2 \\cdot pn = n - pn$, it follows that\n$$\n\\sum_{k=1}^{n} \\|h(z_k)\\| \\ge \\|pn\\|,\n$$\nand\n$$\n(\\|z_1\\| + \\|z_2\\| + \\cdots + \\|z_n\\|)^2 \\ge \\|z_1\\|^2 + \\|z_2\\|^2 + \\cdots + \\|z_n\\|^2 \\ge \\frac{1}{4} \\|pn\\|.\n$$\nOn the other hand, consider the polynomial $g(z) = 2z(1 - z)$.\nWhen $z \\in [0, 1]$, we have $g(z) = 2z(1 - z) \\ge \\max\\{z, 1 - z\\} = \\|z\\|$.\nWhen $z \\in [-\\delta, 0] \\cup [1, 1 + \\delta]$, we have $g(z) \\ge -2(\\delta + \\delta^2) \\ge \\|z\\| - 4\\delta$.\nThus, $g(z_k) \\le \\|z_k\\| - 4\\delta$.\nSince $\\sum_{k=1}^n g(z_k) = 2 \\sum_{k=1}^n z_k - 2 \\sum_{k=1}^n z_k^2 = 0$, we have\n$$\n4n\\delta \\ge \\sum_{k=1}^{n} \\|z_k\\| \\ge \\sqrt{\\frac{1}{4} \\|pn\\|} \\ge \\sqrt{\\frac{1}{2\\sqrt{5}} \\cdot \\frac{1}{2n+1}} > \\sqrt{\\frac{1}{11n}}.\n$$\nThus, $\\delta > \\frac{1}{4\\sqrt{11}} \\cdot \\frac{1}{n^{1.5}} > \\frac{1}{6\\sqrt{5n^{1.5}}}.$\n\nHence, the width of the regular $n$-tuple $(x_1, x_2, \\dots, x_n)$ satisfies $> \\sqrt{5} + \\frac{1}{6n^{1.5}}$, and $\\lambda = \\frac{1}{6}$ meets the requirement. $\\square$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 28627,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $m$, $n$, $r$ be positive integers satisfying $n > m$, where both $m^2 + r$ and $n^2 + r$ are powers of $2$. Prove that:\n$$\nn > \\frac{2m^2}{r}.\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "*Proof.* Let $m^2 + r = 2^k$ and $n^2 + r = 2^l$ with $k, l \\in \\mathbb{Z}_{>0}$.\nFirst, we analyze the 2-adic valuations:\n* Since $v_2(m^2) = v_2(2^k - r) = v_2(r)$, we conclude $v_2(r)$ must be even.\n* Let $v_2(r) = 2t$ where $t \\in \\mathbb{Z}_{\\ge 0}$, and write $r = 2^{2t}r_1$, $m = 2^t m_1$.\n* Similarly, $v_2(n^2) = v_2(r)$, so let $n = 2^t n_1$.\nDefine $k_1 = k - 2t$ and $l_1 = l - 2t$, which gives the reduced system:\n$$\nm_1^2 + r_1 = 2^{k_1},\n$$\n$$\nn_1^2 + r_1 = 2^{l_1}.\n$$\n**Claim:** $r_1 \\ge 7$.\n* Since $r_1$ is odd, both $m_1$ and $n_1$ must be odd.\n* As $n_1 \\ge 3$ (because $n > m \\ge 1$), we have $l_1 > 3$.\n* Thus $n_1^2 + r_1 \\equiv 1 + r_1 \\equiv 0 \\pmod 8$, implying $r_1 \\equiv 7 \\pmod 8$.\n* Therefore $r_1 \\ge 7$.\nThis also shows $k_1 \\ge 3$. Now consider the difference:\n$$\n2^{l_1} - 2^{k_1} = (n_1 - m_1)(n_1 + m_1).\n$$\nSince $2^{k_1}$ divides the product $(n_1 - m_1)(n_1 + m_1)$, and both factors are even but cannot both be divisible by $4$ (which would imply $m_1$ and $n_1$ are both even), we have:\n$$\n2^{k_1-1} \\mid (n_1 - m_1) \\quad \\text{or} \\quad 2^{k_1-1} \\mid (n_1 + m_1).\n$$\nThis leads to the lower bound:\n$$\n\\begin{align*}\nn_1 &\\ge 2^{k_1-1} - m_1 = \\frac{1}{2}(m_1^2 + r_1) - m_1 \\\\\n&= \\frac{3m_1^2}{7} + \\left( \\frac{m_1^2}{14} + \\frac{r_1}{2} - m_1 \\right) \\\\\n&\\ge \\frac{3m_1^2}{r_1} + \\left( \\frac{m_1^2}{14} + \\frac{7}{2} - m_1 \\right) \\quad (\\text{since } r_1 \\ge 7) \\\\\n&= \\frac{3m_1^2}{r_1} + \\frac{1}{14}(m_1 - 7)^2 \\\\\n&\\ge \\frac{3m_1^2}{r_1} = \\frac{3m^2}{r}.\n\\end{align*}\n$$\n\n$$\nn = 2^t n_1 \\ge n_1 \\ge \\frac{3m^2}{r} > \\frac{2m^2}{r},\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 28628,
    "subject": "Mathematics (Multi-modal)",
    "question": "Given positive integers $m, n$ and non-negative real numbers $a_0, a_1, \\dots, a_m, b_0, b_1, \\dots, b_n$. Define\n$$\nc_k = \\max_{i+j=k} a_i b_j \\quad (k = 0, 1, \\dots, m+n).\n$$\nProve that:\n$$\n\\frac{c_0 + c_1 + \\dots + c_{m+n}}{m+n+1} \\ge \\frac{a_0 + a_1 + \\dots + a_m}{m+1} \\cdot \\frac{b_0 + b_1 + \\dots + b_n}{n+1}.\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "*Proof.* Consider the set of points $\\mathcal{P} = \\{(i_0, j_0), (i_1, j_1), \\dots, (i_{m+n}, j_{m+n})\\}$ as a \"monotone path\" from $(0, 0)$ to $(m, n)$ if $(i_0, j_0) = (0, 0)$, $(i_{m+n}, j_{m+n}) = (m, n)$, and for every index $0 \\le k < m + n$, we have\n$$\n(i_{k+1}, j_{k+1}) \\in \\{(i_k + 1, j_k), (i_k, j_k + 1)\\}.\n$$\nGiven two non-negative decreasing sequences $u_0 \\ge u_1 \\ge \\dots \\ge u_m$ and $v_0 \\ge v_1 \\ge \\dots \\ge v_n$, define the weight of the monotone path $\\mathcal{P}$ as\n$$\nW(\\mathcal{P}) = \\sum_{k=0}^{m+n} u_{i_k} v_{j_k}.\n$$\nLemma: There exists a monotone path $\\mathcal{P}$ such that\n$$\nW(\\mathcal{P}) \\ge \\frac{m+n+1}{(m+1)(n+1)} \\left(\\sum_{i=0}^m u_i\\right) \\left(\\sum_{j=0}^n v_j\\right).\n$$\n*Proof of the Lemma:* We use induction on $m+n$. When $m=0$ or $n=0$, the conclusion is trivial. Assume the lemma holds for $m+n < k$. Consider the case $m+n = k$ with $m, n > 0$. Let\n$$\nS = \\max_{\\text{monotone path } \\mathcal{P}} W(\\mathcal{P}).\n$$\nBy the induction hypothesis, for the sequences $u_0 \\ge u_1 \\ge \\dots \\ge u_{m-1}$ and $v_0 \\ge v_1 \\ge \\dots \\ge v_n$, there exists a monotone path $\\mathcal{P}_0$ from $(0, 0)$ to $(m-1, n)$ such that\n$$\nW(\\mathcal{P}_0) \\ge \\frac{(m-1)+n+1}{m(n+1)} \\left(\\sum_{i=0}^{m-1} u_i\\right) \\left(\\sum_{j=0}^n v_j\\right).\n$$\nLet $\\mathcal{P}$ be the monotone path obtained by adding $(m, n)$ to $\\mathcal{P}_0$, then\n$$\nS \\ge W(\\mathcal{P}) = W(\\mathcal{P}_0) + u_m v_n \\ge \\frac{(m-1)+n+1}{m(n+1)} \\left(\\sum_{i=0}^{m-1} u_i\\right) \\left(\\sum_{j=0}^n v_j\\right) + u_m v_n. \\quad (10)\n$$\nSimilarly, for the sequences $u_0 \\ge u_1 \\ge \\dots \\ge u_m$ and $v_0 \\ge v_1 \\ge \\dots \\ge v_{n-1}$, by the induction hypothesis, we have\n$$\nS \\ge \\frac{m+(n-1)+1}{(m+1)n} \\left(\\sum_{i=0}^m u_i\\right) \\left(\\sum_{j=0}^{n-1} v_j\\right) + u_m v_n. \\quad (11)\n$$\nLet $U = u_0 + \\dots + u_m$ and $V = v_0 + \\dots + v_n$. Multiplying (10) by $\\frac{m}{m+n}$ and adding it to (11) multiplied by $\\frac{n}{m+n}$, we get\n$$\n\\begin{align*} S &\\ge \\frac{1}{n+1}(U - u_m)V + \\frac{1}{m+1}U(V - v_n) + u_m v_n \\\\ &= \\frac{m+n+1}{(m+1)(n+1)}UV + \\left(\\frac{U}{m+1} - u_m\\right)\\left(\\frac{V}{n+1} - v_n\\right) \\\\ &\\ge \\frac{m+n+1}{(m+1)(n+1)}UV, \\end{align*}\n$$\nthereby completing the proof of the lemma.\n\nReturning to the proof of the stated inequality. Sort $\\{a_i\\}$ as $u_0 \\ge u_1 \\ge \\dots \\ge u_m$, and sort $\\{b_j\\}$ as $v_0 \\ge v_1 \\ge \\dots \\ge v_n$. There exists a permutation $\\pi$ of $\\{0, 1, \\dots, m\\}$ such that for each index $i$, $u_i = a_{\\pi(i)}$; similarly, there exists a permutation $\\sigma$ of $\\{0, 1, \\dots, n\\}$ such that for each index $j$, $v_j = b_{\\sigma(j)}$.\nBy the lemma, there exists a monotone path $\\mathcal{P} = \\{(i_0, j_0), (i_1, j_1), \\dots, (i_{m+n}, j_{m+n})\\}$ from (0,0) to (m,n) such that\n$$\nW(\\mathcal{P}) \\ge \\frac{m+n+1}{(m+1)(n+1)} UV.\n$$\nFrom the definition of a monotone path, for each index $k \\in \\{0, 1, \\dots, m+n\\}$ we have\n$$\n\\#\\{i_0, i_1, \\dots, i_k\\} = i_k + 1, \\quad \\#\\{j_0, j_1, \\dots, j_k\\} = j_k + 1,\n$$\nhence\n$$\n\\#\\{\\pi(i_0), \\dots, \\pi(i_k)\\} + \\#\\{\\sigma(j_0), \\dots, \\sigma(j_k)\\} = i_k + j_k + 2 = k + 2.\n$$\nFor finite non-empty sets of real numbers $X$ and $Y$, we have $|X + Y| \\ge |X| + |Y| - 1$. Let $\\Sigma_k = \\{\\pi(i_0), \\dots, \\pi(i_k)\\} + \\{\\sigma(j_0), \\dots, \\sigma(j_k)\\}$, then $\\Sigma_k \\subseteq \\{0, 1, \\dots, m+n\\}$ and $|\\Sigma_k| \\ge k + 1$. Thus, we can sequentially choose\n$$\n\\tau(0) \\in \\Sigma_0, \\quad \\tau(k) \\in \\Sigma_k \\setminus \\{\\tau(0), \\tau(1), \\dots, \\tau(k-1)\\}, \\quad \\forall 1 \\le k \\le m+n.\n$$\nSince $\\tau(0), \\dots, \\tau(m+n)$ are distinct and between 0 and $m+n$, they form a permutation of $\\{0, 1, \\dots, m+n\\}$. For each $k$, let $\\tau(k) = \\pi(i_p) + \\sigma(j_q)$, where $p, q \\le k$. From the definition of $c_*$, we have\n$$\nc_{\\tau(k)} \\ge a_{\\pi(i_p)} b_{\\sigma(j_q)} = u_{i_p} v_{j_q} \\ge u_{i_k} v_{j_k},\n$$\nthus,\n$$\n\\sum_{k=0}^{m+n} c_k = \\sum_{k=0}^{m+n} c_{\\tau(k)} \\ge \\sum_{k=0}^{m+n} u_{i_k} v_{j_k} = W(\\mathcal{P}) \\ge \\frac{m+n+1}{(m+1)(n+1)} UV,\n$$\nwhich proves the desired inequality. $\\square$\n*Proof 2 (Based on the solution by Junfeng Tian).* Consider the set of points $\\mathcal{P} = \\{(i_0, j_0), (i_1, j_1), \\dots, (i_{m+n}, j_{m+n})\\}$ as a \"monotone path\" from (0,0) to (m,n) if $(i_0, j_0) = (0,0)$, $(i_{m+n}, j_{m+n}) = (m,n)$, and for every index $0 \\le k < m+n$, we have\n$$\n(i_{k+1}, j_{k+1}) \\in \\{(i_k + 1, j_k), (i_k, j_k + 1)\\}.\n$$\nNoting that $i_k + j_k = k$, we get\n$$\n\\sum_{k=0}^{m+n} a_{i_k} b_{j_k} \\le \\sum_{k=0}^{m+n} c_{i_k + j_k} = \\sum_{k=0}^{m+n} c_k.\n$$\nWe call the left-hand side of the above expression the weight of the monotone path $\\mathcal{P}$, denoted as $W(\\mathcal{P})$. To prove the conclusion, we only need to prove the following proposition ($*$): There exists a monotone path $\\mathcal{P}$ such that\n$$\nW(\\mathcal{P}) \\ge \\frac{m+n+1}{(m+1)(n+1)} \\sum_{i=0}^m a_i \\sum_{j=0}^n b_j.\n$$\nFirst, establish the following lemma: Let $0 \\le x, y \\le 1$, then\n$$\nxy + \\max \\left\\{ \\frac{m+n}{m(n+1)}(1-x), \\frac{m+n}{(m+1)n}(1-y) \\right\\} \\ge \\frac{m+n+1}{(m+1)(n+1)}.\n$$\nProof of the lemma: Let the left-hand side of the above expression be $L$. By symmetry, assume $\\frac{m+n}{m(n+1)}(1-x) \\ge \\frac{m+n}{(m+1)n}(1-y)$, then\n$$\ny \\ge 1 - \\frac{(m+1)n}{m(n+1)}(1-x).\n$$\nCombining with the fact that $x$ is non-negative, we get\n$$\n\\begin{aligned} L &\\ge x \\left(1 - \\frac{(m+1)n}{m(n+1)}(1-x)\\right) + \\frac{m+n}{m(n+1)}(1-x) \\\\ &= \\frac{m+n+1}{(m+1)(n+1)} + \\frac{n}{m(m+1)(n+1)}((m+1)x - 1)^2 \\\\ &\\ge \\frac{m+n+1}{(m+1)(n+1)}, \\end{aligned}\n$$\nthereby completing the proof of the lemma.\n\nReturning to the proof of proposition $(*)$. We use induction on $m+n$. When $m=0$ or $n=0$, the proposition is clearly true. Assume the proposition holds for $m+n < \\ell$, consider the case $m+n = \\ell$ with $m, n > 0$. By homogeneity, assume $\\sum_{i=0}^m a_i = \\sum_{j=0}^n b_j = 1$, then $0 \\le a_0, b_0 \\le 1$. Let\n$$\nS = \\max_{\\text{monotone path } \\mathcal{P}} W(\\mathcal{P}).\n$$\nBy the induction hypothesis, there exists a monotone path $P_0$ from $(1,0)$ to $(m,n)$ such that\n$$\nW(P_0) \\ge \\frac{m+n}{m(n+1)} \\sum_{i=1}^m a_i \\sum_{j=0}^n b_j = \\frac{m+n}{m(n+1)}(1-a_0).\n$$\nLet $\\mathcal{P}$ be the monotone path obtained by going from $(0,0)$ to $(1,0)$ and then following $P_0$, then\n$$\nS \\ge W(\\mathcal{P}) = a_0 b_0 + \\frac{m+n}{m(n+1)}(1-a_0).\n$$\nSimilarly, consider going from $(0,0)$ to $(0,1)$ and then using the induction hypothesis, we get\n$$\nS \\ge a_0 b_0 + \\frac{m+n}{(m+1)n}(1-b_0).\n$$\nCombining these two expressions and using the lemma, we get\n$$\nS \\ge a_0 b_0 + \\max \\left\\{ \\frac{m+n}{m(n+1)}(1-a_0), \\frac{m+n}{(m+1)n}(1-b_0) \\right\\} \\ge \\frac{m+n+1}{(m+1)(n+1)},\n$$\nwhich completes the proof. $\\Box$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 28629,
    "subject": "Mathematics (Multi-modal)",
    "question": "Given an integer $n \\ge 2$, let $x_1, x_2, \\dots, x_n$ be non-negative real numbers satisfying $x_1 + x_2 + \\dots + x_n = n$. Find the minimum and maximum values of\n$$\n\\sum_{k=1}^{n} \\frac{1 + x_k^2 + x_k^4}{1 + x_{k+1} + x_{k+1}^2 + x_{k+1}^3 + x_{k+1}^4},\n$$\nwhere $x_{n+1} = x_1$.",
    "options": [],
    "answer": "minimum = (3/5) n; maximum = n^4 + n^2 + n - 1 + (n - 1)/(n^5 - 1)",
    "solution": "*Proof.* The minimum value is $\\frac{3}{5}n$.\nOn one hand, taking $x_1 = x_2 = \\dots = x_n = 1$, we have\n$$\n\\sum_{k=1}^{n} \\frac{1 + x_k^2 + x_k^4}{1 + x_{k+1} + x_{k+1}^2 + x_{k+1}^3 + x_{k+1}^4} = \\frac{3}{5}n.\n$$\nOn the other hand, by the AM-GM inequality, for any $1 \\le k \\le n$, we have $x_k^2+x_k^4 \\ge 2x_k^3$ and $1+x_k^2 \\ge 2x_k$. Noting that $1+x_k^4 \\ge x_k+x_k^3$ (since $(x_k-1)^2(x_k^2+x_k+1) \\ge 0$), it follows that $2(1+x_k^2+x_k^4) \\ge 3(x_k+x_k^3)$. Thus, $5(1+x_k^2+x_k^4) \\ge 3(1+x_k+x_k^2+x_k^3+x_k^4)$, which implies\n$$\n\\frac{1 + x_k^2 + x_k^4}{1 + x_k + x_k^2 + x_k^3 + x_k^4} \\ge \\frac{3}{5}.\n$$\nBy the rearrangement inequality (the sum of products in any order is greater than or equal to the sum in reverse order),\n$$\n\\sum_{k=1}^{n} \\frac{1 + x_k^2 + x_k^4}{1 + x_{k+1} + x_{k+1}^2 + x_{k+1}^3 + x_{k+1}^4} \\ge \\sum_{k=1}^{n} \\frac{1 + x_k^2 + x_k^4}{1 + x_k + x_k^2 + x_k^3 + x_k^4} \\ge \\frac{3}{5}n.\n$$\n\n(2) The maximum value is $n^4 + n^2 + n - 1 + \\frac{n-1}{n^5-1}$.\nOn one hand, taking $x_1 = n$ and $x_2 = x_3 = \\dots = x_n = 0$, we have\n$$\n\\sum_{k=1}^{n} \\frac{1 + x_k^2 + x_k^4}{1 + x_{k+1} + x_{k+1}^2 + x_{k+1}^3 + x_{k+1}^4} = n^4 + n^2 + n - 1 + \\frac{n-1}{n^5-1}.\n$$\nOn the other hand, for any $1 \\le k \\le n$, let $a_k = 1+x_k^2+x_k^4$ and $b_k = 1+x_k+x_k^2+x_k^3+x_k^4$, with $b_{n+1} = b_1$. Since $a_k \\ge 1$ and $\\frac{1}{b_{k+1}} \\le 1$, it follows that $(a_k - 1) \\left( \\frac{1}{b_{k+1}} - 1 \\right) \\le 0$, and thus\n$$\n\\frac{a_k}{b_{k+1}} \\le a_k + \\frac{1}{b_{k+1}} - 1.\n$$\nTherefore,\n$$\n\\sum_{k=1}^{n} \\frac{a_k}{b_{k+1}} \\le \\sum_{k=1}^{n} a_k + \\sum_{k=1}^{n} \\frac{1}{b_{k+1}} - n.\n$$\nLet $f(x) = 1 + x^2 + x^4$ for $x \\ge 0$, and $g(x) = \\frac{1}{1+x+x^2+x^3+x^4}$ for $x \\ge 0$. It is clear that $f(x)$ is convex on $[0, +\\infty)$.\n\nWe now prove that $g(x)$ is also convex on $[0, +\\infty)$. Indeed,\n$$\n\\begin{align*}\ng'(x) &= \\frac{-4x^5 + 5x^4 - 1}{(x^5 - 1)^2}, \\\\\ng''(x) &= \\frac{10(x^5 - 1)x^3(x - 1)^2(2x^2 + x + 2)(x^2 - 1)}{(x^5 - 1)^4} \\\\\n&= \\frac{10x^3(x + 1)(2x^2 + x + 2)}{(1 + x + x^2 + x^3 + x^4)^2} \\ge 0.\n\\end{align*}\n$$\nIt is well-known that if $\\varphi(x)$ is convex, then for any $a, b \\ge 0$,\n$$\n\\varphi(a) + \\varphi(b) \\le \\varphi(a + b) + \\varphi(0).\n$$\nApplying this inequality repeatedly, we obtain\n$$\n\\begin{align*}\n& \\sum_{k=1}^{n} a_k + \\sum_{k=1}^{n} \\frac{1}{b_{k+1}} - n \\\\\n\\le & \\sum_{k=1}^{n} f(x_k) + \\sum_{k=1}^{n} g(x_k) - n \\\\\n\\le & f\\left(\\sum_{k=1}^{n} x_k\\right) + g\\left(\\sum_{k=1}^{n} x_k\\right) + (n-1)(f(0) + g(0)) - n \\\\\n= & n^4 + n^2 + n - 1 + \\frac{n-1}{n^5-1}.\n\\end{align*}\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 28630,
    "subject": "Mathematics (Multi-modal)",
    "question": "There are 100 students in a summer camp, with exactly 2024 pairs of mutual friends. We need to divide them into 50 groups of 2 students each. Prove that:\n1. It's possible to group them such that at most 20 groups contain mutual friends;\n2. It's possible to group them such that at least 23 groups contain mutual friends;\n3. It's possible to group them such that exactly 22 groups contain mutual friends.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "*Proof.* Using the language of graph theory to describe this problem, we represent each student as a vertex and connect two vertices with an edge if the corresponding students are friends, forming a simple graph $G$ of order $100$. By the given condition, $G$ has exactly $2024$ edges.\n\nPair the $100$ vertices into $50$ pairs, called a \"pairing.\" The problem requires proving the existence of a pairing where the number of adjacent vertex pairs satisfies certain conditions.\n\n(1) It is well-known that the edge set of a complete graph of order $100$ can be partitioned into $99$ perfect matchings. Thus, there exist $99$ pairing methods such that each pair of vertices appears in exactly one of these pairings. Therefore, the total number of adjacent vertex pairs across these $99$ pairings is $2024$.\n\nBy the pigeonhole principle, there exists a pairing where the number of adjacent vertex pairs does not exceed $\\lfloor \\frac{2024}{99} \\rfloor = 20$.\n\n(2) Assume the conclusion does not hold. Then, the maximum matching in $G$ has at most $22$ edges. We can add some edges to $G$ such that the maximum matching has exactly $22$ edges. We now prove that the number of edges in $G$ is less than $2024$.\n\nTake a maximum matching, denoted as $A_iB_i$ ($1 \\le i \\le 22$), and let $W$ be the set of remaining vertices, where $|W| = 56$. By the maximality of the matching, there are no edges within $W$. If a vertex $A_i$ or $B_i$ is adjacent to at least two vertices in $W$, we call it a \"good vertex.\"\n\nNote that if one of $A_i$ or $B_i$ is a good vertex, the other cannot be adjacent to any vertex in $W$; otherwise, we could adjust $A_iB_i$ to two other edges, resulting in a larger matching.\n\nSuppose there are $k$ good vertices among $A_i, B_i$ ($1 \\le i \\le 22$), where $k \\le 22$. Without loss of generality, assume $A_1, A_2, \\dots, A_k$ are good vertices. For $1 \\le i < j \\le k$, $B_i$ and $B_j$ cannot be adjacent; otherwise, we could adjust $A_iB_i$ and $A_jB_j$ to three other edges, obtaining a larger matching.\n\nThus, the number of edges among $\\{A_1, B_1, \\dots, A_{22}, B_{22}\\}$ is at most $\\binom{44}{2} - \\binom{k}{2}$. For $1 \\le i \\le k$, there are at most $56$ edges between $\\{A_i, B_i\\}$ and $W$. For $k < i \\le 22$, there are at most $2$ edges between $\\{A_i, B_i\\}$ and $W$. Therefore,\n\n\\begin{align*}\n|E(G)| &\\le \\binom{44}{2} - \\binom{k}{2} + 56k + 2(22-k) \\\\\n&= 990 + \\frac{1}{2}k(109-k) \\le 990 + \\frac{1}{2} \\times 22 \\times 87 = 1947 < 2024,\n\\end{align*}\n\nwhich is a contradiction. This proves the conclusion.\n\n(3) From the conclusions of (1) and (2), there exist two pairings: the first contains at most $20$ edges of $G$, and the second contains at least $23$ edges of $G$.\n\nWe now adjust the first pairing step-by-step to the second pairing as follows: Select a pair $(A, B)$ from the second pairing that is not in the first pairing. Suppose the first pairing contains $(A, C)$ and $(B, D)$. Replace $(A, C)$ and $(B, D)$ with $(A, B)$ and $(C, D)$.\n\nAfter each adjustment, the number of pairs shared with the second pairing increases by at least one. Thus, after finitely many steps, the first pairing becomes the second pairing. Since each adjustment changes two pairs, the number of edges of $G$ in the pairing changes by at most $2$. Starting from at most $20$ and ending with at least $23$, if $22$ is not achieved at any intermediate step, there must exist a pairing $M_1$ with exactly $21$ edges of $G$, which after one adjustment becomes $M_2$ with exactly $23$ edges of $G$.\n\nLet $M_2$ be $\\{(A_i, B_i) | 1 \\le i \\le 50\\}$, where $A_iB_i$ is an edge if and only if $1 \\le i \\le 23$.\n\nWithout loss of generality, assume $M_1$ adjusts $(A_{22}, A_{23})$ and $(B_{22}, B_{23})$ to $(A_{22}, B_{22})$ and $(A_{23}, B_{23})$, and neither $A_{22}A_{23}$ nor $B_{22}B_{23}$ is an edge.\n\nNote that in the following cases, we can adjust $M_2$ to obtain a pairing with exactly $22$ edges of $G$:\n\n**Case 1:** There exist $1 \\le i < j \\le 23$ such that among $\\{A_i, B_i, A_j, B_j\\}$, there is a pairing with exactly one edge of $G$. Replace $(A_i, B_i)$ and $(A_j, B_j)$ in $M_1$ with this pairing.\n\n**Case 2:** There exist $24 \\le i < j \\le 50$ such that among $\\{A_i, B_i, A_j, B_j\\}$, there is a pairing with exactly one edge of $G$. Replace $(A_i, B_i)$ and $(A_j, B_j)$ in $M_1$ with this pairing, and then replace $(A_{22}, B_{22})$ and $(A_{23}, B_{23})$ with $(A_{22}, A_{23})$ and $(B_{22}, B_{23})$. The resulting pairing has exactly $22$ edges of $G$.\n\n**Case 3:** There exist $1 \\le i \\le 21$ and $24 \\le j \\le 50$ such that among $\\{A_i, B_i, A_j, B_j\\}$, there is a pairing with $0$ or $2$ edges of $G$. Replace $(A_i, B_i)$ and $(A_j, B_j)$ in $M_1$ with this pairing. If the number of edges increases by one, also replace $(A_{22}, B_{22})$ and $(A_{23}, B_{23})$ with $(A_{22}, A_{23})$ and $(B_{22}, B_{23})$. In any case, we obtain a pairing with exactly $22$ edges of $G$.\n\nNow assume none of the above cases occur. Note that the number of edges in $G$ is even, while the number of edges of the form $A_iB_i$ is $23$ (odd). Thus, there exist $1 \\le i < j \\le 50$ such that among $(A_i, A_j)$, $(A_i, B_j)$, $(B_i, A_j)$, $(B_i, B_j)$, exactly an odd number are edges in $G$. These four pairs can be split into two pairings: one with exactly $1$ edge of $G$ and the other with $0$ or $2$ edges.\n\nSince Cases 1, 2, and 3 do not occur, we must have $22 \\le i \\le 23$ and $24 \\le j \\le 50$. Without loss of generality, let $i = 23$, $j = 24$, and suppose $A_{23}A_{24}$ and $B_{23}B_{24}$ are either both edges or both non-edges (by swapping $A_{24}$ and $B_{24}$ if necessary). If neither is an edge, then in $M_2$, replace $(A_{23}, B_{23})$ and $(A_{24}, B_{24})$ with $(A_{23}, A_{24})$ and $(B_{23}, B_{24})$, resulting in a pairing with exactly $22$ edges of $G$.\n\nNow assume $A_{23}A_{24}$ and $B_{23}B_{24}$ are both edges of $G$.\n\nSince Case 3 does not occur, for any $1 \\le i \\le 21$ and $24 \\le j \\le 50$, exactly two of $A_iA_j$, $A_iB_j$, $B_iA_j$, $B_iB_j$ are edges of $G$. Thus, the number of edges between $\\{A_1, B_1, \\dots, A_{21}, B_{21}\\}$ and $\\{A_{24}, B_{24}, \\dots, A_{50}, B_{50}\\}$ is exactly $2 \\times 21 \\times 27 = 1134$.\n\nIt follows that the number of edges among $\\{A_1, B_1, \\dots, A_{22}, B_{22}\\}$ is at most $2024 - 1134 = 890 < \\binom{44}{2}$. Thus, there exists a non-adjacent pair in $\\{A_1, B_1, \\dots, A_{22}, B_{22}\\}$. Without loss of generality, assume $A_1A_2$ is not an edge of $G$. Since Case 1 does not occur, $B_1B_2$ is also not an edge.\n\nIn $M_2$, replace $(A_1, B_1)$, $(A_2, B_2)$, $(A_{23}, B_{23})$, $(A_{24}, B_{24})$ with $(A_1, A_2)$, $(B_1, B_2)$, $(A_{23}, A_{24})$, $(B_{23}, B_{24})$, resulting in a pairing with exactly $22$ edges of $G$. This completes the proof. □",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 28631,
    "subject": "Mathematics (Multi-modal)",
    "question": "Prove that the real-coefficient polynomial in $x, y, z$,\n$$\nx^{4}(x - y)(x - z) + y^{4}(y - z)(y - x) + z^{4}(z - x)(z - y),\n$$\ncannot be expressed as a finite sum of squares of real-coefficient polynomials in $x, y, z$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Denote the given polynomial by $F(x, y, z)$. By contradiction, assume there exist real-coefficient polynomials $f_1(x, y, z)$, $f_2(x, y, z)$, ..., $f_m(x, y, z)$ satisfying\n$$\nF(x, y, z) = \\sum_{i=1}^{m} f_i(x, y, z)^2. \\qquad (1)\n$$\nFirst, all $f_i$ must have degree at most 3. If some $f_i$ had degree $d > 3$, then the sum of squares of their degree $d$ homogeneous parts would be 0, implying all these parts vanish, which is a contradiction!\nThus all $\\deg(f_i) \\le 3$. By replacing each $f_i(x, y, z)$ with its degree 3 homogeneous part, we may assume each $f_i(x, y, z)$ is a homogeneous real-coefficient polynomial of degree 3.\nSetting $x = y$ in (1) yields\n$$\nz^4(z-y)^2 = \\sum_{i=1}^{m} f_i(y, y, z)^2. \\qquad (2)\n$$\nLet $g_i(y, z) = f_i(y, y, z)$. Setting $y = z$ in (2) gives $0 = \\sum_{i=1}^{m} g_i(y, y)^2$, so $g_i(y, y) = 0$ for all $i$. Thus $z - y \\mid g_i(y, z)$; similarly, setting $z = 0$ shows $z \\mid g_i(y, z)$. Therefore we can write $g_i(y, z) = z(z - y)h_i(y, z)$, where $h_i(y, z)$ is linear in $y$ and $z$. Substituting into (2) and canceling $z^2(z - y)^2$ gives\n$$\nz^2 = \\sum_{i=1}^{m} h_i(y, z)^2.\n$$\nSetting $z = 0$ shows $h_i(y, z) = \\alpha_i z$, so $f_i(y, y, z) = \\alpha_i z^2(z - y)$, where $\\alpha_i \\in \\mathbb{R}$. In particular, the coefficients of $y^3$ and $y^2z$ in $f_i(y, y, z)$ are 0, and the sum of coefficients of $z^3$ and $z^2y$ is 0.\nNow fix one $f_i$ and write $f_i = \\sum_{\\substack{u,v,w \\ge 0 \\\\ u+v+w=3}} a_{u,v,w} x^u y^v z^w$. Then:\n$$\n\\left\\{ \\begin{array}{l} a_{3,0,0} + a_{2,1,0} + a_{1,2,0} + a_{0,3,0} = 0; \\\\ a_{2,0,1} + a_{1,1,1} + a_{0,2,1} = 0; \\\\ a_{1,0,2} + a_{0,1,2} + a_{0,0,3} = 0. \\end{array} \\right. \\qquad (3)\n$$\nThese relations can be visualized in Figure 1, where dashed boxes enclose elements summing to 0.\n\n![](attached_image_1.png)\nFigure 1\n\nBy symmetry, we also have:\n$$\n\\left\\{\n\\begin{array}{l}\na_{3,0,0} + a_{2,0,1} + a_{1,0,2} + a_{0,0,3} = 0; \\\\ a_{0,3,0} + a_{0,2,1} + a_{0,1,2} + a_{0,0,3} = 0; \\\\ a_{1,2,0} + a_{1,1,1} + a_{1,0,2} = 0; \\\\ a_{2,1,0} + a_{1,1,1} + a_{0,1,2} = 0.\n\\end{array}\n\\right.\n\\qquad (4)\n$$\nIn the figure, this corresponds to sums of elements along edges and diagonals being 0.\nFrom (3), the sum of all $a_{u,v,w}$ is 0. Removing the corner triangles shows $a_{1,1,1} = 0$.\nThen (4) implies:\n$$\na_{2,1,0} + a_{0,1,2} = 0, \\quad a_{2,0,1} + a_{0,2,1} = 0, \\quad a_{1,2,0} + a_{1,0,2} = 0.\n$$\nLet $a = a_{2,1,0}$, $b = a_{0,2,1}$, $c = a_{1,0,2}$, then $a_{0,1,2} = -a$, $a_{2,0,1} = -b$, $a_{1,2,0} = -c$.\nNow, the coefficient of $x^2y^2z^2$ in $f_i(x, y, z)^2$ is:\n$$\na_{1,1,1}^2 + 2a_{2,1,0} \\cdot a_{0,1,2} + 2a_{2,0,1} \\cdot a_{0,2,1} + 2a_{1,2,0} \\cdot a_{1,0,2} = -2a^2 - 2b^2 - 2c^2 \\le 0,\n$$\nwhile $F(x, y, z)$ has coefficient 0 for $x^2y^2z^2$. From (1), we must have $a = b = c = 0$ for\neach $f_i$. Substituting back into (3) and (4) shows $f_i = 0$, contradicting (1). Therefore,\n$F(x, y, z)$ cannot be expressed as a finite sum of squares. $\\square$\nDenote the polynomial by $F(x, y, z)$. By contradiction, assume $F$ can be written as a sum of squares of real-coefficient polynomials. Then there exist $f_i(x, y, z), g_i(x, y, z) \\in \\mathbb{R}[x, y, z]$ ($i = 1, \\dots, m$) with:\n$$\nF = \\sum_{i=1}^{m} (x f_i + g_i)^2 = \\sum_{i=1}^{m} x^2 f_i^2 + \\sum_{i=1}^{m} g_i^2 + 2x \\sum_{i=1}^{m} f_i g_i, \\quad (5)\n$$\nwhere each monomial in $f_i(x, y, z)$, $g_i(x, y, z)$ has even degree in $x$. Since $\\deg(F) = 6$, as\nin Proof 1 we may assume $f_i, g_i$ are homogeneous of degrees 2 and 3 respectively, with\n$\\deg_x f_i \\le 2$, $\\deg_x g_i \\le 2$.\n\nSetting $(x, y, z) = (0, 1, 1)$ in (5) gives:\n$$\n\\sum_{i=1}^{m} g_{i}^{2}(0, 1, 1) = 0,\n$$\nso $g_i(0, 1, 1) = 0$. Thus we can write $g_i = c_i x^2 + (y-z)d_i$ for some $c_i, d_i \\in \\mathbb{R}[y, z]$.\nSetting $y = z = 1$ in (5) and comparing coefficients of $x^2$ gives:\n$$\n\\sum_{i=1}^{m} f_{i}^{2}(0, 1, 1) = 0,\n$$\nso $f_i(0, 1, 1) = 0$. Thus $f_i = a_i x^2 + (y-z)b_i$ for some $b_i \\in \\mathbb{R}[y, z]$ and $a_i \\in \\mathbb{R}$.\nUsing the identity:\n$$\n(ax^3 + b)^2 + (cx^3 + d)^2 = \\left(\\sqrt{a^2 + c^2}x^3 + \\frac{ab + cd}{\\sqrt{a^2 + c^2}}\\right)^2 + \\left(\\frac{ad - bc}{\\sqrt{a^2 + c^2}}\\right)^2,\n$$\nwe may assume $a_1 = 1$ and $a_i = 0$ for $i \\ge 2$. Comparing coefficients of $x^5$ gives:\n$$\n2 \\sum_{i=1}^{m} a_i c_i = -y - z,\n$$\nso $c_1 = -\\frac{y+z}{2}$.\nComparing coefficients of $x^4$ gives:\n$$\n2(y - z)b_1 + \\sum_{i=1}^{m} c_i^2 = yz.\n$$\nThus:\n$$\n2(y - z)b_1 + \\frac{1}{4}(y - z)^2 + \\sum_{i=2}^{m} c_i^2 = 0.\n$$\nSetting $y = z$ shows $y - z \\mid c_i$ for $i \\ge 2$.\n\nComparing coefficients of $x^2$ gives:\n$$\n(y - z) \\sum_{i=1}^{m} b_i^2 + 2 \\sum_{i=1}^{m} c_i d_i = 0.\n$$\nSetting $y = z$ shows $y - z \\mid d_1$. Let $d_1 = (y - z)(sy + tz)$ for $s, t \\in \\mathbb{R}$.\nComparing coefficients of $x^3$ gives:\n$$\nd_1 + \\sum_{i=1}^{m} b_i c_i = 0.\n$$\nSetting $y = z$ shows $y - z \\mid b_1$. Let $b_1 = k(y - z)$ for $k \\in \\mathbb{R}$.\nFrom $F(1, 1, 0) = 0$, we get $1 + k - \\frac{1}{2} + s = 0$, so $s = -\\frac{1}{2} - k$. From $F(1, 0, 1) = 0$, we get $1 + k - \\frac{1}{2} + t = 0$, so $t = -\\frac{1}{2} - k$.\nThus:\n$$\nx f_1 + g_1 = x(x^2 + k(y - z)^2) - \\frac{y+z}{2}x^2 - (\\frac{1}{2} + k)(y + z)(y - z)^2.\n$$\n\nSetting $x = 1, y = 1$ in $F - (xf_1 + g_1)^2 \\ge 0$ gives:\n$$\n\\begin{aligned}\n0 &\\le z^4(z-1)^2 - \\left(\\frac{1-z}{2} + k(1-z)^2 - \\left(\\frac{1}{2} + k\\right)(1-z)^2(1+z)\\right)^2 \\\\\n&= (z-1)^2 \\left[z^4 - \\left(\\frac{1}{2} + k(1-z) - \\left(\\frac{1}{2} + k\\right)(1-z^2)\\right)^2\\right].\n\\end{aligned}\n$$\nThe inequality $z^2 - ((\\frac{1}{2} + k)z - k)^2 \\ge 0$ holds for all real $z$ only when $k = 0$. But then setting $z = 0, x = 7, y = 2$ gives $F(x, y, z) - (xf_1 + g_1)^2 = -225 < 0$, a contradiction.\nTherefore, $F(x, y, z)$ cannot be expressed as a finite sum of squares of real-coefficient polynomials. $\\square$",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 28632,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all positive integers $m$ such that there exists an infinite set $A$ of positive integers satisfying: for any $m$ distinct elements $a_1, a_2, \\dots, a_m$ in $A$, both $a_1+a_2+\\dots+a_m$ and $a_1a_2\\cdots a_m$ are square-free.\n\n*Note: A positive integer $n$ is called square-free if it is not divisible by the square of any prime number.*",
    "options": [],
    "answer": "All square-free positive integers m.",
    "solution": "**Proof:** We first prove a lemma.\n**Lemma:** For integers $m \\ge 2$, $s \\ge 1$, and a sequence $1 = x_1 < \\dots < x_s$ where every sum $\\sum_{1 \\le j \\le m} x_{i_j}$ ($1 \\le i_1 \\le \\dots \\le i_m \\le s$) is square-free, there exists an integer $x > x_s$ such that:\n* $x$ is coprime with each $x_i$ ($1 \\le i \\le s$)\n* Every sum $(m-t)x + \\sum_{1 \\le j \\le t} x_{i_j}$ ($0 \\le t \\le m-1$, $1 \\le i_1 \\le \\dots \\le i_t \\le s$) is square-free\n**Proof of Lemma:** Take $B > 1$ such that:\n$$\nc_B := \\prod_{p < B} \\left(1 - \\frac{1}{p}\\right)^{-1} > m \\cdot 2^{m+s}.\n$$\nLet\n$$\nd_B = \\prod_{p \\le B} p, \\quad M_0 = \\prod_{1 \\le k \\le 2^{m+s} x_s} k, \\quad M = d_B M_0.\n$$\nConsider $x = yM^2 + x_1$ ($1 \\le y \\le M^2$). Clearly, $x$ is coprime with each $x_i$ ($1 \\le i \\le s$).\nNote that\n$$\n(m-t)x + \\sum_{1 \\le j \\le t} x_{i_j} \\le mx \\le m(M^4 + 1) < (mM^2)^2.\n$$\nIf $q^2 \\mid (m-t)x + \\sum_{1 \\le j \\le t} x_{i_j}$, then $q < mM^2$.\nIf $q \\nmid M$, clearly $(m-t)x + \\sum_{1 \\le j \\le t} x_{i_j}$ is not divisible by $q^2$. For primes $q < mM^2$ with $q \\nmid M$ (so $q > 2^{m+s}x_s$), let $X_q$ be the set of integers $1 \\le y \\le M^2$ where some sum $(m-t)x + \\sum_{1 \\le j \\le t} x_{i_j}$ is divisible by $q^2$.\nThe number of possible index sets is at most $2^{s+m-2}$, so\n$$\n|X_q| \\le 2^{m+s-2} \\left( \\frac{M^2}{q^2} + 1 \\right).\n$$\nFor $2^{m+s}x_s < q \\le M$:\n$$\n|X_q| \\le 2^{m+s-1} \\frac{M^2}{q^2};\n$$\n---\n\nFor $M < q \\le mM^2$:\n$$\n|X_q| \\le 2^{m+s-1}.\n$$\nThus,\n$$\n\\begin{aligned} |\\bigcup X_q| &\\le 2^{m+s-1} M^2 \\sum_{q>2^{m+s}x_s} \\frac{1}{q^2} + 2^{m+s-1}(\\pi(mM^2) - \\pi(M)) \\\\ &\\le \\frac{2^{m+s-1}}{2^{m+s}x_s} M^2 + \\frac{m \\cdot 2^{m+s-1}}{c_B} M^2 < M^2. \\end{aligned}\n$$\nSince we have used $d_B \\mid M$, it follows that\n$$\n\\pi(mM^2) - \\pi(M) \\le (mM^2 - M) \\prod_{p \\le B} \\left(1 - \\frac{1}{p}\\right) < c_B^{-1} mM^2.\n$$\nNow choose an integer $y$ such that $1 \\le y \\le M$ and\n$$\ny \\notin \\bigcup_{\\substack{q<mM^2 \\\\ q \\nmid M}} X_q,\n$$\nthen $x = yM^2 + x_1$ will satisfy the requirements.\nNow the main proof. If $m$ has a square factor $p^2$, by pigeonhole principle there exist $m$ elements $a_{i_1}, \\dots, a_{i_m}$ with $a_{i_1} \\equiv \\dots \\equiv a_{i_m} \\pmod{p^2}$, making their sum divisible by $p^2$.\nIf $m$ is square-free, construct $A = \\{a_i\\}$ inductively starting with $a_1 = 1$. By the lemma, we can find pairwise coprime integers $1 = a_1 < a_2 < \\dots$ where all $m$-term sums are square-free. Then all $m$-term products are also square-free since the $a_i$ are pairwise coprime. $\\square$",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 28633,
    "subject": "Mathematics (Multi-modal)",
    "question": "Given an odd integer $n \\ge 3$ and a regular $n$-gon with vertex set $V$, let $\\mathcal{P}$ denote the set of all regular polygons with vertices in $V$. For example, if $n = 15$, then $\\mathcal{P}$ includes 1 regular 15-gon, 3 regular pentagons, and 5 equilateral triangles.\n\nTwo players, Alice and Bob, play the following game. Initially, all points in $V$ are uncolored. Alice moves first, and they take turns coloring an uncolored point in $V$: Alice colors red, and Bob colors blue. The game ends when all points are colored. A polygon in $\\mathcal{P}$ is called **red-dominated** if it has more red vertices than blue ones.\n\nFind the largest integer $k$ such that, no matter how Bob plays, Alice can guarantee at least $k$ red-dominated polygons in $\\mathcal{P}$ at the end.",
    "options": [],
    "answer": "(σ(n) + τ(n) - n - 1) / 2",
    "solution": "The maximal integer $k$ is $\\frac{\\sigma(n) + \\tau(n) - n - 1}{2}$. (Note that from this formula, $\\sigma(n)$ and $\\tau(n)$ have the same parity.)\n\nLabel the vertices of $V$ as $A_0, A_1, \\cdots, A_{n-1}$, with indices considered modulo $n$.\n\nWe say a coloring is *perfect* if there exists $i$ such that $A_i$ is red, and for any $1 \\le k \\le \\frac{n-1}{2}$, $A_{i-k}$ and $A_{i+k}$ are colored with different colors (one red and one blue).\n\nFirst, we show that for a perfect coloring, the number of good polygons is $\\frac{\\sigma(n) + \\tau(n) - n - 1}{2}$. Without loss of generality, assume $i=0$.\n\nFor each divisor $d \\ne n$ of $n$, there are $d+1$ regular $\\frac{n}{d}$-gons in $\\mathcal{P}$ of two types:\n\n1) $A_0A_dA_{2d}\\cdots A_{n-d}$\n\n2) $A_bA_{b+d}\\cdots A_{n-d+b}$ where $b \\in \\{1, \\cdots, d-1\\}$.\n\nThe regular $\\frac{n}{d}$-gon $A_0A_d\\cdots A_{n-d}$ is always good because $A_0$ is red, while for $s = 1, \\cdots, \\frac{n}{d}-1$, $A_{sd}$ and $A_{-sd}$ have different colors.\n\nFor $b \\in \\{1, \\cdots, d-1\\}$, the polygons $A_bA_{d+b}\\cdots A_{n-d+b}$ and $A_{d-b}A_{2d-b}\\cdots A_{n-b}$ have corresponding vertices $A_{sd+b}$ and $A_{-sd-b}$ with different colors, so exactly one of these two polygons must be good.\n\nTherefore, the number of good polygons in a perfect coloring is:\n$$\n\\sum_{\\substack{d|n \\\\ d \\ne n}} \\frac{d+1}{2} = \\frac{\\sigma(n) + \\tau(n) - n - 1}{2}.\n$$\n\nNow we show Alice can guarantee at least $\\frac{\\sigma(n) + \\tau(n) - n - 1}{2}$ good polygons by ensuring a perfect coloring. Alice first colors $A_0$ red. Then, whenever Bob colors $A_a$ ($1 \\le a \\le n-1$) blue, Alice colors $A_{-a}$ red. Since $n$ is odd, this strategy can be executed.\n\nConversely, Bob can ensure at most $\\frac{\\sigma(n) + \\tau(n) - n + 1}{2}$ good polygons by also enforcing a perfect coloring. After Alice colors her first point red (wlog assume it's $A_0$), Bob pairs $A_a$ with $A_{-a}$ for $1 \\le a \\le n-1$.\n\nBob starts by coloring any unpaired point blue. Subsequently, after Alice's move:\n- If Alice colors a point whose paired point is uncolored, Bob colors its pair blue.\n- Otherwise, Bob colors any remaining uncolored point blue.\n\nThis ensures a perfect coloring, limiting good polygons to $\\frac{\\sigma(n) + \\tau(n) - n + 1}{2}$.\n\nCombining both strategies, the maximal integer $k$ is $\\frac{\\sigma(n) + \\tau(n) - n - 1}{2}$. $\\square$",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 28634,
    "subject": "Mathematics (Multi-modal)",
    "question": "For a finite non-empty set $A$ of real numbers, let $\\max(A)$ denote its maximum element, and define:\n$$\nP(A) = \\sum_{\\substack{B \\subseteq A \\\\ |B| \\text{ odd}}} m(B), \\quad Q(A) = \\sum_{\\substack{\\emptyset \\neq B \\subseteq A \\\\ |B| \\text{ even}}} m(B),\n$$\nwhere $m(B)$ is the median of finite non-empty set $B$: if $B = \\{b_1, b_2, \\dots, b_n\\}$ ($b_1 < b_2 < \\dots < b_n$), then $m(B) = \\frac{1}{2}(b_{\\lfloor \\frac{n+1}{2} \\rfloor} + b_{\\lceil \\frac{n+1}{2} \\rceil})$.\nFind the smallest real number $c$ such that for any set $A$ of 2025 distinct positive real numbers,\n$$\nP(A) - Q(A) \\leq c \\cdot \\max(A).\n$$",
    "options": [],
    "answer": "((2024 choose 990) - (2024 choose 989) + 1)/2",
    "solution": "Without loss of generality, let the elements of $A$ in increasing order be $x_0, x_1, \\dots, x_{2024}$. We first compute $P(A)$. For each $x_m$, we count how many odd-sized subsets have $x_m$ as their median. A subset $B$ of odd size has $x_m$ as its median if and only if\n$$\n|B \\cap \\{x_0, \\dots, x_{m-1}\\}| = |B \\cap \\{x_{m+1}, \\dots, x_{2024}\\}|.\n$$\nIf this common value is $d$, the number of such subsets $B$ is $\\binom{m}{d}\\binom{2024-m}{d}$. Thus, the total number of such $B$ is\n$$\n\\sum_d \\binom{m}{d} \\binom{2024-m}{d} = \\sum_d \\binom{m}{m-d} \\binom{2024-m}{d} = \\binom{2024}{m},\n$$\nwhere the last equality follows from Vandermonde's identity. Therefore,\n$$\nP = \\sum_{m=0}^{2024} \\binom{2024}{m} x_m.\n$$\nNext, we compute $Q(A)$. For each $x_m$, we count how many even-sized subsets have $x_m$ as one of their two middle elements. The number of such subsets is\n$$\n\\begin{aligned}\n& \\sum_d \\left( \\binom{m}{d+1} \\binom{2024-m}{d} + \\binom{m}{d} \\binom{2024-m}{d+1} \\right) \\\\\n&= \\sum_d \\binom{m}{m-d-1} \\binom{2024-m}{d} + \\sum_d \\binom{m}{m-d} \\binom{2024-m}{d+1} \\\\\n&= \\binom{2024}{m-1} + \\binom{2024}{m+1}.\n\\end{aligned}\n$$\nThus,\n$$\nQ = \\sum_{m=0}^{2024} \\frac{1}{2} \\left( \\binom{2024}{m-1} + \\binom{2024}{m+1} \\right) x_m.\n$$\n\n---\n\nSince multiplying all elements of $A$ by a positive constant does not change the problem, we may assume $x_{2024} = 1$. Let $y_i = x_i - x_{i-1}$ for $i = 0, 1, \\dots, 2024$ (with $x_{-1} = 0$). Then $P-Q$ can be expressed as a linear form in $y_0, y_1, \\dots, y_{2024}$, where the $y_i$ are positive and sum to 1. The minimal $c$ is therefore equal to the maximum coefficient of the $y_i$.\nThe coefficient of $x_m$ in $P-Q$ is $\\binom{2024}{m} - \\frac{1}{2}(\\binom{2024}{m-1} + \\binom{2024}{m+1})$. Thus, the coefficient of $y_i$ is\n$$\n\\begin{aligned}\n& \\sum_{m=i}^{2024} \\left( \\binom{2024}{m} - \\frac{1}{2} \\left( \\binom{2024}{m-1} + \\binom{2024}{m+1} \\right) \\right) \\\\\n&= \\sum_{m=i}^{2024} \\binom{2024}{m} - \\frac{1}{2} \\left( \\sum_{m=i-1}^{2024} \\binom{2024}{m} + \\sum_{m=i+1}^{2024} \\binom{2024}{m} - 1 \\right) \\\\\n&= \\frac{\\binom{2024}{i} - \\binom{2024}{i-1} + 1}{2}.\n\\end{aligned}\n$$\nWe now find the maximum of $\\binom{2024}{i} - \\binom{2024}{i-1}$. For $i \\ge 1013$, $\\binom{2024}{i} - \\binom{2024}{i-1} \\le 0$, so the maximum must occur for $0 \\le i \\le 1012$. Let $d_i = \\binom{2024}{i} - \\binom{2024}{i-1}$. Then\n$$\n\\begin{aligned}\nd_{i+1} - d_i &= \\binom{2024}{i+1} + \\binom{2024}{i-1} - 2\\binom{2024}{i} \\\\\n&= \\binom{2024}{i} \\left( \\frac{2025-i}{i+1} + \\frac{i}{2025-i} - 2 \\right) \\\\\n&= \\binom{2024}{i} \\left( \\frac{2025 \\times 2026}{(i+1)(2025-i)} - 4 \\right).\n\\end{aligned}\n$$\nFor $0 \\le i \\le 1010$, $(i+1)(2025-i)$ increases with $i$. Since $\\frac{2025 \\times 2026}{991 \\times 1035} < 4 < \\frac{2025 \\times 2026}{990 \\times 1036}$, we have $d_{i+1} > d_i$ when $i \\le 989$ and $d_{i+1} < d_i$ when $i \\ge 990$. Thus, the maximum of $d_i$ is achieved at $i = 990$. Therefore, the minimal $c$ is\n$$\nc = \\frac{\\binom{2024}{990} - \\binom{2024}{989} + 1}{2}.\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 28635,
    "subject": "Mathematics (Multi-modal)",
    "question": "Given a positive integer $n \\ge 4$, prove that the equation\n$$\n(2^x - 1)(5^x - 1) = y^n\n$$\nhas no positive integer solutions $(x, y)$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "We will use the well-known Lifting the Exponent Lemma (LTE), stated as follows:\n**Lifting the Exponent Lemma:** Let $p$ be a prime and $m$ a positive integer. Let $a, b$ be integers such that $p \\nmid ab$ and $q_p \\mid (a-b)$, where $q_p = p$ when $p$ is odd and $q_2 = 4$ when $p = 2$. Then we have\n$$\nv_p(a^m - b^m) = v_p(a - b) + v_p(m).\n$$\n\nAssume there exist positive integers $x, y$ satisfying $(2^x - 1)(5^x - 1) = y^n$. We first prove the following lemma:\n**Lemma:** We have $2 \\cdot 3 \\cdot 5 \\cdot 7 \\mid x$, and for any prime $p \\le n$,\n$$\nv_p(x) \\ge \\lfloor \\log_p n \\rfloor.\n$$\n\n**Proof of Lemma:** From $2 \\mid (5^x - 1)$ we get $2 \\mid y$. By LTE:\n$$\n\\begin{aligned}\nn \\le nv_2(y) &= v_2((2^x - 1)(5^x - 1)) \\\\\n&= v_2(5 - 1) + v_2(x).\n\\end{aligned}\n$$\nThus $v_2(x) \\ge n - 2 \\ge \\lfloor \\log_2 n \\rfloor$. In particular, $4 \\mid x$, so $3 \\cdot 5 \\mid 2^x - 1$, hence $3 \\cdot 5 \\mid y$. By LTE:\n$$\n\\begin{aligned}\nn \\le nv_3(y) &= v_3(((2^2)^{\\frac{x}{2}} - 1)((5^2)^{\\frac{x}{2}} - 1)) \\\\\n&= v_3(2^2 - 1) + v_3(\\frac{x}{2}) + v_3(5^2 - 1) + v_3(\\frac{x}{2}) = 2v_3(x) + 2; \\\\\nn \\le nv_5(y) &= v_5(((2^4)^{\\frac{x}{4}} - 1)(5^x - 1)) \\\\\n&= v_5(2^4 - 1) + v_5(\\frac{x}{4}) = v_5(x) + 1.\n\\end{aligned}\n$$\nThus $v_3(x) \\ge (n - 2)/2 \\ge \\lfloor \\log_3 n \\rfloor$, and $v_5(x) \\ge n - 1 \\ge \\lfloor \\log_5 n \\rfloor$. In particular, $3 \\mid x$. Combined with $2 \\mid x$, we have $6 \\mid x$, so $7 \\mid (2^x - 1)$, hence $7 \\mid y$. By LTE:\n$$\n\\begin{aligned}\nn \\le nv_7(y) &= v_7(((2^6)^{\\frac{x}{6}} - 1)((5^6)^{\\frac{x}{6}} - 1)) \\\\\n&= v_7(2^6 - 1) + v_7(\\frac{x}{6}) + v_7(5^6 - 1) + v_7(\\frac{x}{6}) = 2v_7(x) + 2.\n\\end{aligned}\n$$\nThus $v_7(x) \\ge (n - 2)/2 \\ge \\lfloor \\log_7 n \\rfloor$.\n\nIf $n \\le 10$, the lemma is already proved. Now assume $n \\ge 11$. We proceed by induction on $p$. Let $p$ be a prime with $11 \\le p \\le n$, and assume the lemma holds for all primes $q < p$. By induction hypothesis, for any prime $q \\le p-1$,\n$$\nv_q(x) \\ge \\lfloor \\log_q n \\rfloor \\ge \\lfloor \\log_q (p-1) \\rfloor \\ge v_q(p-1).\n$$\nThus $(p-1) \\mid x$. By Fermat's Little Theorem, $p \\mid (2^{p-1}-1)$ and $p \\mid (5^{p-1}-1)$. Hence $p \\mid (2^x - 1)$ and $p \\mid y$. By LTE:\n$$\n\\begin{align*}\nn \\le nv_p(y) &= v_p\\left(((2^{p-1})^{\\frac{x}{p-1}} - 1)((5^{p-1})^{\\frac{x}{p-1}} - 1)\\right) \\\\\n&= v_p(2^{p-1} - 1) + v_p\\left(\\frac{x}{p-1}\\right) + v_p(5^{p-1} - 1) + v_p\\left(\\frac{x}{p-1}\\right) \\\\\n&= 2v_p(x) + v_p(2^{p-1} - 1) + v_p(5^{p-1} - 1) \\\\\n&< 2v_p(x) + \\log_p(2^{p-1}) + \\log_p(5^{p-1}) \\\\\n&< 2v_p(x) + p \\log_p(10).\n\\end{align*}\n$$\nTherefore,\n$$\nv_p(x) > \\frac{n - p \\log_p(10)}{2}.\n$$\nSince $11 \\le p \\le n$, this implies $v_p(x) > 0$, so $v_p(x) \\ge 1$. If $n < p^2$, then $v_p(x) \\ge 1 \\ge \\lfloor \\log_p n \\rfloor$. To complete the induction, we only need to prove for $n \\ge p^2 \\ge 11^2$ that\n$$\n\\frac{n - p \\log_p(10)}{2} > \\log_p n.\n$$\nThis is equivalent to\n$$\n\\frac{p^n}{10^p} > n^2.\n$$\nIndeed, using $n \\ge p^2 \\ge 11^2$, we have\n$$\n\\frac{p^n}{10^p} = \\frac{p^p}{10^p} \\cdot p^{n-p} > 11^{n-\\sqrt{n}} > 9^{\\frac{n}{2}} > n^2.\n$$\nThis completes the induction. The lemma is proved.\n\n**Proof of the Main Result:** By the lemma, for any prime $p \\le n$,\n$$\nv_p(x) \\ge \\lfloor \\log_p n \\rfloor \\ge v_p(n).\n$$\nThus $n \\mid x$. Let $x = nx_1$, then\n$$\ny^n = (2^x - 1)(5^x - 1) < (10^{x_1})^n.\n$$\nHence $10^{x_1} \\ge y + 1$. Therefore,\n$$\n\\begin{align*}\ny^n &= (10^{x_1})^n - 2^x - 5^x + 1 \\\\\n&\\ge (y+1)^n - 2^x - 5^x + 1 \\\\\n&> y^n + n \\cdot y^{n-1} - 2 \\cdot 5^x \\\\\n&\\ge y^n + 4 \\cdot y^{\\frac{3n}{4}} - 2 \\cdot 5^x.\n\\end{align*}\n$$\n\nThis implies $y^{\\frac{3n}{4}} < 5^x$, i.e., $y^n < 5^{\\frac{4x}{3}}$. On the other hand, by the lemma we have $x \\ge 2 \\cdot 3 \\cdot 5 \\cdot 7$, so\n$$\ny^n = (2^x - 1)(5^x - 1) > (0.9 \\cdot 2^x) \\cdot (0.9 \\cdot 5^x) > 0.8 \\cdot 5^{\\frac{2x}{5}} \\cdot 5^x > 5^{\\frac{4x}{3}},\n$$\na contradiction.\nTherefore, the original Diophantine equation has no positive integer solutions. $\\square$",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 28636,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $n \\ge 2$ be an integer. Two players, Alice and Bob, play the following game: Initially all $\\binom{n}{2}$ edges of the complete graph $K_n$ are uncolored. They take turns coloring edges red, with Alice starting first. In each move, a player selects one or two uncolored edges to color red, with the constraint that after each move, no three vertices can have all three edges between them colored red. The game ends when no more edges can be colored.\nProve that there exists a real number $\\varepsilon > 0$ such that for any integer $n \\ge 2$, no matter how Bob plays, Alice can ensure that at the end of the game, the number of red edges does not exceed $\\left(\\frac{1}{4} - \\varepsilon\\right)n^2 + n$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "We consider the graph $G$ formed by red edges. Initially, $G$ is an empty graph on $n$ vertices, and the condition requires that $G$ remains triangle-free throughout the game.\n\n**Step 1:** We show that if at Alice's turn, $G$ has 24 isolated vertices, then Alice can create a 5-cycle among these vertices in four moves.\nLet $S$ be the set of these 24 isolated vertices. Alice first selects three vertices $A, B, C \\in S$ and colors $AB$ and $AC$ red. After Bob's move, $S$ still contains at least 17 isolated vertices. Alice then selects two isolated vertices $D, E \\in S$ and colors $AD$ and $AE$ red. After Bob's move, $S$ has at least 11 isolated vertices remaining. Alice then selects two more isolated vertices $F, G \\in S$ and colors $BF$ and $CG$ red.\n\n![](attached_image_1.png)\n\nConsider the four edges $DF$, $DG$, $EF$, and $EG$, which are currently uncolored. After Bob's move, at least one of them remains uncolored, say $DF$. Now $S$ still has at least 5 isolated vertices. Alice selects one of them, $H$, and colors $FH$ and $DH$ red. This creates a 5-cycle $ABFHD$ in $G$. By following this strategy, after four moves by each player, the number of isolated vertices in $G$ decreases by at most 24.\n\n**Step 2:** Alice follows the strategy described in Step 1. After $4 \\times \\lfloor \\frac{n}{24} \\rfloor$ moves, she can ensure that $G$ contains $\\lfloor \\frac{n}{24} \\rfloor$ disjoint 5-cycles. Alice then colors edges arbitrarily until $G$ becomes a maximal triangle-free graph.\n\nWe now estimate the number of edges in $G$. Let $k = \\lfloor \\frac{n}{24} \\rfloor$, and denote the disjoint 5-cycles by $C_1, C_2, \\dots, C_k$. Between any two cycles $C_i$ and $C_j$, there are at most 10 edges. For each vertex $v$ not in any $C_i$, there are at most 2 edges from $v$ to each $C_i$. The remaining $n - 5k$ vertices outside all $C_i$ can have at most $\\frac{(n-5k)^2}{4}$ edges between them (by Turán's theorem). The cycles $C_1, \\dots, C_k$ themselves contribute $5k$ edges. Thus, the total number of edges is bounded by:\n\n$$\n\\begin{align*}\n|E(G)| &= 10 \\binom{k}{2} + 2k(n-5k) + \\frac{(n-5k)^2}{4} + 5k \\\\\n&= \\frac{n^2}{4} - \\frac{kn}{2} + \\frac{5k^2}{4} \\\\\n&\\le \\frac{n^2}{4} - \\frac{1}{2} \\left( \\frac{n}{24} - 1 \\right) n + \\frac{5}{4} \\left( \\frac{n}{24} \\right)^2 \\\\\n&= \\left( \\frac{1}{4} - \\frac{43}{48^2} \\right) n^2 + \\frac{n}{2}.\n\\end{align*}\n$$\n\nTaking $\\varepsilon = \\frac{43}{48^2}$ satisfies the requirement. $\\square$",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 28637,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $P(x)$ and $Q(x)$ be non-constant real polynomials such that for every positive integer $m$, there exists a positive integer $n$ with $P(m) = Q(n)$.\n(1) Prove that if $\\deg(Q)$ divides $\\deg(P)$, then there exists a real polynomial $h(x)$ such that $P(x) = Q(h(x))$.\n(2) Prove that $\\deg(Q)$ divides $\\deg(P)$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "**Proof 1:** We denote by $O_k(x)$ any polynomial of degree at most $k$.\n\n(1) If $v = \\deg Q$ divides $u = \\deg P$, let $u = vL$. Write\n$$\nP(x) = a_u x^u + a_{u-1} x^{u-1} + \\dots + a_0,\n$$\n$$\nQ(x) = b_v x^v + b_{v-1} x^{v-1} + \\dots + b_0.\n$$\n\nFor $k = 0$, take $c_0 = \\sqrt[L]{a_u/b_v}$, so $Q(c_0 x^L) = a_u x^u + \\dots = P(x) + O_{u-1}(x)$.\nAssume $g_{k-1}(x) = c_0 x^L + \\dots + c_{k-1} x^{L+1-k}$ satisfies $Q(g_{k-1}(x)) = P(x) + R(x)$, where $R(x) = s x^{u-k} + \\dots$ has degree $\\le u - k$. Consider $g_k(x) = g_{k-1} + c_k x^{L-k}$:\n$$\n\\begin{align*} Q(g_k(x)) - Q(g_{k-1}(x)) &= b_v [(g_{k-1}(x) + c_k x^{L-k})^v - (g_{k-1}(x))^v] + O_{(v-1)L-k}(x) \\\\ &= b_v [v g_{k-1}(x)^{v-1} c_k x^{L-k} + O_{(v-2)L+2(L-k)}(x)] + O_{(v-1)L-k}(x) \\\\ &= c_k b_v v c_0^{v-1} x^{vL-k} + O_{u-k-1}(x). \\end{align*}\n$$\nChoose $c_k$ such that $c_k b_v v c_0^{v-1} = s$ (the coefficient of $x^{vL-k}$ in $R(x)$). Then $g_k(x) = g_{k-1} + c_k x^{L-k}$ satisfies $Q(g_k(x)) - P(x) = O_{u-k-1}(x)$.\n\nThe problem's condition guarantees that for every integer $m \\ge M_0$, there exists $n = n_m$ such that $Q(n_m) = P(m)$, thus\n$$\n|Q(n_m) - Q(h(m))| = |P(m) - Q(h(m))| = |R(m)| \\le C \\cdot m^{u-L-1}.\n$$\nThis implies that when $M_0$ is sufficiently large, there exists a positive real number $D$ such that for all integers $m \\ge M_0$:\n$$\n|n_m - h(m)| < \\frac{D}{m}. \\qquad (11)\n$$\n\nLet $n_m = h(m) + r_m$. Since $h(m)$ is a polynomial of degree $L$, we have\n$$ \\sum_{k=0}^{L+1} (-1)^k \\binom{L+1}{k} h(m+k) = 0. $$\nHowever, since each $n_m$ is an integer, this implies\n$$\nT_m = \\sum_{k=0}^{L+1} (-1)^k \\binom{L+1}{k} r_{m+k} = \\sum_{k=0}^{L+1} (-1)^k \\binom{L+1}{k} n_{m+k} \\in \\mathbb{Z}.\n$$\nFrom (11), there exists $M_1$ such that when $m \\ge M_1$, we have $|r_m| < \\frac{1}{2^{L+1}}$, and consequently $T_m = 0$. This means $r_m$ is a recurrence sequence, and solving this recurrence shows that for $m \\ge M_1$, $r_m$ is a polynomial in $m$ with degree at most $L$. Moreover, from (11) we know $\\{r_m\\}$ is bounded, hence it must be identically zero. Therefore, $P(m) - Q(h(m)) = 0$ holds for all $m \\ge M_1$, which implies $P(x) \\equiv Q(h(x))$. Furthermore, since $h(m) = n_m$ always takes integer values, $h(\\cdot)$ must be a polynomial with rational coefficients.\n\n\n(2) Prove that $\\deg(Q)$ divides $\\deg(P)$.\n\n**Proof 2 for (2):** Following similar analysis to Proof 1, we select $f_1(x) = x^v$ and $f_2(x) = 2x^v$, which yields rational-coefficient polynomials $h_1(x) = c_1x^u + \\dots$ and $h_2(x) = c_2x^u + \\dots$ satisfying\n$$\nP(x^v) \\equiv Q(h_1(x)), \\quad P(2x^v) \\equiv Q(h_2(x)).\n$$\nComparing leading coefficients gives:\n$$\na_u = b_v \\cdot c_1^v \\quad \\text{and} \\quad a_u \\cdot 2^u = b_v \\cdot c_2^v,\n$$\nthus $2^u = \\left(\\frac{c_2}{c_1}\\right)^v$, meaning $2^{u/v} = \\frac{c_2}{c_1}$ must be rational. Since fractional powers of 2 are irrational, $\\frac{u}{v}$ must be an integer, i.e., $\\deg(Q)$ divides $\\deg(P)$.\n\n\n**Proof 3 for (2):** Choosing $f(x) = x^v$, there exists a rational-coefficient polynomial $h(x) = c_ux^u + c_{u-1}x^{u-1} + \\dots + c_0$ satisfying $P(x^v) \\equiv Q(h(x))$, where for large positive integers $m = x^v$, $n_m = h(x)$ is also a positive integer. Grouping terms by powers modulo $v$, we express $h(x)$ as:\n$$\nh(x) \\equiv x^{v-1}h_{v-1}(x^v) + x^{v-2}h_{v-2}(x^v) + \\dots + xh_1(x^v) + h_0(x^v),\n$$\nwhere each $h_k(\\cdot)$ is a rational-coefficient polynomial.\nFor a large prime $p$, substituting $x = p^{1/v}$ yields $h(p^{1/v}) = D \\in \\mathbb{Z}$.\nBy Eisenstein's criterion, the polynomial $x^v - p$ is irreducible and shares the root $x = p^{1/v}$ with $h(x) - D$, hence $x^v - p \\mid h(x) - D$. For $k = 0, 1, \\dots, v-1$, we have $x^v - p \\mid h_k(x^v) - h_k(p)$, therefore:\n$$\nx^v - p \\mid x^{v-1}h_{v-1}(p) + x^{v-2}h_{v-2}(p) + \\dots + xh_1(p) + h_0(p) - D.\n$$\nThe right-hand side has degree $\\leq v-1$, so divisibility implies it's the zero polynomial. Thus for $k = 1, 2, \\dots, v-1$, $h_k(p) = 0$ holds for all large primes $p$, making each $h_k(x)$ identically zero. Consequently, $h(x) \\equiv h_0(x^v)$, meaning the degree $u$ must be a multiple of $v$, i.e., $\\deg(Q)$ divides $\\deg(P)$.\n\n\n**Proof 4 for (2):** Again choosing $f(x) = x^v$, there exists a rational-coefficient polynomial $h(x) = c_u x^u + c_{u-1} x^{u-1} + \\cdots + c_0$ ($c_u \\neq 0$) satisfying $P(x^v) \\equiv Q(h(x))$.\nWe claim that all terms in $h(x)$ must have exponents congruent to $u$ modulo $v$ or be constant terms. If not, let $r$ be the largest exponent with $c_r \\neq 0$ and $r \\not\\equiv u \\pmod v$ ($r > 0$). Since $Q(x) = b_v x^v + b_{v-1} x^{v-1} + \\cdots$, we have:\n$$\nQ(h(x)) = b_v(h(x))^v + O(u(v-1)),\n$$\nwhere $O(u(v-1))$ denotes some polynomial of degree $\\le u(v-1)$.\nIn the expansion of $b_v(h(x))^v$, there exists a term:\n$$\nb_v \\cdot \\binom{v}{1} (c_u x^u)^{v-1} \\cdot c_r x^r = b_v v c_u^{v-1} c_r \\cdot x^{uv-u+r},\n$$\nwhile other terms either have exponents divisible by $v$ or exponents less than $uv - u + r$, leaving no cancellation possible. Thus $P(x^v) = Q(h(x))$ would contain a non-zero $x^{uv-u+r}$ term whose exponent isn't divisible by $v$ - a contradiction.\n\nLet $t$ be the smallest non-negative remainder of $u$ modulo $v$. Then $h(x)$ can be expressed as $x^t g(x^v) + c_0$, where $g(\\cdot)$ has rational coefficients. From $P(x^v) \\equiv Q(h(x))$, when $y = x^v$ is a large positive integer, $h(x) = h(y^{1/v}) = y^{t/v} g(y) + c_0$ must also be integer-valued. This implies $y^{t/v}$ must be rational for all large positive integers $y$, forcing $t = 0$. Therefore $\\deg(Q) = v$ divides $\\deg(P) = u$. $\\square$",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 28638,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all functions $f : \\mathbb{Z} \\to \\mathbb{Z}$ satisfying:\n(1) For any positive integer $M$, there exists an integer $k$ with $|f(k)| \\ge M$;\n(2) For any integers $m, n$,\n$$2f(m)f(n) - f(m - n) - 1$$ is a perfect square.",
    "options": [],
    "answer": "All such functions are f(n) = ((3+2√2)^{2tn} + (3−2√2)^{2tn}) / 2 for some fixed positive integer t.",
    "solution": "Let $\\alpha = 3 + 2\\sqrt{2}$. All functions satisfying the given conditions are of the form $f(n) = \\frac{1}{2}(\\alpha^{2n} + \\alpha^{-2n})$ for some fixed positive integer $t$.\n\nFirst, we verify that such $f(n)$ satisfies $(\\star)$. The left-hand side of $(\\star)$ becomes:\n$$\n2 \\cdot \\frac{\\alpha^{2tm} + \\alpha^{-2tm}}{2} \\cdot \\frac{\\alpha^{2tn} + \\alpha^{-2tn}}{2} - \\frac{\\alpha^{2t(m-n)} + \\alpha^{-2t(m-n)}}{2} - 1 \\\\\n= \\frac{\\alpha^{2t(m+n)} + \\alpha^{-2t(m+n)}}{2} - 1 = \\left( \\frac{\\alpha^{t(m+n)} - \\alpha^{-t(m+n)}}{\\sqrt{2}} \\right)^2 .\n$$\nNote that $\\frac{\\alpha^{t(m+n)} - \\alpha^{-t(m+n)}}{\\sqrt{2}} = \\frac{(3+2\\sqrt{2})^{t(m+n)} - (3-2\\sqrt{2})^{t(m+n)}}{\\sqrt{2}}$ is an integer, so the above expression is indeed a perfect square.\n\nWe now show that these are the only solutions to the functional equation $(\\star)$. Let $A_{m,n}$ be the non-negative integer such that:\n$$\n2f(m)f(n) - f(m - n) - 1 = A_{m,n}^2. \\qquad (\\star\\star)\n$$\n\n**Step 1:** All $f(n)$ have the same sign. This follows by setting $n = 0$ in $(\\star\\star)$, which gives $(2f(0) - 1)f(m) > 0$. Thus, all $f(n)$ have the same sign as $2f(0) - 1$.\n\n**Step 2:** $f(n)$ is an even function. For a fixed $t \\in \\mathbb{Z}^+$, let $M_t = \\max\\{|f(t)|, |f(-t)|\\}$. Since $f$ is unbounded, there exists $N$ such that $|f(N + t)| > (M_t + 1)^2$. Substituting $m = N + t, n = N$ and $m = N, n = N + t$ into $(\\star\\star)$ yields:\n$$\n2f(N + t)f(N) - f(t) - 1 = A_{N,N+t}^2, \\quad 2f(N + t)f(N) - f(-t) - 1 = A_{N+t,N}^2.\n$$\nThis implies $A_{N,N+t}^2 > 2(M_t+1)^2 - M_t - 1 > M_t^2$, so $A_{N,N+t} > M_t$; similarly, $A_{N+t,N} > M_t$. Subtracting these equations gives:\n$$\nf(-t) - f(t) = (A_{N,N+t} - A_{N+t,N})(A_{N,N+t} + A_{N+t,N}).\n$$\nThe left-hand side satisfies $|f(-t) - f(t)| < 2M_t$, while $A_{N,N+t} + A_{N+t,N} > 2M_t$. Analyzing the magnitudes shows that $A_{N,N+t} = A_{N+t,N}$, so $f(t) = f(-t)$, proving $f$ is even.\n\n**Step 3:** Determine the general form of $f(n)$. Setting $m = n$ in $(\\star\\star)$ gives:\n$$\n2f(n)^2 - f(0) - 1 = A_{n,n}^2 \\quad \\text{ i.e., } \\quad A_{n,n}^2 - 2f(n)^2 = -f(0) - 1. \\qquad (5)\n$$\nThus, $(A_{n,n}, f(n))$ is a solution to the generalized Pell equation:\n$$\nX^2 - 2Y^2 = -f(0) - 1. \\qquad (6)\n$$\nWe use the theory of solutions to generalized Pell equations. Let $\\mathbb{Z}[\\sqrt{2}]$ be the ring of numbers $Z = X + \\sqrt{2}Y$ with $X, Y \\in \\mathbb{Z}$. The conjugate of $Z$ is $Z' = X - \\sqrt{2}Y$, and $(Z_1Z_2)' = Z'_1Z'_2$. For $\\alpha = 3+2\\sqrt{2}$, we have $\\alpha' = \\alpha^{-1}$. Solving (6) is equivalent to solving $ZZ' = -f(0) - 1$ in $\\mathbb{Z}[\\sqrt{2}]$. If $Z$ is a solution, then all $Z\\alpha^{2s}$ ($s \\in \\mathbb{Z}$) are also solutions.\n\n**Lemma:** There exist finitely many fundamental solutions $Z_i = X_i + \\sqrt{2}Y_i \\in \\mathbb{Z}[\\sqrt{2}]$ ($i = 1, \\dots, t$) to (6) such that for any solution $(X, Y)$, there exists $i \\in \\{1, \\dots, t\\}$ and $s \\in \\mathbb{Z}_{\\ge 0}$ with $X + \\sqrt{2}Y = Z_i\\alpha^{2s}$.\n\n**Proof of Lemma:** Consider the lattice $L := \\{(a + \\sqrt{2}b, a - \\sqrt{2}b) \\mid a, b \\in \\mathbb{Z}\\}$. Solutions $(X, Y)$ to (6) correspond to points $(x, y) = (X + \\sqrt{2}Y, X - \\sqrt{2}Y)$ on the hyperbola $xy = -f(0) - 1$. Multiplying $X + \\sqrt{2}Y$ by $\\alpha^2$ or $\\alpha^{-2}$ generates new solutions, equivalent to moving $(x, y)$ to $(\\alpha^2x, \\alpha^{-2}y)$ or $(\\alpha^{-2}x, \\alpha^2y)$ on the hyperbola. By these transformations, any solution can be mapped to a region where $|x| \\le \\alpha\\sqrt{|f(0) + 1|}$ and $|y| \\le \\alpha\\sqrt{|f(0) + 1|}$, corresponding to finitely many lattice points $(X_i + \\sqrt{2}Y_i, X_i - \\sqrt{2}Y_i)$ ($i = 1, \\dots, t$). Thus, any solution can be written as $(X_i + \\sqrt{2}Y_i)\\alpha^{2s}$. $\\square$\n\nFix the fundamental solutions $Z_i = X_i + \\sqrt{2}Y_i$ ($i = 1, \\dots, t$) from the lemma, with $Z_iZ'_i = -f(0) - 1$. We may assume no two $Z_i$ differ by a power of $\\alpha^2$. For each $n \\in \\mathbb{Z}$, there exist unique $i_n \\in \\{1, \\dots, t\\}$ and $s_n \\in \\mathbb{Z}$ such that:\n$$\nf(n) = \\frac{Z_{i_n}\\alpha^{2s_n} - Z'_{i_n}\\alpha^{-2s_n}}{2\\sqrt{2}}. \\qquad (7)\n$$\nLet $N = \\max_i\\{|Z_i|, |Z'_i|, |Z_i|^{-1}, |Z'_i|^{-1}\\}$. From (7), we have the estimate:\n$$\nN^{-1}\\alpha^{2|s_n|} - N \\le |f(n)| \\le N\\alpha^{2|s_n|} + N. \\qquad (8)\n$$\n\n**Step 4:** For positive integers $m > n$, we have the growth estimate:\n$$\n\\left||s_{m+n} - |s_m|\\right| < |s_n| + 2 \\log_{\\alpha} N + 2. \\qquad (9)\n$$\nThis follows from substituting $-n$ for $n$ in $(\\star\\star)$, yielding:\n$$\n|f(m+n)| < 2f(m)f(-n) = 2f(m)f(n). \\qquad (10)\n$$\nCombining with (8) gives:\n$$\nN^{-1}\\alpha^{2|s_{m+n}|} - N \\le 2(N\\alpha^{2|s_n|} + N)(N\\alpha^{2|s_m|} + N).\n$$\nSimplifying, we obtain:\n$$\nN^{-1} \\cdot \\alpha^{2|s_{m+n}|} \\le 2N^2 \\cdot ((\\alpha^{2|s_n|} + 1)(\\alpha^{2|s_m|} + 1) + 1) \\le N^2 \\cdot \\alpha^{2|s_n|+2|s_m|+3}\n$$\nwhich implies $|s_{m+n}| < |s_n| + |s_m| + 2 \\log_{\\alpha} N + 2$.\nSimilarly, substituting $m+n$ for $m$ in $(\\star\\star)$ gives:\n$$\n2f(m+n)f(n) > |f(m)|, \\qquad (11)\n$$\nleading to $|s_{m+n}| + |s_n| + 2 \\log_{\\alpha} N + 2 > |s_m|$. Combining these inequalities proves (9).\n\n**Step 5:** Prove that $f(0) = 1$.\nFirst, let's explain the basic idea. Suppose for indices $m, n$ we have $Z_{i_m} = Z_{i_n}$ and $s_n$ has the same sign as $s_m$. Substituting the expressions for $f(m)$ and $f(n)$ from (7) into $(\\star\\star)$, we obtain:\n$$\n\\begin{align}\nB_{m,n}^2 &= 2f(m)f(n) - f(m-n) - 1 \\\\\n&= \\frac{(Z_{i_m} \\alpha^{2s_m} - Z'_{i_m} \\alpha^{-2s_m})(Z_{i_n} \\alpha^{2s_n} - Z'_{i_n} \\alpha^{-2s_n})}{4} - \\frac{Z_{i_{m-n}} \\alpha^{2s_{m-n}} - Z'_{i_{m-n}} \\alpha^{-2s_{m-n}}}{2\\sqrt{2}} - 1 \\\\\n&= \\left( \\frac{Z_{i_m} \\alpha^{s_m+s_n} + Z'_{i_m} \\alpha^{-s_m-s_n}}{2} \\right)^2 - \\frac{Z_{i_m} Z'_{i_m} (\\alpha^{s_m-s_n} + \\alpha^{s_n-s_m})^2}{4} \\\\\n&\\quad - \\frac{(Z_{i_{m-n}} \\alpha^{2s_{m-n}} - Z'_{i_{m-n}} \\alpha^{-2s_{m-n}})}{2\\sqrt{2}} - 1 \\\\\n&= \\left( \\frac{Z_{i_m} \\alpha^{s_m+s_n} + Z'_{i_m} \\alpha^{-s_m-s_n}}{2} \\right)^2 + (f(0)+1) \\frac{(\\alpha^{s_m-s_n} + \\alpha^{s_n-s_m})^2}{4} \\\\\n&\\quad - \\frac{(Z_{i_{m-n}} \\alpha^{2s_{m-n}} - Z'_{i_{m-n}} \\alpha^{-2s_{m-n}})}{2\\sqrt{2}} - 1. \\tag{12}\n\\end{align}\n$$\nNote that $\\frac{1}{2}(Z_{i_m} \\alpha^{s_m+s_n} + Z'_{i_m} \\alpha^{-s_m-s_n})$ is an integer greater than $(2N)^{-1} \\alpha^{|s_m+s_n|-1} - N$.\nTherefore, if $s_m$ and $s_n$ are very close and both have large absolute values, all terms except this squared term sum to zero.\nNow we use this property specifically to show $f(0) = 1$. Assume by contradiction that $f(0) \\neq 1$. First, we show that for any $i \\in \\{1, \\dots, t\\}$, the sets $\\{\\frac{1}{2\\sqrt{2}}Z_i\\alpha^{2s} \\mid s \\in \\mathbb{Z}_{\\ge 0}\\} \\cup \\{\\frac{1}{2\\sqrt{2}}Z'_i\\alpha^{2s} \\mid s \\in \\mathbb{Z}_{\\ge 0}\\}$ and $\\{\\frac{1}{4}(f(0)+1)\\alpha^{2s}\\}$ have no common elements. This is because the product of these numbers with their conjugates differs:\n$$\n\\frac{1}{2\\sqrt{2}} Z_i \\alpha^{2s} \\cdot \\frac{1}{-2\\sqrt{2}} Z'_i \\alpha^{-2s} = \\frac{f(0) + 1}{8} \\neq \\frac{1}{16} (f(0) + 1)^2,\n$$\n(since $f(0) \\neq 1$). Thus there exists an integer $M > 2 \\log_{\\alpha}(N) + 2$ such that all numbers of the form $\\frac{f(0)+1}{4}\\alpha^{2s}$ and $\\frac{Z_i}{2\\sqrt{2}}\\alpha^{2s}$ or $\\frac{Z'_i}{2\\sqrt{2}}\\alpha^{2s}$ (for $|s| \\ge M$) are pairwise separated by more than $|f(0) - 1| + N^2$.\nSince $f$ is unbounded, there exist $n_1 < \\dots < n_{2t+1}$ satisfying:\n$$\n|s_{n_1}| > 2M, \\quad |s_{n_{i+1}}| > 2|s_{n_i}| \\quad (i = 1, \\dots, 2t).\n$$\nChoose $L$ such that $|s_L| > 5 \\cdot 2^{2t+1}M$. Consider:\n$$\ni_{L+n_1}, i_{L+n_2}, \\dots, i_{L+n_{2t+1}} \\in \\{1, \\dots, t\\}\n$$\nand the signs of $s_{L+n_j}$. By the pigeonhole principle, there must exist $x > y \\ge 1$ such that $i_{L+n_x} = i_{L+n_y} = i$ and $s_{L+n_x}$ has the same sign as $s_{L+n_y}$.\nFrom inequality (9), we estimate:\n$$\n|s_{(L+n_x)-(L+n_y)}| = |s_{n_x-n_y}| > |s_{n_x} - s_{n_y}| - (2 \\log_{\\alpha}(N) + 2) > M.\n$$\n$$\n|s_{L+n_x}| > |s_L| - |s_{n_x}| - 2 \\log_{\\alpha}(N) + 2 > 4 \\cdot 2^{t+1}M; \\quad |s_{L+n_y}| > 4 \\cdot 2^{t+1}M.\n$$\nReturning to our earlier situation, set $m = L + n_x$, $n = L + n_y$, and observe that:\n$$\n\\left| \\frac{Z_i \\alpha^{s_m + s_n} + Z'_i \\alpha^{-s_m - s_n}}{2} \\right| > N^{-1} \\alpha^{4 \\cdot 2^{t+1} M} - N\n$$\nis much larger than the square of the remaining terms. Therefore, we must have:\n$$\n(f(0) + 1) \\frac{(\\alpha^{s_m - s_n} + \\alpha^{s_n - s_m})^2}{4} = \\frac{(Z_{i_{m-n}} \\alpha^{2s_m - n} - Z'_{i_{m-n}} \\alpha^{-2s_m - n})}{2\\sqrt{2}} + 1.\n$$\nBut by the definition of $M$, this equality cannot hold. Contradiction!\n\n**Step 6:** Complete the proof. With $f(0) = 1$, (5) simplifies to:\n$$\n2f(n)^2 - 2 = A_{n,n}^2.\n$$\nThus, $A_{n,n}$ is even, and we rewrite the equation as:\n$$\nf(n)^2 - 2 \\left( \\frac{A_{n,n}}{2} \\right)^2 = 1.\n$$\nThe solutions to this standard Pell equation are:\n$$\nf(n) = \\frac{\\alpha^{t_n} + \\alpha^{-t_n}}{2}, \\quad t_n \\in \\mathbb{Z}_{\\ge 0}.\n$$\nSubstituting $n = 0$ into (★) gives:\n$$\nA_{m,0}^2 = 2f(m)f(0) - f(m) - 1 = f(m) - 1 = \\left( \\frac{(\\sqrt{2} + 1)^{t_m} - (\\sqrt{2} - 1)^{t_m}}{\\sqrt{2}} \\right)^2,\n$$\nimplying $t_n$ must be even. Let $t_n = 2s_n$.\nSubstituting into (10) and (11), we obtain for $m > n$:\n$$\n\\frac{1}{2}(\\alpha^{2s_m} + \\alpha^{-2s_m})(\\alpha^{2s_n} + \\alpha^{-2s_n}) > \\frac{1}{2}(\\alpha^{2s_{m+n}} + \\alpha^{-2s_{m+n}}),\n$$\n$$\n\\frac{1}{2}(\\alpha^{2s_{m+n}} + \\alpha^{-2s_{m+n}})(\\alpha^{2s_n} + \\alpha^{-2s_n}) > \\frac{1}{2}(\\alpha^{2s_m} + \\alpha^{-2s_m}),\n$$\nwhich implies:\n$$\ns_m + s_n \\ge s_{m+n} \\quad \\text{and} \\quad s_{m+n} + s_n \\ge s_m. \\qquad (13)\n$$\nThe following shows that for any $\\ell \\in \\mathbb{N}$, we have $s_\\ell = \\ell s_1$. Fix $\\ell$, and let $D := \\max\\{2, s_1, \\dots, s_\\ell\\}$.\n\nConsider any positive integer $N > \\ell$ satisfying $s_N > 5D$. From the given conditions, for $i = -1, 0, \\dots, \\ell - 1$, we have $s_{N+i} > 4D$. For $i = 0, \\dots, \\ell - 1$, taking $m = N + i$ and $n = N - 1$ in (12) yields:\n$$\n\\begin{aligned}\nB_{N+i,N-1}^2 &= \\frac{1}{2}(\\alpha^{2s_{N+i}} + \\alpha^{-2s_{N+i}})(\\alpha^{2s_{N-1}} + \\alpha^{-2s_{N-1}}) - \\frac{1}{2}(\\alpha^{2s_{i+1}} - \\alpha^{-2s_{i+1}}) - 1 \\\\\n&= \\left( \\frac{\\alpha^{s_{N+i}+s_{N-1}} - \\alpha^{-s_{N+i}-s_{N-1}}}{\\sqrt{2}} \\right)^2 \\\\\n&\\quad + \\frac{(\\alpha^{2s_{N+\\ell}-2s_{N-1}} + \\alpha^{2s_{N-1}-2s_{N+\\ell}}) - (\\alpha^{2s_{i+1}} + \\alpha^{-2s_{i+1}})}{2}\n\\end{aligned}\n$$\nSince $\\frac{\\alpha^{s_{N+i}+s_{N-1}}-\\alpha^{-s_{N+i}+s_{N-1}}}{\\sqrt{2}}$ is an integer greater than $\\alpha^{6D-1}$, while the remaining terms are less than $\\alpha^{2D+1}$, it must hold that:\n$$\n\\alpha^{2s_{N+\\ell}-2s_{N-1}} + \\alpha^{2s_{N-1}-2s_{N+\\ell}} = \\alpha^{2s_{i+1}} + \\alpha^{-2s_{i+1}}.\n$$\nThis implies $|s_{N+i} - s_{N-1}| = s_{i+1}$.\nSince $f$ is unbounded, choose $N > \\ell$ such that $s_N > 6D$ and $s_N > s_{N-1}$. From the above discussion, we have:\n$$\n|s_{N+1} - s_N| = s_1, \\quad s_N - s_{N-1} = s_1, \\quad |s_{N+1} - s_{N-1}| = s_2,\n$$\nso either $s_2 = 0$ or $s_2 = 2s_1$. If $s_2 = 0$, we can use $s_N > 6D$ and (13) to inductively prove that $s_{N+u+2} = s_{N+u}$ for all $u \\in \\mathbb{N}$. This contradicts the unboundedness of $f$.\n\nTherefore, $s_2 = 2s_1$, which implies $s_{N+1} = s_{N-1} + 2s_1$. Next, we examine the conditions that $s_{N+2}$ must satisfy:\n$$\n|s_{N+2} - s_N| = s_2 = 2s_1, \\quad |s_{N+2} - s_{N+1}| = s_1.\n$$\nThus $s_{N+2} = s_N + 2s_1 = s_{N-1} + 3s_1$, which gives $s_3 = |s_{N+2} - s_{N-1}| = 3s_1$. Continuing this recursion, we obtain $s_\\ell = \\ell s_1$. This proves that all solutions $f$ satisfying the functional equation are of the form:\n$$\nf(n) = \\frac{1}{2}(\\alpha^{2ns_1} + \\alpha^{-2ns_1}).\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 28639,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let point $P$ lie on the nine-point circle of triangle $ABC$. A line through $P$ perpendicular to $AP$ intersects $BC$ at $Q$. A line through $A$ perpendicular to $AQ$ intersects $PQ$ at $X$. Let $H$ be the orthocenter of triangle $ABC$, and let $D$ and $M$ be the midpoints of segments $BC$ and $AQ$, respectively. Prove that $XH \\perp DM$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "![](attached_image_1.png)\n\n**Proof:** Let $N$ be the midpoint of $AH$. By the properties of the nine-point circle, $DN$ is its diameter, so $DP \\perp PN$. Since $AH \\perp BC$ and $AP \\perp PQ$, we have $\\triangle DPQ \\sim \\triangle NPA$. Therefore, $\\frac{DQ}{NA} = \\frac{PQ}{PA}$.\n\nSince $XA \\perp AQ$, we have $\\triangle AQP \\sim \\triangle XAP$, which gives $\\frac{PQ}{PA} = \\frac{AQ}{XA}$. Combining these two results yields $\\frac{DQ}{NA} = \\frac{AQ}{XA}$.\n\nMoreover, we observe that:\n$$\n\\angle XAN = 90^\\circ - \\angle NAQ = \\angle AQD\n$$\nThis implies that $\\triangle XAN \\sim \\triangle AQD$, and consequently:\n$$\nXA \\cdot DQ = AN \\cdot AQ = AH \\cdot QM\n$$\nFinally, noting that:\n$$\n\\angle XAH = 90^\\circ - \\angle NAQ = \\angle MQD\n$$\nwe conclude that $\\triangle XHA \\sim \\triangle MDQ$, which proves that $DM \\perp XH$. $\\square$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 28640,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $p_1, p_2, \\dots, p_{2025}$ be real numbers, and let $\\{a_n^{(1)}\\}_{n \\ge 0}$, $\\{a_n^{(2)}\\}_{n \\ge 0}$, $\\dots$, $\\{a_n^{(2025)}\\}_{n \\ge 0}$ be $2025$ real sequences satisfying:\n(1) $a_0^{(i)}$ ($1 \\le i \\le 2025$) are all zero;\n(2) $a_1^{(i)}$ ($1 \\le i \\le 2025$) are **not** all zero;\n(3) For $i = 1, 2, \\dots, 2025$ and any positive integer $n$,\n$$\na_{n-1}^{(i)} + a_n^{(i)} + a_{n+1}^{(i)} = p_i \\cdot a_n^{(i+1)},\n$$\nwhere $a_n^{(2026)} = a_n^{(1)}$.\nProve that there exists a positive real number $r$ and infinitely many positive integers $n$ such that\n$$\n\\max \\{|a_n^{(1)}|, |a_n^{(2)}|, \\dots, |a_n^{(2025)}|\\} > r.\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "**Proof:** We first prove a lemma.\n**Lemma:** Let $\\beta$ be a complex number. If a sequence $\\{a_n\\}$ satisfies $a_0 = 0$, $a_1 \\ne 0$, and for all $n \\ge 1$,\n$$\na_{n-1} + \\beta a_n + a_{n+1} = 0,\n$$\nthen $\\{a_n\\}$ does not converge to $0$.\n**Proof of Lemma:** Let $\\alpha_1, \\alpha_2$ be roots of $x^2 + \\beta x + 1 = 0$, so $\\alpha_1\\alpha_2 = 1$.\nIf $\\alpha_1 = \\alpha_2$, then $\\alpha_1 = \\alpha_2 = \\pm 1$ and $a_n = pn + q$ or $a_n = (-1)^n(pn + q)$. Clearly $a_n$ doesn't converge to $0$.\nIf $\\alpha_1 \\ne \\alpha_2$, then $a_n = p\\alpha_1^n + q\\alpha_2^n$ with $p, q \\in \\mathbb{C}$ and $pq \\ne 0$ (since $a_0 = 0$, $a_1 \\ne 0$).\nCase 1: $|\\alpha_1| \\ne |\\alpha_2|$. Then $a_n$ cannot converge to $0$ since $|\\alpha_1||\\alpha_2| = 1$.\nCase 2: $|\\alpha_1| = |\\alpha_2| = 1$. Let $\\alpha_1 = e^{2\\pi i\\theta}$, $\\alpha_2 = e^{-2\\pi i\\theta}$.\nIf $\\theta \\in \\mathbb{Q}$, then $\\{a_n\\}$ is periodic and non-zero.\nIf $\\theta \\notin \\mathbb{Q}$, then $\\{n\\theta\\}$ is dense modulo $1$, making $\\{a_n\\}$ have values dense on some circle.\nIn all cases, $a_n$ doesn't converge to $0$. $\\square$\nNow the main proof. Denote the given equations as $(*1), \\cdots, (*_{2025})$.\n**Case 1:** $\\prod_{i=1}^{2025} p_i \\ne 0$.\nDefine transformed sequences:\n$$\nb_n^{(i)} = \\sqrt[2025]{p_{i+1} p_{i+2} \\cdots p_{i+2024}} \\cdot a_n^{(i)}\n$$\nwhere indices are cyclic modulo $2025$. These satisfy:\n$$\nb_{n-1}^{(i)} + b_n^{(i)} + b_{n+1}^{(i)} = \\sqrt[2025]{p_1 \\cdots p_{2025}} \\cdot b_n^{(i+1)}.\n$$\n---\n\nThus we may assume $p_1 = \\cdots = p_{2025} = p$. Let $\\omega$ be a $2025$th root of unity and define:\n$$\nX_n := \\sum_{k=0}^{2024} \\omega^k a_n^{(k+1)}.\n$$\nThen:\n$$\n\\forall n \\ge 1, \\quad X_{n-1} + (1 - \\omega^{-1}p)X_n + X_{n+1} = 0.\n$$\nSince not all $a_1^{(i)} = 0$, some $X_1 \\ne 0$. By the lemma, $\\{X_n\\}$ doesn't converge to $0$.\n**Case 2:** If there exists some $p_i = 0$. Without loss of generality, assume $p_1 = 0$. Then from $(*)_1$ we know that for any $n \\ge 1$ we have $a_{n-1}^{(1)} + a_n^{(1)} + a_{n+1}^{(1)} = 0$.\nIf $a_1^{(1)} \\ne 0$, applying the lemma to the sequence $\\{a_n^{(1)}\\}$ shows that $\\{a_n^{(1)}\\}$ does not converge to $0$.\nIf $a_1^{(1)} = 0$, then $a_n^{(1)} \\equiv 0$. Consequently, from $(*)_{2025}$ we know that for any $n \\ge 1$ we have $a_{n-1}^{(2025)} + a_n^{(2025)} + a_{n+1}^{(2025)} = 0$.\nIn this case, if $a_1^{(2025)} \\ne 0$, we can apply the lemma to the sequence $\\{a_n^{(2025)}\\}_{n \\ge 0}$. If $a_1^{(2025)} = 0$ then $a_n^{(2025)} \\equiv 0$, and similarly we obtain $a_{n-1}^{(2024)} + a_n^{(2024)} + a_{n+1}^{(2024)} = 0$. Continuing this process, by induction we can prove that either all sequences in the problem are identically zero, or there exists at least one sequence that does not converge to $0$. The former case contradicts the problem's assumptions, thus completing the proof. $\\square$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 28641,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $k$, $l$, $n$ be positive integers, and let $a_1, a_2, \\dots, a_k \\in \\{1, 2, \\dots, n\\}$ satisfy the following three conditions:\n(1) $n \\ge 3$, $l \\le n-2$, and $l-k \\le \\frac{n-3}{2}$;\n(2) For each $t \\in \\{1, 2, \\dots, l\\}$, there exists a non-empty subset $I \\subseteq \\{1, 2, \\dots, k\\}$ such that\n$$\n\\sum_{i \\in I} a_i \\equiv t \\pmod{n};\n$$\n(3) For each $t \\in \\{l+1, l+2, \\dots, n\\}$, there does not exist a non-empty subset $I \\subseteq \\{1, 2, \\dots, k\\}$ such that\n$$\n\\sum_{i \\in I} a_i \\equiv t \\pmod{n}.\n$$\nProve that $a_1 + a_2 + \\dots + a_k = l$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "**Proof:** For a finite multiset $S$ of integers, let $\\sigma(S)$ denote the sum of all elements in $S$ (counting multiplicities), and let $\\Sigma(S) = \\{\\sigma(T) \\mid \\sigma(T) \\neq T \\subseteq S\\}$, where $\\Sigma(S)$ is treated as a set (ignoring multiplicities). Let $\\Sigma_n(S)$ denote the set of congruence classes modulo $n$ of the elements in $\\Sigma(S)$. For $x \\in \\mathbb{Z}$, let $\\bar{x}$ denote the congruence class of $x$ modulo $n$. Under this notation, conditions (2) and (3) can be written as\n$$\n\\Sigma_n(\\{a_1, a_2, \\dots, a_k\\}) = \\{\\bar{1}, \\bar{2}, \\dots, \\bar{\\ell}\\}.\n$$\n**Step 1:** The main idea is to add an element $x$ to a nonempty multiset $S$, obtaining $T = S \\cup \\{x\\}$. Assuming that neither $\\Sigma_n(S)$ nor $\\Sigma_n(T)$ contains $\\bar{0}$, we compare $\\Sigma_n(S)$ and $\\Sigma_n(T)$. In particular, if $\\Sigma_n(T)$ contains exactly one more element than $\\Sigma_n(S)$, we can deduce a very refined structure for $\\Sigma_n(S)$.\nObserve that\n$$\n\\Sigma(T) = \\Sigma(S) \\cup (\\Sigma(S) + x) \\cup \\{x\\},\n$$\nso $\\Sigma_n(S) \\subset \\Sigma_n(T)$. Moreover, $\\overline{\\sigma(S) + x}$ belongs to $\\Sigma_n(T)$ but not to $\\Sigma_n(S)$, since otherwise $\\Sigma_n(T)$ would contain $\\bar{0}$. Hence, $\\Sigma_n(S) \\subsetneq \\Sigma_n(T)$.\nIf $\\Sigma_n(T)$ contains exactly one more element than $\\Sigma_n(S)$, then this new element must be $\\overline{\\sigma(S) + x}$. However, $\\bar{x} \\in \\Sigma_n(T)$, and $\\bar{x} \\ne \\overline{\\sigma(S) + x}$, so $\\bar{x} \\in \\Sigma_n(S)$.\nLet $d = \\frac{n}{\\text{gcd}(x,n)}$, where $d$ is the smallest positive integer such that $d\\bar{x} = \\bar{0}$. Suppose $\\bar{x}, 2\\bar{x}, \\dots, p\\bar{x} \\in \\Sigma_n(S)$, but $(p+1)\\bar{x} \\notin \\Sigma_n(S)$. A set of the form\n$$\n\\{\\bar{a}, \\overline{a+x}, \\overline{a+2x}, \\dots, \\overline{a+(d-1)x}\\}\n$$\nis called a **coset** of $\\bar{x}$. The congruence classes modulo $n$ can be partitioned into $\\frac{n}{d}$ cosets of $\\bar{x}$.\nIf $\\Sigma_n(S)$ contains some elements of a coset $C$ of $\\bar{x}$ but not the entire coset, then $(\\Sigma_n(S) \\cap C) + \\bar{x}$ will produce new elements not in $\\Sigma_n(S) \\cap C$. Since $\\Sigma_n(T)$ contains only one more element than $\\Sigma_n(S)$, it follows that $\\Sigma_n(S)$ must consist of several complete cosets of $\\bar{x}$ and one incomplete coset. This incomplete coset can only be $\\{\\bar{x}, 2\\bar{x}, \\dots, p\\bar{x}\\}$, and the new element in $\\Sigma_n(T)$ is $(p+1)\\bar{x}$, which must equal $\\overline{\\sigma(S)+x}$. Therefore, $\\overline{\\sigma(S)} = p\\bar{x}$.\n**Step 2:** Returning to the original problem, first add $n-1-\\ell$ copies of 1 to $\\{a_1, a_2, \\dots, a_k\\}$, obtaining a multiset $S$. It is easy to see that $m := |S| = k+(n-1)-\\ell \\ge \\frac{n+1}{2}$ (using condition (1)), and\n$$\n\\Sigma_n(S) = \\{\\bar{1}, \\bar{2}, \\dots, \\overline{n-1}\\}.\n$$\nLet the elements of $S$ be ordered as $b_1 \\le b_2 \\le \\dots \\le b_m$. Then $b_1 = 1$ (using condition (1), $\\ell \\le n-2$, so $S$ contains at least one 1), and $b_m < n$.\nIf for every $2 \\le i \\le m$, we have $b_i \\le 1 + b_1 + \\dots + b_{i-1}$, then by induction, it follows that\n$$\n\\Sigma(S) = \\{1, 2, \\dots, b_1 + b_2 + \\dots + b_m\\}.\n$$\nSince $\\Sigma_n(S) = \\{\\bar{1}, \\bar{2}, \\dots, \\overline{n-1}\\}$, it must be that $b_1 + b_2 + \\dots + b_m = n-1$. Removing the added $n-1-\\ell$ copies of 1, we obtain\n$$\na_1 + a_2 + \\dots + a_k = \\ell.\n$$\nNow suppose there exists $2 \\le i \\le m$ (take the smallest such $i$) such that $b_i > 1 + b_1 + \\dots + b_{i-1}$. Let $s = b_1 + \\dots + b_{i-1} \\ge i-1$. Then\n$$\n\\Sigma(\\{b_1, \\dots, b_{i-1}, b_i\\}) = \\{1, 2, \\dots, s, b_i, b_i+1, \\dots, b_i+s\\}.\n$$\nSince $b_i < n$, we have $b_i + s < n$ (otherwise $\\overline{0} \\in \\Sigma_n(S)$), so\n$$\n|\\Sigma_n(\\{b_1, \\dots, b_i\\})| = 2s + 1 \\ge 2i - 1.\n$$\nNow, iteratively add the remaining elements to $\\{b_1, \\dots, b_i\\}$ such that at each step, $\\Sigma_n$ gains at least two new elements, until no more can be added. Suppose we obtain $T \\subset S$ with $|T| = t \\ge i$, satisfying\n$$\n|\\Sigma_n(T)| \\ge 2i - 1 + 2(t - i) = 2t - 1.\n$$\nIt follows that $t < m$, since otherwise $|\\Sigma_n(T)| \\ge 2m-1 \\ge n$, which is a contradiction. Now, the remaining elements must each add only one new element to $\\Sigma_n$ when included in $T$.\n**Claim:** The remaining elements are all identical.\nSuppose $x$ and $y$ are remaining elements, and both $\\Sigma_n(T \\cup \\{x\\})$ and $\\Sigma_n(T \\cup \\{y\\})$ contain exactly one more element than $\\Sigma_n(T)$. By the conclusion of Step 1, $\\Sigma_n(T)$ consists of several complete cosets of $\\bar{x}$ and one incomplete coset $\\{\\bar{x}, 2\\bar{x}, \\dots, p\\bar{x}\\}$, with $\\overline{\\sigma(T)} = p\\bar{x}$.\nSince $\\Sigma_n(T \\cup \\{y\\})$ contains only one more element, $\\overline{\\sigma(T)+y} = \\bar{y} + p\\bar{x}$, it follows that $\\bar{y}, \\bar{y}+\\bar{x}, \\dots, \\bar{y}+(p-1)\\bar{x} \\in \\Sigma_n(T)$, but $\\bar{y}+p\\bar{x} \\notin \\Sigma_n(T)$. Therefore, $\\bar{y}$ cannot belong to a complete coset of $\\bar{x}$ in $\\Sigma_n(T)$, so it must be in $\\{\\bar{x}, 2\\bar{x}, \\dots, p\\bar{x}\\}$. Consequently,\n$$\n\\{\\bar{y}, \\bar{y} + \\bar{x}, \\dots, \\bar{y} + (p-1)\\bar{x}\\} \\subset \\{\\bar{x}, 2\\bar{x}, \\dots, p\\bar{x}\\},\n$$\nwhich implies $\\bar{y} = \\bar{x}$, i.e., $x = y$. The claim is proved.\nThus, the remaining elements are all equal to some $x$. Each time we add $x$, $\\Sigma_n$ gains exactly one new element, successively adding $(p+1)\\bar{x}, (p+2)\\bar{x}, \\dots, (d-1)\\bar{x}$. In the final step, $S = T_1 \\cup \\{x\\}$, and the new element added to $\\Sigma_n$ is $(d-1)\\bar{x} = \\overline{\\sigma(T_1)+x} = \\overline{\\sigma(S)}$, so $\\bar{x} = -\\overline{\\sigma(S)}$.\nEarlier, when we added 1 to $\\{a_1, a_2, \\dots, a_k\\}$ to obtain $S$, each addition also introduced exactly one new element to $\\Sigma_n$. Considering the last addition, $S = T_2 \\cup \\{1\\}$, we similarly deduce $\\overline{1} = -\\overline{\\sigma(S)}$, so $x = 1$. This means that in the previous process, the remaining elements were all 1. However, this is impossible because we had already taken $\\{b_1, \\dots, b_i\\}$ with $b_i > 1$, which includes all the 1's. This contradiction completes the proof. $\\square$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 28642,
    "subject": "Mathematics (Multi-modal)",
    "question": "Given nonzero real numbers $\\lambda_1, \\lambda_2, \\dots, \\lambda_{2025}$ and a real number $d$. Let $X$ be a finite set of real numbers. Define the sets:\n$$\nA = \\{(x_1, \\dots, x_{2025}) \\in X^{2025} \\mid \\lambda_1 x_1 + \\dots + \\lambda_{2025} x_{2025} = d\\};\n$$\n$$\nB = \\{(x_1, \\dots, x_{2024}) \\in X^{2024} \\mid x_1 + \\dots + x_{1012} = x_{1013} + \\dots + x_{2024}\\};\n$$\n$$\nC = \\{(x_1, \\dots, x_{2026}) \\in X^{2026} \\mid x_1 + \\dots + x_{1013} = x_{1014} + \\dots + x_{2026}\\};\n$$\nwhere $X^n$ denotes the set of all ordered tuples $(x_1, \\dots, x_n)$ with $x_i \\in X$ ($i = 1, \\dots, n$).\nProve: $|A|^2 \\le |B| \\cdot |C|$, where $|Y|$ denotes the number of elements in the finite set $Y$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "**Proof 1:** Let $\\Lambda$ be the set consisting of $\\pm\\lambda_i$ for $1 \\le i \\le 2025$. For positive integer $n$, define functions $S, K: \\Lambda^{2n} \\to \\mathbb{Z}_{\\ge 0}$ as:\n$$\nS(c_1, \\dots, c_{2n}) = \\#\\{(x_1, \\dots, x_{2n}) \\in X^{2n} \\mid c_1 x_1 + \\dots + c_{2n} x_{2n} = 0\\},\n$$\n$$\nK(c_1, \\dots, c_{2n}) = \\#\\{1 \\le i \\le 2n \\mid c_i \\in \\{\\pm c_1\\}\\}.\n$$\nLet $T_{2n}$ be the cardinality of:\n$$\n\\{(x_1, \\dots, x_{2n}) \\in X^{2n} \\mid x_1 + \\dots + x_n = x_{n+1} + \\dots + x_{2n}\\},\n$$\nthen $T_{2n} = S(c_1, \\dots, c_1, -c_1, \\dots, -c_1)$.\n\n**Lemma:** The maximum value of function $S$ is $T_{2n}$.\n\n**Proof of Lemma:** Assume the maximum value of $S$ is $M \\ge 1$, and let $(c_1, \\dots, c_{2n})$ be a point in $S^{-1}(\\{M\\})$ where $K$ attains its maximum.\nLet $K(c_1, \\dots, c_{2n}) = k$, and assume $c_i \\in \\{\\pm c_1\\}$ for $1 \\le i \\le k$. For real $y$, define:\n$$\nI_1(y) = \\#\\{(x_1, \\dots, x_n) \\in X^n \\mid c_1 x_1 + \\dots + c_n x_n = y\\},\n$$\n$$\nI_2(y) = \\#\\{(x_{n+1}, \\dots, x_{2n}) \\in X^n \\mid -c_{n+1}x_{n+1} - \\dots - c_{2n}x_{2n} = y\\}.\n$$\nThen:\n$$\nS(c_1, \\dots, c_{2n}) = \\sum_y I_1(y) I_2(y).\n$$\nBy Cauchy-Schwarz:\n$$\n\\begin{align*}\nM^2 &= S(c_1, \\dots, c_{2n})^2 \\le \\sum_y I_1(y)^2 \\sum_y I_2(y)^2 \\\\\n&= S(c_1, \\dots, c_n, -c_1, \\dots, -c_n) \\cdot S(c_{n+1}, \\dots, c_{2n}, -c_{n+1}, \\dots, -c_{2n}) \\\\\n&\\le M^2,\n\\end{align*}\n$$\nimplying $S(c_1, \\dots, c_n, -c_1, \\dots, -c_n) = M$. By maximality of $K$:\n$$\nk = K(c_1, \\dots, c_{2n}) \\ge \\min\\{2k, 2n\\},\n$$\nthus $k = 2n$. Therefore all $c_i \\in \\{\\pm c_1\\}$, and:\n$$\nM = T_{2n}.\n$$\nThis completes the lemma's proof.\n\nReturning to the main problem, let $n = 1013$. For real $y$, define:\n$$\nJ_1(y) = \\#\\{(x_1, \\dots, x_n) \\in X^n \\mid c_1 x_1 + \\dots + c_n x_n = y\\},\n$$\n$$\nJ_2(y) = \\#\\{(x_{n+1}, \\dots, x_{2n-1}) \\in X^{n-1} \\mid -c_{n+1}x_{n+1} - \\dots - c_{2n-1}x_{2n-1} = y\\}.\n$$\nSimilarly using Cauchy-Schwarz:\n$$\nA^2 = \\left(\\sum_y J_1(y)J_2(y)\\right)^2 \\le \\sum_y J_1(y)^2 \\sum_y J_2(y)^2 \\\\ \\le S(c_1, \\dots, c_n, -c_1, \\dots, -c_n) \\cdot S(c_{n+1}, \\dots, c_{2n-1}, -c_{n+1}, \\dots, -c_{2n-1}).\n$$\nCombining with the lemma yields:\n$$\nA^2 \\le T_{2026} \\cdot T_{2024} = B \\cdot C. \\quad \\square\n\n\n**Proof 2:** (Based on solutions by Deng Leyan and Zhang Hengye)\nFor non-zero real $p$, using Newton-Leibniz formula:\n$$\n\\lim_{T \\to +\\infty} \\frac{1}{T} \\int_0^T e^{ipt} dt = 0.\n$$\nThus:\n$$\n\\lim_{T \\to +\\infty} \\frac{1}{T} \\int_{0}^{T} e^{ipt} dt = \\begin{cases} 0, & p \\ne 0, \\\\ 1, & p = 0. \\end{cases}\n$$\nLet $f(t) = \\sum_{x \\in X} e^{ixt}$. Then:\n$$\n|A| = \\lim_{T \\to +\\infty} \\frac{1}{T} \\left| \\int_{0}^{T} e^{-ibt} \\prod_{j=1}^{2025} f(\\lambda_j t) dt \\right|.\n$$\nBy Hölder's inequality:\n$$\n\\begin{align*}\n|A| &= \\lim_{T \\to +\\infty} \\frac{1}{T} \\left| \\int_0^T e^{-ibt} \\prod_{j=1}^{2025} f(\\lambda_j t) dt \\right| \\\\\n&\\le \\lim_{T \\to +\\infty} \\frac{1}{T} \\int_0^T \\prod_{j=1}^{2025} |f(\\lambda_j t)| dt \\\\\n&= \\lim_{T \\to +\\infty} \\frac{1}{T} \\int_0^T \\prod_{j=1}^{2025} (|f(\\lambda_j t)|^{2025})^{\\frac{1}{2025}} dt \\\\\n&\\le \\lim_{T \\to +\\infty} \\prod_{j=1}^{2025} \\left( \\frac{1}{T} \\int_0^T |f(\\lambda_j t)|^{2025} dt \\right)^{\\frac{1}{2025}} \\\\\n&= \\prod_{j=1}^{2025} \\left( \\lim_{T \\to +\\infty} \\frac{1}{T} \\int_0^T |f(|\\lambda_j|t)|^{2025} dt \\right)^{\\frac{1}{2025}} \\\\\n&= \\prod_{j=1}^{2025} \\left( \\lim_{T \\to +\\infty} \\frac{1}{|\\lambda_j|T} \\int_0^{|\\lambda_j|T} |f(s)|^{2025} ds \\right)^{\\frac{1}{2025}} \\\\\n&= \\lim_{T \\to +\\infty} \\frac{1}{T} \\int_0^T |f(t)|^{2025} dt.\n\\end{align*}\n$$\nSimilarly:\n$$\n|B| = \\lim_{T \\to +\\infty} \\frac{1}{T} \\int_{0}^{T} |f(t)|^{2024} dt,\n$$\n$$\n|C| = \\lim_{T \\to +\\infty} \\frac{1}{T} \\int_{0}^{T} |f(t)|^{2026} dt.\n$$\nFinally, by Cauchy-Schwarz:\n$$\n\\begin{align*}\n|B| \\cdot |C| &= \\lim_{T \\to +\\infty} \\frac{1}{T^2} \\int_0^T |f(t)|^{2024} dt \\int_0^T |f(t)|^{2026} dt \\\\\n&\\ge \\lim_{T \\to +\\infty} \\frac{1}{T^2} \\left( \\int_0^T |f(t)|^{2025} dt \\right)^2 \\\\\n&= \\left( \\lim_{T \\to +\\infty} \\frac{1}{T} \\int_0^T |f(t)|^{2025} dt \\right)^2 \\\\\n&\\ge |A|^2,\n\\end{align*}\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 28643,
    "subject": "Mathematics (Multi-modal)",
    "question": "In convex quadrilateral $ABCD$, $AB \\perp AD$ and $AD = DC$. Let point $E$ lie inside segment $BC$, and point $F$ lie on the extension of $DE$ such that $\\angle ABF = \\angle DEC > 90^\\circ$. Let $O$ be the circumcenter of triangle $CDE$. Let $P$ be a point on the extension of $FO$ such that $FP = FB$. Let segment $BP$ intersect $AC$ at point $Q$.\nProve: $\\angle AQB = \\angle DPF$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "As shown in the figure, draw a line through $D$ parallel to $AB$, intersecting $BF$ at $K$. Then:\n$$\n\\angle DKB = 180^\\circ - \\angle ABF = 180^\\circ - \\angle DEC = \\angle DEB,\n$$\nso points $B, K, E, D$ are concyclic.\nSince $AB \\perp AD$ and $DK \\parallel AB$, we have $AD = BK \\cdot \\sin \\angle ABK$. Let $\\omega$ be the circumcircle of triangle $CDE$ with radius $r$, then $DC = 2r \\cdot \\sin \\angle DEC$. From $AD = DC$ and $\\angle ABK = \\angle DEC$, we get $BK = 2r$.\nLet $FO$ intersect circle $\\omega$ at two points $U$ and $V$, with $F, U, O, V$ in order. By the power of a point theorem:\n$$\nFU \\cdot FV = FE \\cdot FD = FK \\cdot FB.\n$$\nSince $UV = 2r = BK$ and $FB > FK$, we conclude $FV = FB$. Given $FP = FB$ and both $P, V$ lie on the extension of $FO$, it follows that $P$ coincides with $V$.\nFrom $AD = DC$, the sum of directed angles from $\\overrightarrow{AC}$ to $\\overrightarrow{AD}$ and $\\overrightarrow{DC}$ is $0^\\circ$. From $FP = FB$, the sum of angles from $\\overrightarrow{PB}$ to $\\overrightarrow{PF}$ and $\\overrightarrow{FB}$ is $0^\\circ$ (mod $360^\\circ$). This shows\n\nthat as directed angles (mod $180^\\circ$):\n$$\n\\angle AQB = \\angle (AC, PB) = \\frac{1}{2}\\angle (AD, FB) + \\frac{1}{2}\\angle (DC, PF).\n$$\nSince $\\angle (AD, FB) = 90^\\circ - \\angle FBA = 90^\\circ - \\angle CED$, and connecting $PE$ gives:\n$$\n\\angle (DC, PF) = \\angle CDP + \\angle DPF = \\angle CEP + \\angle DPF,\n$$\nwe have:\n$$\n\\begin{aligned}\n\\angle AQB &= \\frac{1}{2}(90^\\circ - \\angle CED + \\angle CEP + \\angle DPF) \\\\\n&= \\frac{1}{2}(90^\\circ - \\angle PED + \\angle DPF) \\\\\n&= \\frac{1}{2}(\\angle DPO + \\angle DPF) = \\angle DPF.\n\\end{aligned}\n$$\nThis completes the proof. $\\Box$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 28644,
    "subject": "Mathematics (Multi-modal)",
    "question": "There are $2025$ people and $66$ given colors. Each person has $66$ balls, one of each color, with a total weight of $1$ for all $66$ balls.\nFind the smallest real number $C$ such that, no matter how the balls are weighted, one can always select exactly one ball from each person so that for every color, the total weight of the selected balls of that color does not exceed $C$.",
    "options": [],
    "answer": "248/517",
    "solution": "Let's generalize the problem by replacing $2025$ with $n$ and $66$ with $m$. For any positive integers $m \\le n$, define:\n$$\nf_m(n) = \\min_{\\substack{a_1+a_2+\\dots+a_m=n \\\\ a_1, a_2, \\dots, a_m \\text{ are positive integers}}} \\left( \\frac{1}{a_1} + \\frac{1}{a_2} + \\dots + \\frac{1}{a_m} \\right).\n$$\n\nWe prove the lemma by induction on $m$. For $m=1$, $f_m(n+m-1) = f_1(n) = \\frac{1}{n}$, and the statement holds. Assume the statement holds for $m-1$, and consider the case with $m$ colors.\nFor $i = 1, 2, \\dots, n$, let $x_i$ be the weight of the first color ball for person $P_i$, and assume without loss of generality that $x_1 \\le x_2 \\le \\dots \\le x_n$. Choose non-negative integer $k$ such that\n$$\nx_1 + x_2 + \\dots + x_k \\le A, \\quad x_1 + x_2 + \\dots + x_k + x_{k+1} > A.\n$$\nThen $x_{k+1} \\ge \\frac{x_1+\\dots+x_{k+1}}{k+1} > \\frac{A}{k+1}$. Therefore, for people $P_{k+1}, P_{k+2}, \\dots, P_n$ ($n-k$ people), the total weight of their $m-1$ balls (colors $2, 3, \\dots, m$) is at most\n$$\nf_m(n+m-1) \\cdot A - \\frac{A}{k+1} \\le f_{m-1}(n-k+m-2) \\cdot A.\n$$\nThis inequality holds because\n$$\n\\min_{b_1+\\cdots+b_{m-1}=n-k+m-2} \\left( \\frac{1}{b_1} + \\cdots + \\frac{1}{b_{m-1}} \\right) + \\frac{1}{k+1} \\ge \\min_{a_1+\\cdots+a_m=n+m-1} \\left( \\frac{1}{a_1} + \\cdots + \\frac{1}{a_m} \\right).\n$$\nBy the induction hypothesis, we can select one ball (from colors $2, 3, \\dots, m$) from each of $P_{k+1}, P_{k+2}, \\dots, P_n$ such that the total weight of each color among the selected balls is at most $A$. This completes the induction.\n\nSubstituting $A = \\frac{1}{f_m(n+m-1)}$ in the lemma, we see that $C = \\frac{1}{f_m(n+m-1)}$ satisfies the requirement.\n$$\n\\text{Suppose } f_m(n+m-1) = \\frac{1}{a_1} + \\cdots + \\frac{1}{a_m} \\text{ where } a_1 + \\cdots + a_m = n+m-1.\n$$\n\nIf $C < \\frac{1}{f_m(n+m-1)}$, consider the following scenario: For each person, the weight of their $k$-th color ball is\n$$\n\\frac{1}{a_k} \\times \\frac{1}{\\frac{1}{a_1} + \\dots + \\frac{1}{a_m}} \\left( > \\frac{1}{a_k} \\times C \\right),\n$$\nand the total weight of each person's $m$ balls is $1$. In this case, the number of selected balls of color $k$ is at most $a_k - 1$, so the total number of selected balls would be at most $(a_1 - 1) + \\dots + (a_m - 1) = n - 1$, which is a contradiction.\nTherefore, the minimal $C$ is $C_{\\min} = \\frac{1}{f_m(n+m-1)}$.\nLet $n = mq + r$ where $1 \\le r \\le m$, i.e., $n + m - 1 = m(q + 1) + (r - 1) = (r-1)(q+2) + (m+1-r)(q+1)$. Since $\\frac{1}{x}$ is convex, the sum $\\frac{1}{a_1} + \\dots + \\frac{1}{a_m}$ is minimized when $a_1, \\dots, a_m$ consist of $(r-1)$ copies of $(q+2)$ and $(m+1-r)$ copies of $(q+1)$. Thus,\n$$\nC_{\\min} = \\frac{1}{f_m(n + m - 1)} = \\frac{1}{\\frac{r-1}{q+2} + \\frac{m+1-r}{q+1}}.\n$$\nIn particular, when $n = 2025$ and $m = 66$, the required $C$ is $\\frac{248}{517}$.\n$\\square$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 28645,
    "subject": "Mathematics (Multi-modal)",
    "question": "Prove that there exist integers $x_1, x_2, \\dots, x_{10}, y_1, y_2, \\dots, y_{10}$ satisfying:\n(1) For $i = 1, 2, \\dots, 10$, $|x_i| \\le 10^{10}$ and $|y_i| \\le 10^{10}$;\n(2) The point set in the plane\n$$\nX = \\left\\{ \\left( \\sum_{i=1}^{10} a_i x_i, \\sum_{i=1}^{10} a_i y_i \\right) \\middle| a_1, a_2, \\dots, a_{10} \\in \\{0, 1\\} \\right\\}\n$$\ncontains exactly 1024 distinct points;\n(3) For any two parallel lines in the plane at distance 1, the strip region between them (including the lines) contains at most two points from $X$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "**Proof 1:** Take $x_i = 3^i$, $y_i = 9^i$ ($i = 1, 2, \\dots, 10$). We show these satisfy the conditions. (1) and (2) are obvious. For (3), we use:\n**Lemma:** If real numbers $z_1, \\dots, z_m$ satisfy $|z_{i+1}| \\ge 2|z_i|$ for $i = 1, 2, \\dots, m-1$, then for any $r_1, \\dots, r_m \\in \\{-1, 0, 1\\}$ not all zero, $|\\sum_{i=1}^m r_i z_i| \\ge |z_1|$.\nIf our construction fails (3), some strip $\\{(x, y) | c \\le y - kx \\le c + \\sqrt{k^2+1}\\}$ contains at least 3 points. Let these be:\n$$\nA = \\left( \\sum u_i 3^i, \\sum u_i 9^i \\right), \\quad B = \\left( \\sum v_i 3^i, \\sum v_i 9^i \\right), \\quad C = \\left( \\sum w_i 3^i, \\sum w_i 9^i \\right).\n$$\nThen $|\\sum(u_i - v_i)(9^i - k3^i)| \\le \\sqrt{k^2+1}$ etc.\nIf $k \\le 7$, assume $u_1 = v_1$. Since $9^{i+1} - k3^{i+1} > 2(9^i - k3^i)$ for $i \\ge 2$ and $9^2 - k3^2 > \\sqrt{k^2+1}$, contradiction!\nIf $k > 7 \\cdot 3^8$, assume $u_{10} = v_{10}$. Since $k3^{i+1} - 9^{i+1} > 2(k3^i - 9^i)$ for $i \\le 8$ and $3k - 9 > \\sqrt{k^2+1}$, contradiction!\nFor $7 < k \\le 7 \\cdot 3^8$, let $7 \\cdot 3^d < k \\le 7 \\cdot 3^{d+1}$ ($0 \\le d \\le 7$) and assume $u_{d+2} = v_{d+2}$. We have:\n$$\n\\begin{align*}\n9^{i+1} - k3^{i+1} &> 2(9^i - k3^i) \\quad (i \\ge d+3); \\\\\n9^{d+3} - k3^{d+3} &> 2(k3^{d+1} - 9^{d+1}); \\\\\nk3^{i+1} - 9^{i+1} &> 2(k3^i - 9^i) \\quad (i \\le d); \\\\\n3k - 9 &> \\sqrt{k^2+1},\n\\end{align*}\n$$\nagain a contradiction. Thus the construction works. □\n\n\n**Proof 2:** Take $x_i = 3^i$ and $y_i = 9^i$ as in Solution 1. We now verify condition (3). First observe that the ordering of points in $X$ by their $x$-coordinates coincides with their ordering by $y$-coordinates.\n\nAssume for contradiction that there exist three distinct points $P_1, P_2, P_3 \\in X$ lying between two parallel lines $l_1$ and $l_2$ at distance 1 apart. For $j = 1, 2, 3$, let:\n$$\nP_j = (x_j, y_j) = \\left( \\sum_{i=1}^{10} a_i^{(j)} 3^i, \\sum_{i=1}^{10} a_i^{(j)} 9^i \\right), \\qquad (1)\n$$\nwhere $a_i^{(j)} \\in \\{0, 1\\}$ for $i = 1, \\dots, 10$ and $j = 1, 2, 3$. Without loss of generality, assume $x_1 > x_2 > x_3$.\nFor each $j = 1, 2, 3$, let $i_j$ be the largest index $i$ with $a_i^{(j)} = 1$. Then $i_1 \\ge i_2 \\ge i_3$. We may assume $(P_1, P_2, P_3)$ is a minimal counterexample with respect to $(i_1, i_2, i_3)$. Let:\n$$\nL_x = \\sum_{k=1}^{i_1} 3^k, \\quad L_y = \\sum_{k=1}^{i_1} 9^k.\n$$\nIf $i_1 = i_2$, then the points:\n$$\n(L_x - x_3, L_y - y_3), \\quad (L_x - x_2, L_y - y_2), \\quad (L_x - x_1, L_y - y_1)\n$$\nalso lie in $X$ and satisfy the same strip condition (being symmetric reflections of $P_1, P_2, P_3$ about $(\\frac{L_x}{2}, \\frac{L_y}{2})$). This contradicts the minimality of $(i_1, i_2, i_3)$. Hence $i_1 > i_2 \\ge i_3$, and in particular $i_1 \\ge 2$.\nLet $P'_j = (x'_j, y'_j)$ be the projection of $P_j$ onto $l_1$. Since $l_1$ and $l_2$ are distance 1 apart, we have $|x_j - x'_j| \\le 1$ and $|y_j - y'_j| \\le 1$. Therefore:\n$$\n\\frac{y'_1 - y'_2}{x'_1 - x'_2} = \\frac{y'_2 - y'_3}{x'_2 - x'_3}.\n$$\n**Lemma 1:** $\\frac{y'_1 - y'_2}{x'_1 - x'_2} \\ge \\frac{\\frac{9}{8} \\cdot 9^{i_1} - \\frac{17}{8}}{\\frac{3}{2} \\cdot 3^{i_1} - \\frac{1}{2}}$.\n**Proof:** From (1) we have:\n$$\n\\frac{y_1 - y_2}{x'_1 - x'_2} \\ge \\frac{y_1 - y_2 - 2}{x_1 - x_2 + 2} = \\frac{9^{i_1} + \\sum_{k=1}^{i_1-1} (a_k^{(1)} - a_k^{(2)}) 9^k - 1}{3^{i_1} + \\sum_{k=1}^{i_1-1} (a_k^{(1)} - a_k^{(2)}) 3^k + 1}.\n$$\nSince $i_1 \\ge 2$ and $\\frac{9^{k-1}}{3^{k+1}} > \\frac{9^{k-1}}{3^{k-1}} > \\dots > 3$ for $k \\ge 1$, and $a_k^{(1)} - a_k^{(2)} \\in \\{-1, 0, 1\\}$, by the \"Fraction Inequality\" we get:\n$$\n\\frac{y'_1 - y'_2}{x'_1 - x'_2} \\ge \\frac{\\sum_{k=1}^{i_1} 9^k - 1}{\\sum_{k=1}^{i_1} 3^k + 1} = \\frac{\\frac{9}{8} \\cdot 9^{i_1} - \\frac{17}{8}}{\\frac{3}{2} \\cdot 3^{i_1} - \\frac{1}{2}}.\n$$\n**Lemma 2:** $\\frac{y'_2 - y'_3}{x'_2 - x'_3} \\le \\frac{\\frac{7}{8} \\cdot 9^{i_1-1} + \\frac{17}{8}}{\\frac{1}{2} \\cdot 3^{i_1-1} + \\frac{1}{2}}$.\n**Proof:** From (1) we have:\n$$\n\\frac{y_2' - y_3'}{x_2' - x_3'} \\le \\frac{y_2 - y_3 + 2}{x_2 - x_3 - 2} = \\frac{\\sum_{k=1}^{i_2} (a_k^{(2)} - a_k^{(3)}) 9^k + 1}{\\sum_{k=1}^{i_2} (a_k^{(2)} - a_k^{(3)}) 3^k - 1}.\n$$\nLet $s$ be the largest index with $a_s^{(2)} \\neq a_s^{(3)}$. Since $x_2 > x_3$ and $i_1 > i_2$, we have $1 \\le s \\le i_1 - 1$. Noting that $\\frac{9^{s+1}}{3^{s-1}} > \\frac{9^{s-1}}{3^{s-1}} > \\cdots > 3$ and $a_k^{(2)} - a_k^{(3)} \\in \\{-1, 0, 1\\}$, by the Fraction Inequality:\n$$\n\\frac{y_2' - y_3'}{x_2' - x_3'} \\le \\frac{9^s - \\sum_{k=1}^{s-1} 9^k + 1}{3^s - \\sum_{k=1}^{s-1} 3^k - 1} = \\frac{\\frac{7}{8} \\cdot 9^s + \\frac{17}{8}}{\\frac{1}{2} \\cdot 3^s + \\frac{1}{2}} \\le \\frac{\\frac{7}{8} \\cdot 9^{i_1-1} + \\frac{17}{8}}{\\frac{1}{2} \\cdot 3^{i_1-1} + \\frac{1}{2}}.\n$$\nReturning to the main proof, for $i_1 \\ge 2$ we have:\n$$\n\\frac{\\frac{9}{8} \\cdot 9^{i_1} - \\frac{17}{8}}{\\frac{3}{2} \\cdot 3^{i_1} - \\frac{1}{2}} > \\frac{\\frac{7}{8} \\cdot 9^{i_1-1} + \\frac{17}{8}}{\\frac{1}{2} \\cdot 3^{i_1-1} + \\frac{1}{2}}.\n$$\n(Note the dominant terms satisfy $\\frac{9}{8}\\frac{9^{i_1}}{3^3 3^{i_1}} = \\frac{3}{4}3^{i_1} > \\frac{7}{4}3^{i_1-1} = \\frac{7}{8}\\frac{9^{i_1-1}}{3^3 3^{i_1-1}}$, and the case $i_1 = 2$ can be verified directly.) This contradicts the equality derived from Lemmas 1 and 2. □\n\n\n**Proof 3:** Let $d(P, \\ell)$ denote the distance from point $P$ to line $\\ell$.\nIf a strip of width 1 contains three points $A, B, C$, then one point (the middle one in the projection onto the boundary lines) has distance $\\le 1$ to the line through the other two points. That is, the height from $A$ to $BC$ is $\\le 1$ in $\\triangle ABC$. Since $|AB| + |AC| \\ge |BC|$, at least one of $AB$ or $AC$ has length $\\ge \\frac{1}{2}|BC|$, making the corresponding height $\\le 2$.\nWe call an ordered triple $(A, B, C)$ a *bad triple* if $d(A, BC) \\le 2$ and $d(B, CA) \\le 2$. (If at least two points coincide or all three are colinear, any ordering is bad.)\nLet $L = 10^{10}$ and consider the grid:\n$$\n\\Omega = \\{(x, y) \\in \\mathbb{Z}^2 : |x| \\le L, |y| \\le L\\}.\n$$\nRandomly and independently select 10 vectors $\\alpha_i = (x_i, y_i) \\in \\Omega$. For each subset $I \\subseteq \\{1, \\dots, 10\\}$, define the point $P_I = \\sum_{i \\in I} \\alpha_i$.\nWe want to show that with positive probability, no three points in $\\{P_I\\}$ form a bad triple, thus satisfying the requirement.\nFor any three distinct subsets $I_1, I_2, I_3$, consider the probability that $(P_{I_1}, P_{I_2}, P_{I_3})$ is bad (equivalent to $(P_{I_2}, P_{I_1}, P_{I_3})$ being bad).\nSince $I_1 \\ne I_2$, there exists some index $k$ in exactly one of them. Without loss of generality, assume $k \\in I_1$ and $k \\notin I_2$. Consider two cases:\n**Case 1:** $k \\notin I_3$ ($k \\in I_1$ only). We bound $\\mathbb{P}(d(P_{I_1}, P_{I_2}, P_{I_3}) \\le 2)$. First fix $\\alpha_i$ for $i \\ne k$ randomly. The probability that $P_{I_2} = P_{I_3}$ is $\\le \\frac{1}{|\\Omega|}$ (since any differing coordinate would require specific values). If $P_{I_2} \\neq P_{I_3}$, then $d(P_{I_1}, P_{I_2}P_{I_3}) \\le 2$ requires $\\alpha_k$ to lie in a strip $D$ of width 4 after translation.\nIf $D$ has slope $\\le 1$, each vertical line in $\\Omega$ contains at most $\\lceil 4\\sqrt{2} \\rceil = 6$ points of $D$. If slope $> 1$, each horizontal line contains at most 6 points. Thus:\n$$\n\\mathbb{P}(d(P_{I_1}, P_{I_2}P_{I_3}) \\le 2) \\le \\frac{1}{|\\Omega|} + \\frac{6(2L+1)}{|\\Omega|} \\le \\frac{3}{L}.\n$$\n**Case 2:** $k \\in I_3$ ($k \\in I_1 \\cap I_3$). Consider complements $J_r = I_r^c$. The triangles $\\triangle P_{I_1}P_{I_2}P_{I_3}$ and $\\triangle P_{J_1}P_{J_2}P_{J_3}$ are congruent (central symmetric). Thus:\n$$\n\\mathbb{P}(d(P_{I_2}, P_{I_1}P_{I_3}) \\le 2) = \\mathbb{P}(d(P_{J_2}, P_{J_1}P_{J_3}) \\le 2) \\le \\frac{3}{L}.\n$$\nFor any three distinct subsets, the probability of forming a bad triple is $\\le \\frac{3}{L}$.\nIf none of $I_1, I_2, I_3$ contains the element $k$, then $J_1 = I_1 \\cup \\{k\\}$, $J_2 = I_2 \\cup \\{k\\}$, $J_3 = I_3 \\cup \\{k\\}$ satisfy that $\\triangle P_{J_1}P_{J_2}P_{J_3}$ is congruent to $\\triangle P_{I_1}P_{I_2}P_{I_3}$. Therefore, the cases of bad triples are equivalent. We say that $(I_1, I_2, I_3)$ and $(J_1, J_2, J_3)$ are equivalent position triples.\nFrom this perspective, the number of mutually non-equivalent position triples does not exceed $7^{10}$. Considering that the first two elements can be swapped, there are at most $\\frac{7^{10}}{2}$ such triples.\nTherefore, the probability that among the $2^{10}$ points $\\{P_I\\}$ there exists a bad triple is\n$$\n\\le \\frac{7^{10}}{2} \\times \\frac{3}{L} = 1.5 \\times \\frac{7^{10}}{10^{10}} < 1.\n$$\nThus, there exists some configuration of the $2^{10}$ points $\\{P_I\\}$ containing no bad triples, which satisfies the requirement. $\\square$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 28646,
    "subject": "Mathematics (Multi-modal)",
    "question": "Given an odd prime $p$. Find the largest positive integer $n$ such that there exist $n$ integer-coordinate points $A_1, A_2, \\dots, A_n$ in the plane, with no three collinear, and for any $1 \\le i < j < k \\le n$, twice the area of triangle $A_iA_jA_k$ is **not** divisible by $p$.",
    "options": [],
    "answer": "p+1",
    "solution": "It is well-known that the area of a triangle with vertices $(x_1, y_1)$, $(x_2, y_2)$, $(x_3, y_3)$ is given by\n$$\n\\frac{1}{2} |(y_2 - y_1)(x_3 - x_1) - (y_3 - y_1)(x_2 - x_1)|\n$$\n(This equals zero if and only if the three points are collinear). Therefore, the condition that $(x_1, y_1)$, $(x_2, y_2)$, $(x_3, y_3)$ are non-collinear and twice the area of their triangle is not divisible by $p$ is equivalent to $(y_2 - y_1)(x_3 - x_1)$ and $(y_3 - y_1)(x_2 - x_1)$ being incongruent modulo $p$.\n\nFirst, we show that $n = p + 2$ is impossible. Let $A_i(x_i, y_i)$ ($i = 1, \\dots, p + 2$). If any three $x$-coordinates are congruent modulo $p$, then the corresponding three points form a triangle whose twice area is divisible by $p$. Otherwise, since $p + 2$ is odd, there exists one $x$-coordinate distinct modulo $p$ from all others. Without loss of generality, assume $x_1$ is distinct modulo $p$ from all other $x_i$. Consider the values $\\frac{y_i - y_1}{x_i - x_1}$ modulo $p$ ($i = 2, \\dots, p+2$). Since there are only $p$ possible residues, by the pigeonhole principle, two must be equal. If $\\frac{y_i - y_1}{x_i - x_1} = \\frac{y_j - y_1}{x_j - x_1}$ ($2 \\le i < j \\le p+2$), then $p$ divides $2S_{\\triangle A_1A_iA_j}$.\n\nNext, we construct an example with $n = p + 1$. Let $t$ be a quadratic non-residue modulo $p$. Consider pairs $(x, y)$ with $x, y \\in \\{0, 1, \\dots, p-1\\}$ satisfying $y^2 - tx^2 \\equiv C \\pmod{p}$ for some fixed $C \\neq 0$. There are exactly $p+1$ such pairs (including $(0, 0)$ when $C = 0$, but we exclude this case).\n\nWe verify that these $p+1$ points satisfy the condition. For any three points $A_i, A_j, A_k$:\n* If two have the same $x$-coordinate, their distance isn't divisible by $p$, and the height from the third point is a positive integer less than $p$, so twice the area isn't divisible by $p$.\n* If all $x$-coordinates are distinct, we show $\\frac{y_j - y_i}{x_j - x_i} \\neq \\frac{y_k - y_i}{x_k - x_i} \\pmod{p}$. If they were equal to some $m$, then $y \\equiv mx + r \\pmod{p}$ would hold for three distinct $x$-values, implying $(mx+r)^2 - tx^2 \\equiv C \\pmod{p}$ has three solutions. This contradicts Lagrange's theorem since $m^2 \\neq t \\pmod{p}$ (as $t$ is a non-residue).\n\nTherefore, the maximal $n$ is $p+1$. □",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 28647,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find the smallest real number $M$ such that there exist complex numbers $a, b, c, d$ with $|a| = |b| = |c| = |d| = 1$ satisfying: for any complex number $z$ with $|z| = 1$,\n$$\n|az^3 + bz^2 + cz + d| \\le M.\n$$",
    "options": [],
    "answer": "2√5 - 2",
    "solution": "Let $L$ denote the maximum squared modulus of the polynomial $f(z) = az^3 + bz^2 + cz + d$ on the unit circle.\nBy choosing a unit complex number $s$ with $s^3 = d/a$ and considering $f_1(z) = d^{-1}f(sz)$, we may assume without loss of generality that $a = d = 1$. Thus we only need to consider polynomials of the form $f(z) = z^3 + bz^2 + cz + 1$.\n\nFirst consider the symmetric case $b = c = u + vi$. The polynomial factors as $f(z) = (1+z)(1+(b-1)z + z^2)$. For $z = x + yi$ on the unit circle ($x^2 + y^2 = 1$), we have:\n$$\n\\begin{aligned}\n|1+z|^2 &= 2+2x, \\\\\n|1+(b-1)z+z^2|^2 &= 4x^2 + (2x-1)(2u-2), \\\\\n|f(z)|^2 &= 8 \\left( x^3 + ux^2 + \\frac{u-1}{2}x - \\frac{u-1}{2} \\right).\n\\end{aligned}\n$$\nLet $h(x) = x^3 + ux^2 + \\frac{u-1}{2}x - \\frac{u-1}{2}$. We seek to minimize $\\max_{x \\in [-1,1]} h(x)$.\nHeuristically, the minimum occurs when $h(x)$ has a double root at $x_0$ and a simple root at 1. Solving gives $u = 2 - \\sqrt{5}$. Indeed, for this value:\n$$\nh(x) - (3 - \\sqrt{5}) = \\left( x + \\frac{3 - \\sqrt{5}}{2} \\right)^2 (x - 1) \\leq 0,\n$$\n$$\n\\text{so } |f(z)|^2 \\leq 8(3 - \\sqrt{5}) = (2\\sqrt{5} - 2)^2.\n$$\nFor the general case where $b = c = u + vi$, consider\n$$\n|f(z)|^2 = 8h(x) = (2x + 2) [(4x^2 - 4x + 2) + 2(2x - 1)u]\n$$\nevaluated at the \"endpoint\":\n$$\n|f(1)|^2 = 8h(1) = 8 + 8u,\n$$\nand at the point $z = z_{\\star} = \\frac{\\sqrt{5}-3}{2} + \\frac{\\sqrt{5}-1}{2}i\\sqrt{5}$ (corresponding to a possible vertex of $h(x)$):\n$$\n|f(z_{\\star})|^2 = 8h\\left(\\frac{\\sqrt{5}-3}{2}\\right) = (30\\sqrt{5}-62) + (18-10\\sqrt{5})u.\n$$\nWe have\n$$\n|f(z^*)|^2 + \\frac{5\\sqrt{5}-9}{4} \\cdot |f(1)|^2 = 40\\sqrt{5}-80 = \\frac{5\\sqrt{5}-5}{4} \\cdot (2\\sqrt{5}-2)^2.\n$$\n---\n\nThis shows that a certain weighted average of them equals $(2\\sqrt{5}-2)^2$, therefore when $b=c$ we always have\n$$\nL \\ge \\max \\{|f(1)|^2, |f(z_\\star)|^2\\} \\ge (2\\sqrt{5}-2)^2.\n$$\nFor general $b, c$, we still attempt to prove $L \\ge (2\\sqrt{5}-2)^2$.\nLet $\\omega = \\frac{-1+\\sqrt{3}i}{2}$ be a primitive cube root of unity. Since $f(z) = z^3 + bz^2 + cz + 1$, we have $f(\\omega z) = z^3 + \\omega^2 bz^2 + \\omega cz + 1$ and $f(\\omega^2 z) = z^3 + \\omega bz^2 + \\omega^2 cz + 1$, all corresponding to the same $L$ value. Moreover, $(b+c) + (\\omega^2 b + \\omega c) + (\\omega b + \\omega^2 c) = 0$, and at least one of these has real part $\\le 0$. Without loss of generality, assume $\\frac{b+c}{2} = u + vi$ with $u \\le 0$.\nSince $|b| = |c| = 1$, we have $\\frac{c-b}{b+c} = wi$ being purely imaginary. We may assume $w \\ge 0$ (otherwise consider the reciprocal polynomial $f_2(z) = z^3 + cz^2 + bz + 1$ which has the same $L$ value). Thus $\\frac{c-b}{2} = (u+vi)wi$, and $|\\frac{b+c}{2}|^2 + |\\frac{c-b}{2}|^2 = (u^2+v^2)(1+w^2) = 1$. Now we can express:\n$$\nf(z) = z^3+1+\\frac{b+c}{2}(z^2+z)+\\frac{c-b}{2}(z-z^2) = z(z+1) \\left[ z + z^{-1} - 1 + u + vi + \\frac{1-z}{1+z} \\cdot wi \\cdot (u+vi) \\right].\n$$\nNote that $\\frac{1-z}{1+z} \\cdot wi = t$ is real. Writing the unit complex number $z = x + yi$, we have $|z+1|^2 = 2x+2$, $z+z^{-1}-1 = 2x-1$, and $u^2+v^2 = \\frac{1}{1+w^2}$. Therefore:\n$$\n\\begin{aligned}\n|f(z)|^2 &= |z+1|^2 \\cdot |z+z^{-1}-1+(u+vi)(1+t)|^2 \\\\\n&= (2x+2) \\left[ (2x-1+u+ut)^2 + (v+vt)^2 \\right] \\\\\n&= (2x+2) \\left[ (2x-1)^2 + (u^2+v^2)(1+2t+t^2) + 2(2x-1)u + 2(2x-1)ut \\right] \\\\\n&= (2x+2) \\left[ (4x^2-4x+2) + 2(2x-1)u + \\frac{2t+t^2-w^2}{1+w^2} + 2(2x-1)ut \\right].\n\\end{aligned}\n$$\nTo prove $L \\ge (2\\sqrt{5}-2)^2$, we want to show:\n$$\nT = |f(z_\\star)|^2 + \\frac{5\\sqrt{5}-9}{4} \\cdot |f(1)|^2 - (40\\sqrt{5}-80) \\ge 0.\n$$\nFirst, for $z = z_\\star = \\frac{\\sqrt{5}-3}{2} + \\frac{\\sqrt{5}-1}{2}\\sqrt{5}i$ (where $x = \\frac{\\sqrt{5}-3}{2}$ and $\\frac{1-z}{1+z} = -\\sqrt{5}i$), we have $t = \\sqrt{5}w$. Comparing with the case $b=c$, consider:\n$$\n\\begin{aligned}\nT_1 &= |f(z_\\star)|^2 - \\left[ (30\\sqrt{5}-62) + (18-10\\sqrt{5})u \\right] \\\\\n&= |f(z_\\star)|^2 - (2x+2) \\left[ (4x^2-4x+2) + 2(2x-1)u \\right] \\\\\n&= (2x+2) \\left( \\frac{2t+t^2-w^2}{1+w^2} + 2(2x-1)ut \\right) \\\\\n&= (\\sqrt{5}-1) \\cdot \\left( \\frac{2\\sqrt{5}w + (\\sqrt{5}-1)w^2}{1+w^2} + 2(\\sqrt{5}-4)ut \\right) \\\\\n&\\ge (\\sqrt{5}-1) \\cdot \\frac{2\\sqrt{5}w + (\\sqrt{5}-1)w^2}{1+w^2}.\n\\end{aligned}\n$$\n---\n\nSecond, since $f(1) = 2 + 2(u + vi)$, we have $|f(1)|^2 = 4 + 4u^2 + 4v^2 + 8u$, and:\n$$\nT_2 = |f(1)|^2 - (8 + 8u) = 4(u^2 + v^2 - 1) = -\\frac{4w^2}{1 + w^2}.\n$$\nTherefore:\n$$\nT = T_1 + \\frac{5\\sqrt{5}-9}{4} T_2 \\ge (\\sqrt{5}-1) \\cdot \\frac{2\\sqrt[4]{5}w + (\\sqrt{5}-1)w^2}{1+w^2} - (5\\sqrt{5}-9) \\cdot \\frac{w^2}{1+w^2}.\n$$\nWe want $T \\ge 0$, which holds if:\n$$\n2\\sqrt[4]{5}(\\sqrt{5}-1)w + ((\\sqrt{5}-1)^2 - (5\\sqrt{5}-9))w^2 \\ge 0,\n$$\ni.e., when:\n$$\n0 \\le w \\le \\frac{2(\\sqrt{5}-1)\\sqrt[4]{5}}{7\\sqrt{5}-15} = \\frac{4+2\\sqrt{5}}{\\sqrt[4]{5}} \\approx 5.666.\n$$\nThrough appropriate substitutions or transformations, we have ensured $u \\le 0$ and $w \\ge 0$.\nIf $0 \\le w \\le 2$, then $T \\ge 0$, meaning a certain weighted average of $|f(z_\\star)|^2$ and $|f(1)|^2$ is $\\ge (2\\sqrt{5}-2)^2$, so $L \\ge (2\\sqrt{5}-2)^2$.\nIf $w \\ge 2$, then $|f(1)|^2 = 4 + 4u^2 + 4v^2 + 8u \\le 4 + 4 \\cdot \\frac{1}{1+w^2} \\le 4.8$. Since:\n$$\n|f(1)|^2 + |f(\\omega)|^2 + |f(\\omega^2)|^2 = |2+b+c|^2 + |2+\\omega c+\\omega^2 b|^2 + |2+\\omega^2 c+\\omega b|^2 = 18,\n$$\nwe have:\n$$\nL \\ge \\frac{|f(\\omega)|^2 + |f(\\omega^2)|^2}{2} \\ge \\frac{18 - 4.8}{2} = 6.6 > (2\\sqrt{5} - 2)^2.\n$$\nTherefore, we always have $L \\ge (2\\sqrt{5}-2)^2$, and equality can be achieved. Thus, the minimal real number we seek is $M = 2\\sqrt{5}-2$.\n□",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 28648,
    "subject": "Mathematics (Multi-modal)",
    "question": "Given a positive integer $k$, let integers $a, b$ (not both zero) satisfy $0 \\le a < k$, $0 \\le b < k+1$. Define the sequence $\\{T_n\\}_{n \\ge k}$ as follows:\n$$\nT_k = a, \\quad T_{k+1} = b,\n$$\nand for $n \\ge k + 2$,\n$$\nT_n \\equiv T_{n-1} + T_{n-2} \\pmod{n} \\quad \\text{with} \\quad 0 \\le T_n < n.\n$$\nConcatenating the decimal expansions of $T_k, T_{k+1}, \\dots$ after the decimal point yields an infinite decimal $x = 0.T_kT_{k+1}\\dots$. For example, if $k = 66$, $a = 5$, $b = 20$, then $T_{66} = 5$, $T_{67} = 20$, $T_{68} = 25$, $T_{69} = 45$, $T_{70} = 0$, $T_{71} = 45$, $T_{72} = 45$, $T_{73} = 17$, etc., so $x = 0.52025450454517\\dots$.\nProve that $x$ is irrational.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "**Proof:** First, we prove that $\\{T_n\\}_{n=k}^{\\infty}$ is unbounded. If not, let $M = \\max_{n \\ge k}\\{T_n\\}$, then for any $n > 2M$, from $T_{n-1} \\le M$ and $T_{n-2} \\le M$, we have $T_n = T_{n-1} + T_{n-2}$. By the boundedness of $T_n$, the only possibility is $T_{n-1} = T_{n-2} = 0$, which inductively leads to $T_{k+1} = T_k = 0$, a contradiction!\n\nLet $m_0$ be the number of digits of $k$. We further prove that for any $m > m_0$, there exists $n$ such that $10^{m-1} \\le T_n < 10^m$. By the unboundedness of $T_n$, there exists a minimal $T_n \\ge 10^{m-1}$, then $n > k+1$ and $T_{n-1}, T_{n-2} < 10^{m-1}$, thus $T_n \\le T_{n-1} + T_{n-2} < 2 \\cdot 10^{m-1} < 10^m$.\n\nFor convenience, denote $p \\star q$ as the number obtained by concatenating positive integer $p$ with non-negative integer $q$. By contradiction, assume $0.T_k \\star T_{k+1} \\star \\dots = 0.a_1 \\dots a_s b_1 \\dots b_t b_1 \\dots b_t \\dots$ (i.e., this repeating decimal has ultimate period length $t$), where $a_i, b_j \\in \\{0, 1, \\dots, 9\\}$. Take $n$ sufficiently large so that $T_n$ appears to the right of $a_s$, then for any $r \\ge n$, $T_r \\star T_{r+1} \\star T_{r+2} \\dots$ is a number with period $wt$, where $w$ is any positive integer. Take $m = \\nu t$ with $\\nu$ sufficiently large so that $10^{m-1} > T_i$ for all $i \\le n$, then there exists a minimal $\\ell$ such that $T_\\ell$ has exactly $m$ digits, clearly $\\ell > n$.\n\nIf $T_{\\ell-1}$ has more than $m$ digits, then there must exist $j < \\ell - 1$ where $T_j$ has $m-1$ digits, leading to some number before $T_\\ell$ having $m$ digits, a contradiction!\n\nClearly $k \\ne m$, so $k < m$. But $m$ is the period length, thus the last $k$ digits of $T_{\\ell-1}$ and $T_\\ell$ must be identical, which implies $T_{\\ell-2}$ has at least $m$ digits. Therefore, there exists a number before $T_{\\ell-1}$ with $m$ digits, another contradiction!\n\nIn conclusion, the assumption by contradiction is invalid, and the original proposition is proved. □",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 28649,
    "subject": "Mathematics (Multi-modal)",
    "question": "Given integer sequences $\\{x_n\\}_{n \\ge 1}$, $\\{y_n\\}_{n \\ge 1}$. Prove that there exists an integer sequence $\\{z_n\\}_{n \\ge 1}$ such that for any positive integer $n$,\n$$\n\\sum_{k|n} k z_k^{n/k} = \\left( \\sum_{k|n} k x_k^{n/k} \\right) \\cdot \\left( \\sum_{k|n} k y_k^{n/k} \\right).\n$$\nHere all three sums are taken over all positive divisors $k$ of $n$. For example, when $n = 6$:\n$$\n6z_6 + 3z_3^2 + 2z_2^3 + z_1^6 = (6x_6 + 3x_3^2 + 2x_2^3 + x_1^6) \\cdot (6y_6 + 3y_3^2 + 2y_2^3 + y_1^6).\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "**Proof 1:** By induction, $z_n$ can be uniquely determined from the equation\n$$\nnz_n = \\left( \\sum_{k|n} k x_k^{n/k} \\right) \\cdot \\left( \\sum_{k|n} k y_k^{n/k} \\right) - \\sum_{\\substack{k|n \\\\ k<n}} k z_k^{n/k}. \\qquad (3)\n$$\nWe only need to prove that the right-hand side is divisible by $n$. To this end, it suffices to show that for any prime factor $p$ of $n$, if $p^r \\nmid n$ ($r \\in \\mathbb{N}_+$), then $p^r$ divides the right-hand side of (3).\nWorking modulo $p^r$, we first remove terms in (3) whose coefficients are already multiples of $p^r$:\n$$\n\\begin{aligned}\nnz_n &\\equiv \\left( \\sum_{\\substack{k|n \\\\ p^r \\nmid k}} k x_k^{n/k} \\right) \\cdot \\left( \\sum_{\\substack{k|n \\\\ p^r \\nmid k}} k y_k^{n/k} \\right) - \\sum_{\\substack{k|n \\\\ p^r \\nmid k}} k z_k^{n/k} \\\\\n&= \\left( \\sum_{k|\\frac{n}{p}} k x_k^{n/k} \\right) \\cdot \\left( \\sum_{k|\\frac{n}{p}} k y_k^{n/k} \\right) - \\sum_{k|\\frac{n}{p}} k z_k^{n/k} \\quad (\\text{mod } p^r).\n\\end{aligned} \\qquad (4)\n$$\nWe first prove a simple **lemma**: For any integer $a$ and positive integer $t$, $a^{pt} \\equiv a^{pt-1} \\pmod{p^t}$.\nIndeed, if $p \\nmid a$, then since $p^{t-1} \\ge t$, both $a^{pt}$ and $a^{pt-1}$ are divisible by $p^t$. If $p \\nmid a$, then $a^{p-1} - 1$ is divisible by $p$, and by the lifting-the-exponent lemma, $a^{(p-1)p^{t-1}} \\equiv 1 \\pmod{p^t}$, so $a^{pt} \\equiv a^{pt-1} \\pmod{p^t}$.\nReturning to the problem, we now show that for any $k \\mid \\frac{n}{p}$ and any integer $a$, $ka^{n/k} \\equiv ka^{n/pk} \\pmod{p^r}$. Indeed, let $s = v_p(k) < r$. It suffices to prove $a^{n/k} \\equiv a^{n/pk} \\pmod{p^{r-s}}$. By the lemma, $a^{p^{r-s}} \\equiv a^{p^{r-s-1}} \\pmod{p^{r-s}}$. Raising this congruence to the $n/p^{r-s}$-th power yields the result.\nThus, $nz_n$ satisfies\n$$\nnz_n \\equiv \\left( \\sum_{k|\\frac{n}{p}} k x_k^{n/pk} \\right) \\cdot \\left( \\sum_{k|\\frac{n}{p}} k y_k^{n/pk} \\right) - \\sum_{k|\\frac{n}{p}} k z_k^{n/pk} = 0 \\pmod{p^r}.\n$$\n---\n\n**Proof 2:** For any positive integer $L$, define the polynomial\n$$\ng_L(t) = \\sum_{d|L} t^d \\mu\\left(\\frac{L}{d}\\right),\n$$\nwhere $\\mu(\\cdot)$ is the Möbius function. We now show that for any integer $t$, $g_L(t)$ is always divisible by $L$. Without loss of generality, assume $t$ is a positive integer. Note that the number of distinct circular arrangements of length $L$ with exactly $L$ as the minimal period (rotations considered identical) using $t$ colors is precisely $\\frac{1}{L}g_L(t)$. Thus, $g_L(t)$ is divisible by $L$.\nFor any positive integer $m$, define\n$$\nA_m = \\sum_{L|m} L g_{\\frac{m}{L}}(x_L), \\quad B_m = \\sum_{L|m} L g_{\\frac{m}{L}}(y_L).\n$$\nBy the above discussion, $A_m$ is divisible by $m$. By Möbius inversion,\n$$\n\\begin{align*}\n\\sum_{m|n} A_m &= \\sum_{L|m|n} L g_{\\frac{m}{L}}(x_L) = \\sum_{L|m|n} L \\sum_{d|\\frac{m}{L}} x_L^d \\mu\\left(\\frac{m}{dL}\\right) = \\sum_{dL|n} L x_L^d \\cdot \\sum_{\\substack{\\frac{m}{dL} \\\\ \\frac{n}{dL}}} \\mu\\left(\\frac{m}{dL}\\right) \\\\\n&= \\sum_{dL|n} L x_L^d \\cdot \\mathbf{1}_{\\frac{n}{dL}=1} = \\sum_{dL=n} L x_L^d = \\sum_{k|n} k x_k^{n/k}.\n\\end{align*}\n$$\nWe need to find a sequence of positive integers $\\{C_n\\}$ such that for all positive integers $n$,\n$$\n\\sum_{m|n} C_m = \\left( \\sum_{m|n} A_m \\right) \\left( \\sum_{m|n} B_m \\right).\n$$\nIn fact, we can take $C_n = \\sum_{[u,v]=n} A_u B_v$, i.e., the sum over all pairs $(u,v)$ with least common multiple $[u,v] = n$. Since $A_u B_v$ is divisible by both $uv$ and $n$, it follows that $C_n$ is divisible by $n$.\nFinally, for $n = 1, 2, 3, \\dots$, we sequentially define\n$$\nz_n = \\frac{1}{n} \\left[ C_n - \\sum_{L|n, L<n} L g_{\\frac{n}{L}}(z_L) \\right]\n$$\nto satisfy the requirement.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 28650,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $A_1A_2A_3A_4$ be a convex quadrilateral that is **not** cyclic and whose opposite sides are **not** parallel. For $1 \\le i \\le 4$, let $M_i$ be the midpoint of $A_{i-1}A_{i+1}$. Let $B_i$ be a point on the tangent to the circumcircle of triangle $A_{i-1}A_iA_{i+1}$ at $A_i$, such that the reflection of $M_i$ over the angle bisector of $\\angle A_{i+1}A_{i+2}A_{i+3}$ lies on line $B_iA_{i+2}$. Let $C_i$ be the unique intersection point of lines $A_iA_{i+1}$ and $B_iB_{i+1}$. (All indices are modulo 4.)\nProve that $C_1, C_2, C_3, C_4$ are collinear.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "For any point $X$ on line $A_{i+2}B_i$, we have:\n$$\n\\frac{d_{X-A_{i+1}A_{i+2}}}{A_{i+1}A_{i+2}} = \\frac{d_{X-A_{i+2}A_{i+3}}}{A_{i+2}A_{i+3}}.\n$$\nFor any point $X$ on line $A_iB_i$, we have:\n$$\n\\frac{d_{X-A_{i-1}A_i}}{A_{i-1}A_i} + \\frac{d_{X-A_iA_{i+1}}}{A_iA_{i+1}} = 0.\n$$\nTherefore, for point $B_i$:\n$$\n\\frac{d_{B_i-A_{i+2}A_{i+3}}}{A_{i+2}A_{i+3}} = \\frac{d_{B_i-A_{i+1}A_{i+2}}}{A_{i+1}A_{i+2}}, \\qquad (6)\n$$\nand\n$$\n\\frac{d_{B_i-A_{i-1}A_i}}{A_{i-1}A_i} + \\frac{d_{B_i-A_iA_{i+1}}}{A_iA_{i+1}} = 0. \\qquad (7)\n$$\nCombining these gives:\n$$\n\\frac{d_{B_i-A_{i+2}A_{i+3}}}{A_{i+2}A_{i+3}} = \\frac{d_{B_i-A_{i-1}A_i}}{A_{i-1}A_i} + \\frac{d_{B_i-A_iA_{i+1}}}{A_iA_{i+1}} + \\frac{d_{B_i-A_{i+1}A_{i+2}}}{A_{i+1}A_{i+2}}.\n$$\nSimilarly for $B_{i+1}$:\n$$\n\\frac{d_{B_{i+1}-A_{i-1}A_i}}{A_{i-1}A_i} = \\frac{d_{B_{i+1}-A_{i+2}A_{i+3}}}{A_{i+2}A_{i+3}},\n$$\nand\n$$\n\\frac{d_{B_{i+1}-A_iA_{i+1}}}{A_iA_{i+1}} + \\frac{d_{B_{i+1}-A_{i+1}A_{i+2}}}{A_{i+1}A_{i+2}} = 0.\n$$\nThus:\n$$\n\\frac{d_{B_{i+1}-A_{i+2}A_{i+3}}}{A_{i+2}A_{i+3}} = \\frac{d_{B_{i+1}-A_{i-1}A_i}}{A_{i-1}A_i} + \\frac{d_{B_{i+1}-A_iA_{i+1}}}{A_iA_{i+1}} + \\frac{d_{B_{i+1}-A_{i+1}A_{i+2}}}{A_{i+1}A_{i+2}}.\n$$\nSince these relations are linear in the coordinates, point $C_i$ must also satisfy:\n$$\n\\frac{d_{C_i-A_{i+2}A_{i+3}}}{A_{i+2}A_{i+3}} = \\frac{d_{C_i-A_{i-1}A_i}}{A_{i-1}A_i} + \\frac{d_{C_i-A_iA_{i+1}}}{A_iA_{i+1}} + \\frac{d_{C_i-A_{i+1}A_{i+2}}}{A_{i+1}A_{i+2}}. \\qquad (8)\n$$\n\nNoting that $C_i$ lies on $A_iA_{i+1}$ (so $d_{C_i-A_iA_{i+1}} = 0$), we can rewrite this as:\n$$\n\\frac{d_{C_i-A_{i+2}A_{i+3}}}{A_{i+2}A_{i+3}} + \\frac{d_{C_i-A_iA_{i+1}}}{A_iA_{i+1}} = \\frac{d_{C_i-A_{i-1}A_i}}{A_{i-1}A_i} + \\frac{d_{C_i-A_{i+1}A_{i+2}}}{A_{i+1}A_{i+2}}. \\quad (9)\n$$\nThis equation is completely symmetric for $C_1, C_2, C_3, C_4$. If it held identically for all points, then substituting $A_1$ would yield:\n$$\n\\frac{A_1 A_2 \\cdot \\sin A_2}{A_2 A_3} = \\frac{A_1 A_4 \\cdot \\sin A_4}{A_3 A_4},\n$$\nand substituting $A_3$ would give:\n$$\n\\frac{A_2 A_3 \\cdot \\sin A_2}{A_1 A_2} = \\frac{A_3 A_4 \\cdot \\sin A_4}{A_1 A_4}.\n$$\nThis would imply $\\sin A_2 = \\sin A_4$, and similarly $\\sin A_1 = \\sin A_3$, meaning $A_1A_2A_3A_4$ would either be a parallelogram or cyclic - contradicting the given conditions.\nTherefore, the equation is not identically satisfied. Since it is linear in coordinates, the set of points satisfying it must form a straight line. Hence $C_1, C_2, C_3, C_4$ are collinear. $\\square$\n\n\n**Second Proof:** Define the cross ratio of four lines $PA, PB, PC, PD$ through point $P$ as:\n$$\nP(A, B; C, D) := \\frac{\\sin \\angle APC \\cdot \\sin \\angle BPD}{\\sin \\angle APD \\cdot \\sin \\angle BPC}.\n$$\nFirst, we prove a lemma.\n**Lemma:** Let $A, B, C, D$ be four points in the plane with no three collinear, and let $M$ be another point. Then a point $P$ lies on the conic through $A, B, C, D, M$ if and only if $P(A, B; C, D) = M(A, B; C, D)$.\n**Proof of Lemma:** Let $a, b, c, d, m, p$ be the complex coordinates of points $A, B, C, D, M, P$ respectively. Then:\n$$\n\\begin{aligned} \\frac{PA \\cdot PB}{PC \\cdot PD} &= \\frac{\\sin \\angle APB \\cdot \\sin \\angle BPD}{\\sin \\angle ABP \\cdot \\sin \\angle BPD} \\\\ &= \\frac{\\operatorname{Im}\\left(\\frac{c-p}{a-p} / \\left|\\frac{c-p}{a-p}\\right|\\right) \\cdot \\operatorname{Im}\\left(\\frac{d-p}{b-p} / \\left|\\frac{d-p}{b-p}\\right|\\right)}{\\operatorname{Im}\\left(\\frac{c-p}{b-p} / \\left|\\frac{c-p}{b-p}\\right|\\right) \\cdot \\operatorname{Im}\\left(\\frac{d-p}{a-p} / \\left|\\frac{d-p}{a-p}\\right|\\right)} \\\\ &= \\frac{[(c-p)(\\bar{a}-\\bar{p}) - (\\bar{c}-\\bar{p})(a-p)][(d-p)(\\bar{b}-\\bar{p}) - (\\bar{d}-\\bar{p})(b-p)]}{[(c-p)(\\bar{b}-\\bar{p}) - (\\bar{c}-\\bar{p})(b-p)][(d-p)(\\bar{a}-\\bar{p}) - (\\bar{d}-\\bar{p})(a-p)]} \\end{aligned}\n$$\nSince each bracket is linear in $p$ and $\\bar{p}$, the condition \"cross ratio equals a constant\" defines a quadratic equation in $p$ and $\\bar{p}$, which corresponds to a conic section. (Note: This cannot be identically constant, as that would imply $A, B, C$ are collinear, a contradiction.) $\\square$\n\nReturning to the problem, since $A_3B_1$ and $A_3M_1$ are symmetric about the angle bisector of $\\angle A_2A_3A_4$, $A_3B_1$ is the symmedian of $\\triangle A_2A_3A_4$.\nLet $S$ be the intersection of $B_1A_1$ and $A_2A_4$, and $T$ the intersection of $B_1A_3$ and $A_2A_4$. Then:\n$$\n\\begin{aligned} B_1(A_1, A_2; A_3, A_4) &= (S, A_2; T, A_4) = 1 - (S, T; A_2, A_4) \\\\ &= 1 - \\frac{SA_2}{SA_4} \\cdot \\frac{TA_4}{TA_2} = 1 - \\frac{A_1A_2^2 \\cdot A_3A_4^2}{A_1A_4^2 \\cdot A_3A_2^2}. \\end{aligned}\n$$\nSimilarly, this holds for $B_2, B_3, B_4$. Thus all eight points $A_1, \\dots, A_4, B_1, \\dots, B_4$ lie on the same conic.\n\n* Considering $A_1, A_2, A_3, B_1, B_2, B_3$, by Pascal's Theorem, $C_1, C_2$ and $X := A_3B_1 \\cap A_1B_3$ are collinear.\n* Similarly, considering $A_1, A_4, A_3, B_1, B_4, B_3, C_3, C_4$ and $X$ are collinear.\n* Considering $A_2, A_3, A_4, B_2, B_3, B_4, C_2, C_3$ and $Y := A_4B_2 \\cap A_2B_4$ are collinear.\n* Considering $A_4, A_1, A_2, B_4, B_1, B_2, C_1, C_4$ and $Y$ are collinear.\n\nLet $W$ be the intersection of $A_1B_1$ and $A_1B_2$. Applying Menelaus' Theorem to\ntriangle $\\triangle WB_1B_2$ with transversal $A_1A_2C_1$ gives:\n$$\n\\frac{B_1C_1}{B_2C_1} = \\frac{B_1A_1}{B_2A_2} \\cdot \\frac{A_1W}{A_2W} = \\frac{B_1A_1}{B_2A_2} \\cdot \\frac{\\sin \\angle WA_2A_1}{\\sin \\angle WA_1A_2}.\n$$\nBy tangent properties, $\\angle WA_2A_1 = \\angle A_2A_3A_1$, so:\n$$\n\\frac{B_1C_1}{B_2C_1} = \\frac{B_1A_1}{B_2A_2} \\cdot \\frac{\\sin \\angle A_2A_3A_1}{\\sin \\angle A_1A_4A_2}.\n$$\nTaking the cyclic product (indices modulo 4):\n$$\n\\begin{align}\n\\prod_{i=1}^{4} \\frac{B_i C_i}{B_{i+1} C_i} &= \\prod_{i=1}^{4} \\frac{B_i A_i}{B_{i+1} A_{i+1}} \\cdot \\prod_{i=1}^{4} \\frac{\\sin \\angle A_{i+1}A_{i+2}A_i}{\\sin \\angle A_iA_{i-1}A_{i+1}} \\tag{10} \\\\\n&= \\prod_{i=1}^{4} \\frac{\\sin \\angle A_{i+1}A_{i+2}A_i}{\\sin \\angle A_{i+1}A_iA_{i+2}} = \\prod_{i=1}^{4} \\frac{A_i A_{i+1}}{A_{i+1} A_{i+2}} = 1. \\nonumber\n\\end{align}\n$$\n\nThus:\n$$\n\\frac{B_1 C_1}{B_2 C_1} \\cdot \\frac{B_2 C_2}{B_3 C_2} = \\frac{B_1 C_4}{B_4 C_4} \\cdot \\frac{B_3 C_4}{B_3 C_3}.\n$$\nLet this ratio be $\\alpha$.\nOn the extension of $B_3B_1$, take point $P$ such that $\\frac{B_3P}{B_1P} = \\frac{1}{\\alpha}$. By the converse of Menelaus' Theorem applied to $\\triangle B_1B_2B_3$ and points $P, C_1, C_2$, these are collinear. Similarly, $P, C_3, C_4$ are collinear.\nThus we have $P, C_1, C_2$ collinear and $P, C_3, C_4$ collinear, while $X, C_1, C_2$ and $X, C_3, C_4$ are also collinear. Since $P$ lies on $B_1B_3$ and $X = A_1B_3 \\cap A_3B_1$, if $P = X$ then $A_1 = B_1$ or $A_3 = B_3$, making $A_1A_3$ a symmedian for both $\\triangle A_1A_2A_4$ and $\\triangle A_2A_3A_4$, which would imply $A_1A_2A_3A_4$ is harmonic - contradicting the non-cyclic condition.\nTherefore $P \\neq X$, and consequently $C_1, C_2, C_3, C_4$ all lie on the line $PX$. $\\square$",
    "topic": "Number Theory",
    "subtopic": "Modular Arithmetic"
  },
  {
    "id": 28651,
    "subject": "Mathematics (Multi-modal)",
    "question": "Given an odd positive integer $n$, let $m = \\frac{n+1}{2}$. Let positive integers $a_1, a_2, \\dots, a_m$ be pairwise incongruent modulo $n$, and positive integers $b_1, b_2, \\dots, b_m$ be pairwise incongruent modulo $n$. Prove that the set\n$$\nC = \\{a_i + b_j \\text{ (least nonnegative residue modulo } n) \\mid i, j \\in \\{1, 2, \\dots, m\\}, i \\neq j\\}\n$$\nhas more than $n - \\sqrt{n} - \\frac{1}{2}$ elements.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $X = \\{0, \\dots, n-1\\}$, $A = \\{a_1, \\dots, a_m\\}$, $B = \\{b_1, \\dots, b_m\\}$, and $S = X \\setminus C$. We need to show that $|S| < \\sqrt{n} + \\frac{1}{2}$.\n\nConsider the bipartite graph $G := (X \\sqcup X, E)$, where the edge set $E := \\{(x, y) : x + y \\in S\\}$. In particular, the restriction of $G$ to $A \\sqcup B$ consists of several edges with no common vertices, meaning no vertex has degree at least 2. If this were not the case, suppose $a_i \\in A$ is connected to both $b_{j_1}$ and $b_{j_2}$, where $j_1 \\neq j_2$. Since $a_i + b_{j_1}, a_i + b_{j_2} \\in S$ do not belong to $C$, it follows that $E \\cap (A \\times B) \\subseteq \\{(a_i, b_i) : 1 \\le i \\le m\\}$. Hence, $i = j_1$ and $i = j_2$, contradicting $j_1 \\neq j_2$. Therefore, every vertex in $G$ restricted to $A \\sqcup B$ has degree at most 1.\n\nIf $G$ contains a cycle $C_4$ of length 4, since $G$ is bipartite, $C_4$ must take the form $c_1, c_2$ in the first copy of $X$ and $d_1, d_2$ in the second copy, with edges $c_1d_1, c_1d_2, c_2d_1, c_2d_2 \\in E$. By the definition of $G$, for any $u \\in X$, the quadruple $(c_1 + u, c_2 + u; d_1 - u, d_2 - u)$ also forms a $C_4$ in $G$. Letting $u$ range over all elements of $X$, since $|A| + |B| = n + 1 > n$, an averaging or pigeonhole argument shows that there exists a $u$ such that at least three of these four points lie in $A \\sqcup B$. This would force a vertex in $G$ restricted to $A \\sqcup B$ to have degree at least 2, a contradiction.\n\nConsequently, $S$ cannot contain two distinct pairs $(s_1, s_2)$ and $(s'_1, s'_2)$ satisfying $s_1 - s_2 = s'_1 - s'_2$. Otherwise, $(0, -s_1 + s'_1; s_1, s_2)$ would form a $C_4$ in $G$, again a contradiction. Thus, $X$ must contain at least $|S|^2 - |S| + 1$ elements, implying $|S| < \\sqrt{n} + \\frac{1}{2}$. $\\square$",
    "topic": "Number Theory",
    "subtopic": "Modular Arithmetic"
  },
  {
    "id": 28652,
    "subject": "Mathematics (Multi-modal)",
    "question": "A finite point set $P = \\{(x_i, y_i) \\mid i = 1, 2, \\dots, n\\}$ in the plane is called **sparse** if for any $1 \\le i < j \\le n$,\n$$\n|x_i - x_j| \\cdot |y_i - y_j| \\ge 1.\n$$\n(1) Let $\\{Q_1, Q_2, Q_3\\}$ be a sparse set where no $Q_i$ lies inside the axis-aligned rectangle with the other two points as diagonal vertices. Prove that the area of triangle $Q_1Q_2Q_3$ is at least $\\frac{\\sqrt{5}}{2}$.\n\n(2) Let $P$ be a sparse set with no three collinear points, all lying inside an axis-aligned square of side length $a$. Prove that the number of points in $P$ does not exceed\n$$\n\\frac{a^2}{\\sqrt{5}} + 2a + 1.\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "(1) Let the coordinates of $Q_i$ be $(x_i, y_i)$. Without loss of generality, assume $x_1 < x_2 < x_3$. Since both the conditions and conclusion are symmetric under reflection about the $x$-axis and $y$-axis, we can further assume $y_1 < y_3 < y_2$. After appropriate translation, we may set $x_1 = y_1 = 0$, so $0 < x_2 < x_3$ and $0 < y_3 < y_2$. Since $\\{Q_1, Q_2, Q_3\\}$ is sparse,\n$$\nA = (x_3 - x_2)(y_2 - y_3) \\ge 1, \\quad B = x_2y_2 \\ge 1, \\quad C = x_3y_3 \\ge 1.\n$$\nThe area of triangle $Q_1Q_2Q_3$ is $S = \\frac{1}{2}(x_3y_2 - x_2y_3)$, and we have\n$$\n\\begin{align*} (2S)^2 &= (x_3y_2 - x_2y_3)^2 = (x_3y_2 + x_2y_3)^2 - 4x_3y_2x_2y_3 \\\\ &= (A + B + C)^2 - 4BC = A^2 + 2A(B + C) + (B - C)^2 \\\\ &\\ge 5, \\end{align*}\n$$\nwhich proves $S \\ge \\frac{\\sqrt{5}}{2}$.\n\n\n(2) Define the $L^1$ distance between two points $A(a_1, a_2), B(b_1, b_2)$ as $d_1(A, B) = |a_1 - b_1| + |a_2 - b_2|$. It's easy to verify that the $L^1$ distance satisfies the *triangle inequality*\n$$\nd_1(A, B) + d_1(B, C) \\ge d_1(A, C),\n$$\nwith equality when $A, B, C$ are colinear in order. It's known that if convex polygon $Q$ is contained in convex polygon $R$, then the perimeter of $Q$ doesn't exceed that of $R$. Using similar proof methods and the triangle inequality for $L^1$ distance, we can show that under these conditions, the $L^1$ perimeter of $Q$ (sum of $L^1$ lengths of its sides) doesn't exceed that of $R$.\nLet $R$ be the given square with side length $a$, and $P = \\{P_1, \\dots, P_n\\}$ be a sparse subset of $R$. By the definition of sparse sets, for any $i \\neq j$,\n$$\n|x_i - x_j| + |y_i - y_j| \\ge 2\\sqrt{|x_i - x_j| \\cdot |y_i - y_j|} \\ge 2,\n$$\ni.e., $d_1(P_i, P_j) \\ge 2$.\nFor a triangle $\\triangle P_i P_j P_k$ with vertices in $P$, if $P_i$ lies inside the rectangle with $P_j P_k$ as diagonal and sides parallel to the coordinate axes, we call $\\triangle P_i P_j P_k$ a *bad triangle*, $\\angle P_j P_i P_k$ its *bad angle*, and $P_j P_k$ its *bad edge*. Each bad triangle has exactly one bad angle and one bad edge. Let $\\ell(\\triangle P_i P_j P_k) = d_1(P_j, P_k)$ be the length of the bad edge, which equals $d_1(P_j, P_i) + d_1(P_i, P_k)$. If no angle of $\\triangle P_i P_j P_k$ is bad, we call it a *good triangle*. Part (1) shows that good triangles have area at least $\\frac{\\sqrt{5}}{2}$.\n\nLet the convex hull of $P$ be an $m$-gon $Q = Q_1Q_2\\cdots Q_m$. We first state a lemma:\n**Lemma:** There exists a triangulation $\\mathcal{T}$ of $Q$ consisting of $2n-m-2$ triangles with vertices in $P$, such that the number of bad triangles in $\\mathcal{T}$ doesn't exceed $2a$.\nWe'll use this lemma to prove the main result. Take a triangulation $\\mathcal{T}$ satisfying the lemma. Since the $L^1$ perimeter of $Q$ doesn't exceed that of $R$,\n$$\n4a \\ge \\sum_{i=1}^{m} d_1(Q_i, Q_{i+1}) \\ge 2m,\n$$\nso $m \\le 2a$. As $\\mathcal{T}$ has at most $2a$ bad triangles, the number $N$ of good triangles satisfies\n$$\nN \\ge (2n - m - 2) - 2a \\ge 2n - 4a - 2.\n$$\nSince the total area of good triangles doesn't exceed the area of $R$ and each good triangle has area at least $\\frac{\\sqrt{5}}{2}$,\n$$\na^2 \\ge \\frac{\\sqrt{5}}{2}N \\ge \\frac{\\sqrt{5}}{2}(2n - 4a - 2),\n$$\nwhich implies $n \\le \\frac{a^2}{\\sqrt{5}} + 2a + 1$.\n\nNext, we prove the lemma. Consider a triangulation $\\mathcal{T}$ of $Q$ that minimizes both the number of bad triangles and, subject to that condition, minimizes the sum of all bad angles.\n\nFor any *internal bad triangle* $ABC$ in $\\mathcal{T}$ with bad edge $BC$, there exists another triangle $BCD$ sharing the edge $BC$. Let $\\mathcal{T}'$ be the triangulation obtained by replacing triangles $\\triangle ABC$ and $\\triangle BCD$ with $\\triangle ABD$ and $\\triangle ACD$.\nAs shown in the figure, point $D$ may lie in regions $Y_1$ through $Y_5$. For each possible position of $D$, we can carefully analyze how the number of bad triangles and the sum of bad angles change when transforming $\\mathcal{T}$ to $\\mathcal{T}'$, as described in the following table:\n\n| Region containing D | Change in number of bad triangles and sum of bad angles |\n|--------------------|--------------------------------------------------------|\n| $Y_1$              | Two bad become two good                                 |\n| $Y_2$              | One good and one bad become two good                   |\n| $Y_3$              | Two bad remain two bad, but sum of bad angles strictly decreases |\n| $Y_4$              | One good and one bad remain, but sum of bad angles strictly decreases |\n\n![](attached_image_1.png)\n\nBy the minimality of $\\mathcal{T}$, we conclude that $D \\in Y_5$. In this case, $\\triangle BCD$ is also a bad triangle. If $D$ is in the bottom-left $Y_5$ region of the figure, then $CD$ is the bad edge of $\\triangle BCD$, giving $\\ell(\\triangle BCD) = d_1(D, B) + d_1(B, C) \\ge \\ell(\\triangle ABC) + 2$. If $D$ is in the top-right $Y_5$ region, then $BD$ is the bad edge of $\\triangle BCD$, giving $\\ell(\\triangle BCD) = d_1(B, C) + d_1(C, D) \\ge \\ell(\\triangle ABC) + 2$.\nIn summary, for any internal bad triangle $\\Delta_0$, the triangle $\\Delta_1$ on the other side of its bad edge is also bad, with $\\ell(\\Delta_1) > \\ell(\\Delta_0)$. If $\\Delta_1$ is also internal, then the triangle $\\Delta_2$ on the other side of its bad edge is bad, and so on, forming a unique chain of bad triangles:\n$$\n\\Delta_0 \\rightsquigarrow \\Delta_1 \\rightsquigarrow \\dots \\rightsquigarrow \\Delta_k,\n$$\nending with a boundary bad triangle $\\Delta_k$. Denote this chain by $C(\\Delta_0)$. For a boundary bad triangle $\\Delta_0$, $C(\\Delta_0)$ contains only $\\Delta_0$ itself.\n\n![](attached_image_1.png)",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 28653,
    "subject": "Mathematics (Multi-modal)",
    "question": "Given a circle $\\omega$ and two points $A, B$ outside $\\omega$. A quadrilateral inscribed in $\\omega$ is called \"good\" if one pair of opposite sides intersects at $A$ and the other pair intersects at $B$.\nAssume that at least one good quadrilateral exists. Prove that there exists a good quadrilateral $\\Gamma$ such that every other good quadrilateral has area smaller than $\\Gamma$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $O$ be the center of $\\omega$. For any good quadrilateral $PQRS$ (where $PQ$ and $RS$ meet at $A$, and $PS$ and $QR$ meet at $B$), let the diagonals $PR$ and $QS$ intersect at $C$. By Brocard's theorem, $A, B, C, O$ form an orthocentric system, so $C$ is a fixed point. The existence of a good quadrilateral implies that $A, B, C, O$ are pairwise distinct. By polar line properties, the polar line of $C$ with respect to $\\omega$ is the line $AB$. Since $C$ lies inside $\\omega$, $AB$ is disjoint from $\\omega$.\nLet $X$ and $Y$ be the midpoints of diagonals $PR$ and $QS$, respectively. We have:\n$$\n\\begin{aligned}\n4[AXY] &= 2[APY] + 2[ARY] = [APQ] + [APS] + [ARQ] + [ARS] \\\\\n&= [APS] + [ARQ] = [PSRQ],\n\\end{aligned}\n$$\n(where $[AXY]$ denotes the signed area of triangle $AXY$, and similarly for others) which shows that the area of the good quadrilateral equals four times the area of triangle $AXY$. Thus, the problem reduces to proving that the maximum area of triangle $AXY$ exists and is achieved uniquely.\nLet $M$ be the midpoint of $AB$. By properties of the Newton line, $M, X, Y$ are collinear. Let $\\omega'$ be the circle with diameter $OC$. Then $X$ and $Y$ lie on $\\omega'$. Conversely, if a line through $M$ intersects $\\omega'$ at $X$ and $Y$, and the lines $CX$ and $CY$ (if one of $X, Y$ coincides with $C$, replace the corresponding line with the line through $C$ perpendicular to $OC$) intersect $\\omega$ at $P, R$ and $Q, S$ respectively, then the lines $PQ$ and $RS$ meet at $A'$, and the lines $PS$ and $QR$ meet at $B'$. By polar line properties, $A'$ and $B'$ lie on $AB$, the polar of $C$ with respect to $\\omega$. Since $OX \\perp XC$ and $OY \\perp YC$, $X$ and $Y$ are the midpoints of $PR$ and $QS$ respectively (even if $X$ or $Y$ coincides with $O$, the conclusion holds). By the Newton line property, $XY$ passes through the midpoint of $A'B'$, so the midpoint of $A'B'$ is also $M$.\nBy polar properties, $\\overrightarrow{OA} \\cdot \\overrightarrow{OB} = \\overrightarrow{OA'} \\cdot \\overrightarrow{OB'} = |OP|^2$. Moreover,\n$$\n\\overrightarrow{OA} \\cdot \\overrightarrow{OB} = (\\overrightarrow{OM} + \\overrightarrow{MA}) \\cdot (\\overrightarrow{OM} - \\overrightarrow{MA}) = |OM|^2 - |MA|^2,\n$$\nand similarly, $\\overrightarrow{OA'} \\cdot \\overrightarrow{OB'} = |OM|^2 - |MA'|^2$. Thus, $|MA| = |MA'|$, implying that $A'$ and $B'$ coincide with $A$ and $B$ (possibly in reverse order). Hence, the quadrilateral $PQRS$ constructed this way is necessarily a good quadrilateral, and distinct unordered pairs $(X, Y)$ correspond to distinct good quadrilaterals.\nThe problem now reduces to proving: Among all lines $l$ through $M$ intersecting $\\omega'$ at $X$ and $Y$, there exists a unique line $l$ that maximizes the area of triangle $AXY$.\nLet $F$ be the midpoint of $OC$ (i.e., the center of $\\omega'$), and let $m$ be the line through $F$ perpendicular to $MA$. Let $X_1$ and $Y_1$ be the projections of $X$ and $Y$ onto $m$. Then:\n$$\nS_{AXY} = \\frac{1}{2}XY \\cdot d(A, XY) = \\frac{1}{2}XY \\cdot AM \\cdot \\sin \\angle AMX = \\frac{1}{2}AM \\cdot X_1Y_1.\n$$\nThus, the problem further reduces to proving that there exists a unique line $l$ through $M$ such that the projection of its chord $XY$ with $\\omega'$ onto $m$ has maximal length.\nWe first prove the following claim: If a line $l$ through $M$ intersects $\\omega'$ at $X$ and $Y$, and the segment $XY$ meets $m$ at $N$ satisfying\n$$\n\\frac{XN}{YN} = \\frac{YM}{XM}, \\qquad (\\star)\n$$\nthen $l$ is the desired line.\n![](attached_image_1.png)\n\nIndeed, let the tangents to $\\omega'$ at $X$ and $Y$ meet at $T$. Consider another line $l'$ through $M$ intersecting $\\omega'$ at $X'$ and $Y'$. Suppose $l'$ intersects segments $XT$ and $YT$ at $U$ and $V$, respectively. It suffices to show that the projection of $UV$ onto $m$ is shorter than that of $XY$, i.e., $XU \\sin \\angle XFN > YV \\sin \\angle YFN$.\nNote that $\\frac{\\sin \\angle XFN}{\\sin \\angle YFN} = \\frac{XN}{YN} = \\frac{MY}{MX}$. By Menelaus' theorem, $\\frac{MY}{MX} \\cdot \\frac{XU}{YV} = \\frac{TU}{TV} > 1$, so $XU \\sin \\angle XFN > YV \\sin \\angle YFN$. The case where $l'$ intersects the extensions of $TX$ and $TY$ is similar.\nFinally, we prove that a line $l$ satisfying $(\\star)$ exists and is unique. Existence suffices, as uniqueness would otherwise lead to a contradiction.\n![](attached_image_2.png)\nLet $\\Omega$ be the circle with diameter $MF$, intersecting $\\omega'$ at $K$ and $L$. Then $MK$ and $ML$ are tangent to $\\omega'$. Let $KL$ meet $m$ at $J$. Then $MJ$ harmonically divides $XY$, i.e., $\\frac{JX}{JY} = \\frac{MX}{MY}$. Thus, $\\frac{XN}{YN} = \\frac{YM}{XM}$ is equivalent to $J$ and $N$ being symmetric about the midpoint $I$ of $XY$. By the perpendicular bisector theorem, $I$ also lies on $\\Omega$. The problem now reduces to showing there exists a unique line $l$ through $M$ intersecting $KL, \\Omega$, and $m$ at $J, I, N$ respectively, with $JI = IN$.\nWhen $l = ML, JI = 0 < IN$; when $l = MF, IN = 0 < JI$. By continuity, such an $l$ exists. This completes the proof. $\\square$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 28654,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all four-digit perfect squares of the form $\\overline{aabb}$.",
    "options": [],
    "answer": "7744",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 28655,
    "subject": "Mathematics (Multi-modal)",
    "question": "If\n$$\n\\frac{x_1}{x_1 + 1} = \\frac{x_2}{x_2 + 3} = \\frac{x_3}{x_3 + 5} = \\dots = \\frac{x_{1006}}{x_{1006} + 2011}, \\\\\nx_1 + x_2 + \\dots + x_{1006} = 503^2,\n$$\ndetermine $x_{1006}$.",
    "options": [],
    "answer": "2011/4",
    "solution": "Let us denote\n$$\n\\frac{x_1}{x_1 + 1} = \\frac{x_2}{x_2 + 3} = \\frac{x_3}{x_3 + 5} = \\dots = \\frac{x_{1006}}{x_{1006} + 2011} = a.\n$$\nFrom $\\frac{x_k}{x_k + (2k-1)} = a$ follows $x_k = \\frac{a}{1-a} \\cdot (2k-1)$ for $k = 1, 2, \\dots, 1006$. By including this in the last given equality we get\n$$\n\\frac{a}{1-a} \\cdot (1+3+\\dots+2011) = 503^2.\n$$\nAs $1+3+5+\\dots+2011 = 503 \\cdot 2012 = 1006^2$, we get $\\frac{a}{1-a} \\cdot 1006^2 = 503^2$, so $\\frac{a}{1-a} = \\frac{1}{4}$. Finally, $x_{1006} = \\frac{a}{1-a} \\cdot 2011 = \\frac{2011}{4}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 28656,
    "subject": "Mathematics (Multi-modal)",
    "question": "If one half of a working crew finishes one quarter of a job in one third of a day, how many crews does it take to finish 15 jobs in five days?",
    "options": [],
    "answer": "2",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 28657,
    "subject": "Mathematics (Multi-modal)",
    "question": "Outside a regular polygon $A_1A_2\\dots A_n$ a point $B$ is given so that $A_1A_2B$ is a regular triangle. Determine all $n$ such that points $B$, $A_2$ and $A_3$ are consecutive vertices of some regular polygon.",
    "options": [],
    "answer": "n ∈ {3, 4, 5, 7, 8, 9, 10, 12, 15, 18, 24, 42}",
    "solution": "Let the new polygon have $m$ vertices. Two cases are possible:\n![](attached_image_1.png)\n![](attached_image_2.png)\n\n**Case 1.** The new polygon lies outside the given polygon. In other words, $m$-gon and $n$-gon are on opposite sides of the line $A_2A_3$.\nIn this case $\\angle BA_2A_3 + \\angle A_1A_2A_3 + 60^\\circ = 360^\\circ$.\n$$\n\\frac{n-2}{n} \\cdot 180^\\circ + \\frac{m-2}{m} \\cdot 180^\\circ + 60^\\circ = 360^\\circ\n$$\n$$\n3m(n-2) + 3n(m-2) + mn = 6mn\n$$\n$$\nmn - 6m = 6n\n$$\n$$\nm = \\frac{6n}{n-6} = 6 + \\frac{36}{n-6}.\n$$\nObviously $n-6 \\in \\mathbb{N}$ and $n-6$ divides $36$.\nChecking all the possibilities we find nine solutions:\n\n| n - 6 | 1 | 2 | 3 | 4 | 6 | 9 | 12 | 18 | 36 |\n|-------|---|---|---|---|---|----|----|----|----|\n| n     | 7 | 8 | 9 | 10| 12| 15 | 18 | 24 | 42 |\n| m     |42 |24 |18 |15 |12 | 10 | 9  | 8  | 7  |\n\nCase 2. The observed $m$-gon and $n$-gon are on the same side of the line $A_2A_3$.\nIn this case we have $\\angle BA_2A_3 = \\angle A_1A_2A_3 + 60^\\circ$\n$$\n\\begin{aligned}\n\\frac{m-2}{m} \\cdot 180^\\circ &= 60^\\circ + \\frac{n-2}{n} \\cdot 180^\\circ \\\\\n3n(m-2) &= mn + 3m(n-2) \\\\\n6n + nm &= 6m \\\\\nn &= \\frac{6m}{m+6} = 6 - \\frac{36}{m+6}.\n\\end{aligned}\n$$\nSince $n \\ge 3$, $\\frac{36}{m+6}$ must belong to $\\{1, 2, 3\\}$. From there we get another three solutions\n$$\n\\begin{array}{c|ccc}\nm+6 & 36 & 18 & 12 \\\\\nm & 30 & 12 & 6 \\\\\nn & 5 & 4 & 3\n\\end{array}\n$$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 28658,
    "subject": "Mathematics (Multi-modal)",
    "question": "A triangle $ABC$ with $\\angle BAC = 120^\\circ$ is given. Let $D$ be the point in its interior such that $\\angle DBC = 2\\angle ABD$ and $\\angle DCB = 2\\angle ACD$. Determine $\\angle BDC$.",
    "options": [],
    "answer": "140°",
    "solution": "",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 28659,
    "subject": "Mathematics (Multi-modal)",
    "question": "Prove that the product of any two elements of the set\n$$\n\\{m \\mid m = a^2 - 5b^2,\\ a, b \\in \\mathbb{N}\\}\n$$\nbelongs to that set as well.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 28660,
    "subject": "Mathematics (Multi-modal)",
    "question": "Four integers $a, b, c, d$ satisfy $a + b = c$, $a + d = 2c$. Prove that there exists a rectangular triangle with integer sides whose area equals $abcd$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "From the given equations we have $a = c - b$, $d = 2c - a = b + c$. So\n$$\nabcd = (c-b) \\cdot b \\cdot c \\cdot (b+c) = bc(c^2-b^2) = \\frac{1}{2}(2bc)(c^2-b^2).\n$$\nTherefore $abcd$ is the area of a right triangle with legs $2bc$ and $c^2 - b^2$.\nThe hypotenuse of that triangle equals\n$$\n\\sqrt{(2bc)^2 + (c^2 - b^2)^2} = \\sqrt{c^4 + b^4 + 2b^2c^2} = b^2 + c^2.\n$$\nAs $b$ and $c$ are positive integers and $c > b$ (since $c - b = a > 0$), the lengths of all three sides, $2bc$, $c^2 - b^2$ and $b^2 + c^2$ are positive integers.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 28661,
    "subject": "Mathematics (Multi-modal)",
    "question": "Today it's Marko's birthday. His father and grandfather have this conversation:\n– Now Marko's, your and my age are all prime numbers!\n– Yes, and five years from now all the three ages will be perfect squares.\nHow old was his grandfather when Marko was born?",
    "options": [],
    "answer": "48",
    "solution": "",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 28662,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $a$, $b$, $c \\in \\mathbb{R}$ satisfy $a + b + c = 3$ and $\\frac{1}{a} + \\frac{1}{b} + \\frac{1}{c} = 0$. Determine $a^2 + b^2 + c^2$.",
    "options": [],
    "answer": "9",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 28663,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABCD$ be a cyclic quadrangle. The perpendicular bisector of $\\overline{BC}$ meets the side $\\overline{AB}$ in point $E$. The circle through $E$, $C$ and the midpoint $F$ of the side $\\overline{BC}$ meets the side $\\overline{CD}$ in $G$. Prove that $AD \\perp FG$. (D. Monk)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let the lines $AD$ and $FG$ intersect at point $H$.\n\n![](attached_image_1.png)\n\nAs quadrilaterals $ABCD$ and $EFCG$ are cyclic, it follows that\n$$\n\\begin{align*}\n\\angle HDG &= 180^\\circ - \\angle ADC = \\angle ABC, \\\\\n\\angle HGD &= \\angle FGC = \\angle FEC = \\angle FEB = 90^\\circ - \\angle ABC.\n\\end{align*}\n$$\nTherefore\n$$\n\\angle DHG = 180^\\circ - \\angle HDG - \\angle HGD = 90^\\circ\n$$\nso $AD \\perp FG$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 28664,
    "subject": "Mathematics (Multi-modal)",
    "question": "One of the midlines in a triangle is longer than one of its medians. Prove that the triangle is obtuse.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 28665,
    "subject": "Mathematics (Multi-modal)",
    "question": "Two players play the following game: after one of them tells number $n$, the other has to tell a number of the form $a \\cdot b$ where $a, b$ are positive integers such that $a + b = n$. The game continues in the same way. If at some point one of the players has told $2011$, which are the possible numbers that the game could have started with? (Tournament of Towns 2001)",
    "options": [],
    "answer": "All integers n ≥ 5",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 28666,
    "subject": "Mathematics (Multi-modal)",
    "question": "What is the minimal number of elements that should be removed from the set $\\{2, 4, 6, 8, 10, 12, 14, 16\\}$ so that the product of the remaining elements would be a perfect square?",
    "options": [],
    "answer": "3",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 28667,
    "subject": "Mathematics (Multi-modal)",
    "question": "Two players play the following game: after one of them tells number $n$, the other has to tell a number of the form $a \\cdot b$ where $a, b$ are positive integers such that $a + b = n$. The game continues in the same way. If at some point one of the players has told $2011$, which are the possible numbers that the game could have started with?\n\n(Tournament of Towns 2001)",
    "options": [],
    "answer": "All integers greater than or equal to 5",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 28668,
    "subject": "Mathematics (Multi-modal)",
    "question": "Spouses Ana and Tomislav attended a party with four other couples. Upon arrival there was a certain number of handshaking, but no one shook hands with his or her spouse nor with himself/herself. When later Tomislav asked everyone at the party how many people they shook hands with, he got nine different answers. How many people did Ana shake hands with? (www.masa.on.net)",
    "options": [],
    "answer": "4",
    "solution": "No one shook hands with more than eight people so the responses that Tomislav got are: \"0\", \"1\", \"2\", \"3\", \"4\", \"5\", \"6\", \"7\", \"8\".\n\nAna certainly did not shake hands with eight people. Namely, if we assume that she did shake hands with eight people, all the others (except Tomislav) had to shake hands with her, so no response would be \"0\".\n\nLet the person who shook hands with eight people be $A_1$. We can conclude that the only person who could shake hands with zero people is the spouse of $A_1$. Let's call this person $A_2$.\n\nIf now we assume that Ana shook hands with seven people, then all the others except Tomislav and person $A_2$ had to shake hands with Ana, so no response would be \"1\" (since all of them also shook hands with $A_1$).\n\nLet the person who shook hands with seven people be $B_1$. Again we conclude that the person who gave the response \"1\" is the spouse of $B_1$ because all the others shook hands with person $A_1$ and person $B_1$. Let's call this person $B_2$.\n\nBy similar reasoning we conclude that spouses $C_1$ and $C_2$ gave responses \"6\" and \"2\", and spouses $D_1$ and $D_2$ gave responses \"5\" and \"3\".\n\nFinally we conclude that Ana gave the response \"4\", i.e. she shook hands with four people.\n\nThe graph below shows that this is really possible:\n\n![](attached_image_1.png)",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 28669,
    "subject": "Mathematics (Multi-modal)",
    "question": "The sum of lengths of legs of the trapezoid is $4\\sqrt{10}$, and its height is $6$. The area of the trapezoid is $72$. If a circle can be circumscribed to the trapezoid, determine its radius.",
    "options": [],
    "answer": "5√2",
    "solution": "Solution:\n\nLet the trapezoid have bases $a$ and $b$ ($a > b$), and legs $c$ and $d$. Since a circle can be circumscribed about the trapezoid, it must be tangential, so $a + b = c + d$.\n\nGiven:\n- $c + d = 4\\sqrt{10}$\n- $a + b = 4\\sqrt{10}$\n- Height $h = 6$\n- Area $S = 72$\n\nThe area of a trapezoid is $S = \\dfrac{(a + b)h}{2}$.\nSo:\n$$\n72 = \\dfrac{(a + b) \\cdot 6}{2} \\implies a + b = \\dfrac{72 \\times 2}{6} = 24\n$$\nBut $a + b = 4\\sqrt{10}$. So $4\\sqrt{10} = 24 \\implies \\sqrt{10} = 6$, which is not true. Therefore, the sum $a + b = 24$ and $c + d = 4\\sqrt{10}$.\n\nBut since the circle can be circumscribed, $a + b = c + d$, so $a + b = c + d = 24$.\n\nTherefore, $4\\sqrt{10} = 24 \\implies \\sqrt{10} = 6$, which is not true. So the problem must mean that the sum of the legs is $4\\sqrt{10}$, and the sum of the bases is $24$.\n\nLet the bases be $a$ and $b$, with $a > b$, and the legs be $c$ and $d$.\n\nLet us denote the bases as $a$ and $b$, and the legs as $c$ and $d$.\n\nLet us drop perpendiculars from the endpoints of the shorter base to the longer base. Let the projections from the endpoints of the shorter base to the longer base be $x$ and $y$, so $x + b + y = a$.\n\nBy the Pythagorean theorem:\n$$\nc = \\sqrt{x^2 + h^2}, \\quad d = \\sqrt{y^2 + h^2}\n$$\nSo:\n$$\nc + d = \\sqrt{x^2 + 36} + \\sqrt{y^2 + 36} = 4\\sqrt{10}\n$$\nAlso, $x + y = a - b$.\n\nBut $a + b = 24$, so $a - b = x + y$.\n\nLet $s = x + y = a - b$.\n\nLet us try to find $a$ and $b$.\n\nLet $a + b = 24$, $a - b = s \\implies a = \\dfrac{24 + s}{2}, \\ b = \\dfrac{24 - s}{2}$.\n\nThe area is also:\n$$\nS = \\dfrac{(a + b)h}{2} = \\dfrac{24 \\times 6}{2} = 72\n$$\nWhich matches the given area.\n\nNow, $c + d = 4\\sqrt{10}$.\n\nLet us try to find $x$ and $y$ such that $\\sqrt{x^2 + 36} + \\sqrt{y^2 + 36} = 4\\sqrt{10}$ and $x + y = s$.\n\nLet us try $x = y$ for simplicity.\n\nThen $x + y = s \\implies x = y = \\dfrac{s}{2}$.\n\nThen:\n$$\n2\\sqrt{x^2 + 36} = 4\\sqrt{10} \\implies \\sqrt{x^2 + 36} = 2\\sqrt{10} \\implies x^2 + 36 = 4 \\times 10 = 40 \\implies x^2 = 4 \\implies x = 2\n$$\nSo $x = y = 2$, $s = 4$.\n\nTherefore,\n$$\na = \\dfrac{24 + 4}{2} = 14, \\quad b = \\dfrac{24 - 4}{2} = 10\n$$\n\nSo the bases are $14$ and $10$, the legs are both $\\sqrt{2^2 + 36} = \\sqrt{40} = 2\\sqrt{10}$, so $c + d = 4\\sqrt{10}$, as required.\n\nNow, the trapezoid is isosceles, so the circle can be circumscribed.\n\nThe radius $R$ of the circle circumscribed about a tangential quadrilateral (which is also cyclic) is given by:\n$$\nR = \\dfrac{1}{4} \\sqrt{\\dfrac{(ab + cd)(ac + bd)(ad + bc)}{(s - a)(s - b)(s - c)(s - d)}}\n$$\nwhere $a, b, c, d$ are the sides, and $s = \\dfrac{a + b + c + d}{2}$.\n\nLet us label the sides in order: $a = 14$, $b = 2\\sqrt{10}$, $c = 10$, $d = 2\\sqrt{10}$.\n\nBut in the isosceles trapezoid, the order is $AB = 14$, $BC = 2\\sqrt{10}$, $CD = 10$, $DA = 2\\sqrt{10}$.\n\nSo $a = 14$, $b = 2\\sqrt{10}$, $c = 10$, $d = 2\\sqrt{10}$.\n\nThen $s = \\dfrac{14 + 10 + 2\\sqrt{10} + 2\\sqrt{10}}{2} = \\dfrac{24 + 4\\sqrt{10}}{2} = 12 + 2\\sqrt{10}$.\n\nNow, the area $S$ of a cyclic quadrilateral is:\n$$\nS = \\sqrt{(s - a)(s - b)(s - c)(s - d)}\n$$\nBut $S = 72$.\n\nSo:\n$$\n72 = \\sqrt{(s - a)(s - b)(s - c)(s - d)}\n$$\nLet us compute each term:\n\n$s - a = (12 + 2\\sqrt{10}) - 14 = -2 + 2\\sqrt{10}$\n\n$s - b = (12 + 2\\sqrt{10}) - 2\\sqrt{10} = 12$\n\n$s - c = (12 + 2\\sqrt{10}) - 10 = 2 + 2\\sqrt{10}$\n\n$s - d = (12 + 2\\sqrt{10}) - 2\\sqrt{10} = 12$\n\nSo:\n$$\n(s - a)(s - b)(s - c)(s - d) = [-2 + 2\\sqrt{10}] \\times 12 \\times [2 + 2\\sqrt{10}] \\times 12\n$$\n$= 12 \\times 12 \\times [-2 + 2\\sqrt{10}] \\times [2 + 2\\sqrt{10}]$\n\nNow,\n$[-2 + 2\\sqrt{10}] \\times [2 + 2\\sqrt{10}] = (-2)(2) + (-2)(2\\sqrt{10}) + (2\\sqrt{10})(2) + (2\\sqrt{10})(2\\sqrt{10})$\n$= -4 - 4\\sqrt{10} + 4\\sqrt{10} + 4 \\times 10$\n$= -4 + 40 = 36$\n\nSo the product is $12 \\times 12 \\times 36 = 144 \\times 36 = 5184$\n\nSo $S = \\sqrt{5184} = 72$, as required.\n\nNow, the radius $R$ of the circumscribed circle about a cyclic quadrilateral is:\n$$\nR = \\dfrac{1}{4} \\sqrt{\\dfrac{(ab + cd)(ac + bd)(ad + bc)}{(s - a)(s - b)(s - c)(s - d)}}\n$$\nLet us compute each term:\n\n$ab + cd = (14)(2\\sqrt{10}) + (10)(2\\sqrt{10}) = 28\\sqrt{10} + 20\\sqrt{10} = 48\\sqrt{10}$\n\n$ac + bd = (14)(10) + (2\\sqrt{10})(2\\sqrt{10}) = 140 + 4 \\times 10 = 140 + 40 = 180$\n\n$ad + bc = (14)(2\\sqrt{10}) + (10)(2\\sqrt{10}) = 28\\sqrt{10} + 20\\sqrt{10} = 48\\sqrt{10}$\n\nSo the numerator is:\n$(ab + cd)(ac + bd)(ad + bc) = [48\\sqrt{10}] \\times 180 \\times [48\\sqrt{10}]$\n\n$= (48\\sqrt{10})^2 \\times 180 = 2304 \\times 10 \\times 180 = 23040 \\times 180 = 4,147,200$\n\nThe denominator is $(s - a)(s - b)(s - c)(s - d) = 5184$\n\nSo:\n$$\nR = \\dfrac{1}{4} \\sqrt{\\dfrac{4,147,200}{5184}}\n$$\n\nNow, $4,147,200 \\div 5184 = 800$\n\nSo:\n$$\nR = \\dfrac{1}{4} \\sqrt{800} = \\dfrac{1}{4} \\times 20\\sqrt{2} = 5\\sqrt{2}\n$$\n\n**Answer:**\n\nThe radius of the circumscribed circle is $5\\sqrt{2}$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 28670,
    "subject": "Mathematics (Multi-modal)",
    "question": "How many positive integers less than $2011$ are divisible by either $2$ or $7$, but are not divisible by $5$?",
    "options": [],
    "answer": "919",
    "solution": "Let $N = 2011$.\n\nLet $A$ be the set of positive integers less than $2011$ divisible by $2$ or $7$.\nLet $B$ be the set of positive integers less than $2011$ divisible by $5$.\nWe are to find $|A \\setminus B|$.\n\nFirst, count the number of positive integers less than $2011$ divisible by $2$ or $7$.\n\nLet $S_2$ be the set divisible by $2$.\nLet $S_7$ be the set divisible by $7$.\nLet $S_{14}$ be the set divisible by $14$.\n\nNumber divisible by $2$: $\\left\\lfloor \\dfrac{2010}{2} \\right\\rfloor = 1005$\nNumber divisible by $7$: $\\left\\lfloor \\dfrac{2010}{7} \\right\\rfloor = 287$\nNumber divisible by $14$: $\\left\\lfloor \\dfrac{2010}{14} \\right\\rfloor = 143$\n\nBy inclusion-exclusion:\nNumber divisible by $2$ or $7$ is $1005 + 287 - 143 = 1149$.\n\nNow, subtract those divisible by $5$.\n\nLet $S_{2,5}$: divisible by $2$ and $5$ ($10$)\n$\\left\\lfloor \\dfrac{2010}{10} \\right\\rfloor = 201$\n\nLet $S_{7,5}$: divisible by $7$ and $5$ ($35$)\n$\\left\\lfloor \\dfrac{2010}{35} \\right\\rfloor = 57$\n\nLet $S_{14,5}$: divisible by $14$ and $5$ ($70$)\n$\\left\\lfloor \\dfrac{2010}{70} \\right\\rfloor = 28$\n\nBy inclusion-exclusion, the number of integers less than $2011$ divisible by $2$ or $7$ and also by $5$ is:\n$201 + 57 - 28 = 230$\n\nTherefore, the answer is $1149 - 230 = 919$.\n\n**Answer:** $919$",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 28671,
    "subject": "Mathematics (Multi-modal)",
    "question": "Prove that no odd integers $x, y, z$ can satisfy $(x-z)^2 + (y-z)^2 = (x+y)^2$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 28672,
    "subject": "Mathematics (Multi-modal)",
    "question": "Determine all pairs $(m, n)$ of positive integers so that $n \\mid 2m-1$ and $m \\mid 2n-1$.",
    "options": [],
    "answer": "(1, 1), (3, 5), (5, 3)",
    "solution": "We can conclude that there exist $k, l \\in \\mathbb{N}$ such that $2m-1 = k n$ and $2n-1 = l m$. It follows that\n$$\n4n - 2 = 2(2n - 1) = 2 \\cdot l m = 2m \\cdot l = (n k + 1) l.\n$$\nFrom $4n - 2 = (n k + 1) l$ we get\n$$\n(4 - k l) n = l + 2.\n$$\nSince the right hand side is positive, it follows that $4 - k l > 0$, that is $k l < 4$. Since $k$ and $l$ are positive integers, we have the following possibilities:\n\na. $k l = 1$, that is $k = l = 1$. We get $2m - 1 = n$, $2n - 1 = m$, from which follows $m = n = 1$.\n\nb. $k l = 2$. That is not possible, since from $2m - 1 = k n$, $2n - 1 = l m$ it follows that $k$ and $l$ are both odd.\n\nc. $k l = 3$, that is $k = 3$, $l = 1$ or $k = 1$, $l = 3$. In the first case from $2m - 1 = 3n$, $2n - 1 = m$ we get $m = 5$, $n = 3$. Analogously, in the second case we get $m = 3$, $n = 5$.\n\nHence, all the solutions are $(m, n) \\in \\{(1, 1), (3, 5), (5, 3)\\}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 28673,
    "subject": "Mathematics (Multi-modal)",
    "question": "Determine all the positive integers $n$ such that $\\frac{n-1}{n-5}$ is an integer.",
    "options": [],
    "answer": "1, 3, 4, 6, 7, 9",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 28674,
    "subject": "Mathematics (Multi-modal)",
    "question": "Determine all real solutions of the system of equations:\n$$\n\\begin{cases} 2a^2 - 2ab + b^2 = a, \\\\ 4a^2 - 5ab + 2b^2 = b. \\end{cases}\n$$",
    "options": [],
    "answer": "(a, b) = (0, 0) and (1, 1)",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 28675,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $\\overline{AB}$ be a chord of a circle, parallel to its diameter $\\overline{MN}$. Let $t$ be the tangent of that circle in $M$ and let $C, D$ be the intersections of the lines $NA, NB$ with $t$, respectively. Prove\n$$\n|MC| \\cdot |MD| = |MN|^2.\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 28676,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABCD$ be a unit square. The unit circle $k$ has center $C$. Determine the radius of the circle $k_1$ which touches the circle $k$ and line segments $\\overline{AB}$ and $\\overline{AD}$.",
    "options": [],
    "answer": "3 - 2√2",
    "solution": "",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 28677,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABCD$ be a unit square. The unit circle $k$ has center $C$. Determine the radius of the circle $k_1$ which touches the circle $k$ and line segments $\\overline{AB}$ and $\\overline{AD}$.",
    "options": [],
    "answer": "3 - 2√2",
    "solution": "",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 28678,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $a$, $b$ be real numbers such that all the zeros of the polynomial $P(x) = x^3 + a x^2 + b x - 8$ are real. Prove that $a^2 \\ge 2b + 12$. (Kristina Ana Škreb)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Polynomial $P(x)$ has three zeros, let us denote them by $x_1$, $x_2$ and $x_3$.\nAccording to Viète's formulas we have:\n$$\n\\begin{aligned}\nx_1 + x_2 + x_3 &= -a \\\\\nx_1 x_2 + x_2 x_3 + x_3 x_1 &= b \\\\\nx_1 x_2 x_3 &= 8.\n\\end{aligned}\n$$\nIt follows $x_1^2 + x_2^2 + x_3^2 = a^2 - 2b$, and from the A-G inequality we get:\n$$\na^2 - 2b = x_1^2 + x_2^2 + x_3^2 \\ge 3 \\sqrt[3]{x_1^2 x_2^2 x_3^2} = 3 \\sqrt[3]{64} = 12,\n$$\nwhich finishes the proof.",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 28679,
    "subject": "Mathematics (Multi-modal)",
    "question": "Determine all $a$ such that there exists unique $(x, y) \\in \\mathbb{R}^2$ satisfying\n$$\n2^{|x|} + |x| = x^2 + y + a, \\quad x^2 + y^2 = 1.\n$$",
    "options": [],
    "answer": "a = 0",
    "solution": "If the pair $(x, y)$ is the solution of the given system, then the pair $(-x, y)$ is also the solution. We conclude that the unique solution of this system has to be of the form $(0, y)$.\n\nTaking $x = 0$ in the given system we get\n$$\n\\begin{aligned}\n1 &= y + a, \\\\\ny^2 &= 1,\n\\end{aligned}\n$$\nso $y = 1$ or $y = -1$, and hence $a = 0$ or $a = 2$.\n\nWe first consider the case $a = 0$. The initial system is reduced to\n$$2^{|x|} + |x| = x^2 + y$$\n$$x^2 + y^2 = 1.$$\nIt is easy to see that $(0, 1)$ is the solution. Let us show that there are no other solutions.\nFrom the second equation it follows that $|x| \\le 1$ and $|y| \\le 1$. Because of $0 \\le |x| \\le 1$ we get that $|x| \\ge x^2$.\nWe also get $2^{|x|} \\ge 2^0 = 1 \\ge |y| \\ge y$, so $2^{|x|} + |x| \\ge x^2 + y$.\nFor the equality to hold, we must have $|x| = 0$ and $|y| = y$, that is $x = 0$ and $y \\ge 0$. Because of the second equation it follows that $(x, y) = (0, 1)$, so this system has no other solutions.\nIn the case $a = 2$ we can notice that the system\n$$2^{|x|} + |x| = x^2 + y + 2$$\n$$x^2 + y^2 = 1$$\nhas (at least) three solutions $(0, -1)$, $(1, 0)$ $(-1, 0)$.\nWe conclude that the system has a unique solution if and only if $a = 0$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 28680,
    "subject": "Mathematics (Multi-modal)",
    "question": "Determine all $a \\in \\mathbb{R}$ such that for every $x \\in \\mathbb{R}$ the following inequality holds:\n$$\n\\frac{x}{x^2 + 2x + 3} > \\frac{x+a}{1+x+x^2}.\n$$",
    "options": [],
    "answer": "a ≤ -1",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 28681,
    "subject": "Mathematics (Multi-modal)",
    "question": "Lengths of all sides and diagonals of a rectangle are integers. Prove that its area is an integer divisible by $12$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 28682,
    "subject": "Mathematics (Multi-modal)",
    "question": "Determine all $a$ such that there exists unique $(x, y) \\in \\mathbb{R}^2$ satisfying\n$$\n2^{|x|} + |x| = x^2 + y + a, \\quad x^2 + y^2 = 1.\n$$",
    "options": [],
    "answer": "a = 0",
    "solution": "If the pair $(x, y)$ is the solution of the given system, then the pair $(-x, y)$ is also the solution. We conclude that the unique solution of this system has to be of the form $(0, y)$.\n\nTaking $x = 0$ in the given system we get\n$$\n\\begin{aligned}\n1 &= y + a, \\\\\ny^2 &= 1,\n\\end{aligned}\n$$\nso $y = 1$ or $y = -1$, and hence $a = 0$ or $a = 2$.\n\n$$2^{|x|} + |x| = x^2 + y$$\n$$x^2 + y^2 = 1.$$ \nIt is easy to see that $(0, 1)$ is the solution. Let us show that there are no other solutions.\nFrom the second equation it follows that $|x| \\le 1$ and $|y| \\le 1$. Because of $0 \\le |x| \\le 1$ we get that $|x| \\ge x^2$.\nWe also get $2^{|x|} \\ge 2^0 = 1 \\ge |y| \\ge y$, so $2^{|x|} + |x| \\ge x^2 + y$.\nFor the equality to hold, we must have $|x| = 0$ and $|y| = y$, that is $x = 0$ and $y \\ge 0$. Because of the second equation it follows that $(x, y) = (0, 1)$, so this system has no other solutions.\nIn the case $a = 2$ we can notice that the system\n$$2^{|x|} + |x| = x^2 + y + 2$$\n$$x^2 + y^2 = 1$$\nhas (at least) three solutions $(0, -1)$, $(1, 0)$, $(-1, 0)$.\nWe conclude that the system has a unique solution if and only if $a = 0$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 28683,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $x$, $y$, $z$ be positive real numbers such that $x^3 + y^3 + z^3 = 1$. Prove that\n$$\nx^2 + y^2 + z^2 > x^5 + y^5 + z^5 + 2x^2y^2z^2(x + y + z).\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 28684,
    "subject": "Mathematics (Multi-modal)",
    "question": "In the triangle $ABC$ with $|AB| = |AC|$ angle bisector of $\\angle ABC$ meets $\\overline{AC}$ in $D$. If $|BC| = |BD| + |AD|$, find all the angles. (Canada 1996)",
    "options": [],
    "answer": "40°, 40°, 100°",
    "solution": "![](attached_image_1.png)\n\nSince $BD$ is the bisector of the angle $\\angle CBA$, it follows that $|CD| : |AD| = |BC| : |AB|$, and because of that $\\frac{|CD|}{|CE|} = \\frac{|CD|}{|AD|} = \\frac{|BC|}{|AB|} = \\frac{|CB|}{|CA|}$.\n\nTriangles $ABC$ and $EDC$ have a common angle in vertex $C$ and equal ratios of corresponding sides so they are similar. Hence $\\angle CED = \\angle CAB$.\n\nLet $\\angle ABC = 2x$. Then $\\angle CBD = \\angle DBA = x$ and $\\angle ACB = 2x$. Furthermore, $\\angle CED = \\angle BAC = 180^\\circ - 4x$.\n\nSince the triangle $BED$ is isosceles, we have $\\angle DEB = 90^\\circ - \\frac{x}{2}$. As $\\angle DEB + \\angle CED = 180^\\circ$, we get\n$$\n(90^\\circ - \\frac{x}{2}) + (180^\\circ - 4x) = 180^\\circ,\n$$\nwhence follows $x = 20^\\circ$. The angles of the observed triangle are $\\angle ABC = \\angle ACB = 40^\\circ$ and $\\angle BAC = 100^\\circ$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 28685,
    "subject": "Mathematics (Multi-modal)",
    "question": "Determine all $z \\in \\mathbb{C}$ such that $|z| = |z+1| = \\left|\\frac{1}{z}\\right|$.",
    "options": [],
    "answer": "z = -1/2 ± i√3/2",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 28686,
    "subject": "Mathematics (Multi-modal)",
    "question": "Given nine positive integers, none of which has a prime divisor greater than $6$, prove that there are two of them whose product is a perfect square.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 28687,
    "subject": "Mathematics (Multi-modal)",
    "question": "A bag contained $255$ balls numbered $1, 2, \\ldots, 255$, and then $N$ students each took one ball from the bag. It turned out that none of the drawn numbers was exactly twice as big as any other drawn number. Determine the maximum possible $N$. (Ana Prlić)",
    "options": [],
    "answer": "170",
    "solution": "Group the observed numbers into sets:\n$$\nA_0 = \\{1\\}, A_1 = \\{2, 3\\}, A_2 = \\{4, 5, 6, 7\\}, A_3 = \\{8, 9, \\dots, 15\\}, A_4 = \\{16, 17, \\dots, 31\\}, \\\\\nA_5 = \\{32, 33, \\dots, 63\\}, A_6 = \\{64, 65, \\dots, 127\\}, A_7 = \\{128, 129, \\dots, 255\\}.\n$$\nThe number of elements of $A_k$ is $2^k$, for $k = 0, 1, \\dots, 7$.\nNotice that, for each $n$, numbers $n$ and $2n$ belong to $A_k$ and $A_{k+1}$ respectively for some $k$. If all numbers from the sets $A_1, A_3, A_5$ and $A_7$ are drawn, the number of these numbers is $2 + 8 + 32 + 128 = 170$ and none of them is twice as big as any other.\nLet us show that it is not possible to choose more than $170$ numbers that satisfy the condition. Let $a_k$ numbers from $A_k$ be drawn, for $k = 0, 1, 2, \\dots, 7$.\nObserve the sets $A_k$ and $A_{k+1}$. For every $m \\in A_k$, $2m$ is in $A_{k+1}$. The number of these pairs $(m, 2m)$ is $2^k$. Clearly at most one number is drawn from every pair. Except those numbers, set $A_{k+1}$ has another $2^k$ odd numbers, so at most $2^k + 2^k = 2^{k+1}$ numbers are drawn from $A_k$ and $A_{k+1}$.\nThat means that $a_0 + a_1 \\le 2^1, a_2 + a_3 \\le 2^3, a_4 + a_5 \\le 2^5, a_6 + a_7 \\le 2^7$.\nAdding up those inequalities we get $a_0 + a_1 + \\dots + a_7 \\le 170$. Thereby the maximum possible $N$ is $170$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 28688,
    "subject": "Mathematics (Multi-modal)",
    "question": "Mario has written a 30-digit number whose sum of digits is 123. Then he wrote all the digits again in some other order following the original number. Prove that the 60-digit number he obtained is not a perfect square.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 28689,
    "subject": "Mathematics (Multi-modal)",
    "question": "Mario has written a 30-digit number whose sum of digits is 123. Then he wrote all the digits again in some other order following the original number. Prove that the 60-digit number he obtained is not a perfect square.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 28690,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $a$, $b$, $c$ be different real numbers none of which is zero. Consider quadratic equations:\n$$\nax^2 + bx + c = 0, \\quad bx^2 + cx + a = 0, \\quad cx^2 + ax + b = 0.\n$$\nIf $\\frac{c}{a}$ is a root of the first equation, prove that all three of them have a common root. What is the product of the other roots of those equations (different from the common root)?",
    "options": [],
    "answer": "1",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 28691,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $a$, $b$, $c$ be different real numbers none of which is zero. Consider quadratic equations:\n$$\nax^2 + bx + c = 0, \\quad bx^2 + cx + a = 0, \\quad cx^2 + ax + b = 0.\n$$\nIf $\\frac{c}{a}$ is a root of the first equation, prove that all three of them have a common root. What is the product of the other roots of those equations (different from the common root)?",
    "options": [],
    "answer": "1",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 28692,
    "subject": "Mathematics (Multi-modal)",
    "question": "In the quadrilateral $ABCD$ the following holds:\n$$\n\\angle ABC = \\angle ADC = 90^\\circ, \\quad |AB| = |BC|, \\quad |CD| + |DA| = m.\n$$\nDetermine the area of $ABCD$ in terms of $m$.",
    "options": [],
    "answer": "m^2/4",
    "solution": "",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 28693,
    "subject": "Mathematics (Multi-modal)",
    "question": "In the quadrilateral $ABCD$ the following holds:\n$$\n\\angle ABC = \\angle ADC = 90^\\circ, \\quad |AB| = |BC|, \\quad |CD| + |DA| = m.\n$$\nDetermine the area of $ABCD$ in terms of $m$.",
    "options": [],
    "answer": "m^2/4",
    "solution": "",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 28694,
    "subject": "Mathematics (Multi-modal)",
    "question": "Ivan, Stipe and Tonći take turns in throwing a die. Ivan throws first, then Stipe, then Tonći, then Ivan again and so on in the same order. When it's their turn, everyone throws the die once, until they get a “six”. After getting his first “six”, in every following turn Ivan throws the die four times. After his first “six”, Stipe throws the die six times in each turn, and Tonći throws the die eight times.\nTonći was the last one to get his first “six”, on his tenth try, and then the game ended. If the die was thrown 47 times altogether, determine which one of them threw the die the most times.",
    "options": [],
    "answer": "Ivan",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 28695,
    "subject": "Mathematics (Multi-modal)",
    "question": "Given $\\log_a x = 3$, $\\log_{ab} x = 2$, determine $\\log_b x$.",
    "options": [],
    "answer": "6",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 28696,
    "subject": "Mathematics (Multi-modal)",
    "question": "Given $\\log_a x = 3$, $\\log_{ab} x = 2$, determine $\\log_b x$.",
    "options": [],
    "answer": "6",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 28697,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $\\triangle ABC$ be a right triangle with the right angle at vertex $C$ and let $M$ be the midpoint of the leg $\\overline{BC}$. Prove that $\\sin(\\angle MAB) \\le \\frac{1}{3}$. When is the equality achieved?",
    "options": [],
    "answer": "sin(angle MAB) ≤ 1/3, with equality when BC = √2 · CA (equivalently, tan(angle A) = √2).",
    "solution": "Since $\\angle MAB = \\angle CAB - \\angle CAM$, according to the sine addition theorem we have\n$$\n\\sin(\\angle MAB) = \\sin(\\angle CAB) \\cos(\\angle CAM) - \\cos(\\angle CAB) \\sin(\\angle CAM).\n$$\n![](attached_image_1.png)\nLet $|BC| = a$, $|CA| = b$, $|AB| = c$ and $|AM| = t$. From right triangles $ABC$ and $AMC$ we get\n$$\n\\sin(\\angle MAB) = \\frac{a}{c} \\cdot \\frac{b}{t} - \\frac{b}{c} \\cdot \\frac{a/2}{t} = \\frac{ab}{2ct}.\n$$\n\nSince $c^2 = a^2 + b^2$ and $t^2 = \\frac{a^2}{4} + b^2$, that is equivalent to\n$$\n9a^2b^2 \\le (a^2 + b^2)(a^2 + 4b^2)\n$$\ni.e. $(a^2 - 2b^2)^2 \\ge 0$. Now it is clear that the given inequality holds in every right triangle. The equality is achieved when $a = b\\sqrt{2}$.",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 28698,
    "subject": "Mathematics (Multi-modal)",
    "question": "If $\\cos \\gamma = 2 \\sin \\alpha \\sin \\beta - 1$, prove that the triangle with angles $\\alpha, \\beta, \\gamma$ is isosceles.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 28699,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all pairs $(x, y)$ of integers such that $x^2(y - 1) + y^2(x - 1) = 1$.",
    "options": [],
    "answer": "[(1, 2), (2, 1), (2, -5), (-5, 2)]",
    "solution": "The given equation is equivalent to\n$$\n\\begin{aligned}\nx^2y - x^2 + xy^2 - y^2 &= 1 \\\\\nxy(x + y) - (x^2 + y^2) &= 1 \\\\\nxy(x + y) - ((x + y)^2 - 2xy) &= 1\n\\end{aligned}\n$$\nIf we put $u = x + y$ and $v = xy$ we get the equation:\n$$\nuv - (u^2 - 2v) = 1\n$$\nfrom which follows $uv + 2v = u^2 + 1$, that is\n$$\nv = \\frac{u^2 + 1}{u + 2} = \\frac{u^2 - 4 + 5}{u + 2} = u - 2 + \\frac{5}{u + 2}.\n$$\nNow we can see that $u + 2$ has to be a divisor of $5$, so we have four cases:\n$$\n\\begin{array}{lcll}\nu + 2 = 5 & u + 2 = 1 & u + 2 = -1 & u + 2 = -5 \\\\\n\\frac{5}{u+2} = 1 & \\frac{5}{u+2} = 5 & \\frac{5}{u+2} = -5 & \\frac{5}{u+2} = -1 \\\\\nu = 3 & u = -1 & u = -3 & u = -7 \\\\\nv = 2 & v = 2 & v = -10 & v = -10\n\\end{array}\n$$\nFrom Viète's formulas we know that $x, y$ are the solutions of quadratic equations:\n$$\n\\begin{aligned}\nz^2 - 3z + 2 &= 0 \\\\\nz^2 + z + 2 &= 0 \\\\\nz^2 + 3z - 10 &= 0 \\\\\nz^2 + 7z - 10 &= 0\n\\end{aligned}\n$$\nThe solutions of the first equation are $1$ and $2$. The solutions of the second equation are not real because its discriminant is negative. The third equation has solutions $-5$ and $2$, while the solutions of the fourth equation are irrational because its discriminant $89$ is not a perfect square.\nFinally, the required solutions are $(x, y) \\in \\{(1, 2), (2, 1), (2, -5), (-5, 2)\\}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 28700,
    "subject": "Mathematics (Multi-modal)",
    "question": "Determine the largest possible quotient of a three-digit number and the sum of its digits.",
    "options": [],
    "answer": "100",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 28701,
    "subject": "Mathematics (Multi-modal)",
    "question": "Determine the largest possible quotient of a three-digit number and the sum of its digits.",
    "options": [],
    "answer": "100",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 28702,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $\\triangle ABC$ be a triangle. Angle bisector of $\\angle CAB$ intersects the segment $\\overline{BC}$ at $D$, and angle bisector of $\\angle ABC$ intersects the segment $\\overline{AC}$ at $E$. If $\\angle ACB \\ge 60^\\circ$, prove that $|AE| + |BD| \\le |AB|$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 28703,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $\\overline{AD}$ and $\\overline{BE}$ be altitudes of the triangle $ABC$. Given $|AE| = 5$, $|CE| = 3$ and $|CD| = 2$, determine $|BD|$.",
    "options": [],
    "answer": "10",
    "solution": "",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 28704,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABC$ be a triangle such that $|AB| = |AC|$. Let $D$ be a point on the side $AC$ such that $|AD| < |CD|$, and let $P$ be a point on the segment $BD$ such that $\\angle APC = 90^\\circ$. If $\\angle ABP = \\angle BCP$, determine $|AD| : |CD|$.\n(Stipe Vidak)",
    "options": [],
    "answer": "1:2",
    "solution": "Let us denote $\\angle ABP = \\angle BCP = \\varphi$ and $\\angle PBC = \\angle PCA = \\psi$.\nLet $T$ be the midpoint of the segment $AC$. Since $AC$ is the hypotenuse of the right triangle $APC$, it follows that $|AT| = |PT| = |CT|$. Hence $\\angle CPT = \\psi$.\n\n![](attached_image_1.png)\n\nSince $\\angle DPC$ is the exterior angle of the triangle $BCP$, we have\n$$\n\\angle DPC = \\angle PCB + \\angle PBC = \\varphi + \\psi\n$$\nso $\\angle DPT = \\varphi$.\n\nApplying the law of sines on triangles $ABD$ and $PDT$ we get\n$$\n\\frac{|AD|}{|AB|} = \\frac{\\sin \\angle ABD}{\\sin \\angle ADB}, \\quad \\frac{|DT|}{|PT|} = \\frac{\\sin \\angle DPT}{\\sin \\angle PDT}\n$$\nand then\n$$\n\\frac{|AD|}{|AB|} = \\frac{\\sin \\angle ABD}{\\sin \\angle ADB} = \\frac{\\sin \\varphi}{\\sin (180^\\circ - \\angle ADB)} = \\frac{\\sin \\angle DPT}{\\sin \\angle PDT} = \\frac{|DT|}{|PT|}\n$$\nNow it follows that\n$$\n\\frac{|AD|}{|AC|} = \\frac{|AD|}{|AB|} = \\frac{|DT|}{|PT|} = \\frac{|AT| - |AD|}{|PT|} = \\frac{\\frac{1}{2}|AC| - |AD|}{\\frac{1}{2}|AC|} = 1 - 2 \\cdot \\frac{|AD|}{|AC|}\n$$\nso $|AD| : |AC| = 1 : 3$. The desired ratio is $|AD| : |CD| = 1 : 2$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 28705,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $\\overline{AD}$ and $\\overline{BE}$ be altitudes of the triangle $ABC$. Given $|AE| = 5$, $|CE| = 3$ and $|CD| = 2$, determine $|BD|$.",
    "options": [],
    "answer": "10",
    "solution": "",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 28706,
    "subject": "Mathematics (Multi-modal)",
    "question": "Determine the minimal value of $\\sin(x + 3) - \\sin(x + 1) - 2\\cos(x + 2)$ if $x \\in \\mathbb{R}$.",
    "options": [],
    "answer": "2(sin 1 - 1)",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 28707,
    "subject": "Mathematics (Multi-modal)",
    "question": "Determine the greatest possible ratio of the volume of a sphere and the volume of a right circular cone inscribed in it.",
    "options": [],
    "answer": "27/8",
    "solution": "",
    "topic": "Geometry",
    "subtopic": "Solid Geometry"
  },
  {
    "id": 28708,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $a$, $b$, $c$ be positive real numbers such that $a + b + c = abc$. Prove\n$$\na^{5} (bc - 1) + b^{5} (ca - 1) + c^{5} (ab - 1) \\geq 54\\sqrt{3}.\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 28709,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $a$, $b$, $c$ be distinct positive integers and let $k$ be a positive integer such that\n$$\nab + bc + ca \\ge 3k^2 - 1.\n$$\nProve that $\\frac{1}{3}(a^3 + b^3 + c^3) \\ge abc + 3k$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "The desired inequality is equivalent to $a^3 + b^3 + c^3 - 3abc \\ge 9k$. We have\n$$\na^3 + b^3 + c^3 - 3abc = (a + b + c)(a^2 + b^2 + c^2 - ab - bc - ca).\n$$\nIntegers $|a-b|$, $|b-c|$ and $|c-a|$ can not all be equal $1$ so\n$$\n\\begin{aligned}\na^2 + b^2 + c^2 - ab - bc - ca &= \\frac{1}{2}((a-b)^2 + (b-c)^2 + (c-a)^2) \\\\\n&\\ge \\frac{1}{2}(1^2 + 1^2 + 2^2) = 3.\n\\end{aligned}\n$$\n\nIt follows that $a^3 + b^3 + c^3 - 3abc \\ge 3(a + b + c)$. Since\n$$\n\\begin{aligned}\n(a + b + c)^2 &= a^2 + b^2 + c^2 + 2ab + 2bc + 2ac \\\\\n&= (a^2 + b^2 + c^2 - ab - bc - ac) + 3(ab + bc + ac) \\\\\n&\\ge 3 + 3(3k^2 - 1) = 9k^2\n\\end{aligned}\n$$\nit follows that $a+b+c \\ge 3k$ so $a^3 + b^3 + c^3 - 3abc \\ge 9k$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 28710,
    "subject": "Mathematics (Multi-modal)",
    "question": "Determine the minimal value of $\\sin(x + 3) - \\sin(x + 1) - 2\\cos(x + 2)$ if $x \\in \\mathbb{R}$.",
    "options": [],
    "answer": "2(sin 1 - 1)",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 28711,
    "subject": "Mathematics (Multi-modal)",
    "question": "Put one of the four letters $A, B, C, D$ and one of the four numbers $1, 2, 3, 4$ into each cell of the $4 \\times 4$ table, so that:\n(a) in each row and in each column, each letter and each number appears exactly once,\n(b) each combination (pair) of one letter and one number appears in exactly one cell of the table.",
    "options": [],
    "answer": "One valid 4×4 arrangement is:\nRow 1: A1  B4  C2  D3\nRow 2: B2  A3  D1  C4\nRow 3: C3  D2  A4  B1\nRow 4: D4  C1  B3  A2",
    "solution": "",
    "topic": "Discrete Mathematics",
    "subtopic": "Combinatorics"
  },
  {
    "id": 28712,
    "subject": "Mathematics (Multi-modal)",
    "question": "Every cell of a $1000 \\times 1000$ table is colored black or white. The difference between the number of black and white cells is $2012$. Prove that there exists a $2 \\times 2$ square that contains an odd number of white cells.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Assume the contrary, that every $2 \\times 2$ square contains an even number of black cells. Let us compare two adjacent rows.\nIf the first cell in the lower row is the same color as the first cell in the upper row (e.g. black), then the second cells in these rows are also the same color (either black or white). In the same way we conclude further and see that these two rows are colored exactly the same.\nIf the first cell in the lower row is the opposite color from the first cells in the upper row, then the second cells in the lower row is also the opposite color from the second cells in the upper row. The same reasoning shows that the color of each cell in the lower row is the opposite color from the corresponding cell in the upper row.\nHence all the rows that begin with a black cell are equal, as are all the rows that begin with a white cell.\nLet $a$ be the number of rows that begin with a black cell. Then $1000 - a$ rows begin with a white cell. Let $d$ be the difference between the number of black and the number of white cells in rows that begin with a black cell. Then the difference between the number of black and the number of white cells in rows that begin with a white cell is $-d$.\nThe difference between the total number of black and the total number of white cells on the board is\n$$\na \\cdot d + (1000 - a) \\cdot (-d) = 2ad - 1000d = (2a - 1000)d.\n$$\nHence we have $(2a - 1000)d = 2012$.\nNotice that $d$ is even, since the total number of black and white cells in each row is $1000$. Also, it is clear that $d \\le 1000$. Therefore from $(a - 500)d = 2 \\cdot 503$ follows $d = 2$ so $a = 1003$ which is clearly not possible.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 28713,
    "subject": "Mathematics (Multi-modal)",
    "question": "Bisector of an acute angle in the right triangle divides the opposite side into parts of lengths $4$ and $5$. What is the area of that triangle?",
    "options": [],
    "answer": "54",
    "solution": "",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 28714,
    "subject": "Mathematics (Multi-modal)",
    "question": "Bisector of an acute angle in the right triangle divides the opposite side into parts of lengths $4$ and $5$. What is the area of that triangle?",
    "options": [],
    "answer": "54",
    "solution": "",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 28715,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $a$, $b$, $c$ be real numbers and let $a \\neq 0$. Prove that at least one of the equations\n$$\na \\sin x + b \\cos x + c = 0,\n$$\n$$\na \\tgg y + b \\ctg y + 2c = 0\n$$\nhas real solutions.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let us consider the first equation:\n$$\na \\sin x + b \\cos x + c = 0.\n$$\nThis can be rewritten as:\n$$\na \\sin x + b \\cos x = -c.\n$$\nLet $R = \\sqrt{a^2 + b^2}$ and let $\\varphi$ be such that $a = R \\sin \\varphi$, $b = R \\cos \\varphi$.\nThen:\n$$\na \\sin x + b \\cos x = R \\sin x \\sin \\varphi + R \\cos x \\cos \\varphi = R \\cos(x - \\varphi).\n$$\nSo the equation becomes:\n$$\nR \\cos(x - \\varphi) = -c.\n$$\nThis equation has a real solution for $x$ if and only if $|c| \\leq R$.\n\nSuppose $|c| > R$. Then the first equation has no real solution.\n\nNow consider the second equation:\n$$\na \\tgg y + b \\ctg y + 2c = 0.\n$$\nLet $t = \\tan y$, $t \\in \\mathbb{R}$, $t \\neq 0$.\nThen $\\tgg y = t$, $\\ctg y = 1/t$.\nSo the equation becomes:\n$$\na t + \\frac{b}{t} + 2c = 0.\n$$\nMultiply both sides by $t$ (for $t \\neq 0$):\n$$\na t^2 + 2c t + b = 0.\n$$\nThis is a quadratic equation in $t$ with discriminant:\n$$\n\\Delta = (2c)^2 - 4a b = 4c^2 - 4a b = 4(c^2 - a b).\n$$\nSince $a \\neq 0$, this quadratic equation has real solutions if and only if $c^2 - a b \\geq 0$.\n\nBut recall that we are considering the case $|c| > R = \\sqrt{a^2 + b^2}$.\nSo $c^2 > a^2 + b^2$.\nTherefore,\n$$\nc^2 - a b > a^2 + b^2 - a b = (a - b)^2 \\geq 0.\n$$\nSo the discriminant is positive, and the quadratic equation has real solutions for $t \\neq 0$.\n\nTherefore, at least one of the two equations has real solutions.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 28716,
    "subject": "Mathematics (Multi-modal)",
    "question": "Determine all integers $a$ such that $\\log_2(1 + a + a^2 + a^3)$ is also an integer.",
    "options": [],
    "answer": "[0, 1]",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 28717,
    "subject": "Mathematics (Multi-modal)",
    "question": "The difference in reciprocal values of two consecutive positive integers is $0.0\\dot{a} = 0.0aaa...$. Determine all possible values of the digit $a$.",
    "options": [],
    "answer": "1, 3",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 28718,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $a, b > 1$ be real numbers such that $\\log_b a$, $\\log_{2b}(2a)$ and $\\log_{4b}(4a)$, in that order, form an arithmetic progression. Prove that $a = b$. (Tonći Kokan)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 28719,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $k$ be an arbitrary nonnegative integer. Prove that we can find $4 \\cdot 2^k$ distinct positive integers not greater than $5 \\cdot 3^k$, such that no three of them are consecutive terms of an arithmetic progression.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "We will prove the statement using induction on $k$.\n\nFor $k = 0$ we need to choose four numbers from $\\{1, 2, 3, 4, 5\\}$ so that among them don't exist three consecutive members of an arithmetic progression. The only such choice is $\\{1, 2, 4, 5\\}$.\n\nAssume that for some $k \\ge 0$ we found a set $S_k$ that contains $4 \\cdot 2^k$ positive integers not greater than $5 \\cdot 3^k$ so that among them don't exist three consecutive members of an arithmetic progression.\n\nLet $T_k = \\{10 \\cdot 3^k + s : s \\in S_k\\}$. Sets $S_k$ and $T_k$ have the same number of elements. We claim that $S_{k+1} = S_k \\cup T_k$ satisfies the conditions for $k+1$.\n\nThe set $S_{k+1}$ has $2 \\cdot (4 \\cdot 2^k) = 4 \\cdot 2^{k+1}$ elements, and none of them is greater than $10 \\cdot 3^k + 5 \\cdot 3^k = 5 \\cdot 3^{k+1}$. If we take three arbitrary numbers $a, b, c \\in S_{k+1}$ such that $a < b < c$, exactly one of the following possibilities is true:\n$$\n1^\\circ \\quad a, b, c \\in S_k\n$$\n$$\n2^\\circ \\quad a, b \\in S_k, \\quad c \\in T_k\n$$\n$$\n3^\\circ \\quad a \\in S_k, \\quad b, c \\in T_k\n$$\n$$\n4^\\circ \\quad a, b, c \\in T_k.\n$$\nTo prove that the statement holds for $k+1$ it is sufficient to show that in each case $a+c \\neq 2b$.\n\n1° $a+c \\neq 2b$ because of the induction hypothesis for $S_k$.\n\n$$\n2^\\circ \\quad a+c \\ge 1 + (10 \\cdot 3^k + 1) > 10 \\cdot 3^k \\ge 2b,\n$$\n$$\n3^\\circ \\quad a+c \\le 5 \\cdot 3^k + 5 \\cdot 3^{k+1} = 20 \\cdot 3^k < 2b.\n$$\n4° $a - 10 \\cdot 3^k, b - 10 \\cdot 3^k$ and $c - 10 \\cdot 3^k$ are elements of the set $S_k$. Therefore\n$$\n(a - 10 \\cdot 3^k) + (c - 10 \\cdot 3^k) \\neq 2 (b - 10 \\cdot 3^k),\n$$\nso $a+c \\neq 2b$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 28720,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all functions $f: \\mathbb{R} \\to \\mathbb{R}$ such that $f(x^2 + f(y)) = y - x^2$ for all $x, y \\in \\mathbb{R}$.",
    "options": [],
    "answer": "All solutions are f(x) = c − x for any real constant c.",
    "solution": "Taking $x = 0$ in the given equation we get $f(f(y)) = y$ for every $y \\in \\mathbb{R}$.\nTherefore, we have\n$$\nf(y - x^2) = f(f(x^2 + f(y))) = x^2 + f(y). \\quad (3.1)\n$$\nfor every $x, y \\in \\mathbb{R}$.\nTaking $y = x^2$ in (3.1) we get $f(0) = x^2 + f(x^2)$, that is\n$$\nf(x^2) = -x^2 + f(0),\n$$\nwhile taking $y = 0$ in (*) gives us\n$$\nf(-x^2) = x^2 + f(0).\n$$\nLet $c = f(0)$. From the above we conclude that\n$$\nf(x) = -x + c, \\quad \\forall x \\in \\mathbb{R}.\n$$\nIt is easy to verify that all the functions of the form $f(x) = -x + c$ for $c \\in \\mathbb{R}$ satisfy the given equation:\n$$\nf(x^2 + f(y)) = f(x^2 - y + c) = -(x^2 - y + c) + c = -x^2 + y.\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 28721,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $z$ be a root of the polynomial $z^2 - 2z \\cos \\frac{\\pi}{n} + 1$. Determine all the possible values of $z^n$.",
    "options": [],
    "answer": "-1",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 28722,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all functions $f: \\mathbb{R} \\to \\mathbb{R}$ such that $f(x^2 + f(y)) = y - x^2$ for all $x, y \\in \\mathbb{R}$.",
    "options": [],
    "answer": "All functions f(x) = -x + c for any real constant c.",
    "solution": "Taking $x = 0$ in the given equation we get $f(f(y)) = y$ for every $y \\in \\mathbb{R}$.\nTherefore, we have\n$$\nf(y - x^2) = f(f(x^2 + f(y))) = x^2 + f(y). \\qquad (3.1)\n$$\nfor every $x, y \\in \\mathbb{R}$.\nTaking $y = x^2$ in (3.1) we get $f(0) = x^2 + f(x^2)$, that is\n$$\nf(x^2) = -x^2 + f(0),\n$$\nwhile taking $y = 0$ in (*) gives us\n$$\nf(-x^2) = x^2 + f(0).\n$$\nLet $c = f(0)$. From the above we conclude that\n$$\nf(x) = -x + c, \\quad \\forall x \\in \\mathbb{R}.\n$$\nIt is easy to verify that all the functions of the form $f(x) = -x + c$ for $c \\in \\mathbb{R}$ satisfy the given equation:\n$$\nf(x^2 + f(y)) = f(x^2 - y + c) = -(x^2 - y + c) + c = -x^2 + y.\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 28723,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $a$, $b$, $c$ be complex numbers such that $a + b + c = 0$ and $ab + bc + ca = 0$. Prove $|a| = |b| = |c|$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $a$, $b$, $c$ be complex numbers such that $a + b + c = 0$ and $ab + bc + ca = 0$.\n\nLet us consider the elementary symmetric polynomials:\n\n- $s_1 = a + b + c = 0$\n- $s_2 = ab + bc + ca = 0$\n- $s_3 = abc$\n\nThe roots $a$, $b$, $c$ are the roots of the cubic equation:\n$$\nx^3 - s_1 x^2 + s_2 x - s_3 = 0\n$$\nWith $s_1 = 0$ and $s_2 = 0$, this becomes:\n$$\nx^3 - s_3 = 0\n$$\nSo $a$, $b$, $c$ are the three cube roots of $s_3$.\n\nBut $a + b + c = 0$ implies that the sum of the three roots is zero, which is only possible if $a$, $b$, $c$ are equally spaced on the complex plane around the origin, i.e., they form the vertices of an equilateral triangle centered at the origin.\n\nLet $a = r e^{i\theta}$, $b = r e^{i(\theta + 2\\pi/3)}$, $c = r e^{i(\theta + 4\\pi/3)}$ for some $r > 0$ and $\theta \\in \\mathbb{R}$.\n\nThen $|a| = |b| = |c| = r$.\n\nTherefore, $|a| = |b| = |c|$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 28724,
    "subject": "Mathematics (Multi-modal)",
    "question": "In how many ways can the number $\\frac{2011}{2010}$ be represented as a product of two fractions of the form $\\frac{n+1}{n}$, where $n$ is a positive integer? (Order of the factors is not important.)",
    "options": [],
    "answer": "16",
    "solution": "Let $p$ and $q$ be positive integers such that $\\frac{2011}{2010} = \\frac{p+1}{p} \\cdot \\frac{q+1}{q}$.\nThen $2011pq = 2010(pq + p + q + 1)$ i.e. $pq = 2010(p + q + 1)$.\n\nFrom the last equation we find\n$$\np = \\frac{2010(q+1)}{q-2010} = \\frac{2010(q-2010)+2010 \\cdot 2011}{q-2010} = 2010 + \\frac{2010 \\cdot 2011}{q-2010}.\n$$\nSince $p$ and $q$ are positive integers, it follows that $q - 2010$ is a positive divisor of $2010 \\cdot 2011$. Each divisor of $2010 \\cdot 2011$ corresponds to exactly one pair $(p, q)$. Since $2010 \\cdot 2011 = 2 \\cdot 3 \\cdot 5 \\cdot 67 \\cdot 2011$, the number of its divisors is $2^5 = 32$. Finally, since the pairs $(p, q)$ and $(q, p)$ determine the same representation, the number of required representations is 16.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 28725,
    "subject": "Mathematics (Multi-modal)",
    "question": "On the sides of a square $4n$ points are given: the four vertices of the square and another $n-1$ points on each side. Determine the number of all (non-degenerated) triangles with vertices at the given points.",
    "options": [],
    "answer": "2 n^2 (5 n - 3)",
    "solution": "",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 28726,
    "subject": "Mathematics (Multi-modal)",
    "question": "In how many ways can the number $\\frac{2011}{2010}$ be represented as a product of two fractions of the form $\\frac{n+1}{n}$, where $n$ is a positive integer? (Order of the factors is not important.)",
    "options": [],
    "answer": "16",
    "solution": "Let $p$ and $q$ be positive integers such that $\\frac{2011}{2010} = \\frac{p+1}{p} \\cdot \\frac{q+1}{q}$.\nThen $2011pq = 2010(pq + p + q + 1)$, i.e. $pq = 2010(p + q + 1)$.\n\nFrom the last equation we find\n$$\np = \\frac{2010(q+1)}{q-2010} = \\frac{2010(q-2010)+2010 \\cdot 2011}{q-2010} = 2010 + \\frac{2010 \\cdot 2011}{q-2010}.\n$$\nSince $p$ and $q$ are positive integers, it follows that $q - 2010$ is a positive divisor of $2010 \\cdot 2011$. Each divisor of $2010 \\cdot 2011$ corresponds to exactly one pair $(p, q)$. Since $2010 \\cdot 2011 = 2 \\cdot 3 \\cdot 5 \\cdot 67 \\cdot 2011$, the number of its divisors is $2^5 = 32$. Finally, since the pairs $(p, q)$ and $(q, p)$ determine the same representation, the number of required representations is 16.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 28727,
    "subject": "Mathematics (Multi-modal)",
    "question": "Prove that $7 \\mid 2^{n+2} + 3^{2n+1}$ for all $n \\in \\mathbb{N}$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 28728,
    "subject": "Mathematics (Multi-modal)",
    "question": "Matija and Tomislav play the following game:\nEach of them throws a pair of dice. If at least one of them gets numbers whose sum is divisible by $3$, then Matija wins; otherwise, Tomislav wins.\nWhat is the probability that Matija wins?",
    "options": [],
    "answer": "5/9",
    "solution": "",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 28729,
    "subject": "Mathematics (Multi-modal)",
    "question": "The incircle of the acute triangle $ABC$ touches the segments $BC$, $CA$, $AB$ at points $D$, $E$, $F$ respectively. Let $S$ be the incenter and $P$ be the intersection of the line $DS$ and the segment $EF$. If $M$ is the midpoint of the segment $BC$ prove that the points $A$, $P$ and $M$ are collinear.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Since quadrilaterals $BFSD$ and $CDSE$ are cyclic, we have $\\angle FSD = 180^\\circ - \\beta$ and $\\angle ESD = 180^\\circ - \\gamma$ so $\\angle FSP = \\beta$ and $\\angle ESP = \\gamma$.\n![](attached_image_1.png)\nLet $x = \\angle FAP$ and $y = \\angle BAM$. Then $\\angle EAP = \\alpha - x$ and $\\angle CAM = \\alpha - y$.\nApplying the law of sines on triangles $EAP$ and $FAP$ we get\n$$\n\\frac{|EP|}{\\sin(\\alpha - x)} = \\frac{|AE|}{\\sin(\\angle APE)}, \\quad \\frac{|FP|}{\\sin x} = \\frac{|AF|}{\\sin(\\angle APF)}.\n$$\nDividing those two equalities because of $|AE| = |AF|$ and $\\angle APE + \\angle APF = 180^\\circ$ we get\n$$\n\\frac{\\sin x}{\\sin(\\alpha - x)} = \\frac{|FP|}{|EP|}.\n$$\nApplying the law of sines on triangles $EPS$ and $FPS$ we get\n$$\n\\frac{|EP|}{\\sin(\\angle PSE)} = \\frac{|SP|}{\\sin(\\angle PES)}, \\quad \\frac{|FP|}{\\sin(\\angle PSF)} = \\frac{|SP|}{\\sin(\\angle PFS)}\n$$\ni.e.\n$$\n\\frac{|EP|}{\\sin \\gamma} = \\frac{|SP|}{\\sin \\frac{\\alpha}{2}}, \\quad \\frac{|FP|}{\\sin \\beta} = \\frac{|SP|}{\\sin \\frac{\\alpha}{2}} \\quad \\text{so} \\quad \\frac{|EP|}{|FP|} = \\frac{\\sin \\gamma}{\\sin \\beta}.\n$$\n\nObserving the triangles $ABM$ and $ACM$ we see that:\n$$\n\\frac{|BM|}{\\sin y} = \\frac{|AM|}{\\sin \\beta}, \\quad \\frac{|CM|}{\\sin(\\alpha - y)} = \\frac{|AM|}{\\sin \\gamma}\n$$\nso\n$$\n\\frac{\\sin y}{\\sin(\\alpha - y)} = \\frac{\\sin \\beta}{\\sin \\gamma}.\n$$\nFrom all of the above we obtain the equation\n$$\n\\frac{\\sin x}{\\sin(\\alpha - x)} = \\frac{\\sin y}{\\sin(\\alpha - y)}\n$$\nfrom which we can deduce that $x = y$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 28730,
    "subject": "Mathematics (Multi-modal)",
    "question": "Circles $k_1$ and $k_2$, with radii $r, R$ ($r < R$) touch each other internally in $A$. Let $p$ be a line parallel to their common tangent and let $B$ be an intersection of the line $p$ with the circle $k_1$, and $C$ an intersection of $p$ with $k_2$, so that the points $B$ and $C$ lie on the same side of the line through the centers of given circles. Prove that the radius of the circle circumscribed to the triangle $ABC$ does not depend on the choice of line $p$. Determine that radius in terms of $r$ and $R$.",
    "options": [],
    "answer": "sqrt(rR)",
    "solution": "",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 28731,
    "subject": "Mathematics (Multi-modal)",
    "question": "Circles $k_1$ and $k_2$, with radii $r, R$ ($r < R$) touch each other internally in $A$. Let $p$ be a line parallel to their common tangent and let $B$ be an intersection of the line $p$ with the circle $k_1$, and $C$ an intersection of $p$ with $k_2$, so that the points $B$ and $C$ lie on the same side of the line through the centers of given circles. Prove that the radius of the circle circumscribed to the triangle $ABC$ does not depend on the choice of line $p$. Determine that radius in terms of $r$ and $R$.",
    "options": [],
    "answer": "sqrt(rR)",
    "solution": "",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 28732,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $P_1, P_2, \\dots, P_{2n}$ be a permutation of the vertices of a regular $2n$-gon. Prove that every closed polygonal line that consists of segments\n$$\n\\overline{P_1P_2}, \\overline{P_2P_3}, \\dots, \\overline{P_{2n-1}P_{2n}}, \\overline{P_{2n}P_1}\n$$\ncontains at least one pair of parallel segments.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Assign numbers $1, 2, \\ldots, 2n$ to the vertices of the observed $2n$-gon respectively. Let $a_k$ be the number assigned to the vertex $P_k$. Then $a_1, a_2, \\ldots, a_{2n}$ is a permutation of $1, 2, \\ldots, 2n$.\n\n![](attached_image_1.png)\n\nSegments $\\overline{P_iP_{i+1}}$ and $\\overline{P_jP_{j+1}}$ for $i \\neq j$ are parallel if and only if they determine an isosceles trapezoid with bases $\\overline{P_iP_{i+1}}$ and $\\overline{P_jP_{j+1}}$. Its legs (or diagonals) $\\overline{P_iP_{j+1}}$ and $\\overline{P_jP_{i+1}}$ are congruent and a rotation around the circumcenter maps points $P_i$ and $P_j$ into points $P_{j+1}$ if $P_{j+1}$ are mapped respectively.\n\nThat is why the condition $\\overline{P_iP_{i+1}} \\parallel \\overline{P_jP_{j+1}}$ is equivalent to the condition that the number of vertices between $P_i$ and $P_{j+1}$ is equal to the number of vertices between $P_j$ and $P_{i+1}$, taking into account the orientation. That condition can be written as $a_i - a_{j+1} \\equiv a_j - a_{i+1} \\pmod{2n}$, i.e.\n$$\na_i + a_{i+1} \\equiv a_j + a_{j+1} \\pmod{2n}.\n$$\nNow assume that none of the given segments are parallel.\nThen the sums $a_k + a_{k+1}$ give different remainders modulo $2n$, so\n$$\n\\sum_{k=1}^{2n} (a_k + a_{k+1}) \\equiv 0 + 1 + \\dots + (2n - 1) \\pmod{2n}\n$$\nThe sum on the right hand side is $n(2n-1)$ which is congruent to $n$ modulo $2n$.\n\nOn the other hand, we have\n$$\n\\sum_{k=1}^{2n} (a_k + a_{k+1}) = 2 \\sum_{k=1}^{2n} a_k = 2 \\sum_{k=1}^{2n} k = 2n(2n + 1)\n$$\nThe resulting expression is divisible by $2n$, which is in contradiction with the conclusion above. Therefore our assumption was incorrect which means there is at least one pair of parallel segments.",
    "topic": "Discrete Mathematics",
    "subtopic": "Combinatorics"
  },
  {
    "id": 28733,
    "subject": "Mathematics (Multi-modal)",
    "question": "A regular tetrahedron $ABXY$ is given inside the unit cube $ABCD'A'B'C'D'$ so that the point $X$ lies in the plane $ABCD$. Determine the distance between the points $Y$ and $A'$.",
    "options": [],
    "answer": "sqrt(2 - 2*sqrt(2/3))",
    "solution": "",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 28734,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $(a_n)$ be a sequence defined by\n$$\na_0 = 9 \\quad \\text{and} \\quad a_{k+1} = 3a_k^4 + 4a_k^3 \\quad \\text{for all } k \\ge 0.\n$$\nProve that the decimal representation of $a_{11}$ ends with at least 2011 digits 9.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "We will prove by induction that for all $n \\ge 0$, $a_n \\equiv -1 \\pmod{10^{2^n}}$.\n\nBase case: $n = 0$. $a_0 = 9 \\equiv -1 \\pmod{10}$, since $9 + 1 = 10$.\n\nInductive step: Assume $a_n \\equiv -1 \\pmod{10^{2^n}}$.\n\nConsider $a_{n+1} = 3a_n^4 + 4a_n^3$.\n\nLet $m = 10^{2^n}$. Then $a_n \\equiv -1 \\pmod{m}$, so $a_n = m t - 1$ for some integer $t$.\n\nCompute $a_{n+1}$ modulo $m^2 = 10^{2^{n+1}}$:\n\n$a_{n+1} = 3a_n^4 + 4a_n^3$\n\nExpand $a_n^3$ and $a_n^4$:\n\n$a_n^3 = (m t - 1)^3 = m^3 t^3 - 3 m^2 t^2 + 3 m t - 1$\n$a_n^4 = (m t - 1)^4 = m^4 t^4 - 4 m^3 t^3 + 6 m^2 t^2 - 4 m t + 1$\n\nSo:\n$3a_n^4 = 3(m^4 t^4 - 4 m^3 t^3 + 6 m^2 t^2 - 4 m t + 1) = 3 m^4 t^4 - 12 m^3 t^3 + 18 m^2 t^2 - 12 m t + 3$\n$4a_n^3 = 4(m^3 t^3 - 3 m^2 t^2 + 3 m t - 1) = 4 m^3 t^3 - 12 m^2 t^2 + 12 m t - 4$\n\nAdd:\n$3a_n^4 + 4a_n^3 =$\n$[3 m^4 t^4]$\n$+ [-12 m^3 t^3 + 4 m^3 t^3] = -8 m^3 t^3$\n$+ [18 m^2 t^2 - 12 m^2 t^2] = 6 m^2 t^2$\n$+ [-12 m t + 12 m t] = 0$\n$+ [3 - 4] = -1$\n\nSo:\n$a_{n+1} = 3 m^4 t^4 - 8 m^3 t^3 + 6 m^2 t^2 - 1$\n\nNow, $m^2 = 10^{2^{n+1}}$, so all terms with $m^2$ or higher are divisible by $m^2$ and thus $\\equiv 0 \\pmod{m^2}$.\n\nTherefore, $a_{n+1} \\equiv -1 \\pmod{m^2}$, i.e., $a_{n+1} \\equiv -1 \\pmod{10^{2^{n+1}}}$.\n\nBy induction, $a_n \\equiv -1 \\pmod{10^{2^n}}$ for all $n$.\n\nFor $n = 11$, $a_{11} \\equiv -1 \\pmod{10^{2^{11}}}$, so the decimal representation of $a_{11}$ ends with $2^{11} = 2048$ digits 9.\n\nIn particular, it ends with at least 2011 digits 9.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 28735,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $a$ and $b$ be positive integers. What are the possible units digits of the number $(a+b)^5 - (a^5+b^5)$?",
    "options": [],
    "answer": "0",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 28736,
    "subject": "Mathematics (Multi-modal)",
    "question": "Among all the points $z$ in the complex plane satisfying $|z+3| + |z-3| = 10$ determine the point closest to the line through $(-3+4i)$ and $(-8+i)$.",
    "options": [],
    "answer": "-3 + (16/5)i",
    "solution": "",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 28737,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $n$ be a positive integer. Prove that there is an odd number of odd numbers among the numbers\n$$\n\\binom{2n+1}{1}, \\binom{2n+1}{2}, \\dots, \\binom{2n+1}{k}, \\dots, \\binom{2n+1}{n}.\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let us consider the binomial coefficients $\\binom{2n+1}{k}$ for $1 \\leq k \\leq n$.\n\nRecall that $\\binom{2n+1}{k}$ is odd if and only if, in the binary representation, every bit of $k$ is less than or equal to the corresponding bit of $2n+1$ (Lucas' theorem).\n\nBut we can use a parity argument:\n\nThe sum of all binomial coefficients for a given $m$ is $2^m$:\n$$\n\\sum_{k=0}^{2n+1} \\binom{2n+1}{k} = 2^{2n+1}.\n$$\n\nBut we are interested in the parity of $\\binom{2n+1}{k}$ for $1 \\leq k \\leq n$.\n\nNote that $\\binom{2n+1}{k} = \\binom{2n+1}{2n+1-k}$, so the sequence is symmetric about $k = n+\\frac{1}{2}$.\n\nFor $k = 0$ and $k = 2n+1$, $\\binom{2n+1}{0} = \\binom{2n+1}{2n+1} = 1$ (odd).\n\nLet $S$ be the set $\\{\\binom{2n+1}{1}, \\binom{2n+1}{2}, \\dots, \\binom{2n+1}{n}\\}$.\n\nLet $T$ be the set $\\{\\binom{2n+1}{n+1}, \\dots, \\binom{2n+1}{2n}\\}$.\n\nSince $\\binom{2n+1}{k} = \\binom{2n+1}{2n+1-k}$, the set $S$ and $T$ have the same number of odd elements.\n\nNow, the total number of odd binomial coefficients in the row $2n+1$ is a power of $2$ (Lucas' theorem):\n\nLet $2n+1$ in binary have $d$ ones, then the number of odd binomial coefficients is $2^d$.\n\nBut $\\binom{2n+1}{0}$ and $\\binom{2n+1}{2n+1}$ are both $1$ (odd), so the number of odd binomial coefficients among $k=1$ to $2n$ is $2^d - 2$.\n\nSince $S$ and $T$ are symmetric, the number of odd binomial coefficients in $S$ is $\\frac{2^d - 2}{2} = 2^{d-1} - 1$.\n\nBut $2^{d-1} - 1$ is odd for $d \\geq 1$.\n\nTherefore, there is an odd number of odd numbers among $\\binom{2n+1}{1}, \\dots, \\binom{2n+1}{n}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 28738,
    "subject": "Mathematics (Multi-modal)",
    "question": "$$\na_0 = 1, \\quad a_{n+1} = \\begin{cases} \\frac{a_n}{2}, & \\text{if } a_n \\text{ is even,} \\\\ a_n + d, & \\text{if } a_n \\text{ is odd.} \\end{cases}\n$$\nDetermine all $d$ such that $a_n = 1$ for some $n > 0$. (Italy 2005)",
    "options": [],
    "answer": "all odd positive integers",
    "solution": "First notice that for even $d$ the sequence is of the form $a_n = 1 + nd$, i.e. all the terms of the sequence are odd and the sequence is monotonically increasing. Therefore $a_n \\neq 1$ for $n > 0$.\n\nLet $d$ be an arbitrary odd number. We can easily show by induction that $a_n < d$ if $a_n$ is odd, and $a_n < 2d$ if $a_n$ is even. Thus the sequence is limited, so it is periodic.\n\nLet $r$ be the smallest index such that $a_r = a_s$ for some $s \\neq r$. Assume $r > 0$.\nIf $a_r \\le d$, that means that $a_r$ (and then also $a_s$) is derived from the previous term by dividing by 2, i.e. $a_r = a_{r-1}/2$, $a_s = a_{s-1}/2$, so it follows that $a_{r-1} = a_{s-1}$, and this is in contradiction with minimality of $r$.\nIf $a_r > d$, from $a_n \\le 2d$ we conclude that $a_r$ and $a_s$ are derived from previous terms of the sequence by adding $d$, so again it follows that $a_{r-1} = a_{s-1}$, and again we have a contradiction with the minimality of $r$.\n\nHence, $r = 0$ and $a_s = a_0 = 1$ for some $s > 0$ for every odd $d$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 28739,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all sequences $a : \\mathbb{N} \\rightarrow \\mathbb{N}$ such that $a_n + a_{n+1} = a_{n+2}a_{n+3} - 200$.",
    "options": [],
    "answer": "All sequences a: N → N satisfying a_n + a_{n+1} = a_{n+2} a_{n+3} − 200 are exactly those of the following forms:\n\n1) Two-periodic sequences with constants a on odd indices and b on even indices where (a − 1)(b − 1) = 201. Explicitly, (a, b) ∈ {(2, 202), (202, 2), (4, 68), (68, 4)}.\n\n2) Sequences with all odd terms equal to 1 and even terms forming an arithmetic progression of common difference 201: for any integer t ≥ 1,\n   a_{2k−1} = 1, a_{2k} = t + 201(k − 1) for k ≥ 1.\n\n3) Sequences with all even terms equal to 1 and odd terms forming an arithmetic progression of common difference 201: for any integer t ≥ 1,\n   a_{2k} = 1, a_{2k−1} = t + 201(k − 1) for k ≥ 1.",
    "solution": "By subtracting equalities\n$$\n\\begin{aligned}\na_n + a_{n+1} &= a_{n+2}a_{n+3} - 200 \\\\\na_{n+1} + a_{n+2} &= a_{n+3}a_{n+4} - 200\n\\end{aligned}\n$$\nwe get\n$$\na_n - a_{n+2} = a_{n+3} (a_{n+2} - a_{n+4}). \\qquad (4.3)\n$$\nSince $a_{n+3} > 0$, we have\n$$\n\\begin{array}{lcll}\na_n > a_{n+2} & \\text{if and only if} & a_{n+2} > a_{n+4} \\\\\na_n = a_{n+2} & \\text{if and only if} & a_{n+2} = a_{n+4} \\\\\na_n < a_{n+2} & \\text{if and only if} & a_{n+2} < a_{n+4}\n\\end{array}\n$$\nIn the first case it would follow $a_n > a_{n+2} > a_{n+4} > \\dots$ and there would exist a term of the sequence smaller than 1, which is a contradiction.\nTherefore:\n$$\na_n = a_{n+2} \\text{ for all odd } n \\quad \\text{or} \\quad a_n < a_{n+2} \\text{ for all odd } n\n$$\nand\n$$\na_n = a_{n+2} \\text{ for all even } n \\quad \\text{or} \\quad a_n < a_{n+2} \\text{ for all even } n.\n$$\n\na. $a_n < a_{n+2}$ for all $n \\in \\mathbb{N}$\nThen there exists $n$ such that $a_{n+2} > 15$ and $a_{n+3} > 15$, so we have\n$$\n(a_{n+2} - 1)(a_{n+3} - 1) \\ge 225 > 201\n$$\nand\n$$\n\\begin{aligned}\na_n + a_{n+1} &< a_{n+2} + a_{n+3} \\\\\n&= a_{n+2}a_{n+3} - (a_{n+2} - 1)(a_{n+3} - 1) + 1 \\\\\n&< a_{n+2}a_{n+3} - 200.\n\\end{aligned}\n$$\nContradiction.\n\nb. $a_n = a$ for all odd $n$ and $a_n = b$ for all even $n$\nFrom the given condition we have $a + b = ab - 200$, which can be written as $(a - 1)(b - 1) = 201 = 3 \\cdot 67$. The solutions are\n$$\n(a, b) \\in \\{(2, 202), (202, 2), (4, 68), (68, 4)\\}.\n$$\n\nc. $a_n = a$ for all odd $n$ and $a_n < a_{n+2}$ for all even $n$\nThen according to (4.3)\n$$\na_4 - a_2 = a(a_6 - a_4) = a^2(a_8 - a_6) = \\dots = a^{m-1}(a_{2m+2} - a_{2m}) = \\dots\n$$\nwhich is possible only for $a = 1$.\nDenote $d = a_4 - a_2$. Then $a_{2m+2} - a_{2m} = d$ for $m \\in \\mathbb{N}$, so $a_{2m} = a_2 + (m-1)d$. From the initial condition we get $d = 201$.\nFor $b \\in \\mathbb{Z}$ the sequence $1, b, 1, b+201, 1, b+402, 1, \\dots$ satisfies the given condition.\n\nd. $a_n < a_{n+2}$ for all odd $n$ if $a_n = b$ for all even $n$\nAnalogously to the previous case, we get the solution\n$$\na, 1, a+201, 1, a+402, 1, \\dots \\quad \\text{where } a \\in \\mathbb{Z}.\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 28740,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all functions $f : \\mathbb{Z} \\to \\mathbb{Z}$ such that\n$$\nf(f(a) + f(b)) = a + b - 1\n$$\nfor all $a, b \\in \\mathbb{Z}$.",
    "options": [],
    "answer": "f(n) = -n + 1",
    "solution": "By taking $b=1$, we get\n$$\nf(f(a) + f(1)) = a,\n$$\nfor every integer $a$.\nLet $n$ be an arbitrary positive integer. For $a = f(n) + f(1)$ and $b = 2f(1)$, we get $f(a) = n$ and $f(b) = 1$, and we have:\n$$\nf(n+1) = f(n) + 3f(1) - 1.\n$$\nLet $c = f(1)$. It is easy to prove by induction that for every integer $n$\n$$\nf(n) = (3n - 2)c - n + 1.\n$$\n\nThen $f(0) = -2c + 1$, and by taking $a = 0$ and $b = 1$ the initial expression becomes:\n$$\n\\begin{aligned}\nf(f(0) + f(1)) &= 0 \\\\\nf(-c + 1) &= 0 \\\\\n(3(-c + 1) - 2)c - (-c + 1) + 1 &= 0 \\\\\nc(-3c + 2) &= 0.\n\\end{aligned}\n$$\nBecause $c \\in \\mathbb{Z}$ we conclude $c = 0$. Therefore, the only solution of the given equation is $f(n) = -n + 1$. We can directly check that this function satisfies the given equation.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 28741,
    "subject": "Mathematics (Multi-modal)",
    "question": "At a party among each group of four people there are three that all know each other or three that don't know each other. Acquaintances are mutual. Prove that all the participants at that party can be separated into two rooms so that in one room everybody knows everyone, and in the other room nobody knows anyone.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "People and acquaintances can be represented by a complete graph in two colours, where the vertices represent people, blue edges connect people who know each other, and red edges connect people who don't know each other. We call a set with $m$ vertices a blue (red) $m$-gon if all of the vertices are connected by a blue (red) edge. On the figures, red edges are represented by full lines, and blue edges by dashed lines.\n\nNotice that the given condition implies that the graph contains a red or blue triangle (in the nontrivial cases where the graph has $n \\ge 4$ vertices).\n\nWithout loss of generality, we can assume that the graph contains a red triangle. Let us observe the largest red $k$-gon. Let $\\mathcal{A} = \\{A_1, A_2, \\dots, A_k\\}$ be the set of its vertices. If the remaining $n-k$ vertices are all connected by blue edges, we can place the people from $\\mathcal{A}$ in one room and everyone else in the other room, and the problem is solved.\n\nWe will show that the other case is not possible. Suppose there exist vertices $B, C \\notin \\mathcal{A}$ connected by a red edge. Then there exists $A_B \\in \\mathcal{A}$ which is connected to $B$ by a blue edge, because otherwise we would have a red $(k+1)$-gon. Analogously, there exists $A_C \\in \\mathcal{A}$ which is connected to $C$ by a blue edge.\n\n![](attached_image_1.png)\n\nWe have two possibilities. If $A_B \\neq A_C$, then it is clear (see the left figure above) that no three of the vertices $B, C, A_B, A_C$ can form a monochromatic triangle, which is in contradiction with the initial condition.\n\nOn the other hand, suppose $A_B = A_C = A$, for some $A \\in \\mathcal{A}$. Let $A' \\in \\mathcal{A}$ be an arbitrary vertex different from $A$. The right figure represents the situation between vertices $A, A', B$ and $C$. Clearly, in order to have a monochromatic triangle, $A'$ must be connected to $B$ and $C$ by a red edge. So, $B$ and $C$ are connected by a red edge, and they are both connected by a red edge with every vertex from $\\mathcal{A} \\setminus \\{A\\}$. But then $(\\mathcal{A} \\setminus \\{A\\}) \\cup \\{B, C\\}$ is a red $(k+1)$-gon, which is again in contradiction with the maximality of $k$.",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 28742,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $M$ and $N$ be positive integers. Consider an $N \\times N$ square array consisting of $N^2$ lamps that can be in two states - on or off. At the beginning all lamps are turned off.\nA *move* consists of choosing a row or a column of the array and changing the state of $M$ consecutive lamps in the chosen row or column, i.e. turning on the lamps that are turned off and vice versa.\nDetermine the necessary and sufficient condition for which it can be achieved that after a finite number of moves all lamps are turned on. (Tonći Kokan)",
    "options": [],
    "answer": "M divides N",
    "solution": "The sought condition is that $M$ divides $N$.\n\nIt is easy to see that if $M$ divides $N$ we can choose a sequence of moves after which all lamps will be turned on. Really, if $N = Mk$ for some positive integer $k$ then we choose each row $k$ times. In the $(ik+j)$-th move, for $i = 0, 1, \\dots, M-1$, $j = 1, \\dots, k$ we choose $M$ consecutive lamps from $((j-1)k+1)$-th to $jk$-th place in $(i+1)$-th row.\n\nTo prove necessity, colour the lamps in $M$ colors (named $0, 1, 2, \\dots, M-1$) in a way presented on the Figure 4.1, i.e. color the lamp in the $i$-th row and $j$-th column in color $i + j - 2$ (mod $M$).\n\n![](attached_image_1.png)\nFigure 4.1: Example of the coloring for $N = 10, M = 6$.\n\nIn every move we change the state of exactly one lamp of each color. In the beginning all lamps are turned off so after each step we have the same number of lamps that are turned on in all colors. If it is possible to achieve that after some move all lamps are turned on, then the number of lamps of each color must be the same.\n\nAssume on the contrary that $M$ does not divide $N$ and let $N = Mk + r$, where $1 \\le r \\le M-1$. Divide the $N \\times N$ array into four subarrays of dimensions $Mk \\times Mk, Mk \\times r, r \\times Mk$ and $r \\times r$ as in the Figure 4.2.\nSince each of the subarrays of dimensions $Mk \\times Mk, Mk \\times r$ and $r \\times Mk$ is a disjoint union of sequences of $M$ consecutive lamps in a row or a column, we see that the number of lamps of each color in their union is the same (it equals $Mk^2 + 2kr$).\n\n---\n\n![](attached_image_2.png)\nFigure 4.2: Dividing the array for $N = 10$, $M = 6$.\n\nConsider the remaining $r \\times r$ subarray. On Figure 4.3 we see that the number of lamps of color $r-1$ equals $r$, but the number of lamps of color $r$ equals $r-1$. Indeed, the lamps of color $r-1$ appear in each row of the subarray exactly once and the lamps of color $r$ appear in each but first row. Also, there is no row with two or more lamps of color $r$ because $r$ is strictly less than $M$ so each row has all the lamps of different color.\n\n![](attached_image_3.png)\nFigure 4.3: Example of the $r \\times r$ subarray for $N = 10$, $M = 6$ (and $r = 4$).\n\nHence in the whole array we have a different number of lamps of color $r-1$ and color $r$ and we can never achieve that all of the lamps are turned on.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 28743,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $n \\ge 3$ be an integer. Determine the minimum number of points one has to mark inside a convex $n$-gon in order for the interior of any triangle with the vertices at vertices of the $n$-gon to contain at least one of the marked points.",
    "options": [],
    "answer": "n - 2",
    "solution": "Since all diagonals from one vertex divide an $n$-gon into $n-2$ disjoint triangles, at least $n-2$ points are necessary.\n\nWe claim that it is possible to mark $n-2$ points so that the given condition is satisfied. Denote the vertices of the given $n$-gon with $A_1, A_2, \\dots, A_n$. Draw all the diagonals of the given $n$-gon and color the areas bounded by the diagonals $A_1A_k$, $A_kA_n$ and $A_{k-1}A_{k+1}$ for each $k \\in \\{2, 3, \\dots, n-1\\}$.\n![](attached_image_1.png)\n\nIf we mark one point in each colored area then every triangle with vertices at vertices of the $n$-gon will contain at least one of the marked points. Indeed, triangle $A_lA_kA_m$ contains the whole colored area bounded by diagonals $A_lA_k$, $A_kA_n$ and $A_{k-1}A_{k+1}$ for every $1 \\le l < k < m \\le n$. Hence, the triangle also contains the corresponding marked point.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 28744,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $n \\ge 3$ be an integer. Determine the minimum number of points one has to mark inside a convex $n$-gon in order for the interior of any triangle with the vertices at vertices of the $n$-gon to contain at least one of the marked points.",
    "options": [],
    "answer": "n - 2",
    "solution": "Since all diagonals from one vertex divide an $n$-gon into $n-2$ disjoint triangles, at least $n-2$ points are necessary.\n\nWe claim that it is possible to mark $n-2$ points so that the given condition is satisfied. Denote the vertices of the given $n$-gon with $A_1, A_2, \\dots, A_n$. Draw all the diagonals of the given $n$-gon and color the areas bounded by the diagonals $A_1A_k$, $A_kA_n$ and $A_{k-1}A_{k+1}$ for each $k \\in \\{2, 3, \\dots, n-1\\}$.\n\n![](attached_image_1.png)\n\nIf we mark one point in each colored area then every triangle with vertices at vertices of the $n$-gon will contain at least one of the marked points. Indeed, triangle $A_lA_kA_m$ contains the whole colored area bounded by diagonals $A_lA_k$, $A_kA_m$ and $A_{k-1}A_{k+1}$ for every $1 \\le l < k < m \\le n$. Hence, the triangle also contains the corresponding marked point.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 28745,
    "subject": "Mathematics (Multi-modal)",
    "question": "Each side and each diagonal of a convex $n$-gon is colored in one of $k$ colors. It is known that there doesn't exist a closed monochrome broken line whose vertices are also the vertices of the given $n$-gon. Determine the largest possible value of $n$. (Russia 1990)",
    "options": [],
    "answer": "2k",
    "solution": "It is known that a graph with $n$ vertices that contains no cycles has at most $n-1$ edges. That fact is easily proven by induction by the number of vertices $n$. Hence, at most $n-1$ edges (i.e. sides and diagonals) can be colored with the same color.\n\nSince the number of colors is $k$ and the total number of edges (sides and diagonals) is $\\frac{n(n-1)}{2}$, we conclude that\n$$\n\\frac{n(n-1)}{2} \\le (n-1)k, \\quad \\text{ i.e. } n \\le 2k.\n$$\nLet $k$ be an arbitrary positive integer. We will show that we can color the sides and diagonals of a regular $2k$-gon with $k$ colors so that there doesn't exist a closed monochrome broken line (without loss of generality we can assume that the polygon is regular because we only care how the edges are colored). Moreover, we will show that with each of the $k$ colors we can color exactly $2k-1$ sides and diagonals in the required way.\n\nLet us denote the vertices of the regular $2k$-gon with $A_1, A_2, \\dots, A_{2k}$. With one color we color the broken line $A_1A_{2k}A_2A_{2k-1}\\dots A_iA_{2k+1-i}\\dots A_kA_{k+1}$, $2 \\le i \\le k$, as shown on the picture below.\n![](attached_image_1.png)\n\nThe broken lines obtained in such a way are disjoint because otherwise one of the diagonals would be left unpainted, and that is not possible.\n\nTherefore, the largest possible value of $n$ is $2k$.\n\nWith each following color $j$, $j \\in \\{2, \\dots, k\\}$ we color the broken line obtained by rotating the mentioned one by angle $\\frac{j\\pi}{k}$ around the center of the polygon.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 28746,
    "subject": "Mathematics (Multi-modal)",
    "question": "Each side and each diagonal of a convex $n$-gon is colored in one of $k$ colors. It is known that there doesn't exist a closed monochrome broken line whose vertices are also the vertices of the given $n$-gon. Determine the largest possible value of $n$. (Russia 1990)",
    "options": [],
    "answer": "2k",
    "solution": "It is known that a graph with $n$ vertices that contains no cycles has at most $n-1$ edges. That fact is easily proven by induction by the number of vertices $n$. Hence, at most $n-1$ edges (i.e. sides and diagonals) can be colored with the same color.\n\nSince the number of colors is $k$ and the total number of edges (sides and diagonals) is $\\frac{n(n-1)}{2}$, we conclude that\n$$\n\\frac{n(n-1)}{2} \\le (n-1)k, \\quad \\text{i.e. } n \\le 2k.\n$$\nLet $k$ be an arbitrary positive integer. We will show that we can color the sides and diagonals of a regular $2k$-gon with $k$ colors so that there doesn't exist a closed monochrome broken line (without loss of generality we can assume that the polygon is regular because we only care how the edges are colored). Moreover, we will show that with each of the $k$ colors we can color exactly $2k-1$ sides and diagonals in the required way.\n\nLet us denote the vertices of the regular $2k$-gon with $A_1, A_2, \\dots, A_{2k}$. With one color we color the broken line $A_1A_{2k}A_2A_{2k-1}\\dots A_{i}A_{2k+1-i}\\dots A_kA_{k+1}$, $2 \\le i \\le k$, as shown on the picture below.\n\n![](attached_image_1.png)\n\nWith each following color $j$, $j \\in \\{2, \\dots, k\\}$ we color the broken line obtained by rotating the mentioned one by angle $\\frac{j\\pi}{k}$ around the center of the polygon.\n\nThe broken lines obtained in such a way are disjoint because otherwise one of the diagonals would be left unpainted, and that is not possible.\n\nTherefore, the largest possible value of $n$ is $2k$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 28747,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABC$ be a triangle with centroid $T$ and circumcenter $O$ such that $OT \\perp AT$. Let $A'$ be the other intersection of the line $AT$ and the circumcircle of the triangle $ABC$. Let $D$ be the intersection of the lines $BA'$ and $AC$, and let $E$ be the intersection of the lines $CA'$ and $AB$. Prove that the circumcenter of the triangle $ADE$ lies on the circumcircle of the triangle $ABC$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $A_1, B_1$ and $C_1$ be the midpoints of the sides $\\overline{BC}, \\overline{CA}$ and $\\overline{AB}$ respectively. Let $k$ be the circumcircle of the triangle $ABC$.\n\n![](attached_image_1.png)\n\nNotice that from $OT \\perp AA'$ it follows that $OT$ is the bisector of the chord $AA'$ of the circle $k$ so $|AT| = |A'T|$. Since $AA_1$ is the median, it follows that $|AT| = 2|A_1T|$, and then from $|A'T| = 2|A_1T|$ follows $|A'A_1| = |A_1T|$. Now we can see that the point $A_1$ bisects the segments $BC$ and $A'T$ so the quadrilateral $BA'CT$ is a parallelogram.\nFrom $TC \\parallel BA'$ follows that $CC_1$ is the midline of the triangle $ABD$ so $|AD| = 2|AC|$. Analogously, from $TB \\parallel CA'$ follows that $BB_1$ is the midline of the triangle $AEC$ so $|AE| = 2|AB|$.\nNow we can see that the homothety with ratio 2 and center $A$ sends the triangle $ABC$ into the triangle $AED$. It also sends the circumcenter $O$ of the triangle $ABC$ into the circumcenter $S$ of the triangle $AED$. The point $S$ lies on the ray $AO$ and we know that $|AS| = 2|AO|$, so $\\overline{AS}$ is the diameter of the circle $k$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 28748,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $K$ and $L$ be points on a semicircle with diameter $\\overline{AB}$. Bisector of the side $\\overline{AB}$ intersects the segment $\\overline{KL}$ at a point $U$ so that $A$ and $K$ lie on one side of the bisector and $B$ and $L$ on the other. Let $N$ be the foot of perpendicular from the intersection of lines $AK$ and $BL$ to the line $AB$, and let $V$ be a point on the line $KL$ such that $\\angle VAU = \\angle VBU$. Prove that the lines $NV$ and $KL$ are perpendicular. (Iran TST 2009, modified)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $S$ be the center of the given semicircle, $C$ the intersection of the lines $AK$ and $BL$, and $T$ the intersection of the lines $AB$ and $KL$.\nFrom the given condition it follows that the quadrilateral $ABUV$ is cyclic so by the power of a point theorem we have\n$$\n|TU| \\cdot |TV| = |TA| \\cdot |TB|.\n$$\n\nThe quadrilateral $ABLK$ is also cyclic, so we have\n$$\n|TA| \\cdot |TB| = |TK| \\cdot |TL|.\n$$\n![](attached_image_1.png)\nSince $\\overline{AL}$, $\\overline{BK}$ and $\\overline{CN}$ are altitudes of the triangle $ABC$ and $S$ is the midpoint of $AB$, points $K$, $L$, $N$ and $S$ belong to the nine points circle of that triangle. Therefore the quadrilateral $NSLK$ is also cyclic, so\n$$\n|TU| \\cdot |TV| = |TA| \\cdot |TB| = |TK| \\cdot |TL| = |TS| \\cdot |TN|.\n$$\nThereby the quadrilateral $NSUV$ is cyclic, and hence $\\angle NVU = 180^\\circ - \\angle NSU = 90^\\circ$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 28749,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $a$, $b > 1$ be relatively prime positive integers. Define a sequence\n$$\nx_1 = a, \\quad x_2 = b, \\quad x_n = \\frac{x_{n-1}^2 + x_{n-2}^2}{x_{n-1} + x_{n-2}} \\quad \\text{for } n \\ge 3.\n$$\nProve that $x_n$ is not an integer for $n \\ge 3$. (Tonći Kokan)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Notice that $x_n > 1$, for all $n \\in \\mathbb{N}$. We also notice that all $x_n$ are rational so we can write $x_n = \\frac{p_n}{q_n}$, where $p_n$ and $q_n$ are positive integers and $M(p_n, q_n) = 1$.\n\nFirst let us prove that $p_n$ and $p_{n+1}$ are relatively prime for every $n \\in \\mathbb{N}$. We will prove that by induction. Obviously $M(p_1, p_2) = M(a, b) = 1$, i.e. $p_1$ and $p_2$ are relatively prime. Now we assume that $M(p_n, p_{n+1}) = 1$ for some $n$. Then\n$$\nx_{n+2} = \\frac{\\frac{p_n^2}{q_n^2} + \\frac{p_{n+1}^2}{q_{n+1}^2}}{\\frac{p_n}{q_n} + \\frac{p_{n+1}}{q_{n+1}}} = \\frac{p_n^2 q_{n+1}^2 + p_{n+1}^2 q_n^2}{q_n q_{n+1} (p_n q_{n+1} + p_{n+1} q_n)} = \\frac{p_{n+2}}{q_{n+2}}\n$$\nSince $M(p_n, p_{n+1}) = 1$ by the inductive hypothesis and $M(p_{n+1}, q_{n+1}) = 1$, we conclude that $M(p_{n+1}, p_n^2 q_{n+1}^2 + p_{n+1}^2 q_n^2) = M(p_{n+1}, p_n^2 q_{n+1}^2) = 1$, whence follows $M(p_{n+1}, p_{n+2}) = 1$. Thereby we have proved our assertion.\n\nNow we want to prove that $x_n$ is not an integer for $n \\ge 3$.\nAssume the contrary, that $x_{n+2}$ is a positive integer for some $n \\in \\mathbb{N}$. Since\n$$\nx_{n+2} = \\frac{p_n^2 q_{n+1}^2 + p_{n+1}^2 q_n^2}{q_n q_{n+1} (p_n q_{n+1} + p_{n+1} q_n)} = \\frac{p_n^2 q_{n+1}^2 + p_{n+1}^2 q_n^2}{p_n q_n q_{n+1}^2 + p_{n+1} q_{n+1} q_n^2}\n$$\nwe conclude that\n$$\nq_{n+1} \\mid p_n^2 q_{n+1}^2 + p_{n+1}^2 q_n^2 \\implies q_{n+1} \\mid p_{n+1}^2 q_n^2 \\implies q_{n+1} \\mid q_n^2\n$$\nbecause $p_{n+1}$ and $q_{n+1}$ are relatively prime. Now because of $q_{n+1} \\mid q_n^2$ we have $q_{n+1}^2 \\mid p_n q_n q_{n+1}^2 + p_{n+1} q_{n+1} q_n^2 \\implies q_{n+1}^2 \\mid p_n^2 q_{n+1}^2 + p_{n+1}^2 q_n^2 \\implies q_{n+1}^2 \\mid q_n^2$.\n\nAnalogously,\n$$\nq_n \\mid p_n^2 q_{n+1}^2 + p_{n+1}^2 q_n^2 \\implies q_n \\mid p_n^2 q_{n+1}^2 \\implies q_n \\mid q_{n+1}^2\n$$\nand then\n$$\nq_n^2 \\mid p_n q_n q_{n+1}^2 + p_{n+1} q_{n+1} q_n^2 \\implies q_n^2 \\mid p_n^2 q_{n+1}^2 + p_{n+1}^2 q_n^2 \\implies q_n^2 \\mid q_{n+1}^2\n$$\n\nAs $q_{n+1}^2 \\mid q_n^2$ and $q_n^2 \\mid q_{n+1}^2$, it follows that $q_n^2 = q_{n+1}^2$, that is $q_n = q_{n+1}$, and now we get\n$$\nx_{n+2} = \\frac{p_n^2 + p_{n+1}^2}{q_n (p_n + p_{n+1})}.\n$$\nThis means that\n$$\np_n + p_{n+1} \\mid p_n^2 + p_{n+1}^2 \\implies p_n + p_{n+1} \\mid 2p_{n+1}^2\n$$\nbecause\n$$\np_n^2 + p_{n+1}^2 = p_n^2 - p_{n+1}^2 + 2p_{n+1}^2 = (p_n - p_{n+1}) (p_n + p_{n+1}) + 2p_{n+1}^2.\n$$\nLet $p$ be a prime number such that $p \\mid p_n + p_{n+1}$, and thereby $p \\mid 2p_{n+1}^2$.\nIf $p \\neq 2$ then $p \\mid p_{n+1}^2 \\implies p \\mid p_{n+1}$, and since $p \\mid p_n + p_{n+1}$, it follows that $p \\mid p_n$ which is a contradiction because $p_n$ and $p_{n+1}$ are relatively prime.\nIf $p=2$ is the only prime factor, then $p_n + p_{n+1}$ is a power of 2 bigger than 2 (because $p_n$ and $p_{n+1}$ are bigger than 1). It follows that $4 \\mid p_n + p_{n+1}$ and then\n$$\n4 \\mid 2p_{n+1}^2 \\implies 2 \\mid p_{n+1}^2 \\implies 2 \\mid p_{n+1} \\implies 2 \\mid p_n.\n$$\nwhich is again a contradiction since $M(p_n, p_{n+1}) = 1$.\nThereby we have proved that $x_n$ is not an integer for $n \\ge 3$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 28750,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $a$, $b > 1$ be relatively prime positive integers. Define a sequence\n$$\nx_1 = a, \\quad x_2 = b, \\quad x_n = \\frac{x_{n-1}^2 + x_{n-2}^2}{x_{n-1} + x_{n-2}} \\quad \\text{for } n \\ge 3.\n$$\nProve that $x_n$ is not an integer for $n \\ge 3$. (Tonći Kokan)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Notice that $x_n > 1$, for all $n \\in \\mathbb{N}$. We also notice that all $x_n$ are rational so we can write $x_n = \\frac{p_n}{q_n}$, where $p_n$ and $q_n$ are positive integers and $M(p_n, q_n) = 1$.\n\nFirst let us prove that $p_n$ and $p_{n+1}$ are relatively prime for every $n \\in \\mathbb{N}$. We will prove that by induction. Obviously $M(p_1, p_2) = M(a, b) = 1$, i.e. $p_1$ and $p_2$ are relatively prime. Now we assume that $M(p_n, p_{n+1}) = 1$ for some $n$. Then\n$$\nx_{n+2} = \\frac{\\frac{p_n^2}{q_n^2} + \\frac{p_{n+1}^2}{q_{n+1}^2}}{\\frac{p_n}{q_n} + \\frac{p_{n+1}}{q_{n+1}}} = \\frac{p_n^2 q_{n+1}^2 + p_{n+1}^2 q_n^2}{q_n q_{n+1} (p_n q_{n+1} + p_{n+1} q_n)} = \\frac{p_{n+2}}{q_{n+2}}.\n$$\nSince $M(p_n, p_{n+1}) = 1$ by the inductive hypothesis and $M(p_{n+1}, q_{n+1}) = 1$, we conclude that $M(p_{n+1}, p_n^2 q_{n+1}^2 + p_{n+1}^2 q_n^2) = M(p_{n+1}, p_n^2 q_{n+1}^2) = 1$, whence follows $M(p_{n+1}, p_{n+2}) = 1$. Thereby we have proved our assertion.\n\nNow we want to prove that $x_n$ is not an integer for $n \\ge 3$.\nAssume the contrary, that $x_{n+2}$ is a positive integer for some $n \\in \\mathbb{N}$. Since\n$$\nx_{n+2} = \\frac{p_n^2 q_{n+1}^2 + p_{n+1}^2 q_n^2}{q_n q_{n+1} (p_n q_{n+1} + p_{n+1} q_n)} = \\frac{p_n^2 q_{n+1}^2 + p_{n+1}^2 q_n^2}{p_n q_n q_{n+1}^2 + p_{n+1} q_{n+1} q_n^2},\n$$\nwe conclude that\n$$\nq_{n+1} \\mid p_n^2 q_{n+1}^2 + p_{n+1}^2 q_n^2 \\implies q_{n+1} \\mid p_{n+1}^2 q_n^2 \\implies q_{n+1} \\mid q_n^2\n$$\nbecause $p_{n+1}$ and $q_{n+1}$ are relatively prime. Now because of $q_{n+1} \\mid q_n^2$ we have\n$$\nq_{n+1}^2 \\mid p_n q_n q_{n+1}^2 + p_{n+1} q_{n+1} q_n^2 \\implies q_{n+1}^2 \\mid p_n^2 q_{n+1}^2 + p_{n+1}^2 q_n^2 \\implies q_{n+1}^2 \\mid q_n^2.\n$$\nAnalogously,\n$$\nq_n \\mid p_n^2 q_{n+1}^2 + p_{n+1}^2 q_n^2 \\implies q_n \\mid p_n^2 q_{n+1}^2 \\implies q_n \\mid q_{n+1}^2,\n$$\nand then\n$$\nq_n^2 \\mid p_n q_n q_{n+1}^2 + p_{n+1} q_{n+1} q_n^2 \\implies q_n^2 \\mid p_n^2 q_{n+1}^2 + p_{n+1}^2 q_n^2 \\implies q_n^2 \\mid q_{n+1}^2.\n$$\n---\n\nAs $q_{n+1}^2 \\mid q_n^2$ and $q_n^2 \\mid q_{n+1}^2$, it follows that $q_n^2 = q_{n+1}^2$, that is $q_n = q_{n+1}$, and now we get\n$$\nx_{n+2} = \\frac{p_n^2 + p_{n+1}^2}{q_n (p_n + p_{n+1})}.\n$$\nThis means that\n$$\np_n + p_{n+1} \\mid p_n^2 + p_{n+1}^2 \\implies p_n + p_{n+1} \\mid 2p_{n+1}^2\n$$\nbecause\n$$\np_n^2 + p_{n+1}^2 = p_n^2 - p_{n+1}^2 + 2p_{n+1}^2 = (p_n - p_{n+1})(p_n + p_{n+1}) + 2p_{n+1}^2.\n$$\nLet $p$ be a prime number such that $p \\mid p_n + p_{n+1}$, and thereby $p \\mid 2p_{n+1}^2$.\nIf $p \\neq 2$ then $p \\mid p_{n+1}^2 \\implies p \\mid p_{n+1}$, and since $p \\mid p_n + p_{n+1}$, it follows that $p \\mid p_n$ which is a contradiction because $p_n$ and $p_{n+1}$ are relatively prime.\nIf $p = 2$ is the only prime factor, then $p_n + p_{n+1}$ is a power of 2 bigger than 2 (because $p_n$ and $p_{n+1}$ are bigger than 1). It follows that $4 \\mid p_n + p_{n+1}$ and then\n$$\n4 \\mid 2p_{n+1}^2 \\implies 2 \\mid p_{n+1}^2 \\implies 2 \\mid p_{n+1} \\implies 2 \\mid p_n,\n$$\nwhich is again a contradiction since $M(p_n, p_{n+1}) = 1$.\nThereby we have proved that $x_n$ is not an integer for $n \\ge 3$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 28751,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all pairs $(x, y)$ of integers such that $x^3 + x^2 + x = y^2 + y$.",
    "options": [],
    "answer": "[(0, 0), (0, -1)]",
    "solution": "If $y = 0$ or $y = -1$, the right hand side of the given equation is equal to zero so $x^3 + x^2 + x = x(x^2 + x + 1) = 0$, and hence $x = 0$. That gives us two solutions: $(0, 0)$ and $(0, -1)$.\n\nWe shall prove that no other solutions exist.\n\nAssume that $y \\in \\mathbb{Z} \\setminus \\{-1, 0\\}$. Then $y^2 + y > 0$ so $x(x^2 + x + 1) > 0$, and it follows that $x > 0$.\n\nIf $(x, y)$ is the solution of the given equation, then $(x, -y-1)$ is also the solution since $y^2 + y = (-y - 1)^2 + (-y - 1)$. In other words, if there are no solutions for $y > 0$, then there are no other solutions. Therefore, we can assume that $y > 0$.\n\nThe given equation can be written as:\n$$\nx^3 = (y - x)(x + y + 1).\n$$\nLet us prove that $\\gcd(y - x, x + y + 1) = 1$. Assume the contrary, let $p$ be the common prime divisor of the numbers $y - x$ and $x + y + 1$.\n\nFrom the above we conclude that $p \\mid x^3$ and hence $p \\mid x$. Now we have:\n$$\np \\mid y - x, \\quad p \\mid x + y + 1, \\quad p \\mid x \\quad \\implies \\quad p \\mid (x + y + 1) - (y - x) - 2x = 1,\n$$\nwhich means that our assumption was wrong, i.e. $y - x$ and $x + y + 1$ are relatively prime numbers whose product is $x^3$ so each of them is a cube of an integer. Furthermore, $x^3 > 0$ and $x + y + 1 > 0$ so $y - x > 0$, which means that $y - x$ and $x + y + 1$ are cubes of positive integers. Let us denote:\n$$\ny - x = a^3 \\\\\nx + y + 1 = b^3.\n$$\nNotice that from $2x + 1 > 0$ it follows that $y - x < x + y + 1$, that is $a < b$, i.e. $b - a \\ge 1$. Subtracting the previous two equations we get $2x + 1 = b^3 - a^3$, and from above it follows $x = ab$. Therefore, $2ab + 1 = b^3 - a^3$. Now:\n$$\n2ab + 1 = b^3 - a^3 = (b - a)(a^2 + ab + b^2) \\ge a^2 + ab + b^2 \\ge 3ab,\n$$\ni.e. $ab \\le 1$, which is clearly not possible since $a$ and $b$ are distinct positive integers. Hence, the only solutions of the given equation are $(0, 0)$ and $(0, -1)$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 28752,
    "subject": "Mathematics (Multi-modal)",
    "question": "$$\nS = \\{0, 1, 1+2, 1+2+3, \\dots, 1+2+3+\\dots+(n-1)\\}.\n$$\n\na) If $n$ is a power of $2$, prove that all the elements of $S$ have different remainders modulo $n$.\n\nb) If $n$ is not a power of $2$, prove that there exist two elements of $S$ with the same remainder modulo $n$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $k$ and $l$ be positive integers such that $0 \\le k < l \\le n-1$. Sums $1+2+\\dots+k$ and $1+2+\\dots+l$ give the same remainder modulo $n$ if and only if\n$$\n\\frac{1}{2} k (k + 1) \\equiv \\frac{1}{2} l (l + 1) \\pmod{n},\n$$\ni.e. if and only if $l(l+1) - k(k+1) = (l-k)(l+k+1)$ is divisible by $2n$.\nNotice that $l-k$ and $l+k+1$ give a different remainder modulo $2$.\n\na) Let $n = 2^m$. Then $2n = 2^{m+1}$, and since $l-k$ and $l+k+1$ give a different remainder modulo $2$, one of them has to be divisible by $2^{m+1}$.\nOn the other hand, clearly $l-k < 2^{m+1}$, and we have $l+k+1 \\le 2l < 2n = 2^{m+1}$, so none of those numbers is divisible by $2^{m+1}$.\nWe conclude that distinct elements of the set $S$ give different remainders modulo $n$.\n\nb) Let $n = 2^m s$ where $s > 1$ is odd.\nWe have to choose $k$ and $l$, $0 \\le k < l \\le n-1$ such that\n$$\n(l-k)(l+k+1) = 2n = 2^{m+1}s.\n$$\nIf $2^{m+1} \\ge s+1$ we will choose them so that $l+k+1 = 2^{m+1}$ and $l-k = s$, and if $2^{m+1} \\le s-1$ then we will choose them so that $l+k+1 = s$ and $l-k = 2^{m+1}$.\nIn the first case solving the system we get $l = 2^m + \\frac{s-1}{2}$ and $k = 2^m - \\frac{s+1}{2}$. Clearly $0 \\le k < l$. Since $2 < s \\le 2^{m+1} - 1$, it follows that\n$$\nl = 2^m + \\frac{s-1}{2} \\le 2^m + \\frac{2^{m+1}-2}{2} = 2^{m+1} - 1 < 2^m s - 1.\n$$\nIn the second case we get $l = \\frac{s-1}{2} + 2^m$ and $k = \\frac{s-1}{2} - 2^m$. Again it is clear that $0 \\le k < l$. Finally, we check\n$$\nl = \\frac{s-1}{2} + 2^m \\le \\frac{s-1}{2} + \\frac{s-1}{2} = s-1 \\le 2^m s - 1.\n$$",
    "topic": "Number Theory",
    "subtopic": "Modular Arithmetic"
  },
  {
    "id": 28753,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find at least one integer $a$ such that for the polynomial $P(x) = x^5 + a x$ the statement\n“if $n \\mid P(k) - P(l)$ then $n \\mid k - l$, for all $k, l \\in \\mathbb{Z}$”\nholds only for finitely many positive integers $n$, including $n = 95$.",
    "options": [],
    "answer": "-95^4",
    "solution": "One such number is $a = -95^4$.\nFor that $a$ we have $P(95) = P(0) = 0$, so $n$ divides $P(95) - P(0)$ for every positive integer $n$, and $n$ does not divide $95 - 0$ if $n$ is not a divisor of $95$. Therefore the statement is valid for only finitely many positive integers $n$.\n\nLet us show that the statement is valid for $n = 95$. Let $k, l \\in \\mathbb{Z}$ such that $95 \\mid P(k) - P(l)$. Since $95 \\mid a$ we conclude that $95 \\mid k^5 - l^5$. We want to show that $95 \\mid k - l$, so it is sufficient to prove that $5 \\mid k - l$ and $19 \\mid k - l$.\n\nBy Fermat's little theorem we have $k^5 \\equiv k \\pmod{5}$ and $l^5 \\equiv l \\pmod{5}$, so it follows that $5 \\mid k - l$ because $5 \\mid k^5 - l^5$.\n\nAlso by Fermat's little theorem, if $k$ and $19$ are relatively prime then $k^{18} \\equiv 1 \\pmod{19}$, so $k^{54} = (k^{18})^3 \\equiv 1 \\pmod{19}$. Therefore $k^{55} \\equiv k \\pmod{19}$ whether $k$ is divisible by $19$ or not. Analogously $l^{55} \\equiv l \\pmod{19}$. We conclude that $k \\equiv (k^5)^{11} \\equiv (l^5)^{11} \\equiv l \\pmod{19}$, i.e. $19 \\mid k - l$, which is what we had to prove.",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 28754,
    "subject": "Mathematics (Multi-modal)",
    "question": "Determine all pairs $(x, y)$ of integers which satisfy\n$$\n6x^{2}y^{2} - 4y^{2} = 2012 - 3x^{2}.\n$$",
    "options": [],
    "answer": "(-2, -10), (-2, 10), (2, -10), (2, 10)",
    "solution": "The given equality is equivalent to\n$$\n\\begin{aligned}\n6x^2y^2 + 3x^2 - 4y^2 &= 2012 \\\\\n3x^2(2y^2 + 1) - 2(2y^2 + 1) &= 2012 - 2 \\\\\n(3x^2 - 2)(2y^2 + 1) &= 2010\n\\end{aligned}\n$$\nSince $2010 = 2 \\cdot 3 \\cdot 5 \\cdot 67$ and the number $2y^2 + 1$ is odd, we conclude that\n$$\n\\begin{aligned}\n2y^2 + 1 &\\in \\{1, 3, 5, 15, 67, 201, 335, 1005\\}, \\\\\n2y^2 &\\in \\{0, 2, 4, 14, 66, 200, 334, 1004\\}, \\\\\ny^2 &\\in \\{0, 1, 2, 7, 33, 100, 167, 502\\}.\n\\end{aligned}\n$$\nThereby $y^2$ must be $0$, $1$ or $100$.\nThere are three cases left:\n$$\n\\left\\{ \n\\begin{array}{l}\n3x^2 - 2 = 2010 \\\\\n2y^2 + 1 = 1\n\\end{array}\n\\right.\n\\quad\n\\left\\{ \n\\begin{array}{l}\n3x^2 - 2 = 670 \\\\\n2y^2 + 1 = 3\n\\end{array}\n\\right.\n\\quad\n\\left\\{ \n\\begin{array}{l}\n3x^2 - 2 = 10 \\\\\n2y^2 + 1 = 201\n\\end{array}\n\\right.\n$$\nWe see that only in the third case there are integer solutions,\n$$\n(x, y) \\in \\{(2, 10), (-2, 10), (2, -10), (-2, -10)\\}.\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 28755,
    "subject": "Mathematics (Multi-modal)",
    "question": "Determine all pairs of integers $(a, b)$ such that $a (a - b) = b$.",
    "options": [],
    "answer": "(0, 0) and (-2, -4)",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 28756,
    "subject": "Mathematics (Multi-modal)",
    "question": "How many perfect squares are there between $4^9$ and $9^4$, excluding those two numbers?",
    "options": [],
    "answer": "430",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 28757,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $a$ be a real number. Determine the sum of all three solutions of the equation\n$$\nx^3 - a^2x + ax - x + a^2 - a = 0.\n$$",
    "options": [],
    "answer": "0",
    "solution": "Let us write the equation as:\n$$\nx^3 - a^2x + ax - x + a^2 - a = 0.\n$$\nGroup the $x$ terms:\n$$\nx^3 + (a - a^2 - 1)x + (a^2 - a) = 0.\n$$\nThis is a cubic equation of the form:\n$$\nx^3 + px + q = 0,\n$$\nwhere $p = a - a^2 - 1$ and $q = a^2 - a$.\n\nBy Vieta's formulas, the sum of the roots of the cubic equation $x^3 + px + q = 0$ is $0$ (since the coefficient of $x^2$ is $0$).\n\nTherefore, the sum of all three solutions is $\\boxed{0}$.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 28758,
    "subject": "Mathematics (Multi-modal)",
    "question": "One of the numbers $x^2$ and $(1-x)^2$ is smaller, and the other is greater than $1$.\nProve that $0 < x^2 - x < 2$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 28759,
    "subject": "Mathematics (Multi-modal)",
    "question": "Prove that for all real numbers $a, b, c$ the following inequality holds\n$$\n\\frac{1}{3}(a+b+c)^2 \\leq a^2 + b^2 + c^2 + 2(a-b+1).\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Transforming the right hand side of the inequality we get\n$$\n\\begin{aligned}\na^2 + b^2 + c^2 + 2(a - b + 1) &= a^2 + 2a + 1 + b^2 - 2b + 1 + c^2 \\\\\n&= (a + 1)^2 + (b - 1)^2 + c^2.\n\\end{aligned}\n$$\nFrom the A–K inequality it follows\n$$\n\\sqrt{\\frac{(a+1)^2 + (b-1)^2 + c^2}{3}} \\ge \\frac{(a+1) + (b-1) + c}{3}\n$$\ni.e.\n$$\n(a + 1)^2 + (b - 1)^2 + c^2 \\ge \\frac{1}{3} (a + b + c)^2.\n$$",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 28760,
    "subject": "Mathematics (Multi-modal)",
    "question": "Each digit of a positive integer $n$ (except the first one) is larger than the digit next left to it. Determine the sum of all digits of the number $9n$. (Russia 1999)",
    "options": [],
    "answer": "9",
    "solution": "Let $n = \\overline{a_{m-1}a_{m-2}\\dots a_1a_0}$. We know that $a_0 > a_1 > \\dots > a_{m-1}$.\n\nSince $9n = 10n - n$, from\n$$\n\\begin{array}{c c c c c c c c c c c c c c c}\n& a_{m-1} & a_{m-2} & a_{m-3} & \\dots & a_1 & a_0 & 0 \\\\\n- & \\multicolumn{1}{c}{a_{m-1}} & a_{m-2} & \\dots & a_2 & a_1 & a_0 \\\\\n\\hline\nb_m & b_{m-1} & b_{m-2} & \\dots & b_2 & b_1 & b_0\n\\end{array}\n$$\nwe conclude $b_0 = 10 - a_0$, $b_1 = a_0 - (a_1 + 1)$, $b_2 = a_1 - a_2$, $b_3 = a_2 - a_3$, ..., $b_{m-1} = a_{m-2} - a_{m-1}$, $b_m = a_{m-1}$. Those are the digits of $9n$, and their sum is $a_{m-1} + (a_{m-2} - a_{m-1}) + \\dots + (a_1 - a_2) + (a_0 - a_1 - 1) + (10 - a_0) = 9$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 28761,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABCD$ be a parallelogram with an acute angle at vertex $A$. $G$ is a point on the line $AB$, different from $B$, such that $|BC| = |CG|$, and $H$ is a point on the line $BC$, different from $B$, such that $|AB| = |AH|$. Prove that the triangle $DGH$ is isosceles.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 28762,
    "subject": "Mathematics (Multi-modal)",
    "question": "If $a + b = 4$ and $a^2 + b^2 = 14$, determine $a^3 + b^3$.",
    "options": [],
    "answer": "52",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 28763,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABCDE$ be a regular pentagon. The lines $BC$ and $DE$ intersect at point $F$. Determine the angles of triangle $BEF$.",
    "options": [],
    "answer": "∠EBF = 108°, ∠BEF = 54°, ∠BFE = 18°",
    "solution": "",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 28764,
    "subject": "Mathematics (Multi-modal)",
    "question": "Determine the number of pairs of integers $(x, y)$ that satisfy the equality:\n$$\n(x + y + 2012)^2 = x^2 + y^2 + 2012^2 .\n$$",
    "options": [],
    "answer": "30",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 28765,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $\\triangle ABC$ be a triangle with an obtuse angle at vertex $B$, let $D$ and $E$ be midpoints of the segments $\\overline{AB}$ and $\\overline{AC}$ respectively, let $F$ be a point on the segment $\\overline{BC}$ such that $\\angle BFE = 90^\\circ$, and let $G$ be a point on the segment $\\overline{DE}$ such that $\\angle BGE = 90^\\circ$. Prove that the points $A, F$ and $G$ are collinear if and only if $2|BF| = |CF|$. (Matija Bašić)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Obviously, $BFEG$ is a rectangle. The triangles $ADE$ and $ABC$ are similar with the coefficient of similarity $2$. The points $A$, $F$ and $G$ are collinear if and only if $|FB| : |GD| = 2 : 1$.\n\nLet $|BC| = a$ and $|BF| = x$. Then $|ED| = \\frac{1}{2}a$, $|EG| = |FB| = x$, $|GD| = |DE| - |EG| = \\frac{1}{2}a - x$, and the equality $|FB| = 2|GD|$ is equivalent to $x = 2\\left(\\frac{1}{2}a - x\\right)$, i.e. $a = 3x$.\n\nTherefore $|FB| = 2|GD|$ is equivalent to $|BC| = 3|BF|$, i.e. $|CF| = 2|BF|$.\n\n![](attached_image_1.png)",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 28766,
    "subject": "Mathematics (Multi-modal)",
    "question": "Azra thought of four real numbers and wrote on the blackboard the sums of all pairs of imagined numbers, and then she deleted one of the sums. There were numbers $-2$, $1$, $2$, $3$ and $6$ left on the blackboard. What numbers did Azra think of? (M. Bašić, M. Bombardelli)",
    "options": [],
    "answer": "−3/2, −1/2, 5/2, 7/2",
    "solution": "Let $a$, $b$, $c$ and $d$ be the numbers Azra thought of. Without loss of generality, we can assume that the deleted sum is $c+d$. Then there are numbers $a+b$, $a+c$, $a+d$, $b+c$ and $b+d$ written on the blackboard, i.e.\n$$\n\\{-2, 1, 2, 3, 6\\} = \\{a+b, a+c, a+d, b+c, b+d\\}.\n$$\nSince\n$$\n(a+c) + (b+d) = (a+d) + (b+c),\n$$\namong the numbers on the blackboard we can choose two pairs of numbers with equal sums. We easily find the only such pairs $-2+6=1+3=4$, and therefore the sum of all numbers Azra thought of equals $4$.\nAmong the numbers on the blackboard, the number $2$ does not appear in the last equality (so $a+b = 2$) and the deleted number $c+d$ equals $4-2=2$.\nLet us now change the notation. Let $a$, $b$, $c$, $d$ be the numbers Azra thought of, such that $a \\le b \\le c \\le d$. We know that $\\{-2, 1, 2, 2, 3, 6\\} = \\{a+b, a+c, a+d, b+c, b+d, c+d\\}$.\nSince $a \\le b \\le c \\le d$, obviously $a+b$ is the smallest sum, and $c+d$ is the largest, so $a+b = -2$, $c+d = 6$. Now it is easy to see that the sum $a+c$ is smaller than all the sums except $a+b$, and analogously $b+d$ is larger than all the sums except $c+d$. So, $a+c = 1$ and $b+d = 3$. Finally, $a+d = b+c = 2$.\nWe see that $2a = (a+b) + (a+c) - (b+c) = -2+1-2 = -3$, i.e. $a = -\\frac{3}{2}$, and further $b = -2 - a = -\\frac{1}{2}$, $c = 1 - a = \\frac{5}{2}$ and $d = 2 - a = \\frac{7}{2}$.\nLet us check that the other equalities are satisfied: $b+c = -\\frac{1}{2} + \\frac{5}{2} = 2$, $b+d = -\\frac{1}{2} + \\frac{7}{2} = 3$, $c+d = \\frac{5}{2} + \\frac{7}{2} = 6$.\nAzra thought of numbers $-\\frac{3}{2}$, $-\\frac{1}{2}$, $\\frac{5}{2}$ and $\\frac{7}{2}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 28767,
    "subject": "Mathematics (Multi-modal)",
    "question": "Eleonora has many cubes with all white sides. First she takes one cube and puts it in an empty box. Then she takes one cube at a time and paints some of its sides green, in such a way that the cube is different from all the other cubes that are already in the box, and then she puts that cube also in the box. At most how many cubes can there be in the box?",
    "options": [],
    "answer": "10",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 28768,
    "subject": "Mathematics (Multi-modal)",
    "question": "The numbers $2009$, $2012$ and $2015$ are written on the board. In each step Željko marks the numbers on the board by $a$, $b$ and $c$ in some order, and then replaces them with the numbers $3a - b$, $3b - c$ and $3c - a$. By repeating this procedure multiple times, can Željko achieve that at some point all the numbers on the board are equal?",
    "options": [],
    "answer": "No",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 28769,
    "subject": "Mathematics (Multi-modal)",
    "question": "An isosceles triangle has the basis of length $a$, legs of length $b$, and the circumradius $R$. Prove that the equality $a^2R^2 + b^4 = 4b^2R^2$ holds, regardless of whether the triangle is acute, right or obtuse.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 28770,
    "subject": "Mathematics (Multi-modal)",
    "question": "Ana is four times as old as Petar was when Ana was as old as Petar is now. When Petar becomes as old as Ana is now, the sum of their ages will be $95$. How old is Ana, and how old is Petar?",
    "options": [],
    "answer": "Ana is 40 years old and Petar is 25 years old.",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 28771,
    "subject": "Mathematics (Multi-modal)",
    "question": "Determine all pairs of prime numbers $p$ and $q$ for which there exists an integer $a$ such that $a^4 = p a^3 + q$.",
    "options": [],
    "answer": "(p, q) = (2, 3)",
    "solution": "",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 28772,
    "subject": "Mathematics (Multi-modal)",
    "question": "Depending on the real parameter $a$, solve the equation\n$$\n(a - 1) (1 + x + x^2)^2 = (a + 1) (1 + x^2 + x^4)\n$$",
    "options": [],
    "answer": "The equation is equivalent to (x^2 + b x + 1)(x^2 + c x + 1) = 0, where b and c are the roots of t^2 + (a − 1)t − a = 0. Explicitly, b, c = (−(a − 1) ± √(a^2 + 6a + 1)) / 2. Hence all real solutions x satisfy either x^2 + ((−(a − 1) + √(a^2 + 6a + 1)) / 2) x + 1 = 0 or x^2 + ((−(a − 1) − √(a^2 + 6a + 1)) / 2) x + 1 = 0.",
    "solution": "First, expand both sides:\n\nLeft side:\n$(a - 1)(1 + x + x^2)^2 = (a - 1)(1 + 2x + 3x^2 + 2x^3 + x^4)$\n\nRight side:\n$(a + 1)(1 + x^2 + x^4)$\n\nBring all terms to one side:\n$$(a - 1)(1 + 2x + 3x^2 + 2x^3 + x^4) - (a + 1)(1 + x^2 + x^4) = 0$$\n\nExpand:\n$(a - 1)(1) + (a - 1)(2x) + (a - 1)(3x^2) + (a - 1)(2x^3) + (a - 1)(x^4)$\n$- (a + 1)(1) - (a + 1)(x^2) - (a + 1)(x^4) = 0$\n\nGroup like terms:\n- Constant: $(a - 1) - (a + 1) = -2$\n- $x$ term: $2(a - 1)$\n- $x^2$ term: $3(a - 1) - (a + 1) = 3a - 3 - a - 1 = 2a - 4$\n- $x^3$ term: $2(a - 1)$\n- $x^4$ term: $(a - 1) - (a + 1) = -2$\n\nSo the equation becomes:\n$$\n-2 + 2(a - 1)x + (2a - 4)x^2 + 2(a - 1)x^3 - 2x^4 = 0\n$$\n\nDivide both sides by $2$:\n$$\n-1 + (a - 1)x + (a - 2)x^2 + (a - 1)x^3 - x^4 = 0\n$$\n\nOr:\n$$\n-x^4 + (a - 1)x^3 + (a - 2)x^2 + (a - 1)x - 1 = 0\n$$\n\nOr, multiplying both sides by $-1$:\n$$\nx^4 - (a - 1)x^3 - (a - 2)x^2 - (a - 1)x + 1 = 0\n$$\n\nLet us factor this quartic. Notice that if $x = 1$:\n\n$x^4 - (a - 1)x^3 - (a - 2)x^2 - (a - 1)x + 1 = 1 - (a - 1) - (a - 2) - (a - 1) + 1$\n$= 1 - a + 1 - a + 2 - a + 1 + 1$\nBut let's compute step by step:\n$1 - (a - 1) = 2 - a$\n$2 - a - (a - 2) = 2 - a - a + 2 = 4 - 2a$\n$4 - 2a - (a - 1) = 4 - 2a - a + 1 = 5 - 3a$\n$5 - 3a + 1 = 6 - 3a$\nSo $x = 1$ is a root if $6 - 3a = 0$, i.e., $a = 2$.\n\nTry $x = -1$:\n$(-1)^4 - (a - 1)(-1)^3 - (a - 2)(-1)^2 - (a - 1)(-1) + 1$\n$= 1 - (a - 1)(-1) - (a - 2)(1) - (a - 1)(-1) + 1$\n$= 1 + (a - 1) - (a - 2) + (a - 1) + 1$\n$= 1 + a - 1 - a + 2 + a - 1 + 1$\n$= (1 - 1 + 2 + 1) + (a - a + a) - 1$\n$= (3) + (a) - 1$\n$= a + 2$\nSo $x = -1$ is a root if $a = -2$.\n\nAlternatively, factor the quartic as follows:\n\nLet us try to factor as $(x^2 + bx + 1)(x^2 + cx + 1)$.\n\nExpand:\n$(x^2 + bx + 1)(x^2 + cx + 1) = x^4 + (b + c)x^3 + (bc + 2)x^2 + (b + c)x + 1$\n\nCompare with $x^4 - (a - 1)x^3 - (a - 2)x^2 - (a - 1)x + 1$:\n\nSo:\n- $x^4$ coefficient: $1$\n- $x^3$ coefficient: $b + c = -(a - 1)$\n- $x^2$ coefficient: $bc + 2 = -(a - 2)$\n- $x$ coefficient: $b + c = -(a - 1)$\n- constant: $1$\n\nSo $b + c = -(a - 1)$\n$bc + 2 = -(a - 2)$\n\nLet $b + c = s = -(a - 1)$\n$bc = -(a - 2) - 2 = -(a) + 2 - 2 = -a$\n\nSo $b$ and $c$ are roots of $t^2 - s t - a = 0$\n\nSo $t^2 + (a - 1)t - a = 0$\n\nThus, the quartic factors as:\n$$\n(x^2 + b x + 1)(x^2 + c x + 1) = 0\n$$\nwhere $b$ and $c$ are roots of $t^2 + (a - 1)t - a = 0$.\n\nTherefore, the solutions are all real $x$ such that\n$$\nx^2 + b x + 1 = 0 \\quad \\text{or} \\quad x^2 + c x + 1 = 0\n$$\nwhere $b$ and $c$ are roots of $t^2 + (a - 1)t - a = 0$.\n\nExplicitly, $b, c = \\frac{-(a - 1) \\pm \\sqrt{(a - 1)^2 + 4a}}{2} = \\frac{-(a - 1) \\pm \\sqrt{a^2 + 2a + 1 + 4a}}{2} = \\frac{-(a - 1) \\pm \\sqrt{a^2 + 6a + 1}}{2}$\n\nSo the solutions are all real $x$ such that\n$$\nx^2 + \\left(\\frac{-(a - 1) + \\sqrt{a^2 + 6a + 1}}{2}\\right)x + 1 = 0\n$$\nor\n$$\nx^2 + \\left(\\frac{-(a - 1) - \\sqrt{a^2 + 6a + 1}}{2}\\right)x + 1 = 0\n$$\n\nThat is, for each $a$, the equation has at most four real solutions, given by the roots of these two quadratics.\n\nSpecial cases:\n- If $a = 2$, then $b + c = -(2 - 1) = -1$, $bc = -2$; so $b$ and $c$ are roots of $t^2 + t - 2 = 0$, i.e., $t = 1, -2$.\n  So the quadratics are $x^2 + x + 1 = 0$ and $x^2 - 2x + 1 = 0$.\n  The first has no real roots, the second is $(x - 1)^2 = 0$, so $x = 1$ is a double root.\n- If $a = -2$, then $b + c = -(-2 - 1) = 3$, $bc = 2$; so $t^2 - 3t - 2 = 0$, $t = \\frac{3 \\pm \\sqrt{9 + 8}}{2} = \\frac{3 \\pm \\sqrt{17}}{2}$.\n  Both quadratics $x^2 + b x + 1 = 0$ and $x^2 + c x + 1 = 0$ have real roots.\n\nIn summary:\n\nFor each real $a$, the solutions are all real $x$ such that\n$$\nx^2 + b x + 1 = 0 \\quad \\text{or} \\quad x^2 + c x + 1 = 0\n$$\nwhere $b$ and $c$ are the roots of $t^2 + (a - 1)t - a = 0$.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 28773,
    "subject": "Mathematics (Multi-modal)",
    "question": "Determine all values of the real parameter $m$ for which the equation\n$$\n(m - 1)x^2 - 2mx + 2 = 0\n$$\nhas no real solutions.",
    "options": [],
    "answer": "no real values of m",
    "solution": "Let us consider the quadratic equation:\n$$\n(m - 1)x^2 - 2mx + 2 = 0.\n$$\nThis equation has no real solutions if and only if its discriminant is negative.\n\nThe discriminant $D$ is:\n$$\nD = [-2m]^2 - 4(m-1) \\cdot 2 = 4m^2 - 8(m-1) = 4m^2 - 8m + 8.\n$$\nWe require:\n$$\n4m^2 - 8m + 8 < 0.\n$$\nDivide both sides by $4$:\n$$\nm^2 - 2m + 2 < 0.\n$$\nBut the quadratic $m^2 - 2m + 2$ has discriminant:\n$$\n(-2)^2 - 4 \\cdot 1 \\cdot 2 = 4 - 8 = -4 < 0.\n$$\nSo $m^2 - 2m + 2$ is always positive for all real $m$.\n\nTherefore, there is **no real value of $m$** for which the equation has no real solutions.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 28774,
    "subject": "Mathematics (Multi-modal)",
    "question": "Prove that for every integer $k \\ge 2$ there exist $k$ positive integers whose sum is equal to their product.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 28775,
    "subject": "Mathematics (Multi-modal)",
    "question": "Ante wrote all integers from $1$ to $40$ in a row without spacings, thus obtaining the multi-digit number $12345\\ldots383940$. Then he decided to erase $60$ digits of this number. What is the largest number Ante can get by doing this?",
    "options": [],
    "answer": "99967383940",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 28776,
    "subject": "Mathematics (Multi-modal)",
    "question": "A triangle with an area of $1.5 \\text{ cm}^2$ is inscribed in a circle of radius $1.25$ cm, with one side of the triangle as the diameter of the circle. What are the lengths of the sides of the triangle?",
    "options": [],
    "answer": "1.5 cm, 2 cm, 2.5 cm",
    "solution": "",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 28777,
    "subject": "Mathematics (Multi-modal)",
    "question": "Determine all positive integers $n$ that satisfy the equality\n$$\n5^n + 2^{n+1} 3^n = 9^n + 4^n.\n$$",
    "options": [],
    "answer": "2",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 28778,
    "subject": "Mathematics (Multi-modal)",
    "question": "An isosceles triangle $ABC$ with $|AB| = |AC|$ is given. Tangents to its circumcircle at points $A$ and $C$ intersect in $D$. If $\\angle DBC = 30^\\circ$, prove that triangle $ABC$ is equilateral.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $BCE$ and $ECF$ be the equilateral triangles such that the points $A$ and $E$ are at the same side of the line $BC$, and $B$ and $F$ at different sides of the line $CE$. We will assume that $E$ and $A$ are different points, otherwise the statement holds.\n\n![](attached_image_1.png)\n\nSince $\\angle FBC = 30^\\circ$, and then $\\angle FBC = \\angle DBC$, we conclude that $B$, $F$ and $D$ are collinear. Triangles $DFC$ and $DFE$ are congruent (SAS), so $|DE| = |DC| = |DA|$. Since $AE$ is perpendicular to $BC$ and then also to $AD$, from the right triangle $DAE$ we see that $|DE| > |DA|$, which is a contradiction.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 28779,
    "subject": "Mathematics (Multi-modal)",
    "question": "For which $x \\in \\mathbb{R}$ is the number $\\sqrt[3]{4+4x}$ greater than the number $1+\\sqrt[3]{x}$?",
    "options": [],
    "answer": "(-1, 1) ∪ (1, ∞)",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 28780,
    "subject": "Mathematics (Multi-modal)",
    "question": "Prove that for all $a, b, c > 0$ such that $a + b + c \\le 3$ the following inequality holds\n$$\n\\frac{a+1}{a(a+2)} + \\frac{b+1}{b(b+2)} + \\frac{c+1}{c(c+2)} \\ge 2.\n$$\n(Tangenta magazine, 2010)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "From the A–H inequality we get\n$$\n\\begin{aligned}\n\\frac{a+1}{a(a+2)} + \\frac{b+1}{b(b+2)} + \\frac{c+1}{c(c+2)} &\\ge \\frac{9}{\\frac{a(a+2)}{a+1} + \\frac{b(b+2)}{b+1} + \\frac{c(c+2)}{c+1}} \\\\\n&= \\frac{9}{\\frac{(a+1)^2-1}{a+1} + \\frac{(b+1)^2-1}{b+1} + \\frac{(c+1)^2-1}{c+1}} \\\\\n&= \\frac{9}{(a+1) + (b+1) + (c+1) - \\left(\\frac{1}{a+1} + \\frac{1}{b+1} + \\frac{1}{c+1}\\right)}\n\\end{aligned}\n$$\nSince $(a+1) + (b+1) + (c+1) = a+b+c+3 \\le 6$ and\n$$\n\\frac{1}{a+1} + \\frac{1}{b+1} + \\frac{1}{c+1} \\ge \\frac{9}{(a+1) + (b+1) + (c+1)} \\ge \\frac{9}{6} = \\frac{3}{2}\n$$\nwe can conclude\n$$\n\\frac{9}{(a+1) + (b+1) + (c+1) - \\left(\\frac{1}{a+1} + \\frac{1}{b+1} + \\frac{1}{c+1}\\right)} \\ge \\frac{9}{6 - \\frac{3}{2}} = 2\n$$\nwhich finishes the proof.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 28781,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $P$ be a point inside the triangle $ABC$. Let $D$, $E$ and $F$ be the feet of the perpendiculars from point $P$ to $BC$, $CA$, and $AB$ respectively. If the quadrilaterals $AEPF$, $BFPD$, and $CDPE$ are circumscribed, prove that $P$ is the incentre of triangle $ABC$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 28782,
    "subject": "Mathematics (Multi-modal)",
    "question": "The figure consisting of five unit squares is called a “plus”. In how many ways can we place a *plus* on a *board* of the same shape that contains $5 \\cdot 5^2$ unit squares, so that the *plus* covers exactly five unit squares of the *board*?\n![](attached_image_1.png)",
    "options": [],
    "answer": "73",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 28783,
    "subject": "Mathematics (Multi-modal)",
    "question": "Determine the number of quadratic functions with coefficients from the set\n$$\nS = \\{1, 2, 4, 8, 16, 32\\},\n$$\nthat have no real roots.",
    "options": [],
    "answer": "135",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 28784,
    "subject": "Mathematics (Multi-modal)",
    "question": "Determine all positive integers smaller than $1000$ that are equal to the sum of squares of their digits.",
    "options": [],
    "answer": "1, 4, 9",
    "solution": "Let $n$ be a positive integer less than $1000$ such that $n$ is equal to the sum of the squares of its digits.\n\nLet $n$ have digits $a$, $b$, $c$ (possibly with leading zeros), so $n = 100a + 10b + c$, where $0 \\leq a \\leq 9$, $0 \\leq b \\leq 9$, $0 \\leq c \\leq 9$, and $n < 1000$.\n\nWe want $n = a^2 + b^2 + c^2$.\n\nLet us check all possible $n$:\n\n- For $n$ with $3$ digits, $n \\geq 100$. The maximum sum of squares of digits is $3 \\times 9^2 = 243$, so $n$ cannot be a $3$-digit number.\n- For $n$ with $2$ digits, $n = 10a + b$, $n \\geq 10$. The maximum sum is $2 \\times 9^2 = 162$, so possible.\n- For $n$ with $1$ digit, $n = a$, $n = a^2$, so $a = 0$ or $1$, but $n$ must be positive, so $n = 1$.\n\nLet us check all $n$ from $1$ to $99$:\n\nFor $n = 1$ to $9$:\n- $n = 1$: $1^2 = 1$ ✓\n- $n = 4$: $2^2 = 4$ ✓\n- $n = 9$: $3^2 = 9$ ✓\n\nFor $n = 10$ to $99$:\nLet $n = 10a + b$, $n = a^2 + b^2$.\nSo $10a + b = a^2 + b^2$.\nRewriting: $a^2 - 10a + b^2 - b = 0$.\nTry $a$ from $1$ to $9$:\n\nFor $a = 1$:\n$1 - 10 + b^2 - b = 0 \\implies b^2 - b - 9 = 0$\nDiscriminant: $1 + 36 = 37$, not a perfect square.\n\nFor $a = 2$:\n$4 - 20 + b^2 - b = 0 \\implies b^2 - b - 16 = 0$\nDiscriminant: $1 + 64 = 65$, not a perfect square.\n\nFor $a = 3$:\n$9 - 30 + b^2 - b = 0 \\implies b^2 - b - 21 = 0$\nDiscriminant: $1 + 84 = 85$, not a perfect square.\n\nFor $a = 4$:\n$16 - 40 + b^2 - b = 0 \\implies b^2 - b - 24 = 0$\nDiscriminant: $1 + 96 = 97$, not a perfect square.\n\nFor $a = 5$:\n$25 - 50 + b^2 - b = 0 \\implies b^2 - b - 25 = 0$\nDiscriminant: $1 + 100 = 101$, not a perfect square.\n\nFor $a = 6$:\n$36 - 60 + b^2 - b = 0 \\implies b^2 - b - 24 = 0$\nAlready checked.\n\nFor $a = 7$:\n$49 - 70 + b^2 - b = 0 \\implies b^2 - b - 21 = 0$\nAlready checked.\n\nFor $a = 8$:\n$64 - 80 + b^2 - b = 0 \\implies b^2 - b - 16 = 0$\nAlready checked.\n\nFor $a = 9$:\n$81 - 90 + b^2 - b = 0 \\implies b^2 - b - 9 = 0$\nAlready checked.\n\nAlternatively, try all $n$ from $10$ to $99$ and check if $n = a^2 + b^2$ for $a, b$ digits.\n\nLet us try $n = 13$:\nDigits: $1, 3$. $1^2 + 3^2 = 1 + 9 = 10 \\neq 13$.\n\nTry $n = 25$:\n$2^2 + 5^2 = 4 + 25 = 29 \\neq 25$.\n\nTry $n = 45$:\n$4^2 + 5^2 = 16 + 25 = 41 \\neq 45$.\n\nTry $n = 82$:\n$8^2 + 2^2 = 64 + 4 = 68 \\neq 82$.\n\nTry $n = 85$:\n$8^2 + 5^2 = 64 + 25 = 89 \\neq 85$.\n\nTry $n = 130$:\n$1^2 + 3^2 + 0^2 = 1 + 9 + 0 = 10 \\neq 130$.\n\nTry $n = 130$ to $999$:\nBut as above, the maximum sum of squares for $3$ digits is $243$, so $n$ cannot be $3$ digits.\n\nNow, try all $n$ from $10$ to $99$:\nLet us check for $n = a^2 + b^2$ where $a$ and $b$ are digits and $n = 10a + b$.\n\nAlternatively, list all possible $a, b$ and compute $n = 10a + b$, and check if $n = a^2 + b^2$.\n\nLet us try $a$ from $0$ to $9$, $b$ from $0$ to $9$:\n\nFor $a = 0$:\n$b^2 = 10 \\times 0 + b \\implies b^2 - b = 0 \\implies b(b - 1) = 0 \\implies b = 0$ or $b = 1$\nSo $n = 0$ or $1$ (but $n$ must be positive), so $n = 1$.\n\nFor $a = 1$:\n$1 + b^2 = 10 + b \\implies b^2 - b = 9 \\implies b^2 - b - 9 = 0$\nDiscriminant: $1 + 36 = 37$, not a perfect square.\n\nFor $a = 2$:\n$4 + b^2 = 20 + b \\implies b^2 - b = 16 \\implies b^2 - b - 16 = 0$\nDiscriminant: $1 + 64 = 65$, not a perfect square.\n\nFor $a = 3$:\n$9 + b^2 = 30 + b \\implies b^2 - b = 21 \\implies b^2 - b - 21 = 0$\nDiscriminant: $1 + 84 = 85$, not a perfect square.\n\nFor $a = 4$:\n$16 + b^2 = 40 + b \\implies b^2 - b = 24 \\implies b^2 - b - 24 = 0$\nDiscriminant: $1 + 96 = 97$, not a perfect square.\n\nFor $a = 5$:\n$25 + b^2 = 50 + b \\implies b^2 - b = 25 \\implies b^2 - b - 25 = 0$\nDiscriminant: $1 + 100 = 101$, not a perfect square.\n\nFor $a = 6$:\n$36 + b^2 = 60 + b \\implies b^2 - b = 24 \\implies b^2 - b - 24 = 0$\nAlready checked.\n\nFor $a = 7$:\n$49 + b^2 = 70 + b \\implies b^2 - b = 21 \\implies b^2 - b - 21 = 0$\nAlready checked.\n\nFor $a = 8$:\n$64 + b^2 = 80 + b \\implies b^2 - b = 16 \\implies b^2 - b - 16 = 0$\nAlready checked.\n\nFor $a = 9$:\n$81 + b^2 = 90 + b \\implies b^2 - b = 9 \\implies b^2 - b - 9 = 0$\nDiscriminant: $1 + 36 = 37$, not a perfect square.\n\nTherefore, the only possible $n$ are $1$, $4$, and $9$.\n\nThus, the positive integers less than $1000$ that are equal to the sum of the squares of their digits are:\n\n$1$, $4$, $9$, $130$, $133$, $155$, $175$, $203$, $222$, $229$, $233$, $262$, $263$, $291$, $292$, $293$, $319$, $320$, $326$, $329$, $346$, $355$, $362$, $365$, $397$, $400$, $466$, $478$, $487$, $490$, $496$, $514$, $518$, $526$, $536$, $556$, $563$, $608$, $617$, $622$, $623$, $632$, $635$, $636$, $637$, $653$, $654$, $655$, $656$, $665$, $671$, $672$, $673$, $674$, $680$, $697$, $700$, $701$, $709$, $713$, $730$, $736$, $748$, $761$, $792$, $793$, $802$, $820$, $833$, $836$, $863$, $874$, $881$, $888$, $899$, $901$, $904$, $907$, $910$, $912$, $913$, $914$, $915$, $916$, $917$, $918$, $919$, $920$, $921$, $922$, $923$, $924$, $925$, $926$, $927$, $928$, $929$, $930$, $931$, $932$, $933$, $934$, $935$, $936$, $937$, $938$, $939$, $940$, $941$, $942$, $943$, $944$, $945$, $946$, $947$, $948$, $949$, $950$, $951$, $952$, $953$, $954$, $955$, $956$, $957$, $958$, $959$, $960$, $961$, $962$, $963$, $964$, $965$, $966$, $967$, $968$, $969$, $970$, $971$, $972$, $973$, $974$, $975$, $976$, $977$, $978$, $979$, $980$, $981$, $982$, $983$, $984$, $985$, $986$, $987$, $988$, $989$, $990$, $991$, $992$, $993$, $994$, $995$, $996$, $997$, $998$, $999$.\n\nHowever, upon checking, only $1$, $4$, and $9$ satisfy the condition for $n < 1000$.\n\n**Final answer:**\n\n$1$, $4$, $9$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 28785,
    "subject": "Mathematics (Multi-modal)",
    "question": "Determine and draw in the complex plane the set of all complex numbers $z$ which satisfy the condition\n$$\n\\text{Re}[(4 + 3i)z^2] \\ge 0.\n$$",
    "options": [],
    "answer": "All z = x + i y such that 4x^2 − 6xy − 4y^2 ≥ 0. Equivalently, arg(z) ∈ [−arctan(2), arctan(1/2)] modulo π; i.e., the closed double cone between the lines y = −2x and y = (1/2)x, including the boundary lines and the origin.",
    "solution": "",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 28786,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $\\triangle ABC$ be an acute triangle, and let $M$ and $N$ be the feet of the altitudes from $A$ and $B$, respectively. If $|AN| = |NM|$, prove that the incentre of $\\triangle ABC$ lies on the altitude $\\overline{BN}$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 28787,
    "subject": "Mathematics (Multi-modal)",
    "question": "Solve $\\sin x \\cdot \\cos 2x \\cdot \\cos 4x = 1$.",
    "options": [],
    "answer": "x = 3π/2 + 2kπ, k ∈ ℤ",
    "solution": "We have $|\\sin x| \\leq 1$, $|\\cos 2x| \\leq 1$, $|\\cos 4x| \\leq 1$. The product $\\sin x \\cdot \\cos 2x \\cdot \\cos 4x$ can only be $1$ if each factor is $1$ or $-1$, and their product is $1$.\n\nBut $|\\sin x| = 1$ only when $\\sin x = 1$ or $\\sin x = -1$.\n\nCase 1: $\\sin x = 1$\nThen $x = \\dfrac{\\pi}{2} + 2k\\pi$, $k \\in \\mathbb{Z}$.\n\nAt $x = \\dfrac{\\pi}{2}$:\n$\\cos 2x = \\cos(\\pi) = -1$\n$\\cos 4x = \\cos(2\\pi) = 1$\nSo $\\sin x \\cdot \\cos 2x \\cdot \\cos 4x = 1 \\cdot (-1) \\cdot 1 = -1 \\neq 1$\n\nCase 2: $\\sin x = -1$\nThen $x = \\dfrac{3\\pi}{2} + 2k\\pi$, $k \\in \\mathbb{Z}$.\n\nAt $x = \\dfrac{3\\pi}{2}$:\n$\\cos 2x = \\cos(3\\pi) = -1$\n$\\cos 4x = \\cos(6\\pi) = 1$\nSo $\\sin x \\cdot \\cos 2x \\cdot \\cos 4x = (-1) \\cdot (-1) \\cdot 1 = 1$\n\nTherefore, all solutions are $x = \\dfrac{3\\pi}{2} + 2k\\pi$, $k \\in \\mathbb{Z}$.\n\nAnswer: $x = \\dfrac{3\\pi}{2} + 2k\\pi$, $k \\in \\mathbb{Z}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 28788,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $D$ be a point on the side $\\overline{BC}$ of triangle $ABC$. Denote $\\alpha_1 = \\angle DAB$ and $\\alpha_2 = \\angle CAD$. Prove the equality\n$$\n\\frac{\\sin(\\alpha_1 + \\alpha_2)}{|AD|} = \\frac{\\sin \\alpha_1}{|AC|} + \\frac{\\sin \\alpha_2}{|AB|}.\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 28789,
    "subject": "Mathematics (Multi-modal)",
    "question": "Prove that there is no integer $n \\ge 2$ such that\n$$\nf(x) = \\cos(x\\sqrt{1}) + \\cos(x\\sqrt{2}) + \\dots + \\cos(x\\sqrt{n})\n$$\nis a periodic function.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Suppose, on the contrary, that the function $f$ is periodic with the period $T$ for some integer $n \\ge 2$. Hence, $f(T) = f(0) = n$.\nNow we have\n$$\nf(T) = \\cos(T\\sqrt{1}) + \\cos(T\\sqrt{2}) + \\dots + \\cos(T\\sqrt{n}) = n,\n$$\nfrom which we conclude that $\\cos(T\\sqrt{1}) = \\cos(T\\sqrt{2}) = \\dots = \\cos(T\\sqrt{n}) = 1$. So, $T = 2k\\pi$ and $T\\sqrt{2} = 2l\\pi$, where $k, l \\in \\mathbb{N}$. Hence, $\\sqrt{2} = l/k \\in \\mathbb{Q}$, which is contradiction. In conclusion, there is no integer $n \\ge 2$ such that $f$ is periodic.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 28790,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABC$ be a right triangle with the right angle at vertex $C$. Let $D$ be a point on the segment $AC$ and let $E$ be a point on the segment $BD$ such that $\\angle ABC = \\angle DAE = \\angle AED$. Prove that $|BE| = 2|CD|$. (Lithuania 2010)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $\\triangle ABC = \\triangle DAE = \\triangle AED = \\beta$.\n![](attached_image_1.png)\nThen\n$$\n\\begin{aligned}\n\\angle BAC &= 90^\\circ - \\beta, \\\\\n\\angle BAE &= (90^\\circ - \\beta) - \\beta = 90^\\circ - 2\\beta, \\\\\n\\angle BEA &= 180^\\circ - \\beta.\n\\end{aligned}\n$$\nApplying the law of sines to the triangle $ABE$:\n$$\n|BE| = \\frac{\\sin \\angle BAE}{\\sin \\angle AEB} \\cdot |AB| = \\frac{\\sin(90^\\circ - 2\\beta)}{\\sin(180^\\circ - \\beta)} \\cdot |AB| = \\frac{\\cos(2\\beta)}{\\sin \\beta} \\cdot |AB|.\n$$\nSince $|CD| = |BC| \\operatorname{ctg} \\angle CDB = |BC| \\operatorname{ctg}(2\\beta)$, we finally obtain\n$$\n\\frac{|BE|}{|CD|} = \\frac{\\frac{\\cos(2\\beta)}{\\sin \\beta} \\cdot |AB|}{|BC| \\operatorname{ctg}(2\\beta)} = \\frac{\\cos(2\\beta)|AB|}{\\sin \\beta \\cdot |BC| \\cdot \\frac{\\cos(2\\beta)}{\\sin(2\\beta)}} = \\frac{\\sin(2\\beta)|AB|}{\\sin \\beta \\cdot |AB| \\cos \\beta} = 2.\n$$",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 28791,
    "subject": "Mathematics (Multi-modal)",
    "question": "Maja wrote all integers from $100$ to $130$ in a row without spacings, thus obtaining the multi-digit number $100101102\\ldots129130$. Then she decided to erase $80$ digits of this number. What is the largest number Maja can get by doing this?",
    "options": [],
    "answer": "9967128129130",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 28792,
    "subject": "Mathematics (Multi-modal)",
    "question": "A sphere of radius $r$ is inscribed to a tetrahedron. If the altitudes of the tetrahedron are $v_1$, $v_2$, $v_3$ and $v_4$, prove\n$$\n\\frac{1}{v_1} + \\frac{1}{v_2} + \\frac{1}{v_3} + \\frac{1}{v_4} = \\frac{1}{r}.\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Geometry",
    "subtopic": "Solid Geometry"
  },
  {
    "id": 28793,
    "subject": "Mathematics (Multi-modal)",
    "question": "For a given prime number $p$ determine all integers $n$ such that $\\sqrt{n^2 + pn}$ is an integer.",
    "options": [],
    "answer": "For any prime p, the solutions are n = 0 and n = -p. If p is odd (p > 2), there are two additional solutions: n = ((p−1)/2)^2 and n = −((p+1)/2)^2.",
    "solution": "Let $\\sqrt{n^2 + pn} = m$, $m \\in \\mathbb{N}_0$. Then $n^2 + pn - m^2 = 0$.\nThe solutions of this quadratic equation are\n$$\nn = \\frac{1}{2} \\left( -p \\pm \\sqrt{p^2 + 4m^2} \\right).\n$$\nThe solutions are integers, so $p^2 + 4m^2 = w^2$, for some $w \\in \\mathbb{N}_0$.\nSince $w + 2m \\ge 0$ and $w + 2m \\ge w - 2m$, from $p^2 = (w - 2m)(w + 2m)$, we obtain the following possibilities:\n$$\n\\begin{cases} w - 2m = 1 \\\\ w + 2m = p^2 \\end{cases} \\quad \\text{or} \\quad \\begin{cases} w - 2m = p \\\\ w + 2m = p \\end{cases}\n$$\nIn the first case we obtain $w = \\frac{p^2 + 1}{2}$, $m = \\frac{p^2 - 1}{4}$, so it follows:\n$$\n\\begin{aligned}\nn &= \\frac{1}{2} \\left( -p \\pm \\sqrt{p^2 + 4 \\left( \\frac{p^2 - 1}{4} \\right)^2} \\right) \\\\\nn_1 &= \\left( \\frac{p-1}{2} \\right)^2, \\quad n_2 = -\\left( \\frac{p+1}{2} \\right)^2.\n\\end{aligned}\n$$\nThese numbers are integers only when $p$ is odd, i.e. only if $p > 2$.\n\nIn the second case we obtain $w = p$, $m = 0$, so it follows that $n = \\frac{1}{2}(-p \\pm p)$ or $n_3 = 0$ and $n_4 = -p$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 28794,
    "subject": "Mathematics (Multi-modal)",
    "question": "Prove that no rational number $x$ satisfies the equality $\\{x^2\\} + \\{x\\} = 1$. Find at least one real number $x$ that satisfies the equality.\nNote: $\\{x\\} = x - \\lfloor x \\rfloor$ where $\\lfloor x \\rfloor$ is the largest integer not greater than $x$.",
    "options": [],
    "answer": "No rational number satisfies the equality. One real solution is x = (sqrt(5) - 1) / 2.",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 28795,
    "subject": "Mathematics (Multi-modal)",
    "question": "Calculate $144^{\\log_5 1000} : 10^{6\\log_5 12}$.",
    "options": [],
    "answer": "1",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 28796,
    "subject": "Mathematics (Multi-modal)",
    "question": "Calculate the sum\n$$\n\\sum_{n=1}^{2012} \\operatorname{tg} n \\operatorname{tg} (n + 1).\n$$",
    "options": [],
    "answer": "cot(1)·(tan(2013) − tan(1)) − 2012",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 28797,
    "subject": "Mathematics (Multi-modal)",
    "question": "The lengths of all sides of a quadrilateral are integers, and each of them is a divisor of the sum of the other three lengths. Prove that at least two of the sides of that quadrilateral have equal lengths.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "On the contrary, let's assume that all sides are of different lengths; i.e. $a < b < c < d$ ($a$, $b$, $c$, $d$ are lengths of the sides, ordered by their length). Each of the lengths is a divisor of $S = a + b + c + d$, by assumption.\nAlso, it must be $a + b + c > d$ or equivalently $S > 2d$ and finally $\\frac{S}{d} > 2$. Since $d$ is a divisor of $S$ it follows that $\\frac{S}{d} \\ge 3$. Now we have $c < d \\le \\frac{S}{3}$ and since $c$ is a divisor of $S$, it follows that $c \\le \\frac{S}{4}$. Analogously we obtain $b \\le \\frac{S}{5}$ and $a \\le \\frac{S}{6}$.\nThen\n$$\nS = a + b + c + d \\le \\frac{S}{6} + \\frac{S}{5} + \\frac{S}{4} + \\frac{S}{3} = S \\left( \\frac{1}{6} + \\frac{1}{5} + \\frac{1}{4} + \\frac{1}{3} \\right) = \\frac{57}{60} S < S.\n$$\nContradiction!",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 28798,
    "subject": "Mathematics (Multi-modal)",
    "question": "$ABCDA_1B_1C_1D_1$ is a cube of edge length $a$. A plane is set through the vertices $A$ and $C_1$ and the midpoint of edge $\\overline{BB_1}$. Determine the area of the intersection of the cube and the plane.",
    "options": [],
    "answer": "a^2*sqrt(6)/2",
    "solution": "",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 28799,
    "subject": "Mathematics (Multi-modal)",
    "question": "A board of size $1000 \\times 1000$ is given. Is it possible to colour exactly $125$ unit squares of the board in such a way that every coloured unit square has an odd number of coloured neighbouring unit squares? Two unit squares are *neighbouring* if they have a common side.",
    "options": [],
    "answer": "No",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 28800,
    "subject": "Mathematics (Multi-modal)",
    "question": "There are some integers written on the blackboard. In each step two numbers $a$ and $b$ are chosen and replaced with the numbers $3a - b$ and $13a - 3b$. If there are numbers $1$, $2$, $3$, $4$, $\\ldots$, $2011$, $2012$ on the blackboard at the beginning, is it possible to get numbers $2$, $4$, $6$, $8$, $\\ldots$, $4022$, $4024$ on the blackboard after a finite number of steps?",
    "options": [],
    "answer": "No",
    "solution": "In each step the sum of all numbers on the blackboard changes by\n$$\n((3a - b) + (13a - 3b)) - (a + b) = 15a - 5b = 5(3a - b).\n$$\nThis difference is divisible by $5$, i.e. the sum of all numbers written on the board in each step gives the same remainder when divided by $5$.\nAt the beginning, the sum of all numbers on the board is\n$$\n1 + 2 + 3 + \\dots + 2012 = \\frac{2012 \\cdot 2013}{2} = 1006 \\cdot 2013\n$$\nand we have\n$$\n1006 \\cdot 2013 \\equiv 1 \\cdot 3 \\equiv 3 \\pmod{5}.\n$$\nOn the other hand,\n$$\n2 + 4 + 6 + \\dots + 4024 = 2 \\cdot (1 + 2 + 3 + \\dots + 2012) \\equiv 2 \\cdot 3 \\equiv 1 \\pmod{5}.\n$$\nIn conclusion, it is impossible to get the numbers $2$, $4$, $6$, $\\ldots$, $4024$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 28801,
    "subject": "Mathematics (Multi-modal)",
    "question": "If the sides of a triangle satisfy the equality $\\frac{a+b}{b+c} = \\frac{c}{a-b}$, determine the largest angle in the triangle.",
    "options": [],
    "answer": "120 degrees",
    "solution": "",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 28802,
    "subject": "Mathematics (Multi-modal)",
    "question": "Determine all values of the real parameter $a$ for which the equation\n$$\n8^a \\sin^2 x = 4 \\cdot 2^{\\cos^2 x}\n$$\nhas exactly one solution in the interval $\\left[-\\frac{\\pi}{6}, \\frac{2\\pi}{3}\\right]$.",
    "options": [],
    "answer": "a = 2/3 or 17/12 − (1/3) log_2 3 < a < 19/12",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 28803,
    "subject": "Mathematics (Multi-modal)",
    "question": "In triangle $ABC$ the side $BC$ is of length $6$, cosine of the angle $\\angle ABC$ equals $\frac{4}{5}$, and the radius of the incircle is $1$. Determine the lengths of the sides $AB$ and $AC$ of the triangle.",
    "options": [],
    "answer": "AB = 15/4, AC = 15/4",
    "solution": "",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 28804,
    "subject": "Mathematics (Multi-modal)",
    "question": "Determine the smallest integer $N$ larger than $1000$ such that exactly half of the numbers from $1$ to $N$ have at least one digit $1$ in their decimal notation.",
    "options": [],
    "answer": "1456",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 28805,
    "subject": "Mathematics (Multi-modal)",
    "question": "Real numbers $x$, $y$ and $z$ are consecutive terms of an arithmetic sequence. If the numbers $\\cos^2 x$, $\\cos^2 y$ and $\\cos^2 z$ are mutually different and consecutive terms of some arithmetic sequence, determine all possible values of $y$.",
    "options": [],
    "answer": "y = π/4 + k·(π/2), for any integer k",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 28806,
    "subject": "Mathematics (Multi-modal)",
    "question": "a) Let $x$ and $y$ be real numbers such that $x + y$, $x^2 + y^2$ and $x^4 + y^4$ are integers. Prove that the number $x^n + y^n$ is an integer for all $n \\in \\mathbb{N}$.\n\nb) Find an example of real numbers $x$ and $y$ that are not integers, such that the numbers $x + y$, $x^2 + y^2$ and $x^4 + y^4$ are all integers.\n\nc) Find an example of real numbers $x$ and $y$ that are not integers, such that the numbers $x + y$, $x^2 + y^2$ and $x^3 + y^3$ are integers, but the number $x^4 + y^4$ is not an integer.\n\n(Neven Elezović)",
    "options": [],
    "answer": "a) For all natural numbers n, x^n + y^n is an integer.\nb) Example: x = sqrt(2), y = −sqrt(2).\nc) Example: x = 1/sqrt(2), y = −1/sqrt(2).",
    "solution": "a) Assume that $a = x + y$, $b = x^2 + y^2$ and $c = x^4 + y^4$ are integers. Then the numbers $a^2 - b = 2xy$ and $b^2 - c = 2x^2 y^2$ are integers as well.\n\nSuppose that $xy$ is not an integer. Then $xy = \\frac{m}{2}$, where $m \\in \\mathbb{Z}$ is odd. But, then $2x^2 y^2 = \\frac{m^2}{2}$ is not an integer. Contradiction! Hence, $xy$ is an integer.\n\nBy induction we can prove that $x^n + y^n$ is an integer for each $n \\in \\mathbb{N}$:\n\nBasis of induction: numbers $x + y$ and $x^2 + y^2$ are integers, by assumption.\n\nLet's suppose that, for some integer $n > 1$, the numbers $x^{n-1} + y^{n-1}$ and $x^n + y^n$ are integers.\n\nThen\n$$\n\\begin{aligned}\nx^{n+1} + y^{n+1} &= (x^n + y^n)(x + y) - (x^n y + x y^n) \\\\\n&= (x^n + y^n)(x + y) - xy(x^{n-1} + y^{n-1}).\n\\end{aligned}\n$$\nNumbers $x + y$ and $xy$ are integers, as well as the numbers $x^{n-1} + y^{n-1}$ and $x^n + y^n$, so, we conclude that the number $x^{n+1} + y^{n+1}$ is an integer, too.\n\nb) $x = \\sqrt{2}$, $y = -\\sqrt{2}$.\n\nc) $x = \\frac{1}{\\sqrt{2}}$, $y = -\\frac{1}{\\sqrt{2}}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 28807,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all solutions of the equation $m! + 2 = n^2$, where $m$ and $n$ are positive integers.",
    "options": [],
    "answer": "(m, n) = (2, 2)",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 28808,
    "subject": "Mathematics (Multi-modal)",
    "question": "Determine the set of all real numbers $x$ that satisfy the equation\n$$\n\\left\\lfloor \\frac{2x-1}{3} \\right\\rfloor + \\left\\lfloor \\frac{4x+1}{6} \\right\\rfloor = \\frac{5x-4}{3},\n$$\nwhere $\\lfloor a \\rfloor$ denotes the largest integer not greater than $a$.",
    "options": [],
    "answer": "x ∈ { -2/5, 1/5, 4/5, 7/5, 2 }",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 28809,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $p_1$ and $q_1$ be integers such that the equation $x^2 + p_1x + q_1 = 0$ has two integer solutions. For all $n \\in \\mathbb{N}$ we define the numbers $p_{n+1}$ and $q_{n+1}$ by the formulas\n$$\np_{n+1} = p_n + 1, \\quad q_{n+1} = q_n + \\frac{1}{2}p_n.\n$$\nProve that there exists an infinite number of positive integers $n$ such that the equation $x^2 + p_nx + q_n = 0$ has two integer solutions.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $D_n$ be a discriminant of quadratic equation $x^2 + p_n x + q_n = 0$, for each $n \\in \\mathbb{N}$, i.e.\n$$D_n = p_n^2 - 4q_n.$$ \nBy assumption we conclude that $D_1$ is a square of an integer.\nFurther, we have:\n$$\nD_{n+1} = p_{n+1}^2 - 4q_{n+1} = (p_n + 1)^2 - 4 \\left( q_n + \\frac{1}{2}p_n \\right) = p_n^2 - 4q_n + 1 = D_n + 1.\n$$\nLet's suppose that the equation $x^2 + p_n x + q_n = 0$ has two integer solutions for some $n$. Then $D_n = k^2$ for some integer $k$, hence, we have:\n$$\nD_{n+2k+1} = D_n + 2k + 1 = k^2 + 2k + 1 = (k+1)^2.\n$$\nMoreover, since $\\frac{1}{2}(-p_n + \\sqrt{D_n})$ and $\\frac{1}{2}(-p_n - \\sqrt{D_n})$ are integers, we conclude that $p_n$ and $D_n$ are of the same parity. But,\n$$\np_{n+2k+1} \\equiv p_n + 2k + 1 \\equiv p_n + 1 \\equiv D_n + 1 \\equiv D_n + 2k + 1 \\equiv D_{n+2k+1} \\pmod{2}\n$$\nso the equation $x^2 + p_{n+2k+1}x + q_{n+2k+1} = 0$ also has two integer solutions.\nHence, we proved that there are infinitely many integers $n$ for which the equation $x^2 + p_n x + q_n = 0$ has two integer solutions.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 28810,
    "subject": "Mathematics (Multi-modal)",
    "question": "Determine and sketch in the Gaussian plane the set of all $z$ that satisfy\n$$\n|z - 1| - |z + 1| = \\sqrt{3}.\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 28811,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $p$ and $q$ be real numbers. The graph of the function $f(x) = x^2 + px + q$ intersects with the coordinate axes in three different points $A$, $B$ and $C$. Prove that the circumcircle of triangle $ABC$ intersects with the $y$-axis at a point with ordinate $1$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Discrete Mathematics",
    "subtopic": "Other"
  },
  {
    "id": 28812,
    "subject": "Mathematics (Multi-modal)",
    "question": "A triangle with the orthocentre $H$ and circumcentre $O$ is given. If one of the angles of the triangle is $60^\\circ$, prove that the angle bisector of that angle is perpendicular to the line $OH$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Without loss of generality, let $\\angle BAC = 60^\\circ$ and $|AB| > |CA|$. Let $D$ be the intersection of the angle bisector of $\\angle BAC$ and the circumcircle of the triangle $ABC$. Point $D$ lies on the perpendicular bisector of $BC$. Let $P$ be the midpoint of $BC$. Since $|OC| = |OD|$ and $\\angle COD = 2\\angle CAD = 60^\\circ$, triangle $COD$ is equilateral. $P$ is the midpoint of $OD$, since $OD \\perp BC$.\n![](attached_image_1.png)\nSince $|AH| = 2|OP|$ we conclude that $|AH| = |OD|$. Also, we notice that the lines $AH$ and $OD$ are parallel, because they are both perpendicular to $BC$. Hence, $AHDO$ is a parallelogram. Since $|AO| = |DO|$, $AHDO$ is a rhombus, therefore, its diagonals $AD$ and $OH$ are perpendicular to each other.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 28813,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find the real number $A$, given that the coefficient with $x^{12}$ in the polynomial\n$$\n(1 + x^4)^{12} + A (x(1 - x^2)^2)^{12}\n$$\nequals 100.",
    "options": [],
    "answer": "-120",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 28814,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $n$ and $d$ be positive integers such that $2n^2$ is divisible by $d$. Prove that $n^2 + d$ is not a perfect square.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $k$ be an integer such that $2n^2 = dk$. Then we have\n$$\n\\begin{aligned}\nk^2(n^2 + d) &= k^2n^2 + k^2d = k^2n^2 + 2kn^2 \\\\\n&= n^2(k^2 + 2k) = n^2((k+1)^2 - 1).\n\\end{aligned}\n$$\nExpression $(k+1)^2 - 1$ is not a square of an integer for $k \\in \\mathbb{N}$. Then, $k^2(n^2 + d)$ is not a square of an integer, so $n^2 + d$ cannot be a square of an integer.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 28815,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $a$, $b$ and $c$ be the lengths of sides of a triangle with area $P$. Prove that $P < \\frac{1}{6}(a^2 + b^2 + c^2)$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 28816,
    "subject": "Mathematics (Multi-modal)",
    "question": "One box contains three blue balls and one red ball, and a second box contains three red and one blue ball. First we randomly choose one ball from the first box and put it in the second one, and then we randomly choose one ball from the second box. What is the probability that the chosen ball is red?",
    "options": [],
    "answer": "13/20",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 28817,
    "subject": "Mathematics (Multi-modal)",
    "question": "For which $n \\in \\mathbb{N}$ exist an angle $\\alpha$ and a convex $n$-gon with angles $\\alpha, 2\\alpha, \\dots, n\\alpha$?",
    "options": [],
    "answer": "n = 3 or 4",
    "solution": "",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 28818,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $n$ be a positive integer. $2^n$ players participated in a tennis tournament. Every player played one match against each of the remaining players. Prove that we can choose $n+1$ players and arrange them in a line in such a way that each of them defeated all players in line behind him.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 28819,
    "subject": "Mathematics (Multi-modal)",
    "question": "We say that two cells of the $10 \\times 10$ table are *friendly* if they have at least one common vertex. Into each cell of the table a positive integer less than or equal to $10$ is written, so that the numbers in friendly cells are relatively prime. Prove that some number appears in the table at least $17$ times.\n\n(St. Petersburg olympiad 2001)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let's divide the given table into $25$ smaller squares $2 \\times 2$. In each of these squares there is at most one even number and at most one number divisible by $3$. Hence, at most $50$ numbers in the table are divisible by $2$ or $3$. At least $50$ numbers remain, and each of them is equal to $1$, $5$ or $7$. By the box principle, at least one of these numbers appears at least $17$ times.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 28820,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all integers greater than $1$ whose all divisors written in an increasing sequence make a geometric sequence.",
    "options": [],
    "answer": "Exactly the prime powers, i.e., all integers of the form p^k with p prime and k a positive integer.",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 28821,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $x$ and $y$ be real numbers such that $x + y \\ge 0$. Prove that\n$$\n2^{n-1} (x^n + y^n) \\ge (x + y)^n \\quad \\text{for all } n \\in \\mathbb{N}.\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 28822,
    "subject": "Mathematics (Multi-modal)",
    "question": "There are $1234$ pebbles on a table. Ratko and Rudi play the following game: first Ratko takes an even number of pebbles, at least two but no more than $100$, and then Rudi takes an odd number of pebbles, at least one but no more than $99$. The players take turns taking pebbles, always following the same conditions. A player wins if he takes all remaining pebbles, or if the other player can't make his move. Who has the winning strategy, i.e. who can win regardless of the opponent's moves?",
    "options": [],
    "answer": "Ratko",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 28823,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $x$, $y$ and $z$ be positive real numbers such that $x + y + z = 18xyz$. Prove the inequality\n$$\n\\frac{x}{\\sqrt{x^2 + 2yz + 1}} + \\frac{y}{\\sqrt{y^2 + 2xz + 1}} + \\frac{z}{\\sqrt{z^2 + 2xy + 1}} \\ge 1.\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "The A-H inequality together with the condition of the problem gives\n$$\nxy + yz + zx = xyz \\left(\\frac{1}{x} + \\frac{1}{y} + \\frac{1}{z}\\right) \\ge xyz \\cdot \\frac{9}{x + y + z} = \\frac{9xyz}{18xyz} = \\frac{1}{2}.\n$$\nUsing $1 \\le 2xy + 2yz + 2zx$ we get\n$$\nx^2 + 2yz + 1 \\le x^2 + 2xy + 2zx + 4yz = (x + 2y)(x + 2z) \\le (x + y + z)^2\n$$\nwhere the last inequality is a consequence of the A-G inequality. Hence $\\frac{x}{\\sqrt{x^2 + 2yz + 1}} \\ge \\frac{x}{x + y + z}$ and\nanalogously $\\frac{y}{\\sqrt{y^2 + 2zx + 1}} \\ge \\frac{y}{x + y + z}$, $\\frac{z}{\\sqrt{z^2 + 2xy + 1}} \\ge \\frac{z}{x + y + z}$.\nAdding these three inequalities finishes the proof.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 28824,
    "subject": "Mathematics (Multi-modal)",
    "question": "Given the sequence of real numbers $x_0 = 1$, $x_1 = 1$,\n$$\nx_n = \\sqrt{\\frac{n}{2} + x_{n-1}x_{n-2}}, \\quad \\text{for } n \\ge 2,\n$$\ndetermine whether there exists a real number $A$ such that $An < x_n < An+1$ for all $n \\in \\mathbb{N}$?",
    "options": [],
    "answer": "A = sqrt(6)/6",
    "solution": "Note that all members of the sequence are positive real numbers. The recursive relation gives\n$$\n2x_n^2 = n + 2x_{n-1}x_{n-2} \\le x_{n-1}^2 + x_{n-2}^2 + n.\n$$\n\n$$\n\\begin{aligned}\n2x_n^2 &\\le x_{n-1}^2 + x_{n-2}^2 + n, \\\\\n2x_{n-1}^2 &\\le x_{n-2}^2 + x_{n-3}^2 + n-1, \\\\\n&\\vdots \\\\\n2x_3^2 &\\le x_2^2 + x_1^2 + 3, \\\\\n2x_2^2 &\\le x_1^2 + x_0^2 + 2.\n\\end{aligned}\n$$\nAdding all these inequalities gives\n$$\n2x_n^2 + x_{n-1}^2 \\le 2x_1^2 + x_0^2 + [n + (n-1) + \\dots + 3 + 2],\n$$\ni.e.\n$$\n2x_n^2 + x_{n-1}^2 \\le \\frac{n^2 + n + 4}{2}. \\quad (*)\n$$\nIf there exists a sought number $A$, the left hand side of this inequality is at least $2A^2n^2 + A^2(n-1)^2$, thus\n$$\n2A^2n^2 + A^2(n-1)^2 \\le \\frac{n^2 + n + 4}{2},\n$$\ni.e. $6A^2n^2 - 4A^2n + 2A^2 \\le n^2 + n + 4$. This inequality holds for every $n \\in \\mathbb{N}$ only if $6A^2 \\le 1$, i.e. $A \\le \\frac{\\sqrt{6}}{6}$.\nWe prove that $A = \\frac{\\sqrt{6}}{6}$ satisfies the conditions of the problem.\nFirst, we show inductively that $x_n > \\frac{n\\sqrt{6}}{6}$, for all $n \\in \\mathbb{N}_0$. Statement is true for $n=0$ and $n=1$. Let us assume $x_{n-1} > \\frac{(n-1)\\sqrt{6}}{6}$ and $x_{n-2} > \\frac{(n-2)\\sqrt{6}}{6}$. Then\n$$\nx_n^2 = \\frac{n}{2} + x_{n-1}x_{n-2} > \\frac{n}{2} + \\frac{(n-1)\\sqrt{6}}{6} \\cdot \\frac{(n-2)\\sqrt{6}}{6} = \\frac{n^2 + 2}{6} > \\frac{n^2}{6},\n$$\ni.e. $x_n > \\frac{n\\sqrt{6}}{6}$, and this finishes the inductive step.\nNext, we prove that $x_n < \\frac{n\\sqrt{6}}{6} + 1$, for all $n \\in \\mathbb{N}$. The statement is true for $n=1$ and $n=2$ by inspection. Inequality $(*)$ and the fact $x_{n-1}^2 > \\frac{(n-1)^2}{6}$ imply\n$$\n2x_n^2 \\le \\frac{n^2 + n + 4}{2} - x_{n-1}^2 < \\frac{n^2 + n + 4}{2} - \\frac{(n-1)^2}{6} = \\frac{2n^2 + 5n + 11}{6}.\n$$\nIt remains to show\n$$\n\\frac{2n^2 + 5n + 11}{6} < 2 \\left( \\frac{n\\sqrt{6}}{6} + 1 \\right)^2,\n$$\nbut this inequality is equivalent to $5n - 1 < 4n\\sqrt{6}$ which holds for $n \\ge 3$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 28825,
    "subject": "Mathematics (Multi-modal)",
    "question": "Determine all functions $f: \\mathbb{R} \\to \\mathbb{R}$ such that for all real numbers $x$ and $y$ holds\n$$\nf(y + f(x)) - f(x + f(y)) = f(x - y)(f(x + y) - 1).\n$$",
    "options": [],
    "answer": "f(x) = 1 for all real x; f(x) = 0 for all real x",
    "solution": "Setting $x = y$ gives $0 = f(0)(f(2x) - 1)$, $\\forall x \\in \\mathbb{R}$.\n\nFirst case: $f(0) \\ne 0$\nWe have $f(2x) - 1 = 0$, $\\forall x \\in \\mathbb{R}$, so $f(x) = 1$, $\\forall x \\in \\mathbb{R}$. We check that this function is a solution.\n\nSecond case: $f(0) = 0$\nSetting $y = 0$ gives $f(f(x)) - f(x) = f(x)(f(x) - 1)$, i.e.\n$$\nf(f(x)) = (f(x))^2, \\quad \\forall x \\in \\mathbb{R}. \\qquad (\\star)\n$$\nInterchanging $x$ and $y$ gives $f(x+f(y)) - f(y+f(x)) = f(y-x)(f(x+y)-1)$, which added to the starting equation gives\n$$\n0 = (f(x+y) - 1)(f(x-y) + f(y-x)), \\quad \\forall x, y \\in \\mathbb{R}.\n$$\nSetting $y = -x$ in thus obtained equation gives $0 = (f(0) - 1)(f(2x) + f(-2x))$, so $f(0) = 0$ implies $f(2x) = -f(-2x)$, $\\forall x \\in \\mathbb{R}$ and hence the function $f$ is odd.\n\nNow we conclude\n$$\nf(f(x)) \\stackrel{*}{=} (f(x))^2 = (-f(-x))^2 = (f(-x))^2 \\stackrel{*}{=} f(f(-x)) = -f(-f(-x)) = -f(f(x)), \\qquad (\\star\\star)\n$$\nfor every $x \\in \\mathbb{R}$\n\nFinally, $(\\star\\star)$ and $(\\star)$ give $2f(f(x)) = 0$, so $2(f(x))^2 = 0$. We check directly that $f(x) = 0$, $\\forall x \\in \\mathbb{R}$ is also a solution.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 28826,
    "subject": "Mathematics (Multi-modal)",
    "question": "Determine all functions $f : \\mathbb{R} \\to \\mathbb{R}$ such that for all real numbers $x$ and $y$ holds\n$$\nf(x^2 + f(y)) = (f(x) + y^2)^2.\n$$",
    "options": [],
    "answer": "f(x) = x^2 for all real x",
    "solution": "Let $a = f(0)$. Setting $x = y = 0$ in the given equation gives $f(a) = a^2$. Setting $y = 0$ gives\n$$\nf(x^2 + a) = (f(x))^2, \\quad \\forall x \\in \\mathbb{R}. \\qquad (\\circ)\n$$\nSetting $x = a$ gives $f(a^2 + a) = (f(a))^2 = a^4$.\nAssume that $a < 0$. Then there exists $b > 0$ such that $a = -b^2$. Setting $x = b$ in (\\circ) gives $a = f(0) = f(b^2 + a) = (f(b))^2 \\ge 0$, a contradiction. Hence $a \\ge 0$.\nSetting $x = \\sqrt{a}$, $y = a$ in the given equation gives $f(a + a^2) = (f(\\sqrt{a}) + a^2)^2$.\nWe have obtained $f(a + a^2) = (f(\\sqrt{a}) + a^2)^2 = a^4$, i.e. $f(\\sqrt{a}) (f(\\sqrt{a}) + 2a^2) = 0$.\nAssume $f(\\sqrt{a}) = -2a^2$. Setting $x = \\sqrt{\\sqrt{a} + 2a^2}$, $y = \\sqrt{a}$ gives\n$$\nf(x^2 + f(y)) = f(\\sqrt{a} + 2a^2 - 2a^2) = f(\\sqrt{a}) = -2a^2 < 0.\n$$\nBut the given equation gives $f(x^2 + f(y)) = (f(x) + y^2)^2 \\ge 0$, a contradiction. Hence $f(\\sqrt{a}) = 0$.\nSetting $x = 0$ in the given equation gives $f(f(y)) = (a + y^2)^2, \\forall y \\in \\mathbb{R}$.\nSetting $y = \\sqrt{a}$ gives $a = f(0) = f(f(\\sqrt{a})) = (a + a)^2 = 4a^2$.\nWe have two cases: $a = \\frac{1}{4}$ or $a = 0$.\nAssume $a = \\frac{1}{4}$. Then $f(\\frac{1}{2}) = 0$. Because $f(f(y)) = (a + y^2)^2$, we have $f(\\frac{1}{4}) = \\frac{1}{16}$. If $x$ and $y$ are real numbers such that $x^2 + f(y) = \\frac{1}{2}$, the given equation gives $f(x) = -y^2$. This is true for, e.g., $x = \\frac{\\sqrt{7}}{4}$ and $y = \\frac{1}{4}$ and we conclude that $f(\\frac{\\sqrt{7}}{4}) = -\\frac{1}{16} < 0$. But setting $x = \\sqrt{\\frac{\\sqrt{7}-1}{4}}$ in (\\circ) gives $f(\\frac{\\sqrt{7}}{4}) \\ge 0$, a contradiction.\nHence $f(0) = a = 0$ and we can simplify obtained identities to $f(x^2) = (f(x))^2$ and $f(f(y)) = y^4$.\nFrom the first identity we have $f(x) \\ge 0$ for all $x \\ge 0$.\nAssume there is $t > 0$ such that $f(t) < t^2$. Setting $x = \\sqrt{t^2 - f(t)}$ and $y = t$ in the given equation gives\n$$\nf(t^2) = \\left( f(\\sqrt{t^2 - f(t)}) + t^2 \\right)^2 \\ge t^4,\n$$\ntherefore $(f(t))^2 \\ge t^4$ which is a contradiction with the assumption $f(t) < t^2$. Hence for all $x \\ge 0$ holds $f(x) \\ge x^2$.\nThis implies that $x^4 = f(f(x)) \\ge (f(x))^2 \\ge x^4$ for all $x > 0$, which is possible only if $f(x) = x^2$ for all $x \\ge 0$.\nLet $w > 0$. Setting $x = -w$ in $f(x^2) = (f(x))^2$ gives $f(-w) = w^2$ or $f(-w) = -w^2$. But if $f(-w) = -w^2$, setting $x = w$, $y = -w$ in the given equation gives $0 = f(0) = f(w^2 - w^2) = f(x^2 + f(y)) = (f(x) + y^2)^2 = 4w^4 > 0$, a contradiction.\nHence the only possible solution is $f(x) = x^2, \\forall x \\in \\mathbb{R}$. We check directly that it really satisfies the given equation.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 28827,
    "subject": "Mathematics (Multi-modal)",
    "question": "One of the numbers $0$ or $1$ is assigned to every vertex of a regular polygon. Rudi divides the polygon to triangles by drawing some diagonals that intersect each other only in the vertices of the polygon, and then inside each triangle writes the sum of the numbers assigned to its vertices. Prove that Rudi can choose the diagonals to draw in such a way that the minimal and the maximal number inside the triangles differ by at most $1$. (Estonia 2010)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Note that if the numbers assigned to all vertices are equal, then Rudi can choose any diagonals.\nHence we assume there is at least one $0$-vertex and at least one $1$-vertex. (We shall call the vertices \"0-vertex\" or \"1-vertex\", according to the number assigned to it.)\nBy the strong induction on the number $n$ of vertices we will prove that Rudi can choose diagonals in such a way that the number written in each triangle is either $1$ or $2$.\nIf $n = 3$, the statement is obviously true.\nThe case $n = 4$ is also easy. If there is a diagonal connecting a $0$-vertex with a $1$-vertex, then Rudi draws that diagonal. Otherwise he draws any diagonal and obtains equal numbers ($1$ or $2$) in both triangles.\nAssume that the statement is true for all polygons with less than $n$ vertices ($n \\ge 5$), such that there are both a $0$-vertex and a $1$-vertex.\nIn a polygon with $n$ vertices, we can choose two adjacent vertices $A$ and $B$ so that $A$ is a $0$-vertex and $B$ a $1$-vertex. Let $P$ be a vertex which is not adjacent to $A$ or $B$ (such vertex exists since $n \\ge 5$). If $P$ is a $0$-vertex, Rudi connects it to $B$, otherwise to $A$. In this way Rudi has divided the polygon into two polygons with smaller number of vertices and each of them has both a $0$-vertex and a $1$-vertex. By the inductive hypothesis each of the two smaller polygons can be divided into triangles such that the numbers $1$ and $2$ are written in all of them, so the same holds for the polygon with $n$ vertices. This finishes the inductive step.",
    "topic": "Discrete Mathematics",
    "subtopic": "Combinatorics"
  },
  {
    "id": 28828,
    "subject": "Mathematics (Multi-modal)",
    "question": "In some country there are $c$ cities and $r$ roads, every road connects two different cities and between any two cities there is at most one road. Roads are named by numbers $1, 2, \\dots, r$. Tonči travels along some roads in such a way that, when he writes down the names of the roads in the order he passes through them, he obtains an ascending sequence of numbers.\nShow that there is a city such that starting from it Tonči can pass through at least $\\frac{2r}{c}$ roads.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "We place one of Tonči's friends in every city. In the step $i$ ($i = 1, 2, \\dots, r$) friends which are at that moment in the cities connected by the road $i$ switch their positions. In each step we have exactly two shifts from one city to another, and all together $2r$ shifts. Hence at least one of the $c$ friends has shifted at least $\\frac{2r}{c}$ times.\nIf Tonči starts in the city where this friend was placed he can satisfy the condition of the problem.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 28829,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $A$ be a subset of the set $\\{1, 2, 3, \\dots, 26\\}$ with seven elements. Prove that there are two distinct nonempty subsets of $A$ such that the sums of their elements are equal.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "We prove a stronger statement that already among subsets of $A$ with at most four elements there are two with equal sums of elements. Set $A$ has\n$$\n\\binom{7}{1} + \\binom{7}{2} + \\binom{7}{3} + \\binom{7}{4} = 98\n$$\nsubsets with at most four elements. The sum of elements of each of these subsets is at least $1$ and at most $26 + 25 + 24 + 23 = 98$. Assume that among these subsets there are no two with equal sums of elements. Then every number between $1$ and $98$ is the sum of elements of exactly one subset of $A$ with at most four elements. In particular, $98$ is the sum of the subset $\\{23, 24, 25, 26\\} \\subset A$. On the other hand, two-element subsets $\\{23, 26\\}$ and $\\{24, 25\\}$ have equal sums, which is a contradiction.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 28830,
    "subject": "Mathematics (Multi-modal)",
    "question": "Along the coast of an island there are 20 villages. Each village has 20 fighters. Every fighter fights all the fighters from all the other villages. No two fighters have equal strength and the stronger fighter wins the fight. We say that the village $A$ is stronger than the village $B$ if in at least $k$ fights among the fighters from $A$ and $B$ a fighter from the village $A$ wins. It turned out that every village is stronger than its neighbour (in the clockwise direction). Show that the maximal possible $k$ is 290.",
    "options": [],
    "answer": "290",
    "solution": "We first show that $k > 290$ is impossible. In every village we rank the fighters (from 1 to 20) according to their strength and we consider the tenth fighter in every village. Let the weakest of these considered fighters comes from the village $A$. Then in any other village $B$ (and in particular in the neighbour of $A$) at least 10 fighters (from the strongest to the tenth by strength) from $B$ has won against 11 fighters (from the tenth to the weakest) from $A$. Hence the number of fights in which the fighters from $A$ have won is at most $20 \\cdot 20 - 10 \\cdot 11 = 290$, a contradiction with the assumption that the village $A$ is stronger than its neighbour.\n\nWe construct an example showing that $k = 290$ can be obtained. Let the 400 fighters be ranked by their strength (1 is the weakest, 400 the strongest), and consider them as 210 weaker fighters (from 1 to 210) and 190 stronger fighters (from 211 to 400). In the village $A_i$ we put $i$ of weaker and $20 - i$ of stronger fighters, more precisely in $A_1$ the fighters ranked 1 and 211–229; in $A_2$ fighters ranked 2–3 and 230–247; etc.\n\nWe show that the village $A_i$ is stronger than the village $A_{i-1}$ for $i = 2, \\dots, 20$. All weaker fighters from $A_i$ have won against all weaker fighters from $A_{i-1}$, and all stronger fighters from $A_i$ have won against all fighters from $A_{i-1}$. Hence the number of fights in which a fighter from $A_i$ has won is $i \\cdot (i-1) + (20-i) \\cdot 20 = i^2 - 21i + 400$. This expression attains its minimum for $i = 10$ or $i = 11$ and then equals 290. In addition, the 19 stronger fighters from $A_1$ have won against all fighters from $A_{20}$, and since $20 \\cdot 19 = 380 > 290$, the village $A_1$ is stronger than the village $A_{20}$. This proves the assertion of the problem.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 28831,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABCD$ be a cyclic quadrilateral such that $|AD| = |BD|$ and let $M$ be the intersection of its diagonals. Let $N$ be the second intersection of the diagonal $AC$ with the circle passing through $B$, $M$ and the incentre of the triangle $BCM$. Prove that $|AN| \\cdot |NC| = |CD| \\cdot |BN|$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "We denote $\\angle BAD = \\angle DBA = \\alpha$ in the isosceles triangle $ABD$. Then $\\angle ACB = \\angle ADB = 180^\\circ - 2\\alpha$.\nLet $I$ be the incentre of the triangle $BCM$. We have\n$$\n\\begin{aligned}\n\\angle MIB &= 180^\\circ - \\frac{1}{2} (\\angle BMC + \\angle CBM) = 180^\\circ - \\frac{1}{2} (180^\\circ - \\angle MCB) \\\\\n&= 180^\\circ - \\frac{1}{2} (180^\\circ - \\angle ACB) = 180^\\circ - \\alpha.\n\\end{aligned}\n$$\nSince the quadrilateral $BIMN$ is cyclic we can prove that $\\angle BNC = \\alpha$ (independently of the position of the point $N$) so the triangle $BCN$ is isosceles and $|BC| = |NC|$.\n\n![](attached_image_1.png)\nNote that the triangles $ABN$ and $DBC$ are similar because $\\angle ANB = \\angle DCB = 180^\\circ - \\alpha$ and $\\angle BAN = \\angle BDC$ (suspended angles over the chord $BC$). Hence\n$$\n\\frac{|NA|}{|BN|} = \\frac{|CD|}{|BC|} = \\frac{|CD|}{|NC|},\n$$\ni.e. $|AN| \\cdot |NC| = |CD| \\cdot |BN|$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 28832,
    "subject": "Mathematics (Multi-modal)",
    "question": "The incircle of a scalene triangle $ABC$ touches the sides $\\overline{AB}$ and $\\overline{CA}$ at the points $M$ and $N$ respectively. The excircles opposite to vertices $B$ and $C$ touch the line $BC$ at points $P$ and $Q$ respectively. Prove that the quadrilateral $MNPQ$ is cyclic if and only if $\\angle CAB = 90^\\circ$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $|BC| = a$, $|CA| = b$ and $|AB| = c$ be the lengths of the sides of the triangle $ABC$ and let $s = \\frac{1}{2}(a+b+c)$ be its semiperimeter. Without loss of generality we may assume $b < c$.\nLet $S$ be the intersection of lines $BC$ and $MN$, and let $X$ be the point where the incircle of the triangle $ABC$ touches the side $\\overline{BC}$.\n\n![](attached_image_1.png)\n\nMenelaus' theorem applied to the line $MN$ and the triangle $ABC$ gives\n$$\n\\frac{|AM|}{|BM|} \\cdot \\frac{|BS|}{|CS|} \\cdot \\frac{|CN|}{|AN|} = 1.\n$$\nSince $|AM| = |AN| = s-a$, $|BM| = |BX| = s-b$ and $|CN| = |CX| = s-c$, we have\n$$\n\\frac{|AM|}{|BM|} \\cdot \\frac{|BX|}{|CX|} \\cdot \\frac{|CN|}{|AN|} = 1,\n$$\nand hence\n$$\n\\frac{|BX|}{|CX|} = \\frac{|BS|}{|CS|} = \\frac{|SX| + |BX|}{|SX| - |CX|}.\n$$\nDenoting $|SX| = d$ gives\n$$\n\\frac{s-b}{s-c} = \\frac{d+(s-b)}{d-(s-c)},\n$$\ni.e. $d(c-b) = 2(s-b)(s-c)$.\nWe also have $|PX| = |CX| + |CP| = (s-c) + (s-a) = b$, analogously $|QX| = c$ and $|SM| \\cdot |SN| = |SX|^2$ (by the power of the point $S$ with respect to the incircle of the triangle $ABC$).\nFinally, the following sequence of equivalent statements finishes the proof:\nThe quadrilateral $MNPQ$ is cyclic.\n$$\n\\begin{align*}\n\\iff |SM| \\cdot |SN| &= |SP| \\cdot |SQ| \\\\\n\\iff |SX|^2 &= (|SX| - |PX|)(|SX| + |QX|) \\\\\n\\iff d^2 &= (d-b)(d+c) \\\\\n\\iff d(c-b) &= bc \\\\\n\\iff 2(s-b)(s-c) &= bc \\\\\n\\iff a^2 - (b-c)^2 &= 2bc \\\\\n\\iff a^2 &= b^2 + c^2 \\\\\n\\iff \\text{The triangle } ABC \\text{ has a right angle at vertex } A.\n\\end{align*}\n$$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 28833,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $\\triangle ABC$ be acute triangle and let $A_1, B_1, C_1$ be points on its sides $\\overline{BC}, \\overline{CA}, \\overline{AB}$ respectively. Prove that the triangles $\\triangle ABC$ and $A_1B_1C_1$ are similar ($\\angle A = \\angle A_1$, $\\angle B = \\angle B_1$, $\\angle C = \\angle C_1$) if and only if the orthocentre of the triangle $A_1B_1C_1$ coincides with the circumcentre of the triangle $\\triangle ABC$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let the triangle $A_1B_1C_1$ be similar to the triangle $ABC$ ($\\angle A = \\angle A_1 = \\alpha$, $\\angle B = \\angle B_1 = \\beta$, $\\angle C = \\angle C_1 = \\gamma$), and let the point $O$ be the orthocentre of the triangle $A_1B_1C_1$. Then $\\angle OB_1C_1 = 90^\\circ - \\gamma$, $\\angle OC_1B_1 = 90^\\circ - \\beta$, so $\\angle B_1OC_1 = 180^\\circ - (90^\\circ - \\gamma) - (90^\\circ - \\beta) = \\beta + \\gamma$. Since $\\angle B_1AC_1 + \\angle B_1OC_1 = \\alpha + \\beta + \\gamma = 180^\\circ$, the quadrilateral $AC_1OB_1$ is cyclic. Hence $\\angle OAB_1 = OC_1B_1 = 90^\\circ - \\beta$ and $\\angle OAC_1 = \\angle OB_1C_1 = 90^\\circ - \\gamma$.\n\nAnalogously, because quadrilaterals $BA_1OC_1$ and $CB_1OA_1$ are cyclic, we get $\\angle OBC_1 = 90^\\circ - \\gamma$, $\\angle OBA_1 = 90^\\circ - \\alpha$ and $\\angle OCA_1 = 90^\\circ - \\alpha$, $\\angle OCB_1 = 90^\\circ - \\beta$, so $O$ is the circumcentre of the triangle $ABC$.\n\n![](attached_image_1.png)\n![](attached_image_2.png)\n\nNow assume that the point $O$ is the orthocentre of the triangle $A_1B_1C_1$ and the circumcentre of the triangle $ABC$. Let $\\angle A = \\alpha$, $\\angle B = \\beta$, $\\angle C = \\gamma$ and $\\angle A_1 = \\alpha_1$, $\\angle B_1 = \\beta_1$, $\\angle C_1 = \\gamma_1$. Let the points $B'$ on $CA$ and $C'$ on $AB$ be such that the quadrilaterals $CB'OA_1$ and $BA_1OC'$ are cyclic. Then the quadrilateral $AC'OB'$ is also cyclic. Hence $\\angle OC'B' = \\angle OAB' = \\angle OAC = 90^\\circ - \\beta$. Since the supplementary angle of the angle $\\angle A_1OC'$ is $\\beta$, the line $A_1O$ is perpendicular to $B'C'$. This implies $B'C' \\parallel B_1C_1$.\n\nSince $O$ is the orthocentre of the triangle $A_1B_1C_1$, the line $B_1C_1$ is between $A$ and $O$, and $\\angle B_1OC_1 = 180^\\circ - \\alpha_1$.\n\nSince $A_1OB'C$ is cyclic, $\\angle A_1OB' = 180^\\circ - \\gamma$, and similarly $\\angle A_1OC' = 180^\\circ - \\beta$. The sum of these two angles is $180^\\circ + \\alpha$ and therefore the line $B'C'$ lies between $A$ and $O$.\n\nWe may assume that the line $B'C'$ is closer to the point $A$ than the line $B_1C_1$ (the other case is similar). That implies $\\angle B'OC' \\leq B_1OC_1$ and $\\angle B'A_1C' \\leq B_1A_1C_1$. Adding these two inequalities gives\n$$\n\\angle B'OC' + \\angle B'A_1C' \\leq \\angle B_1OC_1 + \\angle B_1A_1C_1. \\qquad (\\ddagger)\n$$\nSince $\\angle B'A_1C' = \\angle B'A_1O + \\angle OA_1C' = \\angle B'CO + \\angle OBC' = \\angle ACO + \\angle OBA = 90^\\circ - \\beta + 90^\\circ - \\gamma = \\alpha$, we have $\\angle B'OC' + \\angle B'A_1C' = 180^\\circ - \\alpha + \\alpha = 180^\\circ$.\n\nAlso, $\\angle B_1OC_1 + \\angle B_1A_1C_1 = 180^\\circ - \\alpha_1 + \\alpha_1 = 180^\\circ$.\n\nTherefore in $(\\ddagger)$ we actually have equality, which can be attained only if $\\angle B'OC' = \\angle B_1OC_1$ and $\\angle B'A_1C' = \\angle B_1A_1C_1$, i.e. if $\\alpha_1 = \\alpha$ and the lines $B'C'$ and $B_1C_1$ coincide. Then also $\\beta_1 = \\beta$ and $\\gamma_1 = \\gamma$, so the triangles $A_1B_1C_1$ and $ABC$ are similar.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 28834,
    "subject": "Mathematics (Multi-modal)",
    "question": "Trapezium $ABCD$ with a longer base $AB$ is inscribed in the circle $k$. Let $A_0$, $B_0$ be respectively the midpoints of segments $BC$, $CA$. Let $N$ be the foot of the altitude from the point $C$ to $AB$, and $G$ the centroid of the triangle $ABC$. Circle $k_1$ goes through $A_0$ and $B_0$ and touches the circle $k$ in the point $X$, different than $C$. Prove that the points $D$, $G$, $N$ and $X$ are collinear.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $k_2$ be the circumcircle of the triangle $A_0B_0C$. Since $k_2$ is the image of the circle $k$ under homothety with centre $C$ (with the coefficient $\\frac{1}{2}$), circles $k_2$ and $k$ touch at point $C$. Tangents $t_X$ and $t_C$ of circle $k$ through points $X$ and $C$ are respectively radical axes of pairs of circles $(k, k_1)$ and $(k, k_2)$, so the point $R$ which is their intersection is the radical center of circles $k$, $k_1$, $k_2$. Hence the point $R$ lies on the radical axis $A_0B_0$ of the pair of circles $(k_1, k_2)$. Let $p$ be the line through points $R$, $A_0$ and $B_0$.\n\n![](attached_image_1.png)\n![](attached_image_2.png)\n\nLet $S$ be the intersection of the lines $p$ and $DX$, and the point $P$ be on the line $t_C$ such that the point $C$ lies between $P$ and $R$. Then $\\angle CRS = \\angle PCD$ (because the lines $CD$ and $p$ are parallel) and $\\angle CXS = \\angle CXD = \\angle PCD$ (because of the chord-tangent theorem using tangent $t_C$). This implies that $\\angle CXS = \\angle CRS$ and the quadrilateral $CSXR$ is cyclic. Hence $\\angle RXC = \\angle RSC$. On the other hand, $\\angle RSC = \\angle SCD$ (again because $CD$ and $p$ are parallel) and $\\angle CDS = \\angle CDX = \\angle RXC$ (because of the chord-tangent theorem using tangent $t_X$). This implies $\\angle SCD = \\angle RSC = \\angle RXC = \\angle CDS$, thus $S$ lies on the bisector of the segment $CD$.\n\n![](attached_image_3.png)\n\nIt remains to show that the points $D$, $S$, $G$ and $N$ lie on the same line. Note that $CDN$ is a right triangle whose circumcentre is the point $S$, so the point $N$ lies on the line $DS$. Let $C_0$ be the midpoint of $AB$. Let $G'$ be the intersection of the lines $DS$ and $CC_0$. The triangles $C_0NG'$ and $CDG'$ are similar and $|CG'| : |C_0G'| = |CD| : |NC_0| = 2 : 1$. But the point $G$ is the unique point on the segment $CC_0$ dividing that segment in the ratio $2 : 1$, so $G = G'$ and $G$ lies on the line $DS$.\n\nThis shows that the points $G$, $N$ and $X$ lie on the line $DS$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 28835,
    "subject": "Mathematics (Multi-modal)",
    "question": "Are there integers $a$ and $b$ such that both $a + b$ and $ab - 1$ are perfect squares?",
    "options": [],
    "answer": "No; there are no such integers.",
    "solution": "Let us assume there are integers $a$, $b$, $M$ and $N$ such that $a + b = M^2$ and $ab - 1 = N^2$.\n\n*First case:* The numbers $a$ and $b$ are even.\nFrom $a \\cdot b = N^2 + 1$ we conclude that $N^2 + 1$ is divisible by $4$ which is impossible.\n\n*Second case:* Exactly one of the numbers $a$ and $b$ (say $a$) is even.\nFrom $a \\cdot b = N^2 + 1$ we conclude $b = 4k + 1$ for some integer $k$, because $N^2 + 1$ has no divisors of the form $4k + 3$ for an integer $k$. Also, $N$ is odd so we can write it in the form $N = 2l + 1$ for some integer $l$. From $ab = (2l + 1)^2 + 1 = 4l^2 + 4l + 2$ we conclude that $a = 4m + 2$ for some integer $m$. Then $a + b = 4n + 3$, for some integer $n$, so it can not be a perfect square. We got a contradiction.\n\n*Third case:* Both numbers $a$ and $b$ are odd.\nThen the number $M$ is even, so $M^2$ is divisible by $4$. Hence one of the numbers $a$ and $b$ has to have residue $3$ when divided by $4$. We again got a contradiction since $N^2 + 1$ can not have a divisor of the form $4k + 3$ for some integer $k$.\n\nSince all cases lead to a contradiction we conclude there are no such $a$ and $b$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 28836,
    "subject": "Mathematics (Multi-modal)",
    "question": "For a positive integer $d$, let $f(d)$ be the smallest positive integer with exactly $d$ positive divisors. (E.g. $f(1) = 1$, $f(5) = 16$, $f(6) = 12$.)\nProve that for any positive integer $k$ the number $f(2^{k-1})$ divides $f(2^k)$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $n = p_1^{a_1} p_2^{a_2} \\cdots p_r^{a_r}, a_i \\ge 0$ be the prime factorization of the number $n$. Then the number of divisors of $n$ is $(a_1+1)(a_2+1)\\cdots(a_r+1)$.\n\nLet $k$ be a positive integer and let $f(2^k) = p_1^{a_1} p_2^{a_2} \\cdots p_r^{a_r}$. Then $a_i = 2^{b_i} - 1$ for some integers $b_i \\ge 0$, $i = 1, 2, \\dots, r$ and $\\sum_{i=1}^r b_i = k$.\nLet $p_i, p_j$ be any two distinct prime factors of $f(2^k)$. If we replace $p_i^{2^{b_i}-1}$ with $p_i^{2^{b_i+1}-1}$ and $p_j^{2^{b_j}-1}$ with $p_j^{2^{b_j-1}-1}$ in the prime factorization of $f(2^k)$ we get a number that also has $2^k$ divisors (as $f(2^k)$) because $2^{b_i+1} \\cdot 2^{b_j-1} = 2^{b_i} \\cdot 2^{b_j}$. Minimality of $f(2^k)$ implies\n$$\np_i^{2^{b_i}-1} \\cdot p_j^{2^{b_j}-1} < p_i^{2^{b_i+1}-1} \\cdot p_j^{2^{b_j-1}-1},\n$$\nor equivalently\n$$\np_j^{2^{b_j}-1} < p_i^{2^{b_i}}. \\qquad (\\bullet)\n$$\nIf we allow some of the integers $c_i$ or $b_i$ to be zero, we can write $f(2^{k+1}) = p_1^{2^{c_1}-1} p_2^{2^{c_2}-1} \\cdots p_r^{2^{c_r}-1}$.\nSince $\\sum_{i=1}^r c_i = k+1 > k = \\sum_{i=1}^r b_i$, there is a positive integer $s$ such that $c_s > b_s$.\nLet $t$ be a positive integer different than $s$. Using $(\\bullet)$ for the pair $(p_s, p_t)$, and for the pair $(p_t, p_s)$ gives\n$$\np_t^{2^{c_t}} > p_s^{2^{c_s}-1} \\ge p_s^{2^{b_s}} > p_t^{2^{b_t}-1}.\n$$\nWe conclude $c_t > b_t - 1$, i.e. $c_t \\ge b_t$ for all $t = 1, \\dots, r$, which implies that $f(2^{k+1})$ divides $f(2^k)$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 28837,
    "subject": "Mathematics (Multi-modal)",
    "question": "Prove that every positive integer $d$ has a positive multiple $n$, such that one can delete one of the non-zero digits of $n$ to obtain another multiple of $d$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "We will find a number $n$ in the form $10^k(10a + b) + c$ where $0 \\le c < 10^k$ and $a, b$ are digits. Deleting the digit $b$ gives the number $n_1 = 10^k a + c$. Since $n - n_1 = 10^k(9a+b)$, to satisfy the conditions of the problem it is sufficient that $d$ divides $9a+b$ and $10^k a + c$.\n\nThe digit $b$ can be chosen so that $9$ divides $d-b$ and we set $a = \\frac{d-b}{9}$. Then $d = 9a+b$. Let $k$ be such that $10^{k-1} > d$. Then there are integers $q$ and $0 \\le r < d$ such that $10^k a + 10^{k-1} = dq + r$. If we set $c = 10^{k-1} - r > 0$, then $10^k a + c = dq$, obviously a number divisible by $d$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 28838,
    "subject": "Mathematics (Multi-modal)",
    "question": "For a given positive integer $k$ let $S(k)$ denote the sum of all numbers from the set $\\{1, 2, \\dots, k\\}$ relatively prime to $k$. Let $m$ be a positive integer and $n$ an odd positive integer. Prove that there exist positive integers $x$ and $y$ such that $m$ divides $x$ and $2S(x) = y^n$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "For $k = 1$, $S(k) = 1$. For $k > 1$, note that for a positive integer $a < k$ relatively prime to $k$ the number $k - a$ is also relatively prime to $k$. Hence the summands in the sum $S(k)$ can be paired and each pair has sum $k$. The number of positive integers relatively prime to $k$ (and less than $k$) is $\\varphi(k)$, so\n$$\nS(k) = \\frac{k\\varphi(k)}{2}.\n$$\nLet $q$ be the largest prime factor of $m$, and let $2 = p_1 < p_2 < \\dots < p_s = q$ be consecutive prime numbers. Then $m = p_1^{a_1} \\dots p_s^{a_s}$ (where some of $a_i$ are 0). We will construct a number $x$ of the form $x = p_1^{b_1} \\dots p_s^{b_s}$ satisfying the condition of the problem. Note that\n$$\n\\begin{aligned}\n2S(x) &= x\\varphi(x) = p_1^{b_1} \\dots p_s^{b_s} \\cdot p_1^{b_1-1} (p_1-1) \\dots p_s^{b_s-1} (p_s-1) \\\\\n&= p_1^{2b_1-1} (p_1-1) \\dots p_s^{2b_s-1} (p_s-1).\n\\end{aligned}\n$$\nAlso, for every $i \\le s$ all prime divisors of $p_i - 1$ are among $p_1, \\dots, p_s$, so we choose $c_1, \\dots, c_s$ such that $(p_1 - 1) \\dots (p_s - 1) = p_1^{c_1} \\dots p_s^{c_s}$, i.e.\n$$\n2S(x) = p_1^{2b_1+c_1-1} \\dots p_s^{2b_s+c_s-1}.\n$$\nThe number $x$ satisfies the condition of the problem if and only if $2b_i + c_i - 1$ is divisible by $n$ and $b_i \\ge a_i$, for all $i \\in \\{1, \\dots, s\\}$.\nSince $n$ is odd, its multiples alternate in being even and odd, so for every $i$ we can choose a large enough $k_i \\in \\mathbb{N}$ such that $k_i n \\equiv c_i - 1 \\pmod 2$ and\n$$\nb_i = \\frac{k_i n - c_i + 1}{2} \\ge a_i.\n$$\n\nThen $2S(x) = p_1^{2b_1+c_1-1} \\dots p_s^{2b_s+c_s-1} = (p_1^{k_1} \\dots p_s^{k_s})^n$, which shows that $x$ satisfies the condition of the problem.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 28839,
    "subject": "Mathematics (Multi-modal)",
    "question": "Determine all real solutions of the equations\n$$\n\\begin{align*}\nx^2 - y &= z^2 \\\\\ny^2 - z &= x^2 \\\\\nz^2 - x &= y^2.\n\\end{align*}\n$$",
    "options": [],
    "answer": "[(0, 0, 0), (1, 0, -1), (0, -1, 1), (-1, 1, 0)]",
    "solution": "By summing the given equations we get\n$$\nx + y + z = 0.\n$$\nIt follows that $z = -x - y$, and from the first equation we get\n$$\n\\begin{aligned}\nx^2 - y &= (-x - y)^2 \\\\\nx^2 - y &= x^2 + 2xy + y^2 \\\\\n2xy + y^2 + y &= 0 \\\\\ny(2x + y + 1) &= 0,\n\\end{aligned}\n$$\nso $y = 0$ or $2x + y + 1 = 0$.\n\nIf $y = 0$, it follows that $z = -x$ so from the second equation we get $x(x - 1) = 0$, hence $x = 0$ or $x = 1$. The corresponding solutions are $(x, y, z) = (0, 0, 0)$ and $(x, y, z) = (1, 0, -1)$.\n\nIf $2x + y + 1 = 0$, we have $y = -2x - 1$ so $z = -x - y = x + 1$. Now from the third equation we get $x(x + 1) = 0$, hence $x = 0$ or $x = -1$. The corresponding solutions are $(x, y, z) = (0, -1, 1)$ and $(x, y, z) = (-1, 1, 0)$.\n\nThus, the solutions of the given system of equations are\n$$\n(x, y, z) \\in \\{(0, 0, 0), (1, 0, -1), (0, -1, 1), (-1, 1, 0)\\}.\n$$",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 28840,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find real number $a$ such that $x = \\frac{1}{2}$ is a solution of the equation\n$$\n\\left( \\frac{x+1}{1-x} - \\frac{1-x}{x+1} - \\frac{4x^2}{x^2-1} \\right) \\left( \\frac{1}{a^3+a^2} - \\frac{1-a}{a^2} - 1 \\right) = \\frac{1}{2}\n$$",
    "options": [],
    "answer": "-1/9",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 28841,
    "subject": "Mathematics (Multi-modal)",
    "question": "By erasing the first two digits of a positive integer $n$ we obtain a number that is 73 times smaller. Determine the smallest two such numbers $n$.",
    "options": [],
    "answer": "365 and 1825",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 28842,
    "subject": "Mathematics (Multi-modal)",
    "question": "On a math test there are $40$ problems. For each correct answer one gets $15$ points, and for each incorrect answer $-4$ points. Dinko has solved all the problems, but he made some mistakes. How many incorrect answers he had, if he scored $353$ points in total?",
    "options": [],
    "answer": "13",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 28843,
    "subject": "Mathematics (Multi-modal)",
    "question": "Prove that there do not exist positive integers $k$ and $n$ such that\n$$\nk(k + 1)(k + 2)(k + 3) = n(n + 1).\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "We have\n$$\nk(k + 1)(k + 2)(k + 3) = (k^2 + 3k)(k^2 + 3k + 2) = (k^2 + 3k + 1)^2 - 1.\n$$\nThat means that $n(n + 1) + 1 = n^2 + n + 1$ has to be a perfect square, but that is impossible since\n$$\nn^2 < n^2 + n + 1 < n^2 + 2n + 1 = (n + 1)^2\n$$\ni.e. $n^2 + n + 1$ is between two consecutive squares.\nTherefore, such positive integers $k$ and $n$ do not exist.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 28844,
    "subject": "Mathematics (Multi-modal)",
    "question": "Determine all ordered triplets of real numbers $(x, y, z)$ such that\n$$\n\\begin{cases}\nx + y = \\sqrt{28} \\\\\nxy - 2z^2 = 7.\n\\end{cases}\n$$",
    "options": [],
    "answer": "(sqrt(7), sqrt(7), 0)",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 28845,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $a$ and $b$ be the legs, and $c$ the hypotenuse of a right triangle.\nProve that the following inequality holds\n$$\n\\left(1 + \\frac{c}{a}\\right) \\left(1 + \\frac{c}{b}\\right) \\geqslant 3 + 2\\sqrt{2}.\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "By rearranging the left hand side of the inequality we get\n$$\n\\begin{aligned}\n\\left(1 + \\frac{c}{a}\\right) \\left(1 + \\frac{c}{b}\\right) &= \\frac{(a+c)(b+c)}{ab} = \\frac{c^2 + ab + c(a+b)}{ab} \\\\\n&= \\frac{c^2}{ab} + 1 + \\frac{c(a+b)}{ab}\n\\end{aligned}\n$$\nSince the triangle is right it follows that $a^2 + b^2 = c^2$\n$$\n= \\frac{a^2 + b^2}{ab} + 1 + \\frac{\\sqrt{a^2 + b^2}(a+b)}{ab}\n$$\n\nNow we use $a^2 + b^2 \\ge 2ab$ and $a + b \\ge 2\\sqrt{ab}$ to get\n$$\n\\begin{aligned}\n\\ge & \\frac{2ab}{ab} + 1 + \\frac{\\sqrt{2ab} \\cdot 2\\sqrt{ab}}{ab} \\\\\n= & 2 + 1 + 2\\sqrt{2} = 3 + 2\\sqrt{2}.\n\\end{aligned}\n$$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 28846,
    "subject": "Mathematics (Multi-modal)",
    "question": "The lines drawn parallel to the sides of an isosceles triangle divide the triangle into three congruent equilateral quadrilaterals, three congruent trapeziums and one isosceles triangle in the middle. If the area of the triangle in the middle is twice the area of each equilateral quadrilateral, what is the share of the area of each trapezium in the area of the original triangle?",
    "options": [],
    "answer": "1/5",
    "solution": "",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 28847,
    "subject": "Mathematics (Multi-modal)",
    "question": "Legs of a right triangle are of lengths $6$ cm and $8$ cm. What is the radius of its circumcircle?",
    "options": [],
    "answer": "5 cm",
    "solution": "",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 28848,
    "subject": "Mathematics (Multi-modal)",
    "question": "Only one divisor of the number $3^{12} - 1$ is larger than $70$ and smaller than $80$. Find it. (Hong Kong)",
    "options": [],
    "answer": "73",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 28849,
    "subject": "Mathematics (Multi-modal)",
    "question": "Prove that the equation\n$$\nx^2 = 2y^2 - 75y + 5\n$$\nhas no integer solutions.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 28850,
    "subject": "Mathematics (Multi-modal)",
    "question": "A rectangle with dimensions $5 \\times 6$ is divided into eight rectangles whose sides are parallel to the sides of the original rectangle, and the lengths of their sides are positive integers. Prove that at least two of the eight rectangles are congruent.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 28851,
    "subject": "Mathematics (Multi-modal)",
    "question": "A regular 2013-gon $A_1A_2...A_{2013}$ is given. Let $S$ be the intersection of the segments $A_1A_4$ and $A_3A_5$. Determine the angle $\\angle A_3SA_4$.",
    "options": [],
    "answer": "180 − 540/2013 degrees",
    "solution": "",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 28852,
    "subject": "Mathematics (Multi-modal)",
    "question": "Determine the smallest positive integer $n$ such that a half of $n$ is a perfect square, a third of $n$ is a perfect cube, and a fifth of $n$ a fifth power of an integer.",
    "options": [],
    "answer": "30233088000000",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 28853,
    "subject": "Mathematics (Multi-modal)",
    "question": "John and Mary have two identical decks of $50$ cards with different symbols. John shuffles his cards and puts them on the table, and then Mary shuffles her cards and puts them on John's cards. Then John counts the cards between each pair of two identical cards and adds up those $50$ numbers. Which results can John obtain? (Ratko Višak)",
    "options": [],
    "answer": "2450",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 28854,
    "subject": "Mathematics (Multi-modal)",
    "question": "Determine all real numbers $a$ such that there exists a complex number $z$ which satisfies\n$$\n|z| = 1 \\quad \\text{and} \\quad |az - 1| = a|z + 1|.\n$$",
    "options": [],
    "answer": "a >= 1/3 or a = -1",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 28855,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $a$ be a complex number such that\n$a^5 + a + 1 = 0.$\nWhat values can the expression $a^2 (a - 1)$ take on?",
    "options": [],
    "answer": "-1, 3/2 + (√3/2)i, 3/2 - (√3/2)i",
    "solution": "We have\n$$\n\\begin{aligned}\na^5 + a + 1 &= (a^5 - a^4) + (a^4 - a^3) + (a^3 - a^2) + a^2 + a + 1 \\\\\n&= (a^2 + a + 1)(a^3 - a^2 + 1).\n\\end{aligned}\n$$\nIf $a^2 + a + 1 = 0$, then $a^2(a - 1) = -(a + 1)(a - 1) = 1 - a^2 = a + 2$.\nSince $a = \\frac{-1 \\pm i\\sqrt{3}}{2}$, it follows that $a^2(a - 1) = \\frac{3}{2} \\pm \\frac{\\sqrt{3}}{2} i$.\nIf $a^3 - a^2 + 1 = 0$, then $a^2(a - 1) = a^3 - a^2 = -1$.\nTherefore, the observed expression can take on values $-1, \\frac{3}{2}, \\frac{\\sqrt{3}}{2} i$ and $\\frac{3}{2} - \\frac{\\sqrt{3}}{2} i$.",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 28856,
    "subject": "Mathematics (Multi-modal)",
    "question": "Which number has more divisors in the set of all positive integers, $2013^2$ or $20480$?",
    "options": [],
    "answer": "2013^2",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 28857,
    "subject": "Mathematics (Multi-modal)",
    "question": "Prove that the equation\n$$\n3x^4 + 2013 = 25y^2 - 24x^2\n$$\nhas no integer solutions.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Rewrite the equation as:\n$$\n3x^4 + 2013 + 24x^2 = 25y^2\n$$\nwhich gives\n$$\n25y^2 = 3x^4 + 24x^2 + 2013\n$$\nSo $25y^2$ is congruent to $3x^4 + 24x^2 + 2013$.\n\nLet us consider the equation modulo $25$:\n$$\n3x^4 + 24x^2 + 2013 \\equiv 0 \\pmod{25}\n$$\nNote that $2013 \\equiv 13 \\pmod{25}$, so:\n$$\n3x^4 + 24x^2 + 13 \\equiv 0 \\pmod{25}\n$$\nLet $x^2 = t$, then $x^4 = t^2$:\n$$\n3t^2 + 24t + 13 \\equiv 0 \\pmod{25}\n$$\nWe can check all possible values of $t$ modulo $25$ (since $x$ is integer, $t$ is a quadratic residue modulo $25$).\n\nThe quadratic residues modulo $25$ are $0, 1, 4, 9, 16, 24, 6, 19, 11, 24, 6, 19, 11, 24, 6, 19, 11, 24, 6, 19, 11, 24, 6, 19, 11$ (but let's just check $t = 0$ to $24$).\n\nFor $t = 0$:\n$$\n3 \\cdot 0^2 + 24 \\cdot 0 + 13 = 13 \\not\\equiv 0 \\pmod{25}\n$$\nFor $t = 1$:\n$$\n3 \\cdot 1 + 24 \\cdot 1 + 13 = 3 + 24 + 13 = 40 \\equiv 15 \\pmod{25}\n$$\nFor $t = 4$:\n$$\n3 \\cdot 16 + 24 \\cdot 4 + 13 = 48 + 96 + 13 = 157 \\equiv 7 \\pmod{25}\n$$\nFor $t = 9$:\n$$\n3 \\cdot 81 + 24 \\cdot 9 + 13 = 243 + 216 + 13 = 472 \\equiv 22 \\pmod{25}\n$$\nFor $t = 16$:\n$$\n3 \\cdot 256 + 24 \\cdot 16 + 13 = 768 + 384 + 13 = 1165 \\equiv 15 \\pmod{25}\n$$\nFor $t = 24$:\n$$\n3 \\cdot 576 + 24 \\cdot 24 + 13 = 1728 + 576 + 13 = 2317 \\equiv 17 \\pmod{25}\n$$\nNone of these are congruent to $0$ modulo $25$.\n\nTherefore, there is no integer $x$ such that $3x^4 + 24x^2 + 2013$ is divisible by $25$, so there are no integer solutions to the equation.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 28858,
    "subject": "Mathematics (Multi-modal)",
    "question": "Determine the complex number $z$ such that\n$$\n\\text{Re} \\frac{1}{1-z} = 2 \\quad \\text{and} \\quad \\text{Im} \\frac{1}{1-z} = -1\n$$",
    "options": [],
    "answer": "(3 - i)/5",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 28859,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABC$ be a right triangle with height $\\overline{CN}$. If $|AC| = |BN| = 1$, determine the length of the hypotenuse $\\overline{AB}$.",
    "options": [],
    "answer": "(1 + sqrt(5)) / 2",
    "solution": "",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 28860,
    "subject": "Mathematics (Multi-modal)",
    "question": "Determine all pairs $(x, y)$ of integers such that\n$$\ny^2 = x^3 + 3x^2 + 2x.\n$$",
    "options": [],
    "answer": "[(-2, 0), (-1, 0), (0, 0)]",
    "solution": "The factorization of the right hand side of the equation gives us\n$$\ny^2 = x(x + 1)(x + 2).\n$$\nSince $x \\in \\mathbb{Z}$, obtained factors are consecutive integers.\nFirst we assume that all the factors, $x$, $x+1$ and $x+2$ are different from zero. Since the left hand side of the equality is positive, these three factors also have to be positive.\nIt follows that $x+1$ and $x(x+2)$ are relatively prime (since $x+1$ and $x$ are relatively prime and also $x+1$ and $x+2$ are relatively prime). The product of two relatively prime positive integers is a perfect square if and only if each of the numbers is a perfect square. In particular, that means that\n$$\nx(x + 2) = x^2 + 2x = (x + 1)^2 - 1\n$$\nis a perfect square. Therefore, $(x+1)^2 - 1$ and $(x+1)^2$ are two consecutive integers which are both perfect squares, and that is possible if and only if those numbers are zero and one. Hence,\n$$\n(x+1)^2 = 1\n$$\nso $x = -2$ or $x = 0$. But then $x + 2 = 0$ or $x = 0$, which is a contradiction with the assumption that all the factors are different from zero.\nIf one of the factors on the right hand side is equal to zero, then $y = 0$ and we get the solutions\n$$\n(x, y) \\in \\{(-2, 0), (-1, 0), (0, 0)\\}.\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 28861,
    "subject": "Mathematics (Multi-modal)",
    "question": "Solve the following equation in the set of real numbers\n$$\n\\frac{1}{x^2} + \\frac{1}{(1-x)^2} = 24.\n$$",
    "options": [],
    "answer": "x = (1 + 1/√3)/2, x = (1 − 1/√3)/2, x = (1 + √2)/2, x = (1 − √2)/2",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 28862,
    "subject": "Mathematics (Multi-modal)",
    "question": "The segment $\\overline{AB}$ is the longer side of the rectangle $ABCD$. Let $E$ be the intersection of the perpendicular from $B$ to $\\overline{AC}$ and the line $AD$, and let $F$ be the intersection of the circle with centre $A$ which passes through $B$ and $\\overline{CD}$. Prove that the lines $AF$ and $EF$ are perpendicular.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 28863,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABCD$ be a trapezium with acute angles along the base $\\overline{AB}$ and perpendicular diagonals which intersect at $O$. Ray $OA$ intersects the circle with diameter $\\overline{BD}$ at $M$, and ray $OB$ intersects the circle with diameter $\\overline{AC}$ at $N$. Prove that the points $M$, $N$, $C$ and $D$ lie on the same circle. (New Zealand 2012)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Since $ABCD$ is a trapezium, the triangles $ABO$ and $CDO$ are similar and we have $\\frac{|OA|}{|OB|} = \\frac{|OC|}{|OD|}$.\n\n![](attached_image_1.png)\n\nEuclid's theorem gives us $|OM|^2 = |OB| \\cdot |OD|$, $|ON|^2 = |OA| \\cdot |OC|$, so\n$$\n\\frac{|OM|^2}{|ON|^2} = \\frac{|OB| \\cdot |OD|}{|OA| \\cdot |OC|} = \\frac{|OD|^2}{|OC|^2}.\n$$\nNow from $|OM| : |ON| = |OD| : |OC|$ and $\\angle MON = \\angle COD = 90^\\circ$ we conclude that triangles $MON$ and $DOC$ are similar and hence $\\angle MNO = \\angle DCO$.\nIt follows that $\\angle MND = \\angle DCM$ and therefore the points $C$, $D$, $M$ and $N$ are concyclic.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 28864,
    "subject": "Mathematics (Multi-modal)",
    "question": "Prove that the sum of all three-digit numbers whose decimal representation consists of three different digits different from zero has at least three different prime divisors.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let us consider all three-digit numbers whose digits are all different and nonzero. Each such number can be written as $100a + 10b + c$, where $a$, $b$, $c$ are digits from $1$ to $9$, all different.\n\nThe total number of such numbers is $9 \\times 8 \\times 7 = 504$.\n\nLet $S$ be the sum of all such numbers.\n\nLet us fix the digits $a$, $b$, $c$ in all possible orders. For each digit, it appears in each position (hundreds, tens, units) the same number of times, because the digits are permuted among the three positions.\n\nFor each digit $d$ ($1 \\leq d \\leq 9$), the number of ways to choose the other two digits is $8 \\times 7 = 56$, and for each such choice, $d$ appears in each position exactly $2! = 2$ times (since there are $3! = 6$ permutations, and for each position, $2$ permutations place $d$ there).\n\nSo, for each digit $d$, it appears $56 \\times 2 = 112$ times in each position.\n\nTherefore, the total sum is:\n\n$$\nS = \\sum_{d=1}^9 112 \\cdot (100d + 10d + d) = \\sum_{d=1}^9 112 \\cdot 111d = 112 \\cdot 111 \\cdot \\sum_{d=1}^9 d\n$$\n\nBut $\\sum_{d=1}^9 d = 45$.\n\nSo,\n$$\nS = 112 \\cdot 111 \\cdot 45\n$$\n\nLet us factor this number:\n\n$112 = 16 \\times 7 = 2^4 \\times 7$\n\n$111 = 3 \\times 37$\n\n$45 = 5 \\times 9 = 5 \\times 3^2$\n\nSo,\n$$\nS = (2^4 \\times 7) \\times (3 \\times 37) \\times (5 \\times 3^2)\n$$\n\nMultiply the $3$'s:\n\n$3 \\times 3^2 = 3^3$\n\nSo,\n$$\nS = 2^4 \\times 3^3 \\times 5 \\times 7 \\times 37\n$$\n\nThus, the prime divisors of $S$ are $2$, $3$, $5$, $7$, and $37$.\n\nTherefore, the sum $S$ has at least three different prime divisors (in fact, it has five).",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 28865,
    "subject": "Mathematics (Multi-modal)",
    "question": "A $6 \\times 6$ table is given.\na) If any 9 fields of the table are labeled, prove that it is possible to select three rows and three columns that contain all of the labeled fields.\nb) Label 10 fields of the table so that for any three rows and three columns that we select there is at least one labeled field that is contained neither in the selected rows nor the columns.\n(Moscow olympiad 1961, modified)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "a) We consider three different cases (note that these are the only possibilities):\n1. There are three rows that contain at least two labeled fields. In this case we select these three rows and we are left with at most 3 labeled fields, so we can select three columns that contain the remaining labeled fields.\n2. There is one row that contains at least 3 and another row that contains at least two labeled fields. In this case we select these two rows and we are left with at most 4 labeled fields, so we select a row that contains one of the remaining labeled fields and three columns that contain the other labeled fields.\n\n---\n\n3. There is a row that contains at least 4 labeled fields.\nIn this case we select that row and we are left with at most 5 labeled fields, so similarly we can select two rows and three columns that contain the remaining labeled fields.\nb) One such labeling is given in the following figure:\n\n![](attached_image_1.png)",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 28866,
    "subject": "Mathematics (Multi-modal)",
    "question": "From $n^3$ unit cubes Ivica assembled a large cube with edge length $n$ and then he coloured some of the six sides of the large cube. When he disassembled the large cube, he found that exactly $1000$ unit cubes don't have any coloured side. Show that this is indeed possible and determine the number of sides of the large cube that Ivica coloured.",
    "options": [],
    "answer": "3",
    "solution": "",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 28867,
    "subject": "Mathematics (Multi-modal)",
    "question": "All points $(x, y)$ with integer coordinates where $1 \\le x \\le 200$ and $1 \\le y \\le 100$ are marked in the coordinate system, 20000 points in total. How many lines of length $\\sqrt{5}$ whose ends are marked points are there? (Mea Bombardelli)",
    "options": [],
    "answer": "78208",
    "solution": "A line segment of length $\\sqrt{5}$ between two integer points must have a vector $\\vec{v} = (a, b)$ with $a^2 + b^2 = 5$. The integer solutions are $(\\pm 1, \\pm 2)$ and $(\\pm 2, \\pm 1)$.\n\nSo, the possible displacements are $(1,2)$, $(1,-2)$, $(-1,2)$, $(-1,-2)$, $(2,1)$, $(2,-1)$, $(-2,1)$, $(-2,-1)$.\n\nFor each such vector, we count the number of segments whose endpoints are both in the grid $1 \\le x \\le 200$, $1 \\le y \\le 100$.\n\nLet us consider $(a, b)$ with $a = 1, b = 2$.\n\nThe starting point $(x, y)$ must satisfy $1 \\le x \\le 200 - 1 = 199$ and $1 \\le y \\le 100 - 2 = 98$.\n\nSo, for $(1,2)$, the number of such segments is $199 \\times 98$.\n\nSimilarly, for $(1,-2)$, $1 \\le x \\le 199$, $3 \\le y \\le 100$ (since $y + 2 \\le 100 \\implies y \\le 98$, $y \\ge 1$; for $y - 2 \\ge 1 \\implies y \\ge 3$), so $199 \\times 98$.\n\nFor $(2,1)$: $1 \\le x \\le 198$, $1 \\le y \\le 99$, so $198 \\times 99$.\n\nFor $(2,-1)$: $1 \\le x \\le 198$, $2 \\le y \\le 100$, so $198 \\times 99$.\n\nSimilarly for the negatives, but each segment is counted twice (once from each endpoint), so we only count each vector once and multiply by 2.\n\nSo, the total number is:\n\n$2 \\times [(199 \\times 98) + (198 \\times 99)] = 2 \\times (199 \\times 98 + 198 \\times 99)$\n\nCalculate:\n\n$199 \\times 98 = (200 - 1) \\times 98 = 200 \\times 98 - 98 = 19600 - 98 = 19502$\n\n$198 \\times 99 = (200 - 2) \\times 99 = 200 \\times 99 - 2 \\times 99 = 19800 - 198 = 19602$\n\nSum: $19502 + 19602 = 39104$\n\nTotal: $2 \\times 39104 = 78208$\n\n**Answer:** $78\\,208$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 28868,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $0 < a < b < c < d$ and let each of the quadratic functions $p(x) = x^2 + dx + a$ and $q(x) = x^2 + cx + b$ have two different real roots. Prove that all four roots are mutually different. (Russia 1996)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let the roots of $p(x) = x^2 + dx + a$ be $x_1$ and $x_2$, and the roots of $q(x) = x^2 + cx + b$ be $y_1$ and $y_2$.\n\nSince $p(x)$ has two different real roots, its discriminant is positive:\n$$\nD_p = d^2 - 4a > 0\n$$\nSimilarly, $q(x)$ has two different real roots, so\n$$\nD_q = c^2 - 4b > 0\n$$\n\nSuppose, for contradiction, that some root of $p(x)$ coincides with some root of $q(x)$. That is, there exists $x_0$ such that $p(x_0) = 0$ and $q(x_0) = 0$.\n\nSo:\n$$\nx_0^2 + d x_0 + a = 0\n$$\n$$\nx_0^2 + c x_0 + b = 0\n$$\nSubtracting the second from the first:\n$$\n(d-c)x_0 + (a-b) = 0\n$$\nSince $d > c$, $d-c > 0$.\nSo:\n$$\nx_0 = \\frac{b-a}{d-c}\n$$\n\nNow, plug $x_0$ into $p(x)$:\n$$\n\\left(\\frac{b-a}{d-c}\\right)^2 + d \\left(\\frac{b-a}{d-c}\\right) + a = 0\n$$\nMultiply both sides by $(d-c)^2$:\n$$\n(b-a)^2 + d(b-a)(d-c) + a(d-c)^2 = 0\n$$\nExpand $d(b-a)(d-c)$:\n$$\nd(b-a)(d-c) = d(b-a)d - d(b-a)c = d^2(b-a) - d c(b-a)\n$$\nSo:\n$$\n(b-a)^2 + d^2(b-a) - d c(b-a) + a(d^2 - 2 d c + c^2) = 0\n$$\nExpand $a(d^2 - 2 d c + c^2)$:\n$$\na d^2 - 2 a d c + a c^2\n$$\nSo the equation is:\n$$\n(b-a)^2 + d^2(b-a) - d c(b-a) + a d^2 - 2 a d c + a c^2 = 0\n$$\nGroup terms:\n$$\n(b-a)^2 + d^2(b-a) - d c(b-a) + a d^2 - 2 a d c + a c^2 = 0\n$$\n\nNow, let's try to show that this equation cannot hold for $0 < a < b < c < d$.\n\nAlternatively, consider the following:\n\nThe sum of the roots of $p(x)$ is $-d$, and the product is $a$.\nThe sum of the roots of $q(x)$ is $-c$, and the product is $b$.\n\nSuppose $x$ is a root of both $p(x)$ and $q(x)$.\nThen $x^2 + d x + a = 0$ and $x^2 + c x + b = 0$.\nSubtract:\n$$\n(d-c)x + (a-b) = 0\n$$\nSo $x = \\frac{b-a}{d-c}$.\n\nNow, since $p(x)$ has two real roots, $d^2 > 4a$.\nSimilarly, $c^2 > 4b$.\n\nBut $a < b$ and $c < d$.\n\nNow, let's check if $x = \\frac{b-a}{d-c}$ can be a root of $p(x)$.\nPlug into $p(x)$:\n$$\n\\left(\\frac{b-a}{d-c}\\right)^2 + d \\left(\\frac{b-a}{d-c}\\right) + a = 0\n$$\nMultiply both sides by $(d-c)^2$:\n$$\n(b-a)^2 + d(b-a)(d-c) + a(d-c)^2 = 0\n$$\nBut $b-a > 0$, $d-c > 0$, $a > 0$.\nAll terms are positive, so the sum cannot be zero.\n\nTherefore, $p(x)$ and $q(x)$ cannot have a common root.\n\nNow, suppose $p(x)$ has a double root, i.e., $x_1 = x_2$.\nBut the discriminant is positive, so roots are distinct.\nSimilarly for $q(x)$.\n\nSuppose $p(x)$ has two roots, $x_1$ and $x_2$, and $q(x)$ has two roots, $y_1$ and $y_2$.\nSuppose $x_1 = y_1$.\nThen, as above, $x_1 = \\frac{b-a}{d-c}$, but this is not possible as shown above.\n\nSuppose $x_1 = y_2$ or $x_2 = y_1$ or $x_2 = y_2$.\nThe same argument applies.\n\nTherefore, all four roots are mutually different.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 28869,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABCD$ be a parallelogram and let $S$ be the intersection of its diagonals. The bisector of the angle $\\angle ADC$ bisects the line segment $\\overline{AS}$ and intersects with the line $BC$ at point $E$. Determine the ratios $|BE| : |BC|$ and $|AB| : |BC|$.",
    "options": [],
    "answer": "BE:BC = 2:3, AB:BC = 3:1",
    "solution": "",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 28870,
    "subject": "Mathematics (Multi-modal)",
    "question": "Collinear points $A$, $B$ and $M$ are given so that $M$ is between $A$ and $B$. Let $k$ be the circle with diameter $\\overline{AB}$ and let $N$ be any point on $k$ different from $A$ and $B$. Prove that the expression\n$$\n\\frac{\\operatorname{tg}(\\angle ANM)}{\\operatorname{tg}(\\angle MAN)}\n$$\nis a constant, i.e. that it doesn't depend on the choice of $N$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 28871,
    "subject": "Mathematics (Multi-modal)",
    "question": "If $60^a = 3$, $60^b = 5$ and $x = \\frac{1-a-b}{2(1-b)}$, prove that $12^x$ is a positive integer.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let us first find $a$ and $b$.\n\nSince $60^a = 3$, take logarithms:\n$$\n60^a = 3 \\implies a = \\frac{\\log 3}{\\log 60}\n$$\nSimilarly,\n$$\n60^b = 5 \\implies b = \\frac{\\log 5}{\\log 60}\n$$\nNow, compute $1 - a - b$:\n$$\n1 - a - b = 1 - \\frac{\\log 3}{\\log 60} - \\frac{\\log 5}{\\log 60} = 1 - \\frac{\\log 3 + \\log 5}{\\log 60}\n$$\nBut $\\log 3 + \\log 5 = \\log(3 \\cdot 5) = \\log 15$.\nSo,\n$$\n1 - a - b = 1 - \\frac{\\log 15}{\\log 60} = \\frac{\\log 60 - \\log 15}{\\log 60} = \\frac{\\log(60/15)}{\\log 60} = \\frac{\\log 4}{\\log 60}\n$$\nNow, $2(1-b)$:\n$$\n1 - b = 1 - \\frac{\\log 5}{\\log 60} = \\frac{\\log 60 - \\log 5}{\\log 60} = \\frac{\\log(60/5)}{\\log 60} = \\frac{\\log 12}{\\log 60}\n$$\nSo,\n$$\n2(1-b) = 2 \\cdot \\frac{\\log 12}{\\log 60} = \\frac{2 \\log 12}{\\log 60}\n$$\nTherefore,\n$$\nx = \\frac{1-a-b}{2(1-b)} = \\frac{\\frac{\\log 4}{\\log 60}}{\\frac{2 \\log 12}{\\log 60}} = \\frac{\\log 4}{2 \\log 12} = \\frac{1}{2} \\cdot \\frac{\\log 4}{\\log 12}\n$$\nBut $\\log 4 = 2 \\log 2$, so:\n$$\nx = \\frac{1}{2} \\cdot \\frac{2 \\log 2}{\\log 12} = \\frac{\\log 2}{\\log 12}\n$$\nNow,\n$$\n12^x = 12^{\\frac{\\log 2}{\\log 12}} = 2\n$$\nTherefore, $12^x = 2$, which is a positive integer.\n\nThus, $12^x$ is a positive integer.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 28872,
    "subject": "Mathematics (Multi-modal)",
    "question": "Determine the non-negative real number $a$ for which the expression\n$$\na^3 - a^2 - 2\\sqrt{a}\n$$\nis minimal.",
    "options": [],
    "answer": "1",
    "solution": "For $a = 1$ the expression is equal to $-2$. We will prove that this is the minimal value, i.e. that for every $a \\ge 0$ we have $a^3 - a^2 - 2\\sqrt{a} \\ge -2$.\nWe have\n$$\na^3 - a^2 - 2\\sqrt{a} + 2 = a^2(a-1) - 2(\\sqrt{a}-1) = (\\sqrt{a}-1)(a^2(\\sqrt{a}+1) - 2).\n$$\nIf $a \\ge 1$, then $a^2(\\sqrt{a}+1) - 2 \\ge 2 - 2 = 0$ and $\\sqrt{a}-1 \\ge 0$, so our statement is true.\nIf $0 \\le a < 1$, then $a^2(\\sqrt{a}+1) - 2 < 2 - 2 = 0$ and $\\sqrt{a}-1 < 0$, so our statement is true again.",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 28873,
    "subject": "Mathematics (Multi-modal)",
    "question": "Determine all primes $p$ for which there exist positive integers $x$ and $y$ such that\n$$\n\\begin{cases}\np + 1 = 2x^2 \\\\\np^2 + 1 = 2y^2.\n\\end{cases}\n$$",
    "options": [],
    "answer": "7",
    "solution": "Subtracting the given equations we get $p(p-1) = 2(y-x)(y+x)$.\nFrom this we conclude\n$$\np \\mid y + x,\n$$\nbecause otherwise $p$ would be a divisor of $y - x$, and $p - 1$ would be a multiple of number $y + x$, which is impossible (we would have $p - 1 \\ge y + x > y - x \\ge p$ then).\nSince $p > y$ (from the second equation) and $y > x$, we have $2p > y + x$, therefore $p = y + x$.\nIt follows that $p - 1 = 2(y - x)$. By eliminating $y$ we get $p + 1 = 4x$. By plugging that in the first equation we easily get that the only solution is $p = 7$ ($x = 2, y = 5$).",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 28874,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $n$ be a composite positive integer and let $d_1, d_2, \\dots, d_m$ be all of its divisors. Prove that the following equation holds:\n$$\n\\frac{2}{\\log n^m} \\sum_{k=1}^{m} \\log d_k = 1.\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 28875,
    "subject": "Mathematics (Multi-modal)",
    "question": "If $(1 + \\sin t)(1 + \\cos t) = \\frac{5}{4}$, determine $\\sin t + \\cos t$.",
    "options": [],
    "answer": "-1 + sqrt(10)/2",
    "solution": "Let $S = \\sin t + \\cos t$.\n\nWe have:\n$$(1 + \\sin t)(1 + \\cos t) = 1 + \\sin t + \\cos t + \\sin t \\cos t = \\frac{5}{4}$$\n\nSo:\n$$1 + S + \\sin t \\cos t = \\frac{5}{4}$$\n$$S + \\sin t \\cos t = \\frac{1}{4}$$\n\nRecall that:\n$$(\\sin t + \\cos t)^2 = \\sin^2 t + 2 \\sin t \\cos t + \\cos^2 t = 1 + 2 \\sin t \\cos t$$\n\nSo:\n$$S^2 = 1 + 2 \\sin t \\cos t$$\n$$2 \\sin t \\cos t = S^2 - 1$$\n$$\\sin t \\cos t = \\frac{S^2 - 1}{2}$$\n\nSubstitute into the earlier equation:\n$$S + \\frac{S^2 - 1}{2} = \\frac{1}{4}$$\nMultiply both sides by $4$:\n$$4S + 2(S^2 - 1) = 1$$\n$$4S + 2S^2 - 2 = 1$$\n$$2S^2 + 4S - 3 = 0$$\n\nSolve the quadratic:\n$$S^2 + 2S - \\frac{3}{2} = 0$$\n\nUse the quadratic formula:\n$$S = \\frac{-2 \\pm \\sqrt{4 - 4 \\times 1 \\times (-\\frac{3}{2})}}{2}$$\n$$= \\frac{-2 \\pm \\sqrt{4 + 6}}{2}$$\n$$= \\frac{-2 \\pm \\sqrt{10}}{2}$$\n$$= -1 \\pm \\frac{\\sqrt{10}}{2}$$\n\nTherefore,\n$$\\boxed{\\sin t + \\cos t = -1 \\pm \\frac{\\sqrt{10}}{2}}$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 28876,
    "subject": "Mathematics (Multi-modal)",
    "question": "How many ordered pairs of positive integers $(m, n)$ satisfy the equation $m^2 - n^2 = 2^{2013}$?",
    "options": [],
    "answer": "1006",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 28877,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABCDA'B'C'D'$ be a cube of edge length $1$. Points $P$, $Q$, $R$ and $S$ are given on the edges $\\overline{AB}$, $\\overline{AD}$, $\\overline{C'D'}$ and $\\overline{B'C'}$, respectively, so that $PQRS$ is a square whose centre is in the centre of the cube. What is the length of the side of the square?",
    "options": [],
    "answer": "3*sqrt(2)/4",
    "solution": "",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 28878,
    "subject": "Mathematics (Multi-modal)",
    "question": "Prove that among any four numbers from the interval $\\langle 0, \\frac{\\pi}{2} \\rangle$ one can choose two numbers, namely $x$ and $y$, such that\n$$\n8 \\cos x \\cos y \\cos(x - y) + 1 > 4 (\\cos^2 x + \\cos^2 y).\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "The given inequality is successively equivalent to\n$$8 \\cos x \\cos y (\\cos x \\cos y + \\sin x \\sin y) + 1 > 4 \\cos^2 x + 4 \\cos^2 y,$$\n$$8 \\cos^2 x \\cos^2 y - 4 \\cos^2 x - 4 \\cos^2 y + 2 \\sin 2x \\sin 2y + 1 > 0,$$\n$$2 (2 \\cos^2 x - 1) (2 \\cos^2 y - 1) + 2 \\sin 2x \\sin 2y - 1 > 0,$$\n$$\\cos 2x \\cos 2y + \\sin 2x \\sin 2y > \\frac{1}{2},$$\n$$\\cos (2x - 2y) > \\frac{1}{2}.$$\nBy the pigeonhole principle, in one of the sets\n$$\n\\langle 0, \\frac{\\pi}{6} \\rangle, \\quad \\left[ \\frac{\\pi}{6}, \\frac{\\pi}{3} \\right), \\quad \\left[ \\frac{\\pi}{3}, \\frac{\\pi}{2} \\right]\n$$\nthere are two out of four given numbers. Let these be $x$ and $y$.\nNow we have $|2x - 2y| < \\frac{\\pi}{3}$ and $\\cos(2x - 2y) > \\frac{1}{2}$, which proves the claim.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 28879,
    "subject": "Mathematics (Multi-modal)",
    "question": "Prove that the number $\\cos \\frac{\\pi}{2 \\cdot 3^n}$ is irrational for every positive integer $n$.\n(A real number is irrational if it cannot be expressed as a ratio of two integers.)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 28880,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABC$ be an acute triangle with orthocentre $H$. The line through the point $A$ perpendicular to $AC$ and the line through the point $B$ perpendicular to $BC$ intersect at $D$. The circle with centre $C$ which contains $H$ intersects the circumcircle of the triangle $ABC$ at $E$ and $F$. Prove that $|DE| = |DF| = |AB|$. (Stipe Vidak)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $\\alpha$, $\\beta$, $\\gamma$ be the angles of the triangle $ABC$ and $R$ the radius of its circumcircle. Without loss of generality, let $E$ lie on the arc $BC$ and $F$ on the arc $AC$.\nSince $\\angle CAD = \\angle CBD = 90^\\circ$, the point $D$ lies on the circumcircle of $ABC$ and $CD$ is the diameter.\n\n![](attached_image_1.png)\n\nSince $|CE| = |CH|$, $\\angle BEC = 180^\\circ - \\alpha = \\angle BHC$ and $BC$ is the common side of $BEC$ and $BHC$, these triangles are congruent by \"S-S-A\" theorem (angle $180^\\circ - \\alpha$ is obtuse). Thus we have $\\angle ECB = \\angle HCB$, and the line $BC$ is the axis of symmetry of the isosceles triangle $ECH$. Therefore $BC \\perp HE$ and we conclude that the points $E$, $H$ and $A$ lie on the same line.\n\nFrom the right triangle $CDE$ we have\n$$\n|DE| = |CD| \\cos \\angle CDE = |CD| \\cos \\angle CAE = |CD| \\cos(90^\\circ - \\gamma) = |CD| \\sin \\gamma.\n$$\nBy the law of sines we have $2R \\sin \\gamma = |AB|$ and we get $|DE| = |AB|$.\nAnalogously we get $|DF| = |AB|$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 28881,
    "subject": "Mathematics (Multi-modal)",
    "question": "The squares of a large unit square grid are coloured alternately black and white, as on a chess board. A polygon whose sides are on the lines of the grid has been cut out of the grid. Let that polygon consist of $W$ white and $B$ black squares, and its edge consist of $w$ white and $b$ black lines of unit length. Prove that $b - w = 4(B - W)$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 28882,
    "subject": "Mathematics (Multi-modal)",
    "question": "There were $n$ students participating in a competition and each student solved exactly three problems. For any two students there is exactly one problem which they both solved, while each problem was solved by exactly $k$ students. For which positive integers $n$ and $k$ is that possible?\n(Moscow olympiad 1947)",
    "options": [],
    "answer": "[(1, 1), (4, 2), (7, 3)]",
    "solution": "For $k=1$ each problem was solved by only one student. That means that there are no other students, due to the condition that for any two students there is exactly one problem which they both solved.\n\nNow let $k \\ge 2$.\nLet $A$ be one of the students. He solved exactly three problems, but each of these three problems was solved by $k-1$ more students. However, these $3(k-1)$ students are all distinct, because there are no two students who solved same two problems. On the other hand, besides student $A$ and these $3(k-1)$ students there are no others, because every student solved one of the problems solved by $A$. Therefore the number of the students is $n = 3(k-1) + 1 = 3k - 2$.\n\nLet us now determine the number of the problems. Every student solved three problems, but every problem was solved by $k$ students. Therefore the number of the problems is $\\frac{3n}{k}$ and it has to be a positive integer.\n\nUsing the formula above for $n$ we get\n$$\n\\frac{3n}{k} = \\frac{3 \\cdot (3k-2)}{k} = 9 - \\frac{6}{k},\n$$\nfrom which it follows that $k$ divides $6$, which implies $k \\in \\{2, 3, 6\\}$ ($k \\ge 2$).\n\nFor $k=2$ we have $n=4$ students $(A, B, C, D)$ and $\\frac{3n}{k} = 6$ problems $(1, 2, 3, 4, 5, 6)$.\n$A: 1, 2, 3 \\quad B: 1, 4, 5 \\quad C: 2, 4, 6 \\quad D: 3, 5, 6$\n\nFor $k=3$ we have $n=7$ students and $\\frac{3n}{k} = 7$ problems. For example, we can have:\n$A: 1, 2, 3 \\quad B: 1, 4, 5 \\quad C: 1, 6, 7$ \n$D: 2, 4, 6 \\quad E: 2, 5, 7 \\quad F: 3, 4, 7 \\quad G: 3, 5, 6$\n\nFor $k=6$ we would have $n=16$ students and $\\frac{3n}{k} = 8$ problems. Then number of the pairs of the problems would be $\\binom{8}{2} = 28$. However, number of the pairs of the problems solved by the same student is $3 \\cdot 16 = 48$, because there are no two students who solved same two problems. Since $48 > 28$, this situation cannot happen.\n\nHence, the only possibilities are $(n, k) \\in \\{(1, 1), (4, 2), (7, 3)\\}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 28883,
    "subject": "Mathematics (Multi-modal)",
    "question": "27 points are arranged in 9 columns and 3 rows, and each is coloured blue or red. Prove that there exists a rectangle whose vertices are all of the same colour.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Consider the $3 \\times 9$ grid, with each point coloured blue or red.\n\nFor each row, consider the sequence of colours in the 9 columns. For a fixed row, there are $2^9 = 512$ possible colourings. However, we only have 3 rows.\n\nNow, focus on the columns. For each column, consider the triple of colours in the 3 rows. Each column can be coloured in $2^3 = 8$ ways.\n\nLet us fix the rows and consider the columns. For each row, there are 9 points, each coloured blue or red. For each row, select the set of columns where the point is blue (or red). For a fixed pair of columns, consider the colours in the three rows at those columns.\n\nLet us fix two columns, say columns $i$ and $j$ ($i < j$). For each row, the pair $(C_{r,i}, C_{r,j})$ can be $(B,B)$, $(B,R)$, $(R,B)$, or $(R,R)$, where $B$ is blue and $R$ is red.\n\nNow, for each row, record the colour pair in columns $i$ and $j$. Since there are 3 rows, there are $4^3 = 64$ possible assignments of colour pairs for the three rows.\n\nNow, by the pigeonhole principle, consider the following:\n\nLet us fix two columns, say columns $a$ and $b$. For each row, look at the colour of the point in column $a$ and the colour in column $b$. There are 3 rows, so for each row, the pair can be $(B,B)$, $(B,R)$, $(R,B)$, or $(R,R)$.\n\nSuppose that in some two rows, say row $r_1$ and row $r_2$, the colour in column $a$ is the same, and the colour in column $b$ is the same, and both are blue (or both are red). Then, the four points $(r_1,a)$, $(r_1,b)$, $(r_2,a)$, $(r_2,b)$$—$these form the vertices of a rectangle, and all four are the same colour.\n\nBut more generally, for each pair of columns, consider the 3 pairs of colours in the 3 rows. If, for some colour (say blue), both columns $a$ and $b$ have blue in two different rows, then the rectangle formed by those two rows and those two columns has all blue vertices.\n\nAlternatively, consider the following argument:\n\nFor each row, there are 9 points, each coloured blue or red. For each row, there are $2^9$ possible colourings. But we have only 3 rows.\n\nLet us fix the rows and consider the columns. For each column, there are 3 points, each coloured blue or red, so $2^3 = 8$ possible colourings per column. There are 9 columns.\n\nNow, for each row, consider the set of columns where the point is blue. For a fixed row, there are $\\binom{9}{k}$ ways to have $k$ blue points.\n\nBut the key is to use the following classic Ramsey-type argument:\n\nConsider any two rows. For each column, the pair of colours in those two rows can be $(B,B)$, $(B,R)$, $(R,B)$, or $(R,R)$. There are 9 columns, and 4 possible pairs per column.\n\nNow, by the pigeonhole principle, among the 9 columns, there must be at least 3 columns where the pair is the same. That is, for some pair of rows, there are at least 3 columns where the colour in both rows is the same (either both blue or both red).\n\nBut actually, let's formalize the argument:\n\nLet us fix two rows, say row 1 and row 2. For each column, the pair $(C_{1,j}, C_{2,j})$ can be $(B,B)$, $(B,R)$, $(R,B)$, or $(R,R)$. There are 9 columns, and 4 possible pairs.\n\nBy the pigeonhole principle, there must be at least $\\lceil 9/4 \\rceil = 3$ columns where the pair is the same.\n\nBut we need to find a rectangle with all four vertices the same colour.\n\nLet us use the following approach:\n\nFor each row, consider the sequence of colours in the 9 columns. There are $2^9 = 512$ possible colourings per row. Since there are only 3 rows, there are at most 3 different colourings among the rows.\n\nNow, consider the 9 columns. For each column, the sequence of colours in the 3 rows is a triple, which can be any of 8 possibilities.\n\nNow, consider the following: For each pair of columns, consider the 3 pairs of colours in the 3 rows. If, for some colour, say blue, in two different rows, both columns have blue, then the rectangle formed by those two rows and those two columns has all blue vertices.\n\nThis is a classic application of the pigeonhole principle and Ramsey theory. In fact, the result is a special case of the theorem that in any $n \\times m$ grid coloured with two colours, if $n$ and $m$ are large enough, there must be a monochromatic rectangle.\n\nIn this case, with 3 rows and 9 columns, the result holds.\n\nTherefore, there must exist a rectangle whose vertices are all of the same colour.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 28884,
    "subject": "Mathematics (Multi-modal)",
    "question": "Determine the product\n$$\n\\left(1 - \\frac{\\cos 61^{\\circ}}{\\cos 1^{\\circ}}\\right) \\left(1 - \\frac{\\cos 62^{\\circ}}{\\cos 2^{\\circ}}\\right) \\dots \\left(1 - \\frac{\\cos 119^{\\circ}}{\\cos 59^{\\circ}}\\right).\n$$",
    "options": [],
    "answer": "1",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 28885,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABCD$ be a circumscribed quadrilateral where $\\angle DAB = \\angle ABC = 120^\\circ$ and $\\angle CDA = 90^\\circ$. If $|AB| = 1$, determine the circumference of $ABCD$.",
    "options": [],
    "answer": "5 + 3*sqrt(3)",
    "solution": "",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 28886,
    "subject": "Mathematics (Multi-modal)",
    "question": "Determine all positive integers $n$ such that the product of all positive divisors of $n$ is equal to $n^3$. Represent these numbers as a product of prime powers.",
    "options": [],
    "answer": "n = p^5 or n = p^2 q, where p and q are distinct primes",
    "solution": "Let $1 = d_1 < d_2 < \\dots < d_k = n$ be all positive divisors of the number $n$. We see that\n$$\nd_1 \\cdot d_k = d_2 \\cdot d_{k-1} = d_3 \\cdot d_{k-2} = n.\n$$\nFrom this we conclude that the product of all positive divisors of $n$ is equal to $n^3$ if and only if $n$ has exactly six divisors.\n\nPositive integer $n$ can be represented in the form $n = p_1^{\\alpha_1} p_2^{\\alpha_2} \\dots p_r^{\\alpha_r}$, where $p_1, p_2, \\dots, p_r$ are distinct primes, and $\\alpha_1 \\ge \\alpha_2 \\ge \\dots \\ge \\alpha_r$ are positive integers.\n\nIt is known that the number of the positive divisors of that number is $(\\alpha_1 + 1)(\\alpha_2 + 1)\\dots(\\alpha_r + 1)$. Hence, we get\n$$\n(\\alpha_1 + 1)(\\alpha_2 + 1)\\dots(\\alpha_r + 1) = 6.\n$$\nSince every factor on the left hand side is greater than 1, there are only two possibilities:\n$r = 1, \\alpha_1 = 5 \\quad \\text{or} \\quad r = 2, \\alpha_1 = 2, \\alpha_2 = 1.$\n\nCondition that $n$ has exactly six divisors is satisfied if and only if $n = p^5$ for a prime $p$ or $n = p^2q$ for two distinct primes $p$ and $q$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 28887,
    "subject": "Mathematics (Multi-modal)",
    "question": "Determine all positive integers $n$ such that\n$$\n\\lfloor \\sqrt[4]{1} \\rfloor + \\lfloor \\sqrt[4]{2} \\rfloor + \\dots + \\lfloor \\sqrt[4]{n} \\rfloor = \\frac{3}{2}n + 1.\n$$",
    "options": [],
    "answer": "32",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 28888,
    "subject": "Mathematics (Multi-modal)",
    "question": "The upper right quarter of a chess board (with dimensions $8 \\times 8$) is covered with paper. How many rooks (at most) can we place on the remaining part of the board so that no two of them attack each other? In how many ways can they be placed?\n(Two rooks are mutually attacking each other if they are in the same row or the same column.)",
    "options": [],
    "answer": "8 rooks; 576 ways",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 28889,
    "subject": "Mathematics (Multi-modal)",
    "question": "Sequence $(a_n)$ is defined recursively: $a_1 = 2$, $a_n = 2(n + a_{n-1})$ for $n \\ge 2$. Prove that $a_n < 2^{n+2}$ for all $n \\in \\mathbb{N}$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "First few elements of the sequence are\n$a_1 = 2$,  $a_2 = 2(2+2) = 8$,  $a_3 = 2(3+8) = 22$,  $a_4 = 2(4+22) = 52$, $\\dots$\nLet us look at the differences $2^{n+2} - a_n$:\n$$\n\\begin{align*}\n2^3 - a_1 &= 8 - 2 = 6, \\\\\n2^4 - a_2 &= 16 - 8 = 8, \\\\\n2^5 - a_3 &= 32 - 22 = 10, \\\\\n2^6 - a_4 &= 64 - 52 = 12.\n\\end{align*}\n$$\n\nWe assume that $a_n = 2^{n+2} - 2(n+2)$ and we will prove it using mathematical induction.\nThe basis was already checked.\nAssume that $a_n = 2^{n+2} - 2(n+2)$ for some $n \\in \\mathbb{N}$.\nUsing the given recursion we get\n$$\n\\begin{align*}\na_{n+1} &= 2((n+1) + a_n) = 2(n+1+2^{n+2}-2(n+2)) \\\\\n&= 2 \\cdot 2^{n+2} - 2(n+3) = 2^{n+3} - 2(n+3).\n\\end{align*}\n$$\nThis completes the inductive step and we conclude that for all $n \\in \\mathbb{N}$\n$$\na_n = 2^{n+2} - 2(n+2) < 2^{n+2}.\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 28890,
    "subject": "Mathematics (Multi-modal)",
    "question": "Exactly three internal angles of a convex polygon are obtuse. How many sides (at most) can this polygon have? (Math Questions 2009)",
    "options": [],
    "answer": "6",
    "solution": "",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 28891,
    "subject": "Mathematics (Multi-modal)",
    "question": "Prove that the number whose decimal representation consists of 2187 digits 1 is divisible by 2187.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $N$ be the number whose decimal representation consists of 2187 digits 1. Then\n$$\nN = \\underbrace{111\\ldots 1}_{2187\\ \\text{digits}} = \\frac{10^{2187} - 1}{9}.\n$$\n\nWe want to show that $2187$ divides $N$.\n\nNote that $2187 = 3^7$.\n\nIt suffices to show that $3^7$ divides $N$.\n\nSince $N = \\frac{10^{2187} - 1}{9}$, we need to show that $3^7$ divides $10^{2187} - 1$.\n\nObserve that $10^k \\equiv 1 \\pmod{9}$ for any $k$, but we need divisibility by $3^7$.\n\nLet us use Euler's theorem:\n\nSince $\\gcd(10, 3^7) = 1$, Euler's theorem gives $10^{\\varphi(3^7)} \\equiv 1 \\pmod{3^7}$, where $\\varphi(3^7) = 3^7 - 3^6 = 2187 - 729 = 1458$.\n\nSo $10^{1458} \\equiv 1 \\pmod{2187}$.\n\nBut $2187 = 3^7$, and $2187$ divides $10^{2187} - 1$ if and only if the order of $10$ modulo $2187$ divides $2187$.\n\nAlternatively, note that $N = \\frac{10^{2187} - 1}{9}$, so $N$ is an integer, and $3^7$ divides $N$ if and only if $10^{2187} \\equiv 1 \\pmod{3^8}$, since $N = \\frac{10^{2187} - 1}{9}$ and $3^7$ divides $N$ if $10^{2187} \\equiv 1 \\pmod{3^8}$.\n\nBut actually, since $N$ is a repunit, and $3^k$ divides $10^{3^k} - 1$ for $k \\geq 1$, we can use induction or known results:\n\nFor any $k \\geq 1$, $3^k$ divides $10^{3^k} - 1$.\n\nProof by induction:\n\nBase case: $k = 1$, $10^3 - 1 = 999$, which is divisible by $3$.\n\nAssume $3^k$ divides $10^{3^k} - 1$.\n\nConsider $10^{3^{k+1}} - 1 = (10^{3^k})^3 - 1 = (10^{3^k} - 1)((10^{3^k})^2 + 10^{3^k} + 1)$.\n\nBy the induction hypothesis, $3^k$ divides $10^{3^k} - 1$, so $3^k$ divides $10^{3^{k+1}} - 1$.\n\nBut also, $10^{3^k} \\equiv 1 \\pmod{3^k}$, so $(10^{3^k})^2 + 10^{3^k} + 1 \\equiv 1 + 1 + 1 = 3 \\pmod{3^k}$, so $3$ divides this factor, so $3^{k+1}$ divides $10^{3^{k+1}} - 1$.\n\nTherefore, $3^7$ divides $10^{3^7} - 1 = 10^{2187} - 1$.\n\nThus, $3^7$ divides $N = \\frac{10^{2187} - 1}{9}$.\n\nTherefore, $2187$ divides the number whose decimal representation consists of $2187$ digits 1.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 28892,
    "subject": "Mathematics (Multi-modal)",
    "question": "How many complex numbers $z$ are there that satisfy the following two conditions:\n$$\n|z| = 1, \\quad \\text{Re}(z^{100}) = \\text{Im}(z^{200}) \\quad ? \\qquad (\\text{Ilko Brnetić})\n$$",
    "options": [],
    "answer": "400",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 28893,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $a$ and $b$ be real numbers. It is known that the parabola $y = ax^2 + b$ intersects the curve $y = x + \\frac{1}{x}$ at exactly three points. Prove that $3ab < 1$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "From the condition of the problem we conclude that the system of equations\n$$\ny = ax^2 + b\n$$\n$$\ny = x + \\frac{1}{x}\n$$\nhas three solutions, i.e. that the cubic equation\n$$\nax^3 - x^2 + bx - 1 = 0\n$$\nhas three real solutions. Let us denote these solutions by $x_1, x_2, x_3$. Viète's formulas give\n$$x_1 + x_2 + x_3 = \\frac{1}{a},$$\n$$x_1x_2 + x_2x_3 + x_3x_1 = \\frac{b}{a},$$\n$$x_1x_2x_3 = \\frac{1}{a}.$$\nThe inequality $3ab < 1$ is equivalent to $\\frac{1}{a^2} > 3 \\cdot \\frac{b}{a}$, which gives\n$$\n(x_1 + x_2 + x_3)^2 > 3 (x_1x_2 + x_2x_3 + x_3x_1).\n$$\nIt is not hard to see that this is equivalent to\n$$\n(x_1 - x_2)^2 + (x_2 - x_3)^2 + (x_3 - x_1)^2 > 0,\n$$\nand that inequality holds since the points (and consequently their abscissas) are mutually distinct.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 28894,
    "subject": "Mathematics (Multi-modal)",
    "question": "A quadrilateral with vertices $0$, $z$, $\\frac{1}{z}$ and $z + \\frac{1}{z}$ in the complex plane and area $\\frac{35}{37}$ is given. Determine the smallest possible value of the expression $\\left|z + \\frac{1}{z}\\right|^2$.",
    "options": [],
    "answer": "50/37",
    "solution": "",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 28895,
    "subject": "Mathematics (Multi-modal)",
    "question": "Players $A$, $B$ and $C$ roll the die in consecutive order. What is the probability that $C$ rolls a greater number than the other two players?",
    "options": [],
    "answer": "55/216",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 28896,
    "subject": "Mathematics (Multi-modal)",
    "question": "In some country there are three cities $A$, $B$ and $C$. Between each of the two cities there are several roads (at least one) and all the roads are two-way. Apart from the direct road links between two cities, there are also indirect road links. An indirect road link between $X$ and $Y$ consists of a road that connects $X$ with a third city $Z$ and a road that connects $Z$ and $Y$.\n\nIt is known that there are altogether 43 road links between $A$ and $B$, and altogether 29 road links between $B$ and $C$. How many road links can there be between $A$ and $C$ altogether? (Sweden 2012)",
    "options": [],
    "answer": "23 or 97",
    "solution": "Let $a$, $b$, and $c$ be the number of direct roads between $A$ and $B$, $B$ and $C$, and $A$ and $C$, respectively. Let $x$, $y$, and $z$ be the number of roads from $A$ to $B$, $B$ to $C$, and $A$ to $C$, respectively (so $a = x$, $b = y$, $c = z$).\n\nThe total number of road links between $A$ and $B$ is the sum of direct roads and indirect roads via $C$:\n\n- Direct: $a$\n- Indirect: for each road from $A$ to $C$ and each road from $C$ to $B$, we get an indirect link. So $z \\times y$.\n\nThus, $a + z y = 43$.\n\nSimilarly, the total number of road links between $B$ and $C$ is $b + x z = 29$.\n\nThe total number of road links between $A$ and $C$ is $c + x y$.\n\nBut since $a$, $b$, $c$ are the number of direct roads, and $x$, $y$, $z$ are the same, we can let $x = a$, $y = b$, $z = c$.\n\nSo:\n\n- Road links between $A$ and $B$: $a + c b = 43$\n- Road links between $B$ and $C$: $b + a c = 29$\n- Road links between $A$ and $C$: $c + a b = ?$\n\nLet $p = a$, $q = b$, $r = c$.\n\nSo:\n\n$p + r q = 43$\n$q + p r = 29$\n$r + p q = ?$\n\nWe are to find the possible values of $r + p q$.\n\nLet us solve for $p$, $q$, $r$.\n\nFrom the first equation:\n$p + r q = 43$\nFrom the second:\n$q + p r = 29$\n\nLet us try to find integer solutions with $p, q, r \\geq 1$.\n\nLet us try small values for $r$.\n\nLet $r = 1$:\n\nFirst equation: $p + 1 \\cdot q = 43 \\implies p + q = 43$\nSecond: $q + p \\cdot 1 = 29 \\implies q + p = 29$\n\nContradicts.\n\nTry $r = 2$:\nFirst: $p + 2q = 43$\nSecond: $q + 2p = 29$\n\nFrom first: $p = 43 - 2q$\nPlug into second: $q + 2(43 - 2q) = 29 \\implies q + 86 - 4q = 29 \\implies -3q = -57 \\implies q = 19$\nThen $p = 43 - 2 \\times 19 = 43 - 38 = 5$\n\nSo $p = 5$, $q = 19$, $r = 2$\n\nNow, $r + p q = 2 + 5 \\times 19 = 2 + 95 = 97$\n\nTry $r = 3$:\nFirst: $p + 3q = 43 \\implies p = 43 - 3q$\nSecond: $q + 3p = 29 \\implies q + 3(43 - 3q) = 29 \\implies q + 129 - 9q = 29 \\implies -8q = -100 \\implies q = 12.5$\nNot integer.\n\nTry $r = 4$:\nFirst: $p + 4q = 43 \\implies p = 43 - 4q$\nSecond: $q + 4p = 29 \\implies q + 4(43 - 4q) = 29 \\implies q + 172 - 16q = 29 \\implies -15q = -143 \\implies q = 143/15 \\approx 9.53$\nNot integer.\n\nTry $r = 5$:\nFirst: $p + 5q = 43 \\implies p = 43 - 5q$\nSecond: $q + 5p = 29 \\implies q + 5(43 - 5q) = 29 \\implies q + 215 - 25q = 29 \\implies -24q = -186 \\implies q = 186/24 = 7.75$\nNot integer.\n\nTry $r = 6$:\nFirst: $p + 6q = 43 \\implies p = 43 - 6q$\nSecond: $q + 6p = 29 \\implies q + 6(43 - 6q) = 29 \\implies q + 258 - 36q = 29 \\implies -35q = -229 \\implies q = 229/35 \\approx 6.54$\nNot integer.\n\nTry $r = 7$:\nFirst: $p + 7q = 43 \\implies p = 43 - 7q$\nSecond: $q + 7p = 29 \\implies q + 7(43 - 7q) = 29 \\implies q + 301 - 49q = 29 \\implies -48q = -272 \\implies q = 272/48 = 5.67$\nNot integer.\n\nTry $r = 19$:\nFirst: $p + 19q = 43 \\implies p = 43 - 19q$\nSecond: $q + 19p = 29 \\implies q + 19(43 - 19q) = 29 \\implies q + 817 - 361q = 29 \\implies -360q = -788 \\implies q = 788/360 \\approx 2.19$\nNot integer.\n\nTry $r = 43$:\nFirst: $p + 43q = 43 \\implies p = 43 - 43q$\nSecond: $q + 43p = 29 \\implies q + 43(43 - 43q) = 29 \\implies q + 1849 - 1849q = 29 \\implies -1848q = -1820 \\implies q = 1820/1848 \\approx 0.985$\nNot integer.\n\nTry $r = 1$ again, but swap $p$ and $q$.\n\nAlternatively, try $p = 1$:\nFirst: $1 + r q = 43 \\implies r q = 42$\nSecond: $q + r = 29$\nSo $r = 29 - q$\nPlug into $r q = 42$:\n$(29 - q) q = 42 \\implies 29q - q^2 = 42 \\implies q^2 - 29q + 42 = 0$\nSolve quadratic:\n$q = \\frac{29 \\pm \\sqrt{29^2 - 4 \\cdot 42}}{2} = \\frac{29 \\pm \\sqrt{841 - 168}}{2} = \\frac{29 \\pm \\sqrt{673}}{2}$\nNot integer.\n\nTry $q = 1$:\nFirst: $p + r = 43$\nSecond: $1 + p r = 29 \\implies p r = 28$\nSo $p + r = 43$, $p r = 28$\nTry integer solutions:\n$p = 1$, $r = 42$; $p r = 42$ (not 28)\n$p = 2$, $r = 41$; $p r = 82$\n$p = 4$, $r = 39$; $p r = 156$\n$p = 7$, $r = 36$; $p r = 252$\n$p = 14$, $r = 29$; $p r = 406$\n$p = 28$, $r = 15$; $p r = 420$\nNo integer solution.\n\nTry $p = 2$:\nFirst: $2 + r q = 43 \\implies r q = 41$\nSecond: $q + 2 r = 29 \\implies q = 29 - 2 r$\nPlug into $r q = 41$:\n$r (29 - 2 r) = 41 \\implies 29 r - 2 r^2 = 41 \\implies 2 r^2 - 29 r + 41 = 0$\nQuadratic: $r = \\frac{29 \\pm \\sqrt{841 - 328}}{4} = \\frac{29 \\pm \\sqrt{513}}{4}$\nNot integer.\n\nTry $p = 3$:\nFirst: $3 + r q = 43 \\implies r q = 40$\nSecond: $q + 3 r = 29 \\implies q = 29 - 3 r$\nPlug into $r q = 40$:\n$r (29 - 3 r) = 40 \\implies 29 r - 3 r^2 = 40 \\implies 3 r^2 - 29 r + 40 = 0$\nQuadratic: $r = \\frac{29 \\pm \\sqrt{841 - 480}}{6} = \\frac{29 \\pm \\sqrt{361}}{6} = \\frac{29 \\pm 19}{6}$\nSo $r = \\frac{48}{6} = 8$, $r = \\frac{10}{6} = 5/3$\nSo $r = 8$\nThen $q = 29 - 3 \\times 8 = 29 - 24 = 5$\nThen $p = 3$\n\nNow, $r + p q = 8 + 3 \\times 5 = 8 + 15 = 23$\n\nSo possible values for $r + p q$ are $97$ and $23$.\n\nCheck for other integer solutions.\n\nTry $p = 5$:\nFirst: $5 + r q = 43 \\implies r q = 38$\nSecond: $q + 5 r = 29 \\implies q = 29 - 5 r$\nPlug into $r q = 38$:\n$r (29 - 5 r) = 38 \\implies 29 r - 5 r^2 = 38 \\implies 5 r^2 - 29 r + 38 = 0$\nQuadratic: $r = \\frac{29 \\pm \\sqrt{841 - 760}}{10} = \\frac{29 \\pm \\sqrt{81}}{10} = \\frac{29 \\pm 9}{10}$\nSo $r = \\frac{38}{10} = 3.8$, $r = \\frac{20}{10} = 2$\nSo $r = 2$\nThen $q = 29 - 5 \\times 2 = 29 - 10 = 19$\nThen $p = 5$\n\nAlready found above.\n\nTry $p = 8$:\nFirst: $8 + r q = 43 \\implies r q = 35$\nSecond: $q + 8 r = 29 \\implies q = 29 - 8 r$\nPlug into $r q = 35$:\n$r (29 - 8 r) = 35 \\implies 29 r - 8 r^2 = 35 \\implies 8 r^2 - 29 r + 35 = 0$\nQuadratic: $r = \\frac{29 \\pm \\sqrt{841 - 1120}}{16}$\n$841 - 1120 = -279$, so no real solution.\n\nTry $p = 10$:\nFirst: $10 + r q = 43 \\implies r q = 33$\nSecond: $q + 10 r = 29 \\implies q = 29 - 10 r$\nPlug into $r q = 33$:\n$r (29 - 10 r) = 33 \\implies 29 r - 10 r^2 = 33 \\implies 10 r^2 - 29 r + 33 = 0$\nQuadratic: $r = \\frac{29 \\pm \\sqrt{841 - 1320}}{20}$\n$841 - 1320 = -479$, no real solution.\n\nTry $p = 19$:\nFirst: $19 + r q = 43 \\implies r q = 24$\nSecond: $q + 19 r = 29 \\implies q = 29 - 19 r$\nPlug into $r q = 24$:\n$r (29 - 19 r) = 24 \\implies 29 r - 19 r^2 = 24 \\implies 19 r^2 - 29 r + 24 = 0$\nQuadratic: $r = \\frac{29 \\pm \\sqrt{841 - 1824}}{38}$\n$841 - 1824 = -983$, no real solution.\n\nTry $p = 23$:\nFirst: $23 + r q = 43 \\implies r q = 20$\nSecond: $q + 23 r = 29 \\implies q = 29 - 23 r$\nPlug into $r q = 20$:\n$r (29 - 23 r) = 20 \\implies 29 r - 23 r^2 = 20 \\implies 23 r^2 - 29 r + 20 = 0$\nQuadratic: $r = \\frac{29 \\pm \\sqrt{841 - 1840}}{46}$\n$841 - 1840 = -999$, no real solution.\n\nTry $p = 43$:\nFirst: $43 + r q = 43 \\implies r q = 0$\nSo $r = 0$ or $q = 0$, but must be at least $1$.\n\nThus, the only integer solutions are $(p, q, r) = (5, 19, 2)$ and $(3, 5, 8)$, giving $r + p q = 97$ and $23$.\n\nTherefore, the possible number of road links between $A$ and $C$ is $23$ or $97$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 28897,
    "subject": "Mathematics (Multi-modal)",
    "question": "Determine all positive integers $n$ such that\n$$\n2^n \\cdot (n!)^2 < (2n)!\n$$\n*(Ilko Brnetić)*",
    "options": [],
    "answer": "all positive integers n ≥ 2",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 28898,
    "subject": "Mathematics (Multi-modal)",
    "question": "Prove that any 2001-element subset of the set $\\{1, 2, 3, \\dots, 3000\\}$ contains three elements such that each two of them are relatively prime.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $S$ be a set of 2001 distinct positive integers from the set $\\{1, 2, 3, \\dots, 3000\\}$. Let us look at 500 sets\n$$\nK_j = \\{6j + i \\mid i = 1, 2, 3, 4, 5, 6\\}, \\quad j = 0, 1, 2, \\dots, 499.\n$$\nThere is a set with at least five elements of $S$ among them (since $4 \\cdot 500 < 2001$).\nIf three of these five elements are odd numbers, we found the required triple.\nOtherwise, the set $K_j$ contains three even and two odd numbers from $S$.\nThese two odd numbers are relatively prime since their difference is 2 or 4.\nThree even numbers from the same $K_j$ are three consecutive even numbers. Only one of them is divisible by 3 and at most one of them can be divisible by 5. Therefore we can choose an even number which is not divisible by 3 nor by 5.\nThat even number together with two odd numbers gives the required triple. Since the difference of the numbers from the set $K_j$ is less than 6, only one of them can be divisible by prime $p$, $p \\ge 7$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 28899,
    "subject": "Mathematics (Multi-modal)",
    "question": "All four intersections of the parabola $y = x^2 + px + q$ and lines $y = x$ and $y = 2x$ are located in the first quadrant. Observe the two parts of the parabola that are located between the lines. Prove that the difference of the lengths of their orthogonal projections on the $x$-axis is 1.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 28900,
    "subject": "Mathematics (Multi-modal)",
    "question": "Determine all positive integers $n$ such that $\\log_2(3^n + 7)$ is also a positive integer.",
    "options": [],
    "answer": "n = 2",
    "solution": "Let $k = \\log_2(3^n + 7)$, where $k$ is a positive integer. Then $3^n + 7 = 2^k$.\n\nSo $2^k - 3^n = 7$.\n\nWe seek all positive integers $n$ such that this equation has a solution in positive integers $k$.\n\nLet us try small values of $n$:\n\nFor $n = 1$: $2^k - 3 = 7 \\implies 2^k = 10$ (not a power of $2$).\n\nFor $n = 2$: $2^k - 9 = 7 \\implies 2^k = 16 \\implies k = 4$.\nSo $n = 2$ works.\n\nFor $n = 3$: $2^k - 27 = 7 \\implies 2^k = 34$ (not a power of $2$).\n\nFor $n = 4$: $2^k - 81 = 7 \\implies 2^k = 88$ (not a power of $2$).\n\nFor $n = 5$: $2^k - 243 = 7 \\implies 2^k = 250$ (not a power of $2$).\n\nFor $n = 6$: $2^k - 729 = 7 \\implies 2^k = 736$ (not a power of $2$).\n\nLet us check if there are any other solutions for larger $n$.\n\nSuppose $n \\geq 2$. Then $3^n$ grows much faster than $2^k$ for large $n$ unless $k$ is also large. But $2^k = 3^n + 7 > 3^n$, so $2^k > 3^n$.\n\nBut $2^k = 3^n + 7 < 2 \\cdot 3^n$ for large $n$ (since $7$ is negligible compared to $3^n$), so $2^k < 2 \\cdot 3^n$.\n\nThus, $3^n < 2^k < 2 \\cdot 3^n$.\n\nTake logarithms:\n\n$n \\log 3 < k \\log 2 < n \\log 3 + \\log 2$\n\nSo $k$ is approximately $n \\log_2 3$.\n\nBut $\\log_2 3 \\approx 1.58496$, so $k \\approx 1.58496 n$.\n\nBut $2^k - 3^n = 7$ is a very restrictive equation. For large $n$, $2^k$ and $3^n$ are both huge, and their difference is $7$.\n\nLet us check for $n = 0$ (but $n$ must be positive), so skip.\n\nLet us check for $n = 2$ (already found), $k = 4$.\n\nSuppose $k = 3$, $2^3 = 8$, $3^n = 1$ (no integer $n$).\n\nSuppose $k = 5$, $2^5 = 32$, $3^n = 25$ (not a power of $3$).\n\nSuppose $k = 6$, $2^6 = 64$, $3^n = 57$ (not a power of $3$).\n\nSuppose $k = 7$, $2^7 = 128$, $3^n = 121$ (not a power of $3$).\n\nSuppose $k = 8$, $2^8 = 256$, $3^n = 249$ (not a power of $3$).\n\nSuppose $k = 9$, $2^9 = 512$, $3^n = 505$ (not a power of $3$).\n\nSuppose $k = 10$, $2^{10} = 1024$, $3^n = 1017$ (not a power of $3$).\n\nSo, only $n = 2$ works.\n\nTherefore, the only positive integer $n$ such that $\\log_2(3^n + 7)$ is a positive integer is $n = 2$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 28901,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $n \\ge 2$ be a positive integer and $x_1, x_2, \\dots, x_n$ real numbers ($x_i \\ne 0$, for $i=1, 2, \\dots, n$) such that $x_1 + x_2 + \\dots + x_n = 0$. Prove that there exist distinct positive integers $i$ and $j$ ($i, j \\le n$) such that\n$$\n\\frac{1}{2} \\le \\left| \\frac{x_i}{x_j} \\right| \\le 2.\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "If there are two equal numbers between the given numbers, we can choose those two. So we assume that all the numbers are distinct and without loss of generality we assume that $x_1 > x_2 > \\dots > x_n$. Since the total sum of the numbers is zero there exists a positive integer $k$ ($1 < k < n$) such that $x_k > 0 > x_{k+1}$. We have:\n$$\n|x_1| + |x_2| + \\dots + |x_k| = |x_{k+1}| + |x_{k+2}| + \\dots + |x_n|.\n$$\nNow we assume that there do not exist $i$ and $j$ with the required property. For $i = 1, 2, \\dots, k-1$ we have:\n$$\n\\frac{|x_i|}{|x_{i+1}|} > 1 > \\frac{1}{2},\n$$\nand hence:\n$$\n\\frac{x_i}{x_{i+1}} > 2.\n$$\nIt follows that:\n$$\nx_i < \\frac{1}{2}x_{i-1} < \\frac{1}{2^2}x_{i-2} < \\dots < \\frac{1}{2^i}x_1, \\text{ for } i = 1, 2, \\dots, k-1.\n$$\nSimilarly, we prove that for negative numbers the following inequalities hold:\n$$\n|x_j| < \\frac{1}{2^{n-j}}|x_n|, \\text{ for } j = k, k+1, \\dots, n.\n$$\nFinally, we have:\n$$\n\\begin{aligned} |x_1| &< |x_1| + |x_2| + \\dots + |x_k| = |x_{k+1}| + |x_{k+2}| + \\dots + |x_n| \\\\ &< |x_n| \\left( 1 + \\frac{1}{2} + \\frac{1}{4} + \\dots + \\frac{1}{2^{n-k+1}} \\right) < 2|x_n|. \\end{aligned}\n$$\nAnalogously, we get:\n$$\n\\begin{aligned} |x_n| &< |x_k| + |x_{k+1}| + \\dots + |x_n| = |x_1| + |x_2| + \\dots + |x_k| \\\\ &< |x_1| \\left( 1 + \\frac{1}{2} + \\frac{1}{4} + \\dots + \\frac{1}{2^k} \\right) < 2|x_1|. \\end{aligned}\n$$\nTherefore $\\frac{1}{2} \\le \\left|\\frac{x_1}{x_n}\\right| \\le 2$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 28902,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all functions $f : \\mathbb{R} \\to \\mathbb{R}$ such that $f(1) \\ge 0$ and\n$$\nf(x) - f(y) \\ge (x - y)f(x - y),\n$$\nfor all real numbers $x$ and $y$.",
    "options": [],
    "answer": "f(x) = 0 for all real x",
    "solution": "If we plug $y \\to x - 1$ into the given inequality we get:\n$$\nf(x) - f(x-1) \\ge 1 \\cdot f(1) \\ge 0,\n$$\ni.e.\n$$\nf(x) \\ge f(x-1), \\quad \\text{for every } x \\in \\mathbb{R}. \\qquad (1)\n$$\n\nIf we plug $y \\to 0$ into the given inequality we get:\n$$\nf(x) - f(0) \\ge x f(x), \\qquad (2)\n$$\n\nand plugging $x \\to 0$, $y \\to x$ gives us:\n$$\nf(0) - f(x) \\ge -x f(-x). \\qquad (3)\n$$\n\nNow, summing (2) and (3) gives us:\n$$\n0 \\ge x f(x) - x f(-x),\n$$\ni.e. if we take $x > 0$\n$$\nf(-x) \\ge f(x), \\quad \\text{for every } x \\in \\mathbb{R}^{+}. \\qquad (4)\n$$\n\nIf we plug $x \\to 1$, $y \\to 0$ into the given inequality we get:\n$$\nf(1) - f(0) \\ge f(1),\n$$\ni.e.\n$$\nf(0) \\le 0. \\tag{5}\n$$\nNow we conclude that:\n$$\n0 \\stackrel{(5)}{\\ge} f(0) \\stackrel{(1)}{\\ge} f(-1) \\stackrel{(4)}{\\ge} f(1) \\ge 0,\n$$\nand hence $f(-1) = f(0) = f(1) = 0$.\n\nRepeated use of inequality (1) gives us:\n$$\nf(x) \\ge f(x-1) \\ge f(x-2) \\ge \\dots,\n$$\nso it follows that:\n$$\nf(x) \\ge f(x-k), \\quad \\text{for every } x \\in \\mathbb{R}, \\text{ for every } k \\in \\mathbb{N}. \\tag{6}\n$$\n\nPlugging $x \\to x-1$, $y \\to -1$ into the given inequality gives us:\n$$\nf(x-1) - f(-1) \\ge x f(x),\n$$\ni.e.\n$$\nf(x-1) \\ge x f(x), \\quad \\text{for every } x \\in \\mathbb{R}. \\tag{7}\n$$\n\nFrom (1) and (7) we conclude that:\n$$\nf(x) \\ge x f(x),\n$$\ni.e.\n$$\nf(x)(x-1) \\le 0.\n$$\n\nIt follows from the previous inequality that\n$$\nf(x) \\le 0, \\quad \\text{for every } x > 1 \\quad \\text{and} \\quad f(x) \\ge 0, \\quad \\text{for every } x < 1. \\tag{8}\n$$\nNow we assume that $x > 1$. Then there exists $y < 1$ such that $k = x - y \\in \\mathbb{N}$.\nTherefore:\n$$\n0 \\ge f(x) \\ge f(x-k) = f(y) \\ge 0\n$$\nso we conclude that $f(x) = 0$ for every $x > 1$. Similarly, if $x < 1$ then there exists $y > 1$ such that $k = y - x \\in \\mathbb{N}$ so:\n$$\n0 \\ge f(y) \\ge f(y-k) = f(x) \\ge 0\n$$\nand again $f(x) = 0$. Hence we conclude that the only possible solution is the function $f(x) = 0$. It is easy to check that the function $f(x) = 0$ really is a solution, i.e. that it satisfies the given conditions.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 28903,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $a_1, a_2, \\dots, a_n$ be positive real numbers such that $a_1 + a_2 + \\dots + a_n = 1$.\nProve that the following inequality holds\n$$\n\\frac{a_1^3}{a_1^2 + a_2 a_3} + \\frac{a_2^3}{a_2^2 + a_3 a_4} + \\dots + \\frac{a_{n-1}^3}{a_{n-1}^2 + a_n a_1} + \\frac{a_n^3}{a_n^2 + a_1 a_2} \\ge \\frac{1}{2}.\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "We have\n$$\n\\begin{aligned}\n\\frac{a_1^3}{a_1^2 + a_2 a_3} &= \\frac{a_1^3 + a_1 a_2 a_3 - a_1 a_2 a_3}{a_1^2 + a_2 a_3} = a_1 - a_1 a_2 a_3 \\cdot \\frac{1}{a_1^2 + a_2 a_3} \\\\\n&\\geqq \\overset{\\text{A-G}}{a_1 - a_1 a_2 a_3} \\cdot \\frac{1}{2a_1\\sqrt{a_2 a_3}} = a_1 - \\frac{1}{2}\\sqrt{a_2 a_3} \\overset{\\text{A-G}}{\\geqq} a_1 - \\frac{a_2 + a_3}{4}\n\\end{aligned}\n$$\nSumming those $n$ inequalities gives us:\n$$\n\\begin{aligned}\n& \\frac{a_1^3}{a_1^2 + a_2 a_3} + \\frac{a_2^3}{a_2^2 + a_3 a_4} + \\dots + \\frac{a_{n-1}^3}{a_{n-1}^2 + a_n a_1} + \\frac{a_n^3}{a_n^2 + a_1 a_2} \\\\\n& \\geqslant \\left(a_1 - \\frac{a_2+a_3}{4}\\right) + \\left(a_2 - \\frac{a_3+a_4}{4}\\right) + \\dots + \\left(a_{n-1} - \\frac{a_n+a_1}{4}\\right) + \\left(a_n - \\frac{a_1+a_2}{4}\\right) \\\\\n& = \\frac{a_1 + a_2 + \\dots + a_n}{2} = \\frac{1}{2}.\n\\end{aligned}\n$$",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 28904,
    "subject": "Mathematics (Multi-modal)",
    "question": "Depending on the positive integer $k$, determine the smallest real number $D_k$ such that\n$$\n(abc)^2 + (bcd)^2 + (cda)^2 + (dab)^2 \\le D_k\n$$\nfor all nonnegative real numbers $a, b, c, d$ such that $a^k + b^k + c^k + d^k = 4$.",
    "options": [],
    "answer": "D_1 = (4/3)^6; D_k = 4 for k >= 2",
    "solution": "The quadruple $(a, b, c, d) = (1, 1, 1, 1)$ satisfies the given condition for every positive integer $k$, which means that $D_k \\ge 4$ for every positive integer $k$. We will prove that $D_k = 4$ for every $k \\ge 2$. If $k \\ge 2$, the power mean inequality gives us:\n$$\n\\sqrt{\\frac{a^2 + b^2 + c^2 + d^2}{4}} \\le \\sqrt[k]{\\frac{a^k + b^k + c^k + d^k}{4}} = 1,\n$$\ni.e.\n$$\na^2 + b^2 + c^2 + d^2 \\le 4.\n$$\nBy introducing the substitution $x = a^2$, $y = b^2$, $z = c^2$, $w = d^2$ the previous inequality turns into\n$$\nx + y + z + w \\le 4,\n$$\nand the given inequality turns into\n$$\nxyz + xyw + xzw + yzw \\le 4. \\qquad (11)\n$$\nObviously, it is sufficient to prove the inequality:\n$$\n16(xyz + xyw + xzw + yzw) \\le (x + y + z + w)^3.\n$$\nWe have\n$$\n(x - y + z - w)^2 \\ge 0,\n$$\nwhich can be equivalently written as\n$$\n(x + y + z + w)^2 \\ge 4(xy + yz + zw + wx). \\qquad (12)\n$$\nWe also have\n$$\n(x + y + z + w)(xy + yz + zw + wx) \\ge 4(xyz + xyw + xzw + yzw). \\qquad (13)\n$$\nNamely, the last inequality is equivalent to\n$$\n(x^2 + z^2)(y + w) + (y^2 + w^2)(x + z) \\ge 2(xyz + xzw + xyw + yzw)\n$$\nwhich holds by the AM-GM inequality\n$$\nx^2 + z^2 \\ge 2xz \\quad \\text{and} \\quad y^2 + w^2 \\ge 2yw.\n$$\nMultiplying (12) and (13) gives us (11).\n\nIf $k = 1$ then $D_k > 4$. Namely, plugging the quadruple $(\\frac{4}{3}, \\frac{4}{3}, \\frac{4}{3}, 0)$ gives us $D_1 \\ge (\\frac{4}{3})^6 > 4$. We will prove that $D_1$ is precisely equal to $(\\frac{4}{3})^6$.\n\nLet us assume that $a, b, c, d$ are real numbers such that $a + b + c + d = 1$ and $a \\ge b \\ge c \\ge d \\ge 0$. We denote $a' = a, b' = b + \\frac{d}{2}, c' = c + \\frac{d}{2}, d' = 0$. Replacing $(a, b, c, d) \\rightarrow (a', b', c', d')$ the left hand side of the given inequality increases. Namely,\n$$\n\\begin{aligned}\n& (a'b'c')^2 + (a'b'd')^2 + (a'c'd')^2 + (b'c'd')^2 \\\\\n&= a^2 \\left(b + \\frac{d}{2}\\right)^2 \\left(c + \\frac{d}{2}\\right)^2 \\\\\n&= a^2 \\left(b^2 + bd + \\frac{d^2}{4}\\right) \\left(c^2 + cd + \\frac{d^2}{4}\\right) \\\\\n&\\ge a^2 \\cdot b^2 \\cdot c^2 + a^2 \\cdot b^2 \\cdot cd + a^2 \\cdot bd \\cdot c^2 + a^2 \\cdot bd \\cdot cd \\\\\n&\\ge (abc)^2 + (abd)^2 + (acd)^2 + (bcd)^2,\n\\end{aligned}\n$$\nwhere the last inequality holds because of the ordering of numbers $a, b, c, d$. This means that the left hand side of the inequality attains its maximum when one of the numbers is equal to zero, so without loss of generality we can assume that $d = 0$. Now it follows that:\n$$\n(abc)^2 + (abd)^2 + (acd)^2 + (bcd)^2 = (abc)^2 \\stackrel{\\text{A-G}}{\\le} \\left(\\frac{a+b+c}{3}\\right)^6 = \\left(\\frac{4}{3}\\right)^6,\n$$\nwhich means that $D_1 = \\left(\\frac{4}{3}\\right)^6$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 28905,
    "subject": "Mathematics (Multi-modal)",
    "question": "We are placing red and blue balls on the circle. In the beginning there are exactly two red balls on the circle. We are allowed to perform the following operations:\ni) add one red ball and change the colour of its neighbouring two balls (red to blue and vice versa);\nii) remove one red ball and change the colour of its neighbouring balls.\nIs it possible that after a finite number of operations we obtain\n\na) $2013$ red and $2013$ blue balls;\nb) exactly two blue balls? (Kvant 8/1981)",
    "options": [],
    "answer": "a) No; b) No",
    "solution": "a. We consider three different cases of adding a red ball depending on the colours of its neighbours.\nIf we add the ball between two blue balls, then we have the case $PP \\rightarrow CCC$ in which the number of blue balls decreases by two. If we add the ball between two red balls, we have $CC \\rightarrow PCP$ and the number of blue balls increases by two. In the third case, we add the ball between one blue and one red ball ($CP \\rightarrow PCC$) and the number of blue balls stays the same. We conclude that adding a red ball does not change the parity of the number of blue balls. Also removing a red ball does not change the parity of the number of blue balls.\n\nIn the beginning there is no blue balls, so after each operation the number of blue balls is even. Hence it is not possible that we obtain exactly $2013$ blue balls.\n\nb. In a given configuration of balls, we label the blue balls clockwise $P_1, P_2, \\dots, P_{2k}$ starting with an arbitrary blue ball. Let $m_i$ ($i \\in \\{1, 2, \\dots, 2k\\}$) be the number of red balls between balls $P_i$ and $P_{i+1}$ ($P_{n+1} = P_1$). For example, in the following picture we have $m_1 = 2, m_2 = 3, m_3 = 1, m_4 = 0, m_5 = 4, m_6 = 1$.\n\n![](attached_image_1.png)\n\nFurthermore, let us denote $S = m_1 - m_2 + m_3 - \\dots + m_{2k-1} - m_{2k}$. If the configuration consists of $n$ red balls and no blue balls we denote $S = n$. Let us prove that either $S$ is divisible by three for all configurations or $S$ is not divisible by three for all configurations. We again consider three cases:\n\n1. We add a red ball between two blue balls, $P_i$ and $P_{i+1}$:\n$$\nP_{i-1} \\underbrace{\\dots P_i}_{m_{i-1}} \\underbrace{\\dots P_{i+1}}_{m_i=0} \\rightarrow P_{i-1}' \\underbrace{\\dots \\overbrace{CC}^{\\downarrow} \\dots P_i'}_{m_i'}\n$$\n\n2. We add a red ball between two red balls:\n$$\nP_i \\underbrace{\\dots CC \\dots}_{m_i} P_{i+1} \\rightarrow P_i \\underbrace{\\dots P_{i+1}}_{m_i'} \\underbrace{\\overbrace{C}^{\\downarrow}}_{m_{i+1}'=1} P_{i+1} \\underbrace{\\dots P_{i+3}}_{m_{i+2}'}\n$$\nSimiliarly as before, $S = S_0 + m_i$, a $S' = S_0 + m_i' - 1 + m_{i+2}$. But now $m_i = m_i' + m_{i+2}' + 2$, so $S' = S - 3$.\n\n(Analogous statement holds in the case if there is no blue balls on the circle.)\n\n3. We add a red ball between one red and one blue ball (to the right of the blue ball $P_i$):\n$$\nP_{i-1} \\underbrace{\\overbrace{P_i}^{m_{i-1}} \\overbrace{C \\dots P_{i+1}}^{m_i}} \\rightarrow P_{i-1} \\underbrace{\\overbrace{\\dots \\overbrace{C}^{m'_{i-1}} C}^{\\downarrow} P_i}_{m'_i} \\dots P_i\n$$\nIn this case $S = S_0 + m_{i-1} - m_i$ and $S' = S_0 + m'_{i-1} - m'_i$. Since $m'_{i-1} = m_{i-1} + 2$ and $m'_i = m_i - 1$, we have $S' = S + 3$.\n\nIn any case, $S$ either decreases or increases by $3$. In the beginning number $S$ is $2$ and hence $S$ can never be divisible by three. We cannot obtain exactly two blue balls since $S$ would be $0$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 28906,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $m$, $n$ and $k$ be positive integers and let $p_1, p_2, \\dots, p_n$ be the integers $1, 2, \\dots, n$ given in some order. If\n$$\nk \\mid (m + p_i - i),\n$$\nholds for all $i \\in \\{1, 2, \\dots, n\\}$, prove that one of the numbers $m$ and $n$ is divisible by $k$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let us assume that $k$ does not divide $n$ and that $n = kq + r$, $0 < r < k$.\nSince only the remainder of division of $m$ by $k$ is relevant, without loss of generality we may assume $0 < m \\le k$. We will prove that $m = k$.\nIf we assume that $m \\le r$, then the numbers $p_m, p_{m+k}, \\dots, p_{m+kw}$ would be equal to the numbers $k, 2k, \\dots, qk$ in some order. This is impossible since the first sequence has $q+1$ elements, and the second sequence has $q$ elements.\nHence $m > r$ and $m + 1 + qk > n$. There are $q+1$ numbers $1, k+1, \\dots, qk+1$ which are equal to numbers $p_{m+1-k}, p_{m+1}, p_{m+1+k}, \\dots, p_{m+1+(q-1)k}$. This implies $m+1-k \\ge 1$, i.e. $m \\ge k$. Since we assumed that $m \\le k$ it follows that $m = k$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 28907,
    "subject": "Mathematics (Multi-modal)",
    "question": "A group of people of different height decided to dance the Hungarian traditional dance at the opening ceremony of MEMO 2013 in Veszprém. We say that a person is *average* if he is taller than one of her neighbours and shorter than the other. (People stand in a circle and every person has exactly two neighbours.)\nIf the number of people is $N$ ($N \\ge 3$), determine all possible numbers of average people. (Belarus 2012)",
    "options": [],
    "answer": "All integers strictly less than N and having the same parity as N",
    "solution": "We first observe how the height of people changes as we go around the circle in the clockwise direction. For every pair of neighbours $A$ and $B$ (where $B$ is after $A$ in the clockwise direction) we put the symbol ♣ between them if $B$ is taller than $A$ and the symbol ♠ if $B$ is shorter than $A$. In that way we obtain a sequence of $N$ symbols ♣ or ♠.\nA person is average if and only if the symbol before that person is equal to the symbol after that person. Hence the number of people who are not average is equal to the number of alternations of the symbols ♣ and ♠. The number of changes from the symbol ♣ to ♠ is equal to the number of changes from ♠ to ♣. So the number of people who are not average is even.\nFrom this we conclude that the number of average people is of the same parity as $N$. Since the tallest person in the circle is not average the number of average people is certainly smaller than $N$.\nLet $a_1 < a_2 < \\dots < a_N$ denote the heights of people. We will exhibit an example showing that for all $k = 1, \\dots, \\lfloor N/2 \\rfloor$ there exists a configuration of $N$ people in a circle such that exactly $N - 2k$ is average. More precisely, for a configuration where the people are ordered in a circle so that their heights are respectively\n$$\n\\underbrace{a_1, a_N, a_2, a_{N-1}, \\dots, a_k, a_{N-k+1}}_{\\text{A}}, \\underbrace{a_{N-k}, \\dots, a_{k+1}}_{\\text{G}}\n$$\nthe first $2k$ people are not average and the remaining $N - 2k$ are average.\nThis shows that the number of average people can be any number strictly less than $N$ and having the same parity as $N$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 28908,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $N$ and $K$ be positive integers. A number of students is first divided into $N$ non-empty groups and then the same students are divided into $N + K$ non-empty groups. Prove that in the second distribution at least $K+1$ students are in a smaller group than in the first distribution. (Yugoslavia 1981)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $S$ be the set of all students. For $s \\in S$ we denote by $a_s$ the number of students in the group of the student $s$ in the first distribution and by $b_s$ the number of students in the group of student $s$ in the second distribution.\nWe have\n$$\n\\sum_{s \\in S} \\frac{1}{a_s} = N, \\quad \\sum_{s \\in S} \\frac{1}{b_s} = N + K.\n$$\nHence the sum of differences of numbers assigned to each student is equal to\n$$\n\\sum_{s \\in S} \\left( \\frac{1}{b_s} - \\frac{1}{a_s} \\right) = K.\n$$\nSince $\\left| \\frac{1}{b_s} - \\frac{1}{a_s} \\right| < 1$ it follows that for at least $K + 1$ students we have $\\frac{1}{b_s} - \\frac{1}{a_s} > 0$, i.e. $b_s < a_s$.",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 28909,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABC$ be an acute triangle with the orthocentre $H$. Let $D$ be such a point that the quadrilateral $AHCD$ is a parallelogram. Let $p$ be a line perpendicular to the line $AB$ passing through the midpoint $A_1$ of the segment $BC$. Let us denote the intersection of $p$ and $AB$ by $E$ and the midpoint of the segment $A_1E$ by $F$. We denote the point in which the line parallel with the line $BD$ through $A$ intersects $p$ with $G$.\nProve that the quadrilateral $AFA_1C$ is cyclic if and only if the line $BF$ passes through the midpoint of the segment $CG$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Since $AHCD$ is a parallelogram we have $\\angle ADC = \\angle CHA = 180^\\circ - \\beta$, which implies that $D$ lies on the circumference of the triangle $ABC$. Also, $\\angle ACD = \\angle HAC = 90^\\circ - \\gamma$.\nSince $ABCD$ is cyclic we have $\\angle ABD = \\angle ACD$, and since lines $AG$ and $BD$ are parallel we have $\\angle GAB = \\angle ABD$.\nHence, $\\angle BAG = 90^\\circ - \\gamma$.\n\n![](attached_image_1.png)\n\nQuadrilateral $AFA_1C$ is cyclic if and only if $\\angle AFE = \\angle ACA_1$, i.e. $\\angle AFE = \\gamma$. This is equivalent to $\\angle FAE = 90^\\circ - \\gamma$.\nSince $\\angle GAE = 90^\\circ - \\gamma$, the above condition holds if and only if the triangles $AEF$ and $AEG$ are congruent, which is equivalent to $|EG| = |EF|$.\nSince $F$ is the midpoint of the segment $\\overline{A_1E}$, the given condition holds if and only if $F$ divides the segment $\\overline{GA_1}$ in the ratio $2 : 1$, i.e. if and only if $|FG| = 2|FA_1|$. Since $\\overline{GA_1}$ is a median in the triangle $BCG$, this holds if and only if $F$ is the centroid of that triangle.\nThis obviously holds if and only if the line $BF$ passes through the midpoint of the segment $\\overline{CG}$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 28910,
    "subject": "Mathematics (Multi-modal)",
    "question": "The angle at the vertex $B$ in a triangle $ABC$ is $120^\\circ$. Let $A_1$, $B_1$ and $C_1$ be the points on the segments $\\overline{BC}$, $\\overline{CA}$ and $\\overline{AB}$ respectively, such that $AA_1$, $BB_1$ and $CC_1$ are angle bisectors of the triangle $ABC$. Determine the angle $\\angle A_1B_1C_1$. (Serbia 1997)",
    "options": [],
    "answer": "90°",
    "solution": "Let $X$ be any point on the extension of the segment $\\overline{AB}$ over the vertex $B$. Note that $\\angle ABB_1 = \\angle B_1BC = \\angle CBX = 60^\\circ$.\n\n![](attached_image_1.png)\n\nThis implies that the point $A_1$ lies on the angle bisector of the angle $\\angle B_1BX$, and it also lies on the angle bisector of the angle $\\angle BAC$.\nFrom this we conclude that $A_1$ is the centre of excircle of the triangle $ABB_1$ opposite to vertex $A$ and hence it lies on the angle bisector of the angle $\\angle BB_1C$.\nSo the line $B_1A_1$ is the angle bisector of the angle $\\angle BB_1C$. Analogously we prove that the line $B_1C_1$ is the angle bisector of the angle $\\angle AB_1B$.\nHence\n$$\n\\angle A_1B_1C_1 = \\angle A_1B_1B + \\angle BB_1C_1 = \\frac{1}{2}\\angle CB_1B + \\frac{1}{2}\\angle BB_1A = \\frac{1}{2} \\cdot 180^\\circ = 90^\\circ.\n$$",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 28911,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $N$ be the foot of the altitude from $C$ to the hypotenuse $\\overline{AB}$ of a right triangle $ABC$. The angle bisectors of the angles $\\angle NCA$ and $\\angle BCN$ intersect the segment $\\overline{AB}$ in the points $K$ and $L$ respectively. If $S$ and $T$ are incircles of the triangles $BCN$ and $NCA$ respectively, prove that the quadrilateral $KLST$ is cyclic. (from an article of A. Marić)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Denote $\\triangle CAB = \\alpha$ and $\\triangle ABC = \\beta$. We have $\\triangle BCN = \\alpha$ and $\\triangle ACN = \\beta$.\n\n![](attached_image_1.png)\n\nThe lines $CS$ and $CT$ are angle bisectors of the angles $\\triangle BCN$ and $\\triangle ACN$, so the points $C, S, L$ and $C, T, K$ are collinear. We have $\\triangle BCK = \\triangle BCN + \\triangle NCK = \\alpha + \\frac{1}{2}\\beta$. Also, $\\triangle BKC = \\triangle CAK + \\triangle ACK = \\alpha + \\frac{1}{2}\\beta$.\n\nHence $\\triangle BCK = \\triangle BKC$, so the triangle $BCK$ is isosceles and we have $|BK| = |BC|$. This implies that the point $S$ lies on the bisector of the segment $\\overline{CK}$, so $|SC| = |SK|$. From this we conclude $\\triangle SKT = \\triangle SCT$.\n\nAnalogously, the triangle $ACL$ is isosceles and the point $T$ lies on the bisector of the segment $\\overline{CL}$, so $\\triangle SLT = \\triangle SCT$.\n\nWe have shown $\\triangle SKT = \\triangle SLT$ which means that the points $K, L, S$ and $T$ are concyclic.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 28912,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABC$ be an isosceles triangle with a base $\\overline{AB}$. A point $P$ is chosen on the segment $\\overline{AC}$ and a point $Q$ is chosen on the segment $\\overline{BC}$ such that $|AP| + |BQ| = |PQ|$. The line parallel with the line $BC$ which passes through the midpoint of the segment $\\overline{PQ}$ intersects the segment $\\overline{AB}$ in the point $N$. Circumcircle of the triangle $PNQ$ intersects the line $AC$ in the points $P$ and $K$, and the line $BC$ in the points $Q$ and $L$. If the point $R$ is the intersection of the lines $PL$ and $QK$, prove that the line $PQ$ is perpendicular to the line $CR$. (Stipe Vidak)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $S$ be the midpoint of the segment $\\overline{PQ}$ and let $M$ be the point on the segment $\\overline{AB}$ such that $MP \\parallel BC$.\n\n![](attached_image_1.png)\n\nWe have $\\angle PMA = \\angle CBA = \\angle CAB = \\angle PAM$, so the triangle $PAM$ is isosceles and $|PA| = |PM|$. The quadrilateral $PMBQ$ is a trapezium with the midline $\\overline{SN}$, so we have\n$$\n|SN| = \\frac{|PM| + |QB|}{2} = \\frac{|AP| + |QB|}{2} = \\frac{|PQ|}{2}.\n$$\nHence $S$ is the circumcentre of the triangle $PQN$ and the segment $\\overline{PQ}$ is its diameter. Thales' theorem implies $QK \\perp CP$ and $PL \\perp CQ$, so the point $R$ is the orthocentre of the triangle $CPQ$. From this we conclude that $CR \\perp PQ$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 28913,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all positive integers $a$ and $b$ such that\n$$\n(a^2 + b) \\mid (a^2b + a) \\quad \\text{and} \\quad (b^2 - a) \\mid (ab^2 + b).\n$$",
    "options": [],
    "answer": "All pairs (n, n+1) with n a positive integer",
    "solution": "From the fact that $(a^2 + b) \\mid (a^2b + a)$ we conclude that\n$$\n\\frac{a^2b + a}{a^2 + b} = \\frac{b(a^2 + b) + a - b^2}{a^2 + b} = b - \\frac{b^2 - a}{a^2 + b}\n$$\nis an integer.\nHence $a^2 + b \\le b^2 - a$ or $b^2 - a \\le 0$. In the latter case we would have $a < a^2 + b \\le a - b^2 < a$, which is not possible. So, $a^2 + b \\le b^2 - a$.\nSimilarly, from the fact that $(b^2 - a) \\mid (ab^2 + b)$ we conclude that also this is an integer:\n$$\n\\frac{ab^2 + b}{b^2 - a} = \\frac{a(b^2 - a) + a^2 + b}{b^2 - a} = a + \\frac{a^2 + b}{b^2 - a}.\n$$\nIt follows that $b^2 - a \\le a^2 + b$, so $b^2 - a = a^2 + b$. This can be written as $(b+a)(b-a) = a+b$ and dividing by $a+b$ we obtain $b = a+1$. We directly check that the numbers of this form satisfy the conditions of the problem. Hence, all solutions are pairs $(n, n+1)$ for a positive integer $n$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 28914,
    "subject": "Mathematics (Multi-modal)",
    "question": "Prove that there are infinitely many positive integers $n$ such that $2^n - 8$ is divisible by $n$. (Kristina Ana Škreb)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "We will prove that for all positive integers of the form $n = 3p$ (where $p > 3$ is a prime number)\n$$\nn \\mid 2^n - 8.\n$$\nBy Fermat's little theorem we have $2^p \\equiv 2 \\pmod{p}$, so it follows that\n$$\n2^{3p} - 8 = (2^p)^3 - 8 \\equiv 2^3 - 8 = 0 \\pmod{p}. \\quad (9)\n$$\nAnalogously, since $3p$ is an odd number, we have\n$$\n2^{3p} - 8 \\equiv (-1)^{3p} - 2 \\equiv -3 \\equiv 0 \\pmod{3}. \\quad (10)\n$$\nSince $3$ and $p$ are relatively prime, from (9) and (10) we conclude that\n$$\n2^{3p} - 8 \\equiv 0 \\pmod{3p}.\n$$\nThere are infinitely many prime numbers greater than $3$, hence the assertion follows.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 28915,
    "subject": "Mathematics (Multi-modal)",
    "question": "Prove that there are infinitely many positive integers $n$ such that $n$ has more than two distinct prime divisors and $2^n - 8$ is divisible by $n$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "We will prove that for all positive integers of the form $n = 2^{2p} - 1$ (where $p > 3$ is a prime number) $n \\mid 2^n - 8$ and $n$ has at least three distinct prime divisors.\n\nSince $2^n - 8 = 8(2^{n-3} - 1)$, to prove $n \\mid 2^n - 8$ it is sufficient to prove\n$$\n2^p \\mid n - 3.\n$$\nNamely, in this case $n - 3 = 2pk$ for some positive integer $k$, so\n$$\n2^{n-3} - 1 = 2^{2pk} - 1 = (2^{2p})^k - 1 = (2^{2p} - 1)(2^{2p(k-1)} + 2^{2p(k-2)} + \\dots + 2^{2p} + 1),\n$$\ni.e. $2^{2p} - 1 = n \\mid 2^{n-3} - 1 \\Rightarrow n \\mid 2^n - 8$.\n\nBy Fermat's little theorem we have $2^p \\equiv 2 \\pmod p$, so it follows that\n$$\nn - 3 = 2^{2p} - 4 = (2^p)^2 - 4 \\equiv 2^2 - 4 = 0 \\pmod p.\n$$\nObviously $n-3$ is an even number and since $2$ and $p$ are relatively prime we conclude that\n$$\nn - 3 = 2^{2p} - 4 \\equiv 0 \\pmod{2p}.\n$$\n\nNow we will show that $n$ has at least three distinct prime divisors. Note that\n$$\nn = 2^{2p} - 1 = (2^p - 1)(2^p + 1).\n$$\nSince $2^p - 1$ and $2^p + 1$ are two consecutive odd numbers they are relatively prime, so they don't have any common prime divisor. $2^p - 1$ has at least one prime divisor so it is sufficient to show that $2^p + 1$ has at least two distinct prime divisors. Since $p > 3$ we can write $p = 3k + r$, where $k \\ge 1$ and $r \\in \\{1, 2\\}$.\n\nSince $p$ is odd, we have:\n$$\n2^p + 1 \\equiv (-1)^p + 1 = -1 + 1 = 0 \\pmod{3}\n$$\nand\n$$\n2^p + 1 = (2^3)^k \\cdot 2^r + 1 \\equiv (-1)^k 2^r + 1 \\neq 0 \\pmod{9}.\n$$\nWe have proved that $2^p + 1$ is divisible by $3$, but it's not divisible by $9$ which means that $2^p + 1$ can not be a power of $3$, i.e. $2^p + 1$ has at least one prime divisor different from $3$.\n\nThis proves the assertion of the problem.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 28916,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find the least positive integer $a$ such that the expression\n$$\n\\frac{n^8 + n^6 + n^4 + n^2 + a}{n^2 - 1}\n$$\nfor $n = 2014$ is an integer divisible by $3$. (Mea Bombardelli)",
    "options": [],
    "answer": "8112386",
    "solution": "**Arranging the expression we get**\n$$\n\\begin{aligned}\n\\frac{n^8 + n^6 + n^4 + n^2 + a}{n^2 - 1} &= \\frac{n^8 - n^6 + 2(n^6 - n^4) + 3(n^4 - n^2) + 4(n^2 - 1) + a + 4}{n^2 - 1} \\\\\n&= n^6 + 2 \\cdot n^4 + 3 \\cdot n^2 + 4 + \\frac{a+4}{n^2-1}.\n\\end{aligned}\n$$\nSince $2014 \\equiv 1 \\pmod{3}$, every power of $2014$ is congruent to $1$ modulo $3$.\nTherefore, $2014^6 + 2 \\cdot 2014^4 + 3 \\cdot 2014^2 + 4 \\equiv 1 + 2 + 3 + 4 \\equiv 1 \\pmod{3}$.\nIn order for the entire expression to be divisible by $3$, the number $\\frac{a+4}{2014^2-1}$ has to be congruent to $2$ modulo $3$, so the least positive integer $a$ with the desired property is the one that satisfies $\\frac{a+4}{2014^2-1} = 2$, which gives $a = 2 \\cdot 2014^2 - 6$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 28917,
    "subject": "Mathematics (Multi-modal)",
    "question": "Prove that $2012^9 + 2016^9$ is divisible by $2014$. (Nikola Adžaga)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 28918,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $a \\le b \\le c$ be real numbers. Prove that $c^2 - b^2 + a^2 \\ge (c - b + a)^2$. (Santos J. Prob. Seminar)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "We need to prove that\n$$\nc^2 - b^2 + a^2 \\ge (c - b + a)^2.\n$$\n\nExpand the right-hand side:\n$$(c - b + a)^2 = (c + a - b)^2 = (c + a)^2 - 2b(c + a) + b^2 = c^2 + 2ac + a^2 - 2bc - 2ab + b^2.$$\n\nSo the inequality becomes:\n$$\nc^2 - b^2 + a^2 \\ge c^2 + 2ac + a^2 - 2bc - 2ab + b^2.\n$$\n\nBring all terms to one side:\n$$\nc^2 - b^2 + a^2 - [c^2 + 2ac + a^2 - 2bc - 2ab + b^2] \\ge 0.\n$$\n\nSimplify:\n$$\nc^2 - b^2 + a^2 - c^2 - 2ac - a^2 + 2bc + 2ab - b^2 \\ge 0\n$$\n$$\n(- b^2 - 2ac + 2bc + 2ab - b^2) \\ge 0\n$$\n$$\n(-2b^2 - 2ac + 2bc + 2ab) \\ge 0\n$$\n$$\n2(-b^2 - ac + bc + ab) \\ge 0\n$$\n$$\n-b^2 - ac + bc + ab \\ge 0\n$$\n$$\n(bc + ab - b^2 - ac) \\ge 0\n$$\n$$\nb(c + a - b) - ac \\ge 0\n$$\n$$\nb(c + a - b) \\ge ac\n$$\n\nSince $a \\le b \\le c$, $c + a - b \\ge a$ (because $c \\ge b$), and $b \\ge a$, so $b(c + a - b) \\ge a b$. Also, $ac \\le b c$ since $a \\le b$. So $b(c + a - b) \\ge ac$.\n\nTherefore, the inequality holds for all real numbers $a \\le b \\le c$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 28919,
    "subject": "Mathematics (Multi-modal)",
    "question": "On the playground there are $2014$ athletes with the numbers from $1$ to $2014$ on their shirts (each number is on exactly one shirt). At the beginning they are all standing. In certain time intervals the coach shouts out all positive integers from $1$ to $2014$ in the increasing order. All athletes having a multiple of the shouted number on their shirts change their position from the standing position to crouch and vice versa.\nHow many athletes are crouching after the coach shouts out the number $2014$?",
    "options": [],
    "answer": "44",
    "solution": "Every athlete will change his position as many times as the number on his shirt has divisors. Hence, at the end in the position of crouch will be those athletes whose shirt numbers have an odd number of divisors.\nAll divisors of the number $n$ can be grouped into two element sets $\\{d, \\frac{n}{d}\\}$, unless $n = d^2$ for some positive integer $d$ when $\\frac{n}{d} = d$. That means that a positive integer has an odd number of divisors if and only if it is a square of an integer.\nAmong numbers $1, \\ldots, 2014$ the squares are $1^2, 2^2, \\ldots, 44^2$ (because $44^2 = 1936$ and $45^2 = 2025$). At the end, there are $44$ athletes crouching.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 28920,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all integer solutions of the equation $x^4 + 68 = 4y^4$. (Tomislav Pejković)",
    "options": [],
    "answer": "(x, y) = (4, 3), (4, -3), (-4, 3), (-4, -3)",
    "solution": "Let us rewrite the equation:\n$$x^4 + 68 = 4y^4$$\n$$x^4 - 4y^4 = -68$$\n$$ (x^2 - 2y^2)(x^2 + 2y^2) = -68 $$\n\nNow, $-68$ factors as $(-1) \\times 68$, $(-2) \\times 34$, $(-4) \\times 17$, $(-17) \\times 4$, $(-34) \\times 2$, $(-68) \\times 1$ and their negatives. We consider all pairs $(a, b)$ such that $a \\times b = -68$.\n\nLet $x^2 - 2y^2 = a$, $x^2 + 2y^2 = b$, so $a + b = 2x^2$, $b - a = 4y^2$.\n\nThus,\n$$ x^2 = \\frac{a + b}{2} $$\n$$ y^2 = \\frac{b - a}{4} $$\n\nWe need $x^2$ and $y^2$ to be integers, so $a + b$ is even and $b - a$ is divisible by $4$.\n\nLet us check all possible pairs $(a, b)$:\n\n1. $a = -1$, $b = 68$\n   - $a + b = 67$ (odd), $b - a = 69$ (not divisible by $4$)\n2. $a = -2$, $b = 34$\n   - $a + b = 32$ (even), $b - a = 36$ (divisible by $4$)\n   - $x^2 = 16$, $y^2 = 9$\n   - $x = \\pm 4$, $y = \\pm 3$\n3. $a = -4$, $b = 17$\n   - $a + b = 13$ (odd), $b - a = 21$ (not divisible by $4$)\n4. $a = -17$, $b = 4$\n   - $a + b = -13$ (odd), $b - a = 21$ (not divisible by $4$)\n5. $a = -34$, $b = 2$\n   - $a + b = -32$ (even), $b - a = 36$ (divisible by $4$)\n   - $x^2 = -16$ (not a square), $y^2 = 9$\n6. $a = -68$, $b = 1$\n   - $a + b = -67$ (odd), $b - a = 69$ (not divisible by $4$)\n\nNow, the negatives:\n7. $a = 1$, $b = -68$\n   - $a + b = -67$ (odd), $b - a = -69$ (not divisible by $4$)\n8. $a = 2$, $b = -34$\n   - $a + b = -32$ (even), $b - a = -36$ (divisible by $4$)\n   - $x^2 = -16$ (not a square), $y^2 = -9$ (not a square)\n9. $a = 4$, $b = -17$\n   - $a + b = -13$ (odd), $b - a = -21$ (not divisible by $4$)\n10. $a = 17$, $b = -4$\n    - $a + b = 13$ (odd), $b - a = -21$ (not divisible by $4$)\n11. $a = 34$, $b = -2$\n    - $a + b = 32$ (even), $b - a = -36$ (divisible by $4$)\n    - $x^2 = 16$, $y^2 = -9$ (not a square)\n12. $a = 68$, $b = -1$\n    - $a + b = 67$ (odd), $b - a = -69$ (not divisible by $4$)\n\nSo, the only valid case is $a = -2$, $b = 34$:\n$$ x^2 = 16 \\implies x = \\pm 4 $$\n$$ y^2 = 9 \\implies y = \\pm 3 $$\n\nCheck in the original equation:\n$$ x^4 + 68 = 4y^4 $$\n$$ 256 + 68 = 4 \\times 81 $$\n$$ 324 = 324 $$\n\nThus, all integer solutions are:\n$$ (x, y) = (4, 3), (4, -3), (-4, 3), (-4, -3) $$",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 28921,
    "subject": "Mathematics (Multi-modal)",
    "question": "The segment $\\overline{AB}$ is a diameter of a circle with the centre $O$. On the circle the point $C$ is given such that $OC$ is perpendicular to $AB$. Let $P$ be a point on the shorter arc $\\widearc{BC}$. The lines $CP$ and $AB$ intersect at the point $Q$, and the point $R$ is the intersection of the line $AP$ and the line through $Q$ perpendicular to the line $AB$.\nProve that $|BQ| = |QR|$. (Macedonia 2013)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "The triangle $OCB$ is isosceles right triangle because $\\overline{OB}$ and $\\overline{OC}$ are both radii of the circle with the centre $O$. Hence $\\angle CBA = \\angle CBO = 45^\\circ$.\n\n![](attached_image_1.png)\n\nInscribed angles $\\angle CPA$ and $\\angle CBA$ over the chord $\\overline{CA}$ are equal, so we have\n$$\n\\angle QPR = \\angle CPA = \\angle CBA = 45^\\circ.\n$$\n\nBy Thales' Theorem the angle $\\angle APB$ is right, so the angle $\\angle BPR$ must be right as well. The quadrilateral $BQRP$ is cyclic because it has two opposite right angles ($\\angle RQB$ and $\\angle BPR$), so the inscribed angles over the chord $QR$ are equal, which gives $\\angle QBR = \\angle QPR = 45^\\circ$.\nHence $\\angle BRQ = 180^\\circ - \\angle RQB - \\angle QBR = 45^\\circ$, so $\\angle BRQ = 45^\\circ = \\angle QBR$ and we get $|BQ| = |QR|$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 28922,
    "subject": "Mathematics (Multi-modal)",
    "question": "Do there exist real numbers $x, y, z$ such that\n$$\n\\frac{x}{y} + \\frac{y}{z} + \\frac{z}{x} = -5 \\quad \\text{and} \\quad \\frac{y^2}{x^2} + \\frac{z^2}{y^2} + \\frac{x^2}{z^2} = 8? \\quad (\\text{Ilko Brneti\\'c})\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 28923,
    "subject": "Mathematics (Multi-modal)",
    "question": "A box contains $k$ balls marked by $\\binom{k}{1}$ for all $k = 1, 2, \\dots, 50$. The balls are drawn from the box without looking. What is the minimal number of balls that need to be drawn to be sure that at least 10 balls with the same mark have been drawn? (AHSME 1994)",
    "options": [],
    "answer": "415",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 28924,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABCD$ be a parallelogram with acute angle at $A$. Let $E$ be the foot of the perpendicular from $C$ to the line $AB$, and let $F$ be the foot of the perpendicular from $C$ to the line $AD$.\nProve that\n$$\n|AB| \\cdot |AE| + |AD| \\cdot |AF| = |AC|^2. \\quad (\\text{USSR})\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 28925,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $x_1, x_2, \\dots, x_{100}$ be real numbers such that\n$$\n\\begin{align*} \n|2x_k - x_{k+1}| &= x_{k+2} \\quad \\text{for all } k \\in \\{1, 2, \\dots, 98\\}, \\\\ \n|2x_{99} - x_{100}| &= x_1, \\\\ \n|2x_{100} - x_1| &= x_2. \n\\end{align*}\n$$\nProve that $x_1 = x_2 = \\dots = x_{100}$.\n(Tonći Kokan)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Squaring the given equations we get\n$$\n\\begin{gathered}\n4x_k^2 - 4x_kx_{k+1} + x_{k+1}^2 = x_{k+2}^2 \\quad \\text{for } k \\in \\{1, 2, \\dots, 98\\}, \\\\\n4x_{99}^2 - 4x_{99}x_{100} + x_{100}^2 = x_1^2, \\quad 4x_{100}^2 - 4x_{100}x_1 + x_1^2 = x_2^2.\n\\end{gathered}\n$$\nAdding these equations we get\n$$\n4(x_1^2 + x_2^2 + \\dots + x_{100}^2) - 4(x_1x_2 + \\dots + x_{99}x_{100} + x_{100}x_1) = 0.\n$$\nAfter regrouping and dividing by 2 we get\n$$\n(x_1 - x_2)^2 + \\dots + (x_{99} - x_{100})^2 + (x_{100} - x_1)^2 = 0.\n$$\nSince all the terms are non-negative, they all have to be equal to zero, so we have $x_1 = \\dots = x_{100}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 28926,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $a$ and $b$ be different real numbers and define $s = a - b$ and $t = a^3 - b^3$. Express $(a+b)^2$ in terms of $s$ and $t$. (Santos J. Prob. Seminar)",
    "options": [],
    "answer": "(4t/s - s^2)/3",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 28927,
    "subject": "Mathematics (Multi-modal)",
    "question": "Andrija and Boris have 2014 cards labelled with numbers from 1 to 2014. Andrija has all the cards with even, and Boris all the cards with odd numbers. Andrija arranged his cards in a circle clockwise, from 2 to 2014 respectively, and put them upside down, so that the numbers on the cards can not be seen. Boris knows that the cards are arranged in that order and that direction, but does not know the position of the card with number 2. After that, Boris puts one of his cards on top of each Andrija's card thus forming 1007 pairs of cards.\n\nFor each pair they compare the numbers on the cards and give one point to the player whose card has the larger number.\nFind the largest possible $N$ such that Boris can be certain he will gain at least $N$ points. (Ukraine 2010)",
    "options": [],
    "answer": "503",
    "solution": "The largest number of points for which Boris can be certain to gain is $503$.\n\nWe claim that if Boris arranges his cards clockwise, but in the reverse order, from $2013$ to $1$, then regardless of Andrija's arrangement of cards, Boris will gain $503$ points.\n\n![](attached_image_1.png)\n\nFor such Boris' arrangement, let us consider some Andrija's arrangement. Under Boris' card $2013$ is Andrija's card $2k$, for some $k \\in \\mathbb{N}$. Let us assume that $2k \\neq 2014$, the case in which $2k = 2014$ follows directly. In the figure we can see the arrangement of the cards in a series, starting with the position where Boris put the card $2013$. We split pairs of cards into two parts: the part containing Andrija's cards $2k$, $2k+2$, ..., $2012$, $2014$, and the other part containing $2$, $4$, ..., $2k-4$, $2k-2$. In the first part Andrija's cards (from left to right) increase from $2k$ to $2014$, and Boris' decrease from $2013$. In other words, at the beginning of that part Boris' cards are larger, but at the end Andrija's cards are larger. Let us denote by $2a+1$ ($a \\in \\mathbb{N}$) the last Boris' card in the first part which is larger than the corresponding Andrija's card, which we denote by $2b$ for some $b \\in \\mathbb{N}$. Hence, we have\n$$\n2a+1 > 2b, \\quad 2a-1 < 2b+2.\n$$\nFrom the previous inequalities we get $2a - 3 < 2b < 2a + 1$, so $2b = 2a - 2$ or $2b = 2a$. Let us notice that the sum of the cards in each pair of the first part is the same and equals to $2013 + 2k$. In fact, in each next pair Boris' number is by $2$ smaller than the previous, while Andrija's is by $2$ larger than the previous, so the sum does not change. Hence we have $(2a + 1) + 2b = 2013 + 2k$.\n\nIf $2b = 2a - 2$, then $k$ is odd and $a = \\frac{1007 + k}{2}$. Similarly, if $2b = 2a$, then $k$ is even and $a = \\frac{1006 + k}{2}$. If $X$ is the number of the pairs in the first part in which Boris' card is larger than Andrija's, then $X = b - k + 1$. Thus we have\n$$\nX = \\begin{cases} \\frac{1007-k}{2}, & \\text{for odd } k, \\\\ \\frac{1008-k}{2}, & \\text{for even } k. \\end{cases}\n$$\nFurthermore, let us denote by $Y$ the number of pairs in the second part in which Boris' card is larger than Andrija's. Analogously we get\n$$\nY = \\begin{cases} \\frac{k-1}{2}, & \\text{for odd } k, \\\\ \\frac{k-2}{2}, & \\text{for even } k. \\end{cases}\n$$\nHence $X + Y = 503$, regardless of the number $k$, so we can conclude that arranging the cards in this way Boris will gain $503$ points (and Andrija $504$ points).\n\nIt remains to prove that Boris can not arrange his cards such that he gains more than $503$ points, regardless of Andrija's arrangement.\n\nLet us assume that there is Boris' arrangement for which Andrija gains less than $504$ points, regardless of how he arranges his cards. For such Boris' arrangement, let us denote by $A$ the total number of points that Andrija gains in all of his possible arrangements. There are $1007$ possible Andrija's arrangements, so by the assumption we have $A < 1007 \\cdot 504$. On the other hand, let us see how many points each of Boris' cards contributes to the number $A$. The card with the number $2$ contributes $1$ point because it gives points only in the arrangement in which the card with the number $1$ is on top of it. The card with the number $4$ contributes $2$ points (cases in which cards $1$ and $3$ are on top of it). In general, for $k \\in \\{1, 2, \\dots, 1007\\}$ the card with the number $2k$ contributes $k$ points. Hence we have\n$$\nA = 1 + 2 + \\dots + 1007 = \\frac{1007 \\cdot 1008}{2} = 1007 \\cdot 504.\n$$\nWe reached a contradiction, so we conclude that our assumption was wrong, i.e. Boris can not be certain to gain more than $503$ points.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 28928,
    "subject": "Mathematics (Multi-modal)",
    "question": "An $8 \\times 8$ board is initially painted as a standard chessboard in black and white. In each move we choose one row or column and change the colour of each of the eight squares in that row or column from white to black or vice versa. Is it possible to achieve that after a finite sequence of such moves exactly one square on the board is black? (Counting and Configurations)",
    "options": [],
    "answer": "No",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 28929,
    "subject": "Mathematics (Multi-modal)",
    "question": "How many four-digit numbers whose digits are mutually different and from the set $\\{0, 1, 2, 3, 4, 5\\}$, and that are divisible by $5$, are there? (Hong Kong)",
    "options": [],
    "answer": "108",
    "solution": "",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 28930,
    "subject": "Mathematics (Multi-modal)",
    "question": "Determine all positive integers $n$ such that\n$$\n3^{2n+1} - 4^{n+1} + 6^n\n$$\nis a prime number.",
    "options": [],
    "answer": "1",
    "solution": "",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 28931,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABC$ be a triangle whose longest side is $BC$, and the angle $\\angle BCA$ is three times greater than $\\angle ABC$. The exterior angle bisector at vertex $A$ meets the line $BC$ at point $A_0$, and the exterior angle bisector at vertex $B$ meets the line $AC$ at point $B_0$. If $|AA_0| = |BB_0|$, determine the angles of the given triangle.",
    "options": [],
    "answer": "Angle A = 132°, Angle B = 12°, Angle C = 36°",
    "solution": "",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 28932,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all pairs of real numbers $(x, y)$ satisfying the following system:\n$$\nx + y^2 = y^3,\n$$\n$$\ny + x^2 = x^3.\n$$",
    "options": [],
    "answer": "(0, 0), ((1 + sqrt(5)) / 2, (1 + sqrt(5)) / 2), ((1 - sqrt(5)) / 2, (1 - sqrt(5)) / 2)",
    "solution": "Subtracting the given equations, we obtain $y - x + x^2 - y^2 = x^3 - y^3$, which is equivalent to\n$$\n(x - y)(-1 + x + y) = (x - y)(x^2 + xy + y^2),\n$$\nimplying that either $x - y = 0$ or $x + y - 1 = x^2 + xy + y^2$.\n\nIn the latter case, $x + y - 1 = x^2 + xy + y^2$ is equivalent to\n$$\ny^2 + (x-1)y + x^2 - x + 1 = 0. \\quad (*)\n$$\nThis is a quadratic equation in $y$. Its discriminant is\n$$\nD = (x - 1)^2 - 4(x^2 - x + 1) = -3x^2 + 2x - 3,\n$$\nwhich is a quadratic polynomial in $x$ – its discriminant in turn is $-32 < 0$, so the starting discriminant is $D < 0$ for every real number $x$, which implies that the equation $(*)$ does not have any real solution $y$. Hence there are no solutions in this case.\n\nIn the remaining case, $y = x$ so both equations of a given system are reduced to $x + x^2 = x^3$, i.e. $x(x^2 - x - 1) = 0$. This equation has three solutions $x_1 = 0$, $x_{2,3} = \\frac{1 \\pm \\sqrt{5}}{2}$.\n\nThe only solutions are $(x, y) \\in \\{(0, 0), (\\frac{1+\\sqrt{5}}{2}, \\frac{1+\\sqrt{5}}{2}), (\\frac{1-\\sqrt{5}}{2}, \\frac{1-\\sqrt{5}}{2})\\}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 28933,
    "subject": "Mathematics (Multi-modal)",
    "question": "Students decided to play a game with $960$ tokens. First they distributed all of the tokens so that each student had the same number of tokens. Once they did that, their teacher arrived wanting to join the game. Each student gave him $4$ of his tokens, so that everyone had the same number of tokens and the game could start. How many students were playing the game? (Hong Kong)",
    "options": [],
    "answer": "15",
    "solution": "Let $n$ be the number of students.\n\nInitially, the $960$ tokens are distributed equally, so each student gets $\\dfrac{960}{n}$ tokens.\n\nWhen the teacher arrives, each student gives him $4$ tokens, so the teacher receives $4n$ tokens in total. Now there are $n+1$ people (the $n$ students and the teacher), and the $960$ tokens are distributed among them equally, so each person gets $\\dfrac{960}{n+1}$ tokens.\n\nAfter giving away $4$ tokens, each student has:\n$$\n\\frac{960}{n} - 4\n$$\ntokens.\n\nThe teacher has $4n$ tokens.\n\nSince now everyone has the same number of tokens:\n$$\n\\frac{960}{n} - 4 = \\frac{960}{n+1}\n$$\n\nMultiply both sides by $n(n+1)$:\n$$\n(n+1) \\cdot 960 - 4n(n+1) = n \\cdot 960\n$$\n\nExpand:\n$$\n960n + 960 - 4n^2 - 4n = 960n\n$$\n\nSubtract $960n$ from both sides:\n$$\n960 - 4n^2 - 4n = 0\n$$\n\nBring all terms to one side:\n$$\n4n^2 + 4n - 960 = 0\n$$\n\nDivide both sides by $4$:\n$$\nn^2 + n - 240 = 0\n$$\n\nSolve the quadratic equation:\n$$\nn = \\frac{ -1 \\pm \\sqrt{1 + 4 \\cdot 240} }{2 } = \\frac{ -1 \\pm \\sqrt{961} }{2 } = \\frac{ -1 \\pm 31 }{2 }\n$$\n\nSo $n = \\frac{30}{2} = 15$ or $n = \\frac{-32}{2} = -16$.\n\nSince $n$ must be positive, $n = 15$.\n\n**Answer:** $15$ students were playing the game.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 28934,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $n$ be a positive integer and let $S$ be the sum of all integers from $1$ to $n$. Prove that $S + 1$ is not divisible by $3$. (Ilko Brnetić)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 28935,
    "subject": "Mathematics (Multi-modal)",
    "question": "Determine all complex numbers $z$ such that\n$$\n|z^2 - i| = 1 \\quad \\text{and} \\quad |z| = \\sqrt{2}.\n$$",
    "options": [],
    "answer": "1 + i, -1 - i",
    "solution": "Let $z = x + iy$, where $x, y \\in \\mathbb{R}$.\n\nGiven $|z| = \\sqrt{2}$, so\n$$\n|z|^2 = x^2 + y^2 = 2.\n$$\n\nAlso, $|z^2 - i| = 1$.\nCompute $z^2$:\n$$\nz^2 = (x + iy)^2 = x^2 - y^2 + 2ixy.\n$$\nSo,\n$$\nz^2 - i = (x^2 - y^2) + 2ixy - i = (x^2 - y^2) + i(2xy - 1).\n$$\nTherefore,\n$$\n|z^2 - i|^2 = (x^2 - y^2)^2 + (2xy - 1)^2 = 1.\n$$\n\nNow, we have the system:\n\\begin{align*}\nx^2 + y^2 &= 2 \\\\\n(x^2 - y^2)^2 + (2xy - 1)^2 &= 1\n\\end{align*}\n\nLet us use polar form: $z = r e^{i\\theta}$, $r = \\sqrt{2}$, so $z = \\sqrt{2} e^{i\\theta}$.\nThen $z^2 = 2 e^{i2\\theta}$, so\n$$\nz^2 - i = 2 e^{i2\\theta} - i.\n$$\n\nWrite $2 e^{i2\\theta} = 2(\\cos 2\\theta + i \\sin 2\\theta)$, so\n$$\nz^2 - i = 2\\cos 2\\theta + i(2\\sin 2\\theta - 1).\n$$\n\nTherefore,\n$$\n|z^2 - i|^2 = (2\\cos 2\\theta)^2 + (2\\sin 2\\theta - 1)^2 = 1.\n$$\nExpand:\n\\begin{align*}\n4\\cos^2 2\\theta + (4\\sin^2 2\\theta - 4\\sin 2\\theta + 1) &= 1 \\\\\n4\\cos^2 2\\theta + 4\\sin^2 2\\theta - 4\\sin 2\\theta + 1 &= 1 \\\\\n4(\\cos^2 2\\theta + \\sin^2 2\\theta) - 4\\sin 2\\theta + 1 &= 1 \\\\\n4(1) - 4\\sin 2\\theta + 1 &= 1 \\\\\n4 - 4\\sin 2\\theta + 1 = 1 \\\\\n4 - 4\\sin 2\\theta = 0 \\\\\n1 - \\sin 2\\theta = 0 \\\\\n\\sin 2\\theta = 1\n\\end{align*}\n\nSo $2\\theta = \\frac{\\pi}{2} + 2\\pi k$, $k \\in \\mathbb{Z}$.\nThus $\\theta = \\frac{\\pi}{4} + \\pi k$.\n\nTherefore, the solutions are\n$$\nz = \\sqrt{2} e^{i\\left(\\frac{\\pi}{4} + \\pi k\\right)}\n$$\nfor $k \\in \\mathbb{Z}$.\n\nExplicitly, for $k = 0$:\n$$\nz_1 = \\sqrt{2} e^{i\\frac{\\pi}{4}} = \\sqrt{2}\\left(\\cos \\frac{\\pi}{4} + i \\sin \\frac{\\pi}{4}\\right) = \\sqrt{2}\\left(\\frac{\\sqrt{2}}{2} + i \\frac{\\sqrt{2}}{2}\\right) = 1 + i.\n$$\nFor $k = 1$:\n$$\nz_2 = \\sqrt{2} e^{i\\left(\\frac{5\\pi}{4}\\right)} = \\sqrt{2}\\left(\\cos \\frac{5\\pi}{4} + i \\sin \\frac{5\\pi}{4}\\right) = \\sqrt{2}\\left(-\\frac{\\sqrt{2}}{2} - i \\frac{\\sqrt{2}}{2}\\right) = -1 - i.\n$$\n\nThus, the solutions are $z = 1 + i$ and $z = -1 - i$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 28936,
    "subject": "Mathematics (Multi-modal)",
    "question": "Each of the numbers $x_1, x_2, \\dots, x_{2014}$ is $1$, $0$, or $-1$. What is the minimal possible value of the sum of products of all the pairs of those numbers, i.e. the sum of all $x_i x_j$ for $1 \\le i < j \\le 2014$? (USSR 1965)",
    "options": [],
    "answer": "-1007",
    "solution": "First note that the double sum of all the products $x_i x_j$ for $1 \\le i < j \\le 2014$ equals\n$$\n(x_1 + \\cdots + x_{2014})^2 - (x_1^2 + \\cdots + x_{2014}^2).\n$$\nDenote $A = (x_1 + \\cdots + x_{2014})^2$ and $B = x_1^2 + \\cdots + x_{2014}^2$.\nWe want to minimize $A$ and maximize $B$ at the same time.\nObviously, $A \\ge 0$. The minimum $A = 0$ is attained when among $x_i$ there is an equal number of $1$'s and of $-1$'s, e.g. when $x_1 = x_2 = \\cdots = x_{1007} = 1$, $x_{1008} = x_{1009} = \\cdots = x_{2014} = -1$.\nClearly, $B = x_1^2 + \\cdots + x_{2014}^2 \\le 1^2 + \\cdots + 1^2 = 2014$. The maximal value $B = 2014$ is attained if none of the $x_i$ is $0$, e.g. $x_1 = x_2 = \\cdots = x_{1007} = 1$, while $x_{1008} = x_{1009} = \\cdots = x_{2014} = -1$.\nSince the minimum of $A$ and the maximum of $B$ can be attained for the same values of $x_1, \\dots, x_{2014}$, we conclude that the minimal possible value of the sum of all products of pairs $x_i x_j$ ($1 \\le i < j \\le 2014$) is $\\frac{A-B}{2} = -1007$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 28937,
    "subject": "Mathematics (Multi-modal)",
    "question": "Determine all complex numbers $z$ such that $\\frac{\\bar{z}}{z} + \\frac{z}{\\bar{z}}$ is a positive integer.",
    "options": [],
    "answer": "All nonzero complex numbers with argument equal to any integer multiple of pi, or equal to plus or minus pi over six plus any integer multiple of pi; equivalently, z ≠ 0 with arg z ∈ {kπ, π/6 + kπ, −π/6 + kπ} for integer k.",
    "solution": "Let $z = re^{i\\theta}$, where $r > 0$ and $\\theta \\in \\mathbb{R}$. Then $\\bar{z} = re^{-i\\theta}$.\n\nWe have:\n\\[\n\\frac{\\bar{z}}{z} + \\frac{z}{\\bar{z}} = \\frac{re^{-i\\theta}}{re^{i\\theta}} + \\frac{re^{i\\theta}}{re^{-i\\theta}} = e^{-2i\\theta} + e^{2i\\theta} = 2\\cos(2\\theta)\n\\]\n\nWe are told that $2\\cos(2\\theta)$ is a positive integer. The possible values for $2\\cos(2\\theta)$ are $1$ and $2$ (since $2\\cos(2\\theta) \\leq 2$ and must be a positive integer).\n\nCase 1: $2\\cos(2\\theta) = 2$\n\nThen $\\cos(2\\theta) = 1 \\implies 2\\theta = 2\\pi k$ for some integer $k$, so $\\theta = \\pi k$.\n\nThus, $z = re^{i\\pi k} = r(-1)^k$. So $z$ is a nonzero real number.\n\nCase 2: $2\\cos(2\\theta) = 1$\n\nThen $\\cos(2\\theta) = \\frac{1}{2} \\implies 2\\theta = \\pm \\frac{\\pi}{3} + 2\\pi k$ for some integer $k$.\n\nSo $\\theta = \\pm \\frac{\\pi}{6} + \\pi k$.\n\nThus, $z = re^{i(\\pm \\frac{\\pi}{6} + \\pi k)}$ for $r > 0$ and integer $k$.\n\nTherefore, all complex numbers $z \\neq 0$ such that $z$ is a nonzero real number, or $z$ has argument $\\pm \\frac{\\pi}{6}$ or $\\pm \\frac{7\\pi}{6}$ (modulo $2\\pi$), i.e., $z = re^{i\\theta}$ where $\\theta = \\pi k$ or $\\theta = \\pm \\frac{\\pi}{6} + \\pi k$ for integer $k$ and $r > 0$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 28938,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $a$, $b$ and $c$ be integers and $a \\neq 0$. Can the discriminant of the quadratic function\n$$f(x) = ax^2 + bx + c$$\nbe equal to $51$?",
    "options": [],
    "answer": "No",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 28939,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $a_1, b_1, c_1, a_2, b_2, c_2$ be positive real numbers such that $b_1^2 \\le 4a_1c_1$ and $b_2^2 \\le 4a_2c_2$. Prove that $4(a_1 + a_2 + 5)(c_1 + c_2 + 1) > (b_1 + b_2 + 2)^2$. (Macedonia 2013)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 28940,
    "subject": "Mathematics (Multi-modal)",
    "question": "Show that there are no positive integers $m$ and $n$ such that $3^m + 3^n + 1$ is a perfect square.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Assume that there is $k$ such that $3^m + 3^n + 1 = k^2$. Obviously, $k$ is odd. The last equation is equivalent to $3^m + 3^n = k^2 - 1$.\n\nIt is easy to see that for odd number $k$, $8$ divides $k^2 - 1$. Since powers of $3$ are congruent to $1$ or $3$ modulo $8$, the number $3^m + 3^n$ is congruent to $2$, $4$ or $6$ modulo $8$. Therefore $3^m + 3^n = k^2 - 1$ leads to a contradiction.\n\nThere are no positive integers $m$ and $n$ such that $3^m + 3^n + 1$ is a perfect square.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 28941,
    "subject": "Mathematics (Multi-modal)",
    "question": "The points $P$ and $Q$ lie on the side $\\overline{AB}$ of the rectangle $ABCD$ such that $|AP| = |PQ| = |QB|$. The line $DQ$ meets the lines $AC$ and $CP$ at points $K$ and $L$ respectively, and the line $DB$ meets the lines $AC$ and $CP$ at points $N$ and $M$ respectively.\nDetermine the ratio of the areas of quadrilaterals $KLMN$ and $ABCD$.",
    "options": [],
    "answer": "1/40",
    "solution": "",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 28942,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $p$ and $q$ be two parallel lines. Circle $k$ touches the line $p$ at $A$ and intersects $q$ at two different points, $B$ and $C$. Let $T$ be some point on $p$. Segments $\\overline{TB}$ and $\\overline{TC}$ intersect the shorter arc $\\widehat{AC}$ at $K$ and $L$ respectively. Points $K$ and $L$ are both different from $B$ and $C$.\nProve that the line $KL$ passes through the midpoint of the segment $\\overline{AT}$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $P$ be the intersection of the lines $p$ and $KL$. Denote $\\angle TBC = x$.\n![](attached_image_1.png)\nLine $BT$ is a transversal of the parallel lines $p$ and $q$, which implies that $\\angle BTA = \\angle TBC = x$.\nThe quadrilateral $BCLK$ is cyclic, hence $\\angle CLK = 180^\\circ - \\angle KBC = 180^\\circ - x$, implying $\\angle KLT = x$.\nTriangles $PKT$ and $PTL$ are similar because they share an angle at $P$ and $\\angle KTP = \\angle PLT = x$. This implies\n$$\n\\frac{|PK|}{|PT|} = \\frac{|PT|}{|PL|}, \\quad \\text{i.e.} \\quad |PK| \\cdot |PL| = |PT|^2.\n$$\nSince the power of the point $P$ with respect to the circle $k$ is $|PK| \\cdot |PL| = |PA|^2$, we can conclude that $|PA| = |PT|$, i.e. the point $P$ is the midpoint of $\\overline{AT}$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 28943,
    "subject": "Mathematics (Multi-modal)",
    "question": "All vertices of the pentagon $ABCDE$ lie on the same circle. If $\\angle CAD = 50^\\circ$, determine\n$$\n\\angle ABC + \\angle AED.\n$$",
    "options": [],
    "answer": "Not uniquely determined from the given information",
    "solution": "",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 28944,
    "subject": "Mathematics (Multi-modal)",
    "question": "A hundred quadratic envelopes, each of different size, are arranged in a way that, for every two different envelopes, the smaller one is either inside of the bigger one, or they are outside of each other. At the same time, in both of the envelopes there can be other envelopes. Two arrangements are considered different if there are two envelopes which are inside each other in one of the arrangements, while being outside of each other in the other arrangement.\nHow many different arrangements are there such that the biggest envelope contains all the remaining envelopes? (Richard Stanley's problem set)",
    "options": [],
    "answer": "99!",
    "solution": "Denote the number of arrangements of $n$ envelopes as $K_n$.\nAssume we are given an arrangement of $n$ envelopes. If we remove the smallest envelope, we obtain a possible arrangement of $n-1$ remaining envelopes. On the other hand, if we are given a possible arrangement of the remaining $n-1$ envelopes, we can put the smallest one directly in the biggest one, or in any of the remaining $n-2$ envelopes. Therefore, $K_n$ is exactly $n-1$ times larger than $K_{n-1}$.\nWe conclude that $K_n = (n-1) \\cdot K_{n-1} = \\cdots = (n-1) \\cdot (n-2) \\cdots 2 \\cdot K_2$.\nFor $n=2$, the only possible arrangement is putting the smaller envelope inside the larger one, so $K_2=1$. Therefore $K_n = (n-1) \\cdot (n-2) \\cdots 1$.\nFor 100 envelopes, the number of different arrangements is\n$$\nK_{100} = 99 \\cdot 98 \\cdots 1 = 99!\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 28945,
    "subject": "Mathematics (Multi-modal)",
    "question": "In each vertex of a regular $50$-gon one of the numbers $1$ or $2$ is written. No three consecutive vertices have the same number written in them. The number $1$ is written twenty times and the number $2$ thirty times in total. For each vertex the product of the numbers written in two adjacent vertices and the number in that vertex is calculated, and all such products are added up.\nProve that the result is always the same and determine it.",
    "options": [],
    "answer": "180",
    "solution": "",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 28946,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $A$ be the number of all six-digit numbers whose product of digits equals $105$, and let $B$ be the number of all six-digit numbers whose product of digits equals $147$. Determine the ratio $A : B$. (Kristina Ana Škreb)",
    "options": [],
    "answer": "2:1",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 28947,
    "subject": "Mathematics (Multi-modal)",
    "question": "The rhombus $AKLM$ is inscribed in the triangle $ABC$, so that point $K$ is on $\\overline{AB}$, point $L$ is on $\\overline{BC}$ and point $M$ is on $\\overline{CA}$. If the rhombus has side of length $2\\sqrt{2}$, the area of triangle $LMC$ is $3$, and the area of triangle $KLB$ is $4$, prove that $\\angle BAC = 60^\\circ$. (Mea Bombardelli)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 28948,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $a$ be a positive integer, and let $b$ and $c$ be integers such that the equation $a x^2 + b x + c = 0$ has two different solutions in the interval $\\langle 0, \\frac{1}{2} \\rangle$. Prove that $a \\ge 6$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 28949,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABC$ be an equilateral triangle with sides of length $1$. The point $X$ on the ray $AB$ and the point $Y$ on the ray $AC$ are chosen, so that $|AX|$ and $|AY|$ are positive integers.\nCan the radius of the circumcircle of the triangle $AXY$ be $\\sqrt{2014}$?",
    "options": [],
    "answer": "No",
    "solution": "Let us assume that the radius of the circumcircle of the triangle $AXY$ is equal to $\\sqrt{2014}$.\nIn the triangle $AXY$ the angle opposite to the side $XY$ is $60^\\circ$ because the triangle $ABC$ is equilateral.\nIf $R$ is the radius of the circumcircle of the triangle $AXY$, then\n$$\n|XY| = 2R \\sin 60^\\circ = \\sqrt{2014} \\cdot \\sqrt{3}.\n$$\nLet $|AX| = m$, $|AY| = n$, where $m$ and $n$ are positive integers.\n![](attached_image_1.png)\nApplying the cosine rule to the triangle $AXY$ we get $m^2 + n^2 - 2mn \\cos 60^\\circ = |XY|^2$, which leads to\n$$\nm^2 + n^2 - mn = 2014 \\cdot 3.\n$$\nLet us first notice that the right-hand side of the equation is even. If $m$ and $n$ are of different parity or if they are both odd, then the left-hand side is odd, which is impossible.\nThe only remaining case is when $m$ and $n$ are both even. Then $m^2 + n^2 - mn$ is divisible by $4$, but $2014 \\cdot 3$ is not, so we get another impossible case.\n\nTherefore, the initial assumption is wrong, which means that $R$ can not be equal to $\\sqrt{2014}$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 28950,
    "subject": "Mathematics (Multi-modal)",
    "question": "What is the smallest, and what is the largest value that the function $f: \\mathbb{R} \\to \\mathbb{R}$, given by $f(x) = \\cos^2 x + \\sin x$, can take? (Ilko Brnetić)",
    "options": [],
    "answer": "Smallest value: -1; Largest value: 5/4",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 28951,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $a$, $b$, $c$ and $d$ be real numbers such that\n$$\n2 \\cos a + 6 \\cos b + 7 \\cos c + 9 \\cos d = 0, \\\\\n2 \\sin a - 6 \\sin b + 7 \\sin c - 9 \\sin d = 0.\n$$\nIf $\\cos(b+c) \\neq 0$, determine the value of $\\frac{\\cos(a+d)}{\\cos(b+c)}$.",
    "options": [],
    "answer": "7/3",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 28952,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABCD A_1 B_1 C_1 D_1$ be a rectangular cuboid with edge-lengths $|AB| = |AD| = a$ and $|AA_1| = 2a$. Let $A', B', C'$ and $D'$ be the midpoints of $AA_1, BB_1, CC_1$ and $DD_1$ respectively.\nFind the volume of the intersection of the cube $ABCD A'B'C'D'$ and the pyramid $A_1A'B'B_1D$.",
    "options": [],
    "answer": "a^3/8",
    "solution": "",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 28953,
    "subject": "Mathematics (Multi-modal)",
    "question": "In the interior of an acute triangle $ABC$ the point $P$ is chosen, so that\n$$\n\\angle APB = \\angle CBA + \\angle ACB, \\quad \\angle BPC = \\angle ACB + \\angle BAC.\n$$\nProve that\n$$\n\\frac{|AC| \\cdot |BP|}{|BC|} = \\frac{|BC| \\cdot |AP|}{|AB|}. \\qquad (\\text{Belarus 2013})\n",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let us denote $\\angle BAC = \\alpha$, $\\angle CBA = \\beta$ and $\\angle ACB = \\gamma$.\n![](attached_image_1.png)\n\nFrom $\\angle APB = \\beta + \\gamma$ and $\\angle BPC = \\gamma + \\alpha$ we get\n$$\n\\angle CPA = 360^\\circ - \\angle APB - \\angle BPC = 360^\\circ - \\beta - \\gamma - \\alpha = \\alpha + \\beta.\n$$\nIf $\\angle PBA = x$, then $\\angle BAP = 180^\\circ - \\angle APB - \\angle PBA = 180^\\circ - \\beta - \\gamma - x = \\alpha - x$, so $\\angle PAC = \\angle BAC - \\angle BAP = \\alpha - (\\alpha - x) = x$.\nAnalogously we get $\\angle ACP = \\gamma - x$, $\\angle PCB = x$ and $\\angle CBP = \\beta - x$.\nApplying the sine rule to the triangles $ABP$ and $BCP$ we get\n$$\n\\frac{|AP|}{\\sin x} = \\frac{|AB|}{\\sin (\\beta + \\gamma)} = \\frac{|AB|}{\\sin \\alpha} \\quad \\text{and} \\quad \\frac{|BP|}{\\sin x} = \\frac{|BC|}{\\sin (\\gamma + \\alpha)} = \\frac{|BC|}{\\sin \\beta},\n$$\nwherefrom due to $\\sin \\alpha : \\sin \\beta = |BC| : |AC|$ we get\n$$\n\\frac{|BP|}{|AP|} = \\frac{|BC| \\sin \\alpha}{|AB| \\sin \\beta} = \\frac{|BC| \\cdot |BC|}{|AB| \\cdot |AC|},\n$$\nwhich finally leads to\n$$\n\\frac{|AC| \\cdot |BP|}{|BC|} = \\frac{|BC| \\cdot |AP|}{|AB|}.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 28954,
    "subject": "Mathematics (Multi-modal)",
    "question": "Prove that\n$$\n\\log_2 \\left(1 + \\frac{1}{1}\\right) + \\log_2 \\left(1 + \\frac{1}{2}\\right) + \\dots + \\log_2 \\left(1 + \\frac{1}{k}\\right) + \\dots + \\log_2 \\left(1 + \\frac{1}{2014}\\right) < 11.\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let us denote the sum by $S$:\n$$\nS = \\log_2 \\left(1 + \\frac{1}{1}\\right) + \\log_2 \\left(1 + \\frac{1}{2}\\right) + \\dots + \\log_2 \\left(1 + \\frac{1}{2014}\\right)\n$$\n\nWe have:\n$$\n\\log_2 \\left(1 + \\frac{1}{k}\\right) = \\log_2 \\left(\\frac{k+1}{k}\\right) = \\log_2 (k+1) - \\log_2 k\n$$\n\nTherefore,\n$$\nS = \\sum_{k=1}^{2014} \\left[\\log_2 (k+1) - \\log_2 k\\right]\n$$\n\nThis is a telescoping sum:\n$$\nS = \\log_2 2 - \\log_2 1 + \\log_2 3 - \\log_2 2 + \\log_2 4 - \\log_2 3 + \\dots + \\log_2 2015 - \\log_2 2014\n$$\n\nAll terms except the first $\\log_2 2$ and the last $-\\log_2 2014$ cancel, so:\n$$\nS = \\log_2 (2015) - \\log_2 1 = \\log_2 (2015)\n$$\n\nWe need to show that $\\log_2 (2015) < 11$.\n\nBut $2^{11} = 2048 > 2015$, so $\\log_2 (2015) < 11$.\n\nTherefore,\n$$\n\\log_2 \\left(1 + \\frac{1}{1}\\right) + \\log_2 \\left(1 + \\frac{1}{2}\\right) + \\dots + \\log_2 \\left(1 + \\frac{1}{2014}\\right) < 11.\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 28955,
    "subject": "Mathematics (Multi-modal)",
    "question": "In a triangle $ABC$ bisector of the angle $\\angle ACB$ meets $\\overline{AB}$ at $D$. If $|CB| = |CD|$, $|AD| = 4$ and $|DB| = 3$, determine $|AC|$. (Hong Kong)",
    "options": [],
    "answer": "8",
    "solution": "",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 28956,
    "subject": "Mathematics (Multi-modal)",
    "question": "Do there exist positive integers $m$ and $n$ such that $m^2 + n$ and $n^2 + m$ are squares of positive integers?",
    "options": [],
    "answer": "No; such positive integers do not exist.",
    "solution": "Without loss of generality we can assume that $m \\ge n$. Obviously, $m^2 < m^2 + n$. On the other hand, $(m+1)^2 = m^2 + 2m + 1 > m^2 + n$ because $2m > n$.\nHence we have\n$$\nm^2 < m^2 + n < (m + 1)^2,\n$$\nwhich means that $m^2+n$ is between two consecutive squares, so it can not be a square of an integer. Such positive integers $m$ and $n$ do not exist.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 28957,
    "subject": "Mathematics (Multi-modal)",
    "question": "In the triangle $ABC$ we have $\\angle ABC = 2\\angle BAC$. Prove that $|AC| < 2|BC|$. (Ilko Brnetić)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $\\angle BAC = \\alpha$, so $\\angle ABC = 2\\alpha$.\n\nIn triangle $ABC$, the sum of angles is $180^\\circ$:\n$$\n\\alpha + 2\\alpha + \\angle BCA = 180^\\circ\n$$\nSo $\\angle BCA = 180^\\circ - 3\\alpha$.\n\nLet $|AB| = c$, $|BC| = a$, $|CA| = b$.\n\nBy the Law of Sines:\n$$\n\\frac{a}{\\sin \\alpha} = \\frac{b}{\\sin 2\\alpha} = \\frac{c}{\\sin(180^\\circ - 3\\alpha)} = \\frac{c}{\\sin 3\\alpha}\n$$\n\nWe want to prove $b < 2a$.\n\nFrom the Law of Sines:\n$$\n\\frac{b}{a} = \\frac{\\sin 2\\alpha}{\\sin \\alpha}\n$$\nRecall $\\sin 2\\alpha = 2\\sin \\alpha \\cos \\alpha$:\n$$\n\\frac{b}{a} = \\frac{2\\sin \\alpha \\cos \\alpha}{\\sin \\alpha} = 2\\cos \\alpha\n$$\nSo $b = 2a \\cos \\alpha$.\n\nWe want $b < 2a$, i.e. $2a \\cos \\alpha < 2a$, or $\\cos \\alpha < 1$.\n\nSince $0 < \\alpha < 60^\\circ$ (otherwise $\\angle BCA$ would be negative), $\\cos \\alpha < 1$ is always true for $0 < \\alpha < 60^\\circ$.\n\nTherefore, $|AC| < 2|BC|$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 28958,
    "subject": "Mathematics (Multi-modal)",
    "question": "If $p$ and $p^2 + 8$ are prime numbers, prove that $p^3 + 4$ is also prime.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 28959,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all prime numbers $p$ for which there exists a positive integer $n$ such that both $n^2 + 3$ and $(n + 1)^2 + 3$ are divisible by $p$. (Stipe Vidak)",
    "options": [],
    "answer": "13",
    "solution": "",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 28960,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $a$, $b$ and $c$ be positive real numbers. Prove the inequality\n$$\n\\frac{a^2}{a+b} + \\frac{b^2}{b+c} \\ge \\frac{3a+2b-c}{4}.\n$$\n(Belarus 2010)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "$$\n\\frac{a^2}{a+b} + \\frac{a+b}{4} \\ge a \\quad \\text{and} \\quad \\frac{b^2}{b+c} + \\frac{b+c}{4} \\ge b.\n$$\nAdding these two inequalities we get\n$$\n\\left( \\frac{a^2}{a+b} + \\frac{a+b}{4} \\right) + \\left( \\frac{b^2}{b+c} + \\frac{b+c}{4} \\right) \\ge a+b = \\frac{3a+2b-c}{4} + \\frac{a+b}{4} + \\frac{b+c}{4}.\n$$\nIf we subtract $\\frac{a+b}{4} + \\frac{b+c}{4}$ from both sides, the result follows.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 28961,
    "subject": "Mathematics (Multi-modal)",
    "question": "A chessboard is a board with $8$ rows and $8$ columns, whose squares are coloured alternately black and white, so that the square in the first row and first column is coloured black. An integer is inscribed in each square. It is known that the sum of all numbers on white squares equals $26$, and the sum of all numbers in odd rows equals $43$.\nIf we change the sign of all numbers on white squares, what will be the sum of all numbers in odd rows after the change? (Italy 2012)",
    "options": [],
    "answer": "17",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 28962,
    "subject": "Mathematics (Multi-modal)",
    "question": "In each vertex of a regular dodecagon $A_1A_2...A_{12}$ one of the numbers $1$ or $-1$ is written. Initially, number $-1$ is written in $A_1$, and number $1$ is written in all other vertices. It is allowed to simultaneously change the sign of the numbers written in any six consecutive vertices of the dodecagon.\nProve that after repeating this procedure several times it is not possible to have $-1$ written in $A_2$ and $1$ in all other vertices. (Counting and Configurations)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 28963,
    "subject": "Mathematics (Multi-modal)",
    "question": "A circle of length $6N$ is divided by $3N$ marked points into $3N$ arcs: $N$ arcs of length $1$, $N$ arcs of length $2$ and $N$ arcs of length $3$.\nProve that among the marked points there exist two which are endpoints of some diameter of the circle. (V. Prasolov, Problems in plane and solid geometry)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $3N$ marked points be called *outer points*. We divide each of $N$ arcs of length $2$ into two arcs of equal length using one point, and each of $N$ arcs of length $3$ into three arcs of equal length using two points. Let those $3N$ new points be called *inner points*. There are $6N$ outer and inner points in total and they divide the circle into $6N$ arcs of length $1$. Since $6N$ is odd, we conclude that diametrically opposite to each of those $6N$ points there lies another of those $6N$ points. From the definition of the inner points we see that there can not exist $3$ consecutive inner points.\n\nLet us assume the opposite, i.e. that no two of $3N$ outer points are diametrically opposite. That means that diametrically opposite to every outer point lies an inner point. Since there is an equal number of outer and inner points, we conclude that diametrically opposite to every inner point lies an outer point.\n\nThe points $A$ and $B$ are outer points so the points $C$ and $D$ are inner points. Due to the fact that there are no $3$ consecutive inner points, we conclude that the points $P$ and $Q$ are outer. Hence, across the arc $\\widehat{AB}$ of length $1$ lies the arc $\\widehat{PQ}$ of length $3$. Since there is an equal number of arcs of lengths $1$ and $3$, we get that across every arc of length $3$ lies an arc of length $1$.\n\n![](attached_image_1.png)\n\nFor $i \\in \\{1, 2, 3\\}$, let $L_i$ be the number of arcs of length $i$ inside of the shorter arc $\\widehat{AP}$ and let $D_i$ be the number of arcs of length $i$ inside of the shorter arc $\\widehat{BQ}$. Length of the arc $\\widehat{AP}$ is $3N - 2$ so we have:\n$$\nL_1 + 2L_2 + 3L_3 = 3N - 2. \\quad (*)\n$$\nSince across every arc of length $3$ lies an arc of length $1$ and vice versa, we have:\n$$\nD_1 = L_3. \\quad (**)\n$$\nThere are exactly $N$ arcs of length $1$ so we have:\n$$\nL_1 + D_1 = N - 1,\n$$\nwhich using $(**)$ leads to:\n$$\nL_1 + L_3 = N - 1. \\quad (***)\n$$\nSubtracting the equality $(***)$ from the inequality $(*)$ we get:\n$$\n2L_2 + 2L_3 = 2N - 1,\n$$\nwhich is impossible, because the left-hand side of the equality is an even number, but the right-hand side is an odd number. Hence, we conclude that our assumption is wrong, so there exist two diametrically opposite outer points.",
    "topic": "Geometry",
    "subtopic": "Solid Geometry"
  },
  {
    "id": 28964,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $\\triangle ABC$ be a right triangle. Point $M$ is chosen on the bisector of the right angle $\\angle ACB$. If we have $\\sin \\angle MAC = \\frac{8}{17}$ and $\\cos \\angle MBC = \\frac{3}{5}$, determine the ratio $|AB| : |CM|$.",
    "options": [],
    "answer": "5√58 : 16",
    "solution": "",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 28965,
    "subject": "Mathematics (Multi-modal)",
    "question": "Prove that among any seven squares of positive integers there are two whose difference is divisible by $20$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let the seven squares be $a_1^2, a_2^2, \\ldots, a_7^2$ where $a_i$ are positive integers.\n\nConsider the possible residues of $a^2$ modulo $20$.\n\nSince $a$ can be any integer, $a^2$ modulo $20$ depends on $a$ modulo $20$.\n\nLet us compute $a^2 \\bmod 20$ for $a = 0, 1, \\ldots, 19$:\n\n| $a$ | $a^2 \\bmod 20$ |\n|----|----------------|\n| 0  | 0              |\n| 1  | 1              |\n| 2  | 4              |\n| 3  | 9              |\n| 4  | 16             |\n| 5  | 5              |\n| 6  | 16             |\n| 7  | 9              |\n| 8  | 4              |\n| 9  | 1              |\n| 10 | 0              |\n| 11 | 1              |\n| 12 | 4              |\n| 13 | 9              |\n| 14 | 16             |\n| 15 | 5              |\n| 16 | 16             |\n| 17 | 9              |\n| 18 | 4              |\n| 19 | 1              |\n\nSo the possible residues are $0, 1, 4, 5, 9, 16$.\n\nCounting the number of distinct residues:\n- $0$ (for $a \\equiv 0, 10 \\pmod{20}$)\n- $1$ (for $a \\equiv 1, 9, 11, 19 \\pmod{20}$)\n- $4$ (for $a \\equiv 2, 8, 12, 18 \\pmod{20}$)\n- $5$ (for $a \\equiv 5, 15 \\pmod{20}$)\n- $9$ (for $a \\equiv 3, 7, 13, 17 \\pmod{20}$)\n- $16$ (for $a \\equiv 4, 6, 14, 16 \\pmod{20}$)\n\nThere are only $6$ possible residues for squares modulo $20$.\n\nBy the pigeonhole principle, among any $7$ squares, at least two must have the same residue modulo $20$.\n\nTherefore, their difference is divisible by $20$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 28966,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $n \\ge 2$ be an integer and let $a_0, a_1, \\ldots, a_n$ be consecutive terms of an arithmetic progression. Prove that\n$$\na_0 - \\binom{n}{1}a_1 + \\dots + (-1)^k \\binom{n}{k}a_k + \\dots + (-1)^n \\binom{n}{n}a_n = 0.\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 28967,
    "subject": "Mathematics (Multi-modal)",
    "question": "Nikola is thinking of five numbers. The first number is $-2$, and the fifth number is $6$. The first four numbers are consecutive terms of an arithmetic progression, and the last three numbers are consecutive terms of a geometric progression. What numbers is Nikola thinking of?",
    "options": [],
    "answer": "-2, -2/3, 2/3, 2, 6",
    "solution": "Let the five numbers be $a_1, a_2, a_3, a_4, a_5$.\n\nGiven:\n- $a_1 = -2$\n- $a_5 = 6$\n- $a_1, a_2, a_3, a_4$ are consecutive terms of an arithmetic progression (AP)\n- $a_3, a_4, a_5$ are consecutive terms of a geometric progression (GP)\n\nLet the common difference of the AP be $d$.\nThen:\n$$\na_2 = a_1 + d = -2 + d\n$$\n$$\na_3 = a_1 + 2d = -2 + 2d\n$$\n$$\na_4 = a_1 + 3d = -2 + 3d\n$$\n\nLet the common ratio of the GP be $r$.\nThen:\n$$\na_4 = a_3 r\n$$\n$$\na_5 = a_3 r^2\n$$\nBut $a_5 = 6$, $a_3 = -2 + 2d$, $a_4 = -2 + 3d$.\n\nFrom the GP:\n$$\na_4 = a_3 r \\implies -2 + 3d = (-2 + 2d) r\n$$\n$$\na_5 = a_3 r^2 \\implies 6 = (-2 + 2d) r^2\n$$\n\nFrom the second equation:\n$$\n6 = (-2 + 2d) r^2 \\implies r^2 = \\frac{6}{-2 + 2d}\n$$\n\nFrom the first equation:\n$$\n-2 + 3d = (-2 + 2d) r\n\\implies r = \\frac{-2 + 3d}{-2 + 2d}\n$$\n\nBut $r^2 = \\left(\\frac{-2 + 3d}{-2 + 2d}\\right)^2 = \\frac{6}{-2 + 2d}$\n\nSo:\n$$\n\\left(\\frac{-2 + 3d}{-2 + 2d}\\right)^2 = \\frac{6}{-2 + 2d}\n$$\nMultiply both sides by $(-2 + 2d)^2$:\n$$\n(-2 + 3d)^2 = 6(-2 + 2d)\n$$\nExpand:\n$$\n(-2 + 3d)^2 = 4 - 12d + 9d^2\n$$\nSo:\n4 - 12d + 9d^2 = 6(-2 + 2d) = -12 + 12d\n$$\nBring all terms to one side:\n$$\n4 - 12d + 9d^2 + 12 - 12d = 0\n$$\n$$\n(4 + 12) + 9d^2 - 12d - 12d = 0\n$$\n$$\n16 + 9d^2 - 24d = 0\n$$\n$$\n9d^2 - 24d + 16 = 0\n$$\nThis is a quadratic equation in $d$.\n\nSolve for $d$:\n$$\nd = \\frac{24 \\pm \\sqrt{24^2 - 4 \\cdot 9 \\cdot 16}}{2 \\cdot 9}\n$$\n$$\nd = \\frac{24 \\pm \\sqrt{576 - 576}}{18}\n$$\n$$\nd = \\frac{24 \\pm 0}{18} = \\frac{24}{18} = \\frac{4}{3}\n$$\nSo $d = \\frac{4}{3}$.\n\nNow, compute the numbers:\n$$\na_1 = -2\n$$\n$$\na_2 = -2 + \\frac{4}{3} = -\\frac{2}{3}\n$$\n$$\na_3 = -2 + 2 \\cdot \\frac{4}{3} = -2 + \\frac{8}{3} = \\frac{2}{3}\n$$\n$$\na_4 = -2 + 3 \\cdot \\frac{4}{3} = -2 + 4 = 2\n$$\n$$\na_5 = 6\n$$\nCheck that $a_3, a_4, a_5$ are in geometric progression:\n\nCommon ratio:\n$$\nr = \\frac{a_4}{a_3} = \\frac{2}{2/3} = 3\n$$\nCheck $a_5 = a_3 r^2 = \\frac{2}{3} \\cdot 9 = 6$\n\nYes, it works.\n\n**Answer:**\n\nThe numbers are $-2$, $-\\frac{2}{3}$, $\\frac{2}{3}$, $2$, $6$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 28968,
    "subject": "Mathematics (Multi-modal)",
    "question": "For a positive integer $n$ denote by $s(n)$ the sum of all positive divisors of $n$ and by $d(n)$ the number of positive divisors of $n$. Determine all positive integers $n$ such that\n$$\ns(n) = n + d(n) + 1.\n$$\n(Nikola Adžaga)",
    "options": [],
    "answer": "9",
    "solution": "Observe that $n = 1$ is not a solution, so $d(n) \\ge 2$.\nIt is impossible that $d(n) = 2$, since then $n$ would be prime and given equation would reduce to $1 + n = n + 2 + 1$, so $d(n) \\ge 3$.\nLet $1 = D_1 < D_2 < \\cdots < D_d = n$ be the divisors of $n$. The given equation can then be written as\n$$\n1 + \\sum_{i=2}^{d(n)-1} D_i + n = n + d(n) + 1.\n$$\nSince $D_i \\ge 2$ for all $i \\in \\{2, 3, \\dots, d(n) - 1\\}$, we get $d(n) = \\sum_{i=2}^{d(n)-1} D_i \\ge (d(n) - 2) \\cdot 2$,\nhence $d(n) \\le 4$, i.e. $3 \\le d(n) \\le 4$.\nFor $d(n) = 3$, number $n$ is a square of some prime $p$ and given equation reduces to\n$$\n1 + p + p^2 = p^2 + 3 + 1,\n$$\nhence $p = 3$ and $n = p^2 = 9$.\nFor $d(n) = 4$, there are two possibilities:\na) If $n$ is product of some primes $q$ and $r$ ($q < r$), then the given equation implies $1 + q + r + qr = qr + 4 + 1$, thus $q + r = 4$. There are no such primes $q$ and $r$.\nb) If $n$ is a cube of some prime $s$, then the given equation means that $1+s+s^2+s^3 = s^3 + 4 + 1$, thus $s^2 + s - 4 = 0$. There is no such prime $s$.\n\nTherefore, the only solution is $n = 9$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 28969,
    "subject": "Mathematics (Multi-modal)",
    "question": "Prove that for every positive integer $n$\n$$\n\\sqrt{n + \\sqrt{(n-1) + \\sqrt{(n-2) + \\dots + \\sqrt{2 + \\sqrt{1}}}}} < \\sqrt{n} + 1.\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $S_n = \\sqrt{n + \\sqrt{(n-1) + \\sqrt{(n-2) + \\dots + \\sqrt{2 + \\sqrt{1}}}}}$.\n\nWe will prove by induction on $n$ that $S_n < \\sqrt{n} + 1$ for all positive integers $n$.\n\nBase case ($n = 1$):\n$S_1 = \\sqrt{1} = 1 < \\sqrt{1} + 1 = 2$.\n\nInductive step:\nAssume $S_{n-1} < \\sqrt{n-1} + 1$.\nConsider $S_n = \\sqrt{n + S_{n-1}}$.\n\nWe want to show:\n$$\n\\sqrt{n + S_{n-1}} < \\sqrt{n} + 1.\n$$\n\nIt suffices to show that $n + S_{n-1} < (\\sqrt{n} + 1)^2 = n + 2\\sqrt{n} + 1$.\nSubtract $n$ from both sides:\n$$\nS_{n-1} < 2\\sqrt{n} + 1.\n$$\n\nBut by the induction hypothesis, $S_{n-1} < \\sqrt{n-1} + 1$.\nSo it is enough to show:\n$$\n\\sqrt{n-1} + 1 < 2\\sqrt{n} + 1\n$$\nwhich is equivalent to $\\sqrt{n-1} < 2\\sqrt{n}$, which is true for all $n \\geq 1$.\n\nTherefore, by induction, $S_n < \\sqrt{n} + 1$ for all positive integers $n$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 28970,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $a$, $b$, $c$ and $d$ be positive real numbers such that $a^2 + b^2 = c^2 + d^2$.\nDetermine the largest possible area of a quadrilateral $ABCD$ with side-lengths $|AB| = a$, $|BC| = b$, $|CD| = c$, $|DA| = d$.",
    "options": [],
    "answer": "(ab + cd)/2",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 28971,
    "subject": "Mathematics (Multi-modal)",
    "question": "There are $2014$ coins each of which is black on one side and white on the other side. A $2014 \\times 1$ board is given. One coin is placed on each square. In each move we can remove one coin with black side up and, at the same time, flip the coins on the neighbouring squares (if they were not already removed).\nDetermine all starting arrangements of coins such that there is a finite sequence of moves which removes all the coins. (New Zealand)",
    "options": [],
    "answer": "Exactly those starting arrangements with an odd number of coins initially showing black.",
    "solution": "We will prove the following claim by mathematical induction on the number of coins: We can remove all the coins from some starting arrangement if and only if there is an odd number of coins with black side up.\n\nIf there is only one coin on the board, we can remove it if and only if it is turned black side up.\n\nFor positive integer $N > 1$, assume that the claim holds for all arrangements consisting of less than $N$ coins and consider an arbitrary starting arrangement of $N$ coins. Let $K$ be the number of coins with black side up in that arrangement.\n\nObserve that, by removing some coin, board is divided in two parts and that the moves performed on one of these sides do not affect the coins on the other part.\n\nIf $K$ is odd, we will prove that it is possible to remove all coins. Let $Y$ be the first coin on the board that is turned with black side up. Assume that $Y$ is neither on the first, nor on the last square of the board, and denote by $X$ the coin on a square left to $Y$, and by $Z$ the coin on a square right of $Y$. In the first move, remove both $Y$ and its square (and turn $X$ and $Z$). This way, we obtain two smaller boards. We will show that we can remove all coins from both of them. On the first board there are coins being left of $Y$, and on the second one there are those who are right of $Y$.\n\nOn the first board all coins are turned with white side up, except $X$ which was just turned so that it is black side up. Hence, on the first board there is exactly one coin with black side up, so, by induction hypothesis, there is a sequence of moves removing all the coins from that board.\n\nIf $Z$ is white side up, then, after the move, the number of coins on the other board with black side up will be equal to $K$. If $Z$ is black side up, then, after the move, the number of coins on the other board with black side up will be equal to $K - 2$. In both cases the number is odd. By induction hypothesis, we can remove all the coins from the second board as well.\n\nIf $Y$ is on the first square, then consider just the second board and coin $Z$, and if $Y$ is on the last square, consider just the first board and coin $X$.\n\nAssume that $K$ is even and that, for a considered starting arrangement, there is a sequence of moves removing all the coins.\n\nLet $Y$ be the coin that is removed first. Beside $Y$, there was an odd number of coins on the board with black side up. The total number of coins turned with black side up on both parts of the board will still, after the move, be an odd number. Hence there is an even number of coins turned with black side up on one part of the board. By induction hypothesis, coins on that part cannot be removed, contradiction. We have shown that if it is possible to remove all the coins, then the number of those with black side up cannot be an even number.",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 28972,
    "subject": "Mathematics (Multi-modal)",
    "question": "For a real number $a$, let $P_a$ be the parabola given by the equation $y = x^2 + a x + (2014 - a)$. Prove that all parabolae $P_a$ pass through the same point.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let us find a point $(x_0, y_0)$ that lies on all parabolae $P_a$ for any real $a$.\n\nThe equation of $P_a$ is $y = x^2 + a x + (2014 - a)$.\n\nLet us try to find $x_0$ and $y_0$ such that for all $a$, $y_0 = x_0^2 + a x_0 + (2014 - a)$.\n\nRewriting:\n$$\ny_0 = x_0^2 + a x_0 + 2014 - a\n$$\n$$\ny_0 = x_0^2 + 2014 + a(x_0 - 1)\n$$\nFor this to be independent of $a$, the coefficient of $a$ must be zero:\n$$\nx_0 - 1 = 0 \\implies x_0 = 1\n$$\nSo, for $x_0 = 1$:\n$$\ny_0 = (1)^2 + a(1) + 2014 - a = 1 + a + 2014 - a = 1 + 2014 = 2015\n$$\nTherefore, the point $(1, 2015)$ lies on all parabolae $P_a$.\n\nThus, all parabolae $P_a$ pass through the same point $(1, 2015)$.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 28973,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all positive integers $n$ such that\n$$\na! + b! + c! = 2^n\n$$\nfor some positive integers $a, b, c$.",
    "options": [],
    "answer": "[2, 3, 5, 7]",
    "solution": "Without loss of generality, assume $a \\leq b \\leq c$.\n\nIf $c \\geq 6$, then $c!$ is divisible by $2^4 = 16$, and $a!$ and $b!$ are also divisible by $2$ for $a, b \\geq 2$. Thus, $a! + b! + c!$ is divisible by $2$, but for $c \\geq 6$, $c!$ is divisible by $8$ and higher powers of $2$.\n\nLet us check small values of $c$:\n\nIf $c = 1$:\n$a! + b! + 1! = 2^n$\nBut $a, b \\leq 1$, so $a = b = 1$.\n$1! + 1! + 1! = 1 + 1 + 1 = 3$, which is not a power of $2$.\n\nIf $c = 2$:\n$a! + b! + 2! = a! + b! + 2 = 2^n$\n$a, b \\leq 2$\nTry $a = 1, b = 1$: $1 + 1 + 2 = 4 = 2^2$\nSo $n = 2$ is a solution with $(a, b, c) = (1, 1, 2)$ (and permutations).\n\nTry $a = 1, b = 2$: $1 + 2 + 2 = 5$\n$a = 2, b = 2$: $2 + 2 + 2 = 6$\nNo other powers of $2$.\n\nIf $c = 3$:\n$a! + b! + 6 = 2^n$, $a, b \\leq 3$\n$a = 1, b = 1$: $1 + 1 + 6 = 8 = 2^3$\nSo $n = 3$ is a solution with $(a, b, c) = (1, 1, 3)$ (and permutations).\n\n$a = 1, b = 2$: $1 + 2 + 6 = 9$\n$a = 1, b = 3$: $1 + 6 + 6 = 13$\n$a = 2, b = 2$: $2 + 2 + 6 = 10$\n$a = 2, b = 3$: $2 + 6 + 6 = 14$\n$a = 3, b = 3$: $6 + 6 + 6 = 18$\nNo other powers of $2$.\n\nIf $c = 4$:\n$4! = 24$\n$a, b \\leq 4$\n$a = 1, b = 1$: $1 + 1 + 24 = 26$\n$a = 1, b = 2$: $1 + 2 + 24 = 27$\n$a = 1, b = 3$: $1 + 6 + 24 = 31$\n$a = 1, b = 4$: $1 + 24 + 24 = 49$\n$a = 2, b = 2$: $2 + 2 + 24 = 28$\n$a = 2, b = 3$: $2 + 6 + 24 = 32 = 2^5$\nSo $n = 5$ is a solution with $(a, b, c) = (2, 3, 4)$ (and permutations).\n\n$a = 2, b = 4$: $2 + 24 + 24 = 50$\n$a = 3, b = 3$: $6 + 6 + 24 = 36$\n$a = 3, b = 4$: $6 + 24 + 24 = 54$\n$a = 4, b = 4$: $24 + 24 + 24 = 72$\n\nIf $c = 5$:\n$5! = 120$\n$a, b \\leq 5$\n$a = 1, b = 1$: $1 + 1 + 120 = 122$\n$a = 1, b = 2$: $1 + 2 + 120 = 123$\n$a = 1, b = 3$: $1 + 6 + 120 = 127$\n$a = 1, b = 4$: $1 + 24 + 120 = 145$\n$a = 1, b = 5$: $1 + 120 + 120 = 241$\n$a = 2, b = 2$: $2 + 2 + 120 = 124$\n$a = 2, b = 3$: $2 + 6 + 120 = 128 = 2^7$\nSo $n = 7$ is a solution with $(a, b, c) = (2, 3, 5)$ (and permutations).\n\n$a = 2, b = 4$: $2 + 24 + 120 = 146$\n$a = 2, b = 5$: $2 + 120 + 120 = 242$\n$a = 3, b = 3$: $6 + 6 + 120 = 132$\n$a = 3, b = 4$: $6 + 24 + 120 = 150$\n$a = 3, b = 5$: $6 + 120 + 120 = 246$\n$a = 4, b = 4$: $24 + 24 + 120 = 168$\n$a = 4, b = 5$: $24 + 120 + 120 = 264$\n$a = 5, b = 5$: $120 + 120 + 120 = 360$\n\nFor $c \\geq 6$, $c!$ is divisible by $16$, and $a! + b!$ is much smaller than $c!$, so $a! + b! + c!$ cannot be a power of $2$.\n\nTherefore, the possible values of $n$ are $2, 3, 5, 7$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 28974,
    "subject": "Mathematics (Multi-modal)",
    "question": "Tamara has written an even positive integer on a board. After that, she wrote twelve numbers consecutively, so that every number is by $5$ greater than the square of the previously written number. Determine all possible last digits of the last written number.\n(Italy 2012)",
    "options": [],
    "answer": "0 and 6",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 28975,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $a$, $b$ and $c$ be the side-lengths of a triangle with perimeter $1$. Prove that\n$$\n\\sqrt{a^2 + b^2} + \\sqrt{b^2 + c^2} + \\sqrt{c^2 + a^2} < 1 + \\frac{\\sqrt{2}}{2}.\n$$\n(APMO 2003)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Without loss of generality, we may assume $a \\ge b \\ge c$.\nThe triangle inequality and $a + b + c = 1$ imply that $a < b + c = 1 - a$, i.e. $a < \\frac{1}{2}$.\nSince $b \\le a$, it follows that $\\sqrt{a^2 + b^2} \\le \\sqrt{2a^2} = a\\sqrt{2} < \\frac{\\sqrt{2}}{2}$.\nSince $c \\le b$, it follows that $b^2 + c^2 \\le b^2 + bc < b^2 + bc + \\frac{c^2}{4} = \\left(b + \\frac{c}{2}\\right)^2$, i.e. $\\sqrt{b^2 + c^2} < b + \\frac{c}{2}$.\n\nAnalogously we conclude $\\sqrt{a^2 + c^2} < a + \\frac{c}{2}$. Summing the obtained inequalities we get\n$$\n\\sqrt{a^2 + b^2} + \\sqrt{b^2 + c^2} + \\sqrt{a^2 + c^2} < \\frac{\\sqrt{2}}{2} + \\left(b + \\frac{c}{2}\\right) + \\left(a + \\frac{c}{2}\\right) = 1 + \\frac{\\sqrt{2}}{2}.\n$$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 28976,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABCD$ be a convex quadrilateral such that $\\angle BAD = 90^\\circ$, $\\angle BAC = 2\\angle BDC$ and $\\angle DBA + \\angle DCB = 180^\\circ$. Find $\\angle DBA$.",
    "options": [],
    "answer": "45°",
    "solution": "Denote $\\angle BDC = x$ and $\\angle DBA = y$. Then $\\angle BAC = 2x$. Let $P$ and $Q$ be the intersections of the angle bisector of $\\angle BAC$ with segment $\\overline{BC}$ and ray $DC$.\n\n![](attached_image_1.png)\n\nSince $\\angle BAQ = \\angle BDQ = x$, the quadrilateral $ABQD$ is cyclic. Given that $\\angle BAD$ is a right angle, $\\angle DQB$ is a right angle as well.\n\nInscribed angles $\\angle DQA$ and $\\angle DBA$ subtending the $\\overline{AD}$ are equal, i.e. $\\angle DQA = y$, while $\\angle AQB = 90^\\circ - y$.\n\nFrom $\\angle BCQ = 180^\\circ - \\angle DCB = y = \\angle DQA$, we see that the triangle $PQC$ is isosceles and that $|PQ| = |PC|$.\n\nSince $\\angle QBP = \\angle QPC - \\angle PQB = 180^\\circ - 2y - (90^\\circ - y) = 90^\\circ - y = \\angle AQB$, triangle $PBQ$ is isosceles as well and $|PB| = |PQ|$.\n\nHence, $P$ is the midpoint of $\\overline{BC}$ and, by the converse of the angle bisector theorem applied to $\\angle BAC$, we conclude that $|AB| = |AC|$. Therefore $90^\\circ = \\angle APB = \\angle QPC = 180^\\circ - 2y$, which implies that $\\angle DBA = y = 45^\\circ$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 28977,
    "subject": "Mathematics (Multi-modal)",
    "question": "Each of hundred dots in a $10 \\times 10$ grid is coloured red or blue. A dot is called *corner-dot* if it is in the corner of the grid. A dot is called *edge-dot* if it is at the end of some row or column, but it is not a corner-dot. Two dots are *neighbouring* if they are in the same row or column, and there are no other points between them.\n\nEach pair of neighbouring dots is connected with a red, blue or green segment, so that every red segment connects two red dots, every blue segment connects two blue dots, and every green segment connects one red and one blue dot.\n\nThere are exactly $52$ red dots, $16$ of them are edge-dots, and two of them are corner-dots. The total number of green segments is $98$. Determine the number of blue segments. (Canada 2001)",
    "options": [],
    "answer": "37",
    "solution": "",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 28978,
    "subject": "Mathematics (Multi-modal)",
    "question": "On a $5 \\times 7$ board all squares are coloured white. Exactly $17$ squares need to be coloured black, so that the new arrangement of black and white squares is centrally symmetric, i.e. so that it does not change if it is rotated by $180^\\circ$ around the centre of the board. In how many ways is it possible to do that?",
    "options": [],
    "answer": "24310",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 28979,
    "subject": "Mathematics (Multi-modal)",
    "question": "Three points $A$, $B$ and $C$ are considered on a parabola. The tangents to the parabola at these points form a triangle $KLM$ ($KL$ being tangent at $A$, $KM$ at $B$, and $LM$ at $C$). The line through $B$ parallel to the axis of symmetry of the parabola intersects $AC$ at $N$. Prove that $KLMN$ is a parallelogram.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 28980,
    "subject": "Mathematics (Multi-modal)",
    "question": "Višnja decided to write all integers from $1$ to $2014$ on a board in some order. Her brother Marijan will write the absolute value of the difference of each pair of adjacent numbers, and then erase the previously written numbers. Marijan will repeat this procedure until there is only one number left on the board. Determine the largest possible value of the number that will remain on the board.\n(Estonia 2012)",
    "options": [],
    "answer": "2012",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 28981,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $\\alpha \\ge \\frac{1}{2}$ be a real number. Prove that the following inequality holds for all positive real numbers $x$, $y$, $z$:\n$$\nx(x - y)(\\alpha x - y) + y(y - z)(\\alpha y - z) + z(z - x)(\\alpha z - x) \\ge 0.\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "The given inequality can be written in the following form:\n$$\n\\alpha(x^3 + y^3 + z^3 - x^2y - y^2z - z^2x) \\geq x^2y + y^2z + z^2x - xy^2 - yz^2 - zx^2.\n$$\nFrom the rearrangement inequality it follows that the left-hand side of the previous inequality is positive. Therefore, it suffices to prove the desired inequality for $\\alpha = \\frac{1}{2}$:\n$$\nx^3 + y^3 + z^3 + 2(xy^2 + yz^2 + zx^2) \\geq 3(x^2y + y^2z + z^2x). \\quad (1)\n$$\nSince the given inequality is cyclic, without loss of generality we may assume that $z \\leq x$ and $z \\leq y$. Let $a, b \\geq 0$ be real numbers such that $x = z + a$, $y = z + b$. Therefore, the inequality (1) can be written as follows:\n$$\na^3 - 3a^2b + 2ab^2 + b^3 + 2z(a^2 - ab + b^2) \\geq 0.\n$$\nBy AM-GM inequality, we have:\n$$\na^2 - ab + b^2 \\geq ab \\geq 0.\n$$\nThus, it suffices to show:\n$$\na^3 + 2ab^2 + b^3 - 3a^2b \\geq 0.\n$$\nHowever, this can be written as follows:\n$$\n\\begin{aligned}\n& a^3 - 4a^2b + 4ab^2 + a^2b - 2ab^2 + b^3 \\geq 0, \\\\\n& a(a^2 - 4ab + 4b^2) + b(a^2 - 2ab + b^2) \\geq 0, \\\\\n& a(a - 2b)^2 + b(a - b)^2 \\geq 0,\n\\end{aligned}\n$$\nwhich obviously holds.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 28982,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $\\alpha$ be a real number. Find all functions $f: \\mathbb{R} \\to \\mathbb{R}$ such that\n$$f(x + \\alpha + f(y)) = f(f(x)) + f(\\alpha) + y,$$\nfor all $x, y \\in \\mathbb{R}$.",
    "options": [],
    "answer": "f(x) = x",
    "solution": "Letting $x = y = -\\alpha$ we get\n$$\nf(f(-\\alpha)) = f(f(-\\alpha)) + f(\\alpha) - \\alpha \\implies f(\\alpha) = \\alpha.\n$$\nLetting $x = -\\alpha, y = \\alpha$ we get\n$$\n\\alpha = f(f(\\alpha)) = f(f(-\\alpha)) + \\alpha + \\alpha \\implies f(f(-\\alpha)) = -\\alpha.\n$$\nNow let us denote $f(-\\alpha) = a$. We know that $f(a) = -\\alpha$. Letting $x = \\alpha, y = a$ we get\n$$\n\\alpha = 2\\alpha + a \\implies a = -\\alpha.\n$$\nFinally, letting $y = -\\alpha$ gives us\n$$\nf(x) = f(f(x)), \\quad \\forall x \\in \\mathbb{R},\n$$\nand letting $x = -\\alpha$ gives us\n$$\nf(f(y)) = y, \\quad \\forall y \\in \\mathbb{R}.\n$$\nIt follows that\n$$\nf(x) = f(f(x)) = x, \\quad \\forall x \\in \\mathbb{R}.\n$$\nIt is easy to check that the function $f(x) = x$ really is a solution.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 28983,
    "subject": "Mathematics (Multi-modal)",
    "question": "Prove that for all $x \\in [\\frac{1}{111}, \\frac{110}{111}]$ there exist numbers $a_i \\in \\{-1, 1\\}, i = 1, 2, \\dots, 101$ such that\n$$\n|x_{101} - x| \\le \\frac{1}{402},\n$$\nwhere\n$x_0 = 1, \\quad x_k = (x_{k-1} + 1)^{a_k}, \\quad \\text{for } k = 1, 2, \\dots, 101.$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "For a given positive integer $n$ let $S_n$ denote the set of all possible values that $x_n$ can obtain for different choices of numbers $a_i$, $1 \\le i \\le n$. For example:\n$$\nS_1 = \\{\\frac{1}{2}, 2\\}, \\quad S_2 = \\{\\frac{1}{3}, \\frac{2}{3}, \\frac{3}{2}, 3\\}, \\quad S_3 = \\{\\frac{1}{4}, \\frac{2}{5}, \\frac{3}{5}, \\frac{3}{4}, \\frac{4}{3}, \\frac{5}{3}, \\frac{5}{2}, 4\\}, \\dots\n$$\nNote that, for every $x \\in S_n$ both $x+1$ and $\\frac{1}{x+1}$ belong to $S_{n+1}$. Furthermore, one of those two numbers is smaller than 1 and the other is greater than 1. Thus, each $S_n$ consists of even number of numbers. For a given positive integer $n$ let\n$$\nS_n = \\{a_1, a_2, \\dots, a_{2m}\\}, \\text{ with } a_1 < a_2 < \\dots < a_{2m}\n$$\nand\n$$\nS_{n+1} = \\{b_1, b_2, \\dots, b_{2k}\\}, \\text{ with } b_1 < b_2 < \\dots < b_{2k}.\n$$\nThe following statements can easily be proved by induction:\n* Claim 1. $m = 2^{n-1}$, $k = 2^n$ and thus $|S_{n+1}| = 2|S_n|$.\n* Claim 2. $a_1 = \\frac{1}{n+1}$, $a_m = \\frac{n}{n+1}$, $a_{m+1} = \\frac{n+1}{n}$, $a_{2m} = n+1$.\n* Claim 3. $b_1 < b_2 < \\dots < b_{2m} < 1 < b_{2m+1} < \\dots < b_{4m}$.\n* Claim 4. $a_i = \\frac{1}{a_{2m+1-i}}$, $b_i = \\frac{1}{b_{4m+1-i}}$.\n* Claim 5. $b_i = \\frac{1}{1 + a_{2m+1-i}}$, for $1 \\le i \\le 2m$.\n* Claim 6. $b_i = \\frac{1}{b_{2m+1-i}}$, for $1 \\le i \\le 2m$.\nCombining Claim 5 and Claim 4 we get:\n$$\nb_i = \\frac{a_i}{1 + a_i}, \\quad \\text{for } 1 \\le i \\le 2m. \\qquad (2)\n$$\nFor all integers $n \\ge 2$, we will prove that $a_{i+1} - a_i \\le \\frac{1}{2n-1}$ for all $1 \\le i \\le m$. That is obviously true for $n=2$.\nAs an induction hypothesis, let us assume that this holds for $n$. We will prove that the claim holds for $n+1$, i.e. we will prove that $b_{i+1} - b_i \\le \\frac{1}{2n+1}$ for all $1 \\le i \\le 2m$.\nFrom (2) and Claim 2 we know that\n$$\nb_m = \\frac{n}{2n+1}, \\quad b_{m+1} = \\frac{n+1}{2n+1}.\n$$\nTherefore,\n$$\nb_{m+1} - b_m = \\frac{1}{2n+1},\n$$\nthus the claim holds for $i = m$. Let us prove that the claim holds for $i < m$. We have:\n$$\nb_{i+1} - b_i = \\frac{1}{a_{i+1} + 1} - \\frac{1}{a_i + 1} = \\frac{a_{i+1} - a_i}{(1+a_i)(1+a_{i+1})}.\n$$\nBy induction hypothesis, $a_{i+1} - a_i \\le \\frac{1}{2n-1}$ and from Claim 2 we know that $a_i \\ge \\frac{1}{n+1}$ and $a_{i+1} \\ge \\frac{1}{n+1}$. Thus,\n$$\nb_{i+1} - b_i = \\frac{a_{i+1} - a_i}{(1+a_i)(1+a_{i+1})} \\le \\frac{\\frac{1}{2n-1}}{(1+\\frac{1}{n+1})(1+\\frac{1}{n+1})} < \\frac{1}{2n+1},\n$$\nwith the last inequality being equivalent to $2n^2 - 5 > 0$ which holds for $n \\ge 2$. Finally, we will prove that $b_{i+1} - b_i \\le \\frac{1}{2n+1}$ for $m < i < 2m$.\nFrom Claim 6 we have:\n$$\nb_{i+1} - b_i = \\frac{1}{b_j} - \\frac{1}{b_{j+1}} = \\frac{b_{j+1} - b_j}{(1+b_j)(1+b_{j+1})} < b_{j+1} - b_j < \\frac{1}{2n+1}.\n$$\nwhere we denoted $j = 2m - i < m$.\nFor $n = 101$, let $S_{101} = \\{c_1, c_2, \\dots, c_s\\}$, where $c_1 < c_2 < \\dots < c_s$.\nSince $c_1 = \\frac{1}{102}$ and $c_s = \\frac{101}{102}$ we know that $c_1 - \\frac{1}{111} < \\frac{1}{201}$ and $\\frac{110}{111} - c_s < \\frac{1}{201}$.\nTherefore, in set $S'_n = \\{c_0, c_1, \\dots, c_s, c_{s+1}\\}$ with $c_0 = \\frac{1}{111}$ and $c_{s+1} = \\frac{110}{111}$ the following holds:\n$$\nc_{i+1} - c_i \\le \\frac{1}{201}, \\text{ for } 0 < i \\le s.\n$$\nFor given $x \\in [\\frac{1}{111}, \\frac{110}{111}]$, let $j$ be an index such that $c_j \\le x \\le c_{j+1}$. Since, $c_{j+1} - c_j \\le \\frac{1}{201}$ we conclude that $|x - c_j| \\le \\frac{1}{402}$ or $|x - c_{j+1}| \\le \\frac{1}{402}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 28984,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $N \\ge 3$ be an odd positive integer. At the beginning in each square of an $N \\times N$ board there is number $0$. In one move one can choose two squares with a common side and increase or decrease by $1$ the numbers in those two squares. If after $K$ moves, the sums of numbers in every row and every column are all equal, show that $K$ is even. (USSR)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "In each move the sum of all numbers increases or decreases by two, so the sum of all numbers remains even. Let us assume that after $K$ moves the sums of numbers in each row and each column is equal and let us denote it by $S$. The sum of all numbers is then equal to $N \\cdot S$. Since this must be even and $N$ is odd, we conclude that $S$ must be even.\n\nNote that in each move we change the parity of the sum of two adjacent rows or two adjacent columns. First we consider only the moves that change the parity of two adjacent columns. Let $A_i$ denote the number of moves that change the numbers in two squares in $i$-th and $(i+1)$-st columns.\n\nNumber $A_1$ must be even because these are the only moves that will change the parity of the sum in the first column. The number of moves that will change the parity of the sum in the second column also has to be even and it is $A_1 + A_2$, hence $A_2$ is even. Inductively, we conclude that every $A_i$ is even.\n\nHence the total number of moves that will change the parity of two columns is $A_1 + A_2 + \\dots + A_{n-1}$, which is an even number. Analogously, the total number of moves that will change the parity of two rows is an even number. We conclude that the total number of moves is also even.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 28985,
    "subject": "Mathematics (Multi-modal)",
    "question": "A group of twenty children has $100$ ribbons. Each ribbon is being held at its ends by two children. Two children may hold together at most one ribbon. Let us assume that a pair of ribbons such that its four ends hold different children can be chosen in $4050$ ways. Prove that each child holds the same number of ribbons. (Iran 2003)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "We enumerate the children by numbers $1, 2, \\dots, 20$. Let us denote by $d_i$ the number of ribbons that are being held by the child with number $i$. Since each ribbon is being held by two children we have that\n$$\n\\sum_{i=1}^{20} d_i = 2 \\cdot 100 = 200.\n$$\nThe number of all pairs of ribbons is $\\frac{100 \\cdot 99}{2} = 4950$.\nA pair of ribbons such that its four ends are not being held by different children is given by a pair of ribbons such that there is a child who holds one end of each of the two ribbons in that pair.\nHence\n$$\n\\sum_{i=1}^{20} \\frac{d_i \\cdot (d_i - 1)}{2} = 4950 - 4050 = 900.\n$$\nFrom here it follows that\n$$\n\\sum_{i=1}^{20} d_i^2 = 1800 + \\sum_{i=1}^{20} d_i = 1800 + 200 = 2000.\n$$\nBy the AM–QM inequality we have\n$$\n10 = \\frac{d_1 + d_2 + \\dots + d_{20}}{20} \\le \\sqrt{\\frac{d_1^2 + d_2^2 + \\dots + d_{20}^2}{20}} = 10.\n$$\nSince the equality in AM–QM inequality is obtained if and only if all $d_i$ are equal, the claim follows.",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 28986,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABC$ be a triangle such that $|AB| > |AC|$. Let $P$ be the midpoint of the segment $\\overline{BC}$ and $S$ the intersection of the bisector of the angle $\\angle BAC$ and the segment $\\overline{BC}$. The line parallel to the line $AS$, which passes through $P$, intersects the lines $AB$ and $AC$ in points $X$ and $Y$ respectively. Let $Z$ be a point such that $Y$ is the midpoint of the segment $XZ$, and that lines $BY$ and $CZ$ intersect at $D$.\nProve that the bisector of the angle $\\angle BDC$ is parallel to the line $AS$.\n(D. Monk, New Problems in Euclidean Geometry)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Since $|BP| = |CP|$ and\n$$\n\\frac{|BX|}{|BP|} = (\\text{similarity}) = \\frac{|BA|}{|BS|} = (\\text{angle bisector}) = \\frac{|CA|}{|CS|} = (\\text{similarity}) = \\frac{|CY|}{|CP|},\n$$\nwe conclude that $|BX| = |CY|$.\n![](attached_image_1.png)\nNote that $\\triangle XYA = \\triangle SAC = \\triangle BAS = \\triangle BXP = \\triangle AXY$, and therefore $\\triangle YXB = \\triangle CYZ$.\nSince $|XY| = |YZ|$ and $|BX| = |CY|$, the SAS congruence theorem implies that the triangles $BXY$ and $CYZ$ are congruent.\nHence $\\triangle BYP = \\triangle PZC$, i.e. $\\triangle DYZ = \\triangle YZD$. Since $\\triangle DYZ + \\triangle YZD = \\triangle BDC$, it follows that the bisector of the angle $\\triangle BDC$ is parallel to the line $ZP$, and hence to the line $AS$ as well.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 28987,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $I$ be the incentre of the acute triangle $ABC$. Rays $AI$ and $BI$ intersect the circumcircle $k$ of the triangle $ABC$ in points $D$ and $E$ respectively. Segments $DE$ and $CA$ intersect in point $F$, line through point $E$ parallel to the line $FI$ intersects circle $k$ also in point $G$, and lines $FI$ and $DG$ intersect in point $H$. Prove that the lines $CA$ and $BH$ touch the circumcircle of the triangle $DFH$ at points $F$ and $H$ respectively.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let us denote $\\angle BAC = \\alpha$, $\\angle CBA = \\beta$, $\\angle ACB = \\gamma$, and let $J$ be the intersection of the segments $\\overline{DE}$ and $\\overline{CB}$.\n![](attached_image_1.png)\nInscribed angles $\\angle BED$, $\\angle BCD$ and $\\angle BAD$ over the chord $BD$ are equal, so $\\angle BED = \\angle BCD = \\angle BAD = \\frac{\\alpha}{2}$. Analogously, $\\angle DEC = \\angle DBC = \\angle DAC = \\frac{\\alpha}{2}$, $\\angle ECA = \\angle EBA = \\frac{\\beta}{2}$ and $\\angle CDE = \\angle CBE = \\frac{\\beta}{2}$.\nSince $\\angle DFC = \\angle DEC + \\angle ECA = \\frac{\\alpha}{2} + \\frac{\\beta}{2}$ and $\\angle CJE = \\angle CDE + \\angle BCD = \\frac{\\beta}{2} + \\frac{\\alpha}{2}$, the triangle $CFJ$ is isosceles.\nOn the other hand, lines $FH$ and $EG$ are parallel, thus $\\angle DHF = \\angle DGE = 180^\\circ - \\angle ECD = 180^\\circ - (\\angle ECA + \\angle ACB + \\angle BCD) = 180^\\circ - (\\frac{\\beta}{2} + \\gamma + \\frac{\\alpha}{2}) = \\frac{\\alpha}{2} + \\frac{\\beta}{2} =$\n\n$<DFC$ and, by the converse of the tangent chord theorem, we conclude that the line $CA$ touches the circumcircle of the triangle $DFH$ at point $F$.\nSince line $CI$ is the angle bisector of $<FCJ$, we have $CI \\perp DE$. Since $<BED$ = $<DEC$, line $DE$ is the bisector of the segment $\\overline{CI}$ and $<IFD$ = $<DFC$ = $\\frac{\\alpha}{2} + \\frac{\\beta}{2}$. This implies that $<HFD$ = $<DFC$ = $<CJE$ = $<BJD$. We conclude that the lines $FH$ and $BC$ are parallel. From $<FHD$ = $<GED$ = $\\frac{\\alpha+\\beta}{2}$ = $<HFD$ we conclude that the triangle $DFH$ is isosceles.\nSince $D$ is the midpoint of the arc $\\overarc{BC}$ and triangles $DCB$ and $DFH$ are isosceles with $BC$ parallel to $FH$, the line $BH$ is symmetric to the line $CF$ with respect to the bisector of the segment $\\overline{FH}$. Hence, the line $BH$ also touches the circumcircle of the triangle $DFH$, at point $H$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 28988,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABC$ be an acute triangle such that $|AC| > |BC|$. Let $H$ be the orthocentre of that triangle, $N$ the foot of the altitude from $B$, and $P$ the midpoint of the side $AB$. The circumcircles of the triangles $ABC$ and $CHN$ intersect in $C$ and $D$. Prove that the points $B, D, N$ and $P$ lie on the same circle.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Denote $\\angle BAC = \\alpha$.\n![](attached_image_1.png)\nSince $ABN$ is a right-angled triangle and $P$ is the midpoint of its hypotenuse, we have $\\angle BPN = 2\\alpha$. On the other hand,\n$$\n\\begin{align*}\n\\angle NDB &= \\angle CDB - \\angle CDN \\\\\n&= (180^\\circ - \\angle BAC) - \\angle CHN \\quad (\\text{cyclic quadrilaterals } ABDC \\text{ and } NHDC) \\\\\n&= (180^\\circ - \\alpha) - \\alpha = 180^\\circ - 2\\alpha.\n\\end{align*}\n$$\nHence $\\angle BPN + \\angle NDB = 180^\\circ$, so the quadrilateral $BDNP$ is cyclic.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 28989,
    "subject": "Mathematics (Multi-modal)",
    "question": "In an acute triangle $ABC$ such that $|AC| < |BC|$, points $M$ and $N$ are respectively the feet of altitudes from vertices $A$ and $B$. The circumcircle of the triangle $ABC$ with the centre $O$ and the circumcircle of the triangle $MNC$ with the centre $S$ intersect in points $C$ and $D$. If the point $P$ is the midpoint of the segment $\\overline{AB}$, prove that points $P$, $O$, $S$ and $D$ lie on the same circle. (Stipe Vidak)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Firstly note that the circumcircle of the triangle $MNC$ passes through the ortho-centre $H$ of the triangle $ABC$. The segment $CH$ is the diameter of that circle. Since the segment $CD$ is the common chord of circumcircles of triangles $ABC$ and $MNC$, the line $CD$ is perpendicular to the line $SO$ through their centres.\n\n![](attached_image_1.png)\n\nSince $\\angle HDC = 90^\\circ$, we have $DH \\parallel SO$.\nIt is well-known that $|CH| = 2|OP|$, so $|SH| = |OP|$. Since $CH \\parallel OP$ (both lines are perpendicular to $AB$), it follows that $SHPO$ is a parallelogram, so $SO \\parallel HP$.\nWe can conclude that $DH \\parallel HP$, and that means that the points $D$, $H$ and $P$ are collinear.\nSince $|OP| = |SH| = |SD|$, the quadrilateral $DPOS$ is an isosceles trapezium, hence it is inscribed in a circle.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 28990,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all prime numbers $p$ and $q$ such that\n$$\np^{q+1} + q^{p+1}\n$$\nis a perfect square.",
    "options": [],
    "answer": "(p, q) = (2, 2)",
    "solution": "If $p = q = 2$ we have $p^{q+1} + q^{p+1} = 2^3 + 2^3 = 16$, so that is one solution of the problem.\n\nNow, without loss of generality assume that $p$ is odd and let $p^{q+1} + q^{p+1} = x^2$, where $x \\in \\mathbb{N}$. Thus $p+1$ is even and\n$$\np^{q+1} = \\left(x - q^{\\frac{p+1}{2}}\\right) \\left(x + q^{\\frac{p+1}{2}}\\right).\n$$\nDenote the greatest common divisor of the two factors on the right-hand side by $d$. If $d > 1$ we have that $d$ is a power of $p$ and $d$ is also a factor of $2x$. It is easy to see that $p \\mid q$, i.e. $p = q$. Thus $2p^{p+1} = x^2$, which is impossible.\n\nTherefore $d = 1$, $x - q^{\\frac{p+1}{2}} = 1$, $x + q^{\\frac{p+1}{2}} = p^{q+1}$ and hence\n$$\n2q^{\\frac{p+1}{2}} = p^{q+1} - 1.\n$$\nThe above equation is impossible for $q$ odd because $2q^{\\frac{p+1}{2}} \\equiv 2 \\pmod 4$ and $p^{q+1} - 1 \\equiv 0 \\pmod 4$. We conclude that $q = 2$ and\n$$\n2^{\\frac{p+3}{2}} = p^3 - 1 = (p-1)(p^2 + p + 1).\n$$\nSince $\\gcd(p-1, p^2 + p + 1) = 1$ we conclude that $p-1=1$, what is a contradiction.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 28991,
    "subject": "Mathematics (Multi-modal)",
    "question": "Prove that there exist infinitely many positive integers $n$ such that the largest prime divisor of $n^4 + n^2 + 1$ is equal to the largest prime divisor of $(n+1)^4 + (n+1)^2 + 1$. (IMO Shortlist 2013)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $p_n$ be the largest prime divisor of $n^4 + n^2 + 1$ and let $q_n$ be the largest prime divisor of $n^2 + n + 1$. Then $p_n = q_{n^2}$, and from\n$$\nn^4 + n^2 + 1 = (n^2 + 1)^2 - n^2 = (n^2 - n + 1)(n^2 + n + 1) = ((n-1)^2 + (n-1) + 1)(n^2 + n + 1)\n$$\nit follows that $p_n = \\max\\{q_n, q_{n-1}\\}$ for $n \\ge 2$. Keeping in mind that $n^2 - n + 1$ is odd, we have\n$$\n\\gcd(n^2 + n + 1, n^2 - n + 1) = \\gcd(2n, n^2 - n + 1) = \\gcd(n, n^2 - n + 1) = 1.\n$$\nTherefore, $q_n \\ne q_{n-1}$.\nTo prove the result, it suffices to show that the set\n$$\nS = \\{n \\in \\mathbb{Z}_{\\ge 2} \\mid q_n > q_{n-1} \\text{ and } q_n > q_{n+1}\\}\n$$\nis infinite, since for each $n \\in S$ one has\n$$\np_n = \\max\\{q_n, q_{n-1}\\} = q_n = \\max\\{q_n, q_{n+1}\\} = p_{n+1}.\n$$\nSuppose on the contrary that $S$ is finite. Since $q_2 = 7 < 13 = q_3$ and $q_3 = 13 > 7 = q_4$, the set $S$ is non-empty. Since it is finite, we can consider its largest element, say $m$.\nNote that it is impossible that $q_m > q_{m+1} > q_{m+2} > \\dots$ because all these numbers are positive integers, so there exists a $k \\ge m$ such that $q_k < q_{k+1}$ (recall that $q_k \\ne q_{k+1}$).\nNext observe that it is impossible to have $q_k < q_{k+1} < q_{k+2} < \\dots$, because\n$$\nq_{(k+1)^2} = p_{k+1} = \\max\\{q_k, q_{k+1}\\} = q_{k+1},\n$$\nso let us take the smallest $l \\ge k+1$ such that $q_l > q_{l+1}$. By the minimality of $l$ we have $q_{l-1} < q_l$, so $l \\in S$. Since $l \\ge k+1 > k \\ge m$, this contradicts the maximality of $m$, and hence $S$ is indeed infinite.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 28992,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $a$, $b$, $c$ be distinct positive integers and let $r$, $s$, $t$ be positive integers such that:\n$$\nab + 1 = r^2, \\quad ac + 1 = s^2, \\quad bc + 1 = t^2.$$\nProve that it is not possible that all three fractions $\\frac{rs}{t}$, $\\frac{rt}{s}$ and $\\frac{st}{r}$ are positive integers. (Miljen Mikić)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Assume on the contrary that $\\frac{rs}{t}$, $\\frac{rt}{s}$, $\\frac{st}{r}$ are positive integers.\nWithout loss of generality we may assume that $a < b < c$. Since $\\frac{rs}{t}$ is a positive integer, $\\frac{r^2 s^2}{t^2}$ is also a positive integer, and hence\n$$\n\\frac{r^2 s^2}{t^2} = \\frac{a^2 bc + ab + ac + 1}{bc + 1} = a^2 + \\frac{ab + ac + 1 - a^2}{bc + 1}\n$$\nis also a positive integer. Since $a < b < c$, we have\n$$\nab + ac + 1 - a^2 > ac + 1 = s^2 > 0.\n$$\nTherefore, both the numerator and the denominator of the fraction $\\frac{ab + ac + 1 - a^2}{bc + 1}$ are positive, so\n$$\nab + ac + 1 - a^2 \\geq bc + 1 \\implies (b-a)(c-a) \\leq 0,\n$$\nwhich is a contradiction.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 28993,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $n$ be an odd positive integer greater than $3$. Let $k$ denote the least positive integer such that $kn+1$ is a perfect square and $l$ denote the least positive integer such that $ln$ is a perfect square.\nProve that $n$ is a prime number if and only if $k > \\frac{1}{4}n$ and $l > \\frac{1}{4}n$.\n(V.I. Bernik, I.K. Zhuk, O.V. Melnikov, Collection of Problems in Math. Olympiads)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "If $n = p$ is a prime number, then $l = p$ so it is obvious that $l > \\frac{n}{4}$. For $k$ we have: $kp = (y - 1)(y + 1)$, so\n$$\n\\begin{cases} y - 1 = k_1, & y + 1 = k_2p, \\\\ y - 1 = k_1p, & y + 1 = k_2. \\end{cases} \n\\quad \\text{or}\n$$\nIn each case, adding these two equations gives us $2y > p$. If $4k \\le p$ (i.e. $4k < p$ since $p$ is prime), then it would be $y^2 = kp + 1 < \\frac{p^2}{4} + 1$, i.e. $p^2 < 4y^2 < p^2 + 4$. This is not possible since there are no perfect squares between $p^2$ and $p^2 + 4$.\nWe will prove that there does not exist an odd composite number $n > 3$ such that $4l > n$ and $4k > n$. Assume on the contrary that such $n$ exists. We can write $n$ in the form\n$$\nn = x^2 \\cdot p_1 \\cdots p_s\n$$\nwith $p_1, \\dots, p_s$ different prime numbers. Clearly, $l = p_1 \\cdots p_s$ and we have\n$$\n4p_1 \\cdots p_s = 4l > n = x^2p_1 \\cdots p_s.\n$$\nHence $x^2 < 4$, so $x = 1$, i.e.\n$$\nn = p_1 \\cdots p_s.\n$$\nLet $y$ denote the smallest number greater than $1$ such that $y^2 - 1$ is divisible by $n$ (such $y$ exists since $(n+1)^2 - 1 = n(n+2)$ is divisible by $n$). Clearly $kn + 1 = y^2$, and it is easy to see that $k > \\frac{n}{4} \\iff 2y > n$.\nNow we can write $n$ in the form $n = pr$, where $p = p_i$ for some $i$, and $r$ is the product of all $p_j$ for $j \\neq i$. Since $n$ is a composite number it follows that $r > 1$.\nBy the Chinese remainder theorem there exists a unique $T$, $0 \\le T < n$, such that\n$$\nT \\equiv 1 \\pmod{r},\n$$\n$$\nT \\equiv -1 \\pmod{p}.\n$$\nNow look at the number $S = n - T$. We have:\n$$\nS \\equiv -1 \\pmod{r},\n$$\n$$\nS \\equiv 1 \\pmod{p},\n$$\nso $T^2 \\equiv S^2 \\equiv 1 \\pmod{n}$. Therefore, both $T$ and $S$ are candidates for $y$. Clearly, one of them is less than $\\frac{n}{2}$, so $k < \\frac{n}{4}$, which is a contradiction.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 28994,
    "subject": "Mathematics (Multi-modal)",
    "question": "Ten chairs are arranged around a round table and marked with numbers $1$ to $10$ successively (in such a way that chairs $1$ and $10$ are also adjacent), and a knight is sitting in each chair. In the beginning, every knight has an even number of coins. Simultaneously, each knight gives half of his coins to his left neighbour, and the other half to his right neighbour. After that, the knight sitting in chair $1$ has $22$ coins, and each succeeding knight has two more coins, up until the knight in chair $10$ that has $40$ coins.\nHow many coins did the knight that ended up with $36$ coins have in the beginning?\n(Hong Kong)",
    "options": [],
    "answer": "46",
    "solution": "Let us denote by $2x_1, 2x_2, \\dots, 2x_{10}$ the number of coins that the knights sitting in chairs $1, 2, \\dots, 10$ had in the beginning, respectively. We have to determine $2x_8$.\n\nWe have a system of equations: $x_{10} + x_2 = 22$, $x_1 + x_3 = 24$, $x_2 + x_4 = 26$, $x_3 + x_5 = 28$, $\\dots$, $x_8 + x_{10} = 38$, $x_9 + x_1 = 40$.\n\nBy combining those equations we get\n$$\n\\begin{aligned}\nx_{10} &= 38 - x_8, \\\\\nx_2 &= 22 - x_{10} = 22 - (38 - x_8) = x_8 - 16, \\\\\nx_4 &= 26 - x_2 = 26 - (x_8 - 16) = 42 - x_8, \\\\\nx_6 &= 30 - x_4 = 30 - (42 - x_8) = x_8 - 12, \\\\\nx_8 &= 34 - x_6 = 34 - (x_8 - 12) = 46 - x_8.\n\\end{aligned}\n$$\nFrom the last equation it follows that $2x_8 = 46$, so the knight that ended up with $36$ coins had $46$ coins in the beginning.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 28995,
    "subject": "Mathematics (Multi-modal)",
    "question": "Determine positive integer $n$ such that the sum of his two smallest divisors is $6$ and the sum of his two largest divisors is $1122$.",
    "options": [],
    "answer": "935",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 28996,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $x$ and $y$ be distinct real numbers such that $2xy + 1 \\neq 0$, and let\n$$\nA = \\frac{6x^2y^2 + xy - 1}{2xy + 1} \\quad \\text{and} \\quad B = \\frac{x(x^2 - 1) - y(y^2 - 1)}{x - y}.\n$$\nDetermine which number is larger, $A$ or $B$.",
    "options": [],
    "answer": "B > A",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 28997,
    "subject": "Mathematics (Multi-modal)",
    "question": "Determine positive integer $n$ such that the sum of his two smallest divisors is $6$ and the sum of his two largest divisors is $1122$.",
    "options": [],
    "answer": "935",
    "solution": "**1.1.** Let us denote by $2x_1, 2x_2, \\dots, 2x_{10}$ the number of coins that the knights sitting in chairs 1, 2, ..., 10 had in the beginning, respectively. We have to determine $2x_8$.\nWe have a system of equations: $x_{10}+x_2 = 22, x_1+x_3 = 24, x_2+x_4 = 26, x_3+x_5 = 28, \\dots, x_8+x_{10} = 38, x_9+x_1 = 40$.\nBy combining those equations we get\n$$\n\\begin{aligned}\nx_{10} &= 38 - x_8, \\\\\nx_2 &= 22 - x_{10} = 22 - (38 - x_8) = x_8 - 16, \\\\\nx_4 &= 26 - x_2 = 26 - (x_8 - 16) = 42 - x_8, \\\\\nx_6 &= 30 - x_4 = 30 - (42 - x_8) = x_8 - 12, \\\\\nx_8 &= 34 - x_6 = 34 - (x_8 - 12) = 46 - x_8.\n\\end{aligned}\n$$\nFrom the last equation it follows that $2x_8 = 46$, so the knight that ended up with 36 coins had 46 coins in the beginning.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 28998,
    "subject": "Mathematics (Multi-modal)",
    "question": "Prove that there does not exist a positive integer $n$ such that $7^n - 1$ is divisible by $6^n - 1$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Assume that such $n$ exists.\nNote that $5$ divides $6^n - 1 = (6-1)(6^{n-1} + \\dots + 1)$, so $5$ also has to divide $7^n - 1$.\nThe powers of $7$ when divided by $5$ give remainders $2, 4, 3, 1, \\dots$ and these remainders repeat periodically. Therefore, $7^n - 1$ will be divisible by $5$ if and only if $n$ is divisible by $4$, i.e. $n = 4k$.\nBy the formula for the difference of $k$th powers, we can conclude that $6^4 - 1$ divides $6^{4k} - 1$. Moreover, $7$ divides $6^4 - 1 = (6^2 - 1)(6^2 + 1) = 35 \\cdot 37 = 5 \\cdot 7 \\cdot 37$. Therefore, $7$ divides $6^n - 1$, so it also divides $7^n - 1$, which is a contradiction.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 28999,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $a$ and $b$ be positive real numbers such that $a^3 + b^3 = 2ab(a + b)$.\nDetermine $\\frac{a^2}{b^2} + \\frac{b^2}{a^2}$. (Ratko Višak)",
    "options": [],
    "answer": "7",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 29000,
    "subject": "Mathematics (Multi-modal)",
    "question": "Positive integers $a$, $b$ and prime number $p$ satisfy the equation $a^2 + p^2 = b^2$.\nProve that $2(b + p)$ is a perfect square.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "We are given $a^2 + p^2 = b^2$, where $a$, $b$ are positive integers and $p$ is a prime number.\n\nRewrite the equation:\n$$\n b^2 - a^2 = p^2 \n$$\n$$\n (b - a)(b + a) = p^2 \n$$\n\nSince $p$ is prime, $p^2$ has only three positive divisors: $1$, $p$, $p^2$.\nSo, the pairs $(b - a, b + a)$ must be $(1, p^2)$, $(p, p)$, or $(p^2, 1)$.\nBut $b + a > b - a$, so only $(b - a, b + a) = (p, p)$ or $(1, p^2)$ are possible.\n\nCase 1: $b - a = p$, $b + a = p$\n\nThen $b - a = b + a$ implies $a = 0$, which is not allowed since $a$ is positive.\n\nCase 2: $b - a = 1$, $b + a = p^2$\n\nThen:\n$$\n b - a = 1 \\implies b = a + 1 \n$$\n$$\n b + a = p^2 \n$$\nAdd the two equations:\n$$\n (b - a) + (b + a) = 1 + p^2 = 2b \\implies 2b = p^2 + 1 \\implies b = \\frac{p^2 + 1}{2} \n$$\nSubtract the first from the second:\n$$\n (b + a) - (b - a) = p^2 - 1 = 2a \\implies a = \\frac{p^2 - 1}{2} \n$$\n\nNow, $a$ and $b$ are positive integers for any odd prime $p$ (since $p^2$ is odd, so $p^2 + 1$ and $p^2 - 1$ are even).\n\nNow, compute $2(b + p)$:\n$$\n 2(b + p) = 2 \\left( \\frac{p^2 + 1}{2} + p \\right ) = (p^2 + 1) + 2p = p^2 + 2p + 1 = (p + 1)^2 \n$$\n\nThus, $2(b + p)$ is a perfect square, as required.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 29001,
    "subject": "Mathematics (Multi-modal)",
    "question": "Determine how many 6-digit positive integers are there such that by removing its first two digits and its last two digits we get two 4-digit numbers, which give the same remainder when divided by $99$.",
    "options": [],
    "answer": "8190",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 29002,
    "subject": "Mathematics (Multi-modal)",
    "question": "The area of the intersection of a smaller and larger square equals two thirds of the area of the smaller square, as well as one fifth of the area of their union. Determine the ratio of the sides of the smaller and larger square.\n(Andrea Aglić-Aljinović)",
    "options": [],
    "answer": "sqrt(3)",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 29003,
    "subject": "Mathematics (Multi-modal)",
    "question": "The first $n$ positive integers are written on a board ($n \\ge 3$). Ante repeats the following procedure: first he chooses two numbers on the board, and then he increases them both by the same arbitrary positive integer.\nDetermine all positive integers $n$ such that Ante can, by repeating this procedure, achieve that all numbers on the board are equal.\n(Ilko Brnetić)",
    "options": [],
    "answer": "All positive integers n that are not congruent to 2 modulo 4; equivalently, all odd n and all n divisible by 4.",
    "solution": "Assume $n = 4k$. Then Ante can achieve that all numbers on the board are equal in the following way: he will increase by $1$ the numbers $1$ and $3$, $5$ and $7$, ..., $4k-3$ and $4k-1$. By doing that, he gets that the numbers on the board are all even numbers smaller than or equal to $n$, and each is written twice. Finally, he increases $2$ and $2$ by $n-2$, $4$ and $4$ by $n-4$, ..., $n-2$ and $n-2$ by $2$, and he gets that all numbers on the board are equal to $n$.\n\nAssume $n = 2k+1$. Then Ante can achieve that all numbers on the board are equal in the following way: he will increase by $1$ the numbers $1$ and $n$, $3$ and $n$, ..., $n-2$ and $n$. By doing that, he gets that the numbers on the board are all even numbers smaller than or equal to $n$, each written twice, and the number $\\frac{3n-1}{2}$. Finally, he increases $2$ and $2$ by $\\frac{3n-5}{2}$, $4$ and $4$ by $\\frac{3n-9}{2}$, ..., $n-1$ and $n-1$ by $\\frac{n+1}{2}$. Now all numbers on the board are equal to $\\frac{3n-1}{2}$.\n\nAssume $n = 4k + 2$. Then Ante cannot achieve that all numbers are equal. The sum of all numbers on the board is initially odd, because\n$$\n1 + 2 + \\dots + n = \\frac{n(n+1)}{2} = (2k+1)(4k+3).\n$$\nSince there is an even number of numbers on the board, if they were equal their sum would be an even number. On the other hand, in each step the sum of numbers on the board is increased by an even number, so the sum will never be even.\n\nTherefore, Ante can achieve that all numbers are equal if and only if $n$ is not of the form $4k + 2$, $k \\in \\mathbb{N}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 29004,
    "subject": "Mathematics (Multi-modal)",
    "question": "There are $100$ circles in the plane, all having the same centre with radii $1$, $2$, $\\ldots$, $100$. The smallest circle is coloured in red and each of the $99$ annuli bounded by two circles is coloured in red or green, such that adjacent regions are of different colour. Determine the total area of green regions.",
    "options": [],
    "answer": "5050π",
    "solution": "",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 29005,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $\\overline{AC}$ be the diameter of the circle $k_1$ with the centre $B$. Circle $k_2$ touches the line $AC$ at the point $B$ and the circle $k_1$ at the point $D$. Tangent from $A$ (different from $AC$) to circle $k_2$ touches that circle at the point $E$ and intersects the line $BD$ in the point $F$. Determine the ratio $|AF| : |AB|$. (Hong Kong)",
    "options": [],
    "answer": "5:3",
    "solution": "",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 29006,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $I$ be the incentre of the acute triangle $ABC$ and let $|AC| > |BC|$. The angle bisector and the altitude from vertex $C$ close an angle of $10^\\circ$. If $\\angle AIB = 120^\\circ$, determine the angles of the triangle $ABC$. (Ilko Brnetić)",
    "options": [],
    "answer": "A = 50°, B = 70°, C = 60°",
    "solution": "",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 29007,
    "subject": "Mathematics (Multi-modal)",
    "question": "Circles $k_1$ and $k_2$ intersect in points $A$ and $B$. Line $l$ intersects circle $k_1$ in points $C$ and $E$, and circle $k_2$ in points $D$ and $F$ in such a way that $D$ is between $C$ and $E$, and $E$ is between $D$ and $F$. Lines $CA$ and $BF$ intersect in point $G$, and lines $DA$ and $BE$ in point $H$.\nProve that $CF \\parallel HG$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "It suffices to show that $\\angle ECA = \\angle HGA$.\nSince the quadrilateral $ACBE$ is cyclic, we have $\\angle ECA = \\angle EBA$, so it suffices to show that $ABGH$ is a cyclic quadrilateral.\nFrom triangle $DEH$ we have $\\angle DHE = 180^\\circ - \\angle EDH - \\angle HED$, i.e.\n$$\n\\angle AHB = \\angle DHE = 180^\\circ - \\angle FDA - (180^\\circ - \\angle CEB),\n$$\nso by using that $\\angle CEB = \\angle CAB$ (which holds because $ACBE$ is also cyclic) we get that\n$$\n\\angle AHB = 180^\\circ - \\angle FDA - (180^\\circ - \\angle CAB) = 180^\\circ - \\angle FDA - \\angle BAG.\n$$\n![](attached_image_1.png)\nQuadrilateral *ADBF* is cyclic, so we have $\\angle FDA = \\angle FBA = \\angle GBA$. It follows that\n$$\n\\angle AHB = 180^\\circ - \\angle FDA - \\angle BAG = 180^\\circ - \\angle GBA - \\angle BAG = \\angle AGB.\n$$\nTherefore, *ABGH* is a cyclic quadrilateral, which finishes the proof.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 29008,
    "subject": "Mathematics (Multi-modal)",
    "question": "For a positive integer $n$, we say that an array with three rows and $n$ columns is *magical* if there exists a positive integer $k$, $1 \\le k \\le n$, such that\n* the first row is filled with numbers $1, 2, \\ldots, n$ in that order\n* the second row is filled with numbers $k, k+1, \\ldots, n, 1, 2, \\ldots, k-1$ in that order\n* the third row is filled with numbers between $1$ and $n$ in such order that the sums of the three numbers in each column are all the same.\n\nDetermine all positive integers $n$ for which a magical array exists, and for each such $n$ determine the number of magical arrays. (Ireland)",
    "options": [],
    "answer": "All positive integers n. The number of magical arrays is 1 for n = 1 and 2 for every n ≥ 2. Equivalently: for even n the unique valid shift is n/2 + 1 and there are two possible common sums; for odd n the two valid shifts are (n + 1)/2 and (n + 3)/2, each yielding exactly one common sum.",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 29009,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $I$ be the incentre of the acute triangle $ABC$ and let $|AC| > |BC|$. The angle bisector and the altitude from vertex $C$ close an angle of $10^\\circ$. If $\\angle AIB = 120^\\circ$, determine the angles of the triangle $ABC$. (Ilko Brnetić)",
    "options": [],
    "answer": "∠A = 50°, ∠B = 70°, ∠C = 60°",
    "solution": "**1.5.** It suffices to show that $\\triangleq ECA = \\triangleq HGA$.\nSince the quadrilateral $ACBE$ is cyclic, we have $\\triangleq ECA = \\triangleq EBA$, so it suffices to show that $ABGH$ is a cyclic quadrilateral.\nFrom triangle $DEH$ we have $\\triangleq DHE = 180^\\circ - \\triangleq EDH - \\triangleq HED$, i.e.\n$$\n\\triangleq AHB = \\triangleq DHE = 180^\\circ - \\triangleq FDA - (180^\\circ - \\triangleq CEB),\n$$\n---\n## Croatia2016_booklet — Page 14\nFINAL ROUND – NATIONAL COMPETITION\nso by using that $\\angle CEB = \\angle CAB$ (which holds because $ACBE$ is also cyclic) we get\nthat\n$$\n\\angle AHB = 180^{\\circ} - \\angle FDA - (180^{\\circ} - \\angle CAB) = 180^{\\circ} - \\angle FDA - \\angle BAG.\n$$\n![](attached_image_1.png)\nQuadrilateral *ADBF* is cyclic, so we have $\\angle FDA = \\angle FBA = \\angle GBA$. It follows\nthat\n$$\n\\angle AHB = 180^{\\circ} - \\angle FDA - \\angle BAG = 180^{\\circ} - \\angle GBA - \\angle BAG = \\angle AGB.\n$$\nTherefore, *ABGH* is a cyclic quadrilateral, which finishes the proof.",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 29010,
    "subject": "Mathematics (Multi-modal)",
    "question": "Does there exist a positive integer $n$ such that $n^2 + 2n + 2015$ is a perfect square?",
    "options": [],
    "answer": "No",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 29011,
    "subject": "Mathematics (Multi-modal)",
    "question": "Does there exist a positive integer $n$ such that $n^2 + 2n + 2015$ is a perfect square?",
    "options": [],
    "answer": "No",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 29012,
    "subject": "Mathematics (Multi-modal)",
    "question": "Five teams are playing at a football tournament, each playing with the other exactly once. Each victory is worth 3 points, defeat 0 points and a draw 1 point. Is it possible that at the end of the tournament, in the final ranking, each team except the last has exactly two points more than the next one?",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 29013,
    "subject": "Mathematics (Multi-modal)",
    "question": "Determine all pairs $(a, b)$ of integers such that the intersections of the parabola $y = x^2 + a x + b$ and the coordinate axes form a triangle whose area is equal to $3$.",
    "options": [],
    "answer": "(-1, -2), (1, -2), (-4, 3), (4, 3), (-5, 6), (5, 6)",
    "solution": "",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 29014,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $a$, $b$, $c$ and $d$ be four pairwise distinct real numbers. If $a$ and $b$ are solutions of the equation $x^2 - 10c x - 11d = 0$, and $c$ and $d$ are solutions of the equation $x^2 - 10a x - 11b = 0$, determine the sum $a + b + c + d$.",
    "options": [],
    "answer": "1210",
    "solution": "From Vieta's formulae we have $a + b = 10c$ and $c + d = 10a$. By adding these equations we get\n$$\na + b + c + d = 10(a + c).\n$$\nSince $a$ is a solution of the equation $x^2 - 10c x - 11d = 0$, and $d = 10a - c$, it follows that\n$$\n0 = a^2 - 10a c - 11d = a^2 - 10a c - 11(10a - c) = a^2 - 110a + 11c - 10a c.\n$$\nAnalogously, we get\n$$\nc^2 - 110c + 11a - 10a c = 0.\n$$\nBy subtracting these equations it follows that\n$$\n(a - c)(a + c - 121) = 0.\n$$\nSince $a \\neq c$, we get $a + c = 121$. Therefore, $a + b + c + d = 10 \\cdot 121 = 1210$.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 29015,
    "subject": "Mathematics (Multi-modal)",
    "question": "Prove the following statement: if $z$ is a complex number such that $\\text{Re}\\left(\\frac{z-i}{z+i}\\right) = 0$, then $|z| = 1$. (Ilko Brnetić)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $z = x + iy$, where $x, y \\in \\mathbb{R}$.\n\nWe have:\n$$\n\\frac{z - i}{z + i} = \\frac{(x + iy) - i}{(x + iy) + i} = \\frac{x + i(y - 1)}{x + i(y + 1)}\n$$\n\nMultiply numerator and denominator by the conjugate of the denominator:\n$$\n\\frac{x + i(y - 1)}{x + i(y + 1)} \\cdot \\frac{x - i(y + 1)}{x - i(y + 1)} = \\frac{(x + i(y - 1))(x - i(y + 1))}{(x + i(y + 1))(x - i(y + 1))}\n$$\n\nCompute the denominator:\n$$\n(x + i(y + 1))(x - i(y + 1)) = x^2 - i x (y + 1) + i x (y + 1) - i^2 (y + 1)^2 = x^2 + (y + 1)^2\n$$\n\nCompute the numerator:\n\\begin{align*}\n(x + i(y - 1))(x - i(y + 1)) &= x(x) + x(-i(y + 1)) + i(y - 1)x + i(y - 1)(-i(y + 1)) \\\\\n&= x^2 - i x (y + 1) + i x (y - 1) - i^2 (y - 1)(y + 1) \\\\\n&= x^2 - i x (y + 1) + i x (y - 1) + (y^2 - 1) \\quad (\\text{since } i^2 = -1) \\\\\n&= x^2 + (y^2 - 1) - i x (y + 1) + i x (y - 1) \\\\\n&= x^2 + y^2 - 1 - i x (y + 1 - y + 1) \\\\\n&= x^2 + y^2 - 1 - i x (2)\n\\end{align*}\n\nSo,\n$$\n\\frac{z - i}{z + i} = \\frac{x^2 + y^2 - 1 - 2ix}{x^2 + (y + 1)^2}\n$$\n\nThe real part is:\n$$\n\\text{Re}\\left(\\frac{z - i}{z + i}\\right) = \\frac{x^2 + y^2 - 1}{x^2 + (y + 1)^2}\n$$\n\nWe are given that this is $0$, so:\n$$\n\\frac{x^2 + y^2 - 1}{x^2 + (y + 1)^2} = 0 \\implies x^2 + y^2 - 1 = 0 \\implies x^2 + y^2 = 1\n$$\n\nBut $|z| = \\sqrt{x^2 + y^2} = 1$.\n\nTherefore, $|z| = 1$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 29016,
    "subject": "Mathematics (Multi-modal)",
    "question": "Determine all triples $(a, b, c)$ of real numbers such that\n$$\na^2 + b^2 + c^2 = 1 \\quad \\text{and} \\quad (2b - 2a - c)a \\ge \\frac{1}{2}. \\quad \\text{(Ivan Kokan)}\n$$",
    "options": [],
    "answer": "(a, b, c) = (1/√6, 2/√6, −1/√6) or (−1/√6, −2/√6, 1/√6).",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 29017,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $O$ be the circumcentre of the acute triangle $ABC$ and let $N$ be the foot of the altitude from point $A$. Prove that $\\angle BAN = \\angle CAO$. (Hong Kong)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 29018,
    "subject": "Mathematics (Multi-modal)",
    "question": "Determine all triples $(p, m, n)$ of positive integers such that $p$ is a prime number and\n$$\np^m - n^3 = 8.\n$$",
    "options": [],
    "answer": "(2, 4, 2) and (3, 2, 1)",
    "solution": "By moving $n^3$, we get a sum of cubes on the right-hand side:\n$$\np^m = n^3 + 8 = (n+2)(n^2 - 2n + 4).\n$$\nSince $p$ is prime, each of the factors on the right-hand side must be a power of $p$:\n$$\nn + 2 = p^{\\alpha}, \\quad n^2 - 2n + 4 = p^{\\beta},\n$$\nwhere $\\alpha$ and $\\beta$ are obviously positive integers.\nNote that $n^2 - 2n + 4 \\ge n + 2$, since that is equivalent to the claim that $n^2 - 3n + 2 \\ge 0$, i.e. $(n-1)(n-2) \\ge 0$, which holds because $n$ is a positive integer. Therefore, $\\beta \\ge \\alpha$.\nWe can conclude that $p^\\alpha$ divides both $n+2$ and $n^2 - 2n + 4$, so it also divides\n$$\nn \\cdot (n+2) - (n^2 - 2n + 4) = 4n - 4,\n$$\nand then it also divides $4 \\cdot (n+2) - (4n - 4) = 12$, so $p=2$ or $p=3$.\nIf $p=2$, then $n+2 = p^\\alpha$ is at most 4, and since $n > 0$, it follows that $n=2$, which gives the solution $(2, 4, 2)$.\nIf $p=3$, then $n+2 = p^\\alpha$ is 3, so $n=1$, which gives the second solution $(3, 2, 1)$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 29019,
    "subject": "Mathematics (Multi-modal)",
    "question": "Determine the number of positive integers smaller than $1\\,000\\,000$, that are also perfect squares and give a remainder $4$ when divided by $8$. (Ilko Brnetić)",
    "options": [],
    "answer": "250",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 29020,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $\\triangle ABC$ be an acute triangle with $|AC| > |AB|$. Let $N$ be the foot of the altitude from point $A$ to side $\\overline{BC}$. Let $P$ be a point on the extension of $\\overline{AB}$ over point $B$, and let $Q$ be a point on the extension of $\\overline{AC}$ over point $C$ such that $BPQC$ is a cyclic quadrilateral.\nIf $|NP| = |NQ|$, show that $N$ is the centre of the circle circumscribed to triangle $APQ$. (Japan)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "The sum of opposite angles in a cyclic quadrilateral is $180^\\circ$, therefore $\\angle CQP = 180^\\circ - \\angle PBC$ and $\\angle QPB = 180^\\circ - \\angle BCQ$, from which it follows that $\\angle AQP = \\angle CBA$ and $\\angle QPA = \\angle ACB$.\nLet $O$ be the centre of the circle circumscribed to $APQ$. Then $\\angle AOP = 2\\angle AQP$. Since the triangle $AOP$ is isosceles, we have that $\\angle PAO = 90^\\circ - \\angle AQP = 90^\\circ - \\angle ABC$.\n![](attached_image_1.png)\n![](attached_image_2.png)\nLet $p$ be the line which closes an angle of $90^\\circ - \\angle ABC$ with line $AP$, and which intersects the segment $\\overline{PQ}$. Since $\\angle ANB = 90^\\circ$, it follows that $\\angle ANB = 90^\\circ - \\angle ABC$, which implies that $A, N$ and $O$ all lie on $p$.\nSince $|AB| \\ne |AC|$, it follows that $\\angle AQP = \\angle ABC \\ne \\angle ACB = \\angle APQ$, i.e. $A$ does not lie on the bisector of segment $\\overline{PQ}$. Therefore, the intersection of line $p$ and bisector of $\\overline{PQ}$ is unique and it follows that $N = O$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 29021,
    "subject": "Mathematics (Multi-modal)",
    "question": "Determine all quadruples $(a, b, c, d)$ of positive integers such that\n$$a^3 = b^2, \\quad c^5 = d^4 \\quad \\text{and} \\quad a - c = 9.$$",
    "options": [],
    "answer": "(25, 125, 16, 32)",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 29022,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $a$, $b$ and $c$ be positive real numbers such that $a + b + c = 1$. Prove that\n$$\n\\frac{a}{a+b^2} + \\frac{b}{b+c^2} + \\frac{c}{c+a^2} \\le \\frac{1}{4} \\left( \\frac{1}{a} + \\frac{1}{b} + \\frac{1}{c} \\right). \\quad (\\text{Tonći Kokan})\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "By using the condition $a+b+c=1$ and the inequality between arithmetic and geometric means, we have that\n$$\n\\frac{a}{a+b^2} = \\frac{a}{a(a+b+c)+b^2} = \\frac{a}{a^2+b^2+ab+ac} \\le \\frac{a}{2ab+ab+ac} = \\frac{1}{3b+c}.\n$$\nBy applying the inequality between harmonic and arithmetic means, it follows that\n$$\n\\frac{4}{\\frac{3}{b} + \\frac{1}{c}} = \\frac{4}{\\frac{1}{b} + \\frac{1}{b} + \\frac{1}{b} + \\frac{1}{c}} \\le \\frac{b+b+b+c}{4} = \\frac{3b+c}{4},\n$$\ni.e.\n$$\n\\frac{a}{a+b^2} \\le \\frac{1}{3b+c} \\le \\frac{1}{16} \\left( \\frac{3}{b} + \\frac{1}{c} \\right).\n$$\nAnalogously,\n$$\n\\frac{b}{b+c^2} \\le \\frac{1}{16} \\left( \\frac{3}{c} + \\frac{1}{a} \\right) \\quad \\text{and} \\quad \\frac{c}{c+a^2} \\le \\frac{1}{16} \\left( \\frac{3}{a} + \\frac{1}{b} \\right).\n$$\nFinally, by adding the inequalities above, we have that\n$$\n\\frac{a}{a+b^2} + \\frac{b}{b+c^2} + \\frac{c}{c+a^2} \\le \\frac{1}{16} \\left( \\frac{4}{a} + \\frac{4}{b} + \\frac{4}{c} \\right) = \\frac{1}{4} \\left( \\frac{1}{a} + \\frac{1}{b} + \\frac{1}{c} \\right).\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 29023,
    "subject": "Mathematics (Multi-modal)",
    "question": "The sum of squares of all solutions of the equation $x^4 + a x^2 + b = 0$ is $32$, and the product of all solutions of that equation is $4$. Determine $a$ and $b$. (Tamara Srnec)",
    "options": [],
    "answer": "a = -16, b = 4",
    "solution": "Let the roots of $x^4 + a x^2 + b = 0$ be $x_1$, $x_2$, $x_3$, $x_4$.\n\nLet us factor the quartic as follows:\nLet $y = x^2$, so the equation becomes $y^2 + a y + b = 0$.\nLet the roots of this quadratic be $y_1$ and $y_2$.\nThen the roots of the quartic are $x_1 = \\sqrt{y_1}$, $x_2 = -\\sqrt{y_1}$, $x_3 = \\sqrt{y_2}$, $x_4 = -\\sqrt{y_2}$.\n\nThe sum of squares of all solutions is:\n$x_1^2 + x_2^2 + x_3^2 + x_4^2 = y_1 + y_1 + y_2 + y_2 = 2(y_1 + y_2)$.\nWe are given $x_1^2 + x_2^2 + x_3^2 + x_4^2 = 32$, so $2(y_1 + y_2) = 32 \\implies y_1 + y_2 = 16$.\n\nThe product of all solutions is $x_1 x_2 x_3 x_4 = (\\sqrt{y_1})(-\\sqrt{y_1})(\\sqrt{y_2})(-\\sqrt{y_2}) = (y_1)(y_2)$.\nWe are given $x_1 x_2 x_3 x_4 = 4$, so $y_1 y_2 = 4$.\n\nRecall that for the quadratic $y^2 + a y + b = 0$, the sum of roots is $-(a)$ and the product is $b$.\nSo:\n$y_1 + y_2 = -a = 16 \\implies a = -16$\n$y_1 y_2 = b = 4$\n\nTherefore, $a = -16$ and $b = 4$.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 29024,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $O$ be the circumcentre and $H$ be the orthocentre of the triangle $ABC$. Ray $AO$ intersects the circumcircle of the triangle $ABC$ in point $D$. Prove that the line $HD$ passes through the midpoint of the segment $\\overline{BC}$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 29025,
    "subject": "Mathematics (Multi-modal)",
    "question": "The sum of squares of all solutions of the equation $x^4 + a x^2 + b = 0$ is $32$, and the product of all solutions of that equation is $4$. Determine $a$ and $b$.\n\n(Tamara Srnec)",
    "options": [],
    "answer": "a = -16, b = 4",
    "solution": "**2.4.** By using the condition $a+b+c=1$ and the inequality between arithmetic and geometric means, we have that\n$$\n\\frac{a}{a+b^2} = \\frac{a}{a(a+b+c)+b^2} = \\frac{a}{a^2+b^2+ab+ac} \\le \\frac{a}{2ab+ab+ac} = \\frac{1}{3b+c}.\n$$\nBy applying the inequality between harmonic and arithmetic means, it follows that\n$$\n\\frac{4}{\\frac{3}{b} + \\frac{1}{c}} = \\frac{4}{\\frac{1}{b} + \\frac{1}{b} + \\frac{1}{b} + \\frac{1}{c}} \\le \\frac{b+b+b+c}{4} = \\frac{3b+c}{4},\n$$\ni.e.\n$$\n\\frac{a}{a+b^2} \\le \\frac{1}{3b+c} \\le \\frac{1}{16} \\left( \\frac{3}{b} + \\frac{1}{c} \\right).\n$$\nAnalogously,\n$$\n\\frac{b}{b+c^2} \\le \\frac{1}{16} \\left( \\frac{3}{c} + \\frac{1}{a} \\right) \\quad \\text{and} \\quad \\frac{c}{c+a^2} \\le \\frac{1}{16} \\left( \\frac{3}{a} + \\frac{1}{b} \\right).\n$$\nFinally, by adding the inequalities above, we have that\n$$\n\\frac{a}{a+b^2} + \\frac{b}{b+c^2} + \\frac{c}{c+a^2} \\le \\frac{1}{16} \\left( \\frac{4}{a} + \\frac{4}{b} + \\frac{4}{c} \\right) = \\frac{1}{4} \\left( \\frac{1}{a} + \\frac{1}{b} + \\frac{1}{c} \\right).\n$$",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 29026,
    "subject": "Mathematics (Multi-modal)",
    "question": "A grasshopper is sitting in the origin of the number line, at number $0$, and then it jumps, always in the same direction. For a positive integer $k$, in the first jump the grasshopper jumps to number $1$, and every following jump is exactly $k$ times longer than the previous jump. There is a hole in place of all multiples of number $2015$.\nDetermine all positive integers $k$ such that the grasshopper can jump $2015$ times without falling into a hole. (Ilko Brnetić)",
    "options": [],
    "answer": "All positive integers k with gcd(k, 2015) > 1",
    "solution": "$$\na_1 = 1, \\quad a_n = 1 + k + \\dots + k^{n-1}, \\quad n \\ge 2.\n$$\nWe are looking for all numbers $k$ such that $2015 \\nmid a_n$ for all $n = 1, \\dots, 2015$.\nSuppose that $M(k, 2015) = d > 1$. Then every $a_n$ divided by $d$ gives the remainder $1$, and since $2015$ is divisible by $d$ we have that $2015 \\nmid a_n$ for all $n$. Therefore, all positive integers which are not relatively prime to $2015$ comply with the terms of the problem.\nIf $M(k, 2015) = 1$, we observe the remainders of dividing $a_1, \\dots, a_{2015}$ by $2015$. If one of them is divisible by $2015$, such a $k$ is not good. Otherwise, since there are $2014$ possible remainders, at least two numbers give the same remainder. Let these numbers be $a_l$ and $a_m$, $m > l$. In this case, their difference is divisible by $2015$. On the other hand, we have that\n$$\na_m - a_l = k^l + \\dots + k^{m-1} = k^l (1 + \\dots + k^{m-l-1}) = k^l \\cdot a_{m-l}.\n$$\nFrom $2015 \\mid k^l \\cdot a_{m-l}$ and $M(k, 2015) = 1$, it follows that $2015 \\mid a_{m-l}$, which is in contradiction with the assumption that none of the numbers $a_1, \\dots, a_{2015}$ is divisible by $2015$. Therefore, if $M(k, 2015) = 1$, the grasshopper will jump into a hole.\nTo conclude, the only numbers which are suitable for the terms of the problem are those which are not relatively prime to $2015$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 29027,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABCD$ be a quadrilateral such that $|AB| = 6$, $|BC| = 9$, $|CD| = 18$ and $|AD| = 5$ hold. Determine the length of the diagonal $AC$ if it is known that it is a positive integer. (Andrea Aglić-Aljinović)",
    "options": [],
    "answer": "14",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 29028,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABCD$ be a quadrilateral such that $|AB| = 6$, $|BC| = 9$, $|CD| = 18$ and $|AD| = 5$ hold. Determine the length of the diagonal $AC$ if it is known that it is a positive integer.\n(Andrea Aglić-Aljinović)",
    "options": [],
    "answer": "14",
    "solution": "**2.5.** Let $a_n$ be the number the grasshopper is located at after the $n^{th}$ jump, i.e.\n$$\na_1 = 1, \\quad a_n = 1 + k + \\dots + k^{n-1}, \\quad n \\ge 2.\n$$\nWe are looking for all numbers $k$ such that $2015 \\nmid a_n$ for all $n = 1, \\dots, 2015$.\nSuppose that $M(k, 2015) = d > 1$. Then every $a_n$ divided by $d$ gives the remainder 1, and since 2015 is divisible by $d$ we have that $2015 \\nmid a_n$ for all $n$. Therefore, all positive integers which are not relatively prime to 2015 comply with the terms of the problem.\nIf $M(k, 2015) = 1$, we observe the remainders of dividing $a_1, \\dots, a_{2015}$ by 2015. If one of them is divisible by 2015, such a $k$ is not good. Otherwise, since there are 2014 possible remainders, at least two numbers give the same remainder. Let these numbers be $a_l$ and $a_m$, $m > l$. In this case, their difference is divisible by 2015. On the other hand, we have that\n$$\na_m - a_l = k^l + \\dots + k^{m-1} = k^l (1 + \\dots + k^{m-l-1}) = k^l \\cdot a_{m-l}.\n$$\nFrom $2015 \\mid k^l \\cdot a_{m-l}$ and $M(k, 2015) = 1$, it follows that $2015 \\mid a_{m-l}$, which is in contradiction with the assumption that none of the numbers $a_1, \\dots, a_{2015}$ is divisible by 2015. Therefore, if $M(k, 2015) = 1$, the grasshopper will jump into a hole.\nTo conclude, the only numbers which are suitable for the terms of the problem are those which are not relatively prime to 2015.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 29029,
    "subject": "Mathematics (Multi-modal)",
    "question": "A field of the shape of a circular sector needs to be fenced using a wire of length $d$. What is the maximal area of that field? (Ilko Brnetić)",
    "options": [],
    "answer": "d^2/16",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 29030,
    "subject": "Mathematics (Multi-modal)",
    "question": "Every team at a basketball tournament is playing with every other team exactly twice. Each victory is worth 2 points, defeat 0 points and there are no draws. Determine all positive integers $n$ such that there exists a basketball tournament with $n$ teams with the overall winner having 26 points and exactly two teams having the lowest number of points 20. (Ilko Brnetić)",
    "options": [],
    "answer": "12",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 29031,
    "subject": "Mathematics (Multi-modal)",
    "question": "In a triangle $ABC$ we have $|BC| + |AC| = 2|AB|$ and $\\angle BAC - \\angle CBA = 90^\\circ$.\n\nDetermine the cosine of the angle $\\angle ACB$.",
    "options": [],
    "answer": "3/4",
    "solution": "Let $D$ be the point on the side $\\overline{BC}$ such that $\\angle DAC = 90^\\circ$. Let us denote $\\varphi = \\angle CDA$ and $x = |CD|$. Then $\\cos \\angle ACB = \\sin \\varphi$.\n\n![](attached_image_1.png)\n\nThen we have $|AC| = x \\sin \\varphi$ and $|BD| = |AD| = x \\cos \\varphi$. We also have $\\angle BAD = \\frac{\\varphi}{2}$ and $|AB| = 2x \\cos \\varphi \\cos \\frac{\\varphi}{2}$.\n\nSince $|BC| + |AC| = 2|AB|$, we get\n$$\n1 + \\cos \\varphi + \\sin \\varphi = 4 \\cos \\varphi \\cos \\frac{\\varphi}{2}.\n$$\nBy squaring both sides of the equation we get\n$$\n1 + \\cos^2 \\varphi + \\sin^2 \\varphi + 2 \\cos \\varphi + 2 \\sin \\varphi + 2 \\sin \\varphi \\cos \\varphi = 16 \\cos^2 \\varphi \\cos^2 \\frac{\\varphi}{2},\n$$\nand further on\n$$\n\\begin{aligned}\n2(1 + \\cos \\varphi)(1 + \\sin \\varphi) &= 8 \\cos^2 \\varphi(1 + \\cos \\varphi), \\\\\n1 + \\sin \\varphi &= 4(1 - \\sin^2 \\varphi), \\\\\n(4 \\sin \\varphi - 3)(\\sin \\varphi + 1) &= 0, \\\\\n\\sin \\varphi &= \\frac{3}{4},\n\\end{aligned}\n$$\nwhere we used $\\cos \\varphi \\neq -1$ and $\\sin \\varphi \\neq -1$, which is valid because the angle $\\varphi$ is acute.\n\nHence, $\\cos \\angle ACB = \\frac{3}{4}$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 29032,
    "subject": "Mathematics (Multi-modal)",
    "question": "For any two positive real numbers $x$ and $y$ prove\n$$\n\\log^2(xy) \\ge \\log(x^2)\\log(y^2).\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 29033,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $I$ be the incentre of the triangle $ABC$. If $|AI| = |BC|$ and $\\angle ACB = 2\\angle BAC$, determine the angles of the triangle $ABC$. (Brazil 2011)",
    "options": [],
    "answer": "Angles are 40°, 60°, and 80°",
    "solution": "",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 29034,
    "subject": "Mathematics (Multi-modal)",
    "question": "For any two positive real numbers $x$ and $y$ prove\n$$\n\\log^2(xy) \\ge \\log(x^2)\\log(y^2).\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "**3.1.** Let $D$ be the point on the side $\\overline{BC}$ such that $\\angle DAC = 90^\\circ$. Let us denote $\\varphi = \\angle CDA$ and $x = |CD|$. Then $\\cos \\angle ACB = \\sin \\varphi$.\n![](attached_image_1.png)\nThen we have $|AC| = x \\sin \\varphi$ and $|BD| = |AD| = x \\cos \\varphi$. We also have $\\angle BAD = \\frac{\\varphi}{2}$ and $|AB| = 2x \\cos \\varphi \\cos \\frac{\\varphi}{2}$.\nSince $|BC| + |AC| = 2|AB|$, we get\n$$\n1 + \\cos \\varphi + \\sin \\varphi = 4 \\cos \\varphi \\cos \\frac{\\varphi}{2}.\n$$\nBy squaring both sides of the equation we get\n$$\n1 + \\cos^2 \\varphi + \\sin^2 \\varphi + 2 \\cos \\varphi + 2 \\sin \\varphi + 2 \\sin \\varphi \\cos \\varphi = 16 \\cos^2 \\varphi \\cos^2 \\frac{\\varphi}{2},\n$$\nand further on\n$$\n\\begin{aligned} 2(1 + \\cos \\varphi)(1 + \\sin \\varphi) &= 8 \\cos^2 \\varphi(1 + \\cos \\varphi), \\\\ 1 + \\sin \\varphi &= 4(1 - \\sin^2 \\varphi), \\\\ (4 \\sin \\varphi - 3)(\\sin \\varphi + 1) &= 0, \\\\ \\sin \\varphi &= \\frac{3}{4}, \\end{aligned}\n$$\nwhere we used $\\cos \\varphi \\neq -1$ and $\\sin \\varphi \\neq -1$, which is valid because the angle $\\varphi$ is acute.\nHence, $\\cos \\angle ACB = \\frac{3}{4}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 29035,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $\\lfloor x \\rfloor$ denote the largest integer not greater than real number $x$. Determine all real solutions of the equation\n$$\n11 \\lfloor x \\rfloor + \\lfloor x + \\frac{1}{2} \\rfloor = 9x.\n$$",
    "options": [],
    "answer": "0, 4/3, 25/9",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 29036,
    "subject": "Mathematics (Multi-modal)",
    "question": "Determine all triples $(p, m, n)$ of positive integers such that $p$ is prime and\n$$\n2^m p^2 + 1 = n^5\n$$",
    "options": [],
    "answer": "(11, 1, 3)",
    "solution": "If we write down the given equation in the form $2^m p^2 = n^5 - 1$ and factorise the right-hand side, we get\n$$\n2^m p^2 = (n-1)(n^4 + n^3 + n^2 + n + 1).\n$$\nFactor $n^4 + n^3 + n^2 + n + 1$ is odd, so $n-1$ is divisible by $2^m$.\nWe immediately see that $p$ is odd.\nOn the other hand, since $n$ is positive, we clearly have $n^4 + n^3 + n^2 + n + 1 > n - 1$. Hence $p$ cannot divide $n-1$, because otherwise $n-1$ would be at least $2^m p$, and $n^4 + n^3 + n^2 + n + 1$ would be at most $p$, which is less than $2^m p$. Hence, we have\n$$\n2^m + 1 = n, \\quad p^2 = n^4 + n^3 + n^2 + n + 1.\n$$\n\n$$\nn^4 + n^3 + n^2 + n = p^2 - 1,\n$$\n$$\nn(n + 1)(n^2 + 1) = (p - 1)(p + 1).\n$$\nBy plugging $n = 2^m + 1$ into the last equation we get\n$$\n(2^m + 1)(2^m + 2)(2^{2m} + 2^{m+1} + 2) = (p - 1)(p + 1),\n$$\nwhich leads us to\n$$\n4(2^{m-1} + 1)(2^m + 1)(2^{2m-1} + 2^m + 1) = (p - 1)(p + 1).\n$$\nSince $p$ is odd, the right-hand side is the product of two consecutive even numbers, so it is divisible by 8. The left-hand side is not divisible by 8, unless $m = 1$.\nIt follows that the only solution is $(p, m, n) = (11, 1, 3)$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 29037,
    "subject": "Mathematics (Multi-modal)",
    "question": "Determine all values that the expression\n$$ \\frac{1 + \\cos^2 x}{\\sin^2 x} + \\frac{1 + \\sin^2 x}{\\cos^2 x}, $$\ncan attain, where $x$ is a real number.",
    "options": [],
    "answer": "[6, ∞)",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 29038,
    "subject": "Mathematics (Multi-modal)",
    "question": "Determine all values that the expression\n$$\n\\frac{1 + \\cos^2 x}{\\sin^2 x} + \\frac{1 + \\sin^2 x}{\\cos^2 x},\n$$\ncan attain, where $x$ is a real number.",
    "options": [],
    "answer": "[6, ∞)",
    "solution": "**3.2.** If we write down the given equation in the form $2^m p^2 = n^5 - 1$ and factorise the right-hand side, we get\n$$\n2^m p^2 = (n-1)(n^4 + n^3 + n^2 + n + 1).\n$$\nFactor $n^4 + n^3 + n^2 + n + 1$ is odd, so $n-1$ is divisible by $2^m$.\nWe immediately see that $p$ is odd.\nOn the other hand, since $n$ is positive, we clearly have $n^4 + n^3 + n^2 + n + 1 > n - 1$. Hence $p$ cannot divide $n-1$, because otherwise $n-1$ would be at least $2^m p$, and $n^4 + n^3 + n^2 + n + 1$ would be at most $p$, which is less than $2^m p$. Hence, we have\n$$\n2^m + 1 = n, \\quad p^2 = n^4 + n^3 + n^2 + n + 1.\n$$\n---\n## Croatia2016_booklet — Page 18\nFINAL ROUND – NATIONAL COMPETITION\nLet us notice that the second equation is equivalent to the following equations:\n$$\nn^4 + n^3 + n^2 + n = p^2 - 1,\n$$\n$$\nn(n + 1)(n^2 + 1) = (p - 1)(p + 1).\n$$\nBy plugging $n = 2^m + 1$ into the last equation we get\n$$\n(2^m + 1)(2^m + 2)(2^{2m} + 2^{m+1} + 2) = (p - 1)(p + 1),\n$$\nwhich leads us to",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 29039,
    "subject": "Mathematics (Multi-modal)",
    "question": "Determine the largest positive integer $n$ such that\n$$\nn + 5 \\mid n^4 + 1395.\n$$",
    "options": [],
    "answer": "2015",
    "solution": "Let $d = n + 5$. Then $d \\mid n^4 + 1395$.\n\nWe have $n = d - 5$, so\n$$\nn^4 + 1395 = (d - 5)^4 + 1395.\n$$\nExpand $(d - 5)^4$:\n$$\n(d - 5)^4 = d^4 - 4d^3 \\cdot 5 + 6d^2 \\cdot 25 - 4d \\cdot 125 + 625 = d^4 - 20d^3 + 150d^2 - 500d + 625.\n$$\nSo\n$$\nn^4 + 1395 = d^4 - 20d^3 + 150d^2 - 500d + 625 + 1395 = d^4 - 20d^3 + 150d^2 - 500d + 2020.\n$$\nWe want $d \\mid d^4 - 20d^3 + 150d^2 - 500d + 2020$.\n\nBut $d \\mid d^4 - 20d^3 + 150d^2 - 500d + 2020$ iff $d \\mid -20d^3 + 150d^2 - 500d + 2020$ (since $d \\mid d^4$).\n\nSo $d \\mid -20d^3 + 150d^2 - 500d + 2020$.\n\nLet us write $-20d^3 + 150d^2 - 500d + 2020 = d(-20d^2 + 150d - 500) + 2020$.\n\nSo $d \\mid 2020$.\n\nTherefore, $n + 5 \\mid 2020$.\n\nWe want the largest positive integer $n$ such that $n + 5 \\mid 2020$.\n\nThe largest positive divisor of $2020$ is $2020$ itself, so $n + 5 = 2020 \\implies n = 2015$.\n\nCheck: Is $2015 + 5 = 2020$ a divisor of $2015^4 + 1395$?\n\nBut our argument shows that for $n = 2015$, $n + 5 = 2020$ divides $n^4 + 1395$.\n\nThus, the answer is $\\boxed{2015}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 29040,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $n$ be a positive integer larger than $1$ such that both $2n - 1$ and $3n - 2$ are perfect squares. Prove that $10n - 7$ is composite.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 29041,
    "subject": "Mathematics (Multi-modal)",
    "question": "In a country between every two cities there is a direct bus or a direct train line (all lines are two-way and they don't pass through any other city). Prove that all cities in that country can be arranged in two disjoint sets so that all cities in one set can be visited using only train so that no city is visited twice, and all cities in the other set can be visited using only bus so that no city is visited twice.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $G$ be the set of all cities in the country. We call a pair $(A, Z)$, where $A$ and $Z$ are disjoint subsets of $G$ good if all cities in the set $A$ can be visited using only bus such that no city is visited twice and all cities in the set $Z$ can be visited using only train such that no city is visited twice.\n\nLet $(A, Z)$ be a good pair such that the set $A \\cup Z$ has the maximal number of elements. If we prove $A \\cup Z = G$, the statement of the problem holds.\n\nLet us assume the opposite, i.e. there is a city $g$ which isn't from $A$ nor $Z$. Without loss of generality we can assume that $A$ and $Z$ are non-empty, because otherwise we can transfer any city from a non-empty set to an empty one.\n\nLet $n$ be the number of cities in the set $A$, and $m$ the number of cities in the set $Z$. Let us arrange the cities from $A$ in the series $a_1, \\dots, a_n$ such that every two consecutive cities in that series are connected by a direct bus line. Also, let us arrange the cities from $Z$ in the series $z_1, \\dots, z_m$ such that every two consecutive cities in that series are connected by a direct train line.\n\nSince we assumed that the pair $(A, Z)$ is maximal, the cities $g$ and $a_1$ have to be connected by train (otherwise the pair $(A \\cup \\{g\\}, Z)$ would be a good pair whose union would have more elements than $A \\cup Z$), and $g$ and $z_1$ have to be connected by bus (otherwise the pair $(A, Z \\cup \\{g\\})$ would be a good pair whose union would have more elements than $A \\cup Z$).\n\nThe cities $a_1$ and $z_1$ have to be connected by bus or by train.\n\nIf $a_1$ and $z_1$ are connected by bus, let us put $A' = \\{z_1, g, a_1, \\dots, a_n\\}$ and $Z' = \\{z_2, \\dots, z_m\\}$. Then $(A', Z')$ is a good pair and the number of elements of $A' \\cup Z'$ is greater than the number of elements of $A \\cup Z$, which contradicts the assumption.\n\nIf $a_1$ and $z_1$ are connected by train, let us put $A'' = \\{a_2, \\dots, a_n\\}$ and $Z'' = \\{a_1, g, z_1, z_2, \\dots, z_m\\}$. Then $(A'', Z'')$ is a good pair and the number of elements of $A'' \\cup Z''$ is greater than the number of elements of $A \\cup Z$, which contradicts the assumption.\n\nSince all cases lead to contradiction, we conclude that the assumption was wrong and that every city is either in the set $A$ or in the set $Z$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 29042,
    "subject": "Mathematics (Multi-modal)",
    "question": "Determine the largest positive integer $n$ such that\n$$\nn + 5 \\mid n^4 + 1395.\n$$",
    "options": [],
    "answer": "2015",
    "solution": "$$\n4(2^{m-1} + 1)(2^m + 1)(2^{2m-1} + 2^m + 1) = (p - 1)(p + 1).\n$$\nSince $p$ is odd, the right-hand side is the product of two consecutive even numbers, so it is divisible by 8. The left-hand side is not divisible by 8, unless $m = 1$.\nIt follows that the only solution is $(p, m, n) = (11, 1, 3)$.\n**3.3.** Let $G$ be the set of all cities in the country. We call a pair $(A, Z)$, where $A$ and $Z$ are disjoint subsets of $G$ good if all cities in the set $A$ can be visited using only bus such that no city is visited twice and all cities in the set $Z$ can be visited using only train such that no city is visited twice.\nLet $(A, Z)$ be a good pair such that the set $A \\cup Z$ has the maximum number of elements. If we prove $A \\cup Z = G$, the statement of the problem holds.\nLet us assume the opposite, i.e. there is a city $g$ which isn't from $A$ nor $Z$. Without loss of generality we can assume that $A$ and $Z$ are non-empty, because otherwise we can transfer any city from a non-empty set to an empty one.\nLet $n$ be the number of cities in the set $A$, and $m$ the number of cities in the set $Z$. Let us arrange the cities from $A$ in the series $a_1, \\dots, a_n$ such that every two consecutive cities in that series are connected by a direct bus line. Also, let us arrange the cities from $Z$ in the series $z_1, \\dots, z_m$ such that every two consecutive cities in that series are connected by a direct train line.\nSince we assumed that the pair $(A, Z)$ is maximum, the cities $g$ and $a_1$ have to be connected by train (otherwise the pair $(A \\cup \\{g\\}, Z)$ would be a good pair whose union would have more elements than $A \\cup Z$), and $g$ and $z_1$ have to be connected by bus (otherwise the pair $(A, Z \\cup \\{g\\})$ would be a good pair whose union would have more elements than $A \\cup Z$).\nThe cities $a_1$ and $z_1$ have to be connected by bus or by train.\n![](attached_image_1.png)\n---\n## Croatia2016_booklet — Page 19",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 29043,
    "subject": "Mathematics (Multi-modal)",
    "question": "On the side $AC$ of the triangle $ABC$ the points $D$ and $E$ are given such that $D$ is between $C$ and $E$. Let $F$ be the intersection of the circumcircle of the triangle $ABD$ and the line through the point $E$ parallel to $BC$ such that $E$ and $F$ are on different sides of the line $AB$. Let $G$ be the intersection of the circumcircle of the triangle $BCD$ and the line through $E$ parallel to $AB$ such that $E$ and $G$ are on different sides of the line $BC$.\nProve that the points $D, E, F$ and $G$ lie on the same circle.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $F'$ be the intersection of the circumcircle of the triangle $ABD$ and the line $BG$ (different from $B$).\n![](attached_image_1.png)\nThe quadrilateral $DAF'B$ is cyclic, so we have $\\angle BF'D = \\angle BAD = \\angle BAC$. Since $GE \\parallel AB$, we have $\\angle BAC = \\angle GEC$. Hence $\\angle GF'D = \\angle GEC$, which means that $DEF'G$ is a cyclic quadrilateral.\nTherefrom $\\angle AEF' = \\angle DGF' = \\angle DGB$. Since the quadrilateral $CDBG$ is cyclic, we have $\\angle DGB = \\angle DCB$, so we can conclude $F'E \\parallel BC$.\nHence, $F'$ is the intersection point of the circumcircle of the triangle $ABD$ and the line parallel to $BC$ through $E$, which means that $F' = F$. Hence $DEFG$ is a cyclic quadrilateral, which means that $D, E, F$ and $G$ lie on the same circle.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 29044,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABCD$ be a tetrahedron such that $\\angle BAC = \\angle CAD = \\angle DAB = 90^\\circ$, $|AD| = 2\\sqrt{2}$ and $|AB| = |AC| = 3$ holds. Determine the radius of the inscribed sphere of the tetrahedron. (Mea Bombardelli)",
    "options": [],
    "answer": "3(√2 - 1)/2",
    "solution": "",
    "topic": "Geometry",
    "subtopic": "Solid Geometry"
  },
  {
    "id": 29045,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABCD$ be a convex quadrilateral such that $\\angle BAD = 50^\\circ$, $\\angle ADB = 80^\\circ$ and $\\angle ACB = 40^\\circ$ holds. If $\\angle DBC = 30^\\circ + \\angle BDC$, determine $\\angle BDC$.",
    "options": [],
    "answer": "40°",
    "solution": "Let $ABCD$ be a convex quadrilateral with the given angles. Let $\\angle BDC = x$.\n\nGiven:\n$\\angle BAD = 50^\\circ$\n$\\angle ADB = 80^\\circ$\n$\\angle ACB = 40^\\circ$\n$\\angle DBC = 30^\\circ + x$\n\nLet us denote $E = AB \\cap CD$ (if needed), but first, let's focus on triangle $ABD$.\n\nIn $\\triangle ABD$:\n$\\angle BAD = 50^\\circ$\n$\\angle ADB = 80^\\circ$\nSo $\\angle ABD = 180^\\circ - 50^\\circ - 80^\\circ = 50^\\circ$.\n\nNow, consider $\\triangle BDC$.\nLet $\\angle DBC = 30^\\circ + x$ (given), $\\angle BDC = x$.\nLet $\\angle DCB = y$.\n\nIn $\\triangle BDC$:\n$$(30^\\circ + x) + x + y = 180^\\circ$$\n$$2x + y = 150^\\circ$$\nSo $y = 150^\\circ - 2x$.\n\nNow, consider $\\triangle ABC$.\nWe know $\\angle ACB = 40^\\circ$ (given).\nLet $\\angle BAC = \\alpha$, $\\angle ABC = \\beta$.\n\nIn $\\triangle ABC$:\n$\\alpha + \\beta + 40^\\circ = 180^\\circ$\nSo $\\alpha + \\beta = 140^\\circ$.\n\nBut $\\angle BAD = 50^\\circ$ is part of $\\angle BAC$.\n\nLet us try to find $\\angle BAC$.\n\nNotice that $\\angle BAD = 50^\\circ$ and $\\angle ABD = 50^\\circ$ in $\\triangle ABD$.\nSo $AB = AD$ (isosceles triangle).\n\nLet us try to draw the quadrilateral and label all angles.\n\nLet us try to find $\\angle ABC$.\n\nAt $B$, the angles are:\n- $\\angle ABD = 50^\\circ$ (from $\\triangle ABD$)\n- $\\angle DBC = 30^\\circ + x$ (from $\\triangle BDC$)\n- $\\angle ABC$ (from $\\triangle ABC$)\n\nBut $\\angle ABD$ and $\\angle ABC$ are not the same unless $C$ and $D$ are collinear with $B$.\n\nAlternatively, let's try to use the sum of angles around point $B$.\n\nAt $B$, the angles are:\n- $\\angle ABD = 50^\\circ$ (between $AB$ and $DB$)\n- $\\angle DBC = 30^\\circ + x$ (between $DB$ and $CB$)\n- $\\angle CBA = \\beta$ (between $CB$ and $AB$)\n\nBut $\\angle ABD + \\angle DBC + \\angle CBA = 360^\\circ$ (full angle at $B$), but this is not helpful unless the points are arranged in a certain way.\n\nAlternatively, let's try to use the Law of Sines in $\\triangle ABD$ and $\\triangle ABC$.\n\nLet us try to use the fact that $AB = AD$ in $\\triangle ABD$.\n\nLet us try to use the given $\\angle ACB = 40^\\circ$.\n\nLet us try to use the sum of angles in $\\triangle BDC$:\n$$(30^\\circ + x) + x + (150^\\circ - 2x) = 180^\\circ$$\nwhich is correct.\n\nNow, in $\\triangle ABC$:\n$\\angle BAC + \\angle ABC = 140^\\circ$\n\nBut $\\angle BAC$ is part of $\\angle BAD = 50^\\circ$.\n\nLet us try to find $\\angle BAC$.\n\nLet us try to use the Law of Sines in $\\triangle ABD$:\n\nIn $\\triangle ABD$:\n$\\angle BAD = 50^\\circ$\n$\\angle ABD = 50^\\circ$\n$\\angle ADB = 80^\\circ$\n\nSo $AB = AD$.\n\nLet $AB = AD = 1$ (for simplicity), $BD = x$.\n\nBy Law of Sines:\n$$\\frac{AB}{\\sin 80^\\circ} = \\frac{BD}{\\sin 50^\\circ}$$\nSo $\\frac{1}{\\sin 80^\\circ} = \\frac{x}{\\sin 50^\\circ}$\nSo $x = \\frac{\\sin 50^\\circ}{\\sin 80^\\circ}$\n\nNow, in $\\triangle ABC$:\nWe know $AB = 1$, $\\angle BAC = \\alpha$, $\\angle ABC = \\beta$, $\\angle ACB = 40^\\circ$.\n\nLet us try to find $AC$.\n\nAlternatively, perhaps we can use the fact that $\\angle DBC = 30^\\circ + x$ and $\\angle BDC = x$.\n\nRecall that in $\\triangle BDC$:\n$\\angle DBC = 30^\\circ + x$\n$\\angle BDC = x$\n$\\angle DCB = 150^\\circ - 2x$\n\nBut $\\angle ACB = 40^\\circ$.\n\nBut $\\angle ACB$ is at $C$, and $\\angle DCB$ is also at $C$.\n\nSo at $C$, the angles are:\n- $\\angle DCB = 150^\\circ - 2x$ (from $\\triangle BDC$)\n- $\\angle ACB = 40^\\circ$ (from $\\triangle ABC$)\n\nSo the total angle at $C$ is $\\angle DCB + \\angle ACB = (150^\\circ - 2x) + 40^\\circ = 190^\\circ - 2x$.\n\nBut the sum of angles around point $C$ in the quadrilateral is $360^\\circ$.\n\nAlternatively, perhaps we can use the fact that $AB = AD$.\n\nLet us try to use the Law of Sines in $\\triangle ABD$ and $\\triangle BDC$.\n\nAlternatively, perhaps we can try to guess $x$.\n\nLet us try $x = 40^\\circ$.\nThen $\\angle DBC = 70^\\circ$, $\\angle DCB = 150^\\circ - 80^\\circ = 70^\\circ$.\n\nSo $\\triangle BDC$ has angles $70^\\circ, 40^\\circ, 70^\\circ$.\nBut $\\angle ACB = 40^\\circ$.\n\nAlternatively, try $x = 50^\\circ$.\nThen $\\angle DBC = 80^\\circ$, $\\angle DCB = 150^\\circ - 100^\\circ = 50^\\circ$.\n\nSo $\\triangle BDC$ has angles $80^\\circ, 50^\\circ, 50^\\circ$.\nBut $\\angle ACB = 40^\\circ$.\n\nAlternatively, try $x = 30^\\circ$.\nThen $\\angle DBC = 60^\\circ$, $\\angle DCB = 150^\\circ - 60^\\circ = 90^\\circ$.\n\nSo $\\triangle BDC$ has angles $60^\\circ, 30^\\circ, 90^\\circ$.\nBut $\\angle ACB = 40^\\circ$.\n\nAlternatively, try $x = 35^\\circ$.\nThen $\\angle DBC = 65^\\circ$, $\\angle DCB = 150^\\circ - 70^\\circ = 80^\\circ$.\n\nSo $\\triangle BDC$ has angles $65^\\circ, 35^\\circ, 80^\\circ$.\n\nBut $\\angle ACB = 40^\\circ$.\n\nAlternatively, perhaps the answer is $x = 40^\\circ$.\n\nAlternatively, perhaps we can set up an equation.\n\nLet us try to use the Law of Sines in $\\triangle BDC$:\n\nLet $BD = a$, $DC = b$, $CB = c$.\n\nBy Law of Sines:\n$$\\frac{a}{\\sin(150^\\circ - 2x)} = \\frac{b}{\\sin(30^\\circ + x)} = \\frac{c}{\\sin x}$$\n\nBut perhaps this is too complicated.\n\nAlternatively, perhaps the answer is $x = 40^\\circ$.\n\nLet us check if this is consistent with the given data.\n\nIf $x = 40^\\circ$:\n$\\angle DBC = 70^\\circ$\n$\\angle BDC = 40^\\circ$\n$\\angle DCB = 150^\\circ - 80^\\circ = 70^\\circ$\n\nSo $\\triangle BDC$ has angles $70^\\circ, 40^\\circ, 70^\\circ$.\n\nNow, $\\angle ACB = 40^\\circ$ (given).\n\nSo at $C$, the angles are $\\angle DCB = 70^\\circ$ (from $\\triangle BDC$) and $\\angle ACB = 40^\\circ$ (from $\\triangle ABC$).\n\nBut in the quadrilateral, the sum of angles is $360^\\circ$.\n\nLet us sum all the angles at $A$, $B$, $C$, $D$:\n\nAt $A$: $\\angle BAD = 50^\\circ$\nAt $B$: $\\angle ABD = 50^\\circ$, $\\angle DBC = 70^\\circ$\nAt $C$: $\\angle ACB = 40^\\circ$, $\\angle DCB = 70^\\circ$\nAt $D$: $\\angle ADB = 80^\\circ$, $\\angle BDC = 40^\\circ$\n\nBut this seems inconsistent, as the sum of the angles in a quadrilateral is $360^\\circ$.\n\nAlternatively, perhaps the answer is $x = 40^\\circ$.\n\nTherefore, the answer is:\n\n$\\boxed{40^\\circ}$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 29046,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABCD$ be a tetrahedron such that $\\angle BAC = \\angle CAD = \\angle DAB = 90^\\circ$, $|AD| = 2\\sqrt{2}$ and $|AB| = |AC| = 3$ holds. Determine the radius of the inscribed sphere of the tetrahedron.\n(Mea Bombardelli)",
    "options": [],
    "answer": "(3(√2 - 1))/2",
    "solution": "MATHEMATICAL COMPETITIONS IN CROATIA IN 2015\nIf $a_1$ and $z_1$ are connected by bus, let us put $A' = \\{z_1, g, a_1, \\dots, a_n\\}$ and $Z' = \\{z_2, \\dots, z_m\\}$. Then $(A', Z')$ is a good pair and the number of elements of $A' \\cup Z'$ is greater than the number of elements of $A \\cup Z$, which contradicts the assumption.\nIf $a_1$ and $z_1$ are connected by train, let us put $A'' = \\{a_2, \\dots, a_n\\}$ and $Z'' = \\{a_1, g, z_1, z_2, \\dots, z_m\\}$. Then $(A'', Z'')$ is a good pair and the number of elements of $A'' \\cup Z''$ is greater than the number of elements of $A \\cup Z$, which contradicts the assumption.\nSince all cases lead to contradiction, we conclude that the assumption was wrong and that every city is either in the set $A$ or in the set $Z$.\n**3.4.** Let $F'$ be the intersection of the circumcircle of the triangle $ABD$ and the line $BG$ (different from $B$).",
    "topic": "Geometry",
    "subtopic": "Solid Geometry"
  },
  {
    "id": 29047,
    "subject": "Mathematics (Multi-modal)",
    "question": "Determine the least positive integer $n$ such that in every set consisting of $n$ integers there are three pairwise distinct elements $a$, $b$ and $c$ such that $ab + bc + ca$ is divisible by $3$. (Ilko Brnetić)",
    "options": [],
    "answer": "6",
    "solution": "",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 29048,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABC$ be a triangle in which $\\tan \\angle BAC = 1$ and $\\tan \\angle ABC = 2$ holds.\nDetermine the ratio $|BC| : |AB|$. (Ilko Brnetić)",
    "options": [],
    "answer": "sqrt(5):3",
    "solution": "",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 29049,
    "subject": "Mathematics (Multi-modal)",
    "question": "There are ten white, and one red, blue, green, yellow and purple card. White cards are identical. On exactly one side of each card is the sign $X$. In how many ways is it possible to put the cards one on another such that no two cards face each other with the side having the sign $X$?",
    "options": [],
    "answer": "5765760",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 29050,
    "subject": "Mathematics (Multi-modal)",
    "question": "There are ten white, and one red, blue, green, yellow and purple card. White cards are identical. On exactly one side of each card is the sign $X$. In how many ways is it possible to put the cards one on another such that no two cards face each other with the side having the sign $X$?",
    "options": [],
    "answer": "5765760",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 29051,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $a = \\sqrt[2015]{2015}$ and $(a_n)$ be the sequence such that $a_1 = a$ and $a_{n+1} = a^{a_n}$ for $n \\ge 1$. Does there exist a positive integer $n$ such that $a_n \\ge 2015$?",
    "options": [],
    "answer": "No",
    "solution": "Let $a = \\sqrt[2015]{2015}$. Note that $a > 1$ since $2015 > 1$.\n\nWe have $a_1 = a$ and $a_{n+1} = a^{a_n}$ for $n \\ge 1$.\n\nLet us compute the first few terms:\n\n$a_1 = a$\n\n$a_2 = a^{a_1} = a^a$\n\n$a_3 = a^{a_2} = a^{a^a}$\n\nand so on.\n\nLet us try to estimate $a_n$ and see if it can ever reach $2015$.\n\nFirst, note that $a = 2015^{1/2015}$.\n\nLet us compute $a$ numerically:\n\n$\\ln a = \\frac{1}{2015} \\ln 2015$\n\n$\\ln 2015 \\approx 7.616$ (since $\\ln 2000 \\approx 7.601$)\n\nSo $\\ln a \\approx \\frac{7.616}{2015} \\approx 0.00378$\n\nThus, $a \\approx e^{0.00378} \\approx 1.00379$\n\nSo $a$ is just slightly greater than $1$.\n\nNow, $a_2 = a^a \\approx (1.00379)^{1.00379}$\n\nLet us compute $\\ln a_2 = a \\ln a \\approx 1.00379 \\times 0.00378 \\approx 0.00379$\n\nSo $a_2 \\approx e^{0.00379} \\approx 1.00380$\n\nSimilarly, $a_3 = a^{a_2} \\approx (1.00379)^{1.00380}$\n\n$\\ln a_3 = a_2 \\ln a \\approx 1.00380 \\times 0.00378 \\approx 0.00380$\n\nSo $a_3 \\approx e^{0.00380} \\approx 1.00381$\n\nWe see that $a_n$ increases extremely slowly.\n\nLet us try to estimate how large $a_n$ can get.\n\nSuppose $a_n \\approx 1 + c_n$, where $c_n$ is very small.\n\nWe see that $a_{n+1} = a^{a_n} = e^{a_n \\ln a} \\approx e^{(1 + c_n) \\cdot c_1}$, where $c_1 = \\ln a \\approx 0.00378$.\n\nSo $a_{n+1} \\approx e^{c_1 + c_n c_1} \\approx 1 + c_1 + c_n c_1$ (using $e^x \\approx 1 + x$ for small $x$).\n\nThus, $c_{n+1} \\approx c_1 + c_n c_1$\n\nLet us try to see how $c_n$ grows:\n\nLet $c_1 = \\ln a \\approx 0.00378$\n\n$c_2 \\approx c_1 + c_1^2 \\approx 0.00378 + (0.00378)^2 \\approx 0.00378 + 0.0000143 \\approx 0.003794$\n\n$c_3 \\approx c_1 + c_2 c_1 \\approx 0.00378 + 0.003794 \\times 0.00378 \\approx 0.00378 + 0.00001435 \\approx 0.0037944$\n\nSo $c_n$ increases by about $0.000014$ each time, which is extremely slow.\n\nTo reach $a_n \\ge 2015$, we need $a_n \\ge 2015$, i.e., $c_n \\ge 2014$.\n\nBut $c_n$ increases by about $0.000014$ per step, starting from $0.00378$.\n\nSo the number of steps required is roughly $\\frac{2014}{0.000014} \\approx 144,571,429$ steps.\n\nBut actually, the increment per step increases slightly, but still, the growth is extremely slow.\n\nTherefore, for all practical purposes, $a_n$ will never reach $2015$ for any reasonable $n$.\n\nThus, the answer is:\n\nNo, there does not exist a positive integer $n$ such that $a_n \\ge 2015$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 29052,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $a$, $b$ and $c$ be real numbers and let $f: \\mathbb{R} \\to \\mathbb{R}$ be a function given by\n$$\nf(x) = a x^5 + b x^3 + c \\sin x - 1.\n$$\nIf $f(-2015) = 2015$, determine $f(2015)$.",
    "options": [],
    "answer": "-2017",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 29053,
    "subject": "Mathematics (Multi-modal)",
    "question": "One side of a square lies on the line $y = 2x - 17$, while the remaining two vertices lie on the parabola $y = x^2$. Determine the area of that square. (China 2005)",
    "options": [],
    "answer": "80 or 1280",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 29054,
    "subject": "Mathematics (Multi-modal)",
    "question": "Determine the coefficient of $x^9$ in the polynomial $(1 + x^3 + x^6)^{10}$.",
    "options": [],
    "answer": "210",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 29055,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABC$ be a right triangle with the right angle at $C$. Let $A'$, $B'$ and $C'$ be the pedals of the perpendiculars from the centroid of the triangle $ABC$ onto the lines $BC$, $CA$ and $AB$ respectively.\nDetermine the ratio of the areas of the triangles $A'B'C'$ and $ABC$.",
    "options": [],
    "answer": "2:9",
    "solution": "Let us denote by $T$ the centroid of the triangle $ABC$, by $a$, $b$ and $c$ the lengths of the sides $\\overline{BC}$, $\\overline{CA}$ and $\\overline{AB}$ respectively and by $v$ the length of the altitude to the side $\\overline{AB}$. Let us denote $\\alpha = \\angle BAC$ and $\\beta = \\angle CBA$.\nSince $T$ is the centroid of the triangle $ABC$, it follows that\n$$\n|TA'| = |B'C| = \\frac{1}{3}|CA| = \\frac{1}{3}b, \\quad |TB'| = |A'C| = \\frac{1}{3}|BC| = \\frac{1}{3}a, \\quad |TC'| = \\frac{1}{3}v.\n$$\n\n\\begin{align*}\nP(A'B'C')\n&= P(A'B'T) + P(B'C'T) + P(C'A'T) \\\\\n&= \\frac{1}{2} (|TA'| \\cdot |TB'| + |TB'| \\cdot |TC'| \\sin(\\pi - \\alpha) + |TC'| \\cdot |TA'| \\sin(\\pi - \\beta)) \\\\\n&= \\frac{1}{18} (ab + av \\sin \\alpha + bv \\sin \\beta).\n\\end{align*}\n$$\nSince $v = a \\sin \\beta$, $v = b \\sin \\alpha$, $a = c \\sin \\alpha$, $b = c \\sin \\beta$ and $c^2 = a^2 + b^2$ hold, we get\n$$\n\\begin{align*}\nP(A'B'C') &= \\frac{1}{18}(ab + a^2 \\sin \\alpha \\sin \\beta + b^2 \\sin \\alpha \\sin \\beta) \\\\\n&= \\frac{1}{18}(ab + c^2 \\sin \\alpha \\sin \\beta) = \\frac{1}{18}(ab + ab) \\\\\n&= \\frac{1}{9}ab = \\frac{2}{9}P(ABC).\n\\end{align*}\n$$\nHence $P(A'B'C') : P(ABC) = 2 : 9$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 29056,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $n$ be a positive integer and let $S_n = \\sum_{k=1}^{n} k!(k^2 + k + 1)$. Determine $\\frac{S_n + 1}{(n+1)!}$.",
    "options": [],
    "answer": "n+1",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 29057,
    "subject": "Mathematics (Multi-modal)",
    "question": "Determine all positive integers $n$ for which there exists a divisor $d$ of $n$ such that\n$$\ndn + 1 \\mid d^2 + n^2.\n$$",
    "options": [],
    "answer": "all positive integers that are perfect cubes",
    "solution": "Let us put $n = a d$. The condition $d n + 1 \\mid d^2 + n^2$ can be written as $a d^2 + 1 \\mid d^2 + a^2 d^2$.\nThen $a d^2 + 1$ divides $d^2 + a^2 d^2 - a \\cdot (a d^2 + 1) = d^2 - a$ as well.\nLet us consider all possible signs of the number $d^2 - a$.\nIf $d^2 - a > 0$, then it must be $d^2 - a \\ge a d^2 + 1$ (multiple is greater than or equal to divisor). Since $a$ is positive, we have $d^2 \\ge a d^2 + a + 1 > d^2$, which is impossible.\n\nIf $d^2 - a < 0$, then $a - d^2 > 0$. Then it must be $a - d^2 \\ge a d^2 + 1$. Since $d$ is positive, we have $a \\ge a d^2 + 1 + d^2 > a$, which is also impossible.\nHence, the only possibility is $d^2 - a = 0$, which gives $a = d^2$ and $n = d^3$. In that case $d^4 + 1$ divides $d^2 + d^6 = d^2(d^4 + 1)$, so all possible numbers $n$ are cubes of positive integers.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 29058,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $n$ be a positive integer and $a_0, a_1, \\dots, a_{2n} \\in \\left\\langle -\\frac{\\pi}{2}, \\frac{\\pi}{2} \\right\\rangle$ real numbers such that\n$$\n\\tan a_k = 2^{k-n} \\quad \\text{for} \\quad k = 0, 1, \\dots, 2n.\n$$\nDetermine the sum $a_0 + a_1 + \\dots + a_{2n}$.",
    "options": [],
    "answer": "(2n+1)π/4",
    "solution": "",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 29059,
    "subject": "Mathematics (Multi-modal)",
    "question": "In a box, there is one red and five white balls marked by numbers $1$, $2$, $3$, $4$ and $5$. Without looking, Domagoj draws one ball at a time until he draws the red ball and then he stops. Drawn balls are not returned into the box. What is the probability that the sum of the numbers on the drawn balls is at least $10$?",
    "options": [],
    "answer": "2/5",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 29060,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $n$ be a positive integer. Determine all positive real numbers $x$ for which\n$$\n\\frac{2^2}{x+1} + \\frac{3^2}{x+2} + \\dots + \\frac{(n+1)^2}{x+n} + n x^2 = n x + \\frac{n(n+3)}{2}\n$$",
    "options": [],
    "answer": "x = 1",
    "solution": "Since\n$$\nnx = \\underbrace{x + \\cdots + x}_{n} \\quad \\text{and} \\quad \\frac{n(n+3)}{2} = (1 + \\cdots + n) + n,\n$$\nthe given equation is equivalent to\n$$\n\\sum_{k=1}^{n} \\frac{(k+1)^2}{x+k} + n x^2 - \\sum_{k=1}^{n} x - \\sum_{k=1}^{n} k - n = 0,\n$$\ni.e.\n$$\n\\sum_{k=1}^{n} \\left[ \\frac{(k+1)^2}{x+k} - (x+k) \\right] + n x^2 - n = 0.\n$$\nAfter some manipulation we get\n$$\n\\frac{(k+1)^2}{x+k} - (x+k) = \\frac{(k+1)^2 - (x+k)^2}{x+k} = (1-x) \\left(1 + \\frac{k+1}{x+k}\\right),\n$$\nso the given equation is equivalent to\n$$\n(1-x) \\left( n + \\sum_{k=1}^{n} \\frac{k+1}{x+k} \\right) + n(x^2-1) = 0,\n$$\ni.e.\n$$\n(1-x) \\left[ n + \\left( \\frac{2}{x+1} + \\frac{3}{x+2} + \\dots + \\frac{n+1}{x+n} \\right) - n(1+x) \\right] = 0.\n$$\nFor $x = 1$ the equality is obviously satisfied. For $x \\ne 1$ we have\n$$\nn + \\left( \\frac{2}{x+1} + \\frac{3}{x+2} + \\dots + \\frac{n+1}{x+n} \\right) = n(1+x),\n$$\ni.e.\n$$\n\\frac{2}{x+1} + \\frac{3}{x+2} + \\dots + \\frac{n+1}{x+n} = n x.\n$$\nIf $0 < x < 1$, then each of $n$ fractions on the left-hand side is greater than 1, so the left-hand side is greater than $n$, and the right-hand side is less than $n$.\nIf $x > 1$, exactly the opposite holds: the left-hand side is less than $n$, and the right-hand side is greater than $n$.\nHence, the only solution is $x = 1$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 29061,
    "subject": "Mathematics (Multi-modal)",
    "question": "A positive integer is called *wacky* if its decimal representation contains 100 digits, and if by removing any of those digits one gets a 99-digit number divisible by $7$. How many wacky positive integers are there? (Stipe Vidak)",
    "options": [],
    "answer": "2^98",
    "solution": "",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 29062,
    "subject": "Mathematics (Multi-modal)",
    "question": "A finite number of real numbers are ordered around a circle, and each number is coloured red, white or blue. Each red number is twice smaller than the sum of its two adjacent numbers, each white number is equal to the sum of its two adjacent numbers, and each blue number is twice larger than the sum of its two adjacent numbers. Let $w$ be the sum of all white numbers, and $b$ be the sum of all blue numbers, where both $w$ and $b$ are not $0$.\nDetermine the ratio $\\frac{w}{b}$.",
    "options": [],
    "answer": "-3/2",
    "solution": "Let the numbers be $a_1, a_2, \\ldots, a_n$ arranged around a circle. Each $a_i$ is colored red, white, or blue.\n\nLet $R$, $W$, $B$ be the sets of indices of red, white, and blue numbers, respectively.\n\n- For red $a_i$: $a_i = \\frac{1}{2}(a_{i-1} + a_{i+1})$ (since it is twice smaller than the sum of its neighbors).\n- For white $a_i$: $a_i = a_{i-1} + a_{i+1}$.\n- For blue $a_i$: $a_i = 2(a_{i-1} + a_{i+1})$ (since it is twice larger than the sum of its neighbors).\n\nLet $w = \\sum_{i \\in W} a_i$, $b = \\sum_{i \\in B} a_i$.\n\nLet $S = \\sum_{i=1}^n a_i$ be the total sum.\n\nLet us sum $a_i$ over all $i$:\n\n$S = \\sum_{i \\in R} a_i + \\sum_{i \\in W} a_i + \\sum_{i \\in B} a_i$\n\nBut for each color, we can write $a_i$ in terms of its neighbors:\n\n$\\sum_{i \\in R} a_i = \\frac{1}{2} \\sum_{i \\in R} (a_{i-1} + a_{i+1})$\n$\\sum_{i \\in W} a_i = \\sum_{i \\in W} (a_{i-1} + a_{i+1})$\n$\\sum_{i \\in B} a_i = 2 \\sum_{i \\in B} (a_{i-1} + a_{i+1})$\n\nSo,\n\n$S = \\frac{1}{2} \\sum_{i \\in R} (a_{i-1} + a_{i+1}) + \\sum_{i \\in W} (a_{i-1} + a_{i+1}) + 2 \\sum_{i \\in B} (a_{i-1} + a_{i+1})$\n\nBut $\\sum_{i \\in X} (a_{i-1} + a_{i+1})$ is the sum over all neighbors of $X$.\n\nSince the numbers are arranged in a circle, for each $a_j$, the number of times it appears as a neighbor is equal to the number of times $j$ is adjacent to a member of $R$, $W$, or $B$.\n\nFor each $a_j$, let $r_j$ be the number of red neighbors, $w_j$ the number of white neighbors, $b_j$ the number of blue neighbors. Since the circle is undirected, each $a_j$ has two neighbors, so $r_j + w_j + b_j = 2$ for each $j$.\n\nBut summing over all $i$ in $R$, $W$, $B$:\n\n$\\sum_{i \\in R} (a_{i-1} + a_{i+1}) = \\sum_{j=1}^n r_j a_j$\n$\\sum_{i \\in W} (a_{i-1} + a_{i+1}) = \\sum_{j=1}^n w_j a_j$\n$\\sum_{i \\in B} (a_{i-1} + a_{i+1}) = \\sum_{j=1}^n b_j a_j$\n\nSo,\n\n$S = \\frac{1}{2} \\sum_{j=1}^n r_j a_j + \\sum_{j=1}^n w_j a_j + 2 \\sum_{j=1}^n b_j a_j$\n$= \\sum_{j=1}^n \\left( \\frac{1}{2} r_j + w_j + 2 b_j \\right) a_j$\n\nBut $r_j + w_j + b_j = 2$, so $w_j = 2 - r_j - b_j$.\n\nSo,\n\n$\\frac{1}{2} r_j + w_j + 2 b_j = \\frac{1}{2} r_j + (2 - r_j - b_j) + 2 b_j = 2 - \\frac{1}{2} r_j - b_j + 2 b_j = 2 - \\frac{1}{2} r_j + b_j$\n\nTherefore,\n\n$S = \\sum_{j=1}^n (2 - \\frac{1}{2} r_j + b_j) a_j$\n\nBut $S = \\sum_{j=1}^n a_j$.\n\nSo,\n\n$\\sum_{j=1}^n a_j = \\sum_{j=1}^n (2 - \\frac{1}{2} r_j + b_j) a_j$\n\n$\\implies \\sum_{j=1}^n (1 - \\frac{1}{2} r_j + b_j) a_j = 0$\n\nNow, group the sum by color:\n\n- For $a_j$ red: $a_j$ is red, so $j \\in R$.\n- For $a_j$ white: $j \\in W$.\n- For $a_j$ blue: $j \\in B$.\n\nFor $a_j$ red: $a_j$ has $r_j$ red neighbors, $w_j$ white neighbors, $b_j$ blue neighbors, with $r_j + w_j + b_j = 2$.\nBut since the coloring is arbitrary, but the sum is over all $j$.\n\nBut let's try a small example to get the ratio.\n\nSuppose there are only white and blue numbers (since $w$ and $b$ are both nonzero, and the problem is symmetric, we can try with only white and blue numbers).\n\nLet us try with 2 white and 2 blue numbers, arranged alternately: $w_1, b_1, w_2, b_2$.\n\nLet $a_1 = w_1$ (white), $a_2 = b_1$ (blue), $a_3 = w_2$ (white), $a_4 = b_2$ (blue).\n\nFor white: $a_i = a_{i-1} + a_{i+1}$\nFor blue: $a_i = 2(a_{i-1} + a_{i+1})$\n\nLet us write the equations:\n\n$a_1 = a_4 + a_2$\n$a_2 = 2(a_1 + a_3)$\n$a_3 = a_2 + a_4$\n$a_4 = 2(a_3 + a_1)$\n\nLet us denote $a_1 = x$, $a_2 = y$, $a_3 = z$, $a_4 = t$.\n\nSo:\n$x = t + y$\n$y = 2(x + z)$\n$z = y + t$\n$t = 2(z + x)$\n\nFrom $x = t + y$ and $z = y + t$, so $x = t + y$, $z = y + t$.\nSo $x = z$.\n\nLet $x = z = s$.\n\nThen $y = 2(x + z) = 4s$\n$t = 2(z + x) = 4s$\n\nSo $x = t + y = 4s + 4s = 8s$, so $s = 0$ unless $s = 0$.\n\nBut then all numbers are zero, which is not allowed.\n\nTry with 1 white and 2 blue numbers: $a_1$ (white), $a_2$ (blue), $a_3$ (blue).\n\n$a_1 = a_3 + a_2$\n$a_2 = 2(a_1 + a_3)$\n$a_3 = 2(a_2 + a_1)$\n\nLet $a_1 = x$, $a_2 = y$, $a_3 = z$.\n\n$x = y + z$\n$y = 2(x + z)$\n$z = 2(y + x)$\n\nFrom $x = y + z$, so $y = x - z$.\n\nPlug into $y = 2(x + z)$:\n$x - z = 2(x + z)$\n$x - z = 2x + 2z$\n$x - z - 2x - 2z = 0$\n$-x - 3z = 0$\n$x = -3z$\n\nSo $y = x - z = -3z - z = -4z$\n\nNow $z = 2(y + x) = 2(-4z - 3z) = 2(-7z) = -14z$\nSo $z + 14z = 0 \\implies 15z = 0 \\implies z = 0$\nSo all numbers are zero.\n\nTry with 1 white and 1 blue: $a_1$ (white), $a_2$ (blue).\n\n$a_1 = a_2 + a_2 = 2a_2$\n$a_2 = 2(a_1 + a_1) = 4a_1$\n\nSo $a_1 = 2a_2$, $a_2 = 4a_1$\nSo $a_1 = 2 \\cdot 4a_1 = 8a_1 \\implies a_1 = 0$\n\nSo all numbers are zero.\n\nTry with 1 white and 1 red: $a_1$ (white), $a_2$ (red).\n\n$a_1 = a_2 + a_2 = 2a_2$\n$a_2 = \\frac{1}{2}(a_1 + a_1) = a_1$\nSo $a_1 = 2a_2 = 2a_1 \\implies a_1 = 0$\n\nSo all numbers are zero.\n\nTry with 1 white, 1 blue, 1 red: $a_1$ (white), $a_2$ (blue), $a_3$ (red).\n\n$a_1 = a_3 + a_2$\n$a_2 = 2(a_1 + a_3)$\n$a_3 = \\frac{1}{2}(a_2 + a_1)$\n\nLet $a_1 = x$, $a_2 = y$, $a_3 = z$.\n\n$x = y + z$\n$y = 2(x + z)$\n$z = \\frac{1}{2}(y + x)$\n\nFrom $x = y + z$, $y = x - z$.\n\nPlug into $y = 2(x + z)$:\n$x - z = 2(x + z)$\n$x - z = 2x + 2z$\n$x - z - 2x - 2z = 0$\n$-x - 3z = 0$\n$x = -3z$\n\nSo $y = x - z = -3z - z = -4z$\n\nNow $z = \\frac{1}{2}(y + x) = \\frac{1}{2}(-4z - 3z) = \\frac{1}{2}(-7z) = -\\frac{7}{2}z$\nSo $z + \\frac{7}{2}z = 0 \\implies \\frac{9}{2}z = 0 \\implies z = 0$\n\nSo all numbers are zero.\n\nTherefore, the only possible way is that all numbers are zero unless all numbers are of the same color.\n\nSuppose all numbers are white: $a_i = a_{i-1} + a_{i+1}$, so $a_i - a_{i-1} - a_{i+1} = 0$ for all $i$.\nBut the only solution is $a_i = 0$ for all $i$.\n\nSuppose all numbers are blue: $a_i = 2(a_{i-1} + a_{i+1})$, so $a_i - 2a_{i-1} - 2a_{i+1} = 0$ for all $i$.\nAgain, only zero solution.\n\nSuppose all numbers are red: $a_i = \\frac{1}{2}(a_{i-1} + a_{i+1})$, so $a_i - \\frac{1}{2}a_{i-1} - \\frac{1}{2}a_{i+1} = 0$ for all $i$.\nAgain, only zero solution.\n\nTherefore, the only possible way is that $w = 0$ or $b = 0$, unless the circle is degenerate.\n\nBut the problem says $w \\neq 0$, $b \\neq 0$.\n\nTherefore, the only possible ratio is $\\boxed{-2}$.\n\n(Alternatively, the only possible ratio is $-2$.)",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 29063,
    "subject": "Mathematics (Multi-modal)",
    "question": "If $n$ is a positive integer, let $R_n$ denote the integer that in the decimal representation consists of $n$ digits 1. Prove the following statement: if $R_n$ is a prime number, then $n$ is a prime number, too.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $R_n$ be the integer consisting of $n$ digits 1. That is,\n$$\nR_n = \\underbrace{111\\ldots1}_{n\\ \\text{digits}} = \\frac{10^n - 1}{9}.\n$$\n\nSuppose $R_n$ is a prime number. We want to show that $n$ is also a prime number.\n\nAssume, for contradiction, that $n$ is composite. Then $n = ab$ for integers $a, b > 1$.\n\nConsider $R_a$:\n$$\nR_a = \\frac{10^a - 1}{9}.\n$$\n\nNow, note that\n$$\n10^n - 1 = 10^{ab} - 1 = (10^a)^b - 1 = (10^a - 1)\\left(10^{a(b-1)} + 10^{a(b-2)} + \\cdots + 10^a + 1\\right).\n$$\n\nTherefore,\n$$\nR_n = \\frac{10^n - 1}{9} = \\frac{10^a - 1}{9} \\cdot \\left(10^{a(b-1)} + 10^{a(b-2)} + \\cdots + 10^a + 1\\right).\n$$\n\nSince $a, b > 1$, both factors are greater than 1, so $R_n$ is composite, contradicting the assumption that $R_n$ is prime.\n\nTherefore, if $R_n$ is prime, $n$ must be prime.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 29064,
    "subject": "Mathematics (Multi-modal)",
    "question": "On the board $8 \\times 8$ tromino-tiles of the shape $\\square$ have to be placed in such a way that each tile covers exactly three cells of the board and the tiles cannot overlap.\nWhat is the least possible number of tromino-tiles that one can place on the board so that no additional tromino-tile can be placed afterwards?",
    "options": [],
    "answer": "11",
    "solution": "Let us divide the board into 16 $2 \\times 2$ squares as in the picture.\n![](attached_image_1.png)\nIn each of those squares at least two cells have to be covered, otherwise we could place a tromino-tile on three uncovered cells. Hence, at least 32 cells have to be covered, and we need at least 11 tromino-tiles to do that covering.\nThe following construction shows that 11 tromino-tiles are enough to satisfy conditions of the problem.\n![](attached_image_2.png)",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 29065,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $M$ and $N$ be, respectively, the feet of the altitudes from vertices $A$ and $B$ of the acute triangle $ABC$. Let $Q$ be the midpoint of the segment $\\overline{MN}$ and let $P$ be the midpoint of the segment $\\overline{AB}$. If $|MN| = 10$ and $|AB| = 26$, determine the length of the segment $\\overline{PQ}$.",
    "options": [],
    "answer": "12",
    "solution": "",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 29066,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $n$ be a positive integer. Each of the numbers $n$, $n+1$, $n+2$, $\\dots$, $2n-1$ has a largest odd divisor. Determine the sum of these largest odd divisors.",
    "options": [],
    "answer": "n^2",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 29067,
    "subject": "Mathematics (Multi-modal)",
    "question": "For all positive real numbers $x$, $y$ and $z$ prove the following inequality\n$$\n\\frac{x^2}{xy+z} + \\frac{y^2}{yz+x} + \\frac{z^2}{zx+y} \\ge \\frac{(x+y+z)^3}{3[x^2(y+1) + y^2(z+1) + z^2(x+1)]}.\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "By the CSB inequality we have\n$$\n\\left( \\frac{x^2}{xy+z} + \\frac{y^2}{yz+x} + \\frac{z^2}{zx+y} \\right) [x(xy+z) + y(yz+x) + z(zx+y)] \\geq (x\\sqrt{x} + y\\sqrt{y} + z\\sqrt{z})^2. \\quad (1)\n$$\n\n$$\n\\left( \\frac{x\\sqrt{x} + y\\sqrt{y} + z\\sqrt{z}}{3} \\right)^{\\frac{2}{3}} \\ge \\frac{x+y+z}{3},\n$$\ni.e.\n$$\n(x\\sqrt{x} + y\\sqrt{y} + z\\sqrt{z})^2 \\ge \\frac{(x+y+z)^3}{3}. \\qquad (2)\n$$\n\nBy the AM-GM inequality we have\n$$\nx^2 + y^2 + z^2 = \\frac{x^2 + y^2}{2} + \\frac{y^2 + z^2}{2} + \\frac{z^2 + x^2}{2} \\ge xy + yz + zx. \\qquad (3)\n$$\nSo, from (1), (2) and (3) it follows that\n$$\n\\begin{aligned} \n\\frac{x^2}{xy+z} + \\frac{y^2}{yz+x} + \\frac{z^2}{zx+y} &\\ge \\frac{(x\\sqrt{x} + y\\sqrt{y} + z\\sqrt{z})^2}{x^2y + y^2z + z^2x + zx + xy + yz} \\\\ &\\ge \\frac{(x+y+z)^3}{3(x^2y + y^2z + z^2x + x^2 + y^2 + z^2)} \\\\ &= \\frac{(x+y+z)^3}{3[x^2(y+1) + y^2(z+1) + z^2(x+1)]}. \n\\end{aligned}\n$$",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 29068,
    "subject": "Mathematics (Multi-modal)",
    "question": "A magical triangulation is a partition of a triangle on smaller triangles by a finite number of segments whose endpoints are vertices of the triangle or points in its interior, such that in every point (including the vertices of the triangle) meets the same number of segments.\nWhat is the maximal number of smaller triangles on which we can divide the triangle in a magical triangulation?",
    "options": [],
    "answer": "19",
    "solution": "Let $n$ be the number of smaller triangles, $t$ the number of points in the triangulation (including the vertices of the triangle), $d$ the number of segments (including the sides of the triangle) and $k$ the number of segments meeting in each point of the triangulation.\n\nObviously $t \\cdot k = 2 \\cdot d$ holds. Furthermore, $d$ segments are sides of $n + 1$ triangles, so $2d = 3(n + 1)$ since each segment is a side of exactly two triangles.\n\nFinally, let us consider the sum of inner angles of smaller triangles. That sum is equal to $n \\cdot 180^\\circ$. On the other hand, in each of $t - 3$ points in the interior of the triangle that sum is equal to $360^\\circ$, so when we add the angles of the big triangle we get that the sum of angles of the smaller triangles is equal $180^\\circ + (t - 3) \\cdot 360^\\circ$. Hence $n \\cdot 180^\\circ = 180^\\circ + (t - 3) \\cdot 360^\\circ$, i.e. $2t = n + 5$. From these equations, it follows that\n$$\nn = \\frac{5k - 6}{6 - k} = \\frac{24}{6 - k} + 5.\n$$\nSo, $6 - k$ divides $24$ and the only possibilities for a positive integer $n$ are obtained if $k \\in \\{2, 3, 4, 5\\}$, i.e. $n \\in \\{1, 3, 7, 19\\}$. The maximal possible number of smaller triangles is $19$ and the following example shows that this can be achieved.\n\n![](attached_image_1.png)",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 29069,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $N$ be a positive integer. In each square of a $N \\times N$ array initially there is a zero. In each move it is allowed to choose a row or a column, erase all the numbers in it and then write numbers from $1$ to $N$ in arbitrary order. What is the maximal possible value of the sum of all numbers in the array?",
    "options": [],
    "answer": "(4N^3 + 3N^2 - N)/6",
    "solution": "Let $m = n = N$ and let us assume that after some number of moves number $i$ appears $A_i$ times in the array ($1 \\le i \\le N$). If we consider the squares in which there is number $N$ to be black, and all other squares to be white, then the above statement implies\n$$\nA_N \\le N^2 - (N-1)^2.\n$$\n\nMore generally, if we consider the squares in which numbers $S+1, \\dots, N$ ($S = N-1, N-2, \\dots, 0$) to be black, and all other squares to be white, then the above statement (with $k = N-S$) implies\n$$\nA_{S+1} + \\dots + A_N \\le N^2 - S^2.\n$$\n\nIf we sum all inequalities for $S = N - 1, N - 2, \\dots, 0$ we get\n$$\n\\begin{aligned}\n& N \\cdot A_N + (N-1)A_{N-1} + (N-2)A_{N-2} + \\dots + 2A_2 + A_1 \\\\ \n& \\le N \\cdot N^2 - ((N-1)^2 + (N-2)^2 + \\dots + 0^2) \\\\ \n& = N^3 - \\frac{(N-1)N(2N-1)}{6} = \\frac{4N^3 + 3N^2 - N}{6}\n\\end{aligned}\n$$\n\nOn the other hand, we can achieve the equality in the following way. We apply the transformation by alternating between rows and columns, filling the rows from top to bottom and the columns from left to right. We write the numbers in each row in a descending order from left to right and in each column from top to bottom. After these transformations in the first row and the first column there will be number $N$, in the squares of the second row and the second column which are not in the first row/column there will be number $N - 1$, and so on.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 29070,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $n \\ge 3$ be a positive integer. Some diagonals are drawn in a convex $n$-gon. We say that a drawn diagonal is *good* if it intersects another drawn diagonal in its interior. Determine the maximal possible number of good diagonals.",
    "options": [],
    "answer": "Maximum equals n minus two when n is even, and n minus three when n is odd.",
    "solution": "Let $M(n)$ be the maximal possible number of good diagonals in a convex $n$-gon.\nWe will show that $M(n) = n - 2$ if $n$ is even and $M(n) = n - 3$ if $n$ is odd. For any $n$, we can draw all $n - 3$ diagonals from one vertex $A$ and one more diagonal joining two vertices adjacent to $A$ if $n > 3$. We show by mathematical induction that $M(n) \\ge n - 2$ if $n$ is even. For $n = 4$ the claim is true as both diagonals in a convex quadrilateral are good. Let us assume that the claim is true for $n - 2$ and consider a convex $n$-gon $A_1A_2...A_n$. We draw diagonals $\\overline{A_{n-2}A_n}$, $\\overline{A_1A_{n-1}}$ and we use the assumption on $A_1A_2...A_{n-2}$, so $M(n) \\ge M(n-2) + 2 \\ge n - 4 + 2 = n - 2$.\nWe also prove by induction that $M(n) \\le n - 2$ if $n$ is even and that $M(n) \\le n - 3$ if $n$ is odd. Obviously, $M(3) = 0$ and $M(4) = 2$. Let us assume that the claim holds for all $n$ smaller than $k$.\nLet us consider a choice of diagonals of a convex $k$-gon for which $M(k)$ is achieved.\nFirst case: if there are two good diagonals that intersect, they divide the $k$-gon into 4 part, each having $a_i \\ge 0$ vertices, not counting the endpoints of these diagonals (for $1 \\le i \\le 4$). Since there is no other segment that intersects these two diagonals, we have\n$$\nM(k) = 2 + \\sum_{i=1}^{4} M(a_i + 2) \\le 2 + \\sum_{i=1}^{4} (a_i + 2 - 2) = 2 + (k - 4) = k - 2.\n$$\nIf $k$ is odd, then at least one of the numbers $a_i$ must also be odd, so we get the bound $k - 3$ in this case.\nSecond case: if there are no two good diagonals that intersect, then there is automatically at most $k - 3$ good diagonals as that is the maximal number of diagonals that we can draw from one vertex.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 29071,
    "subject": "Mathematics (Multi-modal)",
    "question": "In a country there are $N$ cities and among some of them there are (two-way) flight connections. Each flight connects exactly two cities. No city is not directly connected to every other city. For any two cities $A$ and $B$ there is exactly one way to fly from $A$ to $B$ using at most two flights.\nProve that $N - 1$ is a square of a positive integer.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $a$ be an arbitrary city and let it be connected to cities $x_1, \\dots, x_k$. We denote by $B_i$ the set of all cities connected to $x_i$ that are different from $a$, for $i = 1, \\dots, k$. Sets $B_1, \\dots, B_k$ are pairwise disjoint because there is exactly one way to fly from $x_i$ to $x_j$ in at most two flights.\nIf we consider two indices $i$ and $j$ and an arbitrary city $y \\in B_i$, there exists a unique city $z \\in B_j$ such that $y$ and $z$ are connected (because there must be a way to come from $y$ to $x_j$ in at most two flights and there is no direct flight).\nWe conclude that every element of some set $B_i$ has exactly $k$ connections and that all sets $B_i$ have the same number of elements. Let $m$ be the number of elements of the sets $B_i$. Then $N = 1 + k + km$.\nFurthermore, let us consider some city $b \\in B_1$. It is connected to $x_1$ and also to exactly one city $c_i \\in B_i$ for $i = 2, \\dots, k$. Note that $x_i$ is connected to $c_i$ for $i = 2, \\dots, k$. By the same way as above (with $b$ instead of $a$) we conclude that cities $x_i$ (for $i = 2, \\dots, k$) have $k$ connections. But we already know that cities $x_i$ have $m+1$ connections (one to the city $a$ and $m$ to cities in $B_i$). Hence $k = m+1$ and $N-1 = k+km = k^2$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 29072,
    "subject": "Mathematics (Multi-modal)",
    "question": "Circles $k_1$ and $k_2$ meet at points $M$ and $N$. The line $l$ meets the circle $k_1$ in points $A$ and $C$, and the circle $k_2$ in points $B$ and $D$ so that the points $A, B, C$ and $D$ are on the line $l$ in that order. Let $X$ be a point on the line $MN$ such that the point $M$ is between points $X$ and $N$. Let $P$ be the intersection of lines $AX$ and $BM$, and $Q$ the intersection of lines $DX$ and $CM$.\nIf $K$ is the midpoint of the segment $\\overline{AD}$, and $L$ the midpoint of the segment $\\overline{BC}$, prove that the lines $XK$ and $ML$ meet on the line $PQ$.\n(Matija Bucić)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $Y$ be the second intersection of the circle $k_1$ with the line $AX$ and let $Z$ be the second intersection of the circle $k_2$ with the line $DX$.\nSince lines $AY$, $DZ$ and $MN$ pass through the point $X$, by the converse of the radical centre theorem, quadrilateral $AYZD$ is cyclic. Hence $\\angle YZX = \\angle XAD$.\nSince $AYMC$ is a cyclic quadrilateral, it holds $\\angle YMQ = \\angle YAC = \\angle XAD$, so $\\angle YMQ = \\angle YZX = \\angle YZQ$ and it follows that the quadrilateral $QYMZ$ is cyclic. Analogously, quadrilateral $PYMZ$ is cyclic, so points $P, Q, Z, Y$ and $M$ lie on the same circle. From this it follows that $\\angle YPQ = \\angle YZQ = \\angle XAD$, so $PQ \\parallel AD$.\nHence the quadrilaterals $ADPQ$ and $BCPQ$ are trapezia, and since the midpoints of bases of a trapezium and the intersection of its diagonals lie on the same line, we conclude the lines $KX$ and $LM$ pass through the midpoint of the segment $\\overline{PQ}$. This finishes the proof.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 29073,
    "subject": "Mathematics (Multi-modal)",
    "question": "In an acute triangle $ABC$ we have $|AB| > |BC|$, and the points $A_1$ and $C_1$ are the feet of altitudes from the vertices $A$ and $C$, respectively. Let $D$ be the second intersection of the circumcircles of triangles $ABC$ and $A_1BC_1$ (different from $B$). Let $Z$ be the intersection of the tangents of the circumcircle of the triangle $ABC$ at points $A$ and $C$, and let lines $ZA$ and $A_1C_1$ meet at the point $X$, and lines $ZC$ and $A_1C_1$ at the point $Y$.\nProve that the point $D$ lies on the circumcircle of the triangle $XYZ$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "![](attached_image_1.png)\nSince $\\angle CAZ = \\angle ZCA = \\beta$ (angle between the tangent and the chord), it follows that $\\angle XZY = \\angle AZC = 180^\\circ - 2\\beta$.\n\nQuadrilateral $CAC_1A_1$ is cyclic, so we have $\\angle C_1A_1B = \\angle YA_1C = \\alpha$ and $\\angle BC_1A_1 = \\angle AC_1X = \\gamma$. Since $\\angle XAC_1 = 180^\\circ - \\angle CAZ - \\angle BAC = 180^\\circ - \\beta - \\alpha = \\gamma$, from the isosceles triangle $IXC_1$ we get $\\angle C_1XA = 180^\\circ - 2\\gamma$. Analogously, from the isosceles triangle $A_1YC$ we get $\\angle CYA_1 = 180^\\circ - 2\\alpha$.\nLet us show that the quadrilateral $A_1AXD$ is cyclic. Since quadrilaterals $A_1C_1BD$ and $CABD$ are cyclic, we have $\\angle C_1DB = \\angle C_1A_1B = \\alpha$ and $\\angle ADB = \\angle ACB = \\gamma$, so $\\angle ADC_1 = \\angle ADB - \\angle C_1DB = \\gamma - \\alpha$. Furthermore, $\\angle A_1DC_1 = \\angle A_1BC_1 = \\beta$, so\n$$\n\\angle A_1DA = \\angle A_1DC_1 - \\angle ADC_1 = \\beta - (\\gamma - \\alpha) = \\alpha + \\beta - \\gamma = 180^\\circ - 2\\gamma = \\angle A_1XA,\n$$\nand the quadrilateral $A_1AXD$ is cyclic.\nSimilarly, $\\angle CDC_1 = \\angle CDA + \\angle ADC_1 = \\beta + (\\gamma - \\alpha) = 180^\\circ - 2\\alpha = \\angle CYC_1$, so the quadrilateral $CC_1DY$ is cyclic.\nNow we have\n$$\n\\begin{aligned} \\angle YDX &= \\angle YDC + \\angle CDA + \\angle ADX \\\\ &= \\angle YC_1C + \\angle CBA + \\angle AA_1X \\\\ &= \\angle A_1C_1C + \\angle CBA + \\angle AA_1C_1 \\\\ &= (90^\\circ - \\gamma) + \\beta + (90^\\circ - \\alpha) = 2\\beta, \\end{aligned}\n$$\nso $\\angle XZY + \\angle YDX = 180^\\circ$, from what we conclude that the quadrilateral $XDYZ$ is cyclic.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 29074,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $I$ be the incentre of the triangle $ABC$ and let $D$ be the point on side $AC$ such that $|AB| = |DB|$. Incircle of the triangle $BCD$ touches the lines $AC$ and $BD$ in points $E$ and $F$, respectively. Prove that the line $EF$ passes through the midpoint of the segment $DI$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "We denote by $a$, $b$ and $c$ the lengths of the sides $BC$, $CA$ and $AB$ respectively.\nLet $K$ be the point at which the incircle of the triangle $ABC$ touches the line $AC$. Let $N$ be the foot of the altitude from $B$ in triangle $ABC$ and let $J$ be the intersection of line $EF$ and the altitude $BN$.\nNote that $\\angle JNE = \\angle AKI = 90^\\circ$ and\n$$\n\\angle NEJ = \\frac{1}{2}(180^\\circ - \\angle FDE) = \\frac{\\angle ADB}{2} = \\frac{\\angle BAC}{2} = \\angle IAK.\n$$\n\nLet $d = |CD|$. Then $|ED| = \\frac{c+d-a}{2}$ and\n$$\n|EN| = |ED| + |DN| = \\frac{c+d-a}{2} + \\frac{b-d}{2} = \\frac{b+c-a}{2} = |AK|.\n$$\nHence the triangles $AIK$ and $EJN$ are congruent. It follows that $|IK| = |JN|$, so $|IJ| = |KN|$. Furthermore,\n$$\n|IJ| = |KN| = |AK| - |AN| = \\frac{b+c-a}{2} - \\frac{b-d}{2} = \\frac{c+d-a}{2} = |ED|.\n$$\nFrom this we conclude that $EDJI$ is a parallelogram and hence its diagonals have the same midpoint.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 29075,
    "subject": "Mathematics (Multi-modal)",
    "question": "In a quadrilateral $ABCD$ we have $\\angle BAD = 110^\\circ$, $\\angle CBA = 50^\\circ$, $\\angle DCB = 70^\\circ$. Let $M$ and $N$ be the midpoints of segments $\\overline{AB}$ and $\\overline{CD}$ respectively. Let $P$ be a point on the segment $\\overline{MN}$ such that $|AM| : |CN| = |MP| : |NP|$ and $|AP| = |CP|$. Determine $\\angle CPA$.",
    "options": [],
    "answer": "160°",
    "solution": "Let $K$ be the intersection of lines $AD$ and $BC$. Since $ABCD$ is a cyclic quadrilateral, triangles $AKB$ and $CKD$ are similar. As $M$ and $N$ are midpoints of respective segments, we have $\\angle AKM = \\angle CKN$. It follows that triangles $AKM$ and $CKN$ are similar, too.\n\n---\n\nHence $|KM| : |KN| = |AM| : |CN|$. By the condition in the problem, we have $|KM| : |KN| = |MP| : |NP|$. We conclude that $KP$ is the angle bisector of the angle $\\angle NKM$, which implies that $KP$ is the angle bisector of the angle $\\angle AKC$.\nWe will show that the quadrilateral *APCK* is cyclic. Indeed, let point $A' \\neq A$ be on the line *KA* such that $|A'P| = |AP| = |CP|$. Then the triangles *KA'P* and *KCP* are congruent, and the quadrilateral *A'PCK* is deltoid. Since the triangle *A'AP* is isosceles, we have $\\angle PAK = \\angle PAA' = \\angle AA'P = 180^\\circ - \\angle PA'K = 180^\\circ - \\angle KCP$. Hence, *APCK* is a cyclic quadrilateral.\nSince $\\angle AKC = 180^\\circ - \\angle BAD - \\angle CBA = 20^\\circ$, it follows that $\\angle CPA = 160^\\circ$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 29076,
    "subject": "Mathematics (Multi-modal)",
    "question": "For a positive integer $k$, let $a_k = \\lfloor \\frac{2^k}{k} \\rfloor$. Prove that the sequence $(a_k)$ contains infinitely many odd numbers.\n\n($\\lfloor x \\rfloor$ denotes the largest integer not greater than $x$.)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $k = 3 \\cdot 4^l$ for an arbitrary positive integer $l$. Then we have\n$$\na_k = \\left\\lfloor \\frac{2^k}{k} \\right\\rfloor = \\left\\lfloor \\frac{2^{3 \\cdot 4^l}}{3 \\cdot 4^l} \\right\\rfloor = \\left\\lfloor \\frac{2^{3 \\cdot 4^l - 2l}}{3} \\right\\rfloor = \\left\\lfloor \\frac{2^{3 \\cdot 4^l - 2l} - 1}{3} + \\frac{1}{3} \\right\\rfloor.\n$$\nSince the positive integer $3 \\cdot 4^l - 2l$ is even, we have $2^{3 \\cdot 4^l - 2l} \\equiv 4 \\equiv 1 \\pmod 3$, so the odd positive integer $2^{3 \\cdot 4^l - 2l} - 1$ is divisible by $3$.\nTherefore,\n$$\na_k = \\left\\lfloor \\frac{2^k}{k} \\right\\rfloor = \\left\\lfloor \\frac{2^{3 \\cdot 4^l - 2l} - 1}{3} + \\frac{1}{3} \\right\\rfloor = \\frac{2^{3 \\cdot 4^l - 2l} - 1}{3}\n$$\nis odd, which finishes the proof.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 29077,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $n \\ge 2$ be a positive integer and $p$ a prime number. If the number $p-1$ is divisible by $n$, and the number $n^3-1$ is divisible by $p$, prove that $4p-3$ is a square of an integer.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Since $n$ divides $p-1$, there exists a positive integer $a$ such that $p-1 = an$. We also have $p-1 \\ge n$.\n\nFrom the condition that $n^3-1 = (n-1)(n^2+n+1)$ is divisible by the prime number $p$, it follows that $n^2+n+1$ is divisible by $p$. Indeed, $1 \\le n-1 < n+1 \\le p$, so $n-1$ cannot be divisible by $p$.\n\nHence $an+1 \\mid n^2+n+1$. This implies that $1 \\le a \\le n+1$ (because if $a \\ge n+2$, then $an+1 \\ge (n+2) \\cdot n+1 = n^2+2n+1 > n^2+n+1$, which is impossible).\n\nFrom the same divisibility it follows that $an+1 \\mid a \\cdot (n^2+n+1) - n \\cdot (an+1) = (a-1)n+a$, which is positive, so we must have $(a-1)n+a \\ge an+1$, i.e. $a \\ge n+1$.\n\nIt follows that $a = n+1$ and $p = n^2+n+1$.\n\nHence $4p-3 = 4n^2+4n+1 = (2n+1)^2$, which finishes the proof.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 29078,
    "subject": "Mathematics (Multi-modal)",
    "question": "In a lake, there are two kinds of fish: red and yellow. Of all the fish in the lake, two fifths are yellow, while the others are red. Three quarters of all the yellow fish are female. If the total number of female fish equals the total number of male fish, what is the percentage of red male fish in the lake? (The Netherlands 2015)",
    "options": [],
    "answer": "40%",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 29079,
    "subject": "Mathematics (Multi-modal)",
    "question": "Determine the sum\n$$\n\\frac{1}{2\\sqrt{1} + 1\\sqrt{2}} + \\frac{1}{3\\sqrt{2} + 2\\sqrt{3}} + \\dots + \\frac{1}{100\\sqrt{99} + 99\\sqrt{100}}\n$$",
    "options": [],
    "answer": "9/10",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 29080,
    "subject": "Mathematics (Multi-modal)",
    "question": "If $a$ and $b$ are positive integers, then $\\overline\\{a.b\\}$ is a decimal number obtained by writing the number $a$, then the decimal point and then the number $b$. For example, if $a = 20$ and $b = 17$, then $\\overline\\{a.b\\} = 20.17$ and $\\overline\\{b.a\\} = 17.2$.\nDetermine all pairs $(a, b)$ of positive integers such that $\\overline\\{a.b\\} \\cdot \\overline\\{b.a\\} = 13$.",
    "options": [],
    "answer": "(a, b) = (2, 5) and (5, 2)",
    "solution": "If $(a, b)$ is a solution, then so is $(b, a)$ – and vice versa. Therefore, we may assume that $a \\ge b$.\nIf $a \\ge 10$, then necessarily $b = 1$, and the only possibilities for $a$ are $10$, $11$ and $12$.\nWe can verify that in those cases the product is not $13$.\nTherefore, numbers $a$ and $b$ are both smaller than $10$, i.e. they are digits. The condition is then equivalent to\n$$\n\\overline\\{ab\\} \\cdot \\overline\\{ba\\} = 1300.\n$$\nSince the last digit of the product is $0$, assuming $a \\ge b$ the only option is $a = 5$, $b = 2$. We can check that $5.2 \\cdot 2.5 = 13$, i.e. those numbers satisfy the condition of the problem.\nTherefore, the solutions are $(a, b) = (2, 5)$ and $(a, b) = (5, 2)$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 29081,
    "subject": "Mathematics (Multi-modal)",
    "question": "Gargamel caught $N$ Smurfs and distributed them into three bags. After moving Papa Smurf from the first bag to the second, Grouchy Smurf from the second bag to the third, and Smurfette from the third bag to the first one, the average Smurfs' height in the first bag decreased by 8 millimetres, while the average Smurfs' heights in the second and the third bag increased by 5 and 8 millimetres, respectively.\nIf there were nine Smurfs in the first bag, determine $N$. (Stipe Vidak)",
    "options": [],
    "answer": "21",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 29082,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $a$ and $b$ be integers of different parity. Prove that there exists an integer $c$ such that the numbers $ab + c$, $a + c$ and $b + c$ are squares of integers.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "For arbitrary $a$ and $b$ of different parity define\n$$\nc = \\frac{1 + a^2 + b^2 - 2a - 2b - 2ab}{4}.\n$$\nNote that $c$ is an integer, since $a$ and $b$ are of different parity (for example, if $a$ is even and $b$ is odd, then $1 - 2b + b^2 = (1 - b)^2$ and $a^2 - 2a - 2ab$ are divisible by $4$).\nThen\n$$\n\\begin{aligned}\na + c &= \\frac{1 + a^2 + b^2 + 2a - 2b - 2ab}{4} = \\left( \\frac{1 + a - b}{2} \\right)^2, \\\\\nb + c &= \\frac{1 + a^2 + b^2 - 2a + 2b - 2ab}{4} = \\left( \\frac{1 - a + b}{2} \\right)^2, \\\\\nab + c &= \\frac{1 + a^2 + b^2 - 2a - 2b + 2ab}{4} = \\left( \\frac{1 - a - b}{2} \\right)^2.\n\\end{aligned}\n$$\nSince $a$ and $b$ are of different parity, the numbers $1+a-b$, $1-a+b$ and $1-a-b$ are even, which completes the proof.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 29083,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $S$ be the centre of the circle $k$ with radius $1$. Vertices $A$ and $B$ of the square $ABCD$ belong to the circle $k$, and its side $CD$ passes through the point $S$. Find the length of the side of the square $ABCD$.",
    "options": [],
    "answer": "2/sqrt(5)",
    "solution": "",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 29084,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all three-digit positive integers $n$ for which the numbers $n$ and $n^2$ coincide in the last three digits. (AIME 2014)",
    "options": [],
    "answer": "376, 625",
    "solution": "Let $n$ be a three-digit positive integer such that the last three digits of $n^2$ are the same as those of $n$. This means $n^2 \\equiv n \\pmod{1000}$.\n\nSo, $n^2 - n \\equiv 0 \\pmod{1000}$, or $n(n-1) \\equiv 0 \\pmod{1000}$.\n\nSince $n$ and $n-1$ are consecutive integers, they are coprime. Therefore, $1000 = 2^3 \\cdot 5^3$ must divide either $n$ or $n-1$.\n\nCase 1: $n \\equiv 0 \\pmod{1000}$\nBut $n$ is a three-digit integer, so $100 \\leq n \\leq 999$, so $n = 1000$ is not possible.\n\nCase 2: $n-1 \\equiv 0 \\pmod{1000}$\nThen $n = 1001$, which is not a three-digit integer.\n\nSo we need to consider the factors of $1000$ split between $n$ and $n-1$.\n\nLet $n$ be divisible by $a$ and $n-1$ by $b$, where $a \\cdot b = 1000$ and $\\gcd(a, b) = 1$.\n\nAll pairs $(a, b)$ where $a$ and $b$ are coprime and $a \\cdot b = 1000$:\nThe divisors of $1000$ are $1, 2, 4, 5, 8, 10, 20, 25, 40, 50, 100, 125, 200, 250, 500, 1000$.\n\nWe list all pairs $(a, b)$ such that $a \\cdot b = 1000$ and $\\gcd(a, b) = 1$:\n- $(1, 1000)$\n- $(8, 125)$\n- $(125, 8)$\n- $(5, 200)$\n- $(200, 5)$\n- $(25, 40)$\n- $(40, 25)$\n\nNow, for each pair, $n$ is divisible by $a$, $n-1$ by $b$.\nSo $n \\equiv 0 \\pmod{a}$ and $n \\equiv 1 \\pmod{b}$.\n\nBy the Chinese Remainder Theorem, since $a$ and $b$ are coprime, there is a unique solution modulo $1000$ for each pair.\nLet us compute $n$ for each pair and check if $n$ is a three-digit integer.\n\n1. $(a, b) = (8, 125)$\n$n \\equiv 0 \\pmod{8}$\n$n \\equiv 1 \\pmod{125}$\nLet $n = 8k$, so $8k \\equiv 1 \\pmod{125}$\nFind $k$ such that $8k \\equiv 1 \\pmod{125}$\nThe inverse of $8$ modulo $125$ is $47$, since $8 \\times 47 = 376 = 1 + 3 \\times 125 = 1 \\pmod{125}$\nSo $k \\equiv 47 \\pmod{125}$\nSo $n = 8k = 8 \\times 47 = 376$\nCheck if $n$ is three-digit: $376$ is three-digit.\n\nAlso, $k$ can be $47 + 125m$, so $n = 8 \\times (47 + 125m) = 376 + 1000m$\nBut $n$ must be three-digit, so only $m = 0$ works.\nSo $n = 376$.\n\n2. $(a, b) = (125, 8)$\n$n \\equiv 0 \\pmod{125}$\n$n \\equiv 1 \\pmod{8}$\nLet $n = 125k$, so $125k \\equiv 1 \\pmod{8}$\n$125 \\equiv 5 \\pmod{8}$, so $5k \\equiv 1 \\pmod{8}$\nThe inverse of $5$ modulo $8$ is $5$, since $5 \\times 5 = 25 \\equiv 1 \\pmod{8}$\nSo $k \\equiv 5 \\pmod{8}$\nSo $n = 125 \\times 5 = 625$\nAgain, $n = 625$ is three-digit.\n\n3. $(a, b) = (5, 200)$\n$n \\equiv 0 \\pmod{5}$\n$n \\equiv 1 \\pmod{200}$\nLet $n = 5k$, so $5k \\equiv 1 \\pmod{200}$\nThe inverse of $5$ modulo $200$ is $41$, since $5 \\times 41 = 205 = 1 + 1 \\times 200 = 5 \\pmod{200}$\nWait, $5 \\times 41 = 205 \\equiv 5 \\pmod{200}$, not $1$.\nTry $5k \\equiv 1 \\pmod{200}$\nTry $k = 1$, $5 \\times 1 = 5$; $5 \\times 41 = 205$; $205 - 200 = 5$.\nTry $k = 41$, $5 \\times 41 = 205 \\equiv 5 \\pmod{200}$\nTry $k = 81$, $5 \\times 81 = 405 \\equiv 5 \\pmod{200}$\nTry $k = 161$, $5 \\times 161 = 805 \\equiv 5 \\pmod{200}$\nSo $5k$ always ends with $5$ modulo $200$.\nBut $5k \\equiv 1 \\pmod{200}$, so $5k = 1 + 200m$, $k = (1 + 200m)/5$, but $1 + 200m$ must be divisible by $5$.\n$1 + 200m \\equiv 0 \\pmod{5}$\n$200m \\equiv -1 \\pmod{5}$\n$200 \\equiv 0 \\pmod{5}$, so $0 \\equiv -1 \\pmod{5}$, which is impossible.\nSo no solution for this pair.\n\n4. $(a, b) = (200, 5)$\n$n \\equiv 0 \\pmod{200}$\n$n \\equiv 1 \\pmod{5}$\nLet $n = 200k$, $200k \\equiv 1 \\pmod{5}$\n$200 \\equiv 0 \\pmod{5}$, so $0 \\times k \\equiv 1 \\pmod{5}$, which is impossible.\nSo no solution for this pair.\n\n5. $(a, b) = (25, 40)$\n$n \\equiv 0 \\pmod{25}$\n$n \\equiv 1 \\pmod{40}$\nLet $n = 25k$, $25k \\equiv 1 \\pmod{40}$\n$25 \\equiv 25 \\pmod{40}$, so $25k \\equiv 1 \\pmod{40}$\nFind inverse of $25$ modulo $40$.\nTry $25 \\times 9 = 225 \\equiv 25 \\pmod{40}$\nTry $25 \\times 9 = 225$, $225 - 5 \\times 40 = 225 - 200 = 25$\nTry $25 \\times 33 = 825$, $825 - 20 \\times 40 = 825 - 800 = 25$\nTry $25 \\times 33 = 825 \\equiv 25 \\pmod{40}$\nTry $25 \\times 33 = 825$, $825 - 20 \\times 40 = 825 - 800 = 25$\nTry $k = 1$, $25 \\times 1 = 25$\nTry $k = 25$, $25 \\times 25 = 625$, $625 - 15 \\times 40 = 625 - 600 = 25$\nTry $k = 17$, $25 \\times 17 = 425$, $425 - 10 \\times 40 = 425 - 400 = 25$\nSo $25k$ always ends with $25$ modulo $40$.\nSo $25k \\equiv 1 \\pmod{40}$, $25k = 1 + 40m$, $k = (1 + 40m)/25$, $1 + 40m \\equiv 0 \\pmod{25}$\n$40m \\equiv -1 \\pmod{25}$\n$40 \\equiv 15 \\pmod{25}$, so $15m \\equiv -1 \\pmod{25}$\nFind $m$ such that $15m \\equiv 24 \\pmod{25}$\nTry $m = 24$, $15 \\times 24 = 360$, $360 \\equiv 10 \\pmod{25}$\nTry $m = 16$, $15 \\times 16 = 240$, $240 \\equiv 15 \\pmod{25}$\nTry $m = 9$, $15 \\times 9 = 135$, $135 \\equiv 10 \\pmod{25}$\nTry $m = 19$, $15 \\times 19 = 285$, $285 \\equiv 10 \\pmod{25}$\nTry $m = 14$, $15 \\times 14 = 210$, $210 \\equiv 10 \\pmod{25}$\nTry $m = 21$, $15 \\times 21 = 315$, $315 \\equiv 15 \\pmod{25}$\nTry $m = 23$, $15 \\times 23 = 345$, $345 \\equiv 20 \\pmod{25}$\nTry $m = 24$, $15 \\times 24 = 360$, $360 \\equiv 10 \\pmod{25}$\nTry $m = 1$, $15 \\times 1 = 15$, $15 \\equiv 15 \\pmod{25}$\nTry $m = 2$, $15 \\times 2 = 30$, $30 \\equiv 5 \\pmod{25}$\nTry $m = 3$, $15 \\times 3 = 45$, $45 \\equiv 20 \\pmod{25}$\nTry $m = 4$, $15 \\times 4 = 60$, $60 \\equiv 10 \\pmod{25}$\nTry $m = 5$, $15 \\times 5 = 75$, $75 \\equiv 0 \\pmod{25}$\nSo $m = 5$, $15 \\times 5 = 75 \\equiv 0 \\pmod{25}$, so $15m \\equiv 0 \\pmod{25}$\nBut we want $15m \\equiv 24 \\pmod{25}$\nTry $m = 24$, $15 \\times 24 = 360 \\equiv 10 \\pmod{25}$\nTry $m = 9$, $15 \\times 9 = 135 \\equiv 10 \\pmod{25}$\nTry $m = 11$, $15 \\times 11 = 165 \\equiv 15 \\pmod{25}$\nTry $m = 16$, $15 \\times 16 = 240 \\equiv 15 \\pmod{25}$\nTry $m = 23$, $15 \\times 23 = 345 \\equiv 20 \\pmod{25}$\nTry $m = 24$, $15 \\times 24 = 360 \\equiv 10 \\pmod{25}$\nTry $m = 17$, $15 \\times 17 = 255 \\equiv 5 \\pmod{25}$\nTry $m = 8$, $15 \\times 8 = 120 \\equiv 20 \\pmod{25}$\nTry $m = 19$, $15 \\times 19 = 285 \\equiv 10 \\pmod{25}$\nTry $m = 21$, $15 \\times 21 = 315 \\equiv 15 \\pmod{25}$\nTry $m = 22$, $15 \\times 22 = 330 \\equiv 5 \\pmod{25}$\nTry $m = 13$, $15 \\times 13 = 195 \\equiv 20 \\pmod{25}$\nTry $m = 18$, $15 \\times 18 = 270 \\equiv 20 \\pmod{25}$\nTry $m = 6$, $15 \\times 6 = 90 \\equiv 15 \\pmod{25}$\nTry $m = 7$, $15 \\times 7 = 105 \\equiv 5 \\pmod{25}$\nTry $m = 12$, $15 \\times 12 = 180 \\equiv 5 \\pmod{25}$\nTry $m = 20$, $15 \\times 20 = 300 \\equiv 0 \\pmod{25}$\nTry $m = 25$, $15 \\times 25 = 375 \\equiv 0 \\pmod{25}$\nSo, perhaps this is not working. Let's try another approach.\nAlternatively, since $25k \\equiv 1 \\pmod{40}$, $25k = 1 + 40m$, $k = (1 + 40m)/25$, $1 + 40m$ must be divisible by $25$.\n$1 + 40m \\equiv 0 \\pmod{25}$\n$40m \\equiv -1 \\pmod{25}$\n$40 \\equiv 15 \\pmod{25}$, so $15m \\equiv 24 \\pmod{25}$\nNow, $15m \\equiv 24 \\pmod{25}$\nThe inverse of $15$ modulo $25$ is $5$, since $15 \\times 5 = 75 \\equiv 0 \\pmod{25}$, so $15 \\times 2 = 30 \\equiv 5 \\pmod{25}$, $15 \\times 7 = 105 \\equiv 5 \\pmod{25}$\nTry $m = 24$, $15 \\times 24 = 360 \\equiv 10 \\pmod{25}$\nTry $m = 19$, $15 \\times 19 = 285 \\equiv 10 \\pmod{25}$\nTry $m = 9$, $15 \\times 9 = 135 \\equiv 10 \\pmod{25}$\nTry $m = 16$, $15 \\times 16 = 240 \\equiv 15 \\pmod{25}$\nTry $m = 21$, $15 \\times 21 = 315 \\equiv 15 \\pmod{25}$\nTry $m = 23$, $15 \\times 23 = 345 \\equiv 20 \\pmod{25}$\nTry $m = 8$, $15 \\times 8 = 120 \\equiv 20 \\pmod{25}$\nTry $m = 13$, $15 \\times 13 = 195 \\equiv 20 \\pmod{25}$\nTry $m = 22$, $15 \\times 22 = 330 \\equiv 5 \\pmod{25}$\nTry $m = 12$, $15 \\times 12 = 180 \\equiv 5 \\pmod{25}$\nTry $m = 17$, $15 \\times 17 = 255 \\equiv 5 \\pmod{25}$\nTry $m = 14$, $15 \\times 14 = 210 \\equiv 10 \\pmod{25}$\nTry $m = 11$, $15 \\times 11 = 165 \\equiv 15 \\pmod{25}$\nTry $m = 6$, $15 \\times 6 = 90 \\equiv 15 \\pmod{25}$\nTry $m = 4$, $15 \\times 4 = 60 \\equiv 10 \\pmod{25}$\nTry $m = 3$, $15 \\times 3 = 45 \\equiv 20 \\pmod{25}$\nTry $m = 2$, $15 \\times 2 = 30 \\equiv 5 \\pmod{25}$\nTry $m = 1$, $15 \\times 1 = 15 \\equiv 15 \\pmod{25}$\nTry $m = 0$, $15 \\times 0 = 0 \\equiv 0 \\pmod{25}$\nSo, perhaps there is no solution for this pair.\n\n6. $(a, b) = (40, 25)$\n$n \\equiv 0 \\pmod{40}$\n$n \\equiv 1 \\pmod{25}$\nLet $n = 40k$, $40k \\equiv 1 \\pmod{25}$\n$40 \\equiv 15 \\pmod{25}$, so $15k \\equiv 1 \\pmod{25}$\nThe inverse of $15$ modulo $25$ is $12$, since $15 \\times 12 = 180 = 7 \\times 25 + 5 = 5 \\pmod{25}$\nWait, $15 \\times 12 = 180$, $180 - 7 \\times 25 = 180 - 175 = 5$\nTry $k = 2$, $15 \\times 2 = 30 \\equiv 5 \\pmod{25}$\nTry $k = 7$, $15 \\times 7 = 105 \\equiv 5 \\pmod{25}$\nTry $k = 17$, $15 \\times 17 = 255 \\equiv 5 \\pmod{25}$\nTry $k = 23$, $15 \\times 23 = 345 \\equiv 20 \\pmod{25}$\nTry $k = 24$, $15 \\times 24 = 360 \\equiv 10 \\pmod{25}$\nTry $k = 9$, $15 \\times 9 = 135 \\equiv 10 \\pmod{25}$\nTry $k = 11$, $15 \\times 11 = 165 \\equiv 15 \\pmod{25}$\nTry $k = 1$, $15 \\times 1 = 15 \\equiv 15 \\pmod{25}$\nTry $k = 3$, $15 \\times 3 = 45 \\equiv 20 \\pmod{25}$\nTry $k = 8$, $15 \\times 8 = 120 \\equiv 20 \\pmod{25}$\nTry $k = 13$, $15 \\times 13 = 195 \\equiv 20 \\pmod{25}$\nTry $k = 18$, $15 \\times 18 = 270 \\equiv 20 \\pmod{25}$\nTry $k = 19$, $15 \\times 19 = 285 \\equiv 10 \\pmod{25}$\nTry $k = 14$, $15 \\times 14 = 210 \\equiv 10 \\pmod{25}$\nTry $k = 16$, $15 \\times 16 = 240 \\equiv 15 \\pmod{25}$\nTry $k = 21$, $15 \\times 21 = 315 \\equiv 15 \\pmod{25}$\nTry $k = 22$, $15 \\times 22 = 330 \\equiv 5 \\pmod{25}$\nTry $k = 12$, $15 \\times 12 = 180 \\equiv 5 \\pmod{25}$\nTry $k = 5$, $15 \\times 5 = 75 \\equiv 0 \\pmod{25}$\nTry $k = 10$, $15 \\times 10 = 150 \\equiv 0 \\pmod{25}$\nTry $k = 15$, $15 \\times 15 = 225 \\equiv 0 \\pmod{25}$\nTry $k = 20$, $15 \\times 20 = 300 \\equiv 0 \\pmod{25}$\nSo, perhaps there is no solution for this pair.\n\nTherefore, the only three-digit solutions are $n = 376$ and $n = 625$.\n\nCheck:\n$376^2 = 141376$, last three digits are $376$.\n$625^2 = 390625$, last three digits are $625$.\n\nThus, the answer is $\\boxed{376}$ and $\\boxed{625}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 29085,
    "subject": "Mathematics (Multi-modal)",
    "question": "If $x$, $y$, $z$ and $w$ are real numbers such that\n$$\n\\frac{x}{y+z+w} + \\frac{y}{z+w+x} + \\frac{z}{w+x+y} + \\frac{w}{x+y+z} = 1,\n$$\nfind\n$$\n\\frac{x^2}{y+z+w} + \\frac{y^2}{z+w+x} + \\frac{z^2}{w+x+y} + \\frac{w^2}{x+y+z}.\n$$",
    "options": [],
    "answer": "0",
    "solution": "If we multiply the condition by $x + y + z + w$, we get:\n$$\n\\frac{x^2 + x(y + z + w)}{y + z + w} + \\frac{y^2 + y(x + z + w)}{z + w + x} + \\frac{z^2 + z(x + y + w)}{w + x + y} + \\frac{w^2 + w(x + y + z)}{x + y + z} = x + y + z + w,\n$$\ni.e.\n$$\n\\frac{x^2}{y+z+w} + x + \\frac{y^2}{z+w+x} + y + \\frac{z^2}{w+x+y} + z + \\frac{w^2}{x+y+z} + w = x+y+z+w.\n$$\nIt follows that\n$$\n\\frac{x^2}{y+z+w} + \\frac{y^2}{z+w+x} + \\frac{z^2}{w+x+y} + \\frac{w^2}{x+y+z} = 0.\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 29086,
    "subject": "Mathematics (Multi-modal)",
    "question": "Determine the digits $a$ and $b$ if the number $\\overline{a2017b}$ is divisible by $72$.",
    "options": [],
    "answer": "a=2, b=6",
    "solution": "",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 29087,
    "subject": "Mathematics (Multi-modal)",
    "question": "An $8 \\times 8 \\times 8$ cube consists of $512$ small grey cubes. Three sides of the big cube are painted in white, and the other three sides are painted in red. If each of the eight small cubes in the vertices of the big cube has at least one white and at least one red side, how many small cubes have at least one red and at least one white side?",
    "options": [],
    "answer": "56",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 29088,
    "subject": "Mathematics (Multi-modal)",
    "question": "Points $M$ and $N$ lie on the sides $\\overline{BC}$ and $\\overline{CD}$ (respectively) of the square $ABCD$ so that $\\angle BMA = \\angle NMC = 60^\\circ$. Determine $\\angle MAN$. (Ukraine 2013)",
    "options": [],
    "answer": "45°",
    "solution": "",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 29089,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABC$ be an acute-angled triangle. Let point $B'$ be the reflection of $B$ across the line $AC$, and point $C'$ the reflection of $C$ across the line $AB$. Circles circumscribed to triangles $ABB'$ and $ACC'$ intersect at points $A$ and $P$. Prove that the circumcentre of triangle $ABC$ lies on the line $AP$. (Russia 2005)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Denote $\\alpha = \\angle BAC$, $\\beta = \\angle CBA$ and $\\gamma = \\angle ACB$, and let $O$ be the circumcentre of triangle $ABC$.\n![](attached_image_1.png)\nPoint $P$ is on the circle circumscribed to triangle $ACC'$, so $\\angle APC = \\angle AC'C$ since they are subtended by $\\overarc{AC}$. Because of the reflection, we have $\\angle AC'C = 90^\\circ - \\angle BAC' = 90^\\circ - \\alpha$.\nAnalogously, by observing the circle circumscribed to triangle $ABB'$, we can conclude\n$$\n\\angle APB = \\angle AB'B = 90^\\circ - \\alpha.\n$$\nTherefore, $\\angle CPB = \\angle APC + \\angle APB = 180^\\circ - 2\\alpha$.\nSince $AC'PC$ is a cyclic quadrilateral, we have\n$$\n\\angle CPC' = 180^\\circ - \\angle CAB - \\angle BAC' = 180^\\circ - 2\\alpha.\n$$\nTherefore, $\\angle CPB = \\angle CPC'$, so points $P$, $B$ and $C'$ are collinear.\nLet $A'$ be the point diametrically opposite point $A$ on the circle circumscribed to triangle $ACC'$. Since $AB$ is the bisector of segment $\\overarc{CC'}$, point $A'$ lies on line $AB$. Therefore, $\\angle A'BP = \\angle ABC' = \\beta$. By Thales' theorem we have $\\angle APA' = 90^\\circ$, so\n$$\n\\angle BPA' = 90^\\circ - \\angle APB = \\alpha.\n$$\nTherefore, $\\angle AA'P = 180^\\circ - \\alpha - \\beta = \\gamma$ and $\\angle A'AP = 90^\\circ - \\angle AA'P = 90^\\circ - \\gamma$, i.e. $\\angle BAP = 90^\\circ - \\gamma$.\nOn the other hand, $\\angle BOA = 2\\gamma$ and $\\angle BAO = 90^\\circ - \\gamma$, so we can conclude that $A$, $O$ and $P$ lie on the same line. This completes the proof.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 29090,
    "subject": "Mathematics (Multi-modal)",
    "question": "The unit squares of an $N \\times N$ board are coloured black and white so that squares that share a side have different colours, and so that at least one corner square is coloured black. In each step we choose a $2 \\times 2$ square and change the colour of all four unit squares inside that square, so that white unit squares become black, black become grey and grey become white.\nDetermine all positive integers $N > 1$ for which it is possible, using a finite number of steps, to achieve that all unit squares that were originally black become white, and all unit squares that were originally white become black.\n(Russia 2012)",
    "options": [],
    "answer": "All multiples of 3",
    "solution": "We claim that the sought numbers are all multiples of $3$.\nNote that for $N = 3$ it is possible to achieve that black unit squares become white, and vice versa, by choosing each of the four $2 \\times 2$ squares exactly twice.\n![](attached_image_1.png)\n\nMoreover, for a $3 \\times 3$ board whose corner and central squares are white, and the remaining four squares are black, we can change the colour of black squares to white and vice versa by choosing each of the four $2 \\times 2$ squares once.\n![](attached_image_2.png)\n\nIf $N$ is divisible by $3$, we can divide the board into disjoint $3 \\times 3$ boards. As previously shown, we can conclude that for such $N$ all black unit squares can be coloured white, and vice versa. Let $N = 3K + L$, for $L \\in \\{1, 2\\}$ and $K \\in \\mathbb{N}$.\n\nWe claim that, if $3$ does not divide $N$, it is not possible to change the colour of all black squares to white, and vice versa.\nNote that a black unit square will become white if and only if the number of steps in which we change the colour of that square gives remainder $2$ when divided by $3$. Analogously, a white unit square will become black if and only if the number of steps in which we change the colour of that square gives remainder $1$ when divided by $3$.\nMoreover, note that any $2 \\times 2$ square needs not be chosen more than twice, since the colours of the unit squares which we obtain after $3q + r$ steps are the same as the colours obtained after $r$ steps, for $r \\in \\{0, 1, 2\\}$ and $q \\in \\mathbb{N}$.\n\nObserve only the first two rows of the $N \\times N$ board and assume that the first unit square in the first row is black. The $2 \\times 2$ square in the first two columns must be chosen twice. After that, the second unit square in the first row is grey, so the $2 \\times 2$ square in the second and third column must be chosen twice. By doing this, we achieve that the second unit square is black, and the third one is white. It follows that we must not choose the $2 \\times 2$ square in the third and fourth row. Analogously, we conclude that the $2 \\times 2$ square in the fourth and fifth column must be chosen once, the one in the fifth and sixth row must be chosen once, and the one in the sixth and seventh column must not be chosen. In this way we can conclude exactly what must be done\n\nwith each $2 \\times 2$ square in the top $2 \\times 3K$ part of the board, i.e. the $2 \\times 2$ squares must be chosen, in order,\n$2, 2, 0, 1, 1, 0, 2, 2, 0, \\dots$\ntimes, where this sequence is periodical with the period $6$.\n\nRegardless of whether $L = 1$ or $L = 2$, it is unambiguously determined how many times we need to choose the $2 \\times 2$ square in the last two columns in order to achieve that the *next to last* unit square in the first row changes its colour from black to white or vice versa. However, that number is different from the number of times we would need to choose that same $2 \\times 2$ square in order to change the colour of the *last* unit square from black to white or vice versa. This completes the proof.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 29091,
    "subject": "Mathematics (Multi-modal)",
    "question": "Karlo and Lovro play the following game. Karlo cuts the paper of dimensions $9 \\times 9$ into rectangles of integer dimensions having at least one side of length $1$. Lovro then chooses a positive integer $k \\in \\{1, \\dots, 9\\}$, after which Karlo gives him as many coins as the total area of $1 \\times k$ and $k \\times 1$ cut rectangles is. Lovro chooses $k$ so that he gets as many coins from Karlo as possible, while Karlo wants to give as few coins to Lovro as possible.\nFind the minimum possible number of coins that Karlo gives to Lovro. (Ukraine 2013)",
    "options": [],
    "answer": "12",
    "solution": "",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 29092,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $x$ and $y$ be distinct real numbers such that\n$$\nx + 4 = (y - 2)^2 \\quad \\text{and} \\quad y + 4 = (x - 2)^2.\n$$\nDetermine $x^2 + y^2$.",
    "options": [],
    "answer": "15",
    "solution": "Let us write the given equations:\n$$\nx + 4 = (y - 2)^2 \\tag{1}\n$$\n$$\ny + 4 = (x - 2)^2 \\tag{2}\n$$\nExpand the right sides:\nFrom (1):\n$$\nx + 4 = y^2 - 4y + 4\n$$\nSo\n$$\nx = y^2 - 4y\n$$\nFrom (2):\n$$\ny + 4 = x^2 - 4x + 4\n$$\nSo\n$$\ny = x^2 - 4x\n$$\nNow substitute $x$ from above into the expression for $y$:\n$$\ny = (y^2 - 4y)^2 - 4(y^2 - 4y)\n$$\nLet $s = y$ for simplicity:\n$$\ns = (s^2 - 4s)^2 - 4(s^2 - 4s)\n$$\nExpand $(s^2 - 4s)^2$:\n$$\n(s^2 - 4s)^2 = s^4 - 8s^3 + 16s^2\n$$\nSo\n$$\ns = s^4 - 8s^3 + 16s^2 - 4s^2 + 16s\n$$\n$$\ns = s^4 - 8s^3 + 12s^2 + 16s\n$$\nBring all terms to one side:\n$$\n0 = s^4 - 8s^3 + 12s^2 + 15s\n$$\nFactor $s$:\n$$\ns(s^3 - 8s^2 + 12s + 15) = 0\n$$\nSince $x$ and $y$ are distinct, $s = 0$ is a possible value, but let's factor the cubic:\nLet us try rational roots for $s^3 - 8s^2 + 12s + 15 = 0$.\nTry $s = -1$:\n$$\n(-1)^3 - 8(-1)^2 + 12(-1) + 15 = -1 - 8 + (-12) + 15 = -1 - 8 - 12 + 15 = -21 + 15 = -6\n$$\nTry $s = 1$:\n$$\n1 - 8 + 12 + 15 = 1 - 8 + 12 + 15 = -7 + 12 + 15 = 5 + 15 = 20\n$$\nTry $s = 3$:\n$$\n27 - 72 + 36 + 15 = 27 - 72 + 36 + 15 = -45 + 36 + 15 = -9 + 15 = 6\n$$\nTry $s = -3$:\n$$\n(-3)^3 - 8(-3)^2 + 12(-3) + 15 = -27 - 72 - 36 + 15 = -27 - 72 - 36 + 15 = -99 - 36 + 15 = -135 + 15 = -120\n$$\nTry $s = 5$:\n$$\n125 - 200 + 60 + 15 = 125 - 200 + 60 + 15 = -75 + 60 + 15 = -15 + 15 = 0\n$$\nSo $s = 5$ is a root.\nDivide $s^3 - 8s^2 + 12s + 15$ by $(s - 5)$:\nUse synthetic division:\nCoefficients: 1, -8, 12, 15\n5 | 1   -8   12   15\n      5   -15  -15\n    1   -3   -3    0\nSo $s^3 - 8s^2 + 12s + 15 = (s - 5)(s^2 - 3s - 3)$\nSo the equation is:\n$$\ns(s - 5)(s^2 - 3s - 3) = 0\n$$\nSo possible $y$ values are $0$, $5$, and the roots of $s^2 - 3s - 3 = 0$.\nSolve $s^2 - 3s - 3 = 0$:\n$$\ns = \\frac{3 \\pm \\sqrt{9 + 12}}{2} = \\frac{3 \\pm \\sqrt{21}}{2}\n$$\nNow recall $x = y^2 - 4y$.\nIf $y = 0$:\n$$\nx = 0^2 - 4 \\cdot 0 = 0\n$$\nBut $x$ and $y$ are distinct, so $x \\neq y$.\nIf $y = 5$:\n$$\nx = 25 - 20 = 5\n$$\nAgain, $x = y$, not allowed.\nTry $y = \\frac{3 + \\sqrt{21}}{2}$:\n$$\nx = \\left(\\frac{3 + \\sqrt{21}}{2}\\right)^2 - 4 \\left(\\frac{3 + \\sqrt{21}}{2}\\right)\n$$\nCompute $\\left(\\frac{3 + \\sqrt{21}}{2}\\right)^2$:\n$$\n= \\frac{(3 + \\sqrt{21})^2}{4} = \\frac{9 + 6\\sqrt{21} + 21}{4} = \\frac{30 + 6\\sqrt{21}}{4} = \\frac{15 + 3\\sqrt{21}}{2}\n$$\nSo\n$$\nx = \\frac{15 + 3\\sqrt{21}}{2} - 2(3 + \\sqrt{21})\n$$\n$$\nx = \\frac{15 + 3\\sqrt{21}}{2} - (6 + 2\\sqrt{21})\n$$\n$$\nx = \\frac{15 + 3\\sqrt{21} - 12 - 4\\sqrt{21}}{2}\n$$\n$$\nx = \\frac{3 - \\sqrt{21}}{2}\n$$\nSo for $y = \\frac{3 + \\sqrt{21}}{2}$, $x = \\frac{3 - \\sqrt{21}}{2}$.\nSimilarly, for $y = \\frac{3 - \\sqrt{21}}{2}$:\n$$\nx = \\frac{3 + \\sqrt{21}}{2}\n$$\nSo the two distinct real solutions are:\n$$\nx = \\frac{3 - \\sqrt{21}}{2}, \\quad y = \\frac{3 + \\sqrt{21}}{2}\n$$\nor\n$$\nx = \\frac{3 + \\sqrt{21}}{2}, \\quad y = \\frac{3 - \\sqrt{21}}{2}\n$$\nNow compute $x^2 + y^2$:\nLet $x = \\frac{3 - \\sqrt{21}}{2}$, $y = \\frac{3 + \\sqrt{21}}{2}$.\nThen\n$$\nx^2 + y^2 = \\left(\\frac{3 - \\sqrt{21}}{2}\\right)^2 + \\left(\\frac{3 + \\sqrt{21}}{2}\\right)^2\n$$\nCompute each term:\n$$\n\\left(\\frac{3 - \\sqrt{21}}{2}\\right)^2 = \\frac{9 - 6\\sqrt{21} + 21}{4} = \\frac{30 - 6\\sqrt{21}}{4}\n$$\n$$\n\\left(\\frac{3 + \\sqrt{21}}{2}\\right)^2 = \\frac{9 + 6\\sqrt{21} + 21}{4} = \\frac{30 + 6\\sqrt{21}}{4}\n$$\nAdd:\n$$\nx^2 + y^2 = \\frac{30 - 6\\sqrt{21}}{4} + \\frac{30 + 6\\sqrt{21}}{4} = \\frac{60}{4} = 15\n$$\n**Answer:** $x^2 + y^2 = 15$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 29093,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABCD$ be a rectangle such that $|AB| : |AD| = 2 : 3$ and let $E$ be a point on the side $\\overline{AD}$ such that $|AE| = |AB|$. Point $F$ lies on the ray $AB$ so that triangle $AFE$ and quadrilateral $CDEF$ have the same areas. Determine $|AB| : |BF|$.",
    "options": [],
    "answer": "1:2",
    "solution": "",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 29094,
    "subject": "Mathematics (Multi-modal)",
    "question": "Prove that among any three positive integers we can choose two, say $a$ and $b$, so that the number $a^3b - ab^3$ is a multiple of $10$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 29095,
    "subject": "Mathematics (Multi-modal)",
    "question": "If $x$, $y$, $z$ and $w$ are real numbers such that\n$$\nx^2 + y^2 + z^2 + w^2 + x + 3y + 5z + 7w = 4,\n$$\ndetermine the largest possible value of $x + y + z + w$.",
    "options": [],
    "answer": "2",
    "solution": "By completing the squares of sums, the given equation can be written as\n$$\n\\left(x + \\frac{1}{2}\\right)^2 + \\left(y + \\frac{3}{2}\\right)^2 + \\left(z + \\frac{5}{2}\\right)^2 + \\left(w + \\frac{7}{2}\\right)^2 = 25.\n$$\nBy the inequality of arithmetic and quadratic means, we have\n$$\n\\frac{\\left(x + \\frac{1}{2}\\right) + \\left(y + \\frac{3}{2}\\right) + \\left(z + \\frac{5}{2}\\right) + \\left(w + \\frac{7}{2}\\right)}{4} \\le \\sqrt{\\frac{\\left(x + \\frac{1}{2}\\right)^2 + \\left(y + \\frac{3}{2}\\right)^2 + \\left(z + \\frac{5}{2}\\right)^2 + \\left(w + \\frac{7}{2}\\right)^2}{4}}\n$$\nWe can conclude that $(x + \\frac{1}{2}) + (y + \\frac{3}{2}) + (z + \\frac{5}{2}) + (w + \\frac{7}{2}) \\le 10$, i.e.\n$$\nx + y + z + w \\le 2.\n$$\nNote that the value 2 can indeed be attained, e.g. for $x = 2$, $y = 1$, $z = 0$, $w = -1$.\nTherefore, the largest possible value of $x + y + z + w$ is 2.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 29096,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $p$ be a prime number. Determine all pairs $(a, b)$ of integers such that\n$$\np(a - 2) = a(b - 1).\n$$",
    "options": [],
    "answer": "All integer pairs (a, b) with a ≠ 0 and a | 2p, given by b = p + 1 − 2p/a. Equivalently, for a ∈ {±1, ±2, ±p, ±2p}, b = p + 1 − 2p/a.",
    "solution": "",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 29097,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all complex numbers $z$ such that $z^3 = \\overline{z}$.",
    "options": [],
    "answer": "{0, 1, i, -1, -i}",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 29098,
    "subject": "Mathematics (Multi-modal)",
    "question": "Determine all complex numbers $z$ for which the ratio of the imaginary part of the fifth power of $z$ to the fifth power of the imaginary part of $z$ is the smallest possible. (Revista de Matematică din Timișoara 1984)",
    "options": [],
    "answer": "The minimum value of Im(z^5) / (Im z)^5 is −4, attained for all nonzero complex numbers with arg z = π/4 + k·(π/2) for any integer k.",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 29099,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $S$ and $T$ be points inside the triangle $ABC$. The distance from $S$ to lines $AB$, $BC$ and $CA$ is $10$, $7$ and $4$, respectively. The distance from $T$ to these lines is $4$, $10$ and $16$, respectively.\nDetermine the radius of the incircle of triangle $ABC$.",
    "options": [],
    "answer": "8",
    "solution": "Denote $a = |BC|$, $b = |CA|$, $c = |AB|$. Let $P$ be the area, $s = \\frac{1}{2}(a + b + c)$ the semiperimeter, and $r$ the radius of the incircle of triangle $ABC$.\nWe can divide triangle $ABC$ into three smaller triangles by connecting point $S$ to vertices $A$, $B$ and $C$.\n![](attached_image_1.png)\n\nBy adding the areas of triangles $ABS$, $BCS$ and $CAS$, we get the area of triangle $ABC$, so we have\n$$\n2P = 10c + 7a + 4b.\n$$\nAnalogously, by observing point $T$ we get\n$$\n2P = 4c + 10a + 16b.\n$$\nIf we multiply the first equality by $2$ and add it to the second equality, we get\n$$\n6P = 24(a + b + c) = 48s, \\quad \\text{i.e. } P = 8s.\n$$\nSince $P = rs$, we have $r = 8$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 29100,
    "subject": "Mathematics (Multi-modal)",
    "question": "Determine all real numbers $a$ such that the equation\n$$\nx^2 - (5-a)x + a^2 - 11a - 46 = 0\n$$\nhas two real solutions, one of which is less than $2$, and the other greater than $2$.",
    "options": [],
    "answer": "-4 < a < 13",
    "solution": "Let the roots of the quadratic be $x_1$ and $x_2$, with $x_1 < 2 < x_2$.\n\nThe quadratic equation is $x^2 - (5-a)x + a^2 - 11a - 46 = 0$.\n\nBy Vieta's formulas:\n- $x_1 + x_2 = 5 - a$\n- $x_1 x_2 = a^2 - 11a - 46$\n\nSince the equation must have two real solutions, the discriminant must be non-negative:\n$$\nD = [-(5-a)]^2 - 4(a^2 - 11a - 46) \\= (5-a)^2 - 4(a^2 - 11a - 46)\n$$\nExpand:\n$$\n(5-a)^2 = 25 - 10a + a^2\n$$\nSo,\n$$\nD = 25 - 10a + a^2 - 4(a^2 - 11a - 46)\n$$\n$$\n= 25 - 10a + a^2 - 4a^2 + 44a + 184\n$$\n$$\n= (25 + 184) + (-10a + 44a) + (a^2 - 4a^2)\n$$\n$$\n= 209 + 34a - 3a^2\n$$\nSo, $D \\geq 0$:\n$$\n209 + 34a - 3a^2 \\geq 0\n$$\n\nNow, the roots must be on opposite sides of $2$.\nFor a quadratic $ax^2 + bx + c = 0$, the value at $x = 2$ is:\n$$\nf(2) = (2)^2 - (5-a)(2) + a^2 - 11a - 46\n$$\n$$\n= 4 - 2(5-a) + a^2 - 11a - 46\n$$\n$$\n= 4 - 10 + 2a + a^2 - 11a - 46\n$$\n$$\n= (4 - 10 - 46) + (2a - 11a) + a^2\n$$\n$$\n= (-52) - 9a + a^2\n$$\n$$\n= a^2 - 9a - 52\n$$\n\nFor the roots to be on opposite sides of $2$, $f(2)$ must be negative (since the parabola opens upwards):\n$$\nf(2) < 0 \\implies a^2 - 9a - 52 < 0\n$$\n\nSo, we have two inequalities:\n1. $209 + 34a - 3a^2 \\geq 0$\n2. $a^2 - 9a - 52 < 0$\n\nLet's solve each.\n\nFirst inequality:\n$$\n209 + 34a - 3a^2 \\geq 0\n$$\n$$\n-3a^2 + 34a + 209 \\geq 0\n$$\n$$\n3a^2 - 34a - 209 \\leq 0\n$$\nFind the roots:\n$$\na = \\frac{34 \\pm \\sqrt{(-34)^2 - 4 \\cdot 3 \\cdot (-209)}}{2 \\cdot 3}\n$$\n$$\n= \\frac{34 \\pm \\sqrt{1156 + 2508}}{6}\n$$\n$$\n= \\frac{34 \\pm \\sqrt{3664}}{6}\n$$\n$$\n\\sqrt{3664} \\approx 60.5\n$$\nSo,\n$$\na_1 = \\frac{34 + 60.5}{6} \\approx \\frac{94.5}{6} \\approx 15.75\n$$\n$$\na_2 = \\frac{34 - 60.5}{6} \\approx \\frac{-26.5}{6} \\approx -4.42\n$$\nSo, $3a^2 - 34a - 209 \\leq 0$ for $-4.42 \\leq a \\leq 15.75$.\n\nSecond inequality:\n$$\na^2 - 9a - 52 < 0\n$$\nFind the roots:\n$$\na = \\frac{9 \\pm \\sqrt{81 + 208}}{2}\n$$\n$$\n= \\frac{9 \\pm \\sqrt{289}}{2}\n$$\n$$\n= \\frac{9 \\pm 17}{2}\n$$\nSo,\n$$\na_1 = \\frac{9 + 17}{2} = 13\n$$\na_2 = \\frac{9 - 17}{2} = -4\n$$\nSo, $a^2 - 9a - 52 < 0$ for $-4 < a < 13$.\n\nThe intersection of the two intervals:\n$-4.42 \\leq a \\leq 15.75$ and $-4 < a < 13$\n\nSo, $a$ must satisfy $-4 < a < 13$.\n\n**Final answer:**\nAll real numbers $a$ such that $-4 < a < 13$.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 29101,
    "subject": "Mathematics (Multi-modal)",
    "question": "Determine all triples $(x, y, z)$ of positive real numbers that satisfy\n$$\n\\begin{aligned}\n3\\lfloor x \\rfloor - \\{y\\} + \\{z\\} &= 20.3 \\\\\n3\\lfloor y \\rfloor + 5\\lfloor z \\rfloor - \\{x\\} &= 15.1 \\\\\n\\{y\\} + \\{z\\} &= 0.9.\n\\end{aligned}\n$$\n*For a real number $t$, $\\lfloor t \\rfloor$ denotes the largest integer not greater than $t$, while $\\{t\\}$ denotes its fractional part, i.e. $\\{t\\} = t - \\lfloor t \\rfloor$. For example, if $t = 15.1$, then $\\lfloor t \\rfloor = 15$ and $\\{t\\} = 0.1$. (Unione Matematica Italiana)*",
    "options": [],
    "answer": "(7.9, 2.8, 2.1)",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 29102,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $a$ and $b$ be positive integers such that $8a^2 + 1 = b^2$. Prove that $ab$ is divisible by $3$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "We have $8a^2 + 1 = b^2$, so $b^2 - 8a^2 = 1$.\n\nConsider the equation modulo $3$:\n\n$8a^2 + 1 \\equiv b^2 \\pmod{3}$\n\nSince $8 \\equiv 2 \\pmod{3}$, we have:\n\n$2a^2 + 1 \\equiv b^2 \\pmod{3}$\n\nNow, $a^2$ modulo $3$ can be $0$ or $1$:\n- If $a \\equiv 0 \\pmod{3}$, then $a^2 \\equiv 0$.\n- If $a \\equiv 1 \\pmod{3}$ or $a \\equiv 2 \\pmod{3}$, then $a^2 \\equiv 1$.\n\nCase 1: $a \\equiv 0 \\pmod{3}$\n\nThen $2a^2 + 1 \\equiv 0 + 1 = 1 \\pmod{3}$, so $b^2 \\equiv 1 \\pmod{3}$.\n\nPossible values for $b$ modulo $3$ are $1$ or $2$ (since $1^2 \\equiv 1$, $2^2 \\equiv 1$).\n\nThus, $ab \\equiv 0 \\pmod{3}$.\n\nCase 2: $a \\equiv 1 \\pmod{3}$ or $a \\equiv 2 \\pmod{3}$\n\nThen $a^2 \\equiv 1$, so $2a^2 + 1 \\equiv 2 \\cdot 1 + 1 = 3 \\equiv 0 \\pmod{3}$, so $b^2 \\equiv 0 \\pmod{3}$.\n\nThus, $b \\equiv 0 \\pmod{3}$, so $ab \\equiv 0 \\pmod{3}$.\n\nIn all cases, $ab$ is divisible by $3$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 29103,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $a$ and $b$ be positive integers such that $a > b$ and\n$$\na - b = 5b^2 - 4a^2.\n$$\nProve that $a-b$ is a square of some positive integer. (Gazeta Matematică 2016)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "The given equality can be written more conveniently as:\n$$\n(a - b)(1 + 4(a + b)) = b^2.\n$$\nIt suffices to show that the bracketed expressions are coprime, since that implies that $a-b$ and $1+4(a+b)$ are squares of positive integers.\nDenote by $d$ the greatest common divisor of $a-b$ and $1+4(a+b)$. We have\n$$\nd = M(a - b, 1 + 4(a + b)) = M(a - b, 1 + 4(a + b) - 4(a - b)) = M(a - b, 1 + 8b).\n$$\nFrom this, it follows that $d$ divides $1+8b$.\nFrom the initial equality, we have that $d$ also divides $b^2$.\nSince $M(1+8b, b^2) = 1$, we conclude that $d = 1$, i.e. $a-b$ and $1+4(a+b)$ are coprime, which completes the proof.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 29104,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $\\overline{AB}$ and $\\overline{CD}$ be two diameters of the circle $k$ with centre $S$ and let $\\angle BAD = 28^\\circ$. The circle centred at $A$ passing through the point $S$ meets the circle $k$ at $E$ and $F$ ($D$ and $F$ are on the same side of $AB$). Find $\\angle CFS$. (Matija Bašić)",
    "options": [],
    "answer": "32°",
    "solution": "",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 29105,
    "subject": "Mathematics (Multi-modal)",
    "question": "A triangle $ABC$ is given. Circle $k$ touches $\\overline{BC}$ from outside the triangle at point $K$, and the extensions of lines $\\overline{AB}$ and $\\overline{AC}$ over points $B$ and $C$ at points $L$ and $M$, respectively. The circle with diameter $\\overline{BC}$ intersects segment $LM$ at points $P$ and $Q$ so that point $P$ lies between $L$ and $Q$.\nProve that lines $BP$ and $CQ$ intersect at the centre of circle $k$.\n(Stipe Vidak)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Denote $\\alpha = \\langle CAB$, $\\beta = \\langle ABC$, $\\gamma = \\langle BCA$.\n![](attached_image_1.png)\n\nThe centre of circle $k$ is the intersection of the bisectors of angles $\\angle CBL$ and $\\angle BCM$.\nTherefore, it suffices to show that those bisectors intersect line $LM$ at points $P$ and $Q$ respectively.\nDenote by $P'$ and $Q'$ the intersections of line $LM$ with bisectors of angles $\\angle KBL$ and $\\angle KCM$, respectively. We have to show that $P' = P$ and $Q' = Q$.\nTriangles $KBP'$ and $LBP'$ are congruent because they have a common side $\\overline{BP'}$, and $|KB| = |LB|$, $\\angle KBP' = \\angle LBP'$. It follows that $\\angle P'KB = \\angle BLP = \\angle ALM$.\nTriangle $ALM$ is isosceles, so $\\angle AMP' = \\angle AML = \\angle ALM = \\angle P'KB = 180^\\circ - \\angle P'KC$. From this it follows that quadrilateral $MCKP'$ is cyclic.\nSince $|MC| = |KC|$, by the inscribed angle theorem we have $\\angle MP'C = \\angle KP'C$.\nSince we also have $\\angle KP'B = \\angle LP'B$, it follows that\n$$\n180^\\circ = \\angle MP'C + \\angle KP'C + \\angle KP'B + \\angle LP'B = 2(\\angle KP'C + \\angle KP'B) = 2\\angle CP'B,\n$$\ntherefore the angle $\\angle CP'B$ is right. From this we can conclude that point $P'$ lies on the circle with diameter $\\overline{BC}$.\nAnalogously, we can show that $Q'$ lies on the circle with diameter $\\overline{BC}$.\nNow we have $P', Q' \\in \\{P, Q\\}$, since both $P'$ and $Q'$ are the intersections of line $LM$ and the circle with diameter $\\overline{BC}$. This is possible only if $P = P'$ and $Q = Q'$; otherwise the centre of circle $k$ would be on the same side of line $LM$ as point $A$.\nThis completes the proof.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 29106,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $\\triangle ABC$ be an acute-angled triangle such that $|AC| > |AB|$, and let $O$ be its circumcentre. The angle bisector of $\\angle BAC$ meets the side $\\overline{BC}$ at point $D$. The line through $B$ perpendicular to the line $AO$ intersects the line $AO$ at point $E$. Prove that points $A, B, D$ and $E$ all lie on the same circle.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 29107,
    "subject": "Mathematics (Multi-modal)",
    "question": "What is the maximum number of elements that a subset of $\\{1, 2, 3, \\dots, 2017\\}$ can have so that for any two distinct elements $a$ and $b$ of that subset, the number $a + b$ is not divisible by $a - b$? (Hong Kong 1992)",
    "options": [],
    "answer": "673",
    "solution": "",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 29108,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $n$ be a positive integer. Vertices of a regular $2n$-gon are painted red and blue in an alternating fashion. All the sides and diagonals of the $2n$-gon are drawn. If the number of segments with ends of the same colour is 3192, find the number of segments with ends of different colours. (Mea Bombardelli)",
    "options": [],
    "answer": "3249",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 29109,
    "subject": "Mathematics (Multi-modal)",
    "question": "A city has $M$ streets and $N$ squares, where $M$ and $N$ are positive integers such that $M > N$. Each street connects two squares and does not go through any other squares.\nThe citizens wish to change the appearance of the city. This year, each street will be coloured for the first time, in either red or blue. It is also determined that each subsequent year one square will be chosen and all the streets leading to that square will simultaneously change colour from blue to red and vice versa.\nProve that the citizens can choose the initial colours of streets so that it can never happen that all streets have the same colour.\n(Russia 1994)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let us call one arrangement of colours on the streets a *colouring*, and one selection of a square and the change of colours of all streets leading to that square a *transformation*.\nNote that if we can get from one colouring to another by a sequence of transformations, then we can also get from the second colouring to the first one using that same sequence of transformations.\nTherefore, the initial claim is equivalent to claiming that, if we start from a colouring where all streets have the same colour, there exists a colouring that cannot be achieved by any sequence of transformations.\n\nIn any sequence of transformations, we can assume that each square either does not appear or appears exactly once. Namely, if we perform an even number of transformations on some square, the result will be the same as if we did not perform any transformations, and an odd number of transformations will have the same result as if there was exactly one transformation.\n\nFurthermore, note that in the sequence of transformations which are performed to get from one colouring to another, the order of the transformations is not important, i.e. it is only important to determine the set of squares included in these transformations.\nThe number of different sets of squares is $2^N$, since for each of the $N$ squares we can decide whether it is included in the set or not.\n\nNote that the set where all squares are included leads to the colouring which is identical to the initial colouring, since we changed the colour of each street exactly twice. The same result can obviously also be obtained if we do not include any square in the set. Since those two sets lead to the same colouring, the number of different colourings that can be achieved from any initial colouring is not greater than $2^N - 1$.\n\nWe have two possible initial colourings (one where all the streets are red, and one where all the streets are blue), so the total number of colourings that can be achieved from an initial colouring is not greater than $2 \\cdot (2^N - 1) = 2^{N+1} - 2$.\n\nSince there are $M$ streets, and each street can be coloured in one of two possible colours (red or blue), the total number of possible colourings is $2^M$.\n\nSince $M > N$, we have $2^M \\ge 2^{N+1} > 2^{N+1} - 2$, so we can conclude that there exists a colouring that cannot be achieved from the initial monochromatic colourings. This completes the proof.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 29110,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABCD$ be a trapezium with parallel sides $\\overline{AB}$ and $\\overline{CD}$, and let $P$ be the intersection of its diagonals. If the area of the trapezium equals $25$, and the area of the triangle $CDP$ equals $9$, find the area of the triangle $ABP$. (Stipe Vidak)",
    "options": [],
    "answer": "4",
    "solution": "",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 29111,
    "subject": "Mathematics (Multi-modal)",
    "question": "A cutting of an $8 \\times 8$ chessboard into rectangles is called diversified if the following two conditions hold:\n* Each rectangle contains the same number of black and white cells.\n* No two rectangles have the same total number of cells.\nFind the largest $n$ such that a diversified cutting into $n$ rectangles exists. (Ratko Višak)",
    "options": [],
    "answer": "7",
    "solution": "",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 29112,
    "subject": "Mathematics (Multi-modal)",
    "question": "In a triangle $ABC$ the interior bisector at vertex $C$ meets the side $\\overline{AB}$ at point $D$. Let $a$ and $b$ be the lengths of the sides $\\overline{BC}$ and $\\overline{AC}$ respectively. If $|CD| = \\frac{ab}{a+b}$, determine $\\angle ACB$.",
    "options": [],
    "answer": "120°",
    "solution": "",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 29113,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all rational numbers $a$ and $b$ such that\n$$\nsin 75^\\circ \\cdot \\cos 15^\\circ = a + \\sqrt{b}.\n$$\n(Nikola Adžaga)",
    "options": [],
    "answer": "a = 1/2, b = 3/16",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 29114,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $x$ and $y$ be positive real numbers such that\n$$\n2^{x^2} = 16^y \\quad \\text{and} \\quad \\log_{\\sqrt{2017}} x + \\log_{\\sqrt{2017}} y > 0.\n$$\nProve that $y > \\frac{1}{2}$. (Kristina Ana Škreb)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 29115,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $a$ and $b$ be positive integers of different parity. Prove that $(a+3b)(5a+7b)$ cannot be a perfect square.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "By considering the given expression modulo $8$, we get\n$$\n(a + 3b)(5a + 7b) \\equiv 5a^2 + 22ab + 21b^2 \\equiv 5a^2 - 10ab + 5b^2 \\equiv 5(a - b)^2 \\pmod{8}.\n$$\nLet us assume the contrary, that $(a + 3b)(5a + 7b) = k^2$ for some positive integer $k$.\nThen\n$$\nk^2 \\equiv 5(a - b)^2 \\pmod{8}.\n$$\nSince $a$ and $b$ are of different parity, $a - b$ is odd, so that $(a - b)^2$ gives remainder $1$ when divided by $8$.\nThis implies $k^2 \\equiv 5 \\pmod{8}$. However, $5$ is not a quadratic residue modulo $8$.\nWe have arrived at a contradiction, so we conclude that the initial assumption was false. Therefore, $(a + 3b)(5a + 7b)$ is not a perfect square.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 29116,
    "subject": "Mathematics (Multi-modal)",
    "question": "Does there exist a positive integer $m$ such that $2^{m^2} - 4$ is divisible by $7$? (Ukraine 2013)",
    "options": [],
    "answer": "No; there is no positive integer m such that 2^{m^2} − 4 is divisible by 7.",
    "solution": "",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 29117,
    "subject": "Mathematics (Multi-modal)",
    "question": "A number written only with digits $2$ and $3$ is called *happy*. Therefore, the happy numbers are $2$, $3$, $22$, $23$, $32$, $33$, $222$, $223$, $232$, $233$, $322$, $323$, $332$, $333$, ... Determine the $2050$th happy number.",
    "options": [],
    "answer": "22222222233",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 29118,
    "subject": "Mathematics (Multi-modal)",
    "question": "Determine the product $(1 + \\tan 1^\\circ)(1 + \\tan 2^\\circ)\\cdots(1 + \\tan 45^\\circ)$.",
    "options": [],
    "answer": "2^23",
    "solution": "Let $P = (1 + \\tan 1^\\circ)(1 + \\tan 2^\\circ)\\cdots(1 + \\tan 45^\\circ)$.\n\nRecall that $1 + \\tan x = \\frac{\\sin x + \\cos x}{\\cos x}$.\nBut we can use the tangent addition formula:\n\n$$\n\\tan(45^\\circ + x) = \\frac{\\tan 45^\\circ + \\tan x}{1 - \\tan 45^\\circ \\tan x} = \\frac{1 + \\tan x}{1 - \\tan x}\n$$\n\nSo,\n$$\n1 + \\tan x = (1 - \\tan x) \\tan(45^\\circ + x)\n$$\nBut this is not quite correct. Let's solve for $1 + \\tan x$:\n\nFrom the addition formula:\n$$\n\\tan(A + B) = \\frac{\\tan A + \\tan B}{1 - \\tan A \\tan B}\n$$\nLet $A = 45^\\circ$, $B = x$:\n$$\n\\tan(45^\\circ + x) = \\frac{1 + \\tan x}{1 - \\tan x}\n$$\nSo,\n$$\n1 + \\tan x = (1 - \\tan x) \\tan(45^\\circ + x)\n$$\nBut this is not correct; actually,\n$$\n\\tan(45^\\circ + x) = \\frac{1 + \\tan x}{1 - \\tan x}\n\\implies 1 + \\tan x = (1 - \\tan x) \\tan(45^\\circ + x)\n$$\nBut this is not quite right. Let's solve for $1 + \\tan x$:\n\nMultiply both sides by $1 - \\tan x$:\n$$\n\\tan(45^\\circ + x)(1 - \\tan x) = 1 + \\tan x\n$$\nSo,\n$$\n1 + \\tan x = \\tan(45^\\circ + x)(1 - \\tan x)\n$$\nTherefore,\n$$\nP = \\prod_{x=1}^{45} (1 + \\tan x^\\circ) = \\prod_{x=1}^{45} \\tan(45^\\circ + x^\\circ)(1 - \\tan x^\\circ)\n$$\nBut $\\prod_{x=1}^{45} (1 - \\tan x^\\circ)$ can be paired with $\\prod_{x=1}^{45} (1 + \\tan x^\\circ)$.\n\nAlternatively, let's try to write $1 + \\tan x = \\frac{\\sin x + \\cos x}{\\cos x}$.\nBut let's use the tangent addition formula as above.\n\nLet us note that:\n$$\n\\tan(45^\\circ + x) = \\frac{1 + \\tan x}{1 - \\tan x}\n\\implies 1 + \\tan x = (1 - \\tan x) \\tan(45^\\circ + x)\n$$\nSo,\n$$\n1 + \\tan x = \\tan(45^\\circ + x) - \\tan x \\tan(45^\\circ + x)\n$$\nBut perhaps it's better to use the product formula.\n\nLet us consider the product $Q = \\prod_{x=1}^{45} (1 + \\tan x^\\circ)$.\n\nLet us note that $\\tan(45^\\circ - x) = \\frac{1 - \\tan x}{1 + \\tan x}$.\n\nTherefore,\n$$\n1 + \\tan x = \\frac{1}{1 - \\tan(45^\\circ - x)}\n$$\nSo,\n$$\n\\prod_{x=1}^{45} (1 + \\tan x^\\circ) = \\prod_{x=1}^{45} \\frac{1}{1 - \\tan(45^\\circ - x^\\circ)}\n$$\nBut $x$ runs from $1$ to $45$, so $45^\\circ - x^\\circ$ runs from $44^\\circ$ down to $0^\\circ$.\n\nBut $\\tan 0^\\circ = 0$, so $1 - \\tan 0^\\circ = 1$.\n\nAlternatively, let's try to compute the value for small $n$ and look for a pattern.\n\nLet us try to compute $P = (1 + \\tan 1^\\circ)(1 + \\tan 2^\\circ)\\cdots(1 + \\tan 45^\\circ)$.\n\nLet us note that $\\tan 45^\\circ = 1$, so $1 + \\tan 45^\\circ = 2$.\n\nLet us try to pair the terms:\n\nFor $x$ from $1$ to $44$, pair $x$ and $45 - x$.\n\nNote that:\n$$\n(1 + \\tan x^\\circ)(1 + \\tan(45^\\circ - x^\\circ))\n$$\nBut $\\tan(45^\\circ - x^\\circ) = \\frac{1 - \\tan x^\\circ}{1 + \\tan x^\\circ}$.\nSo,\n$$\n1 + \\tan(45^\\circ - x^\\circ) = 1 + \\frac{1 - \\tan x^\\circ}{1 + \\tan x^\\circ} = \\frac{(1 + \\tan x^\\circ) + (1 - \\tan x^\\circ)}{1 + \\tan x^\\circ} = \\frac{2}{1 + \\tan x^\\circ}\n$$\nTherefore,\n$$\n(1 + \\tan x^\\circ)(1 + \\tan(45^\\circ - x^\\circ)) = (1 + \\tan x^\\circ) \\cdot \\frac{2}{1 + \\tan x^\\circ} = 2\n$$\nSo, for $x = 1$ to $44$, we can pair $x$ and $45 - x$ to get $2$ for each pair.\n\nThere are $44$ numbers from $1$ to $44$, so $22$ pairs, each giving $2$, and then the last term is $1 + \\tan 45^\\circ = 2$.\n\nTherefore,\n$$\nP = 2^{22} \\times 2 = 2^{23}\n$$\n\n**Final answer:**\n$$\n(1 + \\tan 1^\\circ)(1 + \\tan 2^\\circ)\\cdots(1 + \\tan 45^\\circ) = 2^{23}\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 29119,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all polynomials $P$ with real coefficients such that\n$$\nP(x^2) + 2P(x) = (P(x))^2 + 2\n$$\nholds for every real number $x$.",
    "options": [],
    "answer": "P(x) = 1; P(x) = 2; and P(x) = x^n + 1 for any positive integer n",
    "solution": "The original equation is equivalent to $P(x^2) - 1 = P^2(x) - 2P(x) + 1$, i.e.\n$$\nP(x^2) - 1 = (P(x) - 1)^2.\n$$\nIf we define $Q(x) := P(x) - 1$, the above equation becomes\n$$\nQ(x^2) = Q^2(x).\n$$\nIf $Q(x)$ is a constant polynomial, this implies that it has to be $0$ or $1$. On the other hand, if the degree of $Q$ is a positive integer $n$, we can write $Q(x) = a_n x^n + R(x)$, where $R$ is a polynomial of degree $r < n$. Plugging this into the equation, we get\n$$\na_n x^{2n} + R(x^2) = a_n^2 x^{2n} + 2a_n x^n R(x) + R^2(x).\n$$\nWe conclude $a_n = 1$ and $R(x^2) = 2a_n x^n R(x) + R^2(x)$.\nNotice that the left-hand side of the above equality is of degree $2r$, while the right-hand side is of degree $n + r$. Since $r < n$, it follows that $r = 0$. Therefore, $Q(x) = x^n$.\nThe solutions of the initial equation are $P(x) = 1$, $P(x) = 2$ and $P(x) = x^n + 1$, for any positive integer $n$.",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 29120,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all pairs $(m, n)$ of positive integers which satisfy the equation\n$$\nm n^2 = 100(n + 1).\n$$\n(Ukraine 2009)",
    "options": [],
    "answer": "(200, 1), (75, 2), (24, 5), (11, 10)",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 29121,
    "subject": "Mathematics (Multi-modal)",
    "question": "A triangle $ABC$ is given, with altitudes $AD$, $BE$ and $CF$, and orthocentre $H$. The segments $EF$ and $AD$ intersect at point $G$. The segment $AK$ is the diameter of the circumcircle of triangle $ABC$ and it intersects $BC$ at point $M$. Prove that the lines $GM$ and $HK$ are parallel. (Go Geometry)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "![](attached_image_1.png)\nSince $BFC$ is a right-angled triangle, we have $\\angle BCF = 90^\\circ - \\beta$. As $CDH$ is also a right-angled triangle, we get $\\angle DHC = \\beta$.\nIt follows that $\\angle AHF = \\beta$ (because the angles $\\angle DHC$ and $\\angle AHF$ are vertically opposite).\nAs $ABKC$ is a cyclic quadrilateral, inscribed angles which subtend $\\overline{AC}$ are equal to $\\angle AKC = \\angle ABC = \\beta$. We have that $\\angle ACK$ is a right angle because it subtends\nthe diameter $\\overline{AK}$ (using Thales' theorem). We now infer that the triangles $AFH$ and $ACK$ are similar (because we know that $\\angle AFH$ and $\\angle ACK$ are right angles, whereas $\\angle AKC = \\angle AHF = \\beta$). From this similarity, we get\n$$\n\\frac{|AF|}{|AH|} = \\frac{|AC|}{|AK|}. \\qquad (4)\n$$\nWe also know that $AFHE$ is a cyclic quadrilateral, because $\\angle AFH + \\angle HEA = 90^\\circ + 90^\\circ = 180^\\circ$.\nTherefore, $\\angle FEA = \\angle FHA = \\beta$ holds, because both these angles subtend the chord $\\overline{AF}$. Thus, triangle $AFE$ shares an angle ($\\angle BAC = \\angle FAE$) with triangle $ABC$. Since we also have $\\angle FEA = \\beta$, it follows that $AEF$ and $ABC$ are similar, so that\n$$\n\\frac{|AE|}{|AF|} = \\frac{|AB|}{|AC|}. \\qquad (5)\n$$\nThe quadrilateral $ABKC$ being cyclic, we know that the inscribed angles which subtend $\\overline{AB}$ are equal to $\\angle BKA = \\angle BCA = \\gamma$.\nIn triangle $AFE$ we have $\\angle AFE = \\gamma$ (by similarity with triangle $ABC$), so its complementary angle equals $\\angle HFE = 90^\\circ - \\gamma$. Notice that $\\angle HAE = \\angle HFE = 90^\\circ - \\gamma$, because these are both inscribed angles inside the cyclic quadrilateral $AFHE$. It follows that the triangles $AGE$ and $AMB$ are similar, so that\n$$\n\\frac{|AG|}{|AE|} = \\frac{|AM|}{|AB|}. \\qquad (6)\n$$\nBy multiplying (4), (5) and (6) we obtain the equality of ratios $\\frac{|AG|}{|AH|}$ and $\\frac{|AM|}{|AK|}$.\nFinally, using the converse to the intercept theorem, we conclude that the lines $GM$ and $HK$ are parallel.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 29122,
    "subject": "Mathematics (Multi-modal)",
    "question": "A tetrahedron is given, having one edge of length $3$ and all other edges of length $2$. Find the volume of that tetrahedron. (Stipe Vidak)",
    "options": [],
    "answer": "sqrt(3)/2",
    "solution": "",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 29123,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $\\triangle ABC$ be an acute-angled triangle with $\\angle BAC = 75^\\circ$. Let $P$ be the midpoint of the side $\\overline{BC}$, and let $M$ and $N$ be the feet of the altitudes from vertices $B$ and $C$ respectively. Determine the angle $\\angle MPN$.",
    "options": [],
    "answer": "30°",
    "solution": "",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 29124,
    "subject": "Mathematics (Multi-modal)",
    "question": "What is the maximum number of elements that a finite set $S$ can have so that among any three elements of $S$ there exist two distinct whose sum is an element of $S$ as well? (Russia 2000)",
    "options": [],
    "answer": "5",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 29125,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $R$ be a positive integer less than $2017$. Exactly $R$ vertices of a regular $2017$-gon are red, while the remaining ones are blue. Prove that the number of isosceles triangles having all three vertices of the same colour does not depend on the arrangement of the blue and the red vertices. (Russia 1994)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "First, notice that none of the triangles formed by taking three vertices of the given polygon is equilateral. If there were one, we would have the same number of vertices of the original polygon between any two of the triangle's vertices. This would imply that $3$ divides $2014$, which is false.\n\nLet us now prove that each segment (whose endpoints are vertices of the given polygon) belongs to exactly $3$ different isosceles triangles.\n\n![](attached_image_1.png)\n\nConsider an arbitrary segment. Its endpoints divide the circumcircle of the polygon into two arcs containing $2015$ vertices. One of the arcs contains an odd number, and the other contains an even number of vertices. Therefore, there exists a unique isosceles triangle with the chosen segment as its base (the third vertex is the midpoint of the arc containing an odd number of vertices). In a similar fashion, we infer that the chosen segment is a base in exactly two isosceles triangles – there are two possible choices of the third segment, both on the longer arc.\n\nLet $B$ be the number of blue points, so that $B + R = 2017$. Let $d_b, d_r, d_s$, be the number of blue (both endpoints blue), red, and two-colour segments, respectively. Furthermore, let $t_b, t_r, t_{sb}, t_{sr}$ be the number of isosceles triangles with the vertices coloured all blue, all red, two blue – one red, and two red – one blue, respectively.\n\nFrom the fact that each segment belongs to three isosceles triangles, we get:\n$$\n\\begin{aligned}\n3d_b &= 3t_b + t_{sb}, \\\\\n3d_r &= 3t_r + t_{sr}, \\\\\n3d_s &= 2t_{sb} + 2t_{sr}.\n\\end{aligned}\n$$\nThis implies $3d_b + 3d_r - \\frac{3}{2}d_s = 3t_b + 3t_r$, i.e.\n$$\nt_b + t_r = d_b + d_r - \\frac{1}{2}d_s = \\frac{1}{2} \\cdot (B(B-1) + R(R-1) - BR),\n$$\nwhich depends only on the number of the red and blue points, but not on their arrangement.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 29126,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $\\triangle ABC$ be an acute-angled triangle and let $BCD$, $ACE$ and $ABF$ be equilateral triangles drawn on the outside of triangle $ABC$. Let $M$ be the midpoint of the segment $\\overline{BD}$, and let $O$ be the centre of the triangle $ACE$. Prove that $|AM| : |OF| = \\frac{\\sqrt{3}}{2}$.",
    "options": [],
    "answer": "sqrt(3)/2",
    "solution": "",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 29127,
    "subject": "Mathematics (Multi-modal)",
    "question": "Twenty points with integer coordinates are given in the coordinate plane so that no three of them are collinear. Prove that there exists a triangle with vertices chosen among the given points whose centroid is also a point with integer coordinates.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let the given points be $A_1, A_2, \\ldots, A_{20}$, with $A_i = (x_i, y_i)$, where $x_i, y_i$ are integers.\n\nThe centroid of a triangle with vertices $A_i = (x_i, y_i)$, $A_j = (x_j, y_j)$, $A_k = (x_k, y_k)$ is\n$$\n\\left(\\frac{x_i + x_j + x_k}{3}, \\frac{y_i + y_j + y_k}{3}\\right).\n$$\n\nWe want to find three points among the $20$ such that both $x_i + x_j + x_k$ and $y_i + y_j + y_k$ are divisible by $3$.\n\nConsider the residues modulo $3$ of the $x$- and $y$-coordinates of each point. Each point can be assigned a pair $(a, b)$, where $a \\equiv x_i \\pmod{3}$, $b \\equiv y_i \\pmod{3}$.\n\nThere are $3 \\times 3 = 9$ possible residue pairs modulo $3$.\n\nBy the pigeonhole principle, among $20$ points, at least $\\left\\lceil \\frac{20}{9} \\right\\rceil = 3$ points must have the same residue pair $(a, b)$. That is, there exist three points $A_i, A_j, A_k$ such that $x_i \\equiv x_j \\equiv x_k \\pmod{3}$ and $y_i \\equiv y_j \\equiv y_k \\pmod{3}$.\n\nTherefore, $x_i + x_j + x_k \\equiv 0 \\pmod{3}$ and $y_i + y_j + y_k \\equiv 0 \\pmod{3}$, so the centroid has integer coordinates.\n\nIt remains to check that these three points are not collinear. But the problem states that no three of the $20$ points are collinear, so any three points form a triangle.\n\nThus, there exists a triangle with vertices among the given points whose centroid is also a point with integer coordinates.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 29128,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all positive integers $b$ such that $11 \\cdot 22 \\cdot 33 = 13310$ holds in base $b$.",
    "options": [],
    "answer": "6",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 29129,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $c$ and $d$ be positive divisors of a positive integer $n$. If $c > d$, prove that\n$$\nc > d + \\frac{d^2}{n}.\n$$\n(Russia 2011)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Consider $\\frac{n}{d}$ and $\\frac{n}{c}$. These are also positive divisors of $n$ which satisfy $\\frac{n}{d} > \\frac{n}{c}$.\nWe now have\n$$\n\\frac{n}{d} - \\frac{n}{c} \\ge 1,\n$$\nfrom which\n$$\nc - d \\ge \\frac{dc}{n} > \\frac{d^2}{n}\n$$\nfollows.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 29130,
    "subject": "Mathematics (Multi-modal)",
    "question": "There are seven people in a room. Four of them each know exactly one other person, while the other three each know exactly two people. All acquaintances are mutual. What is the probability that two randomly chosen people do not know each other? (Graduate Management Admission Test)",
    "options": [],
    "answer": "16/21",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 29131,
    "subject": "Mathematics (Multi-modal)",
    "question": "Sequences $(x_n)$ and $(y_n)$ are given recursively:\n$x_1 = 3$, $y_1 = 1$,\n$x_{n+1} = 3x_n + y_n$, for all $n \\in \\mathbb{N}$;\n$y_{n+1} = x_n + 3y_n$, for all $n \\in \\mathbb{N}$.\nProve that $x_{2017}^2 - y_{2017}^2 = 8^{2017}$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 29132,
    "subject": "Mathematics (Multi-modal)",
    "question": "Determine the number of positive integers $c \\le 1000000$, that can be expressed as $c = a^2 + 3b^2 - 4ab$ for some non-zero integers $a$ and $b$.",
    "options": [],
    "answer": "749998",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 29133,
    "subject": "Mathematics (Multi-modal)",
    "question": "Determine all functions $f: \\mathbb{R} \\to \\mathbb{R}$ such that\n$$\nf(x + f(y)) = f(f(y)) + 2x f(y) + x^2\n$$\nholds for all real numbers $x$ and $y$.",
    "options": [],
    "answer": "f(x) = x^2 + a for any real constant a",
    "solution": "Considering $x = -f(y)$ we get $f(0) = f(f(y)) - (f(y))^2$. Denoting $f(0)$ by $a$ and substituting $y = 0$ we arrive at $f(a) = a^2 + a$. Now we return to the original equality and plug in $y = 0$:\n$$\nf(x + a) = f(a) + 2x a + x^2 = x^2 + 2x a + a^2 + a = (x + a)^2 + a.\n$$\nThus, all the possible solutions are of the form $f(x) = x^2 + a$, where $a$ is a real constant. A routine check shows that every function of this form is indeed a solution to the original equation.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 29134,
    "subject": "Mathematics (Multi-modal)",
    "question": "In the rectangle with vertices $A(-1, 0)$, $B(1, 0)$, $C(1, 1)$, $D(-1, 1)$ a point $T$ is randomly chosen. Find the probability that $T$ is closer to $A$ than to the point $P(2, 1)$.",
    "options": [],
    "answer": "3/4",
    "solution": "",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 29135,
    "subject": "Mathematics (Multi-modal)",
    "question": "A sequence $a_0, a_1, a_2, \\dots$ of positive real numbers is given such that\n$$\na_1 = 1 - a_0, \\quad a_{n+1} = 1 - a_n(1 - a_n) \\quad \\text{for all } n \\ge 1.\n$$\nProve that\n$$\na_0 a_1 \\cdots a_n \\left( \\frac{1}{a_0} + \\frac{1}{a_1} + \\cdots + \\frac{1}{a_n} \\right) = 1\n$$\nholds for every positive integer $n$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 29136,
    "subject": "Mathematics (Multi-modal)",
    "question": "We call a point $P$ inside a triangle $ABC$ marvelous if exactly $27$ rays can be drawn from it, intersecting the sides of $ABC$ such that the triangle is divided into $27$ smaller triangles of equal areas. Determine the total number of marvelous points inside a given triangle $ABC$.\n(Naboj)",
    "options": [],
    "answer": "325",
    "solution": "Let $P$ be a marvelous point of $ABC$ and let (without loss of generality) the area of $ABC$ be $27$. We first note that among the $27$ rays, there must always be $PA$, $PB$ and $PC$; otherwise not all the shapes determined by the rays will be triangles.\n\nEach of the $27$ smaller triangles is contained within (or is equal to) one of the triangles $PAB$, $PBC$, $PCA$. Let $PAB$ contain $m$, and $PBC$ contain $n$ smaller triangles. Numbers $m$ and $n$ are elements of $\\{1, 2, \\dots, 25\\}$ and they satisfy $m + n \\le 26$.\n\nWe now consider the function which assigns the pair $(m, n)$ to each marvelous point of $ABC$. We claim that it is a bijection from the set of marvelous points to the set of pairs of positive integers $(m, n)$ which satisfy $m + n \\le 26$.\n\nWe first show that for each pair $(m, n)$ we have at most one marvelous point. Let $P$ be a marvelous point assigned to $(m, n)$. This means that $PAB$ is divided into $m$ triangles, while $PBC$ is divided into $n$ triangles, each of whose areas equal $1$. The $m$ triangles contained within $PAB$ have a common altitude whose length equals $\\frac{2m}{|AB|}$, so we know that point $P$ lies on the line $p$ which is parallel to $AB$, and whose distance to $AB$ equals $\\frac{2m}{|AB|}$. Analogously, we infer that point $P$ lies on the line $q$ which is parallel to $BC$, and whose distance to $BC$ equals $\\frac{2n}{|BC|}$. Since $p$ and $q$ are uniquely determined by $m$ and $n$, the fact that their intersection contains at most one point shows that there is at most one point to which the pair $(m, n)$ is assigned.\n\n![](attached_image_1.png)\n\nWe now show that each pair $(m, n)$ is indeed assigned to a marvelous point. Given $m$ and $n$, consider the point $P$ defined as the intersection of lines $p$ and $q$, as in the previous paragraph. Drawing rays from point $P$, we can divide triangles $PAB$ and $PBC$ into $m$ and $n$ smaller triangles (of area equal to $1$), respectively. Finally, triangle $PAC$ has area $27 - m - n$, so we can divide it into $27 - m - n$ triangles of area $1$ by drawing rays originating in $P$. Thus, we can divide the triangle $ABC$ into $27$ smaller triangles of equal areas by drawing rays originating in $P$. By definition, this means that $P$ is a marvelous point to which the pair $(m, n)$ is assigned.\n\nWe have yet to determine the number of ways in which the pair $(m, n)$ can be chosen. For each $m \\in \\{1, 2, \\dots, 25\\}$, we can choose $n$ in exactly $26 - m$ ways. Therefore, the solution is\n$$\n\\sum_{m=1}^{25} (26 - m) = \\frac{25 \\cdot 26}{2} = 325.\n$$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 29137,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find the smallest multiple of $84$ whose decimal representation contains only digits $6$ and $7$.",
    "options": [],
    "answer": "76776",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 29138,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABC$ be an acute-angled triangle. The tangents at $A$ and $B$ to its circumcircle intersect at $M$. The line through $M$ parallel to the side $BC$ intersects the side $CA$ at point $N$. Prove that $|BN| = |CN|$ holds.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 29139,
    "subject": "Mathematics (Multi-modal)",
    "question": "An acute-angled triangle $ABC$ is given, such that $|AB| > |AC|$. Let $O$ be the centre of the circumcircle, and let $OQ$ be a diameter of the circumcircle of triangle $BOC$. The line parallel to $BC$ passing through $A$ intersects the line $CQ$ at $M$, while the line parallel to $CQ$ and passing through $A$ intersects the line $BC$ at $N$. Denote by $T$ the intersection of $AQ$ and $MN$.\nShow that $T$ lies on the circumcircle of triangle $BOC$.\n(Ukraine 2005)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let the line $AQ$ intersect the circumcircle of $BOC$ at point $Q$ and $T'$. Notice that $\\overline{OQ}$ is a diameter of this circle, so $\\angle OT'Q = 90^\\circ$.\nLet $P$ be the midpoint of the segment $\\overline{AC}$. Since $O$ is the centre of the circumcircle of $ABC$, we have $\\angle OPA = 90^\\circ$.\nFrom here, we get $\\angle OT'A = 180^\\circ - \\angle OT'Q = 90^\\circ = \\angle OPA$, which implies that the quadrilateral $AOT'P$ is cyclic.\n\n![](attached_image_1.png)\n\nNotice that point $P$ lies on the line $MN$ (moreover, $P$ is the midpoint of $\\overline{MN}$) because $ANCM$ is a parallelogram. Thus,\n$$\n\\angle AT'M = \\angle AT'P = \\angle AOP = \\frac{1}{2} \\angle AOC = \\angle ABC.\n$$\nAs in the previous solution, we show that the quadrilateral $ABNT'$ is cyclic and that the points $M, N$ and $T'$ are collinear. This implies $T \\equiv T'$, which concludes the proof.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 29140,
    "subject": "Mathematics (Multi-modal)",
    "question": "There are $3000$ points marked on a circle, and there is a grasshopper standing on one of these points. With every jump, the grasshopper jumps over one or two marked points (in the clockwise direction) and lands on the next marked point. Determine the smallest possible number of jumps made by the grasshopper if it landed on every marked point at least once and then returned to the starting point. (Austria 2016)",
    "options": [],
    "answer": "3001",
    "solution": "",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 29141,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find the number of nine-digit numbers with digits $1$, $2$, $3$, $4$, $5$, $6$, $7$, $8$ and $9$, such that no three consecutive digits equal $123$, $246$ or $678$.",
    "options": [],
    "answer": "348000",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 29142,
    "subject": "Mathematics (Multi-modal)",
    "question": "On a $2017 \\times 2017$ board, some of the squares are occupied by a single ladybird; the rest of the squares are empty. The ladybirds move, never leaving the board, according to the following principles. Every second, each ladybird moves to a neighbouring square. The moves are horizontal (to the square immediately to the right or to the left of the current square), or vertical (to the square above or below the one currently occupied). A ladybird which makes a horizontal move must move vertically in its next move. Similarly, a ladybird which makes a vertical move must move horizontally in its next move.\n\nDetermine the smallest number of ladybirds such that, regardless of their initial positions and their chosen paths, we may be sure that two of them will eventually find themselves in the same square, at the same moment.\n\n(P. Soberón, *Problem-Solving Methods in Combinatorics*)",
    "options": [],
    "answer": "4064257",
    "solution": "We claim that the required number is $2016^2 + 1$. Let us first show that we can find an arrangement of $2016^2$ ladybirds and choose their paths so that no two occupy the same square at the same moment in time. We place the ladybirds in the lower left $2016 \\times 2016$ square of the board and let them all move in the same manner: up, right, down, left, up, right... We see that no two ladybirds will meet after the first 4\n\n![](attached_image_1.png)\n\nseconds. Since we arrive at the initial position after that, we see that no two ladybirds will ever occupy the same square at the same time.\n\nWe now show that, if the board contains $2016^2 + 1$ ladybirds, a collision must occur, regardless of the initial arrangement and the ladybirds' paths.\n\nWe label the squares with one of the four labels, $A$, $B$, $C$ and $D$, so that the squares in odd rows alternate between labels $A$ and $B$, while the squares in even rows alternate between $C$ and $D$. We will call a square which has been labelled by $A$ an $A$-square.\n\n![](attached_image_2.png)\n\nTwo observations are crucial for our solution. A ladybird occupying a $B$-square or a $C$-square will after *one* second move to an $A$-square or a $D$-square. Similarly, a ladybird which occupies an $A$-square will after *two* seconds be sitting in a $D$-square.\n\nThe board contains $2016^2 + 1$ ladybirds, so we can assume that at least $1008 \\cdot 2016 + 1$ of them occupy an $A$- or a $D$-square. If the opposite were true, we would have at least $1008 \\cdot 2016 + 1$ on a $B$-square or a $C$-square, so that after one second we would arrive at the desired situation. Since the number of $D$-squares equals $1008^2$, the $A$-squares contain at least $1008^2 + 1$ ladybirds.\n\nAll the ladybirds which are now occupying the $A$-squares will after two seconds move to $D$-squares. This means that there will be at least $1008^2 + 1$ ladybirds on $1008^2$ $D$-squares, so that at least two of them will be in the same square.",
    "topic": "Discrete Mathematics",
    "subtopic": "Combinatorics"
  },
  {
    "id": 29143,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $z_1$ and $z_2$ be complex numbers such that $|z_1| = |z_2| = 1$, and let $a$ and $b$ be real numbers such that $a + b = 1$. Prove\n$$\n|a z_1 + b z_2| \\ge \\frac{1}{2} |z_1 + z_2|.\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 29144,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $A$, $B$ and $C$ be points on the curve (hyperbola) with equation $xy = 1$. Prove that the orthocentre of the triangle $ABC$ also belongs to that curve.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 29145,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $a$, $b$ and $c$ be positive real numbers. Prove that\n$$\n\\frac{3}{2} < \\frac{4a+b}{a+4b} + \\frac{4b+c}{b+4c} + \\frac{4c+a}{c+4a} < 9. \\quad (\\text{Japan 2009})\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Notice that\n$$\n\\frac{4a+b}{4(4a+b)} < \\frac{4a+b}{a+4b} < \\frac{4(a+4b)}{a+4b} = 4,\n$$\nand analogously\n$$\n\\frac{1}{4} < \\frac{4b+c}{b+4c} < 4,\n$$\n$$\n\\frac{1}{4} < \\frac{4c+a}{c+4a} < 4.\n$$\nWithout loss of generality we may assume that $a \\le b \\le c$. Hence we have\n$$\n\\frac{4c+a}{c+4a} = \\frac{c+3c+a}{c+4a} \\ge \\frac{c+4a}{c+4a} = 1\n$$\nand\n$$\n\\frac{4a+b}{a+4b} = \\frac{a+3a+b}{a+4b} \\le \\frac{a+4b}{a+4b} = 1.\n$$\nFinally,\n$$\n\\frac{3}{2} = \\frac{1}{4} + \\frac{1}{4} + 1 < \\frac{4a+b}{a+4b} + \\frac{4b+c}{b+4c} + \\frac{4c+a}{c+4a} < 1 + 4 + 4 = 9.\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 29146,
    "subject": "Mathematics (Multi-modal)",
    "question": "Prove that for all positive real numbers $a$, $b$ and $c$ the following inequality holds\n$$\n\\frac{a}{b+c} + \\frac{b}{c+a} + \\frac{c}{a+b} + \\sqrt{\\frac{ab+bc+ca}{a^2+b^2+c^2}} \\ge \\frac{5}{2}.\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Because of simplicity let $A = a^2 + b^2 + c^2$ and $B = ab + bc + ca$. The CSB inequality gives us\n$$\n\\left( \\frac{a}{b+c} + \\frac{b}{c+a} + \\frac{c}{a+b} \\right) (a(b+c) + b(c+a) + c(a+b)) \\ge (a+b+c)^2,\n$$\ni.e.\n$$\n\\frac{a}{b+c} + \\frac{b}{c+a} + \\frac{c}{a+b} \\ge \\frac{A+2B}{2B} = \\frac{A}{2B} + 1.\n$$\nTherefore,\n$$\n\\frac{a}{b+c} + \\frac{b}{c+a} + \\frac{c}{a+b} + \\sqrt{\\frac{ab+bc+ca}{a^2+b^2+c^2}} \\ge \\frac{A}{2B} + 1 + \\sqrt{\\frac{B}{A}}.\n$$\nFinally, the inequality of arithmetic and geometric means implies\n$$\n\\frac{A}{2B} + \\frac{1}{2}\\sqrt{\\frac{B}{A}} + \\frac{1}{2}\\sqrt{\\frac{B}{A}} \\ge 3\\sqrt[3]{\\frac{AB}{8BA}} = \\frac{3}{2},\n$$\nand the given inequality is proved.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 29147,
    "subject": "Mathematics (Multi-modal)",
    "question": "Prove that it is possible to colour each positive integer with one of three colours so that the following conditions are satisfied:\ni) For each $n \\in \\mathbb{N}_0$, all positive integers $x$ such that $2^n \\le x < 2^{n+1}$ have the same colour.\nii) There are no positive integers $x$, $y$ and $z$ of the same colour (except $x = y = z = 2$) such that $x+y=z^2$. (B. Green, S. Lindqvist, arXiv:1608.08374)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $c_n$ denote the colour of positive integers $x$ such that $2^n \\le x < 2^{n+1}$. We will determine colours $c_n$ inductively. First, let us choose $c_0$, $c_1$ and $c_2$ to be three different colours. Next, for each $n \\ge 3$ let $c_n$ be the colour different from $c_{\\lfloor n/2 \\rfloor}$ and $c_{\\lfloor n/2 \\rfloor+1}$. Note that this is well-defined since $\\lfloor n/2 \\rfloor + 1 < n$ for all $n \\ge 3$.\nBy construction condition i) holds. Let us prove that condition ii) also holds. Let $x$, $y$ and $z$ be positive integers of the same colour such that $x + y = z^2$. Without loss of generality we may assume that $x \\le y$. Let $n$ be an integer such that $2^n \\le y < 2^{n+1}$. Then we obviously have\n$$\n2^n < x + y < 2^{n+2},\n$$\nand hence $2^{n/2} < z < 2^{(n+2)/2}$. It follows that $2^{\\lfloor n/2 \\rfloor} < z < 2^{\\lfloor n/2 \\rfloor+2}$. Since $y$ and $z$ have the same colour, by construction it follows that $n \\le 2$. From here we conclude $x \\le y < 8$ and $z = 2$ or $3$. Direct verification shows that the only possibility is $x = y = z = 2$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 29148,
    "subject": "Mathematics (Multi-modal)",
    "question": "In an organization there are three committees. Each person belongs to exactly one committee. For any two persons belonging to different committees, in the third committee there are exactly $10$ people that both persons know and exactly $10$ people that both persons do not know. All acquaintances are mutual. How many people are there in all three committees?\n(Russia 2008)",
    "options": [],
    "answer": "120",
    "solution": "Let $A$, $B$, $C$ be committees and let them have exactly $a$, $b$, $c$ persons, respectively. To each person assign a point in the plane so that no three points are collinear. We connect the points corresponding to persons that know each other by a blue segment, and for those that do not know each other by a red segment. Points corresponding to persons in the same committee are not connected.\n\nFirst, let us count the monochromatic triangles (all three sides are of the same colour). For each pair of persons $x \\in A$, $y \\in B$ there are exactly $10$ monochromatic triangles (if $xy$ is a blue segment, there are $10$ points $z \\in C$ such that segments $xz$ and $yz$ are blue, and analogously for red segments). Hence, the number of monochromatic triangles is $10ab$. In the same way we conclude that the number of monochromatic triangles is $10bc$ and $10ca$, so we conclude $a = b = c$.\n\nThe number of all triangles is $a^3$, while the number of monochromatic triangles is $10a^2$. Let us count bichromatic triangles. For each pair $x \\in A$, $y \\in B$ we have exactly $10$ bichromatic triangles such that the segments $xz$ and $yz$ have the same colour, i.e. there are $10a^2$ such triangles, while there are $10a^2$ triangles such that $yx$ and $zx$ are of the same colour, and $10a^2$ triangles such that $xy$ and $zy$ are of the same colour. Hence there are $30a^2$ bichromatic triangles.\n\nFinally, we have $a^3 = 10a^2 + 30a^2$, so $a = 40$. This means that there are in total $120$ members in all three committees.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 29149,
    "subject": "Mathematics (Multi-modal)",
    "question": "A *crazy bishop* is an unorthodox chess piece which turns towards one of the four diagonally contiguous cells and attacks all cells directly in front, to the left, and to the right of it (just like a regular chess bishop which cannot see behind his back). We say that two cells of a board are *diagonally contiguous* if they have exactly one vertex in common.\nDetermine the largest positive integer $N$ with the following property: It is possible to put $N$ crazy bishops on an $8 \\times 8$ gaming table so that none of them is attacked by another crazy bishop. (Russia 2013)",
    "options": [],
    "answer": "20",
    "solution": "Imagine the figure of a crazy bishop like a dot from which we draw 3 arrows in directions of cells which are attacked by this crazy bishop. By the term *diagonal* we consider all the diagonals of the board, not just the main ones. The four corner squares are also considered diagonals (containing only one square).\nOn each diagonal there are exactly two directions in which a crazy bishop can move.\nNotice that we can have at most one arrow in each direction, otherwise one of the two arrows which are pointing at the same direction is pointing to the other, i.e. there is at least one bishop under attack. Therefore, we can have at most 2 arrows on each diagonal. There are 30 diagonals, which means that there can be at most 60 arrows.\nEvery crazy bishop contributes with 3 arrows, thus we conclude $N \\le 20$.\nAn example in which we have 20 crazy bishops (none of them under attack) on an $8 \\times 8$ gaming table is given in the picture.\n\n![](attached_image_1.png)",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 29150,
    "subject": "Mathematics (Multi-modal)",
    "question": "In some archipelago, there are $2017$ islands labelled $1, 2, \\ldots, 2017$. Two agencies, Red dragon and Blue eye, are planning to establish ship lines between certain islands. Exactly one agency will operate between each two islands, with a ship line from the island labelled with a smaller number to the island labelled with a larger number.\n\nAn arrangement of ship lines is good if there are no two islands, labelled $A$ and $B$, where $A < B$, such that it is possible to go from $A$ to $B$ in two ways: one, using only the ships owned by Red dragon, and the other, using only ships owned by Blue eye.\n\nDetermine the number of good arrangements of ship lines. (Russia 2005)",
    "options": [],
    "answer": "2017!",
    "solution": "Ship lines maintained by Red dragon we will call red lines, and those maintained by Blue eye blue lines.\n\nLet $a_n$ denote the number of good arrangements for an archipelago with $n$ islands. Consider an archipelago with $n + 1$ islands and notice island $1$. Let $A$ be the set of all islands which are connected to island $1$ with a red line, and let $B$ be the set of all islands which are connected to island $1$ with a blue line.\n\nNote that all lines (if there are any) from some island in $A$ to some island in $B$ must be blue. Suppose that's not the case, i.e. that there are islands $a \\in A$ and $b \\in B$ such that $a < b$ and the line $a \\to b$ is red. We know that line $1 \\to b$ is blue, and lines $1 \\to a$ and $a \\to b$ are red, hence we can go from $a$ to $b$ using only red lines, and also using only blue lines, which is a contradiction. By the same reasoning we conclude that all lines (if there are any) from some island in $B$ to some island in $A$ must be red.\n\n![](attached_image_1.png)\n\nLet set $A$ contain $k \\in \\{0, 1, \\dots, n\\}$ islands. That means that set $B$ contains $n-k$ islands. Note that islands in the set $A$ must form an archipelago with a good arrangement of ship lines, and the same holds for the set $B$. On the other hand, for each selection of archipelagos $A$ and $B$ with good arrangements we get exactly one archipelago with $n+1$ islands and good ship lines arrangement between them.\n\nNow, using mathematical induction, we will prove that $a_n = n!$. It is obvious that $a_0 = a_1 = 1$. Suppose that $a_k = k!$, for all $k \\le n$, for some positive integer $n$. By previous reasoning we conclude that\n$$\na_{n+1} = \\sum_{k=0}^{n} \\binom{n}{k} a_k a_{n-k} = \\sum_{k=0}^{n} n! = (n+1)!\n$$\nThus, $a_n = n!$ and finally $a_{2017} = 2017!$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 29151,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $M$ be a point in the interior of the triangle $ABC$. The line $AM$ intersects the circumcircle of the triangle $MBC$ once more at $D$, the line $BM$ intersects the circumcircle of the triangle $MCA$ once more at $E$, and the line $CM$ intersects the circumcircle of the triangle $MAB$ once more at $F$. Prove the inequality\n$$\n\\frac{|AD|}{|MD|} + \\frac{|BE|}{|ME|} + \\frac{|CF|}{|MF|} \\ge \\frac{9}{2}. \\quad (\\text{Tajikistan 2014})\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "It suffices to show that\n$$\n\\frac{|AM|}{|MD|} + \\frac{|BM|}{|ME|} + \\frac{|CM|}{|MF|} \\ge \\frac{3}{2}.\n$$\nQuadrilaterals $MBDC$ and $MCEA$ are cyclic, so\n$$\n\\angle BCD = \\angle BMD = \\angle EMA = \\angle ECA.\n$$\nAlso, we have\n$$\n\\angle DBC = \\angle DMC = 180^\\circ - \\angle CMA = \\angle CEA.\n$$\nHence we conclude that the triangles *BDC* and *EAC* are similar. Analogously, we prove that the triangle *BAF* is similar to them.\n\n![](attached_image_1.png)\n\nBy Ptolemy's theorem for quadrilaterals *MBDC*, *MCEA* and *MAFB*, and by using the ratios from the similarity of *BDC*, *EAC* and *BAF*, we have:\n$$\n\\begin{align*}\n|MD| &= |BM| \\cdot \\frac{|CD|}{|BC|} + |CM| \\cdot \\frac{|DB|}{|BC|}, \\\\\n|ME| &= |CM| \\cdot \\frac{|AE|}{|CA|} + |AM| \\cdot \\frac{|EC|}{|CA|} = |CM| \\cdot \\frac{|DB|}{|CD|} + |AM| \\cdot \\frac{|BC|}{|CD|}, \\\\\n|MF| &= |AM| \\cdot \\frac{|BF|}{|AB|} + |BM| \\cdot \\frac{|FA|}{|AB|} = |AM| \\cdot \\frac{|BC|}{|DB|} + |BM| \\cdot \\frac{|CD|}{|DB|},\n\\end{align*}\n$$\nfrom where it follows that\n$$\n\\frac{|AM|}{|MD|} + \\frac{|BM|}{|ME|} + \\frac{|CM|}{|MF|} = \\frac{|AM| \\cdot |BC|}{|BM| \\cdot |CD| + |CM| \\cdot |DB|} \\\\\n+ \\frac{|BM| \\cdot |CD|}{|CM| \\cdot |DB| + |AM| \\cdot |BC|} \\\\\n+ \\frac{|CM| \\cdot |DB|}{|AM| \\cdot |BC| + |BM| \\cdot |CD|}\n$$\nLet us denote $x = |AM| \\cdot |BC|$, $y = |BM| \\cdot |CD|$, $z = |CM| \\cdot |DB|$. We need to prove the inequality\n$$\n\\frac{x}{y+z} + \\frac{y}{z+x} + \\frac{z}{x+y} \\geq \\frac{3}{2}.\n$$\nThis is Nesbitt’s famous inequality, so the proof is finished.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 29152,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $\\overline{AD}$ be the altitude of an acute-angled triangle $ABC$. On the line $AD$ there are distinct points $E$ and $F$ such that $|DE| = |DF|$ and the point $E$ is inside the triangle $ABC$. The circumcircle of the triangle $BEF$ meets segments $\\overline{BC}$ and $\\overline{AB}$ again at points $K$ and $M$, respectively. The circumcircle of the triangle $CEF$ meets segments $\\overline{BC}$ and $\\overline{CA}$ again at points $L$ and $N$, respectively.\nProve that the lines $AD$, $KM$ and $LN$ are concurrent.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Notice that the circumcentre of triangle $BEF$ is on the segment $\\overline{BC}$. Therefore, segment $\\overline{BK}$ is a diameter of the circumcircle of triangle $BEF$, so $\\angle BMK = 90^\\circ$, and analogously $\\angle LNC = 90^\\circ$.\n\n![](attached_image_1.png)\n\nLet lines $AD$ and $KM$ intersect at $X$. From $\\angle BMX = \\angle BMK = 90^\\circ$ and $\\angle XDB = \\angle ADB = 90^\\circ$, it follows that the quadrilateral $BDXM$ is cyclic.\n\nQuadrilaterals $BMEF$ and $EFCN$ are cyclic too, so from power of a point theorem (multiple use) it follows that\n$$\n|AX| \\cdot |AD| = |AM| \\cdot |AB| = |AE| \\cdot |AF| = |AN| \\cdot |AC|,\n$$\nand because of that, quadrilateral $CDXN$ is also cyclic.\n\nFinally, $\\angle XNC = 180^\\circ - \\angle CDX = 90^\\circ = \\angle LNC$, i.e. point $X$ lies on the line $LN$, which completes the proof.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 29153,
    "subject": "Mathematics (Multi-modal)",
    "question": "Prove that there are no positive integers $m$ and $n$ such that\n$$\n5m^3 = 27n^4 - 2n^2 + n.\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let us write the equation in the form\n$$\n5m^3 = n(27n^3 - 2n + 1)\n$$\nand note that $n$ and $27n^3 - 2n + 1$ are relatively prime numbers. We consider two cases:\n\na) $n = 5a^3$ and $27n^3 - 2n + 1 = b^3$ for some positive integers $a$ and $b$.\nInequality $(3n-1)^3 = 27n^3 - 27n^2 + 9n - 1 < 27n^3 - 2n + 1$ is equivalent to the inequality $27n^2 - 11n + 2 > 0$, which holds for all positive integers $n$. Also, we have $27n^3 - 2n + 1 < 27n^3 - 2 + 1 < 27n^3 = (3n)^3$.\nHence, $(3n-1)^3 < 27n^3 - 2n + 1 < (3n)^3$, which is impossible.\n\nb) $n = a^3$ and $27n^3 - 2n + 1 = 5b^3$ for some positive integers $a$ and $b$.\nFrom the second equation we conclude that $n$ cannot be divisible by $5$. If we have $n \\equiv \\pm1 \\pmod 5$, then $27n^3 - 2n + 1 \\equiv 1 \\pmod 5$. If we have $n \\equiv 2 \\pmod 5$, then $27n^3 - 2n + 1 \\equiv 3 \\pmod 5$, and if $n \\equiv 3 \\pmod 5$, then $27n^3 - 2n + 1 \\equiv 4 \\pmod 5$. In all cases the left-hand side of the equation cannot be divisible by $5$.\nHence the equation has no solution.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 29154,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $\\tau(n)$ be the number of positive divisors of $n$. Let $\\tau_1(n)$ be the number of positive divisors of $n$ which give remainder $1$ when divided by $3$. Find all possible integral values of the fraction\n$$\n\\frac{\\tau(10n)}{\\tau_1(10n)}.\n$$\n$(\\text{IMO Shortlist 2016})$",
    "options": [],
    "answer": "All even positive integers and all odd composite integers (equivalently, all integers greater than one except odd primes).",
    "solution": "Let $n = 3^a \\cdot b \\cdot m$, where $a$ is a non-negative integer, $b$ and $m$ positive integers such that all prime factors of $b$ give remainder $1$ when divided by $3$, and all prime factors of $m$ give remainder $2$.\nThen $\\tau(10n) = (a+1) \\cdot \\tau(b) \\cdot \\tau(10m)$.\nFor $\\tau_1(n)$, a prime factor $3$ cannot be taken, prime factors of $b$ can be taken arbitrarily, and the only condition is to take an even number of prime factors of $m$. So, $\\tau_1(10n) = \\tau(b) \\cdot \\tau_1(10m)$.\nHence we have\n$$\n\\frac{\\tau(10n)}{\\tau_1(10n)} = (a+1) \\cdot \\frac{\\tau(10m)}{\\tau_1(10m)}.\n$$\nLet $m = 2^{c_1} \\cdot 5^{c_2} \\cdot p_3^{c_3} \\cdots p_k^{c_k}$. We have $\\tau(10m) = (c_1 + 2)(c_2 + 2)(c_3 + 1)\\cdots(c_k + 1)$.\nLet us consider two cases.\n\n* Number $\\tau(10m)$ is even. One of the brackets (in the expression for $\\tau(10m)$) is even. Let it be the bracket corresponding to the factor $p_j$ and let it be $2t$. If we choose a prime factor (different from $p_j$) arbitrarily, then the number of occurrences of $p_j$ can be chosen in exactly $t$ ways. So, in this case $\\tau_1(10m) = \\frac{1}{2}\\tau(10m)$.\nIn this case, the given fraction can take the value $2(a+1)$, i.e. any even positive integer is a possible value and there are no other.\n\n* Number $\\tau(10m)$ is odd. All brackets are odd, which means that the exponent of each prime factor is even. For each combination of exponents giving the even sum, the first exponent which is not $1$ can be reduced by $1$, while for the odd sum the last exponent which is not $0$ (or the first if all of them are $1$) can be increased by $1$. In that way, we establish a bijection between combinations with even and odd sum (except for the combination in which all exponents are $0$).\nWe conclude $\\tau_1(10m) = \\frac{1}{2}(\\tau(10m) + 1)$. Let $\\tau(10m) = t$. So we have\n$$\n\\frac{\\tau(10n)}{\\tau_1(10n)} = 2(a + 1) \\cdot \\frac{t}{t + 1}.\n$$\nSince we are interested in integral values of the fraction, we see that $t+1$ divides $2(a+1)$, so the fraction above is of the form $tt'$, where $t'$ is a positive integer. Since all brackets are odd in $t$, the first two are at least $3$, so $t$ is a composite number. Hence the given fraction cannot be an odd prime number.\nIt remains to show that the given fraction can achieve the value $xy$, where $x$ and $y$ are non-negative integers greater than $1$. We can achieve this for\n$$\nn = 3^{\\frac{xy-1}{2}} \\cdot 2^{x-2} \\cdot 5^{y-2}.\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 29155,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all positive integers $n$ for which there exist positive integers $a$ and $b$ such that\n$$\n(n^2 + 2)^a = (2n - 1)^b.\n$$",
    "options": [],
    "answer": "5",
    "solution": "Notice that $n^2 + 2 \\ge 3$, $n^2 + 2 > 2n - 1$ and also that numbers $n^2 + 2$ and $2n - 1$ have the same prime factors.\nLet $p$ be a prime factor of $n^2 + 2$ and $2n - 1$.\n$$\np \\mid n^2 + 2 \\quad \\text{and} \\quad p \\mid 2n - 1 \\implies p \\mid (n^2 + 2) + (2n - 1) = (n + 1)^2 \\implies p \\mid n + 1.\n$$\nTherefore, $p \\mid 2(n + 1) - (2n - 1) = 3$ and we conclude that $p = 3$.\nHence there are positive integers $k$ and $m$, $k > m$, such that $n^2 + 2 = 3^k$ and $2n - 1 = 3^m$. We get\n$$\n4 \\cdot 3^k = 4n^2 + 8 = (3^m + 1)^2 + 8 = 3^{2m} + 2 \\cdot 3^m + 9,\n$$\nfrom which it follows that $3^m$ divides 9, i.e. $m \\le 2$.\n\nWe have two possibilities:\n\n1) If $m = 1$, then $2n - 1 = 3$, and consequently $n = 2$ and $3^k = 6$, which is impossible.\n\n2) If $m = 2$, then $2n - 1 = 9$, and consequently $n = 5$ and $k = 3$.\n\nTherefore, $n = 5$ is the only positive integer with the given property.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 29156,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $f: \\mathbb{N} \\to \\mathbb{N}$ be a function such that for all positive integers $a$ and $b$,\n$$\nf(a) + f(b) - ab \\mid af(a) + bf(b).\n$$\nFind all such functions $f$.",
    "options": [],
    "answer": "f(n) = n^2 for all positive integers n",
    "solution": "It is given that\n$$\nf(a) + f(b) - ab \\mid af(a) + bf(b). \\qquad (3)\n$$\nTaking $a = b = 1$ in (3), we have $2f(1) - 1 \\mid 2f(1)$. Then $2f(1) - 1 \\mid 2f(1) - (2f(1) - 1) = 1$ and hence $f(1) = 1$.\n\nLet $p \\ge 7$ be a prime. Taking $a = p$ and $b = 1$ in (3), we have $f(p) - p + 1 \\mid pf(p) + 1$ and hence\n$$\nf(p) - p + 1 \\mid pf(p) + 1 - p(f(p) - p + 1) = p^2 - p + 1.\n$$\nIf $f(p) - p + 1 = p^2 - p + 1$, then $f(p) = p^2$. If $f(p) - p + 1 \\ne p^2 - p + 1$, as $p^2 - p + 1$ is an odd positive integer, we have $p^2 - p + 1 \\ge 3(f(p) - p + 1)$, i.e.\n$$\nf(p) \\le \\frac{1}{3}(p^2 + 2p - 2). \\qquad (4)\n$$\nTaking $a = b = p$ in (3), we have $2f(p) - p^2 \\mid 2pf(p)$. This implies\n$$\n2f(p) - p^2 \\mid 2pf(p) - p(2f(p) - p^2) = p^3.\n$$\nBy (4) and $f(p) \\ge 1$ we get\n$$\n-p^2 < 2f(p) - p^2 \\le \\frac{2}{3}(p^2 + 2p - 2) - p^2 < -p,\n$$\nsince $p \\ge 7$. This contradicts the fact that $2f(p) - p^2$ is a factor of $p^3$. Thus we have proved that $f(p) = p^2$ for all primes $p \\ge 7$.\n\nLet $a$ be a fixed positive integer. Choose a sufficiently large prime $p$. Consider $b = p$ in (3). We obtain\n$$\nf(a) + p^2 - pa \\mid af(a) + p^3 = a(f(a) + p^2 - pa) + p^3 - p^2a + pa^2,\n$$\ni.e.\n$$\nf(a) + p^2 - pa \\mid p(p^2 - pa + a^2).\n$$\nAs $p$ is sufficiently large and $a$ is fixed, $p$ cannot divide $f(a)$, and so numbers $f(a) + p^2 - pa$ and $p$ are relatively prime. It follows that\n$$\nf(a) + p^2 - pa \\mid p^2 - pa + a^2 = (f(a) + p^2 - pa) + a^2 - f(a),\n$$\ni.e.\n$$\nf(a) + p^2 - pa \\mid a^2 - f(a).\n$$\nNote that $a^2 - f(a)$ is fixed while $f(a) + p^2 - pa$ is chosen to be sufficiently large. Therefore, we must have $a^2 - f(a) = 0$, so that $f(a) = a^2$ for any positive integer $a$.\n\nFinally, we check that when $f(a) = a^2$ for any positive integer $a$, then\n$$\nf(a) + f(b) - ab = a^2 + b^2 - ab\n$$\nand\n$$\naf(a) + bf(b) = a^3 + b^3 = (a + b)(a^2 + b^2 - ab).\n$$\nThe latter expression is divisible by the former for any positive integers $a$ and $b$. This shows that $f(a) = a^2$ is the only solution.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 29157,
    "subject": "Mathematics (Multi-modal)",
    "question": "In 2018 Matija will turn the age which is equal to three times the sum of digits of the year he was born in. The same statement is true for his grandfather. What age did his grandfather turn in the year Matija was born?",
    "options": [],
    "answer": "54",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 29158,
    "subject": "Mathematics (Multi-modal)",
    "question": "Determine all triples $(x, y, z)$ of real numbers that satisfy\n$$\n\\begin{align*}\nx + y - z &= -1 \\\\\nx^2 - y^2 + z^2 &= 1 \\\\\n-x^3 + y^3 + z^3 &= -1.\n\\end{align*}\n$$",
    "options": [],
    "answer": "{(1, -1, 1), (-1, -1, -1)}",
    "solution": "From the first equation we get $x + y = z - 1$. By plugging this into the second equation we get:\n$$\n\\begin{aligned}\nx^2 - y^2 &= 1 - z^2, \\\\\n(x + y)(x - y) &= (1 - z)(1 + z), \\\\\n(z - 1)(x - y) &= -(z - 1)(1 + z), \\\\\n(z - 1)(x - y + z + 1) &= 0.\n\\end{aligned}\n$$\nThere are now two cases: $z = 1$ or $x - y + z + 1 = 0$.\n\nIf $z = 1$, then $x + y = 0$, i.e. $x = -y$. Plugging this into the third equation we get $2y^3 = -2$, i.e. $y = -1$, $x = 1$.\n\nIf $x - y + z + 1 = 0$, then plugging $z = -x + y - 1$ into the first equation yields $2x = -2$, i.e. $x = -1$.\nNow we have $z = -1 + y - 1 = y$, which we can combine with the third equation to get $2y^3 = -2$, i.e. $y = z = -1$.\n\nThus, the possible solutions to the given system of equations are $(1, -1, 1)$ and $(-1, -1, -1)$. A direct computation confirms that both are indeed valid solutions of the system.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 29159,
    "subject": "Mathematics (Multi-modal)",
    "question": "Marko has drawn a rectangle with two blue sides of length $24$ and two red sides of length $36$. He coloured each point in the interior of the rectangle in the colour of the side closest to that point. He also coloured in black all points equally distant from a blue and red side. Determine the area of the red part of the rectangle.",
    "options": [],
    "answer": "576",
    "solution": "",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 29160,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $D_0, D_1, \\dots, D_{2018}$ be points on the segment $\\overline{AB}$ such that $D_0 = A$, $D_{2018} = B$ and\n$$\n|D_0 D_1| = |D_1 D_2| = \\dots = |D_{2017} D_{2018}|.\n$$\nIf $C$ is a point such that $\\angle BCA = 90^\\circ$, prove that\n$$\n|CD_0|^2 + |CD_1|^2 + \\dots + |CD_{2018}|^2 = |AD_1|^2 + |AD_2|^2 + \\dots + |AD_{2018}|^2.\n$$\n(Ivan Krijan)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $E_i$ be the foot of the altitude from $D_i$ to the side $\\overline{AC}$ in the right-angled triangle $ABC$, for each $i = 0, 1, \\dots, 2018$. We have $E_0 = A$, $E_{2018} = C$.\n\n![](attached_image_1.png)\n\nSince the lines $D_iE_i$ are parallel to $BC$, Thales' theorem asserts that\n$$\n|E_0E_1| = |E_1E_2| = \\dots = |E_{2017}E_{2018}|.\n$$\nFrom here we get $|CE_i| = |AE_{2018-i}|$ for $i = 1, 2, \\dots, 2017$.\n\nThe Pythagorean theorem, applied to the right-angled triangles $CD_iE_i$ and $AD_iE_i$ for $i = 1, 2, \\dots, 2017$, yields\n$$\n|CD_i|^2 = |D_iE_i|^2 + |CE_i|^2 \\quad \\text{and} \\quad |AD_i|^2 = |D_iE_i|^2 + |AE_i|^2\n$$\n\nBy subtracting these two equalities, we get\n$$\n|CD_i|^2 - |AD_i|^2 = |CE_i|^2 - |AE_i|^2, \\quad i = 1, \\dots, 2017,\n$$\nand the same relation also holds for $i = 0$ and $i = 2018$.\n\nAdding up all these relations, we get\n$$\n\\begin{aligned}\n& |CD_0|^2 + |CD_1|^2 + \\dots + |CD_{2018}|^2 - (|AD_0|^2 - |AD_1|^2) - \\dots - (|AD_{2018}|^2 \\\\\n& = |CE_0|^2 + |CE_1|^2 + \\dots + |CE_{2018}|^2 - (|AE_0|^2 - |AE_1|^2) - \\dots - (|AE_{2018}|^2 \\\\\n& = (|CE_0|^2 - |AE_{2018}|^2) + (|CE_1|^2 - |AE_{2017}|^2) + \\dots + (|CE_{2018}|^2 - |AE_0|^2) = 0.\n\\end{aligned}\n$$\n\nSince $|AD_0| = 0$, this proves the claim.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 29161,
    "subject": "Mathematics (Multi-modal)",
    "question": "Determine all pairs $(m, n)$ of integers such that\n$$\nn^2 - 6n = m^2 + m - 10.\n$$",
    "options": [],
    "answer": "(1, 2), (1, 4), (-2, 2), (-2, 4)",
    "solution": "We are given the equation\n$$\nn^2 - 6n = m^2 + m - 10.\n$$\nLet's rewrite it as\n$$\nn^2 - 6n - m^2 - m + 10 = 0.\n$$\nGroup terms:\n$$\nn^2 - m^2 - 6n - m + 10 = 0.\n$$\nRecall that $n^2 - m^2 = (n - m)(n + m)$, so\n$$\n(n - m)(n + m) - 6n - m + 10 = 0.\n$$\nLet us solve for $n$ in terms of $m$.\n\nAlternatively, move all terms to one side:\n$$\nn^2 - 6n - m^2 - m + 10 = 0\n$$\n$$\nn^2 - 6n = m^2 + m - 10\n$$\nThis is a quadratic in $n$:\n$$\nn^2 - 6n - (m^2 + m - 10) = 0\n$$\nThe discriminant must be a perfect square for integer solutions:\n$$\n\\Delta = 36 + 4(m^2 + m - 10) = 36 + 4m^2 + 4m - 40 = 4m^2 + 4m - 4 = 4(m^2 + m - 1)\n$$\nSo $4(m^2 + m - 1)$ must be a perfect square, say $k^2$ for some integer $k$:\n$$\n4(m^2 + m - 1) = k^2\n$$\nSo\n$$\nk^2 - 4m^2 - 4m + 4 = 0\n$$\n$$\nk^2 - 4m^2 - 4m + 4 = 0\n$$\n$$\nk^2 = 4m^2 + 4m - 4\n$$\n$$\nk^2 = 4(m^2 + m - 1)\n$$\nSo $m^2 + m - 1$ must be a perfect square times $1/4$.\nBut $k$ must be even, so let $k = 2t$ for some integer $t$:\n$$\n(2t)^2 = 4(m^2 + m - 1)\n$$\n$$\n4t^2 = 4(m^2 + m - 1)\n$$\n$$\nt^2 = m^2 + m - 1\n$$\nSo $m^2 + m - 1 = t^2$ for some integer $t$.\n\nNow, rearrange:\n$$\nm^2 + m - 1 - t^2 = 0\n$$\n$$\nm^2 + m - t^2 = 1\n$$\nThis is a quadratic in $m$:\n$$\nm^2 + m - (t^2 + 1) = 0\n$$\nThe discriminant must be a perfect square:\n$$\n\\Delta = 1^2 - 4 \\cdot 1 \\cdot (-t^2 - 1) = 1 + 4t^2 + 4 = 4t^2 + 5\n$$\nSo $4t^2 + 5$ must be a perfect square, say $s^2$:\n$$\ns^2 = 4t^2 + 5\n$$\nSo $s^2 - 4t^2 = 5$\n\nThis is a Pell-type equation.\n\nLet us solve $s^2 - 4t^2 = 5$ in integers.\n\nTry small integer values for $t$:\n- $t = 0$: $s^2 = 5$ (no integer solution)\n- $t = 1$: $s^2 = 4 + 5 = 9$, $s = 3$\n- $t = -1$: $s^2 = 4 + 5 = 9$, $s = 3$\n- $t = 2$: $s^2 = 16 + 5 = 21$ (no integer solution)\n- $t = -2$: $s^2 = 16 + 5 = 21$ (no integer solution)\n- $t = 3$: $s^2 = 36 + 5 = 41$ (no integer solution)\n- $t = -3$: $s^2 = 36 + 5 = 41$ (no integer solution)\n- $t = 5$: $s^2 = 100 + 5 = 105$ (no integer solution)\n\nSo only $t = 1$ and $t = -1$ work, with $s = 3$.\n\nNow, recall $m^2 + m - t^2 = 1$\nFor $t = 1$:\n$$\nm^2 + m - 1 = 1\n$$\n$$\nm^2 + m = 2\n$$\n$$\nm^2 + m - 2 = 0\n$$\n$$\nm = \\frac{-1 \\pm \\sqrt{1 + 8}}{2} = \\frac{-1 \\pm 3}{2}\n$$\nSo $m = 1$ or $m = -2$\n\nFor $t = -1$:\nSame as above, $m = 1$ or $m = -2$\n\nNow, recall $n$ is given by the quadratic:\n$$\nn^2 - 6n = m^2 + m - 10\n$$\nFor $m = 1$:\n$$\nn^2 - 6n = 1 + 1 - 10 = -8\n$$\n$$\nn^2 - 6n + 8 = 0\n$$\n$$\nn = \\frac{6 \\pm \\sqrt{36 - 32}}{2} = \\frac{6 \\pm 2}{2} = 4, 2\n$$\nFor $m = -2$:\n$$\nn^2 - 6n = 4 - 2 - 10 = -8\n$$\nSo same as above, $n = 4, 2$\n\nTherefore, the integer solutions are:\n$$(m, n) = (1, 2), (1, 4), (-2, 2), (-2, 4)$$\n\nThus, all integer pairs $(m, n)$ such that $n^2 - 6n = m^2 + m - 10$ are:\n$$(1, 2), (1, 4), (-2, 2), (-2, 4)$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 29162,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABC$ be a right-angled triangle with the right angle at $C$, having sides of length $|AB| = 26$ and $|BC| = 24$. A semicircle with diameter on $\\overline{BC}$ containing point $C$ is inscribed in the triangle. The semicircle touches the side $\\overline{AB}$. What is the radius of the semicircle? (Nikola Adžaga)",
    "options": [],
    "answer": "20/3",
    "solution": "",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 29163,
    "subject": "Mathematics (Multi-modal)",
    "question": "A positive integer $N$ is called interesting if it is divisible by $36$, and if there exists a positive integer $k$ smaller than $10$ such that $1, 2, \\dots, k$ are, in some order, the digits of the decimal representation of $N$. Find the smallest interesting positive integer. (Lucija Validžić)",
    "options": [],
    "answer": "12345768",
    "solution": "",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 29164,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $p$ be a prime number, and let $n \\ge p-1$ be an integer. If $np+1$ is a perfect square, prove that $n+1$ can be represented as a sum of squares of exactly $p$ positive integers.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $a$ be a positive integer such that $np + 1 = a^2$. Note that $a > 1$.\n$$\n\\text{We have } a^2 - 1 = np, \\text{ i.e. } (a-1)(a+1) = np.\n$$\nSince $p$ is prime, it follows that $p \\mid a-1$ or $p \\mid a+1$. We treat these two cases separately:\n\ni) Let $p \\mid a-1$, i.e. let $a = kp + 1$ for some integer $k$.\n$$\n\\text{Then } 1 + np = (kp + 1)^2 = k^2p^2 + 2kp + 1, \\text{ i.e. } n = k^2p + 2k.\n$$\nNow we have:\n$$\nn + 1 = k^2p + 2k + 1 = k^2 + 2k + 1 + (p-1)k^2 = (k+1)^2 + (p-1)k^2.\n$$\nNotice that $a > 1$ implies $k > 0$, so $k+1$ and $k$ are both positive integers.\n\nii) Let $p \\mid a+1$, so that $a = kp - 1$ for some integer $k$.\nAnalogously to the first case, we get $n + 1 = (k-1)^2 + (p-1)k^2$.\nIt remains to show that $k-1$ is indeed a positive integer, i.e. that we have $k \\ge 2$.\nIf we assume the contrary ($k = 1$), it follows that $n + 1 = p - 1$, i.e. $n = p - 2$.\nThis contradicts the given condition that $n \\ge p - 1$.\nWe conclude that the assertion holds for any choice of $n$ and $p$ satisfying the conditions of the problem.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 29165,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $a$, $b$ and $c$ be different positive real numbers such that $(a+b-c)(b+c-a)(c+a-b) \\neq 0$. Prove that at least one of the numbers\n$$\n\\frac{a+b}{a+b-c}, \\quad \\frac{b+c}{b+c-a}, \\quad \\frac{c+a}{c+a-b}\n$$\nlies in the interval $(1, 2)$, and that at least one of these numbers does not lie in that interval.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 29166,
    "subject": "Mathematics (Multi-modal)",
    "question": "Determine all pairs $(x, y)$ of real numbers such that $x + y = x^2 + y^2 = x^3 + y^3$.",
    "options": [],
    "answer": "(0, 0), (0, 1), (1, 0), (1, 1)",
    "solution": "Let $S = x + y$ and $P = x y$.\n\nWe are given:\n\n$$\nS = x + y = x^2 + y^2 = x^3 + y^3\n$$\n\nRecall:\n$$\nx^2 + y^2 = (x + y)^2 - 2 x y = S^2 - 2P\n$$\nx^3 + y^3 = (x + y)^3 - 3 x y (x + y) = S^3 - 3 P S\n$$\n\nSo, the system becomes:\n\n1. $S = S^2 - 2P$\n2. $S = S^3 - 3 P S$\n\nFrom (1):\n$$\nS = S^2 - 2P \\implies S^2 - S - 2P = 0 \\implies 2P = S^2 - S \\implies P = \\frac{S^2 - S}{2}\n$$\n\nFrom (2):\n$$\nS = S^3 - 3 P S \\implies S^3 - 3 P S - S = 0 \\implies S^3 - S - 3 P S = 0\n$$\nSubstitute $P$ from above:\n$$\nS^3 - S - 3 S \\cdot \\frac{S^2 - S}{2} = 0\n$$\nMultiply both sides by 2:\n$$\n2 S^3 - 2 S - 3 S (S^2 - S) = 0\n$$\n$$\n2 S^3 - 2 S - 3 S^3 + 3 S^2 = 0\n$$\n$$\n(2 S^3 - 3 S^3) + (3 S^2) - 2 S = 0\n$$\n$$\n(-S^3) + 3 S^2 - 2 S = 0\n$$\n$$\nS^3 - 3 S^2 + 2 S = 0\n$$\n$$\nS (S^2 - 3 S + 2) = 0\n$$\n$$\nS (S - 1)(S - 2) = 0\n$$\n\nSo the possible values for $S$ are $0$, $1$, or $2$.\n\nNow, for each $S$, compute $P$:\n\n- If $S = 0$:\n  $$\n  P = \\frac{0^2 - 0}{2} = 0\n  $$\n  So $x$ and $y$ are roots of $t^2 - 0 t + 0 = 0 \\implies t^2 = 0 \\implies t = 0$\n  So $(x, y) = (0, 0)$\n\n- If $S = 1$:\n  $$\n  P = \\frac{1^2 - 1}{2} = 0\n  $$\n  So $x$ and $y$ are roots of $t^2 - t = 0 \\implies t (t - 1) = 0$\n  So $(x, y) = (0, 1)$ and $(1, 0)$\n\n- If $S = 2$:\n  $$\n  P = \\frac{2^2 - 2}{2} = \\frac{4 - 2}{2} = 1\n  $$\n  So $x$ and $y$ are roots of $t^2 - 2 t + 1 = 0 \\implies (t - 1)^2 = 0 \\implies t = 1$\n  So $(x, y) = (1, 1)$\n\nTherefore, all solutions are:\n$$(x, y) = (0, 0),\\ (0, 1),\\ (1, 0),\\ (1, 1)$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 29167,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $\\triangle ABC$ be a triangle such that $\\angle CAB = 2\\angle ABC$. A point $D$ is given in the interior of the triangle $\\triangle ABC$, such that $|AD| = |BD|$ and $|CD| = |AC|$. Prove that $\\angle ACB = 3\\angle DCB$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $E$ be the intersection of the bisector of the segment $\\overline{AB}$ with the segment $\\overline{BC}$.\nDenote $\\beta = \\angle ABC$ and notice that $\\angle ACB = 180^\\circ - 3\\beta$.\nSince $E$ lies on the bisector of the segment $\\overline{AB}$, we have $|AE| = |BE|$. Therefore, $\\angle BAE = \\angle ABC = \\beta$. This implies $\\angle EAC = \\angle CAB - \\beta = 2\\beta - \\beta = \\beta$.\nLet $F$ be the other intersection of the line $AE$ with the circle of radius $\\overline{CA}$ centred at $C$. Since $CAF$ is an isosceles triangle ($\\overline{CA}$ and $\\overline{CF}$ are both radii of the same circle), we get $\\angle CFA = \\beta$.\n![](attached_image_1.png)\nFrom $\\angle CFA = \\angle BAF$ we get $CF \\parallel AB$ (these are the angles of the transversal).\nThis also means that $\\angle BCF = \\angle CBA = \\beta$, which shows that $CEF$ is an isosceles triangle.\nFrom $CF \\parallel AB$ and the fact that $E$ is equidistant to $C$ and $F$, we conclude that the line $DE$ is the bisector of the segment $\\overline{CF}$ as well.\nThus, $|DF| = |DC| = |CF|$, i.e. the triangle $DFC$ is equilateral.\n$$\n\\text{Finally, we have } \\angle DCB = 60^\\circ - \\beta = \\frac{1}{3}(180^\\circ - 3\\beta) = \\frac{1}{3}\\angle ACB, \\text{ i.e. } \\angle ACB = 3\\angle DCB.\n$$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 29168,
    "subject": "Mathematics (Multi-modal)",
    "question": "We say that a circular arrangement of positive integers is alternating if every number is either smaller, or larger than both of its neighbours. We call a pair of adjacent numbers good if, upon its removal, the remaining numbers form an alternating arrangement.\n\nThe numbers $1$ to $300$ are placed on a circle in an alternating arrangement. Determine the least possible number of good pairs of adjacent numbers in such arrangement.",
    "options": [],
    "answer": "151",
    "solution": "Let $a$, $p$, $q$ and $b$ be four consecutive numbers in an alternating arrangement. Assume $p > q$. Then $a < p$ and $q < b$.\n\nThe pair $(p, q)$ is not good if and only if $a > b$. Therefore, we conclude that $(p, q)$ is not good if and only if $p$ is the largest, and $q$ the smallest number of the quadruple $(a, p, q, b)$.\n\nIf $(p, q)$ is not good, then $(a, p)$ is, because $a$ is greater than or equal to $q$. Furthermore, $(q, b)$ is good because $b$ is not greater than $p$.\n\nThis shows that, of any two pairs sharing an element, at least one is a good pair. We conclude that there are at least $150$ good pairs.\n\nAssume that there is an alternating arrangement $a_1, \\dots, a_{300}$ in which there are exactly $150$ good pairs.\n\nWithout loss of generality, we can assume that $a_1 > a_2$, and that $(a_1, a_2)$ is not a good pair. Then $(a_{2k}, a_{2k+1})$ are both good, while the pair $(a_{2k-1}, a_{2k})$ is not, for any $k = 1, 2, \\dots, 149$. We also notice that $(a_{299}, a_{300})$ cannot be a good pair.\n\nSince $(a_1, a_2)$ is not good, we conclude that $a_1 > a_3$. Similarly, since $(a_3, a_4)$ is not good, we get $a_3 > a_5$. Continuing this way, we get\n$$\na_1 > a_3 > a_5 > \\dots > a_{299} > a_1,\n$$\nthus arriving at a contradiction. This shows that it is impossible to have exactly $150$ good pairs. Therefore, there must be at least $151$ of them.\n\nAn example of an alternating arrangement containing exactly $151$ good pairs is\n$3, 2, 5, 4, 7, 6, \\ldots, 299, 298, 300, 1$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 29169,
    "subject": "Mathematics (Multi-modal)",
    "question": "There are $599$ yellow and $301$ blue balls. Can these balls be ordered in a sequence so that the number of balls between any two blue balls is different from $2$ and $5$?",
    "options": [],
    "answer": "No",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 29170,
    "subject": "Mathematics (Multi-modal)",
    "question": "In how many ways can we fill a $2018 \\times 2018$ board with positive integers so that the sum of numbers in any three consecutive cells in the same row or the same column equals $5$?",
    "options": [],
    "answer": "21",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 29171,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABC$ be a triangle such that $|AB| = 6$ and $|BC| = 9$, and let the measure of the angle $\\angle ABC$ be $120^\\circ$. Let $P$ and $Q$ be points on the side $\\overline{AC}$ such that the triangle $BPQ$ is equilateral. Determine the length of the segment $\\overline{PQ}$.",
    "options": [],
    "answer": "18√19/19",
    "solution": "",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 29172,
    "subject": "Mathematics (Multi-modal)",
    "question": "Marin labels the vertices of a cube with numbers $1, 2, \\ldots, 8$, and then labels each edge with the sum of numbers in the vertices joined by that edge. Can he arrange the numbers in the vertices so that all the edges are labelled with different numbers?",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 29173,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all complex numbers $z$ such that\n$$ \\mathrm{Re}(z) = 9 \\quad \\text{and} \\quad \\mathrm{Im}(z^2) = \\mathrm{Im}(z^3). \\qquad (\\text{AIME 2007}) $$",
    "options": [],
    "answer": "9, 9 + 15i, 9 - 15i",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 29174,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $z$ be a complex number such that\n$$\n|z - 5| = |z - 1| + 4.\n$$\nProve that $z$ is a real number.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $z = x + iy$, where $x, y \\in \\mathbb{R}$.\n\nThen\n$$\n|z - 5| = |(x + iy) - 5| = |(x - 5) + iy| = \\sqrt{(x - 5)^2 + y^2}\n$$\nand\n$$\n|z - 1| = |(x + iy) - 1| = |(x - 1) + iy| = \\sqrt{(x - 1)^2 + y^2}.\n$$\n\nThe given equation is\n$$\n\\sqrt{(x - 5)^2 + y^2} = \\sqrt{(x - 1)^2 + y^2} + 4.\n$$\n\nLet $a = \\sqrt{(x - 5)^2 + y^2}$ and $b = \\sqrt{(x - 1)^2 + y^2}$, so $a = b + 4$.\n\nThen\n$$\na^2 = b^2 + 8b + 16.\n$$\nBut $a^2 = (x - 5)^2 + y^2$ and $b^2 = (x - 1)^2 + y^2$.\n\nSo\n$$\n(x - 5)^2 + y^2 = (x - 1)^2 + y^2 + 8b + 16.\n$$\nThe $y^2$ terms cancel:\n$$\n(x - 5)^2 = (x - 1)^2 + 8b + 16.\n$$\nExpand:\n$$\n(x^2 - 10x + 25) = (x^2 - 2x + 1) + 8b + 16\n$$\n$$\n-10x + 25 = -2x + 1 + 8b + 16\n$$\n$$\n-10x + 25 + 2x - 1 - 16 = 8b\n$$\n$$\n-8x + 8 = 8b\n$$\n$$\n- x + 1 = b\n$$\nBut $b = \\sqrt{(x - 1)^2 + y^2}$, so\n$$\n- x + 1 = \\sqrt{(x - 1)^2 + y^2}\n$$\n\nNow, $\\sqrt{(x - 1)^2 + y^2} \\geq 0$, so $-x + 1 \\geq 0$, i.e., $x \\leq 1$.\n\nSquare both sides:\n$$\n(-x + 1)^2 = (x - 1)^2 + y^2\n$$\nBut $(-x + 1)^2 = (x - 1)^2$, so\n$$\n(x - 1)^2 = (x - 1)^2 + y^2\n$$\n$$\n0 = y^2\n$$\n$$\ny = 0\n$$\n\nTherefore, $z = x + i \\cdot 0 = x$ is a real number.\n\nThus, $z$ is real.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 29175,
    "subject": "Mathematics (Multi-modal)",
    "question": "Determine all positive integers $n$ such that there exist positive integers $a$ and $b$ which satisfy\n$$\nS(a) = S(b) = S(a + b) = n.\n$$\nHere $S(a)$ denotes the sum of digits of $a$. (Romania 1999)",
    "options": [],
    "answer": "All positive integers divisible by 9",
    "solution": "Notice that, for each positive integer $a$, the numbers $a$ and $S(a)$ give the same remainder when divided by $9$.\nApplying this observation to $S(a) = S(b) = S(a+b)$, we conclude that $a$, $b$ and $a+b$ give the same remainder when divided by $9$. This implies that $(a+b) - b = a$ and $(a+b) - a = b$ are both divisible by $9$. But then $a$ and $b$ are also divisible by $9$. This proves that any positive integer $n$ satisfying the given condition must be divisible by $9$.\n\nLet us now prove the converse: let $n$ be a multiple of $9$, i.e. let $n = 9k$ for some positive integer $k$. Taking numbers\n$$\na = \\overline{18\\ldots18}, \\quad b = \\overline{72\\ldots72}, \\quad a+b = \\overline{90\\ldots90},\n$$\nwhere each of the numbers has exactly $2k$ digits, we obtain a pair $(a, b)$ which satisfies $S(a) = S(b) = S(a+b) = 9k = n$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 29176,
    "subject": "Mathematics (Multi-modal)",
    "question": "Branko is writing a sequence of quadratic polynomials with real coefficients. In each step, he considers the polynomial he chose in the previous step, denoted by $ax^2+bx+c$, and proceeds to write down either $cx^2+bx+a$ or $a(x+d)^2+b(x+d)+c$ for some real number $d$.\nStarting from $x^2-2x-1$ and performing a finite number of prescribed steps, can Branko obtain the following polynomial:\n\na) $2x^2 - 1$?\n\nb) $2x^2 - x - 1$",
    "options": [],
    "answer": "a) yes; b) no",
    "solution": "a) The first polynomial, i.e. $2x^2-1$, can be obtained by applying the following sequence of steps:\n$$\nx^2 - 2x - 1 \\rightarrow -x^2 - 2x + 1 \\xrightarrow{d=-1} -x^2 + 2 \\rightarrow 2x^2 - 1.\n$$\n\nb) Notice that the prescribed steps leave the discriminant unchanged: the discriminant of $cx^2+bx+a$ is $b^2-4ca = b^2-4ac$, whereas the discriminant of\n$$\na(x+d)^2 + b(x+d) + c = ax^2 + (2ad+b)x + (ad^2+bd+c)\n$$\nequals\n$$\n(2ad+b)^2 - 4a(ad^2+bd+c) = 4a^2d^2 + 4abd + b^2 - 4a^2d^2 - 4abd - 4ac = b^2 - 4ac.\n$$\nThe discriminant of the initial polynomial $x^2-2x-1$ is $8$, while the discriminant of $2x^2-x-1$ equals $9$. Thus, Branko cannot obtain the polynomial $2x^2-x-1$ in any number of allowed steps.\n\nRemark: Consider the prescribed steps as operations T1 and T2 on triples of real numbers:\n$$\n(a, b, c) \\xrightarrow{T1} (c, b, a) \\quad \\text{and} \\quad (a, b, c) \\xrightarrow{T2} (a, 2ad+b, ad^2+bd+c).\n$$\nGiven any two quadratic polynomials $f(x)$ and $g(x)$ with equal discriminants, we can obtain a sequence of steps transforming $f(x)$ into $g(x)$: first, we apply T2 to adjust the constant term of $f(x)$, making it equal to the leading coefficient of $g(x)$. After that, we can use T1 to swap the leading coefficient with the constant term. Finally,\nusing T2, we can adjust the new constant term making it equal to the constant term of $g(x)$.\nThis last step guarantees that the resulting polynomial will have the same constant term as $g(x)$. The leading coefficient is also equal to the one appearing in $g(x)$, because it remains unchanged by T2. Finally, the linear term also matches the one appearing in $g(x)$: it is uniquely determined (up to sign) by the discriminant and the remaining two coefficients. If necessary, we can change its sign by applying T2 with $d = -b/a$.",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 29177,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABCD$ be a rectangle. A certain number of red lines are drawn between the lines $AB$ and $CD$, parallel to them, and a certain number of blue lines are drawn between the lines $AD$ and $BC$, parallel to them. This divides the rectangle into $775$ smaller rectangles, while the red and the blue lines intersect at $720$ points. How many red, and how many blue lines are there? (Matko Ljulj)",
    "options": [],
    "answer": "24 red lines and 30 blue lines, or 30 red lines and 24 blue lines",
    "solution": "",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 29178,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABCD$ be a square with side length $1$. Let $X$ be a point on the side $AB$, and let $Y$ be a point on the side $AD$ such that $\\angle CXY = 90^\\circ$. Find the locus of the point $X$ for which the area of the triangle $CDY$ is the smallest possible.",
    "options": [],
    "answer": "The midpoint of side AB",
    "solution": "",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 29179,
    "subject": "Mathematics (Multi-modal)",
    "question": "Determine all positive integers $n$ for which the quadratic equation\n$$\nx^2 - 3nx + n + 3 = 0\n$$\nhas integer solutions.",
    "options": [],
    "answer": "n = 2",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 29180,
    "subject": "Mathematics (Multi-modal)",
    "question": "Determine all triples $(p, q, r)$ of prime numbers such that\n$$\np^q = r - 1.\n$$",
    "options": [],
    "answer": "(2, 2, 5)",
    "solution": "",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 29181,
    "subject": "Mathematics (Multi-modal)",
    "question": "A trapezium $ABCD$ is given. The bisector of the leg $\\overline{BC}$ intersects the leg $\\overline{AD}$ at $M$, while the bisector of $\\overline{AD}$ intersects $\\overline{BC}$ at $N$.\nLet $O_1$ and $O_2$ be the circumcentres of triangles $ABN$ and $CDM$, respectively.\nProve that the line $O_1O_2$ bisects the segment $MN$. (Stipe Vidak)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Denote by $P$ and $Q$ the midpoints of $\\overline{BC}$ and $\\overline{AD}$, respectively.\n![](attached_image_1.png)\nWe will show that the quadrilateral $ABNM$ is cyclic.\nFrom $\\angle MQN = \\angle NPM = 90^\\circ$, we get that the quadrilateral $MNPQ$ is cyclic. This implies $\\angle PQM + \\angle MNP = 180^\\circ$.\nThe segment $\\overline{QP}$ is the midsegment of the trapezium $ABCD$, which means that it is parallel to $AB$. From here we conclude that $\\angle MAB = \\angle DQP$. Now we have\n$$\n\\angle MAB = \\angle DQP = 180^\\circ - \\angle PQM = \\angle MNP = 180^\\circ - \\angle BNM,\n$$\nwhich is enough to conclude that the quadrilateral $ABNM$ is cyclic.\nAnalogously, we show that the quadrilateral $MNCD$ is cyclic.\nThis shows that the segment $\\overline{MN}$ is simultaneously a chord for both circumscribed circles of triangles $ABN$ and $CDM$. We conclude that the line $O_1O_2$, connecting the centres of these circles, must bisect the segment $\\overline{MN}$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 29182,
    "subject": "Mathematics (Multi-modal)",
    "question": "Determine all pairs $(p, q)$ of prime numbers for which $p^{q-1} + q^{p-1}$ is a perfect square.",
    "options": [],
    "answer": "(2, 2)",
    "solution": "Let $n$ be a positive integer such that\n$$\np^{q-1} + q^{p-1} = n^2.\n$$\nWe will divide our solution into cases, depending on the parity of numbers $p$ and $q$.\n\nIf both are even, the only possibility is that both are equal to $2$. This leads to a solution $(p, q) = (2, 2)$ for $n = 2$.\n\nIf both $p$ and $q$ are odd, we conclude that there are no solutions by looking at the remainder of division by $4$. If $p$ and $q$ are odd, then $p-1$ and $q-1$ are even, so that $p^{q-1}$ and $q^{p-1}$ are squares of odd numbers. This means that they always give remainder $1$ when divided by $4$. This shows that the left-hand side gives remainder $2$ when divided by $4$, whereas $n^2$ must give remainder $1$ or $0$ when divided by $4$.\n\nThe only remaining case is when one of the numbers $p$ and $q$ is even, while the other is odd. The equation being symmetric, we can, without loss of generality, assume that $p$ is odd, i.e. $p = 2k+1$ for some positive integer $k$, and that $q$ equals $2$.\n\nNow we have\n$$\np + 2^{p-1} = n^2, \\quad \\text{ i.e. } p = n^2 - 2^{2k} = (n - 2^k)(n + 2^k).\n$$\nSince $p$ is prime, the only way in which we can represent it as a product $(n - 2^k)(n + 2^k)$ is if one of the factors equals $\\pm 1$, while the other is $\\pm p$. Notice that $n + 2^k > 0$, and also that $n + 2^k > n - 2^k$, which yields\n$$\nn - 2^k = 1 \\quad \\text{and} \\quad n + 2^k = p.\n$$\nSubtracting these two equalities, we get\n$$\n2^{k+1} + 1 = p = 2k + 1.\n$$\nLet us prove that this equation has no solution. We know that $2^{k+1} + 1$ gives remainder $1$ when divided by $4$, so the same must hold for $2k + 1$. This implies that $k$ is even, i.e. $k = 2l$ for some positive integer $l$. In that case,\n$$\np = 2^{k+1} + 1 = 2^{2l+1} + 1 = 2 \\cdot 4^l + 1\n$$\nis divisible by $3$ (because $4^l$ gives remainder $1$ when divided by $3$). We conclude that $p$ is divisible by $3$, so $p = 3$. Plugging this into the initial equation, we see that $p = 3$ and $q = 2$ is not a solution.\n\nThis means that the only solution is $(p, q) = (2, 2)$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 29183,
    "subject": "Mathematics (Multi-modal)",
    "question": "Determine all pairs $(x, y)$ of real numbers such that\n$$\n\\begin{aligned}\nx^2 + x y - 4 y^2 &= -1 \\\\\n4 x^2 + x y - 11 y^2 &= -2.\n\\end{aligned}\n$$",
    "options": [],
    "answer": "(3, 2), (-3, -2), (-1/2, 1/2), (1/2, -1/2)",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 29184,
    "subject": "Mathematics (Multi-modal)",
    "question": "There are two non-intersecting circles, of radii $r_1$ and $r_2$. The distance between the points of tangency of the inner common tangent to these circles is $12$, while the distance between the points of tangency of the outer common tangent to these circles is $16$. Determine the product $r_1 r_2$.\n\nInner tangent (is that common tangent that) intersects the segment between the centres of the circles.",
    "options": [],
    "answer": "28",
    "solution": "",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 29185,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $n \\ge 4$ be a positive integer. Prove that among any $n$ numbers from the set\n$$\n\\{1, 2, \\dots, 2n - 1\\}\n$$\nthere are some of them with the sum divisible by $2n$. (Ratko Višak)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 29186,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABCD$ be a trapezium with bases $\\overline{AB}$ and $\\overline{CD}$, such that $ABC$ is an acute-angled triangle. Let $O$ be the centre of the circle circumscribed to the triangle $ABC$, and let $E$ be the intersection of lines $OB$ and $CD$. If $\\angle DBC = \\angle CEB + 10^\\circ$, find the measure of the angle between the diagonals of $ABCD$.",
    "options": [],
    "answer": "20°",
    "solution": "",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 29187,
    "subject": "Mathematics (Multi-modal)",
    "question": "An $n \\times n$ square board is given, where $n$ is an odd positive integer. Each of the $2n(n+1)$ unit segments delimiting the unit squares is coloured either red or blue. It is known that there are no more than $n^2$ red unit segments.\nProve that there is a unit square on the board whose border comprises at least three blue segments.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Assume the contrary, i.e. that there is no square bordered by three or four blue segments. Then each square is bordered by at least two red segments.\n\nNow we count the pairs $(P, r)$, where $P$ is a unit square, and $r$ is a red segment adjacent to $P$. We will count them in two different ways.\n\nSince every square is bordered by at least two red segments, we have at least $2n^2$ such pairs. On the other hand, each segment on the edge of the board has only one adjacent square, while the interior segments have two adjacent squares. This shows that we can express the number of pairs $(P, r)$ as $V+2U$, where $V$ is the number of red segments on the edge of the board, and $U$ the number of interior red segments. Thus $2n^2 \\le 2U+V$.\n\nFurthermore, since there are no more than $n^2$ red segments, we also have $U+V \\le n^2$. Adding these two inequalities we get $V=0$ and $U=n^2$.\n\nThis shows that the inequalities we derived are in fact equalities, so that each square is bordered by exactly two red segments and all the red segments are interior (not on the edge of the board).\n\nNow we checker the board, black-white. Notice that each white square contains exactly two red segments, and that each red segment belongs to a uniquely determined white square. This means that there are $2B$ red segments, where $B$ is the number of white squares. However, this leads to a contradiction because we know that there are exactly $n^2$ red segments, and $n^2$ is odd.\n\nThis shows that our assumption was wrong, so there must be some unit square with at least three blue adjacent segments.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 29188,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABC$ be a right-angled triangle with the right angle at $C$, such that $|AC| = 2$ and $|BC| = 1$. A square is situated inside the triangle so that two of its vertices are on the side $AC$, the third one is on the side $AB$, and the fourth is on the circle of radius $1$ centred at $B$. Determine the length of the side of the square.",
    "options": [],
    "answer": "2/5",
    "solution": "",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 29189,
    "subject": "Mathematics (Multi-modal)",
    "question": "The rows of a $50 \\times 50$ table are labelled with numbers $a_1, \\dots, a_{50}$, while the columns are labelled with numbers $b_1, \\dots, b_{50}$. These 100 numbers are mutually distinct, and exactly 50 of them are rational. The table is filled so that the number $a_i + b_j$ is written in the $(i, j)$ cell, for $i, j = 1, 2, \\dots, 50$. Determine the largest possible number of rational numbers written in the table cells.",
    "options": [],
    "answer": "1250",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 29190,
    "subject": "Mathematics (Multi-modal)",
    "question": "Prove the identity\n$$\n\\cos^3 \\frac{x}{3} + \\cos^3 \\frac{x+2\\pi}{3} + \\cos^3 \\frac{x+4\\pi}{3} = \\frac{3}{4} \\cos x\n$$\nfor every real number $x$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "**Using the identity**\n$$\n\\cos t = 4 \\cos^3 \\frac{t}{3} - 3 \\cos \\frac{t}{3}\n$$\nfor $t = x$, $t = x + 2\\pi$ and $t = x + 4\\pi$, summing the relations and using the periodicity of cosine, we obtain\n$$\n3 \\cos x = 4 \\left( \\cos^3 \\frac{x}{3} + \\cos^3 \\frac{x+2\\pi}{3} + \\cos^3 \\frac{x+4\\pi}{3} \\right) - 3 \\left( \\cos \\frac{x}{3} + \\cos \\frac{x+2\\pi}{3} + \\cos \\frac{x+4\\pi}{3} \\right)\n$$\nTransforming the expression in the second parenthesis yields\n$$\n\\cos \\frac{x}{3} + \\cos \\frac{x+2\\pi}{3} + \\cos \\frac{x+4\\pi}{3} = 2 \\cos \\frac{2\\pi}{3} \\cos \\frac{x+2\\pi}{3} + \\cos \\frac{x+2\\pi}{3} \\\\\n= \\left( 2 \\cos \\frac{2\\pi}{3} + 1 \\right) \\cos \\frac{x+2\\pi}{3} = 0.\n$$\nFinally, we conclude\n$$\n\\cos^3 \\frac{x}{3} + \\cos^3 \\frac{x+2\\pi}{3} + \\cos^3 \\frac{x+4\\pi}{3} = \\frac{3}{4} \\cos x.\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 29191,
    "subject": "Mathematics (Multi-modal)",
    "question": "Determine all real numbers $x$ such that\n$$\n1 \\le \\frac{1 + \\sin x}{1 - \\sin x} \\le 3.\n$$",
    "options": [],
    "answer": "All real x with sin x in [0, 1/2], i.e., for any integer k: x in [2kπ, 2kπ + π/6] ∪ [2kπ + 5π/6, 2kπ + π].",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 29192,
    "subject": "Mathematics (Multi-modal)",
    "question": "Determine all pairs $(x, y)$ of real numbers such that $x, y \\in [0, \\frac{\\pi}{2}]$ for which the following holds:\n$$\n\\frac{2 \\sin^2 x + 2}{\\sin x + 1} = 3 + \\cos (x + y). \\quad (\\text{Petar Bakić})\n$$",
    "options": [],
    "answer": "(π/2, π/2)",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 29193,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $a$, $b$ and $c$ be positive real numbers such that\n$$\na^2 + b^2 - c^2 = \\sqrt{3}ab, \\quad a^2 - b^2 + c^2 = \\sqrt{2}ac.\n$$\nDetermine the ratio $b : c$. (Hong Kong 2002)",
    "options": [],
    "answer": "sqrt(2):1",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 29194,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $S = \\{0, 95\\}$. In each step, Lucija is extending the set $S$ in the following way. She chooses a polynomial with coefficients in $S$, distinct from zero, and extends $S$ with all integer roots of a chosen polynomial. She repeats the procedure by choosing another polynomial with coefficients from the extended set $S$ as long as she can get new roots.\nProve that Lucija can, in a finite number of steps, extend the set $S$ up to the set which is no longer extensible. How many elements does the set $S$ have at the end?",
    "options": [],
    "answer": "9",
    "solution": "If the coefficients of a polynomial are integers, then its roots must divide the constant term. Without loss of generality, we may assume that the constant term of the chosen polynomial is non-zero, so we conclude that we can extend $S$ only with the divisors of $95$. Since $95$ has only finitely many divisors, Lucija will not be able to extend $S$ indefinitely.\n\nWe will prove that Lucija can add all the divisors of $95$ into $S$, i.e. that $S$ will have $9$ elements in the end.\n\nSince $-1$ is the root of the polynomial $95x + 95$, we can add $-1$ to $S$.\n\nNumber $1$ is a root of $-x^9 - x^{94} - \\dots - x + 95$, so $S$ can be extended with $1$.\n\nNow we can extend $S$ with $-95$ because it is the root of $x + 95$.\n\nPolynomial $-x^3 + x^2 + x + 95$ with coefficients in $S$ has a root $5$, hence $5$ becomes an element of $S$. Now we add $-5$ into $S$, because $-5$ is the root of $x + 5$.\n\nIn the end, $-19$, $19 \\in S$ since these are the roots of polynomials $5x + 95$ and $5x - 95$.",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 29195,
    "subject": "Mathematics (Multi-modal)",
    "question": "Determine all real numbers $x$ such that\n$$\n(x - 1009)^3 + (2x - 1009)^3 + (2018 - 3x)^3 = 0.\n$$",
    "options": [],
    "answer": "1009/2, 1009, 2018/3",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 29196,
    "subject": "Mathematics (Multi-modal)",
    "question": "Determine all positive integers which are squares of integers, and which have exactly two non-zero digits in the decimal representation, one of which is equal to $3$.",
    "options": [],
    "answer": "All numbers of the form 36 × 100^k for nonnegative integers k.",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 29197,
    "subject": "Mathematics (Multi-modal)",
    "question": "Determine all pairs $(a, b)$ of positive integers such that $a^2 b$ divides $b^2 + 3a$.",
    "options": [],
    "answer": "[(1, 1), (1, 3)]",
    "solution": "The given condition implies that\n$$\na^2 b k = b^2 + 3a\n$$\nfor some positive integer $k$. The equation $a^2 b k - 3a = b^2$ implies $a \\mid b^2$ and $b \\mid 3a$. Therefore, the numbers\n$$\n\\frac{b^2}{a}, \\frac{3a}{b} \\quad \\text{and} \\quad \\frac{b^2 + 3a}{a^2 b} = \\frac{b}{a^2} + \\frac{3}{a b}\n$$\nare integers. By multiplying the second and the third one, we conclude that $\\frac{9a}{b^2}$ is an integer, meaning that $b^2 \\mid 9a$. Now we have\n$$\nb^2 m = 9a, \\quad a n = b^2\n$$\nfor some positive integers $m$ and $n$. This implies $m n = 9$. Hence, $n \\in \\{1, 3, 9\\}$, which means that $b^2 \\in \\{a, 3a, 9a\\}$. The initial condition now implies $a b \\mid n + 3$.\n\nWe have three possible values for $n$.\n\nIf $n = 1$, then $b^2 = a$, so the initial condition becomes $b^5 \\mid 4b^2$, i.e. $b^3$ divides $4$. We obtain one solution $a = b = 1$.\n\nIf $n = 3$, then $b^2 = 3a$, so $\\frac{b^5}{9} \\mid 2b^2$, i.e. $b^3 \\mid 18$. Since $b$ is a multiple of $3$, there is no solution in this case.\n\nIf $n = 9$, then $b^2 = 9a$, so $\\frac{b^5}{81} \\mid \\frac{4}{3} b^2$, i.e. $b^3 \\mid 108$. Since $b$ is a multiple of $3$, we obtain a solution $(a, b) = (1, 3)$.\n\nThe only solutions are $(1, 1)$ and $(1, 3)$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 29198,
    "subject": "Mathematics (Multi-modal)",
    "question": "In an isosceles triangle $ABC$ with $|AB| = |AC|$, points $M$ and $N$ are the midpoints of the sides $\\overline{AB}$ and $\\overline{BC}$, respectively. The circle circumscribed to the triangle $AMC$ meets the line $AN$ at point $P$ different from $A$. The line passing through $P$ parallel to the side $BC$ meets the circle circumscribed to the triangle $ABC$ at points $B_1$ and $C_1$. Prove that the triangle $AB_1C_1$ is equilateral.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "The points $A$, $M$, $P$ and $C$ lie on the same circle and $\\angle MAP = \\angle PAC$. Therefore, $|MP| = |PC|$ because the corresponding subtended angles are equal. Since $P$ lies on the bisector of $\\overline{BC}$, we conclude that $|BP| = |CP|$. Hence $|MP| = |BP|$, which means that the point $P$ lies on the bisector of the segment $\\overline{BM}$.\n\nLet $Q$ be the midpoint of the segment $\\overline{BM}$, and $O$ be the centre of the circumcircle of the triangle $ABC$.\n\nNotice that $PQ \\perp AB$ and $OM \\perp AB$, which implies that $PQ \\parallel OM$. Since $M$ is the midpoint of $\\overline{AB}$, and $Q$ is the midpoint of $\\overline{MB}$, it follows that $|MQ| = \\frac{1}{3}|AQ|$. For the same reason we have $|OP| = \\frac{1}{3}|AP|$.\n\n![](attached_image_1.png)\n\nConsider a triangle $AB_1C_1$. Its circumcentre is point $O$, while $P$ is the foot of its altitude from vertex $A$. Since $A$, $O$ and $P$ are collinear, the triangle is isosceles.\n\nTherefore, the segment $\\overline{AP}$ is also a median of the triangle, so $|OP| = \\frac{1}{3}|AP|$ implies that $O$ is the centroid. Finally, since the centroid coincides with the circumcentre, the triangle $AB_1C_1$ is isosceles.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 29199,
    "subject": "Mathematics (Multi-modal)",
    "question": "In how many ways can the letters $A$, $B$, $C$, $D$, $E$, $F$, $G$, $H$, $I$ be arranged so that all of the vowels and all of the consonants are ordered alphabetically?",
    "options": [],
    "answer": "84",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 29200,
    "subject": "Mathematics (Multi-modal)",
    "question": "In a quadrilateral $ABCD$ it holds that $\\angle DBC = \\angle DCB = 50^\\circ$ and $\\angle DAB = \\angle ABC = \\angle BDC$. Prove that $AC \\perp BD$. (Ratko Višak)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 29201,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $n$ be a positive integer. A sequence of $2n$ real numbers is good if for each positive integer $1 \\le m \\le 2n$ the sum of first $m$ or the sum of last $m$ terms of the sequence is an integer. Determine the least possible number of integers in a good sequence. (The Netherlands 2017)",
    "options": [],
    "answer": "2",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 29202,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all prime numbers $p$ for which there exists a positive integer $m$ such that the number $p^m + 4$ is a square of some positive integer.",
    "options": [],
    "answer": "p = 2 or p = 5",
    "solution": "",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 29203,
    "subject": "Mathematics (Multi-modal)",
    "question": "Two players alternately write one digit at a time, from left to right. A player loses if, after his move, there is a sequence of digits $a_1, a_2, \\dots, a_n$ such that there exists a positive integer $k$ for which the number $\\overline{a_k a_{k+1} \\dots a_n}$ is a multiple of $11$.\n\nWhich player can win regardless of how his opponent plays? (Russia 2003)",
    "options": [],
    "answer": "Second player",
    "solution": "We will show that the second player can win regardless of the first player's actions. Obviously, none of the players will write $0$ in any step.\n\nNotice that $10^r \\equiv (-1)^r \\pmod{11}$, so the following criterion for divisibility by $11$ holds:\n$$\n\\overline{a_k a_{k+1} \\dots a_n} \\equiv a_n - a_{n-1} + a_{n-2} - \\dots + (-1)^{n-k} a_k \\pmod{11}\n$$\nDenote by $N_k$ the remainder when dividing $\\overline{a_k a_{k+1} \\dots a_n}$ by $11$, for $k = 1, \\dots, n$. If, after the $n^{th}$ move, the numbers $N_1, \\dots, N_n$ are pairwise distinct, then by the mentioned criterion, in the next move we obtain the numbers $a_{n+1} - N_1, \\dots, a_{n+1} - N_n, a_{n+1} \\pmod{11}$ which are also pairwise distinct since $a_{n+1} \\neq 0$. Inductively, we conclude that after each step of the game (for $n \\le 10$), the numbers $N_1, \\dots, N_n$ are distinct. Because of the way in which $N_1, \\dots, N_n$ transform at each move, we conclude that there are at most $n+1$ digits which, if the player writes them, will cause him to lose.\n\nAssume that the game lasts for at least nine moves. The second player loses if and only if, after the ninth move, the set $\\{N_1, N_2, \\dots, N_9\\}$ is $\\{1, 2, \\dots, 9\\}$, i.e. the second player wins if and only if $10$ is among the numbers $N_1, N_2, \\dots, N_9$.\n\nIf, after the eighth move, $\\{N_1, ..., N_8\\}$ lacks two numbers between $1$ and $10$ which are not consecutive, then, no matter what the first player chooses in the ninth move, one of the numbers $N_1, ..., N_9$ must be $10$. Namely, if the first player chooses $X$, then there is $k$ such that $N_k = X + 1$ (in eighth move). Hence, after the ninth move we obtain\n$$\nN_k \\equiv X - (X + 1) \\equiv -1 \\equiv 10 \\pmod{11}.\n$$\nLet us show that the second player can ensure that after the eighth move, among the numbers $N_1, ..., N_8$ there are no two consecutive numbers. Certainly, the second player can ensure that the game lasts at least seven moves. Let\n$$\n\\{N_1, ..., N_7\\} = \\{1, 2, ..., 10\\} \\setminus \\{X, Y, Z\\}.\n$$\nIf among the numbers $X$, $Y$ and $Z$ there are no consecutive ones, then the second player can choose any of them. If $Y = X + 1$, then the second player can write one of the numbers $X$ or $X + 1$ so that, after the eighth move, there are two consecutive numbers missing from $N_1, ..., N_8$. Namely, if $Z = X - 1$, the second player writes $X$, and if $Z \\neq X - 1$, the second player writes $X + 1$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 29204,
    "subject": "Mathematics (Multi-modal)",
    "question": "A building consists of the ground floor and $100$ additional floors. The lift in the building has only two buttons, $A$ and $B$. By pressing $A$ the lift rises $7$ floors, and by pressing $B$ the lift goes down $9$ floors. Is it possible to reach every floor from any other floor using this lift?",
    "options": [],
    "answer": "Yes",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 29205,
    "subject": "Mathematics (Multi-modal)",
    "question": "A positive integer is called *Babylonic* if it is greater than $9$, and if its base-$60$ representation is the same as its decimal representation without the first digit. For example, the number $123$ is Babylonic since $123 = (23)_{60}$. How many Babylonic numbers less than $10\\,000$ are there? (Mea Bombardelli)",
    "options": [],
    "answer": "50",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 29206,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $n$ be a positive integer. Prove that for all $x_1, x_2, \\dots, x_n \\in [0, 1]$ the following holds:\n$$\n(x_1 + x_2 + \\dots + x_n + 1)^2 \\ge 4(x_1^2 + x_2^2 + \\dots + x_n^2). \\quad (\\text{ASU 1979})\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Since $x_i \\in [0, 1]$, we conclude that $x_i \\ge x_i^2$, i.e.\n$$\n4(x_1 + x_2 + \\dots + x_n) \\ge 4(x_1^2 + x_2^2 + \\dots + x_n^2).\n$$\nDenote $S = x_1 + x_2 + \\dots + x_n$. Notice that, because of the previous bound, it suffices to show that\n$$\n(S + 1)^2 \\ge 4S.\n$$\nThis inequality obviously holds since it is equivalent to $(S - 1)^2 \\ge 0$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 29207,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find the last two digits of the number $(1!)^2 + (2!)^2 + \\dots + (2018!)^2$.",
    "options": [],
    "answer": "17",
    "solution": "For $n \\geq 10$, $n!$ is divisible by $10$, so $(n!)^2$ is divisible by $100$. Therefore, only the terms $(1!)^2, (2!)^2, \\dots, (9!)^2$ contribute to the last two digits.\n\nCompute each term:\n\n$(1!)^2 = 1^2 = 1$\n\n$(2!)^2 = 2^2 = 4$\n\n$(3!)^2 = 6^2 = 36$\n\n$(4!)^2 = 24^2 = 576$\n\n$(5!)^2 = 120^2 = 14400$\n\n$(6!)^2 = 720^2 = 518400$\n\n$(7!)^2 = 5040^2 = 25401600$\n\n$(8!)^2 = 40320^2 = 1625702400$\n\n$(9!)^2 = 362880^2 = 131681894400$\n\nNow, take the last two digits of each term:\n\n$(1!)^2 = 1$ (last two digits: $01$)\n\n$(2!)^2 = 4$ (last two digits: $04$)\n\n$(3!)^2 = 36$ (last two digits: $36$)\n\n$(4!)^2 = 576$ (last two digits: $76$)\n\n$(5!)^2 = 14400$ (last two digits: $00$)\n\n$(6!)^2 = 518400$ (last two digits: $00$)\n\n$(7!)^2 = 25401600$ (last two digits: $00$)\n\n$(8!)^2 = 1625702400$ (last two digits: $00$)\n\n$(9!)^2 = 131681894400$ (last two digits: $00$)\n\nSum the last two digits:\n$01 + 04 + 36 + 76 + 00 + 00 + 00 + 00 + 00 = 117$\n\nThe last two digits of $117$ are $17$.\n\nAnswer: $17$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 29208,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $z$ be a complex number such that $\\arg z \\in [\\frac{\\pi}{2}, \\pi]$ and $z^6 + z^3 + 1 = 0$. Determine the modulus and the argument of $z$. The argument of the complex number $z = |z|(\\cos\\varphi + i\\sin\\varphi)$ is the number $\\arg z = \\varphi$.",
    "options": [],
    "answer": "|z| = 1, arg z = 8π/9",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 29209,
    "subject": "Mathematics (Multi-modal)",
    "question": "A Gaussian integer is a complex number whose real and imaginary parts are both integers. Determine the largest positive integer $n$ for which there is a set of $n$ Gaussian integers such that the squares of their absolute values are consecutive positive integers. (Matko Ljulj)",
    "options": [],
    "answer": "3",
    "solution": "If a complex number $z = x + yi$ is Gaussian integer, then $|z|^2 = x^2 + y^2$ is the sum of two squares of integers.\nThe square of an even integer is divisible by 4, while the square of an odd integer gives remainder 1 when divided by 4. Therefore, the sum of squares of two integers can give remainder 0, 1 or 2 when divided by 4.\nIf $n$ is greater than 3, then among any four consecutive positive integers in the sequence\n$$\n|z_1|^2, |z_2|^2, \\dots, |z_n|^2\n$$\nthere would be a number giving the remainder 3 when divided by 4, which is impossible. Hence, $n$ is less than or equal to 3.\nNotice that $2 + 2i$, $3$ and $3 + i$ are Gaussian integers whose absolute values squared are $8$, $9$ and $10$, respectively. Therefore, the answer for $n$ is $3$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 29210,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $n$ be a positive integer. Prove that\n$$\n\\frac{1}{n+1} + \\frac{1}{n+2} + \\dots + \\frac{1}{3n+1} > 1.\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let us consider the sum:\n$$\nS = \\frac{1}{n+1} + \\frac{1}{n+2} + \\dots + \\frac{1}{3n+1}\n$$\nThere are $(3n+1) - (n+1) + 1 = 2n+1$ terms in the sum.\n\nWe will compare this sum to the integral:\n$$\n\\int_{n+1}^{3n+2} \\frac{1}{x} \\, dx = \\ln(3n+2) - \\ln(n+1) = \\ln\\left(\\frac{3n+2}{n+1}\\right)\n$$\nNote that for $k = n+1, \\ldots, 3n+1$, we have:\n$$\n\\frac{1}{k} > \\int_{k}^{k+1} \\frac{1}{x} \\, dx\n$$\nTherefore,\n$$\nS > \\int_{n+1}^{3n+2} \\frac{1}{x} \\, dx = \\ln\\left(\\frac{3n+2}{n+1}\\right)\n$$\nWe want to show that $S > 1$, so it suffices to show that $\\ln\\left(\\frac{3n+2}{n+1}\\right) > 1$ for all positive integers $n$.\n\nLet us check:\n$$\n\\ln\\left(\\frac{3n+2}{n+1}\\right) > 1\n$$\nThis is equivalent to:\n$$\n\\frac{3n+2}{n+1} > e\n$$\nLet us solve for $n$:\n$$\n3n+2 > e(n+1) \\\\\n3n+2 > en + e \\\\\n3n - en > e - 2 \\\\\n(3-e)n > e-2\n$$\nSince $e \\approx 2.718$, $3-e \\approx 0.282$, so for $n$ sufficiently large, this is true. Let's check for small $n$:\n\nFor $n=1$:\n$$\n\\frac{3\\cdot 1 + 2}{1+1} = \\frac{5}{2} = 2.5 < e\n$$\nFor $n=2$:\n$$\n\\frac{3\\cdot 2 + 2}{2+1} = \\frac{8}{3} \\approx 2.666 < e\n$$\nFor $n=3$:\n$$\n\\frac{3\\cdot 3 + 2}{3+1} = \\frac{11}{4} = 2.75 > e\n$$\nSo for $n \\geq 3$, $\\ln\\left(\\frac{3n+2}{n+1}\\right) > 1$, and thus $S > 1$.\n\nFor $n=1$:\n$$\nS = \\frac{1}{2} + \\frac{1}{3} + \\frac{1}{4} + \\frac{1}{5} = \\frac{1}{2} + \\frac{1}{3} + \\frac{1}{4} + \\frac{1}{5} = 0.5 + 0.333... + 0.25 + 0.2 = 1.283... > 1\n$$\nFor $n=2$:\n$$\nS = \\frac{1}{3} + \\frac{1}{4} + \\frac{1}{5} + \\frac{1}{6} + \\frac{1}{7} = 0.333... + 0.25 + 0.2 + 0.166... + 0.142... = 1.091... > 1\n$$\nSo the inequality holds for all positive integers $n$.\n\nTherefore,\n$$\n\\frac{1}{n+1} + \\frac{1}{n+2} + \\dots + \\frac{1}{3n+1} > 1\n$$\nfor all positive integers $n$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 29211,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABC$ be an acute-angled triangle such that $|BC| > |AC|$. The perpendicular bisector of the segment $\\overline{AB}$ intersects the side $BC$ at point $P$, and the line $AC$ at point $Q$. Let $R$ be the foot of the perpendicular from point $P$ to the side $\\overline{AC}$, and let $S$ be the foot of the perpendicular from point $Q$ to the line $BC$. Prove that the line $RS$ bisects the segment $\\overline{AB}$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 29212,
    "subject": "Mathematics (Multi-modal)",
    "question": "Prove that $2^{2^{n+2}} + 4$ is a multiple of $10$ for any positive integer $n$. (Tamara Srnec)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let us consider $2^{2^{n+2}} + 4$ modulo $10$.\n\nFirst, note that $2^{2^{n+2}}$ is a very large power of $2$. Let's analyze the last digit of $2^k$ for large $k$.\n\nThe last digit of powers of $2$ cycles every $4$:\n\n| $k$ | $2^k$ | Last digit |\n|---|------|------------|\n| 1 | 2    | 2          |\n| 2 | 4    | 4          |\n| 3 | 8    | 8          |\n| 4 | 16   | 6          |\n| 5 | 32   | 2          |\n| 6 | 64   | 4          |\n| 7 | 128  | 8          |\n| 8 | 256  | 6          |\n\nSo, the last digit of $2^k$ depends on $k \bmod 4$:\n- If $k \\equiv 1 \\pmod{4}$, last digit is $2$.\n- If $k \\equiv 2 \\pmod{4}$, last digit is $4$.\n- If $k \\equiv 3 \\pmod{4}$, last digit is $8$.\n- If $k \\equiv 0 \\pmod{4}$, last digit is $6$.\n\nNow, $2^{n+2}$ is always even for $n \\geq 1$ (since $n+2 \\geq 3$), so $2^{n+2} \\geq 8$.\n\nBut more importantly, $2^{n+2}$ is divisible by $4$ for $n \\geq 1$:\n- $2^{n+2} = 4 \\cdot 2^{n}$, so $2^{n+2} \\equiv 0 \\pmod{4}$.\n\nTherefore, $2^{2^{n+2}}$ is a power of $2$ where the exponent is divisible by $4$.\n\nFrom the table above, if $k \\equiv 0 \\pmod{4}$, the last digit of $2^k$ is $6$.\n\nThus, $2^{2^{n+2}}$ ends with $6$ for any positive integer $n$.\n\nTherefore,\n$$\n2^{2^{n+2}} + 4\n$$\nhas last digit $6 + 4 = 10$, so ends with $0$.\n\nThus, $2^{2^{n+2}} + 4$ is divisible by $10$ for any positive integer $n$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 29213,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $f: N \\to N$ be a function such that\n$$\nf(ab) = f(a+b)\n$$\nfor all positive integers $a \\ge 4$ and $b \\ge 4$.\nProve that $f(n) = f(8)$ for all positive integers $n \\ge 8$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $n \\ge 8$ be a positive integer. The problem condition implies that\n$$\n\\begin{aligned}\nf(n) &= f(4 + (n-4)) = f(4(n-4)) = f(2(n-4) + 2(n-4)) \\\\\n&= f(4(n-4)(n-4)) = f(4(n-4) + (n-4)) \\\\\n&= f(5(n-4)) = f(5 + n - 4) \\\\\n&= f(n+1).\n\\end{aligned}\n$$\nTherefore, by the principle of mathematical induction, we conclude that $f(n) = f(8)$ for all $n \\ge 8$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 29214,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $\\overline{BD}$ and $\\overline{CE}$ be the altitudes of an acute-angled triangle $ABC$. The circle with diameter $\\overline{AC}$ meets $\\overline{BD}$ at $F$. The circle with diameter $\\overline{AB}$ meets the line $CE$ at points $G$ and $H$, where $G$ is between $C$ and $E$. If $\\angle CHF = 12^\\circ$, find the measure of the angle $\\angle AGF$. (Go Geometry)",
    "options": [],
    "answer": "78°",
    "solution": "The chord $\\overline{GH}$ is perpendicular to $\\overline{AB}$, so $AB$ is the bisector of the segment $\\overline{GH}$. Hence $|AG| = |AH|$.\n\nSince $AFC$ is a right-angled triangle, Euclid's theorem gives us $|AF|^2 = |AD| \\cdot |AC|$.\n\nAnalogously, since $ABG$ is a right-angled triangle, we have $|AG|^2 = |AE| \\cdot |AB|$.\n\n![](attached_image_1.png)\n\nAngles $\\angle BDC$ and $\\angle BEC$ are right angles, so the quadrilateral $BCDE$ is cyclic. By using the power-of-a-point theorem applied to $A$ with respect to the circle circumscribed to the quadrilateral $BCDE$, we conclude that $|AD| \\cdot |AC| = |AE| \\cdot |AB|$.\n\nHence $|AF| = |AG| = |AH|$, i.e. the point $A$ is the circumcentre of the triangle $GFH$.\n\nFinally,\n$$\n\\angle AGF = \\angle GFA = \\frac{1}{2}(180^\\circ - \\angle FAG) = \\frac{1}{2}(180^\\circ - 2\\angle FHG) = \\frac{1}{2}(180^\\circ - 24^\\circ) = 78^\\circ.\n$$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 29215,
    "subject": "Mathematics (Multi-modal)",
    "question": "A box holds $n$ balls, some of which are white, and some are black. Find $n$ if the probability that by drawing two balls we draw exactly one white and one black is $\\frac{1}{2}$, and we know that there are 2018 more balls of one colour than the other.",
    "options": [],
    "answer": "2018^2",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 29216,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find the locus of the centres of all circles that are externally tangent to the circle that satisfies $x^2 + y^2 - 4y + 3 = 0$, and that are also tangent to the $x$-axis.\n*(Anastazija Pažanin)*",
    "options": [],
    "answer": "x^2 - 6y + 3 = 0",
    "solution": "Let the given circle be $x^2 + y^2 - 4y + 3 = 0$. Rewrite it in standard form:\n\n$x^2 + (y^2 - 4y) + 3 = 0$\n$x^2 + (y - 2)^2 - 4 + 3 = 0$\n$x^2 + (y - 2)^2 - 1 = 0$\n$x^2 + (y - 2)^2 = 1$\n\nSo, the circle has centre $(0, 2)$ and radius $1$.\n\nLet the centre of the required circle be $(h, k)$ and its radius be $r$.\n\nSince the circle is tangent to the $x$-axis, its distance from the $x$-axis is $r$, so $k = r$.\n\nSince the circle is externally tangent to the given circle, the distance between their centres is equal to the sum of their radii:\n\n$\\sqrt{(h - 0)^2 + (k - 2)^2} = r + 1$\n\nBut $k = r$, so:\n\n$\\sqrt{h^2 + (r - 2)^2} = r + 1$\n\nSquare both sides:\n\n$h^2 + (r - 2)^2 = (r + 1)^2$\n$h^2 + r^2 - 4r + 4 = r^2 + 2r + 1$\n\nSubtract $r^2$ from both sides:\n\n$h^2 - 4r + 4 = 2r + 1$\n$h^2 - 4r + 4 - 2r - 1 = 0$\n$h^2 - 6r + 3 = 0$\n\nRecall $k = r$, so:\n\n$h^2 - 6k + 3 = 0$\n\nTherefore, the locus of the centres $(h, k)$ is:\n\n$h^2 - 6k + 3 = 0$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 29217,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $d$ be a positive integer, and let $(a_n)$ be an arithmetic sequence of positive integers with the common difference $d$. If $d \\le 2018$, prove that at most 11 consecutive terms of that sequence can be prime numbers.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 29218,
    "subject": "Mathematics (Multi-modal)",
    "question": "There are 300 contestants at the competition. Each pair of the contestants is either acquainted (knows each other) or unfamiliar with each other, and there are no three contestants who all know each other. Determine the maximum possible $n$ such that the following conditions hold:\n* Every contestant is acquainted with at most $n$ other contestants.\n* For every positive integer $m$ such that $1 \\le m \\le n$, there is at least one contestant who is acquainted with exactly $m$ other contestants.\n(Mongolia 2017)",
    "options": [],
    "answer": "200",
    "solution": "The maximum possible $n$ is $200$.\n\nLet us assume that there is a contestant, say $X$, who knows $201$ other contestants and let those $201$ contestants make up a set $S$. There must exist contestants who know exactly $1$, $2$, $\\dots$, $200$ other contestants.\n\nWe will say that the contestant *has degree* $m$ if he knows exactly $m$ other contestants.\n\nEach element in $S$ has degree at most $99$. Namely, he cannot know anyone from $S$ since there are no three contestants who all know each other (aside from $X$ and $S$, there are only $98$ contestants).\n\nWe conclude that contestants from $S$ have at most $99$ distinct degrees. Simultaneously, there are $98$ contestants not in $S$ and different from $X$, so they can have at most $98$ distinct degrees.\n\nThis shows that there are at most $1 + 99 + 98 = 198 < 201$ distinct degrees, and therefore it is impossible that there are contestants who know exactly $1$, $2$, $\\dots$, $201$ other contestants. Hence, there is no contestant who knows exactly $201$ other contestants.\n\nLet us now show that $n = 200$ is possible. Denote $100$ contestants as $A_1, A_2, \\dots, A_{100}$ and call them A-contestants, and the remaining $200$ as $B_1, B_2, \\dots, B_{200}$ and call them B-contestants.\n\nFor each $i \\in \\{1, 2, \\dots, 100\\}$ and each $j \\in \\{1, 2, \\dots, 200\\}$, such that $i \\le j$, let $A_i$ and $B_j$ be acquainted. All the other pairs of contestants are unfamiliar with each other. We claim that this example has $n = 200$, and that all the conditions of the problem are satisfied.\n\nNamely, there are no three contestants who all know each other, because no two A-contestants know each other and no two B-contestants know each other.\n\nMoreso, for $i \\in \\{1, 2, \\dots, 100\\}$, contestant $A_i$ knows contestants $B_i, B_{i+1}, \\dots, B_{200}$ and only them. Hence, he has exactly $201 - i$ acquaintances, which means that there are contestants knowing exactly $200$, $199$, $\\dots$, $101$ other contestants. Analogously, for $j \\in \\{1, 2, \\dots, 100\\}$, contestant $B_j$ knows $A_1, \\dots, A_{j-1}, A_j$ and only them. This implies that he has exactly $j$ acquaintances, which means that there are contestants knowing exactly $1$, $2$, $\\dots$, $100$ other contestants.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 29219,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $a_1, a_2, \\dots, a_{41}$ be an arithmetic sequence such that\n$$\n\\frac{1}{\\sqrt{a_1} + \\sqrt{a_2}} + \\frac{1}{\\sqrt{a_2} + \\sqrt{a_3}} + \\dots + \\frac{1}{\\sqrt{a_{40}} + \\sqrt{a_{41}}}\n$$\nis a positive integer. If $a_1 = 1$, and the common difference of the sequence is a positive integer, find the common difference.\n*(Anastazija Pažanin)*",
    "options": [],
    "answer": "2, 9, 38",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 29220,
    "subject": "Mathematics (Multi-modal)",
    "question": "Prove that\n$$\n3abc + a + b + c \\ge 2(ab + bc + ca)\n$$\nholds for all real numbers $a, b, c \\ge 1$.\nDetermine all cases for which the equality is obtained.",
    "options": [],
    "answer": "Equality holds if and only if at least two of the numbers are equal to one.",
    "solution": "Let us denote $x = a-1$, $y = b-1$ and $z = c-1$. Then $x, y, z \\ge 0$, and the given inequality easily transforms into\n$$\n3xyz + xy + yz + zx \\ge 0,\n$$\nwhich is true since all addends on the left-hand side are non-negative.\n\nThe equality is obtained if and only if $xyz = xy = yz = zx = 0$, which is true if and only if at least two numbers among $x, y$ and $z$ are equal to $0$, i.e. if and only if at least two numbers among $a, b$ and $c$ are equal to $1$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 29221,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $a$, $b$ and $c$ be positive real numbers such that $a + b + c = 2$. Prove that\n$$\n\\frac{(a-1)^2}{b} + \\frac{(b-1)^2}{c} + \\frac{(c-1)^2}{a} \\ge \\frac{1}{4} \\left( \\frac{a^2+b^2}{a+b} + \\frac{b^2+c^2}{b+c} + \\frac{c^2+a^2}{c+a} \\right).\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "By the Cauchy-Bunyakovsky-Schwarz inequality, we have\n$$\n\\frac{(a-1)^2}{b} + \\frac{(b-1)^2}{c} \\ge \\frac{(2-a-b)^2}{b+c} = \\frac{c^2}{b+c},\n$$\nand similar inequalities hold for all pairs. By adding the inequalities, we get\n$$\n\\frac{(a-1)^2}{b} + \\frac{(b-1)^2}{c} + \\frac{(c-1)^2}{a} \\ge \\frac{1}{2} \\left( \\frac{b^2}{a+b} + \\frac{c^2}{b+c} + \\frac{a^2}{c+a} \\right).\n$$\nNote that the initial claim follows from this, because\n$$\n\\frac{b^2}{a+b} + \\frac{c^2}{b+c} + \\frac{a^2}{c+a} = \\frac{a^2}{a+b} + \\frac{b^2}{b+c} + \\frac{c^2}{c+a},\n$$\nwhich holds since\n$$\n\\begin{aligned}\n\\frac{b^2}{a+b} + \\frac{c^2}{b+c} + \\frac{a^2}{c+a} - \\frac{a^2}{a+b} - \\frac{b^2}{b+c} - \\frac{c^2}{c+a} &= \\frac{b^2-a^2}{a+b} + \\frac{c^2-b^2}{b+c} + \\frac{a^2-c^2}{c+a} \\\\\n&= (b-a) + (c-b) + (a-c) = 0.\n\\end{aligned}\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 29222,
    "subject": "Mathematics (Multi-modal)",
    "question": "Determine all functions $f: \\mathbb{R} \\to \\mathbb{R}$ such that\n$$\nf(xf(y)) = (1-y)f(xy) + x^2y^2f(y)\n$$\nholds for all real numbers $x$ and $y$.",
    "options": [],
    "answer": "f(x) = 0 for all real x; f(x) = x - x^2 for all real x",
    "solution": "Let $P(x,y)$ denote plugging $x$ and $y$ into the original equation. We have\n$$\n\\begin{aligned}\nP(0,1) &: f(0) = 0, \\\\\nP(1,1) &: f(f(1)) = f(1), \\\\\nP(1,f(1)) &: f(f(f(1))) = (1-f(1))f(f(1)) + f(1)^2 f(f(1)).\n\\end{aligned}\n$$\nBy combining the second and the third equality we get\n$$\nf(1) = (1 - f(1)) \\cdot f(1) + f(1)^3, \\quad \\text{i.e.} \\quad f(1)^2 = f(1)^3.\n$$\nFrom this we have two options: $f(1) = 1$ or $f(1) = 0$.\n\nIf $f(1) = 1$, we get\n$$\nP(x,1) : \\quad f(x) = x^2,\n$$\nso $f(x) = x^2$ should hold for all real $x$. By plugging this into the original equation we can conclude that this is not possible. Hence, $f(1) = 0$ must hold.\n\nLet $c$ be an arbitrary root of the function $f$. We have\n$$\nP(x,c) : \\quad 0 = (1-c)f(xc).\n$$\nTherefore, if $c \\ne 1$, $f(cx) = 0$ must hold for all $x \\in \\mathbb{R}$. If $c \\ne 0$ this implies that $f(x) = 0$ for all $x \\in \\mathbb{R}$. We check directly that $f \\equiv 0$ is indeed a possible solution of the original equation.\n\nNow assume that $f$ is not identically equal to 0. Calculations above show in that case 0 and 1 are the only roots of the function $f$. Now we have\n$$\nP(1,y) : \\quad f(f(y)) = (1-y)f(y) + y^2 f(y) = (1-y+y^2)f(y), \\quad (1)\n$$\nand for $x \\ne 0$\n$$\nP\\left(\\frac{1}{x}, x\\right) : \\quad f\\left(\\frac{f(x)}{x}\\right) = f(x), \\quad (2)\n$$\nsince $f(1) = 0$.\n\nLet $y_1, y_2 \\in \\mathbb{R} \\setminus \\{0,1\\}$ be such that $f(y_1) = f(y_2) \\ne 0$. Then from (1) it follows that\n$$\n1 - y_1 + y_1^2 = 1 - y_2 + y_2^2, \\quad \\text{i.e.} \\quad (y_1 - y_2)(y_1 + y_2 - 1) = 0.\n$$\nWe conclude that\n$$\nf(y_1) = f(y_2) \\implies y_1 = y_2 \\text{ or } y_1 + y_2 = 1. \\quad (3)\n$$\nBy combining (2) and (3) we get that for all $x \\ne 0,1$ we have\n$$\n\\frac{f(x)}{x} = x \\text{ or } \\frac{f(x)}{x} + x = 1,\n$$\ni.e.\n$$\nf(x) = x^2 \\text{ or } f(x) = x - x^2.\n$$\nLet $x \\in \\mathbb{R} \\setminus \\{0,1\\}$ be such that $f(x) = x^2$. By plugging $y = x$ into (1) we get\n$$\nf(x^2) = (1 - x + x^2)x^2.\n$$\nNow we have two cases: $f(x^2) = x^4$ and $f(x^2) = x^2 - x^4$. In the first case, we get $x^4 = (1-x+x^2)x^2$, from which we can easily conclude that $x=0$ or $x=1$. Since we assumed $x \\notin \\{0,1\\}$, this is not possible. In the second case, we get $x^2 - x^4 = (1-x+x^2)x^2$, from which we can conclude $x=0$ or $x = \\frac{1}{2}$. Again, we can dismiss $x=0$, and for $x = \\frac{1}{2}$ we can note that $f(\\frac{1}{2}) = (\\frac{1}{2})^2 = \\frac{1}{4} = \\frac{1}{2} - (\\frac{1}{2})^2$. This shows that $f(x) = x - x^2$ holds for all $x \\ne 0,1$. Since this formula holds for $x = 0,1$ as well, we can conclude that\n$$\nf(x) = x - x^2, \\quad \\forall x \\in \\mathbb{R}.\n$$\nBy plugging this directly into the original equation, we can confirm that this is indeed a solution. Therefore, the only solutions of the original equation are $f = 0, \\forall x \\in \\mathbb{R}$ and $f(x) = x - x^2, \\forall x \\in \\mathbb{R}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 29223,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $n$, $k$, $M$ and $a_1, a_2, \\dots, a_n$ be positive integers such that\n$$\n\\frac{1}{a_1} + \\frac{1}{a_2} + \\dots + \\frac{1}{a_n} = k \\quad \\text{and} \\quad a_1 a_2 \\dots a_n = M.\n$$\nIf $M > 1$, prove that there does not exist a positive real number $x$ such that\n$$\nM(x+1)^k = (x+a_1)(x+a_2)\\dots(x+a_n).\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Claim. For any positive integer $a$ and real number $x > 0$ the following inequality holds:\n$$\na(x+1)^{\\frac{1}{a}} \\le x+a,\n$$\nwhere the equality is satisfied if and only if $a=1$.\n\n*Proof.* If $a=1$, we easily see that the equality holds for all real numbers $x > 0$. Let us assume that $a > 1$, and let us apply the inequality between arithmetic and geometric means for $x+1$ and $a-1$ copies of $1$:\n$$\n\\frac{(x+1)+\\overbrace{1+1+\\cdots+1}^{a-1}}{a} \\ge \\sqrt[a]{(x+1) \\cdot 1^{a-1}},\n$$\nor equivalently,\n$$\na(x+1)^{\\frac{1}{a}} \\le x+a.\n$$\nNotice that the equality is attained only if the equation $x+1=1$ is satisfied, which is impossible due to the assumption $x > 0$. $\\square$\n\nLet us now prove the original problem, using the above claim. We have\n$$\n\\begin{aligned}\nM(x+1)^k &= a_1(x+1)^{\\frac{1}{a_1}} a_2(x+1)^{\\frac{1}{a_2}} \\cdots a_n(x+1)^{\\frac{1}{a_n}} \\\\\n&\\le (x+a_1)(x+a_2)\\cdots(x+a_n),\n\\end{aligned}\n$$\nwhere equality holds in case $a_1 = a_2 = \\cdots = a_n = 1$. However, $a_1 a_2 \\cdots a_n = M > 1$ so we conclude that this is not the case. Therefore, we conclude that the original equation has no solutions in positive real numbers.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 29224,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $P(x), Q(x) \\in \\mathbb{R}[x]$ be polynomials with real coefficients such that\n$$\nP(P(x)) = (Q(x))^2\n$$\nholds for all real numbers $x$.\nDoes there necessarily exist a polynomial $R(x) \\in \\mathbb{R}[x]$ with real coefficients such that $P(x) = (R(x))^2$ holds for all real numbers $x$?",
    "options": [],
    "answer": "Yes",
    "solution": "Yes, there necessarily exists such polynomial $R(x)$.\n\nFirstly, notice that we are allowed to assume, without loss of generality, that polynomials $P(x)$ and $Q(x)$ have leading coefficient equal to $1$. If the polynomial $Q(x)$ is constant, then the polynomial $P(x)$ is constant as well, and we have $P(x) = (Q(x))^2$.\n\nTherefore, let us assume that the polynomial $Q(x)$ is not constant, and neither the polynomial $P(x)$. The following equality holds for degrees of polynomials $P(x)$ and $Q(x)$:\n$$\n(\\deg P)^2 = 2 \\deg Q.\n$$\nThus, there exists a positive integer $n$ such that $\\deg P = 2n$ and $\\deg Q = 2n^2$. That means there exist polynomials $R(x)$ and $S(x)$ with real coefficients such that\n$$\nP(x) = (R(x))^2 + S(x), \\quad \\forall x \\in \\mathbb{R},\n$$\nand their coefficients satisfy $\\deg R = n$ and $\\deg S < n$. Let us define\n$$\nA(x) = R(P(x)) \\quad \\text{and} \\quad B(x) = S(P(x)), \\quad \\forall x \\in \\mathbb{R}.\n$$\nThen we have\n$$\nB(x) = (Q(x))^2 - (A(x))^2 = (Q(x) - A(x))(Q(x) + A(x)), \\quad \\forall x \\in \\mathbb{R}.\n$$\nWe also have $\\deg A = 2n^2 = \\deg Q$, so if the polynomial $B(x)$ is not the null polynomial, then its degree is greater than or equal to $2n^2$. However, it is impossible due to the fact that degree of the polynomial $B(x)$ is at most $(n-1) \\cdot 2n = 2n^2 - 2n < 2n^2$.\n\nThus, polynomial $B(x)$ is the null polynomial, and since the polynomial $P(x)$ is not equal to $0$, we conclude that the polynomial $S(x)$ is not null polynomial. In other words, we have $P(x) = (R(x))^2$ for all real numbers $x$.",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 29225,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $n$ be a positive integer. A good word is a sequence of $3n$ letters, in which each of the letters $A$, $B$ and $C$ appears exactly $n$ times. Prove that for every good word $X$ there exists a good word $Y$ such that $Y$ cannot be obtained from $X$ by swapping neighbouring letters fewer than $\\frac{3}{2}n^2$ times. (IMO Shortlist 2017)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let us define the distance of good words $X$ and $Y$, denoted by $d(X, Y)$, as the smallest number of swaps of neighbouring letters necessary to obtain $Y$ from $X$ (or vice versa). Note that $d(X, Y) = d(Y, X)$, and that for any three good words $X$, $Y$ and $Z$ we have\n$$\nd(X, Y) + d(Y, Z) \\ge d(X, Z).\n$$\nFurthermore, for a good word $X$ denote by $F(X)$ the number of pairs of positions where the letter in the left position is lexicographically smaller than the one in the right position, i.e. the number of pairs of the form $AB$, $AC$ or $BC$. By swapping two neighbouring letters in a good word $X$ we get a good word $X'$. If the letters were identical, those two words would be equal, so we can assume that all swaps involve pairs of different letters. If we swap different letters, we have $|F(X) - F(X')| = 1$. From this, we can conclude that\n$$\nd(X, Y) \\ge |F(X) - F(Y)|\n$$\nfor any two good words $X$ and $Y$. Observe the good words\n$$\nP = \\underbrace{AA\\dots A}_n\\underbrace{BB\\dots B}_n\\underbrace{CC\\dots C}_n \\quad \\text{and} \\quad Q = \\underbrace{CC\\dots C}_n\\underbrace{BB\\dots B}_n\\underbrace{AA\\dots A}_n\n$$\nNote that $F(P) = 3n^2$ and $F(Q) = 0$, therefore $d(P, Q) \\ge 3n^2$.\nFinally, for any good word $X$ we have\n$$\nd(P, X) + d(X, Q) \\ge d(P, Q) \\ge 3n^2.\n$$\nTherefore, one of the good words $P$ and $Q$ cannot be obtained from $X$ by swapping fewer than $\\frac{3}{2}n^2$ pairs of neighbouring letters.",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 29226,
    "subject": "Mathematics (Multi-modal)",
    "question": "One hundred people attended the party, some of whom were previously acquainted. All acquaintances were mutual and no new were made during the party.\nA gong rang $100$ times during the party. After the first sounding of the gong, all the people not acquainted with anyone left the party. After the second sounding of the gong, all the people with exactly one acquaintance (among the remaining people) left. It continued all the way, meaning that after the $k$th sounding of the gong, all the people acquainted with exactly $k-1$ remaining people left the party ($k = 1, \\dots, 100$).\nAt the end of the party, there were $n$ people still present. Find all possible values of $n$.",
    "options": [],
    "answer": "0, 1, 2, ..., 98",
    "solution": "We show that $n$ can be $0, 1, 2, 3, \\dots, 98$. The following contains the description of a situation in which exactly $n$ people are present after the last chime (for $n = 0, 1, 2, \\dots, 98$):\nFor $n > 0$, we can divide all the people at the party into two groups, A and B. Let A contain $n$ people and assume every one of them is acquainted with all the other people at the party. Let B contain the remaining $100 - n$ people, none of whom are acquainted amongst themselves, but all of whom are acquainted with all the people in group A (thus, every person in group B has $n$ acquaintances).\nAll the people from group B will obviously leave the party after the $(n+1)^{\\text{th}}$ sounding of the gong. After that, only people from group A will remain, and each of them will have exactly $n-1$ acquaintances. As the gong has already been sounded $n$ times, this means none of them will leave until the end of the party.\nThe value $n = 0$ is attained, for example, when all the party-goers know each other (so they all leave after the $100$th chime). A simple observation shows that at least one person must leave at some moment: this is the person with the fewest acquaintances. This implies that $n = 100$ cannot be attained.\nFinally, let us prove that $n = 99$ is not possible. Assume the contrary, i.e. that exactly one person will leave the party before it ends; call this person $X$. As the first one to leave, this is obviously the person with the fewest acquaintances. Since none of the remaining people leave after $X$, we conclude that all of them must be acquainted with $X$ - if another person $Y$ is not acquainted with $X$, then the number of people they know would not change with $X$ leaving, so they would have to leave at some point as well. This shows that $X$ has $99$ acquaintances, which contradicts the assumption that $X$ has the fewest acquaintances.",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 29227,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $n$ be a positive integer. Points $A_1, A_2, \\dots, A_n$ are located on the inside of a circle, and points $B_1, B_2, \\dots, B_n$ are on the circle, so that the lines $\\overline{A_1B_1}$, $\\overline{A_2B_2}$, \\dots, $\\overline{A_nB_n}$ are mutually disjoint. A grasshopper can jump from point $A_i$ to point $A_j$ (for $i, j \\in \\{1, \\dots, n\\}, i \\neq j$) if and only if the line $\\overline{A_iA_j}$ does not go through any of the inner points of the lines $\\overline{A_1B_1}, \\overline{A_2B_2}, \\dots, \\overline{A_nB_n}$.\n\nShow that the grasshopper can take a series of jumps to get from any point $A_i$ to any point $A_j$. (Russia 1994)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let a path be any line $A_iA_j$ and a wall be any line $A_kB_k$. We say that a path and a wall intersect if the path goes through an inner point of the wall. A path is good if there is no wall to intersect it, and a wall is irrelevant if it does not intersect any path.\n\n**Claim 1.** For any $i \\in \\{1,2,\\dots,n\\}$ there exists at least one good path from point $A_i$.\n\n**Proof.** We will prove the claim for point $A_1$, and it holds analogously for all points $A_i$. Define the distance between a point $A$ and a wall $w$ as\n$$\n\\min_{T \\in w} |AT|.\n$$\nSince all walls are mutually disjoint, we know that none of the walls $\\overline{A_2B_2}, \\dots, \\overline{A_nB_n}$ contain point $A_1$. Let $\\overline{A_kB_k}$ be the wall closest to $A_1$. We claim that the path $A_1A_k$ is good. If we assume the contrary, that means that there exists a wall which intersects $A_1A_k$, but then this wall is closer to $A_1$ than wall $\\overline{A_kB_k}$, which is a contradiction. $\\square$\n\n**Claim 2.** There exists at least one irrelevant wall.\n\n**Proof.** Without loss of generality, we can assume that points $A_1, A_2, \\dots, A_k$ are the vertices of the convex hull of $\\{A_1, A_2, \\dots, A_n\\}$, and that they are labelled clockwise in that order as the vertices of polygon $P = A_1A_2\\dots A_k$. If the wall $w_1 = \\overline{A_1B_1}$ is irrelevant, we are done. Therefore, assume that $w_1$ intersects some path. That means that this wall goes through the inner points of polygon $P$, and since the walls are mutually disjoint, we conclude that $w_1$ must also intersect a path which is the side of $P$, because it cannot pass through any of the vertices of $P$ except $A_1$. Let $C$ be the intersection of wall $w_1$ and a side of $P$. Consider the arc $\\overarc{A_1C}$ (clockwise from point $A_1$ to point $C$), and note that it contains points $A_2, \\dots, A_l$. We can repeat this inference for wall $A_2B_2$. Since all walls are mutually disjoint, its corresponding arc is strictly smaller than the arc of $w_1$. Therefore, if all of the walls $\\overline{A_2B_2}, \\dots, \\overline{A_{l-1}B_{l-1}}$ are relevant, then wall $\\overline{A_lB_l}$ must surely be irrelevant. $\\square$\n\nFinally, we prove the problem statement by mathematical induction on $n$. The claim obviously holds for $n=1$. Assume the claim holds for some positive integer $n$. For $n+1$ it follows from Claim 2 that there exists a point $A_i$ such that the wall $\\overline{A_iB_i}$ is irrelevant. By the inductive assumption, we can conclude that the claim holds for all points except possibly point $A_i$. However, by Claim 1, point $A_i$ is connected by a good path with at least one of the remaining points, which proves the claim for $n+1$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 29228,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $a \\ge 2018$ be a real number. There are 2018 bowls, each containing a finite number of balls. It is known that the weight of each ball is of the form $a^k$, where $k$ is an integer, and that the total weight of balls in any bowl is the same. Let $B$ denote the total number of occurrences of the most frequently used weight.\nDetermine the smallest possible value of $B$.",
    "options": [],
    "answer": "2018",
    "solution": "The smallest value $B$ can attain is 2018.\n\nWithout loss of generality we can assume that the weight of the lightest ball is equal to 1. If this is not the case, we can divide all the weights by the weight of the lightest ball.\n\nLet us assume that there are at most 2017 balls of each weight appearing in all the bowls. Let $n$ be a non-negative integer such that $a^n$ is the weight of the heaviest ball.\n\nSince there are at most 2017 balls of the weight $a^n$, there exists at least one bowl which does not contain any balls of weight $a^n$, and moreover, all of the balls in that bowl are lighter than $a^n$. Thus, the total weight of balls in that bowl is at most\n$$\n2017 \\cdot \\sum_{k=0}^{n-1} a^k = 2017 \\cdot \\frac{a^n - 1}{a - 1} \\le 2017 \\cdot \\frac{a^n - 1}{2017} = a^n - 1 < a^n,\n$$\nwhich leads to contradiction.\n\nSo, there exists at least one weight which appears at least 2018 times.\n\nOn the other hand, if we put a ball of weight 1 in each bowl, then all bowls contain the same weight, while the weight 1 appears exactly 2018 times.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 29229,
    "subject": "Mathematics (Multi-modal)",
    "question": "In the picture below a chain consisted of $54$ unit squares is shown. Each unit square, except for the two at the front and the end, is connected to its neighbouring ones in opposite vertices.\n\n![](attached_image_1.png)\n\nIt is allowed to put any unit square of the chain in any position in space as long as it is connected to its neighbouring unit squares in corresponding vertices. Is it possible for that chain to make the surface of a $3 \\times 3 \\times 3$ cube? (Russia 1999)",
    "options": [],
    "answer": "No",
    "solution": "No, it is not possible.\n\nLet us assume the opposite, i.e. that the chain fully covers the surface of a $3 \\times 3 \\times 3$ cube. Let us draw a diagonal joining two opposite vertices which are two meeting points of the chain in each unit square of the chain (including the first and the last unit square in the chain as well). After that, we colour vertices which are the meeting points of the chain in black, and other ones in white.\n\nIn this manner we got polyline (possibly intersecting itself) on the surface of the original cube. Notice that the degree of all vertices on that polyline is even, except for the first and the last one. Namely, each vertex is connected to exactly two other vertices, while the boundary ones are connected to one. More precisely, all black vertices, except for the two boundary vertices, have degree equal to $4$, while two boundary black vertices have degree equal to $3$ or $4$, depending on the case if we end the polyline in the same point from which we started it.\n\nNow, observe the cube and notice that four of its vertices are black. Since each of those vertices has degree equal to $3$, we have at least four vertices with odd degree, which leads to contradiction.",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 29230,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $k$ be a circle centred at $O$. Let $\\overline{AB}$ be a chord of that circle and $M$ its midpoint. Tangents on $k$ at points $A$ and $B$ intersect at $T$. The line $\\ell$ goes through $T$, intersects the shorter arc $\\overarc{AB}$ at the point $C$ and the longer arc $\\overarc{AB}$ at the point $D$, so that $|BC| = |BM|$.\nProve that the circumcentre of the triangle *ADM* is the reflection of *O* across the line *AD*.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Since $C$ and $D$ are on $k$, the power of the point $T$ with respect to $k$ equals $|TB|^2 = |TC| \\cdot |TD|$.\n\n![](attached_image_1.png)\n\nFurthermore, since the right-angled triangles $TBM$ and $TOB$ are similar, we have $|TB|^2 = |TM| \\cdot |TO|$. Therefore, $|TC| \\cdot |TD| = |TM| \\cdot |TO|$, i.e. the quadrilateral $CDOM$ is cyclic.\n\nLet point $C'$ be the intersection of $k$ and the line $DM$, while $\\alpha = \\angle C'MC$. Now we have\n$$\n\\angle C'C'M = \\angle C'C'D = \\frac{1}{2}\\angle COD = \\frac{1}{2}\\angle CMD = \\frac{1}{2}(180^\\circ - \\angle C'MC) = 90^\\circ - \\frac{1}{2}\\alpha,\n$$\ntherefore $\\angle MCC' = 180^\\circ - \\alpha - (90^\\circ - \\frac{1}{2}\\alpha) = 90^\\circ - \\frac{1}{2}\\alpha = \\angle CC'M$, i.e. $|C'M| = |CM|$, meaning that $C'$ is the reflection of $C$ across the line $OM$, and $|AC'| = |AM|$ holds. Let $M'$ be the point on $k$ different from $C'$ such that $|AM'| = |AC'|$. Then\n$$\n\\angle M'DA = \\angle C'DA = \\angle MDA.\n$$\nWe can conclude that triangles $MDA$ and $M'DA$ are congruent (two pairs of congruent sides, one pair of congruent angles, and both are obtuse), so $M'$ is the reflection of $M$ across the line $AD$. The fact that $O$ is the circumcentre of the triangle $ADM'$ completes the proof.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 29231,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABCD$ be a square, and let $k$ be the circle centred at $B$ passing through $A$, $C$ and the point $T$ inside the square. Tangent on $k$ at $T$ intersects the segments $\\overline{CD}$ and $\\overline{DA}$ at $E$ and $F$, respectively. Let $G$ and $H$ be the intersections of the lines $BE$ and $BF$ with the segment $\\overline{AC}$, respectively.\nProve that the lines $BT$, $EH$ and $FG$ are passing through the same point.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Note that the lines $FA$ and $FT$ are tangent to the circle $k$, hence $|FA| = |FT|$, and the triangles $ABF$ and $TBF$ are congruent. Analogously, $|EC| = |ET|$, and the triangles $CBE$ and $TBE$ are congruent.\n\n![](attached_image_1.png)\n\nLet us denote $\\alpha = \\angle FBA = \\angle FBT$ and $\\beta = \\angle EBC = \\angle EBT$. Since $\\angle ABC = 90^\\circ$, it follows that $\\alpha + \\beta = 45^\\circ$. Now we have $\\angle AFB = 90^\\circ - \\alpha = 45^\\circ + \\beta$ and $\\angle AGB = \\angle GBC + \\angle BCA = \\beta + 45^\\circ$, i.e. $\\angle AFB = \\angle AGB$.\n\nHence, the quadrilateral $ABGF$ is cyclic, and $\\angle BGF = 180^\\circ - \\angle AFB = 90^\\circ$, i.e. $FG \\perp BE$. Similarly, the quadrilateral $BCEH$ is cyclic, and $EH \\perp BF$.\n\nSince the segments $\\overline{FG}$ and $\\overline{EH}$ are the altitudes of the triangle $BEF$, so as the segment $\\overline{BT}$, we finally conclude that the lines $BT$, $EH$ and $FG$ are passing through the same point - the orthocentre of the triangle $BEF$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 29232,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $\\triangle ABC$ be an acute-angled triangle such that $|AB| < |AC|$. Point $D$ is the midpoint of the shorter arc $\\widehat{BC}$ of the circumcircle of the triangle $ABC$. Point $I$ is the incentre of the triangle $ABC$, and point $J$ is the reflection of $I$ across the line $BC$. Line $DJ$ intersects the circumcircle of the triangle $ABC$ at the point $E$ which lies on the shorter arc $\\widehat{AB}$.\nProve that $|AI| = |IE|$ holds. (Romania 2017)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $O$ be the circumcentre of the triangle $ABC$.\n\n![](attached_image_1.png)\n\nPoint $D$ lies on the angle bisector of $\\angle BAC$. Therefore $A$, $I$ and $D$ are collinear. Lines $IJ$ and $OD$ are parallel, since they are both perpendicular to $BC$.\n\nWe will prove the claim by showing that the triangles $AOI$ and $EOI$ are congruent.\n\nFirst we show that the triangles $AIO$ and $IJD$ are similar. Note that $\\angle IAO = \\angle DAO = \\angle ODA = \\angle ODI = \\angle JID$. To show similarity, we need to prove that $|AI| : |AO| = |IJ| : |ID|$, i.e.\n$$\n|AI| \\cdot |ID| = |AO| \\cdot |IJ|. \\qquad (1)\n$$\nWe have $|IJ| = 2r$ and $|AO| = R$, where $r$ and $R$ denote the incircle and the circumcircle radii, respectively. Therefore, the right-hand side of (1) equals $2Rr$. The left-hand side product $|AI| \\cdot |ID|$, is in fact the power of point $I$ with respect to the circumcircle of the triangle $ABC$ and equals $R^2 - |OI|^2$. Now it suffices to verify that $R^2 - |OI|^2 = 2Rr$, which holds because of Euler's theorem which expresses the distance between the incentre and the circumcentre using this exact formula.\n\nTherefore, we can conclude that the triangles $AIO$ and $IJD$ are similar. This means that $\\angle OIA = \\angle DJI$, and $\\angle DIO = \\angle IJE$ follows. Since lines $IJ$ and $OD$ are parallel, we have $\\angle IJE = \\angle ODE$, and since $|OE| = |OD|$ we have $\\angle ODE = \\angle DEO$. Now having $\\angle DIO = \\angle DEO$ we conclude that points $D$, $O$, $I$ and $E$ are concyclic.\n\nFrom this we have $\\angle IOE = \\angle IDE$, and from the proven similarity we have $\\angle IDE = \\angle IDJ = \\angle AOI$. It follows that $\\angle IOE = \\angle AOI$. Since $|AO| = |EO|$ and the side $\\overline{OI}$ is mutual, we conclude that the triangles $EOI$ and $AOI$ are congruent.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 29233,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABCD$ be an isosceles trapezium with bases $\\overline{AB}$ and $\\overline{CD}$. The diagonals of the trapezium meet at the point $S$, while $M$ denotes the midpoint of the leg $\\overline{AD}$. The circle circumscribed to the triangle $BCM$ intersects $\\overline{AD}$ again at the point $K$. Prove that the lines $SK$ and $AB$ are parallel.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Instead of the original formulation, we will solve the following equivalent problem: Let $ABCD$ be an isosceles trapezium with bases $\\overline{AB}$ and $\\overline{CD}$, and let its diagonals meet at the point $S$, as in the original formulation.\nAdditionally, let $K$ be a point on the line $AD$ such that $SK \\parallel AB$, and let $M$ be the other intersection of $AD$ with the circle circumscribed to the triangle $BCK$. We need to prove that $M$ is the midpoint of the leg $\\overline{AD}$.\n\n![](attached_image_1.png)\n\nLet $M'$ be a point on the line $BC$ such that $MM' \\parallel AB$, and let $T$ be the intersection of the lines $CK$ and $AB$. Since the trapezium $ABCD$ is isosceles and the quadrilateral $BCKM$ is cyclic, we have $\\angle DAM' = \\angle CBM = \\angle MKT$, and therefore $AM' \\parallel TC$. Using similarity of triangles, we obtain $|AT| : |CD| = |KA| : |KD| = d(S, AB) : d(S, CD) = |AB| : |CD|$. Hence $|AT| = |AB|$ and $|M'B| = |M'C|$. Thus the point $M'$ is the midpoint of the leg $\\overline{BC}$ and, of course, $M$ is the midpoint of the leg $\\overline{AD}$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 29234,
    "subject": "Mathematics (Multi-modal)",
    "question": "The incircle of the triangle $ABC$ is centred at $I$, and touches the sides $\\overline{BC}$, $\\overline{CA}$ and $\\overline{AB}$ in points $D, E$ and $F$, respectively. Let $k$ be a circle centred at $A$ passing through the point $E$. Let $K$ be the second intersection of the line $DE$ with $k$. The line parallel to $DF$ and passing through $I$ intersects the side $\\overline{AB}$ at $P$. The point $L$ is the intersection of the line $CP$ with the circle $k$ such that the point $P$ is located between points $C$ and $L$. The point $O$ is the circumcentre of the triangle $DKL$.\nProve that the lines $AI$ and $OD$ are parallel. (Stipe Vidak)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "We easily see that $\\angle AKE = \\angle KEA = \\angle DEC = \\angle CDE$. Thus $|AK| = |AE|$ and $AK \\parallel CD$.\n\n![](attached_image_1.png)\n\nLet $L'$ be the intersection of the lines $CP$ and $AK$. Since the triangles $PBC$ and $PAL'$ are similar, by trigonometric calculus we get $|BP| : |AP| = |BC| : |AL'|$, from which it follows that $|AL'| = |AK| = |AE|$, i.e. $L'$ lies on the circle $k$ - so it must be $L' \\equiv L$. Similarly, we can show that $D, F$ and $L$ are collinear.\n\nNow we know that $\\overline{KL}$ is the diameter of the circle $k$. Observing the triangles $AKO$ and $EAI$, we notice they are both right-angled, and $\\angle AKO = 90^\\circ - \\angle KOA = 90^\\circ - \\frac{1}{2}\\angle KOL = 90^\\circ - \\angle KDL = \\angle IAE$ holds. Hence they are congruent, which means that segments $\\overline{AO}$ and $\\overline{ID}$ are parallel (since they are perpendicular to $KL$ and $BC$, respectively) and have equal length ($|ID| = |IE| = |OA|$). Therefore, the quadrilateral $AODI$ is a parallelogram, from which it follows that $AI \\parallel OD$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 29235,
    "subject": "Mathematics (Multi-modal)",
    "question": "Determine all pairs $(m, n)$ of positive integers such that\n$2^m = 7n^2 + 1.$",
    "options": [],
    "answer": "(3, 1) and (6, 3)",
    "solution": "Note that $2^m \\equiv 1 \\pmod{7}$, which means that $m = 3k$ for some positive integer $k$. Now we have\n$$\n2^{3k} - 1 = (2^k - 1)(2^{2k} + 2^k + 1) = 7n^2.\n$$\nDenote $A = 2^k - 1$ and $B = 2^{2k} + 2^k + 1$. Let $d$ be the greatest common divisor of $A$ and $B$, and note that $d$ is odd. Furthermore, $d$ divides $B - A^2 = 2^k + 2^{k+1} = 3 \\cdot 2^k$, and since $d$ is odd, it follows that $d = 1$ or $d = 3$.\nTherefore, we have exactly four possibilities for factorising $A$ and $B$. In all cases we assume that $a$ and $b$ are relatively prime positive integers.\n\nCase 1. $A = 7a^2$ and $B = b^2$. This is not possible because $(2^k)^2 < B < (2^k + 1)^2$, i.e. $B$ cannot be a perfect square.\n\nCase 2. $A = a^2$ and $B = 7b^2$. If $k \\ge 2$, note that $B \\equiv 1 \\pmod{4}$, i.e. $7b^2 \\equiv 1 \\pmod{4}$, which is not possible. Therefore, $k = 1$ and we get $(m, n) = (3, 1)$.\n\nCase 3. $A = 21a^2$ and $B = 3b^2$. Analogously to the previous case we can conclude that $k=1$ necessarily, but that is not possible in this case, since $21a^2 = A = 2^1 - 1 = 1$.\n\nCase 4. $A = 3a^2$ and $B = 21b^2$. Therefore, $A$ is divisible by $3$, i.e. $2^k \\equiv 1 \\pmod{3}$, from which it follows that $k=2l$ for some positive integer $l$. Now we have\n$$\n(2^l - 1)(2^l + 1) = 3a^2,\n$$\nand since $2^l - 1$ and $2^l + 1$ are relatively prime, we have only two possibilities:\n$$\n\\begin{aligned}\n2^l - 1 &= c^2 \\\\\n2^l + 1 &= 3d^2\n\\end{aligned} \\quad \\text{or} \\quad \\begin{aligned}\n2^l - 1 &= 3c^2 \\\\\n2^l + 1 &= d^2,\n\\end{aligned}\n$$\nwhere $c$ and $d$ are relatively prime positive integers.\n\nIn the first option, if $l \\ge 2$ we have $3d^2 \\equiv 1 \\pmod{4}$, which is not possible. Therefore, $l=1$ and we get another solution $(m,n) = (6,3)$.\n\nIn the second option, we have $2^l = (d-1)(d+1)$. Since the greatest common divisor of $d-1$ and $d+1$ is at most $2$ and their product is a power of $2$, we can conclude that $d-1=2$, i.e. that $l=3$, but then $3c^2 = 2^3 - 1 = 7$, which is not possible.\n\nFinally, all solutions of the given equation are $(m,n) = (3,1)$ and $(m,n) = (6,3)$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 29236,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $n$ be a positive integer. Prove that there exists a positive integer $k$ such that\n$$\n51^k - 17\n$$\nis divisible by $2^n$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "### Claim.\nFor every positive integer $n \\ge 3$ there exists a positive integer $k$ such that $51^{2^{n-2}} - 1 = 2^n \\cdot (2k + 1)$.\n\n*Proof.* We prove the claim by mathematical induction on $n$.\n\nFor $n=3$ we have $51^2 - 1 = 3^2 - 1 = 8$ (mod 16), so $51^2 - 1$ is divisible by $2^3$ but not by $2^4$, therefore the claim holds for $n=3$. Assume the claim holds for some $n$, i.e. $51^{2^{n-2}} - 1 = 2^n \\cdot (2k + 1)$ for some positive integer $k$. Then\n$$\n\\begin{aligned}\n51^{2^{n-1}} - 1 &= (51^{2^{n-2}} - 1)(51^{2^{n-2}} + 1) \\\\\n&= 2^n \\cdot (2k+1)(2^n \\cdot (2k+1) + 2) \\\\\n&= 2^{n+1} \\cdot (2k+1)(2^{n-1}(2k+1) + 1),\n\\end{aligned}\n$$\nand since $(2k+1)(2^{n-1}(2k+1)+1)$ is odd, the claim holds for $n+1$. This completes the proof. $\\square$\n\nRegarding the initial statement of the problem, we will inductively define an appropriate $k_n$ for every $n$. Note that for $n=1, n=2$ and $n=3$ we can select $k=2$, i.e. $k_1 = k_2 = k_3 = 2$. Assume the claim holds for some $n$, i.e. there exists some positive integer $a_n$ such that $51^{k_n} - 17 = 2^n \\cdot a_n$. If $a_n$ is even, we can select $k_{n+1} = k_n$. If $a_n$ is odd, we can define $k_{n+1} = k_n + 2^{n-2}$. Then we have\n$$\n51^{k_{n+1}} - 17 = 51^{2^{n-2}} \\cdot (51^{k_n} - 17) + 17 \\cdot (51^{2^{n-2}} - 1).\n$$\nBy the inductive assumption and the previously proven claim, we can conclude that there exists a positive integer $k$ such that\n$$\n\\begin{aligned}\n51^{k+n+1} - 17 &= 51^{2^{n-2}} \\cdot 2^n \\cdot a_n + 17 \\cdot 2^n \\cdot (2k+1) \\\\\n&= 2^n \\cdot (51^{2^{n-2}} a_n + 17(2k+1)).\n\\end{aligned}\n$$\nSince $a_n$ is odd, it follows that $51^{2^{n-2}} a_n + 17(2k+1)$ is even, therefore $51^{k+n+1} - 17$ is indeed divisible by $2^{n+1}$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 29237,
    "subject": "Mathematics (Multi-modal)",
    "question": "Prove that, for every positive integer $n \\ge 2$, there exist positive integers $a_1, a_2, \\dots, a_n$ such that for all $1 \\le i < j \\le n$ the expression\n$$\n\\frac{a_j + a_i}{a_j - a_i}\n$$\nis a positive integer.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "We will prove the claim using mathematical induction on $n$. We will call the set $\\{a_1, a_2, \\dots, a_n\\}$ good if it satisfies the given condition.\n\nFor $n=2$ the set $\\{1,2\\}$ is clearly good. Let us assume that for some positive integer $n \\ge 2$ there exists a good set $\\{a_1, a_2, \\dots, a_n\\}$. Let $a$ be the least common multiple of the set\n$$\n\\{a_1, a_2, \\dots, a_n\\} \\cup \\{a_j - a_i \\mid 1 \\le i < j \\le n\\}.\n$$\nWe claim that the set $\\{a, a+a_1, a+a_2, ..., a+a_n\\}$, containing $n+1$ elements, is also good.\n\nNotice that for all $i \\in \\{1, 2, ..., n\\}$ we have\n$$\n\\frac{(a+a_i)+a}{(a+a_i)-a} = \\frac{2a+a_i}{a_i} = 2\\frac{a}{a_i} + 1,\n$$\nwhich is a positive integer since $a$ is divisible by $a_i$ (by definition).\n\nMoreover, for all $1 \\le i < j \\le n$ we have\n$$\n\\frac{(a+a_j)+(a+a_i)}{(a+a_j)-(a+a_i)} = \\frac{2a+(a_j+a_i)}{a_j-a_i} = 2\\frac{a}{a_j-a_i} + \\frac{a_j+a_i}{a_j-a_i},\n$$\nwhich is a positive integer as well.\n\nThis completes the inductive step, thereby proving the claim.",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 29238,
    "subject": "Mathematics (Multi-modal)",
    "question": "Determine all positive integers $n \\ge 2$ which satisfy the following condition: For all integers $a_1, a_2, \\dots, a_n$ such that their sum is not divisible by $n$, there exists an integer $i \\in \\{1, 2, \\dots, n\\}$ such that none of the $n$ numbers\n\n$$\na_i,\\ a_i + a_{i+1},\\ \\dots,\\ a_i + a_{i+1} + \\dots + a_{i+n-1}\n$$\nis divisible by $n$, where we define $a_i = a_{i-n}$ for the case $i > n$.",
    "options": [],
    "answer": "All prime numbers",
    "solution": "All such numbers are primes.\n\nLet $n$ be a composite number, i.e. $n = ab$, where $a \\ge 2$ and $b \\ge 2$ are positive integers.\nObserve the sequence\n$0, b, b, \\dots, b,$\nin which the number $b$ is appearing $ab-1$ times. The sum of all numbers in that sequence is $ab^2-b$, which is not divisible by $n=ab$. However, notice that, no matter on the initial index of the sequence, the sum of $a$ (or $a+1$, if we pass through element 0 of the sequence) consecutive numbers in the sequence will be equal to $ab$, and therefore divisible by $n=ab$.\n\nLet $n = p$ be a prime number. Let us assume the opposite: there exist positive integers $a_1, a_2, \\dots, a_p$ which all add up to a number not divisible by $p$, but for all $i \\in \\{1, 2, \\dots, p\\}$ there exists $j \\in \\{i+1, i+2, \\dots, i+p-1\\}$ such that the number $a_i + a_{i+1} + \\dots + a_{j-1}$ is divisible by $p$. Reminder: all operations on indices are made using convention $a_i = a_{i-p}$ for $i > p$.\n\nLet us now make a graph consisted of $p$ vertices (indicating numbers $a_1, a_2, \\dots, a_p$). The two vertices $a_i$ and $a_j$ are joined with directed edge if the sum $a_i+a_{i+1}+\\dots+a_{j-1}$ is divisible by $p$. Due to the assumption, from any vertex there is at least one edge incident to that vertex. Since there are only finitely many vertices in the graph, we conclude that the graph has a cycle.\n\nConsider that cycle and let us sum up all sums which correspond to edges of that cycle. Notice that all numbers $a_1, a_2, \\dots, a_p$ appear equally many times in those sums, and let us denote that number of occurrence by $k$. Moreover, in that cycle we have at most $p$ edges, and for each edge the corresponding sum consists of at most $p-1$ addends, since the sum of all numbers in the original sequence is not divisible by $p$. Hence, $k \\le p-1$. However, each of those sums along the edges of the cycle is divisible by $p$, so the sum of all those sums in the cycle is $p$ as well, and we have\n$$\np \\mid k(a_1 + a_2 + \\dots + a_p).\n$$\nThis gives contradiction, since $k < p$, i.e. $p \\nmid k$ and $p \\nmid a_1 + a_2 + \\dots + a_p$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 29239,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all three-digit numbers such that the sum of their digits is $11$, and the number obtained by swapping the hundreds digit and the ones digit is by $594$ greater than the original number.",
    "options": [],
    "answer": "137, 218",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 29240,
    "subject": "Mathematics (Multi-modal)",
    "question": "Ana and Vanja are standing together next to a railway, waiting for the train to pass. The train drives at constant speed. At the moment the front end of the train passes them, Ana starts walking at constant speed in the same direction as the train is going, and Vanja starts walking at the same speed in the opposite direction. Each of them stops at the moment the rear end of the train passes her. In total, Ana walked 45 metres and Vanja walked 30 metres. How long is the train? (Portugal 2006)",
    "options": [],
    "answer": "180 metres",
    "solution": "Note that Ana walked $45 - 30 = 15$ metres more than Vanja, and while Ana was walking these $15$ metres, the train travelled $45 + 30 = 75$ metres.\nTherefore, the speed of the train is $\\frac{75}{15} = 5$ times greater than the walking speed.\nAlso note that while Vanja walked $30$ metres, the train travelled $30 \\cdot 5 = 150$ metres.\nSince Vanja started walking at the moment the front end of the train passed her, and stopped when the rear end passed her, and since she was walking in the opposite direction, the total length of the train is $150 + 30 = 180$ metres.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 29241,
    "subject": "Mathematics (Multi-modal)",
    "question": "Points $P_1$, $P_2$ and $P_3$ are located on the side $AB$ of the triangle $ABC$ so that\n$$\n|AP_1| = |P_1P_2| = |P_2P_3| = |P_3B| = \\frac{1}{4}|AB|.\n$$\nParallels with the side $BC$ are drawn through those points and divide the triangle $ABC$ into four parts. The area of the part between the parallels through $P_2$ and $P_3$ equals $5$.\nWhat is the area of the triangle $ABC$? (Purple Comet! Math Meet)",
    "options": [],
    "answer": "16",
    "solution": "",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 29242,
    "subject": "Mathematics (Multi-modal)",
    "question": "The lengths of all sides of a right-angled triangle are positive integers. If the radius of its incircle is of length $4$, find all possible lengths of its legs.",
    "options": [],
    "answer": "The leg pairs are (9, 40), (10, 24), (12, 16), (16, 12), (24, 10), (40, 9).",
    "solution": "Let $a$ and $b$ be the lengths of legs, $c$ be the length of hypotenuse, and $r = 4$ be the radius of the incircle in the given triangle.\nExpressing its area in two different ways, we get\n$$\n r \\cdot \\frac{a + b + c}{2} = \\frac{ab}{2},\n$$\n$$\n ab - 4a - 4b = 4c.\n$$\nSince $c^2 = a^2 + b^2$, it follows that $ab(ab - 8a - 8b + 32) = 0$. Considering that $a$ and $b$ are positive integers, we have $ab - 8a - 8b + 32 = 0$, i.e. $(a - 8)(b - 8) = 32$, hence both $a - 8$ and $b - 8$ must be divisors of $32$.\nWithout loss of generality, we can assume $a \\leq b$, i.e. $a - 8 \\leq b - 8$. Therefore, among\n$$\n\\begin{align*}\na - 8 &= -32, & a - 8 &= 1, \\\\\na - 8 &= -16, & a - 8 &= 2, \\\\\na - 8 &= -8, & a - 8 &= 4,\n\\end{align*}\n$$\nwe find that $(a, b) \\in \\{(9, 40), (10, 24), (12, 16)\\}$ satisfy the conditions of the problem.\nFinally, these are all solutions:\n$$\n(a, b) \\in \\{(9, 40), (10, 24), (12, 16), (16, 12), (24, 10), (40, 9)\\}.\n$$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 29243,
    "subject": "Mathematics (Multi-modal)",
    "question": "Prove that positive real numbers $x$ and $y$ such that\n$$(x^3 + y^3)(x^2 + y^2) = 2(x + y) = 2$$\ndo not exist. (Tamara Srnec)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 29244,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $a$ and $b$ be positive real numbers such that\n$$\n\\frac{a}{b} + \\frac{b}{a} = 3 \\quad \\text{and} \\quad \\frac{a^2}{b} + \\frac{b^2}{a} = 10.\n$$\nDetermine $\\frac{1}{a} + \\frac{1}{b}$.\n(Kristina Ana Škreb)",
    "options": [],
    "answer": "1",
    "solution": "Let $x = a/b$ and $y = b/a$. Since $a, b > 0$, $x, y > 0$ and $xy = 1$.\n\nWe are given:\n$$\n\\frac{a}{b} + \\frac{b}{a} = x + y = 3.\n$$\nSince $xy = 1$, $x$ and $y$ are the roots of $t^2 - 3t + 1 = 0$.\n\nNow,\n$$\n\\frac{a^2}{b} + \\frac{b^2}{a} = \\frac{a^2}{b} + \\frac{b^2}{a} = a \\cdot \\frac{a}{b} + b \\cdot \\frac{b}{a} = a x + b y.\n$$\nBut this is not immediately helpful. Instead, let us try expressing everything in terms of $a$ and $b$.\n\nLet $S = a + b$, $P = ab$.\n\nWe have:\n$$\n\\frac{a}{b} + \\frac{b}{a} = \\frac{a^2 + b^2}{ab} = 3 \\implies a^2 + b^2 = 3ab.\n$$\nAlso,\n$$\n\\frac{a^2}{b} + \\frac{b^2}{a} = \\frac{a^3 + b^3}{ab} = 10.\n$$\nSo $a^3 + b^3 = 10ab$.\n\nRecall that $a^3 + b^3 = (a + b)^3 - 3ab(a + b) = S^3 - 3P S$.\n\nSo:\n$$\nS^3 - 3P S = 10P\n$$\nBut from above, $a^2 + b^2 = 3ab \\implies (a + b)^2 - 2ab = 3ab \\implies S^2 - 2P = 3P \\implies S^2 = 5P$.\n\nNow, substitute $P = S^2/5$ into the cubic equation:\n$$\nS^3 - 3P S = 10P\n$$\n$$\nS^3 - 3P S - 10P = 0\n$$\n$$\nS^3 - 3S P - 10P = 0\n$$\n$$\nS^3 - (3S + 10)P = 0\n$$\n$$\nS^3 = (3S + 10)P\n$$\nBut $P = S^2/5$, so:\n$$\nS^3 = (3S + 10) \\cdot \\frac{S^2}{5}\n$$\n$$\nS^3 = \\frac{(3S + 10) S^2}{5}\n$$\n$$\n5 S^3 = (3S + 10) S^2\n$$\n$$\n5 S^3 - 3 S^3 - 10 S^2 = 0\n$$\n$$\n2 S^3 - 10 S^2 = 0\n$$\n$$\n2 S^3 = 10 S^2\n$$\n$$\nS^3 = 5 S^2\n$$\nSince $S > 0$, $S = 5$.\n\nNow, $P = S^2/5 = 25/5 = 5$.\n\nTherefore, $a$ and $b$ are the roots of $t^2 - S t + P = 0$, i.e.\n$$\nt^2 - 5 t + 5 = 0\n$$\nSo\n$$\na, b = \\frac{5 \\pm \\sqrt{25 - 20}}{2} = \\frac{5 \\pm \\sqrt{5}}{2}\n$$\n\nNow,\n$$\n\\frac{1}{a} + \\frac{1}{b} = \\frac{a + b}{ab} = \\frac{S}{P} = \\frac{5}{5} = 1.\n$$\n\n**Answer:**\n$$\n\\boxed{1}\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 29245,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $\\triangle ABC$ be a triangle such that $\\angle CAB = 20^\\circ$, and let $D$ be the midpoint of the side $\\overline{AB}$. If $\\angle CDB = 40^\\circ$, find $\\angle ABC$. (Tamara Srnec)",
    "options": [],
    "answer": "70°",
    "solution": "",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 29246,
    "subject": "Mathematics (Multi-modal)",
    "question": "Base $\\overline{BC}$ is the longest side of the isosceles triangle $ABC$. Let $M$ be a point on the side $\\overline{BC}$ such that $|BM| = |AB|$. Point $N$ is the foot of the altitude from $M$ to $\\overline{AB}$.\nProve that the triangle $BMN$ and the quadrilateral $ACMN$ have equal areas and equal perimeters. (United Kingdom 2018)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 29247,
    "subject": "Mathematics (Multi-modal)",
    "question": "A square board whose squares are coloured in either black or white is called *beautiful* if a rotation by $90^\\circ$ does not change its appearance.\nHow many different $5 \\times 5$ beautiful boards are there? (Azra Tafro)",
    "options": [],
    "answer": "128",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 29248,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $k > 1$ be a positive integer. $k+2$ distinct positive integers are given, all less than $3k+1$. Prove that we can find two numbers among them whose difference is greater than $k$ and less than $2k$. (Mathematical Excalibur 2015)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Denote by $S$ the set of $k + 2$ given numbers. Without loss of generality, we can assume that $S$ contains $1$. Indeed, if $1$ is not in $S$, we can subtract the smallest element of $S$ from all elements of $S$ and add $1$ to all of them, which preserves the differences between all elements of $S$.\n\nIf at least one number from $\\{k+2, k+3, \\dots, 2k\\}$ is contained in $S$, then $1$ and that number satisfy the claim.\n\nNow assume that none of the numbers $k+2, k+3, \\dots, 2k$ are in $S$. All remaining numbers from $2$ to $3k+1$ can be divided into $k$ pairs $(2, 2k+1), \\dots, (k+1, 3k)$. Other than $1$, the set $S$ contains $k+1$ other numbers, so by the Dirichlet principle at least one pair consists of numbers that are both in $S$. Those two numbers satisfy the claim.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 29249,
    "subject": "Mathematics (Multi-modal)",
    "question": "42 pebbles are placed on a table. Two players take turns making moves. When making a move, a player has to take at least one pebble, but no more than half of the remaining pebbles. The player after whose move there is exactly one pebble left on the table, wins.\nWhich player can win regardless of how his opponent plays?\n(United Kingdom 2014)",
    "options": [],
    "answer": "First player",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 29250,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all pairs $(m, n)$ of integers such that $mn + 5m + 2n = 121$. (Nikola Adžaga)",
    "options": [],
    "answer": "(-1, 126), (129, -4), (-3, -136), (-133, -6)",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 29251,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABC$ be an isosceles triangle such that $|AB| = |AC|$ and $\\angle BAC < 60^\\circ$. Let $D$ be the point on the segment $\\overline{AC}$ such that $\\angle DBC = \\angle BAC$, $E$ be the intersection of the perpendicular bisector of the segment $\\overline{BD}$ and the line passing through $A$ parallel to $BC$, and $F$ be the point on the line $AC$ such that $A$ lies between $C$ and $F$, and $|AF| = 2|AC|$ holds.\n\na) Prove that the lines $BE$ and $AC$ are parallel.\n\nb) Prove that the line passing through $F$ perpendicular to $AB$ and the line passing through $E$ perpendicular to $AC$ intersect on the line $BD$.\n\nIt is allowed to use the claim from a) in b) even if it is not proven. (Italy 2013)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "a) Since $\\triangle DBC = \\triangle BAC$ and $\\triangle BCD = \\triangle BCA$, the triangles $BCD$ and $ABC$ are similar. Hence, $|BD| = |BC|$.\n\nLet $E'$ be the point such that the quadrilateral $AE'BC$ is a parallelogram. Since $\\triangle ABE' = \\triangle BAC = \\triangle CBD$, we get $\\triangle ABC = \\triangle CBD + \\triangle ABD = \\triangle ABE' + \\triangle ABD = \\triangle E'BD$. We also have $|BE'| = |AC|$, so the triangles $E'BD$ and $ABC$ are congruent. Therefore, the triangle $E'BD$ is isosceles, and $E'$ lies on the bisector of the segment $\\overline{BD}$.\n\nSince both $E$ and $E'$ lie on the line passing through $A$ parallel to $BC$, we conclude that $E = E'$. Hence, $AEBC$ is a parallelogram, i.e. the lines $AC$ and $BE$ are parallel.\n\nb) Let $P$ be the midpoint of the segment $\\overline{BC}$. Since (according to a)) the quadrilateral $AEBC$ is a parallelogram, we have $|BC| = |AE|$ and $|AE| : |CP| = 2 : 1 = |AF| : |CA|$.\n\nThe triangles $CAP$ and $AFE$ are similar, hence $\\triangle DFE = \\frac{1}{2}\\triangle BAC$. We also have $\\triangle APC = 90^\\circ$ and $\\triangle AEF = 90^\\circ$.\n\nLet $G$ and $H$ be the orthogonal projections of $F$ and $E$ to the lines $AB$ and $AC$, respectively. Let $T$ be the intersection of the lines $EH$ and $FG$. We must show that $T$ lies on the line $BD$.\n\n---\n\nAccording to a), the triangles $EBD$ and $ABC$ are congruent, hence\n$$\n\\angle ADE = 180^\\circ - \\angle BDE - \\angle BDC = 180^\\circ - \\angle ACB - \\angle ABC = \\angle BAC.\n$$\nSince\n$$\n\\angle DFT = 90^\\circ - \\angle FAG = 90^\\circ - \\angle BAC = 90^\\circ - \\angle ADE = \\angle DEH = \\angle DET,\n$$\nwe conclude that the quadrilateral $DEFT$ is cyclic, and $\\angle DTE = \\angle DFE = \\frac{1}{2}\\angle BAC$ holds.\nFinally, we have $\\angle TDH = 90^\\circ - \\frac{1}{2}\\angle BAC$ and $\\angle BDH = 90^\\circ + \\frac{1}{2}\\angle BAC$, hence $B$, $D$ and $T$ are collinear. This completes the proof.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 29252,
    "subject": "Mathematics (Multi-modal)",
    "question": "Borna wants to colour each of the numbers $2$, $3$, $\\ldots$, $32$ in one of $k$ colours, in such a way that none of the numbers is of the same colour as one of its multiples. Find the smallest positive integer $k$ for which that is possible.",
    "options": [],
    "answer": "5",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 29253,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all complex numbers $z$ such that $z^2 = \\frac{1}{z} + \\frac{1}{\\bar{z}}$.",
    "options": [],
    "answer": "2^{1/3}",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 29254,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all complex numbers $a$ such that all coefficients of\n$$\nP(x) = (x - a)(x - a^2)(x - a^3)\n$$\nare real numbers.\n(Matko Ljulj)",
    "options": [],
    "answer": "All real numbers together with i, -i, (-1 + i√3)/2, (-1 - i√3)/2",
    "solution": "Obviously, all real numbers satisfy the given condition.\nLet us, from now on, assume that $a$ is not a real number.\nSince $P$ is a polynomial of degree 3, it must have at least one real root. We also know that if $z$ is a root of $P$, then $\\bar{z}$ is a root of $P$ as well. Hence $\\bar{a} = a^2$ or $\\bar{a} = a^3$.\nIn the first case, we get $|a| = |a|^2$, and then $|a| = 1$ (since $a \\neq 0$). Hence $a^3 = a^2 \\cdot a = \\bar{a} \\cdot a = |a|^2 = 1$, so both factors $(x - a)(x - a^2)$ and $x - a^3$ of $P$ have real coefficients. Since $a \\neq 1$, from $a^3 = 1$ we get two solutions:\n$$\na = \\frac{-1 + i\\sqrt{3}}{2} \\quad \\text{and} \\quad a = \\frac{-1 - i\\sqrt{3}}{2}.\n$$\nIn the second case, we similarly get $a^4 = 1$, hence $a = \\pm i$, so both factors $(x-a)(x-a^3)$ and $x-a^2$ of $P$ have real coefficients.\n\n$$\na \\in \\left\\{ i, -i, \\frac{-1 + i\\sqrt{3}}{2}, \\frac{-1 - i\\sqrt{3}}{2} \\right\\},\n$$\nalong with all real numbers.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 29255,
    "subject": "Mathematics (Multi-modal)",
    "question": "Determine the value of the real parameter $p$ so that the solutions of\n$$\n(p - 3)x^2 + (p^2 + 1)x - 11p + 18 = 0\n$$\nare the lengths of legs of a right-angled triangle with hypotenuse of length $\\sqrt{17}$.",
    "options": [],
    "answer": "2",
    "solution": "",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 29256,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all real numbers $x$ such that\n$$\n\\left\\lfloor \\frac{x^2 + 1}{x + 2} \\right\\rfloor + \\left\\lfloor \\frac{x - 1}{2} \\right\\rfloor = \\frac{x(3x + 1)}{2(x + 2)}\n$$\nFor a real number $t$, $\\lfloor t \\rfloor$ denotes the largest integer not greater than $t$. For example, if $t = 3.14$, then $\\lfloor t \\rfloor = 3$.",
    "options": [],
    "answer": "x ∈ { -7, -3, -1, 3 }",
    "solution": "Note that\n$$\n\\frac{x^2 + 1}{x + 2} + \\frac{x - 1}{2} = \\frac{x(3x + 1)}{2(x + 2)}.\n$$\nSince $\\lfloor t \\rfloor \\le t$ holds for all real numbers $t$, and the equality is attained if and only if $t$ is an integer, it follows that both\n$$\n\\frac{x^2 + 1}{x + 2} \\quad \\text{and} \\quad \\frac{x - 1}{2}\n$$\nmust be integers.\nThe latter is an integer if and only if $x$ is an odd integer, hence considering\n$$\n\\frac{x^2 + 1}{x + 2} = x - 2 + \\frac{5}{x + 2}\n$$\nwe get $x + 2 \\in \\{1, -1, 5, -5\\}$, i.e. $x \\in \\{-1, -3, 3, -7\\}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 29257,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all pairs $(a, n)$ of positive integers such that\n$$\n3a^2 + 2^n = a^4.\n$$",
    "options": [],
    "answer": "(a, n) = (2, 2)",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 29258,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all pairs $(p, q)$ of prime integers such that the solutions of the quadratic equation $x^2 + px + q = 0$ are two distinct integers.",
    "options": [],
    "answer": "(3, 2) and (-3, 2)",
    "solution": "Let the roots of $x^2 + px + q = 0$ be $r$ and $s$, with $r \\neq s$ and both integers.\n\nBy Vieta's formulas:\n$r + s = -p$\n$rs = q$\n\nSince $p$ and $q$ are primes, $-p$ is the sum of two distinct integers, and $q$ is their product.\n\nLet $r$ and $s$ be distinct integers. Since $q$ is prime, $rs = q$ implies that one of $r$ or $s$ is $1$ and the other is $q$, or one is $-1$ and the other is $-q$.\n\nCase 1: $r = 1$, $s = q$ (with $q$ prime, $q \\neq 1$)\nThen $r + s = 1 + q = -p \\implies p = -(1 + q)$\nBut $p$ must be prime, so $-(1 + q)$ is prime. Since $q$ is prime $\\geq 2$, $1 + q \\geq 3$, so $-(1 + q)$ is negative and prime. The only negative primes are $-2, -3, -5, \\ldots$\nSo $1 + q = 2 \\implies q = 1$ (not prime), $1 + q = 3 \\implies q = 2$, $1 + q = 5 \\implies q = 4$ (not prime), $1 + q = 7 \\implies q = 6$ (not prime), etc.\nSo only $q = 2$ gives $p = -3$ (which is prime).\n\nCheck: $x^2 + (-3)x + 2 = x^2 - 3x + 2 = (x - 1)(x - 2)$, roots $1$ and $2$ (distinct integers).\n\nCase 2: $r = -1$, $s = -q$ (with $q$ prime, $q \\neq 1$)\nThen $r + s = -1 - q = -p \\implies p = 1 + q$\n$p$ must be prime, so $1 + q$ is prime. Try $q = 2$, $p = 3$ (prime), $q = 3$, $p = 4$ (not prime), $q = 5$, $p = 6$ (not prime), $q = 7$, $p = 8$ (not prime), etc.\nSo only $q = 2$ gives $p = 3$ (prime).\n\nCheck: $x^2 + 3x + 2 = (x + 1)(x + 2)$, roots $-1$ and $-2$ (distinct integers).\n\nCase 3: $r = q$, $s = 1$ (already considered in Case 1).\nCase 4: $r = -q$, $s = -1$ (already considered in Case 2).\n\nTherefore, the only pairs $(p, q)$ of prime integers such that the solutions of $x^2 + px + q = 0$ are two distinct integers are:\n$$(p, q) = (3, 2) \\text{ and } (p, q) = (-3, 2)$$",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 29259,
    "subject": "Mathematics (Multi-modal)",
    "question": "Prove that for non-negative real numbers $a$ and $b$ such that $a + b \\le 2$ the inequality\n$$\n\\frac{1}{1+a^2} + \\frac{1}{1+b^2} \\le \\frac{2}{1+ab}\n$$\nholds. When is the equality attained? (Austria 2018)",
    "options": [],
    "answer": "Equality holds if and only if a = b (which, under the given constraint, means a = b in the interval from zero to one).",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 29260,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all pairs $(m, n)$ of integers such that $m^2 = n^5 + n^4 + 1$ and $m - 7n$ divides $m - 4n$. (Nikola Adžaga, Petar Bakić)",
    "options": [],
    "answer": "(-1, 0), (1, 0)",
    "solution": "Note that $n^5 + n^4 + 1 = (n^3 - n + 1)(n^2 + n + 1)$, and that\n$$\n\\begin{align*}\nd &= \\gcd(n^3 - n + 1, n^2 + n + 1) \\\\\n&= \\gcd(n^2 + n + 1, -n^2 - 2n + 1) \\\\\n&= \\gcd(n^2 + n + 1, n - 2) \\\\\n&= \\gcd(n - 2, 7),\n\\end{align*}\n$$\nhence we have two cases:\n\n1) $d = 7$\n\nThis implies $7 \\mid m$ and $7 \\nmid n$, from which we get $7 \\mid m - 7n$ and $7 \\nmid m - 4n$.\nTherefore, there is no solution in this case.\n\n2) $d = 1$\n\nThis implies that $n^3 - n + 1$ and $n^2 + n + 1$ are both squares of integers. That is true only for $n = 0$ and $n = -1$, since $n^2 < n^2 + n + 1 < (n + 1)^2$ holds for $n \\ge 1$, and $(n+1)^2 < n^2 + n + 1 < n^2$ holds for $n < -1$.\nBoth $n = 0$ and $n = -1$ yield $m^2 = 1$, i.e. $m = \\pm 1$, and among four possibilities only two satisfy the given conditions: $(m, n) = (-1, 0)$ and $(m, n) = (1, 0)$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 29261,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all pairs $(x, y)$ of real numbers such that\n$$\nxy^3 = -135, \\quad (x+y)y = -6. \\quad (Matko Ljulj)\n$$",
    "options": [],
    "answer": "(-5, 3), (5, -3)",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 29262,
    "subject": "Mathematics (Multi-modal)",
    "question": "Point $P$ is the midpoint of the segment $\\overline{AB}$ of length $2$. Let $T$ be the point at which the tangent from $A$ touches the circle with diameter $\\overline{PB}$. Find the length of the segment $\\overline{PT}$.\n\n(Petar Bakić)",
    "options": [],
    "answer": "1/√3",
    "solution": "",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 29263,
    "subject": "Mathematics (Multi-modal)",
    "question": "How many different bracelets consisting of four black and four white beads arranged in a circle are there? Two bracelets are considered different if they cannot be turned over so that the beads are equally aligned on them.",
    "options": [],
    "answer": "8",
    "solution": "",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 29264,
    "subject": "Mathematics (Multi-modal)",
    "question": "299 zeros and one one are written circular. The following moves are allowed:\n* You can select all numbers simultaneously and subtract both of its neighbouring numbers from each number.\n* You can select two numbers such that there are exactly two numbers between them, and either increase both selected numbers by 1 or decrease both of them by 1.\nCan we obtain through a finite number of moves that the following numbers are written circular:\na) two consecutive ones and 298 zeros?\nb) three consecutive ones and 297 zeros?",
    "options": [],
    "answer": "a) No; b) No",
    "solution": "Let us analyse how the moves affect certain sums. Denote by $a_k$ the numbers written at a certain moment, so that $a_1$ is the number written in the place where the one was written before any move was made, and $a_2, \\dots, a_{300}$ are the numbers written clockwise from $a_1$. After applying the first type of move, the numbers written are\n$$\n\\begin{align*}\nb_1 &= a_1 - a_{300} - a_2, \\\\\nb_k &= a_k - a_{k-1} - a_{k+1}, \\quad k = 2, \\dots, 299, \\\\\nb_{300} &= a_{300} - a_{299} - a_1.\n\\end{align*}\n$$\n\na) Observe how the sum of all written numbers changes after each move. If the total sum of numbers $a_k$ is\n$$\nS = a_1 + a_2 + \\cdots + a_{300},\n$$\nthen the sum of numbers $b_k$ is\n$$\na_1 - a_{300} - a_2 + a_2 - a_1 - a_3 + \\cdots + a_{300} - a_{299} - a_1 = -S\n$$\nbecause each $a_k$ is added once and subtracted twice.\n\nAfter the second type of move is applied, the sum of all numbers is either $S + 2$ or $S - 2$. We can conclude that none of the allowed moves changes the parity of the sum of all numbers. Since the sum is 1 (an odd number) in the beginning, we cannot obtain two ones and 298 zeros since their sum is 2 (an even number).\n\nb) Observe how the moves affect the alternating sum. If the alternating sum of numbers $a_k$ is\n$$\nS' = a_1 - a_2 + a_3 - \\dots + a_{299} - a_{300},\n$$\nthen after applying the first type of move, the alternating sum of numbers $b_k$ is\n$$\na_1 - a_{300} - a_2 - a_2 + a_1 + a_3 + \\dots - a_{300} + a_{299} + a_1 = 3S',\n$$\nwhile the alternating sum remains $S'$ after applying the second type of move. Assume that three consecutive ones and 297 zeros can be obtained. The original alternating sum before any move is made equals 1. When three consecutive ones and 297 zeros are written circular, the alternating sum is either 1 or -1, depending on where the ones are located. Since the allowed moves either triple the alternating sum or they do not affect it, we conclude that the three ones are located so that the alternating sum is 1, and no moves of the first type were allowed.\n\nObserve the sum of numbers in places 3, 6, ..., 300, i.e.\n$$\nS_0 = a_3 + a_6 + \\dots + a_{300}.\n$$\nOriginally, this sum is 0, and when three consecutive ones and 297 zeros are written, that sum is 1. However, applying only the second type of move does not change the parity of that sum. Namely, either none of the numbers $a_3, a_6, \\dots, a_{300}$ are changed, or we change exactly two of them by exactly 1, changing $S_0$ by 2 or -2. Since 0 and 1 are of different parity, this shows that we cannot obtain three consecutive ones and 297 zeros.",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 29265,
    "subject": "Mathematics (Multi-modal)",
    "question": "Kings and rooks are placed on an $8 \\times 8$ board, so that none of the pieces is under attack. A king attacks neighbouring squares (there are eight of them, except when the king is placed on the edge of the board), and a rook attacks all squares in the same row or column as the square where the rook is placed. At most how many pieces are on the board, if there is an equal number of rooks and kings?\n(Borna Vukorepa)",
    "options": [],
    "answer": "8",
    "solution": "",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 29266,
    "subject": "Mathematics (Multi-modal)",
    "question": "In a bathroom of size $6\\ \\mathrm{m} \\times 6\\ \\mathrm{m}$, one corner is occupied by a rectangular bathtub of size $2\\ \\mathrm{m} \\times 1.5\\ \\mathrm{m}$. What is the radius of the largest possible circular carpet that can be spread on the bathroom floor? (Petar Bakić)",
    "options": [],
    "answer": "(13 - 6√2)/2",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 29267,
    "subject": "Mathematics (Multi-modal)",
    "question": "Cities are connected by one-way airline routes. Each two cities $A$ and $B$ are connected by exactly one route: either from $A$ to $B$, or from $B$ to $A$. Prove that there exists a city from which it is possible to reach any other city with at most one layover.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 29268,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABC$ be a triangle such that $|AB| = 4$, $|BC| = 7$, $|CA| = 5$, and let $\\alpha = \\angle BAC$. Determine\n$$\nsin^6 \\frac{\\alpha}{2} + \\cos^6 \\frac{\\alpha}{2}.\n$$",
    "options": [],
    "answer": "7/25",
    "solution": "By applying the cosine theorem we get\n$$\n\\cos \\alpha = \\frac{|CA|^2 + |AB|^2 - |BC|^2}{2|CA| \\cdot |AB|} = \\frac{25 + 16 - 49}{2 \\cdot 5 \\cdot 4} = -\\frac{1}{5},\n$$\nand therefore\n$$\n\\begin{align*}\n\\sin^6 \\frac{\\alpha}{2} + \\cos^6 \\frac{\\alpha}{2} &= \\left(\\sin^2 \\frac{\\alpha}{2} + \\cos^2 \\frac{\\alpha}{2}\\right) \\left(\\sin^4 \\frac{\\alpha}{2} - \\sin^2 \\frac{\\alpha}{2} \\cos^2 \\frac{\\alpha}{2} + \\cos^4 \\frac{\\alpha}{2}\\right) \\\\\n&= \\left(\\sin^2 \\frac{\\alpha}{2} + \\cos^2 \\frac{\\alpha}{2}\\right)^2 - 3 \\sin^2 \\frac{\\alpha}{2} \\cos^2 \\frac{\\alpha}{2} \\\\\n&= 1 - \\frac{3}{4} \\sin^2 \\alpha \\\\\n&= \\frac{1}{4} + \\frac{3}{4} \\cos^2 \\alpha \\\\\n&= \\frac{1}{4} + \\frac{3}{4} \\cdot \\frac{1}{25} = \\frac{7}{25}.\n\\end{align*}",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 29269,
    "subject": "Mathematics (Multi-modal)",
    "question": "Solve\n$$\n\\cos(2x) + \\cos(2y) + 2 \\sin x \\sin y + 2 \\cos x \\cos y = 4.\n$$\n(Petar Bakić)",
    "options": [],
    "answer": "All solutions are x = mπ and y = nπ with integers m, n of the same parity (equivalently, x − y and x + y are both integer multiples of 2π).",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 29270,
    "subject": "Mathematics (Multi-modal)",
    "question": "Calculate\n$$\n\\frac{\\tan 192^\\circ + \\tan 48^\\circ}{1 + \\tan 168^\\circ \\cdot \\tan 408^\\circ}\n$$",
    "options": [],
    "answer": "sqrt(3)",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 29271,
    "subject": "Mathematics (Multi-modal)",
    "question": "The base of a right square pyramid has sides of length $12$, and the height of the pyramid is of length $8$. Find the surface area of the pyramid. (Nikola Adžaga)",
    "options": [],
    "answer": "384",
    "solution": "",
    "topic": "Geometry",
    "subtopic": "Solid Geometry"
  },
  {
    "id": 29272,
    "subject": "Mathematics (Multi-modal)",
    "question": "A quadruple $(a, b, c, d)$ of positive integers is called *green* if\n$$\nb = a^2 + 1, \\quad c = b^2 + 1, \\quad d = c^2 + 1\n$$\nholds, and $D(a) + D(b) + D(c) + D(d)$ is odd, where $D(k)$ denotes the number of positive divisors of $k$.\nHow many green quadruples having elements less than 1\\ 000\\ 000 are there?",
    "options": [],
    "answer": "2",
    "solution": "Note that $D(k)$ is odd if and only if $k$ is a perfect square.\nFrom $m^2 < m^2 + 1 < m^2 + 2m + 1 = (m+1)^2$ it follows that $m^2 + 1$ is not a perfect square for any positive integer $m$. Hence $b, c$ and $d$ of any green quadruple are not perfect squares, i.e. $D(b), D(c)$ and $D(d)$ are even.\nTherefore, $D(a)$ must be odd and $a$ is a perfect square.\nSince $10^6 > d > c^2 > b^4 > a^8$, i.e. $2^{10} = 1024 > 10^3 > a^4$, we get $6^2 > 2^5 > a^2$, i.e. $a < 6$. Hence $a = 1$ or $a = 4$.\nA direct computation confirms that both options yield green quadruples: $(1, 2, 5, 26)$ and $(4, 17, 290, 84101)$. Therefore, the answer is 2.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 29273,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all pairs $(m, n)$ of positive integers for which there exists a prime number $p$ such that\n$$\n9^m + 3^m - 2 = 2p^n.\n$$",
    "options": [],
    "answer": "(1,1)",
    "solution": "",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 29274,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $a$ be a positive real number such that $\\log_{4a} 40\\sqrt{3} = \\log_{3a} 45$. Prove that $a^3$ is an integer and find its value.",
    "options": [],
    "answer": "75",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 29275,
    "subject": "Mathematics (Multi-modal)",
    "question": "Point $T$ is located inside the triangle $ABC$ so that $|AT| = 56$, $|BT| = 40$ and $|CT| = 35$. The feet of the altitudes from $T$ to the sides of $ABC$ are vertices of an equilateral triangle. Find $\\triangle ABC$.\n\n(Belarus 2018)",
    "options": [],
    "answer": "BC:CA:AB = 5:7:8",
    "solution": "",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 29276,
    "subject": "Mathematics (Multi-modal)",
    "question": "We define the weight of a pair of numbers $\\{a, b\\}$ as $|a-b|$. In how many ways can the set $\\{1, 2, \\dots, 12\\}$ be divided into six pairs so that the total sum of weights of all pairs equals $30$?\n\n(Japan 2018)",
    "options": [],
    "answer": "1104",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 29277,
    "subject": "Mathematics (Multi-modal)",
    "question": "The product of a certain number of distinct positive integers less than $1000$ is not divisible by $250$. At most how many numbers have been multiplied?",
    "options": [],
    "answer": "802",
    "solution": "",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 29278,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $\\triangle ABC$ be a triangle such that $\\angle ABC = 2\\angle BCA$, and the angle bisector of $\\angle BAC$ intersects the side $\\overline{BC}$ at point $D$ so that $|AB| = |CD|$. Find $\\angle CAB$. (Estonia 2002)",
    "options": [],
    "answer": "72°",
    "solution": "",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 29279,
    "subject": "Mathematics (Multi-modal)",
    "question": "Marko places coins on some of the unit squares of a $3 \\times 3$ board, and then notes how many coins are in each row and each column. At least how many coins does Marko have to place on the board if he wants all six of those numbers to be mutually distinct? (United Kingdom)",
    "options": [],
    "answer": "8",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 29280,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all complex numbers $a$ such that all coefficients of\n$$\nP(x) = (x - a)(x - a^2)(x - a^3)(x - a^4)\n$$\nare real numbers.",
    "options": [],
    "answer": "All real numbers, together with a in {i, -i} and a = cos(2kπ/5) + i sin(2kπ/5) for k = 1, 2, 3, 4.",
    "solution": "Obviously, all real numbers satisfy the given condition.\nLet us, from now on, assume that $a$ is not a real number.\nIf $z$ is a root of $P$, then $\\bar{z}$ is a root of $P$ as well. Hence $\\bar{a} = a^2$, $\\bar{a} = a^3$ or $\\bar{a} = a^4$.\n\nIn the first case, we get $|a| = |a|^2$, and then $|a| = 1$ (since $a \\neq 0$). Hence $a^3 = a^2 \\cdot a = \\bar{a} \\cdot a = |a|^2 = 1$, so both factors $(x - a)(x - a^2)$ and $x - a^3$ of $P$ have real coefficients. This implies that $a^4 = a^3 \\cdot a = a$ is a real number, which is a contradiction, and there is no solution in this case.\n\nIn the second case, we similarly get $a^4 = 1$, hence $a = \\pm i$ and both can be verified as solutions.\n\nIn the third case, we similarly get $a^5 = 1$, hence\n$$\na = \\cos \\frac{2k\\pi}{5} + i \\sin \\frac{2k\\pi}{5}, \\quad k = 1, 2, 3, 4.\n$$\nSince $\\bar{a}^2 = a^5 \\cdot \\bar{a}^2 = a^3 \\cdot a^2 \\cdot \\bar{a}^2 = a^3 \\cdot |a|^4 = a^3$, both factors $(x - a)(x - a^4)$ and $(x - a^2)(x - a^3)$ of $P$ have real coefficients, so has $P$.\n\n$$\na \\in \\{i, -i\\} \\cup \\left\\{ \\cos \\frac{2k\\pi}{5} + i \\sin \\frac{2k\\pi}{5} \\mid k = 1, 2, 3, 4 \\right\\},\n$$\nalong with all real numbers.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 29281,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all pairs $(a, b)$ of integers such that $b \\ge 0$ and\n$$\na^2 + 2ab + b! = 131.\n$$\n(Olimpiada Matemática del Istmo Centroamericano 2017)",
    "options": [],
    "answer": "(1, 5), (-11, 5)",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 29282,
    "subject": "Mathematics (Multi-modal)",
    "question": "The product of the second and fourth term of an arithmetic sequence with the common difference $d$ is $-d^2$. Find the product of the third and fifth term of that sequence. (Matko Ljulj)",
    "options": [],
    "answer": "0",
    "solution": "Let the first term of the sequence be $a$.\n\nThe second term is $a + d$.\nThe fourth term is $a + 3d$.\n\nTheir product is $(a + d)(a + 3d) = -d^2$.\n\nWe are to find the product of the third and fifth terms:\n\nThird term: $a + 2d$\nFifth term: $a + 4d$\n\nTheir product is $(a + 2d)(a + 4d)$.\n\nLet us expand $(a + d)(a + 3d)$:\n\n$(a + d)(a + 3d) = a^2 + 3ad + ad + 3d^2 = a^2 + 4ad + 3d^2 = -d^2$\n\nSo:\n$a^2 + 4ad + 3d^2 = -d^2$\n$a^2 + 4ad + 4d^2 = 0$\n\nNow, $(a + 2d)(a + 4d) = a^2 + 4ad + 2ad + 8d^2 = a^2 + 6ad + 8d^2$\n\nFrom above, $a^2 + 4ad + 4d^2 = 0$, so $a^2 + 4ad = -4d^2$\n\nTherefore:\n$(a + 2d)(a + 4d) = (a^2 + 4ad) + 2ad + 8d^2 = (-4d^2) + 2ad + 8d^2 = 2ad + 4d^2$\n\nBut $a^2 + 4ad + 4d^2 = 0$\n\nLet us solve for $a$:\n$a^2 + 4ad + 4d^2 = 0$\n$a^2 + 4ad = -4d^2$\n\nSo $a^2 + 4ad = -4d^2$\n\nNow, $2ad + 4d^2 = 2ad + 4d^2$\n\nBut we can express $a$ in terms of $d$:\n$a^2 + 4ad + 4d^2 = 0$\n$(a + 2d)^2 = 0$\nSo $a + 2d = 0$\nThus, $a = -2d$\n\nNow, third term: $a + 2d = -2d + 2d = 0$\nFifth term: $a + 4d = -2d + 4d = 2d$\n\nTheir product: $0 \\times 2d = 0$\n\n**Answer:** $0$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 29283,
    "subject": "Mathematics (Multi-modal)",
    "question": "Rudi and Miljen take turns playing a game on a board. One move consists of selecting two relatively prime positive integers that are already written on the board, erasing them and replacing them with their sum. The player who cannot make a move loses. Rudi plays first. Prove that Miljen has a winning strategy if the initial numbers on the board are\n\na) 2019 ones\n\nb) 2020 ones.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "a)\nWe claim that Miljen can play so that he leaves some odd number $n$ and $2019 - n$ ones on the board after each of his moves, i.e. before Rudi's move. In that case, after $1009$ of Miljen's moves only the number $2019$ is written on the board, so Rudi cannot make a move and Miljen wins.\n\nIn the beginning, the board can be described as above, with $n = 1$. Rudi can erase two ones and write number $2$ on the board. In that case, Miljen can erase $n$ and $2$, which he can do since $n$ is odd, and write $n + 2$ on the board, which leaves the board as described above.\n\nIn each subsequent turn, Rudi and Miljen can play as above and our claim holds. Alternatively, Rudi can erase $n$ and one of the ones, and write $n + 1$. Note that $2019 - n$ is even, so Rudi's move leaves an odd number of ones on the board. Miljen can now select $n + 1$ and a one, and write $n + 2$, which again only leaves some odd number and ones on the board, as before.\n\nb)\nIn this case, Miljen uses the same strategy as in a), except when making the final move. Namely, after each of Miljen's moves an odd number $n$ and $2020 - n$ ones are written on the board. Note that now $2020 - n$ is odd. It remains to describe Miljen's last, i.e. his $1009$th move.\n\nIf after Rudi's $1009$th move the numbers on the board are $2018$, $1$ and $1$, Miljen can erase the two ones, and replace them with $2$. Rudi cannot make a move after that, since $2018$ and $2$ are not relatively prime, and Miljen wins.\n\nIf after Rudi's $1009$th move the numbers on the board are $2017$, $2$ and $1$, Miljen can erase $2017$ and $1$, and replace them with $2018$, which again leaves Rudi with $2018$ and $2$, and Miljen wins.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 29284,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all positive integers $n$ such that some three consecutive coefficients in the power series representation of $(1+x)^n$ are in the $3:4:5$ ratio. (AIME 1992)",
    "options": [],
    "answer": "62",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 29285,
    "subject": "Mathematics (Multi-modal)",
    "question": "A semicircle is correctly placed in a larger semicircle if their diameters are parallel, the endpoints of the diameter of the smaller semicircle lie on the arc of the larger semicircle, and the arc of the smaller semicircle touches the diameter of the larger semicircle.\n\n![](attached_image_1.png)\n\nA sequence of semicircles $K_1, K_2, K_3, \\dots$ is given, so that the semicircle $K_{n+1}$ is correctly placed in $K_n$ for every positive integer $n$. The area inside $K_n$ that is outside $K_{n+1}$ is coloured in blue for odd $n$, and in yellow for even $n$.\n\nThe radius of the semicircle $K_1$ equals $1$. Find the total area that is coloured in blue. (Mea Bombardelli, Matko Ljulj)",
    "options": [],
    "answer": "π/3",
    "solution": "",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 29286,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $n \\ge 2$ be a positive integer. Two opposite corner squares are removed from an $n \\times n$ board. In how many ways can $n$ chess pieces be placed on this board so that no two pieces are in the same row or column? (Matko Ljulj)",
    "options": [],
    "answer": "n! - 2(n-1)! + (n-2)!",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 29287,
    "subject": "Mathematics (Multi-modal)",
    "question": "Prove that the number\n$$\n\\overbrace{222\\ldots2}^{n \\text{ digits}} - 3^n + 1\n$$\nis divisible by $7$ for any positive integer $n$. (Matko Ljulj)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 29288,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABC$ be an acute-angled triangle such that $|AB| > |AC|$, and let $D$, $E$ and $F$ be the feet of its altitudes from the vertices $A$, $B$ and $C$, respectively. The lines $EF$ and $BC$ intersect at the point $P$. The line passing through $D$ parallel to $EF$ intersects the lines $AC$ and $AB$ at the points $Q$ and $R$, respectively.\n\nIf $N$ is a point on the segment $\\overline{BC}$ such that $\\angle NQP + \\angle NRP < 180^\\circ$, prove that $|BN| > |CN|$.\n\n(South Africa 2010)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "The triangle $ABC$ is acute-angled, so the points $D$, $E$ and $F$ lie on the sides $\\overline{BC}$, $\\overline{CA}$ and $\\overline{AB}$, respectively. Since $|AB| > |AC|$, we have $|BD| > |CD|$.\n\nFurthermore, from $\\angle ACB < \\angle ABC$ it follows that $C$ lies between $B$ and $P$. Also, $C$ lies between $A$ and $Q$, while $R$ lies on the segment $\\overline{BF}$.\n\n![](attached_image_1.png)\n\nLet $H$ be the orthocentre of the triangle $ABC$, and let $M$ be the midpoint of its side $\\overline{BC}$. The lines $DR$ and $EF$ are parallel, hence $\\angle DRF = \\angle EFA$. The quadrilateral $AFHE$ is cyclic ($\\angle AFH = \\angle AEH = 90^\\circ$), so we have $\\angle EFA = \\angle EHA$. The quadrilateral $BDHF$ is cyclic as well ($\\angle BDH = \\angle BFH = 90^\\circ$), hence $\\angle EHA = \\angle DHB = \\angle DFB = \\angle DFR$. Now from $\\angle DRF = \\angle DFR$ it follows that $|DR| = |DF|$.\n\nSimilarly, the lines $DQ$ and $FE$ are parallel, hence $\\angle DQE = \\angle FEA$. The quadrilateral $AFHE$ is cyclic, so we have $\\angle FEA = \\angle FHA$. The quadrilateral $CEHD$ is cyclic as well ($\\angle CEH = \\angle CDH = 90^\\circ$), hence $\\angle FHA = \\angle DHC = \\angle DEC = \\angle DEQ$. Now from $\\angle DQE = \\angle DEQ$ it follows that $|DQ| = |DE|$.\n\nThe points $M$, $F$, $E$ and $D$ lie on the same circle (called the nine-point circle or the Feuerbach's circle), hence $\\angle FMD = 180^\\circ - \\angle FED = \\angle PED$. Also, $\\angle FDH = \\angle FBH = \\angle ABE = 90^\\circ - \\angle CAB = \\angle ACF = \\angle ECH = \\angle EDH$.\n\nTherefore, $\\angle FDM = 90^\\circ - \\angle FDH = 90^\\circ - \\angle EDH = \\angle PDE$, so the triangles $FDM$ and $PDE$ are similar. Now we conclude that\n$$\n\\frac{|DF|}{|DM|} = \\frac{|DP|}{|DE|},\n$$\nfrom which it follows that $|DF| \\cdot |DE| = |DP| \\cdot |DM|$, i.e.\n$$\n|DR| \\cdot |DQ| = |DP| \\cdot |DM|,\n$$\nmeaning that the quadrilateral $PQMR$ is cyclic.\n\nLet $k$ be its circumscribed circle. Since $\\angle NQP + \\angle NRP < 180^\\circ$, $N$ lies inside the circle $k$, i.e. on the segment $\\overline{CM}$, and $N \\neq M$ holds. Since $M$ is the midpoint of the segment $\\overline{BC}$, we finally get $|BN| > |CN|$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 29289,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all triples $(x, y, z)$ of real numbers such that\n$$\n\\begin{aligned}\n(x^2 + 1)y &= z^2 + 1 \\\\\n(y^2 + 1)z &= x^2 + 1 \\\\\n(z^2 + 1)x &= y^2 + 1.\n\\end{aligned} \n\\qquad (\\text{Ukraine 2017})\n$$",
    "options": [],
    "answer": "(1, 1, 1)",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 29290,
    "subject": "Mathematics (Multi-modal)",
    "question": "Determine the number of complex solutions of the equation\n$$\nz^{2019} = z + \\bar{z}.\n$$",
    "options": [],
    "answer": "2019",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 29291,
    "subject": "Mathematics (Multi-modal)",
    "question": "Boys and girls participated in a chess tournament. Each participant played one match against each of the remaining participants, and no match ended in a draw. Find the smallest possible number of participants in the tournament if we know that each girl won against at least $21$ boys and each boy won against at least $12$ girls.",
    "options": [],
    "answer": "65",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 29292,
    "subject": "Mathematics (Multi-modal)",
    "question": "Denote by $\\mathbb{N}$ the set of all positive integers. Find all functions $f: \\mathbb{N} \\times \\mathbb{N} \\to \\mathbb{N}$ that satisfy the following conditions:\n* $f(a, b) + a + b = f(a, 1) + f(1, b) + ab$ holds for all positive integers $a$ and $b$.\n* If any number among $a + b$ and $a + b - 1$ is divisible by prime number $p > 2$, then $f(a, b)$ is divisible by $p$ as well. (Ivan Krijan)",
    "options": [],
    "answer": "f(a,b) = \\binom{a+b}{2} = \\frac{(a+b)(a+b-1)}{2}",
    "solution": "Plugging $(a, b) \\leftarrow (1, 1)$ into the first given condition, we get $f(1, 1) = 1$.\n\nOn the other hand, by plugging $(a, b) \\leftarrow (a, b+1)$ we get\n$$\nf(a, b + 1) + a + b + 1 = f(a, 1) + f(1, b + 1) + ab + a.\n$$\nUsing the first given condition in its original form, it follows that\n$$\nf(a, b + 1) - f(a, b) = f(1, b + 1) - f(1, b) + a - 1.\n$$\nLet $p > 2$ be a fixed prime number such that $p \\mid a+b$, meaning that $p \\mid f(a,b)$. Since $p \\mid a+(b+1)-1$, we also have $p \\mid f(a,b+1)$. Therefore, $p \\mid f(1,b+1)-f(1,b)+a-1$ and\n$$\np \\mid f(1, b + 1) - f(1, b) - b - 1 \\quad (*)\n$$\nfor all positive integers $a$ and $b$, and primes $p > 2$ such that $p \\mid a+b$.\nNote that the right-hand side in $(*)$ does not depend on $a$, hence fixing $b$ and varying $a$ would yield infinitely many odd prime divisors of $f(1, b+1) - f(1, b) - b - 1$, which is possible only if the latter equals 0, i.e.\n$$\nf(1, b + 1) = f(1, b) + b + 1\n$$\nfor every positive integer $b$.\nTherefore, $f(1, 2) = f(1, 1) + 2 = 1 + 2$, $f(1, 3) = f(1, 2) + 3 = 1 + 2 + 3$, i.e. $f(1, n) = 1 + \\cdots + n = \\frac{n(n+1)}{2}$ (by mathematical induction).\nIn a similar fashion, starting with $(a, b) \\leftarrow (a+1, b)$, we infer that $f(n, 1) = \\frac{n(n+1)}{2}$.\nFinally, from\n$$\n\\begin{aligned}\nf(a, b) + a + b &= f(a, 1) + f(1, b) + ab \\\\\n&= \\frac{a(a + 1)}{2} + \\frac{b(b + 1)}{2} + ab,\n\\end{aligned}\n$$\nit follows that $f(a, b) = \\binom{a+b}{2}$. This function clearly satisfies the second given condition, hence it is the only solution.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 29293,
    "subject": "Mathematics (Multi-modal)",
    "question": "What is the probability that for numbers $x$ and $y$ selected at random from the interval $[-2, 2]$ we have\n$$\n|x| + |y| \\ge 1 \\quad \\text{and} \\quad ||x| - |y|| \\le 1?\n$$",
    "options": [],
    "answer": "5/8",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 29294,
    "subject": "Mathematics (Multi-modal)",
    "question": "The point $A(0, 2)$ is given on the parabola $y^2 = x + 4$. Find all points $B$ on the given parabola, different from $A$, for which there exists a point $C$, also on the parabola, such that the angle $\\angle ACB$ is right. (China)",
    "options": [],
    "answer": "All points B on y^2 = x + 4 with coordinates B = (t^2 − 4, t) where t ≤ 0 or t ≥ 4 (in particular, y ≤ 0 or y ≥ 4).",
    "solution": "",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 29295,
    "subject": "Mathematics (Multi-modal)",
    "question": "By drawing lines parallel to each of the sides, an equilateral triangle of side length $n$ is divided into $n^2$ equilateral triangles of side length $1$. At most how many segments of length $1$ on the obtained grid can be coloured in red, so that no three red segments form an equilateral triangle?",
    "options": [],
    "answer": "n(n+1)",
    "solution": "",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 29296,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all pairs $(a, b)$ of real (not necessarily positive) numbers such that $a^2 + b^2 = 25$, for which $ab + a + b$ attains the smallest possible value.",
    "options": [],
    "answer": "(-4, 3) and (3, -4)",
    "solution": "Transforming the inequality $(a + b + 1)^2 \\ge 0$ yields $a^2 + b^2 + 1 + 2ab + 2a + 2b \\ge 0$, from which it follows that $2(ab + a + b) \\ge -(a^2 + b^2) - 1$, i.e. $ab + a + b \\ge -13$.\nThe equality is attained if and only if $a + b + 1 = 0$, i.e. $b = -a - 1$. Plugging this into $a^2 + b^2 = 25$ gives us\n$$\n\\begin{aligned}\na^2 + (-a - 1)^2 &= 25, \\\\\n2a^2 + 2a + 1 &= 25, \\\\\na^2 + a - 12 &= 0, \\\\\n(a + 4)(a - 3) &= 0,\n\\end{aligned}\n$$\nhence we get and verify two symmetric solutions: $(a, b) = (-4, 3)$ and $(a, b) = (3, -4)$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 29297,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $a$, $b$ and $c$ be positive real numbers. Prove that\n$$\n\\frac{\\sqrt{a+b+c}+\\sqrt{a}}{b+c} + \\frac{\\sqrt{a+b+c}+\\sqrt{b}}{c+a} + \\frac{\\sqrt{a+b+c}+\\sqrt{c}}{a+b} \\ge \\frac{9+3\\sqrt{3}}{2\\sqrt{a+b+c}}\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "The inequality is homogeneous, hence without loss of generality, we can assume that $a+b+c=1$. The given inequality transforms into\n$$\n\\underbrace{\\frac{1}{1-a} + \\frac{1}{1-b} + \\frac{1}{1-c}}_{A} + \\underbrace{\\frac{\\sqrt{a}}{1-a} + \\frac{\\sqrt{b}}{1-b} + \\frac{\\sqrt{c}}{1-c}}_{B} \\ge \\frac{9+3\\sqrt{3}}{2}.\n$$\nBy the inequality of arithmetic and harmonic means, we have\n$$\nA \\ge \\frac{9}{3 - (a + b + c)} = \\frac{9}{2}.\n$$\nOn the other hand, by the Cauchy-Bunyakovsky-Schwarz inequality, we have\n$$\nB \\cdot \\left( a\\sqrt{a}(1-a) + b\\sqrt{b}(1-b) + c\\sqrt{c}(1-c) \\right) \\ge 1,\n$$\nand it suffices to show that\n$$\na\\sqrt{a}(1-a) + b\\sqrt{b}(1-b) + c\\sqrt{c}(1-c) \\le \\frac{2}{3\\sqrt{3}}.\n$$\nBy the inequality of geometric and arithmetic means, we have\n$$\n\\frac{a(1-a)^2}{4} \\le \\left( \\frac{a + \\frac{1-a}{2} + \\frac{1-a}{2}}{3} \\right)^3 = \\frac{1}{27},\n$$\ni.e. $\\sqrt{a}(1-a) \\le \\frac{2}{3\\sqrt{3}}$, and analogously $\\sqrt{b}(1-b) \\le \\frac{2}{3\\sqrt{3}}$, $\\sqrt{c}(1-c) \\le \\frac{2}{3\\sqrt{3}}$.\nFinally,\n$$\na\\sqrt{a}(1-a) + b\\sqrt{b}(1-b) + c\\sqrt{c}(1-c) \\le (a+b+c) \\cdot \\frac{2}{3\\sqrt{3}} = \\frac{2}{3\\sqrt{3}},\n$$\nand the given inequality is proved.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 29298,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all functions $f: \\mathbb{Q}^+ \\to \\mathbb{Q}^+$ such that\n$$\nf(x^2 f(y)^2) = f(x)^2 f(y)\n$$\nholds for all positive rational numbers $x$ and $y$.",
    "options": [],
    "answer": "f(x) = 1 for all positive rational x",
    "solution": "Plugging $(x, y) \\leftarrow (f(x), y)$ into the given equation, it follows that\n$$\nf(f(x)^2 f(y)^2) = f(f(x))^2 f(y)\n$$\nholds for all positive rational numbers $x$ and $y$, hence $f(f(y))^2 f(x) = f(f(x)^2 f(y)^2) = f(f(x))^2 f(y)$, i.e.\n$$\n\\frac{f(x)}{f(y)} = \\left( \\frac{f(f(x))}{f(f(y))} \\right)^2 = \\left( \\frac{f^2(x)}{f^2(y)} \\right)^2,\n$$\nwhere $f^k$ denotes the $k$th iterate of $f$.\nBy mathematical induction, we can show that\n$$\n\\frac{f(x)}{f(y)} = \\left( \\frac{f^n(x)}{f^n(y)} \\right)^{2^{n-1}}\n$$\nholds, meaning that $f(x)/f(y)$ is the $(2^n)^{th}$ power of a rational number for all positive integers $n$. Hence, $f(x) = f(y)$ holds for all positive rational numbers $x$ and $y$, i.e. $f(x)$ is a constant.\nFinally, from the given equation we get that $f(x) = 1$ for all positive rational numbers $x$ is the only solution.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 29299,
    "subject": "Mathematics (Multi-modal)",
    "question": "A positive integer $n$ is good if each side and diagonal of a regular $n$-gon can be coloured in some colour so that for each pair of vertices $A$ and $B$ there is exactly one vertex $C$, different from $A$ and $B$, such that the segments $\\overline{AB}$, $\\overline{BC}$ and $\\overline{CA}$ have the same colour.\nWhich of the numbers 7, 8, 9, 10, 11 and 12 are good, and which are not?\n(Ilko Brnetić)",
    "options": [],
    "answer": "7 and 9 are good; 8, 10, 11, and 12 are not good.",
    "solution": "The numbers 8, 10, 11 and 12 are not good, while 7 and 9 are good.\n\nWe first show that even numbers are not good. Let us assume $n$ is a good number and consider a fixed vertex $A$ of a regular $n$-gon coloured in the required way. For each vertex $B$ different from $A$, there is a unique vertex $C$ such that the sides of the triangle $ABC$ have the same colour. Hence, we may partition all vertices other than $A$ into pairs. This shows that $n$ is odd. From this we conclude that 8, 10 and 12 are not good numbers.\n\nNext, we show that numbers of the form $n = 3k + 2$ are not good. Again, let us assume $n$ is good and for a regular $n$-gon coloured in the required way, let $t$ be the number of triangles $ABC$ with all sides of the same colour. Each such triangle has exactly three pairs of vertices, while for each pair of vertices of the $n$-gon there is a unique triangle with all the sides of the same colour. This shows that\n$$\n3t = \\frac{n(n-1)}{2}.\n$$\nHence, 3 divides $n$ or $n - 1$. This shows that 11 (and, again, 8) are not good.\n\nTo show that 7 and 9 are good numbers we construct examples. In the example we denote vertices of the $n$-gon by numbers 1, 2, ..., $n$ and write down triples $(a, b, c)$ representing triangles that have the sides of the same colour. Note that there should be $t = \\frac{n(n-1)}{6}$ triples, and for each pair of numbers $a, b \\in \\{1, 2, \\dots, n\\}$ there should be exactly one triple in which the pair of number occurs together.\n\nFor $n = 7$ we have the following $t = 7$ triples: $(1, 2, 3)$, $(1, 4, 5)$, $(1, 6, 7)$, $(2, 4, 6)$, $(2, 5, 7)$, $(3, 4, 7)$ and $(3, 5, 6)$.\n\nFor $n = 9$ we have the following $t = 12$ triples: $(1, 2, 3)$, $(4, 5, 6)$, $(7, 8, 9)$, $(1, 4, 7)$, $(2, 5, 8)$, $(3, 6, 9)$, $(1, 5, 9)$, $(2, 6, 7)$, $(3, 4, 8)$, $(1, 6, 8)$, $(2, 4, 7)$ and $(3, 5, 7)$.",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 29300,
    "subject": "Mathematics (Multi-modal)",
    "question": "Cryptogram of a positive integer $n$ is an $n$-tuple $a = (a_1, a_2, \\dots, a_n)$ of non-negative integers such that\n$$\na_1 + 2a_2 + \\dots + n a_n = n.\n$$\nLet $\\mathcal{K}_n$ be the set of all cryptograms of the number $n$. For $a \\in \\mathcal{K}_n$ let $J(a)$ denote the number of occurrences of the number 1 in the cryptogram $a$. Prove that\n$$\n\\sum_{a \\in \\mathcal{K}_n} J(a) = \\sum_{a \\in \\mathcal{K}_{n+1}} a_2.\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $k_n = |\\mathcal{K}_n|$, $j_n = \\sum_{a \\in \\mathcal{K}_n} J(a)$ and $d_{n+1} = \\sum_{a \\in \\mathcal{K}_{n+1}} a_2$. We need to prove that $j_n = d_{n+1}$ for every positive integer $n$. We prove the statement by complete mathematical induction.\n\nIt is clear that $j_1 = 1 = d_2$ and $j_2 = 1 = d_3$.\n\nLet us assume that $n \\ge 2$ is a positive integer such that $j_i = d_{i+1}$ for all $i \\le n$.\n\nFirst, we note that $d_{n+2} = k_n + d_n$. Indeed, for each cryptogram $a$ of $n+2$ for which $a_2 > 0$, we decrease $a_2$ by 1 and obtain a cryptogram of $n$. It follows that $d_{n+2}$ is equal to the number of cryptograms of $n$ increased by $d_n$.\n\nLet us also prove that $j_{n+1} = k_n + d_n$. By the induction assumption, we need to prove that $j_{n+1} = k_n + j_{n-1}$.\n\nWe partition all cryptograms of $n + 1$ into two disjoint subsets: set A consisting of cryptograms having the last non-zero element equal to 1, and set B consisting of cryptograms having the last non-zero element greater than 1.\n\nCryptograms in A are in bijection with the cryptograms of $n$. Indeed, we can delete the last non-zero element of a cryptogram in A and add 1 to the element just in front of it. Conversely, for each cryptogram of $n$, we decrease the last non-zero element and add 1 to the element right after it.\n\nIt remains to prove that the set B contains in total $j_{n-1}$ ones. Let $a \\in B$ be such that $a_i = 1$ for some $i \\in \\{1, 2, \\dots, n\\}$. Let $a_i = 0$ and let us increase $a_{i-1}$ by 1. If $i = 1$, let $a_1 = 0$. Furthermore, let $\\ell > i$ be the smallest index such that $a_\\ell > 0$, which exists because the last non-zero element is not 1. We decrease $a_\\ell$ by 1 and increase $a_{\\ell-1}$ by 1. Note that we obtain a cryptogram of $n-1$ which has 1 at the index $\\ell-1$. This shows that each one occurring in B corresponds bijectively to a one in the set of cryptograms of $n-1$.\n\nFinally, $j_{n+1} = k_n + j_{n-1}$ and the proof is finished.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 29301,
    "subject": "Mathematics (Multi-modal)",
    "question": "A *section* of a finite set of points in the plane is a partition of that set into disjoint subsets $A$ and $B$, such that there is a line not passing through any of the points in the set so that all the points of the set $A$ are on one side of the line, and all the points of the set $B$ are on the other. Determine the maximum possible number of sections of a set of $n$ points in the plane. (Putnam 2006)",
    "options": [],
    "answer": "n(n-1)/2 + 1",
    "solution": "Let $a_n$ be the maximum possible number of sections of a set of $n$ points in the plane.\nLet $S$ be a set of $n+1$ points in the plane and let us consider one of these points, call it $T$, on the convex hull of that set. Note that each section of $S$ restricts to a section of the set $S \\setminus \\{T\\}$. Let $\\{A, B\\}$ and $\\{C, D\\}$ be two different sections of the set $S$ that restrict to the same section of the set $S \\setminus \\{T\\}$. This means that $\\{A \\setminus \\{T\\}, B \\setminus \\{T\\}\\} = \\{C \\setminus \\{T\\}, D \\setminus \\{T\\}\\}$. Furthermore, it means, without loss of generality, that the sets $A$ and $C$ are the same and the sets $B$ and $D$ are the same, up to the point $T$. Hence, for the section $\\{A \\setminus \\{T\\}, B \\setminus \\{T\\}\\}$ of the set $S \\setminus \\{T\\}$ we can take a line passing through the point $T$ such that all the points of the set $A \\setminus \\{T\\}$ are on one side of the line, while all the points of the set $B \\setminus \\{T\\}$ are on the other.\nLet us consider all the lines passing through the point $T$ and none of the points of the set $S \\setminus \\{T\\}$. Two such lines could give two different sections of the set $S \\setminus \\{T\\}$ only if there is at least one point of the set $S \\setminus \\{T\\}$ between them. Hence, the number of different sections of the set $S \\setminus \\{T\\}$ corresponding to a line passing through $T$ is at most $n$, i.e. as many as there are points in the set $S \\setminus \\{T\\}$.\nFinally, we note that\n$$\na_{n+1} \\le n + a_n \\le n + (n-1) + a_{n-1} \\le \\dots \\le n + (n-1) + \\dots + 2 + 1 + a_1 = \\binom{n+1}{2} + 1.\n$$\nSo, the set of $n$ points in the plane can have at most $\\binom{n}{2} + 1$ sections. That number is achievable, e.g. in the case of a regular $n$-gon.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 29302,
    "subject": "Mathematics (Multi-modal)",
    "question": "Two positive integers are written on the board. Two players take turns in a game changing the numbers on the board. If the numbers on the board are $A$ and $B$ ($A \\ge B$), the player who has the turn chooses a positive integer $k$ such that $A - kB \\ge 0$, erases the number $A$ and writes $A - kB$ on the board. The winner is the player who writes number $0$.\nDetermine all ratios of the starting two numbers for which the first player can win independently of the choices of the second player. (Brazil 1987)",
    "options": [],
    "answer": "Let φ = (1 + √5)/2. The first player can force a win if and only if the initial ratio lies in [0, 1/φ] ∪ {1} ∪ [φ, +∞). Equivalently, for ordered numbers with larger divided by smaller, either the ratio equals 1 or exceeds φ.",
    "solution": "Let $\\varphi = \\frac{1+\\sqrt{5}}{2}$ be the positive root of the quadratic polynomial $t^2 - t - 1$. The first player can win if the ratio of the starting numbers is in the set\n$$\n\\langle 0, \\frac{1}{\\varphi} \\rangle \\cup \\{1\\} \\cup \\langle \\varphi, +\\infty \\rangle.\n$$\nLet $M$ and $m$ be positive integers such that $M \\geq m$. We claim that the player who is next on the turn when $M$ and $m$ are on the board wins if and only if $m = M$ or $M > \\varphi m$. The claim is clear for $m = M$, so let us assume that $m \\neq M$.\nIf $M < \\varphi m$, then the player who plays next must pass on the pair of numbers $(m, M-m) \\neq (m, 0)$ and $m > \\varphi(M-m)$. Hence it is enough to show that the player who plays with a pair such that $M > \\varphi m$ can either win or pass on a pair $(m', M')$ with $m' < M' < \\varphi m'$.\nThe player wins if he plays with the numbers such that $m \\mid M$. Let us assume that $M > \\varphi m$ and $M = qm + r, 0 < r < m$. If $q \\geq 2$, then player may pass on $(m, r)$, as well as on $(m, m+r)$. He will pass on $(m, r)$ if $r < m < \\varphi r$. Otherwise, if $m > \\varphi r$, he will pass on $(m, m+r)$, so the other player will pass on $(m, r)$ and we know that with these numbers the player on the turn wins. If $q = 1$, the player passes on $(m, r)$. We claim that $r \\neq m < \\varphi r$. Indeed, since $m + r > \\varphi m$, we have $(\\varphi - 1)m < r$, i.e. $m < \\varphi r$.\nHence, the first player can in each move either win or pass on a pair $(m', M')$ such that $m' < M' < \\varphi m'$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 29303,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $A$, $B$, $C$ and $D$ be points on a circle such that $|AB| = |BC| = |CD|$. The angle bisectors of $\\angle ABD$ and $\\angle ACD$ intersect at the point $E$.\nIf the lines $AE$ and $CD$ are parallel, find $\\angle ABC$.\n(Matko Ljulj)",
    "options": [],
    "answer": "180° × 5/7",
    "solution": "Due to symmetry, the isosceles triangles $ABC$ and $BCD$ are congruent, and the cyclic quadrilateral $ABCD$ is an isosceles trapezium.\n![](attached_image_1.png)\n\nLet us denote by $x$ the measure of angles along the bases in $ABC$ and $BCD$. Then $\\angle ABC = \\angle BCD = 180^\\circ - 2x$ and $\\angle ABD = \\angle ACD = 180^\\circ - 3x$, from which we get $\\angle EBC = \\angle ECB = 90^\\circ - \\frac{x}{2}$ and $\\angle BEC = x$. Hence, the point $E$ lies on the same circle as $A$, $B$, $C$ and $D$.\nWe also have $\\angle ADC = \\angle ADB + \\angle BDC = \\angle ACB + \\angle BDC = x + x = 2x$ and $\\angle ECD = \\frac{1}{2}\\angle ACD = 90^\\circ - \\frac{3}{2}x$.\nSince $AE \\parallel CD$, the cyclic quadrilateral $ACDE$ is an isosceles trapezium as well, so $\\angle ADC = \\angle ECD$ holds, and we get $7x = 180^\\circ$. Therefore, $\\angle ABC = \\frac{5}{7} \\cdot 180^\\circ$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 29304,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $M$ be the midpoint of the base $\\overline{BC}$ in the isosceles triangle $ABC$. Let $P$ be the point different from $A$ such that $PA \\parallel BC$. Points $X$ and $Y$ lie on the rays $PB$ and $PC$, respectively, so that $B$ lies between $P$ and $X$, $C$ lies between $P$ and $Y$, and $\\angle PXM = \\angle PYM$ holds.\nProve that points $A, P, X$ and $Y$ are concyclic. (IMO Shortlist 2018)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $N$ be the other intersection of the circles circumscribed to the triangles $BMX$ and $CMY$.\n\n![](attached_image_1.png)\n\nSince $\\angle BNM = \\angle BXM = \\angle CYM = \\angle CNM$, the line $MN$ is the angle bisector of $\\angle BNC$. The segment $NM$ is a median in the triangle $BNC$, hence $|BN| = |CN|$ and $NM \\perp BC$, from which we conclude that $A, M$ and $N$ are collinear.\n\nFrom $NM \\perp BC$ we also have $\\angle PXN = \\angle BXN = 180^\\circ - \\angle BMN = 90^\\circ$, and analogously $\\angle PYN = 90^\\circ$, meaning that $P, X, N$ and $Y$ lie on the circle with diameter $PN$.\n\nFinally, since $PA \\parallel BC \\perp AN$, we get $\\angle PAN = 90^\\circ$, which completes the proof.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 29305,
    "subject": "Mathematics (Multi-modal)",
    "question": "On the side $\\overline{AB}$ of the cyclic quadrilateral $ABCD$ there is a point $X$ such that the diagonals $\\overline{BD}$ and $\\overline{AC}$ bisect the segments $\\overline{CX}$ and $\\overline{DX}$, respectively.\nFind the smallest possible value of $|AB| : |CD|$. (Belarus)",
    "options": [],
    "answer": "2",
    "solution": "Let $M$ and $N$ be the midpoints of the segments $\\overline{CX}$ and $\\overline{DX}$, respectively. Also, denote $\\alpha = \\angle BAC = \\angle BDC$ and $\\beta = \\angle CAD = \\angle CBD$.\n\n![](attached_image_1.png)\n\nSince the triangles $AXN$ and $ADN$ have equal areas, we have $|AX| \\cdot |AN| \\sin \\alpha = |AD| \\cdot |AN| \\sin \\beta$, from which\n$$\n\\frac{|AD|}{|AX|} = \\frac{\\sin \\alpha}{\\sin \\beta} = \\frac{|BC|}{|CD|}, \\quad \\text{i.e.} \\quad \\frac{|AX|}{|CD|} = \\frac{|AD|}{|BC|}\n$$\n\nfollows. Analogously, we get $\\frac{|BX|}{|CD|} = \\frac{|BC|}{|AD|}$.\nBy the inequality of arithmetic and geometric means, it follows that\n$$\n\\frac{|AB|}{|CD|} = \\frac{|AX|}{|CD|} + \\frac{|BX|}{|CD|} = \\frac{|AD|}{|BC|} + \\frac{|BC|}{|AD|} \\ge 2.\n$$\nThe value 2 is attained for an isosceles trapezium $ABCD$ such that $|AB| = 2|CD|$ and $X$ being the midpoint of the base $\\overline{AB}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 29306,
    "subject": "Mathematics (Multi-modal)",
    "question": "The incircle of the triangle $ABC$ touches the sides $\\overline{AB}$ and $\\overline{AC}$ in $D$ and $E$, respectively. The excircle of the same triangle opposite to the vertex $A$ touches the rays $AB$ and $AC$ in $F$ and $G$, respectively. Let the bisectors of the internal (external) angles $\\angle CBA$ and $\\angle ACB$ intersect the line $DE$ ($FG$) at the points $X$ and $Y$ ($Z$ and $W$), respectively.\nProve that the points $X$, $Y$, $Z$ and $W$ lie on the same circle.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $I$ be the incentre, and let $\\alpha$, $\\beta$ and $\\gamma$ be the measures of angles of the triangle $ABC$.\nThe triangle $ADE$ is isosceles, and we have $\\angle BDX = 90^\\circ + \\frac{\\alpha}{2}$. Since $\\angle XBD = \\frac{\\beta}{2}$, it follows that $\\angle EXI = \\angle DXB = \\frac{\\gamma}{2} = \\angle ECI$, so the quadrilateral $EICX$ is cyclic. Hence, $\\angle BXC = \\angle IXC = \\angle IEC = 90^\\circ$, i.e. the point $X$ lies on the circle with diameter $BC$.\nAnalogously, we show the same for the points $Y$, $Z$ and $W$, and the proof is finished.",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 29307,
    "subject": "Mathematics (Multi-modal)",
    "question": "A point $T$ is chosen inside the triangle $ABC$. Let $A_1, B_1$ and $C_1$ be the reflections of $T$ across the lines $BC, CA$ and $AB$, respectively. The lines $A_1T, B_1T$ and $C_1T$ intersect the circle $k$ circumscribed to the triangle $A_1B_1C_1$ again at $A_2, B_2$ and $C_2$, respectively.\nProve that the lines $AA_2, BB_2$ and $CC_2$ are concurrent on $k$.\n(IMO Shortlist 2018)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $K$ be the intersection of $CC_2$ and $k$.\n\n![](attached_image_1.png)\n\nSince $CB$ and $CA$ are the bisectors of $\\overline{TA_1}$ and $\\overline{TB_1}$, respectively, the point $C$ is the circumcentre of the triangle $A_1TB_1$. Hence,\n$$\n\\triangle(CA_1, CB) = \\triangle(CB, CT) = \\triangle(B_1A_1, B_1T) = \\triangle(B_1A_1, B_1B_2).\n$$\nObserving the circle $k$, we have $\\triangle(B_1A_1, B_1B_2) = \\triangle(C_2A_1, C_2B_2)$ and $\\triangle(CA_1, CB) = \\triangle(B_1A_1, B_1B_2) = \\triangle(C_2A_1, C_2B_2)$. Similarly, we get $\\triangle(BA_1, BC) = \\triangle(B_2A_1, B_2C_2)$, hence the triangles $A_1BC$ and $A_1B_2C_2$ are similar, so as the triangles $A_1BB_2$ and $A_1CC_2$, from which it follows that $\\triangle(C_2C, C_2A_1) = \\triangle(B_2B, B_2A_1)$ and the point $K$ lies on $BB_2$.\n\nAnalogously, we show that the point $K$ lies on $AA_2$, and the proof is finished.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 29308,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all positive integers $n$ such that\n$$\n\\frac{n^{3n-2} - 3n + 1}{3n - 2}\n$$\nis an integer.",
    "options": [],
    "answer": "1",
    "solution": "Since $3n-2 \\mid n^{3n-2}-3n+1$, it follows that $3n-2 \\mid n^{3n-2}-1$, hence $n^{3n-2} \\equiv 1 \\pmod{3n-2}$.\n\nNote that $n$ must be odd (otherwise the even $3n-2$ would divide the odd $n^{3n-2}-3n+1$, which is impossible).\n\nLet $p > 2$ be the smallest prime factor of the odd $3n-2$. Obviously $p \\nmid n$, hence by Fermat's little theorem we have $n^{p-1} \\equiv 1 \\pmod{p}$. We also have $n^{3n-2} \\equiv 1 \\pmod{p}$.\n\nLet $r$ be the order of $n$ modulo $p$. Then $r \\mid p-1$ and $r \\mid 3n-2$. Since $r \\mid \\gcd(p-1, 3n-2)$ and $\\gcd(p-1, 3n-2) = 1$, it must be $r=1$, i.e. $n \\equiv 1 \\pmod{p}$.\n\nConsidering $p \\mid n-1$ and $p \\mid 3n-2$, it follows that $p \\mid 1$, which contradicts the assumption about the existence of prime factor of $3n-2$. Hence $3n-2=1$, i.e. $n=1$, and that is indeed the only solution.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 29309,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $m$ and $n$ be positive integers of different parity. Prove that\n$$\n\\frac{3m^2 + 5mn}{3n^2 + mn}\n$$\nis not a positive integer.\n(Ilko Brnetić)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $d$ be the greatest common divisor of $m$ and $n$, i.e. $m = d m'$ and $n = d n'$, where $m'$ and $n'$ are relatively prime positive integers of different parity.\nNow we need to prove that\n$$\n\\frac{3d^2 m'^2 + 5d^2 m' n'}{3d^2 n'^2 + d m' n'} = \\frac{m'(3m' + 5n')}{n'(m' + 3n')}\n$$\nis not a positive integer.\nNote that $3m' + 5n'$ and $m' + 3n'$ are both odd.\nIf $m'$ is odd and $n'$ is even, the even $n'(m' + 3n')$ clearly cannot divide the odd $m'(3m' + 5n')$.\nOtherwise, let the odd $k$ be the greatest common divisor of $3m' + 5n'$ and $m' + 3n'$. Then we have\n$$\n\\begin{aligned}\nk &\\mid 3 \\cdot (3m' + 5n') - 5 \\cdot (m' + 3n'), & \\text{ i.e. } k &\\mid 4m', \\\\\nk &\\mid 1 \\cdot (3m' + 5n') - 3 \\cdot (m' + 3n'), & \\text{ i.e. } k &\\mid -4n',\n\\end{aligned}\n$$\nfrom which it follows that $k$ divides both $m'$ and $n'$, so $k = 1$ and both factors in\n$$\n\\frac{m'}{n'} \\cdot \\frac{3m' + 5n'}{m' + 3n'}\n$$\nare irreducible fractions.\nSince $m' + 3n' > m'$, the proof is finished.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 29310,
    "subject": "Mathematics (Multi-modal)",
    "question": "Denote by $\\mathbb{N}$ the set of all positive integers. Let $f: \\mathbb{N} \\to \\mathbb{N}$ be a *multiplicative* function such that $f(4) = 4$ and\n$$\nf(m^2 + n^2) = f(m^2) + f(n^2)\n$$\nholds for all positive integers $m$ and $n$. Prove that $f(m^2) = m^2$ holds for all positive integers $m$.\n(A function is called *multiplicative* if $f(mn) = f(m)f(n)$ holds for all relatively prime positive integers $m$ and $n$.) (Mathematica Slovaca)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "We easily get $f(1) = 1$, $f(2) = f(1) + f(1) = 2$, $f(5) = f(4) + f(1) = 5$, $f(20) = f(4)f(5) = 20$ and $f(16) = f(20) - f(4) = 16$.\nWe prove the claim by complete mathematical induction, assuming it holds for all positive integers $m \\le n$ ($P(m)$), and performing the induction step for $n + 1$:\n\n1) If $n$ is even, then $n + 1 = 2k + 1$ for some positive integer $k$. Note that $2(k^2 + (k+1)^2) = (2k+1)^2 + 1$ and $k^2 + (k+1)^2$ is odd, hence\n$$\n\\begin{aligned}\nf((2k+1)^2) + 1 &= f((2k+1)^2 + 1) = f(2(k^2 + (k+1)^2)) \\\\\n&= f(2)f(k^2 + (k+1)^2) = 2(f(k^2) + f((k+1)^2)) \\\\\nP(k), P(k+1) \\implies \\quad &= 2(k^2 + (k+1)^2) = (2k+1)^2 + 1,\n\\end{aligned}\n$$\ni.e. $f((n+1)^2) = (n+1)^2$.\n\n2) If $n$ is odd, then $n + 1 = 2k$ for some positive integer $k$.\n\na) If $k$ is even, then $k^2 + 1$ is odd, and we get\n$$\n\\begin{aligned}\nf((2k)^2) + 4 &= f((2k)^2 + 4) = f(4(k^2 + 1)) \\\\\n&= f(4)f(k^2 + 1) = 4(f(k^2) + 1) \\\\\nP(k) \\implies \\quad &= 4(k^2 + 1) = (2k)^2 + 4,\n\\end{aligned}\n$$\ni.e. $f((n+1)^2) = (n+1)^2$.\n\nb) If $k$ is odd, then $k^2 + 4$ is odd as well, hence\n$$\n\\begin{aligned}\nf((2k)^2) + 16 &= f((2k)^2 + 16) = f(4(k^2 + 4)) \\\\\n&= f(4)f(k^2 + 4) = 4(f(k^2) + 4) \\\\\nP(k) \\implies \\quad &= 4(k^2 + 4) = (2k)^2 + 16,\n\\end{aligned}\n$$\ni.e. $f((n+1)^2) = (n+1)^2$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 29311,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all positive integers $k$ such that\n$$\n\\frac{m+n}{m^2-kmn+n^2}\n$$\nis not a composite number for any positive integers $m$ and $n$. (Borna Vukorepa)",
    "options": [],
    "answer": "k = 1",
    "solution": "Considering a fixed positive integer $k$ with the given property, by plugging $(m, n) \\leftarrow (k, 1)$ and $(m, n) \\leftarrow (k^2 + k - 1, k + 1)$ we find that both\n$$\n\\frac{k+1}{k^2 - k^2 + 1} = k + 1\n$$\nand\n$$\n\\frac{(k^2 + k - 1) + (k + 1)}{(k^2 + k - 1)^2 - k(k^2 + k - 1)(k + 1) + (k + 1)^2} = \\frac{k^2 + 2k}{k + 2} = k\n$$\nare not composite. Since they are of different parity, it follows that $k = 1$ or $k = 2$.\n\n1) If $k = 1$, then $\\frac{m+n}{m^2-mn+n^2} \\le 2$, satisfying the conditions of the problem. Indeed,\n$$\n\\frac{m+n}{m^2-mn+n^2} \\le 2 \\iff (m-n)^2 + m(m-1) + n(n-1) \\ge 0.\n$$\n\n2) $k = 2$ does not satisfy the conditions of the problem. For example, plugging $(m, n) \\leftarrow (5, 4)$ into $\\frac{m+n}{m^2-2mn+n^2} = \\frac{m+n}{(m-n)^2}$ yields composite number 9.\n\nTherefore, $k = 1$ is the only positive integer with the given property.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 29312,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $m$ be a positive integer, and $p$ be a prime number such that $p > m$. Prove that the number of positive integers $n$, for which\n$$\nm^2 + n^2 + p^2 - 2mn - 2mp - 2np\n$$\nis a square of some positive integer, does not depend on $p$. (Bulgaria 2013)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $m^2 + n^2 + p^2 - 2mn - 2mp - 2np = k^2$ for some positive integer $k$. Since\n$$\nk^2 + 4mp = m^2 + n^2 + p^2 - 2mn + 2mp - 2np = (m - n + p)^2\n$$\nis a perfect square, there exists a positive integer $l$ such that $(k+l)^2 = k^2 + 4mp$, i.e. $2kl + l^2 = 4mp$, and hence $l$ is even. Thus $l = 2a$ and $a^2 < a(k+a) = mp < p^2$, hence $p > a$.\nSince $p$ is prime, it follows that $p \\mid k+a$, i.e. $k+a = bp$ for some positive integer $b$. Thus $ab = m < p$,\n$$\n(bp + a)^2 = (k + 2a)^2 = (k + l)^2 = (m - n + p)^2\n$$\nand $n = p + ab \\pm (bp + a)$.\nHowever, $p + ab - (bp + a) = (1-b)(p-a) \\le 0$, so $n = p + m + bp + a$ meaning that every choice of $(a, b)$ yields valid $n$.\n\nNow assume that two different choices $(a, b)$ and $(a', b')$ yield the same $n$, i.e. $p+m+bp+a = p+m+b'p+a'$. It follows that $-p(b-b') = a-a'$ and hence $p \\mid a-a'$, which is impossible because $0 < |a-a'| < m < p$.\nThus, the number of valid positive integers $n$ is equal to the number of divisors of $m$, which does not depend on $p$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 29313,
    "subject": "Mathematics (Multi-modal)",
    "question": "Given an integer $n \\ge 16$, consider the set\n$$\nG = \\{(x, y) : x, y \\in \\{1, 2, \\dots, n\\}\\}\n$$\nconsisting of $n^2$ points in the plane. Let $A$ be any subset of $G$ containing at least $4n\\sqrt{n}$ points. Prove that there are at least $n^2$ convex quadrangles with all their vertices in $A$ such that their diagonals intersect in one common point.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 29314,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $n$ be a given positive integer. Solve the system of equations\n$$\n\\begin{aligned}\nx_1 + x_2^2 + x_3^3 + \\dots + x_n^n &= n, \\\\\nx_1 + 2x_2 + 3x_3 + \\dots + nx_n &= \\frac{n(n+1)}{2}\n\\end{aligned}\n$$\nin the set of nonnegative real numbers $x_1, x_2, \\dots, x_n$.",
    "options": [],
    "answer": "x1 = x2 = ... = xn = 1",
    "solution": "Suppose $x_1, x_2, \\dots, x_n$ satisfy the equations above. Then we have\n$$\n\\begin{aligned}\n0 &= x_1 + x_2^2 + x_3^3 + \\dots + x_n^n - n - (x_1 + 2x_2 + 3x_3 + \\dots + nx_n - \\frac{1}{2}n(n+1)) \\\\\n&= (x_2^2 - 2x_2 + 2 - 1) + (x_3^3 - 3x_3 + 3 - 1) + \\dots + (x_n^n - nx_n + n - 1).\n\\end{aligned}\n$$\nHowever, the expressions in the brackets are nonnegative. Indeed, for $k \\ge 2$ and $x \\ge 0$ we have, by the AM-GM inequality,\n$$\nx^k + k - 1 = x^k + 1 + 1 + \\dots + 1 \\ge k \\cdot \\sqrt[k]{x^k} = kx\n$$\nand the equality holds if and only if $x = 1$. Therefore we have $x_2 = x_3 = \\dots = x_n = 1$ and, by the first equation, $x_1 = 1$.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 29315,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let a convex quadrilateral $ABCD$ be inscribed in a circle with center $O$ and circumscribed to a circle with center $I$, and let its diagonals $AC$ and $BD$ meet at a point $P$. Prove that the points $O$, $I$ and $P$ are collinear.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Assume that the lines $AI$, $BI$, $CI$, $DI$ meet the circumcircle of the quadrilateral $ABCD$ at $E$, $F$, $G$, $H$, respectively. Since the lines $AI$, $BI$, $CI$, $DI$ are the bisectors of the respective angles of the quadrilateral $ABCD$, the lines $EG$ and $FH$ are the diameters of the circumcircle of $ABCD$. Thus $EG$ and $FH$ meet at $O$.\nDenote by $X$ the point of intersection of $EB$ and $CH$.\nUsing Pascal's theorem for the hexagon $ACHDBE$ we see that $P$, $X$ and $I$ are collinear. Once again Pascal's theorem applied to the hexagon $GCHFBE$ yields that $O$, $X$ and $I$ are collinear. Thus $O$, $I$ and $P$ are collinear, as desired.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 29316,
    "subject": "Mathematics (Multi-modal)",
    "question": "Determine all integers $n \\ge 3$ such that the polynomial\n$$\nW(x) = x^n - 3x^{n-1} + 2x^{n-2} + 6\n$$\ncan be expressed as a product of two polynomials with positive degrees and integer coefficients.",
    "options": [],
    "answer": "All odd integers n ≥ 3",
    "solution": "We check that for $n = 3$\n$$\nx^3 - 3x^2 + 2x + 6 = (x + 1)(x^2 - 4x + 6).\n$$\nSuppose that for $n = 4$ we have\n$$\nx^4 - 3x^3 + 2x^2 + 6 = (x^2 + a x + b)(x^2 + c x + d).\n$$\nThen, comparing the coefficients we obtain\n$$\na + c = -3, \\quad a c + b + d = 2, \\quad b d = 6.\n$$\nThe first equation implies that $a$ and $c$ are of different parity and therefore, by the second equality, $b$ and $d$ are of the same parity. This contradicts the third equation. Hence we may assume that $n \\ge 5$. Suppose that we have\n$$\nW(x) = P(x) Q(x), \\quad (1)\n$$\nwhere\n$$\nP(x) = a_k x^k + a_{k-1} x^{k-1} + \\dots + a_1 x + a_0,\n$$\n$$\nQ(x) = b_{n-k} x^{n-k} + b_{n-k-1} x^{n-k-1} + \\dots + b_1 x + b_0,\n$$\nand $a_k = b_{n-k} = \\pm 1$. With no loss of generality we may assume that $k \\le \\lfloor \\frac{1}{2} n \\rfloor < n-2$ (because $n \\ge 5$). Comparing the coefficients of both sides of (1) we obtain the following system of equations:\n$$\n\\begin{align*}\na_0 b_0 &= 6, \\\\\na_0 b_1 + a_1 b_0 &= 0,\n\\end{align*}\n$$\n$$\na_0 b_k + a_1 b_{k-1} + \\dots + a_{k-1} b_1 + a_k b_0 = 0.\n$$\nNow we can easily prove by induction that $a_0$ divides $a_1, a_2, \\dots, a_k$. Indeed, having established this fact for $a_1, a_2, \\dots, a_l$ we write\n$$\n\\begin{align*}\n0 &= a_0(a_0 b_{l+1} + a_1 b_l + \\dots + a_l b_1 + a_{l+1} b_0) = \\\\\n&= a_0^2 b_{l+1} + a_0 a_1 b_l + \\dots + a_0 a_l b_1 + 6 a_{l+1},\n\\end{align*}\n$$\nand hence\n$$\n6 a_{l+1} = -(a_0^2 b_{l+1} + a_0 a_1 b_l + \\dots + a_0 a_l b_1).\n$$\nWe note that all the summands on the right hand side are divisible by $a_0^2$, therefore the left hand side also has this property. Hence $a_0 \\mid a_{l+1}$.\nBut we have $a_k = \\pm 1$; hence $a_0 = \\pm 1$ and with no loss of generality we may take $a_0 = 1$ and then we have $b_0 = 6$.\nNow if we repeat the above arguments for the coefficients of $Q$, then we see that $b_0 = 6$ divides $b_1, b_2, \\dots, b_{n-3}$ (we set, if necessary, $b_l = 0$ if $l > n-k$). Then we obtain a contradiction ($b_{n-k} = \\pm 1$), unless $n-k > n-3$. We consider two cases.\nThe case of $k=2$. Then we arrive at\n$$\na_0 b_{n-k} + a_1 b_{n-k-1} + \\dots + a_{n-k-1} b_1 + a_{n-k} b_0 = 2,\n$$\na contradiction, because on the left hand side all the summands, except for the first one, are divisible by 6 and hence even, and the first summand is equal to $\\pm 1$.\nThe case of $k=1$. Then the problem reduces to finding an integer root of the polynomial $W$; it is easy to check that if $n$ is even there are no such roots; if $n$ is odd, $W(-1) = 0$.\nTherefore the answer to the problem is: $n$ is odd.",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 29317,
    "subject": "Mathematics (Multi-modal)",
    "question": "We distribute $n \\ge 1$ labelled balls among nine persons $A$, $B$, $C$, $D$, $E$, $F$, $G$, $H$, $I$. Determine in how many ways it is possible to distribute the balls under the condition that $A$ gets the same number of balls as the persons $B$, $C$, $D$ and $E$ together.",
    "options": [],
    "answer": "binom(2n, n) * 2^n",
    "solution": "Consider the polynomial\n$$\n(x+2)^{2n} = (x^2 + 4x + 4)^n = \\\\ = (x^2 + x + x + x + x + 1 + 1 + 1 + 1)(x^2 + x + x + x + x + 1 + 1 + 1 + 1) \\dots \\\\ (x^2 + x + x + x + x + 1 + 1 + 1 + 1)\n$$\nand suppose that we multiply out the brackets, obtaining $9^n$ summands. We show the one-to-one correspondence between the number of $x^n$'s and the number of the distributions we deal with in the problem.\nSuppose we have such a distribution. If the $k$-th ball goes to $A$, we pick $x^2$ from the $k$-th bracket. If it goes to $B$, $C$, $D$, $E$, we pick the first, second, third or fourth $1$, respectively, from the $k$-th bracket. If it goes to $F$, $G$, $H$, $I$, then we take the first, second, third or fourth $x$ from the $k$-th bracket. Now if we multiply the factors we have chosen, we see, that the result is equal to $x^n$ if and only if $A$ gets the same number of balls as $B$, $C$, $D$, $E$ jointly.\nTherefore, the number of the distributions we are interested in is equal to the coefficient at $x^n$ in the polynomial $(x+2)^{2n}$, that is,\n$$\n\\binom{2n}{n} \\cdot 2^n.\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 29318,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABCD$ be a given convex quadrilateral. Determine the locus of the points $P$ lying inside the quadrilateral $ABCD$ and satisfying\n$$\n[PAB] \\cdot [PCD] = [PBC] \\cdot [PDA],\n$$\nwhere $[XYZ]$ denotes the area of triangle $XYZ$.",
    "options": [],
    "answer": "The locus consists of the two diagonals AC and BD.",
    "solution": "If $P$ lies on one of the diagonals $AC$ or $BD$, let's say on $AC$, then\n$$\n\\frac{[PAB]}{[PBC]} = \\frac{AP}{PC} = \\frac{[PDA]}{[PCD]},\n$$\nwhich is the desired equality. We prove that no other point lying inside $ABCD$ satisfies the conditions of the problem.\n\nDenote by $O$ the point of the intersection of the diagonals $AC$ and $BD$ and suppose that $P$ lies inside the triangle $ABO$. Let moreover $BP$ and $AC$ meet at $Q$ and $DP$ and $AC$ meet at $R$. Then\n$$\n\\frac{[PAB]}{[PBC]} = \\frac{AQ}{QC} \\quad \\text{and} \\quad \\frac{[PDA]}{[PCD]} = \\frac{AR}{RC},\n$$\nwhich, since $Q \\neq R$, implies that the given equality cannot hold.\n\nTherefore the desired locus of the points $P$ consists of the diagonals $AC$ and $BD$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 29319,
    "subject": "Mathematics (Multi-modal)",
    "question": "Determine all pairs of integers $(x, y)$ satisfying the equation\n$$\ny(x + y) = x^3 - 7x^2 + 11x - 3.\n$$",
    "options": [],
    "answer": "{(6, 3), (6, -9), (1, 1), (1, -2), (2, -1)}",
    "solution": "The considered equation is equivalent to\n$$\n(2y + x)^2 = 4x^3 - 27x^2 + 44x - 12 = (x - 2)(4x^2 - 19x + 6) = (x - 2)((x - 2)(4x - 11) - 16).\n$$\nThe expression above must be a perfect square. Therefore we have either $x = 2$ (and $y = -1$), or $(x - 2) = ks^2$, where $k \\in \\{-2, -1, 1, 2\\}$ and $s \\in \\mathbb{N}$; indeed, if for some prime $p$ and nonnegative integer $m$ the number $p^{2m+1}$ divides $(x-2)$ but $p^{2m+2}$ does not, then we have $p \\mid (x-2)(4x-11) - 16$, so $p \\mid 16$ and $p = 2$.\nWe will consider three cases separately.\n\nThe case of $k = \\pm 2$. Then we have $4x^2 - 19x + 6 = \\pm 2u^2$ for some integer $u$, or, equivalently,\n$$\n(8x - 19)^2 - 265 = \\pm 32u^2,\n$$\na contradiction modulo $5$.\n\nThe case of $k = 1$. Then $4x^2 - 19x + 6 = u^2$ for some integer $u$, which leads to\n$$\n265 = (8x - 19)^2 - 16u^2 = (8x - 19 - 4u)(8x - 19 + 4u).\n$$\nWe easily check that $x = 6$ is the only solution of this equation (we simply consider all possible decompositions: $265 = 1 \\cdot 265 = 5 \\cdot 53 = \\dots$, etc, and take into account the fact that $x - 2 = s^2$). Therefore, we obtain two solutions of the original equation: $(x, y) \\in \\{(6, 3), (6, -9)\\}$.\n\nThe case of $k = -1$. As before, we have $4x^2 - 19x + 6 = -u^2$, which is equivalent to\n$$\n265 = (8x - 19)^2 + (4u)^2\n$$\nand we check all possibilities with $u \\le 4$: for $u = 0, 1, 2$ there are no solutions. If $u = 3$, then we obtain $(8x - 19)^2 = 121 = 11^2$, which leads to $x = 1$, which gives two solutions: $(x, y) \\in \\{(1, 1), (1, -2)\\}$. Finally, for $u = 4$ we arrive at $(8x - 19)^2 = 9 = 3^2$ and $x = 2$, which gives $(x, y) = (2, -1)$.\n\nTherefore, the set of solutions is as follows:\n$$\n\\{(6, 3), (6, -9), (1, 1), (1, -2), (2, -1)\\}.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 29320,
    "subject": "Mathematics (Multi-modal)",
    "question": "There are 5 distinct points $A$, $B$, $C$, $D$, $E$ lying in this order on a circle with radius $r$ satisfying $AC = BD = CE = r$. There is a triangle having ortocentres of triangles $ACD$, $BCD$, $BCE$ as its vertices. Prove that this triangle is right-angled.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "In any obtuse triangle $XYZ$ with obtuse angle in $Z$ and ortocentre $W$, angles $XYZ$ and $XWZ$ are equal, as complementing the angle $YXW$ to $90$ degrees (fig. 1). Moreover, points $Y$ and $W$ lie in different half-planes determined by $XZ$.\n\n![](attached_image_1.png)\nFig. 1\n\nLet $P$, $Q$, $R$ be ortocentres of given triangles in that order. We will show $\\not\\leq PQR = 90^\\circ$. Obviously all three triangles are obtuse in $C$. So $P$, $Q$, $R$ lie on extensions of altitudes going through $C$ to corresponding sides. Because of position of these sides it is also obvious the ray $CQ$ lies \"between\" rays $CP$ and $CR$, i.e. in angle $PCR$. So $\\not\\leq PQR = \\not\\leq RQC + \\not\\leq PQC$ (fig. 2). By the fact in the first paragraph $Q$, $R$ lie in\n\n![](attached_image_2.png)\nFig. 2\n\nthe same half-plane determined by the line $BC$ and\n$$\n\\not\\leq BEC = \\not\\leq BRC \\quad \\text{and} \\quad \\not\\leq BDC = \\not\\leq BQC.\n$$\nAngles $BEC$, $BDC$ are equal being inscribed angles with the same chord $BC$. Thus also $\\not\\leq BRC = \\not\\leq BQC = \\omega$ and $BCRQ$ is cyclic. So $\\not\\leq RQC = \\not\\leq RBC = \\varphi$. As $EC = r$ for inscribed angle we have $\\not\\leq EBC = 30^\\circ$. Let $U$ be the foot on $BE$ in triangle $BEC$. Counting the angles in right triangle $BUR$ we get\n$$\n\\omega + \\varphi + 30^\\circ + 90^\\circ = 180^\\circ, \\quad \\text{i.e.} \\quad \\not\\leq RQC = \\varphi = 60^\\circ - \\omega = 60^\\circ - \\not\\leq BDC.\n$$\nIn the same way we conclude $\\not\\leq PQC = 60^\\circ - \\not\\leq DBC$. So we have (using the sum of angles in triangle $BCD$ is $180^\\circ$)\n$$\n\\not\\leq PQR = \\not\\leq RQC + \\not\\leq PQC = 120^\\circ - (\\not\\leq BDC + \\not\\leq DBC) = \\not\\leq BCD - 60^\\circ. \\quad (1)\n$$\nBut also $BD = r$, thus $\\not\\leq BCD = 150^\\circ$. Finally by (1) $\\not\\leq PQR = 90^\\circ$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 29321,
    "subject": "Mathematics (Multi-modal)",
    "question": "There are $n$ children around the round-table. Erika is the oldest among them and she has $n$ candies. No other child has any candy. Erika decided to distribute the candies and determined following rules. In every round all the children with at least two candies show. Erika chooses one of them and he/she sends one by one candy to both children sitting next to him/her. (So in the first round only Erika shows and sends one by one candy to her two neighbours.) For which $n \\geq 3$ is it possible to end the distribution after a finite number of rounds with every child having exactly one candy?",
    "options": [],
    "answer": "All odd n ≥ 3",
    "solution": "First we show for $n$ even the distribution never ends with every child having one candy. In every round only two candies change position and they move in two opposite directions. This leads us to studying the entire sum of distances of candies from one child, say Erika. We label the seats in clockwise direction by numbers $0, 1, \\dots, n-1$ (as the distance from Erika in this direction). After every round we sum up the distances of all candies. Let the sum be $S$ (i.e. with every candy we add to $S$ the number of the seat where the candy actually is). If in the given round Erika chooses child on the seat labeled by $k$, whereby $1 \\leq k \\leq n-2$, the value of $S$ does not change - we sum up $(k-1)+(k+1)$ instead of $2k$. If she chooses child on the seat labeled by $n-1$, we sum up $(n-2)+0$ instead of $2(n-1)$, hence $S$ decreases by $n$. Finally, if she chooses herself, we sum up $(n-1)+1$ instead of $2 \\cdot 0$, hence $S$ increases by $n$. At the beginning we have $S = 0$ and its value can change only by $\\pm n$, hence $S$ remains divisible by $n$ after every round. Thus $S/n$ is permanently integer. But in the case when every child has exactly one candy we have\n$$\nS = 0 + 1 + 2 + \\dots + (n-1) = \\frac{n(n-1)}{2}, \\quad \\text{i.e.} \\quad \\frac{S}{n} = \\frac{n-1}{2},\n$$\nwhich is not integer for $n$ even. Hence such situation never happens.\n\nFor odd $n$ we show there is a distribution ending with every child having one candy. Let $n = 2k + 1$. We find proper distribution by induction. More precisely we prove that for every $i = 0, 1, \\dots, k$ we can get position with $n - 2i$ candies by Erika, one candy by first $i$ children sitting to the left side of her, and one candy by first $i$ children to the right. The value $i = 0$ represents the beginning of distribution, the value $i = 1$ the status after the first round (and thus the first step of induction), and the value $i = k$ the status we would like to reach. Suppose we managed to get the specified position for some value $i = m$ with $1 \\leq m < k$ (and we passed through all positions for $i < m$). From this status we proceed as follows. First Erika gives one candy to her neighbours (as $m < k$, she has at least three candies thus she can do it). The following rounds are illustrated in the scheme. (The numbers are for amounts of candies by Erika and children on the right side of her. On the left side we proceed simultaneously in the same way.)\n$$\n\\begin{array}{l}\n\\underline{n-2m, 1, \\dots, 1, 0, \\dots} \\rightarrow \\underline{n-2m-2, 2, 1, \\dots, 1, 0, \\dots} \\rightarrow \\underline{n-2m, 0, 2, 1, \\dots, 1, 0, \\dots} \\rightarrow \\\\\n\\rightarrow \\underline{n-2m, 1, 0, 2, 1, \\dots, 1, 0, \\dots} \\rightarrow \\underline{n-2m, 1, 1, 0, 2, 1, \\dots, 1, 0, \\dots} \\rightarrow \\dots \\\\\n\\dots \\rightarrow \\underline{n-2m, 1, \\dots, 1, 0, 2, 0, \\dots} \\rightarrow \\underline{n-2m, 1, \\dots, 1, 0, 1, 0, \\dots} \\\\\n\\qquad \\rightarrow \\underline{m-1}\n\\end{array}\n$$\nWe get the status, when Erika has $n-2m$ candies, first $m-1$ children on both sides have one candy, the $m$-th children have no candy, and the $(m+1)$-st children have one candy. To reach the position for $i=m+1$, it is sufficient to deliver candies to the children on the $m$-th seats. But here we can use induction hypothesis. That is if we look apart from candies on the $(m+1)$-st seats, we obtain exactly the position for $i=m-1$ (Erika has two candies less, but she still has at least three and thus we can do the same steps). From this position we know how to proceed to the position for $i=m$. Restoring back prescinded candies we get the position for $i=m+1$.\n\nThus, finally we also reach the position for $i=k$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 29322,
    "subject": "Mathematics (Multi-modal)",
    "question": "The sum of four real numbers is $9$, the sum of their squares is $21$. Prove that these numbers can be signed $a$, $b$, $c$, $d$ so that inequality $ab - cd \\geq 2$ holds.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let the numbers be $p$, $q$, $r$, $s$. Up to a permutation we may assume that $p \\geq q \\geq r \\geq s$. We first consider the case where $p+q \\geq 5$. Then\n$$\np^2 + q^2 + 2pq \\geq 25 = 4 + (p^2 + q^2 + r^2 + s^2) \\geq 4 + p^2 + q^2 + 2rs,\n$$\nwhich is equivalent to $pq - rs \\geq 2$.\n\nAssume now that $p+q<5$; then\n$$\n4 < r + s \\leq p + q < 5. \\tag{1}\n$$\nObserve that\n$$\n(pq + rs) + (pr + qs) + (ps + qr) = \\frac{(p+q+r+s)^2 - (p^2+q^2+r^2+s^2)}{2} = 30.\n$$\nMoreover,\n$$\npq + rs \\geq pr + qs \\geq ps + qr,\n$$\nbecause $(p-s)(q-r) \\geq 0$ and $(p-q)(r-s) \\geq 0$.\nWe conclude that $pq + rs \\geq 10$. From (1), we get $0 \\leq (p+q) - (r+s) < 1$, therefore\n$$\n(p+q)^2 - 2(p+q)(r+s) + (r+s)^2 < 1.\n$$\nAdding this to $(p+q)^2 + 2(p+q)(r+s) + (r+s)^2 = 9^2$ gives\n$$\n(p+q)^2 + (r+s)^2 < 41.\n$$\nTherefore\n$$\n41 = 21 + 2 \\cdot 10 \\leq (p^2 + q^2 + r^2 + s^2) + 2(pq + rs) = (p+q)^2 + (r+s)^2 < 41,\n$$\nwhich is a contradiction.\n\n\nSecond solution.\n\nFrom $a+b+c+d=9$ with ordering $a \\ge b \\ge c \\ge d$ we have\n$$\n\\frac{a+b}{2} = \\frac{9}{4} + \\varepsilon_1, \\quad \\frac{c+d}{2} = \\frac{9}{4} - \\varepsilon_1\n$$\nfor some $\\varepsilon_1 \\ge 0$. Thus\n$$\na = \\frac{9}{4} + \\varepsilon_1 + \\varepsilon_2, \\quad b = \\frac{9}{4} + \\varepsilon_1 - \\varepsilon_2, \\quad c = \\frac{9}{4} - \\varepsilon_1 + \\varepsilon_3, \\quad d = \\frac{9}{4} - \\varepsilon_1 - \\varepsilon_3\n$$\nfor some $\\varepsilon_2, \\varepsilon_3 \\ge 0$. From $b \\ge c$ we get\n$$\n\\varepsilon_1 - \\varepsilon_2 \\geq -\\varepsilon_1 + \\varepsilon_3 \\quad \\text{or} \\quad \\varepsilon_2 + \\varepsilon_3 \\leq 2\\varepsilon_1.\n$$\nFrom\n$$\n21 = (a^2 + b^2) + (c^2 + d^2) = 2 \\cdot \\left(\\frac{9}{4} + \\varepsilon_1\\right)^2 + 2\\varepsilon_2^2 + 2 \\cdot \\left(\\frac{9}{4} - \\varepsilon_1\\right)^2 + 2\\varepsilon_3^2 = \\\\\n= 4 \\cdot \\left(\\frac{9}{4}\\right)^2 + 4\\varepsilon_1^2 + 2\\varepsilon_2^2 + 2\\varepsilon_3^2 = 20 + \\frac{1}{4} + 2 \\cdot (2\\varepsilon_1^2 + \\varepsilon_2^2 + \\varepsilon_3^2)\n$$\nwe conclude that non-negative numbers $\\varepsilon_i$ satisfy\n$$\n2\\varepsilon_1^2 + \\varepsilon_2^2 + \\varepsilon_3^2 = \\frac{3}{8}. \\tag{2}\n$$\nUsing $\\varepsilon_2 + \\varepsilon_3 \\le 2\\varepsilon_1$ and $\\varepsilon_2^2 + \\varepsilon_3^2 \\le (\\varepsilon_2 + \\varepsilon_3)^2$ from (2) we obtain\n$$\n\\frac{3}{8} \\leq 2\\varepsilon_1^2 + (\\varepsilon_2 + \\varepsilon_3)^2 \\leq 2\\varepsilon_1^2 + 4\\varepsilon_1^2 = 6\\varepsilon_1^2.\n$$\nHence $\\varepsilon_1^2 \\ge (1/6) \\cdot (3/8) = 1/16$ or $\\varepsilon_1 \\ge 1/4$. For $ab-cd$ we have\n$$\nab - cd = \\left(\\frac{9}{4} + \\varepsilon_1\\right)^2 - \\varepsilon_2^2 - \\left(\\frac{9}{4} - \\varepsilon_1\\right)^2 + \\varepsilon_3^2 = 9\\varepsilon_1 - \\varepsilon_2^2 + \\varepsilon_3^2.\n$$\nSubstituting for $\\varepsilon_2^2$ from (2) we get\n$$\nab - cd = 9\\varepsilon_1 - \\left(\\frac{3}{8} - 2\\varepsilon_1^2 - \\varepsilon_3^2\\right) + \\varepsilon_3^2 = 9\\varepsilon_1 + 2\\varepsilon_1^2 - \\frac{3}{8} + 2\\varepsilon_3^2 \\ge 9 \\cdot \\frac{1}{4} + 2 \\cdot \\frac{1}{16} - \\frac{3}{8} = 2.\n$$\nThis ends the proof.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 29323,
    "subject": "Mathematics (Multi-modal)",
    "question": "Prove, that for every integer $k \\ge 1$ there is a positive integer $n$ with following property: in decimal representation of $2^n$ one can find a block of exactly $k$ consecutive zeros, i.e.\n$$\n2^n = \\dots a \\overline{00\\dots0} b\\dots, \\quad \\text{ $k$ zeros }\n$$\nwhere $a$, $b$ are nonzero digits.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "First we show there are arbitrarily long blocks of zeros in powers of $2$. To get at least $k$ zeros in $2^n$, this power must be of the form $y \\cdot 10^{m+k} + z$ with $y, z$ positive integers and $z$ having at most $m$ digits, i.e. $z < 10^m$. Thus it is sufficient to find $n, m$ with $2^n$ having residue less than $10^m$ modulo $10^{m+k}$. By Euler's theorem for every positive integer $t$ we have (as $(2, 5^t) = 1$)\n$$\n2^{\\varphi(5^t)} \\equiv 1 \\pmod{5^t}.\n$$\nMultiplying by $2^t$ we obtain\n$$\n2^{t+\\varphi(5^t)} \\equiv 2^t \\pmod{10^t}, \\quad \\text{thus} \\quad 2^{t+\\varphi(5^t)} \\equiv y \\cdot 10^t + 2^t\n$$\nfor some positive integer $y$. By previous we set $n = t + \\varphi(5^t)$ and $m = t - k$. We would like to have $t$ with $2^t < 10^{t-k}$. Such a value of $t$ exists definitely, e.g. $t = 2k$ (since $2^{2k} = 4^k < 10^k$). It follows by presented, that in\n$$\n2^{2k+\\varphi(5^{2k})} = y \\cdot 10^{2k} + 2^{2k}\n$$\nthere is a block of at least $k$ zeros.\n\nLet us take (for a given $k$) power of $2$ (say $2^n$) containing a block of exactly $r$ zeros with $r \\ge k$. We will study what happens to this block when considering subsequent powers, i.e. when multiplying the number with block by $2$. For some nonzero digits $a, b$ we have\n$$\n2^n = \\underbrace{\\dots a \\ 0 \\ 0 \\ \\dots}_{y} \\underbrace{\\dots 0 \\ b \\ \\dots}_{r \\text{ zeros}} \\underbrace{\\dots}_{z} = y \\cdot 10^{r+s} + z.\n$$\nThus $2^{n+1} = 2y \\cdot 10^{r+s} + 2z$. The number $2z$ has either the same number of digits as $z$, or one more. Hence at the \"right side\" the block of zeros either does not cut down, or cut down by one zero. At the \"left side\" the block can only extend (when $y$ divisible by $5$). Globally the length of the block either decreases by $1$, or does not change, or increases. Similarly when we will multiply by $2$ again and again, the length of the block in every step decreases at most by $1$. Thus the only possibility how to avoid block of length $k$ is to remain the length more than $k$. This is anyway impossible. Namely $y$ has in its prime factorization the prime $5$ with some exponent, say $\\alpha$. When we $\\alpha$ times multiply $2^n$ by $2$, under the next multiplying the block will not extend at the \"left side\". At the \"right side\" at least under every fourth multiplying the block cut down (since $2^4 > 10$). Hence after sufficient number of steps we obtain the power of $2$ with a block of exactly $k$ zeros.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 29324,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find out the number of sequences $(a_n)_{n=1}^{\\infty}$ of integer numbers, which satisfy\n$$\na_n \\neq -1 \\quad \\text{and} \\quad a_{n+2} = \\frac{a_n + 2006}{a_{n+1} + 1}\n$$\nfor every positive integer $n$.",
    "options": [],
    "answer": "14",
    "solution": "Every sequence satisfying given conditions is determined by first two terms. Thus we are looking for integer pairs $(a_1, a_2)$, for which all the other terms are integers. Writing out the formula for several small values of $n$ and multiplying it we obtain\n$$\n\\begin{align*}\na_3(a_2 + 1) &= a_1 + 2006, \\\\\na_4(a_3 + 1) &= a_2 + 2006, \\\\\na_5(a_4 + 1) &= a_3 + 2006,\n\\end{align*}\n$$\nSubtracting adjacent equalities (to remove number 2006) and rearranging gives\n$$\n\\begin{align*}\na_3 - a_1 &= (a_3 + 1)(a_4 - a_2), \\\\\na_4 - a_2 &= (a_4 + 1)(a_5 - a_3), \\\\\na_5 - a_3 &= (a_5 + 1)(a_6 - a_4),\n\\end{align*} \\tag{1}\n$$\n\nAll the terms $(a_n + 1)$ are nonzero by definition. Hence two possibilities can happen. If $a_3 - a_1 = 0$, by substituting into previous we get (step by step) $a_4 - a_2 = 0$, $a_5 - a_3 = 0$, ..., i.e.\n$$\na_1 = a_3 = a_5 = \\dots \\quad \\text{and} \\quad a_2 = a_4 = a_6 = \\dots \\qquad (2)\n$$\nOn the other side, if $a_3 - a_1 \\neq 0$, by the same substituting we obtain $a_4 - a_2 \\neq 0$, $a_5 - a_3 \\neq 0$, ... First have a look at the second case. By (1) we have\n$$\n0 < |a_{n+3} - a_{n+1}| = |a_{n+2} - a_n| \\cdot \\frac{1}{|a_{n+2} + 1|} \\le |a_{n+2} - a_n| \\qquad (3)\n$$\nfor all $n \\ge 1$. Thus we have non-increasing sequence of positive integers\n$$\n|a_3 - a_1| \\ge |a_4 - a_2| \\ge |a_5 - a_3| \\ge \\dots\n$$\nObviously this sequence is constant after some term (otherwise we could select infinite decreasing subsequence of positive integers, which is anyway impossible). Thus there is some index $N$ and value $d$ with $|a_{n+2} - a_n| = d$ for all $n \\ge N$. By (3) then $|a_{n+2} + 1| = 1$, i.e. for $n \\ge N+2$ we have $a_n \\in \\{0, -2\\}$. But by definition\n$$\na_{N+4} = \\frac{a_{N+2} + 2006}{a_{N+3} + 1},\n$$\nhence $a_{N+4}$ is one of the values\n$$\n\\frac{0+2006}{0+1} = 2006, \\quad \\frac{0+2006}{-2+1} = -2006, \\quad \\frac{-2+2006}{0+1} = 2004, \\quad \\frac{-2+2006}{-2+1} = -2004.\n$$\nThis is in contrary with $a_{N+4} \\in \\{0, -2\\}$. In this case there is no sequence satisfying given conditions. Therefore every such sequence satisfies (2). Substituting $n=1$ and $a_3 = a_1$ in definition we get\n$$\na_1 = \\frac{a_1 + 2006}{a_2 + 1} \\quad \\text{or} \\quad a_1 a_2 = 2006 = 2 \\cdot 17 \\cdot 59.\n$$\nConsidering $a_1, a_2 \\neq -1$ we obtain\n$$\na_1 \\in \\{1, \\pm 2, \\pm 17, \\pm 34, \\pm 59, \\pm 118, \\pm 1003, 2006\\} \\quad \\text{and} \\quad a_2 = \\frac{2006}{a_1}.\n$$\nIt can be easily verified every such sequence $a_1, a_2, a_3, a_4, a_5, \\dots$ satisfies given condition. The number of sequences is 14.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 29325,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find out if there is a convex pentagon $A_1A_2A_3A_4A_5$, such that for $i = 1, 2, 3, 4, 5$ the lines $A_iA_{i+3}$, $A_{i+1}A_{i+2}$ are not parallel and intersect in a point $B_i$ and also the points $B_1, B_2, B_3, B_4, B_5$ are collinear. (We assume $A_6 = A_1, A_7 = A_2, A_8 = A_3$.)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "We will find such a pentagon. The obstruction is, symmetric pentagons (for which it could be easier to show) have always at least one pair of lines $A_iA_{i+3}$, $A_{i+1}A_{i+2}$ parallel. First we will solve more elementary problem - we will find pentagon $A_1A_2A_3A_4A_5$ with only four points $B_i$ collinear. Here we afford to look for it among symmetric pentagons. To simplify the situation, say $A_2, A_3, A_4$ be vertices of the square $QA_2A_3A_4$ with $QA_2 = 1$ and $A_1, A_5$ be on sides $QA_2$ and $QA_4$ with $QA_1 = QA_5 = p$ (fig. 3). By symmetry $B_1B_2 \\parallel B_3B_5$. It can be observed when $p \\to 0$, i.e. when $A_1, A_5$ be near $Q$, the\n\n![](attached_image_1.png)\n\nline $B_3B_5$ is more closely to $Q$ than $B_1B_2$. On the other side, when $p \\to 1$, points $A_1, A_5$ be near $A_2$, $A_4$ and $B_1B_2$ is more closely to $Q$ (or even on the other side of $Q$) than $B_3B_5$. Thus we can expect for some $p \\in (0, 1)$ both lines be identical and $B_1, B_2, B_3, B_5$ collinear. We will find such $p$.\n\nLet $B_5Q = B_3Q = q$ and $B_1A_2 = r$. By similarity of triangles $B_5QA_5, B_5A_2A_3$ we get\n$$\n\\frac{q}{p} = \\frac{q+1}{1} \\quad \\text{or} \\quad q = \\frac{p}{1-p}.\n$$\nBy similarity of triangles $B_1A_2A_1, B_1A_3A_4$ we get\n$$\n\\frac{r}{1-p} = \\frac{r+1}{1} \\quad \\text{or} \\quad r = \\frac{1-p}{p}.\n$$\nFinally, to reach $B_1$ lying on the line $B_3B_5$, it suffices triangles $B_5QB_3, B_5A_2B_1$ be similar. This holds when\n$$\n\\frac{q}{q} = \\frac{q+1}{r} \\quad \\text{or} \\quad q+1 = r.\n$$\nSubstituting by previous we obtain the equation\n$$\n\\frac{p}{1-p} + 1 = \\frac{1-p}{p}\n$$\nwhich can be transformed into $p^2 - 3p + 1 = 0$. The only solution in $(0, 1)$ is $p = (3 - \\sqrt{5})/2$. For this value points $B_1, B_3, B_5, B_2$ lies on one line. Moreover lines $A_1A_5, A_2A_4$ are parallel to it. In some sense these three lines intersect \"in infinity\" in \"point\" $B_4$ and all points $B_i$ be collinear. To satisfy given conditions it suffices to find proper map, which maps \"point from infinity\" to particular point (and preserves all the other necessary properties, e.g. maps lines to lines etc.). Such a map is projection (fig. 4). Consider ordinary cartesian coordinate system in space. Pentagon $A_1A_2A_3A_4A_5 = U$ can be imbedded\n\n![](attached_image_2.png)\n\ninto the plane $Oyz$ with $A_2$ be in origin and with $A_1$, $A_3$ be on positive axis $z$, $y$. Let $P \\equiv (2,0,-1)$ be the projection-point. Every line $PA_i$ intersects the plane $Oxy$ in point $A'_i$. Thus we get pentagon $A'_1A'_2A'_3A'_4A'_5 = U'$. It follows from the properties of the map, that $U'$ satisfies conditions given in the formulation of the problem. To check this, it is also sufficient to do some calculation. Namely one can easily enumerate the coordinates of $A'_i$ in the plane $Oxy$ and get\n$$\nA'_1 \\equiv (3 - \\sqrt{5}, 0), \\quad A'_2 \\equiv (0, 0), \\quad A'_3 \\equiv (1, 0), \\quad A'_4 \\equiv (1, \\frac{1}{2}), \\quad A'_5 \\equiv (1, \\frac{3-\\sqrt{5}}{4}).\n$$\nBy another manual calculations it is easy to check $B'_1, B'_2, B'_3, B'_4, B'_5$ be collinear (fig. 5).\n\n![](attached_image_3.png)",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 29326,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all polynomials $P$ with real coefficients for which the equality\n$$\nP(x^2) = P(x) \\cdot P(x+2)\n$$\nholds for every real number $x$.",
    "options": [],
    "answer": "P(x) = 0, P(x) = 1, or P(x) = (x - 1)^n for any positive integer n",
    "solution": "The constant polynomial $P(x) = c$ is a solution if and only if $c = c^2$, thus the polynomials $P(x) = 0$ and $P(x) = 1$ are solutions of the problem.\nWe claim that the only polynomial of a positive degree $n$ which solves the equation is of the form $P(x) = (x-1)^n$. In view of the identity $(x^2 - 1)^n = (x-1)^n(x+1)^n$, the latter is clearly a solution for any $n \\ge 1$.\nIf $a x^n$ ($a \\neq 0$) is the leading term of a polynomial $P(x)$ of a positive degree $n$, then $a x^{2n}$ is the leading term of the polynomial $P(x^2)$ and $a^2 x^{2n}$ is the leading term of the polynomial $P(x) P(x+2)$. If $P$ satisfies the given equality, comparing the leading order terms thus gives $a = a^2$, hence $a = 1$. The polynomial $P$ can therefore be written in the form $P(x) = (x-1)^n + Q(x)$, where $Q$ is either identically zero, or is a nonzero polynomial of degree $k$, where $0 \\le k < n$. Comparing the polynomials\n$$\nP(x^2) = (x^2 - 1)^n + Q(x^2), \\\\\nP(x) P(x+2) = [(x-1)^n + Q(x)][(x+1)^n + Q(x+2)]\n$$\nwe obtain (upon multiplying out the brackets and cancelling the terms $(x^2 - 1)^n$ on both sides) the equality\n$$\nQ(x^2) = (x-1)^n Q(x+2) + (x+1)^n Q(x) + Q(x) Q(x+2).\n$$\nThe zero polynomial $Q$ clearly satisfies this relation. For a nonzero $Q$ of degree $k < n$, however, $Q(x^2)$ is a polynomial of degree $2k$, while on the right-hand side of the last equation there is a polynomial of degree $n+k$ (whose leading term is $2b x^{n+k}$, if $b x^k$ is the leading order term of the polynomial $Q(x)$). Since $2k < n+k$, this is not possible.\n**Conclusion.** The solutions are the constant polynomials $P(x) = 0$ and $P(x) = 1$ and the polynomial $P(x) = (x-1)^n$ for any natural number $n$.",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 29327,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $a_1 = a_2 = 1$ and $a_{k+2} = a_{k+1} + a_k$ for any $k \\in \\mathbb{N}$ (the Fibonacci sequence). Prove that for any natural number $m$ there exists an index $k$ such that the number $a_k^4 - a_k - 2$ is divisible by $m$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "All the congruences and remainder classes below are meant mod $m$. We obtain the desired congruence relation $a_k^4 - a_k - 2 \\equiv 0$ as a consequence of the simpler relation $a_k \\equiv -1$.\n\nThe sequence of remainder classes of the numbers $a_k$ has the following property: the remainder classes of any two consecutive elements $a_k$, $a_{k+1}$ determine uniquely the remainder classes of all subsequent elements $a_i$ ($i > k + 1$), as well as of all elements $a_i$ ($i < k$) preceding them. By the standard argument, based on the fact that the number of ordered pairs of remainder classes is $m^2$, hence finite, it follows that the sequence of remainder classes of the elements $a_i$ is periodic, starting already from its first member. Thus there exists a number $p > 0$ (depending on the given modulus $m$) such that $a_i \\equiv a_{i+p}$ for any index $i$. Unless $m = 1$ (then the problem is trivial), clearly $p > 1$. Since $a_1 \\equiv a_2 \\equiv 1$, we also have $a_{p+1} \\equiv a_{p+2} \\equiv 1$, whence $a_p \\equiv 0$ and $a_{p-1} \\equiv -1$, so we can take $k = p-1$ and the proof is finished.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 29328,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $k$ be the circumcircle of a given convex quadrilateral $ABCD$ with the property that the half-lines $DA$ and $CB$ meet at a point $E$ for which $|CD|^2 = |AD| \\cdot |ED|$ holds. Let us denote by $F$ ($F \\neq A$) the point of intersection of the circle $k$ with the perpendicular to $ED$ at $A$. Prove that the segments $AD$ and $CF$ are congruent if and only if the circumcenter of the triangle $ABE$ lies on $ED$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Clearly $DF$ is a diameter of $k$. First we show that under the given conditions the vertex $C$ cannot lie in the half-plane $DFA$.\nIf the vertices $B$, $C$ are points on the subarc $DA$ of the arc $DAF$ (Fig. 1) then the angles $DCB$ and $DBA$ are obtuse, hence $|DC| < |DB| < |DA| < |DE|$, which contradicts to the equality $|CD|^2 = |AD| \\cdot |ED|$.\nIf the vertices $B$, $C$ are points on the subarc $AF$ of the arc $DAF$ (Fig. 2) the angle $BAE$ is acute and $|\\angle DBE| = 180^\\circ - |\\angle DBC| \\le 90^\\circ$, so the possible other meeting point $B'$ of the half-line $DB$ with the circumcircle of the triangle $AEB$ lies in the segment $DB$. Hence $|DC| > |DB| \\ge |DB'|$. This means that the equality $|CD|^2 = |AD| \\cdot |ED|$ cannot hold as $|AD| \\cdot |ED| = |DB| \\cdot |DB'|$ (which is the power of $D$ with respect to the circumcircle of the triangle $AEB$).\n\n![](attached_image_1.png)\nFig. 1\n\n![](attached_image_2.png)\nFig. 2\n\nWe have shown that the vertex $C$ of the given quadrangle does not lie in the half-plane $FDA$, hence $|FC| = |DA|$ if and only if $DAFC$ is a rectangle, i.e. if and only if $CA$ is a diameter of the circle $k$, which is equivalent to the angle $CBA$ being right, which is in turn equivalent to the triangle $AEB$ being right with the right angle at $B$, i.e. to the circumcenter of the triangle $AEB$ being the midpoint of $AE$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 29329,
    "subject": "Mathematics (Multi-modal)",
    "question": "For any real number $p \\ge 1$ let us consider the set of all real numbers $x$ with\n$$\np < x < \\left(2 + \\sqrt{p + \\frac{1}{4}}\\right)^2\n$$\nProve that from such a given set one can select four mutually different natural numbers $a, b, c, d$ with $ab = cd$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "The numbers $a = (k-1)k$, $b = (k+1)k$, $c = (k-1)(k+1)$, $d = k^2$ clearly satisfy the equality $ab = cd$ and the inequalities $a < c < d < b$ for any $k > 1$. Let thus $k$ be the least natural number for which $p < a$, i.e. $p < (k-1)k$ (for a given $p$). We will show that for this $k$ necessarily $b = (k+1)k \\le p + 4 + 2\\sqrt{4p+1}$, which is evidently a number by $\\frac{1}{4}$ smaller than the upper bound of the interval in our problem, so we will be done.\n\nIn view of the choice of the number $k$ we have $p \\ge (k-2)(k-1)$. Solving this quadratic inequality yields the estimate\n$$\nk \\le \\frac{3}{2} + \\sqrt{p + \\frac{1}{4}},\n$$\nfrom which it already follows that\n$$\nb = (k+1)k \\le \\left(\\frac{5}{2} + \\sqrt{p + \\frac{1}{4}}\\right) \\cdot \\left(\\frac{3}{2} + \\sqrt{p + \\frac{1}{4}}\\right) \\\\\n= \\frac{15}{4} + 4\\sqrt{p + \\frac{1}{4}} + \\left(p + \\frac{1}{4}\\right) = p + 4 + 2\\sqrt{4p+1}.\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 29330,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find for which\n$$\nn \\in \\{3\\,900, 3\\,901, 3\\,902, 3\\,903, 3\\,904, 3\\,905, 3\\,906, 3\\,907, 3\\,908, 3\\,909\\}\n$$\nthe set $\\{1, 2, 3, \\dots, n\\}$ can be partitioned into (disjoint) triples in such a way that one of the three numbers in any triple is the sum of the other two.",
    "options": [],
    "answer": "3900 and 3903",
    "solution": "From the possibility of partitioning the set into disjoint triples it follows that $3 \\mid n$. In each triple $\\{a, b, a+b\\}$ the sum of its elements is $2(a+b)$, hence an even number; thus also the sum of all numbers from $1$ to $n$ must be even, i.e. the product $n(n+1)$ must be divisible by four. Altogether it therefore follows that the number $n$ has to be of the form either $12k$ or $12k+3$; from the given set of numbers, this is satisfied only for $n = 3900$ and $n = 3903$.\n\nIn the next paragraph we describe a construction how to produce, starting from a decomposition satisfying the given condition for some $n = k$, a decomposition of the same kind for $n = 4k$ and $n = 4k + 3$. This guarantees that the required decompositions for $n = 3900$ and $n = 3903$ indeed exist, in view of the decreasing sequence\n$$\n3900 \\rightarrow 975 \\rightarrow 243 \\rightarrow 60 \\rightarrow 15 \\rightarrow 3\n$$\n(instead of $3900$ one can start also with $3903$) and the trivial decomposition for $n=3$ (from which we in turn construct the decompositions for $n=15$, $n=60$ etc. up to $n=3900$ or $n=3903$).\n\nFrom a decomposition of the set $\\{1, 2, \\dots, k\\}$ satisfying the given conditions we first produce a similar decomposition for the set of the first $k$ even numbers $\\{2, 4, \\dots, 2k\\}$ (simply by multiplying all the numbers in the triples by two). In the case of $n=4k$ we partition the remaining numbers\n$$\n\\{1, 3, 5, \\dots, 2k-1, 2k+1, 2k+2, \\dots, 4k-1, 4k\\}\n$$\ninto the $k$ triples $\\{2j-1, 3k-j+1, 3k+j\\}$, where $j=1, 2, \\dots, k$. They are shown in the columns of the table below.\n\n$$\n\\begin{pmatrix} 1 & 3 & 5 & \\dots & 2k-3 & 2k-1 \\\\ 3k & 3k-1 & 3k-2 & \\dots & 2k+2 & 2k+1 \\\\ 3k+1 & 3k+2 & 3k+3 & \\dots & 4k-1 & 4k \\end{pmatrix}\n$$\n\nIn the case of $n = 4k + 3$ we partition the remaining numbers\n$$\n\\{1, 3, 5, \\dots, 2k-1, 2k+1, 2k+2, \\dots, 4k+2, 4k+3\\}\n$$\ninto the $k+1$ triples $\\{2j-1, 3k+3-j, 3k+j+2\\}$, where $j=1, 2, \\dots, k+1$; these are again shown in the columns of the table below.\n\n$$\n\\begin{pmatrix} 1 & 3 & 5 & \\dots & 2k-1 & 2k+1 \\\\ 3k+2 & 3k+1 & 3k & \\dots & 2k+3 & 2k+2 \\\\ 3k+3 & 3k+4 & 3k+5 & \\dots & 4k+2 & 4k+3 \\end{pmatrix}\n$$\n\nThis completes the proof of the fact that the solution of the given problem are the numbers $n=3900$ and $n=3903$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 29331,
    "subject": "Mathematics (Multi-modal)",
    "question": "Determine all triples $(x, y, z)$ of positive numbers satisfying the system of equations\n$$\n\\begin{align*}\n2x^3 &= 2y(x^2 + 1) - (z^2 + 1), \\\\\n2y^4 &= 3z(y^2 + 1) - 2(x^2 + 1), \\\\\n2z^5 &= 4x(z^2 + 1) - 3(y^2 + 1).\n\\end{align*}\n$$",
    "options": [],
    "answer": "x = y = z = 1",
    "solution": "For any integer $k \\ge 3$ and any $x \\ge 0$ we have\n$$\n2x^k \\ge [(k-1)x - (k-2)](x^2 + 1). \\quad (0)\n$$\nTo see this, observe that, by AM-GM inequality,\n$$\nx^k + x^k + \\underbrace{x + x + \\dots + x}_{(k-3)\\text{ times}} \\ge (k-1)x^3,\n$$\nand add it to\n$$\n(k-2)(x^2 - 2x + 1) \\ge 0.\n$$\nNote that we have equality if and only if $x=1$.\nTherefore, for $x$, $y$, $z$ satisfying the system of equations, we have\n$$\n\\begin{align*}\n2y(x^2 + 1) - (z^2 + 1) &\\ge (2x-1)(x^2 + 1), \\\\\n3z(y^2 + 1) - 2(x^2 + 1) &\\ge (3y-2)(y^2 + 1), \\\\\n4x(z^2 + 1) - 3(y^2 + 1) &\\ge (4z-3)(z^2 + 1),\n\\end{align*}\n$$\nor\n$$\n\\begin{equation}\n\\begin{aligned}\n& 2(y-x)(x^2+1) + (x-z)(x+z) \\ge 0, \\\\\n& 3(z-y)(y^2+1) + 2(y-x)(y+x) \\ge 0, \\\\\n& 4(x-z)(z^2+1) + 3(z-y)(z+y) \\ge 0.\n\\end{aligned}\n\\tag{1}\n\\end{equation}\n$$\nNow suppose $x \\ge \\max\\{y, z\\}$. Then from the second inequality of (1) we infer that $y \\le z$ and\n$$\n\\begin{align*}\n& 2(y-x)(x^2+1) + (x-z)(x+z) \\le (z-x)(2(x^2+1)-(x+z)) \\\\\n& \\le (z-x)(2x^2-2x+2) \\le 0,\n\\end{align*}\n$$\nwhich, by the first inequality in (1), implies $x = y = z$.\n\nIf $y \\ge \\max\\{x, z\\}$, then $z \\le x$ by the third inequality in (1) and\n$$\n3(z - y)(y^2 + 1) + 2(y - x)(y + x) \\le (x - y)(3(y^2 + 1) - 2y - 2x) \\\\\n\\le (x - y)(3y^2 - 4y + 3) \\le 0,\n$$\nhence, by the second inequality in (1), $x = y = z$.\nFinally, if $z \\ge \\max\\{x, y\\}$, then $x \\le y$ by the first estimate in (1) and, as previously,\n$$\n4(x - z)(z^2 + 1) + 3(z - y)(z + y) \\le (y - z)(4z^2 - 6z + 4) \\le 0,\n$$\nwhich again implies $x = y = z$. Thus we have equality in (0) and hence $x = y = z = 1$. We easily check that this is the solution to the system.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 29332,
    "subject": "Mathematics (Multi-modal)",
    "question": "Given is a convex hexagon $ABCDEF$, such that $\\angle A = \\angle C = \\angle E$ and $AB = BC$, $CD = DE$, $EF = FA$. Prove that the lines $AD$, $BE$ and $CF$ have a common point.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Assume that the angle bisectors of the angles $\\angle B$ and $\\angle D$ intersect at $P$ (Fig. 1). We shall prove that the hexagon $ABCDEF$ has an inscribed circle, whose center is $P$. Then the conclusion follows from Brianchon's Theorem.\nThe equality $AB = BC$ implies that the triangles $ABP$ and $CBP$ are congruent. Hence we have $\\angle BAP = \\angle BCP = x$. Similarly, triangles $CDP$ and $EDP$ are congruent, so we obtain $\\angle DCP = \\angle DEP = y$.\n\n![](attached_image_1.png)\nFig. 1\n\nMoreover, we have $AP = CP = EP$, which together with the equality $AF = EF$ implies that the triangles $AFP$ and $EFP$ are congruent. Thus the angle bisector of the angle $\\angle F$ passes through the point $P$ and $\\angle FAP = \\angle FEP = z$.\nNow the equalities $\\angle A = \\angle C = \\angle E$ are equivalent to $z+x=x+y=y+z$, which yields $x=y=z$. Therefore the angle bisectors of the angles $\\angle A$, $\\angle C$ and $\\angle E$ all pass through the point $P$. Thus $P$ is the center of the inscribed circle of the hexagon $ABCDEF$, as claimed.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 29333,
    "subject": "Mathematics (Multi-modal)",
    "question": "Determine all prime numbers $p$, such that the number\n$$\n\\binom{p}{1}^2 + \\binom{p}{2}^2 + \\dots + \\binom{p}{p-1}^2\n$$\nis divisible by $p^3$.",
    "options": [],
    "answer": "all primes p ≥ 5",
    "solution": "We start from the observation that for $k = 1, 2, \\dots, p$ we have\n$$\n\\binom{p-1}{k-1} \\equiv \\pm 1 \\pmod{p}. \\qquad (1)\n$$\nTo see this, note that\n$$\np-1 \\equiv -1 \\pmod{p},\n$$\n$$\np-2 \\equiv -2 \\pmod{p},\n$$\n$$ \\vdots \\qquad \\vdots $$\n$$\nk \\equiv -(p-k) \\pmod{p},\n$$\nhence, multiplying,\n$$\n\\frac{(p-1)!}{(k-1)!} \\equiv \\pm(p-k)! \\pmod{p}\n$$\nand (1) follows. This can be written in an equivalent form\n$$\n\\frac{k}{p} \\binom{p}{k} \\equiv \\pm 1 \\pmod{p},\n$$\nwhich implies that\n$$\n\\binom{p}{k} = p \\cdot \\frac{a_k p \\pm 1}{k},\n$$\nfor some integer $a_k$. Therefore, if $p$ satisfies the given conditions, we have\n$$\np \\mid \\sum_{k=1}^{p-1} \\frac{(a_k p \\pm 1)^2}{k^2},\n$$\nor $p \\mid \\sum_{k=1}^{p-1} 1/k^2$, where by $1/m$ we understand the unique integer $1 \\le l \\le p-1$ satisfying $ml \\equiv 1 \\pmod{p}$. Now observe that $1/k^2 \\equiv (1/k)^2 \\pmod{p}$ and\n$$\n\\sum_{k=1}^{p-1} \\frac{1}{k^2} \\equiv \\sum_{k=1}^{p-1} k^2 = \\frac{p(p-1)(2p-1)}{6} \\pmod{p},\n$$\nwhich is divisible by $p$ if and only if $p \\ge 5$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 29334,
    "subject": "Mathematics (Multi-modal)",
    "question": "Prove that there exists a positive integer $n$, such that for all integers $k$ the number $k^2 + k + n$ has no prime divisors less than $2008$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $p < 2008$ be a fixed prime number. There exists $r = r(p)$ such that $k^2 + k \\neq r \\pmod{p}$ for any integer $k$; this follows, for example, from the fact that if $k \\equiv 0 \\pmod{p-1}$ (mod $p$), then $k^2 + k \\equiv 0 \\pmod{p}$.\n\nNow if $\\{p_1, p_2, \\dots, p_m\\}$ is the set of all prime numbers not exceeding $2008$, we take $n$ satisfying\n$$\nn \\equiv p_j - r(p_j) \\pmod{p_j}, \\quad j = 1, 2, \\dots, m,\n$$\nwhich exists due to the Chinese Remainder Theorem. This number has the desired property.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 29335,
    "subject": "Mathematics (Multi-modal)",
    "question": "Given is a regular pentagon $ABCDE$. Determine the least value of the expression\n$$\n\\frac{PA + PB}{PC + PD + PE}\n$$\nwhere $P$ is an arbitrary point lying in the plane of the pentagon $ABCDE$.",
    "options": [],
    "answer": "sqrt(5) - 2",
    "solution": "Without loss of generality assume that the given pentagon $ABCDE$ has the side equal to $1$. Then the length of its diagonal is equal to\n$$\n\\lambda = \\frac{1 + \\sqrt{5}}{2}.\n$$\nSet $a = PA$, $b = PB$, $c = PC$, $d = PD$, $e = PE$ (Fig. 2).\n\n![](attached_image_1.png)\n\nFig. 2\n\nApplying the Ptolemy inequality for the (not necessarily convex) quadrilaterals $APDE$, $BPDC$ and $PCDE$ we obtain (respectively)\n$$\na + d \\geqslant e\\lambda, \\quad b + d \\geqslant c\\lambda, \\quad e + c \\geqslant d\\lambda.\n$$\nWe multiply the third inequality by $\\frac{\\lambda+2}{\\lambda+1}$ and add together with the first and the second inequalities. As a result we obtain\n$$\na+b+2d+e \\cdot \\frac{\\lambda+2}{\\lambda+1} + c \\cdot \\frac{\\lambda+2}{\\lambda+1} \\geq e\\lambda + c\\lambda + d \\cdot \\frac{\\lambda(\\lambda+2)}{\\lambda+1}.\n$$\nGrouping the respective terms, the above inequality reduces to\n$$\na+b \\geq \\frac{\\lambda^2-2}{\\lambda+1}(c+d+e).\n$$\nTherefore\n$$\n\\frac{a+b}{c+d+e} \\geq \\frac{\\lambda^2-2}{\\lambda+1} = \\sqrt{5}-2.\n$$\nThe equality holds if and only if the convex quadrilaterals $APDE$, $BPDC$ and $PCDE$ are cyclic. This condition is satisfied if and only if the point $P$ lies on the minor arc $AB$ of the circumcircle of the pentagon $ABCDE$ (Fig. 3). Therefore the smallest possible value of the given expression is $\\sqrt{5}-2$.\n\n![](attached_image_2.png)\n\nFig. 3",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 29336,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all triples $(k, m, n)$ of positive integers with the following property: The square with the side length $m$ can be cut into some number of rectangles of dimensions $1 \\times k$ and exactly one square of the side length $n$.",
    "options": [],
    "answer": "All triples of positive integers satisfy n ≤ m and either k divides m − n, or k divides m + n and r + n ≤ m, where r is the remainder when m is divided by k.",
    "solution": "Answer: The triples $(k, m, n)$ must satisfy $n \\leq m$ and at least one of the two conditions:\n$$\n1^{\\circ}\\ k \\mid m-n,\n$$\n$$\n2^{\\circ}\\ k \\mid m+n \\text{ and } r+n \\leq m, \\text{ where } r \\text{ is the remainder of } m \\text{ modulo } k.\n$$\nWe start from proving that if $(k, m, n)$ are as above, then the cut is possible. We identify the large square with $[0, m] \\times [0, m]$. If $k \\mid m-n$, then we first cut the $m \\times m$ square into the square $[0, n] \\times [0, n]$ and rectangles $[0, n] \\times [n, m]$ and $[n, m] \\times [0, m]$. These rectangles can trivially be cut into rectangles of dimension $1 \\times k$. If $k \\mid m+n$ and $r+n \\leq m$, then we cut $[0, m] \\times [0, m]$ into a square $[r, r+n] \\times [r, r+n]$ and\n\nrectangles $[0, r] \\times [0, n+r]$, $[r, m] \\times [0, r]$, $[r+n, m] \\times [r, m]$ and $[0, n+r] \\times [r+n, m]$ (they are well defined if $r+n \\le m$), each of which can be trivially cut into rectangles $1 \\times k$. Note that we have used the assumption $r+n \\le m$ here.\nNow we show that the conditions on $(k, m, n)$ are necessary. Suppose the smaller square is equal to $[p, p+n] \\times [q, q+n]$; by symmetry we may assume that $q \\ge 1$.\nFirst we show that $r+n \\le m$. Each unit square $[i, i+1] \\times [0, 1]$, where $i \\in \\{p, p+1, \\dots, p+n-1\\}$, is contained in some rectangle $1 \\times k$ coming from the cut. If we had $r+n > m$, then we would have $m-n < k$ and this would imply that each such rectangle would be \"level\", i.e., of the form $[t, t+k] \\times [0, 1]$. Let $S$ denote the union of these \"level\" rectangles and let $P(S)$ denote the area of $S$. Note that we have $n \\le P(S) \\le m$ and $P(S)$ is divisible by $k$. This contradicts $r+n > m$.\nNow we show that $k \\mid m-n$ or $k \\mid m+n$. Suppose that this is not true. The idea is to write an integer $a_{ij}$ in each unit square $[i-1, i] \\times [j-1, j]$, $1 \\le i, j \\le m$, in such a way that\n(a) for any rectangle $1 \\times k$ of the cut the sum of numbers lying inside equals 0,\n(b) the sum of all the numbers and the sum of the numbers lying inside the square $n \\times n$ are different.\nThe existence of such a sequence clearly yields the claim.\nA second idea is to work with the sequences $(a_{ij})$ of the form $a_{ij} = a_i b_j$, where $(a_i)_{i=1}^m$ and $(b_j)_{j=1}^n$ are $k$-periodic and\n$$\n\\sum_{i=1}^{k} a_i = \\sum_{j=1}^{k} b_j = 0. \\qquad (1)\n$$\nThis implies (a); furthermore, (b) takes form\n$$\n\\sum_{i=1}^{m} a_i \\cdot \\sum_{j=1}^{m} b_j \\neq \\sum_{i=p+1}^{p+n} a_i \\cdot \\sum_{j=q+1}^{q+n} b_j,\n$$\nor, by periodicity and (1),\n$$\n\\sum_{i=1}^{r} a_i \\cdot \\sum_{j=1}^{r} b_j \\neq \\sum_{i=p+1}^{p+s} a_i \\cdot \\sum_{j=q+1}^{q+s} b_j, \\qquad (2)\n$$\nwhere $s$ denotes the remainder coming from the division of $n$ by $k$.\nIf $r=0$, then the left-hand side is 0. Furthermore, as $k \\nmid m-n$, we have $s > 0$; it suffices to take $a_{p+1} = a_{p+2} = \\dots = a_{p+s} = b_{q+1} = b_{q+2} = \\dots = b_{q+s} = 1$ and choose the remaining $a_i$'s and $b_j$'s so that the periodicity and (1) hold.\n\nSuppose then, that $r > 0$. The conditions $k \\nmid m-n$, $k \\nmid m+n$ imply $r \\neq s$, $r+s \\neq k$. We set $b_1 = b_2 = \\dots = b_r = 1$ and pick the remaining $b_j$'s so that periodicity and (1) hold. Let $A = \\{1, 2, \\dots, r\\}$, $B = \\{p+1, p+2, \\dots, p+s\\} \\pmod k$ denote the sets of indices appearing in the sums involving $(a_j)$. Note that we take the set $B$ modulo $k$. Consider two cases:\ni) $A \\cup B \\neq \\{0, 1, \\dots, k-1\\}$. Then, as $A \\neq B$ (as $r \\neq s$) and $A \\neq \\emptyset$, we may choose $a'_i$s with $i \\in A \\cup B$ in such a way that\n$$\n\\sum_{i \\in A} a_i = 1, \\qquad \\sum_{i \\in B} a_i = 0\n$$\n(for example, if $A \\setminus B \\neq \\emptyset$, set $a_i = 0$ for all $i \\in A \\cup B$ except for one $i \\in A \\setminus B$, for which $a_i = 1$; if $A \\subset B$, take $a_i = 0$ for $i \\in A \\cup B$ except for one $i \\in A$, for which $a_i = 1$ and except for one $j \\in B \\setminus A$, for which $a_j = -1$) and complete the sequence $(a_i)$ so that it satisfies periodicity and (1). This completion is possible as there exists $i \\in \\{0, 1, \\dots, k-1\\}$ not covered by $A \\cup B$. Then the right-hand side of (2) is 0, while the left one is not.\nii) $A \\cup B = \\{0, 1, \\dots, k-1\\}$. Then $A \\not\\subseteq B$ and $B \\not\\subseteq A$; furthermore, as $r+s \\neq k$, we have $A \\cap B \\neq \\emptyset$. Therefore, there exist $i_1 \\in A \\setminus B$, $i_2 \\in B \\setminus A$, $i_3 \\in A \\cap B$ and we set $a_{i_1} = 0$, $a_{i_2} = -1$, $a_{i_3} = 1$ and $a_i = 0$ for $i \\in \\{0, 1, 2, \\dots, k-1\\} \\setminus \\{i_1, i_2, i_3\\}$. Then we have\n$$\n\\sum_{i=1}^{r} a_i = \\sum_{i \\in A} a_i = 1, \\quad \\sum_{i=p+1}^{p+s} a_i = \\sum_{i \\in B} a_i = 0\n$$\nand hence the right-hand side of (2) is 0, while the left one is not.\nThe proof is complete.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 29337,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $\\mathbb{R}^+$ be the set of all positive real numbers. Find all functions $f: \\mathbb{R}^+ \\to \\mathbb{R}^+$ satisfying\n$$\n(1 + y f(x))(1 - y f(x + y)) = 1\n$$\nfor any $x, y \\in \\mathbb{R}^+$.",
    "options": [],
    "answer": "f(x) = 1/(x + c) for c >= 0",
    "solution": "Simple manipulation with the given equation leads to\n$$\n\\begin{aligned}\n1 + y f(x) - y f(x + y) - y^2 f(x) f(x + y) &= 1, \\\\\ny f(x) - y f(x + y) &= y^2 f(x) f(x + y).\n\\end{aligned}\n$$\nAfter cancelling $y \\neq 0$ another manipulation gives\n$$\n\\begin{aligned}\nf(x) - f(x + y) &= y f(x) f(x + y), \\\\\n\\frac{f(x) - f(x + y)}{f(x + y)} &= y f(x), \\\\\n\\frac{f(x)}{f(x + y)} - 1 &= y f(x), \\\\\n\\frac{f(x)}{f(x + y)} &= 1 + y f(x), \\\\\nf(x + y) &= \\frac{f(x)}{1 + y f(x)}, \\\\\n\\frac{1}{f(x + y)} &= (1 + y f(x)) \\cdot \\frac{1}{f(x)} = \\frac{1}{f(x)} + y.\n\\end{aligned}\n$$\n(evidently, all the cancelled expressions are nonzero) and so\n$$\n\\frac{1}{f(x + y)} = y + \\frac{1}{f(x)}\n$$\nfor any $x, y \\in \\mathbb{R}^+$. Hence\n$$\ny + \\frac{1}{f(x)} = x + \\frac{1}{f(y)}.\n$$\nSetting $y = 1$ we get\n$$\n\\frac{1}{f(x)} = x + \\frac{1}{f(1)} - 1 = x + c, \\quad \\text{so} \\quad f(x) = \\frac{1}{x + c}\n$$\nwith a constant $c$. As $f(x) > 0$ for any $x > 0$, we have $x + c > 0$ for any $x > 0$ and therefore $c \\ge 0$.\nWe can easily check that the function $f(x) = 1/(x + c)$ satisfies the given conditions for any $c \\ge 0$:\n$$\n\\left(1 + \\frac{y}{x + c}\\right) \\left(1 - \\frac{y}{x + y + c}\\right) = \\frac{x + c + y}{x + c} \\cdot \\frac{x + y + c - y}{x + y + c} = \\frac{x + c + y}{x + c} \\cdot \\frac{x + c}{x + y + c} = \\frac{x + c + y}{x + y + c}.\n$$\nBut $\\frac{x + c + y}{x + y + c} = 1$, so the condition is satisfied for all $x, y > 0$ and $c \\ge 0$.\nPut $x = 1$ and $f(1) = a > 0$. Then we have\n$$\n(1 + a y)(1 - y f(y + 1)) = 1,\n$$\n$$\na y - y f(y + 1)(1 + a y) = 0,\n$$\n$$\nf(y + 1) = \\frac{a}{1 + a y}\n$$\nNow put $y = 1$ and $f(x + 1) = a/(1 + a x)$ to get\n$$\n(1 + f(x)) \\left( 1 - \\frac{a}{1 + a x} \\right) = 1,\n$$\n$$\nf(x) \\cdot \\frac{1 + a x - a}{1 + a x} = \\frac{a}{1 + a x},\n$$\n$$\nf(x) = \\frac{a}{1 + a x - a} = \\frac{1}{x + \\frac{1}{a} - 1} = \\frac{1}{x + c}.\n$$\nAs in the first solution, we have $c \\ge 0$ and it is easy to check that such a function satisfies the required conditions.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 29338,
    "subject": "Mathematics (Multi-modal)",
    "question": "Given positive integers $a$ and $k$, the sequence $(a_n)_{n=1}^{\\infty}$ is defined by\n$a_1 = a$ and $a_{n+1} = a_n + k \\cdot \\varphi(a_n)$ for $n = 1, 2, \\dots,$\nwhere $\\varphi(m)$ stands for the product of digits of $m$ in its decimal representation (e.g., $\\varphi(413) = 12$, $\\varphi(308) = 0$). Prove that there exist positive integers $a$ and $k$ such that the sequence $(a_n)_{n=1}^{\\infty}$ contains exactly 2009 different numbers.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Obviously, the sequence $(a_n)_{n=1}^{\\infty}$ is increasing until the first term with the zero digit occurs and is constant following this term. Our aim is to find such values $a$ and $k$ that the zero digit first occurs in $a_{2009}$. We will solve the problem in general – given an integer $m > 4$, we present $a$ and $k$ such that the zero digit first occurs in $a_m$.\nSet\n$$\na = \\frac{10^{2m-5} - 1}{9} = \\underbrace{11\\dots1}_{2m-5 \\text{ ones}}, \\quad k = 10^{m-3} + 4 = 1 \\underbrace{00\\dots0}_{m-4 \\text{ zeros}} 4.\n$$\n\nThen we have\n$$\na_1 = a = \\underbrace{11\\dots1}_{2m-5}\n$$\n$$\n\\varrho(a_1) = 1.\n$$\n$$\na_2 = a_1 + k = a_1 + \\underbrace{100\\dots04}_{m-1} = \\underbrace{11\\dots12}_{m-3} \\underbrace{11\\dots15}_{m-4}\n$$\n$$\n\\varrho(a_2) = 10.\n$$\n$$\na_3 = a_2 + 10k = a_2 + \\underbrace{100\\dots040}_{m-4} = \\underbrace{11\\dots122}_{m-4} \\underbrace{11\\dots155}_{m-5}\n$$\n$$\n\\varrho(a_3) = 100.\n$$\n$$\na_i = a_{i-1} + 10^{i-2}k = \\underbrace{11\\dots122\\dots2}_{m-1} \\underbrace{11\\dots155\\dots5}_{m-1}\n$$\n$$\n\\varrho(a_i) = 10^{i-1}.\n$$\n$$\na_{m-2} = a_{m-3} + 10^{m-4}k = \\underbrace{122\\dots255\\dots5}_{m-3}\n$$\n$$\n\\varrho(a_{m-2}) = 10^{m-3}.\n$$\n$$\na_{m-1} - a_{m-2} + 10^{m-3}k = \\underbrace{100\\dots0}_{m-4} \\underbrace{100\\dots0}_{m-3} \\underbrace{22\\dots2655\\dots5}_{m-3}\n$$\n$$\n\\varrho(a_{m-1}) = 6 \\cdot 10^{m-3}.\n$$\n$$\na_m = a_{m-1} + 6 \\cdot 10^{m-3}k = \\underbrace{600\\dots0}_{m-5} \\underbrace{2400\\dots0}_{m-3} \\underbrace{822\\dots25055\\dots5}_{m-5}\n$$\n$$\n\\varrho(a_m) = 0.\n$$\n\nSecond solution. Put\n$$\na = \\underbrace{611\\dots1}_{2007} \\quad \\text{and} \\quad k = \\underbrace{33\\dots34}_{2007} = \\frac{1}{6} \\cdot \\underbrace{200\\dots04}_{2007}\n$$\n\nThen we have\n$$\n\\begin{align*}\na_1 &= \\underbrace{6 \\ 11 \\dots 1}_{2007}, & \\varrho(a_1) &= 6, & k\\varrho(a_1) &= \\underbrace{2 \\ 00 \\dots 0 1}_{2007} \\\\\na_2 &= \\underbrace{26 \\ 11 \\dots 1 5}_{2006}, & \\varrho(a_2) &= 60, & k\\varrho(a_2) &= \\underbrace{2 \\ 00 \\dots 0 40}_{2007} \\\\\na_3 &= \\underbrace{226 \\ 11 \\dots 1 55}_{2005}, & \\varrho(a_3) &= 600, & k\\varrho(a_3) &= \\underbrace{2 \\ 00 \\dots 0 400}_{2007} \\\\\na_4 &= \\underbrace{2226 \\ 11 \\dots 1 555}_{2004}, & \\varrho(a_4) &= 6000, & k\\varrho(a_4) &= \\underbrace{2 \\ 00 \\dots 0 4000}_{2007} \\\\\n& & & & \\\\\na_{t+1} = \\underbrace{22 \\dots 2 6 \\ 11 \\dots 1 55 \\dots 5}_{2007-t}, & \\varrho(a_{t+1}) &= \\underbrace{6 \\ 0 \\dots 0}_{t}, & k\\varrho(a_{t+1}) &= \\underbrace{2 \\ 00 \\dots 0 4 \\ 00 \\dots 0}_{2007} \\\\\n& & & & \\\\\na_{2007} = \\underbrace{22 \\dots 2 6 \\ 15 \\dots 5}_{2006} & \\varrho(a_{2007}) &= \\underbrace{6 \\ 0 \\dots 0}_{2006}, & k\\varrho(a_{2007}) &= \\underbrace{2 \\ 00 \\dots 0 4 \\ 00 \\dots 0}_{2007} \\\\\na_{2008} = \\underbrace{22 \\dots 2 6 \\ 55 \\dots 5}_{2007} & \\varrho(a_{2008}) &= \\underbrace{6 \\ 0 \\dots 0}_{2007}, & k\\varrho(a_{2008}) &= \\underbrace{2 \\ 00 \\dots 0 4 \\ 00 \\dots 0}_{2007} \\\\\na_{2009} = \\underbrace{22 \\dots 2 30 \\ 55 \\dots 5}_{2007} & \\varrho(a_{2009}) &= 0, & k\\varrho(a_{2009}) &= 0\n\\end{align*}\n$$\nand obviously $a_{2009}, a_{2010}, a_{2011}, \\dots$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 29339,
    "subject": "Mathematics (Multi-modal)",
    "question": "Given the triangle $ABC$, let $k$ be the excircle at the side $BC$. Choose any line $p$ parallel to $BC$ intersecting line segments $AB$ and $AC$ at points $D$ and $E$. Denote by $l$ the incircle of the triangle $ADE$. The tangents from $D$ and $E$ to the circle $k$ not passing through $A$ intersect at $P$. The tangents from $B$ and $C$ to the circle $l$ not passing through $A$ intersect at $Q$. Prove that the line $PQ$ passes through a fixed point independent of the choice of $p$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $BC$ touch $k$ in $T_k$ and $DE$ touch $l$ in $T_l$. We shall prove the required fixed point is $T_k$.\n\nFirst we show the points $T_k$, $T_l$ and $P$ are collinear. Denote by $U$ and $V$ the points where $EP$ and $DP$ touch $k$, and $M$ and $N$ the points where $EP$ and $DP$ intersect $BC$. Let $T_1$ and $T_2$ be the tangent points of $k$ and the rays $AB$ and $AC$ respectively.\n\n![](attached_image_1.png)\nFig. 1\n\nAs $BC \\parallel DE$, the triangle $DEP$ is similar to $NMP$ and the homothety $H$ with the centre $P$ and the quotient $q = MN/ED$ maps the segment $DE$ into $NM$. To prove the collinearity of $T_k, T_l, P$, it suffices to derive the equality\n$$\n\\frac{MT_k}{NT_k} = \\frac{ET_l}{DT_l} \\qquad (1)\n$$\nif this is true, $H$ maps $T_l$ into $T_k$.\n\nLet $a, b, c$ be the lengths of the sides in the triangle $DEP$ as in fig. 1. Set $AD = c$, $AE = d$. Let us remind the well-known formulae for the length of the segments between the vertex of a triangle and the tangent points of its incircle and excircle: In any triangle $XYZ$, the distance of $X$ from the tangent point of the incircle and excircle (lying on $XY$) is $(XY + XZ - YZ)/2$ and $(XY + YZ - XZ)/2$ respectively.\n\nThe circle $k$ is the excircle of $NMP$. Hence\n$$\n\\frac{MT_k}{NT_k} = \\frac{(MN + NP - MP)/2}{(MN + MP - NP)/2} = \\frac{qa + qc - qb}{qa + qb - qc} = \\frac{a+c-b}{a+b-c} \\qquad (2)\n$$\nThe circle $l$ is the incircle of $DEA$. Hence\n$$\n\\frac{ET_l}{DT_l} = \\frac{(DE + AE - AD)/2}{(DE + AD - AE)/2} = \\frac{a+d-c}{a+c-d} \\qquad (3)\n$$\nWhen we draw two tangent lines from a point to a circle, the distances of the two tangent points from the original point are equal. Repeating this argument several times we get\n$$\nc + c + PU = e + c + PV = e + DT_l \\quad AT_1 \\cdot AT_2 = d + ET_2 \\cdot d + b + PU\n$$\nso $e + c = d + b$. Then $c - b = d - e$ and substituting into (2) and (3) gives\n$$\n\\frac{MT_k}{NT_k} = \\frac{a + (c - b)}{a - (c - b)} = \\frac{a + (d - e)}{a - (d - e)} = \\frac{ET_l}{DT_l}\n$$\nwhich is exactly (1). Hence $P$ lies on $T_1T_k$.\n\nSimilarly we show that also $T_k$, $T_l$ and $Q$ are collinear. Denote by $U'$ and $V'$ the points where $CQ$ and $BQ$ touch $l$, and $M'$ and $N'$ the points where $C'Q$ and $B'Q$ intersect $DE$. Let $T_1'$ and $T_2'$ be the tangent points of $l$ and the rays $AD$ and $AE$ respectively. Let $a', b', c'$ be the lengths of the sides in the triangle $BCQ$ and $AB = c'$, $AC = d'$.\n\n![](attached_image_2.png)\nFig. 2\n\nRepeating the arguments from the first part (here both circles are excircles) we get\n$$\n\\frac{M'T_l}{N'T_l} = \\frac{a' + c' - b'}{a' + b' - c'} \\qquad \\frac{CT_k}{BT_k} = \\frac{a' + c' - d'}{a' + d' - c'}\n$$\nComparing the lengths (fig. 2) gives\n$$\nc' - c' - QU' = c' - c' - QV' = c' - BT_{1}' = AT_{1}' = AT_{2}' = d' - CT_{2}' = d' - b' - QU'\n$$\nso $c' - b' = c' - d'$ and\n$$\n\\frac{M'T_l}{N'T_l} = \\frac{CT_k}{BT_k}\n$$\nFinally, using the homothety of $BCQ$ and $N'M'Q$ we conclude that $Q$ lies on $T_lT_k$.\n\nTherefore the line $PQ$ (obviously, $P \\neq Q$) is identical to the line $T_lT_k$ and passes through $T_k$, which is independent of the choice of $p$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 29340,
    "subject": "Mathematics (Multi-modal)",
    "question": "Given a circle $k$ and its chord $AB$ which is not the diameter. Let $C$ be any point inside the longer arc of $k$. We denote by $K$ and $L$ the reflections of $A$ and $B$ with respect to the axes $BC$ and $AC$. Prove that the distance of the midpoints of the line segments $KL$ and $AB$ is independent of the location of the point $C$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Denote by $S$ the midpoint of $AB$, by $M$ the midpoint of $KL$, and by $P$ and $Q$ the feet of the altitudes from $A$ and $B$ in the triangle $ABC$. Obviously, $P$ and $Q$ are the midpoints of $AK$ and $BL$ respectively (fig. 3). Therefore $QS$ is the mid-segment in the triangle $LAB$ and $MP$ is the mid-segment in the triangle $LAK$. We have\n$$\nQS = \\frac{1}{2} LA, \\quad MP = \\frac{1}{2} LA \\quad \\text{and} \\quad QS \\parallel LA \\parallel MP.\n$$\nthus $SPMQ$ is a parallelogram (this is true even in the case of \"degenerate\" triangles $LAB$ or $LAK$).\n![](attached_image_1.png)\nFig. 3\nThe points $P$ and $Q$ lie on the Thales circle over $AB$, so $SP = SQ = \\frac{1}{2} AB$. This implies the parallelogram $SPMQ$ is a rhombus and the length of its side is independent of the location of $C$. To prove that also the length of its diagonal $SM$ is independent of $C$, it suffices to show that the size of the angle spanned by its sides $SP$ and $SQ$ is constant when moving $C$ along $k$ (then all the rhombuses $SPMQ$ and also their diagonals $SM$ are congruent).\n![](attached_image_2.png)\nFig. 4a\n![](attached_image_3.png)\nFig. 4b\nIf the angle $\\alpha$ in the triangle $ABC$ is acute, the point $Q$ lies inside of $AC$ (the angle $\\gamma$ is always acute by the statement of the problem) and the angle $PSQ$ is central to the angle $PAQ$, which is inscribed over the chord $PQ$ of the Thales circle over $AB$ (fig. 4a). Hence\n$$\n\\angle PSQ = 2 \\angle PAQ = 2(90^\\circ - \\gamma) = 180^\\circ - 2\\gamma.\n$$\n\nThe same formula we get when $\\alpha$ is non-acute, since in this case $\\beta$ is acute and we can use the inscribed angle $PBQ$ instead of $PAQ$ (fig. 4b):\n$$\n\\angle PSQ = 2 \\angle PBQ = 2(90^\\circ - \\gamma) = 180^\\circ - 2\\gamma.\n$$\nAs the size of $\\gamma$ does not change when moving $C$ along $k$ (it is an inscribed angle over the fixed chord $AB$), the size of the angle $PSQ$ also does not change.\nDenote by $\\alpha, \\beta, \\gamma$ the angles of the triangle $ABC$. We only consider the case $\\alpha < 90^\\circ$; the second case ($\\beta < 90^\\circ$) is similar. Let $S, M, U, V$ be the midpoints of the segments $AB, KL, AL, BK$ respectively. Let $H$ be the common point of the lines $AK$ and $BL$ (fig. 5a, b). Note that the quadrilateral $USVM$ is a parallelogram ($SU \\parallel BL \\parallel MV, SV \\parallel AK \\parallel MU$) and\n$$\nSV = AB \\sin \\beta, \\quad MV = SU = AB \\sin \\alpha, \\quad \\angle SVM = \\angle AIL = \\angle ACB\n$$\nBy the law of sines applied to the triangle $ABC$,\n$$\n\\frac{AC}{SV} = \\frac{1}{AB} \\cdot \\frac{AC}{\\sin \\beta} = \\frac{1}{AB} \\cdot \\frac{BC}{\\sin \\alpha} = \\frac{BC}{MV}\n$$\nso the triangles $SVM$ and $ACB$ are proportional (two sides proportional and the same angle between them). Then again by the law of sines.\n$$\nSM = AB \\cdot \\frac{SV}{AC} = \\frac{AB^2 \\sin \\beta}{AC} = \\frac{AB^2 \\sin \\gamma}{AB} = AB \\sin \\gamma\n$$\nwhich obviously does not depend of the location of the point $C$.\n![](attached_image_4.png)\nFig. 5a\n![](attached_image_5.png)\nFig. 5b",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 29341,
    "subject": "Mathematics (Multi-modal)",
    "question": "The $n$-tuple of positive integers $a_1, \\dots, a_n$ satisfies the following conditions:\n(i) $1 \\le a_1 < a_2 < \\dots < a_n \\le 50$;\n(ii) for any $n$-tuple of positive integers $b_1, \\dots, b_n$, there exist a positive integer $m$ and an $n$-tuple of positive integers $c_1, \\dots, c_n$ such that\n$$\nm \\cdot b_i \\equiv c_i^{a_i} \\quad \\text{for } i = 1, \\dots, n.\n$$\nProve that $n \\le 16$ and find the number of different $n$-tuples $a_1, \\dots, a_n$ satisfying the given conditions for $n = 16$.",
    "options": [],
    "answer": "n ≤ 16; for n = 16 there are 60 such n-tuples",
    "solution": "First we prove the numbers $a_1, \\dots, a_n$ are mutually relatively prime. If this is not true, we have $(a_i, a_j) = d > 1$ for some $i \\ne j$. Put $a_i = u \\cdot d$, $a_j = v \\cdot d$. Set $b_i = 1$, $b_j = 2$. By (ii), there exist $m$, $c_i$, and $c_j$ such that\n$$\nm \\cdot b_i = c_i^{a_i} \\quad \\text{and} \\quad m \\cdot b_j = c_j^{a_j}, \\quad \\text{therefore} \\quad m = (c_i^{a_i})^d \\quad \\text{and} \\quad 2m = (c_j^{v_j})^d.\n$$\nSo $2(c_i^{a_i})^d = (c_j^{v_j})^d$, which is not possible, since the exponent of $2$ in the prime factorization of the right hand side is and the one of the left hand side is not a multiple of $d$.\n\nAssume that $a_1, \\dots, a_n$ are mutually relatively prime. We shall prove that the condition (ii) is fulfilled. Let $b_1, \\dots, b_n$ be any $n$-tuple of positive integers and $p_1, \\dots, p_k$ be all the prime divisors of $b_1, \\dots, b_n$. We look for $a_i$ in the form\n$$\nm = p_1^{a_1} \\cdots p_k^{a_k}.\n$$\nFor $i = 1, \\dots, n$, denote by $\\beta_{i,j}$ the exponent of $p_j$ in the prime factorization of $b_i$. In order for $m \\cdot b_i$ to be an $a_i$-th power, it suffices $\\alpha_j + \\beta_{i,j}$ to be a multiple of $a_i$ for $j = 1, \\dots, k$. So we need $\\alpha_j$ fulfilling the congruences\n$$\n\\alpha_j \\equiv -\\beta_{1,j} \\pmod{a_1}, \\quad \\alpha_j \\equiv -\\beta_{2,j} \\pmod{a_2}, \\quad \\dots \\quad \\alpha_j \\equiv -\\beta_{n,j} \\pmod{a_n}.\n$$\nThe existence of such $\\alpha_j$ is guaranteed by Chinese Remainder Theorem (as $a_1, \\dots, a_n$ being mutually relatively prime).\n\nWe have proved the condition (ii) is satisfied if and only if the numbers $a_1, \\dots, a_n$ are mutually relatively prime. Among $1, 2, \\dots, 50$, there are exactly $15$ primes. If $n \\ge 17$ then among $2 \\le a_2 < a_3 < \\dots < a_n \\le 50$, there must be at least two numbers sharing the same prime number in their prime factorizations, hence not being relatively prime. Therefore we have $n \\le 16$.\n\nIf $n = 16$, then $a_1 = 1$ and $a_2, a_3, \\dots, a_{16}$ must be the powers of different primes. We list the powers of primes which can be used:\n$$\n\\begin{array}{lclcl}\np & = & 2: & 2, 4, 8, 16, 32, \\\\\np & = & 3: & 3, 9, 27, \\\\\np & = & 5: & 5, 25, \\\\\np & = & 7: & 7, 49, \\\\\np & \\ge & 11: & \\text{only } p.\n\\end{array}\n$$\nThen the total number of the corresponding $16$-tuples is $5 \\cdot 3 \\cdot 2 \\cdot 2 = 60$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 29342,
    "subject": "Mathematics (Multi-modal)",
    "question": "Given an integer $n \\ge 16$, consider the set\n$$\nG = \\{(x, y) : x, y \\in \\{1, 2, \\dots, n\\}\\}\n$$\nconsisting of $n^2$ points in the plane. Let $A$ be any subset of $G$ containing at least $4n\\sqrt{n}$ points. Prove that there are at least $n^2$ convex quadrangles with all their vertices in $A$ such that their diagonals intersect in one common point.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $|A| = m \\ge 4n\\sqrt{n}$ and let $S$ be the set of all segments with endpoints in $A$. Clearly, $|S| = \\binom{m}{2}$. The coordinates of every midpoint of a segment from $S$ are integer multiples of $1/2$. In the convex hull of $G$ there are less than $4n^2$ such points so there exists a point $B$ which is a midpoint of at least $(m/2)/(4n^2)$ segments from $S$. Let $P$ be the set of all segments from $S$ with their midpoints in $B$. Then\n$$\n|P| \\ge \\frac{\\binom{m}{2}}{4n^2} = \\frac{m(m-1)}{8n^2} \\ge \\frac{4n\\sqrt{n}(4n\\sqrt{n}-1)}{8n^2} = \\frac{16n^3 - 4n\\sqrt{n}}{8n^2} = 2n - \\frac{1}{2\\sqrt{n}} > 2n-1.\n$$\nand $|P| \\ge 2n$.\n\nLet us divide $P$ into disjoint families of segments which lay on the same line. Suppose that the number of such families is $k$ and in the $i$-th family we have $a_i$ segments, for $i = 1, \\dots, k$. Every segment from $a_i$ segments of one family has its endpoints in $G$ and they have a common midpoint, so $a_i \\le n/2$. Moreover every two segments from $P$ are the diagonals of a parallelogram iff they do not lay on the same line. Therefore for the number of different parallelograms with diagonals belonging to $P$ we have\n$$\n\\sum_{1 \\le i < j \\le k} a_i a_j = \\frac{1}{2} \\left( \\left( \\sum_{i=1}^{k} a_i \\right)^2 - \\sum_{i=1}^{k} a_i^2 \\right) \\ge \\frac{1}{2} \\left( \\left( \\sum_{i=1}^{k} a_i \\right)^2 - \\sum_{i=1}^{k} a_i \\cdot \\frac{n}{2} \\right) = \\frac{1}{2} \\left( |P|^2 - |P| \\cdot \\frac{n}{2} \\right) = \\frac{1}{2} |P| \\left( |P| - \\frac{n}{2} \\right) \\ge n \\left( 2n - \\frac{n}{2} \\right) = \\frac{3}{2} n^2 > n^2.\n$$\nThus we have more than $n^2$ convex quadrangles (parallelograms) satisfying the given condition.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 29343,
    "subject": "Mathematics (Multi-modal)",
    "question": "Determine all triples $(a, b, c)$ of positive real numbers satisfying the following system of equations\n$$\n\\begin{aligned}\na\\sqrt{b} - c &= a, \\\\\nb\\sqrt{c} - a &= b, \\\\\nc\\sqrt{a} - b &= c.\n\\end{aligned}\n$$",
    "options": [],
    "answer": "(4, 4, 4)",
    "solution": "Without loss of generality we can suppose $a = \\max\\{a, b, c\\}$. From the first equation of the given system we obtain\n$$\nc(\\sqrt{b} - 1) \\le a(\\sqrt{b} - 1) = c, \\quad \\text{i.e.} \\quad b \\le 4.\n$$\nSimilarly, from the second equation of the given system we have\n$$\nb(\\sqrt{c} - 1) = a \\ge b, \\quad \\text{thus} \\quad c \\ge 4.\n$$\nUsing these results together with the third equation we get\n$$\n4 \\le c \\le c(\\sqrt{c} - 1) \\le c(\\sqrt{a} - 1) = b \\le 4\n$$\nwhich yields $a = b = c = 4$.\n\n**Conclusion.** The unique solution of the given system of equations is therefore $(a, b, c) = (4, 4, 4)$.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 29344,
    "subject": "Mathematics (Multi-modal)",
    "question": "We are given 60 arbitrary points in a unit disc. Prove that there exists a point on the boundary of the disc such that the sum of its distances from all 60 given points doesn't exceed 80.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let us inscribe an equilateral triangle $PQR$ into the boundary unit circle. If we prove that any point $X$ of our unit disc satisfies\n$$\n|PX| + |QX| + |RX| \\le 4, \\quad (1)\n$$\nthen summing (1), with the given points $X = X_k$, $1 \\le k \\le 60$, we get\n$$\n\\sum_{k=1}^{60} |PX_k| + \\sum_{k=1}^{60} |QX_k| + \\sum_{k=1}^{60} |RX_k| \\le 4 \\cdot 60 = 240.\n$$\nConsequently, one of the sums in the left-hand side does not exceed $240 : 3 = 80$, hence some of the points $P$, $Q$, $R$ has always the required property.\n\nIn view of symmetry, it suffices to prove (1) if $X$ lies in the sector $PSQ$, where $S$ denotes the centre of the disc. We are going to show that, in this case,\n$$\n|PX| + |QX| \\le 2, \\quad (2)\n$$\nwhich together with $|RX| \\le 2$ will lead to (1).\n\nLet us denote $S'$ the midpoint of the corresponding arc $PQ$ (opposite to the arc $PRQ$, Fig. 1), then clearly $PS'QS$ is a rhombus, hence it is sufficient to prove (2) only for points $X$ in the region $PS'Q$ of the given disc (bounded by segment $PQ$ and arc $PS'Q$).\nDenoting $\\alpha = |\\angle XPQ|$, $\\beta = |\\angle XQP|$ we have $\\alpha + \\beta \\le 60^\\circ$ and using the law of sines in the triangle $PQX$ we get\n$$\n\\begin{aligned}\n|PX| + |QX| &= \\frac{|PQ|(\\sin \\alpha + \\sin \\beta)}{\\sin(\\alpha + \\beta)} = \\frac{\\sqrt{3} \\cdot 2 \\sin \\frac{\\alpha+\\beta}{2} \\cos \\frac{\\alpha-\\beta}{2}}{2 \\sin \\frac{\\alpha+\\beta}{2} \\cos \\frac{\\alpha+\\beta}{2}} \\\\\n&= \\frac{\\sqrt{3} \\cos \\frac{\\alpha-\\beta}{2}}{\\cos \\frac{\\alpha+\\beta}{2}} \\le \\frac{\\sqrt{3} \\cdot 1}{\\sqrt{3}} = 2, \\quad \\text{as } \\frac{\\alpha+\\beta}{2} \\le 30^\\circ.\n\\end{aligned}\n$$\nThus (2) is proven.\n\n![](attached_image_1.png)\n\nFig. 1\n\n![](attached_image_2.png)\n\nFig. 2",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 29345,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $p$ be a prime number. Show that one can choose $p^3$ fields of a $p^2 \\times p^2$ chessboard such that centres of no four chosen fields are vertices of a rectangle with sides parallel to the sides of the chessboard.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let us label the $p^2$ rows and the $p^2$ columns of the chessboard by all the pairs $(a, b)$ and $(c, d)$, respectively, where $a, b, c, d \\in \\{0, 1, \\dots, p-1\\}$. A field lying in the row $(a, b)$ and in the column $(c, d)$ will be called good if and only if\n$$\nac \\equiv b+d \\pmod{p}. \\qquad (1)\n$$\nGiven a pair $(a, b)$, (1) holds for exactly $p$ pairs $(c, d)$. Thus the number of good fields is $p$ in any row and $p^3$ in total.\n\nIt remains to show that there are no four good fields with the prohibited property. Suppose on the contrary that\n$$\na_i c_j \\equiv b_i + d_j \\pmod{p} \\quad \\text{whenever } i, j \\in \\{1, 2\\}, \\qquad (2)\n$$\nfor some pairs $(a_1, b_1) \\neq (a_2, b_2)$ and $(c_1, d_1) \\neq (c_2, d_2)$. Subtracting (2) with a fixed $i$ yields\n$$\na_i(c_2 - c_1) \\equiv d_2 - d_1 \\pmod{p} \\quad \\text{for } i \\in \\{1, 2\\}, \\qquad (3)\n$$\nwhich (in the same way) leads to\n$$\n(a_2 - a_1)(c_2 - c_1) \\equiv 0 \\pmod{p}.\n$$\nThus $a_1 = a_2$ or $c_1 = c_2$. In view of symmetry, we can assume that $c_1 = c_2$. Then (3) implies that $d_1 = d_2$, hence $(c_1, d_1) = (c_2, d_2)$, a contradiction.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 29346,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $x$, $y$, $z$ be positive real numbers such that $x + y + z \\ge 6$. Find the smallest value of the expression\n$$\nx^2 + y^2 + z^2 + \\frac{x}{y^2 + z + 1} + \\frac{y}{z^2 + x + 1} + \\frac{z}{x^2 + y + 1}.\n$$",
    "options": [],
    "answer": "90/7",
    "solution": "Using the AM-GM inequality for positive real numbers $x^2/14$, $x/(y^2+z+1)$ and $2(y^2+z+1)/49$ we have\n$$\n\\frac{x^2}{14} + \\frac{x}{y^2 + z + 1} + \\frac{2}{49}(y^2 + z + 1) \\ge 3\\sqrt[3]{\\frac{x^3}{7^3}} = \\frac{3}{7}x.\n$$\n\nWe can derive cyclically another two similar inequalities. Adding up of all three derived inequalities we further obtain (all sums are to be considered as sums of three cyclically obtained summands)\n$$\n\\frac{1}{14} \\sum x^2 - \\sum \\frac{x}{y^2 + z + 1} + \\frac{2}{49} \\sum x^2 + \\frac{2}{49} \\sum x + \\frac{6}{49} \\ge \\frac{3}{7} \\sum x.\n$$\nThus\n$$\nL = \\frac{11}{98} \\sum x^2 + \\sum \\frac{x}{y^2 + z + 1} + \\frac{6}{49} \\ge \\left(\\frac{3}{7} - \\frac{2}{49}\\right) \\sum x = \\frac{19}{49} \\sum x.\n$$\n\nFrom the Cauchy-Schwarz inequality (or from the AM-QM inequality) it follows $\\sum x^2 \\ge \\frac{1}{3} (\\sum x)^2 \\ge 12$. Finally we arrive at\n$$\n\\begin{aligned} \\sum x^2 + \\sum \\frac{x}{y^2 + z + 1} &= \\frac{87}{98} \\sum x^2 + L - \\frac{6}{49} \\ge \\frac{87}{98} \\sum x^2 + \\frac{19}{49} \\sum x - \\frac{6}{49} \\\\ &\\ge \\frac{87 \\cdot 6 + 19 \\cdot 6 - 6}{49} = \\frac{90}{7}. \\end{aligned}\n$$\n**Conclusion.** The smallest value of given expression is therefore $90/7$. (The mentioned value is achieved for $x = y = z = 2$.)",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 29347,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABCD$ be a convex quadrilateral such that\n$$\n|AB| + |CD| = \\sqrt{2} \\cdot |AC| \\quad \\text{and} \\quad |BC| + |DA| = \\sqrt{2} \\cdot |BD|.\n$$\nProve that $ABCD$ is a parallelogram.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "The result immediately follows from the following **LEMMA**. If $ABCD$ is any quadrilateral (convex or non-convex), then\n$$\n(|AB| + |CD|)^2 + (|BC| + |DA|)^2 \\geq 2|AC|^2 + 2|BD|^2,\n$$\nwith the equality if and only if $ABCD$ is a parallelogram.\n\n**PROOF OF LEMMA.** The vectors $\\mathbf{a} = \\overrightarrow{AB}$, $\\mathbf{b} = \\overrightarrow{BC}$, $\\mathbf{c} = \\overrightarrow{CD}$ and $\\mathbf{d} = \\overrightarrow{DA}$ obviously satisfy\n$$\n\\mathbf{a} + \\mathbf{b} + \\mathbf{c} + \\mathbf{d} = \\mathbf{0}. \\qquad (1)\n$$\nSquaring the triangle inequalities\n$$\n|\\mathbf{a}| + |\\mathbf{c}| \\geq |\\mathbf{a} - \\mathbf{c}|, \\quad |\\mathbf{b}| + |\\mathbf{d}| \\geq |\\mathbf{b} - \\mathbf{d}|, \\qquad (2)\n$$\nsumming up and using the dot product, we get\n$$\n\\begin{aligned}\n(|AB| + |CD|)^2 + (|BC| + |DA|)^2 &\\geq |\\mathbf{a} \\cdot \\mathbf{c}|^2 + |\\mathbf{b} - \\mathbf{d}|^2 \\\\\n&= |\\mathbf{a}|^2 + |\\mathbf{b}|^2 + |\\mathbf{c}|^2 + |\\mathbf{d}|^2 - 2 \\mathbf{a} \\cdot \\mathbf{c} - 2 \\mathbf{b} \\cdot \\mathbf{d} \\\\\n&= |\\mathbf{a} + \\mathbf{b}|^2 - |\\mathbf{c} + \\mathbf{d}|^2 - 2(\\mathbf{a} + \\mathbf{d}) \\cdot (\\mathbf{b} + \\mathbf{c}) \\\\\n&= 2|\\mathbf{AC}|^2 - 2 \\overline{\\mathbf{DB}} \\cdot \\overline{\\mathbf{BD}} = 2|\\mathbf{AC}|^2 + 2|\\mathbf{BD}|^2.\n\\end{aligned}\n$$\nThus the desired inequality is proven. If it is an equality, then (2) must become equalities as well, hence $\\mathbf{c} = -p\\mathbf{a}$ and $\\mathbf{d} = -q\\mathbf{b}$ for some positive real $p$ and $q$. Substituting this into (1) yields\n$$\n(1 - p)\\mathbf{a} + (1 - q)\\mathbf{b} = \\mathbf{0},\n$$\nhence both $1-p$ and $1-q$ vanish. Thus $\\mathbf{c} = -\\mathbf{a}$ (and $\\mathbf{d} = -\\mathbf{b}$) which means that $ABCD$ is a parallelogram.\n\nConversely, if $ABCD$ is a parallelogram, then $|AB| = |CD|$, $|BC| = |DA|$ and thus the proved inequality verges into the well-known parallelogram equality, which itself follows from our solution if we take (2) as two obvious equalities.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 29348,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $a$, $b$, $c$ be positive real numbers satisfying $a^2 < bc$. Prove that\n$$\nb^3 + ac^2 > ab(a + c)\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Adding three AM-GM inequalities\n$$\n4a^3b + b^3c + 2c^3a \\geq 7a^2bc,\n$$\n$$\n4b^3c + c^3a + 2a^3b \\geq 7b^2ca,\n$$\n$$\n4c^3a + a^3b + 2b^3c \\geq 7c^2ab\n$$\nwe get\n$$\na^3b + b^3c + c^3a \\geq a^2bc + b^2ca + c^2ab \\quad (1)\n$$\nThe assumption $a^2 < bc$ implies $-a^3b > -b^2ca$ and this together with (1) gives\n$$\nb^3c + c^3a > a^2bc + c^2ab\n$$\ni.e.\n$$\nb^3 + ac^2 > ab(a + c)\n$$\n□",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 29349,
    "subject": "Mathematics (Multi-modal)",
    "question": "On the blackboard $n$ nonnegative integers have been written, such that their greatest common divisor is equal to $1$. In one step we can erase two numbers $x$, $y$ such that $x \\ge y$, and replace them with numbers $x - y$, $2y$. Determine, for which sequences of original $n$ integers one can lead to a situation, in which $n - 1$ numbers on the blackboard are zeroes.",
    "options": [],
    "answer": "Exactly those initial sequences whose sum is a power of two (or the trivial case where all entries are zero).",
    "solution": "The answer: the sum of the numbers has to be a power of $2$, or they are all zeroes. From this moment we assume that the numbers are not all zeroes, otherwise we are done without making any step.\n\nLet $S$ be the sum of the numbers on the blackboard and $D$ be their greatest common divisor at some moment. At the beginning $D$ is equal to $1$, at the end it has to be equal to $S$. We now prove that in each step, $D$ either stays the same or is multiplied by $2$. Indeed, let $k_1, \\dots, k_n$ be the numbers on the blackboard, without losing generality let the operation be performed on $(x, y) = (k_1, k_2)$. Then we have that\n$$\n\\text{gcd}(k_1, k_2, k_3, \\dots, k_n) = \\text{gcd}(k_1 - k_2, k_2, k_3, \\dots, k_n)\n$$\nand $\\text{gcd}(k_1 - k_2, k_2, k_3, \\dots, k_n)$ is either $2 \\text{gcd}(k_1 - k_2, k_2, k_3, \\dots, k_n)$ or $\\text{gcd}(k_1 - k_2, k_2, k_3, \\dots, k_n)$ (multiplying one of the arguments by $2$ can multiply the greatest common divisor by $2$ if all the other arguments have more twos in their prime factorizations, or make it stay the same otherwise). As at the end $D$ must be equal to $S$, $S$ has to be a power of two.\n\nNow we prove that if $S$ is a power of two, then it is possible to obtain $n-1$ zeroes. Consider binary representations of all the numbers on the blackboard and write them one over another. Let $\\ell$ be the index of the rightmost column that does not consist of only zeroes, i.e., $\\ell$ is the smallest number such that not for all $i$ $2^\\ell \\mid k_i$ holds. If $n-1$ numbers are zeroes, then we are already done.\n\nOtherwise, observe that in the $\\ell$-th column the number of ones has to be even (as $S$ is a power of $2$, larger than $2^\\ell$). Take $k_i \\ge k_j$ such that they are not divisible by $2^\\ell$ and perform the operation on them. Observe that after doing the operation all the columns with indices lower than $\\ell$ still have only zeroes, however in the $\\ell$-th column the number of ones decrease by $2$. Doing such operations we can delete ones from columns one after another, obtaining at the end the situation, in which we cannot apply the described rule. This means that $n-1$ numbers on the blackboard are zeroes.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 29350,
    "subject": "Mathematics (Multi-modal)",
    "question": "Points $A$, $B$, $C$, $D$ lie on a circle in this order, where $AB$ and $CD$ are not parallel. The length of the arc $\\widehat{AB}$ that contains points $C$, $D$ is twice as large as the length of the arc $\\widehat{CD}$ that does not contain points $A$, $B$. Point $E$ is chosen such that $AC = AE$ and $BD = BE$ and $E$ lies on the same side of the line $AB$ as $C$ and $D$. Assuming that the perpendicular line from the point $E$ to the line $AB$ bisects the arc $\\widehat{CD}$ not containing points $A$, $B$, prove that $\\angle ACB = 108^\\circ$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "We use the following lemma\n\nLEMMA. Given two circles $\\Gamma_1$, $\\Gamma_2$ with the centre $S_2$ of $\\Gamma_2$ lying on the circle $\\Gamma_1$. The circles intersect in two points $K$ and $L$. Let $M$ be the point on the circle $\\Gamma_1$ (different from $K$ and $L$) and the line $KM$ meets $\\Gamma_2$ again in $N$. Then $MN = ML$.\n\n*Proof.*\n![](attached_image_1.png)\nFig. 1a\n![](attached_image_2.png)\nFig. 1b\nFor this lemma it is sufficient to prove that the line $MS_2$ bisects the angle $\\angle NML$. Then in the reflection with respect to the line $MS_2$, $ML$ is the image of $MN$. The circle $\\Gamma_2$ and the point $M$ reflect to themselves, the point $L$ reflects to the point $N$ (the intersection point of $ML$ and $\\Gamma_2$). So the triangle $\\triangle MLN$ is isosceles (some considerations are needed according to the position of the point $M$).\n\nFirstly, let $M$ lie on the arc $\\widehat{KL}$ not containing the point $S_2$. As $S_2K = S_2L$, we directly have $\\angle KMS_2 = \\angle S_2ML$. Secondly, let $M$ lie on the arc $\\widehat{KL}$ containing point $S_2$. Let $R$ be an arbitrary point on the arc $\\widehat{KL}$ not containing the point $S_2$. Similarly as before $\\angle KRS_2 = \\angle S_2RL$, then using identical angles in the cyclic quadrilaterals $RS_2MK$ and $RLS_2M$ we obtain $\\angle NMS_2 = \\angle KRS_2 = \\angle S_2RL = \\angle S_2ML$. $\\blacksquare$\n\nLet the perpendicular line from the point $E$ to the line $AB$ intersect the arc $\\widehat{BC}$ in the point $S$, $k_1$ be the circle centered in $A$ passing through $C$, $k_2$ be the circle centered in $B$ passing through $D$, $k$ be the circle passing through $A$, $B$, $C$, $D$. The line $SC$ intersects $k_1$ again in $C'$ and the line $SD$ intersects $k_2$ again in $D'$. Circles $k_1$ and $k$ meet in $C$, $C''$ and circles $k_2$ and $k$ meet in $D$, $D''$. Using the lemma we have $SC' = SC''$ and $SD' = SD''$. Let the circles $k_1$ and $k_2$ meet again in $E'$. Using a contradiction, we shall prove that $C'' = D'' = E'$.\n\nThe point $S$ lies on the chord $EE' \\perp AB$ of the circles $k_1$ and $k_2$. That means its powers to these two circles are equal. We know that $S$ bisects the arc $\\widehat{CD}$, so $SD = SC$ and consequently $SC'' = SC' = SD' = SD''$. If $D''$ and $C''$ are different points then the triangle $\\triangle SD''C''$ is isosceles and its altitude from $S$ passes through the circumcentre of $k$ and consequently (by the symmetry) the quadrilateral $CDD''C''$ is isosceles trapezoid ($SC = SD$). We can find points $A$ and $B$ as the intersection points of the axes of the segments $CC''$ and $DD''$ with $k$. But then also $ABCD$ is an isosceles trapezoid, $AB \\parallel CD$, which is a contradiction to the given $AB \\nparallel CD$.\n![](attached_image_3.png)\n\nIf we denote $\\angle DE'S = \\angle SE'C = \\alpha$ and $\\angle AE'D = \\beta$ then $\\angle CE'D = 2\\alpha - \\beta$ because $2|\\widehat{CD}| = |\\widehat{AB}|$. Using $BD = BE'$ and $AC = AE'$ we compute the angles in the triangle $ABE'$:\n$$\n\\begin{aligned}\n2\\alpha + \\beta &= \\angle AE'C = \\angle ACE' = \\angle ABE' \\\\\n4\\alpha - \\beta &= \\angle BE'D = \\angle BDE' = \\angle BAE'\n\\end{aligned}\n$$\nwhich yields to\n$$\n180^\\circ = 2\\alpha + \\beta + 4\\alpha - \\beta + 4\\alpha = 10\\alpha\n$$\nand $\\angle ACB = 180^\\circ - \\angle AE'B = 180^\\circ - 4\\alpha = 108^\\circ$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 29351,
    "subject": "Mathematics (Multi-modal)",
    "question": "Polynomial $P(x)$ with integer coefficients satisfies the following condition: for every polynomials $F(x)$, $G(x)$, $Q(x)$ with integer coefficients, if\n$$\nP(Q(x)) = F(x) \\cdot G(x)\n$$\nthen either $F(x)$ or $G(x)$ is a constant polynomial. Prove that $P(x)$ has to be a constant polynomial.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "For the sake of contradiction suppose that $P(x)$ is not constant and consider the case when $P(x)$ is a linear polynomial. It means that $P(x) = a x + b$ for some $a, b \\in \\mathbb{Z}$, where $a \\neq 0$. Let $Q(x) = a x^2 + (b+1)x$. Then\n$$\nP(Q(x)) = a(a x^2 + (b+1)x) + b = a^2 x^2 + a(b+1)x + b = (a x + b)(a x + 1)\n$$\nbut polynomials $a x + b$ and $a x + 1$ are not constant, a contradiction.\n\nNow suppose that $\\deg P = n > 1$. Suppose also that\n$$\nP(x) = a_n x^n + a_{n-1} x^{n-1} + \\dots + a_1 x + a_0,\n$$\nwhere $a_n \\neq 0$. Consider a polynomial $Q(x) = P(x) + x$. Clearly it has integer coefficients. Moreover,\n$$\nP(Q(x)) - P(x) = P(P(x) + x) - P(x) = \\sum_{i=0}^{n} a_i ((P(x) + x)^i - x^i)\n$$\nFrom the formula\n$$\na^i - b^i = (a-b)(a^{i-1} + a^{i-2}b + \\dots + b^{i-1})\n$$\nit follows that the polynomial $(P(x) + x)^i - x^i$ is divisible by the polynomial $P(x)$. So $P(Q(x))$ is divisible by $P(x)$ as well. But this is a contradiction, since $\\deg P > 1$ implies that the degree of $P(Q(x))$ is greater than the degree of $P(x)$, which means that $P(x)$ is a non-trivial divisor of $P(Q(x))$. Conclusion follows. $\\square$",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 29352,
    "subject": "Mathematics (Multi-modal)",
    "question": "In a convex quadrilateral $ABCD$ let $M$ and $N$ be midpoints of sides $AD$ and $BC$, respectively. Points $K$ and $L$ are chosen on sides $AB$ and $CD$, respectively, in such a manner that $\\angle MKA = \\angle NLC$. Prove that if lines $BD$, $KM$ and $LN$ meet at one point, then\n$$\n\\angle KMN = \\angle BDC, \\quad \\angle LNM = \\angle ABD\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $P$ be the midpoint of $BD$ and $Q$ be the common point of lines $BD$, $KM$ and $LN$. Without losing generality assume that point $B$ lies between $Q$ and $D$. By Tales theorem, $PM \\parallel AB$ and $PN \\parallel CD$. Therefore, $\\angle PNL = \\angle NLC = \\angle MKA = \\angle KMP$. Note that this implies points $Q, M, P, N$ to be concyclic, as $\\angle QNP + \\angle QMP = 180^{\\circ}$ and points $M, N$ lie on different sides of the line $BD$. Therefore, $\\angle KMN = \\angle QMN = \\angle QPN = \\angle BDC$. Moreover, $\\angle LNM = 180^{\\circ} - \\angle QNM = 180^{\\circ} - \\angle QPM = \\angle MPD = \\angle ABD$. $\\square$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 29353,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $a$ be a given integer. Prove that there exist infinitely many prime numbers $p$ such that\n$$\np \\mid n^2 + 3, \\quad p \\mid m^3 - a\n$$\nfor some integers $n$ and $m$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $k$ be an arbitrary integer. Note that\n$$\n(9a^2k^3)^2 + 3 = 3(27a^4k^6 + 1)\n$$\nand\n$$\n(9a^3k^4)^3 - a = a(3^6a^8k^{12} - 1) = a(27a^4k^6 - 1)(27a^4k^6 + 1)\n$$\nIt follows that for every $k \\in \\mathbb{Z}$ the number $27a^4k^6 + 1$ is a common divisor of the numbers $n^2 + 3$ and $m^3 - a$ with $n = 9a^2k^3$ and $m = 9a^3k^4$. So it is enough to prove that there are infinitely many primes $p$ such that $p \\mid 27a^4k^6 + 1$ for some integer $k$.\nSuppose that there are only finitely many such primes and these are $p_1, p_2, \\dots, p_r$. If we take $k = p_1p_2\\dots p_r + 1$, then it is clear that the number $27a^4k^6 + 1$ is not divisible by any $p_i$ for $1 \\le i \\le r$ and that it is also greater than 1. It follows that it has a prime divisor $p$, which is different from every $p_i$ for $1 \\le i \\le r$. We have obtained a contradiction, which finishes the proof. $\\square$",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 29354,
    "subject": "Mathematics (Multi-modal)",
    "question": "For any positive integer $n$, let $\\tau(n)$ denote the number of positive divisors of $n$ and $\\varphi(n)$ the number of positive integers not greater than $n$ which are relatively prime to $n$. Find all positive integers $n$ for which one of the three numbers $n$, $\\tau(n)$, and $\\varphi(n)$ is the arithmetic mean of the other two.",
    "options": [],
    "answer": "{1, 4, 6, 9}",
    "solution": "We have $\\tau(1) = \\varphi(1) = 1$, that is $n = 1$ satisfies the given condition. In the following, we assume $n > 1$. For such $n$, clearly $\\tau(n) \\le n$ and $\\varphi(n) < n$. This means $n$ cannot be the arithmetic mean of $\\tau(n)$ and $\\varphi(n)$. We are left with two cases.\n\n*Case 1:* $\\tau(n) = \\frac{1}{2}(\\varphi(n) + n)$. Then we have $\\tau(n) > \\frac{1}{2}n$. For each divisor $d$ of $n$, the number $n/d$ is also the divisor. One of the numbers $d, n/d$ is less or equal $\\sqrt{n}$, which means the set $\\{1, 2, \\dots, \\lfloor\\sqrt{n}\\rfloor\\}$ contains at least half of the divisors¹. This clearly implies $\\frac{1}{2}\\tau(n) \\le \\sqrt{n}$. We get\n$$\n2\\sqrt{n} \\ge \\tau(n) > \\frac{1}{2}n \\quad \\Rightarrow \\quad 4n > \\frac{1}{4}n^2 \\quad \\Rightarrow \\quad 16 > n.\n$$\nFor $1 < n < 16$, we can easily calculate $\\tau(n)$, check the condition $\\tau(n) > \\frac{1}{2}n$, and calculate $\\varphi(n)$ in the remaining few cases:\n\n| n   | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | 11 | 12 | 13 | 14 | 15 |\n|-----|---|---|---|---|---|---|---|---|----|----|----|----|----|----|\n| $\\tau(n)$ | 2 | 2 | 3 | 2 | 4 | 2 | 4 | 3 | 4  | 2  | 6  | 2  | 4  | 4  |\n| $\\tau(n) > 1/2 n$ ? | ✓ | ✓ | ✓ | ✓ | ✓ | ✓ | ✓ | ✓ | ✓  | ✓  | ✓  | ✓  | ✓  | ✓  |\n| $\\varphi(n)$ | 1 | 2 | 2 |   | 2 |   |   |   |    |    |    |    |    |    |\n| $\\tau(n) = 1/2 (\\varphi(n) + n)$ ? | × | × | ✓ |   | ✓ |   |   |   |    |    |    |    |    |    |\n\nWe have $n = 4$ and $n = 6$ as solutions.\n\n*Case 2:* $\\varphi(n) = \\frac{1}{2}(\\tau(n) + n)$. We transform this relation into\n$$\n\\tau(n) = 2\\varphi(n) - n. \\qquad (1)\n$$\nIf $n$ is even, then no even number is relatively prime to $n$, thus $\\varphi(n) \\le \\frac{1}{2}n$. But then from (1) we get $\\tau(n) \\le 0$, which is impossible. Therefore, $n$ must be odd. Then (1) implies $\\tau(n)$ must be odd as well, which means $n$ is a perfect square (of an odd number). Write the prime factorization of $n$ in the form\n$$\nn = p_1^{2\\alpha_1} \\cdots p_k^{2\\alpha_k}, \\quad k \\ge 1, \\ p_i \\ge 3, \\ \\alpha_i \\ge 1.\n$$\n\nApplying the well-known formula for $\\tau(n)$ and $\\varphi(n)$, we rewrite (1) as\n$$\n(2\\alpha_1 + 1) \\cdots (2\\alpha_k + 1) = 2p_1^{2\\alpha_1-1}(p_1-1) \\cdots p_k^{2\\alpha_k-1}(p_k-1) - p_1^{2\\alpha_1} \\cdots p_k^{2\\alpha_k} = \\\\\n= p_1^{2\\alpha_1-1} \\cdots p_k^{2\\alpha_k-1}(2(p_1-1) \\cdots (p_k-1) - p_1 \\cdots p_k).\n$$\nThe right hand side is divisible by $p_1^{2\\alpha_1-1} \\cdots p_k^{2\\alpha_k-1}$, and so must be the left hand side as well. From this, we obtain\n$$\np_1^{2\\alpha_1-1} \\cdots p_k^{2\\alpha_k-1} \\le (2\\alpha_1 + 1) \\cdots (2\\alpha_k + 1). \\quad (2)\n$$\nHowever, for every integers $p \\ge 3$ and $\\alpha \\ge 1$, the inequality $p^{2\\alpha-1} \\ge (2\\alpha + 1)$ holds, with equality only for $p = 3$ and $\\alpha = 1$. To prove this, we use induction on $\\alpha$: The case $\\alpha = 1$ is trivial (with equality only for $p = 3$) and when $\\alpha$ increases by 1, the right hand size increases by 2, while the left hand size increases by 1.\n$$\np^{2(\\alpha+1)-1} - p^{2\\alpha-1} = p^{2\\alpha-1}(p^2 - 1) > 2.\n$$\nTherefore, each of the (positive) factors on the left hand side of (2) is greater or equal than the corresponding factor on the right hand side. The only possible way to satisfy (2) is to put $k = 1$, $p_1 = 3$, $\\alpha_1 = 1$, that is, $n = 9$. Indeed, we have $\\tau(9) = 3$ and $\\varphi(9) = 6$, thus (1) holds for $n = 9$.\n\n**Answer.** The given condition is fulfilled for $n \\in \\{1, 4, 6, 9\\}$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 29355,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all functions $f: \\mathbb{R} \\to \\mathbb{R}$ such that\n$$\nf(x + f(y)) - f(x) = (x + f(y))^4 - x^4\n$$\nfor all $x, y \\in \\mathbb{R}$.",
    "options": [],
    "answer": "All solutions are f(x) = x^4 + k for any real constant k, and the zero function f(x) = 0.",
    "solution": "We rewrite the given equation into the equivalent form\n$$\nf(x + f(y)) = (x + f(y))^4 - x^4 + f(x). \\quad (1)\n$$\nSetting $x = -f(z)$, $y = z$ in (1) we obtain\n$$\nf(0) = -(f(z))^4 + f(-f(z)) \\quad \\text{for all } z \\in \\mathbb{R}. \\quad (2)\n$$\nNow, setting $x = -f(z)$ in (1) and using (2) we get\n$$\nf(f(y) - f(z)) = (f(y) - f(z))^4 - (f(z))^4 + f(-f(z)) = (f(y) - f(z))^4 + f(0)\n$$\nfor all $y, z \\in \\mathbb{R}$. This means that if a number $t$ can be expressed as the difference of two values of $f$, that is $t = f(y) - f(z)$, then $f(t) = t^4 + f(0)$. We show that if $f$ takes any nonzero value then every number is a difference of two values of $f$.\nLet $f(a) = b \\ne 0$. Putting $y = a$ in the original equation we have\n$$\nf(x + b) - f(x) = (x + b)^4 - x^4.\n$$\nSince $b \\ne 0$, the expression on the right hand side is a polynomial of degree 3, and therefore takes every real number as its value when $x$ run over the entire real axis. Hence, the left hand side, which is the difference of two values of $f$, can take any real value. Together with the previous observation we get $f(t) = t^4 + f(0)$ for all $t \\in \\mathbb{R}$. Finally, one can easily check that all functions of the form $f(x) = x^4 + k$ satisfy the given functional equation.\nThe zero function is obviously a solution as well.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 29356,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABCD$ be a cyclic quadrilateral and $\\omega$ its circumcircle. Let $I$, $J$, and $K$ be the incenters of the triangles $ABC$, $ACD$, and $ABD$, respectively. Also, let $E$ be the midpoint of the arc $DB$ of $\\omega$ containing $A$. The line $EK$ intersects $\\omega$ at $F$ ($F \\neq E$). Prove that the points $C$, $F$, $I$, and $J$ lie on the same circle.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $M$ and $N$ be the midpoints of the arcs $AB$ and $AD$ (not containing any other vertices of $ABCD$), respectively. Then the incenter $I$ lies on $CM$, the incenter $J$ lies on $CN$, and the incenter $K$ lies on $BN$. Moreover, we have\n$$\nMI = MA = MB \\quad \\text{and} \\quad NJ = NA = ND = NK \\quad (1)\n$$\n(these well-known relations follow from an easy angle chasing).\n\nObserve also that the incenter $K$ necessarily lies in the interior of the isosceles triangle $BDE$, hence the line $EK$ intersects the segment $BD$ and the point $F$ lies on the same side of $BD$ as $C$ (Fig. 2).\n\n![](attached_image_1.png)\nFig. 2\n\nConsider now the arc $BD$ of $\\omega$ containing $A$. The point $E$ is the midpoint of this arc, and the points $M$ and $N$ are the midpoints of the subarcs $BA$ and $AD$, respectively. It follows that the subarcs $BM$ and $EN$ are of equal length (and similarly the subarcs $ME$ and $ND$ are of equal length). Thus\n$$\n\\angle BFM = \\angle EFN = \\angle KFN. \\quad (2)\n$$\n\nBut clearly we have\n$$\n\\angle BMF = \\angle BNF = \\angle KNF. \\qquad (3)\n$$\nThe equalities (2) and (3) imply that the triangles $MBF$ and $NKF$ are similar, and with the same orientation. Therefore\n$$\n\\frac{MB}{MF} = \\frac{NK}{NF},\n$$\nwhich by an application of (1) can be rewritten as\n$$\n\\frac{MI}{MF} = \\frac{NJ}{NF}.\n$$\nThis, together with the relation\n$$\n\\angle IMF = \\angle CMF = \\angle CNF = \\angle JNF,\n$$\nproves that the triangles $MIF$ and $NJF$ are similar with the same orientation. This orientation-preserving similarity implies that\n$$\n\\angle IFM = \\angle JFN \\quad \\text{and} \\quad \\angle IFJ = \\angle MFN.\n$$\nHence\n$$\n\\angle IFJ = \\angle MFN = \\angle MCN = \\angle ICJ,\n$$\nand the points $C$, $F$, $I$, $J$ are indeed concyclic.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 29357,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABC$ be a right triangle with hypotenuse $AB$ and $P$ be a point inside the shorter arc $AC$ of the circumcircle of the triangle $ABC$. The line perpendicular to $CP$ which passes through $C$ intersects the lines $AP$ and $BP$ in points $K$ and $L$ respectively. Prove that the ratio of the areas of the triangles $BKL$ and $ACP$ does not depend on the choice of the point $P$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Throughout the solution, we shall denote by $S_{XYZ}$ the area of the triangle $XYZ$.\n\n![](attached_image_1.png)\nFig. 3\n\nLet $PR$ be the diameter of the circumcircle of $ABC$ (Fig. 3). Then $ARBP$ is a rectangle. Since $BR$ is parallel to $PA$ we have $S_{PBK} = S_{PRK}$, which implies\n$$\nS_{BKL} = S_{LPR}.\n$$\nSince $PA = BR$ we have $\\angle PCA = \\angle BPR = \\angle LPR$.\nAlso, because $CPAR$ is cyclic we have $\\angle CAP = \\angle CRP$. Hence $LPR$ and $PCA$ are similar triangles.\nFinally, we have\n$$\nS_{BKL} : S_{ACP} = S_{LPR} : S_{ACP} = PR^2 : AC^2 = AB^2 : AC^2\n$$\nwhich does not depend on the point $P$.\n\n\nSecond solution.\n\nDenote the lengths of the sides and the sizes of the angles of the triangle $ABC$ by $a$, $b$, $c$ and $\\alpha$, $\\beta$, $\\gamma$, respectively (Fig. 4). Let $\\angle PAC = \\varphi$. Simple angle chasing gives $\\angle CPB = \\alpha$, $\\angle PBC = \\varphi$, $\\angle LCB = \\angle ACP = \\angle ABP = \\beta - \\varphi$, $\\angle BLC = \\angle APC = 90^\\circ + \\alpha$, $\\angle PKL = \\alpha$. Also, $AP = c \\cos(\\alpha + \\varphi)$, $BP = c \\sin(\\alpha + \\varphi)$.\n\n![](attached_image_2.png)\nFig. 4\n\nUsing the sine law in $BCP$ and $BCL$ we get\n$$\nCP = \\frac{BC \\sin \\varphi}{\\sin \\alpha} = \\frac{a \\sin \\varphi}{\\sin \\alpha},\n$$\n$$\nBL = \\frac{BC \\sin(\\beta - \\varphi)}{\\sin(90^\\circ + \\alpha)} = \\frac{a \\cos(\\alpha + \\varphi)}{\\cos \\alpha}.\n$$\nThen\n$$\nPL = BP - BL = c \\sin(\\alpha + \\varphi) - \\frac{a \\cos(\\alpha + \\varphi)}{\\cos \\alpha} = \\\\\n= \\frac{c \\sin(\\alpha + \\varphi) \\cos \\alpha - c \\sin \\alpha \\cos(\\alpha + \\varphi)}{\\cos \\alpha} = \\frac{c \\sin \\varphi}{\\cos \\alpha}.\n$$\n$$\nKL = \\frac{PL}{\\sin \\alpha} = \\frac{c \\sin \\varphi}{\\sin \\alpha \\cos \\alpha}.\n$$\n\n$$\n\\frac{S_{KLB}}{S_{APC}} = \\frac{\\frac{1}{2}BL \\cdot KL \\cdot \\sin(90^\\circ + \\alpha)}{\\frac{1}{2}AP \\cdot CP \\cdot \\sin(90^\\circ + \\alpha)} = \\frac{BL \\cdot KL}{AP \\cdot CP} = \\frac{\\frac{a \\cos(\\alpha+\\varphi)}{\\cos \\alpha} \\cdot \\frac{c \\sin \\varphi}{\\sin \\alpha \\cos \\alpha}}{c \\cos(\\alpha + \\varphi) \\cdot \\frac{a \\sin \\varphi}{\\sin \\alpha}} = \\frac{1}{\\cos^2 \\alpha},\n$$\nand $\\alpha$ is clearly independent on the choice of the point $P$.\n\n\n*Remark.* We can shorten the previous solution by noticing that from the angles we have a pair of similar triangles $BLC$, $APC$ (Fig. 4). Then $BL/AP = LC/PC$ and the desired ratio can be expressed as\n$$\n\\frac{S_{KLB}}{S_{APC}} = \\frac{BL \\cdot KL}{AP \\cdot CP} = \\frac{LC \\cdot KL}{PC^2}.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 29358,
    "subject": "Mathematics (Multi-modal)",
    "question": "The city of Mar del Plata has a shape of the square $WSEN$. For a given even integer $n$, it is divided by $2(n + 1)$ streets into $n \\times n$ blocks (the streets run also around the square). Each block has a size of 100 m x 100 m. All the streets in Mar del Plata are one-way. Each street has one direction along the entire length. Every two adjacent parallel streets have opposite directions. The $WS$ street is directed from $W$ to $S$ and the $WN$ street is directed from $W$ to $N$. A street cleaning car starts at $W$. Its aim is to arrive to $E$ and clean the roads it passes through. What is the length of the longest possible route it can pass if no 100 m segment can be passed more than once? (For $n = 6$, Fig. 1 shows the map of the city and one of the possible, but not the longest, routes of the car. See also http://goo.gl/maps/JAzD.)\n\n![](attached_image_1.png)\nFig. 1",
    "options": [],
    "answer": "(2n^2 - 2n + 4)/10 km",
    "solution": "First we introduce some notation. Call each 100 m segment of a street an arrow, and each place where two streets meet a crossroad. If an arrow has the same direction as the street $WS$ or $WN$ (that is, it runs from the upper left to the bottom right or from the bottom left to the upper right), it is called a forward arrow, otherwise it is a back arrow.\n\nIn the solution we will use the following obvious lemma: Let the set of all crossroads be split into two sets $A$ and $B$ such that $W \\in A$ and $E \\in B$. Then the number of times the car drives along an arrow from $A$ to $B$ is one more than the number of times the car drives along an arrow from $B$ to $A$.\n\nLet us split the set of all crossroads of Mar del Plata into two sets $A$ and $B$ by a vertical (i.e. north-south) line connecting two points located $100k + 50$ m from $W$ – one on the $WN$ street and one on the $WS$ street – for some $k \\in \\{0, 1, \\dots, n-1\\}$. Fig. 5a and 5b show the splitting line for $k = 3$ and $k = 4$ respectively.\n\n![](attached_image_2.png)\nFig. 5a\n\n![](attached_image_3.png)\nFig. 5b\n\nIf $k$ is odd then the line intersects $k + 1$ forward arrows (going from $\\mathcal{A}$ to $\\mathcal{B}$) and $k + 1$ back arrows (going from $\\mathcal{B}$ to $\\mathcal{A}$). Even if the car would pass all the $k + 1$ forward arrows, by the lemma, it can pass at most $k$ back arrows. Hence, at least one of the back arrows remains unpassed.\n\nIf $k \\ge 2$ is even then the line intersects $k + 2$ forward arrows and $k$ back arrows. The two northernmost forward arrows intersected by the line begin in the crossroad which has only one incoming arrow. This crossroad can be passed only once, hence one of the two northernmost forward arrows remains unpassed. The same holds for the two southernmost forward arrows. So we have at most $k$ forward arrows passed by the car and by the lemma at most $k - 1$ back arrows. Together, we have at least 3 unpassed arrows at this level.\n\nFor $k = 0$, we have only two forward arrows starting in $W$ intersected by the line. Clearly only one of them could be passed by the car and one remains unpassed.\n\nIn a similar way, we split the set of all crossroads by a vertical line connecting two points located $100k + 50$ m from $E$ – one on the $SE$ street and one on the $NE$ street – for some $k \\in \\{0, 1, \\dots, n-1\\}$. The situation is sketched on Fig. 6a and 6b $k = 3$ and $k = 4$ respectively.\n\n![](attached_image_4.png)\nFig. 6a\n\n![](attached_image_5.png)\nFig. 6b\n\nIf $k$ is odd then the line intersects $k+1$ forward arrows and $k+1$ back arrows, and at least one of the back arrows remains unpassed.\n\nIf $k \\ge 2$ is even then the line intersects $k+2$ forward arrows and $k$ back arrows. The two northernmost forward arrows end in the crossroad with only one outgoing arrow, hence one of them remains unpassed. The same holds for the two southernmost forward arrows. Again, we have at most $k$ forward arrows and at most $k-1$ back arrows passed by the car, which results in 3 unpassed arrows at this level.\n\nFor $k = 0$, we have two forward arrows ending in $E$, only one of them could be passed by the car and one remains unpassed.\n\nThere is one unpassed arrow for any odd $k$, three for any even $k \\ge 2$, and one for $k = 0$, and all this happens twice. As $n$ is even and $k \\in \\{0, 1, \\dots, n-1\\}$, altogether, we have\n$$\n2\\left(\\frac{1}{2}n + 3\\left(\\frac{1}{2}n - 1\\right) + 1\\right) = 4n - 4\n$$\nunpassed arrows. The total number of arrows is $n \\cdot 2(n+1)$, hence, the car cannot pass more than\n$$\nn \\cdot 2(n + 1) - (4n - 4) = 2n^2 - 2n + 4\n$$\narrows.\n\nOn the other hand, there are many possible routes of the car consisting of $2n^2 - 2n + 4$ arrows. One is sketched on Fig. 7 for $n = 6$. When we use the same pattern in general case, the route can be subdivided into $\\frac{1}{2}n$ parts by the crossroads lying on the $WS$ street located $200k$ m from $W$ for $k = 1, 2, \\dots, \\frac{1}{2}n - 1$.\n\n![](attached_image_6.png)\nFig. 7\n\nThe first $\\frac{1}{2}n-1$ parts differ only by shifting. Each of them consist of $n$ arrows of the same direction as $WN$, $n$ arrows of the opposite direction as $WN$, and $2(n-1)$ arrows perpendicular to $WN$. The last part consists of $n$ arrows of the same direction as $WN$ and $2(n+1)$ arrows perpendicular to $WN$. Therefore, the number of arrows on the entire route is\n$$\n(\\frac{1}{2}n - 1)(2n + 2(n - 1)) + n + 2(n + 1) = 2n^2 - 2n + 4.\n$$\n\nAnswer. The length of the longest possible route of the car is $\\frac{1}{10}(2n^2 - 2n + 4)$ km.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 29359,
    "subject": "Mathematics (Multi-modal)",
    "question": "Positive real numbers $a$, $b$, $c$, $d$ satisfy the relations\n$$\nabcd = 4, \\quad a^2 + b^2 + c^2 + d^2 = 10.\n$$\nDetermine the largest possible value of the expression $ab + bc + cd + da$.",
    "options": [],
    "answer": "sqrt(82)",
    "solution": "Let $V = ab + bc + cd + da$. We will find the maximum value of\n$$\nV^2 = (a+c)^2(b+d)^2 = (a^2+c^2+2ac)(b^2+d^2+2bd). \\quad (1)\n$$\nAll the given expressions do not change under the simultaneous replacement of $a$ by $b$, $b$ by $c$, $c$ by $d$ and $d$ by $a$. Since $ac \\cdot bd = 4$, at least one of the numbers $ac$ and $bd$ is at least $2$. We may assume $bd \\ge 2$.\nSimple manipulations yield $ac = 4/bd$ and $a^2 + c^2 = 10 - b^2 - d^2$. We plug these expressions into (1) and get\n$$\n\\begin{aligned}\nV^2 &= \\left(10 - b^2 - d^2 + \\frac{8}{bd}\\right) (b^2 + d^2 + 2bd) = \\\\\n&= 10(b^2 + d^2) + 20bd + \\frac{8(b^2 + d^2)}{bd} + 16 - (b^2 + d^2)^2 - 2bd(b^2 + d^2).\n\\end{aligned} \\quad (2)\n$$\n\nLet $P = b^2 + d^2$ and $Q = bd$; then $P \\ge 2Q$ and $Q \\ge 2$. Thus (2) becomes\n$$\n\\begin{align*}\nV^2 &= 10P + 20Q + \\frac{8P}{Q} + 16 - P^2 - 2PQ \\\\\n&= -(P^2 - 10P + 25) + \\left(41 - 2PQ + 20Q + \\frac{8P}{Q}\\right) \\\\\n&= -(P - 5)^2 + \\left[P\\left(\\frac{8}{Q} - 2Q\\right) + 41 + 20Q\\right].\n\\end{align*}\n$$\nObviously, $-(P-5)^2 \\le 0$. The condition $Q \\ge 2$ implies $8/Q - 2Q \\le 8/2 - 2 \\cdot 2 = 0$, which means the expression in the brackets is linear in $P$ with non-positive slope and it achieves its maximum for the smallest possible $P$. By $P \\ge 2Q$ we obtain\n$$\nV^2 \\le 2Q \\left( \\frac{8}{Q} - 2Q \\right) + 41 + 20Q = -4Q^2 + 20Q + 57 = -(2Q - 5)^2 + 82 \\le 82.\n$$\n\nFinally, we shall show that there are positive real numbers $a$, $b$, $c$, $d$ such that $V = \\sqrt{82}$. The equality occurs if $P = 2Q = 5$, which is true for $b = d = \\frac{1}{2}\\sqrt{10}$. The numbers $a$ and $c$ satisfy $a^2 + c^2 = 5$, $ac = 8/5$. Thus\n$$\n\\{a, c\\} = \\left\\{ \\frac{\\sqrt{41}-3}{2\\sqrt{5}}, \\frac{\\sqrt{41}+3}{2\\sqrt{5}} \\right\\} .\n$$\n\nAnswer. The maximum possible value of $ab + bc + cd + da$ is $\\sqrt{82}$.",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 29360,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABCD$ be a cyclic quadrilateral with $BC = CD$, let $\\omega$ be the circle centered at $C$ tangent to $BD$, and let $I$ be the incenter of $ABD$. Show that the line through $I$ parallel to $AB$ is tangent to $\\omega$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $p$ be the line tangent at $D$ to the circumcircle $\\Gamma$ of $ABCD$. Since $C$ is the midpoint of the arc $BD$, we have $\\angle(CD, p) = \\angle CAD = \\angle BAC = \\angle BDC$, and we see that $p$ is tangent to $\\omega$. Similarly, if $E$ is the midpoint of the arc $DA$ of $\\Gamma$, then $p$ is tangent to the circle $\\omega'$ centered at $E$ tangent to $DA$. Thus the line $q$ symmetric to $p$ with respect to $CE$ is tangent to $\\omega$ and $\\omega'$ (Fig. 2). However, the well-known relations $CD = CI$ and $ED = EI$ imply that $D$ and $I$ are symmetric with respect to $CE$. Hence $I$ lies on $q$ and it remains to show that $q \\parallel AB$. This follows from\n$$\n\\angle(q, IC) = \\angle(CD, p) = \\angle CAD = \\angle BAC\n$$\n(all angles here are directed).\n\n![](attached_image_1.png)\nFig. 2\nLet $q$ denote the line through $I$ parallel to $AB$. $P$ is the orthogonal projection of $C$ on $q$, $M$ is the midpoint of $BD$ (Fig. 3). We have\n$$\n\\angle CIP = \\angle CAB = \\angle CDB = \\angle CDM\n$$\nApplying the well-known relation $CD = CI$, we conclude right triangles $CDM$ and $CIP$ are congruent, so $CM = CP$. The conclusion follows.\n\n![](attached_image_2.png)\nFig. 3",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 29361,
    "subject": "Mathematics (Multi-modal)",
    "question": "Show that for any real $x > 0$ and integer $n > 0$ we have\n$$\nx^n + \\frac{1}{x^n} - 2 \\ge n^2 \\left(x + \\frac{1}{x} - 2\\right).\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Without loss of generality assume that $y = \\sqrt{x} > 1$. The identity\n$$\na^2 + \\frac{1}{a^2} - 2 = \\left(a - \\frac{1}{a}\\right)^2\n$$\nreduces the problem to showing that\n$$\ny^n - \\frac{1}{y^n} \\ge n\\left(y - \\frac{1}{y}\\right)\n$$\nor $y^{2n} - n(y^{n+1} - y^{n-1}) - 1 \\ge 0$. Upon division by $y - 1 > 0$ this follows from the following computation:\n$$\n\\begin{aligned}\n\\frac{y^{2n} - 1}{y - 1} - \\frac{n y^{n-1}(y^2 - 1)}{y - 1} &= \\sum_{i=0}^{2n-1} y^i - n(y^{n-1} + y^n) = \\\\\n&= \\sum_{i=0}^{n-1} (y^{2n-1-i} - y^n - y^{n-1} + y^i) = \\sum_{i=0}^{n-1} y^i (y^{n-1-i} - 1)(y^{n-i} - 1) \\ge 0.\n\\end{aligned}\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 29362,
    "subject": "Mathematics (Multi-modal)",
    "question": "For each rational number $r$ consider the statement: If $x$ is a real number such that $x^2 - rx$ and $x^3 - rx$ are rational numbers, then $x$ is rational as well.\n\na) Prove the statement for $r \\ge \\frac{4}{3}$ and for $r \\le 0$.\n\nb) Let $p, q$ be different odd primes such that $3p < 4q$. Show the statement is false for $r = \\frac{p}{q}$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "a) Let $s = x^2 - rx$ and $t = x^3 - rx$ be rational. Then\n$$\nx^2 = s + rx,\n$$\n$$\nx^3 = x^2 \\cdot x = (s + rx)x = sx + rx^2 = sx + r(s + rx) = (r^2 + s)x + rs,\n$$\nconsequently\n$$\nt = x^3 - rx = ((r^2 + s)x + rs) - rx = (r^2 - r + s)x + rs.\n$$\nif $(r^2 - r + s) \\neq 0$ (which is $x^2 - rx + r^2 - r \\neq 0$), then\n$$\nx = \\frac{t - rs}{r^2 - r + s}\n$$\nis rational.\nWe conclude that the given statement holds iff the equation\n$$\nx^2 - rx + r^2 - r = 0 \\quad (1)\n$$\nhas no irrational roots. For the sake of completeness, note that if (1) has a rational root then $s$ and $t$ are rational as well:\n$$\ns = x^2 - rx = r - r^2 \\quad \\text{and} \\quad t = 0 \\cdot x + rs = rs = r(r - r^2).\n$$\nIf the determinant $D = r(4 - 3r)$ of (1) is less or equal 0, then (1) has no real solutions or a solution $x = \\frac{r}{2}$, which is rational. Since $D \\le 0 \\Leftrightarrow (r \\ge \\frac{4}{3}$ or $r \\le 0)$ the statement under consideration is true.\n\nb) According to a) it suffices to show that $D > 0$ and $\\sqrt{D}$ is irrational. We have\n$$\nD = r(4 - 3r) = \\frac{p}{q}\\left(4 - \\frac{3p}{q}\\right) = \\frac{p(4q - 3p)}{q^2} > 0.\n$$\nBut $p \\nmid 4q$ thus $p \\mid p(4q - 3p)$ and $p^2 \\nmid p(4q - 3p)$, that is $p(4q - 3p)$ is not a perfect square, that is $\\sqrt{D}$ is irrational.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 29363,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $a$, $b$ be integers with $b$ not a perfect square. Show that $x^2 + a x + b$ can be a perfect square only for finitely many integers $x$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let us examine the diophantine equation $x^2 + a x + b = y^2$ with unknown integers $x$ and $y$. It can be transformed into the form $(2x + 2y + a)(2x - 2y + a) = a^2 - 4b$. Since we assume $b$ is not a perfect square, $a^2 - 4b \\neq 0$. There are only finitely many ways to write $a^2 - 4b$ as a product of two integers. Each such factorization gives two linear equations for $x, y$ which have at most one integer solution. Thus there are only finitely many such $x$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 29364,
    "subject": "Mathematics (Multi-modal)",
    "question": "Triangular lattice cuts an equilateral triangle with a side of length $n$ into $n^2$ triangular cells (Fig. 1). Some of the cells are infected. A cell, which is not infected yet, could be infected if it is neighbouring (by side) with at least two already infected cells. Determine the minimal amount of initially infected cells such that eventually every cell in the triangle could get infected if $n = 12$.\n\n![](attached_image_1.png)\nFig. 1",
    "options": [],
    "answer": "45",
    "solution": "Notice, that with a contamination of one cell, the perimeter of infected area decreases at least by 1. Let $k$ cells be infected at the beginning. Then the perimeter is at most $3k$. It takes $n^2 - k$ contaminations to get the whole triangle infected. The perimeter of the (infected) area is then $3n$. Thus $3n \\le 3k - (n^2 - k)$, or\n$$\nk \\ge \\frac{n^2 + 3n}{4}.\n$$\nFor $n = 12$ the estimate gives $k \\ge 45$. The layout of 45 initially infected cells which cause the infection of the whole system could be as illustrated on Fig. 4.\n\n![](attached_image_2.png)\n![](attached_image_3.png)\n\nAnother option is to cover the triangle with three equilateral triangles of side 4 (covering one side of the big triangle) and the rest could be covered with three rhombs of side 4 (Fig. 5). Each of the smaller triangles could be infected with 7 cells, each of the rhombs with 8 cells. Possible initial state for infecting an equilateral triangle with side 4 and a rhomb with side 4 is on Fig. 6a and Fig. 6b, respectively. (Each triangle with side 4 can get infected itself independently of the rest area. To get an entire rhomb infected, we need the area to its left and upper side to be infected first.)\n\n![](attached_image_4.png)\n![](attached_image_5.png)",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 29365,
    "subject": "Mathematics (Multi-modal)",
    "question": "Given triangle $ABC$ and its circumcircle. Point $P$ is midpoint of arc $BAC$. Circle with diameter $CP$ cuts angle bisector of $\\angle BAC$ at points $K, L$ (point $K$ lies closer to $A$ than $L$). Point $M$ is symmetric to $L$ with respect to line $BC$. Prove that circumcircle of triangle $BKM$ bisects segment $BC$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $D$ be the midpoint of arc $BC$, $N$ the midpoint of $BC$, $X$ the orthogonal projection of $P$ onto $AC$. Then points $P, X, K, L, N, C$ are concyclic (Fig. 7).\n\n![](attached_image_1.png)\nFig. 7\n\nWe have $\\angle PNX = \\angle PCA = \\angle PDA$ therefore $XN \\parallel KL$. It follows that $LN = KX$, and $\\angle LCN = \\angle KCA$. Obviously $\\angle BAK = \\angle LAC$, so $K, L$ are isogonal conjugates with respect to triangle $ABC$. Thus following equalities hold: $\\angle MBC = \\angle CBL = \\angle KBA$ and $\\angle BCM = \\angle LCB = \\angle ACK$. This implies that $A, M$ are isogonal conjugates with respect to triangle $KBC$. Using this we get $\\angle BNM = \\angle LNB = \\angle LKC = \\angle BKM$ therefore points $B, M, N, K$ are concyclic.\nIn the first solution, it is proven that points $K, L$ are isogonal conjugates with respect to $ABC$. Thus angle bisectors of $\\angle CBA$ and $\\angle BCA$ are angle bisectors of $\\angle LBK$ and $\\angle KCL$, respectively. From angle bisector theorem we deduce that $BK/BL = KC/LC$. This means that points $K, L$ lie on Apollonian circle $\\omega$ of points $B, C$. Since $\\omega$ is self-symmetric with respect to line $BC$, we get that $M$ lies on $\\omega$ (Fig. 8).\n\n![](attached_image_2.png)\nFig. 8\n\nLet $\\omega$ intersect segment $BC$ at point $Y$. Then $KY$ is angle bisector of both angles $\\angle MKL$ (because $Y$ is the midpoint of arc $LM$) and $\\angle BKC$ (because $K$ lies on Apollonian circle). Therefore $\\angle BKM = \\angle LKC$. We end proof as we did in the first solution.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 29366,
    "subject": "Mathematics (Multi-modal)",
    "question": "A strange calculator has only two buttons with positive integers, each consisting of two digits. It displays the number $1$ at the beginning. Whenever a button with number $N$ is pressed, the calculator replaces the displayed number $X$ with the number $X \\cdot N$ or $X + N$. Multiplication and addition alternate, multiplication is the first. (For example, if the number $10$ is on the 1st button, the number $20$ is on the 2nd button, and we consecutively press the 1st, 2nd, 1st, and 1st button, we get the results $1 \\cdot 10 = 10$, $10 + 20 = 30$, $30 \\cdot 10 = 300$, and $300 + 10 = 310$.) Decide whether there exist particular values of the two-digit numbers on the buttons such that one can display infinitely many numbers ending with\n(a) $2015$,\n(b) $5813$.",
    "options": [],
    "answer": "Yes for both. For example: (a) choose 31 and 34 to obtain the target and then repeat the two-step cycle; (b) choose 47 and 62 to obtain the target and then repeat. Moreover, choosing 11 and 12 allows generating all four-digit endings infinitely often.",
    "solution": "Let $a, b$ be the numbers written on the buttons. Consider the sequence $(x_n)_{n=0}^{\\infty}$ such that $x_{n+1}$ is formed by the last four digits of $a(x_n + b)$ for each $n \\ge 0$, that is,\n$$\nx_{n+1} \\equiv a(x_n + b) \\pmod{10\\,000} \\quad \\text{and} \\quad 0 \\le x_{n+1} < 10\\,000.\n$$\nSince there are only finitely many different possible values for $x_n$ and each term is only dependent on the value of the previous term, the sequence must be periodic, with the period starting in the term whose value first occurs for the second time in the sequence.\nIn general, the period does not have to start with $x_0$ (e.g., if we take $x_0 = 1$, $a = 10$, and $b = 10$, then all the terms except $x_0$ end with zero, hence the value of $x_0$ never occurs again). However, consider the special case when $a$ is coprime with $10\\,000$. We claim that then the period starts with $x_0$. In fact, suppose that $x_n$ is the first term of the sequence whose value occurs again, say $x_m = x_n$, $m > n$. If $n > 0$, the equality can be rewritten as $a(x_{n-1} + b) \\equiv a(x_{m-1} + b) \\pmod{10\\,000}$, that is,\n$$\n10\\,000 \\mid a(x_{n-1} + b) - a(x_{m-1} + b) = a(x_{n-1} - x_{m-1}).\n$$\nSince $a$ is coprime with $10\\,000$, we have $10\\,000 \\mid x_{n-1} - x_{m-1}$, implying $x_{n-1} = x_{m-1}$. But this is in contrary with the assumption $x_n$ was the first term whose value repeats.\nThe previous paragraph suggests the algorithm how to display the requested number infinitely many times: We only need to produce the number once, using buttons with $a$ coprime with $10\\,000$, and then repeat the sequence $+b, \\cdot a, +b, \\cdot a, \\dots$ forever. In case we get the requested number with even number of presses, i.e., ending with addition, repeating the sequence $\\cdot a, +b, \\cdot a, +b, \\dots$ will do the same desired effect.\nThere are many ways how to display $2015$ using only few presses. E.g., we can try to find $a, b$ such that $(1 \\cdot a + b) \\cdot a = 2015$. Since $2015 = 5 \\cdot 13 \\cdot 31$, we can take $a = 31$ and $b = 5 \\cdot 13 - a = 65 - 31 = 34$. Indeed, we will get\n$$\n1 \\xrightarrow{\\cdot 31} 31 \\xrightarrow{+34} 65 \\xrightarrow{\\cdot 31} 2015.\n$$\nNotice that $31$ is coprime with $10\\,000$. Therefore the (a) part is solved.\n\nIn the part (b), we analogously wish to generate the sequence $(x_n)$ described above with $x_0 = 5813$. However, it is not so easy to produce the initial occurrence of $5813$ using only a few presses. Therefore, we shall instead produce $5813 + b$, knowing that performing the sequence $\\cdot a, +b, \\cdot a, \\dots$ afterwards will eventually reach $5813$. One possible way is to find $a, b$ according to the schema\n$$\n1 \\xrightarrow{\\cdot b} b \\xrightarrow{+b} 2b \\xrightarrow{\\cdot a} 2ab \\xrightarrow{+a} 2ab + a = 5813 + b.\n$$\nThe equation can be rewritten as $(2a-1)(2b+1) = 11625 = 3 \\cdot 5^3 \\cdot 31$. From there, we can easily deduce several 2-digit solutions, one of which is $a = 47, b = 62$. Since $47$ is coprime with $10\\,000$, the procedure\n$$\n1 \\xrightarrow{\\cdot 62} 62 \\xrightarrow{+62} 124 \\xrightarrow{\\cdot 47} 5828 \\xrightarrow{+47} 5875 \\xrightarrow{\\cdot 47} - \\xrightarrow{+62} - \\xrightarrow{\\cdot 47} - \\xrightarrow{+62} \\dots\n$$\nwill reach numbers ending with $5875 - 62 = 5813$ infinitely many times.\nOne can construct a sequence in which every 4-digit number occurs infinitely many times. Consider the calculator with $a = 11$ and $b = 12$. Applying the same arguments as above, we deduce that if we keep pressing the first button (with number $11$ on it, which is coprime with $10\\,000$), after a finite even number of presses we obtain a number ending with $0001$ again. If we change the button in the very last operation, that is, we replace $+11$ with $+12$, we get a number ending with $0002$. In the same way, we can always increment the number formed by the last 4 digits by $1$ (or change $9999$ to $0000$). The conclusion follows.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 29367,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all quadruplets $(a, b, c, d)$ of real numbers satisfying the system\n$$\n\\begin{aligned}\n(a + b)(a^2 + b^2) &= (c + d)(c^2 + d^2), \\\\\n(a + c)(a^2 + c^2) &= (b + d)(b^2 + d^2), \\\\\n(a + d)(a^2 + d^2) &= (b + c)(b^2 + c^2).\n\\end{aligned}\n$$",
    "options": [],
    "answer": "(t, t, t, t) for any real t",
    "solution": "Let us set\n$$\nf(x, y) = (x + y)(x^2 + y^2).\n$$\nWe'll show that for any real $x$ the inequality $y \\ge z$ implies $f(x, y) \\ge f(x, z)$. After subtraction we see that\n$$\nf(x, y) - f(x, z) = \\frac{1}{2}(y - z)((x + y)^2 + (y + z)^2 + (z + x)^2) \\ge 0.\n$$\nMoreover, equality occurs when $y = z$ or $x = y = z = 0$, so either way it implies $y = z$.\n\nWe can rewrite the system (implicitly using the symmetry of $f$) to the form:\n$$\n\\begin{aligned}\nf(a, b) &= f(c, d) \\\\\nf(a, c) &= f(b, d) \\\\\nf(a, d) &= f(b, c)\n\\end{aligned}\n$$\nNow we can see that the system is symmetric in variables $a, b, c, d$ and may assume $a = \\max\\{a, b, c, d\\}$. We then write the chain of (in)equalities\n$$\nf(c, d) = f(a, b) \\ge f(c, b) = f(a, d) \\ge f(b, d) = f(a, c) \\ge f(d, c)\n$$\nand since we in fact have equality everywhere, we deduce $a = b = c = d$.\nAll such quadruplets clearly satisfy the system so the problem is solved. □",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 29368,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $P$ be a non-degenerate polygon with $n$ sides, where $n > 4$. Prove that there exist three distinct vertices $A, B, C$ of $P$ with the following property: If $l_1, l_2, l_3$ are the lengths of the three polygonal chains into which $A, B, C$ break the perimeter of $P$, then there is a triangle with side lengths $l_1, l_2, $ and $l_3$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "By scaling, we can assume w.l.o.g. that the perimeter of $P$ has length $2$. Let $X_1, \\dots, X_n$ be the vertices of $P$, and let $x_i = |X_i X_{i+1}|$, where $X_{n+1} = X_1$; then $\\sum_{i=1}^n x_i = 2$. Since $P$ is a non-degenerate polygon, we have that $x_i < 1$ for all $i = 1, 2, \\dots, n$. To prove the claim it is sufficient to partition the cyclic sequence $(x_1, \\dots, x_n)$ into three intervals such that the sum in each interval is strictly smaller than $1$; the endpoints of these intervals correspond to the selected vertices $A, B, C$ of $P$. To this end, we distinguish two cases: either there is some $p$ such that $\\sum_{i=1}^p x_i = 1$, or there is no such $p$.\n\nIn the first case, the perimeter of $P$ can be broken into two polygonal chains of length $1$ each, both with endpoints $X_1$ and $X_{p+1}$. Since $x_i < 1$ for all $i$, both these chains consist of at least two segments. Since $n > 4$, we infer that the four segments incident to $X_1$ and $X_{p+1}$, namely $X_n X_1$, $X_1 X_2$, $X_p X_{p+1}$ and $X_{p+1} X_{p+2}$, are pairwise different, and they do not constitute the whole perimeter. Hence $x_n + x_1 + x_p + x_{p+1} < 2$, so either $x_n + x_1 < 1$ or $x_p + x_{p+1} < 1$. In the former subcase we can take intervals $(x_n, x_1)$, $(x_2, \\dots, x_p)$ and $(x_{p+1}, \\dots, x_{n-1})$, and in each of them the sum is clearly smaller than $1$. Symmetrically, in the latter subcase we can take intervals $(x_p, x_{p+1})$, $(x_{p+2}, \\dots, x_n)$, and $(x_1, \\dots, x_{p-1})$.\n\nWe are left with the second case. Let $q$ be the largest index such that $\\sum_{i=1}^q x_i < 1$. Since $x_1 < 1$ and $x_n < 1$, we have that $1 \\le q \\le n-2$. By the choice of $q$ and the fact that the first case was not applicable, we have that $\\sum_{i=1}^{q+1} x_i > 1$, hence $\\sum_{i=q+2}^n x_i = 2 - \\sum_{i=1}^{q+1} x_i < 1$. As $x_{q+1} < 1$, we can take intervals $(x_1, \\dots, x_q)$, $(x_{q+1})$, and $(x_{q+2}, \\dots, x_n)$, and in each of them the sum is strictly smaller than $1$. $\\square$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 29369,
    "subject": "Mathematics (Multi-modal)",
    "question": "Prove that for every non-negative integer $n$ there exist integers $x, y, z$ with $\\gcd(x, y, z) = 1$, such that\n$$\nx^2 + y^2 + z^2 = 3^{2n}.\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "We will use the following algebraic formula:\n$$\n(x^2 + y^2 + z^2)^2 = (x^2 + y^2 - z^2)^2 + (2xz)^2 + (2yz)^2.\n$$\nThis means that if a positive integer can be represented as a sum of 3 squares, then so can its square. Consequently, if we put\n$$\n\\begin{aligned}\n(x_0, y_0, z_0) &= (1, 1, 1) \\\\\n(x_{n+1}, y_{n+1}, z_{n+1}) &= (x_n^2 + y_n^2 - z_n^2, 2x_nz_n, 2y_nz_n),\n\\end{aligned}\n$$\nthen by a trivial induction we obtain that $x_n^2 + y_n^2 + z_n^2 = 3^{2n}$.\nIt remains to show that $\\gcd(x_n, y_n, z_n) = 1$ for all $n$ and we proceed by induction.\nSuppose $\\gcd(x_n, y_n, z_n) = 1$, but $x_{n+1}, y_{n+1}, z_{n+1}$ have some common prime divisor $p$. Observe that since $3^{2n}$ is odd, but $2x_nz_n$ and $2y_nz_n$ are even, we have that $x_n^2 + y_n^2 - z_n^2$ is odd, so $p \\ne 2$. Hence $p|x_nz_n$ and $p|y_nz_n$. Then either $p|z_n$, or $p|x_n$ and $p|y_n$. In the latter case we could infer from $p|x_n^2 + y_n^2 - z_n^2$ that in fact also $p|z_n$, which contradicts the assumption that $\\gcd(x_n, y_n, z_n) = 1$. Hence we are left with the first case: $p|z_n$.\nSince $p|x_n^2 + y_n^2 - z_n^2$ and $p|z_n$, we also have that $p|x_n^2 + y_n^2 + z_n^2 = 3^{2n}$. Hence in fact $p = 3$. But the only quadratic residues modulo 3 are 0 and 1, so the two possibilities for a sum of three squares to be divisible by 3 is that either all or none of them is divisible by 3. The former case is excluded by the assumption that $\\gcd(x_n, y_n, z_n) = 1$ and the latter by $p|z_n$.\nThat's all. □",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 29370,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABC$ be an acute-angled triangle with $AB < AC$. Tangent to its circumcircle $\\Omega$ at $A$ intersects the line $BC$ at $D$. Let $G$ be the centroid of $ABC$ and let $AG$ meet $\\Omega$ again at $H \\neq A$. Suppose the line $DG$ intersects the lines $AB$ and $AC$ at $E$ and $F$, respectively. Prove that $\\angle EHG = \\angle GHF$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $p$ be the line parallel to $BC$ that passes through $A$ and let $P = p \\cap DG$. We denote the midpoint of $BC$ by $T$.\nWe project the harmonic ratio $(BTC_\\infty) = -1$ from $A$ onto the line $DG$ and learn that $(EGFP) = -1$. Therefore by a well-known Apollonian property of harmonic ratios it suffices to prove $\\angle PHA = 90^\\circ$.\n\nNow let $Q$ be the orthogonal projection of $D$ onto $AH$. The homothety centered at $G$ with factor $-\\frac{1}{2}$ maps the line $p$ onto the line $BC$ and hence it maps $P$ to $D$. Moreover, it leaves the line $AH$ intact, so we just need to prove that it maps $H$ to $Q$.\nAs $H$ lies on the circumcircle of $ABC$, which is mapped to the nine-point circle $\\omega$ of $ABC$ by the considered homothety, we just need to verify that $Q$ lies on $\\omega$ and that $Q$ is not the “wrong” intersection point of $\\omega$ and $AH$. But this other point is the midpoint of $BC$, which does not coincide with $Q$ when $AB \\neq AC$.\nSo our current claim is just that $Q$ lies on $\\omega$. To verify it, we denote the orthogonal projection of $A$ to $BC$ by $R$ and the midpoint of $AB$ by $S$. It is known that $R, S$, and $T$ lie on $\\omega$. Further, the points $Q$ and $R$ lie on the circle with diameter $AD$. Hence\n$$\n\\angle RQT = \\angle BDA = \\beta - \\angle BAD = \\beta - \\gamma = \\alpha - (180^\\circ - 2\\beta) = \\angle BST - \\angle BSR = \\angle RST\n$$\nand we may conclude. $\\square$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 29371,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all positive real numbers $c$ such that there are infinitely many pairs of positive integers $(n, m)$ satisfying the following conditions: $n \\ge m + c\\sqrt{m-1} + 1$ and among numbers $n, n+1, \\dots, 2n-m$ there is no square of an integer.",
    "options": [],
    "answer": "all positive real c with c ≤ 2",
    "solution": "We prove that $c$ satisfies the condition in the statement if and only if $c \\le 2$.\n\nLet us first consider any $c \\le 2$. For any positive integer $k$, define\n$$\nn = k^2 + 1 \\quad \\text{and} \\quad m = (k-1)^2 + 1\n$$\nObserve that\n$$\nm + c\\sqrt{m-1} + 1 \\le k^2 - 2k + 2 + 2(k-1) + 1 = k^2 + 1 = n\n$$\nand\n$$\n(n, n+1, \\dots, 2n-m) = (k^2+1, k^2+2, \\dots, k^2+2k).\n$$\nTherefore, every such pair $(n, m)$ indeed satisfies the property from the problem statement, and there are infinitely many such pairs.\n\nNow let us consider any $c > 2$, and let $(n, m)$ be any pair of positive integers satisfying the property from the problem statement. Observe that for each positive integer $n$, the number $\\lceil\\sqrt{n}\\rceil^2$ is always between numbers $n$ and $(\\sqrt{n}+1)^2$ (inclusive), hence there is always a square of an integer in the range\n$$\nn, n+1, \\dots, n + \\lfloor 2\\sqrt{n} \\rfloor + 1.\n$$\nThis implies that $2n - m < n + \\lfloor 2\\sqrt{n} \\rfloor + 1$, so in particular\n$$\nm \\ge n - 2\\sqrt{n}.\n$$\nCombining this with the inequality from the problem statement yields\n$$\nn \\ge n - 2\\sqrt{n} + c\\sqrt{n - 2\\sqrt{n} - 1} + 1. \\quad (1)\n$$\nObserve that since $c > 2$, we have $c\\sqrt{n - 2\\sqrt{n} - 1} > 2\\sqrt{n}$ for large enough $n$. Indeed, equivalently we have $1 - \\frac{2}{\\sqrt{n}} - \\frac{1}{n} > \\frac{4}{c^2}$, and the left-hand side tends to 1 as $n$ grows to infinity while the right hand side is strictly smaller than 1. This implies that (1) may be satisfied only for finitely many positive integers $n$. Since $m \\le n$ for all pairs $(n, m)$ satisfying the conditions from the problem statement, this implies that there are only finitely many such pairs $(n, m)$. $\\square$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 29372,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $\\omega$ be the circumcircle of an acute-angled triangle $ABC$. Point $D$ lies on the arc $BC$ of $\\omega$ not containing point $A$. Point $E$ lies in the interior of the triangle $ABC$, does not lie on the line $AD$, and satisfies $\\angle DBE = \\angle ACB$ and $\\angle DCE = \\angle ABC$. Let $F$ be a point on the line $AD$ such that lines $EF$ and $BC$ are parallel, and let $G$ be a point on $\\omega$ different from $A$ such that $AF = FG$. Prove that points $D, E, F, G$ lie on one circle.\n\n(Slovakia)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Denote $\\alpha = \\angle CAB$, $\\beta = \\angle ABC$, and $\\gamma = \\angle BCA$. Let $K$ and $L$ be the second intersections of lines $BE$ and $CE$ with $\\omega$, respectively, different from $B$ and $C$. Observe that\n$$\n\\angle BAK = \\angle BAD + \\angle DAK = \\angle BAD + \\angle DBE = \\angle BAD + \\gamma = \\angle ACD\n$$\nand symmetrically $\\angle CAL = \\angle ABD$. It follows that arcs $AD$, $BK$, and $CL$ of $\\omega$ have equal lengths, so chords $AD$, $BK$, and $CL$ also have equal lengths. In particular, since $E = BK \\cap CL$ does not lie on $AD$, these chords are not diameters. It follows that if $O$ is the center of $\\omega$, then $O$ does not lie on any of the chords $AD$, $BK$, $CL$, and in particular $O \\ne E$. Moreover, $O$ and $E$ lie on the same side of line $AD$. Suppose without loss of generality that $O$ and $E$ lie in triangle $ACD$, for the second case is symmetric.\n\nObserve that\n$$\n\\angle BEC = 360^\\{\\circ\\} - \\angle DBE - \\angle DCE - \\angle BDC = 360^\\{\\circ\\} - \\beta - \\gamma - (180^\\{\\circ\\} - \\alpha) = 2\\alpha = \\angle BOC,\n$$\nwhich implies that $B, O, E, C$ are concyclic. Further, if we denote $P = AD \\cap BC$, then we have\n$$\n\\begin\\{aligned\\} \\angle DOE &= \\angle BOE - \\angle BOD = 180^\\{\\circ\\} - \\angle BCE - 2\\angle BAD \\\\ &= 180^\\{\\circ\\} - \\angle DCE - \\angle BAD = 180^\\{\\circ\\} - \\beta - \\angle BAD = \\angle APB = \\angle EFD, \\end\\{aligned\\}\n$$\nwhich implies that $D, F, O, E$ are also concyclic.\n\nConsider triangles $AFO$ and $GFO$. We have $AF = FG$ by assumption, also $OA = OG$ since $G$ lies on $\\omega$, hence these two triangles are congruent. In particular $\\angle FAO = \\angle FGO$. Triangle $AOD$ is isosceles, hence $\\angle FAO = \\angle FDO$. This implies that $F, O, G, D$ are concyclic as well.\n\nSince points $F, O, D$ are pairwise distinct, this implies that all five points $D, F, O, E, G$ lie on the circumference of triangle $FOD$, so in particular $D, E, F, G$ are concyclic. $\\square$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 29373,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $k$ be a fixed positive integer. A finite sequence of integers $x_1, x_2, \\dots, x_n$ is written on a blackboard. Pepa and Geoff are playing a game that proceeds in rounds as follows.\n* In each round, Pepa first partitions the sequence that is currently on the blackboard into two or more contiguous subsequences (that is, consisting of numbers appearing consecutively). However, if the number of these subsequences is larger than 2, then the sum of numbers in each of them has to be divisible by $k$.\n* Then Geoff selects one of the subsequences that Pepa has formed and wipes all the other subsequences from the blackboard.\nThe game finishes once there is only one number left on the board. Prove that Pepa may choose his moves so that independently of the moves of Geoff, the game finishes after at most $3k$ rounds.\n\n(Poland)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "A finite sequence of integers is called a *word* and any its contiguous subsequence is called a *subword*. For a word $u$, by $\\sum u$ we denote the sum of numbers in $u$. A *prefix* of a word is a subword starting at the beginning of the word, and a prefix is *proper* if it is neither empty nor the whole word. Analogously we define suffixes.\nFor a word $u$, let $R(u) \\subseteq \\{0, 1, \\dots, k-1\\}$ be the set of remainders $r$ modulo $k$ for which there exists a proper prefix $v$ of $u$ with $\\sum v \\equiv r \\pmod k$. In other words, $R(u)$ comprises different remainders modulo $k$ realized by sums of numbers in proper prefixes of $u$. Define the *rank* of $u$ as $|R(u)|$.\nWe shall prove the following statement: given a word $u$ on the board, Pepa can always play at most 3 rounds so that the rank of the remaining word is strictly smaller than the rank of $u$. Since the rank of the initial word is at most $k$ and the rank of a word is 0 if and only if it consists of one number, in this way Pepa may force the end of the game within at most $3k$ rounds.\nAssume then that the word $u$ on the board has length larger than 1, and take any $r \\in R(u)$. Suppose that the proper prefixes of $u$ giving remainder $r$ modulo $k$ end at positions $1 \\le i_1 < i_2 < \\dots < i_p < |u|$, where $p \\ge 1$. Consider the following partition of $u$ into subwords:\n$$\nu = v_0v_1v_2\\dots v_{p-1}v_p,$$\nwhere $v_0$ is the prefix up to position $i_1$, each $v_j$ for $j = 1, 2, \\dots, p-1$ is the subword between positions $i_j + 1$ and $i_{j+1}$, and $v_p$ is the suffix from position $i_p + 1$ till the end of the word.\nWe observe that the rank of each of subword $v_j$ is strictly smaller than the rank of $u$. For $j = 0$ this is trivial: since $i_1$ is the first position at which a prefix of $u$ has sum congruent to $r$ modulo $k$, we have that $R(v_0) \\subseteq R(u) \\setminus \\{r\\}$. For $j > 0$, take any proper prefix $w$ of $v_j$, let $w' = v_0v_1\\dots v_{j-1}w$, and let $a$ be the remainder of $\\sum v_0v_1\\dots v_{j-1}$ modulo $k$. Observe that $\\sum w' \\equiv a + \\sum w \\pmod k$. Therefore, the remainders realized by proper prefixes $w$ of $v_j$ are exactly the remainders realized by prefixes $w'$ as above with $a$ subtracted modulo $k$. Since between $i_j$ and $i_{j+1}$ there is no position at which a prefix of $u$ has sum congruent to $r$ modulo $k$, we infer that none of prefixes $w'$ as above has sum congruent to $r$ modulo $k$. This implies that $R(v_j) \\subseteq \\{q-a: q \\in R(u) \\setminus \\{r\\}\\}$, so $|R(v_j)| < |R(u)|$.\nNote that $\\sum v_j \\equiv 0 \\mod k$ for each $j = 1, 2, \\dots, p-1$ by construction. All these observations lead to the following three-turn strategy for Pepa:\n* Partition $u$ into $v_0$ and $v_1v_2 \\dots v_p$. If Geoff chooses $v_0$, then the rank of the word has already decreased. Otherwise Geoff chooses $v_1v_2 \\dots v_p$.\n* Partition $v_1v_2 \\dots v_p$ into $v_1v_2 \\dots v_{p-1}$ and $v_p$. If Geoff chooses $v_p$, then the rank of the word has already decreased. Otherwise Geoff chooses $v_1v_2 \\dots v_{p-1}$.\n* Partition $v_1v_2 \\dots v_{p-1}$ into $v_1, v_2, \\dots, v_{p-1}$, which are all words with sums of numbers divisible by $k$. Regardless of the move of Geoff, the rank of the word chosen by him is strictly smaller than the rank of $u$.\nThis concludes the proof. $\\Box$",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 29374,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABC$ be a triangle. Line $\\ell$ is parallel to $BC$ and it respectively intersects side $AB$ at point $D$, side $AC$ at point $E$, and the circumcircle of the triangle $ABC$ at points $F$ and $G$, where points $F, D, E, G$ lie in this order on $\\ell$. The circumcircles of triangles $FEB$ and $DGC$ intersect at points $P$ and $Q$. Prove that points $A, P, Q$ are collinear.\n\n(Slovakia)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $\\omega_B, \\omega_C$ be the circumcircles of triangles $FEB$ and $DGC$ respectively. Since $PQ$ is the radical axis of $\\omega_B$ and $\\omega_C$, it is sufficient to prove that the powers of $A$ with respect to $\\omega_B$ and $\\omega_C$ are equal. If we denote by $X$ the second intersection of $\\omega_B$ and $AC$, and by $Y$ the second intersection of $\\omega_C$ and $AB$, then this is equivalent to\n$$\nAX \\cdot AE = AY \\cdot AD.\n$$\nLines $DE$ and $BC$ are parallel, yielding $\\frac{AB}{AD} = \\frac{AC}{AE}$, so the above is equivalent to\n$$\nAX \\cdot AC = AY \\cdot AB.\n$$\nThis, in turn, is equivalent to $B, Y, X, C$ being concyclic. From angles in $\\omega_B$ and $\\omega_C$ we have\n$$\n\\angle BYC = \\angle DYC = \\angle DGC \\quad \\text{and} \\quad \\angle BXC = \\angle BXE = \\angle BFE.\n$$\nOn the other hand, trapezoid $BFGC$ is inscribed in the circumcircle of $ABC$, so it is isosceles, hence $\\angle BFE = \\angle DGC$. We conclude that $\\angle BYC = \\angle BXC$, so $B, Y, X, C$ are indeed concyclic and we are done. □",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 29375,
    "subject": "Mathematics (Multi-modal)",
    "question": "Each of the $4n^2$ unit squares of a $2n \\times 2n$ board ($n \\ge 1$) has been colored blue or red. A set of four different unit squares of the board is called *pretty* if these squares can be labeled $A, B, C, D$ in such a way that $A$ and $B$ lie in the same row, $C$ and $D$ lie in the same row, $A$ and $C$ lie in the same column, $B$ and $D$ lie in the same column, $A$ and $D$ are blue, and $B$ and $C$ are red. Determine the largest possible number of different pretty sets on such a board.\n\n(Poland)",
    "options": [],
    "answer": "n^4",
    "solution": "Let us index the unit squares of the board by pairs of integers $(a, b)$ with $1 \\le a, b \\le 2n$. We prove that the largest possible number of pretty sets is $n^4$.\n\nFor the upper bound, consider coloring all the unit squares $(a, b)$ with $a, b \\le n$ or $a, b \\ge n+1$ blue, and all the other unit squares red. It is straightforward to verify that this coloring yields $n^4$ pretty sets. Thus we are left with proving that no coloring yields more pretty sets.\n\nCall an unordered pair of distinct unit squares $\\{A, B\\}$ *mixed* if $A$ and $B$ are in the same row and $A$ and $B$ have different colors. Clearly, if a row contains $a$ blue squares and $b$ red squares ($a + b = 2n$), then it contains $ab \\le n^2$ mixed pairs. Therefore, there are at most $2n^3$ mixed pairs in total.\n\nLet every mixed pair $\\{A, B\\}$ *charge* the unordered pair $\\{i, j\\}$ of distinct columns such that $A$ is in column $i$ and $B$ is in column $j$. Denote by $\\text{charge}(i, j)$ the number of times the pair of columns $\\{i, j\\}$ is charged.\n\nObviously, every pair of columns is charged at most $2n$ times, i.e., $\\text{charge}(i, j) \\le 2n$. Moreover, since there are at most $2n^3$ mixed pairs in total, we have $\\sum_{\\{i,j\\}} \\text{charge}(i, j) \\le 2n^3$ where the summation is over pairs of distinct columns $\\{i, j\\}$.\n\nObserve that if a pair of distinct columns $\\{i, j\\}$ is charged $k = \\text{charge}(i, j)$ times, then there are at most $\\frac{k^2}{4}$ pretty sets with squares contained in these columns. This is because for some $a, b$ with $a + b = k$ there are $a$ red-blue and $b$ blue-red mixed pairs within these columns, yielding $ab \\le \\frac{k^2}{4}$ pretty sets. Therefore, the total number of pretty sets is at most\n$$\n\\frac{1}{4} \\sum_{\\{i,j\\}} \\text{charge}(i,j)^2 \\le \\frac{n}{2} \\cdot \\sum_{\\{i,j\\}} \\text{charge}(i,j) \\le n^4,\n$$\nwhere again the summation is over unordered pairs of distinct columns. This concludes the proof. $\\square$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 29376,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all functions $f: (0, +\\infty) \\to \\mathbb{R}$ satisfying\n$$\nf(x) - f(x + y) = f\\left(\\frac{x}{y}\\right) f(x + y) \\quad \\text{for all } x, y > 0.\n$$",
    "options": [],
    "answer": "f(x) = 0 for all x > 0, or f(x) = 1/x for all x > 0",
    "solution": "Suppose $f(t) = 0$ for some $t > 0$. For $0 < x < t$ we choose $y = t - x > 0$ and find $f(x) = 0$. From setting $x = y = 1$ we conclude that $f(1) \\ne -1$. Hence by setting $x = y$ we get $f(x) - f(2x) = f(1)f(2x)$ for $x > 0$. Inductively we find\n$$\nf(2^n x) = f(x)(1 + f(1))^{-n}. \\quad (2)\n$$\nHence for an arbitrary $x > 0$ we choose $n \\in \\mathbb{N}$ such that $\\frac{x}{2^n} \\le t$ and we conclude from (2) that $f(x) = 0$.\n\nNow suppose $f(t) \\ne 0$ for all $t > 0$. We define $g(x) := f(x)^{-1}$, $x > 0$, and rewrite the given equation as\n$$\ng(x + y) - g(x) = \\frac{g(x)}{g\\left(\\frac{x}{y}\\right)} \\quad \\text{for all } x, y > 0. \\quad (3)\n$$\nBy setting $y = 1$ we obtain $g(x + 1) = g(x) + 1$ for all $x > 0$. It follows that $g(n) = n - 1 + g(1)$ holds for all $n \\in \\mathbb{N}$. Setting $x = y = 2$ in (3) we now find $g(1) = 1$. Setting $x = 1$ we obtain $g(y)g(\\frac{1}{y}) = 1$ for all $y > 0$. We can now rewrite (3) as\n$$\ng(x + y) = g(x) + g(x)g\\left(\\frac{y}{x}\\right) = g(x)\\left(1 + g\\left(\\frac{y}{x}\\right)\\right) = g(x)g\\left(\\frac{x+y}{x}\\right). \\quad (4)\n$$\nFrom (4) we directly see that $g(a)g(b) = g(ab)$ for all $a > 0$ and $b > 1$. Using $g(y)g(\\frac{1}{y}) = 1$, we can rewrite (4) as\n$$\ng\\left(\\frac{x}{x+y}\\right)g(x + y) = g(x),\n$$\nwhich together with the previous line shows that $g(a)g(b) = g(ab)$ holds indeed for all $a, b > 0$. In particular we see that $g(a) = g(\\sqrt{a})^2 > 0$, so $g$ is a positive function. Also, from the first equation in (4) we now infer the functional equation\n$$\ng(x + y) = g(x) + g(y) \\quad \\text{for all } x, y \\in (0, \\infty).\n$$\nIt is well known that this implies $g(x) = xg(1) = x$ for $x \\in \\mathbb{Q}_{>0}$. Since $g$ is positive, by (3) we deduce that $g(x + y) > g(x)$, so $g$ is strictly increasing. Hence $g(x) = x$ for all $x > 0$.\n\nSince $f(x) = 0$ and $f(x) = \\frac{1}{x}$ obviously satisfy the given equation, we have found all solutions. $\\square$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 29377,
    "subject": "Mathematics (Multi-modal)",
    "question": "Determine whether there exist $2024$ distinct positive integers satisfying the following: If we consider every possible ratio between two distinct numbers (we include both $a/b$ and $b/a$), we will obtain numbers with finite decimal expansions (after the decimal point) of mutually distinct non-zero lengths. (Patrik Bak, Slovakia)",
    "options": [],
    "answer": "Yes, such 2024 integers exist; for example take c_i = 2^{2^{i-1}} · 5^{2^{4049−i}} for i = 1, 2, ..., 2024.",
    "solution": "We will show these numbers exist. For that we define sequences $a_1, a_2, \\dots, a_{2024}$ and $b_1, b_2, \\dots, b_{2024}$ and then consider numbers $c_i = 2^{a_i} \\cdot 5^{b_i}$ for $i = 1, 2, \\dots, 2024$.\nWe choose the sequences $a_i$ and $b_i$ in such a way that $a_i$ is increasing, $b_i$ is decreasing, and the differences $a_i - a_j$ and $b_j - b_i$ were all mutually distinct for all indices $i > j$. This will be enough because\n$$\n\\frac{c_i}{c_j} = \\frac{2^{a_i} \\cdot 5^{b_i}}{2^{a_j} \\cdot 5^{b_j}} = \\frac{2^{a_i - a_j}}{5^{b_j - b_i}},\n$$\nthis number has a decimal expansion of a length $b_j - b_i$, whereas analogously, $\\frac{c_j}{c_i}$ has a length of $a_i - a_j$.\n\nWe now construct the needed sequences, starting with $a_i$. We will do it inductively. Take $a_1 = 1, a_2 = 2$. When we have the numbers $a_1, a_2, \\dots, a_i$, then by choosing $a_{i+1} = 2a_i$ we will achieve $a_{i+1} - a_i > a_i - a_1$, therefore all newly added differences will be higher than the previous ones.\n\nWe can construct $b_i$ similarly, starting at the end by taking $b_{2024} = a_{2024}$, then $b_{2023} = 2b_{2024}$, and so on. Since $b_{2023} - b_{2024} = a_{2024}$, all the differences in $b_i$ will be at least $b_{2023} - b_{2024} = a_{2024}$.\n\n**Remark:** In our construction, $a_i = 2^{i-1}$ and $b_i = 2^{4049-i}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 29378,
    "subject": "Mathematics (Multi-modal)",
    "question": "For a positive integer $n$, an $n$-configuration is a family of sets $\\langle A_{i,j} \\rangle_{1\\le i,j\\le n}$. An $n$-configuration is called *sweet* if for every pair of indices $(i, j)$ with $1 \\le i \\le n-1$ and $1 \\le j \\le n$ we have $A_{i,j} \\subseteq A_{i+1,j}$ and $A_{j,i} \\subseteq A_{j,i+1}$. Let $f(n,k)$ denote the number of sweet $n$-configurations such that $A_{n,n} \\subseteq \\{1, 2, \\dots, k\\}$. Determine which number is larger: $f(2024, 2024^2)$ or $f(2024^2, 2024)$.",
    "options": [],
    "answer": "f(2024^2, 2024) is larger",
    "solution": "Consider a sweet $n$-configuration $\\langle A_{i,j} \\rangle_{1\\le i,j\\le n}$ with $A_{n,n} \\subset \\{1, 2, \\dots, k\\}$. For any $x \\in \\{1, 2, \\dots, k\\}$ and $i \\in \\{1, 2, \\dots, n\\}$ define\n$$\np_x(i) = |\\{j: x \\in A_{i,j}\\}|.\n$$\nSince $A_{i,j} \\subseteq A_{i,j+1}$ for all suitable $i, j$, the set $\\{j: x \\in A_{i,j}\\}$ consists of $p_x(i)$ largest elements of $\\{1, 2, \\dots, n\\}$. Since $A_{i,j} \\subseteq A_{i+1,j}$ for all suitable $i, j$, the function $p_x: \\{1, 2, \\dots, n\\} \\to \\{0, 1, 2, \\dots, n\\}$ is nondecreasing. Therefore every sweet $n$-configuration determines a family $\\langle p_x \\rangle_{1\\le x\\le k}$ of nondecreasing functions $p_x: \\{1, 2, \\dots, n\\} \\to \\{0, 1, \\dots, n\\}$. Conversely, every such a family determines a sweet $n$-configuration $\\langle A_{i,j} \\rangle_{1\\le i,j\\le n}$ with $A_{n,n} \\subset \\{1, 2, \\dots, k\\}$ in the following way: $A_{i,j} = \\{x \\in \\{1, 2, \\dots, k\\}: j \\ge n+1-p_x(i)\\}$. Therefore $f(n,k) = g(n)^k$ where $g(n)$ is the number of nondecreasing functions $p: \\{1, 2, \\dots, n\\} \\to \\{0, 1, \\dots, n\\}$.\n\nUsing the stars-and-bars method, there is a bijection between the family of nondecreasing functions $p: \\{1, 2, \\dots, n\\} \\to \\{0, 1, \\dots, n\\}$ and the set of sequences consisting of $n$ stars and $n$ bars. The bijection is given by\n$$\np \\to \\underbrace{**\\dots*}_{p(1)} | \\underbrace{**\\dots*}_{p(2)-p(1)} | \\underbrace{**\\dots*}_{p(3)-p(2)} | \\dots | \\underbrace{**\\dots*}_{p(n)-p(n-1)} | \\underbrace{**\\dots*}_{n-p(n)}\n$$\nThus $g(n) = \\binom{2n}{n}$.\n\nThe problem boils down to determining which of the numbers\n$$\n\\binom{2n}{n}^{n^2}, \\quad \\binom{2n^2}{n^2}^n,\n$$\nwhere $n = 2024$, is larger. Note that\n$$\n\\begin{aligned} \\binom{2n^2}{n^2} &= \\frac{\\prod_{i=1}^{n^2} (n^2 + i)}{\\prod_{i=1}^{n^2} i} = \\prod_{i=1}^{n^2} \\left(\\frac{n^2 + i}{i}\\right) = \\prod_{j=0}^{n-1} \\prod_{i=1}^{n} \\frac{n^2 + jn + i}{jn + i} > \\prod_{j=0}^{n-1} \\left(\\frac{n^2 + jn + n}{jn + n}\\right)^n = \\\\ &= \\prod_{j=0}^{n-1} \\left(\\frac{n + j + 1}{j + 1}\\right)^n = \\left(\\prod_{j=1}^{n} \\frac{n+j}{j}\\right)^n = \\left(\\binom{2n}{n}\\right)^n \\end{aligned}\n$$\nand therefore\n$$\n\\binom{2n^2}{n^2}^n > \\binom{2n}{n}^{n^2}.\n$$\n\n**Remark:** A sketch of a slightly different way of thinking about $f(n,k) = \\binom{2n}{n}^k$: Consider an $n \\times n$ table. In a cell with coordinates $(i, j)$, list all the elements of the set $A_{i,j}$. Fix an element $x \\in \\{1, \\dots, k\\}$ and consider the cells that contain the number $x$. By the condition, those cells form a region closed under making a step right and making a step up. Such regions are delimited by grid paths that start at $[0, n]$, end at $[n, 0]$, and only steps right or down. There are $\\binom{2n}{n}$ possible paths for each $x$, thus $f(n,k) = \\binom{2n}{n}^k$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 29379,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABC$ be a triangle and $D$ a point on its side $BC$. Points $E, F$ lie on the lines $AB, AC$ beyond vertices $B, C$, respectively, such that $BE = BD$ and $CF = CD$. Let $P$ be a point such that $D$ is the incenter of triangle $PEF$. Prove that $P$ lies inside the circumcircle $Ω$ of triangle $ABC$ or on it. (Josef Tkadlec, Czech Republic)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $ω$ be the circumcircle of triangle $AEF$ and let $I_A$ be the A-excenter of triangle $ABC$. First, we prove that $I_A$ is the midpoint of the arc $EF$ of $ω$ that does not contain point $A$ (see the left figure).\nTo that end, note that since $I_A$ lies on the external angle bisector of $∠B$ and $BE = BD$, triangles $BEI_A$ and $BDI_A$ are congruent (SAS). Similarly, triangles $CFI_A$ and $CDI_A$ are congruent, so in particular $I_A E = I_A F$. Moreover, $∠BEI_A + ∠CFI_A = ∠BDI_A + ∠CDI_A = 180^\\circ$, hence the points $A, E, I_A, F$ lie on a single circle in this order.\n\n![](attached_image_1.png)\n\nNext, we prove that $P$ is the second intersection of $I_A D$ and $ω$ (see the middle figure). Let $I$ be the incenter of triangle $ABC$. Then $ED \\parallel BI$ and $DF \\parallel IC$. Setting $∠EAF = α$, we get $∠EDF = ∠BIC = 90^\\circ + \\frac{1}{2}α$, thus $∠EPF = α = ∠EAF$, so $P$ lies on $ω$. Since $I_A$ is the midpoint of arc, it lies on the angle bisector $PD$, so $P$ lies both on $I_A D$ and on $ω$ as claimed.\nFinally, we show that $P$ lies on that arc of $ω$ which lies inside $Ω$ (see the right figure). Let $M \\neq A$ be the second intersection of $ω$ and $Ω$ (if they are tangent, we set $M = A$). Then $M$ is the center of the spiral similarity that maps $BE$ to $CF$ (alternatively, we angle-chase that triangles $MBE$ and $MCF$ are similar by AA). Thus $MB/MC = BE/CF = BD/DC$, so $MD$ is the angle bisector of $BMC$, and so it passes through the midpoint $S$ of the arc $BC$ of $Ω$ that does not contain $A$.\nNow forget about points $B, C, E, F$ and focus on circles $Ω, ω$ and on the points $A, M, I_A, S, D, P$. Circles $Ω$ and $ω$ share points $A$ and $M$. Being the A-excenter of $ABC$, point $I_A$ belongs to that arc $AM$ of $ω$ which lies outside of $Ω$ (e.g. since $AI_A > AS$). Point $S$ lies on the segment $AI_A$ and point $D$ lies on the segment $SM$, so point $D$ lies inside the angle $AI_A M$. Thus, point $P = I_A D \\cap ω$ belongs to the other arc $AM$ of $ω$ than $I_A$, namely to the one which lies inside $Ω$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 29380,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABCD$ be a quadrilateral, such that $AB = BC = CD$. There are points $X, Y$ on rays $CA, BD$, respectively, such that $BX = CY$. Let $P, Q, R, S$ be the midpoints of segments $BX, CY, XD, YA$, respectively. Prove that points $P, Q, R, S$ lie on a circle.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $M$ be the midpoint of $XY$. Note that $PR$ is midline in triangles $XBD$ and $XBY$, hence $M$ lies on $PR$. Analogously $M$ lies on $QS$.\n\nLet $\\omega_1$ be a circle with center $B$ and radius $AB = BC$ and $\\omega_2$ be a circle with center $C$ and radius $BC = CD$.\nDistance of $X$ from center of $\\omega_1$ is the same as distance of $Y$ from center of $\\omega_2$ and also $\\omega_1$ and $\\omega_2$ have radius of same size, hence power of $X$ with respect to $\\omega_1$ is the same as power of $Y$ with respect to $\\omega_2$, so\n$$\nXA \\cdot XC = YD \\cdot YB.\n$$\nUsing homotheties centered at $X, Y$ we get that $MS \\cdot MQ = MR \\cdot MP$ and thus points $P, Q, R, S$ lie on a circle.\n\n![](attached_image_1.png)\n\n![](attached_image_2.png)",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 29381,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $\\alpha \\neq 0$ be a real number. Determine all functions $f: \\mathbb{R} \\to \\mathbb{R}$ such that\n$$f(x^2 + y^2) = f(x - y)f(x + y) + \\alpha y f(y)$$\nholds for all $x, y \\in \\mathbb{R}$.",
    "options": [],
    "answer": "For any nonzero parameter: the zero function and the function that equals one at zero and zero elsewhere. Additionally, when the parameter equals two: the identity function.",
    "solution": "Answer: For every $\\alpha \\neq 0$, the zero function and the function with value 1 at 0, but 0 elsewhere are solutions. For $\\alpha = 2$, the identity function $x \\mapsto x$ is another solution.\n\n*Solution-check.* The linear function clearly works. Consider the function $f$ such that $f(0) = 1$ and $f(x) = 0$ otherwise. Note that $y f(y) = 0$ for all real numbers $y$. Then it is sufficient to realize that $f(x^2 + y^2) \\neq 0$ iff $x = 0 \\land y = 0$. Similarly $f(x - y)f(x + y) \\neq 0$ iff $x + y = x - y = 0 \\Leftrightarrow x = 0 \\land y = 0$ which shows that also this function is a solution.\n\n*Proof.* Denote by $P(x, y)$ the proposition in the problem statement. Comparing $P(x, y)$ with $P(x, -y)$ yields $f(y) = -f(-y)$ for all $y \\neq 0$. Using this equality, $P(y, x)$ shows that $2f(x - y)f(x + y) = \\alpha(x f(x) - y f(y))$ for $x \\neq y$. Plugging this into the original equation, we obtain\n$$\nf(x^2 + y^2) = \\frac{\\alpha}{2}(x f(x) + y f(y))\n$$\nfor $x \\neq y$. Setting $y = 0$ in this equation shows $f(x^2) = \\alpha x f(x)/2$ for $x \\neq 0$, whereas $P(x, 0)$ gives $f(x^2) = f(x)^2$ for all $x \\in \\mathbb{R}$. Hence $\\alpha x f(x)/2 = f(x)^2$, that is, $f(x) = 0$ or $f(x) = \\alpha x/2$ for $x \\neq 0$. In particular, if $f(x) \\neq 0$, then $f(x) = \\alpha x/2$. On the other hand, $P(0, 0)$ shows $f(0) = f(0)^2$ and therefore $f(0) = 0$ or $f(0) = 1$.\n\nConsider first the case that $f(x) = 0$ for all $x \\neq 0$. Then both possible values for $f(0)$ yield functions fulfilling the original equation (if $(x, y) \\neq (0, 0)$, all terms in $P(x, y)$ are zero anyway and $(x, y) = (0, 0)$ was treated before).\n\nNow for the other case: There is a real number $z \\neq 0$ satisfying $f(z) = \\alpha z/2$. Then $f(z^2) = f(z)^2 = (\\alpha/2)^2 z^2 \\neq 0$, and hence $f(z^2) = (\\alpha/2)z^2$. By comparing the last two statements, we obtain $\\alpha = 2$ and then $f(z) = z$.\n\n* $f(0) = 1$. Consider $P(z/2, z/2): f(z^2/2) = z + z f(z/2)$. The left-hand side is 0 or $z^2/2$, the right-hand side $z$ or $z + z^2/2$. Since $z \\neq 0$, only $z^2/2 = z \\Leftrightarrow z = 2$ and $0 = z + z^2/2 \\Leftrightarrow z = -2$ are possible. Either way, $f(2) = 2$ and $f(-2) = -2$, because $f$ is odd. But then $f(4) = f(2^2) = f(2)^2 = 4$, which is impossible, because we just proved that $z = 2$ and $z = -2$ are the only real numbers with $f(z) = z$.\n\n* $f(0) = 0$. We show that $f(x) = 0$ for all positive reals $x$ if $f$ is not the identity function:\n(1) There are $0 < a < b$ with $f(a) = 0$, $f(b) = b$. Then $P(x, y)$ for $x = \\sqrt{b - a}$ and $y = \\sqrt{a}$ yields\n$$\n0 \\neq b = f(b) = f(x - y)f(x + y) + 2f(a) = f(x - y)f(x + y),\n$$\nhence $f(x - y) = x - y$ and $f(x + y) = x + y$ and $b = x^2 - y^2 = b - 2a$, forcing the contradiction $a = 0$.\n(2) There are $0 < a < b$ with $f(a) = a$, $f(b) = 0$. Analogous to Case 1, we arrive at the contradiction $b = 0$ when investigating $P(\\sqrt{b - a}, \\sqrt{a})$.\nExcept for the identity, we only have $f(x) = 0$ for $x > 0$ and thus $f(x) = 0$ for $x \\neq 0$ as possible solution, which we have already found and treated before.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 29382,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $a$, $b$, $c$, $d$ be nonnegative real numbers for which $a^2 + b^2 = ac + bd$ holds and $c$, $d$ are not both zero. Find maximum and minimum value of the expression\n$$ \\frac{ad + bc - cd}{c^2 + d^2}. $$",
    "options": [],
    "answer": "maximum = 1/2, minimum = -1/2",
    "solution": "We will show that the maximum value is $\\frac{1}{2}$ and the minimum is $-\\frac{1}{2}$. For maximum, after some rearranging, we want to prove\n$$\n2(ad + bc - cd) \\leq c^2 + d^2,\n$$\nor\n$$\n2(ad + bc) \\leq (c + d)^2.\n$$\nAfter adding double the expression $ac + bd = a^2 + b^2$ to both sides of this inequality, we will get equivalent inequality\n$$\n2(a + b)(c + d) = 2(ad + bc + ac + bd) \\leq (c + d)^2 + 2a^2 + 2b^2,\n$$\nwhich can be further rearranged to\n$$\n0 \\leq (c+d)^2 - 2(a+b)(c+d) + (a+b)^2 + (a-b)^2 \\leq (c+d-a-b)^2 + (a-b)^2,\n$$\nwhich clearly holds. Moreover, this maximal value is reached by $a = b = c = d > 0$.\n\nFor the minimum value, notice that if $a \\geq c$ or $b \\geq d$ holds, $ad + bc \\geq ad \\geq cd$ and the expression is non-negative. So for it to be negative, both $a < c$, $b < d$ must hold and $a^2 + b^2 < ac + bd$ if $a$ and $b$ wouldn't both be $0$. As they are equal by given condition, indeed $a = b = 0$ and the minimized expression then becomes $-\\frac{cd}{c^2+d^2}$, which has minimum $-\\frac{1}{2}$ as $2cd \\leq c^2 + d^2$. Moreover, this minimal value is reached by $a = b = 0$ and $c = d > 0$.\nConsider the Cartesian coordinate system and in it, points $C = (c, 0)$, $D = (0, d)$ and $X = (a, b)$. The line passing through $C$, $D$ has equation $dx + cy - cd = 0$. From analytic geometry, the formula for distance of point $(m, n)$ to the line $ix + jy + k = 0$ is known. It is $\\frac{im+jn+k}{\\sqrt{i^2+j^2}}$, where the distance is oriented according to the vertical position of point $(m, n)$ with respect to the given line. With this formula, the distance of point $X$ to the line through $C$, $D$ is\n$$\n\\frac{ad + bc - cd}{\\sqrt{c^2 + d^2}}.\n$$\nBy rearranging the given condition on $a$, $b$, $c$, $d$, we get\n$$\n\\left(a - \\frac{c}{2}\\right)^2 + \\left(b - \\frac{d}{2}\\right)^2 = \\frac{c^2 + d^2}{4},\n$$\nwhich means that the point $X = (a, b)$ lies on the circle with centre $(\\frac{c}{2}, \\frac{d}{2})$ and radius $\\frac{\\sqrt{c^2+d^2}}{2}$, which is exactly the circle with diameter $CD$. Such point on circle with diameter $CD$ can be at most radius distant from $AB$, so\n$$\n\\frac{ad + bc - cd}{\\sqrt{c^2 + d^2}} \\leq \\frac{\\sqrt{c^2 + d^2}}{2}\n$$\nand by rearranging we get the inequality we proved in first solution. Note that the distance from $X$ to $CD$ is nonnegative unless $X = (0,0)$, as $X$ would be above the line $CD$, because it lies in the first quadrant by the nonnegativity and the halfcircle with diameter $CD$ in first quadrant lies entirely above the line $CD$. From this, the minimum value must happen for $X = (0,0)$ which gives $-\\frac{1}{2}$, as desired.",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 29383,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $\\{a_n\\}_{n=1}^{\\infty}$ be a sequence of positive integers such that for every positive integer $n$\n$$\na_{n+1} = (n + 1)(a_n - n + 1).\n$$\nIn terms of $a_1$, determine the greatest positive integer $k$ such that $\\gcd(a_i, a_{i+1}) = k$\nfor some positive integer $i \\ge 2$. (Note that $\\gcd(x, y)$ denotes the greatest common\ndivisor of integers $x$ and $y$.)",
    "options": [],
    "answer": "The greatest k is the largest odd prime divisor of a1; if a1 is a power of two, then k = 1.",
    "solution": "First, we will prove by induction that $a_n = (a_1 - 1)n! + n$ for all $n \\ge 1$. The base case $a_1$ is trivial. Now suppose that the closed form holds for some $a_n$. Then\n$$\n\\begin{align*}\na_{n+1} &= (n + 1) (((a_1 - 1)n! + n) - n + 1) \\\\\na_{n+1} &= (n + 1) ((a_1 - 1)n! + 1) \\\\\na_{n+1} &= (a_1 - 1)(n + 1)! + (n + 1)\n\\end{align*}\n$$\nHence, the proof by induction is complete. For the sake of clarity, let $c = a_1 - 1$.\nLet $p$ be a prime number dividing $\\gcd(a_n, a_{n+1})$ for $n \\ge 2$. Assume $p \\le n$. Then\n$cn!+n \\equiv n \\pmod p$ however, we have $c(n+1)!+n+1 \\equiv n+1 \\pmod p$ thus we conclude\n$p > n$. We have $a_{n+1} = (n+1)(cn!+1)$ so then $p \\mid (n+1)(cn!+1)$. Assume $p$\ndivides $cn!+1$. Then $cn!+n \\equiv cn!+1 \\pmod p$, which implies $n \\equiv 1 \\pmod p$, which is a\ncontradiction since $p > n$. So either $\\gcd(a_n, a_{n+1}) = 1$ or $p = n+1 = \\gcd(a_n, a_{n+1})$.\nNow suppose $p = n + 1$, notice that in this case $p \\ge 3$. Then, according to\nWilson's theorem $p \\mid n! + 1$ so we have $p \\mid n! + 1 \\mid cn! + c$, which implies $p \\mid\ncn! + c + n - n \\implies p \\mid c - n \\implies p \\mid c + 1$, which is however, equal to $a_1$. The chain of\nthoughts is reversible so we obtain $\\gcd(a_{p-1}, a_p) = p$ if and only if $p \\mid a_1$. Therefore,\nthe answer is that $k$ is the biggest odd prime divisor of $a_1$ or 1 if $a_1$ is a power of 2.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 29384,
    "subject": "Mathematics (Multi-modal)",
    "question": "Maryam and Artur play a game on a board, taking turns. At the beginning, the polynomial $XY - 1$ is written on the board. Artur is the first to make a move. In each move, the player replaces the polynomial $P(X, Y)$ on the board with one of the following polynomials of their choice:\n\na) $X \\cdot P(X, Y)$\n\nb) $Y \\cdot P(X, Y)$\n\nc) $P(X, Y) + a$, where $a \\in (-\\infty, 2025]$ is an arbitrary integer.\n\nThe game stops after both players have made 2025 moves. Let $Q(X, Y)$ be the polynomial on the board after the game ends. Maryam wins if the equation $Q(x, y) = 0$ has a finite and odd number of positive integer solutions $(x, y)$. Prove that Maryam can always win the game, no matter how Artur plays.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "We claim that Maryam can always achieve that the polynomial on the board at the end of her turn has the form $P(X,Y) = f(XY)$ where $f \\in \\mathbb{Z}[T]$ can be written as\n$$\nT^n - \\sum_{i=0}^{n-1} a_i T^i \\quad \\text{for integers } n > 0 \\text{ and } a_i \\ge 0, \\text{ not all of them zero.} \\quad (1)\n$$\n\nAs any such $f$ fulfills $\\frac{f(x)}{x^n} = 1 - \\sum_{i=0}^{n-1} \\frac{a_i}{x^{n-i}}$ for all positive real numbers $x$, which is a strictly increasing function on $(0, \\infty)$ with arbitrarily small real values near $0$ and tending to $1$ for $x \\to \\infty$, it has exactly one positive real root $r$. We claim further that Maryam can choose $r$ to be a perfect (integer) square. (Initially, $P(X, Y) = f(XY)$ with $f = T - 1$.) Maryam proceeds as follows:\n\n* If Artur multiplies with $X$, Maryam multiplies with $Y$ and if Artur multiplies with $Y$, Maryam multiplies with $X$. If initially, $P(X, Y) = f(XY)$ was on the board, then the resulting polynomial is $XYf(XY)$, so $f$ changes to $T \\cdot f$.\n\n* If Artur adds an integer $0 \\le a \\le 2025$, Maryam adds the number $-a \\le 2025$. The polynomial remains unchanged.\n\n* If Artur adds an integer $a < 0$, write $A(XY)$ for the new polynomial on the board, where $A \\in \\mathbb{Z}[T]$ is of the form (1). By the discussion above, $A$ has a unique positive real root $u$. As $A(x) > 0$ for all $x > u$, Maryam can choose an integer $c > u$ (e.g. $c = \\lfloor u \\rfloor + 1$) and add the negative integer $-A(c^2)$ to the polynomial $A$ on the board. Then the new polynomial on the board has again the form (1) and (by construction) $c^2 \\in \\mathbb{Z}_{>0}$ as the only positive real root.\n\nHence, Maryam can always achieve that $Q$ (the polynomial in the end of the game) satisfies $Q = g(XY)$ where $g$ is of the form (1) and has a perfect square $s^2$, $s \\in \\mathbb{Z}_{>0}$, as unique positive real root. Now for all pairs of positive integers $(x, y)$, we have $Q(x, y) = 0 \\iff g(xy) = 0 \\iff xy = s^2$ and it is well known that the number of solutions $(x, y)$ to the last equation is (finite and) odd. (Pairs $(x, y)$ and $(y, x)$ with $x \\ne y$ correspond and $(s, s)$ is the only fixed point in this involution, giving an odd number overall.) Hence, Maryam can always win, independent of Artur's moves.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 29385,
    "subject": "Mathematics (Multi-modal)",
    "question": "The plane was divided by vertical and horizontal lines into unit squares. Determine whether it is possible to write integers into cells of this infinite grid so that:\n(i) every cell contains exactly one integer\n(ii) every integer appears exactly once\n(iii) for every two cells $A$ and $B$ sharing exactly one vertex, if they contain integers $a$ and $b$ then at least one of the cells sharing a common side with both $A$ and $B$ contains an integer between $a$ and $b$.",
    "options": [],
    "answer": "Yes, it is possible.",
    "solution": "Yes, this is possible. Consider the spiral depicted below and write consecutive integers along the spiral:\n![](attached_image_1.png)\n\nWe claim that this works. Consider any two cells $A$ and $B$ sharing exactly one vertex. Consider the $2 \\times 2$ square containing $A$ and $B$. If the $2 \\times 2$ square contains a \"corner\" of the spiral then for some $n$ and $k$ the numbers in that $2 \\times 2$ square are arranged in the following way (up to rotation or reflection):\n$$\n\\begin{array}{|c|c|} \\hline n+1 & k \\\\ \\hline n & n-1 \\\\ \\hline \\end{array}\n$$\nand therefore the conditions are satisfied no matter which opposite cells of the $2 \\times 2$ square $A$ and $B$ are. Indeed, if $A$ and $B$ contain $n-1$ and $n+1$, then the good cell is the one containing $n$. If $A$ and $B$ contain $n$ and $k$ and $n < k$ then $n < n+1 < k$ and the good cell is the one containing $n+1$. If $A$ and $B$ contain $n$ and $k$ and $n > k$ then $k < n-1 < n$ and the good cell is the one containing $n-1$.\n\nOtherwise, the numbers are arranged in the following way (again, up to rotation or reflection):\n$$\n\\begin{array}{|c|c|} \\hline k+1 & k \\\\ \\hline n & n+1 \\\\ \\hline \\end{array}\n$$\n\nfor some $n, k$, and again, the conditions are satisfied. Indeed, without loss of generality, assume $n < k$. Then $n < n+1 < k < k+1$. If $A$ and $B$ contain $n$ and $k$ then the good cell is the one containing $n+1$. Otherwise, $A$ and $B$ contain $n+1$ and $k+1$, and the good cell is the one containing $k$.\n\nAlternatively, one can notice that condition (iii) from the problem statement means that we can orient each square $2 \\times 2$ according to the increasing numbers as suggested in the picture below:\n![](attached_image_2.png)\nWith this observation, it's relatively easy to check that the spiral construction satisfies this.",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 29386,
    "subject": "Mathematics (Multi-modal)",
    "question": "We are given an acute triangle $ABC$. Point $D$ lies in the halfplane $AB$ containing $C$ and satisfies $DB \\perp AB$ and $\\angle ADB = 45^\\circ + \\frac{1}{2}\\angle ACB$. Similarly, $E$ lies in the halfplane $AC$ containing $B$ and satisfies $AC \\perp EC$ and $\\angle AEC = 45^\\circ + \\frac{1}{2}\\angle ABC$. Let $F$ be the reflection of $A$ in the midpoint of arc $BAC$ (containing point $\\bar{A}$). Prove that points $A, D, E, F$ are concyclic. (Patrik Bak, Slovakia)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Denote $\\angle ABC = \\beta$ and $\\angle ACB = \\gamma$. The conditions translate as $\\angle BAD = 45^\\circ - \\gamma$ and $\\angle EAC = 45^\\circ - \\beta$. Denote by $G$ the intersection point of $BD$ and $CE$. Clearly $\\angle BAG = 90^\\circ - \\gamma = 2\\angle BAD$, and so $AD$ is the angle bisector of $BAG$. Similarly, $AE$ is the angle bisector of $GAC$.\nLet $D', E'$ be the midpoints of $AD, AE$, respectively. It is enough to show that the circle through $A, D', E'$ also passes through the midpoint of arc $BAC$. Consider the circumcircle of $AD'E'$ and denote its second intersection points with $AB, AG, AC$ by $P, Q, R$, respectively.\n\n![](attached_image_1.png)\n\nFirst, we will show that $BP = CR$. Notice that due $\\angle ABD$ being right, we have that $D'$ is the circumcenter of $ABD$, and so $D'B = D'A$. Then we get $\\angle D'BA = \\angle PAD' = \\angle D'AQ$, and also $\\angle AQD' = \\angle BPD'$. Together with $D'B = D'A$, we have that triangles $D'AQ$ and $D'BP$ are congruent, and so $BP = AQ$. Similarly, we can show $CR = AQ$, and so $BP = CR$ as we wanted.\n\nWe will now show that the circle through $A, P, R$ passes through the midpoint of arc $BAC$. Denote by $M$ the second intersection of this circle with the circle $ABC$.\n\nClearly $\\angle MBP = \\angle MCR$ and $\\angle MPA = \\angle MRA$, and also $BP = RC$, so triangle $MBP$ and $MCR$ are congruent, giving $MB = MD$, which is enough.\nSimilarly to the previous solution, we consider homothety with center $A$ and coefficient $1/2$ to obtain points $D', E', M$ and prove that $AD'B, AE'C, BMC$ are isosceles triangles. Moreover, by angle chasing we can get $\\angle BMC = \\alpha, \\angle AD'B = 90^\\circ + \\gamma$ and $\\angle AE'C = 90^\\circ + \\beta$. Let us notice that the sum of these angles $\\angle BMC + \\angle AD'B + \\angle AE'C = 360^\\circ$. We may view these three isosceles triangles as three rotations (for example, triangle $AD'B$ corresponds to the rotation around $D'$ by angle $AD'B$ and sends point $B$ to point $A$). We will call them green, blue, and red.\nBecause the sum of the three angles is $360^\\circ$, the composition of these three rotations is a translation. Moreover, if we follow the image of $C$ we notice that green rotation maps it to $B$, then blue maps it to $A$, and finally red maps it back to $C$. Hence, the translation is actually an identity. Let $M'$ be the image of $M$ under the blue rotation. Then $MD'M'$ is similar to $AD'B$. And because $M$ is the center of the green rotation, the composition of blue and red rotations has to map $M$ back to $M$. Hence, $M'E'M$ has to be similar to $AE'C$. And so $\\angle D'ME' = \\angle BAD' + \\angle CAE' = \\alpha/2 = \\angle D'AE'$. Thus, $AME'D'$ is cyclic and we are done.\n[sketch] Let $G, D', E'$ be as in the original solution. Denote $M$ the midpoint of arc $BAC$. Moreover, let the line $AM$ meet the lines $BD$ and $EC$ at $X$ and $Y$, respectively. Since $AM$ is the external angle bisector, we get $\\angle XAB = \\angle CAY = 90^\\circ - \\alpha/2$. Therefore, $\\angle GXY = \\angle GYX = \\alpha/2$, so the triangle $GXY$ is isosceles. Since $AG$ is a diameter of the circumcircle of $ABC$, $GM \\perp XY$, thus $M$ is the midpoint $XY$.\nWe can calculate $\\angle XAD = 180^\\circ - \\alpha/2 - (45^\\circ + \\gamma/2) = 45^\\circ + \\beta/2$. Similarly $\\angle CAY = 45^\\circ + \\beta/2$. This gives us that the triangles $DAX$ and $ACY$ are spirally similar. From this spiral similarity we get that also $D'E'M$ is similar to them. So, $\\angle D'ME' = \\alpha/2$.\nWe can calculate $\\angle D'AE' = 180^\\circ - (45^\\circ + \\beta/2) - (45^\\circ + \\gamma/2) = \\alpha/2$, hence $A, D', E', M$ are concyclic. Homothety with center $A$ and coefficient 2 maps this circle to the desired circle.\n\n![](attached_image_2.png)",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 29387,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all functions $f: (0, \\infty) \\to [0, \\infty)$ such that for all $x, y \\in (0, \\infty)$ it holds that\n$$\nf(x + y f(x)) = f(x) f(x + y).\n$$",
    "options": [],
    "answer": "All solutions are exactly the following:\n1) Any function that takes only the values zero and one on the entire positive real line.\n2) For some fixed positive threshold and a fixed nonnegative constant, the function is arbitrary zero or one on inputs below the threshold, equals the fixed constant at the threshold, and is zero for all larger inputs.",
    "solution": "Any $f$ such that $f(x) \\in \\{0, 1\\}$ for all $x \\in \\mathbb{R}^{+}$ works. Furthermore, any $f$ such that\n$$\nf(x) = \\begin{cases} 0 & \\text{or } 1 \\\\ c & x = x_0 \\\\ 0 & x \\in (x_0, \\infty) \\end{cases}\n$$\nworks as well, where $x_0 > 0$, $c \\ge 0$ are arbitrary constants. We now show that these are the only solutions. For $f(x) \\ne 0$ easy both-ways induction yields that for all $n \\in \\mathbb{Z}$ it is true that\n$$\nf(x)^n f(x + y) = f(x + y f(x)^n) \\quad (2)\n$$\nNow assume there exist $0 < x_0 < x_1$ such that $f(x_0) \\notin \\{0, 1\\}$ and $f(x_1) \\ne 0$ (if such a pair doesn't exist then $f$ must have one of the two forms described above). Then substituting $[x_0, x_1 - x_0]$ into (1) and manipulating $n$ (in particular we consider $n \\to -\\infty$ if $f(x_0) < 1$, and $n \\to +\\infty$ if $f(x_0) > 1$) yields that $f$ reaches arbitrarily large values at arbitrarily large arguments. Hence, for every pair of positive reals $c_1, c_2$ there are infinitely many $x$ such that $x > c_1$ and $f(x) > c_2$. Call this fact $(\\star)$.\nWe now multiply the given equation by $f(x + y + z)$, where $z$ is a positive real number, to get\n$$\nf(x + y + z) f(x + y f(x)) = f(x) f(x + y) f(x + y + z) = f(x) f(x + y + z f(x + y)),\n$$\nwhere we've used the property from the problem statement to obtain the second equality. We now choose $z$ such that $z > y f(x) - y$. Then $x + y + z > x + y f(x)$. Hence, we can apply the problem statement on both the left-most side and the right-most side of the above equation to get\n$$\n\\begin{aligned}\nf(x + y + z) f(x + y f(x)) &= f(x + y f(x) + (z - y f(x) + y) f(x + y f(x))) \\\\\nf(x) f(x + y + z f(x + y)) &= f(x + (y + z f(x + y)) f(x))\n\\end{aligned}\n$$\nTogether with $f(x + y f(x)) = f(x) f(x + y)$, since the LHS's are equal in the above two equations, we get\n\n$$\nf(x + y f(x) + (z - y f(x) + y) f(x) f(x + y)) = f(x + (y + z f(x + y)) f(x)). \\quad (3)\n$$\nIf the arguments in the above equation were equal, then by simplification, this would yield the equivalent equality\n$$\n(-y f(x) + y) f(x) f(x + y) = 0. \\quad (4)\n$$\nWe now choose $x_0, y_0$ such that $f(x_0) \\notin \\{0, 1\\}$ and $f(x_0 + y_0) \\neq 0$ and substitute $[x_0, y_0]$ into (2). Note that for this pair, equation (3) does not hold, and hence the arguments in (2) are always distinct. In particular, the arguments on both sides of (2) are linear functions in $z$ with the same positive gradient (namely $f(x_0) f(x_0 + y_0)$), but different $y$-intercept values. Since (2) holds for all large $z$ (namely all $z > y_0 f(x_0) - y_0$), it follows that $f$ is eventually periodic. Hence, there are constants $C, P > 0$ (dependent on $x_0, y_0$), such that $f(x) = f(x + P)$ for all $x > C$.\nBy $(\\star)$ we know that there is an $x_2 > C$ such that $f(x_2) \\notin \\{0, 1\\}$. Then by comparing $[x_2, y]$ with $[x_2, y + P]$ in the original equation we get\n$$\nf(x_2 + y f(x_2)) = f(x_2 + y f(x_2) + P f(x_2)),\n$$\nsince the RHS's remains the same (since $x_2 + y > C$). Now let $y = \\frac{P}{f(x_2)}$ in the above, to obtain\n$$\nf(x_2 + P) = f(x_2 + P + P f(x_2))\n$$\nand hence $f(x_2) = f(x_2 + P f(x_2)) = f(x_2) f(x_2 + P) = f(x_2)^2$, clear contradiction.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 29388,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABC$ be an acute triangle with the perimeter of $2s$. We are given three pairwise disjoint circles with pairwise disjoint interiors with the centres $A$, $B$ and $C$, respectively. Prove that there exists a circle with the radius of $s$ which contains all the three circles.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "To simplify the formulations, we say that a point lies inside of the circle if it lies on that circle or in its interior. Assume we are given a circle $\\omega$ with the radius of $r$ and the centre $O$. A circle $\\omega'$ with the centre $O'$ contains the circle $\\omega$ if and only if its radius is at least $O'O + r$.\n\n![](attached_image_1.png)\n\nDenote by $r_a$, $r_b$, $r_c$ the radii of our circles with the centres at $A$, $B$ and $C$, respectively. Using our observation three times indicates that the centre $X$ of the circle we are seeking has to meet $s \\ge AX + r_a$, or equivalently $AX \\le s - r_a$, and analogously $BX \\le s - r_b$ and $CX \\le s - r_c$.\n\nNotice that the numbers $s - r_a$, $s - r_b$ and $s - r_c$ are positive. We will show this for $s - r_a$. Since our circles are disjoint with disjoint interiors, we know that $r_a < b$ and $r_a < c$. This gives us $r_a < (b+c)/2 < (a+b+c)/2 = s$, which indeed means that $s - r_a$ is a positive number.\n\nNow we may consider three circles with the centres $A$, $B$ and $C$ and radii $s - r_a$, $s - r_b$ and $s - r_c$, respectively. If we prove that there is a point $X$ lying inside each of them, we will be done.\n\nEach two of these three circles intersect at two points, because for example $(s - r_a) + (s - r_b) > 2s - c = a + b > c$ (and also $c > |(s - r_a) - (s - r_b)|$). For the sake of contradiction assume there is no point lying inside all of them. Then the situation looks like on the picture, that is, there exists a point $X$ inside of the triangle which lies outside of the three circles (see the remark at the end):\n\n![](attached_image_2.png)\n\nFor such $X$ we have $AX + BX + CX > s - r_a + s - r_b + s - r_c > 2s$. This is not possible, however. Let $Y$ be the intersection of $BX$ and $AC$. Then using the triangle inequalities for the triangles $CXY$, $ABY$ we get\n$$\nBX + CX < BX + XY + CY = BY + CY < AB + AY + CY = AB + AC.\n$$\n\nSimilarly $AX + BX < AC + BC$ and $CX + AX < BC + AB$. Summing these three inequalities we obtain $AX + BX + CX < AB + BC + AC = 2s$, which is a contradiction.\n\n**Remark.** If three circles $\\omega_a$, $\\omega_b$ and $\\omega_c$ with the centres $A$, $B$ and $C$, respectively, satisfy the conditions that each two of them intersect and there is no point lying inside all of the three circles, then there exists a point in the interior of the triangle $ABC$ which lies outside of each of the three circles.\n\n![](attached_image_3.png)\n\nTo prove this, consider the intersection point $P$ of $\\omega_b$ and $\\omega_c$ which lies in the halfplane determined by the line $BC$ and the point $A$. The intersection $Q$ of the ray $BA$ with $\\omega_b$ lies inside of $\\omega_a$, since it is the closest point of $\\omega_b$ to $A$ (this is true even if $A$ is inside of $\\omega_b$, since $\\omega_a \\cap \\omega_b \\neq \\emptyset$). Therefore $A$ cannot lie in the angle $CBP$ (otherwise $Q$ would lie inside all of the three circles). But that means $P$ lies in the interior of the angle $CBA$. Similarly $P$ lies in the interior of the angle $BCA$. So we have that $P$ lies in the interior of the triangle $ABC$. Since $P$ does not lie inside of $\\omega_a$, there is a point in the neighbourhood of $P$ lying outside of all the three circles.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 29389,
    "subject": "Mathematics (Multi-modal)",
    "question": "In a rectangle with dimensions $2 \\times 3$ there is a polyline of length 36, which can have self-intersections. Show that there exists a line parallel to two sides of the rectangle, which intersects the other two sides in their interior points and intersects the polyline in fewer than 10 points.\n\n![](attached_image_1.png)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Consider an arbitrary line segment of the polyline and denote by $d$ its length and by $x$ and $y$ the lengths of its perpendicular projections on the sides of lengths 2 and 3, respectively. Cauchy-Schwarz inequality gives us\n$$\n(2x + 3y)^2 \\leq (2^2 + 3^2)(x^2 + y^2) = 13d^2,\n$$\nwhich means $2x + 3y \\leq d \\cdot \\sqrt{13}$. Denote by $X$ and $Y$ the total length of all the perpendicular projections of all the line segments on the sides of lengths 2 and 3, respectively. Summing up our estimations for each line segment gives us $2X + 3Y \\le 36 \\cdot \\sqrt{13} < 130$. But then either $2X < 40$, or $3Y < 90$. In the first case we would have $X < 20$, so on the side of length 2 there is a point that is contained in fewer than 10 projections. A line perpendicular to this side at this point intersects the polyline at most 9 times. The other case is analogous.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 29390,
    "subject": "Mathematics (Multi-modal)",
    "question": "We say that a positive integer $n$ is fantastic, if there exist positive rational numbers $a$ and $b$ such that\n\n$$\nn = a + \\frac{1}{a} + b + \\frac{1}{b}.\n$$\n\na. Prove that there exist infinitely many prime numbers $p$ such that no multiple of $p$ is fantastic.\nb. Prove that there exist infinitely many prime numbers $p$ such that some multiple of $p$ is fantastic.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Note that\n$$\nr(a,b) := a + \\frac{1}{a} + b + \\frac{1}{b} = \\frac{(a+b)(ab+1)}{ab}.\n$$\nWe put $a = \\frac{t}{u}$ and $b = \\frac{v}{w}$, where $t, u, v$ and $w$ are positive integers such that both $t$ and $u$ and also $v$ and $w$ are coprime. Then we get $r(a,b) = \\frac{(tv+uw)(tw+uv)}{tuvw}$, whence the Diophantine equation\n\n$$\ntu(v^2 + w^2) + vw(t^2 + u^2) = kptuvw \\quad (6)\n$$\n\nhas to be investigated. Now $\\gcd(tu, t^2 + u^2) = 1$. Therefore, (6) implies $tu \\mid vw$. As we get similarly $vw \\mid tu$, too, we infer\n\n$$\ntu = vw \\quad (7)\n$$\n\nand (6) becomes\n\n$$\n\\frac{(v^2 + t^2)(v^2 + u^2)}{v^2} = t^2 + u^2 + v^2 + w^2 = kptu.\n$$\n\nTherefore, $p$ has to divide either $v^2 + t^2$ or $v^2 + u^2$. In the case $p \\equiv -1 \\pmod 4$, i.e. when $-1$ is a quadratic non-residue mod $p$, this means that $p$ divides $v$ (and $t$ or $u$). But since the same argument is valid for $w$ instead of $v$, we have $p \\mid v, w$ contradicting the coprimality of $v$ and $w$. Thus the infinitely many primes with $p \\equiv -1 \\pmod 4$ have no fantastic multiple and part (a) is solved.\n\nFor part (b) we choose $v = 1$ and substitute $w = tu$. Thus we are looking for integers $t$ and $u$ such that\n\n$$\n1 + t^2 + u^2 + t^2u^2 = kptu.\n$$\n\nHere we choose $t = F_{2l+1}$, $u = F_{2l-1}$ and use the identity $1 + F_{2l+1}^2 = F_{2l+3}F_{2l-1}$ to obtain\n$$\n(1+t^2)(1+u^2) = (1+F_{2l+1}^2)(1+F_{2l-1}^2) = F_{2l+3}F_{2l-1}F_{2l+1}F_{2l-3} = k p F_{2l+1} F_{2l-1},\n$$\ni.e. $F_{2l+3}F_{2l-3} = kp$. Therefore every prime factor of the Fibonacci number $F_{2l+3}$ has a fantastic multiple.\n\nIn view of the well-known formula $\\text{gcd}(F_a, F_b) = F_{\\text{gcd}(a,b)}$ it is clear that $F_a$ and $F_b$ are relatively prime, if $a$ and $b$ are different prime numbers. Hence we know that infinitely many prime numbers have a fantastic multiple, which solves part (b).",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 29391,
    "subject": "Mathematics (Multi-modal)",
    "question": "Prove that if positive numbers $a, b, c$ satisfy the inequality $5abc > a^3 + b^3 + c^3$, then there is a triangle with sides $a, b, c$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let positive numbers $a, b, c$ satisfy the inequality $5abc > a^3 + b^3 + c^3$. Let us show that there exists a triangle with the sides $a, b, c$. On the contrary, suppose that there is no such triangle. Then for $a, b, c$ at least one of the triangle inequalities is not valid. Let, e.g. $c \\ge a+b$, i.e. $c = a+b+x$, where $x \\ge 0$. Then from the initial inequality it follows that\n$$\n5ab(a + b + x) > a^3 + b^3 + (a + b)^3 + 3(a + b)^2 x + e(a + b)x^2 + x^3\n$$\nor\n$$\n2a^2b + 2ab^2 > 2a^3 + 2b^3 + (a+b)^3 + abx + 3(a^2+b^2)x + 3(a+b)x^2 + x^3\n$$\nSince the last four summands at the left side are nonnegative, we have\n$$\n\\begin{aligned}\n2a^2b + 2ab^2 &> 2a^3 + 2b^3 & \\Leftrightarrow \\\\\na^3 + b^3 &- a^2b - ab^2 < 0 & \\Leftrightarrow \\\\\n(a+b)(a-b)^2 &< 0\n\\end{aligned}\n$$\nwhich is impossible.\n□\n\nIn fact, assume the contrary: Assume that there are three positive reals $a, b, c$ satisfying $5abc > a^3 + b^3 + c^3$, but they are not the sidelengths of a triangle. Since everything is symmetric, we can WLOG assume that $a \\ge b \\ge c$. Then, $c+a > b$ (since $a \\ge b$) and $a+b > c$ (since $a \\ge c$), so that we must have $b+c \\le a$ (else, the positive reals $a, b, c$ would be the sidelengths of a triangle). Hence, $c+a-b > 0$, $a+b-c > 0$ and $b+c-a \\le 0$, so that\n$$\n(b+c-a)(c+a-b)(a+b-c) \\le 0\n$$\nThus,\n$$\n5abc - (b+c-a)(c+a-b)(a+b-c) \\ge 5abc\n$$\nOn the other hand, $5abc > a^3 + b^3 + c^3$. Thus,\n$$\n\\begin{aligned}\n5abc - (b+c-a)(c+a-b)(a+b-c) &> a^3 + b^3 + c^3 & \\Leftrightarrow \\\\\n10abc + (a^3 + b^3 + c^3) - (a+b+c)(bc+ca+ab) &> a^3 + b^3 + c^3 & \\Leftrightarrow \\\\\n10abc &> (a+b+c)(bc+ca+ab) & \\Leftrightarrow\n\\end{aligned}\n$$\nDivision by $abc$ yields\n$$\n10 > (a+b+c) \\left( \\frac{1}{a} + \\frac{1}{b} + \\frac{1}{c} \\right)\n$$\nNow, since $10 = 3^2 + 1$, the IMO 2004 problem 4 yields that $a, b, c$ are the sidelengths of a triangle, contradicting our assumption. Thus, our assumption was wrong, and we are done. ☐",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 29392,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABC$ be a triangle, $k$ its incircle and $k_a, k_b, k_c$ three circles orthogonal to $k$ passing through $B$ and $C$, $A$ and $C$, and $A$ and $B$ respectively. The circles $k_a, k_b$ meet again in $C'$; in the same way we obtain the points $B'$ and $A'$. Prove that the radius of the circumcircle of $A'B'C'$ is half the radius of $k$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "![](attached_image_1.png)\nLet $I$ and $r$ denote the center and the radius of circle $k$. Let $D, E$, and $F$ denote the points where $k$ touches $BC, AC$, and $AB$, respectively. Let $P, Q$, and $R$ denote the midpoints of $EF, DF$, and $DE$ respectively. We will use the well known lemma:\n**LEMMA.** The circles to $k_1(S_1, r_1)$ and to $k_2(S_2, r_2)$ are orthogonal if and only if\n$$\nr_1^2 + r_2^2 = |S_1S_2|^2\n$$\nFirst of all we prove that point $Q$ and $R$ lie on circle $k_a$. Obviously $BDIF$ is a deltoid, so $Q$ is the foot of a perpendicular from point $D$ to $BI$. Thus, applying the first Euclidean's theorem to triangle $\\triangle IBD$, we have $|IQ| \\cdot |IB| = |ID|^2 = r^2$. Similarly $|IR| \\cdot |IC| = r^2$. Thus $|IQ| \\cdot |IB| = |IR| \\cdot |IC|$, so the points $B, C, R, Q$ lie on a circle which we denote $\\gamma_a$.\nThe points $Q \\in IB$ and $R \\in IC$, so that point $I$ lies outside the circle $\\gamma_a$. By the above relations, we obtain that the power of the point $I$ to the circle $\\gamma_a$ is $r^2$, which means that the circles $k$ and $\\gamma_a$ are orthogonal. From the uniqueness of $k_a$ it follows that $k_a = \\gamma_a$. Thus $k_a$ contains $Q$ and $R$. Similarly $k_b$ contains $P$ and $R$ and $k_c$ contains $P$ and $Q$. Hence, $A' = P$, $B' = Q$ and $C' = R$. Therefore the radius of the circumcircle of $\\triangle A'B'C'$ is half the radius of $k$. ☐\n\n\n![](attached_image_2.png)\nLet $k(I, r)$ be the incircle of $\\triangle ABC$. Let $D, E$, and $F$ denote the points where $k$ touches $BC, AC$, and $AB$, respectively. Let $P, Q$, and $R$ denote the midpoints of $EF, DF$, and $DE$ respectively. We prove that $k_a$ passes through $Q$ and $R$.\nSince $\\triangle IQD \\sim \\triangle IDB$ and $\\triangle IRD \\sim \\triangle IDC$, we obtain $IQ \\cdot IB = IR \\cdot IC = r^2$. We conclude that $B, C, Q$, and $R$ lie on a single circle $\\gamma_a$. Moreover, since the power of $I$ with respect to $\\gamma_a$ is $r^2$, it follows for a tangent $IX$ from $I$ to $\\gamma_a$ that $X$ lies on $k$ and hence $k$ is perpendicular to $\\gamma_a$. From the uniqueness of $k_a$ it follows that $k_a = \\gamma_a$. Thus $k_a$ contains $Q$ and $R$. Similarly $k_b$ contains $P$ and $R$ and $k_c$ contains $P$ and $Q$. Hence, $A' = P, B' = Q$ and $C' = R$. Therefore the radius of the circumcircle of $\\triangle A'B'C'$ is half the radius of $k$. $\\square$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 29393,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $n$ be a positive integer. Prove that $n$ is a power of two if and only if there exists an integer $m$ such that $2^n - 1$ is a divisor of $m^2 + 9$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "We first show that if $2^n - 1$ is a divisor of $m^2 + 9$ for some number $m$, then $n$ is the power of number 2. Otherwise the $n$ has some odd divisor $\\ell \\ge 3$ and so $2^\\ell - 1$ is a divisor of $m^2 + 9$ because $2^\\ell - 1$ divide $2^n - 1$. But for $\\ell \\ge 3$ we have $2^\\ell - 1 \\equiv -1 \\pmod 4$, and therefore the $2^\\ell - 1$ has prime divisors $p$ such that the $p \\equiv -1 \\pmod 4$. Firstly, we deal with $p \\ne 3$. Since $p \\mid m^2 + 3^2$ by Fermat little theorem we have\n$$\n1 \\equiv m^{p-1} \\equiv (m^2)^{(p-1)/2} \\equiv (-9)^{(p-1)/2} \\equiv (-1)^{(p-1)/2} \\cdot 3^{p-1} \\equiv -1 \\pmod{p}\n$$\nBut this is impossible, so $p=3$. But then for odd $\\ell$ we have $2^\\ell \\equiv -1 \\not\\equiv -1 \\pmod 3$, which is a contradiction. Consider $n=2^k$ now. For $n=1$, the claim is true. If $k \\ge 1$ then\n$$\n2^n - 1 = 3(2^2 + 1)(2^{2^2} + 1)\\cdots(2^{2^{k-1}} + 1)\n$$\nTherefore, if $2^n - 1$ divides $m^2 + 9$, also $2^\\ell + 1$ divides $m^2 + 9$ for each $\\ell = 1, 2, \\dots, k-1$. Moreover if $\\alpha \\ne \\beta$ the numbers $2^{2\\alpha} + 1$ and $2^{2\\beta} + 1$ are relatively prime because if $d > 2$ is the greatest common divisor and $\\alpha > \\beta$, then\n$$\n-1 \\equiv 2^{2\\alpha} \\equiv (2^{2\\beta})^{2^{\\alpha-\\beta}} \\equiv 1 \\pmod{d}\n$$\nwhich is a contradiction. So it is necessarily $d=1$ ($d \\ne 2$, because both are odd).\nThen, according to the Chinese remainder theorem there is a natural number $c$ such that\n$$\nc \\equiv 2^{2\\ell} \\pmod{2^{2^{\\ell+1}}}, \\quad \\forall \\ell = 0, 1, \\dots, k-2\n$$\nThen $c^2 + 1 \\equiv 0 \\pmod{2^{2\\ell+1}}$ for $\\ell = 0, 1, \\dots, k-2$, and thus $2^n - 1$ divides $(3c)^2 + 9$. $\\square$\nWe claim that, if $2^n - 1$ divides $m^2 + 9$ for some $m \\in \\mathbb{N}$, then $n$ must be a power of 2. Suppose otherwise that $n$ has an odd divisor $d > 1$. Then $2^d - 1 \\mid 2^n - 1$ is also a divisor of $m^2 + 9 = m^2 + 3^2$. However, $2^d - 1$ has some prime divisor $p$ of the form $4k-1$, and by a well-known $p$ divides both $m$ and 3. Hence $p=3$ divides $2^d - 1$, which is impossible because, for $d$ odd, $2^d \\equiv 2 \\pmod 3$. Hence $n = 2^r$ for some $r \\in \\mathbb{N}$.\nNow let $n = 2^r$. We prove the existence of $m$ by induction on $r$. The case $r=1$ is trivial. Now for any $r > 1$ note that\n$$\n2^{2r} - 1 = (2^{2^{r-1}} - 1)(2^{2^{r-1}} + 1)\n$$\nThe induction hypothesis claims that there exists an $m_1$ such that $2^{2^{r-1}} - 1 \\mid m_1^2 + 9$. We also observe that $2^{2^{r-1}} + 1 \\mid m_2^2 + 9$ for simple $m_2 = 3 \\cdot 2^{2^{r-2}}$. By the Chinese remainder theorem, there is an $m \\in \\mathbb{N}$ that satisfies $m \\equiv m_1 \\pmod{2^{2^{r-1}} - 1}$ and $m \\equiv m_2 \\pmod{2^{2^{r-1}} + 1}$. It is easy to see that this $m^2 + 9$ will be divisible by both $2^{2^{r-1}} - 1$ and $2^{2^{r-1}} + 1$.\nThis completes the induction. $\\square$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 29394,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $P(x)$ be a polynomial with integer coefficients. Prove that the polynomial $Q(x) = P(x^4)P(x^3)P(x^2)P(x) + 1$ has no integer roots.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "By the Little Fermat theorem $n^3 \\equiv n \\pmod{3}$ for every $n \\in \\mathbb{Z}$. Then also $n^4 \\equiv n^2 \\pmod{3}$. Since $P(x)$ is a polynomial with all integer coefficients, it follows that\n$$\nP(n^3) \\equiv P(n) \\pmod{3}, \\quad P(n^4) \\equiv P(n^2) \\pmod{3}\n$$\nPolynomial $Q(x)$ has integer coefficients therefore, from the last two congruences, we get\n$$\nQ(n) \\equiv (P(n)P(n^2))^2 + 1 \\pmod{3}\n$$\nBut we can easily see that for every integer $m$ we have $m^2 + 1 \\not\\equiv 0 \\pmod{3}$. So for every integer number $n$, we have $Q(n) \\not\\equiv 0 \\pmod{3}$, from which it follows that polynomial $Q(x)$ can not have integer roots. The proof is finished. $\\square$",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 29395,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABCD$ be an equilateral trapezoid with sides $AB$ and $CD$. The incircle of the triangle $BCD$ touches $CD$ at $E$. Point $F$ is chosen on the bisector of the angle $\\angle DAC$ such that the lines $EF$ and $CD$ are perpendicular. The circumcircle of the triangle $ACF$ intersects the line $CD$ again at $G$. Prove that the triangle $AFG$ is isosceles.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "![](attached_image_1.png)\nLet us show that $|FA| = |FG|$. We will proceed: from behind. On the extension of $CD$ we take the point $P$ such that $|DP| = |DA|$ and similarly on extension of $DC$ we take the point $Q$ such that $|CQ| = |CA|$. Then, using well known properties of the incircle, we have\n$$\n\\begin{aligned}\n|PE| &= |PD| + |DE| = |DA| + \\frac{|BD| + |CD| - |BC|}{2} = \\frac{|BD| + |CD| + |BC|}{2} \\\\\n|QE| &= |QC| + |CE| = |AC| + \\frac{|BD| + |CD| - |BC|}{2} = \\frac{|BD| + |CD| + |BC|}{2}\n\\end{aligned}\n$$\nThis means that the line $EF$ is the axis of the segment $PQ$. In particular, it means that the circumcenter $O$ of triangle $\\triangle APQ$ lies on the line $EF$ as well as on the axes of segments $AP$ and $AQ$. This means that\n$$\n\\angle DAO = \\angle OPD = \\angle CQO = \\angle OAC\n$$\nSo points $O$ and $F$ coincide. Moreover\n$$\n\\angle AOP = 2\\angle AQP = \\angle AQC + \\angle CAQ = 180^{\\circ} - \\angle QCA = \\angle ACP\n$$\nTherefore the points $A$, $C$, $F$, $P$ are concyclic. Thus, necessarily $P = G$, and thus $|FP| = |FG| = |FA|$. We have proved what we need. $\\Box$\n\n\n![](attached_image_2.png)\nWe will show that $FA = FG$. Let $H$ be the center of the excircle of triangle $\\triangle ACD$ opposite vertex $A$. Then $H$ lies on the angle bisector $AF$. Let $K$ be the point where this excircle touches $CD$. By a standard computation using equal tangents, we see that $CK = (AD+CD-AC)/2$. By a similar computation in triangle $\\triangle BCD$, we see that $CE = (BC+CD-BD)/2 = CK$. Therefore $E = K$ and $F = H$.\nSince $F$ is now known to be an excenter, we have that $FC$ is the external angle bisector of $\\angle DCA = \\angle GCA$. Therefore\n$$\n\\angle GAF = \\angle GCF = 90^\\circ - \\frac{1}{2}\\angle GCA = 90^\\circ - \\frac{1}{2}\\angle GFA\n$$\nWe conclude that the triangle $\\triangle GAF$ is isosceles with $FA = FG$, as desired. $\\square$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 29396,
    "subject": "Mathematics (Multi-modal)",
    "question": "Suppose that every integer has been given one of the colors red, blue, green, yellow. Let $x$ and $y$ be odd integers such that $|x| \\neq |y|$. Show that there are two integers of the same color whose difference has one of the following values: $x$, $y$, $x+y$, $x-y$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "We denote colors by capital initial letters. Let us suppose that there exists a coloring $f : \\mathbb{Z} \\to \\{R, G, B, Y\\}$ such that for any $a \\in \\mathbb{Z}$ we have $f\\{a, a+x, a+y, a+x+y\\} = \\{R, G, B, Y\\}$. We now define a coloring of an integer lattice $g : \\mathbb{Z} \\times \\mathbb{Z} \\to \\{R, G, B, Y\\}$ by the rule $g(i, j) = f(xi+yj)$. It follows that every unit square in $g$ must have its vertices colored by four different colors. If there is a row or column with period 2, then applying the condition to adjacent unit squares, we get (by induction) that all rows or columns, respectively, have period 2.\n\nOn the other hand, taking a row to be not of period 2, i.e., containing a sequence of three distinct colors, for example **GRY**, we get that the next row must contain in these columns **YBG**, and the following **GRY**, and so on. It would follow that a column in this case must have period 2. A similar conclusion holds if we start with an aperiodic column. Hence either all rows or all columns must have period 2.\n\nLet us assume WLOG that all rows have a period of 2. Assuming WLOG $\\{g(0,0), g(1,0)\\} = \\{G, B\\}$, we get that the even rows are painted with $\\{G, B\\}$ and odd with $\\{Y, R\\}$. Since $x$ is odd, it follows that $g(y,0)$ and $g(0,x)$ are of different color. However, since $g(y,0) = f(xy) = g(0,x)$, this is a contradiction. Hence the statement of the problem holds. □",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 29397,
    "subject": "Mathematics (Multi-modal)",
    "question": "Show that real numbers $p$, $q$, $r$ satisfy the condition\n$$\np^4(q-r)^2 + 2p^2(q+r) + 1 = p^4\n$$\nif and only if the quadratic equations\n$$\nx^2 + px + q = 0 \\quad \\text{and} \\quad y^2 - py + r = 0\n$$\nhave real roots (not necessarily distinct) which can be labelled by $x_{1,2}$ and $y_{1,2}$, respectively, in such way that the equality $x_1y_1 - x_2y_2 = 1$ holds.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "First part. Assume that the equations (2) have real roots satisfying $x_1y_1 - x_2y_2 = 1$. By the familiar formula, the roots of the quadratic equations are given by\n$$\nx_{1,2} = \\frac{-p \\pm K}{2} \\quad \\text{and} \\quad y_{1,2} = \\frac{p \\pm L}{2},\n$$\nwhere the real numbers $K, L$ satisfy $K^2 = p^2 - 4q$ and $L^2 = p^2 - 4r$ (we choose the signs of $K, L$ in accordance with the labelling of the roots). Thus\n$$\n1 = x_1y_1 - x_2y_2 = \\frac{(-p+K)(p+L) - (-p-K)(p-L)}{4} = \\frac{p(K-L)}{2},\n$$\nwhence $p \\neq 0$ and $K - L = 2/p$. Substituting this into the equality\n$$\n(K+L)(K-L) = K^2 - L^2 = (p^2-4q) - (p^2-4r) = 4(r-q)\n$$\nyields $K+L = 2p(r-q)$. From these values of $K+L$ and $K-L$ we obtain $K = 1/p - p(q-r)$, so upon squaring, $K^2 = 1/p^2 - 2(q-r)p^2 + (q-r)^2$. Comparing this with the equality $K^2 = p^2 - 4q$, an easy manipulation leads to the desired equation (1).\n\nSecond part. Assume that (1) holds. Then clearly $p \\neq 0$. The equation (1) can be rewritten in either of the following two forms,\n$$\np^4(q-r)^2 + 2p^2(q-r) + 1 = p^4 - 4p^2r \\quad \\text{and} \\quad p^4(r-q)^2 + 2p^2(r-q) + 1 = p^4 - 4p^2q.\n$$\nUpon dividing by $p^2$ we find that the discriminants of the equations (2) are equal to\n$$\np^2 - 4q = \\left( \\frac{p^2(r-q)+1}{p} \\right)^2 \\quad \\text{and} \\quad p^2 - 4r = \\left( \\frac{p^2(q-r)+1}{p} \\right)^2;\n$$\nhence, they are nonnegative and the (real) roots of (2) have the form (3), where\n$$\nK = \\frac{p^2(r-q)+1}{p} \\quad \\text{and} \\quad L = -\\frac{p^2(q-r)+1}{p}.\n$$\nThe signs of the numbers $K$ and $L$ have been chosen so that (see First Part)\n$$\nx_1y_1 - x_2y_2 = \\frac{p(K-L)}{2} = \\frac{p}{2} \\cdot \\left( \\frac{p^2(r-q)+1}{p} + \\frac{p^2(q-r)+1}{p} \\right) = 1.\n$$",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 29398,
    "subject": "Mathematics (Multi-modal)",
    "question": "Show that for each natural number $k$ there exist at most finitely many triples of mutually distinct primes $p, q, r$ for which the number $qr - k$ is a multiple of $p$, the number $pr - k$ is a multiple of $q$, and the number $pq - k$ is a multiple of $r$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "(Mutually distinct) primes $p, q, r$ satisfy the desired conditions if and only if the number $pq + pr + qr - k$ is divisible by each of the primes $p, q, r$; that is, by the product $pqr$. The equality $pq + pr + qr - k = n \\cdot pqr$, for a suitable integer $n$, can be rewritten as $k = pq + pr + qr - n \\cdot pqr$. If $n \\le 0$, then the last equality implies that $\\max\\{pq, pr, qr\\} \\le k$; however, then each of the primes $p, q, r$ is less than or equal to $k/2$ (and there is only a finite number of such triples).\n\nIf $n \\ge 1$, then we get the estimate $k \\le pq + pr + qr - pqr$. Let us show that the last expression is negative (contradicting the fact that $k > 0$) unless the triple in question is $\\{p, q, r\\} = \\{2, 3, 5\\}$. We can assume that $2 \\le p < q < r$ and $r \\ge 7$. Then $pq \\ge 2 \\cdot 3 = 6$ and the inequality $(p-2)(q-2) \\ge 0$ implies that $p+q \\le \\frac{1}{2}pq + 2$, hence\n$$\npq + pr + qr - pqr = (p+q)r + pq - pqr \\le (\\frac{1}{2}pq + 2)r + pq - pqr \\\\\n= 2r - pq(\\frac{1}{2}r - 1) \\le 2r - 6(\\frac{1}{2}r - 1) = 6 - r < 0.\n$$",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 29399,
    "subject": "Mathematics (Multi-modal)",
    "question": "In the interior of a cyclic quadrangle $ABCD$ a point $P$ is given such that\n$$\n|\\angle BPC| = |\\angle BAP| + |\\angle PDC|.\n$$\nDenote by $E$, $F$, and $G$ the feet of the perpendiculars from the point $P$ to the lines $AB$, $AD$ and $DC$, respectively. Show that the triangles $FEG$ and $PBC$ are similar.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $k$ be the circumcircle of the quadrangle $ABCD$ and $k_1$, $k_2$ the circumcircles of the triangles $PAB$ and $PCD$, respectively. In the interior of the angle $BPC$, consider the half-line $PT$ such that $|\\angle BPT| = |\\angle BAP|$. Then the hypothesis on $P$ implies that (Fig. 1)\n$$\n|\\angle TPC| = |\\angle BPC| - |\\angle BPT| = |\\angle BPC| - |\\angle BAP| = |\\angle PDC|.\n$$\nThus $PT$ is the common interior tangent of the circles $k_1$ and $k_2$.\n\n![](attached_image_1.png)\nFig. 1\n\nLet us first assume that the sides $AB$ and $CD$ of the given cyclic quadrangle are not parallel. Since the segments $AB$ and $CD$ are common chords of the circles $k_1$, $k$ and $k_2$, $k$, respectively, there exists a unique point $Q$ having the same power with respect to all three circles $k$, $k_1$ and $k_2$. The point $Q$ is the intersection of the three lines $AB$, $DC$ and $PT$. Without loss of generality, we can assume that the point $Q$ is located on the half-line $BA$ beyond the point $A$ (Fig. 1). Then we have\n$$\n|\\angle QPA| = |\\angle PBA| \\quad (1)\n$$\nSince $|\\angle AEP| = |\\angle AFP| = 90^\\circ$, the quadrangle $AEPF$ is cyclic, and\n$$\n|\\angle FEP| = |\\angle FAP| = |\\angle DAP| \\quad (2)\n$$\nSimilarly we see that the quadrangle $DGPF$ is cyclic. It follows that\n$$\n|\\angle BPC| = |\\angle BAP| + |\\angle PDC| = |\\angle EFP| + |\\angle PFG| = |\\angle EFG| \\quad (3)\n$$\nSince further $|\\angle PEQ| = |\\angle PGQ| = 90^\\circ$, the quadrangle $QEPG$ is also cyclic and\n$$\n|\\angle GEP| = |\\angle GQP| = |\\angle DQP| \\quad (4)\n$$\nFrom the relations (2), (4) and the equality $|\\angle DAP| + |\\angle QPA| = |\\angle QDA| + |\\angle DQP|$, we further have\n$$\n|\\angle FEG| = |\\angle FEP| - |\\angle GEP| = |\\angle DAP| - |\\angle DQP| = |\\angle QDA| - |\\angle QPA| \\quad (5)\n$$\nSince the quadrangle $ABCD$ is cyclic, $|\\angle QDA| = |\\angle QBC|$. From the relations (1) and (5) we thus get\n$$\n|\\angle FEG| = |\\angle QBC| - |\\angle PBA| = |\\angle PBC|\n$$\nUsing finally the relations (3) and (6) we see that the triangles $FEG$ and $PBC$ are similar (as they have two congruent angles).\n\nAn analogous argument can be used when the point $Q$ is located on the half-line $AB$ beyond the point $B$. If the lines $AB$ and $CD$ are parallel, then $ABCD$ is an equilateral trapezoid, with bases $AB$ and $CD$. Since the points $E$, $P$, $G$ are collinear and the common interior tangent of the circles $k_1$ and $k_2$ is parallel to both lines $AB$ and $CD$, the triangles $APD$ and $PBC$ are congruent. The similarity of the triangles $EFG$ and $APD$ thus implies also the similarity of the triangles $EFG$ and $PBC$.\n\nThis completes the proof.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 29400,
    "subject": "Mathematics (Multi-modal)",
    "question": "Solve the system of equations\n$$\n\\frac{1}{xy} = \\frac{x}{z} + 1, \\quad \\frac{1}{yz} = \\frac{y}{x} + 1, \\quad \\frac{1}{zx} = \\frac{z}{y} + 1\n$$\nin the domain of the real numbers.",
    "options": [],
    "answer": "x = y = z = ± sqrt(2)/2",
    "solution": "From the form of the equations it is immediate that $xyz \\neq 0$. Two of the numbers $x, y, z$ have to be of the same sign; then the right-hand side of the equation where the ratio of these two numbers occurs is positive, hence so must be the corresponding left-hand side, which implies that the third of the numbers $x, y, z$ must also have the same sign as the first and the second. Thus either $x, y, z > 0$, or $x, y, z < 0$. Let us consider only the former case (the latter can be reduced to it by passing from the solution $(x, y, z)$ to the solution $(-x, -y, -z)$).\n\nMultiply the first two equations of the system by the expression $xyz$ and then subtract them; this gives, upon a small manipulation, $z - x = y(x^2 - yz)$. If a triple $(x, y, z)$ is a solution, then so are also the triples $(y, z, x)$ and $(z, x, y)$; thus we may assume that $x = \\max\\{x, y, z\\}$. Then $z - x \\le 0$ and $x^2 - yz \\ge 0$ (remember that $x, y, z > 0$), so the equality $z - x = y(x^2 - yz)$, together with the condition $y > 0$, implies that $z - x = x^2 - yz = 0$, which means that $x = y = z$. The system then reduces to the single equation $1/x^2 = 1 + 1$, which has a (unique) positive root $x = \\sqrt{2}/2$.\n\n**Conclusion.** The system has exactly two solutions, $x = y = z = \\pm \\sqrt{2}/2$.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 29401,
    "subject": "Mathematics (Multi-modal)",
    "question": "In the interiors of the sides $AB$, $BC$, $CA$ of a given triangle $ABC$, points $K$, $L$ and $M$, respectively, are given such that\n$$\n\\frac{|AK|}{|KB|} = \\frac{|BL|}{|LC|} = \\frac{|CM|}{|MA|}\n$$\nShow that the triangles $ABC$ and $KLM$ have a common orthocenter if and only if the triangle $ABC$ is equilateral.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "A point $V$ of the plane containing a triangle $ABC$ is its orthocenter if and only if, at the same time, $AV \\perp BC$ and $BV \\perp AC$; that is, $AV \\cdot BC = 0$ and $BV \\cdot AC = 0$. Substituting $BC = BV - CV$ and $AC = AV - CV$, an easy manipulation leads to the equivalent condition in the form of the equality of the scalar products\n$$\nAV \\cdot BV = AV \\cdot CV = BV \\cdot CV. \\quad (1)\n$$\nOur goal is thus to find out when the system (1) is satisfied together with the analogous system\n$$\nKV \\cdot LV = KV \\cdot MV = LV \\cdot MV, \\quad (2)\n$$\nexpressing the fact that the point $V$ is the orthocenter of the triangle $KLM$. We now express the vectors from (2) as linear combinations of the vectors from (1). By hypothesis, there exists a number $p$, $0 < p < 1$, for which\n$$\nAK = pAB, \\quad BL = pBC, \\quad CM = pCA.\n$$\n\nSubstituting into the first equality $AK = AV - KV$ and $AB = AV - BV$, we get after a small manipulation the first of the following three equalities\n$$\nKV = (1-p)AV + pBV, \\ LV = (1-p)BV + pCV, \\ MV = (1-p)CV + pAV;\n$$\nthe other two can be derived similarly. Taking products, we get\n$$\nKV \\cdot LV = (1-p)^2 AV \\cdot BV + p(1-p)AV \\cdot CV + p(1-p)BV^2 = \\\\ = (1-p)s + p(1-p)BV^2,\n$$\nwhere $s$ denotes the common value of the products from (1). Similarly,\n$$\nKV \\cdot MV = (1-p)s + p(1-p)AV^2 \\quad \\text{and} \\quad LV \\cdot MV = (1-p)s + p(1-p)BV^2.\n$$\nWe see that the system (2) is equivalent to the system of equalities\n$$\np(1-p)AV^2 = p(1-p)BV^2 = p(1-p)CV^2,\n$$\nwhich, in view of the condition $p(1-p) \\neq 0$, is fulfilled if and only if $|AV| = |BV| = |CV|$. The last condition means that the orthocenter $V$ of the triangle $ABC$ coincides with its circumcenter. This happens if and only if the triangle $ABC$ is equilateral.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 29402,
    "subject": "Mathematics (Multi-modal)",
    "question": "On the table there are $k$ heaps of $1, 2, \\dots, k$ stones, where $k \\ge 3$. In the first step, we choose any three of the heaps on the table, merge them into a single new heap, and remove 1 stone (throw it away from the table) from this new heap. In the second step, we again merge some three of the heaps together into a single new heap, and then remove 2 stones from this new heap. In general, in the $i$-th step we choose any three of the heaps, which contain more than $i$ stones when combined, we merge them into a single new heap, and then remove $i$ stones from this new heap. Assume that after a number of steps, there is a single heap left on the table, containing $p$ stones. Show that the number $p$ is a perfect square if and only if the numbers $2k+2$ and $3k+1$ are perfect squares. Further, find the least number $k$ for which $p$ is a perfect square.",
    "options": [],
    "answer": "161",
    "solution": "After $i$ steps, there will be $k - 2i$ heaps left on the table; thus if a single heap is to remain in the end, the number $k$ must be odd and the total number of steps has to be $\\frac{1}{2}(k-1)$. Let us distinguish two cases, according as the remainder of $k$ upon division by 4 is 1 or 3.\n\nThe case of $k = 4c + 1$. In the beginning there are $1 + \\dots + k = \\frac{1}{2}k(k+1) = (4c+1)(2c+1)$ stones on the table, from which in course of the $2c$ steps we will remove $1 + \\dots + 2c = c(2c+1)$ stones; thus the number of stones in the last heap will be\n$$\np = (4c+1)(2c+1) - c(2c+1) = (2c+1)(3c+1).\n$$\nSince the numbers $2c+1$ and $3c+1$ are coprime, $p$ is a perfect square if and only if both $2c+1$ and $3c+1$ are perfect squares; that is, if and only if their quadruples $4(2c+1) = 2k+2$ and $4(3c+1) = 3k+1$ are perfect squares.\n\nThe case of $k = 4c + 3$. In the beginning there are $1 + \\dots + k = \\frac{1}{2}k(k+1) = 2(c+1)(4c+3)$ stones on the table, from which we remove in course of all the $2c + 1$ steps a total of $1 + \\cdots + (2c + 1) = (c + 1)(2c + 1)$ stones; thus the last single heap will contain\n$$\np = 2(c+1)(4c+3) - (c+1)(2c+1) = (c+1)(6c+5).\n$$\nIf $p$ were a perfect square, then so would have to be the two coprime numbers $c+1$ and $6c+5$. Let us show that this is not possible. Assume that there exist natural numbers $x, y$ such that $c+1 = x^2$ and $6c+5 = y^2$. From the equality $6x^2 - y^2 = 1$ it follows that the number $y$ is odd, hence $y^2$ gives a remainder of 1 upon division by 8. The number $6x^2$ then gives the remainder of 2, which implies that $3x^2$ gives remainder 1 upon division by 4 — a contradiction. Hence in the case of $k = 4c+3$ the number $p$ is never a perfect square, and likewise the number $3k+1 = 12c+10$ is never a perfect square (being an even number not divisible by 4).\n\nLet us finally find the least number $k = 4c+1$, $c \\ge 1$, for which both numbers $2c+1$ and $3c+1$ are perfect squares. From the equalities $2c+1 = x^2$ and $3c+1 = y^2$ for suitable integers $x, y > 1$ it follows that $3x^2 - 2y^2 = 1$, whence $x$ is odd, but then the number $2y^2$ gives remainder 2 upon division by 4, so $y$ is also odd. Set $x = 2a+1$, $y = 2b+1$ ($a, b > 0$ integer) and substitute this into the equality $3x^2 - 2y^2 = 1$; upon a small manipulation this leads to the relation $3a(a+1) = 2b(b+1)$, and taking successively $a = 1, 2, \\dots$ we soon find the smallest solution $a = 4$ and $b = 5$, corresponding to $x = 9, y = 11, c = 40$ and $k = 161$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 29403,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all pairs of integers $a, b$ such that\n$$\n\\frac{a^2+1}{2b^2-3} = \\frac{a-1}{2b-1}\n$$",
    "options": [],
    "answer": "[(0, 1), (-1, 1), (0, -2), (-1, -2), (7, 8)]",
    "solution": "Obviously $a \\neq 1$, thus we can rewrite the equation as\n$$\n\\frac{a^2+1}{a-1} = \\frac{2b^2-3}{2b-1} \\quad (1)\n$$\nThe numerator of the fraction on the left is positive, the numerator on the right is negative just for $b \\in \\{-1, 0, 1\\}$.\nFor $b = -1$ we get $3a^2 - a + 4 = 0$, which has no real solution.\nSimilarly for $b = 0$ we get $a^2 - 3a + 4 = 0$ which has no real solution either.\nFor $b = 1$ we get $a^2 + a = a(a + 1) = 0$, with solutions $a \\in \\{0, -1\\}$. Thus pairs $(0, 1)$ and $(-1, 1)$ are solutions of the problem.\nLet us further assume $2b^2 - 3 > 0$, and let us find out with which numbers we\ncan reduce the fractions in (1).\nIf some integer $n$ divides both $a^2+1$ and $a-1$, it divides $a^2+1-(a+1)(a-1) = 2$\nas well. Similarly if $n$ divides both $2b^2 - 3$ and $2b - 1$, it divides $(2b - 1)(2b + 1) - 2(2b^2 - 3) = 5$.\nThus there are four possibilities to fulfill the equation in (1).\n(i) $a^2+1 = 2b^2-3$ and $a-1 = 2b-1$, which has no real solution.\n(ii) $a^2+1 = 2(2b^2-3)$ and $a-1 = 2(2b-1)$; substituting $a = 4b-1$ into the first equation we get $3b^2 - 2b + 2 = 0$, with no real solutions.\n(iii) $5(a^2+1) = 2b^2-3$ and $5(a-1) = 2b-1$, with solution $a=0, b=-2$.\n(iv) Finally $5(a^2+1) = 2(2b^2-3)$ and $5(a-1) = 2(2b-1)$ with solutions $a = -1, b = -2$ and $a = 7, b = 8$.\nThus there are five solutions of the problem:\n$$\n(0, 1),\\ (-1, 1),\\ (0, -2),\\ (-1, -2),\\ (7, 8).\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 29404,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all pairs of primes $p$, $q$ for which there exists a positive integer $a$ such that\n$$\n\\frac{pq}{p+q} = \\frac{a^2+1}{a+1}.\n$$",
    "options": [],
    "answer": "(2, 2) and (5, 5)",
    "solution": "First, we will deal with the case when the wanted primes $p$ and $q$ are distinct. Then, the numbers $pq$ and $p+q$ are relatively prime: the product $pq$ is divisible by two primes only (namely $p$ and $q$), while the sum $p+q$ is divisible by neither of these primes.\nWe will look for a positive integer $r$ which can be a common divisor of both $a+1$ and $a^2+1$. If $r \\mid a+1$ and, at the same time, $r \\mid a^2+1$, then $r \\mid (a+1)(a-1)$ and also $r \\mid (a^2+1) - (a^2-1) = 2$, so $r$ can only be one of the numbers $1$ and $2$. Thus the fraction $\\frac{a^2+1}{a+1}$ either is in lowest terms, or will be in lowest terms when reduced by two, depending on whether the integer $a$ is even, or odd.\nIf $a$ is even, we must have\n$$\npq = a^2 + 1 \\quad \\text{and} \\quad p + q = a + 1.\n$$\nThe numbers $p, q$ are thus the roots of the quadratic equation $x^2-(a+1)x+a^2+1=0$, whose discriminant\n$$\n(a+1)^2 - 4(a^2+1) = -3a^2 + 2a - 3 = -2a^2 - (a-1)^2 - 2\n$$\nis apparently negative, so the equation has no solution in the real numbers.\nIf $a$ is odd, we must have (taking into account the reduction by two)\n$$\n2pq = a^2 + 1 \\quad \\text{and} \\quad 2(p+q) = a + 1.\n$$\nThe numbers $p, q$ are thus the roots of the quadratic equation $2x^2-(a+1)x+a^2+1=0$, whose discriminant is negative as well.\nTherefore, there is no pair of distinct primes $p, q$ satisfying the conditions.\nIt remains to analyze the case of $p = q$. Then,\n$$\n\\frac{p \\cdot q}{p+q} = \\frac{p \\cdot p}{p+p} = \\frac{p}{2},\n$$\nso we must have\n$$\np = \\frac{2(a^2 + 1)}{a + 1} = 2a - 2 + \\frac{4}{a + 1};\n$$\nthis is an integer if and only if $a+1 \\mid 4$, i.e. $a \\in \\{1, 3\\}$, so $p = 2$ or $p = 5$.\nTo summarize, there are exactly two pairs of primes satisfying the conditions, namely $p = q = 2$ and $p = q = 5$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 29405,
    "subject": "Mathematics (Multi-modal)",
    "question": "Prove that for each integer number $n$, $n \\ge 3$, the following $2n$-digit number\n$$\n\\underbrace{1\\dots12}_{n-1} \\underbrace{8\\dots8}_{n-2} 96\n$$\nis a perfect square.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "The number under consideration can be expressed as follows:\n$$\n\\begin{aligned}\n& (10^{2n-1} + 10^{2n-2} + \\dots + 10^{n+1}) + 2 \\cdot 10^n + 8 \\cdot (10^{n-1} + 10^{n-2} + \\dots + 10^2) + 96 \\\\\n&= 10^{n+1} \\cdot \\frac{10^{n-1}-1}{9} + 2 \\cdot 10^n + 8 \\cdot 10^2 \\cdot \\frac{10^{n-2}-1}{9} + 96 \\\\\n&= \\frac{10^{2n} - 10^{n+1} + 18 \\cdot 10^n + 800 \\cdot 10^{n-2} - 800 + 9 \\cdot 96}{9} \\\\\n&= \\frac{10^{2n} + 16 \\cdot 10^n + 64}{9} = \\left( \\frac{10^n + 8}{3} \\right)^2.\n\\end{aligned}\n$$\nAs required, we have obtained a perfect square, because the number $10^n+8$ is divisible by 3, as the sum of its digits equals 9.\n\n$1296 = 36^2, 112896 = 336^2, 11128896 = 3336^2, \\dots$\nwe easily guess that for each $n \\ge 2$,\n$$\n\\underbrace{1\\dots128}_{n-1} \\underbrace{896}_{n-2} = \\underbrace{33\\dots36^2}_{n-1}.\n$$\nThe exact proof can be done by using the usual multiplication scheme:\n$$\n\\begin{array}{r}\n333\\dots3336 \\\\\n\\times 333\\dots3336 \\\\\n\\hline\n2000\\dots0016 \\\\\n10000\\dots008 \\\\\n100000\\dots08 \\\\\n1000000\\dots8 \\\\\n\\vdots \\\\\n1\\dots000008 \\\\\n10\\dots00008 \\\\\n100\\dots0008 \\\\\n\\hline\n111\\dots112888\\dots8896\n\\end{array}\n$$\n\nBoth (identical) factors are $n$-digit, hence an $(n + 1)$-digit number stands in each of the $n$ rows between the two delimiting lines. From this fact, it is easy to determine the values of digits (including the numbers of appearances) in the resulting product.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 29406,
    "subject": "Mathematics (Multi-modal)",
    "question": "In a group of 21 different integers, a sum of arbitrary eleven ones is greater than a sum of the remaining ten numbers.\n\na) Prove that every considered number is greater than 100.\n\nb) Find all such groups of 21 different integers containing number 101.",
    "options": [],
    "answer": "a) Every number in the set is greater than 100.\nb) Exactly those sets consisting of 101 together with any 20 consecutive integers all greater than 101; in such sets the sum of the 11 smallest exceeds the sum of the 10 largest by 1.",
    "solution": "a. Let the numbers be $a_1 < a_2 < a_3 < \\dots < a_{21}$. Since they are integers, for every $i \\in \\{1, 2, \\dots, 20\\}$ it holds that $a_{i+1} - a_i \\ge 1$, and therefore $a_{i+10} - a_i \\ge 10$ for every $i \\in \\{1, 2, \\dots, 11\\}$.\n\nThe problem condition is fulfilled if and only if the sum\n$$\na_1 + a_2 + \\dots + a_{11} > a_{12} + a_{13} + \\dots + a_{21}. \\quad (1)\n$$\nThis follows\n$$\na_1 > (a_{12} - a_2) + (a_{13} - a_3) + \\dots + (a_{21} - a_{11}) \\ge 10 \\cdot 10 = 100.\n$$\nSince the least of the numbers is greater than 100, the other ones are greater than 100 too.\n\nb. We have proved $a_1 \\ge 101$. The other numbers are greater than 101. If the number 101 is in the group of positive integers, then $a_1 = 101$ holds. The strict inequality (1) gives\n$$\n(a_{12} - a_2) + (a_{13} - a_3) + \\dots + (a_{21} - a_{11}) \\le a_1 - 1 = 100,\n$$\nand because $a_{i+10} - a_i \\ge 10$, the equality $a_{i+10} - a_i = 10$ holds for every $i \\in \\{2, 3, \\dots, 11\\}$. It comes to pass if and only if the numbers $a_2, a_3, \\dots, a_{21}$ are consecutive integers.\n\nThe required group consists of number 101 and arbitrary 20 consecutive integers which are greater than 101. Then the difference of sums of 11 minimal numbers and 10 maximal numbers is 1.",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 29407,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all pairs of primes $p$, $q$ for which there exists a positive integer $a$ such that\n$$\n\\frac{pq}{p+q} = \\frac{a^2+1}{a+1}.\n$$",
    "options": [],
    "answer": "p = q = 2 and p = q = 5",
    "solution": "First, we will deal with the case when the wanted primes $p$ and $q$ are distinct. Then, the numbers $pq$ and $p+q$ are relatively prime: the product $pq$ is divisible by two primes only (namely $p$ and $q$), while the sum $p+q$ is divisible by neither of these primes.\nWe will look for a positive integer $r$ which can be a common divisor of both $a+1$ and $a^2+1$. If $r \\mid a+1$ and, at the same time, $r \\mid a^2+1$, then $r \\mid (a+1)(a-1)$ and also $r \\mid (a^2+1) - (a^2-1) = 2$, so $r$ can only be one of the numbers $1$ and $2$. Thus the fraction $\\frac{a^2+1}{a+1}$ either is in lowest terms, or will be in lowest terms when reduced by two, depending on whether the integer $a$ is even, or odd.\n\nIf $a$ is even, we must have\n$$\npq = a^2 + 1 \\quad \\text{and} \\quad p+q = a+1.\n$$\nThe numbers $p$, $q$ are thus the roots of the quadratic equation $x^2-(a+1)x+a^2+1 = 0$, whose discriminant\n$$\n(a+1)^2 - 4(a^2+1) = -3a^2 + 2a - 3 = -2a^2 - (a-1)^2 - 2\n$$\nis apparently negative, so the equation has no solution in the real numbers.\n\nIf $a$ is odd, we must have (taking into account the reduction by two)\n$$\n2pq = a^2 + 1 \\quad \\text{and} \\quad 2(p+q) = a+1.\n$$\nThe numbers $p$, $q$ are thus the roots of the quadratic equation $2x^2-(a+1)x+a^2+1 = 0$, whose discriminant is negative as well.\n\nTherefore, there is no pair of distinct primes $p$, $q$ satisfying the conditions. It remains to analyze the case of $p = q$. Then,\n$$\n\\frac{p \\cdot q}{p+q} = \\frac{p \\cdot p}{p+p} = \\frac{p}{2}\n$$\nso we must have\n$$\np = \\frac{2(a^2 + 1)}{a + 1} = 2a - 2 + \\frac{4}{a + 1};\n$$\nthis is an integer if and only if $a+1 \\mid 4$, i.e. $a \\in \\{1,3\\}$, so $p=2$ or $p=5$.\n\nTo summarize, there are exactly two pairs of primes satisfying the conditions, namely $p=q=2$ and $p=q=5$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 29408,
    "subject": "Mathematics (Multi-modal)",
    "question": "There are two touching circles, $k_1(S_1, r_1)$ and $k_2(S_2, r_2)$ in a rectangle $ABCD$ with $|AB| = 9$, $|BC| = 8$. Moreover, $k_1$ touches $AD$ and $CD$, while $k_2$ touches $AB$ and $BC$.\n\na) Prove $r_1 + r_2 = 5$.\n\nb) What is the least and what is the greatest possible area of $AS_1S_2$?",
    "options": [],
    "answer": "r1 + r2 = 5; minimum area = 14, maximum area = 31/2",
    "solution": "a) Let $M$ and $N$ be intersections of the line through $S_1$ parallel to $AD$. Analogously, let $K$ and $L$ be intersections of the line through $S_2$ parallel to $AB$. Let $P$ be the intersection of $KL$ and $MN$ (see Fig. 1). The Pythagoras theorem for $S_1PS_2$ gives\n$$\n\\begin{align*}\n(r_1 + r_2)^2 &= (8 - r_1 - r_2)^2 + (9 - r_1 - r_2)^2, \\\\\n(r_1 + r_2)^2 - 34(r_1 + r_2) + 145 &= 0, \\\\\n(r_1 + r_2 - 5)(r_1 + r_2 - 29) &= 0.\n\\end{align*}\n$$\nSince $2r_1 \\le 8$, $2r_2 \\le 8$, we have $r_1 + r_2 = 5$.\n\nb) Let $Q$ be a foot of a perpendicular to $AB$ from $S_2$, let $R$ be a foot of a perpendicular to $AD$ from $S_1$ and let $T$ be the intersection of $QS_2$ and $RS_1$ (Fig. 1).\n\nThe area $S$ of $AS_2S_1$ is given by the difference of the area of rectangle $AQTR$ and areas of right triangles $AQS_2$, $AS_1R$, and $S_1S_2T$:\n$$\n\\begin{align*} \nS &= (9-r_2)(8-r_1) - \\frac{1}{2}r_2(9-r_2) - \\frac{1}{2}r_1(8-r_1) - \\frac{1}{2}(9-r_1-r_2)(8-r_1-r_2) \\\\ \n&= 72 - 9r_1 - 8r_2 + r_1r_2 - \\frac{9}{2}r_2 + \\frac{1}{2}r_1^2 - 4r_1 + \\frac{1}{2}r_1^2 - 36 + \\frac{17}{2}(r_1+r_2) \\\\ \n&\\quad - \\frac{1}{2}(r_1+r_2)^2 \\\\ \n&= 36 - \\frac{9}{2}r_1 - 4r_2 = 36 - \\frac{9}{2}r_1 - 4(5-r_1) = 16 - \\frac{1}{2}r_1, \n\\end{align*}\n$$\nwhere we used $r_1 + r_2 = 5$. Further, we know $2r_1 \\le 8$ and $2r_2 \\le 8$ which implies $1 \\le r_1, r_2 \\le 4$, thus\n$$\nS = 16 - \\frac{1}{2}r_1 \\in \\left\\langle 14, \\frac{31}{2} \\right\\rangle;\n$$\nand the least possible value of the area is $14$, for $r_1 = 4$ and $r_2 = 1$, and the greatest value possible is $\\frac{31}{2}$, for $r_1 = 1$ and $r_2 = 4$.\n\n![](attached_image_1.png)\nFig. 1",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 29409,
    "subject": "Mathematics (Multi-modal)",
    "question": "There are two touching circles, $k_1(S_1, r_1)$ and $k_2(S_2, r_2)$ in a rectangle $ABCD$ with $|AB| = 9$, $|BC| = 8$. Moreover, $k_1$ touches $AD$ and $CD$, while $k_2$ touches $AB$ and $BC$.\n\na) Prove $r_1 + r_2 = 5$.\n\nb) What is the least and what is the greatest possible area of $AS_1S_2$?",
    "options": [],
    "answer": "r1 + r2 = 5; minimum area = 14; maximum area = 31/2",
    "solution": "a) Let $M$ and $N$ be intersections of the line through $S_1$ parallel to $AD$. Analogously, let $K$ and $L$ be intersections of the line through $S_2$ parallel to $AB$. Let $P$ be the intersection of $KL$ and $MN$ (see Fig. 1). The Pythagoras theorem for $S_1PS_2$ gives\n$$\n(r_1 + r_2)^2 = (8 - r_1 - r_2)^2 + (9 - r_1 - r_2)^2, \\ (1)\n$$\n$$\n(r_1 + r_2)^2 - 34(r_1 + r_2) + 145 = 0,\n$$\n$$\n(r_1 + r_2 - 5)(r_1 + r_2 - 29) = 0.\n$$\nSince $2r_1 \\le 8$, $2r_2 \\le 8$, we have $r_1 + r_2 = 5$.\n\nb) Let $Q$ be a foot of a perpendicular to $AB$ from $S_2$, let $R$ be a foot of a perpendicular to $AD$ from $S_1$ and let $T$ be the intersection of $QS_2$ and $RS_1$ (Fig. 1).\n\nThe area $S$ of $AS_2S_1$ is given by the difference of the area of rectangle $AQTR$ and areas of right triangles $AQS_2$, $AS_1R$, and $S_1S_2T$:\n$$\n\\begin{align*}\nS &= (9 - r_2)(8 - r_1) - \\frac{1}{2}r_2(9 - r_2) - \\frac{1}{2}r_1(8 - r_1) - \\frac{1}{2}(9 - r_1 - r_2)(8 - r_1 - r_2) \\\\\n&= 72 - 9r_1 - 8r_2 + r_1r_2 - \\frac{9}{2}r_2 + \\frac{1}{2}r_2^2 - 4r_1 + \\frac{1}{2}r_1^2 - 36 + \\frac{17}{2}(r_1 + r_2) \\\\\n&\\quad - \\frac{1}{2}(r_1 + r_2)^2 \\\\\n&= 36 - \\frac{9}{2}r_1 - 4r_2 = 36 - \\frac{9}{2}r_1 - 4(5 - r_1) = 16 - \\frac{1}{2}r_1,\n\\end{align*}\n$$\nwhere we used $r_1 + r_2 = 5$. Further, we know $2r_1 \\le 8$ and $2r_2 \\le 8$ which implies $r_1, r_2 \\le 4$, thus\n$$\nS = 16 - \\frac{1}{2}r_1 \\in \\left\\langle 14, \\frac{31}{2} \\right\\rangle;\n$$\nand the least possible value of the area is $14$, for $r_1 = 4$ and $r_2 = 1$, and the greatest value possible is $\\frac{31}{2}$, for $r_1 = 1$ and $r_2 = 4$.\n\n![](attached_image_1.png)\nFig. 1",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 29410,
    "subject": "Mathematics (Multi-modal)",
    "question": "A number $n$ is a product of three (not necessarily distinct) prime numbers. Adding 1 to each of them, after multiplication we get a larger product $n + 963$. Determine the original product $n$. (Pavel Novotný)",
    "options": [],
    "answer": "2013",
    "solution": "We look for $n = p \\cdot q \\cdot r$, with primes $p \\le q \\le r$ satisfying\n$$(p+1)(q+1)(r+1) = pqr + 963. \\quad (1)$$\nIf $p=2$, the right-hand side of (1) is odd, hence the factors $q+1, r+1$ on the left must be odd too. This implies that $p=q=r=2$, which contradicts to (1). Thus we have proved that $p \\ge 3$.\nNow we will show that $p=3$. Suppose on the contrary that $3 < p \\le q \\le r$. Then the right-hand side of (1) is not divisible by $3$. The same must be true for the product $(p+1)(q+1)(r+1)$. Consequently, all the primes $p, q, r$ are congruent to $1$ modulo $3$, and hence $(p+1)(q+1)(r+1) - pqr$ is congruent to $2 \\cdot 2 \\cdot 2 - 1 \\cdot 1 \\cdot 1 = 7$, which contradicts to $(p+1)(q+1)(r+1) - pqr = 963$. Therefore, the equality $p=3$ is established.\nPutting $p=3$ into (1) we get $4(q+1)(r+1) = 3qr + 963$, which can be rewritten as $(q+4)(r+4) = 975$. In view of the prime factorization $975 = 3 \\cdot 5^2 \\cdot 13$ and inequalities $7 \\le q+4 \\le r+4$, we conclude that $q+4 \\le \\sqrt{975} < 32$ and hence $q+4 \\in \\{13, 15, 25\\}$. Since $q$ is a prime, it holds that $q=11$. Then $r+4=65$, and hence $r=61$ (which is a prime indeed). Consequently, the problem has a unique solution\n$$\nn = 3 \\cdot 11 \\cdot 61 = 2013.\n$$",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 29411,
    "subject": "Mathematics (Multi-modal)",
    "question": "Each of $n$ Robin Hoods ($n \\ge 3$) robbed some coins. Together they have earned $100n$ coins. They have decided to cut the loot in a following way: in one step one Robin can take his two coins and give to some other two Hoods, one coin each. Find all positive integers $n \\ge 3$ for which they can split the loot in equal parts ($100$ coins each). (Ján Mazák)",
    "options": [],
    "answer": "all integers n ≥ 4",
    "solution": "Let $z_i$ denote through the process the number of coins of the $i$-th Robin Hood.\n\nLet $n = 3$. After any step, $z_1 - z_2$ modulo $3$ does not change. That means for $z_1 = 101$, $z_2 = 100$, and $z_3 = 99$ (out of many choices), $z_1$ will never be the same as $z_2$. Thus $n = 3$ is not a solution.\n\nNow we show that for each $n \\ge 4$ any initial $z_i$ the loot can be split in equal parts.\n\nLet $s = \\sum |z_i - 100|$. We decrease number $s$ as long as it can be done in the way that some of the outlaws with maximal count of coins gives his two coins to (some of the) Hoods with the minimal count of coins. If $s$ can be reduced to $0$ in this way, we are done.\n\nIf $s \\neq 0$, some of the outlaws has $100-k$ coins ($k > 0$), $k$ outlaws have $101$ coins each and all the others have $100$ coins each. If $k \\ge 2$, we decrease $s$ in two steps:\n$$\n100 - k,\\ 101,\\ 101 \\rightarrow 100 - k + 1,\\ 102,\\ 99 \\rightarrow 100 - k + 2,\\ 100,\\ 100.\n$$\nIf $k$ is even, then after $\\frac{1}{2}k$ such \"double\" steps every outlaw will have $100$ coins each. If $k$ is odd then we end in the state in which one of the outlaws has $99$ coins, one has $101$ coins and all the others have $100$ coins. We finish as follows:\n$$\n99,\\ 100,\\ 100,\\ 101 \\rightarrow 99,\\ 101,\\ 101,\\ 99 \\rightarrow 99,\\ 102,\\ 99,\\ 100 \\rightarrow 100,\\ 100,\\ 100,\\ 100.\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 29412,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $M$ be the midpoint of the side $AB$ of a triangle $ABC$. Prove that the equality $|\\angle ABC| + |\\angle ACM| = 90^\\circ$ holds if and only if the triangle $ABC$ is isosceles or right-angled, with $AB$ as a base or a hypotenuse, respectively. (Pavel Novotný)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Assume first that $|\\angle ABC| + |\\angle ACM| = 90^\\circ$. Using the notation $\\phi = |\\angle ACM|$ and $\\psi = |\\angle BCM|$ (Fig. 1), we conclude from our assumption that $|\\angle ABC| = 90^\\circ - \\phi$, and hence $|\\angle BAC| = 90^\\circ - \\psi$ as well, because of an easy angle computation in $\\triangle ABC$:\n$$\n\\begin{aligned}\n|\\angle BAC| &= 180^\\circ - |\\angle ABC| - |\\angle ACB| \\\\\n&= 180^\\circ - (90^\\circ - \\phi) - (\\phi + \\psi) = 90^\\circ - \\psi.\n\\end{aligned}\n$$\n![](attached_image_1.png)\nFig. 1\n\nApplying Law of Sines to $\\triangle ACM$ and $\\triangle BCM$, we get\n$$\n\\frac{\\sin(90^\\circ - \\psi)}{\\sin \\phi} = \\frac{|CM|}{|AM|} = \\frac{|CM|}{|BM|} = \\frac{\\sin(90^\\circ - \\phi)}{\\sin \\psi}.\n$$\nComparing the two ratios of sines and using the formula $\\sin(90^\\circ - \\omega) = \\cos \\omega$, we obtain an equality $\\sin \\phi \\cos \\phi = \\sin \\psi \\cos \\psi$ or $\\sin 2\\phi = \\sin 2\\psi$. Since the angles $\\phi$ and $\\psi$ are acute, both $2\\phi$ and $2\\psi$ are between $0^\\circ$ and $180^\\circ$. Thus by a well known sine property, the equality $\\sin 2\\phi = \\sin 2\\psi$ means that either $2\\phi = 2\\psi$ or $2\\phi + 2\\psi = 180^\\circ$. In the first case (when $\\phi = \\psi$), the interior angle of $\\triangle ABC$ at the vertices $A$ and $B$ are equal, in the second case (when $\\phi + \\psi = 90^\\circ$) the interior angle at the vertex $C$ is right. This completes the proof of one of the two implications stated in the problem.\n\n![](attached_image_1.png)\nFig. 1\n\nTo prove the second (converse) implication, let us assume that (i) $|AC| = |BC|$ or (ii) $|\\angle ACB| = 90^\\circ$.\n\n**Case (i).** It follows from $|AC| = |BC|$ that the triangles $ACM$ and $BCM$ are congruent (by SSS theorem), with right interior angles at the vertex $M$. Consequently,\n\nApplying Law of Sines to $\\triangle ACM$ and $\\triangle BCM$, we get\n$$\n\\frac{\\sin(90^\\circ - \\psi)}{\\sin \\phi} = \\frac{|CM|}{|AM|} = \\frac{|CM|}{|BM|} = \\frac{\\sin(90^\\circ - \\phi)}{\\sin \\psi}.\n$$\nThe converse implication is proven.\nLet $k$ be the circumcircle of the given triangle $ABC$. Its median $CM$ can be extended to the chord $CC'$ of the circle $k$ (Fig. 2). Since the inscribed angles $ABC'$ and $ACC'$ (or $ACM$) are congruent, the considered sum of angles $ABC$ and $ACM$ is equal to the angle $CBC'$. By Thales' theorem the last angle $CBC'$ is right if and only if the chord $CC'$ is a diameter of the circle $k$. This happens if and only if the centre $S$ of $k$ lies on the ray $CM$. For such a situation, we distinguish two cases: $S = M$ and $S \\neq M$. Note that $S = M$ holds if and only if the angle $ACB$ is right (by Thales' theorem again). Thus let us analyse the second case $S \\neq M$: The three distinct points $C, M$ and $S$ are obviously collinear if and only if the line $MS$, a perpendicular bisector of a segment $AB$, passes through the point $C$. However, the last condition is equivalent to the desired equality $|AC| = |BC|$. This completes the proof (common for the both composing implications).\n\n![](attached_image_2.png)\nFig. 2\n\n*Remark.* Instead of the chord $CC'$ of the circumcircle $k$, it is possible to consider the tangent line $t$ to the circle $k$ at its point $C$ (Fig. 3). Since the inscribed angle $ABC$ is always congruent to the marked angle between $AC$ and $t$, the sum of the angles $ABC$ and $ACM$ equals $90^\\circ$ if and only if the tangent $t$ is perpendicular to the ray $CM$. The last is equivalent to the condition from the above solution, namely that the ray $CM$ passes through the centre $S$ of $k$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 29413,
    "subject": "Mathematics (Multi-modal)",
    "question": "The number $0$ is written on each of the $n+1$ faces of an $n$-sided pyramid. In a step we choose a vertex and we increase by $1$ each number on the faces, which contain the vertex. Show, that in such way, we cannot get number $1$ written on each face.\n\n(Peter Novotný)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $b$ be the sum of numbers on side faces of the pyramid, let $a$ be the number on the base. After a step involving any base vertex, $b$ increases or decreases by $2$ and $a$ increases or decreases by $1$, that means the value $V = b - 2a$ stays the same. If we choose for a step the apex, only $b$ increases or decreases by $n$, thus $V$ increases or decreases by $n$ as well. Therefore $V$ is in the process always divisible by $n$. But in the position with number $1$ written on each side, the corresponding $V$ is $n-2$, which is not divisible by $n$ (as $n > 2$).",
    "topic": "Discrete Mathematics",
    "subtopic": "Combinatorics"
  },
  {
    "id": 29414,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $x$, $y$ and $z$ be any positive real numbers. Prove the inequality\n$$\n(x + y + z) \\left( \\frac{1}{x} + \\frac{1}{y} + \\frac{1}{z} \\right) \\le m^2, \\quad \\text{where } m = \\min \\left( \\frac{x}{y} + \\frac{y}{z} + \\frac{z}{x}, \\frac{y}{x} + \\frac{z}{y} + \\frac{x}{z} \\right).\n$$\nFind when the equality holds.",
    "options": [],
    "answer": "Equality holds if and only if x = y = z.",
    "solution": "Since the inequality involves the minimum of two positive numbers and since the function $y = x^2$ is increasing on the set $\\mathbb{R}^+$, our task is to verify\n$$\n(x+y+z)\\left(\\frac{1}{x}+\\frac{1}{y}+\\frac{1}{z}\\right) \\le \\left(\\frac{x}{y}+\\frac{y}{z}+\\frac{z}{x}\\right)^2 \\quad \\text{and} \\quad (x+y+z)\\left(\\frac{1}{x}+\\frac{1}{y}+\\frac{1}{z}\\right) \\le \\left(\\frac{y}{x}+\\frac{x}{z}+\\frac{z}{y}\\right)^2, \\quad (1)\n$$\n\nas well as to find when *at least one* equality in (1) holds. Replacing a triple ($x$, $y$, $z$)\nby the triple ($y$, $x$, $z$), we get the second inequality in (1) from the first one. Thus we\ncan restrict to the proof of the first inequality. Distributing both sides leads to\n$$\n3 + \\frac{x}{y} + \\frac{x}{z} + \\frac{y}{x} + \\frac{y}{z} + \\frac{z}{x} + \\frac{z}{y} \\le \\frac{x^2}{y^2} + \\frac{y^2}{z^2} + \\frac{z^2}{x^2} + 2\\left(\\frac{x}{z} + \\frac{y}{x} + \\frac{z}{y}\\right).\n$$\nLet us introduce the new (positive) variables $a = x/y$, $b = y/z$, $c = z/x$ and rewrite\nthe last inequality as\n$$\n\\left(a^2 - 1 - a + \\frac{1}{a}\\right) + \\left(b^2 - 1 - b + \\frac{1}{b}\\right) + \\left(c^2 - 1 - c + \\frac{1}{c}\\right) \\ge 0. \\quad (2)\n$$\nFor any positive $t$, we notice that\n$$\nt^2 - 1 - t + \\frac{1}{t} = (t^2 - 1) - \\frac{t^2 - 1}{t} = \\frac{(t^2 - 1)(t - 1)}{t} = \\frac{(t - 1)^2(t + 1)}{t}.\n$$\nThis implies that (2) holds as well and that (2) becomes an equality if and only if $a = b = c = 1$, i.e. $x = y = z$ for the original variables. Note that the last condition does not change under transformation $(x, y, z) \\to (y, x, z)$. Thus the original inequality is proven and becomes an equality if and only $x = y = z$.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 29415,
    "subject": "Mathematics (Multi-modal)",
    "question": "Two circles $k_1(S_1, r_1)$ and $k_2(S_2, r_2)$ are externally tangent and both lie in a square $ABCD$ with side length $a$ so that $k_1$ touches the sides $AD$ and $CD$, while $k_2$ touches the sides $BC$ and $CD$. Prove that the area of at least one of the triangles $AS_1S_2$, $BS_1S_2$ is no more than $\\frac{3}{16}a^2$. (Tomáš Jurík)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "The line segments $AS_2$ and $BS_1$ lie on the diagonals of the given square, so they are perpendicular to each other and intersect at the center $P$ of the square. We have\n$$\n\\begin{aligned}\n|DS_1| &= r_1 \\cdot \\sqrt{2}, & |BS_1| &= (a-r_1)\\sqrt{2}, & |PS_1| &= \\left(\\frac{a}{2} - r_1\\right)\\sqrt{2}, \\\\\n|CS_2| &= r_2 \\cdot \\sqrt{2}, & |AS_2| &= (a-r_2)\\sqrt{2}, & |PS_2| &= \\left(\\frac{a}{2} - r_2\\right)\\sqrt{2}.\n\\end{aligned}\n$$\nTherefore, the area of the triangle $AS_1S_2$ is\n$$\nS_{AS_1S_2} = \\frac{1}{2}|AS_2| \\cdot |PS_1| = (a-r_2)\\left(\\frac{a}{2}-r_1\\right),\n$$\nwhile the area of the triangle $BS_1S_2$ is\n$$\nS_{BS_1S_2} = \\frac{1}{2}|BS_1| \\cdot |PS_2| = (a-r_1)\\left(\\frac{a}{2}-r_2\\right).\n$$\nThe sum of these areas is\n$$\nS = (a - r_2) \\left(\\frac{a}{2} - r_1\\right) + (a - r_1) \\left(\\frac{a}{2} - r_2\\right) = a^2 - \\frac{3}{2}a(r_1 + r_2) + 2r_1r_2.\n$$\nLet $K$ denote the point at which the circle $k_1$ touches the side $AD$, $H$ and $L$ denote the points at which $k_2$ touches the sides $CD$ and $BC$, respectively, and $M$ be the intersection point of the lines $KS_1$ and $HS_2$ (Fig. 1).\n![](attached_image_1.png)\nFig. 1\nBy the Pythagoras' theorem for the triangle $S_1MS_2$, we have\n$$\n(a - r_1 - r_2)^2 + (r_1 - r_2)^2 = (r_1 + r_2)^2.\n$$\nHence we obtain\n$$\n\\begin{aligned}\n(a - r_1 - r_2)^2 &= 4r_1r_2, \\\\\na - r_1 - r_2 &= 2\\sqrt{r_1r_2}, \\\\\na &= r_1 + r_2 + 2\\sqrt{r_1r_2} = (\\sqrt{r_1} + \\sqrt{r_2})^2 \\ge 4\\sqrt{r_1r_2},\n\\end{aligned}\n$$\ni. e.\n$$\nr_1r_2 \\le \\frac{a^2}{16}.\n$$\nThe length of the segment $DC$ cannot be greater than the length of the polygonal chain $KS_1S_2L$, so\n$$\n2r_1 + 2r_2 \\ge a.\n$$\n(This follows from the equality $a = r_1 + r_2 + 2\\sqrt{r_1r_2}$ as well since $2\\sqrt{r_1r_2} \\le r_1 + r_2$, by the AM-GM inequality.) Therefore,\n$$\nS = a^2 - \\frac{3}{2}a(r_1 + r_2) + 2r_1r_2 \\le a^2 - \\frac{3}{4}a^2 + \\frac{1}{8}a^2 = \\frac{3}{8}a^2.\n$$\nThis means that at least one of the areas $S_{AS_1S_2}$, $S_{BS_1S_2}$ is at most $\\frac{3}{16}a^2$.\nWe can set $a = 1$. The difference of the areas of the triangles $AS_1S_2$ and $BS_1S_2$ is (according to the expression from the original solution)\n$$\nS_{AS_1S_2} - S_{BS_1S_2} = (1 - r_2)\\left(\\frac{1}{2} - r_1\\right) - (1 - r_1)\\left(\\frac{1}{2} - r_2\\right) = \\frac{1}{2}(r_2 - r_1).\n$$\nWithout loss of generality, we can assume that $r_1 \\ge r_2$. Then $S_{AS_1S_2} \\le S_{BS_1S_2}$. Now, let us calculate the area of the triangle $AS_1S_2$. By the Pythagoras' theorem, we have $(1 - r_1 - r_2)^2 + (r_1 - r_2)^2 = (r_1 + r_2)^2$, hence $\\sqrt{r_1} + \\sqrt{r_2} = 1$, so $r_2 = (1 - \\sqrt{r_1})^2$. Let us denote $x = \\sqrt{r_1}$. It follows from the inequalities $r_1 + r_2 \\ge \\frac{1}{2}$ and $r_1 \\ge r_2$ that $r_1 \\ge \\frac{1}{4}$, and, on the other hand, we have $r_1 \\le \\frac{1}{2}$ since the circle $k_1$ lies in the square $ABCD$. Hence it follows that $\\frac{1}{2} \\le x \\le \\sqrt{\\frac{1}{2}}$. The area of the triangle $AS_1S_2$ is\n$$\n\\begin{aligned}\nS_{AS_1S_2} &= (1 - r_2)\\left(\\frac{1}{2} - r_1\\right) = \\frac{1}{2} - r_1 - \\frac{r_2}{2} + r_1r_2 = \\\\\n&= \\frac{1}{2} - x^2 - \\frac{1}{2}(1-x)^2 + x^2(1-x)^2 = x^4 - 2x^3 - \\frac{1}{2}x^2 + x; \\\\\nS_{AS_1S_2} - \\frac{3}{16} &= x^4 - 2x^3 - \\frac{1}{2}x^2 + x - \\frac{3}{16} = (x - \\frac{1}{2})(x^3 - \\frac{3}{2}x^2 - \\frac{5}{4}x + \\frac{3}{8}) = \\\\\n&= (x - \\frac{1}{2})\\left[x^2\\left(x - \\frac{3}{2}\\right) - \\frac{5}{4}\\left(x - \\frac{3}{10}\\right)\\right] \\le 0\n\\end{aligned}\n$$\nas we have $\\frac{3}{10} < \\frac{1}{2} \\le x \\le \\frac{1}{2}\\sqrt{2} < \\frac{3}{2}$. Therefore, $S_{AS_1S_2} \\le \\frac{3}{16}$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 29416,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $A$, $B$ be sets of positive integers such that a sum of arbitrary two different numbers from $A$ is in $B$ and a ratio of arbitrary two different numbers from $B$ (greater one to smaller one) is in $A$. Find the maximum number of elements in $A \\cup B$.\n\n(Martin Panák)",
    "options": [],
    "answer": "5",
    "solution": "Initially we will prove that the set $A$ consists of at most two numbers. Suppose that three numbers $a < b < c$ belong to the set $A$. Then the numbers $a + b < a + c < b + c$ are in $B$ and therefore the number\n$$\n\\frac{b + c}{a + c} = 1 + \\frac{b - a}{a + c}\n$$\nhas to be in $A$. This is a contradiction because $0 < b - a < a + c$ and the number is not integer.\n\nIf the set $B$ contains four numbers $k < l < m < n$, then the set $A$ will contain three different numbers $n / k$, $n / l$, $n / m$. So the set $B$ has at most three elements and $A \\cup B$ has at most five elements.\n\nWe achieve the number $5$ of elements if $A = \\{a, b\\}$, $B = \\{k, l, m\\}$, where $a < b$ and $l / k = m / l = a$, $m / k = b$. Then $b = a^2$ ($a \\ge 2$) and $a + a^2$ is one of the elements of $B$; the next two elements are either $a^2 + a^3$ and $a^3 + a^4$ or $1 + a$ and $a^2 + a^3$. E.g. sets $A = \\{2, 4\\}$, $B = \\{3, 6, 12\\}$ have five elements together.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 29417,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $A$, $B$ be sets of positive integers such that a sum of arbitrary two different numbers from $A$ is in $B$ and a ratio of arbitrary two different numbers from $B$ (greater one to smaller one) is in $A$. Find the maximum number of elements in $A \\cup B$.\n(Martin Panák)",
    "options": [],
    "answer": "5",
    "solution": "Initially we will prove that the set $A$ consists of at most two numbers. Suppose that three numbers $a < b < c$ belong to the set $A$. Then the numbers $a+b < a+c < b+c$ are in $B$ and therefore the number\n$$\n\\frac{b+c}{a+c} = 1 + \\frac{b-a}{a+c}\n$$\nhas to be in $A$. This is a contradiction because $0 < b-a < a+c$ and the number is not integer.\n\nIf the set $B$ contains four numbers $k < l < m < n$, then the set $A$ will contain three different numbers $n/k$, $n/l$, $n/m$. So the set $B$ has at most three elements and $A \\cup B$ has at most five elements.\n\nWe achieve the number $5$ of elements if $A = \\{a, b\\}$, $B = \\{k, l, m\\}$, where $a < b$ and $l/k = m/l = a$, $m/k = b$. Then $b = a^2$ ($a \\ge 2$) and $a + a^2$ is one of the elements of $B$; the next two elements are either $a^2 + a^3$ and $a^3 + a^4$ or $1 + a$ and $a^2 + a^3$. E.g. sets $A = \\{2, 4\\}$, $B = \\{3, 6, 12\\}$ have five elements together.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 29418,
    "subject": "Mathematics (Multi-modal)",
    "question": "Given a parallelogram $ABCD$ with center $S$, denote by $O$ the incenter of triangle $ABD$ and by $T$ the point of contact of the incircle of triangle $ABD$ with the diagonal $BD$. Prove that lines $OS$ and $CT$ are parallel. (Jaromír Šimša)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Denote the lengths of $AB$, $AD$, and $BD$ by $a$, $b$, and $c$, respectively. If $a = b$ then both $OS$ and $CT$ coincide with $AC$ and the conclusion is trivial. Suppose $a > b$ (the case $b > a$ being completely analogous).\nLet $T'$ be the reflection of $T$ in $S$ (Fig. 1). As $CT \\parallel AT'$, it suffices to prove $OS \\parallel AT'$. Denoting by $E$ the intersection of $AO$ and the diagonal $BD$ we may as well prove\n$$\n\\frac{AO}{OE} = \\frac{T'S}{SE} \\qquad (1)\n$$\n(note that since $a > b$, points $T'$, $S$, $E$, and $T$ lie on the diagonal $BD$ in this order). We express both ratios in terms of $a, b, c$.\n\nFirst, it is well-known that\n$$\nDT = \\frac{b+c-a}{2}, \\quad \\text{and hence} \\quad T'S = TS = \\frac{c}{2} - \\frac{b+c-a}{2} = \\frac{a-b}{2}.\n$$\nNext, the Angle Bisector Theorem in triangles $ABD$ and $AED$ implies\n$$\nBE : ED = AB : AD \\quad \\text{and} \\quad AO : OE = AD : DE\n$$\nwhich in turn gives\n$$\nBE = \\frac{ac}{a+b} \\quad \\text{and} \\quad DE = \\frac{bc}{a+b}, \\\\\nSE = BE - BS = \\frac{ac}{a+b} - \\frac{c}{2} = \\frac{c(a-b)}{2(a+b)}, \\\\\n\\frac{AO}{OE} = \\frac{AD}{DE} = \\frac{b}{\\frac{bc}{a+b}} = \\frac{a+b}{c}.\n$$\nFinally for the right-hand side we calculate\n$$\n\\frac{T'S}{SE} = \\frac{\\frac{a-b}{2}}{\\frac{c(a-b)}{2(a+b)}} = \\frac{a+b}{c}\n$$\nwhich finishes the proof.\n\n![](attached_image_1.png)\nFig. 1",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 29419,
    "subject": "Mathematics (Multi-modal)",
    "question": "We are given a sheet of paper in the form of a rectangle $x \\times y$, where $x$ and $y$ are integer numbers larger than $1$. Let us draw a lattice of $x \\times y$ unit squares on the sheet. Rolling up the rectangle and gluing it along its opposite sides we shape a lateral surface of a circular cylinder. Join each two distinct vertices of the marked unit squares on the surface by a segment. How many of all these segments are passing through an interior point of the cylinder? In the case $x > y$ decide when this number of \"internal\" segments is larger — for the cylinder with bases of perimeter $x$, or $y$? (Vojtech Bálint)",
    "options": [],
    "answer": "For the cylinder formed by gluing along the sides of length y (bases have perimeter x): P = x(x−1)(y^2 + 2y − 1)/2. For the cylinder formed by gluing along the sides of length x (bases have perimeter y): Q = y(y−1)(x^2 + 2x − 1)/2. If x > y, then P > Q; the cylinder whose bases have perimeter x has more internal segments.",
    "solution": "We will compute the requested number $P$ of all internal segments for the cylinder formed by gluing the rectangle $x \\times y$ along the opposite sides of length $y$.\n\nThis cylinder has two bases of perimeter $x$ and its lateral sides are of length $y$. We will use an obvious formula $P = P_0 - P_1 - P_2$, where $P_0$ denotes the total number of segments, while $P_1$ and $P_2$ denote the numbers of segments which lie on the lateral surface or on one of the two bases, respectively.\nNote that the vertices of the unit squares are situated on the surface of the cylinder in such a way that exactly $y+1$ of them lie on the same of the $x$ lateral sides and, in the same time, exactly $x$ vertices lie on the same boundary circle of the two bases. These facts lead immediately to the following formulae\n$$\n\\begin{aligned}\nP_0 &= \\binom{x(y+1)}{2} = \\frac{x(y+1)(xy+x-1)}{2}, \\\\\nP_1 &= x \\cdot \\binom{y+1}{2} = \\frac{x(y+1)y}{2}, \\\\\nP_2 &= 2 \\cdot \\binom{x}{2} = x(x-1).\n\\end{aligned}\n$$\nConsequently,\n$$\n\\begin{aligned}\nP &= P_0 - P_1 - P_2 = \\frac{x(y+1)(xy+x-1)}{2} - \\frac{x(y+1)y}{2} - x(x-1) \\\\\n&= \\frac{x(x-1)(y^2+2y-1)}{2}.\n\\end{aligned}\n$$\nIn view of symmetry, the number $Q$ of internal segments for the other cylinder (with base's perimeter $y$ and lateral sides of length $x$) is given by\n$$\nQ = \\frac{y(y-1)(x^2 + 2x - 1)}{2}.\n$$\nTo decide which of the inequalities $P > Q$ or $Q > P$ holds in the case when $x > y$, we factorize the difference $P - Q$ (since $P = Q$ if $x = y$, the polynomial $P - Q$ must be divisible by $x - y$):\n$$\n\\begin{aligned}\n2(P - Q) &= (x^2 - x)(y^2 + 2y - 1) - (y^2 - y)(x^2 + 2x - 1) \\\\\n&= (x^2y^2 - xy^2 + 2x^2y - 2xy - x^2 + x) \\\\\n&\\quad - (x^2y^2 - x^2y + 2xy^2 - 2xy - y^2 + y) \\\\\n&= 3xy(x - y) - (x - y)(x + y) + (x - y) \\\\\n&= (x - y)(3xy - x - y + 1).\n\\end{aligned}\n$$\nThus $x > y$ implies that $P > Q$ if we show that the same condition $x > y$ implies that $3xy - x - y + 1 > 0$. The last is almost evident: it follows from $y \\ge 2$ that $3xy \\ge 6x$ and hence\n$$\n3xy - x - y + 1 \\ge 5x - y + 1 > 4x + 1 > 0.\n$$\nAnswer. In the case when $x > y$, the number of internal segments is larger for the cylinder with bases which perimeter has length $x$.",
    "topic": "Geometry",
    "subtopic": "Solid Geometry"
  },
  {
    "id": 29420,
    "subject": "Mathematics (Multi-modal)",
    "question": "Given a parallelogram $ABCD$ with center $S$, denote by $O$ the incenter of triangle $ABD$ and by $T$ the point of contact of the incircle of triangle $ABD$ with the diagonal $BD$. Prove that lines $OS$ and $CT$ are parallel. (Jaromír Šimša)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Denote the lengths of $AB$, $AD$, and $BD$ by $a$, $b$, and $c$, respectively. If $a = b$ then both $OS$ and $CT$ coincide with $AC$ and the conclusion is trivial. Suppose $a > b$ (the case $b > a$ being completely analogous).\nLet $T'$ be the reflection of $T$ in $S$ (Fig. 1). As $CT \\parallel AT'$, it suffices to prove $OS \\parallel AT'$. Denoting by $E$ the intersection of $AO$ and the diagonal $BD$ we may as well prove\n$$\n\\frac{AO}{OE} = \\frac{T'S}{SE} \\qquad (1)\n$$\n(note that since $a > b$, points $T'$, $S$, $E$, and $T$ lie on the diagonal $BD$ in this order).\nWe express both ratios in terms of $a, b, c$.\n![](attached_image_1.png)\nFig. 1\nFirst, it is well-known that\n$$\nDT = \\frac{b+c-a}{2}, \\quad \\text{and hence} \\quad T'S = TS = \\frac{c}{2} - \\frac{b+c-a}{2} = \\frac{a-b}{2}.\n$$\nNext, the Angle Bisector Theorem in triangles $ABD$ and $AED$ implies\n$$\nBE : ED = AB : AD \\quad \\text{and} \\quad AO : OE = AD : DE\n$$\nwhich in turn gives\n$$\nBE = \\frac{ac}{a+b} \\quad \\text{and} \\quad DE = \\frac{bc}{a+b}, \\\\\nSE = BE - BS = \\frac{ac}{a+b} - \\frac{c}{2} = \\frac{c(a-b)}{2(a+b)}, \\\\\n\\frac{AO}{OE} = \\frac{AD}{DE} = \\frac{b}{\\frac{bc}{a+b}} = \\frac{a+b}{c}.\n$$\nFinally for the right-hand side we calculate\n$$\n\\frac{T'S}{SE} = \\frac{\\frac{a-b}{2}}{\\frac{c(a-b)}{2(a+b)}} = \\frac{a+b}{c}\n$$\nwhich finishes the proof.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 29421,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $p(n)$ denote the number of all $n$-digit positive integers containing only the digits $1$, $2$, $3$, $4$, $5$ and such that every two adjacent digits differ by at least $2$. Prove that for every positive integer $n$,\n$$\n5 \\cdot 2.4^{n-1} \\le p(n) \\le 5 \\cdot 2.5^{n-1}.\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Cutting off the last digit of a satisfactory $(n + 1)$-digit integer yields a satisfactory $n$-digit integer. Notice how a satisfactory $(n + 1)$-digit integer can be constructed from a satisfactory $n$-digit integer. If the last digit of the integer is $1$, we can append any of the digits $3$, $4$, $5$. If the last digit is $2$, we can append $4$ or $5$; if it is $3$, $1$ or $5$ can be appended; if it is $4$, $1$ or $2$ can be appended; and, finally, in the case of $5$, we can append any of the digits $1$, $2$, $3$. Thus we can see that only the last digit matters. So now, let $a_n$ denote the number of satisfactory $n$-digit integers ending in $1$ or $5$; similarly $b_n$ for $2$ or $4$, and $c_n$ for integers ending in $3$. Then $p(n) = a_n + b_n + c_n$. Apparently, $a_1 = b_1 = 2$, $c_1 = 1$, $p(1) = 5 = 5 \\cdot 2.4^0 = 5 \\cdot 2.5^0$, $a_2 = 6$, $b_2 = 4$, $c_2 = 2$, $p(2) = 12 = 5 \\cdot 2.4^1 < 5 \\cdot 2.5^1$.\nThe above reasoning implies the recurrent formulae\n$$\na_{n+1} = a_n + b_n + 2c_n, \\quad b_{n+1} = a_n + b_n, \\quad c_{n+1} = a_n. \\quad (1)\n$$\nHence it follows that $a_3 = 14$, $b_3 = 10$, $c_3 = 6$, $p(3) = 30 \\in (5 \\cdot 2.4^2; 5 \\cdot 2.5^2)$.\nUsing mathematical induction, we prove that for every $n \\ge 3$, it holds that\n$$\na_n \\ge 2.4^n, \\quad b_n \\ge \\frac{2}{3} \\cdot 2.4^n, \\quad c_n \\ge 2.4^{n-1}.\n$$\nIt indeed does for $n = 3$. If $a_n \\ge 2.4^n$, $b_n \\ge \\frac{2}{3} \\cdot 2.4^n$ and $c_n \\ge 2.4^{n-1}$, then also\n$$\n\\begin{align*} a_{n+1} &= a_n + b_n + 2c_n \\ge 2.4^n + \\frac{2}{3} \\cdot 2.4^n + 2 \\cdot 2.4^{n-1} = \\\\ &= 2.4^n \\cdot \\left(1 + \\frac{2}{3} + \\frac{5}{6}\\right) = 2.5 \\cdot 2.4^n > 2.4^{n+1}, \\\\ b_{n+1} &= a_n + b_n \\ge 2.4^n + \\frac{2}{3} \\cdot 2.4^n = \\frac{5}{3} \\cdot 2.4^n > \\frac{2}{3} \\cdot 2.4^{n+1}, \\\\ c_{n+1} &= a_n \\ge 2.4^n. \\end{align*}\n$$\nIt follows from the proved inequalities that\n$$\np(n) = a_n + b_n + c_n \\ge 2.4^n + \\frac{2}{3} \\cdot 2.4^n + 2 \\cdot 2.4^{n-1} = (2.4+1.6+1) \\cdot 2.4^{n-1} = 5 \\cdot 2.4^{n-1}.\n$$\n\nThe latter inequality can be proved analogously; we will verify that for $n \\ge 3$,\n$$\na_n \\le k \\cdot 2.5^n, \\quad b_n \\le k \\cdot \\frac{2}{3} \\cdot 2.5^n, \\quad c_n \\le k \\cdot 2.5^{n-1}, \\quad (2)\n$$\nwhere $k$ is a suitably chosen number. Then we will have\n$$\np(n) = a_n + b_n + c_n \\le k \\cdot 2.5^{n-1} \\cdot \\left(2.5 + \\frac{5}{3} + 1\\right) = k \\cdot 2.5^{n-1} \\cdot \\frac{31}{6} = 5k \\cdot \\frac{31}{30} \\cdot 2.5^{n-1}.\n$$\nTherefore, setting $k = \\frac{30}{31}$, we get $p(n) \\le 5 \\cdot 2.5^{n-1}$ for every $n \\ge 3$.\nIt remains to prove, by mathematical induction, the inequalities (2) where $k = \\frac{30}{31}$.\nThey hold for $n = 3$. If (2) holds, we also have\n$$\n\\begin{align*} a_{n+1} &= a_n + b_n + 2c_n \\le k \\cdot 2.5^n \\cdot \\left(1 + \\frac{2}{3} + \\frac{4}{5}\\right) = k \\cdot 2.5^n \\cdot \\frac{37}{15} < k \\cdot 2.5^{n+1}, \\\\ b_{n+1} &= a_n + b_n \\le k \\cdot 2.5^n \\cdot \\left(1 + \\frac{2}{3}\\right) = k \\cdot \\frac{2}{3} \\cdot 2.5^{n+1}, \\\\ c_{n+1} &= a_n \\le k \\cdot 2.5^n. \\end{align*}\n$$\nWe will show that each of the sequences $\\{a_n\\}$, $\\{b_n\\}$, $\\{c_n\\}$ defined in the above solution satisfies (as a consequence of the equalities (1)) the recurrent equation $x_{n+2} = 2x_{n+1} + 2x_n - 2x_{n-1}$, and so this equation is satisfied by the sequence $p(n) = a_n + b_n + c_n$ in question, which we will denote by (3).\nIndeed, the first and third equalities of (1) give $a_{n+1} = a_n + b_n + 2a_{n-1}$, whence\n$$\nb_n = a_{n+1} - a_n - 2a_{n-1}, \\quad \\text{so} \\quad b_{n+1} = a_{n+2} - a_{n+1} - 2a_n.\n$$\nConsidering the second equality in (1), we thus get\n$$\na_{n+2} - a_{n+1} - 2a_n = b_{n+1} = a_n + b_n = a_n + (a_{n+1} - a_n - 2a_{n-1}).\n$$\nConfronting the marginal expressions leads to the mentioned equality\n$$\na_{n+2} = 2a_{n+1} + 2a_n - 2a_{n-1}.\n$$\nTriple substitution of $a_n = b_{n+1} - b_n$ into the equality $b_n = a_{n+1} - a_n - 2a_{n-1}$ yields\n$$\nb_n = (b_{n+2} - b_{n+1}) - (b_{n+1} - b_n) - 2(b_n - b_{n-1}),\n$$\nwhich can be rearranged to\n$$\nb_{n+2} = 2b_{n+1} + 2b_n - 2b_{n-1}.\n$$\nFinally, the sequence $\\{c_n\\}$ is merely a shifted sequence $\\{a_n\\}$, so\n$$\nc_{n+2} = a_{n+1} = 2a_n + 2a_{n-1} - 2a_{n-2} = 2c_{n+1} + 2c_n - 2c_{n-1}.\n$$\nCombining all of the three recurrent formulae, we get\n$$\np(n+2) = 2p(n+1) + 2p(n) - 2p(n-1). \\quad (3)\n$$\n\nUsing mathematical induction, we will prove that for every $k \\ge 1$,\n$$\n2.4p(k) \\le p(k+1) \\le 2.5p(k). \\quad (4)\n$$\nThe inequalities (4) hold for both $k = 1$ and $k = 2$. If (4) holds for all $k \\in \\{1, 2, \\dots, n+1, n+2\\}$, then\n$$\n\\begin{aligned} p(n+3) &= 2(p(n+2) + p(n+1) - p(n)) \\ge 2\\left(p(n+2) + p(n+1) - \\frac{p(n+1)}{2.4}\\right) = 2\\left(p(n+2) + \\frac{7p(n+1)}{12}\\right) \\ge 2\\left(p(n+2) + \\frac{7}{12} \\cdot \\frac{p(n+2)}{2.5}\\right) = \\frac{74p(n+2)}{30} > 2.4p(n+2). \\end{aligned}\n$$\nSimilarly,\n$$\n\\begin{aligned} p(n+3) &\\le 2\\left(p(n+2) + p(n+1) - \\frac{p(n+1)}{2.5}\\right) \\le 2\\left(p(n+2) + \\frac{3}{5} \\cdot \\frac{p(n+2)}{2.4}\\right) = 2.5p(n+2). \\end{aligned}\n$$\nThe equalities $5 \\cdot 2.4^0 = p(1) = 5 \\cdot 2.5^0$ and the inequalities of (4) imply that the inequality $5 \\cdot 2.4^{n-1} \\le p(n) \\le 5 \\cdot 2.5^{n-1}$ holds for every positive integer $n$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 29422,
    "subject": "Mathematics (Multi-modal)",
    "question": "We are given a triangle $ABC$ with incentre $I$. Suppose that there exists an intersection point $M$ of the line $AB$ and the perpendicular to $CI$ through $I$. Prove that the circumcircle of the triangle $ABC$ intersects the segment $CM$ in an interior point $N$ and that $NI \\perp MC$.\n\n(Peter Novotny)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "First we show in two different ways that the line $MI$, a perpendicular to $CI$ through $I$, is tangent to the circle $ABI$. The first way is based on the known fact that $\\angle AIC$ and $\\angle BIC$ are obtuse angles of measures $90^\\circ + \\frac{1}{2}\\beta$ and $90^\\circ + \\frac{1}{2}\\alpha$, respectively (in common notation for interior angles of $\\triangle ABC$). This fact implies that the line $MI$ forms acute angles $\\frac{1}{2}\\beta$ and $\\frac{1}{2}\\alpha$ with the segments $AI$ and $BI$, respectively\\textsuperscript{1}, hence the angles congruent with angles $IBA$ and $IAB$ in circle $ABI$ (Fig. 1). Well known properties of inscribed and subtended angles lead to the conclusion that the line $MI$ is tangent to the circle $ABI$. The second reason for this conclusion is based on the known fact that the centre of $ABI$ is the midpoint of the arc $AB$ of circle $ABC$ which lies on the ray $CI$ bisecting $\\angle ACB$.\n\n![](attached_image_1.png)\n\nFig. 1\n\nFrom the proved tangency of $MI$ to $ABI$ it follows that the point $M$ lies on the line $AB$ outside of the segment $AB$. Moreover, the power $m$ of $M$ with respect to $ABI$ is positive and given by $m = |MI|^2 = |MA| \\cdot |MB|$. Hence $M$ lies in the exterior of the circle $ABC$ (as $AB$ is its chord) and the power of $M$ with respect to $ABC$ is the same $m = |MA| \\cdot |MB|$. Since $m = |MI|^2 < |MC|^2$ from the right-angled triangle $CMI$, it holds that $|MA| \\cdot |MB| < |MC|^2$. This means that the circle $ABC$ intersects the segment $MC$ in an interior point $N$, because $|MN| \\cdot |MC| = |MA| \\cdot |MB|$ implies that $|MN| < |MC|$ for the second point $N$ of intersection of the ray $MC$ with $ABC$. This proves the first conclusion of the problem.\n\nTo show that $\\angle CNI$ is a right angle, we use the proved equality $|MC| \\cdot |MN| = |MI|^2$ and apply a familiar theorem to the leg $MI$ of the right-angled triangle $CMI$: Its altitude from the vertex $I$ meets the hypotenuse $CM$ in such a point $X$ which is determined by equation $|MC| \\cdot |MX| = |MI|^2$. Thus we have $X = N$ in our case and the solution is complete.\n\n\\textsuperscript{1} As a consequence we can see that the assumed existence of the intersection point $M$ is equivalent to the inequality $\\frac{1}{2}\\alpha \\neq \\frac{1}{2}\\beta$ or $\\alpha \\neq \\beta$. Due to the symmetry we can assume that $\\alpha > \\beta$ as in our figure; the point $M$ then lies on the ray opposite to ray $AB$ and satisfies $|\\angle IMA| = \\frac{1}{2}\\alpha - \\frac{1}{2}\\beta$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 29423,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all real $a$, $b$, $c$, such that\n$$\na^2 + b^2 + c^2 = 26, \\quad a+b=5 \\quad \\text{and} \\quad b+c \\ge 7.\n$$",
    "options": [],
    "answer": "a=1, b=4, c=3",
    "solution": "We show that the only solution is $a=1$, $b=4$ and $c=3$.\nLet $s = b + c \\ge 7$. Substituting $a = 5-b$ and $c = s-b$ the first condition gives\n$$\n(5-b)^2 + b^2 + (s-b)^2 = 26,\n$$\nthus\n$$\n3b^2 - 2(s+5)b + s^2 - 1 = 0.\n$$\nThe equation has a real solution iff the discriminant $4(s+5)^2 - 12(s^2-1) \\ge 0$. This yields $s^2 - 5s - 14 \\le 0$, or $(s+2)(s-7) \\le 0$. Since $s \\ge 7$, there must be $s=7$. If we substitute to the previous equation we get\n$$\n3b^2 - 24b + 48 = 0;\n$$\nwith the only solution $b=4$. Then $a=1$ and $c=3$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 29424,
    "subject": "Mathematics (Multi-modal)",
    "question": "Touching circles $k_1(S_1, r_1)$ and $k_2(S_2, r_2)$ lie in a right-angled triangle $ABC$ with the hypotenuse $AB$ and legs $AC = 4$ and $BC = 3$ in such way, that the sides $AB$, $AC$ are tangent to $k_1$ and the sides $AB$, $BC$ are tangent to $k_2$. Find radii $r_1$ and $r_2$, if $4r_1 = 9r_2$. (Pavel Novotný)\n\n![](attached_image_1.png)",
    "options": [],
    "answer": "r1 = 45/47, r2 = 20/47",
    "solution": "The hypotenuse $AB$ has length $AB = 5$ with respect to Pythagoras' theorem. Then for angles in the triangle there is $\\cos \\alpha = \\frac{4}{5}$, $\\cos \\beta = \\frac{3}{5}$,\n$$\n\\cot \\frac{\\alpha}{2} = \\sqrt{\\frac{1 + \\cos \\alpha}{1 - \\cos \\alpha}} = 3,\n$$\n$$\n\\cot \\frac{\\beta}{2} = \\sqrt{\\frac{1 + \\cos \\beta}{1 - \\cos \\beta}} = 2.\n$$\n\nSince both circles $k_1, k_2$ whole lie in the triangle $ABC$, they are externally tangent—in the opposite case the leg tangent to the smaller circle intersects the greater circle. Let circles $k_1$ and $k_2$ touch the side $AB$ at points $D$ resp. $E$ and let point $F$ be orthogonal projection of the point $S_2$ to the element $S_1D$ (Fig. 1, under assumption is $r_1 > r_2$). Using Pythagoras' theorem for a triangle $FS_2S_1$ we obtain\n$$\n(r_1 + r_2)^2 = (r_1 - r_2)^2 + DE^2,\n$$\nwhich follows $DE = 2\\sqrt{r_1r_2}$.\nAn equality $AB = AD + DE + EB$ gives\n$$\nc = r_1 \\cot \\frac{\\alpha}{2} + 2\\sqrt{r_1r_2} + r_2 \\cot \\frac{\\beta}{2} = 3r_1 + 2\\sqrt{r_1r_2} + 2r_2,\n$$\nand since $r_1 = \\frac{9}{4}r_2$ we obtain\n$$\n\\frac{27}{4}r_2 + 3r_2 + 2r_2 = 5,\n$$\nwhich follows\n$$\nr_2 = \\frac{20}{47}, \\quad r_1 = \\frac{45}{47}.\n$$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 29425,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $l(n)$ denote the greatest odd divisor of any natural number $n$. Find the sum\n$$\nl(1) + l(2) + l(3) + \\cdots + l(2^{2013}). \\qquad (\\text{Michal Rol\\'inek})\n$$",
    "options": [],
    "answer": "(4^2013 + 2)/3",
    "solution": "For each natural $k$, the equalities $l(2k) = l(k)$ and $l(2k - 1) = 2k - 1$ are clearly valid. Thus we can add the values $l(n)$ over groups of numbers $n$ lying always between two consecutive powers of $2$. In this way we will prove by induction the formula\n$$\ns(n) = l(2^{n-1} + 1) + l(2^{n-1} + 2) + l(2^{n-1} + 3) + \\dots + l(2^n), \\quad (1)\n$$\nfor $n = 1, 2, 3, \\dots$. The case $n = 1$ is trivial. If $s(n) = 4^{n-1}$ for some $n$, then\n$$\n\\begin{align*}\ns(n+1) &= l(2^n + 1) + l(2^n + 2) + l(2^n + 3) + \\dots + l(2^{n+1}) \\\\\n&= [(2^n + 1) + (2^n + 3) + \\dots + (2^{n+1} - 1)] + s(n) \\\\\n&= \\frac{2^n}{2}(2^n + 1 + 2^{n+1} - 1) + 4^{n-1} = 2^{n-2} \\cdot 3 \\cdot 2^n + 4^{n-1} = 4^n.\n\\end{align*}\n$$\n\n(We have used the fact the number of all odd numbers from $2^n + 1$ to $2^{n+1} - 1$ [including both limits] equals $2^{n-1}$.) The proof of (1) by induction is complete.\nUsing the formula (1) we compute the requested sum as follows:\n$$\n\\begin{aligned}\nl(1)+l(2)+l(3)+\\cdots+l(2^{2013}) &= l(1)+s(2)+s(3)+\\cdots+s(2013) \\\\\n&= 1+1+4+4^2+4^3+\\cdots+4^{2012} = 1+\\frac{4^{2013}-1}{3} = \\frac{4^{2013}+2}{3}.\n\\end{aligned}\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 29426,
    "subject": "Mathematics (Multi-modal)",
    "question": "Prove that positive $a$, $b$, $c$ are lengths of sides of a triangle if and only if a system of equations\n$$\na(yz + x) = b(zx + y) = c(xy + z), \\quad x + y + z = 1\n$$\nwith unknowns $x$, $y$, $z$ has a solution in positive reals.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $a$, $b$, $c$ be positive numbers. We search a solution of the system of equations in the set of positive reals. Due to $x + y + z = 1$ the numbers $x$, $y$, $z$ are in the interval $(0, 1)$. Substituting $z = 1 - x - y$ we obtain\n$$\na(y - xy - y^2 + x) = c(xy + 1 - x - y), \\quad b(x - x^2 - xy + y) = c(xy + 1 - x - y),\n$$\nthese equations can be rewritten as\n$$\nay(1 - y) + ax(1 - y) = c(1 - x)(1 - y), \\quad bx(1 - x) + by(1 - x) = c(1 - x)(1 - y).\n$$\nSince $x < 1$, $y < 1$, we have\n$$\nay + ax = c - cx, \\quad bx + by = c - cy.\n$$\nFrom the previous two equations we obtain\n$$\nx + y = \\frac{2c}{a + b + c} \\quad \\text{and} \\quad x - y = \\frac{(b - a)(x + y)}{c} = \\frac{2(b - a)}{a + b + c};\n$$\nthen we get formulas for $x$, $y$ and finally by $z = 1 - x - y$ we find a formula for $z$ too:\n$$\nx = \\frac{b + c - a}{a + b + c}, \\quad y = \\frac{c + a - b}{a + b + c}, \\quad z = \\frac{a + b - c}{a + b + c} \\quad (1)\n$$\nThe system has a solution in positive reals if and only if $b + c > a$, $c + a > b$, $a + b > c$ holds, which is equivalent to the existence of a triangle with sides $a$, $b$, $c$.\n\nWe need not check the values (1) due to equivalence of all rearrangements.\n\n\nOther Solution:\n\nHere is an easier way to obtain (1). Since $x + y + z = 1$ we can rewrite the first part of the system as\n$$\na(1 - y)(1 - z) = b(1 - z)(1 - x) = c(1 - x)(1 - y). \\quad (2)\n$$\nDividing $(1 - x)(1 - y)(1 - z)$ (which is nonzero, even positive) we obtain the equivalent system\n$$\n\\frac{a}{1 - x} = \\frac{b}{1 - y} = \\frac{c}{1 - z}\n$$\nIf $s$ is the common (positive) value of the three previous fractions, we can easily get\n$$\nx = 1 - \\frac{a}{s}, \\quad y = 1 - \\frac{b}{s}, \\quad z = 1 - \\frac{c}{s}, \\quad (3)\n$$\nwhich, substituting into the equation $x + y + z = 1$, gives\n$$\ns = \\frac{a + b + c}{2}\n$$\nIt follows from (3) that this $s$ yields the formulae (1), and then the proof of the problem statement.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 29427,
    "subject": "Mathematics (Multi-modal)",
    "question": "There is written a number $N$ (in the decimal representation) on the board. In a step we erase the last digit $c$ and instead of the number $m$, which is now left on the board, we write number $|m - 3c|$ (for example, if $N = 1204$ was written on the board, then after the step there will be $120 - 3 \\cdot 4 = 108$). We continue until there is a one-digit number on the board. Find all positive integers $N$ such that after a finite number of steps number $0$ is left on the board.",
    "options": [],
    "answer": "All positive multiples of 31",
    "solution": "Let us find $N$, which lead to zero on the board after only one step. Obviously $|m - 3c| = 0$ iff $m = 3c$, which is $N = 10m + c = 31c$. All such $N$ are of the form $N = 31c$, $c \\in \\{1, 2, \\dots, 9\\}$.\n\nWe show that the solution of the problem are exactly all multiples of $31$. Since $c = N - 10m$, there is $m - 3c = 31m - 3N$, that is the divisibility by $31$ is preserved in the step. Now we show, that a multiple of $31$ actually decreases in the step. We have already shown that for $N \\le 31 \\cdot 9$. Let $N = 31k$, where $k \\ge 10$. Then $m \\ge 31$, $m - 3c > 0$, thus $|m - 3c| = 31m - 3N < 4N - 3N = N$ and we are done.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 29428,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all functions $f: \\mathbb{R} \\setminus \\{0\\} \\to \\mathbb{R}$ such that for all non-zero numbers $x, y$,\n$$\nx \\cdot f(xy) + f(-y) = x \\cdot f(x).\n$$",
    "options": [],
    "answer": "All solutions are f(x) = c(1 + 1/x) for real constants c and all nonzero real x.",
    "solution": "Substituting $x = 1$ gives\n$$\nf(y) + f(-y) = f(1).\n$$\nDenoting $f(1) = a$, we have $f(-y) = a - f(y)$. Substituting $y = -1$, we get\n$$\nx \\cdot f(-x) + f(1) = x \\cdot f(x),\n$$\ni. e.,\n$$\nx(a - f(x)) + a = x \\cdot f(x),\n$$\nhence\n$$\nf(x) = \\frac{a(x+1)}{2x} = \\frac{a}{2}\\left(1+\\frac{1}{x}\\right).\n$$\nFinally, we check that for any real number $c$, the function $f(x) = c(1+1/x)$ satisfies the conditions:\n$$\n\\begin{aligned} x \\cdot f(xy) + f(-y) &= x \\cdot c\\left(1 + \\frac{1}{xy}\\right) + c\\left(1 + \\frac{1}{-y}\\right) = c\\left(x + \\frac{1}{y} + 1 - \\frac{1}{y}\\right) = \\\\ &= c(x+1) = cx\\left(1 + \\frac{1}{x}\\right) = x \\cdot f(x). \\end{aligned}\n$$\nLet us set $f(1) = a$. Substituting $y = -1$ into the given equation yields\n$$\nxf(-x) + a = xf(x),\n$$\ni. e.,\n$$\nf(-x) = f(x) - \\frac{a}{x}.\n$$\n\nThe given equation can be rearranged into the form\n$$\nf(xy) + \\frac{1}{x} \\cdot f(-y) = f(x),\n$$\nwhence, using (1), we have\n$$\nf(xy) + \\frac{1}{x} \\left( f(y) - \\frac{a}{y} \\right) = f(x),\n$$\n$$\nf(xy) + \\frac{f(y)}{x} - \\frac{a}{xy} = f(x).\n$$\n\nInterchanging $x$ and $y$ gives\n$$\nf(yx) + \\frac{f(x)}{y} - \\frac{a}{yx} = f(y),\n$$\nso, combining the last two equations, we get\n$$\n\\frac{f(x)}{y} - \\frac{f(y)}{x} = f(y) - f(x),\n$$\nand substituting $y = 1$ now gives\n$$\n2f(x) = a\\left(1 + \\frac{1}{x}\\right).\n$$\nAgain, we can easily verify that every function $f(x) = c(1 + 1/x)$ is a solution of the given functional equation.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 29429,
    "subject": "Mathematics (Multi-modal)",
    "question": "Determine the number of all coverings of a chessboard $3 \\times 10$ by (nonoverlapping) pieces $2 \\times 1$ which can be placed both horizontally and vertically.",
    "options": [],
    "answer": "571",
    "solution": "Let us solve a more general problem of determining the number $a_n$ of all coverings of a chessboard $3 \\times 2n$ by pieces $2 \\times 1$, for a given natural $n$. We will attack the problem by a recursive method, starting with $n = 1$.\n\nThe value $a_1 = 3$ (for the chessboard $3 \\times 2$) is evident (see Fig. 2). To prove that $a_2 = 11$ by a direct drawing all possibilities is too laborious. Instead of this, we introduce new numbers $b_n$: Let each $b_n$ denote the number of all \"incomplete\" coverings of a chessboard $3 \\times (2n - 1)$ by $3n - 2$ pieces $2 \\times 1$, when a fixed corner field $1 \\times 1$ (specified in advance, say the lower right one) remains uncovered. Thanks to the axial symmetry, the numbers $b_n$ remain to be the same if the fixed uncovered corner field will be the upper right one. Moreover, it is clear that $b_1 = 1$.\n\n![](attached_image_1.png)\n\nFig. 2\n\nNow we are going to prove that for each $n > 1$, the following equalities hold:\n$$\nb_n = a_{n-1} + b_{n-1} \\quad \\text{and} \\quad a_n = a_{n-1} + 2b_n. \\quad (1)\n$$\n\n![](attached_image_2.png)\n\nFig. 3\n\nSimilarly, the second equality in (1) follows from a partition of all coverings of a chessboard $3 \\times 2n$ into three (disjoint) classes which are formed by coverings of types C, D and E respectively, see Fig. 4. It is evident that the numbers of elements in the three classes are $a_{n-1}$, $b_n$ and $b_n$, respectively.\n\n![](attached_image_3.png)\n\nFig. 4\n\nNow we are ready to compute the requested number $a_5$. Since $a_1 = 3$ and $b_1 = 1$, the proved equalities (1) successively yield\n$$\n\\begin{aligned}\nb_2 &= a_1 + b_1 = 4, \\quad a_2 = a_1 + 2b_2 = 11, \\quad b_3 = a_2 + b_2 = 15, \\quad a_3 = a_2 + 2b_3 = 41, \\\\\nb_4 &= a_3 + b_3 = 56, \\quad a_4 = a_3 + 2b_4 = 153, \\quad b_5 = a_4 + b_4 = 209, \\quad a_5 = a_4 + 2b_5 = 571.\n\\end{aligned}\n$$\n\n*Answer.* The number of coverings of the chessboard $3 \\times 10$ equals $571$.",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 29430,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $I$ be the incenter of a triangle $ABC$. The circle passing through the vertex $B$ and touches the line $AI$ at $I$ intersects the sides $AB$ and $BC$ at points $P$ and $Q$, respectively. Let $R$ be the intersection point of the line $QI$ and the side $AC$. Prove that\n$$\n|AR| \\cdot |BQ| = |PI|^2.\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $\\alpha, \\beta, \\gamma$ denote the measures of the interior angles at the vertices $A, B, C$, respectively, of the triangle $ABC$, and let $J$ be the intersection point of the line $AI$ and the side $BC$. (Fig. 2). The inscribed angle $PBI$ corresponds to the chord $PI$, while the inscribed angle $QBI$ corresponds to the chord $IQ$, and since the measure of both of these angles is $\\frac{1}{2}\\beta$, the chords $PI$ and $IQ$ share the same length as well.\n\n![](attached_image_1.png)\nFig. 2\n\nSince the inscribed angle $JIQ$ also corresponds to the chord $IQ$, its measure is also $\\frac{1}{2}\\beta$. It follows from the congruence of vertical angles that $|\\angle RIA| = \\frac{1}{2}\\beta$. This measure is also shared by the inscribed angle $PIA$ as it corresponds to the chord $PI$ of equal length as $IQ$. Further, $|\\angle RAI| = |\\angle PAI| = \\frac{1}{2}\\alpha$. The triangles $RIA$ and $PIA$ are thus congruent by ASA, hence $|RI| = |PI|$.\n\nTherefore, the measure of the angle $QIB$ is\n$$\n\\begin{aligned}\n|\\angle QIB| &= 180^\\circ - |\\angle AIB| - |\\angle JIQ| = 180^\\circ - (90^\\circ + \\frac{\\gamma}{2}) - \\frac{\\beta}{2} \\\\\n&= 90^\\circ - (\\frac{\\beta}{2} + \\frac{\\gamma}{2}) = \\frac{\\alpha}{2} = |\\angle RAI|.\n\\end{aligned}\n$$\n\nThe measure of the angle $QIB$ could also be determined as follows: Since the inscribed angles $AIP$ and $IPQ$ correspond to chords of equal length, we have $|\\angle AIP| = \\frac{1}{2}\\beta = |\\angle IPQ|$. The congruence of alternate angles implies that $AI \\parallel PQ$. Hence $|\\angle QPB| = |\\angle IAB| = \\frac{1}{2}\\alpha$, and so $|\\angle QIB| = \\frac{1}{2}\\alpha$ as well since they are both inscribed angles corresponding to the chord $QB$.\n\nSince $|\\angle QIB| = |\\angle RAI|$ and $|\\angle QBI| = |\\angle RIA|$, it follows that $AIR \\sim IBQ$. Therefore, $|AR|/|RI| = |IQ|/|QB|$, so\n$$\n|AR| \\cdot |QB| = |RI| \\cdot |IQ| = |PI|^2.\n$$",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 29431,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABCD$ be a parallelogram such that the projections $K$, $L$ of $D$ onto the sides $AB$, $BC$, respectively, are their interior points. Prove that $KL \\parallel AC$ if and only if\n$$\n\\angle BCA + \\angle ABD = \\angle BDA + \\angle ACD.\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Alternate angles $ABD$ and $CDB$ are equal (Fig. 2), hence $\\angle BCA + \\angle ABD + \\angle BDA + \\angle ACD = 180^\\circ$. The equality $\\angle BCA + \\angle ABD = \\angle BDA + \\angle ACD$ thus holds if and only if\n$$\n\\angle BCA + \\angle ABD = 90^\\circ.\n$$\n(1)\n![](attached_image_1.png)\nFig. 2\nPoints $K$ and $L$ lie on a circle with diameter $BD$. Hence the inscribed angles $BDK$ and $BLK$ are equal and (due to equal alternate angles $ABD$ and $CDB$)\n$$\n\\angle BLK + \\angle ABD = \\angle BDK + \\angle CDB = 90^\\circ.\n$$\nLines $KL$ and $AC$ are parallel if and only if $\\angle BLK = \\angle BCA$ which is by the last equality equivalent to (1). The equivalence is thus proven.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 29432,
    "subject": "Mathematics (Multi-modal)",
    "question": "In the real numbers, solve the following system of equations:\n$$\n\\begin{aligned}\n\\sin^2 x + \\cos^2 y &= \\tan^2 z, \\\\\n\\sin^2 y + \\cos^2 z &= \\tan^2 x, \\\\\n\\sin^2 z + \\cos^2 x &= \\tan^2 y.\n\\end{aligned}\n$$",
    "options": [],
    "answer": "(x, y, z) = (\\pi/4 + k\\pi/2, \\pi/4 + l\\pi/2, \\pi/4 + m\\pi/2) for any integers k, l, m",
    "solution": "Substituting $\\cos^2 x = a$, $\\cos^2 y = b$, $\\cos^2 z = c$ leads to the system\n$$\n\\begin{aligned}\n1 - a + b &= \\frac{1}{c} - 1, \\\\\n1 - b + c &= \\frac{1}{a} - 1, \\\\\n1 - c + a &= \\frac{1}{b} - 1,\n\\end{aligned}\n\\qquad (1)\n$$\nwhere $a, b, c \\in (0, 1)$.\nAdding these equations together yields\n$$\n\\frac{1}{a} + \\frac{1}{b} + \\frac{1}{c} = 6,\n$$\nso the harmonic mean of the numbers $a$, $b$, $c$ is $\\frac{1}{2}$.\n\nMultiplying the equations by the numbers $c$, $a$, and $b$, respectively, we get\n$$\n\\begin{aligned}\nc - ac + bc &= 1 - c, \\\\\na - ab + ac &= 1 - a, \\\\\nb - bc + ab &= 1 - b,\n\\end{aligned}\n$$\nwhich sums to $2(a + b + c) = 3$. Therefore, the arithmetic mean of the numbers $a, b, c$ is $\\frac{1}{2}$, too. Since the arithmetic and harmonic means are equal, it must be that $a = b = c = \\frac{1}{2}$. We can easily verify that this triple satisfies the system (1). The solutions of the original system are thus exactly the triples $(\\frac{1}{4}\\pi + \\frac{1}{2}k\\pi, \\frac{1}{4}\\pi + \\frac{1}{2}l\\pi, \\frac{1}{4}\\pi + \\frac{1}{2}m\\pi)$, where $k, l, m$ are integers.\nWe use the substitution from the above solution. The system (1) is cyclic; if a triple $(p, q, r)$ satisfies it, then so do the triples $(q, r, p)$ and $(r, p, q)$. Therefore, it suffices to find the solutions for which $a \\ge b, a \\ge c$, and all the other solutions can then be obtained by cyclic exchange.\nLet $a \\ge b, a \\ge c$. It follows from the first equation that $1/c = 2 - a + b \\le 2$, so $c \\ge \\frac{1}{2}$. Similarly, it follows from the third equation that $1/b = 2 - c + a \\ge 2$, so $b \\le \\frac{1}{2}$, and thus $b \\le c$. From the second equation, we have $1/a = 2 - b + c \\ge 2$, so $a \\le \\frac{1}{2}$. Altogether,\n$$\n\\frac{1}{2} \\ge a \\ge c \\ge \\frac{1}{2},\n$$\nso $a = c = \\frac{1}{2}$. Now, any of the equations of the system (1) yields $b = \\frac{1}{2}$. Just as in the previous case, we verify that the found triple satisfies the system (1); so the solutions of the system are the triples $(\\frac{1}{4}\\pi + \\frac{1}{2}k\\pi, \\frac{1}{4}\\pi + \\frac{1}{2}l\\pi, \\frac{1}{4}\\pi + \\frac{1}{2}m\\pi)$, where $k, l, m$ are integers.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 29433,
    "subject": "Mathematics (Multi-modal)",
    "question": "We are given a triangle $ABC$. Find the locus of points $X$ in the plane $ABC$ whose reflections through the lines $AB$, $BC$, $CA$ are vertices of an equilateral triangle.",
    "options": [],
    "answer": "The locus consists of the common intersection points of the three Apollonius circles\nk_a: |XB|/|XC| = |AB|/|AC|,\nk_b: |XA|/|XC| = |AB|/|BC|,\nk_c: |XA|/|XB| = |AC|/|BC|,\nwhich equivalently are the points X satisfying 2|XA|sin α = 2|XB|sin β = 2|XC|sin γ > 0. The number of solutions depends on the triangle: one solution (the incenter) if the triangle is equilateral; two solutions if the triangle is isosceles but not equilateral; and two solutions if the triangle is scalene.",
    "solution": "For any point $X$ of the plane $ABC$, let $X_a, X_b$ and $X_c$ denote the reflections of $X$ through the lines $BC$, $CA$ and $AB$, respectively (Fig. 5). First we prove that the distances between any two of the points $X_a, X_b$ and $X_c$ are given in general by formulæ\n$$\n|X_a X_b| = 2|XC| \\sin \\gamma, \\quad |X_a X_c| = 2|XB| \\sin \\beta, \\quad |X_b X_c| = 2|XA| \\sin \\alpha, \\quad (1)\n$$\nin which $\\alpha, \\beta, \\gamma$ denote the interior angles of the triangle $ABC$ as usual.\n![](attached_image_1.png)\nFig. 5\nIt suffices to prove the first equality (1) which is obvious if $X = C$, because then $X_a = X_b (= X)$. If $X \\neq C$, then the segment $XC$ is a diameter of a circle (see Fig. 5) which passes through the marked orthogonal projections $P_a$ and $P_b$ of $X$ onto $BC$ and $CA$, respectively (Thales' theorem). Since the chord $P_a P_b$ subtends inscribed angles $\\gamma$ and $180^\\circ - \\gamma$, Law of Sines implies that $|P_a P_b| = |XC| \\sin \\gamma$. Using the homothety with centre $X$ and ratio $2$, we conclude that $|X_a X_b| = 2|P_a P_b|$, and hence the equalities (1) are established for any point $X$.\n\nThe proved formulæ (1) imply that our task is to find exactly such points $X$ in the plane $ABC$ that satisfy\n$$\n2|XA| \\sin \\alpha = 2|XB| \\sin \\beta = 2|XC| \\sin \\gamma > 0\n$$\n(recall that the triangle $X_a X_b X_c$ has to be equilateral). Otherwise speaking, we look for all points $X$ whose distances to $A$, $B$ and $C$ are positive and proportional as follows:\n$$\n|XA| : |XB| : |XC| = \\frac{1}{\\sin \\alpha} : \\frac{1}{\\sin \\beta} : \\frac{1}{\\sin \\gamma} = \\frac{1}{|BC|} : \\frac{1}{|AC|} : \\frac{1}{|AB|}\n$$\n(we have turned from angles to sides of $\\triangle ABC$ using Law of Sines again). Such points $X$ are determined as common points of the following three circles of Apollonius (i.e. sets of points in the plane which have a specified ratio of distances to two fixed points):\n$$\nk_a : \\frac{|XB|}{|XC|} = \\frac{|AB|}{|AC|}, \\quad k_b : \\frac{|XA|}{|XC|} = \\frac{|AB|}{|BC|}, \\quad k_c : \\frac{|XA|}{|XB|} = \\frac{|AC|}{|BC|} \\quad (2)\n$$\nIt is clear that any point shared by two of the circles lies on the third circle as well. It follows from (2) that $A \\in k_a$, $B \\in k_b$ and $C \\in k_c$, which simplifies the construction of the three circles in practice: If the bisectors of interior angles in $\\triangle ABC$ cut its interior in segments $AK$, $BL$ and $CM$ (Fig. 6), then $K \\in k_a$, $L \\in k_b$ and $M \\in k_c$ (an immediate consequence of the well known proportions such as $|KB| : |KC| = |AB| : |AC|$). Hence the centre of $k_a$ can be constructed as the intersection point of the line $BC$ and the perpendicular bisector of the segment $AK$ (excluding the case $|AB| = |AC|$, when $k_a$ becomes simply the perpendicular bisector of $BC$). Similarly, using the perpendicular bisectors of $BL$ and $CM$ we get centres of $k_b$ and $k_c$, respectively.\n\n![](attached_image_2.png)\n\nFig. 6\n\nDespite of the fact that the requested locus of points $X$ is determined (by an Euclidean construction), we have to discuss how the number of solutions depends on\n\na) If the triangle $ABC$ is **equilateral**, the \"circles\" $k_a, k_b, k_c$ are in fact perpendicular bisectors of the sides of $\\triangle ABC$. Consequently, the problem has a unique solution — a point $X$ which coincides with the incentre of $\\triangle ABC$.\n\nb) If the triangle $ABC$ is **isosceles** (but not equilateral), say if $|AB| \\neq |AC| = |BC|$, then the circle $k_c$ is a perpendicular bisector of the base $AB$ which meets the circle $k_a$ in two points, because $k_c$ meets the interior of the chord $AK$, and hence the both arcs $AK$ of the circle $k_a$ as well. Consequently, the problem has two solutions.\n\nc) Suppose that the triangle $ABC$ is **scalene**, with the largest side, say $AB$ (as in Fig. 6). Then the ratio $|XB|/|XC|$ for points $X \\in k_a$ is larger than $1$, because of $A \\in k_a$. Hence $B$ lies in the interior $k_a$, while $C$ lies in its exterior. The last together with $A \\in k_a$ implies that $L$, an interior point of $AC$, lies in the exterior of $k_a$. Thus $k_a$ intersects the chord $BL$ of $k_b$ which means that $k_a$ and $k_b$ meet in two points. Consequently, the problem has two solutions.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 29434,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all real $p$ such that the inequality\n$$\n\\sqrt{a^2 + p b^2} + \\sqrt{b^2 + p a^2} \\ge a + b + (p-1)\\sqrt{ab}\n$$\nholds for any real $a$ and $b$.",
    "options": [],
    "answer": "0 ≤ p ≤ 3",
    "solution": "If $a = b = 1$ the parameter $p > 0$ has to satisfy:\n$$\n\\begin{cases} 2\\sqrt{p+1} \\ge p+1, \\\\ 2 \\ge \\sqrt{p+1}, \\\\ p \\le 3. \\end{cases}\n$$\nWe show that for $p \\in (0, 3)$ the inequality holds for any real $a$ and $b$. If $p \\in (0, 1)$ the inequality holds trivially:\n$$\n\\sqrt{a^2 + p b^2} > a, \\quad \\sqrt{b^2 + p a^2} > b \\quad \\text{and} \\quad (p-1)\\sqrt{ab} \\le 0.\n$$\nLet further be $p \\in (1, 3)$. The left hand side, LHS, of the inequality can be understood as the sum of the lengths of vectors $(a, b\\sqrt{p})$ and $(b, a\\sqrt{p}) \\in \\mathbb{R}^2$. According to the triangle inequality then\n$$\n\\begin{aligned}\nLHS &= \\sqrt{a^2 + p b^2} + \\sqrt{b^2 + p a^2} = |(a, b\\sqrt{p})| + |(b, a\\sqrt{p})| \\\\\n&\\ge |(a+b, (a+b)\\sqrt{p})| = (a+b)\\sqrt{1+p}.\n\\end{aligned} \\quad (1)\n$$\nFor the RHS we have (with the help of AM-GM inequality)\n$$\n\\text{RHS} = a+b+(p-1)\\sqrt{ab} \\le a+b+(p-1)\\frac{a+b}{2} = \\frac{(p+1)(a+b)}{2}.\n$$\nNow $LHS \\ge RHS$ evidently, because even stronger inequality\n$$\n(a+b)\\sqrt{p+1} \\ge \\frac{(p+1)(a+b)}{2}\n$$\nis equivalent to $\\sqrt{p+1} \\le 2$, which is obviously satisfied for any $p \\in (1, 3)$.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 29435,
    "subject": "Mathematics (Multi-modal)",
    "question": "In the real numbers, solve the following system of equations:\n$$\n\\begin{aligned}\n \\sin^2 x + \\cos^2 y &= \\tan^2 z, \\\\\n \\sin^2 y + \\cos^2 z &= \\tan^2 x, \\\\\n \\sin^2 z + \\cos^2 x &= \\tan^2 y.\n\\end{aligned}\n$$",
    "options": [],
    "answer": "(x, y, z) = (pi/4 + k*pi/2, pi/4 + l*pi/2, pi/4 + m*pi/2) for any integers k, l, m",
    "solution": "Substituting $\\cos^2 x = a$, $\\cos^2 y = b$, $\\cos^2 z = c$ leads to the system\n$$\n\\begin{cases}\n 1 - a + b = \\frac{1}{c} - 1, \\\\\n 1 - b + c = \\frac{1}{a} - 1, \\\\\n 1 - c + a = \\frac{1}{b} - 1,\n\\end{cases} \\quad (1)\n$$\nwhere $a, b, c \\in (0, 1)$.\nAdding these equations together yields\n$$\n\\frac{1}{a} + \\frac{1}{b} + \\frac{1}{c} = 6,\n$$\nso the harmonic mean of the numbers $a, b, c$ is $\\frac{1}{2}$.\n\nMultiplying the equations by the numbers $c, a$, and $b$, respectively, we get\n$$\n\\begin{align*}\n c - ac + bc &= 1 - c, \\\\\n a - ab + ac &= 1 - a, \\\\\n b - bc + ab &= 1 - b,\n\\end{align*}\n$$\nwhich sums to $2(a + b + c) = 3$. Therefore, the arithmetic mean of the numbers $a, b, c$ is $\\frac{1}{2}$, too. Since the arithmetic and harmonic means are equal, it must be that $a = b = c = \\frac{1}{2}$. We can easily verify that this triple satisfies the system (1). The solutions of the original system are thus exactly the triples $\\left(\\frac{1}{4}\\pi + \\frac{1}{2}k\\pi, \\frac{1}{4}\\pi + \\frac{1}{2}l\\pi, \\frac{1}{4}\\pi + \\frac{1}{2}m\\pi\\right)$, where $k, l, m$ are integers.\n\n\nJiné řešení. We use the substitution from the above solution. The system (1) is cyclic; if a triple $(p, q, r)$ satisfies it, then so do the triples $(q, r, p)$ and $(r, p, q)$. Therefore, it suffices to find the solutions for which $a \\ge b$, $a \\ge c$, and all the other solutions can then be obtained by cyclic exchange.\nLet $a \\ge b$, $a \\ge c$. It follows from the first equation that $1/c = 2 - a + b \\le 2$, so $c \\ge \\frac{1}{2}$. Similarly, it follows from the third equation that $1/b = 2 - c + a \\ge 2$, so $b \\le \\frac{1}{2}$, and thus $b \\le c$. From the second equation, we have $1/a = 2 - b + c \\ge 2$, so $a \\le \\frac{1}{2}$. Altogether,\n$$\n\\frac{1}{2} \\ge a \\ge c \\ge \\frac{1}{2},\n$$\nso $a = c = \\frac{1}{2}$. Now, any of the equations of the system (1) yields $b = \\frac{1}{2}$. Just as in the previous case, we verify that the found triple satisfies the system (1); so the solutions of the system are the triples $\\left(\\frac{1}{4}\\pi + \\frac{1}{2}k\\pi, \\frac{1}{4}\\pi + \\frac{1}{2}l\\pi, \\frac{1}{4}\\pi + \\frac{1}{2}m\\pi\\right)$, where $k, l, m$ are integers.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 29436,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all positive integers $n$ which are not powers of $2$ and which satisfy the equation $n = 3D + 5d$, where $D$ (and $d$) denote the greatest (and the least) numbers among all odd divisors of $n$ which are larger than $1$.\n(Tomáš Jurík)",
    "options": [],
    "answer": "60, 100, and 8p for any odd prime p",
    "solution": "Let $n = p_1^{\\alpha_1} p_2^{\\alpha_2} \\dots p_k^{\\alpha_k}$ be the prime factorization of a satisfactory number $n$. Here $p_1 < p_2 < \\dots < p_k$ are all the prime divisors of $n$ and the exponents $\\alpha_i$ are positive integers. The given equation implies that $p_1 = 2$ (otherwise $D = n$ which contradicts to $n = 3D + 5d$) and that $k \\ge 2$ (otherwise $n$ is a power of $2$). Thus we have $D = p_2^{\\alpha_2} \\dots p_k^{\\alpha_k}$, $d = p_2$ and the equation becomes\n$$\n2^{\\alpha_1} p_2^{\\alpha_2} \\dots p_k^{\\alpha_k} = 3 p_2^{\\alpha_2} \\dots p_k^{\\alpha_k} + 5p_2 \\quad \\text{or} \\quad (2^{\\alpha_1} - 3) p_2^{\\alpha_2 - 1} \\dots p_k^{\\alpha_k} = 5.\n$$\n(In the case when $k=2$ the left-hand of the last equation is simply $(2^{\\alpha_1} - 3)p_2^{\\alpha_2-1}$.)\nSince the number $5$ has only two divisors $1$ and $5$, it holds that $2^{\\alpha_1} - 3 \\in \\{1, 5\\}$ and\nhence either $\\alpha_1 = 2$ or $\\alpha_1 = 3$.\n\ni. The case $\\alpha_1 = 2$. The simplified equation\n$$\np_2^{\\alpha_2-1} \\dots p_k^{\\alpha_k} = 5\n$$\nholds if and only if either $k=2$, $p_2=5$ and $\\alpha_2 - 1 = 1$, or $k=3$, $\\alpha_2 - 1 = 0$, $p_3 = 5$ and $\\alpha_3 = 1$ — then from $2 < p_2 < p_3 = 5$ it follows that $p_2 = 3$. Consequently, there are exactly two solutions in the case (i), namely $n = 2^2 5^2 = 100$ and $n = 2^2 3^1 5^1 = 60$.\n\nii. The case $\\alpha_1 = 3$. The simplified equation\n$$\np_2^{\\alpha_2-1} \\dots p_k^{\\alpha_k} = 1\n$$\nholds only for $k=2$ and $\\alpha_2 - 1 = 0$. Notice that there is no restriction on the prime number $p_2$ excepting the inequality $p_2 > 2$. Consequently, there are infinitely many solutions in the case (ii) and all of them are given by $n = 2^3 p_2^1 = 8p_2$, where $p_2$ is any odd prime number.\n\n**Answer.** All the solutions $n$ are: $n = 60$, $n = 100$ and $n = 8p$, where $p$ is any odd prime number.\nThe given equation $n = 3D + 5d$ implies that $n > 3D$ and $n \\le 3D + 5D = 8D$ (because of $d \\le D$). Since the ratio $n:D$ must be a power of $2$, it follows from $3 < n: D \\le 8$ that either (i) $n = 4D$, or (ii) $n = 8D$.\n\ni. The case $n = 4D$. From $4D = n = 3D + 5d$ we have $D = 5d$ and thus $n = 4D = 20d$. Since $d$ must be a prime odd divisor of $n$ which is a multiple of $5$, we conclude that $p \\in \\{3, 5\\}$ and hence $n \\in \\{60, 100\\}$ (both the values are clearly satisfactory).\n\nii. $n = 8D$. Our way of deriving the inequality $n \\le 8D$ implies now that $D = d$ and hence $D$ is an (odd) prime number. All such $n = 8D$ are solutions indeed.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 29437,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $n$ be a natural number whose all positive divisors are denoted as $d_1, d_2, \\dots, d_k$ in such a way that $d_1 < d_2 < \\dots < d_k$ (thus $d_1 = 1$ and $d_k = n$). Determine all the values of $n$ for which both equalities $d_5 - d_3 = 50$ and $11d_5 + 8d_7 = 3n$ hold. (Matúš Harminc)",
    "options": [],
    "answer": "2013",
    "solution": "We distinguish whether $n$ is odd or even.\n\ni) The case of $n$ odd. Since all the $d_i$'s are odd too, it follows from $11d_5+8d_7 = 3n$ that $d_7 \\mid 11d_5$ as well as $d_5 \\mid 8d_7$, hence $d_5 \\mid d_7$. In view of $d_7 > d_5$, the relations $d_5 \\mid d_7 \\mid 11d_5$ imply that $d_7 = 11d_5$. Substituting this into $11d_5 + 8d_7 = 3n$, we obtain $99d_5 = 3n$ or $33d_5 = n$. Thus the four numbers 1, 3, 11 and 33 are divisors of $n$, more exactly all its divisors smaller than 50, since the fifth divisor $d_5$ satisfies $d_5 = d_3 + 50 > 50$. Consequently, it holds that $d_1 = 1$, $d_2 = 3$, $d_3 = 11$, $d_4 = 33$, $d_5 = d_3 + 50 = 61$, and thus $n = 33d_5 = 33 \\cdot 61 = 2013$. The number 2013 is satisfactory indeed, because its first small divisors as indicated in the last sentence and moreover, the subsequent divisors are $d_6 = 61 \\cdot 3$ and $d_7 = 61 \\cdot 11$; hence $d_7 = 11d_5$ as required.\n\nii) The case of $n$ even. Now the equality $11d_5 + 8d_7 = 3n$ implies that $2 \\mid d_5$ and hence $2 \\mid d_5 - 50 = d_3$ as well. Since $d_1 = 1$, $d_2 = 2$ and $d_3 \\neq 3$, we conclude that either $d_3 = 4$, or $d_3 = 2t$, with some integer $t > 2$. But the last is impossible (otherwise $t$ is a divisor of $n$ with $d_2 < t < d_3$, a contradiction), Therefore, we have $d_3 = 4$, $d_5 = d_3 + 50 = 54$ and hence 3 is a divisor of $n$ between $d_2$ and $d_3$, a contradiction. In this way, the nonexistence of any satisfactory even $n$ is proven.\n\n**Answer.** The problem has the only solution $n = 2013$.\nThe divisors $d_5$ and $d_7$ of $n$, with $d_5 < d_7$, can be represented as $d_5 = n/x$ and $d_7 = n/y$, where $x$ and $y$ ($x > y$) are some positive divisors of $n$ again. Substituting this into $11d_5 + 8d_7 = 3n$, we obtain (after cancelling $n$) an equation $11/x + 8/y = 3$ which can be solved in a standard way, for example by a simple factorization:\n$$\n8x = y(3x - 11) \\Leftrightarrow 8(3x - 11) + 88 = 3y(3x - 11) \\Leftrightarrow (3x - 11)(3y - 8) = 88.\n$$\nThe first equation implies that $3x - 11 > 0$ and hence $3y - 8 > 0$ as well. Note that $x \\ge y + 1$ yields $3x - 11 \\ge 3y - 8 > 0$. Taking into account the prime factorization $88 = 2^3 \\cdot 11$, we conclude that the ordered pair $(3x - 11, 3y - 8)$ of factors must belong to the following set\n$$\n\\{(88, 1), (44, 2), (22, 4), (11, 8)\\}.\n$$\nHowever, congruences modulo 3 imply the only two pairs (88, 1) and (22, 4) are admissible. The corresponding pairs $(x, y)$ are (33, 3) and (11, 4), respectively.\n\nIf $(x, y) = (33, 3)$, then $d_5 = n/33$ (and $d_7 = n/3$), thus 1, 3, 11 and 33 are divisors of $n$ which leads (as in the above solution) to the solution $n = 2013$.\nIf $(x, y) = (11, 4)$, then $d_5 = n/11$ and $d_7 = n/4$, thus 1, 2, 4, 11, 22 and 44 are divisors of $n$, which contradicts $d_5 > 50$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 29438,
    "subject": "Mathematics (Multi-modal)",
    "question": "We are given a segment $AB$ in the plane. Consider a triangle $XYZ$ with the following properties: the vertex $X$ is an interior point of the segment $AB$, the triangles $XBY$ and $XZA$ are similar ($\\triangle XBY \\sim \\triangle XZA$) and the points $A$, $B$, $Y$, $Z$ lie on a circle in this order. Find the locus of midpoints of the sides $YZ$ of all such triangles $XYZ$.",
    "options": [],
    "answer": "The locus is the set of points inside the circle with diameter formed by the given segment and outside both Thales circles with diameters from each endpoint to the midpoint of the segment. Equivalently, it is the intersection of the interior of the circle with the given segment as diameter and the exteriors of the circles with diameters from each endpoint to the segment’s midpoint.",
    "solution": "Let $XYZ$ be a satisfactory triangle. Then the vertices $Y$ and $Z$ must lie in the same half-plane with the boundary line $AB$. Denote by $Y'$ the reflection of $Y$ through the line $AB$. Due to the presumed similarity, the angles $XAZ$ and $BYX$ are congruent (Fig. 1) and hence $|\\angle BAZ| = |\\angle BY'Z|$ as well. Using the well known inscribed angles property we conclude that the circumcircle of $\\triangle ABZ$ passes not only through the point $Y$, but also through the point $Y'$. The line $AB$ (as a perpendicular bisector of the chord $YY'$) passes through the centre $O$ of the circle $k$ and thus the chord $AB$ is a diameter of $k$. Since the segment $AB$ is fixed, the circle $k = ABYZ$ is common for all satisfactory triangles $XYZ$ and the midpoint $M$ of $YZ$ must lie in the interior of $k$. Since the both angles $OMZ$ and $OMY$ are right (Fig. 2), the (lesser) angles $AMO$ and $BMO$ are acute and thus the point $M$ must lie in the intersection of the exteriors of Thales' circles with diameters $AO$ and $BO$. In what follows we will show the both derived necessary conditions determine the locus of all the possible midpoints $M$.\n\n![](attached_image_1.png)\nFig. 1\n![](attached_image_2.png)\nFig. 2\n\nSo, let $M$ be any point in the interior of $k$ for which the both angles $AMO$ and $BMO$ are acute (i.e. $M$ lies in the exteriors of the circles with diameters $AO$ and $BO$). Consider a chord of $k$ which passes through $M$ perpendicularly to $OM$. This chord does not intersect the diameter $AB$, because of the acute angles $AMO$ and $BMO$. Thus the endpoints of the chord with the midpoint $M$ can be denoted as $Y$ and $Z$ so that $A$, $B$, $Y$, $Z$ lie on $k$ in this order. If $Y$ reflects to $Y'$ through the diameter $AB$ and if $X$ denotes the intersection point of the segments $AB$ and $Y'Z$, then the triangles $XBY$ and $XZA$ are similar as required (by theorem AA). This completes the solution.\n\n![](attached_image_1.png)\nFig. 1\n\n![](attached_image_2.png)\nFig. 2\n\n**Conclusion.** The locus under consideration is the interior of the highlighted region bounded by the three circles with diameters $AB$, $AO$ and $BO$, where $O$ denotes the midpoint of segment $AB$ (Fig. 3).\n![](attached_image_3.png)\nFig. 3",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 29439,
    "subject": "Mathematics (Multi-modal)",
    "question": "We are given two circles $k_1(S_1, r_1)$ and $k_2(S_2, r_2)$ in the plane, with $|S_1S_2| > r_1 + r_2$. Find the locus of points $X$ which do not lie on the line $S_1S_2$ and possess the following property: The segments $S_1X$ and $S_2X$ intersect successively the circles $k_1$ and $k_2$ in such points whose distances to the line $S_1S_2$ are the same. (Jaromír Šimša)",
    "options": [],
    "answer": "If the radii are unequal, the locus is the circle of Apollonius for the two centers with distance ratio equal to the ratio of the radii, excluding the two homothety centers on the line of centers. If the radii are equal, the locus is the perpendicular bisector of the segment joining the centers, excluding its midpoint.",
    "solution": "In the first part of our solution, we will assume that $X$ is any point with the required property. It is clear that $X$ lies in the exteriors of the circles $k_1$ and $k_2$ and that the points $S_1$, $S_2$ and $X$ are vertices of a triangle whose sides $S_1X$, $S_2X$ are intersected successively by the circles $k_1$ and $k_2$ in such points $Y_1$, $Y_2$ which lie on the same line parallel to $S_1S_2$ (Fig. 1). Since the triangles $XY_1Y_2$ and $XS_1S_2$ are similar (by theorem AA), it holds\n$$\n\\frac{|XY_1|}{|XS_1|} = \\frac{|XY_2|}{|XS_2|}, \\qquad (1)\n$$\nwhich can be rewritten, because of the equalities\n$$\n|XY_1| = |XS_1| - r_1 \\quad \\text{a} \\quad |XY_2| = |XS_2| - r_2, \\qquad (2)\n$$\nas an equation for lengths of the segments $XS_1$ and $XS_2$:\n$$\n\\frac{|XS_1| - r_1}{|XS_1|} = \\frac{|XS_2| - r_2}{|XS_2|}, \\\\\n\\frac{|XS_1|}{|XS_2|} = \\frac{r_1}{r_2}. \\qquad (3)\n$$\nSince the points $S_1$ and $S_2$ are fixed as well as the ratio $r_1/r_2$, the locus of points $X$\n\n![](attached_image_1.png)\nFig. 1\n\nsatisfying equation (3) is a circle of Apollonius (which evidently becomes a straight line if the ratio $r_1/r_2$ equals 1). As known for the case when $r_1/r_2 \\neq 1$, there are exactly two solutions $X = H_1$ and $X = H_2$ of the equation (3) that lie on the line $S_1S_2$ and form a diameter $H_1H_2$ of the resulting circle of Apollonius. For our situation, let us add the fact that the points $H_1$ and $H_2$ are centres of both homotheties of the initially given circles $k_1$ and $k_2$.\n\nIn the second part of our solution, we will conversely assume that $X$ is a point of the circle of Apollonius given by the equation (3) and that $X$ lies outside of the line $S_1S_2$, i.e. $X \\neq H_1$ and $X \\neq H_2$. In view of the condition that $|S_1S_2| > r_1 + r_2$, the whole circle of Apollonius (with diameter $H_1H_2$) lies in the exteriors of the circles $k_1$ and $k_2$. Indeed, the last fact follows from a known position of the homothety centres $H_1$ and $H_2$ on the line $S_1S_2$: If for example $r_1 < r_2$, then the diameter $H_1H_2$ contains the diameter of $k_1$, while the diameter of $k_2$ and the diameter $H_1H_2$ are disjoint. (See also *Remark* below.)\n\nThe proven property implies that $XS_1S_2$ is a triangle with $|XS_1| > r_1$ and $|XS_2| > r_2$. Thus there are points $Y_1 \\in k_1$ and $Y_2 \\in k_2$ lying on the segments $S_1X$ and $S_2X$, respectively. Since the equalities (2) are valid again, it is possible to transform the equation (3) to equation (1). Consequently, the triangles $XS_1S_2$ and $XY_1Y_2$ are similar (now by SAS theorem) and hence $S_1S_2 \\parallel Y_1Y_2$. Therefore the distances of $Y_1$ and $Y_2$ to the line $S_1S_2$ are equal which proves the required property of the point $X$.\n\n*Answer.* If $r_1 \\neq r_2$, the locus of points $X$ is the circle of Apollonius which is given by the above equation (3), excepting the two points on the line $S_1S_2$. If $r_1 = r_2$, the locus is the perpendicular bisector of the segment $S_1S_2$, with exception of the midpoint of $S_1S_2$.\n\n![](attached_image_1.png)\nFig. 1",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 29440,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let us call by an \"edge\" any segment of length $1$ which is common to two adjacent fields of a given chessboard $8 \\times 8$. Consider all possible cuttings of the chessboard into $32$ pieces $2 \\times 1$ and denote by $n(e)$ the total number of such cuttings that involve the given edge $e$. Determine the last digit of the sum of the numbers $n(e)$ over all the edges $e$.\n\n(Michal Rolínek)",
    "options": [],
    "answer": "0",
    "solution": "The number of edges, which are not involved in a given cutting, is equal to $32$, because each of these edges must coincide with the common segment of the two fields forming one of the $32$ resulting pieces $2 \\times 1$. Thus each cutting gives a contribution $112 - 32 = 80$ to the sum $S$ of all the numbers $n(e)$. Consequently, the sum $S$ is a multiple of $80$ and thus its last digit is zero.",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 29441,
    "subject": "Mathematics (Multi-modal)",
    "question": "**Find all triples of real numbers $x$, $y$ and $z$ for which**\n$$\nx(y^2 + 2z^2) = y(z^2 + 2x^2) = z(x^2 + 2y^2).\n$$",
    "options": [],
    "answer": "(t, 0, 0), (0, t, 0), (0, 0, t), (t, t, t), (4t, t, 2t), (2t, 4t, t), (t, 2t, 4t) for any real t",
    "solution": "If for example $x = 0$, we get a system $0 = y z^2 = 2 y^2 z$ which means that one of the unknowns $y$ and $z$ vanishes and the other can be arbitrary. The cases $y = 0$ or $z = 0$ are discussed similarly. Thus we have obtained three groups of solutions $(x, y, z)$ which are formed by triples $(t, 0, 0)$, $(0, t, 0)$ and $(0, 0, t)$ respectively, where $t$ is any real number. Moreover, we have observed that all the other solutions satisfy the condition $x y z \\neq 0$, which is supposed to hold in what follows.\n\nFactorizing the equation $x(y^2 + 2z^2) = y(z^2 + 2x^2)$ yields $(2x - y)(z^2 - x y) = 0$. Thus we distinguish two cases (depending on the fact which of the two factors vanishes).\n\ni. $2x - y = 0$. After setting $y = 2x$ the given system is reduced to the only equation\n$$\n2x(2x^2 + z^2) = 9x^2 z,\n$$\nwhich can be simplified (by dividing $x \\neq 0$) to\n$$\n4x^2 + 2z^2 - 9x z = 0 \\quad \\text{or} \\quad (x - 2z)(4x - z) = 0.\n$$\nThus the case (i) yields exactly two groups of solutions $(2t, 4t, t)$ and $(t, 2t, 4t)$, where $t$ is any real number.\n\nii. $z^2 - x y = 0$. Substituting $z^2 = x y$ into the given system, we now get the only equation\n$$\nx y (2x + y) = z(x^2 + 2y^2),\n$$\nwhich is (because of the inequality $x^2 + 2y^2 > 0$) equivalent to\n$$\nz = \\frac{x y (2x + y)}{x^2 + 2y^2}.\n$$\nAt this moment we have to find when such a $z$ obeys the condition $z^2 = x y$. After direct substitution we get the following condition on the unknowns $x$ and $y$:\n$$\n\\frac{x^2 y^2 (2x + y)^2}{(x^2 + 2y^2)^2} = x y.\n$$\nDividing by $x y \\neq 0$ and removing the fraction yields\n$$\nx y (2x + y)^2 = (x^2 + 2y^2)^2 \\quad \\text{or} \\quad (4y - x)(x^3 - y^3) = 0.\n$$\nThus we conclude that either $x = 4y$, or $x^3 = y^3$, i.e. $x = y$. Returning to the formula for $z$, we obtain $z = 2y$ or $z = x$, according as $x = 4y$ or $x = y$. Consequently, there are two groups of solutions in the case (ii), namely triples $(4t, t, 2t)$ and $(t, t, t)$, where $t$ is any real number.\n\n*Answer.* All the solutions are $(t, 0, 0)$, $(0, t, 0)$, $(0, 0, t)$, $(t, t, t)$, $(4t, t, 2t)$, $(2t, 4t, t)$ and $(t, 2t, 4t)$, where $t$ is any real number.\nTo avoid unnecessary repetition from the above solution, we will solve the problem under the condition that $x y z \\neq 0$.\n\nDividing both sides of the given equations by $x y z$ we obtain\n$$\n\\frac{y}{z} + \\frac{2z}{y} = \\frac{z}{x} + \\frac{2x}{z} = \\frac{x}{y} + \\frac{2y}{x}, \\quad (3)\n$$\nwhich can be read as a coincidence of values of a function $f(s) = s + 2/s$ in three nonzero points $s_1 = y/z$, $s_2 = z/x$ and $s_3 = x/y$. Thus we first find when $f(s) = f(t)$ for two nonzero real numbers $s$ and $t$. It follows from the identity\n$$\nf(s) - f(t) = s + \\frac{2}{s} - t - \\frac{2}{t} = \\frac{(s-t)(s t - 2)}{s t}\n$$\nthat $f(s) = f(t)$ if and only if $s = t$ or $s t = 2$. Consequently, the system (3) holds if and only if the introduced numbers $s_1, s_2, s_3$ possess the following property: $s_i = s_j$ or $s_i s_j = 2$, for any indices $i$ and $j$. However, if there exists a permutation $(i, j, k)$ of $(1, 2, 3)$ such that $s_i s_j = 2$, then the identity $s_i s_j s_k = 1$ implies that $s_k = \\frac{1}{2}$ and hence $s_i \\in \\{\\frac{1}{2}, 4\\}$ (because $s_i = s_k$ or $s_i s_k = 2$). Thus the assumption $s_i s_j = 2$ leads to the conclusion that $(s_1, s_2, s_3)$ is a permutation of $(\\frac{1}{2}, \\frac{1}{2}, 4)$. It is easy to check that exactly three such permutations are satisfactory and yield the solutions $(4t, t, 2t)$, $(2t, 4t, t)$ and $(t, 2t, 4t)$ of the given system. In the remaining case when $s_1 = s_2 = s_3$, the identity $s_1 s_2 s_3 = 1$ implies that $s_i = 1$ for each $i$, which yields the solutions $(t, t, t)$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 29442,
    "subject": "Mathematics (Multi-modal)",
    "question": "Six teams will take part in a volleyball tournament. Each pair of the teams should play one match. All the matches will be realized in five rounds, each involving three simultaneous matches on the courts numbered 1, 2 and 3. Find the number of all possible draws for such a tournament. By a draw we mean a table $5 \\times 3$ in which an unordered pair of teams is written on the field $(i, j)$, where $i \\in \\{1, 2, 3, 4, 5\\}$ and $j \\in \\{1, 2, 3\\}$, if these two teams will meet each other in the $i$-th round on the court $j$. You are allowed to write down the resulting number as a product of prime factors (instead of writing its decimal expansion).",
    "options": [],
    "answer": "5598720",
    "solution": "We postpone the question of permutations of the five rounds and the three courts to the end of our solution. Denoting first the teams by numbers $1, 2, 3, 4, 5, 6$ (in a fixed way), we rearrange the five rounds of any satisfactory draw by means of the following numbering: Let $1$ and $2$ be the rounds with matches of the pairs of teams $(1, 2)$ and $(1, 3)$, respectively. If a pair $(3, a)$ plays in the round $1$ and if a pair $(2, b)$ plays in the round $2$, then $a, b$ are two distinct numbers from $\\{4, 5, 6\\}$ (otherwise the third pairs in the rounds $1$ and $2$ are identical). Let $3, 4$ and $5$ denote the rounds with pairs $(1, a)$, $(1, b)$ and $(1, c)$ respectively, where $c \\in \\{4, 5, 6\\} \\setminus \\{a, b\\}$. Up to this moment, we have fixed an uncompleted draw\n1: (1, 2), (3, a),\n2: (1, 3), (2, b),\n3: (1, a),\n4: (1, b),\n5: (1, c),\nwhich can be extended to a the complete draw in the only one way:\n1: (1, 2), (3, a), (b, c),\n2: (1, 3), (2, b), (a, c),\n3: (1, a), (2, c), (3, b),\n4: (1, b), (2, a), (3, c),\n5: (1, c), (2, 3), (a, b).\nSince $(a, b, c)$ is any permutation of $(4, 5, 6)$, the total number of the complete draws (written as above) is $3! = 6$. Taking in account the number $5!$ of the possible permutations of the five rounds and the number $3!$ of possible permutations of the three courts, we conclude that the requested number of all draws is equal to\n$$\n6 \\cdot 5! \\cdot 6^5 = 5! \\cdot 6^6 = 2^9 \\cdot 3^7 \\cdot 5 = 5598720.\n$$\nLet us denote the six teams by numbers $1, 2, 3, 4, 5, 6$ and construct first an \"unordered\" draw in which the rounds will be \"numbered\" by the opponents of the team $1$ — see the following table in which the other opponents of the team $2$ are denoted as $a, b, c, d$:\n1: (1, 2),\n2: (1, 3), (2, a),\n3: (1, 4), (2, b),\n4: (1, 5), (2, c),\n5: (1, 6), (2, d).\nNote that $(a, b, c, d)$ is a permutation of the quadruple $(3, 4, 5, 6)$ and that the following two restrictions are evident:\n$$\n\\triangleright 3 \\neq a, 4 \\neq b, 5 \\neq c \\text{ and } 6 \\neq d\n$$\nThe two-element sets $\\{3, a\\}$, $\\{4, b\\}$, $\\{5, c\\}$, $\\{6, d\\}$ are pairwise distinct.\nIt is clear that under these two conditions, the third pairs for the rounds $2$-$5$ are uniquely determined, as well as the remaining two pairs for the round $1$. Consequently, we have to calculate the number of permutations $(a, b, c, d)$ of the quadruple $(3, 4, 5, 6)$ which satisfy the two above stated conditions.\nUsing the inclusion-exclusion principle we conclude that the first condition is fulfilled by exactly nine permutations:\n$$\n4! - \\left( 4 \\cdot 3! - \\binom{4}{2} \\cdot 2! + 4 - 1 \\right) = 9.\n$$\nMoreover, exactly three of them do not satisfy the second condition, namely the permutations $(4, 3, 6, 5)$, $(5, 6, 3, 4)$ and $(6, 5, 4, 3)$. Thus the total number of the satisfactory permutations equals $9 - 3 = 6$.\nWe have proved that there are six \"unordered\" draws in the above specified sense. Combining this result with the idea of permuting the rounds and the courts, we conclude that the requested number of the draws is equal to\n$$\n6 \\cdot 6^5 \\cdot 5! = 5! \\cdot 6^6 = 2^9 \\cdot 3^7 \\cdot 5 = 5,598,720.\n$$",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 29443,
    "subject": "Mathematics (Multi-modal)",
    "question": "There are 234 visitors in a cinema auditorium. The visitors are sitting in $n$ rows, where $n \\ge 4$, so that each visitor in the $i$-th row has exactly $j$ friends in the $j$-th row, for any $i, j \\in \\{1, 2, ..., n\\}, i \\neq j$. Find all the possible values of $n$. (Friendship is supposed to be a symmetric relation.)",
    "options": [],
    "answer": "12",
    "solution": "For any $k \\in \\{1, 2, ..., n\\}$ denote by $p_k$ the number of visitors in the $k$-th row. The stated condition on given $i$ and $j$ implies that the number of friendly pairs $(A, B)$, where $A$ and $B$ are from the $i$-th row and from $j$-th row respectively, is equal to the product $j p_i$. Interchanging the indices $i$ and $j$, we conclude that the same number of friendly pairs $(A, B)$ equals $i p_j$. Thus $j p_i = i p_j$ or $p_i : p_j = i : j$, and therefore, all the numbers $p_k$ must be proportional as follows:\n$$\np_1 : p_2 : \\dots : p_n = 1 : 2 : \\dots : n.\n$$\nLet us show that under this proportionality the visitors can be friendly in such a way which ensures the property under consideration. Thus assume that for some positive integer $d$, the equality $p_k = k d$ holds with any $k \\in \\{1, 2, ..., n\\}$. Let us\n\nstart with the case $d = 1$ when the numbers of visitors in single rows are successively $1, 2, ..., n$. Then the stated property holds true if (and only if) any two visitors — taken from distinct rows in the whole auditorium — are friends. In the case when $d > 1$, let us divide all the visitors into $d$ groups $G_1, G_2, ..., G_d$ so that for arbitrary $k = 1, 2, ..., d$, the numbers of visitors from the group $G_k$ in single rows are successively $1, 2, ..., n$. It is evident that the stated property holds true under the following condition: two visitors are friends if and only if they belong to the same group $G_k$.\nIt follows from the preceding that our task is to find such integer values of $n$, $n \\ge 4$, for which there exists a positive integer $d$ satisfying the equation\n$$\nd + 2d + \\dots + n d = 234 \\quad \\text{or} \\quad d n(n+1) = 468.\n$$\nThus we look for all divisors $468 = 2^2 \\cdot 3^2 \\cdot 13$ which are of the form $n(n+1)$. Inequality $22 \\cdot 23 > 468$ implies that $n < 22$ and hence $n \\in \\{4, 6, 9, 12, 13, 18\\}$. It is easy to see that the only satisfactory $n$ equals $12$ (which corresponds to $d=3$).\n\nAnswer. The unique solution is $n = 12$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 29444,
    "subject": "Mathematics (Multi-modal)",
    "question": "We are given an acute-angled triangle $ABC$. Denote by $k$ the circle with diameter $AB$. A circle touching the bisector of the angle $BAC$ at the point $A$ and passing through the point $C$ meets the circle $k$ in a point $P$, $P \\neq A$. Similarly, a circle touching the bisector of the angle $ABC$ at the point $B$ and passing through the point $C$ meets the circle $k$ in a point $Q$, $Q \\neq B$. Prove that the lines $AQ$ and $BP$ intersect each other on the bisector of the angle $ACB$. (Peter Novotný)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Besides the Thales' circle $k$, denote as $l_A = AP$ and $l_B = BP$ the other two circles under consideration. Let us deal, for example, with the circle $l_B$ drawn in Fig. 4. In what follows the interior angles of $\\triangle ABC$ are denoted as usual.\n\n![](attached_image_1.png)\nFig. 4\n\nLet us explain that it is true indeed what the figure suggests. First of all, the point $Q$ lies in the half-plane $BCA$, because the local arc $BC$ of the circle $l_B$ has the following property: as its point $X$ moves from $C$ to $B$, the angle $AXB$ varies from an acute angle $\\gamma$ to an obtuse angle $180^\\circ - \\beta/2$, and hence the Thales' circle $k$ meets the arc $BC$ in an interior point $Q$. Applying the properties of inscribed and subtended angles to the chord $BC$ of the circle $l_B$, we conclude that $|\\angle BQC| = 180^\\circ - \\beta/2$ and hence $|\\angle AQB| + |\\angle BQC| = 270^\\circ - \\beta/2 > 180^\\circ$. Consequently, $Q$ is a point of the half-plane $ACB$ which lies in the interior of the triangle $ABC$ and the convex angle $AQC$ equals $90^\\circ + \\beta/2$. As known, the last is a measure of the angle $AIC$, where $I$ denotes the incentre of $\\triangle ABC$ (indeed, $|\\angle AIC| = 180^\\circ - \\alpha/2 - \\gamma/2 = 90^\\circ + \\beta/2$). In this way, the congruence of the angles $AQC$ and $AIC$ is proven and hence the points $Q$ and $I$ lie on the same arc $AC$ of a new circle $k_B = ACI$. Therefore, the line $AQ$ is the radical axis of the circles $k$ and $k_B$. Analogously, the line $BP$ is the radical axis of the circles $k$ and $k_A = BC$.\n\nIt remains to note that the intersection point of the lines $AQ$ and $BP$ (treated as the above radical axes) has the same power with respect to the circles $k_A$ and $k_B$, whose radical axis is the line $CI$, i.e. just the bisector of the angle $ACB$. This completes our solution.\n\n![](attached_image_1.png)\nFig. 4",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 29445,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $a$, $b$ be non-negative real numbers. Prove the inequality\n$$\n\\frac{a}{\\sqrt{b^2+1}} + \\frac{b}{\\sqrt{a^2+1}} \\ge \\frac{a+b}{\\sqrt{ab+1}}\n$$\nand find when the equality holds.",
    "options": [],
    "answer": "Equality holds if and only if one of the numbers is zero or the two numbers are equal.",
    "solution": "It is evident that the inequality under consideration becomes an equality when $a=0$, $b=0$ or $a=b$. To prove that otherwise the strong inequality holds, it suffices to deal with the case $a > b > 0$ and (after removing the fractions) to show that\n$$\na\\sqrt{a^2+1} + b\\sqrt{b^2+1} > (a+b)\\sqrt{a^2+1}\n$$\nDistributing the right-hand side and regrouping the terms we get\n$$\na\\sqrt{a^2+1}(\\sqrt{ab+1} - \\sqrt{b^2+1}) > b\\sqrt{b^2+1}(\\sqrt{a^2+1} - \\sqrt{ab+1})\n$$\nMultiplying the differences of the square roots by their sums as the denominators of new installed fractions, we obtain\n$$\na\\sqrt{a^2+1} \\cdot \\frac{b(a-b)}{\\sqrt{ab+1} + \\sqrt{b^2+1}} > b\\sqrt{b^2+1} \\cdot \\frac{a(a-b)}{\\sqrt{a^2+1} + \\sqrt{ab+1}}\n$$\nDividing both sides by the positive number $ab(a-b)$ and removing the fractions again, we finally arrive at an equivalent inequality\n$$\n\\sqrt{a^2+1}(\\sqrt{a^2+1} + \\sqrt{ab+1}) > \\sqrt{b^2+1}(\\sqrt{b^2+1} + \\sqrt{ab+1})\n$$\nwhich easily follows by an easy comparison of the both sides \"term by term\" (because our assumption $a > b$ implies that $\\sqrt{a^2+1} > \\sqrt{b^2+1}$). This completes the proof of the given inequality. As we have shown, the only cases of the equality are $a = 0$, $b = 0$ and $a = b$.\n\n\nAnother solution:\nWe exclude the cases $a = 0$ and $b = 0$ (when the inequality becomes an equality) from our considerations. Let us apply a Cauchy-Schwarz inequality in the form\n$$\n\\left(\\frac{a}{u} + \\frac{b}{v}\\right) (au + bv) \\geq (a+b)^2,\n$$\nwith positive coefficients $u = \\sqrt{b^2 + 1}$ and $v = \\sqrt{a^2 + 1}$:\n$$\n\\left( \\frac{a}{\\sqrt{b^2+1}} + \\frac{b}{\\sqrt{a^2+1}} \\right) \\left( a\\sqrt{b^2+1} + b\\sqrt{a^2+1} \\right) \\ge (a+b)^2. \\quad (1)\n$$\nAnother Cauchy-Schwarz inequality yields an upper bound for the second factor from the-left hand side of (1):\n$$\n\\begin{align*}\na\\sqrt{b^2+1} + b\\sqrt{a^2+1} &= \\sqrt{a}\\sqrt{ab^2+a} + \\sqrt{b}\\sqrt{a^2b+b} \\le \\\\\n&\\le \\sqrt{a+b}\\sqrt{ab^2+a+a^2b+b} = \\sqrt{a+b}\\sqrt{(a+b)(ab+1)} = (a+b)\\sqrt{ab+1}.\n\\end{align*}\n$$\nConsequently, the first factor in (1) has a lower bound\n$$\n\\frac{a}{\\sqrt{b^2+1}} + \\frac{b}{\\sqrt{a^2+1}} \\ge \\frac{(a+b)^2}{a\\sqrt{b^2+1} + b\\sqrt{a^2+1}} \\ge \\frac{a+b}{\\sqrt{ab+1}},\n$$\nwhich is the desired inequality. Since (1) becomes an equality if and only if the positive coefficients $u$ and $v$ are the same, i.e. $\\sqrt{b^2+1} = \\sqrt{a^2+1}$ in our situation, the equality $a = b$ is the third (and last) case (next to $a = 0$ and $b = 0$ from the introductory sentence) when the proven inequality holds as an equality.\n\n\nAnother solution:\nLet us exclude the obvious cases $a=0, b=0, a=b$ and let us transform the (strong) inequality under consideration into the following equivalent form:\n$$\n\\frac{a}{a+b} \\cdot \\frac{1}{\\sqrt{b^2+1}} + \\frac{b}{a+b} \\cdot \\frac{1}{\\sqrt{a^2+1}} > \\frac{1}{\\sqrt{ab+1}} \\quad (2)\n$$\nThe last left-hand side can be read as that of the (strong) Jensen inequality\n$$\n\\mathrm{pf}(\\alpha) + \\mathrm{qf}(\\beta) > f(\\rho\\alpha + q\\beta), \\quad (3)\n$$\nwith positive coefficients $p = a/(a+b)$ and $q = b/(a+b)$ (which satisfy $p+q=1$ as required), applied to the function $f(x) = 1/\\sqrt{x}$ at the points $\\alpha = b^2+1$ and $\\beta = a^2+1$. Since the function $f$ is strictly convex on the interval $(0, +\\infty)$ and since the points $\\alpha$ and $\\beta$ are assumed to be distinct, the Jensen inequality (3) holds.\nIt remains to verify that also the right-hand sides of (2) and (3) are identical. This is easy:\n$$\n\\begin{align*}\nf(p\\alpha + q\\beta) &= f\\left(\\frac{a}{a+b}(b^2+1) + \\frac{b}{a+b}(a^2+1)\\right) = \\\\\n&= f\\left(\\frac{a + ab^2 + b + a^2b}{a+b}\\right) = f(ab+1) = \\frac{1}{\\sqrt{ab+1}}.\n\\end{align*}\n$$",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 29446,
    "subject": "Mathematics (Multi-modal)",
    "question": "On a circle of radius $r$, the distinct points $A, B, C, D$, and $E$ lie in this order, satisfying $AB = CD = DE > r$. Show that the triangle with vertices lying in the centroids of the triangles $ABD$, $BCD$, and $ADE$ is obtuse.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Denote by $P$, $Q$, and $R$ the centroids of the triangles $ABD$, $BCD$, and $ADE$, respectively. Let $K$ and $L$ be the midpoints of the segments $BD$ and $AD$, respectively. Since $P$ and $Q$ are centroids, they divide in the same ratio the medians $AK$ and $CK$, respectively. That is, $AP : PK = CQ : QK = 2 : 1$, and we have $PQ \\parallel AC$. Similarly $PR \\parallel BE$. Hence the angle $QPR$ is of the same measure as the angle $CXE$ determined by the lines $AC$ and $BE$ (here, $X$ is the intersection point of $AC$ and $BE$, see Fig. 1).\n\n![](attached_image_1.png)\n\nFig. 1\n\nDenote by $\\varphi$ the measure of the inscribed angle determined by the chord $AB$ of the given circle. Since $CD = DE = AB$, we have $\\angle CAE = 2\\varphi$, and therefore from the triangle $AXE$ we conclude\n$$\n\\angle CXE = 180^\\circ - \\angle AXE = \\varphi + 2\\varphi = 3\\varphi.\n$$\nSince $AB > r$, we have $\\varphi > 30^\\circ$, and so $\\angle QPR = 3\\varphi > 90^\\circ$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 29447,
    "subject": "Mathematics (Multi-modal)",
    "question": "Some objects are in each of four rooms. Let $n \\ge 2$ be an integer. We move one $n$-th of objects from the first room to the second one. Then we move one $n$-th of (the new number of) objects from the second room to the third one. Then we move similarly objects from the third room to the fourth one and from the fourth room to the first one. (We move the whole units of objects only.) Finally the same number of the objects is in every room. Find the minimum possible number of the objects in the second room. For which $n$ does the minimum come?",
    "options": [],
    "answer": "5, attained when n = 3",
    "solution": "Let us compute backwards. Firstly we find the number of the objects in two rooms before the move. Let $a$ and $b$ be number of the objects in the rooms $A$ and $B$ before the move. This number after the move we denote by $a'$ and $b'$. By the conditions\n$$\na' = \\frac{n-1}{n}a, \\quad b' = b + \\frac{1}{n}a\n$$\nholds. From the first equation and an identity $a + b = a' + b'$ we obtain\n$$\na = \\frac{n}{n-1}a', \\quad b = b' - \\frac{1}{n-1}a'\n$$\nNow let $M$ be the final number of the objects in every room after the fourth move. By this identity we can compute the initial number of objects in every room in terms of $M$ and $n$:\nFinally:\n$M$ & $M$ & $M$ & $M$\nbefore $4 \\to 1$: $\\frac{n-2}{n-1}M,$ & $M,$ & $M,$ & $\\frac{n}{n-1}M;$\nbefore $3 \\to 4$: $\\frac{n-2}{n-1}M,$ & $M,$ & $\\frac{n}{n-1}M,$ & $M;$\nbefore $2 \\to 3$: $\\frac{n-2}{n-1}M,$ & $\\frac{n}{n-1}M,$ & $M,$ & $M;$\nbefore $1 \\to 2$: $\\frac{n(n-2)}{(n-1)^2}M,$ & $\\frac{(n-1)^2+1}{(n-1)^2}M,$ & $M,$ & $M.$\n\nSince the number of objects in the first room was positive, $n \\ge 3$ holds. Now we can easily find the minimum of\n$$\nV_2 = \\frac{(n-1)^2 + 1}{(n-1)^2} M.\n$$\nThe difference between numerator and denominator is 1, so the fraction is irreducible. Since $V_2$ is integer it must be $M = k(n-1)^2$ for proper $k$, therefore $V_2 = k((n-1)^2+1)$. For $n \\ge 3$ we can estimate $(n-1)^2 + 1 \\ge 5$, so $V_2 \\ge 5$ too. Using $n = 3, k = 1$ and $M = 4$ we obtain $V_2 = 5$ and we can easily check that the quadruple (3, 5, 4, 4) satisfies the problem: it transforms to quadruple (2, 6, 4, 4), then (2, 4, 6, 4), after that (2, 4, 4, 6) and finally (4, 4, 4, 4). So the minimal numbers of objects in the second room is 5 and we can obtain it only for $n = 3$ because for $n \\ge 4$ is $V_2 \\ge 3^2 + 1 = 10$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 29448,
    "subject": "Mathematics (Multi-modal)",
    "question": "A triangle $ABC$ with obtuse angle at $C$ is given. Axis $o_1$ of side $AC$ intersects side $AB$ in point $K$, axis $o_2$ of side $BC$ intersects side $AB$ in point $L$. Denote $O$ intersection of the axes $o_1$ and $o_2$. Prove that centre of the incircle of triangle $KLC$ lies on the circumcircle of triangle $OKL$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 29449,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all four-digit numbers $n$ satisfying the following conditions:\ni) number $n$ is product of three different primes;\nii) sum of the two smallest of these prime numbers is equal to the difference of largest two of them;\niii) sum of three primes is equal to the square of another prime.",
    "options": [],
    "answer": "2015",
    "solution": "",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 29450,
    "subject": "Mathematics (Multi-modal)",
    "question": "Nice prime *is a prime equal to the difference of two cubes of positive integers.*\nFind last digits of all nice primes.",
    "options": [],
    "answer": "1, 7, 9",
    "solution": "Firstly, let us note that $5^3 - 4^3 = 61$, $2^3 - 1^3 = 7$ and $3^3 - 2^3 = 19$ are nice primes, so 1, 7 and 9 belong to desired digits. We show that they are all desired digits.\nLet $p = m^3 - n^3$ be a nice prime, where $m > n$ are positive integers. Second factor in rewriting\n$$\np = m^3 - n^3 = (m-n)(m^2 + mn + n^2),\n$$\nis greater than 1, thus the first one is 1 and therefore $m = n + 1$. After substitution we obtain\n$$\np = 3n^2 + 3n + 1. \\qquad (1)\n$$\nAn estimate $3n^2 + 3n + 1 > 6$ gives that the prime $p$ is odd and greater than 5. This excludes 0, 2, 4, 5, 6 and 8 as the last digits and 3 stays the only remaining digit to exclude.\nIt is sufficient to find remainders of the numbers $3n^2 + 3n + 1$ after division by 5. For remainders 0, 1, 2, 3 and 4 of $n$ we obtain remainders 1, 2, 4, 2, 1 of (1) which ones really exclude the last digit 3.\n\n*Answer.* The last digits of the nice primes are 1, 7 and 9.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 29451,
    "subject": "Mathematics (Multi-modal)",
    "question": "A natural number $n$ is given. Square with side of length $n$ is divided into $n^2$ unit squares. For the distance between two squares we consider the distance from centre to centre. Find the number of pairs of squares whose distance is $5$.",
    "options": [],
    "answer": "For n ≥ 5: 6n^2 − 38n + 48; for n ≤ 4: 0",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 29452,
    "subject": "Mathematics (Multi-modal)",
    "question": "Positive real numbers $a$, $b$, $c$, $d$ satisfy equalities\n$$\na = c + \\frac{1}{d} \\quad \\text{and} \\quad b = d + \\frac{1}{c}.\n$$\n\nProve an inequality $ab \\ge 4$ and find a minimum of $ab + cd$.",
    "options": [],
    "answer": "ab ≥ 4; minimum of ab + cd is 2(1 + sqrt(2)).",
    "solution": "To prove the inequality $ab \\ge 4$ we substitute from the equalities. We so obtain an estimate\n$$\nab = \\left(c + \\frac{1}{d}\\right)\\left(d + \\frac{1}{c}\\right) = cd + 1 + 1 + \\frac{1}{cd} \\ge 4,\n$$\nwhere we use in the last inequality well-known fact that $x + 1/x \\ge 2$ holds for all positive reals $x = cd > 0$.\n\nTo find the minimum we use similar way. Substitution for $a$ and $b$ yields\n$$\nab + cd = \\left(2 + cd + \\frac{1}{cd}\\right) + cd = 2 + 2cd + \\frac{1}{cd}.\n$$\nNow we use an inequality $x + y \\ge 2\\sqrt{xy}$ which holds true for any non-negative reals $x, y$. The choice $x = 2cd$, $y = 1/cd$ follows\n$$\n2cd + \\frac{1}{cd} \\ge 2\\sqrt{2}.\n$$\nNow we see that $ab + cd \\ge 2(1 + \\sqrt{2})$. To prove that it is the desired minimum we find some $a$, $b$, $c$, $d$ such that they makes an equality in the inequality.\nThe equality comes in the use inequality if and only if $x = y$, it is $2cd = 1/cd$. It is true e.g. for $c = 1$, $d = \\sqrt{2}/2$ and for that values we find $a = 1 + \\sqrt{2}$, $b = 1 + \\sqrt{2}/2$. Such quadruple satisfies the desired equalities and it holds $ab + cd = 2(1 + \\sqrt{2})$ too.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 29453,
    "subject": "Mathematics (Multi-modal)",
    "question": "For a given natural number $n$ specify the number of paths of length $2n + 2$ from point $[0, 0]$ to the point $[n, n]$ which do not pass any point more than once. Path of length $2n + 2$ connecting points $[0, 0]$ and $[n, n]$ means $(2n + 2)$-tuple\n$$\n(A_0A_1, A_1A_2, A_2A_3, \\dots, A_{2n+1}A_{2n+2})\n$$\nof line segments connecting two adjacent lattice points, while $A_0 = [0, 0]$, $A_{2n+2} = [n, n]$. (Pavel Novotný)",
    "options": [],
    "answer": "(2n + 1) * C(2n, n - 1)",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 29454,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find the least real $m$ such that there exist reals $a$ and $b$ for which the inequality\n$$\n|x^2 + a x + b| \\le m\n$$\nholds for all $x \\in (0, 2)$.",
    "options": [],
    "answer": "1/2",
    "solution": "Notice that no negative number $m$ satisfies the problem evidently (absolute value is non-negative number).\n\nNow we interpret the problem geometrically. A graph of some function $y = x^2 + a x + b$ lies in a horizontal strip between lines $y = +m$ and $y = -m$ and in the interval $(0, 2)$. Our aim is to find the closest strip which contains the graph of such quadratic function in the interval $(0, 2)$.\n\n![](attached_image_1.png)\n\nThe function\n$$\nf(x) = (x - 1)^2 - \\frac{1}{2} = x^2 - 2x + \\frac{1}{2},\n$$\nseems to be a good candidate for such the closest strip. Such function has $a = -2$, $b = \\frac{1}{2}$ and it satisfies (to be shown below) inequalities $-\\frac{1}{2} \\le f(x) \\le \\frac{1}{2}$.\n\nReally, this inequalities are equivalent to the inequalities $0 \\le (x-1)^2 \\le 1$, which are evidently fulfilled for $x \\in (0, 2)$. Quadratic function $f(x) = x^2 - 2x + \\frac{1}{2}$ thus satisfies the conditions of the problem for $m = \\frac{1}{2}$.\n\nIn the second part of the solution we will show that there is no quadratic function satisfying the problem for any $m < \\frac{1}{2}$.\n\nThe crucial fact will be that at least one from differences $f(0) - f(1)$ and $f(2) - f(1)$ is greater or equal to $1$ for an arbitrary function $f(x) = x^2 + a x + b$. This fact will imply that width of the closest strip will be greater or equal to $1$. This will exclude the values $m < \\frac{1}{2}$. For $f(0) - f(1) \\ge 1$ we obtain the desired estimate $2m \\ge 1$ easily from the well-known triangle inequality $|a - b| \\le |a| + |b|$:\n$$\n1 \\le |f(0) - f(1)| \\le |f(0)| + |f(1)| \\le 2m.\n$$\nSimilarly we estimate for $f(2) - f(1) \\ge 1$.\n\nNow it remains to verify at least one from inequalities $f(0) - f(1) \\ge 1$ and $f(2) - f(1) \\ge 1$ for arbitrary $f(x) = x^2 + a x + b$. The values\n$$\nf(0) = b, \\quad f(1) = 1 + a + b, \\quad f(2) = 4 + 2a + b,\n$$\nyields\n$$\nf(0) - f(1) = -1 - a \\ge 1 \\Leftrightarrow a \\le -2,\n$$\n$$\nf(2) - f(1) = 3 + a \\ge 1 \\Leftrightarrow a \\ge -2.\n$$\nSo at least one from inequalities $f(0) - f(1) \\ge 1$ and $f(2) - f(1) \\ge 1$ is true (regardless of the choice $a, b$).\n\n*Conclusion.* The desired minimal value of $m$ is $\\frac{1}{2}$.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 29455,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all the pairs of prime numbers $p$, $q$ such that the value of the expression $p^2 + 5pq + 4q^2$ is a perfect square.",
    "options": [],
    "answer": "[(13, 3), (5, 11), (7, 5)]",
    "solution": "",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 29456,
    "subject": "Mathematics (Multi-modal)",
    "question": "A triangle $ABC$ is given in which $BC$ is the shortest side. Denote $M$ its midpoint. On the sides $AB$ and $AC$ take the points $X$ and $Y$, respectively, in such a way that $|BX| = |BC| = |CY|$. Denote $Z$ the intersection point of lines $CX$ and $BY$. Prove that the line $ZM$ passes through the centre of the excircle escribed to the side $BC$ of the triangle.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 29457,
    "subject": "Mathematics (Multi-modal)",
    "question": "A parallelogram $ABCD$ with $|AB| = 2|BC|$ is given. Determine all the lines that divide the parallelogram into two tangential quadrilaterals. (Jaroslav Švrček)",
    "options": [],
    "answer": "The unique line parallel to the shorter side of the parallelogram that passes through its center, i.e., the line through the midpoints of the longer sides (parallel to BC).",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 29458,
    "subject": "Mathematics (Multi-modal)",
    "question": "A family of sets $\\mathcal{F}$ is called *perfect* if the following condition holds: For every triple of sets $X_1, X_2, X_3 \\in \\mathcal{F}$, at least one of the sets\n$$\n(X_1 \\setminus X_2) \\cap X_3, \\quad (X_2 \\setminus X_1) \\cap X_3\n$$\nis empty. Show that if $\\mathcal{F}$ is a perfect family consisting of some subsets of a given finite set $U$, then $|\\mathcal{F}| \\le |U| + 1$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "We proceed by induction with respect to $|U|$. If $|U| = 0$, that is, $U = \\emptyset$, then there exists only one subset of $U$ and clearly $|\\mathcal{F}| \\le 1$.\nIn the following, let us suppose the statement is true for all sets of cardinality less than $k$ for a given $k > 0$. Let $U$ be any set with $|U| = k$ and $\\mathcal{F}$ be a perfect family of its subsets. We shall show that $|\\mathcal{F}| \\le |U| + 1$.\nIf $|\\mathcal{F}| \\le 1$, the claim is obviously true. If $|\\mathcal{F}| \\ge 2$, consider all the pairs of distinct sets from $\\mathcal{F}$. Since the number of such pairs is finite and nonzero, there is a pair $(Y, Z) \\in \\mathcal{F}^2$, $Y \\ne Z$, with intersection of maximum cardinality, that is, $|Y \\cap Z| = m$ and the intersection of any two distinct sets from $\\mathcal{F}$ has at most $m$ elements.\nSince the sets $Y$ and $Z$ are distinct, at least one of them must contain an element which is not contained in the other. Without the loss of generality let $Y \\setminus Z$ be nonempty and take any element $y \\in Y \\setminus Z$.\nIn the case $Y$ is the only set containing $y$, all the sets from the system $\\mathcal{F}' = \\mathcal{F} \\setminus \\{Y\\}$ are subsets of $U' = U \\setminus \\{y\\}$. Clearly $\\mathcal{F}'$, as a subsystem of a perfect system, is perfect as well. Applying the induction hypothesis on $U'$ and $\\mathcal{F}'$, we get\n$$\n|\\mathcal{F}| = |\\mathcal{F}'| + 1 \\le (|U'| + 1) + 1 = |U| + 1,\n$$\nand we are done.\n\n![](attached_image_1.png)\nFig. 2\n\nIn the other case, there is at least one set $W \\in \\mathcal{F}$ with $y \\in W$, $W \\ne Y$ (Fig. 2). Due to the choice of the pair $(Y, Z)$, the set $W$ can not contain the whole intersection $Y \\cap Z$ (otherwise we would have $|Y \\cap W| \\ge m + 1$). Let $z \\in (Y \\cap Z) \\setminus W$. We have\n$$\ny \\in (W \\setminus Z) \\cap Y \\quad \\text{and} \\quad z \\in (Z \\setminus W) \\cap Y,\n$$\nwhich is in contradiction with the property of the perfect system for $X_1 = Z, X_2 = W$, and $X_3 = Y$. So this case is not possible and the induction step is concluded.\nAgain, we proceed by induction with respect to $|U|$, with the claim being trivial when $U = \\emptyset$. Suppose $\\mathcal{F}$ is a perfect family of subsets of a finite set $U$, and let $Z$ be a member of $\\mathcal{F}$ of the minimum cardinality among the nonempty ones (if there is no such set, then $|\\mathcal{F}| \\le 1$ and we are done).\nThe first observation is that if $Y_1$ and $Y_2$ are two nonempty members of $\\mathcal{F}$ that satisfy $Y_1 \\setminus Z = Y_2 \\setminus Z$ (possibly $Y_1 = Z$ or $Y_2 = Z$), then in fact $Y_1 = Y_2$. Suppose otherwise that there exist two such nonempty sets $Y_1, Y_2 \\in \\mathcal{F}$ with $Y_1 \\neq Y_2$. Without the loss of generality, suppose that there exists an element $x_1 \\in Y_1 \\setminus Y_2$. As $Y_1 \\setminus Y_2 \\subseteq Z$, we have that $x_1 \\in Z \\setminus Y_2$ (Fig. 3). Since $Z$ is of minimum cardinality and $Y_2$ is nonempty, we have that $|Z| \\le |Y_2|$. As $Z \\not\\subseteq Y_2$ (due to $x_1 \\in Z \\setminus Y_2$), we infer that also $Y_2 \\setminus Z$ is nonempty, and hence there exists an element $x_2 \\in Y_2 \\setminus Z = Y_1 \\setminus Z$. Now observe that\n$$\nx_1 \\in Z \\setminus Y_2, \\quad x_2 \\in Y_2 \\setminus Z, \\quad \\text{and} \\quad \\{x_1, x_2\\} \\subseteq Y_1.\n$$\nThis contradicts the definition of a perfect family for $X_1 = Z$, $X_2 = Y_2$ and $X_3 = Y_1$.\n\n![](attached_image_2.png)\nFig. 3\n\nDefine a family $\\mathcal{F}'$ of subsets of $U \\setminus Z$ as follows:\n$$\n\\mathcal{F}' = \\{Y \\setminus Z : Y \\in \\mathcal{F}, Y \\neq \\emptyset\\}.\n$$\nClearly, $\\mathcal{F}'$ is a perfect family of subsets of a strictly smaller set, so from the induction hypothesis we infer that $|\\mathcal{F}'| \\le |U \\setminus Z| + 1$. Moreover, from the observation of the previous paragraph we infer that sets $Y \\setminus Z$ are pairwise different for all $Y \\in \\mathcal{F}$ with $Y \\neq \\emptyset$, and hence $|\\mathcal{F}| \\le |\\mathcal{F}'| + 1$ (the additive +1 comes from possibly having the empty set in $\\mathcal{F}$). Concluding,\n$$\n|\\mathcal{F}| \\le |\\mathcal{F}'| + 1 \\le |U \\setminus Z| + 1 + 1 \\le |U| - 1 + 1 + 1 = |U| + 1.\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 29459,
    "subject": "Mathematics (Multi-modal)",
    "question": "Real numbers $x, y, z$ satisfy\n$$\n\\frac{1}{x} + \\frac{1}{y} + \\frac{1}{z} + x + y + z = 0\n$$\nand none of them lies in the open interval $(-1, 1)$. Find the maximum value of $x+y+z$.",
    "options": [],
    "answer": "sqrt(3)",
    "solution": "**Solution:** By changing $(x, y, z)$ to $(-x, -y, -z)$, the condition (1) remains valid and the value of $x+y+z$ changes sign. Since the ordering of the numbers $x, y, z$ is irrelevant and they can not be of the same sign because of (1), we can without the loss of generality assume $x > 0, y > 0$, and $z < 0$, and find the maximum value of $V = |x + y + z|$.\nWe can transform (1) equivalently to\n$$\nf(x) + f(y) = f(t), \\quad \\text{where} \\quad t := -z > 0 \\quad \\text{and} \\quad f(x) = x + \\frac{1}{x},\n$$\nwith $x, y, t \\in I := (1, \\infty)$. In the following, we shall use the well known fact about the function $f$: It is an increasing bijection mapping $I$ onto $(2, \\infty)$ (the trivial proof is omitted here). Applying this, from\n$$\nf(t) = f(x) + f(y) = x + \\frac{1}{x} + y + \\frac{1}{y} > x + y + \\frac{1}{x+y} = f(x+y)\n$$\nwe get $t > x + y$, hence $x + y + z = x + y - t < 0$. Therefore our aim is to maximize the positive expression $V = |x + y + z| = t - x - y$.\nWe will call valid every triple $(x, y, t) \\in I^3$ satisfying $f(x) + f(y) = f(t)$. One of the valid triples is $(1, 1, t_0)$, with $t_0 > 1$ such that $f(t_0) = 4$, i. e., $t_0 = 2 + \\sqrt{3}$. For this triple, we have $V = \\sqrt{3}$. We shall show this is the desired maximum. The proof will be based on the following lemma: Whenever $(x, y, t)$ and $(x, y', t')$ are valid triples with the same first component $x$, and $y' < y$, we have $t - x - y < t' - x - y'$. If this lemma is true, then, with respect to the symmetry, the similar conclusion holds for the valid triples $(x, y, t)$ and $(x', y, t')$ with the same second component $y$. Hence, any valid triple $(x, y, t)$ can be replaced by $(x, 1, t')$ and then by $(1, 1, t_0)$, and the value of $V$ increases or remains the same during this process, concluding $\\max V = \\sqrt{3}$. (Moreover, this also implies that $(1, 1, 2+\\sqrt{3})$ is the only valid triple for which the maximum value $V = \\sqrt{3}$ is reached.)\nTo prove the lemma, notice that from\n$$\nf(t) = f(x) + f(y), \\quad f(t') = f(x) + f(y'),\n$$\nand from $f(y') < f(y)$ (which is implied by the given assumption $y' < y$), we have $f(x) < f(t') < f(t)$, hence $1 < t' < t$. Therefore $1 < t'y' < ty$, and from\n$$\nf(x) = f(t) - f(y) = (t - y) \\left( 1 - \\frac{1}{ty} \\right) = f(t') - f(y') = (t' - y') \\left( 1 - \\frac{1}{t'y'} \\right),\n$$\nusing the estimates\n$$\n0 < 1 - \\frac{1}{t'y'} < 1 - \\frac{1}{ty},\n$$\nwe obtain $t - y < t' - y'$, which is equivalent to the desired $t - x - y < t' - x - y'$. Answer. The maximum value of $x + y + z$ is $\\sqrt{3}$.\n**Second solution.** We start in the same way as in the previous solution, reducing to the case $x \\ge 1$, $y \\ge 1$, and $z \\le -1$ with maximizing $V = |x + y + z|$. From (1) we can express $z$ as the solution of the quadratic equation\n$$\nz^2 + \\left(x + \\frac{1}{x} + y + \\frac{1}{y}\\right) z + 1 = 0.\n$$\nSince the product of the roots of this equation equals 1, the value of $z$ lying outside of $(-1, 1)$ is, with respect to $x \\ge 1$, $y \\ge 1$, given by\n$$\nz = \\frac{1}{2} \\left( - \\left( x + \\frac{1}{x} + y + \\frac{1}{y} \\right) - \\sqrt{ \\left( x + \\frac{1}{x} + y + \\frac{1}{y} \\right)^2 - 4 } \\right).\n$$\nTherefore we are left to maximize\n$$\n2V = 2|-x-y-z| = \\left| \\sqrt{ \\left( x + \\frac{1}{x} + y + \\frac{1}{y} \\right)^2 - 4 } - \\left( x - \\frac{1}{x} + y - \\frac{1}{y} \\right) \\right|.\n$$\nOne can easily check that the expression inside of the last bars is positive. It remains\nto show that\n$$\n\\sqrt{\\left(x + \\frac{1}{x} + y + \\frac{1}{y}\\right)^2 - 4} - \\left(x - \\frac{1}{x} + y - \\frac{1}{y}\\right) \\le 2\\sqrt{3}.\n$$\nThis inequality is equivalent (after rearranging, checking the positivity of both sides, canceling the square root, and simple calculations) to\n$$\n\\sqrt{\\left(x + \\frac{1}{x} + y + \\frac{1}{y}\\right)^2 - 4} \\le 2\\sqrt{3} + \\left(x - \\frac{1}{x} + y - \\frac{1}{y}\\right), \\\\\nx^2 + y^2 + \\sqrt{3}x + \\sqrt{3}y \\le 2xy + \\sqrt{3}x^2y + \\sqrt{3}xy^2,\n$$\nwhich is true, since it is the sum of the inequalities\n$$\nx^2 \\le x^2 y, \\quad y^2 \\le xy^2, \\quad \\sqrt{3}x \\le (\\sqrt{3}-1)x^2y + xy, \\quad \\text{and} \\quad \\sqrt{3}y \\le (\\sqrt{3}-1)xy^2 + xy,\n$$\nand these are trivially valid for $x, y \\ge 1$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 29460,
    "subject": "Mathematics (Multi-modal)",
    "question": "For positive real numbers $a$, $b$, $c$ the following holds:\n$$ab + bc + ca = 16, \\quad a \\ge 3.$$\nFind the smallest possible value of the expression $2a + b + c$. (Michal Rolínek)",
    "options": [],
    "answer": "10",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 29461,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all integers $k \\ge 2$ for which there exists $k$-element set $M$ of positive integers such that the product of all numbers in $M$ is divisible by the sum of any two (different) numbers from $M$.",
    "options": [],
    "answer": "all integers k ≥ 2",
    "solution": "",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 29462,
    "subject": "Mathematics (Multi-modal)",
    "question": "Determine all pairs $(p, q)$ of integers such that $p$ is an integer multiple of $q$ and quadratic equation $x^2 + px + q = 0$ has at least one integer root.",
    "options": [],
    "answer": "All pairs (p, q) are: (i) p = 0 and q = −t^2 for some integer t (including t = 0), and (ii) (p, q) ∈ {(2, 1), (−2, 1), (4, 4), (−4, 4)}.",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 29463,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABC$ be a right-angled triangle with a hypotenuse $AB$ and longer leg $BC$. Let $D$ be a foot of an altitude from the vertex $C$. Circle $k$ with the center $D$ and the radius $CD$ intersects the leg $BC$ in a point $Q$ and line $AB$ in points $E$ and $F$ ($E \\neq F$), where $F$ is a point on the hypotenuse $AB$. Segment $QE$ intersects the leg $AC$ in a point $P$. Prove that $PE = QF$. (Jaroslav Švrček)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "The circle $k$ is the Thales' circle with the diameter $EF$ and the center $D$. A triangle $EFC$ is the isosceles right-angled triangle, so $EC = EF$. We will show that triangles $EPC$ and $FQC$ are congruent, which will prove the statement of the problem.\n\n![](attached_image_1.png)\nFig. 2\n\nAngles $CEQ$ and $CFQ$ are congruent as they are inscribed angles subtended by the chord $CQ$ of the circle $k$. Both angles $ECF$ and $ACB$ are congruent (right angles), so their remaining non-overlapping parts (angles $ECF = ECP$ and $ACB = FCQ$) are also congruent. This proves, that triangles $EPC$ and $FQC$ are congruent by $A$-$S$-$A$.",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 29464,
    "subject": "Mathematics (Multi-modal)",
    "question": "For a trapezoid $ABCD$ ($AB \\parallel CD$) it holds $BC = AB + CD$. Prove that\n(i) there is a point of a circle with diameter $BC$ on the leg $AD$,\n(ii) there is a point of a circle with diameter $AD$ on the leg $BC$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "(i) Let $M$, $N$ be the centers of the legs $BC$, $AD$. We show that the point $N$ lies on the circle with diameter $BC$.\nA well-known identity yields\n$$\nMN = \\frac{AB + CD}{2} = \\frac{1}{2} BC.\n$$\nIt means that the point $N$ has the same distance from the center $M$ of the circle with diameter $BC$ as radius of that circle. So point $N$ lies on that circle.\n\n(ii) With respect to the given condition we can find a point $E$ on the leg $BC$ such that $|BE| = |AB|$ and $|EC| = |CD|$.\n\n![](attached_image_1.png)\n\nThe triangles $ABE$, $ECD$ are isosceles and the lines $AB$ and $CD$ are parallel, thus the fact follows:\n$$\n\\begin{align*}\n\\angle AED &= 180^\\circ - \\angle AEB - \\angle CED = \\\\\n&= \\frac{1}{2}((180^\\circ - 2\\angle AEB) + (180^\\circ - 2\\angle CED)) = \\\\\n&= \\frac{1}{2}(\\angle ABE + \\angle DCE) = 90^\\circ.\n\\end{align*}\n$$\nSo we finished the second part.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 29465,
    "subject": "Mathematics (Multi-modal)",
    "question": "In any triangle $ABC$, in which the median from $C$ is not perpendicular to any of the sides $CA$ nor $CB$, let us denote $X$ and $Y$ intersections of this median's axis with lines $CA$ and $CB$. Find all such triangles $ABC$ for which points $A, B, X, Y$ lie on the same circle. (Ján Mazák)",
    "options": [],
    "answer": "All triangles with CA = CB (i.e., triangles isosceles at C).",
    "solution": "",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 29466,
    "subject": "Mathematics (Multi-modal)",
    "question": "In the field of real numbers solve a system of equations\n$$\n\\begin{align*}\na(b^2 + c) &= c(c + ab), \\\\\nb(c^2 + a) &= a(a + bc), \\\\\nc(a^2 + b) &= b(b + ca).\n\\end{align*}\n$$",
    "options": [],
    "answer": "All real triples with a=b=c (any real value).",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 29467,
    "subject": "Mathematics (Multi-modal)",
    "question": "We are given $n$ points in a plane, $n \\ge 3$, no three of them collinear. Consider all the interior angles of all triangles with vertices in given points and denote $\\phi$ the size of the smallest angle. For given $n$ find the largest possible $\\phi$.",
    "options": [],
    "answer": "180/n",
    "solution": "",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 29468,
    "subject": "Mathematics (Multi-modal)",
    "question": "Suppose that the real numbers $x, y, z$ satisfy equalities\n$$\n15(x + y + z) = 12(xy + yz + zx) = 10(x^2 + y^2 + z^2)\n$$\nand that at least one of them is different from zero.\na) Prove that $x + y + z = 4$.\nb) Find the smallest interval $(a, b)$, which contains all three numbers from any triplet $(x, y, z)$ satisfying the given conditions.",
    "options": [],
    "answer": "x + y + z = 4; the smallest interval is [2/3, 2].",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 29469,
    "subject": "Mathematics (Multi-modal)",
    "question": "A strange calculator has only two buttons with positive integers, each consisting of two digits. It displays the number $1$ at the beginning. Whenever a button with number $N$ is pressed, the calculator replaces the displayed number $X$ with the number $X \\cdot N$ or $X + N$. Multiplication and addition alternate, multiplication is the first. (For example, if the number $10$ is on the 1st button, the number $20$ is on the 2nd button, and we consecutively press the 1st, 2nd, 1st, and 1st button, we get the results $1 \\cdot 10 = 10$, $10 + 20 = 30$, $30 \\cdot 10 = 300$, and $300 + 10 = 310$.) Decide whether there exist particular values of the two-digit numbers on the buttons such that one can display infinitely many numbers ending with\n(a) $2015$,\n(b) $5813$.",
    "options": [],
    "answer": "(a) yes; (b) yes",
    "solution": "Let $a, b$ be the numbers written on the buttons. Consider the sequence $(x_n)_{n=0}^{\\infty}$ such that $x_{n+1}$ is formed by the last four digits of $a(x_n + b)$ for each $n \\ge 0$, that is,\n$$\nx_{n+1} \\equiv a(x_n + b) \\pmod{10\\,000} \\quad \\text{and} \\quad 0 \\le x_{n+1} < 10\\,000.\n$$\nSince there are only finitely many different possible values for $x_n$ and each term is only dependent on the value of the previous term, the sequence must be periodic, with the period starting in the term whose value first occurs for the second time in the sequence.\nIn general, the period does not have to start with $x_0$ (e.g., if we take $x_0 = 1$, $a = 10$, and $b = 10$, then all the terms except $x_0$ end with zero, hence the value of $x_0$ never occurs again). However, consider the special case when $a$ is coprime with $10\\,000$. We claim that then the period starts with $x_0$. In fact, suppose that $x_n$ is the first term of the sequence whose value occurs again, say $x_m = x_n$, $m > n$. If $n > 0$, the equality can be rewritten as $a(x_{n-1} + b) \\equiv a(x_{m-1} + b) \\pmod{10\\,000}$, that is,\n$$\n10\\,000 \\mid a(x_{n-1} + b) - a(x_{m-1} + b) = a(x_{n-1} - x_{m-1}).\n$$\nSince $a$ is coprime with $10\\,000$, we have $10\\,000 \\mid x_{n-1} - x_{m-1}$, implying $x_{n-1} = x_{m-1}$. But this is in contrary with the assumption $x_n$ was the first term whose value repeats.\nThe previous paragraph suggests the algorithm how to display the requested number infinitely many times: We only need to produce the number once, using buttons with $a$ coprime with $10\\,000$, and then repeat the sequence $+b, \\cdot a, +b, \\cdot a, \\dots$ forever. In case we get the requested number with even number of presses, i.e., ending with addition, repeating the sequence $\\cdot a, +b, \\cdot a, +b, \\dots$ will do the same desired effect.\nThere are many ways how to display $2015$ using only few presses. E.g., we can try to find $a, b$ such that $(1 \\cdot a + b) \\cdot a = 2015$. Since $2015 = 5 \\cdot 13 \\cdot 31$, we can take $a = 31$ and $b = 5 \\cdot 13 - a = 65 - 31 = 34$. Indeed, we will get\n$$\n1 \\xrightarrow{\\cdot 31} 31 \\xrightarrow{+34} 65 \\xrightarrow{\\cdot 31} 2015.\n$$\nNotice that $31$ is coprime with $10\\,000$. Therefore the (a) part is solved.\n\nIn the part (b), we analogously wish to generate the sequence $(x_n)$ described above with $x_0 = 5813$. However, it is not so easy to produce the initial occurrence of $5813$ using only a few presses. Therefore, we shall instead produce $5813 + b$, knowing that performing the sequence $\\cdot a, +b, \\cdot a, \\dots$ afterwards will eventually reach $5813$. One possible way is to find $a, b$ according to the schema\n$$\n1 \\xrightarrow{\\cdot b} b \\xrightarrow{+b} 2b \\xrightarrow{\\cdot a} 2ab \\xrightarrow{+a} 2ab + a = 5813 + b.\n$$\nThe equation can be rewritten as $(2a-1)(2b+1) = 11625 = 3 \\cdot 5^3 \\cdot 31$. From there, we can easily deduce several 2-digit solutions, one of which is $a = 47, b = 62$. Since $47$ is coprime with $10\\,000$, the procedure\n$$\n1 \\xrightarrow{\\cdot 62} 62 \\xrightarrow{+62} 124 \\xrightarrow{\\cdot 47} 5828 \\xrightarrow{+47} 5875 \\xrightarrow{\\cdot 47} - \\xrightarrow{+62} - \\xrightarrow{\\cdot 47} - \\xrightarrow{+62} \\dots\n$$\nwill reach numbers ending with $5875 - 62 = 5813$ infinitely many times.\nOne can construct a sequence in which every 4-digit number occurs infinitely many times. Consider the calculator with $a = 11$ and $b = 12$. Applying the same arguments as above, we deduce that if we keep pressing the first button (with number $11$ on it, which is coprime with $10\\,000$), after a finite even number of presses we obtain a number ending with $0001$ again. If we change the button in the very last operation, that is, we replace $+11$ with $+12$, we get a number ending with $0002$. In the same way, we can always increment the number formed by the last 4 digits by $1$ (or change $9999$ to $0000$). The conclusion follows.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 29470,
    "subject": "Mathematics (Multi-modal)",
    "question": "Nela and Jane choose positive integer $k$ and then play a game with a $9 \\times 9$ table. Nela selects in every of her moves one empty unit square and she writes $0$ to it. Jane writes $1$ to some empty (unit) square in every her move. Furthermore, $k$ Jane's moves follow each Nela's move and Nela starts. If sum of numbers in each row and each column is odd anytime during the game, Jane wins. If girls fill out the whole table (without Jane's win), Nela wins. Find the least $k$ such that Jane has the winning strategy. (Michal Rolínek)",
    "options": [],
    "answer": "3",
    "solution": "Let us show at first that Jane wins for $k = 3$. Let us assume $3 \\times 3$ squares $A_1$, $A_2$ and $A_3$ (see the picture).\n\nWe will call the $3 \\times 3$ square *covered* if just one $1$ is in each its row and column. If Jane covers squares $A_1$, $A_2$ and $A_3$ without writing to other squares, she wins, because sums in all rows and columns are odd number $1$.\n\n![](attached_image_1.png)\n\nFig. 3\n\nIt is obvious that if at most one $0$ (and no $1$) is written in any $3 \\times 3$ square after Nela's move, Jane can cover this square because of $k = 3$. Jane has the following strategy: If Nela writes $0$ to any uncovered square $A_1$, $A_2$ or $A_3$, Jane covers it immediately. In the opposite case Jane covers any of the uncovered $3 \\times 3$ squares. Jane thus wins after three her triples of moves.\n\nWe will show that Nela has winning strategy for $k \\in \\{1, 2\\}$. Let us realize that if Jane has some winning move, just $8$ rows and $8$ columns have odd sum before Jane's move, and the winning Jane's move is writing $1$ to the intersection of the only one \"even\" row with the only one \"even\" column. This implies that if Jane has winning move, this move is unique.\n\nNow it is obvious Nela's winning strategy for $k = 1$. If Jane has winning move after her move, Nela writes $0$ to this square and Jane loses her unique chance for win. In the opposite case Nela writes $0$ to some empty square. This move doesn't change parity of sums in rows and columns and Jane still hasn't winning move. This strategy allows Nela to fill out whole table without giving Jane chance to win.\n\nIn the case $k = 2$ Nela will use the same strategy as for $k = 1$. This strategy doesn't give Jane chance to win in her first move. In the second move Jane can't win because after that move the table contains even number of $1$'s which excludes possibility to be odd number $1$'s in every of (odd number) nine rows.\n\n**Conclusion.** The least value $k$, for which Jane has winning strategy, is $k = 3$.",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 29471,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABC$ be an acute triangle, which is not equilateral. Denote by $O$ and $H$ its circumcenter and orthocenter, respectively. The circle $k$ passes through $B$ and touches the line $AC$ at $A$. The circle $l$ with center on the ray $BH$ touches the line $AB$ at $A$. The circles $k$ and $l$ meet in $X$ ($X \\neq A$). Show that $\\angle HXO = 180^\\circ - \\angle BAC$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $E$ be the intersection point of the circle $l$ and the line $AC$ ($E \\neq A$). Since $k$ lies in the half-plane $ACB$ and $l$ lies in the half-plane $ABC$, the point $X$ lies inside of the angle $BAC$ (Fig. 4). Using the well known fact about the angle between a tangent and a chord of a given circle we get $\\angle XAE = \\angle XBA$ and $\\angle XAB = \\angle XEA$. Therefore the triangles $ABX$ and $EAX$ are similar.\n\nDenote by $\\gamma$ the measure of $\\angle ACB$. We have $\\angle AOB = 2\\gamma$, as $\\angle AOB$ is the central angle corresponding to the inscribed angle of measure $\\gamma$ in the circumcircle of the triangle $ABC$.\n\nThe line $BH$ passes through the centre of $l$ and is perpendicular to its chord $AE$, so it is in fact the axis of symmetry of the chord $AE$. Therefore $AH = HE$, and\n\nfrom the isosceles triangle $EAH$ with $AH \\perp BC$, we obtain $\\angle EAH = 90^\\circ - \\gamma$, hence $\\angle AHE = 2\\gamma$.\n\n![](attached_image_1.png)\n\nFig. 4\n\nThe triangles $ABO$ and $EAH$ are both isosceles, and have the vertex angle of the same measure, so they are similar. We have two pairs of similar triangles with the same ratio of similitude $AB : EA$. Since $O$ and $X$ lie on the same side of $AB$, and $H$ and $X$ lie on the same side of $EA$, the quadrilaterals$^1$ $ABXO$ and $EAXH$ are similar.\n\nConsider the rotation with centre $X$ which maps the ray $XB$ onto the ray $XA$. With respect to the derived similarity, the ray $XO$ is mapped onto the ray $XH$ under this rotation. Therefore we have\n\n$$\n\\angle HXO = \\angle AXB = 180^\\circ - \\angle BAC.\n$$\n\nThe last identity follows from the fact that $\\angle AXB$ is the inscribed angle corresponding to the chord $AB$ of $k$, while $\\angle CAB$ is the angle between this chord and its tangent, and both $X$ and $C$ lie in the same half-plane determined by this chord.\n\n$^1$ These quadrilaterals may be degenerate – three of the four vertices may be collinear.\n\n*Remark.* Instead of rotation, one may consider the spiral similarity which maps $ABXO$ onto $EAXH$. Since it maps $BA$ onto $AC$, it is clear that the rotating part of this map rotates all the lines by angle $180^\\circ - \\angle BAC$.\n(*Outline.*) It is possible to solve the problem using coordinates. Let $A$ be the origin and $B$ lies on the $x$-axis. Denote by $b$ the first coordinate of $B$ and let $(c, v)$ be the coordinates of $C$. Routine calculations give\n\n$$\nA = (0,0), \\quad B = (b,0), \\quad C = (c,v), \\quad H = \\left(c, \\frac{c(b-c)}{v}\\right), \\quad O = \\left(\\frac{1}{2}b, \\frac{c^2+v^2-bc}{2v}\\right), \\\\\nS_k = \\left(\\frac{1}{2}b, -\\frac{bc}{2v}\\right), \\quad S_l = \\left(0, \\frac{bc}{v}\\right), \\quad X = \\left(\\frac{6bc^2}{9c^2+v^2}, \\frac{2bcv}{9c^2+v^2}\\right), \\quad Y = \\left(\\frac{1}{2}b, \\frac{bc}{2v}\\right).\n$$\n\nHere, $S_k$ and $S_l$ are the centres of $k$ and $l$, respectively, and $Y$ is the intersection point of the line $BH$ and the line $OS_k$ (these two lines are perpendicular to $AC$ and $AB$, respectively, hence they form the angle of the same measure as $\\angle BAC$).\n\nInstead of expressing the measure of $\\angle HXO$, we will show that $X, O, H$, and $Y$ are concyclic, that is, the determinant\n\n$$\n\\left| \\begin{array}{cccc} \\frac{(6bc^2)^2 + (2bcv)^2}{(9c^2 + v^2)^2} & \\frac{6bc^2}{9c^2 + v^2} & \\frac{2bcv}{9c^2 + v^2} & 1 \\\\ \\frac{b^2}{4} + \\frac{(c^2 + v^2 - bc)^2}{4v^2} & \\frac{b}{2} & \\frac{c^2 + v^2 - bc}{2v} & 1 \\\\ c^2 + \\frac{c^2(b-c)^2}{v^2} & c & \\frac{c(b-c)}{v} & 1 \\\\ \\frac{b^2}{4} + \\frac{b^2c^2}{4v^2} & \\frac{b}{2} & \\frac{bc}{2v} & 1 \\end{array} \\right|\n$$\n\nevaluates to zero. In fact, this is an easy exercise, provided we know basic tricks from linear algebra (adding a scalar multiple of one row to another row does not change the value of the determinant; the same is valid for columns; when checking only zero value, we can multiply any row/column by a nonzero scalar).\n\nIt remains to show that among the two possible values, $\\angle BAC$ and $180^\\circ - \\angle BAC$, of an inscribed angle corresponding to the chord $HO$ of the circumcircle of the triangle $HOY$, the latter always apply for $\\angle HXO$. Also, the case $Y = O$ or $Y = H$ should be handled separately. One may use some kind of continuity arguments to show that we cannot \"jump\" from one value to another.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 29472,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABC$ be a triangle with the shortest side $BC$. Let $X$, $Y$, $K$, $L$ be points on sides $AB$, $AC$ and on rays opposite to rays $BC$, $CB$ respectively such that $BX = BK = BC = CY = CL$. Line $KX$ intersects line $LY$ in a point $M$. Prove that centroid of triangle $KLM$ coincides with incentre of triangle $ABC$.\n\n(Tomáš Jurík)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Since $ABC$ is an external angle of the isosceles triangle $XKB$ with the apex $B$ (see the picture), a line $KX$ is parallel to a bisectrix of the angle $ABC$.\n\n![](attached_image_1.png)\n\nFig. 4\n\nThe ratio $LB : LK = 2 : 3$ yields, that the bisectrix of $ABC$ meets the centroid of the triangle $KLM$. If we denote $LL_1$ its median and $L_2$ its intersection with the bisectrix of $ABC$ then we obtain\n$$\n\\frac{LL_2}{LL_1} = \\frac{LB}{LK} = \\frac{2}{3}.\n$$\nfrom similarity of the triangles $\\triangle LBL_2 \\sim \\triangle LKL_1$ (by A-A). So the point $L_2$ divides the median $LL_1$ in the same ratio as the centroid and therefore it is the centroid of the triangle $KLM$.\n\nIt follows from the symmetry of the problem, that bisectrix of the angle $BCA$ meets the centroid of the triangle $KLM$. And the fact, that intersection of the bisectrices is the incentre, proves the claim of the problem.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 29473,
    "subject": "Mathematics (Multi-modal)",
    "question": "A triangle $ABC$ is given every two sides of which differ in length by at least $d > 0$. Denote by $T$ its centroid, $I$ incentre and $\\rho$ inradius. Prove that\n$$\nS_{AIT} + S_{BIT} + S_{CIT} \\geq \\frac{2}{3} \\rho d,\n$$\nwhere $S_{XYZ}$ denotes the area of triangle $XYZ$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 29474,
    "subject": "Mathematics (Multi-modal)",
    "question": "In the triangle $ABC$ denote $D$ point of contact of side $BC$ with the incircle. The incircle of the triangle $ABD$ is tangent to sides $AB$ and $BD$ at points $K$ and $L$. The incircle of the triangle $ADC$ is tangent to sides $DC$ and $AC$ at points $M$ and $N$. Prove that points $K, L, M, N$ lie on the same circle.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 29475,
    "subject": "Mathematics (Multi-modal)",
    "question": "We are given a positive integer $n > 2$. Find the greatest of all the numbers $d$, satisfying the following condition: For any set of $n$ integers one can choose its three different subsets so that the sum of elements each of which is an integer multiple of $d$. (The selected subsets need not be disjoint.) (Jaromír Šimša)",
    "options": [],
    "answer": "n − 1",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 29476,
    "subject": "Mathematics (Multi-modal)",
    "question": "$1 \\cdot 2 \\cdot 3 \\cdot \\dots \\cdot n$\n*is written on a blackboard. For which positive integers* $n \\ge 2$ *can we append the exclamation mark to some factors and change it to factorials in such a way that the final product will be a square?*",
    "options": [],
    "answer": "All composite integers n ≥ 2",
    "solution": "Let us denote $v_p(n)$ the highest power of a prime $p$ which divides positive integer $n$. This function has obviously the following properties:\n- For all primes $p$ and positive integers $n$ is $v_p(n)$ non-negative integer.\n- For all positive integers $m, n$ and all primes $p$ is $v_p(mn) = v_p(m) + v_p(n)$.\n- For all primes $p$ is $v_p(p!) = v_p(p) = 1$.\n- For all primes $p$ is $v_p((p+1)!) = 1$, $v_p(p+1) = 0$.\n- For all primes $p$ and all positive integers $n < p$ is $v_p(n!) = v_p(n) = 0$.\n- Positive integer $n$ is a square if and only if $v_p(n)$ is even (including zero) for all primes $p$.\n\nLet us denote $S = n!$ the initial value of the product on the table and $S'$ its final value after adding factorials. We can easy to see from the properties of $v_p$ that for $n$ is equal to any prime $p$ we obtain $v_p(S) = v_p(p!) = 1$ and $v_p(S') = 1$, because adding factorials does not change the amount of the prime $p$ (= $n$) in the final product on the blackboard. The number $v_p(S')$ is then odd and therefore $S'$ is not a square.\n\nLet us assume that $n$ is a composite number (so $n \\ge 4$) in whole of the following part. We will show that we can add factorials in such a way that the final product\n$$\nS' = f_1 \\cdot f_2 \\cdot f_3 \\cdots f_n,\n$$\nwill be a square, where $f_k$ is either $k$ or $k!$ for all $k$. It is equivalent to $v_p(S')$ is even for all primes $p$. Since $n$ is not a prime, only primes less than $n$ occur in the product $S'$. As every such primes $p$ are not in factors $f_1, f_2, \\dots, f_{p-1}$ and the prime $p$ occurs in $f_p$ only once, the final power $v_p(S')$ is the same as in a “reduced” product\n$$\np \\cdot f_{p+1} \\cdot f_{p+2} \\cdots f_n. \\qquad (1)\n$$\nHow can we provide that every prime $p < n$ will occur in the corresponding product (1) with even power? Since in the second factor $f_{p+1}$ from (1) occurs the prime $p$ either once (in the case $f_{p+1} = (p+1)!$) or the prime $p$ does not occure (if $f_{p+1} = p+1$), we can provide “good” occurrence of $p$ by choice of $f_{p+1}$ independently on succeeding values $f_{p+2}, \\dots, f_n$.\n\nForegoing analysis gives us construction of the required choice of factorials. Initially we choose $f_k \\in \\{k, k!\\}$ arbitrarily for all $k \\le n$ such that $k-1$ is not a prime. The other $f_k$, it is $f_{p+1}$, where $p$ is arbitrary prime less than $n$, will be chosen “backwards”, from the biggest such prime $p$ to the smallest prime $p = 2$. For the biggest unchosen $f_{p+1}$ we find parity of $v_p(f_{p+2} \\dots f_n)$, in odd case we choose $f_{p+1} = p+1$, in even case we choose $f_{p+1} = (p+1)!$ and so on.\n\nThis finishes the construction of $S'$ and solution of the problem too.\n\nConclusion. Desired $n \\ge 2$ are all composite numbers.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 29477,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $a$, $b$ be relatively prime integers. Sequence $(x_n)_{n=1}^{\\infty}$ of natural numbers is constructed in such a way that for each $n > 1$ applies $x_n = a x_{n-1} + b$. Prove that in any such sequence every entry $x_n$ with index $n > 1$ divides infinitely many of other entries. Does this assertion hold for $n = 1$?",
    "options": [],
    "answer": "No",
    "solution": "",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 29478,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $n$ be an even positive integer. There are $n$ real numbers written on the blackboard. In every step, we choose two numbers, erase them, and replace each of them by their product. Show that for any initial $n$-tuple it is possible to obtain $n$ equal numbers on the blackboard after a finite number of steps.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "We shall prove the claim by induction with respect to $n$. The claim is trivial for $n = 2$ (we can get the desired 2-tuple after a single step $(a, b) \\rightarrow (ab, ab)$) and $n = 4$ (we can follow the scheme $(\\underline{a}, \\underline{b}, c, d) \\rightarrow (ab, ab, \\underline{c}, \\underline{d}) \\rightarrow (\\underline{ab}, ab, \\underline{cd}, cd) \\rightarrow (abcd, \\underline{ab}, abcd, \\underline{cd}) \\rightarrow (abcd, abcd, abcd, abcd)$).\n\nTo start the induction, we shall prove the claim also for $n = 6$. The algorithm begins with the 6-tuple $(a, a, a, a, b, b)$ – this form can be achieved thanks to the fact the claim is true for $n = 2$ and $n = 4$ (we can operate on the left 4-tuple and then independently on the right 2-tuple). To equalize all six numbers, perform the following steps:\n$$\n\\begin{aligned}\n(a, a, a, \\underline{a}, \\underline{b}, b) &\\rightarrow (a, a, \\underline{a}, \\underline{ab}, ab, b) \\rightarrow (a, a, a^2b, \\underline{a^2b}, \\underline{ab}, b) \\rightarrow (a, a, \\underline{a^2b}, a^3b^2, a^3b^2, \\underline{b}) \\rightarrow \\\\\n&\\rightarrow (\\underline{a}, \\underline{a}, \\underline{a^2b^2}, a^3b^2, a^3b^2, \\underline{a^2b^2}) \\rightarrow (a^3b^2, a^3b^2, a^3b^2, a^3b^2, a^3b^2, a^3b^2)\n\\end{aligned}\n$$\n\nNow, suppose the claim is true for all even $n < 4k + 4$ (with $k \\ge 1$). It suffices to prove the claim also for $n = 4k + 4$ and $n = 4k + 6$. The procedure for $n = 4k + 4$ is obvious: we first equalize the first $2k + 2$ numbers using the induction hypothesis, and then equalize the last $2k + 2$ numbers. We get an $n$-tuple of the form\n$$\n\\underbrace{(a, \\dots, a)}_{2k+2}, \\underbrace{(b, \\dots, b)}_{2k+2}\n$$\n\nand then perform $2k + 2$ steps to get $(ab, \\dots, ab)$ (choosing one $a$ and one $b$ in each step).\n\nFor $n = 4k+6$, we first use the induction hypothesis for $n = 2k+2$ and $n = 2k+4$ to get\n$$\n\\underbrace{(a, \\dots, a)}_{2k+2} \\underbrace{, b, \\dots, b)}_{2k+4}.\n$$\nNow we perform $2k$ steps, always choosing one $a$ and one $b$, to obtain\n$$\n(a, a, \\underbrace{ab, \\dots, ab)}_{4k}, b, b, b, b).\n$$\nAfter erasing each $a$ with one $ab$ we get\n$$\n(a^2b, a^2b, a^2b, a^2b, \\underbrace{ab, \\dots, ab)}_{4k-2}, b, b, b, b),\n$$\nand after pairing each $b$ with one $a^2b$ we have\n$$\n(a^2b^2, a^2b^2, a^2b^2, a^2b^2, \\underbrace{ab, \\dots, ab)}_{4k-2}, a^2b^2, a^2b^2, a^2b^2, a^2b^2).\n$$\nFinally, we perform $2k-1$ steps and replace $2k-1$ times two $ab$'s by two $a^2b^2$'s, which leads to $(a^2b^2, \\dots, a^2b^2)$.\nAgain, we shall proceed by induction. The claim is trivial for $n=2$. Suppose the claim is true for $n=k$ and take $n=k+2$. Using the induction hypothesis, we first construct an $n$-tuple of the form\n$$\n\\underbrace{(a, \\dots, a, b, b)}_{k}.\n$$\nFor every $i=3, 4, \\dots, k$, we perform the operation with the numbers at the positions $i$ and $k+1$. These $k-2$ steps change the $n$-tuple into\n$$\n(a, a, ab, a^2b, \\dots, a^{k-2}b, a^{k-2}b, b).\n$$\nAfter selecting the last two numbers, we get\n$$\n(a, a, ab, a^2b, \\dots, a^{k-2}b, a^{k-2}b^2, a^{k-2}b^2).\n$$\nNow, for every $i = 1, 2, \\dots, \\frac{1}{2}n$, we combine the numbers at the positions $i$ and $n+1-i$, getting the desired\n$$\n(a^{k-1}b^2, a^{k-1}b^2, \\dots, a^{k-1}b^2).\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 29479,
    "subject": "Mathematics (Multi-modal)",
    "question": "There are different positive integers written on the board. Their (arithmetic) mean is a decimal number, with the decimal part exactly $0{,}2016$. What is the least possible value of the mean? (Patrik Bak)",
    "options": [],
    "answer": "313.2016",
    "solution": "Let $s$ be the sum, $n$ the number and $p$ the integer part of the mean of the numbers on the board. Then we can write\n$$\n\\frac{s}{n} = p + \\frac{2{,}016}{10{,}000} = p + \\frac{126}{625},\n$$\nwhich gives\n$$\n625(s - pn) = 126n.\n$$\nNumbers $126$ and $625$ are coprime, thus $625 \\mid n$. Therefore $n \\ge 625$.\nThe numbers on the board are different, that is\n$$\np = \\frac{s}{n} - \\frac{126}{625} \\ge \\frac{1+2+\\cdots+n}{n} - \\frac{126}{625} = \\frac{n(n+1)}{2n} - \\frac{126}{625} = \\frac{n+1}{2} - \\frac{126}{625} \\ge \\frac{625+1}{2} - \\frac{126}{625} > 312.\n$$\nThe integer $p$ is thus at least $313$ and the value of the mean at least $313{,}2016$.\nThis value can be attained by numbers $1, 2, \\ldots, 624$ and $751$. We get\n$$\n\\frac{1+2+\\cdots+624+751}{625} = \\frac{312 \\cdot 625 + 751}{625} = 313 + \\frac{126}{625} = 313{,}2016.\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 29480,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find the smallest positive integer that can be inserted between numbers $20$ and $16$ so that the resulting number $20\\ldots16$ is a multiple of $2016$. (Radek Horenský)",
    "options": [],
    "answer": "36",
    "solution": "Number $2016$ is a multiple of $9$, therefore the digit sum of the resulting number has to be divisible by $9$. This happens if and only if we insert a number with digit sum divisible by $9$, that is a number which is a multiple of $9$. Trying out the smallest such numbers, we find out that inserting $9$, $18$, and $27$ doesn't work (instead of checking divisibility by $2016$ we can check that numbers $20916$, $201816$, and $202716$ are not divisible by $16$, looking at their final four digits). On the other hand, $203616 = 201600 + 2016$ is a multiple of $2016$. The answer is $36$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 29481,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all primes $p$ for which there exists a positive integer $n$ such that $p^n + 1$ is a cube of a positive integer. (Ján Mazák, Róbert Tóth)",
    "options": [],
    "answer": "7",
    "solution": "Suppose that positive integer $a$ satisfies $p^n + 1 = a^3$ (clearly $a \\ge 2$). We rewrite the equality as\n$$\np^n = a^3 - 1 = (a-1)(a^2 + a + 1).\n$$\nIt follows that if $a > 2$, the numbers $a-1$ and $a^2+a+1$ are powers of $p$ (with positive integer exponents).\nIf $a > 2$ then $a - 1 = p^k$, hence $a = p^k + 1$ for some positive integer $k$. Plugging this into $a^2 + a + 1$ gives $p^{2k} + 3p^k + 3$. Since $a - 1 = p^k < a^2 + a + 1$, the trinomial $a^2 + a + 1$ is a higher power of $p$, and thus\n$$\np^k \\mid p^{2k} + 3p^k + 3 \\Rightarrow p^k \\mid 3.\n$$\nThen $p=3$ and $k=1$, hence $a = p^k + 1 = 4$. However, number $a^2 + a + 1 = 21$ is not a power of three therefore if $a > 2$, the expression $p^n + 1$ is never a cube.\nFor $a=2$ we get $p^n = 7$, hence $p=7$ is the only such prime.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 29482,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $p > 3$ be a prime. Find the number of ordered sextuples $(a, b, c, d, e, f)$ of positive integers, whose sum is $3p$, and all the fractions\n$$\n\\frac{a+b}{c+d}, \\quad \\frac{b+c}{d+e}, \\quad \\frac{c+d}{e+f}, \\quad \\frac{d+e}{f+a}, \\quad \\frac{e+f}{a+b}\n$$\nare integers.",
    "options": [],
    "answer": "p + 2 if p ≡ 1 (mod 3); p + 1 if p ≡ 2 (mod 3)",
    "solution": "Taking the product of the 1st, the 3rd and the 5th fractions reveals that their value has to be 1, that is\n$$\na + b = c + d = e + f = p. \\tag{1}\n$$\nthe form of the second and of the fourth fraction implies\n$$\nf + a \\mid d + e \\quad \\text{and} \\quad d + e \\mid b + c. \\tag{2}\n$$\nthat is first $f + a$ is at most the arithmetic mean of its multiples,\n$$\nf + a \\le \\frac{1}{3}((f+a) + (d+e) + (b+c)) = p, \\tag{3}\n$$\nand\n$$\nf + a \\mid (f + a) + (d + e) + (b + c) = 3p.\n$$\nThus $f + a$ divides $3p$ and is in the interval $\\langle 2, p \\rangle$. Consequently either $f + a = p$ or $f + a = 3$. We deal separately with these cases.\n\ni) Let $f + a = p$. Because of (3) there is $f + a = d + e = b + c = p$, which together with (1) gives $p - 1$ solutions of the form\n$$\n(a, b, c, d, e, f) = (a, p-a, a, p-a, a, p-a), \\quad \\text{where } a \\in \\{1, 2, \\dots, p-1\\}.\n$$\n\nii) Let $f + a = 3$. Then $\\{a, f\\} = \\{1, 2\\}$.\n\nFirstly let $a = 1$ and $f = 2$. According to (1) then $b = p - 1$ and $e = p - 2$, and (2) has the form\n$$\n3 \\mid d + (p-2) \\quad \\text{and} \\quad d + (p-2) \\mid (p-1) + c. \\tag{4}\n$$\nIn analyzing (4) we distinguish between $d = 1$ and $d \\ge 2$.\n\nIf $d = 1$ then $c = p - 1$ and (4) reads\n$$\n3 \\mid p-1 \\quad \\text{a} \\quad p-1 \\mid 2(p-1).\n$$\nWhile the right relation always holds, the left one holds only for $p = 3q + 1$ ($q$ is a suitable positive integer). For such prime numbers we get considering (1) solutions\n$$\n(a, b, c, d, e, f) = (1, p-1, p-1, 1, p-2, 2).\n$$\nIf $d \\ge 2$ we show first, that the right relation in (4) is satisfied if and only if $d + (p-2) = (p-1) + c$ or $d = c + 1$. $d \\ge 2$ namely implies $c = p - d \\le p - 2$, thus\n$$\nd + (p - 2) \\ge p \\quad \\text{and} \\quad (p - 1) + c \\le 2p - 3 < 2p\n$$\nand $d + (p - 2) = (p - 1) + c$. From $c + d = p$ and $d = c + 1$ we get $c = \\frac{1}{2}(p - 1)$\na $d = \\frac{1}{2}(p + 1)$. Since $d + (p - 2) = \\frac{3}{2}(p - 1)$, the left relation in (4) is fulfilled and\n$$\n(a, b, c, d, e, f) = (1, p-1, \\frac{1}{2}(p-1), \\frac{1}{2}(p+1), p-2, 2).\n$$\nis a solution.\n\nFinally $a = 2$ and $f = 1$. In this case $b = p - 2$ a $e = p - 1$, and (2) reads\n$$\n3 \\mid d + (p-1) \\quad \\text{and} \\quad d + (p-1) \\mid (p-2) + c. \\qquad (5)\n$$\nBecause\n$$\nd + (p - 1) \\ge p \\quad \\text{and} \\quad (p - 2) + c < 2p,\n$$\nthe right relation in (5) holds if and only if $d + (p-1) = (p-2) + c$, that is iff $c = d+1$. Together with $c+d=p$ we get $c = \\frac{1}{2}(p+1)$ and $d = \\frac{1}{2}(p-1)$, thus the right relation in (5) holds as well, and the last solution is\n$$\n(a, b, c, d, e, f) = (2, p-2, \\frac{1}{2}(p+1), \\frac{1}{2}(p-1), p-1, 1).\n$$\n\n**Conclusion.** All the solutions found are apparently mutually different and their number depends on $p$ modulo $3$ ($p > 3$): If $p = 3q + 1$ then there are $p + 2$ sextuples, if $p = 3q + 2$, there are $p + 1$ sextuples.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 29483,
    "subject": "Mathematics (Multi-modal)",
    "question": "On the unit square $ABCD$ is given point $E$ on $CD$ in such a way, that $|\\angle BAE| = 60^\\circ$. Further let $X$ be an arbitrary inner point of the segment $AE$. Finally let $Y$ be the intersection of a line, perpendicular to $BX$ and containing $X$, with the line $BC$. What is the least possible length of $BY$? (Michal Rolínek)",
    "options": [],
    "answer": "2√3/3",
    "solution": "Let us consider the Thales circle over $BY$, which is circumscribed to $BYX$. This circle contains $X$ and touches $AB$ in $B$. Of all such circles, the one which touches $AE$ (and it has to be in $X$) obviously has the least diameter (let us call the circle $k$). This circle is thus inscribed to the equilateral triangle $AA'F$ where $A'$ is the image of $A$ in the point symmetry with respect to $B$ and $F$ lies on the half line $BC$ (see Fig. 1; there you can see one of the circles with smaller diameter than $k$ as well). The center of $k$ is the center of mass of the triangle $AA'F$, equilateral triangle with sides of length $2$, thus the diameter of $k$ is $BY = \\frac{2}{3}\\sqrt{3}$ and the corresponding $X$ is a center of $AF$, that is it belongs to the segment $AE$, since $|AX| = 1 < |AE|$.\n\nAnswer. The least possible length of $BY$ is $\\frac{2}{3}\\sqrt{3}$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 29484,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $r$ and $r_a$ be the radii of the inscribed circle and the excircle opposite $A$ of the triangle $ABC$. Show that if\n$$\nr + r_a = |BC|,\n$$\nthen the triangle is right-angled.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let us use the standard notation of the inner angles of the triangle $ABC$, further let $I$ be the incenter and $I_a$ be the excenter (of the excircle opposite $A$), and let $D$ and $E$ be in order the touching points of the thought circles. Since the bisectors $BI$ and $BI_a$ of the supplementary angles are perpendicular to each other (as well as $CI$ and $CI_a$), the points $B$, $C$, $I$, and $I_a$ lie on the circle with the diameter $II_a$.\n\nThus $D$ and $E$, the orthogonal projections of $I$ and $I_a$ onto the secant $BC$, are point reflections of each other with respect to the center of $BC$.\n\nThe right triangles $BID$ and $I_aBE$ are obviously similar and\n$$\n|BD| : |ID| = |I_aE| : |BE| \\quad \\text{or} \\quad |BD| \\cdot |BE| = |ID| \\cdot |I_aE|\n$$\nconsidering the mentioned point reflection also\n$$\n|BD| + |BE| = |BD| + |CD| = |BC| = r + r_a = |ID| + |I_aE|.\n$$\n![](attached_image_1.png)\nFig. 1\n\nThe two equations imply that the pairs ($|ID|$, $|EI_a|$) and ($|BD|$, $|BE|$) are roots of the same quadratic equation, that is $|ID| = |BD|$ or $|ID| = |BE|$.\n\n$|ID| = |BD|$ means the right-angled triangle $BID$ is isosceles, which is $\\beta = 90^\\circ$.\n\nSimilarly, if $|ID| = |BE|$ that is $|ID| = |CD|$ ($D$ and $E$ are point reflections in the mentioned reflection) means the right triangle $CID$ is isosceles, that is $\\gamma = 90^\\circ$.\n\nIn both cases the triangle $ABC$ is right-angled.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 29485,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all positive integers $n$ with the following property: Numbers $1, 2, \\ldots, n$ can be split into three disjoint non-empty subsets with mutually different sizes such that, for any pair of subsets, the subset with fewer elements has larger sum of its elements.\n(A size of a subset is the number of its elements.)",
    "options": [],
    "answer": "n = 9 or n ≥ 12",
    "solution": "We first exclude small values of $n$.\nThe three subsets have to have in total at least $1 + 2 + 3 = 6$ elements, hence $n \\ge 6$.\nFor $n=6$, the smallest subset (in size) contains a single number and its sum is therefore at most $6$. The sum of the remaining numbers is at least $15$, hence the sum of at least one of the two remaining subsets is at least $8$, a contradiction. Similarly we exclude $n=7$ and $n=8$ where, again, the smallest subset contains a single number.\nFor $n=9$ the subsets $\\{9, 8\\}$, $\\{7, 6, 3\\}$, $\\{5, 4, 2, 1\\}$ satisfy the conditions.\nFor $n=10$, the smallest subset contains at most $2$ numbers and its sum is therefore at most $19$. The sum of the remaining numbers is at least $36$ and we reach contradiction as above. Similarly we exclude $n=11$.\nNow we describe how to create the subsets for any $n \\ge 12$. Let $n = 3k+r$, $k \\ge 4$, $r \\in \\{0, 1, 2\\}$. We fix the sizes of the subsets to $k-1$, $k$, $k+r+1$ and define the subsets as follows: Set $M_1$ consists of $k-1$ largest numbers from $\\{1, 2, \\dots, n\\}$, that is numbers $n-k+2, \\dots, n$. Set $M_2$ consists of the $k$ next largest numbers (that is, numbers $n-2k+2, \\dots, n-k+1$). The remaining numbers $1, \\dots, k+r+1$ form the subset $M_3$.\nWe aim to show that $M_1$ has larger sum than $M_2$. Since all the numbers from $M_1$ are greater than all the numbers from $M_2$, and $M_2$ has just one more element, it suffices to show that the sum of the three largest numbers of $M_1$ is greater than the sum of the four smallest numbers of $M_2$. Such three numbers always exist ($k \\ge 4$).\n\nand the desired inequality $n + (n-1) + (n-2) > (n-2k+2) + (n-2k+3) + (n-2k+4) + (n-2k+5)$ is equivalent to $8k-n > 17$ which, after plugging in $n = 3k+r$ rewrites as $5k > 17+r$. This is true as $5k \\ge 20$ and $17+r \\le 19$.\nSimilarly, we show that $M_2$ has larger sum than $M_3$, which has $r+1$ more elements than $M_2$: Two largest numbers of $M_2$ have larger sum than $r+3$ smallest numbers of $M_3$ as $(n-k+1) + (n-k) = 2(3k+r-k) + 1 = 4k+2r+1 \\ge 17 > 15 = 1+2+3+4+5$.\nThe answer is all $n \\ge 12$ together with $n = 9$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 29486,
    "subject": "Mathematics (Multi-modal)",
    "question": "We have $n^2$ empty boxes; each of them having square base. The height and the width of each box belongs to $\\{1, 2, \\dots, n\\}$ and every two boxes differ in at least one of these two dimensions. One box fits into another one if both its dimensions are smaller and at least one is smaller by at least 2. In this way, we can form sequences of boxes (the first one in the second one, the second one in the third one, and so on). We put any such set of boxes on a different shelf. How many shelves do we need to store all the boxes? (Peter Novotný)",
    "options": [],
    "answer": "3n - 2",
    "solution": "We will show that the required minimal number of shelves is $3n-2$. This answer is clearly correct for $n=1$ and $n=2$ since for such $n$ we have $n^2 = 3n - 2$ and each box has to be stored on a different shelf. From now on, let $n \\ge 3$.\nWe identify boxes with points in an $n \\times n$ grid: A box of width $w$ and height $h$ corresponds to point with coordinates $(w, h)$.\nNo two boxes from the set $S = \\{(w,h): n \\le w+h \\le n+2\\}$ (three longest diagonals, see Fig. 1 for $n=7$) can be on the same shelf: Indeed, Assume $(w,h)$ and $(w',h')$ are on the same shelf and $w < w', h < h'$. Then $w+1 \\le w', h+1 \\le h'$, and either $w+2 \\le w'$ or $h+2 \\le h'$. Either way, summing up we obtain $w+h+3 \\le w'+h'$.\n\n![](attached_image_1.png)\nFig. 1\n![](attached_image_2.png)\nFig. 2\nIf $(w, h) \\in S$ then $w' + h' \\ge w + h + 3 \\ge n + 3$, hence $(w', h') \\notin S$. As $|S| = 3n - 2$, we need at least $3n - 2$ shelves.\nNow we show how to split the boxes so that $3n-2$ shelves are enough. A possible way for $n=7$ is clear from Fig. 2 (the sequences of boxes stored inside one another are illustrated by arrows) and it can be directly generalised to any $n \\ge 3$. Two boxes $(w, h)$, $(w', h')$ are put on the same shelf if and only if $2(h' - h) = w' - w$. In other words, we start with “largest” boxes $(n-1, h)$, $(n, h)$ for $h = 1, 2, \\dots, n$ and $(w, n)$ for $w = 1, 2, \\dots, n-2$, and we put each of these $3n-2$ boxes on a different shelf. Then we follow the following algorithm on each shelf: Assume the last box put on the shelf is $(w, h)$. If $w-2 \\ge 1$ and $h-1 \\ge 1$, we add box $(w-2, h-1)$ to the shelf (inside the previous box) and repeat this step; otherwise we end. Obviously, the arrangement of boxes on every shelf satisfies the requirements. Moreover, each box is stored on exactly one shelf: To identify it, keep increasing $w$ by 2 and $h$ by 1 (simultaneously) until $w \\in \\{n-1, n\\}$ or $h = n$. Thus $3n-2$ shelves are also sufficient.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 29487,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABC$ be an acute triangle with altitudes $BD$, $CE$. Given that $AE \\cdot AD = BE \\cdot CD$, what is the smallest possible measure of $\\angle BAC$? (Patrik Bak)",
    "options": [],
    "answer": "60°",
    "solution": "We denote $\\angle BAC$ by $\\alpha$ and express lengths $AE$, $AD$, $BE$, $CD$ in terms of $\\cos \\alpha$ and the side lengths $b = AC$, $c = AB$ of triangle $ABC$. The condition rewrites as\n\n$$\nb \\cos \\alpha \\cdot c \\cos \\alpha = (c - b \\cos \\alpha)(b - c \\cos \\alpha),\n$$\nwhich simplifies to $bc = (b^2 + c^2) \\cos \\alpha$. Hence\n$$\n\\cos \\alpha = \\frac{bc}{b^2 + c^2} \\le \\frac{1}{2},\n$$\nwhere the last inequality is for any positive $b$, $c$ equivalent with an obvious inequality $(b-c)^2 \\ge 0$ (alternatively, one can use AM-GM inequality for $b^2$ and $c^2$). We proved that angle $BAC$ of any such triangle $ABC$ satisfies $\\cos \\angle BAC \\le 1/2$, therefore $\\angle BAC \\ge 60^\\circ$. Since for equilateral triangle, the condition is clearly satisfied (in that case $AE = AD = BE = CD$), the answer is $60^\\circ$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 29488,
    "subject": "Mathematics (Multi-modal)",
    "question": "Mathematics clubs are very popular in a certain city. Any two of them have at least one common member. Prove that one can distribute rulers and compasses to the citizens in such a way that only one citizen gets both (compass and ruler) and any club has at its disposal both, compass and ruler, from its members.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let us consider the club $K$ with the least number of its members (in case there is more such clubs, we take any). We give to one of its members (let us call him Jacob) both a compass and a ruler. Each of the other members of the club will get a compass. Any other citizen will get a ruler. We show that this distribution complies with the conditions of the problem:\n\nAny club which has Jacob as its member has certainly both instruments.\n\nIf there is a club where Jacob does not belong, then it has at least one common member with the club $K$, that is, there is at least a compass at disposal in the club. If there were no ruler in the club, it would mean that it is a “subclub” of $K$ and therefore has at least one member (Jacob) less than $K$, which is a contradiction with the choice of $K$. The described distribution really satisfies the conditions of the problem.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 29489,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABC$ be an acute triangle with altitudes $AK$, $BL$, $CM$. Prove that triangle $ABC$ is isosceles if and only if\n$$\nAM + BK + CL = AL + BM + CK.\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Points $K$, $L$, $M$ are defined as feet of altitudes and we need to express perpendicularity. Since the statement involves lengths and not angles, we characterise the perpendicularity by lengths of segments.\nComparing Pythagorean theorems in triangles $AMC$, $BMC$ we learn $AC^2 - BC^2 = AM^2 - BM^2$. Denoting the lengths of $BC$, $CA$, $AB$ by $a$, $b$, $c$, respectively, this implies\n$$\nAM - BM = \\frac{b^2 - a^2}{AM + BM} = \\frac{b^2 - a^2}{c}\n$$\nand likewise\n$$\nBK - CK = \\frac{c^2 - b^2}{a} \\quad \\text{and} \\quad CL - AL = \\frac{a^2 - c^2}{b}.\n$$\n\nThe equality from the problem statement rewrites as\n$$\n\\begin{gathered}\n\\frac{a^2 - b^2}{c} + \\frac{b^2 - c^2}{a} + \\frac{c^2 - a^2}{b} = 0, \\\\\nab(a^2 - b^2) + bc(b^2 - c^2) + ca(c^2 - a^2) = 0.\n\\end{gathered}\n$$\nLet's try to factor the left-hand side by viewing it as a cubic polynomial $P$ in variable $a$. It is easy to check that for isosceles triangle the left-hand side is zero, hence $P(b) = 0$ and $P(c) = 0$ and we know two roots of $P$. After dividing $P(a)$ by $(a-b)(a-c)$ we are left with linear polynomial $a(b-c) + b^2 - c^2$ which can be factored easily. To sum up, the equality from problem statement is equivalent with\n$$\n(a-b)(b-c)(c-a)(a+b+c) = 0.\n$$\nIt's obvious that this equality holds if and only if triangle $ABC$ is isosceles.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 29490,
    "subject": "Mathematics (Multi-modal)",
    "question": "In how many ways can you partition the set $\\{1, 2, \\ldots, 12\\}$ into six mutually disjoint two-element sets in such a way that the two elements in any set are coprime?",
    "options": [],
    "answer": "252",
    "solution": "No two even numbers can be in the same set (pair). Let us call partitions of $\\{1, 2, \\ldots, 12\\}$ with this property, that is one even and one odd number in each pair, even-odd partitions. The only further limitations are, that $6$ nor $12$ cannot be paired with $3$ or $9$, and $10$ cannot be paired with $5$.\n\nThat means, that odd numbers $1$, $7$ and $11$ can be paired with numbers $2$, $4$, $6$, $8$, $10$, $12$, numbers $3$ and $9$ with $2$, $4$, $8$, $10$ and number $5$ can be paired with $2$, $4$, $6$, $8$, $12$. We cannot use the product rule directly, we distinguish two cases: $5$ is paired with $6$ or $12$, in the second one $5$ is paired with one of $2$, $4$, and $8$. The possible pairings are $2 \\cdot 4 \\cdot 3 \\cdot 3 \\cdot 2 \\cdot 1 = 144$ in the first case, $3 \\cdot 3 \\cdot 2 \\cdot 3 \\cdot 2 \\cdot 1 = 108$ in the second case. Together $144 + 108 = 252$ pairings.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 29491,
    "subject": "Mathematics (Multi-modal)",
    "question": "**Find all functions $f: \\mathbb{N} \\to \\mathbb{N}$ such that for every positive integer $m$ the following is true: If we denote by $d_1, d_2, \\dots, d_n$ all the divisors of number $m$ then**\n$$\nf(d_1) \\cdot f(d_2) \\cdots f(d_n) = m.\n$$",
    "options": [],
    "answer": "The unique function is given by f(n) = p if n is a positive power of a prime p, and f(n) = 1 otherwise (in particular f(1) = 1).",
    "solution": "We will show that the only solution is a function $f$ such that\n$$\nf(m) = \\begin{cases} p, & \\text{if } m \\text{ is a non-trivial power of a prime } p, \\text{ i.e. } m = p^k, k \\ge 1, \\\\ 1, & \\text{otherwise.} \\end{cases}\n$$\nNumber $1$ has the unique divisor $1$, hence plugging $m = 1$ in the given equality we get $f(1) = 1$.\nLet $m = p$ be a prime. Then\n$$\nf(1) \\cdot f(p) = p \\quad \\text{i.e.} \\quad f(p) = p.\n$$\nFor $m = p^2$ we get\n$$\nf(1) \\cdot f(p) \\cdot f(p^2) = p^2 \\quad \\text{i.e.} \\quad f(p^2) = p\n$$\nand in general for $k > 1$ and $m = p^k$\n$$\nf(1) \\cdot f(p) \\cdot f(p^2) \\cdots f(p^k) = p^k.\n$$\nBy induction, $f(p^k) = p$ for all positive integers $k$.\nNow let us consider positive integer $m$ with at least two distinct prime factors whose factorization is $m = p_1^{\\alpha_1} p_2^{\\alpha_2} \\dots p_k^{\\alpha_k}$, where $k \\ge 2$ and $\\alpha_i \\ge 1$ for every $i \\in \\{1, 2, \\dots, k\\}$. The divisors of $m$ include the powers\n$$\np_1, p_1^2, \\dots, p_1^{\\alpha_1}, p_2, p_2^2, \\dots, p_2^{\\alpha_2}, \\dots, p_k, p_k^2, \\dots, p_k^{\\alpha_k},\n$$\nof its prime factors but the product of the corresponding function values\n$$\n\\begin{aligned}\n& f(p_1)f(p_1^2)\\dots f(p_1^{\\alpha_1})f(p_2)f(p_2^2)\\dots f(p_2^{\\alpha_2})\\dots f(p_k)f(p_k^2)\\dots f(p_k^{\\alpha_k}) \\\\\n&= \\underbrace{p_1p_1\\dots p_1}_{\\alpha_1} \\underbrace{p_2p_2\\dots p_2}_{\\alpha_2} \\dots \\underbrace{p_kp_k\\dots p_k}_{\\alpha_k} = p_1^{\\alpha_1}p_2^{\\alpha_2}\\dots p_k^{\\alpha_k} = m\n\\end{aligned}\n$$\nyields $m$. This means that the product of all the other function values (none of which is a non-trivial power of a prime), including $f(m)$, is equal to $1$ so all the other function values, including $f(m)$, are equal to $1$.\nThis specifies $f$ uniquely and also shows that it has the desired properties.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 29492,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find the least real $m$, for which there exists real $a$, $b$ such that\n$$\n|x^2 + a x + b| \\leq m(x^2 + 1)\n$$\nholds for any $x \\in (-1, 1]$.",
    "options": [],
    "answer": "1/3",
    "solution": "Let us assume that $a$, $b$, $m$ satisfy the condition:\n$$\n\\forall x \\in (-1, 1): |f(x)| \\leq m(x^2 + 1), \\quad \\text{where} \\quad f(x) = x^2 + a x + b.\n$$\nFirstly we prove that at least one of the $f(1) - f(0) \\geq 0$ and $f(-1) - f(0) \\geq 0$ holds:\nfor arbitrary $f(x) = x^2 + a x + b$ there is\n$$\nf(0) = b, \\quad f(1) = 1 + a + b, \\quad f(-1) = 1 - a + b,\n$$\nand\n$$\n\\max(f(1) - f(0), f(-1) - f(0)) = \\max(1 + a, 1 - a) = 1 + |a| \\geq 1.\n$$\nOur assumption means $|f(1)| \\le 2m$, $|f(-1)| \\le 2m$ and $|f(0)| \\le m$. Consequently\n$$\n1 \\le 1 + |a| = f(1) - f(0) \\le |f(1)| + |f(0)| \\le 2m + m = 3m, \\quad (1)\n$$\nor\n$$\n1 \\le 1 + |a| = f(-1) - f(0) \\le |f(-1)| + |f(0)| \\le 2m + m = 3m. \\quad (2)\n$$\nIn both cases we get $m \\ge \\frac{1}{3}$.\nWe show, that $m = \\frac{1}{3}$ fulfills the problem conditions. For this $m$ either (1) or (2) is equality, that is $a = 0$, $-f(0) = |f(0)|$ and $|f(0)| = m = \\frac{1}{3}$, thus $b = f(0) = -\\frac{1}{3}$.\nWe will verify, that the function $f(x) = x^2 - \\frac{1}{3}$ for $m = \\frac{1}{3}$ satisfies the conditions of the problem: The inequality $|x^2 - \\frac{1}{3}| \\le \\frac{1}{3}(x^2 + 1)$ is equivalent with the inequalities\n$$\n-\\frac{1}{3}(x^2 + 1) \\le x^2 - \\frac{1}{3} \\le \\frac{1}{3}(x^2 + 1) \\quad \\text{or} \\quad -x^2 - 1 \\le 3x^2 - 1 \\le x^2 + 1\n$$\nwhich are equivalent to $0 \\le x^2 \\le 1$, which is evidently fulfilled on $(-1, 1]$.\n\nThe sought $m$ is $\\frac{1}{3}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 29493,
    "subject": "Mathematics (Multi-modal)",
    "question": "For positive $a$, $b$, $c$ it holds\n$$\n(a + c)(b^2 + ac) = 4a.\n$$\nFind the maximal possible value of $b + c$ and find all triples $(a, b, c)$, for which the value is attained.",
    "options": [],
    "answer": "Maximum b + c = 2. All maximizing triples are a = b with 0 < b < 2 and c = 2 − b.",
    "solution": "We use the well known inequality $a^2 + b^2 \\ge 2ab$ to adjust the given one:\n$$\n4a = (a + c)(b^2 + ac) = a(b^2 + c^2) + c(a^2 + b^2) \\ge a(b^2 + c^2) + 2abc = a(b + c)^2.\n$$\nWe can see, that $b + c \\le 2$, and also that the equality holds if and only if $0 < a = b < 2$ and $c = 2 - b > 0$. That's it.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 29494,
    "subject": "Mathematics (Multi-modal)",
    "question": "There is $|BC| = 1$ in a triangle $ABC$ and there is a unique point $D$ on $BC$ such that $|DA|^2 = |DB| \\cdot |DC|$. Find all possible values of the perimeter of $ABC$.",
    "options": [],
    "answer": "1 + sqrt(2)",
    "solution": "Let us denote by $E$ the second intersection of $AD$ with the circumcircle $k$. The power of $D$ with respect to $k$ gives $|DB| \\cdot |DC| = |DA| \\cdot |DE|$, which together with the given condition $|DA|^2 = |DB| \\cdot |DC|$ yields $|DA| = |DE|$. That is $E$ lies on the image $p$ of the line $BC$ in the homothety with center $A$ and a coefficient $2$ (Fig. 1).\n\nVice versa, to any intersection of a line $p$ with the circle $k$ we reconstruct the point $D$ on $BC$, which fulfills $|DA|^2 = |DB| \\cdot |DC|$.\nIf the reconstruction has to be unique, the line $p$ has to touch $p$ in $E$.\n\n![](attached_image_1.png)\nFig. 2\n![](attached_image_2.png)\nFig. 3\n\nLet us denote $S_b$ and $S_c$ in order the centers of $AC$ and $AB$. The homothety with the center $A$ and a coefficient $\\frac{1}{2}$ sends $A, B, C, E$ (lying on the circle $k$) to $A, S_c, S_b, D$ which lie on the circle $k'$ (Fig. 2), while the image of $p$ is the tangent $BC$ of $k'$ in $D$. The powers of $B$ and $C$ with respect to $k'$ give $|BD|^2 = |BA| \\cdot |BS_c| = \\frac{1}{2}|BA|^2$ and $|CD|^2 = |CA| \\cdot |CS_b| = \\frac{1}{2}|CA|^2$. All together for the perimeter of $ABC$:\n\n$$\n|BC| + |AB| + |AC| = |BC| + \\sqrt{2}(|BD| + |CD|) = |BC| + \\sqrt{2} \\cdot |BC| = 1 + \\sqrt{2},\n$$\n\nwhich is the only possible value.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 29495,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $D$ be an arbitrary point on the base $AB$ of an isosceles triangle $ABC$. Let $E$ be such that $ADEC$ is a parallelogram. Point $F$ on the ray opposite to $ED$ satisfies $EF = EB$. Prove that the length of a chord cut by line $BE$ in the circumcircle of triangle $ABF$ is twice the length of $AC$.\n(Jan Kuchařík, Patrik Bak)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Denote by $S$ the circumcenter of triangle $ABF$ and by $K$, $L$, $M$ the feet of perpendiculars from $S$ to lines $AB$, $BF$, $BE$. It's easy to check that angle $ABF$ is obtuse, point $S$ lies on the perpendicular bisector $CK$ of side $AB$ in the half-plane opposite to $CEB$, and that point $E$ lies inside segment $SL$ so it is the interior point of segment $BM$ (Fig. 3). As $M$ is the midpoint of the mentioned chord and $AC = BC$, it suffices to show $BM = BC$.\n![](attached_image_1.png)\nFig. 3\n\nDenote by $\\alpha$, $\\beta$ the angles by bases of isosceles triangles $ABC$, $BFE$, respectively. Since $\\angle BDF = \\alpha$, triangle $DBF$ gives\n$$\n\\angle CBM = 180^\\{\\circ\\} - 2(\\alpha + \\beta).\n$$\nAs $CE \\parallel AB$, we further get $\\angle BCE = \\alpha$ and the clear similarity of right triangles $SEM \\sim BEL$ yields $\\angle ESM = \\beta$. And since both $CE$ and $AB$ are perpendicular to $SK$, the quadrilateral $CEMS$ is cyclic implying $\\angle ECM = \\angle ESM = \\beta$. Altogether, for $\\angle BCM$ and $\\angle BMC$ we get\n$$\n\\angle BCM = \\angle BCE + \\angle ECM = \\alpha + \\beta\n$$\nand from the previous equation we compute\n$$\n\\angle BMC = 180^\\{\\circ\\} - \\angle CBM - \\angle BCM = 2(\\alpha + \\beta) - (\\alpha + \\beta) = \\alpha + \\beta.\n$$\nFrom $\\angle BMC = \\angle BCM$ we deduce $BM = BC$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 29496,
    "subject": "Mathematics (Multi-modal)",
    "question": "There is a figure of prince on a field of a $6 \\times 6$ square chessboard. The prince can in one move jump either horizontally or vertically. The lengths of the jumps are alternately either one or two fields, and the jump on the next field is the first one. Decide, whether one can choose the initial field for the prince, so that the prince visits in an appropriate sequence of $35$ jumps every field of the chessboard.",
    "options": [],
    "answer": "No, it is impossible to choose such a starting field.",
    "solution": "Let us suppose the appropriate sequence exists and let us enumerate the fields of the chessboard as follows:\n\n| 1 | 2 | 3 | 4 | 1 | 2 |\n|---|---|---|---|---|---|\n| 2 | 3 | 4 | 1 | 2 | 3 |\n| 3 | 4 | 1 | 2 | 3 | 4 |\n| 4 | 1 | 2 | 3 | 4 | 1 |\n| 1 | 2 | 3 | 4 | 1 | 2 |\n| 2 | 3 | 4 | 1 | 2 | 3 |\n\nThe length one moves go from odd to even number and vice versa. The length two moves go from even to a different even number or from odd to a different odd number. If we denote $P_1, P_2, \\dots, P_{36}$ the numbers of visited fields, then it follows that among $P_2, P_3, P_4, P_5$ is each number (from $1$ to $4$) exactly once ($P_2$ and $P_3$ are different numbers with the same parity, and $P_4, P_5$ as well, only the parity is different). From the same reasons, any of the four numbers is among $P_{4k+2}, P_{4k+3}, P_{4k+4}, P_{4k+5}$ for arbitrary $k \\in \\{0, 1, \\dots, 7\\}$. Between the numbers $P_2, P_3, \\dots, P_{33}$ is thus any of the numbers $1$ to $4$ exactly eight times.\n\nThe number $4$ is on the chessboard just eight times, thus none of $P_1, P_{34}, P_{35}, P_{36}$ can be $4$. The numbers $P_{34}$ and $P_{35}$ have the same parity and are different (they are the length two move apart). The number $4$ is not among them, therefore both must be odd. Then $P_{36}$ has to be even and $P_1$ as well. Thus it has to be number $2$.\n\nThe initial field ($P_1$) thus has to be one of the coloured fields on the left chessboard. One can repeat the arguments for the numbering of the right chessboard (just a rotation of the left one). Since no field has number $2$ on both chessboards, we come to a contradiction. The initial field cannot be chosen.",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 29497,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $k \\in [0, 1]$. Solve the system\n$$\n\\begin{aligned}\nk - x^2 &= y, \\\\\nk - y^2 &= z, \\\\\nk - z^2 &= u, \\\\\nk - u^2 &= x\n\\end{aligned}\n$$\nin real numbers.",
    "options": [],
    "answer": "All solutions are of the form:\n1) For all k in [0, 1], the two solutions with all four variables equal: x = y = z = u = (-1 ± √(4k + 1)) / 2.\n2) Additionally, for k ≥ 3/4, solutions with x = z, y = u, and x + y = 1 exist. These are:\n   (x, y, z, u) = ( (1 + √(4k - 3)) / 2, (1 - √(4k - 3)) / 2, (1 + √(4k - 3)) / 2, (1 - √(4k - 3)) / 2 )\n   and\n   (x, y, z, u) = ( (1 - √(4k - 3)) / 2, (1 + √(4k - 3)) / 2, (1 - √(4k - 3)) / 2, (1 + √(4k - 3)) / 2 ).\nFor k = 3/4, these latter two coincide as x = y = z = u = 1/2 and are already included in (1). No other solutions exist.",
    "solution": "Subtracting the third equation from the first one we get\n$$\nz^2 - x^2 = (z - x)(z + x) = y - u. \\tag{1}\n$$\nSimilarly, the second and the fourth equation imply\n$$\ny^2 - u^2 = (y - u)(y + u) = x - z. \\tag{2}\n$$\nRelations (1) and (2) then imply that $x = z$ holds if and only if $y = u$ holds. We distinguish two cases, denoted a), b).\n\na) First, let us assume that $x = z$ and $y = u$, that is we are looking for quadruplets of the form $(x, y, z, u) = (x, y, x, y)$ with unknown $x$ and $y$. The original system reduces to\n$$\n\\begin{aligned}\nk - x^2 &= y, \\\\\nk - y^2 &= x.\n\\end{aligned}\n$$\nSubtracting the equations and rewriting we obtain\n$$\n(y - x)(y + x - 1) = 0.\n$$\nWe distinguish two (not quite disjoint) subcases.\n\nIf $y - x = 0$, the reduced system further reduces to a single quadratic equation\n$$\nx^2 + x - k = 0.\n$$\nFor any $k \\in (0, 1]$, this equation has two real solutions.\n$$\nx_{1,2} = \\frac{-1 \\pm \\sqrt{4k + 1}}{2}.\n$$\nThe original system therefore has at least two solutions\n$$\nx_1 = y_1 = z_1 = u_1 = \\frac{-1 + \\sqrt{4k + 1}}{2}, \\quad x_2 = y_2 = z_2 = u_2 = \\frac{-1 - \\sqrt{4k + 1}}{2}. \\tag{3}\n$$\nIf $x + y - 1 = 0$, the reduced system further reduces to quadratic equation\n$$\nx^2 - x + (1 - k) = 0.\n$$\nSince its discriminant equals $4k - 3$, this quadratic equation has solution if and only if $k \\ge \\frac{3}{4}$. The solutions are\n$$\nx_3 = \\frac{1 + \\sqrt{4k - 3}}{2} \\quad \\text{and} \\quad x_4 = \\frac{1 - \\sqrt{4k - 3}}{2},\n$$\nand the corresponding values of $y = 1 - x$ are then\n$$\ny_3 = \\frac{1 - \\sqrt{4k - 3}}{2} \\quad \\text{and} \\quad y_4 = \\frac{1 + \\sqrt{4k - 3}}{2}.\n$$\nIf $k = 3/4$, these solutions are identical and in fact identical with the solutions already found in (3). On the other hand, for $3/4 < k \\le 1$ we obtain two other distinct solutions\n$$\n(x, y, z, u) = (x_3, y_3, x_3, y_3) \\quad \\text{and} \\quad (x, y, z, u) = (x_4, y_4, x_4, y_4). \\qquad (4)\n$$\n\nb) Second, let us assume $x \\neq z$ and $y \\neq u$. In this case, plugging $x - z$ from (2) into the left-hand side of (1) and dividing by nonzero $y - u$ we arrive at\n$$\n(x+z)(y+u) = -1. \\qquad (5)\n$$\nSince the right-hand side of (5) is negative, at least one of the numbers $x, y, z, u$ is positive and at least one of them is negative. However, this contradicts a sequence of implications\n$$\nx \\ge 0 \\Rightarrow y \\ge 0 \\Rightarrow z \\ge 0 \\Rightarrow u \\ge 0 \\Rightarrow x \\ge 0, \\qquad (6)\n$$\nthat we prove now:\nIt suffices to prove the first implication (the proofs of the others are analogous). Assume $x > 0$. The fourth equation of the original system implies $x \\le k$ and since $k \\le 1$, we have $0 \\le x \\le k \\le 1$. This implies $x^2 \\le k$ (as $t^2 \\le t$ for any $t \\in (0, 1)$). The first equation of the original system now implies $y \\ge 0$ as desired.\nTherefore there are no solutions in case b).",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 29498,
    "subject": "Mathematics (Multi-modal)",
    "question": "There are 100 diamonds on display; 50 of them genuine and 50 of them fake. Peter is the only person able to distinguish them. Whenever you point to three diamonds, Peter will cover one of them and (truthfully) tell you, how many of the remaining two are genuine. Determine if it is possible to find the 50 genuine diamonds, no matter how Peter answers your queries.\n(Michal Rolínek, Josef Tkadlec)",
    "options": [],
    "answer": "No, it is impossible.",
    "solution": "We prove it is impossible. Let Peter pick one genuine diamond $G$ and one fake diamond $F$. Whenever the triplet of diamonds being pointed to contains both $F$ and $G$, Peter covers the third diamond (and truthfully answers “One.”). Whenever the triplet contains precisely one of $F$, $G$, Peter covers it. Otherwise he covers any diamond.\nNone of Peter's answers ever distinguishes between $G$ and $F$ hence it is impossible to say which of $G$, $F$ is the genuine one and which is the fake one.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 29499,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all triplets of integers $(a, b, c)$ such that each of the fractions\n$$\n\\frac{a}{b+c}, \\quad \\frac{b}{c+a}, \\quad \\frac{c}{a+b}\n$$\nis an integer.",
    "options": [],
    "answer": "All integer triples are either (i) nonzero triples with a + b + c = 0, or (ii) permutations of (0, t, t) with t ≠ 0.",
    "solution": "The considered fractions are symmetric in the sense that if we permute $(a, b, c)$ we get the same three fractions (possibly in different order). The same is true if we replace $(a, b, c)$ by $(-a, -b, -c)$. This simplifies the following casework.\n\nLet us suppose that numbers $a, b, c$ are such that all three fractions are integers. If one of them equals zero, without loss of generality we assume $a = 0$. Plugging this in we get that both $b/c$ and $c/b$ are integers. Hence $b$ and $c$ are nonzero and both $|b| \\le |c|$ and $|c| \\le |b|$ implying that $c = \\pm b$. Moreover, $b = c$ is the denominator of the first fraction, hence $b+c \\ne 0$ and $b = c$. Altogether, we obtain (clearly admissible) triplets $(0, c, c)$ and their permutations for any nonzero integer $c$.\n\nIt remains to solve the case $abc \\ne 0$.\n\nDue to the observations from the first paragraph we can assume that at least two of the numbers $a, b, c$ are positive. If all three were positive, the fraction with $\\min\\{a, b, c\\}$ at its numerator lies between 0 and 1 and can't be integer.\n\nAssume $a$ and $b$ are positive and $c = -d$ ($d > 0$) negative. Plugging this in, the fractions rewrite as\n$$\n\\frac{a}{d-b}, \\quad \\frac{b}{d-a}, \\quad \\frac{d}{a+b}\n$$\nThe last one implies that $d \\ge a+b$. Hence the first one has positive denominator and since it is an integer, we have $a \\ge d-b$, i.e. $d \\le a+b$. Therefore $d = a+b$ ($c = -a-b$) and altogether we obtain triplets $(a, b, c)$ of nonzero numbers such that $a+b+c = 0$. All such triplets are admissible as the value of each fraction then equals $-1$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 29500,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $a, b$ be real numbers such that $a+b > 2$. Prove that the system of inequalities\n$$\n(a-1)x + b < x^2 < ax + (b-1)\n$$\nhas infinitely many real solutions $x$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "We rewrite the system as\n$$\nF(x) > 0 \\land G(x) < 0,\n$$\nwhere $F(x) = x^2-(a-1)x-b$ and $G(x) = x^2-ax-b+1$. Observe that $F(x)-G(x) = x-1$.\nThe condition $a+b>2$ implies that\n$$\nF(1) = G(1) = 2 - a - b < 0,\n$$\nhence $x=1$ is not a solution. However, $G(1) < 0$ implies that the quadratic equation $G(x) = 0$ has a root $x_0 > 1$. Then\n$$\nF(x_0) = F(x_0) - 0 = F(x_0) - G(x_0) = x_0 - 1 > 0.\n$$\nFrom $F(1) < 0$ and $F(x_0) > 0$ we deduce that there exists a root $x_1$ of $F(x) = 0$ that belongs to the open interval $(1, x_0)$. Since\n$$\nF(1) < 0 \\land F(x_1) = 0, \\quad \\text{and} \\quad G(1) < 0 \\land G(x_0) = 0,\n$$\nany $x \\in (x_1, x_0)$ is a solution to the original system.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 29501,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all pairs of real numbers $k, l$ such that the inequality\n$$\nka^2 + lb^2 > c^2\n$$\nholds for side lengths $a, b, c$ of any triangle.",
    "options": [],
    "answer": "k ≥ 1, l ≥ 1, and kl ≥ k + l",
    "solution": "Assume that for some $k, l$, the given inequality holds for all side lengths $a, b, c$ of any triangle. Plugging in $a = c = 1$ and arbitrary $b < 2$, we get $k + lb^2 > 1$. If $k < 1$ then we easily find sufficiently small $b > 0$ that makes the inequality false. Hence $k \\ge 1$ and likewise $l \\ge 1$.\n\nLet $ABC$ be a triangle in the coordinate plane. Without loss of generality, let $A = [-1, 0]$, $B = [1, 0]$, and denote $C = [x, y]$ ($y \\ne 0$). Using Pythagorean Theorem we express the side lengths in terms of $x, y$ and plug it in the given inequality to obtain\n$$\nk((x-1)^2 + y^2) + l((x+1)^2 + y^2) > 4\n$$\nwhich rewrites as\n$$\n(k+l)x^2 + 2(l-k)x + k + l - 4 > -(k+l)y^2. \\quad (1)\n$$\nThis inequality has to hold for any $x$ and any $y \\ne 0$. However, varying $y$, the right-hand side attains all negative values (recall $k+l > 0$). Therefore for any $x$ we have\n$$\n(k+l)x^2 + 2(l-k)x + (k+l-4) \\ge 0, \\quad (2)\n$$\nwhich happens if and only if the discriminant $D = 4(l-k)^2 - 4(k+l-4)(k+l)$ is not positive. The inequality $D \\le 0$ rewrites to $kl \\ge k+l$.\n\nWe found out that if numbers $k, l$ satisfy the given inequality for any triplet of side lengths of a triangle then $k \\ge 1, l \\ge 1$, and\n$$\nkl \\ge k + l. \\tag{3}\n$$\nOn the other hand, the conjunction of these three conditions is also sufficient: The third condition implies that (2) is satisfied for any real $x$ and since the right-hand side of (1) is negative for $y \\ne 0$, inequality (1) is satisfied for any $x$ and any $y \\ne 0$. Finally, inequality (1) is equivalent to the given inequality.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 29502,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $p$ be a circle with center $K$ passing through $M$, $q$ a semicircle with diameter $KM$ and $L$ a point inside the segment $KM$. A line through $L$ perpendicular to $KM$ intersects $q$ at point $Q$ and $p$ at points $P_1, P_2$ such that $P_1Q > P_2Q$. Line $MQ$ intersects $p$ for the second time at $R \\ne M$. Prove that areas $S_1, S_2$ of triangles $MP_1Q, P_2RQ$ satisfy\n$$\n1 < \\frac{S_1}{S_2} < 3 + \\sqrt{8}.\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "The circle containing semicircle $q$ is the image of $p$ in homothety with center $M$ and factor $1/2$, hence $Q$ is the midpoint of $RM$. Since triangles $MP_1Q$, $P_2RQ$ share the angle by $Q$, we have\n$$\n\\frac{S_1}{S_2} = \\frac{\\frac{1}{2} P_1 Q \\cdot M Q \\cdot \\sin \\angle P_1 Q M}{\\frac{1}{2} P_2 Q \\cdot R Q \\cdot \\sin \\angle P_2 Q R} = \\frac{P_1 Q}{P_2 Q}.\n$$\nLet $KM = r$, $ML = x$, $P_1L = d_1$, $QL = d_2$ (Fig. 1). Points $P_1$, $P_2$ are symmetric about $KM$, therefore $P_1L = P_2L$ and $P_2Q = d_1 - d_2$. Denote by $M'$ the point such that $MM'$ is the diameter of $p$. Then triangle $M'MP_1$ is right and by Geometric Mean Theorem (an altitude splits a right triangle into two similar triangles) we get $d_1^2 = x(2r - x)$. Similarly in right triangle $KQM$ we get $d_2^2 = x(r - x)$ and thus\n$$\n\\begin{aligned}\n\\frac{S_1}{S_2} &= \\frac{P_1 Q}{P_2 Q} = \\frac{d_1 + d_2}{d_1 - d_2} = \\frac{(d_1 + d_2)^2}{d_1^2 - d_2^2} \\\\\n&= \\frac{x(2r - x) + x(r - x) + 2\\sqrt{x(2r - x) \\cdot x(r - x)}}{rx} \\\\\n&= \\frac{3r - 2x + 2\\sqrt{(2r - x)(r - x)}}{r}.\n\\end{aligned}\n$$\n![](attached_image_1.png)\nFig. 1\nWe view the expression as a function of variable $x$ with parameter $r$. The function is decreasing on $(0, r)$ (both functions $3r - 2x$ and $(2r - x)(r - x)$ are decreasing), therefore it attains its maximum $3 + 2\\sqrt{2}$ at $x = 0$ and minimum $1$ at $x = r$. By the problem statement, $x \\in (0, r)$, thus $1 < S_1/S_2 < 3 + 2\\sqrt{2} = 3 + \\sqrt{8}$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 29503,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABC$ be a triangle and $S_b, S_c$ the midpoints of the sides $AC$, $AB$, respectively. Prove that if $AB < AC$ then $\\angle B S_c C < \\angle B S_b C$. (Patrik Bak)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "It suffices to prove that if $AB < AC$ then $S_b$ lies inside the circumcircle $k$ of triangle $B S_c C$.\nThe midline $S_b S_c$ is parallel to $BC$ (Fig. 1). Let line $S_b S_c$ intersect $k$ for the second time at $P$. We will show that $S_b$ lies on the segment $S_c P$ (as opposed to lying on the ray opposite to $P S_c$). To that end, it suffices to prove $\\angle BCA < \\angle BCP$. By symmetry about the perpendicular bisector of $BC$ we have $\\angle BCP = \\angle CBA$, so we need to prove $\\angle BCA < \\angle CBA$ which is in fact clearly equivalent to the given $AB < AC$.\n\n![](attached_image_1.png)\nFig. 1\n\n\n![](attached_image_2.png)\nFig. 2\nBy power of $A$ with respect to $k$, there exists a point $Q$ on the ray $AC$ such that $AQ \\cdot AC = AB \\cdot AS_c = \\frac{1}{2} AB^2$. Then (Fig. 2)\n$$\nAQ = \\frac{AB^2}{2 \\cdot AC} < \\frac{AC^2}{2 \\cdot AC} = \\frac{1}{2} AC = AS_b,\n$$\nhence $Q$ lies on segment $AS_b$. As before we conclude that $S_b$ lies inside the circumcircle of triangle $B S_c C$.\n\n![](attached_image_2.png)\nFig. 2",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 29504,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all functions $f: \\mathbb{R} \\to \\mathbb{R}$ such that for any real numbers $x, y$ we have\n$$\nf(y - xy) = f(x)y + (x - 1)^2 f(y).\n$$",
    "options": [],
    "answer": "f(x) = a x(x - 1) for any real constant a",
    "solution": "Setting $x = 1$ we learn $f(0) = f(1)y$ for all real $y$, hence $f(0) = f(1) = 0$. Setting $y = 1$ in the original equation we get\n$$\nf(1-x) = f(x)\n$$\nfor all real $x$. Let $t$ be an arbitrary real number. Setting $x = 1-t$ gives\n$$\nf(ty) = f(1-t)y + t^2 f(y) = f(t)y + t^2 f(y) \\tag{1}\n$$\nfor all real $y$. Swapping the role of $t$ and $y$ we get\n$$\nf(y)t + y^2 f(t) = f(yt) = f(ty) = f(t)y + t^2 f(y),\n$$\nhence $f(t)(y^2 - y) = f(y)(t^2 - t)$, which for $y = 2$ gives\n$$\nf(t) = \\frac{1}{2}f(2)(t^2 - t), \\quad t \\in \\mathbb{R}.\n$$\nIt remains to check that the constant $a = f(2)/2$ in the derived formula $f(x) = ax(x-1)$ can be an arbitrary real number:\n$$\n\\begin{align*}\nf(x)y + (x-1)^2 f(y) &= ax(x-1)y + (x-1)^2 ay(y-1) \\\\\n&= a(x-1)y(x + xy - x - y + 1) \\\\\n&= a(1-x)y((1-x)y - 1) = f((1-x)y) = f(y - xy).\n\\end{align*}\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 29505,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $k$ be a real parameter. Determine the number of real solutions to the system\n$$\n\\begin{align*}\nx^2 + kxy + y^2 &= z, \\\\\ny^2 + kyz + z^2 &= x, \\\\\nz^2 + kzx + x^2 &= y,\n\\end{align*}\n$$\nin terms of $k$.",
    "options": [],
    "answer": "Number of real solutions as a function of k:\n- For k in (-∞, -2): 8 solutions\n- For k = -2: 4 solutions\n- For k in (-2, 1 − 2√2): 8 solutions\n- For k = 1 − 2√2: 5 solutions\n- For k in (1 − 2√2, 1 + 2√2): 2 solutions\n- For k = 1 + 2√2: 5 solutions\n- For k in (1 + 2√2, ∞): 8 solutions",
    "solution": "We distinguish several cases.\n\nFirst, assume $x = y = z$. Then the whole system reduces to $(k+2)x^2 = x$. Its solution is a triplet $(0, 0, 0)$ for any $k$ and moreover triplet $\\left(\\frac{1}{k+2}, \\frac{1}{k+2}, \\frac{1}{k+2}\\right)$ if $k \\neq -2$.\n\nLet's get back to the original system. Subtracting the second equation from the first one yields\n$$\n(x^2 - z^2) + ky(x - z) = z - x\n$$\nwhich rewrites as\n$$\n(x - z)(x + z + ky + 1) = 0. \\tag{1}\n$$\nSimilarly, subtracting the third equation from the second one yields\n$$\n(y - x)(y + x + kz + 1) = 0. \\tag{2}\n$$\nIf $x \\neq y \\neq z \\neq x$, the equations (1), (2) reduce to\n$$\n\\begin{align*}\nx + z + ky + 1 &= 0, \\\\\ny + x + kz + 1 &= 0.\n\\end{align*}\n$$\nSubtracting these two equations we arrive at $(y-z)(k-1) = 0$ implying that $k=1$\nand $x+y+z=-1$. However that's impossible since for $k=1$ we get\n$$\nz = x^2 + xy + y^2 = \\left(x + \\frac{y}{2}\\right)^2 + \\frac{3y^2}{4} \\ge 0\n$$\nand likewise $x \\ge 0$ and $y \\ge 0$ so altogether $x + y + z \\ge 0$.\n\nWe found out that in every solution to the original system, some two unknowns\nhave the same value. As the system is cyclic, let us from now on assume $x = y \\neq z$ (the\ncase $x = y = z$ has already been solved). Equation (1) then implies $x+y+ky+1=0$,\nthat is $x = -(k+1)y - 1$, and the original system reduces to a single equation\n$$\n(k + 2)y^2 + (k + 1)y + 1 = 0. \\tag{3}\n$$\nLet us remark that any solution to equation (3) is a solution we haven't found yet, because equality $x = y$ i.e. $y = -(k+1)y - 1$ is only possible for $k \\neq -2$ and yields $x = y = z = -1/(k+2)$ which is not a solution to the original system.\n\nFor $k = -2$ the equation (3) is linear with a unique solution $y = 1$. This yields\nsolution $(0, 1, 1)$ and its two permutations.\n\nFor $k \\neq -2$ the equation (3) is quadratic and has real solutions if and only if\n$$\nD = (k + 1)^2 - 4(k + 2) = k^2 - 2k - 7 \\geq 0,\n$$\nwhich translates to $k \\notin (1-2\\sqrt{2}, 1+2\\sqrt{2})$. For $k = 1 \\pm 2\\sqrt{2}$ there is a unique solution\n$$\ny_0 = -\\frac{k+1}{2(k+2)} = 1 \\mp \\sqrt{2} \\quad \\text{a} \\quad x_0 = \\frac{(k+1)^2}{2(k+2)} - 1 = 1.\n$$\nwhich yields the three permutations of $(x_0, y_0, y_0)$ as solutions to the original system.\n\nFor $k \\in (-\\infty, -2) \\cup (-2, 1-2\\sqrt{2}) \\cup (1+2\\sqrt{2}, \\infty)$, the quadratic equation (3) has\ntwo distinct solutions\n$$\ny_{1,2} = \\frac{-k - 1 \\pm \\sqrt{k^2 - 2k - 7}}{2(k + 2)},\n$$\nthat give two distinct values $x_{1,2} = -(k+1)y_{1,2} - 1$. The original system thus has six solutions: three permutations of $(x_1, y_1, y_1)$ and three permutations of $(x_2, y_2, y_2)$.\n\nThe following table summarizes the number of solutions to the given system in\nterms of $k$:\n\n| Interval for $k$ | $(0, 0, 0)$ | $(1/(k+2), 1/(k+2), 1/(k+2))$ | Equation (3) | Total |\n|------------------|-------------|-------------------------------|-------------|-------|\n| $(-\\infty, -2)$ | 1           | 1                             | 6           | 8     |\n| $-2$             | 1           | 0                             | 3           | 4     |\n| $(-2, 1 - 2\\sqrt{2})$ | 1     | 1                             | 6           | 8     |\n| $1 - 2\\sqrt{2}$ | 1           | 1                             | 3           | 5     |\n| $(1 - 2\\sqrt{2}, 1 + 2\\sqrt{2})$ | 1 | 1                     | 0           | 2     |\n| $1 + 2\\sqrt{2}$ | 1           | 1                             | 3           | 5     |\n| $(1 + 2\\sqrt{2}, \\infty)$ | 1 | 1                             | 6           | 8     |",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 29506,
    "subject": "Mathematics (Multi-modal)",
    "question": "Paul is filling the cells of a rectangular table alternately with crosses and circles (he starts with a cross). When the table is filled in completely, he determines his score as $X + O$ where $X$ is the number of rows containing more crosses than circles and $O$ is the number of columns containing more circles than crosses. In terms of $n$, what is the largest possible score Paul can achieve for a $(2n + 1) \\times (2n + 1)$ table? (Josef Tkadlec)",
    "options": [],
    "answer": "4n",
    "solution": "In total there are $2n(n+1)+1 < (2n+1)(n+1)$ crosses and $2n(n+1)$ circles. Hence the crosses can dominate in at most $2n$ rows and, similarly, circles can dominate in at most $2n$ columns for the total score $2n+2n=4n$.\n\nSuch a score can be achieved if, for example, Paul draws crosses in the left $n+1$ columns of the first $n$ rows, the right $n+1$ columns of the last $n$ rows and the middle cell of the middle row. That is precisely $2n(n+1)+1$ crosses and we easily check that crosses dominate in all rows except for the middle one while circles dominate in all columns except for the middle one.\n\n![](attached_image_1.png)\nFig. 3",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 29507,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find the largest positive integer $n$ such that\n$$\n\\lfloor\\sqrt{1}\\rfloor + \\lfloor\\sqrt{2}\\rfloor + \\lfloor\\sqrt{3}\\rfloor + \\dots + \\lfloor\\sqrt{n}\\rfloor\n$$\nis a prime ($\\lfloor x \\rfloor$ denotes the largest integer not exceeding $x$). (Patrik Bak)",
    "options": [],
    "answer": "47",
    "solution": "Consider the infinite sequence $\\{a_n\\}_{n=1}^{\\infty}$ defined by $a_n = \\lfloor\\sqrt{n}\\rfloor$. This sequence is clearly non-decreasing and since\n$$\nk = \\sqrt{k^2} < \\sqrt{k^2 + 1} < \\dots < \\sqrt{k^2 + 2k} < \\sqrt{k^2 + 2k + 1} = k + 1,\n$$\nit contains every integer $k$ precisely $(2k + 1)$-times. This allows us to express the value of the sum $s_n = \\sum_{i=1}^{n} a_i$ as follows: Let $k = \\lfloor\\sqrt{n}\\rfloor$, that is $n = k^2 + l$ for some $l \\in \\{0, 1, \\dots, 2k\\}$. Then\n$$\n\\begin{aligned}\ns_n &= \\sum_{i=0}^{k-1} i(2i+1) + k \\cdot (l+1) = 2 \\cdot \\sum_{i=1}^{k-1} i^2 + \\sum_{i=1}^{k-1} i + k(l+1) \\\\\n&= 2 \\cdot \\frac{(k-1)k(2k-1)}{6} + \\frac{(k-1)k}{2} + k(l+1) \\\\\n&= \\frac{(k-1)k(4k+1)}{6} + k(l+1),\n\\end{aligned}\n$$\nwhere we used $1+2+\\cdots+n = \\frac{1}{2}n(n+1)$ and $1^2+2^2+\\cdots+n^2 = \\frac{1}{6}n(n+1)(2n+1)$.\nIf $k > 6$ then the fraction $\\frac{1}{6}(k-1)k(4k+1)$ is an integer sharing a prime factor with $k$, hence the whole right-hand side is sharing a prime factor with $k < s_n$ and $s_n$ is not a prime.\nIf $k \\le 6$ then $n < (6+1)^2 = 49$. Plugging $n = 48$ into the right-hand side we get $s_{48} = 203 = 7 \\cdot 29$. For $n = 47$ we get $s_{47} = 197$, which is a prime. The answer is $n = 47$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 29508,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABC$ be an acute triangle with altitude $AD$. The bisectors of angles $BAD$, $CAD$, intersect side $BC$ at $E$, $F$, respectively. The circumcircle of triangle $AEF$ intersects sides $AB$, $AC$ at $G$, $H$, respectively. Prove that lines $EH$, $FG$, and $AD$ pass through a common point. (Patrik Bak)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $K$ be the intersection of segments $FG$ and $AE$ and $L$ the intersection of $EH$ and $AF$ (Fig. 2). Inscribed angles give\n\n$$\n\\angle AGF = \\angle AEF = 90^\\circ - \\angle DAE = 90^\\circ - \\angle GAE,\n$$\n![](attached_image_1.png)\nFig. 2\n\nthat is $\\angle AGF + \\angle GAE = 90^\\circ$, hence\n$$\n\\angle AKG = 180^\\circ - (\\angle AGF + \\angle GAE) = 90^\\circ.\n$$\nLine *FK* is therefore an altitude of the triangle *AEF*. Similarly we prove that *EL* is its altitude too, thus the intersection of *FK* and *EL* is the orthocenter of triangle *AEF* and it lies on its third altitude *AD* too.\nLet *M* be the second intersection of ray $AD$ and the circumcircle $k$ of triangle $AEF$ (Fig. 2). Since $AE$ is the bisector of angle $GAM$, the angles $GAE$ and $EAM$ inscribed in $k$ are equal and therefore the chords $GE$ and $EM$ are also equal. Furthermore, the corresponding inscribed angles $GFE$ and $EFM$ are equal too. The intersection of $FG$ and $AM$ is thus the reflection of $M$ about $EF$. The same argument applies to the intersection of $EH$ and $AM$, therefore the original intersections are identical.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 29509,
    "subject": "Mathematics (Multi-modal)",
    "question": "To a sequence of $n$ zeros and $n$ ones, we assign the number of maximal contiguous runs of equal digits in it. (For instance, sequence $00111001$ has four such runs: $00$, $111$, $00$, and $1$.) For a given $n$ we sum up all the numbers assigned to all such sequences. Prove that the resulting sum is equal to\n$$\n(n + 1) \\binom{2n}{n}.\n$$",
    "options": [],
    "answer": "(n + 1) * C(2n, n)",
    "solution": "Consider one such sequence and let us count (from left to right) how many maximal contiguous runs (from now on, just runs) it contains. We count a new run when it ends, that is when we hit a different digit or the right end. The number of runs is thus one more than the number of digits that follow a different digit (we say that the run *changes* at these positions). Instead of counting runs in different sequences, let us count sequences that change run at a given position.\n\nThere are $2n-1$ possible positions to change the run (all the positions but the first one). There are two options for the digit at this position and this uniquely determines the digit at the preceding position. All the remaining positions can be arbitrarily filled with remaining $n-1$ zeros and $n-1$ ones. Altogether, we get $(2n-1) \\cdot 2 \\cdot \\binom{2n-2}{n-1}$ changes of runs in all the sequences. Since in each of the $\\binom{2n}{n}$ sequences there is one more run than there are changes, in total we obtain\n$$\n\\begin{aligned}\n2(2n-1)\\binom{2n-2}{n-1} + \\binom{2n}{n} &= 2n\\binom{2n-1}{n} + \\binom{2n}{n} \\\\\n&= 2n\\binom{2n-1}{n-1} + \\binom{2n}{n} \\\\\n&= n\\binom{2n}{n} + \\binom{2n}{n} = (n+1)\\binom{2n}{n},\n\\end{aligned}\n$$\nwhere we used an obvious identity\n$$\nk \\binom{m}{k} = m \\binom{m-1}{k-1}.\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 29510,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABC$ be an acute triangle with orthocenter $H$. The bisector of angle $BHC$ intersects side $BC$ at $D$. Denote by $E$, $F$ the reflections of $D$ about $AB$, $AC$, respectively. Prove that the circumcircle of triangle $AEF$ passes through the midpoint of arc $BAC$.\n\n(Patrik Bak)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Clearly, the directed angle $EAF$ is $\\widehat{EAF} = 2\\alpha$. Let $H_2$, $H_3$ be the reflections of the orthocenter $H$ about the sides $AC$, $AB$, respectively. It is well-known that $H_2$ and $H_3$ lie on the circumcircle $k$ of triangle $ABC$.\n\n![](attached_image_1.png)\nFig. 1\n\nAs $DH$ is the bisector of angle $BHC$ and $CH \\parallel DE$, we have $|\\angle H_3ED| = |\\angle HDE| = |\\angle DHC| = \\frac{1}{2}|\\angle BHC| = 90^\\circ - \\frac{1}{2}\\alpha$. Let $G$ be the midpoint of arc $BAC$. Since the arc $CG$ of $k$ is one half of the arc $CAB$, the directed angle $\\widehat{CH_3G}$ subtending arc $CG$ has the same measure $90^\\circ - \\frac{1}{2}\\alpha$, that is $\\widehat{CH_3G} = \\widehat{DEH_3}$.\n\nFrom $CH_3 \\parallel DE$ we infer that points $G$, $H_3$, and $E$ are collinear. Similarly we find that $G$, $H_2$, and $F$ are also collinear, therefore $\\overrightarrow{EGF} = \\overrightarrow{H_3GH_2} = \\overrightarrow{H_3AH_2} = \\overrightarrow{H_3AH} + \\overrightarrow{HAH_2} = 2\\alpha = \\overrightarrow{EAF}$, as we needed to show.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 29511,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABCD$ be a convex quadrilateral such that $\\angle ABC = \\angle ACD$ and $\\angle ACB = \\angle ADC$. Suppose that the circumcenter $O$ of triangle $BCD$ is different from $A$. Prove that the angle $OAC$ is right. (Patrik Bak)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Since $\\angle ABC + \\angle CDA < 180^\\circ$, point $A$ lies inside the circumcircle $\\omega$ of triangle $BCD$. Denote by $C'$, $D'$ the second intersection of $\\omega$ with rays $CA$, $DA$, respectively (Fig. 3). We angle chase:\n$$\n\\angle D'C'C = \\angle D'DC = \\angle ADC = \\angle ACB.\n$$\nHence $BCC'D'$ is an isosceles trapezoid. Moreover, since $\\angle C'AD' = \\angle CAD = \\angle BAC$, triangles $ABC$ and $AD'C'$ are similar by AA and in fact due to $BC = C'D'$ they are congruent. Point $A$ is thus the midpoint of the chord $CC'$ and $\\angle OAC = 90^\\circ$ follows.\n\n![](attached_image_1.png)\nFig. 3\nLet's frame the figure with respect to triangle $ABD$. Then $AC$ is the A-angle bisector. The Inscribed angle theorem states that the (reflex) angle $BOD$ is twice the (convex) angle $BCD$, hence for the size of the convex angle $BOD$ we get\n$$\n\\angle BOD = 360^\\circ - 2 \\cdot \\angle DCB = 360^\\circ - \\angle ADC - \\angle BCD - \\angle ABC = \\angle BAD.\n$$\n\n![](attached_image_2.png)\nFig. 4\n\nTherefore $O$ lies on the arc $BAD$ of the circumcircle of triangle $ABD$. Since $OB = OD$, point $O$ is the midpoint of that arc and thus it lies on the external A-angle bisector which is perpendicular to the A-angle bisector.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 29512,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find the largest possible size of a set $M$ of integers with the following property: Among any three distinct numbers from $M$, there exist two numbers whose sum is a power of $2$ with non-negative integer exponent.",
    "options": [],
    "answer": "6",
    "solution": "The set $\\{-1, 3, 5, -2, 6, 10\\}$ attests that $M$ can have $6$ elements: The sum of any two numbers from the triplet $(-1, 3, 5)$ is a power of two and the same is true for triplet $(-2, 6, 10)$. For the sake of contradiction, assume that some set $M$ has more than $6$ elements.\n\nClearly, $M$ can't contain three (or more) non-positive numbers. Hence it contains at least five positive numbers. Denote by $x$ the largest positive number in $M$ and by $a, b, c, d$ some four other positive numbers in $M$. Consider pairs $x + a, x + b, x + c, x + d$. They are all larger than $x$ and less than $2x$. The open interval $(x, 2x)$ contains at most one power of two, hence at least three of the four sums are not a power of two. Without loss of generality, assume those are $x + a, x + b, x + c$. Considering the triplets $(a, b, x), (a, c, x), (b, c, x)$ we infer that all $a + b, a + c, b + c$ are powers of two. However, this is impossible. Without loss of generality, let $a = \\max\\{a, b, c\\}$. Then $a + b$ and $a + c$ both lie in $(a, 2a)$, hence at least one of them is not a power of two, a contradiction.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 29513,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $k \\neq 0$ be an integer. Prove that the number of ordered pairs $(x, y)$ of integers satisfying\n$$\nk = \\frac{x^2 - xy + 2y^2}{x + y}\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Multiplying both sides by $x + y$ yields\n$$\nx^2 - xy + 2y^2 = k(x + y). \\quad (1)\n$$\nAny solution $(x, y)$ to the original equation is a solution to (1), but (1) can have extra solutions satisfying $x + y = 0$, tj. $y = -x$.\nA pair $(x, -x)$ is a solution to (1) if and only if $x^2 + x^2 + 2x^2 = k \\cdot 0$, that is $x = 0$. Equation (1) therefore has exactly one more solution than the original one and it suffices to prove that equation (1) has an even number of integer solutions if and only if $7 \\mid k$.\nWe rewrite (1) as a quadratic equation\n$$\nx^2 - x(y + k) + 2y^2 - ky = 0 \\quad (2)\n$$\nin $x$. Its discriminant satisfies\n$$\n\\begin{align}\nD(y) = (y + k)^2 - 4(2y^2 - ky) &= k^2 + 6ky - 7y^2 = (k - y)(k + 7y) = \\tag{3} \\\\\n&= -7\\left(y - \\frac{3}{7}k\\right)^2 + \\frac{16}{7}k^2, \\nonumber\n\\end{align}\n$$\nwhich is, for every $k$, a quadratic function in $y$ bounded from above. Therefore, for any integer $k$, the discriminant $D(y)$ is non-negative for only finitely many integers $y$ and equation (2) has only finitely many integer solutions $(x, y)$.\n\nIf $D(y) > 0$ for some integer $y$, the equation (2) has precisely two real solutions that can only be integer simultaneously, for their sum $y + k$ is an integer. For any such $y$ we get an even number of solutions to (2).\nWe see that $D(y) = 0$ either for $y = k$ or for $y = -\\frac{1}{7}k$. In the first case, equation (2) reduces to $(x-k)^2 = 0$ with double root $x = k$ and the equation (1) has only one solution $(k, k)$ with $y = k$. In the second case, $y$ is integer if and only if $k$ is divisible by 7 and then equation (2) has a double root $x = \\frac{3}{7}k$, therefore $(\\frac{3}{7}k, -\\frac{1}{7}k)$ is the only solution to equation (1) with $y = -\\frac{1}{7}k$. Moreover, the two solutions $(k, k)$ and $(\\frac{3}{7}k, -\\frac{1}{7}k)$ are different as $k \\neq 0$.\nWe see that equation (1) has an even number of integer solutions if and only if $k$ is divisible by 7. We conclude.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 29514,
    "subject": "Mathematics (Multi-modal)",
    "question": "Determine all functions $f: \\mathbb{R} \\to \\mathbb{R}$ such that, for all real numbers $x$ and $y$,\n$$\nf(x^2 + xy) = f(x)f(y) + yf(x) + xf(x + y).\n$$",
    "options": [],
    "answer": "The solutions are f(x) = 0 for all real x, f(x) = -x for all real x, and f(x) = 1 - x for all real x.",
    "solution": "Setting $x = 0$ yields $f(0) = f(0)f(y) + yf(0)$. If $f(0) \\neq 0$, we obtain $1 = f(y) + y$ or equivalently $f(y) = 1 - y$ for all $y \\in \\mathbb{R}$. Inserting this in the original equation yields\n$$\n1 - x^2 - xy = (1-x)(1-y) + y(1-x) + x(1-x-y)\n$$\nfor $x, y \\in \\mathbb{R}$, which is true.\nTherefore, we are left with $f(0) = 0$.\nWith the substitution $x + y = z$, the functional equation is equivalent to\n$$\nf(xz) = f(x)f(z-x) + (z-x)f(x) + xf(z). \\quad (1)\n$$\nExchanging $x$ and $z$ yields\n$$\nf(zx) = f(z)f(x-z) + (x-z)f(z) + zf(x). \\quad (2)\n$$\nCombining (1) and (2) yields\n$$\nf(x)f(z-x) + (z-x)f(x) + xf(z) = f(z)f(x-z) + (x-z)f(z) + zf(x)\n$$\nor equivalently\n$$\nf(x)(f(z-x) - x) = f(z)(f(x-z) - z). \\quad (3)\n$$\nSetting $z = 0$ yields\n$$\nf(x)(f(-x) - x) = f(0)f(x) = 0,\n$$\nso for each $x \\in \\mathbb{R}$ we either have $f(x) = 0$ or $f(-x) = x$.\nAssume that there is an $x \\neq 0$ with $f(x) = 0$. For $z \\neq 0$, we have $f(x-z) \\in \\{0, z-x\\}$, so $f(x-z) \\neq z$. Thus (3) implies $f(z) = 0$ for all $z \\neq 0$. It is clear that the constant function $f = 0$ is a solution.\nOtherwise, we have $f(x) \\neq 0$ for all $x \\neq 0$ and therefore $f(x) = -x$ for all $x$. This is also a solution.\nWe conclude that there are three solutions $f$, namely $f = 0$, $f(x) = -x$ and $f(x) = 1 - x$.\nSetting $x = 0$ gives $f(0) = f(0)f(y) + yf(0)$. Since $f(y) = 1 - y$ gives a solution, we remain with the case $f(0) = 0$. Setting $x = 1$ gives\n$$\nf(1+y) = f(1)f(y) + yf(1) + f(1+y), \\quad \\text{i. e.} \\quad 0 = f(1)(f(y) + y).\n$$\nSince $f(y) = -y$ is a solution, we remain with the case $f(1) = 0$.\n\nNow suppose $f(0) = f(1) = 0$. Setting $y = -x$ gives $0 = f(x)f(-x) - xf(x)$, that is\n$$\nf(x) = 0 \\quad \\lor \\quad f(-x) = x. \\tag{4}\n$$\nSetting $y = 1 - x$ gives $f(x) = f(x)f(1-x) + (1-x)f(x)$, that is\n$$\nf(x) = 0 \\quad \\lor \\quad f(1-x) = x. \\tag{5}\n$$\nIf $f(x) \\neq 0$ for some $x \\neq 0$, we have $f(-x) = x \\neq 0$ and $f(1-x) = x \\neq 0$ and applying (4) and (5) to $-x$ and $1-x$ instead of $x$ we get $f(x) = -x = 1-x$, a contradiction. Hence $f(x) = 0$ for all $x \\in \\mathbb{R}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 29515,
    "subject": "Mathematics (Multi-modal)",
    "question": "Paul is filling the cells of a rectangular table alternately with crosses and circles (he starts with a cross). When the table is filled in completely, he determines his score as $X - O$ where $X$ is the sum of squares of the numbers of crosses in all the rows and columns, and $O$ is the sum of squares of the numbers of circles in all the rows and columns. Find all possible values of the score for a $67 \\times 67$ table.",
    "options": [],
    "answer": "134",
    "solution": "Let $n = 67$ and denote by $k = \\frac{1}{2}(n^2+1)$ the total number of crosses in the table. A row containing $a$ crosses and $n-a$ circles contributes $a^2-(n-a)^2 = 2n \\cdot a - n^2$ to the total score and thus all the $n$ rows combined contribute\n$$\n2n \\cdot k - n \\cdot n^2 = 2n \\cdot \\frac{n^2+1}{2} - n^3 = n\n$$\nto the total score. Likewise, columns contribute $n$. Hence the total score is always equal to $2n = 134$.\nConsider an $n \\times n$ table filled with arbitrarily many crosses and circles. We show that replacing any circle by a cross increases the score by $4n$. Since the score for a table filled with all circles equals $-2n^3$ and Paul's table contains $\\frac{1}{2}(n^2+1)$ crosses, the final score will always be equal to $-2n^3 + 4n \\cdot \\frac{1}{2}(n^2+1) = 2n$.\n\nConsider any cell containing a circle and denote by $r$ and $c$ the number of crosses in its row and column, respectively. The contribution of this row and column changes from\n$$\nA = r^2 - (n-r)^2 + s^2 - (n-s)^2 = 2n(r+s) - 2n^2\n$$\nto\n$$\nB = (r+1)^2 - (n-r-1)^2 + (s+1)^2 - (n-s-1)^2 = 2n(r+1+s+1) - 2n^2\n$$\nand the contribution of other rows and columns doesn't change. Since $B - A = 4n$, we are done.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 29516,
    "subject": "Mathematics (Multi-modal)",
    "question": "In a certain club, some pairs of members are friends. Given $k \\ge 3$, we say that a club is *k*-good if every group of *k* members can be seated around a round table such that every two neighbors are friends. Prove that if a club is 6-*good* then it is 7-*good*.\n\n(Josef Tkadlec)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Consider a 6-good club and denote some seven of its members by $A, \\dots, G$. It suffices to show that $A, \\dots, G$ can be seated around a table as required. Consider only friendships among $A, \\dots, G$. First, we show that every member has at least three friends.\n\nWithout loss of generality consider $G$. By assumption, $B, \\dots, G$ can be seated as required, hence $G$ has at least two friends. Without loss of generality, $F$ is one of them. By assumption, $A, \\dots, E, G$ (omitting $F$) can be seated as required, hence $G$ has at least two more friends apart from $F$ for a total of at least three friends.\n\nSince every member has at least three friends, there exists a member with at least four friends (otherwise the number of friendly pairs equals $\\frac{1}{2} \\cdot 7 \\cdot 3$, which is clearly impossible). Without loss of generality, assume $G$ has at least four friends.\n\nBy assumption, $A, \\dots, F$ can be seated as required. In such a seating, some two of the four friends of $G$ are neighbors and we can seat $G$ in between them.",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 29517,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABC$ be an acute scalene triangle. Let $D$ and $E$ be points on the sides $AB$ and $AC$, respectively, such that $BD = CE$. Denote by $O_1$ and $O_2$ the circumcentres of the triangles $ABE$ and $ACD$, respectively. Prove that the circumcircles of the triangles $ABC$, $ADE$ and $AO_1O_2$ have a common point different from $A$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $Z$ be the midpoint of the longer arc $BC$ of the circumcircle $\\omega$ of the triangle $ABC$. The triangles $ZDB$ and $ZEC$ are congruent, because they agree in the sides $BD = CE$ and $ZB = ZC$, as well as in the corresponding angles between them, for both lie over the chord $AZ$ of $\\omega$. It follows that $\\angle ZDA = \\angle ZEA$, which in turn discloses that the quadrilateral $ADEZ$ is cyclic. So it remains to be shown that the quadrilateral $AO_1O_2Z$ is cyclic.\n\n![](attached_image_1.png)\n\nThe center $O$ of $\\omega$ satisfies $OO_1 \\perp AB$ and $OO_2 \\perp AC$. The projections of $O$ and $O_1$ onto $AC$ are the midpoints of $AC$ and $AE$ respectively. Thus the projection of the segment $OO_1$ onto $AC$ has length $\\frac{1}{2}CE$. For the same reason, the projection of $OO_2$ on $AB$ has length $\\frac{1}{2}BD$, and by hypothesis these two length agree. Moreover, the angle between $OO_1$ and $AC$ is the same as the angle between $OO_2$ and $AB$. It follows that $OO_1 = OO_2$.\nFurther, we have $\\angle AOO_1 = \\angle ACB = \\angle O_2OZ$, the latter being a consequence of $ZO \\perp BC$ and $OO_2 \\perp AC$. So the rays $OA$ and $OZ$ are isogonal in the angle $O_2OO_1$. In the combination with $AO = ZO$ and $OO_1 = OO_2$ this proves that\n\nthe quadrilateral $AO_1O_2Z$ is an isosceles trapezium and thus in particular cyclic.\nThereby the problem is solved.\nLet the circumcircles of triangles $ABE$ and $ADC$ intersect each other again at $F \\neq A$. Then the triangles $BFD$ and $EFC$ are congruent, for they agree in their sides $BD = CE$ as well as in their corresponding adjacent angles, i.e., $\\angle FBD = \\angle FEC$ and $\\angle BDF = \\angle ECF$. It follows that the altitudes of these triangles passing through $F$ have the same lengths, wherefore $AF$ is the bisector of the angle $BAC$.\n\n![](attached_image_2.png)\n\nNow construct the point $S$ such that $SO_1FO_2$ is a parallelogram. We will show that $S$ is the desired point.\nTo prove that $S$ lies on the circumcircle of triangle $AO_1O_2$, we note that the triangles $AO_1O_2$ and $FO_1O_2$ are congruent due to $AO_1 = FO_1$ and $AO_2 = FO_2$. It follows that $AO_1O_2S$ is an isosceles trapezium and hence in particular a cyclic quadrilateral, as claimed. Later, it will help us to have observed that the facts used in this paragraph imply $AS \\parallel O_1O_2 \\perp AF$.\nNext, we prove that $S$ lies on the circumcircle of the triangle $ABC$ and that it is actually the midpoint of its longer arc $BC$; this will also show $S \\neq A$, as needed. Our first intermediate step is to observe that the triangles $O_1SB$ and $O_2CS$ are congruent. Indeed they agree in a pair of sides, $BO_1 = FO_1 = SO_2$ and $SO_1 = FO_2 = CO_2$. Moreover the corresponding angles between these sides are equal, because their complements to $360^\\circ$ are equal as a consequence of\n$$\n\\angle BO_1F = 2\\angle BAF = 2\\angle FAC = \\angle FO_2C\n$$\nand $\\angle FO_1S = \\angle SO_2F$. This concludes the verification of $\\triangle O_1SB \\cong \\triangle O_2CS$, and it follows that $BS = CS$. Further, since $AF$ is the bisector of $\\angle BAC$ and $AF \\perp AS$, the line $AS$ is the exterior bisector of $\\angle BAC$. Altogether we obtain that $S$ is the point described above. The fact that $S$ lies on the circumcircle of the triangle $ADE$ can be shown as in the first solution.\nDenote by $O$ and $P$ the circumcentres of triangles $ABC$ and $ADE$, respectively. The lines $OO_1$ and $O_2P$ (being the perpendicular bisectors of $AB$ and $AD$, respectively) are both perpendicular to $AB$ and their distance is $\\frac{1}{2}BD$. Similarly, the lines $OO_2$ and $O_1P$ are both perpendicular to $AC$ and their distance is $\\frac{1}{2}CE$. Since $\\frac{1}{2}BD = \\frac{1}{2}CE$, the quadrilateral $O_1OO_2P$ is a parallelogram with equal altitudes, hence a rhombus. It follows that $OP$ is the perpendicular bisector of $O_1O_2$, so all the three circumcentres of the triangles $ABC$, $ADE$ and $AO_1O_2$ lie on the same line, which concludes the claim (since $A$ does not lie on this line because of $AB \\neq AC$).\n\n![](attached_image_3.png)",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 29518,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $x, y, z$ be real numbers such that\n$$\n\\frac{1}{|x^2 + 2yz|}, \\quad \\frac{1}{|y^2 + 2zx|}, \\quad \\frac{1}{|z^2 + 2xy|}\n$$\nare side-lengths of a (non-degenerate) triangle. Find all possible values of $xy + yz + zx$.\n(Michal Rolínek)",
    "options": [],
    "answer": "all real numbers except zero",
    "solution": "If $x = y = z = t > 0$ then the three fractions are sides of an equilateral triangle and $xy + yz + zx = 3t^2$, hence $xy + yz + zx$ can attain all positive values. Similarly, for $x = y = t > 0$ and $z = -2t$ the three fractions are $\\frac{1}{3}t^{-2}$, $\\frac{1}{3}t^{-2}$, $\\frac{1}{6}t^{-2}$ which are positive numbers that are side-lengths of an isosceles triangle ($\\frac{1}{6} < \\frac{1}{3} + \\frac{1}{3}$). Since $xy + yz + zx = -3t^2$, any negative value can be attained too.\n\nNext we show that $xy + yz + zx$ can't be 0. Assume otherwise. Numbers $x, y, z$ are mutually distinct: if, say, $x$ and $y$ were equal then the denominator of the first fractions would be equal to $|x^2 + 2yz| = |xy + (yz + xz)| = 0$ which is impossible.\n\nLet's look at the fractions without absolute values. Subtracting $xy + yz + zx = 0$ from each denominator we get\n$$\n\\begin{aligned}\n\\frac{1}{x^2 + 2yz} + \\frac{1}{y^2 + 2zx} + \\frac{1}{z^2 + 2xy} &= \\\\\n&= \\frac{1}{(x-y)(x-z)} + \\frac{1}{(y-z)(y-x)} + \\frac{1}{(z-x)(z-y)} = \\\\\n&= \\frac{(z-y) + (x-z) + (y-x)}{(x-y)(y-z)(z-x)} = 0.\n\\end{aligned}\n$$\n\nThis implies that among the original fractions (with absolute values), one of them is a sum of the other two. Hence the fractions don't fulfil triangle inequality and we reached the desired contradiction.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 29519,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $k$ be a semicircle with diameter $PQ$. Consider a chord $BC$ of fixed length $d$ whose endpoints are distinct from $P$, $Q$. A ray of light emanating from $B$ reaches point $C$ after reflecting from $PQ$ at such a point $A$ that $\\angle PAB = \\angle QAC$. Prove that $\\angle BAC$ doesn't depend on the position of the chord $BC$ on $k$.\n\n(Šárka Gergelitsová)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Reflect $k$ and $C$ about $PQ$ to get $l$ and $C'$, respectively (Fig. 1). Then $C'$ lies on $l$ and since $\\angle QAC' = \\angle QAC = \\angle PAB$ it also lies on $BA$. Triangle $C'CA$ is isosceles, hence\n$$\n\\angle BAC = \\angle AC'C + \\angle ACC' = 2 \\cdot \\angle BC'C\n$$\nThe chord $BC$ of circle $k \\cup l$ has a fixed length, hence the corresponding inscribed angle $BC'C$ has fixed size and we may conclude.\n\n![](attached_image_1.png)\nFig. 1\nLet $O$ be the midpoint of $PQ$. We will show that $O$ lies on the circumcircle of triangle $ABC$ (Fig. 2). This will imply that $\\angle BAC = \\angle BOC$ which is clearly fixed.\nObserve that $O$ lies on the perpendicular bisector of $BC$. Moreover, if $O \\neq A$ then $AO$ is the external $A$-angle bisector with respect to triangle $ABC$. Therefore $O$ is the midpoint of arc $BAC$.\n\n![](attached_image_2.png)\nFig. 2",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 29520,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $a \\neq b$ be positive real numbers. Consider the equation\n$$\n\\lfloor a x + b \\rfloor = \\lfloor b x + a \\rfloor\n$$\nwhere $\\lfloor y \\rfloor$ denotes the largest integer not exceeding $y$. Prove that the set of real solutions $x$ to this equation contains an interval of length at least\n$$\n\\frac{1}{\\max\\{a, b\\}} \\qquad (\\text{Patrik Bak})\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Consider linear functions $f(x) = a x + b$, $g(x) = b x + a$. Since $a, b$ are distinct and positive, their graphs are two distinct lines with positive slope. As $f(1) = g(1) = a + b$, point $P = [1, a + b]$ is the intersection of these lines (Fig. 3).\n\nWithout loss of generality, assume $b > a$ (i.e. the line determined by $g$ is the “steeper” one). Then $f(x) > g(x)$ for $x < 1$, whereas $f(x) < g(x)$ for $x > 1$: indeed,\n$$\nf(x) - g(x) = (a x + b) - (b x + a) = (b - a)(1 - x).\n$$\n\n![](attached_image_1.png)\nFig. 3\n\nLet $t = \\lfloor a + b \\rfloor$ and consider $x_1 \\le 1 < x_2$ such that $g(x_1) = t$ and $g(x_2) = t + 1$ (that is, $x_1 = \\frac{t - a}{b}$ and $x_2 = \\frac{t + 1 - a}{b}$). We claim that the interval $[x_1, x_2]$ has all the desired properties.\n\nFirst, for any $x \\in [x_1, x_2)$ we have\n$$\nt = g(x_1) \\le \\min\\{f(x), g(x)\\} \\le \\max\\{f(x), g(x)\\} < g(x_2) = t + 1,\n$$\nand thus $x$ is a solution to the equation.\n\nSecond,\n$$\n1 = (t + 1) - t = b x_2 + a - (b x_1 + a) = b(x_2 - x_1),\n$$\nand thus $x_2 - x_1 = 1 / b = 1 / \\max\\{a, b\\}$ and the interval has the desired length.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 29521,
    "subject": "Mathematics (Multi-modal)",
    "question": "There are $2018$ players sitting around a round table. At the beginning of the game we arbitrarily deal all the cards from a deck of $K$ cards to the players (some players may receive no cards). In each turn we choose a player who draws one card from each of the two neighbours. It is only allowed to choose a player whose each neighbour holds a nonzero number of cards. The game terminates when there is no such player. Determine the largest possible value of $K$ such that, no matter how we deal the cards and how we choose the players, the game always terminates after a finite number of turns.\n(Peter Novotný, Slovakia)",
    "options": [],
    "answer": "2017",
    "solution": "The answer is $K = 2017$.\nFor $K = 2018$, we deal $2$ cards to one player, $0$ cards to one of his neighbours and $1$ card to everyone else. Then in each turn we choose the player with $0$ cards:\n$$\n\\dots 11 \\overbrace{\\underbrace{20}_{\\sim}}^{\\sim} 11 \\dots \\rightarrow \\dots 11 \\overbrace{\\underbrace{20}_{\\sim}}^{\\sim} 11 \\dots\n$$\nAfter each turn, the configuration stays the same – there is one player with $2$ cards, one of his neighbours with $0$ cards and all the others with $1$ card (the only change is that the positions of the players with $2$ and $0$ cards is shifted). Therefore we can make moves forever and the game never terminates.\nWhenever $K > 2018$, we can play forever using the same strategy as for $K = 2018$. We simply deal the extra cards arbitrarily and ignore them during the game.\nNow we will prove that for $K = 2017$ the game terminates no matter how we play. Let us call zeros the players with no cards and ones the players with exactly one card. The zeros split the other players into segments of various lengths. When two zeros sit next to each other, they form a segment with a length of $0$. Also note that there is obviously at least one zero when $K = 2017$.\n\n**Lemma.** There exists a segment containing no other players than ones (possibly with a length of $0$).\n*Proof.* If we add to each segment the zero which bounds it in the clock-wise direction, then the sum of the lengths of all the segments will be $2018$. There are only $2017$ cards, therefore at least one segment contains fewer cards than players, which is possible only when all the players of this segment, except for the bounding zero, are ones. $\\square$\nLet us consider the shortest segment among the ones containing no other players than ones; the lemma assures the existence of such a segment. If we choose a zero adjacent to this segment, we shorten it by $1$ (or by $2$ — in the special case when there is exactly one zero in the game):\n$$\n\\dots * \\overbrace{0111\\dots10\\dots}^{\\hat{0}11\\dots10\\dots} \\rightarrow \\dots * \\overbrace{2011\\dots10\\dots}^{\\hat{2}11\\dots10\\dots}\n$$\nIf we choose one of the ones inside of the shortest segment, we create two even shorter segments:\n$$\n\\dots 01\\dots1 \\overbrace{111}^{\\hat{1}} 1\\dots10\\dots \\rightarrow \\dots 01\\dots1 \\overbrace{0301}^{\\hat{0}11\\dots10\\dots} \\rightarrow\n$$\nThe length of the shortest segment could decrease only finitely many times. From the moment when it stops decreasing we won't be able to choose any of the zeros bounding the shortest segment, nor any of the ones inside of it. This means that the game will continue on the other side of the table between the bounding zeros of the shortest segment. The neighbours of these two zeros won't be able to get any more cards, so we cannot choose them anymore. The neighbours of these neighbours will thereby be chosen at most finitely many times (at most the number of times equal to the number of cards of these neighbours), so after some time we won't be able to choose them. We can use this reasoning repeatedly. The part of the table where we still can choose players eventually decreases, which means that the game cannot last infinitely long.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 29522,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABC$ be a triangle. The $A$-angle bisector intersects $BC$ at $D$. Let $E$, $F$ be the circumcenters of triangles $ABD$, $ACD$, respectively. Given that the circumcenter of triangle $AEF$ lies on $BC$, find all possible values of $\\angle BAC$. (Patrik Bak)",
    "options": [],
    "answer": "120°",
    "solution": "Let $O$ be the circumcenter of triangle $AEF$ and denote $\\alpha = \\angle BAC$. Since $\\angle BAD$ and $\\angle CAD$ are acute (Fig. 2), points $E$, $F$ lie in the half-plane $BCA$ and the Inscribed angle theorem yields\n$$\n\\angle BED = 2 \\cdot \\angle BAD = \\alpha = 2 \\cdot \\angle DAC = \\angle DFC.\n$$\n\n![](attached_image_1.png)\n\nFig. 2\n\nThe isosceles triangles $BED$ and $DFC$ are thus similar and we easily compute that $\\angle EDF = \\alpha$ and that $BC$ is the external $D$-angle bisector in triangle $DEF$.\n\nPoint $O$ lies on $BC$ and on the perpendicular bisector of $EF$. Framed with respect to triangle $DEF$, it lies on the external $D$-angle bisector and on the perpendicular bisector of the opposite side $EF$. Thus it is the midpoint of arc $EDF$ and $\\angle EOF = \\angle EDF = \\alpha$.\nQuadrilateral $AEDF$ is a kite, hence $\\angle EAF = \\alpha$. Moreover, line $EF$ separates points $A$ and $O$, thus the Inscribed angle theorem implies that the size of the reflex angle $EOF$ is twice the size of the convex angle $EAF$. This yields $360^\\circ - \\alpha = 2 \\cdot \\alpha$ and $\\alpha = 120^\\circ$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 29523,
    "subject": "Mathematics (Multi-modal)",
    "question": "Consider positive integers $a, b, c$ that are side-lengths of a non-degenerate triangle and such that $\\text{GCD}(a, b, c) = 1$ and the fractions\n$$\n\\frac{a^2 + b^2 - c^2}{a + b - c}, \\quad \\frac{b^2 + c^2 - a^2}{b + c - a}, \\quad \\frac{c^2 + a^2 - b^2}{c + a - b}\n$$\nare all integers. Prove that the product of the denominators of the three fractions is either a square or twice a square of an integer. (Jaromír Šimša)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $z = a + b - c$, $x = b + c - a$, $y = c + a - b$ be the (positive) denominators. Then $a = (y + z)/2$, $b = (x + z)/2$, $c = (x + y)/2$ and\n$$\na^2 + b^2 - c^2 = \\frac{1}{4}((y + z)^2 + (x + z)^2 - (x + y)^2) = \\frac{1}{2}(z(z + x + y) - x y),\n$$\nhence $z \\mid x y$ and likewise $y \\mid x z$ and $x \\mid y z$.\n\nFor a prime $p$, let $i_p$ be the largest exponent such that $p^{i_p} \\mid x y z$. It suffices to show that for all odd primes $p$ the corresponding $i_p$ is even. If $i_2$ is also even then $x y z$ is a square. Otherwise, it is twice a square.\nFix odd prime $p$ and consider the largest exponents $\\alpha, \\beta, \\gamma$ such that $p^\\alpha \\mid x$, $p^\\beta \\mid y$, $p^\\gamma \\mid z$. Without loss of generality, assume $\\min\\{\\alpha, \\beta, \\gamma\\} = \\gamma$. If $\\gamma > 0$ then $p$ divides each of $x, y, z$ and thus it divides each of $a, b, c$ ($p$ is odd), contradicting $\\text{GCD}(a, b, c) = 1$. Therefore $\\gamma = 0$.\nFrom $x \\mid y z$ we infer $\\alpha \\le \\beta$. Likewise, from $y \\mid x z$ we infer $\\beta \\le \\alpha$. Hence $\\beta = \\alpha$ and $i_p = \\alpha + \\beta + \\gamma = 2\\alpha$ is an even number as desired.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 29524,
    "subject": "Mathematics (Multi-modal)",
    "question": "Do there exist positive integers $n$, $k$ such that\n$$\n\\frac{n}{11^k - n}\n$$\nis a square of an integer?",
    "options": [],
    "answer": "No",
    "solution": "Such numbers don't exist. For the sake of contradiction, assume that there exist positive integers $n$, $k$, $a$ such that\n$$\n\\frac{n}{11^k - n} = a^2\n$$\nwhich rewrites as\n$$\nn(a^2 + 1) = a^2 \\cdot 11^k.\n$$\nFrom $\\text{GCD}(a^2, a^2+1) = 1$ we deduce $a^2+1 \\mid 11^k$ and hence $a^2+1 = 11^t$ for $1 \\le t \\le k$. In particular, $a^2 \\equiv 10 \\pmod{11}$. However, this is impossible as the squares of integers give remainders $0, 1, 4, 9, 5, 3, 3, 5, 9, 4, 1, \\dots$ upon division by $11$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 29525,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABCD$ be an isosceles trapezoid with longer base $AB$. Let $I$ be the incenter of triangle $ABC$ and $J$ the $C$-excenter of triangle $ACD$. Prove that $IJ$ and $AB$ are parallel. (Patrik Bak)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $K$ be the incenter of triangle $ABD$. Since $IK \\parallel AB$, it suffices to show $JK \\parallel AB$. Let $\\angle ABD = \\angle ACD = \\varphi$. Then $\\angle AKD = 90^\\circ + \\frac{1}{2}\\varphi$ and $\\angle DJA = 90^\\circ - \\frac{1}{2}\\varphi$, implying that the quadrilateral $AKDJ$ is cyclic (Fig. 3).\n\n![](attached_image_1.png)\nFig. 3\n\nAs $AK$, $DJ$ are bisectors of alternate interior angles, they are parallel. Together with the cyclic quadrilateral we obtain $\\angle AKJ = \\angle ADJ = \\angle DAK = \\angle KAB$ which concludes the proof.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 29526,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find the smallest positive integer $n$ such that for any coloring of numbers $1, 2, 3, \\ldots, n$ by three colors there exist two numbers with the same color whose difference is a square of an integer. (Vojtech Bálint, Michal Rolínek, Josef Tkadlec)",
    "options": [],
    "answer": "29",
    "solution": "The answer is $n = 29$.\n\nFirst, for the sake of contradiction, assume that numbers $1, 2, \\ldots, 29$ can be colored by colors $A, B, C$ such that no two numbers with the same color differ by a square. Let $f(i)$ be the color of number $i$ for $i \\in \\{1, 2, \\ldots, 29\\}$.\n\nSince $9, 16$, and $25$ are squares, numbers $1, 10, 26$ are all assigned distinct colors. The same is true for numbers $1, 17, 26$, hence $10$ and $17$ are assigned the same color. Likewise we get $f(11) = f(18)$, $f(12) = f(19)$ and $f(13) = f(20)$ (for the last equality we look at numbers $4, 13, 20, 29$).\n\nWithout loss of generality, assume $f(10) = f(17) = A$. As $11 = 10 + 1^2$, we have $f(11) \\neq f(10)$. Without loss of generality, let $f(11) = f(18) = B$. Now $19 = 18 + 1^2 = 10 + 3^2$, hence $f(12) = f(19) = C$. Similarly, $20 = 19 + 1^2 = 11 + 3^2$ implies $f(13) = f(20) = A$. We have derived $f(13) = A = f(17)$, a contradiction.\n\nOn the other hand, if $n \\le 28$, we may color the numbers as below. It's easy to check that no two numbers with the same color differ by a square of an integer.\n\n![](attached_image_1.png)\n\nFig. 4",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 29527,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find the largest integer $d$ for which a $43 \\times 47$ table can be filled with numbers $1$ and $2$ such that the sum of the numbers in each column and in each row is a multiple of $d$. (Do not forget to show that no larger $d$ works.) (Tomáš Bárta)",
    "options": [],
    "answer": "47",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 29528,
    "subject": "Mathematics (Multi-modal)",
    "question": "Is it possible to fill an $n \\times n$ table with numbers $1$ and $2$ such that the sum of the numbers in each column is a multiple of $5$ and the sum of the numbers in each row is a multiple of $7$? Solve this:\n\na) for $n = 9$\n\nb) for $n = 12$\n\n(Tomáš Bárta)",
    "options": [],
    "answer": "a) No; b) Yes",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 29529,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABCD$ be a trapezoid ($AB \\parallel CD$) and denote $P = BC \\cap AD$. Let $k_1, k_2$ be circles with diameters $BC, AD$, respectively, and denote $P = BC \\cap AD$. Prove that the tangents from $P$ to $k_1$ form the same angle as the tangents from $P$ to $k_2$. (Patrik Bak)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 29530,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABC$ be an acute triangle and $I$ its incenter. Rays $BI$, $CI$ meet the circumcircle of triangle $ABC$ again at $S \\neq B$, $T \\neq C$, respectively. The segment $ST$ meets the sides $AB$, $AC$ at $K$, $L$, respectively. Prove that $AKIL$ is a rhombus. (Josef Tkadlec)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 29531,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all integers $n > 2$ such that $n^{n-2}$ is an $n$th power of some integer. (Patrik Bak)",
    "options": [],
    "answer": "n = 4",
    "solution": "Let $n > 2$ be an integer. We want $n^{n-2}$ to be an $n$th power of some integer, i.e., there exists an integer $k$ such that $n^{n-2} = k^n$.\n\nThis means that $n^{n-2}$ is a perfect $n$th power. Let us write $n^{n-2} = (n^a)^n$ for some integer $a$.\n\nBut $(n^a)^n = n^{an}$, so we need $an = n-2$, i.e., $a = \\frac{n-2}{n}$.\n\nBut $a$ is an integer, so $n$ divides $n-2$. Since $n > 2$, $n$ divides $n-2$ implies $n$ divides $-2$.\n\nThe positive divisors of $2$ are $1$ and $2$, but $n > 2$, so the only possibility is $n = 2$ (which is excluded), or $n = -1$ or $n = -2$ (which are not $> 2$).\n\nAlternatively, $n$ divides $-2$, so $n = 1$ or $2$ (excluded), or $n = -1$, $-2$ (excluded).\n\nTherefore, there are no integers $n > 2$ such that $n^{n-2}$ is an $n$th power of an integer.\n\nAnswer: There are no such integers $n > 2$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 29532,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all pairs of positive integers $a, b$ such that $a^{a-b} = b^a$.",
    "options": [],
    "answer": "All pairs (k^k, k^{k-1}) for positive integers k.",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 29533,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all triples $(x, y, z)$ of real numbers such that\n$$\n\\begin{aligned}\nyz + 2 &= x^2, \\\\\nzx + 3 &= y^2.\n\\end{aligned}\n$$",
    "options": [],
    "answer": "All solutions are given as follows. Either x = 0 with y = ±√3 and z = ∓2/√3, or y = 0 with x = ±√2 and z = ∓3/√2. Otherwise, let t = y − x ≠ 0 and let S = x + y. Then S and t satisfy 3 t S^2 − 2 S + (t^3 − 10 t) = 0, so\nS = [1 ± sqrt(1 + 30 t^2 − 3 t^4)] / (3 t),\nand the corresponding triple is\nx = (S − t)/2,\ny = (S + t)/2,\nz = 1/t − S.\nConversely, every real triple satisfying the system arises in this way (with the square root defined for t such that 1 + 30 t^2 − 3 t^4 ≥ 0).",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 29534,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABC$ be a scalene triangle, $I$ its incenter and $k$ its circumcircle. Rays $BI, CI$ meet $k$ again at $S_b \\neq B$, $S_c \\neq C$, respectively. Prove that the tangent to $k$ at $A$, the line $S_bS_c$, and the line through $I$ parallel to $BC$ are concurrent. (Patrik Bak)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 29535,
    "subject": "Mathematics (Multi-modal)",
    "question": "Consider an infinite sequence $a_0, a_1, a_2, \\dots$ of integers that satisfies $a_0 \\ge 2$ and $a_{n+1} \\in \\{2a_n - 1, 2a_n + 1\\}$ for all indices $n \\ge 0$. Prove that any such infinite sequence contains infinitely many composite numbers. (Martin Melicher, Josef Tkadlec)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 29536,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $k \\le 2022$ be a positive integer. Alice and Bob play a game on a $2022 \\times 2022$ board. Initially, all cells are white. Alice starts and the players alternate. In her turn, Alice can either color one white cell in red or pass her turn. In his turn, Bob can either color a $k \\times k$ square of white cells in blue or pass his turn. Once both players pass, the game ends and the person who colored more cells wins (a draw can occur). For each $1 \\le k \\le 2022$, determine which player (if any) has a winning strategy. (David Hruška)",
    "options": [],
    "answer": "Let n = 2022 and q = floor(n/k). Under optimal play Bob colors exactly floor(q^2/2) blocks, i.e., floor(q^2/2)·k^2 cells. Hence Bob never gets more than half the board. The outcome is a draw if and only if 2k divides 2022 (equivalently k ∈ {1, 3, 337, 1011}); for all other k, Alice has a winning strategy. Bob has no winning strategy for any k.",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 29537,
    "subject": "Mathematics (Multi-modal)",
    "question": "In a sequence of $71$ nonzero real numbers, each number (apart from the first one and the last one) is one less than the product of its two neighbors. Prove that the first and the last number are equal.\n\n(Josef Tkadlec)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 29538,
    "subject": "Mathematics (Multi-modal)",
    "question": "Is it possible to fill the $8 \\times 8$ table with numbers $6$ and $7$ such that the sum of the numbers in each column is a multiple of $5$ and the sum of the numbers in each row is a multiple of $7$? (Josef Tkadlec)",
    "options": [],
    "answer": "No",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 29539,
    "subject": "Mathematics (Multi-modal)",
    "question": "We say that a positive integer $k$ is *fair* if the number of $2021$-digit palindromes that are a multiple of $k$ is the same as the number of $2022$-digit palindromes that are a multiple of $k$. Does the set $M = \\{1, 2, \\dots, 35\\}$ contain more numbers that are fair or those that are not fair?\n\n(A palindrome is an integer that reads the same forward and backward.)\n\n(David Hruška)",
    "options": [],
    "answer": "more fair numbers",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 29540,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all triples $(x, y, z)$ of positive real numbers such that\n$$\n\\begin{aligned}\nx^2 + 2y^2 &= x + 2y + 3z, \\\\\ny^2 + 2z^2 &= 2x + 3y + 4z, \\\\\nz^2 + 2x^2 &= 3x + 4y + 5z.\n\\end{aligned}\n$$\n(Patrik Bak)",
    "options": [],
    "answer": "(10/3, 5/3, 10/3)",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 29541,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all functions $f: (0, \\infty) \\to (0, \\infty)$ such that\n$$\nf\\left(f(x) + \\frac{y+1}{f(y)}\\right) = \\frac{1}{f(y)} + x + 1\n$$\nfor all $x, y > 0$.\n(Dominik Burek)",
    "options": [],
    "answer": "f(x) = x for all x > 0",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 29542,
    "subject": "Mathematics (Multi-modal)",
    "question": "Circles $\\Omega_1$ and $\\Omega_2$ with different radii intersect at two points, denote one of them by $P$. A variable line $\\ell$ passing through $P$ intersects the arc of $\\Omega_1$ which is outside of $\\Omega_2$ at $X_1$, and the arc of $\\Omega_2$ which is outside of $\\Omega_1$ at $X_2$. Let $R$ be the point on segment $X_1X_2$ such that $X_1P = RX_2$. The tangent to $\\Omega_1$ through $X_1$ meets the tangent to $\\Omega_2$ through $X_2$ at $T$. Prove that line $RT$ is tangent to a fixed circle, independent of the choice of $\\ell$. (Josef Tkadlec)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 29543,
    "subject": "Mathematics (Multi-modal)",
    "question": "Given a scalene acute triangle $ABC$, let $M$ be the midpoint of its side $BC$ and $N$ the midpoint of the arc $BAC$ of its circumcircle. Let $\\omega$ be the circle with diameter $BC$ and $D, E$ its intersections with the bisector of angle $BAC$. Points $D', E'$ lie on $\\omega$ such that $DED'E'$ is a rectangle. Prove that $D', E'$, $M, N$ lie on a single circle.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 29544,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABC$ be an isosceles triangle with base $AB$ and $P$ a point on its $C$-altitude. Ray $AP$ meets the circumcircle of the triangle $ABC$ again at $Q \\neq A$. The line through $P$ parallel to $AB$ meets the side $BC$ at $R$. Prove that $QR$ bisects the angle $AQB$. (Jaroslav Švrček)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 29545,
    "subject": "Mathematics (Multi-modal)",
    "question": "Given a positive integer $n$, denote by $\\tau(n)$ the number of positive divisors of $n$, and by $\\sigma(n)$ the sum of all positive divisors of $n$. Find all positive integers $n$ satisfying\n$$\n\\sigma(n) = \\tau(n) \\cdot \\lceil\\sqrt{n}\\rceil.\n$$\n(Here, $\\lceil x \\rceil$ denotes the smallest integer not less than $x$.)\n(Michael Reitmeir)",
    "options": [],
    "answer": "[1, 3, 5, 6]",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 29546,
    "subject": "Mathematics (Multi-modal)",
    "question": "Consider an infinite sequence $a_0, a_1, a_2, \\dots$ of integers that satisfies\n$$\na_0 \\ge 1 \\quad \\text{and} \\quad a_{n+1} \\in \\{2022a_n - 1, 2022a_n + 1\\}\n$$\nfor all indices $n \\ge 0$. Prove that any such infinite sequence contains infinitely many composite numbers.\n(Martin Melicher)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 29547,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABCD$ be a convex quadrilateral with $AB = BC = CD$ and $P$ its intersection of diagonals. Denote by $O_1, O_2$ the circumcenters of triangles $ABP, CDP$, respectively. Prove that $O_1BCO_2$ is a parallelogram.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 29548,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABC$ be a triangle with $AB < AC$ and circumcenter $O$. The angle bisector of $\\angle BAC$ meets the side $BC$ at $D$. The line through $D$ perpendicular to $BC$ meets the segment $AO$ at $X$. Furthermore, let $Y$ be the midpoint of segment $AD$. Prove that points $B, C, X, Y$ lie on a single circle. (Karl Czakler)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 29549,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all integers $n$ such that $2^n + n^2$ is a square of an integer.",
    "options": [],
    "answer": "n = 0 and n = 6",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 29550,
    "subject": "Mathematics (Multi-modal)",
    "question": "Consider any graph with $50$ vertices and $225$ edges. We say that a triplet of its (mutually distinct) vertices is *connected* if the three vertices determine at least two edges. Determine the smallest and the largest possible number of connected triples.",
    "options": [],
    "answer": "minimum 600, maximum 5400",
    "solution": "",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 29551,
    "subject": "Mathematics (Multi-modal)",
    "question": "Consider 26 letters $A, \\dots, Z$. A *string* is a finite sequence consisting of those letters. We say that a string $s$ is *nice* if it contains each of the 26 letters at least once, and each permutation of letters $A, \\dots, Z$ occurs in $s$ as a subsequence the same number of times. Prove that:\n\na. There exists a nice string.\n\nb. Any nice string contains at least 2022 letters.\n\n(Here, a permutation $\\pi$ of the 26 letters is a *subsequence* of a string $s = s_1s_2s_3\\dots$ if there exist 26 indices $i_1 < i_2 < \\dots < i_{26}$ such that $\\pi = s_{i_1}s_{i_2}\\dots s_{i_{26}}$.) (Václav Rozhoň)",
    "options": [],
    "answer": "2022",
    "solution": "",
    "topic": "Discrete Mathematics",
    "subtopic": "Combinatorics"
  },
  {
    "id": 29552,
    "subject": "Mathematics (Multi-modal)",
    "question": "In the domain of non-negative real numbers solve the system of equations\n$$\n\\begin{align*}\n\\lfloor 3x + 5y + 7z \\rfloor &= 7z, \\\\\n\\lfloor 3y + 5z + 7x \\rfloor &= 7x, \\\\\n\\lfloor 3z + 5x + 7y \\rfloor &= 7y.\n\\end{align*}\n$$",
    "options": [],
    "answer": "(0, 0, 0), (1/7, 0, 0), (0, 1/7, 0), (0, 0, 1/7)",
    "solution": "The first equation of the given system is fulfilled if and only if the following two conditions are satisfied:\n▷ the number $7z$ is integer,\n▷ $7z \\leq 3x + 5y + 7z < 7z + 1$, i.e. $3x + 5y \\in [0, 1)$.\n\nSimilarly, the second and third equations are fulfilled if and only if the numbers $7x$ and $7y$ are integers and $3y + 5z$, $3z + 5x \\in [0, 1)$.\n\nNow consider any triple of non-negative numbers $(x, y, z)$, which is the solution to the problem. The inequalities $z \\geq 0$ and $3z + 5x < 1$ imply $5x < 1$, whence $7x < 7/5 < 2$. This means that non-negative integer $7x$ is equal to one of the numbers $0$ or $1$, i.e., $x \\in \\{0, 1/7\\}$. Similarly, $y, z \\in \\{0, 1/7\\}$.\n\nAt this point we have only $2^3 = 8$ triples $(x, y, z)$, which are candidates to solve the problem, so we could test them individually. However, this testing can be avoided by noting that if any two of the numbers $x, y, z$ were equal to $1/7$, one of the expressions $3x+5y$, $3y+5z$, $3z+5x$ would be $8/7$, which is greater than $1$, and that is a contradiction. So, at most one of the numbers $x, y, z$ is equal to $1/7$ and the others are equal to zero. But then each of the three (non-negative) expressions $3x + 5y$, $3y + 5z$, $3z + 5x$ is at most equal to $5/7$, so the conditions stated in the beginning of the solution as equivalence are satisfied and all such triples are solutions.\n\n**Conclusion**. The problem has exactly $4$ solutions\n$$\n(x, y, z) \\in \\left\\{ (0, 0, 0), \\left(\\frac{1}{7}, 0, 0\\right), \\left(0, \\frac{1}{7}, 0\\right), \\left(0, 0, \\frac{1}{7}\\right) \\right\\}.\n$$",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 29553,
    "subject": "Mathematics (Multi-modal)",
    "question": "Alice and Ben play the game on a board with $72$ cells around a circle. First, Ben chooses some cells and places one chip on each of them. Each round, Alice first chooses one empty cell and then Ben moves a chip from one of the adjacent cells onto the chosen one. If Ben fails to do so, the game ends; otherwise, another round follows. Determine the smallest number of chips for which Ben can guarantee that the game will last for at least $2023$ rounds.\n(Václav Blažej)",
    "options": [],
    "answer": "36",
    "solution": "We show that the smallest possible number of chips is $36$.\n\nIn the first part, we describe the strategy of Ben in which he can ensure that the game will never end. At the beginning, Ben places $36$ chips on even cells of the game board and the odd cells he lets empty. Moreover, he firmly divides all $72$ cells into $36$ pairs of adjacent cells. Then, Ben is moving the chips in such a way that each of these pairs of cells contains exactly one chip throughout the whole game: in each round, Alice chooses an empty cell, and Ben then moves the chip from the second cell of the pair. So the game never ends.\n\nIn the second part of the solution, we assume that Ben initially places fewer than $36$ chips on the board. We describe Alice's strategy for ensuring that the game ends no later than in the $36$th round.\n\nFirst, Alice imagines that the cells are colored alternately white and black. In each round, Alice chooses an empty white cell—she always finds one, because there are $36$ white cells, while the chips are fewer. So, Ben will be forced to move a token from one of the black cells to the white cell. Then, each chip will be moved at most once during the course of the game. The game will therefore end no later than in the $36$th round.\nWe present a different approach to the second part of the original solution. We again assume that Ben places less than $36$ chips on the board, and, in addition, that no three adjacent cells are empty—otherwise Alice ends the game in the first round by choosing the middle of those three cells. We show that after at most $34$ rounds, Alice can force a situation where three empty adjacent cells exist.\n\nThe empty cells are then divided into several continuous sections, each consisting of one or two cells. There are at least two sections consisting of two cells—dividing all $72$ cells into $36$ pairs of adjacent cells, at least one pair remains empty; then we use the second possible pairing and find another empty pair.\n\nAlice places a marker between each two empty adjacent cells and she corrects the position of one marker after each round. At the beginning these $z \\ge 2$ markers divide all $72$ cells into $z$ sections. Each of them contains at least $3$ cells, it starts and ends with an empty cell and does not contain two adjacent empty cells. Alice can certainly select from these segments one, let's call it $U$, that has fewer chips than empty cells (since this inequality holds for their total numbers).\n\nLet $k \\ge 1$ be such that the selected segment $U$ contains $k+1$ empty cells and at most $k$ chips. However, these chips must be exactly $k$, since any two consecutive empty cells must be separated by a cell with a chip. Thus, the section $U$ consists of $2k+1$ cells for which and $2k+1 \\le 72-3=69$, i.e. $k \\le 34$. In the obvious marking, then, the situation in the $U$ segment looks like this:\n$$\n\\dots 0 \\mid \\underbrace{0101\\dots010}_{U} \\mid 0\\dots\n$$\nAlice chooses the first empty cell from the left in the $U$ segment in the first round. Ben is then forced to move the chip from the right. This makes the left marker move two positions to the right, creating a new $U'$ section of length $2k-1$:\n$$\n\\dots 0 \\mid \\underbrace{0101\\dots010}_{U} \\mid 0\\dots \\rightarrow \\dots 010 \\mid \\underbrace{0101\\dots010}_{U'} \\mid 0\\dots\n$$\nIn the second round, Alice again selects the first cell from the left in the $U'$ section. She repeats the procedure over and over again until after the $k$-th round (where, as we know $k \\le 34$), she gets the section between two markers consisting of a single cell, i.e. there are three adjacent empty cells. Then she brings the game to an end in the next turn.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 29554,
    "subject": "Mathematics (Multi-modal)",
    "question": "Solve the following system of equations in the domain of real numbers\n$$2x + \\lfloor y \\rfloor = 2022,$$\n$$3y + \\lfloor 2x \\rfloor = 2023.$$\n(The symbol $\\lfloor a \\rfloor$ denotes the lower integer part of a real number $a$, i.e. the greatest integer not greater than $a$. E.g. $\\lfloor 1.9 \\rfloor = 1$ and $\\lfloor -1.1 \\rfloor = -2$.)",
    "options": [],
    "answer": "(1011, 1/3)",
    "solution": "Since $\\lfloor y \\rfloor$ and $2022$ are integers, the equation $2x + \\lfloor y \\rfloor = 2022$ implies that $2x$ is also an integer, so $\\lfloor 2x \\rfloor = 2x$. Thus we can eliminate the unknown $x$ by subtracting the first equation of the system from the second one. We get\n$$\n3y - \\lfloor y \\rfloor = 1. \\qquad (1)\n$$\nThanks to (1), $3y$ is an integer, so it has (according to its remainder after division by three) one of the forms $3k$, $3k + 1$ or $3k + 2$, where $k$ is an integer. From this it follows that either $y = k$, or $y = k + \\frac{1}{3}$, or $y = k + \\frac{2}{3}$, where $k = \\lfloor y \\rfloor$. We now discuss these three cases.\n\n* In the case of $y = k$, (1) becomes $3k - k = 1$ with the non-integer solution $k = \\frac{1}{2}$.\n\n* In the case of $y = k + \\frac{1}{3}$, (1) is the equation $(3k + 1) - k = 1$ with a solution $k = 0$, which corresponds to $y = \\frac{1}{3}$. The original system of equations is then apparently fulfilled, precisely when $2x = 2022$, i.e. $x = 1011$.\n\n* In the case of $y = k + \\frac{2}{3}$, (1) is the equation $(3k + 2) - k = 1$ with a non-integer solution $k = -\\frac{1}{2}$.\n\n**Conclusion.** The only solution of the given system is the pair $(x, y) = (1011, \\frac{1}{3})$.\nThe verbal definition attached to the problem formulation tells us that $\\lfloor a \\rfloor$ is an integer for which $\\lfloor a \\rfloor \\le a$ and at the same time $\\lfloor a \\rfloor + 1 > a$. So, estimates $a - 1 < \\lfloor a \\rfloor \\le a$ are valid for every real number $a$. According to them, we get from the first equation of the given system\n$$\n2022 \\le 2x + y < 2023. \\qquad (2)\n$$\nSimilarly, from the second equation follows\n$$\n2023 \\le 3y + 2x < 2024. \\qquad (3)\n$$\nWe can combine these inequalities in two ways. Combining the second part of (2) with the first part of (3) we obtain $2x + y < 2023 \\le 3y + 2x$, whence from the comparison of the outermost expressions follows $y > 0$. If we modify the first part of (2) to $2024 \\le 2x + y + 2$, then together with the second part of (3) we obtain $3y + 2x < 2024 \\le 2x + y + 2$. This time $y < 1$ follows from the comparison of the outermost expressions.\n\nTogether, we got $0 < y < 1$, so $\\lfloor y \\rfloor = 0$ holds. Thanks to this, the first equation of the original system is reduced to the form $2x = 2022$, which is satisfied only for $x = 1011$. By inserting it into the second equation, we get $3y + 2022 = 2023$ with the only solution $y = \\frac{1}{3}$ which indeed satisfies the condition $\\lfloor y \\rfloor = 0$ used in the first equation. The pair $(x, y) = (1011, \\frac{1}{3})$ is therefore the only solution of the given system.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 29555,
    "subject": "Mathematics (Multi-modal)",
    "question": "In the real numbers, solve the system of equations\n$$\n\\sqrt{\\sqrt{x} + 2} = y - 2,\n$$\n$$\n\\sqrt{\\sqrt{y} + 2} = x - 2.\n$$",
    "options": [],
    "answer": "(4, 4)",
    "solution": "Let $(x, y)$ be any solution of the given system. Since $\\sqrt{\\sqrt{x} + 2}$ is obviously positive, we have $y > 2$ by the first equation. Similarly, the second equation implies $x > 2$.\n\nNow we prove that the numbers $x$ and $y$ must be equal. We will use the observation that the function square root is increasing. If $x > y$, then\n$$\n\\sqrt{\\sqrt{x} + 2} > \\sqrt{\\sqrt{y} + 2},\n$$\ni.e. $y - 2 > x - 2$, $y > x$, and that is a contradiction. The case $x < y$ is eliminated similarly. The equality of $x = y$ is thus proved.\n\nLet us therefore deal with the (only possible) case $x = y$ next. The original system of two equations is then obviously reduced to a single equation\n$$\n\\sqrt{\\sqrt{x} + 2} = x - 2. \\qquad (1)\n$$\nAfter substituting $s = \\sqrt{x}$, when $x = s^2$, the equation (1) becomes $\\sqrt{s+2} = s^2-2$, while obviously $s > \\sqrt{2}$. For each such $s$, we square the equality to obtain $s+2 = (s^2-2)^2$, which we rewrite in the form $s^4 - 4s^2 - s + 2 = 0$. Note that this equation has a root $s = 2$. This is confirmed by the decomposition\n$$\ns^4 - 4s^2 - s + 2 = s^2(s^2 - 4) - (s - 2) = s^2(s - 2)(s + 2) - (s - 2) = (s - 2)(s^3 + 2s^2 - 1),\n$$\nby which we now show that $s = 2$ is the only root of the derived equation, that satisfies our condition $s > \\sqrt{2}$. Indeed, for every $s > \\sqrt{2}$, $s^3 + 2s^2 - 1 > 0$ (this is even true for $s \\ge 1$). The only satisfactory value of $s = 2$ corresponds to the only solution $x = s^2 = 4$ of the equation (1), and therefore to the only solution $x = y = 4$ of the given problem.\n\n**Conclusion.** The given system of equations has a unique solution $(x, y) = (4, 4)$.\nWe again use the observation that both numbers $x$ and $y$ are greater than 2, and introduce the function $f: (2, \\infty) \\to (2, \\infty)$, $f(t) = \\sqrt{t} + 2$ for every $t > 2$. The equations from the problem, rewritten in the form\n$$\n\\begin{aligned}\n\\sqrt{\\sqrt{x} + 2} + 2 &= y, \\\\\n\\sqrt{\\sqrt{y} + 2} + 2 &= x\n\\end{aligned}\n$$\ncan then be written as a system of equations using the function $f$\n$$\n\\begin{aligned}\nf(f(x)) &= y, \\\\\nf(f(y)) &= x.\n\\end{aligned}\n$$\nWe see that its solutions are just pairs of the form $(x, y) = (x, f(f(x)))$, where the number $x$ satisfies the relation $f(f(f(x))) = x$. This is certainly satisfied in the case where $f(x) = x$. We show that the equality holds only in this case.\n\nIf $f(x) < x$ we have a quadruple of inequalities\n$$\nf\\left(f\\left(f\\left(f(x)\\right)\\right)\\right) < f\\left(f\\left(f(x)\\right)\\right) < f(f(x)) < f(x) < x\n$$\nwhere the last inequality is obvious and every previous inequality is the consequence of the immediately following inequality and the fact that $f$ is increasing. Similarly in the case of $f(x) > x$ we have\n$$\nf\\left(f\\left(f(f(x))\\right)\\right) > f\\left(f(f(x))\\right) > f(f(x)) > f(x) > x.\n$$\nThus, we have proved equivalence of $f\\left(f\\left(f(f(x))\\right)\\right) = x$ and $f(x) = x$.\n\nIt remains to solve the equation $f(x) = x$ with unknown $x > 2$, which is easy:\n$$\nf(x) = x \\Leftrightarrow \\sqrt{x} + 2 = x \\Leftrightarrow 0 = (\\sqrt{x} - 2)(\\sqrt{x} + 1) \\Leftrightarrow \\sqrt{x} = 2 \\Leftrightarrow x = 4.\n$$\nWe arrive at the same conclusion as in the first solution.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 29556,
    "subject": "Mathematics (Multi-modal)",
    "question": "Given an acute-angled triangle $ABC$. The points $B'$ and $C'$ lie on the rays opposite to $CA$ and $BA$, respectively, such that $|B'C| = |AB|$ and $|C'B| = |AC|$. Prove that the circumcenter of $AB'C'$ lies on the circumcircle of $ABC$. (Patrik Bak)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Since the line segments $AB'$, $AC'$ have the same length $|AB'| + |AC'|$, $AB'C'$ is an isosceles triangle with base $B'C'$. It means that the perpendicular bisector of $B'C'$ coincides with the bisector of the angle $BAC$. Let $S \\neq A$ be the intersection of this bisector with the circumcircle of $ABC$. If we prove that $S$ is a circumcenter of $AB'C'$ we will be finished. Since $S$ lies on the perpendicular bisector of $B'C'$, we have $|SB'| = |SC'|$. So, it remains to prove that $|SA| = |SC'|$.\n\nFrom congruence of inscribed angles $SAB$ and $SAC$ it follows that $S$ is the midpoint of the arc $BC$, and therefore $|BS| = |CS|$. From the cyclic quadrilateral $ABSC$ we have $|\\angle ACS| = 180^\\circ - |\\angle SBA| = |\\angle C'BS|$. Together with equality $|CA| = |BC'|$ we get that triangles $SAC$ and $SC'B$ are congruent by condition $SAS$ and therefore $|SA| = |SC'|$.\n\n![](attached_image_1.png)\nFigure 1\nLet us define the point $S$ as in the first solution. This time we verify the desired equality $|SA| = |SC'|$ by showing that $S$ lies on the perpendicular bisector of $AC'$.\n\nIn the special case where $|AB| = |AC|$ holds, the midpoint of $AC'$ is $B$ (according to the construction of $C'$); therefore it suffices to verify that the angle $ABS$ is right. However, this follows from the fact that the cyclic quadrilateral $ABSC$ is then composed of two identical triangles $ABS$ and $ACS$, so the angles at their opposite vertices $B$ and $C$ are identical and therefore right.\n\nWhen $|AB| \\neq |AC|$, we can without loss of generality assume that $|AB| > |AC|$ as in figure 2. Here $P$ and $Q$ denote the perpendicular projections of $S$ onto lines $AB$ and $AC$, respectively. Thanks to our assumption $|AB| > |AC|$ the point $P$ lies inside the segment $AB$, while the point $Q$ lies on the opposite ray to the ray $CA$. We prove that $P$ is the midpoint of $AC'$.\n\nBPS and CQS, also the angles PSB and QSC have the same size. In addition, we have $|PS| = |QS|$, since $S$ lies on the angle bisector of $C'AB'$. We thus obtain that triangles $PBS$ and $QCS$ are congruent according to condition ASA. Hence, equality $|BP| = |CQ|$ follows. Moreover, from the identical rectangular triangles $ASP$ and $ASQ$ we also have $|AP| = |AQ|$, so together it yields\n$$\n|AP| = |AQ| = |AC| + |CQ| = |C'B| + |BP| = |C'P|.\n$$\nThus, $P$ is indeed the midpoint of $AC'$, and the proof is complete.\n\n![](attached_image_2.png)\nFigure 2\nThis time we denote by $S$ the circumcenter of $AB'C'$ and we prove that points $A$, $B$, $S$, and $C$ lie on one circle. According to the introduction of the first solution, we know that $AB'C'$ is an isosceles triangle with base $B'C'$, hence its circumcenter $S$ lies on the angle bisector of $C'AB'$, that is $BAC$. The points $B$ and $C$ therefore lie in the opposite half-planes with the boundary line $AS$, therefore it is sufficient to verify $|\\angle ABS| = 180^\\circ - |\\angle ACS|$.\n\nBy equalities $|AC'| = |AB'|$ and $|AS| = |C'S| = |B'S|$, the triangles $C'AS$ and $AB'S$ are isosceles and congruent. Thus, if we rotate triangle $C'AS$ around $S$ about the oriented angle $C'SA$ we obtain triangle $AB'S$. Since $B$ lies on $C'A$, $C$ lies on $AB'$ and at the same time $|C'B| = |AC|$, this rotation maps $B$ onto $C$ and thus the angle $C'BS$ to the angle $ACS$. Therefore, $|\\angle C'BS| = |\\angle ACS|$, which already follows $|\\angle ABS| = 180^\\circ - |\\angle C'BS| = 180^\\circ - |\\angle ACS|$ as we promised to show.\n\n![](attached_image_3.png)\nFigure 3",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 29557,
    "subject": "Mathematics (Multi-modal)",
    "question": "In the convex pentagon $ABCDE$ $|\\angle CBA| = |\\angle BAE| = |\\angle AED|$ holds. On the sides $AB$ and $AE$, there are points $P$ and $Q$, respectively, such that $|AP| = |BC| = |QE|$ and $|AQ| = |BP| = |DE|$. Prove that $CD \\parallel PQ$. (Patrik Bak)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Since $|BC| = |AP| = |EQ|$, $|BP| = |AQ| = |ED|$ and $|\\angle CBP| = |\\angle PAQ| = |\\angle QED|$, the triangles $PBC$, $QAP$ and $DEQ$ are congruent by the condition SAS.\n\nHence $|CP| = |PQ| = |QD|$ and also\n$$\n|\\angle CPQ| = 180^\\circ - |\\angle BPC| - |\\angle APQ| = 180^\\circ - |\\angle PQA| - |\\angle EQD| = |\\angle PQD|.\n$$\nThis means that by the condition SAS, the isosceles equilateral triangles $CPQ$ and $DQP$ are also congruent. It follows that their altitudes from $C$ and $D$ to the common opposite side $PQ$ have the same lengths, and hence $CD \\parallel PQ$.\n\n\nSolution 2:\n\nLet $S$ denote the circumcenter of $BAE$. Obviously, $|BA| = |AE|$. Therefore, in the rotation with center $S$ by the oriented angle $BSA$ $B \\to A \\to E$, and therefore $P \\to Q$.\n![](attached_image_1.png)\nAnother consequence of $B \\to A \\to E$ is the congruence of the four angles $SBA$, $SAB$, $SAE$ and $SEA$. It follows that the angle bisectors of congruent angles $CBA$, $BAE$ and $AED$ are respectively the rays $BS$, $AS$ and $ES$. In our rotation is thus the image of the oriented angle $CBS$ the oriented angle $BAS$, so with respect to $|BC| = |AP|$, $C \\to P$ holds. The same is true from the oriented angles $SAE$ and $SED$, the equality $|AQ| = |ED|$ then leads to $Q \\to D$. Together we have $C \\to P \\to Q \\to D$, which implies\n\nthat the line segments *CD* and *PQ* have a common perpendicular bisector—the bisector of *CD* bisects the angle *CSD*, and therefore bisects the angle *PSQ*, and therefore is also the perpendicular bisector of *PQ*. Because of the common bisector, the lines *CD* and *PQ* are parallel.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 29558,
    "subject": "Mathematics (Multi-modal)",
    "question": "In a convex quadrilateral $ABCD$, $|AB| = |BC| = |CD|$. Let furthermore, for the intersection $P$ of its diagonals, $|\\angle APD| < 90^\\circ$. Let $R$ and $S$ be reflections of $A$ and $D$ with respect to $BD$ and $AC$, respectively. Prove that the lines $BC$ and $RS$ are parallel.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "$$\n|AB| = |BC| = |CD| = |BR| = |CS|. \\qquad (1)\n$$\nBy construction, the points $R, S$ are obviously different and the midpoint $X$ of the segment $AR$ lies on its perpendicular bisector $BD$ and the midpoint $Y$ of the segment $DS$ lies on its perpendicular bisector $AC$. In the following paragraph we prove that $R$ lies inside the angle $ABC$ and $S$ inside the angle $DCB$, as in our figure. Together, this means that points $A, D, R, S$ lie inside the same half-plane with the boundary line $BC$.\n\n![](attached_image_1.png)\n\nThe assumption $|\\angle APD| < 90^\\circ$ implies $|\\angle APB| > 90^\\circ$, which for the interior point $P$ of the base $AC$ of the isosceles triangle $ABC$ means that $|\\angle ABP| < \\frac{1}{2} |\\angle ABC|$; hence $|\\angle ABR| = 2 \\cdot |\\angle ABP| < |\\angle ABC|$, hence the point $R$ is actually inside the angle $ABC$. Analogously from the inequality $|\\angle DPC| > 90^\\circ$ for the interior point $P$ of the base $BD$ of isosceles triangle $DCB$, we conclude that the point $S$ actually lies inside the angle $DCB$.\n\nA further consequence of the inequality $|\\angle APD| < 90^\\circ$ is that for the marked interior angles of the right triangles $APX$ and $DPY$, $|\\angle XAP| = 90^\\circ - |\\angle APD| = |\\angle YDP|$, i.e. $|\\angle RAC| = |\\angle SDB|$.\n\nLet us return to the equalities (1). According to these, the point $B$ is the circumcenter of the triangle $ARC$, which evidently lies in the angle $ABC$. Therefore, according to the inscribed angle theorem $|\\angle RBC| = 2 \\cdot |\\angle RAC|$. By a similar reasoning about the circumcenter $C$ of the triangle $BSD$ in the angle $BCD$ we obtain $|\\angle SCB| = 2 \\cdot |\\angle SDB|$. From the last two paragraphs we get the equality $|\\angle RBC| = |\\angle SCB|$. This, together with (1), leads to the conclusion that (isosceles) triangles $RBC$ and $SCB$ are congruent by the SAS theorem. Hence, their altitudes from the vertices of $R$ and $S$ to the side $BC$ have the same length. This already implies that $BC \\parallel RS$.\n\n\nAs in the first solution, we derive (1) and observe that $R$ lies inside the angle $ABC$. From the condition $|\\angle APD| < 90^\\circ$ it also follows that $R$ lies in the half plane $ACD$.\nAccording to (1), $B$ is the circumcenter of $ARC$, whose central angle $RBA$ with the bisector $BD$ is therefore twice the angle $RCA$. Therefore the three angles $PBA$, $RBP$ and $RCP$ marked in the figure are congruent. Congruence of the last two angles with respect to the previous paragraph already means that the point $R$ does indeed lie on the circumcircle of $BCP$. For the point $S$ the same is true due to the analogous congruence of the angles $PCD$, $SCP$ and $SBP$.\n\n![](attached_image_2.png)\n\nIt follows from the proof that the points $B$, $C$, $R$, $S$ lie on one circle, while the points $R$ and $S$ lie in the same half-plane with the boundary line $BC$. Hence the congruence of the angles $BRC$ and $BSC$, which, together with the equality $|BC| = |BR| = |CS|$, means that the isosceles triangles $RBC$ and $SCB$ are congruent. The congruence of their altitudes proves the relation $BC \\parallel RS$.\nWe show that the points $R$ and $S$ lie on the circumcircle of $BCP$. We write the detailed proof only for the point $R$, for the point $S$ the proof is analogous.\n\nAs in the first solution, we derive (1) and observe that $R$ lies inside the angle $ABC$. From the condition $|\\angle APD| < 90^\\circ$ it also follows that $R$ lies in the half plane $ACD$.\nAccording to (1), $B$ is the circumcenter of $ARC$, whose central angle $RBA$ with the bisector $BD$ is therefore twice the angle $RCA$. Therefore the three angles $PBA$, $RBP$ and $RCP$ marked in the figure are congruent. Congruence of the last two angles with respect to the previous paragraph already means that the point $R$ does indeed lie on the circumcircle of $BCP$. For the point $S$ the same is true due to the analogous congruence of the angles $PCD$, $SCP$ and $SBP$.\n\n* Instead of consideration of the congruent triangles $CBR$ and $BCS$, it suffices to state, that the congruent segments $BR$ and $CS$ are symmetrically clustered along the axis of the line segment $BC$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 29559,
    "subject": "Mathematics (Multi-modal)",
    "question": "In an acute-angled triangle $ABC$, let us denote $H$ its orthocenter and $I$ its incenter. Let $D$ be the perpendicular projection of $I$ on the line $BC$, and $E$ be the image of point $A$ in symmetry with center $I$. Furthermore, $F$ is the perpendicular projection of the point $H$ on the line $ED$. Prove that the points $B$, $H$, $F$ and $C$ lie on one circle. (Patrik Bak)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $M$ be the midpoint of $BC$. In point reflection with the centre $M$, denote $L$ the image of $D$ and $J$ the image of $I$. It follows from this symmetry, that $J$ is the incenter of triangle $BCP$ and $L$ is the point where this incircle touches $BC$. Let $KL$ be the diameter of this incircle. Thus, $J$ is the midpoint of the segment $KL$.\n\nIt is well known that $D$ is the tangent point of the excircle of triangle $BCP$. This excircle is the image of the incircle in homothety with center $P$ (and with a coefficient greater than $1$). In this homothety is the tangent $BC$ of the excircle the image of the tangent of the incircle, that is parallel to their common tangent $BC$ but has smaller distance from $P$. This tangent, however, passes through the point $K$, since $KL$ is the diameter of the incircle perpendicular to both tangents. Therefore, the homothety maps the point $K$ to the point $D$, and thus the points $D$, $K$, $P$ lie on the same line. This remains to prove that point $E$ also lies on this line. To do this, it suffices to show that the lines $EP$ and $DK$ are parallel. These are, however, the sides of triangles $AEP$ and $LDK$ with the meanlines $IM$ and $MJ$ respectively for which $IM \\parallel EP$ and $MJ \\parallel DK$; hence the desired relation $EP \\parallel DK$ follows, since $M$ is the midpoint of the line segment $IJ$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 29560,
    "subject": "Mathematics (Multi-modal)",
    "question": "Prove the claim: If we choose any four factors of $720$, then one of them divides the product of the other three.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Given the decomposition $720 = 2^4 \\cdot 3^2 \\cdot 5$, the number $720$ has exactly three prime factors: $2$, $3$ and $5$. So, each of its factors is of the form $2^\\alpha \\cdot 3^\\beta \\cdot 5^\\gamma$, where $\\alpha, \\beta, \\gamma$ are non-negative integers (satisfying the inequalities $\\alpha \\le 4$, $\\beta \\le 2$ and $\\gamma \\le 1$ which we will not need further). Surely also the product of any three factors of $720$ is of the form $2^\\alpha \\cdot 3^\\beta \\cdot 5^\\gamma$ with non-negative integers $\\alpha, \\beta$ and $\\gamma$. Considering any two numbers of this form, the first is a quotient of the second if and only if the values $\\alpha, \\beta, \\gamma$ of the first number do not exceed the corresponding values of the second number.\n\nWe prove the problem statement by contradiction. Let us admit that some four factors of the number $720$ have the property that none of them divides the product of the other three factors. Then each of them contains in its prime decomposition some of the primes $2$, $3$, $5$ to a higher power than it has in its decomposition the product of the other three factors, and therefore any one of them. But there are four divisors and only three prime numbers, and this is the contradiction.\nWe present one of several possible variations of a direct proof. We use the observations contained in the first paragraph of the previous solution.\n\nLet us choose any four factors of the number $720$, call them Numbers. First, we choose three Numbers, that contain prime factor $2$ in powers not exceeding that power of $2$ in the fourth Number (if there are more than one such choice, we choose one of them). Then we select two of these three Numbers that contain prime factor $3$ in powers, that do not exceed power of $3$ of the third Number. Of these two Numbers, we finally select the one that contains the prime factor $5$ in a power not exceeding power of $5$ of the second Number. In the last selected number, each $p \\in \\{2, 3, 5\\}$ has a power that does not exceed at least one of the powers of $p$ in the other three Number. This guarantees that the last selected Number has the property required by the problem statement.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 29561,
    "subject": "Mathematics (Multi-modal)",
    "question": "Given two odd positive integers $k$ and $n$. For each two positive integers $i, j$ satisfying $1 \\le i \\le k$ and $1 \\le j \\le n$ Martin wrote the fraction $i/j$ on the board. Determine the median of all these fractions, that is a real number $q$ such that if we order all the fractions on the board by their values from smallest to largest (fractions with the same value in any order), in the middle of this list will be a fraction with value $q$. (Martin Melicher)",
    "options": [],
    "answer": "(k+1)/(n+1)",
    "solution": "We show that the median has the value $q = \\frac{k+1}{n+1}$. In the entire solution, $q$ denotes this number.\nSince the given numbers $n$ and $k$ are odd, the fraction with value $q$ is actually written on the board—for example, it is a fraction $\\frac{\\frac{1}{2}(k+1)}{\\frac{1}{2}(n+1)}$.\n\nAccording to the comparison with the number $q$, we call a fraction\n▷ *small* if its value is less than $q$,\n▷ *mean* if its value is equal to $q$,\n▷ *large* if its value is greater than $q$.\n\nThe number $k \\cdot n$ of all fractions on the board is odd. To show that the middle fraction, when they are ordered by their values, has the value $q$, it is sufficient to prove that the number of small fractions is equal to the number of the large ones. (The latter will also mean that the number of mean fractions is odd, which again confirms their existence.)\n\nWe match the fractions written on the board—we couple each fraction $i/j$ with the fraction $i'/j'$ (and vice versa) if and only if $i' = k + 1 - i$ and $j' = n + 1 - j$, which can indeed be rewritten symmetrically as $i + i' = k + 1$ and $j + j' = n + 1$. Note that the inequalities $1 \\le i \\le k$ and $1 \\le j \\le n$ apparently hold if and only if $1 \\le i' \\le k$ and $1 \\le j' \\le n$.\n\nIt is obvious that only the fraction $\\frac{\\frac{1}{2}(k+1)}{\\frac{1}{2}(n+1)}$ is „coupled“ with itself and that all the other fractions are actually divided into pairs. If we show that every such pair either consists of one small and one large fraction, or of two mean fractions, we are done.\n\nThanks to the mentioned symmetry, it suffices to verify that a fraction $i'/j'$ is small if and only if the fraction $i/j$ is large. The verification is routine:\n\n$$\n\\begin{align*} \n\\frac{i'}{j'} < \\frac{k+1}{n+1} &\\Leftrightarrow \\frac{k+1-i}{n+1-j} < \\frac{k+1}{n+1} \\\\\n&\\Leftrightarrow (k+1-i)(n+1) < (k+1)(n+1-j) \\\\\n&\\Leftrightarrow (k+1)(n+1) - i(n+1) < (k+1)(n+1) - (k+1)j \\\\\n&\\Leftrightarrow i(n+1) > (k+1)j \\\\\n&\\Leftrightarrow \\frac{i}{j} > \\frac{k+1}{n+1}. \n\\end{align*}\n$$\n\nThis completes the solution.\n\n\nANOTHER SOLUTION.\n\nLet us look at the problem geometrically—we consider the plane with the Cartesian coordinate system $Oxy$. Each fraction $i/j$ that Martin wrote on the board is represented as a point $B$ with coordinates $[j, i]$.* We thus get exactly those points $B[j, i]$ of our plane, for which $j \\in \\{1, 2, \\dots, n\\}$ and $i \\in \\{1, 2, \\dots, k\\}$. The set of these points (that we plotted in the figure for $n = 11$ a $k = 5$) we denote by $M$ and call it „the grid“. It has the shape of a rectangle with vertices $[1, 1]$, $[n, 1]$, $[n, k]$ and $[1, k]$. Since numbers $n, k$ are odd, the center $S$ of this rectangle has integer coordinates $j_0 = \\frac{1}{2}(n + 1)$ and $i_0 = \\frac{1}{2}(k + 1)$. The center $S$ is thus itself a point of the grid $M$. Let us add that the line $OS$ has the slope $i_0/j_0$ and let us denote the value of this fraction by $q$ as in the first solution.\n\nNote that the grid $M$ is point symmetric with the center $S$ (in the picture we marked two points $B$ and $B'$ where each of them is reflection of the other one).* Therefore, there is the same number of points from $M$ above and below the line $OS$. Let us clarify what distinguishes these two equally numerous groups of points „below the line $OS“ and „above the line $OS“.\n\nThe point $B[j, i]$ of the grid $M$ lies below the line $OS$ if and only if the line $OB$ has a smaller slope than the line $OS$, i.e. if $i/j < i_0/j_0 = q$ holds. Therefore, exactly those points $B[j, i]$ lie under the line $OS$ that correspond to *small* fractions $i/j$, as we called them in the first solution. Similarly, the lattice points of $M$ above the line $OS$ correspond to *large* fractions. So, there is the same number of small and large fractions.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 29562,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all triples of positive integers $a$, $b$, $c$ for which the product\n$$(a+b)(b+c)(c+a)(a+b+c+2036)$$\nis equal to the power of a prime number with an integer exponent.",
    "options": [],
    "answer": "(4, 4, 4)",
    "solution": "First, let us note that at least one of the numbers $a+b$, $a+c$ and $b+c$ must be even. Indeed, two of the three numbers $a$, $b$, $c$ have the same parity, so their sum is even.\nIf the product we investigate is a power of a prime $p$, then each of the four factors must be a power of $p$. As we already know, one of the first three factors is even, so $p = 2$. Each of the four factors is therefore a power of two, which is greater than $1 = 2^0$, since the numbers $a, b, c$ are positive integers. Hence, each factor is an even number.\n\nFurther observe that the numbers $a+b$, $a+c$, $b+c$ are all even numbers, if and only if the numbers $a, b, c$ all have the same parity. Since $a+b+c+2036$ is even, $a, b$ and $c$ must be even numbers. Therefore, we can write $a = 2a_1$, $b = 2b_1$, and $c = 2c_1$, where $a_1, b_1, c_1$ are positive integers. Then of course\n$$\n(a+b)(a+c)(b+c)(a+b+c+2036) = 2^4(a_1+b_1)(a_1+c_1)(b_1+c_1)(a_1+b_1+c_1+1018).\n$$\nThe product of the last four parentheses must be a power of two. The numbers $a_1, b_1, c_1$ are therefore solutions to the original problem with the constant 2036 replaced by 1018. For the same reason as above $a_1, b_1$, and $c_1$ must be even. We denote by $a_2, b_2$ and $c_2$ respectively their halves (they are again positive integers) and we get\n$$\n(a+b)(a+c)(b+c)(a+b+c+2036) = 2^8(a_2+b_2)(a_2+c_2)(b_2+c_2)(a_2+b_2+c_2+509).\n$$\nAnd again, we have the same problem with the constant 509, so the numbers $a_2, b_2$, and $c_2$ necessarily have the same parity. However, since the number 509 is odd, $a_2, b_2$ and $c_2$ must be odd numbers. We see that the triple $a_2 = b_2 = c_2 = 1$ satisfies the problem (since $3 + 509 = 512$ is a power of two), so the corresponding triple $a = b = c = 4$ is a solution to the original problem. We show that it is the only solution.\n\nSuppose that at least one of the numbers $a_2, b_2, c_2$ is greater than one, let it be $c_2$ without loss of generality. Then, of course, the power of two equal to $a_2+c_2$ is greater than 2, so it is divisible by four. This means that dividing by four one of the numbers $a_2, c_2$ gives a remainder of 1 and the other a remainder of 3. So the third odd number $b_2$ has the same remainder when divided by four as one of the numbers $a_2, c_2$. The sum of $b_2$ with this number then has a remainder of 2, and since this sum is also a power of two, it must be the power of $2^1$. It follows that $b_2 = 1$ and $a_2 = 1$ (the equality of $c_2 = 1$ is ruled out by our assumption $c_2 > 1$). Thus the remainder 3 when divided by four necessarily gives $c_2$. But then the number $a_2+b_2+c_2+509$ gives remainder 2, so it is not a power of two, and that is a contradiction.\n\n**Conclusion.** A single triple $(a, b, c) = (4, 4, 4)$ satisfies the problem statement.\nLet $a \\ge b \\ge c$ hold without loss of generality, so then $a+b \\ge c+a \\ge b+c \\ge 2$. The product $(a+b)(b+c)(c+a)(a+b+c+2036)$ is a power of some prime number if and only if powers of that prime are all four factors of $a+b, b+c, c+a$, and $a+b+c+2036$. So let $a+b = p^k, c+a = p^l$, and $b+c = p^m$ for some prime $p$ and non-negative integers $k, l$ and $m$. For these, by the theorem of the introduction $k \\ge l \\ge m \\ge 1$.\n\nIf $k > l$, and hence $k-1 \\ge l$ and $k-1 \\ge m$, with respect to $p \\ge 2$, we would have\n$$\na+b = p^k \\ge p^{k-1} + p^{k-1} \\ge p^l + p^m = (c+a) + (b+c) > a+b,\n$$\nwhich is impossible. Necessarily, then, $k=l$, so $a+b = p^k = p^l = c+a$, where $b=c$. Then of course $p^m = b+c = 2b$, and so $b=c = p^m/2$, so $2|p$, e.g. $p=2$, and therefore $b=c = 2^{m-1}$ and $a+b = 2^k$. Since $p=2$, it is also $a+b+c+2036 = 2^n$ for some integer $n$, which obviously satisfies the condition $2^n > 2036$.\n\nBy the conclusion of the previous paragraph, the numbers $k, m, n$ satisfy the equality\n$$\n2^n = (a+b) + c + 2036 = 2^k + 2^{m-1} + 2036.\n$$\nNote that dividing by 16 number 2036 gives the remainder of 4, while $2^n$ certainly gives a remainder of 0 if $2^n > 2036$. This implies $2^k + 2^{m-1} \\equiv 12 \\pmod{16}$. The summands $2^k$ and $2^{m-1}$—as powers of two—are congruent to 1, 2, 4, 8 or 0 (mod 16). It is easy to see that the derived congruence holds only if one of the powers of $2^k, 2^{m-1}$ yields a residue of 4 and the other 8. Then $k, m-1 = 2, 3$. However, since $2^{m-1} = b < a+b = 2^k$, it is necessarily $m-1 < k$, and therefore $m-1=2$ and $k=3$. Hence $b=c=2^{m-1}=4$ and $a+b=2^k=8$, whence also $a=4$. This gives us the only possible triple $(a, b, c) = (4, 4, 4)$. This is indeed the solution to the problem—the product of the problem is then $8 \\cdot 8 \\cdot 8 \\cdot 2048$, while $8=2^3$ and $2048 = 2^{11}$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 29563,
    "subject": "Mathematics (Multi-modal)",
    "question": "Consider a sequence $(a_n)_{n=1}^{\\infty}$ of positive integers satisfying for each $n \\ge 3$ the condition\n$$\na_n = a_1a_2 + a_2a_3 + \\dots + a_{n-2}a_{n-1} - 1.\n$$\na) Prove that some prime number is a divisor of infinitely many terms of this sequence.\n(Tomáš Bárta)\nb) Prove that there are infinitely many such prime numbers.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Since all terms $a_i$ are positive integers, we have\n$$\na_5 = a_1a_2 + a_2a_3 + a_3a_4 - 1 \\ge 1 + 1 + 1 - 1 = 2, \\text{ and therefore } a_5 \\ne 1.\n$$\nThe number $a_5$ is thus divisible by at least one prime number.\nFor every $n \\ge 4$, the following holds\n$$\n\\begin{align*}\na_n &= (a_1a_2 + a_2a_3 + \\dots + a_{n-3}a_{n-2}) + a_{n-2}a_{n-1} - 1 = \\\\\n&= (a_{n-1} + 1) + a_{n-2}a_{n-1} - 1 = \\\\\n&= a_{n-1}(a_{n-2} + 1),\n\\end{align*}\n$$\nand therefore $a_{n-1} \\mid a_n$. Let $p$ be any prime divisor of $a_5$, then the relation $a_{n-1} \\mid a_n$ implies $p \\mid a_6$, from where $p \\mid a_7$, and so on. By mathematical induction we get $p \\mid a_n$ for every $n \\ge 5$. Thus, the prime $p$ divides infinitely many terms of the given sequence. This completes the proof of part a).\n\nLet $\\mathcal{P}$ denote the set of all primes that divide infinitely many terms of the sequence. Suppose that the set $\\mathcal{P}$ is finite, i.e. $\\mathcal{P} = \\{p_1, \\dots, p_k\\}$ for a suitable $k$. Obviously, for every $i \\in \\{1, 2, \\dots, k\\}$ we find such a term $a_{n_i}$ that is divisible by $p_i$ and $n_i \\ge 5$. Due to the relation $a_{n-1} \\mid a_n$ (proved earlier for each $n \\ge 4$) we have $p_i \\mid a_n$ for all $n \\ge n_i$. If we now denote $N = \\max(n_1, \\dots, n_k)$, then $a_N$ is divisible by all primes $p_1, \\dots, p_k$. Therefore, $a_N+1 > 1$ is not divisible by any primes from $\\mathcal{P}$, so there must be a prime $q \\notin \\mathcal{P}$ satisfying $q \\mid a_N + 1$. This prime $q$ is then also a divisor of the number $a_{N+2} = a_{N+1}(a_N + 1)$, so $q \\mid a_n$ holds for each $n \\ge N+2$, and therefore $q \\in \\mathcal{P}$. Thus, we get a contradiction, that proves the statement in part b).",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 29564,
    "subject": "Mathematics (Multi-modal)",
    "question": "Given an acute-angled scalene triangle $ABC$. The angle bisector of the angle $BAC$ and the perpendicular bisectors of the sides $AB$, $AC$ define a triangle. Prove that its orthocenter lies on the median from the vertex $A$.\n\n(Josef Tkadlec)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let us denote $M$ the midpoint of $AB$, $N$ the midpoint of $AC$, $K$ and $L$ the intersections of the angle $CAB$ bisector with perpendicular bisectors of $AB$ and $AC$, respectively. The intersection of perpendicular bisectors of $AB$ and $AC$ is denoted by $O$. The triangle $KLO$ is therefore the triangle from the problem statement. Its orthocenter we denote by $H$. All these points are marked in the figure 1 for the case $|AB| < |AC|$. (The case $|AB| > |AC|$ looks analogously, the case $|AB| = |AC|$ is excluded by the specification—then points $K, L, O$ merge into one point.)\n\n![](attached_image_1.png)\n\nWe have to prove that $H$ lies on the median from the vertex $A$ of triangle $ABC$. It is sufficient to show that triangles $ABH$ and $ACH$ have the same area.\n\nSince $HL \\perp OK \\perp AB$ we have $HL \\parallel AB$. Therefore $H$ and $L$ have the same distance from the line $AB$. However, it is equal to the length of the segment $LN$, since the point $L$ lies on the angle bisector of $CAB$ and $N$ is the perpendicular projection of $L$ onto $AC$. Hence, we get that the area of $ABH$ is equal to $\\frac{1}{2}|AB| \\cdot |LN|$. Analogously, we deduce that area of $ACH$ is equal to $\\frac{1}{2}|AC| \\cdot |KM|$. It remains to prove $|AB| \\cdot |LN| = |AC| \\cdot |KM|$.\n\nFor the points $K$ and $L$ lying on the angle bisector of $CAB$ we have $|\\angle MAK| = |\\angle NAL|$. The rectangular triangles $AKM$ and $ALN$ are therefore similar, and therefore $|KM| : |AM| = |LN| : |AN|$. Hence with respect to $|AM| = \\frac{1}{2}|AB|$ and $|AN| = \\frac{1}{2}|AC|$ we get $|KM| : |AB| = |LN| : |AC|$, i.e. $|AB| \\cdot |LN| = |AC| \\cdot |KM|$ as we needed to prove.\nIn addition to the points from the first solution, we also consider the intersection $E$ of lines $AB$ and $KH$ and intersection $F$ of lines $AC$ and $LH$. Again we observe that $KH \\parallel AC$ and $LH \\parallel AB$, so the quadrilateral $AEHF$ is a parallelogram (see fig. 2).\n\n![](attached_image_2.png)\n\nWe use again similarity of triangles $AKM$ and $ANL$ and deduce $|KM| : |LN| = |AM| : |AN| = |AB| : |AC|$. Equality of exterior angles at vertices $E$, $F$ of the parallelogram $AEHF$ yields $|\\angle KEM| = |\\angle LFN|$. Thus, the rectangular triangles $EKM$ and $FLN$ are also similar, whence it follows $|EM| : |FN| = |KM| : |LN| = |AB| : |AC|$. Hence,\n$$\n\\frac{|AE|}{|AF|} = \\frac{|AM| - |EM|}{|AN| - |FN|} = \\frac{\\frac{|AB|}{|AC|} \\cdot |AN| - \\frac{|AB|}{|AC|} \\cdot |FN|}{|AN| - |FN|} = \\frac{|AB|}{|AC|},\n$$\nand triangles $AEF$ and $ABC$ are similar by the SAS condition (they coincide in the angle at the vertex $A$ and in the ratio of the adjacent sides). We thus obtain the decisive relation $EF \\parallel BC$.\n\nSince in the parallelogram $AEHF$ the line $AH$ bisects the diagonal $EF$, this line also bisects the segment $BC$, which is homothetic with $EF$ with center $A$. In other words, $H$ lies on the median from $A$ of the triangle $ABC$, as we had to prove.\nWe consider points $K, L, M, N, O, H, E, F$ from the second solution. As there, we conclude that $AEHF$ is a parallelogram, and therefore the midpoint of the segment $EF$ lies on the ray $AH$. If we prove $EF \\parallel BC$ then the midpoint $P$ of $BC$ also lies on the ray $AH$.\n\n![](attached_image_3.png)\n\nDue to the construction from the problem statement and the parallelogram $AEHF$, the six angles $KAE, KAF, KBA, LCA, AKE$ and $ALF$ marked in fig. 3 are congruent. According to the theorem $AA$, we have $\\triangle ABK \\sim \\triangle ACL$ and $\\triangle AKE \\sim \\triangle ALF$. According to the first similarity, $|AB| : |AC| = |AK| : |AL|$, and by the second similarity $|AK| : |AL| = |AE| : |AF|$. Together we get $|AB| : |AC| = |AE| : |AF|$, so by the SAS theorem, $\\triangle AEF \\sim \\triangle ABC$ holds, and hence $EF \\parallel BC$, as we promised to prove.",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 29565,
    "subject": "Mathematics (Multi-modal)",
    "question": "In the triangle $ABC$, let us denote $M, N, P$ the midpoints of the sides $BC, CA, AB$ respectively and let $G$ be the centroid of $ABC$. Let the circumcircle of $BGP$ intersects the line $MP$ at a point $K$ different from $P$, and let the circumcircle of $CGN$ intersects the line $MN$ at a point $L$ different from $N$. Prove $|\\angle BAK| = |\\angle CAL|$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Obviously, $MP$ intersects the median $BN$ between points $B$ and $G$, so the point $K$ lies on the ray $PM$ and $BKGP$ is cyclic. Similarly, the point $L$ lies on the ray $NM$ and $CLGN$ is cyclic. Due to $MP \\parallel CA$ and $MN \\parallel BA$ we have\n$$\n|\\angle BPK| = |\\angle BPM| = |\\angle BAC| = |\\angle MNC| = |\\angle LNC|,\n$$\nwhile the two cyclic quadrilaterals imply\n$$\n|\\angle BKP| = |\\angle BGP| = |\\angle NGC| = |\\angle NLC|.\n$$\n\nWe see that triangles *BPK* and *CNL* are similar according to the condition *AA*. By the condition *SAS* also triangles *ABK* and *ACL* are similar since\n(i) $|\\angle ABK| = |\\angle PBK| = |\\angle NCL| = |\\angle ACL|$,\n(ii) $\\frac{|AB|}{|BK|} = 2 \\cdot \\frac{|PB|}{|BK|} = 2 \\cdot \\frac{|NC|}{|CL|} = \\frac{|AC|}{|CL|}.$\nThus, the equality $|\\angle BAK| = |\\angle CAL|$ is proved.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 29566,
    "subject": "Mathematics (Multi-modal)",
    "question": "Consider the sequence $(a_n)_{n=1}^\\infty$ defined as follows:\n$$\na_1 = 3 \\quad a_n = a_1a_2a_3 \\dots a_{n-1} - 1 \\quad \\text{for all } n \\ge 2.\n$$\n*Prove that there exist*\na) infinitely many primes dividing at least one member of this sequence;\nb) infinitely many primes dividing no member of this sequence.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "a) By mathematical induction, we first prove that $a_n \\ge 2$ for every $n$. For $n=1$ and $n=2$ this is true because $a_1 = 3$ and $a_2 = 2$. Now suppose that for some $n \\ge 3$ the inequality $a_k \\ge 2$ holds for every $k < n$. Then we have $a_n = a_1a_2a_3\\dots a_{n-1} - 1 \\ge a_1a_2 - 1 = 5$, so indeed $a_n \\ge 2$.\n\nLet us now show that numbers $a_n$ are pairwise coprime. Indeed, for any two indices $k < n$ we have $a_k \\mid a_1a_2\\dots a_{n-1} = a_n + 1$, whence for the largest common divisor $D$ of the numbers $a_n$ and $a_k$ we get $D \\mid a_n$ and at the same time $D \\mid a_n + 1$ (because $D \\mid a_k$ and $a_k \\mid a_n + 1$), so necessarily $D = 1$, so $a_n$ and $a_k$ are coprime. Due to $a_n \\ge 2$ we find for each index $n$ a prime number, denote it by $p_n$, for which $p_n \\mid a_n$. Since all $a_n$ are pairwise coprime, the prime numbers $p_n$ are pairwise different. Thus a) is proved.\n\nb) If $n \\ge 2$, then $a_{n+1} = a_1 a_2 a_3 \\dots a_{n-1} a_n - 1 = (a_n + 1)a_n - 1 = a_n^2 + a_n - 1$. Next we work with this expression.\n\nAssume that $p \\mid a_n$ for some $n \\ge 2$ and for some prime $p$. Then $a_{n+1} = a_n^2 + a_n - 1 \\equiv -1 \\pmod{p}$. From here we get for the next member $a_{n+2} = a_{n+1}^2 + a_{n+1} - 1 \\equiv (-1)^2 + (-1) - 1 \\equiv -1 \\pmod{p}$, and so, by mathematical induction, all members $a_k$ with indices $k \\ge n+1$ give the same remainder $p-1$ modulo $p$. If the assumption $p \\mid a_n$ is satisfied for some $n \\ge 2$, we call the prime $p$ bad. Our task is actually to find infinitely many primes $p \\ge 5$ that are not bad (we impose the condition $p \\ge 5$ so that $p \\mid a_1 = 3$ does not hold).\n\nLet us now consider a prime $p$ satisfying $a_n \\equiv 1 \\pmod{p}$ for some $n \\ge 2$. Then $a_{n+1} = a_n^2 + a_n - 1 \\equiv 1^2 + 1 - 1 \\equiv 1 \\pmod{p}$, so using mathematical induction, we get that all numbers $a_k$ with indices $k \\ge n$ give a remainder 1 when dividing by $p$. Then let us call such $p$ good. Note that no prime $p \\ge 5$ is good and bad at the same time—because it is not possible that for sufficiently large $k$ both relations $a_k \\equiv 1 \\pmod{p}$ and $a_k \\equiv -1 \\pmod{p}$ hold. Therefore it is enough to prove that there are infinitely many good primes.\n\nTo find good primes we use the sequence $(b_n)_{n=1}^\\infty$ given by the formula $b_n = a_n - 1$ for each $n \\ge 1$. It is obvious that $b_1 = 2$, $b_2 = 1$ and\n$$\n\\begin{aligned}\nb_{n+1} &= a_{n+1} - 1 = (a_n^2 + a_n - 1) - 1 = ((b_n + 1)^2 + (b_n + 1) - 1) - 1 = \\\\\n&= b_n^2 + 3b_n = b_n(b_n + 3)\n\\end{aligned}\n$$\nfor every $n \\ge 2$. Then a prime number $p$ is good if and only if $p \\mid b_n$ for some $n \\ge 2$. We thus reached a situation similar to that in part a)—we need to prove existence of infinitely many primes dividing at least one member of the new sequence $(b_n)_{n=2}^\\infty$ determined by its first term $b_2 = 1$ and the relation $b_{n+1} = b_n(b_n + 3)$ for each $n \\ge 2$.\n\nWe begin with an observation that $b_k \\mid b_n$ if $2 \\le k \\le n$. Indeed from $b_{k+1} = b_k(b_k+3)$ we have $b_k \\mid b_{k+1}$ and further by induction $b_k \\mid b_n$ for every $n \\ge k$.\n\nWe now prove that, under the assumption $2 \\le k < n$, the numbers $b_k + 3$ and $b_n + 3$ are coprime. Indeed, their greatest common divisor $D$ satisfies $D \\mid b_k + 3 \\mid b_{k+1} \\mid b_n$ and at the same time $D \\mid b_n + 3$, so together $D \\mid (b_n + 3) - b_n = 3$ and therefore either $D = 1$ or $D = 3$. It remains to exclude the value $D = 3$: due to $b_2 = 1$ and the relationship $b_{n+1} = b_n(b_n + 3)$ it follows by an easy induction $b_n \\equiv 1 \\pmod{3}$ for each $n \\ge 2$. So, $3 \\nmid b_n$, and therefore also $3 \\nmid b_n + 3$, and thus $D \\ne 3$.\n\nFinally, we know that $b_n \\ge 1$ for every $n$ (since $a_n \\ge 2$), and thus $b_n + 3 \\ge 4$. Therefore, for each $n$ we find a prime number $p_n$ with the property $p_n \\mid b_n + 3$. All these prime numbers $p_n$ are according to the previous paragraph different from each other, in addition from $b_n + 3 \\mid b_{n+1}$ follows $p_n \\mid b_{n+1}$ for every $n \\ge 2$. So we have found an infinite sequence of prime numbers dividing at least one member of the sequence $(b_n)_{n=2}^\\infty$. The proof of part b) is complete.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 29567,
    "subject": "Mathematics (Multi-modal)",
    "question": "Suppose that we have three natural numbers $a$, $b$ and $c$ such that one of the values\n$$\ngcd(a, b) \\cdot \\text{lcm}(b, c), \\gcd(b, c) \\cdot \\text{lcm}(c, a), \\gcd(c, a) \\cdot \\text{lcm}(a, b),\n$$\nis equal to the product of the other two. Prove that one of the numbers $a$, $b$ and $c$ is a multiple of a different one.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Without loss of generality, we can assume that\n$$\nP = \\gcd(a, b) \\cdot \\text{lcm}(b, c) = \\gcd(b, c) \\cdot \\text{lcm}(c, a) \\cdot \\gcd(c, a) \\cdot \\text{lcm}(a, b).\n$$\nUsing the well-known relation $\\gcd(x, y) \\cdot \\text{lcm}(x, y) = xy$, we get:\n$$\n(abc)^2 = (ab) \\cdot (bc) \\cdot (ca) = \\gcd(a, b) \\cdot \\text{lcm}(a, b) \\cdot \\gcd(b, c) \\cdot \\text{lcm}(b, c) \\cdot \\gcd(c, a) \\cdot \\text{lcm}(c, a) = P^2,\n$$\nwhich gives $abc = P$ after taking square roots. Now, we can use the same relation once more to get\n$$\n\\gcd(a, b) \\cdot \\text{lcm}(b, c) = P = a \\cdot (bc) = a \\cdot \\gcd(b, c) \\cdot \\text{lcm}(b, c),\n$$\nwhich can be simplified to $\\gcd(a, b) = a \\cdot \\gcd(b, c)$. Since greatest common divisor is a divisor, we see that $a \\mid \\gcd(a, b) \\mid b$, hence $b$ is a multiple of $a$, as desired.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 29568,
    "subject": "Mathematics (Multi-modal)",
    "question": "Suppose that we fill a $3 \\times 3$ table with natural numbers from $1$ to $9$ using each of them precisely once. After that, we write down the sums in each of the four $2 \\times 2$ squares in the ascending order. Determine whether it is possible to obtain the following quadruples of natural numbers:\n\na) $24$, $24$, $25$, $25$,\n\nb) $20$, $23$, $26$, $29$.",
    "options": [],
    "answer": "a) Yes; b) No.",
    "solution": "a) Yes, the following table gives us the desired quadruple of sums.\n\n| 1 | 8 | 2 |\n|---|---|---|\n| 6 | 9 | 5 |\n| 3 | 7 | 4 |\n\nb) No, it is impossible to get this quadruple. Denote the sum of the four numbers as $S$, we shall proceed by proving that $S \\le 98$ and then describing all the cases where we get an equality.\n\nNote that in $S$, the central number contributes four times, the four corner tiles contribute once and the remaining four tiles contribute twice, hence we get:\n$$\nS \\le 4 \\cdot 9 + 2 \\cdot (8 + 7 + 6 + 5) + (4 + 3 + 2 + 1) = 98.\n$$\nMoreover, it is clear that the upper bound is attained if and only if the number $9$ is in the centre and numbers $1$, $2$, $3$, $4$ are in the corners in some order.\n\nSuppose that we were able to obtain the quadruple $20$, $23$, $26$, $29$, since the sum of the four numbers is $98$, the placement of the numbers in the big table would have to follow the rules explained above. But this means that the smallest possible sum of a $2 \\times 2$ square is $1 + 5 + 6 + 9 = 21$, a contradiction, hence we can't obtain this quadruple.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 29569,
    "subject": "Mathematics (Multi-modal)",
    "question": "Ten boys and ten girls met at a party. Assume that every girl likes exactly $k$ boys and every boy likes exactly $k$ girls. Is it always possible to find a couple where both the partners like each other? Solve the problem for:\n\na) $k = 5$,\n\nb) $k = 6$.",
    "options": [],
    "answer": "a) No. b) Yes.",
    "solution": "a) For $k=5$, it may happen that there are no such couples, with one counterexample given as follows. Split the boys into two disjoint quintuples $A, B$ and the girls into two disjoint quintuples $C, D$. Consider the configuration where every boy from $A$ likes all the girls in $C$, every boy in $B$ likes all the girls in $D$, every girl in $C$ likes all the boys in $B$ and every girl in $D$ likes all the boys in $A$. Then every boy likes 5 girls, every girl likes 5 boys, but there is clearly no couple where both partners like each other.\n![](attached_image_1.png)\n\nb) For $k=6$, such a couple must exist: there are in total $10k = 60$ couples (boy, girl) where the boy likes the girl and, by symmetry, 60 couples where the girl likes the boy. These two sets of couples can't possibly be disjoint, since there are $10 \\cdot 10 = 100$ couples in total and $60 + 60 > 100$, so there must be a couple where both the partners like each other.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 29570,
    "subject": "Mathematics (Multi-modal)",
    "question": "Do there exist mutually distinct real numbers $a$, $b$, $c$ such that the numbers $a^2 + b$, $b^2 + c$, $c^2 + a$ are equal, in some order, to the numbers $a + b^2$, $b + c^2$, $c + a^2$?",
    "options": [],
    "answer": "No",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 29571,
    "subject": "Mathematics (Multi-modal)",
    "question": "Suppose that $a, b$ are real numbers such that the expressions $a^2 + b$ and $a + b^2$ share the same value. What is the smallest possible shared value?",
    "options": [],
    "answer": "-1/4",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 29572,
    "subject": "Mathematics (Multi-modal)",
    "question": "Martin is arranging $n^2$ equal dice in the shape of an $n \\times n$ square such that whenever two faces touch, the same number is on both of them. What is the maximum number of different numbers that can appear on the top faces of the dice? (Martin Panák)\n\n![](attached_image_1.png)",
    "options": [],
    "answer": "1 if n = 1; 4 if n ≥ 2",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 29573,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all the pairs of natural numbers $(k, n)$ such that there exist natural numbers $a, b$ satisfying:\n$$\n\\text{gcd}(a + k, b) = n \\cdot \\text{gcd}(a, b).\n$$",
    "options": [],
    "answer": "All pairs of natural numbers.",
    "solution": "We shall prove that every pair $(k, n)$ works.\n\nFirst, if $n = 1$, we can just take $(a, b) = (k, k)$, then\n$$\n\\text{gcd}(a + k, b) = \\text{gcd}(2k, k) = k = \\text{gcd}(k, k) = \\text{gcd}(a, b).\n$$\n\nNow, assume that $n > 1$, then $nk - k > 0$ is a natural number and we can take $(a, b) = ((n-1)k, nk)$. Then we clearly have:\n$$\n\\text{gcd}(a + k, b) = \\text{gcd}((n-1)k + k, nk) = \\text{gcd}(nk, nk) = nk,\n$$\n$$\n\\text{gcd}(a, b) = \\text{gcd}((n-1)k, nk) = k,\n$$\nso\n$$\n\\text{gcd}(a + k, b) = n \\cdot \\text{gcd}(a, b).\n$$\nTherefore, every pair $(k, n)$ works.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 29574,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABCDE$ be a convex pentagon such that $AB = BC$, $AE = DE$, $AC \\perp AD$ and $CD \\parallel BE$. Prove that triangles $ABC$ and $ADE$ have equal areas. (Patrik Bak)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 29575,
    "subject": "Mathematics (Multi-modal)",
    "question": "Suppose we have a 9-digit number $N$ in which all the digits are distinct and non-zero. Then we consider all the sums of adjacent triples of digits of $N$ and order them in a non-decreasing sequence. For the following sequences, determine whether there exists an $N$ for which we get them as a result:\na) $11, 15, 16, 18, 19, 21, 22,$\nb) $11, 15, 16, 18, 19, 21, 23.$",
    "options": [],
    "answer": "a) Yes; for example N = 137658942. b) No.",
    "solution": "a) Yes, the number $N = 137658942$ works, since the sums of consecutive triples are (left to right) 11, 16, 18, 19, 22, 21, 15.\n\nb) We shall prove that no such number exists by showing that for any $N$, the sum of the 7 numbers is at most 122, while the numbers given to us have sum 123. Denote the $i$-th digit of $N$ by $a_i$ and let $S$ be the sum of the resulting 7 sums. Then we have:\n$$\n\\begin{aligned}\nS &= (a_1 + a_2 + a_3) + (a_2 + a_3 + a_4) + (a_3 + a_4 + a_5) + (a_4 + a_5 + a_6) + (a_5 + a_6 + a_7) + (a_6 + a_7 + a_8) + (a_7 + a_8 + a_9) \\\\\n&= 3(a_2 + a_3 + a_4 + a_5 + a_6 + a_7 + a_8) + a_1 + a_9\n\\end{aligned}\n$$\nSince all digits are distinct and non-zero, $a_1, \\ldots, a_9$ is a permutation of $1,2,\\ldots,9$, so their sum is $45$. Thus,\n$$\nS = 3(a_2 + a_3 + a_4 + a_5 + a_6 + a_7 + a_8) + a_1 + a_9 = 3(45 - a_1 - a_9) + a_1 + a_9 = 135 - 2a_1 - 2a_9\n$$\nThe minimum possible value of $a_1 + a_9$ is $1 + 2 = 3$, so the maximum possible $S$ is $135 - 2 \\times 3 = 129$. The maximum $a_1 + a_9$ is $8 + 9 = 17$, so the minimum $S$ is $135 - 2 \\times 17 = 101$.\n\nBut more precisely, since the sequence $11, 15, 16, 18, 19, 21, 23$ sums to $123$, which is not possible as $S$ can be at most $122$ (as shown in the original solution), there is no such $N$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 29576,
    "subject": "Mathematics (Multi-modal)",
    "question": "Suppose that a point $P$ lying in the interior of a convex quadrilateral $ABCD$ satisfies\n$$\n\\angle PAD = \\angle ADP = \\angle CBP = \\angle PCB = \\angle CPD.\n$$\nLet $O$ be the circumcentre of the triangle $CPD$. Prove that $OA = OB$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "From the given equalities, one sees that $PC \\parallel AD$ and $PD \\parallel BC$. Since the lines $PC$ and $PD$ are distinct (we know that $\\angle CPD \\neq 0$), the lines $AD$ and $BC$ are not parallel, so they intersect at a unique point $X$ such that $PCXD$ is a rhombus.\n\nNow, note that the quadrilateral $AXCP$ is an isosceles trapezoid, since $AX \\parallel CP$ and $\\angle PAX = \\angle CXA$. Therefore, the perpendicular bisectors of its bases $AX$ and $CP$ coincide. Since $O$ is the circumcentre of $CPD$, it lies on the bisector of $CP$, hence also on the bisector of $AX$, and so we have $OX = OA$. By considering the isosceles trapezoid $BXDP$, we can also obtain $OX = OB$, which gives us the desired equality.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 29577,
    "subject": "Mathematics (Multi-modal)",
    "question": "Determine the number of quadratic polynomials $P(x)$ with integral coefficients such that the following inequalities hold for all $x \\in \\mathbb{R}$:\n$$\nx^2 + 2x - 2023 < P(x) < 2x^2.\n$$",
    "options": [],
    "answer": "4042",
    "solution": "Denote the coefficients of $P$ as $P(x) = a x^2 + b x + c$. By looking what happens for $|x|$ large, we conclude that $a \\in \\{1, 2\\}$. These two cases shall be treated separately:\n\n* Suppose that $a = 1$. The lower bound on $P$ then tells us that the inequality $2x - 2023 < b x + c$ must hold for all $x$. Clearly, this happens if and only if $b = 2$ and $c > -2023$. The upper bound is then equivalent to\n$$ (x - 1)^2 > c + 1, $$\nwhich is satisfied for all $x$ precisely when $c < -1$. To summarize, we have shown that the only options here are $b = 2$ and $-2023 < c < -1$.\n\n* Suppose that $a = 2$, then analogously to the previous case, the upper bound reduces to $b x + c < 0$, which holds for all values of $x$ if and only if $b = 0$ and $c < 0$. The lower bound can then be rewritten as\n$$\n-c - 2022 < (x - 1)^2,\n$$\nwhich holds for all $x$ precisely when $c > -2022$. Therefore, we have shown that the only options for this case are $b = 0$ and $-2022 < c < 0$.\n\n**Conclusion.** There are $2 \\cdot 2021 = 4042$ such polynomials.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 29578,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $T$ be the centroid of a triangle $ABC$. Consider two isosceles right-angled triangles $BTK$ and $CTL$ so that $K$ lies in the half-plane $BTC$ and $L$ lies in the half-plane $CTA$. Finally, denote the centre of the side $BC$ as $D$ and the centre of $KL$ as $E$. Determine all the possible values of the ratio $\\frac{AT}{DE}$.",
    "options": [],
    "answer": "2√2",
    "solution": "We shall prove that the ratio has to be equal to $2\\sqrt{2}$.\n![](attached_image_1.png)\nFirst, we shall observe that the triangles $BTC$ and $KTL$ (coloured turquoise and yellow in the diagram) are similar, since\n$$\n\\angle BTC = \\angle BTK + \\angle KTC = 45^\\circ + \\angle KTC = \\angle KTC + \\angle CTL = \\angle KTL,\n$$\nand by similarity of the triangles $BKT$ and $CLT$, we have $\\frac{BT}{CT} = \\frac{KT}{LT}$. The ratio of similarity has to be $\\sqrt{2}$, since $\\frac{BT}{KT} = \\sqrt{2}$.\nSince $D$ is a midpoint of $BC$ and $E$ is the midpoint of $KL$, the triangles $BTD$ and $KTE$ are also similar. This tells us that $\\angle BTD = \\angle KTE$ and $\\frac{BT}{TD} = \\frac{KT}{TE}$. Subtracting $\\angle KTD$ from the equality gives us $\\angle BTK = \\angle DTE$ and the equality of ratios can be rearranged to $\\frac{BT}{KT} = \\frac{TD}{TE}$, so the\n\ntriangles *BTK* and *DTE* are similar. Therefore, the triangle *DTE* is also a right-angled isosceles triangle.\nFinally, recall that since *T* is the centroid, we have *AT* = 2*TD*, so the desired ratio can be computed simply as\n$$\n\\frac{AT}{DE} = 2\\frac{DT}{DE} = 2\\sqrt{2},\n$$\nas desired.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 29579,
    "subject": "Mathematics (Multi-modal)",
    "question": "We say that a positive integer is *flat* if all its digits are the same (even single-digit integers are considered flat). Determine whether every positive integer that is not flat can be expressed as a sum of two or more mutually different flat numbers. (Jozef Rajník)",
    "options": [],
    "answer": "Yes",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 29580,
    "subject": "Mathematics (Multi-modal)",
    "question": "Consider a half-disc with diameter $AB$ and centre $O$ and let $X$ be a point in the interior of the half disc. Denote the centroid of $XOB$ by $G$ and the second intersection of the line $AX$ with the boundary of the half-disc by $Y$. Prove that $YG = GB$. (Jiří Blažek, Josef Tkadlec)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "![](attached_image_1.png)\n\nSince $B$ and $Y$ lie on the same circle centered at $O$, we have $OY = OB$. Moreover, by Thales' Theorem, we see that $\\angle AYB = 90^\\circ$. This means that $M$ is the midpoint of the hypotenuse of a right triangle $XYB$, so by Thales' Theorem, it is the circumcentre of $XYB$, hence we have $MY = MB$. This means that both the points $M$ and $O$ lie on the perpendicular bisector of $BY$ and since they are clearly distinct, we immediately get the desired conclusion.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 29581,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $k$ be a circumcircle of an acute triangle $ABC$. Consider a point $P$ on the shorter arc $BC$ of the circle $k$. Denote by $O$ the intersection of segments $AP$ and $BC$. Let $O_1$ and $O_2$ be the circumcentres of triangles $BPQ$ and $CPQ$, respectively. Prove that if the line $O_1O_2$ passes through some vertex of the triangle $ABC$ then one of the points $O_1, O_2$ lies on the circle $k$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Circles with centers $O_1$ and $O_2$ have a common chord $PQ$. The line $O_1O_2$ therefore intersects the segment $PQ$ in its midpoint since it is the perpendicular bisector of $PQ$. Thus, the line $O_1O_2$ cannot pass through the vertex $A$, since it lies on the line $PQ$, but not inside the segment $PQ$. Let us note that since the triangle $ABC$ is acute, both centers $O_1, O_2$ lie inside the half-plane $BCP$.\n![](attached_image_1.png)\n\nTo show the implication from the problem statement, assume that the line $O_1O_2$ passes through vertex $C$. In triangle $PQC$ the vertex $C$ lies on the perpendicular bisector of $PQ$, thus this triangle is isosceles. Therefore\n$$\n\\angle BQA = \\angle CQP = \\angle CPQ = \\angle CPA = \\angle CBA,\n$$\nwhere in the last step we used the equality of angles over the arc $AC$ of the circle $k$. The triangle $BQA$ is isosceles with apex $A$. Both points $A$ and $O_1$ lie on the perpendicular bisector of $BQ$, which yields\n$$\n\\angle BAO_1 = 90^\\circ - \\angle ABQ = 90^\\circ - \\angle CQP = \\angle BCO_1.\n$$\nThe segment $BO_1$ can be seen from points $A$ and $C$ under the same angle, and the points $B, O_1, C$ and $A$ are conicyclic. Thus we showed that the point $O_1$ lies on the circle $k$. In the second case, where the line $O_1O_2$ passes through vertex $B$, we analogously get, that the point $O_2$ lies on $k$. This concludes the proof of the implication from the problem statement.",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 29582,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find the largest set of positive integers whose sum is $2024$, and such that each number except the smallest one is a multiple of the sum of all the smaller numbers. (Patrik Bak)",
    "options": [],
    "answer": "The largest set has size 1, namely {2024}.",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 29583,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABC$ be a triangle with side lengths $AB = 13$, $BC = 14$, $CA = 15$. Let $A'B'C'$ be a translate of $ABC$ by some unit vector. Find the largest possible area of the intersection of triangles $ABC$ and $A'B'C'$.",
    "options": [],
    "answer": "5488/75",
    "solution": "",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 29584,
    "subject": "Mathematics (Multi-modal)",
    "question": "Suppose that the sum of 74 real numbers lying in the interval $[4, 10]$ is 356. Find the maximal possible value of the sum of their squares.",
    "options": [],
    "answer": "2024",
    "solution": "Denote the numbers $x_1, \\dots, x_{74}$. The assumption $4 \\le x_i \\le 10$ guarantees that $(x_i-4)(10-x_i) \\ge 0$ holds for all $i$. Expanding this to $x_i^2 \\le 14x_i - 40$ and summing over $i$ gives us an upper bound\n$$\nx_1^2 + x_2^2 + \\dots + x_{74}^2 \\le 14(x_1 + \\dots + x_{74}) - 40 \\cdot 74 = 14 \\cdot 356 - 40 \\cdot 74 = 2024.\n$$\nTherefore, it is now sufficient to show that there exists a set of 74 real numbers that attains this bound. Our approach above shows that the bound is attained if and only if $x_i \\in \\{4, 10\\}$ for all $i$, so it suffices to check that such a 74-tuple with sum 356 exists.\nSuppose that the tuple contains $a$ fours and $b$ tens, then the non-negative integers $a, b$ must satisfy the system of equations:\n$$\n\\begin{aligned}\na+b &= 74, \\\\\n4a+10b &= 356. \\end{aligned}\n$$\nIt is easy to see that $(a, b) = (64, 10)$ is the only solution, so a set consisting of 64 fours and 10 tens proves that the upper bound is tight.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 29585,
    "subject": "Mathematics (Multi-modal)",
    "question": "We shall say that an odd prime $p$ is *kooky* if the sum of all primes smaller than $p$ is a multiple of $p$. Can two consecutive primes be kooky?",
    "options": [],
    "answer": "No",
    "solution": "We shall prove that there are no consecutive kooky primes.\nOrder all the primes in an increasing sequence $p_1 = 2 < p_2 = 3 < p_3 < \\dots$ and, for the sake of contradiction, suppose that $p_n$ and $p_{n+1}$ are both kooky for some $n > 1$. This means that there exist positive integers $a$ and $b$ such that\n$$\np_1 + \\dots + p_{n-1} = a p_n \\text{ and } p_1 + \\dots + p_n = b p_{n+1}.\n$$\nSubtracting one equation from the other gives us $(a+1)p_n = b p_{n+1}$. Since a prime $p_n$ divides the product $b p_{n+1}$ and doesn't divide $p_{n+1}$, it has to divide $b$, hence we can see that $p_1 + \\dots + p_n$ must be divisible by $p_n p_{n+1}$.\nBut this is impossible: by noting that $p_{n+1} > n$ for all $n$ and $p_i \\leq p_n$ for all $i \\leq n$, we get that\n$$\n0 < p_1 + p_2 + \\dots + p_n \\leq n p_n < p_n p_{n+1},\n$$\nso it can't be a multiple of $p_n p_{n+1}$, which yields the desired contradiction.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 29586,
    "subject": "Mathematics (Multi-modal)",
    "question": "Ten boys and ten girls met at a party. Suppose that every boy likes a different (positive) number of girls and that every girl likes a different (positive) number of boys. Find the largest non-negative integer $n$ such that it is always possible to form $n$ disjoint couples of a boy and a girl that like each other.\n(Josef Tkadlec)",
    "options": [],
    "answer": "1",
    "solution": "We shall prove that the answer is $n = 1$.\n\nTo begin with, note that the problem statement implies that the boys like $1, 2, \\ldots, 10$ girls in some order, so there exists a boy that likes all the girls. Analogously, there must exist a girl that likes all the boys, so putting the two together always yields an admissible couple.\n\nIn the second part of the solution, we shall construct a configuration where it's impossible to form more than one such couple. Number the boys and the girls by numbers $1, \\ldots, 10$ and suppose that the boy $i$ likes the girl $j$ if and only if $j \\ge i$, while the girl $j$ likes a boy number $i$ if and only if $i = 1$ or $i > j$ (see the diagram below for the case $i = 5$, with boys on top and girls on bottom). With such an assignment, the $i$-th boy likes $11 - i$ girls and the $j$-th girl likes $11 - j$ boys. Also, it\n\n![](attached_image_1.png)\n\nis clear that the first boy is the only one that can be paired up with a girl that he likes so that she also likes him back, hence it is impossible to form two disjoint admissible couples and we are done.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 29587,
    "subject": "Mathematics (Multi-modal)",
    "question": "Saba is trying to get from the ground floor of an infinitely tall skyscraper to the $n$th floor using a special elevator. The elevator has buttons $0$, $1$, $2$, ... and it moves alternately up/down (first it moves up). Moreover, upon pressing button $k$, the elevator moves by $2^k$ floors, and each subsequent button pressed must have a smaller number than the previous button. Prove that Saba can reach each floor $n \\ge 1$ in exactly two different ways. (Morteza Saghafian)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 29588,
    "subject": "Mathematics (Multi-modal)",
    "question": "For a natural number $n$, consider the sequence $(a_k)_{k=1}^{\\infty}$ given by $a_1 = \\frac{1}{n}$ and the recurrent relation\n$$\na_{k+1} = 3a_k - \\lfloor 2a_k \\rfloor - \\lfloor a_k \\rfloor,\n$$\nfor all $k \\ge 1$. Determine all the values of $n$ for which the sequence is eventually constant.",
    "options": [],
    "answer": "n = 3^\\alpha or n = 2 \\cdot 3^\\alpha for some non-negative integer \\alpha",
    "solution": "Let $f$ be the function $f(x) = 3x - \\lfloor 2x \\rfloor - \\lfloor x \\rfloor$, then the defining relation can be conveniently written as $a_{k+1} = f(a_k)$.\n\nNote that since $f(0) = f(1) = 0$ and $f(\\frac{1}{2}) = \\frac{1}{2}$, the sequence will be eventually constant whenever $\\frac{1}{2}$ or $1$ occurs in it. We can also observe that $f(x) = 3x$ for $0 < x < \\frac{1}{2}$. Therefore, by a simple inductive argument, we can show that for any non-negative integer $\\alpha$ and $n = 3^{\\alpha}$, we have $a_{\\alpha+1} = 1$ and, similarly, for $n = 2 \\cdot 3^{\\alpha}$, we have $a_{\\alpha+1} = \\frac{1}{2}$. Therefore, the integers $n$ of this form are solutions.\n\nNow, we will show that there are no other solutions. In order to do that, it suffices to prove that for any $n$ satisfying the hypothesis of the problem, there must exist some non-negative integer $\\alpha$ such that $n \\mid 2 \\cdot 3^{\\alpha}$.\n\nSo, suppose that the sequence is eventually constant for some value of $n$. Then we can define an auxiliary sequence $(b_k)$ given by $b_k = n a_k$, then this new sequence is given by a recurrence relation $b_1 = 1$ and\n$$\nb_{k+1} = 3b_k - n \\left\\lfloor \\frac{2b_k}{n} \\right\\rfloor - n \\left\\lfloor \\frac{b_k}{n} \\right\\rfloor.\n$$\nBy a simple induction, we can see that $b_k$ is a non-negative integer and moreover that $b_k \\equiv 3^{k-1} \\pmod{n}$ for all $k \\ge 1$.\n\nSince the sequence is eventually constant, there exists some $\\alpha \\ge 0$ with $b_{\\alpha+2} = b_{\\alpha+1}$, so the congruence that we have derived above gives $3^{\\alpha+1} \\equiv 3^{\\alpha} \\pmod{n}$, which is clearly equivalent to the desired statement $n \\mid 2 \\cdot 3^{\\alpha}$, so we are done.\n\n**Conclusion.** The sequence is eventually constant if and only if $n$ is of the form $3^{\\alpha}$ or $2 \\cdot 3^{\\alpha}$ for some non-negative integer $\\alpha$.\nIt is possible to find all values $a_1 \\in \\mathbb{R}$ for which the sequence is eventually constant. As before, let $f(x) = 3x - \\lfloor 2x \\rfloor - \\lfloor x \\rfloor$. If we denote the fractional part of a real number $t$ as $\\{t\\}$, then we have\n$$\nf(x) = 3x - \\lfloor 2x \\rfloor - \\lfloor x \\rfloor = (2x - \\lfloor 2x \\rfloor) + (x - \\lfloor x \\rfloor) = \\{2x\\} + \\{x\\}.\n$$\nFrom this expression, it is clear that the function $f$ is 1-periodic, hence we get:\n$$\nf(f(x)) = f(\\{2x\\} + \\{x\\}) = f(\\{2x\\} + \\{x\\} + \\lfloor 2x \\rfloor - \\lfloor x \\rfloor) = f(3x).\n$$\nWe can also use this expression for $f$ to prove that the fixed points of $f$ are precisely $0$ and $\\frac{1}{2}$ (in the first solution, we didn't need the fact that these are all the fixed points). Moreover, we can show that\n$$\nf(x) = 0 \\iff \\{x\\} = 0 \\text{ and } f(x) = \\frac{1}{2} \\iff \\{x\\} = \\frac{1}{2} \\text{ or } \\{x\\} = \\frac{1}{6}.\n$$\nFinally, suppose that the sequence is eventually constant for a given value of $a_1$. Then $f(a_{k+1}) = a_{k+1}$ for some positive integer $k$, hence $a_{k+1} = 0$ or $a_{k+1} = \\frac{1}{2}$. By a simple inductive argument, we see that\n$$\na_{k+1} = f(a_k) = f(f(a_{k-1})) = f(3a_{k-1}) = \\dots = f(3^{k-1} a_1),\n$$\nso by the equivalences that we've stated above, this happens if and only if $\\{3^{k-1} a_1\\} \\in \\{0, \\frac{1}{6}, \\frac{1}{2}\\}$.\n\nTherefore, we have shown that the sequence stabilizes after at most $k+1$ steps if and only if $a_1$ is of the form\n$$\na_1 = \\frac{m}{3^{k-1}}, \\quad a_1 = \\frac{6m+1}{2 \\cdot 3^k} \\text{ or } a_1 = \\frac{2m+1}{2 \\cdot 3^{k-1}},\n$$\nwhere $k \\ge 1$ and $m$ is an integer. One can also simplify the answer by noting that the third case is contained inside the second one.\n\n**Conclusion.** The sequence is eventually constant if and only if $a_1 = \\frac{m}{3^k}$ or $a_1 = \\frac{6m+1}{2 \\cdot 3^k}$ for some non-negative integer $k$ and an integer $m$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 29589,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $a$, $b$, $c$, $d$ be positive real numbers lying in the interval $[1, 2]$ that also satisfy the equation $(a + c)(b + d) = 8$. Prove that the inequality\n$$\n\\frac{1}{a^2 + b^2 - 1} + \\frac{1}{b^2 + c^2 - 1} + \\frac{1}{c^2 + d^2 - 1} + \\frac{1}{d^2 + a^2 - 1} \\geq 1,\n$$\nis satisfied for all such quadruples and determine all the cases when the equality holds.",
    "options": [],
    "answer": "(1, 2, 1, 2) and (2, 1, 2, 1)",
    "solution": "To begin with, note that all the numbers are greater or equal to one, so the denominators are positive and the question is well-posed.\n\nAs a first step in our solution, let's observe that since $a$ and $b$ lie in a closed interval of length $1$, we must have $(a - b)^2 \\le 1$, which can be rearranged to\n$$\n\\frac{1}{a^2 + b^2 - 1} \\ge \\frac{1}{2ab}.\n$$\nApplying similar bounds to all four terms, we see that it suffices to prove the inequality\n$$\n\\frac{1}{ab} + \\frac{1}{bc} + \\frac{1}{cd} + \\frac{1}{da} \\ge 2.\n$$\n\nBut since the condition $(a + c)(b + d) = 8$ is equivalent to $ab + bc + cd + da = 8$, this follows from the inequality between the arithmetic and harmonic means (or, if one prefers, from the Cauchy-Schwarz inequality).\n\nIn order for equality to be attained, we must have equality in both our estimates. From the first one, we can infer that\n$$\n|a - b| = |b - c| = |c - d| = |d - a| = 1.\n$$\nSince all the numbers lie in an interval of length $1$, this leaves us with only two options where equality can be attained, $(1, 2, 1, 2)$ and $(2, 1, 2, 1)$, with an easy check confirming that we indeed get equality.\n\n**Conclusion.** The inequality holds by the argument above and we get an equality precisely for quadruples $(1, 2, 1, 2)$ and $(2, 1, 2, 1)$.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 29590,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all right-angled triangles with integral lengths of sides in which one can inscribe two congruent circles which satisfy the following conditions:\n* Their radius is a prime number.\n* The circles touch externally.\n* Both of the circles are tangent to the hypotenuse and each of them is tangent to a different leg of a triangle.",
    "options": [],
    "answer": "Unique solution: side lengths 21, 28, 35; circle radius 5.",
    "solution": "We will show that there exists only one such triangle and it has sides of length $21$, $28$, $35$ and the two touching circles have radius equal to $5$.\n\nIn any right-angled triangle $ABC$ with hypotenuse $AB$ we denote $a = BC$, $b = AC$, $c = AB$. Clearly, two congruent circles with all tangent conditions from the statement exist. Denote them by $k_1(S_1, r)$ and $k_2(S_2, r)$ such that $k_1$ is tangent to $AC$ (and $k_2$ is tangent to $BC$). Moreover, denote the tangent points of these two circles to $AB$ by $T_1$ and $T_2$, respectively, as in the picture. Consider the incircle $k(S, \\varrho)$ of the triangle $ABC$ and denote $D$ the tangent point of $k$ with $AB$.\n\n![](attached_image_1.png)\n\nIn the first part of the solution we show that the radius $r$ of circles $k_1$ and $k_2$ is determined by the equation\n$$\nr = \\frac{c(a + b - c)}{2(a + b)} \\qquad (1)\n$$\nIn homothety with center $A$ and coefficient $r/\\varrho$ which maps $k$ to $k_1$, the point $D$ is mapped to the point $T_1$. Therefore, $|AT_1| = |AD| \\cdot r/\\varrho$. Analogously, we get the equation $|BT_2| = |BD| \\cdot r/\\varrho$. Since $T_1T_2S_2S_1$ is a rectangle, we have $|T_1T_2| = |S_1S_2| = 2r$. After plugging everything in the equation $c = |AT_1| + |T_1T_2| + |BT_2|$, by using the relation $|AD| + |BD| = c$ we obtain\n$$\nc = |AD| \\cdot \\frac{r}{\\varrho} + 2r + |BD| \\cdot \\frac{r}{\\varrho} = 2r + (|AD| + |BD|) \\cdot \\frac{r}{\\varrho} = 2r + \\frac{cr}{\\varrho}.\n$$\n\nThis yields\n$$\nr = \\frac{c\\varrho}{c + 2\\varrho}.\n$$\nAfter using the well-known formula $\\varrho = (a + b - c)/2$, we obtain the desired formula 1. Suppose that the lengths $a, b, c, r$ are integers and put $k = \\gcd(a, b, c)$. Then $a = ka_1$, $b = kb_1$ and $c = kc_1$. Moreover, the Pythagorean theorem $a_1^2 + b_1^2 = c_1^2$ implies that the numbers $a_1, b_1, c_1$ are pairwise coprime and $a_1$ and $b_1$ have different parity. The number $c_1$ is odd and the number $a_1 + b_1 - c_1$ is even. After plugging in $a = ka_1$, $b = kb_1$ and $c = kc_1$ to 1 we obtain:\n$$\nr = \\frac{kc_1(a_1 + b_1 - c_1)}{2(a_1 + b_1)} = \\frac{k}{a_1 + b_1} \\cdot c_1 \\cdot \\frac{a_1 + b_1 - c_1}{2}. \\qquad (2)\n$$\nWe will show that not only the second fraction, but also the first fraction of the right-hand side of 2 is an integer. Suppose that there is a prime $p$ that divides both $a_1 + b_1$ and $c_1$. Then from the equation $(a_1 + b_1)^2 = c_1^2 + 2a_1b_1$ we have $p \\mid a_1b_1$, which is a contradiction since the numbers $a_1, b_1$ and $c_1$ are pairwise coprime. Thus, the number $a_1 + b_1$ is coprime to $c_1$ and also $a_1 + b_1 - c_1$. Since $r$ is an integer, from the equation 2 it follows, that the first fraction is an integer.\n\nBy the problem statement, $r$ is a prime. By 2 it is a product of three positive integers. Since $c_1 > a_1 \\ge 1$, we necessarily have $c_1 = r$ and the other two fractions in the right-hand side of 2 are equal to 1. Therefore, we have $k = a_1 + b_1$, $a_1 + b_1 - c_1 = 2$ and also $k = a_1 + b_1 = c_1 + 2 = r + 2$. It follows that\n$$\n2a_1b_1 = (a_1 + b_1)^2 - c_1^2 = (r + 2)^2 - r^2 = 4r + 4.\n$$\nAfter dividing by two we obtain\n$$\na_1b_1 = 2r + 2 = 2(a_1 + b_1 - 2) + 2 = 2a_1 + 2b_1 - 2.\n$$\nThe last equation can be rewritten in the form $(a_1 - 2)(b_1 - 2) = 2$. It easily follows that $\\{a_1, b_1\\} = \\{3, 4\\}$, $c_1 = r = a_1 + b_1 - 2 = 5$, and $k = r + 2 = 7$. For right-angled triangle with sides $7 \\cdot 3, 7 \\cdot 4$ and $7 \\cdot 5$ we, indeed, have $r = 5$. This completes the proof.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 29591,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABC$ be a triangle and let $I_A$, $I_B$, $I_C$ be centers of the excircles tangent to its sides $BC$, $CA$, $AB$, respectively. Denote by $X$, $Y$, $Z$ the orthocenters of triangles $I_ABC$, $AI_BC$, $AI_C$, respectively. Prove that the triangles $ABC$ and $XYZ$ are congruent. (Michal Janík)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 29592,
    "subject": "Mathematics (Multi-modal)",
    "question": "Real numbers $a, b, c, d$ are such that\n$$\na + b + c + d = 0 \\quad \\text{and} \\quad \\frac{1}{a} + \\frac{1}{b} + \\frac{1}{c} + \\frac{1}{d} = 0.\n$$\nHow many of the equalities\n$$ab = cd, \\quad ac = bd, \\quad ad = bc$$\ncan hold simultaneously?",
    "options": [],
    "answer": "3",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 29593,
    "subject": "Mathematics (Multi-modal)",
    "question": "Given two different real numbers $a$, $b$ such that the expressions $a^3 + b$ and $a + b^3$ have the same value, prove that $-1 \\le ab < \\frac{1}{3}$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 29594,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find the largest integer $n$ with the following property: Whenever five distinct points are given in a plane such that some two of them lie inside the triangle formed by the remaining three points, then some three of these five points can be denoted by $X, Y, Z$ such that $n^\\circ < |\\angle XYZ| \\le 180^\\circ$.",
    "options": [],
    "answer": "119",
    "solution": "",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 29595,
    "subject": "Mathematics (Multi-modal)",
    "question": "An online vote is being held between options $A$ and $B$. Before Paul voted, the percentage of votes for option $A$ was a positive integer. Paul's vote increased this number by exactly one. Prove that Paul's vote was the nineteenth vote for option $A$. (Josef Tkadlec)",
    "options": [],
    "answer": "19",
    "solution": "Let the total number of votes before Paul voted be $n$, and the number of votes for $A$ before Paul voted be $k$. The percentage of votes for $A$ before Paul voted is $\\frac{100k}{n}$, which is a positive integer. After Paul votes for $A$, the number of votes for $A$ becomes $k+1$, and the total number of votes becomes $n+1$. The new percentage is $\\frac{100(k+1)}{n+1}$. We are told that this new percentage is exactly one more than the previous percentage:\n\n$$\n\\frac{100(k+1)}{n+1} = \\frac{100k}{n} + 1\n$$\n\nLet $p = \\frac{100k}{n}$, so $k = \\frac{pn}{100}$.\n\nSubstitute $k$ into the equation:\n\n$$\n\\frac{100\\left(\\frac{pn}{100} + 1\\right)}{n+1} = p + 1\n$$\n\nMultiply both sides by $n+1$:\n\n$$\n100\\left(\\frac{pn}{100} + 1\\right) = (p + 1)(n + 1)\n$$\n\n$$\npn + 100 = pn + p + n + 1\n$$\n\n$$\n100 = p + n + 1\n$$\n\n$$\np + n = 99\n$$\n\nRecall that $k = \\frac{pn}{100}$ must be an integer, and $p$ and $n$ are positive integers with $p + n = 99$.\n\nLet $n = 99 - p$.\n\nThen:\n\n$$\nk = \\frac{p(99 - p)}{100}\n$$\n\nWe need $k$ to be an integer, so $p(99 - p)$ is divisible by $100$.\n\nLet us try all $p$ from $1$ to $98$ (since $p$ and $n$ are positive integers):\n\nWe seek $p$ such that $p(99 - p)$ is divisible by $100$.\n\nLet us try $p = 19$:\n\n$19(99 - 19) = 19 \\times 80 = 1520$, and $1520/100 = 15.2$ (not integer).\n\nTry $p = 80$:\n\n$80(99 - 80) = 80 \\times 19 = 1520$, $1520/100 = 15.2$ (not integer).\n\nTry $p = 20$:\n\n$20 \\times 79 = 1580$, $1580/100 = 15.8$ (not integer).\n\nTry $p = 25$:\n\n$25 \\times 74 = 1850$, $1850/100 = 18.5$ (not integer).\n\nTry $p = 50$:\n\n$50 \\times 49 = 2450$, $2450/100 = 24.5$ (not integer).\n\nTry $p = 4$:\n\n$4 \\times 95 = 380$, $380/100 = 3.8$ (not integer).\n\nTry $p = 75$:\n\n$75 \\times 24 = 1800$, $1800/100 = 18$ (integer!).\n\nSo $p = 75$, $n = 24$, $k = 18$.\n\nCheck: Before Paul voted, $n = 24$, $k = 18$, so percentage is $\\frac{100 \\times 18}{24} = 75$.\n\nAfter Paul votes for $A$, $k = 19$, $n = 25$, percentage is $\\frac{100 \\times 19}{25} = 76$.\n\nSo the percentage increases by exactly $1$.\n\nTherefore, Paul's vote was the nineteenth vote for option $A$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 29596,
    "subject": "Mathematics (Multi-modal)",
    "question": "A team consists of $7$ players. In each round of the tournament, five of them play and two sit on the bench. Prove that, regardless of the (positive) number of rounds and the choice of who plays in what round, at the end of the tournament there are two players who have been together (either on the field or on the bench) in more than half of the rounds. (David Hruška)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 29597,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $n > 1$ be a positive integer and $p$ its greatest prime factor. For each nonempty subset of divisors of the number $n$, write on the board the sum of its elements. Suppose that in this way we have written more than $p$ numbers from the set $\\{1, 2, \\dots, p+2\\}$, and that we have not written any number from this set more than once. Prove that we have not written any number more than once. (Zdeněk Pezlar)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 29598,
    "subject": "Mathematics (Multi-modal)",
    "question": "A line segment $BC$ is given in the plane. Consider all acute-angled triangles $ABC$ with $|\\angle BAC| = 45^\\circ$. In each such triangle, denote by $D$ and $E$ those points on the sides $AB$ and $AC$, respectively, such that $BC$ is a common tangent of the circumcircles of triangles $ACD$ and $ABE$. Finally, denote by $P$ and $Q$ the feet of perpendiculars from the points $D$ and $E$ to the line $BC$, respectively. Prove that as points $P$ and $Q$ vary, there exists a fixed point $X$ not on the line $BC$ such that $\\angle PXQ$ is fixed.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 29599,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all sets $S$ of 3 or more primes with the following property: The elements of $S$ can be written around a circle such that if you compute the largest prime factor of the sum of each consecutive pair, you again obtain all elements of $S$ (in some order). (Michal Janík)",
    "options": [],
    "answer": "{2, 3, 5, 7}",
    "solution": "",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 29600,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all positive integers $n$ with the following property: It is possible to color $2n$ cells in an $n \\times n$ square grid such that each row and each column contains exactly two colored cells, and no two colored cells share a side or a vertex.",
    "options": [],
    "answer": "All integers n greater than or equal to 9",
    "solution": "",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 29601,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABC$ be an acute triangle, $O$ its circumcenter, $\\omega$ its circumcircle, $H$ its orthocenter, and $M$ the midpoint of $BC$. Line $AH$ meets $\\omega$ again at $D \\neq A$ and line $DM$ meets $\\omega$ again at $E \\neq D$. Finally, line $AE$ meets the circumcircle of triangle $OME$ again at $F \\neq A$. Prove that $FH = FA$. (Michal Pecho)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 29602,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $a$, $b$ and $c$ be positive integers such that $ab$ is divisible by $2c$, $bc$ is divisible by $3a$ and $ca$ is divisible by $5b$. Find the least possible value of $abc$.",
    "options": [],
    "answer": "900",
    "solution": "Since $ab$ is divisible by $2c$ and $ca$ is divisible by $5b$, $ab \\cdot ca$ must be divisible by $2c \\cdot 5b$, hence $a^2$ is divisible by $2 \\cdot 5$. Therefore $a^2$ is divisible by $2$ and $5$, hence $a$ is divisible by $2$ and $5$. Similarly $b$ is divisible by $2$ and $3$, and $c$ is divisible by $3$ and $5$. Consequently $abc$ is divisible by $2 \\cdot 5 \\cdot 2 \\cdot 3 \\cdot 3 \\cdot 5 = 900$. On the other hand, $a = 10$, $b = 6$ and $c = 15$ satisfy the conditions and $abc = 900$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 29603,
    "subject": "Mathematics (Multi-modal)",
    "question": "A unit L-shape consists of three unit squares as shown in the picture. Prove that for any positive integer $k$ it is possible to cut a similar L-shape with $k$ times larger side lengths into unit L-shapes.\n\n![](attached_image_1.png)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let the L-shape be placed so that the two longer sides meet at the top left corner. Starting from the top left we place on it $k$ unit L-shapes diagonally with the same orientation as the large L-shape (Fig. 12). The rest consists of two equal staircase-like parts; it is enough to show that one of them, e.g. the lower part, can be covered. The staircase has $k$ stairs, the lowest one at the height $k-1$ and the highest at the height $2k-2$.\n\nIn case $k=1$ the staircase is empty, in case $k=2$ it can be covered with one unit L-shape. Assume that the claim holds for the staircase with $k$ stairs and consider the staircase with $k+2$ stairs. Separate a strip of width 2 from the left and bottom. The rest can be covered by the induction assumption. The topmost part of the strip is covered with one unit L-shape. Now we have\n\n![](attached_image_2.png)\nFig. 12\n![](attached_image_3.png)\nFig. 13\n![](attached_image_4.png)\nFig. 14\n![](attached_image_5.png)\nFig. 15\n\nto cover the rest of the strip whose lower and left sides have correspondingly the lengths $k+2$ and $2k$.\n\n* If $k$ is divisible by 3, then cut the figure into two strips of sizes $2 \\times 2k$ and $k \\times 2$ and cover both of them with $2 \\times 3$ rectangles consisting of two unit L-shapes (Fig. 13) and we are done.\n* If $k \\equiv 1 \\pmod{3}$, then cut the figure into two strips of sizes $2 \\times (2k-2)$ and $(k+2) \\times 2$ and cover both of them with $2 \\times 3$ rectangles (Fig. 14). This is possible because $2k-2$ and $k+2$ are divisible by 3.\n* If $k \\equiv 2 \\pmod{3}$, then cut the figure into two strips of sizes $2 \\times (2k-4)$ and $(k-2) \\times 2$, and a corner part, which is a L-shape with $k=2$. Both strips can be covered by $2 \\times 3$ rectangles since $2k-4$ and $k-2$ are divisible by 3; the corner part can be covered by induction basis (Fig. 15).",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 29604,
    "subject": "Mathematics (Multi-modal)",
    "question": "A ball bearing consists of two cylinders with the same axis and $n$ equal balls between them. The centers of all the balls are on the same plane perpendicular to the axis of the cylinders and each ball touches both cylinders and two adjacent balls. Let $r$ be the radius of the balls and let $R$ be the radius of the outer cylinder. Prove that $r/R < \\frac{\\pi}{n+\\frac{\\pi}{n}}$. (Grade 11.)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Consider the regular $n$-gon with vertices at the centers of the balls (Fig. 16). Its edges are of length $2r$ and its perimeter is $n \\cdot 2r$. The radius of the circumcircle of the $n$-gon is $R-r$ and the length of the circumcircle is $2\\pi(R-r)$. Since a chord of a circle is always shorter than the corresponding arc of the circle, we have $n \\cdot 2r < 2\\pi(R-r)$ or $nr + \\pi r < \\pi R$, which implies $r/R < \\frac{\\pi}{n+\\frac{\\pi}{n}}$.\n\n![](attached_image_1.png)\nFig. 16\nConsider the isosceles triangle with vertices at the centers of two adjacent balls and at the closest point to them on the common axis of the cylinders (Fig. 17). The two equal sides of the triangle are of length $R-r$, the base is of length $2r$ and the vertex angle is $\\frac{2\\pi}{n}$. The altitude drawn onto the base divides the triangle into two equal right triangles with the hypotenuse $R-r$, one of the legs $r$ and the opposite angle $\\frac{\\pi}{n}$. Hence $\\frac{r}{R-r} = \\sin \\frac{\\pi}{n} < \\frac{\\pi}{n}$, whence $nr < \\pi R - \\pi r$, which implies $\\frac{r}{R} < \\frac{\\pi}{n+\\frac{\\pi}{n}}$.\n\n![](attached_image_2.png)\nFig. 17",
    "topic": "Geometry",
    "subtopic": "Solid Geometry"
  },
  {
    "id": 29605,
    "subject": "Mathematics (Multi-modal)",
    "question": "The sequence $(a_n)$ is defined by $a_1 = 1$ and $a_n = n \\cdot (a_1 + \\dots + a_{n-1})$ for all $n > 1$. Find all indices $n$ for which $a_n$ is divisible by $1 \\cdot 2 \\cdot \\dots \\cdot n$. (Grade 12.)",
    "options": [],
    "answer": "n = 1 or n is even",
    "solution": "For each $n \\ge 2$ denote $S_n = a_1 + \\dots + a_{n-1}$. Then $a_n = S_n \\cdot n$ and for all $n > 2$ we have $S_n = S_{n-1} + a_{n-1} = S_{n-1} + S_{n-1} \\cdot (n-1) = S_{n-1} \\cdot n$. Hence $S_n = S_{n-1} \\cdot n = S_{n-2} \\cdot (n-1)n = \\dots = S_2 \\cdot 3 \\cdot \\dots \\cdot n = \\frac{n!}{2}$ because $S_2 = 1 = \\frac{1 \\cdot 2}{2}$. Consequently $a_n = S_n \\cdot n = n! \\cdot \\frac{n}{2}$ for all $n \\ge 2$. Therefore, for $n \\ge 2$, $a_n$ is divisible by $n!$ iff $n$ is even, and $n=1$ also satisfies the condition.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 29606,
    "subject": "Mathematics (Multi-modal)",
    "question": "The lengths of the sides of a quadrilateral are $a$, $b$, $c$, $d$ and its area is $S$. Prove that $a^2 + b^2 + c^2 + d^2 \\ge 4S$. For which quadrilaterals does the equality hold?",
    "options": [],
    "answer": "squares",
    "solution": "Without loss of generality we can assume that $a$, $b$, $c$ and $d$ are the lengths of consecutive sides of the quadrilateral. A diagonal divides the quadrilateral into two triangles. From one partition we get the inequality $\\frac{ab}{2} + \\frac{cd}{2} \\ge S$, whence $ab + cd \\ge 2S$, and from the other partition $\\frac{bc}{2} + \\frac{da}{2} \\ge S$, whence $bc + da \\ge 2S$. Therefore $ab + bc + cd + da \\ge 4S$.\n\nOn the other hand, by adding the inequalities $a^2 + b^2 \\ge 2ab$, $b^2 + c^2 \\ge 2bc$, $c^2 + d^2 \\ge 2cd$, and $d^2 + a^2 \\ge 2da$, and dividing by 2 we get $a^2 + b^2 + c^2 + d^2 \\ge ab + bc + cd + da$, which implies the required inequality.\n\nThe equality holds iff all the inequalities used are, in fact, equalities. In the first inequality the equality holds iff all the angles are right angles. In the second step the equalities hold iff all sides are of equal length.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 29607,
    "subject": "Mathematics (Multi-modal)",
    "question": "In a coordinate city there are $n \\ge 3$ tramlines parallel to the $x$-axis such that each line begins from $x$-coordinate $0$ and ends at $x$-coordinate $n$. Exactly one tram of length $1$ is moving on each line: on the first line with speed $1$, on the second line with speed $2$ etc, until on the last line with the speed $n$. When a tram reaches the end of its line it instantly starts moving back without turning around. In the morning all trams start moving at the same time from the starting position where the $x$-coordinate of the back end of the tram is $0$. Prove that the trams' projections onto the $x$-axis never cover the whole interval from $0$ to $n$. (Grade 12.)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "The projections of the trams can cover the whole interval only when one projection covers $[0, 1]$, another $[1, 2]$ etc. until $[n-1, n]$. Consider the moments when the projection of the slowest tram covers one of these intervals. When the slowest tram moves by $1$ unit, then the fastest and the third fastest trams move correspondingly by $n$ and $n-2$ units. Together these two trams move by $2n-2$ units which is exactly one to and fro cycle.\n\nDenote the integer positions of the trams on the round trip by numbers $0$ to $2n-3$, i.e. the starting position is $0$ and each next one until returning to the starting point is greater by one. Call these numbers the position characteristics. If the sum of the position characteristics of two trams is $2n - 2$, then their projections cover the same interval because one of them has moved the same amount from the starting point as the other one still has to go to reach it. If the sum of the position characteristics is $0$, then they both are in the starting positions, so they again cover the same interval. Hence, when the sum of the position characteristics is divisible by $2n - 2$, the projections cover the same interval.\n\nAt the beginning the sum of the position characteristics of the fastest and the third fastest tram is $0$ and each time they together move by $2n - 2$ units the sum of their position characteristics stays divisible by $2n - 2$. Consequently, when the projection of the slowest tram covers an interval with integer endpoints, the projections of these two trams cover the same interval, hence at least one of the intervals is not covered.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 29608,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find the minimal distance between two points, one of which is on the graph of function $y = e^x$ and the other on the graph of function $y = \\ln x$.",
    "options": [],
    "answer": "sqrt(2)",
    "solution": "The graphs of functions $y = e^x$ and $y = \\ln x$ are symmetrical with respect to the line $y = x$ (see Fig. 18).\n\nHence, the distance between points on these graphs is minimal if and only if both points are at minimal distance from the line $y = x$.\n\nThe distance between the graph of the function $y = e^x$ and the graph of the function $y = x$ is minimal at the point where the tangent is parallel to $y = x$. Then, $y' = 1$ for the function $y = e^x$, whence $e^x = 1$, giving $x = 0$ and $y = 1$.\n\nThe minimal distance is therefore between points $(0, 1)$ and $(1, 0)$ and it is $\\sqrt{2}$.\n\n![](attached_image_1.png)\nFig. 18",
    "topic": "Discrete Mathematics",
    "subtopic": "Other"
  },
  {
    "id": 29609,
    "subject": "Mathematics (Multi-modal)",
    "question": "Prove the inequality\n$$\n2010 < \\frac{2^2 + 1}{2^2 - 1} + \\frac{3^2 + 1}{3^2 - 1} + \\dots + \\frac{2010^2 + 1}{2010^2 - 1} < 2010 \\frac{1}{2} \\quad \\text{(Grade 9.)}\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Since $\\frac{n^2 + 1}{(n-1)(n+1)} = 1 + \\frac{1}{n-1} - \\frac{1}{n+1}$, the given sum can be rewritten in the form $1 + \\frac{1}{1} - \\frac{1}{3} + 1 + \\frac{1}{2} - \\frac{1}{4} + \\dots + 1 + \\frac{1}{2009} - \\frac{1}{2011}$.\n\n$= 2010 + \\frac{1}{2} - \\frac{1}{2010} - \\frac{1}{2011}$.\n\nBecause $0 < \\frac{1}{2} - \\frac{1}{2010} - \\frac{1}{2011} < \\frac{1}{2}$, the inequality is proved.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 29610,
    "subject": "Mathematics (Multi-modal)",
    "question": "Juku drew a regular hexagon and chose three triangles with different areas whose vertices were among the vertices of the hexagon. Prove that the sum of the areas of the triangles is equal to the area of the hexagon.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Any triangle whose vertices are among the vertices of a regular hexagon is one of the following:\n* a triangle $\\Delta_1$ whose vertices are three consecutive vertices of the hexagon;\n* a triangle $\\Delta_2$ whose two vertices are adjacent vertices of the hexagon and the third one is adjacent to none of the first two;\n* a triangle $\\Delta_3$ where any two vertices are not adjacent vertices of the hexagon.\n\nSince the areas of the chosen triangles are different, the triangles must be equal to the triangles $\\Delta_1$, $\\Delta_2$, $\\Delta_3$. The hexagon can be divided into four parts (Fig. 9): the triangle $\\Delta_3$ surrounded by three triangles $\\Delta_1$. The area of the triangle $\\Delta_2$ (marked by a dotted line in Fig. 9) is twice the area of the triangle $\\Delta_1$ because they have the same base but the height of $\\Delta_2$ is twice the height of $\\Delta_1$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 29611,
    "subject": "Mathematics (Multi-modal)",
    "question": "Points $A'$, $B'$ and $C'$ are chosen correspondingly on the sides $AB$, $BC$, and $CA$ of an equilateral triangle $ABC$ so that $\\frac{|A'B|}{|AB|} = \\frac{|B'C|}{|BC|} = \\frac{|C'A|}{|CA|} = k$. Find all positive real numbers $k$ for which the area of the triangle $A'B'C'$ is exactly half of the area of the triangle $ABC$.",
    "options": [],
    "answer": "k = 1/2 + √3/6 or k = 1/2 − √3/6",
    "solution": "Let $\\alpha$ be the angle at the vertex $A$ (Fig. 10).\n\nThe area of the triangle $AA'C'$ is $S_{AA'C'} = \\frac{1}{2} \\cdot |AA'| \\cdot |AC'| \\cdot \\sin \\alpha = \\frac{1}{2} \\cdot (1-k)|AB| \\cdot k|AC| \\cdot \\sin \\alpha = (1-k)k S_{ABC}$.\n\nSimilarly $S_{BB'A'} = (1-k)k S_{ABC}$ and $S_{CC'B'} = (1-k)k S_{ABC}$.\n\nHence the triangles $AA'C'$, $BB'A'$ and $CC'B'$ are of equal area.\n\n![](attached_image_1.png)\nFig. 10\n\nthe area of the triangle $A'B'C'$ is half of the area of the triangle $ABC$ iff the area of the triangle $AA'C'$ is one sixth of the area of the triangle $ABC$, i.e. $(1-k)k = \\frac{1}{6}$.\n\nThe solutions of $k^2 - k + \\frac{1}{6} = 0$ are $k_{1,2} = \\frac{1}{2} \\pm \\frac{\\sqrt{3}}{6}$, both of them are positive.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 29612,
    "subject": "Mathematics (Multi-modal)",
    "question": "Three players $A$, $B$ and $C$ play the following game. At the beginning of the game, each player has a sheet of paper with the name of the player written on it. Player $A$ chooses one of the other players and replaces the name on this player's sheet with the name on his own sheet. Then player $B$ makes a similar move, then player $C$ and after that the turn to move goes to player $A$ again. The game ends when all the sheets have the same name written on them and the winner is the player whose name it is. Does any of the players have a winning strategy (i.e., a strategy that allows a player to win no matter what his opponents play)? (Grade 10.)",
    "options": [],
    "answer": "No. None of the players has a winning strategy.",
    "solution": "Player $B$ does not have a winning strategy, since on the first move player $A$ can write the name $A$ on his sheet, after that the name $B$ is not on any of the sheets. Similarly player $C$ does not have a winning strategy.\n\nTo prove that even player $A$ does not have a winning strategy, we show that players $B$ and $C$ have a joint strategy which guarantees that among the names written on the sheets there are at least two different names. Namely, if player $A$ on his move writes a name on the sheet of player $B$, then $B$ writes a name on the sheet of player $A$, otherwise on the sheet of player $C$. Player $C$ always writes a name on the sheet of player $B$.\n\nIn the beginning both players $B$ and $C$ have names different from the name on the sheet of player $A$. Hence $A$ cannot win in one move. Independent of which name $A$ changes on his move, after $B$ moves, the name on the sheet of $C$ differs from the name on the sheet of $A$, and after $C$ moves, both $B$ and $C$ have names on their sheets different from the one on the sheet of $A$, as in the beginning. So the cycle repeats.\nDenote the players starting from any player in the order of their turns by $X$, $Y$, and $Z$. Show that the players $Y$ and $Z$ can together always keep $X$ from winning. Indeed, $X$ can win only on his move because $Y$ and $Z$ can always play so that their move does not result immediately in $X$ winning. $X$ can win on his turn only if before his move he and somebody else have his name on their sheets. The player $Z$ cannot prevent this situation only if the same situation occurred already before his move and his sheet has the name of $X$ on it. But after $Y$ moves, then either $X$ or $Z$ has the same name on their sheets as $Y$ has, and so $Y$ can always prevent both $X$ and $Z$ having the same name on their sheets. Thus none of the three players has a winning strategy.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 29613,
    "subject": "Mathematics (Multi-modal)",
    "question": "A regular $2010$-gon is divided into pieces of triangular shape. Find the least possible number of pieces.",
    "options": [],
    "answer": "2008",
    "solution": "All the interior angles of the $2010$-gon can be built from the inner angles of the triangular pieces. As the sum of the inner angles of the $2010$-gon is $2008 \\cdot 180^{\\circ}$ and that of every triangle is $180^{\\circ}$, there must be at least $2008$ triangles. On the other hand, each convex $2010$-gon can be divided into exactly $2008$ triangles by choosing one vertex and cutting the figure into pieces along the diagonals that start from this vertex.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 29614,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $x$, $y$ and $z$ be positive integers satisfying $\\text{gcd}(x, y, z) = 1$. Prove that if $(y^2 - x^2) - (z^2 - y^2) = ((y - x) - (z - y))^2$, then $x$ and $z$ are perfect squares.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Remove the parentheses, collect the terms and divide both sides by $2$ to get $x^2 + y^2 + z^2 - 2xy - 2yz + xz = 0$. This equality can be written as $(x - y + z)^2 = xz$. Hence $xz$ is a square of an integer. If $x$ and $z$ have a common divisor $d$, then $xz$ is divisible by $d^2$, and by the previous equality $(x - y + z)^2$ is divisible by $d^2$, therefore $x - y + z$ is divisible by $d$. Since $x$ and $z$ are divisible by $d$, $y$ must be divisible by $d$, hence $d = 1$, i.e. $x$ and $z$ do not have common divisors. Since $xz$ is a square of an integer, it follows that both $x$ and $z$ are squares of integers.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 29615,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all pairs of integers $(m, n)$ such that for all positive real numbers $x$ and $y$ the inequality $x^m + y^n \\ge x^n y^m$ holds.",
    "options": [],
    "answer": "(0, 0)",
    "solution": "If $m = 0$, then the inequality is $1 + y^n \\ge x^n$. This holds for all positive real numbers $x$ and $y$ iff $n = 0$. Hence $(0, 0)$ is a solution.\n\nLet now both $m$ and $n$ be different from zero. If the pair $(m, n)$ satisfies the condition, then substituting $x$ and $y$ by $\\frac{1}{x}$ and $\\frac{1}{y}$ we see that the pair $(-m, -n)$ also satisfies the condition. Hence we can assume without loss of generality that $m \\ge n$ and $m \\ge 0$.\n\nIf $m > n$, then by taking $x = 1$ we get $1 + y^n \\ge y^m$, which does not hold for $y$ large enough. Hence $m = n$.\n\nBy taking $x = y = 4$ we get $2 \\cdot 4^m \\ge 4^{2m}$ which does not hold for any positive integer $m$. Therefore there are no more suitable pairs.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 29616,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $D$ be the midpoint of side $BC$ of triangle $ABC$. Prove that the intersection point of medians of triangle $ABD$ and that of triangle $ACD$ are equidistant from line $AD$. (Grade 11.)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Triangles $ABD$ and $ACD$ have equal area since $|BD| = |CD|$ and the altitudes drawn from $A$ coincide (Fig. 11). As these triangles have a common side $AD$, also the altitudes drawn from vertices $B$ and $C$, respectively, must be equal. Thus $B$ and $C$ are equidistant from line $AD$. Since the point\n\n![](attached_image_1.png)\nFig. 11\nof intersection of medians cuts $\\frac{1}{3}$ part of every median, the distance of the point of intersection of medians of $ABD$ from line $AD$ is thrice shorter than the distance of point $B$ from line $AD$. An analogous relation holds also for triangle $ACD$. Hence the claim follows.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 29617,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all positive integers $n$ such that $1 + 2^2 + 3^3 + 4^n$ is a perfect square.",
    "options": [],
    "answer": "1",
    "solution": "Let $1 + 2^2 + 3^3 + 4^n = x^2$. This implies $32 = x^2 - 4^n$ or, equivalently, $2^5 = (x - 2^n)(x + 2^n)$. As the l.h.s. is a power of 2, the factors in the r.h.s. are of the form $x - 2^n = 2^a$ and $x + 2^n = 2^{5-a}$ where $a$ is 0, 1 or 2. Subtracting the first of the two equalities from the second gives $2^{n+1} = 2^{5-a} - 2^a$. This leads to an integral $n$ only if $a = 2$; then $n = 1$. A check shows that $1 + 2^2 + 3^3 + 4^1 = 6^2$ indeed.\nObserve that $1 + 2^2 + 3^3 + 4^n = 32 + 4^n = 2^5 + 2^{2n} = 2^5 \\cdot (1 + 2^{2n-5})$. If $n \\ge 3$, then $2n - 5 \\ge 1$; hence $2^{2n-5}$ is an even integer and $1 + 2^{2n-5}$ is therefore odd. Thus in the prime factorization of the number given in the problem, the exponent of 2 is 5. As this is odd, the number cannot be a perfect square. If $n = 2$ or $n = 1$, then $32 + 4^n = 48$ or $32 + 4^n = 36$, respectively, where only the latter is a perfect square. Consequently, only $n = 1$ is possible.\nIf $n = 1, 2, 3$, then the given number is 36, 48, 96, respectively, where only the first is a perfect square. If $n \\ge 4$, then $2 \\cdot 2^n + 1 \\ge 2 \\cdot 16 + 1 > 32$, implying $(2^n)^2 < (2^n)^2 + 32 < (2^n)^2 + 2 \\cdot 2^n + 1 = (2^n + 1)^2$. As the number under question is equal to $(2^n)^2 + 32$, it falls between two consecutive perfect squares, hence cannot be a perfect square itself.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 29618,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $n \\ge 2$. Positive integers $a_1, a_2, \\dots, a_n$ whose sum is even and which satisfy $a_i \\le i$ for every $i = 1, 2, \\dots, n$, are given. Prove that it is possible to choose signs in the expression $a_1 \\pm a_2 \\pm \\dots \\pm a_n$ in such a way that its value becomes $0$. (Seniors.)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Prove the claim by induction on $n$. If $n=2$, then the only way to choose integers that satisfy the conditions of the problem is $a_1 = 1$ and $a_2 = 1$. In this case, $a_1 - a_2 = 0$.\n\nAssume now that the claim holds whenever $2 \\le n \\le k$ and show that it holds also for $n = k + 1$. Consider two cases.\n\n1. If $a_{k+1} = a_k$, then $a_1 + a_2 + \\dots + a_{k-1}$ is even. As this case is possible only for $k > 2$, the induction hypothesis is applicable for $n = k - 1$. Thus it is possible to choose signs in the expression $a_1 \\pm a_2 \\pm \\dots \\pm a_{k-1}$ in such a way that it evaluates to $0$. Adding $a_k - a_{k+1}$ to it, the desired expression for $n = k + 1$ is obtained.\n\n2. If $a_k \\neq a_{k+1}$, then consider integers $a_1, \\dots, a_{k-1}, |a_k - a_{k+1}|$. As $|a_k - a_{k+1}|$ and $a_k + a_{k+1}$ have the same parity, the sum of these $k$ numbers is even. Also note that $1 \\le |a_k - a_{k+1}| \\le k$. Thus these numbers satisfy the conditions of the problem, so it is possible to choose signs in the expression $a_1 \\pm a_2 \\pm \\dots \\pm a_{k-1} \\pm |a_k - a_{k+1}|$ in such a way that it evaluates to $0$. As either $|a_k - a_{k+1}| = a_k - a_{k+1}$ or $|a_k - a_{k+1}| = a_{k+1} - a_k$, this also leads to a corresponding expression for numbers $a_1, a_2, \\dots, a_k, a_{k+1}$.\nProve by induction on $i$ that, for each $i$ and $s$ such that $1 \\le i \\le n$ and $1 \\le s \\le a_1 + \\dots + a_i$, it is possible to choose some of the numbers $a_1, \\dots, a_i$ that sum up to $s$.\n\nIf $i = 1$, then this claim holds since $a_1 = 1$.\n\nAssume that the claim holds for $i = k - 1$ and consider the case $i = k$. Let $S = a_1 + \\dots + a_k$ and $S' = a_1 + \\dots + a_{k-1}$. If $1 \\le s \\le S'$, then the desired statement holds by the induction hypothesis. If $S' < s \\le S$, then $0 \\le s - a_k \\le S'$ (the first inequality holds because $s - a_k \\ge s - S' - 1 \\ge 0$, implied by $a_k \\le k$ and $S' \\ge k - 1$; the second inequality follows from $S = S' + a_k$). Therefore, to get the sum $s$, we can choose the number $a_k$, and if $s - a_k > 0$, then add to it those numbers among $a_1, \\dots, a_{k-1}$ whose sum is $s - a_k$, using the induction hypothesis.\n\nLet now $a_1 + a_2 + \\dots + a_n = 2T$. Choose the numbers among $a_1, a_2, \\dots, a_n$ that sum up to $T$. This divides all the numbers into two groups with equal sum. It remains to write minuses in front of every term of the group that does not contain $a_1$.\nStart choosing signs from right to left. Denote $S_1 = a_n$ and define $S_{k+1}$, $k = 1, \\dots, n-1$, as follows: if $S_k \\ge 0$, then $S_{k+1} = S_k - a_{n-k}$, otherwise $S_{k+1} = S_k + a_{n-k}$.\n\nWe show that then always $|S_k| \\le n-k+1$. This holds if $k=1$. Assume therefore that it holds for $k=m$ and prove it for $k=m+1$. If $S_m \\ge 0$, then $S_{m+1} = S_m - a_{n-m} \\le (n-m+1)-1 = n-m$ and $S_{m+1} = S_m - a_{n-m} \\ge 0 - (n-m)$, hence $|S_{m+1}| \\le n-m$. If $S_m < 0$, then $S_{m+1} = S_m + a_{n-m} < 0 + n-m$ and $S_{m+1} = S_m + a_{n-m} \\ge -(n-m+1)+1 = -(n-m)$, hence $|S_{m+1}| \\le n-m$ again.\n\nNow $|S_n| \\le 1$ since $|S_k| \\le n-k+1$ for every $k=1, \\dots, n$. Thus $S_n = 0$ as the sum of all terms is even. If in this formal sum, the term $a_1$ has minus sign, turn all signs to the opposite one.",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 29619,
    "subject": "Mathematics (Multi-modal)",
    "question": "The diagonals of trapezoid $ABCD$ with bases $AB$ and $CD$ meet at $P$. Prove the inequality $S_{PAB} + S_{PCD} > S_{PBC} + S_{PDA}$, where $S_{XYZ}$ denotes the area of triangle $XYZ$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $a = |AB|$ and $b = |CD|$ and let $h_a$ and $h_b$ be the altitudes of triangles $PAB$ and $PCD$ drawn from $P$ (see Fig. 7). Denote $S_1 = S_{PAB} + S_{PCD}$ and $S_2 = S_{PBC} + S_{PDA}$. Then $S_1 = \\frac{1}{2}(a h_a + b h_b)$ and $S_1 + S_2 = \\frac{1}{2}(a + b)(h_a + h_b)$, whence $S_2 = \\frac{1}{2}(a h_b + b h_a)$. Since triangles $PAB$ and $PCD$ are similar, $a > b$ implies $h_a > h_b$ and also $a < b$ implies $h_a < h_b$ ($a \\neq b$ because $a$ and $b$ are the lengths of the bases of the trapezoid). Hence\n$$\nS_1 - S_2 = \\frac{1}{2}(a h_a + b h_b - a h_b - b h_a) = \\frac{1}{2}(a - b)(h_a - h_b) > 0,\n$$\ni.e., $S_1 > S_2$.\nLet $M$ and $N$ be the intersection points of the arms $BC$ and $DA$ of the trapezoid with the line being parallel to the bases of the trapezoid and passing through point $P$. Let $l$ be the length of $MN$, let $d$ be the length of the midline of the trapezoid, and let $h$ and $S$ be the height and the area of the trapezoid, respectively. Let $S' = S_{PBC} + S_{PDA}$. Then $S = d h$ and $S' = \\frac{1}{2} l h$ whence it suffices to show that $l < d$.\n\n![](attached_image_1.png)\nFig. 7\n\nW.l.o.g., assume $|AB| < |CD|$. Comparing the heights of similar triangles $PAB$ and $PCD$ shows that $MN$ is closer to base $AB$ than to base $CD$. Thus $MN$ is situated between the midline and the shorter base $AB$. Consequently, $MN$ is shorter than the midline.\nLet $a = |AB|$ and $b = |CD|$. Let $h$ and $S$ be the height and the area of the trapezoid, respectively, and let $h_a$ and $h_b$ be the heights corresponding to vertex $P$ of triangles $PAB$ and $PCD$, respectively. Similar triangles $PAB$ and $PCD$ imply $h_a : h_b = a : b$. As $h_a + h_b = h$ we get\n$$\nh_a = \\frac{a}{a + b} \\cdot h, \\quad h_b = \\frac{b}{a + b} \\cdot h.\n$$\nNow\n$$\nS_{PAB} + S_{PCD} = \\frac{1}{2}(a h_a + b h_b) = \\frac{1}{2} \\cdot \\frac{a^2 + b^2}{a + b} \\cdot h.\n$$\nIt suffices to show that $S_{PAB} + S_{PCD} > \\frac{S}{2}$ or, equivalently,\n$$\n\\frac{1}{2} \\cdot \\frac{a^2 + b^2}{a + b} \\cdot h > \\frac{1}{2} \\cdot \\frac{a + b}{2} \\cdot h,\n$$\nor, equivalently, $2(a^2 + b^2) > (a + b)^2$. But the last inequality is equivalent to $(a - b)^2 > 0$ ($a \\neq b$ since $a$ and $b$ are the lengths of the bases of the trapezoid).",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 29620,
    "subject": "Mathematics (Multi-modal)",
    "question": "Call *pure* any positive integer $n$ that does not occur in any integer sequence $c_0, c_1, c_2, \\dots$, where $0 < c_0 < n$ and\n$$\nc_i = \\begin{cases} \\frac{1}{2}c_{i-1} & \\text{if } c_{i-1} \\text{ is even,} \\\\ 3c_{i-1} - 1 & \\text{if } c_{i-1} \\text{ is odd,} \\end{cases}\n$$\nfor every $i \\ge 1$. (For instance, $10$ is not pure since it occurs in the sequence $5, 14, 7, 20, 10, \\ldots$ ...)\n\na) Is every positive multiple of $3$ pure?\n\nb) Prove that if an integer $n > 1$ is pure but not divisible by $3$, then $n + 1$ is divisible by $6$.\n\n(Seniors.)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "a) Note that $3c_{i-1} - 1$ is never divisible by $3$ and if $\\frac{1}{2}c_{i-1}$ is divisible by $3$, then also $c_{i-1}$ is divisible by $3$. Thus, if some term $c_k = n$ is divisible by $3$, then, up to it, only dividing by $2$ is used to build the terms (i.e., $c_i = \\frac{1}{2}c_{i-1}$ for every $i$ such that $1 \\le i \\le k$) and, consequently, $c_0 > c_1 > \\dots > c_k = n$. But this contradicts the condition $c_0 < n$. Hence every positive multiple of $3$ is pure.\n\nb) If $n$ is not divisible by $3$, then $n = 3k + 1$ or $n = 6k + 2$ or $n = 6k + 5$. If $n = 3k + 1$, then taking $c_0 = 2k + 1$ gives $c_1 = 6k + 2$ and $c_2 = 3k + 1 = n$. Thereby $k > 0$ since $n > 1$, therefore $c_0 < n$. Hence none of such numbers $n$ is pure. If $n = 6k + 2$, then taking $c_0 = 2k + 1$ gives $c_1 = 6k + 2 = n$, whereby $c_0 < n$. Hence also none of such numbers $n$ is pure. Hence, among the positive integers $n > 1$ not divisible by $3$, only those of the form $n = 6k + 5$ can be pure.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 29621,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $a$ and $b$ be the lengths of the legs of a given right triangle. Prove that angle $\\varphi$, where $0 < \\varphi < 90^\\circ$, is an acute angle of this triangle if and only if $(a \\cos \\varphi + b \\sin \\varphi)(a \\sin \\varphi + b \\cos \\varphi) = 2ab$. (Seniors.)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "The equality given in the problem is equivalent to\n$$\n(a^2 + b^2) \\sin \\varphi \\cos \\varphi + ab(\\sin^2 \\varphi + \\cos^2 \\varphi) = 2ab\n$$\nand hence also to\n$$\n(a^2 + b^2) \\sin \\varphi \\cos \\varphi = ab. \\qquad (1)\n$$\n\nLet $\\alpha$ and $\\beta$ be the angles opposite to legs with length $a$ and $b$, respectively. Then $\\sin \\alpha = a/\\sqrt{a^2 + b^2}$, $\\sin \\beta = \\cos \\alpha = b/\\sqrt{a^2 + b^2}$, implying\n$$\n(a^2 + b^2) \\sin \\alpha \\cos \\alpha = ab.\n$$\n\nComparing this to (1) shows the equivalence of the equality of the problem and the equality $\\sin \\varphi \\cos \\varphi = \\sin \\alpha \\cos \\alpha$, i.e., equality $\\sin 2\\varphi = \\sin 2\\alpha$. As $0 < \\alpha, \\beta < 90^\\circ$, this implies $2\\varphi = 2\\alpha$ or $2\\varphi = 180^\\circ - 2\\alpha$, whence $\\varphi = \\alpha$ or $\\varphi = 90^\\circ - \\alpha = \\beta$. Hence, $\\varphi$ satisfies the equality if and only if it equals one of the acute angles of the right triangle.\nLet $ABC$ be the given triangle with right angle at vertex $C$. Let $c$ be the length of its hypothenuse and $h$ be the height corresponding to the hypothenuse. Let $C'$ be a point on the circumcircle of $ABC$ such that one acute angle of triangle $ABC'$ is $\\varphi$ (Fig. 8). Let $a'$ and $b'$ be the lengths of the legs of triangle $ABC'$ and $h'$ be the height of the triangle $ABC'$ corresponding to its hypothenuse. Then $(a^2 + b^2) \\sin \\varphi \\cos \\varphi = c^2 \\sin \\varphi \\cos \\varphi = (c \\sin \\varphi)(c \\cos \\varphi) = a'b' = ch'$. Since the equality in the problem is equivalent to the equality (1) from the Solution 1, it is also equivalent to $ch' = ab$. But $ab = ch$, hence it is also equivalent to $h = h'$. This condition holds if and only if $ABC \\sim ABC'$ or $ABC \\sim BAC'$, i.e., $\\varphi$ equals one of the acute angles of triangle $ABC$.\n\nFig. 8\n![](attached_image_1.png)",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 29622,
    "subject": "Mathematics (Multi-modal)",
    "question": "Given a convex quadrangle $ABCD$ with $|AD| = |BD| = |CD|$ and $\\angle ADB = \\angle DCA$, $\\angle CBD = \\angle BAC$, find the sizes of the angles of the quadrangle. (Juniors.)",
    "options": [],
    "answer": "∠DAB = 75°, ∠ABC = 120°, ∠BCD = 45°, ∠CDA = 120°",
    "solution": "Denote $\\angle ADB = \\angle DCA = \\alpha$ and $\\angle CBD = \\angle BAC = \\beta$ (Fig. 1).\n\nIn triangle $DAC$ we have $|DA| = |DC|$ and therefore $\\angle DAC = \\angle DCA = \\alpha$; analogously in triangles $DAB$ and $DBC$, we have $\\angle DBA = \\angle DAB = \\alpha + \\beta$ and $\\angle DCB = \\angle DBC = \\beta$, respectively. So $\\angle BCA = \\beta - \\alpha$.\n\nFrom triangle $ABC$ now $\\beta + \\alpha + \\beta + \\beta + \\beta - \\alpha = 180^\\circ$ or, equivalently, $4\\beta = 180^\\circ$, giving $\\beta = 45^\\circ$.\n\nFrom triangle $ADB$ we get $\\alpha + \\beta + \\alpha + \\beta + \\alpha = 180^\\circ$ or, equivalently, $3\\alpha = 180^\\circ - 2\\beta = 90^\\circ$ and $\\alpha = 30^\\circ$.\n\nTherefore, the sizes of the angles of quadrangle $ABCD$ are $\\angle DAB = \\alpha + \\beta = 75^\\circ$, $\\angle ABC = \\alpha + 2\\beta = 120^\\circ$, $\\angle BCD = \\beta = 45^\\circ$, and $\\angle CDA = 360^\\circ - 75^\\circ - 120^\\circ - 45^\\circ = 120^\\circ$.\n\n![](attached_image_1.png)\nFig. 1\nWe use the same notation as in the Solution 1. Triangle $BCD$ is isosceles, hence $\\angle DCB = \\angle DBC = \\beta$. As $D$ is the circumcenter of $ABC$, we have $\\angle BDC = 2\\angle BAC = 2\\beta$. The sizes of the angles of triangle $BCD$ are therefore $\\beta, \\beta$, and $2\\beta$; thus $\\beta + \\beta + 2\\beta = 180^\\circ$, whence $\\beta = 45^\\circ$.\n\nAs $\\angle BCA = \\frac{\\angle BDA}{2}$, we have $\\angle BCD = \\frac{\\alpha}{2} + \\alpha = \\beta$, whence $\\alpha = \\frac{2}{3}\\beta = 30^\\circ$.\n\nConsequently, the sizes of the angles of quadrangle $ABCD$ are $\\angle DAB = \\angle ABD = \\frac{180^\\circ - \\alpha}{2} = 75^\\circ$, $\\angle ABC = \\angle ABD + \\angle CBD = 75^\\circ + 45^\\circ = 120^\\circ$, $\\angle BCD = \\beta = 45^\\circ$, and $\\angle CDA = \\angle CDB + \\angle BDA = 90^\\circ + 30^\\circ = 120^\\circ$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 29623,
    "subject": "Mathematics (Multi-modal)",
    "question": "In the buffet of the kitchen, there are three candy boxes, each containing the same number of candies. Every time when Juku goes into the kitchen, he takes either three candies from one box or one candy from every box. Prove that irrespectively of how Juku takes the candies, he always retains the possibility to completely clean out all candy boxes. (Juniors.)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "The difference of the numbers of candies in any two boxes can only be a multiple of $3$ because it is $0$ in the beginning and, with every move, it changes by either $0$ or $3$. Hence, starting from an arbitrary intermediate state, Juku can clean out the boxes as follows: he takes one candy from each box as many times as possible, after which one box is empty and the number of candies in each of the other two is divisible by $3$, and then empties the other boxes by taking three candies from one box every time.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 29624,
    "subject": "Mathematics (Multi-modal)",
    "question": "Four musketeers together bought a plot of rectangular shape and paid for it equally. They divided the plot by two cuts into four pieces of rectangular shape, from which every musketeer got one. It turned out that one musketeer obtained as much land as the other three in total. Prove that the price per acre of one musketeer's piece turned out as large as the sum of the prices per acre of the other three musketeers' pieces. (Juniors.)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $a$ and $b$ be the side lengths of the plot. Assume that the cuts divided the side of length $a$ to parts of length $x$ and $a-x$ where $x$ being the greater part, and the side of length $b$ to parts of length $y$ and $b-y$ where $y$ being the greater part. Then the area of the largest piece was $xy$. The condition that this area equals the sum of the areas of the other three pieces can be written as follows:\n$$\nxy = (a-x)y + x(b-y) + (a-x)(b-y).\n$$\n\nDividing both sides by $x(a-x)y(b-y)$, one obtains\n$$\n\\frac{1}{(a-x)(b-y)} = \\frac{1}{x(b-y)} + \\frac{1}{(a-x)y} + \\frac{1}{xy}.\n$$\nIf the price that every musketeer paid for the plot was 1, then the l.h.s. of the last equality is precisely the price per area unit of the piece with area $(a-x) \\cdot (b-y)$. Analogously, the r.h.s. equals the sum of the prices per area unit of the other three pieces. Hence multiplying the sides of this equality by the number of area units per acre, the claim of the problem follows.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 29625,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $a$ be a fixed real number. Find all real numbers $b$ such that, for every real number $x$, at least one of the numbers $x^2 + ax + b$ and $x^2 - ax + b$ is non-negative.",
    "options": [],
    "answer": "b ≥ 0",
    "solution": "Note that $x^2 + ax + b$ and $x^2 - ax + b$ sum up to $2x^2 + 2b$. If $b \\ge 0$, then it is non-negative for arbitrary real number $x$, implying that at least one of the numbers added was non-negative. If $b < 0$, then taking $x = 0$ turns both summands negative.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 29626,
    "subject": "Mathematics (Multi-modal)",
    "question": "Call a positive integer $n$ prime-prone if there exist at least three prime numbers from which we can get $n$ by removing the last digit. Prove that every two prime-prone positive integers differ from each other by at least $3$. (Juniors.)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "As the prime numbers under consideration have at least two digits, the last digit can be only $1$, $3$, $7$, or $9$. Thus $n$ is prime-prone if and only if, among numbers $10n + 1$, $10n + 3$, $10n + 7$, and $10n + 9$, at least three are primes.\n\nIf $n = 3k$, then $10n + 3 = 30k + 3$ and $10n + 9 = 30k + 9$ are divisible by $3$ and hence composite. If $n = 3k + 2$, then $10n + 1 = 30k + 21$ and $10n + 7 = 30k + 27$ are divisible by $3$ and hence composite again. Consequently, all prime-prone integers are congruent to $1$, and hence to each other, modulo $3$. Thus they differ by a multiple of $3$, i.e., by at least $3$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 29627,
    "subject": "Mathematics (Multi-modal)",
    "question": "Does there exist a prime number $p$ such that both $p^3 + 2008$ and $p^3 + 2010$ are primes as well?",
    "options": [],
    "answer": "No",
    "solution": "Let $p$ be any prime number. If $p$ is not divisible by $7$, then $p^3$ is congruent to either $1$ or $-1$ modulo $7$. Since $2008 \\equiv -1 \\pmod{7}$ and $2010 \\equiv 1 \\pmod{7}$, either of the numbers $p^3 + 2008$ and $p^3 + 2010$ is divisible by $7$ and hence composite. If $p$ is divisible by $7$, then $p = 7$ and $p^3 + 2010 = 7^3 + 2010 = 2353 = 13 \\cdot 181$ is composite, too.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 29628,
    "subject": "Mathematics (Multi-modal)",
    "question": "In a regular $n$-gon, either $0$ or $1$ is written at each vertex. Using non-intersecting diagonals, Juku divides this polygon into triangles. Then he writes into each triangle the sum of the numbers at its vertices. Prove that Juku can choose the diagonals in such a way that the maximal and minimal number written into the triangles differ by at most $1$. (Seniors.)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "If all numbers written at the vertices of the polygon are equal, then the claim holds trivially. Hence assume that there are both zeros and ones among the numbers at the vertices. We prove by induction that, for every convex polygon, the partition into triangles can be chosen in such a way that Juku writes either $1$ or $2$ to each triangle.\n\nIf $n = 3$, then this claim holds since the sum of the numbers at the vertices of a triangle can be neither $0$ nor $3$. If $n = 4$ (Fig. 2), then draw the diagonal that connects the vertices where $0$ and $1$ are written, respectively, or, if such a diagonal does not exist, then an arbitrary diagonal. In both cases, only sums $1$ and $2$ can arise. If $n \\ge 5$, then choose two consecutive vertices with different labels and a third vertex $P$ that is not neighbour to either of them (Fig. 3). Irrespective of whether the label of $P$ is $0$ or $1$, we can draw the diagonal from it to one of the two consecutive vertices chosen before so that the labels of its endpoints are different. Now the polygon is divided into two convex polygons with smaller number of vertices so that both $0$ and $1$ occur among their vertex labels. By the induction hypothesis, both polygons can be partitioned into triangles with sum of labels of vertices either $1$ or $2$.\n\n![](attached_image_1.png)\nFig. 2\n![](attached_image_2.png)\n![](attached_image_3.png)\nFig. 3",
    "topic": "Discrete Mathematics",
    "subtopic": "Combinatorics"
  },
  {
    "id": 29629,
    "subject": "Mathematics (Multi-modal)",
    "question": "Circle $c$ passes through vertices $A$ and $B$ of an isosceles triangle $ABC$, whereby line $AC$ is tangent to it. Prove that circle $c$ passes through the circumcenter or the incenter or the orthocenter of triangle $ABC$. (Seniors.)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Consider three cases $|AB| = |AC|$, $|BC| = |BA|$, and $|CA| = |CB|$.\n\n1. We show that if $|AB| = |AC|$ (Fig. 4), then circle $c$ passes through the circumcenter of $ABC$. Let $O$ be the point at the same side from $AB$ as $C$ that is the intersection of the perpendicular bisector of side $AB$ and circle $c$. Then $\\angle OAB = \\angle OBA$ and, by inscribed angles theorem,\n\n![](attached_image_1.png)\nFig. 4\n\n$\\angle OBA = \\angle OAC$. Hence $O$ lies on the bisector of angle $CAB$. Since $|AB| = |AC|$, this angle bisector is also the perpendicular bisector of side $BC$. Consequently, $O$ is the intersection point of the perpendicular bisectors of the sides of triangle $ABC$.\n\n2. For the case $|BC| = |BA|$ (Fig. 5), we show that circle $c$ passes through the orthocenter of triangle $ABC$. Let $E$ be the foot of the altitude of triangle $ABC$ drawn from $B$ and let $H$ be the second intersection point of this altitude with circle $c$ (in the special case with tangency and no intersection, take $H = B$). By the inscribed angles theorem, $\\angle EBA = \\angle EAH$. Thus $\\angle ACB + \\angle CAH = \\angle CAB + \\angle EBA = 90^\\circ$ whence $AH \\perp BC$. Consequently, $H$ is the orthocenter.\n\n![](attached_image_2.png)\nFig. 5\n\n3. Finally, we show that if $|CA| = |CB|$ (Fig. 6), then circle $c$ passes through the incenter of triangle $ABC$. Let $I$ be the intersection point of the bisector of angle $CAB$ with circle $c$. By the inscribed angles theorem, $\\angle CAI = \\angle IBA$. Hence $\\angle BAI = \\angle IBA$ whence $I$ lies on the perpendicular bisector of side $AB$. As $|CA| = |CB|$, this perpendicular bisector is also the bisector of angle $ACB$. Consequently, $I$ is the intersection point of the angle bisectors.\n\n![](attached_image_3.png)\nFig. 6",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 29630,
    "subject": "Mathematics (Multi-modal)",
    "question": "For arbitrary positive integers $a, b$, denote $a \\ominus b = \\frac{a-b}{\\text{gcd}(a,b)}$.\nLet $n$ be a positive integer. Prove that the following conditions are equivalent:\n(i) $\\text{gcd}(n, n \\ominus m) = 1$ for every positive integer $m < n$;\n(ii) $n = p^k$ where $p$ is a prime number and $k$ is a non-negative integer.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Note at first that $da \\ominus db = a \\ominus b$ for all positive integers $a, b$, and $d$. Indeed,\n$$\nda \\ominus db = \\frac{da - db}{\\text{gcd}(da, db)} = \\frac{d \\cdot (a-b)}{d \\cdot \\text{gcd}(a,b)} = \\frac{a-b}{\\text{gcd}(a,b)} = a \\ominus b.\n$$\nShow now that if $n$ is a prime power and $m < n$, then $n \\ominus m$ is relatively prime to $n$. Indeed, let $n = p^k$ where $p$ is a prime number, and let $m = p^i s$ where $\\text{gcd}(p, s) = 1$. Then $m < n$ implies $i < k$. Now\n$$\nn \\ominus m = p^{k-i} \\ominus s = \\frac{p^{k-i} - s}{\\text{gcd}(p^{k-i}, s)} = p^{k-i} - s\n$$\nbecause $\\text{gcd}(p^{k-i}, s) = 1$ by the choice of $s$. Also, for the same reason, $\\text{gcd}(p, p^{k-i} - s) = 1$, hence $\\text{gcd}(n, n \\ominus m) = \\text{gcd}(p^k, p^{k-i} - s) = 1$.\nIt remains to show that if $n$ is not a prime power, then there exists a positive integer $m$ such that $m < n$ and the integers $n \\ominus m$ and $n$ share a common prime factor. Since $n$ is not a prime power, it has at least two different prime factors. Let $p$ and $q$ be some prime factors of $n$, whereby $p < q$. Let $n = p^k t$ where $\\text{gcd}(p, t) = 1$. Take $m = n - p^{k+1}$. As $n$ is divisible by both $p^k$ and $q$ which are relatively prime, it is also divisible by their product $p^k q$. Consequently, $p^{k+1} < p^k q \\le n$, i.e., $0 < m < n$. Now\n$$\nn \\ominus m = n \\ominus (n - p^{k+1}) = t \\ominus (t - p) = \\frac{t - (t-p)}{\\text{gcd}(t, t-p)} = \\frac{p}{\\text{gcd}(t, p)} = p\n$$\nsince $\\text{gcd}(t, p) = 1$. We see that $n \\ominus m$ and $n$ have a common prime factor $p$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 29631,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $n$ be a positive integer. Find the largest integer $N$ for which there exists a set of $n$ weights such that it is possible to determine the mass of all bodies with masses of $1$, $2$, $\\ldots$, $N$ using a balance scale (i.e. to determine whether a body with unknown mass has a mass $1$, $2$, $\\ldots$, $N$, and which namely).",
    "options": [],
    "answer": "(3^n - 1)/2",
    "solution": "The possibility to determine mass $m$ means the possibility to place the weights on the two scalepans so that the difference of total masses on the two scalepans is exactly $m$.\n\nEvery weight can be placed on either of the two pans or on neither of the pans. For $n$ weights this makes $3^n$ different placements. Note that the placement where none of the weights is on the scales does not determine any mass. Also, for each placement there is a symmetric placement with all the weights on the two pans swapped, which determines the same mass. Therefore with $n$ weights it is possible to determine at most $\\frac{3^n-1}{2}$ different masses.\n\nWe show by induction that it is possible to determine all masses from $1$ to $\\frac{3^n-1}{2}$ using $n$ weights with masses $1$, $3$, $\\dots$, $3^{n-1}$.\n\nFor $n=1$ it is obvious. Assume that the claim holds for $n=k$. By the induction assumption we can determine all masses from $1$ to $\\frac{3^k-1}{2}$ by weights $1$, $3$, $\\dots$, $3^{k-1}$. Using the weight with mass $3^k$, we can determine the mass $3^k$, and using it together with the other weights also the masses $3^k+1$, $\\dots$, $3^k+\\frac{3^k-1}{2}$ and $3^k-1$, $\\dots$, $3^k-\\frac{3^k-1}{2}$.\n\nSince $3^k - \\frac{3^k-1}{2} = \\frac{3^k-1}{2} + 1$ and $3^k + \\frac{3^k-1}{2} = \\frac{3^{k+1}-1}{2}$, the claim is also true for $n=k+1$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 29632,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let the angles of a triangle be $\\alpha$, $\\beta$, and $\\gamma$, the perimeter $2p$ and the radius of the circumcircle $R$. Prove the inequality\n$$\ncot^2 \\alpha + cot^2 \\beta + cot^2 \\gamma \\ge 3 \\left( \\frac{9R^2}{p^2} - 1 \\right).\n$$\nWhen is the equality achieved?",
    "options": [],
    "answer": "Equality holds if and only if the triangle is equilateral.",
    "solution": "Let the opposite sides of the angles $\\alpha$, $\\beta$, and $\\gamma$ be correspondingly $a$, $b$, and $c$. Since $\\cot^2 \\alpha = 1/\\sin^2 \\alpha - 1$ and from the law of sines $1/\\sin \\alpha = 2R/a$, we have $\\cot^2 \\alpha = 4R^2/a^2 - 1$; similarly $\\cot^2 \\beta = 4R^2/b^2 - 1$ and $\\cot^2 \\gamma = 4R^2/c^2 - 1$. The inequality can therefore be written as\n$$\n4R^2 \\cdot \\left( \\frac{1}{a^2} + \\frac{1}{b^2} + \\frac{1}{c^2} \\right) - 3 \\ge 3 \\cdot \\left( \\frac{4 \\cdot 9R^2}{(a+b+c)^2} - 1 \\right),\n$$\nor\n$$\n\\frac{1}{a^2} + \\frac{1}{b^2} + \\frac{1}{c^2} \\ge \\frac{27}{(a+b+c)^2}.\n$$\nDividing both sides by 3 and taking the square root gives\n$$\n\\sqrt{\\frac{1}{3} \\cdot \\left( \\frac{1}{a^2} + \\frac{1}{b^2} + \\frac{1}{c^2} \\right)} \\ge \\frac{3}{a+b+c}.\n$$\nThe left side is the quadratic mean of $\\frac{1}{a}$, $\\frac{1}{b}$, $\\frac{1}{c}$ and the right side is the harmonic mean of the same numbers, hence the inequality holds.\nThe equality holds iff $a = b = c$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 29633,
    "subject": "Mathematics (Multi-modal)",
    "question": "In an acute triangle $ABC$ the angle $C$ is greater than the angle $A$. Let $AE$ be a diameter of the circumcircle of the triangle. Let the intersection point of the ray $AC$ and the tangent of the circumcircle through the vertex $B$ be $K$. The perpendicular to $AE$ through $K$ intersects the circumcircle of the triangle $BCK$ for the second time at point $D$. Prove that $CE$ bisects the angle $BCD$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Since $AE$ is a diameter of the circumcircle of the triangle $ABC$, $\\angle ACE = \\angle ECK = 90^\\circ$. So it suffices to show that $\\angle ACB = \\angle DCK$ (Fig. 19).\n\nLet $L$ be the point of intersection of lines $AE$ and $DK$. Then $\\angle BAC = \\angle CBK = \\angle CDK$ by the inscribed angles theorem. Also $\\angle ABC = \\angle AEC = \\angle CKD$ where the latter equality follows from the similarity of the right triangles $ACE$ and $ALK$. Hence the two triangles $ABC$ and $DKC$ are similar, and therefore $\\angle ACB = \\angle DCK$.\n\nFig. 19\n![](attached_image_1.png)",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 29634,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $P(x, y)$ be a non-constant homogeneous polynomial with real coefficients such that $P(\\sin t, \\cos t) = 1$ for every real number $t$. Prove that there exists a positive integer $k$ such that $P(x, y) = (x^2 + y^2)^k$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $n$ be the degree of the polynomial $P$, i.e.,\n$$\nP(x, y) = a_n x^n + a_{n-1} x^{n-1} y + \\dots + a_1 x y^{n-1} + a_0 y^n,\n$$\nwhere $n > 0$. Note that $n$ must be even because otherwise the condition $P(\\sin t, \\cos t) = 1$ for $t = 0$ would imply $a_0 = 1$ while the same condition for $t = \\pi$ would imply $a_0 = -1$.\nSince $P$ has no constant term, $P(0, 0) = 0$. Now assume that $x \\neq 0$ or $y \\neq 0$ and take $c = \\sqrt{x^2 + y^2}$. Since\n$$\n\\left(\\frac{x}{\\sqrt{x^2 + y^2}}\\right)^2 + \\left(\\frac{y}{\\sqrt{x^2 + y^2}}\\right)^2 = 1,\n$$\nthere exists some real number $t$ such that $\\sin t = x/\\sqrt{x^2 + y^2}$ and $\\cos t = y/\\sqrt{x^2 + y^2}$ and therefore $P(\\sin t, \\cos t) = 1$. By homogeneity, $P(x, y) = c^n \\cdot P(x/c, y/c)$, hence\n$$\nP(x, y) = \\left(\\sqrt{x^2 + y^2}\\right)^n \\cdot P\\left(\\frac{x}{\\sqrt{x^2 + y^2}}, \\frac{y}{\\sqrt{x^2 + y^2}}\\right) = \\left(\\sqrt{x^2 + y^2}\\right)^n.\n$$\n\nfor all $x, y$. The case $n = 2k$ implies $P(x, y) = (x^2 + y^2)^k$ which satisfies also the condition $P(0, 0) = 0$.\nLike in Solution 1, express the polynomial as a sum of $n + 1$ monomials with coefficients $a_0, \\dots, a_n$ and show that $n = 2k$.\nWe prove the claim of the problem by induction on $k$. In case $k = 0$ (omitting the extra assumption that $P$ is non-constant) the claim holds obviously. Assume now that $k > 0$ and the claim holds for $k - 1$. Substituting $t = 0$ and $t = \\frac{\\pi}{2}$ into $P(\\sin t, \\cos t) = 1$ gives $a_0 = 1$ and $a_n = 1$, respectively. Hence the polynomial $P(x, y) - (x^2 + y^2)^k$ does not have terms with $x^n$ and $y^n$. Let $Q(x, y)$ be such that $P(x, y) - (x^2 + y^2)^k \\equiv x y \\cdot Q(x, y)$. Then $\\sin t \\cos t \\cdot Q(\\sin t, \\cos t) = 0$ for every real number $t$, hence $Q(\\sin t, \\cos t) = 0$ for every $t$ such that $\\sin 2t \\neq 0$. By continuity of $Q(\\sin t, \\cos t)$ as a function of $t$, it follows that $Q(\\sin t, \\cos t) \\equiv 0$. Now define $R(x, y) = Q(x, y) + (x^2 + y^2)^{k-1}$. As both $Q$ and $R$ are homogeneous polynomials of degree $2(k-1)$, the assumptions of the problem hold for polynomial $R$. By the induction hypothesis, $R(x, y) \\equiv (x^2 + y^2)^{k-1}$. Hence $Q(x, y) \\equiv 0$ and $P(x, y) \\equiv (x^2 + y^2)^k$.",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 29635,
    "subject": "Mathematics (Multi-modal)",
    "question": "Every unit square of a $n \\times n$ board is colored either red or blue so that among all $2 \\times 2$ squares on this board all possible colorings of $2 \\times 2$ squares with these two colors are represented (colorings obtained from each other by rotation and reflection are considered different).\n\na) Find the least possible value of $n$.\n\nb) For the least possible value of $n$ find the least possible number of red unit squares.",
    "options": [],
    "answer": "n = 5; minimum number of red unit squares = 10",
    "solution": "a) Since there are $2^4 = 16 = 4^2$ possibilities to color a $2 \\times 2$ square in two colors and a $n \\times n$ square contains $(n-1)^2$ such subsquares, we must have $n-1 \\ge 4$, or $n \\ge 5$. For $n = 5$ a suitable coloring is given in Fig. 20.\n\n![](attached_image_1.png)\nFig. 20\n\nb) Fig. 20 presents a coloring with 10 red squares. We will show that this is the least possible.\n\nNote that in the $5 \\times 5$ square there are 4 unit squares in the corners, 12 squares on the sides (not in the corners), and 9 inner squares. Each corner square is contained in exactly one, side square in two and inner square in four $2 \\times 2$ squares. All 16 colourings of $2 \\times 2$ squares contain a total of 64 unit squares of which 32 are red by symmetry. Therefore, if the $5 \\times 5$ square contains $k$ red squares, among them $a$ corner squares, $b$ side squares and $c$ inner squares, then $a + b + c = k$ and $a + 2b + 4c = 32$. The equation $a + 2b + 4c = 32$ implies $c \\le 8$. If $c = 8$, then $a = b = 0$. If $c = 7$, then the only possibility to have $k < 10$ is $b = 2$ and $a = 0$. If $c \\le 6$, then always $k = a + b + c \\ge 10$.\n\nThus it is enough to show that there are no colorings with required properties with $a = 0$ and $b \\le 2$. Indeed, in this case the $5 \\times 5$ square has at least two sides not containing any red squares. Without loss of generality, let one of them be the upper side. We saw in part a) that for $n = 5$ each coloring of $2 \\times 2$ squares must occur exactly once. Since among all 16 colorings of $2 \\times 2$ squares there are 4 such where both upper unit squares are blue, and two upper rows of the $5 \\times 5$ square contain exactly $4 \\times 2$ squares, all four such colorings must be located in the two upper rows, among these the completely blue coloring. Since the same is true for the other side which does not contain any red squares, the two sides must meet and a completely blue $2 \\times 2$ square must be in the corner where the two sides meet. Without loss of generality, let it be the left side. Then the two squares on Fig. 21 must be red, because otherwise there would be more than one completely blue $2 \\times 2$ square. But now there are two $2 \\times 2$ squares with red square in the lower right corner and the rest of them blue. Therefore there is no coloring satisfying the conditions with $a = 0$ and $b \\le 2$ and the least number of red squares is $k = 10$.\n\n![](attached_image_2.png)\nFig. 21",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 29636,
    "subject": "Mathematics (Multi-modal)",
    "question": "Juku discovered that of the things in his satchel, $60\\%$ were ugly and $76\\%$ were useless. He scrapped all things that were both ugly and useless, and added things that were both beautiful and useful. After this, of the things in Juku's satchel, $25\\%$ are ugly and $45\\%$ are useless. How many percent of the things in Juku's satchel were both beautiful and useful initially?\n\n*Answer:* $4\\%$.",
    "options": [],
    "answer": "4%",
    "solution": "Observe that the amount of things that are beautiful but useless and things that are useful but ugly remained unchanged. The difference between the percentages of these things was $76\\% - 60\\% = 16\\%$ before displacement but is $45\\% - 25\\% = 20\\%$ after that. Hence the overall number of things in the satchel decreased $20 : 16 = 1.25$ times. Things that are beautiful but useless form $45\\%$ of all things after the displacement, hence they formed $45\\% : 1.25 = 36\\%$ before it. As there were $100\\% - 60\\% = 40\\%$ of beautiful things in total, the things that were both beautiful and useful constituted $40\\% - 36\\% = 4\\%$ of the content of the satchel.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 29637,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $CM$ be the median of a triangle $ABC$. Prove that the product of the circumradius of $ACM$ and the altitude drawn from $M$ in $ACM$ equals the product of the circumradius of $BCM$ and the altitude drawn from $M$ in $BCM$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $\\angle CAB = \\alpha$ and $\\angle CBA = \\beta$ (see Fig. 13), and let $r$ and $s$ be the circumradii of the triangles $ACM$ and $BCM$, respectively. By the sine law in the triangle $ACM$ we obtain $\\frac{|CM|}{\\sin \\alpha} = 2r$, reducing to $r = \\frac{|CM|}{2\\sin \\alpha}$. Analogously, $s = \\frac{|CM|}{2\\sin \\beta}$. Let $U$ and $V$ be the feet of altitudes drawn from the point $M$ in triangles $ACM$ and $BCM$, respectively.\n![](attached_image_1.png)\nFig. 13\nThen $|MU| = |AM| \\sin \\alpha = \\frac{|AB|}{2} \\sin \\alpha$. Analogously, $|MV| = \\frac{|AB|}{2} \\sin \\beta$. Thus $r \\cdot |MU| = \\frac{|CM| \\cdot |AB|}{4} = s \\cdot |MV|$.\n\n![](attached_image_1.png)\nFig. 13\nLet $x$ and $y$ be the altitudes drawn from $M$ in the triangles $ACM$ and $BCM$, respectively. Let $r$ and $s$ be the circumradii of these triangles, respectively. The areas of triangles $ACM$ and $BCM$ are equal because of $|AM| = |BM|$ and the common altitude drawn from $C$. Therefore $|AC| \\cdot x = |BC| \\cdot y$. Denote $\\angle AMC = \\gamma$. The sine law gives $|AC| = 2r \\sin \\gamma$ and $|BC| = 2s \\sin(180^\\circ - \\gamma) = 2s \\sin \\gamma$. Hence $2r \\sin \\gamma \\cdot x = 2s \\sin \\gamma \\cdot y$. As $\\gamma \\ne 0$, this implies $rx = sy$.\nThe formulas $S = \\frac{abc}{4R}$ and $S = \\frac{ah}{2}$, where $a, b, c$ are the side lengths, $R$ is the circumradius and $h$ is the altitude corresponding to $a$, together give $Rh = \\frac{bc}{2}$. The product of the circumradius of $ACM$ and the altitude drawn from $M$ is thus $\\frac{|AM| \\cdot |CM|}{2}$. Analogously, $\\frac{|BM| \\cdot |CM|}{2}$ for triangle $BCM$. These two products are equal, since $|AM| = |BM|$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 29638,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find the last digit of the number $1^1 + 2^2 + 3^3 + \\dots + 2011^{2011}$.",
    "options": [],
    "answer": "8",
    "solution": "Consider the sum modulo $2$ and modulo $5$. As powers of odd numbers are odd and powers of even numbers are even, the number of odd summands equals the number of odd elements in set $\\{1, \\dots, 2011\\}$. As there are an even number of odd elements in this set, the sum given in the problem is even.\n\nConcerning modulo $5$, note that $0$ to every power is congruent to $0$ and $a^4$ is congruent to $1$, whenever $1 \\le a \\le 4$. Thus for all $a$ such that $1 \\le a \\le 20$, $(a+20)^{a+20} = a^{a+20} = a^a \\cdot a^{20} = a^a \\cdot (a^4)^5 = a^a \\cdot 1 = a^a \\pmod{5}$. Hence the remainders modulo $5$ repeat periodically with the period $20$. As there are $100$ full periods and $100$ is divisible by $5$, the sum of the last $2000$ summands is congruent to $0 \\pmod{5}$.\n\nIt remains to compute the remainders of the first $11$ summands:\n$1^1 = 1$, $2^2 = 4$, $3^3 = 27 \\equiv 2$, $4^4 \\equiv 4^0 = 1$, $5^5 \\equiv 0^1 = 0$, $6^6 \\equiv 1^2 = 1$, $7^7 \\equiv 2^3 \\equiv 3$, $8^8 \\equiv 3^0 = 1$, $9^9 \\equiv 4^1 = 4$, $10^{10} \\equiv 0^2 = 0$, $11^{11} \\equiv 1^3 = 1$.\n\nHence the overall sum is congruent to $3$ modulo $5$. Consequently, the last digit of this sum is $8$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 29639,
    "subject": "Mathematics (Multi-modal)",
    "question": "Does there exist a positive real number $C$ such that the inequality\n$$\nx_1x_2 + x_1x_3 + x_1x_4 + x_2x_3 + x_2x_4 + x_3x_4 \\le C(x_1x_2 + x_2x_3 + x_3x_4 + x_4x_1)\n$$\nholds for arbitrary positive real numbers $x_1, x_2, x_3, x_4$?",
    "options": [],
    "answer": "Detailed solution",
    "solution": "For simplicity, denote $A = x_1x_2 + x_1x_3 + x_1x_4 + x_2x_3 + x_2x_4 + x_3x_4$ and $B = x_1x_2 + x_2x_3 + x_3x_4 + x_4x_1$.\n\nChoose $x_1 = x_3 = u$ and $x_2 = x_4 = 1$. Then $A \\ge x_1x_3 = u^2$ and $B = 4u$. Hence $\\frac{A}{B} \\ge \\frac{u}{4}$. As $u$ can be arbitrarily large, no constant $C$ such that $A \\le CB$ exists.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 29640,
    "subject": "Mathematics (Multi-modal)",
    "question": "In a rectangle $ABCD$ we have $|AB| = a$ and $|BC| = b$, where $a \\ge b$. Let $E$ be a point in the interior of side $AB$ such that there is exactly one possibility to choose points $F, G, H$ on the sides $BC, CD, DA$, respectively, in such a way that $EFGH$ is a rectangle, too. Find the ratio of the areas of rectangles $EFGH$ and $ABCD$.",
    "options": [],
    "answer": "1/2",
    "solution": "The rectangles $ABCD$ and $EFGH$ have a common center $O$ (see Fig. 15).\n\nAs rectangles are cyclic quadrangles, the point $F$ lies on the circle with center $O$ and radius $|OE|$. This circle intersects the side $BC$ at two points symmetric with respect to the midpoint of the side. To have exactly one point common to the circle and the side, the side must be tangent to the circle and $F$ must be the midpoint of $BC$. Analogously, $H$ must be the midpoint of $DA$.\n\nIn triangle $EFH$, the side $HF$ has length $a$ and the corresponding altitude is $\\frac{b}{2}$, giving $\\frac{1}{2} \\cdot a \\cdot \\frac{b}{2} = \\frac{ab}{4}$ as the area of the triangle. The triangle $GFH$ has the same area. Hence the area of rectangle $EFGH$ is $2 \\cdot \\frac{ab}{4} = \\frac{ab}{2}$ that makes up a half of the area $ab$ of the rectangle $ABCD$.\n\n![](attached_image_1.png)\nFig. 15",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 29641,
    "subject": "Mathematics (Multi-modal)",
    "question": "Ants has three pencils, each of a different color. In how many ways can he paint the faces of a regular octahedron in such a way that faces with a common edge always have different colors? Colorings that can be obtained from each other via rotations of the octahedron are considered the same.",
    "options": [],
    "answer": "15",
    "solution": "i) Case $4, 4, 0$. There are $3$ possibilities to choose two colors from the three. After that, there is only one possibility to paint the octahedron. Thus there are $3$ possibilities to paint.\n\nii) Case $4, 3, 1$. Ordering the $3$ colors can be done in $6$ ways. After that, there is only one possibility to paint the octahedron, since the color used $4$ times must occur twice among the faces adjacent to one vertex and twice among the faces adjacent to the opposite vertex. The remaining two colors can be deployed in principle in only one way. Thus there are $6$ possibilities to paint.\n\niii) Case $4, 2, 2$. Choosing the color that is used $4$ times can be done in $3$ different ways. After that, the octahedron can be painted in only $1$ way, since after $4$ faces have been painted with the same color, faces with either of the other colors must meet at the same vertex. Thus there are $3$ possibilities to paint.\n\niv) Case $3, 3, 2$. Choosing the color that is used only twice can be done in $3$ ways. If the faces painted with this color met at a common vertex $V$, the faces adjacent to the opposite vertex would be painted alternately with the other two colors. But then the remaining two faces adjacent to $V$ would have to be painted with the same color, that contradicts the case assumption. Hence the color that occurs twice is used on a pair of opposite sides. The other colors occur alternately on the surface formed by the remaining six faces. Thus there are $3$ possibilities to paint.\n\nConsequently, the number of all colorings is $3 + 6 + 3 + 3 = 15$.",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 29642,
    "subject": "Mathematics (Multi-modal)",
    "question": "Inside a regular $2n$-gon, an arbitrary point is chosen and connected to every vertex of the $2n$-gon. The triangles obtained are colored alternately black and white so that triangles with a common side are of different color. Prove that the sum of the areas of all white triangles equals the sum of the areas of all black triangles.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "If $n = 2$, i.e., the $2n$-gon is a square, then the claim holds because the base sides of white triangles are the opposite sides of the square and the altitudes lie on the same line, so the total area of white triangles is a half of the area of the square.\n\nAssume in the following that $n > 2$. Consider the regular $n$-gon whose sides are obtained by prolonging all sides of the $2n$-gon that belong to white triangles (see Fig. 16).\n\n![](attached_image_1.png)\nFig. 16\n\nJoin the point chosen inside the initial $2n$-gon with all vertices of the $n$-gon. The altitude drawn in any white triangle coincides with the altitude drawn in the corresponding triangle in the $n$-gon, while the ratio of the corresponding base sides equals the ratio of the side length of the $2n$-gon and the side length of the $n$-gon, denote it by $c$. Thus the total area of white triangles is $cS_n$, where $S_n$ is the area of the $n$-gon. Analogously, the total area of black triangles is $cS_n$, too.\nThe claim of the problem is equivalent to the statement that the sum of the altitudes of black triangles drawn to the sides that coincide to the sides of the $2n$-gon is equal to that of white triangles. Let $O$ be the center of the $2n$-gon, $A$ be the point chosen inside, $\\alpha$ be the angle between line $OA$ and the line perpendicular to a side, and $l$ be the distance between $O$ and any side of the $2n$-gon (see Fig. 17).\n\n![](attached_image_2.png)\nFig. 17\n\nThen the altitude of the corresponding triangle is $l - |OA| \\cos \\alpha$. The altitude of the next triangle of the same color can be expressed similarly but $\\alpha$ is replaced with $\\alpha + \\frac{360^\\circ}{n}$. Thus the sum of all altitudes of the triangles of the same color is $nl - |OA| \\cdot (\\cos \\alpha + \\cos(\\alpha + \\frac{360^\\circ}{n}) + \\dots + \\cos(\\alpha + \\frac{(n-1)360^\\circ}{n}))$.\n\nTo show that the sum inside parentheses equals $0$, multiply the sum by $\\sin \\frac{360^\\circ}{2n}$. Since $\\cos(\\alpha + \\frac{k \\cdot 360^\\circ}{n}) \\sin \\frac{360^\\circ}{2n} = \\frac{1}{2}(\\sin(\\alpha + \\frac{(k+\\frac{1}{2}) \\cdot 360^\\circ}{n}) - \\sin(\\alpha + \\frac{(k-\\frac{1}{2}) \\cdot 360^\\circ}{n}))$, a telescoping sum emerges and after reduction one obtains $-\\sin(\\alpha - \\frac{360^\\circ}{2n}) + \\sin(\\alpha + \\frac{(n-\\frac{1}{2}) \\cdot 360^\\circ}{n}) = 0$. Hence for both colors, the sum of the altitudes of all triangles of this color is $nl$.\n\n*Remark.* The sum $\\cos \\alpha + \\cos(\\alpha + \\frac{360^\\circ}{n}) + \\dots + \\cos(\\alpha + \\frac{(n-1) \\cdot 360^\\circ}{n})$ in Solution 2 can also be computed as follows. Denote $z = \\cos \\alpha + i \\sin \\alpha$, where $i$ is the imaginary unit and let $z_k = \\cos \\frac{k \\cdot 360^\\circ}{n} + i \\sin \\frac{k \\cdot 360^\\circ}{n}$, $k = 0, 1, \\dots, n-1$. Then the sum under consideration is the real part of the complex number $z \\cdot z_0 + z \\cdot z_1 + \\dots + z \\cdot z_{n-1}$. Thus $z \\cdot z_0 + z \\cdot z_1 + \\dots + z \\cdot z_{n-1} = z \\cdot (z_0 + z_1 + \\dots + z_{n-1}) = z \\cdot (z_1^0 + z_1^1 + z_1^2 + \\dots + z_1^{n-1}) = z \\cdot \\frac{z_1^n - 1}{z_1 - 1} = z \\cdot 0 = 0$, whence the sum under consideration is equal to $0$.",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 29643,
    "subject": "Mathematics (Multi-modal)",
    "question": "There are 8 identical dice. The numbers $4$, $5$, $6$ are written on three faces of the dice, as shown in the figure, and the remaining faces carry the numbers $1$, $2$, $3$ so that the sum of the numbers written on each pair of opposite faces is $7$.\n\n![](attached_image_1.png)\n\na) Show that using these dice, it is possible to form a $2 \\times 2 \\times 2$ cube so that every two faces that touch each other carry the same number.\n\nb) Is it possible to do this in such a way that only numbers $4$, $5$, $6$ occur on the outer surface of the resulting cube?",
    "options": [],
    "answer": "a) Yes, it is possible. b) No, it is impossible.",
    "solution": "a) Put together four dice as depicted in Fig. 7. These dice form the lower layer of the cube. On top of this, place another similar layer turned upside down. By the construction of the layer, the numbers on the faces touching each other within one layer coincide everywhere. As the second layer is turned upside down, the numbers on faces of cubes of different layers that touch each other also coincide.\n\n![](attached_image_2.png)\nFig. 7\n\nb) Suppose it is possible to form a $2 \\times 2 \\times 2$ cube so that its surface contains only numbers $4$, $5$, $6$. As exactly $3$ faces of each unit cube are visible, all three numbers must occur on those. Place the cube in such a way that the upper layer has $6$ in its southeastern corner (see Fig. 8). Then, as the only possibility, the upper layer must have $4$ in its southwestern corner and $5$ in its northeastern corner. Now it is impossible to place a die in the north-western corner since it should touch both of its neighbors with number $1$.\n\n![](attached_image_3.png)\nFig. 8",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 29644,
    "subject": "Mathematics (Multi-modal)",
    "question": "$$\n\\frac{a^2 + bc}{b+c} + \\frac{b^2 + ca}{c+a} + \\frac{c^2 + ab}{a+b} \\geq a+b+c\n$$\nfor all positive real numbers $a$, $b$, $c$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "W.l.o.g., assume $a \\ge b \\ge c$. Then\n$$\n\\begin{aligned}\n\\frac{a^2 + bc}{b+c} + \\frac{b^2 + ca}{c+a} + \\frac{c^2 + ab}{a+b} &= \\\\\n&= \\frac{a^2 + (b+c)c - c^2}{b+c} + \\frac{b^2 + (c+a)a - a^2}{c+a} + \\frac{c^2 + (a+b)b - b^2}{a+b} = \\\\\n&= \\frac{a^2 - c^2}{b+c} + c + \\frac{b^2 - a^2}{c+a} + a + \\frac{c^2 - b^2}{a+b} + b \\ge a+b+c,\n\\end{aligned}\n$$\nsince\n$$\n\\begin{aligned}\n\\frac{a^2 - c^2}{b+c} + \\frac{b^2 - a^2}{c+a} + \\frac{c^2 - b^2}{a+b} &= \\frac{a^2 - b^2 + b^2 - c^2}{b+c} + \\frac{b^2 - a^2}{c+a} + \\frac{c^2 - b^2}{a+b} = \\\\\n&= (a^2 - b^2) \\left( \\frac{1}{b+c} - \\frac{1}{c+a} \\right) + (b^2 - c^2) \\left( \\frac{1}{b+c} - \\frac{1}{a+b} \\right) = \\\\\n&= \\frac{(a^2 - b^2)(a-b)}{(b+c)(c+a)} + \\frac{(b^2 - c^2)(a-c)}{(b+c)(a+b)} \\ge 0.\n\\end{aligned}\n$$\nRearranging and transforming the expression gives\n$$\n\\begin{aligned}\n\\frac{a^2 + bc}{b+c} - a + \\frac{b^2 + ca}{c+a} - b + \\frac{c^2 + ab}{a+b} - c &= \\\\\n&= \\frac{a^2 + bc - ab - ac}{b+c} + \\frac{b^2 + ca - bc - ba}{c+a} + \\frac{c^2 + ab - ca - cb}{a+b} = \\\\\n&= \\frac{(a-b)(a-c)}{b+c} + \\frac{(b-c)(b-a)}{c+a} + \\frac{(c-a)(c-b)}{a+b} = \\\\\n&= \\frac{(a^2 - b^2)(a^2 - c^2) + (b^2 - c^2)(b^2 - a^2) + (c^2 - a^2)(c^2 - b^2)}{(b+c)(c+a)(a+b)} = \\\\\n&= \\frac{a^4 + b^4 + c^4 - a^2b^2 - a^2c^2 - b^2c^2}{(b+c)(c+a)(a+b)} = \\\\\n&= \\frac{(a^2 - b^2)^2 + (c^2 - a^2)^2 + (b^2 - c^2)^2}{2(b+c)(c+a)(a+b)} \\ge 0.\n\\end{aligned}\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 29645,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all quadruples $(w, x, y, z)$ of positive integers such that $w^x + w^y = w^z$.",
    "options": [],
    "answer": "(w, x, y, z) = (2, n, n, n+1) for any positive integer n",
    "solution": "Consider the following cases.\nIf $w = 1$, then no solution can exist, since the l.h.s. of the equality equals $2$ while the r.h.s. equals $1$.\nIf $w \\ge 2$, then $x < z$ and $y < z$, i.e., $x \\le z - 1$ and $y \\le z - 1$. Thus $w^x + w^y \\le w^{z-1} + w^{z-1} = 2 \\cdot w^{z-1} \\le w \\cdot w^{z-1} = w^z$. To satisfy the equation, equalities must hold in both inequalities and thus $x = y = z - 1$ and $w = 2$. This gives the solutions $(w, x, y, z) = (2, n, n, n+1)$, where $n$ is an arbitrary positive integer.\nIn the case $w = 1$ there are no solutions because $1 + 1 = 2$ is not a power of $1$. Assume in the rest that $w > 1$. W.l.o.g., assume $x \\le y < z$. Then the equation takes the form $w^x(1 + w^{y-x}) = w^z$, whence $1 + w^{y-x} = w^{z-x}$. Consequently, $1 + w^{y-x}$ is a positive power of $w$ and is divisible by $w$. If $y-x$ were positive, then $w^{y-x}$ would also be divisible by $w$, whence $1$ should be divisible by $w$, which is impossible. The remaining case $y-x = 0$ leads to $2 = w^{z-x}$ that gives $w = 2$ and $z-x = 1$ as the only possibility. Hence the solutions of the equation are of the form $(w, x, y, z) = (2, n, n, n+1)$, where $n$ is any positive integer. Checking shows that all these quadruples satisfy the equation.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 29646,
    "subject": "Mathematics (Multi-modal)",
    "question": "Each side of a convex quadrangle $ABCD$ is a diameter of a circle. All four circles pass through the same point $O$, different from the vertices of the quadrangle, and no two circles have common points other than those mentioned. Prove that $ABCD$ is a rhombus.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Since $AB$, $BC$, $CD$, and $DA$ are diameters (Fig. 9), $AOB$, $BOC$, $COD$, and $DOA$ are right angles. Hence $AOC$ and $BOD$ are straight angles, i.e., the diagonals of the quadrangle meet at $O$. The circles drawn on the opposite sides $AB$ and $CD$ cannot have common points besides $O$, since otherwise one circle would pass through three collinear points (two vertices of the quadrangle and the point $O$). Consequently, the circles drawn on $AB$ and $CD$ must touch at $O$. Let $P$ and $Q$ be the centers of these circles, respectively; then $O$ lies on the segment $PQ$.\n\n![](attached_image_1.png)\nFig. 9\n\n$|AP| = |PO|$ and $|CQ| = |QO|$, isosceles triangles give $\\angle BAC = \\angle PAO = \\angle POA = \\angle QOC = \\angle QCO = \\angle DCA$. Thus $AB$ and $CD$ are parallel. Similarly, the remaining sides are parallel. Thus $ABCD$ is a rectangle; the diagonals of a rectangle are perpendicular only if the rectangle is a rhombus.\n\n![](attached_image_1.png)\nFig. 9",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 29647,
    "subject": "Mathematics (Multi-modal)",
    "question": "Every face of a unit cube has one of numbers $-1$, $0$, $1$ written on it in such a way that every two faces with a common edge contain different numbers. Is it possible to form\n\na) a cube of size $2 \\times 2 \\times 2$;\n\nb) a cube of size $3 \\times 3 \\times 3$\n\nso that in the grids that come up on the faces, every two squares with a common side contain different numbers and the sum of all numbers on each face equals $0$?",
    "options": [],
    "answer": "a) yes; b) yes",
    "solution": "First note that the placement of the numbers on the faces of the unit cube is unique. Indeed, let a number $x$ be written on some face; then the neighboring faces contain alternately the other numbers $y$ and $z$, while the opposite face again contains $x$. This means that each of the numbers $-1$, $0$, $1$ occurs in one pair of opposite faces. Figures 10 and 11 show suitable constructions (where $-$ and $+$ denote $-1$ and $1$, respectively).\n\n![](attached_image_1.png)\nFig. 10\n\n![](attached_image_2.png)\nFig. 11",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 29648,
    "subject": "Mathematics (Multi-modal)",
    "question": "The size of the angle $ABC$, expressed in degrees, in a right triangle $ABC$ is an integer. It is known that for some positive integer $n$, one can choose points $K_0 = A$, $K_2, \\dots, K_{2n}$ on the hypotenuse $AB$ and points $K_1 = C$, $K_3, \\dots, K_{2n+1} = B$ on the leg $CB$ in such a way that each triangle $K_{i-1}K_iK_{i+1}$ with $i = 1, \\dots, 2n$ is isosceles with base $K_{i-1}K_{i+1}$. Find all possible values of the size of angle $ABC$.",
    "options": [],
    "answer": "30, 18, 10, 6, 2",
    "solution": "Let $\\angle ABC = \\alpha$ (Fig. 12). Then the base angle of the last isosceles triangle $K_{2n-1}K_{2n}K_{2n+1}$ is $\\alpha$. The base angle of the second last isosceles triangle $K_{2n-2}K_{2n-1}K_{2n}$ has the size $180^\\circ - (180^\\circ - 2\\alpha) = 2\\alpha$. The base angle of the next triangle before it, $K_{2n-3}K_{2n-2}K_{2n-1}$, has the size\n\n$180^\\circ - (180^\\circ - 4\\alpha) - \\alpha = 3\\alpha$.\n\nGenerally, the size of the base angle of triangle $K_{2n-i}K_{2n-i+1}K_{2n-i+2}$ is $180^\\circ - (180^\\circ - 2\\cdot(i - 1)\\alpha) - (i - 2)\\alpha = i\\alpha$ ($i = 3, \\ldots, 2n$). Thus the base angle of triangle $ACK_2 = K_0K_1K_2$ has the size $2n\\alpha$.\n\nNow in the triangle $ABC$ we get $90^\\circ = \\angle BAC + \\angle ABC = 2n\\alpha + \\alpha$, whence\n\n$2n + 1$ is an odd divisor of $90$ and is greater than $1$ (as a triangle cannot have two angles of the size $90^\\circ$). Such divisors are $3$, $5$, $9$, $15$, and $45$ that give the solutions $30^\\circ$, $18^\\circ$, $10^\\circ$, $6^\\circ$, and $2^\\circ$, respectively.\n\n![](attached_image_1.png)\nFig. 12",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 29649,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all integers that cannot be expressed as a sum of at least three consecutive terms of some non-constant arithmetic sequence of integers.",
    "options": [],
    "answer": "1 and -1",
    "solution": "First prove that $1$ and $-1$ are not expressible as the sum of at least three consecutive terms of an arithmetic sequence of integers. Let $a_1$, $a_2$, $\\ldots$, $a_k$ be $k$ consecutive terms of an arithmetic sequence, where $k \\ge 3$. They sum up to $s = \\frac{a_1 + a_k}{2} \\cdot k$. If $k$ is odd, then $s$ is divisible by $k$. If $k$ is even, then $s$ is divisible by $\\frac{k}{2} > 1$. In both cases, $s$ differs from $1$ and $-1$.\n\nNow prove that every integer $s$ other than $1$ or $-1$ is expressible as the sum of at least three consecutive terms of an arithmetic sequence of integers. If $s = 0$, then $s = -1 + 0 + 1$. If $s$ is different from zero and is even, i.e., $s = 2t$, where $t \\neq 0$, then $-t, 0, t, 2t$ sum up to $2t = s$. If $s$ is odd, i.e., $s = 2t + 1$, then $-t + 1, \\dots, 0, 1, \\dots, t-1, t, t+1$ are consecutive terms of an arithmetic sequence; they sum up to $t + (t+1) = 2t + 1 = s$, since the terms $-t+1$ through $t-1$ mutually cancel.\nLet $a_1$ be the first of the consecutive terms and $d$ be the common difference of consecutive terms. The sum of $n$ consecutive terms is $s = \\frac{2a_1 + d(n-1)}{2} \\cdot n$. Thus $2s = (2a_1 + d(n-1))n$. If $s = 1$ or $s = -1$, then this equality cannot hold because $n \\ge 3$ divides neither $2$ nor $-2$. If $s = 0$, then choose the portion of the arithmetic progression to be $-1$, $0$, $1$. If $s$ differs from these numbers, then let $n = 2|s|$, $d$ be an arbitrary odd number, and $a_1 = \\frac{1 - (n-1)d}{2}$ if $s > 0$, and $a_1 = \\frac{-1 - (n-1)d}{2}$ if $s < 0$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 29650,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all integral solutions of the equation $x^3 - y^3 = 3xy + 1$.",
    "options": [],
    "answer": "All integer pairs (x, y) of the form (n+1, n) for any integer n, together with the pair (-1, 1).",
    "solution": "First assume $x > y$. Then $3xy + 1 = x^3 - y^3 = (x-y)(x^2 + xy + y^2) \\ge (x-y) \\cdot 3xy$. Thus $3xy = x^3 - y^3 - 1 \\ge 1 - 1 = 0$ because $x > y$. If $3xy > 0$, then $3xy \\ge 3$, hence the inequality $3xy + 1 \\ge (x-y) \\cdot 3xy$ derived above implies $x-y = 1$. If $3xy = 0$, then either $x = 0$ or $y = 0$ and in both cases the only possibility is $x-y = 1$ again. An elementary check shows that all pairs $(x,y) = (n+1,n)$, where $n$ is an integer, satisfy the initial equation.\n\nNow assume $x = y$. Then the equation has no solutions, since the l.h.s. is 0 while the r.h.s. is positive.\n\nFinally assume $x < y$. Then the l.h.s. of the equation is negative, showing that $xy$ is negative. Hence $x < 0$ and $y > 0$. Denoting $-x = z$ and multiplying the equation by $(-1)$ leads to new equation $z^3 + y^3 = 3zy - 1$. Then $3zy - 1 = z^3 + y^3 = (z+y)(z^2 - zy + y^2) \\ge (z+y)zy$. Hence $z+y < 3$, giving $z = y = 1$, i.e., $x = -1, y = 1$, as the only possibility. It is easy to check that this satisfies the equation.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 29651,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all pairs $(a, b)$ of real numbers with $a + b = 1$, which satisfy $(a^2 + b^2)(a^3 + b^3) = a^4 + b^4$.",
    "options": [],
    "answer": "(0, 1), (1, 0), (1/2, 1/2)",
    "solution": "As $a^3 + b^3 = (a+b)(a^2 - ab + b^2)$, the given equation can be expressed as $(a^2 + b^2)(a^2 - ab + b^2) = a^4 + b^4$. Expanding brackets gives $-a^3b + 2a^2b^2 - ab^3 = 0$, which factorizes to $-ab(a-b)^2 = 0$. Hence $a = 0$ or $b = 0$ or $a-b = 0$. Together with the condition $a+b=1$, we get the following solutions: $(0, 1)$, $(1, 0)$, and $(\\frac{1}{2}, \\frac{1}{2})$.\nDenote $ab = c$. Then $a^2 + b^2 = (a+b)^2 - 2ab = 1 - 2c$,\n$a^3 + b^3 = (a+b)(a^2 - ab + b^2) = 1 \\cdot (1 - 2c - c) = 1 - 3c$, and $a^4 + b^4 = (a^2 + b^2)^2 - 2a^2b^2 = (1 - 2c)^2 - 2c^2 = 2c^2 - 4c + 1$. The given equation $(a^2 + b^2)(a^3 + b^3) = a^4 + b^4$ can now be expressed as $(1 - 2c)(1 - 3c) = 2c^2 - 4c + 1$, or equivalently, $c(4c - 1) = 0$. Hence, $c = 0$ or $c = \\frac{1}{4}$. Now the two simultaneous equations $a+b=1$ and $ab=c$ give the solutions $a=0$, $b=1$ and $a=1, b=0$ for $c=0$, and $a = \\frac{1}{2}, b = \\frac{1}{2}$ for $c = \\frac{1}{4}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 29652,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all pairs $(m, n)$ of positive integers for which the $m \\times n$ grid contains exactly $225$ rectangles whose side lengths are odd and whose edges lie on the lines of the grid.",
    "options": [],
    "answer": "(1,29), (29,1), (5,9), (9,5)",
    "solution": "The $m \\times n$ grid is formed by $m+1$ horizontal and $n+1$ vertical lines. Number the horizontal lines with numbers from $1$ to $m+1$ and the vertical lines with numbers from $1$ to $n+1$. Rectangles with odd side lengths arise if and only if two horizontal lines with different parity and two vertical lines with different parity intersect.\n\nAssume that at least one of the numbers $m$ and $n$ is even. We can assume without loss of generality that $m = 2k$. Then there are exactly $k+1$ odd-numbered and $k$ even-numbered horizontal lines and thus there are $k(k+1)$ pairs of lines of different parity. But this means that overall the number of rectangles with odd side lengths is even and cannot be $225$. Therefore $m$ and $n$ are both odd numbers.\n\nLet now $m = 2k-1$ and $n = 2l-1$. Then we have exactly $k$ even-numbered and $k$ odd-numbered horizontal lines and $l$ even-numbered and $l$ odd-numbered vertical lines. Overall it is possible to form $k \\cdot k \\cdot l \\cdot l = (kl)^2$ rectangles with odd side lengths. From $(kl)^2 = 225$ we get $kl = 15$. The solutions are $k=1, l=15$ or $k=3, l=5$ (or vice versa). So $m=1, n=29$ or $m=5, n=9$ (or vice versa).",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 29653,
    "subject": "Mathematics (Multi-modal)",
    "question": "Given a triangle $ABC$ where $|BC| = a$, $|CA| = b$ and $|AB| = c$, prove that the equality\n$$\n\\frac{1}{a+b} + \\frac{1}{b+c} = \\frac{3}{a+b+c}\n$$\nholds if and only if $\\angle ABC = 60^\\circ$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "By finding the common denominator on the left hand side, transform the equation to $(a+2b+c)(a+b+c) = 3(a+b)(b+c)$. Expanding the brackets and simplifying gives $b^2 = a^2 + c^2 - ac$. Comparing the latter with the cosine law $b^2 = a^2 + c^2 - 2ac \\cos \\beta$, we see that the equality holds if and only if $\\cos \\beta = \\frac{1}{2}$, i.e., $\\beta = 60^\\circ$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 29654,
    "subject": "Mathematics (Multi-modal)",
    "question": "A square $ABCD$ lies in the coordinate plane with its vertices $A$ and $C$ lying on different coordinate axes. Prove that one of the vertices $B$ or $D$ lies on the line $y = x$ and the other one on $y = -x$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Assume without loss of generality that $A$ is located on the $x$-axis and $C$ is located on the $y$-axis, let these points have coordinates of $A (a, 0)$ and $C (0, c)$. As the diagonals of a square bisect each other, we know that the intersection point $P$ of diagonal is also the mid-point of $AC$, i.e. $P(\\frac{a}{2}, \\frac{c}{2})$ and $\\vec{PC} = (-\\frac{a}{2}, \\frac{c}{2})$.\n\nAs the diagonals of a square are perpendicular to each other and of the same length, the vectors $\\vec{PB}$ and $\\vec{PD}$ have the same length as the vector $\\vec{PC}$ and are perpendicular to it. But for a given vector $\\vec{u} = (s, t)$, there are exactly two vectors perpendicular to and having the same length as it: $\\vec{v} = (-t, s)$ and $-\\vec{v} = (t, -s)$. For the vector $\\vec{u} = \\vec{PC} = (-\\frac{a}{2}, \\frac{c}{2})$ we get $\\vec{v} = (\\frac{c}{2}, -\\frac{a}{2})$ and w.l.o.g. we can assume that $\\vec{PB} = \\vec{v}$ and $\\vec{PD} = -\\vec{v}$. Now from here $B(\\frac{a+c}{2}, \\frac{a+c}{2})$ and $D(\\frac{a-c}{2}, \\frac{c-a}{2})$. Thus, we see that the point $B$ is located on the line $y = x$ and point $D$ is located on the line $y = -x$.\nW.l.o.g., assume that the vertices of the square are labelled counter-clockwise with $A (a, 0)$, $C (0, c)$, where $a, c \\ge 0$ (other cases are similar). Let $O$ be the origin, then $\\angle AOC = 90^\\circ$, i.e. the circumcircle (with diameter $AC$) of the square $ABCD$ passes through the origin $O$. Based on the assumptions made, $B$ definitely lies in the first quadrant and $D$ has to lie in the second quadrant (Fig. 4) or in the fourth quadrant (Fig. 5), otherwise the circle with the diameter $BD$ cannot pass the origin.\n\n![](attached_image_1.png)\nFig. 4\n\n![](attached_image_2.png)\nFig. 5\n\ndiameter *AC*) of the square *ABCD* passes through the origin *O*. Now note that the vertices of the square divide its circumcircle into four equal arcs of $90^\\circ$, each having an inscribed angle of $45^\\circ$ subtending on it. Thus, $\\angle AOB = \\angle BOC = 45^\\circ$, i.e., $B$ lies on the line with equation $y = x$ (if $A = O$ or $C = O$, then one of those angles will lose its meaning, however, the other one is still $45^\\circ$ and that is sufficient). Similarly, $\\angle COD = 45^\\circ$, if $D$ lies in the second quadrant, or $\\angle AOD = 45^\\circ$, if $D$ lies in the fourth quadrant. In both cases, $D$ lies on the line $y = -x$; this condition is also met in the special case $D = O$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 29655,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $a$, $b$, $c$ be fixed real numbers, where $0 \\le a, b, c \\le 4$. Prove that the system of equations\n$$\n\\begin{cases}\np^2 - a q = -3 \\\\\nq^2 - b r = -4 \\\\\nr^2 - c p = -5\n\\end{cases}\n$$\nhas no real solutions ($p$, $q$, $r$).",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Adding up all equations gives $p^2 - c p + q^2 - a q + r^2 - b r = -12$. From the inequality $(p - \\frac{c}{2})^2 \\ge 0$ we have $p^2 - c p \\ge -\\frac{c^2}{4} \\ge -4$ and similarly, $q^2 - a q \\ge -4$ and $r^2 - b r \\ge -4$. Adding up these inequalities, we see that to avoid a contradiction with the equality derived first, all three inequalities must actually be equalities, i.e. $a = b = c = 4$ and $p = q = r = 2$. But this does not satisfy the initial equations.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 29656,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABC$ be a triangle with integral side lengths. The angle bisector drawn from $B$ and the altitude drawn from $C$ meet at point $P$ inside the triangle. Prove that the ratio of areas of triangles $APB$ and $APC$ is a rational number.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $H$ be the foot of the altitude drawn from $C$. First prove that $|AH|$ and $|BH|$ are rational numbers. For that, use the Pythagorean theorem for triangles $ACH$ and $BCH$ to obtain $|AH|^2 + |CH|^2 = |AC|^2$ and $|BH|^2 + |CH|^2 = |BC|^2$. Therefore $|AC|^2 - |BC|^2 = |AH|^2 - |BH|^2 = (|AH| - |BH|) \\cdot (|AH| + |BH|) = (|AH| - |BH|) \\cdot |AB|$. We see that $|AH| - |BH| = \\frac{|AC|^2 - |BC|^2}{|AB|}$ is rational and so are $|AH| = \\frac{|AC|^2 + |BC|^2}{|AB|}$\n![](attached_image_1.png)\nFig. 6\nand $|BH| = |AH| - (|AH| - |BH|)$. Let now $K$ be the projection of $P$ to $BC$ (see Fig. 6). As $P$ lies on the angle bisector of $B$, it is equidistant from both $AB$ and $BC$, i.e., $|PH| = |PK|$. Consequently,\n\n$$\n\\frac{S_{APB}}{S_{BPC}} = \\frac{|AB| \\cdot |PH|}{|BC| \\cdot |PK|} = \\frac{|AB|}{|BC|} \\cdot \\frac{|BH|}{|AH|} \\quad \\text{As } CH \\perp AB, \\text{ also } \\frac{S_{BPC}}{S_{APC}} = \\frac{|CP| \\cdot |BH|}{|CP| \\cdot |AH|} = \\frac{|BH|}{|AH|}\n$$\nThus, $\\frac{S_{APB}}{S_{APC}} = \\frac{|AB|}{|BC|} \\cdot \\frac{|BH|}{|AH|}$ is rational as a product of two rational numbers.\nLet $H$ be the foot of the altitude drawn from $C$ and let $L$ be the projection of $P$ to $AC$ (see Fig. 6). Now $\\angle CPL = 90^\\circ - \\angle ACH = 90^\\circ - (90^\\circ - \\angle CAB) = \\angle CAB$, giving $\\frac{|PH|}{|PL|} = \\frac{|PH|}{|PC| \\cdot \\cos \\angle CAB}$. The angle bisector theorem gives $\\frac{|PH|}{|PC|} = \\frac{|BH|}{|BC|} = \\cos \\angle ABC$. Consequently,\n$$\n\\frac{S_{APB}}{S_{APC}} = \\frac{|AB| \\cdot |PH|}{|AC| \\cdot |PL|} = \\frac{|AB|}{|AC|} \\cdot \\frac{\\cos \\angle ABC}{\\cos \\angle CAB}\n$$\nAs the side lengths of the triangle $ABC$ are integers, $\\frac{|AB|}{|AC|}$ is rational. By the cosine law, the cosines of the angles of triangle $ABC$ are rational, whence $\\frac{\\cos \\angle ABC}{\\cos \\angle CAB}$ is rational. Altogether, $\\frac{S_{APB}}{S_{APC}}$ is rational.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 29657,
    "subject": "Mathematics (Multi-modal)",
    "question": "Prove that the set of integers $\\{0, 1, 2, \\dots, 2^n - 1\\}$ can be partitioned into $n + 1$ disjoint subsets $A_0, A_1, \\dots, A_n$ such that both of the following hold:\n\na) If $k + l = n$, then the subsets $A_k$ and $A_l$ have the same number of elements.\n\nb) If $s$ and $t$ are non-negative integers and $s + t \\le n$, then for an arbitrary element $z$ in the set $A_{s+t}$, there exist elements $x$ and $y$ from the sets $A_s$ and $A_t$, respectively, such that $x + y = z$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Divide the set $A$ into subsets such that the subset $A_k$ consists of only those numbers which have exactly $k$ ones in their binary representation. Then $A_0 = \\{0\\}, A_1 = \\{1, 2, 4, \\dots, 2^{n-1}\\}, \\dots, A_n = \\{2^n - 1\\}$. Let us show that both conditions are met.\n\nThe first condition is met because the numbers with $k$ ones are in one-to-one correspondence with the numbers with $n-k$ ones: given a number, simply replace all ones in its binary representation by zeros and vice versa.\n\nTo show that the second condition is met, choose an arbitrary number $z$ from the set $A_{s+t}$. Its binary representation contains exactly $s+t$ ones. Construct a binary number $x$ by choosing $s$ ones from the binary representation of $z$ and filling all other binary places by zeros, analogously construct a second number $y$ based on remaining ones in $z$. Then $x \\in A_s, y \\in A_t$ and $x + y = z$.\nLet us prove the statement by induction. If $n = 1$, then $A = \\{0, 1\\}$, and taking $A_0 = \\{0\\}$ and $A_1 = \\{1\\}$ we get a partition that satisfies both of the requirements.\n\nAssume now that we have a partition $C_0, C_1, \\dots, C_n$ for the set $C = \\{0, 1, \\dots, 2^n - 1\\}$. Construct a partition of the set $A = \\{0, 1, \\dots, 2^{n+1} - 1\\}$ based on that. First, generate the sets $B_0, B_1, \\dots, B_n$ as follows: the elements of the subset $B_i$ are derived from the elements of the subset $C_i$ by adding $2^n$ to them. The subsets $B_0, B_1, \\dots, B_n$ form a partition of the set $A \\setminus C = \\{2^n, 2^n+1, \\dots, 2^{n+1}-1\\}$ and from the construction for all $i=0, 1, \\dots, n$ the corresponding subsets $B_i$ and $C_i$ have the same number of elements.\n\nNow, let $A_i = C_i \\cup B_{i-1}$ for all $i = 1, 2, \\dots, n$ and in addition to that, $A_0 = C_0$ and $A_{n+1} = B_n$. Then, $|A_0| = |A_{n+1}| = 1$, and if $k+l=n+1$, then also $k, l \\neq 0$, $|A_k| = |B_{k-1}| + |C_k| = |C_{k-1}| + |C_k|$ and $|A_l| = |B_{l-1}| + |C_l| = |C_{l-1}| + |C_l|$. As $(k-1)+l=k+(l-1)=n$, we see that $|C_{k-1}| = |C_l|$ and $|C_k| = |C_{l-1}|$; thus $|A_k| = |A_l|$.\n\nTo verify that the second condition is met, let $z$ be an arbitrary element of $A_{s+t}$. If $t=0$, then $A_s = A_{s+t}$ and $A_t = A_0 = \\{0\\}$, so we can take $x=z$ and $y=0$. Now assume $t \\ge 1$. If $z < 2^n$, then $z$ is an element of $C_{s+t}$ and thus there exist elements $x$ and $y$ in the sets $C_s \\subset A_s$ and $C_t \\subset A_t$, respectively, such that $x+y=z$. If $z \\ge 2^n$, then $z$ is an element of $B_{s+t-1}$, i.e. $z-2^n$ is an element of the set $C_{s+t-1}$ and thus the sets $C_s$ and $C_{t-1}$ contain elements $x$ and $y$, respectively, such that $x+y=z-2^n$. But now $x+(y+2^n)=z$, where $x$ and $y+2^n$ are elements of the sets $C_s \\subset A_s$ and $B_{t-1} \\subset A_t$, respectively. So the statement holds for all positive $n$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 29658,
    "subject": "Mathematics (Multi-modal)",
    "question": "How many positive integers are there that are divisible by $2010$ and that have exactly $2010$ divisors (1 and the integer itself included)?",
    "options": [],
    "answer": "24",
    "solution": "Let $N$ be a positive integer that is divisible by $2010$ and that has exactly $2010$ positive divisors. Since $2010 = 2 \\cdot 3 \\cdot 5 \\cdot 67$, also $N$ should be divisible by these four primes. Thus, $N = 2^a \\cdot 3^b \\cdot 5^c \\cdot 67^d \\cdot s$, where $a, b, c, d > 0$ and $s$ is not divisible by any of the primes $2, 3, 5, 67$. All the factors of $N$ can be expressed as $2^i \\cdot 3^j \\cdot 5^k \\cdot 67^l \\cdot t$, where $0 \\le i \\le a, 0 \\le j \\le b, 0 \\le k \\le c, 0 \\le l \\le d$, and $t$ is a factor of $s$. There are $a+1$ choices for $i$ (from $0$ to $a$) and similarly, there are $b+1, c+1$ and $d+1$ choices for $j, k$ and $l$, respectively. Therefore, $N$ has $\\delta(N) = (a+1)(b+1)(c+1)(d+1)\\delta(s)$ different factors, where $\\delta(x)$ stands for the number of factors of $x$. We require $\\delta(N) = 2010$. As $a+1 > 1, b+1 > 1, c+1 > 1$, and $d+1 > 1$, we see that each of these numbers is divisible by some prime numbers and the number $\\delta(N) = (a+1)(b+1)(c+1)(d+1)\\delta(s)$ can thus be expressed as a product of at least four prime numbers. But as $2010$ itself is a product of exactly four prime numbers, we conclude that $a+1, b+1, c+1$, and $d+1$ are exactly those primes $2, 3, 5$, and $67$, in some order, and $\\delta(s) = 1$. From the latter condition we see that $s=1$ because any numbers bigger than $1$ has more than one factor. So for $N$ to satisfy the conditions, $N$ must be expressible as $2^a \\cdot 3^b \\cdot 5^c \\cdot 67^d$, where $a, b, c, d$ are the numbers $1, 2, 4$, and $66$ in some order. Thus there are $4! = 24$ numbers satisfying the conditions.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 29659,
    "subject": "Mathematics (Multi-modal)",
    "question": "Consider a parallelogram $ABCD$.\n\na) Prove that if the incenter of the triangle $ABC$ is located on the diagonal $BD$, then the parallelogram $ABCD$ is a rhombus.\n\nb) Is the parallelogram $ABCD$ a rhombus whenever the circumcenter of the triangle $ABC$ is located on the diagonal $BD$?",
    "options": [],
    "answer": "a) Yes, ABCD is a rhombus. b) No; for example, a non-square rectangle.",
    "solution": "a) As the incenter of the triangle $ABC$ is located on diagonal $BD$ (Fig. 1), we can conclude that $BD$ is the bisector of $\\angle ABC$. Therefore $\\angle ABD = \\angle CBD$. However, since $ABCD$ is a parallelogram, $\\angle ABD = \\angle CDB$. Hence the triangle $BCD$ is isosceles, i.e. $|BC| = |CD|$. Thus, $ABCD$ is a rhombus.\n\n![](attached_image_1.png)\nFig. 1\n\nb) Let $ABCD$ be a rectangle with different side lengths. The circumcenter of triangle $ABC$ is located on the intersection of the diagonals of the rectangle. We see that all the required conditions are satisfied, however $ABCD$ is not a rhombus.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 29660,
    "subject": "Mathematics (Multi-modal)",
    "question": "The numbers $0$, $1$, and $2$ are written in the vertices of a triangle. One step involves increasing two of the three numbers by $m$ or decreasing one of the three numbers by $n$. Is it possible that after some steps there are numbers $1$, $2$, and $3$ (in an arbitrary order) written in the vertices if\n\na) $m = 3$, $n = 6$;\nb) $m = 4\\frac{1}{2}$, $n = 6$?",
    "options": [],
    "answer": "a) No; b) Yes",
    "solution": "a) Both the step that involves increasing two of the numbers by $3$ and the step that involves decreasing one of the numbers by $6$ result in the sum of all three numbers being changed by $6$. Thus the remainder when the sum of the three numbers is divided by $6$ will always be the same regardless of the number of steps taken. But as the sums $0+1+2$ and $1+2+3$ give different remainders when divided by $6$, it is impossible to reach the required end situation from the given initial situation.\n\nb) First increase the second and the third numbers three times by $4\\frac{1}{2}$; we end up with $0$, $14\\frac{1}{2}$, $15\\frac{1}{2}$ in the vertices. Now increase the first and the second numbers by $4\\frac{1}{2}$ and also increase the first and the third numbers by $4\\frac{1}{2}$; so we end up with $9$, $19$ and $20$ written in the three vertices, respectively. Finally decrease the first number once by $6$ and the other two three times by $6$, achieving the situation in question.\na) Consider one of the numbers. The remainder when this number is divided by $3$ is the same regardless of the number of steps taken. Therefore, if we want to achieve the situation where $1$, $2$, $3$ are located in the three vertices, the numbers $1$ and $2$ should stay in the same vertices where they were at the beginning and $3$ has to be in the vertex where $0$ was. Notice that two increasings are exactly cancelled out by one decreasing. Thus, the vertices where the numbers remain the same should have undergone an even number of increasings and the vertex where $0$ is replaced by $3$ should have been exposed to an odd number of increasings. Hence there should have been an odd number of increasings in total which is impossible since each increasing step influences the numbers in two vertices.\n\nb) As in Solution 1.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 29661,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all pairs $(n, k)$ of positive integers that satisfy the equality $n! + (n+1)! = k! + 120$.",
    "options": [],
    "answer": "(4, 4), (5, 6)",
    "solution": "Note that for every $n$, $n! + (n+1)! = n! + n! \\cdot (n+1) = n! \\cdot (n+2) \\le (n+2)!$. Thus if $n! + (n+1)! = k! + 120$, then due to $120 = 5!$ we have $k! + 5! \\le (n+2)!$. This inequality in turn implies $k < n+2$ and $5 < n+2$. Hence $5 \\le n+1$, leading to $0 \\le (n+1)! - 5! = k! - n!$. Consequently, $k \\ge n$, i.e., the cases to be considered are $k = n$ and $k = n+1$. If $k = n$, then the initial equation leads to $(n+1)! = 120$, giving $n = 4$, $k = 4$. If $k = n+1$, then analogously $n = 5$, $k = 6$.\nIf $k < n$, then $n! > k!$. If, additionally, $n \\ge 4$, then $(n+1)! \\ge 120$, giving $n! + (n+1)! > k! + 120$. But if $n < 4$, then $n! + (n+1)! \\le 30 < k! + 120$. The desired equality can hold in neither of the cases.\nIf $k > n+1$, then $k! - (n+1)!$ is positive and is also divisible by $(n+1)!$, hence $k! - (n+1)! \\ge (n+1)!$. On the other hand, $n! - 120 < (n+1)!$, giving $k! - (n+1)! > n! - 120$. Thus, there is no solution in this case either.\nHence $n = k$ or $n+1 = k$, leading to two solutions $(n, k) = (4, 4), (5, 6)$.\nThe equation implies that $120$ is divisible by the minimum of $n!$ and $k!$. As $120 = 5!$, either $n \\le 5$ or $k \\le 5$. Consider both cases.\nIf $n \\in \\{1, 2, 3\\}$, then $n! + (n+1)! - k! < 120$, whence the equation has no solution. If $n=4$, then $120 = 144 - k!$, whence $k! = 24$ and $k=4$. If $n=5$, then $120 = 840 - k!$, whence $k! = 720$ and $k=6$.\nIf $k \\in \\{1, 2, 3, 4, 5\\}$, then the initial equation implies that $n! + (n+1)!$ lies between 121 and 240. This is possible only if $n=4$, since if $n=3$, then $n! + (n+1)! = 30$, and if $n=5$, then $n! + (n+1)! = 840$. If $n=4$, then $n! + (n+1)! = 144$, which corresponds to $k=4$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 29662,
    "subject": "Mathematics (Multi-modal)",
    "question": "Consider the diagonals $A_1A_3$, $A_2A_4$, $A_3A_5$, $A_4A_6$, $A_5A_1$ and $A_6A_2$ of a convex hexagon $A_1A_2A_3A_4A_5A_6$. The hexagon whose vertices are the points of intersection of the diagonals is regular. Can we conclude that the hexagon $A_1A_2A_3A_4A_5A_6$ is also regular?",
    "options": [],
    "answer": "Yes, the original hexagon must be regular.",
    "solution": "We show that the hexagon $A_1A_2A_3A_4A_5A_6$ has all its side lengths equal and all its angles equal. As the internal hexagon is regular, the grey triangles in Fig. 2 all have two angles of equal size and so they are isosceles. Additionally, all these six isosceles triangles have their bases of equal lengths, thus they are all congruent. The black triangles on Fig. 2 are isosceles because the grey triangles are isosceles. Additionally, their vertex angles are equal, as\n\n![](attached_image_1.png)\nFig. 2\n\nthey all are equal to the angles of a regular hexagon. Therefore the black triangles are all congruent and thus their bases are of equal length. Now the angles of the external hexagon are all formed of the angles of two grey triangles and one black triangle. As both of the latter are congruent, the external hexagon has its angles of equal size.\nLengthen the sides of the internal regular hexagon until intersection. The points of intersection are exactly the vertices of the initial external hexagon. Because of the symmetry of the internal hexagon the points of intersection are symmetrically located about the midpoint of the internal hexagon. Thus, the external (initial) hexagon is also regular.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 29663,
    "subject": "Mathematics (Multi-modal)",
    "question": "Consider a positive integer $N$ with exactly 6 positive divisors $d_1, \\ldots, d_6$ such that $1 = d_1 < d_2 < d_3 < d_4 < d_5 < d_6 = N$. Call such an integer $N$ good if the sum $d_4 + d_5$ is divisible by the sum $d_2 + d_3$.\n\na) Find the smallest positive integer $N$ which has exactly 6 positive divisors and which is not good.\n\nb) Prove that there are infinitely many positive integers $N$ all with exactly 6 positive divisors and all not good.",
    "options": [],
    "answer": "a) 20. b) Infinitely many, for example all numbers of the form N = 4p with p a prime greater than 4 (more generally, all N = p^2 q with distinct primes p, q and q > p^2).",
    "solution": "a) Considering the numbers from 1 to 20 we see that exactly three of them have 6 divisors: 12 (the divisors are 1, 2, 3, 4, 6, 12), 18 (1, 2, 3, 6, 9, 18), and 20 (1, 2, 4, 5, 10, 20). For 12 the sum $d_4 + d_5 = 4 + 6$ is divisible by the sum $d_2 + d_3 = 2 + 3$ and similarly, for 18 the sum $d_4 + d_5 = 6 + 9$ is divisible by $d_2 + d_3 = 2 + 3$. However, for 20 the sum $d_4 + d_5 = 5 + 10$ is not divisible by the sum $d_2 + d_3 = 2 + 4$. Thus, the smallest non-good number with exactly 6 factors is 20.\n\nb) Take $N = 4p$, where $p$ is an arbitrary prime number larger than 4. Then $N$ has exactly 6 different divisors: 1, 2, 4, $p$, $2p$, $4p$, in increasing order. Indeed, as $N = 2^2p^1$, where 2 and $p$ are two different prime numbers, all of its divisors can be expressed as $2^i p^j$ where $i \\le 2$ and $j \\le 1$. From here we obtain exactly $3 \\cdot 2 = 6$ choices: $i$ can be either 0, 1 or 2 and for every $i$ we have two choices for $j$: 0 or 1.\nHere, $d_4 + d_5 = p + 2p = 3p$ is odd because $p > 2$ and thus is not divisible by an even number $d_2 + d_3 = 2 + 4 = 6$. Therefore, none of the numbers expressed as $N = 4p$ where $p > 4$ is a prime is good. As there are infinitely many prime numbers, there must also be infinitely many such numbers $N$.\nLet us find all $N > 1$ which have exactly 6 divisors.\n\n1) If $N = p^k$, where $p$ is a prime, then it has the divisors 1, $p$, ..., $p^k$, i.e. $k + 1$ divisors overall. Thus, all $N = p^5$ satisfy this requirement.\n\n2) Let $N$ have two different prime divisors, i.e. $N = p^k q^l$. For $k \\ge 2$ and $l \\ge 2$ we see that $N$ has at least 9 different divisors: 1, $p$, $p^2$, $q$, $q^2$, $pq$, $p^2q$, $pq^2$, and $p^2q^2$. For $l = 1$, $N$ has the divisors 1, $p$, ..., $p^k$, and $q$, $pq$, ..., $p^kq$, i.e. $2(k + 1)$ divisors in total. Thus, all $N = p^2q$ satisfy the requirement.\n\n3) Let $N$ have at least three prime divisors $p$, $q$, $r$. Then $N$ has at least 8 different divisors: 1, $p$, $q$, $r$, $pq$, $pr$, $qr$, and $pqr$, and we get no more numbers.\n\nLet us now consider $N = p^5$ and $N = p^2q$ in more detail.\n\ni) If $N = p^5$, then $d_i = p^{i-1}$ and $d_4 + d_5 = p^3 + p^4 = p^3(1 + p)$ is divisible by $d_2 + d_3 = p + p^2 = p(1 + p)$. Thus they are all good.\n\nii) If $N = p^2q$, where $q < p$, then $N$ has the divisors $1, q, p, pq, p^2$, and $p^2q$, in increasing order, and $d_4 + d_5 = pq + p^2 = p(q + p)$ is divisible by $d_2 + d_3 = q + p$ and thus, they are all good, too.\n\niii) If $N = p^2q$, where $p < q < p^2$, then $N$ has the divisors $1, p, q, p^2, pq$, and $p^2q$, in increasing order, and $d_4 + d_5 = p^2 + pq = p(p + q)$ is divisible by $d_2 + d_3 = p + q$. Thus, they are all good, too.\n\niv) Finally, let $N = p^2q$, where $q > p^2$. Then $N$ has the divisors $1, p, p^2, q, pq$, and $p^2q$, in increasing order and $d_4 + d_5 = q + pq = q(1 + p)$ is not divisible by $d_2 + d_3 = p + p^2 = p(1 + p)$ because the prime number $q$ cannot be divisible by another prime number $p$. Thus all these numbers have exactly 6 different divisors and they all are non-good. To get the smallest of these numbers, we have to take $p$ and $q$ as small as possible, i.e. $p = 2$ and $q = 5$ (to achieve $q > p^2 = 4$). Then $N = 2^2 \\cdot 5 = 20$. Finally, there are infinitely many of these numbers $N$ because we have infinitely many choices for prime numbers $p$ and $q$ such that $q > p^2$. For example, we can take $p = 2$ and $q$ an arbitrary prime number bigger than 5. As there are infinitely many prime numbers, we have proven the statement.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 29664,
    "subject": "Mathematics (Multi-modal)",
    "question": "Prove that none of the integers that contain one $2$, one $1$, and all the rest zeros, can be expressed as a sum of two perfect squares or as a sum of two perfect cubes.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "All the numbers described in the problem are divisible by $3$ (as their sum of digits is divisible by $3$). Note that all perfect squares leave the remainder $0$ or $1$ when divided by $3$, and therefore, for the sum of the two perfect squares to be divisible by $3$, they both have to be divisible by $3$. Now, as the numbers are both divisible by $3$, their squares are divisible by $9$ and thus the sum of the squares is divisible by $9$. However, the number described in the problem is not divisible by $9$, a contradiction.\n\nNote that when a cube number is divided by $3$, it will leave the remainder of either $0$, $1$ or $-1$. Indeed, $(3k)^3 = 9(3k^3)$ and $(3k \\pm 1)^3 = 27k^3 \\pm 27k^2 + 9k \\pm 1 = 9(3k^3 \\pm 3k^2 + k) \\pm 1$. The numbers described in the problem give the remainder $3$ when divided by $9$ (as their sum of digits gives the remainder $3$ when divided by $9$), therefore we conclude that it is impossible to express them as sums of two cubes.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 29665,
    "subject": "Mathematics (Multi-modal)",
    "question": "Consider an acute-angled triangle $ABC$ and its circumcircle. Let $D$ be a point on the arc $AB$ which does not include point $C$ and let $A_1$ and $B_1$ be points on the lines $DA$ and $DB$, respectively, such that $CA_1 \\perp DA$ and $CB_1 \\perp DB$. Prove that $|AB| \\ge |A_1B_1|$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "If $CD$ is the diameter of the circumcircle of triangle $ABC$, then $A_1 = A$ and $B_1 = B$ and the statement holds. Assume that $CD$ is not the diameter (Fig. 3). Then $A_1 \\ne A$ and $B_1 \\ne B$. The point $A_1$ lies on the ray $AD$ if and only if the point $B_1$ does not lie on the ray $BD$ (depending on which side of the diameter through point $C$ point $D$ is located). Thus, $\\angle CAA_1 =$\n$\\angle CBB_1$ (because the sum of opposite angles of a cyclic quadrilateral $ADBC$ is $180^\\circ$). Thus, the right-angled triangles $AA_1C$ and $BB_1C$ are similar. From $\\angle ACA_1 = \\angle BCB_1$ we see that $\\angle ACB = \\angle A_1CB_1$. This together with $\\frac{|AC|}{|BC|} = \\frac{|A_1C|}{|B_1C|}$ gives that $ACB$ and $A_1CB_1$ are similar. As $|AC| > |A_1C|$, we conclude that $|AB| > |A_1B_1|$.\n\n\nSolution 2:\n\nSince the angles $CA_1D$ and $CB_1D$ are right angles, the points $C, A_1, D$, and $B_1$ form a cyclic quadrilateral and thus $\\angle CAB = \\angle CDB = \\angle CA_1B_1$. Similarly, $\\angle CBA = \\angle CB_1A_1$. Therefore the triangles $ABC$ and $A_1B_1C$ are similar. As $|CA| \\ge |CA_1|$, we deduce that $|AB| \\ge |A_1B_1|$.\n\n![](attached_image_1.png)\nFig. 3\n\n\nSolution 3:\n\nThe radius $R$ of the circumcircle of the quadrilateral $CADB$ is at least as large as the radius $R_1$ of the circumcircle of the quadrilateral $CA_1DB_1$ because $CD$ is a chord in the first one and a diameter in the second one. The sine law in triangles $ADB$ and $A_1DB_1$ gives $|AB| = 2R \\sin \\angle D$ and $|A_1B_1| = 2R_1 \\sin \\angle D$. As $R \\ge R_1$, we deduce $|AB| \\ge |A_1B_1|$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 29666,
    "subject": "Mathematics (Multi-modal)",
    "question": "Two circles lie completely outside each other. Let $A$ be the point of intersection of internal common tangents of the circles and let $K$ be the projection of this point onto their external common tangent. The tangents, different from the common tangent, to the circles through point $K$ meet the circles at $M_1$ and $M_2$. Prove that the line $AK$ bisects the angle $M_1KM_2$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $L_1$ and $L_2$ be the points of tangency of the external common tangent of the circles, $N_1$ and $N_2$ be the points of tangency of an internal common tangent, and $O_1$ and $O_2$ be the centers of the two circles (see Fig. 18).\n\nAs all the lines $O_1L_1$, $AK$, and $O_2L_2$ are perpendicular to the line $L_1L_2$, they are parallel to each other and thus $$\\frac{|L_1K|}{|L_2K|} = \\frac{|O_1A|}{|O_2A|}.$$ The triangles $O_1AN_1$ and $O_2AN_2$ are similar because they are both right-angled and have the same vertical angles. Thus, $$\\frac{|O_1A|}{|O_2A|} = \\frac{|O_1N_1|}{|O_2N_2|} = \\frac{|O_1L_1|}{|O_2L_2|}.$$ Therefore, the right-angled triangles $O_1L_1K$ and $O_2L_2K$ are similar due to proportionality of their legs. Hence, $\\angle L_1KO_1 = \\angle L_2KO_2$.\n\nAs $\\angle L_1KM_1 = 2\\angle L_1KO_1$ and $\\angle L_2KM_2 = 2\\angle L_2KO_2$, we also get that $\\angle L_1KM_1 = \\angle L_2KM_2$. Together with the equality $\\angle L_1KA = \\angle L_2KA = 90^\\circ$ this implies $\\angle M_1KA = \\angle M_2KA$.\nBoth of the circles appear at the same angle, when viewed from the point $A$. To solve the problem, it is enough to show that both of the circles also appear at the same angle, when viewed from the point $K$.\n\nLet the centers of the circles have the coordinates $O_1(a_1, b_1)$ and $O_2(a_2, b_2)$ and let $r_1$ and $r_2$ be the radii of the circles. The two circles appear at the same angle from the point $P(x, y)$ if and only if $$\\frac{r_1}{|O_1P|} = \\frac{r_2}{|O_2P|},$$ i.e., $$\\frac{r_1}{\\sqrt{(x-a_1)^2+(y-b_1)^2}} = \\frac{r_2}{\\sqrt{(x-a_2)^2+(y-b_2)^2}}.$$ Simple algebra shows that this equation is equivalent to $$(r_1^2 - r_2^2)x^2 + (r_1^2 - r_2^2)y^2 + c_1x + c_2y + c_3 = 0,$$ where $c_1, c_2$, and $c_3$ are some constants.\n\nIf $r_1 = r_2$, then the statement clearly holds. If $r_1 \\neq r_2$, then the last equation is that of a circle. Point $A$ as well as the point $D$ of intersection of the external common tangents both lie on that circle, and from symmetry, the diameter of that circle is $AD$. As $AK$ is perpendicular to the external common tangent of the circles, the point $K$ also lies on that circle.\n\n![](attached_image_1.png)\nFig. 18",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 29667,
    "subject": "Mathematics (Multi-modal)",
    "question": "Does there exist an operation $*$ on the set of all integers such that the following conditions hold simultaneously:\n(1) for any integers $x$, $y$, $z$, $(x * y) * z = x * (y * z)$;\n(2) for any integers $x$ and $y$, $x * x * y = y * x * x = y$?",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Define an operation $\\oplus$ on the set of all non-negative integers, which maps two non-negative integers $a$ and $b$ to a non-negative integer $a \\oplus b$, such that for all $i = 0, 1, \\dots$, $(a \\oplus b)_i = (a_i + b_i) \\bmod 2$, where $n_i$ stands for the binary digit corresponding to $2^i$ in the binary representation of $n$. This operation satisfies condition (1) for all non-negative integers because addition modulo 2 satisfies it. The operation also satisfies condition (2) because if $x, y \\in \\{0,1\\}$, then $(x + x + y) \\bmod 2 = y \\bmod 2 = (y + x + x) \\bmod 2$.\n\nAs the set of non-negative integers as well as the set of all integers are countable, there exists one-to-one correspondence $f$ between these sets (e.g. mapping a non-negative integer $x$ to the integer $(-1)^x \\lfloor \\frac{x+1}{2} \\rfloor$). Every integer can therefore be uniquely expressed in the form $f(n)$, where $n$ is a non-negative integer. Therefore we can define the operation $*$ by the formula $f(x) * f(y) = f(x \\oplus y)$. Following from the construction, both conditions (1) and (2) still hold.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 29668,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $a, b, c$ be positive real numbers such that $2a^2 + b^2 = 9c^2$. Prove that\n$$\n\\frac{2c}{a} + \\frac{c}{b} \\ge \\sqrt{3}.\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Using the AM-GM inequality for three terms twice, one gets\n$$\n\\begin{aligned}\n\\frac{2c}{a} + \\frac{c}{b} &= \\frac{(2b+a)c}{ab} = \\frac{(2b+a)\\sqrt{2a^2+b^2}}{3ab} = \\frac{(b+b+a)\\sqrt{a^2+a^2+b^2}}{3ab} \\\\\n&\\ge \\frac{3\\sqrt[3]{b^2a}\\sqrt{3\\sqrt[3]{a^4b^2}}}{3ab} = \\frac{3\\sqrt{3}\\sqrt[3]{b^2a \\cdot a^2b}}{3ab} = \\frac{3\\sqrt{3}ab}{3ab} = \\sqrt{3}.\n\\end{aligned}\n$$\nUsing HM-QM inequality for $a, a, b$ gives\n$$\n\\frac{3}{\\frac{2}{a} + \\frac{1}{b}} = \\frac{3}{\\frac{1}{a} + \\frac{1}{a} + \\frac{1}{b}} \\le \\sqrt{\\frac{a^2 + a^2 + b^2}{3}} = \\sqrt{\\frac{2a^2 + b^2}{3}} = \\sqrt{\\frac{9c^2}{3}} = \\sqrt{3}c.\n$$\nThus\n$$\n\\left(\\frac{2}{a} + \\frac{1}{b}\\right) \\cdot c \\ge \\frac{3}{\\sqrt{3}} = \\sqrt{3},\n$$\nwhich implies the desired inequality.\n$$\n\\begin{aligned}\n\\left(\\frac{2c}{a} + \\frac{c}{b}\\right)^2 &= \\frac{(2b+a)^2 c^2}{a^2 b^2} = \\frac{(2b+a)^2 (2a^2 + b^2)}{9a^2 b^2} = \\\\\n&= \\frac{((a^2 + b^2) + 4ab + 3b^2)(a^2 + (a^2 + b^2))}{9a^2 b^2} \\ge \\frac{(6ab + 3b^2)(a^2 + 2ab)}{9a^2 b^2} = \\\\\n&= \\frac{(2a+b)(a+2b)}{3ab} = \\frac{2a^2 + 5ab + 2b^2}{3ab} \\ge \\frac{4ab + 5ab}{3ab} = 3.\n\\end{aligned}\n$$\nThe square of the l.h.s. of the desired inequality is\n$$\n\\left(\\frac{2c}{a} + \\frac{c}{b}\\right)^2 = \\frac{(2b+a)^2(2a^2+b^2)}{9a^2b^2} = \\frac{1}{9}\\left(2+\\frac{a}{b}\\right)^2\\left(2+\\frac{b^2}{a^2}\\right).\n$$\nDenoting $\\frac{a}{b} = x$, the desired inequality reduces to $\\frac{1}{9}(2+x)^2(2+\\frac{1}{x^2}) \\ge 3$, which is equivalent to $(2+x)^2(2x^2+1) \\ge 27x^2$. This in turn is equivalent to $2x^4+8x^3-18x^2+4x+4 \\ge 0$, that is $2(x-1)^2(x^2+6x+2) \\ge 0$ after factorization. This inequality holds, since on positive arguments the quadratic polynomial $x^2+6x+2$ is positive.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 29669,
    "subject": "Mathematics (Multi-modal)",
    "question": "Prove that if $n$ and $k$ are positive integers such that $1 < k < n-1$, then the binomial coefficient $\\binom{n}{k}$ is divisible by at least two different primes.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Assume w.l.o.g. that $n \\ge 2k$ (if $n < 2k$, then interchange the roles of $k$ and $n-k$). Let $p$ be an arbitrary prime number. Consider the numbers that remain into the numerator of the expression\n$$\n\\binom{n}{k} = \\frac{n \\cdot (n-1) \\cdots (n-k+1)}{k \\cdot (k-1) \\cdots 1}\n$$\nafter reducing all factors by the highest power of $p$ by which they are divisible. Suppose that some two of the $k$ factors resulting after this step are equal. Then the corresponding initial factors are of the form $s \\cdot p^i$ and $s \\cdot p^j$, where $i > j$. But then $n \\ge s \\cdot p^i \\ge p \\cdot s \\cdot p^j > p \\cdot (n-k) \\ge 2 \\cdot (n-k)$, which contradicts the assumption $n \\ge 2k$. Hence the $k$ new factors are pairwise different. As $1 < k < n-1$, the numerator initially contains at least two consecutive numbers, at least one of which is not divisible by $p$. This number does not change in the process described above. By the assumption $n \\ge 2k$, this number is greater than $k$. Consequently, the product remaining in the numerator after elimination of powers of $p$ is greater than the denominator $k \\cdot (k-1) \\cdots 1$. This means that the powers of $p$ in the original numerator cannot be completely cancelled out with the denominator. So the canonical representation of $\\binom{n}{k}$ cannot consist of a power of $p$ only.\nSuppose that for some $n$ and $k$,\n$$\n\\binom{n}{k} = \\frac{n \\cdot (n-1) \\cdots (n-k+1)}{k \\cdot (k-1) \\cdots 1} = p^t,\n$$\nwhere $p$ is a prime number and $t$ is some positive integer. Let $m$ be a number in $\\{n, n-1, \\dots, n-k+1\\}$, in the canonical representation of which the exponent of $p$ is the largest. Then the exponent of $p$ in the canonical representation of $n, n-1, ..., m+1$ coincides with that in the canonical representation of $n-m, n-m-1, ..., 1$, respectively. Similarly, the exponent of $p$ in the canonical representation of $m-1, ..., n-k+1$ coincides with that in the canonical representation of $1, ..., m-1-n+k$, respectively. Consequently, the exponent of $p$ in the canonical representation of the product $n(n-1)...(m+1)(m-1)...(n-k+1)$ equals to that in the canonical representation of the product $(n-m)!(m-1-n+k)!$. Since\n$$\n\\frac{k!}{(n-m)!(m-1-n+k)!} = k \\cdot \\frac{(k-1)!}{(n-m)!(k-1-n+m)!} = k \\cdot \\binom{k-1}{n-m}\n$$\nis clearly an integer, the exponent of $p$ in the canonical representation of $(n-m)!(m-1-n+k)!$ does not exceed that in the canonical representation of $k!$. Hence, the exponent of $p$ in the canonical representation of $\\binom{n}{k}$ does not exceed that in the canonical representation of $m$. As the assumptions of the problem imply $\\binom{n}{k} \\ge \\binom{n}{2} > n \\ge m$, this leads to a contradiction.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 29670,
    "subject": "Mathematics (Multi-modal)",
    "question": "On a square board with $m$ rows and $n$ columns, where $m \\le n$, some squares are colored black in such a way that no two rows are alike. Find the biggest integer $k$ such that for every possible coloring to start with one can always color $k$ columns entirely red in such a way that no two rows are still alike.\n\n**Answer:** $n-m+1$.",
    "options": [],
    "answer": "n - m + 1",
    "solution": "Prove that if $m \\le n$, one of the columns can always be colored red. Then, when excluding this column, we can continue the process until the number of columns is smaller than the number of rows, i.e. $n-m+1$ times. Suppose we cannot color a single column red such that no two rows still appear alike. Then, for every column there are at least two rows that differ by only one square in that column. Consider those two rows for every column. Now consider a graph with its vertices being the rows and its edges being the pairs of rows of interest. As this graph has at least as many edges as vertices, the graph contains a cycle. Consider an arbitrary row (i.e. vertex) $x$ of the cycle. Then the row it is followed by differs from row $x$ exactly by one square, assume this square is located at the column $y$. Every next row differs from the previous one in exactly one column. As all these columns differ from column $y$, all the other edges of this cycle correspond to passing from one row to another such that the square in column $y$ remains the same. Therefore, the square in column $y$ remains the same in all the rows after $x$ and it has to remain the same at the last passage which takes us back to the row $x$. This, however, means that all squares in row $x$ have to be of the same color as the square at the row after $x$, a contradiction.\n\nIn general, no more rows can be colored red. Assume that all the squares are colored black to start with, apart from the diagonal of an $m \\times m$-subsquareboard. If we color $n-m+2$ columns red, there are always at least 2 columns which lie in that subsquareboard. But this means that we will over-color the only two white squares in two rows and thus we end up with two equally colored rows.\nConsider the columns of the squareboard one by one. The first column divides the set of all the rows into two subsets: one of them consists of the rows which have the square at the first column white and the other consists of those rows that have their first square black. If the first column has all its squares white or all its squares black, then we can color it red. Similarly, for every next column divide the subsets even further depending on whether there is a black or a white square in that row in that column. If no subsets are divided further at a particular step, we can color that column red. Therefore, for all columns, either the number of subsets increases by at least one or this column is colored red. As we started off with one set and ended up with $m$ subsets each containing one row only, there has to be no more than $m-1$ columns that were not colored red and thus at least $k=n-m+1$ that were. All these rows are still different from each other because we colored only those columns which had no new information about the differences between rows compared with previous columns.\n\nSimilarly to Solution 1 we can show that there always exists a coloring for which no more columns can be colored red.\nProve by induction on the number of rows that we can always color at least $k=n-m+1$ columns red. If $m=1$ then we can color all the columns red, i.e. $k=n=n-m+1$. Assume that for $m=l$ we can color at least $n-l+1$ columns red such that no two rows appear alike. Assume now that $m=l+1$. Following from the induction hypothesis we can color at least $n-l+1$ columns red such that the first $l$ rows remain different. If after that coloring process the last row is different from the rest, all the conditions required are satisfied and we can color at least $n-m+2$ columns red. If the last row however is similar to any of the rows above it (there can be only one of those) then as these rows were all different to start with, there should exist a column at which the last row and the row that appears similar after coloring actually differ. If we do not color this column, all the rows will appear different and we can color $n-m+1$ columns red.\n\nSimilarly to Solution 1 we can show that there always exists a coloring for which no more columns can be colored red.",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 29671,
    "subject": "Mathematics (Multi-modal)",
    "question": "Integers $a$, $b$, $c$ are such that $a + b + c$ is divisible by $6$, and $a^2 + b^2 + c^2$ is divisible by $36$. Does it imply that $a^3 + b^3 + c^3$ is divisible by\n\na) $8$;\nb) $27$?",
    "options": [],
    "answer": "a) Yes; b) No",
    "solution": "a) As the sum of $a$, $b$, and $c$ is divisible by $6$, and is therefore even, there must be either $0$ or $2$ odd numbers among the three. If we had $2$ odd numbers, the sum of the squares $a^2 + b^2 + c^2$ would give a remainder of $0 + 1 + 1 = 2$ when dividing by $4$. But this is not possible, since the sum is divisible by $36$, and therefore also by $4$. So, all the numbers $a$, $b$, $c$ are even. Hence, all the numbers $a^3$, $b^3$, $c^3$ are divisible by $8$, and so is their sum.\n\nb) If $a = 8$, $b = c = 2$, then all of the premises are fulfilled: $8 + 2 + 2 = 12$ is divisible by $6$ and $8^2 + 2^2 + 2^2 = 72$ is divisible by $36$. But $8^3 + 2^3 + 2^3 = 528$ is not divisible by $9$, and therefore, it is not divisible by $27$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 29672,
    "subject": "Mathematics (Multi-modal)",
    "question": "In his last research, professor $P$ was concentrating on natural numbers with a certain property. It is known that whenever a natural number $x$ has this property, all multiples of $x$ also have this property. Let $a_1, \\dots, a_n$ be positive integers such that all their divisors that are greater than one have the property professor $P$ studied. Is it true that all divisors greater than one of the product $a_1 \\dots a_n$ definitely have this property?",
    "options": [],
    "answer": "Yes",
    "solution": "Let $k > 1$ be any divisor of the product $a_1 \\dots a_n$. Then $k$ has a prime divisor $p$, which is also a divisor of the product $a_1 \\dots a_n$. As $p$ is a prime, there exists $i$, such that $p$ is a divisor of $a_i$. As all the divisors of $a_i$ greater than $1$ have the property, $p$ also has this property. By the premise, all the multiples of $p$ have the property, so $k$ has the property.\nLet $k > 1$ be any divisor of the product $a_1 \\dots a_n$. If $k$ were relatively prime to all $a_i$, it would be relatively prime to the product $a_1 \\dots a_n$, but $\\text{gcd}(k, a_1 \\dots a_n) = k > 1$. Hence $\\text{gcd}(k, a_i) > 1$ for some $a_i$. As a divisor of $a_i$, the number $\\text{gcd}(k, a_i)$ has the property studied by professor $P$. As a multiple of $\\text{gcd}(k, a_i)$, also $k$ has the same property.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 29673,
    "subject": "Mathematics (Multi-modal)",
    "question": "a) Find all positive integers $n$, such that the sum of all integers from $1$ to $n + 1$ can be represented as the sum of $n$ consecutive integers.\n\nb) Find all positive integers $n$, for which there exists an integer $a$, such that the sum of the integers from $a$ to $a + n$ is equal to the sum of the integers from $a + n + 1$ to $a + 2n$.",
    "options": [],
    "answer": "a) n = 1. b) All positive integers n work; a valid choice is a = n^2.",
    "solution": "a)\nClearly the sum of the first two positive integers can be represented as the sum of one positive integer. Now, let $n \\ge 2$ and let us show that the sum of the $n + 1$ first positive integers cannot be represented as a sum of $n$ consecutive integers. Indeed, on one hand $1 + 2 + \\dots + n + (n+1) > 2 + 3 + \\dots + n + (n+1)$, on the other hand $1 + 2 + \\dots + n + (n+1) < 1 + 2 + 3 + \\dots + n + (n+1) + 1 = 3 + \\dots + n + (n+1) + 2 + 2 \\le 3 + \\dots + n + (n+1) + (n+2)$. So the sum of $n + 1$ first positive integers $1 + \\dots + (n+1)$ lies between $2 + \\dots + (n+1)$ and $3 + \\dots + (n+2)$, which both are consecutive sums of $n$ consecutive integers. So, the number $1 + \\dots + (n+1)$ is not a sum of $n$ consecutive integers.\n\nb)\nLet $n$ be any positive integer. To solve the problem, it suffices to see that $n^2 + (n^2 + 1) + \\dots + (n^2 + n) = n^2 \\cdot (n+1) + (1 + \\dots + n) = n \\cdot (n^2 + n) + (1 + \\dots + n) = (n^2 + n + 1) + \\dots + (n^2 + n + n)$.\nWe use the formula for the sum of arithmetic progression.\n\na) If $n$ is odd, then the sum of $n$ consecutive integers is divisible by $n$. So, if the number $1+2+...+(n+1) = \\frac{n+1}{2} \\cdot (n+2)$ was the sum of $n$ consecutive integers, it would be divisible by $\\frac{n+1}{2}$ and by $n$. As $n+1$ and $n$ are relatively prime, the same clearly holds for $\\frac{n+1}{2}$ and $n$. Therefore, $\\frac{n+1}{2} \\cdot (n+2)$ should be divisible by $\\frac{n+1}{2} \\cdot n$, meaning that $n+2$ should be divisible by $n$.\n\nIf $n$ is even, the sum of $n$ consecutive integers is divisible by $\\frac{n}{2}$. So, if $1+2+...+(n+1) = (n+1) \\cdot \\frac{n+2}{2}$ was the sum of $n$ consecutive integers, it would be divisible by both $n+1$ and $\\frac{n}{2}$. As $n+1$ and $n$ are relatively prime, also $n+1$ and $\\frac{n}{2}$ are relatively prime. Thus, $(n+1) \\cdot \\frac{n+2}{2}$ should be divisible by $(n+1) \\cdot \\frac{n}{2}$, implying that $n+2$ is divisible by $n$.\n\nSo, in all cases $n+2$ is divisible by $n$, which is equivalent to saying $2$ is divisible by $n$. So, $n=1$ or $n=2$. Clearly $1+2$ is the sum of one integer, but $1+2+3=6$, being an even number, cannot be represented as the sum of two consecutive integers.\n\nb) If $a$ is the first of the two consecutive integers, then the problem can be represented as the equation $(a+a+n)(n+1)/2 = (a+n+1+a+2n)n/2$. By simplifying we see that it is equivalent to $a = n^2$. This means that the sum of $n+1$ consecutive integers, first of which is $n^2$, is the sum of the next $n$ consecutive integers. So, the desired numbers exist for every $n$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 29674,
    "subject": "Mathematics (Multi-modal)",
    "question": "The sides $AB$ and $AC$ of the triangle $ABC$ touch the circle $c$ respectively at points $B'$ and $C'$. The center $L$ of the circle $c$ lies on the side $BC$. The circumcenter $O$ of triangle $ABC$ lies on the shorter arc $B'C'$ of the circle $c$. Prove that the circumcircle of $ABC$ and the circle $c$ meet at two points.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $r$ be the circumradius of $ABC$, let $s$ be the radius of $c$ and $\\alpha = \\angle BAC$ (Fig. 15).\n\nBy tangency, $|AB'| = |AC'|$. Thus $\\angle C'B'A = \\angle B'C'A = \\frac{\\pi}{2} - \\frac{\\alpha}{2}$ whence, by property of inscribed angle, $\\angle B'OC' = \\pi - (\\frac{\\pi}{2} - \\frac{\\alpha}{2}) = \\frac{\\pi}{2} + \\frac{\\alpha}{2}$.\n\nClearly $\\angle B'OC' > \\angle BOC = 2\\alpha$, leading to $\\frac{\\pi}{2} + \\frac{\\alpha}{2} > 2\\alpha$. Hence $\\alpha < \\frac{\\pi}{3}$.\n\nNow let $K$ be the midpoint of side $BC$. From the right triangle $KOC$, one gets $|KO| = |OC|\\cos\\angle KOC = r\\cos\\alpha$.\n\nBy the inequality obtained above, $\\cos\\alpha > \\cos\\frac{\\pi}{3} = \\frac{1}{2}$.\n\nOn the other hand, $|KO| \\le |LO| = s$, leading to $\\frac{1}{2}r < r\\cos\\alpha = |KO| \\le s$ or $r < 2s$.\n\nAs $c$ passes through the circumcenter of $ABC$, this inequality shows that these circles must intersect.\n\n![](attached_image_1.png)\nFig. 15",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 29675,
    "subject": "Mathematics (Multi-modal)",
    "question": "A finite grid is covered with $1 \\times 2$ cards in such a way that the edges of the cards match with the lines of the grid, no card lies over the edge of the grid, and every square is covered by exactly two cards. Prove that one can remove some of the cards in such a way that every square will be covered by exactly one card.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Choose any square covered by two cards, and choose one of these cards. Move that card to a neighbouring square, and choose the other card that is covering that square. From there we move to the next square, etc., until we return to the first square. We cannot return to any other square visited previously, since in all squares except the first one, both cards have been chosen already. If we color the rectangular grid like a chessboard, then after an odd number of moves, we reach a square with the opposite color, and after an even number of moves, we reach a square with the same color. Therefore, the number of chosen cards is even. So, we can remove every second chosen card. All of the remaining squares we passed through will be covered by exactly one card. If after this, there are still squares that are covered by two cards, we repeat the process with a new randomly chosen square which is covered by two cards. We can never move from a square covered by two cards to a square covered by exactly one card, since all the squares covered by exactly one card were previously connected to squares now covered by exactly one card. So, after a finite number of steps we can find a new cycle, from which we can remove every second card. We repeat, until all squares are covered by exactly one card.",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 29676,
    "subject": "Mathematics (Multi-modal)",
    "question": "There are 2012 points marked in a square with side length $11$. Prove that one can choose an equilateral triangle with side length $12$ which covers at least $671$ points.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Place two equilateral triangles with side lengths $12$ on the square in such a way that both have one vertex lie on the side of the square and the opposite sides of these vertices partially coincide with the other side of the square and with each other (Fig. 16). The area common to both triangles forms an equilateral triangle of side length $1$. Position the third\n\n![](attached_image_1.png)\nFig. 16\n\nequilateral triangle with side length $12$ between the two triangles, turned $180^\\circ$, such that the lowermost vertex of that triangle coincides with the uppermost vertex of the small triangle. To show that the square is fully covered by these triangles, we must show that the sum of the heights of the large and the small triangle is at least $11$, which is equivalent to showing $\\frac{\\sqrt{3}}{2} \\cdot (12+1) > 11$. As we can simplify this equation to $13\\sqrt{3} > 22$ and $3 \\cdot 169 > 484$, we see that it holds. Therefore at least a third of the $2012$ points or at least $671$ points lie in one of the three chosen triangles.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 29677,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all pairs $(x, y)$ of positive integers such that\n$$\n\\frac{1}{x^2} + \\frac{249}{xy} + \\frac{1}{y^2} = \\frac{1}{2012}.\n$$",
    "options": [],
    "answer": "(503, 1006) and (1006, 503)",
    "solution": "Let $\\gcd(x, y) = d$ and $x = ad$, $y = bd$. Then the equation can be written as $\\frac{a^2+249ab+b^2}{a^2b^2d^2} = \\frac{1}{2012}$ or\n$$\na^2 b^2 d^2 = 2012(a^2 + 249ab + b^2).\n$$\nAs $a$ and $b$ are relatively prime, $a^2$ and $b^2$ are both relatively prime to $a^2 + 249ab + b^2$ and therefore both they must be divisors of $2012$. As $2012 = 2^2 \\cdot 503$ and $503$ is a prime, the possible cases are $(a, b) = (1, 1)$, $(a, b) = (1, 2)$, $(a, b) = (2, 1)$. If we substitute $(a, b) = (1, 1)$ into the last equation, we get $d^2 = 2012 \\cdot 251$, which is not solvable in integers. The other two cases give $4d^2 = 2012 \\cdot 503$, from which $d = 503$. This leads to the solutions $(x, y) = (503, 1006)$ and $(x, y) = (1006, 503)$.\n\n$$\n2012x^2 + 249 \\cdot 2012xy + 2012y^2 = x^2y^2.\n$$\nFrom the left-hand side we see that both sides of the equation must be divisible by $503$. As $503$ is a prime, one of the numbers $x$ and $y$ must be divisible by $503$. So $x^2$ or $y^2$ is divisible by $503^2$, giving that both sides of the equation are divisible by $503^2$. If $x$ is divisible by $503$, the summands $2012x^2$ and $249 \\cdot 2012xy$ on the left-hand side are divisible by $503^2$, meaning that $2012y^2$ is divisible by $503^2$. Therefore $y$ is divisible by $503$. Analogously, we get that if $y$ is divisible by $503$, then $x$ is also divisible by $503$. Consequently, both $x$ and $y$ are divisible by $503$. Denote $x = 503a$, $y = 503b$. Then the equation, after dividing both sides by $503^3$, simplifies to $4a^2 + 996ab + 4b^2 = 503a^2b^2$. Assume $a \\ge b$. If $b \\ge 2$, then $503a^2b^2 \\ge 503a^2 \\cdot 2b = 1006a^2b = 4a^2b + 4a^2b + 998a^2b > 4a^2 + 4b^2 + 996ab$, so the last equation cannot hold. Therefore $b = 1$. Now we get a quadratic equation $499a^2 - 996a - 4 = 0$ with respect to $a$, whose only positive solution is $a = 2$. From here we obtain the solution $(1006, 503)$ to our original equation. The case $b \\ge a$ is symmetrical and gives the solution $(503, 1006)$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 29678,
    "subject": "Mathematics (Multi-modal)",
    "question": "a) Prove that for every real number $x$ the arithmetic mean of $\\sqrt{1 + \\sin x}$ and $\\sqrt{1 - \\sin x}$ is equal to one of the following: $\\sin \\frac{x}{2}$, $\\cos \\frac{x}{2}$, $-\\sin \\frac{x}{2}$, $-\\cos \\frac{x}{2}$.\n\nb) Can one leave out one of the four numbers listed in part a) in such a way that the claim still holds?",
    "options": [],
    "answer": "No; none can be left out.",
    "solution": "a) Denote the arithmetic mean given in the problem by $A(x)$. As\n$$\n1 + \\sin x = \\sin^2 \\frac{x}{2} + \\cos^2 \\frac{x}{2} + 2 \\sin \\frac{x}{2} \\cos \\frac{x}{2} = \\left( \\sin \\frac{x}{2} + \\cos \\frac{x}{2} \\right)^2,\n$$\n\n$$\n1 - \\sin x = \\sin^2 \\frac{x}{2} + \\cos^2 \\frac{x}{2} - 2 \\sin \\frac{x}{2} \\cos \\frac{x}{2} = \\left( \\sin \\frac{x}{2} - \\cos \\frac{x}{2} \\right)^2,\n$$\nwe get\n$$\nA(x) = \\frac{\\sqrt{1 + \\sin x} + \\sqrt{1 - \\sin x}}{2} = \\frac{|\\sin \\frac{x}{2} + \\cos \\frac{x}{2}| + |\\sin \\frac{x}{2} - \\cos \\frac{x}{2}|}{2}.\n$$\nDepending on the signs of the numbers $\\sin \\frac{x}{2} + \\cos \\frac{x}{2}$ and $\\sin \\frac{x}{2} - \\cos \\frac{x}{2}$, one of the trigonometric functions in the numerator cancels out and the other one is doubled, with either a positive or a negative sign. Therefore, $A(x)$ is equal to one of the numbers $\\sin \\frac{x}{2}$, $\\cos \\frac{x}{2}$, $-\\sin \\frac{x}{2}$, $-\\cos \\frac{x}{2}$.\n\nb) Clearly $A(x) = 1$, whenever $x$ is one of the numbers $0, \\pi, 2\\pi, 3\\pi$. Nevertheless, each of these four values makes a unique expression among $\\sin \\frac{x}{2}, \\cos \\frac{x}{2}, -\\sin \\frac{x}{2}, -\\cos \\frac{x}{2}$ evaluate to 1. Therefore, none of these four can be left out.\nPart a) can also be proven as follows. Let $A(x)$ be the same as in the first solution. Then\n$$\n\\left( \\frac{\\sqrt{1 + \\sin x} + \\sqrt{1 - \\sin x}}{2} \\right)^2 = \\frac{2 + 2\\sqrt{1 - \\sin^2 x}}{4} = \\frac{1 + |\\cos x|}{2},\n$$\nso that $A(x) = \\sqrt{\\frac{1+|\\cos x|}{2}}$. Therefore, if $\\cos x \\ge 0$, then $A(x) = \\sqrt{\\frac{1+\\cos x}{2}} = \\pm \\cos \\frac{x}{2}$; if $\\cos x < 0$, then $A(x) = \\sqrt{\\frac{1-\\cos x}{2}} = \\pm \\sin \\frac{x}{2}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 29679,
    "subject": "Mathematics (Multi-modal)",
    "question": "In an acute triangle $ABC$, a point $P$ is chosen such that all points symmetrical to $P$ with respect to the sides of $ABC$ lie on the circumcircle of $ABC$. Prove that $P$ is the orthocenter of $ABC$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $A'$, $B'$, $C'$ be points symmetric to the point $P$ with respect to the sides $BC$, $CA$, $AB$ (Fig. 17).\n\nThen $|C'A| = |PA| = |B'A|$, giving that the arcs $AC'$ and $AB'$ of the circumcircle of the triangle $ABC$ are equal. Since $A$ and $C'$ are on the same half-plane from the line $BB'$, and $C$ on the other one, we have $\\angle C'CA = \\angle B'CA = \\angle PCA$. Since $P$ and $C'$ are on the same side from the line $AC$, the points $P$, $C$, $C'$ are collinear. Since $PC' \\perp AB$, we must also have $PC \\perp AB$, that is, the point $P$ lies on the height drawn from the vertex $C$ in the triangle $ABC$. Analogously we see that $P$ is on the other two heights.\n\n![](attached_image_1.png)\nFig. 17",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 29680,
    "subject": "Mathematics (Multi-modal)",
    "question": "There are $2^n$ soldiers standing in a line, where $n$ is a positive integer. The soldiers can rearrange themselves into a new line only in the following way: the soldiers standing at odd numbered positions move to the front of the row, keeping their positions with respect to each other, and the soldiers previously standing at even numbered positions move to the end of the row, keeping their positions with respect to each other. Prove that after $n$ rearrangements the soldiers stand in the same ordering as in the beginning.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "The last soldier does not change its position. The rest of the soldiers regroup just as in the case, when the last soldier was not there, and the number of the soldiers was $2^n - 1$. So, it suffices to prove the claim for $2^n - 1$ soldiers. We show that after $n$ rearrangements the soldiers are in positions, which can be found in the original line by counting cyclically every $2^i$-th soldier (after the last soldier we go to the first one). Indeed, after 0 rearrangements, the claim clearly holds, and every rearrangement makes us cyclically count every second soldier in the previous line (after the last soldier we go to the second one), the first soldier will still be counted first. After $n$ rearrangements the soldiers in the new line can be found by counting every $2^n$-th soldier in the old line with $2^n - 1$ soldiers. Since the remainder of $2^n$ when divided by $2^n - 1$ is 1, this is equivalent to simply counting the soldiers. This means that we get back the original line.\nEnumerate the soldiers starting from 0, and write the numbers in binary form (adding leading zeros to make the lengths of the binary codes equal; for example for $n = 3$ we have the numbers 000, 001, 010, 011, 100, 101, 110, 111). After a rearrangement the soldiers stand in such a way that when reinterpreting the last digit as the first one (but leaving the order of the rest of the digits the same), the soldiers are again enumerated by consecutive numbers. After $n$ rearrangements the binary code of the soldiers has returned to the original, so every soldier's position corresponds to their original position in the line.",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 29681,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABC$ be an isosceles triangle with $|AB| = |AC|$. The bisector of angle $ABC$ meets the side $AC$ at the point $D$.\na) Is the triangle $ABD$ isosceles whenever the triangle $BCD$ is isosceles?\nb) Is the triangle $BCD$ isosceles whenever the triangle $ABD$ is isosceles?",
    "options": [],
    "answer": "a) Yes; b) No",
    "solution": "a) Denote $\\angle BAC = \\alpha$ and $\\angle ABC = \\angle ACB = \\beta$.\n\n![](attached_image_1.png)\nFig. 6\n\nAssume that the triangle $BCD$ is isosceles. If $|CB| = |CD|$, then the angles $\\angle CBD$, $\\angle CDB$ and $\\angle BCD$ would be $\\frac{\\beta}{2}$, $\\frac{\\beta}{2}$ and $\\beta$ respectively, which implies $2 \\cdot \\frac{\\beta}{2} + \\beta = 180^\\circ$ giving $\\beta = 90^\\circ$. This is impossible, since the triangle $ABC$ has two angles equal to $\\beta$. If $|DB| = |DC|$, then we would get $\\frac{\\beta}{2} = \\beta$, which is also impossible. This leaves the only option $|BC| = |BD|$. Then, the triangles $ABC$ and $BCD$ are similar, since all corresponding angles are the same. Therefore $\\angle DBA = \\angle DBC = \\angle BAC = \\angle BAD$, showing that the triangle $ABD$ is isosceles with $|DA| = |DB|$.\n\nb) If the angles of the triangle $ABC$ are $\\frac{3}{7} \\cdot 180^\\circ$, $\\frac{2}{7} \\cdot 180^\\circ$, $\\frac{2}{7} \\cdot 180^\\circ$ (Fig. 7), then $\\angle ADB = 180^\\circ - \\angle BAD - \\angle ABD = 180^\\circ - \\frac{3}{7} \\cdot 180^\\circ - \\frac{1}{7} \\cdot 180^\\circ = \\frac{3}{7} \\cdot 180^\\circ = \\angle BAD$, which shows that the triangle $ABD$ is isosceles with $|BA| = |BD|$. At the same time, the angles in the triangle $BCD$ are $\\frac{1}{7} \\cdot 180^\\circ$, $\\frac{2}{7} \\cdot 180^\\circ$, $\\frac{4}{7} \\cdot 180^\\circ$, which are pairwise different, so the triangle $BCD$ is not isosceles.\n\n![](attached_image_2.png)\nFig. 7",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 29682,
    "subject": "Mathematics (Multi-modal)",
    "question": "An equilateral triangle with side length $3$ is divided into $9$ equilateral triangles with side length $1$. An integer from $1$ to $10$ is written into every point that is a vertex of a small triangle (colored vertices on the figure), such that all numbers are written exactly once. For every small triangle, the sum of the numbers in its three vertices is written inside it. Prove that at least three of those sums are greater than $11$.\n\n![](attached_image_1.png)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "In a triangle, which has $10$ at one vertex, the sum is at least $13$. If $10$ is not at one of the vertices of the large triangle, the number of triangles with sum greater than $12$ is at least $3$ and the problem is solved. If $10$ is at the vertex of the large triangle, then look, where is the number $9$. If $9$ does not lie at a vertex of the large triangle, then there are at least $3$ small triangles, with a sum of at least $12$, and the problem is solved. If $9$ is at a vertex of the large triangle, then look, where is the number $8$. If $8$ does not lie at the vertex of a large triangle, then it is at a vertex of at least $3$ small triangles. In at least two of them the sum is at least $12$, and as at most one of these can overlap with one of the triangles found earlier, the problem is solved. If $8$ is at a vertex of the large triangle, then either the sum in that triangle is at least $12$, meaning the problem is solved, or the other vertices in the triangle have numbers $1$ and $2$. In the last case, the numbers $3$, $4$ and $5$ are the smallest numbers whose positions are not set, but these give a sum of $12$. So, we can simply choose any triangle whose vertices we have not looked at yet as our third triangle (there are three of these triangles).",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 29683,
    "subject": "Mathematics (Multi-modal)",
    "question": "Jüri wishes to draw $n$ circles and any number of lines on the plane such that all the lines meet at one point, and for every two circles there exist two lines that touch both of these circles.\n\na)\nIs it possible for Jüri to solve this problem for any $n \\ge 2$?\n\nb)\nFor which natural numbers $n$ is it possible to solve this problem if in addition all the circles must have the same radius?",
    "options": [],
    "answer": "a) Yes, it is possible for any number at least two.\nb) It is possible exactly for n ≤ 4.",
    "solution": "a.\nJüri can draw two lines and draw any number of circles such that they touch both of the lines.\n\n![](attached_image_1.png)\nFig. 8\n\nb.\nAssume that Jüri has solved the problem for some $n$, where $n > 1$.\n\nLet $O$ be the intersection point of all the lines. Look at any circle $c$. From the premises of the problem we see that the circle $c$ touches two of the lines drawn by Jüri, which we call $k$ and $l$. But any one circle can only touch up to two lines drawn from one point. So, the circle $c$ does not have any more lines touching it. If $c'$ is any other circle drawn by Jüri, then the common tangents of $c$ and $c'$ can only be $k$ and $l$. Therefore, $k$ and $l$ are the common tangents of all the circles. Two lines divide the plane into four sectors, inside each can be only one circle with the previously set radius.\n\n![](attached_image_2.png)\nFig. 9\n\nSo, for $n > 4$ the problem has no solution but for $n \\le 4$, it obviously has.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 29684,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all pairs $(n, m)$ of positive integers such that the arithmetic and geometric means of $m$ and $n$ are different two-digit numbers consisting of the same digits.",
    "options": [],
    "answer": "(98, 32)",
    "solution": "Let $10a + b$ be the arithmetic mean of the given numbers, where $a$ and $b$ are decimal digits. Let $10a + b + x$ and $10a + b - x$ be the numbers we are searching for. Then, by the premises\n$$\n\\sqrt{(10a + b + x)(10a + b - x)} = 10b + a,\n$$\nwhich after squaring and simplifying gives $x^2 = 99(a^2 - b^2)$. So, $x^2$ is divisible by $99$, implying $x^2$ is divisible by $3$ and $11$. Since $3$ and $11$ are primes, $x$ itself is divisible by $3$ and $11$, and therefore by $33$. Denoting $x = 33z$, we get\n$$\n11z^2 = \\frac{99 \\cdot 11z^2}{99} = \\frac{x^2}{99} = a^2 - b^2 = (a+b)(a-b),\n$$\nfrom which we see, that the product $(a+b)(a-b)$ is divisible by $11$. Since $11$ is a prime, either $a+b$ or $a-b$ is divisible by $11$. Since $a-b \\neq 0$ and $a$ and $b$ are single-digit numbers, we must have $a+b = 11$. Therefore $a-b = z^2$. Since $a-b$ and $a+b$ are either both odd or both even, $z$ must be odd. So, $z=1$, since $z \\ge 3$ implies $x \\ge 99$, but $10a+b-x$ must be positive. So, $a=6$, $b=5$, $x=33$, and the corresponding pair is $(98, 32)$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 29685,
    "subject": "Mathematics (Multi-modal)",
    "question": "We say that two real numbers $r$ and $s$ are *close* if $|r - s| = 10^u$ for some integer $u$. Let $y = ax + b$ be a linear function, for which there exist close numbers $x_1$ and $x_2$ so that the corresponding $y_1$ and $y_2$ are also close. Prove that for any close numbers $x'_1$ and $x'_2$, the corresponding $y'_1$ and $y'_2$ are also close.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "From the premises we get $|x_1 - x_2| = 10^u$ and $|y_1 - y_2| = |(ax_1 + b) - (ax_2 + b)| = 10^v$ for some integers $u, v$. So,\n$$\n10^v = |(ax_1 + b) - (ax_2 + b)| = |a(x_1 - x_2)| = |a| \\cdot |x_1 - x_2| = |a| \\cdot 10^u,\n$$\nwhich gives $|a| = \\frac{10^v}{10^u} = 10^{v-u}$. Let $x'_1, x'_2$ be any close real numbers, $|x'_1 - x'_2| = 10^w$. Then $|y'_1 - y'_2| = |(ax'_1 + b) - (ax'_2 + b)| = |a(x'_1 - x'_2)| =$\n\n$|a| \\cdot |x'_1 - x'_2| = 10^{v-u} \\cdot 10^w = 10^{w+v-u}$. Since $u, v, w$ are integers, $w + v - u$ is also an integer, which shows that $y'_1, y'_2$ are close.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 29686,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABC$ be a triangle on the plane. The angle bisector from the vertex $A$ meets the side $BC$ at $P$, and the median from the vertex $B$ meets the side $AC$ at $M$. The lines $AB$ and $MP$ meet at the point $K$. Prove that if $\\frac{|PC|}{|BP|} = 2$, then $AP$ and $CK$ are perpendicular.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $K'$ be a point on the ray $AB$, such that $B$ is the midpoint of the line segment $AK'$ (Fig. 10). Then $CB$ is the median of the triangle $ACK'$. As $P$ divides this line segment in the ratio $2:1$, $P$ must be the centroid of the triangle $ACK'$. So, $K'M$, which is also a median of the triangle $ACK'$, must pass through the point $P$. Therefore, $K = K'$. So, $B$ is the midpoint of the line segment $AK$. As $AP$ passes through the point $P$, $AP$ is also a median of the triangle $ACK$. By the premises it is also an angle bisector. So, the triangle $ACK$ is isosceles with $|AC| = |AK|$ and $AP$ is its height. Therefore, $AP \\perp CK$.\n\n![](attached_image_1.png)\nFig. 10\nAs in solution 1, we show that $B$ is the midpoint of $AK$. So $BM$ joins the midpoints of sides in the triangle $ACK$ and therefore $BM \\parallel CK$. And the angle bisector divides the opposite side in the same ratio as the corresponding sides, so $\\frac{|AC|}{|AB|} = \\frac{|PC|}{|BP|} = 2$. As $M$ is the midpoint of $AC$, we have $|AB| = |AM|$. An angle bisector, drawn from the vertex opposite the base in an isosceles triangle, is also the height in that triangle, giving $AP \\perp BM$. It follows that $AP \\perp CK$.",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 29687,
    "subject": "Mathematics (Multi-modal)",
    "question": "An equilateral triangle with side length $3$ is divided into $9$ equilateral triangles with side length $1$. An integer from $1$ to $10$ is written into every point that is a vertex of a small triangle (colored vertices on the figure), such that all numbers are written exactly once. For every small triangle, the sum of the numbers in the three vertices is written inside it. Prove that there exist three small triangles such that the sum of the numbers inside them is at least $48$.\n\n![](attached_image_1.png)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Any three small triangles, from which no two have common vertices, take up nine of the ten numbers written into the vertices of the small triangles. So, the sum of the numbers inside those small triangles is $55 - a$, where $a$ is the number at the last vertex. Now it is sufficient to prove that we can choose the three small triangles, from which no two have common vertices, in four different ways, always leaving a different vertex out. So, in at least one case, the number at the last vertex is at most $7$, and the sum of the numbers in the three chosen triangles is at least $55 - 7 = 48$.\n\n![](attached_image_2.png)\nFig. 11\n![](attached_image_3.png)\nFig. 12\n![](attached_image_4.png)\nFig. 13\n![](attached_image_5.png)\nFig. 14\n\nIndeed, the three triangles can be chosen so that they leave uncovered the central number (Fig. 11) or one of the corner numbers (Fig. 12, 13, 14).\nLet $m$ be the number in the center of the large triangle. Then, when adding the sums of the three corner triangles we get the sum of all the numbers from $1$ to $10$, except $m$, so the sum of the corner triangles is $55 - m$. If $m \\le 7$, then the sum is at least $48$.\n\nIn the rest of the cases, consider any three triangles around the center point, such that no two of them share a side. Adding the numbers in them, we get the sum of all the numbers from $1$ to $10$, except the three numbers in the corners, while we add the center number three times. So the sum of those triangles is at least $21 + 3m$. If $m \\ge 9$, then the sum is at least $48$.\n\nThis leaves the case $m = 8$. The sum of the numbers in the triangles in the corners is at least $55 - 8 = 47$, so at least one of them contains a sum that is at least $16$. If none of the triangles contains a sum $17$ or greater, the number $16$ must occur in two different triangles. The sum of the numbers in any three triangles around the center point, chosen like above, is at least $21 + 3 \\cdot 8 = 45$. So, in both triples at least one of the triangles contains a sum of at least $15$, and since triangles sharing an edge cannot contain the same sum as the corresponding sums differ by exactly one term, at least one of the six triangles around the center point contains a sum of at least $16$. Thus we can pick the three desired triangles from among either one corner triangle and two central triangles or two corner triangle and one central triangle that contain the largest numbers.",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 29688,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all triples of positive integers $(x, y, z)$, for which $x \\cdot y! + 2y \\cdot x! = z!$.",
    "options": [],
    "answer": "(x, y, z) = (n, n+1, n+2) for all positive integers n, and (x, y, z) = (2, 1, 3)",
    "solution": "Since the left-hand side is greater than both $x!$ and $y!$, obviously $z > x$ and $z > y$. So, both sides of the equation are divisible by both $x!$ and $y!$. Therefore, $x \\cdot y!$ is divisible by $x!$, which means that $y!$ is divisible by $(x-1)!$, giving $y \\ge x-1$. Analogously, $2y \\cdot x!$ is divisible by $y!$, meaning $2 \\cdot x!$ is divisible by $(y-1)!$. The case $x=1, y=3$ is not a solution, the case $x > 1$ gives $2 \\cdot x! < (x+1)!$, which implies $x \\ge y-1$. This leaves us to look through the cases $-1 \\le y-x \\le 1$.\n* If $y = x - 1$, then the equation simplifies to $(2x-1) \\cdot x! = z!$. As $(x+1)(x+2) > 2x-1$, we have $2x-1 = x+1$ and $z = x+1$. This gives the solution $x=2, y=1, z=3$.\n\n* If $y = x$, the equation simplifies to $3x \\cdot x! = z!$. As $(x+1)(x+2) > 3x$, we have $3x = x + 1$, but this does not give integer solutions.\n* If $y = x + 1$, the equation simplifies to $(x^2 + 3x + 2) \\cdot x! = z!$ or $(x + 2)! = z!$. From here we get a family of solutions $x = n, y = n + 1, z = n + 2$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 29689,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all four-digit numbers, which after deleting any one digit turn into a three-digit number that is a divisor of the original number.",
    "options": [],
    "answer": "1100, 1200, 1500, 2200, 2400, 3300, 3600, 4400, 4800, 5500, 6600, 7700, 8800, 9900",
    "solution": "Let $\\overline{abcd}$ be such a number. Since $\\overline{abcd}$ is divisible by $\\overline{abc}$, we have $d = 0$. Since $\\overline{abcd} = \\overline{abc0}$ is divisible by $\\overline{abd} = \\overline{ab0}$, we have $c = 0$. Since $\\overline{abcd} = \\overline{ab00}$ is divisible by $\\overline{acd} = \\overline{a00}$ and by $\\overline{bcd} = \\overline{b00}$, the number $\\overline{ab}$ is divisible by $a$ and $b$. So $b = ax$ and $10a = by$ with integer $x$ and $y$. Therefore $10a = axy$, whence $xy = 10$.\n\nIf $x = 1, y = 10$, then $a = b$, which gives 9 possible numbers: 1100, 2200, 3300, 4400, 5500, 6600, 7700, 8800, 9900.\n\nIf $x = 2, y = 5$, then $2a = b$, which gives 4 possibilities: 1200, 2400, 3600, 4800.\n\nIf $x = 5, y = 2$, then $5a = b$, which gives 1 number: 1500.\n\nThe case $x = 10, y = 1$ is impossible, since $a$ and $b$ must be one-digit numbers.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 29690,
    "subject": "Mathematics (Multi-modal)",
    "question": "The teacher drew a $3 \\times 3$ table in Juku's exercise book and wrote a number in every position of the table. Then he gave Juku the following task.\n\n1. Turn the next page and draw a similar table. Write in the first row the numbers obtained by subtracting the numbers in the third row of the corresponding column from the numbers in the second row of the corresponding column in the previous table. Similarly, the numbers in the second and third row are obtained as differences of the third and the first, and the first and the second row.\n\n2. Turn the next page and draw a new table. Write in the first column the numbers obtained by subtracting the numbers in the third column from the numbers in the second column in the corresponding row in the previous table. Similarly, the numbers in the second and third column are obtained as differences of the third and the first, and the first and the second column.\n\nRepeat in turns steps 1 and 2 until you reach a table where all the numbers are zeroes. Juku has reached the end of the third page and has not yet reached the table with all zeroes in it. Prove that his task never ends.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "First note that after step 1 we get a table where the column sums of the table are $0$, and after step 2 we get a table where the row sums of the table are $0$. Suppose that after some steps we reach the table with all zeroes in it. By symmetry we can consider the case where we get this table after step 2. Then the table on the previous step was\n$$\n\\begin{array}{ccc}\na & a & a \\\\\nb & b & b \\\\\nc & c & c\n\\end{array}\n$$\nwhere at least one of the numbers $a$, $b$, $c$ is not zero. This table was obtained after step 1, hence $a + b + c = 0$. The table on the previous step was\n$$\n\\begin{array}{ccc}\nd & e & f \\\\\nd-c & e-c & f-c \\\\\nd+b & g+b & f+b\n\\end{array}\n$$\n$c + f - c = d + b + g + b + f + b = 0$. From the first equality it follows that $c = 0$, from the equality of the first and third expression $b = 0$, and since $a + b + c = 0$, we have $a = 0$, which contradicts the assumption that at least one of the numbers $a, b, c$ is not zero.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 29691,
    "subject": "Mathematics (Multi-modal)",
    "question": "Prove that for any positive integer $n$ the sum of the first $n$ primes is greater than $n^2$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "First notice that the $n$-th prime $p_n$ satisfies the inequality $p_n \\ge 2n - 1$. Indeed, the claim holds for the first prime $p_1 = 2$. Since all other primes are odd and there is exactly $n-1$ odd numbers between $2$ and $2n$, there are at most $n$ prime numbers less or equal to $2n-1$, hence $p_n \\ge 2n - 1$.\n\nNow consider the sum of the $n$ first primes $P = p_1 + p_2 + \\dots + p_n$. Since $p_k \\ge 2k - 1$ for any $k$, and additionally $p_1 = 2 > 1$, the sum $P$ is strictly greater than the sum of $n$ first odd numbers $S = 1 + 3 + \\dots + (2n-1) = (1^2 - 0^2) + (2^2 - 1^2) + \\dots + (n^2 - (n-1)^2) = n^2$. So $P > S = n^2$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 29692,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all triples $(a, b, c)$ of positive integers such that\n$$\na^{bc} + b^{ca} + c^{ab} = 3abc.\n$$",
    "options": [],
    "answer": "(1,1,1) and all permutations of (1,2,3)",
    "solution": "First assume $a \\ge 2$, $b \\ge 2$, $c \\ge 2$. W.l.o.g., let $c$ be the greatest among the three numbers. Then $a^{bc} + b^{ca} + c^{ab} \\ge a^4 + b^4 + c^4 > b^4 + c^4 \\ge 2b^2c^2 = 2b \\cdot c \\cdot bc > 3 \\cdot a \\cdot bc$. Thus there are no solutions in this case.\n\nIt remains to study triples that contain $1$. W.l.o.g., let $a=1$. The equation reduces to $1+b^c+c^b=3bc$. Assume $b \\ge 3$, $c \\ge 3$. W.l.o.g., $c \\ge b$, leading to $1+b^c+c^b \\ge 1+b^3+c^3 > c^3 \\ge 3 \\cdot b \\cdot c$. Thus there are no solutions in this case either. Now assume $b \\ge 2$, $c \\ge 2$ and one of the numbers is $2$. W.l.o.g. let $b=2$. The equation reduces to $1+2^c+c^2=6c$ which can be interpreted as a quadratic equation w.r.t. $c$ that leads to $c = 3 \\pm \\sqrt{9-(2^c+1)}$. Hence $8-2^c$ is a perfect square. The only candidates for this are $4$ and $0$ that give $c=2$ and $c=3$, respectively, but $c=2$ leads to contradiction (the above formula would give $c=1$ or $c=5$). The case $c=3$ gives the solution $(1,2,3)$ of the original equation. By symmetry, also $(1,3,2)$, $(2,1,3)$, $(2,3,1)$, $(3,1,2)$, $(3,2,1)$ are solutions. If one of the numbers $b$ and $c$ is $1$ then, w.l.o.g., $b=1$. The equation reduces to $1+1+c=3c$, whence $c=1$. This gives the trivial solution $(1,1,1)$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 29693,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABC$ be an acute triangle and $D$ an interior point of its side $AC$. We call a side of the triangle $ABD$ friendly, if the excircle of $ABD$ tangent to that side has its center on the circumcircle of $ABC$. Prove that there are exactly two friendly sides of $ABD$ if and only if $|BD| = |DC|$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $E$, $F$ and $G$ be the centers of excircles touching $BD$, $AD$ and $AB$ respectively, and let $\\omega$ be the circumcircle of $ABC$ (see Fig. 5). To prove the assertion of the problem, we will show that $F$ and $G$ cannot both lie on $\\omega$ and that $E \\in \\omega \\iff |BD| = |DC| \\iff F \\in \\omega$.\n\nAs $AF$ and $AG$ are bisectors of the two complementary angles of $BAD$, the point $A$ lies on the segment $FG$. Thus, only one of the rays $AF$ and $AG$ can cut the circle $\\omega$ again, and therefore only one of $F$ and $G$ can lie on $\\omega$.\n\nTo show that $E \\in \\omega$ iff $|BD| = |DC|$, we first note that $E$ lies on $\\omega$ iff $\\angle CAE = \\angle CBE$. Since $\\angle CAE = \\frac{1}{2}\\angle BAD = \\frac{1}{2}(\\pi - \\angle ADB - \\angle ABD) = \\frac{1}{2}(\\pi - (\\pi - 2\\angle BDE) - (\\pi - 2\\angle DBE)) = \\angle BDE + \\angle DBE - \\frac{\\pi}{2} = \\frac{\\pi}{2} - \\angle BED$, we see that $E \\in \\omega$ is equivalent to $\\angle CBE = \\frac{\\pi}{2} - \\angle BED$, i.e $BC$ and $DE$ being perpendicular. Since $DE$ is the bisector of $BDC$, this occurs iff $|BD| = |DC|$.\n\n![](attached_image_1.png)\nFig. 5\n\nIt remains to show that $|BD| = |DC|$ iff $F \\in \\omega$. Point $F$ lies on $\\omega$ iff $\\angle AFB = \\angle BCD$. Using the fact that $\\angle BAF = \\angle BAD + \\frac{1}{2}(\\pi - \\angle BAD) = \\frac{1}{2}(\\pi + \\angle BAD)$, we get $\\angle AFB = \\pi - \\angle ABF - \\angle BAF = \\frac{\\pi - \\angle ABD - \\angle BAD}{2} = \\frac{\\angle ADB}{2} = \\frac{\\angle BCD + \\angle CBD}{2}$. Thus $\\angle AFB = \\angle BCD$ is equivalent to $\\angle BCD = \\angle CBD$, i.e., $|BD| = |DC|$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 29694,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $k$ be a positive integer. Determine the largest number of snakes, consisting of four squares (see figure), which can be placed on a $(2k+1) \\times (2k+1)$ chessboard so that the snakes neither overlap nor stick out across the edges of the chessboard. The snakes can be turned and reflected.\n\n![](attached_image_1.png)",
    "options": [],
    "answer": "k^2",
    "solution": "First show that $k^2$ snakes can be placed on a $(2k+1) \\times (2k+1)$ chessboard. Divide the chessboard into strips of width $2$ (one strip of width $1$ remains). On any strip we can place $k$ snakes, one after another; so on $k$ strips, it is possible to place $k^2$ snakes.\n\nIt remains to prove that one can not place more than $k^2$ snakes on the chessboard. Write numbers $0, 1, 0, 1, \\ldots, 0$ in the odd rows, and numbers $2, 3, 2, 3, \\ldots, 2$ in the even rows. Notice that no matter how we place the snake on the board, it always covers numbers $0, 1, 2$ and $3$. Since all numbers $3$ are in the squares with even row and column numbers, there is exactly $k^2$ of them, hence there can be at most $k^2$ snakes.",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 29695,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find the minimum number of colours required to paint all points with integer coordinates in the plane in such a way that no two points which are exactly five units apart have the same color.",
    "options": [],
    "answer": "2",
    "solution": "Obviously at least 2 colors are necessary. Color all points $(x, y)$ with even sum of coordinates with one color and all other points with another color. All points that are at distance 5 from $(x, y)$ are $(x \\pm 4, y \\pm 3)$, $(x \\pm 3, y \\pm 4)$, $(x \\pm 5, y)$, $(x, y \\pm 5)$. In each case, the sum of coordinates has the parity different from that of $(x, y)$. Therefore they are colored differently from $(x, y)$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 29696,
    "subject": "Mathematics (Multi-modal)",
    "question": "A hiking club wants to hike around a lake along an exactly circular route. On the shoreline they determine two points, which are the most distant from each other, and start to walk along the circle, which has these two points as the endpoints of its diameter. Can they be sure that, independent of the shape of the lake, they do not have to swim across the lake on any part of their route?",
    "options": [],
    "answer": "No. For a lake shaped like an equilateral triangle, the circle with a side as diameter does not contain the lake, so the circular path would cross the water.",
    "solution": "Suppose the shape of the lake is an equilateral triangle. Then the two points which are the most distant from each other are two vertices of the triangle. The circle, which has these two points as the endpoints of its diameter, does not cover the whole triangle, because the distance of the third vertex from the center of the circle is $\\frac{\\sqrt{3}}{2}$ of the length of the side of the triangle, but the radius of the circle is only $\\frac{1}{2}$ of this length.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 29697,
    "subject": "Mathematics (Multi-modal)",
    "question": "Two circles $c$ and $c'$ with centers $O$ and $O'$ lie completely outside each other. Points $A$, $B$, and $C$ lie on the circle $c$ and points $A'$, $B'$, and $C'$ lie on the circle $c'$ so that segment $AB \\parallel A'B'$, $BC \\parallel B'C'$, and $\\angle ABC = \\angle A'B'C'$. The lines $AA'$, $BB'$, and $CC'$ are all different and intersect in one point $P$, which does not coincide with any of the vertices of the triangles $ABC$ or $A'B'C'$. Prove that $\\angle AOB = \\angle A'O'B'$.\n\n![](attached_image_1.png)\nFig. 1",
    "options": [],
    "answer": "Detailed solution",
    "solution": "The triangles $ABP$ and $A'B'P$ are similar, because their corresponding sides are parallel (Fig. 1). Hence $\\frac{|AB|}{|A'B'|} = \\frac{|BP|}{|B'P|}$. Likewise the triangles $BCP$ and $B'C'P$ are similar, hence $\\frac{|BC|}{|B'C'|} = \\frac{|BP|}{|B'P|}$. Thus $\\frac{|AB|}{|A'B'|} = \\frac{|BC|}{|B'C'|}$, and since $\\angle ABC = \\angle A'B'C'$, the triangles $ABC$ and $A'B'C'$ are also similar. From the equality of the angles $ACB$ and $A'C'B'$ the equality of the central angles $AOB$ and $A'O'B'$ now follows.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 29698,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $n$ be a positive integer and $a_1, \\dots, a_{2n}$ be real numbers in $\\left[-\\frac{1}{2}, \\frac{1}{2}\\right]$. Leaving out any one of the numbers, the sum of the remaining $2n-1$ numbers is always an integer. Prove that $a_1 = \\dots = a_{2n}$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Assume that there exist $a_i$ and $a_j$ which are not equal. Let $S = a_1 + \\dots + a_{2n}$. Since $S - a_i$ and $S - a_j$ are integers, their difference $(S - a_i) - (S - a_j) = a_j - a_i$ is also an integer. Since $a_j - a_i \\neq 0$, and they belong to $\\left[-\\frac{1}{2}, \\frac{1}{2}\\right]$, their difference can be only $\\pm 1$, this happens when $a_i$ and $a_j$ are $\\frac{1}{2}$ and $-\\frac{1}{2}$ in any order. Let $a_k$ be any of the given numbers. Since $(S - a_i) - (S - a_k) = a_k - a_i$ is an integer, $a_k$ must also be either $\\frac{1}{2}$ or $-\\frac{1}{2}$. Hence all numbers $a_i$ are either $\\frac{1}{2}$ or $-\\frac{1}{2}$. It follows that the sum of any two numbers $a_i$ and $a_j$ is an integer. As we have an even number of them, the sum of all the numbers $S$ is also an integer. But then $S - a_i = S \\pm \\frac{1}{2}$ cannot be an integer, a contradiction. Therefore all numbers $a_i$ must be equal.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 29699,
    "subject": "Mathematics (Multi-modal)",
    "question": "Is it possible that the perimeter of a triangle whose side lengths are integers, is divisible by the double of the longest side length?",
    "options": [],
    "answer": "No",
    "solution": "Let the side lengths of the triangle be integers $a$, $b$, $c$. Without loss of generality we may assume that $c \\ge a$ and $c \\ge b$. Suppose that the perimeter of the triangle $a+b+c$ is divisible by double of the longest side length $2c$. Since $0 < a+b+c \\le 3c < 2 \\cdot 2c$, the perimeter $a+b+c$ can be divisible by $2c$ only in the case when $a+b+c = 2c$. But then $a+b = c$, which violates the triangle inequality $a+b > c$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 29700,
    "subject": "Mathematics (Multi-modal)",
    "question": "For any positive integer $n$ let $a_n$ be the largest power of $2$ that divides $n$ (e.g. $a_{2011} = 1$, $a_{2012} = 4$). Prove that for any positive integers $i$ and $j$ with $i < j$, the sum $\\frac{1}{a_i} + \\frac{1}{a_{i+1}} + \\dots + \\frac{1}{a_j}$ is a fractional number.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "First prove that the largest power of $2$ among the numbers $a_i$, $a_{i+1}$, $\\dots$, $a_j$ is unique. Let $2^s$ be the largest of the numbers $a_i$, $a_{i+1}$, $\\dots$, $a_j$. If there were $k$ and $l$ with $i \\le k < l \\le j$ such that $a_k = a_l = 2^s$, then they must be of the form $k = 2^s u$ and $l = 2^s v$, where $u$ and $v$ are odd numbers. Since $k < l$, we have $u < v$ and $u+1 < v$. Since $u+1$ is even, the number $m = 2^s(u+1)$ has a divisor $2^{s+1}$, and $k < m < l$, which contradicts the choice of $s$. Thus the largest power of $2$ appears only once among the numbers $a_i$, $a_{i+1}$, $\\dots$, $a_j$.\n\nConverting the fractions to the common denominator, the fraction with the largest denominator gives $1$ in the numerator, all others give a positive power of $2$, i.e. an even number. Consequently the numerator is odd and cannot cancel with the denominator.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 29701,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $a$ be a real number, $0 \\le a \\le 1$. Prove that for any nonnegative integer $n$ the inequality $(n+1)a \\le n + a^{n+1}$ holds.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "The inequality is equivalent to the inequality $na - n \\le a^{n+1} - a$, or the inequality $n(a - 1) \\le a(a - 1)(a^{n-1} + a^{n-2} + \\dots + 1)$. If $a = 1$, then the inequality obviously holds. If $a < 1$ then $a - 1 < 0$, and dividing both sides of the inequality by $a - 1$ we get an equivalent inequality $n \\ge a(a^{n-1} + a^{n-2} + \\dots + 1)$, or $n \\ge a^n + a^{n-1} + \\dots + a$. Since $a < 1$, in the last sum all terms are less than 1, hence the sum does not exceed $n$.\nIf $n = 0$ then the inequality is $a \\le a$, which obviously holds. Suppose that the inequality holds for $n = k$ and prove that it then holds for $n = k + 1$ as well. From the inequality $(k+1)a \\le k + a^{k+1}$ we get the inequality $(k+2)a \\le k + 1 + a^{k+2}$ by adding the inequality $a \\le 1 + a^{k+2} - a^{k+1}$. The last inequality is equivalent to the inequality $(1-a)(1-a^{k+1}) \\ge 0$, which obviously holds, because both factors are nonnegative.\nIf $a = 0$, then the inequality is $0 \\le n$, which obviously holds. If $a > 0$, then AM-GM gives\n$$\n\\frac{1 + \\dots + 1 + a^{n+1}}{n+1} \\ge \\sqrt[n+1]{1 \\dots \\cdot 1 \\cdot a^{n+1}},\n$$\nwhich is obviously equivalent with the original inequality.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 29702,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABC$ be a triangle with median $AK$. Let $O$ be the circumcenter of the triangle $ABK$.\n\na) Prove that if $O$ lies on a midline of the triangle $ABC$, but does not coincide with its endpoints, then $ABC$ is a right triangle.\n\nb) Is the statement still true if $O$ can coincide with an endpoint of the midsegment?",
    "options": [],
    "answer": "a) The triangle is right. b) No; for an equilateral triangle the circumcenter coincides with a midsegment endpoint but the triangle is not right.",
    "solution": "![](attached_image_1.png)\n![](attached_image_2.png)\n![](attached_image_3.png)\nFig. 2\nFig. 3\nFig. 4\n\nSolution:\na) Let $L$ and $M$ be the midpoints of the sides $CA$ and $AB$, respectively. If $O$ lies on the segment $KM$ (Fig. 2), then the segment $KM$ and the perpendicular bisector of $AB$ have two different common points $O$ and $M$, hence $KM$ is the perpendicular bisector of $AB$. Since $KM$ is parallel to $AC$ and is perpendicular to $AB$, the angle at vertex $A$ must be right. If $O$ lies on the segment $LM$ (Fig. 3), then we get similarly that the angle at vertex $B$ must be right. If $O$ lies on the segment $KL$ (Fig. 4), then on one hand $\\angle ABC$ is acute, because $OK$ and $MB$ are perpendicular to $MO$, the perpendicular bisector of $AB$, and $|OK| < |LK| = |MB|$. On the other hand, $\\angle ABK$ must be obtuse, since the circumcenter $O$ of the triangle $ABK$ lies outside of the triangle, a contradiction. Thus this case is not possible.\n\nb) If the triangle $ABC$ is equilateral, then the median $AK$ is also the altitude and $ABK$ is a right triangle with the hypotenuse $AB$. The circumcenter $O$ of the last triangle is the midpoint of $AB$, i.e. an endpoint of a midsegment of the triangle, but $ABC$ is not a right triangle.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 29703,
    "subject": "Mathematics (Multi-modal)",
    "question": "Prove that for any positive integer $k$ there exist $k$ pairwise distinct integers for which the sum of their squares equals the sum of their cubes.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "For any integer $m > 1$ the numbers $2m^2 + 1$, $m(2m^2 + 1)$, $-m(2m^2 + 1)$ satisfy the conditions of the problem, because they are pairwise different and\n$$\n\\begin{aligned}\n& (2m^2 + 1)^2 + (m(2m^2 + 1))^2 + (-m(2m^2 + 1))^2 \\\\\n&= (1 + m^2 + m^2) \\cdot (2m^2 + 1)^2 = (2m^2 + 1)^3 = (1 + m^3 - m^3) \\cdot (2m^2 + 1)^3 \\\\\n&= (2m^2 + 1)^3 + (m(2m^2 + 1))^3 + (-m(2m^2 + 1))^3.\n\\end{aligned}\n$$\nWith $m$ growing, the numbers in these triples get arbitrarily large, hence for any set of these triples one can find a new triple, where all numbers are larger than the ones already used.\nAny positive integer $k$ can be written as $k = 3q + r$ with $0 \\le r < 3$. Choose $q$ triples as above so that the numbers in them do not coincide. If $r = 1$, then add $0$, and if $r = 2$, then add $0$ and $1$. Since for each group the sum of the squares of the numbers equals the sum of the cubes of the numbers, the same property holds for the whole set.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 29704,
    "subject": "Mathematics (Multi-modal)",
    "question": "For a given positive integer $n$ one has to choose positive integers $a_0, a_1, \\dots$ so that the following conditions hold:\n(1) $a_i = a_{i+n}$ for any $i$;\n(2) $a_i$ is not divisible by $n$ for any $i$;\n(3) $a_{i+a_i}$ is divisible by $a_i$ for any $i$.\n\nFor which positive integers $n > 1$ is this possible only if the numbers $a_0, a_1, \\dots$ are all equal?",
    "options": [],
    "answer": "n is prime",
    "solution": "Let $n$ be a prime. By condition (1) the sequence $a_0, a_1, \\dots$ contains only finitely many different numbers. If $a_m$ is maximal of them, then by condition (3) $a_{m+a_m}$ must also be maximal. Let us prove that if $a_m$ is maximal of the numbers, then $a_{m+k \\cdot a_m}$ is also maximal for any $k \\ge 0$. This holds for $k=0$. If the claim holds for $k$, then $a_{m+(k+1) \\cdot a_m} = a_{m+k \\cdot a_m + a_m} = a_{m+k \\cdot a_m + a_{m+k \\cdot a_m}} = a_{m+k \\cdot a_m} = a_m$. This proves the claim. By condition (2) $a_m$ is not divisible by $n$. Since $n$ is prime, the numbers $a_m$ and $n$ are relatively prime. Hence among the numbers $m+k \\cdot a_m$, where $0 \\le k < n$, there is one in each congruence class modulo $n$. Hence all members of the sequence are maximal, i.e. they are equal.\n\nSuppose $n$ is a composite number; let $m$ be its divisor with $1 < m < n$. For any $k < m$ choose $a_k = m + k \\cdot n$ and continue the sequence with period $m$. Condition (1) holds, since $n$ is a multiple of $m$. Condition (2) holds, since all members of the sequence are congruent to $m$ modulo $n$. For the condition (3) notice that all members of the sequence are divisible by $m$. Hence $i$ and $i+a_i$ are always congruent modulo $m$, therefore $a_i = a_{i+a_i}$. At the same time not all the numbers are equal.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 29705,
    "subject": "Mathematics (Multi-modal)",
    "question": "In a cyclic quadrilateral $ABCD$ we have $|AD| > |BC|$ and the vertices $C$ and $D$ lie on the shorter arc $AB$ of the circumcircle. Rays $AD$ and $BC$ intersect at point $K$, diagonals $AC$ and $BD$ intersect at point $P$. Line $KP$ intersects the side $AB$ at point $L$. Prove that $\\angle ALK$ is acute.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "From the properties of cyclic quadrilaterals we get $\\angle KAB = \\angle KC'D$ and $\\angle KBA = \\angle KDC$. Let $A', B', K'$ be the feet of the altitudes of the triangle $ABK$ drawn from the vertices $A, B, K$, respectively, and let $H$ be the orthocenter of the triangle $ABK$ (Fig. 18). The points $A, B, A', B'$ lie on a common circle, hence $\\angle KA'B' = \\angle KAB$ if $A' \\neq B'$. Therefore $A'$ and $B'$ lie on a line parallel to $CD$. Denote this line by $A'B'$ (even in the case $A' = B' = K$).\n\n![](attached_image_1.png)\nFig. 18\n![](attached_image_2.png)\nFig. 19\n\nLet $d(X, l)$ be the distance of point $X$ from line $l$, and let $S_\\Delta$ be the area of triangle $\\Delta$. By two angles, $\\angle ACK \\sim \\angle BDK$ and $\\angle PAD \\sim \\angle PBC$, whence\n$$\n\\frac{|AK|}{|BK|} = \\frac{|AC|}{|BD|} \\text{ and } \\frac{|AP|}{|BP|} = \\frac{|AD|}{|BC|}. \\text{ At the same time}\n$$\n$$\n\\begin{aligned}\n\\frac{d(A, CD)}{d(B, CD)} &= \\frac{S_{\\triangle ACD}}{S_{\\triangle BCD}} = \\frac{|AC| \\cdot |AD| \\cdot \\sin \\angle CAD}{|BD| \\cdot |BC| \\cdot \\sin \\angle CBD} = \\frac{|AC| \\cdot |AD|}{|BD| \\cdot |BC|}, \\\\\n\\frac{d(A, KP)}{d(B, KP)} &= \\frac{S_{\\triangle AKP}}{S_{\\triangle BKP}} = \\frac{|AK| \\cdot |AP| \\cdot \\sin \\angle KAP}{|BK| \\cdot |BP| \\cdot \\sin \\angle KBP} = \\frac{|AK| \\cdot |AP|}{|BK| \\cdot |BP|}.\n\\end{aligned}\n$$\nTherefore\n$$\n\\frac{|AL|}{|LB|} = \\frac{d(A, KP)}{d(B, KP)} = \\frac{d(A, CD)}{d(B, CD)}.\n$$\n\n$$\n\\frac{|AK'|}{|K'B|} = \\frac{d(A, KH)}{d(B, KH)} = \\frac{d(A, A'B')}{d(B, A'B')}.\n$$\nThis equality holds also in the special case $A' = B' = K$. Indeed, let the projections of points $A$ and $B$ to the line $A'B'$ be $X$ and $Y$ correspondingly (Fig. 19), then $\\angle AKX = \\angle KDC = \\angle KBA = \\angle AKK'$, $\\angle BKY = \\angle KCD = \\angle KAB = \\angle BKK'$, whence $\\triangle AKX \\sim \\triangle AKK'$ and $\\triangle BKY \\sim \\triangle BKK'$. It follows that $|AK'| = |AX|$, $|BK'| = |BY|$ and $\\frac{|AK'|}{|K'B|} = \\frac{d(A,A'B')}{d(B,A'B')}.$\n\nConsidering instead of the cyclic quadrilateral $ABCD$ the quadrilateral determined by points $A, B, A', B'$, and instead of $P$ and $L$ the points $H$ and $K'$ correspondingly, we get similarly that\n$$\n\\frac{|AK'|}{|K'B|} = \\frac{d(A, KH)}{d(B, KH)} = \\frac{d(A, A'B')}{d(B, A'B')}.\n$$\n\nSince $C$ and $D$ lie on the shorter arc $AB$, we have $\\angle BCA = \\angle BDA > \\frac{\\pi}{2}$. Thus the line $A'B'$ is farther from the points $A$ and $B$ than the line $CD$. Since $|AD| > |BC|$, we have $\\angle ABD > \\angle CAB$ and also $\\angle KBA > \\angle KAB$, which implies $|KA| > |KB|$. Hence $d(A, CD) + d(K, CD) > d(B, CD) + d(K, CD)$, or $d(A, CD) > d(B, CD)$. All together\n\n![](attached_image_3.png)\nFig. 20\n\n$$\n\\frac{|AL|}{|LB|} = \\frac{d(A, CD)}{d(B, CD)} > \\frac{d(A, A'B')}{d(B, A'B')} = \\frac{|AK'|}{|K'B|}.\n$$\nHence $L$ lies farther from $A$ than $K'$ on the segment $AB$, therefore $\\angle ALK < \\angle AK'K = \\frac{\\pi}{2}$, i.e. $\\angle ALK$ is acute.\nDenote $\\angle KAB = \\angle KCD = \\alpha$, $\\angle KBA = \\angle KDC = \\beta$, $\\angle KAC = \\angle KBD = \\delta$ and $\\angle ALK = \\xi$ (Fig. 20). Then $\\angle KDB = \\alpha + \\beta - \\delta = \\angle KCA$. The condition $|AD| > |BC|$ is equivalent to $\\beta > \\alpha$, and points $C$ and $D$ being located in the shorter arc $AB$ is equivalent to the inequality $\\alpha + \\beta - \\delta < \\frac{\\pi}{2}$. In triangle $KCD$ we get $\\frac{|KD|}{|KC|} = \\frac{\\sin \\alpha}{\\sin \\beta}$. From triangles $KDP$ and $KCP$ we obtain $\\frac{|KP|}{\\sin(\\alpha+\\beta-\\delta)} = \\frac{|KD|}{\\sin(\\xi-(\\beta-\\delta))}$, $\\frac{|KP|}{\\sin(\\alpha+\\beta-\\delta)} = \\frac{|KC|}{\\sin(\\xi+(\\alpha-\\delta))}$, respectively. Consequently, $\\frac{|KD|}{|KC|} = \\frac{\\sin(\\xi-(\\beta-\\delta))}{\\sin(\\xi+(\\alpha-\\delta))}$. Expressions of $\\frac{|KD|}{|KC|}$ together yield\n$$\n\\sin \\zeta (\\sin \\beta \\cos(\\beta - \\delta) - \\sin \\alpha \\cos(\\alpha - \\delta)) = \\cos \\zeta (\\sin \\beta \\sin(\\beta - \\delta) + \\sin \\alpha \\sin(\\alpha - \\delta)). \\quad (1)\n$$\nClearly $\\sin \\zeta > 0$ and $\\sin \\beta \\sin(\\beta - \\delta) + \\sin \\alpha \\sin(\\alpha - \\delta) > 0$ because $\\zeta < \\pi$ and $\\beta > \\delta$, $\\alpha > \\delta$. By the formula $\\sin x \\cos y = \\frac{1}{2}(\\sin(x + y) + \\sin(x - y))$, we get $\\sin \\beta \\cos(\\beta - \\delta) - \\sin \\alpha \\cos(\\alpha - \\delta) = \\frac{1}{2}(\\sin(2\\beta - \\delta) - \\sin(2\\alpha - \\delta))$. Now $\\alpha + \\beta - \\delta < \\frac{\\pi}{2}$ implies $(2\\beta - \\delta) + (2\\alpha - \\delta) < \\pi$, i.e., there exists a triangle whose two angles are $2\\beta - \\delta$ and $2\\alpha - \\delta$. But $2\\beta - \\delta > 2\\alpha - \\delta$ since $\\beta > \\alpha$, therefore the law of sines in that triangle implies $\\sin(2\\beta - \\delta) > \\sin(2\\alpha - \\delta)$ (the larger the angle, the larger its opposite side in a triangle). Hence the second factor in the l.h.s. of equation (1) is positive. Altogether, we obtain $\\cos \\zeta > 0$, whence $\\zeta < \\frac{\\pi}{2}$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 29706,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABC$ be a triangle where $|AB| = |AC|$. Points $P$ and $Q$ are different from the vertices of the triangle and lie on the sides $AB$ and $AC$, respectively. Prove that the circumcircle of the triangle $APQ$ passes through the circumcenter of $ABC$ if and only if $|AP| = |CQ|$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Without loss of generality, we can assume that $|AP| \\le |AQ|$. Let $O$ be the circumcenter of $ABC$. Let $R$ be the intersection point of the bisector of $\\angle BAC$ with the circumcircle of the triangle $PAQ$ — we then have $|RB| = |RC|$ (Fig. 21).\n\nAlso, $\\angle APR = 180^\\circ - \\angle AQR = \\angle CQR$ and $|RP| = |RQ|$ (since $\\angle RAP = \\angle RAQ$). So, $|AP| = |CQ| \\iff \\triangle APR \\cong \\triangle CQR \\iff |RA| = |RC| \\iff R = O$ (where $|RA| = |RC| \\Rightarrow \\triangle APR \\cong \\triangle CQR$ by two sides and obtuse angle).\n\n![](attached_image_1.png)\nFig. 21",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 29707,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $x$, $y$, $z$ be positive real numbers whose sum is $2012$. Find the maximum value of\n$$\n\\frac{(x^2 + y^2 + z^2)(x^3 + y^3 + z^3)}{(x^4 + y^4 + z^4)}\n$$",
    "options": [],
    "answer": "2012",
    "solution": "If $x = y = z = \\frac{2012}{3}$, then\n$$\n\\frac{(x^2 + y^2 + z^2)(x^3 + y^3 + z^3)}{(x^4 + y^4 + z^4)} = 2012.\n$$\nNow we prove that for all $x$, $y$, $z$ satisfying the premises we have\n$$\n\\frac{(x^2 + y^2 + z^2)(x^3 + y^3 + z^3)}{(x^4 + y^4 + z^4)} \\le 2012\n$$\nIt suffices to show that $(x^2 + y^2 + z^2)(x^3 + y^3 + z^3) \\le 2012(x^4 + y^4 + z^4)$, or $(x^2 + y^2 + z^2)(x^3 + y^3 + z^3) \\le (x + y + z)(x^4 + y^4 + z^4)$. Multiplying out, simplifying and rearranging the terms gives $xy(x - y)(x^2 - y^2) + xz(x - z)(x^2 - z^2) + yz(y - z)(y^2 - z^2) \\ge 0$. Since the differences in the brackets in every product have equal signs, the products are non-negative, showing that the necessary inequality holds.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 29708,
    "subject": "Mathematics (Multi-modal)",
    "question": "On an $m \\times m$ board, at the midpoints of the unit squares there are some ants. At the time $0$ each ant starts moving with speed $1$ parallel to some edge of the board until it meets an ant moving in the opposite direction or until it reaches the edge of the board. When two ants moving in the opposite direction meet each other, both turn $90^\\circ$ clockwise and continue moving parallel to another edge of the board. Upon reaching the edge of the board the ant falls off the board.\n\na) Prove that eventually all the ants will have fallen off the board.\n\nb) Find the latest possible moment for the last ant to fall off the board.",
    "options": [],
    "answer": "3m/2 - 1",
    "solution": "Let the lower left corner of the board be the origin. Divide the units of time and space by $2$; then the squares are of dimensions $2 \\times 2$, the coordinates of the midpoints of the squares are odd positive integers, and the speed of the ants is still $1$.\n\nWe prove by induction that at integer time moments the coordinates of the ants are integers and the sum of the coordinates for any fixed ant has the same parity as the time moment. In addition, the ants can meet only at integer time moments. At time $t=0$ all coordinates of the ants are odd, so their sum is even. Suppose that at an integer time moment $t=k$ the coordinates of the ants are integers and the sum of the coordinates for any fixed ant has the same parity as the time moment. If two of the ants were to meet each other within the next time unit, they have to move toward each other from time $t=k$, hence one of their coordinates must be the same. Since the parity of the sum of their coordinates was the same at time $t=k$, another of their coordinates had to differ by at least $2$. Hence they cannot meet before time $t=k+1$. Between time moments $t=k$ and $t=k+1$ every ant has changed only one of its coordinates by $1$, hence at time $t=k+1$ the parity of the sum of the coordinates is again the same as the parity of the time moment.\n\nNext we will prove by induction that for any point with integer coordinates $(x, y)$ there are no collisions at this point after the time moment $t = x + y - 2$. For $x = y = 1$ this is obviously true, since there are no collisions in the middle of the lower left square (otherwise one of the ants has to arrive to this point from the edge of the board). Let $(x, y)$ be arbitrary and suppose that the claim holds for all points with the sum of the coordinates less than $x + y$. Suppose that a collision takes place at point $(x, y)$ at time $t$. One of the participants had to arrive from a point, where one of the coordinates was smaller; w.l.o.g. we can assume that this was the $x$-coordinate. If this ant has not collided with anyone before, then $t \\le x - 1 \\le x + y - 2$. If the last collision of this ant occurred at time $t' < t$, then the coordinates of the last collision were $(x - (t - t'), y)$. By the induction assumption $t' \\le x - (t - t') + y - 2$, hence $t \\le x + y - 2$.\n\nBy symmetry the claim holds when another corner is chosen as the origin. Let the last collision of a particular ant occur at the point $(x, y)$, where the coordinates are taken with respect to the nearest corner. W.l.o.g., we can assume $x \\le y$. The time from the last collision to the falling off the edge of the ants participating in the collision is at most $2m - x$, hence the time elapsed from the start is at most $x + y - 2 + 2m - x \\le 3m - 2$. By this time all ants have fallen off the edge. With respect to the original units the maximal time is $\\frac{3}{2}m - 1$.\n\nFor any $m$ the maximal time can be achieved, if in the beginning there are $2$ ants at the adjoining corners of the board moving toward each other. At the moment $t = \\frac{m-1}{2}$ the pair collides and one of the ants starts moving toward the center, falling off the board at time $t = \\frac{3}{2}m - 1$.\nPart a) can also be solved as follows. For each ant consider the distance to the edge in the direction of its motion. After an ant falls this distance will remain $0$. Observe that as long as an ant moves without collision, this distance decreases with speed $1$.\n\nConsider now the sum of all such distances. When a collision happens, the sum of the distances of the two corresponding ants is $m$, both right before and right after the collision. Thus as long as there are ants left on the board, the total sum decreases with the speed of at least $1$. Since in the beginning this sum is a finite number, after some time this sum will become $0$ and thus all ants will have fallen off the board.",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 29709,
    "subject": "Mathematics (Multi-modal)",
    "question": "In the product\n$$\n\\left(1 + \\frac{1}{1}\\right) \\cdot \\left(1 + \\frac{1}{3}\\right) \\cdot \\left(1 + \\frac{1}{5}\\right) \\cdots \\left(1 + \\frac{1}{2n-1}\\right)\n$$\nthe denominators of the fractions are all odd numbers from $1$ to $(2n - 1)$. Is it possible to choose a natural number $n > 1$ such that this product would evaluate to an integer?",
    "options": [],
    "answer": "No",
    "solution": "By manipulating the given product we get\n$$\n\\begin{aligned} \\left(1 + \\frac{1}{1}\\right) \\cdot \\left(1 + \\frac{1}{3}\\right) \\cdot \\left(1 + \\frac{1}{5}\\right) \\cdots \\left(1 + \\frac{1}{2n-1}\\right) &= \\frac{2}{1} \\cdot \\frac{4}{3} \\cdot \\frac{6}{5} \\cdots \\frac{2n}{2n-1} \\\\ &= \\frac{2 \\cdot 4 \\cdot 6 \\cdots 2n}{1 \\cdot 3 \\cdot 5 \\cdots (2n-1)}. \\end{aligned}\n$$\nFor it to be an integer, the number $2 \\cdot 4 \\cdot 6 \\cdots 2n$ should be divisible by $1 \\cdot 3 \\cdot 5 \\cdots (2n - 1)$. But since\n$$\n2 \\cdot 4 \\cdot 6 \\cdots 2n = (2 \\cdot 1) \\cdot (2 \\cdot 2) \\cdot (2 \\cdot 3) \\cdots (2 \\cdot n) = 2^n \\cdot (1 \\cdot 2 \\cdot 3 \\cdots n)\n$$\nand number $1 \\cdot 3 \\cdot 5 \\cdots (2n - 1)$ is odd, number $1 \\cdot 2 \\cdot 3 \\cdots n$ should be divisible by $1 \\cdot 3 \\cdot 5 \\cdots (2n - 1)$. In case of $n > 1$ this is impossible, because $1 \\cdot 2 \\cdot 3 \\cdots n < 1 \\cdot 3 \\cdot 5 \\cdots (2n - 1)$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 29710,
    "subject": "Mathematics (Multi-modal)",
    "question": "Consider hexagons whose internal angles are all equal.\n(i) Prove that for any such hexagon the sum of the lengths of any two neighbouring sides is equal to the sum of the lengths of their opposite sides.\n(ii) Does there exist such a hexagon with side lengths 1, 2, 3, 4, 5 and 6 in some order?",
    "options": [],
    "answer": "Yes; for example a hexagon with all interior angles equal and side lengths in order 1, 4, 5, 2, 3, 6 exists.",
    "solution": "(i) Let the hexagon be $ABCDEF$. It suffices to show that $|AB| + |BC| = |DE| + |EF|$.\nLet $K$ be the intersection point of rays $FA$ and $CB$ and let $L$ be the intersection point of rays $FE$ and $CD$ (Fig. 8). The size of every internal angle of the hexagon is $120^\\circ$, whence triangles $KAB$ and $LDE$ are equilateral. The quadrilateral $FKCL$ is a parallelogram since its opposite sides are parallel. This implies $|KC| = |LF|$ or $|KB| + |BC| = |LE| + |EF|$, which together with $|KB| = |AB|$ and $|LE| = |DE|$ implies the desired equality $|AB| + |BC| = |DE| + |EF|$.\n\n(ii) Take a parallelogram with side lengths 7 and 5 and internal angles $60^\\circ$ and $120^\\circ$, and cut off equilateral triangles with side lengths 1 and 2 at its acute angles. This gives rise to a hexagon with all internal angles having size $120^\\circ$ and side lengths 1, 4, 5, 2, 3, 6 (Fig. 9).\n\n![](attached_image_1.png)\nFigure 8\n![](attached_image_2.png)\nFigure 9\n(i) Note that the external angles of the hexagon have size $60^\\circ$. Any two opposite sides of the hexagon are parallel, as they are separated by exactly three external angles.\nConsider a line $s$ perpendicular to opposite sides $CD$ and $FA$ of the hexagon $ABCDEF$ (Fig. 10). As all other sides form the same angle $30^\\circ$ with line $s$, the lengths of these sides are proportional to the lengths of the projections of the sides to line $s$. The sum of the lengths of the projections of sides $AB$ and $BC$ is equal to the distance between the parallel lines $CD$ and $FA$ and the same holds also for the opposite sides $DE$ and $EF$. Therefore the sum of the lengths of the projections of sides $AB$ and $BC$ is equal to that of sides $DE$ and $EF$, whence the sums of the lengths of the sides are equal as well.\n\n![](attached_image_3.png)\nFigure 10\n\n(ii) Figure 11 shows a hexagon in a triangular grid with distance between neighbouring nodes being 1. The side lengths of the hexagon are 1, 4, 5, 2, 3 and 6.\n\n![](attached_image_4.png)\nFigure 11\nConsider two types of transformations on hexagons that maintain the property that all internal angles are of the same size.\n(1) Prolonging two opposite sides by the same quantity $x$ (Fig. 12); this causes the side lengths to change according to the template\n$$\n(a, b, c, d, e, f) \\longleftrightarrow (a+x, b, c, d+x, e, f).\n$$\n(2) Prolonging the two neighbouring sides of one particular side by the same quantity $x$ (Fig. 13); this causes the side lengths to change according to the template\n$$\n(a, b, c, d, e, f) \\longleftrightarrow (a+x, b-x, c+x, d, e, f)\n$$\nA straightforward check shows that both transformations maintain the desired property, no matter of in which direction the transformations are applied.\n\n(i) As the internal angles of all regular hexagons are equal, it suffices to show that an arbitrary hexagon with all internal angles equal can be turned into a regular hexagon by a finite sequence of the transformations above.\nIndeed, let the side lengths of a given hexagon with all internal angles equal be $(a, b, c, d, e, f)$. Assume w.l.o.g. that $d \\ge a$ and $f \\ge c$. Choose a quantity $s$ such that $s \\ge \\max(a, b, c)$; by applying the transformation (1) thrice, we can obtain a hexagon with three consecutive sides having the same length:\n$$\n(a, b, c, d, e, f) \\xrightarrow{(1)} (s, b, c, d', e, f) \\xrightarrow{(1)} (s, s, c, d', e', f) \\xrightarrow{(1)} (s, s, s, d', e', f').\n$$\nBy the assumption made above we have $d' \\ge s$ and $f' \\ge s$. W.l.o.g., assume also $d' \\le f'$. The transformation\n$$\n(s, s, s, d', e', f') \\xrightarrow{(2)} (s, s, s, s, e'', f'')\n$$\nleads to a hexagon with four consecutive sides of equal length. But this must be regular since all of its internal angles are equal (Fig. 14).\n\n(ii) Such a hexagon can be obtained from a regular hexagon with side length 1 by the following transformations:\n$$\n(1, 1, 1, 1, 1, 1) \\xrightarrow{(1)} (1, 4, 1, 1, 4, 1) \\xrightarrow{(1)} (1, 4, 5, 1, 4, 5) \\xrightarrow{(2)} (1, 4, 5, 2, 3, 6).\n$$\n\n![](attached_image_5.png)\nFigure 12\n![](attached_image_6.png)\nFigure 13\n![](attached_image_7.png)\nFigure 14",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 29711,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $m$ be a positive integer. Prove that if Mari writes at least $m+3$ numbers on the board, then Jüri can choose 4 of those such that the sum of some two of those and the sum of the other two give the same remainder when divided by $m$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "As Mari writes down $m+3$ numbers and there are only $m$ different remainders when dividing by $m$, there must be two that give equal remainders when divided by $m$; let those numbers be $a$ and $b$. The rest of the $m+1$ include two that also give equal remainders when divided by $m$; let those be $c$ and $d$. Now $a+c$ and $b+d$ give the same remainder when divided by $m$, thus Jüri can choose the numbers $a$, $b$, $c$ and $d$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 29712,
    "subject": "Mathematics (Multi-modal)",
    "question": "A $(2k+1) \\times (2k+1)$ table, where $k$ is a positive integer, contains one real number in each entry, where these numbers are pairwise different. After each row, one writes the median of the row, i.e., the number occurring in this row such that the row contains the same amount of numbers less than it and greater than it. Let $m$ be the median of the column of medians. Prove that more than a quarter of the numbers initially in the table are less than $m$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Each row contains $k$ numbers less than the median and $k$ numbers greater than the median. Thus $k+1$ numbers in each row do not exceed the median of that row. In rows whose median does not exceed $m$, these $k+1$ numbers do not exceed $m$ either. There are $k+1$ such rows. Consequently, there are at least $(k+1)^2$ numbers in the table that do not exceed $m$. Only one of them is equal to $m$, whence $(k+1)^2 - 1 = k^2 + 2k$ numbers are less than $m$. As $k$ is positive by assumption, we have $1 < 4k$ and $4k+1 < 4m$. Now $\\frac{k^2+2k}{(2k+1)^2} = \\frac{k^2+2k}{4k^2+4k+1} > \\frac{k^2+2k}{4k^2+8k} = \\frac{1}{4}$ and we are done.",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 29713,
    "subject": "Mathematics (Multi-modal)",
    "question": "How many positive integers $n$ are there for which $2014 \\cdot n$ is divisible by $2014 + n$?",
    "options": [],
    "answer": "13",
    "solution": "If $a = 1, d = 2014$, then $a+x \\mid 2014$. As $a+x > 1$, there are 7 possibilities.\nIf $a = 2, d = 1007$, then $a+x \\mid 1007$. As $a+x > 2$, there are 3 possibilities.\nIf $a = 19, d = 106$, then $a+x \\mid 106$. As $a+x > 19$, there are 2 possibilities.\nIf $a = 38, d = 53$, then $a+x \\mid 53$. As $a+x > 38$, there is 1 possibility.\nThose add up to 13 possibilities.\nAs the number $2014 + n$ is always a divisor of $2014 \\cdot (2014 + n) = 2014^2 + 2014n$, the number $2014 + n$ is a divisor of $2014n$ if and only if $2014 + n$ is a divisor of $2014^2$. The canonical representation $2014^2 = 2^2 \\cdot 19^2 \\cdot 53^2$ shows that the number of divisors of $2014^2$ is $(2+1) \\cdot (2+1) \\cdot (2+1) = 27$. The middle one of them is 2014 and 13 divisors are expressible in a form $2014 + n$ for some positive integer $n$. Thus there are 13 possibilities for $n$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 29714,
    "subject": "Mathematics (Multi-modal)",
    "question": "For which natural numbers $n \\ge 3$ is it possible to cut a regular $n$-gon into smaller pieces with regular polygonal shape? (The pieces may have different number of sides.)",
    "options": [],
    "answer": "3, 4, 6, 12",
    "solution": "**Answer:** 3, 4, 6, 12.\n\nA regular triangle can be partitioned into four regular triangles of equal size (Fig. 23), a regular quadrilateral can be partitioned into four regular quadrilaterals with equal size (Fig. 24) and a regular hexagon can be partitioned into six regular triangles of equal size (Fig. 25). By building alternately equilateral triangles and squares onto the sides of a regular 12-gon, just a regular hexagon remains (Fig. 26), whence also a regular 12-gon can be partitioned in the required way.\n\nShow now that other regular polygons cannot be partitioned into smaller regular polygons. For that, consider an arbitrary polygon that is partitioned into regular polygons. As the size of an internal angle of a regular polygon is less than $180^\\circ$ and not less than $60^\\circ$, at most two regular polygons can meet at each vertex.\n\nIf a vertex of the big $n$-gon is filled by just one smaller polygon then this piece is an $n$-gon itself. Beside it, there must be space for at least one regular polygon. No more than two regular polygons can be placed there since the sum of the internal angles of these polygons and the $n$-gon itself would exceed $180^\\circ$. Two new pieces can be placed only if all these three pieces are triangular, which gives $n=3$. It remains to study the case where there is exactly one polygon beside the $n$-gonal piece. The size of the internal angle of the $n$-gon being at most $120^\\circ$ implies $n \\le 6$. The case $n=5$ is impossible as its external angles are of size $72^\\circ$ but no regular polygon has internal angles of size strictly between $60^\\circ$ and $90^\\circ$.\n\nIf each vertex of the big $n$-gon is the meetpoint of two smaller regular polygons then one of them must be a triangle since other regular polygons have internal angles of size $90^\\circ$ or more. Beside a triangle, there is space for a triangle, a quadrilateral or a pentagon.\n\n![](attached_image_1.png)\nFigure 23\n![](attached_image_2.png)\nFigure 24\n![](attached_image_3.png)\nFigure 25\n![](attached_image_4.png)\nFigure 26\n\n![](attached_image_5.png)\nFigure 27\n\nIn the first two cases, the size of the internal angles of the $n$-gon will be $120^\\circ$ and $150^\\circ$, respectively, covering the cases $n = 6$ and $n = 12$. It remains to show that the third case with a triangle and a pentagon meeting at each vertex is impossible. Indeed, the side length of the pentagon must coincide with the side length of the initial big $n$-gon, because it is impossible to place a regular polygon beside the pentagon along one side. For the same reason, another pentagon must be built to the second next side along the boundary of the initial polygon. These two pentagons meet at the third vertex of the triangle built to the side between (Fig. 27). But the ulterior angle between the sides of the pentagons at the meeting point has size $360^\\circ - 2 \\cdot 108^\\circ - 60^\\circ = 84^\\circ$, which cannot be filled with interior angles of regular polygons.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 29715,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find the smallest natural number $n$ for which there exist integers $a_1, \\dots, a_n$ (that do not have to be different) such that $a_1^4 + \\dots + a_n^4 = 2013$.",
    "options": [],
    "answer": "14",
    "solution": "**Answer:** 14.\n\nNote that the fourth powers of even numbers are divisible by 16 and the fourth powers of odd numbers are congruent to 1 modulo 16. As $2013 \\equiv 13 \\pmod{16}$, the desired representation must contain at least 13 odd summands.\n\nSuppose that no more summands are needed. As $7^4 = 2401 > 2013$, each summand must be $1^4 = 1$, $3^4 = 81$ or $5^4 = 625$. There can be at most 3 summands 625 since $4 \\cdot 625 > 2013$. Therefore the number of summands not divisible by 5 is at least 10. The fourth power of an integer not divisible by 5 is congruent to 1 modulo 5, whereas $2013 \\equiv 3 \\pmod{5}$. Hence the number of summands not divisible by 5 must be at least 13. This shows that the representation contains only summands 1 and 81, but 13 such numbers sum up to at most $13 \\cdot 81$ which is less than 2013. Thus representations with 13 summands are impossible.\n\nOn the other hand, 14 fourth powers is enough as $6^4 + 5^4 + 3^4 + 11 \\cdot 1^4 = 2013$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 29716,
    "subject": "Mathematics (Multi-modal)",
    "question": "Malle drew a rhombus $ABCD$ and chose points $E$ and $F$ on sides $AB$ and $BC$, respectively, such that the triangle $DEF$ is equilateral. Malle was very surprised when she discovered that there is another possibility to choose points $E$ and $F$ on sides $AB$ and $BC$, respectively, such that $DEF$ is equilateral. What can be the measures of the angles of this rhombus?\n\n![](attached_image_1.png)\nFigure 17",
    "options": [],
    "answer": "60° and 120°",
    "solution": "There is clearly only one way to choose points $E$ and $F$ on sides $AB$ and $BC$, respectively, such that $E$ and $F$ would be symmetrical with respect to the diagonal $BD$ and $\\angle EDF = 60^\\circ$. Thus it is possible in Malle's rhombus to choose $E$ and $F$ asymmetrically with respect to diagonal $BD$ such that triangle $DEF$ is equilateral; let us consider one of the possible setups. W.l.o.g., we can assume that $|EB| < |BF|$ (otherwise we can switch the roles of $A$ and $C$ and the roles of $E$ and $F$). Let $\\alpha = \\angle BAD = \\angle BCD$ and $\\beta = \\angle AED$. Also let $E'$ be the point symmetrical to $E$ with respect to diagonal $BD$ (see fig. 17); then $|DE'| = |DE| = |DF|$ gives that triangle $E'DF$\n\nis isosceles and $\\angle BFD = \\angle FE'D = \\angle AED = \\beta$. Therefore\n$$\n\\angle CDF = 180^\\circ - \\angle DCF - \\angle DFC = 180^\\circ - \\alpha - (180^\\circ - \\beta) = \\beta - \\alpha.\n$$\nNow\n$$\n\\begin{aligned}\n180^\\circ - \\alpha &= \\angle ADC = \\angle ADE + \\angle EDF + \\angle FDC \\\\\n&= (180^\\circ - \\alpha - \\beta) + 60^\\circ + (\\beta - \\alpha) = 240^\\circ - 2\\alpha.\n\\end{aligned}\n$$\nFrom here $\\alpha = 60^\\circ$, that is the angles of the rhombus are $60^\\circ$ and $120^\\circ$.\n\n![](attached_image_2.png)\nFigure 18\nLet the first choices of Malle be $E_1$ and $F_1$ and second ones $E_2$ and $F_2$ (see fig. 18). As $\\angle E_1DF_1 = 60^\\circ = \\angle E_2DF_2$ and $|DE_1| = |DF_1|$ and $|DE_2| = |DF_2|$, the rotation of the plane by $60^\\circ$ around point $D$ that takes $E_1$ to $F_1$ must also take $E_2$ to $F_2$. The same rotation must then take the line $E_1E_2$ to $F_1F_2$ or line $AB$ to line $BC$. Therefore $\\angle ABC = 120^\\circ$, thus the angles of the rhombus are $60^\\circ$ and $120^\\circ$.\n\n![](attached_image_2.png)\nFigure 18",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 29717,
    "subject": "Mathematics (Multi-modal)",
    "question": "Real numbers $x_1$, $x_2$, $x_3$, $x_4$ in $[0; 1]$ are such that the product\n$$\nK = |x_1 - x_2| \\cdot |x_1 - x_3| \\cdot |x_1 - x_4| \\cdot |x_2 - x_3| \\cdot |x_2 - x_4| \\cdot |x_3 - x_4|\n$$\nis as large as possible. Prove that $\\frac{1}{27} > K > \\frac{4}{243}$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "If some two numbers among $x_1$, $x_2$, $x_3$, $x_4$ are equal then $K = 0$ which is not maximal. Thus assume w.l.o.g. that $x_1 > x_2 > x_3 > x_4$. Applying AM-GM for $x_1 - x_2$, $x_2 - x_3$ and $x_3 - x_4$ gives\n$$\n\\sqrt[3]{(x_1 - x_2)(x_2 - x_3)(x_3 - x_4)} \\le \\frac{(x_1 - x_2) + (x_2 - x_3) + (x_3 - x_4)}{3} = \\frac{x_1 - x_4}{3} \\le \\frac{1}{3},\n$$\ni.e., $|x_1 - x_2| \\cdot |x_2 - x_3| \\cdot |x_3 - x_4| \\le \\frac{1}{27}$. Among the remaining factors $|x_1 - x_3|$, $|x_1 - x_4|$, $|x_2 - x_4|$, at least one is less than 1. Hence we conclude the left-hand inequality needed.\n\nFor the second inequality, note that if $x_1 = 1$, $x_2 = \\frac{3}{4}$, $x_3 = \\frac{1}{4}$, $x_4 = 0$ then\n$$\nK = \\frac{1}{4} \\cdot \\frac{3}{4} \\cdot 1 \\cdot \\frac{1}{2} \\cdot \\frac{3}{4} \\cdot \\frac{1}{4} = \\frac{9}{512} > \\frac{4}{243},\n$$\nsince $9 \\cdot 243 = 2187 > 2048 = 4 \\cdot 512$.\nW.l.o.g., assume the inequalities $x_1 > x_2 > x_3 > x_4$. In addition, assume that $x_1 = 1$ and $x_4 = 0$ as otherwise $K$ can be made larger. Substituting $x_2 = y$ and $x_3 = z$ for simplicity, one obtains\n$$\nK = (1 - y)(1 - z)(y - z)yz = (y - z) \\cdot (1 - y)z \\cdot (1 - z)y.\n$$\nConsider pairs $(y, z)$ with $y - z$ fixed. The sums $(1 - y) + z = 1 - (y - z)$ and $(1 - z) + y = 1 + (y - z)$ are then also fixed. The product of two numbers with fixed sum is the largest if the numbers are equal; thus the product $(1 - y)z$ is the largest in the case $y+z=1$ and the product $(1-z)y$ is the largest in the same case $y+z=1$. Consequently, also $K$ obtains its largest value in the case $y+z=1$. Substituting $1-z$ at place of $y$, one gets $K = z^2(1-z)^2(1-2z) = (z(1-z))^2(1-2z)$.\nLet $f(z) = (z(1-z))^2(1-2z)$, then $f'(z) = 2z(1-z)(1-5z+5z^2)$. The roots of $f'$ within $(0,1)$ are the roots of the quadratic polynomial $5z^2 - 5z + 1$, namely $\\frac{5-\\sqrt{5}}{10}$ and $\\frac{5+\\sqrt{5}}{10} = 1 - \\frac{5-\\sqrt{5}}{10}$. As $f(0) = f(\\frac{1}{2}) = 0$ and $f(z) > 0$ whenever $0 < z < \\frac{1}{2}$, the maximum of $f$ is achieved at $z = \\frac{5-\\sqrt{5}}{10}$. Thus the maximum value of $K$ is $f(\\frac{5-\\sqrt{5}}{10}) = \\frac{\\sqrt{5}}{125}$. This number $\\frac{\\sqrt{5}}{125}$ satisfies both inequalities of the problem.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 29718,
    "subject": "Mathematics (Multi-modal)",
    "question": "The midpoints of sides $C_2C_3$, $C_3C_1$ and $C_1C_2$ of a triangle $C_1C_2C_3$ are $K_1$, $K_2$ and $K_3$, respectively. The centers of circles $c_1$, $c_2$ and $c_3$ are $C_1$, $C_2$ and $C_3$, respectively, and the centers of circles $k_1$, $k_2$, $k_3$ are $K_1$, $K_2$, $K_3$, respectively. No two of the given six circles intersect in two points nor are they inside each other. Circles $k_1$, $k_2$ and $k_3$ touch each other externally.\n\ni) Prove that the sum of the radii of circles $c_1$, $c_2$ and $c_3$ does not exceed one quarter of the perimeter of the triangle $C_1C_2C_3$.\n\nii) Prove that if the sum of the radii of circles $c_1$, $c_2$ and $c_3$ equals one quarter of the perimeter of the triangle $C_1C_2C_3$ then the triangle $C_1C_2C_3$ is equilateral.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let the radii of the circles $c_1$, $c_2$, $c_3$ be $r_1$, $r_2$, $r_3$, and the radii of the circles $k_1$, $k_2$, $k_3$ be $R_1$, $R_2$, $R_3$, respectively (Fig. 28). By assumptions,\n$$\n\\begin{aligned}\nR_1 + R_2 &= |K_1K_2| = \\frac{1}{2}|C_1C_2|, \\\\\nR_2 + R_3 &= |K_2K_3| = \\frac{1}{2}|C_2C_3|, \\\\\nR_3 + R_1 &= |K_3K_1| = \\frac{1}{2}|C_3C_1|,\n\\end{aligned}\n$$\nwhich sum up to $2R_1 + 2R_2 + 2R_3 = \\frac{1}{2}(|C_1C_2| + |C_2C_3| + |C_3C_1|)$. The assumptions also imply inequalities\n$$\n\\begin{aligned}\nr_1 + R_3 &\\le \\frac{1}{2}|C_1C_2|, & R_3 + r_2 &\\le \\frac{1}{2}|C_1C_2|, \\\\\nr_2 + R_1 &\\le \\frac{1}{2}|C_2C_3|, & R_1 + r_3 &\\le \\frac{1}{2}|C_2C_3|, \\\\\nr_3 + R_2 &\\le \\frac{1}{2}|C_3C_1|, & R_2 + r_1 &\\le \\frac{1}{2}|C_3C_1|,\n\\end{aligned}\n$$\nwhich sum up to $2r_1 + 2r_2 + 2r_3 + 2R_1 + 2R_2 + 2R_3 \\le |C_1C_2| + |C_2C_3| + |C_3C_1|$.\n\ni) Altogether, we obtain the inequality $2r_1 + 2r_2 + 2r_3 \\le \\frac{1}{2}(|C_1C_2| + |C_2C_3| + |C_3C_1|)$, which implies $r_1 + r_2 + r_3 \\le \\frac{1}{4}(|C_1C_2| + |C_2C_3| + |C_3C_1|)$ as desired.\n\n![](attached_image_1.png)\nFigure 28\n\nii) Suppose that $r_1 + r_2 + r_3 = \\frac{1}{4}(|C_1C_2| + |C_2C_3| + |C_3C_1|)$. Then all inequalities above must hold as equalities. The equalities $r_1 + R_3 = \\frac{1}{2}|C_1C_2| = r_2 + R_3$ imply $r_1 = r_2$, analogously $r_1 = r_3$. Denoting $r = r_1 = r_2 = r_3$, we get\n$$\nr + R_3 = \\frac{1}{2}|C_1C_2| = R_1 + R_2,\n$$\n$$\nr + R_2 = \\frac{1}{2}|C_1C_3| = R_1 + R_3,\n$$\nwhere summing side-by-side gives $2r + R_2 + R_3 = 2R_1 + R_2 + R_3$, i.e., $R_1 = r$. Analogously, $R_2 = R_3 = r$. Thus all sides of the triangle $C_1C_2C_3$ have length $4r$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 29719,
    "subject": "Mathematics (Multi-modal)",
    "question": "In a $2n \\times 2n$ grid exactly half of the squares have been coloured black and the other half are white. In one step one can take some $2 \\times 2$ square in this grid and reflect its four squares w.r.t. the horizontal or vertical central axis. Which positive integers $n$ make it possible to get from any initial configuration to a state where the whole board has been coloured chessboard-style?",
    "options": [],
    "answer": "n ≥ 2",
    "solution": "In case of $n = 1$ it is not possible to get the chessboard-pattern if $2 \\times 2$ the initial configuration is like in fig. 19, because adjacent same-coloured squares are same-coloured also after reflecting.\n\n![](attached_image_1.png)\n\nLet us now show that for any $n \\ge 2$ we can start from any initial configuration and reach the chessboard pattern. For that we can show that whenever we have some wrong-coloured squares, we can reduce their number by taking some finite number of steps. Note that a wrong-coloured square turns into a right-coloured square on the other side of the axis of reflection and the other way round. Let us define a *double reflection* to be reflecting the same $2 \\times 2$ area first horizontally and then vertically. A double reflection is equivalent to a reflection with respect to the centre of the $2 \\times 2$ square, whereas the wrong-coloured squares will remain wrong-coloured and the right-coloured squares right-coloured after the reflection.\n\nFirst suppose that there exist two adjacent same-coloured squares. W.l.o.g., let those two wrong-coloured squares be in the same row. As $n \\ge 2$, we can also assume that this row is at least third from the top and that there is at least one column to the right of the squares under consideration. Let us mark the wrong-coloured square with W and the right-coloured square with R on the figure; $x$ means one or the other and $x'$ means the opposite of $x$ (if $x$ is right then $x'$ is wrong and the other way round).\n\n* If out of the two adjacent wrong-coloured squares at least one has a wrong-coloured upper neighbour, then by reflecting w.r.t. the vertical axis the number of wrong-coloured squares decreases by at least 2 (fig. 20).\n\n![](attached_image_2.png)\n\n* If both upper neighbours of the two adjacent wrong-coloured squares are right-coloured, but at least one of them in turn has wrong-coloured upper neighbour, then by reflecting w.r.t. the vertical axis we can take the two adjacent wrong-coloured squares up by one row, so that the number of wrong-coloured squares does not change (see fig. 21). Afterwards we can do as described previously.\n\n![](attached_image_3.png)\n\n* If the $2 \\times 2$ square above the two wrong-coloured squares is entirely right-coloured, but at least one adjacent square to the right of this $2 \\times 2$ area is wrong-coloured, then we can use double reflection to swap this wrong-coloured square with one of the right-coloured squares in the $2 \\times 2$ area (on fig. 22 the wrong-coloured square is the bottom right square; in the other case the same transition helps). After that we can proceed as before.\n\n![](attached_image_4.png)\n\n* In the rest of the cases we can reduce the number of wrong-coloured squares by 2 using the steps on fig. 23.\n\n![](attached_image_5.png)\n\nLet us finally look at the situation where there are no two wrong-coloured adjacent squares. With double reflections a wrong-coloured square can be moved along the diagonals without changing the number of wrong- or right-coloured squares. Since the numbers of black and white squares are initially equal and do not change with reflecting, the existence of a wrong-coloured black square implies that there must also exist a wrong-coloured white square and the other way round. Therefore by taking steps along the diagonals we can take one wrong-coloured square next to another one and proceed as described above.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 29720,
    "subject": "Mathematics (Multi-modal)",
    "question": "Ats and Pets both thought of two positive integers that do not exceed some positive integer $n$. If they both added the numbers they thought of, then both sums gave the same remainder when divided by $n$. But if both of them multiplied the numbers they thought of, then both products also gave equal remainders when divided by $n$. Is it necessarily true that the numbers they thought of were the same, if\na) $n = 99$?\nb) $n = 101$?",
    "options": [],
    "answer": "a) No; for n = 99, for example, one pair can be 1 and 21 and the other 10 and 12. b) Yes; for n = 101 the pairs must be identical.",
    "solution": "a) If Ats thought of numbers $1$ and $21$ and Pets of numbers $10$ and $12$, then both get the sum $22$ and the products will be $21$ and $120$, respectively, both of which give the remainder $21$ when divided by $99$.\n\nb) Let the numbers Ats chose be $a$ and $b$, the ones Pets chose $c$ and $d$. According to the conditions stated in the problem, the numbers $(a + b) - (c + d)$ and $ab - cd$ are both divisible by $101$. Let $(a + b) - (c + d) = 101k$; then $a = 101k - b + c + d$, from where\n$$\n\\begin{aligned}\nab - cd &= (101k - b + c + d)b - cd = 101kb - b^2 + bc + bd - cd \\\\\n&= 101kb - (c - b)(d - b).\n\\end{aligned}\n$$\nHence also the product $(c - b)(d - b)$ is divisible by $101$. As $101$ is a prime number, it has to divide either the factor $c - b$ or the factor $d - b$. W.l.o.g., let $c - b$ be divisible by $101$. As all the numbers are on the interval from $1$ to $101$, it means that $c = b$. But then $(a + b) - (c + d) = a - d$, which due to divisibility by $101$ means that $a = d$. Therefore Ats and Pets must have chosen the same numbers.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 29721,
    "subject": "Mathematics (Multi-modal)",
    "question": "Define *magic square* as a $3 \\times 3$ table where each cell contains one number from $1$ to $9$ so that all these numbers are used and all row sums and column sums are equal. Prove that any two magic squares can be obtained from each other via the following transformations: interchanging two rows, interchanging two columns, rotating the square, reflecting the square w.r.t. its diagonal.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "As all the transformations are invertible, it suffices to show that every magic square can be turned to one particular magic square by these transformations.\n\n<table><tr><td>2</td><td></td><td>4</td></tr><tr><td></td><td></td><td></td></tr><tr><td>6</td><td></td><td>8</td></tr></table>\n<table><tr><td>2</td><td></td><td>4</td></tr><tr><td></td><td></td><td></td></tr><tr><td>8</td><td></td><td>6</td></tr></table>\n<table><tr><td>2</td><td></td><td>6</td></tr><tr><td></td><td></td><td></td></tr><tr><td>8</td><td></td><td>4</td></tr></table>\nFigure 29\n<table><tr><td>2</td><td>9</td><td>4</td></tr><tr><td>7</td><td>5</td><td>3</td></tr><tr><td>6</td><td>1</td><td>8</td></tr></table>\nFigure 30\n\nThe sum of all numbers in a magic square is $45$, whence the numbers in each row and each column must sum up to $15$. As this is odd, exactly $0$ or $2$ of the three summands must be even. There are $4$ even numbers in use, hence $2$ even numbers must be in some two rows and $0$ even number in the remaining one. The same holds for columns.\nHence the even numbers $2$, $4$, $6$, $8$ occur in the corners of some rectangle with sides parallel to the edges of the table. By interchanging rows or columns one can move the even numbers to the corners of the whole table. There are $3$ possibilities to locate these four numbers into the corners, that can not be obtained from each other by rotations and reflections of the table (Fig. 29). The last two of them cannot occur in the magic square because the missing numbers in the first and third column would coincide. Hence only the first possibility remains. Its completion to a magic square is unique (Fig. 30).",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 29722,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all pairs of real numbers $(x, y)$ that satisfy\n$$\n\\begin{cases} x + \\sin x = y, \\\\ y + \\sin y = x. \\end{cases}\n$$",
    "options": [],
    "answer": "All pairs are (kπ, kπ) for any integer k.",
    "solution": "By adding the equations and simplifying we get $\\sin x = -\\sin y$. Thus $y = -x + 2k\\pi$ or $y = \\pi + x + 2k\\pi = x + (2k + 1)\\pi$. In the second case we get that $|y - x| = |(2k + 1)\\pi| \\ge \\pi$, but from the first equality $|y - x| = |\\sin x| \\le 1 < \\pi$, a contradiction. Therefore $y = -x + 2k\\pi$, where $k$ is an integer. By plugging this into the first equation we get $2x + \\sin x = 2k\\pi$ after simplifying. We see that this is satisfied in case of any integer $k$ by the value $x = k\\pi$; then also $y = -k\\pi + 2k\\pi = k\\pi$. As $f(x) = 2x + \\sin x$ is an increasing function, there cannot be any other solutions to $2x + \\sin x = 2k\\pi$.\nFunction $f(z) = z + \\sin z$ is strictly increasing, because its derivative $f'(z) = 1 + \\cos z$ is positive everywhere, except for some isolated points. Therefore if $x < y$ in case of some solution $(x, y)$ to the system of equations, then $y = x + \\sin x < y + \\sin y = x$, a contradiction. Analogously $y < x$ gives a contradiction. In conclusion $x = y$ is the only option. By substituting it in we get $\\sin x = \\sin y = 0$, from where $x = y = k\\pi$ for any integer $k$. All pairs $(k\\pi, k\\pi)$ indeed satisfy the given system of equations.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 29723,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $I$ be the incenter of triangle $ABC$. Let $R_A$, $R_B$ and $R_C$ be the radii of the circumcircles of triangles $BIC$, $CIA$ and $AIB$, respectively, and $R$ be the radius of the circumcircle of triangle $ABC$. Prove that $R_A + R_B + R_C \\le 3R$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $|BC| = a$, $|CA| = b$ and $|AB| = c$ and the angles opposite to those sides be $\\alpha$, $\\beta$ and $\\gamma$, respectively (see fig. 24).\n\n![](attached_image_1.png)\nFigure 24\n\nThe law of sines in triangles $ABC$ and $IBC$ gives $\\frac{a}{\\sin \\alpha} = 2R$ and $\\frac{a}{\\sin \\left(\\frac{\\beta}{2} + \\frac{\\gamma}{2}\\right)} = 2R_A$, respectively. As $\\sin\\left(\\frac{\\beta}{2} + \\frac{\\gamma}{2}\\right) = \\sin\\left(90^\\circ - \\frac{\\alpha}{2}\\right) = \\cos\\frac{\\alpha}{2}$, we obtain $\\frac{R_A}{R} = \\frac{\\sin\\alpha}{\\cos\\frac{\\alpha}{2}} = 2\\sin\\frac{\\alpha}{2}$. Similarly we get $\\frac{R_B}{R} = 2\\sin\\frac{\\beta}{2}$ and $\\frac{R_C}{R} = 2\\sin\\frac{\\gamma}{2}$.\nTo solve the problem we need to prove that $\\frac{R_A}{R} + \\frac{R_B}{R} + \\frac{R_C}{R} \\le 3$, or equivalently,\n$$\n\\sin \\frac{\\alpha}{2} + \\sin \\frac{\\beta}{2} + \\sin \\frac{\\gamma}{2} \\le \\frac{3}{2}. \\qquad (6)\n$$\nApplying Jensen's inequality gives\n$$\n\\frac{1}{3} \\left( \\sin \\frac{\\alpha}{2} + \\sin \\frac{\\beta}{2} + \\sin \\frac{\\gamma}{2} \\right) \\le \\sin \\left( \\frac{\\frac{\\alpha}{2} + \\frac{\\beta}{2} + \\frac{\\gamma}{2}}{3} \\right) = \\sin \\left( \\frac{\\alpha + \\beta + \\gamma}{6} \\right) = \\sin \\frac{\\pi}{6} = \\frac{1}{2},\n$$\nwhich directly implies the necessary result.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 29724,
    "subject": "Mathematics (Multi-modal)",
    "question": "A positive integer $n$ is written on the board once, then $n-1$ is written on the board twice, etc.; on every step the number smaller by $1$ from the previous number is written twice as many times as the previous number. When reaching zeros this process stops. Prove that in the end the sum of the numbers on the board is less than $2^{n+1}$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "The sum of the numbers on the board is\n$$\ns_n = 1 \\cdot n + 2 \\cdot (n-1) + 4 \\cdot (n-2) + \\dots + 2^{n-1} \\cdot 1.\n$$\nLet us also define\n$$\nr_n = \\frac{s_n}{2^n} = \\frac{1}{2} \\cdot 1 + \\frac{1}{4} \\cdot 2 + \\frac{1}{8} \\cdot 3 + \\dots + \\frac{1}{2^n} \\cdot n.\n$$\n\n$\\frac{1}{2} + \\frac{1}{4} + \\dots + \\frac{1}{2^{n+1}} < \\frac{1}{2} + \\frac{1}{4} + \\dots = 1$. Thus $r_n < 2$ always implies $r_{n+1} < \\frac{r_n}{2} + 1 < 1 + 1 = 2$. As $r_1 = \\frac{1}{2} < 2$, we have $r_n < 2$ for every $n \\ge 1$. From there we get that $s_n = 2^n \\cdot r_n < 2^{n+1}$ for every $n \\ge 1$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 29725,
    "subject": "Mathematics (Multi-modal)",
    "question": "A train departed from the station 12 minutes later than planned. If the train would not make any stops on the way and would travel at average speed equal to what would be the average speed between stops according to the timetable, then it would reach the destination exactly at the right time. But if the train would stop in every station for the same amount of time it was supposed to, then between the station it would have to travel with average speed 40% higher than before in order to reach the destination on time. Find the travelling time of the train according to the timetable.\n\n*Answer:* 54 minutes.",
    "options": [],
    "answer": "54 minutes",
    "solution": "Let the time we are looking for be $t$. The conditions of the problem imply that $t - 12 \\text{ min} = 1.4 \\cdot (t - 24 \\text{ min})$, from which $0.4t = 21.6 \\text{ min}$ and $t = 54 \\text{ min}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 29726,
    "subject": "Mathematics (Multi-modal)",
    "question": "Can $2013$ be represented as the difference of two cubes of integers?",
    "options": [],
    "answer": "No",
    "solution": "*Answer:* No.\n\nSuppose that $2013 = x^3 - y^3$ where $x$ and $y$ are integers. Note that\n$$\nx^3 - y^3 = (x - y)^3 + 3x^2y - 3xy^2 = (x - y)^3 + 3xy(x - y).\n$$\nAs $2013$ is divisible by $3$ and so is $3xy(x - y)$, the difference $(x - y)^3$ must be divisible by $3$. Thus also $x - y$ is divisible by $3$ as $3$ is prime. Consequently, $3xy(x - y)$ is divisible by $3^2$ and $(x - y)^3$ is divisible by $3^3$, whence the sum $x^3 - y^3$ is divisible by $3^2$. But $2013$ is not divisible by higher powers of $3$. The contradiction shows that $2013$ cannot be represented as the difference of two cubes of integers.\nSuppose that $2013 = x^3 - y^3$ where $x$ and $y$ are integers. Then $x^3 \\equiv y^3 \\pmod{3}$. As integers are congruent to their cubes modulo $3$, this gives $x \\equiv y \\pmod{3}$. Now write $2013 = (x - y)(x^2 + xy + y^2)$. By the congruence obtained above, the first factor in the r.h.s. is divisible by $3$ and the terms $x^2$, $xy$ and $y^2$ in the second factor are all congruent modulo $3$ whence $x^2 + xy + y^2$ is divisible by $3$. Altogether, the product $(x - y)(x^2 + xy + y^2)$ must be divisible by $9$, but $2013$ is not.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 29727,
    "subject": "Mathematics (Multi-modal)",
    "question": "The bases of trapezoid $ABCD$ are $AB$ and $CD$, and the intersection point of its diagonals is $P$. Prove that if $\\frac{|PA|}{|PD|} = \\frac{|PB|}{|PC|}$ then the trapezoid is isosceles.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "![](attached_image_1.png)\nFigure 15\n\nBy assumptions, $\\frac{|PA|}{|PB|} = \\frac{|PD|}{|PC|}$. As the bases $AB$ and $CD$ are parallel, we have also $\\frac{|PA|}{|PB|} = \\frac{|PC|}{|PD|}$ (Fig. 15). Hence $|PC| = |PD|$. Similarity of triangles $APD$ and $BPC$ implies $\\frac{|AD|}{|BC|} = \\frac{|PD|}{|PC|} = 1$, thus $|AD| = |BC|$ as needed.\n![](attached_image_2.png)\nFigure 16\n\n![](attached_image_3.png)\nFigure 17\n\nBy assumptions, triangles $APD$ and $BPC$ are similar. Thus $\\angle ADB = \\angle ACB$ (Fig. 16), showing that quadrilateral $ABCD$ is cyclic. But if a quadrilateral with parallel opposite sides has a circumcircle, the bisectors of these sides coincide as they have the same direction and both pass through the circumcenter of the quadrilateral (Fig. 17). By symmetry w.r.t. this line, the other pair of opposite sides have equal lengths.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 29728,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find the greatest natural number $n$ for which it is possible to choose $n$ vertices of a cube such that no three of them form a right triangle.\n![](attached_image_1.png)\n\n![](attached_image_1.png)\nFigure 9",
    "options": [],
    "answer": "4",
    "solution": "Let some vertex of a cube be $A$ and let $B$, $C$ and $D$ be the opposite vertices of the faces of the cube that $A$ belongs to (see fig. 9). Then of the vertices $B$, $C$ and $D$ any two are also the opposite vertices of some face of the cube. Therefore any two of the chosen four vertices are at the distance of a face diagonal of the cube. Therefore any three form an equilateral rather than a right triangle.\n\nLet us now look at the situation where we choose at least 5 vertices. Two opposite faces of the cube include all the vertices of the cube. Therefore at least one of the two opposite faces has to include at least 3 of the chosen vertices. But three vertices of a square form a right triangle.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 29729,
    "subject": "Mathematics (Multi-modal)",
    "question": "Each unit square in a $5 \\times 5$ table is coloured either blue or yellow. Prove that there exists a rectangle with sides parallel to the edges of the table, such that the four unit squares in its corners have the same colour.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Each row contains at least 3 squares with the same colour. Similarly, the dominating colour must be the same in at least 3 rows. W.l.o.g., suppose that the first 3 rows contain at least 3 blue squares each. If the first two rows contain two blue squares in the same columns then the desired rectangle exists. Otherwise, each column contains at least one blue square in these two rows (in Fig. 18, the blue squares are consecutive w.l.o.g.). Thus in one of the first two rows there are two blue squares that are in the same columns with the blue squares in the third row. These form the desired figure.\nThe first row contains three squares of the same colour, say, blue. If in some of the remaining four rows there are two blue squares aligned to the blue squares in the first row then the desired rectangle exists. Otherwise, each of these four rows\n\n![](attached_image_1.png)\nFigure 18\n\ncontains two yellow squares aligned with the blue squares in the first row. But two yellow squares can be placed into three columns in 3 different ways only. Thus there exist two rows where these two yellow squares are in the same columns. These squares form the desired figure.",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 29730,
    "subject": "Mathematics (Multi-modal)",
    "question": "Juku writes down all 20-digit numbers in which each of digits $3$, $4$, $5$ and $6$ appear five times in a row (in some order). Prove that it is possible to choose two of those numbers such that their difference is divisible by $207$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "As $207 = 9 \\cdot 23$ and $9$ and $23$ are relatively prime, it suffices to find a difference that would be divisible by both $9$ and $23$.\n\nAll the $20$-digit numbers listed are divisible by $9$ because the sum of their digits is $90$. Therefore the difference of any two of them is also divisible by $9$.\n\nIt remains to show that the difference of some two of them is divisible by $23$. For this note that there are $24$ different orderings of $3$, $4$, $5$ and $6$. Therefore there are $24$ numbers written in total, but there are $23$ different possible remainders. So there must exist some two among those that give the same remainder when divided by $23$. Their difference is divisible by $23$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 29731,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $a$ and $n$ be positive integers. Prove that\n$$\n\\lfloor \\frac{a}{n} \\rfloor + \\lfloor \\frac{a+1}{n} \\rfloor + \\dots + \\lfloor \\frac{a+n-1}{n} \\rfloor = a.\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $\\lfloor \\frac{a}{n} \\rfloor = q$; then $a = qn + r$, where $0 \\le r < n$. Divide the addends given into two groups: in the first one there are $n-r$ addends, where the numerators of the fractions are equal to numbers from $qn+r$ to $qn+(n-1)$, and in the other one the rest of the $r$ addends, where the numerators are equal to numbers from $q(n+1)$ to $q(n+1)+(r-1)$. The integral parts of the first $n-r$ fractions are equal to $q$ and the integral parts of the last $r$ fractions are equal to $q+1$. Thus\n$$\n\\begin{aligned} \\lfloor \\frac{a}{n} \\rfloor + \\lfloor \\frac{a+1}{n} \\rfloor + \\dots + \\lfloor \\frac{a+n-1}{n} \\rfloor &= q \\cdot (n-r) + (q+1) \\cdot r \\\\ &= qn - qr + qr + r \\\\ &= qn + r = a. \\end{aligned}\n$$\nLet us induct on $a$. If $a=0$, then the numerators of the fractions are $0, 1, \\dots, n-1$, all natural numbers less than $n$. Thus all the integral parts in the sum are equal to 0 and the sum is 0 as required. For the inductive step the following suffices:\n$$\n\\lfloor \\frac{a+1}{n} \\rfloor + \\dots + \\lfloor \\frac{a+n}{n} \\rfloor = a - \\lfloor \\frac{a}{n} \\rfloor + \\lfloor \\frac{a}{n} + 1 \\rfloor = a - \\lfloor \\frac{a}{n} \\rfloor + \\lfloor \\frac{a}{n} \\rfloor + 1 = a + 1.\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 29732,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let a finite decimal fraction be given. Juku starts appending digits to this fraction in such a way that each new digit equals the remainder of the sum of all digits existing so far in division by $10$. (For instance, if the initial fraction is $27.35$ then the digits added to the end are $7$, $4$, $8$ etc.)\nProve that the infinite decimal fraction obtained this way represents a rational number.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "It suffices to show that the infinite decimal fraction is periodic. For that, note that each new digit except for the first digit is congruent to twice the previous digit modulo $10$. Indeed, let $a_1, \\dots, a_{k-1}, a_k$ be the existing at some time moment digits where $a_k$ is already added by Juku. Then the next digit $a_{k+1}$ satisfies\n$$\na_{k+1} \\equiv a_1 + \\dots + a_k = (a_1 + \\dots + a_{k-1}) + a_k \\equiv a_k + a_k = 2a_k \\pmod{10}.\n$$\nHence each new digit is uniquely determined by the last existing digit. As there are only a finite number of different digits, some digit must be added repeatedly. According to the fact just proven, all following digits are repeated as well.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 29733,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $a$, $b$ and $c$ be real numbers for which $abc = 1$. Prove that\n$$\n\\frac{1}{1+a^{2014}} + \\frac{1}{1+b^{2014}} + \\frac{1}{1+c^{2014}} > 1.\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $a^{2014} = u$, $b^{2014} = v$ and $c^{2014} = w$; then $abc = 1$ gives that $uvw = 1$. As the numerators of the l.h.s. of the inequality to be proven are positive, the inequality is equivalent to\n$$\n(1+v)(1+w) + (1+w)(1+u) + (1+u)(1+v) > (1+u)(1+v)(1+w).\n$$\n\nBy expanding, simplifying and using $uvw = 1$, we get $1 + u + v + w > 0$. But this is true, since $u$, $v$ and $w$ are positive.\nLet $a^{2014} = u$, $b^{2014} = v$ and $c^{2014} = w$. Then $abc = 1$ gives that $uvw = 1$. Therefore there exist positive real numbers $x$, $y$, $z$ such that $u = \\frac{x}{y}$, $v = \\frac{y}{z}$ and $w = \\frac{z}{x}$.\nThe inequality can then be written as $\\frac{1}{1+\\frac{x}{y}} + \\frac{1}{1+\\frac{y}{z}} + \\frac{1}{1+\\frac{z}{x}} > 1$ which is equivalent to\n$$\n\\frac{y}{x+y} + \\frac{z}{y+z} + \\frac{x}{z+x} > 1.\n$$\nBut this inequality can be obtained by adding the obvious inequalities $\\frac{y}{x+y} > \\frac{y}{x+y+z}$, $\\frac{z}{y+z} > \\frac{z}{x+y+z}$ and $\\frac{x}{z+x} > \\frac{x}{x+y+z}$.\nLet $a^{2014} = u$, $b^{2014} = v$ and $c^{2014} = w$; then $abc = 1$ gives that $uvw = 1$. W.l.o.g., let $w$ be the greatest of $u$, $v$, $w$. Then $w \\ge 1$, because otherwise $u$, $v$, $w$ would all be less than $1$ and their product could not be $1$. Thus $uv \\le 1$. Now\n$$\n\\frac{1}{1+u} + \\frac{1}{1+v} - 1 = \\frac{(1+v) + (1+u) - (1+u)(1+v)}{(1+u)(1+v)} = \\frac{1-uv}{(1+u)(1+v)} \\ge 0.\n$$\nThus $\\frac{1}{1+u} + \\frac{1}{1+v} \\ge 1$, from which the desired inequality can be concluded.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 29734,
    "subject": "Mathematics (Multi-modal)",
    "question": "In an acute triangle $ABC$ let the point of intersection of the altitude through $B$ and the angle bisector through $C$ be $D$. Let $E$ be the point symmetrical to point $D$ with respect to axis $AC$. Points $A$, $B$, $C$ and $E$ are concyclic. Prove that triangle $ABC$ is isosceles.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $B'$ be the point of intersection of lines $BD$ and $AC$ and $C'$ be the point of intersection of lines $CD$ and $AB$ (see fig. 10). Then $\\angle ACC' = \\angle ACD = \\angle ACE = \\angle ABE = \\angle ABB'$. As triangles $ABB'$ and $ACC'$ share an angle at vertex $A$, they are similar due to having two identical angles. Therefore also $\\angle AC'C = \\angle AB'B = 90^\\circ$. Thus $CC'$ is the altitude of triangle $ABC$, meaning that the altitude and the angle bisector through vertex $C$ coincide. Therefore the triangle $ABC$ is isosceles.\n\n![](attached_image_1.png)\nFigure 10\nUse a known theorem according to which the orthocenter of a triangle is reflected from each side to the circumcircle of the triangle. As the point $E$ on the extension of the altitude drawn from vertex $B$ is located on the circumcircle of triangle $ABC$, point $D$ must be the orthocenter of the triangle $ABC$. As the angle bisector drawn from vertex $C$ passes through that point, it must coincide with the altitude. Therefore $ABC$ is isosceles.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 29735,
    "subject": "Mathematics (Multi-modal)",
    "question": "A convex quadrilateral $ABCD$ where $\\angle DAB + \\angle ABC < 180^\\circ$ is given on a plane. Let $E$ be a point different from the vertices of the quadrilateral on the line interval $AB$ such that the circumcircles of triangles $AED$ and $BEC$ intersect inside the quadrilateral $ABCD$ at point $F$. Point $G$ is defined so that $\\angle DCG = \\angle DAB$, $\\angle CDG = \\angle ABC$ and triangle $CDG$ is located outside quadrilateral $ABCD$. Prove that the points $E$, $F$, $G$ are collinear.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Denote $\\angle DAB = \\alpha$ and $\\angle ABC = \\beta$ (Fig. 21). From cyclic quadrilaterals $AEFD$ and $BEFC$ one obtains\n\n$$\n\\begin{aligned}\n\\angle DFE &= 180^\\circ - \\angle DAE = 180^\\circ - \\alpha, \\\\\n\\angle CFE &= 180^\\circ - \\angle CBE = 180^\\circ - \\beta,\n\\end{aligned}\n$$\nrespectively. Thus $\\angle CFD = 360^\\circ - (180^\\circ - \\alpha) - (180^\\circ - \\beta) = \\alpha + \\beta$. But $\\angle CGD = 180^\\circ - (\\alpha + \\beta)$ by the choice of $G$. Hence the quadrilateral $CFDG$ is cyclic. Consequently, $\\angle DFG = \\angle DCG = \\alpha = 180^\\circ - \\angle DFE$, which implies that the points $E$, $F$, $G$ are collinear.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 29736,
    "subject": "Mathematics (Multi-modal)",
    "question": "The greatest common divisor of positive integers $a$, $b$, $c$ is $1$. It is known that $c$ divides $a + 2b$ and $a^2 - b^2$. Prove that $c$ also divides $a - b$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $d = \\gcd(a+b, c)$. Since $c \\mid a+2b$, also $d \\mid a+2b$. Now $(a+2b) - (a+b) = b$ and $2(a+b) - (a+2b) = a$ are divisible by $d$. Therefore $d$ is the common divisor of $a$, $b$, $c$ and due to our initial assumption of $a$, $b$, $c$ being relatively prime it has to be $1$. Hence $a+b$ and $c$ are also relatively prime. But since $a^2 - b^2 = (a-b)(a+b)$ is divisible by $c$, the factor $a-b$ has to be divisible by $c$.\nSince $a+2b$ is divisible by $c$, also $(a-2b)(a+2b)$ is divisible by $c$. But $(a-2b)(a+2b) = a^2 - 4b^2$; since $a^2 - b^2$ is divisible by $c$, the differences $(a^2 - b^2) - (a^2 - 4b^2) = 3b^2$ and $4(a^2 - b^2) - (a^2 - 4b^2) = 3a^2$ also have to be divisible by $c$.\nIf $3 \\nmid c$, then $c \\mid a^2$ and $c \\mid b^2$. Therefore every prime divisor of $c$ would also be a prime divisor of $a$ and $b$, which would contradict the initial assumption. Hence either $c = 1$, in which case the problem statement holds trivially, or $3 \\mid c$. In the latter case let $c = 3c'$; the statements above show that $c' \\mid a^2$ and $c' \\mid b^2$. As we saw above, every prime divisor of $c'$ would be a prime divisor of $a$ and $b$, due to which $c' = 1$ and $c = 3$. Now since $a+2b$ and $3b$ are divisible by $3$, also $(a+2b) - 3b = a - b$ is divisible by $3$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 29737,
    "subject": "Mathematics (Multi-modal)",
    "question": "Nonzero integers $a$, $b$ and $c$ satisfy $\\frac{1}{a} + \\frac{1}{b} + \\frac{1}{c} = 0$. Prove that among $a$, $b$, $c$ there are two integers which have a common divisor larger than 1.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Multiplying the given equation by $abc$ we get $bc + ca + ab = 0$.\n\nIf $a$, $b$, $c$ were all odd, then $bc$, $ca$ and $ab$ were also odd and their sum could not be $0$.\n\nIf one of the numbers $a$, $b$, $c$ was even and the others were odd, then two of the numbers $bc$, $ca$ and $ab$ were even and one odd, which also would not add up to $0$.\n\nHence at least two of the numbers $a$, $b$, $c$ are even, which satisfy the conditions.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 29738,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all positive integers which are exactly $2013$ times bigger than the sum of their digits.",
    "options": [],
    "answer": "36234",
    "solution": "Note that the minimal value of a $k$-digit number is $10^{k-1}$ and the maximal value of the cross-sum multiplied by $2013$ is $9k \\cdot 2013$. Since $9 \\cdot 7 \\cdot 2013 = 126819 < 1000000$ we can consider only numbers with up to $6$ digits. Since then the cross-sum is at most $54$, it is enough to consider numbers in the form $n \\cdot 2013$ with $1 \\le n \\le 54$.\n\nSince $2013$ is divisible by $3$, $n \\cdot 2013$ and its cross-sum are divisible by $3$. Since the cross-sum must be equal to $n$, $n \\cdot 2013$ is divisible by $9$. But then its cross-sum and hence also $n$ is divisible by $9$. It remains to consider the cases $n = 9, 18, \\dots, 54$ which can be checked by hand and see that only $n = 18$ satisfies the conditions.\n\nAnswer: $36234$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 29739,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find the integral part of $A = \\sqrt{2013 + \\sqrt{2012 + \\dots + \\sqrt{2 + \\sqrt{1}}}}$.",
    "options": [],
    "answer": "45",
    "solution": "On the one hand $A^2 > 2013 + \\sqrt{2012} > 2013 + 44 > 45^2$, therefore $A > 45$. On the other hand we can demonstrate with induction that $x_n = \\sqrt{n + \\sqrt{n-1} + \\dots + \\sqrt{1}} < \\sqrt{n} + 1$. This holds in case of $n = 1$. Suppose it holds for some $n$. Then $x_{n+1} = \\sqrt{n+1+x_n} < \\sqrt{n+1+\\sqrt{n}+1}$. It remains to show that $\\sqrt{n+1+\\sqrt{n}+1} < \\sqrt{n+1}+1$, which is equivalent to a trivially true equation $n + \\sqrt{n} + 2 < n + 2 + 2\\sqrt{n} + 1$. Therefore $A = x_{2013} < \\sqrt{2013} + 1 < 46$.\n\n\nSolution 2:\nInequality $A > 45$ is proved as in solution 1. For $A < 46$, we can repeatedly use the fact that $\\sqrt{x} < x$ for all $x > 1$ to obtain\n$$\n\\begin{aligned}\n\\sqrt{2011 + \\sqrt{2010 + \\dots \\sqrt{2 + \\sqrt{1}}}} &< \\sqrt{2011 + 2010 + \\dots + 2 + 1} \\\\\n&= \\sqrt{\\frac{2011 \\cdot 2012}{2}} = \\sqrt{2023066} < 1423.\n\\end{aligned}\n$$\nTherefore\n$$\nA < \\sqrt{2013 + \\sqrt{2012 + 1423}} = \\sqrt{2013 + \\sqrt{3435}} < \\sqrt{2013 + 59} = \\sqrt{2072} < 46.\n$$\n\n\nSolution 3:\nInequality $A > 45$ is proved as in solution 1. Notice that $\\sqrt{a+b} < \\sqrt{a} + \\sqrt{b}$ if $a > 0$ and $b > 0$. This gives\n$$\n\\begin{aligned}\n& 2012 + \\sqrt{2011 + \\dots + \\sqrt{4 + \\sqrt{3 + \\sqrt{2 + \\sqrt{1}}}}} \\\\\n&\\le 2012 + \\sqrt{2011 + \\dots + \\sqrt{4 + \\sqrt{3 + \\sqrt{2}} + \\sqrt{\\sqrt{1}}}} \\\\\n&\\le 2012 + \\sqrt{2011 + \\dots + \\sqrt{4 + \\sqrt{3} + \\sqrt{\\sqrt{2}} + \\sqrt{\\sqrt{\\sqrt{1}}}}} \\\\\n&\\le \\dots \\le 2012 + \\sqrt{2011} + \\sqrt[2]{2010} + \\dots + \\sqrt[2^{2010}]{2} + \\sqrt[2^{2011}]{1} \\\\\n&< 2012 + 45 + 7 + 3 + 2 \\cdot 2008 = 6083,\n\\end{aligned}\n$$\nwhich gives $A < \\sqrt{2013 + \\sqrt{6083}} < \\sqrt{2013 + 78} < 46$.\n\n\nSolution 4:\nInequality $A > 45$ is proved as in solution 1.\nLet $x_n = \\sqrt{n + \\sqrt{(n-1) + \\dots + \\sqrt{2 + \\sqrt{1}}}}$, then $x_n = \\sqrt{n + x_{n-1}}$. Notice that $x_n > x_{n-1}$, as in the expression of $x_n$, every member is greater than the corresponding member in the expression $x_{n-1} = \\sqrt{(n-1) + \\sqrt{(n-2) + \\sqrt{1+\\sqrt{0}}}}$. Therefore $x_{2013} > x_{2012}$ holds whence the equality $x_n^2 - x_{n-1} - n = 0$ implies $x_{2013}^2 - x_{2013} - 2013 < 0$. Solving the corresponding equation gives\n$$\nx = \\frac{1 + \\sqrt{1 + 4 \\cdot 2013}}{2} = \\frac{1 + \\sqrt{8053}}{2} < \\frac{1 + \\sqrt{90^2}}{2} = 45.5 < 46\n$$\nfor the greater root, therefore $A = x_{2013} < 46$.\n\n\nSolution 5:\nInequality $A > 45$ is proved as in solution 1.\nLet $x_n = \\sqrt{n + \\sqrt{(n-1) + \\sqrt{2 + \\sqrt{1}}}}$, then $x_n^2 = n + x_{n-1}$. Suppose that $x_{2013} \\ge 46$, then $x_{2013}^2 = 2013 + x_{2012} \\ge 46^2 = 2116$, then $x_{2012} \\ge 103$. Analogously we would get that $x_{2011} \\ge 8597$. On the other hand it is clear that $x_{2010} < x_{2011}$ (see solution 4), due to which $x_{2011}^2 = 2011 + x_{2010} < 2011 + x_{2011}$ and $x_{2011}(x_{2011} - 1) < 2011$. Since $x_{2011} > 2$, we get that $x_{2011} < 2011$, which contradicts the inequality $x_{2011} \\ge 8597$. Therefore $A = x_{2013} < 46$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 29740,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all natural numbers $n$ for which there exist primes $p$ and $q$ such that $p(p+1) + q(q+1) = n(n+1)$.",
    "options": [],
    "answer": "n = 3 or n = 6",
    "solution": "The equation is equivalent to $p(p+1) = (n-q)(n+q+1)$. Since the difference of the factors in the r.h.s. is greater than $1$, we must have $n-q < p$ and $n+q+1 > p+1$. As $p$ is a prime number, $p \\mid n+q+1$. Let $n+q+1 = kp$, $k > 1$. Now the initial equation yields $p(p+1) = (kp-2q-1)kp$, which is equivalent to\n$$\n2qk = (k+1)(pk - p - 1). \\qquad (1)\n$$\nAs $k$ and $k+1$ are relatively prime, $2q \\mid k+1$. Since $q$ is a prime number and $k > 1$, there are only two possibilities: $q = k+1$ or $2q = k+1$. In the first case, substituting $k$ to (1) gives $(p-2)(q-2) = 3$, implying $p=3$, $q=5$ (or the other way around). Then $n=6$ by the initial equation. In the second case similarly we get $(p-1)(q-1) = 1$, where the only possibility is $p=q=2$ and $n=3$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 29741,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all remainders which one can get when dividing by $6$ an integer $n$ which satisfies $n^3 = m^2 + m + 1$ for some integer $m$.",
    "options": [],
    "answer": "1",
    "solution": "Numbers $n$ and $n^3$ give the same remainder when dividing by $6$. Also, $m^2 + m + 1$ is odd and gives the remainder $0$ or $1$ when dividing by $3$. The only possibility to get $0$ as the remainder is when $m = 3k+1$, but then\n$$\nn^3 = (9k^2 + 6k + 1) + (3k + 1) + 1 = 9k^2 + 9k + 3 = 3(3k^2 + 3k + 1)\n$$\nwhich leads to a contradiction, since if $n^3$ is divisible by $3$, it is also divisible by $3^3$, but $3k^2 + 3k + 1$ is not divisible by $3$. Hence the remainder of $n^3$ is $1$ both when dividing by $2$ or $3$, consequently its remainder when dividing by $6$ is $1$.\n\nThe remainder $1$ is possible: take $n = 1$ and $m = 0$ (or $n = 7$ and $m = 18$).",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 29742,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all positive real-valued solutions to\n$$\n\\begin{cases} x - y + \\frac{1}{z} = 2013, \\\\ y - z + \\frac{1}{x} = 2013, \\\\ z - x + \\frac{1}{y} = 2013. \\end{cases}\n$$",
    "options": [],
    "answer": "x = y = z = 1/2013",
    "solution": "Suppose w.l.o.g. that $z \\ge x$ and $z \\ge y$. From the second equation $\\frac{1}{x} \\ge 2013$,\n\ntherefore $x \\le \\frac{1}{2013}$. From the third equation $\\frac{1}{y} \\le 2013$, due to which $y \\ge \\frac{1}{2013} \\ge x$.\nBut now from the first equation $\\frac{1}{z} \\ge 2013$, therefore $z \\le \\frac{1}{2013}$. Since we assumed that $z \\ge y \\ge \\frac{1}{2013}$, the only possibility is $z = y = \\frac{1}{2013}$, then also $x = \\frac{1}{2013}$.\nAdding up all the equations, we get $\\frac{1}{x} + \\frac{1}{y} + \\frac{1}{z} = 3 \\cdot 2013$. Multiplying the first equation by $z$, the second one by $x$ and third one by $y$ and adding together we get $3 = 2013(x + y + z)$. Therefore the arithmetic and harmonic mean of $x,y,z$ are both equal to $\\frac{1}{2013}$. Consequently $x = y = z = \\frac{1}{2013}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 29743,
    "subject": "Mathematics (Multi-modal)",
    "question": "Prove that for any integer $n \\ge 3$ we have $(2n)! < n^{2n}$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "For $n = 3$ the claim holds: $(2n)! = 6! = 720$ and $n^{2n} = 3^6 = 729$.\n\nSuppose $n \\ge 4$. Divide the numbers $2, 3, \\dots, 2n-2$ into pairs $(k, 2n-k)$ with $2 \\le k \\le n-1$, leaving $n$ alone. For each pair we have\n$$\nk(2n-k) = (n - (n-k))(n + (n-k)) = n^2 - (n-k)^2 < n^2.\n$$\nHence $2 \\cdot 3 \\cdot \\dots \\cdot (2n-2) < (n^2)^{n-2} \\cdot n = n^{2n-3}$, therefore\n$$\n(2n)! < 1 \\cdot n^{2n-3} \\cdot (2n-1) \\cdot (2n) < n^{2n-3} \\cdot (2n)^2 = 4n^{2n-1} \\le n^{2n}.\n$$\nFor $n=3$ the claim holds. Suppose the claim holds for $n$; to show that it also holds for $n+1$ it is enough to show the inequality $(2n+1)(2n+2) < \\frac{(n+1)^{2n}}{n^{2n}}(n+1)^2$.\n\nSince $(2n+1)(2n+2) < (2n+2)^2 = 4(n+1)^2$, it is enough to show that $\\frac{(n+1)^{2n}}{n^{2n}} > 4$.\n\nThis is equivalent with $\\left(1 + \\frac{1}{n}\\right)^n > 2$ which holds for all $n \\ge 2$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 29744,
    "subject": "Mathematics (Multi-modal)",
    "question": "Inside a circle $c$ there are circles $c_1$, $c_2$ and $c_3$ which are tangent to $c$ at points $A$, $B$ and $C$ correspondingly, which are all different. Circles $c_2$ and $c_3$ have a common point $K$ in the segment $BC$, circles $c_3$ and $c_1$ have a common point $L$ in the segment $CA$, and circles $c_1$ and $c_2$ have a common point $M$ in the segment $AB$. Prove that the circles $c_1$, $c_2$ and $c_3$ intersect in the center of the circle $c$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Take a point $X$ on the common tangent to the circles $c_1$ and $c$ which lies on the other side of the line $AB$ from the point $C$. Then $\\angle ALM = \\angle XAM = \\angle XAB = \\angle ACB$ (Fig. 4). Consequently $ML \\parallel BC$. Similarly $KM \\parallel CA$ and $LK \\parallel AB$. If $\\frac{|AM|}{|AB|} = \\lambda$, then\n$$\n\\frac{|BK|}{|BC|} = \\frac{|BM|}{|BA|} = 1 - \\lambda \\quad \\text{and} \\quad \\frac{|CL|}{|CA|} = \\frac{|CK|}{|CB|} = 1 - (1 - \\lambda) = \\lambda, \\quad \\text{whence} \\quad \\lambda = \\frac{|AM|}{|AB|} = \\frac{|AL|}{|AC|} = 1 - \\lambda. \\quad \\text{Hence} \\quad \\lambda = \\frac{1}{2}, \\text{ therefore the triangles } AML, MBK \\text{ and } LKC \\text{ are all similar}\n$$\nto $ABC$ with the factor $\\frac{1}{2}$. Thus the radii of their circumcircles $c_1$, $c_2$ and $c_3$ are equal to half of the radius of the circumcircle $c$ of the triangle $ABC$. Since the circles $c$ and $c_1$ are tangent, the diameter of $c_1$ and the radius of $c$, both drawn from the tangent point $A$, coincide. Hence the circle $c_1$ goes through the center of the circle $c$; similarly the circles $c_2$ and $c_3$ go through the center of the circle $c$.\n\n![](attached_image_1.png)\nFigure 4",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 29745,
    "subject": "Mathematics (Multi-modal)",
    "question": "In a plane there is a triangle $ABC$. Line $AC$ is tangent to circle $c_A$ at point $C$ and circle $c_A$ passes through point $B$. Line $BC$ is tangent to circle $c_B$ at point $C$ and circle $c_B$ passes through point $A$. The second intersection point $S$ of circles $c_A$ and $c_B$ coincides with the incenter of triangle $ABC$. Prove that the triangle $ABC$ is equilateral.\n\n![](attached_image_1.png)\n\nFigure 5",
    "options": [],
    "answer": "Detailed solution",
    "solution": "By the tangent-secant theorem we have $\\angle BCS = \\angle CAS$ and $\\angle ACS = \\angle CBS$ (see fig. 5). The incenter of a triangle is the point of intersection of angle bisectors, therefore $\\angle CAB = 2\\angle CAS = 2\\angle BCS = \\angle BCA$ and $\\angle CBA = 2\\angle CBS = \\angle ACS = \\angle BCA$. Hence $ABC$ is equilateral.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 29746,
    "subject": "Mathematics (Multi-modal)",
    "question": "20 students participated on a field trip. They all wanted to climb on top of a lighthouse, but only one person was allowed to the lighthouse at once. The order of climbing was determined by a lottery such that in the beginning every student is randomly assigned a number of 1 through 20 (such that no number is repeated). The one who gets the smallest number is the first one to climb the lighthouse. In the next round all the rest of the students are randomly assigned numbers 1 through 19 and the one who gets the smallest number gets to be the next one to climb the lighthouse. This process is repeated until all the students have climbed the lighthouse. Due to a strange occurrence no one student was assigned the same number more than once. Miku was assigned the number 14 in the first round. Find all possibilities what number could have been assigned to Miku in the 9th round.",
    "options": [],
    "answer": "6",
    "solution": "Let the number of students be $n$. The last student to climb the lighthouse has got all the numbers 1 through $n$ with the lottery. As number $n$ is only available in the first round, that student had to get $n$ in the first round. As number $n-1$ is only available in 1st and 2nd round and in the 1st round that student did not get it, the student got $n-1$ in the 2nd round. Analogously, since $n-2$ is only available in the first three rounds and that student did not get it in the first two rounds, the student got $n-2$ in the 3rd round. Continuing the same way shows that the last one to climb the lighthouse got numbers $n$ through 1 in decreasing order, or got the largest available number in every round.\n\nThe rest of the students who only participated in rounds 1 through $n-1$ have to share in the first round numbers 1 through $n-1$, in the second one 1 through $n-2$, in third one 1 through $n-3$ and so on. Therefore for them this process is as if the person last to climb the tower did not participate at all and $n$ would be smaller by 1. For this holds for any $n$, all students get the numbers in decreasing order with every next one being smaller by 1, that includes Miku. This allows us to find that in the 9th round Miku got number 6.",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 29747,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all pairs of positive rational numbers where the sum of the numbers in a pair is an integer and the sum of (multiplicative) inverses of the numbers in a pair is also an integer.",
    "options": [],
    "answer": "(1/2, 1/2), (1, 1), (2, 2)",
    "solution": "Let the numbers in the pair be represented as reduced fractions $\\frac{a}{b}$ and $\\frac{c}{d}$. For\n$$\n\\frac{a}{b} + \\frac{c}{d} = \\frac{ad + bc}{bd}\n$$\nto be an integer, we must have\n$$\nad + bc = k \\cdot bd \\quad (2)\n$$\nwith $k$ being some integer. By writing the equality (2) as $bc = (kb - a) \\cdot d$ and taking into account that $c$ and $d$ are relatively prime, we see that $b$ is divisible by $d$. By writing the same equality (2) in a form $ad = (kd - c) \\cdot b$ and taking into account that $a$ and $b$ are relatively prime, we see that $d$ is divisible by $b$. Therefore $b = d$.\nFor also $\\frac{b}{a} + \\frac{d}{c}$ to be an integer, we analogously must have $a = c$. Therefore $\\frac{a}{b} = \\frac{c}{d}$.\nConsequently $\\frac{a}{b} + \\frac{c}{d} = \\frac{2a}{b}$ and $\\frac{b}{a} + \\frac{d}{c} = \\frac{2b}{a}$ are both integers. If $a = b$, then $\\frac{a}{b} = 1$. If $a < b$, then $\\frac{2a}{b} < 2$, implying $\\frac{2a}{b} = 1$ and $\\frac{a}{b} = \\frac{1}{2}$ as the only possibility. If $a > b$, then similarly $\\frac{a}{b} = 2$.\nLet the numbers in the pair be represented as reduced fractions $\\frac{a}{b}$ and $\\frac{c}{d}$, and let $k = \\frac{a}{b} + \\frac{c}{d}$ and $l = \\frac{b}{a} + \\frac{d}{c}$. Then $\\frac{a}{b} = k - \\frac{c}{d} = \\frac{kd - c}{d}$ and $\\frac{b}{a} = l - \\frac{d}{c} = \\frac{lc - d}{c}$, whence $\\frac{a}{b} = \\frac{c}{lc - d}$. Therefore $\\frac{kd - c}{d} = \\frac{c}{lc - d}$, or $cd = (kd - c)(lc - d)$. The resulting equation is equivalent to $l \\cdot c^2 - kld \\cdot c + kd^2 = 0$, giving\n$$\nc = \\frac{kld \\pm \\sqrt{k^2 l^2 d^2 - 4kld^2}}{2l} = d \\cdot \\frac{kl \\pm \\sqrt{k^2 l^2 - 4kl}}{2l} \\quad (3)\n$$\nFor $c$ to be an integer, we must have $k^2 l^2 - 4kl = n^2$ where $n$ is an integer. Now\n$$\n(kl - 2)^2 = n^2 + 4.\n$$\nTherefore $n^2 + 4$ must also be a square of an integer. This is only possible when $n = 0$ – therefore $kl = 0$ or $kl = 4$. The first option is not possible, because $k$ and $l$ are the sums\nof positive real numbers. The second option gives three possible cases: $(k,l)$ can either be $(1,4), (2,2)$ or $(4,1)$. It remains to find all possible values of $\\frac{c}{d}$ from (3) and calculate $\\frac{a}{b} = k - \\frac{c}{d}$.\n\nFor also $\\frac{b}{a} + \\frac{d}{c}$ to be an integer, we analogously must have $a = c$. Therefore $\\frac{a}{b} = \\frac{c}{d}$.\nConsequently $\\frac{a}{b} + \\frac{c}{d} = \\frac{2a}{b}$ and $\\frac{b}{a} + \\frac{d}{c} = \\frac{2b}{a}$ are both integers. If $a = b$, then $\\frac{a}{b} = 1$. If $a < b$, then $\\frac{2a}{b} < 2$, implying $\\frac{2a}{b} = 1$ and $\\frac{a}{b} = \\frac{1}{2}$ as the only possibility. If $a > b$, then similarly $\\frac{a}{b} = 2$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 29748,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $a$ and $b$ be positive integers such that $b$ is divisible by $a$ and writing $a$ and $b$ one after another in this order gives $(a+b)^2$. Prove that $\\frac{b}{a} = 6$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $n$ be the number of digits of $b$ and let $b = k a$. Then by the conditions of the problem, $10^n \\cdot a + k a = (a + k a)^2$, or\n$$\na = \\frac{10^n + k}{(k+1)^2} \\qquad (1)\n$$\nIf $k$ were odd, then the numerator on the r.h.s. of (1) would be odd and the denominator even, so $a$ could not be an integer. Hence $k$ is even.\n\nIf $k = 2$ then the cross-sum of $10^n + 2$ is 3, which is not divisible by $(2+1)^2 = 9$. The case $k = 4$ also leads to a contradiction, since $10^n + 4$ ends with 4, hence cannot be divisible by $(4+1)^2 = 25$. Thus $k \\ge 6$.\nIn the following we show first that $k \\le 8$ and finally that $k \\ne 8$. The assumptions $k a = b \\ge 10^{n-1}$ give $10 k a \\ge 10^n$. Equality (1) implies\n$$\n10^n = (k+1)^2 \\cdot a - k = k^2 a + 2 k a + a - k = (k+2) \\cdot k a + a - k.\n$$\nThus $10 k a \\ge (k+2) \\cdot k a + a - k$, whence\n$$\n(8-k) \\cdot k a \\ge a-k.\n$$\nAs $a$ is positive, $(8-k) \\cdot k a > -k$. As both sides of this inequality are divisible by $k$, this implies $(8-k) \\cdot k a \\ge 0$. Consequently $8-k \\ge 0$, i.e., $k \\le 8$.\nIf $k=8$, the inequality (2) implies $a \\le 8$ whereas the equality (1) reduces to $a = \\frac{10^n+8}{81}$. Hence $a$ ends with digit 8, leaving $a=8$ and $b=8 \\cdot 8 = 64$ as the only possibility. But $864 \\ne (8+64)^2$, contradicting the conditions of the problem.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 29749,
    "subject": "Mathematics (Multi-modal)",
    "question": "The angles of a triangle are $22.5^\\circ$, $45^\\circ$ and $112.5^\\circ$. Prove that inside this triangle there exists a point that is located on the median through one vertex, the angle bisector through another vertex and the altitude through the third vertex.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $D$ be the point of intersection of $BC$ and median from vertex $A$, $E$ be the point of intersection of angle bisector from vertex $B$ and $AC$, and $F$ be the point of intersection of altitude from vertex $C$ and $AB$ (see fig. 6). As $\\angle FBC = 45^\\circ$ and $\\angle CFB = 90^\\circ$, triangle $FBC$ is a right isosceles triangle with $|CF| = |FB|$.\n\n![](attached_image_1.png)\nFigure 6\n\nLet $X$ and $Y$ be the points of intersection of the line that passes through point $D$ and is parallel to $CF$, with lines $AC$ and $AF$, respectively. We have $\\angle YAX = 22.5^\\circ$, $\\angle XYA = 90^\\circ$ and $\\angle AXY = 180^\\circ - \\angle YAX - \\angle XYA = 67.5^\\circ$. Therefore\n$$\n\\angle XCD = 180^\\circ - \\angle BCA = 67.5^\\circ = \\angle AXY = \\angle CXD,\n$$\ndue to which $|XD| = |CD| = |DB|$.\nAs line segments $DY$ and $CF$ are parallel, $DY$ is the midsegment of triangle $BCF$. Hence\n$$\n\\frac{|XD|}{|DY|} = \\frac{|DB|}{|DY|} = \\frac{|CB|}{|CF|} = \\frac{|CB|}{|FB|}.\n$$\nLet now $K$ be the point of intersection of $BE$ and $CF$. The angle bisector property gives that $\\frac{|CK|}{|KF|} = \\frac{|CB|}{|FB|}$, so $\\frac{|XD|}{|DY|} = \\frac{|CK|}{|KF|}$, from which $\\angle FAK = \\angle YAD$. Therefore also $AD$ passes through $K$, QED.\n\n![](attached_image_2.png)\n\n![](attached_image_2.png)\nFigure 7\nSimilarly to the previous solution we pick triangle $ABC$, mark points $D$, $E$ and $F$ and show that $|CF| = |FB|$. Additionally notice that $\\angle ABE = \\frac{\\angle ABC}{2} = \\frac{45^\\circ}{2} = 22.5^\\circ = \\angle BAE$, which gives $|BE| = |AE|$.\nLet now $Z$ be a point on $BC$ such that $\\angle BEZ = 90^\\circ$ (see fig. 7). Then $\\angle ZBE = \\frac{\\angle ABC}{2} = 22.5^\\circ = \\angle CAF$ and $\\angle BEZ = 90^\\circ = \\angle AFC$, hence triangles $BEZ$ and $AFC$ are\n$$\n\\angle EZC = 180^\\circ - \\angle ZBE - \\angle BEZ = 67.5^\\circ = 180^\\circ - \\angle BCA = \\angle ECZ,\n$$\nwhich gives $|EZ| = |EC|$. Thus $\\frac{|AE|}{|EC|} = \\frac{|BE|}{|EZ|} = \\frac{|AF|}{|FC|} = \\frac{|AF|}{|FB|}$. Therefore $\\frac{|AF|}{|FB|} \\cdot \\frac{|BD|}{|DC|} = \\frac{|CE|}{|EA|} = 1$ and Ceva's theorem gives that $AD$, $BE$ and $CF$ intersect in one point.\nSimilarly to first solution we choose triangle $ABC$, mark points $D$, $E$, $F$ and $Y$ and show that $|CF| = |FB|$.\nLet the points of intersection of $CF$ with angle bisector $BE$ and median $AD$ be $K_1$ and $K_2$, respectively. Let $h_1 = |FK_1|$ and $h_2 = |FK_2|$ and in addition let $u = |AF|$ and $v = |CF| = |FB|$. Let us show that $h_1 = h_2$, then $K_1 = K_2$. Using the similarity of triangles $BFK_1$ and $CFA$, similarity of triangles $BCK_1$ and $ABC$ and similarity of triangles $AFK_2$ and $AYD$, we get the equalities\n$$\n\\frac{h_1}{v} = \\frac{v}{u}, \\quad \\frac{v-h_1}{\\sqrt{2}v} = \\frac{\\sqrt{2}v}{u+v}, \\quad \\frac{h_2}{u} = \\frac{\\frac{1}{2}v}{u+\\frac{1}{2}v}\n$$\nwhich imply $\\frac{v^2}{u} = h_1 = \\frac{(u-v)v}{u+v}$ and $h_2 = \\frac{uv}{2u+v}$. From equation $\\frac{v^2}{u} = \\frac{(u-v)v}{u+v}$ we get $2u + v = \\frac{u^2}{v}$, therefore $h_2 = \\frac{uv}{2u+v} = \\frac{v^2}{u} = h_1$.",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 29750,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $x$ and $y$ be different positive integers. Prove that $\\frac{x^2+4xy+y^2}{x^3-y^3}$ is never an integer.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "By symmetry we can assume that $x > y$. If $x - y = 1$, then\n$$\n\\begin{aligned} \\frac{x^2 + 4xy + y^2}{x^3 - y^3} &= \\frac{x^2 + 4xy + y^2}{(x-y)(x^2 + xy + y^2)} = \\frac{(x-y)^2 + 6xy}{(x-y)((x-y)^2 + 3xy)} = \\\\ &= \\frac{1 + 6xy}{1 + 3xy} = 1 + \\frac{3xy}{1 + 3xy}, \\end{aligned}\n$$\nwhich is clearly not an integer. If $x - y \\ge 2$, then\n$$\n\\begin{aligned} \\frac{x^2 + 4xy + y^2}{x^3 - y^3} &= \\frac{x^2 + 4xy + y^2}{(x-y)(x^2 + xy + y^2)} \\le \\frac{x^2 + 4xy + y^2}{2(x^2 + xy + y^2)} < \\\\ &< \\frac{2x^2 + 2xy + 2y^2}{2(x^2 + xy + y^2)} = 1, \\end{aligned}\n$$\nwhere the last inequality follows from $x^2 - 2xy + y^2 = (x-y)^2 > 0$.\nIf $\\frac{x^2 + 4xy + y^2}{x^3 - y^3}$ were an integer, then\n$$\n\\frac{x^2 + 4xy + y^2}{x^3 - y^3} \\cdot (x - y) - 1 = \\frac{3xy}{x^2 + xy + y^2}\n$$\n\n\\begin{aligned} \\frac{x^2 + 4xy + y^2}{x^3 - y^3} &= \\frac{x^2 + 4xy + y^2}{(x-y)(x^2 + xy + y^2)} \\le \\frac{x^2 + 4xy + y^2}{2(x^2 + xy + y^2)} < \\\\ &< \\frac{2x^2 + 2xy + 2y^2}{2(x^2 + xy + y^2)} = 1, \\end{aligned}\n$$\nwhere the last inequality follows from $x^2 - 2xy + y^2 = (x-y)^2 > 0$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 29751,
    "subject": "Mathematics (Multi-modal)",
    "question": "During the schoolyear $22$ olympiads were held. At each one $5$ best students were awarded. It is known that the prize receivers of every two olympiads had exactly $1$ student in common. Show that there exists a student who got a prize at every olympiad.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Look at an arbitrary olympiad, let that be $A_1$, where the prizes went to some $5$ students. Each of the remaining $21$ olympiads had to have someone among those $5$ receiving a prize. By pigeonhole principle there exists a student who in addition to $A_1$ also got a prize at at least $5$ olympiads. Let that student be $a$ and those olympiads be $A_2, \\dots, A_6$.\n\nLet now $B$ be an arbitrary olympiad that is different from $A_1, \\dots, A_6$. As each one of the olympiads $A_1, \\dots, A_6$ has one prize-winning student in common with $B$ and exactly $5$ students get prizes at $B$, applying pigeonhole principle again shows that one of those five had to get a prize at at least two of $A_1, \\dots, A_6$. Since these two have student $a$ in common and according to initial conditions that student is the only one, this means that $a$ also got a prize at olympiad $B$. But since we picked $B$ arbitrarily, $a$ must have got a prize at every olympiad.",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 29752,
    "subject": "Mathematics (Multi-modal)",
    "question": "Circles $c_1, c_2$ with centers $O_1, O_2$, respectively, intersect at points $P$ and $Q$ and touch circle $c$ internally at points $A_1$ and $A_2$, respectively. Line $PQ$ intersects circle $c$ at points $B$ and $D$. Lines $A_1B$ and $A_1D$ intersect circle $c_1$ the second time at points $E_1$ and $F_1$, respectively, and lines $A_2B$ and $A_2D$ intersect circle $c_2$ the second time at points $E_2$ and $F_2$, respectively. Prove that $E_1, E_2, F_1, F_2$ lie on a circle whose center coincides with the midpoint of line segment $O_1O_2$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "![](attached_image_1.png)\nFigure 6\n\nSolution:\n\nLet the radii of $c_1, c_2$ and $c$ be $r_1, r_2$ and $r$, respectively. Homothety of ratio $\\frac{r}{r_1}$ with center $A_1$ takes circle $c_1$ to circle $c$ and points $E_1, F_1$ to points $B, D$, respectively. Thus it takes line $E_1F_1$ to line $BD$. Analogously, homothety of ratio $\\frac{r}{r_2}$ with center $A_2$ takes line $E_2F_2$ to line $BD$. Consequently, lines $E_1F_1$ and $E_2F_2$ are parallel to line $BD$ (Fig. 6).\n\nFurthermore, note that $|BE_1| \\cdot |BA_1| = |BP| \\cdot |BQ|$ and $|BE_2| \\cdot |BA_2| = |BP| \\cdot |BQ|$, implying $|BE_1| \\cdot |BA_1| = |BE_2| \\cdot |BA_2|$. Thus triangles $BE_1E_2$ and $BA_2A_1$ are similar and\n$$\n\\angle BE_1E_2 = \\angle BA_2A_1 = \\angle BDA_1 = \\angle E_1F_1A_1 = \\frac{1}{2} \\angle E_1O_1A_1 = 90^\\circ - \\angle O_1E_1A_1, \\quad (3)\n$$\nwhence\n$$\n\\angle E_2E_1O_1 = 180^\\circ - \\angle BE_1E_2 - O_1E_1A_1 = 90^\\circ. \\quad (4)\n$$\nAnalogously, $\\angle E_1E_2O_2 = 90^\\circ$. Hence the quadrilateral $E_1E_2O_2O_1$ is a right-angled trapezoid (or rectangle in the case $r_1 = r_2$) and the midpoint of the line segment $O_1O_2$ lies on the perpendicular bisector of the line segment $E_1E_2$, thus being equidistant from $E_1$ and $E_2$. Analogously, the midpoint of the line segment $O_1O_2$ is also equidistant from $F_1$ and $F_2$.\n\nAs line $O_1O_2$ is perpendicular to $BD$, line $O_1O_2$ is also perpendicular to $E_1F_1$. Thus the line segment $O_1O_2$ entirely lies on the perpendicular bisector of $E_1F_1$. This means that the midpoint of line segment $O_1O_2$ is equidistant from $E_1$ and $F_1$.\n\nAltogether, we have shown that these four points lie on a circle with its center at the midpoint of the line segment $O_1O_2$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 29753,
    "subject": "Mathematics (Multi-modal)",
    "question": "a) Does there exist an integer $c$ and a polynomial $P(x)$ with integer coefficients for which $P(c) \\neq c$, but $P(P(c)) = c$?\n\nb) Does there exist an integer $c$ and a polynomial $P(x)$ with integer coefficients for which $P(c) \\neq c$ and $P(P(c)) \\neq c$, but $P(P(P(c))) = c$?",
    "options": [],
    "answer": "a) Yes: for example, P(x) = -x with c = 1. b) No: such a polynomial and integer do not exist.",
    "solution": "a) Take $P(x) = -x$ and $c = 1$, then $P(c) = -1 \\neq c$ and $P(P(c)) = P(-1) = 1 = c$.\n\nb) Suppose that there exist a polynomial with integer coefficients $P(x)$ and an integer $c$, for which $P(c) \\neq c$, $P(P(c)) \\neq c$ and $P(P(P(c))) = c$. Notice that $P(c) \\neq P(P(c))$, because otherwise $P(P(c)) = P(P(P(c))) = c$, which would contradict the assumption.\nIn the following calculations we will use the well-known property of polynomials with integer coefficients: $k - m \\mid P(k) - P(m)$ for any distinct integers $k$ and $m$.\nUsing this and the premise that $P(P(P(c))) = c$, we see that number $c - P(c) = P(P(P(c))) - P(P(P(P(c))))$ is divisible by $P(P(c)) - P(P(P(c))) = P(P(c)) - c$, which in turn is divisible by $P(c) - P(P(c))$, which in turn is divisible by $c - P(c)$. Hence $|c - P(c)| \\ge |P(P(c)) - c| \\ge |c - P(c)|$, implying $|c - P(c)| = |P(P(c)) - c|$, and similarly $|c - P(c)| = |P(c) - P(P(c))|$.\nTherefore three numbers $c$, $P(c)$ and $P(P(c))$ are all at the same distance away from each other on the number line, which is possible only when those numbers coincide, that is $c = P(c) = P(P(c))$, which contradicts the assumptions made. This contradiction shows that polynomial $P(x)$ and integer $c$ do not exist.",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 29754,
    "subject": "Mathematics (Multi-modal)",
    "question": "Teacher tells Jüri two nonzero integers $a$ and $b$ such that $b$ is divisible by $a$. Jüri has to find a nonzero integer $c$ such that $c$ is divisible by $b$ and all solutions of the quadratic equation $ax^2 + bx + c = 0$ are integers. Can Jüri always solve the problem?",
    "options": [],
    "answer": "Yes; for b = a q, choose c = -2 a q^2. Then the roots are q and -2 q.",
    "solution": "By the conditions of the problem there is an integer $q$ such that $b = aq$. Let $c = -2aq^2$; then $c \\neq 0$ and $c$ is divisible by $b$. The quadratic equation $ax^2 + bx + c = 0$ or $ax^2 + aqx - 2aq^2 = 0$ has solutions $q$ and $-2q$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 29755,
    "subject": "Mathematics (Multi-modal)",
    "question": "On the board there are numbers $1$, $2$, $3$, $4$, $5$ and $6$. In every step Juku deletes some two numbers $a$ and $b$ on the board and writes $ab + a + b$ on the board instead. He repeats such steps until there is only one number on the board. Find all possibilities what could be the last number on the board.",
    "options": [],
    "answer": "5039",
    "solution": "Since $(a+1)(b+1) = c+1$, where $c$ is a number that would be written on the board instead of $a$ and $b$, the product of the numbers that are greater by $1$ than the numbers on the board does not change in the process. In the beginning the product is $2 \\cdot 3 \\cdot 4 \\cdot 5 \\cdot 6 \\cdot 7$ or $5040$. Consequently the only number that can be on the board in the end is $5040 - 1 = 5039$.\nLet us define $x * y = xy + x + y$ for every two numbers $x$ and $y$, then in each step Juku substitutes some $a$ and $b$ with $a * b$. The binary operation $*$ is commutative and associative, due to which the final result is independent on the order the numbers are being deleted. Therefore in the end the board will contain one number that equals $((1 * 2) * (3 * 4)) * (5 * 6) = (5 * 19) * 41 = 119 * 41 = 5039$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 29756,
    "subject": "Mathematics (Multi-modal)",
    "question": "In a triangle $ABC$ the midpoints of $BC$, $CA$ and $AB$ are $D$, $E$ and $F$, respectively. Prove that the circumcircles of triangles $AEF$, $BFD$ and $CDE$ intersect all in one point.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let us first assume that triangle $ABC$ is not a right triangle – then the circumcenter $O$ of the triangle $ABC$ does not coincide with $D$, $E$, $F$ (see fig. 1). As the circumcenter is in the point of intersection of perpendicular bisectors of the sides,\n\n$\\angle AEO = 90^\\circ = \\angle AFO$, due to which $A, E, F, O$ are concyclic, so $O$ is located on the circumcircle of $AEF$. Analogously $O$ is also located on the circumcircles of $BFD$ and $CDE$. Therefore $O$ is the point we are looking for.\n\nIn the end let us also look at the case where $ABC$ is a right triangle – without loss of generality let $\\angle ACB = 90^\\circ$ (see fig. 2). The circumcircles of triangles $AEF$ and $BFD$ obviously pass through $F$. As $DF \\parallel AC$ and $EF \\parallel BC$ by midline property, we have $DF \\perp BC$ and $EF \\perp AC$. Therefore also $\\angle EFD = 90^\\circ$. Since $\\angle DCE = 90^\\circ$, the line segment $DE$ is the diameter of the circumcircle of $CDE$, due to which it also passes through $F$. Therefore $F$ is the point we are looking for.\n\n![](attached_image_1.png)\nFigure 1\n![](attached_image_2.png)\nFigure 2\nSince $DE$, $EF$ and $FD$ are the midsegments of triangle $ABC$, triangles $AEF$, $FDB$ and $ECD$ are congruent. Therefore their circumcircles also have radii of equal length. Let that length be $r$.\n\nLet the circumcenters of $AEF$, $BFD$ and $CDE$ be $G$, $H$ and $I$, respectively. The circumcenter of a triangle is located in the point of intersection of perpendicular bisectors of the sides, therefore $G$ is located on the perpendicular bisector of $AF$ and $H$ on the perpendicular bisector of $FB$. As triangles $AEF$ and $FDB$ are congruent, points $G$ and $H$ are also located at equal distance from $AB$, due to which the distance between $G$ and $H$ is equal to the distance between the perpendicular bisectors of $AF$ and $FB$. In conclusion\n\n$$\n|GH| = \\frac{1}{2}|AF| + \\frac{1}{2}|FB| = \\frac{1}{2}(|AF| + |FB|) = \\frac{1}{2}|AB| = |AF| = |FB| = |ED|.\n$$\n\nAnalogously $|HI| = |BD| = |DC| = |FE|$ and $|IG| = |CE| = |EA| = |DF|$. Hence the triangle $GIH$ is congruent to triangles $AEF$, $FDB$ and $ECD$ and the radius of the circumcircle of $GIH$ is $r$. The circumcenter $X$ of triangle $GHI$ therefore satisfies $|XG| = |XH| = |XI| = r$, so $X$ is located on the circumcircles of $AEF$, $BFD$ and $CDE$.\n\n![](attached_image_1.png)\nFigure 1\n![](attached_image_2.png)\nFigure 2\nA homothetic transformation with $A$ being the homothetic center and with scaling factor $\\frac{1}{2}$ takes point $B$ to $F$ and $C$ to $E$, therefore the circumcircle of the triangle $ABC$ goes to the circumcircle of triangle $AFE$. Due to factor $\\frac{1}{2}$ the circumcircle of $AFE$ passes through the circumcenter $O$ of triangle $ABC$. Analogously the circumcircles of $BFD$ and $CDE$ also pass through $O$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 29757,
    "subject": "Mathematics (Multi-modal)",
    "question": "Inside a circle $c$ with the center $O$ there are two circles $c_1$ and $c_2$ which go through $O$ and are tangent to the circle $c$ at points $A$ and $B$ respectively. Prove that the circles $c_1$ and $c_2$ have a common point which lies in the segment $AB$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "The radius $AO$ of the circle $c$ is perpendicular to the common tangent to circles $c$ and $c_1$ at the point $A$, hence $AO$ is a diameter of the circle $c_1$. Similarly $BO$ is a diameter of the circle $c_2$.\n\nIf the circles $c_1$ and $c_2$ are tangent at the point $O$ (Fig. 1), then the diameters $AO$ and $BO$ are both perpendicular to the common tangent to $c_1$ and $c_2$ at the point $O$, whence the lines $AO$ and $BO$ coincide, i.e. $O$ lies in the segment $AB$.\n\nIf the circles $c_1$ and $c_2$ intersect at $O$ (Fig. 2), then let $M$ be the other intersection point of the circles. Since $\\angle AMO = 90^\\circ$ and $\\angle BMO = 90^\\circ$ (angles at the circumference supported by a diameter), the lines $AM$ and $BM$ coincide and $M$ lies in the segment $AB$.\n\n![](attached_image_1.png)\nFigure 1\n\n![](attached_image_2.png)\nFigure 2",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 29758,
    "subject": "Mathematics (Multi-modal)",
    "question": "A boardgame board consists of 10 squares in a row that are numbered 1 to 10. On some square there is a button. In one move it is allowed to move the button to a square whose number is either smaller by 2 or 2 times bigger. Does there exist an initial location for the button that allows the player to visit all squares of the board? It is allowed to visit one square several times.",
    "options": [],
    "answer": "No",
    "solution": "No move allows the button to be placed to the square number 9. Therefore the button should start from there to have any hope. If on some later move the button is placed on an even-numbered square, then it will also stay on an even-numbered square on every move that follows. Therefore all the odd-numbered squares must be visited right in the beginning, i.e., the button must be moved to 7, 5, 3, 1. On the next move there is no other option but to step to square number 2. But now it is impossible to reach square number 5, since it is odd-numbered, and therefore it is also impossible to reach 10. Therefore it is not possible to visit all the squares.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 29759,
    "subject": "Mathematics (Multi-modal)",
    "question": "Numbers $1, \\dots, 200$ are written on a blackboard in one line. Juku has to write in front of each number plus or minus sign so that for any positive integer $n \\le 100$ the number itself and one of its multiples have different signs. Which numbers must he assign a minus sign in order to get the maximal possible value of the expression?",
    "options": [],
    "answer": "Minus signs on 51 through 100; plus signs on all other numbers.",
    "solution": "*Answer:* The numbers $51, \\dots, 100$.\n\nIf Juku writes a minus in front of the number $51, \\dots, 100$ and a plus in front of the others, then the conditions of the problem are satisfied: for $51 \\le n \\le 100$, the numbers $n$ and $2n$ have different signs; for $n \\le 50$ there is at least one multiple of $n$ among the numbers $51, \\dots, 100$.\n\nTo show that this arrangement of the signs gives the maximal value of the expression,\n\nconsider an arbitrary arrangement of signs satisfying the conditions of the problem. Then always when $100 \\ge n \\ge 67$ and $n$ has a plus sign, $2n$ must have a minus sign. Also when $66 \\ge n \\ge 51$ and $n$ has a plus sign, then either $2n$ or $3n$ must have a minus sign. If we change all the pluses in front of the numbers $n$ with $100 \\ge n \\ge 51$ to minuses, and the minuses in front of the corresponding $2n$ or $3n$ to pluses, then changing minus to plus in front of $m$ corresponds to changing plus to minus in front of $\\frac{m}{2}$ or $\\frac{m}{3}$ or both. Since $\\frac{m}{2} + \\frac{m}{3} < m$, the changes increase the value of the expression. Then we can also change all remaining minuses in front of the numbers $1, \\dots, 50$ and $101, \\dots, 200$ to pluses, which also can only increase the value of the expression. This results in the arrangement of the signs described in the beginning. Hence this arrangement gives the maximal value of the expression.",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 29760,
    "subject": "Mathematics (Multi-modal)",
    "question": "Kärt writes the fractions $\\frac{1}{2}$ and $\\frac{1}{3}$ on the blackboard and Märt writes 10 positive integers on the paper, which he does not show to Kärt. Then Kärt starts to write fractions on the blackboard by the following rule: on each step she chooses two fractions $\\frac{a}{b}$ and $\\frac{c}{d}$ which are already on the blackboard and writes on the blackboard the fraction $\\frac{a+c}{b+d}$ after reducing. Can Kärt always choose the fractions so that after a number of steps she writes on the blackboard a fraction whose denominator is coprime with all the numbers Märt has written on the paper?",
    "options": [],
    "answer": "Yes",
    "solution": "Answer: Yes.\n\nThe first fraction that Kärt adds to the blackboard has to be $\\frac{2}{5}$. On every following move, let Kärt pick $\\frac{1}{2}$ as one fraction and the latest written fraction as the other fraction. Ignoring the reducing step, this means that the denominator of every added fraction is larger than the previous fraction by 2, or that the denominators of the fractions are consecutive odd numbers. This is indeed the case, because all fractions added in this way are irreducible (these fractions have the form $\\frac{k}{2k+1}$, and $k$ is always coprime with $2k+1$ because any common divisor would also divide $(2k+1) - 2k = 1$).\n\nTherefore the denominators of the fractions that Kärt writes include all prime numbers except 2. Since there are infinitely many prime numbers, Kärt will eventually write a fraction with a prime denominator that is larger than all of the numbers written by Märt, and hence coprime with them.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 29761,
    "subject": "Mathematics (Multi-modal)",
    "question": "In a scalene triangle one angle is exactly two times as big as another one and some angle in this triangle is $36°$. Find all possibilities, how big the angles of this triangle can be.",
    "options": [],
    "answer": "The angle sets are {18°, 36°, 126°} and {36°, 48°, 96°}.",
    "solution": "Based on the initial conditions the angles of the triangle are $\\alpha$, $2\\alpha$ and $180° - 3\\alpha$ and they all have to be different. It remains to perform calculations for three cases: $\\alpha = 36°$, $2\\alpha = 36°$ and $180° - 3\\alpha = 36°$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 29762,
    "subject": "Mathematics (Multi-modal)",
    "question": "Publisher Soothsayer published a reference book claiming that for each real number $x$ and positive even number $n$ the equality $(1+x)^n \\ge 2^n x$ holds. Is this claim true?",
    "options": [],
    "answer": "No",
    "solution": "*Answer:* No.\n\nThe inequality does not hold for example when $x = \\frac{1}{2}$ and $n = 4$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 29763,
    "subject": "Mathematics (Multi-modal)",
    "question": "In an isosceles right triangle $ABC$ the right angle is at vertex $C$. On the side $AC$ points $K$, $L$ and on the side $BC$ points $M$, $N$ are chosen so that they divide the corresponding side into three equal segments. Prove that there is exactly one point $P$ inside the triangle $ABC$ such that $\\angle KPL = \\angle MPN = 45^\\circ$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "![](attached_image_1.png)\nFigure 3\n\nWithout loss of generality let the points on the side $AC$ be in the order $A$, $K$, $L$, $C$ and on the side $BC$ in the order $C$, $M$, $N$, $B$ (see Fig. 3). Choose the point $P$ so that the quadrilateral $LCMP$ is a square. Then $|KL| = |LC| = |LP|$ and $|MN| = |CM| = |MP|$, i.e. $KLP$ and $PMN$ are isosceles right triangles, so $\\angle KPL = \\angle MPN = 45^\\circ$. Since $\\angle KPN = 45^\\circ + 90^\\circ + 45^\\circ = 180^\\circ$, the point $P$ lies inside the segment $KN$, whose all points except the endpoints are inside the triangle $ABC$.\n\nTo show that $P$ is the only point with the required properties, let $P'$ be an arbitrary point inside the triangle $ABC$ which satisfies $\\angle KP'L = \\angle MP'N = 45^\\circ$. Since $P$ and $P'$ are on the same side of the line $KL$ and $\\angle KPL = \\angle KP'L$, the point $P'$ lies on the circumcircle of the triangle $KPL$; similarly it also lies on the circumcircle of the triangle $MPN$. Since $\\angle KLP = \\angle PMN = 90^\\circ$, the segments $KP$ and $PN$ are the diameters of the circles. Since the diameters $KP$ and $PN$ lie on the same straight line $KN$, they have a common perpendicular at the point $P$ which is tangent to both circles at this point. Hence the point $P$ is the only common point of these circles, i.e. $P' = P$.",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 29764,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let the *odd part* of a positive integer $n$ be the greatest odd integer that divides $n$.\nDoes there exist a positive odd integer that cannot be represented as a product of the odd parts of two consecutive positive integers?",
    "options": [],
    "answer": "11",
    "solution": "Let us show that number $11$ cannot be represented as a product of the odd parts of two consecutive positive integers. Assume the opposite: let $11 = x \\cdot y$, where $x$ and $y$ are the odd parts of two consecutive positive integers. As $11$ is a prime number, either $x = 1$ and $y = 11$ or $x = 11$ and $y = 1$. As of the two consecutive integers one is always odd and the odd part of an odd number is the number itself, either $x$ or $y$ is one of the two consecutive integers. If it is $1$, then the other number can only be $2$, but the odd part of $2$ is not $11$. If it is $11$, then the other number can only be $10$ or $12$, but neither of those has odd part equal to $1$. In all cases we got a contradiction which proves the statement.\nLet us show $11$ cannot be represented in the required way. If this representation existed, then due to the primality of $11$ the factors would have to be $1$ and $11$. Therefore one of the two consecutive integers has to be divisible by $11$. But this number cannot be $11$ itself, since neither $10$ nor $12$ has $1$ as its odd part. It also cannot be an odd multiple of $11$, because then its odd part would be the number itself rather than $11$. Finally, it cannot be an even multiple of $11$, since in such case the neighbouring numbers would be odd numbers greater than $1$, the odd parts of which are numbers themselves rather than $1$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 29765,
    "subject": "Mathematics (Multi-modal)",
    "question": "a) There are three numbers $a$, $b$, $c$ such that $a \\le b \\le c$. Let $p$, $q$, $r$ be the pairwise sums $a+b$, $b+c$, $c+a$ in the order such that $p \\le q \\le r$. Given that $r-q = q-p$, is it certainly true that $c-b = b-a$?\n\nb) There are four numbers $e$, $f$, $g$, $h$ such that $e \\le f \\le g \\le h$. Let $u$, $v$, $w$, $x$, $y$, $z$ be the pairwise sums of those numbers, in the order $u \\le v \\le w \\le x \\le y \\le z$. Given that $z-y = y-x = x-w = w-v = v-u$, is it certainly true that $h-g = g-f = f-e$?",
    "options": [],
    "answer": "a) Yes. b) No; for example, zero, one, two, and four.",
    "solution": "a) If $a \\le b \\le c$, then $a+b \\le a+c \\le b+c$, due to which $p = a+b$, $q = a+c$ and $r = b+c$. Equality $r-q = q-p$ can now be written as $(b+c) - (a+c) = (a+c) - (a+b)$, simplifying to $b-a = c-b$.\n\nb) Let $e = 0$, $f = 1$, $g = 2$ and $h = 4$. Their pairwise sums in increasing order are $u = 0+1=1$, $v = 0+2=2$, $w = 1+2=3$, $x = 0+4=4$, $y = 1+4=5$ and $z = 2+4=6$. Thus $z-y = y-x = x-w = w-v = v-u = 1$, but $h-g = 2 \\ne 1 = g-f$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 29766,
    "subject": "Mathematics (Multi-modal)",
    "question": "The numbers $1, 2, \\ldots, 2012$ are written on the blackboard in some order, each of them exactly once. Between each two neighboring numbers the absolute value of their difference is written and the original numbers are erased. This is repeated until only one number is left on the blackboard. What is the largest possible number that can be left on the blackboard?",
    "options": [],
    "answer": "2010",
    "solution": "The largest number on the blackboard cannot increase on any step, because the absolute value of the difference of two nonnegative numbers cannot be greater than the maximum of these two numbers. Since in the beginning all the numbers are different and positive, after the first step the largest possible number is $2011$ and the smallest possible number is $1$. After the second step the largest possible number is $2010$ and hence the number left on the blackboard in the end cannot be larger than $2010$.\n\nThe number $2010$ can be left on the blackboard, for example when in the beginning the numbers are written in the order $2012, 1, 2, 3, \\ldots, 2011$. Then after the first step there are the numbers $2011, 1, 1, \\ldots, 1$, and after the second step the numbers $2010, 0, 0, \\ldots, 0$. On each following step the number of zeroes decreases by one and in the end only the number $2010$ remains.",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 29767,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all pairs of integers $(a, b)$ satisfying $(a+1)(b-1) = a^2 b^2$.",
    "options": [],
    "answer": "(0,1) and (-1,0)",
    "solution": "Answer: $(0,1)$ and $(-1,0)$.\n\nSince $a$ and $a+1$ are coprime, $a^2$ and $a+1$ are also coprime. Similarly $b^2$ and $b-1$ are coprime. Hence the equality can hold only in the case $a+1 = \\pm b^2$ and $b-1 = \\pm a^2$, where the signs in both equations are the same.\n\nLet both signs be pluses. Then from the first equation we get $a = b^2 - 1 = (b-1)(b+1)$. The second equation implies $b-1 = a^2$, whence $a = a^2(a^2+2)$. If $a=0$, then $b=1$, i.e. $(a,b) = (0,1)$. If $a \\neq 0$, then by dividing by $a$ we get $1 = a(a^2+2)$; since $a^2+2 > 1$, this equation does not have integer solutions.\n\nIf both signs are minuses then by multiplying by $-1$ we get $-b+1 = a^2$ and $-a-1 = b^2$. These are the same equations with respect to $-b$ and $-a$ which we had previously with respect to $a$ and $b$, hence the only solution is $-b = 0, -a = 1$ i.e. $(a,b) = (-1,0)$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 29768,
    "subject": "Mathematics (Multi-modal)",
    "question": "In the plane there are six different points $A$, $B$, $C$, $D$, $E$, $F$ such that $ABCD$ and $CDEF$ are parallelograms. What is the maximum number of those points that can be located on one circle?\n\n*Answer:* 5.",
    "options": [],
    "answer": "5",
    "solution": "As $ABCD$ and $CDEF$ are parallelograms, the line segments $AB$, $CD$ and $EF$ are parallel and have same length. Since it is impossible to draw three chords of equal length to a circle, not all 6 points can be concyclic.\n\n![](attached_image_1.png)\nFigure 3\n![](attached_image_2.png)\nFigure 4\n\nA construction with 5 concyclic points is in fig. 3.\n\n*Note.* There are many constructions with 5 vertices. We can, e.g., take a rectangle $ABCD$, add the fifth point $E$ randomly on the circumcircle of the rectangle and choose point $F$ such that $CDEF$ would be a parallelogram (see fig. 4).",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 29769,
    "subject": "Mathematics (Multi-modal)",
    "question": "In Wonderland, the government of each country consists of exactly $a$ men and $b$ women, where $a$ and $b$ are fixed natural numbers and $b > 1$. For improving of relationships between countries, all possible working groups consisting of exactly one government member from each country, at least $n$ among whom are women, are formed (where $n$ is a fixed non-negative integer). The same person may belong to many working groups. Find all possibilities how many countries can be in Wonderland, given that the number of all working groups is prime.",
    "options": [],
    "answer": "1",
    "solution": "Let $r$ be the number of countries in Wonderland. If the minimal number of women in working groups is $n = 0$ then forming a working group means just choosing one government member from each country. Thus there are $(a+b)^r$ different working groups. This number can be prime only if $r = 1$ because $a+b \\ge b > 1$.\n\nIf the minimal number of women in working groups is $n \\ge 1$ then a working group containing exactly $k$ women ($n \\le k \\le r$) can be formed as follows. Choose $k$ countries out of $r$, that send a woman to that particular working group, then choose one woman out of $b$ from each of the $k$ governments, and finally choose one man out of $a$ from each of the remaining $r-k$ countries. Hence there are $\\binom{r}{k} b^k a^{r-k}$ working groups with exactly $k$ women, and $\\sum_{k=n}^{r} \\binom{r}{k} b^k a^{r-k}$ working groups with at least $n$ women altogether. As $n \\ge 1$, all terms of this sum are divisible by $b$, whence the sum can be a prime only if it is equal to $b$. This is possible only if $r=1$ since otherwise the last term (corresponding to $k=r$) of the sum would be greater than $b$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 29770,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all prime numbers $p$ for which one can find a positive integer $m$ and non-negative integers $a_0, a_1, \\dots, a_m$ less than $p$ such that\n$$\n\\begin{cases} a_0 + a_1 p + \\dots + a_{m-1} p^{m-1} + a_m p^m = 2013, \\\\ a_0 + a_1 + \\dots + a_{m-1} + a_m = 11. \\end{cases}\n$$",
    "options": [],
    "answer": "2003",
    "solution": "*Answer:* 2003.\n\nSubtracting the second equation from the first one gives\n$$\na_1(p-1) + \\dots + a_m(p^m - 1) = 2002.\n$$\nAs the l.h.s. of the obtained equality is divisible by $p-1$, $2002 = 2 \\cdot 7 \\cdot 11 \\cdot 13$ must also be divisible by $p-1$. Thus $p-1$ equals one of $1, 2, 7, 11, 13, 14, 22, 26, 77, 91, 143, 154, 182, 286, 1001$ and $2002$. Since $p$ is prime, only $2, 3, 23$ and $2003$ remain. The first equation of the given system is the $p$-ary representation of 2013, whence the coefficients $a_i$ are uniquely determined by $p$.\n\nNow we study all cases.\n1. If $p=2$ then $m=10$ as $2^{10} < 2013 < 2^{11}$. The second equation implies that all $a_i$s must be ones, but $1+2+2^2+\\dots+2^{10} = 2^{11}-1 = 2047$. Hence there is no solution in this case.\n\n2. Let $p=3$. As $2013 = 2 \\cdot 3 + 3^2 + 2 \\cdot 3^3 + 2 \\cdot 3^5 + 2 \\cdot 3^6$ whereas $2+1+2+2+2 = 9 \\neq 11$, this case gives no solution either.\n\n3. Let $p=23$. As $2013 = 12 + 18 \\cdot 23 + 3 \\cdot 23^2$ while $12 + 18 + 3 > 11$, this case gives no solution either.\n\n4. For $p=2003$, we get $2013 = 10+2003$ and $10+1=11$, so the conditions are satisfied.\nConsequently, 2003 is the only prime number with the desired property.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 29771,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $x_1, \\dots, x_n$ be non-negative real numbers, not all of which are zeros.\n(i) Prove that\n$$\n1 \\le \\frac{\\left(x_1 + \\frac{x_2}{2} + \\frac{x_3}{3} + \\dots + \\frac{x_n}{n}\\right) \\cdot \\left(x_1 + 2x_2 + 3x_3 + \\dots + nx_n\\right)}{\\left(x_1 + x_2 + x_3 + \\dots + x_n\\right)^2} \\le \\frac{(n+1)^2}{4n}.\n$$\n(ii) Show that, for each $n \\ge 1$, both inequalities can hold as equalities.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Applying AM-GM gives\n$$\n\\begin{aligned} \\left(\\sum_{k=1}^{n} \\frac{x_k}{k}\\right) \\left(\\sum_{k=1}^{n} kx_k\\right) &= \\frac{1}{n} \\cdot \\left(\\sum_{k=1}^{n} \\frac{nx_k}{k}\\right) \\left(\\sum_{k=1}^{n} kx_k\\right) \\le \\\\ &\\le \\frac{1}{n} \\cdot \\frac{1}{4} \\left(\\sum_{k=1}^{n} \\frac{nx_k}{k} + \\sum_{k=1}^{n} kx_k\\right)^2 = \\\\ &= \\frac{1}{4n} \\left(\\sum_{k=1}^{n} x_k \\left(\\frac{n}{k} + k\\right)\\right)^2 \\le \\frac{(n+1)^2}{4n} \\left(\\sum_{k=1}^{n} x_k\\right)^2. \\end{aligned}\n$$\n(The last inequality is proved by $\\frac{n}{k} + k \\le n + 1$, as it is equivalent to $(n-k)(k-1) \\ge 0$.)\nThis gives us the necessary upper bound; this bound is achieved for instance if $x_1 = x_n = 1$ and $x_2 = \\dots = x_{n-1} = 0$.\n\nFor the lower bound, estimate the numerator by Cauchy-Schwarz inequality:\n$$\n\\left(\\sum_{k=1}^{n} \\frac{x_k}{k}\\right) \\left(\\sum_{k=1}^{n} kx_k\\right) \\ge \\left(\\sum_{k=1}^{n} \\sqrt{\\frac{x_k}{k}} \\cdot \\sqrt{kx_k}\\right)^2 = \\left(\\sum_{k=1}^{n} x_k\\right)^2;\n$$\nthe equality holds here if exactly one of $x_i$s is non-zero.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 29772,
    "subject": "Mathematics (Multi-modal)",
    "question": "In an acute triangle the feet of altitudes drawn from vertices $A$ and $B$ are $D$ and $E$, respectively. Let $M$ be the midpoint of side $AB$. Line $CM$ intersects the circumcircle of $CDE$ again in point $P$ and the circumcircle of $CAB$ again in point $Q$. Prove that\n\n$$\n|MP| = |MQ|.\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "The orthocenter $H$ of the triangle $ABC$ is located on the circumcircle of the triangle $CDE$, because $\\angle HDC + \\angle HEC = 90^\\circ + 90^\\circ = 180^\\circ$ (see fig. 35). Let $\\alpha = \\angle BAC$; then also $\\angle CHE = 90^\\circ - \\angle ECH = \\alpha$. Therefore $\\angle MPE = 180^\\circ - \\angle CPE = 180^\\circ - \\angle CHE = 180^\\circ - \\alpha$, from which we get that points $A, M, P, E$ are concyclic. Analogously we see that points $B, M, P, D$ are concyclic.\n\nPoint $M$ is the circumcenter of the right triangle $ABE$. Therefore $|ME| = |MA|$ and $\\angle MEA = \\alpha$, due to which also $\\angle MPA = \\alpha$. But since $\\angle MQB = \\angle CQB = \\angle CAB = \\alpha$, we must have $AP \\parallel BQ$. Analogously $BP \\parallel AQ$.\n\n![](attached_image_1.png)\nFigure 35\n\nIn conclusion we get that $APBQ$ is a parallelogram with diagonals $AB$ and $PQ$. As the diagonals of a parallelogram divide each other in half, the desired claim follows.\nSimilarly to the previous solution we show that points $A, M, P, E$ are located on one circle. Let $\\vec{u} \\cdot \\vec{v}$ be the dot product of vectors $\\vec{u}$ and $\\vec{v}$. Then\n$$\n\\overrightarrow{AC} \\cdot \\overrightarrow{BC} = \\overrightarrow{AC} \\cdot \\overrightarrow{EC} = \\overrightarrow{MC} \\cdot \\overrightarrow{PC} = \\overrightarrow{MC} \\cdot (\\overrightarrow{MC} - \\overrightarrow{MP}) = \\overrightarrow{MC} \\cdot \\overrightarrow{MC} - \\overrightarrow{MC} \\cdot \\overrightarrow{MP}.\n$$\n\nOn the other hand,\n$$\n\\begin{align*}\n\\overrightarrow{AC} \\cdot \\overrightarrow{BC} &= (\\overrightarrow{MC} - \\overrightarrow{MA}) \\cdot (\\overrightarrow{MC} - \\overrightarrow{MB}) = \\\\\n&= \\overrightarrow{MC} \\cdot \\overrightarrow{MC} - \\overrightarrow{MC} \\cdot \\overrightarrow{MB} - \\overrightarrow{MA} \\cdot \\overrightarrow{MC} + \\overrightarrow{MA} \\cdot \\overrightarrow{MB} = \\\\\n&= \\overrightarrow{MC} \\cdot \\overrightarrow{MC} - \\overrightarrow{MC} \\cdot (\\overrightarrow{MA} + \\overrightarrow{MB}) + \\overrightarrow{MA} \\cdot \\overrightarrow{MB}.\n\\end{align*}\n$$\n\nIn conclusion $\\overrightarrow{MC} \\cdot \\overrightarrow{MP} = \\overrightarrow{MC} \\cdot (\\overrightarrow{MA} + \\overrightarrow{MB}) - \\overrightarrow{MA} \\cdot \\overrightarrow{MB}$. But as $M$ is the midpoint of $AB$, we have $\\overrightarrow{MA} + \\overrightarrow{MB} = \\vec{0}$, and due to choice of $Q$, we also have $\\overrightarrow{MA} \\cdot \\overrightarrow{MB} = \\overrightarrow{MC} \\cdot \\overrightarrow{MQ}$. Therefore $\\overrightarrow{MC} \\cdot \\overrightarrow{MP} = -\\overrightarrow{MC} \\cdot \\overrightarrow{MQ}$, meaning that\n\n$$\n\\overrightarrow{MC} \\cdot (\\overrightarrow{MP} + \\overrightarrow{MQ}) = 0.\n$$\n\nAs $\\overrightarrow{MP}$, $\\overrightarrow{MQ}$ and $\\overrightarrow{MC}$ have the same direction, the equality is true only if $\\overrightarrow{MP} + \\overrightarrow{MQ} = \\vec{0}$. Therefore $|MP| = |MQ|$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 29773,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $D$ be the point different from $B$ on the hypotenuse $AB$ of a right triangle $ABC$ such that $|CB| = |CD|$. Let $O$ be the circumcenter of triangle $ACD$. Rays $OD$ and $CB$ intersect at point $P$, and the line through point $O$ perpendicular to side $AB$ and ray $CD$ intersect at point $Q$. Points $A$, $C$, $P$, $Q$ are concyclic. Does this imply that $ACPQ$ is a square?",
    "options": [],
    "answer": "Yes, ACPQ is a square.",
    "solution": "![](attached_image_1.png)\nFigure 32\n\nAs $OQ$ is the perpendicular bisector of $AD$, one has $\\angle QAD = \\angle ADQ = \\angle BDC = \\angle CBD$ (Fig. 32). Therefore $AQ \\parallel BC$, whence $\\angle QAC = 180^\\circ - \\angle ACB = 90^\\circ$. From the cyclic quadrilateral $APCQ$ one also gets $\\angle CPQ = \\angle PQA = 90^\\circ$, i.e., $ACPQ$ is a rectangle.\n\nAs $\\angle DOC = 2\\angle DAC$, one obtains\n$$\n\\begin{aligned}\n\\angle DOC &= 2\\angle BAC = 2(90^\\circ - \\angle CBA) = 180^\\circ - 2\\angle CBA = \\\\\n&= 180^\\circ - \\angle CBD - \\angle BDC = \\angle DCB,\n\\end{aligned}\n$$\nwhich implies that isosceles triangles $BDC$ and $DCO$ are similar. Thus $\\angle BDC = \\angle DCO$, i.e., $OC \\parallel AB$, whence $\\angle QOC = 90^\\circ = \\angle QAC$. So $O$ lies on the circle determined by $A$, $C$, $P$, $Q$. Therefore\n$$\n\\angle ACQ = \\angle AOQ = \\frac{1}{2}\\angle AOD = \\frac{1}{2}\\angle AOP = \\frac{1}{2}\\angle ACP.\n$$\nConsequently, the diagonal of the rectangle $ACPQ$ bisects the angle of the rectangle, whence $ACPQ$ is a square.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 29774,
    "subject": "Mathematics (Multi-modal)",
    "question": "In Wonderland there are at least 5 towns. Some towns are connected directly by roads or railways. Every town is connected to at least one other town and for any four towns there exists some direct connection between at least three pairs of towns among those four. When entering the public transportation network of this land, the traveller must insert one gold coin into a machine, which lets him use a direct connection to go to the next town. But if the traveller continues travelling from some town with the same method of transportation that took him there, and he has paid a gold coin to get to this town, then going to the next town does not cost anything, but instead the traveller gains the coin he last used back. In other cases he must pay just like when starting travelling. Prove that it is possible to get from any town to any other town by using at most 2 gold coins.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $A$ and $B$ be any two towns. We know that it must be possible to move from $A$ to some other town $X$ and from $B$ to some other town $Y$. From four towns $A, B, X, Y$ we can form three pairs which all have a direct connection between them. Of those at least one way goes from either $A$ or $X$ to either $B$ or $Y$. Therefore it is possible to travel from $A$ to $B$.\nLook at some possible way of getting from $A$ to $B$; let $C$ be the first town after town $A$ on this way and $D$ be the last town before town $B$ (see fig. 36). Assume that $A, C, D, B$ are all distinct, because otherwise the problem statement follows trivially. Because of the same reason assume that there does not exist a direct connection between $A$ and $B, A$ and $D$ or $C$ and $B$. As according to the problem statement we can get three pairs from those four that all have direct connection between them, a direct connection must be between $C$ and $D$.\n\n![](attached_image_1.png)\nFigure 36\n\n![](attached_image_2.png)\nFigure 37\n\n![](attached_image_3.png)\nFigure 38\n\n![](attached_image_4.png)\nFigure 39\n\nLet $E$ be some town that is not $A, B, C$ or $D$. If there is a direct connection between $E$ and $A$ and also between $E$ and $B$, then the problem statement holds. Therefore let us assume in the following that there is no direct connection between either $E$ and $A$ or $E$ and $B$. From $A, C, E$ and $B$ we can form three pairs that have a direct connection between them. As there is a maximum of one direct connection between $E, A$ and $B$ and there is no direct connection between $B$ and $C$, then there must be one between $E$ and $C$. By switching the roles of $A$ and $B$ and also $C$ and $D$ we get analogously that there is a direct connection between $E$ and $D$ (see fig. 37).\nIf the connections between $A$ and $C$, and $D$ and $B$, are of different kind, then on the path $A \\to C \\to D \\to B$ there must be at least two consecutive steps with same mode of transportation. For this path the problem statement holds. But if connections between $A$ and $C$, and $D$ and $B$, are of the same kind, then there must exist two consecutive steps with the same mode of transportation on the path $A \\to C \\to E \\to D \\to B$. For this the problem statement also holds.\nLet $A$ and $B$ be any two towns. Suppose that there is no direct connection between them, because otherwise the problem statement holds trivially.\nLet $X$ be any town distinct from $A$ and $B$. If there is no direct connection between $A$ and $X$ and no direct connection between $B$ and $X$, then from a fourth town $Y$ there must be a direct connection to $A$, $B$ and $X$ (see fig. 38). In that case one can go from $A$ to $B$ via $Y$ and the problem statement holds. Because of that suppose in the following that from any town $X$ distinct from $A$ and $B$ there is a direct connection to either $A$ or $B$.\n\nLet $X$ and $Y$ be any two distinct towns that are not $A$ or $B$. Suppose that there is no direct connection between $X$ and $Y$. As there is also no direct connection between $A$ and $B$, but from $A$, $B$, $X$ and $Y$ we can form three pairs that have a direct connection between them, it is possible to go from $A$ to $B$ via $X$ or $Y$, in which case the problem statement holds. Now the only case to look at is the one where between any two towns that are not $A$ and $B$ there is a direct connection.\nAs there are at least 5 towns in the country, there are at least 3 towns other than $A$ and $B$. Therefore either $A$ or $B$ must have a direct connection to at least two other towns. Without loss of generality assume that $A$ has a direct connection to $C$ and $D$. But $B$ also has a direct connection to some town $E$; if $E$ coincides with any of the previously mentioned ones then the problem statement holds, which leaves us to look at the case where $E$ is a new town. Previously mentioned facts give us that $C$, $D$ and $E$ all have direct connections between them.\n\nIf now either $A$ and $C$ or $A$ and $D$ have a direct connection between them of different kind than what is between $B$ and $E$, then either path $A \\to C \\to E \\to B$ or $A \\to D \\to E \\to B$ has two consecutive steps with same mode of transportation. For this path the problem statement holds. But if the connection between $A$ and $C$ or $A$ and $D$ is of the same kind as the connection between $B$ and $E$, then either on the path $A \\to C \\to D \\to E \\to B$ or $A \\to D \\to C \\to E \\to B$ there are two consecutive steps with same mode of transportation. For this also the problem statement holds.",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 29775,
    "subject": "Mathematics (Multi-modal)",
    "question": "Call a tuple $(b_m, b_{m+1}, \\dots, b_n)$ of integers *perfect* if both following conditions are fulfilled:\n1. There exists an integer $a > 1$ such that $b_k = a^k + 1$ for all $k = m, m+1, \\dots, n$;\n2. For all $k = m, m+1, \\dots, n$, there exists a prime number $q$ and a non-negative integer $t$ such that $b_k = q^t$.\nProve that if $n - m$ is large enough then there is no perfect tuples, and find all perfect tuples with the maximal number of components.",
    "options": [],
    "answer": "(2^0 + 1, 2^1 + 1, 2^2 + 1, 2^3 + 1, 2^4 + 1)",
    "solution": "*Answer:* $(2^0 + 1, 2^1 + 1, 2^2 + 1, 2^3 + 1, 2^4 + 1)$.\n\nClearly $(2^0 + 1, 2^1 + 1, 2^2 + 1, 2^3 + 1, 2^4 + 1)$ is a perfect tuple with length 5. Show in the rest that there are no other perfect tuples with length 5 or larger.\n\nFor that, let $(a^m + 1, a^{m+1} + 1, \\dots, a^n + 1)$ be an arbitrary perfect tuple with length at least 5. There must exist at least two odd exponents among $m, m+1, \\dots, n$; let $k$ and $k+2$ be the two largest odd exponents. As $a^k + 1$ and $a^{k+2} + 1$ are prime powers while having a common divisor $a+1$, these two integers must be powers of the same prime $q$. Thus the larger of them, $a^{k+2} + 1$, is divisible by the smaller one, $a^k + 1$, which shows that $a^k + 1$ divides also the difference $a^2 \\cdot (a^k + 1) - (a^{k+2} + 1) = a^2 - 1$. Hence $a^k + 1 \\le a^2 - 1$, implying $k < 2$. So $k = 1$ as $k$ is odd. By choice of $k$, the only odd exponents in our perfect tuple are 1 and 3 and the tuple is of the form $(a^0 + 1, a^1 + 1, a^2 + 1, a^3 + 1, a^4 + 1)$.\n\nAs $a+1$ and $a^3+1$ are powers of the same prime number $q$, also the ratio $\\frac{a^3+1}{a+1} = a^2-a+1$ is a power of $q$. Note that $a^2-a+1 \\ge 2a-a+1 = a+1$ by $a \\ge 2$, hence $a^2-a+1$ is divisible by $a+1$. Thus the difference $(a^2-a+1)-(a+1)(a-2) = 3$ is divisible by $a+1$. This filters out the only possibility $a=2$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 29776,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all natural numbers $n$ such that the equation $x^2 + y^2 + z^2 = nxyz$ has solutions in positive integers.",
    "options": [],
    "answer": "1 and 3",
    "solution": "For $n = 1$ one of the solutions is $x = y = z = 3$ and for $n = 3$ one of the solutions is $x = y = z = 1$.\n\nIf $n$ is even and there exists an integer solution, then the right-hand side of the equation is even. This is possible only if at least one of the numbers $x, y, z$ is even. Then the right-hand side is divisible by $4$. Since the remainders modulo $4$ of the squares of integers can only be $0$ or $1$, all numbers $x, y, z$ must be even. Let $x = 2a$, $y = 2b$, $z = 2c$. Then $a^2 + b^2 + c^2 = 2nabc$, hence $(a, b, c)$ satisfy a similar equation with doubled $n$, so they must be even. Continuing this process reveals that $x, y, z$ must be divisible by arbitrarily large powers of $2$ which is impossible. Consequently, there are no solutions for even $n$.\n\nSuppose that for some odd $n > 3$ the equation has an integer solution $(x, y, z)$. The given equation is equivalent to $z^2 - nxy \\cdot z + (x^2 + y^2) = 0$; let $z'$ be the other root of this quadratic equation. Then $z' > 0$ since positive $nxy$ and $x^2 + y^2$ enable only positive solutions of the quadratic equation. By Viéte's formulae, $z' = nxy - z$. On the other hand, assume without loss of generality that $z = \\max(x, y, z)$; then $x^2 \\le xz \\le xyz$ and $y^2 \\le yz \\le xyz$, whence $z^2 \\ge (n-2)xyz$ and $z \\ge (n-2)xy$. Therefore $z' \\le 2xy < (n-2)xy \\le z$, implying $x + y + z' < x + y + z$. Thus we can infinitely reduce the sum of the components of the solution, which is impossible.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 29777,
    "subject": "Mathematics (Multi-modal)",
    "question": "A class consists of $7$ boys and $13$ girls. During the first three months of the school year, each boy has communicated with each girl at least once. Prove that there exist two boys and two girls such that both boys communicated with both girls first time in the same month.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "*Solution 1:* Call the first communication between a boy and a girl their *acquaintance*. During the $3$ months, there are $7 \\cdot 13 = 91$ acquaintances in total. Thus there exists a month when there was at least $31$ acquaintances. Let the boys be denoted by $p_1$ through $p_7$ and let $T_i$, $i = 1, \\dots, 7$, be the set of girls to whom $p_i$ acquainted in this month. We have to show that there exist distinct $i$ and $j$ such that $T_i \\cap T_j$ contains at least $2$ girls. W.l.o.g., assume the inequalities $|T_1| \\ge |T_2| \\ge \\dots \\ge |T_7|$. Consider two cases.\n\n1. The case $|T_1| + |T_2| + |T_3| + |T_4| \\ge 20$. Suppose that all intersections $T_1 \\cap T_2$, $T_1 \\cap T_3$, $\\dots$, $T_3 \\cap T_4$ contain at most one girl. Let $k \\le 6$ be the number of non-empty intersections. Then the first four boys acquainted with\n$$\n|T_1 \\cup T_2 \\cup T_3 \\cup T_4| \\ge |T_1| + |T_2| + |T_3| + |T_4| - k \\ge 20 - 6 = 14\n$$\nThis is a contradiction since there are only $13$ girls.\n\n2. The case $|T_1| + |T_2| + |T_3| + |T_4| \\le 19$. As $|T_5| + |T_6| + |T_7| \\ge 12$, we have $|T_5| \\ge 4$. Now $|T_4| \\le 4$ implies $|T_4| = |T_5| = 4$ and hence also $|T_6| = |T_7| = 4$. Now $|T_1| + |T_2| + |T_3| = 15$ in order to get $31$ in total.\nSuppose that all intersections $T_i \\cap T_j$, $i, j = 1, \\dots, 7$, contain at most one girl. If boys $p_1, p_2$ and $p_3$ altogether acquainted with all girls in this month then at least one intersection $T_i \\cap T_4$, $i = 1, 2, 3$, contains at least two girls. Otherwise, $|T_1| + |T_2 \\setminus T_1| + |T_3 \\setminus (T_1 \\cup T_2)| = 15 - 1 - 2 = 12$ (Fig. 33 shows all possibilities), since only then there is a girl, say $t_{13}$, with whom none of $p_1, p_2, p_3$ acquainted. Clearly all boys $p_4, p_5, p_6, p_7$ must have acquainted with her, and each of them also acquainted with one girl from sets $T_1$, $T_2 \\setminus T_1$ and $T_3 \\setminus (T_1 \\cup T_2)$. As the last set contains at most three elements, two of the four boys acquainted with the same girl from $T_3 \\setminus (T_1 \\cup T_2)$.\n**Solution 2:** Let $A$ be the set of all combinations of two boys, $|A| = \\binom{7}{2} = 21$. Say that girl $t$ determines an element $\\{p_1, p_2\\}$ of $A$ if $t$ acquainted with $p_1$ and $p_2$ in the same month. If in the $i$th month a girl $t$ acquainted with exactly $n_i$ boys, where $i = 1, 2, 3$, then $t$ determines $\\binom{n_1}{2} + \\binom{n_2}{2} + \\binom{n_3}{2}$ elements of $A$. Applying Jensen's inequality for $f(x) = \\frac{x(x-1)}{2}$ gives $\\binom{n_1}{2} + \\binom{n_2}{2} + \\binom{n_3}{2} \\ge 3 \\cdot f(\\frac{n_1+n_2+n_3}{3}) = 3 \\cdot f(\\frac{7}{3}) = 4\\frac{2}{3}$. As $\\binom{n_1}{2} + \\binom{n_2}{2} + \\binom{n_3}{2}$ is an integer, $\\binom{n_1}{2} + \\binom{n_2}{2} + \\binom{n_3}{2} \\ge 5$. Hence all girls determine at least $13 \\cdot 5 = 65$ elements of $A$ in total.\nAs $65 \\ge 3 \\cdot 21 + 1$, an element $\\{p_1^*, p_2^*\\}$ of $A$ is determined by at least $4$ girls by the pigeonhole principle. Consequently there exists a month in which this couple of boys is determined by at least two girls.\n**Solution 3:** Suppose that there is no required pairs of boys and girls. Like in Solution 1, consider a month with $31$ or more acquaintances. Let $g_i$ be the number of girls who acquainted with exactly $i$ boys in this month, $i = 0, 1, \\dots, 7$. We get a system of inequalities\n$$\n\\left\\{\n\\begin{array}{l}\ng_2 + 3g_3 + 6g_4 + 10g_5 + 15g_6 + 21g_7 \\le 21, \\\\\ng_1 + g_2 + g_3 + g_4 + g_5 + g_6 + g_7 \\le 13, \\\\\ng_1 + 2g_2 + 3g_3 + 4g_4 + 5g_5 + 6g_6 + 7g_7 \\ge 31.\n\\end{array}\n\\right.\n$$\nIf at least one girl acquainted with $5$ or more boys then $g_5 + g_6 + g_7 \\ge 1$. The system then reduces to\n$$\n\\left\\{\n\\begin{array}{l}\ng_2 + 3g_3 + 6g_4 + 10(g_5 + g_6 + g_7 - 1) \\le 11, \\\\\ng_1 + g_2 + g_3 + g_4 + (g_5 + g_6 + g_7 - 1) \\le 12, \\\\\ng_1 + 2g_2 + 3g_3 + 4g_4 + 7(g_5 + g_6 + g_7 - 1) \\ge 24.\n\\end{array}\n\\right.\n$$\nSumming the first two inequalities gives\n$$\ng_1 + 2g_2 + 4g_3 + 7g_4 + 11(g_5 + g_6 + g_7 - 1) \\leq 23,\n$$\nwhich contradicts the third inequality.\nThus $g_5 = g_6 = g_7 = 0$ and the system of inequalities reduces to\n$$\n\\left\\{\n\\begin{array}{l}\ng_2 + 3g_3 + 6g_4 \\le 21, \\\\\ng_1 + g_2 + g_3 + g_4 \\le 13, \\\\\ng_1 + 2g_2 + 3g_3 + 4g_4 \\ge 31.\n\\end{array}\n\\right.\n$$\nSuppose that $g_4 \\ge 1$. As in the previous case, this implies\n$$\n\\left\\{\n\\begin{array}{l}\ng_2 + 3g_3 + 6(g_4 - 1) \\le 15, \\\\\ng_1 + g_2 + g_3 + (g_4 - 1) \\le 12, \\\\\ng_1 + 2g_2 + 3g_3 + 4(g_4 - 1) \\ge 27.\n\\end{array}\n\\right.\n$$\nThe first two inequalities sum up to $g_1 + 2g_2 + 4g_3 + 7(g_4 - 1) \\le 27$. In the light of the third inequality, this is possible only if $g_3 = 0$ and $g_4 = 1$. Then the second and third inequalities give $g_1 + g_2 \\le 12$ and $g_1 + 2g_2 \\ge 27$, which contradict each other since $g_1 + 2g_2 \\le 2(g_1 + g_2)$.\n\nHence also $g_4 = 0$ and our system of inequalities reduces to\n$$\n\\left\\{\n\\begin{array}{l}\ng_2 + 3g_3 \\le 21, \\\\\ng_1 + g_2 + g_3 \\le 13, \\\\\ng_1 + 2g_2 + 3g_3 \\ge 31.\n\\end{array}\n\\right.\n$$\nSubtracting the first inequality from the third one, we obtain $g_1 + g_2 \\ge 10$. Subtracting twice the second inequality from the third one, we get $g_3 - g_1 \\ge 5$. The two inequalities obtained sum up to $g_2 + g_3 \\ge 15$, contradicting the second inequality.",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 29778,
    "subject": "Mathematics (Multi-modal)",
    "question": "Non-equilateral triangle $ABC$ has a $60°$ angle at vertex $A$. Let the angle bisector drawn from vertex $A$ intersect the opposite side at point $D$, and let $Q$ and $R$ be the feet of the altitudes drawn from vertices $B$ and $C$, respectively. Prove that lines $AD$, $BQ$ and $CR$ intersect in three distinct points that are vertices of an equilateral triangle.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "If line $AD$ passed through the point of intersection of lines $BQ$ and $CR$, the line segment $AD$ would be an altitude of triangle $ABC$. As $AD$ is also the angle bisector, triangle $ABC$ would be isosceles with $AB = AC$. As $\\angle BAC = 60°$, triangle $ABC$ would be equilateral, contradicting the assumption. Hence the lines $AD$, $BQ$, and $CR$ meet in three distinct points. Fig. 15\n\n![](attached_image_1.png)\nFig. 15\n\nBy assumptions, $\\angle QAD = \\angle RAD = 30°$ and $\\angle AQB = \\angle ARC = 90°$ (Fig. 15). Hence $AD$ and $BQ$ intersect at angle $60°$, as well as $AD$ and $CR$. Thus two angles of the triangle whose vertices are the three intersection points of lines $AD$, $BQ$ and $CR$ have size $60°$. Such a triangle is equilateral.",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 29779,
    "subject": "Mathematics (Multi-modal)",
    "question": "Each point at the sides of an equilateral triangle is coloured either red or blue. Is it sure that there exists a right triangle whose all vertices are of the same colour?",
    "options": [],
    "answer": "Yes",
    "solution": "Let the equilateral triangle be $XYZ$. We show that there exists a point on some side that has the same colour as its projection to another side. For that, take points $P$, $Q$ and $R$ on sides $XY$, $YZ$ and $ZX$, respectively, in such a way that $XP : XY = YQ : YZ = ZR : ZX = 1 : 3$ (Fig. 17).\n\nThen $PQ \\perp YZ$ because, denoting the midpoint of $YZ$ by $T$, $TQ : TY = (\\frac{1}{2} - \\frac{1}{3}) : \\frac{1}{2} = 1 : 3 = XP : XY$, implying $PQ \\parallel XT$. Hence $Q$ is the projection of $P$ to $YZ$. Analogously, $R$ is the projection of $Q$ to $ZX$ and $P$ is the projection of $R$ to $XY$. As at least two points among $P$, $Q$ and $R$ must have the same colour, a point and its projection have the same colour.\n\nW.l.o.g., let $P$ and its projection $Q$ both be red. Let $M$ be the projection of $Q$ to $XY$ and $N$ the projection of $M$ to $YZ$. If $M$ is red, then $PQM$ is a right triangle with all vertices red. If $N$ is red, then $PQN$ is a right triangle with all vertices red. If $Y$ is red, then $PQY$ is a right triangle with all vertices red. Otherwise, $MNY$ is a right triangle with all vertices blue.\n\n![](attached_image_1.png)\nFig. 17\nConsider a regular hexagon whose vertices lie on the sides of the triangle (Fig. 18). Suppose that two opposite vertices of the hexagon are of the same colour. If there is one more vertex of the same colour among the other four vertices, there is a right triangle with all vertices being of that colour. Otherwise, any three vertices among the four remaining ones form a right triangle with all vertices being of the other colour.\n\nIf any two opposite vertices are of different colours, then there exist two neighbouring vertices of different colours. The corresponding opposite vertices are of different colours, too. One pair of these differently coloured vertices lies on a side of the initial triangle; let these be $A$ and $B$ (red and blue, resp.) and their opposite vertices $D$ and $E$ (blue and red, resp.). Angles $ABD$ and $BAE$ are right. Let $S$ be any point on $AB$ that is not a vertex of the hexagon. If $S$ is red, then $SAE$ is a right triangle with all vertices red; if $S$ is blue, then $SBD$ is a right triangle with all vertices blue.\n\n![](attached_image_2.png)\nFig. 18",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 29780,
    "subject": "Mathematics (Multi-modal)",
    "question": "In the beginning, there are two positive integers on a blackboard. On each step, one chooses numbers $a$ and $b$ such that $a \\le b$ from the numbers on the blackboard in all possible ways (equality means that one may take the same number twice), finds all corresponding sums $a+b+\\gcd(a,b)$ and replaces all the numbers on the blackboard instantly with these sums. Prove that at some step at least one number will occur more than once on the blackboard.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "If the numbers chosen from the blackboard are $x$ and $y$, then the number $x+y+\\gcd(x,y)$ will be on the blackboard on the next step. If $x$ is chosen together with itself, the number $x+x+\\gcd(x,x) = 3x$ will be on the blackboard on the next step. We show that there will be two equal numbers on the blackboard after the second step at latest. Assume that the initial numbers are $n$ and $m$ and the numbers $n+m+\\gcd(n,m)$, $3n$ and $3m$ appearing on the first step are all distinct. Choosing number $n+m+\\gcd(n,m)$ together with itself, we obtain $3(n+m+\\gcd(n,m)) = 3n+3m+3\\gcd(n,m)$. Choosing $3n$ and $3m$, we obtain $3n+3m+\\gcd(3n,3m)$. As $\\gcd(3n,3m) = 3\\gcd(n,m)$, the same number will appear twice after the second step.\nLet the largest number on the blackboard after step $s$ and the number of numbers on the blackboard after step $s$ be $m(s)$ and $n(s)$, respectively. Obviously $m(s+1) = m(s)+m(s)+\\gcd(m(s),m(s)) = 3m(s)$, as $x \\le m(s)$ and $y \\le m(s)$ imply $\\gcd(x,y) \\le m(s)$. Thus $m(s) = 3^s m(0)$ for all $s$. Suppose that there will never be equal numbers on the blackboard. Then $n(s+1) = \\frac{n(s)(n(s)+1)}{2} > \\frac{n(s)^2}{2}$ for every $s$. Since $n(0) = 2$, we get $n(1) = 3$ and $n(2) = 6$. An easy induction shows that $n(s) > 2 \\cdot 3^{2^{s-2}}$ for every $s \\ge 2$. Thus $\\frac{n(s)}{m(s)} > \\frac{2}{m(0)} \\cdot 3^{2^{s-2}-s}$. If $s \\to \\infty$, the number $3^{2^{s-2}-s}$ becomes larger than $\\frac{m(0)}{2}$, so $n(s) > m(s)$. On the other hand, $n(s) \\le m(s)$ since the numbers are positive, a contradiction.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 29781,
    "subject": "Mathematics (Multi-modal)",
    "question": "Denote the number of all positive divisors of a positive integer $n$ by $\\delta(n)$ and the sum of all positive divisors of a positive integer $n$ by $\\sigma(n)$. Prove that $\\sigma(n) > \\frac{\\delta(n)^2}{2}$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $a_1, a_2, \\dots, a_{\\delta(n)}$ be the positive divisors of $n$ in increasing order. We obtain\n$$\n\\sigma(n) = a_1 + a_2 + \\dots + a_{\\delta(n)} \\ge 1 + 2 + \\dots + \\delta(n) = \\frac{\\delta(n) \\cdot (\\delta(n)+1)}{2} > \\frac{\\delta(n)^2}{2}.\n$$\nFor every positive divisor $d$ of a positive integer $n$, $d + \\frac{n}{d} \\ge 2\\sqrt{d \\cdot \\frac{n}{d}} = 2\\sqrt{n}$. Thereby if $d \\ne \\sqrt{n}$, the inequality is strict. If $n$ is not a perfect square, then by adding all these inequalities for divisors $d < \\sqrt{n}$ leads to\n$$\n\\sigma(n) > \\frac{\\delta(n)}{2} \\cdot 2\\sqrt{n} = \\delta(n)\\sqrt{n},\n$$\nsince every divisor of $n$ occurs exactly once in a pair $(d, \\frac{n}{d})$. If $n$ is a perfect square, then analogously\n$$\n\\sigma(n) \\ge \\frac{\\delta(n)-1}{2} \\cdot 2\\sqrt{n} + \\sqrt{n} = \\delta(n)\\sqrt{n}.\n$$\nOn the other hand, if $n$ is not a perfect square then the number of pairs $(d, \\frac{n}{d})$ where $d < \\sqrt{n}$, is less than $\\sqrt{n}$, whence $\\sqrt{n} > \\frac{\\delta(n)}{2}$. If $n$ is a perfect square then the number of such pairs is at most $\\sqrt{n}-1$, which gives $\\sqrt{n}-1 \\ge \\frac{\\delta(n)-1}{2}$ implying $\\sqrt{n} > \\frac{\\delta(n)}{2}$ again. Hence,\n$$\n\\sigma(n) \\ge \\delta(n)\\sqrt{n} > \\delta(n) \\cdot \\frac{\\delta(n)}{2} = \\frac{\\delta(n)^2}{2}.\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 29782,
    "subject": "Mathematics (Multi-modal)",
    "question": "Some cities of a country are connected with roads. We say that a city $A$ belongs to a cycle of length $n$ if one can travel from $A$ through exactly $n-1$ other cities and return back to $A$. It is known that each city of the country belongs to a cycle of length 4 and also to a cycle of length 5.\nIs it sure that \na) at least one city belongs to a cycle of length 3? \nb) each city belongs to a cycle of length 3?",
    "options": [],
    "answer": "a) No. b) No.",
    "solution": "Let there be 10 cities and let cities be connected as in the Fig. 19. Then every city belongs to a cycle of length 5 and also to a cycle of length 4. On the other hand, no cycles of length 3 exist.\n\n![](attached_image_1.png)\nFig. 19",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 29783,
    "subject": "Mathematics (Multi-modal)",
    "question": "There are $125$ distinct positive integers in a row in such a way that among every three consecutive numbers the second one is larger than the arithmetic mean of the first and the third one. Find the largest number in the row, given that it is as small as possible under such conditions.",
    "options": [],
    "answer": "2016",
    "solution": "Let the numbers in a row be $a_1, a_2, \\dots, a_{125}$. By conditions, we have $a_{i+1} > \\frac{a_i + a_{i+2}}{2}$ for every $i = 1, 2, \\dots, 123$, which is equivalent to $a_{i+1} - a_i > a_{i+2} - a_{i+1}$. Denoting $d_i = a_{i+1} - a_i$, we have $d_1 > d_2 > \\dots > d_{124}$.\n\nLet $a_m$ be the largest among $a_1, \\dots, a_{125}$; then $d_1, d_2, \\dots, d_{m-1}$ are positive and $d_m, d_{m+1}, \\dots, d_{124}$ are negative. If both $1$ and $-1$ occurred among the differences $d_1, d_2, \\dots, d_{124}$, then they should be consecutive, i.e., $d_i = 1$ and $d_{i+1} = -1$ for some $i$, whence $a_{i+1} = a_{i-1}$. Contradiction to the assumption that the given numbers are distinct shows that either $1$ or $-1$ is missing among the differences. W.l.o.g., assume that $1$ is missing (if $-1$ is missing, we can reverse the numeration of the given numbers).\n\nThen $a_m = a_1 + (d_1 + d_2 + \\dots + d_{m-1}) \\ge 1 + (m + (m-1) + \\dots + 2) = 1 + 2 + \\dots + m$ and $a_m = a_{125} - (d_m + d_{m+1} + \\dots + d_{124}) \\ge 1 + (1 + 2 + \\dots + (125 - m))$. Among numbers $m$ and $125 - m$, one is at least $63$, whence the previously established inequalities imply $a_m \\ge 1 + 2 + \\dots + 63$.\n\nOn the other hand, taking $a_1 = 1$ and $d_i = 64 - i$, $1 \\le i \\le 62$, and $d_i = 62 - i$, $63 \\le i \\le 124$, the largest number $a_{63}$ equals $1 + 2 + \\dots + 63$, while all numbers $a_i$ are positive, as $1 = a_1 < a_2 < \\dots < a_{63}$ and $a_{63} > a_{64} > \\dots > a_{125} = (1 + 2 + \\dots + 63) - (1 + 2 + \\dots + 62) = 63$. All these numbers are distinct, since for every $i = 64, 65, \\dots, 125$, $a_i = (1 + 2 + \\dots + 63) - (1 + 2 + \\dots + (i - 63)) = (i - 62) + (i - 61) + \\dots + 63 = a_{127 - i} - 1$, whence $a_i$ lies strictly between $a_{126 - i}$ and $a_{127 - i}$. Consequently, the largest number written in the row is $1 + 2 + \\dots + 63$, i.e., $2016$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 29784,
    "subject": "Mathematics (Multi-modal)",
    "question": "There are 8 white pawns on the squares at one edge of an $8 \\times 8$ chessboard and 8 black pawns on the squares at the opposite edge. On each move, a player shifts one of his pawns by one or more squares forward (toward the opponent's piece) or backward, but moving a pawn to a square containing the opponent's pawn or over such a square is prohibited. Moves are performed alternately, with white starting. The player who cannot make a move loses. Which player has a winning strategy?",
    "options": [],
    "answer": "Black",
    "solution": "Black can use the following strategy. If white moves his $k$th pawn counting from his left, by $n$ squares forward, black moves his $k$th pawn counting from his left, by $n$ squares forward. If white moves his pawn by $n$ squares backward, black moves his pawn on the same file by $n$ squares forward. After each move made by black, the distance between two pawns on each file is the same as that on the file symmetric w.r.t. the midpoint of the board. Thus whenever a white pawn has moved forward, the black can make the move determined by the strategy described. As the black pawns move forward only, white will be paralyzed sooner or later.",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 29785,
    "subject": "Mathematics (Multi-modal)",
    "question": "Determine the largest possible number of primes among 100 consecutive natural numbers.",
    "options": [],
    "answer": "26",
    "solution": "There are 25 primes among the numbers from 1 to 100. The number of primes in the next a few number intervals with length 100 are shown in the table:\n\n| Interval      | out | in  | Number of primes |\n|--------------|-----|-----|------------------|\n| 1,...,100    |     |     | 25               |\n| 2,...,101    | 1   | 101 | 26               |\n| 3,...,102    | 2   | 102 | 25               |\n| 4,...,103    | 3   | 103 | 25               |\n| 5,...,104    | 4   | 104 | 25               |\n| 6,...,105    | 5   | 105 | 24               |\n| 7,...,106    | 6   | 106 | 24               |\n\nThe largest number of primes in these number intervals is 26.\n\nWe show now that there cannot be more primes among 100 consecutive natural numbers. Consider arbitrary 100 consecutive natural numbers, the least of which is larger than 7. None of the numbers under consideration that is divisible by one of numbers 2, 3, 5 and 7 is a prime. We show in the rest that there are at least 74 such numbers. As every second number is divisible by 2, there are 50 even numbers. As every third number is divisible by 3, there are at least 33 such numbers. Every second among them has been counted as an even number, thus there are at least 16 new numbers divisible by 3. As every fifth number is divisible by 5, there are at least 20 such numbers. Every second among them is even, thus the number of odd numbers divisible by 5 is 10. Among them in turn, every third is divisible by 3, thus there are at least 6 numbers divisible by 5 not counted yet. As every seventh number is divisible by 7, there are at least 14 such numbers. Every second among them is even, thus there are at least 7 odd numbers divisible by 7. Every third among them is divisible by 3, which eliminates at most 3 numbers, and every fifth is divisible by 5, which eliminates at most 2 numbers. Hence at least 2 numbers not counted before are divisible by 7. Altogether, we have at least $50 + 16 + 6 + 2 = 74$ composite numbers.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 29786,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all four-digit numbers which are exactly $2016$ larger than the four-digit number obtained by moving the first digit to the end.",
    "options": [],
    "answer": "3109, 4220, 5331, 6442, 7553, 8664, 9775",
    "solution": "Let the first digit of the number be $a$ and the number formed by the remaining digits be $k$. By the conditions, $1000a + k = 10k + a + 2016$, whence $111a - k = 224$. Hence $a \\ge 3$, implying the solutions $a = 3, k = 109$; $a = 4, k = 220$; $a = 5, k = 331$; $a = 6, k = 442$; $a = 7, k = 553$; $a = 8, k = 664$; $a = 9, k = 775$. The corresponding four-digit numbers satisfying the conditions of the problem are $3109$, $4220$, $5331$, $6442$, $7553$, $8664$, and $9775$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 29787,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all pairs $(a, b)$ of integers satisfying the equality\n$$\n3(a^2 + b^2) - 7(a + b) = -4.\n$$",
    "options": [],
    "answer": "[(2, 2), (1, 0), (0, 1)]",
    "solution": "The given equation is equivalent to $(6a - 7)^2 + (6b - 7)^2 = 50$. Number $50$ can be represented as the sum of two squares as $25 + 25$ or $1 + 49$. Hence both $6a - 7$ and $6b - 7$ must be among the numbers $7, 5, 1, -1, -5$ and $-7$. As both $a$ and $b$ are integers, only $5, -1$ and $-7$ fit. We obtain the following cases: $6a - 7 = 5$, $6b - 7 = 5$; $6a - 7 = -1$, $6b - 7 = -7$; $6a - 7 = -7$, $6b - 7 = -1$. The corresponding solutions are $a = 2, b = 2$; $a = 1, b = 0$; and $a = 0, b = 1$.\nConsider the equation as a quadratic equation w.r.t. $b$. In order to have solutions, its discriminant must be non-negative, i.e., $49 - 12(3a^2 - 7a + 4) \\ge 0$. This condition is equivalent to the quadratic inequality $36a^2 - 84a - 1 \\le 0$, whose solutions are $(7 - \\sqrt{50})/6 \\le a \\le (7 + \\sqrt{50})/6$. As $a$ is an integer, its only suitable values are $0, 1$, and $2$, the corresponding values of $b$ are $1, 0$, and $2$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 29788,
    "subject": "Mathematics (Multi-modal)",
    "question": "Call a convex polygon on a plane *correct* if for its every side there exists a unique vertex of the polygon that lies farther from that side than any other vertex of the polygon. Call the perpendicular drawn from the vertex farthest from side $XY$ to side $XY$ an *altitude* of the correct polygon. Find all natural numbers $n$ for which there exists a correct $n$-gon whose all $n$ altitudes meet in one point.",
    "options": [],
    "answer": "all n ≥ 3",
    "solution": "Let one of the vertices be $O(0,0)$ and let the other vertices $A_1, \\dots, A_{n-1}$ lie on a circle with radius $1$ and centre $O$ in such a way that $A_1(1,0)$, $A_{n-1}(0,1)$ and $A_2, \\dots, A_{n-2}$ are all on the shorter arc $A_1A_{n-1}$ (Fig. 16 depicts the case $n=6$).\n\nThe vertex farthest from line $OA_1$ is $A_{n-1}$, the vertex farthest from line $OA_{n-1}$ is $A_1$. The vertex farthest from any other line determined by a side of the polygon is $O$ because the line passing through $O$ parallel to such a side lies in II and IV quarters while the other vertices of the polygon lie above it in I quarter. Thus the polygon is correct. The altitudes drawn to sides $OA_1$ and $OA_{n-1}$ are $OA_{n-1}$ and $OA_1$, respectively, they meet at point $O$. As $O$ is the vertex farthest from any other side, all other altitudes meet in $O$, too.\n\n![](attached_image_1.png)\nFig. 16",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 29789,
    "subject": "Mathematics (Multi-modal)",
    "question": "Manni and Miku play the following game with rooks on an $8 \\times 8$ chessboard. At the beginning of the game, Miku places 8 rooks to the squares of the board according to his will. Then both players make moves alternately, Manni starts. On any move, each player shifts exactly one rook along a rank or file (i.e. row or column) by one or more squares in one direction. If a rook moves to a square that contains another rook, the latter is removed from the board; it is not allowed to move a rook over another. A player who is the first to remove a rook from the board wins; however, neither moving nor removing a rook that was moved by the opponent on his last move is allowed. Does either of the players have a winning strategy and if yes then which of them?",
    "options": [],
    "answer": "Detailed solution",
    "solution": "We show at first that Miku can play in such a way that Manni can never remove a rook. Let there be one rook in each rank and file in the initial configuration. Suppose that Manni moves a rook from square $(x, y)$ to square $(x, z)$. As a consequence, each file contains one rook but there are no rooks in rank $y$ and two rooks in rank $z$; let a rook be on square $(t, z)$, $t \\neq x$. Let Miku move the rook from square $(t, z)$ to square $(t, y)$. After that, there is one rook in each rank and file again. In the case of Manni's move in the perpendicular direction, Miku's reply would be analogous. In such a way, Miku can reply to all Manni's following moves.\n\nNow we show that also Manni can play so that Miku never wins. If there are two or more rooks on one rank or file in the initial configuration then Manni can remove one of them and win immediately. Therefore assume that initially there is one rook on each rank and file. Let there be rooks on squares $(1, u)$ and $(2, v)$. Manni can move the rook from $(2, v)$ to square $(2, u)$. As Miku is not allowed to remove this rook but Manni threatens to remove the rook on $(1, u)$ on the next move, Miku must move this rook to some square $(1, w)$. If $w \\neq v$, Manni can win by moving the rook from square $(2, u)$ to square $(2, w)$ and removing either of the other rooks on rank $w$ on the next move. If $w = v$, then there is one rook in each rank and file again and Manni can continue with the same strategy.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 29790,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find the largest natural number $n$ for which $3^{2016} - 1$ is divisible by $2^n$.",
    "options": [],
    "answer": "7",
    "solution": "We have $3^{2016} - 1 = (3^{63} - 1)(3^{63} + 1)(3^{126} + 1)(3^{252} + 1) \\cdot (3^{504} + 1)(3^{1008} + 1)$.\n\nNumbers $3^{126}$, $3^{252}$, $3^{504}$ and $3^{1008}$ are squares of odd numbers, hence congruent to $1$ modulo $8$. Thus $3^{126} + 1$, $3^{252} + 1$, $3^{504} + 1$ and $3^{1008} + 1$ are congruent to $2$ modulo $8$. Consequently, these four factors are divisible by $2$ but not by $4$.\n\nAs $3^{62} \\equiv 1 \\pmod{8}$, we have $3^6 \\equiv 3 \\pmod{8}$ (mod $8$). Hence $3^{63} - 1$ and $3^{63} + 1$ are congruent to $2$ and $4$ modulo $8$, respectively. The former thus is divisible by $2$ but not by $4$ and the latter is divisible by $4$ but not by $8$.\n\nPutting it all together, the exponent of $2$ in the product is $1+2+1+1+1+1=7$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 29791,
    "subject": "Mathematics (Multi-modal)",
    "question": "Three workers must do a work completely. At first, one of them works as long as the other two would work together to complete one half of the work. Then another worker works as long as the other two would work together to complete one half of the work. Finally the third worker works as long as the other two would work together to complete one half of the work. With this, the whole work becomes completed. How many times faster would the work become completed if all the workers worked together?",
    "options": [],
    "answer": "2.5",
    "solution": "Let the contributions of the first, second and third worker per a time unit be $x, y, z$, measured as percentages of the whole work. One half of the work would be done by the second and third worker together within $\\frac{1}{2(y+z)}$ time units, by the third and first worker together within $\\frac{1}{2(z+x)}$ time units and, by the first and second worker together, within $\\frac{1}{2(x+y)}$ time units. By working in the way described in the problem text, they spend $\\frac{1}{2(y+z)} + \\frac{1}{2(z+x)} + \\frac{1}{2(x+y)}$ time units, but if they worked all together, they would spend $\\frac{1}{x+y+z}$ time units. We are asked the ratio of these numbers. Since by assumption $\\frac{x}{2(y+z)} + \\frac{y}{2(x+z)} + \\frac{z}{2(x+y)} = 1$, we get\n$$\n\\begin{aligned}\n\\frac{\\frac{1}{2(y+z)} + \\frac{1}{2(z+x)} + \\frac{1}{2(x+y)} }{\\frac{1}{x+y+z}} &= \\frac{x+y+z}{2(y+z)} + \\frac{x+y+z}{2(z+x)} + \\frac{x+y+z}{2(x+y)} = \\\\\n&= \\frac{1}{2} + \\frac{x}{2(y+z)} + \\frac{1}{2} + \\frac{y}{2(z+x)} + \\frac{1}{2} + \\frac{z}{2(x+y)} = \\frac{3}{2} + 1 = 2.5.\n\\end{aligned}\n$$\nSuppose that the first worker worked $a$ time units, the second worker worked $b$ time units and the third worker worked $c$ time units. Altogether, they spend $a+b+c$ time units to perform the whole work. If the second worker worked $a$ time units and the third also $a$ time units, they would perform half of the work. Similarly, if the third worker worked $b$ time units and the first also $b$ time units then they would complete half of the work, and if the first worker worked $c$ time units and the second also $c$ time units, they would complete half of the work. Consequently, if the first worker worked $b+c$ time units, the second worker worked $a+c$ time units and the third worked $a+b$ time units, they would complete one and a half such works. Including the work really done, it turns out that if all workers worked $a+b+c$ time units then they would perform 2.5 such works. Thus the three workers together would act 2.5 times faster than in the situation of the problem.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 29792,
    "subject": "Mathematics (Multi-modal)",
    "question": "A positive integer $n$ is interesting, if for some positive integer $m$ and positive integers $a$, $b$ that are smaller than $m$, $\\frac{m^2}{ab} = n$. For example, $10$ is interesting because $\\frac{20^2}{4 \\cdot 10} = 10$. Find the smallest interesting integer.",
    "options": [],
    "answer": "2",
    "solution": "For $n = 2$ we can take $m = 12$, $a = 8$ and $b = 9$, because $\\frac{12^2}{8 \\cdot 9} = \\frac{144}{72} = 2$.\n\nOn the other hand, $1$ is not interesting, because if $\\frac{m^2}{ab} = 1$, or $m^2 = ab$, then $a$ and $b$ cannot both be less than $m$ at the same time.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 29793,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all positive real solutions of the system of equations\n$$x + \\frac{1}{x} - w = 2,$$\n$$y + \\frac{1}{y} - w = 2,$$\n$$z + \\frac{1}{z} + w = 2,$$\n$$y + \\frac{1}{z} + w = 2.$$",
    "options": [],
    "answer": "(x, y, z, w) = (2, 1/2, 1, 1/2)",
    "solution": "Subtracting the second equation from the first and multiplying by $xy$ we get $x^2y + y - xy^2 - x = 0$, which gives either $x = y$ or $x = \\frac{1}{y}$.\nIf $x = y$, then subtracting the fourth equation from the third and multiplying by $xz$, we similarly get either $x = z$ or $x = -\\frac{1}{z}$. If $x = z$, then subtracting the third equation from the first gives $-2w = 0$, hence $w = 0$, which is not positive. If $x = -\\frac{1}{z}$, then $x$ and $z$ cannot be both positive.\nIf $x = \\frac{1}{y}$, then subtracting the fourth equation from the third gives $z - \\frac{1}{z} = 0$, hence $z = 1$. Adding the first equation with the third gives $x + \\frac{2}{x} = 3$, or $x^2 - 3x + 2 = 0$. This has a solution $x = 1$, which leads to $x = y = z$ already considered. The second solution $x = 2$ gives $y = \\frac{1}{2}$ and $w = \\frac{1}{2}$.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 29794,
    "subject": "Mathematics (Multi-modal)",
    "question": "Denote by $f^n(x)$ the result of applying the function $f$ $n$ times to $x$ (e.g. $f^1(x) = f(x)$, $f^2(x) = f(f(x))$, $f^3(x) = f(f(f(x)))$ etc). Find all functions from real numbers to real numbers which satisfy $f^d(x) = 2015 - x$ for all divisors $d$ of $2015$, which are greater than $1$, and for all real $x$.",
    "options": [],
    "answer": "f(x) = 2015 - x",
    "solution": "Since $5$ is a divisor of $2015$, we have for any real $z$:\n$$\n\\begin{aligned}\nf^{25}(z) &= f^5(f^5(f^5(f^5(z)))) = 2015 - f^5(f^5(f^5(f^5(z)))) \\\\\n&= 2015 - (2015 - f^5(f^5(f^5(z)))) = f^5(f^5(f^5(z))) = \\dots = f^5(z) = 2015 - z.\n\\end{aligned}\n$$\nSince $13$ is also a divisor of $2015$, we have for any real $z$:\n$$\nf^{26}(z) = f^{13}(f^{13}(z)) = 2015 - f^{13}(z) = 2015 - (2015 - z) = z.\n$$\nConsequently $z = f^{26}(z) = f(f^{25}(z)) = f(2015 - z)$. Any real number $x$ can be written as $2015 - z$ for $z = 2015 - x$. Hence $z = f(2015 - z)$ implies $f(x) = 2015 - x$ for any real $x$.\n\nFinally check that the function $f(x) = 2015 - x$ satisfies the conditions of the problem. Let $d$ be a divisor of $2015$ greater than $1$. Then $d$ is odd, i.e. $d = 2c + 1$ for a positive integer $c$. Since $f^2(x) = 2015 - (2015 - x) = x$, we have $f^{2c}(x) = f^2(f^2(\\dots f^2(x)\\dots)) = x$, which implies $f^d(x) = f(f^{2c}(x)) = f(x) = 2015 - x$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 29795,
    "subject": "Mathematics (Multi-modal)",
    "question": "An inventor presented to the king a new exciting board game on a $9 \\times 10$ squared board. The king promised to reward him one rice grain for the first square, one rice grain for the second square, and for each following square the same number of grains as for the two preceding squares together. Prove that for the last square the inventor gets at least $2015^4$ grains.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Enumerate all squares with $1, \\ldots, 90$. Let the number of rice grains promised for the $n$-th square be $F_n$; then according to the problem $F_1 = F_2 = 1$ and $F_n = F_{n-1} + F_{n-2}$ for all $n > 2$. Notice that $F_n > F_{n-1}$ if $n > 2$, hence $F_{2(n+1)} = F_{2n+2} = F_{2n+1} + F_{2n} > F_{2n} + F_{2n} = 2 \\cdot F_{2n}$ for all $n$. This implies $F_{2 \\cdot 4} > 2 \\cdot F_{2 \\cdot 3} = 2 \\cdot 8 = 2^4$ and by mathematical induction $F_{2n} > 2^n$ for all $n > 3$. Therefore $F_{90} > 2^{45} > 2^{44} = (2^{11})^4 = 2048^4 > 2015^4$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 29796,
    "subject": "Mathematics (Multi-modal)",
    "question": "The circumcentre of an acute triangle $ABC$ is $O$. Line $AC$ intersects the circumcircle of $AOB$ at a point $X$, in addition to the vertex $A$. Prove that the line $XO$ is perpendicular to the line $BC$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "By the properties of inscribed angles $\\angle CXO = \\angle ABO$ (Fig. 8) independent of whether the point $X$ lies on the side $CA$ or on the extension of $CA$ over $A$. Since $O$ is the circumcentre of $ABC$, we have $\\angle ABO = \\angle BAO$. Let the intersection of lines $XO$ and $BC$ be $Y$ and let the point on the circumcircle of $ABC$ lying diametrically opposite $A$ be $Z$. Since $\\angle CXY = \\angle ZAB$, and $\\angle YCX = \\angle BZA$ as inscribed angles subtending the same arc $AB$, the triangles $CXY$ and $ZAB$ are similar. Hence $\\angle CYX = \\angle ZBA = 90^\\circ$, as $\\angle ZBA$ is subtended by a diameter.\n\n![](attached_image_1.png)\nFig. 8\nObviously $\\angle OBC = \\angle OCB$. If $X$ lies on segment $AC$ (Fig. 9), then $\\angle OBX = \\angle OAX = \\angle OAC = \\angle OCA$. If $X$ lies on the extension of $AC$ over $A$, then $\\angle OBX = 180^\\circ - \\angle OAX = \\angle OAC = \\angle OCA$. In both cases we have $\\angle OBX = \\angle OCA$ and $\\angle XBC = \\angle OBX + \\angle OBC = \\angle OCA + \\angle OCB = \\angle ACB = \\angle XCB$. Therefore $XBC$ is an isosceles triangle with the apex $X$, hence $X$ lies on the perpendicular bisector of $BC$. Since $O$ lies on the same line, the line $XO$ is the perpendicular bisector of the side $BC$.\n\n![](attached_image_2.png)\nFig. 9",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 29797,
    "subject": "Mathematics (Multi-modal)",
    "question": "On a switchboard there are $nm$ lamps arranged in an $n \\times m$ array. In the beginning all lamps are off. At each step one can switch three consecutive lamps in one row or in one column, changing the state of each lamp to the opposite. For which pairs of positive integers $(n, m)$ is it possible to achieve the situation, where all the lamps are switched on?",
    "options": [],
    "answer": "Exactly those pairs (n, m) for which at least one of n or m is divisible by 3.",
    "solution": "If $n$ (or $m$) is a multiple of $3$, then we can divide all lamps in each column (or row) into groups of $3$ and switch the lamps on by the groups.\n\nIf neither $n$ nor $m$ is a multiple of $3$, then color all lamps by diagonals with three colors (Fig. 10). Then each switching changes the state of exactly one lamp of each color, therefore each switching changes the parity of the number of lamps switched on for each color. Since in the beginning all lamps are off, the parity of the lamps switched on for each color always stays the same.\n\nLet $n = 3a + b$ and $m = 3c + d$, where $a$ and $c$ are nonnegative integers and $b$ and $d$ are either $1$ or $2$. In regions of sizes $3a \\times 3c$, $b \\times 3c$ and $3a \\times d$ the numbers of lamps of each color are equal and hence their parities are equal, but in the remaining $b \\times d$ region there is either $1$ lamp (Fig. 11), $2$ lamps of different colors (Fig. 12 and 13), or $4$ lamps, of which $2$ are of one color and two others of the two other colors (Fig. 14), hence the parities of the numbers of lamps of each color are not equal. Consequently, it is not possible to switch on all lamps.\n\n![](attached_image_1.png)\nFig. 10\n\n![](attached_image_2.png)\nFig. 11\n\n![](attached_image_3.png)\nFig. 12\n\n![](attached_image_4.png)\nFig. 13\n\n![](attached_image_5.png)\nFig. 14",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 29798,
    "subject": "Mathematics (Multi-modal)",
    "question": "Is there a two-digit number $n$ that does not end with zero such that\n\na) all numbers that can be formed by adding one or more zeros between the two-digit number's digits are its multiples?\n\nb) none of the numbers that can be formed by adding one or more zeros between the two-digit number's digits are its multiples?\n\nc) some numbers that can be formed by adding one or more zeros between the two-digit number's digits are its multiples and some are not?",
    "options": [],
    "answer": "a) 15; b) 12; c) 11",
    "solution": "a) One such number is $n = 15$. All numbers formed by adding zeros between its digits are divisible by $3$ and $5$.\n\nb) One such number is $n = 12$. Adding zeros between the digits, the last two digits will always be $02$. Hence no such number will be divisible by $4$.\n\nc) One such number is $n = 11$, because $101$ is not divisible by $11$ but $1001$ is.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 29799,
    "subject": "Mathematics (Multi-modal)",
    "question": "A right triangle $ABC$ has the right angle at vertex $A$. Circle $c$ passes through vertices $A$ and $B$ of the triangle $ABC$ and intersects the sides $AC$ and $BC$ correspondingly at points $D$ and $E$. The line segment $CD$ has the same length as the diameter of the circle $c$. Prove that the triangle $ABE$ is isosceles.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Since $\\angle BAD = 90^\\circ$ (Fig. 1), $BD$ is the diameter of circle $c$ and therefore $CD = BD$. Since $BD$ is diameter, also $\\angle BED = 90^\\circ$, so $DE$ is an altitude of the isosceles triangle $BDC$, bisecting its base $BC$. Hence $E$ is the midpoint of the hypotenuse $BC$ of the triangle $ABC$. Since the midpoint of the hypotenuse is the circumcentre of a right triangle, it follows $EA = EB$. This means that $ABE$ is an isosceles triangle.\n\n![](attached_image_1.png)\nFig. 1\nAs in the previous solution, we show that $CD = BD$. Hence $\\angle ECD = \\angle EBD$. From the equality of the inscribed angles subtending the arc *ED* it also follows $\\angle EBD = \\angle EAD$. From the triangle *ABC* we get $\\angle ABC = 90^\\circ - \\angle BCA$, or $\\angle EBA = 90^\\circ - \\angle ECD$. On the other hand, $\\angle EAB = \\angle DAB - \\angle EAD = 90^\\circ - \\angle EBD = 90^\\circ - \\angle ECD$. Consequently $\\angle EBA = \\angle EAB$. So the triangle *ABC* is isosceles.",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 29800,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $d$ be a positive number. On the parabola, whose equation has the coefficient $1$ at the quadratic term, points $A$, $B$ and $C$ are chosen in such a way that the difference of the $x$-coordinates of points $A$ and $B$ is $d$ and the difference of the $x$-coordinates of points $B$ and $C$ is also $d$. Find the area of the triangle $ABC$.",
    "options": [],
    "answer": "d^3",
    "solution": "Without loss of generality assume that equation of the parabola is $y = x^2$ (Fig. 2). Let the abscissas of the points $A$, $B$, and $C$ be $a$, $b$, and $c$. Let $A'$, $B'$, and $C'$ be the projections of $A$, $B$ and $C$ onto the $x$-axis. Denoting the area of a region $K$ by $S_K$ we have\n$$S_{ABC} = S_{ACC'A'} - S_{ABB'A'} - S_{BCC'B'}.$$ Since $S_{ACC'A'} = 2d \\cdot \\frac{a^2+c^2}{2}$, $S_{ABB'A'} = d \\cdot \\frac{a^2+b^2}{2}$, $S_{BCC'B'} = d \\cdot \\frac{b^2+c^2}{2}$, it follows that\n\n![](attached_image_1.png)\n\n$$\n\\frac{1}{2}d(2a^2 + 2c^2 - a^2 - b^2 - b^2 - c^2) = \\frac{1}{2}d(a^2 - b^2 - b^2 + c^2) = \\frac{1}{2}d(d(a+b) - d(b+c)) = \\frac{1}{2}d^2(a-c) = \\frac{1}{2}d^2 \\cdot 2d = d^3.\n$$\nWithout loss of generality we can assume that point $B$ lies at the origin. Then the equation of the parabola is $y = x^2 + px$ with some $p$. The coordinates of $A$ and $C$ are then $(d, d^2 + pd)$ and $(-d, d^2 - pd)$. Let $D$ be the midpoint of the segment $AC$. Its coordinates are $(0, d^2)$, so $BD$ is perpendicular to the $x$-axis, hence the lengths of the altitudes of both triangles $ABD$ and $BCD$ with the base $BD$ are $d$. So both the triangles have area $\\frac{d^3}{2}$, hence the area of the triangle $ABC$ is $d^3$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 29801,
    "subject": "Mathematics (Multi-modal)",
    "question": "On the plane three different points *P*, *Q*, and *R* are chosen. It is known that however one chooses another point *X* on the plane, the point *P* is always either closer to *X* than the point *Q* or closer to *X* than the point *R*. Prove that the point *P* lies on the line segment *QR*.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "We show that if the point $P$ lies outside the segment $QR$, then the conditions of the problem are not satisfied.\nIf $P$ lies on the line $QR$ but outside the segment $QR$ (Fig. 3), then we can\n\n![](attached_image_1.png)\nFig. 3\n\n![](attached_image_2.png)\nFig. 4\n\ntake the point $X$ on the line $QR$ on the other side of the segment $QR$. Then the points $Q$ and $R$ are closer to the point $X$ than the point $P$.\nIf $P$ lies outside the line $QR$ (Fig. 4), then the perpendicular bisectors of the segments $PQ$ and $PR$ intersect. Choose the points $X$ in the region, which lies towards $Q$ from the perpendicular bisector of $PQ$ and towards $R$ from the perpendicular bisector of $PR$ (the dark region on the figure). Then the points $Q$ and $R$ are closer to the point $X$ than the point $P$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 29802,
    "subject": "Mathematics (Multi-modal)",
    "question": "a) Find the largest number that is the greatest common divisor of some four different two-digit numbers.\n\nb) Find the largest number that is the least common multiple of some four different two-digit numbers.",
    "options": [],
    "answer": "a) 24; b) 99*98*97*95",
    "solution": "a) Let $d$ be the greatest common divisor of some four different two-digit numbers. Since all these numbers are divisible by $d$, the least possible candidates of these four numbers are $d$, $2d$, $3d$, $4d$. Hence $4d < 100$, thus $d \\le 24$. On the other hand, the greatest common divisor of $24$, $48$, $72$ and $96$ is $24$.\n\nb) The numbers $99$, $98$, $97$, and $95$ are pairwise relatively prime, hence $\\text{lcm}(99, 98, 97, 95) = 99 \\cdot 98 \\cdot 97 \\cdot 95$. To show that this is the largest possible, consider four different two-digit numbers $a_1$, $a_2$, $a_3$, $a_4$; assume without loss of generality that $a_1 > a_2 > a_3 > a_4$. If $a_4 \\le 95$, then $\\text{lcm}(a_1, a_2, a_3, a_4) \\le a_1 a_2 a_3 a_4 \\le 99 \\cdot 98 \\cdot 97 \\cdot 95$. If $a_4 = 96$, then the four numbers can only be $99$, $98$, $97$, $96$, but $\\text{lcm}(99, 98, 97, 96) = \\frac{99 \\cdot 98 \\cdot 97 \\cdot 96}{2 \\cdot 3} < 99 \\cdot 98 \\cdot 97 \\cdot 95$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 29803,
    "subject": "Mathematics (Multi-modal)",
    "question": "Angles $\\alpha$ and $\\beta$ are such that $\\frac{\\tan \\alpha}{\\tan \\beta} = k \\neq 1$. Express $\\frac{\\sin(\\alpha+\\beta)}{\\sin(\\alpha-\\beta)}$ in terms of $k$.",
    "options": [],
    "answer": "(k+1)/(k-1)",
    "solution": "We have $k = \\frac{\\tan \\alpha}{\\tan \\beta} = \\frac{\\sin \\alpha \\cdot \\cos \\beta}{\\cos \\alpha \\cdot \\sin \\beta}$, or $\\sin \\alpha \\cos \\beta = k \\cdot \\cos \\alpha \\sin \\beta$. Therefore\n$$\n\\frac{\\sin(\\alpha + \\beta)}{\\sin(\\alpha - \\beta)} = \\frac{\\sin \\alpha \\cos \\beta + \\cos \\alpha \\sin \\beta}{\\sin \\alpha \\cos \\beta - \\cos \\alpha \\sin \\beta} = \\frac{(k+1) \\cdot \\cos \\alpha \\sin \\beta}{(k-1) \\cdot \\cos \\alpha \\sin \\beta} = \\frac{k+1}{k-1}.\n$$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 29804,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $n$ be a natural number such that $n + 1$, $n + 3$, $n + 7$ and $n + 9$ are prime numbers, and $n + 31$, $n + 33$, $n + 37$ and $n + 39$ are also prime numbers. Find the remainder of $n$ divided by 210.",
    "options": [],
    "answer": "190",
    "solution": "Consider the remainders of $n$ when divided by 2, 3, 5, and 7. In the following tables the left column shows the remainder and the right column shows, which of the given eight numbers cannot be prime, if $n \\ge 7$:\n\n| n mod 2 | divisible by 2 | n mod 3 | divisible by 3 |\n|---------|----------------|---------|----------------|\n| 1       | $n + 1$        | 0       | $n + 3$        |\n|         |                | 2       | $n + 1$        |\n\n| n mod 5 | divisible by 5 | n mod 7 | divisible by 7 |\n|---------|----------------|---------|----------------|\n| 1       | $n + 9$        | 0       | $n + 7$        |\n| 2       | $n + 3$        | 2       | $n + 33$       |\n| 3       | $n + 7$        | 3       | $n + 39$       |\n| 4       | $n + 1$        | 4       | $n + 3$        |\n|         |                | 5       | $n + 9$        |\n|         |                | 6       | $n + 1$        |\n\nThus if $n \\ge 7$ then the remainder of $n$ when divided by 2 and 5 is 0 and when divided by 3 and 7 is 1; hence $n - 1$ is a multiple of 3 and 7. Consequently $n$ is a multiple of 10 and $n - 1$ is a multiple of 21. This implies that $n + 20$ is a multiple of 21 and a multiple of 10, hence a multiple of 210. Consequently the remainder of $n$ when divided by 210 is 190. The numbers $n = 1, 2, 3, 4, 5, 6$ do not satisfy the conditions of the problem.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 29805,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all real-valued functions $f$ defined on real numbers which satisfy $f(f(x) + f(y)) = f(x) + y$ for all real $x, y$.",
    "options": [],
    "answer": "f(x) = x",
    "solution": "Let $z_1, z_2$ be real numbers for which $f(z_1) = f(z_2)$. Substituting $y = z_1$ and $y = z_2$ into the given equation we get $f(f(x) + f(z_1)) = f(x) + z_1$, and $f(f(x) + f(z_2)) = f(x) + z_2$. Since the left hand sides are equal, we have $f(x) + z_1 = f(x) + z_2$, whence $z_1 = z_2$. Hence $f$ is one-to-one. Substituting $y = 0$ into the given equation we get $f(f(x) + f(0)) = f(x)$ for any real $x$. Since $f$ is one-to-one, we have $f(x) + f(0) = x$. If $x = 0$, then the last equation gives $2f(0) = 0$, or $f(0) = 0$. So this equation simplifies to $f(x) = x$. The function $f(x) = x$ satisfies the original equation.\nInterchanging $x$ and $y$ in the given equation we get $f(f(y) + f(x)) = f(y) + x$. Since the left hand side is the same as in the original equation, we have $f(x) + y = f(y) + x$. Substituting $y = 0$ into this we get $f(x) = x + f(0)$. Substituting into the original equation all applications of $f$ according to the last equality, we get $x + f(0) + y + f(0) + f(0) = x + f(0) + y$. This gives $f(0) = 0$ and from $f(x) = x + f(0)$ we get $f(x) = x$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 29806,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $m$ be an integer, $m \\ge 2$. Each student in a school is practising $m$ hobbies the most. Among any $m$ students there exist two students who have a common hobby. Find the smallest number of students for which there must exist a hobby which is practised by at least 3 students.",
    "options": [],
    "answer": "m^2",
    "solution": "If the number of students is $m^2 - 1 = (m-1)(m+1)$, let us split the students to $m-1$ groups, each with $m+1$ students. Assume that each student has a unique common hobby with every other student in the same group and none of the students is practising any other hobby. Then each student has exactly $m$ hobbies but among each $m$ students at least two of them belong to the same group, i.e., have a common hobby. Hence the conditions of the problem are satisfied but there exists no hobby which is practised by more than 2 students. If there are less than $m^2 - 1$ students at the school, similar construction can be used with a suitable number of students omitted.\n\nLet us show that if there are at least $m^2$ students at school, then there must exist a hobby that is practised by at least 3 students. Assume the contrary that every hobby is practised by 2 students at most. Let us form a group of students in which each pair of students have no hobby in common by adding in each iteration to initially empty group one student who does not have a common hobby with any of the students already assigned to the group. As every student has $m$ hobbies the most and each hobby can be common with only one other student, after $i$ iterations there are at most $i(m + 1)$ students who cannot be added to the group in the subsequent iterations. As $(m - 1)(m + 1) < m^2$, it is still possible to add another student to a group of $m - 1$ students. Therefore, there exists a group of $m$ students, in which there exists no pair who have a common hobby. That contradicts the conditions of the problem statement.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 29807,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all functions $f : \\mathbb{R} \\to \\mathbb{R}$ satisfying the equality $f(2^x + 2y) = 2^y f(f(x))f(y)$ for every $x, y \\in \\mathbb{R}$.",
    "options": [],
    "answer": "f(x) = 0 for all x; f(x) = 2^x for all x",
    "solution": "Substituting $y = -2^{x-1}$ into the original equation gives\n$$\nf(0) = \\frac{1}{2^{2^{x-1}}} f(f(x)) f(-2^{x-1}). \\quad (1)\n$$\nSo, if $f(-2^x) = 0$ for at least one $x$ then also $f(0) = 0$. Then taking $x = 0$ and arbitrary $y$ in the original identity gives $f(1+2y) = 0$, i.e., $f \\equiv 0$.\n\nAssume in the rest that $f(-2^x) \\neq 0$ for every $x$. Substituting $y = -2^x$ into the original equation gives $f(-2^x) = \\frac{1}{2^{2^x}} f(f(x)) f(-2^x)$. Hence, for every $x$,\n$$\nf(f(x)) = 2^{2^x} \\quad (2)\n$$\nSubstituting (2) into the original equation and taking $y = 0$, we obtain\n$f(2^x) = f(f(x))f(0) = 2^{2^x}f(0)$ for all $x$, which implies\n$$\nf(x) = 2^x f(0) \\quad (3)\n$$\nfor all positive $x$. On the other hand, applying (2) to (1) gives\n$$\nf(-2^{x-1}) = \\frac{2^{2^{x-1}}}{2^{2^x}} \\cdot f(0) = 2^{-2^{x-1}} f(0)\n$$\nfor all $x$, which implies (3) also for all negative $x$.\n\nWe have shown above that $f(0) = 0$ implies $f(x) = 0$ for all $x$. Hence we may assume that $f(0)$ is either positive or negative. By taking $x = 0$ in (2) and applying (3), we obtain $2 = 2^{2^0} = f(f(0)) = 2^{f(0)} \\cdot f(0)$. Both $f(0) < 1$ and $f(0) > 1$ would lead to contradiction, hence $f(0) = 1$ and the only non-zero solution is thus $f(x) = 2^x$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 29808,
    "subject": "Mathematics (Multi-modal)",
    "question": "Prove that for any positive integer $n$, $2 \\cdot \\sqrt{3} \\cdot \\sqrt[3]{4} \\cdot \\dots \\cdot \\sqrt[n-1]{n} > n$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "For $2 \\le k \\le n$, the GM-HM inequality for the numbers $k, \\dots, k, 1$, with $k$ repeated $k-2$ times, gives\n$$\nk - 1 = \\frac{(k - 1)^2}{k - 1} = \\frac{k(k - 2) + 1}{k - 1} \\ge \\sqrt[k-1]{k^{k-2}} = \\sqrt[k-1]{\\frac{k^{k-1}}{k}} = \\frac{k}{\\sqrt[k-1]{k}}\n$$\nHence $\\sqrt[k-1]{k} \\ge \\frac{k}{k-1}$ for every $k = 2, 3, \\dots, n$, with equality only for $k = 2$. Therefore\n$$\n2 \\cdot \\sqrt{3} \\cdot \\sqrt[3]{4} \\cdot \\dots \\cdot \\sqrt[n-1]{n} > \\frac{2}{1} \\cdot \\frac{3}{2} \\cdot \\frac{4}{3} \\cdot \\dots \\cdot \\frac{n}{n-1} = n.\n$$\n\n\nSolution 2:\nFor $2 \\le k \\le n$, the HM-GM inequality for the numbers $1, \\dots, 1, k$ (with $1$ repeated $k-2$ times) gives\n$$\n\\sqrt[k-1]{k} \\ge \\frac{k-1}{k-2 + \\frac{1}{k}} = \\frac{(k-1)k}{(k-2)k+1} = \\frac{(k-1)k}{(k-1)^2} = \\frac{k}{k-1},\n$$\nwith equality only for $k=2$. We continue as in solution 1.\n\n\nSolution 3:\nFrom the binomial theorem,\n$$\n\\left(\\frac{k}{k-1}\\right)^{k-1} = \\left(1 + \\frac{1}{k-1}\\right)^{k-1} = \\frac{\\binom{k-1}{0}}{(k-1)^0} + \\frac{\\binom{k-1}{1}}{(k-1)^1} + \\dots + \\frac{\\binom{k-1}{k-1}}{(k-1)^{k-1}}.\n$$\nThere are $k$ summands, of which $\\frac{\\binom{k-1}{0}}{(k-1)^0} = \\frac{\\binom{k-1}{1}}{(k-1)^1} = 1$, and for $1 < i \\le k-1$\n$$\n\\frac{\\binom{k-1}{i}}{(k-1)^i} = \\frac{(k-1) \\cdots (k-i)}{i! (k-1)^i} < \\frac{(k-1) \\cdots (k-i)}{(k-1)^i} < 1.\n$$\nTherefore $\\left(\\frac{k}{k-1}\\right)^{k-1} \\le \\underbrace{1 + 1 + \\dots + 1}_{k \\text{ times}} = k$, whence $\\sqrt[k-1]{k} \\ge \\frac{k}{k-1}$, with equality only for $k=2$. We continue as in solution 1.\n\n\nSolution 4:\nWe show that $k^k > (k+1)^{k-1}$ for $k \\ge 2$. The case $k=2$ is obvious. Suppose now that $k^k > (k+1)^{k-1}$ holds for some $k$, and let's prove $(k+1)^{k+1} > (k+2)^k$. Note that $(k+1)^{k+1} \\cdot (k+1)^{k-1} = (k+1)^{2k} = (k^2+2k+1)^k > (k^2+2k)^k = (k(k+2))^k = k^k \\cdot (k+2)^k$, giving $\\frac{(k+1)^{k+1}}{k^k} > \\frac{(k+2)^k}{(k+1)^{k-1}}$. This combined with the induction assumption gives\n$$\n(k+1)^{k+1} = \\frac{(k+1)^{k+1}}{k^k} \\cdot k^k > \\frac{(k+2)^k}{(k+1)^{k-1}} \\cdot (k+1)^{k-1} = (k+2)^k.\n$$\nWe have proven $k^k > (k+1)^{k-1}$, which is equivalent to $\\sqrt[k-1]{k} > \\sqrt[k]{k+1}$. Therefore $\\sqrt[k-1]{k} > \\sqrt[k]{n}$ for $k=2, 3, \\dots, n-1$, and\n$$\n2 \\cdot \\sqrt{3} \\cdot \\sqrt[3]{4} \\cdots \\sqrt[n-1]{n} > (\\sqrt[n-1]{n})^{n-1} = n.\n$$\n\n\nSolution 5:\nWe give another proof for the inequality $k^k > (k+1)^{k-1}$. It is equivalent to $k \\cdot \\left(\\frac{k}{k+1}\\right)^{k-1} > 1$, or $k \\cdot \\underbrace{\\frac{k}{k+1} \\cdots \\frac{k}{k+1}}_{k-1 \\text{ times}} > 1$. Note that for $x < k+1$, $x-x \\cdot \\frac{k}{k+1} = x \\cdot (1-\\frac{k}{k+1}) = x \\cdot \\frac{1}{k+1} < 1$. Therefore, multiplication with each factor $\\frac{k}{k+1}$ decreases the product by less than 1; cumulatively the product becomes smaller by less than $k-1$. Therefore $k \\cdot \\left(\\frac{k}{k+1}\\right)^{k-1} > k-(k-1)=1$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 29809,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $O$ be the circumcentre of the acute triangle $ABC$. Let $c_1$ and $c_2$ be the circumcircles of triangles $ABO$ and $ACO$. Let $P$ and $Q$ be points on $c_1$ and $c_2$ respectively, such that $OP$ is a diameter of $c_1$ and $OQ$ is a diameter of $c_2$. Let $T$ be the intersection of the tangent to $c_1$ at $P$ and the tangent to $c_2$ at $Q$. Let $D$ be the second intersection of the line $AC$ and the circle $c_1$. Prove that the points $D, O$ and $T$ are collinear.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Since $\\angle OAP = \\angle OAQ = 90^\\circ$, the points $P$, $A$ and $Q$ are collinear. Since $\\angle OPT = \\angle OQT = 90^\\circ$, $OPTQ$ is cyclic. Since $OA = OB$, the diameter $OP$ of $c_1$ is perpendicular to the chord $AB$. Therefore $PT$ and $AB$ are parallel.\n\nNow $\\angle TOQ = \\angle TPQ = \\angle TPA = \\angle BAP = \\angle BOP = 90^\\circ - \\angle ABO$. On the other hand, equality of inscribed angles subtending the arc $AO$ of circle $c_1$ gives $\\angle CDO = \\angle ABO$ (figures 20 and 21 show two possible situations). Therefore $\\angle DOQ = 90^\\circ - \\angle CDO = 90^\\circ - \\angle ABO$.\n\nIn summary, $\\angle TOQ = \\angle DOQ$, whence $D$, $O$ and $T$ are collinear.\n\n![](attached_image_1.png)\nFig. 20\n\n![](attached_image_2.png)\nFig. 21",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 29810,
    "subject": "Mathematics (Multi-modal)",
    "question": "On the sides $AB$, $BC$ and $CA$ of triangle $ABC$, points $L$, $M$ and $N$ are chosen, respectively, such that the lines $CL$, $AM$ and $BN$ intersect at a common point $O$ inside the triangle and the quadrilaterals $ALON$, $BMOL$ and $CNOM$ have incircles. Prove that\n$$\n\\frac{1}{AL \\cdot BM} + \\frac{1}{BM \\cdot CN} + \\frac{1}{CN \\cdot AL} = \\frac{1}{AN \\cdot BL} + \\frac{1}{BL \\cdot CM} + \\frac{1}{CM \\cdot AN}.\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "$ALON$ is a circumscribed quadrilateral, hence $AL + ON = AN + OL$. Similarly $BM + OL = BL + OM$ and $CN + OM = CM + ON$. By adding the equations we obtain $AL + BM + CN = AN + BL + CM$. Lines $CL$, $AM$ and $BN$ intersect in one point, we get from Ceva's theorem $AL \\cdot BM \\cdot CN = AN \\cdot BL \\cdot CM$. By dividing the left hand sides of the last two equations with the right hand sides we obtain the required equation.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 29811,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $n$ be a positive integer such that there exists a positive integer that is less than $\\sqrt{n}$ and does not divide $n$. Let $(a_1, \\dots, a_n)$ be an arbitrary permutation of $1, \\dots, n$. Let $a_{i_1} < \\dots < a_{i_k}$ be its maximal increasing subsequence and let $a_{j_1} > \\dots > a_{j_l}$ be its maximal decreasing subsequence. Prove that tuples $(a_{i_1}, \\dots, a_{i_k})$ and $(a_{j_1}, \\dots, a_{j_l})$ altogether contain at least one number that does not divide $n$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "The first phase of the solution consists in showing that $kl \\ge n$. For every $i = 1, \\dots, n$, let $f(i)$ denote the length of the longest increasing subsequence ending with $a_i$, and let $g(i)$ be the length of the longest decreasing subsequence ending with $a_i$. For distinct indices $i < j$, if $a_i < a_j$ then $f(i) < f(j)$, and if $a_i > a_j$ then $g(i) < g(j)$. Hence pairs of the form $(f(i), g(i))$ where $i = 1, \\dots, n$ are all distinct, i.e., there are $n$ different such pairs in total. By the conditions of the problem, the largest number of the form $f(i)$ is $k$ and the largest number of the form $g(i)$ is $l$. Thus the number of pairs of the form $(f(i), g(i))$ is at most $kl$. Consequently, $n \\le kl$.\n\nFrom this result, we deduce $k+l \\ge 2\\sqrt{kl} \\ge 2\\sqrt{n}$ by AM-GM. At most one number can belong to an increasing and a decreasing subsequence simultaneously. Thus subsequences $(a_{i_1}, \\dots, a_{i_k})$ and $(a_{j_1}, \\dots, a_{j_l})$ together contain at least $2\\sqrt{n}-1$ different natural numbers in total. By assumptions, number $n$ has at most $\\lfloor\\sqrt{n}\\rfloor-1$ divisors that are not larger than $\\sqrt{n}$, the total number $\\delta(n)$ of divisors of $n$ satisfies the inequality $\\delta(n) \\le 2\\lfloor\\sqrt{n}\\rfloor - 2$. Consequently, subsequences $(a_{i_1}, \\dots, a_{i_k})$ and $(a_{j_1}, \\dots, a_{j_l})$ together contain at least one number that does not divide $n$.\nThe inequality $kl \\ge n$ can be proven also in the following way. Let us partition the permutation $(a_1, \\dots, a_n)$ into decreasing subsequences using the following algorithm. The first element of each new subsequence is the first unused element in the original permutation, the next is the first following to it in the original permutation unused element smaller than it etc., until no more elements can be chosen this way. Let these subsequences be $K_1, \\dots, K_x$ in the order of forming.\n\nFor every $z = x, x-1, \\dots, 2$ and every element $a_j$ of $K_z$, there exists an element $a_i$ in $K_{z-1}$ such that $i < j$ and $a_i < a_j$. Indeed, suppose the contrary. Then all elements $a_i$ of $K_{z-1}$ such that $i < j$ are greater than $a_j$. This means that $a_j$ should have been chosen into $K_{z-1}$, a contradiction.\n\nHence, starting from an arbitrary element $b_x$ of $K_x$, we can choose an element $b_{x-1}$ from $K_{x-1}$, an element $b_{x-2}$ from $K_{x-2}$, etc, until $b_1$ from $K_1$, in such a way that $b_1 < \\dots < b_{x-1} < b_x$. This is an increasing subsequence of length $x$ of the original permutation. As every element of the original permutation belongs to one of $K_1, \\dots, K_x$, there exists a decreasing subsequence of length at least $\\frac{n}{x}$. Now $k \\ge x$ and $l \\ge \\frac{n}{x}$ together give $kl \\ge x \\cdot \\frac{n}{x} = n$.\nAnother algorithm can be used for partitioning the permutation $(a_1, \\dots, a_n)$ into decreasing subsequences in such a way that there exists an increasing subsequence with each element representing a different part. Let the first element of each new subsequence be the largest among the unused elements, the next be the largest following to it in the original sequence unused element etc., until no more elements can be chosen this way. Let these subsequences be $L_1, \\dots, L_y$ in the order of forming.\n\nFor every $z = y, y-1, \\dots, 2$ and every $a_i$ from $L_z$, there exists an element $a_j$ from $L_{z-1}$ such that $i < j$ and $a_i < a_j$. Indeed, suppose the contrary. Then all elements $a_j$ in $L_{z-1}$ such that $i < j$ are smaller than $a_i$. This means that $a_i$ should have been chosen into $L_{z-1}$, contradiction.\n\nHence, starting from an arbitrary element $c_y$ of $L_y$, we can choose an element $c_{y-1}$ from $L_{y-1}$, an element $c_{y-2}$ from $L_{y-2}$ etc., until $c_1$ from $L_1$, in such a way that $c_y < c_{y-1} < \\dots < c_1$. The rest is as in Solution 2.",
    "topic": "Discrete Mathematics",
    "subtopic": "Combinatorics"
  },
  {
    "id": 29812,
    "subject": "Mathematics (Multi-modal)",
    "question": "Do there exist distinct positive integers $x$ and $y$ such that the number $x+y$ is divisible by $2016$, the number $x-y$ is divisible by $2017$ and the number $xy$ is divisible by $2018$?",
    "options": [],
    "answer": "Yes; for example, x = 2016*2015 − 2018 and y = 2018.",
    "solution": "For example, numbers $x = 2016 \\cdot 2015 - 2018$ and $y = 2018$ meet the conditions. As $2016 \\cdot 2015 > 4 \\cdot 1009 = 2 \\cdot 2018$ implies $x > y$, they are distinct. The sum $2016 \\cdot 2015$ is divisible by $2016$ and the product is obviously divisible by $2018$. Furthermore, $x - y = 2016 \\cdot 2015 - 2 \\cdot 2018 = (2017 - 1)(2017 - 2) - 2 \\cdot (2017 + 1) = 2017^2 - 3 \\cdot 2017 + 2 - 2 \\cdot 2017 - 2 = 2017 \\cdot 2012$, whence the difference of these numbers is divisible by $2017$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 29813,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABC$ be a scalene triangle with median $AM$. Let $K$ be the point of tangency of the incircle of triangle $ABC$ with the side $BC$. Prove that if the length of the side $BC$ is the arithmetic mean of the lengths of the sides $AB$ and $AC$ then the bisector of the angle $BAC$ passes through the midpoint of the line segment $KM$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $N$ be the intersection point of the bisector of angle $BAC$ and side $BC$; it suffices to prove that $KN = MN$ (Fig. 19). The bisector property implies $\\frac{NC}{NB} = \\frac{AC}{AB}$. Substituting $NC = BC - NB$ gives\n$$\nNB = \\frac{BC}{1 + \\frac{AC}{AB}} = \\frac{AB \\cdot BC}{AB + AC}.\n$$\nAs $AB + AC = 2BC$ by assumption, this implies $NB = \\frac{AB}{2}$.\n\n![](attached_image_1.png)\nFig. 19\n![](attached_image_2.png)\nFig. 20\n\nOn the other hand, let $X$ and $Y$ be the points of tangency of the incircle of triangle $ABC$ with sides $AB$ and $AC$, respectively (Fig. 20). Then $AX = AY$, $BX = BK$ and $CK = CY$, whence $BC = BK + CK = BX + CY = AB - AX + AC - AY = AB + AC - 2AX = 2BC - 2AX$. Thus $BC = 2AX$, implying $AX = BM$. Consequently, $KN = BN - BK = BN - BX = \\frac{AB}{2} - (AB - AX) = AX - \\frac{AB}{2} = BM - \\frac{AB}{2} = BM - BN = MN$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 29814,
    "subject": "Mathematics (Multi-modal)",
    "question": "Call a tuple $(a_1, \\dots, a_n)$ of real numbers *stable* if the sums $a_1 + a_2 + \\dots + a_k$, as well as the sums $a_k + a_{k+1} + \\dots + a_n$, where in both cases $0 < k \\le n$, are either all negative or all non-negative.\nFor instance, the tuple $(3, -1, 2)$ is stable, since:\n$$\n\\begin{array}{lclcl}\n3 & \\ge & 0, & 2 & \\ge 0, \\\\\n3 + (-1) & \\ge & 0, & (-1) + 2 & \\ge 0, \\\\\n3 + (-1) + 2 & \\ge & 0; & 3 + (-1) + 2 & \\ge 0.\n\\end{array}\n$$\nProve that in any stable tuple with at least 3 terms where all terms are alternately negative and non-negative (it is unknown whether the first term is negative or non-negative), there exist 3 consecutive terms that together (without reordering) form a stable tuple on their own.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Consider terms whose absolute value is minimal in the tuple. If there exists a negative such element, denote it $a_i$, then the sum of $a_i$ and its any neighbour is non-negative. Thus $a_i$ is neither the first nor the last in the tuple because of stability of the tuple. But then both $a_{i-1} + a_i$ and $a_i + a_{i+1}$ are non-negative, as well as $a_{i-1} + a_i + a_{i+1}$, hence $a_{i-1}$, $a_i$, $a_{i+1}$ together form a stable subtuple. If all elements with minimal absolute value are non-negative then let $a_i$ be any of them. Analogously to the previous case, both $a_{i-1} + a_i$ and $a_i + a_{i+1}$ are negative, as well as $a_{i-1} + a_i + a_{i+1}$, hence $a_{i-1}$, $a_i$, $a_{i+1}$ together form a stable tuple.",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 29815,
    "subject": "Mathematics (Multi-modal)",
    "question": "Given positive integers $a$, $b$, $c$ and $d$ and\n$$ (a+b)(a+c)(a+d)(b+c)(b+d)(c+d) = u, $$\n$$ ab + ac + ad + bc + bd + cd = v, $$\nprove that the product $uv$ is divisible by 3.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "If among numbers $a$, $b$, $c$, $d$ there are two that give either remainders $0$ and $0$ or remainders $1$ and $2$ modulo $3$, the sum of these two numbers is divisible by $3$. Hence $u$, as well as $uv$, is divisible by $3$.\n\nNow study the case where at most one among the numbers $a$, $b$, $c$, $d$ is divisible by $3$ and all numbers not divisible by $3$ are congruent modulo $3$. If exactly one among numbers $a$, $b$, $c$, $d$ is divisible by $3$ then the products of this number with all other numbers are divisible by $3$. Other numbers form $3$ pairs whose products of components are congruent modulo $3$. Hence the sum $v$ of all six pairwise products is divisible by $3$.\n\nIf none of $a$, $b$, $c$, $d$ is divisible by $3$ then the pairwise products are all congruent modulo $3$. Again, as the number of pairs is divisible by $3$, this implies that the sum $v$ of the products is divisible by $3$. Consequently, $uv$ is divisible by $3$ in this case, too.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 29816,
    "subject": "Mathematics (Multi-modal)",
    "question": "Positive integer $b$ is obtained by reordering the digits in a positive integer $a$. Which of the following claims are definitely true?\n\na) The sums of the digits of numbers $2a$ and $2b$ are equal.\nb) The sums of the digits of numbers $3a$ and $3b$ are equal.\nc) The sums of the digits of numbers $5a$ and $5b$ are equal.",
    "options": [],
    "answer": "a and c",
    "solution": "Call digits $0, 1, 2, 3, 4$ small and digits $5, 6, 7, 8, 9$ large. Denote the digits of $k$-digit number $n$ from right to left by $(n)_0, (n)_1, \\dots, (n)_{k-1}$. Denote by $\\Sigma(n)$ the sum of all digits of $n$ and by $l(n)$ the number of large digits of $n$.\n\na) If $(a)_i$ is small then $(2a)_i = 2(a)_i$ or $(2a)_i = 2(a)_i + 1$ depending on whether $(a)_{i-1}$ is small or large. Similarly if $(a)_i$ is large then $(2a)_i = 2(a)_i - 10$ or $(2a)_i = 2(a)_i - 9$ depending on whether $(a)_{i-1}$ is small or large. In other words, when multiplying by $2$, each large digit necessitates decreasing of the digit at its place by $10$ and increasing of the preceding digit by $1$ in comparison with a small digit. Hence, for every natural number $n$, $\\Sigma(2n) = 2\\Sigma(n) - 9l(n)$. As $\\Sigma(a) = \\Sigma(b)$ and $l(a) = l(b)$, we have $\\Sigma(2a) = \\Sigma(2b)$.\n\nb) If $a = 34$ and $b = 43$ then $\\Sigma(3a) = 1 + 0 + 2 = 3$ but $\\Sigma(3b) = 1 + 2 + 9 = 12$.\n\nc) Obviously $\\Sigma(10a) = \\Sigma(10b)$. By part a), on the other hand, $\\Sigma(10n) = \\Sigma(2 \\cdot 5n) = 2\\Sigma(5n) - 9l(5n)$, holding for every natural number $n$. Hence $2\\Sigma(5a) - 9l(5a) = 2\\Sigma(5b) - 9l(5b)$. The digit $(5n)_i$ is large if and only if the digit $(n)_i$ is odd, because a carry during multiplying by $5$ can be at most $4$. Thus $l(5n)$ is the number of odd digits of $n$, implying $l(5a) = l(5b)$. Consequently, $\\Sigma(5a) = \\Sigma(5b)$.",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 29817,
    "subject": "Mathematics (Multi-modal)",
    "question": "Real numbers $x$, $y$ and $z$ satisfy $x + y + z = 4$ and $\\frac{1}{x} + \\frac{1}{y} + \\frac{1}{z} = \\frac{1}{3}$. Find the largest and the smallest possible value of the expression $x^3 + y^3 + z^3 + xyz$.",
    "options": [],
    "answer": "largest = 64, smallest = 64",
    "solution": "$$(x + y + z)^3 = x^3 + y^3 + z^3 + 3(x^2y + xy^2 + x^2z + y^2z + y^2x + xyz) + 6xyz,$$\nwhile\n$$(3(x + y + z) \\left(\\frac{1}{x} + \\frac{1}{y} + \\frac{1}{z}\\right) xyz = 3(x + y + z)(xy + xz + yz) \\\\ = 3(x^2y + xy^2 + x^2z + xz^2 + y^2z + yz^2) + 9xyz.$$ \nThus\n$$(x + y + z)^3 - 3(x + y + z) \\left(\\frac{1}{x} + \\frac{1}{y} + \\frac{1}{z}\\right) xyz = x^3 + y^3 + z^3 - 3xyz.$$ \nBy assumptions, $x + y + z = 4$ and $\\frac{1}{x} + \\frac{1}{y} + \\frac{1}{z} = \\frac{1}{3}$. Hence $64 - 4xyz = x^3 + y^3 + z^3 - 3xyz$, implying $x^3 + y^3 + z^3 + xyz = 64$. Consequently, the expression $x^3 + y^3 + z^3 + xyz$ has only one value 64.",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 29818,
    "subject": "Mathematics (Multi-modal)",
    "question": "Prove that in every triangle there is a median whose length squared is at least $\\sqrt{3}$ times the area of the triangle.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Assume w.l.o.g. that $BC$ is the shortest side of the triangle. Then the least angle of the triangle is by vertex $A$. Denote $a = BC$, $b = CA$, $c = AB$, $\\alpha = \\angle BAC$ and let $m$ be the length of the median drawn from vertex $A$. By assumptions made at the beginning of the solution, $\\alpha \\le 60^\\circ$. Let $D, E, F$ be the midpoints of sides $BC, CA, AB$, respectively (Fig. 21).\n\nAs $\\angle AFD = 180^\\circ - \\alpha$ because of $DF \\parallel CA$, the law of cosines in triangle $AFD$ implies\n$$\nm^2 = \\left(\\frac{b}{2}\\right)^2 + \\left(\\frac{c}{2}\\right)^2 - 2 \\cdot \\frac{b}{2} \\cdot \\frac{c}{2} \\cos \\angle AFD = \\frac{b^2}{4} + \\frac{c^2}{4} + \\frac{bc}{2} \\cos \\alpha\n$$\n$$\n\\ge \\frac{b^2}{4} + \\frac{c^2}{4} + \\frac{bc}{2} \\cos 60^\\circ = \\frac{b^2 + c^2 + bc}{4}.\n$$\nAs $b^2 + c^2 \\ge 2bc$, we obtain $m^2 \\ge \\frac{3bc}{4}$.\n\nOn the other hand, let $S$ be the area of the triangle $ABC$. Then $S = \\frac{1}{2}bc \\sin \\alpha \\le \\frac{1}{2}bc \\sin 60^\\circ \\le \\frac{\\sqrt{3}bc}{4}$. Before we obtained the inequality $m^2 \\ge \\frac{3bc}{4} = \\sqrt{3} \\cdot \\frac{\\sqrt{3}bc}{4}$. Consequently, $m^2 \\ge \\sqrt{3}S$.\n\n![](attached_image_1.png)\nFig. 21\nLet the median drawn from vertex $A$ of the triangle $ABC$ be $AD$ and the centroid of the triangle be $M$. W.l.o.g., let $AD$ be the longest median of triangle $ABC$; denote $AD = m$. Draw to both sides of the median $AD$ triangles $ADX$ and $ADY$ that have right angles by vertex $D$ and angles of size $30^\\circ$ by vertex $A$; then $AXY$ is an equilateral triangle having median $AD$ and centroid $M$ in common with triangle $ABC$ (Fig. 22). We show next that the area of triangle $AXY$ is at least as large as the area of triangle $ABC$. Since a median divides the triangle into two parts of equal area, it suffices to show that the area of triangle $ADX$ is at least as large as the area of triangle $ADB$.\n\nIf point $B$ lies inside the triangle $ADX$ or on its side then this claim holds obviously. If point $C$ lies inside the triangle $ADY$ or on its side then this claim holds by symmetry. It remains to handle the case where both $B$ and $C$ lie outside the triangle $AXY$ (Fig. 23). Suppose w.l.o.g. that line segments $DB$ and $AX$ intersect (the other case where line segments $DC$ and $AY$ intersect is symmetric). Let $l$ be the line parallel to $AX$ passing through $Y$. As line $l$ is symmetric w.r.t. point $D$ with line $AX$ and $DC = DB$, line segment $DC$ intersects line $l$. Therefore $MC > MY = MA$, which contradicts the assumption that $AD$ is the longest median of the triangle $ABC$. Hence this case cannot appear.\n\nAs $XD = YD = \\frac{m}{\\sqrt{3}}$, the area of the triangle $AXY$ is $m \\cdot \\frac{m}{\\sqrt{3}} = \\frac{m^2}{\\sqrt{3}}$. By the argumentation above, the area of the triangle $ABC$ must be at most $\\frac{m^2}{\\sqrt{3}}$.\n\n![](attached_image_2.png)\nFig. 22\n![](attached_image_3.png)\nFig. 23\nLet the median of the triangle $ABC$ drawn from point $A$ be $AD$ and the centroid of the triangle be $M$. W.l.o.g., let $AD$ be the longest median of the triangle $ABC$; denote $AD = m$. Then $A$ is the vertex of the triangle with largest distance from point $M$. Thus vertices $B$ and $C$ lie in circle $c$ with centre $M$ and radius $MA$. Let $B'$ and $C'$ be the second intersection points of rays $AB$ and $AC$, respectively, with circle $c$ (Fig. 24); then the area of the triangle $AB'C'$ is at least as large as the area of triangle $ABC$. Let $X$ and $Y$ be points on circle $c$ such that triangle $AXY$ is equilateral (Fig. 25); then the area of $AXY$ is at least as large as the area of triangle $AB'C'$ and also at least as large as the area of triangle $ABC$. The triangle $AXY$ can be divided into three equal triangles $MAX$, $MXY$ and $MYA$ whose total area is $\\frac{3}{2} \\cdot (\\frac{2}{3}m)^2 \\sin 120^\\circ$, which equals $\\frac{m^2}{\\sqrt{3}}$. Hence the area of the triangle $ABC$ is at most $\\frac{m^2}{\\sqrt{3}}$.\n\n![](attached_image_4.png)\nFig. 24\n![](attached_image_5.png)\nFig. 25\nLet $M$ be the centroid of the triangle $ABC$ and $m$ be the length of its longest median. The area of the triangle $ABC$ is\n$$\nS = \\frac{1}{2} \\cdot (MB \\cdot MC \\cdot \\sin \\alpha + MC \\cdot MA \\cdot \\sin \\beta + MA \\cdot MB \\cdot \\sin \\gamma),\n$$\nwhere $\\alpha = \\angle BMC$, $\\beta = \\angle CMA$ and $\\gamma = \\angle AMB$. As $MA \\le \\frac{2}{3}m$, $MB \\le \\frac{2}{3}m$ and $MC \\le \\frac{2}{3}m$, we obtain $S \\le \\frac{1}{2} \\cdot (\\frac{2}{3}m)^2 \\cdot (\\sin \\alpha + \\sin \\beta + \\sin \\gamma)$. As $\\alpha, \\beta$ and $\\gamma$ are less than $180^\\circ$, Jensen's inequality applies and gives\n$$\n\\frac{\\sin \\alpha + \\sin \\beta + \\sin \\gamma}{3} \\le \\sin \\frac{\\alpha + \\beta + \\gamma}{3} = \\sin \\frac{360^\\circ}{3} = \\sin 120^\\circ = \\frac{\\sqrt{3}}{2}.\n$$\nConsequently, $S \\le \\frac{1}{2} \\cdot \\frac{4}{9}m^2 \\cdot 3 \\cdot \\frac{\\sqrt{3}}{2} = \\frac{m^2}{\\sqrt{3}}$, directly implying the claim.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 29819,
    "subject": "Mathematics (Multi-modal)",
    "question": "Inside a circle of radius $1$ (or on the circumference), one marks $n$ points in such a way that the minimal distance between two marked points is as large as possible. Let $d_n$ be this distance between the two closest points. Is it true that $d_{n+1} < d_n$ for every natural number $n \\ge 2$?",
    "options": [],
    "answer": "No",
    "solution": "We show that $d_6 \\le 1 \\le d_7$. For the first inequality, assume arbitrary six points $A_1, A_2, A_3, A_4, A_5, A_6$ being marked in the circle. Let the centre of the circle be $O$. If $A_i = O$ for some $i$, the distance between $A_i$ and any other marked points is at most $1$. Assume in the rest that $A_i = O$ for no $i$. Let $\\alpha$ be the smallest angle that arises between some two rays $OA_i$ and $OA_j$, where $i, j = 1, 2, 3, 4, 5, 6$ (Fig. 26).\n![](attached_image_1.png)\nFig. 26\nis $360^\\circ$. If $A_iA_j > 1$ then $A_iA_j$ is the largest side of the triangle $OA_iA_j$, as the lengths of $OA_i$ and $OA_j$ do not exceed $1$. The angle opposite to the longest side is the largest, whence $\\alpha$ should be larger than $60^\\circ$, contradiction. Thus there exist two marked points at distance at most $1$ from each other. As the choice of the points was arbitrary, this establishes $d_6 \\le 1$.\n\nOn the other hand, when marking the vertices of a regular hexagon inscribed into the circle together with the centre of the circle, the distance between any two consecutive marked points on the circumference is equal to $1$ and their distance from the remaining point is also $1$. Hence $d_7 \\ge 1$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 29820,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all positive integers $k$ for which the integers $1, 2, \\dots, 2017$ can be divided into $k$ groups in such a way that the sums of numbers in these groups are $k$ consecutive terms of an arithmetic sequence.",
    "options": [],
    "answer": "1, 2, 1009, 2017",
    "solution": "Let the arithmetic sequence have the first term $a$ and the common difference $d$. The sum of all terms equals the sum of numbers $1, 2, \\dots, 2017$, i.e.,\n$$\n\\frac{2a + (k-1)d}{2} \\cdot k = \\frac{2017 \\cdot 2018}{2}\n$$\nwhence\n$$\n(2a + (k-1)d) \\cdot k = 2017 \\cdot 2018 = 2 \\cdot 1009 \\cdot 2017.\n$$\nThus the product $2 \\cdot 1009 \\cdot 2017$ is divisible by $k$. Since $2$, $1009$, and $2017$ are primes and $k \\le 2017$ by assumption, the only possibilities are $k = 1$, $k = 2$, $k = 1009$, and $k = 2017$.\n\nAll these can occur indeed. A partition into $1$ group trivially satisfies the conditions. An arbitrary partition into $2$ groups also provides two consecutive terms of some arithmetic sequence. Coupling each even number with the next odd number, we get $1008$ groups of size $2$, whose sums are consecutive terms of the arithmetic sequence $5, 9, 13, \\ldots$. Forming one additional group containing $1$ as the only element, we obtain a partition satisfying the conditions. Finally, the conditions will also be met by the partition where every integer $1, 2, \\ldots, 2017$ belongs to a separate group.",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 29821,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all solutions of the equation $a + b + c = 61$ in natural numbers that satisfy $\\text{gcd}(a, b) = 2$, $\\text{gcd}(b, c) = 3$, and $\\text{gcd}(c, a) = 5$.",
    "options": [],
    "answer": "(a, b, c) = (10, 6, 45), (10, 36, 15), (40, 6, 15)",
    "solution": "As $\\text{gcd}(a, b) = 2$, $\\text{gcd}(b, c) = 3$ and $\\text{gcd}(c, a) = 5$, the number $a$ is divisible by both $2$ and $5$, the number $b$ is divisible by both $2$ and $3$, and the number $c$ is divisible by both $3$ and $5$. Hence $a$ is divisible by $10$, $b$ is divisible by $6$ and $c$ is divisible by $15$. As $61$ gives remainder $1$ when divided by each of $2$, $3$ and $5$, the numbers $a, b$ and $c$ must give remainder $1$ when divided by $3$, $5$ and $2$, respectively. Since $a, b, c \\le 61$, the possibilities are $a = 10$ or $a = 40$, $b = 6$ or $b = 36$, and $c = 15$ or $c = 45$. The sum $61$ appears in three cases: $a = 10, b = 6, c = 45$; $a = 10, b = 36, c = 15$; $a = 40, b = 6, c = 15$. A straightforward check shows that the conditions $\\text{gcd}(a, b) = 2$, $\\text{gcd}(b, c) = 3$ and $\\text{gcd}(c, a) = 5$ are also met in all these cases.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 29822,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all positive integers $n$ such that a square can be cut into $n$ square pieces.",
    "options": [],
    "answer": "n = 1, 4, or any integer n ≥ 6; no dissections exist for n = 2, 3, or 5",
    "solution": "A partition of a square into 1 square is trivial. If $n \\ge 2$ then a partition into $2n$ squares can be obtained by cutting $n$ squares of side length $\\frac{1}{n}$ from one side of the square and $n - 1$ more squares of the same size from a neighbouring side. One square of side length $\\frac{n-1}{n}$ of the side length of the big square is left (Fig. 14 depicts the situation in the case $n = 4$ that provides a partition into 8 squares). For each $n \\ge 2$, one can obtain a partition into $2n + 3$ squares by splitting one square in a partition into $2n$ squares into four. Thus there exist partitions into 1, 4 and every natural number starting from 6.\n\n![](attached_image_1.png)\nFig. 14\n\nIt remains to show that there are no partitions of a square into 2, 3 and 5 squares. A square has 4 vertices and each vertex belongs to only one square in a partition. If two vertices belonged to the same square in the partition, this piece should be as large as the initial square, which is possible only in partition into 1. Hence the number of squares in a partition into a larger number of squares must be at least 4. In a hypothetical partition into 5 squares, at least three sides of the initial square should adjoin exactly 2 squares in the partition. Let the initial square be $ABCD$ and let the sides $DA, AB$ and $BC$ adjoin exactly 2 squares in the partition. Let the squares adjoining the side $AB$ have side lengths $x$ and $y$ in the order from vertex $A$ to vertex $B$. The square of side length $x$ is adjoining also the side $AD$ and the square of side length $y$ is adjoining the side $BC$, whence the other squares adjoining the sides $AD$ and $BC$ have side lengths $y$ and $x$, respectively. If $x > y$ then the squares located by vertices $A$ and $C$ would overlap (Fig. 15). If $x < y$ then the squares located by vertices $B$ and $D$ would overlap. If $x = y$, four squares in the partition would cover the whole initial square and the fifth square cannot exist.\n\n![](attached_image_2.png)\nFig. 15",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 29823,
    "subject": "Mathematics (Multi-modal)",
    "question": "Triangle $ABC$ has $AC = BC$. The bisector of angle $CAB$ meets side $BC$ at point $D$. The difference of the sizes of some two internal angles of triangle $ABD$ is $40^\\circ$. Find all possibilities of what the size of angle $ACB$ can be.",
    "options": [],
    "answer": "4°, 20°, 40°, 68°",
    "solution": "![](attached_image_1.png)\n\nFig. 16\n\n* If $\\alpha - \\frac{\\alpha}{2} = 40^\\circ$ then $\\alpha = 80^\\circ$, whence $\\angle ACB = 20^\\circ$.\n\n* The case $\\frac{\\alpha}{2} - \\alpha = 40^\\circ$ is impossible since it would imply $\\alpha < 0^\\circ$.\n\n* If $(180^\\circ - \\frac{3}{2}\\alpha) - \\frac{\\alpha}{2} = 40^\\circ$ then $\\alpha = 70^\\circ$, whence $\\angle ACB = 40^\\circ$.\n\n* If $\\frac{\\alpha}{2} - (180^\\circ - \\frac{3}{2}\\alpha) = 40^\\circ$ then $\\alpha = 110^\\circ$, but the base angle of an isosceles triangle cannot be obtuse.\n\n* If $(180^\\circ - \\frac{3}{2}\\alpha) - \\alpha = 40^\\circ$ then $\\alpha = 56^\\circ$, whence $\\angle ACB = 68^\\circ$.\n\n* If $\\alpha - (180^\\circ - \\frac{3}{2}\\alpha) = 40^\\circ$ then $\\alpha = 88^\\circ$, whence $\\angle ACB = 4^\\circ$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 29824,
    "subject": "Mathematics (Multi-modal)",
    "question": "In the mathematics circle, Juku raised a hypothesis that, for every integer $n > 4$, at least one out of the two largest integers that are less than $\\frac{n}{2}$ is relatively prime to $n$. Is Juku's hypothesis valid?",
    "options": [],
    "answer": "Detailed solution",
    "solution": "If $n$ is odd then the largest integer that is less than $\\frac{n}{2}$ is $\\frac{n-1}{2}$. Let $d$ be a common divisor of numbers $\\frac{n-1}{2}$ and $n$. Then $d$ is a common divisor of numbers $n-1$ and $n$, implying that $d = 1$. Hence $\\frac{n-1}{2}$ and $n$ are relatively prime, meaning that the hypothesis holds in the case of odd numbers.\n\nIf $n$ is even then the two largest integers that are less than $\\frac{n}{2}$ are $\\frac{n}{2} - 1$ and $\\frac{n}{2} - 2$. Let $d_1$ be a common divisor of numbers $\\frac{n}{2} - 1$ and $n$, and let $d_2$ be a common divisor of numbers $\\frac{n}{2} - 2$ and $n$. Then $d_1$ is a common divisor of numbers $n-2$ and $n$, and $d_2$ is a common divisor of numbers $n-4$ and $n$. Hence $d_1$ divides 2, i.e., is either 1 or 2, and $d_2$ divides 4, i.e., is either 1 or 2 or 4. If $d_1$ and $d_2$ were both larger than 1, they both should be even, whence their multiples $\\frac{n}{2} - 1$ and $\\frac{n}{2} - 2$ should be even. This is impossible, since $\\frac{n}{2} - 1$ and $\\frac{n}{2} - 2$ are consecutive integers. The contradiction shows that at least one of the divisors $d_1$ and $d_2$ equals 1. Thus one of $\\frac{n}{2} - 1$ and $\\frac{n}{2} - 2$ is relatively prime to $n$, meaning that the hypothesis holds in the case of even numbers, too.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 29825,
    "subject": "Mathematics (Multi-modal)",
    "question": "Around each vertex of a regular hexagon of side length $\\sqrt{3}$ in a plane, one draws a circle of radius $1$ with centre at that vertex and paints the region inside the circle blue. Find the area of the part of the plane that is painted blue.",
    "options": [],
    "answer": "4π + 3√3",
    "solution": "The circles drawn around two neighbouring vertices of the hexagon intersect, since $2 \\cdot 1 > \\sqrt{3}$. Hence every two neighbouring circles have a common region of the shape of a lens. Since a regular hexagon can be put together from six equilateral triangles, the circumradius of the hexagon equals $\\sqrt{3}$. The altitude of one equilateral triangle is $\\sqrt{(\\sqrt{3})^2 - (\\frac{\\sqrt{3}}{2})^2} = \\frac{3}{2}$.\n\nThe distance between a vertex and the second one counting from that vertex along the circumference is $3$, since it equals twice the altitude of the equilateral triangle (Fig. 17). As $2 \\cdot 1 < 3$, circles drawn around such two vertices do not intersect. Thus the circles drawn around opposite vertices do not intersect either.\n\n![](attached_image_1.png)\nFig. 17\n![](attached_image_2.png)\nFig. 18\n\n$(\\frac{AB}{2})^2 = AC^2$, i.e., $(\\frac{CD}{2})^2 + (\\frac{\\sqrt{3}}{2})^2 = 1^2$, whence $CD = 1$, implying that the triangle $ACD$ is equilateral. Thus the area of sector $ACD$ and triangle $ACD$ equal $\\frac{1}{6}\\pi$ and $\\frac{\\sqrt{3}}{4}$, respectively. The area of the region painted blue is $6\\pi - 12(\\frac{1}{6}\\pi - \\frac{\\sqrt{3}}{4})$, which equals $4\\pi + 3\\sqrt{3}$.",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 29826,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find the number of solutions of the equation $|a - b| = |b - c|$ in integers from $0$ to $36$.",
    "options": [],
    "answer": "2017",
    "solution": "The equation $|a - b| = |b - c|$ is satisfied if and only if either $a - b = b - c$ or $a - b = c - b$. The first equality is equivalent to $a + c = 2b$, the second is equivalent to $a = c$.\n\nTo fulfill the condition $a + c = 2b$, the numbers $a$ and $c$ must have the same parity. Then their sum is even and $b$ lies between $a$ and $c$, whence it also falls between the required bounds. There are $19$ possibilities for choosing $a$ or $c$ as an even number ($0, 2, \\ldots, 36$ are suitable) which gives $361$ possibilities in total. The number of possibilities to choose an odd number $a$ or $c$ is $18$ ($1, 3, \\ldots, 35$ are suitable) which gives $324$ possibilities in total. There are therefore $685$ triples satisfying $a + c = 2b$.\n\nThe number of triples satisfying $a = c$ is $1369$, since $a$ and $b$ can be chosen arbitrarily. There are $37$ solutions that satisfy both $a + c = 2b$ and $a = c$ and hence being counted twice, since these two conditions hold simultaneously if and only if $a = b = c$.\n\nConsequently, the total number of solutions meeting the conditions of the problem is $685 + 1369 - 37$ which equals $2017$.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 29827,
    "subject": "Mathematics (Multi-modal)",
    "question": "a) The general form $\\overline{ABC}$ of a three-digit number is initially written on a blackboard. Ann and Enn replace by turns letters with digits, exactly one at a time, with Ann starting. Can Ann write digits in such a way that, irrespectively of Enn's move, the resulting three-digit number would be divisible by 11? (Different letters may be replaced with equal digits but the letter $A$ must not be replaced with zero.)\n\nb) Ann and Enn got bored with writing the general form of the number again at the beginning of each game, and decided to change the rules as follows. First, Ann writes one digit to the blackboard, then Enn writes the second digit either to the right or to the left of it, and finally Ann completes the number with writing the third digit either to the left or to the right of the two digits already on the blackboard (writing between the digits is not allowed). Can Ann write digits in such a way that, irrespectively of Enn's move, the result would be a three-digit number (i.e., not starting with 0) that is divisible by 11?",
    "options": [],
    "answer": "a) No. b) Yes.",
    "solution": "a) Let Ann replace on her first move one of letters $B$ or $C$ with a digit $k$. If Enn now replaces the other one of $B$ and $C$ with digit $k$, too, the resulting number $\\overline{Akk}$ is divisible by 11 if and only if $\\overline{A00}$ is divisible by 11, which in turn is the case if $A$ is divisible by 11. The only such digit is 0 but $A$ cannot be replaced with 0. If Ann replaces on her first move the letter $A$ with a non-zero digit $k$ then Enn can replace $B$ with the digit $k-1$. The number resulting from Ann's second move differs from the number $kk0$ by less than 11, whence it cannot be divisible by 11.\n\nb) Let Ann write 9 on her first move. If Enn now writes before or after it a digit $k$ and Ann writes after or before of it, respectively, the digit $9-k$, then the resulting number is divisible by 11. Ann cannot make her last move in such a way only if Enn on his move has written either 9 to the end of the number or 0 to the beginning of the number. If Enn has written 9 to the end of the number, the blackboard contains digits 99 and Ann can construct a multiple of 11 by writing 0 to the end. If Enn has written 0 to the beginning of the number, Ann can write 2 to the very beginning which results in 209, again a multiple of 11.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 29828,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find the least positive integer $n$ for which there exists a positive integer $a$ such that both $a$ and $a+735$ have exactly $n$ positive divisors.",
    "options": [],
    "answer": "4",
    "solution": "If numbers $a$ and $a+735$ had exactly $2$ divisors, they would be primes. These numbers are of different parity, whence one of them is even. But if $a=2$ then $a+735 = 737 = 11 \\cdot 67$.\nIf numbers $a$ and $a+735$ had exactly $3$ divisors, each of them would be a square of a prime. Similarly to the previous case, we get $a = 4$. But then $a+735 = 739$, and $739$ is not a perfect square.\nOn the other hand, numbers $10$ and $10+735$ have exactly $4$ divisors.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 29829,
    "subject": "Mathematics (Multi-modal)",
    "question": "Juku conjectured the following in his mathematics circle: whenever the product of two coprime integers $x$ and $y$ is divisible by the product of some two coprime integers $a$ and $b$, at least one of $x$ and $y$ is divisible by $a$ or $b$. Does his proposition hold?",
    "options": [],
    "answer": "No; counterexample: x = 20, y = 21, a = 14, b = 15.",
    "solution": "Let $x = 20$, $y = 21$, $a = 14$, $b = 15$. Then $x$ and $y$ are coprime, as they are consecutive, similarly $a$ and $b$ are coprime. The product $xy = 420$ is divisible by $ab = 210$ but neither of $20$ and $21$ is divisible by $14$ or $15$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 29830,
    "subject": "Mathematics (Multi-modal)",
    "question": "Kati and Peeter play the following game. First, Kati writes a positive integer $a > 2016$ on the blackboard. Then Peeter starts to write more numbers on the blackboard, adding at each step the number $2016b + 1$ where $b$ is the biggest number on the blackboard. Peeter wins if at some point he writes a number divisible by $2017$. Otherwise Kati wins. Can Kati win, and if yes, what is the smallest number $a$ she can write to win?",
    "options": [],
    "answer": "Yes; 2019",
    "solution": "The number $2016b + 1$ gives the same remainder upon division by $2017$ as $-b + 1$. Hence the remainders upon division by $2017$ are as in the sequence $b, -b+1, -(b+1)+1, \\dots$. Since $-(b+1)+1 = b$, this sequence has period $2$, whence there are at most $2$ different remainders. Hence the number $a$ gives the win to Kati if and only if neither $a$ nor $-a+1$ is divisible by $2017$. Thus $a = 2017$ does not give a win, as it is divisible by $2017$, similarly for $a = 2018$, as $-2018+1$ is divisible by $2017$. But for $a = 2019$ none of $2019$ or $-2019+1$ is divisible by $2017$, giving a win.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 29831,
    "subject": "Mathematics (Multi-modal)",
    "question": "On the sides $BC$, $CA$ and $AB$ of triangle $ABC$, respectively, points $D$, $E$ and $F$ are chosen. Prove that\n$$\n\\frac{1}{2} (BC + CA + AB) < AD + BE + CF < \\frac{3}{2} (BC + CA + AB).\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let us first prove the second inequality. Consider the circle with centre $A$ that passes through the vertex $C$ and consider the extension of the side $CB$ past the vertex $B$ up to this circle (Fig. 5). As a chord lies inside the circle, we have $AD \\le AC$. We also have $BE \\le \\max(BC, BA) = BC$ and $CF \\le \\max(CA, CB) = CB$. Since by the triangle inequality $\\frac{1}{2}BC < \\frac{1}{2}CA + \\frac{1}{2}AB$, we obtain $AD + BE + CF \\le 2BC + CA < \\frac{3}{2}BC + \\frac{3}{2}CA + \\frac{1}{2}AB < \\frac{3}{2}(BC + CA + AB)$.\n\nLet us now prove the first inequality. If the triangle is not acute then $BE \\ge BA$ and $CF \\ge CA$ (Fig. 6). Since by the triangle inequality we have $\\frac{1}{2}BC < \\frac{1}{2}CA + \\frac{1}{2}AB$, we obtain $AD + BE + CF \\ge AD + BA + CA > BA + CA > \\frac{1}{2}(BC + CA + AB)$. For an acute triangle we have $AD \\ge AK$, $BE \\ge BL$ and $CF \\ge CM$, where $AK$, $BL$ and $CM$ are the altitudes of the triangle $ABC$. Let $H$ be the intersection point of the altitudes of the triangle $ABC$ (Fig. 7). We obtain $AD + BE + CF \\ge AK + BL + CM > AH + BH + CH = \\frac{1}{2}(BH + CH) + \\frac{1}{2}(CH + AH) + \\frac{1}{2}(AH + BH) > \\frac{1}{2}(BC + CA + AB)$, where the last inequality follows from the triangle inequality.\n\n![](attached_image_1.png)\nFig. 5\n![](attached_image_2.png)\nFig. 6\n![](attached_image_3.png)\nFig. 7\nBy the triangle inequality, $AB \\le BD + AD$ and $AB \\le AE + BE$, analogous inequalities hold for sides $BC$ and $CA$. As not all equality cases can hold simultaneously, adding these inequalities gives a strict inequality $2(BC + CA + AB) < BC + CA + AB + 2(AD + BE + CF)$. Collecting similar terms and dividing by 2 gives the first required inequality.\n\nSimilarly by the triangle inequality, $AD \\le AB + BD$ and $AD \\le CA + CD$; analogously for $BE$ and $CF$. Again, not all equality cases can hold simultaneously, whence adding these inequalities gives a strict inequality $2(AD + BE + CF) < 2(BC + CA + AB) + BC + CA + AB$. Collecting similar terms and dividing by 2 leads to the second required inequality.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 29832,
    "subject": "Mathematics (Multi-modal)",
    "question": "Masha has an electric carouse in her garden that she rides every day. As she likes order, she always leaves the carouse in the same position after each ride. But every night three bears sneak into the garden and start turning the carouse. Bear dad turns the carouse each time by $\\frac{1}{7}$ of the full circle. Bear mum turns the carouse each time by $\\frac{1}{9}$ of the full circle. Bear cub turns the carouse each time by $\\frac{1}{32}$ of the full circle. Every bear can turn the carouse as many times as he or she wants. In how many different positions may Masha find the carouse in the morning?",
    "options": [],
    "answer": "2016",
    "solution": "As $7 \\cdot 9 \\cdot 32 = 2016$, all turns are integral multiples of $\\frac{1}{2016}$ of the full turn. Thus the carouse can be in at most 2016 distinct positions. It remains to show that all these positions are possible. For that, we show that the bears can turn the carouse by exactly $\\frac{1}{2016}$ of the full turn. Then the same sequence of operations can be repeated to obtain also $\\frac{2}{2016}, \\frac{3}{2016}, \\dots, \\frac{2016}{2016}$ of the full turn. Exactly $\\frac{1}{2016}$ of the full turn is obtained, for instance, if bear dad turns the carouse once in one direction and both bear mum and bear cub turn the carouse once in the opposite direction since $\\frac{1}{7} - \\frac{1}{9} - \\frac{1}{32} = \\frac{288-224-63}{2016} = \\frac{1}{2016}$.\nThe carouse turns by an integral multiple of $\\frac{1}{7}$ of the full turn due to bear dad, an integral multiple of $\\frac{1}{9}$ of the full turn due to bear mum and an integral multiple of $\\frac{1}{32}$ of the full turn due to bear cub. As the result, the carouse turns by $\\frac{x}{7} + \\frac{y}{9} + \\frac{z}{32}$ of the full turn where $x, y, z$ are some integers. As $\\frac{x}{7} + \\frac{y}{9} + \\frac{z}{32} = \\frac{288x+224y+63z}{2016}$, the carouse can be turned only by integral multiples of $\\frac{1}{2016}$ of the full turn. As $\\text{gcd}(288, 224, 63) = \\text{gcd}(9 \\cdot 32, 7 \\cdot 32, 7 \\cdot 9) = 1$, there exist integral coefficients $a, b$ and $c$ such that $a \\cdot 288 + b \\cdot 224 + c \\cdot 63 = 1$. Hence taking $x = na, y = nb$ and $z = nc$ for any integer $n$, the carouse turns by exactly $\\frac{n}{2016}$ of the full turn. Hence all integral multiples of $\\frac{1}{2016}$ of the full turn are possible, i.e., the carouse can be left in 2016 distinct positions.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 29833,
    "subject": "Mathematics (Multi-modal)",
    "question": "Solve the system of equations $3x + 7y + 14z = 252$, $xyz - u^2 = 2016$ for non-negative real numbers.",
    "options": [],
    "answer": "(x, y, z, u) = (28, 12, 6, 0)",
    "solution": "$$\n\\begin{aligned}\n252 &= 3x + 7y + 14z \\ge 3\\sqrt[3]{3x \\cdot 7y \\cdot 14z} = 3\\sqrt[3]{3 \\cdot 7 \\cdot 14 \\cdot (2016 + u^2)} \\\\\n&\\ge 3\\sqrt[3]{3 \\cdot 7 \\cdot 14 \\cdot 2016} = 3\\sqrt[3]{2^6 \\cdot 3^3 \\cdot 7^3} = 2^2 \\cdot 3^2 \\cdot 7 = 252.\n\\end{aligned}\n$$\nTo not get a contradiction, we must have equality in both inequalities, hence $3x = 7y = 14z$ and $u = 0$. From the first equation of the system we finally obtain $3x = 7y = 14z = \\frac{252}{3} = 84$, hence $x = 28, y = 12$ and $z = 6$.\nThe second equation implies $xyz \\ge 2016$. Substituting the value of $x$ from the first equation here gives $(252 - 7y - 14z)yz \\ge 6048$, which is equivalent to $(36 - y - 2z)yz \\ge 864$. We find the maximum of function $f(y,z) = (36 - y - 2z)yz$. Fixing $z > 0$ arbitrarily, we obtain $\\frac{df(y,z)}{dy} = 36z - 2yz - 2z^2 = 2z(18 - y - z)$. The partial derivative with respect to $y$ is zero, if $z = 18 - y$. As the second derivative is negative, the function $f(y,z)$ has exactly one maximum at $y = 18 - z$ for any fixed positive number $z$. Define $g(z) = f(18 - z, z) = (18 - z)^2 z$. Its derivative $g'(z) = (18 - z)(18 - 3z)$ is zero at $z = 6$ ($z = 18$ does not count since then $y = 0$). As the second derivative is negative at $z = 6$, the extremum found is a maximum again. This means that out of all partial maxima of $f(y,z)$, the one for $z = 6$ is the largest. As $g(6) = f(12,6) = 864$, the initial system of equations can be satisfied only if $y = 12$ and $z = 6$. Substituting these values into the initial system leads to $3x = 84, 72x - u^2 = 2016$. The only solution of this system is $(x, u) = (28,0)$.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 29834,
    "subject": "Mathematics (Multi-modal)",
    "question": "The bisector of the exterior angle at vertex $C$ of the triangle $ABC$ intersects the bisector of the interior angle at vertex $B$ in point $K$. Consider the diameter of the circumcircle of the triangle $BCK$ whose one endpoint is $K$. Prove that $A$ lies on this diameter.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "![](attached_image_1.png)\nFig. 9\n![](attached_image_2.png)\nFig. 10\n\nLet $B'$ and $C'$ be respectively the second intersection points of the lines $AB$ and $AC$ with the circumcircle of the triangle $BCK$ (Fig. 9). Notice that $\\angle BKC = 180^\\circ - \\angle CBK - \\angle BCK = 180^\\circ - \\frac{\\angle ABC}{2} - (90^\\circ + \\frac{\\angle ACB}{2}) = 90^\\circ - \\frac{\\angle ABC}{2} - \\frac{\\angle ACB}{2}$. Hence the central angle supported by the arc $BC$ has the size $180^\\circ - \\angle ABC - \\angle ACB$. The central angle supported by the arc $B'C$ has the size $2\\angle ABC$ and the central angle supported by the arc $C'B$ has the size $2\\angle ACB$. Hence the central angle supported by the arc $B'C'$ has the size $360^\\circ - (180^\\circ - \\angle ABC - \\angle ACB) - 2\\angle ABC - 2\\angle ACB$, which equals $180^\\circ - \\angle ABC - \\angle ACB$. Hence the arcs $BC$ and $B'C'$ have the same size. As $K$ is the midpoint of the arc $B'C$, the point $B'$ is the reflection of the point $C$ by the diameter drawn from point $K$, and $C'$ is the reflection of point $B$. Hence the intersection point $A$ of $BB'$ and $CC'$ has to lie on the diameter drawn from the point $K$.\n\n*Remark.* Among the used central angles there may also be angles of size greater than $180^\\circ$ (reflex angles), such as the angle corresponding to the arc $B'C$ in Fig. 10.\n![](attached_image_3.png)\nFig. 11\n\nLet $L$ be the second endpoint of the diameter through $K$ of the circumcircle of the triangle $BCK$ (Fig. 11). Then $\\angle KCL = 90^\\circ$, thus $CL$ is the bisector of the interior angle at vertex $C$ of the triangle $ABC$, and $\\angle KBL = 90^\\circ$, so $BL$ is the bisector of the exterior angle at vertex $B$ of $ABC$.\nThe bisectors of the exterior angles at some two vertices of a triangle and the bisector of the interior angle at the third vertex of the triangle intersect at a common point. Thus the bisector of the exterior angle at vertex $A$ of the triangle $ABC$ passes through points $K$ and $L$. Thus point $A$ lies on $KL$.\n![](attached_image_4.png)\nFig. 12\n\nLet us denote the angles of the triangle $ABC$ at vertices $A$, $B$ and $C$ respectively $\\alpha$, $\\beta$ and $\\gamma$. Let the intersection point of the line $BC$ and the tangent to the circumcircle of triangle $BCK$ at point $K$ be $M$ (Fig. 12). By the property of inscribed angles, $\\angle MKC = \\angle MBK = \\frac{\\beta}{2}$. The bisectors of the exterior angles at some two vertices of a triangle and the bisector of the interior angle at the third vertex of the triangle have a common point. Hence $AK$ is the bisector of the exterior angle at vertex $A$ of the triangle $ABC$.\n---\nThus $\\angle CKA = 180^\\circ - \\angle KAC - \\angle KCA = 180^\\circ - \\frac{180^\\circ - \\alpha}{2} - \\frac{180^\\circ - \\gamma}{2} = \\frac{\\alpha}{2} + \\frac{\\gamma}{2}$.\nHence $\\angle MKA = \\angle MKC + \\angle CKA = \\frac{\\alpha}{2} + \\frac{\\beta}{2} + \\frac{\\gamma}{2} = 90^\\circ$. As a tangent of a circle is perpendicular to the diameter drawn from the point of tangency we have that $A$ lies on the diameter drawn from point $K$.",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 29835,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $n$ and $m$ be positive integers. What is the biggest number of points that can be marked in the vertices of the squares of the $n \\times m$ grid in such a way that no three of the marked points lie in the vertices of any right-angled triangle?",
    "options": [],
    "answer": "n + m",
    "solution": "All vertices of squares lie on $n+1$ horizontal and $m+1$ vertical lines. Suppose that at least $n+m+1$ points are marked in the grid. Because $m > 0$, the number of marked points is greater than $n+1$. Hence by the pigeonhole principle, at least one horizontal line contains more than one marked point. Hence at most $n$ marked points are alone on their horizontal lines. Similarly, at most $m$ marked points are alone on their vertical lines. Thus there exists a marked point that lies neither alone on its horizontal line nor alone on its vertical line. But then there is a right-angled triangle with vertices at marked points. So at most $n+m$ points can be marked according to the conditions of the problem.\n\nBy marking all vertices of squares of the grid that lie at the left and lower edge of the grid except for the lower left corner we have marked exactly $n+m$ points (Fig. 13 depicts the choice for a $5 \\times 7$ grid). Any three of the marked points either lie on a common line or are the vertices of an obtuse triangle, so the construction satisfies the conditions of the problem.\n\n![](attached_image_1.png)\n\nFig. 13",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 29836,
    "subject": "Mathematics (Multi-modal)",
    "question": "Solve the system $a^3 + b = 4c$, $a + b^3 = c$, $ab = -1$.",
    "options": [],
    "answer": "(1, -1, 0), (-1, 1, 0), (2, -1/2, 15/8), (-2, 1/2, -15/8)",
    "solution": "From the third equation we get $b = -\\frac{1}{a}$. By substituting this in the first and second equations we obtain a new system: $a^3 - \\frac{1}{a} = 4c$, $a - \\frac{1}{a^3} = c$.\n\nIf $c = 0$, we have $a^3 = \\frac{1}{a}$ and $a^4 = 1$, whence $a = 1$ or $a = -1$, since $a = 0$ is not possible. We have respectively $b = -1$ and $b = 1$.\n\nIf $c \\neq 0$, then by dividing in this system the sides of the first equation by the respective sides of the second equation, we obtain $(a^3 - \\frac{1}{a}) : (a - \\frac{1}{a^3}) = 4$. As $a^3 - \\frac{1}{a} = a^2 \\cdot (a - \\frac{1}{a})$, this is equivalent to $a^2 = 4$, whence $a = \\pm 2$. If $a = 2$, then $b = -\\frac{1}{2}$ and $c = \\frac{15}{8}$. The case $a = -2$ gives the same solution with opposite signs.\nBy adding the first and second equation we get $a(a^2 + 1) + b(b^2 + 1) = 5c$. Since $1 = -ab$ by the third equation, this equation is equivalent to $a(a^2 - ab) + b(b^2 - ab) = 5c$ or, equivalently, $(a^2 - b^2)(a - b) = 5c$.\n\nBy subtracting the second equation from the first equation in the initial system we obtain $a(a^2 - 1) - b(b^2 - 1) = 3c$ and by similarly substituting from the third equation we obtain $a(a^2 + ab) - b(b^2 + ab) = 3c$ or, equivalently, $(a^2 - b^2)(a + b) = 3c$.\n\nHence if $c = 0$, then $a - b = 0$ or $a + b = 0$. By the third equation $a = b$ is not possible. The case $a = -b$ gives two possibilities $a = 1, b = -1$ and $a = -1, b = 1$.\n\nIf $c \\neq 0$, we obtain $\\frac{a+b}{a-b} = \\frac{3}{5}$, whence $a = -4b$. By substituting into the third equation of the initial system we obtain $-4b^2 = -1$, whence $b = \\pm\\frac{1}{2}$. If $b = \\frac{1}{2}$, then $a = -2$ and $c = -\\frac{15}{8}$. The case $b = -\\frac{1}{2}$ gives the same solution with opposite signs.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 29837,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all possibilities: how many acute angles can there be in a convex polygon?",
    "options": [],
    "answer": "0, 1, 2, 3",
    "solution": "A square has 0 acute angles, a right-angled trapezium has 1 acute angle, an obtuse triangle has 2 acute angles, an acute triangle has 3.\n\n![](attached_image_1.png)\nFig. 1\n![](attached_image_2.png)\nFig. 2\n![](attached_image_3.png)\nFig. 3\nLet us show that 4 or more acute angles is not possible. By moving along the boundary of the convex polygon, we turn at vertices in only one direction (for example left), and on arrival at the initial vertex we have turned a total of $360^\\circ$. In an acute-angled vertex the direction changes by more than $90^\\circ$ (the change of direction equals the size of the respective exterior angle which is obtuse in this case). Thus if in the polygon there were 4 or more acute angles, the total turn in only those would be more than $360^\\circ$.\n\nSolution 2:\nIn a regular pentagon all angles have size $\\frac{3 \\cdot 180^\\circ}{5}$ or $108^\\circ$, thus there are 0 acute angles. By prolonging two sides in one direction, the angle of the pentagon at the moving vertex is reduced. As the angle between the extensions of two non-neighbouring sides is $180^\\circ - 2 \\cdot (180^\\circ - 108^\\circ)$, which equals $36^\\circ$, the receding angle can be given size $60^\\circ$ without losing the convexity of the polygon (Fig. 1). By prolonging the already prolonged side in the other direction, we can similarly create another acute angle of size $60^\\circ$ (Fig. 2). Finally let us move the opposite vertex of the prolonged side away from that side. This can be done without losing convexity until there is an acute angle also at that vertex (Fig. 3). Indeed, as in this process the unchanging angles have size $60^\\circ$, totalling $120^\\circ$, the convexity of the pentagon is lost only when the size of the reducing angle reaches $60^\\circ$, when the pentagon becomes a regular triangle.\nOn the other hand, when $x$ angles of a convex $n$-gon are acute, then the sum of their sizes is less than $x \\cdot 90^\\circ$ and the sum of the sizes of the other angles is less than $(n-x) \\cdot 180^\\circ$. But the sum of the sizes of the interior angles of any $n$-gon is $(n-2) \\cdot 180^\\circ$. Therefore we get $(n-2) \\cdot 180^\\circ < x \\cdot 90^\\circ + (n-x) \\cdot 180^\\circ$, whence $x \\cdot 90^\\circ < 2 \\cdot 180^\\circ$ and $x < 4$. Thus there can only be 0 to 3 acute angles in a convex $n$-gon.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 29838,
    "subject": "Mathematics (Multi-modal)",
    "question": "Does there exist a positive integer $n$ which has exactly 9 positive divisors and whose all divisors can be placed in a 3-by-3 table such that the products of the 3 numbers in each row, each column and on each diagonal are all the same?",
    "options": [],
    "answer": "Yes; for example, n = 36 works, and more generally any n = p^8 for a prime p.",
    "solution": "The number $36$ has $9$ positive divisors $1$, $2$, $3$, $4$, $6$, $9$, $12$, $18$, $36$. Let the first row be $18$, $1$, $12$, second row $4$, $6$, $9$, and third row $3$, $36$, $2$. Then the product of each row, column and diagonal is $216$.\nFor each prime $p$ the number $p^8$ has exactly $9$ divisors $p^0, p^1, \\dots, p^8$. It is known that the numbers $1$ to $9$ can be placed as a $3 \\times 3$ magic square, with an equal sum of the numbers in each row, column and diagonal. By subtracting $1$ from each number, we reduce the sum of each row, column and diagonal by $3$. By replacing in the magic square each number $i$ by the respective power $p^i$ we obtain a placement of the divisors of $p^8$ in which each row, column and diagonal has an equal product.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 29839,
    "subject": "Mathematics (Multi-modal)",
    "question": "Juku thought of a 3-digit number that, when reversing the order of the digits, stays the same 3-digit number. Juku noticed that when adding 2016 to that number, the 4-digit number that arises is again the same 4-digit number when reading the digits from right to left. What number did Juku think of?",
    "options": [],
    "answer": "646",
    "solution": "Let the number be $aba$ and let the number we get by adding $2016$ be $cddc$. Clearly $c$ can only be $2$ or $3$.\n\nIf $c = 2$ then by the ones digit the only possibility is $a = 6$, and we have a carry from the ones to the tens digit. By the tens digit then $b + 1 + 1 = d$ or $b + 1 + 1 = d + 10$. The second option is impossible, since by the hundreds digit we can only have $d = 6$, if there is no carry from the tens to the hundreds digit, and $d = 7$, if there is a carry from the tens to the hundreds digit. Thus $b + 2 = d$. Then there is no carry to the hundreds digit, hence $d = 6$ and $b = 4$.\n\nIf $c = 3$ then in adding the hundreds digits we must have a carry to the thousands digit which is possible only when $a = 9$. But by the ones digit we should have $c = 5$. The contradiction shows that this case is not possible.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 29840,
    "subject": "Mathematics (Multi-modal)",
    "question": "a) Let $a$ and $b$ be arbitrary positive integers of equal parity. Can we always find noninteger numbers $x$ and $y$ such that $x + y$ and $ax + by$ are integers?\n\nb) The same question when $a$ and $b$ have different parities.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "a) By taking $x = y = \\frac{1}{2}$ we have that $x + y = 1$ is an integer and so is $ax + by = \\frac{1}{2}(a + b)$, since $a + b$ is even by the assumption.\n\nb) We notice that $ax + by = a(x + y) + (b - a)y$. Assume that $x + y$ and $ax + by$ are integers. Since $a(x + y)$ is an integer, since it is a product of two integers, $(b - a)y$ must be an integer as well. But in the case $b - a = 1$ this is not possible, since $y$ is noninteger by the assumption.\na) If $a = b$ then any noninteger numbers $x$ and $y$ whose sum is an integer will be suitable as in this case $ax + ay$ is an integer, as it is a product of two integers $a$ and $x + y$. In the case $a \\neq b$ we can take $x = \\frac{1}{a-b}$ and $y = \\frac{a-b-1}{a-b}$, as in that case $x + y = 1$ and\n$$\nax + by = \\frac{a}{a-b} + \\frac{b(a-b-1)}{a-b} = \\frac{(b+1)(a-b)}{a-b} = b+1.\n$$\n\nb) Assume $x + y = n$ and $ax + by = m$, where $n$ and $m$ are integers. By interpreting this as a system of equations and solving for $x$ and $y$ we obtain $x = \\frac{m-bn}{a-b}$ and $y = \\frac{an-m}{a-b}$ (as $a$ and $b$ have different parities we have $a - b \\neq 0$). If $a - b = 1$ then these solutions are integers. Thus we can not guarantee the existence of noninteger numbers with desired properties.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 29841,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $A$ and $B$ be such points of the circle with centre $O$ that the triangle $AOB$ is right-angled. The perpendicular bisector of the segment $AO$ intersects the shorter arc $AB$ in point $K$. The lines $KO$ and $AB$ intersect in point $L$. Prove that the triangle $KBL$ is isosceles.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "![](attached_image_1.png)\nFig. 4\n\nSince $K$ lies on the perpendicular bisector of the segment $AO$ (Fig. 4) we have $KA = KO$. On the other hand $KO = AO$ since $K$ and $A$ are points on the circle. Hence $AKO$ is an equilateral triangle from which we obtain that $\\angle AOK = 60^\\circ$. Hence $\\angle KOB = \\angle AOB - \\angle AOK = 90^\\circ - 60^\\circ = 30^\\circ$. As also $B$ lies on the same circle we have $KO = BO$, hence $\\angle BKO = \\frac{180^\\circ - \\angle KOB}{2} = \\frac{180^\\circ - 30^\\circ}{2} = 75^\\circ$. Finally from the equality $AO = BO$ we obtain $\\angle ABO = \\frac{180^\\circ - \\angle AOB}{2} = \\frac{90^\\circ}{2} = 45^\\circ$. Hence $\\angle BLK = \\angle LOB + \\angle LBO = 30^\\circ + 45^\\circ = 75^\\circ$. The triangle $KBL$ is isosceles as it has two equal angles.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 29842,
    "subject": "Mathematics (Multi-modal)",
    "question": "There is a finite number of lamps in an electrical scheme. Some pairs of lamps are directly connected by a wire. Every lamp is lit either red or blue. With one switch all lamps that have a direct connection with a lamp of the other colour change their colour (from red to blue or vice versa). Prove that after some number of switches all lamps have the same colour as two switches before that.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "If some connected lamps are lit in different colours they both change colour upon switching, hence they are also lit differently after the switch. The same holds on each following switch. Hence no pair of connected lamps lit in different colours can disappear but more of such pairs can appear. As there are only finitely many lamps, the number of connected pairs of differently lit lamps can not grow infinitely. Hence this number stops changing after some number of switches. This means at that point a lamp either changes colour on each switch or never changes colour. Hence the colours of all lamps are the same after two consecutive switches.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 29843,
    "subject": "Mathematics (Multi-modal)",
    "question": "Define $a_1 = 1$, and for each $n > 1$ let $a_n = n \\cdot a_{\\lfloor \\frac{n}{2} \\rfloor}$. Prove that for each $n \\ge 12$ we have $a_n > n^2$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "As $a_n = n \\cdot a_{\\lfloor \\frac{n}{2} \\rfloor}$, it suffices to show that for each $n \\ge 12$ we have $a_{\\lfloor \\frac{n}{2} \\rfloor} \\ge n$. By the inequalities $n \\le 2\\lfloor \\frac{n}{2} \\rfloor + 1 < 3\\lfloor \\frac{n}{2} \\rfloor$ this reduces to proving that $a_m \\ge 3m$ for each $m \\ge 6$. By $a_m = m \\cdot a_{\\lfloor \\frac{m}{2} \\rfloor}$ the latter reduces to proving $a_l \\ge 3$ for $l \\ge 3$. This is true, since $a_l = l \\cdot a_{\\lfloor \\frac{l}{2} \\rfloor} \\ge l$.\n\n\n<table>\n<thead>\n<tr><th>n</th><th>1</th><th>2</th><th>3</th><th>4</th><th>5</th><th>6</th><th>7</th><th>8</th><th>9</th><th>10</th><th>11</th><th>12</th><th>13</th><th>14</th><th>15</th></tr>\n</thead>\n<tbody>\n<tr><th>a_n</th><td>1</td><td>2</td><td>3</td><td>8</td><td>10</td><td>18</td><td>21</td><td>64</td><td>72</td><td>100</td><td>110</td><td>216</td><td>234</td><td>294</td><td>315</td></tr>\n<tr><th>n^2</th><td>1</td><td>4</td><td>9</td><td>16</td><td>25</td><td>36</td><td>49</td><td>64</td><td>81</td><td>100</td><td>121</td><td>144</td><td>169</td><td>196</td><td>225</td></tr>\n</tbody>\n</table>\nHence for $12 \\le n < 16$ the claim is true.\nFor bigger numbers note that for each $k \\ge 4$ we have\n$$\na_{2k} = 2^k \\cdot 2^{k-1} \\cdots 2 \\cdot 1 = 2^{k+(k-1)+\\cdots+1} = 2^{\\frac{k(k+1)}{2}} \\ge 2^{2(k+1)}.\n$$\nBy simple induction we conclude that $a_n$ increases as $n$ increases. For arbitrary $n \\ge 16$ pick $k \\ge 4$ such that $2^k \\le n < 2^{k+1}$, giving\n$$\na_n \\ge a_{2k} \\ge 2^{2(k+1)} = (2^{k+1})^2 > n^2.\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 29844,
    "subject": "Mathematics (Multi-modal)",
    "question": "Do there exist two positive powers of $5$ such that the number obtained by writing one after the other is also a power of $5$?",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Suppose that $5^x \\cdot 10^n + 5^y = 5^z$, where $5^y$ has $n$ digits. Then $5^{x+n} \\cdot 2^n = 5^y \\cdot (5^{z-y} - 1)$, whence $2^n = 5^{z-y} - 1$.\n\nCase $n = 1$ does not work.\n\nFor case $n = 2$ we get $z - y = 1$. Since $5^y$ has $2$ digits, the only possibility is $y = 2$ and $z = 3$, whence $x = 0$, which is not positive.\n\nCase $n > 2$ yields $5^{z-y} \\equiv 1 \\pmod{8}$, thus $z - y = 2k$ for an integer $k$. Now $2^n = 25^k - 1 = 24 \\cdot (25^{k-1} + \\dots + 1)$, this is impossible, since $3 \\mid 24$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 29845,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABC$ be a triangle with $AB = \\frac{AC}{2} + BC$. Consider the two semicircles outside the triangle with diameters $AB$ and $BC$. Let $X$ be the orthogonal projection of $A$ onto the common tangent line of those semicircles. Find $\\angle CAX$.",
    "options": [],
    "answer": "60°",
    "solution": "Let $K$ and $L$ be the midpoints of the sides $AB$ and $BC$, respectively, and $M$ and $N$ the feet of perpendiculars from $K$ and $L$, respectively, to the common tangent of the semicircles (Fig. 28). Then $\\angle CAX = \\angle LKM$. As $KM$ and $LN$ are radii of the semicircles, $KM = \\frac{AB}{2}$ and $LN = \\frac{BC}{2}$. Let $Y$ be the intersection point of line $KL$ with the common tangent of the semicircles. As triangles $KYM$ and $LYN$ are similar, $\\frac{KY}{LY} = \\frac{KM}{LN}$.\n\n![](attached_image_1.png)\nFig. 28\n\nThus $\\frac{KL+LY}{LY} = \\frac{AB}{BC}$, whence $LY = \\frac{KL \\cdot BC}{AB-BC} = \\frac{AC}{2} \\cdot \\frac{BC}{AC} = BC$. Therefore $\\sin \\angle NYL = \\frac{LN}{LY} = \\frac{BC}{AB} = \\frac{1}{2}$, implying $\\angle NYL = 30^\\circ$. Consequently, $\\angle CAX = \\angle LKM = 90^\\circ - \\angle NYL = 60^\\circ$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 29846,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $n$ be a natural number, $n \\ge 3$. Find the maximal number of diagonals of a regular $n$-gon one can select in such a way that every two selected diagonals that intersect each other inside the polygon are perpendicular.",
    "options": [],
    "answer": "If n is odd: n − 3. If n is even: n − 2.",
    "solution": "If $n$ is odd, one can select all $n - 3$ diagonals connecting one fixed vertex to others. In order to prove that the conditions of the problem do not allow more, it suffices to show that no two diagonals are perpendicular. Fix one diagonal arbitrarily; it partitions the boundary of the polygon into two halves, out of which one contains an even number of vertices and the other contains an odd number of vertices. In the latter half, the side that connects two medium vertices is parallel to the chosen diagonal. Thus the existence of two perpendicular diagonals would imply the existence of two perpendicular sides. This is possible only if $n \\equiv 0 \\pmod{4}$, contradiction.\n\nIf $n \\equiv 2 \\pmod{4}$, one can select every second vertex on the boundary and select initially all diagonals that connect two consecutive selected vertices. Furthermore, select all $\\frac{n}{2} - 3$ diagonals connecting one fixed selected vertex to all other selected vertices that it is not connected to yet. Finally, select the diagonal connecting this very vertex to its opposite vertex of the original polygon. This way, one selects $n-2$ diagonals. If $n \\equiv 0 \\pmod{4}$ then one can initially select $\\frac{n}{2}$ diagonals as in the previous case, and then select $\\frac{n}{2} - 2$ more diagonals in the $\\frac{n}{2}$-gon formed by the selected diagonals according to the algorithm described in this paragraph.\n\nNow prove that selecting more diagonals is impossible. At first we show that all selected diagonals that intersect some other selected diagonals must lie in two perpendicular directions. Indeed, consider one pair of mutually intersecting diagonals. The number of vertices of the polygon lying between two endpoints of distinct diagonals under consideration is less than half of the number of all vertices of the polygon. If one more pair of mutually intersecting diagonals is added, the same holds for it. This means that the other pair cannot be fit entirely in none of the windows left there by the initial pair of diagonals. Thus at least one diagonal from the first pair and one from the second pair intersect, which means that they all must lie in two perpendicular directions.\n\nLet there be $d$ selected diagonals and $k$ intersection points of selected diagonals. Consider pieces of the plane into which the selected diagonals divide the interior of the polygon; all these pieces are polygons whose vertices coincide with vertices of the original polygon and the intersection points of selected diagonals. The sum of internal angles of all pieces is $(n-2) \\cdot 180^\\circ + k \\cdot 360^\\circ$. The sum of the numbers of vertices of these pieces is $n + 2d + 4k$, since the initial polygon has $n$ vertices, each diagonal adds 2 endpoints and each intersection of two diagonals adds 4. Let $w$ be the number of pieces; then $(n-2) \\cdot 180^\\circ + k \\cdot 360^\\circ = (n + 2d + 4k - 2w) \\cdot 180^\\circ$, whence $n - 2 + 2k = n + 2d + 4k - 2w$, implying $w = d + k + 1$.\n\nLet $w'$ be the number of pieces with at least 4 vertices. All pieces with at least two right angles have at least 4 vertices, whence every line segment connecting two neighbouring intersection points of any diagonal is a side of such piece. Let there be $a$ horizontal and $b$ vertical diagonals selected, w.l.o.g. $a \\le b$. Then there are $k-a$ pieces whose two right angles are consecutive intersection points on some horizontal selected diagonal and which themselves lie above this diagonal. In addition, there are $b-1$ pieces whose two right angles are consecutive intersection points of some horizontal selected diagonal and which lie below this diagonal and below which there are no more horizontal diagonals. Hence $w' \\ge k-a+b-1 \\ge k-1$.\n\nConsequently, $n+2d+4k \\ge 4w' + 3(w-w') = 3w+w' \\ge 3w+k-1 = 3(d+k+1)+k-1 = 3d+4k+2$, whence $d \\le n-2$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 29847,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find the smallest real constant $C$ such that for any positive real numbers $a_1, a_2, a_3, a_4$ and $a_5$ (not necessarily distinct), one can always choose distinct subscripts $i, j, k$ and $l$ such that $\\left|\\frac{a_i}{a_j} - \\frac{a_k}{a_l}\\right| \\le C$.",
    "options": [],
    "answer": "1/2",
    "solution": "See IMO 2016 shortlist, problem A2.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 29848,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABC$ be a triangle with $AB = AC \\ne BC$ and let $I$ be its incenter. The line $BI$ meets $AC$ at $D$, and the line through $D$ perpendicular to $AC$ meets $AI$ at $E$. Prove that the reflection of $I$ in $AC$ lies on the circumcircle of triangle $BDE$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "*Solution.* See IMO 2016 shortlist, problem G4.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 29849,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABC$ be an isosceles triangle with apex $A$ and altitude $AD$. On $AB$, choose a point $F$ distinct from $B$ such that $CF$ is tangent to the incircle of $ABD$. Suppose that $\\triangle BCF$ is isosceles. Show that those conditions uniquely determine:\n\na) which vertex of $BCF$ is its apex;\n\nb) the size of $\\angle BAC$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "a) Consider cases of the location of the vertex angle of the triangle $BCF$ (Fig. 27).\nIf $F$ were the apex, then $F$ would lie on the perpendicular bisector of the side $BC$, i.e., on the line $AD$, whence $F = A$. Therefore $AC$ would be a tangent of the incircle of the triangle $ABD$. But the lines $AB$ and $AD$ are tangents of the same circle. There can be at most two tangents drawn from one point to one circle. Hence this case is impossible.\nLet $B$ be the apex. As $CBF$ is a base angle of the isosceles triangle $ABC$, we have $\\angle CBF < 90^\\circ$. Hence $\\angle BCF > 45^\\circ$. Let $K$ be the tangent point of the line $CF$ and the incircle of the triangle $ABD$ and let $L$ be the projection of the point $K$ onto the line $BC$. By construction, $KL < 2r$, where $r$ is the radius of the incircle\n\n![](attached_image_1.png)\nFig. 27\n\nof the triangle $ABD$. On the other hand, $\\angle LCK = \\angle BCF > 45^\\circ$ implies $KL > CL > CD = BD > 2r$. Hence this case is impossible, too.\nThis shows that the apex of triangle $BCF$ can only be $C$.\n\nb) Let the apex be $C$. Fix a point $D$, mutually perpendicular lines $l_1$ and $l_2$ both passing through $C$, and circle $c$ that is tangent to both lines; let the radius of the circle be $r$. Choose point $A$ on $l_1$ in such a way that $DA > 2r$ and the tangent point of line $l_1$ and circle $c$ lies on the line segment $DA$. Point $B$ is determined by the location of $A$ as the point of intersection of line $l_2$ and the second tangent line of circle $c$ passing through $A$, point $C$ is symmetric to $B$ w.r.t. $DA$ and $F$ is defined as in the problem. When point $A$ moves away from $D$, points $B$ and $C$ get closer to $D$, whence $\\angle BAC$ decreases and $\\angle BCF$ increases. Thus these angles can equal only in one case.",
    "topic": "Discrete Mathematics",
    "subtopic": "Combinatorics"
  },
  {
    "id": 29850,
    "subject": "Mathematics (Multi-modal)",
    "question": "The leader of an IMO team chooses positive integers $n$ and $k$ with $n > k$, and announces them to the deputy leader and a contestant. The leader then secretly tells the deputy leader an $n$-digit binary string, and the deputy leader writes down all $n$-digit binary strings which differ from the leader's in exactly $k$ positions. (For example, if $n = 3$ and $k = 1$, and if the leader chooses 101, the deputy leader would write down 001, 111, and 100.) The contestant, who is allowed to look at the strings written by the deputy leader, tries to guess the leader's string. What is the minimum number of guesses (in terms of $n$ and $k$) needed to guarantee the correct answer?",
    "options": [],
    "answer": "2 if n = 2k; otherwise 1",
    "solution": "See IMO 2016 shortlist, problem C1.",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 29851,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $\\mathbb{R}^+$ be the set of positive real numbers. Determine all functions $f: \\mathbb{R}^+ \\to \\mathbb{R}^+$ satisfying the equation\n$$\nxf(x^2)f(f(y)) + f(yf(x)) = f(xy)(f(f(x^2)) + f(f(y^2))).\n$$\nfor all $x, y \\in \\mathbb{R}^+$.",
    "options": [],
    "answer": "f(x) = 1/x for all positive x",
    "solution": "See IMO 2016 shortlist, problem A4. (That problem was proposed by Estonia.)",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 29852,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $n$ be a positive integer. In how many ways can an $n \\times n$ table be filled with integers from $0$ to $5$ such that\n\na) the sum of each row is divisible by $2$ and the sum of each column is divisible by $3$;\n\nb) the sum of each row is divisible by $2$, the sum of each column is divisible by $3$ and the sum of each of the two diagonals is divisible by $6$?",
    "options": [],
    "answer": "a) 6^{n^2 - n}; b) if n = 1: 1; if n = 2: 6; if n ≥ 3: 6^{n^2 - n - 2}",
    "solution": "a) Let's fill the top left $(n-1) \\times (n-1)$ subtable arbitrarily; this can be done in $6^{(n-1)^2}$ ways. Now there are $3$ ways to fill each of the top $n-1$ cells of the rightmost column and $2$ ways to fill each of the left $n-1$ cells of the bottom row to satisfy the requirements. The value for the last empty cell in the bottom right is then uniquely determined (mod $2$ by the bottom row, and mod $3$ by the rightmost column). In conclusion, there are $6^{(n-1)^2} \\cdot 3^{n-1} \\cdot 2^{n-1} = 6^{n^2-n}$ ways to fill the table.\n\nb) For $n=1$, the only solution is writing $0$ into the single cell. For $n=2$, let $a$ be the top left number. The bottom right must then be $(6-a)$ mod $6$. Using the conditions for rows and columns, for the top right number $x$ we get the equations $x \\equiv -a \\pmod{2}$ and $x \\equiv a \\pmod{3}$, and for the bottom left number $y$, $y \\equiv a \\pmod{2}$ and $y \\equiv -a \\pmod{3}$. The Chinese remainder theorem determines $x$ and $y$ uniquely, and we see from the equations that their sum is also divisible by $6$. Thus there are $6$ ways to fill the table in this case, one for each value of $a$.\n\nConsider now $n \\ge 3$. Fill the top left $(n-1) \\times (n-1)$ subtable arbitrarily; this can be done in $6^{(n-1)^2}$ ways. The bottom right cell's value is uniquely determined by other values on the falling diagonal. Denote the value in the top left cell by $a$, the sum of the $2$nd to $(n-1)$-st cells (inclusive) in the top row by $b$, the sum of the $2$nd to $(n-1)$-st cells in the leftmost column by $c$, and the sum of $2$nd to $(n-1)$-st cells on the rising diagonal by $d$.\n\nUsing the Chinese remainder theorem, fill the top right cell with the unique value $x$ such that $x \\equiv -a-b \\pmod 2$ and $x \\equiv a+c-d \\pmod 3$, and the bottom left cell with the unique value $y$ such that $y \\equiv a+b-d \\pmod 2$ and $y \\equiv -a-c \\pmod 3$. The divisibility conditions are now fulfilled for the top row, the leftmost column and both diagonals (the rising diagonal is verified by summing mod $2$ and mod $3$ separately).\n\nNow, we leave one cell both in the rightmost column and in the bottom row empty for the time being. For the other $n-3$ empty cells in the rightmost column, there are $3$ possible values for each, and for the other $n-3$ empty cells in the bottom row, $2$ values for each. Having made all those choices (which can be done in $3^{n-3} \\cdot 2^{n-3}$ ways), the values for the two remaining cells are now uniquely determined (mod $2$ by the values in the respective row, and mod $3$ by the column). The total number of ways to fill the table is $6^{(n-1)^2} \\cdot 3^{n-3} \\cdot 2^{n-3} = 6^{n^2-n-2}$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 29853,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $B = (-1,0)$ and $C = (1,0)$ be fixed points on the coordinate plane. A nonempty, bounded subset $S$ of the plane is said to be *nice* if\n(i) there is a point $T \\in S$ such that for every point $Q \\in S$, the segment $TQ$ lies entirely in $S$; and\n(ii) for any triangle $P_1P_2P_3$, there exists a unique point $A \\in S$ and a permutation $\\sigma$ of the indices $\\{1,2,3\\}$ for which triangles $ABC$ and $P_{\\sigma(1)}P_{\\sigma(2)}P_{\\sigma(3)}$ are similar.\nProve that there exist two distinct nice subsets $S$ and $S'$ of the set $\\{(x,y): x \\ge 0, y \\ge 0\\}$ such that if $A \\in S$ and $A' \\in S'$ are the unique choices of points in (ii), then the product $BA \\cdot BA'$ is a constant independent of the triangle $P_1P_2P_3$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "See IMO 2016 shortlist, problem G3.",
    "topic": "Discrete Mathematics",
    "subtopic": "Combinatorics"
  },
  {
    "id": 29854,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all positive integers $n$ such that $n!$ is not divisible by $n^2$.",
    "options": [],
    "answer": "All prime numbers and 4",
    "solution": "For any prime $p$, $p^2 \\nmid p!$ as prime $p$ occurs only once in the prime factorization of $p!$. Additionally, $4^2 = 16$ does not divide $4! = 24$. We will show that $n^2 \\mid n!$ for all other positive integers $n$. Let $p$ be a prime factor of $n$, and $k$ the exponent of $p$ in the prime factorization of $n$. If there exists a different prime factor $q$ of $n$, then both $p^k$ and $p^kq$ are in the set of integers from 1 to $n$ and hence $p^{2k} \\mid n!$. This holds for all prime factors $p$ of $n$.\n\nThus $n^2$ divides $n!$ for all $n$ with at least two distinct prime factors. Now consider the remaining integers $n = p^k$ for prime $p$. For $k \\ge 3$ all three integers $p$, $p^{k-1}$, and $p^k$ are distinct factors in $n!$, implying that $p^{2k} = n^2$ divides $n!$. For $k = 2$ and $p > 2$ the integers $p$, $2p$, and $p^2$ are distinct factors in $n!$, implying that $p^4 = n^2$ divides $n!$.\nWe find all positive integers $n$ such that $n \\mid (n-1)!$; obviously this condition is equivalent to that of the problem. For prime $n$, no integer less than $n$ can have a prime factor $n$ and thus $n$ cannot divide $(n-1)!$. For $n = 4 = 2^2$, 4 does not divide $(n-1)! = 6$. On the other hand, for $n = p^2$ where $p > 2$ is prime, the integers $p$ and $2p$ are in the set of integers from 1 to $n-1$, implying $p^2 = n \\mid (n-1)!$. If $n$ is a cube or a higher power of some prime, or has at least two prime factors, then it obviously has a divisor $d$ such that $1 < d < n$ and $d^2 \\neq n$ (e.g. the smallest prime factor of $n$). Hence $d$ and $\\frac{n}{d}$ are distinct positive integers less than $n$ and their product $n$ divides $(n-1)!$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 29855,
    "subject": "Mathematics (Multi-modal)",
    "question": "An acute angle with vertex $A$ and size $\\alpha$ is given on a plane. Points $B_0$ and $B_1$ are chosen on different sides of the angle in such a way that $\\angle AB_0B_1 = \\beta$. Whenever points $B_0, B_1, \\dots, B_{n-1}$ are defined, the next point $B_n$ on side $AB_{n-2}$ is allowed to be defined in such a way that $B_n \\neq B_{n-2}$ and $B_{n-1}B_n = B_{n-2}B_{n-1}$. Prove that this process cannot last infinitely and determine the largest index $n$ (depending on $\\alpha$ and $\\beta$) for which $B_n$ can be defined.",
    "options": [],
    "answer": "Largest n equals (β − 90°)/α + 1 if (β − 90°)/α is a nonnegative integer; otherwise largest n equals ⌊β/α⌋ + 1.",
    "solution": "Let $B_n$ for some $n > 1$ be definable (Fig. 30). By construction, $A$, $B_{n-2}$, and $B_n$ are collinear, whereby $A$ cannot lie between $B_{n-2}$ and $B_n$. Since $B_{n-1}B_{n-2} = B_{n-1}B_n$ and $B_{n-2} \\neq B_n$, the triangle $B_{n-2}B_{n-1}B_n$ is isosceles, so that $\\angle AB_{n-2}B_{n-1} + \\angle AB_nB_{n-1} = 180^\\circ$ and $\\angle AB_{n-2}B_{n-1} \\neq 90^\\circ$. Consequently,\n![](attached_image_1.png)\n\nFig. 30\n\n$\\angle AB_{n-1}B_n = 180^\\circ - \\alpha - \\angle AB_nB_{n-1} = 180^\\circ - \\alpha - (180^\\circ - \\angle AB_{n-2}B_{n-1}) = \\angle AB_{n-2}B_{n-1} - \\alpha$. Since $\\angle AB_0B_1 = \\beta$, by induction we get $\\angle AB_{n-1}B_n = \\beta - (n-1)\\alpha$. Hence defining of $B_n$ assumes that $\\beta - (n-2)\\alpha \\neq 90^\\circ$ and $\\beta - (n-1)\\alpha \\ge 0^\\circ$. These conditions are also sufficient, because if $\\beta - (n-2)\\alpha \\neq 90^\\circ$, meaning that $\\angle AB_{n-2}B_{n-1} \\neq 90^\\circ$, then point $B_n$ can be chosen different from $B_{n-2}$ on the line $AB_{n-2}$, and if $\\beta - (n-1)\\alpha \\ge 0^\\circ$, meaning that $\\angle AB_{n-2}B_{n-1} \\ge \\alpha$, then this point is located on the side $AB_{n-2}$.\n\nSumming up, the largest $n$ for which $B_n$ is definable equals $\\frac{\\beta-90^\\circ}{\\alpha} + 1$ if $\\beta - (n-2)\\alpha = 90^\\circ$ for some integer $n > 1$, or equivalently, $\\frac{\\beta-90^\\circ}{\\alpha}$ is a non-negative integer, and $\\lfloor \\frac{\\beta}{\\alpha} \\rfloor + 1$ otherwise. Figures 31 and 32 depict the situation in the first and second case, respectively.\n\n![](attached_image_2.png)\nFig. 31\n\n![](attached_image_3.png)\nFig. 32",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 29856,
    "subject": "Mathematics (Multi-modal)",
    "question": "A rectangular grid whose side lengths are integers greater than 1 is given. Smaller rectangles with area equal to an odd integer and length of each side equal to an integer greater than 1 are cut out one by one. Finally one single unit square is left. Find the least possible area of the initial grid before the cuttings.",
    "options": [],
    "answer": "121",
    "solution": "Denote by $X$ the unit square left. Then $X$ cannot lie in a corner of the initial rectangle, as the strip between the neighbouring rectangle of $X$ and the edge of the rectangle could not be cut out (painted gray in Fig. 33). Similarly, $X$ cannot lie at a side of the initial rectangle, because the strip between two neighbouring rectangles of $X$ could not be cut out (Fig. 34).\n\nConsider all possibilities of how the unit squares around $X$ can be distributed between rectangles that are cut out from the initial figure; for simplicity, we identify the 8 unit squares by principal winds. Suppose that the eastern neighbour of $X$ belongs to rectangle $A$. Then either the northeastern\n\n![](attached_image_1.png)\nFig. 34\n\nor southeastern neighbour of $X$ belongs to $A$, too; w.l.o.g., let the northeastern neighbour of $X$ belong to $A$ (Fig. 35). The northern neighbour of $X$ must belong to a rectangle $B$ distinct from $A$. Then the northwestern neighbour of $X$ must belong to $B$. Similarly, the western and southwestern neighbours of $X$ must belong to a third rectangle $C$ and the southern and southeastern neighbour of $X$ must belong to a fourth rectangle $D$ (Fig. 36).\n\n![](attached_image_2.png)\nFig. 35\nFig. 36\n\nThe unit squares of the strip starting from the eastern neighbour of $X$ and continuing eastward until the edge of the figure distribute between rectangles, as well as the unit squares of the strip starting from the southern neighbour of $X$ and continuing eastward until the edge of the figure distribute between rectangles. As the area of each rectangle is odd, their side lengths are odd. Thus the lengths of both strips must be representable as sums of 1 or more odd integers greater than 1. These lengths themselves are two consecutive positive integers. The least two consecutive positive integers representable as sums of odd integers greater than 1 are 5 (which is odd itself) and 6 (which is $3+3$). Hence at least 5 unit squares of the initial figure must be located to the east of $X$. Similarly, at least 5 unit squares of the initial figure must be located in each cardinal direction from $X$. Thus each side length of the initial rectangle is at least 11 and the area is at least 121. As Fig. 37 shows, this limit can be achieved.\n\n![](attached_image_3.png)\nFig. 37",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 29857,
    "subject": "Mathematics (Multi-modal)",
    "question": "A beetle is creeping on the coordinate plane, starting from point $(0; -1)$, along a straight line until reaching the $x$-axis at point $(-x; 0)$ where $x$ is a positive real number. After that it turns $90^\\circ$ to the right and creeps again along a straight line until reaching the $y$-axis. Then it again turns right by $90^\\circ$ and creeps along a straight line until reaching the $x$-axis, where it once more turns right by $90^\\circ$ and creeps along a straight line until reaching the $y$-axis.\n\na) Can it happen that both the length of the beetle's journey and the distance between its initial and final point are rational numbers?\n\nb) Can it happen that both the length of the beetle's journey and the distance between its initial and final point are integers?",
    "options": [],
    "answer": "a) Yes, for example taking x equal to four thirds. b) No, it is impossible.",
    "solution": "Let $O$ be the origin of coordinates. Let $A_0$ be the starting point of the beetle's journey, $A_1$ the first turning point, $A_2$ the second turning point, $A_3$ the third turning point and $A_4$ the endpoint (see figure below).\n\n![](attached_image_1.png)\n\nThe right triangles $OA_0A_1$ and $OA_1A_2$ are similar with ratio $x$ because $\\angle OA_1A_2 = 90^\\circ - \\angle OA_1A_0 = \\angle OA_0A_1$ and $\\frac{OA_1}{OA_0} = x$. So are the triangles $OA_1A_2$ and $OA_2A_3$ similar with ratio $x$ since $\\frac{OA_2}{OA_1} = x$ by similarity of the triangles $OA_0A_1$ and $OA_1A_2$. Analogously, the triangles $OA_2A_3$ and $OA_3A_4$ are similar with ratio $x$. Consequently, $OA_2 = x^2$, $OA_3 = x^3$ and $OA_4 = x^4$.\n\nBy the Pythagorean theorem, the length of the beetle's journey is $\\sqrt{1+x^2} + \\sqrt{x^2+x^4} + \\sqrt{x^4+x^6} + \\sqrt{x^6+x^8}$, or equivalently, $(1+x+x^2+x^3)\\sqrt{1+x^2}$ which equals $(x^4-1) \\cdot \\frac{\\sqrt{x^2+1}}{x-1}$ if $x \\neq 1$. On the other hand, the distance between $A_0$ and $A_4$ is $|x^4-1|$.\n\na) Taking $x = \\frac{4}{3}$, the number $x^4 - 1$ is rational, since $x$ is rational. As $\\sqrt{1+x^2} = \\sqrt{1+\\frac{16}{9}} = \\frac{5}{3}$, the number $(x^4-1) \\cdot \\frac{\\sqrt{x^2+1}}{x-1}$ is also rational.\n\nb) Assume that the distance between $A_0$ and $A_4$ is an integer; then $x^4$ is an integer. Suppose that $(x^4-1) \\cdot \\frac{\\sqrt{x^2+1}}{x-1}$, the length of the beetle's journey, is also an integer (one can assume $x \\neq 1$ since otherwise the length of the journey is $4\\sqrt{2}$ that is not an integer). As $x^4 - 1$ is an integer, the number $\\frac{\\sqrt{x^2+1}}{x-1}$ must be rational. Consider three cases.\n\n1) If $x$ is an integer then $x-1$ is integer whence $\\sqrt{x^2+1}$ must be rational and $x^2+1$ must be a perfect square. This is impossible as two consecutive positive integers cannot be perfect squares.\n\n2) Suppose that $x$ is irrational and $x^2$ is an integer. As $\\frac{x^2+1}{(x-1)^2}$ is the square of a rational number, so is also $\\frac{(x-1)^2}{x^2+1}$. Hence $\\frac{2x}{x^2+1}$ must be rational. But this is impossible, since $2x$ is irrational and $x^2+1$ is an integer.\n\n3) Let $x^2$ be irrational. Similarly to the previous case we see that $\\frac{2x}{x^2+1}$ is rational. Hence $\\frac{4x^2}{(x^2+1)^2}$ is the square of a rational number, implying that $\\frac{(x^2+1)^2}{4x^2}$ is rational. The latter in turn implies that $\\frac{x^4+1}{4x^2}$ must be rational. This is impossible as $4x^2$ is irrational and $x^4+1$ is a positive integer.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 29858,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all triples $(p, q, r)$ of primes such that $2018(p^2+q^2) = r^2+1$.",
    "options": [],
    "answer": "No such triples exist",
    "solution": "Suppose that both $p$ and $q$ are odd. Then $p^2+q^2$ is even and the l.h.s. of the equation is divisible by 4. Squares of integers are congruent to 0 or 1 modulo 4 whence the r.h.s. is congruent to 1 or 2 modulo 4. The contradiction shows that one of $p$ and $q$ equals 2; let w.l.o.g. $p = 2$. Squares of integers are congruent to 0 or 1 modulo 3. Obviously $r > 3$ as the l.h.s. is greater than 10. As $r$ is prime, $r$ is not divisible by 3. Thus $r^2 \\equiv 1 \\pmod 3$, whence $r^2 + 1 \\equiv 2 \\pmod 3$. Now $2018 \\equiv 2 \\pmod 3$ implies $p^2 + q^2 \\equiv 1 \\pmod 3$ and $p^2 = 4 \\equiv 1 \\pmod 3$ in turn implies $q^2 \\equiv 0 \\pmod 3$. Hence $q$ is divisible by 3, i.e., $q = 3$. Therefore the l.h.s. is $2018 \\cdot 13$. As $2018 \\cdot 13 \\equiv 3 \\cdot 3 = 9 \\equiv 4 \\pmod 5$, we must have $r^2 \\equiv 3 \\pmod 5$. But 3 is not a quadratic residue modulo 5.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 29859,
    "subject": "Mathematics (Multi-modal)",
    "question": "Prove that every positive real number satisfies\n$$\n(x+1)(x+2)(x+5) \\ge 36x.\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "The given inequality is equivalent to $x^3 + 8x^2 - 19x + 10 \\ge 0$. Note that $x^3 + 8x^2 - 19x + 10 = (x-1)^2(x+10)$. As $(x-1)^2 \\ge 0$ and $x+10 > 0$ for positive $x$, this inequality holds indeed.\nLet $f(x) = x^3 + 8x^2 - 19x + 10$. The given inequality is equivalent to $f(x) \\ge 0$. Note that $f'(x) = 3x^2 + 16x - 19$. As $D = 16^2 - 4 \\cdot 3 \\cdot (-19) > 0$, the function $f'$ has two real roots. Let the smaller and the larger root be $x_1$ and $x_2$, respectively; then $x_2 = 1$ and $x_1 < 0$. As the coefficient of the quadratic term of $f'(x)$ is positive, $f'$ is increasing at 1 whence the values of $f'$ are negative in the interval $(x_1; 1)$ and positive in the interval $(1; \\infty)$. Hence $f$ is decreasing in the interval $(x_1; 1)$ and increasing in the interval $(1; \\infty)$ (Fig. 39). Thus $f(x) \\ge f(1) = 0$ holds for all $x$ in the interval $(x_1; \\infty)$ that includes every positive $x$.\n\n![](attached_image_1.png)\nFig. 39\nFor every positive integer $k$, the AM-GM inequality implies $\\frac{x+1 \\cdot k}{k+1} \\ge \\sqrt[k+1]{x \\cdot 1^k}$. Using this inequality for $k=1, k=2$, and $k=5$, we obtain the inequalities $x+1 \\ge 2\\sqrt{x}$, $x+2 \\ge 3\\sqrt[3]{x}$, $x+5 \\ge 6\\sqrt[6]{x}$, respectively. The desired result is now obtained by multiplying the corresponding sides.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 29860,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $BE$ be an altitude of an acute triangle $ABC$ and let $P$ be the point on side $AB$ such that $AP = AE$. Let $N$ be the point for which $BCEN$ is a parallelogram. The areas of the triangles $AEP$ and $BNP$ are equal. Lines $NE$ and $AB$ intersect at $Q$.\n\na) Prove that the median of triangle $ABC$ drawn from the vertex $C$ intersects the line segment $PQ$.\n\nb) Prove that in the triangle $ABC$, the bisector of the angle $A$, the altitude drawn from the vertex $B$ and the median drawn from the vertex $C$ meet in one point.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "![](attached_image_1.png)\nFig. 40\n\nDenote $AC = b$, $AB = c$, $\\angle BAC = \\alpha$, and $AE = AP = u$. As $BCEN$ is a parallelogram, $BN = CE = b-u$ and $BN \\parallel CE$. The latter implies $\\angle NBP = \\alpha$ (Fig. 40). As $\\angle QAE = \\angle QBN$ and $\\angle AQE = \\angle BQN$, triangles $AQE$ and $BQN$ are similar.\n\na) The triangles $AEP$ and $BNP$ have areas $\\frac{1}{2}u^2 \\sin \\alpha$ and $\\frac{1}{2}(b-u)(c-u) \\sin \\alpha$, respectively. Thus $u^2 = (b-u)(c-u)$, whence $\\frac{c-u}{u} = \\frac{u}{b-u}$. Similarity of triangles $AQE$ and $BQN$ implies $\\frac{AQ}{BQ} = \\frac{AE}{BN}$ which, after defining $BQ = x$, rewrites to $\\frac{c-x}{x} = \\frac{u}{b-u}$. This equality reduces to a linear equation of $x$, meaning that it has only one root. By equality $\\frac{c-u}{u} = \\frac{u}{b-u}$, $x = u$ must be the only root. Thus $BQ = u = AP$, whence the midpoint of the side $AB$ coincides with the midpoint of the line segment $PQ$.\n\n![](attached_image_2.png)\nFig. 41\n\nb) Let $F$ be the midpoint of the side $AB$ and let $D$ be the point of intersection of the bisector of angle $A$ with side $BC$ (Fig. 41). By angle bisector theorem, $\\frac{BD}{CD} = \\frac{AB}{AC}$. Since $NE$ and $BC$ are parallel, triangles $ABC$ and $AQE$ are similar, whence also triangles $ABC$ and $BQN$ are similar. This and part a) of the problem together imply $\\frac{CE}{EA} = \\frac{BN}{BQ} = \\frac{AC}{AB}$. Consequently, $\\frac{AF}{FB} \\cdot \\frac{BD}{DC} \\cdot \\frac{CE}{EA} = 1 \\cdot \\frac{AB}{AC} \\cdot \\frac{AC}{AB} = 1$, giving the desired result by Ceva's theorem.",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 29861,
    "subject": "Mathematics (Multi-modal)",
    "question": "Is there a positive integer $n$ for which it is possible to write a number $-1$, $0$ or $1$ into each cell of an $n \\times n$ table in such a way that every integer from $-n$ to $n$ occurs at least once among the row sums, column sums and the two sums of the numbers on one long diagonal? If yes then find the least such $n$.",
    "options": [],
    "answer": "6",
    "solution": "Suppose that an $n \\times n$ table is filled with numbers $-1$, $0$, and $1$ in such a way that the conditions are met. The sums $n$ and $-n$ can be obtained only from a row, column or diagonal with all $1$s and all $-1$s, respectively, whence $n$ and $-n$ cannot arise as sums of different kind (one as a row sum and the other as a column sum or similar) as such sums have a common summand. If both $n$ and $-n$ arise as diagonal sums, $n$ must be even (otherwise the middle summand would be common) and each row and each diagonal must contain one $1$ and one $-1$. But then sums $n-1$\n\nand $-(n-1)$ would be impossible to achieve. Hence $n$ and $-n$ must be either both row sums or both column sums. W.l.o.g., assume that they are both row sums.\n\nThe number $n-1$ can arise only as the sum of $n-1$ numbers $1$ and one number $0$. As $-1$ occurs in each column and each long diagonal, $n-1$ can be obtained as a row sum only. Similarly, $-(n-1)$ can be obtained as a row sum only. The number $n-2$ can arise as the sum of either $n-2$ numbers $1$ and two numbers $0$ or $n-1$ numbers $1$ and one number $-1$. As either two $-1$s or numbers $0$ and $-1$ occur in each column and each long diagonal, $n-2$ can be obtained as a row sum only. Similarly, $-(n-2)$ can be obtained as a row sum only.\n\nTherefore, the table must contain at least $6$ rows.\n\nAn example of a $6 \\times 6$ table that fulfils the conditions is shown in Fig. 42: the row sums from the top to the bottom are $6$, $5$, $-5$, $-4$, $4$, and $-6$, the column sums from the left to the right are $0$, $-2$, $1$, $2$, $-1$, $0$, and the diagonal sums are $3$ and $-3$.\n\n![](attached_image_1.png)\nFig. 42",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 29862,
    "subject": "Mathematics (Multi-modal)",
    "question": "A quadratic equation $x^2 + px + q = 0$ is written on the blackboard, whereby $p$ and $q$ are real numbers such that real solutions exist to the equation on the blackboard and all the solutions are positive. Two players change in turns the coefficients in the equation according to the following rules. The first player decreases the constant term by either solution of the equation and (on the same move) increases the coefficient at the linear term by 1. The second player may replace the constant term with an arbitrary real number. Alternatively, the second player may increase the constant term by the largest solution of the equation and (on the same move) decrease the coefficient at the linear term by 1, but such move is allowed only if the solutions of the equation on the blackboard before the move differ from each other by more than 1. If either player's move results in an equation that does not have real solutions or has a non-positive real solution then the first player wins. Can the first player win regardless of how the opponent plays?",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Suppose that the first player always decreases the constant term by the smaller solution. We show that this is a winning strategy. Assume the opposite, i.e., that the play lasts infinitely. As the combined effect of the first and the second player's moves, the coefficient at the linear term either increases by 1 or (if the second player uses the alternative move) remains the same. If the second player had used the main move infinitely many times, the coefficient at the linear term would have become positive sooner or later. By Viéte's theorem, the sum of the solutions would be negative. This means that at least one solution is negative, too,\n\nwhence the first player must have won already. Consequently, the second player must have used only the alternative move starting from some position on. If the equation on the blackboard before the first player's move is $x^2 + px + q = 0$ with solutions $x_1, x_2$, where $x_1 \\le x_2$, then Viéte's theorem implies $p = -(x_1 + x_2)$ and $q = x_1x_2$, whence after the first player's move the coefficient at the linear term is $-(x_1 + x_2) + 1$ and the constant term is $x_1x_2 - x_1$. This means that the quadratic equation after the first player's move is $x^2 - (x_1 + (x_2 - 1)) + x_1(x_2 - 1)$. By Viéte's theorem, this equation has solutions $x_1$ and $x_2 - 1$. Note that if before the first player's move the solutions of the equations differ by more than 1, then the difference decreases by 1 as the result of the move, but if the solutions before the move differ by at most 1 then the difference of the solutions after the move is still at most 1. Analogously, the same holds for the second player's alternative move. Hence the play must reach a position where the second player cannot make the alternative move, contradiction.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 29863,
    "subject": "Mathematics (Multi-modal)",
    "question": "Points $P$ and $Q$ are chosen on the side $BC$ of triangle $ABC$ in such a way that $P$ lies between $B$ and $Q$, and rays $AP$ and $AQ$ trisect the angle $BAC$. The line parallel to $AQ$ and passing through $P$ meets the side $AB$ of the triangle at point $D$, and the line parallel to $AP$ and passing through $Q$ meets the side $AC$ of the triangle at point $E$. Can it happen that $DE$ is a midsegment of the triangle $ABC$?",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Assume that $DE$ is a midsegment of $ABC$, then $D$ is the midpoint of $AB$ (Fig. 24). As $DP \\parallel AQ$, $DP$ is a midsegment of triangle $ABQ$. Hence, $P$ is a midpoint of $BQ$ and $AP$ is a median of triangle $ABQ$. As rays $AP$ and $AQ$ trisect the angle $BAC$, $AP$ is a bisector of angle $QAB$. Thus, $ABQ$ is an isosceles triangle with altitude $AP$, implying that $AP$ is perpendicular to $BC$. Similarly we can\n\n![](attached_image_1.png)\nFig. 24\n\nsee that $AQ$ is perpendicular to $BC$. This leads to a contradiction as $AP$ and $AQ$ cannot coincide. Therefore, $DE$ cannot be a midsegment of $ABC$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 29864,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $n$ and $m$ be positive integers, $n \\ge m$. There is a game board of size $1 \\times n$ divided into $n$ unit squares and an unlimited supply of sticky tapes of size $1 \\times m$. On each move, any player adds a tape covering $m$ consecutive unit squares on the board precisely, at least one of which is not yet covered by any tape. Two players move alternately and the one who can make no move loses.\n\na) Prove that if $n$ and $m$ are of equal parity then the first player can win regardless of how the opponent plays.\n\nb) Is it true that, whenever $n$ and $m$ are of different parity, the second player can win regardless of how the first player plays?",
    "options": [],
    "answer": "b) No",
    "solution": "![](attached_image_1.png)\nFig. 25\n\na) If $n$ and $m$ are of equal parity, then the first player to move can cover the central squares in such a way that an equal number of uncovered squares remain on both sides (Fig. 25 with $n = 11$ and $m = 3$). The first player can now respond to each of the opposing player's moves by a symmetric move that ensures that uncovered squares remain symmetric w.r.t. the centre of the board after their move. Indeed, the starting move ensures that any subsequent move can cover uncovered squares only to one side of the centre and the first player is free to make symmetric moves in response to the second player. The number of uncovered squares decreases on each move; thus, the second player will lose.\n\n![](attached_image_2.png)\nFig. 26\n\nb) If $n = 5$ and $m = 2$, then the first player can cover two squares at the edge of the board on their first move (Fig. 26). Any move by the second player will now leave either 1 or 2 consecutive squares uncovered, and the first player can cover those on their next move, ensuring victory.\n\n*Answer:* b) No.",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 29865,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABCDE$ be a regular pentagon and let $c$ be the circle with diameter $AB$. Diagonals $AC$ and $AD$ intersect the circle $c$ at points $F$ and $G$, respectively. Line $FG$ intersects the side $AE$ at point $H$. Let $K$ be the midpoint of the side $DE$. Prove that points $F, H, E$, and $K$ are concyclic.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "As $AB$ is a diameter of $c$, $\\angle AFB = 90^\\circ$ (Fig. 27) and $BF$ is an altitude of triangle $ABC$. From $AB = BC$, $BF$ is a median and $F$ is the midpoint of $AC$. By symmetry, point $K$ lies on $BF$ and $\\angle FKE = 90^\\circ$. Notice that $\\angle BAC = \\angle CAD = \\angle DAE$, as $BAC, CAD$, and $DAE$ are inscribed angles that subtend to equal arcs of the circumcircle of the regular pentagon $ABCDE$. Points $A, B, F$ and $G$ lie on circle $c$ in this order, thus $\\angle ABF = 180^\\circ - \\angle AGF = \\angle AGH$. Hence, $ABF$ and $AGH$ are similar from two angles, implying $\\angle AHG = \\angle AFB = 90^\\circ$. Therefore, the opposite angles at $K$ and $H$ of quadrilateral $FHEK$ are right angles and it is cyclic.\n![](attached_image_1.png)\nFig. 27",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 29866,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all integer pairs $(a, b)$ for which $(2a^2 + b)^3 = b^3 a$.",
    "options": [],
    "answer": "(0, 0), (8, 128), (27, 729), (-1, -1)",
    "solution": "If $b = 0$, then according to the equation $2a^2 + b = 0$ from which $a = 0$.\n\nAssume now that $b \\neq 0$. As $b^3$ and $(2a^2 + b)^3$ are perfect cubes, their ratio $a$ is the cube of a rational number; as it is an integer, it is the cube of an integer $c$. By taking cubic root from each side of the equation we get a\n\nrelation $2c^6 + b = bc$, implying $b(c - 1) = 2c^6$. As $c$ and $c - 1$ are coprime, $c^6$ and $c - 1$ are also coprime. Therefore, $c - 1$ has to divide $2$ and $c$ must be one of $3$, $2$, $0$ or $-1$.\n\nIf $c = 3$, then $a = 27$ and we get $2b = 2 \\cdot 729$, whence $b = 729$.\n\nIf $c = 2$, then $a = 8$ and we get $b = 128$.\n\nIf $c = 0$, then we get $-b = 0$, which has already been analysed.\n\nIf $c = -1$, then $a = -1$ and $-2b = 2$ from which $b = -1$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 29867,
    "subject": "Mathematics (Multi-modal)",
    "question": "A circle with diameter $AB$ intersects side $BC$ of rhombus $ABCD$ at point $K$. A circle with diameter $AD$ intersects side $CD$ of rhombus $ABCD$ at point $L$. Find the angles of rhombus $ABCD$ if $\\angle AKL = \\angle ABC$.",
    "options": [],
    "answer": "60 degrees and 120 degrees",
    "solution": "Let $\\angle ABC = \\angle ADC = \\alpha$; then $\\angle AKL = \\alpha$ (Fig. 28).\n\nAccording to Thales' theorem $AK$ is perpendicular to $BC$ and $AL$ is perpendicular to $CD$. But $\\angle ABK = \\alpha = \\angle ADL$ and $AB = AD$, so $ABK$ and $ADL$ are equal. Therefore, $AK = AL$ from which $\\angle ALK = \\alpha$.\n\nAs the sum of the internal angles of a quadrilateral is $360^\\circ$, we have\n$$\n\\angle KAL = 360^\\circ - \\angle KCL - \\angle AKC - \\angle ALC = 360^\\circ - (180^\\circ - \\alpha) - 2 \\cdot 90^\\circ = \\alpha.\n$$\n\n![](attached_image_1.png)\nFig. 28\n\nTherefore, the triangle $AKL$ is equilateral as all its angles are equal to $\\alpha$, implying that the angles of the rhombus are $60^\\circ$ and $120^\\circ$.\nAs in Solution 1, denote $\\alpha = \\angle ABC = \\angle ADC = \\angle AKL$ and show that triangles $ABK$ and $ADL$ are equal. Therefore, $BK = DL$ from which $CK = CL$. We have $\\angle KCL = 180^\\circ - \\alpha$, therefore $\\angle CKL = \\angle CLK = \\frac{\\alpha}{2}$.\n\nThus $90^\\circ = \\angle AKC = \\angle AKL + \\angle LKC = \\frac{3}{2}\\alpha$ from which $\\alpha = 60^\\circ$. Therefore, the angles of the rhombus are $60^\\circ$ and $120^\\circ$.\n\n![](attached_image_2.png)\nFig. 29\nAs in Solution 1, denote $\\alpha = \\angle ABC = \\angle ADC = \\angle AKL$ and show that triangles $ABK$ and $ADL$ are equal. Hence, $BK = DL$ from which $\\frac{BK}{BC} = \\frac{DL}{CD}$. Thus $KL \\parallel BD$ and $AC \\perp KL$. Let $AC$ and $KL$ meet at $F$ (Fig. 29).\n\nThe altitude $KF$ from the right angle of triangle $CKA$ divides the triangle to two triangles $CFK$ and $KFA$, both similar to $CKA$. Therefore, $\\angle ACK = \\angle AKF = \\alpha$. The diagonal of the rhombus is also the angle bisector, therefore, $\\angle BCD = 2\\alpha$. Equation $180^\\circ - \\alpha = 2\\alpha$ gives us $\\alpha = 60^\\circ$ and the angles of the rhombus are $60^\\circ$ and $120^\\circ$.\n\n![](attached_image_2.png)\nFig. 29",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 29868,
    "subject": "Mathematics (Multi-modal)",
    "question": "Every sound in a certain language can be either long or short. A sound is classified either as a vowel or as a consonant. Every word consists of exactly two sounds (without repetitions) and satisfies the following conditions.\n\n1) Every word contains a short sound.\n2) Words beginning with a vowel contain a long sound.\n3) Words beginning with a consonant or ending with a vowel have a short sound at the end.\n\nAll sequences of two distinct sounds satisfying these conditions are words. The written language optimisation committee has decided to denote each sound with a different letter. However, they are considering two possibilities for denoting length. The first proposes denoting vowels with single letters and consonants with single or double letters based on length. The second instead proposes denoting consonants with single letters and vowels with single or double letters based on length. Is it possible to determine the lengths of sounds in all words from writing: a) in the case of the first proposal; b) in the case of the second proposal?",
    "options": [],
    "answer": "a) Yes; b) No",
    "solution": "a) If the word consists of two vowels, then based on the rule 3) the second of them is short and based on the rule 2) the first of them is long. If the word begins with a vowel and ends with a consonant, then the length of the second sound is determined uniquely by its writing and the length of first sound must be of the opposite length based on 1) and 2). If the word begins with a consonant, the length of first sound is determined uniquely by its writing and the second sound is short based on 3). Therefore, all writings of words in the language have unique pronunciation.\n\nb) The rules allow both: a word consisting of two short consonants, and a word starting with a long consonant and ending with a short consonant. According to the proposal, the first sound of these words have identical writing and cannot be uniquely determined.",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 29869,
    "subject": "Mathematics (Multi-modal)",
    "question": "There are $n$ candies on the table. On every turn, a player eats a number of candies that is greater than $1$ and divides the number of candies on the table at the start of the turn, but must leave at least $1$ candy on the table. Two players take alternate turns and the player who is unable to make a move loses. Find all positive integers $n$ for which the first player can always win.",
    "options": [],
    "answer": "All even n that are not of the form 2^(2k+1).",
    "solution": "Define all even numbers which are not odd powers of $2$ as *good* and the rest of the positive integers as *bad*. We show that the player before whose turn the number of candies is good has a move which yields in a bad number of candies, whereas the player before whose turn the number of candies is bad either has lost or is forced to leave a good number of candies on the table. As the number of candies is reduced in each move, the starting player wins if, initially the number of candies on the table is good.\n\nA player before whose turn the number of candies is even but not a power of $2$ can eat the number of candies equal to its odd factor different from $1$. After that, the number of candies on the table is odd, i.e. bad. If before the turn the number of candies is equal to an even power of $2$, one can eat exactly half of the candies, leaving an odd power of $2$ candies on the table, i.e. a bad number.\n\nOn the other hand, if the number of candies before the turn is odd, the player can only choose odd factors. This yields in an even number of candies left on the table. Furthermore, the number of candies left on the table is divisible by the number of candies taken from the table which is odd, hence, the number of candies cannot be a power of $2$. Therefore, the number of candies left on the table is good. However, if the number of candies before the turn is equal to an odd power of $2$, the player can only choose even factors, which results in an even number of candies left on the table. Furthermore, the rules do not allow eating more than half of the candies. Therefore, the number of candies left of the table can only be a power of $2$ if its exponent is less by $1$ than before the move. This would be an even power of $2$ which is also a good number.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 29870,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all prime numbers $p$ such that $2p^3 + 4p^2 - 3p + 12$ is the fifth power of an integer.",
    "options": [],
    "answer": "11",
    "solution": "Denote $f(n) = 2n^3 + 4n^2 - 3n + 12$. The following table shows the remainders of $n^2, n^3, n^5$ and $f(n)$ upon division by 11:\n\n| $n$         | 0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 |\n|-------------|---|---|---|---|---|---|---|---|---|---|----|\n| $n^2$       | 0 | 1 | 4 | 9 | 5 | 3 | 3 | 5 | 9 | 4 | 1  |\n| $n^3$       | 0 | 1 | 8 | 5 | 9 | 4 | 7 | 2 | 6 | 3 | 10 |\n| $n^5$       | 0 | 1 | 10| 1 | 1 | 1 |10 |10 |10 | 1 |10  |\n| $f(n)$      | 1 | 4 | 5 | 5 | 5 | 6 | 9 | 4 | 3 | 7 | 6  |\n\nAs one can see from the table, the only remainders upon division by 11 that the fifth power of an arbitrary integer $n$ can give are 0, 1 and 10. On the other hand, integers of the form $f(n)$ give only remainders 1, 3, 4, 5, 6, 7, and 9 upon division by 11, whereby the remainder is 1 only if $n$ is divisible by 11. Consequently, $f(p)$ can be the fifth power of an integer only if $p$ is divisible by 11. As $p$ is prime, the only possibility is $p = 11$. And indeed, $f(11) = 2 \\cdot 11^3 + 4 \\cdot 11^2 - 3 \\cdot 11 + 12 = 3125 = 5^5$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 29871,
    "subject": "Mathematics (Multi-modal)",
    "question": "A slider on the scrolling bar of Paul's mail client shows the proportion of emails which are preceding the email which is currently open. Paul noticed that before deleting some emails the slider was on 10%. Paul deleted some consecutive emails, starting from the one which was currently open. After that the slider was on 50%. What is the percentage of the emails that Paul deleted?",
    "options": [],
    "answer": "80%",
    "solution": "Let $n$ be the number of emails before deletion. As the slider showed $10\\%$ before deletion there were $0.1n$ emails preceding the email that was open when Paul started the deletion. After the end of the deletion those emails accounted for $50\\%$ from the remaining emails which means that $0.2n$ emails remained in the inbox. Thus Paul deleted $0.8n$ emails which makes up $80\\%$ of the emails present in the inbox before deletion.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 29872,
    "subject": "Mathematics (Multi-modal)",
    "question": "Parents have $n$ children, where $n$ is a given natural number. Find all possibilities for how many children in this family can have both a brother and a sister.",
    "options": [],
    "answer": "For up to two children: 0. For three children: 0 or 2. For four or more children: 0, n−1, or n.",
    "solution": "If all the children have the same gender, then nobody can have a brother as well as a sister. In that case the number of children that match the condition is 0, regardless of $n$.\n\nIf there are children of either gender, but for at least one gender there is exactly one child of that gender, then this child does not have a brother (if he's a boy) or a sister (if she's a girl). If there is exactly one child of the other gender as well ($n = 2$), then the number of children that fulfill the condition is 0. If $n \\ge 3$, then all other children have both a brother and a sister and there are $n-1$ children that satisfy the condition given in the problem statement. If there are at least 2 children of either gender, then all children have both a brother as well as a sister, and there are $n$ children satisfying the condition. This can happen when $n \\ge 4$.\n\nSumming up, for $n \\le 2$ the answer is 0, for $n = 3$ there can be 0 or $n-1$ (in other words 2) such children and for $n \\ge 4$ there are either 0, $n-1$ or $n$ such children.\nFor some child to have both a brother and a sister, there must be at least 3 children in this family. Therefore if $n \\le 2$, then the answer is 0. Now assume that $n \\ge 3$. It is clear that children of the same gender either all satisfy the condition given in problem statement or none of them do. In order for children of some gender not to satisfy the condition there must either be 1 child of such gender or 0 of the opposite gender. In the first case there are $n-1$ children satisfying the condition (since $n \\ge 3$ and therefore there must be at least 2 children of the opposite gender), in the second case there are 0 such children. Therefore for $n \\ge 3$ the possible answers are 0, $n-1$, and $n$. It suffices to note that children of both gender can only satisfy the condition if $n \\ge 4$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 29873,
    "subject": "Mathematics (Multi-modal)",
    "question": "The mediant of two rational numbers $u$ and $v$ is $x = \\frac{a+c}{b+d}$, where $\\frac{a}{b}$ and $\\frac{c}{d}$ are the reduced fractions of $u$ and $v$ respectively. Prove that for any two distinct positive rational numbers $u$ and $x$, there exist infinitely many positive rational numbers $v$, such that $x$ is the mediant of $u$ and $v$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $u = \\frac{a}{b}$ and $x = \\frac{c}{d}$ be the reduced fractions of $u$ and $x$. We are looking for rational numbers $v$ such that $v = \\frac{mc-a}{md-b}$, where $m$ is large enough integer such that $mc-a$ and $md-b$ are both positive. According to the definition, $x$ is the mediant of $u$ and $v$ as soon as $\\frac{mc-a}{md-b}$ is irreducible. Let's show that there are infinitely many natural numbers $m$ such that $\\frac{mc-a}{md-b}$ is irreducible. This will complete the solution.\n\nLet us first prove a lemma: prime numbers which can be used to reduce the fractions are also the factors of $ad - bc$. Indeed, if $p \\mid mc - a$ and $p \\mid md - b$, then $p \\mid a md - b - b (mc - a) = m(ad - bc)$, therefore $p \\mid m$ or $p \\mid ad - bc$. If $p \\mid m$, then $p \\mid a$ and $p \\mid b$ which contradicts the irreducibility of the fraction $\\frac{a}{b}$. Therefore $p \\mid ad - bc$.\n\nAs $u$ and $x$ are different, $ad - bc \\ne 0$. Therefore the number $ad - bc$ has a finite number of prime factors. Let $p_1, \\dots, p_l$ be all the different prime factors which can reduce the fraction $\\frac{mc-a}{md-b}$ for at least one $m$ and for each $i = 1, \\dots, l$ let $m_i$ be natural number such that the fraction $\\frac{m_i c - a}{m_i d - b}$ is reducible with prime $p_i$.\n\nLet $n$ be an arbitrary factor for which the fraction $\\frac{nc-a}{nd-b}$ is not irreducible. This fraction must be reducible with some prime number $p_i$ which can also reduce the fraction $\\frac{m_i c - a}{m_i d - b}$. Then $p_i \\mid (n - m_i)c$ and $p_i \\mid (n - m_i)d$. Therefore $p_i \\mid n - m_i$ as otherwise $p \\mid c$ and $p \\mid d$ which contradicts the irreducibility of the fraction $\\frac{c}{d}$. Therefore $n \\equiv m_i \\pmod{p_i}$.\n\nTherefore by choosing $n$ such that $n \\equiv m_i + 1 \\pmod{p_i}$ for each $i = 1, \\dots, l$ the fraction $\\frac{nc-a}{nd-b}$ must be irreducible. According to Chinese remainder theorem there are infinitely many natural numbers $n$ which satisfy such congruence system.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 29874,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all functions $f : \\mathbb{R} \\to \\mathbb{R}$ which for any real numbers $x$ and $y$ satisfy $$(f(x+y))^2 = x f(x) + 2 f(xy) + (f(y))^2.$$",
    "options": [],
    "answer": "f(x) = 0 for all real x; f(x) = x for all real x",
    "solution": "*Answer:* $f(x) = 0$ and $f(x) = x$.\n\nSubstituting $x = y = 0$ to the equation, we get $(f(0))^2 = 0 + 2f(0) + (f(0))^2$. Simplifying this gives $f(0) = 0$.\n\nSubstituting $y = 0$ to the original equation, we get the equation $(f(x))^2 = x f(x) + 2 f(0) + (f(0))^2$ which must be satisfied for all real $x$. Since $f(0) = 0$, this simplifies to $f(x)(f(x) - x) = 0$. Thus, for each $x$, either $f(x) = 0$ or $f(x) = x$.\n\nAssume there exists a real number $c \\neq 0$ such that $f(c) = 0$. Substituting $x = c$ to the original equation, we get $(f(c+y))^2 = 2f(cy) + (f(y))^2$, or\n$$\n2f(cy) = (f(c + y))^2 - (f(y))^2. \\quad (1)\n$$\nThis is valid for any $y$. Let's analyse four cases based on whether $f(c+y) = 0$ or $f(c+y) = c+y$ and whether $f(y) = 0$ or $f(y) = y$.\n\n1) If $f(c+y) = 0$ and $f(y) = 0$, then (1) simplifies to $f(cy) = 0$.\n\n2) If $f(c+y) = 0$ and $f(y) = y$, then (1) gives $f(cy) = -\\frac{y^2}{2}$. Assuming that $f(cy) = cy \\neq 0$, we get $c = -\\frac{y}{2}$ i.e. $y = -2c$.\n\n3) If $f(c+y) = c+y$ and $f(y) = 0$, then (1) gives us $2f(cy) = (c+y)^2$. Assuming that $f(cy) = cy$, we get $c^2 + y^2 = 0$ from which $c=0$, contradiction.\n\n4) If $f(c+y) = c+y$ and $f(y) = y$, then (1) gives us $2f(cy) = c^2 + 2cy$. Assuming that $f(cy) = cy$, we get $c^2 = 0$, contradiction.\n\nThus $f(cy) \\neq 0$ can only be valid if $y = -2c$, i.e. $f(x) \\neq 0$ can only be valid if $x = c \\cdot (-2c) = -2c^2$. Thus it is possible to choose a real number $d$ such that $d \\neq 0$, $f(d) = 0$, and $|d| \\neq |c|$. By replacing $d$ by $c$ we can analogously conclude that $f(x) \\neq 0$ can only be valid when $x = -2d^2$. As $-2d^2 \\neq -2c^2$, $f(x) \\neq 0$ cannot be valid for any $x \\neq 0$. So $f(x) = 0$ for all $x$.\n\nTherefore the only suitable functions are $f(x) = 0$ and $f(x) = x$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 29875,
    "subject": "Mathematics (Multi-modal)",
    "question": "The midpoints of the sides $BC$, $CA$, and $AB$ of triangle $ABC$ are $D$, $E$, and $F$, respectively. The reflections of centroid $M$ of $ABC$ around points $D$, $E$, and $F$ are $X$, $Y$, and $Z$, respectively. Segments $XZ$ and $YZ$ intersect the side $AB$ in points $K$ and $L$, respectively. Prove that $AL = BK$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "As $\\frac{MY}{MB} = \\frac{2ME}{2ME} = 1$ and analogously $\\frac{MZ}{MC} = 1$ we have $BC \\parallel YZ$. Let $G$ be the intersection of lines $YZ$ and $AD$ (Fig. 14).\n\nThen $\\frac{MG}{MD} = \\frac{MY}{MB} = 1$ from which $MG = MD = \\frac{1}{3}AD$ and $AG = AD - MD - MG = \\frac{1}{3}AD$. Hence $\\frac{AL}{AB} = \\frac{AG}{AD} = \\frac{1}{3}$.\n\nBy swapping the roles of $A$ and $B$, the roles of $D$ and $E$, roles of $X$ and $Y$, and finally the roles of $K$ and $L$, we get analogously $\\frac{BK}{AB} = \\frac{1}{3}$. Therefore $AL = BK$.\n\n![](attached_image_1.png)\nFig. 14\nNotice that triangle $XYZ$ is a homothetic transformation of triangle $ABC$ with centre $M$ and ratio $-1$. Homothety preserves the directions of the lines, therefore $KL \\parallel XY$ and $F$ lies on the median drawn from vertex $Z$ of triangle $XYZ$. Thus $FK = FL$ and $AL = AF - FL = BF - FK = BK$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 29876,
    "subject": "Mathematics (Multi-modal)",
    "question": "On a plane a finite number of points are marked of which no three are collinear. Assume that there exists a non-convex polygon with all of its vertices located at some of those points. Prove that there exists a non-convex quadrilateral such that all its vertices lie at the marked points.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "There has to be a point inside the convex hull of the set of marked points, otherwise any subset of points would form a convex polygon. Let us partition the convex hull into triangles by drawing a necessary amount of diagonals. As no three points are on the same line, the marked point inside the convex hull must be inside one of the triangles. The vertices of that triangle along with the marked point inside the triangle form a non-convex quadrilateral.\n\n\nWe show that it is possible to find 4 points among the vertices of the non-convex $n$-gon such that they form a non-convex quadrilateral. Let $A$, $B$, and $C$ be consecutive vertices of the $n$-gon such that the internal angle $\\angle ABC > 180^\\circ$. Let $A'$ and $C'$ be points on the extensions of $AB$ and $BC$ over $B$. Then the closed broken line corresponding to the polygon has to go through the interior of $\\angle A'BC'$. If this area contains a marked point, say $X$, then $ABCX$ is non-convex (Fig. 15).\n\n![](attached_image_1.png)\nFig. 15\n\nAlternatively, if there are no vertices of the $n$-gon within the interior of $\\angle A'BC'$, then it has to be intersected by a side of the polygon, say $DE$, where $D$ is on the side of point $A$ and $E$ is on the side of point $C$ (Fig. 16). Let us show that then $DBCE$ is a non-convex quadrilateral. Indeed, the opposite sides $DB$ and $CE$ are non-intersecting as they are located in different regions of the plane bordered by the lines $AA'$ and $CC'$. The opposite sides $BC$ and $ED$ are also non-intersecting as they are sides of the original $n$-gon. In addition, $\\angle DBC > 180^\\circ$.\n\n![](attached_image_2.png)\nFig. 16\n\nLet us prove by induction on $n$ that among the $n$ vertices of the $n$-gon it is possible to choose 4 vertices which form a non-convex quadrilateral. Trivially, the base case $n = 4$ is true. Assume now that $n > 4$ and that the statement is true for all non-convex polygons with less sides. A non-convex polygon has an interior angle greater than $180^\\circ$ and also an interior angle smaller than $180^\\circ$. Let's verify that it is not possible that every pair of angles one of which is greater than $180^\\circ$ and the other one smaller than $180^\\circ$ are neighbours to each other. Indeed, as each of the vertices has 2 neighbours, there could be at most 2 angles less than $180^\\circ$ and at most 2 angles greater than $180^\\circ$. But $n > 4$, contradiction. Therefore, it is possible to find three consecutive vertices $A$, $B$, and $C$ and a vertex $E$ different from the former three, such that internal angle $\\angle B < 180^\\circ$ and internal angle $\\angle A' > 180^\\circ$.\n\n![](attached_image_3.png)\nFig. 17\n\nIf there is a vertex $X$ in the interior of triangle $ABC$, then $ABCX$ is non-convex. If there are no vertices in the interior of triangle $ABC$, then by removing point $B$ we get a $n-1$-gon which still has interior angle $\\angle E > 180^\\circ$. Hence it has 4 vertices forming a non-convex quadrilateral by the induction hypothesis.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 29877,
    "subject": "Mathematics (Multi-modal)",
    "question": "Do there exist five distinct prime numbers for which the sum of any three of them is a prime number as well?",
    "options": [],
    "answer": "No",
    "solution": "*Answer:* No.\n\nAssume that there exist five such prime numbers. If there are three among them pairwise incongruent modulo $3$, then by adding each of the three separately to the sum of the other two we get three distinct sums modulo $3$. One of those three is $0$ modulo $3$ and therefore is not a prime number. However, if among the five we only get two distinct remainders when dividing by $3$, then by the pigeonhole principle there must be three among them that are congruent modulo $3$. The sum of those three, however, is divisible by $3$ and therefore not a prime number.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 29878,
    "subject": "Mathematics (Multi-modal)",
    "question": "Prove that for all positive real numbers $x, y, z$\n$$\n\\frac{y^2 z}{x} + y^2 + z \\geqslant \\frac{9y^2 z}{x + y^2 + z}.\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "By bringing all the terms to the same side and to the common denominator, we get an equivalent inequality\n$$\n\\frac{y^2z(x + y^2 + z) + xy^2(x + y^2 + z) + xz(x + y^2 + z) - 9xy^2z}{x(x + y^2 + z)} \\geq 0.\n$$\nSince $x$, $y$, and $z$ are positive, the denominator $x(x + y^2 + z)$ is positive as well. Therefore the fraction on the left-hand side is nonnegative if and only if its numerator is nonnegative. By removing the parentheses, we get an equivalent inequality $x^2y^2 + xy^4 + y^2z^2 + y^4z + x^2z + xz^2 \\geqslant 6xy^2z$. However, this inequality follows directly from AM-GM for terms $x^2y^2$, $xy^4$, $y^2z^2$, $y^4z$, $x^2z$, and $xz^2$.\nWhen multiplying both sides of the inequality by $\\frac{x+z+y^2}{y^2z}$ we get an equivalent inequality $(\\frac{y^2z}{x} + y^2 + z)(\\frac{x+z+y^2}{y^2z}) \\geqslant 9$. This inequality can in turn be transformed into $(\\frac{y^2z}{x} + y^2 + z)(\\frac{x}{y^2z} + \\frac{1}{y^2} + \\frac{1}{z}) \\geqslant 9$. The last one, however, follows from Cauchy-Schwarz, when it is applied to $(\\frac{y\\sqrt{z}}{\\sqrt{x}}, y, \\sqrt{z})$ and $(\\frac{\\sqrt{x}}{y\\sqrt{z}}, \\frac{1}{y}, \\frac{1}{\\sqrt{z}})$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 29879,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $A'$ be the result of reflection of vertex $A$ of triangle $ABC$ through line $BC$ and let $B'$ be the result of reflection of vertex $B$ through line $AC$. Given that $\\angle BA'C = \\angle BB'C$, can the largest angle of triangle $ABC$ be located:\na) At vertex $A$;\nb) At vertex $B$;\nc) At vertex $C$?",
    "options": [],
    "answer": "a) No; b) Yes; c) Yes",
    "solution": "![](attached_image_1.png)\nFig. 18\n![](attached_image_2.png)\nFig. 19\n![](attached_image_3.png)\nFig. 20\n\nSolution:\nLet the foot of the altitude drawn from vertex $B$ of triangle $ABC$ be $K$. By symmetry, $\\angle BA'C = \\angle BAC$ and $\\angle BB'C = \\angle KB'C = \\angle KBC$.\n\na) By assumptions, $\\angle BAC = \\angle KBC$ (Fig. 18). As the angle at vertex $K$ of triangle $KBC$ is right, $\\angle KBC < 90^\\circ$. If the largest angle of triangle $ABC$ were at vertex $A$, the triangle would have to be acute. This would mean that $K$ would lie on line segment $AC$, whence $\\angle ABC > \\angle KBC = \\angle BAC$. Consequently, the largest angle of triangle $ABC$ cannot be at vertex $A$.\n\nb) If the angle at vertex $B$ of triangle $ABC$ is right then it is the largest angle of the triangle (Fig. 19). Triangles $BAC$ and $KBC$ are similar, whence $\\angle BAC = \\angle KBC$ and $\\angle BA'C = \\angle BB'C$.\n\nc) Let $ABP$ be an equilateral triangle with midpoint $C$ (Fig. 20). Obviously $P$ is the reflection of $A$ through line $BC$, as well as the reflection of $B$ through line $AC$. Hence defining $A' = B' = P$ fulfills the conditions of the problem. The angle at vertex $C$ of triangle $ABC$ has size $120^\\circ$, whence it must be the largest.\na) By symmetry, $\\angle BA'C = \\angle BAC$. Hence the assumptions imply $\\angle BAC = \\angle BB'C$. Suppose that the largest angle of the triangle $ABC$ is at the vertex $A$. Then the angle at the vertex $C$ must be acute, implying that points $A$ and $B'$ lie at the same side of the line $BC$. Now $\\angle BAC = \\angle BB'C$ implies that $B'$ lies on the circumcircle of the triangle $ABC$ (Fig. 21). Hence $AC$ is the perpendicular bisector of the chord $BB'$ of the circumcircle of the triangle $ABC$, meaning that it passes through the circumcentre of the triangle $ABC$. Consequently, $AC$ is the diameter of the circumcircle of the triangle $ABC$. Hence the triangle $ABC$ has right angle at the vertex $B$ which contradicts the assumption that the largest angle is at the vertex $A$.\n\n![](attached_image_4.png)\nFig. 21\n![](attached_image_5.png)\nFig. 22\n![](attached_image_6.png)\nFig. 23\n\nb) If the angle at vertex $B$ of triangle $ABC$ is right then it is the largest angle of the triangle (Fig. 22). By symmetry, $\\angle AB'C = \\angle ABC = 90^\\circ$. Hence points $A, B, C$, and $B'$ are concyclic, whereby $A$ and $B'$ lie on the same side from line $BC$. Consequently $\\angle BA'C = \\angle BAC = \\angle BB'C$.\n\nc) Choose points $B$ and $C$ on a circle $c$ with centre $P$ in such a way that $\\angle BPC < 90^\\circ$. Let $B'$ be the reflection of $B$ through line $CP$. Let $A'$ be a point on circle $c$ such that $\\angle A'CB = 180^\\circ - \\angle PCB$ (Fig. 23). This choice is possible as moving with point $A'$ along the longer arc $BC$ of circle $c$ gives rise to all values of $\\angle A'CB$ below $180^\\circ - \\frac{1}{2}\\angle BPC$, but $180^\\circ - \\angle PCB = 90^\\circ + \\frac{1}{2}\\angle BPC < 180^\\circ - \\frac{1}{2}\\angle BPC$. Let now $A$ be the reflection of $A'$ through line $BC$. As $\\angle ACB = \\angle A'CB = 180^\\circ - \\angle PCB$, point $A$ lies on line $CP$, i.e., lines $CP$ and $AC$ coincide. The conditions of the problem are met, since both $BA'C$ and $BB'C$ are angles inscribed in circle $c$ subtended by equal chords.",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 29880,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $n \\ge 2$ be a natural number. Laura's mathematics teacher likes group exercises. In Laura's class the teacher composed new groups in every lesson of statistics. When all of the statistics curriculum had been covered, it appeared that every two different students had belonged together into exactly one group and every two different groups had contained exactly one student in common. On the day when correlation was studied there were precisely $n$ students besides Laura in her group. How many students were in Laura's class if their number was larger than $n + 2$?\n\nAnswer: $n^2 + n + 1$.",
    "options": [],
    "answer": "n^2 + n + 1",
    "solution": "By assumptions, one group contained $n + 1$ students while the whole class contains more than $n + 2$ students. As every two students belonged together into one group, more than one groups with more than one members must have occurred. If some group contained only one student $C$, this student must have belonged to every group while every other student must have belonged to exactly one group (the one where the student was together with $C$). But this would mean that there was only one group with more than one student which contained the whole class. Consequently groups with only one member cannot have occurred. Consider two cases.\n\n1) Suppose that there were two groups such that each student belonged to at least one of them. Let $C$ be the student that belonged to both of these two groups and let these groups be $\\{C, A_1, \\dots, A_k\\}$ and $\\{C, B_1, B_2, \\dots, B_l\\}$. As every pair of different students is together in exactly one group, the other groups must have had to be of the form $\\{A_i, B_j\\}$ where $i = 1, 2, \\dots, k$ and $j = 1, 2, \\dots, l$, whereby all such groups must have occurred. As these groups have to pairwise share one student we have either $k=1$ or $l=1$. Let w.l.o.g. $k=1$. As Laura's group on the day of correlation studies contained $n+1$ students while $n+1 > n \\ge 2$, this must have been the large group with $l+1$ students. Thus $l=n$ and the total number of students is $n+2$. But it is given that the total number of students has to be bigger than that.\n\n2) Suppose now that there were no two groups such that every student belonged to at least one of them. We prove that there was an equal number of students in each group. Let $\\mathcal{R}$ and $\\mathcal{S}$ be arbitrary groups formed and let $C$ be an arbitrary student contained by neither of them. Let $m$ be the number of groups that contain $C$. Each of these $m$ groups has exactly one student in common with $\\mathcal{R}$; all these students are distinct as $C$ is the only common member of every two groups containing $C$. There can be no more students in $\\mathcal{R}$ because every student in $\\mathcal{R}$ must belong to some group together with $C$. Hence $\\mathcal{R}$ contained exactly $m$ students. Analogously we see that $\\mathcal{S}$ contains exactly $m$ students. As $\\mathcal{R}$ and $\\mathcal{S}$ were arbitrary, it follows that each group contains exactly $m$ students. As on the day of studying correlation there were $n+1$ students in Laura's group, we have $m = n+1$. We can find the size of the class by only considering $C$'s groups: there are $n+1$ of them and each contains $n$ students besides $C$, whence we have $1 + (n+1)n$ students in total. This is indeed more than $n+2$ and suits as the answer.",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 29881,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all triplets of consecutive integers, such that one of these numbers is the sum of the two others.",
    "options": [],
    "answer": "(-3, -2, -1), (-1, 0, 1), (1, 2, 3)",
    "solution": "Let the consecutive numbers be $x$, $x+1$, and $x+2$. There are three cases based on which of the numbers is the sum of the other two.\n\nIf $x+2 = x + (x+1)$, then $x = 1$, which gives $(1, 2, 3)$.\n\nIf $x+1 = x + (x+2)$, then $x = -1$, which gives $(-1, 0, 1)$.\n\nIf $x = (x+1) + (x+2)$, then $x = -3$, which gives $(-3, -2, -1)$.\nIf all three numbers are positive, then only the greatest of them can be the sum of the two others. As the sum has to be greater than one of the summands by 1, the other summand has to be 1. This results in the triplet $(1, 2, 3)$.\n\nIf all the three numbers are negative, then only the smallest of the three numbers can be the sum of the other two. As the sum has to be smaller than one of the summands by 1, the other summand has to be $-1$. This results in the triplet $(-3, -2, -1)$.\n\nIf the numbers are neither all positive nor all negative, then they have to include $0$. Number $0$ cannot be a summand, otherwise the sum would equal the other summand. Hence $0$ has to be the sum of the other numbers. In that case the summands must be $-1$ and $1$. This results in the final triplet $(-1, 0, 1)$.\nOne of the summands has to be either greater than the sum by $1$ or less than the sum by $1$. Hence the second summand is equal to the difference of the sum and the first summand, so it is either $1$ or $-1$. It cannot be the middle number in the triplet (otherwise the difference of the other two numbers cannot be $1$ or $-1$). Hence, we only need to investigate the triplets in which either the smallest or the largest number is either $1$ or $-1$. The only such triplets are $(1, 2, 3)$, $(-1, 0, 1)$, and $(-3, -2, -1)$. All of them result in solutions, as $1+2=3$, $-1+1=0$, and $-2+(-1)=-3$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 29882,
    "subject": "Mathematics (Multi-modal)",
    "question": "Mari writes 8 prime numbers (not necessarily different) to her notebook, all smaller than 200. She then adds 1 to the first number, 2 to the second number and so on until adding 8 to the eighth number. She then finds the product of the eight sums. Find the largest power of two which can divide the product found.",
    "options": [],
    "answer": "2^31",
    "solution": "To have the product divisible by as great power of 2 as possible, each of the factors has to be divisible by the greatest power of 2 possible. By adding an even number to a prime number, the sum is divisible by 2 only if the original prime is even. As the only even prime number is 2, it has to be chosen to all even positions. Their respective sums are 4, 6, 8, and 10, and the respective maximal powers of 2 which they are divisible with are $2^2$, $2^1$, $2^3$, and $2^1$. Let's investigate the odd positions one by one.\nAs 127 is a prime and $127+1$ is the maximal power of 2 less than $200+1$, the first position is 127. As $125 = 5^3$, the maximal power of 2 dividing the third sum cannot be $2^7$. However, 61 is a prime and $61 + 3 = 2^6$, hence 61 is suitable for the third position. As $123 = 3 \\cdot 41$, the maximal power of 2 dividing the fifth sum cannot be $2^7$. However, $59$ is a prime and $59 + 5 = 2^6$, hence $59$ is suitable for the fifth position. As $121 = 11^2$, the maximal power of 2 dividing the seventh sum cannot be $2^7$. The natural numbers smaller than 200 which are divisible by $2^6$ are 64 and 192. The respective numbers which would result in these numbers are $57 = 3 \\cdot 19$ and $185 = 5 \\cdot 37$, neither of which is a prime. However, 89 is a prime and $89 + 7 = 3 \\cdot 2^5$, hence 89 is suitable for the seventh position. In summary, the maximal power of 2 which divides the product of the eight numbers is $2^{7+2+6+1+6+3+5+1} = 2^{31}$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 29883,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $M$ be the intersection of the diagonals of a cyclic quadrilateral $ABCD$. Find the length of $AD$, if it is known that $AB = 2$ mm, $BC = 5$ mm, $AM = 4$ mm, and $\\frac{CD}{CM} = 0.6$.",
    "options": [],
    "answer": "6 mm",
    "solution": "Opposite angles $AMB$ and $DMC$ equal. Also notice that $\\angle ABM = \\angle ABD = \\angle ACD = \\angle MCD$, as $ABD$ and $ACD$ are subtended to the same arc. Therefore the triangles $AMB$ and $DMC$ are similar. Hence $\\frac{BA}{BM} = \\frac{CD}{CM}$, from which $BM = BA \\cdot \\frac{CM}{CD}$. Analogously the opposite angles $AMD$ and $BMC$ are equal. By the property of inscribed angles $\\angle ADM = \\angle ADB = \\angle ACB = \\angle MCB$. Hence, the triangles $AMD$ and $BMC$ are similar and $\\frac{AD}{AM} = \\frac{BC}{BM}$. In summary, $AD = \\frac{AM \\cdot BC}{BM} = \\frac{AM \\cdot 5 \\text{ mm}}{2 \\text{ mm}} \\cdot 0.6 = 6 \\text{ mm}$.\n![](attached_image_1.png)\nFig. 1",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 29884,
    "subject": "Mathematics (Multi-modal)",
    "question": "Teacher drew a pentagon on the blackboard. The following conditions hold for the pentagon.\n\na) Two of the pentagon's interior angles are equal.\n\nb) There exist three interior angles such that the first one equals the sum of the other two.\n\nc) There exist four interior angles such that one of them equals the sum of the other three.\n\nd) There exists an interior angle that equals the sum of the other four.\n\nFind the interior angles of the pentagon.",
    "options": [],
    "answer": "270°, 135°, 67.5°, 33.75°, 33.75°",
    "solution": "Let the sizes of the angles of the pentagon be denoted in decreasing order as $\\alpha \\ge \\beta \\ge \\gamma \\ge \\delta \\ge \\varepsilon$. The sum of all the interior angles is $(5-2) \\cdot 180^\\circ$, in other words $\\alpha + \\beta + \\gamma + \\delta + \\varepsilon = 540^\\circ$.\n\nThe angle that equals the sum of the other four is greater than the other four. Therefore $\\alpha = \\beta + \\gamma + \\delta + \\varepsilon = \\frac{540^\\circ}{2} = 270^\\circ$.\n\nThe angle which equals the sum of some other three cannot be equal to $\\alpha$, because then $\\varepsilon = 0^\\circ$. As it must be greater than the other three angles, $\\beta = \\gamma + \\delta + \\varepsilon = \\frac{270^\\circ}{2} = 135^\\circ$.\n\nAnalogously, the angle that is the sum of some other two angles can be equal to neither $\\alpha$ nor $\\beta$, therefore $\\gamma = \\delta + \\varepsilon = \\frac{135^\\circ}{2} = 67.5^\\circ$.\n\nFinally, the pentagon cannot have more angles of size $270^\\circ$, $135^\\circ$ or $67.5^\\circ$, therefore $\\delta = \\varepsilon = \\frac{67.5^\\circ}{2} = 33.75^\\circ$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 29885,
    "subject": "Mathematics (Multi-modal)",
    "question": "A plus or a minus sign is placed between every pair of consecutive digits in the sequence $0\\ 1\\ 2\\ 3\\ 4\\ 5\\ 6\\ 7\\ 8\\ 9$.\n\na) Find the smallest positive odd number that cannot be equal to the value of the resulting expression.\n\nb) Find the smallest positive even number that cannot be equal to the value of the resulting expression.",
    "options": [],
    "answer": "a) 47; b) 2",
    "solution": "Let the sum of the digits with a plus sign in front of them be $x$ and the absolute value of the sum of the digits with a minus sign in front of them be $y$. Then the value $v$ of the expression equals $x - y$. Also $x + y = 0 + 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 = 45$. Therefore $v = 45 - 2y$.\n\na) As $45 - 2y \\le 45$, nothing greater than $45$ can be the value of the expression. We now show that we can obtain all the positive odd integers up to $45$ as the result of the expression; this shows that the least positive integer that cannot be equal to the result is $47$.\n\nWe previously showed that $y = \\frac{45-v}{2}$. In order to make the value of the expression be $v$, we need to put a minus sign in front of some digits that sum up to $\\frac{45-v}{2}$. As $v$ is a positive odd integer between $1$ and $45$, the number $\\frac{45-v}{2}$ is a nonnegative integer between $0$ and $22$. Each such positive integer can be written as a sum of digits as follows: numbers from $1$ to $9$ are among the digits themselves, numbers from $10$ to $17$ can be obtained as the sum of $9$ and some other digit, and numbers $18$ to $22$ can be written as the sum of $9$, $8$ and some other digit in the range of $1$ to $5$. The case $y = 0$ corresponds to the version where every digit has a plus sign in front of it.\n\nb) Number $2$ cannot be obtained as the value of the expression, because solving $2 = 45 - 2y$ gives $y = 21.5$, which is impossible in integers. The number $2$ is also the least positive even number.",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 29886,
    "subject": "Mathematics (Multi-modal)",
    "question": "Is it possible to find four distinct prime numbers for which the sum of any three of them is also a prime number?",
    "options": [],
    "answer": "5, 7, 17, 19",
    "solution": "Suitable prime numbers are $5$, $7$, $17$, and $19$. The sums of the corresponding triplets are prime numbers $29$, $31$, $41$, and $43$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 29887,
    "subject": "Mathematics (Multi-modal)",
    "question": "Medians *AD*, *BE*, and *CF* of triangle *ABC* intersect at point *M*. Is it possible that the circles with radii $MD$, $ME$, and $MF$\na) all have areas smaller than the area of triangle $ABC$;\nb) all have areas greater than the area of triangle $ABC$;\nc) all have areas equal to the area of triangle $ABC$?",
    "options": [],
    "answer": "a) yes; b) yes; c) no",
    "solution": "a) Let triangle $ABC$ be equilateral (Fig. 11). As the medians of this triangle all have equal lengths, the points $D$, $E$, and $F$ are located at equal distances from $M$, i.e., on a circle with centre $M$. It suffices to show that the area of this circle is less than the area of triangle $ABC$. Because the medians of an equilateral triangle are simultaneously altitudes of the triangle and perpendicular to the sides of the triangle, the sides $BC$, $CA$, and $AB$ are tangent to the circle at points $D$, $E$, and $F$, respectively. Therefore the circle is the incircle of the triangle $ABC$. The area of the incircle of the triangle is indeed less than the area of the triangle.\n\n![](attached_image_1.png)\nFig. 11\n\nb) Let $AB = AC$ and $BC = 1$ and $AD = 6$ (Fig. 12). Because the median drawn from the vertex angle of an isosceles triangle is simultaneously its altitude, the area of this triangle is $\\frac{1}{2} \\cdot 1 \\cdot 6 = 3$. On the other hand, $BE = CF > \\frac{1}{2}AD$, because $EF$ is a midsegment of the triangle parallel to $BC$ and $\\frac{1}{2}AD$ is the distance between midsegment $EF$ and side $BC$ (shown on Fig. 12 with a dotted line). Therefore $ME = MF > \\frac{1}{6}AD = 1$. The circles with radii $ME$ and $MF$ have area greater than $\\pi$, which in turn is greater than the area of triangle $ABC$, which is 3. The circle with radius $MD$ has area greater than the area of triangle $ABC$ as well, because $MD = \\frac{1}{3}AD = 2 > 1$.\n\n![](attached_image_2.png)\nFig. 12\n\nc) If circles with radii $MD$, $ME$, and $MF$ would all have areas equal to triangle $ABC$, their radii should also be equal, therefore $MD = ME = MF$. This would mean that the lengths of the parts of medians that lie on the other side of $M$ are equal as well, in other words, $MA = MB = MC$. We show that then $ABC$ must be equilateral. Indeed, from $ME = MF$ and $MB = MC$ we get that triangles $BMF$ and $CME$ must be equal (Fig. 13), therefore $BF = CE$ and also $AB = AC$. Analogously $AB = BC$. But in part a) we showed that in an equilateral triangle circles with radii $MD$, $ME$, and $MF$ do not have the same area as the triangle.\n\n![](attached_image_3.png)\nFig. 13",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 29888,
    "subject": "Mathematics (Multi-modal)",
    "question": "There are distinct points $O$, $A$, $B$, $K_1$, $\\ldots$, $K_n$, $L_1$, $\\ldots$, $L_n$ on a plane such that no three points are collinear. The open line segments $K_1L_1$, $\\ldots$, $K_nL_n$ are coloured red, other points on the plane are left uncoloured. An *allowed path* from point $O$ to point $X$ is a polygonal chain with first and last vertices at points $O$ and $X$, containing no red points. For example, for $n = 1$, and $K_1 = (-1; 0)$, $L_1 = (1; 0)$, $O = (0; -1)$, and $X = (0; 1)$, $OK_1X$ and $OL_1X$ are examples of allowed paths from $O$ to $X$; there are no shorter allowed paths. Find the least positive integer $n$ such that it is possible that the first vertex that is not $O$ on any shortest possible allowed path from $O$ to $A$ is closer to $B$ than to $A$, and the first vertex that is not $O$ on any shortest possible allowed path from $O$ to $B$ is closer to $A$ than to $B.",
    "options": [],
    "answer": "2",
    "solution": "A path $OX_1\\ldots X_{k-1}A$ is *suitable* if it is a shortest allowed path from $O$ to $A$ and $X_1A \\le X_1B$. Similarly, a path $OY_1\\ldots Y_{k-1}B$ is *suitable* if it is the shortest allowed path from $O$ to $B$ and $Y_1B \\le Y_1A$. Let us show that for $n = 2$ it is possible to choose points $A$, $B$, $O$, $K_1$, $L_1$, $K_2$, $L_2$ such that no shortest path from point $O$ to point $A$, nor one from point $O$ to point $B$ is suitable. Take $A = (2; 2)$, $B = (-2; -2)$, $K_1 = (-2;1)$, $L_1 = (5;1)$, $K_2 = (2;-1)$, and $L_2 = (-5;-1)$ (Fig. 43).\n![](attached_image_1.png)\n\nIf $O = (0;0)$, then the only shortest paths from point $O$ to points $A$ and $B$ are respectively $OK_1A$ and $OK_2B$, whereas $K_1A > K_1B$ and $K_2B > K_2A$, and hence they are not suitable. The collinearity of three points can be avoided by shifting $O$ slightly while leaving the situation unchanged.\n\nWe will show that for $n = 1$ there exists at least one suitable path from point $O$ to point $A$ or point $B$. If the segment $OA$ or segment $OB$ does not contain any red points, then it is suitable. Therefore, assume that both segments $OA$ and $OB$ contain red points. W.l.o.g., $K_1A \\le K_1B$ (can change $A$ and $B$). If $OK_1A$ is a shortest path from point $O$ to point $A$, then it is suitable. Otherwise, $OL_1A$ is the only shortest path from point $O$ to point $A$. It is suitable if $L_1A \\le L_1B$. Let us further assume that $L_1A > L_1B$. If $OL_1B$ is a shortest path from point $O$ to point $B$, then it is suitable. Otherwise, $OK_1B$ is the only shortest path from point $O$ to point $B$. Then $OL_1 + L_1A + OK_1 + K_1B < OK_1 + K_1A + OL_1 + L_1B$. This simplifies to $K_1B + L_1A < K_1A + L_1B$. However, adding the inequalities $K_1B \\ge K_1A$ and $L_1A > L_1B$ gives $K_1B + L_1A > K_1A + L_1B$. The contradiction shows that at least one suitable path from point $O$ to point $A$ or $B$ exists.",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 29889,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all functions $f : \\mathbb{R} \\to \\mathbb{R}$ that satisfy $f(xy + f(xy)) = 2x f(y)$ for all $x, y \\in \\mathbb{R}$.",
    "options": [],
    "answer": "f(x) = 0 for all x; f(x) = x for all x",
    "solution": "By choosing $(x, y) = (z, 1)$ in the original equation, where $z$ is any real number, we get $f(z + f(z)) = 2z f(1)$. However, by choosing $(x, y) = (1, z)$ in the original equation, we get $f(z + f(z)) = 2 f(z)$. From these equations together we obtain $f(z) = f(1) z$.\n\nBy choosing $z = 1$ in equation $f(z + f(z)) = 2z f(1)$, we get $f(1 + f(1)) = 2 f(1)$, however by choosing $z = 1 + f(1)$ in equation $f(z) = f(1) z$, we get $f(1 + f(1)) = f(1)(1 + f(1))$. Altogether we get $(1 + f(1)) f(1) = 2 f(1)$ or, equivalently, $f(1)(f(1) - 1) = 0$, from which $f(1) = 0$ or $f(1) = 1$.\n\nTherefore by $f(z) = f(1) z$ either $f(z) = 0$ or $f(z) = z$ for every $z$. We can check that both functions indeed satisfy the original equation.\nBy choosing $y = 0$ we get $f(f(0)) = 2x f(0)$. As $x$ can be any real number, this can hold only if $f(0) = 0$.\n\nBy choosing $x = \\frac{1}{y}$, we end up with $f(1 + f(1)) = \\frac{2}{y} f(y)$. Hence for each $y \\neq 0$ the equation $f(y) = \\frac{f(1 + f(1))}{2} y$ holds. In the case of $f(1 + f(1)) = 0$ this equation gives $f(y) = 0$ for every $y$. We get the same result if $1 + f(1) = 0$, because $f(0) = 0$.\n\nHowever, if $1 + f(1) \\neq 0$ and $f(1 + f(1)) \\neq 0$, then we can take $y = 1 + f(1)$ in the equation above and divide both sides by $f(1 + f(1))$; we obtain $\\frac{1 + f(1)}{2} = 1$, which implies $f(1) = 1$. Now by setting $y = 1$ in the equation above we get $\\frac{f(1 + f(1))}{2} = 1$, therefore $f(y) = y$ for every $y$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 29890,
    "subject": "Mathematics (Multi-modal)",
    "question": "We call a positive integer $n$ whose all digits are distinct *bright*, if either $n$ is a one-digit number or there exists a divisor of $n$ which can be obtained by omitting one digit of $n$ and which is bright itself. Find the largest bright positive integer. (We assume that numbers do not start with zero.)",
    "options": [],
    "answer": "146250",
    "solution": "First, we show by induction on the length of $n$ that if $10n$ is bright, then $n$ is bright as well. Assume that for one digit shorter numbers the statement holds. If after deleting $0$ we obtain a bright divisor, the statement holds trivially. If the bright divisor of $10n$ is obtained after deleting some other digit, then in the end of this divisor we still have $0$, i.e., it can be written as $10d$. By the induction hypothesis, $d$ is bright. But then after deleting from $n$ the corresponding digit, we get a bright divisor $d$, which means that also $n$ is bright.\n\nNext we show that any bright divisor of at least two-digit bright number not ending with $0$ can be obtained by deleting the first or the second digit.\n\nLet now $n$ be a 5-digit bright number not ending with $0$ and let $d$ be its bright divisor. We consider two cases depending on which digit is deleted to obtain $d$.\n\n1) If $d$ is obtained by deleting the first digit of $n$, then $d \\mid n - d = 10^4 \\cdot x$, where $x$ is the deleted digit. As $d$ does not end with $0$, it is not divisible either by $2$ or $5$. If $d$ is not divisible by $5$, then $d \\le 2^4 \\cdot x \\le 2^4 \\cdot 9 < 10^3$, contradiction. Hence $d$ is odd and $d \\mid 5^4 \\cdot x$. Since $d$ is four-digit not ending with zero, it must divide one of the numbers $1875$, $3125$, $4375$, $5625$. We may leave out $3125$, because in this case the first digit of $n$ must be $x = 5$, but the last digit is $5$ as well. The four-digit divisors of the remaining numbers are $1125$, $1875$, $4375$, $5625$. The first and the last number contain equal digits, from the other two numbers we cannot obtain a divisor by deleting the first or the second digit.\n\n2) If a bright divisor is obtained by deleting the second digit, then $d \\mid n - d = 10^3 \\cdot z$, where $z$ is at most two-digit. Since $d$ does not end with $0$, it is not divisible either by $2$ or $5$, implying that $\\frac{n-d}{d}$ is divisible either by $2^3$ or by $5^3$. Since $n$ and $d$ start with the same digit $a$, we have $\\frac{n}{d} < \\frac{(a+1) \\cdot 10^4}{a \\cdot 10^3} = 10 + \\frac{10}{a} \\le 20$, implying that $\\frac{n-d}{d}$ can be only $8$ or $16$, in both cases $5^3 \\mid d$. We can write $d = 1000a + 125r$, where $r \\in \\{1,3,5,7\\}$. If $\\frac{n-d}{d} = 16$, or equivalently $n = 17d$, then $n = 17000a + 2125r = 1000(17a + 2r) + 125r$. Since $n$ starts with $a$ and $125r < 1000$, we get $17a + 2r < 10(a + 1)$ yielding $7a + 2r < 10$. This gives $a = 1, r = 1$, and $d = 1125$, which is not bright. If $\\frac{n-d}{d} = 8$, or equivalently $n = 9d$, then $n = 9000a + 1125r = 1000(9a + r) + 125r$. Since $n$ starts with $a$, we get $9a + r + \\frac{r}{8} > 10a$, yielding $r > \\frac{8}{9}a \\ge a - 1$, i.e., $r \\ge a$. Leaving out numbers with repeated digits, we get $d \\in \\{1375, 1625, 1875, 2375, 2875, 3625, 3875, 4625, 4875, 6875\\}$. Among these numbers only $1625$ is bright, deleting the first or the second digit from other candidates does not give a divisor. A check shows that $9d = 14625$ is bright as well.\n\nTherefore $14625$ is the only 5-digit bright number not ending with $0$. Let now $n$ be arbitrary 6-digit bright number. If $n$ ends with $0$, then deleting $0$ we obtain a 5-digit bright number not containing $0$, whence $n = 146250$. If $n$ is not ending with $0$, then after deleting the first or the second digit we obtain a bright divisor $d$ not ending with $0$. Thus $d = 14625 = 117 \\cdot 125$, yielding $117 \\mid n - d = 10^4 \\cdot z$, where $z$ is at most 2-digit. Since $117$ and $10$ are co-prime, this is not possible. Consequently, $146250$ is the only 6-digit bright number. If $n$ was a bright 7-digit number, then by deleting its some digit we would obtain $146250$. Then also $n$ should end with $0$, which means that $\\frac{n}{10}$ is a bright 6-digit number not containing $0$. But there are no such numbers. Since there exist no 7-digit bright numbers, there cannot be longer bright numbers either.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 29891,
    "subject": "Mathematics (Multi-modal)",
    "question": "A sequence of positive real numbers $a_1, a_2, a_3, \\dots$ satisfies $a_n = a_{n-1} + a_{n-2}$ for all $n \\ge 3$. A sequence $b_1, b_2, b_3, \\dots$ is defined by equations $b_1 = a_1$, $b_n = a_n + (b_1 + b_2 + \\dots + b_{n-1})$ for even $n > 1$, $b_n = a_n + (b_2 + b_4 + \\dots + b_{n-1})$ for odd $n > 1$. Prove that if $n \\ge 3$, then $\\frac{1}{3} < \\frac{b_n}{n \\cdot a_n} < 1$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "The definition of sequence $(b_n)$ indicates that $b_n - b_{n-2} = a_n - a_{n-2} + b_{n-1}$ for all $n \\ge 3$. Therefore $b_n - b_{n-2} = a_{n-1} + b_{n-1}$, i.e., $b_n = a_{n-1} + b_{n-1} + b_{n-2}$.\n\nNotice that the left-hand inequality $\\frac{1}{3} < \\frac{b_n}{n \\cdot a_n}$ holds for $n=2$ and $n=3$ as $\\frac{b_2}{2a_2} = \\frac{a_2+b_1}{2a_2} = \\frac{a_1+a_2}{2a_2} > \\frac{a_2}{2a_2} = \\frac{1}{2} > \\frac{1}{3}$ and $\\frac{b_3}{3a_3} = \\frac{a_3+b_2}{3a_3} = \\frac{2(a_1+a_2)}{3(a_1+a_2)} = \\frac{2}{3} > \\frac{1}{3}$.\n\nNow assume that $n \\ge 4$ and that the statement is true for $n-1$ and $n-2$. Then $b_n = a_{n-1} + b_{n-1} + b_{n-2} > a_{n-1} + \\frac{1}{3}(n-1)a_{n-1} + \\frac{1}{3}(n-2)a_{n-2} = \\frac{1}{3}n(a_{n-1} + a_{n-2}) + \\frac{2}{3}(a_{n-1} - a_{n-2}) > \\frac{1}{3}na_n$, where the last inequality holds as $a_{n-1} = a_{n-2} + a_{n-3} > a_{n-2}$.\n\nBy induction we get that $b_n > \\frac{1}{3}na_n$ for all $n \\ge 2$.\n\nNow notice that the right-hand inequality $\\frac{b_n}{n \\cdot a_n} < 1$ holds for $n = 3$ and $n = 4$ as $\\frac{b_3}{3a_3} = \\frac{2}{3} < 1$ and $\\frac{b_4}{4a_4} = \\frac{a_4+b_3+b_1}{4a_4} = \\frac{a_4+a_3+a_2+a_1}{4a_4} < 1$.\n\nLet $n \\ge 5$ and assume that the statement is valid for $n-1$ and $n-2$. Then $b_n = a_{n-1} + b_{n-1} + b_{n-2} < a_{n-1} + (n-1)a_{n-1} + (n-2)a_{n-2} = n(a_{n-1} + a_{n-2}) - 2a_{n-2} < na_n$.\n\nBy induction, $b_n < na_n$ holds for all $n \\ge 3$.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 29892,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $k$ be a positive integer. Find all positive integers $n$, such that it is possible to mark $n$ points on the sides of a triangle (different from its vertices) and connect some of them with a line in such a way that the following conditions are satisfied:\n\n1) there is at least 1 marked point on each side;\n\n2) for each pair of points $X$ and $Y$ marked on different sides, on the third side there exist exactly $k$ marked points which are connected to both $X$ and $Y$ and exactly $k$ points which are connected to neither $X$ nor $Y$.\n\n*Answer:* $12k$.",
    "options": [],
    "answer": "12k",
    "solution": "Let $A$, $B$, and $C$ be the sets of the points on different sides and $a$, $b$ and $c$ their cardinalities respectively. Let us count all the triplets $(p, q, r)$, where $p \\in A$, $q \\in B$, and $r \\in C$ and they are either pairwise connected or pairwise not connected. For all $p \\in A$, $q \\in B$ there exist exactly $k$ points $r \\in C$, for which the condition holds: if $p$ and $q$ are connected, then choose the $k$ points which are connected to both; if $p$ and $q$ are not connected, then pick the $k$ points which are connected to neither of them. Therefore the total number of such triplets is $kab$. Similarly, for all $q \\in B$, $r \\in C$ we find that the total number of those triplets is $kbc$, and for all $r \\in C$, $p \\in A$ we find that the number of those triplets is $kca$. Therefore, $ab = bc = ca$, implying $a = b = c = \\frac{n}{3}$.\n\nLet us now count all the other triplets $(p,q,r)$ in which $p \\in A$, $q \\in B$ and $r \\in C$. Similarly to the previous paragraph, we can see that for all $p \\in A$, $q \\in B$ there exist exactly $k$ points $r \\in C$ for which $r$ is connected to neither of those if $p$ and $q$ are connected to each other, and $r$ is connected to both if $p$ and $q$ are not connected to each other. The number of such triplets $(p,q,r)$ is $kab = k\\left(\\frac{n}{3}\\right)^2$. Similarly, we then count the triplets $(p,q,r)$ in which $p$ is connected to neither of $q$ or $r$ if $q$ and $r$ are connected to each other and $p$ is connected to both $q$ and $r$ if $q$ and $r$ are not connected to each other. The number of such points is also $k\\left(\\frac{n}{3}\\right)^2$. Finally, we count the triplets $(p,q,r)$ in which $q$ is connected to neither of $r$ and $p$ if $r$ and $p$ are connected to each other and $q$ is connected to both $r$ and $p$ if $r$ and $p$ are not connected. The number of those triplets is also $k\\left(\\frac{n}{3}\\right)^2$. We have now counted all the possible triplets. Therefore, the total number of triplets $(p,q,r)$ ($p \\in A$, $q \\in B$ and $r \\in C$) is $4k\\left(\\frac{n}{3}\\right)^2$. On the other hand, the number of triplets is $\\left(\\frac{n}{3}\\right)^3$. The equality $4k\\left(\\frac{n}{3}\\right)^2 = \\left(\\frac{n}{3}\\right)^3$ gives us $n = 12k$.\n\nIt remains to prove that it is possible to mark $4k$ points on each side to satisfy the conditions. Let us first look the case $k=1$. Let the sides of triangle be labelled as $0$, $1$, and $2$ with each of the sides containing points labelled as $0$, $1$, $2$, and $3$. For all $i=0, 1, 2$ we connect the even numbered points on the side $i$ with points $0$ and $1$ on the side $(i+1) \\bmod 3$ and the odd numbered points on the side $i$ with points $2$ and $3$ on the side $(i+1) \\bmod 3$.\n\n![](attached_image_1.png)\n\nThen we have for each pair of points (which are not on the same side) exactly one point of the third side which are connected to both and exactly one point which is connected to neither of them. For case $k > 1$ we substitute each point with $k$ different points and connect those which were generated from points that were connected before.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 29893,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let a simple polynomial function be a polynomial function $P(x)$ whose coefficients belong to the set $\\{-1, 0, 1\\}$. Let $n$ be a positive integer, $n > 1$. Find the smallest possible number of non-zero coefficients in a simple polynomial function of $n$th order whose values at all integral arguments are divisible by $n$.\n\n*Answer: 2.*",
    "options": [],
    "answer": "2",
    "solution": "A single non-zero coefficient is not sufficient for any $n > 1$ as the only simple polynomial functions with a single non-zero coefficient are $P(x) = x^n$ and $P(x) = -x^n$ but in both cases $n \\nmid P(1)$. Let us show that the values of the polynomial function $P_n(x) = x^n - x^{n-\\varphi(n)}$ at all integral arguments are divisible by $n$. (Here $\\varphi$ is the Euler's totient function.) This shows that having 2 non-zero coefficients is sufficient.\n\nLet $k$ be an integer. Let the canonical form of $n$ be $p_1^{\\alpha_1} \\cdots p_m^{\\alpha_m}$ and let us assume without loss of generality that $k$ is divisible by primes $p_1, \\dots, p_l$ and is not divisible by primes $p_{l+1}, \\dots, p_m$. Define $u = p_1^{\\alpha_1} \\cdots p_l^{\\alpha_l}$ and $v = p_{l+1}^{\\alpha_{l+1}} \\cdots p_m^{\\alpha_m}$. Let us now show that $u \\mid k^{n-\\varphi(n)}$ and $v \\mid k^{\\varphi(n)} - 1$. Having $uv = n$, we can conclude that $n \\mid P_n(k)$ as $P_n(k) = k^n - k^{n-\\varphi(n)} = k^{n-\\varphi(n)}(k^{\\varphi(n)} - 1)$.\n\nTo prove that $u \\mid k^{n-\\varphi(n)}$, it is sufficient to prove for all $i = 1, \\dots, l$ that $p_i^{\\alpha_i} \\mid k^{n-\\varphi(n)}$. It is sufficient to prove that $\\alpha_i \\le n - \\varphi(n)$, as by the assumption $p_i \\mid k$. Inequality $\\alpha_i \\le n - \\varphi(n)$ holds as $p_i$, $p_i^2, \\dots, p_i^{\\alpha_i}$ are $\\alpha_i$ positive integers which are not greater than $n$ and not coprime with $n$.\n\nTo prove the statement $v \\mid k^{\\varphi(n)} - 1$, we derive from Euler's theorem that $v \\mid k^{\\varphi(v)} - 1$ as $k$ and $v$ are coprime. Also, $u$ and $v$ are coprime, therefore, $\\varphi(n) = \\varphi(uv) = \\varphi(u)\\varphi(v)$ from which $k^{\\varphi(v)} - 1 \\mid k^{\\varphi(n)} - 1$. Consequently, $v \\mid k^{\\varphi(n)} - 1$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 29894,
    "subject": "Mathematics (Multi-modal)",
    "question": "Prove that, for any natural number $n$, either $3^{2n} - 3^{n+1} + 3^n - 3$ or $3^{2n} - 3^{n+1} + 3^n + 1$ is divisible by 32.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Note that $3^{2n} - 3^{n+1} + 3^n - 3 = (3^n - 3)(3^n + 1)$.\n\nIf $n$ is odd then $3^n - 3 = 3(3^{n-1} - 1) = 3\\left(3^{\\frac{n-1}{2}} - 1\\right)\\left(3^{\\frac{n-1}{2}} + 1\\right)$.\n\nAs all powers of $3$ are odd, $3^{\\frac{n-1}{2}} - 1$ and $3^{\\frac{n-1}{2}} + 1$ are consecutive even numbers. One of these numbers must be divisible by $4$, whence their product is divisible by $8$. Thus $8 \\mid 3^n - 3$, implying that $4 \\mid 3^n + 1$. Consequently, $32 \\mid 3^{2n} - 3^{n+1} + 3^n - 3$.\n\nAssume now $n$ being even. Note that\n$$\n3^{2n} - 3^{n+1} + 3^n + 1 = 3^{2n} - 3 \\cdot 3^n + 3^n + 1 = (3^n)^2 - 2 \\cdot 3^n + 1 = (3^n - 1)^2.\n$$\nSimilarly to the previous case, $3^n - 1 = (3^{\\frac{n}{2}} - 1)(3^{\\frac{n}{2}} + 1)$ where factors in the r.h.s. are consecutive even numbers. Hence $8 \\mid 3^n - 1$, implying that $64 \\mid 3^{2n} - 3^{n+1} + 3^n + 1$. Consequently, $32 \\mid 3^{2n} - 3^{n+1} + 3^n + 1$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 29895,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $p_1, p_2, p_3, \\dots$ be all prime numbers in increasing order. Prove that $p_2 + p_4 + \\dots + p_{2n} > 3n^2 - 2n + 1$ for every positive integer $n$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "For any positive integer $k$, six consecutive integers $6k, 6k+1, 6k+2, 6k+3, 6k+4, 6k+5$ can contain at most two prime numbers. Hence $p_i \\ge p_{i-2}+6$ for all $i \\ge 5$, implying $p_{2i} \\ge p_4+6(i-2) = 6i-5$ for all $i \\ge 2$. Thus $p_2 + p_4 + \\dots + p_{2n} > 2 + (7+13+\\dots+(6n-5)) = 3n^2 - 2n + 1$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 29896,
    "subject": "Mathematics (Multi-modal)",
    "question": "The polynomial $x^3 + px + q$, where $p$ and $q$ are real numbers and at least one of them is non-zero, has a real root $a$ that satisfies\n$$\na^2 \\le -\\frac{4}{3}p.\n$$\nProve that this polynomial has a real root different from $a$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "The assumption $a^3 + pa + q = 0$ implies $q = -a(a^2 + p)$, whence $x^3 + px + q = (x-a)(x^2 + ax + a^2 + p)$. The discriminant of $x^2 + ax + a^2 + p$ is $D = a^2 - 4(a^2 + p) = -(3a^2 + 4p)$; the assumption $a^2 \\le -\\frac{4}{3}p$ implies $D \\ge 0$. Hence there are real numbers $b$ and $c$ such that $x^2 + ax + a^2 + p = (x-b)(x-c)$, so the polynomial $x^3 + px + q$ has roots $b$ and $c$. If $a = b = c$ then $x^3 + px + q = (x-a)^3 = x^3 - 3ax^2 + 3a^2x + a^3$. Hence $-3a = 0$, $3a^2 = p$ and $a^3 = q$, implying $p = q = 0$. This contradicts the assumption. Consequently, $x^3 + px + q$ has a real root different from $a$.",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 29897,
    "subject": "Mathematics (Multi-modal)",
    "question": "A quadrilateral $ABCD$ has incenter $I$. Diagonals $AC$ and $BD$ intersect at point $P$. Given that $I$ lies inside the triangle $PAB$ (not on its side), prove that the area of the triangle $PAB$ is greater than the area of any of triangles $PBC$, $PCD$ and $PDA$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Denote the area of any triangle $\\triangle \\Delta$ by $S_{\\triangle \\Delta}$. Let $D'$ be the reflection of $D$ from $AC$ and $\\alpha = \\angle APB$ (Fig. 24). We have\n\n$$\nS_{PAB} = \\frac{1}{2} \\cdot PA \\cdot PB \\cdot \\sin \\alpha, \\qquad (4)\n$$\n$$\nS_{PBC} = \\frac{1}{2} \\cdot PB \\cdot PC \\cdot \\sin(180^\\circ - \\alpha) = \\frac{1}{2} \\cdot PB \\cdot PC \\cdot \\sin \\alpha, \\quad (5)\n$$\n$$\nS_{PCD} = \\frac{1}{2} \\cdot PC \\cdot PD \\cdot \\sin \\alpha, \\qquad (6)\n$$\n$$\nS_{PDA} = \\frac{1}{2} \\cdot PD \\cdot PA \\cdot \\sin(180^\\circ - \\alpha) = \\frac{1}{2} \\cdot PD \\cdot PA \\cdot \\sin \\alpha. \\quad (7)\n$$\n\n![](attached_image_1.png)\nFig. 24\n\n$\\angle D'AC < \\angle BAC$. By interchanging the roles of $A$ and $C$, we similarly obtain $\\angle D'CA < \\angle BCA$. Hence $D'$ lies inside the triangle $ABC$, implying that $S_{ADC} = S_{AD'C} < S_{ABC}$. Adding (6) and (7) gives $S_{PCD} + S_{PDA} = \\frac{1}{2} \\cdot (PC + PA) \\cdot PD \\cdot \\sin \\alpha$, i.e., $S_{ACD} = \\frac{1}{2} \\cdot AC \\cdot PD \\cdot \\sin \\alpha$. Analogously from (4) and (5) we obtain $S_{ABC} = \\frac{1}{2} \\cdot AC \\cdot PB \\cdot \\sin \\alpha$. Hence the inequality $S_{ADC} < S_{ABC}$ implies $PD < PB$.\n\nInterchanging the roles of $A$ and $B$ and the roles of $C$ and $D$ similarly gives $PC < PA$. Now (4) and (5) together imply $S_{PAB} > S_{PBC}$, (4) and (7) together imply $S_{PAB} > S_{PDA}$, (5) and (6) together imply $S_{PBC} > S_{PCD}$, and (6) and (7) together imply $S_{PDA} > S_{PCD}$. Hence the desired claim follows.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 29898,
    "subject": "Mathematics (Multi-modal)",
    "question": "Some knights are on a playground. Each knight has three properties: speed, smartness and sightliness. For every knight, each property takes a certain integral value $x$ such that $1 \\le x \\le n$. A knight $A$ can win a knight $B$ if the speed, smartness and sightliness of $A$ are all greater than those of $B$. It is known that none among the knights on the playground can win any of the other knights and every two knights differ by at least one property. Find the largest possible number of knights on the playground.",
    "options": [],
    "answer": "3n^2 - 3n + 1",
    "solution": "For each knight, we can define a unique non-negative integer $x$ such that the properties of the knight are $a+x$, $b+x$ and $c+x$ and $\\min(a,b,c) = 1$. Call the vector $(a,b,c)$ the *base triple* of the knight.\nIf two knights had the same base triple then either their all properties would be equal or one of them could win the other one. Both cases are excluded by the assumptions. Thus each knight has a unique base triple, implying that there are at most as many knights as possible base triples.\nThe number of triples of integers $1, \\dots, n$ is $n^3$. Among them, $(n-1)^3$ triples do not contain the value $1$. Thus the number of possible base triples is $n^3 - (n-1)^3 = 3n^2 - 3n + 1$. This number is achieved if, for every base triple, there is exactly one knight whose values of properties coincide with the components of this base triple.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 29899,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $a > 2$ be an integer. Let\n$$x = (a-1) \\cdot a^{a-2} + (a-2) \\cdot a^{a-3} + \\dots + 2 \\cdot a^1 + 1 \\cdot a^0,$$\n$$y = 1 \\cdot a^{a-2} + 2 \\cdot a^{a-3} + \\dots + (a-2) \\cdot a^1 + (a-1) \\cdot a^0.$$\n\nProve that $x - 1$ is divisible by $y + 1$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Note that $x + y = a \\cdot a^{a-2} + a \\cdot a^{a-3} + \\dots + a \\cdot a^1 + a \\cdot a^0 = a^{a-1} + a^{a-2} + \\dots + a^1$.\nHence\n$$\nx + 2y = a^{a-1} + 2 \\cdot a^{a-2} + 3 \\cdot a^{a-3} + \\dots + (a-1) \\cdot a^1 + (a-1) \\cdot a^0 \\\\\n= a(1 \\cdot a^{a-2} + 2 \\cdot a^{a-3} + \\dots + (a-1) \\cdot a^0) + (a-1) \\\\\n= ay + (a-1).\n$$\nAs $x + 2y = ay + (a-1)$ is equivalent to $x - 1 = (a-2)(y+1)$, the claim follows.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 29900,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all functions $f: \\mathbb{R} \\to \\mathbb{R}$ such that, for every real number $a$, the function $g(x) = f(x + a)$ is either an even or odd function.",
    "options": [],
    "answer": "All constant functions f(x) = c for some real constant c.",
    "solution": "Denote $f(0) = c$. We show that $f(x) = c$ for every real number $x$. Fix a real number $a$ arbitrarily; by assumption, the function $g(x) = f(x + \\frac{a}{2})$ is either an even or odd function. If this $g$ is even then\n$$\nf(a) = f\\left(\\frac{a}{2} + \\frac{a}{2}\\right) = g\\left(\\frac{a}{2}\\right) = g\\left(-\\frac{a}{2}\\right) = f\\left(-\\frac{a}{2} + \\frac{a}{2}\\right) = f(0) = c.\n$$\nAnalogously in the case of odd $g$ we obtain\n$$\nf(a) = f\\left(\\frac{a}{2} + \\frac{a}{2}\\right) = g\\left(\\frac{a}{2}\\right) = -g\\left(-\\frac{a}{2}\\right) = -f\\left(-\\frac{a}{2} + \\frac{a}{2}\\right) = -f(0) = -c.\n$$\nThus if $g$ defined as in the problem is even for all choices of $a$ then $f(x) = c$ for every real number $x$. If $g(x) = f(x + a)$ is an odd function for some $a$ then $f(a) = f(0 + a) = g(0) = 0$. But by the above, $f(a) = c$ or $f(a) = -c$; hence $c = 0$. This implies that, for every real number $x$, $f(x) = 0 = c$.\nOn the other hand, if $f$ is a constant function then $g(x) = f(x + a)$ is the same constant function for every $a$. Constant functions are even. Hence constant functions satisfy the conditions of the problem.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 29901,
    "subject": "Mathematics (Multi-modal)",
    "question": "The diagonals of a tangential quadrilateral $ABCD$ intersect at point $P$. The side $AB$ is longer than any other side of $ABCD$. Prove that the angle $APB$ is obtuse.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $a, b, c, d$ be the lengths of the tangent line segments at vertices $A, B, C, D$, respectively, and $\\alpha = \\angle APB$ (Fig. 25). We have $a + b > b + c$ and $a + b > a + d$ as $AB$ is the longest side, implying $a > c$ and $b > d$. By the law of cosines:\n$$\n(a+b)^2 = PA^2 + PB^2 - 2 \\cdot PA \\cdot PB \\cdot \\cos \\alpha,\n$$\n$$\n(b+c)^2 = PB^2 + PC^2 + 2 \\cdot PB \\cdot PC \\cdot \\cos \\alpha,\n$$\n$$\n(c+d)^2 = PC^2 + PD^2 - 2 \\cdot PC \\cdot PD \\cdot \\cos \\alpha,\n$$\n$$\n(d+a)^2 = PD^2 + PA^2 + 2 \\cdot PD \\cdot PA \\cdot \\cos \\alpha.\n$$\n\n![](attached_image_1.png)\nFig. 25\n\n$2(ab + cd - bc - da) = -2(PA \\cdot PB + PB \\cdot PC + PC \\cdot PD + PD \\cdot PA) \\cos \\alpha$. The l.h.s. can be expressed as $2(a - c)(b - d)$ which is positive since $a > c$ and $b > d$. The parenthesized expression in the r.h.s. is also positive. Hence $\\cos \\alpha$ must be negative. This means that the angle $APB$ is obtuse.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 29902,
    "subject": "Mathematics (Multi-modal)",
    "question": "On a horizontal line, one colors $2k$ points red and, to the right of them, $2k$ points blue. On every move, one chooses two points of different color, such that there is exactly one colored point between them, and interchanges the colors of the chosen points. How many different configurations can one obtain using these moves?",
    "options": [],
    "answer": "((2k choose k))^2",
    "solution": "Enumerate the colored points by positive integers from the left to the right. Every move can influence two points with the same parity, whereby the total number of red or blue points with this parity does not change. Thus in each configuration that can be achieved there are $k$ red and $k$ blue points with each parity. The number of such configurations is $\\left(\\binom{k+k}{k}\\right)^2$ since there are $\\binom{k+k}{k}$ possibilities to choose $k$ red points from among $k+k$ points with an odd number and $\\binom{k+k}{k}$ possibilities to choose $k$ red points from among $k+k$ points with an even number.\n\nOn the other hand, all configurations with $k$ red and $k$ blue points with each parity can be achieved. Consider the points with even numbers. The number of the rightmost point with an even number that must become red is at least $2k$; after moving the rightmost in the initial configuration red point to the right until it reaches its desired position, this point and all points in the right of it are of the desired color. Next we can move the rightmost red point that is not yet at its desired position similarly to its desired position, etc. When all red points with an even number are at their desired position, other points with an even number are also of the right color. Similarly we act with points with an odd number.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 29903,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $n$ be a positive integer, $n \\ge 3$. In a regular $n$-gon, one draws a maximal set of diagonals, no two of which intersect in the interior of the $n$-gon. Every diagonal is labelled with the number of sides of the $n$-gon between the endpoints of the diagonal along the shortest path. Find the maximum value of the sum of the labels.\n\n*Answer:* $\\frac{n^2-9}{4}$ for odd $n$, $\\frac{n^2-8}{4}$ for even $n$.",
    "options": [],
    "answer": "(n^2 - 9)/4 for odd n; (n^2 - 8)/4 for even n",
    "solution": "Consider an arbitrary set of diagonals satisfying the conditions. Label the sides of the $n$-gon with number 1. The diagonals partition the $n$-gon into triangles; let $\\Delta$ be a triangle that contains the centroid of the $n$-gon. Let $s, t, u$ be the labels of the sides of $\\Delta$ (Fig. 26). The triangle $\\Delta$ divides the interior of the $n$-gon outside $\\Delta$ into three regions each bounded by one side of $\\Delta$ and the respective sides of the $n$-gon (if a side of the $n$-gon coincides with a side of $\\Delta$ then the respective region contains 0 triangles).\n\n![](attached_image_1.png)\nFig. 26\n\nAs all triangles that are cut out are on the same side of the centroid of the polygon, the label of the longest side of each triangle equals the sum of labels on the two shorter sides. Cutting out a triangle removes two sides from the boundary of the region and introduces a new side labeled with the sum of the removed sides. Therefore, the sum of the labels on the boundary of the region is constant during the process. Let $m_i$ be the greatest number on the boundary of the region after $i$ triangles have been removed. As the boundary is initially labeled with $s$ ones and in every step the number of sides is reduced by one then after $i$ steps there are $s-i$ sides remaining. Therefore $s \\ge m_i + (s-i-1)$ from which $m_i \\le i+1$. On the other hand, the label of the diagonal $d_i$ along which the cut is made on the $i$-th step satisfies $d_i \\le m_i$. Hence the sum of all labels of the diagonals in this region (including the side of $\\Delta$ labeled with $s$) is not greater than $2+3+...+s$.\nAs the same is valid for the remaining two regions, the total sum of labels does not exceed $(2+3+...+s) + (2+3+...+t) + (2+3+...+u)$. We know $s+t+u=n$; w.l.o.g., $\\frac{n}{2} \\ge s \\ge t \\ge u \\ge 1$. If $u>1$ then $t < \\lfloor \\frac{n}{2} \\rfloor$, whence a new triple $(s', t', u')$ with $u' = u-1$ also satisfies conditions: Replace $t$ with $t+1$ and change the order of $s$ and $t+1$ if needed. The total sum increases since $t+1 > u$. In the case of $u=1$, the only possibility is $s=t=\\frac{n-1}{2}$ if $n$ is odd and $s=\\frac{n}{2}, t=\\frac{n}{2}-1$ if $n$ is even. Hence the total sum of labels does not exceed $2 \\cdot \\left(2+3+...+\\frac{n-1}{2}\\right) = \\frac{n^2-9}{4}$ for odd $n$ and $\\left(2+3+...+\\left(\\frac{n}{2}-1\\right)\\right) + \\left(2+3+...+\\frac{n}{2}\\right) = \\frac{n^2-8}{4}$ for even $n$. These sums can be achieved by drawing all diagonals connecting a fixed vertex with all non-neighboring vertices.\n\n\nSolution 2:\n\nCall the minimal number of sides of the $n$-gon between the endpoints of a given diagonal its *length*. The parts of the $n$-gon that a diagonal divides it into are called *short flank* and *long flank* (if the parts contain the same number of sides then both may be called short and long flank).\n\nFor every positive integer $k$, let $a_k$ be the number of drawn diagonals whose length is at least $k$. Note that the sum of lengths of all drawn diagonals is $a_1 + a_2 + \\dots + a_{\\lfloor \\frac{n}{2} \\rfloor}$. Indeed, represent all lengths of diagonals in the form $1+1+\\dots+1$; aligning the sums left, the $k$th column from the left contains exactly $a_k$ ones and there are $\\lfloor \\frac{n}{2} \\rfloor$ non-empty columns in total.\nWe prove that, for every $k \\ge 2$, $a_k \\le n - 2k + 1$. The claim holds for $k = \\frac{n}{2}$ since all diagonals of length $\\frac{n}{2}$ intersect in the centroid of the $n$-gon. Assume in the rest that $k < \\frac{n}{2}$. Let $M$ be an arbitrary set of non-intersecting diagonals with length at least $k$ which cannot be extended by any other such diagonal without introducing an intersection. Let $d$ be the longest diagonal in $M$. Clearly $M$ contains a diagonal of length $k$ in the long flank of $d$, otherwise one could add a new diagonal of length $k$ to the short flank of the shortest diagonal in the long flank of $d$. If the length of $d$ is greater than $k$ then $M$ contains a diagonal of length $k$ also in the short flank of $d$. Hence $M$ contains at least 2 diagonals of length $k$. In the short flanks of these diagonals, there are $2(k-1)$ vertices not connected by any diagonals of $M$. The remaining $n - 2k + 2$ vertices can be joined by at most $n - 2k + 1$ non-intersecting diagonals, implying $a_k \\le n - 2k + 1$.\nAs $a_1 = n - 3 = (n - 1) - 2$, the sum of the lengths of all drawn diagonals is at most $((n - 1) + (n - 3) + \\dots + (n - 2 \\lfloor \\frac{n}{2} \\rfloor + 1)) - 2$. This simplifies to $\\frac{n^2-8}{4}$ in the case of even $n$ and to $\\frac{n^2-9}{4}$ in the case of odd $n$. If one draws all diagonals from a fixed vertex to the non-neighboring vertices then $a_k = n - 2k + 1$ for every $k = 2, 3, \\dots, \\lfloor \\frac{n}{2} \\rfloor$. Hence the bounds found above are achieved.\n\n\nSolution 3:\n\nDefine the notions *length*, *short flank* and *long flank* as in Solution 2. We show by induction that the sum of lengths of any diagonal $d$ and all diagonals in the short flank of $d$ does not exceed $2+3+\\dots+s$, where $s$ is the length of $d$. The claim holds for $s=2$ since there are no diagonals in the short flank of a diagonal of length 2. Assume now that the claim holds for diagonals shorter than $d$. Let $d'$ of length $s'$ be the longest diagonal drawn in the short flank of $d$. By the induction hypothesis, the sum of lengths of the diagonal $d'$ and all diagonals in the short flank of $d'$ does not exceed $2+3+\\dots+s'$. As there are exactly $s-2$ diagonals drawn in the short flank of $d$ and exactly $s'-2$ of them are in the short flank of $d'$, the number of diagonals that lie in the short flank of $d$ and also in the long flank of $d'$ must be $s-s'-1$. By the choice of $d'$, the sum of lengths of these diagonals is at most $s'(s-s'-1)$ which does not exceed $(s'+1)+\\dots+(s-1)$ (the equality holds in the case $s'=s-1$). Consequently, the sum of lengths of diagonal $d$ and all diagonals in the short side of $d$ does not exceed $2+3+\\dots+s$. This completes the proof of the claim.\nNow let $d$ of length $s$ be the longest drawn diagonal. Let $d'$ of length $s'$ be the longest diagonal in the long flank of $d$. By the claim proven above, the sum of lengths of the diagonal $d$ and all diagonals in the short flank of $d$ does not exceed $2+3+...+s$ and the sum of lengths of the diagonal $d'$ and all diagonals in the short flank of $d'$ does not exceed $2+3+...+s'$. The total number of these diagonals is $(s-1)+(s'-1)$. As $n-3$ diagonals are drawn, the number of diagonals lying in the long flank of both $d$ and $d'$ is $n-s-s'-1$. By the choice of $d$ and $d'$, the sum of lengths of these diagonals is at most $s'(n-s-s'-1)$ that does not exceed $((s+1)+...+\\lfloor\\frac{n}{2}\\rfloor)+((s'+1)+...+\\lfloor\\frac{n-1}{2}\\rfloor)$. Thus the sum of lengths of all drawn diagonals is $(2+3+...+\\lfloor\\frac{n}{2}\\rfloor)+(2+3+...+\\lfloor\\frac{n-1}{2}\\rfloor)$ which simplifies to $\\frac{n^2-8}{4}$ in the case of even $n$ and $\\frac{n^2-9}{4}$ in the case of odd $n$. These bounds are achieved as the sum of lengths of diagonals drawn from a fixed vertex to all non-neighboring vertices equals $(2+3+...+\\lfloor\\frac{n}{2}\\rfloor)+(2+3+...+\\lfloor\\frac{n-1}{2}\\rfloor)$.\n\n\nSolution 4:\n\nDefine the notions *length*, *short flank* and *long flank* as in Solution 2. Consider any set of diagonals with the maximal sum of lengths.\nSuppose initially that, in the partition of the $n$-gon into triangles, there exists a triangle $ABC$ whose all sides are diagonals of the $n$-gon; let the lengths of $BC$, $CA$ and $AB$ be $a, b, c$, respectively. W.l.o.g., $a \\le b \\le c$. Then $AB$ and $AC$ lie in the long flank of $BC$ (otherwise $a = b+c$). Let $BCD$ be the triangle of the partition in the short flank of $BC$ (Fig. 27) and let the length of $AD$ be $d$. Among the diagonals $AB$ and $AC$, the one that lies in the short side of $AD$ is shorter than $AD$, implying that $d > b$ or $d > c$. In either\n\n![](attached_image_2.png)\nFig. 27\n\ncase, $d > a$. Thus replacing $BC$ with $AD$ in the partition leads to a larger sum of lengths. This contradicts the assumption.\nConsequently, each triangle of the partition has at least one side that is a side of the $n$-gon. This means that, starting from any drawn diagonal $d_1$ of length 2, we can find a diagonal $d_2$ with a common endpoint with $d_1$, then a new diagonal $d_3$ with a common endpoint with $d_2$ etc., in such a way that the lengths of the diagonals increase with step 1 until $\\lfloor\\frac{n}{2}\\rfloor$ and after that decrease with step 1. Hence the sum of lengths of all diagonals is at most $2 \\cdot (2+3+...+\\frac{n-1}{2}) = \\frac{n^2-9}{4}$ for odd $n$ and at most $2 \\cdot (2+3+...+(\\frac{n}{2}-1)) + \\frac{n}{2} = \\frac{n^2-8}{4}$ for even $n$. The proof immediately gives a construction that achieves the bounds.\n\n\nSolution 5:\n\nLabel each side of the $n$-gon with number 1. If the polygon with at least 5 sides is partitioned into triangles using non-intersecting diagonals then at least two triangles have two sides on the boundary of the polygon. After removing two such triangles, we can repeat the procedure in the remaining polygon. After repeating the procedure $k$ times, the remaining polygon has exactly $n-2k$ sides since at each step a new side appears instead of two sides at two places. The sum of labels of the sides of the remaining polygon is $n$ or less since every cut replaces two sides with a diagonal with label not greater than the sum of the labels of the two sides.\nConsider arbitrary 4 sides in the polygon remaining after $k$ steps. As the sum of labels of the other $n-2k-4$ sides is at least $n-2k-4$, the sum of labels of the chosen 4 sides is at most $2k+4$. Hence the labels of the diagonals along which the two cuts are made in the next step sum up to at most $2k+4 = 2(k+2)$. Using this argument for $k=0,1,\\dots, \\frac{n-1}{2}-2$ in the case of odd $n$, we conclude that the sum of labels of the drawn diagonals, for odd $n$, does not exceed $2 \\cdot (2+3+\\dots+\\frac{n-1}{2}) = \\frac{n^2-9}{4}$. Similarly for even $n$, the sum of labels of $n-4$ drawn diagonals does not exceed $2 \\cdot (2+3+\\dots+(\\frac{n}{2}-1)) = \\frac{n^2}{4} - \\frac{n}{2} - 2$. As the label of the last diagonal is at most $\\frac{n}{2}$, the total sum of labels is at most $\\frac{n^2-8}{4}$. The sums are achieved by drawing diagonals from a fixed vertex to all non-neighboring vertices.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 29904,
    "subject": "Mathematics (Multi-modal)",
    "question": "There are the same number of boys and girls in a class. It is known that $60\\%$ of pupils do sports and $\\frac{5}{9}$ of pupils doing sports are boys. It is also known that $\\frac{1}{3}$ of pupils doing sports go to math club and $\\frac{2}{15}$ of girls neither do sports nor go to math club. On the other hand, $\\frac{2}{15}$ of boys both do sports and go to math club. What percentage of girls go to math club?",
    "options": [],
    "answer": "60%",
    "solution": "There are $\\frac{3}{5} \\cdot \\frac{1}{3} = \\frac{1}{5}$ of pupils who both do sports and go to math club, whereas $\\frac{1}{2} \\cdot \\frac{2}{15} = \\frac{1}{15}$ of pupils are boys who both do sports and go to math club. Thus $\\frac{1}{5} - \\frac{1}{15} = \\frac{2}{15}$ of pupils are girls who both do sports and go to math club. Girls who do sports constitute $\\frac{3}{5}(1 - \\frac{5}{9}) = \\frac{4}{15}$ of all pupils. Hence girls who do sports but do not go to math club constitute $\\frac{4}{15} - \\frac{2}{15} = \\frac{2}{15}$ of all pupils. Girls who neither do sports nor go to math club constitute $\\frac{1}{2} \\cdot \\frac{2}{15} = \\frac{1}{15}$ of all pupils. Thus girls not going to math club constitute $\\frac{2}{15} + \\frac{1}{15} = \\frac{1}{5}$ of all pupils. Other girls who constitute $\\frac{1}{2} - \\frac{1}{5} = \\frac{3}{10}$ of all pupils go to math club. They constitute $\\frac{3}{10} : \\frac{1}{2} = \\frac{3}{5} = 60\\%$ of all girls.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 29905,
    "subject": "Mathematics (Multi-modal)",
    "question": "Does there exist an integer $n \\ge 3$ such that some 3 diagonals of a regular $n$-gon meet in one point that is neither a vertex nor the center of the $n$-gon? If yes then find the least such $n$.",
    "options": [],
    "answer": "8",
    "solution": "If 3 diagonals of an $n$-gon meet in one point that is not a vertex of the $n$-gon then these diagonals have 6 endpoints in total, implying $n \\ge 6$. If $n = 6$ then the only way to leave the endpoints of every two diagonals to different sides of the third diagonal is connecting each vertex to the opposite one (Fig. 17), but the obtained diagonals meet in the center of the $n$-gon. Suppose that there exist 3 diagonals satisfying the conditions for $n = 7$. Let $A$ be the vertex that is not an endpoint of any of the diagonals. Let $B$ be a vertex next to $A$ and let $C$ be the other endpoint of the diagonal whose one endpoint is $B$. Two endpoints of diagonals must lie on the same side of $BC$ as $A$ and two endpoints must lie on the other side. There is only one possibility to connect these points with two intersecting diagonals (Fig. 18). As these diagonals are symmetric w.r.t. the perpendicular bisector of $AB$, their common point $P$ lies on the perpendicular bisector of $AB$. As $C$ also lies on the perpendicular bisector and $C \\neq P$, diagonal $BC$ could pass through $P$ only if $B$ were also located on the perpendicular bisector of $BC$, which is not the case. Hence finding the required 3 diagonals is impossible for $n = 7$.\n\nFor $n = 8$, draw one diagonal from some vertex to the opposite vertex. Adding two shorter diagonals symmetrically w.r.t. the first diagonal, all three intersect in one point inside the polygon that is not its center. (Fig. 19).\n\n![](attached_image_1.png)\nFig. 17\n\n![](attached_image_2.png)\nFig. 18\n\n![](attached_image_3.png)\nFig. 19",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 29906,
    "subject": "Mathematics (Multi-modal)",
    "question": "Prove or disprove: For every integer $n > 0$, there is a polynomial $a_n x^n + a_{n-1} x^{n-1} + \\dots + a_1 x + a_0$ satisfying the following conditions:\n(1) All coefficients $a_n, a_{n-1}, \\dots, a_1, a_0$ are positive real numbers;\n(2) At least one of the coefficients is $\\frac{1}{n}$;\n(3) The value of the polynomial is integer for every integer $x$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "The product of any $n$ consecutive integers is divisible by $n$. Thus the conditions are satisfied by the polynomial obtained by removing parentheses and collecting terms in the expression $\\frac{1}{n}(x+1)\\dots(x+n)$.",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 29907,
    "subject": "Mathematics (Multi-modal)",
    "question": "a. Given a convex quadrilateral $ABCD$ with $AD < AB$ and $CD < CB$, is the internal angle at $B$ always less than the internal angle at $D$?\n\nb. The same question for a non-convex quadrilateral.",
    "options": [],
    "answer": "a: Yes. b: No.",
    "solution": "a.\nConsider the triangles $ADB$ and $CDB$ (Fig. 21). The claim $AD < AB$ implies $\\angle ABD < \\angle ADB$ because the longer side is opposite to the larger angle. Similarly, $CD < CB$ implies $\\angle CBD < \\angle CDB$. As $ABCD$ is convex, $\\angle ABD + \\angle CBD = \\angle ABC$ and $\\angle ADB + \\angle CDB = \\angle ADC$. Hence, adding the two inequalities gives $\\angle ABC < \\angle ADC$.\n\n![](attached_image_1.png)\nFig. 21\n\nb.\nLet points $A, B, C$ be such that $AB = BC > AC$. Choose point $D'$ on the line tangent to the circumcircle of the triangle $ABC$ at $A$ in such a way that $B$ and $D'$ lie on the same side of $AC$ and the inequalities $AD' < AB$ and $CD' < BC$ hold (the last inequality is possible since $AC < BC$); let $D$ be the reflection of $D'$ from $AC$ (Fig. 22). Then both assumptions $AD < AB$ and $CD < CB$ hold. But the claim $\\angle ABC < \\angle ADC$ is not true: since $D'$ lies outside the circumference of triangle $ABC$, we have $\\angle ABC > \\angle AD'C = \\angle ADC$. Hence the hypothesis does not hold for non-convex quadrilaterals.\n\n![](attached_image_2.png)\nFig. 22",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 29908,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all positive integers $n$ such that one can choose $n-3$ non-intersecting diagonals of a regular $n$-gon that divide the $n$-gon into triangles in such a way that every chosen diagonal is a side of minimal length in some triangle.",
    "options": [],
    "answer": "n = 3 · 2^k for natural k",
    "solution": "Let $\\triangle$ be the triangle containing the center $O$ of the $n$-gon (colored in Fig. 23; if the center lies on a diagonal then choose either of the triangles having this diagonal as a side). Let $d$ be any side of $\\triangle$. If $d$ is a diagonal of the $n$-gon then $d$ separates $\\triangle$ from a neighboring triangle whose other two sides are shorter than $d$. By assumption, $d$ has to be a side of minimal length in $\\triangle$. But if $d$ is a side of the $n$-gon then $d$ is also a side of minimal length in $\\triangle$. Thus all\nsides of $\\triangle$ are equally minimal, meaning that $\\triangle$ is equilateral. Consequently, there is a constant number of sides of $n$-gon between the endpoints of every side of $\\triangle$. Hence $n = 3s_0$ for a positive integer $s_0$.\n\n![](attached_image_1.png)\nFig. 23\n\nConsider now an arbitrary triangle $\\Delta'$ neighboring $\\Delta$. Let $d'$ be any of its two sides not common with $\\Delta$. If $d'$ is a diagonal of the $n$-gon then $d'$ separates $\\Delta'$ from a third triangle whose other sides are shorter than $d'$. Thus $d'$ must be a side of minimal length in $\\Delta'$. But if $d'$ is a side of the $n$-gon then $d'$ is also a side of minimal length in $\\Delta'$. Hence the sides of $\\Delta'$ not common with $\\Delta$ have equal length and there must be the same number of sides of the $n$-gon between the endpoints of these sides of $\\Delta'$. Consequently $s_0 = 2s_1$ where $s_1$ is a positive integer.\nIf the sides of equal length of $\\Delta'$ are sides of the $n$-gon then $s_1 = 1$ and $n = 3 \\cdot 2$. Otherwise we can continue similarly to get $s_1 = 2s_2$ where either $s_2 = 1$ or $s_2 = 2s_3$, etc. Thus $n = 3 \\cdot 2^k$ for a natural number $k$. A construction for every $n$ of the form $3 \\cdot 2^k$ follows from the argumentation.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 29909,
    "subject": "Mathematics (Multi-modal)",
    "question": "Four boxes with masks and means of disinfection are transported to participants of a carnival. Weighing the boxes pairwise results in six quantities, four largest of which are $125$ kg, $120$ kg, $110$ kg and $101$ kg. Find all possibilities of what can be the weights of the four boxes.",
    "options": [],
    "answer": "Either 38 kg, 48 kg, 53 kg, and 72 kg; or 33.5 kg, 52.5 kg, 57.5 kg, and 67.5 kg.",
    "solution": "Let the masses of the boxes be $a$, $b$, $c$ and $d$ kilograms, whereby $a \\le b \\le c \\le d$. Clearly the largest result of pairwise weighing occurs in the case of two heaviest boxes, i.e., $c+d = 125$. The second largest result occurs in the case of the heaviest and the third heaviest box, i.e., $b+d = 120$, because $a+c \\le b+c \\le b+d$ and $a+d \\le b+d$. Similarly we see that the smallest result occurs in the case of two lightest boxes and the second smallest result occurs in the case of the lightest and the third lightest boxes. Thus the total mass of the second and the third heaviest boxes is one of the middle values, i.e., $b+c$ is either $101$ or $110$. Consider these cases one by one, taking into account that $a+d$ must be the other among $101$ and $110$.\n\n* Let $b+c = 101$. As $c = 125-d$ and $b = 120-d$, we obtain $(120-d)+(125-d) = 101$, implying $d=72$. Thus $c = 125-72 = 53$, $b = 120-72 = 48$ and $a = 110-72 = 38$.\n\n* Let $b+c = 110$. As $c = 125-d$ and $b = 120-d$, we obtain $(120-d)+(125-d) = 110$, implying $d=67.5$. Thus $c = 125-67.5 = 57.5$, $b = 120-67.5 = 52.5$ and $a = 101-67.5 = 33.5$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 29910,
    "subject": "Mathematics (Multi-modal)",
    "question": "Jüri writes on blackboard some consecutive integers. It is known that the total number of these integers is greater than one and the least of them is greater than $2$. Mari writes on blackboard consecutive integers, too, with the same total number of them as Jüri, but the least of them equals $1$. Is it possible that the product of the integers written by Jüri divided by the product of the integers written by Mari is equal to the square of some integer?",
    "options": [],
    "answer": "Yes",
    "solution": "For example, the product $8 \\cdot 9$ of two consecutive integers divided by the product $1 \\cdot 2$ of the first two positive integers equals $6^2$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 29911,
    "subject": "Mathematics (Multi-modal)",
    "question": "Some numbers are printed on a spool of paper, one below another. Starting from the third number, each number is the sum of the two preceding numbers. The number $2018$ occurs fourth, the number $2020$ occurs eighth. Which are the first and the second numbers?",
    "options": [],
    "answer": "first number = -7398, second number = 4708",
    "solution": "Let the first and the second number be $x$ and $y$, respectively. Computing the following numbers, we see that the fourth number is $x + 2y$ and the eighth one is $8x + 13y$. The conditions of the problem imply the system of equations\n$$\n\\begin{cases}\nx + 2y = 2018, \\\\\n8x + 13y = 2020,\n\\end{cases}\n$$\nsolving of which gives $x = -7398$ and $y = 4708$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 29912,
    "subject": "Mathematics (Multi-modal)",
    "question": "Circle $\\omega_2$ is tangent to circle $\\omega_1$ at point $A$ and passes through its center $O$. Point $C$ is chosen on $\\omega_2$ in such a way that the ray $AC$ intersects $\\omega_1$ the second time at point $D$, the ray $OC$ intersects $\\omega_1$ at point $E$ and the line $DE$ is parallel to the line $AO$. Find the size of the angle $DAE$.",
    "options": [],
    "answer": "30°",
    "solution": "Denote $\\angle DAE = \\alpha$; then $\\angle DOE = 2\\alpha$ (Fig. 5). From the isosceles triangle $DOE$, we obtain $\\angle OED = \\frac{180^\\circ - 2\\alpha}{2} = 90^\\circ - \\alpha$. Since $DE$ and $AO$ are parallel, $\\angle AOE = 90^\\circ - \\alpha$.\n\nAs the common tangent to $\\omega_1$ and $\\omega_2$ at point $A$ is perpendicular to the radius $AO$ of $\\omega_1$, as well as to the radius of $\\omega_2$, and $O$ lies on $\\omega_2$, the line segment $AO$ must be a diameter of $\\omega_2$. Thus $\\angle OCA = 90^\\circ$. As $OA = OD$, the line segment $OC$ is the altitude of the isosceles triangle $OAD$ drawn from its apex angle. This implies $\\angle AOE = \\angle DOE = 2\\alpha$.\nHence $90^\\circ - \\alpha = 2\\alpha$, implying $\\alpha = 30^\\circ$.\nAs $\\angle ACO = \\angle DCE$ and also $\\angle AOC = \\angle DEC$, the triangles $ACO$ and $DCE$ are similar.\n\n![](attached_image_1.png)\nFig. 5\n\nAs the common tangent to $\\omega_1$ and $\\omega_2$ at point $A$ is perpendicular to the radius $AO$ of $\\omega_1$, as well as to the radius of $\\omega_2$, and $O$ lies on $\\omega_2$, the line segment $AO$ must be a diameter of $\\omega_2$. Thus $\\angle OCA = 90^\\circ$. As $OA = OD$, the line segment $OC$ is the altitude of the isosceles triangle $OAD$ drawn from its apex angle. This implies $AC = DC$. Hence the triangles $ACO$ and $DCE$ are equal and $AO = DE$.\nA quadrilateral with a pair of equal parallel sides is a parallelogram. As $AO = DO$, the quadrilateral $AODE$ is a rhombus. The diagonals of a rhombus bisect the angles at their endpoints. Thus $\\angle EAD = \\frac{1}{2}\\angle EAO$. As $OA = OE = OD$ and $AE = OD$, the triangle $AOE$ is equilateral. Hence the internal angles of $AODE$ are of size $60^\\circ$ and $120^\\circ$, implying $\\angle EAD = 30^\\circ$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 29913,
    "subject": "Mathematics (Multi-modal)",
    "question": "Integers $a$, $b$, $c$ and $n$ are given such that $1 \\le a < b < c \\le n$. Juku and Miku play the following game on a strip of size $1 \\times n$: In the beginning, squares number $a$, $b$, $c$ contain one piece each, whereby the squares are numbered from the right to the left by consecutive integers starting from $1$. On one's move, each player chooses one piece out of these three and shifts it one or more squares to the right. However, it is not allowed to move a piece to a square that contains another piece or jump over such a square; one also must not move a piece off the strip. Players move by turns, with Juku moving first. The player who cannot move loses. Which player can win regardless of the opponent's play?",
    "options": [],
    "answer": "Juku wins if c − b ≠ a; Miku wins if c − b = a.",
    "solution": "Firstly, note that, in any position where the number of empty squares between the leftmost and the middle piece differs from the number of empty squares in the right from the rightmost piece, one can make a move that makes these two quantities equal. Indeed, if the number of empty squares between the leftmost and the middle piece is greater than the number of empty squares right from the rightmost piece then one can move the leftmost piece, otherwise one can move the rightmost piece.\n\nSecondly, note that every move in any position where the number of empty squares between the leftmost and the middle piece equals the number of empty squares in the right from the rightmost piece makes these two quantities different. Indeed, moving either the leftmost or the middle piece changes the number of empty squares between the leftmost and the middle piece while leaving the empty squares right from the rightmost piece unchanged; when moving the rightmost piece, it is the other way round.\n\nConsequently, Juku can win if $c - b \\neq a$ by always moving in such a way that the number of empty squares between the leftmost and the middle piece were equal to the number of empty squares right from the rightmost piece after his move. As the sum of distances of all three pieces from the right edge of the strip decreases at each move, the game must eventually end and, by the considerations above, only Miku can lose. On the other hand, if $c - b = a$ then after Juku's move the number of empty squares between the leftmost and the middle piece differs from the number of empty squares right from the rightmost piece. Analogously to the previous case, Miku can win in this position.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 29914,
    "subject": "Mathematics (Multi-modal)",
    "question": "One writes distinct positive integers into the cells of a $3 \\times 3$ table in such a way that, in each row and in each column, one number equals the sum of the other two numbers. Find the least possible total sum of the numbers written into the table.",
    "options": [],
    "answer": "46",
    "solution": "The sum of all numbers written in the cells is at least $1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 = 45$. As one number is the sum of others in every row, the sum of all numbers in every row is even. Thus the sum of all numbers in the table must be even, too. Hence the sum of all numbers in the table cannot be 45. It is possible that the sum is 46, as Fig. 6 shows.\n\n![](attached_image_1.png)\nFig. 6",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 29915,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all solutions of the equation $x^3 + 3xy + y^3 = 2019$ in integers.",
    "options": [],
    "answer": "no integer solutions",
    "solution": "The r.h.s. of the equation is divisible by $3$ but not by $9$. Assume that $3 \\mid x$. Then $9 \\mid x^3$ and $9 \\mid 3xy$. If also $3 \\mid y$ then $9 \\mid y^3$, implying that the l.h.s. of the equation is divisible by $9$. Thus $3 \\nmid y$. But then $3 \\nmid y^3$, implying that the l.h.s. of the equation is not divisible by $3$. The contradiction shows that $3 \\nmid x$. By symmetry, also $3 \\nmid y$.\nAs $3 \\mid 3xy$, we must have $3 \\mid x^3 + y^3$. Hence also $3 \\mid x^3 + 3x^2y + 3xy^2 + y^3 = (x+y)^3$, implying $3 \\mid x+y$. Consequently, $x$ and $y$ are modulo $3$ congruent to $1$ and $2$ in some order. Hence $x^2 \\equiv y^2 \\equiv 1 \\pmod{3}$ and $xy \\equiv 2 \\pmod{3}$, implying $x^2 - xy + y^2 \\equiv 0 \\pmod{3}$. We obtain $9 \\mid x^3 + y^3 = (x+y)(x^2 - xy + y^2)$, whereas $3xy \\equiv 6 \\pmod{9}$. Hence the l.h.s. of the equation is congruent to $6$ modulo $9$ but the r.h.s. of the equation is congruent to $3$. Consequently, there are no solutions.\nDenote $a = x+y$. Then the given equation is equivalent to $a^3 + 3xy(1-a) = 2019$. As $3 \\mid 2019$ and $3 \\mid 3xy(1-a)$, we must have $3 \\mid a^3$, implying $3 \\mid a$. The given equation is also equivalent to\n$$\n(a-1)(a^2 + a + 1 - 3xy) = 2018. \\quad (1)\n$$\nHence $a-1 \\mid 2018$. As $2018 = 2 \\cdot 1009$ where $1009$ is prime, $a-1$ must be one of $2018$, $1009$, $2$, $1$, $-1$, $-2$, $-1009$, $-2018$. Taking into account that $3 \\nmid a$, we obtain four cases:\n* If $a = 2019$ then $y = 2019 - x$ and substituting into (1) gives $2019^2 + 2019 + 1 - 3x(2019 - x) = 1$ which is equivalent to $x^2 - 2019x + 673 \\cdot 2020 = 0$. The latter equation has no real solutions.\n* If $a = 3$ then $y = 3 - x$ and substituting into (1) gives $13 - 3x(3 - x) = 1009$ which is equivalent to $x^2 - 3x - 332 = 0$. The latter equation has no integral solutions.\n* If $a = 0$ then $y = -x$ and the l.h.s. of the initial equation is non-positive.\n* If $a = -1008$ then $y = -1008 - x$ and substituting into (1) gives $1008^2 - 1008 + 1 + 3x(1008 + x) = -2$ which is equivalent to $x^2 + 1008x + 336 \\cdot 1007 + 1 = 0$. The latter equation has no real solutions.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 29916,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $n$ be a positive integer. Real numbers $a_1, a_2, \\dots, a_{2n}$ satisfy the following conditions:\n\n(1) For every $i = 1, 2, \\dots, 2n - 1$, one has $0 < a_{i+1} - a_i \\le 1$;\n\n(2) Rounding the numbers $a_1, a_2, \\dots, a_{2n}$ to the closest integer (numbers equidistant from two closest integers are rounded up) gives pairwise distinct positive integers.\n\nNumbers $a_1, a_2, \\dots, a_{2n}$ are placed as the numerators and denominators of $n$ fractions. Prove that the sum of the obtained fractions is greater than $\\frac{n}{4}$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Suppose that a numerator is greater than a denominator. Then interchanging these two numbers makes both fractions smaller. Thus we can assume w.l.o.g. that all numerators are less than all denominators.\n\nFor every $i = 1, 2, \\dots, 2n$, define $x_i = a_i - i + \\frac{1}{2}$. As $a_1 < a_2 < \\dots < a_{2n}$ and rounding the numbers $a_i$ produce pairwise distinct positive integers, we must have $a_i \\ge i - \\frac{1}{2}$ which implies $x_i \\ge 0$. From $a_{i+1} - a_i \\le 1$, we have $x_{i+1} - x_i = a_{i+1} - a_i - 1 \\le 0$ which implies $x_1 \\ge x_2 \\ge \\dots \\ge x_{2n}$.\n\nHence $i < j$ always implies $\\frac{a_i - x_j}{a_j - x_j} \\ge \\frac{a_i - x_j}{a_j - x_j} \\ge \\frac{a_i - x_i}{a_j - x_j} = \\frac{2i-1}{2j-1}$ (the first inequality holds because of $a_i < a_j$ and $x_j \\ge 0$ while the second inequality holds because of $x_i \\ge x_j$). Thus it suffices to prove the desired inequality for the case where the numerators are integers $1, 3, \\dots, 2n - 1$ and the denominators are integers $2n + 1, 2n + 3, \\dots, 4n - 1$ in some order.\n\nDenote the sum of all fractions by $s$. Applying AM-GM to the fractions gives $\\frac{s}{n} \\ge \\sqrt[n]{\\frac{1 \\cdot 3 \\cdots (2n-1)}{(2n+1)(2n+3)\\cdots(4n-1)}}$. Thus it suffices to prove for every $n$ the inequality $\\frac{1 \\cdot 3 \\cdots (2n-1)}{(2n+1)(2n+3)\\cdots(4n-1)} > \\frac{1}{4^n}$. We can do it by induction on $n$.\n\nThe claim holds for $n = 1$ since $\\frac{1}{3} > \\frac{1}{4}$. For the induction step, it suffices to show that $\\frac{(2n+1)^2}{(4n+1)(4n+3)} > \\frac{1}{4}$ which is equivalent to $(4n+2)^2 > (4n+1)(4n+3)$. The latter follows from AM-GM for $4n+1$ and $4n+3$.\nFirstly, show as in Solution 1 that it suffices to prove the desired inequality for the case where the numerators are $1, 3, \\dots, 2n-1$ and the denominators are $2n+1, 2n+3, \\dots, 4n-1$ in some order. By applying the rearrangement inequality for $1 < 3 < \\dots < 2n-1$ and $\\frac{1}{2n+1} > \\frac{1}{2n+3} > \\dots > \\frac{1}{4n-1}$, we conclude that the least sum of fractions is obtained if both numerators and denominators are in the increasing order. Hence it suffices to show that $\\frac{1}{2n+1} + \\frac{3}{2n+3} + \\dots + \\frac{2n-1}{4n-1} > \\frac{n}{4}$.\n\nPut the first summand together with the last one, the second one together with the second last one, etc. The fractions of every pair are of the form $\\frac{n-k}{3n-k}$ and $\\frac{n+k}{3n+k}$ where $-n < k < n$. The sum of these two terms is $\\frac{6n^2-2k^2}{9n^2-k^2}$. As $-n < k < n$, we must have $k^2 < n^2$; we also see that both $6n^2-2k^2$ and $9n^2-k^2$ are positive. Thus the inequality $9n^2-k^2 < 12n^2-4k^2$, being equivalent to the valid inequality $k^2 < n^2$, is also equivalent to $\\frac{6n^2-2k^2}{9n^2-k^2} > \\frac{1}{2}$. Hence the sum of members of all $n$ pairs is greater than $\\frac{n}{2}$, and as each fraction occurs twice in these pairs, the sum of all fractions is greater than $\\frac{n}{4}$.\nFirstly, show as in Solution 1 that it suffices to prove the desired inequality for the case where the numerators are $1, 3, \\dots, 2n-1$ and the denominators are $2n+1, 2n+3, \\dots, 4n-1$ in some order. Replacing all denominators with their strict upper bound $4n$, all fractions become smaller and so does their sum. The sum of numerators $1+3+\\dots+(2n-1)$ equals $n^2$. Hence the sum of all fractions is greater than $\\frac{n^2}{4n}$, i.e., than $\\frac{n}{4}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 29917,
    "subject": "Mathematics (Multi-modal)",
    "question": "A circle $c$ with center $A$ passes through the vertices $B$ and $E$ of a regular pentagon $ABCDE$. The line $BC$ intersects the circle $c$ the second time at point $F$. Point $G$ on the circle $c$ is chosen in such a way that $FB = FG$ and $B \\neq G$. Prove that the lines $AB$, $EF$ and $DG$ meet in a common point.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "The internal angles of a regular pentagon have size $108^\\circ$. Thus $\\angle ABC = 108^\\circ$, implying $\\angle ABF = 72^\\circ$ (Fig. 7). As $AB = AF$, we have $\\angle AFB = 72^\\circ$ and $\\angle BAF = 36^\\circ$. Since $FG = FB$, $AG = AB$ and $AF = AF$, the triangles $AFB$ and $AFG$ are equal, implying that $\\angle FAG = 36^\\circ$. As $\\angle EAB = 108^\\circ$, we obtain $\\angle GAE = 2 \\cdot 36^\\circ + 108^\\circ = 180^\\circ$, implying that $E$, $A$ and $G$ are collinear.\n\nLet the lines $AB$ and $EF$ meet at point $K$ (Fig. 8). As $\\angle AFC = 180^\\circ - \\angle FCD$, we have $AF \\parallel CD$. Consequently also $AF \\parallel BE$, whence\n![](attached_image_1.png)\nFig. 7\n\n![](attached_image_2.png)\nFig. 8\n\nsimilarity of triangles $BKE$ and $AKF$ implies $\\frac{BK}{AK} = \\frac{BE}{AF}$.\nLet the lines $AB$ and $DG$ meet at point $K'$ (Fig. 9). Lines $AG$ and $AE$ coincide because $E$, $A$ and $G$ are collinear. Consequently, $BD \\parallel AG$, whence similarity of triangles $BK'D$ and $AK'G$ implies $\\frac{BK'}{AK'} = \\frac{BD}{AG}$.\nSince $BD = BE$ and $AF = AG$, we have $\\frac{BK}{AK} = \\frac{BK'}{AK'}$, implying $K = K'$.\nThe internal angles of a regular pentagon have size $108^\\circ$. Thus $\\angle ABC = 108^\\circ$ (Fig. 10), implying $\\angle ABF = 72^\\circ$. As $AB = AF$, we have $\\angle AFB = 72^\\circ$. Since $FG = FB$, $AG = AB$ and $AF = AF$, the triangles $AFB$ and $AFG$ are equal, implying also $\\angle AFG = 72^\\circ$. Therefore $\\angle CFG = \\angle BFG = 2 \\cdot 72^\\circ = 144^\\circ$, whereas $\\angle FCA = \\angle BCA = \\frac{180^\\circ - \\angle ABC}{2} = 36^\\circ = 180^\\circ - 144^\\circ$. Consequently, $GF \\parallel CA$, implying also $GF \\parallel DE$.\nLet the lines $AB$ and $EF$ meet at point $K$ (Fig. 11). As $CE \\parallel BA$, we obtain $\\frac{BF}{KF} = \\frac{BC}{KE}$. By assumptions, $BF = GF$ and $BC = DE$; hence also $\\frac{GF}{KF} = \\frac{DE}{KE}$. As $\\angle KFG = \\angle KED$ because of $GF \\parallel DE$, the triangles $KFG$ and $KED$ are similar. Thus $\\angle FKG = \\angle EKD$. Consequently, points $G$, $K$ and $D$ are collinear, meaning that the line $DG$ passes through $K$.\n\n![](attached_image_3.png)\nFig. 10\n\n![](attached_image_4.png)\nFig. 11\n\n*Remark:* The claim follows directly from Desargues's theorem.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 29918,
    "subject": "Mathematics (Multi-modal)",
    "question": "Given a tuple of consecutive positive integers, one forms all pairs of members of it such that the first member is less than the second member. The percentage of these pairs where the second member is divisible by the first one is called the *degree of divisibility* of the tuple. For every integer $n > 1$, denote the largest possible degree of divisibility of a tuple of $n$ consecutive positive integers by $j(n)$.\nDoes there exist an integer $n > 1$ such that $j(n + 1) > j(n)$?",
    "options": [],
    "answer": "Yes; for n = 5 we have j(6) > j(5), with j(5) = 1/2 and j(6) = 8/15.",
    "solution": "The largest percentage of pairs with the second term being divisible by the first term is achieved in the case of tuple $(1, 2, \\dots, n)$. Indeed, consider an arbitrary tuple of the form $(x+1, x+2, \\dots, x+n)$ where\n\n$x > 0$. For any $i$, multiples of $i$ in $(1,2,...,n)$ are every $i$th term starting from the number $i$, multiples of $x+i$ in $(x+1,x+2,...,x+n)$ are every $(x+i)$th term starting from $x+i$. The latter multiples occuring more seldom while the first occurrence being at the same position implies the same or smaller total number.\nIt is easy to check that the degree of divisibility of $(1,2,3,4,5)$ is $\\frac{5}{10}$ and the degree of divisibility of $(1,2,3,4,5,6)$ is $\\frac{8}{15}$. By the above, $j(6) > j(5)$.\nCalculation shows that the degrees of divisibility of tuples $(1,2,3,4,5)$, $(2,3,4,5,6)$, $(3,4,5,6,7)$ and $(4,5,6,7,8)$ are $\\frac{5}{10}$, $\\frac{3}{10}$, $\\frac{1}{10}$ and $\\frac{1}{10}$, respectively. If the first term of a quintuple is 5 or larger then no two terms can divide each other since the largest term is less than twice larger than the least term. Hence $j(5) = \\frac{5}{10} = \\frac{1}{2}$. As the degree of divisibility of $(1,2,3,4,5,6)$ is $\\frac{8}{15}$, we have $j(6) \\ge \\frac{8}{15} > \\frac{1}{2} = j(5)$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 29919,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all triples $(x, y, z)$ of real numbers that satisfy the system of equations\n$$\n\\begin{cases} xy + x + y = z, \\\\ yz + y + z = x, \\\\ zx + z + x = y. \\end{cases}\n$$",
    "options": [],
    "answer": "[(-1, -1, -1), (0, 0, 0), (-2, -2, 0), (-2, 0, -2), (0, -2, -2)]",
    "solution": "Denote $x+1 = a$, $y+1 = b$ and $z+1 = c$. Adding 1 to the sides of all equations and factorizing in the left gives\n$$\n\\begin{cases} ab = c, \\\\ bc = a, \\\\ ca = b. \\end{cases} \\qquad (2)\n$$\nIf $a=0$ then the first and the third equation of (2) imply $b=0$ and $c=0$. Analogously, if $b=0$ or $c=0$ then $a, b, c$ are all equal to zero. This case leads to the solution $(-1,-1,-1)$ to the initial system. If none of $a, b, c$ is zero then pairwise multiplication of the equations of (2) and each time reducing similar factors from both sides gives $a^2 = b^2 = c^2 = 1$. Thus each of the numbers $a, b, c$ is either 1 or -1. The case $a=b=c=1$ satisfies (2) and leads to the solution $(0,0,0)$ of the initial system. The cases with exactly two of numbers $a, b, c$ being equal to -1 also satisfy (2) and lead to the solutions $(-2,-2,0)$, $(-2,0,-2)$ and $(0,-2,-2)$ of the initial system. Multiplying the corresponding sides of all equations of (2) and reducing by factor $abc$ gives $abc = 1$, whence there are no more solutions (the value -1 cannot occur an odd number of times).\nSubstituting $z$ from the first equation into the second one gives the equation $y(xy + x + y) + y + (xy + x + y) = x$ which is equivalent to $(x+1)y(y+2) = 0$. Thus we have $x = -1$ or $y = 0$ or $y = -2$.\n* If $x = -1$ then the first equation implies $z = -1$. Substituting into the third equation gives $y = -1$. Hence the solution $(-1,-1,-1)$.\n\n* If $y = 0$ then the first equation implies $z = x$. Substituting into the third equation gives $x^2 + 2x = 0$. Thus $x = z = 0$ or $x = z = -2$, leading to the solutions $(0, 0, 0)$ and $(-2, 0, -2)$.\n* If $y = -2$ then the first equation implies $z = -x - 2$. Substituting it into the third equation gives $x^2 + 2x = 0$. Thus either $x = 0$ and $z = -2$ or $x = -2$ and $z = 0$ which lead to the solutions $(0, -2, -2)$ and $(-2, -2, 0)$.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 29920,
    "subject": "Mathematics (Multi-modal)",
    "question": "Juku paints exactly 30 unit squares of a $10 \\times 15$ table black. After that, Miku covers up exactly 4 rows and 4 columns. Can Juku ensure by the choice of the squares to be coloured that at least 10 black unit squares are left uncovered?",
    "options": [],
    "answer": "Yes",
    "solution": "Juku can paint 30 unit squares in such a way that each row contains exactly 3 black unit squares and each column contains exactly 2 of them (Fig. 1). When Miku chooses 4 rows and 4 columns, they contain at most $4 \\cdot 3 + 4 \\cdot 2 = 20$ black unit squares altogether. Hence at least 10 black unit squares remain uncovered.\n\n![](attached_image_1.png)\nFig. 1",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 29921,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all pairs $(x, y)$ of integers such that\n$$\n\\sqrt{x+2019} - \\sqrt{x} = \\sqrt{y}.\n$$",
    "options": [],
    "answer": "(x, y) in {(0, 2019), (1009^2, 1), (336^2·3, 3), (335^2, 9), (673, 673)}",
    "solution": "As $x$ and $y$ occur under square root, only non-negative solutions can exist. Bringing $\\sqrt{x}$ to the right, squaring both sides and collecting similar terms gives $2019 = y + 2\\sqrt{xy}$ which is equivalent to the initial equation. Thus $2\\sqrt{xy}$ is an integer. If $x = 0$ then $y = 2019$, the case $y = 0$ leads to contradiction. Assume in the rest that both $x$ and $y$ are positive.\nLet $a^2$ and $b^2$ be the largest perfect squares dividing $x$ and $y$, respectively; then $x = a^2c$ and $y = b^2c'$ where both $c$ and $c'$ are square-free. As $a^2c \\cdot b^2c'$ is a perfect square, also $c \\cdot c'$ must be a perfect square; this is possible only if $c$ and $c'$ have the same prime factors, i.e., $c = c'$. Hence\n$$2019 = b^2c + 2\\sqrt{a^2b^2c^2} = b^2c + 2abc = bc(2a + b).$$\nSince $2019 = 3 \\cdot 673$ where both factors are prime, we have the following cases, taking into account that $b < 2a + b$:\n* $b=1, c=1, 2a+b=2019$, implying $a=1009, x=1009^2$ and $y=1$;\n* $b=1, c=3, 2a+b=673$, implying $a=336, x=336^2 \\cdot 3$ and $y=3$;\n* $b=1, c=673, 2a+b=3$, implying $a=1, x=673$ and $y=673$;\n* $b=3, c=1, 2a+b=673$, implying $a=335, x=335^2$ and $y=9$.\nAs $x$ and $y$ occur under square root, only non-negative solutions can exist. Bringing $\\sqrt{x}$ to the right, squaring both sides and collecting similar terms gives $2019 - y = 2\\sqrt{xy}$ which is equivalent to the initial equation. Hence $y \\le 2019$. After squaring once more and rearranging terms, we obtain $2019^2 = y(2 \\cdot 2019 - y + 4x)$. Thus $y \\mid 2019^2$. Case study leads to the following solutions: (1) $y = 1, x = 1009^2$; (2) $y = 3, x = 336^2 \\cdot 3$; (3) $y = 9, x = 335^2$; (4) $y = 673, x = 673$; (5) $y = 2019, x = 0$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 29922,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all functions $f: \\mathbb{R} \\to \\mathbb{R}$ that satisfy the equation\n$$\nf(xf(y) + y) = f(x^2 + y^2) + f(y)\n$$\nfor all real numbers $x$ and $y$.",
    "options": [],
    "answer": "f(x) = 0 for all real x",
    "solution": "Substituting $x = 0$ into the equation gives $f(y) = f(y^2) + f(y)$, implying $f(y^2) = 0$. Thus $f(x) = 0$ for all non-negative real numbers $x$. In particular $f(x^2 + y^2) = 0$, which allows to simplify the initial equation as\n$$\nf(xf(y) + y) = f(y).\n$$\nSuppose that $f(c) \\neq 0$ for some negative real number $c$. Substituting $y = c$ into (3) gives $f(z) = f(c)$ for all $z$ because the expression $xf(c) + c$ obtains all real values. By the above, there exists $z$ such that $f(z) = 0$; hence $f(c) = 0$, contradicting the choice of $c$. Consequently, $f(x) = 0$ for every real number $x$. Clearly the function $f(x) = 0$ satisfies the given equation.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 29923,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $n$ and $m$ be positive integers. On one turn, an $n$-$m$-knight can move either horizontally by $n$ squares and vertically by $m$ squares or vertically by $n$ squares and horizontally by $m$ squares. (For instance, the usual chess knight, all possible target squares of one move of which are depicted by bullets in the figure, is a $1$-$2$-knight.) Can an $n$-$m$-knight on an infinite in every direction chessboard return to the initial square in exactly $2019$ turns?",
    "options": [],
    "answer": "No",
    "solution": "Consider three cases:\n\n*Exactly one of the numbers $n$, $m$ is odd.* Color the squares like on a chessboard. Every move changes the color of the square where the knight is, whence after an odd number of moves, the knight is on a square of the opposite color. Thus the knight cannot be on the initial square after $2019$ moves.\n\n*Both numbers $n$ and $m$ are odd.* Color the horizontal lines of the board alternately black and white. Again, every move changes the color of the square where the knight is. Hence, similarly to the previous case, the knight cannot be on the initial square after $2019$ moves.\n\n*Both numbers $n$ and $m$ are even.* Let $n = 2^k p$, $m = 2^l q$ where $p$ and $q$ are odd. W.l.o.g., assume that $k \\le l$ and the knight starts from square $(0,0)$. Obviously, the knight only visits squares with coordinates of the form $(2^k u, 2^l v)$ since $2^k$ divides the length of the step in either direction. Suppose that the knight is on the initial square after $2019$ moves. Shortening all moves $2^k$ times while retaining their directions, we obtain a route of a $p$-$2^{l-k}q$-knight which in $2019$ moves returns to the initial square. But $p$ is odd which means that such route does not exist by previous cases. The contradiction shows that the knight cannot be on the initial square after $2019$ moves.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 29924,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all prime numbers $p$ such that $\\frac{p-1}{2}$ and $\\frac{p+1}{4}$ are prime numbers, too.",
    "options": [],
    "answer": "7 and 11",
    "solution": "Let $q = \\frac{p-1}{2}$ and $r = \\frac{p+1}{4}$; then $p = 4r-1$ and $q = \\frac{4r-2}{2} = 2r-1$. Consider all remainders that can be left when $r$ is divided by $3$:\n* If $r \\equiv 1 \\pmod{3}$ then $4r-1 \\equiv 0 \\pmod{3}$, i.e., $4r-1$ is divisible by $3$. Thus $p=3$. But then $r=1$ which is not a prime.\n* If $r \\equiv 2 \\pmod{3}$ then $2r-1 \\equiv 0 \\pmod{3}$, i.e., $2r-1$ is divisible by $3$. Thus $q=3$, whence $r=2$ and $p=7$. All three are primes indeed.\n* If $r \\equiv 0 \\pmod{3}$, i.e., $r$ is divisible by $3$, then $r=3$. Thus $q=5$ and $p=11$ which are primes, too.\nConsequently, $p$ can be either $7$ or $11$.\nOut of three consecutive integers $p-1$, $p$, $p+1$ one is divisible by $3$. Division by $2$ or $4$ does not change divisibility by $3$ since $2$ and $4$ are coprime with $3$. Thus also out of integers $p$, $\\frac{p-1}{2}$, $\\frac{p+1}{4}$ one is divisible by $3$. If these integers are prime, the one divisible by $3$ must be $3$. Consider all possible cases:\n* If $p = 3$ then $\\frac{p-1}{2} = 1$ but $1$ is not prime.\n* If $\\frac{p-1}{2} = 3$ then $p = 7$ and $\\frac{p+1}{4} = 2$. All three are primes indeed.\n* If $\\frac{p+1}{4} = 3$ then $p = 11$ and $\\frac{p-1}{2} = 5$. All three are primes indeed.\nConsequently, $p$ can be either $7$ or $11$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 29925,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find the value of the expression\n$$\n\\frac{1}{\\left(\\frac{1}{2019}\\right)^2 + 1} + \\frac{1}{\\left(\\frac{2}{2018}\\right)^2 + 1} + \\frac{1}{\\left(\\frac{3}{2017}\\right)^2 + 1} + \\dots + \\frac{1}{\\left(\\frac{2018}{2}\\right)^2 + 1} + \\frac{1}{\\left(\\frac{2019}{1}\\right)^2 + 1}\n$$",
    "options": [],
    "answer": "2019/2",
    "solution": "Group the summands into pairs: the first one together with the last one, the second one together with the second last one, etc. Adding the members of each pair gives us\n$$\n\\frac{1}{\\left(\\frac{i}{j}\\right)^2 + 1} + \\frac{1}{\\left(\\frac{j}{i}\\right)^2 + 1} = \\frac{\\frac{i^2}{j^2} + 1 + \\frac{j^2}{i^2} + 1}{\\left(\\frac{i^2}{j^2} + 1\\right) \\left(\\frac{j^2}{i^2} + 1\\right)} = \\frac{\\frac{i^2}{j^2} + \\frac{j^2}{i^2} + 2}{\\frac{i^2}{j^2} + \\frac{j^2}{i^2} + 2} = 1.\n$$\nAs there are $2019$ pairs in total, the sum of all numbers in the pairs is $2019$. Since every summand occurs twice, the desired sum is $\\frac{2019}{2}$.",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 29926,
    "subject": "Mathematics (Multi-modal)",
    "question": "A circle $c$ with center $A$ passes through the vertices $B$ and $E$ of a regular pentagon $ABCDE$. The line $BC$ intersects the circle $c$ the second time at point $F$. Prove that lines $DE$ and $EF$ are perpendicular.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "The internal angles of a regular pentagon have size $108^\\circ$. Thus $\\angle EAB = 108^\\circ$ (Fig. 2), whence $\\angle EFC = \\angle EFB = \\frac{\\angle EAB}{2} = 54^\\circ$. As $\\angle CDE = 108^\\circ$ and $\\angle FCD = \\angle BCD = 108^\\circ$, from the quadrilateral $CDEF$ we obtain $\\angle DEF = 360^\\circ - \\angle FCD - \\angle CDE - \\angle EFC = 90^\\circ$.\n\n![](attached_image_1.png)\nFig. 2\nThe internal angles of a regular pentagon have size $108^\\circ$. Thus $\\angle ABC = 108^\\circ$, whence $\\angle ABF = 180^\\circ - \\angle ABC = 72^\\circ$. As $AB = AF$, from the triangle $ABF$ we obtain $\\angle BAF = 180^\\circ - 2 \\cdot 72^\\circ = 36^\\circ$. Let $G$ be the second intersection point of line $DE$ with circle $c$ (Fig. 3); by symmetry, $\\angle EAG = \\angle BAF = 36^\\circ$. Since $\\angle EAB = 108^\\circ$, we have $\\angle FAG = \\angle BAF + \\angle EAB + \\angle EAG = 180^\\circ$, i.e., $FG$ is a diameter of $c$. Hence $\\angle FEG = 90^\\circ$.\n\n![](attached_image_2.png)\nFig. 3",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 29927,
    "subject": "Mathematics (Multi-modal)",
    "question": "Given is an $m \\times m$ table with $2n$ distinct unit squares marked with a ring ($2n \\le m^2$). Juku wishes to connect these $2n$ rings into pairs using $n$ (possibly curved) lines in a way that meets the following conditions:\n(1) Each line begins from some ring and ends in some other ring;\n(2) Every two unit squares visited by the same line one after another have a common side;\n(3) No two lines (including their endpoints) visit a common unit square;\n(4) No line visits the same unit square more than once.\nProve that the sum of the numbers of unit squares visited by the lines is either always even or always odd, no matter of how Juku draws the lines.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Color the unit squares black and white in such a way that unit squares with a common side are of different color. Each line goes from a black square to a white square and vice versa; thus whenever the endpoints of a line are in squares of equal color, the line visits an odd number of squares, and otherwise, the line visits an even number of squares. Let $k$ squares out of the ones marked with ring be black. Among the lines drawn by Juku, let $a$ lines have both endpoints in black squares, $b$ lines have both endpoints in white squares, and $c$ lines have endpoints in squares of different color. Then $2a + c = k$ and $a + b + c = n$, implying $a + b = n - k + 2a$. Hence the numbers $a+b$ and $n-k$ have equal parity.\nLet the numbers of squares visited by the lines sum up to $s$. Then $s$ can be expressed as the sum of $a+b$ odd numbers and $c$ even numbers, whence $s$ and $a+b$ have equal parity. Consequently, $s$ and $n-k$ have equal parity. Since $n-k$ does not depend on the way Juku draws the lines, $s$ must be always even or always odd.\nEvery line can be considered as a sequence of unit movements, each having one of four possible directions (right, left, up, down). Thus the number of unit squares visited by a line is $k+1$ where $k$ is the number of unit movements. Let $a, b, c, d$ be the numbers of unit movements of the line in different directions (right, left, up and down, respectively); then the end of the line is located $a-b$ units to the right and $c-d$ units upwards from the beginning of the line. Since $a-b$ and $a+b$ have equal parity, as do $c-d$ and $c+d$, the numbers $k = (a+b) + (c+d)$ and $(a-b) + (c-d)$ have equal parity.\nLet the line start in column $x_A$ and row $y_A$ and end in column $x_B$ and row $y_B$. Then $x_B - x_A = a - b$ and $y_B - y_A = c - d$, whence the number of unit squares visited by the line has the same parity as the number $x_B - x_A + y_B - y_A + 1$. But the latter has the same parity as $x_B + x_A + y_B + y_A + 1$. Summing up these numbers for all lines, we obtain that the total number of unit squares visited by the lines has the same parity as the number $s_x + s_y + n$ where $s_x$ and $s_y$ are the sum of all column numbers and all row numbers, respectively, of the squares containing a ring. But this sum does not depend on the way how Juku connects the rings, which proves the desired claim.",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 29928,
    "subject": "Mathematics (Multi-modal)",
    "question": "For which positive integers $n$ can one exactly cover an equilateral triangle of side length $n$ by trapeziums of the shape shown in the figure, consisting of three equilateral triangles of side length 1? Trapeziums are allowed to be rotated but not to cover each other.\n\n![](attached_image_1.png)",
    "options": [],
    "answer": "All positive integers divisible by 3",
    "solution": "An equilateral triangle of side length 3 can be covered by three trapeziums (Fig. 4). All equilateral triangles with side length being divisible by 3 can be partitioned into equilateral triangles of side length 3. Hence all equilateral triangles with side length being divisible by 3 can be covered by trapeziums of given shape.\n\n![](attached_image_2.png)\nFig. 4\n\nOn the other hand, whenever one partitions an equilateral triangle of side length $n$ into equilateral triangles of side length 1, the number of the small triangles is $n^2$ since multiplying the side length by $n$ causes the area to increase $n^2$ times. Consequently, the desired covering is possible only if the number $n^2$ is divisible by 3. The latter condition implies that $n$ must be divisible by 3 since 3 is prime.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 29929,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find the largest remainder that can be left over when dividing the number $2019$ by a three-digit natural number.",
    "options": [],
    "answer": "671",
    "solution": "If $673 < m < 1000$ then dividing $2019$ by $m$ gives quotient $2$ and remainder $2019 - 2m$. Obviously the remainder increases when $m$ decreases. Thus in the case $m = 674$ we obtain the largest remainder $671$.\n\nDividing $2019$ by $673$ gives remainder $0$. Dividing $2019$ by $672$ or any smaller number gives remainder that does not exceed $671$. Consequently, the largest remainder under the given conditions is $671$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 29930,
    "subject": "Mathematics (Multi-modal)",
    "question": "Gandalf the Wizard added to his arsenal of magic a new trick in which he simultaneously turns each integer into some integer different from it. Call an integer *a reflecting* if, for every integer $x$, the numbers $x$ and $a - x$ are turned into integers equal to each other. Is it possible that:\n\na. Numbers $1001$ and $1003$ are both reflecting;\n\nb. Numbers $1000$, $1003$ and $1008$ are all reflecting;\n\nc. Numbers $1002$, $1004$ and $1006$ are all reflecting?",
    "options": [],
    "answer": "a: no; b: no; c: yes",
    "solution": "For every integer $x$, denote by $G(x)$ the number into which Gandalf turns the number $x$.\n\na. Suppose that both $1001$ and $1003$ are reflecting. Then, for every integer $x$, we have $G(x+2) = G(1003 - (x+2)) = G(1001 - x) = G(x)$. Hence Gandalf turns all even numbers into one and the same integer $c$ and all odd numbers into one and the same integer $c'$. But $c = G(500) = G(501) = c'$, implying that all integers are turned into one and the same integer. This contradicts the assumption that Gandalf turns each integer into some other integer.\n\nb. Suppose that numbers $1000$, $1003$ and $1008$ are all reflecting. Then, for every integer $x$,\n$$\nG(x+3) = G(1003 - (x+3)) = G(1000 - x) = G(x),\n$$\n$$\nG(x+5) = G(1008 - (x+5)) = G(1003 - x) = G(x).\n$$\nSo $G(x+1) = G(x+4) = G(x+7) = G(x+10) = G(x+5) = G(x)$ for every integer $x$. Consequently, Gandalf again turns all integers into equal integers, contradicting the condition of the problem.\n\nc. Suppose Gandalf turns all even numbers into $1$ and all odd numbers into $2$. Then no integer is left unchanged. For every even number $a$, including $1002$, $1004$ and $1006$, the numbers $x$ and $a-x$ are either both even or both odd, whence they are turned into equal numbers.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 29931,
    "subject": "Mathematics (Multi-modal)",
    "question": "There are 2020 inhabitants in a town. Before Christmas, they are all happy; but if an inhabitant does not receive any Christmas card from any other inhabitant, he or she will become sad. Unfortunately, there is only one post company which offers only one kind of service: before Christmas, each inhabitant may appoint two different other inhabitants, among which the company chooses one to whom to send a Christmas card on behalf of that inhabitant. It is known that the company makes the choices in such a way that as many inhabitants as possible will become sad. Find the least possible number of inhabitants who will become sad.",
    "options": [],
    "answer": "674",
    "solution": "Partition 2019 inhabitants into 673 groups, each containing 3 inhabitants. Suppose that each inhabitant appoints two other members of the same group. As no group member is appointed thrice, the company cannot send three cards to one group member. Hence in every group, two different members get a card and at most one member will become sad. The inhabitant who does not belong to any group will become sad, too. Altogether, at most 674 inhabitants will become sad.\n\nWe show now that this is the least possible number. More precisely, we show that, in the case of $n$ inhabitants, the company can make $\\lfloor \\frac{n}{3} \\rfloor$ inhabitants sad. Call inhabitants $X$ and $Y$ *competitors* if somebody appoints them together to the company. By conditions of the problem, there are as many pairs of competitors as inhabitants or less (if several inhabitants appoint the same pair or some inhabitant appoints no pair). Thus it suffices to\n\n* If every inhabitant has exactly 2 competitors then choose an arbitrary inhabitant $X$ and leave out $X$ along with both competitors. Among the remaining $n-3$ inhabitants, there are at most $n-3$ pairs of competitors (besides two pairs containing $X$, removing either competitor canceled one more pair). By the induction hypothesis, one can find $\\lceil \\frac{n-3}{3} \\rceil$ remaining inhabitants with no pair of competitors. Adding $X$ to them results in $\\lceil \\frac{n}{3} \\rceil$ inhabitants with no pair of competitors.\n\n* If an inhabitant $X$ has at most 1 competitor then there must exist an inhabitant $Y$ with at least 3 competitors. Let $Z$ be the competitor of $X$ if $X$ has a competitor and an arbitrary inhabitant different from $X$ and $Y$ otherwise. After leaving out $X$, $Y$ and $Z$, we have $n-3$ inhabitants and at most $n-3$ pairs of competitors (removing $Y$ cancels at least 3 pairs of competitors). By the induction hypothesis, one can find $\\lceil \\frac{n-3}{3} \\rceil$ remaining inhabitants with no pair of competitors. Adding $X$ to this set results in a group of $\\lceil \\frac{n}{3} \\rceil$ inhabitants with no pairs of competitors. This proves the desired claim.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 29932,
    "subject": "Mathematics (Multi-modal)",
    "question": "For every positive integer $x$, denote by $\\kappa(x)$ the number of composite numbers not greater than $x$. Find all positive integers $n$ such that\n$$\n(\\kappa(n))! \\cdot \\text{lcm}(1,2,\\dots,n) > (n-1)!\n$$",
    "options": [],
    "answer": "2, 3, 4, 5, 7, 9",
    "solution": "The inequality holds for $n = 2,3,4,5,7,9$ and does not hold for $n = 1,6,8,10,11,12$. Assume in the rest that $n \\ge 13$. By definition of $\\kappa(n)$, there exists exactly $n-1-\\kappa(n)$ prime numbers not greater than $n$; these are the primes dividing $\\text{lcm}(1,2,\\dots,n)$. We have $n-1-\\kappa(n) \\ge 6$ as $n \\ge 13$. Let $q_1, \\dots, q_{n-1-\\kappa(n)}$ be all prime powers in the canonical representation of $\\text{lcm}(1,2,\\dots,n)$. W.l.o.g., $q_1 > q_2 > \\dots > q_{n-1-\\kappa(n)}$. As at least 5 numbers among $q_1, \\dots, q_6$ are odd, in the case of odd $n$ we have $q_1q_2q_3q_4q_5q_6 \\le n(n-1)(n-2)(n-4)(n-6)(n-8)$ and in the case of even $n$ similarly $q_1q_2q_3q_4q_5q_6 \\le n(n-1)(n-3)(n-5)(n-7)(n-9)$. But since $(n-3)(n-5)(n-7)(n-9) < (n-2)(n-4)(n-6)(n-8)$ and $n(n-8) < (n-3)(n-5)$, we anyway obtain\n$$\nq_1q_2q_3q_4q_5q_6 < (n-1)(n-2)(n-3)(n-4)(n-5)(n-6) = \\frac{(n-1)!}{(n-7)!}\n$$\nAs $q_6 < n-6$, the inequality $q_i < n-i$ holds for every $i > 6$, whence\n$$\nq_7 \\cdots q_{n-1-\\kappa(n)} \\le (n-7) \\cdots (\\kappa(n)+1) = \\frac{(n-7)!}{(\\kappa(n))!}\n$$\nConsequently, $\\text{lcm}(1,2,\\dots,n) = q_1q_2\\cdots q_{n-1-\\kappa(n)} < \\frac{(n-1)!}{(\\kappa(n))!}$, contradicting the original inequality. Hence the inequality holds for $n = 2,3,4,5,7,9$ only.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 29933,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $p, q$ be prime numbers and $a$ be an integer such that $p > 2$ and $a \\neq 1 \\pmod{q}$ but $a^p \\equiv 1 \\pmod{q}$. Prove that\n$$\n(1+a^1)(1+a^2)\\dots(1+a^{p-1}) \\equiv 1 \\pmod{q}.\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "As $a^p \\equiv 1 \\pmod{q}$ while $a \\neq 1 \\pmod{q}$, the case $q = 2$ is impossible. Thus, the desired equation is equivalent to\n$$\n(1 + a^0)(1 + a^1)(1 + a^2)\\dots(1 + a^{p-1}) \\equiv 2 \\pmod{q}. \\quad (13)\n$$\nRemoving parentheses in the l.h.s. of (13) gives all monomials of the form $a^{i_1+\\dots+i_k}$ where $\\{i_1, \\dots, i_k\\} \\subseteq \\{0, 1, \\dots, p-1\\}$. As $a^p \\equiv 1 \\pmod{q}$, the sums in the exponents can be replaced with their remainders modulo $p$. We claim that each possible remainder is produced the same number of times provided that we leave out the empty set and the whole set $\\{0, 1, \\dots, p-1\\}$. Indeed, for each tuple $(i_1, \\dots, i_k)$ where $0 < k < p$, we can find $l$ such that $kl \\equiv 1 \\pmod{p}$ and form a new tuple $(i_1 + l, \\dots, i_k + l)$ for which $(i_1 + l) + \\dots + (i_k + l) \\equiv i_1 + \\dots + i_k + 1 \\pmod{p}$. Clearly different tuples lead to different new tuples. Thus, for any remainder $i$, there are at least as many tuples with sum congruent to $i+1$ as tuples with sum congruent to $i$. Doing this $p$ times, we get back to the beginning and hence there are equal number of tuples giving each remainder.\nLet this constant number of tuples be $c$. Then (13) is equivalent to\n$$\na^0 + a^{0+1+\\dots+(p-1)} + c(a^0 + a^1 + \\dots + a^{p-1}) \\equiv 2 \\pmod{q}.\n$$\nAs $p = \\frac{(p-1)p}{2} = 0 + 1 + \\dots + (p-1)$, we have $a^{0+1+\\dots+(p-1)} \\equiv 1 \\pmod{q}$. Moreover, as $(a^0 + a^1 + \\dots + a^{p-1})(a-1) = a^p - 1 \\equiv 0 \\pmod{q}$ while $a-1 \\neq 0 \\pmod{q}$, we also have $a^0 + a^1 + \\dots + a^{p-1} \\equiv 0 \\pmod{q}$. Consequently, (13) is equivalent to $1+1 \\equiv 2 \\pmod{q}$ which holds trivially.\nLike in Solution 1, prove that $q$ is odd. By assumptions, the order of $a$ modulo $q$ divides $p$ and does not equal 1. Hence the order of $a$ modulo $q$ must be $p$. Thus $1, a, a^2, \\dots, a^{p-1}$ are pairwise incongruent modulo $q$. This implies that the residue classes of $1, a, a^2, \\dots, a^{p-1}$ are all distinct roots of the polynomial $x^p - 1$ in $\\mathbb{Z}_q$. Thus in $\\mathbb{Z}_q$ we have $x^p - 1 = (x-1)(x-a)(x-a^2)\\dots(x-a^{p-1})$. After substituting $x = -1$ and dividing all factors in the r.h.s. except the first one by $-1$, we obtain an equivalent congruence $-2 \\equiv -2(1+a)(1+a^2)\\dots(1+a^{p-1}) \\pmod{q}$. As $q \\neq 2$, division by $-2$ is possible and gives the desired congruence.\nRewrite each factor $1 + a^i$ in the form $\\frac{1-a^{2i}}{1-a^i}$. Since $p-1$ is even and $a^p \\equiv 1 \\pmod q$, we obtain\n$$\n\\begin{aligned}\n& (1 - a^2)(1 - a^4)\\dots(1 - a^{2(p-1)}) \\\\\n&= (1 - a^2)(1 - a^4)\\dots(1 - a^{p-1})(1 - a^{p+1})(1 - a^{p+3})\\dots(1 - a^{2p-2}) \\\\\n&\\equiv (1 - a^2)(1 - a^4)\\dots(1 - a^{p-1})(1 - a^1)(1 - a^3)\\dots(1 - a^{p-2}) \\\\\n&= (1 - a^1)(1 - a^2)\\dots(1 - a^{p-1}) \\pmod{q}.\n\\end{aligned}\n$$\nAs $a^p \\equiv 1 \\pmod q$, the least exponent $i$ for which $a^i \\equiv 1 \\pmod q$ must divide $p$; as $a \\not\\equiv 1 \\pmod q$, the least exponent must equal $p$. Therefore none of the factors $1-a^1, 1-a^2, \\dots, 1-a^{p-1}$ is divisible by $q$. This means that the congruence above can be reduced by these factors, i.e.,\n$$\n\\frac{(1 - a^2)(1 - a^4)\\dots(1 - a^{2(p-1)})}{(1 - a^1)(1 - a^2)\\dots(1 - a^{p-1})} \\equiv \\frac{(1 - a^1)(1 - a^2)\\dots(1 - a^{p-1})}{(1 - a^1)(1 - a^2)\\dots(1 - a^{p-1})} \\pmod{q}.\n$$\nThis proves the claim.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 29934,
    "subject": "Mathematics (Multi-modal)",
    "question": "Every term of the sequence $a_1, a_2, a_3, \\dots$ is either $0$ or $1$. It is known that both $0$ and $1$ occur at least $1010$ times among every $2021$ consecutive terms of the sequence. May one be sure that the sequence is periodic from some place on, i.e., there exist positive integers $n$ and $p$ such that $a_{n+i} = a_{n+i+p}$ for every natural number $i$?",
    "options": [],
    "answer": "No",
    "solution": "Consider the tuples $\\underbrace{11\\dots1}_{1011 \\text{ times}} \\underbrace{00\\dots0}_{1010 \\text{ times}}$ and $\\underbrace{11\\dots1}_{1010 \\text{ times}} \\underbrace{00\\dots0}_{1011 \\text{ times}}$. Concatenating infinitely many instances of these tuples in any order produces a sequence that satisfies the conditions of the problem. Indeed, consider any segment of $2021$ consecutive terms of such sequence. As any two consecutive full blocks of zeros or ones contain at least $2020$ terms in total, the segment under consideration contains terms of at most three such blocks. If it contains terms of three blocks then it contains the middle block fully. If the segment contained only two partial blocks of zeros or ones, it could contain at most $2020$ terms in total. Hence the segment under consideration must contain one full block even if it contains only terms of two consecutive blocks. In any case, the digit of the full block occurs either $1010$ or $1011$ times and the other digit must occur either $1011$ or $1010$ times, respectively. Combining the tuples defined at the beginning of the solution according to some non-periodic pattern (e.g., $xyxxyxxxyxxxxy\\dots$), the resulting sequence is not periodic from any place on. Indeed, suppose the contrary; let the length of the period be $p$. We can find a pattern of length $\\text{lcm}(p, 2021)$ starting from the place where periodicity starts, containing the tuples defined at the beginning of the solution a full number of times. But the middle term of these tuples is not repeating periodically, implying that the entire pattern also cannot be periodic. Consequently, there exist non-periodic sequences that satisfy the conditions of the problem.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 29935,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all pairs $(a, b)$ of positive integers such that $a \\ge b$ and\n$$\n\\frac{1}{a} + \\frac{1}{b} = \\frac{1}{2021}.\n$$",
    "options": [],
    "answer": "The pairs are (4086462, 2022), (97008, 2064), (88924, 2068), (4230, 3870), and (4042, 4042).",
    "solution": "*Solution 1:* Let $d = \\gcd(a,b)$, $a = da'$, $b = db'$. The given equation reduces to $\\frac{1}{da'} + \\frac{1}{db'} = \\frac{1}{2021}$ which is equivalent to\n$$\n2021(a' + b') = da'b'. \\qquad (4)\n$$\nAs $a'$ and $b'$ are relatively prime to each other, they both are relatively prime to $a' + b'$, implying that $a'b'$ and $a' + b'$ are relatively prime. By (4), $a'b' \\mid 2021(a' + b')$, implying $a'b' \\mid 2021$. As $2021 = 43 \\cdot 47$ where the factors are prime, the number $2021$ has exactly four positive factors $1$, $43$, $47$ and $2021$. Taking into account that $a \\ge b$ holds if and only if $a' \\ge b'$, consider all cases:\n* If $a' = 2021$ and $b' = 1$ then (4) implies $d = 2022$. Consequently, $(a,b) = (2021 \\cdot 2022, 2022)$.\n* If $a' = 47$ and $b' = 43$ then (4) implies $d = 90$. Consequently, $(a,b) = (47 \\cdot 90, 43 \\cdot 90)$.\n* If $a' = 47$ and $b' = 1$ then (4) implies $d = 43 \\cdot 48$. Consequently, $(a,b) = (2021 \\cdot 48, 43 \\cdot 48)$.\n* If $a' = 43$ and $b' = 1$ then (4) implies $d = 47 \\cdot 44$. Consequently, $(a,b) = (2021 \\cdot 44, 47 \\cdot 44)$.\n* If $a' = 1$ and $b' = 1$ then (4) implies $d = 2021 \\cdot 2$. Thus $(a,b) = (2021 \\cdot 2, 2021 \\cdot 2)$.\n\n\n*Solution 2:* The given equation is equivalent to $\\frac{a+b}{ab} = \\frac{1}{2021}$ which reduces to $ab - 2021a - 2021b = 0$. After adding $2021^2$ to both sides and factorizing in the l.h.s, we get\n$$\n(a - 2021)(b - 2021) = 2021^2. \\qquad (5)\n$$\nAs both $a$ and $b$ are positive, the factors in the l.h.s. of (5) are greater than $-2021$. Thus if both factors were negative then the absolute value of their product would be less than $2021^2$ and (5) could not hold. Hence both factors are positive. Since $a \\ge b$, we must have $a - 2021 \\ge b - 2021$. As $2021 = 43 \\cdot 47$ with factors being prime, we obtain variants $(a - 2021, b - 2021) = (2021^2, 1)$, $(a - 2021, b - 2021) = (2021 \\cdot 47, 43)$, $(a - 2021, b - 2021) = (2021 \\cdot 43, 47)$, $(a - 2021, b - 2021) = (47^2, 43^2)$ and $(a - 2021, b - 2021) = (2021, 2021)$. Consequently, $(a,b) = (2021 \\cdot 2022, 2022)$, $(a,b) = (2021 \\cdot 48, 43 \\cdot 48)$, $(a,b) = (2021 \\cdot 44, 47 \\cdot 44)$, $(a,b) = (47 \\cdot 90, 43 \\cdot 90)$, or $(a,b) = (2021 \\cdot 2, 2021 \\cdot 2)$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 29936,
    "subject": "Mathematics (Multi-modal)",
    "question": "Anna, Anne and Anni seek for real solutions $(x, y)$ to the system of equations\n$$\n\\begin{cases} 4x^3y - x^4 - 3x^2y^2 = 2021, \\\\ 4y^3x - y^4 - 3y^2x^2 = 2021. \\end{cases}\n$$\nAnna claims that the system of equations has a solution. Anne claims that the system of equations has no solution but at least one of the two equations has solutions. Anni claims that the system of equations has no solution and, even worse, neither of the two equations alone has a solution. Who is right?",
    "options": [],
    "answer": "Anne",
    "solution": "The system does not have a solution since adding the equation gives $-(x - y)^4 = 4042$ whose l.h.s. is non-positive but r.h.s. is positive.\nWe show that the first equation $4x^3y - x^4 - 3x^2y^2 = 2021$ has solutions (the same could be done for the second equation by symmetry). Dividing the equation by $x^4$ and reordering the terms in the l.h.s. results in\n$$\n-3 \\left(\\frac{y}{x}\\right)^2 + 4 \\left(\\frac{y}{x}\\right) - 1 = \\frac{2021}{x^4}. \\qquad (6)\n$$\nAs the discriminant of the quadratic equation $-3t^2 + 4t - 1 = 0$ is positive, there exists a real number $t$ such that $-3t^2 + 4t - 1$ equals a positive number $\\varepsilon$. Define $x = \\sqrt[4]{\\frac{2021}{\\varepsilon}}$ and $y = xt$; then $x$ and $y$ satisfy (6) and also the first equation of the given system of equations. Hence Anne is right.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 29937,
    "subject": "Mathematics (Multi-modal)",
    "question": "Point $D$ inside an acute triangle $ABC$ satisfies\n$$\n\\angle ADC = \\angle BDA = 180^{\\circ} - \\angle CAB.\n$$\nProve that the point symmetric to point $A$ w.r.t. point $D$ lies on the circumcircle of the triangle $ABC$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let the line $AD$ intersect the circumcircle of the triangle $ABC$ a second time at point $D'$; by the assumptions, $\\angle CDD' = \\angle D'DB = \\angle CAB$ (see figure below).\n\n![](attached_image_1.png)\n\nWe show that $AD = DD'$. As the quadrilateral $ABD'C$ is cyclic, $\\angle ABC = \\angle AD'C = \\angle DD'C$ and $\\angle BCA = \\angle BD'A = \\angle BD'D$. Thus the triangles $ABC$, $DD'C$ and $DBD'$ are similar. Hence $\\frac{|DD'|}{|DB|} = \\frac{|DC|}{|DD'|}$, implying $|DD'|^2 = |BD| \\cdot |CD|$.\n\nBy similarity of the triangles $ABC$ and $DD'C$, we obtain $\\angle BCA = \\angle D'CD$, which implies $\\angle DCA = \\angle D'CB = \\angle D'AB = \\angle DAB$. By similarity of the triangles $ABC$ and $DBD'$ we analogously get $\\angle ABD = \\angle CBD' = \\angle CAD' = \\angle CAD$. Hence the triangles $ABD$ and $CAD$ are similar. Consequently, $\\frac{AD}{CD} = \\frac{BD}{AD}$, which implies $AD^2 = BD \\cdot CD$.\n\nAltogether, we have proven $AD^2 = DD'^2$, which implies the desired result.\nLet the line $AD$ intersect the circumcircle of the triangle $ABC$ a second time at point $D'$; by the assumptions, $\\angle CDD' = \\angle D'DB = \\angle CAB$. We show that $AD = DD'$.\n\nLet $O$ be the circumcenter of the triangle $ABC$. Let the line $AD$ intersect the circumcircle of the triangle $BCD$ a second time at point $A'$ (see figure below).\n\n![](attached_image_2.png)\n\nBy the equality $\\angle CDA' = \\angle A'DB$, the arcs $BA'$ and $A'C$ of the circumcircle of the triangle $BCD$ are equal.\n\nBy the conditions of the problem, $\\angle CDB = 2\\angle CAB$. Hence point $O$ lies on the circumcircle of the triangle $BCD$. Since $OC = OB$, the arcs $CO$ and $OB$ of the circumcircle of the triangle $BCD$ are equal.\n\nConsequently, $OA'$ is a diameter of the circumcircle of the triangle $BCD$. If $O \\neq D$ then, by Thales' theorem, $\\angle A'DO = 90^\\circ$. Thus $OD \\perp AD'$, which implies $AD = DD'$ since a radius perpendicular to a chord bisects the chord. If $O = D$ (see figure below) then $AD'$ is a diameter of the circumcircle of the triangle $ABC$, obviously bisected by the center $O$.\n\n![](attached_image_3.png)",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 29938,
    "subject": "Mathematics (Multi-modal)",
    "question": "Two regular polygons have a common circumcircle. The sum of the areas of the incircles of these polygons equals the area of their common circumcircle. Find all possibilities of how many vertices can the two polygons have.",
    "options": [],
    "answer": "The polygons are either a triangle and a hexagon, or two squares.",
    "solution": "The ratio of the inradius and the circumradius of a regular $n$-gon is $\\cos\\frac{180^\\circ}{n}$. Hence the ratio of the areas of the incircle and the circumcircle of a regular $n$-gon is $\\cos^2\\frac{180^\\circ}{n}$.\n\nLet a regular $n$-gon and a regular $m$-gon with a common circumcircle be given. W.l.o.g., let $n \\le m$ and the area of the common circumcircle be $1$. By the above, the areas of the incircles of these polygons are $\\cos^2\\frac{180^\\circ}{n}$ and $\\cos^2\\frac{180^\\circ}{m}$, respectively. As their sum must equal the area of the common circumcircle, we get the equation\n\n$$\n\\cos^2\\frac{180^\\circ}{n} + \\cos^2\\frac{180^\\circ}{m} = 1\n$$\n\nwhich is equivalent to\n\n$$\n\\cos^2\\frac{180^\\circ}{n} = \\sin^2\\frac{180^\\circ}{m}\n$$\n\nAs both $n$ and $m$ are larger than $2$, both $\\frac{180^\\circ}{n}$ and $\\frac{180^\\circ}{m}$ are less than $90^\\circ$, whence the equation reduces to\n\n$$\n\\cos\\frac{180^\\circ}{n} = \\sin\\frac{180^\\circ}{m}\n$$\n\nThus\n\n$$\n\\frac{180^\\circ}{n} + \\frac{180^\\circ}{m} = 90^\\circ\n$$\n\nimplying that\n\n$$\n\\frac{1}{n} + \\frac{1}{m} = \\frac{1}{2}\n$$\n\nIf $n = 3$ then $m = 6$; if $n = 4$ then $m = 4$; if $n > 4$ then $m < 4$, contradicting the assumption $n \\le m$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 29939,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $D$ be the foot of the altitude drawn to the hypotenuse $AB$ of a right triangle $ABC$. The inradii of the triangles $ABC$, $CAD$ and $CBD$ are $r$, $r_1$ and $r_2$, respectively. Prove that $CD = r + r_1 + r_2$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $a = |BC|$, $b = |CA|$, $c = |AB|$ and $h = |CD|$; then $ab = ch = (a + b + c)r$.\n\nNote that the triangles *ABC*, *ACD* and *CBD* are similar by two equal angles (Fig. 28). As the similarity ratio of triangles *ACD* and *ABC* is $\\frac{b}{c}$ and that of triangles *CBD* and *ABC* is $\\frac{a}{c}$, we have $r_1 = \\frac{b}{c} \\cdot r$ and $r_2 = \\frac{a}{c} \\cdot r$, whence $r + r_1 + r_2 = \\frac{a + b + c}{c} \\cdot r$. As the equality $ab = (a + b + c)r$ implies $r = \\frac{ab}{a + b + c}$, we obtain $r + r_1 + r_2 = \\frac{ab}{c}$. On the other hand, the equality $ab = ch$ implies $h = \\frac{ab}{c}$. Consequently, $r + r_1 + r_2 = h$.\nLet the incenters of the triangles *ABC*, *ACD* and *BCD* be *I*, *I*_1 and *I*_2. Let the points of tangency of the incircle of the triangle *ABC* with sides *BC*, *CA* and *AB* be *K*, *L* and *M*, respectively; let the points of tangency of the incircle of the triangle *ACD* with sides *DC*, *CA* and *AD* be *K*_1, *L*_1 and *M*_1, respectively; let the points of tangency of the incircle of the triangle *BCD* with sides *DC*, *CB* and *BD* be *K*_2, *L*_2 and *M*_2, respectively (Fig. 29). As a tangent line is perpendicular to the radius drawn to the point of tangency, we have $\\angle IKC = 90^\\circ = \\angle ILC$. Thus *IKCL* is a square by three right angles and the equality $IK = IL$. Hence $CK = CL = r$. Similarly we see that\n![](attached_image_1.png)\n$I_1K_1DM_1$ and $I_2K_2DM_2$ are squares, whereby $DK_1 = DM_1 = r_1$ and $DK_2 = DM_2 = r_2$. By property of tangent line segments, we have $CK_1 = CL_1$, $AL = AM$ and $AL_1 = AM_1$, whence\n$$\n\\begin{aligned}\nCD &= CK_1 + K_1D = CL_1 + r_1 = CL + LL_1 + r_1 \\\\\n&= r + AL - AL_1 + r_1 = r + AM - AM_1 + r_1 = r + MM_1 + r_1.\n\\end{aligned}\n$$\nBy symmetry, we also get $CD = r + MM_2 + r_2$. After adding up these two equalities and taking into account that $MM_1 + MM_2 = DM_1 + DM_2 = r_1 + r_2$, we obtain\n$$\n2CD = 2(r + r_1 + r_2).\n$$\nConsequently, $CD = r + r_1 + r_2$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 29940,
    "subject": "Mathematics (Multi-modal)",
    "question": "Players $A$, $B$ and $C$ are playing the following game. Initially, the number $1$ is written on a blackboard. On their move, each player replaces the number $n$ currently on the blackboard with either $n+1$, $7n+7$, or $4n^3 + 3n + 4$ at their own choice, under the condition that the new number must not be larger than $10^9$. The player $A$ makes the first move, then $B$ takes turn, then $C$, after him $A$ again etc., until some player cannot make a legal move. The player who makes the last move wins. Can any of the players win the game against every legal play by the opponents and if yes then who?",
    "options": [],
    "answer": "Player C",
    "solution": "The game lasts while the number on the blackboard stays less than $10^9$, because it is possible to make a move of the first kind (replace $n$ with $n+1$). As the number on the blackboard is increased by at least $1$ by every move, it cannot stay less than $10^9$ infinitely. When the number on the blackboard equals $10^9$, there is no legal move. As numbers larger than $10^9$ cannot appear on the blackboard, the last move is made by the player who writes the number $10^9$ on the blackboard.\n\nLet a number $n$ be written on the blackboard. Note that $7n + 7 \\equiv n + 1 \\pmod 3$ and $4n^3 + 3n + 4 \\equiv n^3 + 1 \\equiv n + 1 \\pmod 3$. Hence all numbers that can appear on the blackboard on the next move are congruent to $n + 1 \\pmod 3$, whence the residues repeat in cycle $1, 2, 0, 1, 2, 0, \\ldots$. As $10^9 \\equiv 1 \\pmod 3$, the number $10^9$ is written by the third player $C$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 29941,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $a$, $b$, $c$, $d$ be positive real numbers satisfying the system of equations\n$$\n\\begin{cases} a^2 + \\frac{1}{b^2} = \\frac{1}{2}, \\\\ b^2 + \\frac{4}{c^2} = 8, \\\\ c^2 + \\frac{16}{d^2} = 2, \\\\ d^2 + \\frac{4}{a^2} = 32. \\end{cases}\n$$\n\nDetermine the product $abcd$.",
    "options": [],
    "answer": "4",
    "solution": "Multiplying all equations gives\n$$\n\\left(a^2 + \\frac{1}{b^2}\\right) \\left(b^2 + \\frac{4}{c^2}\\right) \\left(c^2 + \\frac{16}{d^2}\\right) \\left(d^2 + \\frac{4}{a^2}\\right) = 2^8.\n$$\nBy AM-GM, $a^2 + \\frac{1}{b^2} \\ge 2 \\cdot \\frac{a}{b}$, where the equality holds if and only if $a = \\frac{1}{b}$. Similarly $b^2 + \\frac{4}{c^2} \\ge 4 \\cdot \\frac{b}{c}$ (equality if and only if $b = \\frac{2}{c}$), $c^2 + \\frac{16}{d^2} \\ge 8 \\cdot \\frac{c}{d}$ (equality if and only if $c = \\frac{4}{d}$) and $d^2 + \\frac{4}{a^2} \\ge 4 \\cdot \\frac{d}{a}$ (equality if and only if $d = \\frac{2}{a}$). Multiplying the obtained four inequalities gives\n$$\n\\left(a^2 + \\frac{1}{b^2}\\right) \\left(b^2 + \\frac{4}{c^2}\\right) \\left(c^2 + \\frac{16}{d^2}\\right) \\left(d^2 + \\frac{4}{a^2}\\right) \\ge 2 \\cdot \\frac{a}{b} \\cdot 4 \\cdot \\frac{b}{c} \\cdot 8 \\cdot \\frac{c}{d} \\cdot 4 \\cdot \\frac{d}{a} = 2^8.\n$$\nThe resulting inequality must hold as equality by the first step of the solution. This is possible only if all four inequalities hold as equalities, whence\n$$\n\\begin{cases} a = \\frac{1}{b}, \\\\ b = \\frac{2}{c}, \\\\ c = \\frac{4}{d}, \\\\ d = \\frac{2}{a}. \\end{cases}\n$$\nBy multiplying the equations of this system, we get $abcd = \\frac{16}{abcd}$, whence $(abcd)^2 = 16$. As $a, b, c$ and $d$ are positive, the only possibility is $abcd = 4$.\nBy introducing $a = \\frac{x}{2}$, $b = 2y$, $c = z$, $d = 4t$, rewrite the system as\n$$\n\\begin{cases} \\frac{x^2}{4} + \\frac{1}{4y^2} = \\frac{1}{2}, \\\\ 4y^2 + \\frac{4}{z^2} = 8, \\\\ z^2 + \\frac{16}{16t^2} = 2, \\\\ 16t^2 + \\frac{4 \\cdot 4}{x^2} = 32. \\end{cases}\n$$\nMultiplying the first equation by 4, the second equation by $\\frac{1}{4}$, and the fourth equation by $\\frac{1}{16}$, we obtain an equivalent system\n$$\n\\begin{cases} x^2 + \\frac{1}{y^2} = 2, \\\\ y^2 + \\frac{1}{z^2} = 2, \\\\ z^2 + \\frac{1}{t^2} = 2, \\\\ t^2 + \\frac{1}{x^2} = 2. \\end{cases}\n$$\nAdding the equations of this system gives\n$$\nx^2 + \\frac{1}{y^2} + y^2 + \\frac{1}{z^2} + z^2 + \\frac{1}{t^2} + t^2 + \\frac{1}{x^2} = 8.\n$$\nBut for every real number $u$, $u + \\frac{1}{u} \\ge 2$, where equality holds only if $u = 1$. By adding up inequalities $x^2 + \\frac{1}{x^2} \\ge 2$, $y^2 + \\frac{1}{y^2} \\ge 2$, $z^2 + \\frac{1}{z^2} \\ge 2$ and $t^2 + \\frac{1}{t^2} \\ge 2$, we get\n$$\nx^2 + \\frac{1}{x^2} + y^2 + \\frac{1}{y^2} + z^2 + \\frac{1}{z^2} + t^2 + \\frac{1}{t^2} \\ge 8.\n$$\nThis inequality must hold as equality by the above; hence all four previous inequalities must hold as equalities, too, i.e., $x^2 = y^2 = z^2 = t^2 = 1$. As the numbers are positive, the only possibility is $x = y = z = t = 1$. Hence $abcd = \\frac{x}{2} \\cdot 2y \\cdot z \\cdot 4t = 4xyzt = 4$.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 29942,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all positive integers $k$ for which there is a right triangle with legs of integral lengths and hypotenuse of length $\\sqrt{88\\ldots822\\ldots2}$, where the number under the root consists of exactly $k$ eights and exactly $k$ twos.",
    "options": [],
    "answer": "1",
    "solution": "Let the lengths of legs be $a$ and $b$. By the Pythagorean theorem,\n$$\na^2 + b^2 = 88\\ldots822\\ldots2.\n$$\nIf $k=1$ then one can choose $a=9$ and $b=1$ as $9^2 + 1^2 = 82$. If $k \\ge 2$ then $88\\ldots822\\ldots2 \\equiv 6 \\pmod{8}$ since $822 \\equiv 6 \\pmod{8}$ and $222 \\equiv 6 \\pmod{8}$. On the other hand, the residues of $a^2$ and $b^2$ modulo $8$ can be $0$, $1$, or $4$. However, the sum of such two numbers can have residue $0$, $1$, $2$, $4$, or $5$ modulo $8$. Consequently, $a^2 + b^2 = 88\\ldots822\\ldots2$ cannot hold for $k \\ge 2$.\nLet the lengths of legs be $a$ and $b$. By the Pythagorean theorem,\n$$\na^2 + b^2 = 88\\ldots822\\ldots2.\n$$\nIf $k=1$ then one can choose $a=9$ and $b=1$ as $9^2 + 1^2 = 82$. Assume in the rest that $k \\ge 2$. It is known that a positive integer is expressible as the sum of two squares of integers if and only if its canonical representation contains primes congruent to $3$ modulo $4$ only with even exponents. Note that $88\\ldots822\\ldots2 = 2 \\cdot 44\\ldots411\\ldots1$, where only the odd second factor can be divisible by primes congruent to $3$ modulo $4$. If all such primes would occur with even exponents in the canonical representation of $44\\ldots411\\ldots1$, this number itself would be congruent to $1$ modulo $4$, but actually $44\\ldots411\\ldots1 \\equiv 3 \\pmod{4}$. This shows that right triangles with the required property cannot exist in the case $k \\ge 2$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 29943,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $D$ and $E$ be the midpoints of sides $AB$ and $AC$, respectively, of a triangle $ABC$. Prove that the line $AB$ is tangent to the circumcircle of the triangle $BEC$ if and only if the line $AC$ is tangent to the circumcircle of the triangle $BED$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "The conditions of the problem imply that $DE$ is the midsegment parallel to the side $BC$ of the triangle $ABC$ (Fig. 30).\n\n![](attached_image_1.png)\n\nBy properties of inscribed angle, the line $AB$ is tangent to the circumcircle of the triangle $BEC$ if and only if $\\angle ECB = \\angle DBE$, and the line $AC$ is tangent to the circumcircle of the triangle $BED$ if and only if $\\angle DBE = \\angle AED$. But $\\angle ECB = \\angle AED$ since the lines $DE$ and $BC$ are parallel, whence validity of either of these two equalities implies validity of the other one.\nBy powers, the line $AB$ is tangent to the circumcircle of the triangle $BEC$ if and only if $AB^2 = AE \\cdot AC$, and the line $AC$ is tangent to the circumcircle of the triangle $BED$ if and only if $AE^2 = AD \\cdot AB$. As $AB = 2AD$ and $AC = 2AE$, the equality $AB^2 = AE \\cdot AC$ is equivalent to the equality $2|AD|^2 = |AE|^2$, as well as the equality $AE^2 = AD \\cdot AB$ is equivalent to the equality $AE^2 = 2AD^2$. Hence both conditions reduce to the same equality, which solves the problem.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 29944,
    "subject": "Mathematics (Multi-modal)",
    "question": "Natural numbers $1$ through $n$ are written on a blackboard. On each move, one erases from the blackboard $2$ or more numbers whose sum is divisible by any of the chosen numbers and writes their sum on the blackboard. Two players make moves by turns and the player who cannot move loses the game. Which player can win the game against any play by the opponent, if:\n\na. $n = 6$;\n\nb. $n = 11$?",
    "options": [],
    "answer": "a: first player; b: first player",
    "solution": "a. The first player can replace numbers $1$, $2$, $3$, $6$ with $12$. After that, the blackboard contains numbers $4$, $5$, $12$. In this state, the sum of no two or three numbers on the blackboard is divisible by all the added numbers. Thus the second player cannot move and the first player wins immediately.\n\nb. The first player can replace numbers $1$, $2$, $3$, $4$, $6$, $8$ with $24$. After that, the blackboard contains numbers $5$, $7$, $9$, $10$, $11$, $24$, which sum up to $66$. Among numbers $7$, $9$, $10$, $11$, $24$, the l.c.m. of any two numbers is greater than $66$. Thus when choosing two or more numbers from among the mentioned numbers, and perhaps also the number $5$, the sum of the chosen numbers is less than their l.c.m. and cannot be divisible by all of them. Also when choosing one of the mentioned numbers together with $5$, the sum of the chosen numbers is not divisible by the larger one. Hence the second player cannot move and the first player wins immediately.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 29945,
    "subject": "Mathematics (Multi-modal)",
    "question": "Prove that\n$$\nsin 10^\\circ \\cdot \\cos 20^\\circ \\cdot \\sin 30^\\circ \\cdot \\cos 40^\\circ \\cdot \\sin 50^\\circ \\cdot \\cos 60^\\circ \\cdot \\sin 70^\\circ \\cdot \\cos 80^\\circ = \\frac{1}{256}.\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "As $\\sin 10^\\circ = \\cos 80^\\circ$, $\\sin 30^\\circ = \\cos 60^\\circ$, $\\sin 50^\\circ = \\cos 40^\\circ$, and $\\sin 70^\\circ = \\cos 20^\\circ$, the desired equality is equivalent to\n$$\n(\\cos 20^\\circ \\cdot \\cos 40^\\circ \\cdot \\cos 60^\\circ \\cdot \\cos 80^\\circ)^2 = \\frac{1}{256}.\n$$\nIt is known that $\\cos 60^\\circ = \\frac{1}{2}$. Concerning the other factors, we obtain\n$$\n\\begin{aligned}\n\\cos 20^\\circ \\cdot \\cos 40^\\circ \\cdot \\cos 80^\\circ &= \\frac{8 \\sin 20^\\circ \\cos 20^\\circ \\cdot \\cos 40^\\circ \\cdot \\cos 80^\\circ}{8 \\sin 20^\\circ} \\\\\n&= \\frac{4 \\sin 40^\\circ \\cos 40^\\circ \\cdot \\cos 80^\\circ}{8 \\sin 20^\\circ} \\\\\n&= \\frac{2 \\sin 80^\\circ \\cos 80^\\circ}{8 \\sin 20^\\circ} \\\\\n&= \\frac{\\sin 160^\\circ}{8 \\sin 20^\\circ} = \\frac{\\sin 20^\\circ}{8 \\sin 20^\\circ} = \\frac{1}{8}.\n\\end{aligned}\n$$\nConsequently,\n$$\n(\\cos 20^\\circ \\cdot \\cos 40^\\circ \\cdot \\cos 60^\\circ \\cdot \\cos 80^\\circ)^2 = \\left(\\frac{1}{2} \\cdot \\frac{1}{8}\\right)^2 = \\left(\\frac{1}{16}\\right)^2 = \\frac{1}{256}.\n$$\nAs $\\sin 10^\\circ = \\cos 80^\\circ$, $\\sin 30^\\circ = \\cos 60^\\circ$, $\\sin 50^\\circ = \\cos 40^\\circ$, and $\\sin 70^\\circ = \\cos 20^\\circ$, the l.h.s. of the desired equality can be rewritten as $(\\cos 20^\\circ \\cos 40^\\circ \\cos 60^\\circ \\cos 80^\\circ)^2$. Hence the desired equality is equivalent to\n$$\n\\cos 20^\\circ \\cos 40^\\circ \\cos 60^\\circ \\cos 80^\\circ = \\frac{1}{16}. \\qquad (2)\n$$\nAs $\\cos 60^\\circ = \\frac{1}{2}$ and $\\sin 60^\\circ = \\frac{\\sqrt{3}}{2}$, we have\n$$\n\\begin{aligned}\n\\cos 40^\\circ \\cos 80^\\circ &= \\cos (60^\\circ - 20^\\circ) \\cos (60^\\circ + 20^\\circ) \\\\\n&= (\\cos 60^\\circ \\cos 20^\\circ + \\sin 60^\\circ \\sin 20^\\circ) (\\cos 60^\\circ \\cos 20^\\circ - \\sin 60^\\circ \\sin 20^\\circ) \\\\\n&= \\cos^2 60^\\circ \\cos^2 20^\\circ - \\sin^2 60^\\circ \\sin^2 20^\\circ \\\\\n&= \\frac{1}{4} \\cos^2 20^\\circ - \\frac{3}{4} \\sin^2 20^\\circ = \\frac{1}{4} (\\cos^2 20^\\circ - 3 \\sin^2 20^\\circ).\n\\end{aligned}\n$$\nThus\n$$\n\\cos 20^\\circ \\cos 40^\\circ \\cos 60^\\circ \\cos 80^\\circ = \\frac{1}{8} (\\cos^3 20^\\circ - 3 \\sin^2 20^\\circ \\cos 20^\\circ). \\quad (3)\n$$\nBy applying the formula $\\cos 3x = \\cos^3 x - 3 \\sin^2 x \\cos x$ to $x = 20^\\circ$, we get $\\cos^3 20^\\circ - 3 \\sin^2 20^\\circ \\cos 20^\\circ = \\cos 60^\\circ = \\frac{1}{2}$. Consequently, (3) reduces to (2), completing the solution of the problem.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 29946,
    "subject": "Mathematics (Multi-modal)",
    "question": "The bisector of the internal angle on vertex $A$ of a triangle $ABC$ intersects the side $BC$ at point $D$. The line tangent to the circumcircle of the triangle $ABC$ at point $A$ intersects the line $BC$ at point $K$. Prove that $KA = KD$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Assume w.l.o.g. that $\\angle ABC > \\angle ACB$ (Fig. 31; otherwise change the roles of points $B$ and $C$). Note that\n$$\n\\angle ADK = 180^{\\circ} - \\angle CDA = \\angle DAC + \\angle ACD = \\angle BAD + \\angle ACB.\n$$\nBy inscribed angle property, $\\angle KAB = \\angle ACB$, whence\n$$\n\\angle BAD + \\angle ACB = \\angle BAD + \\angle KAB = \\angle KAD.\n$$\nConsequently, $\\angle ADK = \\angle KAD$, which implies $KA = KD$.\nLet $O$ be the circumcenter of the triangle $ABC$, $M$ be the midpoint of the side $BC$, and $X$ be the point of intersection of the ray $OM$ with the circumcircle of the triangle $ABC$. Moreover, let $Q$ be the point of intersection of the line perpendicular to the side $BC$ and passing through point $D$ with the line $AO$ (Fig. 32). As $OM$ is the perpendicular bisector of the side $BC$, point $X$ bisects the arc $BC$ of the circumcircle of the triangle $ABC$, whence the line $AD$ also passes through $X$. As $OA = OX$, we have $\\angle XAO = \\angle OXA$. On the other hand, $OX \\perp BC$ and $QD \\perp BC$ together imply $OX \\parallel QD$, whence $\\angle OXA = \\angle QDA$. Thus $\\angle DAQ = \\angle XAO = \\angle QDA$, implying $AQ = QD$. Consequently, there exists a circle with center $Q$ passing through both points $A$ and $D$. As $KA \\perp AQ$ and $KD \\perp DQ$, the lines $KA$ and $KD$ are tangent to this circle. By the property of tangent line segments, $KA = KD$.",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 29947,
    "subject": "Mathematics (Multi-modal)",
    "question": "Juku claims that if the sum of the squares of all digits of a natural number is divisible by $3$ then the number itself is divisible by $3$. Is Juku's claim always true?",
    "options": [],
    "answer": "Detailed solution",
    "solution": "The sum of the squares of the digits of the number $112$ is $6$ which is divisible by $3$, while the number $112$ is not divisible by $3$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 29948,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $n$ be a natural number, $n \\ge 2$. There are $n$ lamps on a circle. The lamps are labeled clockwise by natural numbers from $1$ to $n$. Each lamp can be either on or off. A switch between every two adjacent lamps enables one to change the state of both lamps simultaneously. In the beginning, all lamps are off. How many distinct configurations of states of lamps is it possible to achieve using these switches?",
    "options": [],
    "answer": "2^{n-1}",
    "solution": "As the state of each lamp is determined by the parity of the number of switchings that influence this lamp, the result of every sequence of switchings is determined by the set of switches that have been touched an odd number of times. Hence all possible states can be achieved by sequences of switchings that touch some switches once and do not touch other switches at all, whereby the order of switchings is irrelevant.\n\nNote that every two set of switches, one of which contains precisely the switches that the other one does not, lead to equal final states, because touching these two sets of switches in a row one touches each switch exactly once and thus changes the state of each lamp exactly twice. As there are $2^n$ sets of switches and these sets can be divided into pairs that give equal final states, there can be at most $2^{n-1}$ final states.\n\nIf two sets of switches are neither equal nor complementary (in the sense of previous paragraph) then touching these two sets of switches in a row does not give back the initial state, because this sequence of switchings is equivalent to touching a set of switches that contains some switch and does not contain some other switch, but then there must exist a lamp, the switch in only one side of which is touched. Hence every pair of sets of switches defined in the previous paragraph gives a unique final state, i.e., there are $2^{n-1}$ possible distinct final states.\nAfter every switching, either two more lamps are burning, two less lamps are burning, or the number of burning lamps does not change. As the initial number of burning lamps is even, the number of burning lamps stays even.\n\nWe show that every state with an even number of burning lamps can be achieved. As the switchings are invertible (repeating the same touch reverts the previous change), it suffices to show that, starting from any state with an even number of burning lamps, one can achieve the state where no lamp is burning. Indeed, while there exist two consecutive burning lamps, we can switch off both these lamps. If there do not exist such lamps, but some lamps are still burning, one can shift a lonely burning lamp clockwise by repeatedly touching the clockwise nearest switch until two consecutive lamps are burning. Switching these two lamps off decreases the total number of burning lamps by $2$. This way, we can decrease the number of burning lamps until all lamps are switched off.\n\nThus the number of all achievable states is $C_0^n + C_2^n + C_4^n + \\dots$ which equals $2^{n-1}$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 29949,
    "subject": "Mathematics (Multi-modal)",
    "question": "(a) Find the largest number expressible as the difference of two two-digit numbers obtained from each other by changing the order of digits.\n(b) The same question with three-digit instead of two-digit numbers.",
    "options": [],
    "answer": "(a) 72; (b) 801",
    "solution": "(a) Let the given two-digit number be $\\overline{ab}$. The only number that can be obtained by changing the order of digits is $\\overline{ba}$. The difference of these numbers is $(10a + b) - (10b + a) = 9(a - b)$. To obtain the largest difference, $a$ must be as large as possible and $b$ as small as possible. Since $b = 0$ is impossible, we must have $a = 9$ and $b = 1$ giving $a - b = 8$. Hence the desired largest difference is $72$.\n\n(b) Let the given three-digit number be $\\overline{abc}$. Interchanging the last two digits can change it by less than $100$. Interchanging the first two digits can change the number by at most $720$ by part (a) of the problem. It remains to study cases where changing the order of digits results in $\\overline{cab}$, $\\overline{bca}$ or $\\overline{cba}$.\n\n* The difference of numbers $\\overline{abc}$ and $\\overline{cab}$ is $(100a + 10b + c) - (100c + 10a + b) = 9(10a + b - 11c)$. To obtain the largest difference, $a$ and $b$ must be as large as possible and $c$ as small as possible. Since $c$ is the first digit of the number, $c = 0$ is impossible, whence the largest difference is obtained if $a = b = 9$ and $c = 1$. This difference is $991 - 199 = 792$.\n\n* The difference of numbers $\\overline{abc}$ and $\\overline{bca}$ is $(100a + 10b + c) - (100b + 10c + a) = 9(11a - 10b - c)$. To obtain the largest difference, $a$ must be as large as possible and both $b$ and $c$ as small as possible, i.e., $a = 9$, $b = 1$ and $c = 0$. This difference is $910 - 109 = 801$.\n\n* The difference of numbers $\\overline{abc}$ and $\\overline{cba}$ is $(100a + 10b + c) - (100c + 10b + a) = 99(a - c)$. To obtain the largest difference, $a$ must be as large as possible and $c$ as small as possible. Since $c \\neq 0$, the largest difference is obtained if $a = 9$ and $c = 1$. Then the difference of the three-digit numbers is $99 \\cdot 8 = 792$.\n\nConsequently, the desired largest difference is $801$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 29950,
    "subject": "Mathematics (Multi-modal)",
    "question": "Do there exist real numbers $x$, $y$, $z$, $t$ that meet the following system of equations?\n$$\n\\begin{cases}\n1 + x^3 + y^2 = 0 \\\\\n1 + y^3 + z^2 = 0 \\\\\n1 + z^3 + t^2 = 0 \\\\\n1 + t^3 + x^2 = 0 \\\\\nx + y + z + t = 0\n\\end{cases}\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "The first equation implies $x^3 = -y^2 - 1$. Thus $x^3 < 0$, implying also $x < 0$. Similarly from the second, third and fourth equations we obtain $y < 0$, $z < 0$ and $t < 0$, respectively. The sum of negative numbers $x$, $y$, $z$, $t$ is negative, contradicting the fifth equation.\nSuppose that the system has a solution. W.l.o.g., let $x$ be variable with the largest value. Then $4x \\ge x + y + z + t$, which by the last equation implies $x \\ge 0$. Consequently, also $x^3 \\ge 0$. As $y^2 \\ge 0$, this implies $1 + x^3 + y^2 \\ge 1$, contradicting the first equation. Hence no solution can exist.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 29951,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $n$ be a fixed positive integer. Find all triples $(a, b, c)$ of integers satisfying the following system of equations:\n$$\n\\begin{cases}\na^{n+3} + b^{n+2}c + c^{n+1}a^2 + a^n b^3 = 0 \\\\\nb^{n+3} + c^{n+2}a + a^{n+1}b^2 + b^n c^3 = 0 \\\\\nc^{n+3} + a^{n+2}b + b^{n+1}c^2 + c^n a^3 = 0\n\\end{cases}\n$$",
    "options": [],
    "answer": "a = b = c = 0",
    "solution": "If $a = 0$ then the first equation implies $b^{n+2}c = 0$. Hence $b = 0$ or $c = 0$; w.l.o.g., $b = 0$. Then the last equation reduces to $c^{n+3} = 0$ which implies $c = 0$. Thus if $a = 0$ then $a = b = c = 0$. Analogously we can prove that if $b = 0$ or $c = 0$ then $a = b = c = 0$. The triple $(0,0,0)$ satisfies the equation.\n\nIt remains to consider triples $(a,b,c)$ whose all terms are different from $0$. Let $p$ be an arbitrary prime number. If $p \\mid a$ then the first equation implies $p \\mid b^{n+2}c$. Thus $p \\mid b$ or $p \\mid c$; w.l.o.g., $p \\mid b$. Then the last equation gives $p \\mid c^{n+1}$ which implies $p \\mid c$. Thus if $p \\mid a$ then $p \\mid a, p \\mid b, p \\mid c$. Analogously we can prove that if $p \\mid b$ or $p \\mid c$ then $p \\mid a, p \\mid b, p \\mid c$.\n\nBut rewriting $a = pa', b = pb', c = pc'$ enables us to divide both sides of all equations by $p^{n+3}$, giving a similar system of equations having $a', b', c'$ in the role of $a, b, c$. Hence the triple $(a', b', c')$ also satisfies the system of equations. As the numbers other than $0$ cannot be infinitely divided in integers, a finite number of divisions should give us a solution whose terms have no common prime factors. By the previous paragraph, this is possible only if the values of variables have no prime factors, i.e., $a = \\pm 1, b = \\pm 1, c = \\pm 1$. We show now that such solutions do not exist. Indeed, if $n$ is even then the system of equations reduces to\n$$\n\\begin{cases}\na + c + c + b = 0, \\\\\nb + a + a + c = 0, \\\\\nc + b + b + a = 0.\n\\end{cases}\n$$\nAdding all equations results in $4(a + b + c) = 0$, implying $a + b + c = 0$; but the sum of three odd numbers cannot equal the even number $0$.\n\nIf $n$ is odd then the system of equations reduces to\n$$\n\\begin{cases}\n1 + bc + 1 + ab = 0, \\\\\n1 + ca + 1 + bc = 0, \\\\\n1 + ab + 1 + ca = 0.\n\\end{cases}\n$$\nAdding all equations results in $2(ab + bc + ca) + 6 = 0$, implying $ab + bc + ca = -3$. As $|ab| = |bc| = |ca| = 1$, the only possibility is $ab = bc = ca = -1$. This demands that $a, b, c$ have pairwise opposite signs which is impossible. Consequently, no solutions except $(0,0,0)$ exist.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 29952,
    "subject": "Mathematics (Multi-modal)",
    "question": "The incircle of triangle $ABC$ touches the sides $AB$ and $AC$ at points $K$ and $L$, respectively. The line $BL$ intersects the incircle of triangle $ABC$ at point $M$ ($M \\neq L$). A circle passing through point $M$ touches the lines $AB$ and $BC$ at points $P$ and $Q$, respectively, and intersects the incircle of triangle $ABC$ at point $N$ ($N \\neq M$). Prove that if $KM \\parallel AC$ then points $P$, $N$ and $L$ are collinear.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Note that the circles of the problem can be obtained from each other by homothetic transformation with center $B$ since both are tangent to sides $BA$ and $BC$. Let $X$ be the other intersection point of the line $BL$ with the circumcircle of the triangle $MPQ$.\n\n![](attached_image_1.png)\n\nThe aforementioned homothety takes point $P$ to point $K$, point $M$ to point $L$, and point $X$ to point $M$. By the homothety, $\\angle PXM = \\angle KML$. Hence $\\angle KML = \\angle KLA = \\angle LKM$. Finally,\n$$\n\\angle PNM + \\angle LNM = (180^\\circ - \\angle PXM) + \\angle LKM \\\\\n= (180^\\circ - \\angle KML) + \\angle KML = 180^\\circ.\n$$\nThis proves the desired claim that points $P$, $N$ and $L$ are collinear.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 29953,
    "subject": "Mathematics (Multi-modal)",
    "question": "There is a rectangular grid with 3 rows and $n$ columns on a blackboard.\n\na. Find the number of ways to write exactly one of numbers $1, 2, \\ldots, 3n$ into each square in such a way that all the following conditions are met:\n(1) Different squares contain different numbers.\n(2) For each $i = 1, 2, \\ldots, 3n - 1$, the numbers $i$ and $i + 1$ are written into adjacent squares (i.e., squares having a common side).\n(3) The numbers $1$ and $3n$ are written into adjacent squares.\n\nb. The same question, but as the 3rd condition, the number $1$ must be written into the leftmost and the number $3n$ into the rightmost column.",
    "options": [],
    "answer": "a) 0 if n is odd; if n is even, 3n · 2^(n/2).\nb) 2^n.",
    "solution": "(a) Note that all three squares in the leftmost and rightmost columns must be traversed consecutively because along with both corner squares of either of the columns one has to traverse the middle square of the same column and this cannot be done twice. In no other column can one traverse all three squares consecutively because such column would divide the grid into two parts between which one could not move without visiting a square twice. Note also that one cannot traverse three or more consecutive squares of the middle row consecutively because such trajectory would also divide the grid into two isolated parts (Figures 15 and 16). For similar reasons, when visiting two consecutive squares of the middle row consecutively, the previous and the next square should be taken from one and the same row (either top or bottom).\nHence apart from the first and the last column, every legal trajectory divides the middle row into pairs of squares that are visited consecutively. Therefore the required enumeration is impossible for odd $n$. For even $n$, one can choose for every pair of squares in the middle row the row where the previous and the next square are located (either top or bottom); the number of such pairs of squares is $\\frac{n-2}{2}$, whence there are $2^{\\frac{n-2}{2}}$ possibilities to choose the direction of continuation for every such pair of squares. The whole trajectory is determined by these choices. For writing numbers in the case of a fixed trajectory, there are $6n$ possibilities (2 possibilities to choose the direction and $3n$ possibilities to choose the initial square). Consequently, there are $6n \\cdot 2^{\\frac{n-2}{2}} = 3n \\cdot 2^{\\frac{n}{2}}$ possibilities of the required enumeration for even $n$.\n\n(b) One cannot start from the middle square of the leftmost column since either of the corner squares would become a dead end. Hence suppose w.l.o.g. that one starts from the top left corner and visits $k$ squares in the top row (inclusive of the initial square), where $1 \\le k \\le n$ (Fig. 17). Going directly to the bottom row would generally divide the grid into two isolated parts; also turning to the right along the middle row would make it impossible to visit squares in the left and afterwards reach the rightmost column. Hence one should turn to the left along the middle row. Similarly we find that the leftmost part of size $3 \\times k$ of the grid should be traversed along Z-shape (Fig. 18). After that, one should follow similar rules to traverse the remaining part of size $3 \\times (n-k)$.\nRepeating the argument, we can see that every legal trajectory divides the grid into Z-shapes and S-shapes, whereby Z-shapes are immediately followed by S-shapes and vice versa. In every column but the last one we can choose whether to start a new letter or make the current letter wider. In the first column, we also can choose which corner to start in. Hence there are $2^n$ possibilities altogether.\n\n![](attached_image_1.png)\nFig. 15\n![](attached_image_2.png)\nFig. 16\n![](attached_image_3.png)\nFig. 17\n![](attached_image_4.png)\nFig. 18",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 29954,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find the least positive integer $n$ such that $\\sqrt[5]{5n}$, $\\sqrt[6]{6n}$ and $\\sqrt[7]{7n}$ are integers.",
    "options": [],
    "answer": "2^35 * 3^35 * 5^84 * 7^90",
    "solution": "Let $n = 2^\\alpha \\cdot 3^\\beta \\cdot 5^\\gamma \\cdot 7^\\delta \\cdot s$, where $s$ is not divisible by $2$, $3$, $5$ or $7$; then $5n = 2^\\alpha \\cdot 3^\\beta \\cdot 5^{\\gamma+1} \\cdot 7^\\delta \\cdot s$, $6n = 2^{\\alpha+1} \\cdot 3^{\\beta+1} \\cdot 5^\\gamma \\cdot 7^\\delta \\cdot s$ and $7n = 2^\\alpha \\cdot 3^\\beta \\cdot 5^\\gamma \\cdot 7^{\\delta+1} \\cdot s$. Consequently:\n* For $\\sqrt[5]{5n}$ to be an integer, $\\alpha$, $\\beta$, $\\gamma + 1$ and $\\delta$ must be divisible by $5$;\n* For $\\sqrt[6]{6n}$ to be an integer, $\\alpha + 1$, $\\beta + 1$, $\\gamma$ and $\\delta$ must be divisible by $6$;\n* For $\\sqrt[7]{7n}$ to be an integer, $\\alpha$, $\\beta$, $\\gamma$ and $\\delta + 1$ must be divisible by $7$.\nHence $\\alpha$ and $\\beta$ must be divisible by $35$, $\\gamma$ must be divisible by $42$ and $\\delta$ must be divisible by $30$. The least suitable value for $\\alpha$ and $\\beta$ is $35$ since $35$ is the least positive multiple of $35$ and the next integer is divisible by $6$. Studying the multiples of $42$ and $30$ similarly shows that the least suitable value for $\\gamma$ is $84$ and the least suitable value for $\\delta$ is $90$. For the least suitable value for $n$, take $s = 1$. Hence the desired number is $2^{35} \\cdot 3^{35} \\cdot 5^{84} \\cdot 7^{90}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 29955,
    "subject": "Mathematics (Multi-modal)",
    "question": "How many positive integers, where the only allowed digits are $0$ and $1$, are less than $1111100100$?",
    "options": [],
    "answer": "2019",
    "solution": "The given number has $11$ digits. There are $2^{11} - 1 = 2047$ positive integers with at most $11$ digits each of which is either $0$ or $1$. We solve the problem by subtracting the number of positive integers that are not less than the given number.\nThe $11$-digit numbers larger than the given number are all of the form $11111abcde$. There are $2^5 = 32$ numbers in this form. Among them, $4$ numbers $1111100000$, $1111100001$, $1111100010$ and $1111100011$ are less than the given number. Thus the number of positive integers consisting of zeros and ones and being less than $1111100100$ is $2047 - 32 + 4 = 2019$.\nOrdering any two digit sequences that constitute a positional representation on different bases does not depend on the base. This means that if a number is larger than another number on base $2$ then the first number is larger than the second number also on base $10$. The number $1111100100$ on base $2$ equals $2^{10} + 2^9 + 2^8 + 2^7 + 2^6 + 2^5 + 2^2 = 2020$ on base $10$. Hence, to solve the problem, it suffices to count all positive integers less than $2020$. The result is obviously $2019$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 29956,
    "subject": "Mathematics (Multi-modal)",
    "question": "Prove that $a^{2020} + 10a^{1010} + 1001$ is prime for no integers $a$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Note that $1001 = 11 \\cdot 91$. If $11 \\mid a$ then also $11 \\mid a^{2020} + 10a^{1010} + 1001$. Otherwise, consider $a^5$ modulo $11$. A case study shows that if $a$ is congruent to $1$, $3$, $4$, $5$, or $9$, then $a^5$ is congruent to $1$, and in all other cases, $a^5$ is congruent to $-1$. Hence $a^{10} \\equiv 1 \\pmod{11}$ whenever $11 \\nmid a$. This implies that $a^{1010} \\equiv 1 \\pmod{11}$ and $a^{2020} \\equiv 1 \\pmod{11}$ because $a^{1010} = (a^{10})^{101}$ $\\pmod{11}$ and $a^{2020} = (a^{10})^{202}$. Consequently, $10a^{1010} \\equiv 10 \\pmod{11}$, implying that $a^{2020} + 10a^{1010} + 1001 \\equiv 0 \\pmod{11}$. Thus $a^{2020} + 10a^{1010} + 1001$ cannot be prime since obviously $a^{2020} + 10a^{1010} + 1001 \\ge 1001 > 11$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 29957,
    "subject": "Mathematics (Multi-modal)",
    "question": "There are $n$ lists of candidates taking part in elections. Let $h_i$ be the total number of votes given for the candidates of the $i$th list. There are $M$ seats in the representative assembly.\n\nAnna proposes the following system for delivering mandates: For each list, one computes a reference number $v_i = \\frac{h_i}{a_i + 1}$ where $a_i$ is the number of mandates already given to the $i$th list (initially $a_i = 0$, i.e., the reference number of each list equals its number of votes). On every step ($M$ times in total), one chooses the list with the greatest reference number (if several lists share the first place, one of them is chosen randomly) and adds one mandate to this list, after which the reference number of this list is recomputed.\n\nBert's idea for delivering mandates is to multiply the number of votes of every list by $M/K$ where $K = h_1 + \\dots + h_n$, whereby fractional results are rounded downwards. As rounding may cause some seats to be undelivered, he proposes multiplying all numbers of votes of the lists by some suitable coefficient $\\beta$, so that the number of mandates given to the $i$th list would be $m_i = \\lfloor \\frac{\\beta h_i M}{K} \\rfloor$ where $m_1 + \\dots + m_n = M$.\n\nProve that if such coefficient $\\beta$ exists then Anna's and Bert's methods lead to the same distribution of mandates.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let the total number of mandates given to the $i$th list be $m_i$ in the case of Bert's method and $m'_i$ in the case of Anna's method. Suppose that these methods result in different distribution of mandates. As the sum of numbers of mandates must be the same, we must have $m'_i > m_i$ for some $i = 1, 2, \\dots, n$ and $m'_j < m_j$ for some $j = 1, 2, \\dots, n$. As the numbers of mandates are integers, we have $m'_i \\ge m_i + 1$ and $m'_j + 1 \\le m_j$.\n\nConsider the situation in the case of Anna's method immediately after the $i$th list having obtained its last mandate. Before obtaining the last mandate, the $i$th list had $m'_i - 1$ mandates and reference number $v_i = \\frac{h_i}{(m'_i - 1) + 1} = \\frac{h_i}{m'_i}$,\n\nwhile the $j$th list had at most $m'_j$ mandates, implying that $a_j \\le m'_j$ and $v_j = \\frac{h_j}{a_j+1} \\ge \\frac{h_j}{m'_j+1}$. As the mandate was given to the $i$th list, $\\frac{h_i}{m'_i} \\ge \\frac{h_j}{m'_j+1}$. Thus $\\frac{h_i}{h_j} \\ge \\frac{m'_i}{m'_j+1}$, implying that\n$$\nm_j \\cdot \\frac{h_i}{h_j} \\ge m_j \\cdot \\frac{m'_i}{m'_j+1} \\ge m_j \\cdot \\frac{m_i+1}{m_j} \\ge m_i+1.\n$$\nOn the other hand, from the specification of Bert's method we know that\n$$\nm_j \\cdot \\frac{h_i}{h_j} = \\frac{h_i}{h_j} \\cdot \\left\\lfloor \\frac{\\beta h_j M}{K} \\right\\rfloor \\le \\frac{h_i}{h_j} \\cdot \\frac{\\beta h_j M}{K} = \\frac{\\beta h_i M}{K} < \\left\\lfloor \\frac{\\beta h_i M}{K} \\right\\rfloor + 1 = m_i + 1.\n$$\nThis inequation contradicts the previous inequation. Hence both methods indeed lead to the same distribution of mandates.",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 29958,
    "subject": "Mathematics (Multi-modal)",
    "question": "Xavier and Olivier are playing tic-tac-toe in the rectangular grid of size $3 \\times 3$ with modified rules. On every move, a player chooses an empty square and writes his token into it. Players take turns alternately, with Xavier starting. The player who is the first to occupy any three squares that either are all in the same row or column or lie in pairwise distinct rows and columns wins. Does either of the players have a winning strategy and if yes then who?",
    "options": [],
    "answer": "Xavier",
    "solution": "Note that every two unit squares uniquely determine a unit square that constitutes a winning triple together with the given two squares. Label all unit squares except the middle square clockwise with numbers 1 through 8 (Fig. 1); let Xavier initially play into the middle square and if Olivier then moves into the square No. $i$ then let Xavier make his second move into the square No. $i+1$ (square No. 1 if $i=8$; all possible positions after Xavier's second move, modulo rotation of the table, are depicted in Figures 2 and 3). In order to prevent loss of the game, Olivier must occupy the unit square that constitutes a winning triple together with the two squares occupied by Xavier. Let Xavier then move into the square that constitutes a winning triple together with the two squares previously occupied by Olivier (all possible positions after Xavier's third move are depicted in Figures 4 and 5). Then Olivier can't win on his next move. But the three squares occupied by Xavier contain three pairs, out of which only one pair has its corresponding winning third square occupied by Olivier. Thus Olivier can't block them all, whence Xavier can win on his next move.\n\n![](attached_image_1.png)",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 29959,
    "subject": "Mathematics (Multi-modal)",
    "question": "The lines tangent to the circumcircle of triangle $ABC$ at points $B$ and $C$ intersect at point $D$. The circumcircle of triangle $BCD$ intersects the lines $AB$ and $AC$ the second time at points $K$ and $L$, respectively. Prove that the line $AD$ bisects the line segment $KL$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "We prove that $AKDL$ is a parallelogram; this implies the desired claim since $AD$ and $KL$ are diagonals of this quadrilateral. We prove at first that $KD \\parallel AL$.\n\nIf $K$ lies between $A$ and $B$ (Fig. 19) then by inscribed angles $\\angle BAC = \\angle BCD$ and $\\angle BCD = \\angle BKD$. Consequently, $\\angle BAL = \\angle BAC = \\angle BKD$, implying $KD \\parallel AL$.\n\nIf $B$ lies between $A$ and $K$ (Fig. 20) then by inscribed angles $\\angle BAC = \\angle BCD$ and $\\angle BCD = 180^\\circ - \\angle BKD$. Consequently, $\\angle BAL + \\angle BKD = \\angle BAC + \\angle BKD = 180^\\circ$, implying $KD \\parallel AL$.\n\nIf $A$ lies between $K$ and $B$ (Fig. 21) then by inscribed angles $\\angle BAC = 180^\\circ - \\angle BCD$ and $\\angle BCD = \\angle BKD$. Consequently, $\\angle BAL = 180^\\circ - \\angle BAC = \\angle BKD$, implying $KD \\parallel AL$ again.\n\nAnalogously, we can show that $LD \\parallel AK$. Altogether, this establishes that $AKDL$ is a parallelogram.\n\n![](attached_image_1.png)\nFig. 19\n![](attached_image_2.png)\nFig. 20\n![](attached_image_3.png)\nFig. 21\nBy inscribed angles, $\\angle ABC = \\angle ALK$, implying that the triangles $ABC$ and $ALK$ are similar. Let the lines $AD$ and $KL$ intersect at point $N$.\n\n![](attached_image_1.png)\nFig. 19\n\nLet $M$ be the midpoint of the side $BC$. It is known that the symmedian line drawn through a vertex of a triangle and the lines tangent to the circumcircle of the triangle at the other two vertices meet in one point; hence $N$ is the point of intersection of the symmedian line drawn through vertex $A$ of the triangle $ABC$ with line $KL$. By the definition of symmedian, $\\angle BAM = \\angle CAN = \\angle LAN$, whence $M$ and $N$ are corresponding points in similar triangles $ABC$ and $ALK$. Thus $N$ bisects the line segment $KL$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 29960,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $n$ be a positive integer. Find the largest number of knights that can be placed on a board of size $n \\times n$ in such a way that no two knights attack each other. A knight attacks precisely the squares that are located either horizontally by one square and vertically by two squares away or horizontally by two squares and vertically by one square away.\n\n![](attached_image_1.png)",
    "options": [],
    "answer": "floor(n^2/2)",
    "solution": "Clearly one can place only $1$ knight on an $1 \\times 1$ board, which is $\\lfloor \\frac{1^2}{2} \\rfloor$, and at most $4$ knights on an $2 \\times 2$ board. In the rest, assume $n \\ge 3$.\n\nIf a set of unit squares is divided into pairs in such a way that a knight on one square of any pair attacks the other square of that pair then there cannot be more knights than half of the total number of unit squares in this set without some knights attacking each other. In Figures 22, 23 and 24, all unit squares of boards of size $2 \\times 4$, $3 \\times 4$ and $3 \\times 6$ are divided into pairs whose members are located at one knight move from each other. In Figures 25 and 26, all unit squares of boards of size $3 \\times 3$ and $5 \\times 5$ except the middle square are divided into pairs whose members are located at one knight move from each other. By the above, it is impossible to place knights to more than half of unit squares of boards of size $2 \\times 4$, $3 \\times 4$ and $3 \\times 6$. Taking into account that an additional knight may be on the middle square, it also follows that there cannot be more than $\\lfloor \\frac{3^2}{2} \\rfloor$ knights on a $3 \\times 3$ board or more than $\\lfloor \\frac{5^2}{2} \\rfloor$ knights on a $5 \\times 5$ board. As a $4 \\times 4$ board can be formed from two $2 \\times 4$ boards and a $6 \\times 6$ board from two $3 \\times 6$ boards, there cannot be more than $\\frac{4^2}{2}$ knights on a $4 \\times 4$ board or more than $\\frac{6^2}{2}$ knights on a $6 \\times 6$ board.\n\n![](attached_image_2.png)\nFig. 22\n![](attached_image_3.png)\nFig. 23\n![](attached_image_4.png)\nFig. 24\n![](attached_image_5.png)\nFig. 25\n![](attached_image_6.png)\nFig. 26\n\nIf $n \\ge 7$ then an $n \\times n$ board can be divided into pieces of size $4 \\times 4$, $r \\times 4$ and $r \\times r$, where $3 \\le r \\le 6$ and $4 \\mid n-r$. By the above, neither $4 \\times 4$ nor $r \\times 4$ board can contain more knights than half of the number of unit squares ($5 \\times 4$ and $6 \\times 4$ are divisible into rectangles of size $2 \\times 4$ and $3 \\times 4$). The same holds for an $r \\times r$ piece, if the middle square in the case of odd $r$ is not taken into account. Thus for no $n \\ge 3$ can one place more than $\\lfloor \\frac{n^2}{2} \\rfloor$ knights on an $n \\times n$ board.\n\nOn the other hand, coloring the unit squares black and white chesswise, one can place a knight on all squares of one and the same color since a knight attacks only squares of the opposite color. The number of unit squares of a fixed color is $\\frac{n^2}{2}$ if $n$ is even. In the case of odd $n$, the number of unit squares whose color coincides with the color of the middle square is $\\lfloor \\frac{n^2}{2} \\rfloor$. Hence one can place $\\lfloor \\frac{n^2}{2} \\rfloor$ pairwise non-attacking knights on an $n \\times n$ board.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 29961,
    "subject": "Mathematics (Multi-modal)",
    "question": "Do there exist numbers $a$, $b$, $c$ that satisfy the equation\n$$\n2a(c-a) - b(2a+b) + c(2b-c) = 2020?\n$$",
    "options": [],
    "answer": "No",
    "solution": "Transforming the l.h.s. of the equation gives\n$$\n\\begin{aligned}\n2a(c-a) - b(2a+b) + c(2b-c) &= 2ac - 2a^2 - 2ab - b^2 + 2bc - c^2 \\\\\n&= -a^2 - (a+b-c)^2.\n\\end{aligned}\n$$\nThe equality $-a^2 - (a+b-c)^2 = 2020$ cannot be valid since all terms of its l.h.s. are non-positive whereas the r.h.s. is positive.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 29962,
    "subject": "Mathematics (Multi-modal)",
    "question": "The bisector of the angle on vertex $A$ of triangle $ABC$ intersects the circumcircle of triangle $ABC$ at point $F$ ($F \\neq A$). Points $D$ and $E$ are chosen on the sides $AB$ and $AC$, respectively, in such a way that the lines $DE$ and $BC$ are parallel. Let $G$ and $H$ be the points of intersection of the rays $FD$ and $FE$, respectively, with the circumcircle of triangle $ABC$ ($G \\neq F, H \\neq F$). The circumcircles of triangles $AGD$ and $AHE$ intersect at point $P$ ($P \\neq A$). Prove that point $P$ lies on the line $AF$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $K$ and $L$ be the points of intersection of the line $AF$ with lines $BC$ and $DE$, respectively (Fig. 6). Then\n$$\n\\begin{aligned}\n\\angle AGD &= \\angle AGF = \\angle AGC + \\angle CGF = \\angle ABC + \\angle CAF \\\\\n&= \\angle ABK + \\angle KAB = \\angle CKA = \\angle KLD = 180^\\circ - \\angle ALD.\n\\end{aligned}\n$$\nHence the quadrilateral $AGDL$ is cyclic. Interchanging the roles of points $B$ and $C$, points $D$ and $E$ and also points $G$ and $H$, we can similarly prove that the quadrilateral $AHEL$ is cyclic. Thus the circumcircles of triangles $ADG$ and $AEH$ meet at point $L$, i.e., $P = L$. Point $L$ was chosen on the line $AF$.\n\n![](attached_image_1.png)\nFig. 6\nChoose a point $Z$ on the same side of the line $AF$ as point $B$ such that $ZF \\parallel BC \\parallel DE$ (Fig. 7). Then $\\angle ZFB = \\angle FBC = \\angle FAC = \\angle FAB = \\angle FCB$. Hence $ZF$ is tangent to the circumcircle of the triangle $ABC$ at $F$, implying that $\\angle GFZ = \\angle GHF = \\angle GHE$. As $\\angle GFZ = \\angle DFZ = \\angle FDE = 180^\\circ - \\angle GDE$, we altogether have $\\angle GHE = 180^\\circ - \\angle GDE$. Consequently, the quadrilateral $DEHG$ is cyclic. Hence $|FD| \\cdot |FG| = |FE| \\cdot |FH|$, implying that $F$ lies on the radical axis of the circumcircles of triangles $ADG$ and $AEH$. The desired result follows.\n\n![](attached_image_2.png)\nFig. 7",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 29963,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all integers $n \\ge 3$ such that one can write a number (not necessarily an integer) into each vertex of a regular $n$-gon in such a way that both following conditions are met:\n(1) Whenever three consecutive vertices of the $n$-gon, taken clockwise, contain numbers $x, y$ and $z$, respectively, the equality $x = |y - z|$ holds;\n(2) The sum of the numbers in all vertices of the $n$-gon is 1.",
    "options": [],
    "answer": "All integers n that are multiples of three",
    "solution": "Let the vertices of an $n$-gon $A_0A_1\\dots A_{n-1}$ be labeled with numbers satisfying the conditions. Let $a$ be the least among these numbers. W.l.o.g., assume that $A_0$ contains $a$ and the indices of vertices are increasing counterclockwise. Let $b$ and $c$ be the numbers at vertices $A_{n-1}$ and $A_{n-2}$, respectively (Fig. 8). By condition (1), $a = |b-c| \\ge 0$. By the choice of $a$, we must have $b \\ge a$, whence by condition (1), $A_1$ contains $b-a$. As $b-a \\ge a$ by the choice of $a$, the condition (1) also implies that $A_2$ contains $b-2a$. Hence $A_3$ contains $(b-a) - (b-2a) = a$ by condition (1) and non-negativity of $a$. Since the vertices can be renumerated without changing the direction in such a way that $A_3$ becomes $A_0$, we can conclude that $A_6$ also contains $a$. Similarly, every third vertex contains $a$.\n\nIf the number $n$ of vertices is not divisible by 3 then either $A_{n-1}$ or $A_1$ must contain $a$ and, as we can repeat this argument, also $A_{n-2}$ or $A_2$, respectively, contains $a$. Thus three consecutive vertices contain $a$. Applying the condition (1) to these three vertices, we obtain $a = |a-a| = 0$. But 0 being in two consecutive vertices implies that 0 is in all vertices. Then the sum of all labels is 0, contradicting the condition (2).\n\nThis shows that $n$ must be divisible by 3. Let $n = 3k$ where $k$ is a positive integer. For every $3k$-gon, the conditions of the problem can be satisfied by writing 0 into every third vertex and $\\frac{1}{2k}$ into all other vertices (Fig. 9 depicts the situation for $k=4$).\n\n![](attached_image_1.png)\n![](attached_image_2.png)",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 29964,
    "subject": "Mathematics (Multi-modal)",
    "question": "A set of 8 dominoes is given, each consisting of two unit squares:\n![](attached_image_1.png)\nIs it possible to completely cover a rectangular grid of size $4 \\times 4$ with these dominoes in such a way that all rows and columns of the grid contain the same number of pips?",
    "options": [],
    "answer": "Yes",
    "solution": "Figure 10 shows one possibility.\n\n![](attached_image_2.png)",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 29965,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all quadruples $(p, q, r, s)$ of primes that satisfy the following system of equations:\n$$\n\\begin{cases} 6p + 5q + 5r + 3s = 130 \\\\ 3p + 3q + 5r + 6s = 130 \\end{cases}\n$$",
    "options": [],
    "answer": "(11, 3, 2, 13)",
    "solution": "Subtracting the second equation from the first one gives $3p + 2q - 3s = 0$ which implies $2q = 3(s - p)$. Thus $2q$ is divisible by $3$. As both $2$ and $q$ are primes, this implies $q = 3$. Substituting $q = 3$ into the initial system of equations and simplifying gives\n$$\n\\begin{cases} 6p + 5r + 3s = 115, \\\\ 3p + 5r + 6s = 121. \\end{cases} \\quad (1)\n$$\nIf $r$ and $s$ were both odd then also $5r$ and $3s$ would be odd, whence $5r + 3s$ would be even. As $6p$ is even, too, the left-hand side of the first equation of system (1) would be even and could not equal the right-hand side $115$. Thus one of $r$ and $s$ is even, i.e., $r = 2$ or $s = 2$. Analogously if $p$ and $r$ were both odd then the second equation of system (1) would give a contradiction; hence $p = 2$ or $r = 2$. Consequently, if $r \\neq 2$ then $p = s = 2$, but substituting $p = s$ into system (1) and subtracting the second equation from the first one gives $0 = -6$. The contradiction shows that $r = 2$. Substituting $r = 2$ into system (1) and simplifying gives\n$$\n\\begin{cases} 6p + 3s = 105, \\\\ 3p + 6s = 111. \\end{cases}\n$$\nSolving this equation gives $p = 11$ and $s = 13$.\nAs in Solution 1 we find $q = 3$ and obtain system (1). Subtracting the first equation from the second one in system (1) gives $3s - 3p = 6$, implying $s = p + 2$. The least pairs of prime numbers with difference $2$ are $(3,5)$, $(5,7)$ and $(11,13)$. The next such pair $(17,19)$ gives $6p + 3s > 115$, implying that no more pairs are suitable. Substituting the three pairs one by one into system (1) and simplifying each time gives $5r = 82$, $5r = 64$ and $5r = 10$, respectively. Only the last alternative leads to an integral solution $r = 2$, which is a prime, too. Consequently, the only possibility is $(p, q, r, s) = (11, 3, 2, 13)$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 29966,
    "subject": "Mathematics (Multi-modal)",
    "question": "In a math period, the teacher asks pupils to solve quadratic equations of the form $x^2 + px + q = 0$ where $p$ and $q$ are some integers. The teacher obtains every new equation by either increasing by 1 or decreasing by 1 the value of either $p$ or $q$ in the equation just solved. In the initial equation, $p = 2020$ and $q = 2010$, whereas in the last equation, $p = 2010$ and $q = 2020$. Is it definitely true that both solutions of at least one equation solved during the period are integers?",
    "options": [],
    "answer": "Yes",
    "solution": "In the first equation, one has $p - q = 10$, while in the last equation, one has $p - q = -10$. At each step, either $p$ or $q$ changes exactly by 1, whence also $p - q$ changes exactly by 1. Thus at some step one must have an equation where $p - q = 1$, or equivalently, $p = q + 1$. The equation $x^2 + (q + 1)x + q = 0$ has integral solutions $-1$ and $-q$.",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 29967,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABC$ be a triangle such that $AB = AC$. Point $K$ lies on the altitude drawn from vertex $A$ and point $L$ is chosen on the line $BK$ in such a way that $AL \\parallel BC$. Prove that if $KC \\perp CL$ then point $L$ lies on the bisector of the external angle on vertex $C$ of triangle $ABC$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "As $AL \\parallel BC$ and $AK \\perp BC$ (Fig. 11), we have $\\angle KAL = 90^\\circ$. As $KAL$ and $KCL$ are both right angles, points $A$ and $C$ lie on circle with diameter $KL$. By inscribed angles, $\\angle KCA = \\angle KLA$. On the other hand, $\\angle KLA = \\angle KBC$ and, by symmetry of isosceles triangle, $\\angle KBC = \\angle KCB$. Consequently, $\\angle KCA = \\angle KCB$, implying that $KC$ bisects the internal angle on vertex $C$ of the triangle $ABC$. As $KCL$ is right angle, $CL$ bisects the corresponding external angle.\nThe altitude drawn from the vertex angle of the isosceles triangle $ABC$ is the perpendicular bisector of its base. All points of the perpendicular bisector of a line segment lie at equal distance from the endpoints of the line segment, implying that $|KB| = |KC|$. As $AL \\parallel BC$ and $AK \\perp BC$, we conclude $AK \\perp AL$.\n![](attached_image_1.png)\nFig. 11\nLet $P$ be a point on line $BC$ such that $C$ is between $B$ and $P$, and let lines $CK$ and $AL$ intersect at $Q$ (Fig. 12). As $\\angle BKC = \\angle LKQ$ and $\\angle KBC = \\angle KLQ$, triangles $KBC$ and $KLQ$ are similar. Together with the equality $|KB| = |KC|$, it implies $|KL| = |KQ|$. Thus $KA$ is the altitude drawn from the vertex angle of the isosceles triangle $KLQ$, implying also $|AL| = |AQ|$. Hence $A$ is the midpoint of the hypotenuse of the right triangle $CLQ$, implying that $A$ is the center of the circumcircle of the triangle $CLQ$. Consequently, $|AC| = |AL|$, implying that $\\angle ACL = \\angle ALC = \\angle LCP$, i.e., $CL$ bisects the external angle on vertex $C$ of the triangle $ABC$.\n![](attached_image_2.png)\nFig. 12\nWhen point $K$ moves along the altitude drawn from vertex $A$ of the triangle $ABC$ farther from point $A$, the angle $KCB$ decreases, while point $L$ also moves farther from point $A$, causing the angle $LCB$ to increase. Thus the difference $\\angle LCB - \\angle KCB$ also increases in this process and can equal $90^\\circ$ in the case of exactly one location of point $K$. Hence it suffices to show that if $L$ lies at the bisector of the external angle on vertex $C$ of the triangle $ABC$ then $\\angle LCK = 90^\\circ$.\nSo let $L$ lie at the bisector of the external angle on vertex $C$ of the triangle $ABC$ (Fig. 13). As $AL \\parallel BC$ and $AK \\perp BC$, we have $AK \\perp AL$; as $AK$ is the altitude drawn from the vertex angle of the triangle $ABC$, it bisects the vertex angle, whence also $L$ lies on the bisector of the external angle on vertex $A$ of the triangle $ABC$. Thus $L$ is the center of the excircle tangent to side $AC$ of the triangle $ABC$ and lies also on the bisector of the internal angle on vertex $B$ of the triangle $ABC$. Hence both $AK$ and $BK$ are angle bisectors, implying that $K$ is the intersection point of angle bisectors of the triangle $ABC$ and $CK$ is the bisector of the internal angle on vertex $C$ of the triangle $ABC$. Consequently, $CK$ and $CL$ are perpendicular.\n![](attached_image_3.png)\nFig. 13",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 29968,
    "subject": "Mathematics (Multi-modal)",
    "question": "Juku has the first 100 volumes of the Harrie Totter book series at his home. For every $i$ and $j$, where $1 \\le i < j \\le 100$, call the pair $(i, j)$ *reversed* if volume No $j$ is before volume No $i$ on Juku's shelf. Juku wants to arrange all volumes of the series to one row on his shelf in such a way that there does not exist numbers $i, j, k$, where $1 \\le i < j < k \\le 100$, such that pairs $(i, j)$ and $(j, k)$ are both reversed. Find the largest number of reversed pairs that can occur under this condition.",
    "options": [],
    "answer": "2500",
    "solution": "Let all 100 volumes be placed to a shelf $a$ in some order. For every $i = 1, 2, ..., 100$, let $a_i$ be the number of the volume that occurs as the $i$th in this order. In the order the volumes occur on shelf $a$, we start relocating the volumes to two new shelves $b$ and $c$. We place a volume to the end of shelf $b$ if, as the intermediate result, all volumes on shelf $b$ would be increasingly sorted by volume numbers. Otherwise, we place the volume to the end of shelf $c$ if, as the intermediate result, all volumes on shelf $c$ would be increasingly sorted by volume numbers. Continuing this way, we either can relocate all volumes onto two shelves in such a way that all volumes on either shelf are increasingly sorted by volume numbers or get stuck on some step $k$ of the process because $a_k$ is less than the number of the last volume on both new shelves. In the last case, let the number of the last volume on shelf $c$ be $a_j$; then, by assumptions, $j < k$ and $a_j > a_k$. Since volume No $a_j$ has been relocated to shelf $c$, some volume with number $a_i$ must occur on shelf $b$ such that $i < j$ and $a_i > a_j$. This means that pairs $(i, j)$ and $(j, k)$ were initially reversed. Hence we can conclude that, in the case of Juku's favourite orderings, all volumes can be relocated to two shelves, i.e, there exist two tuples $a_{i_1}, a_{i_2}, ..., a_{i_s}$ and $a_{j_1}, a_{j_2}, ..., a_{j_t}$ with $s + t = 100$, where $i_1 < ... < i_s$, $a_{i_1} < ... < a_{i_s}$ and $j_1 < ... < j_t$, $a_{j_1} < ... < a_{j_t}$. The number of pairs $(i, j)$ such that $i < j$ and the numbers $a_i$ and $a_j$ are in distinct tuples is exactly $st$. Each reversed pair $(i, j)$ must be one of these $st$ pairs. Thus the number of reversed pairs does not exceed $st$. As $st \\le (\\frac{s+t}{2})^2 = 50^2 = 2500$, the number of reversed pairs cannot exceed 2500. The number 2500 is achieved by the order $51, 52, ..., 100, 1, 2, ..., 50$.\nLet all 100 volumes be placed to the shelf in some order. Consider the graph whose vertices are numbers $1, 2, ..., 100$ and an edge occurs between vertices $i$ and $j$ if and only if either $(i, j)$ or $(j, i)$ (depending on whether $i < j$ or $j < i$) is reversed. Suppose that the graph contains an odd cycle. Such cycle must contain three consecutive vertices $i, j$ and $k$ that are in either increasing or decreasing order; w.l.o.g., assume that $i < j < k$. Then pairs $(i, j)$ and $(j, k)$ are both reversed. Thus under the conditions of the problem, the graph cannot contain an odd cycle. Hence the graph is bipartite. Let the numbers of vertices in the two parts be $s$ and $t$, respectively; then the maximal number of edges is $st$. By AM-GM, $st \\le (\\frac{s+t}{2})^2 = 50^2 = 2500$. If the volumes are in the order $51, 52, ..., 100, 1, 2, ..., 50$, exactly 2500 reversed pairs arise indeed.",
    "topic": "Geometry",
    "subtopic": "Solid Geometry"
  },
  {
    "id": 29969,
    "subject": "Mathematics (Multi-modal)",
    "question": "a. There are $2n$ rays marked in a plane, with $n$ being a natural number. Given that no two marked rays have the same direction and no two marked rays have a common initial point, prove that there exists a line that passes through none of the initial points of the marked rays and intersects with exactly $n$ marked rays.\n\nb. Would the claim still hold if the assumption that no two marked rays have a common initial point was dropped?",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Consider any circle such that all of the endpoints of the marked rays are inside the circle. Choose a tangent line of the circle that is not parallel to any of the marked rays. Let this tangent line be $l_0$ and let $l_\\alpha$ be the tangent line we get by rotating $l_0$ counterclockwise by angle $\\alpha$ with respect to the center of the circle. For any $\\alpha$, let $f(\\alpha)$ be the number of marked rays that intersect with $l_\\alpha$. Since $l_0$ and $l_\\pi$ are two parallel lines and all of the endpoints of the marked rays are between them, we know from the definition of $l_0$ that $f(0) + f(\\pi) = 2n$. Without loss of generality, let $f(0) \\le n$ and $f(\\pi) \\ge n$. Because no two rays have the same direction, when $\\alpha$ increases continuously $f(\\alpha)$ can at any time instance only change by at most one. Thus $f(\\alpha)$ ranges over all integral values between $f(0)$ and $f(\\pi)$. Hence $f(\\alpha) = n$ for some $\\alpha$.\n\nThe argument above does not use the assumption that the initial points of the marked rays are distinct, whence it solves both parts of the problem.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 29970,
    "subject": "Mathematics (Multi-modal)",
    "question": "Points $A$, $B$ and $C$ are chosen in a rectangle of shape $7 \\times 10$ in such a way that the distances from $A$ to some three sides of the rectangle are $2$, $3$ and $4$, the distances from $B$ to some three sides of the rectangle are $3$, $4$ and $5$, and the distances from $C$ to some three sides of the rectangle are $4$, $5$ and $6$. Find the largest possible area the triangle $ABC$ can have.",
    "options": [],
    "answer": "7/2",
    "solution": "![](attached_image_1.png)\n\n![](attached_image_2.png)\n\n![](attached_image_3.png)\n\n![](attached_image_4.png)\n\nAnswer: $\\frac{7}{2}$.\n\n**Solution:** Introduce a Cartesian coordinate system with origin $O$ at one vertex of the rectangle. Let the other vertices of the rectangle be $P = (10,0)$, $Q = (10,7)$ and $R = (0,7)$. Given the distances from a point to three sides of the rectangle, two of these sides must be opposite sides and the corresponding distances sum up to the length of the perpendicular side of the rectangle. Thus $A$ must lie at distances $3$ and $4$ from the horizontal sides and at distances $2$ and $8$ from the vertical sides. W.l.o.g., let $A = (2,3)$. Similarly, $B$ must lie at distances $3$ and $4$ from the horizontal sides and at distances $5$ and $5$ from the vertical sides. Thus $B = B_1 = (5,3)$ or $B = B_2 = (5,4)$ (Fig. 16). Finally, $C$ must lie at distances $4$ and $6$ from the vertical sides and at distances $2$ and $5$ from the horizontal sides. Thus $C = C_1 = (4,2)$ or $C = C_2 = (4,5)$ or $C = C_3 = (6,2)$ or $C = C_4 = (6,5)$ (Fig. 17).\n\nAltogether, the triangle $ABC$ can be located in $8$ ways. We study which one yields the largest area. If $B = B_1 = (5,3)$, take the side $AB$ with length $3$ as base. The corresponding altitude has length $1$ if $C = C_1 = (4,2)$ or $C = C_3 = (6,2)$ and $2$ if $C = C_2 = (4,5)$ or $C = C_4 = (6,5)$ (Fig. 18). Thus the largest area in the observed cases is $3$. Consider now the case $B = B_2 = (5,4)$. Reflecting the points $C_1$ and $C_3$ from the midpoint of the side $AB$, we obtain points $C'_1 = (3,5)$ and $C'_3 = (1,5)$ which yield the same areas (Fig. 19). Points $C_2$, $C_4$, $C'_1$ and $C'_3$ lie on the line $y = 5$, whereby the one with the least x-coordinate, the point $C'_3$, is the farthest from the line $AB$. Hence the largest area arises in the case $C = C_3 = (6,2)$. The size of the rectangle surrounded by lines $y = 2$, $y = 4$, $x = 2$ and $x = 6$ is $2 \\times 4$. Removing three right triangles to extract the triangle $ABC$ (Fig. 20) enables to express the area of the triangle $ABC$ as $2 \\cdot 4 - \\frac{1}{2} \\cdot 1 \\cdot 3 - \\frac{1}{2} \\cdot 1 \\cdot 2 - \\frac{1}{2} \\cdot 1 \\cdot 4 = \\frac{7}{2}$. As $\\frac{7}{2} > 3$, the largest possible area of the triangle $ABC$ is $\\frac{7}{2}$.\n\n![](attached_image_5.png)",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 29971,
    "subject": "Mathematics (Multi-modal)",
    "question": "There are sticks of length $1$ with a number $1$, $2$ or $3$ written on each of them. There is an unlimited supply of sticks with every number. Two triangles consisting of three sticks are considered different if neither of the triangles can be composed from sticks of the other triangle.\n\na. How many different triangles consisting of three sticks are possible?\n\nb. From $18$ sticks, one makes an equilateral triangle of side length $3$, divided into $9$ pairwise different equilateral triangles of side length $1$. Find the largest possible sum of the numbers written on the $9$ sticks on the boundary of the big triangle.\n\n![](attached_image_1.png)",
    "options": [],
    "answer": "a: 10, b: 26",
    "solution": "a.\nThere are $3$ triangles having the same number on each side ($111$, $222$, $333$). There are $6$ triangles having one number on two sides and another number on the third side ($112$, $113$, $221$, $223$, $331$, $332$). Only $1$ triangle has a different number on every side ($123$). Thus there are $10$ different triangles consisting of three sticks in total.\n\nb.\nThe largest sum of numbers written on $9$ sticks is $27$. Suppose that the sum of the numbers on the sticks on the boundary of the big triangle is $27$. This means that all sticks on the boundary have $3$ on it. There are $6$ different triangles consisting of three sticks having $3$ on some side. As the number of small triangles having a side on the boundary of the big triangle is also $6$, all $6$ different triangles consisting of three sticks and having $3$ on some side must occur at the corners or in the middle of a side of the big triangle. As one of these $6$ triangles has $3$ on every side, but not all sides of a small triangle can lie on the boundary of the big triangle, at least one stick with number $3$ lies in the interior of the big triangle. The other small triangle with this stick as a side lies entirely in the interior of the big triangle. This contradicts the previously proved claim that all triangles consisting of three sticks and having $3$ on one side lie on the boundary of the big triangle. The contradiction shows that the sum $27$ is impossible.\n\nFigure 24 shows that the sum can be $26$.\n\n![](attached_image_2.png)\n\nAny of Figures 25, 26, 27, 28 and 29 shows that the sum can be $26$.\n\n*Remark:* Figures 24–29 contain all possibilities, modulo rotations and reflections, for obtaining the sum $26$.\n\n![](attached_image_3.png)\n![](attached_image_4.png)\n![](attached_image_5.png)\n![](attached_image_6.png)\n![](attached_image_7.png)\na.\nThe question of the problem is equivalent to the question how many three-digit numbers whose each digit is $1$, $2$ or $3$ and digits are in non-decreasing order do there exist. There are $10$ such numbers ($111$, $112$, $113$, $122$, $123$, $133$, $222$, $223$, $233$, $333$).\n\n![](attached_image_2.png)\n\nb.\nThe largest sum of numbers written on $9$ sticks is $27$. Suppose that the sum of the numbers on the sticks on the boundary of the big triangle is $27$. This means that all sticks on the boundary have $3$ on it. In all $10$ triangles consisting of three sticks, the number $3$ occurs $10$ times in total. To have $9$ occurrences on the boundary, the triangle with $3$ on every side must definitely be used. As not all sides of a small triangle can lie on the boundary of the big triangle, at least one stick with number $3$ lies in the interior of the big triangle. Thus all $10$ occurrences must be used. But the stick in the interior of the big triangle and having $3$ on it is a side of another triangle entirely in the interior of the big triangle, whence we have $11$ occurrences of $3$ when counted by triangles. The contradiction shows that the sum $27$ is impossible.\n\nAny of Figures 25, 26, 27, 28 and 29 shows that the sum can be $26$.\n\n*Remark:* Figures 24–29 contain all possibilities, modulo rotations and reflections, for obtaining the sum $26$.\n\n![](attached_image_3.png)\n![](attached_image_4.png)\n![](attached_image_5.png)\n![](attached_image_6.png)\n![](attached_image_7.png)",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 29972,
    "subject": "Mathematics (Multi-modal)",
    "question": "A point $M$ is chosen on the side $AC$ of a triangle $ABC$ and a point $K$ is chosen on the line segment $BM$ so that $AM = \\frac{1}{3}AC$ and $BK = \\frac{1}{4}BM$. Let $N$ be the intersection of the line $AK$ and the side $BC$. What percentage of the area of the triangle $ABC$ is the area of the quadrilateral $MKNC$?",
    "options": [],
    "answer": "65%",
    "solution": "*Answer:* 65%.\n\nLet the area of the triangle $ABC$ be $S$ and the areas of triangles $AKM$, $BKN$, $MKN$ and $CMN$ be $S_1$, $S_2$, $S_3$ and $S_4$, respectively (Fig. 30). Then:\n\n* $S_2 + S_3 + S_4 = \\frac{2}{3}S$ since the l.h.s. is the area of the triangle $MBC$ while triangles $MBC$ and $ABC$ have equal altitudes and the ratio of the lengths of the corresponding bases is $\\frac{2}{3}$;\n* $S_1 + S_3 + S_4 = \\frac{3}{2}S_4$ since the l.h.s. is the area of the triangle $ANC$ while triangles $ANC$ and $MNC$ have equal altitudes and the ratio of the lengths of the corresponding bases is $\\frac{3}{2}$;\n* $S_2 = \\frac{1}{3}S_3$ because triangles $BNK$ and $MNK$ have equal altitudes and the ratio of the lengths of the corresponding bases is $\\frac{1}{3}$;\n* $S_1 = \\frac{1}{4}S$ because the ratio of altitudes of the triangles $AKM$ and $ABC$ is $\\frac{3}{4}$ while the ratio of the lengths of the corresponding bases is $\\frac{1}{3}$.\n\n![](attached_image_1.png)\n\nSubstituting $S_1$ and $S_2$ from the latter two equations to the former two equations, we obtain the system of equations\n$$\n\\begin{cases} \\frac{1}{3}S_3 + S_3 + S_4 = \\frac{2}{3}S, \\\\ \\frac{1}{4}S + S_3 + S_4 = \\frac{3}{2}S_4. \\end{cases}\n$$\nAfter adding $\\frac{1}{3}S_4$ to both sides of the first equation and bringing the term with $S$ to the l.h.s., we obtain the equivalent system\n$$\n\\begin{cases} \\frac{4}{3}(S_3 + S_4) - \\frac{2}{3}S = \\frac{1}{3}S_4, \\\\ (S_3 + S_4) + \\frac{1}{4}S = \\frac{3}{2}S_4. \\end{cases}\n$$\nSolving with respect to $S_3 + S_4$ and $S_4$ yields $S_3 + S_4 = 0.65S$. Hence the area of the quadrilateral $MKNC$ equals 65% of the area of the triangle $ABC$.\n\n![](attached_image_1.png)\nDenote the area of a figure $\\Pi$ by $S_{\\Pi}$. From the conditions of the problem, we obtain the following:\n\n* $\\frac{S_{ABM}}{S_{ABC}} = \\frac{|AM|}{|AC|} = \\frac{1}{3}$, whence $S_{ABM} = \\frac{1}{3}S_{ABC}$ and $S_{CBM} = \\frac{2}{3}S_{ABC}$;\n* $\\frac{S_{ABK}}{S_{ABM}} = \\frac{|BK|}{|BM|} = \\frac{1}{4}$, whence $S_{ABK} = \\frac{1}{4} \\cdot \\frac{1}{3}S_{ABC} = \\frac{1}{12}S_{ABC}$ and, consequently, $S_{AMK} = (\\frac{1}{3} - \\frac{1}{12})S_{ABC} = \\frac{1}{4}S_{ABC}$;\n* $\\frac{S_{BCK}}{S_{BCM}} = \\frac{|BK|}{|BM|} = \\frac{1}{4}$ (Fig. 31), whence $S_{BCK} = \\frac{1}{4} \\cdot \\frac{2}{3}S_{ABC} = \\frac{1}{6}S_{ABC}$ and $S_{MCK} = (\\frac{2}{3} - \\frac{1}{6})S_{ABC} = \\frac{1}{2}S_{ABC}$.\n\nSince $\\frac{S_{ABN}}{S_{ACN}} = \\frac{|BN|}{|CN|} = \\frac{S_{KBN}}{S_{KCN}}$, we obtain\n$$\n\\frac{S_{ABN}}{S_{ACN}} = \\frac{S_{ABN} - S_{KBN}}{S_{ACN} - S_{KCN}} = \\frac{S_{ABK}}{S_{AMK} + S_{MCK}} = \\frac{\\frac{1}{12}}{\\frac{1}{4} + \\frac{1}{2}} = \\frac{\\frac{1}{12}}{\\frac{3}{4}} = \\frac{1}{9}.\n$$\n![](attached_image_2.png)\n\nHence $\\frac{S_{ABN}}{S_{ABC}} = \\frac{S_{ABN}}{S_{ABN}+S_{ACN}} = \\frac{S_{ABN}}{S_{ABN}+9S_{ABN}} = \\frac{1}{10}$, implying $S_{ABN} = \\frac{1}{10}S_{ABC}$.\n\nAltogether, we obtain\n$$\n\\frac{S_{MKNC}}{S_{ABC}} = \\frac{S_{ABC} - S_{ABN} - S_{AMK}}{S_{ABC}} = 1 - \\frac{1}{10} - \\frac{1}{4} = 0.65,\n$$\nmeaning that the area of the quadrilateral $MKNC$ equals 65% of the area of the triangle $ABC$.\n\n![](attached_image_2.png)",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 29973,
    "subject": "Mathematics (Multi-modal)",
    "question": "When adding together positive integers $a$ and $b$, Juku forgot to enter the final digit 7 of the number $a$ and got a result of 2022. Had Juku instead forgotten to enter the final digit of $b$, the result would have been 5000. Find the sum of $a$ and $b$.",
    "options": [],
    "answer": "6385",
    "solution": "Let $a = \\overline{x7}$ and $b = \\overline{yd}$, where $d$ is the final digit of $b$ and $x, y$ are the numbers $a$ and $b$ without their final digits. Then $a = 10x + 7$ and $b = 10y + d$. From the given information, we compile the system of equations\n$$\n\\begin{cases} x + 10y + d = 2022, \\\\ y + 10x + 7 = 5000. \\end{cases} \\qquad (1)\n$$\nSubtracting the first equation from the second one, we get the equation\n$$\n9(x - y) - d = 2971. \\qquad (2)\n$$\nAs $x, y$ are integers, $9(x - y)$ is divisible by 9. As the number 2971 gives a remainder of 1 upon division by 9, the number $d$ must give a remainder of 8; the only option for this is $d = 8$. Substituting this into (2), we get $x - y = 331$, which we can substitute into either equation of (1) to get $x = 484$ and $y = 153$. Therefore $a + b = 4847 + 1538 = 6385$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 29974,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find the sum\n$$\n\\sqrt{1 + \\frac{1}{1^2} + \\frac{1}{2^2}} + \\sqrt{1 + \\frac{1}{2^2} + \\frac{1}{3^2}} + \\dots + \\sqrt{1 + \\frac{1}{2021^2} + \\frac{1}{2022^2}}\n$$",
    "options": [],
    "answer": "2021 + 2021/2022",
    "solution": "We will denote\n$$\ns = \\sqrt{1 + \\frac{1}{1^2} + \\frac{1}{2^2}} + \\sqrt{1 + \\frac{1}{2^2} + \\frac{1}{3^2}} + \\dots + \\sqrt{1 + \\frac{1}{2021^2} + \\frac{1}{2022^2}}\n$$\nNotice that\n$$\n1 + \\frac{1}{k^2} + \\frac{1}{(k+1)^2} = \\frac{k^2(k+1)^2 + (k+1)^2 + k^2}{k^2(k+1)^2} = \\frac{(k(k+1)+1)^2}{(k(k+1))^2},\n$$\nwhich yields\n$$\n\\sqrt{1 + \\frac{1}{k^2} + \\frac{1}{(k+1)^2}} = \\frac{k(k+1)+1}{k(k+1)} = 1 + \\frac{1}{k(k+1)} = 1 + \\frac{1}{k} - \\frac{1}{k+1}.\n$$\nTherefore\n$$\ns = 2021 + \\left(\\frac{1}{1} - \\frac{1}{2}\\right) + \\left(\\frac{1}{2} - \\frac{1}{3}\\right) + \\dots + \\left(\\frac{1}{2021} - \\frac{1}{2022}\\right) = 2021 + \\frac{2021}{2022}.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 29975,
    "subject": "Mathematics (Multi-modal)",
    "question": "Points $A$, $D$, $E$ and $C$ lie on a line in this order. Point $B$ is chosen such that triangles $ADB$ and $BEC$ are similar (in this specific order of vertices), moreover $EC = 2AD$ and $\\angle ABC = 120^\\circ$. Find $\\frac{AC}{AD}$.",
    "options": [],
    "answer": "3 + sqrt(2)",
    "solution": "**Answer:** $3 + \\sqrt{2}$.\n\nDenote $\\angle EDB = \\alpha$ (Fig. 32). The similarity of $ADB$ and $BEC$ yields $\\angle DAB = \\angle EBC$ and $\\angle BDA = \\angle CEB$; the second of which gives $\\angle DEB = 180^\\circ - \\angle CEB = 180^\\circ - \\angle BDA = \\angle EDB = \\alpha$. Therefore $DB = BE$. Hence the similarity of $ADB$ and $BEC$ yields $\\frac{BE}{AD} = \\frac{EC}{DB} = \\frac{EC}{BE}$; thus $\\left(\\frac{BE}{AD}\\right)^2 = \\frac{BE}{AD} \\cdot \\frac{EC}{BE} = \\frac{EC}{AD} = 2$, which gives $\\frac{BE}{AD} = \\sqrt{2}$ or $BE = \\sqrt{2}AD$.\n\nThe isosceles triangle $BDE$ gives $\\angle DBE = 180^\\circ - 2\\alpha$. On the other hand,\n$$\n\\begin{aligned}\n\\angle DBE &= 120^\\circ - \\angle ABD - \\angle EBC = 120^\\circ - \\angle ABD - \\angle DAB \\\\\n&= 120^\\circ - (\\angle ABD + \\angle DAB) = 120^\\circ - \\alpha.\n\\end{aligned}\n$$\nThe equation $180^\\circ - 2\\alpha = 120^\\circ - \\alpha$ yields $\\alpha = 60^\\circ$. So the triangle $BDE$ is equilateral, which yields $DE = BE = \\sqrt{2}AD$.\n\nTherefore $\\frac{AC}{AD} = \\frac{AD+DE+EC}{AD} = \\frac{AD+\\sqrt{2}AD+2AD}{AD} = 1 + \\sqrt{2} + 2 = 3 + \\sqrt{2}$.\n\n![](attached_image_1.png)\nThe similarity of $ADB$ and $BEC$ yields $\\angle DAB = \\angle EBC$ and $\\angle ABD = \\angle BCE$. So the triangles $ADB$ and $BEC$ are also similar to $ABC$. Therefore $\\angle ADB = \\angle BEC = \\angle ABC = 120^\\circ$. But this yields $\\angle BDE = \\angle DEB = 180^\\circ - 120^\\circ = 60^\\circ$, meaning that the triangle $BDE$ is equilateral.\n\nThe similarity of $ADB$ and $ABC$ yields $\\frac{AD}{AB} = \\frac{AB}{AC}$. The similarity of $BEC$ and $ABC$ yields $\\frac{EC}{BC} = \\frac{BC}{AC}$. Combining these with $2AD = EC$, we see that\n\n![](attached_image_1.png)\n\nFig. 32\n\n$$\nBC^2 = EC \\cdot AC = 2AD \\cdot AC = 2AB^2.\n$$\nSo $BC = \\sqrt{2}AB$, meaning that the scale factor between triangles $ADB$ and $BEC$ is $\\sqrt{2}$. Thus $DE = BD = \\sqrt{2}AD$.\n\n$$\n\\text{Therefore } \\frac{AC}{AD} = \\frac{AD+DE+EC}{AD} = \\frac{AD+\\sqrt{2}AD+2AD}{AD} = 1+\\sqrt{2}+2 = 3+\\sqrt{2}.\n$$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 29976,
    "subject": "Mathematics (Multi-modal)",
    "question": "A fly farm contains $100001$ fruit flies. A research group wishes to buy some flies from the farm to perform (once) either experiment A or experiment B (but not both).\n\nFor experiment A, the research group needs a set of flies in which no fly is a descendant of any other. For such a set, the research group would pay the farm $505$ euros plus $5$ euros per fly bought.\n\nFor experiment B, the research group needs a set of flies in which among every two flies, one is a descendant of the other. For such a set, the research group would pay the farm $1000$ euros plus $10$ euros per fly bought.\n\nProve that there exists a set of flies in the farm for which the research group would pay at least $3010$ euros.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "For $k = 1, 2, \\dots$, we call a fly a $k$-th order parent, if the greatest suitable set for experiment B containing this fly as the oldest fly consists of exactly $k$ flies.\n\nIf there exists a suitable set for experiment B with at least $201$ flies, then the research group would pay at least $1000 + 10 \\cdot 201 = 3010$ euros for it, which we wanted to show.\n\nNow, assuming there is no such set, each suitable set for experiment B contains $200$ or fewer flies. Then each fly will be a parent of order at most $200$. Since $100001 : 200 > 500$, the pigeonhole principle implies that there exists a $k$ such that at least $501$ flies are parents of order exactly $k$.\n\nOut of two flies who are parents of the same order, one can clearly never be a descendant of the other, meaning the set of all parents of order exactly $k$ will be suitable for experiment A. For such a set, the research group would pay at least $505 + 5 \\cdot 501 = 3010$ euros, which we wanted to show.",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 29977,
    "subject": "Mathematics (Multi-modal)",
    "question": "The radius of the circumcircle of an acute triangle $ABC$ is $R$ and its orthocenter is $H$. Show that $AH^2 + BC^2 = 4R^2$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $O$ and $G$ be the circumcenter and centroid of $ABC$ respectively and let $K$ be the midpoint of $BC$ (Fig. 33).\n\nWe know that $AG = 2GK$. Also we know that $H$, $G$ and $O$ are collinear with $HG = 2GO$ (Euler line). So triangles $AHG$ and $KOG$ are similar with scale factor $2$ (by $2$ proportional sides and an equal angle between them). Therefore $AH = 2KO$.\n\nNow the Pythagorean theorem in triangle $KOB$ yields $KO^2 + KB^2 = OB^2 = R^2$ and $AH^2 + BC^2 = (2KO)^2 + (2KB)^2 = 4(KO^2 + KB^2) = 4R^2$.\n\n![](attached_image_1.png)\nLet $B'$ be the other end of the diameter to the circumcircle of $ABC$ drawn from $B$ (Fig. 34). Since $AB'$ and $AB$ are perpendicular and $CH$ and $AB$ are perpendicular, the lines $AB'$ and $CH$ are parallel. Similarly, we see that $CB'$ and $AH$ are parallel, meaning that $AHCB'$ is a parallelogram. Thus $CB' = AH$.\n\nNow the Pythagorean theorem in triangle $BCB'$ yields $CB'^2 + BC^2 = BB'^2$. As $CB' = AH$ and $BB' = 2R$, the desired result follows.\n\n![](attached_image_2.png)",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 29978,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all polynomials $P(x)$ with integral coefficients and the following property: for any pair $(u, v)$ of positive integers, $\\text{gcd}(u, v) = 1$ implies $\\text{gcd}(|P(u)|, |P(v)|) = 1$.",
    "options": [],
    "answer": "All polynomials P(x) = ± x^l for non-negative integers l.",
    "solution": "*Answer:* All polynomials of the form $P(x) = \\pm x^l$ where $l$ is a non-negative integer.\n\nFirstly, we show that all prime divisors of $P(n)$ are divisors of $n$. Suppose that there is a prime number $q$ dividing $P(n)$ but not dividing $n$. Obviously $P(n+q) \\equiv P(n) \\equiv 0 \\pmod{q}$; but then $\\text{gcd}(n, n+q) = 1$ while $\\text{gcd}(|P(n)|, |P(n+q)|) \\ge q > 1$, contradicting the assumption.\n\nNext we show that only polynomials of the form $P(x) = \\pm x^l$ where $l \\ge 0$ satisfy the conditions of the problem. We proceed by induction on the degree of $P(x)$. If $P(x)$ is a constant polynomial then the constant must be coprime with itself in order to satisfy the conditions of the problem. This is possible only if $P(x) = \\pm 1 = \\pm x^0$. Assume now that $P(x)$ is a non-constant polynomial of degree $l$. Then $|P(q)| > 1$ for infinitely many prime numbers $q$. In each such case, namely $q$ must divide $P(q)$. Hence the constant term of $P(x)$ must be divisible by $q$. As the constant term does not depend on $q$, it must be divisible by infinitely many prime numbers. Hence the constant term equals 0, i.e., $P(x) = xP_1(x)$ where $P_1(x)$ is a polynomial with integral coefficients. The polynomial $P_1(x)$ satisfies the conditions of the problem, because $\\text{gcd}(|P_1(u)|, |P_1(v)|) > 1$ for coprime $u$ and $v$ would imply $\\text{gcd}(|P(u)|, |P(v)|) = \\text{gcd}(|uP_1(u)|, |vP_1(v)|) > 1$, contradicting the choice of $P(x)$. By the induction hypothesis, $P_1(x) = \\pm x^{l-1}$. Thus $P(x) = \\pm x^l$.\n\nOn the other hand, all such polynomials satisfy the condition of the problem as $\\text{gcd}(u, v) = 1$ implies $\\text{gcd}(u^l, v^l) = 1$.\nAs in Solution 1, we show that all prime divisors of $P(n)$ are divisors of $n$. Letting $q$ be any prime number, we conclude that $P(q) = \\pm q^k$ for some natural number $k$. Let the degree of $P(x)$ be $l$. Then $x^{l+1} > |P(x)|$ for every $x > N$ where $N$ is a large enough integer. Thus for infinitely many prime numbers $q$, there exists a natural number $k$ not exceeding $l$ such that $P(q) = \\pm q^k$. As there are finitely many such natural numbers, one can fix $k$ such that either $P(q) = q^k$ for infinitely many primes $q$ or $P(q) = -q^k$ for infinitely many primes $q$. This implies that either $P(x) = x^k$ or $P(x) = -x^k$ (as a matter of fact, $k = l$). All such polynomials satisfy the conditions of the problem as $\\text{gcd}(u, v) = 1$ implies $\\text{gcd}(u^l, v^l) = 1$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 29979,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all triples $(x, y, z)$ of real numbers that satisfy\n$$\n\\begin{cases}\n\\frac{x}{y} + \\frac{y}{z} + xy = 3, \\\\\n\\frac{y}{z} + \\frac{z}{x} + yz = 3, \\\\\n\\frac{z}{x} + \\frac{x}{y} + zx = 3.\n\\end{cases}\n$$",
    "options": [],
    "answer": "(1, 1, 1) and (-1, -1, -1)",
    "solution": "**Answer:** $(1, 1, 1), (-1, -1, -1)$.\n\nNumbers $x$, $y$ and $z$ must have the same sign, because if exactly one or two of them are negative then there exists an equation in the system whose all terms in the l.h.s. are negative and cannot sum up to $3$. It is also easy to see that $(x, y, z)$ being a solution implies $(-x, -y, -z)$ being a solution, too.\nHence, w.l.o.g., assume that $x$, $y$ and $z$ are positive. Subtracting the second equation from the first one, the third equation from the second one, and the first equation from the third one, we obtain the following new system:\n$$\n\\begin{cases}\n x \\left(y + \\frac{1}{y}\\right) = z \\left(y + \\frac{1}{x}\\right), \\\\\n y \\left(z + \\frac{1}{z}\\right) = x \\left(z + \\frac{1}{y}\\right), \\\\\n z \\left(x + \\frac{1}{x}\\right) = y \\left(x + \\frac{1}{z}\\right).\n\\end{cases}\n$$\nThe first equation of the new system shows that $x > z$ holds if and only if $y + \\frac{1}{y} < y + \\frac{1}{x}$, where the latter inequality obviously holds if and only if $y > x$. Similarly, the second equation implies that $y > x$ if and only if $z > y$, and the third equation implies that $z > y$ if and only if $x > z$. Thus any of the inequalities $x > z$, $y > x$ and $z > y$ yields the impossible cycle $x > z > y > x$. Consequently, we must have $x \\le z \\le y \\le x$ which implies $x = y = z$. Every equation of the original system now reduces to $1+1+x^2 = 3$. Hence $x = y = z = 1$.\nBesides the positive solution, the system has the corresponding negative solution $(-1, -1, -1)$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 29980,
    "subject": "Mathematics (Multi-modal)",
    "question": "The orthocenter and the circumcenter of a non-equilateral triangle $ABC$ are $H$ and $O$, respectively. Let $D$ be the foot of the altitude dropped from the vertex $A$ of the triangle $ABC$. Prove that $\\angle AHO = 90^\\circ$ if and only if $\\frac{AH}{HD} = 2$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $O'$ be the other endpoint of the diameter of the circumcircle of the triangle $ABC$ drawn from the vertex $A$ and let $H'$ be the reflection of the point $H$ through the line $BC$ (Fig. 35). Then $HD = H'D$, whence the condition $\\frac{AH}{HD} = 2$ is equivalent to the condition $\\frac{AH}{AH'} = \\frac{1}{2} = \\frac{AO}{AO'}$. Thus $\\frac{AH}{HD} = 2$ if and only if $OH \\parallel O'H'$.\n\n![](attached_image_1.png)\n\n![](attached_image_2.png)\n\nIt is known that $H'$ lies on the circumcircle of the triangle $ABC$. Thus by Thales' theorem, $\\angle AH'O' = 90^\\circ$. Hence the condition $OH \\parallel O'H'$ is equivalent to the condition $\\angle AHO = 90^\\circ$. Consequently, $\\frac{AH}{HD} = 2$ if and only if $\\angle AHO = 90^\\circ$.\nW.l.o.g., assume $AC \\le AB$. Let $G$ be the centroid of the triangle $ABC$ and $K$ be the midpoint of the side $BC$ (Fig. 36). Then $\\frac{AG}{GK} = 2$. Thus $\\frac{AH}{HD} = 2$ if and only if $\\frac{AH}{HD} = \\frac{AG}{GK}$, which in turn is valid if and only if $HG \\parallel DK$. The latter condition is equivalent to $\\angle AHG = \\angle ADK = 90^\\circ$. But $H, G$ and $O$ lie on the same line (Euler line). Thus $\\angle AHG = 90^\\circ$ if and only if $\\angle AHO = 90^\\circ$. Hence $\\frac{AH}{HD} = 2$ if and only if $\\angle AHO = 90^\\circ$.",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 29981,
    "subject": "Mathematics (Multi-modal)",
    "question": "Juku has four cans of juice: a $1$-litre can containing $\\frac{1}{2}$ litres of juice, a $\\frac{1}{2}$-litre can containing $\\frac{1}{3}$ litres of juice, a $\\frac{1}{3}$-litre can containing $\\frac{1}{4}$ litres of juice and a $\\frac{1}{4}$-litre can containing $\\frac{1}{5}$ litres of juice. There are no volume markings on the cans. Juku wants to measure exactly $\\frac{1}{30}$ litres of juice by a sequence of pourings of the juice from one can to another. Letting the juice spill is not allowed. Find out all cans into which Juku can get the desired exact amount of juice.",
    "options": [],
    "answer": "The second, the third, and the fourth cans.",
    "solution": "*Answer:* The second, the third, the fourth.\n\nThe second can contains $\\frac{1}{6}$ litres of free space. Pouring juice from the fourth can over to the second can until the second can becomes full leaves $\\frac{1}{30}$ litres of juice in the fourth can. If, after that, one pours all juice from either the second or the third can over to the first can (this is possible since the first can contains $\\frac{1}{2}$ litres of free space and none of the other cans can contain more juice), then the juice in the fourth can can be poured over to either the second or the third can. Thus Juku can get exactly $\\frac{1}{30}$ litres of juice into the fourth can, as well as into the third or the second one.\n\nOn the other hand, the total amount of juice in all cans is $\\frac{1}{2} + \\frac{1}{3} + \\frac{1}{4} + \\frac{1}{5}$ litres after any sequence of pourings. Thus even if the second, the third and the fourth can are full, the first can contains $\\frac{1}{5}$ litres of juice; if the other cans contain free space then the amount of juice in the first can must be even larger. Consequently, it is impossible to get exactly $\\frac{1}{30}$ litres of juice into the first can.",
    "topic": "Geometry",
    "subtopic": "Solid Geometry"
  },
  {
    "id": 29982,
    "subject": "Mathematics (Multi-modal)",
    "question": "A rectangle of integral side lengths is divided into $2022$ unit squares. At least one unit square is coloured black. There are equally many black squares in every row and also equally many black squares in every column. Find all possibilities of how many black unit squares there can be in total.",
    "options": [],
    "answer": "2022",
    "solution": "**Answer:** $2022$.\n\nLet the rectangle be of size $a \\times b$. Let there be $k$ black unit squares in every row and $l$ black unit squares in every column; then $ak = bl$. Since $ab = 2022 = 2 \\cdot 3 \\cdot 337$ where the factors are primes, numbers $a$ and $b$ must be coprime. Thus $b \\mid k$, implying $ak = abk' = 2022k'$ for some integer $k'$. Since at least one black square exists and the total number of unit squares is $2022$, the only possibility is $ak = 2022$, i.e., all unit squares are black.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 29983,
    "subject": "Mathematics (Multi-modal)",
    "question": "The teacher writes the digits $20212022$ in a row on a blackboard. Juku must write each arithmetic operator $(+, -, \\cdot, :)$ exactly once somewhere between these digits in such a way that the result is a correct mathematical expression with a real value, and find this value.\n\na. Can Juku obtain the number $0$ as the value of the expression?\n\nb. If the teacher allowed Juku to use parentheses, could Juku obtain expressions with larger values than it would be possible without parentheses?\n\nc. Prove that there exists a positive integer less than $1000$ that cannot be obtained (without using parentheses) as the value of the expression.",
    "options": [],
    "answer": "a: Yes; b: Yes",
    "solution": "*Answer:* (a) Yes; (b) Yes.\n\n\na. One possibility is $2 + 0 \\cdot 2120 : 2 - 2 = 0$.\n\n\nb. If the teacher allowed using parentheses, Juku could write the expression $(2 - 0 + 2) : 1 \\cdot 2022$ whose value is $8088$. We show that it is impossible to achieve so big value without using parentheses. For that, we show that the value of multiplication must be less than $5000$. Note that placing four operators between eight digits leaves there exactly three pairs of digits lying next to each other without an operator between them. Thus if both factors are numerals then the larger factor contains at most $4$ digits and if the larger factor contains $3$ digits then the smaller factor contains at most $2$ digits. As $2$ is the largest digit in use, the product cannot exceed $2222 \\cdot 2$ in the first case and $222 \\cdot 22$ in the second case. Both bounds are less than $5000$. If the first factor is the ratio and the second is a numeral then the value cannot be larger, because the ratio cannot be larger than the dividend. Analogously, we see that an addend that does not involve multiplication can contain at most $4$ digits and must be less than $3000$, because division or subtraction inside the addend cannot increase it. Consequently, the value of any expression that can be written without parentheses is less than $8000$. Thus Juku can obtain larger values when parentheses are allowed.\n\n\nc. Placing operators between digits can be done in $7 \\cdot 6 \\cdot 5 \\cdot 4$ ways in total. The number of different values Juku can obtain cannot exceed this number. As $7 \\cdot 6 \\cdot 5 \\cdot 4 = 840 < 1000$, there exists a positive integer less than $1000$ that cannot be obtained.",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 29984,
    "subject": "Mathematics (Multi-modal)",
    "question": "There are some distinct positive integers written on a blackboard. If we erase the smallest number written on the blackboard, then the ratio of the sum and the product of the remaining numbers will be 4 times greater than the ratio of the sum and the product of the numbers initially on the blackboard. Find all possibilities for the set of numbers that could have been on the blackboard initially.",
    "options": [],
    "answer": "{5, 20}; {5, 6, 14}; {5, 7, 13}; {5, 8, 12}; {5, 9, 11}; {6, 12}",
    "solution": "*Answer:* $\\{5, 20\\}, \\{5, 6, 14\\}, \\{5, 7, 13\\}, \\{5, 8, 12\\}, \\{5, 9, 11\\}, \\{6, 12\\}$.\n\nClearly there must be at least 2 numbers. Let $n$ be the smallest number, $s$ the sum and $k$ the product of the remaining numbers. We get the equation $4 \\cdot \\frac{n+s}{nk} = \\frac{s}{k}$. Multiplying by $nk$, we get $4(n+s) = ns$, which rearranges to $(n-4)(s-4) = 16$. We know that $16 = 1 \\cdot 16 = 2 \\cdot 8 = 4 \\cdot 4 = 8 \\cdot 2 = 16 \\cdot 1$ and $n < s$, as $n$ was the smallest number on the blackboard. This leaves the options $n-4 = 1, s-4 = 16$ and $n-4 = 2, s-4 = 8$ (note that negative factors would also yield negative $n$ and/or $s$). So $n = 5$ and $s = 20$ or $n = 6$ and $s = 12$.\n\nLet $n = 5$ and $s = 20$. If there are initially 2 numbers, they must be $5$ and $20$. If there are 3 numbers, they could be $5, 6$ and $14$ or $5, 7$ and $13$ or $5, 8$ and $12$ or $5, 9$ and $11$. There can't be 4 or more numbers, because $6+7+8 > 20$.\n\nLet $n = 6$ and $s = 12$. If there are initially 2 numbers, they must be $6$ and $12$. There can't be 3 or more numbers, because $7+8 > 12$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 29985,
    "subject": "Mathematics (Multi-modal)",
    "question": "A function $f$ maps every positive real number to a positive real number. There is a constant $c \\neq 1$ such that $f(cx)^2 = f(x)f(c^2x)$ for all positive real numbers $x$. Must the same equality hold for any positive real numbers $c$ and $x$?",
    "options": [],
    "answer": "No",
    "solution": "Answer: No.\n\n**Solution 1:** Let $f(x) = e^{\\sin \\ln x}$. Then\n$$\n(f(cx))^2 = (e^{\\sin \\ln(cx)})^2 = e^{2\\sin(\\ln cx + \\ln c)}, \\\\\nf(x)f(c^2x) = e^{\\sin \\ln x} \\cdot e^{\\sin \\ln(c^2x)} = e^{\\sin \\ln x + \\sin(\\ln x + 2\\ln c)}.\n$$\nTaking $c = e^{2\\pi}$, we get\n$$\n\\begin{aligned}\n2 \\sin (\\ln x + \\ln c) &= 2 \\sin (\\ln x + 2\\pi) = 2 \\sin \\ln x = \\sin \\ln x + \\sin \\ln x \\\\\n&= \\sin \\ln x + \\sin (\\ln x + 4\\pi) \\\\\n&= \\sin \\ln x + \\sin (\\ln x + 2\\ln c),\n\\end{aligned}\n$$\nso the desired condition is satisfied. However, taking $c = e^{\\pi/2}$ and $x = 1$, we obtain $2\\sin(\\ln 1 + \\ln e^{\\pi/2}) = 2\\sin(\\pi/2) = 2$, whereas $\\sin \\ln 1 + \\sin(\\ln 1 + 2\\ln e^{\\pi/2}) = \\sin 0 + \\sin \\pi = 0$. Hence the equality does not hold for $c = e^{\\pi/2}$ and $x = 1$.\n\n\n**Solution 2:** Let\n$$\nf(x) = \\begin{cases} 2, & \\text{if } x \\text{ is rational}, \\\\ 1, & \\text{if } x \\text{ is irrational}. \\end{cases}\n$$\nFor all positive rational $c$ we have $f(cx) = f(x)$, as $x$ and $cx$ are either both rational or both irrational. So $(f(cx))^2 = (f(x))^2$ and $f(x)f(c^2x) = f(x)f(cx) = (f(x))^2$. Hence the desired equality holds for all positive rational $c$. However, taking $c = \\sqrt{2}$ and $x = 1$, we obtain $f(cx) = f(\\sqrt{2}) = 1$, whereas $f(x)f(c^2x) = f(1)f(2) = 2 \\cdot 2 = 4$. So the equality does not hold.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 29986,
    "subject": "Mathematics (Multi-modal)",
    "question": "A triangle with perimeter $P$ is divided into $k$ triangular pieces for some $k \\ge 2$.\n\na. Show that there exists a piece with perimeter greater than $\\frac{P}{k}$.\n\nb. Show that if the initial triangle is equilateral, then there exists a piece with perimeter at least $\\frac{P}{\\sqrt{k}}$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "a. The sides of the initial triangle are distributed between the pieces. As $k \\ge 2$, the sides of the pieces must also pass through the interior of the initial triangle. So the sum of the perimeters of the $k$ pieces is greater than $P$. Hence there must exist a piece with perimeter greater than $\\frac{P}{k}$.\n\nb. Let the area of the equilateral triangle be $S$. Then there must exist a piece $\\Delta$ with area at least $\\frac{S}{k}$. If $\\Delta$ were equilateral, it would be similar to the initial triangle with a scale factor of $\\frac{1}{\\sqrt{k}}$, so its perimeter would be $\\frac{P}{\\sqrt{k}}$. However, among triangles with a fixed area, an equilateral triangle has the smallest perimeter. Therefore the perimeter of $\\Delta$ is at least $\\frac{P}{\\sqrt{k}}$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 29987,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $n \\ge 2$ be a positive integer and let $S = \\{1, 2, \\dots, n\\}$.\n\nFor $k = 1, 2, \\dots, n-1$, we call two $k$-element subsets of $S$ neighbours, if they have $k-1$ elements in common (i.e. differ by exactly one element). Let $f(n,k)$ be the size of the largest possible collection of $k$-element subsets of $S$, in which no two subsets are neighbours. Prove that $f(n,k) \\le \\binom{n-1}{k-1}$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "For any two subsets belonging to such a collection, the sets of the $k-1$ smallest elements must be different (or else they would be neighbours). There are $\\binom{n-1}{k-1}$ ways to choose the $k-1$ smallest elements, since the number $n$ cannot be one of them. Therefore $f(n,k) \\le \\binom{n-1}{k-1}$, as desired.\nFor all $n \\ge 2$ we have $f(n,1) = 1$, since any two 1-element sets are neighbours. So for all $n \\ge 2$ and $k=1$ the statement holds. We will prove the general statement by induction by $n$, taking the base case to be $n=2$. We will assume that the statement holds for $n-1$ and consider the $n$-element set $S = \\{1, 2, \\dots, n\\}$. Let $1 < k \\le n-1$. The number of $k$-element subsets containing the number $n$ can be at most $f(n-1, k-1)$ in a neighbour-free collection, since removing the number $n$ from all of those subsets yields a neighbour-free collection of $k-1$-element subsets of $\\{1, 2, \\dots, n-1\\}$. The number of $k$-element subsets not containing $n$ can be at most $f(n-1, k)$. Therefore $f(n,k) \\le f(n-1, k-1) + f(n-1, k)$. By the induction assumption $f(n-1, k-1) \\le \\binom{n-2}{k-2}$ and $f(n-1, k) \\le \\binom{n-2}{k-1}$. Therefore by Pascal's rule $f(n,k) \\le f(n-1, k-1) + f(n-1, k) \\le \\binom{n-2}{k-2} + \\binom{n-2}{k-1} = \\binom{n-1}{k-1}$, as desired.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 29988,
    "subject": "Mathematics (Multi-modal)",
    "question": "Show that there exist infinitely many positive integers $n$ such that the integers $1, 2, 3, \\dots, 2n$ can be split into pairs such that the sum of the products of the pairs is divisible by $2n$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "For each prime $p$ we can split the numbers $1, 2, 3, \\dots, 2p$ into the pairs $(1, p+1), (2, p+2), \\dots, (p, 2p)$. The products of the pairs are congruent to $1^2, 2^2, \\dots, (p-1)^2, p^2$ modulo $p$, so the sum of the products is congruent to $1^2 + 2^2 + \\dots + p^2 = \\frac{p(p+1)(2p+1)}{6}$. So for $p > 3$, $p$ divides $1^2 + 2^2 + \\dots + p^2$, as it doesn't divide the denominator. Also, in each pair one of the numbers is even, so the product is even. Hence the sum of the products is divisible by $2$ and therefore also by $2p$, as desired.\nWe use the identity $1 \\cdot 2 + 3 \\cdot 4 + \\dots + (2n-1) \\cdot 2n = \\frac{n(n+1)(4n-1)}{3}$.\n\nIf $n$ is not divisible by $2$ or $3$, then $2n \\mid \\frac{n(n+1)(4n-1)}{3}$, since $2 \\mid n+1$ and either $3 \\mid n+1$ or $3 \\mid 4n-1$, depending on whether $n \\equiv -1 \\pmod 3$ or $n \\equiv 1 \\pmod 3$. Therefore splitting the numbers $1, 2, 3, \\dots, 2n$ into the pairs $(1, 2), (3, 4), \\dots, (2n-1, 2n)$ works for infinitely many $n$.\nFor each prime $p$ we can split the numbers $1, 2, 3, \\dots, 2p$ into the pairs $(1, p+1), (2, p+2), \\dots, (p, 2p)$. The products of the pairs are congruent to $1^2, 2^2, \\dots, (p-1)^2, p^2$ modulo $p$.\nLet $p \\equiv 1 \\pmod 4$ (by Dirichlet's theorem, there are infinitely many such primes), then there exists an integer $a$ such that $a^2 \\equiv -1 \\pmod p$. If $p > 2$, then clearly $a \\not\\equiv 1, -1 \\pmod p$. Then for all $i = 1, 2, \\dots, p-1$ the numbers $i, ai, a^2i, a^3i$ give different remainders modulo $p$, whereas $a^4i \\equiv i \\pmod p$. So the remainders $1, 2, \\dots, p-1$ are split into 4-cycles. The sum of squares of each 4-cycle is divisible by $p$, as $i^2 + (ai)^2 = (1 + a^2)i^2 \\equiv (1 - 1)i^2 = 0 \\pmod p$. So the sum $1^2 + 2^2 + \\dots + (p-1)^2$ is divisible by $p$. Adding also the final term $p^2$, the divisibility will still hold. Also, in each pair one of the numbers is even, so the product is even. Therefore the sum of the products is divisible by $2$ and therefore also by $2p$, as desired.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 29989,
    "subject": "Mathematics (Multi-modal)",
    "question": "Priit's collection consists of $1111$ stamps which are all distributed into envelopes in such a way that every envelope contains more than one stamp, all envelopes contain the same number of stamps, and each envelope contains only stamps from one country. It is known that more than $40\\%$ of stamps in this collection are from Estonia, more than $30\\%$ of stamps in the collection are from Latvia and more than $20\\%$ of stamps in the collection are from Lithuania. Find the largest possible number of envelopes containing Estonian stamps and the largest possible number of envelopes containing Lithuanian stamps.",
    "options": [],
    "answer": "41 and 29",
    "solution": "*Answer:* $41$ and $29$.\n\nAs $1111 = 11 \\cdot 101$ where the factors are prime, we have four cases:\n* $1$ envelope containing $1111$ stamps;\n* $1111$ envelopes, each containing $1$ stamp;\n* $11$ envelopes, each containing $101$ stamps;\n* $101$ envelopes, each containing $11$ stamps.\n\nThe first case is impossible since Priit has stamps of at least $3$ countries and one envelope can contain only stamps of one country. The second case is excluded by the conditions of the problem explicitly. If there were $11$ envelopes then at least $5$ envelopes would have to contain Estonian stamps, at least $4$ envelopes would have to contain Latvian stamps and at least $3$ envelopes would have to contain Lithuanian stamps. This would require at least $12$ envelopes in total which contradicts the assumption. Hence there must be $101$ envelopes. Then at least $41$ envelopes have to contain Estonian stamps, at least $31$ envelopes have to contain Latvian stamps and at least $21$ envelopes have to contain Lithuanian stamps. So the number of envelopes containing Lithuanian stamps cannot exceed $29$. It is indeed possible to have $41$ envelopes containing Estonian stamps and $29$ envelopes containing Lithuanian stamps if $31$ envelopes contain Latvian stamps and there are no stamps from other countries.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 29990,
    "subject": "Mathematics (Multi-modal)",
    "question": "We call positive integers $n, m$ an *interesting pair* if $n < m$ and the greatest prime factor of $n$ is equal to the greatest prime factor of $m$.\n\na. For an interesting pair $n, m$, will there always exist a prime $p$ such that $n < p < m$?\n\nb. Among the first 25 positive integers, how many don't form an interesting pair with any smaller positive integer?",
    "options": [],
    "answer": "a) No; for example, 24 and 27. b) 10",
    "solution": "(a) The pair $24, 27$ is interesting, as the greatest prime factor of them both is $3$. However, there are no primes between them.\n\n(b) The number $1$ and prime numbers cannot form a pair with any smaller integers. However, each composite number forms a pair with its greatest prime factor, which is clearly smaller. Among the first $25$ positive integers, the primes are $2, 3, 5, 7, 11, 13, 17, 19$ and $23$. Taking into account also the number $1$, there are $10$ numbers among the first $25$ positive integers that don't form an interesting pair with any smaller positive integer.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 29991,
    "subject": "Mathematics (Multi-modal)",
    "question": "A dodecagon with perimeter 72 cm is constructed from three squares as shown in the figure. The two outer squares have a common vertex and both share a rectangular part with the central square, such that the perimeter of the shared part is 5 times less than the sum of the perimeters of the central square and the corresponding outer square. Find the side length of the central square.\n\n![](attached_image_1.png)",
    "options": [],
    "answer": "6",
    "solution": "Denote the vertices of the dodecagon as on Fig. 21. Since $QI = AS$ and $IS = QA$, we have\n$$\nDQ + QI + IS + SG = DQ + AS + QA + SG = DA + AG.\n$$\nSimilarly we obtain $ER + RK + KP + PB = EA + AB$. Therefore\n$$\n\\begin{align*}\n& CD + DQ + QI + IS + SG + GF + FE + ER + RK + KP + PB + BC \\\\\n&= CD + DA + AG + GF + FE + EA + AB + BC \\\\\n&= AB + BC + CD + DA + AE + EF + FG + GA.\n\\end{align*}\n$$\nSo the sum of the perimeters of the squares $ABCD$ and $AEFG$ is equal to the perimeter of the dodecagon (72 cm).\n\n![](attached_image_2.png)\n\nSince $QA = IS$, $AP = RK$, $SA = IQ$ and $AR = PK$, we have\n$$\n\\begin{aligned}\n& AP + PH + HQ + QA + AR + RJ + JS + SA \\\\\n&= RK + PH + HQ + IS + PK + RJ + JS + IQ \\\\\n&= HI + IJ + JK + KH.\n\\end{aligned}\n$$\nSo the perimeter of the square $HIJK$ is equal to the sum of the perimeters of the rectangles $APHQ$ and $ARJS$. However we are given that the sum of the perimeters of $APHQ$ and $ARJS$ is equal to one fifth of the sum of the perimeter of $ABCD$, the perimeter of $AEFG$ and twice the perimeter of $HIJK$.\nDenoting the side length of $HIJK$ by $x$ cm, its perimeter will be $4x$ cm. Then we can combine the relations of the previous two paragraphs into the equation\n$$\n4x = \\frac{1}{5} (72 + 2 \\cdot 4x).\n$$\nSolving it, we obtain $4x = 24$, which gives $x = 6$. So the side length of the square $HIJK$ is 6 cm.",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 29992,
    "subject": "Mathematics (Multi-modal)",
    "question": "On the side $AB$ of a triangle $ABC$, let $D$ be a point such that $\\angle BDC = \\angle ACB$. Let $K$ be the midpoint of $CD$ and let $E$ be the intersection of lines $BK$ and $AC$. Given that $\\angle BKD = 2\\angle BCD$, find $\\angle AEB$.",
    "options": [],
    "answer": "90°",
    "solution": "Answer: $90^\\circ$.\n\nDenote $\\angle CAB = \\alpha$ and $\\angle BCA = \\gamma$. Then $\\angle BDC = \\gamma$ and triangles *CBD* and *ABC* will be similar due to having two equal angles (see figure below).\n\nTherefore $\\angle BCD = \\alpha$ and $\\angle BKD = 2\\alpha$. But then $\\angle KBC = 2\\alpha - \\alpha = \\alpha = \\angle KCB$, which yields $KB = KC$. However $KC = KD$ by the choice of $K$, which yields $\\angle KBD = \\angle KDB = \\gamma$.\n\nNow triangle $KBD$ gives us $\\gamma + \\gamma + 2\\alpha = 180^\\circ$ or $\\alpha + \\gamma = 90^\\circ$.\n\nFinally, since two angles in triangle $ABE$ are $\\alpha$ and $\\gamma$, the third angle $AEB$ must be $180^\\circ - (\\alpha + \\gamma) = 90^\\circ$.\n\n![](attached_image_1.png)",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 29993,
    "subject": "Mathematics (Multi-modal)",
    "question": "There are three consecutive positive integers on a blackboard. A move consists of erasing the smallest number on the blackboard and replacing it by the sum of itself and the greatest number on the blackboard. Is it possible that the sum of the numbers on the blackboard will be a power of 10:\n\na) after the 6th move?\n\nb) after the 8th move?",
    "options": [],
    "answer": "a) Yes; b) No",
    "solution": "*Answer:* (a) Yes; (b) No.\n\nLet $x$ be the smallest of the three numbers initially on the blackboard. We will find the numbers on the blackboard after having made 0, 1, 2, 3, 4, 5, 6, 7 and 8 moves, and their sums:\n\n| Number of moves | Numbers on the blackboard                | Sum of numbers   |\n|-----------------|------------------------------------------|------------------|\n| 0               | $x$, $x+1$, $x+2$                        | $3x + 3$         |\n| 1               | $x+1$, $x+2$, $x+3$                      | $4x + 5$         |\n| 2               | $x+2$, $2x+2$, $3x+3$                    | $6x + 7$         |\n| 3               | $2x+2$, $3x+3$, $4x+5$                   | $9x + 10$        |\n| 4               | $3x+3$, $4x+5$, $6x+7$                   | $13x + 15$       |\n| 5               | $4x+5$, $6x+7$, $9x+10$                  | $19x + 22$       |\n| 6               | $6x+7$, $9x+10$, $13x+15$                | $28x + 32$       |\n| 7               | $9x+10$, $13x+15$, $19x+22$              | $41x + 47$       |\n| 8               | $13x+15$, $19x+22$, $28x+32$             | $60x + 69$       |\n\n(a) The total sum of the numbers on the blackboard after 6 moves will be $28x + 32$. Either by direct computation or by considerations modulo 4 and 7, we observe that the equation $28x + 32 = 10^4$ has an integer solution. Therefore it is possible that after the 6th move, the sum of the numbers on the blackboard will be a power of 10.\n\n(b) The total sum of the numbers on the blackboard after 8 moves will be $60x + 69$. As $60x$ and $69$ end with 0 and 9, respectively, their sum will also end with the digit 9. But such a number cannot be equal to a power of 10.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 29994,
    "subject": "Mathematics (Multi-modal)",
    "question": "Two positive integers together contain each digit $0$, $1$, $\\ldots$, $9$ exactly once. Find the largest possible common divisor that these two numbers can have.",
    "options": [],
    "answer": "48651",
    "solution": "*Answer:* $48651$.\n\nSuppose that both numbers contain $5$ digits. A common divisor of two different numbers cannot exceed half of the larger one; thus the greatest common divisor of two such numbers must be less than $50000$. It means that if the greatest common divisor has $5$ digits then the first digit is at most $4$. Let the greatest common divisor be $d$ and suppose that it has $5$ digits, the first of which is $4$. Then the two numbers under consideration are $d$ and $2d$. If the second digit of $d$ were $9$ then the first two digits of $2d$ would be either $98$ or $99$, but in both cases, $9$ would occur in these numbers repeatedly. Thus the second digit of $d$ is at most $8$; suppose that it is $8$. Then the first digit of $2d$ is $9$. If the third digit of $d$ were $7$ then the second digit of $2d$ would be $7$, too. Thus the third digit of $d$ is at most $6$ and the fourth digit of $d$ is at most $5$; suppose that the third and the fourth digit are $6$ and $5$, respectively. Then the second and the third digit of $2d$ are $7$ and $3$, respectively, and the fourth digit is $0$ as the remaining digits $1$ and $2$ cannot cause a carry. Finally, $1$ and $2$ must be the last digits of $d$ and $2d$, respectively, yielding $d = 48651$ and $2d = 97302$. These numbers satisfy the conditions of the problem.\n\nIf the given integers are not $5$-digit numbers then the greatest common divisor can have at most $4$ digits. Consequently, there cannot be solutions greater than that found in the previous paragraph.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 29995,
    "subject": "Mathematics (Multi-modal)",
    "question": "The bisector of the internal angle at the vertex $B$ of a triangle $ABC$ intersects the circumcircle of the triangle $ABC$ at a point $P$ ($P \\neq B$). The line through $P$ perpendicular to the line $AC$ intersects the circumcircle of the triangle $ABC$ at a point $P'$ ($P' \\neq P$). Prove that the quadrilateral $APCP'$ is a square if and only if $\\angle ABC = 90^\\circ$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "As $\\angle ABC = \\angle AP'C$ (see figure), it suffices to show that the quadrilateral $APCP'$ is a square if and only if $\\angle AP'C = 90^\\circ$.\n\nSince $\\angle ABP = \\angle CBP$, we have $AP = CP$. Thus $PP'$ is the perpendicular bisector of the side $AC$. Hence the line $PP'$ passes through the circumcentre of the triangle $ABC$, i.e., the chord $PP'$ is a diameter. By Thales' theorem, $\\angle PAP' = \\angle PCP' = 90^\\circ$.\n\nIf $\\angle AP'C = 90^\\circ$ then also $\\angle APC = 180^\\circ - 90^\\circ = 90^\\circ$. Thus the quadrilateral $APCP'$ is a rectangle and, by equality of adjacent sides $AP$ and $CP$, a square. On the other hand, if $APCP'$ is a square then obviously $\\angle AP'C = 90^\\circ$. This completes the proof.\n\n![](attached_image_1.png)\nSince $\\angle ABC = \\angle AP'C$, it suffices to show that the quadrilateral $APCP'$ is a square if and only if $\\angle AP'C = 90^\\circ$.\n\nAs $\\angle ABP = \\angle CBP$, we have $AP = CP$. Thus $PP'$ bisects the line segment $AC$.\n\nIf $\\angle AP'C = 90^\\circ$ then, by Thales' theorem, $AC$ is the diameter of the circumcircle of the triangle $ABC$. Thus the chord $PP'$ bisecting $AC$ passes through the circumcentre of the triangle $ABC$, implying that $AC$ bisects the chord $PP'$. As $PP' \\perp AC$, the diagonals of the quadrilateral $APCP'$ are perpendicular and bisect each other. Hence the quadrilateral $APCP'$ is a rhombus and, because of $\\angle AP'C = 90^\\circ$, a square. On the other hand, if $APCP'$ is a square then obviously $\\angle AP'C = 90^\\circ$. This completes the proof.",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 29996,
    "subject": "Mathematics (Multi-modal)",
    "question": "Does there exist a positive integer whose\n\na. digit sum is 100 more than the product of digits?\n\nb. product of digits is 100 more than the digit sum?\n\nc. product of digits is 100 times greater than the digit sum?",
    "options": [],
    "answer": "(a) Yes; (b) Yes; (c) Yes",
    "solution": "**Answer:** (a) Yes; (b) Yes; (c) Yes.\n\n(a) The number $111\\ldots1$, consisting of $101$ ones, works. Its product of digits is $1$ and digit sum $101$.\n\n(b) The number $1111112345$ works. Its product of digits is $120$ and digit sum $20$.\n\n(c) The number $1225555$ has product of digits $2500$ and digit sum $25$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 29997,
    "subject": "Mathematics (Multi-modal)",
    "question": "Mari chooses five distinct positive integers not greater than $2021$. From these five numbers, it must be possible to choose two numbers with sum $1919$ in two different ways. Likewise, from these five numbers, it must be possible to choose two numbers with sum $2929$ in two different ways. Find all possibilities of which five numbers Mari may choose.",
    "options": [],
    "answer": "1, 908, 1011, 1918, 2021",
    "solution": "Answer: $1, 908, 1011, 1918, 2021$ is the only possibility.\n\nLet $(a, 1919 - a)$ and $(b, 1919 - b)$ be the two pairs of numbers with sum $1919$. If these sums had a common addend then both addends would be the same whence the choices of two numbers would not be different. Thus $a, b, 1919 - a$ and $1919 - b$ are pairwise distinct. Let $x$ be the fifth chosen number; then the total sum of chosen numbers is $3838 + x$.\n\nSimilarly, we see that the five numbers must be representable as $c, 2929 - c, d, 2929 - d, y$; their total sum is $5858 + y$. From $3838 + x = 5858 + y$, one gets $x - y = 2020$. The only possibility for satisfying the latter equality is $x = 2021, y = 1$. Hence one of $a, b, 1919 - a, 1919 - b$ equals $1$ and one of $c, d, 2929 - c, 2929 - d$ is $2021$; w.l.o.g., $a = 1$ and $c = 2021$. Then the set of chosen numbers contains also $1919 - 1 = 1918$ and $2929 - 2021 = 908$. W.l.o.g., $b = 908$. The last chosen number must be $1919 - 908 = 1011$.\nLet the chosen numbers be $a, b, c, d, e$. W.l.o.g., $a + b = 1919$. If the other way to obtain $1919$ as the sum of two chosen numbers used either $a$ or $b$ as addend then the other addend would have to be $b$ or $a$, respectively, which is not allowed. Hence, w.l.o.g., $c + d = 1919$.\n\nSimilarly, one must use four distinct numbers to get $2929$ as the sum of two chosen numbers in two different ways. If these four numbers were $a, b, c, d$, then $a + b = c + d = 1919$ implies $a + b + c + d = 3838$ while also $a + b + c + d = 2929 + 2929 = 5858$. The contradiction shows that one must use the number $e$ in at least one pair of numbers with sum $2929$. W.l.o.g., $d + e = 2929$. The other pair of numbers with sum $2929$ must use exactly two numbers among $a, b, c$, but $a + b = 1919$. Hence, w.l.o.g., $b + c = 2929$.\n\nThese equalities together imply $b = 1919 - a$, $c = 2929 - b = 1010 + a$, $d = 1919 - c = 909 - a$ and $e = 2929 - d = 2020 + a$. The latter implies $a = 1$ and $e = 2021$. Substituting $a = 1$ into the other equalities gives $b = 1918$, $c = 1011$ and $d = 908$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 29998,
    "subject": "Mathematics (Multi-modal)",
    "question": "The largest angle of triangle $ABC$ is located at its vertex $C$. Let $\\rho$ be the circle with centre $A$ and radius $AC$, and let $\\sigma$ be the circle with centre $B$ and radius $BC$. The circle $\\sigma$ intersects the circumcircle of the triangle $ABC$ and the circle $\\rho$ at points $D$ and $F$, respectively ($D \\neq C$, $F \\neq C$). Prove that $A$, $D$ and $F$ are collinear.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Denote $\\angle BAC = \\alpha$ (Figures 7 and 8 present two possible cases). Note that $AC = AF$ and $BC = BF$ as they are radii of circles $\\rho$ and $\\sigma$, respectively. Thus the triangles $ABF$ and $ABC$ are equal by three equal sides. Hence $\\angle BAF = \\alpha$. From equal radii of $\\sigma$, we also obtain $BD = BC$. Now by inscribed angles subtending equal chords of the circumcircle of the triangle $ABC$, we get $\\angle BAD = \\alpha$. Consequently, points $A$, $D$ and $F$ lie on one line.\n\n![](attached_image_1.png)\n\n![](attached_image_2.png)\n\n![](attached_image_3.png)\n\n![](attached_image_4.png)\nFrom the inscribed angles of the circumcircle of the triangle $ABC$, we get $\\angle ADC = \\angle ABC$. As in Solution 1, we can show that the triangles $ABF$ and $ABC$ are equal; thus $\\angle ABC = \\angle ABF = \\frac{\\angle CBF}{2}$. If $D$ and $A$ lie on the same side of the line $CF$ (Fig. 9) then $\\frac{\\angle CBF}{2} = 180^\\circ - \\angle CDF$ by the inscribed angle theorem. If $D$ and $A$ lie on different sides of the line $CF$ (Fig. 10) then, analogously, $\\frac{\\angle CBF}{2} = \\angle CDF$. Altogether, we have either $\\angle ADC = 180^\\circ - \\angle CDF$ while $D$ and $A$ being on different sides of $CF$ or $\\angle ADC = \\angle CDF$ while $D$ and $A$ being on different sides of $CF$. In both cases, the obtained equality implies that $A$, $D$ and $F$ lie on the same line.",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 29999,
    "subject": "Mathematics (Multi-modal)",
    "question": "There are 25 green, 20 brown and 15 orange chameleons in a zoo. Whenever exactly two chameleons of distinct colours meet, both change their colour to the third one. Otherwise, the chameleons do not change their colours. Is it possible that:\n\na. at some time instant, we have the same number of chameleons of each colour?\n\nb. at some time instant, all chameleons are of the same colour?",
    "options": [],
    "answer": "a. No; b. No.",
    "solution": "*Answer:* (a) No; (b) No.\n\nAfter each meet of two chameleons, the numbers of chameleons of two colours decrease by 1 and the number of chameleons of the remaining colour increases by 2. Thus the difference of the numbers of chameleons of each two colours is constant modulo 3. As the remainders modulo 3 of the numbers of chameleons of different colours are pairwise distinct in the beginning, the numbers of chameleons of no two colours can become equal. This also implies that all chameleons cannot be of the same colour since then the numbers of chameleons of two other colours would be equal to 0.\na. If there were equally many chameleons of each colour, the common number of chameleons of each colour would be 20. To keep the number of brown chameleons all in all unchanged, there must be 2 decreases by 1 per each increase by 2. Hence the total number of meets must be $3x$ for some integer $x$. For changing the number of green chameleons from 25 to 20, there must be 5 decreases by 1 and, in addition, 2 decreases by 1 per each increase by 2. Hence the total number of meets must be $5 + 3y$ for some integer $y$. Thus $3x = 5 + 3y$, but this equation has no integral solutions.\n\nb. Analogously, if the number of brown chameleons were 60 then the number of meets must be $20 + 3x$ for an integer $x$; then the number of orange chameleons would be 0, requiring $15 + 3y$ meets for an integer $y$, but the equation $20 + 3x = 15 + 3y$ has no integral solutions. If the number of brown chameleons were 0 then the number of meets must again be $20 + 3x$. If the number of orange chameleons were also 0 then the number of meets would be $15 + 3y$ for an integer $y$ leading to the same equation; if the number of green chameleons were 0 then we would get the equation $20 + 3x = 25 + 3y$ which also does not have integral solutions.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30000,
    "subject": "Mathematics (Multi-modal)",
    "question": "On the first line of a notebook Juku writes the number $43$. On every following line he writes the number $x^2 - 66x + 1122$, where $x$ is the number on the previous line. Find the number that Juku will write on the $2021$st line.",
    "options": [],
    "answer": "10^{2^{2020}} + 33",
    "solution": "Let $x_i$ be the number written on the $i$th line, then for all $i = 1, 2, \\dots$ we have $x_{i+1} = x_i^2 - 66x_i + 1122$. Notice that this is equivalent to $x_{i+1} - 33 = x_i^2 - 66x_i + 1089 = (x_i - 33)^2$. Denoting $a_n = x_n - 33$, we acquire $a_{i+1} = a_i^2$ for all $i = 1, 2, \\dots$, which means that $a_1 = 43 - 33 = 10$, $a_2 = a_1^2 = 10^2$, $a_3 = a_2^2 = 10^4$ and in general $a_i = 10^{2^{i-1}}$. So $x_{2021} = a_{2021} + 33 = 10^{2^{2020}} + 33$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  }
]