[
  {
    "id": 12501,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all prime numbers $p$ such that $\\frac{p-1}{2}$ and $\\frac{p+1}{4}$ are prime numbers, too.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $q = \\frac{p-1}{2}$ and $r = \\frac{p+1}{4}$. Then $p = 4r - 1$ and $q = 2r - 1$.\n\nConsider all possible remainders when $r$ is divided by $3$:\n\n- If $r \\equiv 1 \\pmod{3}$, then $4r - 1 \\equiv 0 \\pmod{3}$, so $p$ is divisible by $3$. Thus $p = 3$, but then $r = 1$, which is not prime.\n- If $r \\equiv 2 \\pmod{3}$, then $2r - 1 \\equiv 0 \\pmod{3}$, so $q = 3$. Thus $r = 2$ and $p = 7$. All three are primes.\n- If $r \\equiv 0 \\pmod{3}$, then $r = 3$. Thus $q = 5$ and $p = 11$, which are also primes.\n\nConsequently, the possible values for $p$ are $7$ and $11$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 12502,
    "subject": "Mathematics (Olympiad)",
    "question": "Given that $\\frac{c}{a}$ is a root of the equation $ax^2 + bx + c = 0$, prove that all three quadratic equations\n\n$$ax^2 + bx + c = 0$$\n$$bx^2 + cx + a = 0$$\n$$cx^2 + ax + b = 0$$\nhave a common zero.",
    "options": [],
    "answer": "See solution",
    "solution": "Applying Vieta's formulas to the first equation, the product of the roots is $x_1 x_2 = \\frac{c}{a}$, where $x_1, x_2$ are the roots. Since $x_1 = \\frac{c}{a}$ is given, it follows that $x_2 = 1$. Substituting $x = 1$ into the first equation yields $a + b + c = 0$, which implies that $x = 1$ is a root of all three equations.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 12503,
    "subject": "Mathematics (Olympiad)",
    "question": "a) Let $A$ be a matrix in $M_2(\\mathbb{C})$, with $A \\neq a I_2$ for any $a \\in \\mathbb{C}$. Prove that a matrix $X$ in $M_2(\\mathbb{C})$ commutes with $A$ (i.e., $AX = XA$) if and only if there exist complex numbers $\\alpha$ and $\\alpha'$ such that $X = \\alpha A + \\alpha' I_2$.\n\nb) Let $A$, $B$, and $C$ be matrices in $M_2(\\mathbb{C})$ such that $AB \\neq BA$, $AC = CA$, and $BC = CB$. Prove that $C$ commutes with all matrices in $M_2(\\mathbb{C})$.",
    "options": [],
    "answer": "See solution",
    "solution": "a) Clearly, if $X = \\alpha A + \\alpha' I_2$, then $X$ and $A$ commute. Conversely, let\n$$\nA = \\begin{pmatrix} a_1 & a_2 \\\\ a'_1 & a'_2 \\end{pmatrix}, \\quad X = \\begin{pmatrix} x_1 & x_2 \\\\ x'_1 & x'_2 \\end{pmatrix}.\n$$\nThe equality $AX = XA$ implies\n$$\na_2 x'_1 = a'_1 x_2, \\quad (1)\n$$\n$$\n(a_1 - a'_2)x_2 + a_2(x'_2 - x_1) = 0, \\quad (2)\n$$\n$$\n(a_1 - a'_2)x'_1 + a'_1(x'_2 - x_1) = 0. \\quad (3)\n$$\nSince $A \\neq a I_2$, either one of $a_2, a'_1$ is nonzero, or $a_2 = a'_1 = 0$ and $a_1 \\neq a'_2$.\n\nIn the first case, if $a_2 \\neq 0$, we obtain\n$$\nX = \\frac{x_2}{a_2} A + \\left( x_1 - \\frac{a_1}{a_2} x_2 \\right) I_2.\n$$\nIn the second case,\n$$\nX = \\frac{x_1 - x'_2}{a_1 - a'_2} A + \\frac{a_1 x'_2 - a'_2 x_1}{a_1 - a'_2} I_2.\n$$\n\nb) We prove that $C = \\gamma I_2$ for some $\\gamma \\in \\mathbb{C}$, hence $C$ commutes with all matrices in $M_2(\\mathbb{C})$. Suppose the contrary. Since $A$ and $C$ commute, there exist $\\alpha, \\alpha'$ such that $A = \\alpha C + \\alpha' I_2$. Similarly, there exist $\\beta, \\beta'$ such that $B = \\beta C + \\beta' I_2$. Therefore, $AB = BA$, a contradiction.",
    "topic": "Algebra",
    "subtopic": "Linear Algebra"
  },
  {
    "id": 12504,
    "subject": "Mathematics (Olympiad)",
    "question": "For any positive real numbers $a_1 \\le a_2 \\le a_3 \\le a_4 \\le a_5$, consider the five fractions:\n$$\n\\frac{a_1}{a_2}, \\frac{a_3}{a_4}, \\frac{a_1}{a_5}, \\frac{a_2}{a_3}, \\frac{a_4}{a_5}.\n$$\nWhat is the smallest possible value of $C$ such that, among these fractions, there must exist two with distinct indices whose difference is less than $C$?",
    "options": [],
    "answer": "See solution",
    "solution": "Each fraction lies in the interval $[0, 1]$. By the Pigeonhole Principle, at least three of them must lie in $[0, \\frac{1}{2}]$ or $[\\frac{1}{2}, 1]$ simultaneously. Thus, there must be two consecutive terms within an interval of length $\\frac{1}{2}$, so their difference is less than $\\frac{1}{2}$. Since the indices are distinct, we conclude $C \\le \\frac{1}{2}$. Therefore, $C = \\frac{1}{2}$ is the smallest possible choice.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 12505,
    "subject": "Mathematics (Olympiad)",
    "question": "設 $n$ 是大於或等於 $1$ 的整數。在集合 $\\{1, 2, \\dots, n\\}$ 中，最多可以選出幾對數字，滿足：\n\n- 每一對都是相異的兩個數；\n- 每一對中的兩數之和與其他對的和皆不同；\n- 每一對的和都不超過 $n$？",
    "options": [],
    "answer": "See solution",
    "solution": "考慮集合 $\\{1, 2, 3, \\dots, n\\}$ 中的 $x$ 對這樣的數字。這些 $2x$ 個數字的總和 $S$ 至少是 $1+2+\\cdots+2x$，因為這些選出來的數字兩兩互異。另一方面，$S \\le n + (n-1) + \\cdots + (n-x+1)$，因為每對數字的和均不相同，並且都不超過 $n$。於是給出下列不等式：\n\n$$\n\\frac{2x(2x+1)}{2} \\le nx - \\frac{x(x-1)}{2}\n$$\n\n所以推得 $x \\le \\frac{2n-1}{5}$。所以最多可以有 $\\left\\lfloor \\frac{2n-1}{5} \\right\\rfloor$ 對滿足題設的數字。\n\n以下建構 $\\left\\lfloor \\frac{2n-1}{5} \\right\\rfloor$ 對滿足題設的數字。首先考慮 $n = 5k + 3$ ($k \\ge 0$) 的情形，此時 $\\left\\lfloor \\frac{2n-1}{5} \\right\\rfloor = 2k + 1$。這些對數字可以用下列的表格表示：\n\n![](path/to/file.png)\n\n以上的 $2k+1$ 對數字使用了 $1$ 到 $4k+2$ 的所有數字；各對數字的和包含 $3k+3$ 到 $5k+3$ 的所有數字。當 $n = 5k + 4$ 或 $5k + 5$ ($k \\ge 0$) 時可以用同樣的建構法。在這些情形下要達成的對數 $\\left\\lfloor \\frac{2n-1}{5} \\right\\rfloor$ 仍然是 $2k+1$，而上表中所有的和也都不超過 $5k+3$。\n\n在 $n = 5k + 2$ ($k \\ge 0$) 的情形下，只需要 $2k$ 對數字；此時可以將上表中的最後一行刪除（於是移除了和是 $5k+3$ 的那一對數）。最後，當 $n = 5k + 1$ ($k \\ge 0$) 時也需要 $2k$ 對數字。此時只要將上表中的最後一行刪除，並且將第一列的各個數字減 $1$ 即可。",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 12506,
    "subject": "Mathematics (Olympiad)",
    "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 $n$.",
    "options": [],
    "answer": "See solution",
    "solution": "We look for $n = p \\cdot q \\cdot r$, with primes $p \\leq q \\leq r$ satisfying\n\n$$\n(p+1)(q+1)(r+1) = pqr + 963.\n$$\n\nIf $p=2$, the right-hand side is odd, so $q+1$ and $r+1$ must be odd too. This implies $p=q=r=2$, which contradicts the equation. Thus, $p \\geq 3$.\n\nNow, we show $p=3$. Suppose $3 < p \\leq q \\leq r$. Then $pqr + 963$ is not divisible by 3, so $(p+1)(q+1)(r+1)$ must also not be divisible by 3. This means $p, q, r \\equiv 1 \\pmod{3}$, so $(p+1)(q+1)(r+1) - pqr \\equiv 8 - 1 = 7 \\pmod{3}$, which contradicts $(p+1)(q+1)(r+1) - pqr = 963$. Therefore, $p=3$.\n\nSubstituting $p=3$ into the equation:\n\n$$\n4(q+1)(r+1) = 3qr + 963\n$$\n\nwhich can be rewritten as\n\n$$\n(q+4)(r+4) = 975.\n$$\n\nSince $975 = 3 \\cdot 5^2 \\cdot 13$ and $q$ is prime, possible values for $q+4$ are $13, 15, 25$. Thus, $q=11$ and $r+4=65$, so $r=61$ (which is prime). Therefore, the original product is\n\n$$\nn = 3 \\cdot 11 \\cdot 61 = 2013.\n$$",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 12507,
    "subject": "Mathematics (Olympiad)",
    "question": "In the acute-angled triangle $ABC$, with $AB \\neq AC$, $D$ is the foot of the angle bisector of angle $A$, and $E$, $F$ are the feet of the altitudes from $B$ and $C$, respectively. The circumcircles of triangles $DBF$ and $DCE$ intersect for the second time at $M$. Prove that $ME = MF$.\n\n![](images/RMC2013_final_p66_data_e689d6a121.png)",
    "options": [],
    "answer": "See solution",
    "solution": "Triangles $AEF$ and $ABC$ are similar, therefore $AF \\cdot AB = AE \\cdot AC$. It follows that point $A$ is on the radical axis of the two circumcircles, hence $M \\in AD$. We have that $m(\\angle EMF) = 360^\\circ - (180^\\circ - m(\\angle FBD)) - (180^\\circ - m(\\angle ECD)) = m(\\angle B) + m(\\angle C) = 180^\\circ - m(\\angle A)$; it follows that the quadrilateral $AEMF$ is cyclic. This means that $\\angle MEF \\equiv \\angle FAM$ and $\\angle MFE \\equiv \\angle EAM$, i.e., triangle $MEF$ is isosceles, with $ME = MF$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 12508,
    "subject": "Mathematics (Olympiad)",
    "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": "See solution",
    "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 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 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 on 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": 12509,
    "subject": "Mathematics (Olympiad)",
    "question": "Find the minimal positive integer $k$ for which there exist pairwise distinct positive integers $a$, $b$, and $c$ such that all three numbers:\n\n$$\n4a^2 + k b + c,\n$$\n$$\n4b^2 + k c + a,\n$$\n$$\n4c^2 + k a + b\n$$\nare perfect squares.\n\nLet $k$ be the minimal positive integer for which there exist pairwise distinct positive integers $a$, $b$, and $c$ such that all three numbers above are perfect squares. Find all possible triples $(a, b, c)$ for such $k$.",
    "options": [],
    "answer": "See solution",
    "solution": "**Answer:** $k = 4$ and $(a, b, c)$ are cyclic permutations of $(120, 85, 141)$.\n\nSince the expressions $4a^2 + k b + c$, $4b^2 + k c + a$, and $4c^2 + k a + b$ are invariant under cyclic permutations of $a$, $b$, and $c$, we can assume $c = \\max(a, b, c)$. For $k \\leq 3$:\n\n$$\n(2c)^2 < 4c^2 + k a + b \\leq 4c^2 + 3c + c < 4c^2 + 4c + 1 = (2c + 1)^2,\n$$\nso $4c^2 + k a + b$ is not a perfect square if $k \\leq 3$.\n\nFor $k = 4$, one can find the triple $(a, b, c) = (120, 85, 141)$, considering the expressions modulo $20$.\n\nFor the alternative version, we find all triples $(a, b, c)$ which satisfy the problem conditions for $k = 4$.\n\nAssuming $c = a + \\frac{b - 1}{4}$, we obtain $3a \\leq 4.25b - 1.25$. Consider $4b^2 + 4c + a$:\n\n$$\n\\begin{aligned}\n4b^2 + 4c + a &= 4b^2 + 5a + b - 1 \\\\\n&\\leq 4b^2 + \\frac{5}{3}(4.25b - 1.25) + b - 1 \\\\\n&= 4b^2 + 8.083...b - 2.083... \\\\\n&< 4b^2 + 12b + 9 = (2b + 3)^2.\n\\end{aligned}\n$$\n\nTherefore $(2b + 1)^2 < 4b^2 + 4c + a < (2b + 3)^2$, so $4b^2 + 4c + a = (2b + 2)^2$, i.e., $8b + 4 = 4c + a$.\n\nNote that $8a + 4 = 2(4c + 1 - b) + 4 = 8c + 6 - 2b > 4b + c$, hence\n\n$$\n(2a)^2 < 4a^2 + 4b + c < 4a^2 + 8a + 4 = (2a + 2)^2,\n$$\nwhich yields $4a^2 + 4b + c = (2a + 1)^2$, i.e., $4b + c = 4a + 1$.\n\nThus, the triple $(a, b, c)$ satisfies the system:\n\n$$\n\\begin{cases}\n4a + b = 4c + 1 \\\\\n8b + 4 = 4c + a \\\\\n4b + c = 4a + 1\n\\end{cases}\n$$\n\nThe solution to this system is $(120, 85, 141)$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 12510,
    "subject": "Mathematics (Olympiad)",
    "question": "Prove that for all positive integers $n \\ge 2$ we have\n\n$$\n\\frac{1}{2} + \\sqrt{\\frac{1}{2}} + \\sqrt[3]{\\frac{2}{3}} + \\dots + \\sqrt[n]{\\frac{n-1}{n}} < \\frac{n^2}{n+1}.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "We use proof by induction.\n\nFor $n = 2$ we have\n\n$$\n\\frac{1}{2} + \\sqrt{\\frac{1}{2}} < \\frac{4}{3} \\Leftrightarrow \\sqrt{\\frac{1}{2}} < \\frac{5}{6} \\Leftrightarrow \\frac{1}{2} < \\frac{25}{36},\n$$\n\nwhich is true. For $n = 2$ the inequality holds.\n\nNow assume that the inequality holds for $n$ and let us prove it for $n + 1$. We wish to show that\n\n$$\n\\frac{1}{2} + \\sqrt{\\frac{1}{2}} + \\sqrt[3]{\\frac{2}{3}} + \\dots + \\sqrt[n]{\\frac{n-1}{n}} + \\sqrt[n+1]{\\frac{n}{n+1}} < \\frac{(n+1)^2}{n+2}. \\quad (1)\n$$\n\nBy the induction hypothesis we can estimate\n\n$$\n\\frac{1}{2} + \\sqrt{\\frac{1}{2}} + \\sqrt[3]{\\frac{2}{3}} + \\dots + \\sqrt[n]{\\frac{n-1}{n}} + \\sqrt[n+1]{\\frac{n}{n+1}} < \\frac{n^2}{n+1} + \\sqrt[n+1]{\\frac{n}{n+1}}.\n$$\n\nIt therefore suffices to show that\n\n$$\n\\begin{aligned}\n\\frac{n^2}{n+1} + \\sqrt[n+1]{\\frac{n}{n+1}} &\\le \\frac{(n+1)^2}{n+2} \\\\\n\\Leftrightarrow \\sqrt[n+1]{\\frac{n}{n+1}} &\\le \\frac{(n+1)^2}{n+2} - \\frac{n^2}{n+1} = \\frac{n^2 + 3n + 1}{(n+2)(n+1)} = 1 - \\frac{1}{(n+2)(n+1)}.\n\\end{aligned}\n$$\n\nUsing the Arithmetic-Geometric Mean Inequality we get\n\n$$\n\\sqrt[n+1]{\\frac{n}{n+1}} = \\sqrt[n+1]{\\frac{n}{n+1} \\cdot 1 \\cdot 1 \\cdots 1} \\le \\frac{\\frac{n}{n+1} + 1 + \\cdots + 1}{n+1} = \\frac{n^2 + 2n}{(n+1)^2} = 1 - \\frac{1}{(n+1)^2}.\n$$\n\nFrom here it follows that\n\n$$\n\\sqrt[n+1]{\\frac{n}{n+1}} \\le 1 - \\frac{1}{(n+1)^2} < 1 - \\frac{1}{(n+2)(n+1)},\n$$\n\nwhich proves the inequality (1) and completes the induction step.\n\n*Remark: The inequality*\n\n$$\n\\sqrt[n+1]{\\frac{n}{n+1}} \\le 1 - \\frac{1}{(n+1)^2}\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 12511,
    "subject": "Mathematics (Olympiad)",
    "question": "Given a sequence of real numbers $ (x_n) $:\n\n$$\nx_1 = 3 \\quad \\text{and} \\quad x_n = \\frac{n+2}{3n}(x_{n-1} + 2), \\quad \\forall n \\ge 2.\n$$\n\nProve that the sequence has a finite limit as $ n \\to \\infty $ and calculate this limit.",
    "options": [],
    "answer": "See solution",
    "solution": "For $ n \\ge 1 $, we have\n\n$$\nx_{n+1} - x_n = \\left(\\frac{n+3}{3(n+1)} - 1\\right)x_n + \\frac{2(n+3)}{3(n+1)} = \\frac{2}{3(n+1)}(n+3 - nx_n). \\quad (1)\n$$\n\nWe first prove that\n\n$$\nx_n > 1 + \\frac{3}{n} \\quad \\forall n \\ge 2. \\quad (2)\n$$\n\nThe proof proceeds by induction on $ n $. For $ n=2 $ we have\n\n$$\nx_2 = \\frac{4}{6}(3+2) = \\frac{10}{3} > 1 + \\frac{3}{2}.\n$$\n\nSuppose that $ x_k > 1 + \\frac{3}{k} $ for some $ k \\ge 2 $, then we have\n\n$$\nx_{k+1} = \\frac{k+3}{3(k+1)}(x_k + 2) > \\frac{k+3}{3(k+1)}\\left(1 + \\frac{3}{k} + 2\\right) = 1 + \\frac{3}{k+1}.\n$$\n\nThus, (2) is proved. From (1) and (2), we have\n\n$$\nx_n > x_{n+1} > 1, \\quad \\forall n \\ge 2.\n$$\n\nTherefore, $ (x_n) $ is a decreasing sequence for $ n \\ge 2 $ and is bounded by 1. This implies that $ (x_n) $ has a finite limit. Now, we can easily find this limit by solving the equation\n\n$$\nx = \\frac{1}{3}(x+2),\n$$\n\nwhich implies that the limit is 1.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 12512,
    "subject": "Mathematics (Olympiad)",
    "question": "Prove that for every positive integer $n$ there exists a (not necessarily convex) polygon with no three collinear vertices, which admits exactly $n$ different triangulations. A triangulation is a dissection of the polygon into triangles by interior diagonals which have no common interior points with each other nor with the sides of the polygon.",
    "options": [],
    "answer": "See solution",
    "solution": "The left figure below shows an example of a polygon admitting a unique triangulation: the only diagonals lying inside the polygon come from $A$, so all of them must be drawn. (Notice that the \"exterior\" polygon $B_1B_2\\ldots B_n$ is convex.)\n\nNow we prove that the right figure below shows a polygon $A_1A_2B_1B_2\\ldots B_n$ with exactly $n$ triangulations. Indeed, any triangulation must contain a triangle with side $A_1A_2$, and there are $n$ possible such, namely $A_1A_2B_i$ for $i = 1, 2, \\ldots, n$. After such a triangle has been chosen, the rest of the polygon splits into two (or one) polygons admitting a unique triangulation. Hence the result.\n\n![](images/RMC_2019_var_3_p96_data_97197a145c.png)",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 12513,
    "subject": "Mathematics (Olympiad)",
    "question": "We are given a rectangular $m \\times n$ chessboard with $m$ unit squares in each row and $n$ unit squares in each column. We are going to assign an integer to each unit square. A rectangle $R$ consisting of one or more unit squares is called a *shelf* if there is an integer $h$ satisfying the following two conditions:\n\n1. The number in each unit square in $R$ is larger than $h$.\n2. The number in a unit square out of $R$ sharing an edge or a point with $R$ is at most $h$.\n\n(We assume that a shelf $R$ contains all interior unit squares in the rectangle.)\n\nWhat is the number of shelves if we assign integers to maximize the number of shelves?",
    "options": [],
    "answer": "See solution",
    "solution": "The answer is $\\left\\lfloor \\frac{(n+1)(m+1)}{2} \\right\\rfloor - 1$.\n\nFor a shelf $R$, let $\\tilde{R}$ be the rectangular area by extending $R$ by 1 row to the top and 1 column to the left. Let us also extend the initial chessboard by 1 row to the top and 1 column to the left. We assign $-\\infty$ to each of those new unit squares so that the number of shelves is preserved. For a shelf $R$, let us call the integer $h$ satisfying two conditions the *height* of $R$.\n\n**Claim 1:** If two shelves $R_1$ and $R_2$ do not intersect, then $\\tilde{R}_1 \\cap \\tilde{R}_2 = \\emptyset$.\n\n*Proof of Claim 1:* Let $h_1, h_2$ be the heights of $R_1, R_2$ respectively. Without loss of generality, let us assume $h_1 \\le h_2$. If $\\tilde{R}_1 \\cap \\tilde{R}_2 \\ne \\emptyset$, then some unit square $x$ in $R_2$ is adjacent to $R_1$ and therefore the integer in $x$ is less than or equal to $h_1$. Since $x \\in R_2$, the integer in $x$ is larger than $h_2$, contradictory to the assumption that $h_1 \\le h_2$. This proves Claim 1.\n\n**Claim 2:** If $R_1$ and $R_2$ are distinct maximal shelves smaller than the whole chessboard, then $R_1 \\cap R_2 = \\emptyset$.\n\n*Proof of Claim 2:* Let $h_1, h_2$ be the heights of $R_1, R_2$ respectively. Without loss of generality, let us assume $h_1 \\le h_2$. Suppose that $R_1 \\cap R_2 \\ne \\emptyset$. Suppose a unit square $x$ in $R_2$ shares at least one point with $R_1$. If $x$ does not belong to $R_1$, then the integer in $x$ should be at most $h_1$ but since $x \\in R_2$, the integer in $x$ should be larger than $h_2$, contradictory to the assumption that $h_1 \\le h_2$. So such $x$ belongs to $R_1$ and therefore $R_2 \\subseteq R_1$. However, this contradicts the assumption that $R_2$ is maximal. So Claim 2 is proved.\n\n**Claim 3:** The number of shelves in the $n \\times m$ chessboard is less than or equal to $\\frac{(n+1)(m+1)}{2} - 1$.\n\n*Proof of Claim 3:* We proceed by induction on $n+m$. If $n+m=2$, then $n=m=1$ and the number of shelves is 1.\n\nNow let us assume $n+m > 2$. Let $R_1, R_2, \\dots, R_k$ be the maximal shelves strictly smaller than the whole $n \\times m$ chessboard.\n\nBy Claims 1 and 2, $\\tilde{R}_i \\cap \\tilde{R}_j = \\emptyset$. By the induction hypothesis, the number of shelves contained in $R_i$ for each $i$ is at most $|\\tilde{R}_i|/2 - 1$. Therefore, the number of all shelves is at most\n\n$$\n\\sum_{i=1}^{k} \\left( \\left\\lfloor \\frac{|\\tilde{R}_i|}{2} \\right\\rfloor - 1 \\right) \\le \\frac{(n+1)(m+1)}{2} - k - 1.\n$$\n\nSo Claim 3 is proved if $k > 1$. We may now assume that $k=1$. Then it is enough to show that\n\n$$\n\\max \\left( \\frac{m(n+1)}{2}, \\frac{(m+1)n}{2} \\right) - 1 + 1 \\le \\frac{(n+1)(m+1)}{2} - 1.\n$$\n\nIt is easy to see this because $n \\ge 1$ is equivalent to the inequality $\\frac{m+n-1}{2} \\le \\frac{(n+1)(m+1)}{2} - 1$. This proves Claim 3.\n\nNow it remains to show that there is an assignment of integers so that the number of shelves is exactly $\\left\\lceil \\frac{(n+1)(m+1)}{2} \\right\\rceil - 1$. We proceed by induction on $n+m$. It is trivial if $\\max(n,m) \\le 2$.\n\nNow by symmetry let us assume that $m > 2$. Let us write 1 in each unit square on the second row. For the first row, we write integers larger than 1 obtained by the induction hypothesis on a $1 \\times n$ chessboard. For the remaining rows, we write integers larger than 1 obtained by the induction hypothesis on a $(m-2) \\times n$ chessboard.\n\nNow the number of shelves in this assignment is\n\n$$\n1 + \\left\\lfloor \\frac{2(n+1)}{2} - 1 \\right\\rfloor + \\left\\lceil \\frac{(m-1)(n+1)}{2} - 1 \\right\\rceil\n$$\n\nby the induction hypothesis. This completes the proof. $\\Box$",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 12514,
    "subject": "Mathematics (Olympiad)",
    "question": "A doubly infinite sequence $a_n$, for $n \\in \\mathbb{Z}$, has each $a_n$ equal to either $0$ or $1$. Prove that there exist numbers $p$ and $q > 1$ such that $a_{p+k} = a_{p+q+k}$ for $k = 0, 1, \\dots, q-1$.",
    "options": [],
    "answer": "See solution",
    "solution": "Suppose, for contradiction, that no such $p$ and $q > 1$ exist; that is, the sequence contains no two identical consecutive blocks of length $q > 1$.\n\n(a) The run $010$ must extend to $00100$, since $1010$ and $0101$ are excluded.\n\n(b) The run $000$ must extend to $10001$, since $0000$ is excluded.\n\n(c) The run $000$ must extend to $1100011$. For $000$ extends to $10001$ by (b). If this were continued to $100010$, it would extend further to $10001000$ by (a), which is impossible. Hence we must have $100011$, and, by symmetry, the same argument works on the left.\n\n(d) The sequence does not contain $010$. If it did, we would have $00100$ by (a), which cannot continue to $001001$, so it must be $001000$, which contradicts (c).\n\n(e) The sequence does not contain $101$. This follows from (d) and the apparent symmetry between $0$ and $1$.\n\n(f) The run $01$ must extend to $0011$ by (d) and (e).\n\n(g) The run $10$ must extend to $1100$ by (d) and (e).\n\n(h) The sequence does not contain $001100$. Suppose it does. If it continues on the right as $0011001$, it would extend to $00110011$ by (f) and give a contradiction. Hence it continues on the right as $0011000$. By symmetry, it will be continued on the left as $00011000$. By (c), we get $0001100011$ and again a contradiction.\n\n(i) The sequence does not contain $110011$ by (h) and symmetry.\n\n(j) The run $000$ must extend to $111000111$. Perforce, we already have $1100011$ by (c). A continuation $11000110$ on the right would extend by (g) into $110001100$, contradicting (h). Hence we get $11000111$, and by symmetry also $111000111$.\n\n(k) The run $111$ must extend to $000111000$ by (j) and symmetry.\n\n(l) The sequence does not contain $000$. For by (j), this would extend to $111000111$, which must continue as $111000111000$ by (k).\n\n(m) The sequence does not contain $10$. For this would extend to $1100$ by (g), then $11001$ by (l) and $110011$ by (f), contradicting (i).\n\n(n) The sequence does not contain $01$ by (m) and symmetry.\n\n(o) By (m) and (n), the sequence can contain neither $01$ nor $10$, and so must be constant, which is absurd. $\\blacktriangledown$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 12515,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all pairs $(a, b)$ of real numbers such that\n\n$$\na \\cdot \\lfloor b n \\rfloor = b \\cdot \\lfloor a n \\rfloor\n$$\n\nfor all positive integers $n$.",
    "options": [],
    "answer": "See solution",
    "solution": "The solutions are all pairs $(a, b)$ with $a = 0$, $b = 0$, $a = b$, or both $a$ and $b$ integers.\n\nLet $a_0 = \\lfloor a \\rfloor$ and $a_i$ be the binary digits of the fractional part of $a$ such that $a = a_0 + \\sum_{i=1}^{\\infty} \\frac{a_i}{2^i}$ with $a_0 \\in \\mathbb{Z}$ and $a_i \\in \\{0, 1\\}$ for $i \\ge 1$. Similarly, let $b = b_0 + \\sum_{i=1}^{\\infty} \\frac{b_i}{2^i}$ with $b_0 \\in \\mathbb{Z}$ and $b_i \\in \\{0, 1\\}$ for $i \\ge 1$. In the case of a non-unique binary expansion, we choose the expansion ending on infinitely many zeros.\n\nNow choose $n = 2^k$ and $m = 2^{k-1}$ in the given equation. We get the equations\n\n$$\n\\begin{aligned}\na\\left(2^k b_0 + \\sum_{i=1}^k b_i 2^{k-i}\\right) &= b\\left(2^k a_0 + \\sum_{i=1}^k a_i 2^{k-i}\\right), \\\\\na\\left(2^{k-1} b_0 + \\sum_{i=1}^{k-1} b_i 2^{k-i-1}\\right) &= b\\left(2^{k-1} a_0 + \\sum_{i=1}^{k-1} a_i 2^{k-i-1}\\right).\n\\end{aligned}\n$$\n\nThe first equation for $k = 0$ and the difference of the first equation and the doubled second equation for $k \\ge 1$ yields\n\n$$\nab_k = b a_k \\tag{1}\n$$\n\nfor $k \\ge 0$.\n\nNow, we consider three cases. If one or both of $a$ and $b$ are zero, then the original equation is clearly satisfied. If both fractional parts are zero, then both numbers are integers and again, the original equation is satisfied. So, finally, we consider the case that $a, b \\ne 0$ and that there is a $k \\ge 1$ with $a_k = 1$. The equation (1) shows that $b_k$ cannot be zero, so we get $b_k = 1$ and thus from the same equation $a = b$. This clearly satisfies the original equation. (Of course, $b_k = 1$ leads to the same conclusion.) Therefore, the solutions are exactly the pairs listed above.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 12516,
    "subject": "Mathematics (Olympiad)",
    "question": "Determine all triples $(x, y, z)$ of positive integers with $x > y > z > 0$, such that\n$$\nx^2 = y \\cdot 2^x + 1.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "We first note that the right-hand side of the equation is odd, so $x^2$ is odd, and therefore $x$ is odd. One of the neighbors of $x$ must be divisible by $4$, so we have $x = 2^p a \\pm 1$ with $p > 1$ and $a$ odd.\n\n**Case 1:** $x = 2^p a + 1$\n\nIf $a > 1$, then $x^2 - 1 = 2^{p+1} a (2^{p-1} a + 1)$, so $y$ must contain all odd factors of $x^2 - 1$:\n$$\ny \\geq 3(2^{p-1} a + 1) > 2^p a + 1 = x,\n$$\nwhich contradicts $x > y$. Thus, $a = 1$, so $x = 2^p + 1$. The only possible value for $y$ is $2^{p-1} + 1$, since $b(2^{p-1} + 1) > 2^p + 1 = x$ for any $b > 1$. Thus, possible solutions are $x = 2^p + 1$, $y = 2^{p-1} + 1$, $z = p + 1$. For $p = 2$, $x = 5$, $y = z = 3$, which contradicts $y > z$. For $p > 2$, $x > y > z$ holds, since $2^{p-1} + 1 > p + 1$ for $p > 2$. So all triples $(2^p + 1, 2^{p-1} + 1, p + 1)$ for $p > 2$ are solutions.\n\n**Case 2:** $x = 2^p a - 1$\n\nIf $a > 1$, $x^2 - 1 = 2^{p+1} a (2^{p-1} a - 1)$, so\n$$\ny \\geq 3(2^{p-1} a + 1) = 2^p a - 1 + 2^{p-1} a - 2 > 2^p a - 1 + 2^p - 2 > x,\n$$\ncontradicting $x > y$. Thus, $a = 1$. If $x = 2^p - 1$, possible $y$ are $2^{p-1} - 1$ or $2(2^{p-1} - 1)$, since $b(2^{p-1} - 1) > 2^p - 1$ for $b > 2$. Thus, two further groups of solutions:\n\n- $x = 2^p - 1$, $y = 2^{p-1} - 1$, $z = p + 1$. For $p = 2$, $x = 3$, $y = 1$, $z = 3$ (contradicts $y > z$). For $p = 3$, $x = 7$, $y = 3$, $z = 4$ (again $y < z$). For $p \\geq 4$, $y > z$ holds, since $2^{p-1} - 1 > p + 1$ for $p > 3$. So all triples $(2^p - 1, 2^{p-1} - 1, p + 1)$ for $p > 3$ are solutions.\n\n- $x = 2^p - 1$, $y = 2^p - 2$, $z = p$. For $p = 2$, $x = 3$, $y = z = 2$ (contradicts $y > z$). For $p > 2$, $y > z$ holds, since $2^p - 2 > p$ for $p > 2$. So all triples $(2^p - 1, 2^p - 2, p)$ for $p > 2$ are solutions.\n\n**Summary:**\n\n$$\n\\begin{array}{l}\n(2^p + 1,\\ 2^{p-1} + 1,\\ p + 1) \\quad \\text{for } p > 2, \\\\\n(2^p - 1,\\ 2^{p-1} - 1,\\ p + 1) \\quad \\text{for } p > 3, \\\\\n(2^p - 1,\\ 2^p - 2,\\ p) \\quad \\text{for } p > 2.\n\\end{array}\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 12517,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $n$ be a positive integer. Initially, a bishop is placed in each square of the top row of a $2^n \\times 2^n$ chessboard; those bishops are numbered from 1 to $2^n$, from left to right. A jump is a simultaneous move made by all bishops such that the following conditions are satisfied:\n\n- Each bishop moves diagonally, in a straight line, some number of squares, and\n- At the end of the jump, the bishops all stand in different squares of the same row.\n\nFind the total number of permutations $\\sigma$ of the numbers $1, 2, \\ldots, 2^n$ with the following property: There exists a sequence of jumps such that all bishops end up on the bottom row arranged in the order $\\sigma(1), \\sigma(2), \\ldots, \\sigma(2^n)$, from left to right.",
    "options": [],
    "answer": "See solution",
    "solution": "The required number is $2^{n-1}$.\n\n**Step 1.** We show that the length of any jump is of the form $2^d$ for some integer $d \\le n-1$. Assign each bishop the number of the column it is situated on before the jump. Let $k$ be the length of the jump; then each bishop's column number either increases by $k$, or decreases by $k$ in the jump.\n\nThus, bishops $1, 2, \\ldots, k$ should move to columns $k+1, k+2, \\ldots, 2k$, as they cannot move leftwards. On the other hand, after the jump, columns $1, 2, \\ldots, k$ should be filled by the bishops $k+1, k+2, \\ldots, 2k$. So the leftmost $2k$ bishops still fill the columns $1, 2, \\ldots, 2k$ after the jump.\n\nRepeating the argument shows that the next $k$ bishops move rightwards, and the next $k$ bishops beyond move leftwards, and so on. Finally, the bishops all split into contiguous groups of length $2k$, and in each group the leftmost $k$ bishops move rightwards, whereas the rightmost $k$ bishops move leftwards. Hence $2k \\mid 2^n$, so $k$ is indeed of the form $2^d$ with $d \\le n-1$.\n\n**Step 2.** To make a more explicit description of the column change during the jump, assign each column the $n$-digit binary expansion of one less than its number, augmented with zeroes leftwards if necessary. It is then easily seen that a jump of length $2^d$ just switches the $d$-th digit from the left, $0$ to $1$ and vice versa.\n\nThus, the resulting permutation also has the following form: For every $d = 0, 1, \\ldots, n-1$, the $d$-th digit is either swapped for all bishops, or it is preserved for them all.\n\nMoreover, notice that the total length of all jumps is odd, so there will be an odd number of jumps of length $1$. Hence the $0$-th (the rightmost) digit will be switched anyway. This leaves room for $2^{n-1}$ possible permutations.\n\n**Step 3.** It remains to show that all $2^{n-1}$ permutations are indeed possible. Let us show how to reach any of them.\n\nStart by getting to the bottom row by downward jumps of lengths $1, 2, 4, \\ldots, 2^{n-1}$ that will switch all $n$ digits.\n\nNow, if we want to switch the $i$-th digit back, $1 \\le i \\le n-1$, make two upward jumps of length $2^{i-1}$, followed by a downward jump of length $2^i$. Combine such modifications for all possible digit combinations to get all desired permutations.",
    "topic": "Discrete Mathematics",
    "subtopic": "Combinatorics"
  },
  {
    "id": 12518,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $\\triangle ABC$ be a triangle with $|AB| < |AC| < |BC|$ and with circumcircle $\\Gamma$ having centre $O$. Let $\\omega_1$ be the circle with centre $B$ and radius $|AC|$ and let $\\omega_2$ be the circle with centre $C$ and radius $|AB|$. The circles $\\omega_1$ and $\\omega_2$ intersect in a point $E$ such that $A$ and $E$ lie on opposite sides of the line $BC$. The circles $\\Gamma$ and $\\omega_1$ intersect in a point $F$ and the circles $\\Gamma$ and $\\omega_2$ intersect in a point $G$ such that $F$ and $G$ lie on the same side of the line $BC$ as $E$.\n\nProve that the antipode $K$ of $A$ relative to $\\Gamma$ is the circumcentre of $\\triangle EFG$.\n\n![](images/NLD_ABooklet_2023_p25_data_addd855757.png)",
    "options": [],
    "answer": "See solution",
    "solution": "Since $|BE| = |AC|$ and $|CE| = |AB|$, $ABEC$ is a parallelogram. Therefore $BE \\parallel AC$. On the other hand, $ABGC$ is a cyclic quadrilateral with $|CG| = |AB|$. It follows that $ABGC$ is an isosceles trapezoid with $BG \\parallel AC$. Therefore $B$, $G$, and $E$ are collinear. Completely analogously, we see that $C$, $F$, and $E$ are collinear.\n\nSince $K$ is the antipode of $A$ relative to $\\Gamma$, we have $\\angle ACK = 90^\\circ$. As $AC \\parallel BE$ and $G$ lies on the line $BE$, it follows that $CK$ is perpendicular to $GE$. But we also know that $|CE| = |AB| = |CG|$, so $\\triangle ECG$ is isosceles and $CK$ is the perpendicular bisector of $EG$. In the same way, we see that $BK$ is the perpendicular bisector of $EF$. So $K$ lies on the perpendicular bisectors of $EG$ and $EF$. We conclude that $K$ is the circumcentre of $\\triangle EFG$. $\\square$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 12519,
    "subject": "Mathematics (Olympiad)",
    "question": "a) The longest sides of an isopentagon are those meeting at the $60^\text{∘}$ vertex. How many isopentagons have longest side 5 cm?\n\n![](images/2018-Australian-Scene-W2_p39_data_ea5ead0a0a.png)\n\nb) For the isopentagon $(5, 3, 2, 3, 5)$ (bottom-left pentagon in part a), what is its $t$-number (number of internal triangles)?\n\nc) For isopentagons with longest side less than 5 cm:\n- Are there any with longest side 1 cm?\n- How many with longest side 2 cm, and what is their $t$-number?\n- How many with longest side 3 cm, and what are their $t$-numbers?\n\nList all isopentagons with $t$-number less than 20.",
    "options": [],
    "answer": "See solution",
    "solution": "a) There are 4 isopentagons with longest side 5 cm.\n\nb) The $(5, 3, 2, 3, 5)$ isopentagon has $t$-number 46.\n\nc) There are no isopentagons with longest side 1 cm.\n\nThere is only one isopentagon with longest side 2 cm, $(2, 1, 1, 1, 2)$, which has $t$-number 7.\n\nThere are two isopentagons with longest side 3 cm, $(3, 1, 2, 1, 3)$ and $(3, 2, 1, 2, 3)$, which have $t$-numbers 14 and 17, respectively.\n\nIf the longest side of an isopentagon is 4 cm or more, then it contains an equilateral triangle of side length 4 cm, which in turn contains 16 grid triangles. The isopentagon also contains a trapezium which has at least another 7 grid triangles, a total of at least 23.\n\nSo the only isopentagons with $t$-number less than 20 are:\n- $(2, 1, 1, 1, 2)$ with $t$-number 7\n- $(3, 1, 2, 1, 3)$ with $t$-number 14\n- $(3, 2, 1, 2, 3)$ with $t$-number 17",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 12520,
    "subject": "Mathematics (Olympiad)",
    "question": "We call a divisor $d$ of a positive integer $n$ *special* if $d + 1$ is also a divisor of $n$. Prove that at most half the positive divisors of a positive integer can be special. Determine all positive integers for which exactly half the positive divisors are special.",
    "options": [],
    "answer": "See solution",
    "solution": "We prove that no positive divisor $d$ of $n$ that is greater or equal to $\\sqrt{n}$ can be special: if $d$ is special, then $d + 1$ is also a divisor, so $\\frac{n}{d}$ and $\\frac{n}{d + 1}$ are both integers, which means that their difference is at least 1. Thus\n\n$$\n\\frac{n}{d} \\geq \\frac{n}{d+1} + 1,\n$$\n\nwhich is equivalent to $n \\ge d(d+1)$. But since $d(d+1) > d^2 \\ge n$, this is a contradiction. Thus only divisors less than $\\sqrt{n}$ can be special. Since divisors come in pairs $(a$ and $n/a)$ such that one of them is less than $\\sqrt{n}$ and one greater than $\\sqrt{n}$ (when $n$ is a square, $\\sqrt{n}$ is paired with itself), this means that at most half the divisors can be special.\n\nIf precisely half the divisors are special, then $n$ cannot be a square, and every divisor less than $\\sqrt{n}$ has to be special. Thus 1 has to be a special divisor, meaning that 2 is a divisor (and thus also special), so 3 is a divisor, and so on, up to the greatest integer $k$ that is less than $\\sqrt{n}$. Finally, $k$ is special, so $k + 1$ has to be a divisor as well. Since $k$ is the greatest divisor less than $\\sqrt{n}$ and $k + 1$ the least divisor greater than $\\sqrt{n}$, their product must be $n$, so $n = k(k + 1) = k^2 + k$. Moreover, $k - 1$ is also a divisor of $n = k^2 + k$ (unless $k = 1$), so it also divides $n - (k - 1)(k + 2) = k^2 + k - (k^2 + k - 2) = 2$.\n\nThis leaves us with $k = 1$, $k = 2$ and $k = 3$ as the only possibilities, giving us $n = 2$, $n = 6$ or $n = 12$. In all these cases, exactly half the divisors are special.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 12521,
    "subject": "Mathematics (Olympiad)",
    "question": "Determine\n\na) the smallest number\nb) the biggest number\n\nfor $n \\ge 3$ of non-negative integers $x_1, x_2, \\dots, x_n$ having the sum $2011$ and satisfying:\n\n$$\n\\begin{aligned}\n&x_1 \\leq |x_2 - x_3|,\\\\\n&x_2 \\leq |x_3 - x_4|,\\\\\n&\\dots,\\\\\n&x_{n-2} \\leq |x_{n-1} - x_n|,\\\\\n&x_{n-1} \\leq |x_n - x_1|,\\\\\n&x_n \\leq |x_1 - x_2|.\n\\end{aligned}\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "Let $x_1, x_2, \\dots, x_n$ be non-negative integers satisfying the given conditions. Arrange them on a circle in this order. Let $M$ be the largest among $x_1, \\dots, x_n$. Without loss of generality, suppose $x_1 = M$. Clearly, $M \\neq 0$.\n\nFrom $x_1 \\leq |x_2 - x_3|$ and the choice of $x_1$, it follows that $\\{x_2, x_3\\} = \\{M, 0\\}$. Therefore, any $M$ is followed by an $M$ and a $0$ on the circle, either as $M$-$M$-$0$ or $M$-$0$-$M$. By induction, we find that only $0$ and $M$ appear on the circle.\n\n- There are no neighboring zeros (otherwise, all numbers would be zero, contradicting the sum $2011$).\n- There are no three consecutive $M$'s.\n\nSince $2011$ is prime and $x_1 + x_2 + \\dots + x_n = kM$, where $k$ is the number of $M$'s, we get $M = 1$ and $k = 2011$.\n\na) The smallest $n$ occurs when the number of zeros is minimized. Out of any three consecutive numbers, at least one is $0$, so there are at least $1006$ zeros, giving at least $3017$ numbers. A possible configuration: $x_{3k} = 0$ for $k = 1, 2, \\dots, 1005$, $x_{3017} = 0$, and the rest $x_j = 1$.\n\nb) The largest $n$ occurs when the number of zeros is maximized. Since there can be no neighboring zeros, at most $2011$ zeros are possible. This happens if $0$ and $1$ alternate around the circle, so the largest $n$ is $4022$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 12522,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $n > 3$ be a positive integer. Suppose that $n$ children are arranged in a circle, and $n$ coins are distributed between them (some children may have no coins). At every step, a child with at least 2 coins may give 1 coin to each of their neighbours on the right and left. Determine all initial distributions of coins from which it is possible that, after a finite number of steps, each child has exactly one coin.",
    "options": [],
    "answer": "See solution",
    "solution": "The answer is: all distributions where $\\sum_{i=1}^n i c_i \\equiv \\frac{n(n+1)}{2} \\pmod n$, where $c_i$ denotes the number of coins the $i$-th child starts with.\n\nEncode the sequence $c_i$ as the polynomial $p(x) = \\sum_i a_i x^i$. The cyclic nature of the problem makes it natural to work modulo $x^n - 1$. Child $i$ performing a step is equivalent to adding $x^i (x-1)^2$ to the polynomial, and we want to reach the polynomial $q(x) = 1 + x + \\dots + x^{n-1}$.\n\nSince we only add multiples of $(x-1)^2$, this is only possible if $p(x) = q(x)$ modulo the ideal generated by $x^n - 1$ and $(x-1)^2$, i.e.\n\n$$\n(x^n - 1, (x-1)^2) = (x-1) \\left( \\frac{x^n - 1}{x-1}, (x-1) \\right) = (x-1) \\cdot (n, (x-1))\n$$\n\nThis is equivalent to $p(1) = q(1)$ (which simply translates to the condition that there are $n$ coins) and $p'(1) = q'(1) \\pmod n$, which translates to the invariant. We can also show that this condition is sufficient. $\\square$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 12523,
    "subject": "Mathematics (Olympiad)",
    "question": "給定凸五邊形 $ABCDE$，令點 $A_1$ 為直線 $BD$ 與 $CE$ 的交點，點 $B_1$ 為直線 $CE$ 與 $DA$ 的交點，並依此類推定義 $C_1, D_1, E_1$ 等點。又令點 $A_2$ 為三角形 $ABD_1$ 的外接圓與三角形 $AEC_1$ 的外接圓的另一個交點，點 $B_2$ 為三角形 $BCE_1$ 的外接圓與三角形 $BAD_1$ 的外接圓的另一個交點，並依此類推定義 $C_2, D_2, E_2$ 等點。證明直線 $AA_2, BB_2, CC_2, DD_2, EE_2$ 共點。",
    "options": [],
    "answer": "See solution",
    "solution": "解：對 $A$ 作反演，反演後的點以撇號表示。\n\n![](images/19-1J_p36_data_47f64004d1.png)\n\n令 $P \\equiv AC'_1 \\cap \\odot(AB'B'_2)$，$Q \\equiv AD'_1 \\cap \\odot(AE'E'_2)$，$R \\equiv B'D'_1 \\cap \\odot(AB'D')$，$S \\equiv E'C'_1 \\cap \\odot(AC'E')$。顯然 $E'_1 \\in \\odot(AB'D')$，$B'_2 \\in \\odot(AB'C')$。\n\n$B'D'_1$ 和 $B', C', E'_1, B'_2$ 共圓，因此由 Reim 定理得 $AR \\parallel C'B'_2$。同理可證 $AS \\parallel D'E'_2$。\n\n注意 $A, B', E', C'_1, D'_1$ 共圓，因此由 Reim 定理得 $C'_1D'_1 \\parallel B'_2P \\parallel E'_2Q \\parallel D'R \\parallel C'S$，因此\n\n$$\n\\left\\{ \\begin{array}{l} \\frac{AD'_1}{C'D'_1} = \\frac{D'_1R}{D'_1B'_2} = \\frac{D'C'_1}{C'_1P} \\\\ \\\\ \\frac{AC'_1}{D'C'_1} = \\frac{C'_1S}{C'_1E'_2} = \\frac{C'D'_1}{D'_1Q} \\end{array} \\right. \\Longrightarrow \\frac{AC'_1}{C'_1P} = \\frac{AD'_1}{D'_1Q} \\Longrightarrow C'_1D'_1 \\parallel PQ,\n$$\n\n這表示 $P, Q, B'_2, E'_2$ 共線且 $C'_1D'_1 \\parallel B'_2E'_2$。\n\n由 Reim 定理得 $B', E', B'_2, E'_2$ 共圓，因此 $B, E, B_2, E_2$ 在同一圓 $\\Gamma$ 上，這表示 $AA_2, BB_2, EE_2$ 交於 $\\Gamma$、$\\odot(ABD_1)$ 和 $\\odot(AEC_1)$ 的根心 $T$。同理可證 $T$ 在 $CC_2$ 和 $DD_2$ 上，因此 $AA_2, BB_2, CC_2, DD_2, EE_2$ 共點。",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 12524,
    "subject": "Mathematics (Olympiad)",
    "question": "What is the largest number of rectangles of size $1 \\times 3$ that can be colored on a board of size $20 \\times 13$ such that no two colored rectangles share any points?\n\n![](images/Ukrajina_2013_p31_data_33f58fad4b.png)",
    "options": [],
    "answer": "See solution",
    "solution": "Let's mark 30 squares of size $2 \\times 2$ and 5 rectangles of size $1 \\times 2$ (see the figure). Each rectangle of size $1 \\times 3$ shares points with exactly one of the marked rectangles. Therefore, there can be at most 35 rectangles of size $1 \\times 3$. The example shows that this number can be achieved.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 12525,
    "subject": "Mathematics (Olympiad)",
    "question": "Does there exist a polynomial $f(x)$ such that\n$$\begin{aligned}\nf(a) &= bc, \\\\\nf(b) &= ca, \\\\\nf(c) &= ab?\n\ndisplaymath$$",
    "options": [],
    "answer": "See solution",
    "solution": "Yes.\n\nThe polynomial $f(x) = (a+b+c)x^2 - (ab+bc+ca)x + abc$ fits. Alternatively, $f(x) = x^3 - (x-a)(x-b)(x-c)$ also satisfies the conditions.",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 12526,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $\\overline{a_1a_2\\ldots a_n}$ be an $n$-digit number. The sum of this number and the number obtained by reversing its digits is a palindrome. How many such $n$-digit numbers are there?",
    "options": [],
    "answer": "See solution",
    "solution": "We consider two cases separately:\n\n* If the sum is an $n$-digit number, then obviously $a_1 + a_n < 10$. We show that no carry occurs when adding $\\overline{a_1a_2\\ldots a_n}$ and $\\overline{a_n a_{n-1}\\ldots a_1}$. Suppose, to the contrary, that a carry occurs. Let the result of the addition be $\\overline{b_1b_2\\ldots b_n}$, with the first carry occurring in the $i$-th position from the end. Then $a_n + a_1 = b_n$, $a_{n-1} + a_2 = b_{n-1}$, $\\ldots$, $a_{n+1-(i-1)} + a_{i-1} = b_{n+1-(i-1)}$ and $a_{n+1-i} + a_i \\ge 10$. Due to the last inequality, a carry also occurs in the $(n+1-i)$-th position from the end. Therefore, $b_{i-1} = a_{i-1} + a_{n+1-(i-1)} + 1 = b_{n+1-(i-1)} + 1$. But for $\\overline{b_1b_2\\ldots b_n}$ to be a palindrome, the equality $b_{i-1} = b_{n+1-(i-1)}$ must hold, which is a contradiction.\n\n* If the sum is a $(n+1)$-digit number, its first digit must be 1; let the remaining digits of the sum be $\\overline{b_1b_2\\ldots b_n}$. We will show that for each $i=1, \\ldots, n$, either $a_{n+1-i} = a_i = 0$ or $a_{n+1-i} + a_i = 11$. The statement is obviously true for $i=1$, since $b_n=1$, but $a_n+a_1 \\ne 1$ due to $a_n>0$, $a_1>0$. Now let $i>1$ and assume that the statement holds for $i-1$. Consider the case $a_{n+1-(i-1)} + a_{i-1} = 11$. Depending on whether or not there is a carry in the $(n+1-i)$-th position from the end, $b_{i-1}=2$ or $b_{i-1}=1$. Accordingly, $b_{n-(i-1)}=2$ or $b_{n-(i-1)}=1$, that is, $b_{n+1-i}=2$ or $b_{n+1-i}=1$. Since there is a carry in the $(i-1)$-th position from the end, the last digit of $a_{n+1-i}+a_i$ must be 1 or 0, respectively.\n\nSimilarly, in the case $a_{n+1-(i-1)} = a_{i-1} = 0$, depending on whether or not there is a carry in the $(n+1-i)$-th position from the end, $b_{i-1}=1$ or $b_{i-1}=0$. Accordingly, $b_{n-(i-1)}=1$ or $b_{n-(i-1)}=0$, that is, $b_{n+1-i}=1$ or $b_{n+1-i}=0$. Since there is no carry in the $(i-1)$-th position from the end, the last digit of $a_{n+1-i}+a_i$ is 1 or 0, respectively.\n\nIn conclusion, we have shown that if there is a carry in the $(n+1-i)$-th position from the end, the last digit of $a_{n+1-i}+a_i$ is 1, and if there is no carry in the $(n+1-i)$-th position from the end, the last digit of $a_{n+1-i}+a_i$ is 0. In the first case, $a_{n+1-i}+a_i = 11$, and in the second case, $a_{n+1-i} = a_i = 0$. This completes the induction step and proves the desired statement.\n\nAll described numbers trivially satisfy the problem's condition. Let's count them.\n\nIf $n$ is even, the number of $n$-digit numbers for which no carry occurs when adding the number and the number obtained by reversing its digits is $$36 \\cdot 55^{\\frac{n-2}{2}}$$ because the number of pairs of digits $(a, b)$ whose sum is less than 10 is $8+7+\\ldots+1 = 36$ (when zero is not allowed, as in the first and last positions), and $10+9+\\ldots+1 = 55$ (when zero is allowed, as in the other positions).\n\nThe number of $n$-digit numbers for which the sum of the first and last digits is 11 and the sum of the digits equidistant from the middle is either 0 or 11 is $$8 \\cdot 9^{\\frac{n-2}{2}}$$.\n\nSimilarly, if $n$ is odd, the number of $n$-digit numbers for which no carry occurs when adding the number and the number obtained by reversing its digits is $$36 \\cdot 55^{\\frac{n-3}{2}} \\cdot 5$$.\n\nThe number of $n$-digit numbers for which the sum of the first and last digits is 11 and the sum of the digits equidistant from the middle is either 0 or 11 is $$8 \\cdot 9^{\\frac{n-3}{2}}$$.\n\nThus, the total number of such numbers is\n$$36 \\cdot 55^{\\frac{n-2}{2}} + 8 \\cdot 9^{\\frac{n-2}{2}}$$\nif $n$ is even, and\n$$36 \\cdot 55^{\\frac{n-3}{2}} \\cdot 5 + 8 \\cdot 9^{\\frac{n-3}{2}}$$\nif $n$ is odd.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 12527,
    "subject": "Mathematics (Olympiad)",
    "question": "a) Does there exist a set $A$ of 2016 pairwise different positive integers such that for every non-empty subset $B \\subset A$ with $B \\neq A$ and every non-empty subset $C \\subset (A \\setminus B)$, the sum of the elements of $B$ is not divisible by the sum of the elements of $C$?\n\nb) Does there exist a set $A$ of 2016 pairwise different positive integers such that for every non-empty subset $B \\subset A$ with $B \\neq A$ and $|B| \\geq 2$, and every non-empty subset $C \\subset (A \\setminus B)$, the product of the elements of $B$ is divisible by the sum of the elements of $C$?",
    "options": [],
    "answer": "See solution",
    "solution": "a) Let us first choose the numbers as $2, 3, 5, \\ldots, p_{2016}$, where $p_i$ is the $i$-th prime number in increasing order. Consider all possible pairs of subsets $B$ and $C$ as described; their number is finite, say $N$. We will iteratively modify the elements of $A$ to construct a set that satisfies the condition for each pair. Denote the elements as $a_1, a_2, \\ldots, a_{2016}$ (initially $a_i = p_i$). For each pair $(B, C)$, let $p = p_{2016+k}$ for the $k$-th pair. Multiply all elements of $C$ by $p$, and in $B$, multiply all but the smallest-indexed element by $p$. Thus, the sum of $B$ is not divisible by $p$, while the sum of $C$ is divisible by $p$, ensuring the sum of $B$ is not divisible by the sum of $C$. Repeating this for all $N$ pairs, we obtain a set $A$ satisfying the required property. This construction works for any finite number of pairwise distinct elements.\n\nb) We construct such a set for any $n$. Let $N = \\left(\\frac{n(n+1)}{2}\\right)!$, and define $a_i = i \\cdot N$ for $i = 1, \\ldots, n$. For any non-empty subset $C$, its sum $L$ satisfies $L \\leq \\frac{n(n+1)}{2} N$, so $L = K N$ for some $K \\leq \\frac{n(n+1)}{2}$. The product of any two elements of $A$ is divisible by $N^2$, and thus by $L$, since $N$ is divisible by any integer up to $\\frac{n(n+1)}{2}$. Therefore, the product of the elements of $B$ is divisible by the sum of the elements of $C$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 12528,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $ABC$ be a triangle and let $C_0$, $A_0$, $B_0$ be points on the sides $AB$, $BC$, $CA$ respectively. Show that\n$$\nR(AB_0C_0) + R(BC_0A_0) + R(CA_0B_0) \\ge 6r(A_0B_0C_0).\n$$\nHere $R(XYZ)$ is the circumradius and $r(XYZ)$ is the inradius of a triangle $XYZ$.",
    "options": [],
    "answer": "See solution",
    "solution": "First, we prove the following well-known lemma, due to Fejes Tóth:\n\n**Lemma.** For any point $P$ inside a triangle $ABC$ with inradius $r$, we have $PA + PB + PC \\ge 6r$.\n\n*Proof.* Assume that $\\angle C > 120^\\circ$. Consider the triangle $A_1B_1C_1$ such that $A \\in B_1C_1$, $B \\in C_1A_1$, $C \\in A_1B_1$, and $AC \\perp B_1C_1$, $BC \\perp A_1C_1$, and $A_1B_1$ is perpendicular to the bisector of the angle $C$. Then $\\angle AC_1B = 180^\\circ - \\angle ACB < 60^\\circ$. Hence $A_1B_1 < B_1C_1 = A_1C_1$. Let $Q$ be an arbitrary point inside the triangle $A_1B_1C_1$. Let $A'$, $B'$, $C'$ be the bases of perpendiculars from $Q$ to the sides $B_1C_1$, $C_1A_1$, $A_1B_1$ respectively. We observe that\n$$\nB_1C_1 \\cdot QA' + C_1A_1 \\cdot QB' + A_1B_1 \\cdot QC' = 2S_{A_1B_1C_1}.\n$$\nHence,\n$$\n(QA' + QB' + QC') \\cdot B_1C_1 = 2S_{A_1B_1C_1} + (B_1C_1 - A_1B_1) \\cdot QC.\n$$\nSince $A_1B_1 < B_1C_1$, the sum $QA' + QB' + QC'$ takes the minimum value if and only if $Q \\in A_1B_1$. Thus, for an arbitrary point $P$ inside the triangle $ABC$, the sum $PA+PB+PC$ takes the minimum value if and only if $P \\equiv C$. Thus $PA+PB+PC \\ge AC+BC$.\n\nLet $BC = a$, $AC = b$, $AB = c$ and $h_c$ is the altitude from vertex $C$ of the triangle $ABC$. Then it is obvious that $a \\ge h_c$ and $b \\ge h_c$. Hence,\n$$\n6r(a+b+c) = 12S_{ABC} \\le 4ab+4S = 4ab+2ch_c \\le (a+b)^2+c(a+b) = (a+b)(a+b+c).\n$$\nTherefore $a+b \\ge 6r$. Thus $PA+PB+PC \\ge 6r$.\n\nNow assume that $\\angle A$, $\\angle B$, $\\angle C \\le 120^\\circ$. Let $B'$, $C'$, $P'$ be the images of the points $B$, $C$, $P$ under the rotation by $60^\\circ$ (clockwise) at the point $A$. Then $PA+PB+PC = PP'+PB+PC'$ and it takes the minimum value if and only if the points $B$, $P$, $P'$, $C'$ are collinear. Hence $\\angle APB = \\angle AP'C' = APC = 120^\\circ$. Let $AB = c$, $AC = b$, $\\angle A = \\alpha$. We conclude that\n$$\n(PA + PB + PC)^2 \\ge BC'^2 = b^2 + c^2 - 2bc \\cos(\\alpha + 60^\\circ) = b^2 + c^2 - bc \\cos \\alpha + bc \\sqrt{3} \\sin \\alpha \\tag{1}\n$$\nSince\n$$\n(p \\cdot r)^2 = S^2 = p(p-a)(p-b)(p-c) \\le p \\left( \\frac{p-a+p-b+p-c}{3} \\right)^3 = \\frac{p^4}{27},\n$$\n$$\nr \\le \\frac{p}{3\\sqrt{3}}. \\text{ Hence } S = p \\cdot r \\ge 3\\sqrt{3}r^2.\n$$\nOn the other hand,\n$$\nS \\le \\frac{p^2}{3\\sqrt{3}} = \\frac{(a+b+c)^2}{4 \\cdot 3\\sqrt{3}} \\le \\frac{a^2+b^2+c^2}{4\\sqrt{3}}.\n$$\nTherefore, from (1),\n$$\n(PA + PB + PC)^2 \\ge 2\\sqrt{3}S_{ABC} + 2\\sqrt{3}S_{ABC} = 4\\sqrt{3}S_{ABC} \\ge 36r^2.\n$$\nThus $PA+PB+PC \\ge 6r$.\n\n![](images/MNG_ABooklet_2015_p33_data_e3bce1bc5d.png)\n\nNow we prove our problem. Let $\\omega(XYZ)$ be the circumcircle of a triangle $XYZ$. By Miquel's theorem, the circles $\\omega(AB_0C_0)$, $\\omega(BC_0A_0)$, and $\\omega(CA_0B_0)$ pass through a common point, say $M$. Let $O_1$, $O_2$, and $O_3$ be the centres of the circles $\\omega(AB_0C_0)$, $\\omega(BC_0A_0)$, and $\\omega(CA_0B_0)$, respectively. For the triangle $O_1O_2O_3$ and the point $M$, we use the Erdős-Mordell inequality and get that\n$$\nMO_1 + MO_2 + MO_3 = R(AB_0C_0) + R(BC_0A_0) + R(CA_0B_0) \\ge MA_0 + MB_0 + MC_0.\n$$\nBy the lemma above, $MA_0 + MB_0 + MC_0 \\ge 6r(A_0B_0C_0)$. This completes the proof.\n\nFrom the solution it follows that equality holds if and only if the triangles $ABC$ and $A_0B_0C_0$ are equilateral.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 12529,
    "subject": "Mathematics (Olympiad)",
    "question": "Solve the equation\n\n$$\n(x + 1)\\sqrt{x^2 + 2x + 2} + x\\sqrt{x^2 + 1} = 0.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "Let $a = \\sqrt{x^2 + 2x + 2} > 0$ and $b = \\sqrt{x^2 + 1} > 0$. Therefore,\n\n$$\nx = \\frac{(x^2 + 2x + 2) - (x^2 + 1) - 1}{2} = \\frac{a^2 - b^2 - 1}{2}\n$$\n$$\nx + 1 = \\frac{a^2 - b^2 + 1}{2}.\n$$\n\nThe equation is equivalent to:\n\n$$\n\\begin{aligned}\n& \\frac{a^2 - b^2 + 1}{2} \\cdot a + \\frac{a^2 - b^2 - 1}{2} \\cdot b = 0 \\\\\n& (a^2 - b^2)a + a + (a^2 - b^2)b - b = 0 \\\\\n& (a-b)\\left((a+b)^2 + 1\\right) = 0.\n\\end{aligned}\n$$\n\nHence, $a = b$ and $x = -\\frac{1}{2}$.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 12530,
    "subject": "Mathematics (Olympiad)",
    "question": "Find $a_3, a_4, \\dots, a_{2008}$, such that $a_i = \\pm 1$ for $i = 3, \\dots, 2008$ and\n$$\n\\sum_{i=3}^{2008} a_i 2^i = 2008,\n$$\nand show that the numbers $a_3, a_4, \\dots, a_{2008}$ are uniquely determined by these conditions.",
    "options": [],
    "answer": "See solution",
    "solution": "*Existence:* Dividing both sides by $8$, we require\n$$\n251 = \\sum_{i=0}^{2005} a_{i+3} 2^i.\n$$\nNow $251 = 2^0 + 2^1 + 2^2 + 2^3 + 2^4 + 2^5 + 2^6 + 2^7$. Also, $-(1+2+\\dots+2^{m-1}) + 2^m = 1$, for $m = 1, 2, \\dots$\n\nSo\n$$\n251 = 2^0 + 2^1 + 2^2 + 2^3 + 2^4 + 2^5 + 2^6 + 2^7 = (-1 - 2 - \\dots - 2^{1997} + 2^{1998}).\n$$\nSo $a_3 = a_5 = a_6 = a_7 = a_8 = a_9 = a_{2008} = +1$ and $a_4 = a_{10} = a_{11} = a_{12} = \\dots = a_{2007} = -1$ gives a solution.\n\n*Uniqueness:* More generally, for $n \\geq 1$, each odd integer $m$ with $-2^n < m < 2^n$ has a unique expression as\n$$\nm = \\sum_{i=0}^{n-1} a_i 2^i, \\quad \\text{where } a_i = \\pm 1, \\text{ for each } i.\n$$\nWe prove this by induction on $n$. The base case $n = 1$ is just the statement that $1 = 2^0$ and $-1 = -2^0$.\n\nLet $n > 1$ and assume the result for $n-1$. Then there is a unique integer $a_0 = \\pm 1$ such that $m - a_0 \\equiv 2 \\pmod{4}$. Clearly $-2^n < m - a_0 < 2^n$. So $-2^{n-1} < (m - a_0)/2 < 2^{n-1}$ and $(m - a_0)/2$ is odd. The inductive hypothesis implies that\n$$\n\\frac{m - a_0}{2} = \\sum_{i=1}^{n-1} a_i 2^{i-1}, \\quad \\text{for unique } a_i = \\pm 1.\n$$\nThis gives the expression for $m$. The uniqueness of the expression is apparent from the construction. It is also a consequence of the fact that there are $2^n$ odd integers $m$ with $-2^n < m < 2^n$, but only $2^n$ expressions $\\sum_{i=0}^{n-1} a_i 2^i$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 12531,
    "subject": "Mathematics (Olympiad)",
    "question": "We will call _smalls_ the positive integers not larger than 2025.\n\na) Find the number of smalls which are perfect squares.\n\nb) Find the number of smalls which are perfect squares and leave remainder 0 when divided by 45.\n\nc) Find the number of smalls which neither are perfect squares, nor leave remainder 0 when divided by 45.",
    "options": [],
    "answer": "See solution",
    "solution": "a) Since $45^2 = 2025$, the smalls which are perfect squares are $1^2, 2^2, 3^2, \\dots, 45^2$, that is, 45 such numbers.\n\nb) The smalls which are perfect squares and are divisible by 45 are $15^2$, $4 \\cdot 15^2$, and $9 \\cdot 15^2$—there are 3 of them.\n\nc) The smalls divisible by 45 are $45 \\cdot 1, 45 \\cdot 2, \\dots, 45 \\cdot 45$—there are 45 such numbers. The number of smalls which are perfect squares or are divisible by 45 is $45 + 45 - 3 = 87$. So, the number of smalls which are neither perfect squares, nor leave remainder 0 when divided by 45 is $2025 - 87 = 1938$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 12532,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all positive integers $x$, $y$, $p$, $n$, $k$ such that:\n\n$$\n\\begin{cases}\n5x + y = p^k, \\\\\n5y + x = p^{k+n}.\n\\end{cases}\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "*Answer:* For all $k \\ge 3$, $x = 2^{k-3}$, $y = 3 \\cdot 2^{k-3}$, $p = 2$, and $n = 1$.\n\n*Solution.* Rewrite the first equation as $25x + 5y = 5p^k$ and subtract the second equation: $24x = p^k(5 - p^n)$. Thus, $5 - p^n > 0$, so only the following cases are possible:\n\n*Case 1.* $p = 1$, $n \\in \\mathbb{N}$. Since $5x + y \\ge 6$, there are no solutions.\n\n*Case 2.* $p = 2$, $n = 1$. Then $24x = 2^k \\cdot 3$, so $x = 2^{k-3}$, with $k \\ge 3$. From the first equation, $y = 2^k - 5 \\cdot 2^{k-3} = 3 \\cdot 2^{k-3}$. These values also satisfy the second equation. Thus, for any $k \\ge 3$, the solution is $x = 2^{k-3}$, $y = 3 \\cdot 2^{k-3}$, $p = 2$, $n = 1$.\n\n*Case 3.* $p = 2$, $n = 2$ or $p = 4$, $n = 1$. Then $24x = 2^k$, so there are no solutions.\n\n*Case 4.* $p = 3$, $n = 1$. Then $24x = 3^k \\cdot 2$, so there are no solutions.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 12533,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all finite sets $A$ of positive real numbers with at least two elements, such that for every $a, b \\in A$ with $a \\neq b$, $a^2 + b^2 \\in A$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $a_1 < a_2 < \\dots < a_n$ be the elements of $A$. Then $a_1^2 + a_2^2, a_1^2 + a_3^2, \\dots, a_1^2 + a_n^2, a_2^2 + a_n^2, \\dots, a_{n-1}^2 + a_n^2$ all belong to $A$, and\n\n$$\na_1^2 + a_2^2 < a_1^2 + a_3^2 < \\dots < a_1^2 + a_n^2 < a_2^2 + a_n^2 < \\dots < a_{n-1}^2 + a_n^2\n$$\n\nwhich implies $2n - 3 \\leq n$. Thus, $A$ has at most 3 elements.\n\nIf $n = 3$ and $A = \\{a, b, c\\}$ with $a < b < c$, then $a^2 < b^2 < c^2$, so $a^2 + b^2 < a^2 + c^2 < b^2 + c^2$.\n\nSince $A = \\{a, b, c\\} = \\{a^2 + b^2, a^2 + c^2, b^2 + c^2\\}$, we have\n\n$$\na^2 + b^2 = a, \\quad a^2 + c^2 = b, \\quad b^2 + c^2 = c. \\tag{1}\n$$\n\nFrom the first and last equations in (1), $a^2 - c^2 = a - c$, so $a + c = 1$. Substituting into the second equation of (1):\n\n$$\nb = a^2 + (1 - a)^2 = 2a^2 - 2a + 1 = 2(a^2 - a) + 1 = -2b^2 + 1,\n$$\n\nso $(2b - 1)(b + 1) = 0$, hence $b = \\frac{1}{2}$. Then $a^2 - a + \\frac{1}{4} = 0$ leads to $a = \\frac{1}{2}$, which contradicts $a < b$.\n\nTherefore, $n = 2$, and $A = \\{a, b\\}$ with $a^2 + b^2 = a$. For $a \\in (0, 1]$, $b = \\sqrt{a - a^2}$. In conclusion, the desired sets are $\\{a, \\sqrt{a - a^2}\\}$, with $a \\in (0, 1]$, $a \\neq \\frac{1}{2}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 12534,
    "subject": "Mathematics (Olympiad)",
    "question": "From the natural numbers $2, 3, 4, \\ldots, 2019$, one constructs $1009$ fractions and chooses the maximum of these. What is the minimum possible value of this maximum fraction?",
    "options": [],
    "answer": "See solution",
    "solution": "First, note that the minimum possible value for the maximum fraction can be achieved by choosing the following fractions:\n\n$$\n\\frac{1}{1011}, \\frac{2}{1012}, \\frac{3}{1013}, \\ldots, \\frac{1010}{2019}.\n$$\n\nSuppose, for contradiction, that a smaller value for the maximum fraction can be obtained in some other way. Clearly, $2019$ cannot be a numerator, so consider the fraction with denominator $2019$. Its numerator must be less than $1010$; denote it as $a < 1010$.\n\nThere are at least $1009$ numbers in the set\n\n$$\nM = \\{a, a+1, \\ldots, 1010, 1011, \\ldots, 2018\\}.\n$$\n\nBy the pigeonhole principle, some of these numbers must form one of the remaining $1008$ fractions (excluding $\\frac{a}{2019}$). If we denote them $b < c$, then $\\frac{1010}{2019} < \\frac{b}{2019} < \\frac{b}{c}$, which is a contradiction.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 12535,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all prime numbers $p$ such that $p^3 - 4p + 9$ is a perfect square.",
    "options": [],
    "answer": "See solution",
    "solution": "It is easy to check that $p = 2$ is a solution and $p = 3$ is not. We suppose that $p > 3$.\n\nFrom $p^3 - 4p + 9 = n^2$, we have $$(p-2)p(p+2) = (n-3)(n+3).$$ As one of the three factors on the left-hand side is a multiple of $3$, we get that $n$ is a multiple of $3$. We suppose now that $n = 3k$, $k \\ge 1$, and thus $$(p-2)p(p+2) = 9(k-1)(k+1).$$ Because $p \\ne 0$, the products on both sides are nonzero. The prime number $p > 3$ divides $9(k-1)(k+1)$ and thus it divides exactly one of the two factors $k-1$ or $k+1$.\n\nSuppose $k-1 = lp$. Then $$(p-2)(p+2) = 9l(lp+2).$$ Reducing modulo $p$, we get $$18l + 4 \\equiv 0 \\pmod{p}.$$ As $p$ is odd, $p$ divides $9l + 2$. In particular, $p \\le 9l + 2$, $p-2 \\le 9l$, and thus $(p-2)(p+2) = 9l(lp + 2)$ implies $p+2 \\ge lp+2$. But this is possible only for $l = 1$ and $p = 9l+2 = 11$. Thus, $p = 11$ is a solution.\n\nSimilarly, if $k+1 = lp$, we get $p = 7$.\n\nTherefore, the solutions are $p = 2, 7, 11$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 12536,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all real numbers $r$ such that there is exactly one real number $a$ satisfying the following system of inequalities:\n\n$$\na + r \\geq 0 \\qquad (1)\n$$\n\n$$\nar \\geq -1 \\qquad (2)\n$$\n\n$$\na \\geq 0 \\qquad (3)\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "**Case 1** $r < -1$\n\nThen (1) implies $a > 1$ while (2) implies $a < 1$, which is a contradiction. So this case does not occur.\n\n**Case 2** $r = -1$\n\nThen (1) implies $a \\geq 1$, while (2) implies $a \\leq 1$. So $a = 1$, and this also satisfies (3). Hence $r = -1$ is a solution.\n\n**Case 3** $-1 < r < 0$\n\nWe may write $r = -s$ where $0 < s < 1$. Thus (1), (2) and (3) become\n\n$$\na \\geq s \\qquad (1')\n$$\n\n$$\nas \\leq 1 \\qquad (2')\n$$\n\n$$\na \\geq 0 \\qquad (3')\n$$\n\nSince $s < 1$, any $a$ satisfying $s < a < 1$ also satisfies (1'), (2') and (3'). Since there are infinitely many such $a$, this case does not occur.\n\n**Case 4** $r \\geq 0$\n\nAny $a \\geq 0$ satisfies (1), (2) and (3). So this case does not occur. $\\square$",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 12537,
    "subject": "Mathematics (Olympiad)",
    "question": "Given $d = 1$, let $k = 15$ and consider the sequence:\n\n8, 1, 15, 10, 6, 3, 13, 12, 4, 5, 11, 14, 2, 7, 9.\n\nTwo neighbouring numbers in this sequence always add up to 9, 16, or 25. For square $d > 1$, take $k = 15$ and the same sequence as above, but multiply all numbers by $d$. Then, two neighbouring numbers in this sequence always add up to $9d$, $16d$, or $25d$, which are all perfect squares.\n\nNow, consider a non-square $d$. Does there exist a $k$ and a sequence $a_1d, a_2d, \\dots, a_kd$, where $\\{a_1, a_2, \\dots, a_k\\} = \\{1, 2, \\dots, k\\}$, such that every sum of two neighbouring terms is a perfect square? For which $d$ is this possible?",
    "options": [],
    "answer": "See solution",
    "solution": "Suppose such a sequence exists for a non-square $d$. Write $d = c m^2$, where $m$ is a positive integer and $c$ is square-free ($c > 1$). For all $i$ with $1 \\leq i \\leq k-1$, $a_i d + a_{i+1} d$ is a perfect square, so $d \\mid a_i d + a_{i+1} d$, i.e., $c m^2 \\mid (a_i + a_{i+1}) d$. Thus, $c \\mid a_i + a_{i+1}$, so $a_{i+1} \\equiv -a_i \\pmod{c}$, and $a_{i+2} \\equiv -a_{i+1} \\equiv a_i \\pmod{c}$. Therefore, only two residue classes modulo $c$ can occur among the $a_i$. But since $\\{a_1, \\dots, a_k\\} = \\{1, \\dots, k\\}$ and $k \\geq 3$, we must have $c \\leq 2$. Since $d$ is not a square, $c = 1$ is impossible, so $c = 2$. But then $a_{i+1} \\equiv -a_i \\equiv a_i \\pmod{2}$, so only one residue class modulo 2 occurs, which is a contradiction.\n\nTherefore, the only possible $d$ are perfect squares.",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 12538,
    "subject": "Mathematics (Olympiad)",
    "question": "令 $p$ 為一奇質數，並設 $N = \\frac{1}{4}(p^3 - p) - 1$。我們將數字 $1, 2, \\dots, N$ 分別塗成紅色或藍色中的一個顏色。對於任何正整數 $n \\leq N$，令 $R(n)$ 為 $\\{1, 2, \\dots, n\\}$ 中紅色數字的個數，並令 $r(n) = R(n)/n$。試證明存在正整數 $a \\in \\{1, 2, \\dots, p-1\\}$，使得 $r(n) \\ne a/p$ 對於所有 $n = 1, 2, \\dots, N$ 都成立。",
    "options": [],
    "answer": "See solution",
    "solution": "假設原命題不成立，則對於每個 $a \\in \\{1, 2, \\dots, p-1\\}$，存在 $n_a$ 使得 $r(n_a) = a/p$，從而 $R(n_a) = a n_a / p$。這表示 $p \\mid n_a$，因此 $m_a = n_a / p \\in \\mathbb{N}$ 且 $R(n_a) = a m_a$。不失一般性假設 $m_1 < m_{p-1}$（否則只要把紅藍兩色完全對調即可）。\n\n注意到此時，我們必然有\n\n$$\nm_a \\leq \\frac{N}{p} < \\frac{p^2 - 1}{4} \\quad (1)\n$$\n\n對於 $a = 1, \\dots, p-1$ 皆成立。但同時，我們有以下性質：\n\n*引理：* 若 $m_a < m_b$，其中 $a, b \\in \\{1, 2, \\dots, p-1\\}$，\n\n$$\nm_b \\geq \\frac{a}{b} m_a \\quad \\text{且} \\quad m_b \\geq \\frac{p-a}{p-b} m_b.\n$$\n\n*證明：* 基於 $n_b = p m_b > p m_a = n_a$，我們有 $b m_b = R(n_b) \\geq R(n_a) = a m_a$，故第一部分得證。將兩色對調便得第二部分。□\n\n讓我們證明引理與 (1) 矛盾，從而原命題必須成立。考慮 $q = (p-1)/2$。\n\n**Case 1.** $m_1, m_2, \\dots, m_q$ 皆小於 $m_{p-1}$\n\n取 $m_a = \\max\\{m_1, \\dots, m_q\\}$，則基於 $m_i$ 互異，必有 $m_a \\geq q \\geq a$。但由引理，我們有\n\n$$\nm_{p-1} \\geq \\frac{p-a}{p-(p-1)} m_a \\geq (p-q) q = \\frac{p^2 - 1}{4},\n$$\n\n與 (1) 矛盾。\n\n**Case 2.** 存在 $k \\leq q$ 使得 $m_k > m_p$\n\n令 $k$ 為滿足 $m_k > m_p$ 中最小者。我們自然地有 $1 < k \\leq q < p-1$。此外，同 Case 1 定義最大值 $m_a$，則我們有 $a \\leq k-1 \\leq m_a < m_{p-1}$。因此，由引理，\n\n$$\n\\begin{aligned}\nm_k &\\geq \\frac{p-1}{k} m_{p-1} \\geq \\frac{p-1}{k} \\frac{p-a}{p-(p-1)} m_a \\\\\n&\\geq \\frac{p-1}{k} (p-k+1)(k-1) \\geq \\frac{k-1}{k} (p-1)(p-q) \\geq \\frac{1}{2} \\frac{p^2 - 1}{2},\n\\end{aligned}\n$$\n\n與 (1) 矛盾。\n\n綜以上，不論任何狀況皆會導致矛盾，故原命題成立。",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 12539,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $a, b, c > 0$. Find the greatest $x > 0$ such that there exist positive numbers $p, q, r$ with $p + q + r = 1$ and\n$$\nx \\leq a \\frac{p}{q}, \\quad x \\leq b \\frac{q}{r}, \\quad x \\leq c \\frac{r}{p}.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "The greatest such $x$ is $\\sqrt[3]{abc}$.\n\nMultiplying the three inequalities gives\n$$\nx^3 \\leq a b c \\frac{p}{q} \\frac{q}{r} \\frac{r}{p} = abc,\n$$\nso $x \\leq \\sqrt[3]{abc}$.\n\nTo show $x = \\sqrt[3]{abc}$ is attainable, consider the system\n$$\n\\begin{cases}\n\\sqrt[3]{abc} = a \\frac{p}{q} \\\\\n\\sqrt[3]{abc} = b \\frac{q}{r} \\\\\np + q + r = 1\n\\end{cases}\n$$\nFrom the first two equations:\n$$\np = \\frac{yq}{a}, \\quad r = \\frac{bq}{y}, \\quad \\text{where } y = \\sqrt[3]{abc}.\n$$\nSubstitute into $p + q + r = 1$:\n$$\n\\frac{yq}{a} + q + \\frac{bq}{y} = 1 \\implies q = \\frac{1}{\\frac{y}{a} + 1 + \\frac{b}{y}}.\n$$\nThus, $p, q, r > 0$ exist, so $x = \\sqrt[3]{abc}$ is possible.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 12540,
    "subject": "Mathematics (Olympiad)",
    "question": "Find the smallest positive integer $k$ with the following property: for any three points with integer coordinates on a plane,\n\n$$\nL_{\\max} - L_{\\min} > \\frac{1}{\\sqrt{k} \\cdot L_{\\max}},\n$$\n\nwhere $L_{\\max}$ and $L_{\\min}$ are respectively the maximum and minimum distances among them.",
    "options": [],
    "answer": "See solution",
    "solution": "The answer is $k = 4$.\n\nFirst, note that $L_{\\max} > L_{\\min}$ because a triangle formed by three integer-coordinate points cannot be equilateral. If it were, the area would be $S = \\frac{a^2 \\sqrt{3}}{4}$, which is irrational for rational $a^2$, while the area of a triangle with integer vertices must be rational.\n\nSince $L_{\\max}^2$ and $L_{\\min}^2$ are integers, $L_{\\max}^2 - L_{\\min}^2 \\ge 1$, so\n\n$$\nL_{\\max} - L_{\\min} \\ge \\frac{1}{L_{\\max} + L_{\\min}} > \\frac{1}{2L_{\\max}} = \\frac{1}{\\sqrt{4} L_{\\max}}.\n$$\n\nTo show that $k \\le 3$ does not work, consider an isosceles triangle with base and height both equal to $2$. The side length is $\\sqrt{5}$, so\n\n$$\nL_{\\max} - L_{\\min} = \\sqrt{5} - 2 < \\frac{1}{\\sqrt{3} \\sqrt{5}}.\n$$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 12541,
    "subject": "Mathematics (Olympiad)",
    "question": "Prove that a triangle $ABC$ is right-angled if and only if\n\n$$\nsin^2 A + sin^2 B + sin^2 C = 2.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "Suppose one of $\\angle A, \\angle B, \\angle C$ is a right angle. Say $\\angle A$ is. Then $\\angle B + \\angle C = \\pi/2$ and so\n\n$$\n\\sin^2 A + \\sin^2 B + \\sin^2 C = 1 + \\sin^2 B + \\sin^2(\\pi/2 - B) = 1 + \\sin^2 B + \\cos^2 B = 2.\n$$\n\nSuppose the identity holds. By the Sine Rule,\n\n$$\n\\frac{\\sin A}{a} = \\frac{\\sin B}{b} = \\frac{\\sin C}{c} = \\frac{1}{2R} \\equiv \\lambda.\n$$\n\nHence\n\n$$\n2 = \\lambda^2(a^2 + b^2 + c^2), \\quad \\lambda^2 = \\frac{2}{a^2 + b^2 + c^2}.\n$$\n\nso that\n\n$$\n\\sin^2 A = \\frac{2a^2}{a^2 + b^2 + c^2}, \\quad \\cos^2 A = \\frac{b^2 + c^2 - a^2}{a^2 + b^2 + c^2} = \\frac{2bc \\cos A}{a^2 + b^2 + c^2}.\n$$\n\nSimilarly,\n\n$$\n\\cos^2 B = \\frac{2ca \\cos B}{a^2 + b^2 + c^2}, \\quad \\cos^2 C = \\frac{2ab \\cos C}{a^2 + b^2 + c^2}.\n$$\n\nHence $\\cos A, \\cos B, \\cos C$ are nonnegative. Suppose all are positive, then\n\n$$\n\\cos A = \\frac{2bc}{a^2 + b^2 + c^2}, \\cos B = \\frac{2ca}{a^2 + b^2 + c^2}, \\cos C = \\frac{2ab}{a^2 + b^2 + c^2}.\n$$\n\nHence, for instance,\n\n$$\n(b^2 + c^2 - a^2)(b^2 + c^2 + a^2) = (2bc)^2, \\quad (b^2 + c^2)^2 - a^4 = 4b^2c^2,\n$$\n\nor\n\n$$\n(b^2 - c^2)^2 - a^4 = 0, \\quad (b^2 - c^2 - a^2)(b^2 - c^2 + a^2) = 0,\n$$\n\ni.e., $\\cos B \\cos C = 0$, a contradiction. Hence, one of $\\cos A, \\cos B, \\cos C$ is zero.\n\n**Second solution:** Let $S$ be the circumcircle of $ABC$ and suppose that $S$ has diameter $d$. Using the sine rule we have\n\n$$\n\\sin^2 A + \\sin^2 B + \\sin^2 C = \\frac{1}{d^2}(a^2 + b^2 + c^2).\n$$\n\nSuppose that $ABC$ is not a right-angled triangle. There are two cases to consider:\n\n**Case I:** One of the angles is obtuse. Without loss of generality, suppose that $A > \\frac{\\pi}{2}$. It is easy to see from a diagram that $a < d$ and that $b^2 + c^2 < d^2$. Thus\n\n$$\n\\frac{1}{d^2}(a^2 + b^2 + c^2) < 2\n$$\n\nin this case.\n\n**Case II:** $ABC$ is an acute-angled triangle (i.e., all angles $< \\frac{\\pi}{2}$). Then let $X$ be the point on $S$ that is diametrically opposite to $A$ (so $|AX| = d$). Let $e = |BX|$ and $f = |CX|$. Pythagoras' Theorem implies that\n\n$$\nb^2 + e^2 + c^2 + f^2 = 2d^2\n$$\n\nHowever, $\\angle BXC > \\frac{\\pi}{2}$, so $e^2 + f^2 < a^2$ (a simple application of the cosine rule to the triangle $BXC$). Therefore, in this case we have\n\n$$\n\\frac{1}{d^2}(a^2 + b^2 + c^2) > 2.\n$$\n\n**Third solution:** Assume\n\n$$\n2 = \\sin^2 A + \\sin^2 B + \\sin^2 C = \\frac{4\\Delta^2}{b^2c^2} + \\frac{4\\Delta^2}{c^2a^2} + \\frac{4\\Delta^2}{a^2b^2},\n$$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 12542,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $\\mathbb{N}$ be the set of positive integers. Let $f: \\mathbb{N} \\to \\mathbb{N}$ be a function satisfying the following two conditions:\n\n(a) $f(m)$ and $f(n)$ are relatively prime whenever $m$ and $n$ are relatively prime.\n\n(b) $n \\leq f(n) \\leq n + 2012$ for all $n$.\n\nProve that for any natural number $n$ and any prime $p$, if $p$ divides $f(n)$ then $p$ divides $n$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $g(n)$ be the smallest prime factor of $f(n)$. (Since $f(n) \\geq n$, $f(n)$ must have a prime factor unless $n=1$ and $f(1)=1$. In this case, we define $g(1)=1$.)\n\nFirst, we show that for any prime $p$ and any $k \\geq 1$, $f(p^k)$ is a power of $g(p)$.\n\nSuppose for the sake of contradiction that $f(p^k)$ is not a power of $g(p)$ for some $p$ and $k$. Choose $M$ sufficiently large so that $p^k < M \\cdot 2013!$, and let $P$ be the set of primes less than or equal to $M \\cdot 2013! + 1$. For any $q \\in P$, we have\n\n$$\ng(q) \\leq f(q) \\leq q + 2012 \\leq M \\cdot 2013! + 2013,\n$$\n\nand furthermore the numbers $M \\cdot 2013! + i$ are composite for $2 \\leq i \\leq 2013$. It follows that $g(q) \\in P$, and so $g$ can be treated as a function from $P$ to $P$. Clearly, $g$ is injective because $f(q_1)$ and $f(q_2)$ are relatively prime for any distinct $q_1, q_2 \\in P$. Then, $g$ is bijective and in particular surjective.\n\nNow we can apply the same reasoning we used on primes in $P$ to $p^k$. We have\n\n$$\nf(p^k) \\leq p^k + 2012 \\leq M \\cdot 2013! + 2013,\n$$\n\nand so $f(p^k)$ has only prime factors in $P$. Suppose for the sake of contradiction that there is some prime $q \\in P$ not equal to $g(p)$ which divides $f(p^k)$. Then, by the surjectivity of $g$, there exists $r \\in P$ such that $g(r) = q$ (and with $r \\neq p$). We then find that $p^k$ and $r$ are relatively prime, but $q$ divides both $f(r)$ and $f(p^k)$, contradicting the first condition in the problem. It follows that $f(p^k)$ is a power of $g(p)$, as claimed.\n\nNext, we show that in fact $g(p) = p$. Indeed, suppose instead that $g(p) = q \\neq p$. Then, choose $\\ell$ sufficiently large so that $q^\\ell > 2013$, and letting $P = p^{q^\\ell - q^{\\ell-1}}$, choose $k$ sufficiently large so that $P^k > q^\\ell$. We find that\n\n$$\nP^k \\equiv 1^k \\equiv 1 \\pmod{q^\\ell}.\n$$\n\nBy the second condition of the problem, the only possible residues of $f(P^k)$ modulo $q^\\ell$ are the numbers between 1 and 2013, so in particular, $f(P^k)$ cannot be divisible by $q^\\ell$. On the other hand, $f(P^k) \\geq P^k > q^\\ell$, so $f(P^k)$ cannot be any power of $q$, contradicting the fact that since $P^k$ is a power of $p$, $f(P^k)$ must be a power of $g(p) = q$. This proves our claim that $g(p) = p$.\n\nFinally, let $p$ be any prime, and let $n$ be any positive integer relatively prime to $p$. Then, $f(p)$ and $f(n)$ are relatively prime, and since $f(p)$ is a power of $p$, it follows that $f(n)$ is not divisible by $p$. This is the contrapositive of the desired result, so we are done.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 12543,
    "subject": "Mathematics (Olympiad)",
    "question": "A number $r_X > 0$ is assigned to any point $X$ in the plane such that $2|r_X - r_Y| \\leq |XY|$ for any two points $X$ and $Y$. A cricket can jump from $X$ to $Y$ if $r_X = |XY|$. Prove that for any two points $X$ and $Y$, the cricket can move from $X$ to $Y$ by a finite number of jumps.",
    "options": [],
    "answer": "See solution",
    "solution": "Denote by $D(Z, r)$ and $S(Z, r)$ the open disc and the circle with center $Z$ and radius $r$, respectively.\n\nLet $A = A_X$ be the set of points that can be reached from $X$ by a finite number of jumps (possibly zero). We have to show that $A = \\mathbb{R}^2$.\n\nFirst, we shall prove that $D(X, 2r_X) \\subset A$. We may assume that $X = O$ and $r_0 = 1$. Set $a_n = 2^{-n}$, $b_n = 2 - a_n$, and $A_n = \\{X \\in \\mathbb{R}^2 : a_n \\leq |OX| \\leq b_n\\}$ for $n \\in \\mathbb{N}_0$. It is enough to show that $A_n \\subset A$.\n\nWe proceed by induction. For $n=0$, this follows from the given condition. Suppose that $A_k \\subset A$. It suffices to prove that $Y \\in A$ for $Y \\in [a_{k+1}, a_k) \\cup (b_k, b_{k+1}]$ and then apply rotation. Let $f(Z) = f_Y(Z) = |YZ| - r_Z$. By the condition of the problem, $1 - a_{k+1} \\leq r_{a_k} \\leq 1 + a_{k+1}$ and $a_{k+1} \\leq r_{b_k} \\leq b_{k+1}$. So, if $Y \\in [a_{k+1}, a_k)$, then $f(a_k) < a_k - 1 \\leq 0 < f(-b_k)$, and if $Y \\in (b_k, b_{k+1}]$, then $f(b_k) < 0 < 1 - a_{k+1} < f(-a_k)$. Since $A_k$ is a linearly connected set and $f$ is a continuous function, we get that $f(Z) = 0$ for some $Z \\in A_k$ and hence $Y \\in A$.\n\nAssume now that $A \\neq \\mathbb{R}^2$. Then $\\sup\\{r : D(X, r) \\subset A\\} = R < \\infty$. Let $m = \\inf\\{r_Y : Y \\in D(X, R)\\}$. Since $r$ is a continuous function, $m > 0$ and, by the above, $D(Y, 2m) \\subset A$ for $Y \\in D(X, R)$. Hence $D(X, R + 2m) \\subset A$, a contradiction.\n\n_Remark._ The statement of the problem remains true under the much weaker conditions that $r$ is a continuous function and\n\n$$\n\\limsup_{X \\to \\infty} (r_X - |OX|) < +\\infty, \\quad \\liminf_{X \\to \\infty} (r_X - |OX|) = -\\infty.\n$$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 12544,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $a$ and $n$ be positive integers such that:\n\n- $a^{2^n} - a$ is divisible by $n$,\n- $\\sum_{k=1}^{n} k^{2024} a^{2^k}$ is not divisible by $n$.\n\nProve that $n$ has a prime factor *smaller* than $2024$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $n$ be a positive integer. Call a positive integer $a$ *good* for $n$ if $n \\mid a^{2^n} - a$.\n\nDefine $d_n(a) = d$ to be the smallest positive integer such that $n \\mid a^{2^d} - a$. Then $a^{2^{d+k}} \\equiv a^{2^k} \\pmod{n}$ holds for $k \\ge 0$, so $a^{2^i} \\pmod{n}$ is periodic with period $d$. The minimality of $d$ implies that it is the smallest period. Therefore $d \\mid n$. So $d \\mid a^{2^d} - a$, which implies $a$ is good for $d$ as well.\n\nAssume $n > 1$. If $\\gcd(a, n) > 1$, then the sequence $a^{2^i} \\pmod{n}$ for $i = 1, 2, \\dots, n$ does not contain any residues relatively prime to $n$. Otherwise, the sequence doesn't contain the residue $0$. In any case, the sequence cannot contain all distinct elements, so $d_n(a) < n$.\n\n**Lemma A1.** If $a$ is good for $n$, then the numbers $a^{2^i} + i$ for $i = 1, \\dots, n$ form a complete residue system modulo $n$.\n\n*Proof.* We will prove this by induction on $n$. The base case $n = 1$ is true since there's only one number. Consider $n > 1$ for the induction step. Assume for contradiction that there exist distinct $i, j \\le n$ such that $a^{2^i} + i \\equiv a^{2^j} + j \\pmod{n}$. Let $d = d_n(a)$. Then $d \\mid n$, $d < n$, and $a$ is good for $d$ as well. So $a^{2^i} + i \\equiv a^{2^j} + j \\pmod{d}$. Since $a$ is good for $d$, reducing $i, j$ modulo $d$ does not change the values of $a^{2^i} + i \\pmod{d}$ and $a^{2^j} + j \\pmod{d}$. Therefore, by the induction hypothesis, $i \\equiv j \\pmod{d}$.\n\nBut then $a^{2^i} \\equiv a^{2^j} \\pmod{n}$ implies $i \\equiv j \\pmod{n}$. Contradiction! Hence the numbers are pairwise distinct modulo $n$, and so form a complete residue system. $\\square$\n\n**Lemma A2.** If $a$ is good for $n$, then so is $a^t$ for every $t \\in \\mathbb{N}$.\n\n*Proof.* $n \\mid a^{2^n} - a$ implies $n \\mid a^{t \\cdot 2^n} - a^t$.\n\nWe return to the main problem. Clearly we must have $n > 1$. Assume for contradiction that all the prime factors of $n$ are $\\ge 2024$. We will prove by induction on $i$ that\n\n$$\n\\sum_{k=1}^{n} k^i a^{2^k}\n$$\n\nis divisible by $n$ for any good $a$ and $0 \\le i \\le 2024$, which will give us the desired contradiction.\n\nFor the base case, $i = 0$. By Lemma A1,\n\n$$\n\\sum_{k=1}^{n} (a^{2^k} + k) \\equiv \\sum_{k=1}^{n} k \\pmod{n} \\implies \\sum_{k=1}^{n} a^{2^k} \\equiv 0 \\pmod{n}.\n$$\n\nAssume that the hypothesis holds for all $i < m$ for some $m \\le 2024$. Then by Lemma A1,\n\n$$\n\\sum_{k=1}^{n} (a^{2^k} + k)^{m+1} \\equiv \\sum_{k=1}^{n} k^{m+1} \\pmod{n} \\implies \\sum_{i=0}^{m} \\binom{m+1}{i} \\sum_{k=1}^{n} k^i a^{(m+1-i)2^k} \\equiv 0 \\pmod{n}.\n$$\n\nBy Lemma A2 and the induction hypothesis, all the sums of the form\n\n$$\n\\sum_{k=1}^{n} k^i a^{(m+1-i)2^k}\n$$\n\nfor $0 \\le i \\le m - 1$ are divisible by $n$. Hence we get\n\n$$\nn \\mid (m+1) \\sum_{k=1}^{n} k^m a^{2^k}\n$$\n\nBut $m + 1 \\le 2025$, and $2024, 2025$ are not primes. So all prime factors of $m + 1$ are $< 2024$, which implies $\\gcd(n, m + 1) = 1$. Therefore\n\n$$\nn \\mid \\sum_{k=1}^{n} k^m a^{2^k}\n$$\n\nwhich completes the induction. $\\square$",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 12545,
    "subject": "Mathematics (Olympiad)",
    "question": "Prove that among any 20 consecutive positive integers there exists a number $d$ such that for each positive integer $n$ we have the inequality\n\n$$\nn\\sqrt{d}\\{n\\sqrt{d}\\} > \\frac{5}{2}\n$$\n\nwhere $\\{x\\}$ denotes the fractional part of the real number $x$.",
    "options": [],
    "answer": "See solution",
    "solution": "Among the given numbers, there is a number of the form $20k + 15 = 5(4k + 3)$. We shall prove that $d = 5(4k + 3)$ satisfies the statement's condition. Since $d \\equiv -1 \\pmod{4}$, it follows that $d$ is not a perfect square, and thus for any $n \\in \\mathbb{N}$ there exists $a \\in \\mathbb{N}$ such that $a + 1 > n\\sqrt{d} > a$, that is, $(a+1)^2 > n^2d > a^2$. Actually, we are going to prove that $n^2d \\ge a^2 + 5$. Indeed:\n\nIt is known that each positive integer of the form $4s+3$ has a prime divisor of the same form. Let $p \\mid 4k+3$ and $p \\equiv -1 \\pmod{4}$. Because of the form of $p$, the numbers $a^2+1^2$ and $a^2+2^2$ are not divisible by $p$, and since $p \\mid n^2d$, it follows that $n^2d \\ne a^2+1, a^2+4$. On the other hand, $5 \\mid n^2d$, and since $5 \\nmid a^2+2, a^2+3$, we conclude $n^2d \\ne a^2+2, a^2+3$. Since $n^2d > a^2$ we must have $n^2d \\ge a^2+5$ as claimed. Therefore,\n\n$$\nn\\sqrt{d}\\{n\\sqrt{d}\\} = n\\sqrt{d}(n\\sqrt{d}-a) \\ge a^2+5 - a\\sqrt{a^2+5} > a^2+5 - \\frac{a^2+(a^2+5)}{2} = \\frac{5}{2},\n$$\n\nwhich was to be proved.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 12546,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $S$ be a square with sides of length $2$ and $R$ be a rhombus with sides of length $2$ and angles measuring $60^\\circ$ and $120^\\circ$. These quadrilaterals are arranged to have the same centre, and the diagonals of the rhombus are parallel to the sides of the square. Calculate the area of the region on which the figures overlap.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $S$ be the square $ABCD$, with centre $O$. Let $R$ be the rhombus $EFGH$, with $EG$ the short diagonal, and $O$ the midpoint of $EG$. Since the diagonals of $R$ bisect the angles of $R$, we have that $\\angle OFE = 30^\\circ$, so that $\\sin 30^\\circ = \\frac{1}{2}$ forces $EG$ to have length $2$. We may therefore assume that $E$ is the midpoint of $AD$ and $G$ is the midpoint of $BC$.\n\n![](images/s3s2020_p1_data_f9580e1e81.png)\n\nConsequently, $\\angle AEJ = 30^\\circ$ so that $\\tan 30^\\circ = \\frac{1}{\\sqrt{3}}$ implies that $AJ = \\frac{1}{\\sqrt{3}}$. The area of the region where $R$ and $S$ overlap is therefore, by symmetry, equal to the area of $S$ minus four times the area of triangle $AJE$, i.e.,\n\n$$\n4 - 4 \\cdot \\frac{1}{2} \\cdot 1 \\cdot \\frac{1}{\\sqrt{3}} = 4 \\left(1 - \\frac{1}{2\\sqrt{3}}\\right).\n$$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 12547,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $f(x)$ be a polynomial with complex coefficients such that its leading coefficient is rational. If for some positive integer $k$ we have $f(x)^k \\in \\mathbb{Z}[X]$, then show that $f(x) \\in \\mathbb{Z}[X]$.",
    "options": [],
    "answer": "See solution",
    "solution": "Assume $f(x) = a_n x^n + \\cdots + a_0$. Then $f(x)^k = b_k x^{kn} + \\cdots + b_1 x + b_0$. Assume inductively that $a_n, a_{n-1}, \\ldots, a_{n-s}$ are rational; we must prove $a_{n-s-1}$ is rational. Comparing the coefficient of $x^{kn-s-1}$, we have $b_{kn-s-1} = a_{n-s-1} a_n^{k-1} + S$ where $S$ is rational, so $a_{n-s-1}$ is rational. Thus, $f(x)$ has rational coefficients.\n\nNow, to show $f(x)$ has integer coefficients: If $f(x)^k = f(x)^{k-1} \\cdot f(x) \\in \\mathbb{Z}[X]$, by Gauss's lemma, there exists a rational $q$ such that $q f(x)$ and $q^{-1} f(x)^{k-1}$ have integer coefficients. Let $q = \\frac{a}{b}$ with $\\gcd(a, b) = 1$. Then $a f(x)$ and $b f(x)^{k-1}$ have integer coefficients, so $a^{k-1} f(x)^{k-1}$ has integer coefficients. Since there exist integers $s, t$ with $a^{k-1} s + b t = 1$, the polynomial $a^{k-1} s f(x)^{k-1} + b t f(x)^{k-1} = f(x)^{k-1}$ has integer coefficients. Continuing, we see $f(x)$ has integer coefficients.",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 12548,
    "subject": "Mathematics (Olympiad)",
    "question": "Suppose that\n\n$$\n2^n + 7^n = m^2\n$$\n\nfor some positive integers $n$ and $m$. Clearly, $n = 1$ is a solution. Show that there is no solution for $n > 1$.",
    "options": [],
    "answer": "See solution",
    "solution": "**Case 1:** $n$ is odd and $n > 1$\n\nLet $n = 2k + 1$, where $k$ is a positive integer. Equation (1) can be rewritten as\n\n$$\n\\begin{aligned}\n2 \\cdot 2^{2k} + 7 \\cdot 7^{2k} &= m^2 \\\\\n2(2^{2k} - 7^{2k}) &= m^2 - 9 \\cdot 7^{2k} \\\\\n&= (m - 3 \\cdot 7^k)(m + 3 \\cdot 7^k)\n\\end{aligned}\n$$\n\nThe left-hand side is even, so the right-hand side must be even too. This implies that $m$ is odd. Both factors on the right are even, so the right-hand side is a multiple of 4. However, the left-hand side is not a multiple of 4. So there are no solutions in this case.\n\n**Case 2:** $n$ is even\n\nLet $n = 2k$, where $k$ is a positive integer. Equation (1) can be rewritten as\n\n$$\n\\begin{aligned}\n2^{2k} &= m^2 - 7^{2k} \\\\\n&= (m - 7^k)(m + 7^k)\n\\end{aligned}\n$$\n\nHence there exist non-negative integers $r < s$ such that\n\n$$\nm - 7^k = 2^r\n$$\n$$\nm + 7^k = 2^s\n$$\n\nSubtracting the first equation from the second yields\n\n$$\n2^r(2^{s-r} - 1) = 2 \\cdot 7^k\n$$\n\nHence $r = 1$, so $m = 7^k + 2$. Substituting this into equation (1) yields\n\n$$\n\\begin{aligned}\n2^{2k} + 7^{2k} &= (7^k + 2)^2 \\\\\n&= 7^{2k} + 4 \\cdot 7^k + 4 \\\\\n4^k &= 4 \\cdot 7^k + 4\n\\end{aligned}\n$$\n\nHowever, this is impossible because the left-hand side is smaller than the right-hand side.\n\nHaving covered all possible cases, the proof is complete. $\\Box$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 12549,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $ABC$ be an acute triangle and let $I$ be the incentre of triangle $ABC$. The lines $AI$ and $BI$ meet the circumcircle of triangle $ABC$ again at $A_1$ and $B_1$. The segments $A_1B_1$ and $CI$ meet at $C_1$. Let $S$ be the circumcentre of triangle $IA_1C_1$. The lines $SC_1$ and $BI$ meet the segment $AC$ at $D$ and $E$. Find the angles of triangle $ABC$ if $|CD| = |DE|$ and $\\angle CBA = 2\\angle ACB$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let us write $\\angle ACB = \\gamma$. Then $\\angle CBA = 2\\gamma$ and $\\angle BAC = \\pi - 3\\gamma$. If we draw a figure where $\\angle CBA = 2\\angle ACB$, we notice that $IA_1C_1$ is a right triangle. Let us prove this.\n\nSince $ABA_1B_1$ is a cyclic quadrilateral, we have $\\angle C_1A_1I = \\angle B_1A_1A = \\angle B_1BA = \\frac{\\angle BCA}{2} = \\gamma$. The angle $\\angle A_1IC$ is equal to $\\pi - \\angle CIA$. Since $\\angle CIA = \\pi - \\angle CAI - \\angle ACI$, we get\n\n$$\n\\angle A_1IC = \\angle CAI + \\angle ACI = \\frac{\\pi - 3\\gamma}{2} + \\frac{\\gamma}{2} = \\frac{\\pi}{2} - \\gamma.\n$$\n\n![](images/Slovenija_2010_p28_data_a2c8d3398b.png)\n\nSo, $\\angle A_1IC + \\angle C_1A_1I = \\frac{\\pi}{2}$ and $\\angle IC_1A_1 = \\frac{\\pi}{2}$. Thales' theorem states that the circumcentre of triangle $IA_1C_1$ (i.e., the point $S$) is the midpoint of the segment $IA_1$. The central angle is twice the inscribed angle, so $\\angle C_1SI = 2\\angle C_1A_1I = 2\\frac{\\angle CBA}{2} = 2\\gamma$. The quadrilateral $ABA_1C_1$ is cyclic, so $\\angle CA_1A = \\angle CBA = 2\\gamma$. Hence, $CA_1$ is parallel to $DS$. But $D$ is the midpoint of the segment $CE$ and $S$ is the midpoint of the segment $IA_1$, so $DS$ is the midsegment of the quadrilateral $SIA_1C_1$. Since $DS$ is parallel to $CA_1$, it must also be parallel to $EI$. So, $\\angle EIA = \\angle DSA = 2\\gamma$.\n\nFor the inner angles of triangle $ABI$ we have $\\angle BAI = \\angle BAC = \\frac{\\pi-3\\gamma}{2}$, $\\angle IBA = \\angle CBA = \\frac{\\pi-3\\gamma}{2}$. Since the sum of the inner angles in an arbitrary triangle is equal to $\\pi$, we get $\\frac{\\pi-3\\gamma}{2} + \\gamma + \\pi - 2\\gamma = \\pi$ or $\\gamma = \\frac{\\pi}{5}$.\n\nThe angles of triangle $ABC$ measure $\\angle BAC = \\angle CBA = \\frac{2\\pi}{5}$ and $\\angle ACB = \\frac{\\pi}{5}$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 12550,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $G$ be a connected graph with $n \\ge 2$ vertices. Prove that we can color its vertices in two colors so that if $x$ and $y$ are the numbers of “multicolored” and “single colored” edges, respectively, then $x - y \\ge \\lfloor \\frac{n}{2} \\rfloor$.\n\nFurthermore, for an arbitrary connected graph $G$ with $n \\ge 3$ vertices and $m$ edges, show that deleting at most $\\frac{1}{2} \\cdot \\left( m - \\left\\lfloor \\frac{n}{2} \\right\\rfloor \\right)$ edges is sufficient to make the graph bipartite, and that this bound is tight.",
    "options": [],
    "answer": "See solution",
    "solution": "We proceed by induction on $n$.\n\nFor $n = 2, 3$, the statement is easily verified.\n\nAssume $n \\ge 4$ and $G$ is a connected graph with $n$ vertices. Let $u$ and $v$ be two vertices as in the lemma. Remove $u$ and $v$, and color $G \\setminus \\{u, v\\}$ by the induction hypothesis. We can color $u$ and $v$ so that the difference $x - y$ increases by at least $1$. This is clear if $u$ and $v$ are leaves; otherwise, by considering the parity of their neighbors in $G \\setminus \\{u, v\\}$, such a coloring is always possible. Thus, the statement holds by induction.\n\nNow, for any connected graph $G$ with $n \\ge 3$ vertices and $m$ edges, color as above. We have $x - y \\ge \\lfloor \\frac{n}{2} \\rfloor$ and $x + y = m$, so\n\n$$\ny \\le \\frac{1}{2} \\left( m - \\left\\lfloor \\frac{n}{2} \\right\\rfloor \\right)\n$$\n\nDeleting $y$ edges suffices to make the graph bipartite, so $k \\le \\frac{1}{2}$.\n\nTo show $k \\ge \\frac{1}{2}$, consider the complete graph on $n$ vertices. Making it bipartite requires deleting all edges within each part. For $n = 2n_1$, the minimum is $2 \\binom{n_1}{2} = n_1^2 - n_1$ edges. For $n = 2n_1 + 1$, the minimum is $\\binom{n_1+1}{2} + \\binom{n_1}{2} = n_1^2$ edges. In both cases, this matches the upper bound, so $k = \\frac{1}{2}$ is tight.",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 12551,
    "subject": "Mathematics (Olympiad)",
    "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$$\n(n + 1) \\binom{2n}{n}.\n$$",
    "options": [],
    "answer": "See solution",
    "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$$\n\\begin{aligned}\n2(2n-1)\\binom{2n-2}{n-1} + \\binom{2n}{n} &= 2n\\binom{2n-1}{n} + \\binom{2n}{n} \\\\\n&= 2n\\binom{2n-1}{n-1} + \\binom{2n}{n} \\\\\n&= n\\binom{2n}{n} + \\binom{2n}{n} = (n+1)\\binom{2n}{n},\n\\end{aligned}\n$$\n\nwhere we used the identity\n\n$$\nk \\binom{m}{k} = m \\binom{m-1}{k-1}.\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 12552,
    "subject": "Mathematics (Olympiad)",
    "question": "Suppose there is a game where two players take turns drawing either a vertical or a horizontal line on a sheet of paper. After $n$ lines have been drawn, the sheet is divided into rectangles. The winner is determined by the parity (odd or even) of the number of rectangles formed. Who wins the game, depending on whether $n$ is odd or even, and who starts?",
    "options": [],
    "answer": "See solution",
    "solution": "If $n$ is odd, the winner is always Sara, regardless of who starts. If $n$ is even, the winner is the player who does not start.\n\nSuppose at the end there are $p$ vertical and $r$ horizontal lines, with $p + r = n$. The sheet is divided into $(p+1)(r+1)$ rectangles.\n\n- If $n$ is odd, one of $p$ or $r$ is odd, so $(p+1)(r+1)$ is even. Thus, Sara wins no matter who starts.\n- If $n$ is even, after $n-1$ moves there are $s$ vertical and $t$ horizontal lines, $s + t = n-1$ (odd), so $s$ and $t$ have different parities. The last move can make the number of rectangles odd or even, so the last player (the one who does not start) can always win.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 12553,
    "subject": "Mathematics (Olympiad)",
    "question": "Обозначим числа на окружности через $a_1, \\dots, a_{2009}$, и положим $a_{n+2009} = a_n = a_{n-2009}$. Пусть $N = 100400$.\n\nПусть разрешено выбрать любую пару соседних чисел $(a_i, a_{i+1})$ и увеличить оба на 1. Какое наименьшее $k$ гарантирует, что, начиная с любых неотрицательных $a_1, \\dots, a_{2009}$, если каждую пару $(a_i, a_{i+1})$ можно увеличить не более $k$ раз, то можно сделать все числа равными?",
    "options": [],
    "answer": "See solution",
    "solution": "1. Пусть $a_2 = a_4 = \\dots = a_{2008} = 100$ и $a_1 = a_3 = \\dots = a_{2009} = 0$. Пусть мы сумели сделать все числа равными при каком-то значении $k$. Рассмотрим сумму $S = (a_2 - a_3) + (a_4 - a_5) + \\dots + (a_{2008} - a_{2009})$. Эта сумма увеличивается на 1 при прибавлении единицы к паре $(a_1, a_2)$, уменьшается на 1 при прибавлении к паре $(a_{2009}, a_1)$ и не изменяется при всех остальных операциях. Поскольку исходное значение $S$ равно $S_0 = 100 \\cdot 1004 = N$, а конечное должно быть нулем, то пара $(a_{2009}, a_1)$ увеличивалась хотя бы $N$ раз. Это значит, что $k \\ge N$.\n\n2. Осталось показать, что при $k = N$ требуемое всегда возможно. Рассмотрим произвольный набор чисел $a_i$. Увеличим каждую пару $(a_i, a_{i+1})$ ровно $s_i = a_{i+2} + a_{i+4} + \\dots + a_{i+2008}$ раз. Тогда число $a_i$ превратится в\n\n$$a_i + s_{i-1} + s_i = a_i + (a_{i+1} + a_{i+3} + \\dots + a_{i+2007}) + (a_{i+2} + a_{i+4} + \\dots + a_{i+2008}) = a_1 + \\dots + a_{2009},$$\n\nто есть все числа станут равными. С другой стороны, $s_i \\le 1004 \\cdot 100 = N$, что и требовалось.\n\n*Ответ:* $k = 100400$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 12554,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $a$ and $n$ be positive integers such that $a^{2n} - a$ is divisible by $n$.\n\nProve that $n$ has a prime factor *smaller* than $2024$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $n$ be a positive integer. Call a positive integer $a$ *good* for $n$ if $n \\mid a^{2n} - a$.\n\nDefine $d_n(a) = d$ to be the smallest positive integer such that $n \\mid a^{2d} - a$. Then $a^{2d+k} \\equiv a^{2k} \\pmod{n}$ holds for $k \\ge 0$, so $a^{2i} \\pmod{n}$ is periodic with period $d$. The minimality of $d$ implies that it is in fact the smallest period. Therefore $d \\mid n$. So $d \\mid n \\mid a^{2d} - a$, which implies $a$ is *good* for $d$ as well.\n\nAssume $n > 1$. If $\\gcd(a, n) > 1$, then the sequence $a^{2i} \\pmod{n}$ for $i = 1, 2, \\dots, n$ does not contain any residues relatively prime to $n$. Otherwise, the sequence doesn't contain the residue $0$. In any case, the sequence cannot contain all distinct elements, so $d_n(a) < n$.\n\n**Lemma A1.** If $a$ is *good* for $n$, then the numbers $a^{2i} + i$ for $i = 1, \\dots, n$ form a complete residue system modulo $n$.\n\n*Proof.* We will prove this by induction on $n$. Base case $n = 1$ is true since there's only one number. Consider $n > 1$ for the induction step. Assume for the sake of contradiction that there exist distinct $i, j \\le n$ such that $a^{2i} + i \\equiv a^{2j} + j \\pmod{n}$. Let $d = d_n(a)$. Then $d \\mid n$, $d < n$ and $a$ is *good* for $d$ as well. So $a^{2i} + i \\equiv a^{2j} + j \\pmod{d}$. Note that, since $a$ is *good* for $d$, reducing $i, j$ modulo $d$ does not change the values of $a^{2i} + i \\pmod{d}$ and $a^{2j} + j \\pmod{d}$. Therefore, by induction hypothesis, $i \\equiv j \\pmod{d}$.\n\nBut then $a^{2i} \\equiv a^{2j} \\pmod{n}$ implies $i \\equiv j \\pmod{n}$. Contradiction! Hence the numbers are pairwise distinct modulo $n$, and so form a complete residue system. $\\square$\n\n**Lemma A2.** If $a$ is *good* for $n$, then so is $a^t$ for every $t \\in \\mathbb{N}$.\n\n*Proof.* $n \\mid a^{2n} - a$ implies $n \\mid a^{t \\cdot 2n} - a^t$.\n\nWe return to the main problem. Clearly we must have $n > 1$. Assume for the sake of contradiction that all the prime factors of $n$ are $\\ge 2024$. We will prove using induction on $i$ that\n\n$$\n\\sum_{k=1}^{n} k^i a^{2k}\n$$\n\nis divisible by $n$ for any *good* $a$ and $0 \\le i \\le 2024$, which will give us the desired contradiction. For the base case, $i = 0$. By Lemma A1,\n\n$$\n\\sum_{k=1}^{n} (a^{2k} + k) \\equiv \\sum_{k=1}^{n} k \\pmod{n} \\implies \\sum_{k=1}^{n} a^{2k} \\equiv 0 \\pmod{n}.\n$$\n\nAssume that the hypothesis holds for all $i < m$ for some $m \\le 2024$. Then by Lemma A1,\n\n$$\n\\sum_{k=1}^{n} (a^{2k} + k)^{m+1} \\equiv \\sum_{k=1}^{n} k^{m+1} \\pmod{n} \\implies \\sum_{i=0}^{m} \\binom{m+1}{i} \\sum_{k=1}^{n} k^i a^{(m+1-i)2k} \\equiv 0 \\pmod{n}.\n$$\n\nBy Lemma A2 and the induction hypothesis, all the sums of the form\n\n$$\n\\sum_{k=1}^{n} k^i a^{(m+1-i)2k}\n$$\nfor $0 \\le i \\le m-1$ are divisible by $n$. Hence we get\n\n$$\nn \\mid (m+1) \\sum_{k=1}^{n} k^m a^{2k}\n$$\n\nBut $m+1 \\le 2025$, and $2024, 2025$ are not primes. So all prime factors of $m+1$ are $< 2024$, which implies $\\gcd(n, m+1) = 1$. Therefore\n\n$$\nn \\mid \\sum_{k=1}^{n} k^m a^{2k}\n$$\n\nwhich completes the induction. $\\square$",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 12555,
    "subject": "Mathematics (Olympiad)",
    "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 $\\frac{3}{16}a^2$.",
    "options": [],
    "answer": "See 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$$\n\\begin{aligned}\n|DS_1| &= r_1 \\sqrt{2}, & |BS_1| &= (a - r_1) \\sqrt{2}, & |PS_1| &= \\left( \\frac{a}{2} - r_1 \\right) \\sqrt{2}, \\\\\n|CS_2| &= r_2 \\sqrt{2}, & |AS_2| &= (a - r_2) \\sqrt{2}, & |PS_2| &= \\left( \\frac{a}{2} - r_2 \\right) \\sqrt{2}.\n\\end{aligned}\n$$\n\nTherefore, the area of the triangle $AS_1S_2$ is\n\n$$\nS_{AS_1S_2} = \\frac{1}{2} |AS_2| \\cdot |PS_1| = (a - r_2) \\left( \\frac{a}{2} - r_1 \\right),\n$$\n\nwhile the area of the triangle $BS_1S_2$ is\n\n$$\nS_{BS_1S_2} = \\frac{1}{2} |BS_1| \\cdot |PS_2| = (a - r_1) \\left( \\frac{a}{2} - r_2 \\right).\n$$\n\nThe sum of these areas is\n\n$$\nS = (a - r_2) \\left( \\frac{a}{2} - r_1 \\right) + (a - r_1) \\left( \\frac{a}{2} - r_2 \\right) = a^2 - \\frac{3}{2} a (r_1 + r_2) + 2 r_1 r_2.\n$$\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 $HS_2$ (see figure below).\n\n![](images/Czech_and_Slovak_booklet_2013_p2_data_2792213c9c.png)\n\nBy the Pythagorean theorem for the triangle $S_1MS_2$, we have\n\n$$\n(a - r_1 - r_2)^2 + (r_1 - r_2)^2 = (r_1 + r_2)^2.\n$$\n\nHence we obtain\n\n$$\n\\begin{aligned}\n(a - r_1 - r_2)^2 &= 4 r_1 r_2, \\\\\na - r_1 - r_2 &= 2 \\sqrt{r_1 r_2}, \\\\\na &= r_1 + r_2 + 2 \\sqrt{r_1 r_2} = (\\sqrt{r_1} + \\sqrt{r_2})^2 \\ge 4 \\sqrt{r_1 r_2},\n\\end{aligned}\n$$\n\ni.e.\n\n$$\nr_1 r_2 \\le \\frac{a^2}{16}.\n$$\n\nThe length of the segment $DC$ cannot be greater than the length of the polygonal chain $KS_1S_2L$, so\n\n$$\n2 r_1 + 2 r_2 \\ge a.\n$$\n\n(This follows from the equality $a = r_1 + r_2 + 2 \\sqrt{r_1 r_2}$ as well since $2 \\sqrt{r_1 r_2} \\le r_1 + r_2$, by the AM-GM inequality.) Therefore,\n\n$$\nS = a^2 - \\frac{3}{2} a (r_1 + r_2) + 2 r_1 r_2 \\le a^2 - \\frac{3}{4} a^2 + \\frac{1}{8} a^2 = \\frac{3}{8} a^2.\n$$\n\nThis means that at least one of the areas $S_{AS_1S_2}$, $S_{BS_1S_2}$ is at most $\\frac{3}{16} a^2$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 12556,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $f : [0, 1] \\to \\mathbb{R}$ be a differentiable function with\n$$\nf(0) = f(1), \\quad \\int_{0}^{1} f(x) \\, dx = 0, \\quad \\text{and} \\quad f'(x) \\neq 1 \\text{ for every } x \\in [0, 1].\n$$\n\na) Prove that the function $g : [0, 1] \\to \\mathbb{R}$ given by $g(x) = f(x) - x$ is decreasing.\n\nb) Prove that for any integer $n \\ge 1$ the following inequality holds:\n$$\n\\left| \\sum_{k=0}^{n-1} f \\left( \\frac{k}{n} \\right) \\right| < \\frac{1}{2}.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "a) Since $f'$ has the intermediate value property, it follows that $f'(x) < 1$ for every $x \\in [0, 1]$, or $f'(x) > 1$ for every $x \\in [0, 1]$. But in the second case, $f$ would be strictly increasing, which contradicts $f(0) = f(1)$. So $f'(x) < 1$ for every $x \\in [0, 1]$, whence $g(x) = f(x) - x$ has $g'(x) = f'(x) - 1 < 0$ for all $x \\in [0, 1]$, that is, $g$ is strictly decreasing.\n\nb) Notice that\n$$\ns_{\\Delta}(g) < \\int_{0}^{1} g(x) \\, dx < S_{\\Delta}(g),\n$$\nwhere $s_{\\Delta}(g)$ and $S_{\\Delta}(g)$ are the lower and upper Darboux sums of $g$ for the partition $\\Delta = \\{0, \\frac{1}{n}, \\dots, \\frac{n-1}{n}, 1\\}$. Then\n$$\n\\frac{1}{n} \\sum_{k=0}^{n-1} \\left( f\\left(\\frac{k+1}{n}\\right) - \\frac{k+1}{n} \\right) < \\int_{0}^{1} (f(x) - x) \\, dx < \\frac{1}{n} \\sum_{k=0}^{n-1} \\left( f\\left(\\frac{k}{n}\\right) - \\frac{k}{n} \\right).\n$$\nSince $f(0) = f(1)$ and $\\int_{0}^{1} f(x) \\, dx = 0$, this yields\n$$\n\\sum_{k=0}^{n-1} f\\left(\\frac{k}{n}\\right) - \\frac{n+1}{2} < -\\frac{n}{2} < \\sum_{k=0}^{n-1} f\\left(\\frac{k}{n}\\right) - \\frac{n-1}{2},\n$$\nwhence the required inequality.\n\n**Alternative Solution (b).** Let $G : [0, 1] \\to \\mathbb{R}$, $G(x) = F(x) - \\frac{1}{2}x^2$, where $F$ is a primitive of $f$. The function $G'(x) = f(x) - x$ is strictly decreasing, because $G''(x) = f'(x) - 1 < 0$ for every $x \\in [0, 1]$.\n\nNow, by the mean value theorem applied to $G$ on the intervals $[\\frac{k}{n}, \\frac{k+1}{n}]$, $k = 0, \\dots, n-1$, we have\n$$\nf\\left(\\frac{k+1}{n}\\right) - \\frac{k+1}{n} < n\\left(F\\left(\\frac{k+1}{n}\\right) - F\\left(\\frac{k}{n}\\right) - \\frac{2k+1}{2n^2}\\right) < f\\left(\\frac{k}{n}\\right) - \\frac{k}{n}.\n$$\nThese inequalities add up to\n$$\n\\sum_{k=0}^{n-1} f\\left(\\frac{k+1}{n}\\right) - \\frac{n+1}{2} < n\\left(F(1) - F(0) - \\frac{1}{2}\\right) < \\sum_{k=0}^{n-1} f\\left(\\frac{k}{n}\\right) - \\frac{n-1}{2}.\n$$\nSince $F(1) - F(0) = \\int_{0}^{1} f(x) \\, dx = 0$ and $f(0) = f(1)$, the required inequality follows immediately.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 12557,
    "subject": "Mathematics (Olympiad)",
    "question": "Prove that there exist infinitely many natural numbers $n$ such that all prime factors of $n^2 + 1$ are less than $n$.",
    "options": [],
    "answer": "See solution",
    "solution": "This holds for all numbers $n = 2a^2$ where $a > 1$ and $a \\equiv 1 \\pmod{5}$.\n\nIf $n = 2a^2$, then\n$$\nn^2 + 1 = 4a^4 + 1 = (2a^2 + 2a + 1)(2a^2 - 2a + 1).\n$$\nSince $2a^2 - 2a + 1 < 2a^2$, it remains to ensure that all prime factors of $2a^2 + 2a + 1$ are less than $2a^2$. If $a \\equiv 1 \\pmod{5}$, then $2a^2 + 2a + 1 \\equiv 0 \\pmod{5}$, so $2a^2 + 2a + 1 = 5 \\cdot \\frac{2a^2 + 2a + 1}{5}$, and both these factors are less than $2a^2$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 12558,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $ABCDEF$ be an inscribed hexagon such that $AB = BC$, $CD = DE$, and $EF = FA$. Show that the lines $AD$, $BE$, and $CF$ intersect at exactly one point.",
    "options": [],
    "answer": "See solution",
    "solution": "Equality $AB = BC$ implies that $\\angle AEB = \\angle BEC$, hence $EB$ is a bisector of $\\triangle ACE$. Similarly, one can conclude the same about $AD$ and $CF$. Therefore, they intersect at one point.\n\n![](images/Ukraine_2020_booklet_p6_data_3b25ff8160.png)",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 12559,
    "subject": "Mathematics (Olympiad)",
    "question": "a) Does there exist an increasing bijection between open intervals of rational numbers $(a, b)$ and $(c, d)$? What about between $A = \\mathbb{Q} \\cap (0, \\sqrt{2})$ and $B = \\mathbb{Q} \\cap (0, \\sqrt{3})$?\n\nb) Does there exist an increasing bijection between two countable dense subsets $A, B \\subset \\mathbb{R}$?\n\nc) Does there exist an increasing bijection $f: \\mathbb{R} \\to B$ where $B = \\mathbb{R} \\times \\mathbb{R}$ with the lexicographic order?\n\nd) Does there exist an increasing bijection between the sets\n$$\nA = \\left\\{ (2^{-n}, 1/2) \\mid n \\in \\mathbb{N} \\right\\}, \\quad B = A \\cup \\{(0, 1/2)\\}\n$$\nwith the lexicographic order?\n\ne) Let $Y = \\{2^{-n} \\mid n \\in \\mathbb{N}\\} \\cup \\{0\\}$ and define\n$$\nA = Y \\cup (Y + 1) \\cup (Y + 2) \\cup \\dots, \\quad B = A \\cup \\{-1\\}\n$$\nDoes there exist an increasing bijection from $A$ to $B$?",
    "options": [],
    "answer": "See solution",
    "solution": "a) Yes, such a function exists. For rational numbers $a, b, c, d$, the function\n$$\nf(x) = \\frac{d-c}{b-a}(x-a) + c\n$$\ndefines an increasing bijection between $(a, b)$ and $(c, d)$. For $A = \\mathbb{Q} \\cap (0, \\sqrt{2})$ and $B = \\mathbb{Q} \\cap (0, \\sqrt{3})$, consider strictly increasing sequences $\\{p_n\\}_{n \\ge 0}$ and $\\{q_n\\}_{n \\ge 0}$ with $p_0 = q_0 = 0$, $p_n \\to \\sqrt{2}$, $q_n \\to \\sqrt{3}$. Define\n$$\nf(x) = \\begin{cases} x & x \\le 0 \\\\ \\frac{q_{i+1}-q_i}{p_{i+1}-p_i}(x-p_i) + q_i & x \\in [p_i, p_{i+1}] \\end{cases}\n$$\nThis gives an increasing bijection from $A$ to $B$.\n\nb) Yes, such a function exists. Since $A$ and $B$ are countable and dense in $\\mathbb{R}$, enumerate $A = \\{a_1, a_2, \\dots\\}$ and $B = \\{b_1, b_2, \\dots\\}$. Define $f(a_1) = b_1$, and inductively assign $f(a_k)$ to an unused $b$ in the appropriate interval to maintain strict increase, and similarly for the inverse. Repeating this process ensures $f$ is a strictly increasing bijection from $A$ to $B$.\n\nc) No, such a function does not exist. Suppose $f: \\mathbb{R} \\to B$ is an increasing bijection. Consider $C = \\{0\\} \\times \\mathbb{R} \\subseteq B$ and $D = f^{-1}(C) \\subseteq \\mathbb{R}$. $D$ must be an open interval, but $C$ has no maximum or minimum, so $D$ cannot have one either. This leads to a contradiction when considering the preimage of endpoints, so such $f$ cannot exist.\n\nd) No, such a function does not exist. In the ordered set $A$, the set of end elements is\n$$\n\\mathbf{End}(A) = \\left\\{ (\\frac{1}{2}, \\frac{1}{2}), (\\frac{1}{4}, \\frac{1}{2}), (\\frac{1}{8}, \\frac{1}{2}), \\dots \\right\\}\n$$\nwhile for $B$ it is\n$$\n\\mathbf{End}(B) = \\left\\{ (\\frac{1}{2}, \\frac{1}{2}), (\\frac{1}{4}, \\frac{1}{2}), (\\frac{1}{8}, \\frac{1}{2}), \\dots, (0, \\frac{1}{2}) \\right\\}\n$$\n$(0, \\frac{1}{2})$ is the minimum of $\\mathbf{End}(B)$, but $\\mathbf{End}(A)$ has no minimum. Thus, an increasing bijection cannot exist.\n\ne) Define $Y = \\{2^{-n} \\mid n \\in \\mathbb{N}\\} \\cup \\{0\\}$, $A = Y \\cup (Y + 1) \\cup (Y + 2) \\cup \\dots$, $B = A \\cup \\{-1\\}$. Define\n$$\nf(x) = \\begin{cases} -1 & x < 1 \\\\ x - 1 & x \\ge 1 \\end{cases}, \\quad g(x) = \\begin{cases} 0 & x = -1 \\\\ x & x \\ne -1 \\end{cases}\n$$\nBoth $f: A \\to B$ and $g: B \\to A$ are increasing and surjective. However, there is no increasing bijection from $A$ to $B$ because $0$ is the minimum of $A$ and would have to map to $-1$, but $0$ has no successor in $A$, while $-1$ does in $B$, leading to a contradiction.",
    "topic": "Discrete Mathematics",
    "subtopic": "Other"
  },
  {
    "id": 12560,
    "subject": "Mathematics (Olympiad)",
    "question": "Two circles $\\omega_1$ and $\\omega_2$ intersect at points $A$ and $B$. A line through $B$ intersects $\\omega_1$ at point $C$ and $\\omega_2$ at point $D$. The line $AD$ intersects $\\omega_1$ at point $E$ (different from $A$), and the line $AC$ intersects $\\omega_2$ at point $F$ (different from $A$). If $O$ is the circumcenter of triangle $AEF$, prove that $OB \\perp CD$.\n\n![](images/MNG_ABooklet_2017_p18_data_89861cb308.png)",
    "options": [],
    "answer": "See solution",
    "solution": "We draw the circumcircle of $AEF$, and denote $\\angle EAC = \\alpha$. So $\\angle EAF = 180^\\circ - \\alpha$, and from here we get $\\angle EOF = 2\\alpha$. On the other hand, $\\alpha = \\angle EBC = \\angle EAC = \\angle FAD = \\angle FBD$, so $\\angle EBF = 180^\\circ - 2\\alpha$. From here we have $\\angle EOF + \\angle EBF = 180^\\circ$, hence $EOFB$ is cyclic. Therefore, from $EO = FO$ we have $\\angle EBO = \\angle OBF$. From here we have $\\angle OBC = \\angle CBE + \\angle EBO = \\angle OBF + \\angle FBD = \\angle OBD$. We know $\\angle OBC + \\angle OBD = 180^\\circ$, so $OB \\perp CD$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 12561,
    "subject": "Mathematics (Olympiad)",
    "question": "The sequence of real numbers $\\{x_n\\}_{n \\ge 1}$ is given by\n$$\nx_{n+1} = \\left|x_n - \\frac{1}{n}\\right|, \\quad \\text{for } n \\ge 1.\n$$\nProve that it has a finite limit, and calculate it.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $N$ be an arbitrary positive integer. There exists a least $k \\in \\mathbb{N}$ such that $x_N < \\frac{1}{N} + \\frac{1}{N+1} + \\dots + \\frac{1}{N+k}$, since the harmonic series is divergent. Then\n$$\nx_{N+k} = x_N - \\left(\\frac{1}{N} + \\dots + \\frac{1}{N+k-1}\\right) < \\frac{1}{N+k} \\le \\frac{1}{N}.\n$$\nAlso, $x_n < \\frac{1}{N}$ for $n \\ge N+k$. The proof is by simple induction, since $x_{n+1} = \\left|x_n - \\frac{1}{n}\\right|$, and if $x_n > \\frac{1}{n}$, then $x_{n+1} = x_n - \\frac{1}{n} < x_n < \\frac{1}{N}$, while if $x_n \\le \\frac{1}{n}$, then $x_{n+1} = \\frac{1}{n} - x_n \\le \\frac{1}{n} < \\frac{1}{N}$.\n\nTherefore, for any $N$ there exists $N_1$ such that $0 \\le x_n < \\frac{1}{N}$ for $n \\ge N_1$, which means the sequence converges to $0$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 12562,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $a$, $b$, $c$, $d$ be positive real numbers such that $a + b + c + d = 4$. Prove that\n\n$$\na\\sqrt{a+8} + b\\sqrt{b+8} + c\\sqrt{c+8} + d\\sqrt{d+8} \\geq 12.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "For any positive $x$, applying the AM-GM inequality twice, we find that\n\n$$\nx\\sqrt{x+8} \\geq x\\sqrt{9\\sqrt[9]{x}} = 3x\\sqrt[18]{x} = \\frac{1}{6}\\left(18\\sqrt[18]{x^{19}} + 1 - 1\\right) \\geq \\frac{1}{6}(19x - 1).\n$$\n\nThus,\n\n$$\n\\sum a\\sqrt{a+8} \\geq \\sum \\frac{19a-1}{6} = \\frac{19 \\cdot 4 - 4}{6} = 12.\n$$\n\nEquality occurs when $a = b = c = d = 1$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 12563,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $s = x^2 + y^2$ and $t = x + y$. Show that\n\n$$\ns^3 \\ge 8t(3s - t^2)(t^2 - 2t - s)\n$$\n\nfor $2s \\ge t^2$ and $t \\ge 0$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $s = rt$. This transforms the inequality to\n\n$$\nr^3 \\ge 8(3r - t)(t - 2 - r)\n$$\n\nfor $2r \\ge t \\ge 0$.\n\nSince $r^3 - 8(3r - t)(t - 2 - r) = 8(t - (2r + 1))^2 + r^3 - 8r^2 + 16r - 8 \\ge r^3 - 8r^2 + 16r = r(r - 4)^2 \\ge 0$ for $2r \\ge t \\ge 0$, the inequality holds.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 12564,
    "subject": "Mathematics (Olympiad)",
    "question": "Intercept the extension of the line $B_1A$ at $P_1$. Draw the arc $\\overarc{P_1Q_1}$ with center $B_1$ and radius $B_1P_1$, intercepting the extension of the line $B_1C_0$ at $Q_1$. Draw the arc $\\overarc{Q_1P'_0}$ with center $C_0$ and radius $C_0Q_1$, intercepting the extension of the line $AB_0$ at $P'_0$.\n\nProve that:\n\n1. $P'_0$ and $P_0$ are coincident, and the arcs $\\overarc{P_0Q_0}$ and $\\overarc{P_0Q_1}$ are tangent to each other at $P_0$.\n2. The points $P_0$, $Q_0$, $Q_1$, $P_1$ are concyclic.",
    "options": [],
    "answer": "See solution",
    "solution": "(1) From the properties of an ellipse, we know\n\n$$\nB_1C_0 + C_0B_0 = B_1C_1 + C_1B_0.\n$$\n\nAlso, it is obvious that\n\n$$\nB_0P_0 = B_0Q_0, \\quad C_1B_0 + B_0Q_0 = C_1P_1,\n$$\n\n$$\nB_1C_1 + C_1P_1 = B_1C_0 + C_0Q_1, \\quad C_0Q_1 = C_0B_0 + B_0P'_0.\n$$\n\nAdding these equations, we get $B_0P_0 = B_0P'_0$.\n\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 $\\overarc{P_0Q_0}$) lie 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 $\\overarc{Q_1P_0}$ 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, 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, draw a common tangent line $R_1S_1$ through $Q_1$, and suppose it intercepts $P_0T$ and $P_1T$ at points $R_1$ 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$$\n\\begin{aligned}\n\\angle P_0Q_1P_1 &= \\pi - \\angle P_0Q_1R_1 - \\angle P_1Q_1S_1 \\\\\n&= \\pi - (\\angle P_1P_0T - \\angle Q_1P_0P_1) \\\\\n&\\quad - (\\angle P_0P_1T - \\angle Q_1P_1P_0).\n\\end{aligned}\n$$\n\nSince\n\n$$\n\\pi - \\angle P_0Q_1P_1 = \\angle Q_1P_0P_1 + \\angle Q_1P_1P_0,\n$$\n\nwe obtain\n\n$$\n\\angle P_0Q_1P_1 = \\pi - \\frac{1}{2}(\\angle P_1P_0T + \\angle P_0P_1T).\n$$\n\nIn the same way, we can prove that\n\n$$\n\\angle P_0Q_0P_1 = \\pi - \\frac{1}{2}(\\angle P_1P_0T + \\angle P_0P_1T).\n$$\n\nIt implies that points $P_0$, $Q_0$, $Q_1$, $P_1$ are concyclic.\n\n![](images/Mathematical_Olympiad_in_China_2007-2008-Problems_and_Solutions_p56_data_3d16556b89.png)",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 12565,
    "subject": "Mathematics (Olympiad)",
    "question": "Consider the functions $f, g : \\mathbb{R} \\to \\mathbb{R}$, such that $g(x) = 2f(x) + f(x^2)$ for all $x \\in \\mathbb{R}$.\n\na) Prove that if $f$ is locally bounded at the origin and $g$ is continuous at the origin, then $f$ is continuous at the origin.\n\nb) Give an example of a function $f$, discontinuous at the origin, such that $g$ is continuous at the origin.",
    "options": [],
    "answer": "See solution",
    "solution": "a) Let $\\varepsilon > 0$ be arbitrary. By hypothesis, there exist $\\delta_1, M > 0$ such that $|f(x)| < M$ for all $x \\in (-\\delta_1, \\delta_1)$. Since $g$ is continuous at $0$, there exists $\\delta_2 > 0$ (depending on $\\varepsilon$) such that $|g(x) - g(0)| < \\frac{\\varepsilon}{2}$ for any $x \\in (-\\delta_2, \\delta_2)$. Let $\\delta = \\min\\{\\delta_1, \\delta_2, 1\\}$. For $x \\in (-\\delta, \\delta)$, $|x| < \\delta \\le 1$ implies $|x|^2 < \\delta$.\n\nFor $x \\in (-\\delta, \\delta)$:\n\n$$\n\\begin{aligned}\n|f(x) - f(0)| &= \\frac{|2f(x) - 2f(0)|}{2} \\\\\n&\\le \\frac{|2f(x) + f(x^2) - 3f(0)| + |f(x^2) - f(0)|}{2} \\\\\n&= \\frac{|g(x) - g(0)|}{2} + \\frac{|f(|x|^2) - f(0)|}{2} \\\\\n&< \\frac{\\varepsilon}{4} + \\frac{|f(|x|^2) - f(0)|}{2}.\n\\end{aligned}\n$$\n\nInductively,\n\n$$\n|f(x) - f(0)| < \\varepsilon \\left( \\frac{1}{2^2} + \\frac{1}{2^3} + \\dots + \\frac{1}{2^{n+1}} \\right) + \\frac{|f(|x|^{2^n}) - f(0)|}{2^n}, \\quad n \\in \\mathbb{N}^*\n$$\n\nLet $p \\in \\mathbb{N}^*$ such that $2^p > \\frac{4M}{\\varepsilon}$. Then $|x|^{2p} < \\delta \\le \\delta_1$, and\n\n$$\n\\begin{aligned}\n|f(x) - f(0)| < \\frac{\\varepsilon \\left(1 - \\frac{1}{2^p}\\right)}{2} + \\frac{|f(|x|^{2p})| + |f(0)|}{2^p} \\le \\frac{\\varepsilon \\left(1 - \\frac{1}{2^p}\\right)}{2} + \\frac{2M}{2^p} < \\frac{\\varepsilon}{2} + \\frac{\\varepsilon}{2} = \\varepsilon.\n\\end{aligned}\n$$\n\nb) For $a \\in (0, 1)$ and $A = \\{\\pm a^{2^n} : n \\in \\mathbb{Z}\\}$, define\n\n$$\nf: \\mathbb{R} \\to \\mathbb{R}, \\quad f(x) = \\begin{cases} (-1)^n 2^n, & |x| = a^{2^n},\\ n \\in \\mathbb{Z} \\\\ 0, & x \\in \\mathbb{R} \\setminus A \\end{cases}\n$$\n\nBecause $\\lim_{k \\to \\infty} a^{2^{2k}} = 0$ and $\\lim_{k \\to \\infty} f(a^{2^{2k}}) = \\infty$, $f$ is discontinuous at the origin.\n\nFor $x \\in \\mathbb{R} \\setminus A$, $x^2 \\in \\mathbb{R} \\setminus A$, so $g(x) = 0$. For $x \\in A$, there is $n \\in \\mathbb{Z}$ such that $|x| = a^{2^n}$, so\n\n$$\ng(x) = 2f(a^{2^n}) + f(a^{2^{n+1}}) = 2^{n+1}[(-1)^n + (-1)^{n+1}] = 0.\n$$\n\nTherefore, $g$ is the zero function.\n\nAlternative solution for a):\n\nLet\n$$\nL := \\limsup_{x \\to 0} f(x) = \\lim_{\\varepsilon \\searrow 0} \\sup\\{f(x) : x \\in (-\\varepsilon, \\varepsilon) \\setminus \\{0\\}\\}\n$$\nand\n$$\n\\ell := \\liminf_{x \\to 0} f(x) = \\lim_{\\varepsilon \\searrow 0} \\inf\\{f(x) : x \\in (-\\varepsilon, \\varepsilon) \\setminus \\{0\\}\\}\n$$\nAs $f$ is locally bounded, $\\ell, L \\in \\mathbb{R}$, with $\\ell \\le L$.\n\nConsider sequences $(x_n)_{n \\ge 1}$ and $(y_n)_{n \\ge 1}$, with nonzero elements, converging to $0$, such that $\\lim_{n \\to \\infty} f(x_n) = L$ and $\\lim_{n \\to \\infty} f(y_n) = \\ell$. By continuity of $g$ at $0$ and properties of $\\limsup$ and $\\liminf$:\n\n$$\n\\begin{aligned}\ng(0) &= \\lim_{n \\to \\infty} g(x_n) = \\lim_{n \\to \\infty} (2f(x_n) + f(x_n^2)) \\\\\n&= \\liminf_{n \\to \\infty} (2f(x_n) + f(x_n^2)) \\\\\n&\\ge 2 \\liminf_{n \\to \\infty} f(x_n) + \\liminf_{n \\to \\infty} f(x_n^2) \\\\\n&= 2L + \\ell\n\\end{aligned}\n$$\n\nand\n\n$$\n\\begin{aligned}\ng(0) &= \\lim_{n \\to \\infty} g(y_n) = \\lim_{n \\to \\infty} (2f(y_n) + f(y_n^2)) \\\\\n&= \\limsup_{n \\to \\infty} (2f(y_n) + f(y_n^2)) \\\\\n&\\le 2 \\limsup_{n \\to \\infty} f(y_n) + \\limsup_{n \\to \\infty} f(y_n^2) \\\\\n&= 2\\ell + L\n\\end{aligned}\n$$\n\nSince $\\ell \\le L$, the inequality $2L + \\ell \\le g(0) \\le 2\\ell + L$ implies $L = \\ell = \\frac{g(0)}{3}$. So $f$ is continuous at the origin and $\\lim_{x \\to 0} f(x) = \\frac{g(0)}{3} = f(0)$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 12566,
    "subject": "Mathematics (Olympiad)",
    "question": "In the square with diagonal $A(0, 0)B(10, 10)$, a spaceship needs to deliver a letter from $A$ to $B$. The spaceship can move as follows: from a point $(n_1, m_1)$, where $n_1$ and $m_1$ are integers of the same parity, it can move to any point $(n_2, m_2)$ where $n_2 > n_1$, $m_2 > m_1$, and $n_2, m_2$ are also integers of the same parity. Enemies have set up location systems. The zone covered by the location system is shown in the figure below by arrows. The spaceship is detected if its trajectory intersects an arrow (the endpoint does not belong to the covered zone).\n\n![](images/Ukrajina_2011_p8_data_e2374835ce.png)\n\nWhich trajectories will result in the spaceship being detected or undetected?",
    "options": [],
    "answer": "See solution",
    "solution": "Trajectories for which the spaceship is undetected are those that do not intersect the covered zones.\n\nWe can add extra vertical segments to those that belong to the covered zone to form 'stairs' as shown in the figure below. These stairs are between the lines $x + y = 10$ and $x + y = 11$. Each trajectory from $A$ to $B$ must intersect these stairs exactly once, since at each step the sum of the coordinates increases by at least $2$.\n\nA trajectory is called *dangerous* if the spaceship is detected on it, and *safe* otherwise. Consider two groups of trajectories: those that pass through the point $(5, 5)$ and those that do not. All trajectories passing through $(5, 5)$ are safe.\n\n![](images/Ukrajina_2011_p9_data_888c5d02cc.png)\n\nFor any trajectory $f(x)$, consider its symmetric trajectory $g(x)$ with respect to the line $AB$. If one trajectory from the second group is safe (dangerous), then its symmetric is dangerous (safe), so the number of safe and dangerous trajectories in this group is equal. Combining this with the previous observation, we conclude that the number of safe trajectories equals the number of dangerous ones.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 12567,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all positive integers $n$ for which there exists a polynomial $P(x)$ with integer coefficients such that for every positive divisor $d$ of $n$, $P(d) = \\left(\\frac{n}{d}\\right)^2$.",
    "options": [],
    "answer": "See solution",
    "solution": "Obviously, such a polynomial exists for $n = 1$. In this case, the only condition is that $P(1) = 1$, and the polynomial $P(x) = x$ has this property.\n\nIf $n$ is a prime, then its only two divisors are $1$ and $n$. The polynomial $P$ must satisfy $P(1) = n^2$ and $P(n) = 1$. Let us write $P(x) = ax + b$ and solve the resulting system of equations. We get $P(x) = (-n - 1)x + n^2 + n + 1$.\n\nAssume that $n = k \\cdot l$ is not a prime and $k, l > 1$. We have $P(1) = n^2$, $P(l) = k^2$, $P(k) = l^2$, and $P(n) = 1$. We know that for arbitrary integers $a$ and $b$, the number $P(a) - P(b)$ is divisible by $a - b$. So, $n - k = k(l - 1)$ divides $P(n) - P(k) = 1 - l^2 = (1 - l)(1 + l)$. This implies that $k$ divides $l + 1$. Similarly, we show that $n - l$ divides $P(n) - P(l)$. So $l(k - 1)$ divides $(1 - k)(1 + k)$ and $l$ divides $k + 1$. Hence, $kl$ divides $(k + 1)(l + 1)$ and therefore it also divides $(k + 1)(l + 1) - kl = k + l + 1$. We must have $kl \\leq k + l + 1$, which implies that $kl - k - l + 1 \\leq 2$ or $(k - 1)(l - 1) \\leq 2$. We may assume that $k \\leq l$. The only possible case is $k = 2$ and $l = 3$, whence $n = 6$.\n\nLet us find a polynomial $P$, satisfying the conditions $P(1) = 36$, $P(2) = 9$, $P(3) = 4$, and $P(6) = 1$. For the polynomial $Q(x) = P(x) - 1$ we have $Q(1) = 35$, $Q(2) = 8$, $Q(3) = 3$, and $Q(6) = 0$, so $6$ is a root of $Q$. Let us write $Q(x) = (x - 6)R(x)$. For $R(x)$ we have $R(1) = -7$, $R(2) = -2$, and $R(3) = -1$, so $3$ is a root of the polynomial $R(x) + 1$ and $R(x) = 1 + (x - 3)S(x)$. For the polynomial $S(x)$ we have $S(1) = 4$ and $S(2) = 3$. We see that $S(x) = 5 - x$, whence $P(x) = 1 + (x - 6)(1 + (x - 3)(5 - x))$.",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 12568,
    "subject": "Mathematics (Olympiad)",
    "question": "Suppose $a_1, a_2, \\ldots, a_n$ are positive real numbers such that $\\max(a_1, a_2, \\ldots, a_n) \\le n \\cdot \\min(a_1, a_2, \\ldots, a_n)$. What is the smallest integer $n$ such that among any such $n$ numbers, there must exist three that are the side-lengths of an acute triangle?",
    "options": [],
    "answer": "See solution",
    "solution": "First, we prove that any $n \\ge 13$ is a solution. Suppose $a_1, a_2, \\ldots, a_n$ satisfy $\\max(a_1, a_2, \\ldots, a_n) \\le n \\cdot \\min(a_1, a_2, \\ldots, a_n)$, and that we cannot find three that are the side-lengths of an acute triangle. Assume $a_1 \\le a_2 \\le \\ldots \\le a_n$. Then $a_{i+2}^2 \\ge a_i^2 + a_{i+1}^2$ for all $i \\le n-2$.\n\nLet $(F_n)$ be the Fibonacci sequence, with $F_1 = F_2 = 1$ and $F_{n+1} = F_n + F_{n-1}$. It is easy to check that $F_n < n^2$ for $n \\le 11$, $F_{12} = 12^2$, and $F_n > n^2$ for $n > 12$. The inequality $a_{i+2}^2 \\ge a_i^2 + a_{i+1}^2$ and the ordering imply $a_i^2 \\ge F_i \\cdot a_1^2$ for all $i \\le n$. Hence, if $n \\ge 13$, we obtain $a_n^2 > n^2 \\cdot a_1^2$, contradicting the hypothesis. This shows that any $n \\ge 13$ is a solution.\n\nBy taking $a_i = \\sqrt{F_i}$ for $1 \\le i \\le n$, we have $\\max(a_1, a_2, \\ldots, a_n) \\le n \\cdot \\min(a_1, a_2, \\ldots, a_n)$ for any $n < 13$, but no three $a_i$ can be the side-lengths of an acute triangle. Hence, the answer is: all $n \\ge 13$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 12569,
    "subject": "Mathematics (Olympiad)",
    "question": "An integer $N$ is determined by its first (leftmost) six digits and must satisfy the following conditions:\n\n- The first occurrence of either the string $2014$ or its reverse, $4102$, must start at position 1, 2, or 3 among the digits.\n- In each case, there are two other digits to be chosen.\n- These digits can be chosen arbitrarily if $2014$ or $4102$ starts at position 1, but the first digit must be non-zero in the other cases.\n\nHow many such integers $N$ are there?",
    "options": [],
    "answer": "See solution",
    "solution": "There are three possible starting positions for the string $2014$ or $4102$ (positions 1, 2, or 3):\n\n- If the string starts at position 1, the two remaining digits can be any digits: $10 \\times 10 = 100$ choices.\n- If the string starts at position 2 or 3, the first digit (to the left) must be non-zero, so $9$ choices for the first digit and $10$ for the other: $9 \\times 10 = 90$ choices for each case.\n- There are two strings ($2014$ and $4102$), so multiply the total by $2$.\n\nTotal: $2 \\times (100 + 90 + 90) = 2 \\times 280 = 560$ numbers.",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 12570,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $a_1, a_2, \\dots, a_{n+1}$ be a sequence of real numbers such that $a_1 \\ge 1$ and $a_{k+1} - a_k \\ge 1$ for all $k \\ge 1$. Prove that\n\n$$\n\\sum_{k=1}^{n+1} a_k^3 \\ge \\left(\\sum_{k=1}^{n+1} a_k\\right)^2.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "We will prove the statement by induction.\n\nFirst, for $n=1$, $a_1^3 \\ge a_1^2$ since $a_1 \\ge 1$.\n\nSuppose the statement holds for $n$ terms:\n$$\n\\sum_{k=1}^{n} a_k^3 \\ge \\left(\\sum_{k=1}^{n} a_k\\right)^2.\n$$\nFor $n+1$ terms:\n$$\n\\begin{aligned}\n\\sum_{k=1}^{n+1} a_k^3 &= a_{n+1}^3 + \\sum_{k=1}^{n} a_k^3 \\\\\n&\\ge a_{n+1}^3 + \\left(\\sum_{k=1}^{n} a_k\\right)^2 \\\\\n&= \\left(\\sum_{k=1}^{n+1} a_k\\right)^2 + a_{n+1}^3 - a_{n+1}^2 - 2a_{n+1} \\sum_{k=1}^{n} a_k.\n\\end{aligned}\n$$\nTo complete the induction, we show $a_{n+1}^3 - a_{n+1}^2 - 2a_{n+1} \\sum_{k=1}^{n} a_k \\ge 0$.\n\nSince $a_{k+1} - a_k \\ge 1$, we have $a_{k+1}^2 - a_k^2 \\ge a_{k+1} + a_k$. Summing over $k=1, \\dots, n$ and using $a_1^2 - a_1 \\ge 0$, we get\n$$\na_{n+1}^2 - a_1^2 \\ge a_{n+1} + 2 \\sum_{k=1}^{n} a_k - a_1.\n$$\nThis implies\n$$\na_{n+1}^3 - a_{n+1}^2 - 2a_{n+1} \\sum_{k=1}^{n} a_k \\ge 0.\n$$\nThus, the inequality holds for all $n$ by induction.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 12571,
    "subject": "Mathematics (Olympiad)",
    "question": "Determine all positive integers $n$ such that the number $N$ is also an integer, where\n\n$$\nN = \\frac{2 \\cdot 1}{\\sqrt{1^2 + 1 + 4 + \\sqrt{1^2 - 1 + 4}}} + \\frac{2 \\cdot 2}{\\sqrt{2^2 + 2 + 4 + \\sqrt{2^2 - 2 + 4}}} + \\dots + \\frac{2n}{\\sqrt{n^2 + n + 4 + \\sqrt{n^2 - n + 4}}}\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "**Answer:** $n = 3$.\n\nLet us note that\n\n$$\n\\frac{2k}{\\sqrt{k^2 + k + 4} + \\sqrt{k^2 - k + 4}} = \\frac{(\\sqrt{k^2 + k + 4} - \\sqrt{k^2 - k + 4})2k}{(k^2 + k + 4) - (k^2 - k + 4)} = \\sqrt{k^2 + k + 4} - \\sqrt{k^2 - k + 4} = \\sqrt{k^2 + k + 4} - \\sqrt{(k-1)^2 + (k-1) + 4}.\n$$\n\nAnd thus\n\n$$\nN = (\\sqrt{6} - \\sqrt{4}) + (\\sqrt{10} - \\sqrt{6}) + (\\sqrt{16} - \\sqrt{10}) + \\dots + (\\sqrt{n^2 + n + 4} - \\sqrt{(n-1)^2 + (n-1) + 4}) = \\sqrt{n^2 + n + 4} - \\sqrt{4} = \\sqrt{n^2 + n + 4} - 2.\n$$\n\nIt means that $N$ is integer if and only if $n^2 + n + 4$ is a perfect square.\n\nSince $n^2 < n^2 + n + 4 < n^2 + 4n + 4 = (n + 2)^2$, then for every positive integer $n$ we obtain $n^2 + n + 4 = (n + 1)^2 = n^2 + 2n + 1$. And thus $n = 3$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 12572,
    "subject": "Mathematics (Olympiad)",
    "question": "Find the minimum and the maximum of the sum $S = \\frac{a}{b} + \\frac{c}{d}$ where $a, b, c, d \\in \\mathbb{N}$ satisfy $a + c = 2020$, $b + d = 2020$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $a + c = 2020$ and $b + d = 2020$. Consider $S = \\frac{a}{b} + \\frac{c}{d}$ for $a, b, c, d \\in \\mathbb{N}$.\n\nBy symmetry, assume $b \\leq d$. Then $1 \\leq b \\leq 1010$.\n\nFor the maximum, set $a = 2019$, $c = 1$, $b = 1$, $d = 2019$:\n$$\nS_{\\max} = \\frac{2019}{1} + \\frac{1}{2019} = 2019 + \\frac{1}{2019}\n$$\n\nFor the minimum, set $a = 1$, $c = 2019$, $b = 1010$, $d = 1010$:\n$$\nS_{\\min} = \\frac{1}{1010} + \\frac{2019}{1010} = \\frac{2020}{1010} = 2\n$$\n\nThus, the minimum value of $S$ is $2$, and the maximum value is $2019 + \\frac{1}{2019}$, under the given constraints.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 12573,
    "subject": "Mathematics (Olympiad)",
    "question": "Find the smallest number $n$ with the following property: if among the numbers from 1 to 1000 we choose $n$ numbers such that no two of them are divisible by the square of the same prime number, then at least one of these numbers is necessarily the square of a prime number.",
    "options": [],
    "answer": "See solution",
    "solution": "The answer is $617$.\n\nFirst, let's construct a set of $616$ numbers from $1$ to $1000$ such that there are no prime squares and no two numbers are divisible by the square of the same prime. Take the $8$ numbers of the form $2p^2$ not exceeding $1000$, where $p$ is a prime:\n\n$$\n2 \\cdot 2^2,\\ 2 \\cdot 3^2,\\ 2 \\cdot 5^2,\\ 2 \\cdot 7^2,\\ 2 \\cdot 11^2,\\ 2 \\cdot 13^2,\\ 2 \\cdot 17^2,\\ 2 \\cdot 19^2\n$$\n\nand all natural numbers from $1$ to $1000$ that are square-free (not divisible by the square of any prime). There are $608$ such numbers. To verify this, use the inclusion-exclusion principle:\n\n$$\n1000 - \\sum_{1 \\le i \\le 11} \\left\\lfloor \\frac{1000}{p_i^2} \\right\\rfloor + \\sum_{1 \\le i < j \\le 11} \\left\\lfloor \\frac{1000}{p_i^2 p_j^2} \\right\\rfloor - \\sum_{1 \\le i < j < k \\le 11} \\left\\lfloor \\frac{1000}{p_i^2 p_j^2 p_k^2} \\right\\rfloor,\n$$\n\nwhere $p_1 = 2,\\ p_2 = 3,\\ p_3 = 5,\\ p_4 = 7,\\ p_5 = 11,\\ p_6 = 13,\\ p_7 = 17,\\ p_8 = 19,\\ p_9 = 23,\\ p_{10} = 29,\\ p_{11} = 31$. Calculating:\n\n$$\n\\begin{aligned}\n1000 &- (250 + 111 + 40 + 20 + 8 + 5 + 3 + 2 + 1 + 1 + 1) \\\\\n&+ (27 + 10 + 5 + 2 + 1 + 4 + 2) - 1 \\\\\n&= 1000 - 442 + 51 - 1 = 608.\n\\end{aligned}\n$$\n\nNow, suppose we select $617$ numbers from $1$ to $1000$ with no squares of primes among them. Since there are only $608$ square-free numbers, at least $9$ of the chosen numbers are divisible by the square of a prime. Denote these by $a_1, a_2, \\dots, a_9$, and for each $a_i$, let $q_i^2$ be a prime square dividing $a_i$. If all $q_i^2$ are distinct, then some $q_k^2 \\geq 23^2 = 529$. Since $a_k \\neq q_k^2$, $a_k \\geq 2q_k^2 = 1058$, which exceeds $1000$—a contradiction. Therefore, at least two numbers are divisible by the square of the same prime.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 12574,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $ABCD$ be a cyclic quadrangle such that none of the triangles $BCD$ and $CDA$ is equilateral. Prove that, if the Simson line of $A$ in the triangle $BCD$ and the Euler line of this triangle are perpendicular, then so are the Simson line of $B$ in the triangle $CDA$ and the Euler line of this triangle.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $O$ be the center of the circle $\\gamma$ circumscribed about quadrilateral $ABCD$. Denote by $H_a$ and $H_b$ the orthocenters of triangles $BCD$ and $ACD$, respectively. Let the perpendicular from $A$ to the line $CD$ meet the circle $\\gamma$ again at $A_1$, and define $B_1$ similarly.\n\nNotice that $ABB_1A_1$ is a trapezoid, since the chords $AA_1$ and $BB_1$ are parallel; and so is $H_aH_bB_1A_1$, since $CD$ is the common perpendicular bisector of the line segments $H_bA_1$ and $H_aB_1$. Consequently, $AH_bH_aA$ is a parallelogram. Let $M$ be the common midpoint of the diagonals $AH_a$ and $BH_b$.\n\nIt is a fact that $M$ lies on the Simson line $\\ell_a$ of $A$ in triangle $BCD$. Notice that $\\ell_a$ is a midline in triangle $H_bA_1B$, so $\\ell_a$ is parallel to $A_1B$. Similarly, the Simson line $\\ell_b$ of $B$ in triangle $ACD$ is parallel to $AB_1$.\n\nAt this point, we are to prove that the line $OH_b$ — the Euler line of triangle $ACD$ — is perpendicular to $AB_1$, provided that the lines $OH_a$ and $BA_1$ are perpendicular. To this end, notice that $OBH_aA_1$ is either a kite or a chevron, since $OB = OA_1$ and its diagonals are perpendicular, so $BH_a = A_1H_a$. Notice that $BH_a = AH_b$ and $A_1H_a = B_1H_b$ to deduce that $AH_b = B_1H_b$. Since $OA = OB_1$, the line $OH_b$ is the perpendicular bisector of the line segment $AB_1$, whence the conclusion.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 12575,
    "subject": "Mathematics (Olympiad)",
    "question": "In the acute-angled triangle $ABC$, consider the altitudes $BB'$ and $CC'$. The half-line $C'B'$ intersects the circumcircle of the triangle at $B''$, and denote by $\\alpha_A$ the angle $\\widehat{ABB''}$. In a similar way, define the angles $\\alpha_B$ and $\\alpha_C$. Prove the inequality\n\n$$\n\\sin \\alpha_A \\sin \\alpha_B \\sin \\alpha_C \\le \\frac{3\\sqrt{6}}{32}.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "Let $C''$ be the second intersection point of the line $B'C'$ with the circumcircle of $\\triangle ABC$.\n\n![](images/shortlistBMO2015_p11_data_1532328027.png)\n\nConsider the intersection point $A_1$ of the diameter $AA_2$ with the line $B'C'$. Since $BC'B'C$ is cyclic, we have $\\widehat{ABC} = \\widehat{AA_2C} = \\frac{\\pi}{2} - \\widehat{A_1AB'} = \\widehat{AB'A_1}$, thus $AA_1 \\perp B'C'$, i.e., $AA_1 \\perp B''C''$.\n\nBecause the chord $B''C''$ is perpendicular to the diameter $AA_2$, it follows that $\\triangle AB''C''$ is isosceles. Clearly, we have $\\widehat{ABB''} = \\widehat{AC''B''} = \\widehat{AB''C''}$.\n\nIn $\\triangle AB'C'$, we have $AB' = c \\cdot \\cos A$, where $a, b, c$ denote the side lengths of $\\triangle ABC$. It follows $AA_1 = c \\cdot \\cos A \\sin B$. Let $R$ be the circumradius of triangle $ABC$. We get $A_1B''^2 + (R - AA_1)^2 = R^2$, hence $A_1B''^2 = 2R \\cdot AA_1 - AA_1^2$. Then we obtain $AB''^2 = AA_1^2 + A_1B''^2 = 2R \\cdot AA_1 = 2R \\cdot \\cos A \\sin B$, therefore\n\n$$\n\\sin \\alpha_A = \\sin \\widehat{ABB''} = \\sin \\widehat{AB''A_1} = \\frac{AA_1}{AB''} = \\frac{\\cos A \\sin B}{\\sqrt{2R \\cos A \\sin B}} = \\sqrt{\\frac{c}{2R}} \\cos A \\sin B = \\sqrt{\\cos A \\sin B \\sin C}\n$$\n\nand similar relations hold for $\\sin \\alpha_B, \\sin \\alpha_C$.\n\nIt follows\n\n$$\n(\\sin \\alpha_A \\sin \\alpha_B \\sin \\alpha_C)^2 = (\\cos A \\cos B \\cos C)(\\sin A \\sin B \\sin C)^2.\n$$\n\nSince the functions $f(x) = \\ln \\cos x$ and $g(x) = \\ln \\sin x$ are concave on the interval $(0, \\frac{\\pi}{2})$, we obtain $\\cos A \\cos B \\cos C \\le \\frac{1}{8}$ and $\\sin A \\sin B \\sin C \\le \\frac{3\\sqrt{3}}{8}$. Therefore\n\n$$\n\\sin \\alpha_A \\sin \\alpha_B \\sin \\alpha_C \\le \\frac{1}{2\\sqrt{2}} \\cdot \\frac{3\\sqrt{3}}{8} = \\frac{3\\sqrt{6}}{32},\n$$\n\nas desired. The equality holds if and only if the triangle $ABC$ is equilateral. $\\square$\n\n**Remark (P.S.C.):** We can avoid reference to functions since it is well known that for acute-angled triangles $ABC$ it holds $0 \\le \\cos A \\cos B \\cos C \\le \\frac{1}{8}$ and that for any triangle $ABC$ it holds $0 \\le \\sin A \\sin B \\sin C \\le \\frac{3\\sqrt{3}}{8}$, with both of these relations admitting plenty of other proofs, for example, trigonometric.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 12576,
    "subject": "Mathematics (Olympiad)",
    "question": "Tenemos 2021 colores y 2021 fichas de cada color. Colocamos las $2021^2$ fichas en fila. Se dice que una ficha, $F$, es \"mala\" si a cada lado de $F$ quedan un número impar de las $2020 \\times 2021$ fichas que no comparten color con $F$.\n\n(a) Determina cuál es el mínimo número posible de fichas malas.\n\n(b) Si se impone la condición de que cada ficha ha de compartir color con al menos una ficha adyacente, ¿cuál es el mínimo número posible de fichas malas?",
    "options": [],
    "answer": "See solution",
    "solution": "(a) Como $2020 \\times 2021$ es par, para decidir si una ficha es mala es suficiente comprobar que el número de fichas a su izquierda con las que no comparte color es impar. Sea $A$ el conjunto de fichas que ocupan una posición impar en la fila, es decir, las fichas que tienen un número par de fichas a su izquierda. Es claro que $|A| = \\frac{2021^2+1}{2}$.\n\nSea $B$ el conjunto de fichas que, entre las de su color, ocupan una posición impar. Es decir, hay un número par de fichas de su color a su izquierda. Hay 1011 fichas de cada color en $B$, así que $|B| = 1011 \\times 2021$.\n\nPor construcción, las fichas que están en $B$ pero no en $A$ son malas (tienen un número impar de fichas a su izquierda, de las cuales un número par comparten color con ella). Por lo tanto, hay al menos $|B| - |A| = 2021 \\times 1011 - \\frac{2021^2+1}{2} = 1010$ fichas malas.\n\nEste número se puede conseguir de la siguiente forma: colocamos 2020 fichas de color 1, luego 2020 fichas de color 2, etc., hasta tener 2020 fichas de cada color; luego colocamos las fichas restantes por orden de color. Las únicas fichas malas en esta configuración son las que son última de cada color par. En la imagen se ve esta configuración para 5 colores y 5 fichas de cada color.\n\n![](images/OME2021_booklet_p4_data_67d6605c61.png)\n\n(b) Sean $X = A \\cap B$, $Y = A \\cap B^c$ y $Z = A^c \\cap B$, donde $A$ y $B$ son los conjuntos descritos en el apartado anterior. Las fichas malas son precisamente $Y \\cup Z$ (contando las fichas a la izquierda que no comparten color con la ficha dada, como antes), por lo que tenemos que minimizar el tamaño de este conjunto.\n\nObsérvese que si la ficha $F$ es adyacente a una ficha $F' \\in Z$ del mismo color, entonces $F \\in Y$. Como cada ficha de $Z$ es adyacente a al menos una ficha de $Y$, y cada ficha de $Y$ es adyacente a como mucho dos fichas de $Z$, tenemos que $|Z| \\le 2|Y|$.\n\nAdemás, tenemos que $|Z| - |Y| = (|X| + |Z|) - (|X| + |Y|) = |B| - |A| = 1010$.\nY concluimos que\n\n$$\n|Y| + |Z| = 3(|Z| - |Y|) - 2(|Z| - 2|Y|) \\geq 3 \\times 1010 - 2 \\times 0 = 3030.\n$$\n\n(La conclusión del último párrafo también se puede obtener haciendo:\n$$|Z| = 1010 + |Y| \\geq 1010 + \\frac{1}{2}|Z|, \\text{ luego } |Z| \\geq 2020 \\text{ y } |Z| + |Y| \\geq \\frac{3}{2}|Z| \\geq 3030.$$)\n\nUna forma de tener exactamente 3030 fichas malas es colocar primero 2018 fichas de color 1, luego 2018 de color 2, etc., y después colocar las tres fichas restantes de color 1, luego las tres de color 2, etc. Las únicas fichas malas son las tres últimas de cada color par.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 12577,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $f : [a, b] \\to \\mathbb{R}$ be a Riemann integrable function with the property $(\\mathcal{P})$:\n\nFor any $n \\in \\mathbb{N}^*$ and any $x \\in [a, b]$, the following equality holds:\n\n$$\nf(x) = \\frac{1}{2^n} \\sum_{k=0}^{2^n-1} f\\left(\\frac{x + (2^n - 1 - k)a + kb}{2^n}\\right).\n$$\n\nShow that any function with property $(\\mathcal{P})$ is constant on $[a, b]$, and conversely, that any constant function satisfies $(\\mathcal{P})$.",
    "options": [],
    "answer": "See solution",
    "solution": "For any $t \\in \\mathbb{R}$, there is a unique function with property $(\\mathcal{P})$, namely $f : [a, b] \\to \\mathbb{R}$ defined by $f(x) = \\frac{t}{b-a}$ for all $x \\in [a, b]$, so that $\\int_a^b f(x) dx = t$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 12578,
    "subject": "Mathematics (Olympiad)",
    "question": "How many ways are there to distribute integers into the boxes of a $3 \\times 3$ grid so that the sum of the numbers in each row and each column is a multiple of $3$?",
    "options": [],
    "answer": "See solution",
    "solution": "Let us replace every number $n$ in the grid by $p = 0, 1, 2$ such that $n \\equiv p \\pmod{3}$. The condition that the sum of the numbers in each row and column is a multiple of $3$ is preserved after this replacement.\n\nAmong the triples $(a, b, c)$ with $a, b, c \\in \\{0, 1, 2\\}$, only $(0, 0, 0)$, $(1, 1, 1)$, $(2, 2, 2)$, and $(0, 1, 2)$ (and its permutations) have their sum divisible by $3$. Thus, each row and column must be either all the same or all different.\n\nIf we represent the remainders $0, 1, 2$ by the letters A, B, C (in any order), the only possible placements that satisfy the requirements are the four shown below:\n\n![](images/Japan_2015_p4_data_3461d5f408.png)\n![](images/Japan_2015_p4_data_cffea6da1d.png)\n![](images/Japan_2015_p4_data_f67edc5ad3.png)\n![](images/Japan_2015_p4_data_790853145d.png)\n\nFor each placement, there are $3!$ ways to assign the three remainders to the letters, so $4 \\times 3! = 24$ possible configurations of remainders $p$.\n\nFor each configuration, there are $3$ choices for $n$ for each $p$, so $(3!)^3 = 216$ ways to assign integers for each configuration.\n\nTherefore, the total number of ways is $24 \\times 216 = 5184$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 12579,
    "subject": "Mathematics (Olympiad)",
    "question": "Determine all polynomials $P \\in \\mathbb{R}[x, y]$ such that\n$$\nP(a, b^2 - ac) + P(b, c^2 - ab) + P(c, a^2 - bc) = 0\n$$\nfor all $a, b, c \\in \\mathbb{R}$.",
    "options": [],
    "answer": "See solution",
    "solution": "We use properties of polynomials:\n\n**Fact 1.** For $P \\in \\mathbb{R}[x, y]$, if $P(c, y) = 0$ for infinitely many $y$, then $x - c$ divides $P(x, y)$.\n\n**Fact 2.** If $P(c, y) = 0$ for infinitely many $y$ and infinitely many $c$, then $P$ is the zero polynomial.\n\n**Fact 3.** If $P(x, y, z) = 0$ for all $x, y, z \\in \\mathbb{R}$, then $P$ is the zero polynomial.\n\nSubstitute $a = b = c$:\n$$\nP(a, 0) = 0 \\implies y \\text{ divides } P(x, y).\n$$\nSubstitute $a = b = 0$:\n$$\nP(0, c^2) = 0 \\implies x \\text{ divides } P(x, y).\n$$\nSo $P(x, y) = xyQ(x, y)$ for some polynomial $Q$.\n\nSubstitute $a = 0$:\n$$\nP(b, c^2) = -P(c, -bc) \\implies Q(b, c^2) = Q(c, -bc).\n$$\nSubstitute $c = 0$:\n$$\nQ(b, 0) = Q(0, 0) \\implies y \\text{ divides } Q(x, y) - Q(0, 0).\n$$\nLet $Q(x, y) = yR(x, y) + k_0$.\n\nNow,\n$$\nc^2 R(b, c^2) = -bc R(c, -bc) \\implies c R(b, c^2) = -b R(c, -bc).\n$$\nSubstitute $b = 0$:\n$$\nc R(0, c^2) = 0 \\implies R(x, y) = xP_1(x, y).\n$$\nSo $P(x, y) = xy(k_0 + x y P_1(x, y))$.\n\nIterating this process, we find\n$$\nP(x, y) = \\sum_{i=0}^{\\infty} k_i (x y)^{2i+1}.\n$$\nSubstituting into the original equation, only the term $k_0 x y$ survives, so\n$$\nP(x, y) = k x y\n$$\nfor some $k \\in \\mathbb{R}$.\n\nThus, all solutions are $P(x, y) = k x y$.",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 12580,
    "subject": "Mathematics (Olympiad)",
    "question": "Si $n$ es un número natural, el $n$-ésimo número triangular es $T_n = 1 + 2 + \\cdots + n$. Hallar todos los valores de $n$ para los que el producto de los 16 números triangulares consecutivos $T_n T_{n+1} \\cdots T_{n+15}$ es un cuadrado perfecto.",
    "options": [],
    "answer": "See solution",
    "solution": "Como $T_n = \\dfrac{n(n+1)}{2}$, el producto de los 16 números triangulares es\n$$\nP_n = \\frac{n(n+1)(n+2)\\cdots(n+15)(n+1)(n+2)\\cdots(n+16)}{2^{16}}\n$$\nObservando que $(n+1)(n+2)\\cdots(n+15)$ aparece dos veces, $P_n = \\dfrac{n(n+16)\\left[(n+1)(n+2)\\cdots(n+15)\\right]^2}{2^{16}}$. El factor $\\left[(n+1)(n+2)\\cdots(n+15)\\right]^2$ es un cuadrado perfecto, así que $P_n$ es un cuadrado perfecto si y sólo si $n(n+16)$ es un cuadrado perfecto.\n\nComo $n$ y $n+16$ son coprimos salvo posiblemente por el 2, para que $n(n+16)$ sea un cuadrado perfecto, ambos deben ser cuadrados o el doble de un cuadrado. Escribimos $n = 2^a m^2$ y $n+16 = 2^b t^2$ con $a, b = 0$ o $1$.\n\nComo $n$ y $n+16$ tienen la misma paridad, sólo son posibles $a = b = 0$ o $a = b = 1$.\n\n- Si $a = b = 0$: $n = m^2$, $n+16 = t^2$, así que $t^2 - m^2 = 16 \\implies (t-m)(t+m) = 16$. Las soluciones enteras positivas son $t-m = 2$, $t+m = 8$ $\\implies$ $t = 5$, $m = 3$, así que $n = 9$.\n- Si $a = b = 1$: $n = 2m^2$, $n+16 = 2t^2$, así que $2t^2 - 2m^2 = 16 \\implies t^2 - m^2 = 8$. Las soluciones enteras positivas son $t-m = 2$, $t+m = 4$ $\\implies$ $t = 3$, $m = 1$, así que $n = 2$.\n\nPor lo tanto, los valores de $n$ son $n = 2$ y $n = 9$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 12581,
    "subject": "Mathematics (Olympiad)",
    "question": "令 $n$ 為某個大於 $3$ 的整數且 $S = \\{1, 2, \\dots, n\\}$。令 $A_1, A_2, \\dots, A_n$ 為 $S$ 的子集，其中對所有 $i$ 都有 $|A_i| \\ge 2$。假設對於任意兩個元素的子集 $S' \\subseteq S$，都存在唯一的 $i$ 使得 $S' \\subseteq A_i$，試證：對於任意 $1 \\le i < j \\le n$，都有 $A_i \\cap A_j \\ne \\emptyset$。",
    "options": [],
    "answer": "See solution",
    "solution": "由題意得\n\n$$\n\\sum_{i=1}^{n} \\binom{|A_i|}{2} = \\binom{n}{2}.\n$$\n\n令 $d_i = |\\{k \\mid i \\in A_k\\}|$。明顯地，\n\n$$\n\\sum_{i=1}^{n} d_i = \\sum_{k=1}^{n} |A_k|.\n$$\n\n由「存在與唯一」性質，得到\n\n$$\n\\sum_{i=1}^{n} \\binom{d_i}{2} = \\sum_{1 \\le i < j \\le n} |A_i \\cap A_j|.\n$$\n\n再加上 $|A_i \\cap A_j| \\le 1$（否則違反題意的「唯一」性質）。要證明 $A_i \\cap A_j \\ne \\emptyset$，等於是證明 $|A_i \\cap A_j| = 1$，這等價於\n\n$$\n\\sum_{i=1}^{n} \\binom{d_i}{2} = \\binom{n}{2}.\n$$\n\n由於上述等式以及二項式係數的定義 $\\binom{x}{2} = (x^2 - x)/2$，上面的等式又等價於\n\n$$\n\\sum_{i=1}^{n} d_i^2 = \\sum_{k=1}^{n} |A_k|^2.\n$$\n\n以下考慮有序對 $(i, k)$，其中 $i \\notin A_k$。假設 $A_k = \\{j_1, j_2, \\dots, j_t\\}$，則 $\\{i, y_1\\}, \\{i, y_2\\}, \\dots, \\{i, y_t\\}$ 是相異的 $t$ 個 $A_j$ 的子集（因為二元集 $\\{y_a, y_b\\}$ 是唯一包含於 $A_k$ 中），所以 $d_i \\ge |A_k|$，也會有\n\n$$\n\\frac{d_i}{n - d_i} \\ge \\frac{|A_k|}{n - |A_k|}.\n$$\n\n將上面的不等式對於所有要求的 $(i, k)$ 做和，得到\n\n$$\n\\begin{aligned}\n\\sum_{i=1}^{n} d_i &= \\sum_{i=1}^{n} \\sum_{k\\mid i \\notin A_k} \\frac{d_i}{n - d_i} \\\\\n&\\geq \\sum_{i=1}^{n} \\sum_{k\\mid i \\notin A_k} \\frac{|A_k|}{n - |A_k|} \\\\\n&= \\sum_{k=1}^{n} \\sum_{i\\mid i \\notin A_k} \\frac{|A_k|}{n - |A_k|} \\\\\n&= \\sum_{k=1}^{n} |A_k|.\n\\end{aligned}\n$$\n\n由等式，上述應該是等式，最後就得到 $d_i = |A_k|$ 當 $i \\notin A_k$ 時。接著考慮等式\n\n$$\n\\begin{aligned}\n\\sum_{i=1}^{n} (n - d_i) d_i &= \\sum_{i=1}^{n} \\sum_{k\\mid i \\notin A_k} d_i \\\\\n&= \\sum_{i=1}^{n} \\sum_{k\\mid i \\notin A_k} |A_k| \\\\\n&= \\sum_{k=1}^{n} \\sum_{i\\mid i \\notin A_k} |A_k| \\\\\n&= \\sum_{k=1}^{n} (n - |A_k|) |A_k|.\n\\end{aligned}\n$$\n\n加上等式推得上述等式成立。證明完畢！",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 12582,
    "subject": "Mathematics (Olympiad)",
    "question": "A jury of 3366 film critics are judging the Oscars. Each critic makes a single vote for their favorite actor, and a single vote for their favorite actress. It turns out that for every integer $n \\in \\{1, 2, 3, \\ldots, 100\\}$, there is an actor or actress who has been voted for exactly $n$ times. Show that there are two critics who voted for the same actor and the same actress.",
    "options": [],
    "answer": "See solution",
    "solution": "Assume, for contradiction, that every critic votes for a different pair of actor and actress. \n\nDefine a *double-vote* as a critic's choice of both an actor and an actress, and a *single-vote* as each individual choice. Each double-vote corresponds to two single-votes.\n\nFor each $n = 34, 35, \\ldots, 100$, select one actor or actress who received exactly $n$ votes, and let $S$ be the set of these movie stars. Let $a$ and $b$ be the number of men and women in $S$, so $a + b = 67$.\n\nLet $S_1$ be the set of double-votes with exactly one single-vote in $S$, and $S_2$ the set with both single-votes in $S$. Let $s_1$ and $s_2$ be their respective counts. The total number of double-votes with at least one single-vote in $S$ is $s_1 + s_2$, and those with both in $S$ is $s_2 \\leq ab$.\n\nThe total number of single-votes in $S$ is $s_1 + 2s_2 = 34 + 35 + \\ldots + 100 = 4489$. Thus, $s_1 + s_2 = (s_1 + 2s_2) - s_2 \\geq 4489 - ab$.\n\nThe maximum value of $ab$ for $a + b = 67$ is $33 \\times 34 = 1122$ (since $ab = \\frac{(a+b)^2 - (a-b)^2}{4}$ is maximized when $|a-b| = 1$).\n\nTherefore, there must be at least $4489 - 1122 = 3367$ critics, which contradicts the given number of 3366. Thus, there must be two critics who voted for the same actor and the same actress.\n\n**Remark:** The choice of 34 is motivated by maximizing the lower bound for the number of critics, using similar calculations for other values of $k$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 12583,
    "subject": "Mathematics (Olympiad)",
    "question": "Prove that for every nonnegative integer $n$, the number $7^{7^n} + 1$ is the product of at least $2n + 3$ (not necessarily distinct) primes.",
    "options": [],
    "answer": "See solution",
    "solution": "The proof is by induction.\n\n**Base case ($n = 0$):**\n$7^{7^0} + 1 = 7^1 + 1 = 8 = 2^3$, which is the product of $3$ primes.\n\n**Inductive step:**\nAssume the statement holds for $n = k$. For $n = k + 1$, let $x = 7^{2m-1}$ for some positive integer $m$. We show that $\\dfrac{x^7 + 1}{x + 1}$ is composite, so $x^7 + 1$ has at least two more prime factors than $x + 1$.\n\nObserve:\n$$\n\\begin{aligned}\n\\frac{x^7 + 1}{x + 1} &= \\frac{(x + 1)^7 - ((x + 1)^7 - (x^7 + 1))}{x + 1} \\\\\n&= (x + 1)^6 - \\frac{7x(x^5 + 3x^4 + 5x^3 + 5x^2 + 3x + 1)}{x + 1} \\\\\n&= (x + 1)^6 - 7x(x^4 + 2x^3 + 3x^2 + 2x + 1) \\\\\n&= (x + 1)^6 - 7^{2m}(x^2 + x + 1)^2 \\\\\n&= \\left((x + 1)^3 - 7^m(x^2 + x + 1)\\right)\\left((x + 1)^3 + 7^m(x^2 + x + 1)\\right)\n\\end{aligned}\n$$\n\nEach factor exceeds $1$. For the smaller factor, $\\sqrt{7x} \\le x$ gives:\n$$\n\\begin{aligned}\n(x + 1)^3 - 7^m(x^2 + x + 1) &= (x + 1)^3 - \\sqrt{7x}(x^2 + x + 1) \\\\\n&\\geq x^3 + 3x^2 + 3x + 1 - x(x^2 + x + 1) \\\\\n&= 2x^2 + 2x + 1 \\geq 113 > 1.\n\\end{aligned}\n$$\n\nHence $\\dfrac{x^7 + 1}{x + 1}$ is composite, and the proof is complete.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 12584,
    "subject": "Mathematics (Olympiad)",
    "question": "Given a regular nonagon $ABCDEFGHI$ (all sides equal, each angle $\\frac{7 \\cdot 180^\\circ}{9} = 140^\\circ$) and a regular hexagon $ABJKLM$ (all sides equal, each angle $120^\\circ$), prove that $\\angle KEL = \\angle LGK = \\angle HMG$.\n\n% IMAGE: ![](images/Argentina2022_p7_data_b66a0e53f4.png)",
    "options": [],
    "answer": "See solution",
    "solution": "Because $HM$ and $GL$ are parallel, $\\angle HMG = \\angle LGK = \\angle KEL$, and we are done.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 12585,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all pairs of integers $a, b$ such that\n\n$$\n\\frac{a^2+1}{2b^2-3} = \\frac{a-1}{2b-1}\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "Obviously $a \\neq 1$, thus we can rewrite the equation as\n\n$$\n\\frac{a^2+1}{a-1} = \\frac{2b^2-3}{2b-1} \\quad (1)\n$$\n\nThe numerator of the fraction on the left is positive, the numerator on the right is negative only for $b \\in \\{-1, 0, 1\\}$.\n\nFor $b = -1$ we get $3a^2 - a + 4 = 0$, which has no real solution.\n\nFor $b = 0$ we get $a^2 - 3a + 4 = 0$, which has no real solution either.\n\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.\n\nNow assume $2b^2 - 3 > 0$, and consider possible reductions of the fractions in (1).\n\nIf some integer $n$ divides both $a^2+1$ and $a-1$, it divides $a^2+1-(a+1)(a-1) = 2$ as well. Similarly, if $n$ divides both $2b^2 - 3$ and $2b - 1$, it divides $(2b - 1)(2b + 1) - 2(2b^2 - 3) = 5$.\n\nThus, there are four possibilities to fulfill equation (1):\n\n1. $a^2+1 = 2b^2-3$ and $a-1 = 2b-1$, which has no real solution.\n2. $a^2+1 = 2(2b^2-3)$ and $a-1 = 2(2b-1)$; substituting $a = 4b-1$ into the first equation gives $3b^2 - 2b + 2 = 0$, with no real solutions.\n3. $5(a^2+1) = 2b^2-3$ and $5(a-1) = 2b-1$, with solution $a=0, b=-2$.\n4. $5(a^2+1) = 2(2b^2-3)$ and $5(a-1) = 2(2b-1)$, with solutions $a = -1, b = -2$ and $a = 7, b = 8$.\n\nThus, the five solutions are:\n\n$$\n(0, 1),\\ (-1, 1),\\ (0, -2),\\ (-1, -2),\\ (7, 8)\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 12586,
    "subject": "Mathematics (Olympiad)",
    "question": "Find the greatest integer $k \\le 2023$ such that, no matter how Alice colors exactly $k$ numbers among $\\{1, 2, \\dots, 2023\\}$ in red, Bob can color some of the remaining uncolored numbers in blue so that the sum of the red numbers equals the sum of the blue numbers.",
    "options": [],
    "answer": "See solution",
    "solution": "The answer is $592$.\n\nFor $k \\ge 593$, Alice can color the largest $593$ numbers $1431, 1432, \\dots, 2023$ and any other $(k-593)$ numbers, so their sum $s$ satisfies\n\n$$\ns \\ge \\frac{2023 \\cdot 2024}{2} - \\frac{1430 \\cdot 1431}{2} > \\frac{1}{2} \\cdot \\left( \\frac{2023 \\cdot 2024}{2} \\right),\n$$\n\nso Bob cannot match the sum with the remaining numbers.\n\nFor $k = 592$, let $s$ be the sum of Alice's $592$ numbers; note $s < \\frac{1}{2} \\cdot \\left(\\frac{2023 \\cdot 2024}{2}\\right)$. Bob can always find a subset of the remaining $1431$ numbers whose sum is either $s$ or $\\frac{2023 \\cdot 2024}{2} - 2s$, both $\\le \\frac{1}{3} \\cdot \\left(\\frac{2023 \\cdot 2024}{2}\\right)$.\n\n**Case 1:** $s_0 \\ge 2024$. Let $s_0 = 2024a + b$, $0 \\le b \\le 2023$. Bob can find two uncolored numbers with sum $b$ or $2024 + b$, and $a$ (or $a-1$) pairs with sum $2024$. There are enough such pairs to guarantee at least one is uncolored, so Bob can always achieve the sum $s_0$.\n\n**Case 2:** $s_0 \\le 2023$. Since $s \\ge 1 + 2 + \\dots + 592 > 2023$, $s_0 = \\frac{2023 \\cdot 2024}{2} - 2s$. If $s_0 > 2 \\cdot 593$, among the $593$ pairs $(1, s_0-1), (2, s_0-2), \\dots, (593, s_0-593)$, at least one is uncolored. If $s_0 \\le 2 \\cdot 593$, then Alice cannot have colored any of $1, 2, \\dots, 838$, so Bob can choose numbers from these to sum to $s_0$.\n\n**Remark:** For general $n \\ge 100$, the answer is $k = \\left\\lfloor \\frac{(2n + 1) - \\sqrt{n^2 + (n + 1)^2}}{2} \\right\\rfloor$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 12587,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $f : \\mathbb{R} \\to \\mathbb{R}$ be a function satisfying the following properties:\n\n1. For all $x, y \\in \\mathbb{R}$, $f(xy + x) = f(x)f(y) + f(x) + f(x + y - 1) + 1$.\n2. For all $x, y \\in \\mathbb{R}$, $f(xy + y) = f(x)f(y + 1) + f(x + y)$.\n3. $f(1) = 0$.\n4. $f(0) = -1$.\n\nFind all such functions $f$.",
    "options": [],
    "answer": "See solution",
    "solution": "Define $g(x) = f(x) + 1$ for each $x \\in \\mathbb{R}$. Then, from property (1), we have\n\n$$\ng(t + x) = g(t) + g(x)\n$$\nfor every $t, x \\in \\mathbb{R}$, so $g$ is additive.\n\nFurthermore, from property (2),\n$$\n\\begin{aligned}\ng(xy) - 1 &= (g(x) - 1)(g(y) - 1) + g(x + y - 1) - 1 \\\\\n&= g(x)g(y) - g(x) - g(y) + g(x + y - 1) \\\\\n&= g(x)g(y) - 1.\n\\end{aligned}\n$$\nThis implies that $g$ is multiplicative.\n\nAn additive and multiplicative function on $\\mathbb{R}$ is either identically zero or the identity function. Since $g$ is not identically zero, we get $g(x) = x$ for every $x \\in \\mathbb{R}$, so $f(x) = x - 1$ for every $x \\in \\mathbb{R}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 12588,
    "subject": "Mathematics (Olympiad)",
    "question": "In the Cartesian plane $Oxy$ we consider the points $A_1(40,1)$, $A_2(40,2)$, ..., $A_{40}(40,40)$, as well as the line segments $OA_1, OA_2, ..., OA_{40}$. A point of the Cartesian plane will be called \"good\" if its coordinates are integers and it lies in the interior of a line segment $OA_i$, $i = 1, 2, 3, ..., 40$. Also, a line segment from $OA_1, OA_2, ..., OA_{40}$ will be called \"good\" if it contains at least one \"good\" point. Determine the number of \"good\" points and the number of \"good\" segments.",
    "options": [],
    "answer": "See solution",
    "solution": "![](images/Hellenic_Mathematical_Competitions_2011_booklet_p7_data_b1612f7d3d.png)\n\nA point $M(k, l)$ will belong to the interior of the line segment $OA_i$ if and only if $\\overrightarrow{OM}$ and $\\overrightarrow{OA_i}$ have the same slope (with $k, l$ integers and $0 < k \\leq 40$), that is,\n\n$$\n\\frac{i}{40} = \\frac{l}{k} \\quad (\\text{with } k, l \\text{ integers and } 0 < k \\leq 40).\n$$\n\nFor the line segment $OA_i$ to be \"good,\" it is necessary and sufficient that the fraction $\\frac{i}{40}$ is reducible (then we have a fraction $\\frac{l}{k}$ with integer terms equivalent to $\\frac{i}{40}$, and its terms give the coordinates of the \"good\" point $M(k, l)$).\n\nHence, if $\\gcd(40, i) > 1$, then the line segment $OA_i$ is \"good\" and we have $\\gcd(40, i) - 1$ \"good\" points on the segment $OA_i$. For example, for $A_2(40, 2)$, the corresponding \"good\" segment $OA_2$ contains the good point $(20, 1)$. For $A_4(40, 4)$, the corresponding \"good\" segment $OA_4$ contains the \"good\" points $(10, 1)$, $(20, 2)$, $(30, 3)$. Working in this way, we finally find 24 \"good\" segments and 140 \"good\" points.\n\nAn easy solution can be given using Euler's totient function $\\phi$. It is known that the number of positive integers less than or equal to $n$ and not relatively prime to $n$ is $n - \\phi(n)$. Since $40 = 5 \\cdot 2^3$, we have\n\n$$\n\\phi(40) = 40 \\left(1 - \\frac{1}{2}\\right) \\left(1 - \\frac{1}{5}\\right) = 40 \\cdot \\frac{1}{2} \\cdot \\frac{4}{5} = 16.\n$$\n\nHence, the number of good segments is $40 - \\phi(40) = 24$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 12589,
    "subject": "Mathematics (Olympiad)",
    "question": "Determine the largest possible number of primes among 100 consecutive natural numbers.",
    "options": [],
    "answer": "See solution",
    "solution": "The largest possible number of primes among 100 consecutive natural numbers is $26$.\n\nThere are $25$ primes among the numbers from $1$ to $100$. The number of primes in the next few intervals of length $100$ are shown in the table:\n\n| Interval      | Out | In  | Number of primes |\n|--------------|-----|-----|------------------|\n| $1,\\ldots,100$   |     |     | $25$              |\n| $2,\\ldots,101$   | $1$ | $101$ | $26$              |\n| $3,\\ldots,102$   | $2$ | $102$ | $25$              |\n| $4,\\ldots,103$   | $3$ | $103$ | $25$              |\n| $5,\\ldots,104$   | $4$ | $104$ | $25$              |\n| $6,\\ldots,105$   | $5$ | $105$ | $24$              |\n| $7,\\ldots,106$   | $6$ | $106$ | $24$              |\n\nThe largest number of primes in these intervals is $26$.\n\nTo show that there cannot be more than $26$ primes among $100$ consecutive natural numbers, consider any $100$ consecutive natural numbers, the least of which is greater than $7$. None of the numbers divisible by $2$, $3$, $5$, or $7$ is prime. We show that there are at least $74$ such numbers:\n\n- Every second number is divisible by $2$, so there are $50$ even numbers.\n- Every third number is divisible by $3$, so there are at least $33$ such numbers. Every second among them has already been counted as even, so there are at least $16$ new numbers divisible by $3$.\n- Every fifth number is divisible by $5$, so there are at least $20$ such numbers. Every second among them is even, so the number of odd numbers divisible by $5$ is $10$. Among them, every third is divisible by $3$, so there are at least $6$ numbers divisible by $5$ not counted yet.\n- Every seventh number is divisible by $7$, so there are at least $14$ such numbers. Every second among them is even, so there are at least $7$ odd numbers divisible by $7$. Every third among them is divisible by $3$ (eliminating at most $3$ numbers), and every fifth is divisible by $5$ (eliminating at most $2$ numbers). Hence, at least $2$ numbers not counted before are divisible by $7$.\n\nAltogether, we have at least $50 + 16 + 6 + 2 = 74$ composite numbers.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 12590,
    "subject": "Mathematics (Olympiad)",
    "question": "The natural numbers $m$ and $n$ are given such that $mn$ is divisible by $9$ but not by $27$. The $m \\times n$ rectangle is divided into corners of three cells each. Depending on the orientation of these corners, they are divided into four types, as shown in the figure below. Could it happen that the numbers of corners of each type are exact squares of some natural numbers?\n\n![](images/Ukraine2022-23_p16_data_0c70280611.png)\n\n*Fig. 8*",
    "options": [],
    "answer": "See solution",
    "solution": "**Answer:** No.\n\nWe will place the corner in a $2 \\times 2$ square. Depending on which cell of the square is not occupied, we will denote four types of corners. Let type (1) be the corners where the upper left cell is absent, type (2) the upper right one, type (3) the lower right one, and type (4) the lower left cell. Let $a_i$ be the number of corners of type $(i)$.\n\nNumber the columns from $1$ to $n$, and the rows from $1$ to $m$ (let the lower left cell of the board be in the first row and the first column), and write the sum of its column and row in each cell. Since $mn$ is divisible by $3$, the board can be divided into $1 \\times 3$ rectangles, each of which has a sum of numbers divisible by $3$, so the sum of numbers in all cells is divisible by $3$.\n\nNote that the sum of numbers in corners of types (1) and (3) is divisible by $3$. The sum of numbers in corners of type (2) gives a remainder of $2$ when divided by $3$, and type (4) a remainder of $1$. Thus, the sum of numbers on the board is congruent to $a_4 - a_2$ modulo $3$, so $a_2 \\equiv a_4 \\pmod{3}$. Similarly, $a_1 \\equiv a_3 \\pmod{3}$, since we can make a numbering where the lower right cell is in the first row and the first column.\n\nSuppose not all $a_i$ are divisible by $3$. Without loss of generality, let $a_1$ not be divisible by $3$. If $a_1$ is a square, then $a_1 \\equiv a_3 \\equiv 1 \\pmod{3}$. Since $mn$ is divisible by $9$, the number of corners is divisible by $3$. Therefore, $a_1 + a_2 + a_3 + a_4$ is divisible by $3$, which means $a_2 \\equiv a_4 \\equiv 2 \\pmod{3}$, so $a_2$ is not a square.\n\nIf all $a_i$ are divisible by $3$, and they are squares, then they are all divisible by $9$, so the number of corners is divisible by $9$, hence $mn$ is divisible by $27$, which contradicts the condition. Therefore, such a situation is impossible.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 12591,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $m, n$ be positive integers and $a, b$ be positive real numbers different from $1$. Suppose that $m > n$ and\n$$\n\\frac{a^{m+1}-1}{a^m-1} = \\frac{b^{n+1}-1}{b^n-1} = c.\n$$\nShow that $a^m c^n > b^n c^m$.",
    "options": [],
    "answer": "See solution",
    "solution": "Note that\n$$\n\\frac{b^{n+1}-1}{b^n-1} = \\frac{b^n + b^{n-1} + \\dots + 1}{b^{n-1} + \\dots + 1}\n$$\nHence,\n$$\nc(b^{n-1} + \\dots + 1) = b^n + b^{n-1} + \\dots + 1 = b(b^{n-1} + \\dots + 1) + 1 \\Rightarrow c > b.\n$$\nWe also have\n$$\nc(b^{n-1} + \\dots + 1) = b^n + b^{n-1} + \\dots + 1 \\Rightarrow (c-1)(b^{n-1} + \\dots + 1) = b^n \\Rightarrow (c-1)(b^n + \\dots + 1) = c b^n.\n$$\nThen, $c > 1$ and defining $z = (c b^n)^{1/(n+1)}$, we see that $z > b$ and $z^{n+1} = (c-1)(b^n + \\dots + 1) < (c-1)(z^n + \\dots + 1)$. In other words, for the polynomial $P(x) = x^{n+1} - (c-1)(x^n + \\dots + 1)$, we have $P(z) < 0$. Since the leading coefficient of $P(x)$ is positive, there is a number $b_1$ with $b_1 > z$ and $P(b_1) = 0$. For this number,\n$$\nb_1^{n+1} = (c-1)(b_1^n + \\dots + 1) \\Rightarrow b_1^{n+1} + b_1^n + \\dots + 1 = c(b_1^n + \\dots + 1) \\Rightarrow \\frac{b_1^{n+2}-1}{b_1^{n+1}-1} = c\n$$\nand\n$$\nb_1^{n+1} > z^{n+1} = c b^n.\n$$\nSet $b = b_0$. This number satisfies $\\frac{b_0^{n+1}-1}{b_0^n-1} = c$ and the arguments above define a number $b_1$ from $b_0$ satisfying $\\frac{b_1^{n+2}-1}{b_1^{n+1}-1} = c$ and $b_1^{n+1} > c b_0^n = c b^n$. Use the same arguments to define $b_2$ from $b_1$, so that $\\frac{b_2^{n+3}-1}{b_2^{n+2}-1} = c$ and $b_2^{n+2} > c b_1^{n+1} > c^2 b^n$. Inductively, use the same arguments to define a number $b_k$ from $b_{k-1}$ so that $\\frac{b_k^{n+k}-1}{b_k^{n+k}-1} = c$ and $b_k^{n+k} > c^k b^n$. If we prove $b_{m-n} = a$, the proof will be complete.\n\nWe have $b_{m-n}^m = (c-1)(b_{m-n}^{m-1} + \\dots + 1)$ and $a^m = (c-1)(a^{m-1} + \\dots + 1)$. Hence,\n$$\n\\frac{1}{c-1} = \\frac{1}{b_{m-n}} + \\dots + \\frac{1}{b_{m-n}^m} = \\frac{1}{a} + \\dots + \\frac{1}{a^m}.\n$$\nThis clearly shows that neither of $b_{m-n}$ and $a$ is greater than the other, so they are equal. $\\square$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 12592,
    "subject": "Mathematics (Olympiad)",
    "question": "a) Show that if $n$ is an odd perfect number, then $n$ must have the form $n = p^{\\alpha} m^2$, where $p$ is a prime congruent to $1$ modulo $4$ and $\\alpha$ is congruent to $1$ modulo $4$.\n\nb) Prove that if $\\frac{n(n+1)}{2}$ is an odd perfect number, then $n = 7$ is the only possibility.",
    "options": [],
    "answer": "See solution",
    "solution": "Then $4 \\mid n$, a contradiction. Thus, $n$ is odd.\n\nRecall a well-known lemma: even perfect numbers have the form $2^s (2^{s+1} - 1)$, where $2^{s+1} - 1$ is prime.\n\nApplying the lemma, there exists a positive integer $s$ such that $n = 2^s (2^{s+1} - 1) + 1$ and $2^{s+1} - 1$ is prime. Hence,\n\n$$\n\\frac{n(n+1)}{2} = [2^s(2^{s+1}-1)+1][2^{s-1}(2^{s+1}-1)+1]\n$$\n\nis an odd perfect number, so\n\n$$\n[2^s(2^{s+1}-1)+1][2^{s-1}(2^{s+1}-1)+1] = p_2^{s_2} m_2^2,\n$$\n\nwhere $p_2$ is a prime congruent to $1$ modulo $4$, $s_2 \\equiv 1 \\pmod{4}$, and $m_2$ is an odd positive integer not divisible by $p_2$. Since $\\gcd(2^s(2^{s+1}-1)+1, 2^{s-1}(2^{s+1}-1)+1) = 1$, at least one of these numbers must be a square.\n\nWe will prove that neither can be a square when $s \\ge 2$. If $s \\ge 2$, these numbers are odd. Suppose $2^s(2^{s+1}-1)+1 = b^2$ for some odd $b > 1$. Then\n\n$$\n2^s (2^{s+1} - 1) = (b-1)(b+1),\n$$\n\nand $2^{s+1} - 1$ is prime, so $2(2^{s+1} - 1)$ divides $b-1$ or $b+1$. But $2(b-1) \\ge b+1$, so\n\n$$\n(b-1)(b+1) \\ge 2(2^{s+1}-1)^2 > 2^s(2^{s+1}-1) = (b-1)(b+1).\n$$\n\nThis is a contradiction.\n\nThe case $2^{s-1}(2^{s+1}-1)+1$ is similar. Thus, $s=1$ and $n=7$ is the only possibility. $\\square$",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 12593,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $n$ be a positive integer greater than or equal to 3. Determine all those $n$ for which there exists an $n$-gon with all the sides having the same length and with all its interior angles having either $120^\\circ$ or $240^\\circ$.",
    "options": [],
    "answer": "See solution",
    "solution": "We will show that $n = 6$ and $n = 2k$, where $k$ is an integer greater than or equal to 5, satisfy the condition of the problem and that there are no other $n$ satisfying the requirement.\n\nFirst, let us show that $n = 6$ and $n = 2k$ with $k \\ge 5$ satisfy the condition of the problem.\n\n![](images/Japan_2010_p33_data_1eade6d2c6.png)\n\nLet us consider figures obtained by piecing together a number of regular hexagons as shown in the diagram above. The figure on the left in the diagram above is a $(6+4m)$-gon, and the one on the right is a $(12+4m)$-gon. So, we see that $n = 6+4m$, $12+4m$, where $m$ is a non-negative integer, satisfy the condition of the problem. Since the numbers 6 and $2k$ with $k$ being an integer greater than or equal to 5 can be expressed in the form $6+4m$ or $12+4m$, we see that these numbers satisfy the condition of the problem.\n\nWe next show that no other numbers satisfy the condition of the problem. Let us suppose that a number $n$ greater than or equal to 3 satisfies the condition of the problem, and denote by $\\ell$ the number (greater than or equal to 0) of inner angles with $240^\\circ$ of the $n$-gon satisfying the condition of the problem. Then, since the sum of all the inner angles of an $n$-gon is $(n-2) \\times 180^\\circ$, we have\n\n$$\n(n - 2) \\times 180 = (n - \\ell) \\times 120 + \\ell \\times 240,\n$$\n\nfrom which we conclude that $n = 2\\ell + 6$. This means that if $n$ is an odd integer or is equal to 4, then it does not satisfy the condition of the problem. Therefore, it remains only to show that if $n = 8$, then the condition is not satisfied. So, let us suppose that $n = 8$. It then follows that $\\ell = 1$, and thus the octagon which satisfies the condition must have 7 inner angles with $120^\\circ$ each, and 1 inner angle with $240^\\circ$. It must be the case, then, that the 7 inner angles with $120^\\circ$ would appear consecutively in the figure of this octagon, but this is impossible, since 6 inner angles with $120^\\circ$ appearing consecutively will form a regular hexagon because of the assumption that the length of each side must be the same. Thus, we conclude that $n = 8$ cannot satisfy the condition of the problem.\n\nThis establishes our claim that $n$ must equal 6 or be of the form $2k$ with $k$ being greater than or equal to 5 in order to satisfy the condition of the problem.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 12594,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $m$ be a positive integer greater than one. Prove that the number $m^4 + 1$ does not have a divisor between $m^2$ and $(m+1)^2$.",
    "options": [],
    "answer": "See solution",
    "solution": "Assume that there exists a divisor $n$ of $m^4 + 1$ such that $m^2 < n < (m+1)^2$. Let $n = m^2 + x$ where $1 \\leq x \\leq 2m$. Then $m^4 + 1 = (m^2 + x)(m^2 - y)$ for some $y \\in \\mathbb{Z}_{\\geq 1}$. Hence $1 + xy = m^2(x - y)$ and it is obvious that $x > y \\geq 1$.\n\n1. If $x - y = 1$, then $1 + (y+1)y = m^2 \\iff 3 + (2y+1)^2 = (2m)^2 \\iff 3 = (2m+2y-1)(2m-2y-1)$. Since $2m+2y-1 \\geq 2m+1 \\geq 5$, there is no solution for $m$ and $y$.\n\n2. If $x-y=2$, then $1+y^2+2y=2m^2 \\implies (y+1)^2=2m^2$. This has no positive integer solution as $\\sqrt{2}$ is irrational.\n\n3. If $x-y=3$, then $1+y^2+3y=3m^2 \\implies y^2+1=3(m^2-y)$. But $y^2+1 \\not\\equiv 0 \\pmod{3}$, which is a contradiction.\n\n4. If $x-y \\geq 4$, then $1+x^2 > 1+xy=m^2(x-y) \\geq 4m^2 = (2m)^2$. Since $x \\leq 2m$, we get that $x=2m$. But $m^2+2m$ does not divide $m^4+1$, which is a contradiction.\n\nConsequently, $m^4+1$ has no divisor $n$ such that $m^2 < n < (m+1)^2$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 12595,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $n = 8$. Determine whether there exists a labelling of an $8 \\times 8$ chessboard such that the following condition is fulfilled: the difference of any two rook products is always divisible by $65$.\n\nLet $n = 10$. Determine whether there exists a labelling of a $10 \\times 10$ chessboard such that the following condition is fulfilled: the difference of any two rook products is always divisible by $101$.",
    "options": [],
    "answer": "See solution",
    "solution": "(a) No, there is no such labelling.\n\nOn the contrary, we show that for every labelling there exist two rook products whose difference is not divisible by $65$. Suppose that an $8 \\times 8$ chessboard is labelled with the numbers $1, 2, \\ldots, 64$ such that no number is used twice.\n\nWe can construct a rook product that is divisible by $13$ by placing a rook on the square with the label $13$ and the other seven rooks non-attackingly, but otherwise arbitrarily.\n\nWe can construct a rook product that is not divisible by $13$ as follows. Notice that only four labels are divisible by $13$, namely $13, 26, 39,$ and $52$. These four labels are located in at most four rows; we denote the index set of these rows $R \\subseteq \\{1, \\ldots, 8\\}$. Similarly, there are at least four columns that do not contain any of these four labels; we denote the index set of these columns $C \\subseteq \\{1, \\ldots, 8\\}$. Since $|R| \\leq |C|$, it is possible to place non-attacking rooks in rows $R$ using only columns from $C$. The remaining rooks are placed in the remaining rows non-attackingly, but otherwise arbitrarily. The resulting rook product is not divisible by $13$ since the rooks avoid the squares whose labels are divisible by $13$.\n\nThe difference of the two rook products is not divisible by $13$, since one rook product is divisible by $13$ whereas the other one is not. Hence the difference is not divisible by $65$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 12596,
    "subject": "Mathematics (Olympiad)",
    "question": "The sides and diagonals of a regular $n$-gon are colored in $k \\geq 3$ 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.",
    "options": [],
    "answer": "See 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.\n\nDenote 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 $1$ connectivity components 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 $2$.\n\nIf $A_1B$ 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; it 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.\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": 12597,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $n \\ge 2$ be a positive integer. Let $a_1, a_2, \\dots, a_n$ be a sequence of positive integers such that\n$$\n\\gcd(a_1, a_2), \\gcd(a_2, a_3), \\dots, \\gcd(a_{n-1}, a_n)\n$$\nis a strictly increasing sequence. Find, in terms of $n$, the maximum possible value of\n$$\n\\frac{1}{a_1} + \\frac{1}{a_2} + \\dots + \\frac{1}{a_n}.\n$$\nover all such sequences.",
    "options": [],
    "answer": "See solution",
    "solution": "We claim the maximum possible value is $2$.\n\nTo see that this is achievable, let the sequence $(a_i)_{i=1}^n$ be $1, 2, 4, \\dots, 2^{n-2}, 2^{n-2}$. Then $\\gcd(a_i, a_{i+1}) = 2^{i-1}$, which is an increasing sequence, and it is easy to check that $\\sum \\frac{1}{a_i} = 2$.\n\nWe now show this is the maximum.\n\nLet $d_i = \\gcd(a_i, a_{i+1})$. Since $d_{i-1} < d_i$, we have that $\\frac{a_i}{d_{i-1}} > \\frac{a_i}{d_i}$. But these are both integers, so we get that\n$$\n1 \\le \\frac{a_i}{d_{i-1}} - \\frac{a_i}{d_i} \\implies \\frac{1}{a_i} \\le \\frac{1}{d_{i-1}} - \\frac{1}{d_i}.\n$$\nAdding everything up, we get that\n$$\n\\sum_{i=1}^{n} \\frac{1}{a_i} \\le \\frac{1}{a_1} + \\frac{1}{d_1} - \\frac{1}{d_{n-1}} + \\frac{1}{a_n}.\n$$\nBut $d_{n-1} \\mid a_n$, so $d_{n-1} \\le a_n$, and $\\frac{1}{d_{n-1}} \\ge \\frac{1}{a_n}$. Thus, this sum is at most $\\frac{1}{a_1} + \\frac{1}{d_1} \\le 2$.\n\n$\\square$",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 12598,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $x, y, z$ be nonnegative real numbers. Then for any $r > 0$,\n\n$$\nx^r(x-y)(x-z) + y^r(y-z)(y-x) + z^r(z-x)(z-y) \\geq 0.\n$$\n\nEquality holds if and only if $x = y = z$ or if two of $x, y, z$ are equal and the third is equal to $0$.",
    "options": [],
    "answer": "See solution",
    "solution": "The proof of the inequality is rather simple. Since the inequality is symmetric in the three variables, we may assume without loss of generality that $x \\geq y \\geq z$. Then the given inequality may be rewritten as\n\n$$\n(x - y)[x^r(x - z) - y^r(y - z)] + z^r(x - z)(y - z) \\geq 0,\n$$\n\nand every term on the left-hand side is clearly nonnegative. The first term is positive if $x > y$, so equality requires $x = y$, as well as $z^r(x-z)(y-z) = 0$, which gives either $x = y = z$ or $z = 0$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 12599,
    "subject": "Mathematics (Olympiad)",
    "question": "Each point in the plane is labelled with a real number. For each cyclic quadrilateral $ABCD$ in which the line segments $AC$ and $BD$ intersect, the sum of the labels at $A$ and $C$ equals the sum of the labels at $B$ and $D$.\n\nProve that all points in the plane are labelled with the same number.",
    "options": [],
    "answer": "See solution",
    "solution": "For any point $P$ in the plane, let $f(P)$ denote its label. Consider two points $A$ and $B$, and construct any cyclic pentagon $ABPQR$ whose vertices lie in that order.\n\nThen we have $f(A) + f(Q) = f(P) + f(R) = f(B) + f(Q)$.\n\nThis implies that the arbitrarily chosen points $A$ and $B$ satisfy $f(A) = f(B)$.\n\nSo it is necessarily true that all points in the plane are labelled with the same number.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 12600,
    "subject": "Mathematics (Olympiad)",
    "question": "A dodecahedron and a vertex $X$ of the dodecahedron are given. An ant started from $X$, walked along the edges of the dodecahedron, passing each of the vertices of the dodecahedron except $X$ just once, and returned to $X$. How many such routes exist? We consider a route and its reversal to be different.",
    "options": [],
    "answer": "See solution",
    "solution": "$60$",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 12601,
    "subject": "Mathematics (Olympiad)",
    "question": "The country has $n \\geq 3$ airports, some pairs of which are connected by bidirectional flights. Every day, the government closes the airport from which the largest number of flights is flying. What is the maximum number of days this can continue?",
    "options": [],
    "answer": "See solution",
    "solution": "For $n = 3$, the process lasts $1$ day; for $n \\geq 4$, it lasts $n - 3$ days.\n\nWe interpret the problem using graph theory, considering the complement graph for convenience. Each day, the vertex of strictly lowest degree is removed.\n\nWhen only $2$ vertices remain, no further removals are possible. For $n = 3$, this bound is achieved by connecting two of the three vertices with an edge.\n\nWe show that if at some point there were $4$ vertices, there will always be at least $3$ days of removals. Suppose the vertices are $A$, $B$, $C$, $D$, and first $D$ is removed, then $C$. Then $D$ cannot be connected to $A$ or $B$. Also, in a graph of three vertices $A$, $B$, $C$, the vertex $C$ cannot be connected to $A$ or $B$. Thus, in the graph on these four vertices, the degree of $D$ was at least that of $C$, which is a contradiction.\n\nFor example, consider the following graph: a chain of vertices $A_1, A_2, A_3, \\ldots, A_n$, where every two adjacent vertices are connected, and in addition $A_{n-2}$ is connected to $A_n$:\n\n![](images/Ukraine2022-23_p33_data_73062726f0.png)\n\nIt is clear that the vertices will be removed one by one, in order $A_1, A_2, A_3, \\ldots, A_{n-3}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 12602,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all polynomials $P(x)$ with integer coefficients and all integers $d$ such that there exist infinitely many pairs of integers $(x, y)$ with $x \\ne y$ satisfying both $x - y = a$ (for some fixed integer $a$) and $P(x) - P(y) = d$.",
    "options": [],
    "answer": "See solution",
    "solution": "Suppose there are infinitely many pairs $(x, y)$ such that $x - y = a$ and $P(x) - P(y) = d$. Note that\n\n$$\nP(x) - P(x - a) - d\n$$\n\nis a polynomial. The only way it can have infinitely many zeros is if it is identically zero. Thus, $P(x) = \\frac{d}{a}x + b$ for some integer $b$.\n\nThis gives the first set of solutions: arbitrary integer $d$ and polynomial $P(x) = \\frac{d}{a}x + b$, where $a$ divides $d$ and $b$ is arbitrary. This includes the case $d = 0$ and $P(x)$ constant.\n\nFor the rest, assume $d = 0$. If the degree of $P(x)$ is odd, then for large $x$, $P(x)$ is strictly monotonic, so $P(x) - P(y) = 0$ only for finitely many $(x, y)$ with $x \\ne y$.\n\nThus, $P(x)$ must be of even degree. Now, for large $x$, $P(x)$ is increasing for $x \\ge A$ and decreasing for $x \\le B$, or vice versa. Infinitely many solutions $P(x) - P(y) = 0$ with $x \\ne y$ can only occur if $x + y$ is constant. Let\n\n$$\nP(x) = a x^n + b x^{n-1} + \\dots\n$$\n\nAssume $a > 0$. For large $x$, $P(x) - P(y)$ is positive unless $an(x + y) = -2b$. In this case,\n\n$$\nP(x) - P\\left(\\frac{-2b}{an} - x\\right)\n$$\n\nis a polynomial with infinitely many zeros, so it is constant. Thus,\n\n$$\nP(x) = P\\left(\\frac{-2b}{an} - x\\right)\n$$\n\nfor all $x$, which implies $Q(x) = P\\left(\\frac{-b}{an} + x\\right)$ is even: $Q(x) = Q(-x)$. Therefore, $Q(x)$ contains only even powers of $x$, i.e., $Q(x) = R(x^2)$. This gives the second set of solutions: $P(x) = R((x-c)^2)$, where $2c$ is integer, and $d = 0$.",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 12603,
    "subject": "Mathematics (Olympiad)",
    "question": "For each positive integer $x$, let $\\sigma(x)$ be the sum of its positive divisors and $\\tau(x)$ be the number of positive divisors of $x$.\n\nFind all positive integers $n$ such that $\\sigma(\\tau(n)) = n$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $\\tau(n) = k$. For any divisor $d$ of $n$ with $d \\leq \\sqrt{n}$, $\\frac{n}{d}$ is also a divisor and $\\frac{n}{d} \\geq \\sqrt{n}$, so $\\tau(n) \\leq 2\\sqrt{n}$ for any $n \\in \\mathbb{N} \\setminus \\{0\\}$.\n\nIf $1 = d_1 < d_2 < \\dots < d_k = n$ are the divisors of $n$, then\n\n$$\n2\\sigma(n) = (d_1 + d_k) + (d_2 + d_{k-1}) + \\dots + (d_k + d_1) = \\left(d_1 + \\frac{n}{d_1}\\right) + \\dots + \\left(d_k + \\frac{n}{d_k}\\right).\n$$\n\nIf $d$ divides $n$, then $1 + n - d - \\frac{n}{d} = \\frac{n(d-1)}{d} - (d-1) = \\frac{(n-d)(d-1)}{d} \\geq 0$. Thus, $\\sigma(n) \\leq \\frac{1}{2}\\tau(n)(n+1)$. Therefore,\n\n$$\nn = \\sigma(\\tau(n)) \\leq \\frac{1}{2}\\tau(\\tau(n)) (\\tau(n)+1) \\leq \\frac{1}{2} \\cdot 2\\sqrt{\\tau(n)} (2\\sqrt{n}+1) \\leq \\sqrt{2\\sqrt{n}} (2\\sqrt{n}+1).\n$$\n\nThis leads to\n\n$$\nn^2 \\leq 2\\sqrt{n}(4n + 4\\sqrt{n} + 1) \\implies n\\sqrt{n} \\leq 2(4n + 4\\sqrt{n} + 1) \\implies \\sqrt{n} \\leq 8 + \\frac{8}{\\sqrt{n}} + \\frac{2}{n}.\n$$\n\nSuppose $n \\geq 81$. Then $\\sqrt{n} \\geq 9$ and $8 + \\frac{8}{9} + \\frac{2}{81} < 9 \\leq \\sqrt{n}$, a contradiction. Hence, $n \\leq 80$.\n\nThus, $n$ can only have one of the following forms: $1, p, p^2, p^3, p^4, p^5, pq, p^2q, p^3q, p^4q, p^2q^2, p^3q^2, pqr, p^2qr$, where $p, q, r$ are primes.\n\nEach such $n$ has at most 12 divisors. Checking all $\\tau(n) \\in \\{1, 2, \\dots, 12\\}$, the only solutions are $n = 1, 3, 4, 12$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 12604,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all functions $f : \\mathbb{R} \\to \\mathbb{R}$ satisfying the following conditions:\n\n1. There is at most one number $a$ such that $f(a) = 0$.\n2. $f(f(x+y)f(x-y) + 25xy + y^2) = 24y f(x) + x f(x+y)$ for all real numbers $x, y$.",
    "options": [],
    "answer": "See solution",
    "solution": "First, set $y = 0$:\n\n$$f(f(x)^2) = x f(x).$$\n\nSet $x = 0$:\n\n$$f(f(y)f(-y) + y^2) = 24y f(0).$$\n\nNow, replace $y$ by $-y$; the left side does not change, so:\n\n$$24y f(0) = -24y f(0) \\implies f(0) = 0.$$ \n\nFrom $f(f(y)f(-y) + y^2) = 0$, we get $f(y)f(-y) = -y^2$.\n\nSet $y = -x$:\n\n$$f(-24x^2) = -24x f(x).$$\n\nChanging $x$ to $-x$ shows $f(x)$ is odd. Thus $f(y)^2 = y^2$, so $f(x) = x$ or $f(x) = -x$ for all $x$.\n\nNow, consider possible mixtures. Let $A = \\{a \\in \\mathbb{R} \\setminus \\{0\\} : f(a) = a\\}$ and $B = \\{b \\in \\mathbb{R} \\setminus \\{0\\} : f(b) = -b\\}$. Suppose both are non-empty. Take $a \\in A$, $b \\in B$.\n\nLet $x = a$, $y = b - a$:\n\n- $f(x + y) = f(b) = -b$\n- $f(x - y) = f(2a - b)$\n\nSo:\n\n$$\n\\text{LHS} = f(-b f(2a - b) + 2a(b - a) + (b - a)^2)\n$$\n\n$$\n\\text{RHS} = (b - a)a + a(-b) = -a^2\n$$\n\nFor each pair $a, b$, there are 4 sign choices, so for fixed $a$, at most 8 choices for $b$, implying $|B| \\leq 8$. Similarly, $|A| \\leq 8$. But $A \\cup B = \\mathbb{R} \\setminus \\{0\\}$, a contradiction.\n\nHence, the only solutions are $f(x) = x$ for all $x$ and $f(x) = -x$ for all $x$. $\\square$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 12605,
    "subject": "Mathematics (Olympiad)",
    "question": "How many ways are there to color an $n \\times 2$ board with four colors such that every $2 \\times 2$ square contains all four colors?",
    "options": [],
    "answer": "See solution",
    "solution": "The answer is $3 \\cdot 2^{n+2} - 24$.\n\nSuppose there are at least three different colors $A, B, C$ in the first row. Then they occur consecutively, say in the order $ABC$. Then the cell below the $B$ has the fourth color $D$, and all other cells are determined by the first row, because three cells in a $2 \\times 2$ square determine the other one, and we can fill the board from right to left and left to right, beginning at the $D$. Notice that the cells below $ABC$ are $CDA$, so the next row always has three different colors and can be filled. Finally, notice that the colors alternate in the columns in between $A, C$ and $B, D$, so all columns have two different colors. There are $\\binom{4}{2} = 6$ ways to choose the two colors for the top two cells in the first column and $2^n$ ways to choose the two colors for the top two cells in each next column: if we chose $\\{A, B\\}$ for the first column, then we can choose $(A, B)$ or $(B, A)$ for the first column, $(C, D)$ or $(D, C)$ for the next column, $(A, B)$ or $(B, A)$ for the next column, and so on. We only need to exclude the cases where there are only two colors in the first row: in this case, we still have 6 choices for the first column, and 2 choices for each of the first two columns orders. The other orders are determined to be the same as the preceding ones, so to repeat the pattern. So we must exclude $6 \\cdot 2 \\cdot 2 = 24$ cases.\n\nNow we deal with the case in which there are only two colors in the first row. There are $4 \\cdot 3$ ways to choose the colors in the first row. Then each following row has the other two colors, alternated in one of two ways. So in this case we have $12 \\cdot 2^{n-1}$ colorings.\n\nSo the grand total is $6 \\cdot 2^n - 24 + 12 \\cdot 2^{n-1} = 3 \\cdot 2^{n+2} - 24$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 12606,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all functions $f: \\mathbb{R} \\to \\mathbb{R}$ satisfying, for all real $x$ and $y$,\n\n$$\nf(2^x + 2y) = 2^y f(f(x)) f(y).\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "The solutions are $f(x) = 0$ and $f(x) = 2^x$.\n\nSubstitute $y = -2^x$:\n\n$$\nf(-2^x) = \\frac{1}{2^{2x}} f(f(x)) f(-2^x).\n$$\n\nThus, for each $x$, either $f(-2^x) = 0$ or $f(f(x)) = 2^{2^x}$.\n\nSubstitute $y = -2^{x-1}$:\n\n$$\nf(0) = \\frac{1}{2^{2^{x-1}}} f(f(x)) f(-2^{x-1}).\n$$\n\nSo, if $f(-2^z) = 0$ for some $z$, then $f(0) = 0$. Taking $x = 0$ in the original equation gives $f(1 + 2y) = 0$, so $f \\equiv 0$.\n\nAssume now that $f(-2^z) \\neq 0$ for all $z$. Then\n\n$$\nf(f(x)) = 2^{2^x}\n$$\n\nfor all $x$. Taking $y = 0$ in the original equation and applying the above,\n\n$$\nf(2^x) = f(f(x)) f(0) = 2^{2^x} f(0),\n$$\n\nso\n\n$$\nf(z) = 2^z f(0)\n$$\n\nfor all positive $z$. Similarly, applying the above to $f(-2^{x-1})$ gives\n\n$$\nf(-2^{x-1}) = 2^{-2^{x-1}} f(0),\n$$\n\nso $f(z) = 2^z f(0)$ for all $z$.\n\nIf $f(0) = 0$, then $f(x) = 0$ for all $x$. Otherwise, taking $x = 0$ in $f(f(0)) = 2^{2^0}$ and using $f(0) = 1$ gives\n\n$$\n2 = f(f(0)) = 2^{f(0)} \\cdot f(0),\n$$\n\nwhich is only possible for $f(0) = 1$. Thus, the only nonzero solution is $f(x) = 2^x$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 12607,
    "subject": "Mathematics (Olympiad)",
    "question": "We call a polynomial *mixed* if it has both positive and negative coefficients (we do not consider zero to be either positive or negative). Is it true that the product of two mixed polynomials is always a mixed polynomial?",
    "options": [],
    "answer": "See solution",
    "solution": "Here is a counterexample:\n\n$$\nf(x) = x^5 + x^4 - x^3 - x^2 + x + 1 = (x^4 - x^2 + 1)(x + 1)\n$$\n\n$$\ng(x) = x^4 - x^3 + 2x^2 - x + 1 = (x^2 - x + 1)(x^2 + 1)\n$$\n\nBoth of these polynomials are mixed, but their product is not mixed, since\n\n$$\n\\begin{aligned}\nf(x)g(x) &= (x^4 - x^2 + 1)(x + 1)(x^2 - x + 1)(x^2 + 1) \\\\\n&= (x^6 + 1)(x^3 + 1) \\\\\n&= x^9 + x^6 + x^3 + 1.\n\\end{aligned}\n$$\n\nThe product $f(x)g(x)$ has only positive coefficients, so it is not mixed.",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 12608,
    "subject": "Mathematics (Olympiad)",
    "question": "Sea $ABC$ un triángulo, con $AB < AC$, y sea $\\Gamma$ su circuncírculo. Sean $D$, $E$ y $F$ los puntos de tangencia del incírculo con $BC$, $CA$ y $AB$, respectivamente. Sea $R$ el punto de $EF$ tal que $DR$ es una altura del triángulo $DEF$ y sea $S$ el punto de corte de la bisectriz exterior del ángulo $\\angle BAC$ con $\\Gamma$. Probar que $AR$ y $SD$ se cortan sobre $\\Gamma$.",
    "options": [],
    "answer": "See solution",
    "solution": "Sea $X$ el segundo punto de corte de $AR$ con $\\Gamma$. Por comodidad, pongamos $\\beta = \\angle ABC$ y $\\gamma = \\angle BCA$. Se tiene entonces que\n\n$$\n\\frac{BX}{XC} = \\frac{\\sin \\angle BAX}{\\sin \\angle CAX} = \\frac{FR}{RE} = \\frac{FD \\sin(\\gamma/2)}{DE \\sin(\\beta/2)} = \\frac{BD \\sin \\beta \\sin(\\gamma/2) \\cos(\\gamma/2)}{DC \\sin(\\beta/2) \\cos(\\beta/2) \\sin \\gamma} = \\frac{BD}{DC},\n$$\n\ndonde en la primera igualdad se ha usado el teorema del seno en los triángulos $ABX$ y $ACX$; en la segunda que $AFE$ es isósceles; en la tercera el teorema del seno en $FDR$ y $DRE$; y en la cuarta nuevamente el teorema del seno en $BDF$ y $CDE$.\n\nSea ahora $Y$ el segundo punto de corte de $SD$ con $\\Gamma$. Tenemos que $YD$ es una bisectriz del ángulo $\\angle BYC$ y por tanto podemos aplicar el teorema de la bisectriz y deducir que\n\n$$\n\\frac{BY}{YC} = \\frac{BD}{DC}.\n$$\n\nPor tanto, tenemos que $X$ e $Y$ son dos puntos sobre el arco de $BC$ que no contiene a $A$ y que satisfacen $\\frac{BX}{XC} = \\frac{BY}{YC}$, lo que implica necesariamente que $X = Y$ (dado que $X$ e $Y$ están sobre el círculo de Apolonio de $B$ y $C$, que corta en un único punto al arco $BC$ que no contiene a $A$). Esto demuestra lo que pedía el enunciado. $\\square$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 12609,
    "subject": "Mathematics (Olympiad)",
    "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\n$$\nx_1 + x_2,\\ x_2 + x_3,\\ \\dots,\\ x_{n-1} + x_n,\\ x_n + x_1\n$$\n\nare all in the set $A$.",
    "options": [],
    "answer": "See solution",
    "solution": "**Solution**\n\nLet $m_1 = x_1 + x_2$, $m_2 = x_2 + x_3$, $\\dots$, $m_{n-1} = x_{n-1} + x_n$, $m_n = x_n + x_1$.\n\nFirst, note that $m_1 \\neq m_2$, otherwise $x_1 = x_3$, which contradicts the fact that the $x_i$ are distinct. Similarly, $m_i \\neq m_{i+1}$ for $i = 1, 2, \\dots, n$, where $m_{n+1} = m_1$. It follows that $k \\ge 2$.\n\nFor $k = 2$, let $A = \\{a, b\\}$, where $a \\neq b$. It follows that\n\n$$\n\\begin{cases}\nx_1 + x_2 = a, \\\\\nx_2 + x_3 = b, \\\\\n\\vdots \\\\\nx_{n-1} + x_n = a, \\\\\nx_n + x_1 = b,\n\\end{cases}\n\\quad \\text{(if $n$ is even)}\n$$\n\nor\n\n$$\n\\begin{cases}\nx_1 + x_2 = a, \\\\\nx_2 + x_3 = b, \\\\\n\\vdots \\\\\nx_{n-1} + x_n = b, \\\\\nx_n + x_1 = a,\n\\end{cases}\n\\quad \\text{(if $n$ is odd)}\n$$\n\nFor the odd case, we have $x_n = x_2$, which is possible. For the even case, it follows that\n\n$$\n\\begin{aligned}\n\\frac{n}{2}a &= (x_1 + x_2) + (x_3 + x_4) + \\dots + (x_{n-1} + x_n) \\\\\n&= (x_2 + x_3) + (x_4 + x_5) + \\dots + (x_n + x_1) \\\\\n&= \\frac{n}{2}b,\n\\end{aligned}\n$$\n\nand hence $a = b$, which is impossible since $a \\neq b$. It follows that $k \\ge 3$.\n\nFor $k = 3$, one can construct a valid example as follows:\n\ndefine $x_{2k-1} = k$ ($k \\ge 1$) and $x_{2k} = n+1-k$ ($k \\ge 1$). When $n$ is even,\n\n$$\nx_i + x_{i+1} =\n\\begin{cases}\nn+1, & \\text{if } i \\text{ is odd} \\\\\nn+2, & \\text{if } i \\text{ is even and } i < n \\\\\n\\frac{n}{2} + 2, & \\text{if } i = n, \\text{ where } x_{n+1} = x_n\n\\end{cases}\n$$\n\nWhen $n$ is odd,\n\n$$\nx_i + x_{i+1} =\n\\begin{cases}\nn+1, & \\text{if } i \\text{ is odd and } i < n \\\\\nn+2, & \\text{if } i \\text{ is even} \\\\\n\\frac{n-1}{2} + 2, & \\text{if } i = n, \\text{ where } x_{n+1} = x_n\n\\end{cases}\n$$\n\nTherefore, the smallest positive integer $k$ is $3$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 12610,
    "subject": "Mathematics (Olympiad)",
    "question": "定義 $f_k(n)$ 為正整數 $n$ 的所有正因數之 $k$ 次方總和，也就是\n\n$$\nf_k(n) := \\sum_{m \\mid n,\\ m>0} m^k.\n$$\n\n試找出所有正整數對 $(a, b)$，使得 $f_a(n) \\mid f_b(n)$ 對於所有正整數 $n$ 均成立。",
    "options": [],
    "answer": "See solution",
    "solution": "$a = b$ 為所有可能。\n\n首先帶入 $n = 2$，必須有 $1 + 2^a \\mid 1 + 2^b$。設 $b = aq + r$，其中 $0 \\leq r < a$，可知\n\n$$\n1 + 2^b = 1 + 2^{aq + r} \\equiv 1 + (-1)^q \\times 2^r \\pmod{1 + 2^a}.\n$$\n\n由於 $|1 + (-1)^q \\times 2^r| < 1 + 2^a$，必須有 $1 + (-1)^q \\times 2^r = 0$，故 $q$ 為奇數且 $r = 0$。換言之，$b$ 必須為 $a$ 的奇數倍。\n\n以下證明 $q = 1$。不妨設 $q$ 有奇質數 $p$，並令 $q = lp$，則 $b = lpa$。取 $n = 2^{p-1}$，則必須有\n\n$$\n\\frac{2^{pa} - 1}{2^a - 1} = 1 + 2^a + \\cdots + 2^{(p-1)a} = f_a(n) \\mid f_b(n) = 1 + 2^b + \\cdots + 2^{(p-1)b}.\n$$\n\n然而\n\n$$\n1 + 2^b + \\cdots + 2^{(p-1)b} \\equiv 1 + 1 + \\cdots + 1 = p \\pmod{2^{pa} - 1},\n$$\n\n而 $0 < p < 1 + 2^a + \\cdots + 2^{(p-1)a}$，故上述整除是不可能的，矛盾。\n\n故 $q$ 不能有奇質數，也就是 $q = 1$。證畢。",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 12611,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all functions $f : \\mathbb{Z} \\to \\mathbb{Z}$ that satisfy the following equation for any integers $a, b, c$:\n\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": "See solution",
    "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 $2f(2) - 2f(1) = (f(1))^2 + (f(-1))^2 = 2f(1)$. Hence, $f(2) = 2f(1)$.\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 $g(0) = g(1) = 0$.\n\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$$\n2(f(k) - f(-k)) = (f(k+1))^2 - (f(k-1))^2\n$$\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 the above equation, 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\n$$\nf(2k + 1) = \\frac{1}{2}(f(k + 1))^2 - \\frac{1}{2}(f(k - 1))^2 + 1\n$$\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$$\n2f(2k) - 2f(-2k) = (f(k + 2))^2 - (f(k - 2))^2\n$$\n\nBy substituting $k = 1$ into the previous equation, we have $f(3) = 3$.\n\nBy substituting $k = 2$ into the above, we obtain $(f(4))^2 = 2(f(4) - f(-4))$. If $f(4) = 0$, then substituting $k = 3$ into the earlier equation 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\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 $k = 0, 1, 2, \\dots, m-1$.\n\nFor $m = 2k + 1 \\geq 5$, using the previous equation, we obtain $f(2k + 1) = \\frac{1}{2}((f(k+1))^2 - (f(k-1))^2) + 1 = 2k + 1$.\n\nFor $m = 2k \\geq 6$, utilizing the earlier equation, 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 the earlier equation gives us\n\n$$\n2(f(m) - f(-m)) = (f(m + 1))^2 - (f(m - 1))^2 = 4m, \\Rightarrow f(-m) = -m.\n$$\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": 12612,
    "subject": "Mathematics (Olympiad)",
    "question": "On the set $A = [0, \\infty)$ of all nonnegative real numbers, consider three functions $f, g, h : A \\to A$ and the binary operation $*: A \\times A \\to A$ defined by\n\n$$\nx * y = f(x) + g(y) + h(x) \\cdot |x - y|, \\quad \\text{for any } x, y \\ge 0.\n$$\n\nIf $(A, *)$ is a commutative monoid:\n\na) Show that the function $h$ is continuous on $A$.\n\nb) Determine the functions $f, g, h$.",
    "options": [],
    "answer": "See solution",
    "solution": "a) Let $e$ be the unit element of the monoid $(A, *)$. Then\n\n$$\nf(0) + g(e) + h(0) \\cdot e = 0 * e = 0 \\quad \\text{and} \\quad f(e) + g(0) + h(e) \\cdot e = e * 0 = 0,\n$$\n\nso $f(e) = g(e) = f(0) = g(0) = h(e) \\cdot e = h(0) \\cdot e = 0$, whence $e = e * e = f(e) + g(e) = 0$.\n\nThen $0 * x = x$ and $x * 0 = x$ for any $x \\ge 0$, and we obtain\n\n$$\nf(0) + g(x) + h(0) \\cdot x = x \\quad \\text{and} \\quad f(x) + g(0) + h(x) \\cdot x = x,\n$$\n\nso $f(x) = x(1 - h(x))$ and $g(x) = x(1 - h(0))$ for any $x \\ge 0$. Since $f(x), g(x) \\ge 0$, it follows that $h(x) \\in [0, 1]$ for all $x \\ge 0$.\n\nThen\n\n$$\nx * y = x + y - x h(x) - y h(0) + h(x) |x - y|, \\quad \\forall x, y \\ge 0.\n$$\n\nFrom commutativity of $*$,\n\n$$\nx h(x) - y h(y) = h(0)(x - y) + (h(x) - h(y)) |x - y|, \\quad \\forall x, y \\ge 0.\n$$\n\nSince $h$ is bounded, $\\lim_{x \\to y} x h(x) = y h(y)$ for any $y \\ge 0$, so the function $p(x) = x h(x)$ is continuous. Thus $h$ is continuous on $(0, \\infty)$.\n\nAlso, for any $y > 0$,\n\n$$\n\\lim_{x \\to y} \\frac{p(x) - p(y)}{x - y} = h(0),\n$$\n\nso there are $a = h(0)$ and $b \\ge 0$ such that $p(y) = a y + b = h(0) y + b$ for all $y > 0$. Then $b = \\lim_{y \\to 0} p(y) = p(0) = 0$. Thus $y h(y) = p(y) = y h(0)$ for any $y > 0$, so $h(y) = h(0)$ for all $y > 0$. The function $h$ is thus constant, hence continuous.\n\nb) Let $k = h(0)$. Then $h(x) = k$ and $f(x) = g(x) = x(1 - k)$ for any $x \\ge 0$, and\n\n$$\nx * y = (x + y)(1 - k) + k |x - y|, \\quad \\forall x, y \\ge 0.\n$$\n\nThen $(1 * 1) * 2 = 1 * (1 * 2) \\implies k(4k - 2) = 0$, so $k \\in \\{0, \\frac{1}{2}\\}$.\n\nFor $k = 0$, $f = g = \\text{id}_A$ and $x * y = x + y$ for all $x, y \\ge 0$.\n\nFor $k = \\frac{1}{2}$, $f(x) = g(x) = \\frac{x}{2}$ for all $x \\ge 0$ and $x * y = \\frac{x + y}{2} + \\frac{|x - y|}{2} = \\max(x, y)$ for all $x, y \\ge 0$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 12613,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $I$ be the incenter of $\\triangle ABC$ and let $H_a$, $H_b$, and $H_c$ be the orthocenters of $\\triangle BIC$, $\\triangle CIA$, and $\\triangle AIB$, respectively. The line $H_aH_b$ meets $AB$ at $X$ and the line $H_aH_c$ meets $AC$ at $Y$. If the midpoint $T$ of the median $AM$ of $\\triangle ABC$ lies on $XY$, prove that the line $H_aT$ is perpendicular to $BC$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let the lines through $B$ and $C$ parallel to $XY$ meet $AM$ at $P$ and $Q$, respectively. Since $BM = MC$, we have $PM = MQ$ and\n\n$$\n\\frac{BX}{XA} + \\frac{CY}{YA} = \\frac{PT}{TA} + \\frac{QT}{TA} = \\frac{PT + QT}{TA} = \\frac{2MT}{TA} = 2.\n$$\n\n![](images/BMO_Short_list_2014_p29_data_b6dada0683.png)\n\nLet $I_a$, $I_b$, and $I_c$ be the excenters of $\\triangle ABC$ opposite $A$, $B$, and $C$, respectively. Since the figures $BH_aCI_a$, $CH_bAI_b$, and $AH_cBI_c$ are parallelograms, we have\n\n$$\n\\begin{aligned}\n2 &= \\frac{BX}{XA} + \\frac{CY}{YA} = \\frac{BH_a}{H_bA} + \\frac{CH_a}{H_cA} = \\frac{I_aC}{CI_b} + \\frac{I_aB}{BI_c} \\\\\n&= \\frac{r_a}{r_b} + \\frac{r_a}{r_c} = \\frac{S/(p-a)}{S/(p-b)} + \\frac{S/(p-a)}{S/(p-c)} = \\frac{p-b}{p-a} + \\frac{p-c}{p-a},\n\\end{aligned}\n$$\n\nwhich is equivalent to $2a = b + c$.\n\n![](images/BMO_Short_list_2014_p29_data_111cce7eef.png)\n\nOn the other hand, $H_aT \\perp BC \\Leftrightarrow BH_a^2 - H_aC^2 = BT^2 - TC^2$. Let $U$ be the tangency point of the ex-circle opposite $A$ and $BC$. Then\n\n$$\nBH_a^2 - H_aC^2 = CI_a^2 - I_aB^2 = CU_a^2 - UB^2 = (p-b)^2 - (p-c)^2 = a(c-b).\n$$\n\n![](images/BMO_Short_list_2014_p30_data_5c9af9d5d5.png)\n\nAlso, since $BT$ and $CT$ are medians in $\\triangle ABM$ and $\\triangle ACM$, respectively, we have\n\n$$\n\\begin{aligned}\nBT^2 - TC^2 &= \\frac{1}{2}(BA^2 + BM^2) - \\frac{1}{4}AM^2 - \\frac{1}{2}(CA^2 + CM^2) + \\frac{1}{4}AM^2 \\\\\n&= \\frac{1}{2}(BA^2 - CA^2) = \\frac{1}{2}(c-b)(c+b).\n\\end{aligned}\n$$\n\nIf $b = c$, then $H_aT \\perp BC$ by symmetry. If $b \\neq c$, then the above implies that\n\n$$\nH_aT \\perp BC \\Leftrightarrow BH_a^2 - H_aC^2 = BT^2 - TC^2 \\Leftrightarrow a = \\frac{1}{2}(b + c),\n$$\n\nas needed.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 12614,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $ABC$ be an acute-angled triangle with $AB < AC$, incenter $I$, and let $M$ be the midpoint of the major arc $BAC$. Suppose the perpendicular line from $A$ to segment $BC$ meets lines $BI$, $CI$, and $MI$ at points $P$, $Q$, and $K$ respectively. Prove that the $A$-median line in $\\triangle AIK$ passes through the circumcentre of $\\triangle PIQ$.",
    "options": [],
    "answer": "See solution",
    "solution": "Observe that $\\angle PIQ = 90 - \\frac{\\angle A}{2}$, $\\angle IQP = 90 - \\frac{\\angle C}{2}$, and $\\angle IPQ = 90 - \\frac{\\angle B}{2}$. Thus, if $\\triangle DEF$ is the orthic triangle of $\\triangle IPQ$, then it is similar to $\\triangle ABC$.\n\nLet $\\ell$ be the $I$-midline in $\\triangle IPQ$ and $O$ be the circumcenter. Now, let $X = \\ell \\cap AO$ and $K'$ be the reflection of $I$ across $X$. Then, we just want that $K'$ and $K$ coincide, or equivalently that $I$, $M$, $X$ are collinear.\n\nNow, with respect to triangle $IPQ$, $AI$ is tangent to the circumcircle of $IPQ$ as $\\angle AIQ = 90 - \\frac{\\angle B}{2}$ as required.\n\nLet $H$ be the orthocenter of $\\triangle IPQ$ and $N$ be the midpoint of $PQ$. Then, we have that $\\angle AIM = \\angle NHD$, so we just want $\\angle NHD + \\angle AIX = 180^\\circ$. But $\\angle AIX = 90^\\circ + \\angle OIX$ and $\\angle NHD = 90^\\circ - \\angle HND$. Thus, we just want that $\\angle HND = \\angle OIX$.\n\nTaking homothety with dilation factor $+2$ from $I$, we have $X$ going to $K'$, which is now the intersection of $PQ$ and the line through the antipode of $I$ in $IPQ$ and the point $R$ on $(IPQ)$ such that $AR$ is tangent to $(IPQ)$.\n\nNow let $HN \\cap (IPQ) = S_1, S_2$ where $S_2$ is the antipode of $I$ and let $H'$ be the reflection of $H$ in $PQ$.\n\nNow, we just want $\\angle HND = \\angle HS_2H' = \\angle S_1IH$. Thus, we want $\\angle S_1IH = \\angle OIK'$, or that $IK'$ and $IS_1$ are isogonal in $\\angle QIP$.\n\nNow, performing $\\sqrt{bc}$ and reflection in $\\triangle IPQ$, we get that $S_1$ and $K'$ interchange. Thus, $IK'$ and $IS_1$ are isogonal in $\\angle QIP$ as required. Thus, we are done. $\\square$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 12615,
    "subject": "Mathematics (Olympiad)",
    "question": "We have 3366 film critics who sent their preferences for the best actor and best actress for the Oscars. It turns out that for every integer $n \\in \\{1, 2, \\dots, 100\\}$ there is an actor or an actress who has been voted exactly $n$ times. Show that there are two critics who voted in exactly the same manner.",
    "options": [],
    "answer": "See solution",
    "solution": "Assume, for contradiction, that every critic votes for a different pair of actor and actress.\n\nCall the vote of each critic (their choice of actor and actress) a *double-vote*, and each individual choice (actor or actress) a *single-vote*. Thus, each double-vote corresponds to two single-votes.\n\nFor each $n = 34, 35, \\dots, 100$, pick one actor or actress who has been voted by exactly $n$ critics (i.e., appears in exactly $n$ single-votes), and let $S$ be the set of these movie stars. Let $a$ and $b$ be the number of men and women in $S$, so $a + b = 67$.\n\nLet $S_1$ be the set of double-votes with exactly one of its two single-votes in $S$, and $S_2$ the set with both single-votes in $S$. Let $s_1 = |S_1|$ and $s_2 = |S_2|$. The number of double-votes with at least one single-vote in $S$ is $s_1 + s_2$, and those with both in $S$ is $s_2 \\leq ab$.\n\nThe total number of single-votes in $S$ is $s_1 + 2s_2 = 34 + 35 + \\dots + 100 = 4489$. Thus, $s_1 + s_2 = (s_1 + 2s_2) - s_2 \\geq 4489 - ab$.\n\nSince all double-votes are distinct, there must be at least $s_1 + s_2$ critics. The maximum value of $ab$ for $a + b = 67$ is $33 \\cdot 34 = 1122$ (since $ab = \\frac{(a+b)^2 - (a-b)^2}{4}$, maximized when $|a-b|=1$).\n\nTherefore, there are at least $4489 - 1122 = 3367$ critics, which contradicts the given number of 3366. Thus, there must be two critics who voted in exactly the same manner. $\\square$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 12616,
    "subject": "Mathematics (Olympiad)",
    "question": "On the circle $\\gamma$ there are points $A$ and $B$. The circle $\\omega$ is tangent to the segment $AB$ at the point $K$ and intersects the circle $\\gamma$ at points $M$ and $N$. The points lie on the circle $\\gamma$ in the following order: $A, M, N, B$. Prove that $\\angle AMK = \\angle KNB$.\n\n![](images/Ukrajina_2013_p38_data_377a4b5aae.png)",
    "options": [],
    "answer": "See solution",
    "solution": "Let us draw the segment $MN$ and denote $\\angle AMK = x$, $\\angle KNB = y$. As $\\angle AMN + \\angle NBA = 180^\\circ$, we can see that $x + \\beta + 180^\\circ - \\beta - y = 180^\\circ$. The result is that $x = y$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 12617,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $a$, $b$, $c$, $d$ be real numbers such that $ab(c+d) = cd(a+b)$. Prove that\n\n$$\n\\frac{a+1}{a^2+3} + \\frac{b+1}{b^2+3} \\ge \\frac{c-1}{c^2+3} + \\frac{d-1}{d^2+3}.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "Replace $(c, d)$ with $(-c, -d)$ to rephrase the problem as: subject to the constraint $ab(c+d) + cd(a+b) = 0$, prove that $\\sum \\frac{a+1}{a^2+3} \\ge 0$.\n\nSuppose one of the numbers equals $0$ and notice that (at least) another is $0$ as well – let them be $c$ and $d$. The claim rewrites as\n\n$$\n\\frac{a+1}{a^2+3} + \\frac{b+1}{b^2+3} \\ge -\\frac{2}{3},\n$$\n\nwhich holds true as $\\frac{x+1}{x^2+3} \\ge -\\frac{1}{3}$ for all $x \\in \\mathbb{R}$.\n\nSuppose now that $a, b, c, d \\ne 0$ and set $x = \\frac{1}{a}$, $y = \\frac{1}{b}$, $z = \\frac{1}{c}$, $t = \\frac{1}{d}$. Once again, the problem is restated:\n\n$$\n x + y + z + t = 0 \\implies \\sum \\frac{x^2 + x}{3x^2 + 1} \\ge 0.\n$$\n\nLet $|t| = \\max\\{|x|, |y|, |z|, |t|\\}$. The inequality is equivalent to\n\n$$\n\\sum \\left( \\frac{6x^2 + 6x}{3x^2 + 1} + 1 \\right) \\ge 4 \\iff \\sum \\frac{(3x+1)^2}{3x^2 + 1} \\ge 4\n$$\n\n$$\n\\iff \\frac{(3x+1)^2}{3x^2+1} + \\frac{(3y+1)^2}{3y^2+1} + \\frac{(3z+1)^2}{3z^2+1} \\ge 4 - \\frac{(3t+1)^2}{3t^2+1} = \\frac{3(1-t)^2}{3t^2+1}.\n$$\n\nNotice that $x^2+y^2+z^2 \\le 3t^2$ and apply Cauchy-Schwarz inequality to conclude:\n\n$$\n\\frac{(3x+1)^2}{3x^2+1} + \\frac{(3y+1)^2}{3y^2+1} + \\frac{(3z+1)^2}{3z^2+1} \\ge \\frac{3(x+y+z+1)^2}{x^2+y^2+z^2+1} \\ge \\frac{3(1-t)^2}{3t^2+1}.\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 12618,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $p$ be a prime and $m$ a positive integer. Determine all pairs $(p, m)$ satisfying the equation:\n\n$$\np(p + m) + p = (m + 1)^3.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "The given equation can be rewritten as\n\n$$\np(p + m + 1) = (m + 1)^3. \\tag{1}\n$$\n\nSince $p$ is a prime, $p$ must divide $(m + 1)^3$. Thus, $p \\mid (m + 1)$, so there exists a positive integer $k$ such that $m + 1 = k p$. Substituting into (1):\n\n$$\np(p + k p) = (k p)^3 \\implies p^2 (k + 1) = k^3 p^3.\n$$\nDividing both sides by $p^2$ (since $p \\neq 0$):\n\n$$\nk + 1 = k^3 p.\n$$\nSo $k^3$ divides $k + 1$, which is only possible if $k = 1$.\n\nTherefore, $k = 1$, so $m + 1 = p$, which gives $m = p - 1$. Substituting $k = 1$ into $k + 1 = k^3 p$ gives $2 = p$, so $p = 2$ and $m = 1$.\n\nThus, the only solution is $(p, m) = (2, 1)$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 12619,
    "subject": "Mathematics (Olympiad)",
    "question": "找出所有的實係數多項式 $P(x)$，使得對滿足 $2xyz = x + y + z$ 的非零實數，皆有\n\n$$\n\\frac{P(x)}{yz} + \\frac{P(y)}{zx} + \\frac{P(z)}{xy} = P(x - y) + P(y - z) + P(z - x)\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "定義\n\n$$\nQ(x, y, z) = xP(x) + yP(y) + zP(z) - xyz[P(x - y) + P(y - z) + P(z - x)]\n$$\n\n則 $Q(x, y, z)$ 也是實係數多項式，且當 $xyz \\neq 0$ 且 $2xyz = x + y + z$ 時，有 $Q(x, y, z) = 0$。\n\n這個性質可延伸到複數上，即 $x, y, z$ 也可用複數帶入。當 $(x, y, z) = (t, -t, 0)$ 時，帶入得 $P(t) = P(-t)$，知 $P(x)$ 是偶函數。\n\n又帶入\n\n$$\n(x, y, z) = \\left(x, \\frac{i}{\\sqrt{2}}, -\\frac{i}{\\sqrt{2}}\\right)\n$$\n\n得到\n\n$$\n\\begin{aligned}\n& xP(x) + \\frac{i}{\\sqrt{2}}\\left(P\\left(\\frac{i}{\\sqrt{2}}\\right) - P\\left(-\\frac{i}{\\sqrt{2}}\\right)\\right) \\\\\n&= \\frac{1}{2}x\\left(P\\left(x - \\frac{i}{\\sqrt{2}}\\right) + P\\left(x + \\frac{i}{\\sqrt{2}}\\right) + P(\\sqrt{2}i)\\right)\n\\end{aligned}\n$$\n\n推出\n\n$$\nP\\left(x + \\frac{i}{\\sqrt{2}}\\right) + P\\left(x - \\frac{i}{\\sqrt{2}}\\right) - 2P(x) = P(\\sqrt{2}i)\n$$\n\n由此可知 $\\deg P(x) \\leq 2$，所以 $P(x)$ 的一般形式是 $ax^2 + b$。",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 12620,
    "subject": "Mathematics (Olympiad)",
    "question": "Determine the maximum value of the function\n\n$$\nf_k(x, y) = (x + y) - (x^{2k+1} + y^{2k+1})\n$$\n\nover all real numbers $x$ and $y$ satisfying $x^2 + y^2 = 1$, for all positive integers $k$.",
    "options": [],
    "answer": "See solution",
    "solution": "Since $x^2 + y^2 = 1$, we have $|x| \\leq 1$ and $|y| \\leq 1$. Define $g_k(x) := x - x^{2k+1}$; note that $g_k(-x) = -g_k(x)$. The function can be written as $f_k(x, y) = g_k(x) + g_k(y)$, so $f_k(x, y) \\leq f_k(|x|, |y|)$. Since $x^2 + y^2 = 1$ implies $|x|^2 + |y|^2 = 1$, we may assume $x, y \\geq 0$.\n\nThe quadratic mean is\n\n$$\nm_2(x, y) = \\sqrt{\\frac{x^2 + y^2}{2}} = \\frac{\\sqrt{2}}{2},\n$$\n\nand\n\n$$\nx + y = 2m_1(x, y) \\leq 2m_2(x, y).\n$$\n\nAlso,\n\n$$\n-(x^{2k+1} + y^{2k+1}) \\leq -2m_2^{2k+1}(x, y).\n$$\n\nThus, $x + y \\leq \\sqrt{2}$ and\n\n$$\n-(x^{2k+1} + y^{2k+1}) \\leq \\frac{2}{\\sqrt{2}^{2k+1}} = \\frac{\\sqrt{2}}{2^k},\n$$\n\nso\n\n$$\nf_k(x, y) \\leq \\frac{2^k - 1}{2^k} \\sqrt{2}\n$$\n\nwith equality for $x = y = \\frac{\\sqrt{2}}{2}$.\n\nqed",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 12621,
    "subject": "Mathematics (Olympiad)",
    "question": "令 $\\mathbb{R}$ 代表所有實數所成的集合。確定所有單射函數 $f: \\mathbb{R} \\to \\mathbb{R}$ 使得\n\n$$\n(f(a) - f(b))(f(b) - f(c))(f(c) - f(a)) = f(ab^2 + bc^2 + ca^2) - f(a^2b + b^2c + c^2a)\n$$\n\n對所有實數 $a, b, c$ 都成立。",
    "options": [],
    "answer": "See solution",
    "solution": "$f(x) = \\alpha x + \\beta$ 或 $f(x) = \\alpha x^3 + \\beta$，其中 $\\alpha \\in \\{-1, 0, 1\\}$，$\\beta \\in \\mathbb{R}$。\n\n這些函數滿足題目中的等式。設 $f(x)$ 滿足該等式，記為 $E(a, b, c)$。顯然 $f(x) + C$ 也滿足，因此可假設 $f(0) = 0$。\n\n由 $E(a, b, 0)$ 得：\n\n$$\nf(a)f(b)(f(a) - f(b)) = f(a^2b) - f(ab^2). \\quad (1)\n$$\n\n設 $\\kappa := f(1)$，由單射性 $\\kappa = f(1) \\neq f(0) = 0$。令 $b = 1$ 代入 (1)：\n\n$$\n\\kappa f(a)(f(a) - \\kappa) = f(a^2) - f(a). \\quad (2)\n$$\n\n將 $a$ 換成 $-a$ 並相減：\n\n$$\n\\kappa(f(a) - f(-a))(f(a) + f(-a) - \\kappa) = f(-a) - f(a).\n$$\n\n若 $a \\neq 0$，由單射性 $f(a) - f(-a) \\neq 0$，因此\n\n$$\nf(a) + f(-a) = \\kappa - \\kappa^{-1} =: \\lambda \\quad (3)\n$$\n\n因此\n\n$$\nf(a) - f(b) = f(-b) - f(-a)\n$$\n\n對所有非零 $a, b$ 成立。將 (1) 中 $a, b$ 換成 $-a, -b$，兩式相加，利用 (3)：\n\n$$\n(f(a) - f(b))(f(a)f(b) - f(-a)f(-b)) = 0,\n$$\n即 $f(a)f(b) = f(-a)f(-b) = (\\lambda - f(a))(\\lambda - f(b))$，對所有非零 $a \\neq b$ 成立。若 $\\lambda \\neq 0$，則 $f(a) + f(b) = \\lambda$，與單射性矛盾。因此 $\\lambda = 0$，$\\kappa = \\pm 1$，$f$ 為奇函數。若必要可將 $f$ 換成 $-f$，可假設 $f(1) = 1$。\n\n由 (2) 得 $f(a^2) = f^2(a)$。將 (1) 於 $(a, b)$ 和 $(a, -b)$ 相加，得 $-2f(a)f^2(b) = -2f(ab^2)$，即 $f(a)f(b^2) = f(ab^2)$。令 $b = \\sqrt{x}$，對所有 $x \\ge 0$，$f(ax) = f(a)f(x)$。由奇性，對所有 $a, x$ 成立。又 $f(a^2) = f^2(a)$，故 $f(x) \\ge 0$ 對 $x \\ge 0$。由單射性 $f(x) > 0$ 對 $x > 0$。\n\n若 $f(x)$ 對 $x > 0$ 不是 $f(x) = x^\\tau$，則 $f$ 的圖形在 $(0, \\infty)^2$ 上稠密。可取 $b < 1/10$ 使 $f(b) > 1$。若 $f(x) = x^\\tau$ 且 $\\tau < 0$，同理。則 $x^2 + xb^2 + b \\ge 0$，由 $E(1, b, x)$ 得：\n\n$$\nf(b^2+bx^2+x) = f(x^2+xb^2+b) + (f(b)-1)(f(x)-f(b)(f(x)-1)) \\ge -\\frac{(f(b)-1)^3}{4}\n$$\n\n即 $f$ 在 $(b^2 - \\frac{1}{4b}, +\\infty)$ 下有下界，由奇性在 $(0, \\frac{1}{4b} - b^2)$ 上有上界。這與 $f(x) = x^\\tau$，$\\tau < 0$ 或稠密性矛盾。\n\n因此 $f(x) = x^\\tau$，$x > 0$，$\\tau > 0$。將 $E(a, b, c)$ 除以 $(a-b)(b-c)(c-a) = (ab^2 + bc^2 + ca^2) - (a^2b + b^2c + c^2a)$，令 $a, b, c \\to 1$，得 $\\tau^3 = \\tau \\cdot 3^{\\tau-1}$，即 $\\tau^2 = 3^{\\tau-1}$，$F(\\tau) := 3^{\\tau/2-1/2} - \\tau = 0$。$F$ 嚴格凸，至多兩根，得 $\\tau \\in \\{1, 3\\}$。$\\square$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 12622,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $a$, $b$, $c$, $d$ be positive integers satisfying $ab = cd$. Is it possible that $a + b + c + d$ is a prime number?",
    "options": [],
    "answer": "See solution",
    "solution": "It follows from the problem statement that $\\frac{ab}{c}$ is a positive integer. Then there should exist positive integers $m, n, x, y$ such that $c = mn$, $a = mx$, $b = ny$. This implies that $d = \\frac{ab}{c} = xy$, and so\n$$\na + b + c + d = mx + ny + mn + xy = (n + x)(m + y),\n$$\nwhich is, obviously, not prime.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 12623,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $H$ be the orthocenter of triangle $ABC$. Let $M$ and $N$ be the midpoints of sides $AB$ and $AC$, respectively. Assume that $H$ lies inside the quadrilateral $BMNC$ and that the circumcircles of triangles $BMH$ and $CNH$ are tangent to each other. The line through $H$ parallel to $BC$ intersects the circumcircles of triangles $BMH$ and $CNH$ at points $K$ and $L$, respectively. Let $F$ be the intersection point of $MK$ and $NL$, and let $J$ be the incenter of triangle $MHN$. Prove that $FJ = FA$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $Y$ and $Z$ be the feet of the perpendiculars from $B$ and $C$ to $AC$ and $AB$, respectively. Hence $BZYC$ is cyclic due to $\\angle BZC = 90^\\circ = \\angle BYC$. So we may let\n\n$$\n\\angle HBZ = \\angle YCH = x.\n$$\n\nOur plan is to calculate many angles in terms of $x$.\n\n![](images/2018-Australian-Scene-W2_p121_data_de69fa9c39.png)\n\nSince $M$ and $N$ are the midpoints of $AB$ and $AC$, respectively, it follows that $MN \\parallel BC \\parallel KL$. Using this and cyclic $BKMH$ yields\n\n$$\n\\angle NMF = \\angle LKF = \\angle HKM = \\angle HBM = x.\n$$\n\nSimilarly,\n\n$$\n\\angle FNM = \\angle FLK = x.\n$$\n\nThe above angle equalities imply $FK = FL$ and $FM = FN$. Hence\n\n$$\nFM \\cdot FK = FN \\cdot FL.\n$$\n\nThus $F$ has equal power with respect to circles $BKMH$ and $CHNL$. Hence $F$ lies on the radical axis of these two circles. Since the two circles are tangent, it follows that $F$ lies on the common tangent at $H$. Hence $FH$ is tangent to the two circles at $H$. Using the alternate segment theorem we deduce\n\n$$\n\\angle FHM = \\angle HBM = x \\quad \\text{and} \\quad \\angle NHF = \\angle NCH = x.\n$$\n\nIn particular, $FH$ bisects $\\angle NHM$, and so $F$, $J$, and $H$ are collinear. Another consequence is that $FMHN$ is cyclic due to $\\angle NMF = x = \\angle NHF$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 12624,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $A$ be a set containing 2000 integers and $B$ be a set containing 2016 integers. Denote $K$ to be the number of pairs $(m, n)$ satisfying\n$$\n\\begin{cases}\n  m \\in A, \\, n \\in B, \\\\\n  |m - n| \\le 1000.\n\\end{cases}\n$$\nFind the maximum value of $K$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $a_1 < a_2 < \\dots < a_{2000}$ be all elements in $A$. For every positive integer $1 \\leq k \\leq 2000$, denote $x_k$ to be the number of elements $m$ in $B$ such that $|a_k - m| \\leq 1000$.\n\nWe will prove that\n$$\nx_k + x_{2001-k} \\leq \\min\\{4002, 2016 + 2k\\}, \\text{ for all } 1 \\leq k \\leq 1000. \\quad (\\spadesuit)\n$$\n\nFirstly, it is clear that $x_k \\leq 2001$ for all $1 \\leq k \\leq 2000$, hence $x_k + x_{2001-k} \\leq 4002$. Let $X_k$ be the set of numbers $m$ in $B$ such that $|a_k - m| \\leq 1000$. It is obvious that\n$$\nX_k \\subset B_k = \\{a_k - 1000, \\dots, a_k, \\dots, a_k + 1000\\}.\n$$\n\nSimilarly,\n$$\nX_{2001-k} \\subset B_{2001-k} = \\{a_{2001-k}-1000, \\dots, a_{2001-k}, \\dots, a_{2001-k}+1000\\}.\n$$\n\nThus, $X_k \\cap X_{2001-k} \\subset B_k \\cap B_{2001-k}$. On the other hand, we have $a_k + 2001 - 2k \\leq a_{2001-k}$, which implies $|B_k \\cap B_{2001-k}| \\leq 2k$ for all $k$. Therefore,\n$$\n\\begin{aligned}\nx_k + x_{2001-k} &= |B_k| + |B_{2001-k}| \\\\\n  &= |B_k \\cup B_{2001-k}| + |B_k \\cap B_{2001-k}| \\\\\n  &\\leq 2016 + 2k.\n\\end{aligned}\n$$\n\nSo $(\\spadesuit)$ is proved. Thus,\n$$\n\\begin{aligned}\nK &= \\sum_{k=1}^{1000} x_k \\leq \\sum_{k=1}^{1000} \\min\\{4002, 2016 + 2k\\} \\\\\n  &= \\sum_{k=1}^{993} (2016 + 2k) + 7 \\cdot 4002 = 3016944.\n\\end{aligned}\n$$\n\nSo the maximum value of $K$ is $3016944$ and the following two sets\n$$\nA = \\{9, \\dots, 2008\\}, \\quad B = \\{1, \\dots, 2016\\}\n$$\nsatisfy the conditions. $\\Box$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 12625,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $k$ be a fixed odd integer with $k > 3$. Prove that there exist infinitely many positive integers $n$ such that there are two positive integers $d_1, d_2$ satisfying $d_1, d_2 \\mid \\frac{n^2+1}{2}$ and $d_1 + d_2 = n + k$.",
    "options": [],
    "answer": "See solution",
    "solution": "Consider the Diophantine equation\n\n$$\n((k-2)^2 + 1)xy = (x + y - k)^2 + 1\n$$\n\nWe prove that this equation has infinitely many positive odd solutions $(x, y)$.\n\nObviously, $(1, 1)$ is one positive odd solution. Let $(x_1, y_1) = (1, 1)$. Assume $(x_i, y_i)$ is a positive odd solution with $x_i \\le y_i$. Define $x_{i+1} = y_i$ and $y_{i+1} = (k-1)(k-3)y_i + 2k - x_i$.\n\nSince the equation can be rewritten as\n\n$$\nx^2 - ((k-1)(k-3)y + 2k)x + (y-k)^2 + 1 = 0,\n$$\n\nby Vieta's theorem, $(x_{i+1}, y_{i+1})$ is also an integer solution. Since $x_i, y_i$, and $k$ are all positive odd integers and $k \\ge 5$, $x_{i+1}$ is a positive odd integer, and\n\n$$\ny_{i+1} = (k-1)(k-3)y_i + 2k - x_i \\equiv -x_i \\equiv 1 \\pmod{2},\n$$\n\nso $y_{i+1}$ is also odd. Moreover, $y_{i+1} > y_i > 0$, so $(x_{i+1}, y_{i+1})$ is a positive odd solution, and $x_i + y_i < x_{i+1} + y_{i+1}$.\n\nBy starting from $(x_1, y_1)$ and iterating, we obtain an infinite sequence of positive odd solutions $(x_i, y_i)$ with strictly increasing $x_i + y_i$.\n\nFor any $i$ with $x_i + y_i > k$, let $n = x_i + y_i - k$, $d_1 = x_i$, $d_2 = y_i$. Then $n$ is a positive odd integer, $d_1 + d_2 = n + k$, and since $((k-2)^2 + 1)$ is even, $d_1, d_2$ both divide $\\frac{n^2 + 1}{2}$.\n\nThus, there exist infinitely many positive integers $n$ satisfying the required conditions.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 12626,
    "subject": "Mathematics (Olympiad)",
    "question": "An acute-angled triangle $ABC$ has the side $BC > AB$, and the bisector $BL = AB$. On the segment $BL$ there exists a point $M$ for which $\\angle AML = \\angle BCA$. Prove that $AM = LC$.\n\n![](images/ukraine_2015_Booklet_p7_data_ea71bc6a76.png)",
    "options": [],
    "answer": "See solution",
    "solution": "On the segment $BC$, place a point $D$ such that $BD = BL$. Then $\\triangle ABL \\cong \\triangle BLD$, so $\\angle LAB = \\angle BLA = \\angle BLD = \\angle BDL$. On the segment $BD$, choose a point $K$ such that $\\angle ALM = \\angle DLK$. Then $\\triangle ALM \\cong \\triangle DLK$ because $LD = AL$, and thus $\\angle AML = \\angle LKD = \\angle BCA$. Therefore, $KL = AM = LC$, which is exactly what we have to prove.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 12627,
    "subject": "Mathematics (Olympiad)",
    "question": "A necklace contains 2016 pearls, each of which has one of the colours black, green, or blue. In each step, we replace simultaneously each pearl with a new pearl, where the colour of the new pearl is determined as follows:\n\n- If the two original neighbours were of the same colour, the new pearl has their colour.\n- If the neighbours had two different colours, the new pearl has the third colour.\n\n(a) Is there such a necklace that can be transformed with such steps to a necklace of blue pearls if half of the pearls were black and half of the pearls were green at the start?\n\n(b) Is there such a necklace that can be transformed with such steps to a necklace of blue pearls if a thousand of the pearls were black at the start and the rest green?\n\n(c) Is it possible to transform a necklace that contains exactly two adjacent black pearls and 2014 blue pearls to a necklace that contains one green pearl and 2015 blue pearls?\n",
    "options": [],
    "answer": "See solution",
    "solution": "(a) Since 2016 is divisible by 4, we can alternate two black and two green pearls. In the first step, all pearls are already replaced by blue pearls.\n\n(b) Assign to each blue pearl the number 0, to each green pearl the number 1, and to each black pearl the number 2. In each step, the new colour of a pearl modulo 3 is equal to the negative sum of its two original neighbours. The new total sum of all colours modulo 3 can be calculated by multiplying the old total sum by 2 and changing the sign. But modulo 3, multiplication by $-2$ is equivalent to multiplication by 1, so the total sum always remains the same modulo 3.\n\nFor a necklace with only blue pearls, the total sum is 0. But for 1000 black and 1016 green pearls, it is $2000 + 1016 \\equiv 1 \\pmod{3}$. Therefore, there does not exist an arrangement of 1000 black and 1016 green pearls that can be transformed into a necklace with only blue pearls using such steps.\n\n(c) Using the same assignment of numbers modulo 3, in each step the sum of all colours in even positions becomes the sum of the colours in odd positions, and vice versa. If these sums are $A$ and $B$ in the beginning, then at the end we still have these same two sums modulo 3, maybe with switched positions.\n\nBut in the beginning, we have sums 2 and 2 modulo 3, because both among the even and among the odd positions there is exactly one black pearl with value 2, and otherwise only blue pearls with value 0. However, at the end we are supposed to have sums 1 and 0 because one of the two sums is determined only by blue pearls with value 0, and the other by exactly one green pearl with value 1 and only blue pearls with value 0 otherwise. Therefore, it is not possible.\n\n$\\Box$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 12628,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $f(n)$ denote the minimum number of square tiles needed to tile an $n$-staircase, where an $n$-staircase is a shape consisting of $n$ rows, with the $i$-th row containing $i$ unit squares aligned to the left. \n\n(a) For which positive integers $n$ does $f(n) = n$? \n\n(b) For which positive integers $n$ does $f(n) = n + 1$?\n\n![](images/Kanada_2010_p3_data_3f32d0db15.png)\n![](images/Kanada_2010_p3_data_26651aab5a.png)\n![](images/Kanada_2010_p4_data_ec97d01873.png)\n![](images/Kanada_2010_p4_data_1566e643bf.png)\n![](images/Kanada_2010_p4_data_37ed5d5b91.png)\n![](images/Kanada_2010_p4_data_70e0e47f6f.png)",
    "options": [],
    "answer": "See solution",
    "solution": "Consider the left-most unit square in the fourth row. The only square tile that can cover this unit square and a diagonal square is a $4 \\times 4$ square tile.\n\nContinuing this construction, the side lengths of the square tiles encountered are $1, 2, 4, \\ldots, 2^k$ for some nonnegative integer $k$. Therefore, $n$, the height of the $n$-staircase, is $1 + 2 + 4 + \\cdots + 2^k = 2^{k+1} - 1$. Alternatively, $n = 2^k - 1$ for some positive integer $k$. Let $p(k) = 2^k - 1$.\n\nConversely, a $p(k)$-staircase can be tiled with $p(k)$ square tiles recursively: $p(1) = 1$, and a 1-staircase can be tiled with 1 square tile. Assume a $p(k)$-staircase can be tiled with $p(k)$ square tiles for some $k$. For a $p(k+1)$-staircase, place a $2^k \\times 2^k$ square tile in the bottom left corner, covering a diagonal square. Then $p(k+1) - 2^k = 2^{k+1} - 1 - 2^k = 2^k - 1 = p(k)$, leaving two $p(k)$-staircases. These can be tiled with $2p(k)$ square tiles, so $2p(k) + 1 = p(k+1)$ square tiles.\n\nTherefore, $f(n) = n$ if and only if $n = 2^k - 1 = p(k)$ for some positive integer $k$. In other words, the binary representation of $n$ consists of all 1s, with no 0s.\n\n(b) Let $n$ be a positive integer such that $f(n) = n + 1$, and consider a minimal tiling of an $n$-staircase. Since there are $n$ diagonal squares, every square tile except one covers a diagonal square. The square tile that covers the bottom-left unit square must be the one that does not cover a diagonal square.\n\nIf $n$ is even, this is obvious, as the square tile covering the bottom-left unit square cannot cover any diagonal square. If $n$ is odd, let $n = 2m + 1$ with $m \\ge 1$. Suppose the square tile covering the bottom-left unit square also covers a diagonal square; its side length must be $m + 1$. After placing this $(m+1) \\times (m+1)$ square tile, two $m$-staircases remain. Thus, $f(n) = 2f(m) + 1$, which is odd, while $n + 1 = 2m + 2$ is even, so $f(n) \\neq n + 1$, a contradiction. Therefore, the square tile covering the bottom-left unit square does not cover a diagonal square.\n\nLet $t$ be the side length of this square tile. Every other square tile must cover a diagonal square, so $n = 1 + 2 + 4 + \\cdots + 2^{k-1} + t = 2^k + t - 1$ for some $k$. The top $p(k) = 2^k - 1$ rows must be tiled as in the minimal tiling of a $p(k)$-staircase. The horizontal line between rows $p(k)$ and $p(k)+1$ and the vertical line between columns $t$ and $t+1$ are *fault lines*, partitioning two $p(k)$-staircases.\n\nIf these two $p(k)$-staircases do not overlap, $t = p(k)$, so $n = 2p(k)$. If they overlap, their intersection is a $[p(k) - t]$-staircase, which is tiled as the top $p(k) - t$ rows of a minimal $p(k)$-staircase, so $p(k) - t = p(l)$ for some $l < k$, so $t = p(k) - p(l)$. Then\n$$\nn = t + p(k) = 2p(k) - p(l).\n$$\nSince $p(0) = 0$, $n$ must be of the form\n$$\nn = 2p(k) - p(l) = 2^{k+1} - 2^l - 1,\n$$\nwhere $k$ is a positive integer and $l$ is a nonnegative integer. If $n$ is of this form, an $n$-staircase can be tiled with $n + 1$ square tiles.\n\nFinally, $n$ is of this form if and only if the binary representation of $n$ contains exactly one 0:\n$$\n2^{k+1} - 2^l - 1 = \\underbrace{11\\dots1}_{k-l \\ 1s} \\ 0 \\ \\underbrace{11\\dots1}_{l \\ 1s}.\n$$\n\n$\\Box$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 12629,
    "subject": "Mathematics (Olympiad)",
    "question": "Determine all continuous functions $f : \\mathbb{R} \\to \\mathbb{R}$ such that\n\n$$\n\\begin{aligned}\n&f(1) = e, \\\\\n&f(x + y) = e^{3xy} \\cdot f(x) \\cdot f(y), \\quad \\text{for all } x, y \\in \\mathbb{R}.\n\\end{aligned}\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "Let $f : \\mathbb{R} \\to \\mathbb{R}$ satisfy the given conditions.\n\nSuppose there exists $a \\in \\mathbb{R}$ such that $f(a) = 0$. Then for any $x \\in \\mathbb{R}$:\n$$\nf(x) = f(a + (x - a)) = e^{3a(x-a)} f(a) f(x-a) = 0,\n$$\nwhich would imply $f(x) = 0$ for all $x$, contradicting $f(1) = e \\neq 0$. Thus, $f(x) \\neq 0$ for all $x$.\n\nAlso,\n$$\nf(x) = e^{3x^2/4} f^2(x/2) \\geq 0,\n$$\nso $f(x) > 0$ for all $x$.\n\nTake logarithms:\n$$\n\\ln f(x + y) = 3xy + \\ln f(x) + \\ln f(y).\n$$\nRewriting,\n$$\n\\ln f(x + y) - \\frac{3(x + y)^2}{2} = [\\ln f(x) - \\frac{3x^2}{2}] + [\\ln f(y) - \\frac{3y^2}{2}].\n$$\nDefine $g(x) = \\ln f(x) - \\frac{3x^2}{2}$. Then $g(x + y) = g(x) + g(y)$, so $g$ is additive and continuous, hence linear: $g(x) = cx$.\n\nFrom $f(1) = e$:\n$$\ng(1) = \\ln f(1) - \\frac{3}{2} = 1 - \\frac{3}{2} = -\\frac{1}{2},\n$$\nso $g(x) = -\\frac{x}{2}$.\n\nTherefore,\n$$\n\\ln f(x) - \\frac{3x^2}{2} = -\\frac{x}{2} \\implies \\ln f(x) = \\frac{3x^2 - x}{2} \\implies f(x) = e^{\\frac{x(3x-1)}{2}}.\n$$\n\nThis function satisfies the original conditions.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 12630,
    "subject": "Mathematics (Olympiad)",
    "question": "Find the maximum value of the positive real number $k$ such that the inequality\n\n$$\n\\frac{1}{kab + c^2} + \\frac{1}{kbc + a^2} + \\frac{1}{kca + b^2} \\geq \\frac{k+3}{a^2 + b^2 + c^2}\n$$\n\nholds for all positive real numbers $a, b, c$ such that\n\n$$\na^2 + b^2 + c^2 = 2(ab + bc + ca).\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "Let $a \\to 0^+$ and $b = c = 1$, one can get\n\n$$\n2 + \\frac{1}{k} \\geq \\frac{k+3}{2}\n$$\n\nso $k \\leq 2$. For $k = 2$, we need to prove that\n\n$$\n\\frac{1}{2ab + c^2} + \\frac{1}{2bc + a^2} + \\frac{1}{2ca + b^2} \\geq \\frac{5}{a^2 + b^2 + c^2}\n$$\n\nis true for all positive triples $(a, b, c)$ satisfying $a^2 + b^2 + c^2 = 2(ab + bc + ca)$. First, we will prove that\n\n$$\n\\frac{1}{a^2 + 2bc} + \\frac{1}{b^2 + 2ac} + \\frac{1}{c^2 + 2ab} \\geq \\frac{2}{ab + bc + ac} + \\frac{1}{a^2 + b^2 + c^2}\n$$\n\nis true for all positive real numbers $a, b, c$. Indeed, the above inequality can be rewritten as\n\n$$\n\\frac{a^2 + b^2 + c^2}{a^2 + 2bc} + \\frac{a^2 + b^2 + c^2}{b^2 + 2ac} + \\frac{a^2 + b^2 + c^2}{c^2 + 2ab} \\geq \\frac{2(a^2 + b^2 + c^2)}{ab + bc + ca} + 1,\n$$\n\nthus\n\n$$\n\\frac{(b-c)^2}{a^2+2bc} + \\frac{(c-a)^2}{b^2+2ca} + \\frac{(a-b)^2}{c^2+2ab} \\geq \\frac{(a-b)^2 + (b-c)^2 + (c-a)^2}{ab+bc+ac}.\n$$\n\nIf $(a-b)(b-c)(c-a) = 0$ then the above inequality is true. Now consider the case $(a-b)(b-c)(c-a) \\neq 0$, then\n\n$$\n\\begin{aligned}\n\\text{LHS} &\\geq \\frac{\\left[\\sum (b-c)^2\\right]^2}{\\sum (a^2+2bc)(b-c)^2} = \\frac{\\left[\\sum (b-c)^2\\right]^2}{(ab+bc+ca)\\left[\\sum (b-c)^2\\right]} \\\\\n&= \\frac{(a-b)^2 + (b-c)^2 + (c-a)^2}{ab+bc+ac} = \\text{RHS}.\n\\end{aligned}\n$$\n\nBack to the original problem, applying the above inequality, combined with $a^2 + b^2 + c^2 = 2(ab + bc + ca)$, we get\n\n$$\n\\begin{aligned}\n\\frac{1}{a^2+2bc} + \\frac{1}{b^2+2ac} + \\frac{1}{c^2+2ab} &\\geq \\frac{2}{ab+bc+ac} + \\frac{1}{a^2+b^2+c^2} \\\\\n&= \\frac{5}{a^2+b^2+c^2}.\n\\end{aligned}\n$$\n\nSo the maximum positive real number $k$ is $2$. $\\square$",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 12631,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $a, b, c, d$ be positive integers such that:\n\n1. $a + b + c + d = 2023$\n2. $2023 \\mid ab - cd$\n3. $2023 \\mid a^2 + b^2 + c^2 + d^2$\n\nShow that $a, b, c, d$ are all multiples of $17$.",
    "options": [],
    "answer": "See solution",
    "solution": "Remark 1. Taking the original problem modulo $7$ by the same reasoning gives $a^2 + b^2 \\equiv c^2 + d^2 \\equiv 0 \\pmod{7}$, which forces $a \\equiv b \\equiv c \\equiv d \\pmod{7}$. Therefore, the assumption that $a, b, c, d$ are divisible by $7$ is not needed.\n\nRemark 2. The numbers $(a, b, c, d) = (7, 266, 2016, 1757)$ satisfy all of the conditions of the problem except that $a + b + c + d = 2 \\cdot 2023$ rather than $2023$. None of these are multiples of $17$. It follows that there is no solution to the original problem using only arithmetic modulo $2023$. At some point, inequalities have to be applied to exclude such cases.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 12632,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $k$ be a real number such that the system\n\n$$\n\\begin{aligned}\n|25 + 20i - z| &= 5 \\\\\n|z - 4 - k| &= |z - 3i - k|\n\\end{aligned}\n$$\n\nhas exactly one complex solution $z$. The sum of all possible values of $k$ can be written as $\\frac{m}{n}$, where $m$ and $n$ are relatively prime positive integers. Find $m+n$. Here $i = \\sqrt{-1}$.",
    "options": [],
    "answer": "See solution",
    "solution": "Geometrically, $|a-b|$ represents the distance between complex numbers $a$ and $b$ in the complex plane. Thus $|25+20i-z| = 5$ means that the distance from $25+20i$ to $z$ is 5. The set of solutions for $z$ is then a circle with radius 5 and center $25+20i$.\n\nSimilarly, $|z-(4+k)| = |z-(k+3i)|$ means that the distance from $z$ to $4+k$ equals the distance from $z$ to $k+3i$. Geometrically, the set of points equidistant from two fixed points $A$ and $B$ is the perpendicular bisector of $\\overline{AB}$. Thus the solution set for this equation is the line that is the perpendicular bisector of the segment connecting $4+k$ and $k+3i$.\n\nAny intersection of the circle from the first equation and the line from the second equation is a solution to the system of equations. For the system to have exactly one complex solution, the line and the circle must be tangent. There are two such lines, $\\ell_1$ and $\\ell_2$, as shown below.\n\n![](images/2025AIME_I_Solutions_p5_data_ab79025a53.png)\n\nSwitching to Cartesian coordinates, for any $k$, the slope of the line between $(k, 3)$ and $(4+k, 0)$ equals $-\\frac{3}{4}$, and hence the slopes of lines $\\ell_1$ and $\\ell_2$ are both equal to $\\frac{4}{3}$. Because line $\\ell_1$ passes through $(2+k_1, \\frac{3}{2})$, the equation of this line is\n\n$$\ny - \\frac{3}{2} = \\frac{4}{3}(x - (2 + k_1)),\n$$\n\nand the $x$-coordinate of its $x$-intercept is $k_1 + \\frac{7}{8}$. Similarly, the $x$-coordinate of the $x$-intercept of $\\ell_2$ is $k_2 + \\frac{7}{8}$. The line $\\ell_3$, parallel to lines $\\ell_1$ and $\\ell_2$, whose $x$-intercept is the midpoint of the $x$-intercepts of lines $\\ell_1$ and $\\ell_2$, passes through the center of the circle, which is at $(25, 20)$. Thus the equation of line $\\ell_3$ is\n\n$$\ny - 20 = \\frac{4}{3}(x - 25),\n$$\n\nand the $x$-coordinate of its $x$-intercept is 10.\n\nTherefore\n\n$$\n\\frac{k_1 + k_2}{2} + \\frac{7}{8} = 10,\n$$\n\nfrom which $k_1 + k_2 = \\frac{73}{4}$. The requested sum is $73 + 4 = 77$.\n\n**Note:** The two values of $k$ are $2\\frac{7}{8}$ and $15\\frac{3}{8}$, which correspond to solutions $z = 21 + 23i$ and $z = 29 + 17i$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 12633,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $ABC$ be a triangle with a point $D$ on $AB$ and a point $E$ on $BC$ such that $AD = CE$ and $2DE = AC$. Show that the circumradius of $BDE$ is equal to half the circumradius of $ABC$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let us choose a point $S$ on the circumcircle of $BDE$ such that $SD = SE$. Since $\\angle BES = \\angle BDS$, we see that $\\angle SDA = \\angle SEC$. Because $AD = EC$ and $SD = SE$, we have $\\triangle SDA = \\triangle SEC$. It follows that $\\angle ASC = \\angle DSE = \\angle DBE = \\angle ABC$. Hence $S$ must lie on the circumcircle of $ABC$. Thus $\\triangle DSE \\sim \\triangle ASC$. From the congruency, we have $\\frac{DE}{AC} = \\frac{1}{2}$. Hence the ratio of the circumradii of $BDE$ and $ABC$ is $\\frac{1}{2}$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 12634,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $S = \\{1, 2, \\dots, 2017\\}$.\n\nFind the maximal $n$ with the property that there exist $n$ distinct subsets of $S$ such that for no two subsets their union equals $S$.",
    "options": [],
    "answer": "See solution",
    "solution": "The answer is $n = 2^{2016}$.\n\n**Proof:**\n\nThere are $2^{2016}$ subsets of $S$ which do not contain $2017$. The union of any two such subsets does not contain $2017$ and is thus a proper subset of $S$. Thus, $n \\ge 2^{2016}$.\n\nTo show the other direction, we group the subsets of $S$ into $2^{2016}$ pairs so that every subset forms a pair with its complement. If $n > 2^{2016}$, then the $n$ subsets would contain such a pair, and their union would be $S$, a contradiction.\n\nThus, $n = 2^{2016}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 12635,
    "subject": "Mathematics (Olympiad)",
    "question": "Con baldosas cuadradas de lado un número exacto de unidades se ha podido embaldosar una habitación de superficie $18144$ unidades cuadradas de la siguiente manera: el primer día se puso una baldosa, el segundo dos baldosas, el tercero tres, etc. ¿Cuántas baldosas fueron necesarias?",
    "options": [],
    "answer": "See solution",
    "solution": "Supongamos que fueron necesarias $n$ baldosas y que su tamaño es $k \\times k$. Entonces $n k^2 = 18144 = 2^5 \\times 3^4 \\times 7$. Hay nueve casos posibles para $n$, a saber, $2 \\times 7$, $2^3 \\times 7$, $2^5 \\times 7$, $2 \\times 3^2 \\times 7$, $2^3 \\times 3^2 \\times 7$, $2^5 \\times 3^2 \\times 7$, $2 \\times 3^4 \\times 7$, $2^3 \\times 3^4 \\times 7$, $2^5 \\times 3^4 \\times 7$. Además, este número tiene que poderse expresar en la forma $1+2+3+\\cdots+N = N(N+1)/2$ y esto sólo es posible en el caso sexto: $2^5 \\times 3^2 \\times 7 = 63 \\times 64/2 = 2016$. Para descartar los otros casos rápidamente observamos que $N$ y $N+1$ son números primos entre sí. Si por ejemplo $N(N+1)/2 = 2^3 \\times 7$, tendría que ser $N+1 = 2^4$ y $N = 7$, que es imposible, etc. Por tanto, se necesitaron $2016$ baldosas.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 12636,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $a_1, a_2, \\dots, a_n$ be an arbitrary arrangement of the numbers $1, 2, \\dots, n$ on a circle. Find\n$$\n\\min \\sum_{j=1}^{n} |a_j - a_{j+1}| \\quad \\text{and} \\quad \\max \\sum_{j=1}^{n} |a_j - a_{j+1}|,\n$$\nwhere $a_{n+1} = a_1$, and the extrema are taken over all possible arrangements of $1, 2, \\dots, n$.",
    "options": [],
    "answer": "See solution",
    "solution": "**Minimum:**\n\nConsider $1$ and $n$ on the circle. They divide the circle into two arcs. The sum of the absolute differences on either arc is at least $n-1$. Suppose, for example, the numbers $1 = b_1, b_2, \\dots, b_k = n$ appear on one of the arcs between $1$ and $n$, in that order. Then the sum of absolute differences of adjacent numbers on this arc is\n$$\n|1 - b_2| + |b_2 - b_3| + \\dots + |b_{k-1} - n| \\geq |1 - n| = n - 1.\n$$\nSimilarly, the least sum of absolute differences on the other arc is also $n-1$. Hence,\n$$\n\\sum_{j=1}^{n} |a_{j+1} - a_{j}| \\geq 2(n-1).\n$$\nThis is achieved by the permutation $(a_1, a_2, \\dots, a_n)$, where $a_j = j$ for $1 \\leq j \\leq n$.\n\n**Maximum:**\n\nWe have\n$$\n\\sum_{j=1}^{n} |a_{j+1} - a_{j}| = \\sum_{j=1}^{n} \\pm (a_{j+1} - a_{j}).\n$$\nEach of the numbers $1, 2, \\dots, n$ appears in the sum twice. To maximize the sum, assign positive signs to larger numbers in both occurrences and negative signs to smaller numbers. Thus, $n, n-1, n-2, \\dots, \\left\\lfloor \\frac{n}{2} \\right\\rfloor$ should get positive signs, and $1, 2, 3, \\dots, \\left\\lfloor \\frac{n}{2} \\right\\rfloor - 1$ should get negative signs. This happens when:\n\n- For even $n$, the arrangement is $1, n, 2, n-1, 3, n-2, \\dots, \\frac{n}{2}, \\frac{n}{2} + 1$;\n- For odd $n$, the arrangement is $1, n, 2, n-1, 3, n-2, \\dots, \\left\\lfloor \\frac{n}{2} \\right\\rfloor + 2, \\left\\lfloor \\frac{n}{2} \\right\\rfloor + 1$.\n\nThe corresponding sums are:\n$$\n\\frac{n^2}{2} \\quad \\text{when } n \\text{ is even,} \\qquad \\frac{n^2-1}{2} \\quad \\text{when } n \\text{ is odd.}\n$$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 12637,
    "subject": "Mathematics (Olympiad)",
    "question": "設 $n$ 為正整數。一條東西向的路上從東到西有 $n$ 個城鎮。每個城鎮都派出兩頭犀牛，一隻從城鎮往東出發，另一隻從城鎮往西出發（犀牛不會轉向），而這 $2n$ 頭犀牛的大小都不一樣。當兩頭犀牛面對面相撞時，大隻的會把小隻的撞出道路；但如果一隻犀牛從背後撞向另一頭犀牛，不論犀牛的大小，從背後被撞的犀牛都會被撞出道路。\n\n假設有兩個城鎮 $A$ 和 $B$，其中 $B$ 在 $A$ 的東邊。如果 $A$ 的東進犀牛可以一路抵達 $B$ 而把其間的犀牛全部撞飛，則稱 $A$ 城輾過 $B$ 城。反之，如果 $B$ 的西進犀牛可以一路抵達 $A$ 並把其間的犀牛全部撞飛，則稱 $B$ 城輾過 $A$。\n\n證明：恰有一個城鎮不會被任何其他城鎮輾過。\n\nLet $n$ be a positive integer. There are $n$ towns arranged on an East-West road. Each town has two rhinoceroses, one heading East from the town, and the other heading West from the town (rhinoceroses cannot turn into another direction). These $2n$ rhinoceroses all have different sizes. When two rhinoceroses confront face to face, the larger one knocks the smaller one out of the road. However, if a rhinoceros is bumped by another from its rear end, the bumped rhinoceros is knocked out of the road, regardless of their sizes.\n\nLet $A$ and $B$ be two towns, with $B$ being East to $A$. We say that $A$ tramples $B$ if the East-heading rhinoceros of $A$ can reach $B$ by knocking off all rhinoceroses in between. Similarly, we say that $B$ tramples $A$ if the West-heading rhinoceros of $B$ can reach $A$ by knocking off all rhinoceroses in between.\n\nProve that there is exactly one town that would not be trampled by any other town.",
    "options": [],
    "answer": "See solution",
    "solution": "對 $n$ 歸納。當 $n=1$ 時顯然成立。\n\n假設原命題在 $n \\leq N$ 時都成立。當 $n = N + 1$ 時，除去最西邊的西進犀牛和最東邊的東進犀牛（牠們沒有功能），考慮剩餘的 $2N$ 隻犀牛中最大者。不失一般性，假設最大隻的犀牛為從西邊數來第 $k$ 座城鎮的東進犀牛，其中 $k < N + 1$（因為我們不考慮最東邊城鎮的東進犀牛）。\n\n顯然，在第 $k$ 座城鎮以東的城鎮都會被第 $k$ 座城鎮輾過；且第 $k$ 座城鎮以東的城鎮都被這頭最大犀牛擋住，而不能輾過第 $k$ 座城鎮及其西邊的任何城鎮。所以可以丟棄第 $k$ 座城鎮以東的諸城鎮。基於 $k \\leq N$，由歸納假設，剩下的 $k$ 座城中，恰有一個不會被所有其他城鎮輾過，故證畢。",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 12638,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $\\mathbb{N}$ be the set of all positive integers. A subset $A$ of $\\mathbb{N}$ is *sum-free* if, whenever $x$ and $y$ are (not necessarily distinct) members of $A$, their sum $x + y$ does not belong to $A$.\n\nDetermine all *surjective* functions $f: \\mathbb{N} \\to \\mathbb{N}$ such that, for each sum-free subset $A$ of $\\mathbb{N}$, the image $\\{f(a): a \\in A\\}$ is again sum-free.",
    "options": [],
    "answer": "See solution",
    "solution": "The identity is the only surjection of the positive integers onto themselves sending every sum-free set onto a sum-free set.\n\nTo prove this, fix a function $f$ satisfying the conditions in the statement, and proceed in several steps.\n\n**Step 1.** Notice that a 2-element set $\\{x, y\\}$, where $x < y$, is *not* sum-free if and only if $y = 2x$.\n\nChoose any $a \\in \\mathbb{N}$, and for any $i \\ge 0$ choose some $x_i$ such that $f(x_i) = 2^i a$. The set $f(\\{x_i, x_{i+1}\\})$ is not sum-free, so neither is $\\{x_i, x_{i+1}\\}$, whence $x_i = 2x_{i+1}$ or $x_{i+1} = 2x_i$. Since the $x_i$ are all distinct, the same option should hold for all $i$. The former option yields $x_i = x_0 2^{-i}$ which cannot hold for large enough $i$. So $x_{i+1} = 2x_i$ for all $i$.\n\nTherefore, $f(2x) = 2f(x)$ for all $x$, and, moreover, $x$ is the only argument $t$ with $f(t) = f(2x)/2$. Therefore, $f$ is injective (and hence bijective).\n\n**Step 2.** Say that a 3-element set $\\{a, b, c\\}$ is good if it is not sum-free, but each of its 2-element subsets is (in other words, no element is twice another). It is easily seen that a set $\\{a, b, c\\}$, where $a < b$, is good only if $c = b \\pm a$. Notice that the pre-image of a good set is also a good set, due to Step 1.\n\nNow let $f(1) = a$. We show that $f(n) = an$ by induction on $n$. The base cases are $n = 1, 2, 3, 4, 5$; for $n = 1, 2, 4$ the result follows from Step 1.\n\nSet $t = f^{-1}(3a)$ and $s = f^{-1}(5a)$. The sets $\\{a, 4a, 3a\\}$ and $\\{a, 4a, 5a\\}$ are good, hence so are $\\{1, 4, t\\}$ and $\\{1, 4, s\\}$. Therefore, $\\{s, t\\} = \\{3, 5\\}$. But the set $\\{a, 5a, 6a\\}$ is also good, so the pair $\\{1, s\\}$ is contained in one more good set, which is not the case if $s = 3$, since $\\{1, 3\\}$ is contained in one single good set, namely, $\\{1, 4, 3\\}$. Thus $t = 3$ and $s = 5$, which establishes the base.\n\nFor the induction step, assume that $f(k) = ak$ for all $k \\le n$, where $n \\ge 5$. Choose $t = f^{-1}((n+1)a)$. Then the pair $\\{a, na\\}$ is contained in two good sets, namely, $\\{a, na, (n-1)a\\}$ and $\\{a, na, (n+1)a\\}$. Their pre-images, $\\{1, n, n-1\\}$ and $\\{1, n, t\\}$, are also good, and injectivity of $f$ forces $t = n+1$. This completes the induction step.\n\nFinally, since $f$ is surjective, $1 = f(n) = an$ for some positive integer $n$, so $a = 1 = n$. Consequently, $f$ is the identity, as claimed at the beginning of the solution.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 12639,
    "subject": "Mathematics (Olympiad)",
    "question": "Real numbers $x_1, x_2, x_3, x_4$ in $[0, 1]$ are such that the product\n\n$$\nK = |x_1 - x_2| \\cdot |x_1 - x_3| \\cdot |x_1 - x_4| \\cdot |x_2 - x_3| \\cdot |x_2 - x_4| \\cdot |x_3 - x_4|\n$$\n\nis as large as possible. Prove that $\\frac{1}{27} > K > \\frac{4}{243}$.",
    "options": [],
    "answer": "See 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 without loss of generality that $x_1 > x_2 > x_3 > x_4$.\n\nApplying the AM-GM inequality to $x_1 - x_2$, $x_2 - x_3$, and $x_3 - x_4$ gives\n\n$$\n\\sqrt[3]{(x_1 - x_2)(x_2 - x_3)(x_3 - x_4)} \\leq \\frac{(x_1 - x_2) + (x_2 - x_3) + (x_3 - x_4)}{3} = \\frac{x_1 - x_4}{3} \\leq \\frac{1}{3},\n$$\n\ni.e., $|x_1 - x_2| \\cdot |x_2 - x_3| \\cdot |x_3 - x_4| \\leq \\frac{1}{27}$. 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\n$$\nK = \\frac{1}{4} \\cdot \\frac{3}{4} \\cdot 1 \\cdot \\frac{1}{2} \\cdot \\frac{3}{4} \\cdot \\frac{1}{4} = \\frac{9}{512} > \\frac{4}{243},\n$$\n\nsince $9 \\cdot 243 = 2187 > 2048 = 4 \\cdot 512$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 12640,
    "subject": "Mathematics (Olympiad)",
    "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 symmetric to $D$ with respect to the axis $AC$. Points $A$, $B$, $C$, and $E$ are concyclic. Prove that triangle $ABC$ is isosceles.",
    "options": [],
    "answer": "See 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 the figure below). 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, $\\angle AC'C = \\angle AB'B = 90^\\circ$. Thus, $CC'$ is the altitude of triangle $ABC$, meaning that the altitude and the angle bisector through vertex $C$ coincide. Therefore, triangle $ABC$ is isosceles.\n\n![](images/prob1314_p16_data_72e756b08f.png)\n\nFigure 10",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 12641,
    "subject": "Mathematics (Olympiad)",
    "question": "Determine all prime numbers $p < q < r$ so that $A = (r-p)(r-q)(q-p)+1$ and $B = 3p+5q$ equal the same prime number.",
    "options": [],
    "answer": "See solution",
    "solution": "**Answer:** $p=2$, $q=5$, $r=7$.\n\n**Solution.** Let $p, q, r$ be the prime numbers that satisfy the conditions of the problem. If $p > 2$, then all $p, q, r$ are odd, thus the number $A = (r-p)(r-q)(q-p)+1$ is odd and the number $B = 3p + 5q$ is even, which contradicts the conditions of the problem. Therefore, $p=2$. Thus:\n\n$$\nA = (r-2)(r-q)(q-2)+1\n$$\nand\n$$\nB = 3p + 5q = 6 + 5q.\n$$\n\nSince $p < q < r$, we have $r-2 > q-2$ and $r-q \\ge 2$. Therefore, $6 + 5q > 2(q-2)^2 + 1$. Solving this inequality, we find $q < 7$. The prime numbers less than $7$ and greater than $2$ are $q=3$ and $q=5$.\n\nIf $q=3$, then $B = 3p + 5q = 21$, which is not prime.\n\nIf $q=5$, then $B = 3p + 5q = 31$, which is prime. This implies $A = (r-2)(r-5)(5-2)+1 = (r-2)(r-5) \\cdot 3 + 1 = 31$. Setting $(r-2)(r-5) \\cdot 3 + 1 = 31$ gives $(r-2)(r-5) = 10$. The only integer solution for $r$ with $r > 5$ is $r=7$.\n\nThus, the only solution is $p=2$, $q=5$, $r=7$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 12642,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $N$ be a positive integer. Determine the number of pairs $(a, b)$ of positive integers $a$ and $b$ such that the number\n\n$$\n\\frac{ab}{a+b}\n$$\n\nis a divisor of $N$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $d(n)$ denote the number of positive divisors of a positive integer $n$. We show that the number of pairs is $d(N)^2$.\n\nLet $m$ be a fixed positive divisor of $N$. The number of pairs $(a, b)$ such that $\\frac{ab}{a+b} = m$ can be found by rewriting the equation as\n\n$$\n(a - m)(b - m) = m^2.\n$$\n\nBoth $a - m$ and $b - m$ are positive divisors of $m^2$, so for each $m \\mid N$, there are $d(m^2)$ such pairs. Therefore, the total number of pairs is\n\n$$\n\\sum_{m \\mid N} d(m^2).\n$$\n\nAssume $N$ has prime factorization $p_1^{s_1} \\cdots p_t^{s_t}$. Since $d(kl) = d(k)d(l)$ when $\\gcd(k, l) = 1$,\n\n$$\n\\begin{aligned}\n\\sum_{m \\mid N} d(m^2) &= \\prod_{i=1}^{t} (d(1) + d(p_i^2) + \\cdots + d(p_i^{2s_i})) \\\\\n&= \\prod_{i=1}^{t} (1 + 3 + \\cdots + (2s_i + 1)) \\\\\n&= \\prod_{i=1}^{t} (s_i + 1)^2 \\\\\n&= d(N)^2.\n\\end{aligned}\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 12643,
    "subject": "Mathematics (Olympiad)",
    "question": "Probar que\n\n$$\n2014^{2013} - 1013^{2013} - 1001^{2013}\n$$\n\nes múltiplo de\n\n$$\n2014^3 - 1013^3 - 1001^3\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "Se tiene que $2014 = 1013 + 1001$. Sea $a = 2014$ y $b = 1013$, entonces deberemos probar que\n\n$$\na^{2013} - b^{2013} - (a-b)^{2013}\n$$\n\nes múltiplo de\n\n$$\na^3 - b^3 - (a-b)^3 = 3ab(a-b).\n$$\n\nAhora bien, por el binomio de Newton, se obtiene\n\n$$\na^{2013} - b^{2013} - (a-b)^{2013} = \\sum_{n=1}^{2012} \\binom{2013}{n} (-1)^n a^{2013-n} b^n,\n$$\n\ny agrupando por parejas simétricas resulta\n\n$$\n\\begin{aligned}\na^{2013} - b^{2013} - (a-b)^{2013} &= \\sum_{n=1}^{1006} \\binom{2013}{n} (-1)^n (a^{2013-n} b^n - a^n b^{2013-n}) \\\\\n&= \\sum_{n=1}^{1006} \\binom{2013}{n} (-1)^n a^n b^n (a^{2013-2n} - b^{2013-2n}).\n\\end{aligned}\n$$\n\nTeniendo en cuenta que\n\n$$\na^k - b^k = (a-b)(a^{k-1} + a^{k-2}b + \\dots + ab^{k-2} + b^{k-1}) = (a-b)p_k(a,b),\n$$\n\npodemos escribir\n\n$$\na^{2013} - b^{2013} - (a-b)^{2013} = ab(a-b) \\sum_{n=1}^{1006} \\binom{2013}{n} (-1)^n a^{n-1} b^{n-1} p_n(a,b).\n$$\n\nFinalmente, observemos que $2014^{2013} - 1013^{2013} - 1001^{2013}$ es múltiplo de 3, pero no $a = 2014$ ni $b = 1013$ ni $a - b = 1001$, de donde se concluye que\n\n$$\na^{2013} - b^{2013} - (a-b)^{2013} \\text{ es múltiplo de } 3ab(a-b).\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 12644,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $ABC$ be a right triangle with hypotenuse $AB$, and let $P$ be a point inside the shorter arc $AC$ of the circumcircle of triangle $ABC$. The line perpendicular to $CP$ passing through $C$ intersects the lines $AP$ and $BP$ at points $K$ and $L$, respectively. Prove that the ratio of the areas of triangles $BKL$ and $ACP$ does not depend on the choice of the point $P$.\n\n![](images/Cesko-Slovacko-Poljsko_2012_p5_data_03c61078a1.png)",
    "options": [],
    "answer": "See solution",
    "solution": "Throughout the solution, we denote by $S_{XYZ}$ the area of triangle $XYZ$.\n\nLet $PR$ be the diameter of the circumcircle of $ABC$. Then $ARBP$ is a rectangle. Since $BR$ is parallel to $PA$, we have $S_{PBK} = S_{PRK}$, which implies\n\n$$\nS_{BKL} = S_{LPR}.\n$$\n\nSince $PA = BR$, we have $\\angle PCA = \\angle BPR = \\angle LPR$.\n\nAlso, because $CPAR$ is cyclic, we have $\\angle CAP = \\angle CRP$. Hence, $LPR$ and $PCA$ are similar triangles.\n\nFinally,\n\n$$\nS_{BKL} : S_{ACP} = S_{LPR} : S_{ACP} = PR^2 : AC^2 = AB^2 : AC^2\n$$\n\nwhich does not depend on the point $P$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 12645,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $a_0$ be a positive integer, and define a sequence $(a_i)$ of positive rational numbers by the following rule: for each $i \\geq 0$, either\n\n$$\na_{i+1} = 2a_i + 1 \\quad \\text{or} \\quad a_{i+1} = \\frac{a_i}{a_i + 2}.\n$$\n\nSuppose that for some $n$, $a_n = 2014$. What is the smallest possible value of $n$?",
    "options": [],
    "answer": "See solution",
    "solution": "Let $a_i = \\frac{p_i}{q_i}$ in lowest terms, with $p_i, q_i > 0$ and $\\gcd(p_i, q_i) = 1$. The two recurrence options are:\n\n$$\n\\frac{p_{i+1}}{q_{i+1}} = \\frac{2p_i + q_i}{q_i} \\quad \\text{or} \\quad \\frac{p_{i+1}}{q_{i+1}} = \\frac{p_i}{p_i + 2q_i}.\n$$\n\nIf $p_i$ and $q_i$ are both odd, so are $2p_i + q_i$, $q_i$, $p_i$, and $p_i + 2q_i$. Thus, all $p_k, q_k$ remain odd, contradicting $a_n = 2014$ (an integer). Therefore, $p_i$ and $q_i$ must always have opposite parity for $i = 0, 1, \\dots, n$.\n\nIf $p_i$ is odd, only the first recurrence is possible, and $q_i$ is even. Then:\n\n$$\np_{i+1} = p_i + \\frac{q_i}{2}, \\quad q_{i+1} = \\frac{q_i}{2}.\n$$\n\nIf $p_i$ is even, only the second recurrence is possible, and $p_i$ is divisible by $2$:\n\n$$\np_{i+1} = \\frac{p_i}{2}, \\quad q_{i+1} = \\frac{p_i}{2} + q_i.\n$$\n\nIn both cases, $p_{i+1} + q_{i+1} = p_i + q_i$, so $p_i + q_i = 2015$ for all $i$ (since $p_n + q_n = 2015$ when $a_n = 2014$).\n\nBy induction, $(p_n, q_n) \\equiv \\left(\\frac{p_0}{2^n}, \\frac{q_0}{2^n}\\right) \\pmod{2015}$. Since $q_0 = 1$, we require $2^n \\equiv 1 \\pmod{2015}$.\n\nSince $2015 = 5 \\cdot 13 \\cdot 31$, we need $4 \\mid n$, $12 \\mid n$, and $5 \\mid n$. The least common multiple is $60$, so the smallest $n$ is $60$.\n\n**Answer:** $n = 60$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 12646,
    "subject": "Mathematics (Olympiad)",
    "question": "a) Consider a differentiable and convex function $f: [0, \\infty) \\to [0, \\infty)$ such that $f(x) \\leq x$ for $x \\geq 0$. Prove that $f'(x) \\leq 1$ for all $x \\geq 0$.\n\nb) Determine all differentiable and convex functions $f: [0, \\infty) \\to [0, \\infty)$ with the properties $f(0) = 0$ and $f'(x) \\cdot f(f(x)) = x$ for all $x \\geq 0$.",
    "options": [],
    "answer": "See solution",
    "solution": "a) Suppose the contrary: there exists $a \\geq 0$ such that $f'(a) > 1$. Since $\\lim_{x \\to a} \\frac{f(x) - f(a)}{x - a} > 1$, there is $b > a$ such that $\\frac{f(b) - f(a)}{b - a} > 1$. For any $x > b$, by the convexity of $f$, we have $\\frac{f(x) - f(b)}{x - b} \\geq \\frac{f(b) - f(a)}{b - a} = m > 1$. Thus, $f(x) \\geq m x - m b + f(b)$, so $f(x) > x$ for sufficiently large $x$, a contradiction.\n\nb) We will prove that $f(x) = x$ for all $x \\geq 0$. Since $f'(x) = \\frac{x}{f(f(x))} > 0$ for $x > 0$, $f$ is increasing. The function $f$ is convex and differentiable, so $f'$ is non-decreasing.\n\nSuppose, for contradiction, that $f(a) < a$. Then $f(f(a)) < f(a) < a$, so $f'(a) > 1$. By part (a), there is $b > a$ such that $f(b) = b$. Then $f(f(b)) = b$, which gives $f'(b) = 1 < f'(a)$, contradicting the monotonicity of $f'$.\n\nTherefore, $f(x) \\geq x$ for $x \\geq 0$. Then $f(f(x)) \\geq f(x) \\geq x$, so $f'(x) \\leq 1$ for $x \\geq 0$. By the Mean Value Theorem, $f(x) - f(0) = x f'(c_x) \\leq x$ for $x > 0$, giving $f(x) = x$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 12647,
    "subject": "Mathematics (Olympiad)",
    "question": "Suppose that a point $P$ lying in the interior of a convex quadrilateral $ABCD$ satisfies\n\n$$\n\\angle PAD = \\angle ADP = \\angle CBP = \\angle PCB = \\angle CPD.\n$$\n\nLet $O$ be the circumcentre of the triangle $CPD$. Prove that $OA = OB$.\n\n![](images/CZE_ABooklet_2024_p10_data_79f3da4f63.png)",
    "options": [],
    "answer": "See 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\n![](images/CZE_ABooklet_2024_p10_data_79f3da4f63.png)\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": 12648,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $a_1, a_2, \\dots, a_n$ be non-negative integers such that for all real numbers $x_1 > x_2 > x_3 > \\dots > x_n > 0$ with $x_1 + x_2 + \\dots + x_n < 1$ it holds that $\\sum_{k=1}^n a_k x_k^3 < 1$.\n\nShow that\n\n$$\nna_1 + (n-1)a_2 + \\dots + (n-j+1)a_j + \\dots + a_n \\le \\frac{n^2(n+1)^2}{4}.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "The assertion can be rewritten as\n\n$$\n\\sum_{k=1}^{n} \\sum_{j=1}^{k} a_j \\le \\sum_{k=1}^{n} k^3.\n$$\n\nIt therefore suffices to prove\n\n$$\n\\sum_{j=1}^{k} a_j \\le k^3 \\quad (6)\n$$\nfor every $k = 1, \\dots, n$.\n\nIn order to prove (6) we fix $k$ and set\n\n$$\nx_i = \\frac{1}{k} - \\frac{i}{N} \\quad \\text{for } i = 1, \\dots, k\n$$\n\nand\n\n$$\nx_i = \\frac{n+1-i}{N^2} \\quad \\text{for } i = k+1, \\dots, n\n$$\n\nfor some integer $N > 0$ to be determined as follows:\n\nThe conditions $x_1 > x_2 > \\dots > x_n > 0$ are certainly fulfilled in the case $k = n$ and for $k < n$ the only non-trivial relation is $x_k > x_{k+1}$, that is, $\\frac{1}{k} - \\frac{k}{N} > \\frac{n-k}{N^2}$. Hence we choose $N > kn$, in order to have $\\frac{1}{k} - \\frac{k}{N} > \\frac{n-k}{N} \\ge \\frac{n-k}{N^2}$.\n\nThe condition $x_1 + \\dots + x_n < 1$ means $1 - \\frac{k(k+1)}{N} + \\frac{(n-k)(n-k+1)}{N^2} < 1$ and will be satisfied for\n\n$$\nN > \\frac{(n-k)(n-k+1)}{k(k+1)}\n$$\n\nFor $N > \\max\\{kn, \\frac{(n-k)(n-k+1)}{k(k+1)}\\}$ the numbers $x_1, \\dots, x_n$ fulfill the relevant conditions and we conclude\n\n$$\n\\sum_{i=1}^{n} a_i x_i^3 < 1.\n$$\n\nBy taking the limit $N \\to \\infty$ we get $\\sum_{i=1}^{n} a_i \\lim_{N \\to \\infty} x_i^3 \\le 1$, which gives $\\sum_{i=1}^{k} a_i \\frac{1}{k^3} \\le 1$, hence the desired estimate (6). $\\square$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 12649,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $ABC$ be an acute triangle and $D$ a variable point on side $AC$. Point $E$ is on $BD$ such that\n\n$$\nBE = \\frac{BC^2 - CD \\cdot CA}{BD}.\n$$\n\nAs $D$ varies on side $AC$, prove that the circumcircle of $ADE$ passes through a fixed point other than $A$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let the circumcircle of triangle $CED$ intersect $BC$ at point $G$. From the power of a point, we have\n\n$$\nBG \\cdot BC = BE \\cdot BD.\n$$\n\nCombining this with the given condition,\n\n$$\n\\frac{BG \\cdot BC}{BD} = BE = \\frac{BC^2 - CD \\cdot CA}{BD},\n$$\n\nso\n\n$$\nCD \\cdot CA = BC(BC - BG) = BC \\cdot CG.\n$$\n\nThis implies that $D$, $A$, $B$, $G$ are concyclic as well. Thus,\n\n$$\n\\angle BEC = \\angle BGD = 180^\\circ - \\angle BAD = 180^\\circ - \\angle CAB.\n$$\n\nNow let the circumcircle of $ADE$ and $BEC$ intersect again at $X$. Since\n\n$$\n\\angle XCB = \\angle XEB = 180^\\circ - \\angle XED = \\angle XAD = \\angle XAC,\n$$\n\nand\n\n$$\n\\angle BXC = \\angle BEC = 180^\\circ - \\angle BAC,\n$$\n\nwe have that $X$ is on the unique circle through $A$ and $C$ tangent to side $BC$ at point $C$ and the circumcircle of $BHC$, where $H$ is the orthocenter of triangle $ABC$. This intersection is unique, and we are done.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 12650,
    "subject": "Mathematics (Olympiad)",
    "question": "Which regular $n$-gons have a triangulation consisting of isosceles triangles?",
    "options": [],
    "answer": "See solution",
    "solution": "Call $n$ *good* if the regular $n$-gon can be triangulated with isosceles triangles. By *segments* we mean the sides and the diagonals of the $n$-gon; the sides are the shortest among all segments.\n\nLet $n$ be good and $T$ an isosceles triangulation of the regular $n$-gon $P$. Suppose that the base of a triangle $\\Delta \\in T$ is a side $a$ of $P$. Then the vertex of $\\Delta$ opposing $a$ is on the perpendicular bisector of $a$, which passes through the center of $P$, and also the circumcircle of $P$. Hence $n$ is odd and the center of $P$ is interior to $\\Delta$, implying that such a triangle $\\Delta$ is unique.\n\nLet $n$ be even. Then sides are the shortest segments and none of them is a base of a triangle of $T$. So all of them are divided into pairs of consecutive ones, and each pair contains the equal sides of a triangle of $T$. Deleting these $\\frac{n}{2}$ isosceles triangles leaves a regular $\\frac{n}{2}$-gon which therefore also admits of an isosceles triangulation. It follows that an even $n \\ge 6$ is good if and only if so is $\\frac{n}{2}$.\n\nLet $n$ be odd. Then the sides cannot be paired up like in the even case; one of them must be a base of a triangle $\\Delta$ from $T$ as explained earlier. The equal sides of $\\Delta$ are diagonals of $P$ (longest ones). Removing $\\Delta$ leaves two congruent polygons which must have isosceles triangulations. Let $P_1$ be one of them. It has $n_1 = \\frac{1}{2}(n(n+1))$ sides; thus $n_1$ is good. One of the sides is a diagonal $d_1$, the rest are sides of $P$, hence shorter. So $d_1$ is a base of a triangle $\\Delta_1$ of $T$, and its opposite vertex divides the remaining $n_1 - 3$ vertices into two equal halves. It follows that $n_1$ is odd. Remove $\\Delta_1$ from $P_1$ and denote by $P_2$ one of the two obtained congruent polygons with $n_2 = \\frac{1}{2}(n_1 + 1)$ sides; $n_2$ is good. The same argument applies to $P_2$ because one of its sides is a diagonal $d_2$, and the rest are sides of $P$.\n\nWe conclude that $n_2$ is odd, then define $n_3 = \\frac{1}{2}(n_2 + 1)$, and so on. Thus each of the numbers $n > n_1 > n_2 > \\dots$ is odd and good, as long as it is $\\ge 3$. Let $k$ be such that $n_k \\ge 3 > n_{k+1}$. If $n_{k+1} = 1$ then $n_k = 1$ which is false. So $n_{k+1} = 2$ and so $n_k = 3$. Write $n_k = 3 = 2^1 + 1$ and backwards to obtain $n_{k-1} = 2^2 + 1$ and likewise $n_{k-2} = 2^3 + 1$, ..., $n_1 = 2^k + 1$, $n = 2^{k+1} + 1$. Therefore $n-1$ is a power of 2. In addition, the steps of the argument imply a construction showing that the converse is also true.\n\nTo sum up, consider two cases for a general $n$. If $n \\ge 4$ is a power of 2, $n = 2^m$ with $m \\ge 2$, then it is good if and only if so are $2^{m-1}, 2^{m-2}, \\dots, 2^2 = 4$. Since the square has an isosceles triangulation, the powers of 2 are good. If $n \\ge 3$ is not a power of 2 then $n = 2^m k$ with $k \\ge 3$ odd and $m \\ge 0$. By the above, $n$ is good if and only if so is $k$, and the latter holds if and only if $k$ is of the form $k = 2^l + 1$ with $l \\ge 1$. Hence $n = 2^u + 2^v$ with $u > v \\ge 0$. In conclusion, the good numbers are $2^m$ with $m \\ge 2$ and $2^u + 2^v$ with $u > v \\ge 0$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 12651,
    "subject": "Mathematics (Olympiad)",
    "question": "Determine all positive integers $n$ for which there exists a polynomial $f(x)$ with real coefficients, with the following properties:\n\n1. For each integer $k$, the number $f(k)$ is an integer if and only if $n$ does not divide $k$.\n2. The degree of $f$ is less than $n$.",
    "options": [],
    "answer": "See solution",
    "solution": "We will show that such a polynomial exists if and only if $n = 1$ or $n$ is a power of a prime.\n\n**Lemma 1.** If $p^a$ is a power of a prime and $k$ is an integer, then\n$$\n\\frac{(k-1)(k-2)\\cdots(k-p^a+1)}{(p^a-1)!}\n$$\nis divisible by $p$ if and only if $k$ is not divisible by $p^a$.\n\n*Proof.*\n- If $p^a \\mid k$, then in\n$$\n\\frac{(k-1)(k-2)\\cdots(k-p^a+1)}{(p^a-1)!} = \\frac{k-1}{p^a-1} \\cdot \\frac{k-2}{p^a-2} \\cdots \\frac{k-p^a+1}{1},\n$$\n$p$ has the same maximal exponent in numerator and denominator for each fraction, so the product is an integer not divisible by $p$.\n- If $p^a \\nmid k$, then\n$$\n\\frac{(k-1)(k-2)\\cdots(k-p^a+1)}{(p^a-1)!} = \\frac{p^a}{k} \\cdot \\frac{k(k-1)\\cdots(k-p^a+1)}{(p^a)!},\n$$\nand $\\frac{p^a}{k}$ is not an integer, so the whole expression is divisible by $p$.\n\n**Lemma 2.** If $g(x)$ is a polynomial with degree less than $n$, then\n$$\n\\sum_{\\ell=0}^{n} (-1)^{\\ell} \\binom{n}{\\ell} g(x + n - \\ell) = 0.\n$$\n*Proof.* Induction on $n$.\n- For $n = 1$, $g(x)$ is constant and $g(x+1) - g(x) = 0$.\n- Assume true for $n-1$. Let $h(x) = g(x+1) - g(x)$. Then\n$$\n\\sum_{\\ell=0}^{n-1} (-1)^{\\ell} \\binom{n-1}{\\ell} h(x+n-1-\\ell) = 0\n$$\nExpanding and rearranging gives\n$$\n\\sum_{\\ell=0}^{n} (-1)^{\\ell} \\binom{n}{\\ell} g(x+n-\\ell) = 0.\n$$\n\n**Lemma 3.** If $n$ has at least two distinct prime divisors, then the greatest common divisor of $\\binom{n}{1}, \\binom{n}{2}, \\ldots, \\binom{n}{n-1}$ is $1$.\n\n*Proof.* Suppose $p$ divides all $\\binom{n}{k}$ for $1 \\leq k \\leq n-1$. Then $p \\mid n$. Let $a$ be the exponent of $p$ in $n$. Since $n$ has at least two prime divisors, $1 < p^a < n$. Thus $p$ divides both $\\binom{n}{p^a}$ and $\\binom{n}{p^a-1}$, so $p$ divides $\\binom{n}{p^a} - \\binom{n}{p^a-1} = \\binom{n-1}{p^a-1}$, contradicting Lemma 1.\n\n**Construction:**\n- For $n=1$, $f(x) = \\frac{1}{2}$ works.\n- If $n = p^a$ (prime power), let\n$$\nf(x) = \\frac{1}{p} \\binom{x-1}{p^a-1} = \\frac{1}{p} \\cdot \\frac{(x-1)(x-2)\\cdots(x-p^a+1)}{(p^a-1)!}.\n$$\nThe degree is $n-1$. By Lemma 1, $f(k)$ is integer if and only if $n$ does not divide $k$.\n\n**Impossibility for composite $n$ with at least two distinct primes:**\nIf $n$ has at least two distinct prime divisors, Lemma 3 shows no such $f(x)$ can exist.\n\n**Conclusion:**\nAll such $n$ are $n=1$ or $n$ a power of a prime.",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 12652,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $Q_n$ be the set of $n$-tuples $x = (x_1, \\ldots, x_n)$, where $x_i \\in \\{0, 1, 2\\}$ for $i = 1, \\ldots, n$.\n\nA triad $(x, y, z)$, where $x = (x_1, \\ldots, x_n)$, $y = (y_1, \\ldots, y_n)$, $z = (z_1, \\ldots, z_n)$, of distinct elements of $Q_n$ is called *good* if there exists at least one $i \\in \\{1, \\ldots, n\\}$ such that $\\{x_i, y_i, z_i\\} = \\{0, 1, 2\\}$.\n\nA subset $A$ of $Q_n$ is called *good* if every three elements of $A$ form a good triad.\n\nProve that every good subset of $Q_n$ has at most $2\\left(\\frac{3}{2}\\right)^n$ elements.",
    "options": [],
    "answer": "See solution",
    "solution": "We use induction on $n$.\n\nFor $n = 1$, the claim is obvious.\n\nAssume every good subset of $Q_{n-1}$ has at most $2\\left(\\frac{3}{2}\\right)^{n-1}$ elements.\n\nDefine\n$$\nA_0 = \\{(x_1, \\ldots, x_n) \\in A : x_n \\neq 0\\}, \\quad A_1 = \\{(x_1, \\ldots, x_n) \\in A : x_n \\neq 1\\}, \\quad A_2 = \\{(x_1, \\ldots, x_n) \\in A : x_n \\neq 2\\}.\n$$\n\nEach $A_j$ is a good subset. For $A_0$, deleting the last coordinate from each element gives $A_0'$, a good subset of $Q_{n-1}$, and $|A_0'| = |A_0|$ if $|A_0| \\geq 3$.\n\nIf $|A_0'| \\neq |A_0|$, then there exist $x, y \\in A_0$ with the same first $n-1$ coordinates and last coordinates $1$ and $2$. For any $z \\in A_0$, its last coordinate cannot be $0$, so $x, y, z$ cannot form a good triad—a contradiction.\n\nBy induction,\n$$\n|A_0| \\leq \\max\\{2, |A_0'|\\} \\leq 2\\left(\\frac{3}{2}\\right)^{n-1}.\n$$\nSimilarly, $|A_1|, |A_2| \\leq 2\\left(\\frac{3}{2}\\right)^{n-1}$.\n\nEach element of $A$ appears in exactly two of $A_0, A_1, A_2$, so\n$$\n|A| = \\frac{1}{2}(|A_0| + |A_1| + |A_2|) \\leq 2\\left(\\frac{3}{2}\\right)^n.\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 12653,
    "subject": "Mathematics (Olympiad)",
    "question": "Suppose that at instant 0 the traffic lights turn green and the car passes through the first light at instant $t_0 \\ge 0$ (time is measured in seconds). The traffic lights will stay green between the time instants $150k + 90$ and $150(k+1)$, for each integer $k$. The car will pass through the lights at the instants $t_0 + \\frac{1500}{v} r$, for each nonnegative integer $r$.\n\nWhat are the possible values for the speed $v$ of the car (in m/s) that allow a non-stopping trip through all the lights?",
    "options": [],
    "answer": "See solution",
    "solution": "A necessary and sufficient condition for a non-stopping trip is that, for every integer $r$, $\\frac{t_0}{150} + \\frac{10}{v} r$ equals an integer plus a number between $0$ and $\\frac{3}{5}$. This is possible if $\\frac{10}{v}$ is an integer (with any $t_0$ between $0$ and $90$) or if $\\frac{10}{v}$ is half of an odd integer (with any $t_0$ between $0$ and $15$).\n\nTo show these are the only possibilities: If $\\frac{10}{v}$ is irrational, then by Kronecker's theorem, the fractional part of $\\frac{t_0}{150} + \\frac{10}{v} r$ is dense in $(0,1)$ and will eventually fall between $\\frac{3}{5}$ and $1$, violating the condition. If $\\frac{10}{v} = \\frac{p}{q}$ with $\\gcd(p, q) = 1$, then the fractional part cycles through $\\frac{s}{q}$ for $0 \\leq s \\leq q-1$. The gap between consecutive values is $\\frac{1}{q}$, so to avoid the interval $\\left(\\frac{3}{5}, 1\\right)$, we need $\\frac{1}{q} \\geq 1 - \\frac{3}{5} = \\frac{2}{5}$, so $q \\leq 2$. Thus, $\\frac{10}{v}$ is either an integer or half of an odd integer.\n\nTherefore, the possible values for the speed of the car are $v = \\frac{20}{k}$ m/s, for every positive integer $k$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 12654,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $ABCDEF$ be a regular hexagon of side length $2$. Through the vertices and the midpoints of the sides, we construct lines parallel to the sides, which divide the hexagon into $24$ congruent equilateral triangles, whose vertices are called *nodes*.\n\nA *sheet* is any (non-degenerate) equilateral triangle whose vertices are nodes. A *trio* of a node $X$ is the figure formed by three adjacent sheets such that their intersection is only $X$ and they are not congruent in pairs.\n\n**a)** Find the maximal possible area of a trio.\n\n**b)** Show that there exists a node whose trios can cover the whole hexagon, and a node whose trios cannot cover the whole hexagon.\n\n**c)** Determine the total number of trios associated to the hexagon.\n\n![](images/RMC_2023_v2_p68_data_86e0aac227.png \"Figura 1\")\n\n![](images/RMC_2023_v2_p68_data_c94aa6a6b4.png \"Figura 2\")",
    "options": [],
    "answer": "See solution",
    "solution": "**a)** Consider the regular hexagon $ABCDEF$, centered at $O$, and let $T$, $U$, $V$, $X$, $Y$, $Z$ be the midpoints of the sides $AB$, $BC$, $CD$, $DE$, $EF$, $FA$, respectively.\n\nNotice that the nodes situated on the sides of the hexagon cannot have trios, so a node that admits a trio is either $O$, or situated at distance $1$ from $O$. Denote these latter nodes by $M$, $N$, $P$, $Q$, $R$, $S$, as shown in Figure 1.\n\nThe side length of an equilateral triangle with nodes as vertices may be $1$, $\\sqrt{3}$, $2$, $\\sqrt{7}$, $3$, or $2\\sqrt{3}$. A sheet of side length $3$ or $2\\sqrt{3}$ cannot be part of a trio, because the vertices of this sheet would be situated on the hexagon sides.\n\nThere are no sheets of side $\\sqrt{7}$ that have $O$ as a vertex. Also, there are only two sheets of side $\\sqrt{7}$ that have $R$ as a vertex, namely $RAU$ and $RCT$. Moreover, the intersection of a side $2$ sheet and a side $\\sqrt{7}$ sheet at $R$ is non-empty, so we can't form trios with such sheets.\n\nConsequently, the sheets $RAU$, $RDP$, and $REY$ (see Figure 1) form a trio with maximal possible area, and the maximal area is:\n\n$$\n\\frac{\\sqrt{3}}{4} + \\frac{3\\sqrt{3}}{4} + \\frac{7\\sqrt{3}}{4} = \\frac{11\\sqrt{3}}{4}\n$$\n\n**b)** Considering the colored trio from Figure 2, we notice that $OAB$ covers one sixth of the hexagon. Rotating, $OBC$, $OCD$, $ODE$, $OEF$, and $OFA$ are sheets for five other trios of $O$. All these six trios cover the whole hexagon.\n\nThe node $R$ has exactly two sheets of maximal side length $\\sqrt{7}$, namely $RAU$ and $RCT$. Since $RB = 3 > \\sqrt{7}$, it follows that not all the points of the segment $RB$ can be covered with the trios of $R$.\n\n**c)** We say that a trio is $(x, y, z)$-type if the side lengths of its sheets are $x, y, z$.\n\nThe center $O$ has only $(1, \\sqrt{3}, 2)$-type trios, as indicated in Figure 2. Now, we count how many trios have $OQR$ as one of their sheets. The only side $2$ sheets that can be part of a trio are $OAF$, $OAB$, and $OBC$. For each of these side $2$ sheets, we can choose the side $\\sqrt{3}$ sheet in two ways, so there are exactly $6$ trios which contain $OQR$. Since $O$ is the vertex of six side $1$ sheets, the node $O$ has exactly $36$ trios.\n\nThe node $R$ admits only $(1, \\sqrt{3}, 2)$ and $(1, \\sqrt{3}, \\sqrt{7})$-type trios.\n\nConsider the sheets $RDP$ and $RNZ$. To complete a trio, there are two choices for the side-$1$ sheet, specifically $REX$ and $REY$. The same applies for the sheets $RFM$ and $RNV$. Therefore, the node $R$ has exactly four $(1, \\sqrt{3}, 2)$-type trios.\n\nLikewise, the sheets $RAU$ and $RDP$ can form a trio together with one of the following three triangles of side length $1$: $REX$, $REY$, and $RSY$. Also, there are three side $1$ sheets that form a trio with $RCT$ and $RFM$. Consequently, $R$ has six $(1, \\sqrt{3}, \\sqrt{7})$-type trios.\n\nIt follows that there are $10$ trios associated to the node $R$, so there are $60$ trios associated to nodes other than $O$. Adding all up, there are $96$ trios associated to the hexagon.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 12655,
    "subject": "Mathematics (Olympiad)",
    "question": "A positive integer $n$ is *good* if every positive divisor of $n$, when increased by $1$, is a divisor of $n+1$. Find all good positive integers.",
    "options": [],
    "answer": "See solution",
    "solution": "$1$ together with all odd primes.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 12656,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $A$ be a set with $|A| = 225$, meaning that $A$ has 225 elements. Suppose further that there are eleven subsets $A_1, \\dots, A_{11}$ of $A$ such that $|A_i| = 45$ for $1 \\le i \\le 11$ and $|A_i \\cap A_j| = 9$ for $1 \\le i < j \\le 11$. Prove that $|A_1 \\cup A_2 \\cup \\dots \\cup A_{11}| \\ge 165$, and give an example for which equality holds.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $S$ be the complement of $A_1 \\cup A_2 \\cup \\dots \\cup A_{11}$ in $A$; we wish to prove that $|S| \\le 60$. For $\\ell \\ge 0$, define\n\n$$\n\\theta(\\ell) = \\left(1 - \\frac{\\ell}{2}\\right) \\left(1 - \\frac{\\ell}{3}\\right) = 1 - \\frac{2}{3}\\ell + \\frac{1}{3}\\binom{\\ell}{2}.\n$$\n\nNote that $\\theta(0) = 1$ and $\\theta(\\ell) \\ge 0$ for any integer $\\ell > 0$. For $n \\in A$, let $\\ell(n)$ be the number of sets among $A_1, \\dots, A_{11}$ containing $n$. Since $S$ is the intersection of the complements of the $A_i$, we see that\n\n$$\n|S| \\le \\sum_{n \\in A} \\theta(\\ell(n)).\n$$\n\nOn the other hand, we have\n\n$$\n\\sum_{n \\in A} \\theta(\\ell(n)) = \\sum_{n \\in A} \\left(1 - \\frac{2}{3}\\ell(n) + \\frac{1}{3}\\binom{\\ell(n)}{2}\\right) = |A| - \\frac{2}{3}\\sum_{i} |A_i| + \\frac{1}{3}\\sum_{i<j} |A_i \\cap A_j|.\n$$\n\nPutting these two equations together, we obtain\n\n$$\n|S| \\le 225 - \\frac{2}{3} \\cdot 11 \\cdot 45 + \\frac{1}{3} \\cdot \\binom{11}{2} \\cdot 9 = 60,\n$$\n\nand therefore $|A_1 \\cup A_2 \\cup \\dots \\cup A_{11}| \\ge 165$.\n\nIt remains to give an example showing that this lower bound is best possible. Let $p_1, p_2, \\dots, p_{11}$ be a set of 11 distinct primes, and let $A'$ denote the set of all products of three of these primes. Let $A'' = \\{q_1, q_2, q_3, \\dots, q_{60}\\}$ be a set of 60 distinct positive integers that are all coprime to $p_1, \\dots, p_{11}$. Set $A = A' \\cup A''$, and define\n\n$$\nA_i = \\{n \\in A' : p_i \\mid n\\}.\n$$\n\nThen $|A_i| = \\binom{10}{2} = 45$, $|A_i \\cap A_j| = \\binom{9}{1} = 9$, and\n\n$$\n|A_1 \\cup A_2 \\cup \\dots \\cup A_{11}| = |A'| = \\binom{11}{3} = 165.\n$$\n\nFinally, $|A| = |A'| + |A''| = 165 + 60 = 225$, so this $A$ and $A_1, \\dots, A_{11}$ give a valid example.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 12657,
    "subject": "Mathematics (Olympiad)",
    "question": "$n$ students with different names are forming a circle and standing toward the center. Initially, $n$ name tags are distributed to them at random. The students repeat the following process:\n\nEvery student having their own name tag gets out of the circle, and the remaining students pass their name tags to the right.\n\nFind the number of initial distributions of name tags such that some students still remain after 4 processes.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $A$ be a finite set of positive integers and $\\pi$ a permutation (bijection) on $A$. A pair $(x, \\pi(x)) \\in A^2$ is called an \"up\" if $x < \\pi(x)$, and $\\text{up}(\\pi)$ is the number of ups in $\\pi$.\n\nGiven a permutation $\\pi$ on $A = \\{x_1 < x_2 < \\dots < x_m\\}$, define $\\pi'$ by\n\n$$\n\\pi' = x_2 x_3 \\cdots x_1 \\circ \\pi,\n$$\n\nwhere $\\circ$ denotes functional composition. Then:\n\n*Lemma.* If $\\pi$ is a permutation on $A$ without fixed points, then\n$$\n\\text{up}(\\pi') = \\text{up}(\\pi) - 1.\n$$\n\nLet $S_0 = \\{s_1, \\dots, s_n\\}$ be the students and $T_0 = \\{t_1, \\dots, t_n\\}$ the name tags. A distribution is a bijection $\\tau_0: T_0 \\to S_0$, which we can view as a permutation on $n$. Let $F_1$ be the set of fixed points of $\\tau_0$. Define $S_1 = S_0 - \\{s_i \\mid i \\in F_1\\}$ and $T_1 = T_0 - \\{t_i \\mid i \\in F_1\\}$. Then $\\tau_1: T_1 \\to S_1$ is the state after students with their own name tag leave. It is clear that $\\text{up}(\\tau_0) = \\text{up}(\\tau_1)$.\n\nPassing name tags to the right transforms $\\tau_1$ to $\\tau_1'$, and by the lemma:\n$$\n\\text{up}(\\tau_1') = \\text{up}(\\tau_1) - 1 = \\text{up}(\\tau_0) - 1.\n$$\n\nThus, each process decreases the number of ups by 1. If the initial distribution $\\tau_0$ has at most 3 ups, then all students will have their own name tag within 3 processes, i.e., all will leave by the 4th process. Therefore, the answer is:\n$$\nn! - |\\{\\tau_0 \\in \\text{Sym}_n \\mid \\text{up}(\\tau_0) \\le 3\\}|,\n$$\nwhere $\\text{Sym}_n$ is the set of permutations on $\\{1, \\dots, n\\}$.\n\nLet $B(n, k) = \\{\\pi \\in \\text{Sym}_n \\mid \\text{up}(\\pi) \\le k\\}$ and $b(n, k) = |B(n, k)|$. By considering the position of $n$ in $\\pi \\in B(n, k)$, we get the recurrence:\n$$\nb(n, k) = (k + 1) b(n - 1, k) + (n - k - 1) b(n - 1, k - 1), \\quad n \\ge 2\n$$\nwith $b(n, 0) = 1$ and $b(n, k) = 0$ for $k \\ge n$.\n\nFor $k = 1$:\n$$\nb(n, 1) = 2 b(n - 1, 1) + (n - 2), \\quad b(1, 1) = 1.\n$$\nThis yields $b(n, 1) = 2^n - n$.\n\nFor $k = 2$:\n$$\nb(n, 2) = 3 b(n - 1, 2) + (n - 3)(2^{n-1} - n + 1), \\quad b(1, 2) = 1.\n$$\nThis yields $b(n, 2) = 3^n - n 2^n + \\binom{n}{2}$.\n\nFor $k = 3$:\n$$\nb(n, 3) = 4 b(n - 1, 3) + (n - 4) \\left( 3^{n-1} - (n-1) 2^{n-1} + \\binom{n-1}{2} \\right), \\quad b(1, 3) = 1.\n$$\nThis yields $b(n, 3) = 4^n - n 3^n + \\binom{n}{2} 2^n - \\binom{n}{3}$.\n\nTherefore, the answer is:\n$$\nn! - b(n, 3) = n! - 4^n + n 3^n - \\binom{n}{2} 2^n + \\binom{n}{3}.\n$$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 12658,
    "subject": "Mathematics (Olympiad)",
    "question": "A group of 16 people will be partitioned into 4 indistinguishable 4-person committees. Each committee will have one chairperson and one secretary. The number of different ways to make these assignments can be written as $3^r M$, where $r$ and $M$ are positive integers and $M$ is not divisible by 3. What is $r$?\n\n(A) 5  (B) 6  (C) 7  (D) 8  (E) 9",
    "options": [],
    "answer": "See solution",
    "solution": "The 16 people can be partitioned into the 4 committees, each of size 4, in\n\n$$\n\\frac{16!}{(4!)^5}\n$$\n\nways; four of the $4!$ factors come from permuting the members of the committees and one $4!$ factor comes from permuting the committees. Then there are $4^4$ ways to choose the four chairpersons and $3^4$ ways to choose the four secretaries. This gives a total of\n\n$$\n\\frac{16! \\cdot 4^4 \\cdot 3^4}{(4!)^5}\n$$\n\nassignments. The numerator has 1 factor of 3 in each of 15, 12, 6, and 3; it has 2 factors of 3 in 9 and 4 factors in $3^4$, a total of 10. The denominator has 5 factors of 3. Thus $r = 10 - 5 = 5$.\n\nAlternatively,\n\nThere are $16 \\cdot 15$ ways to choose the chairperson and secretary of the first committee, $14 \\cdot 13$ ways for the second, $12 \\cdot 11$ for the third, and $10 \\cdot 9$ for the fourth. There are then $\\frac{8 \\cdot 7}{2} = 28$ ways to choose the remaining members of the first committee, $\\frac{6 \\cdot 5}{2} = 15$ for the second, $\\frac{4 \\cdot 3}{2} = 6$ for the third, and $\\frac{2 \\cdot 1}{2} = 1$ for the fourth. There are $4!$ ways to account for the indistinguishability of the committees. This gives a total of\n\n$$\n\\frac{16 \\cdot 15 \\cdot 14 \\cdot 13 \\cdot 12 \\cdot 11 \\cdot 10 \\cdot 9 \\cdot 28 \\cdot 15 \\cdot 6 \\cdot 1}{4!}\n$$\n\nassignments. There are $1 + 1 + 2 + 1 + 1 = 6$ factors of 3 in the numerator and 1 factor of 3 in the denominator, so $r = 6 - 1 = 5$.\n\n*Note:* The number of ways to make the assignments is $54,486,432,000 = 3^5 \\cdot 224,224,000$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 12659,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all integers $n \\ge 2$ with the property that there exists a permutation $(a_1, a_2, \\ldots, a_n)$ of the set $\\{1, 2, \\ldots, n\\}$ such that the numbers\n\n$$\na_1 + a_2 + \\cdots + a_k, \\quad k = 1, 2, \\ldots, n,\n$$\n\nleave different remainders when divided by $n$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $n$ be such a number. For every $k = 1, \\ldots, n$, let $r_k$ be the remainder obtained from dividing $s_k = a_1 + a_2 + \\cdots + a_k$ by $n$.\n\nFirst, we prove that $a_1 = n$. Suppose not; then there would be $k \\ge 2$ such that $a_k = n$. Since $s_k = s_{k-1} + n$, we have $n \\mid s_k - s_{k-1}$, so $r_k = r_{k-1}$, a contradiction. Thus, $a_1 = n$ and $r_1 = 0$.\n\nSince $s_n = \\frac{n(n+1)}{2}$, if $n$ were odd, then $n \\mid s_n$, so $r_n = 0 = r_1$, a contradiction. Therefore, $n$ must be even. We will show that every even number has the required property.\n\nIndeed, consider $n$ even and, for $k = 1, 2, \\ldots, n$, define\n\n$$\na_k = \\begin{cases} n+1-k, & k \\text{ odd} \\\\ k-1, & k \\text{ even} \\end{cases}.\n$$\n\nFor each $k \\in \\{1, 2, \\ldots, \\frac{n}{2}\\}$, we have\n\n$$\ns_{2k-1} = a_1 + a_2 + \\cdots + a_{2k-1} = k(n-1) + 1,\n$$\n\n$$\ns_{2k} = a_1 + a_2 + \\cdots + a_{2k} = k(n+1),\n$$\n\nso\n\n- $r_{2k-1} = k(n-1) + 1 \\pmod{n}$,\n- $r_{2k} = k(n+1) \\pmod{n}$.\n\nIf $a, b \\in \\{1, 2, \\ldots, n-1\\}$ have the same parity, then $r_a \\ne r_b$. If there exist $k, j \\in \\{1, 2, \\ldots, \\frac{n}{2}\\}$ such that $r_{2k-1} = r_{2j}$, then $j = 0$ or $j = n-k+1$, which is impossible.\n\nTherefore, all even $n \\ge 2$ satisfy the required property.",
    "topic": "Discrete Mathematics",
    "subtopic": "Combinatorics"
  },
  {
    "id": 12660,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all functions $f: \\mathbb{Z}_{>0} \\to \\mathbb{Z}_{>0}$ for which $f(n) \\mid f(m) - n$ if and only if $n \\mid m$ for all natural numbers $m$ and $n$.",
    "options": [],
    "answer": "See solution",
    "solution": "Substituting $m = n$ gives $f(n) \\mid f(n) - n$, so for all natural numbers $n$ we have $f(n) \\mid n$. Applying this to the original condition, it follows that $f(n) \\mid f(m)$ if and only if $n \\mid m$.\n\nWe show that $f(n) = n$ by induction on the number of prime factors of $n$. The base of the induction is the case $n = 1$. In this case, we have $f(1) \\mid 1$, therefore $f(1) = 1$.\n\nSuppose that $f(k) = k$ for all natural numbers $k$ with fewer prime factors than $n$, and suppose for a contradiction that $f(n) \\mid n$ is a strict divisor of $n$. Then there exists a prime number $p$ such that $f(n) \\mid \\frac{n}{p} = f\\left(\\frac{n}{p}\\right)$, by the induction hypothesis. But $n$ does not divide $\\frac{n}{p}$, which contradicts our assumption that $f(n) \\neq n$. Therefore $f(n) = n$, and this concludes the induction.\n\nNote that $f(n) = n$ is indeed a solution, since $n \\mid m$ holds if and only if $n \\mid m - n$.\n\n![](images/NLD_ABooklet_2022_p20_data_fa089821ec.png)",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 12661,
    "subject": "Mathematics (Olympiad)",
    "question": "A cube of size $n \\times n \\times n$ loses 104 small cubes at the first step when $n = 10$. Increasing the cube's dimension by 1 increases the number of lost cubes by 12, one for each edge. For what value of $n$ does the cube lose 200 small cubes at the first step?",
    "options": [],
    "answer": "See solution",
    "solution": "Let $n$ be the cube's dimension. Each increment increases the lost cubes by 12. Since $200 - 104 = 96$ and $96 / 12 = 8$, the required cube is $10 + 8 = 18$. Thus, the cube that loses 200 small cubes at the first step is $18 \\times 18 \\times 18$.",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 12662,
    "subject": "Mathematics (Olympiad)",
    "question": "Adrian teaches a class of six pairs of twins. He wishes to set up teams for a quiz, but wants to avoid putting any pair of twins into the same team. Subject to this condition:\n\n1. In how many ways can he split them into two teams of six?\n2. In how many ways can he split them into three teams of four?",
    "options": [],
    "answer": "See solution",
    "solution": "1. When Adrian is picking Team 1, he starts with a choice of 12 pupils to pick first. Each time he picks someone for team 1, he has 2 fewer people to pick from next time (i.e., he cannot pick the same person again, nor can he pick their twin). He can pick the 6 pupils in $6!$ orders, so there are\n\n$$\n\\frac{12 \\times 10 \\times 8 \\times 6 \\times 4 \\times 2}{6!} = \\frac{2^6 \\times 6!}{6!} = 2^6\n$$\n\nways of picking team 1. Each way of picking team 1 gives exactly one way of picking team 2, which must consist of all remaining pupils. Each pair of twins has one twin in team 1, and thus team 2 will indeed be a valid team. Adrian can pick team 1 and team 2 in $2!$ orders, so the total number of ways of picking teams is $64/2 = 32$.\n\n2. As before, when picking team 1, Adrian will have two fewer people to choose from after each choice, and can pick the 4 people in $4!$ orders, and so this gives\n\n$$\n\\frac{12 \\times 10 \\times 8 \\times 6}{4!} = 240\n$$\n\nways of picking team 1.\n\nThen there are two pairs of twins and four others to split between teams 2 and 3. The two pairs of twins cannot have both twins in the same team, so Adrian must pick one twin from each pair to go in team 2. He must then pick two of the four singletons to fill the remaining places in team 2. So there are\n\n$$\n2 \\times 2 \\times \\frac{4 \\times 3}{2!} = 24\n$$\n\nways to pick team 2, having chosen team 1. The remaining four people do not contain any twins, so they then make up team 3. The three teams can be picked in $3!$ orders, and so there are in total\n\n$$\n\\frac{240 \\times 24 \\times 1}{3!} = 960\n$$\n\nways of picking the three teams.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 12663,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $k$ be a circle with center $M$. $T$ is a point on $k$ and $t$ is the tangent to $k$ at $T$. $P$ is a point on $t$ with $P \\neq T$, and $g$ is a line through $P$ with $g \\neq t$. The line $g$ intersects $k$ at points $U$ and $V$ ($U \\neq V$), and $S$ is the midpoint of the arc $UV$ not containing $T$. $Q$ is the point symmetric to $P$ with respect to $TS$. Prove that $QTUV$ is a trapezoid.\n\n![](images/AustriaMO2011_p2_data_37468398d7.png)",
    "options": [],
    "answer": "See solution",
    "solution": "Let $R$ be the intersection of $t$ and the tangent $s$ to $k$ at $S$. Since $S$ is the midpoint of the arc $UV$, $s$ is parallel to $g$ (and not to $t$). Since $MR$ is perpendicular to $TS$ and bisects $\\angle SRT$, $QP$ bisects $\\angle UPT$. Let $W$ be the intersection of $PQ$ and $TS$. By symmetry, $PQ \\perp TS$, so triangle $PWT$ is right-angled.\n\nWe have\n\n$$\n\\begin{align*}\n\\angle WTP + \\angle WPT &= 90^\\circ \\\\\n\\Leftrightarrow 2 \\cdot \\angle WTP + 2 \\cdot \\angle WPT &= 180^\\circ \\\\\n\\Leftrightarrow \\angle QTP + \\angle UPT &= 180^\\circ,\n\\end{align*}\n$$\n\nso *UV* is parallel to *QT*. Thus, *QTUV* is a trapezoid, as claimed. QED.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 12664,
    "subject": "Mathematics (Olympiad)",
    "question": "Given the equation:\n\n$$\nn! + 8 = 2^k\n$$\n\nFind all integer solutions $(n, k)$.",
    "options": [],
    "answer": "See solution",
    "solution": "Case 1: $n \\geq 6$\n\nWe have $2^4 \\mid n!$ and $2^3 \\nmid 8$. Thus $2^3 \\nmid n! + 8$. Hence $2^3 \\nmid 2^k$, so $k = 3$. But this implies $n! = 0$, which is impossible. So there are no solutions in this case.\n\nCase 2: $n \\leq 3$\n\nWe have $2^3 \\nmid n!$ and $2^3 \\nmid 8$. Hence $2^3 \\nmid 2^k$, so $k < 3$. It follows that $n! < 0$, which is impossible. So there are no solutions in this case.\n\nCase 3: $n = 4$ or $n = 5$\n\nIf $n = 4$ then $k = 5$, and if $n = 5$ then $k = 7$.\n\n$\\square$\n\n$^1$ For a prime number $p$ and integers $k \\geq 0$ and $N \\geq 1$, the notation $p^k \\mid N$ means that $p^k$ divides $N$ but $p^{k+1}$ does not divide $N$. Put another way, it means that the exponent of $p$ in the prime factorisation of $N$ is $k$.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 12665,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all odd integer pairs $\\{a, b\\}$, with $a, b > 1$, such that\n\n$$\n7\\varphi^2(a) - \\varphi(ab) + 11\\varphi^2(b) = 2(a^2 + b^2),\n$$\n\nwhere $\\varphi(n)$ denotes the number of positive integers less than $n$ and relatively prime to $n$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $p^\\alpha \\mid b$ mean that $p^\\alpha$ is the highest power of $p$ dividing $b$ (i.e., $p^\\alpha \\mid b$ but $p^{\\alpha+1} \\nmid b$).\n\nSuppose $\\{a, b\\}$ is a solution. If $p$ is a prime and $p^2 \\mid a, p^2 \\mid b$, then $\\left\\{\\frac{a}{p}, \\frac{b}{p}\\right\\}$ is also a solution. Thus, for each prime $p$, at least one of $a, b$ is coprime to $p$ or divisible exactly by $p$.\n\nSince $a, b > 2$, $2 \\mid \\varphi(a)$ and $2 \\mid \\varphi(b)$. Then:\n\n$$\n7\\varphi^2(a) \\ge 7 \\left(\\frac{2}{3}\\right)^2 a^2 > 2.5a^2,\n$$\n$$\n11\\varphi^2(b) \\ge 11 \\left(\\frac{2}{3}\\right)^2 b^2 > 2.5b^2.\n$$\n\nAlso, $\\varphi(ab) < ab \\le 0.5a^2 + 0.5b^2$, so the left side of the equation is larger than the right, and equality cannot hold. Thus, either $4 \\mid \\varphi(a)$ or $4 \\mid \\varphi(b)$.\n\nAssume $4 \\mid \\varphi(a)$. If $16 \\mid \\varphi(ab)$, taking the equation modulo 16 gives $2(a^2 + b^2) \\equiv 11\\varphi^2(b) \\pmod{16}$. But $11\\varphi^2(b) = 12 \\pmod{16}$, a contradiction. The same applies if $4 \\mid \\varphi(b)$.\n\nThus, either $4 \\mid \\varphi(ab)$ or $8 \\mid \\varphi(ab)$. In any case, $ab$ has at most three distinct prime factors, and if three, all are of the form $4k+3$.\n\n**Case 1:** $ab$ is a prime power, say $p^k$. Then $p$ must be $4k+1$, and $p \\mid a, p \\mid b$. But then\n$$\n7\\varphi^2(a) > 2.5a^2, \\quad 11\\varphi^2(b) > 2.5b^2,\n$$\nso the equation cannot hold.\n\n**Case 2:** $ab$ has two distinct prime factors, $p$ and $q$. At least one must be 3, otherwise the above bound holds. $a$ and $b$ cannot be coprime, or else each has only one prime factor and the bound holds. If $pq \\mid a$ and $pq \\mid b$, then $p=3$, $q=7$ (since $q>7$ fails the bound). If $q^2 \\mid a$ or $q^2 \\mid b$, this leads to a contradiction. If $a = 3^\\alpha \\cdot 7$, $b = 3^\\beta \\cdot 7$, then (i) $\\alpha=\\beta$ leads to $17\\varphi^2(a) = 4a^2$, impossible; (ii) $\\alpha > \\beta$ or $\\alpha < \\beta$ also yield contradictions. Thus, no solutions in this case.\n\nIf $a$ and $b$ are not coprime and $pq$ does not divide both, suppose $a$ and $b$ share $p$. If $pq \\mid b$, $a$ has only one prime factor, and the bound holds. Thus, $pq \\mid a$, $p \\mid b$, $(q, a) = 1$, and $p=3$, $q=5$ or $7$.\n\n- If $q=5$, $3 \\nmid (q-1)$. If $3^2 \\mid a$ or $3^2 \\mid b$, contradiction. So $3\\mid a, 3\\mid b$, $d=3$. If $5^2 \\mid a$, contradiction. Thus, $5\\mid a$, $a=15$. After checking, $(a, b) = (15, 3)$ is a solution.\n- If $q=7$, $3^2 \\mid a$ leads to contradiction. Let $a=3 \\cdot 7^\\beta$, $b=3^\\alpha$. If $\\alpha \\ge 3$, contradiction. Thus, $b=3$ or $9$, but the equation does not balance, so no solutions.\n\n**Case 3:** $ab$ has three distinct prime factors, all $4k+3$. Then $8 \\mid \\varphi(ab)$, so $4 \\nmid \\varphi(a)$ or $4 \\nmid \\varphi(b)$, i.e., $a$ or $b$ has only one prime factor. If $a$ has one, then\n$$\n7\\varphi^2(a) \\ge 7 \\left(\\frac{2}{3}\\right)^2 a^2 > 2.5a^2,\n$$\n$$\n11\\varphi^2(b) \\ge 11 \\left(\\frac{2}{3} \\cdot \\frac{6}{7} \\cdot \\frac{10}{11}\\right)^2 b^2 > 2.5b^2,\n$$\nso the bound holds. If $b$ has one prime factor $p$, and $a$ has the other two, $3 \\mid a$ or else the bound holds. If $a$ and $b$ are coprime, $3 \\nmid b$, $p \\ge 7$, and\n$$\n11\\varphi^2(b) \\ge 11 \\cdot \\frac{36}{49} b^2 > 8b^2.\n$$\nMeanwhile, $a$ has two prime factors, so\n$$\n7\\varphi^2(a) \\ge 7 \\left(\\frac{2}{3} \\cdot \\frac{36}{49}\\right)^2 a^2 > 2.25a^2.\n$$\nThus,\n$$\n7\\varphi^2(a) + 11\\varphi^2(b) > 2.25a^2 + 8b^2 > 2a^2 + 2b^2 + \\varphi(ab),\n$$\ncontradiction. If $(a, b) \\ne 1$, $p \\mid a$. If $p^2 \\mid a$, by descent, $p = b$, but then $p^2 \\nmid 11\\varphi(b)$, impossible. Thus, $p \\nmid a$. Taking the equation modulo $p$ twice gives $p \\mid 7\\varphi^2(b)$, so $p=7$ or $p$ divides $q-1$ or $r-1$.\n\n**Conclusion:** The only solution is $(a, b) = (15, 3)$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 12666,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $P_n$ be the point with coordinates $(n, n^3 - 2014n^2)$. Show that for any distinct integers $a, b, c$, the points $P_a$, $P_b$, and $P_c$ are collinear if and only if $a + b + c = 2014$.",
    "options": [],
    "answer": "See solution",
    "solution": "We claim that defining $P_n$ to be the point with coordinates $(n, n^3 - 2014n^2)$ will satisfy the conditions of the problem. Recall that points $(x_1, y_1)$, $(x_2, y_2)$, and $(x_3, y_3)$ are collinear if and only if\n\n$$\n\\begin{vmatrix} x_1 & y_1 & 1 \\\\ x_2 & y_2 & 1 \\\\ x_3 & y_3 & 1 \\end{vmatrix} = 0.\n$$\n\nTherefore, we examine the determinant\n\n$$\n\\begin{vmatrix} a & a^3 - 2014a^2 & 1 \\\\ b & b^3 - 2014b^2 & 1 \\\\ c & c^3 - 2014c^2 & 1 \\end{vmatrix} = \\begin{vmatrix} a & a^3 & 1 \\\\ b & b^3 & 1 \\\\ c & c^3 & 1 \\end{vmatrix} - 2014 \\begin{vmatrix} a & a^2 & 1 \\\\ b & b^2 & 1 \\\\ c & c^2 & 1 \\end{vmatrix}.\n$$\n\nThe first determinant on the right is a homogeneous polynomial of degree four divisible by $(a-b)(b-c)(c-a)$. The remaining factor has degree one, is symmetric, and yields an $ab^3$ term when the product is expanded, hence must be $(a+b+c)$. The second determinant is a homogeneous polynomial of degree three divisible by $(a-b)(b-c)(c-a)$, and comparing coefficients of the $ab^2$ term we see that this is the desired polynomial. Thus\n\n$$\n\\begin{vmatrix} a & a^3 - 2014a^2 & 1 \\\\ b & b^3 - 2014b^2 & 1 \\\\ c & c^3 - 2014c^2 & 1 \\end{vmatrix} = (a-b)(b-c)(c-a)(a+b+c-2014).\n$$\n\nIt follows that for distinct $a, b$, and $c$, this expression will equal zero if and only if $a + b + c = 2014$, as desired.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 12667,
    "subject": "Mathematics (Olympiad)",
    "question": "Kevoff numbers are numbers whose digit sum is divisible by the number of digits. Are there any pairs of consecutive Kevoff numbers with the same number of digits? If so, for which digit lengths do such pairs exist, and can you give examples?",
    "options": [],
    "answer": "See solution",
    "solution": "For 2-digit and 3-digit numbers, there are no pairs of consecutive Kevoff numbers. This is because increasing a number by 1 either changes the digit sum by an amount not divisible by the number of digits, or the digit sums do not align for both numbers to be Kevoff. \n\nFor 4-digit numbers, consecutive Kevoff pairs do exist. For example, $2419$ and $2420$ are both Kevoff numbers:\n- $2419$: digit sum $2+4+1+9=16$, $16/4=4$\n- $2420$: digit sum $2+4+2+0=8$, $8/4=2$\n\nBoth are divisible by $4$. Other examples include $3539$ and $3540$, $4659$ and $4660$, $5779$ and $5780$.\n\nFor 8-digit numbers, similar pairs exist, such as $22210009$ and $22210010$.\n\nThus, consecutive Kevoff numbers with the same number of digits exist for 4-digit and 8-digit numbers, but not for 2- or 3-digit numbers.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 12668,
    "subject": "Mathematics (Olympiad)",
    "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": "See solution",
    "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 $n \\ge 6$.\n\nIf $n = 6$, the only way to leave the endpoints of every two diagonals to different sides of the third diagonal is by connecting each vertex to the opposite one, but the obtained diagonals meet in the center of the $n$-gon.\n\nSuppose 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. As these diagonals are symmetric with respect to 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 with respect to the first diagonal, all three intersect in one point inside the polygon that is not its center.\n\nThus, the least such $n$ is $8$.",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 12669,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $\\triangle ABC$ be an acute-angled triangle with circumcircle $\\omega$. A circle $\\Gamma$ is internally tangent to $\\omega$ at $A$ and also tangent to $BC$ at $D$. Let $AB$ and $AC$ intersect $\\Gamma$ at $P$ and $Q$ respectively. Let $M$ and $N$ be points on line $BC$ such that $B$ is the midpoint of $DM$ and $C$ is the midpoint of $DN$. Lines $MP$ and $NQ$ meet at $K$ and intersect $\\Gamma$ again at $I$ and $J$ respectively. The ray $KA$ meets the circumcircle of triangle $IJK$ at $X \\neq K$. Prove that $\\angle BXP = \\angle CXQ$.",
    "options": [],
    "answer": "See solution",
    "solution": "![](images/Saudi_Arabia_booklet_2024_p45_data_1a25eb3d9c.png)",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 12670,
    "subject": "Mathematics (Olympiad)",
    "question": "Jerry writes all the integers from $1$ to $2024$ on the whiteboard. He repeatedly chooses four numbers, erases them, and replaces them with either their sum or their product. After several such operations, all remaining numbers on the whiteboard are odd. What is the maximum possible number of integers on the whiteboard at that time?\n\n(A) 1010 (B) 1011 (C) 1012 (D) 1013 (E) 1014",
    "options": [],
    "answer": "See solution",
    "solution": "Each time Jerry performs the operation, the number of even integers on the whiteboard decreases by at most $3$. Initially, there are $\\frac{2024}{2} = 1012$ even integers. Since $\\frac{1011}{3} = 337$ and $\\frac{1014}{3} = 338$, Jerry needs at least $338$ operations to eliminate all even numbers. After $338$ operations, there are $2024 - (338 \\times 3) = 1010$ numbers left on the whiteboard.\n\nTo achieve this, Jerry could, for $0 \\leq n \\leq 336$, choose $6n + 1$, $6n + 2$, $6n + 4$, and $6n + 6$ and replace them with their sum. For the final operation, he could erase $3$, $5$, $9$, and $2024$ and replace them with their sum. After these $338$ operations, all $1010$ numbers remaining are odd.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 12671,
    "subject": "Mathematics (Olympiad)",
    "question": "Нехай біла частина шахівниці розрізана на $p$ прямокутників, а чорна — на $q$ прямокутників. Аналогічно тому, як зображено на рисунку для $n=8$, на білій частині відмітимо $n-1$ клітинку, а на чорній — $n$ клітинок. Кожному з прямокутників розрізання відповідної частини належить не більше однієї відміченої в цій частині клітинки. Доведіть, що $p \\ge n-1$, $q \\ge n$, і що з урахуванням рівності $p+q=2n$ можливі лише такі варіанти: $p = n-1$, $q = n+1$ або $p = n$, $q = n$. Для кожного $n \\ge 3$ наведіть розрізання із $q=n$ та із $q=n+1$.",
    "options": [],
    "answer": "See solution",
    "solution": "Відповідь: на $n$ прямокутників або ж на $n+1$ прямокутник.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 12672,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all positive integers $a, b$ such that $$(a-b)^{a+b} = a^a.$$",
    "options": [],
    "answer": "See solution",
    "solution": "Let $d = \\gcd(a, b)$. Write $a = d x$ and $b = d y$. Then the equation reduces to $d^b (x - y)^{a+b} = x^a$. Since $\\gcd(x, x-y) = 1$, it follows that $x - y = \\pm 1$ and $d^b = x^a$. Using $b = d x$ and $a = d y$, this further reduces to $d y = x^x$ or\n\n$$\nd^{x \\pm 1} = x^x\n$$\n\nNote that $d = 1 = x$ does not lead to any solution. Let $p$ be a prime dividing $d$ and hence $x$. Let $p^\\alpha \\mid d$, $p^\\beta \\mid x$. Considering the powers of $p$, we see that\n\n$$\n\\alpha(x \\pm 1) = \\beta x.\n$$\n\nThis implies that $x \\nmid \\alpha$. Putting $\\alpha = x \\gamma$, we obtain $\\gamma(x \\pm 1) = \\beta$. If $\\gamma(x+1) = \\beta$, then we have $x+1 = k \\gamma^\\beta + 1 > \\beta$ and the relation is not tenable. If $\\gamma(x-1) = \\beta$ and $\\beta > 1$, then again we see that\n\n$$\nx - 1 = k \\gamma^\\beta - 1 \\ge 2^\\beta - 1 > \\beta,\n$$\n\nso that the relation is again not valid. The only possibility is $\\beta = 1$. This gives $\\gamma(kp - 1) = 1$, which leads to $p = 2, k = 1, \\gamma = 1$. In turn, we get $x = 2, d = 4, y = 1, a = 8, b = 4$. Observe $$(8 - 4)^{8+4} = 4^{12} = 8^8.$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 12673,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $p, n$ be positive integers such that $p$ is prime and $p < n$. If $p$ divides $n + 1$ and $\\left(\\left\\lfloor\\frac{n}{p}\\right\\rfloor, (p-1)!\\right) = 1$, then prove that $p \\cdot \\left\\lfloor\\frac{n}{p}\\right\\rfloor^2$ divides $\\binom{n}{p} - \\left\\lfloor\\frac{n}{p}\\right\\rfloor$. (Here $\\lfloor x \\rfloor$ represents the integer part of the real number $x$.)",
    "options": [],
    "answer": "See solution",
    "solution": "Since $p \\mid n + 1$, there exists $k \\in \\mathbb{N}$ such that $n = k p + p - 1$ and $\\left\\lfloor \\frac{n}{p} \\right\\rfloor = k$.\n\n$$\n\\begin{aligned}\n\\binom{n}{p} - \\left\\lfloor \\frac{n}{p} \\right\\rfloor &= \\binom{k p + p - 1}{p} - k \\\\\n&= \\frac{(k p + p - 1)(k p + p - 2) \\dots (k p + 1)(k p)}{p!} - k \\\\\n&= \\frac{k (k p + 1)(k p + 2) \\dots (k p + p - 1) - k (p - 1)!}{(p - 1)!}\n\\end{aligned}\n$$\n\n$$\n= \\frac{k (k \\cdot p \\cdot r + (p-1)! ) - k (p-1)!}{(p-1)!} = \\frac{k^2 \\cdot p \\cdot r}{(p-1)!} \\in \\mathbb{N}\n$$\n\nSince $\\left( \\left\\lfloor \\frac{n}{p} \\right\\rfloor, (p-1)! \\right) = 1$, $(p-1)!$ divides $r$ and therefore $\\binom{n}{p} - \\left\\lfloor \\frac{n}{p} \\right\\rfloor$ is divisible by $p \\cdot \\left\\lfloor \\frac{n}{p} \\right\\rfloor^2$ as required. $\\square$",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 12674,
    "subject": "Mathematics (Olympiad)",
    "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 $DM \\cdot KE = DN \\cdot KF$.",
    "options": [],
    "answer": "See solution",
    "solution": "It is easy to see that points $I$, $D$, $E$, and $B$ are concyclic and\n\n$$\n\\angle AID = 90^\\circ - \\angle IAD, \\\\\n\\angle MED = \\angle FDA = 90^\\circ - \\angle IAD.\n$$\n\nSo $\\angle AID = \\angle MED$, thus points $I$, $D$, $E$, and $M$ are concyclic.\n\nHence, five points $I$, $D$, $B$, $E$, $M$ are concyclic and $\\angle IMB = \\angle IEB = 90^\\circ$, that is $AM \\perp BM$.\n\nSimilarly, points $I$, $D$, $A$, $N$, and $F$ are concyclic and $BN \\perp AN$.\n\nLet lines $AN$ and $BM$ intersect at point $G$. We see point $I$ is the\n\n![](images/Mathematical_Olympiad_in_China_2011-2014_p302_data_72fbfdbae3.png)\n\northocenter of $\\triangle GAB$ and $ID \\perp AB$, so points $G$, $I$, and $D$ are collinear.\n\nSince points $G$, $N$, $D$, and $B$ are concyclic, we see that $\\angle ADN = \\angle G$.\n\nSimilarly, $\\angle BDM = \\angle G$. So $DK$ bisects $\\angle MDN$, thus\n\n$$\n\\frac{DM}{DN} = \\frac{KM}{KN}. \\qquad \\textcircled{1}\n$$\n\nSince points $I$, $D$, $E$, and $M$ are concyclic, and points $I$, $D$, $N$, and $F$ are concyclic, we see that\n\n$$\nKM \\cdot KE = KI \\cdot KD = KF \\cdot KN.\n$$\n\nTherefore,\n\n$$\n\\frac{KM}{KN} = \\frac{KF}{KE}. \\qquad \\textcircled{2}\n$$\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": 12675,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $f \\in \\mathbb{Q}[x]$ be a polynomial. Define $f^\\omega(Q)$ as the set of points in $Q$ whose forward orbits under $f$ are eventually periodic. Prove that for any sequence $(x_n)_{n \\ge 1}$ of rational numbers such that $x_n = f(x_{n+1})$, the sequence is periodic.",
    "options": [],
    "answer": "See solution",
    "solution": "Since $f(f^\\omega(Q)) = f^\\omega(Q)$, the restriction of $f$ to $f^\\omega(Q)$ is a permutation of this finite set. Thus, any sequence $(x_n)$ with $x_n = f(x_{n+1})$ must have all its terms in $f^\\omega(Q)$, which is finite. Therefore, the sequence is eventually periodic.",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 12676,
    "subject": "Mathematics (Olympiad)",
    "question": "Show that $\\sum_{k=0}^{n} (-1)^k \\binom{2n+1}{2k+1} 2008^k$ is not divisible by $19$ for every positive integer $n$.",
    "options": [],
    "answer": "See solution",
    "solution": "Observe that $-2008 \\equiv 6 \\equiv 5^2 \\pmod{19}$. Thus,\n\n$$\n\\begin{aligned}\n2 \\sum_{k=0}^{n} \\binom{2n+1}{2k+1} (-2008)^k &\\equiv 2 \\sum_{k=0}^{n} \\binom{2n+1}{2k+1} 5^{2k} \\pmod{19} \\\\\n&\\equiv (1+5)^{2n+1} - (1-5)^{2n+1} \\pmod{19} \\\\\n&\\equiv 6^{2n+1} + 4^{2n+1} \\\\\n&\\equiv 2^{2n+1} (3^{2n+1} + 2^{2n+1}) \\pmod{19}.\n\\end{aligned}\n$$\n\nSince\n\n$$\n3^{2n+1} + 2^{2n+1} \\equiv (-16)^{2n+1} + 2^{2n+1} \\equiv 2^{2n+1} (1 - 2^{6n+3}) \\pmod{19}\n$$\n\nand $2^{18} \\equiv 1 \\pmod{19}$. We can see that\n\n$$\n2^{6(n+3)+3} \\equiv 2^{6n+3} \\pmod{19}\n$$\n\nfor each $n = 0, 1, 2, \\dots$.\n\nTherefore, it suffices to consider the divisibility of $3^{2n+1} + 2^{2n+1}$ by $19$ when $n = 0, 1, 2$. We now verify that\n\n$$\n3^1 + 2^1 \\equiv 5 \\pmod{19}\n$$\n\n$$\n3^3 + 2^3 \\equiv 35 \\equiv 16 \\pmod{19}\n$$\n\n$$\n3^5 + 2^5 \\equiv 275 \\equiv 9 \\pmod{19}\n$$\n\nHence, $19 \\nmid \\sum_{k=0}^{n} \\binom{2n+1}{2k+1} (-2008)^k$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 12677,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all triples $ (a, b, c) $ of real numbers such that $ ab + bc + ca = 1 $ and\n\n$$\na^2b + c = b^2c + a = c^2a + b.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "First, we find all solutions in which at least one of the terms is zero.\n\nIf, without loss of generality, $a = 0$, then $bc = 1$ and $c = b^2c = b$.\n\nThus, $b = c$ and $bc = 1$, so either $b = c = 1$ (giving $(a, b, c) = (0, 1, 1)$) or $b = c = -1$ (giving $(a, b, c) = (0, -1, -1)$).\n\nBy symmetry, there are six solutions in total in which at least one of the terms is zero:\n\n$$\n(a, b, c) = (0, 1, 1),\\ (1, 0, 1),\\ (1, 1, 0),\\ (0, -1, -1),\\ (-1, 0, -1),\\ (-1, -1, 0).\n$$\n\nNow assume that none of $a, b, c$ are zero.\n\nWe have $a^2b + c = b^2c + a \\implies b^2c - c + a(1 - ab) = 0$, and since $1 - ab = bc + ca$,\n\n$$\nb^2c - c + abc + a^2c = 0. \\quad (1)\n$$\n\nSince $c \\neq 0$, $ab + a^2 + b^2 - 1 = 0$.\n\nSimilarly, $bc + b^2 + c^2 - 1 = 0$ and $ca + c^2 + a^2 - 1 = 0$.\n\nAdding these three equations gives $2(a^2 + b^2 + c^2) + ab + bc + ca - 3 = 0$. Substituting $ab + bc + ca = 1$ yields\n\n$$\na^2 + b^2 + c^2 = 1. \\quad (2)\n$$\n\nCombining (1) and (2):\n\n$$\nabc = c(1 - a^2 - b^2) \\implies abc = c^3.\n$$\n\nSimilarly, $abc = a^3$ and $abc = b^3$, so\n\n$$\na^3 = b^3 = c^3 \\implies a = b = c.\n$$\n\nSubstituting $a = b = c$ into $ab + bc + ca = 1$ gives $3a^2 = 1$, so $a = \\pm\\frac{1}{\\sqrt{3}} = b = c$. This gives two further solutions:\n\n$$\n(a, b, c) = \\left(\\frac{1}{\\sqrt{3}}, \\frac{1}{\\sqrt{3}}, \\frac{1}{\\sqrt{3}}\\right),\\ \\left(-\\frac{1}{\\sqrt{3}}, -\\frac{1}{\\sqrt{3}}, -\\frac{1}{\\sqrt{3}}\\right).\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 12678,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $n > 1$ be a positive integer and $a_1, a_2, \\dots, a_n$ be a sequence of $n$ positive integers. Define\n\n$$\nb_i = \\left[ \\frac{a_1 + \\dots + a_{i-1} + a_{i+1} + \\dots + a_n}{n-1} \\right], \\quad 1 \\leq i \\leq n.\n$$\n\nLet $f$ be a mapping such that $f(a_1, a_2, \\dots, a_n) = (b_1, b_2, \\dots, b_n)$.\n\na) Let the function $g: \\mathbb{N} \\rightarrow \\mathbb{N}$ be defined such that $g(1)$ is the number of different elements in the sequence $f(a_1, a_2, \\dots, a_n)$, and $g(m)$ is the number of different elements in the sequence $f^m(a_1, a_2, \\dots, a_n)$, where $f^m$ denotes $f$ applied $m$ times, $m > 1$. Prove that there is a positive integer $k_0$ such that for $m \\geq k_0$, the function $g(m)$ is periodic.\n\nb) Prove that\n$$\n\\sum_{m=1}^{k} \\frac{g(m)}{m(m+1)} < C\n$$\nfor any positive integer $k$, where the constant $C$ does not depend on $k$.",
    "options": [],
    "answer": "See solution",
    "solution": "a) Let $n > 2$. We will show that for sufficiently large $m$, $g(m) = 1$.\n\nLet $a_1, a_2, \\dots, a_n$ be a sequence of positive integers. Then\n\n$$\nf(a_1, a_2, \\dots, a_n) = \\left( \\left[ \\frac{a_2 + a_3 + \\dots + a_n}{n-1} \\right], \\left[ \\frac{a_1 + a_3 + \\dots + a_n}{n-1} \\right], \\dots, \\left[ \\frac{a_1 + a_2 + \\dots + a_{n-1}}{n-1} \\right] \\right),\n$$\n\nwhere some of the elements may be equal, so $g(1) \\leq n$. Similarly, $g(m) \\leq n$ for every positive integer $m$. Let $S_r$ be the sum of the elements in the sequence $f^r(a_1, a_2, \\dots, a_n)$. For the sum of the elements of the sequences $(b_1, \\dots, b_n)$ and $f(b_1, \\dots, b_n)$, we have\n\n$$\nS_{r+1} = \\sum_{i=1}^n \\left[ \\frac{S_r - b_i}{n-1} \\right] \\leq \\sum_{i=1}^n \\frac{S_r - b_i}{n-1} = S_r.\n$$\n\nThus, $0 \\leq S_{r+1} \\leq S_r$ for every $r$. Therefore, there is a positive integer $k_0$ such that for $m \\geq k_0$, $S_m = S_{m+1} = \\dots$ is constant. Equality holds only if the numbers in the sequence $f^m(a_1, a_2, \\dots, a_n)$ are all equal for large $m$.\n\nWe will prove that $S_{r+1} = S_r$ implies $d_1 = d_2 = \\dots = d_n$, where $f^m(a_1, a_2, \\dots, a_n) = (d_1, \\dots, d_n)$ for large $m$. For equality, it is necessary that $n-1 \\mid S_r - b_i$ for all $i$, so\n\n$$\nb_1 \\equiv b_2 \\equiv \\dots \\equiv b_n \\pmod{n-1}.\n$$\n\nFrom this, $\\frac{S_r - b_i}{n-1}$ are integers for all $i$. Also,\n\n$$\n|c_i - c_j| = \\left| \\frac{b_i - b_j}{n-1} \\right| < |b_i - b_j|.\n$$\n\nThus, the differences decrease at each step, so after finitely many steps, all $d_i$ are equal. Therefore, there is a $k_0$ such that for $m \\geq k_0$, the elements of $f^m(a_1, a_2, \\dots, a_n)$ are all equal, so $g(m) = 1$.\n\nFor $n = 2$, it is clear that $(a_1, a_2) = f(a_1, a_2)$, so $g(m) \\leq 2$ for all $m$.\n\nb) For $n > 2$ and any positive integer $k$,\n\n$$\n\\sum_{m=1}^{k} \\frac{g(m)}{m(m+1)} < \\sum_{m=1}^{\\infty} \\frac{g(m)}{m(m+1)} = \\sum_{m=1}^{k_0} \\frac{g(m)}{m(m+1)} + \\sum_{m=k_0+1}^{\\infty} \\frac{1}{m(m+1)}.\n$$\n\nSince $g(m) = 1$ for $m > k_0$, the sum converges and is bounded by a constant $C$ independent of $k$.\n\nFor $n = 2$, $\\sum_{m=1}^{k} \\frac{g(m)}{m(m+1)} < 2 \\sum_{m=1}^{\\infty} \\frac{1}{m(m+1)} = 2$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 12679,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $a$, $b$, and $c$ be the lengths of the sides opposite vertices $A$, $B$, and $C$ of a triangle, respectively. Let $D$ be the intersection of $BC$ and the angle bisector of $\\angle A$. Let $p = BD$ and $q = CD$. Prove the following inequality involving the lengths $\\ell_A$, $\\ell_B$, $\\ell_C$ (defined as the lengths from $D$ to the feet of the perpendiculars from $D$ to $AB$ and $AC$, and similarly for the other vertices):\n\n$$\n\\frac{\\ell_A \\ell_B \\ell_C}{\\ell^3} \\le \\frac{(a+b-c)^2 (b+c-a)^2 (c+a-b)^2}{64a^2b^2c^2}\n$$\n\nwhere $\\ell$ is the length of the angle bisector $AD$.",
    "options": [],
    "answer": "See solution",
    "solution": "Since $AD$ is the angle bisector of $\\angle A$, we have $bp = cq$, which leads to\n$$\np = \\frac{ac}{b+c}, \\quad q = \\frac{ab}{b+c}.\n$$\nBy the Law of Cosines and the given geometric configuration,\n$$\n\\frac{x^2 + p^2 - c^2}{2px} + \\frac{x^2 + q^2 - b^2}{2qx} = 0,\n$$\nwhere $x = AD$. Solving, we get\n$$\nx^2 = bc - pq = bc \\left[ 1 - \\left( \\frac{a}{b+c} \\right)^2 \\right] = \\frac{bc(b+c-a)}{(b+c)^2} \\ell.\n$$\nLet $E$ and $F$ be the feet of the perpendiculars from $D$ to $AB$ and $AC$. Using the Sine Law and area $T$ of $\\triangle ABC$,\n$$\n\\ell_A = x \\sin A = \\frac{2xT}{bc} = \\frac{2T}{(b+c)\\sqrt{bc}} \\sqrt{(b+c-a)\\ell}.\n$$\nSimilarly,\n$$\n\\ell_B = \\frac{2T}{(c+a)\\sqrt{ca}} \\sqrt{(c+a-b)\\ell},\n$$\n$$\n\\ell_C = \\frac{2T}{(a+b)\\sqrt{ab}} \\sqrt{(a+b-c)\\ell}.\n$$\nMultiplying,\n$$\n\\ell_A \\ell_B \\ell_C = \\frac{8T^3 \\ell \\sqrt{(a+b-c)(b+c-a)(c+a-b)\\ell}}{abc(a+b)(b+c)(c+a)} = \\frac{\\ell^3 (a+b-c)^2 (b+c-a)^2 (c+a-b)^2}{8abc(a+b)(b+c)(c+a)}.\n$$\nBy the AM-GM inequality,\n$$\n\\frac{\\ell_A \\ell_B \\ell_C}{\\ell^3} \\le \\frac{(a+b-c)^2 (b+c-a)^2 (c+a-b)^2}{64a^2b^2c^2}.\n$$\nAlso,\n$$\n0 < (a+b-c)(c+a-b) \\le a^2, \\quad 0 < (a+b-c)(b+c-a) \\le b^2, \\quad 0 < (b+c-a)(c+a-b) \\le c^2,\n$$\nso\n$$\n0 < (a-b+c)^2(b-a+c)^2(c-a+b)^2 \\le a^2b^2c^2.\n$$\nThus, the required inequality follows.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 12680,
    "subject": "Mathematics (Olympiad)",
    "question": "There are 100 diamonds on display; 50 of them are genuine and 50 are 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.",
    "options": [],
    "answer": "See solution",
    "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.\n\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": 12681,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all positive integers $p$ such that $p^2 - p - 1$ is a perfect cube.",
    "options": [],
    "answer": "See solution",
    "solution": "We have $p^2 - p - 1 = n^3$. First, note that $p^2 > n^3 \\ge n^2 \\Rightarrow p \\ge n + 1$. If $p = n + 1$, then the equation becomes $n^3 - n^2 - n + 1 = 0 \\Rightarrow n = 1, p = 2$, which is a solution of the problem.\n\nLet now $p > n + 1$. Rewrite the equation as $p(p-1) = (n+1)(n^2-n+1)$. So, $(n+1)(n^2-n+1) \\nmid p$. Since $n+1 < p$, we have $n^2-n+1 \\nmid p$, hence\n\n$$\nn^2 - n + 1 = pk, \\quad k \\in \\mathbb{N}. \\qquad (1)\n$$\n\nThus $p-1 = k(n+1)$ or\n\n$$\np = kn + k + 1. \\qquad (2)\n$$\n\nSubstituting $p$ into (1) gives\n\n$$\nn^2 - (k^2 + 1)n - (k^2 + k - 1) = 0. \\qquad (3)\n$$\n\nThe discriminant of this equation is $D = k^4 + 6k^2 + 4k - 3$ which must be a square of an integer. But if $k \\ge 4$, then it is easy to see that $(k^2 + 3)^2 < D < (k^2 + 4)^2$, a contradiction.\n\nIt remains to inspect the cases $k = 1, 2, 3$, which gives the answer: $p = 2$ and $p = 37$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 12682,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all positive integer solutions of the equation\n\n$$\n(x^2 - 1)^2 (y^2 - 1)^2 + 16x^2 y^2 = z^2.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "Let $a = x^2 - 1$ and $b = y^2 - 1$. Then the equation becomes\n\n$$\na^2 b^2 + 16(a+1)(b+1) = z^2.\n$$\n\nWhen $a, b > 13$, it can be shown that\n\n$$\n(ab + 8)^2 < z^2 < (ab + 9)^2,\n$$\n\nso there is no solution in this case. If at least one of $a$ or $b$ is not greater than $13$ (i.e., at least one of $x$ or $y$ is not greater than $3$), checking all possibilities shows that the solutions are $(x, y, z) = (1, n, 4n)$ or $(x, y, z) = (n, 1, 4n)$, where $n \\in \\mathbb{Z}_{\\ge 1}$.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 12683,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $ABC$ be a scalene triangle with incentre $I$ and circumcircle $\\omega$. The lines $AI$, $BI$, $CI$ intersect $\\omega$ for the second time at the points $D$, $E$, $F$, respectively. The lines through $I$ parallel to the sides $BC$, $AC$, $AB$ intersect the lines $EF$, $DF$, $DE$ at the points $K$, $L$, $M$, respectively. Prove that the points $K$, $L$, $M$ are collinear.",
    "options": [],
    "answer": "See solution",
    "solution": "First we will prove that $KA$ is tangent to $\\omega$.\n\nIndeed, it is a well-known fact that $FA = FB = FI$ and $EA = EC = EI$, so $FE$ is the perpendicular bisector of $AI$. It follows that $KA = KI$ and\n\n$$\n\\angle KAF = \\angle KIF = \\angle FCB = \\angle FEB = \\angle FEA,\n$$\n\nso $KA$ is tangent to $\\omega$. Similarly, we can prove that $LB$, $MC$ are tangent to $\\omega$ as well.\n\n![](images/Greek2015_booklet_p9_data_076bd54cd8.png)\n\nLet $A'$, $B'$, $C'$ be the intersections of $AI$, $BI$, $CI$ with $BC$, $CA$, $AB$ respectively. From Pascal's Theorem on the cyclic hexagon $AACDEB$ we get $K$, $C'$, $B'$ collinear. Similarly, $L$, $C'$, $A'$ are collinear and $M$, $B'$, $A'$ are collinear.\n\nThen from Desargues' Theorem for $\\triangle DEF$, $\\triangle A'B'C'$ which are perspective from the point $I$, we get that points $K$, $L$, $M$ of the intersection of their corresponding sides are collinear as wanted.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 12684,
    "subject": "Mathematics (Olympiad)",
    "question": "Prove that if $a + \\frac{b}{a} - \\frac{1}{b}$ is an integer, then it is a perfect square, where $a, b \\in \\mathbb{N}$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $a, b \\in \\mathbb{N}$ and suppose $a + \\frac{b}{a} - \\frac{1}{b} = k \\in \\mathbb{Z}$. \n\nFrom $a + \\frac{b}{a} - \\frac{1}{b} = k$, multiply both sides by $a$ to get:\n\n$$a^2 + b - \\frac{a}{b} = k a$$\n\nMultiply both sides by $b$:\n\n$$a^2 b + b^2 - a = k a b$$\n\nSo $b$ divides $a$, let $a = bq$ for some $q > 0$, $q \\in \\mathbb{Z}$. Substitute:\n\n$$b^3 q^2 + b^2 - bq = k b^2 q$$\n\nDivide both sides by $b$ ($b > 0$):\n\n$$b^2 q^2 + b - q = k b q$$\n\nSo $q$ divides $b$, let $b = q t$ for some $t > 0$, $t \\in \\mathbb{Z}$. Substitute:\n\n$$q^4 t^2 + q t - q = k q^2 t$$\n\nDivide both sides by $q$ ($q > 0$):\n\n$$q^3 t^2 + t - 1 = k q t$$\n\nSo $t$ divides $1$ and $t > 0$, so $t = 1$. Thus $b = q$ and $a = b q = q^2$.\n\nNow, substitute back:\n\n$$k = a + \\frac{b}{a} - \\frac{1}{b} = q^2 + \\frac{q}{q^2} - \\frac{1}{q}$$\n\nSince $q \\in \\mathbb{N}$, $\\frac{q}{q^2} - \\frac{1}{q} = \\frac{1}{q} - \\frac{1}{q} = 0$, so $k = q^2$.\n\nTherefore, $k$ is a perfect square.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 12685,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $x$, $y$, $z$ be positive real numbers. Prove that\n\n$$\n\\sqrt{\\frac{xy}{x^2 + y^2 + 2z^2}} + \\sqrt{\\frac{yz}{y^2 + z^2 + 2x^2}} + \\sqrt{\\frac{zx}{z^2 + x^2 + 2y^2}} \\leq \\frac{3}{2}.\n$$\n\nWhen does equality hold?",
    "options": [],
    "answer": "See solution",
    "solution": "We have\n\n$$\n\\begin{aligned}\n& \\sqrt{\\frac{xy}{x^2 + y^2 + 2z^2}} + \\sqrt{\\frac{yz}{y^2 + z^2 + 2x^2}} + \\sqrt{\\frac{zx}{z^2 + x^2 + 2y^2}} \\\\\n\\leq & \\sqrt{\\frac{xy}{xy + yz + zx + z^2}} + \\sqrt{\\frac{yz}{xy + yz + zx + x^2}} + \\sqrt{\\frac{zx}{xy + yz + zx + y^2}} \\\\\n= & \\sqrt{\\frac{xy}{(z + x)(y + z)}} + \\sqrt{\\frac{yz}{(x + y)(z + x)}} + \\sqrt{\\frac{zx}{(y + z)(x + y)}} \\\\\n\\leq & \\frac{\\frac{x}{z + x} + \\frac{y}{y + z}}{2} + \\frac{\\frac{y}{x + y} + \\frac{z}{z + x}}{2} + \\frac{\\frac{z}{y + z} + \\frac{x}{x + y}}{2} \\\\\n= & \\frac{\\frac{x + y}{x + y} + \\frac{y + z}{y + z} + \\frac{z + x}{z + x}}{2} = \\frac{3}{2}.\n\\end{aligned}\n$$\n\nEquality holds if and only if $x = y = z$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 12686,
    "subject": "Mathematics (Olympiad)",
    "question": "Determine all functions $f : \\mathbb{R} \\to \\mathbb{R}$ such that for every pair of real numbers $x$ and $y$,\n\n$$\nf(x + y^2) = f(x) + |y f(y)|.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "We first claim that $f$ is nondecreasing. Let $r \\geq s$ be any two real numbers, and let $t = \\sqrt{r-s}$. Then we have\n\n$$\nf(r) = f(s + t^2) = f(s) + |t f(t)| \\geq f(s),\n$$\n\nas desired.\n\nNow, setting $x = 0$ and rearranging the original equation yields\n\n$$\nf(y^2) - f(0) = |y f(y)|. \\qquad (1)\n$$\n\nDefine $g(a) = f(a) - f(0)$ for all real $a$. By (1), we get $g(y^2) = |y f(y)|$ for all real $y$. Since $f(x + y^2) - f(x) = g(x + y^2) - g(x)$, we have $g(x + y^2) - g(x) = |y f(y)| = g(y^2)$, whence\n\n$$\ng(z + t) = g(z) + g(t)\n$$\n\nholds for all nonnegative reals $z, t$. Thus, $g$ satisfies the Cauchy functional equation for nonnegative reals. Since $g$ must also be nondecreasing (because $f$ is nondecreasing), $g(x)$ is linear over the nonnegative reals, that is, there is some $c$ such that $g(x) = c x$ for all $x \\geq 0$. Consequently, $f(x) = c x + f(0)$ for all $x \\geq 0$.\n\nSubstituting into (1), we obtain $c y^2 = |c y^2 + f(0) y|$ for all $y \\geq 0$. Setting $y = 1$ gives\n\n$$\nc = |c + f(0)|. \\qquad (2)\n$$\n\nSetting $y = 2$ and dividing by $4$ gives $c = |c + f(0)/2|$. Equating the expressions in absolute values, we see that either $f(0) = f(0)/2$, which implies $f(0) = 0$, or $2c + 3f(0)/2 = 0$, which implies $f(0) = -4c/3$. In the latter case, substituting $f(0)$ into (2) yields $c = |c - 4c/3|$, whence $c = 0$, which implies $f(0) = 0$. In either case, we have $f(0) = 0$, so $f(x) = c x$ for all $x \\geq 0$.\n\nWe now claim that $f(y) = c y$ for all real $y$. Since $f(0) = 0$, we have $f(y^2) = |y f(y)|$ for all real $y$. Thus, for any $y < 0$, we have\n\n$$\nc y^2 = f(y^2) = |y f(y)|,\n$$\n\nwhich implies $|f(y)| = -c y$. Thus, $f(y) = c y$ or $f(y) = -c y$. If $f(y) = c y$, we are done. Otherwise, since $y \\leq 0$, $f(y) \\leq f(0) = 0 \\leq -c y$, $f(y) = -c y$ would imply $0 = -c y$, hence $c = 0$, hence $|f(y)| = 0$ and $f(y) = c y$. Thus, we see that $f(y) = c y$ for all real $y$.\n\nFinally, since $f(1^2) = c = |1 \\cdot f(1)| \\geq 0$, we must have $c \\geq 0$. Conversely, for any $c \\geq 0$, we have\n\n$$\nf(x + y^2) = c x + c y^2 = f(x) + |y f(y)|,\n$$\n\nso the solutions are $f(x) = c x$ for any $c \\geq 0$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 12687,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $a_1, a_2, \\dots$ be a sequence of integers with infinitely many positive and infinitely many negative terms. Suppose that for each positive integer $n$, the numbers $a_1, a_2, \\dots, a_n$ leave distinct remainders upon division by $n$. Prove that every integer occurs exactly once in the sequence.",
    "options": [],
    "answer": "See solution",
    "solution": "The conditions of the problem can be reformulated by saying that for every positive integer $n$, the numbers $a_1, a_2, \\dots, a_n$ form a complete set of residues modulo $n$. We proceed with the proof as follows:\n\n1. **Distinctness:** The sequence consists of distinct integers; that is, if $1 \\leq i < j$, then $a_i \\neq a_j$. Otherwise, the set $\\{a_1, a_2, \\dots, a_j\\}$ would contain at most $j-1$ distinct residues modulo $j$, violating the condition.\n\n2. **Closeness:** If $1 \\leq i < j \\leq n$, then $|a_i - a_j| \\leq n - 1$. If $m = |a_i - a_j| \\geq n$, then the set $\\{a_1, a_2, \\dots, a_m\\}$ would contain two numbers congruent modulo $m$, again violating the condition.\n\n3. **Consecutive Block:** The set $\\{a_1, a_2, \\dots, a_n\\}$ contains a block of consecutive numbers. For every $n$, let $i_n$ and $j_n$ be the indices such that $a_{i_n}$ and $a_{j_n}$ are the smallest and largest among $a_1, a_2, \\dots, a_n$. By (2), $a_{j_n} - a_{i_n} \\leq n - 1$. By (1), $\\{a_1, a_2, \\dots, a_n\\}$ consists of all integers between $a_{i_n}$ and $a_{j_n}$ (inclusive).\n\n4. **Completeness:** Every integer appears in the sequence. Let $x$ be arbitrary. Since there are infinitely many positive and negative terms and all terms are distinct, there exist $i$ and $j$ such that $a_i < x < a_j$. For $n \\geq \\max\\{i, j\\}$, by (3), every number between $a_i$ and $a_j$, including $x$, is in $\\{a_1, a_2, \\dots, a_n\\}$. Thus, every integer occurs exactly once in the sequence.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 12688,
    "subject": "Mathematics (Olympiad)",
    "question": "Во равенството $25! = 15\\ 511 \\times 10\\ 043\\ 330\\ x85\\ 984\\ y00\\ 000$ определи ги цифрите $x$, $y$ и $z$ за да тоа е точно.",
    "options": [],
    "answer": "See solution",
    "solution": "По дефиниција, $25! = 1 \\cdot 2 \\cdot 3 \\cdot 4 \\cdot 5 \\cdots 25$. Ако овој број го разложиме на прости множители (канонична факторизација), се добива:\n\n$$\n25! = 2^{22} \\cdot 3^{10} \\cdot 5^6 \\cdot 7^3 \\cdot 11^2 \\cdot 13 \\cdot 17 \\cdot 23 = 10^6 \\cdot 2^{16} \\cdot 3^{10} \\cdot 7^3 \\cdot 11^2 \\cdot 13 \\cdot 17 \\cdot 23\n$$\n\nЗначи $25!$ завршува на 6 нули, па следува дека $z=0$.\n\nБројот $25!$ е делив со 9, следува дека $x + y$ е таков што $61 + x + y$ е делив со 9. Значи:\n\n$$\nx + y = 2 \\text{ или } x + y = 11\n$$\n\n(бидејќи $x$ и $y$ се едноцифрени броеви).\n\nБројот $25!$ е делив со 11. Критериумот за деливост со 11 гласи: бројот $a_n...a_5a_4a_3a_2a_1a_0$ е делив со 11 ако бројот $(a_0 + a_2 + a_4 +...) - (a_1 + a_3 + a_5 +...)$ е делив со 11. Па, следува дека $(34 + x) - (27 + y) = 7 + x - y$ е делив со 11. Значи:\n\n$$\n-x + y = 7 \\text{ или } x - y = 4\n$$\n\n(бидејќи $x$ и $y$ се едноцифрени броеви).\n\nОд горните равенки ги формираме следниве системи:\n\n$$\n\\begin{cases} x + y = 2 \\\\ -x + y = 7 \\end{cases} \\qquad \\begin{cases} x + y = 2 \\\\ x - y = 4 \\end{cases} \\qquad \\begin{cases} x + y = 11 \\\\ -x + y = 7 \\end{cases} \\qquad \\begin{cases} x + y = 11 \\\\ x - y = 4 \\end{cases}\n$$\n\nЦелобројни решенија се добиваат само кај вториот и третиот систем, т.е. $x = 2$, $y = 9$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 12689,
    "subject": "Mathematics (Olympiad)",
    "question": "You may pick any three piles and take 2 stones out of the first pile, 3 stones out of the second pile, and 4 stones out of the third pile, provided that each pile has enough stones. Is it possible, after a finite number of such operations, to get exactly $3^{1005}$ stones in each pile?",
    "options": [],
    "answer": "See solution",
    "solution": "No.\n\nAfter each operation, the total number of stones changes by a number divisible by $9$. At the end, the total number is $2012 \\cdot 3^{1005}$, which is divisible by $9$. However, at the starting moment, the total number is $2^0 + 2^1 + 2^2 + \\ldots + 2^{2011} = 2^{2012} - 1$, which is not divisible by $9$. Indeed, $2012 = 335 \\cdot 6 + 2$, $2^6 = 64 \\equiv 1 \\pmod{9}$, therefore\n\n$$\n2^{2012}-1 = (2^6)^{335} \\cdot 2^2 - 1 = 4 - 1 = 3 \\pmod{9}.\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 12690,
    "subject": "Mathematics (Olympiad)",
    "question": "En un grupo de 2022 estudiantes, algunos son amigos entre sí, y la amistad es siempre recíproca. Sabemos que cualquier subconjunto de esos estudiantes tiene la siguiente propiedad: siempre existe un estudiante del subconjunto que es amigo de, a lo sumo, 100 estudiantes del mismo.\n\n(a) Determina el menor entero positivo $N$ que nos asegura que se cumple la siguiente propiedad: es posible dividir a los estudiantes en $N$ grupos (no necesariamente del mismo tamaño), de manera que dos estudiantes que están en el mismo grupo nunca son amigos entre sí.\n\n(b) Numeramos a los estudiantes del 1 al 2022. Sea $c_i$ el número de amigos del estudiante $i$. Determina el máximo valor que puede tomar la suma\n\n$$\nc_1 + c_2 + \\dots + c_{2022}.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "Para la primera parte, seguimos la siguiente estrategia: tomamos un estudiante con a lo sumo 100 amigos (que existe por hipótesis). Diremos que es el estudiante 1. Sacamos a ese estudiante y en el subconjunto resultante, existirá un estudiante con a lo sumo 100 amigos en dicho subconjunto; este será el estudiante 2, y así sucesivamente.\n\nVamos a probar entonces que la respuesta es $N = 101$. Empezamos a asignar grupo por el final. Al estudiante 2022 le asignamos un grupo cualquiera. En general, al estudiante $i$ le asignaremos un grupo que no haya sido usado en ninguno de los estudiantes ya asignados de los que sea amigo. Por construcción, sabemos que es amigo de como máximo 100 estudiantes con un número mayor, así que con tener 101 grupos es suficiente para completar esta asignación.\n\nPara ver que $N = 101$ es la mejor opción, consideremos la siguiente configuración: tomamos 100 estudiantes de los 2022 y hacemos que esos 100 sean amigos de todos (tanto entre ellos como con los 1922 restantes); entre los demás no hay más relaciones de amistad. Si ahora tomamos esos 100 estudiantes y uno de los otros tenemos que hay todas las relaciones de amistad posibles, con lo que 100 grupos, o cualquier cantidad inferior, no sería suficiente.\n\nPara el segundo apartado seguimos esta misma estrategia. Tomamos un estudiante con lo sumo 100 amigos y lo eliminamos. Cada vez que hacemos este proceso, si quedaban por lo menos 101, estamos eliminando a lo sumo 100 relaciones de amistad. Cuando quedan $i$ estudiantes, con $i \\leq 100$, estamos sacando como máximo $i-1$ relaciones de amistad. En total, no sacamos más de\n\n$$\n100 \\cdot 1922 + \\sum_{i=1}^{99} i = 192200 + \\binom{100}{2}\n$$\n\nrelaciones de amistad. Teniendo en cuenta que, por el enunciado del apartado (b), cada relación de amistad ha de contarse dos veces, la suma pedida será menor o igual que $384400 + 9900 = 394300$. Está claro que podemos conseguir la cota. Para ello, al igual que antes, tomamos 100 de los 2022 y hacemos que esos 100 sean amigos de todos (tanto entre ellos como con los 1922 restantes). En este caso, el número de amistades es el proporcionado por la cota.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 12691,
    "subject": "Mathematics (Olympiad)",
    "question": "Consider the system\n\n$$\n(x + 20)(y + 12)(z - 4) = n\n$$\n\n$$\n(x + 20)(y - 12)(z + 4) = n\n$$\n\n$$\n(x - 20)(y + 12)(z + 4) = n.\n$$\n\n(a) For $n = 1215$, find all real solutions $(x, y, z)$ to the system.\n\n(b) Find the smallest integer $n > 1215$ for which the system has a real solution $(x, y, z)$.",
    "options": [],
    "answer": "See solution",
    "solution": "Subtracting the second equation from the first gives:\n\n$$\n(x + 20) \\left[(y + 12)(z - 4) - (y - 12)(z + 4)\\right] = 0.\n$$\n\nSimplifying yields $y = 3z$.\n\nSimilarly, subtracting the third equation from the first gives $x = 5z$.\n\nSubstituting $x$ and $y$ into the first equation:\n\n$$\nn = 15(z + 4)^2(z - 4)\n$$\n\n(a) For $n = 1215$, $(z+4)^2(z-4) = 81$. The only real solution with $z > 4$ is $z = 5$.\n\nThus, $x = 25$, $y = 15$, $z = 5$ is the only solution.\n\n(b) Since $(z+4)^2(z-4)$ is strictly increasing for $z \\ge 4$, the smallest integer $n > 1215$ is for $z = 6$:\n\n$$\nn = 15(6 + 4)^2(6 - 4) = 3000.\n$$\n\nFor $z = 6$, $y = 18$, $x = 30$; these values satisfy the system.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 12692,
    "subject": "Mathematics (Olympiad)",
    "question": "Given an isosceles triangle $ABC$ with $AB = AC$, let point $D$ satisfy $AD \\parallel BC$ and $DB > DC$. Point $E$ is on the arc $\\widearc{BC}$ of the circumcircle of $\\triangle ABC$ that does not contain $A$, with $EB < EC$. Let $F$ be a point on the extension of $BC$ such that $\\angle DFE = \\angle ADE$. Let the extension of $FD$ intersect the extension of $BA$ at point $X$ and the extension of $CA$ at point $Y$.\n\nProve that $\\angle XEY$ is a fixed value.\n\n![](images/China-TST-2024B_p24_data_328fa52a8b.png)",
    "options": [],
    "answer": "See solution",
    "solution": "*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$$\n\\frac{AB^2}{DF^2} = \\frac{AP \\cdot AE}{DQ \\cdot DE} = \\frac{AE^2}{DE^2},\n$$\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$$\n\\angle XEY = \\angle XTY = \\angle TXD - \\angle TYD = \\frac{1}{2}\\angle AXD - \\frac{1}{2}\\angle AYD = \\frac{1}{2}\\angle XAY = \\frac{1}{2}\\angle BAC\n$$\n\nis a fixed value. $\\square$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 12693,
    "subject": "Mathematics (Olympiad)",
    "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": "See solution",
    "solution": "Let $\\angle DAE = \\alpha$. Then $\\angle DOE = 2\\alpha$.\n\nFrom the isosceles triangle $DOE$, $\\angle OED = \\frac{180^\\circ - 2\\alpha}{2} = 90^\\circ - \\alpha$. Since $DE$ and $AO$ are parallel, $\\angle AOE = 90^\\circ - \\alpha$.\n\nThe common tangent to $\\omega_1$ and $\\omega_2$ at $A$ is perpendicular to the radius $AO$ of $\\omega_1$ and to the radius of $\\omega_2$. Since $O$ lies on $\\omega_2$, the segment $AO$ must be a diameter of $\\omega_2$, so $\\angle OCA = 90^\\circ$. As $OA = OD$, the segment $OC$ is the altitude of isosceles triangle $OAD$ from its apex. Thus, $\\angle AOE = \\angle DOE = 2\\alpha$.\n\nTherefore, $90^\\circ - \\alpha = 2\\alpha$, so $\\alpha = 30^\\circ$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 12694,
    "subject": "Mathematics (Olympiad)",
    "question": "We call a *contour* a set of segments of length 1, as for a $1 \\times 1$ square. A machine can color one contour in one operation. What is the minimal number of operations required to color all segments (internal and external) of a $2010 \\times 2011$ board?\n\nContours can have common points and sides, and can also have points that lie outside the board.",
    "options": [],
    "answer": "See solution",
    "solution": "It is clear that each segment on the border of the board must be colored, so we need at least $2 \\cdot 2011 + 2 \\cdot 2010 - 4 = 8038$ operations.\n\nWe also need to color the internal $2008 \\times 2009$ squares, possibly without their borders. We split the internal square into $1 \\times 2$ dominos. To color the segment in the middle of a domino, one must color at least one square forming this domino.\n\n![](images/Ukrajina_2011_p7_data_7b65e1bc2b.png)\n\nIt is easy to see that if we perform an operation on each black square of each domino, then all segments of the board are colored. The total number of dominos is $\\frac{2008 \\times 2009}{2} = 2017036$. Hence, we need at least $2017036 + 8038 = 2025074$ operations.\n\n![](images/Ukrajina_2011_p7_data_b164fbc9ec.png)",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 12695,
    "subject": "Mathematics (Olympiad)",
    "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 triangle $XOB$ by $G$, and let $Y$ be the second intersection of the line $AX$ with the boundary of the half-disc. Prove that $YG = GB$.\n\n![](images/CZE_ABooklet_2024_p6_data_05e4e64cd2.png)",
    "options": [],
    "answer": "See solution",
    "solution": "Denote the centre of the segment $XB$ by $M$. We shall prove that the median $OM$ of triangle $OXB$ coincides with the perpendicular bisector of segment $BY$. Since the point $G$ lies on the line $OM$, this clearly implies the desired equality.\n\nSince $B$ and $Y$ lie on the same circle centered at $O$, we have $OY = OB$. Moreover, by Thales' Theorem, $\\angle AYB = 90^\\circ$. This means that $M$ is the midpoint of the hypotenuse of right triangle $XYB$, so by Thales' Theorem, it is the circumcentre of $XYB$, hence $MY = MB$. This means that both points $M$ and $O$ lie on the perpendicular bisector of $BY$, and since they are clearly distinct, we immediately get the desired conclusion.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 12696,
    "subject": "Mathematics (Olympiad)",
    "question": "Positive real numbers $x, y, z$ satisfy $xyz + xy + yz + zx = x + y + z + 1$. Prove that\n\n$$\n\\frac{1}{3} \\left( \\sqrt{\\frac{1+x^2}{1+x}} + \\sqrt{\\frac{1+y^2}{1+y}} + \\sqrt{\\frac{1+z^2}{1+z}} \\right) \\leq \\left( \\frac{x+y+z}{3} \\right)^{5/8}.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "By the given condition, we have\n\n$$\n1 = \\frac{xyz + xy + yz + xz}{x + y + z + 1}.\n$$\n\nWe may therefore compute\n\n$$\n\\begin{aligned}\n1 + x^2 &= \\frac{xyz + xy + yz + xz}{x + y + z + 1} + x^2 \\\\\n&= \\frac{xyz + xy + yz + xz + x^3 + x^2y + x^2z + x^2}{x + y + z + 1} \\\\\n&= \\frac{(xyz + yz) + (x^2z + xz) + (x^2y + xy) + (x^3 + x^2)}{x + y + z + 1} \\\\\n&= \\frac{(1+x)(yz + xz + xy + x^2)}{x + y + z + 1} \\\\\n&= \\frac{(1+x)(x+y)(x+z)}{x + y + z + 1}.\n\\end{aligned}\n$$\n\nBy the AM-GM inequality, this implies that\n\n$$\n\\frac{1}{3}\\sqrt{\\frac{1+x^2}{1+x}} = \\frac{1}{3}\\sqrt{\\frac{(x+y)(x+z)}{x+y+z+1}} \\leq \\frac{x+y+x+z}{6\\sqrt{x+y+z+1}}.\n$$\n\nAdding the last inequality to its cyclic analogues yields\n\n$$\n\\frac{1}{3} \\left( \\sqrt{\\frac{1+x^2}{1+x}} + \\sqrt{\\frac{1+y^2}{1+y}} + \\sqrt{\\frac{1+z^2}{1+z}} \\right) \\leq \\frac{2(x+y+z)}{3\\sqrt{x+y+z+1}}.\n$$\n\nIt thus suffices to show that\n\n$$\n\\frac{2(x + y + z)}{3\\sqrt{x + y + z + 1}} \\leq \\left(\\frac{x + y + z}{3}\\right)^{5/8}.\n$$\n\nWriting $s = x + y + z$, this is equivalent to\n\n$$\n\\left(\\frac{s}{3}\\right)^{3/4} \\leq \\frac{1}{4}(s+1) = \\frac{1}{4}\\left(\\frac{s}{3} + \\frac{s}{3} + \\frac{s}{3} + 1\\right),\n$$\n\nwhich holds by the AM-GM inequality, completing the proof.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 12697,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $\\mathbb{Z}[x]$ denote the set of single-variable polynomials in $x$ with integer coefficients.\n\nFind all functions $\\theta : \\mathbb{Z}[x] \\to \\mathbb{Z}$ such that for any polynomials $p, q \\in \\mathbb{Z}[x]$:\n\n- $\\theta(p + 1) = \\theta(p) + 1$\n- If $\\theta(p) \\neq 0$ then $\\theta(p)$ divides $\\theta(p \\cdot q)$.",
    "options": [],
    "answer": "See solution",
    "solution": "The answer is $\\theta : p \\mapsto p(c)$ for each choice of $c \\in \\mathbb{Z}$. Obviously, these work, so we prove these are the only ones. In what follows, $x \\in \\mathbb{Z}[x]$ is the identity polynomial, and $c = \\theta(x)$.\n\n**First solution (Merlijn Staps)**\n\nConsider an integer $n \\neq c$. Because $x - n \\mid p(x) - p(n)$, we have\n\n$$\n\\theta(x-n) \\mid \\theta(p(x)-p(n)) \\implies c-n \\mid \\theta(p(x))-p(n).\n$$\n\nOn the other hand, $c-n \\mid p(c)-p(n)$. Combining the previous two gives $c-n \\mid \\theta(p(x))-p(c)$, and by letting $n$ be large we conclude $\\theta(p(x))-p(c) = 0$, so $\\theta(p(x)) = p(c)$.\n\n**Second solution**\n\nFirst, we settle the case $\\deg p = 0$. In that case, from the second property, $\\theta(m) = m + \\theta(0)$ for every integer $m \\in \\mathbb{Z}$ (viewed as a constant polynomial). Thus $m + \\theta(0) \\mid 2m + \\theta(0)$, hence $m + \\theta(0) \\mid -\\theta(0)$, so $\\theta(0) = 0$ by taking $m$ large. Thus $\\theta(m) = m$ for $m \\in \\mathbb{Z}$.\n\nNext, we address the case of $\\deg p = 1$. We know $\\theta(x+b) = c+b$ for $b \\in \\mathbb{Z}$. Now for each particular $a \\in \\mathbb{Z}$, we have\n\n$$\nc+k \\mid \\theta(x+k) \\mid \\theta(ax+ak) = \\theta(ax) + ak \\implies c+k \\mid \\theta(ax) - ac.\n$$\n\nfor any $k \\neq -c$. Since this is true for large enough $k$, we conclude $\\theta(ax) = ac$. Thus $\\theta(ax+b) = ac+b$.\n\nWe now proceed by induction on $\\deg p$. Fix a polynomial $p$ and assume it's true for all $p$ of smaller degree. Choose a large integer $n$ (to be determined later) for which $p(n) \\neq p(c)$. We then have\n\n$$\n\\frac{p(c)-p(n)}{c-n} = \\theta\\left(\\frac{p-p(n)}{x-n}\\right) \\mid \\theta(p-p(n)) = \\theta(p)-p(n).\n$$\n\nSubtracting off $c-n$ times the left-hand side gives\n\n$$\n\\frac{p(c)-p(n)}{c-n} \\mid \\theta(p)-p(c).\n$$\n\nThe left-hand side can be made arbitrarily large by letting $n \\to \\infty$, since $\\deg p \\geq 2$. Thus $\\theta(p) = p(c)$, concluding the proof.",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 12698,
    "subject": "Mathematics (Olympiad)",
    "question": "Find the smallest positive integer which has exactly $2015$ positive divisors.",
    "options": [],
    "answer": "See solution",
    "solution": "The number of positive divisors of an integer $N$ with the prime factorization $p_1^{a_1} p_2^{a_2} \\cdots p_n^{a_n}$ is $(a_1 + 1)(a_2 + 1) \\cdots (a_n + 1)$. Since $2015 = 1 \\cdot 2015 = 5 \\cdot 403 = 13 \\cdot 155 = 31 \\cdot 65 = 5 \\cdot 13 \\cdot 31$, the required number has one of the following forms:\n\n- $a = 2^{2014}$\n- $b = 2^{402} \\cdot 3^4$\n- $c = 2^{154} \\cdot 3^{12}$\n- $d = 2^{64} \\cdot 3^{30}$\n- $e = 2^{30} \\cdot 3^{12} \\cdot 5^4$\n\nWe claim that the smallest number is $2^{30} \\cdot 3^{12} \\cdot 5^4$. Indeed, $a > e$ since $2^{1984} > 3^{12} \\cdot 5^4$; $b > e$ since $2^{372} > 3^8 \\cdot 5^4$; $c > e$ for $2^{124} > 5^4$ and $d > e$ for $2^{34} \\cdot 3^{18} > 5^4$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 12699,
    "subject": "Mathematics (Olympiad)",
    "question": "Suppose two sandwich numbers differ by 82. A sandwich number is a four-digit number where the middle two digits equal the product of the first and last digits. Neither number can start or end with 0 or 1. List all pairs of sandwich numbers that differ by 82.",
    "options": [],
    "answer": "See solution",
    "solution": "Let the sandwich numbers be $a b_1 b_2 c$ and $a d_1 d_2 (c+2)$, where $a$ is the first digit, $c$ is the last digit of the smaller number, and $c+2$ is the last digit of the larger. The middle two digits are $a \\times c$ and $a \\times (c+2)$. Their difference is $a \\times (c+2) - a \\times c = a \\times 2 = 8$, so $a = 4$. The possible values for $c$ are $3, 4, 5, 6, 7$ (since neither number can end in 0 or 1, and $c+2 \\leq 9$). The pairs of sandwich numbers that differ by 82 are:\n\n- $4123$ and $4205$\n- $4164$ and $4246$\n- $4205$ and $4287$\n- $4246$ and $4328$\n- $4287$ and $4369$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 12700,
    "subject": "Mathematics (Olympiad)",
    "question": "Determine the greatest positive integer that has pairwise distinct digits and is divisible by each of its digits.",
    "options": [],
    "answer": "See solution",
    "solution": "First, note that $0$ cannot be one of the digits, since division by $0$ is undefined. If $5$ is one of the digits, the number must end with $5$ to be divisible by $5$, but then it cannot be divisible by $2$, $4$, $6$, or $8$. Thus, omitting $5$ allows for more digits.\n\nIf all digits except $0$ and $5$ are present, the number is not divisible by $3$, $6$, or $9$. Therefore, we must omit another digit to ensure divisibility by $3$. The possible cases are:\n\n**Case 1:** If either $1$ or $7$ is omitted, the number is divisible by $3$ and $6$, but not by $9$, so $9$ is not present and the number has $6$ digits.\n\n**Case 2:** If $4$ is omitted, it is possible to arrange the remaining digits so that each digit divides the number. Thus, the number has $7$ digits: $1, 2, 3, 6, 7, 8, 9$.\n\nOur goal is to form the largest such number. Since the number must be divisible by $9$ and $3$ (which is always true for these digits), we focus on the divisibility by $8$ and $7$.\n\nThe last three digits must form a number divisible by $8$. After checking possible arrangements, $9867312$ is the largest number with distinct digits, each of which divides the number, and it is divisible by $7$.\n\n**Answer:** $9867312$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 12701,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $p$ be a prime number. For how many positive integers $n$ is $\\dfrac{n^3 + np + 1}{n + p + 1}$ an integer?",
    "options": [],
    "answer": "See solution",
    "solution": "Recall that $x^3 + y^3 + 1 - 3xy$ is divisible by $x + y + 1$. Writing\n\n$$\n\\begin{aligned}\n\\frac{n^3 + np + 1}{n + p + 1} &= \\frac{n^3 + p^3 + 1 - 3pn}{n + p + 1} - \\frac{p(p^2 - 4n)}{n + p + 1} \\\\\n&= \\frac{n^3 + p^3 + 1 - 3pn}{n + p + 1} + 4p - \\frac{p(p + 2)^2}{n + p + 1},\n\\end{aligned}\n$$\n\nwe see that $\\frac{n^3 + np + 1}{n + p + 1}$ is an integer if and only if $\\frac{p(p+2)^2}{n + p + 1}$ is an integer. Let $f(p)$ denote the number of positive integers $n$ for which $n + p + 1$ divides $p(p + 2)^2$. We will prove that $f(p) \\equiv 1 \\pmod{3}$ for any prime $p$.\n\nIt is clear that $f(2) = 4$, so assume $p$ is odd. Let $\\psi(m)$ denote the number of positive integer divisors of $m$. Note that for any integer $k$, the number of positive integer divisors of $k^2$ that are smaller than $k$ and larger than $k$ are equal. By noting that $\\gcd(p, p+2) = 1$ and $n + p + 1 \\ge p + 2$, we obtain\n\n$$\nf(p) = (\\psi((p+2)^2) - 1) + \\frac{\\psi((p+2)^2) + 1}{2} = \\frac{3 \\cdot \\psi((p+2)^2) - 1}{2},\n$$\n\nwhich completes the proof.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 12702,
    "subject": "Mathematics (Olympiad)",
    "question": "Consider a vertical or horizontal $1 \\times 4$ shape. The two end cells are called \"bad cells.\" If the letter \"Ц\" is written in a bad cell, it is a losing move. Let $M_{max}$ be the maximum possible number of bad cells. Find the value of $M_{max}$.\n\n% IMAGE: ![](images/2013-ilovepdf-compressed_p43_data_114212e242.png)\n\nFor any such shape, the number of bad cells is $2$ ($M=2$). For a specific case, the maximum number of adjacent bad cells is $2$.\n\n% IMAGE: ![](images/2013-ilovepdf-compressed_p43_data_3a90574bc7.png)",
    "options": [],
    "answer": "See solution",
    "solution": "We have $M_{max} \\leq 2k$.\n\n$$\n2k + k + k + 1 \\geq M_{max} + k + k + 1 > 64 \\rightarrow 4k > 63,\n$$\n\n$$\nk > \\frac{63}{4} \\rightarrow k_{min} \\geq 16.\n$$\n\nThus, the minimum value of $k$ is $16$. Now, let's draw a configuration where $k_{min} = 16$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 12703,
    "subject": "Mathematics (Olympiad)",
    "question": "Three identical square sheets of paper, each with side length 6, are stacked on top of each other. The middle sheet is rotated clockwise $30^\\circ$ about its center, and the top sheet is rotated clockwise $60^\\circ$ about its center, resulting in the 24-sided polygon shown in the figure below. The area of this polygon can be expressed in the form $a - b\\sqrt{c}$, where $a$, $b$, and $c$ are positive integers, and $c$ is not divisible by the square of any prime. What is $a + b + c$?\n\n![](images/2021_AMC10B_Solutions_Fall_p9_data_d1a87ccf39.png)\n\n(A) 75 (B) 93 (C) 96 (D) 129 (E) 147",
    "options": [],
    "answer": "See solution",
    "solution": "Let $O$ be the center of the polygon, and label $11$ points as shown in the figure. Let $a = AB = BC$.\n\n![](images/2021_AMC10B_Solutions_Fall_p9_data_5b754c3631.png)\n\nTriangle $BCK$ is a $30$-$60$-$90\\degree$ triangle, so $BK = 2a$ and $CK = KG = a\\sqrt{3}$. Then $AG = 3a + a\\sqrt{3} = 6$, so $a = 3 - \\sqrt{3}$. The area of the $24$-sided polygon can be computed as $12$ times the area of kite $OBCD$. The longer diagonal of this kite is $OC$, half of a diagonal of the square, so $OC = 3\\sqrt{2}$. The shorter diagonal of the kite is $BD$, the hypotenuse of isosceles right triangle $BCD$ with leg $a = 3 - \\sqrt{3}$. The area of a kite is half the product of the lengths of its diagonals, so the area of the $24$-sided polygon is\n\n$$\n12 \\cdot \\frac{1}{2} \\cdot 3\\sqrt{2} \\cdot (3 - \\sqrt{3}) \\sqrt{2} = 108 - 36\\sqrt{3}.\n$$\n\nTherefore, $a + b + c = 108 + 36 + 3 = 147$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 12704,
    "subject": "Mathematics (Olympiad)",
    "question": "Дараалал $\\{x_n\\}$ нь $x_0 = a$, $x_1 = 2$ ба\n\n$$\nr_n = 2x_n - x_{n-2} - x_{n-1} - x_{n-2} - 1\n$$\nтомёогоор өгөгдсөн бол $2x_{3n} - 1$ тоо бүхэл тооны квадрат байх бүх $a$ тоог ол.",
    "options": [],
    "answer": "See solution",
    "solution": "$$\n2x_n - 1 = 2(2x_{n-1}x_{n-2} - x_{n-1} - x_{n-2} + 1) - 1 \\\\\n= (2x_{n-1} - 1)(2x_{n-2} - 1)\n$$\n\n$$\na_n = 2x_n - 1 \\text{ гэе. } a_n = a_{n-1} \\cdot a_{n-2},\\ a_1 = 3,\\ a_0 = 2a - 1.\n$$\n\n$$\na_0 = (2k + 1)^2 = 4k^2 + 4k + 1 = 2a - 1 \\Rightarrow a = 2k^2 + 2k + 1\n$$\n\n$$\na_{3n} = k^2,\\ a_{3n+1} = 3m^2 \\text{ бол } a_{3n+2} = k^2 \\cdot 3m^2 = 3(mk)^2,\\ a_{3n+1} = 3k^2\n$$\n\n$$\n\\text{ба } a_{3n+2} = 3m^2 \\text{ бол } a_{3n+3} = (3km)^2,\\ a_{3n+2} = 3k^2,\\ a_{3(n+1)} = m^2\n$$\n\n$$\n\\text{ба } a_{3(n+1)+1} = 3(mk)^2.\\ \\text{Иймд } a = 2k^2 + 2k + 1,\\ \\forall k \\in \\mathbb{Z} \\text{ бол } 2x_{3n} - 1 \\text{ нь бүтэн квадрат болно.}\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 12705,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $N$ be the product of all the positive integer divisors of $42$. What is the units digit of $N$?\n\n(A) 0  (B) 2  (C) 4  (D) 6  (E) 8",
    "options": [],
    "answer": "See solution",
    "solution": "The prime factorization of $42$ is $2 \\cdot 3 \\cdot 7$. A divisor of $42$ is determined by choosing whether or not to include each prime as a factor, so there are $2^3 = 8$ divisors. The divisors can be paired so that each pair multiplies to $42$: $1 \\cdot 42$, $2 \\cdot 21$, $3 \\cdot 14$, and $6 \\cdot 7$. Thus, $N = 42^4$. The units digit of $N$ is the same as the units digit of $2^4$, which is $6$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 12706,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $x$, $y$, and $z$ be positive real numbers. Show that\n$$\nx^2 + x y^2 + x y z^2 \\ge 4 x y z - 4.\n$$",
    "options": [],
    "answer": "See 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$$\nx^2 + x y^2 + x y z^2 \\ge (4x - 4) + x(4y - 4) + x y (4z - 4) = 4 x y z - 4.\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 12707,
    "subject": "Mathematics (Olympiad)",
    "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": "See solution",
    "solution": "By the properties of inscribed angles, $\\angle CXO = \\angle ABO$ (see figure) independent of whether the point $X$ lies on the side $CA$ or on the extension of $CA$ beyond $A$. 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 $ABC$ 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![](images/prob1516_p12_data_aaed0f68b8.png)\n\nFig. 8",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 12708,
    "subject": "Mathematics (Olympiad)",
    "question": "Show that the sum of the numbers in any $3 \\times 5$ or $5 \\times 3$ rectangle of consecutive integers is divisible by $15$.",
    "options": [],
    "answer": "See solution",
    "solution": "Consider a $3 \\times 5$ rectangle of consecutive integers, with the center number $x$:\n\n$$\n\\begin{array}{ccccc}\nx - 12 & x - 11 & x - 10 & x - 9 & x - 8 \\\\\nx - 2 & x - 1 & x & x + 1 & x + 2 \\\\\nx + 8 & x + 9 & x + 10 & x + 11 & x + 12\n\\end{array}\n$$\n\nThe sum is $15x$, which is divisible by $15$.\n\nSimilarly, for a $5 \\times 3$ rectangle:\n\n$$\n\\begin{array}{ccc}\nx - 21 & x - 20 & x - 19 \\\\\nx - 11 & x - 10 & x - 9 \\\\\nx - 1 & x & x + 1 \\\\\nx + 9 & x + 10 & x + 11 \\\\\nx + 19 & x + 20 & x + 21\n\\end{array}\n$$\n\nThe sum is again $15x$, divisible by $15$.\n\nAlternatively, since both rectangles have an odd number of rows and columns, the average of all $15$ numbers is the center number, so the sum is $15$ times the center, always divisible by $15$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 12709,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $n$ ($n \\geq 1$) be a positive integer and $U = \\{1, \\dots, n\\}$. Let $S$ be a nonempty subset of $U$ and let $d$ ($d \\neq 1$) be the smallest common divisor of all elements of the set $S$. Find the smallest positive integer $k$ such that for any subset $T$ of $U$, consisting of $k$ elements, with $S \\subset T$, the greatest common divisor of all elements of $T$ is equal to $1$.",
    "options": [],
    "answer": "See solution",
    "solution": "We will show that $k_{\\min} = 1 + \\lfloor \\frac{n}{d} \\rfloor$ (where $\\lfloor \\cdot \\rfloor$ denotes the integer part).\n\nObviously, the number of elements of $S$ is not greater than $\\lfloor \\frac{n}{d} \\rfloor$, i.e., $|S| \\leq \\lfloor \\frac{n}{d} \\rfloor$, and $S \\neq U$.\n\nIf $S \\subset T$ and the greatest common divisor of elements of $T$ is equal to $1$, then $|T| \\geq |S| + 1$.\n\n1) Assume that $|S| < \\lfloor \\frac{n}{d} \\rfloor$. Let $T$ be the subset of $U$ consisting of all multiples of $d$ in $U$. Thus, $|T| = \\lfloor \\frac{n}{d} \\rfloor$ and $S \\subset T$. Therefore, the greatest common divisor of all elements of $T$ is $d > 1$. Thus, $k \\geq 1 + \\lfloor \\frac{n}{d} \\rfloor$.\n\n2) Assume $|S| = \\lfloor \\frac{n}{d} \\rfloor$. Let $T$ be any subset of $U$ with $S \\subset T$, $S \\neq T$. Therefore, $|T| \\geq 1 + \\lfloor \\frac{n}{d} \\rfloor$. Let $q$ be the greatest common divisor of all elements of $T$. Assume that $q > 1$. Therefore, $q$ is a common divisor of all elements of $S$ as well. Hence, $q \\geq d$. It follows that $|T| \\leq \\lfloor \\frac{n}{q} \\rfloor \\leq \\lfloor \\frac{n}{d} \\rfloor$, a contradiction. Hence, $q = 1$.\n\nTherefore, the minimal possible value of $k$ is $1 + \\lfloor \\frac{n}{d} \\rfloor$.\n\n$\\boxed{1 + \\lfloor \\frac{n}{d} \\rfloor}$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 12710,
    "subject": "Mathematics (Olympiad)",
    "question": "A cube of side length $n$ is made up of $n^3$ smaller unit cubes. Some of the six faces of the large cube are fully painted. When the large cube is taken apart, 245 smaller cubes do not have any paint on them.\n\nDetermine the value(s) of $n$ and how many faces of the large cube were painted.",
    "options": [],
    "answer": "See solution",
    "solution": "Let us first determine how the number of painted faces influences the formula for the total number of unpainted unit cubes.\n\nIf zero faces are painted, the number of cubes is $n^3$, and if all six are painted, the number of unpainted cubes is $(n - 2)^3$. Since 245 is not a perfect cube, both of these are ruled out.\n\nIf one face is painted, the number of unpainted unit cubes is $n^3 - n^2$.\n\nIf two faces are painted, the number of unpainted unit cubes is either $n^2(n - 2)$ (if the painted faces are opposite each other) or $n(n - 1)^2$ (if they share an edge).\n\nIf three faces are painted, the number of unpainted unit cubes is either $(n - 1)^3$ (if they share a vertex) or $n(n - 1)(n - 2)$.\n\nIf four faces are painted, the number of unpainted unit cubes is either $n(n - 2)^2$ (if the two unpainted faces are opposite each other) or $(n - 1)^2(n - 2)$ (otherwise).\n\nIf five faces are painted, the number of unpainted unit cubes is $(n - 1)(n - 2)^2$.\n\nNow we factorise $245 = 5 \\times 7 \\times 7$. Since these are three prime factors, the only possibility from the expressions above is if $5 \\cdot 7^2 = (n - 2)n^2$.\n\nThus $n = 7$ and two faces opposite each other are painted.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 12711,
    "subject": "Mathematics (Olympiad)",
    "question": "For any prime number $p \\ge 3$, show that for all $x \\in \\mathbb{N}$ sufficiently large, one of the integers $x+1, x+2, \\dots, x + \\frac{p+3}{2}$ has a prime factor larger than $p$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $q = \\frac{p+3}{2}$.\n\nAssume, for contradiction, that none of the integers $x+1, x+2, \\dots, x+q$ has a prime factor larger than $p$.\n\nLet $p_i^{\\alpha_i}$ be the highest prime power dividing $x+i$ (i.e., $p_i^{\\alpha_i+1} \\nmid x+i$). The number of primes $\\le p$ is at most $q-1$. Thus, for $x \\ge (q-1)^{q-1}$, we have $p_i^{\\alpha_i} > x^{1/(q-1)} \\ge q-1$ for each $i=1,2,\\dots,q$.\n\nSince there are $q$ numbers but at most $q-1$ primes $\\le p$, by the pigeonhole principle, there exist distinct $i_1, i_2 \\in \\{1,2,\\dots,q\\}$ with $p_{i_1} = p_{i_2}$. Then $(x+i_1, x+i_2) \\ge \\min(p_{i_1}^{\\alpha_{i_1}}, p_{i_2}^{\\alpha_{i_2}}) > q-1$. But $(x+i_1, x+i_2) = (x+i_1, i_2-i_1) \\le |i_2-i_1| \\le q-1$, a contradiction. Therefore, the statement holds for all $x \\ge (q-1)^{q-1}$.\n\n$\\square$",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 12712,
    "subject": "Mathematics (Olympiad)",
    "question": "A *snake of length* $k$ is an animal which occupies an ordered $k$-tuple $(s_1, \\dots, s_k)$ of cells in an $n \\times n$ grid of square unit cells. These cells must be pairwise distinct, and $s_i$ and $s_{i+1}$ must share a side for $i = 1, \\dots, k-1$. If the snake is currently occupying $(s_1, \\dots, s_k)$ and $s$ is an unoccupied cell sharing a side with $s_1$, the snake can *move* to occupy $(s, s_1, \\dots, s_{k-1})$ instead. The snake has *turned around* if it occupied $(s_1, s_2, \\dots, s_k)$ at the beginning, but after a finite number of moves occupies $(s_k, s_{k-1}, \\dots, s_1)$ instead.\n\nDetermine whether there exists an integer $n > 1$ such that one can place some snake of length at least $0.9n^2$ in an $n \\times n$ grid which can turn around.",
    "options": [],
    "answer": "See solution",
    "solution": "The answer is yes (and $0.9$ is arbitrary).\n\nThe following solution is due to Brian Lawrence. For illustration reasons, we give below a figure of a snake of length $89$ turning around in an $11 \\times 11$ square (which generalizes readily to odd $n$). We will see that a snake of length $(n-1)(n-2)-1$ can turn around in an $n \\times n$ square, so this certainly implies the problem.\n\n![](images/sols-TST-IMO-2019_p8_data_7d3b36e82d.png)\n\nUse the obvious coordinate system with $(1, 1)$ in the bottom left. Start with the snake as shown in Figure 1, then have it move to $(2, 1)$, $(2, n)$, $(n, n-1)$ as in Figure 2. Then, have the snake shift to the position in Figure 3; this is possible since the snake can just walk to $(n, n)$, then start walking to the left and then follow the route; by the time it reaches the $i$th row from the top its tail will have vacated by then. Once it achieves Figure 3, move the head of the snake to $(3, n)$ to achieve Figure 4.\n\nIn Figure 5 and 6, the snake begins to “deform” its loop continuously. In general, this deformation by two squares is possible in the following way. The snake walks first to $(1, n)$ then retraces the steps left by its tail, except when it reaches $(n-1, 3)$ it makes a brief detour to $(n-2, 3)$, $(n-2, 4)$, $(n-1, 4)$ and continues along its way; this gives the position in Figure 5. Then it retraces the entire loop again, except that when it reaches $(n-4, 4)$ it turns directly down, and continues retracing its path; thus at the end of this second revolution, we arrive at Figure 6.\n\nBy repeatedly doing perturbations of two cells, we can move all the “bumps” in the path gradually to protrude from the right; Figure 7 shows a partial application of the procedure, with the final state as shown in Figure 8.\n\nIn Figure 9, we stretch the bottom-most bump by two more cells; this shortens the “tail” by two units, which is fine. Doing this for all $(n-3)/2$ bumps arrives at the situation in Figure 10, with the snake’s head at $(3, n)$. We then begin deforming the turns on the bottom-right by two steps each as in Figure 11, which visually will increase the length of the head. Doing this arrives finally at the situation in Figure 12. Thus the snake has turned around.",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 12713,
    "subject": "Mathematics (Olympiad)",
    "question": "Given a positive integer $n$, consider a triangular array with entries $a_{ij}$ where $i$ ranges from $1$ to $n$ and $j$ ranges from $1$ to $n - i + 1$. The entries of the array are all either $0$ or $1$, and, for all $i > 1$ and any associated $j$, $a_{ij}$ is $0$ if $a_{i-1,j} = a_{i-1,j+1}$, and $a_{ij}$ is $1$ otherwise.\n\nLet $S$ denote the set of binary sequences of length $n$, and define a map $f: S \\to S$ via\n$$\nf: (a_{11}, a_{12}, \\dots, a_{1n}) \\mapsto (a_{n1}, a_{n-1,2}, \\dots, a_{1n}).\n$$\nDetermine the number of fixed points of $f$.",
    "options": [],
    "answer": "See solution",
    "solution": "For convenience, denote $b_k = a_{1, n-k}$ and $c_k = a_{k+1, n-k}$ for every $k = 0, 1, \\dots, n-1$. Our aim is to find the set of relations for $(b_k)$ which are equivalent to the relation $(b_k) = (c_k)$. All calculations are in $\\mathbb{Z}_2$.\n\nThe definition of $a_{ij}$ is equivalent to\n$$\na_{ij} = a_{i-1, j} + a_{i-1, j+1}.\n$$\nA straightforward check shows that\n$$\na_{ij} = \\sum_{\\ell=0}^{i-1} \\binom{i-1}{\\ell} a_{1, \\ell + j}.\n$$\nFor two nonnegative integers $k$ and $\\ell$, write $\\ell \\preceq k$ if the binary representation of $\\ell$ can be obtained from that of $k$ by replacing some ones by zeroes (leading zeroes allowed; thus $0 \\preceq k$ and $k \\preceq k$ for every $k$). Write $\\ell \\prec k$ if $\\ell < k$ and $\\ell \\preceq k$. By Lucas' theorem, $\\binom{k}{\\ell}$ is odd if and only if $\\ell \\preceq k$. Thus,\n$$\nc_k = a_{k+1, n-k} = \\sum_{\\ell=0}^{k} \\binom{k}{\\ell} a_{1, \\ell + n - k} = \\sum_{\\ell=0}^{k} \\binom{k}{\\ell} b_{k-\\ell} = \\sum_{\\ell=0}^{k} \\binom{k}{\\ell} b_{\\ell} = \\sum_{\\ell \\preceq k} b_{\\ell}.\n$$\nNow, the conditions $(b_k) = (c_k)$ rewrite as the set of equations\n$$\n0 = \\sum_{\\ell \\prec k} b_{\\ell} \\qquad (*)_k\n$$\nfor all $k = 0, 1, \\dots, n-1$.\n\nLet $T$ be the set of all strings $(b_k)$ such that $(*)_k$ are satisfied for all odd $k \\le n-1$. Each string in this set is determined uniquely by the values of $b_{2i-1}$ ($2i < n$) and $b_{n-1}$: the values of $b_{2i}$ (for $2i < n-1$) are found inductively from $(*)_{2i+1}$. Thus,\n$$\n|T| = 2^{\\lceil n/2 \\rceil} = 2^{\\lfloor (n+1)/2 \\rfloor}.\n$$\nNow we claim that $T$ is exactly the desired set of fixed points; in fact, all the relations $(*)_d$ for even $d$ follow from the relations $(*)_k$ for odd $k$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 12714,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $x$, $y$, $z$ be positive real numbers with $x + y + z \\ge 3$. Prove that\n$$\n\\frac{1}{x+y+z^2} + \\frac{1}{y+z+x^2} + \\frac{1}{z+x+y^2} \\le 1.\n$$\n\n*When does equality hold?*",
    "options": [],
    "answer": "See solution",
    "solution": "By Cauchy's inequality, we have\n$$\n(x + y + z^2)(x + y + 1) \\ge (x + y + z)^2,\n$$\nhence\n$$\n\\frac{1}{x+y+z^2} \\le \\frac{x+y+1}{(x+y+z)^2}.\n$$\nThus it suffices to show that\n$$\n\\sum_{\\text{cyc}} \\frac{x+y+1}{(x+y+z)^2} = \\frac{2(x+y+z)+3}{(x+y+z)^2} \\le 1.\n$$\nThis is equivalent to the inequality\n$$\n(x + y + z)^2 - 2(x + y + z) - 3 \\ge 0,\n$$\nwhich holds for $x + y + z \\ge 3$.\n\nEquality in the Cauchy step holds if and only if $(x, y, z^2)$ and $(x, y, 1)$ are collinear, i.e., $z^2 = 1$ or, equivalently, $z = 1$. Cyclic permutation shows that equality holds if and only if $x = y = z = 1$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 12715,
    "subject": "Mathematics (Olympiad)",
    "question": "Mari writes 8 prime numbers (not necessarily different), all smaller than 200, in her notebook. She then adds 1 to the first number, 2 to the second, and so on, adding 8 to the eighth number. She then finds the product of these eight sums. Find the largest power of two that can divide the resulting product.",
    "options": [],
    "answer": "See solution",
    "solution": "*Answer:* $2^{31}$.\n\nTo maximize the power of 2 dividing the product, each sum should be divisible by the highest possible power of 2. Adding an even number to a prime yields an even sum only if the prime is 2 (the only even prime). Thus, 2 must be chosen for all even positions (positions 2, 4, 6, 8), giving sums 4, 6, 8, and 10, which are divisible by $2^2$, $2^1$, $2^3$, and $2^1$ respectively.\n\nFor the odd positions:\n- 1st: $127$ is prime and $127+1=128=2^7$.\n- 3rd: $61$ is prime and $61+3=64=2^6$.\n- 5th: $59$ is prime and $59+5=64=2^6$.\n- 7th: $89$ is prime and $89+7=96=3 \\times 2^5$.\n\nSo, the maximal powers of 2 dividing the sums are $2^7$, $2^2$, $2^6$, $2^1$, $2^6$, $2^3$, $2^5$, $2^1$.\n\nMultiplying the exponents: $7+2+6+1+6+3+5+1=31$.\n\nThus, the largest power of two dividing the product is $2^{31}$.\n\n*Remark:* This set of primes is unique under the problem's constraints, as other candidates do not yield higher powers of 2 in the sums.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 12716,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $f: \\mathbb{R}^+ \\to \\mathbb{R}^+$ be a function satisfying\n\n$$\n\\frac{1}{x} + \\frac{1}{f(y)} = f\\left(\\frac{1}{y} + f\\left(\\frac{1}{x}\\right)\\right)\n$$\nfor all $x, y > 0$. Find all such functions $f$.",
    "options": [],
    "answer": "See solution",
    "solution": "We can rewrite the given equation as\n\n$$\n\\frac{1}{x} + \\frac{1}{f(y)} = f\\left(\\frac{1}{y} + f\\left(\\frac{1}{x}\\right)\\right).\n$$\n\nReplacing $x$ with $\\frac{1}{x}$, we get\n\n$$\nP(x, y): x + \\frac{1}{f(y)} = f\\left(\\frac{1}{y} + f(x)\\right).\n$$\n\nSuppose there exists $\\alpha > 0$ with $\\alpha > f(\\alpha)$. Then\n\n$$\nP\\left(\\alpha, \\frac{1}{\\alpha - f(\\alpha)}\\right) \\implies \\alpha = f(\\alpha) - \\frac{1}{f(\\alpha - f(\\alpha))} < f(\\alpha),\n$$\nwhich is a contradiction. So $x \\leq f(x)$ for all $x > 0$.\n\nBut $P(x, y)$ also gives\n\n$$\n\\begin{align*}\nx + \\frac{1}{f(y)} &= f\\left(\\frac{1}{y} + f(x)\\right) \\geq \\frac{1}{y} + f(x) \\geq x + \\frac{1}{y} \\\\\n\\implies \\frac{1}{f(y)} \\geq \\frac{1}{y} \\implies y \\geq f(y), \\forall y > 0\n\\end{align*}\n$$\n\nCombining, we get $f(x) = x$ for all $x > 0$, which is indeed a solution.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 12717,
    "subject": "Mathematics (Olympiad)",
    "question": "$$\n\\begin{array}{|c|c|c|}\n\\hline\n+ & ? & ? \\\\\n\\hline\n? & s & s+1 \\\\\n\\hline\n? & s+2 & s+3 \\\\\n\\hline\n\\end{array}\n$$\n\nSuppose you have four numbers placed on counters, and you form sums by adding pairs from different counters. Show that either one or three of the four numbers on the counters are even.",
    "options": [],
    "answer": "See solution",
    "solution": "Since $s$ is either even or odd, we have two cases:\n\n$$\n\\begin{array}{|c|c|c|}\n\\hline\n+ & x & ? \\\\\n\\hline\n? & \\text{even} & \\text{odd} \\\\\n\\hline\n? & \\text{even} & \\text{odd} \\\\\n\\hline\n\\end{array}\n$$\n\n$$\n\\begin{array}{|c|c|c|}\n\\hline\n+ & x & ? \\\\\n\\hline\n? & \\text{odd} & \\text{even} \\\\\n\\hline\n? & \\text{odd} & \\text{even} \\\\\n\\hline\n\\end{array}\n$$\n\nSince $x$ is either even or odd, we can complete each of these tables in two ways:\n\n$$\n\\begin{array}{|c|c|c|}\n\\hline\n+ & \\text{even} & \\text{odd} \\\\\n\\hline\n\\text{even} & \\text{even} & \\text{odd} \\\\\n\\hline\n\\text{even} & \\text{even} & \\text{odd} \\\\\n\\hline\n\\end{array}\n$$\n\n$$\n\\begin{array}{|c|c|c|}\n\\hline\n+ & \\text{even} & \\text{odd} \\\\\n\\hline\n\\text{odd} & \\text{odd} & \\text{even} \\\\\n\\hline\n\\text{odd} & \\text{odd} & \\text{even} \\\\\n\\hline\n\\end{array}\n$$\n\n$$\n\\begin{array}{|c|c|c|}\n\\hline\n+ & \\text{odd} & \\text{even} \\\\\n\\hline\n\\text{odd} & \\text{even} & \\text{odd} \\\\\n\\hline\n\\text{odd} & \\text{even} & \\text{odd} \\\\\n\\hline\n\\end{array}\n$$\n\n$$\n\\begin{array}{|c|c|c|}\n\\hline\n+ & \\text{odd} & \\text{even} \\\\\n\\hline\n\\text{even} & \\text{odd} & \\text{even} \\\\\n\\hline\n\\text{even} & \\text{odd} & \\text{even} \\\\\n\\hline\n\\end{array}\n$$\n\nIn each case, either one or three of the four numbers on the counters are even.\n\n**Alternative:**\n\nIf all four numbers were even or all were odd, all sums would be even and not consecutive. If two numbers were even and two odd, and both evens were on the same counter, all sums would be odd and not consecutive. Thus, each counter must have an even and an odd number. If the two smaller numbers were odd, the two larger would be even, so the lowest and highest sums would both be even and not consecutive. Similarly, if the two smaller numbers were even. If one smaller number was odd and the other even, then one larger would be even and the other odd, so the lowest and highest sums would both be odd and not consecutive. Therefore, either one or three of the four numbers on the counters are even.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 12718,
    "subject": "Mathematics (Olympiad)",
    "question": "On a table, there are ten thousand matches, two of which are inside a bowl.\n\nAnna and Bernd play the following game: They alternate taking turns, with Anna going first. On each turn, the player counts the matches in the bowl, chooses a proper divisor $d$ of this number, and adds $d$ matches to the bowl. The game ends when there are more than $2024$ matches in the bowl. The person who played the last turn wins.\n\nProve that Anna can win regardless of how Bernd plays.",
    "options": [],
    "answer": "See solution",
    "solution": "Anna's strategy is to always add a single match to the bowl while there are fewer than $1350$ matches inside.\n\nWith this strategy, she always changes an even number of matches to an odd number, so Bernd is forced to choose an odd divisor and give her an even number of matches again.\n\nBernd will have at most $1350$ matches and can add at most a third, since there is no larger odd proper divisor. Since $$1350 + \\frac{1}{3} \\cdot 1350 = 1350 + 450 = 1800 < 2024,$$ he cannot reach more than $2024$ matches in this phase of the game.\n\nTherefore, there will come a turn where Anna starts with an even number of at least $1350$ matches. She can add half of them and obtains at least $$1350 + \\frac{1}{2} \\cdot 1350 = 1350 + 675 = 2025$$ matches and has won.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 12719,
    "subject": "Mathematics (Olympiad)",
    "question": "Hallar las soluciones enteras de la ecuación\n\n$$\nx^{4} + y^{4} = 3x^{3}y.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "Supongamos, en primer lugar, que $x = 0$. En este caso se tiene $y^4 = 0$, por lo que $y$ también tiene que ser 0. Así pues, una solución es $x = y = 0$.\n\nSi $x \\neq 0$, dividimos toda la ecuación por $x^4$, quedando\n\n$$\n1 + \\left(\\frac{y}{x}\\right)^{4} = 3\\frac{y}{x}.\n$$\n\nSea $t = y/x$, entonces, las soluciones enteras de la ecuación $x^4 + y^4 = 3x^3y$ dan lugar a soluciones racionales de la ecuación $t^4 - 3t + 1 = 0$. Sin embargo, esta ecuación no tiene soluciones racionales pues, de tenerlas, el numerador de la fracción debería ser un divisor del término independiente y el denominador un divisor del coeficiente de $t^4$. Es decir, las posibles soluciones racionales solo pueden ser $1$ o $-1$, pero ninguna de ellas verifica $t^4 - 3t + 1 = 0$. Por tanto, como no hay soluciones racionales, no hay soluciones enteras de $x^4 + y^4 = 3x^3y$ con $x \\neq 0$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 12720,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $ABC$ be an acute triangle with altitudes $AK$, $BL$, $CM$. Prove that triangle $ABC$ is isosceles if and only if\n\n$$\nAM + BK + CL = AL + BM + CK.\n$$",
    "options": [],
    "answer": "See 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 characterize the perpendicularity by lengths of segments.\n\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\n$$\nAM - BM = \\frac{b^2 - a^2}{AM + BM} = \\frac{b^2 - a^2}{c}\n$$\n\nand likewise\n\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$$\n\\begin{gathered}\n\\frac{a^2 - b^2}{c} + \\frac{b^2 - c^2}{a} + \\frac{c^2 - a^2}{b} = 0, \\\\\nab(a^2 - b^2) + bc(b^2 - c^2) + ca(c^2 - a^2) = 0.\n\\end{gathered}\n$$\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 an 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 the problem statement is equivalent with\n\n$$\n(a-b)(b-c)(c-a)(a+b+c) = 0.\n$$\n\nIt's obvious that this equality holds if and only if triangle $ABC$ is isosceles.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 12721,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all natural numbers $x$, $n$, and prime $p$ such that\n$$\nx^3 + 3x + 14 = 2p^n.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "It is easy to see that $x^3 + 3x + 14 = (x+2)(x^2 - 2x + 7)$, so the initial equality can be rewritten as\n$$\n(x+2)(x^2 - 2x + 7) = 2p^n.\n$$\n\nNote that $x^2 - 2x + 7 > x+2$ for all $x \\in \\mathbb{N}$. Consider two cases:\n\n1. $x+2 = 2p^k$, $x^2 - 2x + 7 = p^{n-k}$.\n2. $x+2 = p^k$, $x^2 - 2x + 7 = 2p^{n-k}$.\n\nIn both cases, $n-k \\geq k \\geq 0$. For both, $2(x^2 - 2x + 7)$ does not divide $(x+2)$. Thus, $2(x^2 - 2x + 7)/(x+2)$ must be integer, which implies $30/(x+2)$ is integer. So $x+2$ must be a divisor of $30$.\n\nFrom the equation, $x+2$ can have at most two prime divisors, one of which is $2$. Thus, possible values for $x+2$ are $3, 5, 6, 10$, i.e., $x = 1, 3, 4, 8$.\n\n- For $x = 1$: $(x+2)(x^2 - 2x + 7) = 3 \\cdot 6 = 18 = 2 \\cdot 3^2$, so $p = 3$, $n = 2$.\n- For $x = 3$: $(x+2)(x^2 - 2x + 7) = 5 \\cdot 10 = 50 = 2 \\cdot 5^2$, so $p = 5$, $n = 2$.\n- For $x = 4$: $(x+2)(x^2 - 2x + 7) = 6 \\cdot 15 = 90 = 2 \\cdot 3^2 \\cdot 5$, which cannot be $2p^n$ for any $p$ and $n$.\n- For $x = 8$: $(x+2)(x^2 - 2x + 7) = 10 \\cdot 55 = 550 = 2 \\cdot 5^2 \\cdot 11$, which cannot be $2p^n$ for any $p$ and $n$.\n\n**Answer:** $(x, n, p) = (1, 2, 3)$ and $(x, n, p) = (3, 2, 5)$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 12722,
    "subject": "Mathematics (Olympiad)",
    "question": "Determine all triples of real numbers $ (x, y, z) $ such that the equation\n\n$$\n4x^4 - x^2 (4y^4 + 4z^4 - 1) - 2xyz + y^8 + 2y^4z^4 + y^2z^2 + z^8 = 0\n$$\n\nholds.",
    "options": [],
    "answer": "See solution",
    "solution": "We first note that\n\n$$\n\\begin{aligned}\n& 4x^4 - x^2(4y^4 + 4z^4 - 1) - 2xyz + y^8 + 2y^4z^4 + y^2z^2 + z^8 \\\\\n& = (4x^4 + y^8 + z^8 - 4x^2y^4 - 4x^2z^4 + 2y^4z^4) + (x^2 - 2xyz + y^2z^2) \\\\\n& = (2x^2 - y^4 - z^4)^2 + (x - yz)^2.\n\\end{aligned}\n$$\n\nThe given equation is therefore equivalent to\n\n$$\n(2x^2 - y^4 - z^4)^2 + (x - yz)^2 = 0.\n$$\n\nIt therefore follows that both $x = yz$ and $2x^2 - y^4 - z^4 = 0$ must hold.\n\nSubstituting $x = yz$ in the second of these equations, we obtain $2y^2z^2 - y^4 - z^4 = 0$, which is equivalent to $-(y^2 - z^2)^2 = 0$. We see that $z = \\pm y$ must hold. For $z = y = t$, we obtain $x = t^2$, and for $-z = y = t$, we obtain $x = -t^2$. It therefore follows that the set of all solutions is\n\n$$\n\\{ (t^2, t, t) \\mid t \\in \\mathbb{R} \\} \\cup \\{ (-t^2, t, -t) \\mid t \\in \\mathbb{R} \\}.\n$$",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 12723,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $n, k$ be positive integers. Julia and Florian play a game on a $2n \\times 2n$ board. Julia has secretly tiled the entire board with invisible dominos. Florian now chooses $k$ cells. All dominos covering at least one of these cells then turn visible. Determine the minimal value of $k$ such that Florian has a strategy to always deduce the entire tiling.",
    "options": [],
    "answer": "See solution",
    "solution": "The minimal value of $k$ with this property is $n^2$.\n\nWe first show that in order for Florian to be able to deduce the entire tiling, we must have $k \\geq n^2$. If Julia picks an independent tiling on each of $n^2$ disjoint $2 \\times 2$ regions, then Florian needs to reveal at least one domino from each region to deduce the entire tiling. Hence $k \\geq n^2$.\n\nNow color the squares of the board with 4 different colors, such that the coloring is periodic with period 2 squares in both horizontal and vertical directions. We show that if Florian reveals the dominos covering the $n^2$ squares of one color class, then there is at most one tiling of the board that contains this arrangement of revealed dominos.\n\nLet red be one of the 4 colors and assume that we have two distinct tilings $A$ and $B$ of the board that agree on all the dominos covering a red square. We call a square of the board *augmented* if it is covered in a different way by $A$ and $B$. Given an augmented square $s$, let $a(s)$ and $b(s)$ be the two squares covered by the same domino in the tiling $A$ and $B$, respectively. By definition, we have $a(s) \\neq b(s)$ and $a(s)$ and $b(s)$ must both be augmented as well. Repeating this argument, we find distinct augmented squares $s_1, s_2, s_3, \\dots, s_m$ with\n\n$$\n(a(s_k), b(s_k)) = (s_{k+1}, s_{k-1}),\n$$\n\nwhere indices are taken modulo $m$. Hence, there is a closed path $P$ of orthogonally neighboring squares that does not contain a red square and that is tiled by the restriction of both tilings $A$ and $B$.\n\nNote, however, that any orthogonal, closed path that avoids red squares must enclose an orthogonally connected interior of odd size. This is easily shown by induction on the number of red squares enclosed: Consider the top-most, left-most red square $x$ in the interior of the closed path. Of its four orthogonal neighbors, either two, three, or all four are in the path. In the last case the interior is a single square; in the others, we can locally alter the path to exclude $x$ from the interior while reducing the interior by an even number of squares.\n\nNow, since the tilings $A$ and $B$ tile the path $P$, they must also tile the odd-sized region enclosed by $P$. This is impossible since every domino covers exactly 2 squares, contradicting our initial assumption. We conclude that if two tilings agree on the red squares, they must in fact be the same tiling. Hence, if Florian knows how the red squares are tiled, there is a unique way to complete this tiling to the entire board, which he can deduce by searching through all the finitely many tilings of the board.",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 12724,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $a_{ij}$, $i = 1, 2, \\dots, m$ and $j = 1, 2, \\dots, n$, be positive real numbers. Prove that\n\n$$\n\\sum_{i=1}^{m} \\left( \\sum_{j=1}^{n} \\frac{1}{a_{ij}} \\right)^{-1} \\leq \\left( \\sum_{j=1}^{n} \\left( \\sum_{i=1}^{m} a_{ij} \\right)^{-1} \\right)^{-1}.\n$$\n\nWhen does equality hold?",
    "options": [],
    "answer": "See solution",
    "solution": "We will use the following lemma:\n\n**Lemma.** If $a_1, a_2, \\dots, a_n, b_1, b_2, \\dots, b_n$ are positive real numbers, then\n\n$$\n\\frac{1}{\\sum_{j=1}^{n} \\frac{1}{a_j}} + \\frac{1}{\\sum_{j=1}^{n} \\frac{1}{b_j}} \\leq \\frac{1}{\\sum_{j=1}^{n} \\frac{1}{a_j + b_j}}.\n$$\n\nEquality holds when $\\frac{a_1}{b_1} = \\frac{a_2}{b_2} = \\dots = \\frac{a_n}{b_n}$.\n\n*Proof.* Set $x_j = \\frac{1}{a_j}$ and $y_j = \\frac{1}{b_j}$ for each $j = 1, 2, \\dots, n$. Then we have to prove that\n\n$$\n\\frac{1}{\\sum_{j=1}^{n} x_j} + \\frac{1}{\\sum_{j=1}^{n} y_j} \\leq \\frac{1}{\\sum_{j=1}^{n} \\frac{x_j y_j}{x_j + y_j}}\n$$\n\nor equivalently,\n\n$$\n\\sum_{j=1}^{n} \\frac{x_j y_j}{x_j + y_j} \\leq \\frac{\\left(\\sum_{j=1}^{n} x_j\\right) \\left(\\sum_{j=1}^{n} y_j\\right)}{\\sum_{j=1}^{n} x_j + \\sum_{j=1}^{n} y_j}.\n$$\n\nSubtract $\\sum_{j=1}^{n} x_j$ from both sides, and we have to prove\n\n$$\n\\sum_{j=1}^{n} \\left( x_j - \\frac{x_j y_j}{x_j + y_j} \\right) \\geq \\sum_{j=1}^{n} x_j - \\frac{\\left( \\sum_{j=1}^{n} x_j \\right) \\left( \\sum_{j=1}^{n} y_j \\right)}{\\sum_{j=1}^{n} x_j + \\sum_{j=1}^{n} y_j}\n$$\n\nor\n\n$$\n\\sum_{j=1}^{n} \\left( \\frac{x_j^2}{x_j + y_j} \\right) \\geq \\frac{\\left( \\sum_{j=1}^{n} x_j \\right)^2}{\\sum_{j=1}^{n} x_j + \\sum_{j=1}^{n} y_j}.\n$$\n\nThe last inequality follows from the Cauchy-Schwarz inequality, so the lemma is proved.\n\nNow, by repeatedly applying the lemma, we obtain the desired inequality. For example, if $a_1, a_2, \\dots, a_n, b_1, b_2, \\dots, b_n, c_1, c_2, \\dots, c_n$ are positive reals, then by applying the lemma twice:\n\n$$\n\\frac{1}{\\sum_{j=1}^{n} \\frac{1}{a_j}} + \\frac{1}{\\sum_{j=1}^{n} \\frac{1}{b_j}} + \\frac{1}{\\sum_{j=1}^{n} \\frac{1}{c_j}} \\leq \\frac{1}{\\sum_{j=1}^{n} \\frac{1}{a_j + b_j}} + \\frac{1}{\\sum_{j=1}^{n} \\frac{1}{c_j}} \\leq \\frac{1}{\\sum_{j=1}^{n} \\frac{1}{a_j + b_j + c_j}}.\n$$\n\nBy induction, we can prove that\n\n$$\n\\sum_{i=1}^{m} \\left( \\sum_{j=1}^{n} \\frac{1}{a_{ij}} \\right)^{-1} = \\sum_{i=1}^{m} \\frac{1}{\\sum_{j=1}^{n} \\frac{1}{a_{ij}}} \\leq \\frac{1}{\\sum_{j=1}^{n} \\frac{1}{\\sum_{i=1}^{m} a_{ij}}} = \\left( \\sum_{j=1}^{n} \\left( \\sum_{i=1}^{m} a_{ij} \\right)^{-1} \\right)^{-1},\n$$\n\nwhich is the desired result.\n\nEquality holds if and only if\n\n$$\n\\frac{a_{i1}}{a_{11}} = \\frac{a_{i2}}{a_{12}} = \\dots = \\frac{a_{in}}{a_{1n}}\n$$\n\nfor all $i = 1, 2, \\dots, m$. $\\square$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 12725,
    "subject": "Mathematics (Olympiad)",
    "question": "Sequence $\\left(x_1, x_2, \\dots\\right)$ is defined as\n\n$$\nx_1 = 20, \\quad x_2 = 12, \\quad x_{n+2} = x_n + x_{n+1} + 2\\sqrt{x_n x_{n+1} + 121},\n$$\nfor $n \\ge 1$.\n\n1) Compute $x_{10}$.\n\n2) Determine with justification if every term in the sequence is an integer.",
    "options": [],
    "answer": "See solution",
    "solution": "It is clear that $x_3 = 20 + 12 + 2 \\cdot 19 = 70$.\n\nWe note that\n\n$$\n\\begin{aligned}\nx_{n+3} &= x_{n+1} + x_{n+2} + 2\\sqrt{x_{n+1}x_{n+2} + 121} \\\\\n&= x_{n+1} + x_{n+2} + 2\\sqrt{x_{n+1}(x_n + x_{n+1} + 2\\sqrt{x_n x_{n+1} + 121}) + 121} \\\\\n&= x_{n+1} + x_{n+2} + 2\\sqrt{x_{n+1}^2 + 2x_{n+1}\\sqrt{x_n x_{n+1} + 121} + x_n x_{n+1} + 121} \\\\\n&= x_{n+1} + x_{n+2} + 2(x_{n+1} + \\sqrt{x_n x_{n+1} + 121}) \\\\\n&= 3x_{n+1} + x_{n+2} + 2\\sqrt{x_n x_{n+1} + 121} \\\\\n&= 3x_{n+1} + x_{n+2} + x_{n+2} - x_n - x_{n+1} \\\\\n&= 2x_{n+2} + 2x_{n+1} - x_n.\n\\end{aligned}\n$$\n\nTherefore, $(x_1, x_2, \\dots)$ is an integer sequence and it is straightforward to find $x_4 = 144$, $x_5 = 416$, $x_6 = 1050$, $x_7 = 2788$, $x_8 = 7260$, $x_9 = 19046$, and $x_{10} = 49824$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 12726,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $\\Gamma_A$ be the circle centered at $A$ and passing through points $C$ and $L$. Let $\\Gamma_B$ be the circle centered at $B$ and passing through points $C$ and $K$. Let $L_1$ be the second intersection point of line $BL$ with $\\Gamma_A$. Similarly, let $K_1$ be the second intersection point of line $AK$ with $\\Gamma_B$. Let $C_1$ be the second intersection point of $\\Gamma_A$ with $\\Gamma_B$.\n\n![](images/Australian_Scene_2012_-_AMT_Publishing_-_273p_p192_data_3a3e893813.png)\n\nFrom $AC \\perp BC$, we know that $\\Gamma_A$ is tangent to $BC$ at $C$. Considering the power of $B$ with respect to $\\Gamma_A$ yields $BC^2 = BL \\cdot BL_1$.\n\nHowever, since $BC = BK$, we have $BK^2 = BL \\cdot BL_1$. Thus $\\Gamma$ is tangent to $BK$ at point $K$. Similarly, $\\Gamma$ is tangent to $AL$ at point $L$.\n\nShow that $MK = ML$ where $M$ is a point from which tangents $MK$ and $ML$ are drawn to $\\Gamma$.",
    "options": [],
    "answer": "See solution",
    "solution": "Since $X$ lies on $CC_1$, by power of a point first in $\\Gamma_A$ and then in $\\Gamma_B$, we have $XL \\cdot XL_1 = XC \\cdot XC_1 = XK \\cdot XK_1$. Thus points $L_1, K, L, K_1$ are cyclic. Let $\\Gamma$ be the circle containing points $L_1, K, L, K_1$.\n\nFrom $AC \\perp BC$, we know that $\\Gamma_A$ is tangent to $BC$ at $C$. Considering the power of $B$ with respect to $\\Gamma_A$ yields $BC^2 = BL \\cdot BL_1$.\n\nHowever, since $BC = BK$, we have $BK^2 = BL \\cdot BL_1$. Thus $\\Gamma$ is tangent to $BK$ at point $K$. Similarly, $\\Gamma$ is tangent to $AL$ at point $L$.\n\nHence $MK$ and $ML$ are the two tangents from $M$ to $\\Gamma$ and therefore, $MK = ML$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 12727,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $m, n$ be positive integers with $m \\ge n$, and let $S$ be the set of all ordered $n$-tuples $(a_1, a_2, \\dots, a_n)$ of positive integers such that $a_1 + a_2 + \\dots + a_n = m$. Show that\n$$\n\\sum_{S} 1^{a_1} 2^{a_2} \\cdots n^{a_n} = \\binom{n}{n} n^m - \\binom{n}{n-1} (n-1)^m + \\cdots + (-1)^{n-2} \\binom{n}{2} 2^m + \\cdots + (-1)^{n-1} \\binom{n}{1}.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "Let $m = k + n$, and let $T$ be the set of all $n$-term sequences of nonnegative integers $(b_1, b_2, \\dots, b_n)$ such that $b_1 + b_2 + \\dots + b_n = k$. It suffices to show that\n\n$$\nn! \\sum_{T} 1^{b_1} 2^{b_2} \\cdots n^{b_n} = \\binom{n}{n} n^{k+n} - \\binom{n}{n-1} (n-1)^{k+n} + \\cdots + (-1)^{n-2} \\binom{n}{2} 2^{k+n} + \\cdots + (-1)^{n-1} \\binom{n}{1}.\n$$\n\nWe claim that both sides of the desired equation count the number of ways to color $k+n$ objects with $n$ colors such that each color is used at least once. For $1 \\le t \\le n$, the number of ways to color $k+n$ objects with $t$ colors is $t^{k+n}$, so the principle of inclusion-exclusion shows exactly that the right hand side counts these colorings.\n\nIt suffices to show that the left hand side also counts these colorings. Label the objects $1, 2, \\dots, k+n$ in some order, and, for $1 \\le i \\le n$, let $c_i$ be the smallest object of color $i$. Take $b_i = c_{i+1} - c_i$, where we let $c_{n+1} = n+k+1$. Then, notice that any such coloring is specified uniquely by the following data:\n\n(a) the order in which the colors first appear,\n\n(b) the number of objects between $c_i$ and $c_{i+1}$ for each $i$, where $c_{n+1} = n+k+1$,\n\n(c) the colors of the objects between $c_i$ and $c_{i+1}$.\n\nWe now count how many ways these data can be chosen. There are $n!$ choices for datum (a), and it is independent of (b) and (c). Datum (b) is specified uniquely by a choice of $(b_1, b_2, \\dots, b_n)$ such that $b_1 + b_2 + \\dots + b_n = k$, that is, an element of $T$. Given such a choice of $(b_1, b_2, \\dots, b_n)$, there are $1^{b_1} 2^{b_2} \\dots n^{b_n}$ ways to color the intermediate objects, as each of the $b_i$ between $c_i$ and $c_{i+1}$ admits $i$ choices of color since only $i$ colors appear before $c_{i+1}$. Summing over all choices of (a), (b), and (c), we see that the number of colorings of this type is\n\n$$\nn! \\sum_{T} 1^{b_1} 2^{b_2} \\dots n^{b_n},\n$$\n\nwhich matches the left hand side of the desired equation, completing the proof.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 12728,
    "subject": "Mathematics (Olympiad)",
    "question": "A total of $2^n$ coins are distributed among several children. If one of the children has at least half of the coins, the coins are redistributed: coins are transferred from that child to each of the other children so that each of them gets as many coins as they had. If one child possesses all the coins, no redistribution is possible. What is the greatest number of consecutive redistributions?\n\nFor example, if 32 coins are distributed among 6 children as follows: 17, 2, 9, 1, 2, 1, then after one redistribution the children will have: 2, 4, 18, 2, 4, 2 coins, respectively; in this example, that number is 2.\n\nExplain your answer!",
    "options": [],
    "answer": "See solution",
    "solution": "At most $n$ consecutive redistributions are possible.\n\n**Example:**\nLet $2^n$ coins be distributed among 3 children as follows: $1$, $2^{n-1} + 2^{n-2} + \\dots + 2$, $1$.\n\nThe successive redistributions (a total of $n$) will be:\n\n$$\n2^1,\\ 2^{n-1} + 2^{n-2} + \\dots + 2^2,\\ 2^1\n$$\n$$\n2^2,\\ 2^{n-1} + 2^{n-2} + \\dots + 2^3,\\ 2^2\n$$\n$$\n2^{n-2},\\ 2^{n-1},\\ 2^{n-2}\n$$\n$$\n2^{n-1},\\ 0,\\ 2^{n-1}\n$$\n$$\n0,\\ 0,\\ 2^n\n$$\n\n**Proof of maximality:**\nAfter each redistribution, the number of coins each child has becomes divisible by a higher power of 2: after the first, by 2; after the second, by 4; and so on. After the $n$-th redistribution, each child has a number divisible by $2^n$. Since the total is $2^n$, the only possible distribution is $2^n, 0, 0, \\dots, 0$, so no further redistribution is possible. Thus, $n$ is the maximal number.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 12729,
    "subject": "Mathematics (Olympiad)",
    "question": "Suppose there are 207 balls, each colored either red or white, with exactly 100 red and 107 white balls. Zehra can ask for the color relationship between pairs of balls by writing pairs $(i, j)$ on the board, and Asli answers whether the two balls have the same color or not. What is the minimum number $N$ of pairs Zehra must choose so that, regardless of Asli's answers, Zehra can guarantee to determine the color of every ball?",
    "options": [],
    "answer": "See solution",
    "solution": "Let us show that Zehra can guarantee to determine the colors of all balls for $N = 205$.\n\nZehra writes the pairs $(1,2), (1,3), \\dots, (1,206)$ on the board. Assign a number $i$ to the ball in box $i$. Suppose that among balls numbered $2,3,\\dots,206$, the number of balls same-colored with $1$ is $a$ and the number of differently colored balls is $b = 205 - a$. Then the colors of all balls can be determined:\n\nOne of $a$ and $b$ is at least $103$. If $a \\ge 103$, then ball $1$ is white. $a+1$ is either $106$ or $107$. In the first case, ball $207$ is white; in the second case, ball $207$ is red. If $b \\ge 103$, then ball $1$ is red. $b$ is either $106$ or $107$. In the first case, ball $207$ is white; in the second case, ball $207$ is red.\n\nNow, we show that if $N \\le 204$, Zehra cannot guarantee to determine the colors of all balls. Suppose $N$ pairs $(i, j)$ are already chosen. Define a graph $G$ with $207$ vertices representing balls, where vertices $i$ and $j$ are connected if $(i, j)$ is on the board. Since $N \\le 204$, $G$ has at least $3$ connected components. Asli divides these into $3$ non-empty groups $C_1, C_2, C_3$ with sizes $|C_1| \\ge |C_2| \\ge |C_3|$. Then $|C_2| + |C_3| \\le 138$. At least one of $C_2$, $C_3$, or $C_2 \\cup C_3$ contains an even number of vertices. Asli can divide this group into two parts $A_1$ and $A_2$ with $|A_1| = |A_2| < 100$. Since Zehra must guarantee to determine ball colors, Asli can color all balls in $A_1$ red and $A_2$ white, or vice versa, and color the rest to ensure $100$ red and $107$ white balls in total. In both cases, Asli's answers are the same, so Zehra cannot determine the colors of this group containing an even number of boxes.\n\nDone.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 12730,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $x$, $y$, and $z$ be positive real numbers. Show that\n$$\nx^2 + x y^2 + x y z^2 \\ge 4 x y z - 4.\n$$",
    "options": [],
    "answer": "See 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$$\nx^2 + x y^2 + x y z^2 \\ge (4x - 4) + x(4y - 4) + x y (4z - 4) = 4 x y z - 4.\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 12731,
    "subject": "Mathematics (Olympiad)",
    "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?\n\n![](path/to/file.png)",
    "options": [],
    "answer": "See solution",
    "solution": "We shall prove that there are no consecutive kooky primes.\n\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\n$$\np_1 + \\dots + p_{n-1} = a p_n \\quad \\text{and} \\quad p_1 + \\dots + p_n = b p_{n+1}.\n$$\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 $p_n p_{n+1}$.\n\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\n$$\n0 < p_1 + p_2 + \\dots + p_n \\leq n p_n < p_n p_{n+1},\n$$\n\nso it can't be a multiple of $p_n p_{n+1}$, which yields the desired contradiction.\n\n**Remark.** All the kooky primes known to date (OEIS A007506) are\n\n5, 71, 369119, 415074643, 55691042365834801.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 12732,
    "subject": "Mathematics (Olympiad)",
    "question": "Asha, Bree, and Cala are three robots programmed to run athletic track races. When Asha runs a 400 m race, she catches Bree at the finish line if Bree starts 20 m ahead of Asha. Asha catches Cala at the finish line of a 1500 m race if Cala has a 246 m head start. How many metres must Cala start ahead of Bree in an 800 m race if they are to finish at the same time? Assume each of the three robots runs at a constant speed.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $a$, $b$, $c$ denote the running speeds of Asha, Bree, and Cala, respectively.\n\n**Method 1**\n\nAsha runs 400 m in the same time as Bree runs $400 - 20 = 380$ m.\n\nSo $\\frac{400}{a} = \\frac{380}{b}$. Hence $\\frac{a}{b} = \\frac{20}{19}$.\n\nAsha runs 1500 m in the same time as Cala runs $1500 - 246 = 1254$ m.\n\nSo $\\frac{1500}{a} = \\frac{1254}{c}$. Hence $\\frac{c}{a} = \\frac{209}{250}$.\n\nSuppose Bree gives Cala a $h$ metre head start in the 800 m race and they finish at the same time. Then $\\frac{800}{b} = \\frac{800 - h}{c}$. Therefore\n$$\n\\frac{800-h}{800} = \\frac{c}{b} = \\left(\\frac{c}{a}\\right)\\left(\\frac{a}{b}\\right) = \\frac{209}{250} \\times \\frac{20}{19} = \\frac{22}{25}.\n$$\nSo $800-h = 800 \\times \\frac{22}{25} = 704$ and $h = 96$.\n\n**Method 2**\n\nSuppose Bree gives Cala a $h$ metre head start in the 800 m race and they finish at the same time.\n\nIn the time $T_1$ that Asha runs 400 m, Bree runs $400 - 20 = 380$ m. In the time $T_2$ that Bree runs 800 m, Cala runs $800 - h$ m. In the time $T_3$ that Asha runs 1500 m, Cala runs $1500 - 246 = 1254$ m.\n\nSo Asha's speed is $\\frac{400}{T_1} = \\frac{1500}{T_3}$, Bree's is $\\frac{380}{T_1} = \\frac{800}{T_2}$ and Cala's is $\\frac{800-h}{T_2} = \\frac{1254}{T_3}$. Hence\n$$\n\\frac{800-h}{1254} = \\frac{T_2}{T_3} = \\frac{800T_1}{380} \\times \\frac{400}{1500T_1} = \\frac{80}{38} \\times \\frac{4}{15} = \\frac{32}{57}.\n$$\nSo $h = 800 - 1254 \\times \\frac{32}{57} = 800 - 704 = 96$.\n\n**Method 3**\n\nSuppose Asha, Bree, and Cala start side-by-side. After running 400 m, Asha would be 20 m ahead of Bree. Therefore, after running 1500 m, she would be $20 \\times \\frac{1500}{400} = 75$ m ahead of Bree.\n\nThus Bree would be $1500 - 75 = 1425$ m from the start and $246 - 75 = 171$ m ahead of Cala. Hence, at 800 m from the start, Bree would be $171 \\times \\frac{800}{1425} = 171 \\times \\frac{32}{57} = 96$ m ahead of Cala.\n\nSo Bree will catch Cala at the finish of an 800 m race if Bree gives Cala a head start of **96 m**.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 12733,
    "subject": "Mathematics (Olympiad)",
    "question": "Suppose that the excircle of $\\triangle ABC$ opposite the vertex $A$ touches the side $BC$ at the point $A_1$. Define the points $B_1$ on $CA$ and $C_1$ on $AB$ analogously by using the excircles opposite to $B$ and $C$, respectively. Suppose that the circumcentre of $\\triangle A_1B_1C_1$ lies on the circumcircle of\n![](images/Mathematical_Olympiad_in_China_2011-2014_p349_data_a7aa4f65aa.png \"Fig. 3. 1\")\n$\\triangle ABC$. Prove that $\\triangle ABC$ is right-angled.\n\n(The excircle of $\\triangle ABC$ opposite the vertex $A$ is the circle that touches the line segment $BC$, the ray $AB$ beyond $B$, and the ray $AC$ beyond $C$. The excircles opposite $B$ and $C$ are similarly defined.)",
    "options": [],
    "answer": "See solution",
    "solution": "Denote the circumcircles of $\\triangle ABC$ and $\\triangle A_1B_1C_1$ by $\\Omega$ and $\\Gamma$, respectively. Let $A_0$ be the midpoint of arc $\\widehat{BC}$ on $\\Omega$ containing point $A$. Points $B_0$ and $C_0$ are defined analogously. Let $Q$ be the centre of circle $\\Gamma$, then $Q$ is on $\\Omega$ by the hypothesis of the problem. First, we give the following lemma.\n\n*Lemma*. $A_0B_1 = A_0C_1$. Points $A$, $A_0$, $B_1$ and $C_1$ are concyclic.\n\nIf points $A_0$ and $A$ coincide, then $\\triangle ABC$ is isosceles, thus $AB_1 = AC_1$. Otherwise, $A_0B = A_0C$ by the definition of $A_0$. It is evident that\n\n$$\nBC_1 = CB_1 \\left( = \\frac{1}{2} (b + c - a) \\right),\n$$\n\nand\n\n$$\n\\angle C_1BA_0 = \\angle ABA_0 = \\angle ACA_0 = \\angle B_1CA_0.\n$$\n\nThus, $\\triangle A_0BC_1 \\cong \\triangle A_0CB_1$. $\\textcircled{1}$\n\nSo, we have $A_0B_1 = A_0C_1$.\n\nAlso, by $\\textcircled{1}$, we have $\\angle A_0C_1B = \\angle A_0B_1C$, hence $\\angle A_0C_1A = \\angle A_0B_1A$. Thus, points $A$, $A_0$, $B_1$ and $C_1$ are concyclic.\n\nObviously, points $A_1$, $B_1$ and $C_1$ are on a semi-arc of $\\Gamma$, thus $\\triangle A_1B_1C_1$ is an obtuse-angled triangle. Without loss of generality, we may suppose that $\\angle A_1B_1C_1$ is obtuse, then points $Q$ and $B_1$ are on different sides of $A_1C_1$. Obviously, so",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 12734,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $k$ be a positive integer and $m$ an odd integer. Show that there exists a positive integer $n$ for which the number $n^n - m$ is divisible by $2^k$.",
    "options": [],
    "answer": "See solution",
    "solution": "We prove the assertion by induction on $k$.\n\n**Base case ($k=1$):** Let $n=1$. Then $n^n - m = 1 - m$. Since $m$ is odd, $1 - m$ is even, so divisible by $2$.\n\n**Inductive step:** Suppose the assertion holds for $k = t$, i.e., there exists $n_0$ such that $n_0^{n_0} \\equiv m \\pmod{2^t}$. Since $n_0^{n_0}$ is odd, $n_0$ is odd.\n\nIf $n_0^{n_0} \\equiv m \\pmod{2^{t+1}}$, then $n_0$ works for $k = t+1$.\n\nOtherwise, $n_0^{n_0} \\equiv m + 2^t \\pmod{2^{t+1}}$. Let $n = n_0 + 2^t$ (which is odd). By Euler's theorem, $n^2 \\equiv 1 \\pmod{2^{t+1}}$, so\n\n$$\nn^n = n^{n_0 + 2^t} \\equiv n^{n_0} \\cdot n^{2^t} \\equiv n^{n_0} \\pmod{2^{t+1}}.\n$$\n\nExpanding $n^{n_0} = (n_0 + 2^t)^{n_0}$ by the binomial theorem, all terms with $i \\ge 2$ are divisible by $2^{t+1}$, so\n\n$$\nn^{n_0} \\equiv n_0^{n_0} + n_0 C_1 \\cdot 2^t n_0^{n_0-1} \\pmod{2^{t+1}}.\n$$\n\nThis simplifies to\n\n$$\nn^{n_0} \\equiv (m + 2^t) + 2^t n_0^{n_0-1} n_0 \\pmod{2^{t+1}}.\n$$\n\nSince $n_0^{n_0-1} n_0 = n_0^{n_0}$, we have\n\n$$\nn^{n_0} \\equiv m + 2^t (n_0^{n_0} + 1) \\pmod{2^{t+1}}.\n$$\n\nSince $n_0^{n_0}$ is odd, $n_0^{n_0} + 1$ is even, so $2^t (n_0^{n_0} + 1)$ is divisible by $2^{t+1}$. Thus,\n\n$$\nn^n \\equiv m \\pmod{2^{t+1}}.\n$$\n\nThis completes the induction.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 12735,
    "subject": "Mathematics (Olympiad)",
    "question": "Миколка та Андрійко по черзі (Миколка починає першим) вписують натуральні числа $a_1, a_2, \\ldots, a_{2006}$ так, що $a_1 = 1$, $a_n \\leq a_{n+1} \\leq 3a_n$ для $1 \\leq n \\leq 2005$. Якщо числа $a_1 + a_2 + \\ldots + a_{2005}$ і $a_1 + a_2 + \\ldots + a_{2006}$ є взаємно простими, то переможцем є Андрійко, інакше — Миколка. Хто з гравців може забезпечити собі перемогу незалежно від дій суперника? Обґрунтуйте відповідь.",
    "options": [],
    "answer": "See solution",
    "solution": "Відповідь: Андрійко може забезпечити собі перемогу.\n\nДостатньо показати, що Андрійко може записати $a_{2006} = M - 1$, де $M = \\sum_{k=1}^{2005} a_k$. Очевидно, $M - 1 \\geq a_{2005}$. Доведемо, що Андрійко може забезпечити й виконання нерівності $a_{2006} = M - 1 \\leq 3a_{2005}$.\n\nНехай Миколка на своєму ході записує $a_{2k-1}$, тоді Андрійко відповідає $a_{2k} = 3a_{2k-1}$ для $1 \\leq k \\leq 1002$. Оскільки $a_{2k+1} \\geq a_{2k} = 3a_{2k-1}$, то, додавши нерівності $a_3 \\geq 3a_1$, $a_5 \\geq 3a_3$, \\ldots, $a_{2005} \\geq 3a_{2003}$, отримаємо $a_{2005} \\geq 1 + 2A$, де $A = a_1 + a_3 + \\ldots + a_{2003}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 12736,
    "subject": "Mathematics (Olympiad)",
    "question": "No landmine can fully contain the red arc $XY$.\n\n% IMAGE: ![](images/2025-SL-a_p21_data_c3984ff75d.png)\n\nFurthermore, if the distance $OA$ is still greater than $\\frac{1}{\\sqrt{2}}$, i.e., $\\frac{1}{\\sqrt{2}} < OA \\le 1$, consider the circles $(O, \\frac{1}{\\sqrt{2}})$ and $(A, 1)$. Show that no landmine can fully contain the red arc $PQ$.\n\n% IMAGE: ![](images/2025-SL-a_p21_data_780c8f6537.png)",
    "options": [],
    "answer": "See solution",
    "solution": "If a landmine were to exist, consider its center. Since it is closer to $X$ and $Y$ than $A$, it must be in the blue region bounded by the perpendicular bisectors of $XA$ and $YA$. Since it is also closer to $X$ and $Y$ than $O$, it must be in the green region bounded by the perpendicular bisectors of $XO$ and $YO$.\n\nNotice that the two regions are disjoint, so the center of the landmine cannot exist. $\\Box$\n\nHence, the rabbit can hop onto the arc. The maximum distance from the arc to $O$ is at $XO = YO = \\sqrt{d^2-1}$. Thus, the squared distance between the rabbit and $O$ decreases by at least $1$ every move, so the rabbit can eventually achieve a distance of at most $1$ from $O$.\n\nNow, if $OA > \\frac{1}{\\sqrt{2}}$, then $\\frac{1}{\\sqrt{2}} < OA \\le 1$. Consider the circles $(O, \\frac{1}{\\sqrt{2}})$ and $(A, 1)$. Since $OP^2 + OA^2 > 1 = AP^2$, $\\angle POA$ is acute, so $O$ and $A$ lie on opposite sides of the line $PQ$. Therefore, as before, the perpendicular bisectors form two disjoint regions, so no landmine fully contains the red arc $PQ$.\n\nTherefore, the rabbit can hop onto the red arc and will be of distance at most $\\frac{1}{\\sqrt{2}}$ from $O$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 12737,
    "subject": "Mathematics (Olympiad)",
    "question": "A quadrilateral $ABCD$ satisfies $AB = 5$, $BC = 7$, $CD = 6$. The diagonals $AC$ and $BD$ are perpendicular to each other. Find the length of $DA$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $P$ be the intersection point of $AC$ and $BD$. By the Pythagorean theorem:\n\n- $AB^2 = AP^2 + BP^2$\n- $BC^2 = BP^2 + CP^2$\n- $CD^2 = CP^2 + DP^2$\n- $DA^2 = DP^2 + AP^2$\n\nAdding the first and third equations and subtracting the second gives:\n\n$$DA^2 = AB^2 + CD^2 - BC^2 = 5^2 + 6^2 - 7^2 = 25 + 36 - 49 = 12$$\n\nSo $DA = 2\\sqrt{3}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 12738,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all triples $(p, q, r)$ of primes such that $$2018(p^2 + q^2) = r^2 + 1.$$",
    "options": [],
    "answer": "See solution",
    "solution": "Suppose both $p$ and $q$ are odd. Then $p^2 + q^2$ is even, so the left-hand side is divisible by 4. Squares of integers are congruent to 0 or 1 modulo 4, so the right-hand side is congruent to 1 or 2 modulo 4. This is a contradiction, so one of $p$ and $q$ must be 2; let $p = 2$ without loss of generality. Squares of integers are congruent to 0 or 1 modulo 3. Clearly $r > 3$ since the left-hand side is greater than 10. As $r$ is prime, $r$ is not divisible by 3, so $r^2 \\equiv 1 \\pmod{3}$, which means $r^2 + 1 \\equiv 2 \\pmod{3}$. Now $2018 \\equiv 2 \\pmod{3}$, so $p^2 + q^2 \\equiv 1 \\pmod{3}$, and since $p^2 = 4 \\equiv 1 \\pmod{3}$, we have $q^2 \\equiv 0 \\pmod{3}$. Thus $q$ is divisible by 3, i.e., $q = 3$. Therefore, the left-hand side is $2018 \\cdot 13$. Since $2018 \\cdot 13 \\equiv 4 \\pmod{5}$, we must have $r^2 \\equiv 3 \\pmod{5}$. But 3 is not a quadratic residue modulo 5. Thus, there are no such triples.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 12739,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $n \\ge 5$ be an integer. Consider $n$ squares with side lengths $1, 2, \\dots, n$, respectively. The squares are arranged in the plane with their sides parallel to the $x$ and $y$ axes. Suppose that no two squares touch, except possibly at their vertices. Show that it is possible to arrange these squares in a way such that every square touches exactly two other squares.",
    "options": [],
    "answer": "See solution",
    "solution": "Set aside the squares with side lengths $n-3, n-2, n-1$, and $n$ and suppose we can split the remaining squares into two sets $A$ and $B$ such that the sum of the side lengths of the squares in $A$ is 1 or 2 units larger than the sum of the side lengths of the squares in $B$.\n\nString the squares of each set $A, B$ along two parallel diagonals, one for each diagonal. Now use the four largest squares along two perpendicular diagonals to finish the construction: one will have side lengths $n$ and $n-3$, and the other, side lengths $n-1$ and $n-2$. If the sum of the side lengths of the squares in $A$ is 1 unit larger than the sum of the side lengths of the squares in $B$, attach the squares with side lengths $n-3$ and $n-1$ to the $A$-diagonal, and the other two squares to the $B$-diagonal. The resulting configuration, in which the $A$ and $B$-diagonals are represented by unit squares, and the side lengths $a_i$ of squares from $A$ and $b_j$ of squares from $B$ are indicated within each square, follows:\n\n![](images/2023_Australian_Scene_p86_data_76ddbac0dc.png)\n\nSince $(a_1 + a_2 + \\dots + a_k)\\sqrt{2} + \\frac{((n-3)+(n-2))\\sqrt{2}}{2} = (b_1 + b_2 + \\dots + b_\\ell + 2)\\sqrt{2} + \\frac{(n+(n-1))\\sqrt{2}}{2}$, this case is done.\n\nIf the sum of the side lengths of the squares in $A$ is 1 unit larger than the sum of the side lengths of the squares in $B$, attach the squares with side lengths $n-3$ and $n-2$ to the $A$-diagonal, and the other two squares to the $B$-diagonal. The resulting configuration follows:\n\n![](images/2023_Australian_Scene_p86_data_291cc08910.png)\n\nSince $(a_1 + a_2 + \\dots + a_k)\\sqrt{2} + \\frac{((n-3)+(n-1))\\sqrt{2}}{2} = (b_1 + b_2 + \\dots + b_\\ell + 1)\\sqrt{2} + \\frac{(n+(n-2))\\sqrt{2}}{2}$, this case is also done.\n\nIn both cases, the distance between the A-diagonal and the B-diagonal is $\\frac{((n-3)+n)\\sqrt{2}}{2} = \\frac{(2n-3)\\sqrt{2}}{2}$.\n\nSince $a_i, b_j \\le n-4$, $\\frac{(a_i+b_j)\\sqrt{2}}{2} < \\frac{(2n-4)\\sqrt{2}}{2} < \\frac{(2n-3)\\sqrt{2}}{2}$, and therefore the A- and B-diagonals do not overlap.\n\nFinally, we prove that it is possible to split the squares of side lengths 1 to $n-4$ into two sets $A$ and $B$ such that the sum of the side lengths of the squares in $A$ is 1 or 2 units larger than the sum of the side lengths in $B$. One can do that in several ways; we present two possibilities:\n\n* **Direct construction:** Split the numbers from 1 to $n-4$ into several sets of four consecutive numbers $\\{t, t+1, t+2, t+3\\}$, beginning with the largest numbers; put squares of side lengths $t$ and $t+3$ in $A$ and squares of side lengths $t+1$ and $t+2$ in $B$. Notice that $t + (t+3) = (t+1) + (t+2)$. In the end, at most four numbers remain.\n    - If only 1 remains, put the corresponding square in $A$, so the sum of the side lengths of the squares in $A$ is one unit larger than those in $B$;\n    - If 1 and 2 remain, put the square of side length 2 in $A$ and the square of side length 1 in $B$ (the difference is 1);\n    - If 1, 2, and 3 remain, put the squares of side lengths 1 and 3 in $A$, and the square of side length 2 in $B$ (the difference is 2);\n    - If 1, 2, 3, and 4 remain, put the squares of side lengths 2 and 4 in $A$, and the squares of side lengths 1 and 3 in $B$ (the difference is 2).\n* **Indirect construction:** Starting with $A$ and $B$ as empty sets, add the squares of side lengths $n-4, n-3, \\dots, 2$ to either $A$ or $B$ in that order such that at each stage the difference between the sum of the side lengths in $A$ and the sum of the side lengths of $B$ is minimized. By induction it is clear that after adding an integer $j$ to one of the sets, this difference is at most $j$. In particular, the difference is 0, 1 or 2 at the end. Finally adding the final 1 to one of the sets can ensure that the final difference is 1 or 2. If necessary, flip $A$ and $B$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 12740,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $P$ be an arbitrary cycle in a graph with length greater than $4$, such that no vertex is inside $P$. Show that the number of vertices of $P$ with no $90^\text{\\circ}$ turns (denoted $A$) is given by:\n\n$$\nA = 2B - 2C + 2,\n$$\n\nwhere $B$ is the number of squares inside $P$ with only two opposite sides in $P$, and $C$ is the number of squares inside $P$ with no edge in $P$.",
    "options": [],
    "answer": "See solution",
    "solution": "We use induction on $k$, the number of squares inside $P$ (with $k > 1$).\n\nFor $k = 2$, the assertion is trivial. Suppose $k > 2$. Consider a new graph whose nodes are the centers of the squares inside $P$, and two nodes are adjacent if and only if their corresponding squares share a common edge (not necessarily in $P$). This graph is connected and acyclic (otherwise, a vertex would be inside $P$), so it is a tree and has a leaf. Thus, there is a square $Q$ inside $P$ with three sides in $P$.\n\nConsider the induction hypothesis for the cycle obtained by deleting these three edges and adding the remaining side of $Q$:\n\n$$\nA' = 2B' - 2C' + 2.\n$$\n\nLet $xyzt$ be the path in $P$ with the vertices of $Q$. Consider the following cases (see images):\n\n* There are $90^\text{\\circ}$ turns at both $x$ and $t$. Then $A = A' - 2$. Also, $B = B' - 1$, $C = C'$, or $B = B'$, $C = C' + 1$ depending on $Q$'s adjacent square inside $P$.\n\n![](images/2012_p32_data_895af738d7.png)\n\n* There is a $90^\text{\\circ}$ turn at only one of $x, t$. Then $A = A'$, $B = B'$, $C = C'$.\n\n![](images/2012_p32_data_a918727cde.png)\n\n* There is no $90^\text{\\circ}$ turn at $x$ and $t$. Then $A = A' + 2$, $B = B' + 1$, $C = C'$.\n\n![](images/2012_p32_data_50a09920d1.png)\n\nThe induction claim follows in each case. $\\Box$\n\nUsing this lemma, we get\n\n$$\nA = 2B' - 2C' + 2, \\quad (3)\n$$\n\nwhere $B', C'$ are the number of squares inside $P$ with the same properties as $B, C$. Let $R$ be the inner $(m-2) \\times (n-2)$ rectangle and draw its boundary together with the edges of $P$. These edges partition the square into regions. The boundary of each region (other than the inside of $P$ and the outermost region) is a cycle containing one edge in the boundary of $R$ and a path in $P$. Use the lemma for each region to get\n\n$$\nA_i = 2B_i - 2C_i + 2, \\quad 1 \\leq i \\leq t, \\quad (4)\n$$\n\nwhere $t$ is the number of regions. It is easily seen that\n\n$$\n\\begin{align*}\nB &= B' + \\sum_{i} B_i + 1, \\\\\nC &= C' + \\sum_{i} C_i + t + 4, \\\\\nA &= \\frac{1}{2}(A' + \\sum_{i} A_i + A''),\n\\end{align*}\n$$\n\nwhere $A''$ is the number of vertices without $90^\text{\\circ}$ turns on the boundary of $R$. The assertion follows by substituting equations (3) and (4) into $A - B + C$. $\\Box$",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 12741,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $n$ be a positive integer and $A$ a set containing $8n + 1$ positive integers, each co-prime with $6$ and less than $30n$. Prove that there exist two distinct numbers $a, b \\in A$ such that $a$ divides $b$.",
    "options": [],
    "answer": "See solution",
    "solution": "In the set $\\{1, 2, \\dots, 30n\\}$, there are $8n$ numbers co-prime with $30$.\n\nIndeed, let us denote:\n- $B_1 = \\{1 \\le x \\le 30n \\mid 2 \\text{ divides } x\\}$\n- $B_2 = \\{1 \\le x \\le 30n \\mid 3 \\text{ divides } x\\}$\n- $B_3 = \\{1 \\le x \\le 30n \\mid 5 \\text{ divides } x\\}$\n\nWe have $B = \\{1, 2, \\dots, 30n\\} \\setminus (B_1 \\cup B_2 \\cup B_3)$, and since\n$$\n\\text{card}(B_1 \\cup B_2 \\cup B_3) = 15n + 10n + 6n - 5n - 3n - 2n + n = 22n,\n$$\nwe get\n$$\n\\text{card } B = 30n - 22n = 8n.\n$$\n(Alternatively, there are $n$ numbers for each of the remainders $1, 7, 11, 13, 17, 19, 23, 29$ modulo $30$.)\n\nDenote $x_1, x_2, \\dots, x_{8n}$ as the elements of $B$. Partition the set $A$ into $8n$ classes $C_i = \\{x_i \\cdot 5^k \\in A\\}$. Since $A$ has $8n + 1$ numbers and there are $8n$ classes, by the Pigeonhole Principle, one class $C_i$ must contain at least two elements $a = x_i \\cdot 5^s$ and $b = x_i \\cdot 5^t$ with $s \\ne t$. Then $a \\mid b$ or $b \\mid a$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 12742,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $n$ be a positive integer. Find the smallest possible value of $m$ such that there exists a sequence of positive integers $a_1, a_2, \\dots, a_n$ with $a_1 < a_2 < \\cdots < a_n = m$ and for each $k = 1, 2, \\dots, n-1$, the number $\\frac{a_k^2 + a_{k+1}^2}{2}$ is a perfect square.",
    "options": [],
    "answer": "See solution",
    "solution": "We will show that the smallest possible value for $m$ is $2n^2 - 1$.\n\nIf we let $a_k = 2k^2 - 1$ for $k = 1, 2, \\dots, n$, then\n$$\n\\frac{a_k^2 + a_{k+1}^2}{2} = \\frac{(2k^2 - 1)^2 + (2(k+1)^2 - 1)^2}{2} = (2k^2 + 2k + 1)^2\n$$\nare all perfect squares for each $k = 1, 2, \\dots, n-1$, and $m = a_n = 2n^2 - 1$ in this case.\n\nNow, we show that if there exists a sequence $a_1, a_2, \\dots, a_n$ satisfying the two conditions, then $m \\ge 2n^2 - 1$ must hold. It suffices to show $a_k \\ge 2k^2 - 1$ for each $k = 1, 2, \\dots, n$.\n\n**Lemma.** Let $k$ be a positive integer. For any pair of positive integers $x, y$ with $2k^2 - 1 \\le x < y < 2(k+1)^2 - 1$, the number $\\frac{x^2 + y^2}{2}$ is not a perfect square.\n\n*Proof.* If $x$ and $y$ have different parity, $\\frac{x^2 + y^2}{2}$ is not an integer. If they have the same parity:\n$$\n\\frac{x^2 + y^2}{2} - \\left(\\frac{x+y}{2}\\right)^2 = \\left(\\frac{y-x}{2}\\right)^2 > 0\n$$\nAlso,\n$$\ny - x \\le (2(k+1)^2 - 2) - (2k^2 - 1) = 4k + 1\n$$\nSince $y-x$ is even, $y-x \\le 4k$. Thus,\n$$\n\\left(\\frac{x+y}{2}+1\\right)^2 - \\frac{x^2+y^2}{2} = x+y+1 - \\left(\\frac{y-x}{2}\\right)^2 \\ge (2k^2-1) + (2k^2+1) + 1 - (2k)^2 = 1 > 0\n$$\nSo,\n$$\n\\left(\\frac{x+y}{2}\\right)^2 < \\frac{x^2+y^2}{2} < \\left(\\frac{x+y}{2}+1\\right)^2\n$$\nThus, $\\frac{x^2+y^2}{2}$ lies strictly between two consecutive perfect squares, so it cannot be a perfect square.\n\nNow, we prove $a_k \\ge 2k^2 - 1$ by induction. The base case $k=1$ is clear. Suppose $a_\\ell \\ge 2\\ell^2 - 1$. If $a_{\\ell+1} < 2(\\ell+1)^2 - 1$, then taking $x = a_\\ell$ and $y = a_{\\ell+1}$ in the Lemma gives a contradiction, since $\\frac{a_\\ell^2 + a_{\\ell+1}^2}{2}$ would not be a perfect square. Thus, $a_{\\ell+1} \\ge 2(\\ell+1)^2 - 1$, completing the induction.\n\nTherefore, $a_k \\ge 2k^2 - 1$ for all $k$, and in particular, $m = a_n \\ge 2n^2 - 1$. This shows that the minimum value of $m$ is $2n^2 - 1$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 12743,
    "subject": "Mathematics (Olympiad)",
    "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 centre on the circumcircle of $ABC$. Prove that there are exactly two friendly sides of $ABD$ if and only if $BD = CD$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $E$, $F$, and $G$ be the centres of excircles touching $BD$, $AD$, and $AB$ respectively, and let $\\Gamma$ be the circumcircle of $ABC$. To prove the assertion, we will show that $F$ and $G$ cannot both lie on $\\Gamma$ and that\n\n$$\nBD = CD \\iff E \\in \\Gamma \\iff F \\in \\Gamma.\n$$\n\n![](images/Baltic_Way_SHL_2009-11_13-16_p121_data_be2d911094.png)\n\n1. As $AF$ and $AG$ are bisectors of the two complementary angles of $BAD$, the point $A$ lies on the segment $FG$. Thus, only one of the two rays $AF$ and $AG$ can cut the circle $\\Gamma$ again, and therefore only one of $F$ and $G$ can lie on $\\Gamma$.\n\n2. To show that $BD = CD$ iff $E \\in \\Gamma$, we note that $E$ lies on $\\Gamma$ iff $\\angle EAC = \\angle EBC$. Since\n\n$$\n\\begin{aligned}\n\\angle EAC &= \\frac{1}{2}\\angle BAD = \\frac{1}{2}(180^\\circ - \\angle ADB - \\angle DBA) \\\\\n&= \\frac{1}{2}(180^\\circ - (180^\\circ - 2\\angle BDE) - (180^\\circ - 2\\angle EBD)) \\\\\n&= \\angle BDE + \\angle EBD - 90^\\circ = 90^\\circ - \\angle DEB,\n\\end{aligned}\n$$\n\nwe see that $E \\in \\Gamma$ is equivalent to $\\angle EBC = 90^\\circ - \\angle DEB$, i.e., $BC$ and $DE$ being perpendicular. Since $DE$ is the bisector of $BDC$, this occurs iff $BD = CD$.\n\n3. It remains to show that $BD = CD$ iff $F \\in \\Gamma$. The point $F$ lies on $\\Gamma$ iff $\\angle AFB = \\angle DCB$. Using the fact that\n\n$$\n\\angle BAF = \\angle BAD + \\frac{1}{2}(180^\\circ - \\angle BAD) = \\frac{1}{2}(180^\\circ + \\angle BAD),\n$$\n\nwe compute\n\n$$\n\\begin{aligned}\n\\angle AFB &= 180^\\circ - \\angle FBA - \\angle BAF \\\\\n&= \\frac{1}{2}(180^\\circ - \\angle DBA - \\angle BAD) \\\\\n&= \\frac{1}{2}\\angle ADB = \\frac{1}{2}(\\angle DCB + \\angle CBD).\n\\end{aligned}\n$$\n\nThus $\\angle AFB = \\angle DCB$ is equivalent to $\\angle DCB = \\angle CBD$, or $BD = CD$.\n\n_Remark._ The claim holds in the case of right or obtuse triangles, too. The problem with the above proof is that if $\\angle ABC > 90^\\circ$, then the point $E$ may fall inside triangle $ABC$, whence the equality $\\angle CAE = \\angle CBE$ is no longer equivalent to $E \\in \\Gamma$. Nevertheless, one can show that if $BD = CD$, then $E$ must lie outside triangle $ABC$, extending the validity of the claim to the obtuse case, too.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 12744,
    "subject": "Mathematics (Olympiad)",
    "question": "Find the greatest positive integer $k$ such that the following inequality holds for all positive real numbers $a, b, c$ satisfying $abc = 1$:\n\n$$\n\\frac{1}{a} + \\frac{1}{b} + \\frac{1}{c} + \\frac{k}{a+b+c+1} \\geq 3 + \\frac{k}{4}\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "Let $a = b = x$ and $c = \\frac{1}{x^2}$ for $0 < x \\neq 1$. Substituting into the inequality gives:\n\n$$\n\\frac{1}{x} + \\frac{1}{x} + x^2 + \\frac{k}{2x + \\frac{1}{x^2} + 1} \\geq 3 + \\frac{k}{4}\n$$\n\nor\n\n$$\n\\frac{k}{4} \\leq x^2 + 2x + \\frac{2}{x} - \\frac{3}{2x+1}\n$$\n\nFor $x = \\frac{2}{3}$, we get $k \\leq \\frac{880}{63} < 14$. Since $k$ is a positive integer, $k \\leq 13$. We now prove that $k = 13$ satisfies the requirement.\n\n$$\n\\frac{1}{a} + \\frac{1}{b} + \\frac{1}{c} + \\frac{13}{a+b+c+1} \\geq \\frac{25}{4}\n$$\n\nDefine\n\n$$\nf(a, b, c) = \\frac{1}{a} + \\frac{1}{b} + \\frac{1}{c} + \\frac{13}{a+b+c+1}\n$$\n\nAssume without loss of generality that $a = \\max\\{a, b, c\\}$. Then:\n\n$$\n\\begin{aligned}\nf(a, b, c) - f(\\sqrt{a}, \\sqrt{b}, \\sqrt{c}) &= \\left(\\frac{1}{\\sqrt{b}} + \\frac{1}{\\sqrt{c}} - \\frac{2}{\\sqrt{bc}}\\right) + 13 \\left(\\frac{1}{a+b+c+1} - \\frac{1}{a+2\\sqrt{bc}+1}\\right) \\\\\n&= (\\sqrt{b} - \\sqrt{c})^2 \\left[ \\frac{1}{bc} - \\frac{13}{(a+b+c+1)(a+2\\sqrt{bc}+1)} \\right]\n\\end{aligned}\n$$\n\nSince $a = \\max\\{a, b, c\\}$ and $abc = 1$, $bc \\le 1$, so $\\frac{1}{bc} \\ge 1$. By the AM-GM inequality,\n\n$$\n\\frac{13}{(a+b+c+1)(a+2\\sqrt{bc}+1)} \\le \\frac{13}{(3\\sqrt[3]{abc}+1)^2} = \\frac{13}{16}\n$$\n\nThus, $f(a, b, c) \\ge f(\\sqrt{a}, \\sqrt{b}, \\sqrt{c})$. It suffices to prove\n\n$$\nf\\left(\\frac{1}{x^2}, x, x\\right) \\ge \\frac{25}{4} \\quad \\text{where} \\quad x = \\sqrt{bc},\\ 0 < x \\le 1.\n$$\n\nFor $x = 1$, equality holds. For $0 < x < 1$, the inequality becomes\n\n$$\n\\frac{(x+2)(2x^3 + x^2 + 1)}{x(2x+1)} \\ge \\frac{13}{4}\n$$\n\nor\n\n$$\n8x^4 + 20x^3 - 18x^2 - 9x + 8 \\ge 0.\n$$\n\nThe left side can be written as\n\n$$\n\\begin{aligned}\n& (8x^4 - 8x^2 + 2) + (20x^3 - 20x^2 + 5x) + (10x^2 - 14x + 6) \\\\\n&= 2(2x^2 - 1)^2 + 5x(2x - 1)^2 + 2(5x^2 - 7x + 3) > 0.\n\\end{aligned}\n$$\n\nTherefore, $k = 13$ is the desired value. $\\square$",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 12745,
    "subject": "Mathematics (Olympiad)",
    "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": "See solution",
    "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$$\n\\frac{1}{a} + \\frac{2}{b} + \\frac{3}{c} \\geq \\frac{36}{a + 2b + 3c} \\geq 6,\n$$\nand the equal sign holds when $a = b = c = 1$.\n\nTherefore, the minimum value of $\\frac{1}{a} + \\frac{2}{b} + \\frac{3}{c}$ is $6$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 12746,
    "subject": "Mathematics (Olympiad)",
    "question": "Two circles $\\Gamma_1$ and $\\Gamma_2$ have common external tangents $l_1$ and $l_2$ meeting at $T$. Suppose $l_1$ touches $\\Gamma_1$ at $A$ and $l_2$ touches $\\Gamma_2$ at $B$. A circle $\\Omega$ through $A$ and $B$ intersects $\\Gamma_1$ again at $C$ and $\\Gamma_2$ again at $D$, such that quadrilateral $ABCD$ is convex.\n\nSuppose lines $AC$ and $BD$ meet at point $X$, while lines $AD$ and $BC$ meet at point $Y$. Show that $T$, $X$, $Y$ are collinear.",
    "options": [],
    "answer": "See solution",
    "solution": "We present four solutions.\n\n**First solution, elementary (original)** We have $\\triangle YAC \\sim \\triangle YBD$, from which it follows\n\n$$\n\\frac{d(Y, AC)}{d(Y, BD)} = \\frac{AC}{BD}.\n$$\n\nMoreover, if we denote by $r_1$ and $r_2$ the radii of $\\Gamma_1$ and $\\Gamma_2$, then\n\n$$\n\\frac{d(T, AC)}{d(T, BD)} = \\frac{T A \\sin \\angle (AC, l_1)}{T B \\sin \\angle (BD, l_2)} = \\frac{2 r_1 \\sin \\angle (AC, l_1)}{2 r_2 \\sin \\angle (BD, l_2)} = \\frac{AC}{BD}\n$$\n\nthe last step by the law of sines.\n\n![](images/sols-TST-IMO-2020_p2_data_047bfade40.png)\n\nThis solves the problem up to configuration issues: we claim that $Y$ and $T$ both lie inside $\\angle AXB \\equiv \\angle CXD$. WLOG $TA < TB$.\n\n* The former is since $Y$ lies outside segments $BC$ and $AD$, since we assumed $ABCD$ was convex.\n* For the latter, we note that $X$ lies inside both $\\Gamma_1$ and $\\Gamma_2$ in fact on the radical axis of the two circles (since $X$ was an interior point of both chords $AC$ and $BD$). In particular, $X$ is contained inside $\\angle ATB$, and moreover $\\angle ATB < 90^\\circ$, and this is enough to imply the result.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 12747,
    "subject": "Mathematics (Olympiad)",
    "question": "a) Solve the equation in the set of real numbers:\n$$\n[x]^2 - x = -0.99\n$$\n\nb) Show that, for every $a \\leq -1$, the equation\n$$\n[x]^2 - x = a\n$$\nhas no real solutions.",
    "options": [],
    "answer": "See solution",
    "solution": "a) The equation can be rewritten as $[x]^2 - [x] = \\{x\\} - 0.99$, so $\\{x\\} - 0.99 \\in \\mathbb{Z}$.\n\nSince $0 \\leq \\{x\\} < 1$, we have $-0.99 \\leq \\{x\\} - 0.99 < 0.01$, which implies $\\{x\\} - 0.99 = 0$, so $\\{x\\} = 0.99$.\n\nAlso, $[x]^2 - [x] = 0$ gives $[x] = 0$ or $[x] = 1$. Thus, $x \\in \\{0.99,\\ 1.99\\}$.\n\nb) The equation can be rewritten as $[x]^2 - [x] = \\{x\\} + a$. Assume, for contradiction, that there exists $a \\leq -1$ for which the equation has real solutions. Since $\\{x\\} < 1$, $[x]^2 - [x] = \\{x\\} + a < 0$.\n\nLet $[x] = y \\in \\mathbb{Z}$. Then $y(y - 1) < 0$, so $y \\in (0, 1)$, which is impossible for integer $y$. Thus, there are no real solutions.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 12748,
    "subject": "Mathematics (Olympiad)",
    "question": "Given positive numbers $a$, $b$ and a segment $AB$ of length $a$ on a plane, suppose two moving points $C$, $D$ in this plane satisfy that $ABCD$ is a non-degenerate convex quadrilateral with $BC = CD = b$ and $DA = a$. It is easy to see that there exists a circle $I$ tangent to all four sides of quadrilateral $ABCD$. Find the trajectory of the center $I$.",
    "options": [],
    "answer": "See solution",
    "solution": "Denote the center of the inscribed circle as $I$. Then $I$ lies on $AC$ such that\n\n$$\n\\frac{AI}{AC} = \\frac{a}{a+b}.\n$$\n\nHence, the trajectory of $I$ is the homothetic image of the trajectory of $C$ with respect to point $A$. (This can also be calculated by the coordinate method, based on the two geometric facts that $I$ is on $AC$ and $I$ is equidistant from $AB$ and $BC$.)\n\nSince $BC = b$ is fixed, the position of $C$ is on the circle $\\omega_1$ with center $B$ and radius $b$.\n\nTherefore, $I$ must be on a circle $\\omega_2$ with center $O$ on segment $AB$, satisfying $\\frac{AO}{AB} = \\frac{a}{a+b}$ or $AO = \\frac{a^2}{a+b}$, and radius $AO = \\frac{ab}{a+b}$.\n\nFor $ABCD$ to be a convex quadrilateral, both $\\angle ACB$ and $\\angle CAB$ must be acute in $\\triangle ABC$.\n\nWe discuss the following cases:\n\n- If $a > b$, the range of $C$ is two open arcs (both ends excluded, symmetric about the line containing $AB$) on $\\omega_1$. Each arc has one end at $C'$ on the extension of $AB$ with $BC' = b$, and the other at $C''$ on the circle with\n  $$\n  \\cos \\angle ABC'' = \\frac{b}{a}.\n  $$\n  The trajectory of $I$ is two arcs on $\\omega_2$ with\n  $$\n  \\angle AOI \\in \\left( \\arccos \\frac{b}{a}, \\pi \\right).\n  $$\n\n- If $a = b$, then $\\omega_2$ is a circle with $AB$ as its diameter, excluding only the points $A$ and $B$.\n\n- If $a < b$, the range of $C$ is two open arcs (both ends excluded, symmetric about the line containing $AB$) on $\\omega_1$. Each arc has one end at $C'$ on the extension of $AB$ with $BC' = b$, and the other at $C''$ on the circle with\n  $$\n  \\cos \\angle ABC'' = \\frac{a}{b}.\n  $$\n  The trajectory of $I$ is two arcs on $\\omega_2$ with\n  $$\n  \\angle AOI \\in \\left( \\arccos \\frac{a}{b}, \\pi \\right).\n  $$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 12749,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $G$ be a finite group of order $n$. Define the set\n$$\nH = \\{x : x \\in G \\text{ and } x^2 = e\\},\n$$\nwhere $e$ is the neutral element of $G$. Let $p = |H|$ be the cardinality of $H$. Prove that:\n\na) $|H \\cap xH| \\ge 2p - n$, for any $x \\in G$, where $xH = \\{xh : h \\in H\\}$.\n\nb) If $p > \\frac{3n}{4}$, then $G$ is commutative.\n\nc) If $\\frac{n}{2} < p \\le \\frac{3n}{4}$, then $G$ is non-commutative.",
    "options": [],
    "answer": "See solution",
    "solution": "a) Since $G$ is a group, $|xH| = p$. Therefore,\n$$\nn = |G| \\ge |H \\cup xH| = |H| + |xH| - |H \\cap xH| = 2p - |H \\cap xH|,\n$$\nwhence $|H \\cap xH| \\ge 2p - n$.\n\nb) Let $x \\in H$ and $y \\in H \\cap xH$. Then $y = y^{-1}$ and $y = xh$ for some $h \\in H$. Since $xy = x^2h = h \\in H$, it follows that $xy = (xy)^{-1} = y^{-1}x^{-1} = yx$. Therefore, $x$ commutes with all elements of $H \\cap xH$.\n\nSince $|H \\cap xH| \\ge 2p - n > \\frac{3n}{2} - n = \\frac{n}{2}$, the subgroup of elements that commute with $x$ has at least $\\frac{n}{2}$ elements. By Lagrange's theorem, $x$ commutes with all elements of $G$. Therefore, $H \\subseteq Z(G)$, the center of $G$. It follows that $|Z(G)| \\ge |H| = p > \\frac{3n}{4} > \\frac{n}{2}$, so $Z(G) = G$.\n\nc) Assume $G$ is commutative.\n\nIf $x, y \\in H$, then $(xy)^2 = x^2y^2 = e$, so $xy \\in H$. Since $H$ is finite, $H$ is a subgroup of $G$.\n\nThe condition $|H| > \\frac{n}{2}$ forces $H = G$, so $\\frac{3n}{4} \\ge |H| = |G| = n$, a contradiction.",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 12750,
    "subject": "Mathematics (Olympiad)",
    "question": "Determine, with proof, all possible values of $n$ such that, after finishing all the matches in a round-robin tournament among $n$ players, every player is not out-performed by any other player. (A player $A$ is not out-performed by player $B$ if at least one of player $A$'s losers is not a loser of $B$.)",
    "options": [],
    "answer": "See solution",
    "solution": "The answer is $n = 3$ or $n \\geq 5$.\n\n1. **Case $n = 3$:**\n   Let $A$, $B$, and $C$ be the players. Suppose $A$ beats $B$, $B$ beats $C$, and $C$ beats $A$. Each player is not out-performed by any other, as required.\n\n2. **Case $n = 4$:**\n   Assume the condition holds. No player can win all three matches, or else the others would be out-performed. Similarly, no player can lose all three. Thus, each player wins one or two matches. Suppose $A$ beats $B$ and $D$, but loses to $C$. Then both $B$ and $D$ must beat $C$; otherwise, they would be out-performed by $A$. The loser of $B$ vs. $D$ only beats $C$, so that player is out-performed by the winner. Thus, $n = 4$ is impossible.\n\n3. **Case $n = 6$ (and higher even $n$):**\n   One can construct a tournament as shown below:\n\n![](images/Mathematical_Olympiad_in_China_2011-2014_p242_data_4169f865f2.png)\n\n4. **Inductive Step:**\n   Suppose the condition holds for $n$ players $A_1, \\dots, A_n$. Add two new players $M$ and $N$, and define the results:\n\n$$\nA_i \\to M, \\quad M \\to N, \\quad N \\to A_i \\quad \\text{for all } i = 1, \\dots, n\n$$\n\nand keep the original results among the $A_i$. For any $G \\in \\{A_1, \\dots, A_n\\}$, consider $G, M, N$; this reduces to the $n = 3$ case, so the property holds.\n\n![](images/Mathematical_Olympiad_in_China_2011-2014_p243_data_647cb8a5ca.png)\n\nThus, by induction, the condition holds for all odd $n \\geq 3$ and all even $n \\geq 6$.\n\nTherefore, all possible values are $n = 3$ and $n \\geq 5$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 12751,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all positive integers $n$ such that the number of digits of $n$ is equal to the number of its distinct prime divisors, and the sum of the distinct prime divisors is equal to the sum of their exponents in the prime factorization of $n$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $n = p_1^{a_1} p_2^{a_2} \\dots p_k^{a_k}$. From the problem's condition:\n$$\np_1 + p_2 + \\dots + p_k = a_1 + a_2 + \\dots + a_k.\n$$\nConsider the number of digits of $n$. If $n$ has 4 digits, then it must have 4 distinct prime divisors. Then $n \\geq 2^{14} \\cdot 3 \\cdot 5 \\cdot 7 > 10^4$, which is not possible. If $n$ has $k > 4$ digits, then\n$$\nn \\geq 2^{14} \\cdot 3 \\cdot 5 \\cdot 7 \\cdot 2^{p_5 + \\dots + p_k - (k-1)} \\cdot p_5 \\dots p_k > 10^4 \\cdot 10^{k-1} = 10^k,\n$$\nwhich is again not possible.\n\nSo, $n$ can have at most 3 digits.\n\nIf $n$ has 3 digits, then $n = p_1^{a_1} p_2^{a_2} p_3^{a_3}$. If $5 \\mid n$, then $n \\geq 2^8 \\cdot 3 \\cdot 5 > 10^3$, which is not possible. The only primes $\\leq 3$ are 2 and 3, but we need 3 distinct primes, which is a contradiction.\n\nIf $n$ has 2 digits, then $n = p_1^{a_1} p_2^{a_2}$. If $5 \\mid n$, then $n \\geq 2^6 \\cdot 5 > 10^2$. So, $n = 2^{a_1} 3^{a_2}$ with $a_1 + a_2 = 5$. Checking possible values, we get $n = 2^4 \\cdot 3 = 48$ and $n = 2^3 \\cdot 3^2 = 72$ as solutions.\n\nIf $n$ has 1 digit, only $n = 2^2 = 4$ fulfills the condition.\n\n![](images/Macedonia_2017_p2_data_fec3af2d64.png)",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 12752,
    "subject": "Mathematics (Olympiad)",
    "question": "On the board there are numbers $1, 2, 3, 4, 5,$ and $6$. In every step, Juku deletes two numbers $a$ and $b$ from the board and writes $ab + a + b$ on the board instead. He repeats such steps until there is only one number left on the board. Find all possible values for the last number on the board.",
    "options": [],
    "answer": "See solution",
    "solution": "Notice that $(a+1)(b+1) = ab + a + b + 1 = c + 1$, where $c$ is the number written on the board instead of $a$ and $b$. Thus, the product of all numbers on the board, each increased by $1$, remains invariant throughout the process. Initially, the product is $2 \\times 3 \\times 4 \\times 5 \\times 6 \\times 7 = 5040$. In the end, only one number $n$ remains, so $n + 1 = 5040$, which gives $n = 5039$. Therefore, the only possible final number is $5039$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 12753,
    "subject": "Mathematics (Olympiad)",
    "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\n(A) For any positive divisor $m$ of $n$,\n\n$$\nn \\mid \\sum_{d \\mid m} f(d) \\binom{n/d}{m/d}.\n$$\n\n(B) For any positive divisor $k$ of $n$,\n\n$$\nk \\mid \\sum_{d \\mid k} f(d).\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "Define $g: D \\to \\mathbb{Z}$ by\n\n$$\ng(k) = \\sum_{d \\mid k} f(d), \\quad \\forall k \\in D.\n$$\n\nBy Möbius inversion,\n\n$$\nf(k) = \\sum_{d \\mid k} \\mu\\left(\\frac{k}{d}\\right) g(d), \\quad \\forall k \\in D.\n$$\n\nTherefore,\n\n$$\n\\begin{align}\n\\sum_{d \\mid m} f(d) \\binom{n/d}{m/d} &= \\sum_{d \\mid m} \\sum_{x \\mid d} \\mu\\left(\\frac{d}{x}\\right) g(x) \\binom{n/d}{m/d} \\\\\n&= \\sum_{x \\mid m} g(x) \\left( \\sum_{x \\mid d,\\ d \\mid m} \\mu\\left(\\frac{d}{x}\\right) \\binom{n/d}{m/d} \\right) \\\\\n&= \\sum_{x \\mid m} g(x) \\left( \\sum_{s \\mid \\frac{m}{x}} \\mu(s) \\binom{n/(xs)}{m/(xs)} \\right).\n\\end{align}\n$$\n\n**Lemma:** If $b \\mid a$, then\n\n$$\n\\sum_{k \\mid b} \\mu(k) \\binom{a/k}{b/k} \\equiv 0 \\pmod{a}.\n$$\n\nAssuming the lemma, we show (A) and (B) are equivalent.\n\n**(B) $\\Rightarrow$ (A):** If $x \\mid g(x)$ for all $x \\in D$, the lemma gives\n\n$$\n\\frac{n}{x} \\mid \\sum_{s \\mid \\frac{m}{x}} \\mu(s) \\binom{n/(xs)}{m/(xs)},\n$$\n\nso (A) follows from the above sum.\n\n**(A) $\\Rightarrow$ (B):** Assume (A). By induction, suppose $k \\mid g(k)$ for all $k < m$. For $k = m$,\n\n$$\n\\begin{aligned}\n0 &\\equiv \\sum_{d \\mid m} f(d) \\binom{n/d}{m/d} \\\\\n&\\equiv \\sum_{x \\mid m} g(x) \\left( \\sum_{s \\mid \\frac{m}{x}} \\mu(s) \\binom{n/(xs)}{m/(xs)} \\right) \\\\\n&\\equiv g(m) \\cdot \\frac{n}{m} + \\sum_{x \\mid m,\\ x < m} g(x) \\left( \\sum_{s \\mid \\frac{m}{x}} \\mu(s) \\binom{n/(xs)}{m/(xs)} \\right) \\\\\n&\\equiv g(m) \\cdot \\frac{n}{m} \\pmod{n},\n\\end{aligned}\n$$\n\nso $m \\mid g(m)$, completing the induction.\n\n**Proof of the lemma:**\n\n$$\n\\sum_{k \\mid b} \\mu(k) \\binom{a/k}{b/k} = \\frac{a}{b} \\sum_{k \\mid b} \\mu(k) \\binom{\\frac{a}{k}-1}{\\frac{b}{k}-1}.\n$$\n\nLet $b = \\prod_{i=1}^t p_i^{\\beta_i}$. It suffices to show\n\n$$\n\\sum_{k \\mid b} \\mu(k) \\binom{\\frac{a}{k}-1}{\\frac{b}{k}-1} \\equiv 0 \\pmod{p_i^{\\beta_i}}\n$$\n\nfor each $i$. For $u, v$ multiples of $p^\\beta$,\n\n$$\n\\binom{u-1}{v-1} - \\binom{\\frac{u}{p}-1}{\\frac{v}{p}-1} \\equiv 0 \\pmod{p^\\beta}.\n$$\n\nThis follows by expanding the binomial coefficients and noting divisibility by $p^\\beta$.\n\n**Alternative proof:**\n\nLet $X$ be the set of $b$-element subsets of $\\mathbb{Z}/a$, and $h: X \\to X$ by $h(S) = S+1$. For $k \\mid a$, the number of fixed points is $\\binom{k}{b/(a/k)}$. By orbit counting,\n\n$$\n\\sum_{k \\mid a} \\mu(k) \\binom{a/k}{b/k} \\equiv 0 \\pmod{a}.\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 12754,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $n \\in \\mathbb{N}$, $n \\ge 2$, and let $a_1, a_2, \\dots, a_n$ be real numbers. Denote\n$$\nS = \\sum_{1 \\le i < j \\le n} |a_j - a_i| \\quad \\text{and} \\quad d = \\max\\{|a_j - a_i| \\mid 1 \\le i, j \\le n\\}.\n$$\nProve that\n$$\n(n-1)d \\le S \\le \\frac{n^2 d}{4}\n$$\nand specify when equality holds in each of the two inequalities.",
    "options": [],
    "answer": "See solution",
    "solution": "Without loss of generality, assume $a_1 \\le a_2 \\le \\dots \\le a_n$, and let $d_k = a_{k+1} - a_k$ for $k = 1, 2, \\dots, n-1$. Then:\n\n- $d_1 + d_2 + \\dots + d_{n-1} = d$;\n- $|a_j - a_i| = d_i + d_{i+1} + \\dots + d_{j-1}$ for $i < j$.\n\nLet $n_k$ be the number of pairs $(i, j)$ with $1 \\le i < j \\le n$ such that $[a_k, a_{k+1}] \\subset [a_i, a_j]$. Then $i \\le k < k+1 \\le j$, so $n_k = k(n-k)$. Thus,\n$$\nS = \\sum_{k=1}^{n-1} n_k d_k = \\sum_{k=1}^{n-1} k(n-k) d_k. \\qquad (1)\n$$\nFor any $k \\in \\{1, \\dots, n-1\\}$, $k(n-k) \\ge n-1$.\n\nIndeed, $k(n-k) \\ge n-1 \\iff (k-1)(n-k-1) \\ge 0$, which holds for all $k$ in the range.\n\nFrom (1),\n$$\nS \\ge \\sum_{k=1}^{n-1} (n-1)d_k = (n-1)d,\n$$\nproving the left inequality.\n\nEquality holds if and only if $a_1 \\le a_2 = a_3 = \\dots = a_{n-1} \\le a_n$.\n\nBy AM-GM, $k(n-k) \\le \\frac{n^2}{4}$ for all $k$, so\n$$\nS = \\sum_{k=1}^{n-1} k(n-k)d_k \\le \\sum_{k=1}^{n-1} \\frac{n^2}{4}d_k = \\frac{n^2}{4}d, \\qquad (2)\n$$\nproving the right inequality.\n\nEquality in $k(n-k) \\le \\frac{n^2}{4}$ holds when $k = n-k$ (i.e., $n$ even). Thus:\n\n- If $d_k = 0$ for all $k$, i.e., $a_1 = a_2 = \\dots = a_n$, equality holds for any $n \\ge 2$.\n- If some $d_t \\ne 0$:\n    - If $n$ is odd, the right inequality is strict.\n    - If $n$ is even, equality holds only if $t = n-t$ and $d_t = d$, i.e., $n = 2t$ and $a_1 = \\dots = a_t \\le a_{t+1} = \\dots = a_{2t}$.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 12755,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $AB\\Gamma$ be an acute-angled triangle with $AB < A\\Gamma < B\\Gamma$. Let $c$ be its circumcircle, and let $\\Delta$ and $E$ be the midpoints of $AB$ and $A\\Gamma$, respectively. Draw externally two semicircles with diameters $AB$ and $A\\Gamma$, which intersect $E\\Delta$ at $M$ and $N$, respectively. The lines $MB$ and $M\\Gamma$ intersect the circumcircle at $T$ and $\\Sigma$, respectively. If the lines $MB$ and $M\\Gamma$ intersect at $H$, prove that:\n\n(a) The point $H$ lies on the circumcircle of triangle $AMN$.\n\n(b) The lines $AH$ and $T\\Sigma$ intersect perpendicularly at the point $Z$, and $Z$ is the center of the circumcircle of triangle $AMN$.",
    "options": [],
    "answer": "See solution",
    "solution": "(a) The angles $\\widehat{AMB}$ and $\\widehat{AN\\Gamma}$ are right since they subtend the diameters $AB$ and $A\\Gamma$. Therefore, the quadrilateral $AMHN$ is cyclic, which is the desired result.\n\n(b) The quadrilateral $B\\Gamma\\Sigma T$ is inscribed in the circle $c(O, R)$ so:\n\n![](images/Greece_olympiad_2018_p22_data_157b41a3c6.png)\n\n$$\n\\hat{T}_1 = \\hat{I}_1 \\quad (1).\n$$\n\nThe line $E\\Delta$ connects the midpoints of $AB$ and $A\\Gamma$, so $MN$ is parallel to $B\\Gamma$. Therefore:\n\n$$\n\\hat{r}_1 = \\hat{N}_1 \\quad (2).\n$$\n\nFrom the quadrilateral $AMHN$ we have:\n\n$$\n\\hat{H}_1 = \\hat{N}_2 \\quad (3).\n$$\n\nFrom the relations (1), (2), (3) we have:\n\n$$\n\\hat{T}_1 + \\hat{H}_1 = \\hat{N}_1 + \\hat{N}_2 = A\\hat{N}H = 90^\\circ.\n$$\n\n![](images/Greece_olympiad_2018_p23_data_dd2e4f7193.png)\n\nFrom the quadrilateral $AB\\Sigma\\Gamma$ we have:\n\n$$\nA\\hat{B}\\Gamma = A\\hat{\\Sigma}\\Gamma = \\hat{B} \\quad \\text{and from } E\\Delta \\parallel B\\Gamma \\text{ we have: } \\hat{E}_1 = \\hat{B}.\n$$\n\nFrom the equality $A\\hat{\\Sigma}\\Gamma = A\\hat{\\Sigma}N$ in combination with the previous ones, we get that the quadrilateral $AE\\Sigma N$ is cyclic.\n\nThe quadrilateral $AZ\\Sigma N$ is also cyclic, since $A\\hat{N}\\Sigma = A\\hat{Z}\\Sigma = 90^\\circ$. To this end, we have that $\\hat{Z}_1 = \\hat{N}_2 = \\hat{H}_1$ and so $BH \\parallel EZ$. Since $E$ is the midpoint of $AB$, we conclude that $Z$ is the midpoint of $AH$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 12756,
    "subject": "Mathematics (Olympiad)",
    "question": "Real numbers $x$ and $y$ are such that\n$$\nx^4 y^2 + y^4 + 2x^3 y + 6x^2 y + x^2 + 8 \\le 0.\n$$\nProve that $x \\ge -\\frac{1}{6}$.",
    "options": [],
    "answer": "See solution",
    "solution": "By removing $y^4$ from the left-hand side of the inequality, we obtain\n$$\nx^4 y^2 + 2x^2(x+3)y + x^2 + 8 \\le 0.\n$$\nThis is a quadratic inequality in $y$ whose discriminant is $4x^4(6x+1)$. If $x < -\\frac{1}{6}$, then the discriminant is negative, so the left-hand side is always positive for real $y$, which is a contradiction.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 12757,
    "subject": "Mathematics (Olympiad)",
    "question": "Suppose there are $n$ people at a party. For each person $P_i$ at the party, let\n\n$$\nS_i = \\{j : P_i \\text{ knows } P_j\\}\n$$\n\ni.e., $S_i$ is the set of people that $P_i$ knows.\n\nEach $P_i$ knows exactly 22 people. For each $P_i$, count the number of distinct pairs $(j, k)$ such that $j \\in S_i$ and $k \\in S_j$. If the total number of such pairs is 484, how many people $n$ are at the party?",
    "options": [],
    "answer": "See solution",
    "solution": "Fix $i$. Each $P_i$ knows 22 people, so $|S_i| = 22$.\n\nThe number of pairs $(j, k)$ with $j \\in S_i$ and $k \\in S_j$ is $22^2 = 484$.\n\nThere are 22 such pairs with $k = i$ (since for each $j \\in S_i$, $i \\in S_j$ if and only if $P_j$ knows $P_i$).\n\nSuppose $k \\neq i$. Then $P_k$ is one of the $n - 22 - 1$ people different from $P_i$ that $P_i$ does not know. For each such $k$, there are 6 corresponding $j$ for which $(j, k)$ is counted.\n\nSo:\n$$\n484 = 22 + 6(n - 23)\n$$\nSolving for $n$:\n$$\n484 - 22 = 6(n - 23) \\\\\n462 = 6(n - 23) \\\\\nn - 23 = 77 \\\\\nn = 100\n$$\nThus, there are $\\boxed{100}$ people at the party.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 12758,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $p$ be a prime number. Prove that there exists a prime number $q$ such that for every integer $n$, the number $n^p - p$ is not divisible by $q$.",
    "options": [],
    "answer": "See solution",
    "solution": "We approach indirectly by assuming that such $q$ does not exist. Then for any fixed prime $q$, there is a positive integer $n$ such that $n^p - p$ is divisible by $q$, that is\n\n$$\nn^p \\equiv p \\pmod{q}. \\quad (*)\n$$\n\nIf $q$ divides $n$, then $q$ divides $p$, and so $q = p$. We further assume that $q \\ne p$. Hence $q$ does not divide $n$. We start with the following well-known fact.\n\n*Lemma.* Let $q$ be a prime, and let $n$ be a positive integer relatively prime to $q$. Denote by $d_n$ the order of $n$ modulo $q$, that is, $d_n$ is the smallest positive integer such that $n^{d_n} \\equiv 1 \\pmod{q}$. Then for any positive integer $m$ such that $n^m \\equiv 1 \\pmod{q}$, $d_n$ divides $m$.\n\n*Proof.* By the minimality of $d_n$, we can write $m = d_n k + r$ where $k$ and $r$ are integers with $1 \\leq k$ and $0 \\leq r < d_n$. Then\n\n$$\n1 \\equiv n^m \\equiv n^{d_n k + r} \\equiv n^{d_n k} \\cdot n^r = n^r \\pmod{q}.\n$$\n\nBy the minimality of $d_n$, $r=0$, that is, $d_n$ divides $m$.\n\nBy *Fermat's Little Theorem*, $n^{q-1} \\equiv 1 \\pmod{q}$. Thus, by the Lemma, $d_n$ divides $q-1$. For the positive integer $n$, because $n^p \\equiv p \\pmod{q}$, we have $n^{p d_p} \\equiv p^{d_p} \\equiv 1 \\pmod{q}$. Thus, by the Lemma, $d_n$ divides both $q-1$ and $p d_p$, implying that $d_n$ divides \\text{gcd}$(q-1, p d_p)$.\n\nNow we pick a prime $q$ such that (a) $q$ divides $\\frac{p^p-1}{p-1} = 1 + p + \\cdots + p^{p-1}$, and (b) $p^2$ does not divide $q-1$. First we show that such a $q$ does exist. Note that $1 + p + \\cdots + p^{p-1} \\equiv 1 + p \\not\\equiv 1 \\pmod{p^2}$. Hence there is a prime divisor of $1 + p + \\cdots + p^{p-1}$ that is not congruent to $1 \\pmod{p^2}$, and we can choose that prime to be our $q$.\n\nBy (a), $p^p \\equiv 1 \\pmod{q}$ (and $p \\ne q$). By the Lemma, $d_p$ divides $p$, and so $d_p = p$ or $d_p = 1$.\n\nIf $d_p = 1$, then $p \\equiv 1 \\pmod{q}$.\n\nIf $d_p = p$, then $d_n$ divides \\text{gcd}$(p^2, q-1)$. By (b), the possible values of $d_n$ are $1$ and $p$, implying that $n^p \\equiv 1 \\pmod{q}$. By relation $(*)$, we conclude $p \\equiv 1 \\pmod{q}$.\n\nThus, in any case, we have $p \\equiv 1 \\pmod{q}$. But then by (a), $0 \\equiv 1 + p + \\cdots + p^{p-1} \\equiv p \\pmod{q}$, implying that $p = q$, which is a contradiction. Therefore our original assumption was wrong, and there is a $q$ such that for every integer $n$, the number $n^p - p$ is not divisible by $q$.\n\n*Note.* The proof can be shortened by starting directly with the definition of $q$ as in the second half of the above proof. But we think the argument in the first part provides motivation for the choice of this particular $q$.\n\nMany students were able to apply Fermat's Little Theorem to realize that $n^{p d_p} \\equiv p^{d_p} \\equiv 1 \\pmod{q}$. It is also not difficult to see that there are integers $n$ such that $n^{d_p} \\ne 1 \\pmod{q}$, because of the existence of primitive roots modulo $q$. By the minimality of $d_p$, we conclude that $d_p = p k$, where $k$ is some divisor of $d_p$. Consequently, we have $p k \\mid (q-1)$, implying that $q \\equiv 1 \\pmod{p}$. This led people to think about various applications of *Dirichlet's Theorem*, which is a very popular but fatal approach to this problem. However, a solution with advanced mathematics background is available. It involves a powerful prime density theorem. The prime $q$ satisfies the required condition if and only if $q$ remains a prime in the field $k = \\mathbb{Q}(\\sqrt[p]{p})$. By applying Chebotarev's density theorem to the Galois closure of $k$, we can show that the set of such $q$ has density $\\frac{1}{p}$, implying that there are infinitely many $q$ satisfying the required condition. Of course, this approach is far beyond the knowledge of most IMO participants.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 12759,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $A$ to $J$ be the numbers in the vacant squares as shown.\n\n![](images/Australian_Scene_2012_-_AMT_Publishing_-_273p_p46_data_3297e3213c.png)\n\nThe table has the following structure:\n\n| $A$ | $\\times$ | $B$ | $\\times$ | $C$ | $\\times$ | $D$ |\n|-----|----------|-----|----------|-----|----------|-----|\n| $E$ | $\\times$ | $6$ | $\\times$ | $F$ | $\\times$ | $G$ |\n| $H$ | $\\times$ | $I$ | $\\times$ | $J$ | $\\times$ | $9$ |\n| $4$ |          | $16$|          | $9$ |          | $25$|\n\nFill in the values of $A$ to $J$ so that the products and sums in the table are correct.",
    "options": [],
    "answer": "See solution",
    "solution": "Since the only allowable products ending in 1 are $81$ and $21$, $B = 2$ or $8$. Hence $I = 8$ or $2$ respectively. Since $84$ is not an allowable product, $I$ must be $2$ and $B = 8$.\n\nThe first column adds to $4$, so $A, E, H$ must each be less than $3$. As $22$ and $26$ are not allowable products, $E$ must be $1$ and $H$ must be $1$. Then $A = 2$ and we get:\n\n![](images/Australian_Scene_2012_-_AMT_Publishing_-_273p_p46_data_3297e3213c.png)\n\nThere are other possibilities for some of the multiplications: $16 = 2 \\times 8 = 4 \\times 4$, $12 = 2 \\times 6 = 3 \\times 4$, $24 = 4 \\times 6 = 3 \\times 8$, $18 = 2 \\times 9 = 3 \\times 6$. The entries in the addition columns are unique.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 12760,
    "subject": "Mathematics (Olympiad)",
    "question": "A $+1$ or $-1$ is written at each vertex of a regular $n$-gonal prism so that the product of numbers on each face is $-1$. For which $n \\geq 3$ is this possible?",
    "options": [],
    "answer": "See solution",
    "solution": "Call *vertical* the $n$ edges that join corresponding vertices of the two bases. A vertical edge is *odd* if it has different numbers at its endpoints and *even* otherwise. Take two adjacent vertical edges $e_1$ and $e_2$. They determine a lateral face $F$, which is a rectangle with opposite sides $e_1$ and $e_2$. The product of the numbers at the vertices of $F$ is known to be $-1$. It is also the product of the numbers at the endpoints of $e_1$ multiplied by the respective product for $e_2$. Hence $e_1$ and $e_2$ are of different kinds: one is odd and the other is even. Thus, odd and even vertical edges alternate, which is possible only if $n$ is even.\n\nLet $k_1$ and $k_2$ be the numbers of $-1$ at the vertices of the two bases. Both $k_1$ and $k_2$ are odd by hypothesis, hence the total number $k = k_1 + k_2$ of $-1$ is even. On the other hand, $k$ has the same parity as the number of odd vertical edges, which by the above equals $n/2$. Hence $n/2$ is even, meaning that $n$ is divisible by $4$.\n\nSo such an assignment of $+1$ and $-1$ is possible only if $n$ is a multiple of $4$. Conversely, let $n = 4k$, $k \\geq 1$. Let the two bases be $A_1A_2\\ldots A_{4k}$ and $B_1B_2\\ldots B_{4k}$. Assign a $-1$ to each of the vertices $A_1, A_2, \\ldots, A_{4k-3}$ ($2k-1$ of them), and also to $B_{4k-1}$. Write a $+1$ at all remaining vertices. The product of numbers on each face is $-1$. A vertical edge $A_iB_i$ is odd or even according as $i$ is odd or even, hence the product of numbers on each lateral face is\n\n![](images/Argentina_2015_Booklet_p4_data_e15214cce6.png)\n\nalso $-1$. So the assignment has the necessary properties. In conclusion, the answer to the question is: for all $n$ divisible by $4$.",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 12761,
    "subject": "Mathematics (Olympiad)",
    "question": "There are 6 distinct lines chosen in space. Find the largest possible number of points where at least 3 of the chosen lines intersect.",
    "options": [],
    "answer": "See solution",
    "solution": "*Solution 1:* Call a point *rich* if at least 3 chosen lines intersect there. Let the number of rich points be $n$. We count pairs $(P, s)$ where $P$ is a rich point and $s$ is a chosen line that passes through $P$. The number of these pairs is at least $3n$ since each of the $n$ rich points occurs in at least 3 pairs. On the other hand, after fixing a chosen line, the other lines passing through distinct rich points on the fixed line are distinct (two lines cannot intersect in more than one point). Hence there can be at most $\\lfloor \\frac{6-1}{2} \\rfloor = 2$ rich points on every\n\n![](images/EST_ABooklet_2021_p38_data_e40b2aa44e.png)\n\nFig. 39\n\n![](images/EST_ABooklet_2021_p38_data_a1b4c3e6e5.png)\n\nFig. 40\n\nchosen line. This implies that the number of pairs under consideration does not exceed $6 \\cdot 2$, i.e., 12. Altogether, we have established $3n \\le 12$, which implies $n \\le 4$.\n\nA configuration with 4 rich points can be obtained by choosing 6 lines determined by the 6 edges of a tetrahedron; those lines intersect at 4 vertices of the tetrahedron, 3 in each one (Figure 39).",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 12762,
    "subject": "Mathematics (Olympiad)",
    "question": "設三角形 $ABC$ 中有 $AB < AC$，並設 $I_a$ 為位於 $\\angle A$ 內的旁心。令 $D$ 為 $I_a$ 到 $BC$ 的投影點。設 $X$ 為 $AI_a$ 與 $BC$ 的交點，並於直線 $AC, AB$ 上分別取點 $Y, Z$ 使得 $X, Y, Z$ 落在一條與 $AI_a$ 垂直的直線上。設三角形 $AYZ$ 的外接圓與 $AI_a$ 再交於點 $U$。已知過 $A$ 並與 $ABC$ 的外接圓相切的直線交 $BC$ 於點 $T$，而線段 $TU$ 與 $ABC$ 的外接圓交於點 $V$。\n\n證明：$\\angle BAV = \\angle DAC$。",
    "options": [],
    "answer": "See solution",
    "solution": "設 $\\Gamma$ 為 $ABC$ 的外接圓，$\\Gamma_I$ 為內切圓，$\\Gamma_{I_a}$ 為 $A$ 旁切圓。設 $\\Gamma_1$ 和 $\\Gamma_2$ 為分別與 $AB, AC$ 及 $\\Gamma$ 相切的兩圓，其中 $\\Gamma_1$ 在 $\\Gamma$ 內部，$\\Gamma_2$ 在外部。設 $\\Gamma_1$ 與 $\\Gamma$ 相切於 $P$，$\\Gamma_2$ 與 $\\Gamma$ 相切於 $Q$。關鍵觀察是 $P, Q, T, U$ 共線。這可分為兩部分證明：\n\n**引理 1. $P, Q, U$ 共線。**\n\n*證明*：$P$ 是 $\\Gamma$ 與 $\\Gamma_1$ 的外部相似中心，$Q$ 是 $\\Gamma$ 與 $\\Gamma_2$ 的內部相似中心。由 Monge 定理，$PQ$ 通過 $\\Gamma_1$ 與 $\\Gamma_2$ 的內部相似中心。\n\n注意 $XI/XI_a = AI/AI_a$，這是 $\\Gamma_I$ 與 $\\Gamma_{I_a}$ 半徑的比。因此 $X$ 是 $\\Gamma_I$ 與 $\\Gamma_{I_a}$ 的內部相似中心。考慮以 $A$ 為中心、比為 $AU/AX$ 的放射變換，$I$ 會被送到 $\\Gamma_1$ 的圓心，$I_a$ 會被送到 $\\Gamma_2$ 的圓心。這說明 $\\Gamma_I$ 被送到 $\\Gamma_1$，$\\Gamma_{I_a}$ 被送到 $\\Gamma_2$。因為 $X$ 被送到 $U$，所以 $U$ 是 $\\Gamma_1$ 與 $\\Gamma_2$ 的內部相似中心，因此 $PQ$ 通過 $U$。$\\square$\n\n**引理 2. $P, Q, T$ 共線。**\n\n有多種證明方法，以下列舉：\n\n*證明*：考慮以 $A$ 為中心、半徑 $\\sqrt{AB \\cdot AC}$ 的反演，然後對 $AI_a$ 作對稱。則 $B \\to C$，$C \\to B$，$\\Gamma_1 \\to \\Gamma_{I_a}$，$\\Gamma_2 \\to \\Gamma_I$。設 $X \\to X'$，$P \\to P'$，$Q \\to Q'$，則 $\\Gamma_I$ 在 $BC$ 切於 $Q$，$\\Gamma_{I_a}$ 在 $BC$ 切於 $P$。$X'$ 在 $\\Gamma$ 上，且 $\\angle CAX' = \\angle XAB = \\angle ACB$，所以 $AX'$ 平行於 $BC$。只需證明 $AX'P'Q'$ 共圓。因 $BP' = CQ'$，$AX'P'Q'$ 為等腰梯形，得證。$\\square$\n\n*其他證明略。*\n\n由兩引理，$V$ 為 $P$ 或 $Q$。接下來證明 $V = P$。因 $AB < AC$，$TB < TC$（$TB/TC = (AB/AC)^2$）。又 $\\angle C < \\angle B$，所以 $\\angle AXB = 90^\\circ + \\frac{1}{2}(\\angle B - \\angle C) > 90^\\circ$，$D$ 在 $X$ 與 $C$ 之間，$\\angle DAC < \\frac{1}{2}\\angle A$。已知 $\\angle BAP = \\angle DAC$，所以 $\\angle BAP < \\frac{1}{2}\\angle A = \\angle BAU$。同理 $\\angle BAQ > \\angle BAU$。因此 $T, P, U, Q$ 順序排列，$V = P$。所以 $\\angle BAV = \\angle BAP = \\angle DAC$，得證。",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 12763,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $p$ and $q$ be distinct odd primes. For which products $pq$ is $pq$ **not** super-deficient? (A number $n$ is called super-deficient if $2$ times the sum of its proper divisors is less than $n$.)",
    "options": [],
    "answer": "See solution",
    "solution": "Let the proper factors of $pq$ be $1$, $p$, and $q$ (since $p$ and $q$ are distinct primes). We may assume $p < q$.\n\nThe product $pq$ is not super-deficient if and only if\n$$\n2(1 + p + q) \\ge pq.\n$$\nThis can be rearranged as\n$$\npq - 2p - 2q + 4 \\le 6,\n$$\nwhich is equivalent to\n$$\n(p - 2)(q - 2) \\le 6.\n$$\nSince $p - 2 < q - 2$ and $p$ is a prime, $p - 2 = 1$ or $2$, so $p = 3$ or $4$. But $p$ must be an odd prime, so $p = 3$.\n\nThen $q - 2 \\le 6$, so $q = 5$ or $7$ (since $q$ is an odd prime and $q > p$).\n\nTherefore, the only products $pq$ that are not super-deficient are $3 \\times 5 = 15$ and $3 \\times 7 = 21$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 12764,
    "subject": "Mathematics (Olympiad)",
    "question": "Johan and Quintijn play the following game.\n\nBefore the start of the game, the integers $1, 2, \\ldots, 2024$ are written on a board. The players then each take turns, starting with Johan. On their turn, a player must wipe out two integers $a$ and $b$ from the board and write their (possibly negative) difference $a - b$ on the board. The game ends when only one integer is left on the board. If this integer is divisible by $3$, Johan wins; otherwise, Quintijn wins.\n\nDetermine which of the two players has a winning strategy.",
    "options": [],
    "answer": "See solution",
    "solution": "We show that Quintijn has a winning strategy.\n\nObserve that each move reduces the number of integers by exactly one, so at the start of Johan's turn the number is always even and at the start of Quintijn's turn the number is always odd. Moreover, the number must be at least $2$, otherwise the game would have already ended. In particular, at the start of Quintijn's turn, the number of integers is always at least $3$.\n\nFor any residue class $i$ modulo $3$, denote the number of integers on the board inside $i$ with $x_i$. If $i, j, k$ are residue classes modulo $3$ with $i - j \\equiv k$, then we denote any move removing integers $a$ and $b$ with $a \\equiv i \\pmod{3}$ and $b \\equiv j \\pmod{3}$ by $(i, j) \\to k$.\n\nNote Johan cannot reach a position in which $x_1 = x_2 = 0$ unless at the start of his turn either $x_1 = 2$ and $x_2 = 0$, or $x_1 = 0$ and $x_2 = 2$ (or $x_1 = 0$ and $x_2 = 0$) hold. In particular, if at the start of his turn at least one of $x_1$ and $x_2$ is odd, then he cannot reach a position in which $x_1 = x_2 = 0$.\n\nOn the other hand, if $x_1$ and $x_2$ are not both $0$, Quintijn can always make a move such that at least one of $x_1$ and $x_2$ becomes odd, as we will show below.\n\nFirst suppose that $x_1$ and $x_2$ are both even. Then at least one of $x_1$ and $x_2$ must be at least $2$, say $x_1 \\ge 2$. Since Quintijn always has an odd number of integers left at the start of his turn, there must also be an integer on the board that is divisible by $3$. With $(0, 1) \\to 2$, Quintijn ensures that $x_1$ decreases by $1$ and $x_2$ increases by $1$, so $x_1$ and $x_2$ both become odd.\n\nNow suppose that at least one of $x_1$ and $x_2$ is odd. If there exists an $i$ with $x_i \\ge 2$, then performing any move $(i, i) \\to 0$ will not change the parity of $x_1$ and $x_2$, so at least one of them remains odd. Otherwise all $x_i \\le 1$ so there are at most $3$ integers left, and in fact we must have equality here as there are always at least $3$ integers left at the start of Quintijn's turn. So $x_0 = x_1 = x_2 = 1$, hence Quintijn can perform the move $(1, 0) \\to 1$.\n\nTherefore, Quintijn can always make a move that causes at least one of $x_1$ and $x_2$ to be odd. Now the strategy of making such a move is winning for Quintijn.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 12765,
    "subject": "Mathematics (Olympiad)",
    "question": "Prove that the inequality\n\n$$\nx^2 \\sqrt{1 + 2y^2} + y^2 \\sqrt{1 + 2x^2} \\ge xy(x + y + \\sqrt{2})\n$$\n\nholds for any two real numbers $x$ and $y$.",
    "options": [],
    "answer": "See solution",
    "solution": "By the quadratic-arithmetic mean inequality, we have\n\n$$\n1 + 2t^2 \\ge \\frac{1}{2}(1 + t\\sqrt{2})^2.\n$$\n\nIt follows that\n\n$$\n\\begin{align*}\nx^2 \\sqrt{1 + 2y^2} + y^2 \\sqrt{1 + 2x^2} &\\ge x^2 \\cdot \\frac{1 + y\\sqrt{2}}{2} + y^2 \\cdot \\frac{1 + x\\sqrt{2}}{2} \\\\\n&= \\frac{x^2 + y^2 + xy(x + y)\\sqrt{2}}{\\sqrt{2}} \\\\\n&\\ge \\frac{2xy + xy(x + y)\\sqrt{2}}{\\sqrt{2}} = xy(x + y + \\sqrt{2}).\n\\end{align*}\n$$\n\nThis is what we need to prove. $\\square$",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 12766,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all functions $f : \\mathbb{R}^{+} \\to \\mathbb{R}^{+}$ such that\n$$\nf(xf(x + y)) = y f(x) + 1\n$$\nholds for all $x, y \\in \\mathbb{R}^{+}$.",
    "options": [],
    "answer": "See solution",
    "solution": "We will show that for every $x \\in \\mathbb{R}^{+}$, $f(x) = \\frac{1}{x}$. It is easy to check that this function satisfies the equation.\n\nLet $P(x, y)$ denote the assertion $f(xf(x + y)) = y f(x) + 1$.\n\n**Injectivity:**\nAssume $f(x_1) = f(x_2)$ and take any $x < x_1, x_2$. Then $P(x, x_1 - x)$ and $P(x, x_2 - x)$ give\n$$\n(x_1 - x) f(x) + 1 = f(xf(x_1)) = f(xf(x_2)) = (x_2 - x) f(x) + 1\n$$\nso $x_1 = x_2$.\n\n**Surjectivity for $z > 1$:**\nFor every $z > 1$, there is $x$ such that $f(x) = z$. Indeed, $P(x, \\frac{z-1}{f(x)})$ gives\n$$\nf\\left(x f\\left(x + \\frac{z-1}{f(x)}\\right)\\right) = z.\n$$\nNow, given $z > 1$, take $x$ such that $f(x) = z$. Then $P(x, \\frac{z-1}{z})$ gives\n$$\nf\\left(x f\\left(x + \\frac{z-1}{z}\\right)\\right) = \\frac{z-1}{z} f(x) + 1 = z = f(x).\n$$\nSince $f$ is injective, $f\\left(x + \\frac{z-1}{z}\\right) = 1$.\n\nSo there is $k \\in \\mathbb{R}^{+}$ such that $f(k) = 1$. Since $f$ is injective, this $k$ is unique. Therefore, $x = k + \\frac{1}{z} - 1$. That is, for every $z > 1$,\n$$\nf\\left(k + \\frac{1}{z} - 1\\right) = z.\n$$\nWe must have $k + \\frac{1}{z} - 1 \\in \\mathbb{R}^{+}$ for each $z > 1$. Taking the limit as $z \\to \\infty$, we deduce $k \\ge 1$. Set $r = k - 1$.\n\nNow $P\\left(r + \\frac{1}{6}, \\frac{1}{3}\\right)$ gives\n$$\nf\\left(\\left(r + \\frac{1}{6}\\right) f\\left(r + \\frac{1}{6} + \\frac{1}{3}\\right)\\right) = \\frac{1}{3} f\\left(r + \\frac{1}{6}\\right) + 1 = \\frac{6}{3} + 1 = 3 = f\\left(r + \\frac{1}{3}\\right).\n$$\nBut\n$$\nf\\left(\\left(r + \\frac{1}{6}\\right) f\\left(r + \\frac{1}{6} + \\frac{1}{3}\\right)\\right) = f\\left(\\left(r + \\frac{1}{6}\\right) f\\left(r + \\frac{1}{2}\\right)\\right) = f\\left(2r + \\frac{1}{3}\\right).\n$$\nThe injectivity of $f$ now shows $r = 0$, i.e., $f(1) = k = 1$.\n\nThis shows $f\\left(\\frac{1}{z}\\right) = z$ for every $z > 1$, i.e., $f(x) = \\frac{1}{x}$ for every $x < 1$. Now for $x > 1$, consider $P(1, x-1)$ to get $f(f(x)) = (x-1) f(1) + 1 = x = f\\left(\\frac{1}{x}\\right)$. Injectivity of $f$ shows $f(x) = \\frac{1}{x}$.\n\nThus, for all $x \\in \\mathbb{R}^{+}$, $f(x) = \\frac{1}{x}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 12767,
    "subject": "Mathematics (Olympiad)",
    "question": "Does there exist positive real numbers $a$ and $b$ such that $\\lfloor an + b \\rfloor$ is a prime for each positive integer $n$?",
    "options": [],
    "answer": "See solution",
    "solution": "No, there are no such numbers.\n\nConsider the sequence $x_i = \\lfloor a i + b \\rfloor$. We can estimate the difference\n\n$$\nx_{i+1} - x_i = \\lfloor a(i+1) + b \\rfloor - \\lfloor a i + b \\rfloor < a(i+1) + b - (a i + b - 1) = a + 1\n$$\n\nand see that it is bounded. As there are arbitrarily long sequences of consecutive composite numbers (for example, $k! + 2, k! + 3, \\dots, k! + k$) and $\\{x_i\\}$ is unbounded, then $x_i$ cannot all be primes.\n\n![](images/Baltic_Way_SHL_2009-11_13-16_p280_data_898b43a865.png)",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 12768,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $ABCD$ be a cyclic quadrilateral and let the lines $CD$ and $BA$ meet at $E$. The line through $D$ which is tangent to the circle $ADE$ meets the line $CB$ at $F$. Prove that the triangle $CDF$ is isosceles.\n\n![](images/British_Booklet_2016_p3_data_5a3fe50110.png)",
    "options": [],
    "answer": "See solution",
    "solution": "By angles in a cyclic quadrilateral,\n\n$$\n\\angle DCF = 180^\\circ - \\angle DAB = \\angle EAD\n$$\n\nBy the alternate segment theorem,\n\n$$\n\\angle EAD = \\angle XDE = \\angle CDF,\n$$\n\nso\n\n$$\n\\angle DCF = \\angle EAD = \\angle CDF\n$$\n\nso $\\triangle CDF$ is isosceles.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 12769,
    "subject": "Mathematics (Olympiad)",
    "question": "Find the sum\n$$\n\\sum_{k \\in A} \\frac{1}{k-1}\n$$\nif $A = \\{m^n : m, n \\in \\mathbb{Z},\\ m, n \\ge 2\\}$.",
    "options": [],
    "answer": "See solution",
    "solution": "The sum is $1$.\n\nThis can be seen as follows:\n\n$$\n\\sum_{k \\in A} \\frac{1}{k-1} = \\sum_{k \\in A} \\left( \\frac{1}{k} \\cdot \\frac{1}{1-1/k} \\right) = \\sum_{k \\in A} \\sum_{i \\ge 0} \\left( \\frac{1}{k} \\cdot \\frac{1}{k^i} \\right) = \\sum_{k \\in A} \\sum_{i \\ge 1} \\frac{1}{k^i}\n$$\n\n*Claim:* For $x \\in \\mathbb{Z}$, the sets $\\{(m, n) : x = m^n \\text{ for some } m, n \\in \\mathbb{Z} \\text{ with } m, n \\ge 2\\}$ and $\\{(k, i) : x = k^i \\text{ for some } k \\in A,\\ i \\in \\mathbb{Z} \\text{ with } i \\ge 1\\}$ have the same number of elements.\n\n*Proof of the claim:* Let $x = p_1^{a_1} \\cdots p_t^{a_t}$ be the prime factorization of $x$. Let $a = \\gcd(a_1, \\dots, a_t)$, $a_j = a b_j$ for $1 \\le j \\le t$ and $y = p_1^{b_1} \\cdots p_t^{b_t}$. Then $x = m^n$ if and only if $n \\mid a$ and $m = y^{a/n}$. So both of the sets above have $\\tau(a) - 1$ elements, where $\\tau(a)$ stands for the number of positive divisors of $a$.\n\nTherefore:\n\n$$\n\\sum_{k \\in A} \\sum_{i \\ge 1} \\frac{1}{k^i} = \\sum_{m \\ge 2} \\sum_{n \\ge 2} \\frac{1}{m^n} = \\sum_{m \\ge 2} \\frac{1/m^2}{1 - 1/m} = \\sum_{m \\ge 2} \\frac{1}{m(m-1)}\n$$\n\nFinally, as $N \\to \\infty$,\n\n$$\n\\sum_{m=2}^{N} \\frac{1}{m(m-1)} = \\sum_{m=2}^{N} \\left( \\frac{1}{m-1} - \\frac{1}{m} \\right) = 1 - \\frac{1}{N} \\to 1.\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 12770,
    "subject": "Mathematics (Olympiad)",
    "question": "We will say that the positive integers $m$ and $n$ have property $\\mathcal{P}$ if for every divisor $d_1$ of $m$ and every divisor $d_2$ of $n$, the number $d_1 + d_2$ is a prime.\n\n(a) Prove that if $m$ and $n$ have property $\\mathcal{P}$ and are different, then $m + n$ is odd.\n\n(b) Find all the pairs $(m, n)$ of positive integers with $m \\leq n$ having property $\\mathcal{P}$.",
    "options": [],
    "answer": "See solution",
    "solution": "**(a)** Without loss of generality, suppose $1 \\leq m < n$.\n\nIf $m$ and $n$ are both odd, then $m \\geq 1$ and $n \\geq 3$. Taking $d_1 = 1$ and $d_2 = n$ gives $d_1 + d_2 = n + 1$, which is even and at least $4$, so it is composite—a contradiction.\n\nIf $m$ and $n$ are both even, taking $d_1 = 2$ and $d_2 = 2$ gives $d_1 + d_2 = 4$, which is not prime—a contradiction.\n\nTherefore, $m$ and $n$ must have different parities, so $m + n$ is odd.\n\n**(b)** If $m = n$, then $d_1 = m$ and $d_2 = n$ gives $d_1 + d_2 = 2m$, which must be prime, so $m = n = 1$.\n\nIf $m < n$, then $m \\neq n$ and, from (a), $m$ and $n$ have different parities.\n\n- If $m$ is even and $n$ is odd, then $m \\geq 2$ and $n \\geq 3$. Taking $d_1 = 1$ and $d_2 = n$ gives $d_1 + d_2 = n + 1 \\geq 4$, which is composite—a contradiction.\n\n- If $m$ is odd and $n$ is even, then $m \\geq 1$ and $n \\geq 2$. Write $n = 2^a b$ with $b$ odd.\n\n    - If $a \\geq 3$, then $8$ divides $n$ and taking $d_1 = 1$, $d_2 = 8$ gives $d_1 + d_2 = 9$, which is not prime—a contradiction.\n\n    - If $a = 1$, then $n = 2b$ with $b$ odd.\n        - If $b \\geq 3$, take $d_1 = 1$, $d_2 = b$ to get $d_1 + d_2 = b + 1 \\geq 4$, which is composite—a contradiction.\n        - If $b = 1$, then $n = 2$ and since $m < n$ and $m$ is odd, $m = 1$. The pair $(1, 2)$ is a solution.\n\n    - If $a = 2$, then $n = 4b$ with $b$ odd.\n        - If $b \\geq 3$, $d_1 = 1$, $d_2 = b$ gives $d_1 + d_2 = b + 1 \\geq 4$, which is composite—a contradiction.\n        - If $b = 1$, then $n = 4$ and $m < n$ with $m$ odd, so $m \\in \\{1, 3\\}$. Only $(1, 4)$ satisfies the statement.\n\nThe solutions are $(1, 1)$, $(1, 2)$, and $(1, 4)$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 12771,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $ABCD$ be a quadrilateral with $AB$ not parallel to $CD$. The circle with diameter $AB$ is tangent to the side $CD$ at $X$. The circle with diameter $CD$ is tangent to the side $AB$ at $Y$.\n\nProve that the quadrilateral $BCXY$ is cyclic.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $M$ and $N$ be the midpoints of $AB$ and $CD$, respectively. Then $MX$ is perpendicular to $CD$, since $MX$ is the radius of a circle to which $CD$ is tangent. Similarly, $NY$ is perpendicular to $AB$. It follows that the points $M, N, X, Y$ lie on a circle.\n\n![](images/Australian-Scene-2017_p80_data_d7e68a14b2.png)\n\nLet $\\angle ABX = x$ and note that $\\angle AMX = 2x$, since it is the angle subtended at the centre of the circumcircle of triangle $ABX$. It follows that $\\angle YNX = \\angle YMX = \\angle AMX = 2x$, where we have used the fact that $MNXY$ is a cyclic quadrilateral.\n\nSo $\\angle CNY = 180^\\circ - \\angle YNX = 180^\\circ - 2x$. However, note that triangle $CNY$ is isosceles with $CN = NY$. Therefore, $\\angle NCY = \\angle NYC = x$. Since $\\angle XCY = \\angle XBY = x$, we have deduced that the quadrilateral $BCXY$ is cyclic.\n\nA second case arises when $X$ and $Y$ are on different sides of the line $MN$, in which case we have the equality $\\angle YNX = 180^\\circ - \\angle YMX$ rather than $\\angle YNX = \\angle YMX$. This can be handled in an analogous manner or with the use of directed angles.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 12772,
    "subject": "Mathematics (Olympiad)",
    "question": "Prove that $a^3 + b^3 + 3abc > c^3$, where $a$, $b$, $c$ are the sides of a triangle.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $a$, $b$, $c$ be the sides of a triangle. From the triangle inequality $a + b > c$, we have\n\n$$\n\\begin{aligned}\na^3 + b^3 + 3abc &= (a+b)(a^2-ab+b^2) + 3abc \\\\\n&> c \\cdot (a^2-ab+b^2) + 3abc \\\\\n&= c \\cdot (a^2 - ab + b^2 + 3ab) \\\\\n&= c \\cdot (a^2 + 2ab + b^2) \\\\\n&= c \\cdot (a+b)^2 \\\\\n&> c \\cdot c^2 = c^3\n\\end{aligned}\n$$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 12773,
    "subject": "Mathematics (Olympiad)",
    "question": "In a cyclic quadrilateral $ABCD$, let $\\Gamma_1$ and $\\Gamma_2$ be the circles tangent to $AB$ at $E$ and to $AD$ at $F$, respectively, and both tangent to $BC$ at $G$ and to $CD$ at $H$. Prove that $EG \\perp FH$.",
    "options": [],
    "answer": "See solution",
    "solution": "If the quadrilateral $ABCD$ has a pair of parallel sides, the result follows from the symmetry of the construction. So we suppose that $ABCD$ is neither a trapezoid nor a rectangle.\n\nLet $X = BC \\cap AD$. Since $\\Gamma_2$ touches $AD$ at point $F$, we have $XF^2 = XB \\cdot XC$.\n\n![](images/Blr2012_p17_data_78cc276a22.png)\n\nSimilarly, $XH^2 = XA \\cdot XD$. Since $ABCD$ is cyclic, we have $XB \\cdot XC = XA \\cdot XD$, so $XF = XH$, and hence $\\triangle XHF$ is isosceles. Thus $\\angle BHF = 0.5(180^\\circ - \\angle CXD) = 0.5(\\angle C + \\angle D)$. Similarly, $\\angle BGE = 0.5(\\angle A + \\angle D)$. Let $GE \\cap HF = Y$. Then $\\angle GYH = 360^\\circ - (\\angle B + 0.5(\\angle C + \\angle D) + 0.5(\\angle A + \\angle D)) = 360^\\circ - (\\angle B + \\angle D) - 0.5(\\angle A + \\angle C)$. Since $\\angle A + \\angle C = \\angle B + \\angle D = 180^\\circ$, we have $\\angle GYH = 360^\\circ - 180^\\circ - 90^\\circ = 90^\\circ$, so $EG \\perp FH$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 12774,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $u_j = \\binom{2j}{j}$ for $j = 0, 1, 2, \\dots$. For fixed $n$ and $j$, let $m = n - j$ and define $f(j) = u_j u_{n-j} = u_j u_m$.\n\n(a) Show that $f(j)$ decreases for $0 \\leq j < n/2$.\n\n(b) Show that $f(j)$ is convex, i.e., $f(j+1) + f(j-1) > 2f(j)$ for all $j$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $f(j) = u_j u_{n-j}$ and $m = n - j$.\n\n(a) Since\n\n$$\n\\begin{align*}\nu_{j+1} &= \\frac{2(2j+1)}{j+1} u_j, \\\\\nu_m &= \\frac{2(2m-1)}{m} u_{m-1},\n\\end{align*}\n$$\n\nwe have\n\n$$\n\\begin{align*}\nf(j) - f(j+1) &= u_j u_m - u_{j+1} u_{m-1} \\\\\n&= 2 \\left( \\frac{2m-1}{m} - \\frac{2j+1}{j+1} \\right) u_{j+1} u_{m-1} \\\\\n&= 2 \\left( \\frac{(2m-1)(j+1) - (2j+1)m}{m(j+1)} \\right) u_{j+1} u_{m-1} \\\\\n&= 2 \\left( \\frac{m-j-1}{m(j+1)} \\right) u_{j+1} u_{m-1} \\\\\n&\\ge 0,\n\\end{align*}\n$$\n\nif $0 \\leq j < m = n-j$, i.e., $0 \\leq j < n/2$. Thus, $f(j)$ decreases for $0 \\leq j < n/2$.\n\nSince $f(n-j) = f(j)$, $f$ is symmetric and thus increases after $j = n/2$. Therefore,\n\n$$\n\\min\\{f(j) : j = 0, 1, \\dots, n\\} = \\begin{cases} u_{n/2}^2, & \\text{if } n \\text{ is even} \\\\ u_{(n+1)/2}u_{(n-1)/2}, & \\text{if } n \\text{ is odd} \\end{cases}\n$$\n\n(b) To show convexity, note that\n\n$$\n\\begin{align*}\nf(j+1) + f(j-1) - 2f(j) &= u_{j+1}u_{m-1} + u_{j-1}u_{m+1} - 2u_j u_m \\\\\n&= \\frac{2(2j+1)}{j+1}u_{j-1}u_{m-1} + u_{j-1} \\frac{2(2m+1)}{m+1}u_m - 2u_j u_m \\\\\n&= \\frac{4(4j^2-1)}{j(j+1)}u_{j-1}u_{m-1} + u_{j-1} \\frac{4(4m^2-1)}{m(m+1)}u_{m-1} \\\\\n&\\quad - \\frac{8(2j-1)(2m-1)}{j(m)}u_{j-1}u_{m-1} \\\\\n&= 4u_{j-1}u_{m-1} \\left( \\frac{(4j^2-1)}{j(j+1)} + \\frac{(4m^2-1)}{m(m+1)} - \\frac{2(2j-1)(2m-1)}{jm} \\right) \\\\\n&= 4u_{j-1}u_{m-1} \\frac{3m^2+m+3j^2+j-2jm-2}{j(j+1)m(m+1)} \\\\\n&= 4u_{j-1}u_{m-1} \\frac{(m+1)(2m-1)+(j+1)(2j-1)+(j-m)^2}{j(j+1)m(m+1)} \\\\\n&> 0.\n\\end{align*}\n$$\n\nHence, $f(j)$ is convex.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 12775,
    "subject": "Mathematics (Olympiad)",
    "question": "Diagonals of a convex quadrilateral $ABCD$ intersect at $E$. The four points of tangency of the circles $(ABE)$ and $(CDE)$ with their external common tangents lie on a circle $\\omega$. Analogously, the four points of tangency of the circles $(ADE)$ and $(BCE)$ with their external common tangents lie on a circle $\\gamma$. Prove that the centers of $\\omega$ and $\\gamma$ coincide.",
    "options": [],
    "answer": "See solution",
    "solution": "Let us denote the centers of the circumscribed circles of triangles $ABE$, $BCE$, $CDE$, $ADE$ by $O_{AB}$, $O_{BC}$, $O_{CD}$, $O_{AD}$ respectively. Let $T_1, T_2$ be the points of tangency of one of the common tangents with the circumscribed circles of triangles $ABE$ and $CDE$, respectively; denote by $O$ and $T$ the midpoints of segments $O_{AB}O_{CD}$ and $T_1T_2$, respectively (see figure 1). Then, in the right trapezoid $O_{AB}T_1T_2O_{CD}$, the line $OT$ is the midline, hence it is the perpendicular bisector of segment $T_1T_2$. Note that the circle $\\omega$ is symmetric with respect to the line $O_{AB}O_{CD}$, on which the point $O$ also lies, meaning that $O$ is the center of $\\omega$.\n\n![](images/2025-01_p3_data_4312274881.png)\n\nSimilarly, we find that the midpoint of segment $O_{AD}O_{BC}$ is the center of $\\gamma$. Therefore, the statement of the problem is equivalent to the fact that $O_{AB}O_{BC}O_{CD}O_{AD}$ is a parallelogram. To prove this, it suffices to note that $O_{AB}O_{BC}$ and $O_{CD}O_{AD}$ are the perpendicular bisectors of segments $EB$ and $ED$, respectively, hence $O_{AB}O_{BC} \\parallel O_{CD}O_{AD}$; similarly, $O_{AB}O_{AD} \\parallel O_{BC}O_{CD}$, from which the required conclusion follows.\n\n![](images/2025-01_p3_data_dd6c425e1f.png)",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 12776,
    "subject": "Mathematics (Olympiad)",
    "question": "Let there be a $20 \\times 20$ grid where an operation consists of changing the sign of all entries in a selected row or column. If the operation is applied an even number of times to any row or column, it is equivalent to not applying it at all. Thus, we can assume the operation is applied exactly once to some rows and columns, and not at all to the others. Let the operation be applied to $x$ rows and $y$ columns. Then the total number of cells in which the signs are changed is $c = 20x + 20y - 2xy$. In particular, $c$ is even. We can rewrite this as:\n\n$$\n|x - 10| \\cdot |y - 10| = |100 - c/2|.\n$$\n\nGiven that $c$ can take the values $40, 42, 44, 46, 48, 50, 52, 54$ depending on the number of initial minuses changed to pluses, determine which values of $c$ are possible under these constraints.",
    "options": [],
    "answer": "See solution",
    "solution": "If $c$ can be $40, 42, 44, 46, 48, 50, 52, 54$, then the right-hand side of the equation $|x - 10| \\cdot |y - 10| = |100 - c/2|$ becomes $80, 79, 78, 77, 76, 75, 74, 73$ respectively. Since $|x - 10|$ and $|y - 10|$ are both at most $10$, the only possible product is $80 = 8 \\cdot 10$. This corresponds to $c = 40$, which means all initial $7$ minuses keep their positions. Thus, only $c = 40$ is possible.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 12777,
    "subject": "Mathematics (Olympiad)",
    "question": "Let us say that a binary sequence of length $n$ is *acceptable* if it does not contain two consecutive 0s. Let $A_n$ be the number of acceptable binary sequences of length $n$.\n\nLet $B_n$ be the number of *superb* binary sequences of length $n$, where a superb sequence is defined as a binary sequence such that:\n\n1. The second and the second last digits are both 1s.\n2. It contains no subsequence of the form $0, x, 0$ (where $x$ is any digit).\n\nFind the smallest $n$ such that $20$ divides $B_n$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let us say that a binary sequence of length $n$ is *acceptable* if it does not contain two consecutive 0s. Let $A_n$ be the number of acceptable binary sequences of length $n$.\n\n**Lemma**\n\nThe sequence $A_1, A_2, \\dots$ satisfies $A_1 = 2$, $A_2 = 3$, and\n\n$$\nA_{n+1} = A_n + A_{n-1} \\quad \\text{for } n = 2, 3, 4, \\dots\n$$\n\n**Proof**\n\nIt is easy to verify by inspection that $A_1 = 2$ and $A_2 = 3$.\n\nConsider an acceptable sequence $S$ of length $n+1$, where $n \\ge 2$.\n\nIf $S$ starts with a 1, then this 1 can be followed by any acceptable sequence of length $n$. Hence there are $A_n$ acceptable sequences in this case.\n\nIf $S$ starts with a 0, then the next term must be a 1, and the remaining terms can be any acceptable sequence of length $n-1$. Hence there are $A_{n-1}$ acceptable sequences in this case.\n\nPutting it all together yields $A_{n+1} = A_n + A_{n-1}$. $\\square$\n\nReturning to the problem at hand, observe that any superb sequence satisfies the following two properties:\n\n1. The second and the second last digits are both 1s.\n2. It contains no subsequence of the form $0, x, 0$.\n\nMoreover, these two properties completely characterise superb sequences. Property (ii), in particular, motivates us to look at every second term of a superb sequence.\n\n**Case 1:** $n = 2m$ for $m \\ge 2$\n\nFrom (i), the superb sequence is $a_1, 1, a_3, a_4, \\dots, a_{2m-3}, a_{2m-2}, 1, a_{2m}$.\n\nFrom (ii), the subsequences $a_1, a_3, a_5, \\dots, a_{2m-3}$ and $a_4, a_6, \\dots, a_{2m}$ are both acceptable. The first subsequence has $m-1$ terms as does the second. Hence from the lemma we have $B_{2m} = A_{m-1}^2$.\n\n**Case 2:** $n = 2m + 1$ for $m \\ge 3$\n\nFrom (i), the superb sequence is $a_1, 1, a_3, a_4, \\dots, a_{2m-2}, a_{2m-1}, 1, a_{2m+1}$.\n\nFrom (ii), the subsequences $a_1, a_3, \\dots, a_{2m+1}$ and $a_4, a_6, \\dots, a_{2m-2}$ are both acceptable. The first subsequence has $m+1$ terms and the second has $m-2$ terms. Hence from the lemma we have $B_{2m+1} = A_{m+1}A_{m-2}$.\n\nUsing the lemma, it is a simple matter to compute the values of $A_1, A_2, \\dots$ modulo 20 and enter them into the following table. Then the values of $B_{2m}$ and $B_{2m+1}$ are calculated modulo 20 using the two formulas above.\n\n![]( <table><thead><tr><th>m</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>...</th></tr></thead><tbody><tr><th>A<sub>m</sub></th><td>2</td><td>3</td><td>5</td><td>8</td><td>13</td><td>1</td><td>14</td><td>15</td><td>9</td><td>4</td><td>13</td><td>17</td><td>...</td></tr><tr><th>B<sub>2m</sub></th><td>(1)</td><td>4</td><td>9</td><td>5</td><td>4</td><td>9</td><td>1</td><td>16</td><td>5</td><td>1</td><td>16</td><td>9</td><td>...</td></tr><tr><th>B<sub>2m+1</sub></th><td>(3)</td><td>(5)</td><td>16</td><td>19</td><td>5</td><td>12</td><td>15</td><td>9</td><td>16</td><td>15</td><td>13</td><td>0</td><td>...</td></tr></tbody></table> )\n\nThus from the table, $n = 25$ is the smallest $n$ such that $20 \\mid B_n$. $\\square$\n\nHere we are forgetting about whether or not a sequence is superb for the time being.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 12778,
    "subject": "Mathematics (Olympiad)",
    "question": "A computer program reads the numbers displayed, identifies the selected ones, and does one of the following actions:\n\n- If A is pressed, every selected number is changed to its successor.\n- If B is pressed, every selected number is changed to its triple.\n\nOn Andrei and Bogdan's computer screens are displayed the numbers $1, 3, 3^2, 3^3, \\ldots, 3^{19}$.\n\n**a)** Andrei will select 5 numbers and then press A (this is a type A step).\n\nDetermine if there is a succession of type A steps in order to obtain the sum of all displayed numbers to be equal to $2019^{2020}$.\n\n**b)** Bogdan will select 5 numbers and then press B (this is a type B step).\n\nWhat is the minimum number of type B steps needed in order to obtain all the displayed numbers to be equal?",
    "options": [],
    "answer": "See solution",
    "solution": "**a)** At every step of type A, the sum of the displayed numbers increases by 5. The remainder modulo 5 of the sum is invariant.\n\nThe sum of four consecutive powers of 3 is $3^n + 3^{n+1} + 3^{n+2} + 3^{n+3} = 3^n(1 + 3 + 9 + 27) = 40 \\cdot 3^n$, therefore initially we have the remainder zero modulo 5 and it is impossible to obtain $2019^{2020}$.\n\n**b)** At every step, the product of the displayed numbers is multiplied by $3^5$.\n\nInitially, the product is $3^{1+2+\\ldots+19} = 3^{190}$ and after $n$ type B steps the product will be $3^{190+5n}$.\n\nSuppose that in $n$ steps the numbers are equal to a power of 3, at least $3^{19}$. The product of the 20 numbers will be $3^{20p}$, where $p \\geq 19$.\n\nWe must have $3^{20p} = 3^{190+5n}$, so $n + 38 = 4p$. Since $p \\geq 19$, we have $n \\geq 38$.\n\nNext, we provide a succession of 38 type B steps to display in the end equal numbers.\n\nIn 15 steps, the numbers $1, 3, 3^2, 3^3, 3^4$ become $3^{15}, 3^{16}, 3^{17}, 3^{18}, 3^{19}$. In the next 10 steps $3^5, 3^6, 3^7, 3^8, 3^9$ become $3^{15}, 3^{16}, 3^{17}, 3^{18}, 3^{19}$ and the numbers $3^{10}, 3^{11}, 3^{12}, 3^{13}, 3^{14}$ become $3^{15}, 3^{16}, 3^{17}, 3^{18}, 3^{19}$ in 5 steps.\n\nThus in 30 steps, numbers $3^{15}, 3^{16}, 3^{17}, 3^{18}, 3^{19}$ are displayed each 4 times.\n\nNumbers $3^{15}, 3^{15}, 3^{15}, 3^{15}, 3^{16}$ become $3^{18}, 3^{18}, 3^{18}, 3^{18}, 3^{19}$ after three steps.\n\nNumbers $3^{16}, 3^{16}, 3^{16}, 3^{17}, 3^{17}$ become $3^{18}, 3^{18}, 3^{18}, 3^{19}, 3^{19}$ in 2 steps.\n\nNumbers $3^{17}, 3^{17}, 3^{18}, 3^{18}, 3^{18}$ become $3^{18}, 3^{18}, 3^{19}, 3^{19}, 3^{19}$ in one step.\n\nTherefore, after 36 steps on Bogdan's computer screen, there are 10 of $3^{19}$ and 10 of $3^{18}$.\n\nIn the last 2 steps, we make all numbers equal to $3^{19}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 12779,
    "subject": "Mathematics (Olympiad)",
    "question": "En el cuadrilátero convexo $ABCD$, se tiene $\\angle ABC = \\angle CDA = 90^\\circ$. La perpendicular a $BD$ desde $A$ corta a $BD$ en el punto $H$. Los puntos $S$ y $T$ están en los lados $AB$ y $AD$, respectivamente, y son tales que $H$ está dentro del triángulo $SCT$ y\n\n$$\n\\angle CHS - \\angle CSB = 90^\\circ,\n$$\n\n$$\n\\angle THC - \\angle DTC = 90^\\circ.\n$$\n\nDemostrar que la recta $BD$ es tangente a la circunferencia circunscrita del triángulo $TSH$.",
    "options": [],
    "answer": "See solution",
    "solution": "Claramente, $B$ y $D$ están en la circunferencia de diámetro $AC$. Luego la altura $AH$ del triángulo $ABD$ mide $AH = \\frac{AB \\cdot AD}{AC}$.\n\nConsideremos la circunferencia circunscrita a $CSH$, con centro en $O_C$. Por ángulo central, tenemos que\n\n$$\n\\angle CO_CS = 2(180^{\\circ} - \\angle CHS) = 180^{\\circ} - 2\\angle CSB.\n$$\n\nComo $CO_CS$ es isósceles en $O_C$, $\\angle CO_CS = 180^{\\circ} - 2\\angle CSO_C$, es decir, $O_C$ está en la semirrecta por $B$ con origen en $S$. La mediatriz de $SH$ es también la bisectriz de $\\angle SO_CH$ por ser $SO_CH$ isósceles en $O_C$, y corta a la recta $AP$ en el punto $O$. Definamos análogamente el circuncentro $O_B$ de $CHT$, que está en la semirrecta por $D$ con origen en $T$, y el punto $O'$ que es el corte de la mediatriz de $HT$ y bisectriz de $TO_BH$ con la recta $AP$. Claramente, el problema es equivalente a demostrar que $O = O'$, pues en ese caso el circuncentro de $HST$ estaría en la perpendicular $AH$ a $BD$, con lo que $BD$ sería en efecto la tangente a la circunferencia circunscrita a $HST$ en $H$. Por el teorema de la bisectriz, este resultado es a su vez equivalente a\n\n$$\n\\frac{OA}{HC} = \\frac{OB}{HB}.\n$$\n\nComo $\\angle OAO_C = \\angle HAO_C = \\angle HAB$, y $\\cos \\angle HAB = \\frac{AH}{AB}$ por ser $AHB$ rectángulo en $H$, tenemos por el teorema del coseno que\n\n$$\nHO_C^2 = AH^2 + AO_C^2 - 2AH^2 \\cdot \\frac{AO_C}{AB}.\n$$\n\nAl mismo tiempo, por ser $ABC$ rectángulo en $B$, se tiene que $\\angle CAO_C = \\angle CAB$ y $\\cos \\angle CAB = \\frac{AB}{AC}$, luego nuevamente por el teorema del coseno, se tiene\n\n$$\nCO_C^2 = AC^2 + AO_C^2 - 2AB \\cdot AO_C.\n$$\n\nDenotando como es habitual $AB = a$, $DA = d$, y llamando además $AH = h$, $AC = 2R$ por ser el diámetro de la circunferencia circunscrita a $ABD$, y al ser $CO_C = HO_C$ por definición, se tiene\n\n$$\nAO_C = \\frac{16R^4 - a^2d^2}{2a(4R^2 - d^2)},\n$$\n\ndonde además se ha usado que $2Rh = ad$. Entonces,\n\n$$\nCO_C^2 = \\frac{(16R^4 - a^2d^2)^2 - 4a^2d^2(4R^2 - a^2)(4R^2 - d^2)}{4a^2(4R^2 - d^2)^2},\n$$\n\ncon lo que\n\n$$\n\\left(\\frac{O_C H}{O_C A}\\right)^2 = \\frac{(16R^4 - a^2d^2)^2 - 4a^2d^2(4R^2 - a^2)(4R^2 - d^2)}{(16R^4 - a^2d^2)^2}.\n$$\n\nNótese que esta expresión es simétrica en $a$, $d$, luego análogamente obtendremos la misma expresión para $\\frac{O_B H}{O_B A}$, como queríamos demostrar.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 12780,
    "subject": "Mathematics (Olympiad)",
    "question": "In an unopened pack, there are red, green, and blue candies that weigh $2\\ \\mathrm{g}$, $5\\ \\mathrm{g}$, and $25\\ \\mathrm{g}$ respectively. There are the same number of candies of each color. After Mary eats some candies from that pack, the remaining candies weigh exactly $787\\ \\mathrm{g}$ in total. Find the least number of candies that Mary could have eaten.",
    "options": [],
    "answer": "See solution",
    "solution": "Let there be $n$ candies of each color in the unopened pack. Since one red, one green, and one blue candy weigh $2 + 5 + 25 = 32$ grams in total, the total weight of all candies in the unopened pack is $32n$ grams. Therefore, Mary eats $32n - 787$ grams of candies.\n\nWe want to minimize the number of candies eaten. Consider possible values for $n$:\n\n- If $n = 25$, then $32 \\times 25 = 800$ grams, so Mary ate $800 - 787 = 13$ grams. The minimum number of candies to reach $13$ grams is $1$ green ($5$ g) and $4$ red ($2 \\times 4 = 8$ g), totaling $5$ candies.\n- If $n = 26$, then $32 \\times 26 = 832$ grams, so Mary ate $832 - 787 = 45$ grams. The minimum number of candies to reach $45$ grams is $1$ blue ($25$ g) and $4$ green ($5 \\times 4 = 20$ g), totaling $5$ candies.\n- If $n = 27$, then $32 \\times 27 = 864$ grams, so Mary ate $864 - 787 = 77$ grams. The minimum number of candies to reach $77$ grams is $3$ blue ($25 \\times 3 = 75$ g) and $1$ red ($2$ g), totaling $4$ candies.\n- If $n = 28$, then $32 \\times 28 = 896$ grams, so Mary ate $896 - 787 = 109$ grams. Any $4$ candies weigh at most $4 \\times 25 = 100$ grams, so at least $5$ candies are needed.\n\nFor $n > 28$, the total weight eaten increases, so more candies are needed.\n\nThus, the least number of candies that Mary could have eaten is $4$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 12781,
    "subject": "Mathematics (Olympiad)",
    "question": "Consider an acute triangle $ABC$ with area $S$. Let $CD \\perp AB$ ($D \\in AB$), $DM \\perp AC$ ($M \\in AC$), and $EN \\perp BC$ ($N \\in BC$). Denote by $H_1$ and $H_2$ the orthocentres of the triangles $MNC$ and $MND$ respectively. Find the area of the quadrilateral $AH_1BH_2$ in terms of $S$.\n\n![](images/MACEDONIAN_MATHEMATICAL_SOCIETY_p14_data_fa2ee95289.png)",
    "options": [],
    "answer": "See solution",
    "solution": "Let $O$, $P$, $K$, $R$, and $T$ be the midpoints of the segments $CD$, $MN$, $CN$, $CH_1$, and $MH_1$, respectively. From $\\triangle MNC$, we have $\\overline{PK} = \\frac{1}{2}\\overline{MC}$ and $PK \\parallel MC$.\n\nAnalogously, from $\\triangle MH_1C$, we have $\\overline{TR} = \\frac{1}{2}\\overline{MC}$ and $TR \\parallel MC$. Consequently, $\\overline{PK} = \\overline{TR}$ and $PK \\parallel TR$. Also, $\\overline{OK} \\parallel \\overline{DN}$ (from $\\triangle CDN$), and since $\\overline{DN} \\perp \\overline{BC}$ and $\\overline{MH_1} \\perp \\overline{BC}$, it follows that $\\overline{TH_1} \\parallel \\overline{OK}$. Since $O$ is the circumcenter of $\\triangle CMN$, $\\overline{OP} \\perp \\overline{MN}$. Thus, $CH_1 \\perp MN$ implies $\\overline{OP} \\parallel CH_1$. We conclude $\\triangle TRH_1 \\cong \\triangle KPO$ (they have parallel sides and $\\overline{TR} = \\overline{PK}$), hence $\\overline{RH_1} = \\overline{PO}$, i.e., $\\overline{CH_1} = 2\\overline{PO}$ and $CH_1 \\parallel PO$.\n\nAnalogously, $\\overline{DH_2} = 2\\overline{PO}$ and $DH_2 \\parallel PO$. From $\\overline{CH_1} = 2\\overline{PO} = \\overline{DH_2}$ and $CH_1 \\parallel PO \\parallel DH_2$, the quadrilateral $CH_1H_2D$ is a parallelogram, thus $\\overline{H_1H_2} = \\overline{CD}$ and $H_1H_2 \\parallel CD$. Therefore, the area of the quadrilateral $AH_1BH_2$ is $$\\frac{\\overline{AB} \\cdot \\overline{H_1H_2}}{2} = \\frac{\\overline{AB} \\cdot \\overline{CD}}{2} = S.$$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 12782,
    "subject": "Mathematics (Olympiad)",
    "question": "There are 2025 people and 66 given colors. Each person has 66 balls, one of each given color, and the total weight of these 66 balls is 1.\n\nFind the minimal real number $C$ satisfying the following condition: Regardless of how the balls are weighted, one can always select exactly one ball from each person such that among all selected 2025 balls, the total weight of balls of each color does not exceed $C$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let us generalize the problem by replacing 2025 with $n$ and 66 with $m$. For any positive integers $m \\leq n$, define:\n\n$$\nf_m(n) = \\min_{\\substack{a_1 + a_2 + \\dots + a_m = n \\\\ a_1, a_2, \\dots, a_m > 0}} \\left( \\frac{1}{a_1} + \\frac{1}{a_2} + \\dots + \\frac{1}{a_m} \\right).\n$$\n\n**Lemma:** For $n$ people $P_1, P_2, \\dots, P_n$, if for each person the total weight of their $m$ balls is at most $f_m(n + m - 1) \\cdot A$, then one can select one ball from each person such that among the selected $n$ balls, the total weight of balls of each color is at most $A$.\n\nWe prove the lemma by induction on $m$.\n\n- For $m = 1$, $f_1(n) = \\frac{1}{n}$, and the statement holds.\n- Assume the statement holds for $m-1$, and consider the case with $m$ colors.\n\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 $x_1 \\leq x_2 \\leq \\dots \\leq x_n$. Choose $k$ such that\n\n$$\nx_1 + x_2 + \\dots + x_k \\leq A, \\quad x_1 + x_2 + \\dots + x_k + x_{k+1} > A.\n$$\n\nThen $x_{k+1} > \\frac{A}{k+1}$. Therefore, for people $P_{k+1}, \\dots, P_n$ ($n-k$ people), the total weight of their $m-1$ balls (colors 2, 3, ..., $m$) is at most\n\n$$\nf_m(n + m - 1) \\cdot A - \\frac{A}{k+1} \\leq f_{m-1}(n - k + m - 2) \\cdot A.\n$$\n\nThis holds because\n\n$$\n\\min_{b_1 + \\cdots + b_{m-1} = n - k + m - 2} \\left( \\frac{1}{b_1} + \\cdots + \\frac{1}{b_{m-1}} \\right) + \\frac{1}{k+1} \\geq \\min_{a_1 + \\cdots + a_m = n + m - 1} \\left( \\frac{1}{a_1} + \\cdots + \\frac{1}{a_m} \\right).\n$$\n\nBy the induction hypothesis, we can select one ball (from colors 2, 3, ..., $m$) from each of $P_{k+1}, \\dots, P_n$ 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\nSuppose $f_m(n + m - 1) = \\frac{1}{a_1} + \\cdots + \\frac{1}{a_m}$ where $a_1 + \\cdots + a_m = n + m - 1$.\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$$\n\\frac{1}{a_k} \\times \\frac{1}{\\frac{1}{a_1} + \\dots + \\frac{1}{a_m}} > \\frac{1}{a_k} \\times C,\n$$\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.\n\nTherefore, the minimal $C$ is $C_{\\min} = \\frac{1}{f_m(n + m - 1)}$.\n\nLet $n = mq + r$ where $1 \\leq r \\leq m$, 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\n$$\nC_{\\min} = \\frac{1}{f_m(n + m - 1)} = \\frac{1}{\\frac{r-1}{q+2} + \\frac{m+1-r}{q+1}}.\n$$\n\nIn particular, when $n = 2025$ and $m = 66$, the required $C$ is $\\frac{248}{517}$.\n\n$\\boxed{\\dfrac{248}{517}}$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 12783,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all triples of non-negative integers $a, b, c$ for which the number\n\n$$\n\\frac{(a+b)^4}{c} + \\frac{(b+c)^4}{a} + \\frac{(c+a)^4}{b}\n$$\n\nis an integer and $a + b + c$ is prime.",
    "options": [],
    "answer": "See solution",
    "solution": "The possible triples are $(1, 1, 1)$, $(1, 2, 2)$, and $(2, 3, 6)$.\n\nLet $p = a + b + c$, so $a + b = p - c$, $b + c = p - a$, and $c + a = p - b$. The expression becomes:\n\n$$\n\\frac{(p-c)^4}{c} + \\frac{(p-a)^4}{a} + \\frac{(p-b)^4}{b}\n$$\n\nExpanding, we see that $p^4\\left(\\frac{1}{a} + \\frac{1}{b} + \\frac{1}{c}\\right)$ must also be an integer. Since $a, b, c$ are not divisible by $p$ (as $a + b + c = p$ is prime), $\\frac{1}{a} + \\frac{1}{b} + \\frac{1}{c}$ must be a non-negative integer. This only occurs for $(1, 1, 1)$, $(1, 2, 2)$, and $(2, 3, 6)$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 12784,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all functions $f: \\mathbb{R} \\to \\mathbb{R}$ satisfying\n\n$$\nf(x + y f(x^2)) = f(x) + x f(xy)\n$$\n\nfor all real numbers $x$ and $y$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $P(x, y)$ be the assertion $f(x + y f(x^2)) = f(x) + x f(xy)$.\n\n$P(1, 0)$ yields $f(0) = 0$.\n\nIf there exists $x_0 \\neq 0$ with $f(x_0^2) = 0$, then $P(x_0, y)$ gives $f(x_0 y) = 0$ for all $y \\in \\mathbb{R}$. Since $x_0 \\neq 0$, any $c \\in \\mathbb{R}$ can be written as $x_0 y$, so $f(c) = 0$. Thus, the zero function is a solution.\n\nNow assume $f(x^2) \\neq 0$ for all $x \\neq 0$.\n\nBy $P(1, y)$:\n$$\nf(1 + y f(1)) = f(1) + f(y).\n$$\nIf $f(1) \\neq 1$, there exists $y$ with $1 + y f(1) = y$, so $f(1) = 0$, contradicting $f(x^2) \\neq 0$ for $x \\neq 0$. Thus, $f(1) = 1$.\n\nConsider $P(x, -x / f(x^2))$ for $x \\neq 0$:\n$$\nf(x) = -x f\\left(-\\frac{x^2}{f(x^2)}\\right) \\quad \\forall x \\neq 0. \\qquad (1)\n$$\nReplacing $x$ by $-x$ in (1), we get $f(x) = -f(-x)$ for $x \\neq 0$. Since $f(0) = 0$, $f$ is odd: $f(x) = -f(-x)$ for all $x$.\n\n$P(x, -y)$ gives\n$$\nf(x - y f(x^2)) = f(x) - x f(xy)\n$$\nAdding $P(x, y)$ and $P(x, -y)$:\n$$\nf(x + y f(x^2)) + f(x - y f(x^2)) = 2 f(x) \\quad \\forall x, y.\n$$\nLet $y = x / f(x^2)$ for $x \\neq 0$:\n$$\nf(2x) = 2 f(x) \\quad \\forall x.\n$$\nSo,\n$$\nf(x + y f(x^2)) + f(x - y f(x^2)) = f(2x) \\quad \\forall x, y.\n$$\nFor any $u, v$ with $u \\neq -v$, choose\n$$\nx = \\frac{u + v}{2}, \\quad y = \\frac{v - u}{2 f\\left(\\left(\\frac{u + v}{2}\\right)^2\\right)}\n$$\nThen\n$$\nf(u) + f(v) = f(u + v). \\qquad (2)\n$$\nSince $f$ is odd, (2) holds for $u = -v$ too, so\n$$\nf(x) + f(y) = f(x + y) \\quad \\forall x, y.\n$$\nThus, $f$ is additive.\n\nFrom $P(x, y)$:\n$$\nf(y f(x^2)) = x f(xy) \\quad \\forall x, y.\n$$\nSo,\n$$\nf(f(x^2)) = x f(x) \\quad \\forall x. \\qquad (3)\n$$\nAnd\n$$\nf(x f(x^2)) = x f(x^2) \\quad \\forall x. \\qquad (4)\n$$\nUsing (2) and (3):\n$$\n\\begin{aligned}\nx f(x) + y f(y) + x f(y) + y f(x) &= (x + y)(f(x) + f(y)) \\\\\n&= f(f((x + y)^2)) \\\\\n&= f(f(x^2) + 2 x y + y^2) \\\\\n&= f(f(x^2)) + f(y^2) + f(2 x y) \\\\\n&= f(x^2) + f(y f(y)) + f(f(2 x y)) \\\\\n&= x f(x) + y f(y) + f(f(2 x y)) \\\\\n&= x f(x) + y f(y) + 2 f(f(x y))\n\\end{aligned}\n$$\nSo,\n$$\n2 f(f(x y)) = x f(y) + y f(x) \\quad \\forall x, y. \\qquad (5)\n$$\nUsing (5):\n$$\n2 f(f(x)) = x + f(x) \\quad \\forall x. \\qquad (6)\n$$\nUsing (3) and (6):\n$$\n2 x f(x) = 2 f(f(x^2)) = x^2 + f(x^2) \\quad \\forall x. \\qquad (7)\n$$\nLet $y = f(x^2)$ in (5):\n$$\n2 f(f(x f(x^2))) = x f(f(x^2)) + f(x^2) f(x) \\quad \\forall x. \\qquad (8)\n$$\nUsing (4), $f(f(x f(x^2))) = x f(x^2)$, and by (3) and (8):\n$$\n2 x f(x^2) = x^2 f(x) + f(x^2) f(x) \\quad \\forall x. \\qquad (9)\n$$\nFrom (7), $f(x^2) = 2 x f(x) - x^2$. Substitute into (9):\n$$\n2 x (2 x f(x) - x^2) = x^2 f(x) + (2 x f(x) - x^2) f(x)\n$$\nwhich simplifies to\n$$\n2 x (x - f(x))^2 = 0 \\quad \\forall x.\n$$\nThus, $f(x) = x$ for all $x$, which satisfies the original equation.\n\nTherefore, all solutions are $f(x) = 0$ and $f(x) = x$ for all $x \\in \\mathbb{R}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 12785,
    "subject": "Mathematics (Olympiad)",
    "question": "It is given that the bookshelf can fit 9 of the same thick books, but the 10th one will not fit anymore. Similarly, it can hold 15 of the same thin books, but the 16th will not fit anymore. Is it possible for that shelf to hold simultaneously:\n\n- a) 7 thick and 5 thin books?\n- b) 6 thick and 6 thin books?",
    "options": [],
    "answer": "See solution",
    "solution": "a) See solution to **Problem 2b) 7 grade**.\n\nb) Let us denote the length of the shelf by $S$, the width of the thick book by $x$, and the width of the thin book by $y$. Then, we have the conditions:\n\n$$\n9x \\leq S < 10x \\quad \\text{and} \\quad 15y \\leq S < 16y\n$$\n\nwhich implies\n\n$$\n\\frac{1}{10}S < x \\leq \\frac{1}{9}S \\quad \\text{and} \\quad \\frac{1}{16}S < y \\leq \\frac{1}{15}S.\n$$\n\nWe can re-write these as:\n\n$$\n\\frac{72}{720}S < x \\leq \\frac{80}{720}S \\quad \\text{and} \\quad \\frac{45}{720}S < y \\leq \\frac{48}{720}S.\n$$\n\nSelect $x = \\frac{73}{720}S$ and $y = \\frac{46}{720}S$. Then,\n\n$$\n6x + 6y = \\frac{438}{720}S + \\frac{276}{720}S = \\frac{714}{720}S < S.\n$$\n\nThus, with these values, 6 thick and 6 thin books will fit on the shelf.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 12786,
    "subject": "Mathematics (Olympiad)",
    "question": "Solve the inequality\n\n$$\n\\log_2(x^{12} + 3x^{10} + 5x^8 + 3x^6 + 1) < 1 + \\log_2(x^4 + 1).\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "As\n\n$$\n1 + \\log_2(x^4 + 1) = \\log_2(2x^4 + 2),\n$$\n\nand $\\log_2 y$ is monotonically increasing over $(0, +\\infty)$, the given inequality is equivalent to\n\n$$\nx^{12} + 3x^{10} + 5x^8 + 3x^6 + 1 < 2x^4 + 2\n$$\n\nor\n\n$$\nx^{12} + 3x^{10} + 5x^8 + 3x^6 - 2x^4 - 1 < 0.\n$$\n\nIt can be rewritten as\n\n$$\n\\begin{aligned}\n& x^{12} + x^{10} - x^8 \\\\\n& \\quad + 2x^{10} + 2x^8 - 2x^6 \\\\\n& \\quad + 4x^8 + 4x^6 - 4x^4 \\\\\n& \\quad + x^6 + x^4 - x^2 \\\\\n& \\quad + x^4 + x^2 - 1 < 0.\n\\end{aligned}\n$$\n\nThat is to say,\n\n$$\n(x^8 + 2x^6 + 4x^4 + x^2 + 1)(x^4 + x^2 - 1) < 0.\n$$\n\nThen we have $x^4 + x^2 - 1 < 0$. It follows that $x^2 < \\frac{-1+\\sqrt{5}}{2}$, i.e.\n\n$$\n-\\sqrt{\\frac{-1+\\sqrt{5}}{2}} < x < \\sqrt{\\frac{-1+\\sqrt{5}}{2}}.\n$$\n\nSo the solution set is $\\left(-\\sqrt{\\frac{-1+\\sqrt{5}}{2}}, \\sqrt{\\frac{-1+\\sqrt{5}}{2}}\\right)$.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 12787,
    "subject": "Mathematics (Olympiad)",
    "question": "The bisector of the internal angle at vertex $B$ of triangle $ABC$ and the line through point $C$ perpendicular to the side $BC$ intersect at point $D$. Let $M$ and $N$ be the midpoints of the line segments $BC$ and $BD$, respectively. Given that $N$ lies on the side $AC$ and $\\frac{AM}{BC} = \\frac{CD}{BD}$, find all possibilities of what the sizes of the angles of the triangle $ABC$ can be.",
    "options": [],
    "answer": "See solution",
    "solution": "By assumptions, $MN$ is the midsegment of triangle $BCD$ such that $MN \\parallel CD$ (see figure below). Thus $\\angle BMN = \\angle BCD = 90^\\circ$, implying $NB = NC$. Denote $\\gamma = \\angle ABN = \\angle NBC$. The assumption $\\frac{AM}{BC} = \\frac{CD}{BD}$ is equivalent to the assertion that the length of $AM$ equals the height drawn from the right angle of the right triangle $BCD$. Let $E$ be the foot of this altitude; let $K$ be the projection of $M$ to $BD$ and $L$ the point of intersection of lines $MK$ and $AB$ (see figure below). As $M$ bisects $BC$ and $MK \\parallel CE$, $MK$ is the midsegment of triangle $BEC$, implying $CE = 2MK$. As $BK$ is an altitude and an angle bisector of triangle $BML$, it is also a median, i.e., $ML = 2MK$. Thus $MA = CE = ML$, whence either $L = A$ or the triangle $AML$ is isosceles with apex angle at $M$.\n\nConsider the case $L = A$ (see figure below). As $BD$ is the axis of symmetry of triangle $BML$, we have $\\angle BAC = \\angle BAN = \\angle BLN = \\angle BMN = 90^\\circ$. From the right triangle $ABC$, we get $3\\gamma = 90^\\circ$, implying $\\gamma = 30^\\circ$. The sizes of internal angles of the triangle $ABC$ are $30^\\circ$, $60^\\circ$, $90^\\circ$.\n\nIt remains to study the case $L \\neq A$, $ML = MA$; as the angle $BLM$ is acute, $A$ lies on line segment $BL$. Let $J$ be the foot of the altitude drawn from vertex $M$ of the triangle $MCN$ (see figure below). Since the triangle $BNC$ is isosceles, triangles $BMN$ and $CNM$ are equal, whence $JM = KM = \\frac{1}{2}LM = \\frac{1}{2}AM$. In the right triangle $AJM$, leg is twice shorter than the hypotenuse, implying $\\angle JAM = 30^\\circ$. On the other hand, $\\angle JAM = \\angle LAM - \\angle LAC = \\angle ALM - \\angle LAC = \\angle KMB - (\\angle ABC + \\angle BCA) = (90^\\circ - \\gamma) - 3\\gamma = 90^\\circ - 4\\gamma$. Hence $90^\\circ - 4\\gamma = 30^\\circ$, implying $\\gamma = 15^\\circ$. The sizes of internal angles of the triangle $ABC$ are $15^\\circ$, $30^\\circ$, $135^\\circ$.\n\n![](images/EST_ABooklet_2020_p16_data_e872c8ab5a.png)\n![](images/EST_ABooklet_2020_p17_data_cdc7488b44.png)\n![](images/EST_ABooklet_2020_p17_data_7a103e2e43.png)\n![](images/EST_ABooklet_2020_p17_data_c204397780.png)\n![](images/EST_ABooklet_2020_p17_data_32392c212c.png)",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 12788,
    "subject": "Mathematics (Olympiad)",
    "question": "For an integer $n > 1$, let $gpf(n)$ denote the greatest prime factor of $n$. A *strange pair* is an unordered pair of distinct primes $p$ and $q$ such that $\\{p, q\\} = \\{gpf(n), gpf(n + 1)\\}$ for no integer $n > 1$. Prove that there exist infinitely many strange pairs.",
    "options": [],
    "answer": "See solution",
    "solution": "We show that there are infinitely many strange pairs of the form $\\{2, q\\}$ where $q$ is an odd prime.\n\nThe lemma below provides a sufficient condition for such a pair to be strange. For an odd prime $q$, let $ord_q(2)$ denote the multiplicative order of $2$ modulo $q$, i.e., the least positive integer $s$ satisfying $q \\mid 2^s - 1$.\n\n**Lemma.** If some primes $2 < q_1 < q_2$ satisfy $ord_{q_1}(2) = ord_{q_2}(2)$, then $\\{2, q_1\\}$ is a strange pair.\n\n*Proof.* Arguing indirectly, suppose first that $2 = gpf(n)$ and $q_1 = gpf(n+1)$; in particular, $n = 2^k$ for some positive integer $k$, and $q_1 \\mid 2^k+1$. This yields $q_1 \\mid 2^{2k}-1$, so $ord_{q_2}(2) = ord_{q_1}(2) \\mid 2k$. Therefore, $q_2 \\mid 2^{2k}-1 = (2^k-1)(2^k+1)$, but $q_2 \\nmid 2^k-1$, hence $q_2 \\mid 2^k+1$. So $gpf(n+1) \\ge q_2$, which is a contradiction.\n\nSimilarly, but easier, if $2 = gpf(n+1)$ and $q_1 = gpf(n)$, then $n+1 = 2^k$, so $ord_{q_2}(2) = ord_{q_1}(2) \\mid k$ and hence $q_2 \\mid 2^k-1$. Therefore, $gpf(n+1) \\ge q_2$, a contradiction. $\\Box$\n\nIt remains to show that there exist infinitely many disjoint pairs of primes $q_1 < q_2$ satisfying the conditions in the lemma.\n\nLet $p = 2r - 1 > 5$ be a prime, and let $N = 2^{2p} + 1$. We prove that:\n\n1. $N$ has at least two distinct prime factors greater than $5$;\n2. $ord_q(2) = 4p$ for every prime factor $q > 5$ of $N$.\n\nThus, every prime $p > 5$ provides a pair of odd primes satisfying the conditions in the lemma. Moreover, (2) shows that distinct primes $p > 5$ provide disjoint such pairs, whence the conclusion.\n\nTo prove (1), notice that $3 \\nmid N$, and write $N = (4+1)(4^{p-1}-4^{p-2}+\\cdots+1) \\equiv 5p \\pmod{25}$, to infer that $25 \\nmid N$.\n\nNext, write $N = (2^p+1)^2 - 2^{p+1} = (2^p - 2^r + 1)(2^p + 2^r + 1)$. The two factors are coprime (since they are odd, and their difference is $2^{r+1}$), and each is larger than $5$. Hence each has a prime factor greater than $5$. This establishes (1).\n\nTo prove (2), consider a prime factor $q > 5$ of $N$, and notice that $ord_q(2) \\mid 4p$, since $q \\mid N \\mid 2^{4p} - 1$. If $ord_q(2) < 4p$, then either $ord_q(2) \\mid 2p$ or $ord_q(2) \\mid 4$. The former is impossible due to $2^{2p} - 1 = N - 2 \\equiv -2 \\pmod q$, the latter — due to $q \\nmid 15 = 2^4 - 1$. This establishes (2) and completes the proof.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 12789,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $a$, $b$, $c$, $d$ be positive numbers. Prove that\n$$\n(a^2 + b^2 + c^2 + d^2)^2 \\\\geq (a+b)(b+c)(c+d)(d+a).\n$$\n\nWhen does equality hold?",
    "options": [],
    "answer": "See solution",
    "solution": "By the inequality between the arithmetic and geometric means, we have\n$$\n(a + b)(b + c)(c + d)(d + a) \\leq \\left( \\frac{(a + b) + (b + c) + (c + d) + (d + a)}{4} \\right)^4 = 2^4 \\left( \\frac{a + b + c + d}{4} \\right)^4.\n$$\n\nBy the inequality between the quadratic and arithmetic means, we have\n$$\n2^4 \\left( \\frac{a+b+c+d}{4} \\right)^4 \\leq 2^4 \\left( \\frac{a^2+b^2+c^2+d^2}{4} \\right)^2 = (a^2+b^2+c^2+d^2)^2,\n$$\n\nas required.\n\nIn the second inequality, equality holds if and only if $a = b = c = d$, and in that case, equality also holds in the original inequality. Therefore, equality holds if and only if $a = b = c = d$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 12790,
    "subject": "Mathematics (Olympiad)",
    "question": "For each rational number $r$, consider the statement:\n\nIf $x$ is a real number such that $x^2 - rx$ and $x^3 - rx$ are rational numbers, then $x$ is rational as well.\n\n1. Prove the statement for $r \\geq \\frac{4}{3}$ and for $r \\leq 0$.\n2. Let $p, q$ be different odd primes such that $3p < 4q$. Show the statement is false for $r = \\frac{p}{q}$.",
    "options": [],
    "answer": "See solution",
    "solution": "a) Let $s = x^2 - rx$ and $t = x^3 - rx$ be rational. Then\n\n$$\nx^2 = s + rx,\n$$\n\n$$\nx^3 = x^2 x = (s + rx)x = sx + rx^2 = sx + r(s + rx) = (r^2 + s)x + rs,\n$$\n\nso\n\n$$\nt = x^3 - rx = ((r^2 + s)x + rs) - rx = (r^2 - r + s)x + rs.\n$$\n\nIf $(r^2 - r + s) \\neq 0$ (i.e., $x^2 - rx + r^2 - r \\neq 0$), then\n\n$$\nx = \\frac{t - rs}{r^2 - r + s}\n$$\n\nis rational.\n\nThus, the statement holds if the equation\n\n$$\nx^2 - rx + r^2 - r = 0\n$$\n\nhas no irrational roots. If (1) has a rational root, then $s$ and $t$ are rational as well:\n\n$$\ns = x^2 - rx = r - r^2, \\quad t = rs = r(r - r^2).\n$$\n\nThe discriminant $D = r(4 - 3r)$. If $D \\leq 0$, (1) has no real solutions or a rational solution $x = \\frac{r}{2}$. Since $D \\leq 0$ if and only if $r \\geq \\frac{4}{3}$ or $r \\leq 0$, the statement is true in these cases.\n\nb) By (a), it suffices to show $D > 0$ and $\\sqrt{D}$ is irrational. For $r = \\frac{p}{q}$:\n\n$$\nD = r(4 - 3r) = \\frac{p}{q}\\left(4 - \\frac{3p}{q}\\right) = \\frac{p(4q - 3p)}{q^2} > 0.\n$$\n\nSince $p \\nmid 4q$, $p(4q - 3p)$ is not a perfect square, so $\\sqrt{D}$ is irrational.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 12791,
    "subject": "Mathematics (Olympiad)",
    "question": "From the digits 3, 7, 1, 9, 0, and 4, Mila formed the largest and the smallest six-digit numbers using each digit exactly once in each number. Then she divided their difference by 9. Which number did she get?",
    "options": [],
    "answer": "See solution",
    "solution": "The largest six-digit number that can be formed from these digits is $974310$, and the smallest is $103479$. Their difference is $$974310 - 103479 = 870831.$$ Dividing this difference by $9$ gives $$\\frac{870831}{9} = 96759.$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 12792,
    "subject": "Mathematics (Olympiad)",
    "question": "The inscribed circle of an acute triangle $ABC$ touches sides $AB$ and $BC$ at points $C_1$ and $A_1$, respectively. Let $M$ be the midpoint of side $AC$, and $N$ be the midpoint of arc $ABC$ of the circumcircle of $ABC$. Also, let $P$ be the projection of point $M$ onto the segment $A_1C_1$. Prove that the points $I$, $P$, and $N$ lie on the same line.\n\n![](images/UkraineMO2019_booklet_p37_data_231fb47b95.png)",
    "options": [],
    "answer": "See solution",
    "solution": "Let the second intersection point of the line $BI$ with the circumcircle of $\\triangle ABC$ be $W$.\n\nClearly, $W$ is the midpoint of the smaller arc $AC$, and points $M$, $W$, and $N$ lie on the same line.\n\nIt is clear that $A_1C_1 \\perp BI$, implying $PM \\parallel BW$ and $\\angle PMN = \\angle BWN$.\n\nHence, it is enough to prove that triangles $PMN$ and $IWN$ are similar (which will imply that $\\angle PNM = \\angle INW$ and the desired collinearity). In order to prove the similarity, we are going to show that $\\frac{NM}{NW} = \\frac{PM}{IW}$.\n\nLet $\\angle B = 2\\beta$, then $\\angle MNA = \\beta$ and\n\n$$\n\\frac{NM}{NW} = \\frac{PM}{IW} \\cdot \\frac{NA}{NW} = \\cos^2 \\beta.\n$$\n\nConsidering segment $PM$ yields that $PM$ is a midline of trapezoid $AXYC$, where $X$, $Y$ are the projections of $A$ and $C$ onto the line $A_1C_1$, respectively. Going further, we get\n\n$$\n2PM = CY + XA = CA_1 \\sin YA_1C + C_1 \\sin XC_1A = (CA_1 + AC_1)\\cos \\beta = AC\\cos \\beta,\n$$\nbecause $\\angle YA_1C = \\angle XC_1A = 90^\\circ - \\beta$, and $CA_1 + AC_1 = AC$, since $A_1$ and $C_1$ are the points where the inscribed circle touches the sides of $\\triangle ABC$.\n\nFinally, from the trillium theorem, $IW = CW$, meaning that\n\n$$\n\\frac{PM}{IW} = \\frac{AC}{2CW} \\cos \\beta = \\frac{CM}{CW} \\cos \\beta = \\cos^2 \\beta,\n$$\n\nfinishing the proof.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 12793,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all pairs of positive integers $ (a, b) $ such that $ a^p - b^p - 1 $ is divisible by $43$ for every prime $p \\ge 5$.",
    "options": [],
    "answer": "See solution",
    "solution": "If $p = 43$, then by Fermat's little theorem, $a \\equiv b + 1 \\pmod{43}$. Thus, if $a$ is divisible by $43$, then $b \\equiv -1 \\pmod{43}$, and if $b$ is divisible by $43$, then $a \\equiv 1 \\pmod{43}$. For such pairs $(a, b)$, $a^p - b^p - 1$ is divisible by $43$ for every prime $p$.\n\nAssume $ab$ is not divisible by $43$. Since $a^{41} - b^{41} - 1$ is divisible by $43$, we have:\n\n$$\nb \\equiv b a^{42} \\equiv a b^{42} + a b \\equiv a + a b \\pmod{43}.\n$$\n\nThus, $b^2 + b + 1 \\equiv 0 \\pmod{43}$, so $b \\equiv 6 \\pmod{43}$ and $a \\equiv 7 \\pmod{43}$. To see that $a^{41} - b^{41} - 1$ is divisible by $43$ for any $p \\ge 5$, it suffices to check that $(b+1)^p - b^p - 1 \\equiv 0 \\pmod{b^2 + b + 1}$, which follows from $(-b^2)^p - b^p - 1 \\equiv 0 \\pmod{b^2 + b + 1}$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 12794,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $I$ be the incenter of triangle $ABC$. Let $D$ be a point on side $BC$, and $E$ be a point on ray $BC$ such that $C$ lies between $E$ and $D$ and $\\frac{BD}{DC} = \\frac{BE}{EC}$. Let $H$ be the foot of the perpendicular from $D$ to line $IE$. Prove that $\\angle AHE = \\angle IDE$.\n\n![](images/MNG_ABooklet_2017_p25_data_5546a9298c.png)",
    "options": [],
    "answer": "See solution",
    "solution": "Let $\\omega$ be the incircle of triangle $ABC$, and let $A_1$, $B_1$, $C_1$ be the points where $\\omega$ touches $BC$, $AC$, and $AB$, respectively. Let $\\omega'$ be the circle with diameter $EI$, and let $K$ be the intersection point of $\\omega$ and $\\omega'$, different from $A_1$. Let $M$ be the intersection point of $AK$ and $\\omega'$. We claim that $I$, $M$, and $D$ are collinear. For this, it suffices to show that $E$, $C$, $D'$, and $B$ are harmonic, where $D'$ is the intersection point of $IM$ and $BC$, since $D$ is uniquely determined by the given ratio.\n\nSince $\\angle MLA_1 = \\angle KC_1A_1 = \\angle KIE = \\angle KME$, we get $LA_1 \\parallel EM$, so $LA_1 \\perp MI$, i.e., $D'L$ is tangent to the circle $\\omega$. Thus,\n\n$$\n\\sin(\\angle D'IB) = \\sin\\left(\\frac{\\angle LIC_1}{2}\\right), \\quad \\sin(\\angle CID') = \\sin\\left(\\frac{\\angle B_1IL}{2}\\right). \\quad (*)\n$$\n\nClearly, the points $K$, $B_1$, $L$, $C_1$ are harmonic, and the cross-ratio is defined as a ratio of sines. Therefore, we can conclude that $E$, $C$, $D'$, and $B$ are harmonic since $\\sin(\\angle EIB) = \\sin\\left(\\frac{\\angle KIC_1}{2}\\right)$, $\\sin(\\angle EIC) = \\sin\\left(\\frac{\\angle KIB_1}{2}\\right)$, and $(*)$. Hence $D = D'$, implying the claim.\n\nLet $H'$ be the intersection point of $EI$ and $AK$. Since $\\angle KEI = \\angle KMI = \\angle IEA_1$, the points $H'$, $E$, $M$, and $D$ lie on a circle. Thus $H = H'$ and $\\angle AH'E = \\angle ID'E$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 12795,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $n$ be a positive integer. A regular hexagon with side length $n$ is divided into equilateral triangles with side length $1$ by lines parallel to its sides.\n\nFind the number of regular hexagons all of whose vertices are along the vertices of the equilateral triangles.",
    "options": [],
    "answer": "See solution",
    "solution": "By a lattice hexagon we will mean a regular hexagon whose sides run along edges of the lattice. Given any regular hexagon $H$, we construct a lattice hexagon whose edges pass through the vertices of $H$, as shown in the figure, which we will call the enveloping lattice hexagon of $H$. Given a lattice hexagon $G$ of side length $m$, the number of regular hexagons whose enveloping lattice hexagon is $G$ is exactly $m$.\n\nAlso, there are precisely $3(n-m)(n-m+1)+1$ lattice hexagons of side length $m$ in our lattice: they are those with centers lying at most $n-m$ steps from the centre of the lattice. In particular, the total number of regular hexagons is\n\n$$\nN = \\sum_{m=1}^{n} \\left[3(n-m)(n-m+1)+1\\right]m = (3n^2+3n) \\sum_{m=1}^{n} m - 3(2n+1) \\sum_{m=1}^{n} m^2 + 3 \\sum_{m=1}^{n} m^3.\n$$\n\nSince $\\sum_{m=1}^{n} m = \\frac{n(n+1)}{2}$, $\\sum_{m=1}^{n} m^2 = \\frac{n(n+1)(2n+1)}{6}$, and $\\sum_{m=1}^{n} m^3 = \\left(\\frac{n(n+1)}{2}\\right)^2$, it is easily checked that\n\n$$N = \\left(\\frac{n(n+1)}{2}\\right)^2.$$\n\n![](images/Macedonia_2014_p13_data_9a17a4d582.png)",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 12796,
    "subject": "Mathematics (Olympiad)",
    "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": "See solution",
    "solution": "Without loss of generality, assume $a < b < c$. Let $x = a+b$, $y = a+c$, $z = b+c$. Then $x < y < z$, $x+y > z$, and $x+y+z$ is even. Conversely, if there exist $x, y, z \\in A$ such that $x < y < z$, $x+y > z$, and $x+y+z$ is even, set:\n\n$$\na = \\frac{x+y-z}{2}, \\quad b = \\frac{x+z-y}{2}, \\quad c = \\frac{y+z-x}{2}\n$$\n\nIt is clear that $a, b, c$ are pairwise distinct elements of $S$, and $x = a+b$, $y = a+c$, $z = b+c$.\n\nThus, the required property is equivalent to: for any $k$-element subset $A$ of $S$, there exist $x, y, z \\in A$ such that\n\n$$\nx < y < z, \\quad x + y > z, \\quad \\text{and} \\quad x + y + z \\text{ is even.} \\tag{*}\n$$\n\nConsider $A = \\{1, 2, 3, 5, 7, \\dots, 2011\\}$, which has $|A| = 1007$. This $A$ does not contain three elements satisfying $(*)$. Therefore, $k \\geq 1008$.\n\nNow, we show that any $1008$-element subset of $S$ contains three elements satisfying $(*)$.\n\nWe prove a general statement: For any integer $n \\geq 4$, any $(n+2)$-element subset of $\\{1, 2, \\dots, 2n\\}$ contains three elements satisfying $(*)$. We proceed by induction on $n$.\n\n**Base case ($n=4$):** 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 $4, 6, 8 \\in A$ satisfy $(*)$. If it contains exactly two even numbers, then it contains two odd numbers. For any two odd numbers $x, y$ from $\\{3, 5, 7\\}$, two of $(4, x, y)$, $(6, x, y)$, $(8, x, y)$ would satisfy $(*)$, so one of them is in $A$. If $A \\cap \\{3, 4, 5, 6, 7, 8\\}$ contains exactly one even number $x$, then it contains all three odd numbers, and $(x, 5, 7)$ satisfies $(*)$. Thus, the result holds for $n=4$.\n\n**Inductive step:** Assume the result holds for $n \\geq 4$. Consider $n+1$. Let $A$ be a $(n+3)$-element subset of $\\{1, 2, \\dots, 2n+2\\}$. If $|A \\cap \\{1, 2, \\dots, 2n\\}| \\geq n+2$, the result follows by induction. If $|A \\cap \\{1, 2, \\dots, 2n\\}| = n+1$ and $2n+1, 2n+2 \\in A$, then:\n- If $A$ contains an odd $x$ in $\\{1, 2, \\dots, 2n\\}$, then $x, 2n+1, 2n+2$ satisfy $(*)$.\n- If not, then $A = \\{1, 2, 4, 6, \\dots, 2n, 2n+1, 2n+2\\}$, and $4, 6, 8 \\in A$ satisfy $(*)$.\n\nTherefore, the smallest $k$ with the required property is $1008$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 12797,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $a$, $b$, $c$, and $d$ be four integers such that\n\n$$\n7a + 8b = 14c + 28d.\n$$\n\nProve that $a \\cdot b$ is a multiple of $14$.",
    "options": [],
    "answer": "See solution",
    "solution": "We consider the equation modulo $2$ and modulo $7$, respectively, and obtain\n\n$$\na \\equiv 0 \\pmod{2}.\n$$\n\n$$\nb \\equiv 0 \\pmod{7}.\n$$\n\nWe conclude that $a$ is even and $b$ is a multiple of $7$. Therefore, $ab$ is divisible by $2 \\cdot 7$. $\\square$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 12798,
    "subject": "Mathematics (Olympiad)",
    "question": "Between (and including) 98 and 200, how many integers are multiples of 2 or 3?",
    "options": [],
    "answer": "See solution",
    "solution": "Between 98 and 200, there are 51 multiples of 2. Between 98 and 199, there are 34 multiples of 3. Between 102 and 198, there are 17 multiples of 6 (i.e., numbers divisible by both 2 and 3). By the inclusion-exclusion principle, the total number is:\n\n$$\n51 + 34 - 17 = 68\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 12799,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $\\triangle ABC$ be an acute triangle such that $|AB| > |AC|$. Let $D$ be a point different from $C$ on the segment $BC$, such that $|AC| = |AD|$. Let $H$ denote the orthocentre of the triangle $ABC$, and let $A_1, B_1$ be the feet of the altitudes from $A$ and $B$, respectively. The line $DH$ intersects the line $AC$ at $E$ and the line $A_1B_1$ at $F$. Let $G$ be the intersection of the lines $AF$ and $BH$. Prove that $EG$, $CH$ and $AD$ meet at the same point.",
    "options": [],
    "answer": "See solution",
    "solution": "Denote the intersection of the lines $EF$ and $AG$ by $I$ and the intersection of the lines $EF$ and $AH$ by $J$.\n\nNow, $EF$ is parallel to $BD$ and $AH$ is parallel to $CE$, so the quadrilateral $EGHJ$ is a parallelogram. Furthermore, $AG$ and $CF$ are parallel, so $FIGH$ is also a parallelogram. Hence, $|FH| = |IG|$, $|HJ| = |GE|$ and $\\angle FHJ = \\angle IGE$, which means that the triangles $FHW$ and $IGE$ are congruent. Thus, $|FJ| = |IE|$.\n\nSince $EF$ is parallel to $BD$, there are three pairs of similar triangles: $EAF$ and $BAD$, $EAI$ and $BAG$, $JAF$ and $HAD$. Hence, $\\frac{|EA|}{|BA|} = \\frac{|FA|}{|DA|} \\cdot \\frac{|EA|}{|BA|} = \\frac{|EI|}{|BG|}$ and $\\frac{|FA|}{|DA|} = \\frac{|JF|}{|DH|}$. We see that $\\frac{|EI|}{|BG|} = \\frac{|JF|}{|DH|}$ and together with $|FJ| = |IE|$ this implies $|BG| = |DH|$.\n\n![](images/Slovenija_2009_p16_data_da2c6eef8e.png)\n\nLet $S$ be the midpoint of the segment $AC$. Since $AGCH$ is a parallelogram, $S$ is also the midpoint of the segment $GH$. The equality $|BG| = |DH|$ implies that $S$ is the midpoint of $BD$ as well. The segments $AC$ and $BD$ bisect one another, so $ABCD$ is a parallelogram.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 12800,
    "subject": "Mathematics (Olympiad)",
    "question": "Consider factorizations of numbers $EFGH$ with all digits different, and identify all factorizations consisting of two 2-digit numbers. Start with the smallest possible number 1023 and stop when a solution is found.\n\n$$\n\\begin{align*}\n1023 &= 3 \\cdot 11 \\cdot 31 = 11 \\cdot 93 = 31 \\cdot 33 \\\\\n1024 &= 2^{10} = 16 \\cdot 64 = 32 \\cdot 32 \\\\\n1025 &= 5^2 \\cdot 41 = 25 \\cdot 41 \\\\\n1026 &= 2 \\cdot 3^3 \\cdot 19 = 18 \\cdot 57 = 19 \\cdot 54 = 27 \\cdot 38 \\\\\n1027 &= 13 \\cdot 79 \\\\\n1028 &= 2^2 \\cdot 257 \\\\\n1029 &= 3 \\cdot 7^3 = 21 \\cdot 49 \\\\\n1032 &= 2^3 \\cdot 3 \\cdot 43 = 12 \\cdot 86 = 24 \\cdot 43 \\\\\n1034 &= 2 \\cdot 517 \\\\\n1035 &= 3^2 \\cdot 5 \\cdot 23 = 15 \\cdot 69 = 23 \\cdot 45 \\\\\n1036 &= 2^2 \\cdot 7 \\cdot 37 = 14 \\cdot 74 = 28 \\cdot 37 \\\\\n1037 &= 17 \\cdot 61 \\\\\n1038 &= 2 \\cdot 3 \\cdot 173 \\\\\n1039 & \\quad \\text{prime}\n\\end{align*}\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "$$\n\\begin{align*}\n1042 &= 2 \\cdot 521 \\\\\n1043 &= 7 \\cdot 149 \\\\\n1045 &= 5 \\cdot 11 \\cdot 19 = 11 \\cdot 95 = 19 \\cdot 55 \\\\\n1046 &= 2 \\cdot 523 \\\\\n1047 &= 3 \\cdot 349 \\\\\n1048 &= 2^3 \\cdot 131 \\\\\n1049 &= \\text{prime} \\\\\n1052 &= 2^2 \\cdot 263 \\\\\n1053 &= 3^4 \\cdot 13 = 13 \\cdot 81 = 27 \\cdot 39 \\\\\n1054 &= 2 \\cdot 17 \\cdot 31 = 17 \\cdot 62 = 31 \\cdot 34 \\\\\n1056 &= 2^5 \\cdot 3 \\cdot 11 = 11 \\cdot 96 = 12 \\cdot 88 = 16 \\cdot 66 = 22 \\cdot 48 = 24 \\cdot 44 = 32 \\cdot 33 \\\\\n1057 &= 7 \\cdot 151 \\\\\n1058 &= 2 \\cdot 23^2 = 23 \\cdot 46 \\quad \\text{Eureka!}\n\\end{align*}\n$$\n\nSome simple observations help in reducing cases. Without loss of generality, we will assume throughout $AB < CD$. We cannot have $B = 0$ or $D = 0$, as this would imply $H = 0$. Similarly, we cannot have $B = 1$ or $D = 1$, as this would imply $H = B$ or $H = D$. We cannot have $A = 1$, as this would imply $E = 1$, since $AB \\times CD < 20 \\cdot 98 = 1960 < 2000$.\n\nIf $AB = 21$ then $AB \\times CD \\le 21 \\cdot 98 = 2058$ and $E \\in \\{1, 2\\}$ which is impossible. Thus the smallest possible value of $AB$ is 23.\n\nIf $AB = 23$, the smallest possible value of $CD$ is 45. But $23 \\cdot 45 = 1035$, however $23 \\cdot 46 = 1058$ yields a solution.\n\nWe need to show that there is no solution with a smaller value of $AB \\times CD$. We only need to consider these possibilities: $AB = 24, 25, 26, 27, 28, 29, 32$, because $32^2 = 1024 < 1058 < 1089 = 33^2$. In each case we keep in mind that we wish to achieve $1000 < AB \\times CD < 1058$.\n\nFor $AB = 24$ there are no possibilities for $CD$, since $CD$ cannot contain the digit 4, and $24 \\cdot 39 < 1000$ while $24 \\cdot 50 > 1058$. Therefore no solutions exist in this case.\n\nFor $AB = 25$, the possibilities for $CD$ are 40, 41, 42. None of these work.\nFor $AB = 26$, the possibilities for $CD$ are 39, 40. Neither works.\nFor $AB = 27$, the possibilities for $CD$ are 38, 39. Neither works.\nFor $AB = 28$, the possibilities for $CD$ are 36, 37. Neither works.\nFor $AB = 29$, the possibilities for $CD$ are 35, 36. Neither works.\nFor $AB = 32$, the possibilities for $CD$ are 32, 33. Neither works.\nHence $23 \\cdot 46 = 1058$ is the smallest solution.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 12801,
    "subject": "Mathematics (Olympiad)",
    "question": "Given positive integers $a$ and $k$, the sequence $(a_n)_{n=1}^{\\infty}$ is defined by\n\n$$\na_1 = a \\quad \\text{and} \\quad a_{n+1} = a_n + k \\cdot \\varphi(a_n) \\quad \\text{for } n = 1, 2, \\dots,$$\n\nwhere $\\varphi(m)$ stands for the product of digits of $m$ in its decimal representation (e.g., $\\varphi(413) = 12$, $\\varphi(308) = 0$).\n\nProve that there exist positive integers $a$ and $k$ such that the sequence $(a_n)_{n=1}^{\\infty}$ contains exactly 2009 different numbers.",
    "options": [],
    "answer": "See solution",
    "solution": "Obviously, the sequence $(a_n)_{n=1}^{\\infty}$ is increasing until the first term with a 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 $a_m$.\n\nSet\n\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\nThen we have\n\n$$\na_1 = a = \\underbrace{11\\dots1}_{2m-5}$$\n\n$$\\varphi(a_1) = 1.$$  \n\n$$a_2 = a_1 + k = a_1 + \\underbrace{100\\dots04}_{m-1} = \\underbrace{11\\dots12}_{m-3}\\underbrace{11\\dots15}_{m-4}$$\n\n$$\\varphi(a_2) = 10.$$  \n\n$$a_3 = a_2 + 10k = a_2 + \\underbrace{100\\dots040}_{m-4} = \\underbrace{11\\dots122}_{m-4}\\underbrace{11\\dots155}_{m-5}$$\n\n$$\\varphi(a_3) = 100.$$  \n\n$$a_i = a_{i-1} + 10^{i-2}k = \\underbrace{11\\dots122\\dots2}_{m-1}\\underbrace{11\\dots155\\dots5}_{m-1}$$\n\n$$\\varphi(a_i) = 10^{i-1}.$$  \n\n$$a_{m-2} = a_{m-3} + 10^{m-4}k = \\underbrace{122\\dots255\\dots5}_{m-3}$$\n\n$$\\varphi(a_{m-2}) = 10^{m-3}.$$  \n\n$$a_{m-1} = a_{m-2} + 10^{m-3}k = \\underbrace{100\\dots0}_{m-4}\\underbrace{100\\dots0}_{m-3}\\underbrace{22\\dots2655\\dots5}_{m-3}$$\n\n$$\\varphi(a_{m-1}) = 6 \\cdot 10^{m-3}.$$  \n\n$$a_m = a_{m-1} + 6 \\cdot 10^{m-3}k = \\underbrace{600\\dots0}_{m-5}\\underbrace{2400\\dots0}_{m-3}\\underbrace{822\\dots25055\\dots5}_{m-5}$$\n\n$$\\varphi(a_m) = 0.$$  \n\n*Conclusion.* The sequence contains exactly 2009 different numbers for $a = \\frac{1}{9}(10^{4013} - 1)$ and $k = 10^{2006} + 4$.\n\nSecond solution. Put\n\n$$a = \\underbrace{611\\dots1}_{2007} \\quad \\text{and} \\quad k = \\underbrace{33\\dots34}_{2007} = \\frac{1}{6} \\cdot \\underbrace{200\\dots04}_{2007}$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 12802,
    "subject": "Mathematics (Olympiad)",
    "question": "Given a regular triangular prism, the length of each edge is 3. Then the volume of its circumscribed sphere is ____.",
    "options": [],
    "answer": "See solution",
    "solution": "As shown in the figure below, let the centroids of faces *ABC* and *A₁B₁C₁* be *O* and *O₁*, respectively. Denote the midpoint of segment *OO₁* 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![](images/Mathematical_Olympiad_In_China_2021_2022___Problems_and_Solutions_2024_p30_data_13a66d1b8c.png)\n\nIt is easy to find that $PO \\perp AO$, and thus\n\n$$\nPA = \\sqrt{PO^2 + AO^2} = \\sqrt{\\left(\\frac{3}{2}\\right)^2 + (\\sqrt{3})^2} = \\frac{\\sqrt{21}}{2}.\n$$\n\nTherefore, the volume of the circumscribed sphere is\n$$\n\\frac{4}{3}\\pi\\left(\\frac{\\sqrt{21}}{2}\\right)^3 = \\frac{7\\sqrt{21}}{2}\\pi.\n$$\n$\\square$",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 12803,
    "subject": "Mathematics (Olympiad)",
    "question": "Juku and Miku play the following game on a grid of dimensions $n \\times m$:\n\nIn the beginning, all unit squares are white. Each player, on their turn, paints one white unit square either red or blue of their choice, but no two unit squares with a common side or a common vertex can be painted the same color. The players take turns and Juku starts. A player who cannot make an allowed move loses.\n\nIs it possible for Juku to win the game regardless of how Miku plays if:\n\n(a) $n = 2023$ and $m = 2023$;\n\n(b) $n = 2023$ and $m = 2024$;\n\n(c) $n = 2024$ and $m = 2024$?",
    "options": [],
    "answer": "See solution",
    "solution": "If $n$ and $m$ are odd, then there is a middle square on the grid. Let Juku paint the middle square any color on the first move. From now on, each of Juku's moves should mirror Miku's last move relative to the center of the grid. If before Miku's move the position is symmetrical with respect to the center of the grid, then Juku can certainly respond symmetrically, and before Miku's next move the position is again symmetrical with respect to the center of the grid. Therefore, Juku always wins.\n\nIf $n$ or $m$ is even, then Miku can mirror Juku's last move with respect to the center of the grid in each of his moves, but with the opposite color. Then, before each move by Juku, the unit squares symmetrical to the center are of the opposite color. So if Juku can make a move, Miku can respond symmetrically to the center. Consequently, with this strategy, Miku wins. It follows that on a $2023 \\times 2023$ grid Juku can win for any moves of Miku, but on $2023 \\times 2024$ and $2024 \\times 2024$ grids it is not always possible.",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 12804,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $A, B, C \\in \\mathcal{M}_n(\\mathbb{R})$ be such that $ABC = 0_n$ and $\\operatorname{rank} B = 1$. Prove that $AB = O_n$ or $BC = O_n$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $K \\in \\mathcal{M}_{n,1}(\\mathbb{C})$ and $L \\in \\mathcal{M}_{1,n}(\\mathbb{C})$ be such that $KL = B$.\n\nThen $O_n = (AK)(LC)$, with $AK \\in \\mathcal{M}_{n,1}(\\mathbb{C})$, $LC \\in \\mathcal{M}_{1,n}(\\mathbb{C})$.\n\nLet $AK = {}^t(a_1, a_2, \\dots, a_n)$ (where ${}^t X$ denotes the transpose of $X$), and let $LC = (b_1, b_2, \\dots, b_n)$. We infer\n\n$$\nO_n = \\begin{pmatrix} a_1 b_1 & a_1 b_2 & \\cdots & a_1 b_n \\\\ \\vdots & \\vdots & \\cdots & \\vdots \\\\ a_n b_1 & a_n b_2 & \\cdots & a_n b_n \\end{pmatrix}\n$$\n\nIf $LC \\neq O_{1,n}$, then there exists $1 \\le i \\le n$ with $b_i \\neq 0$.\n\nIt follows $a_1 = a_2 = \\cdots = a_n = 0$, hence $AK = O_{n,1}$, and so at least one of the matrices $AK$ and $LC$ is null.\n\nIn the former case, $AB = (AK)L = O_{n,1}L = O_n$, while in the latter, $BC = K(LC) = K O_{1,n} = O_n$.\n\n**Remarks:** The matrices $A, B, C$ may well be taken non-square.\n\nAn alternative solution, based on rank calculation (using Frobenius' theorem), may also be presented.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 12805,
    "subject": "Mathematics (Olympiad)",
    "question": "Determine the greatest possible value of $M$ for which:\n\n$$\n\\frac{x}{1+\\frac{yz}{x}} + \\frac{y}{1+\\frac{zx}{y}} + \\frac{z}{1+\\frac{xy}{z}} \\ge M,\n$$\n\nfor all real numbers $x, y, z > 0$ satisfying the equation $xy + yz + zx = 1$.",
    "options": [],
    "answer": "See solution",
    "solution": "The inequality is equivalent to\n\n$$\n\\frac{x^2}{x+yz} + \\frac{y^2}{y+zx} + \\frac{z^2}{z+xy} \\ge M.\n$$\n\nSince $x, y, z > 0$, by the Cauchy-Schwarz inequality we have\n\n$$\n\\left( \\frac{x^2}{x+yz} + \\frac{y^2}{y+zx} + \\frac{z^2}{z+xy} \\right) (x+yz+y+zx+z+xy) \\ge (x+y+z)^2\n$$\nwhich implies\n$$\n\\frac{x^2}{x+yz} + \\frac{y^2}{y+zx} + \\frac{z^2}{z+xy} \\ge \\frac{(x+y+z)^2}{x+y+z+xy+yz+zx}.\n$$\nGiven $xy + yz + zx = 1$, this becomes\n$$\n\\frac{x^2}{x+yz} + \\frac{y^2}{y+zx} + \\frac{z^2}{z+xy} \\ge \\frac{(x+y+z)^2}{x+y+z+1}.\n$$\n\nEquality holds when $x = y = z$, and $xy + yz + zx = 1$ gives $x = y = z = \\frac{\\sqrt{3}}{3}$.\n\nSince $(x+y+z)^2 \\ge 3(xy+yz+zx)$, we have $x+y+z \\ge \\sqrt{3}$, with equality at $x = y = z = \\frac{\\sqrt{3}}{3}$.\n\nConsider $f(u) = \\frac{u^2}{u+1}$ for $u \\ge \\sqrt{3}$, which is strictly increasing. Thus,\n$$\n\\frac{(x+y+z)^2}{x+y+z+1} \\ge \\frac{3}{\\sqrt{3}+1}.\n$$\n\nTherefore,\n$$\n\\frac{x^2}{x+yz} + \\frac{y^2}{y+zx} + \\frac{z^2}{z+xy} \\ge \\frac{3}{\\sqrt{3}+1}\n$$\nfor all $x, y, z > 0$ with $xy + yz + zx = 1$. Thus, the greatest possible value is\n$$\nM = \\frac{3}{\\sqrt{3}+1}.\n$$",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 12806,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $r$ be the radius of a circle. Two chords of lengths $24$ and $32$ are drawn in the circle. Let $x$ and $y$ be the distances from the center of the circle to these chords, respectively (see the figure).\n\nWhat is the length of a chord that is equidistant from the center as the average of $x$ and $y$?\n\n![](images/Hong_Kong_2015_Booklet_p8_data_3a75df7a2f.png)",
    "options": [],
    "answer": "See solution",
    "solution": "Let $r$ be the radius of the circle. Denote by $x$ and $y$ the distances from the two chords to the center of the circle. Since the perpendicular from the center to a chord bisects the chord, by the Pythagorean theorem:\n\n$$\n\\left(\\frac{24}{2}\\right)^2 + x^2 = r^2 \\implies 12^2 + x^2 = r^2\n$$\nSimilarly,\n$$\n16^2 + y^2 = r^2\n$$\nCombining the two equations:\n$$\n12^2 + x^2 = 16^2 + y^2 \\implies x^2 - y^2 = 16^2 - 12^2 = 256 - 144 = 112\n$$\nSo,\n$$\n(x - y)(x + y) = 112\n$$\nThere are two possibilities:\n\n* If the two chords lie on the same side of the center, $x - y = 14$, so $x + y = 8$, $y = -3$ (impossible).\n* If the two chords lie on opposite sides, $x + y = 14$, $x - y = 8$, so $x = 11$, $y = 3$.\n\nThus, $r = \\sqrt{11^2 + 12^2} = \\sqrt{121 + 144} = \\sqrt{265}$.\n\nThe desired chord is at a distance of $\\frac{11 - 3}{2} = 4$ from the center. Using the Pythagorean theorem again, its length is:\n$$\n2 \\times \\sqrt{r^2 - 4^2} = 2 \\sqrt{265 - 16} = 2\\sqrt{249}\n$$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 12807,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $O$ be the center of a square with side length $8$. Let $A$, $B$, $C$, and $D$ be the midpoints of the sides of the square. Let $E$ be the intersection of lines $AB$ and $OC$. Find the area of quadrilateral $BCDE$.",
    "options": [],
    "answer": "See solution",
    "solution": "(a) The area of trapezoid $OABC$ is:\n\n$$\n\\frac{AB + OC}{2} \\cdot OA = \\frac{4 + 2}{2} \\cdot 2 = 6.\n$$\n\n(b) Let $A'$, $B'$, $C'$, and $D'$ be the reflections of $A$, $B$, $C$, and $D$ across $O$, respectively. Since $O$ is the center of the square, $B'$ and $D'$ lie on the sides of the square. The square is divided into four congruent (non-convex) polygons, each with area $\\frac{64}{4} = 16$. Thus, $BCDE$ has area $16 - 6 = 10$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 12808,
    "subject": "Mathematics (Olympiad)",
    "question": "The points $A_1$ and $C_1$ are chosen on the sides $BC$ and $AB$ of triangle $ABC$ so that the segments $AA_1$ and $CC_1$ are equal and perpendicular. Prove that if $\\angle ABC = 45^\\circ$, then $AC = AA_1$.",
    "options": [],
    "answer": "See solution",
    "solution": "First, we show that the triangle is acute. Suppose $\\angle BAC \\geq 90^\\circ$. Then $\\angle ACB \\leq 45^\\circ = \\angle ABC < \\angle AA_1C$. In $\\triangle AA_1C$, we get $AC > AA_1$. On the other hand, $\\angle C_1AC \\geq 90^\\circ$, so in $\\triangle AC_1C$, the side $CC_1$ is the longest, in particular, $CC_1 > AC$. Hence $CC_1 > AC > AA_1$, a contradiction.\n\nLet $AA_1$ and $CC_1$ intersect at $O$. Consider a circle with diameter $AC$; then $O$ lies on it. This circle intersects $AB$ and $BC$ at $P$ and $Q$, respectively. The points $P$ and $Q$ lie inside the sides of $\\triangle ABC$ because the triangle is acute. Also, since $AA_1$ and $CC_1$ are cevians, $O$ lies inside $\\triangle ABC$, that is, on the arc $PQ$ of the circle.\n\nWe will prove that $\\angle A_1AQ = \\angle QAC$. If true, $\\triangle CAA_1$ is isosceles, because $AQ$ is both an angle bisector and altitude, so $A_1A = AC$. Suppose $\\angle A_1AQ = \\beta > \\alpha = \\angle QAC$. Then $A_1A > AC$. Clearly, $\\angle PBC = \\angle BAQ = 45^\\circ$. Then\n\n$$\n\\angle PCC_1 = 45^\\circ - \\angle OCQ = 45^\\circ - \\angle OAQ = 45^\\circ - \\beta.\n$$\n\nAlso, $\\angle PCA = 90^\\circ - \\angle PAC = 45^\\circ - \\alpha$. Since $\\beta > \\alpha$, we have $45^\\circ - \\beta < 45^\\circ - \\alpha$, so $\\angle C_1CP < \\angle PCA$, hence $CC_1 < AC$. But $A_1A > AC$, so $CC_1 < AA_1$—contradiction. The case $\\beta < \\alpha$ is similar.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 12809,
    "subject": "Mathematics (Olympiad)",
    "question": "Find the minimum value of $\\dfrac{a}{b} + \\dfrac{b}{a}$ for positive real numbers $a$ and $b$.",
    "options": [],
    "answer": "See solution",
    "solution": "It is easy to see that if $a = b$, then the expression is equal to $2$, and a few trials will suggest that $2$ is the smallest value. A proof is that\n\n$$\n\\frac{a}{b} + \\frac{b}{a} = \\frac{a^2 + b^2}{ab} = 2 + \\frac{a^2 - 2ab + b^2}{ab} = 2 + \\frac{(a-b)^2}{ab} \\geq 2,\n$$\n\nsince $(a - b)^2$ is a perfect square and $ab > 0$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 12810,
    "subject": "Mathematics (Olympiad)",
    "question": "Two circles $O_1$ and $O_2$ intersect at two distinct points $P$ and $Q$. The tangent line to the circle $O_1$ at the point $P$ intersects the circle $O_2$ at $R$, different from $P$, and the tangent line to the circle $O_2$ at the point $Q$ intersects the circle $O_1$ at $S$, different from $Q$. Let $X$ be the point of intersection of the two lines $PR$ and $QS$. If $XR = 9$ and $XS = 2$, what is the value of the ratio $\\frac{r_1}{r_2}$, where $r_1$ and $r_2$ are the radii of the circles $O_1$ and $O_2$, respectively? Here $YZ$ denotes the length of the line segment $YZ$.",
    "options": [],
    "answer": "See solution",
    "solution": "$$\n\\sqrt[3]{\\frac{2}{9}}\n$$\n\nIn view of a well-known theorem on angles subtended by arcs on a circle, we have $\\angle PSQ = \\angle QPR$, and $\\angle SQP = \\angle PRQ$. This implies that the triangles $PSQ$ and $QPR$ are similar. Since the circles $O_1$ and $O_2$ are circumcircles of the triangles $PSQ$ and $QPR$, respectively, the ratio $\\frac{r_1}{r_2}$ of the radii of these circles must be the same as the similarity ratio $\\frac{PQ}{QR}$ of these triangles. The same theorem also tells us that $\\angle XPS = \\angle XQP = \\angle XRQ$. Since the angle $\\angle X$ is common to all three triangles $XPS$, $XQP$, and $XRQ$, these triangles are similar. Hence, $\\frac{XS}{XP} = \\frac{XP}{XQ} = \\frac{XQ}{XR}$.\n\nConsequently,\n$$\n\\left(\\frac{XQ}{XR}\\right)^3 = \\frac{XQ}{XR} \\cdot \\frac{XP}{XQ} \\cdot \\frac{XS}{XP} = \\frac{XS}{XR} = \\frac{2}{9}.\n$$\n\nFinally, from the similarity of the triangles $XQP$ and $XRQ$, we also get $\\frac{PQ}{QR} = \\frac{XQ}{XR}$, which enables us to conclude that\n$$\n\\frac{r_1}{r_2} = \\frac{PQ}{QR} = \\frac{XQ}{XR} = \\sqrt[3]{\\frac{2}{9}}\n$$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 12811,
    "subject": "Mathematics (Olympiad)",
    "question": "Барон Мюнхгаузен стверджує, що можна розмістити 2006 точок на колі так, щоб жодні дві ланки ламаної, утвореної послідовним з'єднанням цих точок, не були паралельними. Чи правий барон?",
    "options": [],
    "answer": "See solution",
    "solution": "Ні, барон помиляється. Позначимо 2006 точок числами $0, 1, 2, \\ldots, 2005$, записаними послідовно. Ланки ламаної будемо позначати номерами її кінців. Нескладно перевірити, що якщо $i + j \\equiv k + l \\pmod{2006}$, то ланки $i$ і $j$ та $k$ і $l$ будуть паралельними.\n\nПрипустимо, що барон Мюнхгаузен правий. Тоді існує така перестановка $n_1, n_2, \\ldots, n_{2006}$ чисел $0, 1, 2, \\ldots, 2005$, що всі суми $n_1 + n_2, n_2 + n_3, \\ldots, n_{2005} + n_{2006}, n_{2006} + n_1$ попарно неконгруентні за модулем 2006. Тоді цей набір також є деякою перестановкою чисел $0, 1, 2, \\ldots, 2005$. Таким чином,\n\n$$\n(n_1 + n_2) + (n_2 + n_3) + \\dots + (n_{2006} + n_1) \\equiv 0 + 1 + \\dots + 2005 \\equiv 1003 \\pmod{2006}\n$$\n\nАле, з іншого боку,\n\n$$\n2n_1 + 2n_2 + 2n_3 + \\dots + 2n_{2006} \\equiv 2(0 + 1 + 2 + \\dots + 2005) \\equiv 0 \\pmod{2006}.\n$$\n\nОдержана суперечність і доводить, що барон Мюнхгаузен помиляється.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 12812,
    "subject": "Mathematics (Olympiad)",
    "question": "Consider the following system of 10 equations in 10 real variables $v_1, \\dots, v_{10}$:\n\n$$\nv_i = 1 + \\frac{6 v_i^2}{v_1^2 + v_2^2 + \\dots + v_{10}^2} \\quad (i = 1, \\dots, 10).\n$$\n\nFind all 10-tuples $(v_1, v_2, \\dots, v_{10})$ that are solutions of this system.",
    "options": [],
    "answer": "See solution",
    "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\n$$\nv_i = 1 + \\frac{6v_i^2}{s} \\implies 6v_i^2 - sv_i + s = 0.\n$$\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).\n\nIf all the $v_i$ are equal, then\n\n$$\nv_i = 1 + \\frac{6}{10} = \\frac{8}{5}\n$$\n\nfor all $i$. Otherwise, let $5+k$ of the $v_i$ be $a$, and $5-k$ of the $v_i$ be $b$, where $0 < k \\leq 4$. By the AM-GM inequality,\n\n$$\n6ab = s = (5 + k)a^2 + (5 - k)b^2 \\geq 2ab\\sqrt{25 - k^2}.\n$$\n\nFrom the given equations, $v_i \\geq 1$ for all $i$, so $a$ and $b$ are positive.\nThen $\\sqrt{25 - k^2} \\leq 3 \\implies 25 - k^2 \\leq 9 \\implies k^2 \\geq 16 \\implies k = 4$. Hence,\n$6ab = 9a^2 + b^2 \\implies (b - 3a)^2 = 0 \\implies b = 3a$.\n\nAdding all ten equations, we get\n\n$$\nv_1 + v_2 + \\dots + v_{10} = 16.\n$$\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": 12813,
    "subject": "Mathematics (Olympiad)",
    "question": "For a function $P(a, b, c)$ of three variables $a, b, c$, define\n$$\n\\sum_{cyc} P(a, b, c) = P(a, b, c) + P(b, c, a) + P(c, a, b).\n$$\n\nConsider the inequality:\n$$\n\\sum_{cyc} \\frac{a}{1 + kbc + 4(b-c)^2} \\geq \\frac{1}{2}\n$$\nfor $a, b, c \\geq 0$ with $a + b + c = 1$. Find the maximum value of $k$ for which the inequality holds for all such $a, b, c$.",
    "options": [],
    "answer": "See solution",
    "solution": "If we substitute $a = 0$, $b = c = \\frac{1}{2}$, the inequality becomes\n$$\n\\frac{\\frac{1}{2}}{1 + \\frac{1}{4}k} + \\frac{\\frac{1}{2}}{1 + \\frac{1}{4}k} \\geq \\frac{1}{2},\n$$\nwhich implies $k \\leq 4$.\n\nWe now show the inequality holds for $k = 4$.\n\nBy the Cauchy-Schwarz inequality,\n$$\n\\left( \\sum_{cyc} \\frac{a}{1 + 9bc + 4(b-c)^2} \\right) \\left( \\sum_{cyc} a(1 + 9bc + 4(b-c)^2) \\right) \\geq (a+b+c)^2 = 1.\n$$\n\nAlso,\n$$\n\\sum_{cyc} a(1 + 9bc + 4(b-c)^2) = a + b + c + 3abc + 4(a^2b + a^2c + b^2a + c^2a + c^2b).\n$$\nBy Schur's inequality,\n$$\n3abc + 4(a^2b + a^2c + b^2a + c^2a + c^2b) \\leq (a+b+c)^3.\n$$\nThus,\n$$\n\\sum_{cyc} a(1 + 9bc + 4(b-c)^2) \\leq a + b + c + (a+b+c)^3 = 2.\n$$\n\nCombining the above,\n$$\n\\sum_{cyc} \\frac{a}{1 + 9bc + 4(b-c)^2} \\geq \\frac{1}{2}.\n$$\nTherefore, the maximum value of $k$ is $4$.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 12814,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $S$ be a subset of $\textbf{N} \times \textbf{N}$ such that for each $x \\in \\mathbb{N}$, there are only finitely many elements of $S$ with first coordinate $x$, and similarly for the second coordinate. Suppose $S$ satisfies a non-empty set of equations of the form $P(x) = Q(y)$, where $P$ and $Q$ are polynomials. Let $P_0(x) = Q_0(y)$ be the equation of minimal degree among those satisfied by an infinite subset $S_0 \\subset S$. Show that any other such equation $P(x) = Q(y)$ satisfied by $S$ is derived from $P_0(x) = Q_0(y)$ in the sense that $P(x) = F(P_0(x))$ and $Q(y) = F(Q_0(y))$ for some polynomial $F$ with rational coefficients. \n\nb) Let $P(x) \\in \\mathbb{Z}[x]$ be a monic polynomial and $d$ a positive integer dividing $\\deg P$. Prove that there exist $N \\in \\mathbb{N}$ and polynomials $T(x), R(x) \\in \\mathbb{Z}[x]$ such that\n\n$$\nNP(x) = (T(x))^d + R(x),\n$$\n\nand for sufficiently large $x$,\n\n$$\n(T(x))^d \\leq NP(x) \\leq (T(x) + 1)^d.\n$$\n\nUsing this, show that if $d = \\gcd(\\deg P, \\deg Q) > 1$ for polynomials $P, Q$, then the equation $P(x) = Q(y)$ is not primitive.",
    "options": [],
    "answer": "See solution",
    "solution": "a) For any equation $P(x) = Q(y)$ satisfied by $S$, we can write $P(x)$ and $Q(y)$ in terms of $P_0(x)$ and $Q_0(y)$ using the division algorithm. After clearing denominators, we obtain equations with integer coefficients:\n\n$$\nNP(x) = A'(x)P_0(x) + B'(x), \\quad NQ(y) = C'(y)Q_0(y) + D'(y)\n$$\n\nFor $(x_0, y_0) \\in S_0$, subtracting gives\n\n$$\n(A'(x_0) - C'(y_0))b = B'(x_0) - D'(y_0)\n$$\n\nFor large $|x_0|, |y_0|$, the right side is small compared to $b$, so $A'(x_0) = C'(y_0)$ and $B'(x_0) = D'(y_0)$. Since $B'(x) = D'(y)$ has infinitely many solutions, $B', D'$ must be constant, say $c$. Thus,\n\n$$\nA'(x) = \\frac{NP(x) - c}{P_0(x)}, \\quad C'(y) = \\frac{NQ(y) - c}{Q_0(y)}\n$$\n\nRepeating this process, we see $P(x) = F(P_0(x))$ and $Q(y) = F(Q_0(y))$ for some rational polynomial $F$.\n\nb) *Lemma 1.* Let $P(x) \\in \\mathbb{Z}[x]$ be monic, $d | \\deg P$. Then there exist $N \\in \\mathbb{N}$, $T(x), R(x) \\in \\mathbb{Z}[x]$ such that\n\n$$\nNP(x) = (T(x))^d + R(x),\n$$\n\nand for large $x$,\n\n$$\n(T(x))^d \\leq NP(x) \\leq (T(x) + 1)^d\n$$\n\n*Proof.* Write $\\deg P = dm$. Construct $T_1(x) \\in \\mathbb{Q}[x]$ so $P(x) - (T_1(x))^d$ has degree $<(m-1)d$. Choose $T_1$ recursively, clear denominators to get $T_2(x) = nT_1(x)$, $N = n^d$, $S_2(x) = NP(x) - (T_2(x))^d$. Adjust $T(x)$ so $R(x)$ has desired properties.\n\nNow, for $d = \\gcd(\\deg P, \\deg Q) > 1$, apply the lemma to $P$ and $Q$ to get\n\n$$\nNP(x) = (T(x))^d + R(x), \\quad NQ(y) = (U(y))^d + V(y)\n$$\n\nFor large $x, y$, every solution of $P(x) = Q(y)$ is a solution of $T(x) = U(y)$, which has lower degree. Thus, $P(x) = Q(y)$ is not primitive unless $d = 1$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 12815,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all positive integers $n < 2027$ that satisfy the following conditions:\n\n1. For every positive divisor $d$ of $n$, the numbers $1^d, 2^d, \\dots, 2026^d$ all have distinct remainders when divided by $2027$.\n2. $\\tau(n)^2 \\mid n$, where $\\tau(n)$ is the number of positive divisors of $n$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let us call the numbers satisfying the given conditions \"good\" numbers.\n\nFirst, suppose $n$ is even. Then $d = 2$ is a divisor of $n$, but $1^2 \\equiv 2026^2 \\equiv 1 \\pmod{2027}$, so the remainders are not all distinct. Thus, $n$ must be odd.\n\nWe see that $n = 1$ satisfies both conditions.\n\nNow consider $n > 1$ and write $n = p_1^{a_1} p_2^{a_2} \\dots p_k^{a_k}$, where $p_i$ are distinct odd primes and $a_i > 0$. Since $\\tau(n)^2 \\mid n$, $\\tau(n)$ must be odd, so all $a_i$ are even and $n$ is a perfect square.\n\nIf $k \\geq 3$, then $n \\geq (3 \\cdot 5 \\cdot 7)^2 = 11025 > 2027$, so $k \\leq 2$.\n\n**Case 1:** $k = 1$. Then $n = p^{2m}$ for some odd prime $p$ and $m \\geq 1$.\n- If $2m = 2$, $p = 3$, $n = 9$.\n- If $2m = 4$, $p = 5$, $n = 625$.\n- If $2m \\geq 6$, $n \\geq 7^6 = 117649 > 2027$.\nSo possible values: $n = 9, 625$.\n\n**Case 2:** $k = 2$. Then $n = p^{2a} q^{2b}$ for distinct odd primes $p, q$ and $a, b \\geq 1$.\n- If $2a, 2b \\geq 4$, $n \\geq (3 \\cdot 5)^4 = 50625 > 2027$.\n- So at least one of $2a, 2b = 2$.\n- If $q = 3$, $n = 9p^{2a}$. If $2a = 2$, $\\tau(n) = 9$, but $9^2 \\nmid n$. If $2a \\geq 4$, $n \\geq 9 \\cdot 5^4 = 5625 > 2027$.\n- If $p = 3$, $n = 3^{2a} q^2$. If $2a = 4$, $q = 5$, $n = 3^4 \\cdot 5^2 = 2025$.\n\nSo possible values: $n = 2025$.\n\nNow, check that $n = 1, 9, 625, 2025$ all satisfy condition (i):\n\nFor any positive divisor $d$ of $n$, $d$ is odd. Suppose $u, v \\in \\{1, 2, \\dots, 2026\\}$ with $u \\neq v$ and $u^d \\equiv v^d \\pmod{2027}$. Let $s$ be the inverse of $v$ modulo $2027$, and set $t = us$. Then $t^d \\equiv 1 \\pmod{2027}$. Let $h = \\operatorname{ord}_{2027}(t)$. Since $2027$ is prime, $\\gcd(d, 2026) = 1$, so $h = 1$ and $t \\equiv 1 \\pmod{2027}$, so $u \\equiv v \\pmod{2027}$, a contradiction.\n\nTherefore, all positive divisors $d$ of these $n$ satisfy condition (i).\n\n**Final answer:**\n\nAll such $n$ are $1, 9, 625, 2025$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 12816,
    "subject": "Mathematics (Olympiad)",
    "question": "The line through $C$ perpendicular to $AB$ meets line $BE$ at point $F$. Line $AF$ meets $DE$ at point $G$. The line through $A$ parallel to $BG$ meets $DE$ at $H$. Prove that $GE = GH$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $K$ be the intersection of $AB$ and $DE$, and $M$ be the intersection of $AB$ and $CF$. Join $FK$, $AE$, and $ME$. In $\\triangle ABC$, it follows from $CM \\perp AB$ that $BM \\cdot BA = BC^2 = BE^2$, so $\\triangle BEM \\sim \\triangle BAE$, and $\\angle BEM = \\angle BAE$.\n\n![](images/Mathematical_Olympiad_in_China_2011-2014_p241_data_4741127875.png)\n\nAs $\\angle FMK = \\angle FEK = 90^\\circ$, $MFEK$ is cyclic, and $\\angle BEM = \\angle FKM$. It follows from $\\angle BAE = \\angle BEM = \\angle FKM$, so $FK \\parallel AE$, and hence\n\n$$\n\\frac{KA}{KB} = \\frac{EF}{BF}, \\quad \\text{i.e., } \\frac{KA}{KB} \\cdot \\frac{BF}{FE} = 1. \\qquad \\textcircled{1}\n$$\n\nAs the line $EGA$ intersects $\\triangle EBK$, we have\n\n$$\n\\frac{EG}{GK} \\cdot \\frac{KA}{AB} \\cdot \\frac{BF}{FE} = 1 \\qquad \\textcircled{2}\n$$\n\nAs $BG \\parallel AH$, we have $\\frac{HK}{KG} = \\frac{AK}{KB}$, so\n\n$$\n\\frac{HG}{GK} = \\frac{AB}{BK}. \\qquad \\textcircled{3}\n$$\n\nFrom (1)–(3), we have $\\frac{EG}{HG} = 1$, and so $EG = HG$. $\\square$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 12817,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all numbers $p$, $q$, and $r$ such that $p$ and $r$ are prime, $q$ is a positive integer, and they satisfy the equation:\n\n$$\n(p + q + r)^2 = 2p^2 + 2q^2 + r^2.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "After simplifying the equation, we get $2r(p + q) = (p - q)^2$. Since $r$ is prime, it follows that $r$ divides $p - q$, so $r^2$ divides the right-hand side. This implies $r$ divides $2(p + q)$. If $r > 2$, then $r$ divides $p + q$, so $r$ must divide both $p$ and $q$. Since $p$ is prime, this is possible only if $p = r$ and $q = s r$. After simplification, $2(1 + s) = (s - 1)^2$, so $s^2 - 4s - 1 = 0$, which has no integer solutions. Thus, no solution in this case.\n\nIf $r = 2$, then $p$ and $q$ have the same parity. The case $p = 2$ is impossible, and $p = r$ was already considered, so $p$ must be odd. Let $a \\neq 2$ be a prime divisor of $p + q$; then $a$ must also divide $p - q$, so $a$ divides both $p$ and $q$, which is possible only if $p = a$ and $q = s a$. In this case, $4(1 + s) = a(s - 1)^2$, leading to $a^2 - (2a + 4)s + (a - 4) = 0$, with solutions $\\frac{a + 2 \\pm 2\\sqrt{2a + 1}}{a}$. If $\\sqrt{2a + 1}$ is integer, then $2a + 1 = 4b^2 + 4b + 1$, so $a = 2b(b + 1)$, which cannot be prime. Therefore, $p + q$ and $p - q$ must be powers of $2$, i.e., $p - q = 2^k$ and $p + q = 2^{2k - 2}$. Then $2p = 2^k + 2^{2k - 2}$ and $2q = 2^{2k - 2} - 2^k$, and since $p$ and $q$ are odd, $k = 1$, but then $p + q = 1$, which is impossible.\n\nIt follows that the equation has no prime number solutions.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 12818,
    "subject": "Mathematics (Olympiad)",
    "question": "Twelve students are seated at the vertices of a regular 12-gon. Each student is to be given a red or blue hat. No three students whose positions form an equilateral triangle may all receive hats of the same color, and no four students whose positions form a square may all receive hats of the same color. How many possible colorings are there?",
    "options": [],
    "answer": "See solution",
    "solution": "First, divide the 12 students into four groups of 3, each forming an equilateral triangle. For each group, there are $2^3 - 2 = 6$ possible colorings (excluding monochromatic triangles), so there are $6^4 = 1296$ colorings in total.\n\nNext, subtract the colorings where at least one square is monochromatic. The 12 students can be divided into three groups of 4, each forming a square.\n\n- For one particular square to be monochromatic: $2 \\times 3^4 = 162$ colorings (2 color choices for the square, and for the remaining 8 vertices, paired as in the triangle grouping, each pair has 3 valid colorings).\n- For two particular squares to be monochromatic: If both are the same color, the rest must be the other color; if different, each remaining vertex can be either color. This gives $2 + 2 \\times 2^4 = 34$ colorings.\n- For all three squares to be monochromatic: $2^3 - 2 = 6$ colorings (cannot all be the same color due to the triangle restriction).\n\nBy inclusion-exclusion:\n$$\n3 \\times 162 - 3 \\times 34 + 6 = 390\n$$\n\nSo, the number of valid colorings is:\n$$\n1296 - 390 = 906\n$$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 12819,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all functions $f: \\mathbb{N} \\to \\mathbb{N}$ such that\n$$\nn + f(m) \\mid f(n) + n f(m)\n$$\nfor any $m, n \\in \\mathbb{N}$.",
    "options": [],
    "answer": "See solution",
    "solution": "We consider two cases: whether the range of $f$ is infinite or finite.\n\n*Case 1.* $f$ has infinite range. Fix $n \\in \\mathbb{N}$ and let $m$ be arbitrary. By the condition,\n$$\nn + f(m) \\mid f(n) + n f(m).\n$$\nRewrite:\n$$\nf(n) + n f(m) = f(n) - n^2 + n(f(m) + n),\n$$\nso\n$$\nn + f(m) \\mid f(n) - n^2.\n$$\nSince $f$ has infinite range, we can choose $m$ so that $n + f(m) > |f(n) - n^2|$, forcing $f(n) = n^2$ for all $n$. Checking:\n$$\nn + f(m) = n + m^2,\n$$\n$$f(n) + n f(m) = n^2 + n m^2 = n(n + m^2),\n$$\nso $n + f(m) \\mid f(n) + n f(m)$ holds.\n\n*Case 2.* $f$ has finite range. Then there exists $k$ such that $1 \\leq f(n) \\leq k$ for all $n$. There is some $s$ with $f(n) = s$ for infinitely many $n$. For such $n, m$ with $f(n) = f(m) = s$:\n$$\nn + s \\mid s + n s = s - s^2 + s(n + s),\n$$\nso\n$$\nn + s \\mid s^2 - s.\n$$\nFor large $n$, $n + s > s^2 - s$ forces $s^2 = s$, so $s = 1$. Thus, $f(n) = 1$ for infinitely many $n$.\n\nFix $m$ and let $n$ with $f(n) = 1$:\n$$\nn + f(m) \\mid 1 + n f(m) = 1 - (f(m))^2 + f(m)(n + f(m)),\n$$\nso\n$$\nn + f(m) \\mid (f(m))^2 - 1.\n$$\nFor large $n$, $n + f(m) > (f(m))^2 - 1$ forces $f(m) = 1$ for all $m$. Checking:\n$$\nn + f(m) = n + 1,\n$$\n![](images/Macedonia_2017_p11_data_41ab04db02.png)\n$$f(n) + n f(m) = 1 + n,\n$$\nso $n + f(m) \\mid f(n) + n f(m)$ holds.\n\nThus, all solutions are $f(n) = n^2$ for all $n \\in \\mathbb{N}$ or $f(n) = 1$ for all $n \\in \\mathbb{N}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 12820,
    "subject": "Mathematics (Olympiad)",
    "question": "Players A and B play a game as follows. Initially, A arranges the numbers $1, 2, \\dots, n$ in a row as he wishes, where $n$ is a given positive integer. Next, B chooses one number and puts a stone on it. Then A moves the stone to an adjacent number, B does the same, and so on. The stone can be placed on number $k$ at most $k$ times for $k = 1, \\dots, n$; the initial move of B is counted. The one who cannot move loses. For each $n$, determine who has a winning strategy.",
    "options": [],
    "answer": "See solution",
    "solution": "Player A has a winning strategy if $n \\equiv 0$ or $n \\equiv -1 \\pmod{4}$; otherwise, B has one.\n\nPutting the stone on a number can be viewed as subtracting $1$ from it. We may assume that B chooses a number in A's arrangement and subtracts $1$ from it; then A must subtract $1$ from an adjacent number, and so on. Operating on a number (subtracting $1$) is allowed only if the number is positive. Note that each player always moves at positions with the same parity.\n\nLet $a_1, \\dots, a_n$ be an arrangement of $n$ nonnegative integers, not necessarily distinct. We call it *balanced* if there exist nonnegative integers $x_0, x_1, \\dots, x_n$ such that\n\n$$\nx_0 = x_n = 0 \\quad \\text{and} \\quad a_k = x_{k-1} + x_k \\quad \\text{for } k = 1, \\dots, n.\n$$\n\nWe show that $A$ can win if and only if his initial arrangement is balanced. This applies not only to $1, \\dots, n$ but to any given collection of nonnegative integers (zeros and repetitions are allowed).\n\nSuppose that $B$ has a move in a balanced arrangement $a_1, \\dots, a_n$, subtracting $1$ from $a_k = x_{k-1} + x_k$. Then $x_{k-1} > 0$ or $x_k > 0$ as the move is possible, say $x_k > 0$. So $A$ is able to respond: he can subtract $1$ from $a_{k+1}$ since $a_{k+1} = x_k + x_{k+1} \\ge x_k > 0$. Moreover, the resulting arrangement is balanced. Only $a_k$ and $a_{k+1}$ have changed, replaced by $a'_k = x_{k-1} + x'_k$ and $a'_{k+1} = x'_k + x_{k+1}$ with $x'_k = x_k - 1$, and $x'_k \\ge 0$ due to $x_k > 0$. Hence, if $B$ has a move in a balanced arrangement, then $A$ has an answering move leading to a balanced arrangement again. Since the game always terminates, it will be $B$ who ends up without a legal move.\n\nSuppose next that $A$'s initial arrangement $a_1, \\dots, a_n$ is not balanced. Then $B$ can win by reducing the game to a balanced case like above where he plays the winning role. Define\n\n$$\nx_0 = 0 \\quad \\text{and} \\quad x_k = a_k - x_{k-1} \\quad \\text{for } k = 1, \\dots, n.\n$$\n\nSet $a_{n+1} = 0$ and observe that $x_k > a_{k+1}$ for some $k = 1, \\dots, n$. Indeed, let $x_j \\le a_{j+1}$ for all $1 \\le j \\le n-1$. Then $x_1, x_2, \\dots, x_n \\ge 0$ by the definition of the $x_j$. Now notice that $x_n \\ne 0$. Otherwise, the equalities above would hold with nonnegative $x_j$'s and the arrangement would be balanced. In conclusion, $x_n > 0 = a_{n+1}$.\n\nLet $B$ start at the first position $k$ such that $x_k > a_{k+1}$, that is, $a_k > x_{k-1} + a_{k+1}$. Note that $x_j \\ge 0$ for $j < k$ by the minimum choice of $k$. Since $a_k \\ge x_{k-1} + a_{k+1}$ holds after the opening move, $B$ can play at position $k$ at least $x_{k-1} + a_{k+1}$ more times, regardless of $A$'s moves on $a_{k-1} = x_{k-2} + x_{k-1}$ or $a_{k+1}$. (For $k=1$ assume $a_{k-1} = x_{k-1} = x_{k-2} = 0$.) So let $B$ keep moving at $k$ until $A$ has to move at $k-1$ for the $(x_{k-1} + 1)$st time. Call such a move of $A$ *move M*. It is forced since $A$ has at most $a_{k+1}$ moves at $k+1$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 12821,
    "subject": "Mathematics (Olympiad)",
    "question": "Given triangle $ABC$, let $J$ be the center of the excircle opposite vertex $A$. This excircle is tangent to side $BC$ at $M$, and to lines $AB$ and $AC$ at $K$ and $L$, respectively. Lines $LM$ and $BJ$ meet at $F$, and lines $KM$ and $CJ$ meet at $G$. Let $S$ be the intersection of lines $AF$ and $BC$, and let $T$ be the intersection of lines $AG$ and $BC$. Prove that $M$ is the midpoint of $ST$.\n\n(The excircle of $ABC$ opposite vertex $A$ is tangent to $BC$, to the ray $AB$ beyond $B$, and to the ray $AC$ beyond $C$.)",
    "options": [],
    "answer": "See solution",
    "solution": "Set $\\angle CAB = 2x$, $\\angle ABC = 2y$, and $\\angle BCA = 2z$. Hence $x + y + z = 90^\\circ$. It is not difficult to see that $\\angle BAJ = \\angle CAJ = x$, $\\angle KBJ = \\angle KBM = x + z$, $\\angle MCJ = \\angle LCJ = x + y$, $\\angle BMK = \\angle BKM = y$, and $\\angle CML = \\angle CLM = z$.\n\n![](images/USA_IMO_2013-2014_p72_data_d4c16fcd0d.png)\n\n**Solution 1.** Notice that $\\angle KAJ = x$ and that\n\n$$\n\\angle KGJ = \\angle MCJ - \\angle GMC = \\angle MCJ - \\angle KMB = x + y - y = x.\n$$\n\nThis implies that $\\angle KAJ = \\angle KGJ$, hence $AKJG$ is cyclic. In particular, $\\angle AGC = \\angle AKJ = 90^\\circ$, meaning that $AG \\perp GJ$, so $AG \\parallel ML$. Now, $CML$ is an isosceles triangle with altitude $CJ$, so because $AT \\parallel ML$, $ACT$ is isosceles with altitude $LG$. Similarly, $ABS$ is isosceles with altitude $BF$.\n\nNotice now that $\\angle SAT = 2x + \\angle SAB + \\angle CAT = 2x + y + z$, where we have used the fact that $SAB$ and $CAT$ are isosceles. On the other hand, $\\angle MGT = 90^\\circ + \\angle KGJ = 90^\\circ + \\angle KAJ = 90^\\circ + x$, which implies that $\\angle MGT = \\angle SAT$, hence $SA \\parallel MG$. Because $G$ is the midpoint of $AT$, this implies that $MG$ is the midline of triangle $AST$, so $M$ is the midpoint of $ST$.\n\n**Solution 2.** Applying Menelaus' theorem to triangle $ASC$ and line $FML$ yields\n\n$$\n1 = \\frac{SF}{FA} \\cdot \\frac{AL}{LC} \\cdot \\frac{CM}{MS}.\n$$",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 12822,
    "subject": "Mathematics (Olympiad)",
    "question": "Prove that the edges of a planar finite simple graph can be oriented so that the outdegree of each vertex is at most three.",
    "options": [],
    "answer": "See solution",
    "solution": "Assign an arbitrary orientation to each edge of the graph. The key ingredient is the lemma below.\n\n**Lemma.** *There is a directed path from any vertex which is not a sink (i.e., of positive outdegree) to some vertex whose outdegree is at most 2.*\n\nAssume the lemma for the moment and let $x$ be a vertex whose outdegree exceeds 3 (if any). By the lemma, there is a directed path from $x$ to some vertex $y$ of outdegree at most 2. Reversal of the orientation of every edge along this path decreases the outdegree of $x$ by 1 and increases that of $y$ by 1, so the latter does not exceed 3. Iteration of this procedure eventually yields an orientation having the desired property.\n\nTo prove the lemma, notice that every directed planar graph $G = (V, E)$ has a vertex of outdegree at most 2, for\n\n$$\n\\sum_{x \\in V} \\mathrm{outdeg} x = |E| \\leq 3(|V| - 2).\n$$\n\nNow let $x \\in V$ have a positive outdegree and let $V'$ be the set of all vertices to which there exists a directed path from $x$. Clearly, the edges connecting $V'$ and $V \\setminus V'$ must enter $V'$, so deletion of these edges makes $V'$ into the vertex set of a directed subgraph $G'$ each vertex of which has the same outdegree as before. Since $G'$ is planar, it has a vertex of outdegree at most 2 and we are through.",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 12823,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all functions $f: (0, +infty) \\to \\mathbb{R}$ satisfying\n\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": "See solution",
    "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\n$$\nf(2^n x) = f(x)(1 + f(1))^{-n}.\n$$\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 above that $f(x) = 0$.\n\nNow suppose $f(t) \\ne 0$ for all $t > 0$. We define $g(x) := f(x)^{-1}$ for $x > 0$, and rewrite the given equation as\n\n$$\ng(x + y) - g(x) = \\frac{g(x)}{g\\left(\\frac{x}{y}\\right)} \\quad \\text{for all } x, y > 0.\n$$\n\nBy setting $y = 1$ we obtain $g(x + 1) = g(x) + 1$ for all $x > 0$. It follows that $g(n) = n - 1 + g(1)$ for all $n \\in \\mathbb{N}$. Setting $x = y = 2$ we find $g(1) = 1$. Setting $x = 1$ we obtain $g(y)g\\left(\\frac{1}{y}\\right) = 1$ for all $y > 0$. We can now rewrite the equation as\n\n$$\ng(x + y) = g(x) + g(x)g\\left(\\frac{y}{x}\\right) = g(x)\\left(1 + g\\left(\\frac{y}{x}\\right)\\right) = g(x)g\\left(\\frac{x+y}{x}\\right).\n$$\n\nFrom this, we see that $g(a)g(b) = g(ab)$ for all $a > 0$ and $b > 1$. Using $g(y)g\\left(\\frac{1}{y}\\right) = 1$, we can rewrite as\n\n$$\ng\\left(\\frac{x}{x+y}\\right)g(x + y) = g(x),\n$$\n\nwhich together with the previous line shows that $g(a)g(b) = g(ab)$ holds for all $a, b > 0$. In particular, $g(a) = g(\\sqrt{a})^2 > 0$, so $g$ is positive. Also, from the first equation above we infer the functional equation\n\n$$\ng(x + y) = g(x) + g(y) \\quad \\text{for all } x, y \\in (0, \\infty).\n$$\n\nIt is well known that this implies $g(x) = xg(1) = x$ for $x \\in \\mathbb{Q}_{>0}$. Since $g$ is positive and strictly increasing, $g(x) = x$ 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": 12824,
    "subject": "Mathematics (Olympiad)",
    "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 extension of $AB_0$, draw arc $\\overarc{P_0Q_0}$ with center $B_0$ and radius $B_0P_0$, intercepting the extension of $C_1B_0$ at $Q_0$. Draw arc $\\overarc{Q_0P_1}$ with center $C_1$ and radius $C_1Q_0$, intercepting the extension of $B_1A$ at $P_1$. Draw arc $\\overarc{P_1Q_1}$ with center $B_1$ and radius $B_1P_1$, intercepting the extension of $B_1C_0$ at $Q_1$. Draw arc $\\overarc{Q_1P'_0}$ with center $C_0$ and radius $C_0Q_1$, intercepting the extension of $AB_0$ at $P'_0$.\n\n![](images/Mathematical_Olympiad_in_China_2007-2008-Problems_and_Solutions_p55_data_cbd9b900ec.png)\n\nProve that:\n\n1. $P'_0$ and $P_0$ are coincident, and arcs $\\overarc{P_0Q_0}$ and $\\overarc{Q_1P_0}$ are tangent to each other at $P_0$.\n2. Points $P_0, Q_0, Q_1, P_1$ are concyclic.",
    "options": [],
    "answer": "See 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 $B_0P_0 = B_0P'_0$.\n\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 collinear, we know that $\\overarc{Q_1P_0}$ and $\\overarc{P_0Q_0}$ are tangent at $P_0$.\n\n(2) 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, 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, draw a common tangent line $R_1S_1$ through $Q_1$, and suppose it intercepts $P_0T$ and $P_1T$ at points $R_1$ 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 follows that points $P_0, Q_0, Q_1, P_1$ are concyclic.\n\n![](images/Mathematical_Olympiad_in_China_2007-2008-Problems_and_Solutions_p56_data_3d16556b89.png)",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 12825,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $a_1, a_2, \\dots, a_n$ be nonzero integers such that\n\n$$\na_1 a_2 \\dots a_n \\left( \\frac{1}{a_1^2} + \\frac{1}{a_2^2} + \\dots + \\frac{1}{a_n^2} \\right)\n$$\n\nis an integer. Prove that $a_k^2 \\mid a_1 a_2 \\dots a_n$ for all $k = 1, 2, \\dots, n$.",
    "options": [],
    "answer": "See solution",
    "solution": "For each $k = 1, 2, \\dots, n$, denote $b_k = \\frac{a_1 a_2 \\dots a_n}{a_k^2}$ and consider the following polynomial:\n\n$$\nP(x) = (x - b_1)(x - b_2) \\dots (x - b_n) = x^n + c_{n-1}x^{n-1} + \\dots + c_1x + c_0.\n$$\n\nBy Vieta's theorem, all coefficients $c_k$ are integers because each $b_{i_1} b_{i_2} \\dots b_{i_k}$ is an integer. Since $P(x)$ is monic with integer coefficients, all its roots $b_1, b_2, \\dots, b_n$ are integers. Therefore,\n\n$$\n\\frac{a_1 a_2 \\dots a_n}{a_1^2}, \\frac{a_1 a_2 \\dots a_n}{a_2^2}, \\dots, \\frac{a_1 a_2 \\dots a_n}{a_n^2} \\in \\mathbb{Z}.\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 12826,
    "subject": "Mathematics (Olympiad)",
    "question": "A point $(x, y)$ in the plane is a *lattice point* if $x$ and $y$ are both integers. Let $n$ be a positive integer. Prove that there exists a disk in the plane containing exactly $n$ lattice points in its interior.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $P$ be the point $(\\sqrt{2}, \\frac{1}{4})$. We shall prove that no two lattice points are equidistant from $P$. Suppose to the contrary that lattice points $Q(a, b)$ and\n![alt](path \"title\")\n$Q'(a', b')$ are equidistant from $P$. Thus,\n$$\n\\begin{aligned}\n(a - \\sqrt{2})^2 + \\left(b - \\frac{1}{4}\\right)^2 &= (a' - \\sqrt{2})^2 + \\left(b' - \\frac{1}{4}\\right)^2 \\\\\n(a - \\sqrt{2})^2 - (a' - \\sqrt{2})^2 &= \\left(b' - \\frac{1}{4}\\right)^2 - \\left(b - \\frac{1}{4}\\right)^2 \\\\\n(a - a') (a + a' - 2\\sqrt{2}) &= (b' - b) \\left(b' + b - \\frac{1}{2}\\right)\n\\end{aligned}\n$$\nThe right-hand side is rational, so $a = a'$. We must then have $b' = b$ and $(a, b) = (a', b')$.\n\nLet $r_1 < r_2 < r_3 < \\dots$ be the distances from $P$ to the lattice points. The disk centered at $P$ with radius $r$ where $r_n < r \\le r_{n+1}$ will have exactly $n$ lattice points in its interior.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 12827,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $ABCD$ be a cyclic quadrilateral. Suppose that there exists a line $l \\parallel BD$, which is tangent to the inscribed circles of triangles $ABC$ and $CDA$. Prove that the line $l$ contains the incenter of one of $\\triangle ABC$ and $\\triangle DAB$.",
    "options": [],
    "answer": "See solution",
    "solution": "Without loss of generality, suppose that this common tangent lies on the same side of $BD$ as $A$. Let's show that $BC = CD$.\n\nSuppose it's false; without loss of generality, let $BC < CD$. Let $CK$ and $CL$ be the angle bisectors in triangles $ABC$ and $CDA$, and let $B'$ and $D'$ be the points symmetric to $B$ and $D$ with respect to $CK$ and $CL$, respectively (see the figure below).\n\n![](images/Ukraine_2021-2022_p31_data_01af233597.png)\n\nDenote by $d(X)$ the oriented distance from point $X$ to the line $BD$, where $d(A) > 0$.\n\nNote that $\\angle DBC > \\frac{1}{2}\\angle BAD$, implying\n\n$$\n\\begin{aligned}\n\\angle AKB' &= 180^\\circ - 2\\angle BKC \\\\\n&= 2\\angle ABC + \\angle ACB - 180^\\circ \\\\\n&> 2\\angle ABD + \\angle BAD + \\angle ADB \\\\\n&\\geq -180^\\circ = \\angle ABD,\n\\end{aligned}\n$$\n\nso $d(B') < d(K)$. Similarly, $\\angle ALD' < \\angle ADB$, so $d(D') > d(L)$. Also note that\n\n$$\n\\frac{AK}{KB} = \\frac{AC}{CB} > \\frac{AC}{CD} = \\frac{AL}{LD} \\implies d(K) < d(L).\n$$\n\nConsider the line $k \\parallel BD$, with respect to which the points $K$ and $L$ lie on different sides. Then the inscribed circle of $\\triangle ABC$ lies on the same side of $k$ as the segment $BD$ (since this circle lies in the quadrilateral $BCB'K$, which is on this side with respect to $k$), and the inscribed circle of $\\triangle CDA$ has a point lying on the opposite side with respect to $k$ (as it's tangent to the segment $LD'$, which is on the opposite side with respect to $k$). So these inscribed circles can't have a common tangent parallel to $BD$ on the same side as $A$; this contradiction completes the proof.\n\nAlso, note that when $C$ is the midpoint of arc $BD$, we have $CB' = CB = CD = CD'$, so $B' = D'$ is the incenter of $\\triangle DAB$ (see the figure below). Furthermore, $\\angle AKB' = 180^\\circ - 2\\angle BKC = 2\\angle ABC + \\angle ACB - 180^\\circ = 2\\angle ABD + \\angle BAD + \\angle ADB - 180^\\circ = \\angle ABD$, so $KB' \\parallel BD$. Similarly, $LD' \\parallel BD$, so the line $KL$ is the common tangent.\n\n![](images/Ukraine_2021-2022_p31_data_a33e16399e.png)",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 12828,
    "subject": "Mathematics (Olympiad)",
    "question": "Пусть $N$ — такое натуральное число, что сумма $N$ и его наибольшего делителя, меньшего $N$, равна степени десяти: $N + m = 10^k$, где $m$ — наибольший делитель $N$, меньший $N$, и $k$ — натуральное число. Найдите все такие $N$.",
    "options": [],
    "answer": "See solution",
    "solution": "Ответ: $N = 75$.\n\nПусть $m$ — наибольший делитель числа $N$, меньший, чем $N$. Тогда $N = m p$, где $p$ — наименьший простой делитель числа $N$. Имеем $N + m = 10^k$, то есть $m(p + 1) = 10^k$. Число $10^k$ не делится на $3$, поэтому $p > 2$. Следовательно, $N$ нечётное, и $m$ нечётно. Так как $10^k$ делится на $m$, получаем $m = 5^s$. Если $m = 1$, то $N = p = 10^k - 1$, что невозможно, так как $10^k - 1$ делится на $9$ и не является простым. Значит, $s \\ge 1$, $N$ делится на $5$, и $p \\le 5$. Если $p = 3$, то $4 \\cdot 5^s = 10^k$, откуда $k = 2$, $m = 25$, $N = 75$. Если $p = 5$, то $p + 1 = 6$, и $10^k$ делится на $3$, что невозможно.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 12829,
    "subject": "Mathematics (Olympiad)",
    "question": "The diagonals $AC$ and $BD$ of the convex quadrilateral $ABCD$ intersect at point $O$. The points $A_1$, $B_1$, $C_1$, and $D_1$ on 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 $\\triangle AOB$ and $\\triangle COD$; $N$ be the second intersection point of the circumcircles of $\\triangle AOD$ and $\\triangle BOC$; $P$ be the second intersection point of the circumcircles of $\\triangle A_1OB_1$ and $\\triangle C_1OD_1$; and $Q$ be the second intersection point of the circumcircles of $\\triangle A_1OD_1$ and $\\triangle B_1OC_1$. Prove that the points $M$, $N$, $P$, and $Q$ are concyclic.",
    "options": [],
    "answer": "See solution",
    "solution": "It follows from the conditions that $\\angle MAC = \\angle MBD$ and $\\angle MCA = \\angle MDB$. Therefore, $\\triangle MAC \\sim \\triangle MBD$. 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 $\\triangle OXY$. 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 same argument for the quadrilateral $A_1B_1C_1D_1$ yields that $P$ and $Q$ also belong to the circumcircle of $\\triangle OXY$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 12830,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all triples $x, y, z$ of real numbers such that the following system holds:\n\n$$\n\\begin{cases}\nx^3 + y = z^2 \\\\\ny^3 + z = x^2 \\\\\nz^3 + x = y^2\n\\end{cases}\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "The only triple satisfying the given system of equations is $(x, y, z) = (0, 0, 0)$.\n\nFirst, consider the case $x, y, z \\ge 0$. Note that for all $t \\ge 0$ we have $t^3 + t \\ge 2t^2$. Summing up the three equations, we obtain $x^3 + x + y^3 + y + z^3 + z = x^2 + y^2 + z^2$. By our observation, the left-hand side is greater than or equal to $2(x^2 + y^2 + z^2)$. This yields $x^2 + y^2 + z^2 \\le 0$, which holds only for $x = y = z = 0$. The triple $(x, y, z) = (0, 0, 0)$ clearly satisfies the system.\n\nSecond, consider the case when at least two of $x, y, z$ are negative. Because of cyclicity, we may assume that $x, y < 0$. Then $z^2 = x^3 + y < 0$, which is impossible.\n\nWe are left with the case when exactly one of $x, y, z$ is negative, say $x$. Then $y, z \\ge 0$. Note that $y > y + x^3 = z^2$ and $z^3 > z^3 + x = y^2$, in particular both $y, z$ are positive. We have $z^3 > y^2 > y z^2$ which gives $z > y$. Further, $y > z^2 > y^2$, hence $1 > y$.\n\nOn the other hand, the first equation multiplied by $x$ gives $x^4 + x y = x z^2$, and the third equation multiplied by $y$ gives $y z^3 + x y = y^3$. Subtracting the obtained equalities yields $x^4 - y z^3 = x z^2 - y^3$. Rearranging terms leads to $x^4 + y^3 = x z^2 + y z^3$. Hence $0 < x^4 + y^3 = z^2(x + y z)$ which implies $x + y z > 0$.\n\nFinally, note that $x^2 = y^3 + z^2 > x^3 + y^2 = z^2$. Keeping in mind that $z > 0 > x$, we obtain $-x > z$. Therefore $0 > x + z > x + y z > 0$, a contradiction.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 12831,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $AB\\Gamma$ be an acute-angled scalene triangle with $AB < A\\Gamma < B\\Gamma$. Let $\\Delta$, $E$, and $Z$ be the midpoints of the sides $B\\Gamma$, $A\\Gamma$, and $AB$, respectively, and let $BK$ and $G\\Lambda$ be altitudes. On the extension of $\\Delta Z$ beyond $Z$, consider a point $M$ such that the parallel from $M$ to $K\\Lambda$ intersects the extensions of $\\Gamma A$, $BA$, and $\\Delta E$ at points $\\Sigma$, $T$, and $N$, respectively. If the circumcircle of triangle $MB\\Delta$, denoted $(c_1)$, intersects the line $\\Delta N$ at point $P$, and the circumcircle of triangle $N\\Gamma\\Delta$, denoted $(c_2)$, intersects the line $\\Delta M$ at point $\\Pi$, prove that $\\Sigma T \\parallel \\Pi P$.\n\n![](images/GreekMO2014_booklet_p19_data_29558f1272.png)",
    "options": [],
    "answer": "See solution",
    "solution": "Since $\\Delta$, $E$, and $Z$ are the midpoints of the sides $B\\Gamma$, $A\\Gamma$, and $AB$, respectively, the quadrilaterals $AE\\Delta Z$, $ZE\\Delta B$, and $ZE\\Gamma\\Delta$ are parallelograms.\n\nFrom $\\Delta M \\parallel \\Gamma \\Sigma$ and $MN \\parallel K\\Lambda$, we conclude that $\\angle M = \\angle \\Sigma_1$ and $\\angle \\Sigma_1 = \\angle K$.\n\nFrom the cyclic quadrilateral $B\\Lambda K\\Gamma$ (since $\\angle B\\Lambda\\Gamma = \\angle B K \\Gamma = 90^\\circ$), we get $\\angle K = \\angle B$.\n\nFrom the last three equalities of angles, we arrive at $\\angle M = \\angle B$, and from this we conclude that the points $M$, $B$, $\\Delta$, $T$ are concyclic.\n\nSimilarly, since $\\Delta N \\parallel BT$ and $MN \\parallel K\\Lambda$, we have $\\angle N = \\angle T_1$ and $\\angle T_1 = \\angle \\Lambda$. Also, from the cyclic quadrilateral $B\\Lambda K\\Gamma$ (since $\\angle B\\Lambda\\Gamma = \\angle B K \\Gamma = 90^\\circ$), we have $\\angle \\Lambda = \\angle \\Gamma$.\n\nFrom the last three equalities of angles, we have $\\angle N = \\angle \\Gamma$, and from this we conclude that the points $N$, $\\Gamma$, $\\Delta$, $\\Sigma$ are concyclic.\n\n![](images/GreekMO2014_booklet_p19_data_29558f1272.png)\n\nSince the points $B$, $\\Delta$, $P$, $T$, $M$ belong to the circle $(c_1)$, from the cyclic quadrilateral $MTP\\Delta$ we have $\\angle T_2 + \\angle \\Delta = 180^\\circ$.\n\nSince the points $\\Gamma$, $\\Delta$, $\\Pi$, $\\Sigma$, $N$ belong to the circle $(c_2)$, from the cyclic quadrilateral $\\Delta\\Pi\\Sigma N$ we have $\\angle \\Sigma_2 + \\angle \\Delta = 180^\\circ$.\n\nFrom the last two equalities we get:\n\n$$\n\\angle T_2 = \\angle \\Sigma_2 = 180^\\circ - \\angle \\Delta = 180^\\circ - \\angle A.\n$$\n\nSince $\\angle T_2 + \\angle E = 180^\\circ - \\angle A + \\angle A = 180^\\circ$, the quadrilateral $\\Sigma TPE$ is cyclic; let $(c)$ be its circumcircle. Next, we will prove that the quadrilateral $TPEZ$ is cyclic, which means that the point $Z$ belongs to the circle $(c)$. For that, we are going to prove $\\angle TZE + \\angle PTE = 180^\\circ$ ($\\angle PTE = \\angle P\\Delta$).\n\nThe quadrilateral $MTP\\Delta$ is inscribed in the circle $(c_1)$, and hence:\n\n$$\n\\angle PTE = \\angle P\\Delta = 180^\\circ - \\angle M\\Delta = 180^\\circ - \\angle B\\Gamma = 180^\\circ - \\angle B.\n$$\n\nAlso, $\\angle TZE = \\angle TB\\Gamma = \\angle B$. Summing up the last two equalities, we have:\n\n$$\n\\angle TZE + \\angle PTE = 180^{\\circ} - \\angle B + \\angle B = 180^{\\circ}.\n$$\n\nFinally, we will prove that the point $\\Pi$ belongs to $(c)$.\n\nThe angle $\\angle \\Sigma_2 = 180^{\\circ} - \\angle A$ is an external angle of triangle $\\Sigma\\Pi M$, and so:\n\n$$\n\\angle \\Sigma\\Pi Z = \\angle \\Sigma\\Pi M = \\angle \\Sigma_2 - \\angle M = 180^{\\circ} - \\angle A - \\angle B = \\angle \\Gamma = \\angle \\Sigma E Z.\n$$\n\nThe quadrilateral $\\Sigma\\Pi\\Gamma\\Pi$ is inscribed in the circle $(c)$, and since $\\angle T_2 = \\angle \\Sigma_2$, it is an isosceles trapezium, so $\\Sigma T \\parallel \\Pi P$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 12832,
    "subject": "Mathematics (Olympiad)",
    "question": "Suppose there are 100 players, and each pair of players must meet exactly once in a tournament. What is the minimum number of days required to schedule the tournament if, on each day, each player is either assigned to be 'white', 'black', or 'idle', and all 'white' players play all 'black' players that day (no other matches occur)?",
    "options": [],
    "answer": "See solution",
    "solution": "The answer is 99.\n\nClearly, the tournament can be run in 99 days if we assign, on day $k$ (where $1 \\leq k \\leq 99$):\n- player $k$ to be *white*;\n- players $k+1, k+2, \\ldots, 100$ to be *black*;\n- all other players to be *idle*.\n\nIn this way, player $i$ and player $j$ (where $1 \\leq i < j \\leq 100$) would have met exactly once on day $i$.\n\nNow suppose the number of days is less than 99, and we derive a contradiction. Assign a real number $x_i$ to each player $i$, such that:\n\n1. At least one $x_i$ is nonzero;\n2. The sum of all $x_i$ is zero;\n3. The sum of the $x_i$'s of all 'white' players on each day is zero.\n\nThis is possible because (2) and (3) represent a homogeneous system of fewer than 100 linear equations in 100 unknowns, so a nonzero solution exists.\n\nIt follows that\n\n$$\n0 = (x_1 + x_2 + \\cdots + x_{100})^2 = (x_1^2 + x_2^2 + \\cdots + x_{100}^2) + 2 \\sum_{1 \\leq i < j \\leq 100} x_i x_j. \\quad (4)\n$$\n\nSince any two players have met exactly once,\n\n$$\n\\sum_{1 \\leq i < j \\leq 100} x_i x_j = \\sum_k \\left( \\sum_{w \\in W_k} x_w \\sum_{b \\in B_k} x_b \\right)\n$$\n\nwhere $W_k$ and $B_k$ are the sets of 'white' and 'black' players on day $k$. Since the sum of the $x_i$'s of all 'white' players on each day is zero, the above sum is zero. On the other hand, $x_1^2 + x_2^2 + \\cdots + x_{100}^2 > 0$ as at least one $x_i$ is nonzero. This contradicts (4). Thus, at least 99 days are required.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 12833,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $a_n$ be a sequence defined for positive integers $n$, and let $b_1 = a_1 = 1$ and $b_n = a_n - a_{n-1}$ for $n > 1$. Suppose that for all $n \\geq 1$,\n\n$$\n\\sum_{i|n} b_i = n^{10} - (n-1)^{10}.\n$$\n\nLet $\\mu(n)$ denote the Möbius function, and $\\phi(n)$ Euler's totient function. Prove the following:\n\n1. For all $n \\geq 1$, $\\phi(n)$ divides $b_n$.\n2. For all $n \\geq 1$, $b_n > 0$, and thus $a_n \\geq n$.\n3. For any integer $c > 1$ and $n > 1$, show that $n$ divides $c^{a_n} - c^{a_{n-1}}$.",
    "options": [],
    "answer": "See solution",
    "solution": "We start by noting that for $n > 1$,\n\n$$\n\\sum_{i=1}^{n-1} \\left( a_{\\lfloor n/i \\rfloor} - a_{\\lfloor (n-1)/i \\rfloor} \\right) + a_1 = n^{10} - (n-1)^{10}.\n$$\n\nDefining $b_1 = a_1 = 1$ and $b_n = a_n - a_{n-1}$ for $n > 1$, we have\n\n$$\n\\sum_{i|n} b_i = n^{10} - (n-1)^{10}.\n$$\n\nBy Möbius inversion, for any sequence $y_n = \\sum_{i|n} x_i$, we have\n\n$$\nx_n = \\sum_{i|n} \\mu(i) y_{n/i}.\n$$\n\nApplying this to our sequence,\n\n$$\nb_n = \\sum_{i|n} \\mu(i) \\left( \\left( \\frac{n}{i} \\right)^{10} - \\left( \\frac{n}{i} - 1 \\right)^{10} \\right).\n$$\n\nExpanding, we get\n\n$$\nb_n = 10 \\sum_{i|n} \\mu(i) \\left( \\frac{n}{i} \\right)^9 - 45 \\sum_{i|n} \\mu(i) \\left( \\frac{n}{i} \\right)^8 + \\dots + 10 \\sum_{i|n} \\mu(i) \\left( \\frac{n}{i} \\right) - \\sum_{i|n} \\mu(i).\n$$\n\nFor $n > 1$, $\\sum_{i|n} \\mu(i) = 0$. For each $r \\geq 1$,\n\n$$\n\\sum_{i|n} \\mu(i) \\left( \\frac{n}{i} \\right)^r = n^r \\prod_{p|n} \\left(1 - \\frac{1}{p^r}\\right) = \\phi(n) n^{r-1} \\prod_{p|n} \\left(1 + \\frac{1}{p} + \\dots + \\frac{1}{p^{r-1}}\\right).\n$$\n\nTherefore,\n\n$$\n\\frac{b_n}{\\phi(n)} = 10 n^8 \\prod_{p|n} \\left(1 + \\frac{1}{p} + \\dots + \\frac{1}{p^8}\\right) - 45 n^7 \\prod_{p|n} \\left(1 + \\frac{1}{p} + \\dots + \\frac{1}{p^7}\\right) + \\dots - 45 n \\prod_{p|n} \\left(1 + \\frac{1}{p}\\right) + 10.\n$$\n\n**(1) $\\phi(n) \\mid b_n$:**\n\nSince $b_n$ is an integer and the right side is $\\phi(n)$ times an integer, $\\phi(n)$ divides $b_n$.\n\n**(2) $b_n > 0$ and $a_n \\geq n$:**\n\nDirect computation for small $n$ shows $b_1 = 1 > 0$, $b_2 = 2^{10} - 2 > 0$, $b_3 = 3^{10} - 2^{10} - 1 > 0$, $b_4 = 4^{10} - 3^{10} - 2^{10} + 1 > 0$. For $n \\geq 5$, the formula above shows $b_n > 0$. Thus, $a_n = a_{n-1} + b_n \\geq n$ by induction.\n\n**(3) $n \\mid c^{a_n} - c^{a_{n-1}}$ for $n > 1$:**\n\nLet $n = m s$, where $m$ is the product of prime divisors of $n$ dividing $c$, and $s$ is coprime to $c$. Then $m \\mid c^{n-1} \\mid c^{a_{n-1}}$ (since $a_{n-1} \\geq n-1$), and $s \\mid c^{\\phi(s)-1} \\mid c^{b_n-1}$ (since $\\phi(n) \\mid b_n$). Therefore, $n \\mid c^{a_{n-1}} (c^{b_n-1}) = c^{a_n} - c^{a_{n-1}}$.",
    "topic": "Number Theory",
    "subtopic": "Number-Theoretic Functions"
  },
  {
    "id": 12834,
    "subject": "Mathematics (Olympiad)",
    "question": "The number $2019$ is written on the board. Katia and Mykola play the following game: one by one (starting with Katia), they choose any divisor $d$ of the number $N$ currently on the board and change $N$ to $N - (4d - 1)$, provided the result is a positive integer. Whoever cannot make a move loses. Who will win this game and what is the winning strategy, assuming both play optimally?\n\n![](path/to/file.png)",
    "options": [],
    "answer": "See solution",
    "solution": "Mykola wins.\n\nWith each turn, the number on the board decreases, and its parity changes. Since Katia starts, after her turn the number is even; after Mykola's turn, it is odd. A move is impossible if the number is $1$, $2$, or $3$. For all other numbers, choosing $d = 1$ works, so $N$ becomes $N - 3$.\n\nWhen Mykola plays, the number is even. He can only lose if he faces $2$. For Katia to leave $2$, she must choose a divisor $d$ of $N = dD$ such that:\n\n$$\ndD - (4d - 1) = d(D - 4) + 1 = 2 \\implies d(D - 4) = 1 \\implies d = 1,\\ D = 5 \\implies N = 5.\n$$\n\nSo, if $5$ appears after Mykola's turn, Katia can win. Let's check when $5$ can appear:\n\n$$\nd(D - 4) + 1 = 5 \\implies d(D - 4) = 4.\n$$\n\nPossible cases:\n\n1. $d = 1$, $D = 8$ ($N = 8$): Mykola chooses $d = 2$, so $N - (4d - 1) = 8 - 7 = 1$.\n2. $d = 2$, $D = 6$ ($N = 12$): Mykola chooses $d = 3$, so $N - (4d - 1) = 12 - 11 = 1$.\n3. $d = 4$, $D = 5$ ($N = 20$): Mykola chooses $d = 5$, so $N - (4d - 1) = 20 - 19 = 1$.\n\nThus, Mykola can always avoid leaving $5$ on the board. If after Katia's turn the number is $8$, $12$, or $20$, Mykola replaces it with $1$. Otherwise, he can make any valid move. In this way, $5$ will never appear, and Mykola wins.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 12835,
    "subject": "Mathematics (Olympiad)",
    "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 for how many vertices the two polygons can have.",
    "options": [],
    "answer": "See solution",
    "solution": "The ratio of the inradius to the circumradius of a regular $n$-gon is $\\cos\\frac{180^\\circ}{n}$. Thus, the ratio of the areas of the incircle and circumcircle is $\\cos^2\\frac{180^\\circ}{n}$.\n\nLet a regular $n$-gon and a regular $m$-gon share a common circumcircle. Without loss of generality, let $n \\leq m$ and let the area of the common circumcircle be $1$. Then the areas of the incircles are $\\cos^2\\frac{180^\\circ}{n}$ and $\\cos^2\\frac{180^\\circ}{m}$, respectively. Since their sum equals the area of the circumcircle, we have:\n\n$$\n\\cos^2\\frac{180^\\circ}{n} + \\cos^2\\frac{180^\\circ}{m} = 1\n$$\n\nThis is equivalent to:\n\n$$\n\\cos^2\\frac{180^\\circ}{n} = \\sin^2\\frac{180^\\circ}{m}\n$$\n\nSince $n, m > 2$, both $\\frac{180^\\circ}{n}$ and $\\frac{180^\\circ}{m}$ are less than $90^\\circ$, so:\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\nwhich implies:\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 $m < 4$, which contradicts $n \\leq m$.\n\n**Answer:** The possible pairs are $(n, m) = (3, 6)$ and $(4, 4)$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 12836,
    "subject": "Mathematics (Olympiad)",
    "question": "Which positive integers $d$ have the property that the set $A_d$ of all positive integers consisting only of the digit $1$ (in base $10$) and divisible by $d$ is finite?",
    "options": [],
    "answer": "See solution",
    "solution": "All positive integers which are coprime with $10$.\n\nIf a number is divisible by either $2$ or $5$, then it cannot divide any number of the form $11\\ldots1$. Therefore, for all numbers which are not coprime with $10$, the set $A_d$ is not finite.\n\nNow let $d$ be a positive integer which is coprime with $10$ and $b > d$. Let us show that any number $a_b a_{b-1} \\dots a_1$ is not in the set $A_d$. Consider the numbers $a_1, a_2 a_1, a_3 a_2 a_1, \\dots, a_b a_{b-1} \\dots a_1 \\pmod d$. Since $d > b$, there are $i$ and $j$, $i > j$, such that $a_i a_{i-1} \\dots a_1 \\equiv a_j a_{j-1} \\dots a_1 \\pmod d$. Therefore, $a_i a_{i-1} \\dots a_{j-1} \\cdot 10^j \\equiv 0 \\pmod d$. Since $d$ is coprime with $10$, we get $d \\mid a_i a_{i-1} \\dots a_{j-1}$ and hence $a_b a_{b-1} \\dots a_1$ is not in $A_d$. Thus, all numbers in $A_d$ have at most $d$ digits and consequently $A_d$ is finite.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 12837,
    "subject": "Mathematics (Olympiad)",
    "question": "There is an infinite sequence of letters $a$ and $b$. In this sequence, one can perform a substitution $abb \\rightarrow baa$. It is known that regardless of the order of these substitutions, one can only make a finite number of them. Prove that, in that case, substitutions $aabb \\rightarrow bbaa$ can also be made only a finite number of times.",
    "options": [],
    "answer": "See solution",
    "solution": "Since only a finite number of substitutions can be made, there exists a number $N$ such that, starting from position $N$, no further substitutions are possible. This means there are no pairs of consecutive $b$'s beyond $N$, because otherwise, with an $a$ to the left, a substitution would still be possible. Therefore, the second type of substitution ($aabb \\rightarrow bbaa$) is also impossible beyond $N$, since it requires a pair of $b$'s. All other pairs of $b$'s are moved leftward by each second-type substitution, and since there are only finitely many such pairs, only finitely many such substitutions can occur.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 12838,
    "subject": "Mathematics (Olympiad)",
    "question": "Suppose for a triangle $ABC$ the following condition is satisfied:\n\nA circle $O$ going through the vertices $B, C$ intersects the line segments $AB$ and $AC$ (excluding the end-points) at the points $D$ and $E$, respectively, and $AD + AE = BC$ is satisfied.\n\nLet $I$ be the incenter of the triangle $ABC$ and suppose that the lines $BI$ and $CI$ intersect the circle $O$ at $P, Q$ (different from $B, C$), respectively. Prove that the points $A, I, P, Q$ lie on the same circumference of a circle.\n\nHere, $XY$ denotes the length of the line segment $XY$.",
    "options": [],
    "answer": "See solution",
    "solution": "From $AD + AE = BC$ it follows that we can choose a point $F$ on the side $BC$ such that $AD = CF$ and $AE = BF$. We have $DP = CP$, since the angles the chords $DP$ and $CP$ subtend on the circle $O$ are equal. We also have $\\angle PDA = \\angle PCB = \\angle PCF$, since the points $P, C, B, D$ lie on the circumference of the same circle. These facts, together with $AD = FC$, imply that the triangles $PDA$ and $PCF$ are congruent. Therefore, we have",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 12839,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $n \\ge 2$ be an integer. Call an *interval* a subset $A \\subseteq \\{1, 2, \\dots, n\\}$ for which there exist integers $1 \\le a < b \\le n$ such that $A = \\{a, a+1, \\dots, b-1, b\\}$. Let $\\mathcal{A}$ be a family of subsets $A_i \\subseteq \\{1, 2, \\dots, n\\}$, with $1 \\le i \\le N$, such that for any $1 \\le i < j \\le N$, the intersection $A_i \\cap A_j$ is an interval. Prove that $N \\le \\lfloor n^2/4 \\rfloor$, and that this bound is sharp.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $A_i \\subseteq B_i = [\\min A_i, \\max A_i] \\cap \\{1, 2, \\dots, n\\}$ for $1 \\le i \\le N$, so $B_i$ are intervals. If $B_i = B_j$ for some $i \\ne j$, then $A_i = A_j$, since $A_i \\cap A_j$ is an interval. Since $A_i \\cap A_j \\subseteq B_i \\cap B_j$, we have $|B_i \\cap B_j| \\ge |A_i \\cap A_j| \\ge 2$, so the family $\\mathcal{B}$ of the subsets $B_i$ also has the required property.\n\nBut $\\bigcap_{i=1}^{N} B_i$ contains an interval of length at least $1$; let that be $[a+1, a+2]$. Then there are at most\n\n$$\n(a+1)(n-a-1) = \\left(\\frac{n}{2}\\right)^2 - \\left(a - \\frac{n-2}{2}\\right)^2\n$$\n\npossibilities, so $N \\le \\lfloor \\left(\\frac{n}{2}\\right)^2 \\rfloor = \\lfloor n^2/4 \\rfloor$.\n\nAn example of such a maximal family is the family of all intervals containing the interval $\\lfloor n/2 \\rfloor, \\lfloor n/2 \\rfloor + 1$, so the bound is sharp.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 12840,
    "subject": "Mathematics (Olympiad)",
    "question": "Suppose a number $n$ exists with the property that $3n$ and $7n$ have only even digits. Let $N$ be the smallest such number. Is it possible for such an $N$ to exist?",
    "options": [],
    "answer": "See solution",
    "solution": "If $N$ is divisible by $10$, then $\\frac{N}{10}$ would be a smaller such number, so $N$ cannot end in $0$. Thus, $N$ must end in $2$, $4$, $6$, or $8$.\n\nConsider $3N + 7N = 10N$. The unit digit of $N$ is even, so the unit digit of $10N$ is $0$, and the tens digit of $10N$ is the unit digit of $N$. However, the sum $3N + 7N$ must have only even digits, but the carry from the unit digits would make the tens digit odd, which is a contradiction. Therefore, such an $N$ cannot exist.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 12841,
    "subject": "Mathematics (Olympiad)",
    "question": "設 $P(x) = x^2 + 1$。設 $P(x) = a x^2 + b$，對足夠大的 $x$，取 $y = \\sqrt{a x}$。若 $|y^2 - P(x)| < 2x$，求 $a$ 與 $b$ 的值。",
    "options": [],
    "answer": "See solution",
    "solution": "由 $y = \\sqrt{a x}$，有 $y^2 = a x$，所以\n\n$$\n|a x - (a x^2 + b)| < 2x\n$$\n\n整理得：\n\n$$\n|a x - a x^2 - b| < 2x\n$$\n\n考慮主項，令 $x$ 很大，則 $a x^2$ 必須與 $x^2$ 的係數相同，否則不等式不成立。故 $a = 1$。\n\n再設 $y = x + 1$，則 $y^2 = (x + 1)^2 = x^2 + 2x + 1$，所以\n\n$$\n|x^2 + 2x + 1 - (x^2 + 1 + b)| < 2x\n$$\n\n即\n\n$$\n|2x + 1 - b| < 2x\n$$\n\n這表示 $-2x < 2x + 1 - b < 2x$，即 $-2x < 2x + 1 - b$ 和 $2x + 1 - b < 2x$。\n\n第一個不等式：\n\n$$\n-2x < 2x + 1 - b \\implies b < 4x + 1\n$$\n\n第二個不等式：\n\n$$\n2x + 1 - b < 2x \\implies 1 - b < 0 \\implies b > 1\n$$\n\n所以 $b = 1$。\n\n因此，$a = 1$，$b = 1$。",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 12842,
    "subject": "Mathematics (Olympiad)",
    "question": "You are given a triangle $ABC$, $\\omega$ is its circumscribed circle, $I$ is its incenter. Let $K$ be any point on the arc $AC$ of $\\omega$, not containing point $B$. Point $P$ is symmetrical to the point $I$ with respect to the point $K$. Point $T$ on the arc $AC$ of the circle $\\omega$, which contains point $B$, is such that $\\angle KCT = \\angle PCI$. Prove that the bisectors of angles $AKC$ and $ATC$ meet at the line $CI$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $W$ be the midpoint of the arc $AC$ of circle $\\omega$ which doesn't contain point $B$, $M$ be the midpoint of $CI$, and the line $TW$ meet $CI$ at the point $Q$. As $TW$ is the bisector of $\\angle ATC$, it's enough to show that $Q$ lies on the bisector of $\\angle AKC$.\n\n![](images/Ukraine_2021-2022_p26_data_a12b59a2b8.png)\n\nNote that $KM$ is the midline of triangle $PIC$, and therefore we get $\\angle QMK = \\angle ICP = \\angle TCK = 180^\\circ - \\angle TWK = 180^\\circ - \\angle QWK$. (We can do this in oriented angles as well.) So, points $Q, M, W, K$ are concyclic. By the trilliim theorem, we get $WI = WC$, so $\\angle WMC = 90^\\circ$, so $WQ$ is the diameter of the circle. Then $\\angle QKW = 90^\\circ$, and as $KQ \\perp WK$—the external bisector of $\\angle AKC$—$Q$ has to lie on the internal bisector of angle $AKC$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 12843,
    "subject": "Mathematics (Olympiad)",
    "question": "Suppose we have a necklace of $n$ beads. Each bead is labeled with an integer and the sum of all these labels is $n-1$. Prove that we can cut the necklace to form a string whose consecutive labels $x_1, x_2, \\dots, x_n$ satisfy\n\n$$\n\\sum_{i=1}^{k} x_i \\leq k-1\n$$\n\nfor any $k = 1, \\dots, n$.",
    "options": [],
    "answer": "See solution",
    "solution": "Number the beads $1, 2, \\dots, n$ starting from some arbitrary position, and let $z_i$ be the label of bead $i$, with the convention that $z_{n+i} = z_i$.\n\nLet $S_j = z_1 + z_2 + \\dots + z_j - \\frac{j(n-1)}{n}$. We have $S_j = S_{n+j}$. Then the sum of the labels on beads $m+1, m+2, \\dots, m+k$ is $S_{m+k} - S_m + \\frac{k(n-1)}{n}$.\n\nNow choose $m$ such that $S_m$ is maximal and cut between $m$ and $m+1$. If we do so, we have\n\n$$\nS_{m+k} - S_m + \\frac{k(n-1)}{n} \\leq k - \\frac{k}{n},\n$$\n\nbut the left side is an integer, so we can replace the right side by $k-1$.",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 12844,
    "subject": "Mathematics (Olympiad)",
    "question": "A Pattano coin is a coin which has a blue side and a yellow side. A positive integer not exceeding 100 is written on each side of every coin (the sides may have different integers).\n\nTwo Pattano coins are *identical* if the number on the blue side of both coins are equal and the number on the yellow side of both coins are equal.\n\nTwo Pattano coins are *pairable* if the number on the blue side of both coins are equal or the number on the yellow side of both coins are equal.\n\nGiven 2559 Pattano coins such that no two coins are identical, show that at least one Pattano coin is pairable with at least 50 other coins.",
    "options": [],
    "answer": "See solution",
    "solution": "We represent each Pattano coin by an ordered pair $ (i, j) $, where $ i $ and $ j $ are the numbers written on the blue and yellow side respectively. Let $ C $ be the set of non-identical 2559 Pattano coins. Let $ b_i $ denote the number of Pattano coins in $ C $ with $ i $ on its blue side, and $ y_j $ denote the number of Pattano coins in $ C $ with $ j $ on its yellow side. For a Pattano coin $ (i, j) $ in $ C $, let $ p_{i,j} $ be the number of coins in $ C $ that are pairable with $ (i, j) $ (apart from itself).\n\nIt is easy to see that\n\n$$\np_{i,j} = b_i + y_j - 2,\n$$\n\nthus\n\n$$\n\\sum_{(i,j) \\in C} p_{i,j} = \\sum_{(i,j) \\in C} b_i + \\sum_{(i,j) \\in C} y_j - 5118.\n$$\n\nNote that $ b_i $ appears in the sum $ \\sum_{(i,j) \\in C} b_i $ exactly $ b_i $ times, and $ y_j $ appears in the sum $ \\sum_{(i,j) \\in C} y_j $ exactly $ y_j $ times. Hence\n\n$$\n\\sum_{(i,j) \\in C} b_i = \\sum_{i=1}^{100} b_i^2 \\quad \\text{and} \\quad \\sum_{(i,j) \\in C} y_j = \\sum_{j=1}^{100} y_j^2.\n$$\n\nApplying the Cauchy-Schwarz inequality to the vectors $ (b_1, b_2, \\dots, b_{100}) $, $ (1, 1, \\dots, 1) $ and $ (y_1, y_2, \\dots, y_{100}) $, $ (1, 1, \\dots, 1) $ we obtain,\n\n$$\n\\left(\\sum_{i=1}^{100} b_i\\right)^2 \\leq 100 \\sum_{i=1}^{100} b_i^2, \\qquad \\left(\\sum_{j=1}^{100} y_j\\right)^2 \\leq 100 \\sum_{j=1}^{100} y_j^2.\n$$\n\nCombining the above yields\n\n$$\n\\sum_{(i,j) \\in C} p_{i,j} \\geq \\frac{1}{100} \\left[ \\left( \\sum_{i=1}^{100} b_i \\right)^2 + \\left( \\sum_{j=1}^{100} y_j \\right)^2 \\right] - 5118.\n$$\n\nBut $ \\sum_{i=1}^{100} b_i = \\sum_{j=1}^{100} y_j = 2559 $, thus\n\n$$\n\\sum_{(i,j) \\in C} p_{i,j} \\geq \\frac{1}{100} (2559^2 + 2559^2) - 5118 = \\frac{2559^2}{50} - 5118.\n$$\n\nFinally, by the pigeonhole principle, there exists $ (i,j) \\in C $ such that $ p_{i,j} \\geq \\left\\lfloor \\frac{2559}{50} - 2 \\right\\rfloor = 50 $. That is, there exists a Pattano coin which is pairable with at least 50 other coins.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 12845,
    "subject": "Mathematics (Olympiad)",
    "question": "Given 50 points in the plane, with no three collinear, each point is colored with one of four colors. Prove that there exists a color such that there are at least 130 scalene triangles whose vertices are all of that color.",
    "options": [],
    "answer": "See solution",
    "solution": "Since $50 = 4 \\times 12 + 2$, by the pigeonhole principle, at least one color is used for at least 13 points. With these 13 points, we can form $\\binom{13}{3} = 286$ triangles, since no three are collinear.\n\nWe claim that there are at most $12 \\times 13 = 156$ isosceles triangles among these. There are $\\binom{13}{2} = 78$ distinct line segments, and each can be the base of at most two isosceles triangles (since no three points are collinear). Thus, there are at most $78 \\times 2 = 156$ isosceles triangles. Therefore, the number of scalene triangles is at least $286 - 156 = 130$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 12846,
    "subject": "Mathematics (Olympiad)",
    "question": "La sucesión $a_1, a_2, \\dots$ de números reales positivos verifica\n\n$$\na_{k+1} \\ge \\frac{k a_k}{a_k^2 + (k-1)}\n$$\n\npara todo entero positivo $k$. Demostrar que\n\n$$\na_1 + a_2 + \\dots + a_n \\ge n\n$$\n\npara todo $n \\ge 2$.",
    "options": [],
    "answer": "See solution",
    "solution": "Observemos que de la condición $a_{k+1} \\geq \\frac{k a_k}{a_k^2 + (k-1)}$ se sigue que\n\n$$\n\\frac{k}{a_{k+1}} \\leq a_k + \\frac{k-1}{a_k}\n$$\n\nluego\n\n$$\na_k \\geq \\frac{k}{a_{k+1}} - \\frac{k-1}{a_k}\n$$\n\nSumando las desigualdades anteriores para $k = 1, 2, \\dots, m$ tenemos\n\n$$\na_1 + a_2 + \\dots + a_m \\geq \\left(\\frac{1}{a_2} - 0\\right) + \\left(\\frac{2}{a_3} - \\frac{1}{a_2}\\right) + \\dots + \\left(\\frac{m}{a_{m+1}} - \\frac{m-1}{a_m}\\right) = \\frac{m}{a_{m+1}}\n$$\n\nDemostraremos el resultado deseado por inducción sobre $n$:\n\nPara $n = 2$, la condición del enunciado para $k = 1$ asegura\n\n$$\na_1 + a_2 \\geq a_1 + \\frac{1}{a_1} \\geq 2\n$$\n\nSuponiendo que el resultado es cierto para $n \\geq 2$, y que $a_{n+1} \\geq 1$, la hipótesis de inducción lleva a\n\n$$\n(a_1 + a_2 + \\dots + a_n) + a_{n+1} \\geq n + a_{n+1} \\geq n + 1\n$$\n\nPor último, si fuera $a_{n+1} < 1$ resultaría\n\n$$\n(a_1 + a_2 + \\dots + a_n) + a_{n+1} \\geq \\frac{n}{a_{n+1}} = \\frac{n-1}{a_{n+1}} + \\left(\\frac{1}{a_{n+1}} + a_{n+1}\\right) > (n-1) + 2 = n + 1\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 12847,
    "subject": "Mathematics (Olympiad)",
    "question": "If $x$, $y$, and $z$ are real numbers such that $x^2 + y^2 + z^2 = 2$, prove that $x + y + z \\leq xyz + 2$.",
    "options": [],
    "answer": "See solution",
    "solution": "Notice that $2xy \\le x^2 + y^2 \\le x^2 + y^2 + z^2 = 2$, therefore $xy \\le 1$. Similarly, $xz \\le 1$, and $yz \\le 1$. We also have $(x + y)^2 = x^2 + y^2 + 2xy \\le 4$, so $x + y \\le |x + y| \\le 2$. Also, $x + z \\le 2$, and $y + z \\le 2$.\n\nIf one of the numbers is negative, let's say $z$, then the conclusion follows from the inequalities $x + y \\le 2$ and $z \\le xyz$.\n\nIn the case $x, y, z \\in [0, 1]$, we have $z(1 - xy) \\le 1 - xy \\le 2 - x - y$ (the last inequality is equivalent to $(1 - x)(1 - y) \\ge 0$).\n\nWe may have at most one number greater than $1$, otherwise the product of two such numbers would be greater than $1$. We are left with the case $x, y \\in [0, 1]$ and $z > 1$. In this case $(1 - x)(1 - y)(z - 1) \\ge 0 \\Leftrightarrow xyz + 2 \\ge xy + yz + xz + 3 - (x + y + z) \\ge x + y + z$, the last inequality being equivalent to $2(xy + yz + zx) + 6 = (x + y + z)^2 + 4 \\ge 4(x + y + z)$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 12848,
    "subject": "Mathematics (Olympiad)",
    "question": "Suppose $a$, $b$, $c$, $d$ are distinct nonzero real numbers such that $ac = bd$ and\n$$\n\\frac{a}{b} + \\frac{b}{c} + \\frac{c}{d} + \\frac{d}{a} = 4.\n$$\nDetermine the largest possible value of\n$$\n\\frac{a}{c} + \\frac{b}{a} + \\frac{c}{d} + \\frac{d}{b}.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "Plugging $d = \\frac{ac}{b}$ into the second relation yields\n$$\n\\frac{a}{b} + \\frac{b}{c} + \\frac{b}{a} + \\frac{c}{b} = 4.\n$$\nSet $x = \\frac{a}{b}$ and $y = \\frac{b}{c}$. We seek the maximum of\n$$\nxy + \\frac{1}{xy} + \\frac{x}{y} + \\frac{y}{x} = \\left(x + \\frac{1}{x}\\right)\\left(y + \\frac{1}{y}\\right)\n$$\nsubject to $x + \\frac{1}{x} + y + \\frac{1}{y} = 4$.\n\nLet $A = x + \\frac{1}{x}$, $B = y + \\frac{1}{y}$, so $A + B = 4$. If $x, y > 0$, then $A, B \\ge 2$, so $A = B = 2$, which gives $x = y = 1$. This leads to $a = b = c = d$, a contradiction.\n\nWithout loss of generality, assume $A < 0$ and $B > 0$, i.e., $x < 0$ and $y > 0$. Then $A \\le -2$, so $B \\ge 6$. We need to maximize $AB = (4 - B)B$ with $B \\ge 6$. Since $B - 4 \\ge 2$, $B(B - 4) \\ge 12$, so $B(4 - B) \\le -12$. This value is achieved when $A = -2$, $B = 6$, i.e., $x = -1$, $y = 3 \\pm 2\\sqrt{2}$, which means $b = -a$, $c = -a(3 \\mp 2\\sqrt{2})$, $d = a(3 \\mp 2\\sqrt{2})$.\n\n**In conclusion, the required maximum is $-12$.**",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 12849,
    "subject": "Mathematics (Olympiad)",
    "question": "Сколько существует хороших раскрасок вершин $n$-угольника в два цвета (чёрный и белый), если раскраска считается хорошей, если для любого треугольника, образованного вершинами многоугольника, его вершины не все одного цвета?\n\nНазовём раскраску вершин *упорядоченной*, если все чёрные вершины на границе многоугольника идут подряд (то есть есть ровно две разноцветные стороны).\n\nДокажите, что хорошая раскраска возможна тогда и только тогда, когда она упорядочена.",
    "options": [],
    "answer": "See solution",
    "solution": "Ввиду леммы, осталось лишь посчитать число упорядоченных раскрасок $n$-угольника. Для каждого возможного количества чёрных вершин (от $1$ до $n-1$) можно $n$ способами выбрать расположение их блока среди всех $n$ вершин, то есть число способов равно $n(n-1)$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 12850,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $k > 1$ be an integer. A set of natural numbers $S$ is called *good* if all positive integers can be painted in $k$ colors such that no element of $S$ is a sum of two distinct numbers having one and the same color. Find the largest positive integer $t$ for which the set\n\n$$\nS = \\{a+1, a+2, a+3, \\dots, a+t\\}\n$$\n\nis good for all positive integers $a$.",
    "options": [],
    "answer": "See solution",
    "solution": "We show that the desired number equals $t = 2k - 2$.\n\nConsider the set $S = \\{3, 4, \\dots, 2k, 2k+1\\}$. The sum of any two distinct numbers from $1, 2, \\dots, k+1$ is an element of $S$. Since among $1, 2, \\dots, k+1$ there exist two numbers having one and the same color, we conclude that $S$ is not good. Now $|S| = 2k-1$ implies $t \\le 2k-2$.\n\nIt remains to prove that the set $S = \\{a+1, a+2, \\dots, a+2k-2\\}$ is good for any $a$.\n\n1. Let $a$ be an odd number. Color the numbers $1, 2, \\dots, \\frac{a+1}{2}$ in the first color and each of the numbers $\\frac{a+2s-1}{2}$ for $s = 2, 3, \\dots, k$ in color $s$. Let all numbers greater than $\\frac{a+2k-1}{2}$ be of color $k$. It is easy to see that the sum of any two numbers of one and the same color is not an element of $S$.\n\n2. Let $a$ be an even number. Color the numbers $1, 2, \\dots, \\frac{a}{2}$ in the first color and each of the numbers $\\frac{a+2s-2}{2}$ for $s = 2, 3, \\dots, k$ in color $s$. Let all numbers greater than $\\frac{a+2k-2}{2}$ be of color $k$. It is easy to see that the sum of any two numbers of one and the same color is not an element of $S$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 12851,
    "subject": "Mathematics (Olympiad)",
    "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": "See solution",
    "solution": "**Proof** For any given positive integer $k$, choose a positive integer $m$ sufficiently large so 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 greater than 1. For 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 any positive integer. For any fixed integer $i$ ($1 \\le i \\le k$), we have $2^{n_t} \\equiv 2^m \\pmod{p_i}$. If $p_i = 2$, the result is clear. If $p_i \\ne 2$, by Fermat's little theorem,\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, $3^{n_t} \\equiv 3^m \\pmod{p_i}$. Thus,\n$$\n2^{n_t} + 3^{n_t} - i \\equiv 2^m + 3^m - i \\equiv 0 \\pmod{p_i},\n$$\n$$\n2^{n_t} + 3^{n_t} - i > 2^m + 3^m - i.\n$$\nTherefore, $2^{n_t} + 3^{n_t} - i$ is composite.\n\nThus, $n_t$ is a positive integer $n$ such that\n$$\n2^n + 3^n - 1,\\ 2^n + 3^n - 2,\\ \\dots,\\ 2^n + 3^n - k\n$$\nare all composite. Since $t$ is arbitrary, there are infinitely many such positive integers.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 12852,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $P_1P_2 \\cdots P_{100}$ be a cyclic 100-gon, and let $P_i = P_{i+100}$ for all $i$. Define $Q_i$ as the intersection of diagonals $\\overline{P_{i-2}P_{i+1}}$ and $\\overline{P_{i-1}P_{i+2}}$ for all integers $i$.\n\nSuppose there exists a point $P$ satisfying $\\overline{PP_i} \\perp \\overline{P_{i-1}P_{i+1}}$ for all integers $i$. Prove that the points $Q_1, Q_2, \\dots, Q_{100}$ are concyclic.",
    "options": [],
    "answer": "See solution",
    "solution": "We let $\\overline{PP_2}$ and $\\overline{P_1P_3}$ intersect (perpendicularly) at point $K_2$, and define $K_\\bullet$ cyclically.\n\n![](images/sols-TST-IMO-2020_p11_data_cf84b6e35d.png)\n\n**Claim** — The points $K_\\bullet$ are concyclic, say with circumcircle $\\gamma$.\n\n*Proof*. Note that $PP_1 \\times PK_1 = PP_2 \\times PK_2 = \\dots$ so the result follows by inversion at $P$. $\\square$\n\nLet $E_i$ be the second intersection of line $\\overline{P_{i-1}K_iP_{i+1}}$ with $\\gamma$; then it follows that the perpendiculars to $\\overline{P_{i-1}P_{i+1}}$ at $E_i$ all concur at a point $E$, which is the reflection of $P$ across the center of $\\gamma$.\n\nWe let $H_2 = \\overline{P_1P_3} \\cap \\overline{P_2P_4}$ denote the orthocenter of $\\triangle PP_2P_3$ and define $H_\\bullet$ cyclically.\n\n**Claim** — We have\n\n$$\n\\overline{EH_2} \\perp \\overline{P_1P_4} \\parallel \\overline{K_2K_3} \\text{ and } \\overline{PH_2} \\perp \\overline{E_2E_3} \\parallel \\overline{P_2P_3}.\n$$\n\n*Proof*. Both parallelisms follow by Reim's theorem through $\\angle E_2H_2E_3 = \\angle K_2H_2K_3$, so we need to show the perpendicularities.\n\nNote that $\\overline{H_2P}$ and $\\overline{H_2E}$ are respectively circum-diameters of $\\triangle H_2K_2K_3$ and $\\triangle H_2E_2E_3$. As $\\overline{K_2K_3}$ and $\\overline{E_2E_3}$ are anti-parallel, it follows $\\overline{H_2P}$ and $\\overline{H_2E}$ are isogonal and we derive both perpendicularities. $\\square$\n\n**Claim** — The points $E, Q_3, E_3$ are collinear.\n\n*Proof.* We use the previous claim. The parallelisms imply that\n\n$$\n\\frac{E_3 H_2}{E_3 P_2} = \\frac{E_2 H_2}{E_2 P_3} = \\frac{E_4 H_3}{E_4 P_3} = \\frac{E_3 H_3}{E_3 P_4}.\n$$\n\nNow consider a homothety centered at $E_3$ sending $H_2$ to $P_2$ and $H_3$ to $P_4$. Then it should send the orthocenter of $\\triangle EH_2H_3$ to $Q_3$, proving the result. $\\square$\n\nFrom all this it follows that $\\triangle EQ_2Q_3 \\sim \\triangle PK_2K_3$ as the opposite sides are all parallel. Repeating this we actually find a homothety of 100-gons\n\n$$\nQ_1 Q_2 Q_3 \\cdots \\sim K_1 K_2 K_3 \\cdots\n$$\n\nand that concludes the proof.\n\n**Remark.** The proposer remarks that in fact, if one lets $s$ be an integer and instead defines $R_i = P_i P_{i+s} \\cap P_{i+1} P_{i+s+1}$, then the $R_\\bullet$ are concyclic. The present problem is the case $s=3$. We comment on a few special cases:\n\n- There is nothing to prove for $s = 1$.\n- If $s = 0$, this amounts to proving that poles of $\\overline{P_i P_{i+1}}$ are concyclic; by inversion this is equivalent to showing the midpoints of the sides are concyclic. This is an interesting problem but not as difficult.\n- The problem for $s = 2$ is to show that our $H_\\bullet$ are concyclic, which uses the $s = 0$ case as a lemma.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 12853,
    "subject": "Mathematics (Olympiad)",
    "question": "Given a positive integer $n$, let $D$ denote the set of all positive divisors of $n$. Let $A$ and $B$ be subsets of $D$ such that for any $a \\in A$ and $b \\in B$, we have $a \\nmid b$ and $b \\nmid a$. Prove that\n\n$$\n\\sqrt{|A|} + \\sqrt{|B|} \\le \\sqrt{|D|}.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "Decompose $D$ into the following disjoint unions: $D = X \\sqcup Y \\sqcup Z \\sqcup W$, where\n\n$$\nX = \\{x \\in D : \\exists a \\mid x,\\ \\exists b \\mid x\\}, \\quad Y = \\{x \\in D : \\exists a \\mid x,\\ \\nexists b \\mid x\\},\n$$\n$$\nZ = \\{x \\in D : \\nexists a \\mid x,\\ \\exists b \\mid x\\}, \\quad W = \\{x \\in D : \\nexists a \\mid x,\\ \\nexists b \\mid x\\}.\n$$\n\nThe assumption implies $A \\subseteq Y$ and $B \\subseteq Z$. 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 is equivalent to\n\n$$\n|Y| + |Z| + 2\\sqrt{|Y| \\cdot |Z|} \\le |D| = |X| + |Y| + |Z| + |W| \\iff 2\\sqrt{|Y| \\cdot |Z|} \\le |X| + |W|.\n$$\n\nThis is implied by $|Y| \\cdot |Z| \\le |X| \\cdot |W|$, which can be rewritten as\n\n$$\n(|X|+|Y|)(|X|+|Z|) = |X|(|X|+|Y|+|Z|)+|Y|\\cdot|Z| \\le |X|(|X|+|Y|+|Z|)+|X|\\cdot|W| = |X|\\cdot|D|.\n$$\n\nLet $U = X \\cup Y$ and $V = X \\cup Z$. Then the inequality becomes $|U| \\cdot |V| \\le |U \\cap V| \\cdot |D|$.\n\nNote that $U = \\{x \\in D : \\exists a \\mid x\\}$ is upward-closed: if $x \\in U$ and $x \\mid x'$, then $x' \\in U$. Similarly, $V = \\{x \\in D : \\exists b \\mid x\\}$ is also upward-closed.\n\nWe now prove: for any two nonempty upward-closed subsets $U, V$ of $D$, $|U| \\cdot |V| \\le |U \\cap V| \\cdot |D|$.\n\nLet $n = p_1^{\\alpha_1} \\cdots p_k^{\\alpha_k}$ be the prime factorization of $n$, and proceed by induction on $k$. Let $p = p_k$, $\\alpha = \\alpha_k$, and $n = p^{\\alpha} n'$. Define $D_k = \\{x \\in D : v_p(x) = k\\}$, $U_k = U \\cap D_k$, $V_k = V \\cap D_k$. For $k = 0, 1, \\dots, \\alpha - 1$, for any $x \\in U_k$, the upward-closure of $U$ implies $p x \\in U_{k+1}$, so $|U_k| \\le |U_{k+1}|$; similarly for $V_k$.\n\nNote that $\\frac{1}{p^k}U_k$ and $\\frac{1}{p^k}V_k$ are upward-closed subsets of $D(n')$. By induction,\n\n$$\n|\\left(\\frac{1}{p^k}U_k\\right) \\cap \\left(\\frac{1}{p^k}V_k\\right)| \\ge \\frac{1}{|D(n')|} \\cdot |\\frac{1}{p^k}U_k| \\cdot |\\frac{1}{p^k}V_k|.\n$$\n\nSo $|U_k \\cap V_k| \\ge \\frac{1+\\alpha}{|D|} |U_k| \\cdot |V_k|$. By the rearrangement inequality,\n\n$$\n\\begin{aligned}\n|U \\cap V| &= \\sum_{k=0}^{\\alpha} |U_k \\cap V_k| \\ge \\frac{1+\\alpha}{|D|} \\sum_{k=0}^{\\alpha} |U_k| \\cdot |V_k| \\\\\n&\\ge \\frac{1+\\alpha}{|D|} \\cdot \\frac{1}{1+\\alpha} \\left( \\sum_{k=0}^{\\alpha} |U_k| \\right) \\cdot \\left( \\sum_{k=0}^{\\alpha} |V_k| \\right) \\\\\n&= \\frac{1}{|D|} |U| \\cdot |V|.\n\\end{aligned}\n$$\n\nThus, the problem is proved.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 12854,
    "subject": "Mathematics (Olympiad)",
    "question": "Given a positive integer $k$, let $r = k + \\frac{1}{2}$. Define $f^{(1)}(r) = f(r) = r \\lfloor r \\rfloor$, and for $l \\geq 2$, $f^{(l)}(r) = f(f^{(l-1)}(r))$. Prove that there exists a positive integer $m$ such that $f^{(m)}(r)$ is an integer.\n\n(Here $\\lceil x \\rceil$ denotes the minimum integer not less than $x$; for example, $\\lfloor \\frac{1}{2} \\rfloor = 1$, $\\lceil 1 \\rceil = 1$.)",
    "options": [],
    "answer": "See solution",
    "solution": "Define $v_2(n)$ as the exponent of $2$ in the positive integer $n$. We will prove that $f^{(m)}(r)$ is an integer for $m = v_2(k) + 1$ by induction on $v_2(k) = v$.\n\n**Base case ($v = 0$):**\nIf $k$ is odd, then $k + 1$ is even. Then\n$$\nf(r) = f^{(1)}(r) = \\left\\lceil k + \\frac{1}{2} \\right\\rceil \\left\\lfloor k + \\frac{1}{2} \\right\\rfloor = \\left\\lceil k + \\frac{1}{2} \\right\\rceil (k + 1)\n$$\nis an integer.\n\n**Inductive step:**\nAssume the proposition is true for $v - 1$ ($v \\geq 1$). For $v \\geq 1$, write\n$$\nk = 2^v + \\alpha_{v+1} 2^{v+1} + \\alpha_{v+2} 2^{v+2} + \\dots\n$$\nwhere $\\alpha_i \\in \\{0, 1\\}$ for $i = v + 1, v + 2, \\dots$. Then\n$$\n\\begin{align*}\nf(r) &= \\left\\lceil k + \\frac{1}{2} \\right\\rceil \\left\\lfloor k + \\frac{1}{2} \\right\\rfloor = \\left\\lceil k + \\frac{1}{2} \\right\\rceil (k + 1) \\\\\n&= \\frac{1}{2} + \\frac{k}{2} + k^2 + k \\\\\n&= \\frac{1}{2} + 2^{v-1} + (\\alpha_{v+1} + 1) 2^v + (\\alpha_{v+1} + \\alpha_{v+2}) 2^{v+1} + \\dots + 2^{2v} + \\dots \\\\\n&= k' + \\frac{1}{2}, \\tag*{\\text{①}}\n\\end{align*}\n$$\nwhere\n$$\nk' = 2^{v-1} + (\\alpha_{v+1} + 1) 2^v + (\\alpha_{v+1} + \\alpha_{v+2}) 2^{v+1} + \\dots + 2^{2v} + \\dots\n$$\nClearly, $v_2(k') = v - 1$. Let $r' = k' + \\frac{1}{2}$. By the induction hypothesis, $f^{(v)}(r')$ is an integer, which equals $f^{(v+1)}(r)$ by (①). Thus, the proof is complete.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 12855,
    "subject": "Mathematics (Olympiad)",
    "question": "The sequence $\\{x_n\\}$ is defined by $x_1 = 2$, $x_2 = 12$, and for $n = 1, 2, \\dots$, $$x_{n+2} = 6x_{n+1} - x_n.$$ 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": "See solution",
    "solution": "It is easy to see\n\n$$\nx_n = \\frac{1}{2\\sqrt{2}}\\left((3+2\\sqrt{2})^n - (3-2\\sqrt{2})^n\\right), \\quad n = 1, 2, \\dots\n$$\n\nLet $a_n, b_n$ be positive integers such that $a_n + b_n\\sqrt{2} = (3+2\\sqrt{2})^n$. Then\n\n$$\na_n - b_n\\sqrt{2} = (3 - 2\\sqrt{2})^n,\n$$\n\nso $x_n = b_n$, and $a_n^2 - 2b_n^2 = 1$ for $n = 1, 2, \\dots$\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\n$$\na_n + b_n \\sqrt{2} = (3 + 2\\sqrt{2})^{du} \\equiv (a_d + b_d\\sqrt{2})^u \\pmod{q},\n$$\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$$\n\\begin{aligned}\na_n &= (3 + 2\\sqrt{2})^n = (3 + 2\\sqrt{2})^{du} \\cdot (3 + 2\\sqrt{2})^r \\\\\n&\\equiv (a_d + b_d\\sqrt{2})^u (a_r + b_r\\sqrt{2}) \\pmod{q},\n\\end{aligned}\n$$\n\nwe have\n\n$$\na_d^u b_r \\equiv 0 \\pmod{q}. \\qquad (1)\n$$\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 $q \\mid b_r$, which contradicts the definition of $d$. So $r = 0$, and the lemma is proven.\n\nNow, as $q$ is a prime, $q \\mid \\binom{q}{i}$ for $i = 1, 2, \\dots, q-1$.\n\nUsing Fermat's little theorem, we have\n\n$$\n3^q \\equiv 3 \\pmod{q}, \\quad 2^q \\equiv 2 \\pmod{q}.\n$$\n\nAs $q \\neq 2$, $2^{\\frac{q-1}{2}} \\equiv \\pm 1 \\pmod{q}$, so\n\n$$\n\\begin{aligned}\n(3+2\\sqrt{2})^q &= \\sum_{i=0}^{q} \\binom{q}{i} \\cdot 3^{q-i} (2\\sqrt{2})^i \\\\\n&\\equiv 3^q + (2\\sqrt{2})^q \\\\\n&= 3^q + 2^q \\cdot 2^{\\frac{q-1}{2}} \\sqrt{2} \\\\\n&\\equiv 3 \\pm 2\\sqrt{2} \\pmod{q}.\n\\end{aligned}\n$$\n\nBy the same argument,\n\n$$\n(3+2\\sqrt{2})^{q^2} \\equiv (3 \\pm 2\\sqrt{2})^q \\equiv 3+2\\sqrt{2} \\pmod{q}.\n$$\n\nSo\n\n$$\n(a_{q^2-1} + \\sqrt{2}b_{q^2-1})(3 + 2\\sqrt{2}) \\equiv 3 + 2\\sqrt{2} \\pmod{q}.\n$$\n\nThus,\n\n$$\n\\begin{cases}\n3a_{q^2-1} + 4b_{q^2-1} \\equiv 3 \\pmod{q}, \\\\\n2a_{q^2-1} + 3b_{q^2-1} \\equiv 2 \\pmod{q}.\n\\end{cases}\n$$\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 $d \\in \\{1, p\\}$. 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": 12856,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $\\triangle ABC$ be a right triangle with $\\angle A = 90^\\circ$ and $BC = 38$. There exist points $K$ and $L$ inside the triangle such that\n\n$$\nAK = AL = BK = CL = KL = 14.\n$$\n\nThe area of the quadrilateral $BKLC$ can be expressed as $n\\sqrt{3}$ for some positive integer $n$. Find $n$.",
    "options": [],
    "answer": "See solution",
    "solution": "Because $AL = CL = KL$, the circumcenter of $\\triangle ACK$ is $L$. Similarly, the circumcenter of $\\triangle ABL$ is $K$.\nBecause $\\triangle AKL$ is equilateral, $\\angle ALK = \\angle AKL = 60^\\circ$, and it follows that $\\angle ACK = 30^\\circ = \\angle ABL$.\nBecause\n\n$$\n\\angle BAK = \\angle BAC - \\angle KAL - \\angle LAC = 30^\\circ - \\angle LAC = 30^\\circ - \\angle LCA = \\angle ACK - \\angle LCA = \\angle KCL,\n$$\n\nit follows that the isosceles triangles $\\triangle ABK$ and $\\triangle CKL$ are congruent. Thus $AB = CK$. Because $\\angle KBL = \\angle ABL - \\angle ABK = \\angle ACK - \\angle LCK$, it is also true that $\\triangle BLK \\cong \\triangle ACL$ and $AC = BL$.\nIt follows by SAS that $\\triangle ABL \\cong \\triangle KCA$.\n\n![](images/2025AIME_II_Solutions_p12_data_45906458e2.png)\n\nTherefore\n\n$$\n\\begin{align*}\n\\text{Area}(BKLC) &= \\text{Area}(\\triangle ABC) - \\text{Area}(\\triangle ALK) - \\text{Area}(\\triangle ACL) - \\text{Area}(\\triangle ABK) \\\\\n&= \\text{Area}(\\triangle ABC) - \\text{Area}(\\triangle ALK) - \\text{Area}(\\triangle BLK) - \\text{Area}(\\triangle ABK) \\\\\n&= \\text{Area}(\\triangle ABC) - \\text{Area}(\\triangle ABL).\n\\end{align*}\n$$\n\nSet $c = AB = CK$ and $b = AC = BL$. Then\n\n$$\n\\text{Area}(\\triangle ABL) = \\frac{bc \\sin(\\angle ABL)}{2} = \\frac{bc}{4} \\quad \\text{and} \\quad \\text{Area}(\\triangle ABC) = \\frac{bc}{2},\n$$\n\nfrom which $\\text{Area}(BKLC) = \\frac{bc}{4}$. Applying the Law of Cosines to $\\triangle ABL$ gives\n\n$$\n14^2 = AL^2 = b^2 + c^2 - bc\\sqrt{3} = 38^2 - bc\\sqrt{3}.\n$$\n\nThus $bc\\sqrt{3} = 38^2 - 14^2 = 52 \\cdot 24$, implying that $\\text{Area}(BKLC) = 52 \\cdot 2\\sqrt{3} = 104\\sqrt{3}$. The requested coefficient of $\\sqrt{3}$ is 104.\n\nAlternatively,\n\nAs in the first solution, $AC = BL$. Observe that a $60^\\circ$ rotation around the center of $\\triangle AKL$ moves $\\triangle AKB$ onto $\\triangle KLC$ and $\\triangle KLB$ onto $\\triangle LAC$. Hence there exists a point $P$ such that $\\triangle AKL$ and $\\triangle PBC$ are concentric equilateral triangles, as shown below.\n\n![](images/2025AIME_II_Solutions_p13_data_bfe5c96db0.png)\n\nOnce $\\triangle AKL$ is removed from $\\triangle PBC$, the area of the quadrilateral $BKLC$ is $\\frac{1}{3}$ of the area of the remaining shape. Thus\n\n$$\n\\text{Area}(BKLC) = \\frac{1}{3}(\\text{Area}(\\triangle PBC) - \\text{Area}(\\triangle AKL)) = \\frac{1}{12}(38^2 - 14^2)\\sqrt{3} = 104\\sqrt{3},\n$$\n\nas in the first solution.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 12857,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $\\triangle ABC$ be a triangle with $\\angle BCA = 90^\\circ$, and let $D$ be the foot of the altitude from $C$. Let $X$ be a point in the interior of the segment $CD$. Let $K$ be the point on the segment $AX$ such that $BK = BC$. Similarly, let $L$ be the point on the segment $BX$ such that $AL = AC$. Let $M$ be the point of intersection of $AL$ and $BK$. Show that $MK = ML$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let points $C'$ and $C$ be symmetric over the line $AB$, and $\\omega_1$ and $\\omega_2$ be circles with centres $A$ and $B$, radius $AL$ and $BK$, respectively. Since $AC' = AC = AL$ and $BC' = BC = BK$, points $C$ and $C'$ are both on circles $\\omega_1$ and $\\omega_2$. Since $\\angle BCA = 90^\\circ$, lines $AC$ and $BC$ are tangent to circles $\\omega_2$ and $\\omega_1$, respectively, at point $C$. Let $K_1$ be another intersection point of line $AX$ and circle $\\omega_2$ different to $K$, and $L_1$ be another intersection point of line $BX$ and circle $\\omega_1$ different to $L$.\n\n![](images/Mathematical_Olympiad_in_China_2011-2014_p341_data_0acbfa3330.png)\n\nBy the Circle-Power Theorem, we have\n\n$$\nXK \\cdot XK_1 = XC \\cdot XC' = XL \\cdot XL_1,\n$$\n\nhence $K_1, L, K, L_1$ are four concyclic points at circle denoted by $\\omega_3$.\n\nBy applying Circle-Power Theorem to circle $\\omega_2$, we have\n\n$$\nAL^2 = AC^2 = AK \\cdot AK_1.\n$$\n\nThis implies the line $AL$ is tangent to circle $\\omega_3$ at point $L$. By the similar argument, line $BK$ is tangent to circle $\\omega_3$ at point $K$.\n\nTherefore, $MK$ and $ML$ are two tangent lines from point $M$ to circle $\\omega_3$, thus $MK = ML$. $\\square$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 12858,
    "subject": "Mathematics (Olympiad)",
    "question": "In a soccer tournament, each team plays exactly one game with all others. The winner gets 3 points, the loser gets 0, and each team gets 1 point in case of a draw.\n\nIt is known that $n$ teams ($n \\ge 3$) took part in a tournament and the final classification is given by an arithmetic progression of points, with the last team having only 1 point.\n\n(a) Prove that this is not possible in the Championship of the Republic of Moldova (with $n = 12$).\n\n(b) Find all values of $n$ and all configurations when this is possible.",
    "options": [],
    "answer": "See solution",
    "solution": "a) The total number of matches is $\\frac{n(n-1)}{2}$. Let $w$ be the number of games ended with a victory and $e$ the number of games ended in a draw ($e \\ge 1$ due to the last team). Thus, $w + e = \\frac{n(n-1)}{2}$. If $r$ is the step of the arithmetic progression, the total number of points in the final classification is\n\n$$\n\\frac{(2 + (n-1)r)n}{2} = 3w + 2e = w + 2(w + e) = w + n(n-1),\n$$\n\nor\n\n$$\n2w - 2n = n(n-1)(r-2).\n$$\n\nThe case $r=0$ is impossible (each team would have only 1 point). For $r=1$ we get $2w = 3n - n^2$, a contradiction (for $n=3$ one has 3 games, $w=0$, all games ended in a draw, but the last team has only 1 point; for $n \\ge 3$ this implies $w < 0$). If $r \\ge 3$, then $2w \\ge 2n + n(n-1) = n(n+1)$, which contradicts the total number of matches $\\frac{n(n-1)}{2}$.\n\nThus, the only possible value is $r=2$. In this case $w = n$ and the number of points of each team in decreasing order is the sequence $2n-1, 2n-3, 2n-5, \\dots, 1$.\n\nLet $w_i$ and $e_i$ be the number of victories and draws of the $i$-th classified team (in decreasing order). Note that $w_i + e_i \\le n-1$. Considering the number of points obtained by the first three teams, $2n-1, 2n-3, 2n-5$, we get $2n-1 = 3w_1 + e_1 = 2w_1 + (w_1 + e_1) \\le 2w_1 + n-1$, so $w_1 \\ge n/2$. Similarly, $w_2 \\ge n/2 - 1$ and $w_3 \\ge n/2 - 2$.\n\nFor $n \\ge 7$ we obtain $w_1 + w_2 + w_3 \\ge 3n/2 - 3 > n$, a contradiction with the fact that the total number of victories is $n$. Thus, $n \\le 6$ and it is impossible to have $n=12$.\n\nb) Such a configuration does not exist for $n=3$ ($w=3$, $e = \\frac{n(n-1)}{2} - w = 0$).\n\nThe case $n=4$ is possible, where the points $7, 5, 3, 1$ in the final classification are realized by the following results of the matches of teams:\n- T1–T2: draw\n- T1–T3: T1 won\n- T1–T4: T1 won\n- T2–T3: T2 won\n- T2–T4: draw\n- T3–T4: T3 won\n\nFor $n=5$ we get the configuration $9, 7, 5, 3, 1$. The only way to write 9 as a sum of at most four numbers of 3's and 1's is $9 = 3+3+3$, for 7 is $7 = 3+3+1$. In this case, the number of all victories $w = n = 5$ is reached by the first two teams. It follows that we have to write 5 points of the third as a sum of at most four numbers of 1's, which is impossible.\n\nFor $n=6$ we get the configuration $11, 9, 7, 5, 3, 1$. The only way to write 11 as a sum of at most five numbers of 3's and 1's is $11 = 3+3+3+1+1$, for 9 is $9 = 3+3+3$. Again, the number of all victories $w = n = 6$ is reached by the first two teams. But in this case, we have to write 7 points of the third as a sum of at most five numbers of 1's, which is impossible.\n\nTherefore, the only possibility is $n=4$, with the configuration described above.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 12859,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $I$ be the incenter of triangle $ABC$ and let $k$ be a circle through the points $A$ and $B$. This circle intersects\n\n- the line $AI$ in points $A$ and $P$,\n- the line $BI$ in points $B$ and $Q$,\n- the line $AC$ in points $A$ and $R$, and\n- the line $BC$ in points $B$ and $S$,\n\nwith none of the points $A, B, P, Q, R,$ and $S$ coinciding and such that $R$ and $S$ are interior points of the line segments $AC$ and $BC$, respectively.\n\nProve that the lines $PS$, $QR$, and $CI$ meet in a single point.",
    "options": [],
    "answer": "See solution",
    "solution": "![](images/bwf2015englishSolutions_p4_data_da2f940dd7.png)\n\nWe define angles $\\alpha = \\angle BAC$ and $\\beta = \\angle CBA$ as usual. Since points $A$, $B$, $S$, and $R$ lie on a common circle, we have $\\angle BSR = 180^\\circ - \\alpha$, and therefore $\\angle RSC = \\alpha$. Similarly, $\\angle CRS = \\beta$ also holds.\n\nIf $P$ lies in the interior of $ABC$, we have $\\angle RSP = \\angle RAP = \\alpha/2$. This means that $PS$ bisects the angle $\\angle CSR$.\n\nIf $Q$ is outside of $ABC$, we have $\\angle QRA = \\angle QBA = \\beta/2$, and in this case $QR$ also bisects the angle $\\angle SRC$.\n\nIndependent of the positioning of $Q$ and $R$ with respect to the triangle, we therefore see that $QR$, $PS$, and $CI$ are the bisectors of the interior angles of $CRS$, and they therefore meet in the incenter of this triangle, as claimed.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 12860,
    "subject": "Mathematics (Olympiad)",
    "question": "假設 $a, b, c, d$ 為滿足 $(a+c)(b+d) = ac+bd$ 的正實數。試確定\n\n$$\nS = \\frac{a}{b} + \\frac{b}{c} + \\frac{c}{d} + \\frac{d}{a}\n$$\n\n的最小可能值。",
    "options": [],
    "answer": "See solution",
    "solution": "最小可能值是 $8$。\n\n**解法 1.** 要證明 $S \\ge 8$，可兩次應用 AM-GM 不等式如下：\n\n$$\n\\begin{aligned}\n\\left(\\frac{a}{b} + \\frac{c}{d}\\right) + \\left(\\frac{b}{c} + \\frac{d}{a}\\right) &\\ge 2\\sqrt{\\frac{ac}{bd}} + 2\\sqrt{\\frac{bd}{ac}} \\\\\n&= \\frac{2(ac + bd)}{\\sqrt{abcd}} = \\frac{2(a + c)(b + d)}{\\sqrt{abcd}} \\\\\n&\\ge 2 \\cdot \\frac{2\\sqrt{ac} \\cdot 2\\sqrt{bd}}{\\sqrt{abcd}} = 8.\n\\end{aligned}\n$$\n\n當 $a = c$ 且 $b = d$ 時，上述不等式取等號。此時條件 $(a+c)(b+d) = ac+bd$ 可化為 $4ab = a^2 + b^2$，即 $a/b = 2 \\pm \\sqrt{3}$。因此，$S$ 可取到 $8$，例如 $a = c = 1$，$b = d = 2 + \\sqrt{3}$。",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 12861,
    "subject": "Mathematics (Olympiad)",
    "question": "For $a, b, c, d, e, f \\in \\mathbb{R}$, suppose that the 6 distinct roots of the following equation\n\n$$\n(x^2 + a x + b)(x^2 + c x + d)(x^2 + e x + f) = 0\n$$\n\nare 6 consecutive integers. Find the minimum value of\n\n$$\nT = a^2 + c^2 + e^2 - 2(b + d + f).\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "Using Vieta's theorem, we have\n\n$$\n\\begin{aligned}\nT &= (a^2 - 2b) + (c^2 - 2d) + (e^2 - 2f) \\\\\n  &= x_1^2 + x_2^2 + \\dots + x_6^2 \\\\\n  &= m^2 + (m + 1)^2 + \\dots + (m + 5)^2 \\\\\n  &= 6m^2 + 30m + 55.\n\\end{aligned}\n$$\n\nNote that $(m + 2)(m + 3) \\ge 0$ for all $m \\in \\mathbb{Z}$, so $m^2 + 5m \\ge -6$. Thus,\n\n$$\nT = -6(m^2 + 5m) + 55 \\ge 55 - 36 = 19.\n$$\n\nThe equality holds when $m = -2$ or $m = -3$.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 12862,
    "subject": "Mathematics (Olympiad)",
    "question": "The medians $AA'$, $BB'$, $CC'$ of triangle $ABC$ meet the nine-point circle at $D$, $E$, $F$, respectively. The points $L$, $M$, $N$ are the feet of the altitudes of $ABC$ ($L$ lies on $AA'$, etc). The tangents to the nine-point circle at $D$, $E$, $F$ meet the lines $MN$, $LN$, and $LM$ at points $P$, $Q$, $R$. Show that points $P$, $Q$, and $R$ are collinear.",
    "options": [],
    "answer": "See solution",
    "solution": "1) Triangles $PMD$ and $PDN$ are similar (indeed, $\\widehat{PDM}$ is half inscribed, $\\widehat{DNP}$ is inscribed, and both subtend the same arc in the Euler circle; moreover, $\\hat{P}$ is the same in both triangles). Therefore\n\n$$\n\\frac{PM}{PD} = \\frac{DM}{DN} = \\frac{PD}{PN} \\implies \\frac{PM}{PN} = \\left(\\frac{DM}{DN}\\right)^2. \\quad (*)\n$$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 12863,
    "subject": "Mathematics (Olympiad)",
    "question": "Calculate the angles in triangle $ABC$, given that the angle between the altitude from $C$ and the bisector of angle $ACB$ is $90^\\circ$, and the angle between the bisectors of the exterior angles at vertices $A$ and $B$ is $61^\\circ$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $\\alpha$, $\\beta$, and $\\gamma$ be the angles at vertices $A$, $B$, and $C$ of triangle $ABC$, and $\\alpha_1$, $\\beta_1$, $\\gamma_1$ the corresponding exterior angles. Let $D$ be the foot of the altitude from $C$, $E$ the foot of the bisector of $\\angle ACB$, and $F$ the intersection of the bisectors of the exterior angles at $A$ and $B$.\n\n![](images/Makedonija_2008_p12_data_1401b8a6fb.png)\n\nWe have $\\angle DCE = 90^\\circ$ and $\\angle AFB = 61^\\circ$. The bisectors split the exterior angles: $\\angle XAY = \\frac{\\alpha_1}{2}$, $\\angle NBM = \\frac{\\beta_1}{2}$, $\\angle ACE = \\frac{\\gamma_1}{2}$.\n\nIn triangle $ABF$, $\\angle FAB = \\frac{\\alpha_1}{2}$, $\\angle ABF = \\frac{\\beta_1}{2}$, and $\\angle AFB = 61^\\circ$. Thus,\n\n$$\n\\frac{\\alpha_1}{2} + \\frac{\\beta_1}{2} + 61^\\circ = 180^\\circ\n$$\nso $\\alpha_1 + \\beta_1 = 238^\\circ$. Since $\\alpha_1 = 180^\\circ - \\alpha$ and $\\beta_1 = 180^\\circ - \\beta$,\n\n$$\n180^\\circ - \\alpha + 180^\\circ - \\beta = 238^\\circ \\implies \\alpha + \\beta = 122^\\circ\n$$\n\nTherefore,\n\n$$\n\\gamma = 180^\\circ - (\\alpha + \\beta) = 58^\\circ\n$$\n\nFrom the right triangle $ADC$,\n\n$$\n\\alpha = 90^\\circ - \\angle ACD = 90^\\circ - \\left(\\frac{\\gamma}{2} - 9^\\circ\\right) = 90^\\circ - \\left(29^\\circ - 9^\\circ\\right) = 90^\\circ - 20^\\circ = 70^\\circ\n$$\n\nThus,\n\n$$\n\\beta = 122^\\circ - 70^\\circ = 52^\\circ\n$$\n\nThe angles are $\\alpha = 70^\\circ$, $\\beta = 52^\\circ$, $\\gamma = 58^\\circ$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 12864,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $f: \\mathbb{R} \\to \\mathbb{R}$ be a function such that $f(x + y + xy) = f(x) + f(y) + f(xy)$ for all $x, y \\in \\mathbb{R}$. Prove that $f$ satisfies $f(x + y) = f(x) + f(y)$ for all $x, y \\in \\mathbb{R}$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $x = y = 0$. Then $f(0) = 3f(0)$, so $f(0) = 0$.\n\nTaking $y = -1$, we get $f(-1) = f(x) + f(-1) + f(-x)$, so $f(-x) = -f(x)$ for all $x \\in \\mathbb{R}$.\n\nTaking $y = 1$, we get $f(2x + 1) = 2f(x) + f(1)$ for all $x \\in \\mathbb{R}$.\n\nReplace $x$ by $u + v + uv$ in this, we get\n\n$$\nf(2(u + v + uv) + 1) = 2f(u + v + uv) + f(1) = 2f(u) + 2f(v) + 2f(uv) + f(1),\n$$\nfor all $u, v \\in \\mathbb{R}$.\n\nTaking $x = u$, $y = 2v + 1$ in the original equation, we get\n\n$$\nf(u + 2v + 1 + 2uv + u) = f(u) + f(2v + 1) + f(2uv + u),\n$$\nwhich reduces to\n\n$$\nf(2(u + v + uv) + 1) = f(u) + f(2v + 1) + f(2uv + u),\n$$\nfor all $u, v \\in \\mathbb{R}$.\n\nComparing the two expressions and using $f(2x + 1) = 2f(x) + f(1)$, we get\n\n$$\nf(2uv + u) = f(u) + 2f(uv),\n$$\nfor all $u, v \\in \\mathbb{R}$.\n\nPut $v = -\\frac{1}{2}$ in this to get\n\n$$\n0 = f(0) = f(u) + 2f\\left(-\\frac{u}{2}\\right) = f(u) - 2f\\left(\\frac{u}{2}\\right).\n$$\n\nThis shows that $f\\left(\\frac{u}{2}\\right) = \\frac{f(u)}{2}$, or $f(2u) = 2f(u)$ for all $u \\in \\mathbb{R}$.\n\nThus,\n\n$$\nf(2uv + u) = f(2uv) + f(u),\n$$\nfor all $u, v \\in \\mathbb{R}$.\n\nGiven any $x \\neq 0$ and $y$, we can find $u, v$ such that $2uv = y$ and $u = x$. Thus $f(x + y) = f(x) + f(y)$ for all $x \\neq 0$ and $y$. This also holds for $x = 0$. Therefore, $f(x + y) = f(x) + f(y)$ for all $x, y \\in \\mathbb{R}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 12865,
    "subject": "Mathematics (Olympiad)",
    "question": "Consider the following two operations, A and B:\n\n- Operation A: Cut a cube into 8 equal cubes, each with half the dimensions of the original.\n- Operation B: Cut a cube into 27 equal cubes, each with one third the dimensions of the original.\n\nOperation A increases the total number of cubes by 7, and operation B increases it by 26. Thus, by performing operation A $x$ times and operation B $y$ times, one can obtain $1 + 7x + 26y$ cubes.\n\nGiven that $\\gcd(7, 26) = 1$ and $x, y$ are non-negative integers, what is the smallest integer $n_0$ such that any number of cubes greater than or equal to $n_0$ can be obtained?",
    "options": [],
    "answer": "See solution",
    "solution": "Since $\\gcd(7, 26) = 1$, the set of numbers of the form $1 + 7x + 26y$ (with $x, y \\geq 0$) covers all integers greater than $1 + 7 \\cdot 26 - 7 - 26$. Therefore, the minimal such $n_0$ is:\n\n$$\n1 + 7 \\cdot 26 - 7 - 26 + 1 = 1 + 182 - 7 - 26 + 1 = 1 + 182 - 33 + 1 = 151\n$$\n\nSo, $n_0 = 151$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 12866,
    "subject": "Mathematics (Olympiad)",
    "question": "Given 2017 positive real numbers $a_1, a_2, \\dots, a_{2017}$. For every $n > 2017$, define\n\n$$\na_n = \\max\\left\\{ a_{i_1} a_{i_2} a_{i_3} \\mid i_1 + i_2 + i_3 = n,\\ 1 \\leq i_1 \\leq i_2 \\leq i_3 \\leq n-1 \\right\\}.\n$$\n\nProve that there exists a positive integer $m \\leq 2017$ and $N > 4m$ such that $a_n a_{n-4m} = a_{n-2m}^2$ for every $n > N$.",
    "options": [],
    "answer": "See solution",
    "solution": "For every $n > 0$, let $b_n = \\ln a_n$. The problem reduces to: Given 2017 real numbers $b_1, b_2, \\dots, b_{2017}$. For every $n > 2017$, define\n\n$$\nb_n = \\max\\left\\{ b_{i_1} + b_{i_2} + b_{i_3} \\mid i_1 + i_2 + i_3 = n,\\ 1 \\leq i_1 \\leq i_2 \\leq i_3 \\leq n-1 \\right\\}.\n$$\n\nProve there exist positive integers $m \\leq 2017$ and $N > 4m$ such that $b_n + b_{n-4m} = 2b_{n-2m}$ for every $n > N$.\n\nLet $\\ell$ ($1 \\leq \\ell \\leq 2017$) be such that $\\frac{b_\\ell}{\\ell} = \\max\\left\\{ \\frac{b_i}{i} \\mid 1 \\leq i \\leq 2017 \\right\\}$. **Claim 1:** For every $n \\in \\mathbb{Z}^+$,\n\n$$\n\\frac{b_n}{n} \\leq \\frac{b_\\ell}{\\ell}.\n$$\n\n*Proof.* Induction on $n$. True for $n \\leq 2017$ by definition. For $n > 2017$, by induction hypothesis and the definition of $b_n$, there exist $j_1, j_2, j_3$ with $j_1 + j_2 + j_3 = n$ such that\n\n$$\nb_n = b_{j_1} + b_{j_2} + b_{j_3} \\leq (j_1 + j_2 + j_3) \\cdot \\frac{b_\\ell}{\\ell} = n \\cdot \\frac{b_\\ell}{\\ell}.\n$$\n\nThus $\\frac{b_n}{n} \\leq \\frac{b_\\ell}{\\ell}$.\n\nDefine $c_n = n b_\\ell - \\ell b_n$. From Claim 1, $c_n \\geq 0$ for all $n$. For $n \\geq 2017$,\n\n$$\n\\begin{aligned}\nc_{n+2\\ell} &= (n+2\\ell) b_\\ell - \\ell b_{n+2\\ell} \\\\\n&\\leq (n+2\\ell) b_\\ell - \\ell (b_n + 2b_\\ell) \\\\\n&= n b_\\ell - \\ell b_n = c_n.\n\\end{aligned}\n$$\n\nSo $c_{n+2k\\ell} \\leq c_{n+2(k-1)\\ell} \\leq \\dots \\leq c_n$ for $n \\geq 2017$, $k \\geq 1$.\n\nLet $x$ be the smallest positive integer such that $2x\\ell > 2017$ and set\n\n$$\nM = \\max\\left\\{ c_i \\mid 1 \\leq i \\leq 4x\\ell - 1 \\right\\}.\n$$\n\nFor every $n > 2x\\ell$, write $n = 2k x\\ell + r$ ($0 \\leq r < 2x\\ell$), then\n\n$$\nc_n \\leq c_{r+2(k-1)x\\ell} \\leq \\dots \\leq c_{r+2x\\ell} \\leq M.\n$$\n\n**Claim 2:** For every $n$, there exist natural numbers $s_1, \\dots, s_{2017}$ such that\n\n$$\nc_n = s_1 c_1 + \\dots + s_{2017} c_{2017}.\n$$\n\n*Proof.* Induction on $n$. True for $n \\leq 2017$. For $n > 2017$, from the definition of $b_n$ and the induction hypothesis, $c_n$ is a sum of $c_{j_1}, c_{j_2}, c_{j_3}$, each of which is a linear combination of $c_1, \\dots, c_{2017}$.\n\n**Claim 3:** The sequence $(c_n)$ takes only finitely many values.\n\n*Proof.* Follows from boundedness and Claim 2.\n\nSince $c_n$ takes only finitely many values and $c_{n+2kx\\ell} \\leq \\dots \\leq c_n$, there exists $N_1$ such that $c_n = c_{n-2x\\ell}$ for all $n > N_1$. Thus,\n\n$$\nn b_\\ell - \\ell b_n = (n-2x\\ell) b_\\ell - \\ell b_{n-2x\\ell}, \\quad \\forall n > N_1,\n$$\nwhich gives\n\n$$\nb_n = 2b_\\ell + b_{n-2x\\ell}, \\quad \\forall n > N_1.\n$$\n\nSimilarly,\n\n$$\nb_{n-2x\\ell} = 2b_\\ell + b_{n-4x\\ell}, \\quad \\forall n > N_1 + 2x\\ell.\n$$\n\nCombining,\n\n$$\nb_n + b_{n-4x\\ell} = 2b_{n-2x\\ell}, \\quad \\forall n > N_1 + 2x\\ell.\n$$\n\nChoose $N = N_1 + 2x\\ell$ and $m = x\\ell$, as desired.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 12867,
    "subject": "Mathematics (Olympiad)",
    "question": "For every real numbers $a \\ne b$, solve the system:\n\n$$\n\\begin{cases}\n3x + z = 2y + (a + b), \\\\\n3x^2 + 3xz = y^2 + 2(a + b)y + ab, \\\\\nx^3 + 3x^2z = y^2(a + b) + 2yab.\n\\end{cases}\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "The solutions are $x = y = a,\\ z = b$ and $x = y = b,\\ z = a$.\n\nLet $(x, y, z)$ be a solution of the system. Consider polynomials\n\n$$\n\\begin{aligned}\nP(t) &= (t - x)^3 (t - z) = t^4 + p_1 t^3 + q_1 t^2 + r_1 t + s_1, \\\\\nQ(t) &= (t - y)^2 (t - a)(t - b) = t^4 + p_2 t^3 + q_2 t^2 + r_2 t + s_2,\n\\end{aligned}\n$$\n\nand let\n\n$$\nf(t) = P(t) - Q(t) = (p_1 - p_2)t^3 + (q_1 - q_2)t^2 + (r_1 - r_2)t + (s_1 - s_2).\n$$\n\nNumbers $x, x, x, z$ are roots of $P(t)$ and $y, y, a, b$ are roots of $Q(t)$. Using Vieta's formulas, we get:\n\n$$\n\\begin{aligned}\np_1 &= -(3x + z) = -(2y + (a + b)) = p_2, \\\\\nq_1 &= 3x^2 + 3xz = y^2 + 2(a + b)y + ab = q_2, \\\\\nr_1 &= -(x^3 + 3x^2z) = -(y^2(a + b) + 2yab) = r_2.\n\\end{aligned}\n$$\n\nHence $f(t) = (s_1 - s_2) = C = \\text{const}$. Then\n\n$$\nP'(t) = (t - x)^2 (4t - 3z - x), \\qquad Q'(t) = (t - y) (4t^2 - (3a + 3b + 2y)t + 2ab + y(a + b)).\n$$\n\nConsider $R(t) = 4t^2 - (3a + 3b + 2y)t + 2ab + y(a + b)$ and take $t_0 = \\frac{1}{2}(a + b)$. We obtain\n\n$$\nR(t_0) = (a + b)^2 - 3(a + b)\\frac{a + b}{2} - 2y\\frac{a + b}{2} + 2ab + y(a + b) = -\\frac{(a + b)^2}{2} + 2ab = -\\frac{(a - b)^2}{2} < 0.\n$$\n\nThus $R(t)$ has two different roots and the discriminant of $R(t)$ is positive. But we have $P'(t) \\equiv Q'(t)$ and $P'(t) \\ne (t - x)^2$. Then $Q'(t) \\ne (t - x)^2$. Since $R(t)$ has two different roots, exactly one of them equals $x$ and $y = x$. Moreover, $R(y) = 0$, that is\n\n$$\nR(y) = 4y^2 - (3a + 3b + 2y)y + 2ab + y(a + b) = 2(y^2 - (a + b)y + ab) = 0.\n$$\n\nBut this is possible only if $y = a$ or $y = b$. Finally, we obtain\n\n$$\nf(t) = P(t) - Q(t) = (t - x)^3 (z - c) = C,\n$$\n\nwhere $c = a$ or $c = b$. But this is possible only if $z = c$. Therefore $P$ equals $Q$ and we have two solutions:\n\n$$\nx = y = a,\\ z = b, \\quad \\text{or} \\quad x = y = b,\\ z = a.\n$$\n\nThe second part can also be done using Rolle's theorem. Consider polynomial $Q(t) = (t - y)^2 (t - a)(t - b)$. Suppose that $y, a, b$ are pairwise distinct. Then derivative $Q'(t)$ has three distinct roots. One of them is $t = y$ and two others belong to the intervals between $y, a, b$. But $P'(t) = Q'(t)$ and thus $P'(t)$ also has three distinct roots. But $P'(t) = (t - x)^2$. Contradiction. Thus some of the numbers $y, a, b$ are equal. Since $a \\ne b$, it follows that $y = a$ or $y = b$. Consider the case $y = a$. Then $Q(t) = (t - y)^3 (t - b)$ and $Q'(t) = (t - x)^2$. Using that $Q'(t) = P'(t) = (t - x)^2$, we get $x = y$, and thus:\n\n$$\nC = (t - x)^3 (t - z) - (t - x)^3 (t - b) = (t - x)^3 (z - b),\n$$\n\nwhich is only possible if $z = b$. Thus we obtain the solution $x = y = a,\\ z = b$. Similarly, in the second case we get the solution $x = y = b,\\ z = a$.\n\n**Alternative solution.** From the first and second equations we get:\n\n$$\n\\begin{cases}\na + b = 3x + z - 2y \\\\\nab = 3x^2 + 3y^2 - 6xy + 3xz - 2yz.\n\\end{cases}\n\\qquad (1)\n$$\n\nIf we combine this with the third equation, we obtain:\n\n$$\nx^3 - 4y^3 - 6x^2y + 3x^2z + 9xy^2 - 6xyz + 3y^2z = 0.\n$$\n\nNote that $x^3 - 4y^3 - 6x^2y + 3x^2z + 9xy^2 - 6xyz + 3y^2z = (x - y)^2 (x - 4y + 3z)$. Hence we have two cases:\n\n1) $x = y$. In this case, (1) becomes:\n\n$$\n\\begin{cases}\na + b = x + z \\\\\nab = xz.\n\\end{cases}\n$$\n\nAnd we obtain the solutions $(a, a, b)$ and $(b, b, a)$.\n\n2) $x = 4y - 3z$. In this case, (1) becomes:\n\n$$\n\\begin{cases}\na + b = 10y - 8z \\\\\nab = 27y^2 - 44yz + 18z^2.\n\\end{cases}\n$$\n\nThen $a$ and $b$ are the solutions of the equation\n\n$$\nt^2 - (10y - 8z)t + 27y^2 - 44yz + 18z^2\n$$\n\nwith discriminant\n\n$$\n\\frac{1}{4}D = (5y - 4z)^2 - (27y^2 - 44yz + 18z^2) = -2(y - z)^2 \\le 0.\n$$\n\nAnd we must have $D = 0$. This means that $a = b$ and we get a contradiction.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 12868,
    "subject": "Mathematics (Olympiad)",
    "question": "Circles $\\Gamma_1$ and $\\Gamma_2$, with centres $O_1$ and $O_2$, are externally tangent at the point $D$ and are internally tangent to a circle $\\Gamma$ at points $E$ and $F$ respectively. Line $l$ is the common tangent to $\\Gamma_1$ and $\\Gamma_2$ at $D$. Let $AB$ be the diameter of $\\Gamma$ perpendicular to $l$ so that $A, E, O_1$ are on the same side of the line $l$. Prove that $AO_1$, $BO_2$, and $EF$ are concurrent.",
    "options": [],
    "answer": "See solution",
    "solution": "**Solution.** First, we prove a lemma:\n\n*Lemma:* Let $AB$ and $PR$ be two parallel lines. Let $Q$ be a point on the segment $PR$ such that $AP \\perp BQ$ and $AQ \\perp BR$. Let $U$ and $V$ be the midpoints of $PQ$ and $QR$ respectively; let $S$ and $T$ be the feet of perpendiculars from $B$ onto $AP$ and $AQ$ respectively. Then $AU$, $BV$, and $ST$ are concurrent.\n\n*Proof of Lemma:* Let $AU$ and $BV$ meet $ST$ at $X$ and $Y$ respectively. Let $ST$ meet $PQ$ at $M$ and $AB$ at $N$.\n\n![](images/Indija_TS_2007_p5_data_dc9dfcb70b.png)\n\nConsidering the projections from $T$, we have\n\n$$\n\\frac{QM}{MR} = \\frac{AN}{NB}.\n$$\n\nSimilarly, the projections from $S$ give\n\n$$\n\\frac{PM}{MQ} = \\frac{AN}{NB}.\n$$\n\nThus $\\frac{PM}{MQ} = \\frac{QM}{MR}$. Using $PU = UQ$ and $QV = VR$, we get\n\n$$\n\\frac{\\frac{PM}{MQ}}{\\frac{PU}{UQ}} = \\frac{\\frac{QM}{MR}}{\\frac{QV}{VR}}.\n$$\n\nThus the cross-ratios of the pencils $AP, AU, AQ, AM$ and $BQ, BV, BR, BM$ are equal. Observe that $S, X, T, M$ lie on the first pencil and $S, Y, T, M$ lie on the second pencil. Thus\n\n$$\n\\frac{SX}{XT} \\div \\frac{SM}{MT} = \\frac{SY}{YT} \\div \\frac{SM}{MT}.\n$$\n\nIt follows that $XT = YT$ and hence $X = Y$.\n\nNow we proceed to prove the result given in the problem. Observe that $O_1O_2$ is perpendicular to the line $l$, which is perpendicular to $AB$. Thus $O_1O_2$ is parallel to $AB$. Thus $DO_2$ is parallel to $AO_1$. Hence the homothety that carries $\\Gamma_2$ to $\\Gamma$ moves $D$ to $A$. Hence the point $R$, which is obtained by the intersection of $\\Gamma_2$ and the extension of $DO_2$, corresponds to $B$ under the above homothety. It follows that $A, D, F$ and $B, R, F$ are collinear.\n\nNote that $PR$ is parallel to $AB$, $AP \\perp BD$, and $AD \\perp BR$; $O_1, O_2$ are the midpoints of $PD$ and $DR$ respectively. Moreover, $E$ and $F$ are the feet of perpendiculars from $B$ onto $AP$ and $AD$ respectively. By the lemma, $AO_1$, $BO_2$, and $EF$ are concurrent.\n\n![](images/Indija_TS_2007_p5_data_775c670572.png)",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 12869,
    "subject": "Mathematics (Olympiad)",
    "question": "Провери ја точноста на равенството:\n\n$$\n\\sum_{k=1}^{n} \\frac{(-1)^{k+1}}{k} \\binom{n}{k} = 1 + \\frac{1}{2} + \\frac{1}{3} + \\dots + \\frac{1}{n},\n$$\n\nкаде $n \\in \\mathbb{N}$.",
    "options": [],
    "answer": "See solution",
    "solution": "Равенството ќе го докажеме со принципот на математичка индукција.\n\nКе воведеме ознака\n\n$$\nx_n = \\sum_{k=1}^{n} \\frac{(-1)^{k+1}}{k} \\binom{n}{k}\n$$\n\nЈасно е дека $x_{n-1} = \\sum_{k=1}^{n-1} \\frac{(-1)^{k+1}}{k} \\binom{n-1}{k}$.\n\nТврдењето е точно за $n=1$. Навистина\n\n$$\nx_1 = \\sum_{k=1}^{1} \\frac{(-1)^{k+1}}{k} \\binom{1}{k} = (-1)^{1+1} \\binom{1}{1} = 1.\n$$\n\nНека тврдењето е точно за $n-1$, т.е. нека е точно равенството\n\n$$\nx_{n-1} = \\sum_{k=1}^{n-1} \\frac{(-1)^{k+1}}{k} \\binom{n-1}{k} = 1 + \\frac{1}{2} + \\dots + \\frac{1}{n-1}.\n$$\n\nОд равенството $\\binom{n}{k} = \\binom{n-1}{k} + \\binom{n-1}{k-1}$, за $k=1,2,\\ldots,n-1$, за природниот број $n$, заради индуктивната претпоставка имаме\n\n$$\n\\begin{align*}\nx_n &= \\sum_{k=1}^{n} \\frac{(-1)^{k+1}}{k} \\binom{n}{k} \\\\\n&= \\sum_{k=1}^{n-1} \\frac{(-1)^{k+1}}{k} \\binom{n}{k} + \\frac{(-1)^{n+1}}{n} \\binom{n}{n} \\\\\n&= \\sum_{k=1}^{n-1} \\frac{(-1)^{k+1}}{k} \\left[ \\binom{n-1}{k} + \\binom{n-1}{k-1} \\right] + \\frac{(-1)^{n+1}}{n} \\cdot 1 \\\\\n&= \\sum_{k=1}^{n-1} \\frac{(-1)^{k+1}}{k} \\binom{n-1}{k} + \\sum_{k=1}^{n-1} \\frac{(-1)^{k+1}}{k} \\binom{n-1}{k-1} + \\frac{(-1)^{n+1}}{n} \\\\\n&= x_{n-1} + \\sum_{k=1}^{n-1} \\frac{(-1)^{k+1}}{k} \\binom{n-1}{k-1} + \\frac{(-1)^{n+1}}{n}\n\\end{align*}\n$$\n\nСо промена на индексот $j = k-1$ во вториот збир, добиваме\n\n$$\n\\sum_{k=1}^{n-1} \\frac{(-1)^{k+1}}{k} \\binom{n-1}{k-1} = \\sum_{j=0}^{n-2} \\frac{(-1)^{j+2}}{j+1} \\binom{n-1}{j} = -\\sum_{j=0}^{n-2} \\frac{(-1)^{j+1}}{j+1} \\binom{n-1}{j}\n$$\n\nЗатоа\n\n$$\nx_n = x_{n-1} - \\sum_{j=0}^{n-2} \\frac{(-1)^{j+1}}{j+1} \\binom{n-1}{j} + \\frac{(-1)^{n+1}}{n}\n$$\n\nНо, соодветно уредување и користење на биномната формула, се добива дека\n\n$$\nx_n = x_{n-1} + \\frac{1}{n}\n$$\n\nЗатоа, по индукција,\n\n$$\nx_n = 1 + \\frac{1}{2} + \\frac{1}{3} + \\dots + \\frac{1}{n}\n$$\n\n(да забележиме дека збирот $\\sum_{k=0}^{n} (-1)^k \\binom{n}{k} = 0$ според биномната формула, односно\n\n$$\n\\sum_{k=0}^{n} (-1)^k \\binom{n}{k} = (1 - 1)^n = 0.\n$$\n\nСега, според принципот на математичка индукција добиваме дека равенството\n\n$$\n\\sum_{k=1}^{n} \\frac{(-1)^{k+1}}{k} \\binom{n}{k} = 1 + \\frac{1}{2} + \\frac{1}{3} + \\dots + \\frac{1}{n}\n$$\n\nе точно за секој природен број $n$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 12870,
    "subject": "Mathematics (Olympiad)",
    "question": "An integer-valued function $f(n)$ is multiplicative (that is, $f(ab) = f(a)f(b)$ for all coprime $a$ and $b$) and satisfies the equation\n\n$$\nf(m + k - 3) = f(m) + f(k) - f(3) \\quad \\text{for all primes } m \\text{ and } k.\n$$\n\nFind $f(11)$.",
    "options": [],
    "answer": "See solution",
    "solution": "The possible values for $f(11)$ are $1$ or $11$.\n\nNote that $f(1) = 1$ and the following equalities hold:\n\n$$\n\\begin{aligned}\n  f(1) &= f(2 + 2 - 3) = f(2) + f(2) - f(3), \\\\\n  f(7) &= f(5 + 5 - 3) = f(5) + f(5) - f(3), \\\\\n  f(10) &= f(2)f(5) = f(11 + 2 - 3) = f(11) + f(2) - f(3), \\\\\n  f(11) &= f(7 + 7 - 3) = f(7) + f(7) - f(3), \\\\\n  f(15) &= f(3)f(5) = f(11 + 7 - 3) = f(11) + f(7) - f(3).\n\\end{aligned}\n$$\n\nLet $a = f(2)$, $b = f(3)$, $c = f(5)$, $d = f(7)$, $e = f(11)$. Solving these equations, we obtain the quadratic $a^2 - 3a + 2 = 0$, which has solutions $a = 1$ or $a = 2$. This leads to two possible functions: $f(k) = 1$ for $k = 2, 3, 5, 7, 11$ and $f(k) = k$ for $k = 2, 3, 5, 7, 11$. These correspond to $f(x) \\equiv 1$ and $f(x) = x$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 12871,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $P(x)$ be a polynomial such that for all real numbers $a, b, c$ satisfying $ab + bc + ca = 0$, the following holds:\n\n$$\nP(a - b) + P(b - c) + P(c - a) = 2P(a + b + c).\n$$\n\nFind all such polynomials $P(x)$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $Q(x) = P(\\sqrt{x})$. The given equation becomes:\n\n$$\nQ(x^2) + Q(y^2) + Q((x + y)^2) = 2Q(x^2 + xy + y^2).\n$$\n\nSetting $x = y$ gives:\n\n$$\n2Q(x^2) + Q(4x^2) = 2Q(3x^2). \\qquad (\\dagger')\n$$\n\nFrom $x = a - b = b - c$, we have $a = b + x$ and $c = b - x$. Substituting into $ab + bc + ca = 0$ gives $b = \\frac{|x|}{\\sqrt{3}}$. Thus, for every real $x$, there is a triple\n\n$$\n(a, b, c) = \\left( \\frac{|x|}{\\sqrt{3}} + x, \\frac{|x|}{\\sqrt{3}}, \\frac{|x|}{\\sqrt{3}} - x \\right)\n$$\n\nsuch that $x = a - b = b - c$, so $(\\dagger')$ holds for all real $x$. Since $Q(x)$ is a polynomial, $2Q(x) + Q(4x) = 2Q(3x)$ holds for all real $x$.\n\nLet $Q(x) = \\sum_{i=1}^n p_i x^i$. Then $2p_i + 4^i p_i = 2 \\cdot 3^i p_i$, so $p_i(4^i + 2 - 2 \\cdot 3^i) = 0$ for all $i$. For $i \\ge 3$, $p_i = 0$. Thus, $Q(x) = c_1 x + c_2 x^2$, so $P(x) = Q(x^2) = c_1 x^2 + c_2 x^4$.\n\nAlternatively, let $P(x) = x^2 f(x^2)$. The condition becomes:\n\n$$\n(a-b)^2 f((a-b)^2) + (b-c)^2 f((b-c)^2) + (c-a)^2 f((c-a)^2) = 2(a+b+c)^2 f((a+b+c)^2).\n$$\n\nPlugging $(a, b, c) = [(1 - \\sqrt{3})b, b, (1 + \\sqrt{3})b]$ gives:\n\n$$\n6b^2 f(3b^2) + 12b^2 f(12b^2) = 18b^2 f(9b^2).\n$$\n\nThis leads to:\n\n$$\n\\frac{f(12b^2) - f(9b^2)}{3b^2} = \\frac{f(9b^2) - f(3b^2)}{6b^2}.\n$$\n\nIf $f$ is degree $\\ge 2$, $f$ is convex for large $x$, leading to a contradiction. Thus, $f$ is at most linear: $f(x) = c_1 + c_2 x$, so $P(x) = c_1 x^2 + c_2 x^4$.\n\nTherefore, all polynomials of the form $P(x) = c_1 x^2 + c_2 x^4$ (with real $c_1, c_2$) are solutions.",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 12872,
    "subject": "Mathematics (Olympiad)",
    "question": "a) Find the largest possible value of the number\n$$\nx_1x_2 + x_2x_3 + \\dots + x_{n-1}x_n,\n$$\nif $x_1, x_2, \\dots, x_n$ ($n \\ge 2$) are non-negative integers and their sum is $2011$.\n\nb) Find the numbers $x_1, x_2, \\dots, x_n$ for which the maximum value determined at **a)** is obtained.",
    "options": [],
    "answer": "See solution",
    "solution": "a) Let $x_1, x_2, \\dots, x_n$ be non-negative integers satisfying the conditions from the statement. Let $M = \\max_{1 \\le i \\le n} x_i$. If $x_j = M$, then\n$$\nx_1x_2 + x_2x_3 + \\dots + x_{n-1}x_n \\le x_1x_j + x_2x_j + \\dots + x_{j-1}x_j + x_jx_{j+1} + x_jx_{j+2} + \\dots + x_jx_n = x_j(2011 - x_j) = M(2011 - M) \\le 1005 \\cdot 1006.\n$$\nIndeed, the last inequality comes from $(M - 1005)(M - 1006) \\ge 0$, which is true for any integer $M$. The largest possible value is $1005 \\cdot 1006$ because this value can be obtained by choosing, for example, $x_1 = 1005$, $x_2 = 1006$, and $x_k = 0$ for $k \\ge 3$.\n\nb) For $n = 2$, we have $x_1x_2 = 1005 \\cdot 1006 \\Leftrightarrow x_1(2011 - x_1) = 1005 \\cdot 1006 \\Leftrightarrow (x_1 - 1005)(x_1 - 1006) = 0 \\Leftrightarrow (x_1, x_2) \\in \\{(1005, 1006), (1006, 1005)\\}$.\n\nFor $n = 3$, $x_1x_2 + x_2x_3 = 1005 \\cdot 1006 \\Leftrightarrow x_2(x_1 + x_3) = 1005 \\cdot 1006$, which, as above, gives $x_2 = 1005$, $x_1 + x_3 = 1006$ or $x_2 = 1006$, $x_1 + x_3 = 1005$. We obtain\n$$(x_1, x_2, x_3) \\in \\{(k, 1005, 1006 - k) \\mid k = 0, 1, \\dots, 1006\\} \\cup \\{(k, 1006, 1005 - k) \\mid k = 0, 1, \\dots, 1005\\}.$$\n\nFor $n \\ge 4$, let $j$ be the smallest index for which $x_j > 0$. Then, replacing $x_j$ by $0$ and $x_{j+2}$ by $x_{j+2} + x_j$ increases the value of the sum by $x_jx_{j+3}$. Using this, it is easy to see that if $x_1x_2 + x_2x_3 + \\dots + x_{n-1}x_n = 1005 \\cdot 1006$, then at most three of the terms can be non-zero. We obtain\n$$(x_1, \\dots, x_n) \\in \\{(0, \\dots, 0, k, 1005, 1006 - k, 0, \\dots, 0) \\mid k = 0, 1, \\dots, 1006\\} \\cup \\{(0, \\dots, 0, k, 1006, 1005 - k, 0, \\dots, 0) \\mid k = 0, 1, \\dots, 1005\\},$$\nwhere the group of the three non-zero components can be located anywhere.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 12873,
    "subject": "Mathematics (Olympiad)",
    "question": "Prove that there exist infinitely many non-isosceles triangles with rational side lengths, rational lengths of altitudes, and perimeter equal to $3$.",
    "options": [],
    "answer": "See solution",
    "solution": "If the lengths $a$, $b$, and $c$ of the sides are rational, then since $a h_a = b h_b = c h_c = 2A$, where $h_a$, $h_b$, and $h_c$ denote the lengths of the altitudes of the triangle, it is enough to find an infinite number of triangles with rational area. From Heron's formula:\n\n$$\nA = \\sqrt{\\frac{3}{2} \\left(\\frac{3}{2} - a\\right) \\left(\\frac{3}{2} - b\\right) \\left(\\frac{3}{2} - c\\right)} = \\frac{1}{4} \\sqrt{3(3 - 2a)(3 - 2b)(3 - 2c)}.\n$$\n\nHence, for the area to be rational for infinitely many choices of sides, it is enough that the quantity under the radical is a square of a rational number. Therefore, it suffices to find rational numbers $x$, $y$, and $z$ such that\n\n$$\n3 - 2a = 3x^2, \\quad 3 - 2b = 3y^2, \\quad 3 - 2c = 3z^2.\n$$\n\nThis is feasible by putting\n\n$$\nx = \\frac{2uv}{u^2 + v^2 + w^2}, \\quad y = \\frac{2uw}{u^2 + v^2 + w^2}, \\quad z = \\frac{-u^2 + v^2 + w^2}{u^2 + v^2 + w^2},\n$$\n\nwhere $u$, $v$, and $w$ are rational. It is easily checked that for these values of $x$, $y$, and $z$, we have $x^2 + y^2 + z^2 = 1$, and therefore, there exists a triangle of side lengths $a$, $b$, and $c$ with perimeter $3$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 12874,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $S$ be a finite set and let $\\mathcal{P}(S)$ be its power set, i.e., the set of all subsets of $S$, including the empty set and $S$ itself.\n\nIf $\\mathcal{A}$ and $\\mathcal{B}$ are non-empty subsets of $\\mathcal{P}(S)$, define\n\n$$\n\\mathcal{A} \\vee \\mathcal{B} = \\{X : X \\subseteq \\mathcal{A} \\cup \\mathcal{B},\\ A \\in \\mathcal{A},\\ B \\in \\mathcal{B}\\}.\n$$\n\nGiven a non-negative integer $n \\leq |S|$, determine the minimal size $|\\mathcal{A} \\vee \\mathcal{B}|$ may have, where $\\mathcal{A}$ and $\\mathcal{B}$ are non-empty subsets of $\\mathcal{P}(S)$ such that $|\\mathcal{A}| + |\\mathcal{B}| > 2^n$.",
    "options": [],
    "answer": "See solution",
    "solution": "The required minimum is $2^n$, which is achieved, for example, if $\\mathcal{A} = \\mathcal{B} = \\mathcal{P}(T)$, where $T$ is an $n$-element subset of $S$.\n\nWe show by induction on $n$ that $2^n$ is a global lower bound. For $n=0$, this is clear. Assume the claim holds for all non-negative integers less than $n$.\n\nLet $\\mathcal{A}$ and $\\mathcal{B}$ be subsets of $\\mathcal{P}(S)$ such that $|\\mathcal{A}| + |\\mathcal{B}| > 2^n$.\n\nEnlarge $\\mathcal{A}$ and $\\mathcal{B}$ to $\\mathcal{A}^- = \\{X : X \\subseteq A,\\ A \\in \\mathcal{A}\\}$ and $\\mathcal{B}^- = \\{X : X \\subseteq B,\\ B \\in \\mathcal{B}\\}$, respectively. Notice that $\\mathcal{A}^- \\vee \\mathcal{B}^- = \\mathcal{A} \\vee \\mathcal{B}$, so we may assume $\\mathcal{A} = \\mathcal{A}^-$ and $\\mathcal{B} = \\mathcal{B}^-$. In particular, if a non-empty set is a member of $\\mathcal{A}$ (respectively, $\\mathcal{B}$), then so is every singleton subset of that set.\n\nIf $\\mathcal{A} \\cap \\mathcal{B} = \\{\\emptyset\\}$, then $|\\mathcal{A} \\cap \\mathcal{B}| = 1$, and\n$$\n|\\mathcal{A} \\vee \\mathcal{B}| \\geq |\\mathcal{A} \\cup \\mathcal{B}| = |\\mathcal{A}| + |\\mathcal{B}| - 1 \\geq 2^n.\n$$\n\nOtherwise, $\\mathcal{A} \\cap \\mathcal{B}$ contains some singleton $\\{x\\}$, $x \\in S$. Let $\\mathcal{A}' = \\{X : X \\in \\mathcal{A},\\ x \\notin X\\}$, $\\mathcal{A}'' = \\{X : X \\in \\mathcal{A}',\\ X \\cup \\{x\\} \\in \\mathcal{A}\\}$, and note $\\mathcal{A} \\setminus \\mathcal{A}' = \\{X \\cup \\{x\\} : X \\in \\mathcal{A}''\\}$, so $|\\mathcal{A}'| + |\\mathcal{A}''| = |\\mathcal{A}|$. Similarly for $\\mathcal{B}$.\n\nThus,\n$$\n(|\\mathcal{A}'| + |\\mathcal{B}''|) + (|\\mathcal{A}''| + |\\mathcal{B}'|) = |\\mathcal{A}| + |\\mathcal{B}| > 2^n,\n$$\nso either $|\\mathcal{A}'| + |\\mathcal{B}''| > 2^{n-1}$ or $|\\mathcal{A}''| + |\\mathcal{B}'| > 2^{n-1}$. By symmetry, assume $|\\mathcal{A}'| + |\\mathcal{B}''| > 2^{n-1}$, so $|\\mathcal{A}' \\vee \\mathcal{B}''| \\geq 2^{n-1}$ by induction.\n\nTo complete the induction, $\\mathcal{A}' \\vee \\mathcal{B}''$ and $\\mathcal{C} = \\{X \\cup \\{x\\} : X \\in \\mathcal{A}' \\vee \\mathcal{B}''\\}$ are disjoint isomorphic subsets of $\\mathcal{A} \\vee \\mathcal{B}$, so\n$$\n|\\mathcal{A} \\vee \\mathcal{B}| \\geq |\\mathcal{A}' \\vee \\mathcal{B}''| + |\\mathcal{C}| = 2|\\mathcal{A}' \\vee \\mathcal{B}''| \\geq 2^n.\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 12875,
    "subject": "Mathematics (Olympiad)",
    "question": "Given an integer $k \\ge 3$ and a sequence $\\{a_n\\}$ that satisfies $a_k = 2k$ and for each $n > k$,\n\n$$\na_n = \\begin{cases} a_{n-1} + 1, & \\text{if } a_{n-1} \\text{ and } n \\text{ are coprime,} \\\\ 2n, & \\text{otherwise.} \\end{cases}\n$$\n\nProve that $a_n - a_{n-1}$ is a prime for infinitely many $n$.",
    "options": [],
    "answer": "See solution",
    "solution": "Suppose that $a_l = 2l$, $l \\ge k$. Let $p$ be the least prime divisor of $l-1$. Then\n$$(l-1, i) = \\begin{cases} 1, & 1 \\le i < p, \\\\ p, & i = p. \\end{cases}$$\nThus,\n$$(2l + i - 2, l + i - 1) = \\begin{cases} 1, & 1 \\le i < p, \\\\ p, & i = p. \\end{cases}$$\nFrom the recurrence,\n$$a_{l+i-1} = \\begin{cases} 2l + i - 1, & 1 \\le i < p, \\\\ 2l + 2p - 2, & i = p. \\end{cases}$$\nThen $a_{l+p-1} - a_{l+p-2} = (2l+2p-2) - (2l+p-2) = p$ is a prime number, and $a_{l+p-1} = 2(l+p-1)$. From the discussion above, 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": 12876,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $f(r)$, $g(r)$, and $h(r)$ be functions such that\n\n$$\n\\frac{1}{f(r)g(r)h(r)} = \\frac{1}{4} \\left( 2\\sqrt{r} - \\sqrt{r-1} - \\sqrt{r+1} \\right).\n$$\n\nShow that\n\n$$\nS = \\sum_{r=1}^{2020} \\frac{1}{f(r)g(r)h(r)} = \\frac{1}{4} \\left( \\sqrt{2020} + 1 - \\sqrt{2021} \\right).\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "To see that $S = 0.25$ to two decimal places, we note that:\n\n$$\n0 \\le \\frac{1}{4} - S = \\frac{\\sqrt{2021} - \\sqrt{2020}}{4} = \\frac{1}{4(\\sqrt{2020} + \\sqrt{2021})} < \\frac{1}{8\\sqrt{1600}} = \\frac{1}{320}.\n$$\n\nThus the difference does not affect the second decimal place.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 12877,
    "subject": "Mathematics (Olympiad)",
    "question": "Suppose $a_0, \\dots, a_{100}$ are positive real numbers. For each $k$ in $\\{0, 1, \\dots, 100\\}$, consider the polynomial:\n\n$$\na_{100+k}x^{100} + 100a_{99+k}x^{99} + a_{98+k}x^{98} + a_{97+k}x^{97} + \\dots + a_{2+k}x^2 + a_{1+k}x + a_k\n$$\n\nwhere indices are taken modulo $101$, i.e., $a_{100+i} = a_{i-1}$ for any $i$ in $\\{1, 2, \\dots, 100\\}$. Show that it is impossible for each of these $101$ polynomials to have all roots real.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $n = 50$. Assume, for contradiction, that each polynomial has all real roots; these roots must be negative. Let\n\n$$\n-\\alpha_{1,k}, -\\alpha_{2,k}, \\dots, -\\alpha_{2n,k}\n$$\n\nbe the roots of the polynomial\n\n$$\na_{2n+k}x^{2n} + 2n a_{2n-1+k}x^{2n-1} + a_{2n-2+k}x^{2n-2} + a_{2n-3+k}x^{2n-3} + \\dots + a_{2+k}x^2 + a_{1+k}x + a_k\n$$\n\n(indices modulo $2n+1$). By Vieta's formulas:\n\n$$\n\\sum_{j=1}^{2n} \\alpha_{j,k} = 2n \\cdot \\left( \\frac{a_{k-2}}{a_{k-1}} \\right), \\quad \\prod_{j=1}^{2n} \\alpha_{j,k} = \\frac{a_k}{a_{k-1}}\n$$\n\nSince the $\\alpha_{j,k}$ are positive, by AM-GM:\n\n$$\n\\left(\\frac{a_{k-2}}{a_{k-1}}\\right)^{2n} \\ge \\frac{a_k}{a_{k-1}}\n$$\n\nfor each $k$. Multiplying over all $k$:\n\n$$\n\\prod_{k=0}^{2n} \\left( \\frac{a_{k-2}}{a_{k-1}} \\right)^{2n} \\ge \\prod_{k=0}^{2n} \\frac{a_k}{a_{k-1}}\n$$\n\nBoth sides equal $1$, so all AM-GM inequalities are equalities, implying $\\alpha_{1,k} = \\dots = \\alpha_{2n,k} = \\frac{a_{k-2}}{a_{k-1}}$. For $n \\ge 2$, Vieta gives:\n\n$$\n\\frac{a_{k-3}}{a_{k-1}} = \\binom{2n}{2} \\left(\\frac{a_{k-2}}{a_{k-1}}\\right)^2\n$$\n\nwhich simplifies to\n\n$$\n\\binom{2n}{2} a_{k-2}^2 = a_{k-1} a_{k-3}\n$$\n\nfor each $k$. Multiplying all these equations yields\n\n$$\n\\left( \\binom{2n}{2}^{2n+1} - 1 \\right) \\left( \\prod_{k=0}^{2n} a_k \\right)^2 = 0\n$$\n\nwhich forces some $a_k = 0$, contradicting positivity. $\\square$",
    "topic": "Number Theory",
    "subtopic": "Modular Arithmetic"
  },
  {
    "id": 12878,
    "subject": "Mathematics (Olympiad)",
    "question": "Нека $\\Delta ABC$ е рамностран. На страната $AB$ се избрани точки $C_1$ и $C_2$, на $AC$ се избрани точки $B_1$ и $B_2$, и на страната $BC$ се избрани точки $A_1$ и $A_2$, при што важи: $\\overline{A_1A_2} = \\overline{B_1B_2} = \\overline{C_1C_2}$. Нека пресечните точки на правите $A_2B_1$ и $B_2C_1$, $B_2C_1$ и $C_2A_1$, $C_2A_1$ и $A_2B_1$ се $E, F, G$ соодветно. Покажи дека триаголникот формиран од отсечките $B_1A_2$, $A_1C_2$ и $C_1B_2$ е сличен со триаголникот $\\Delta EFG$.\n\n![](images/Makedonija_2009_p26_data_33b9b4e82a.png)",
    "options": [],
    "answer": "See solution",
    "solution": "Нека триаголникот формиран од отсечките $B_1A_2$, $A_1C_2$ и $C_1B_2$ го означиме со $\\Delta A_3B_3C_3$. Нека $P$ е точка во внатрешноста на $\\Delta EFG$ така што $C_1C_2PB_2$ е паралелограм.\n\nТогаш $\\Delta B_2PB_1$ е рамностран.\n\nЗначи $PA_1A_2B_1$ е паралелограм. Од претходните разгледувања имаме $PC_2 \\parallel EF$ и $PA_1 \\parallel EG$.\n\nЗатоа $\\Delta PC_2A_1 \\sim \\Delta EFG$. Од друга страна $\\Delta PC_2A_1 \\cong \\Delta A_3B_3C_3$. Значи $\\Delta A_3B_3C_3 \\sim \\Delta EFG$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 12879,
    "subject": "Mathematics (Olympiad)",
    "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\n(a) 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\n(b) Can it happen that both the length of the beetle's journey and the distance between its initial and final point are integers?\n\n![](images/prob1718_p24_data_ac1683df5b.png)",
    "options": [],
    "answer": "See solution",
    "solution": "**Answer:** (a) Yes; (b) No.\n\nLet $O$ be the origin. Let $A_0$ be the starting point, $A_1$ the first turning point, $A_2$ the second, $A_3$ the third, and $A_4$ the endpoint. The 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 $\\frac{OA_1}{OA_0} = x$. Similarly, $OA_2 = x^2$, $OA_3 = x^3$, and $OA_4 = x^4$.\n\nBy the Pythagorean theorem, the length of the beetle's journey is:\n\n$$\n\\sqrt{1 + x^2} + \\sqrt{x^2 + x^4} + \\sqrt{x^4 + x^6} + \\sqrt{x^6 + x^8}\n$$\n\nor equivalently,\n\n$$\n(1 + x + x^2 + x^3) \\sqrt{1 + x^2} = (x^4 - 1) \\cdot \\frac{\\sqrt{x^2 + 1}}{x - 1}, \\quad x \\neq 1.\n$$\n\nThe distance between $A_0$ and $A_4$ is $|x^4 - 1|$.\n\n(a) Take $x = \\frac{4}{3}$. Then $x^4 - 1$ is rational, and $\\sqrt{1 + x^2} = \\sqrt{1 + \\frac{16}{9}} = \\frac{5}{3}$, so $(x^4 - 1) \\cdot \\frac{\\sqrt{x^2 + 1}}{x - 1}$ is also rational.\n\n(b) Suppose the distance between $A_0$ and $A_4$ is an integer, so $x^4$ is integer. If $(x^4 - 1) \\cdot \\frac{\\sqrt{x^2 + 1}}{x - 1}$ is also integer (with $x \\neq 1$), then $\\frac{\\sqrt{x^2 + 1}}{x - 1}$ must be rational. Consider three cases:\n\n1. If $x$ is integer, $x^2 + 1$ must be a perfect square, which is impossible for consecutive integers.\n2. If $x$ is irrational and $x^2$ is integer, $\\frac{2x}{x^2 + 1}$ must be rational, but $2x$ is irrational.\n3. If $x^2$ is irrational, $\\frac{2x}{x^2 + 1}$ must be rational, so $\\frac{4x^2}{(x^2 + 1)^2}$ is rational, implying $\\frac{x^4 + 1}{4x^2}$ is rational, which is impossible as $4x^2$ is irrational and $x^4 + 1$ is integer.\n\nThus, both quantities cannot be integers.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 12880,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $ABCD$ be a quadrilateral inscribed in the circle $c(O, R)$ and let $K, L, M, N, S, T$ be the midpoints of the segments $AB$, $BC$, $CD$, $AD$, $AC$, and $BD$, respectively. Prove that the centers of the circumcircles of the triangles $KLS$, $LMT$, $MNS$, and $NKT$ form a cyclic quadrilateral similar to $ABCD$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $c_1, c_2, c_3, c_4$ be the circumcircles of the triangles $KLS$, $LMT$, $MNS$, and $NKT$, respectively. It is easy to prove that the quadrilaterals $KLMN$, $TLSN$, and $KSMT$ are parallelograms, so $KM$, $NL$, and $TS$ will pass through the same point $G$. Therefore, the points $K$, $S$, $L$ are symmetric to the points $M$, $T$, $N$, respectively, with respect to a symmetry centered at $G$.\n\nLet $c'_1$ be the circumcircle of the triangle $MTN$. Then $c'_1$ passes through the center $O$ of the circle $c(O, R)$ and has diameter $OD = R$ (since $OMD = OND = 90^\\circ$).\n\nHowever, the circles $c_1$ and $c'_1$ are symmetric with respect to $G$ (since the triangles defining the two circles are point-to-point symmetric with respect to $G$).\n\nHence, the circle $c_1$ has radius $\\frac{R}{2}$ and passes through $O'$ (the symmetric point of $O$ with respect to $G$).\n\n![](images/Hellenic_Mathematical_Competitions_2011_booklet_p24_data_f09b0e47a5.png)\n\nSimilarly, we prove that the circles $c_2$, $c_3$, $c_4$ have radius $\\frac{R}{2}$ and pass through $O'$. Therefore, the centers of the circles $c_1$, $c_2$, $c_3$, $c_4$ belong to the circle with center $O'$ and radius $\\frac{R}{2}$.\n\nThe centers of the circles $c'_1$, $c'_2$, $c'_3$, and $c'_4$ (which are the midpoints of the segments $OA$, $OB$, $OC$, and $OD$, respectively) define a quadrilateral with sides parallel to the corresponding sides of $ABCD$. Hence, the centers of the circles $c_1$, $c_2$, $c_3$, $c_4$ (symmetric to the centers of the circles $c'_1$, $c'_2$, $c'_3$, $c'_4$) define a quadrilateral similar to $ABCD$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 12881,
    "subject": "Mathematics (Olympiad)",
    "question": "In triangle $ABC$, $AL$, $BM$, and $CN$ are medians. Prove that $\\angle ANC = \\angle ALB$ if and only if $\\angle ABM = \\angle LAC$.",
    "options": [],
    "answer": "See solution",
    "solution": "Lines $LN$ and $AC$ are parallel, hence $\\angle NLA = \\angle LAC$. We have to show the following implication:\n\n$$\n\\angle ANC = \\angle ALB \\Leftrightarrow \\angle ABM = \\angle NLA.\n$$\n\nLet $G$ be the centroid. Then we have the following equivalences:\n\n$\\angle ANC = \\angle ALB \\Leftrightarrow \\angle BNC + \\angle ALB = \\pi \\Leftrightarrow BNLG$ is cyclic $\\Leftrightarrow \\angle ABM = \\angle NLA$, since they share the common segment. The statement is proved.\n\n![](images/Ukrajina_2011_p9_data_3dfa027cff.png)",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 12882,
    "subject": "Mathematics (Olympiad)",
    "question": "Given a set $S$ of $n$ variables, a binary operation $\\times$ on $S$ is called *simple* if it satisfies $$(x \\times y) \\times z = x \\times (y \\times z)$$ for all $x, y, z \\in S$ and $x \\times y \\in \\{x, y\\}$ for all $x, y \\in S$.\n\nGiven a simple operation $\\times$ on $S$, any string of elements in $S$ can be reduced to a single element, such as $xyz \\rightarrow x \\times (y \\times z)$. A string of variables in $S$ is called *full* if it contains each variable in $S$ at least once, and two strings are equivalent if they evaluate to the same variable regardless of which simple $\\times$ is chosen. For example, $xxx$, $xx$, and $x$ are equivalent, but these are only full if $n = 1$.\n\nSuppose $T$ is a set of strings such that any full string is equivalent to exactly one element of $T$. Determine the number of elements of $T$.",
    "options": [],
    "answer": "See solution",
    "solution": "**Solution.** The answer is $n!^2$.\n\nLet $s_1 \\sim s_2$ denote that strings $s_1$ and $s_2$ are equivalent. If $x_1x_2 \\cdots x_k$ is any string in $S$, then\n\n$$\nx_1x_2\\cdots x_kx_1x_2\\cdots x_k = (x_1\\cdots x_k)(x_1\\cdots x_k) \\sim x_1\\cdots x_k\n$$\nbecause $xx \\in \\{x\\}$.\n\nAlso, if $a_1a_2 \\cdots a_i$, $b_1b_2 \\cdots b_j$ are strings in $S$ and $x \\in S$, then\n\n$$\nxa_1a_2 \\cdots a_ixb_1b_2 \\cdots b_jx \\sim xa_1a_2 \\cdots a_ib_1b_2 \\cdots b_jx\n$$\n\nTo prove this, it suffices to show $xaxbx \\sim xabx$. Suppose for some $\\times$, these did not evaluate to the same element of $S$. Then $a \\times x = x$, since otherwise $x(a \\times x)bx = xabx$. Similarly, $x \\times b = x$. This means\n\n$$\nxaxbx = x(a \\times x)bx = xxbx = x(x \\times b)x = xxx = x\n$$\n\nso by assumption $xabx \\neq x$. But if $xabx = a$, then $xa = a$, $ab = a$, and $ax = a$ must all hold, the last of which is a contradiction. Similarly, if $xabx = b$ we reach a contradiction. Thus, this equivalence holds.\n\nGiven any full string $w = x_1x_2 \\cdots x_k$, we claim it is equivalent to some string consisting of two concatenated permutations of $S$, i.e., $\\sigma_1\\sigma_2$ where $\\sigma_1$ and $\\sigma_2$ are permutations of the $n$ variables of $S$. Note that $w \\sim ww = x_1x_2 \\cdots x_kx_1x_2 \\cdots x_k$. This string may be reduced by our previous observation so that, in any case where an element $y$ appears more than twice, all but the outermost instances of $y$ are removed. Because $w$ was full, the new string $\\tilde{w} \\sim w$ will have each element of $S$ occurring exactly twice. The first occurrences of each variable all occur in the first instance of $w$, and the last occurrences all occur in the second instance of $w$ (in $ww$), so $\\tilde{w}$ consists of two concatenated permutations of $S$, of which there are $n!^2$ possibilities.\n\nIt remains to show that two double-permutations $\\sigma_1\\sigma_2$ and $\\sigma'_1\\sigma'_2$ are not equivalent. Suppose otherwise. Without loss of generality, assume $\\sigma_1 \\neq \\sigma'_1$; otherwise, reverse the order of the operation $\\times$ found below. Choose two elements $a, b$ such that $a$ occurs before $b$ in $\\sigma_1$ but their order is reversed in $\\sigma'_1$. Then choose an operation $\\times$ such that for any $x, y \\in S$,\n\n$$\nx \\times y = \\begin{cases} y & \\text{if } x \\notin \\{a, b\\},\\ y \\in \\{a, b\\}, \\\\ x & \\text{otherwise.} \\end{cases}\n$$\n\nThis operation satisfies $x \\times y \\in \\{x, y\\}$, and some easy casework on which of $x, y, z$ are in $\\{a, b\\}$ shows that $(x \\times y) \\times z = x \\times (y \\times z)$ for all $x, y, z \\in S$. Hence, the operation is simple. Furthermore, $\\sigma_1\\sigma_2$ evaluates to $a$ under this operation, while $\\sigma'_1\\sigma'_2$ evaluates to $b$. Thus, the two strings are not equivalent, finishing the solution.",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 12883,
    "subject": "Mathematics (Olympiad)",
    "question": "Four circles with a common center are drawn in a plane, and the distances between adjacent circles are equal. Prove that it is not possible to draw a square with each vertex lying on a different circle.\n\n![](images/Baltic_Way_SHL_2009-11_13-16_p208_data_d9c91af88e.png)",
    "options": [],
    "answer": "See solution",
    "solution": "First, we prove the following lemma:\n\n**Lemma 1.** If $ABCD$ is a square, then for any point $E$,\n\n$$\nEA^2 + EC^2 = EB^2 + ED^2.\n$$\n\n*Proof.* Let $A'$, $C'$ be the projections of $E$ onto sides $AB$ and $CD$, respectively. Then\n\n$$\nEA^2 + EC^2 = (EA'^2 + AA'^2) + (EC'^2 + CC'^2)\n$$\n$$\nEB^2 + ED^2 = (EA'^2 + A'B^2) + (EC'^2 + C'D^2)\n$$\n\nSince $A'B = CC'$ and $AA' = C'D$, it follows that $EA^2 + EC^2 = EB^2 + ED^2$. Note that $E$ does not have to be inside the square; this holds for any point. $\\square$\n\nNow, let $O$ be the common center of the circles and $ABCD$ a square with each vertex on a different circle. Assume $A$ lies on the largest circle. If $a$ is the radius of the smallest circle and $p$ is the distance between circles, then the radii are $a$, $a + p$, $a + 2p$, and $a + 3p$. These are also the distances from $O$ to the vertices of the square $ABCD$, with $OA = a + 3p$. Consider the value $OA^2 + OC^2 - OB^2 - OD^2$. Its smallest possible value is attained when $OC = a$, so\n\n$$\nOA^2 + OC^2 - OB^2 - OD^2 \\geq (a + 3p)^2 + a^2 - (a + p)^2 - (a + 2p)^2 = 4p^2 > 0,\n$$\n\nwhich contradicts the lemma.",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 12884,
    "subject": "Mathematics (Olympiad)",
    "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 and minimum values of $f(x, y, z)$.",
    "options": [],
    "answer": "See solution",
    "solution": "First, prove that $f \\le \\frac{1}{7}$; when $x = y = z = \\frac{1}{3}$, we have $f = \\frac{1}{7}$.\n\nSince\n$$\n\\begin{aligned}\nf &= \\sum \\frac{x(x+3y-1)}{1+x+3y} \\\\\n  &= 1 - 2\\sum \\frac{x}{1+x+3y}\n\\end{aligned}\n$$\nBy Cauchy's inequality,\n$$\n\\sum \\frac{x}{1+x+3y} \\ge \\frac{(\\sum x)^2}{\\sum x(1+x+3y)} = \\frac{1}{\\sum x(1+x+3y)}\n$$\nand\n$$\n\\sum x(1+x+3y) - \\sum x(2x+4y+z) = 2 + \\sum xy \\le \\frac{7}{3}.\n$$\nSo $\\sum \\frac{x}{1+x+3y} \\ge \\frac{3}{7}$, thus $f \\le 1 - 2 \\times \\frac{3}{7} = \\frac{1}{7}$.\nTherefore, $f_{\\max} = \\frac{1}{7}$; when $x = y = z = \\frac{1}{3}$, $f = \\frac{1}{7}$.\n\nSecond, prove that $f \\ge 0$; when $x = 1$, $y = z = 0$, we have $f = 0$.\n\nIn fact,\n$$\n\\begin{aligned}\nf(x, y, z) &= \\frac{x(2y-z)}{1+x+3y} + \\frac{y(2z-x)}{1+y+3z} + \\frac{z(2x-y)}{1+z+3x} \\\\\n&= xy \\left( \\frac{2}{1+x+3y} - \\frac{1}{1+y+3z} \\right) \\\\\n&\\quad + xz \\left( \\frac{2}{1+z+3x} - \\frac{1}{1+x+3y} \\right) \\\\\n&\\quad + yz \\left( \\frac{2}{1+y+3z} - \\frac{1}{1+z+3x} \\right) \\\\\n&= \\frac{7xyz}{(1+x+3y)(1+y+3z)} \\\\\n&\\quad + \\frac{7xyz}{(1+z+3x)(1+x+3y)} \\\\\n&\\quad + \\frac{7xyz}{(1+y+3z)(1+z+3x)} \\\\\n&\\ge 0.\n\\end{aligned}\n$$\nSo $f_{\\min} = 0$ and $f_{\\max} = \\frac{1}{7}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 12885,
    "subject": "Mathematics (Olympiad)",
    "question": "假設 $ABC$ 是一個不等邊的銳角三角形，其垂心為 $H$。令 $l_a$ 為通過 $B$ 對 $CH$ 的對稱點和 $C$ 對 $BH$ 的對稱點之直線。類似地，我們定義 $l_b$ 和 $l_c$。若 $l_a, l_b$ 和 $l_c$ 的交點形成一個三角形，試證明該三角形的垂心與外心和 $H$ 共線。",
    "options": [],
    "answer": "See solution",
    "solution": "![](images/2024-TWN_p111_data_6fc7896821.png)\n\n設 $A_b, A_c$ 分別為 $A$ 關於 $BH$ 和 $CH$ 的對稱點。$B_c, B_a$ 及 $C_a, C_b$ 亦同理定義。依定義，$l_a = B_cC_b,\\ l_b = C_aA_c,\\ l_c = A_bB_a$。設 $A_1 = l_b \\cap l_c,\\ B_1 = l_c \\cap l_a,\\ C_1 = l_a \\cap l_b$，$H_1, O_1$ 分別為 $\\triangle A_1B_1C_1$ 的垂心與外心。\n\n*Claim 1.* $\\triangle AA_bA_c \\cong \\triangle ABC$。\n\n*證明.* 設 $P = BH \\cap AC,\\ Q = CH \\cap AB$，已知 $\\triangle APQ \\cong \\triangle ABC$。以 $A$ 為中心、比例 $2$ 的放大將 $\\triangle APQ$ 映至 $\\triangle AA_bA_c$，故 $\\triangle AA_bA_c \\cong \\triangle ABC$。\n\n*Claim 2.* $\\triangle AA_bA_c \\cong \\triangle AB_aC_a$ 且 $A_1$ 在以 $H$ 為圓心的 $\\triangle AA_bA_c$ 外接圓上。\n\n*證明.* 因 $B_a, C_a$ 為 $B, C$ 關於 $AH$ 的對稱點，$\\triangle AB_aC_a \\cong \\triangle ABC$。結合 *Claim 1*，$\\triangle AA_bA_c \\cong \\triangle AB_aC_a$，$A$ 為相似中心。故 $\\angle A_c A_1 A_b = \\angle A_c AA_b$，即 $A_1$ 在 $\\odot AA_bA_c$ 上。對稱性知 $HA_b = HA = HA_c$，故 $H$ 為此圓心。\n\n*Claim 3.* $\\triangle A_1B_1C_1 \\cong \\triangle ABC$。\n\n*證明.* 由 *Claim 2*，\n\n$$\n\\angle C_1A_1B_1 = \\angle A_cA_1A_b = \\angle A_cAA_b = -\\angle CAB\n$$\n\n同理 $\\angle A_1B_1C_1 = -\\angle ABC,\\ \\angle B_1C_1A_1 = -\\angle BCA$，故 $\\triangle A_1B_1C_1 \\cong \\triangle ABC$。\n\n設 $\\triangle A_1B_1C_1$ 與 $\\triangle ABC$ 的相似比為 $\\lambda (= \\frac{B_1C_1}{BC})$，則\n\n$$\n\\lambda = \\frac{H_1A_1}{HA} = \\frac{H_1B_1}{HB} = \\frac{H_1C_1}{HC}.\n$$\n\n由 *Claim 2*，$HA = HA_1$，同理 $HB = HB_1,\\ HC = HC_1$，故\n\n$$\n\\lambda = \\frac{H_1A_1}{HA_1} = \\frac{H_1B_1}{HB_1} = \\frac{H_1C_1}{HC_1}.\n$$\n\n因此，圓 $A_1B_1C_1$ 為 $HH_1$ 段的阿波羅尼斯圓，比例為 $\\lambda$，所以 $HH_1$ 通過 $O_1$。",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 12886,
    "subject": "Mathematics (Olympiad)",
    "question": "Azra thought of four numbers $a$, $b$, $c$, and $d$. She wrote on the blackboard all possible sums of two of these numbers, but then erased one of the sums. The remaining sums on the blackboard are:\n\n$$\n\\{-2, 1, 2, 3, 6\\}\n$$\n\nWhat are the four numbers Azra thought of?",
    "options": [],
    "answer": "See solution",
    "solution": "Let $a, b, c, d$ be the numbers Azra thought of. Without loss of generality, assume the deleted sum is $c+d$. The sums on the blackboard are $a+b, a+c, a+d, b+c, b+d$, i.e.\n\n$$\n\\{-2, 1, 2, 3, 6\\} = \\{a+b, a+c, a+d, b+c, b+d\\}.\n$$\n\nSince\n$$\n(a+c) + (b+d) = (a+d) + (b+c),\n$$\nthere must be two pairs of numbers among those on the blackboard with equal sums. The only such pairs are $-2+6=1+3=4$, so the sum of all numbers Azra thought of is $4$.\n\nThe number $2$ does not appear in the last equality, so $a+b=2$, and the deleted number $c+d=4-2=2$.\n\nNow, let $a \\leq b \\leq c \\leq d$. The set of all pairwise sums is:\n$$\n\\{-2, 1, 2, 2, 3, 6\\} = \\{a+b, a+c, a+d, b+c, b+d, c+d\\}.\n$$\n\nSince $a+b$ is the smallest and $c+d$ is the largest, $a+b=-2$, $c+d=6$. The next smallest is $a+c=1$, and the next largest is $b+d=3$. The remaining sums are $a+d=b+c=2$.\n\nNow,\n$$\n2a = (a+b) + (a+c) - (b+c) = -2 + 1 - 2 = -3 \\implies a = -\\frac{3}{2}\n$$\nThen,\n$$\nb = -2 - a = -2 + \\frac{3}{2} = -\\frac{1}{2} \\\\\nc = 1 - a = 1 + \\frac{3}{2} = \\frac{5}{2} \\\\\nd = 2 - a = 2 + \\frac{3}{2} = \\frac{7}{2}\n$$\n\nCheck:\n- $b+c = -\\frac{1}{2} + \\frac{5}{2} = 2$\n- $b+d = -\\frac{1}{2} + \\frac{7}{2} = 3$\n- $c+d = \\frac{5}{2} + \\frac{7}{2} = 6$\n\nThus, Azra thought of the numbers $-\\frac{3}{2}$, $-\\frac{1}{2}$, $\\frac{5}{2}$, and $\\frac{7}{2}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 12887,
    "subject": "Mathematics (Olympiad)",
    "question": "Given the set of keys $\\{1, 2, 3, 4, 5, 6\\}$, each director of a company is given a set of 3 different keys. No two directors have the same set of keys. No two directors together can open the safe (i.e., the union of their keys is not the full set $\\{1,2,3,4,5,6\\}$). What is the maximum number of directors the company can have?",
    "options": [],
    "answer": "See solution",
    "solution": "Let $\\mathcal{K} = \\{1, 2, 3, 4, 5, 6\\}$ be the set of all keys. Each director receives a subset of 3 keys, i.e., an element of $\\binom{\\mathcal{K}}{3}$. Two directors together can open the safe if their sets $A$ and $B$ satisfy $A \\cup B = \\mathcal{K}$. Construct a graph $G$ where the vertices are all 3-element subsets of $\\mathcal{K}$, and two vertices are connected if their union is $\\mathcal{K}$. An independent set in this graph corresponds to a set of directors such that no two together can open the safe. The largest independent set corresponds to the maximum number of directors. The graph $G$ is a disjoint union of $K_2$'s (pairs of complementary 3-element subsets), so the independence number is $10$. Thus, the maximum number of directors is $10$.\n\n![](images/Slovenija_2015_p32_data_c30a987f43.png)",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 12888,
    "subject": "Mathematics (Olympiad)",
    "question": "We start with a square with side length $1$. During the first minute, small squares with side length $\\frac{1}{3}$ grow on the middle of the vertical sides. During the next minute, on the middle of each vertical line segment in the new figure, a new small square grows, whose sides have length $\\frac{1}{3}$ of these line segments. Below you can see the situation after 0, 1, and 2 minutes.\n\n![](images/NLD_ABooklet_2023_p8_data_a53d5df15b.png)\n\nThis process continues like this. Each minute, on the middle of each vertical line segment a new square grows, whose sides are $\\frac{1}{3}$ of the length of that line segment. After one hour this process of new squares growing on the figure has happened 60 times.\n\nWhat is the circumference of the figure after one hour?",
    "options": [],
    "answer": "See solution",
    "solution": "$84$",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 12889,
    "subject": "Mathematics (Olympiad)",
    "question": "Given a starting positive integer $n$ on a blackboard, at each step you may replace the current number with the sum of two positive integers whose sum equals the current number. What is the smallest number that can be obtained after a finite number of such steps?",
    "options": [],
    "answer": "See solution",
    "solution": "If $n < 5$, the smallest number obtainable is $n$ itself, since no operation can decrease it further. If $n \\geq 5$, the smallest number obtainable is $5$. This is because any number $m > 5$ can be reduced step by step to $5$, but not below, as shown by considering the possible decompositions and replacements.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 12890,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all prime numbers $p, q, r$ such that\n\n$$\np(p+1) + q(q+1) = r(r+1).\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "**Answer:** $p=2$, $q=2$, $r=3$.\n\n**Solution.** Without loss of generality, let $p \\leq q$. Clearly, $r > p$. We can write:\n\n$$\np(p+1) = (r-q)(r+q+1). \\quad (1)\n$$\n\nSince $p$ is prime, either $r-q$ or $r+q+1$ is divisible by $p$. If $p$ divides $r-q$, then $r-q \\geq p$,\n\n$$\np(p+1) \\leq (r-q)(r-q+1) < (r-q)(r+q+1),\n$$\n\nwhich contradicts (1).\n\nSuppose $r+q+1$ is divisible by $p$. If $p=2$, then $r+q+1$ is even, so $r$ is odd, hence $q$ is even, thus $q=2$ and $r=3$.\n\nIf $p > 2$, let $r+q+1 = kp$. Then $r$ and $q$ are odd and $k$ is odd, $k > 1$. Then $p+1 = k(r-q)$ and we get $k^2(r-q) = kp + k$ or $r+q+1+k = k^2 r - k^2 q$, which is equivalent to\n\n$$\n(k^2+1)q = (k^2-1)r - (k+1).\n$$\n\nThe right-hand side is divisible by $(k+1)$, so the left-hand side is as well. Since $k$ is odd, $\\frac{k^2+1}{2}$ is divisible by $\\frac{k+1}{2}$. Observe that\n\n$$\n\\left(\\frac{k^2+1}{2}, \\frac{k+1}{2}\\right) = \\left(\\frac{k^2+1}{2} - \\frac{(k-1)(k+1)}{2}, \\frac{k+1}{2}\\right) = \\left(1, \\frac{k+1}{2}\\right) = 1.\n$$\n\nThus, $\\frac{k^2+1}{2}$ and $\\frac{k+1}{2}$ are coprime, so $\\frac{k+1}{2}$ is divisible by $q$. Since $k > 1$, $\\frac{k+1}{2} > 1$ and $q = \\frac{k+1}{2}$.\n\nWe get $r = kp - q - 1$ and $q = \\frac{k+1}{2}$. Plugging this into $p+1 = k(r-q)$ gives $p+1 = (kp - k - 2)k$. But this is impossible because $k \\geq 3$, $p \\geq 3$ implies\n\n$$\n(kp - k - 2)k > kp - k - 2 = k(p-1) - 2 \\geq 3(p-1) - 2 = p + (2p-5) \\geq p+1.\n$$\n\nThis completes the proof.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 12891,
    "subject": "Mathematics (Olympiad)",
    "question": "Show that there is no sequence $\\{a_n\\}$ of positive integers with $N > 5$ members satisfying\n$$\na_{n+2} = a_{n+1} + \\sqrt{a_{n+1} + a_n}\n$$\nfor all $n = 1, 2, \\dots, N-2$. Additionally, describe all such sequences with exactly five members.",
    "options": [],
    "answer": "See solution",
    "solution": "Since each $a_i$ is a positive integer, from the recurrence we have $a_{n+1} + a_n = k^2$ for some integer $k$. The recurrence gives $(a_{n+2} - a_{n+1})^2 = a_{n+1} + a_n$, leading to a quadratic in $a_{n+1}$:\n$$\na_{n+1}^2 - (2a_{n+2} + 1)a_{n+1} + a_{n+2}^2 - a_n = 0.\n$$\nThe discriminant $D = 4(a_n + a_{n+2}) + 1$ must be a perfect square, say $D = (2m+1)^2$, so $a_n + a_{n+2} = m(m+1)$. Adding $a_n$ to both sides of the original recurrence and using these results, we get $m(m+1) = k^2 + k$, so $m = k$. Thus,\n$$\n\\begin{cases}\na_n + a_{n+1} = k^2 \\\\\na_n + a_{n+2} = k^2 + k\n\\end{cases}\n$$\nfor some integer $k$. Writing these for $n=2$ and $n=3$ and solving, we find\n$$\na_2 = \\frac{2k^2 - l^2 + k}{2}, \\quad a_3 = \\frac{l^2 - k}{2}, \\quad a_4 = \\frac{l^2 + k}{2}, \\quad a_5 = \\frac{l^2 + 2l + k}{2}\n$$\nfor integers $k < l$ of the same parity. For $a_6$, we get $a_6 = a_5 + \\sqrt{a_5 + a_4} = a_5 + \\sqrt{l^2 + l + k}$, which cannot be integer for $k < l$, so no such sequence exists for $N > 5$. All sequences with five members are given by the above formulas for $k < l$ of the same parity, with $a_1 = (a_3 - a_2)^2 - a_2$, and $a_1, a_2 > 0$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 12892,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $k$ and $m$ be integers with $1 < k < m$. For a positive integer $i$, let $L_i$ be the least common multiple of $1, 2, \\dots, i$. Prove that $k$ is a divisor of\n\n$$\nL_i \\cdot \\left[ \\binom{m}{i} - \\binom{m-k}{i} \\right]\n$$\n\nfor all $i \\ge 1$.\n\nHere, $\\binom{n}{i} = \\frac{n!}{i!(n-i)!}$ denotes a binomial coefficient, and $\\binom{n}{i} = 0$ if $n < i$.",
    "options": [],
    "answer": "See solution",
    "solution": "We prove the statement by induction on $m$.\n\n**Base case ($m = k$):**\n\n$$\nL_i \\left[ \\binom{k}{i} - \\binom{0}{i} \\right] = L_i \\binom{k}{i} = L_i \\cdot \\frac{k!}{i!(k-i)!} = k \\cdot \\frac{L_i}{i} \\cdot \\frac{(k-1)!}{(i-1)!(k-i)!} = k \\cdot \\frac{L_i}{i} \\cdot \\binom{k-1}{i-1}.\n$$\n\nSince $\\frac{L_i}{i}$ and $\\binom{k-1}{i-1}$ are integers, $k$ divides the expression.\n\n**Induction step:**\n\nAssume the statement holds for $m$. Using $\\binom{m+1}{i} = \\binom{m}{i} + \\binom{m}{i-1}$:\n\n$$\n\\begin{aligned}\nL_i \\left[ \\binom{m+1}{i} - \\binom{m+1-k}{i} \\right] &= L_i \\left[ \\binom{m}{i} - \\binom{m-k}{i} \\right] + L_i \\left[ \\binom{m}{i-1} - \\binom{m-k}{i-1} \\right] \\\\\n&= L_i \\left[ \\binom{m}{i} - \\binom{m-k}{i} \\right] + \\frac{L_i}{L_{i-1}} \\cdot L_{i-1} \\left[ \\binom{m}{i-1} - \\binom{m-k}{i-1} \\right].\n\\end{aligned}\n$$\n\nBy the induction hypothesis, $k$ divides both $L_i[\\binom{m}{i} - \\binom{m-k}{i}]$ and $L_{i-1}[\\binom{m}{i-1} - \\binom{m-k}{i-1}]$. Since $L_{i-1}$ divides $L_i$, $k$ also divides the sum, completing the proof.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 12893,
    "subject": "Mathematics (Olympiad)",
    "question": "For a chessboard of size $2008 \\times 2008$, assign to each square (all of different colors) one of the letters $C$, $G$, $M$, or $O$. If every $2 \\times 2$ square contains all four letters, we call it a *harmonic chessboard*. How many different harmonic chessboards are there?",
    "options": [],
    "answer": "See solution",
    "solution": "There are $12 \\times 2^{2008} - 24$ harmonic chessboards.\n\nWe first prove the following claim:\n\nIn every harmonic chessboard, at least one of the following two situations occurs: (1) each row is composed of just two letters, in an alternating way; (2) each column is composed of just two letters, in an alternating way.\n\nSuppose that one row is not composed of two letters; then there must be three consecutive squares containing different letters. Without loss of generality, assume these three letters are $C$, $G$, $M$, as shown in Fig. 1. We then get $X_2 = X_5 = O$, $X_1 = X_4 = M$, and $X_3 = X_6 = C$, as shown in Fig. 2.\n\n![](images/Mathematical_Olympiad_in_China_2009-2010_p129_data_2077dba06f.png)\n\nFig. 1\n\n![](images/Mathematical_Olympiad_in_China_2009-2010_p129_data_6bdff9b792.png)\n\nFig. 2\n\nThe same argument shows that each of these three columns is composed of two letters in an alternating way, and so is every column.\n\nNow we calculate the total number of different harmonic chessboards. If the leftmost column is composed of two letters (e.g., $C$ and $M$), then 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; it is easy to check that this forms a harmonic chessboard. Therefore, there are $\\binom{4}{2} = 6$ 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 where each column is composed of two letters in an alternating way. Similarly, there are $6 \\times 2^{2008}$ configurations where each row is composed of two letters in an alternating way.\n\nWe need to subtract the configurations that are counted twice, i.e., those that are alternating on both rows and columns. Any such configuration is in one-to-one correspondence with the $2 \\times 2$ square at the upper-left corner, which gives $4! = 24$ different ways. Thus, the total number is $12 \\times 2^{2008} - 24$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 12894,
    "subject": "Mathematics (Olympiad)",
    "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.",
    "options": [],
    "answer": "See 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": 12895,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $b_n$ be the number of ways of covering a $3 \\times (2n+1)$ chessboard with one corner square removed, using $3n+1$ dominoes. What is the value of $b_{1010}$ modulo $19$?",
    "options": [],
    "answer": "See solution",
    "solution": "Suppose the corner square is removed from the first column of the $(2n+1) \\times 3$ chessboard.\n\nIf $n \\ge 1$, the dominoes that cover the squares of the last column can be placed in three different ways:\n\n![](images/IRL_ABooklet_2021_p46_data_d43593a808.png)\n\nIn type 1, the squares of the second last column are already covered. In types 2 and 3, there are two possibilities for how the remaining two squares in the second last column can be covered, provided that $n \\ge 2$:\n\n![](images/IRL_ABooklet_2021_p46_data_c1664faa96.png)\n\nWhen $n=1$ and the bottom square of the first column is removed, type 3b is not possible. Using the obvious $b_0=1$, we see that $b_1=4$; the four possibilities are obtained from types 1, 2a, 2b, and 3a.\n\nTo express $b_{n+1}$ in terms of $b_n$ and $b_{n-1}$, note that there are $b_n$ possibilities that the squares of the last two columns of the $(2(n+1)+1) \\times 3$ chessboard are covered as in type 1. The same is true for types 2a and 3a.\n\nTypes 2b and 3b can only occur when there is space for a vertical domino in the third last column. This corresponds to types 2 and 3, respectively, for the $(2n+1) \\times 3$ chessboard obtained by removing the last two columns. Each type 2 tiling of a $(2n+1) \\times 3$ chessboard gives rise to exactly one type 2b tiling of a $(2(n+1)+1) \\times 3$ chessboard, and similarly for types 3 and 3b.\n\nTherefore, the contribution to $b_{n+1}$ from types 2b and 3b is equal to the number of possibilities to tile the $(2n+1) \\times 3$ chessboard so that the last two columns are not of type 1. The number of possibilities to tile a $(2n+1) \\times 3$ chessboard with the last two columns of type 1 is $b_{n-1}$, so the contribution to $b_{n+1}$ from types 2b and 3b together is $b_n - b_{n-1}$. Thus,\n\n$$\nb_{n+1} = 3b_n + b_n - b_{n-1} = 4b_n - b_{n-1}.\n$$\n\nStarting with $b_0 = 1$ and $b_1 = 4$, we can now calculate $b_n$ modulo $19$:\n\n| $n$ | 0 | 1 | 2 | 3 | 4 | 5 | 6 |\n|-----|---|---|----|----|----|---|---|\n| $b_n$ (mod $19$) | 1 | 4 | -4 | -1 | 0 | 1 | 4 |\n\nIt follows that $b_n$ modulo $19$ repeats with period $5$ and $b_{5k} \\equiv 1 \\pmod{19}$. In particular, $b_{1010} \\equiv 1 \\pmod{19}$.",
    "topic": "Number Theory",
    "subtopic": "Modular Arithmetic"
  },
  {
    "id": 12896,
    "subject": "Mathematics (Olympiad)",
    "question": "The sequence $a_1, a_2, \\dots$ is defined by the equalities $a_1 = 2$, $a_2 = 12$, and $a_{n+1} = 6a_n - a_{n-1}$ for every positive integer $n \\ge 2$. Prove that no member of this sequence is equal to a perfect power (greater than one) of a positive integer.",
    "options": [],
    "answer": "See solution",
    "solution": "We shall use the following assertion.\n\n**Lemma.** Let $k \\ge 2$ be a positive integer. Then the equation $2x^{2k} + 1 = y^2$ does not have solutions in positive integers.\n\n*Proof.* Assume that $x, y$ and $k \\ge 2$ are positive integers such that $2x^{2k} + 1 = y^2$ and $x$ is minimum possible. It is obvious that $x$ is even and $y$ is odd. Let us denote $x = 2a$ and $y = 2b + 1$. Then $2^{2k-1}a^{2k} = b(b+1)$ and $(b, b+1) = 1$. There are two possibilities:\n\n- If $b = x_1^{2k}$ and $b+1 = 2^{2k-1}x_2^{2k}$, $x_1, x_2 \\in \\mathbb{N}$, $x_1x_2 = a$, then $2^{2k-1}x_2^{2k} - x_1^{2k} = 1$, which gives a contradiction modulo $4$.\n- If $b = 2^{2k-1}x_1^{2k}$ and $b+1 = x_2^{2k}$, $x_1, x_2 \\in \\mathbb{N}$, $x_1x_2 = a$, then $x_2^{2k} - 2^{2k-1}x_1^{2k} = 1$, which leads to the equation $y_1^2 = 2^{2k-1}x_1^{2k} + 1$, $y_1 = x_2^k$, where we notice that $x_1 < x$.\n\nIt is clear that the above argument of decreasing the degrees of $2$ can be continued until we have degree at most $5$. Therefore, we reach the equation $y_0^2 = 8x_0^{2k} + 1$, where $x_0 < x$ and $y_0 = y_2^k$, $y_2 \\in \\mathbb{N}$. Clearly, $y_0$ is odd and we set $y_0 = 2c+1$. We obtain $c(c+1) = 2x_0^{2k}$, where $(c, c+1) = 1$. We have again two possibilities:\n\n- If $c = x_3^{2k}$ and $c+1 = 2x_4^{2k}$, $x_3, x_4 \\in \\mathbb{N}$, $x_3x_4 = x_0$, then $4x_4^{2k} = 2c+2 = y_2^k+1$, whence $(2x_4^k-1)(2x_4^k+1) = y_2^k$. This leads to $2x_4^k-1 = y_3^k$, $2x_4^k+1 = y_4^k$, $y_3, y_4 \\in \\mathbb{N}$, $y_3y_4 = y_2$, and finally $y_4^k - y_3^k = 2$, which is impossible.\n- If $c = 2x_3^{2k}$ and $c+1 = x_4^{2k}$, $x_3, x_4 \\in \\mathbb{N}$, $x_3x_4 = x_0$, then $2x_3^{2k}+1 = (x_4^k)^2$, which contradicts the choice of $x$ as minimal.\n\nThis completes the proof of the lemma.\n\nThe roots of the characteristic equation $t^2 - 6t + 1 = 0$ of our sequence are $t_{1,2} = 3 \\pm 2\\sqrt{2}$. Therefore, we find (using the conditions $a_1 = 2$ and $a_2 = 12$):\n\n$$\na_n = \\frac{(3 + 2\\sqrt{2})^n - (3 - 2\\sqrt{2})^n}{2\\sqrt{2}}.\n$$\n\nDenote $(3 + 2\\sqrt{2})^n = \\alpha_n + \\beta_n\\sqrt{2}$, $\\alpha_n, \\beta_n \\in \\mathbb{N}$. Then $(3 - 2\\sqrt{2})^n = \\alpha_n - \\beta_n\\sqrt{2}$, $\\alpha_n = \\beta_n$ and $\\alpha_n^2 - 2\\beta_n^2 = 1$. Now, if $a_n$ is a perfect power for some $n$, then the last two equalities give a contradiction with the lemma.\n\n*Remark.* The last calculation can be replaced by the following argument. Let $b_n$ be the sequence with the same recurrence relation and initial conditions $b_1 = 3$ and $b_2 = 17$. Then it is easy to prove by induction that $2a_n^2 + 1 = b_n^2$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 12897,
    "subject": "Mathematics (Olympiad)",
    "question": "We are given a triangle $ABC$ and a point $P$ in its interior. The lines through $P$ and parallel to the sides of the triangle divide the triangle into three parallelograms and three triangles.\n\n**a)** If $P$ is the incenter of $ABC$, show that the perimeter of each of the three small triangles is equal to the length of the adjacent side.\n\n**b)** For a given triangle $ABC$, determine all inner points $P$ such that the perimeter of each of the three small triangles equals the length of the adjacent side.\n\n**c)** For which inner point does the sum of the areas of the three small triangles attain a minimum?",
    "options": [],
    "answer": "See solution",
    "solution": "**a)** Let $I$ be the incenter of $ABC$. Let $X$ be the intersection of $AB$ with the line through $I$ parallel to $CA$, and $Y$ be the intersection of $CA$ with the line through $I$ parallel to $AB$. $AXIY$ is a parallelogram, and since $I$ is the incenter of $ABC$, we have $\\angle IAX = \\angle IAY$. Since $\\angle IAX = \\angle AIX$ must also hold in the parallelogram $AXIY$, we see that $\\angle IAX = \\angle AIX$ holds. The triangle $AIX$ is therefore isosceles with $XA = XI$. If $Z$ denotes the intersection of $AB$ with the line through $I$ parallel to $BC$, we similarly obtain $ZB = ZI$, and it therefore follows that\n\n$$\nXI + XZ + ZI = XA + XZ + ZB = AB\n$$\n\nas claimed.\n\n**b)** Assume that a point $P \\neq I$ with this property exists. Such a point must lie between one of the sides of the triangle and the line parallel to this side through $I$. Without loss of generality, assume it lies between $AB$ and $YI$. The triangle $PX'Z'$ is similar to $IXZ$, and since $P$ is closer to $AB$ than $I$ is, the perimeter of $PX'Z'$ is smaller than that of $IXZ$, which is equal to the length of $AB$. $P$ therefore does not fulfill the required condition. Thus, $I$ is the only point with this property.\n\n**c)** The point $P$ determines three triangles $A_1B_1C_1$, $A_2B_2C_2$, and $A_3B_3C_3$ (with $P = C_1 = A_2 = B_3$) as shown.\n\nThe sum of the areas of the triangles is given by\n\n$$\n\\frac{1}{2}a_1b_1 + \\frac{1}{2}a_2b_2 + \\frac{1}{2}a_3b_3.\n$$\n\n![](images/Austria_2010_p7_data_2720dca159.png)\n\nSince $b_1 + b_2 + b_3 = c = |AB|$ and $a_1 + a_2 + a_3 = h_c$, and all three triangles are similar to $ABC$, we have\n\n$$\na_1 : a_2 : a_3 = b_1 : b_2 : b_3 = t_1 : t_2 : t_3 \\quad \\text{with} \\quad t_1 + t_2 + t_3 = 1.\n$$\n\nIt follows that\n\n$$\n\\begin{aligned}\n\\frac{1}{2}a_1b_1 + \\frac{1}{2}a_2b_2 + \\frac{1}{2}a_3b_3 &= \\frac{1}{2}h_c \\cdot c \\cdot (t_1^2 + t_2^2 + t_3^2) \\\\\n&\\geq h_c \\cdot c \\cdot \\left( \\frac{t_1 + t_2 + t_3}{2} \\right)^2 \\\\\n&= \\frac{h_c \\cdot c}{4},\n\\end{aligned}\n$$\n\nwith equality if and only if $t_1 = t_2 = t_3 = \\frac{1}{3}$. The sum of the areas is therefore minimized if the distance of $P$ from each of the sides is equal to one third of each altitude. This is the case for the centroid of $ABC$, so this is the required point.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 12898,
    "subject": "Mathematics (Olympiad)",
    "question": "Consider a grid of cards where each border between two adjacent cards is assigned a value: $+1$ if the cards have the same colour, $-1$ if they have different colours. The *monochromaticity* of the arrangement is the sum of these values over all borders.\n\n(a) What are the three largest possible values of monochromaticity for such a grid?\n\n% IMAGE: ![](images/NLD_ABooklet_2022_p14_data_52f638a0e0.png)\n\n(b) If $x$ is a possible value for monochromaticity, what other value must also be possible?\n\n(c) What is the smallest possible positive even value for monochromaticity, and how can it be achieved?\n\n% IMAGE: ![](images/NLD_ABooklet_2022_p15_data_715b197b4.png)\n\n% IMAGE: ![](images/NLD_ABooklet_2022_p15_data_9d6f391db4.png)",
    "options": [],
    "answer": "See solution",
    "solution": "(a) There are $3 \\cdot 4 = 12$ horizontal borders and 12 vertical borders, for a total of 24 borders. If $k$ borders count as $-1$, then $24 - k$ borders count as $+1$. The monochromaticity is $(24 - k) \\cdot (+1) + k \\cdot (-1) = 24 - 2k$, which is always even.\n\nIf all cards have the same colour, monochromaticity is 24. Monochromaticity 22 is impossible: if only one border is $-1$, then in any $2 \\times 2$ square, you must cross an even number of borders where the colour changes, contradicting the assumption of only one such border. Thus, 22 cannot occur.\n\nThe next largest possible values are 20 and 18, which can be achieved by appropriate colourings:\n\n% IMAGE: ![](images/NLD_ABooklet_2022_p14_data_52f638a0e0.png)\n\nThe three largest numbers on Niek's list are 24, 20, and 18.\n\n(b) If the monochromaticity is $x$, then by turning half the cards in a chessboard pattern, all border signs are reversed, yielding monochromaticity $-x$. Thus, $-x$ is also possible. Therefore, the three smallest numbers are $-24$, $-20$, and $-18$.\n\n(c) The monochromaticity is always even. The smallest possible positive even value is 2, which can be achieved by having 13 borders between squares of the same colour and 11 between squares of different colours. Examples:\n\n% IMAGE: ![](images/NLD_ABooklet_2022_p15_data_715b197b4.png)\n\n% IMAGE: ![](images/NLD_ABooklet_2022_p15_data_9d6f391db4.png)",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 12899,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all monic polynomials $f$ with integer coefficients satisfying the following condition: there exists a positive integer $N$ such that $p$ divides $2(f(p)) + 1$ for every prime $p > N$ for which $f(p)$ is a positive integer.\n\nA monic polynomial has leading coefficient equal to 1.",
    "options": [],
    "answer": "See solution",
    "solution": "Suppose $f$ is a constant polynomial. Then for $p \\geq 5$, $f(p) = 1$, so $p$ does not divide $2(f(p)) + 1 = 3$.\n\nFrom the divisibility $p \\mid 2(f(p)) + 1$, we see that $f(p) < p$ for all primes $p > N$ (otherwise $p \\mid f(p)$, so $p \\mid 1$, which is impossible).\n\nIf $\\deg f = m > 1$, then $f(x) = x^m + Q(x)$ with $\\deg Q \\leq m-1$, so for large $p$, $f(p) > p$, contradicting the above. Thus, $\\deg f = 1$ and $f(x) = x - a$ for some integer $a$.\n\nNow, the condition becomes $p \\mid 2(p - a) + 1$ for large $p$ with $p - a > 0$.\n\nUsing Wilson's theorem, for $a = 3$:\n\n$$\np \\mid 2(p - 3) + 1\n$$\n\nFor $p > (a - 1)!$, only $a = 3$ works, so $f(x) = x - 3$ is the only solution.\n\nThus, the only monic polynomial satisfying the condition is $f(x) = x - 3$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 12900,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $A_1, A_2, \\dots, A_6$ be selected subsets and let $A = A_1 \\cup \\dots \\cup A_6$. For any coloring of $A$ with two colors, show that there exists a subset $A_i$ containing elements of both colors.",
    "options": [],
    "answer": "See solution",
    "solution": "We prove the statement by considering the maximum number of times an element of $A$ appears among the $A_i$.\n\n- If $1$ appears $6$ times, color $1$ red and $A \\setminus \\{1\\}$ green. Each set contains $1$, so each has both colors.\n\n- If $1$ appears $5$ times, color $1$ red. Assume $A_6$ does not contain $1$; color one element of $A_6$ red, the others green. All sets have both colors.\n\n- If $1$ appears $4$ times, and $A_5, A_6$ do not contain $1$, consider intersections and use the pigeonhole principle to find elements $a \\in A_5$, $b \\in A_6$ such that $\\{1, a, b\\}$ is not selected. Color $1, a, b$ red, others blue; all sets have both colors.\n\n- If $1$ appears $3$ times, not in $A_4, A_5, A_6$, analyze the element with highest multiplicity among $A_4, A_5, A_6$ and reduce to previous cases, or use tuples to ensure a coloring with both colors in each set.\n\n- If $1$ appears $2$ times, belongs to $A_1, A_2$, let $2$ be the element with highest multiplicity in $A_3 \\cup \\dots \\cup A_6$ and reduce to previous cases.\n\n- If $1$ appears once, every element appears once. Color one element from each set red, others blue.\n\nThus, the minimum number $k_n$ such that for any coloring of $A$ with two colors, there exists a subset $A_i$ containing both colors is $k_3 = 1$, $k_4 = 4$, $k_5 = k_6 = 9$, and $k_n = 6$ for all $n \\ge 7$. $\\square$",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 12901,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all pairs of natural numbers $p > q$ such that\n\n$$\n\\frac{p+1}{p} \\cdot \\frac{q+1}{q} = \\frac{2013}{2011}.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "Since $2011$ is a prime, either $p$ or $q$ must be divisible by $2011$. Assume $2011 \\mid p$, so $p = 2011k$ for some $k$. Substituting, we get:\n\n$$\n\\frac{2011k + 1}{2011k} \\cdot \\frac{q + 1}{q} = \\frac{2013}{2011},\n$$\nwhich leads to\n$$\n\\frac{q+1}{q} = \\frac{2013k}{2011k+1} \\implies \\frac{1}{q} = \\frac{2k-1}{2011k+1} \\implies q = \\frac{2011k+1}{2k-1}.\n$$\n\nSince $q$ is a natural number, $2k-1$ must divide $2011k+1$. Consider $2q = \\frac{4022k+2}{2k-1}$, so\n$$\n2k - 1 \\mid 4022k + 2 \\implies 2k - 1 \\mid 4022k + 2 - 2011(2k - 1) = 2013.\n$$\n\nThus, $2k - 1$ is a positive divisor of $2013 = 3 \\cdot 11 \\cdot 61$. All 8 positive divisors yield solutions:\n\n$$\n\\begin{aligned}\n2k - 1 &= 1 &\\implies p &= 2012, &q &= 2011 \\\\\n2k - 1 &= 3 &\\implies p &= 4022, &q &= 1341 \\\\\n2k - 1 &= 11 &\\implies p &= 12066, &q &= 1097 \\\\\n2k - 1 &= 33 &\\implies p &= 34187, &q &= 1036 \\\\\n2k - 1 &= 61 &\\implies p &= 62341, &q &= 1022 \\\\\n2k - 1 &= 183 &\\implies p &= 185012, &q &= 1011 \\\\\n2k - 1 &= 671 &\\implies p &= 675696, &q &= 1007 \\\\\n2k - 1 &= 2013 &\\implies p &= 2025077, &q &= 1006\n\\end{aligned}\n$$\n\nBecause $p > q$, in the first solution the values of $p$ and $q$ should be interchanged.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 12902,
    "subject": "Mathematics (Olympiad)",
    "question": "The closure (interior and boundary) of a convex quadrangle is covered by four closed discs, each centered at a vertex of the quadrangle. Show that three of these discs cover the closure of the triangle determined by their centers.",
    "options": [],
    "answer": "See solution",
    "solution": "Suppose, for contradiction, that the conclusion does not hold. Then no three discs meet, and each disc contains points of the closure of the triangle determined by the centers of the other three discs, not covered by the latter.\n\nAmong the four discs, choose one, say $\\Delta_0$, containing the point $O$ where the diagonals of the quadrangle cross. Let $A_0$ be the center of $\\Delta_0$, and label the other three centers in circular order as $A_1, A_2, A_3$, so that the opposite angles $A_0OA_1$ and $A_2OA_3$ are not obtuse. Let $\\Delta_i$ denote the disc centered at $A_i$.\n\nBefore proceeding, we state a useful lemma (proof postponed for clarity):\n\n**Lemma.** Let $ABCD$ be a convex quadrangle, let $\\Delta$ be a disc centered at $A$, and let $E$ be the point where the ray $AC$ from $A$ crosses the boundary of $\\Delta$. If the orthogonal projection of $B$ onto $AC$ falls on the closed ray $EA$ from $E$, then $\\text{dist}(B, [ACD] \\setminus \\Delta) \\ge BE$, where $[ACD]$ is the closure of triangle $ACD$.\n\nApply the lemma to show that $O$ is also covered by $\\Delta_1$. Let $B_0$ be the point where the ray $A_0O$ from $A_0$ crosses the boundary of $\\Delta_0$. Since $\\Delta_0$ contains $O$, $O$ lies on the closed segment $A_0B_0$, and since the angle $A_0OA_1$ is not obtuse, $A_1O \\leq A_1B_0$. On the other hand, $\\Delta_1$ contains points of the closure $[A_0A_2A_3]$ not covered by $\\Delta_0$, so the radius of $\\Delta_1$ is at least $\\text{dist}(A_1, [A_0A_2A_3] \\setminus \\Delta_0) \\geq A_1B_0$ by the lemma. Thus, $O$ is covered by $\\Delta_1$.\n\nSince no three discs meet, neither $\\Delta_2$ nor $\\Delta_3$ contains $O$. For $i = 2, 3$, the open segment $A_iO$ crosses the boundary of $\\Delta_i$ at some point $B_i$. The open segments $A_2B_3$ and $A_3B_2$ cross, so $r_2 + r_3 = A_2B_2 + A_3B_3 < A_2B_3 + A_3B_2$, where $r_i$ is the radius of $\\Delta_i$.\n\nWe show that $r_2 \\geq A_2B_3$ and $r_3 \\geq A_3B_2$, reaching a contradiction. Consider $r_2 \\geq A_2B_3$; the argument for $r_3$ is similar. Since the angle $A_2OA_3$ is not obtuse, the orthogonal projection $A'_2$ of $A_2$ onto $A_1A_3$ falls on the closed ray $OA_3$ from $O$. If $A'_2$ fell on the closed segment $B_3O$, then rotating the line $A_2A'_2$ slightly about its midpoint would separate $\\Delta_3$ and the closure $[A_0A_1A_2]$, contradicting the earlier remark. Thus, $A'_2$ lies on the open ray $B_3A_3$ from $B_3$, so $\\text{dist}(A_2, [A_0A_1A_3] \\setminus \\Delta_3) \\geq A_2B_3$ by the lemma. Since $\\Delta_2$ covers points in $[A_0A_1A_3] \\setminus \\Delta_3$, $r_2 \\geq A_2B_3$.\n\n*Proof of the lemma.* Since $ABCD$ is convex, all points lie on one side of the line $AB$, say $\\mathcal{H}$. Let $F$ be the point where the ray $AD$ from $A$ crosses the boundary of $\\Delta$, let $\\alpha$ be the arc $EF$ of $\\Delta$'s boundary in $\\mathcal{H}$, and let $r$ be the ray from $E$ along $AC$, not containing $A$.\n\nIf $X$ is in $[ACD] \\setminus \\Delta$, then $BX$ meets either $\\alpha$ or $r$ (this fails if $ABCD$ is not convex at $D$), so consider only $X$ in $\\alpha \\cup r$.\n\nAs $X$ traces $\\alpha$ from $E$ to $F$, $BX$ increases by the cosine law in $ABX$ (this fails if $ABCD$ is not convex at $A$ or $B$), so $BX \\geq BE$.\n\nFinally, since the orthogonal projection of $B$ onto $AC$ is not interior to $r$, $BX$ increases as $X$ runs along $r$ away from $E$, so $BX \\geq BE$ again. This completes the lemma and the solution.",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 12903,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $p$ be a prime. Prove that any complete graph with $1000p$ vertices, whose edges are labeled with integers, has a simple cycle (with no repetitions of vertices or edges) whose sum of labels is divisible by $p$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $G$ be our graph. Define a *gate* to be a subgraph of $G$ with five vertices $A, B, C, D, E$ where we have edges from $A$ and $E$ to each of $B, C, D$, as well as the edges $BC$, $CD$, $DB$. The vertices $A, E$ will be referred to as the *endpoints* of the gate, and the vertices $B, C, D$ as *inner vertices*. It is not difficult to check that, in a gate, either the sum of the weights of the edges $BC$, $CD$, $DB$ is congruent to $0 \\pmod{p}$, or there are two different paths from $A$ to $E$ whose total weights are different modulo $p$. In the former case, we are done, so suppose the latter is true for any gate.\n\nWe now \"chain\" $p$ gates together in a loop. More precisely, fix vertices $A_1, \\dots, A_p$, and let $A_{p+1} = A_1$. For $i = 1, 2, \\dots, p$, construct a gate $G_i$ with endpoints $A_i, A_{i+1}$ and inner vertices $B_i, C_i, D_i$, where the $A_i, B_i, C_i, D_i$ are all chosen to be distinct from each other. Now, consider a path through this loop of gates. By the above, in each gate $G_i$, we have two choices for a path through $G_i$ from $A_i$ to $A_{i+1}$ which yield distinct weights modulo $p$.\n\nIt is now a consequence of the Cauchy-Davenport Theorem (the proof is also not difficult by more elementary means) that, for any residue class $c$ (mod $p$), there exists a choice of paths through the gates $G_i$ with total weight congruent to $c$ (mod $p$). In particular, we may take $c = 0$, so the problem is solved.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 12904,
    "subject": "Mathematics (Olympiad)",
    "question": "A triangle $AB\\Gamma$ is given with $\\hat{A} = 105°$ and $\\hat{\\Gamma} = \\frac{\\hat{B}}{4}$.\n\n(a) Determine the measures of the angles $\\hat{B}$ and $\\hat{\\Gamma}$.\n\n(b) If $O$ is the center of the circumcircle of the triangle $AB\\Gamma$ and $\\Delta$ is the antipodal of $B$, prove that the distance of $\\Gamma$ from $B\\Delta$ is $\\frac{B\\Delta}{4}$.",
    "options": [],
    "answer": "See solution",
    "solution": "(a) Since $\\hat{A} + \\hat{B} + \\hat{\\Gamma} = 180°$ and $\\hat{A} = 105°$, $\\hat{\\Gamma} = \\frac{\\hat{B}}{4}$, we have\n$$\n105° + \\hat{B} + \\frac{\\hat{B}}{4} = 180° \\implies \\hat{B} = 60°, \\text{ and hence } \\hat{\\Gamma} = 15°.\n$$\n\n(b) Since $OB = O\\Delta = O\\Gamma$, it follows that $\\angle B\\Gamma\\Delta = 90°$. Moreover, we have\n$$\n\\angle BO\\Gamma = 360° - 2 \\cdot 105° = 150°.\n$$\nTherefore,\n$$\n\\angle O\\Gamma B = \\frac{180° - 150°}{2} = 15°.\n$$\nHence $\\Gamma O\\Delta = 30°$ and $\\Gamma E = \\frac{O\\Gamma}{2} = \\frac{B\\Delta}{4}$.\n\n![](images/Greek_07_Booklet_p11_data_bd1bb78a73.png)\n\nFigure 4",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 12905,
    "subject": "Mathematics (Olympiad)",
    "question": "Given a circumcircle $(O)$ and two fixed points $B$, $C$ on $(O)$ ($BC$ is not the diameter of $(O)$). A point $A$ moves on $(O)$ such that $ABC$ is an acute triangle. Let $E$, $F$ be the feet of the altitudes from $B$, $C$ of triangle $ABC$. Let $(I)$ be an arbitrary circle passing through $E$ and $F$.\n\na) Let $(I)$ touch $BC$ at $D$. Prove that $\\frac{DB}{DC} = \\sqrt{\\frac{\\cot B}{\\cot C}}$.\n\nb) Suppose that $(I)$ intersects $BC$ at $M$ and $N$. Let $H$ be the orthocenter of triangle $ABC$ and $P$, $Q$ be the intersections of $(I)$ and the circumcircle of triangle $HBC$. Let $(K)$ be the circle passing through $P$, $Q$ and touching $(O)$ at $T$ ($T$ is on the same side as $A$ with respect to $PQ$). Prove that the interior angle bisector of $\\angle MTN$ passes through a fixed point.",
    "options": [],
    "answer": "See solution",
    "solution": "*Solution.* In the case $AB = AC$, it is obvious that $DB = DC$ and the bisector of $\\angle MTN$ passes through the midpoint of the minor arc $BC$ of $(O)$ (which is the fixed point revealed in part b), so we only need to consider the case where $AB \\neq AC$.\n\na) Let $R$, $S$ be the second intersections of $(I)$ with $AB$, $AC$ respectively. Since $E$, $F$ lie on a circle with diameter $BC$, and quadrilateral $EFRS$ is cyclic, we conclude $BC \\parallel RS$ by Reim's theorem.\n\n![](images/Vietnamese_mathematical_competitions_p110_data_a27938630c.png)\n\nBy assumption, $(I)$ touches $BC$ at $D$ then\n\n$$\n\\frac{BD^2}{CD^2} = \\frac{BF \\cdot BR}{CE \\cdot CS} = \\frac{BF \\cdot AB}{CE \\cdot AC} = \\frac{BF \\cdot BE}{CE \\cdot CF} = \\frac{\\cot B}{\\cot C}\n$$\n\nwhich means $\\frac{BD}{CD} = \\sqrt{\\frac{\\cot B}{\\cot C}}$.\n\nb) Denote by $G$ the intersection of $EF$ and $BC$. We have:\n\n* The radical axis of $(BHC)$ and $(I)$ is $PQ$,\n* The radical axis of $(I)$ and $(BFCE)$ is $EF$,\n* The radical axis of $(BHC)$ and $(BFCE)$ is $BC$.\n\n![](images/Vietnamese_mathematical_competitions_p111_data_40ad3ead0f.png)\n\nHence $G$ is the radical center of these circles, which means $G$ lies on $PQ$. On the other hand, we have:\n\n* Let the radical axis of $(O)$ and $(TPQ)$ be $d$,\n* The radical axis of $(TPQ)$ and $(BHC)$ is $PQ$,\n* The radical axis of $(BHC)$ and $(O)$ is $BC$.\n\nThen $d$, $PQ$ and $BC$ concur at $G$, which means $GT$ is the radical axis of $(O)$ and $(TPQ)$, hence $GT$ touches $(TPQ)$ and $(O)$. Since $PQMN$ is a cyclic quadrilateral,\n\n$$\n\\overline{GM} \\cdot \\overline{GN} = \\overline{GP} \\cdot \\overline{GQ} = GT^2,\n$$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 12906,
    "subject": "Mathematics (Olympiad)",
    "question": "A finite collection $C$ of (not necessarily distinct) real numbers is *suitable* if it contains two numbers $a$ and $b$ such that $a + b \\neq s + 1$, where $s$ is the sum of all numbers in $C$; such numbers $a$ and $b$ form an *eligible* pair.\n\nFix an integer $n \\geq 2$. A number of $n$ pairwise distinct real numbers are written on a board. A step consists in choosing an eligible pair $(a, b)$ amongst numbers on the board (if any), crossing them out and replacing them by the number\n\n$$\n\\frac{(a + b)(s + 1) - a^2 - a b - b^2}{s - a - b + 1}.\n$$\n\na) Prove that there is a sequence of $n - 1$ steps such that at each stage the numbers on the board form a suitable collection.\n\nb) Determine the final number on the board in terms of the initial numbers.",
    "options": [],
    "answer": "See solution",
    "solution": "a) Two real numbers (not necessarily distinct) always form a suitable collection. Let $n \\geq 3$ and consider the initial collection. Note that any $a$ can be paired off with some $b \\neq a$ to form an eligible pair: Otherwise, $a + b = s + 1 = a + c$ for distinct $b, c \\neq a$, so $b = c$, contradicting the fact that the initial numbers are pairwise distinct. Hence the initial collection is suitable.\n\nChoose an eligible pair $a_1, a_2$ and use the formula to replace them by $b_1$. As the remaining numbers are pairwise distinct, there is at most one that cannot be paired with $b_1$ to form an eligible pair, so there are at least $n - 2$ candidates to form an eligible pair with $b_1$. Let $a_3$ be one such and use the formula to replace $b_1$ and $a_3$ by $b_2$. Repeat the argument to replace $b_2$ and some $a_4$ by $b_3$ and so on and so forth all the way down to some $b_{n-2}$ and $a_n$ (possibly, $b_{n-2} = a_n$). These latter form an eligible pair, so they can be replaced by a single number.\n\nb) Let $c_1, c_2, \\dots, c_n$ be the initial numbers. The final number is $\\sum_i c_i + \\sum_{i < j} c_i c_j$. This is clearly the case if $n = 2$, so let $n \\geq 3$.\n\nConsider a generic stage $x_1, x_2, \\dots, x_m$ and let $s = x_1 + x_2 + \\dots + x_m$. We will prove that $s + \\sum_{1 \\leq i < j \\leq m} x_i x_j$ does not change upon passing to the next stage. Let $x$ be the number obtained by replacing an eligible pair $(x_k, x_\\ell)$. Then $s$ changes by $x - x_k - x_\\ell$ and the other sum changes by $-x_k x_\\ell - (x_k + x_\\ell)(s - x_k - x_\\ell) + x(s - x_k - x_\\ell)$, so the overall change is\n\n$$\nx - x_k - x_\\ell - x_k x_\\ell - (x_k + x_\\ell)(s - x_k - x_\\ell) + x(s - x_k - x_\\ell).\n$$\n\nFinally, express $x$ in terms of $s$, $x_k$ and $x_\\ell$ and carry out calculations to show that the overall change vanishes, whence the desired invariance.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 12907,
    "subject": "Mathematics (Olympiad)",
    "question": "For any integer $n$ with $n > 1$, let $n = p_1^{a_1} \\cdots p_t^{a_t}$ be its standard factorization. Define:\n$$\n\\omega(n) = t, \\quad \\Omega(n) = a_1 + \\cdots + a_t.\n$$\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": "See solution",
    "solution": "The answer is YES.\n\nFrom the definition of $\\omega$ and $\\Omega$, we have\n$$\n\\omega(ab) \\leq \\omega(a) + \\omega(b),\n$$\n$$\n\\Omega(ab) = \\Omega(a) + \\Omega(b),\n$$\nfor any positive integers $a, b$. Given a fixed positive integer $k$ and positive real numbers $\\alpha, \\beta$, take a positive integer $m > (\\omega(k) + 1)\\alpha$. Since there are infinitely many prime numbers, choose a sufficiently large prime $p$ such that $\\frac{\\Omega(k)+1}{p^m} + \\log_p 2 < \\beta$, and select $m$ pairwise distinct prime numbers $q_1, q_2, \\dots, q_m$ 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 $n_1 = \\frac{n+k}{k} = 2^{q_1 q_2 \\cdots q_m} + 1$. Since $q_1, q_2, \\dots, q_m$ are all odd primes, $2^{q_i} + 1$ divides $n_1$ for $1 \\leq i \\leq m$, and $d_i = \\frac{2^{q_i} + 1}{3}$ is an integer greater than 1.\n\n![alt](path \"title\")\n\nNote that\n\n$(2^r - 1, 2^s - 1) = 2^{(r,s)} - 1$ for all positive integers $r, s$,\nand $(q_i, q_j) = 1$ for $i \\neq j$, so\n$$\n\\begin{aligned}\n(d_i, d_j) &= \\frac{1}{3}(2^{q_i} + 1, 2^{q_j} + 1) \\leq \\frac{1}{3}(2^{2q_i} - 1, 2^{2q_j} - 1) \\\\\n&= \\frac{2^{(2q_i, 2q_j)} - 1}{3} = \\frac{2^2 - 1}{3} = 1.\n\\end{aligned}\n$$\n\nThus, $d_1, d_2, \\dots, d_m$ are pairwise coprime factors of $n_1$, each greater than 1. Hence, $\\omega(n_1) \\geq m$. From the above and the choice of $m$,\n$$\n\\frac{\\omega(n+k)}{\\omega(n)} \\geq \\frac{\\omega(n_1)}{\\omega(n)} \\geq \\frac{\\omega(n_1)}{\\omega(k)+1} \\geq \\frac{m}{\\omega(k)+1} > \\alpha.\n$$\n\nNext, we prove $\\frac{\\Omega(n+k)}{\\Omega(n)} < \\beta$. Since $q_1 q_2 \\cdots q_m$ is odd and not divisible by 3, $n_1 = 2^{q_1 q_2 \\cdots q_m} + 1 \\equiv \\pm 3 \\pmod{9}$, so $3 \\nmid n_1$. Suppose $q$ is a prime factor of $\\frac{n_1}{3}$ and $q \\leq p$, then\n$$\n2^{2q_1 q_2 \\cdots q_m} - 1 = (2^{q_1 q_2 \\cdots q_m} - 1) \\cdot n_1 \\equiv 0 \\pmod{q}.\n$$\nBy Fermat's Little Theorem, $2^{q-1} \\equiv 1 \\pmod{q}$. Thus, $q \\mid 2^{(2q_1 q_2 \\cdots q_m, q-1)} - 1$. Since $(q-1, 2q_1 q_2 \\cdots q_m) = (q-1, 2) \\leq 2$, and $q-1 < p < q_i$ for all $i$, we get $q \\mid 2^2 - 1$, so $q = 3$. 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 $\\frac{n_1}{3} > p^{\\Omega(n_1/3)}$.\n\nFrom above and the choice of $p$ and $q_1, \\dots, q_m$,\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$$\n\\frac{\\Omega(n+k)}{\\Omega(n)} < \\frac{\\Omega(k) + 1 + q_1 q_2 \\cdots q_m \\log_p 2}{q_1 q_2 \\cdots q_m} < \\frac{\\Omega(k) + 1}{p^m} + \\log_p 2 < \\beta. \\quad \\square\n$$",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 12908,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $a_1, a_2, \\dots, a_n$ and $b_1, b_2, \\dots, b_n$ be real numbers. Find the minimum value of the polynomial\n\n$$\np(x) = a_1(x - b_1)^2 + a_2(x - b_2)^2 + \\dots + a_n(x - b_n)^2\n$$\n\nin terms of $a_i$ and $b_i$.\n\n% IMAGE: ![](path/to/file.png)",
    "options": [],
    "answer": "See solution",
    "solution": "*Proof.* Expanding and completing the square gives\n\n$$\np(x) = \\left(\\sum_{i=1}^{n} a_i\\right) x^2 - 2 \\left(\\sum_{i=1}^{n} a_i b_i\\right) x + \\sum_{i=1}^{n} a_i b_i^2 = \\sum_{i=1}^{n} a_i \\left(x - \\frac{\\sum_{i=1}^{n} a_i b_i}{\\sum_{i=1}^{n} a_i}\\right)^2 + \\sum_{i=1}^{n} a_i b_i^2 - \\frac{\\left(\\sum_{i=1}^{n} a_i b_i\\right)^2}{\\sum_{i=1}^{n} a_i}\n$$\n\nfrom which the desired result follows. $\\square$",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 12909,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $\\mathbb{Z}_{>0} = \\{1, 2, 3, \\dots\\}$ be the set of all positive integers. Determine all functions $f : \\mathbb{Z}_{>0} \\to \\mathbb{Z}_{>0}$ such that, for each positive integer $n$:\n\ni) $\\sum_{k=1}^{n} f(k)$ is a perfect square;\n\nii) $f(n)$ divides $n^3$.",
    "options": [],
    "answer": "See solution",
    "solution": "We use induction on $n$ to show that $f(n) = n^3$ for all positive integers $n$.\n\nIt is easy to check that this $f$ satisfies the given conditions. The base case, $n=1$, is clear.\n\nAssume $n \\ge 2$ and that $f(m) = m^3$ for all $m < n$. Then\n$$\n\\sum_{k=1}^{n-1} f(k) = \\frac{n^2(n-1)^2}{4}.\n$$\nBy the first condition,\n$$\nf(n) = \\sum_{k=1}^{n} f(k) - \\sum_{k=1}^{n-1} f(k) = \\left( \\frac{n(n-1)}{2} + k \\right)^2 - \\frac{n^2(n-1)^2}{4} = k(n^2 - n + k),\n$$\nfor some positive integer $k$.\n\nThe divisibility condition implies $k(n^2 - n + k) \\le n^3$, or $(n-k)(n^2+k) \\ge 0$, so $k \\le n$.\n\nAlso, $n^2 - n + k$ must divide $n^3$. If $k < n$, then\n$$\nn < \\frac{n^3}{n^2 - 1} \\le \\frac{n^3}{n^2 - n + k} \\le \\frac{n^3}{n^2 - n + 1} < \\frac{n^3 + 1}{n^2 - n + 1} = n + 1,\n$$\nso $\\frac{n^3}{n^2 - n + k}$ cannot be an integer.\n\nThus, $k = n$, so $f(n) = n^3$. This completes the induction and the proof.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 12910,
    "subject": "Mathematics (Olympiad)",
    "question": "Show that $$\\frac{9}{2} < 3 \\log_2 \\pi < 5.$$",
    "options": [],
    "answer": "See solution",
    "solution": "The inequality $\\frac{9}{2} < 3 \\log_2 \\pi$ is equivalent to $3 < \\log_2 \\pi^2$ or $2^3 < \\pi^2$. This last inequality holds since $\\pi^2 > 3^2 = 9 > 8 = 2^3$.\n\nNow, let us prove the second inequality, $3 \\log_2 \\pi < 5$. This one is equivalent to $\\pi^3 < 2^5$. Again, we see that $\\pi^3 < 3.15^2 \\cdot 3.2 < 10 \\cdot 3.2 = 32 = 2^5$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 12911,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $m$ and $n$ be positive integers of different parity. Show that\n\n$$\n(m + 3n)(5m + 7n)(7m + 5n)(3m + n)\n$$\n\ncan never be a perfect square.",
    "options": [],
    "answer": "See solution",
    "solution": "Assume there exist numbers $m$ and $n$, one even and one odd, such that $A = (m + 3n)(5m + 7n)(7m + 5n)(3m + n)$ is a perfect square. Let $d = \\gcd(m, n)$, and write $m = d m_1$, $n = d n_1$ so that $m_1$ and $n_1$ are coprime. Then\n\n$$\nA = d^4 (m_1 + 3n_1)(5m_1 + 7n_1)(7m_1 + 5n_1)(3m_1 + n_1).\n$$\n\nLet $B = (m_1 + 3n_1)(5m_1 + 7n_1)(7m_1 + 5n_1)(3m_1 + n_1)$. Since $A$ is a perfect square, so is $B$.\n\nLet $p$ be a positive integer dividing both $m_1 + 3n_1$ and $5m_1 + 7n_1$. Since $m_1$ and $n_1$ have different parity, both $m_1 + 3n_1$ and $5m_1 + 7n_1$ are odd, so $p$ is odd. Also,\n\n$$\np \\mid 5(m_1 + 3n_1) - (5m_1 + 7n_1) = 8n_1.\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 12912,
    "subject": "Mathematics (Olympiad)",
    "question": "Show that the maximum size $m$ of a set $S$ of permutations of $\\{1,2,\\dots,n\\}$ such that, for any three numbers $1 \\leq a < b < c \\leq n$, there is no permutation in $S$ where $c$ lies between $a$ and $b$, satisfies $m \\leq 2^{n-1}$. Also, construct such a set $S$ of size $2^{n-1}$.",
    "options": [],
    "answer": "See solution",
    "solution": "We first show that $m \\leq 2^{n-1}$ by induction on $n$.\n\n*Base case*: For $n=3$, the statement is trivial.\n\n*Inductive step*: Assume the statement holds for $n=k$. For $n=k+1$, consider a set $S_{k+1}$ satisfying the problem's conditions. If we delete $k+1$ from each permutation in $S_{k+1}$, we obtain a set $S_k$ for $n=k$. For each permutation $r$ in $S_k$, at most two permutations in $S_{k+1}$ can map to $r$ (otherwise, the forbidden configuration would occur). Thus, $|S_{k+1}| \\leq 2|S_k| \\leq 2^{k} = 2^{n-1}$.\n\nTo achieve $m = 2^{n-1}$, construct permutations inductively: place 1 first; then, for each $l=2$ to $n$, place $l$ either to the left or right of all previously placed numbers. This gives $2^{n-1}$ permutations. In any such permutation, for any $1 \\leq a < b < c \\leq n$, $c$ never lies between $a$ and $b$. Thus, the bound is tight.",
    "topic": "Discrete Mathematics",
    "subtopic": "Combinatorics"
  },
  {
    "id": 12913,
    "subject": "Mathematics (Olympiad)",
    "question": "Consider the set $A = \\{1, 2, 3, \\ldots, 2n-1\\}$, where $n \\geq 2$ is a positive integer. We remove from the set $A$ at least $n-1$ elements such that:\n\n- If $a \\in A$ has been removed, then remove $2a$ as long as $2a \\in A$.\n- If $a, b \\in A$ ($a \\neq b$) have been removed, then remove $a+b$ as long as $a+b \\in A$.\n\nWhich numbers are to be removed in order to maximize the sum of the remaining numbers?",
    "options": [],
    "answer": "See solution",
    "solution": "Denote by $e_1, \\dots, e_m$ the removed numbers, $1 \\leq e_1 < e_2 < \\dots < e_m \\leq 2n-1$, and let $E$ be their set.\n\nWe claim that $e_i + e_{m-i+1} \\geq 2n$ for all $1 \\leq i \\leq m$. Suppose not. Then the following numbers are among the removed ones:\n\n$$\ne_i + e_1 < \\dots < e_i + e_{m-i} < e_i + e_{m-i+1}.\n$$\n\nNotice that the above sequence contains $m-i+1$ numbers, all greater than $e_i$, while there are only $m-i$ numbers greater than $e_i$ that have been removed, namely $e_{i+1} < \\dots < e_m$, a contradiction.\n\nConsequently,\n\n$$\n2n(n-1) \\leq 2nm = \\sum_{i=1}^{m} 2n \\leq \\sum_{i=1}^{m} (e_i + e_{m-i+1}) = \\sum_{i=1}^{m} e_i + \\sum_{i=1}^{m} e_{m-i+1} = 2 \\sum_{e \\in E} e,\n$$\n\nhence\n\n$$\n\\sum_{a \\in A \\setminus E} a = \\sum_{k=1}^{2n-1} k - \\sum_{e \\in E} e \\leq n(2n-1) - n(n-1) = n^2.\n$$\n\nThe maximum value of the sum of the remaining numbers is $n^2$ as shown by the following obvious model:\n\n$$\nE_0 = \\{2, 4, \\dots, 2n-2\\}, \\text{ with } \\sum_{a \\in A \\setminus E_0} a = 1 + 3 + \\dots + (2n-1) = n^2.\n$$\n\nTo end, we prove that this is the only possible model. Indeed, the equality is reached when $m = n-1$ and $e_i + e_{n-i} = 2n$ for all $1 \\leq i \\leq n-1$. To get $m = n-1$, we need $e_1 + e_i = e_{i+1}$ for all $1 \\leq i \\leq n-2$, hence $e_i = i e_1$ for all $1 \\leq i \\leq n-1$. It follows that $2n - e_1 = e_{n-1} = (n-1)e_1$, hence $e_1 = 2$, which leads to the model $E_0$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 12914,
    "subject": "Mathematics (Olympiad)",
    "question": "Given an integer $n \\ge 2$, show that there exist $n+1$ pairwise distinct numbers $x_1, x_2, \\dots, x_n, x_{n+1}$ in $\\mathbb{Q} \\setminus \\mathbb{Z}$ such that\n$$\n\\{x_1^3\\} + \\{x_2^3\\} + \\dots + \\{x_n^3\\} = \\{x_{n+1}^3\\},\n$$\nwhere $\\{x\\}$ is the fractional part of the real number $x$.",
    "options": [],
    "answer": "See solution",
    "solution": "Notice that, if $w_1 < w_2 < \\dots < w_{n+1} < w_{n+2}$ are positive integers such that\n$$\nw_1^3 + w_2^3 + \\dots + w_{n+1}^3 = w_{n+2}^3, \\quad (*)\n$$\nthen the numbers $x_k = w_k/w_{n+2}$ for $k = 1, 2, \\dots, n$, and $x_{n+1} = -w_{n+1}/w_{n+2}$ meet the required conditions.\n\nWe now show by induction on $n \\ge 2$ that there exist integers $3 = w_1 < w_2 < \\dots < w_{n+1} < w_{n+2}$ satisfying $(*)$.\n\nThe equalities $3^3 + 4^3 + 5^3 = 6^3$ and $3^3 + 15^3 + 21^3 + 36^3 = 39^3$ settle the cases $n=2$ and $n=3$, respectively.\n\nFor the induction step $n \\mapsto n+2$, notice that if $3 < w_2 < \\dots < w_{n+1} < w_{n+2}$ are $n+2$ integers satisfying $(*)$, then $3 < 4 < 5 < 2w_2 < \\dots < 2w_{n+1} < 2w_{n+2}$ are $n+4$ integers satisfying the corresponding condition.\n\n**Remarks.** By Andreescu, T., Andrica, D., and Cucurezeanu, I., _An Introduction to Diophantine Equations_, Birkhauser, 2010, Example 5, p. 44, if $n$ is an integer greater than or equal to 412, there exist positive integers $a_1, \\dots, a_n$ such that $1/a_1^3 + \\dots + 1/a_n^3 = 1$. Now, given an integer $n \\ge 412$ and a positive real number $\\varepsilon$, choose a positive integer $a > 1/\\varepsilon$ and set $x_k = 1/(aa_k)$ for $k = 1, \\dots, n$, and $x_{n+1} = 1/a$, to obtain $n+1$ positive rational numbers less than $\\varepsilon$ such that $x_1^3 + \\dots + x_n^3 = x_{n+1}^3$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 12915,
    "subject": "Mathematics (Olympiad)",
    "question": "а) 68 натурал тоонуудаас гуравынх нь нийлбэр дөрөвдэх тоотойгоо тэнцүү байх 4-н тоог ямагт сонгон авч болох уу?\n\nб) 69 натурал тоонуудаас гуравынх нь нийлбэр дөрөвдэх тоотойгоо тэнцүү байх 4-н тоог ямагт сонгон авч болох уу?",
    "options": [],
    "answer": "See solution",
    "solution": "а) $\\{33, 34, \\ldots, 100\\}$ гэсэн 68 тоог авахад хамгийн бага 3-н тоо буюу $33 + 34 + 35 = 102$ байх тул үүнээс нөхцөллийг хангах 4-н тоо олдохгүй.\n\nб) $1 \\leq a_1 < a_2 < \\dots < a_{69} \\leq 100$ тоонууд сонгогдсон гэе. Дирихлейн зарчмаар $\\{1, 2, \\ldots, 32\\}$-оос дор хаяж 1 тоо сонгогдсон байх тул $a_1 \\leq 32$ байна.\n\n$$\n1 < a_3 + a_1 < a_4 + a_1 < a_5 + a_1 < \\dots < a_{69} + a_1 \\leq 100 + 32 = 132\n$$\n\n$$\n1 \\leq a_3 - a_2 < a_4 - a_2 < a_5 - a_2 < \\dots < a_{69} - a_2 < 100 < 132\n$$\n\nДирихлейн зарчим ёсоор $a_i + a_1 = a_j - a_2$ байх тул $a_1 + a_2 + a_i = a_j$ гэсэн 4-н тоо олдоно.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 12916,
    "subject": "Mathematics (Olympiad)",
    "question": "As shown in the figure, in a trapezoid $ABCD$ with $AB \\parallel CD$, circle $\\odot O_1$ is tangent to the segments $DA$, $AB$, and $BC$, and circle $\\odot O_2$ is tangent to the segments $CD$, $BC$, and $DA$. Let $P$ be the point of tangency of $\\odot O_1$ with $AB$, and $Q$ be the point of tangency of $\\odot O_2$ with $CD$. Prove that $AC$, $BD$, and $PQ$ are concurrent.\n\n![alt](images/Mathematical_Olympiad_in_China_2011-2014_p224_data_f9c4a20dec.png)\n![alt](images/Mathematical_Olympiad_in_China_2011-2014_p224_data_50fd51dbdf.png)",
    "options": [],
    "answer": "See solution",
    "solution": "Let $R$ be the intersection of lines $AC$ and $BD$, and join $O_1A$, $O_1B$, $O_1P$, $O_2C$, $O_2D$, $O_2Q$, $PR$, and $QR$.\n\nAs $BA$ and $BC$ are tangent lines of $\\odot O_1$,\n\n$$\n\\angle PBO_1 = \\angle CBO_1 = \\frac{1}{2} \\angle ABC.\n$$\n\nSimilarly, $\\angle QCO_2 = \\frac{1}{2} \\angle BCD$.\n\nSince $AB \\parallel CD$, $\\angle ABC + \\angle BCD = 180^{\\circ}$.\nTherefore, $\\angle PBO_1 + \\angle QCO_2 = 90^{\\circ}$, and as right triangles $O_1BP$ and $CO_2Q$ are similar, we have $\\frac{O_1P}{BP} = \\frac{CQ}{O_2Q}$.\n\nSimilarly, $\\frac{AP}{O_1P} = \\frac{O_2Q}{DQ}$. Multiplying these two identities gives $\\frac{AP}{BP} = \\frac{CQ}{DQ}$, which implies $\\frac{AP}{AP + BP} = \\frac{CQ}{CQ + DQ}$, i.e., $\\frac{AP}{AB} = \\frac{CQ}{CD}$.\n\nAgain, since $AB \\parallel CD$, triangles $ABR$ and $CDR$ are similar, so $\\frac{AR}{AB} = \\frac{CR}{CD}$. Comparing with $\\frac{AP}{AB} = \\frac{CQ}{CD}$, we have $\\frac{AR}{AP} = \\frac{CR}{CQ}$. Meanwhile, triangles $PAR$ and $QCR$ are similar as $\\angle PAR = \\angle QCR$. Thus $\\angle PRA = \\angle QRC$, so $P$, $R$, $Q$ are collinear. Therefore, $AC$, $BD$, and $PQ$ are concurrent. $\\square$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 12917,
    "subject": "Mathematics (Olympiad)",
    "question": "Prove that the following inequality holds for any positive real numbers $a$, $b$, and $c$:\n\n$$\n\\frac{a}{2a+b+c} + \\frac{b}{a+2b+c} + \\frac{c}{a+b+2c} \\le \\frac{3}{4}.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "Set $S = a + b + c$ to rewrite the left-hand side as\n\n$$\n\\frac{a}{S+a} + \\frac{b}{S+b} + \\frac{c}{S+c} = 3 - S \\left( \\frac{1}{S+a} + \\frac{1}{S+b} + \\frac{1}{S+c} \\right).\n$$\n\nWe need to prove that\n$$\n4S \\left( \\frac{1}{S+a} + \\frac{1}{S+b} + \\frac{1}{S+c} \\right) \\ge 9.\n$$\n\nSince\n$$\n4S = (S+a) + (S+b) + (S+c)\n$$\nand for all $x, y, z > 0$,\n$$\n(x+y+z) \\left( \\frac{1}{x} + \\frac{1}{y} + \\frac{1}{z} \\right) \\ge 9,\n$$\nthe inequality holds.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 12918,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $a_1, a_2, \\dots, a_n$ be $n$ even integers. Show that there exists a subsequence $i_1 < i_2 < \\dots < i_k$ of $1, 2, \\dots, n$ such that $a_{i_1} + \\dots + a_{i_k}$ is divisible by $2n$.",
    "options": [],
    "answer": "See solution",
    "solution": "Proof:\n\nConsider the following integers:\n\n$$\nS_i = \\frac{1}{2}(a_1 + a_2 + \\dots + a_i), \\quad 1 \\leq i \\leq n\n$$\n\nIf there is some $i$ such that $n \\mid S_i$, we are done. Otherwise, since there are $n$ numbers and $n$ possible residues modulo $n$, by the pigeonhole principle, there exist $i < j$ such that $S_i \\equiv S_j \\pmod{n}$. This implies:\n\n$$\nS_j - S_i = \\frac{1}{2}(a_{i+1} + \\dots + a_j) \\equiv 0 \\pmod{n}\n$$\n\nSo $n \\mid \\frac{1}{2}(a_{i+1} + \\dots + a_j)$, which means $2n \\mid a_{i+1} + \\dots + a_j$. Thus, the proof is complete. $\\square$\n\nNow for the main problem, we have two cases:\n\n* If $n$ is odd, the answer is $k = 2n$. For $k < 2n$, if we set $a_1 = a_2 = \\dots = a_k = 1$, we cannot choose an even number of $a_i$'s with sum divisible by $n$. On the other hand, if $a_1, a_2, \\dots, a_{2n}$ are $2n$ integers, define $S_i = a_1 + \\dots + a_{2i}$ for $i = 1, 2, \\dots, n$. By similar arguments as in the lemma, there is some $i$ such that $n \\mid S_i$ or there exists $i < j$ such that $n \\mid S_j - S_i$. In both cases, we have found an even number of $a_i$'s with sum divisible by $n$.\n\n* If $n = 2m$ is even, we claim that $k = n + 1 = 2m + 1$ is the answer. For $k \\leq 2m$, if we set $a_1 = a_2 = \\dots = a_{k-1} = 1$ and $a_k = 0$, it is not possible to select an even number of $a_i$'s with sum divisible by $2m$. On the other hand, suppose $a_1, a_2, \\dots, a_{2m+1}$ are arbitrary integers. Let $b_1, b_2, \\dots, b_s$ be the even numbers among the $a_i$'s and $c_1, c_2, \\dots, c_r$ the odd numbers. Since $r + s = 2m + 1$, exactly one of $r$ or $s$ is odd and the other is even. Suppose $r$ is odd and $s$ is even (the other case is similar). Consider the $m$ numbers $a_1 + a_2, a_3 + a_4, \\dots, a_{r-2} + a_{r-1}, b_1 + b_2, \\dots, b_{s-1} + b_s$. All these sums are even, so by the lemma, we can select some of them with sum divisible by $2m = n$. Since each is the sum of two $a_i$'s, we have found an even number of $a_i$'s with sum divisible by $n$, as desired.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 12919,
    "subject": "Mathematics (Olympiad)",
    "question": "Suppose $\\triangle ABC$ with angles $A$, $B$, and $C$, and the corresponding sides $a$, $b$, and $c$ satisfies the equation\n\n$$\na \\cos B - b \\cos A = \\frac{3}{5}c.\n$$\n\nThen the value of $\\frac{\\tan A}{\\tan B}$ is \\_\\_\\_\\_\\_.",
    "options": [],
    "answer": "See solution",
    "solution": "By the given condition and the Law of Cosines, we have\n\n$$\na \\cdot \\frac{c^2 + a^2 - b^2}{2ca} - b \\cdot \\frac{b^2 + c^2 - a^2}{2bc} = \\frac{3}{5}c\n$$\n\nor $a^2 - b^2 = \\frac{3}{5}c^2$. Therefore,\n\n$$\n\\frac{\\tan A}{\\tan B} = \\frac{\\sin A \\cos B}{\\sin B \\cos A} = \\frac{a \\cdot \\frac{c^2 + a^2 - b^2}{2ca}}{b \\cdot \\frac{b^2 + c^2 - a^2}{2bc}}\n$$\n$$\n= \\frac{c^2 + a^2 - b^2}{b^2 + c^2 - a^2} = \\frac{\\frac{8}{5}c^2}{\\frac{2}{5}c^2} = 4\n$$\n\nThe answer is $4$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 12920,
    "subject": "Mathematics (Olympiad)",
    "question": "A quadrilateral $ABCD$ is given with $AD \\parallel BC$. The midpoints of $AD$ and $BC$ are denoted by $M$ and $N$, respectively. The line $MN$ intersects the diagonals $AC$ and $BD$ at points $K$ and $L$, respectively. Prove that the circumcircles of the triangles $AKM$ and $BNL$ have a common point on the line $AB$.",
    "options": [],
    "answer": "See solution",
    "solution": "Denote by $T$ the intersection point of the lines $AD$ and $BC$, and by $O$ the intersection point of the diagonals $AC$ and $BD$. Let the second intersection point of the circumcircles of $\\triangle AOD$ and $\\triangle BOC$ be $P$. It is $P \\neq O$ as $AD \\parallel BC$.\n\nWe have $\\angle PAD = \\angle POD = \\angle PCB$ from the circles of $A, P, O, D$ and $B, C, O, P$ as in the schema (or $\\angle PAD = \\angle POB = \\angle PCB$ depending on the relative positions of $A, B, C, D$). Analogously, $\\angle ADP = \\angle PBC$. Therefore, $\\triangle ADP \\sim \\triangle CBP$. Since $PM$ and $PN$ are corresponding medians in these triangles, we obtain $\\angle PMD = \\angle PNB$. Then the points $M, P, N, T$ are concyclic. Then $\\angle PAK = \\angle PAC = \\angle PTC = \\angle PTN = \\angle PMN = \\angle PMK$, i.e., $K$ belongs to the circumcircle of $\\triangle APM$. Analogously, $L$ belongs to the circumcircle of $\\triangle BPN$.\n\n![](images/shortlistBMO2015_p15_data_30104e398b.png)\n\nLet these two circles intersect also at $Q$. Then $\\angle AQP = \\angle PMD = \\angle PNB$ (by the concyclicity of $A, B, P, M$ and the similarity of $\\triangle ADP, \\triangle CBP$) and $\\angle BQP = \\angle PNB$ (by the concyclicity of $B, Q, N, P$), thus $\\angle AQP = \\angle BQP$, i.e., $Q \\in AB$ (or depending on the relative position of $A, B, Q$ but not shown in the given figures it could be $\\angle AQP + \\angle BQP = \\angle PKO + \\angle PLO = 180^\\circ$, giving again $Q \\in AB$). $\\square$\n\n![](images/shortlistBMO2015_p15_data_6fedba7c6d.png)\n\n**Remark.** After showing $\\triangle ADP \\sim \\triangle CBP$ and similarly that $\\triangle ACP \\sim \\triangle DBP$, one can argue as follows:\n$$\n\\frac{AK}{CK} = \\frac{S_{AMN}}{S_{CMN}} = \\frac{S_{DMN}}{S_{BMN}} = \\frac{DL}{BL}\n$$\nand this implies, because of the similarity of $\\triangle ACP, \\triangle DBP$, that $\\angle AKP = \\angle DLP$. Then $K, P, L, O$ are concyclic and $\\angle PKL = \\angle POL = \\angle PCB = \\angle PAD$, i.e., $P$ belongs to the circumcircle of $\\triangle AMK$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 12921,
    "subject": "Mathematics (Olympiad)",
    "question": "Find the difference between the largest and the next largest odd factors of $2016$.",
    "options": [],
    "answer": "See solution",
    "solution": "2016 is $2^5 \\times 3^2 \\times 7$. The largest odd factor is $3^2 \\times 7 = 63$; and the next largest odd factor is $3 \\times 7 = 21$. $63 - 21 = 42$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 12922,
    "subject": "Mathematics (Olympiad)",
    "question": "In a triangle $\\triangle ABC$, points $L$, $P$, and $Q$ lie on the segments $AB$, $AC$, and $BC$, respectively, such that $PCQL$ is a parallelogram. The circle with center at the midpoint $M$ of segment $AB$ and radius $CM$, and the circle with diameter $CL$, intersect for the second time at point $T$. Prove that the lines $AQ$, $BP$, and $LT$ intersect at a single point.",
    "options": [],
    "answer": "See 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 points $X$ such that $\\triangle AXQ$ and $\\triangle PXB$ are oriented in the same direction and have equal areas is a straight line $\\ell$ passing through the intersection point of $AQ$ and $BP$. Therefore, $\\ell \\equiv KL$ and $AQ$, $BP$, and $KL$ intersect at a point.\n\nLet $N$ be the midpoint of $CL$. The line $KL$ is the image of $MN$ under a homothety with 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": 12923,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $a_1, a_2, \\dots, a_n$ ($n \\ge 3$) be real numbers satisfying $a_1 + a_2 + \\dots + a_n = 0$, and\n\n$$\n2a_k \\le a_{k-1} + a_{k+1}, \\quad \\text{for } k = 2, 3, \\dots, n-1.\n$$\n\nDetermine the smallest $\\lambda(n)$ such that for any $k \\in \\{1, 2, \\dots, n\\}$, one has\n\n$$\n|a_k| \\le \\lambda(n) \\cdot \\max\\{|a_1|, |a_n|\\}.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "$$\n\\lambda(n)_{\\min} = \\frac{n+1}{n-1}.\n$$\n\nFirst, define a sequence $\\{a_k\\}$ as follows: $a_1 = 1$, $a_2 = -\\frac{n+1}{n-1}$, and\n\n$$\na_k = -\\frac{n+1}{n-1} + \\frac{2n(k-2)}{(n-1)(n-2)}, \\quad \\text{for } k = 3, 4, \\dots, n.\n$$\n\nThen $a_1 + a_2 + \\dots + a_n = 0$, and $2a_k \\le a_{k-1} + a_{k+1}$ for $k = 2, 3, \\dots, n-1$. In this case, it is easy to check that $\\lambda(n) \\ge \\frac{n+1}{n-1}$.\n\nNext, suppose that $\\{a_k\\}$ is a sequence satisfying the two conditions stated in the question. We will prove that the following inequalities hold:\n\n$$\na_k \\le \\frac{n+1}{n-1} \\max\\{|a_1|, |a_n|\\}, \\quad \\text{for all } k \\in \\{1, 2, \\dots, n\\}.\n$$\n\nAs $2a_k \\le a_{k-1} + a_{k+1}$, so $a_{k+1} - a_k \\ge a_k - a_{k-1}$, which implies\n\n$$\na_n - a_{n-1} \\ge a_{n-1} - a_{n-2} \\ge \\dots \\ge a_2 - a_1.\n$$\n\nUsing a telescoping sum, we have\n\n$$\n\\begin{aligned}\n(k-1)(a_n - a_1) &= (k-1)[(a_n - a_{n-1}) + (a_{n-1} - a_{n-2}) + \\dots + (a_2 - a_1)] \\\\\n&\\ge (n-1)[(a_k - a_{k-1}) + (a_{k-1} - a_{k-2}) + \\dots + (a_2 - a_1)] \\\\\n&= (n-1)(a_k - a_1),\n\\end{aligned}\n$$\n\nand hence\n\n$$\n\\begin{aligned}\na_k &\\le \\frac{k-1}{n-1}(a_n - a_1) + a_1 \\\\\n&= \\frac{1}{n-1}[(k-1)a_n + (n-k)a_1].\n\\end{aligned}\n\\qquad (1)\n$$\n\nSimilarly, for any fixed $k$ ($k \\notin \\{1, n\\}$), and for any $j \\in \\{1, 2, \\dots, k\\}$,\n\n$$\na_j \\le \\frac{1}{k-1}[(j-1)a_k + (k-j)a_1].\n$$\n\nFor $j \\in \\{k, k+1, \\dots, n\\}$,\n\n$$\na_j \\le \\frac{1}{n-1}[(j-k)a_n + (n-j)a_k].\n$$\n\nConsequently,\n\n$$\n\\begin{aligned}\n\\sum_{j=1}^{k} a_j &\\le \\frac{1}{k-1} \\sum_{j=1}^{k} [(j-1)a_k + (k-j)a_1] \\\\\n&= \\frac{k}{2}(a_1 + a_k),\n\\end{aligned}\n$$\n\n$$\n\\begin{aligned}\n\\sum_{j=k}^{n} a_j &\\le \\frac{1}{n-k} \\sum_{j=k}^{n} [(j-k)a_n + (n-j)a_k] \\\\\n&= \\frac{n+1-k}{2}(a_k + a_n).\n\\end{aligned}\n$$\n\nSumming these two inequalities,\n\n$$\n\\begin{aligned}\na_k &= \\sum_{j=1}^{k} a_j + \\sum_{j=k}^{n} a_j \\le \\frac{k}{2}(a_1 + a_k) + \\frac{n+1-k}{2}(a_k + a_n) \\\\\n&= \\frac{k}{2}a_1 + \\frac{n+1}{2}a_k + \\frac{n+1-k}{2}a_n.\n\\end{aligned}\n$$\n\nTherefore,\n\n$$\na_k \\ge -\\frac{1}{n-1}[k a_1 + (n+1-k)a_n]. \\qquad (2)\n$$\n\nFrom (1) and (2), for $k=2, 3, \\dots, n-1$,\n\n$$\n\\begin{aligned}\n|a_k| &\\le \\max\\left\\{\\frac{1}{n-1} |(k-1)a_n + (n-k)a_1|, \\\\ \n&\\qquad \\frac{1}{n-1} |k a_1 + (n+1-k)a_n|\\right\\} \\\\\n&\\le \\frac{n+1}{n-1} \\max\\{|a_1|, |a_n|\\}.\n\\end{aligned}\n$$\n\nConsequently, it follows that $\\lambda(n)_{\\min} = \\frac{n+1}{n-1}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 12924,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $ABC$ be a triangle and a circle $\\omega$ passing through $C$ and its incenter $I$. The circle $\\omega$ meets $CA$ and $CB$ at $P$ and $Q$, respectively. The circumcircles $(CPQ)$ and $(ABC)$ meet at $L$. The angle bisector of $\\angle ALB$ meets $AB$ at $K$. Show that, as $\\omega$ varies, $\\angle PKQ$ is constant.",
    "options": [],
    "answer": "See solution",
    "solution": "Clearly, $IP = IQ$ since $CI$ is a bisector in $\\omega$. Our aim is to show that $IK = IP$, as it would follow that $I$ is the circumcenter of triangle $PKQ$, whence $\\angle PKQ = \\frac{1}{2}\\angle PIQ = 90^\\circ - \\frac{1}{2}\\angle ACB$.\n\nLet the angle bisectors $CI$ and $LK$ intersect at the midpoint $T$ of the arc $\\widehat{AB}$ from $k$, not containing $C$. We have $\\angle TAK = \\angle TAB = \\angle BCT = \\angle ACT = \\angle ALT$, thus $\\triangle AKT \\sim \\triangle LAT$ and $TA^2 = TK \\cdot TL$. Since $TA = TI$ from the trillium lemma, we deduce $TI^2 = TK \\cdot TL$ and $\\triangle IKT \\sim \\triangle LIT$. Hence\n\n$$\nIK = IT \\cdot \\frac{LI}{LT} = AT \\cdot \\frac{LI}{LT}.\n$$\n\nOn the other hand, the circumcircles $k$ and $\\omega$ yield $\\angle LPI = 180^\\circ - \\angle LCI = \\angle LAT$ and $\\angle PLI = \\angle PCI = \\angle ALT$. Hence $\\triangle LAT \\sim \\triangle LPI$ and so\n\n$$\n\\frac{LI}{LT} = \\frac{PI}{AT}\n$$\n\nwhich completes the proof. $\\square$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 12925,
    "subject": "Mathematics (Olympiad)",
    "question": "The tangents through $S$ to the circumcircle of the trapezoid meet the circumcircle in $E$ and $F$, respectively, where $E$ is on the same side of the line $CD$ as $A$.\n\nCharacterize good trapezoids $ABCD$ (in terms of the side lengths and/or angles of the trapezoid) for which the angles $\\angle BSE$ and $\\angle FSC$ are equal. The characterization should be as simple as possible.\n\n![](images/Austria2019_booklet_p12_data_27cad3b606.png)",
    "options": [],
    "answer": "See solution",
    "solution": "The angles $\\angle BSE$ and $\\angle FSC$ are equal if and only if $\\angle BAD = 60^\\circ$ or $AB = AD$.\n\nWe denote the circumcircle of the trapezoid by $u$, the second intersection point of the line $SB$ with $u$ by $T$, and the centre of $u$ by $M$ (see Figure 4).\n\nAs the trapezoid is inscribed, it is isosceles. Since $ABSD$ is a parallelogram by construction, we have $BS = AD = BC$ and $DS = AB$.\n\nConsider the reflection across the line $MS$. It clearly maps $E$ and $F$ to each other and maps $u$ to itself. We say that the trapezoid meets the *angle condition* if $\\angle BSE = \\angle FSC$.\n\nThe trapezoid meets the angle condition if and only if the reflection maps the rays $SB$ and $SC$ to each other. Equivalently, the intersection points of these rays with $u$ are mapped to each other corresponding to the order of the points on the rays.\n\nWe first consider the case that $B$ is between $S$ and $T$ (see Figure 4). Then the trapezoid meets the angle condition if and only if the reflection maps $B$ and $C$ to each other. Equivalently, the triangle $BSC$ is isosceles with axis of symmetry $SM$. As $M$ lies on the perpendicular bisector of $BC$ in any case, this is equivalent to $CS = BS$. As $BS = BC$, this is in turn equivalent to the triangle $BSC$ being equilateral. Again by $BS = BC$, this is equivalent to $\\angle CSB = 60^\\circ$. As $ABSD$ is a parallelogram, the trapezoid meets the angle condition in this case if and only if $\\angle BAD = 60^\\circ$.\n\nWe now consider the case that $T$ lies between $S$ and $B$ (see Figure 5). Then the above considerations show that the trapezoid meets the angle condition if and only if the reflection maps $B$ and $D$ to each other. Equivalently, the triangle $BSD$ is isosceles with axis of symmetry $MS$. By the same argument as in the first case, this is equivalent to $SB = SD$. This is equivalent to $AB = AD$.\n\n$\\square$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 12926,
    "subject": "Mathematics (Olympiad)",
    "question": "Draw a semicircle with radius $1$, and let $A$ and $B$ be the two endpoints of the arc. Take two points $C$ and $D$ on arc $AB$ so that $A$, $C$, $D$, and $B$ are on the arc in this order.\n\nLet $2a = AC$, $2b = AD$, $\\angle CAB = \\alpha$, and $\\angle DAB = \\beta$. Express $a$ in terms of $\\cos \\alpha$ and $\\beta$, and find the area of triangle $ACD$.\n\nNext, by taking points $A_1, A_2, \\dots, A_n$ on the arc $AB$ with $AA_i = 2a_i$ and considering the sum of the areas of triangles $AA_iA_{i+1}$, show that the left side of the given inequality does not exceed $\\pi/4$, half the area of the semicircle.\n\nFinally, prove that $c$ cannot be smaller than $\\pi/4$ by dividing the arc $AB$ into $2^n$ equal parts and considering the area not covered by the polygon $A_0A_1 \\cdots A_{2^n}$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $2a = AC$, $2b = AD$, $\\angle CAB = \\alpha$, $\\angle DAB = \\beta$. Then $a = \\cos \\alpha, \\beta$ and the area of triangle $ACD$ is\n\n$$\n\\frac{1}{2}(2a)(2b) \\sin(\\alpha - \\beta) = 2ab(\\sin \\alpha \\cos \\beta - \\cos \\alpha \\sin \\beta)\n= 2ab(b\\sqrt{1-a^2} - a\\sqrt{1-b^2}) = 2f(a, b).\n$$\n\nBy taking points $A_1, A_2, \\dots, A_n$ on the arc $AB$ with $AA_i = 2a_i$ and summing the areas of triangles $AA_iA_{i+1}$, the total area does not exceed $\\pi/4$, half the area of the semicircle.\n\nTo show $c$ cannot be smaller than $\\pi/4$, divide the arc $AB$ into $2^n$ equal parts: $A = A_0, A_1, \\dots, A_{2^n} = B$, with $AA_i = 2a_i$. Let $S_n$ be the part of the semicircle not covered by polygon $A_0A_1 \\cdots A_{2^n}$, which consists of $2^n$ congruent figures $P_n$. Since $2|P_{n+1}| < \\frac{1}{2}|P_n|$, we have $|S_{n+1}| < \\frac{1}{2}|S_n|$, so $|S_n| < \\frac{1}{2^{n-1}}|S_1|$. Thus, if $c < \\pi/4$, for large enough $n$ the uncovered area is less than $\\pi/4 - c$, which is a contradiction. Therefore, the minimum value of $c$ is $\\pi/4$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 12927,
    "subject": "Mathematics (Olympiad)",
    "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.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $a$ and $b$ be the side lengths of the plot. Assume that the cuts divided the side of length $a$ into parts of length $x$ and $a - x$ (with $x$ being the greater part), and the side of length $b$ into parts of length $y$ and $b - y$ (with $y$ being the greater part). Then the area of the largest piece is $xy$. The condition that this area equals the sum of the areas of the other three pieces can be written as:\n\n$$\nxy = (a - x)y + x(b - y) + (a - x)(b - y)\n$$\n\nDividing both sides by $x(a - x)y(b - y)$, we obtain:\n\n$$\n\\frac{1}{(a - x)(b - y)} = \\frac{1}{x(b - y)} + \\frac{1}{(a - x)y} + \\frac{1}{xy}\n$$\n\nIf the price that every musketeer paid for the plot was 1, then the left-hand side of the last equality is precisely the price per area unit of the piece with area $(a - x)(b - y)$. Analogously, the right-hand side equals the sum of the prices per area unit of the other three pieces. Hence, multiplying both sides of this equality by the number of area units per acre, the claim of the problem follows.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 12928,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $G$ be the point of intersection of the lines $EF$ and $DC$.\n\n% IMAGE: ![](images/Turska_2011_p15_data_83206e7d7d.png)\n\nGiven that $\\angle CFG = \\angle BFE = \\angle CDF$, prove that the lines $DF$ and $EG$ are perpendicular. Also, if $BA \\cdot BE = BT^2 = BF^2$, show that triangles $BAF$ and $BFE$ are similar, $BF = CF$, $EF = GF$, and $\\angle EDF = \\angle GDF = \\angle CDF$.",
    "options": [],
    "answer": "See solution",
    "solution": "Since $\\angle CFG = \\angle BFE = \\angle CDF$, the lines $DF$ and $EG$ are perpendicular.\n\nOn the other hand, $BA \\cdot BE = BT^2 = BF^2$ implies that the triangles $BAF$ and $BFE$ are similar and $\\angle BAF = \\angle BFE = \\angle CDF$. Therefore $BF = CF$, $EF = GF$ and $\\angle EDF = \\angle GDF = \\angle CDF$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 12929,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $A$ and $C$ be two points on a circle $C$ such that $AC$ is not a diameter, and let $P$ be a point on the line segment $AC$, other than its midpoint. Circles $c_1$ and $c_2$ are interiorly tangent to the circle $C$ at $A$ and $C$, respectively. Both pass through $P$ and intersect again at $Q$. The line $PQ$ intersects circle $C$ at $B$ and $D$. Circle $c_1$ intersects line segments $AB$ and $AD$ at $K$ and $N$, respectively, while circle $c_2$ intersects line segments $CB$ and $CD$ at $L$ and $M$, respectively. Prove that:\n\na) The quadrilateral $KLMN$ is an isosceles trapezoid.\n\nb) $Q$ is the midpoint of the line segment $BD$.",
    "options": [],
    "answer": "See solution",
    "solution": "a) Point $B$ lies on the radical axis $BD$ of circles $c_1$ and $c_2$, so $BK \\cdot BA = BL \\cdot BC$, which shows that quadrilateral $AKLC$ is cyclic. Similarly, $ACMN$ is cyclic. It follows that $\\angle LKB = \\angle ACB$. The angle between the tangent at $A$ to circle $c_1$ (also tangent to $C$) and line $AB$ subtends arcs $AK$ of $c_1$ and $AB$ of $C$, so $\\angle ANK = \\angle ADB$. Thus, $NK \\parallel BD$ and, similarly, $ML \\parallel BD$. (This also follows from the homotheties that transform circles $c_1$ and $c_2$ into $C$.) Finally,\n\n$$\n\\angle NKL = 180^\\circ - \\angle AKN - \\angle LKB = 180^\\circ - \\angle ABD - \\angle LKB = 180^\\circ - \\angle ACD - \\angle ACB = \\angle BAD.\n$$\n\nSimilarly, $\\angle MNK = \\angle BAD$, which leads to the conclusion. Alternatively, one could notice that line segments $ML$, $PQ$, and $NK$ share the same perpendicular bisector.\n\nb) The radical axes of circles $c_1$, $c_2$, and $C$ (one for each pair), i.e., the tangent line to $C$ at $A$, the tangent line to $C$ at $C$, and the line $BD$, are not all parallel, so they are concurrent at the radical center. Diagonal $BD$ is then a symmedian of triangle $ABC$, which means quadrilateral $ABCD$ is harmonic. (This fact can also be used to give a different proof for part a).)\n\nAs $NPQK$ is an isosceles trapezoid, it follows that $\\angle NAP = \\angle QAK$, which means rays $AP$ and $AQ$ are isogonal with respect to angle $\\angle DAB$. Since $ABCD$ is harmonic, $AP$ is a symmedian of triangle $DAB$, so $AQ$, its isogonal, is the median.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 12930,
    "subject": "Mathematics (Olympiad)",
    "question": "在三角形 $ABC$ 中，$\\angle A = 60^\\circ$。設點 $O$、$H$ 分別為 $\\triangle ABC$ 的外心及垂心。在 $BH$ 線段上取一點 $M$，並在直線 $CH$ 上取一點 $N$，使得 $H$ 位於 $C$、$N$ 之間，且 $BM = CN$。試求\n\n$$\n\\frac{MH + NH}{OH}\n$$\n\n的所有可能值。",
    "options": [],
    "answer": "See solution",
    "solution": "$$\n\\frac{MH + NH}{OH} = \\sqrt{3}.\n$$\n\n在 $BH$ 線段上取 $K$ 點使得 $BK = CH$。連接 $OK$、$OB$、$OC$ 等線段。\n\n![](images/17-3J_p27_data_4fea9f0724.png)\n\n因為 $O$ 是 $\\triangle ABC$ 的外心，所以 $\\angle BOC = 2\\angle A = 120^\\circ$。又因為 $H$ 是 $\\triangle ABC$ 的垂心，所以 $\\angle BHC = 180^\\circ - \\angle A = 120^\\circ$。因此 $\\angle BOC = 120^\\circ = \\angle BHC$，故 $B$、$O$、$H$、$C$ 四點共圓。於是有 $\\angle OBH = \\angle OCH$。\n\n注意到 $OB = OC$ 及 $BK = CH$，再加上 $\\angle OBK = \\angle OBH = \\angle OCH$，故 $\\triangle BOK$ 與 $\\triangle COH$ 全等。所以有 $\\angle BOK = \\angle COH$ 以及 $OK = OH$。因此，\n\n$$\n\\angle KOH = \\angle BOC = 120^\\circ, \\quad \\angle OKH = \\angle OHK = 30^\\circ.\n$$\n\n$\\triangle OKH$ 為以 $KH$ 為底邊的 $120^\\circ-30^\\circ-30^\\circ$ 的等腰三角形，所以 $KH = \\sqrt{3}OH$。由於 $BM = CN$ 及 $BK = CH$，知 $KM = NH$。所以\n\n$$\n\\frac{MH + NH}{OH} = \\frac{MH + KM}{OH} = \\frac{KH}{OH} = \\sqrt{3}.\n$$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 12931,
    "subject": "Mathematics (Olympiad)",
    "question": "Consider two disjoint finite sets of positive integers, $A$ and $B$, with $n$ and $m$ elements, respectively. It is known that every $k$ in $A \\cup B$ satisfies at least one of the conditions: $k + 17 \\in A$ or $k - 31 \\in B$. Prove that $17n = 31m$.",
    "options": [],
    "answer": "See solution",
    "solution": "We construct sequences in $A \\cup B$ as follows: Start with any $x_1 \\in A \\cup B$. If $x_1 + 17 \\in A$, set $x_2 = x_1 + 17$; otherwise, set $x_2 = x_1 - 31 \\in B$. Continue: for each $x_j$, if $x_j + 17 \\in A$, set $x_{j+1} = x_j + 17$; otherwise, $x_{j+1} = x_j - 31 \\in B$. Since $A \\cup B$ is finite, eventually $x_{j+1}$ repeats a previous value. Suppose $x_{j+1} = x_k$ for some $k < j+1$. By minimality, $x_{j+1} = x_1$.\n\nLet $a$ be the number of steps where $x_{m+1} = x_m + 17$ (i.e., $x_{m+1} \\in A$), and $b = j - a$ the number where $x_{m+1} = x_m - 31$ (i.e., $x_{m+1} \\in B$). Summing the increments, $x_1 = x_1 + 17a - 31b$, so $17a = 31b$. Thus, in each such cycle, the ratio of $A$ to $B$ elements is $a/b = 31/17$.\n\nIf $A \\cup B$ is not yet exhausted, repeat the process with an unused element, forming more cycles. Each cycle has the same $A$ to $B$ ratio. When all elements are covered, $n/a = m/b = k$ for some integer $k$, so $n = ka$, $m = kb$, and $17n = 31m$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 12932,
    "subject": "Mathematics (Olympiad)",
    "question": "Даден е остроаголен триаголник $ABC$ таков што аголот во темето $C$ е најголем. Нека $E$ и $G$ се пресечните точки на висината спуштена од $A$ кон $BC$ со опишаната кружница на триаголникот $ABC$ и со $BC$ соодветно, и центарот $O$ на опишаната кружница лежи на нормалата спуштена од $A$ кон $BE$. Точките $M$ и $F$ се подножјата на висините спуштени од $E$ кон $AC$ и $AB$ соодветно. Докажи дека $P_{MFE} < P_{FVEG}$.",
    "options": [],
    "answer": "See solution",
    "solution": "Нека пресекот на $EM$ со $BC$ е $V$ (пресекот секогаш ќе постои бидејќи аголот во $C$ е остар). Од теорема на Симсон следува дека точките $M$, $G$ и $F$ се колинеарни. Да забележиме дека $\\angle EAC = \\angle EBC$ како агли над ист кружен лак. Уште $\\angle CAE = \\angle BAO$. Четириаголникот $FBEG$ е тетивен. Па добиваме $\\angle GBE = \\angle GFE$. Уште $\\angle GAO = \\angle GBE$ како агли со нормални краци. Добивме $\\angle CAE = \\angle GAO = \\angle BAO$. Па добиваме дека $AO$ е симетрала на аголот и нормала во триаголникот $ABE$. Следува дека $ABE$ е рамнокрак, од каде $GF$ е паралелна со $BE$. Правите $AO$, $BG$ и $EF$ се сечат во една точка ($EF$ и $BG$ се висини во триаголникот $ABE$). Да забележиме дека $AGMV$ е тетивен. Имаме $\\angle MVG = \\angle GAM$. Уште добиваме $\\angle MAV = \\angle VGM = \\angle FGB$. Па $AM$ е висина и симетрала на аголот во триаголникот $EAV$. Следува дека триаголникот $EAV$ е рамнокрак и $M$ е средина на страната $VE$. Уште $\\angle AVE \\cong \\angle AEB$. Јасно е дека $\\angle EGV \\cong \\angle EGB$ и бидејќи и двата се правоаголни $P_{GEM} = \\frac{1}{2}P_{GEB}$. Од друга страна $P_{GFE} = P_{GFB}$, од каде се добива бараното неравенство.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 12933,
    "subject": "Mathematics (Olympiad)",
    "question": "Triangle $ABC$ is given. Let $M$ be the midpoint of segment $AB$ and $T$ be the midpoint of arc $BC$ not containing $A$ of the circumcircle of $ABC$. Let $K$ be a point inside triangle $ABC$ such that $MATK$ is an isosceles trapezoid with $AT \\parallel MK$. Show that $AK = KC$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $TK$ intersect the circumcircle of $ABC$ at points $T$ and $S$. Then $\\angle ABS = \\angle ATS = \\angle BAT$, so $ASBT$ is a trapezoid. Thus $MK \\parallel AT \\parallel SB$, and $M$ is the midpoint of $AB$, so $K$ is the midpoint of $TS$. But $\\angle TAC = \\angle BAT = \\angle ATS$, so $ACTS$ is an inscribed trapezoid, which is isosceles. Therefore, $KA = KC$ as $K$ is the midpoint of $TS$. $\\blacktriangleleft$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 12934,
    "subject": "Mathematics (Olympiad)",
    "question": "Реши ја равенката\n\n$$\n\\left(\\frac{x^3 + x}{3}\\right)^3 + \\frac{x^3 + x}{3} = 3x.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "Ако воведеме смена $\\frac{x^3+x}{3} = y$, тогаш $y^3 + y = 3x$ и $x^3+x=3y$. Според тоа, ако $x_0$ е решение на равенката и $\\frac{x_0^3+x_0}{3} = y_0$, тогаш $(x_0, y_0)$ е решение на системот равенки\n\n$$\n\\begin{cases}\nx^3 + x = 3y \\\\\ny^3 + y = 3x\n\\end{cases}.\n$$\n\nАко од првата равенка на системот ја одземеме втората равенка добиваме\n\n$$\n\\begin{aligned}\n& (x^3 - y^3) + (x - y) = 3(y - x) \\\\\n& (x - y)(x^2 + xy + y^2 + 1) = 3(y - x) \\\\\n& (x - y)(x^2 + xy + y^2 + 4) = 0\n\\end{aligned}\n\\qquad (1)\n$$\n\nБидејќи\n\n$$\nx^2 + xy + y^2 + 4 = x^2 + 2x\\frac{y}{2} + \\frac{y^2}{4} + \\frac{3}{4}y^2 + 4 = \\left(x + \\frac{y}{2}\\right)^2 + \\frac{3y^2}{4} + 4 \\ge 4,\n$$\n\nза било кои $x, y \\in \\mathbb{R}$, од равенката (1) имаме $x - y = 0$, односно $x = y$. Според тоа $\\frac{x^3+x}{3} = x$, од каде добиваме дека $x(x^2-2)=0$. Решенија на последната равенка се $x_1 = 0$, $x_2 = \\sqrt{2}$, $x_3 = -\\sqrt{2}$. Не е тешко да се провери дека истите се решенија и на почетната равенка.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 12935,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $\\Gamma$ be the circumcircle of $\\triangle ABC$. Let $D$ be a point on the side $BC$. The tangent to $\\Gamma$ at $A$ intersects the parallel line to $BA$ through $D$ at point $E$. The segment $CE$ intersects $\\Gamma$ again at $F$. Suppose $B, D, F, E$ are concyclic. Prove that $AC$, $BF$, $DE$ are concurrent.",
    "options": [],
    "answer": "See solution",
    "solution": "From the conditions, we have\n\n![](images/2020_Australian_Scene_W_p114_data_9849eccaaa.png)\n\n$$\n\\begin{aligned}\n\\angle CBA &= 180^\\circ - \\angle EDB = 180^\\circ - \\angle EFB \\\\\n&= 180^\\circ - \\angle EFA - \\angle AFB \\\\\n&= 180^\\circ - \\angle CBA - \\angle ACB = \\angle BAC.\n\\end{aligned}\n$$\n\nLet $P$ be the intersection of $AC$ and $BF$. Then we have\n\n$$\n\\angle PAE = \\angle CBA = \\angle BAC = \\angle BFC.\n$$\n\nThis implies $A$, $P$, $F$, $E$ are concyclic. It follows that\n\n$$\n\\angle FPE = \\angle FAE = \\angle FBA,\n$$\n\nand hence $AB$ and $EP$ are parallel. So $E$, $P$, $D$ are collinear, and the result follows.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 12936,
    "subject": "Mathematics (Olympiad)",
    "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\n(a) Numbers 1001 and 1003 are both reflecting;\n\n(b) Numbers 1000, 1003 and 1008 are all reflecting;\n\n(c) Numbers 1002, 1004 and 1006 are all reflecting?",
    "options": [],
    "answer": "See solution",
    "solution": "*Answer:* a) No; b) No; c) Yes.\n\n*Solution:* For every integer $x$, denote by $G(x)$ the number into which Gandalf turns the number $x$.\n\n(a) 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\n(b) Suppose that numbers 1000, 1003 and 1008 are all reflecting. Then, for every integer $x$,\n\n$$\nG(x+3) = G(1003 - (x+3)) = G(1000 - x) = G(x),\n$$\n\n$$\nG(x+5) = G(1008 - (x+5)) = G(1003 - x) = G(x).\n$$\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\n(c) 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": 12937,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all functions $f : \\mathbb{R} \\setminus \\{1\\} \\to \\mathbb{R} \\setminus \\{0\\}$ which satisfy the equations $f(0) = 1$ and\n$$\nf(f(xy)) = 1 - \\frac{1}{y f(f(x))}\n$$\nfor any real numbers $x$ and $y$ such that $xy \\neq 0$, $xy \\neq 1$, and $x \\neq 1$.",
    "options": [],
    "answer": "See solution",
    "solution": "Note that if $x \\neq 0$ and $x \\neq 1$, then $f(x) \\neq 0$ and $f(x) \\neq 1$. The difference from $0$ is given in the problem, and if we suppose $f(x) = 1$, taking $y = 1$ would yield $f(f(xy)) = f(f(x)) = f(1)$, meaning that $f(f(xy))$ would be undefined and could not satisfy the equation. Hence, one can apply $f$ infinitely to any real number $x \\notin \\{0, 1\\}$.\n\nPlugging $x = 2$, $y = \\frac{z}{2}$, where $z \\notin \\{0, 1\\}$, into the original equation yields\n$$\nf(f(z)) = 1 - \\frac{2}{z f(f(f(2)))}.\n$$\nWe see that $f(f(z))$ obtains all real values except $1$ and $1 - \\frac{2}{f(f(f(2)))}$, when $z$ obtains all real values except $0$ and $1$. As $f(f(z))$ cannot obtain the value $0$, the only option is $1 - \\frac{2}{f(f(f(2)))} = 0$, i.e., $f(f(z))$ obtains all real values except $0$ and $1$.\n\nNow substituting $x = z$ and $y = 1$, where still $z \\notin \\{0, 1\\}$, into the original equation yields\n$$\nf(f(z)) = 1 - \\frac{1}{f(f(f(z)))}.\n$$\nDenoting $f(f(z)) = x$, we obtain $x = 1 - \\frac{1}{f(x)}$, i.e.,\n$$\nf(x) = \\frac{1}{1-x}\n$$\nThis holds for any real number $x$ except $0$ and $1$. By the condition of the problem, $f(0) = 1$, which implies the equality also in the case $x = 0$.\n\nAn easy check shows that $f(x) = \\frac{1}{1-x}$ satisfies all conditions of the problem.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 12938,
    "subject": "Mathematics (Olympiad)",
    "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.",
    "options": [],
    "answer": "See 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, $\\angle BAG = 90^\\circ - \\gamma = 2\\angle BAD$, so $AD$ is the angle bisector of $BAG$. Similarly, $AE$ is the angle bisector of $GAC$.\n\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![](images/2025CZPS_p4_data_cc2fd1498f.png)\n\nFirst, we will show that $BP = CR$. Notice that since $\\angle ABD$ is right, $D'$ is the circumcenter of $ABD$, so $D'B = D'A$. Then $\\angle D'BA = \\angle PAD' = \\angle D'AQ$, and also $\\angle AQD' = \\angle BPD'$. Together with $D'B = D'A$, triangles $D'AQ$ and $D'BP$ are congruent, so $BP = AQ$. Similarly, $CR = AQ$, so $BP = CR$ as desired.\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 triangles $MBP$ and $MCR$ are congruent, giving $MB = MD$, which is enough.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 12939,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $\\mathbb{N}$ be the set of positive integers. Find all functions $f : \\mathbb{N} \\to \\mathbb{N}$ that satisfy the equation\n\n$$\nf^{abc-a}(abc) + f^{abc-b}(abc) + f^{abc-c}(abc) = a + b + c\n$$\n\nfor all $a, b, c \\ge 2$.\n\n(Here $f^1(n) = f(n)$ and $f^k(n) = f(f^{k-1}(n))$ for every integer $k$ greater than 1.)",
    "options": [],
    "answer": "See solution",
    "solution": "Clearly, if $f(n) = n - 1$ for $n > 1$, the desired identity will be satisfied. In fact, one can easily check that the value of $f(2)$ is also irrelevant, so any $f$ such that $f(n) = n - 1$ for all $n > 2$ will work. We show that these are the only such functions.\n\nPlug in $a = b = c$ to obtain $f^{a^3-a}(a^3) = a$ for all $a$. Using this twice, $f^{a^9-a}(a^9) = a$. Then, $b = c = a^4$ gives $f^{a^9-a^4}(a^9) = a^4$. Meanwhile, $b = a^3$, $c = a^5$ gives $f^{a^9-a^5}(a^9) = a^5$; consequently, we get $f^{a^5-a^4}(a^5) = a^4$.\n\nNext, for any given $b, c$, plug in $a = (bc)^4$, and after simplifying using the above identities, obtain\n\n$$\nf^{bc-b}(bc) + f^{bc-c}(bc) = b + c.\n$$\n\nNow fix a large number $N$, and for all divisors $d \\mid N$ with $1 < d < N$, define $g(d)$ by $g(d) = f^{N-d}(N) - d$. We claim that $g(bc) = g(b) + g(c)$ whenever $bc$ is still a proper divisor of $N$. Proof: put $a = N / bc$, and write\n\n$$\n\\begin{aligned}\ng(bc) &= f^{abc-bc}(abc) - bc \\\\\n&= (a + bc - f^{abc-a}(abc)) - bc \\quad (\\text{by above}) \\\\\n&= a - f^{abc-a}(abc) \\\\\n&= f^{abc-b}(abc) - b + f^{abc-c}(abc) - c \\\\\n&= g(b) + g(c).\n\\end{aligned}\n$$\n\nThis holds for any $N$. In particular, fix any $b \\ge 2$, and put $N = (b^r(b+1)^s)^3$, where $r, s$ are any relatively prime integers both greater than $b+1$. Repeatedly using the multiplicative property of $g$ defined above, we get\n\n$$\nr g(b) + s g(b+1) = g(b^r(b+1)^s) = 0.\n$$\n\nThe only solution to this with integers $g(b), g(b+1)$ satisfying $g(b) \\ge -b$ and $g(b+1) \\ge -(b+1)$ is $g(b) = g(b+1) = 0$. Hence,\n\n$$\nf^{N-b}(N) = b, \\quad f^{N-(b+1)}(N) = b+1\n$$\n\nand comparing gives $f(b+1) = b$, which is what we set out to prove.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 12940,
    "subject": "Mathematics (Olympiad)",
    "question": "Suppose the domain of the function $f(x)$ is $D = (-\\infty, 0) \\cup (0, +\\infty)$ and $f(x) = \\frac{f(1) \\cdot x^2 + f(2) \\cdot x - 1}{x}$ for any $x \\in D$. Then the sum of all the zeros of $f(x)$ is ______.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $f(1)$ and $f(2)$ be unknowns. We have:\n\n$$\n\\begin{align*}\nf(1) &= \\frac{f(1) \\cdot 1^2 + f(2) \\cdot 1 - 1}{1} = f(1) + f(2) - 1, \\\\\nf(2) &= \\frac{f(1) \\cdot 4 + f(2) \\cdot 2 - 1}{2} = 2f(1) + f(2) - \\frac{1}{2}.\n\\end{align*}\n$$\n\nSolving these equations:\n\n- $f(1) = f(1) + f(2) - 1 \\implies f(2) = 1$\n- $f(2) = 2f(1) + f(2) - \\frac{1}{2} \\implies 2f(1) = \\frac{1}{2} \\implies f(1) = \\frac{1}{4}$\n\nSo,\n\n$$\nf(x) = \\frac{1}{x} \\left( \\frac{1}{4}x^2 + x - 1 \\right), \\quad x \\neq 0.\n$$\n\nSet $f(x) = 0$:\n\n$$\n\\frac{1}{4}x^2 + x - 1 = 0\n$$\n\nMultiply both sides by $4$:\n\n$$\nx^2 + 4x - 4 = 0\n$$\n\nThe sum of the roots is $-4$.\n",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 12941,
    "subject": "Mathematics (Olympiad)",
    "question": "Non-negative integers are written in some cells of a $100 \\times 100$ table. For each $k$, $1 \\leq k \\leq 100$, the $k$-th row of the table contains the numbers from $1$ to $k$ written in increasing order (from left to right), but not necessarily in consecutive cells. The empty cells are filled with zeroes. Prove that there exist two columns such that the sum of numbers in one of them is at least $19$ times greater than the sum in the second column.",
    "options": [],
    "answer": "See solution",
    "solution": "Observe that the sum of numbers in the first column is at most $1 \\cdot 100 = 100$. The sum in the first and second columns is at most $1 \\cdot 100 + 2 \\cdot 99$, the sum in the first, second, and third columns is at most $1 \\cdot 100 + 2 \\cdot 99 + 3 \\cdot 98$, etc. The sum of all nonzero numbers equals $\\sum_{i=1}^{100} i(101 - i)$. Therefore, the sum in the columns from $31$-st to $100$-th is at least\n\n$$\n\\sum_{i=31}^{100} i(101-i) = \\sum_{i=1}^{70} i(101-i) = 101 \\sum_{i=1}^{70} i - \\sum_{i=1}^{70} i^2.\n$$\n\nCalculating:\n\n$$\n\\sum_{i=1}^{70} i = \\frac{70 \\cdot 71}{2} = 2485,\n$$\n$$\n\\sum_{i=1}^{70} i^2 = \\frac{70 \\cdot 71 \\cdot 141}{6} = 116045.\n$$\n\nSo,\n$$\n101 \\cdot 2485 - 116045 = 251985 - 116045 = 135940.\n$$\n\nTherefore, one of these columns has a sum at least $\\frac{135940}{70} \\approx 1942$. Therefore, the ratio of sums in this column and in the first one is more than $19$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 12942,
    "subject": "Mathematics (Olympiad)",
    "question": "Let the centres of the five circles be $P, Q, R, S$ and $T$ and $A, B, C, D$ and $E$ the intersections of adjacent circles. Let $F, G, H, K$ and $L$ be the intersections of alternate circles as shown in the drawing.\n\n![](images/Irska_2009_p41_data_700a0d29b4.png)\n\nFind the value of $r_1 r_2$, where $r_1 = |OA|$ and $r_2 = |OF|$ as defined in the figure.",
    "options": [],
    "answer": "See solution",
    "solution": "OPAQ is a rhombus since all the radii are equal, so $PQ$ bisects the common chord $AO$ at right angles. Because $PQRST$ is a regular pentagon, $\\angle POQ = 72^\\circ$. Since the two isosceles triangles $PAO$ and $QAO$ are congruent, $\\angle POA = 36^\\circ$. This implies\n\n$$\nr_1 = |OA| = 2|OP| \\cos \\angle POA = 2 \\cos 36^\\circ.\n$$\n\nSince $TQ$ bisects $OF$ at right angles,\n$$\nr_2 = |OF| = 2|OQ| \\cos \\angle POQ = 2 \\cos 72^\\circ\n$$\nand so\n$$\nr_1 r_2 = 4 \\cos 36^\\circ \\cos 72^\\circ = 4(1 - 2 \\sin^2 18^\\circ) \\sin 18^\\circ.\n$$\n\nNow, since $\\sin 36^\\circ = \\cos 54^\\circ$, we obtain\n$$\n2 \\sin 18^\\circ \\cos 18^\\circ = 4 \\cos^3 18^\\circ - 3 \\cos 18^\\circ.\n$$\nThis implies\n$$\n2 \\sin 18^\\circ = 4 \\cos^2 18^\\circ - 3 = 4(1 - \\sin^2 18^\\circ) - 3.\n$$\nAbbreviating $x = \\sin 18^\\circ$, this can be rewritten as\n$$\n4x^2 + 2x - 1 = 0,\n$$\nfrom which we get the two equalities $2x(2x+1) = 1$ and $2x+1 = 2(1-2x^2)$. Substituting the second into the first and using the expression for $r_1 r_2$ obtained before, we get\n$$\nr_1 r_2 = 4x(1-2x^2) = 1.\n$$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 12943,
    "subject": "Mathematics (Olympiad)",
    "question": "Given a non-isosceles triangle $ABC$ with incenter $I$, show that the circumcircle $k$ of $\\triangle AIB$ is not tangent to the lines $CA$ or $CB$.\n\nLet the second intersection point of $k$ with $CA$ be $P$, and the second intersection with $CB$ be $Q$. Prove that the points $A$, $B$, $P$, and $Q$ (not necessarily in this order) are the vertices of a trapezoid.",
    "options": [],
    "answer": "See solution",
    "solution": "It is well known that the angle bisector at $C$ and the bisector of side $AB$ intersect at a point $M$ on the circumcircle of $\\triangle ABC$. This point $M$ is equidistant from $A$ and $B$.\n\nNow consider $\\triangle AIM$. Let the angles at $A$, $B$, and $C$ be $\\alpha$, $\\beta$, and $\\gamma$, respectively. We have:\n\n$$\n\\angle MAI = \\frac{\\alpha}{2} + \\angle BAM = \\frac{\\alpha}{2} + \\angle BCM = \\frac{\\alpha}{2} + \\frac{\\gamma}{2}.\n$$\n\nFurthermore,\n$$\n\\angle IMA = \\angle CMA = \\angle CBA = \\beta,\n$$\nso\n$$\n\\angle MIA = 180^\\circ - \\angle IMA - \\angle AIM = 180^\\circ - \\beta - \\left(\\frac{\\alpha}{2} + \\frac{\\gamma}{2}\\right) = \\frac{\\alpha}{2} + \\frac{\\gamma}{2}.\n$$\n\nThus, $\\triangle AIM$ is isosceles and $MA = MI$, so $M$ is the midpoint of arc $AB$ on $k$.\n\n![](images/AustriaMO2011_p4_data_6654e77008.png)\n\nLet $U$ and $V$ be the midpoints of $AP$ and $BQ$, respectively. Triangles $MUC$ and $MVC$ are congruent, as they both have angles $90^\\circ$ and $\\frac{\\gamma}{2}$ and share hypotenuse $CM$. Thus, $MU = MV$. Since $MA = MP = MB = MQ$, triangles $AMP$ and $BMQ$ are both isosceles with equal side lengths and altitudes, so they are congruent, implying $AP = BQ$.\n\nTherefore, quadrilateral $AQBP$ is cyclic and has two sides of equal length, so it is a trapezoid.\n\nFinally, neither $AC$ nor $BC$ can be tangent to $k$. If either were tangent, the congruent triangles $AMP$ and $BMC$ would degenerate to segments perpendicular to $AC$ and $BC$, respectively, making both lines tangent to $k$. The tangent segments would then be equal, implying $ABC$ is isosceles, contradicting the assumption. This completes the proof.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 12944,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $f_1 \\in \\mathbb{R}[x]$ be a quadratic polynomial with positive leading coefficient. Set $f_{n+1} = f_1 \\circ f_n$ for $n \\geq 1$. It is known that the polynomial $f_2$ has four non-positive different zeroes. Prove that the polynomial $f_n$ has $2^n$ different real zeroes.",
    "options": [],
    "answer": "See solution",
    "solution": "Note that if $x_1, \\dots, x_{2^n}$ are the zeroes of $f_n$, then the zeroes of $f_{n+1}$ are the solutions of the equations $f_1(x) = x_k$, $1 \\leq k \\leq 2^n$. Moreover, the equation $f_1(x) = a$ has two different real roots if and only if $a > m := \\min f_1$.\n\nAssume that $f_2$ has four non-positive different zeroes. Then it is easy to see that $f_1$ has two zeroes $x_1 < x_2 \\leq 0$ and $x_1 > m$. Using the argument from the beginning, it follows by induction on $n$ that all the zeroes of $f_{n+1}$ are different and belong to the interval $(x_1, x_2]$.",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 12945,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $a$, $b$, $c$ be positive real numbers satisfying $a^2 < bc$. Prove that\n\n$$\nb^3 + ac^2 > ab(a + c)\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "Adding three AM-GM inequalities:\n\n$$\n4a^3b + b^3c + 2c^3a \\geq 7a^2bc,\n$$\n\n$$\n4b^3c + c^3a + 2a^3b \\geq 7b^2ca,\n$$\n\n$$\n4c^3a + a^3b + 2b^3c \\geq 7c^2ab\n$$\n\nwe get\n\n$$\na^3b + b^3c + c^3a \\geq a^2bc + b^2ca + c^2ab \\quad (1)\n$$\n\nThe assumption $a^2 < bc$ implies $-a^3b > -b^2ca$, and this together with (1) gives\n\n$$\nb^3c + c^3a > a^2bc + c^2ab\n$$\n\ni.e.\n\n$$\nb^3 + ac^2 > ab(a + c)\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 12946,
    "subject": "Mathematics (Olympiad)",
    "question": "Eha and Koit play the following game. At the beginning, each vertex of a square has an empty box. On each turn, a player can either:\n\n1. Add one stone to any box, or\n2. Move each box clockwise to the next vertex of the square.\n\nKoit starts, and they alternate turns for a total of 2012 steps (each player takes 1006 turns). After these moves, Koit marks one vertex of the square and allows Eha to make one more move. Koit wins if, after Eha's final move, some box contains more stones than the box at the marked vertex; otherwise, Eha wins.\n\nWhich player has a winning strategy?",
    "options": [],
    "answer": "See solution",
    "solution": "Eha can ensure that, before Koit's marking, the boxes at opposite corners always have equal numbers of stones. Initially, all boxes are empty, so this holds. Suppose before Koit's move the boxes have $(a, b, a, b)$ stones. If Koit adds a stone to a box, Eha adds a stone to the opposite box; if Koit rotates the boxes, Eha also rotates them. Thus, after 2012 steps, the boxes have $(a, b, a, b)$ stones.\n\nAssume $a \\geq b$. If Koit marks a vertex with $a$ stones, Eha adds a stone to that box and wins. If Koit marks a vertex with $b$ stones, Eha rotates the boxes, so the marked box now has $a$ stones, and Eha wins. Therefore, Eha has a winning strategy.",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 12947,
    "subject": "Mathematics (Olympiad)",
    "question": "How many three-digit palindromes are there?",
    "options": [],
    "answer": "See solution",
    "solution": "There are 9 ways to select the nonzero hundreds digit, 1 way to select the units digit (same as the hundreds digit), and 10 ways to select the tens digit. Thus, there are $9 \\times 1 \\times 10 = 90$ three-digit palindromes.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 12948,
    "subject": "Mathematics (Olympiad)",
    "question": "Construct a point $E$ on the extension of $AC$ such that $CD = CE$.",
    "options": [],
    "answer": "See solution",
    "solution": "Using the condition, we find that\n\n$$\n\\begin{aligned}\nMB \\cdot MD &= MA \\cdot MC + MA \\cdot CD \\\\\n&= MA(MC + CE) \\\\\n&= MA \\cdot ME.\n\\end{aligned}\n$$\n\nThis implies $A, B, E, D$ are concyclic.\n\nNow, since $\\angle CDE = \\angle CED = \\frac{1}{2} \\angle DCA = \\angle KCA$, we obtain\n\n$$\n\\angle BKC = \\angle BAC - \\angle KCA = \\angle BDE - \\angle CDE = \\angle BDC\n$$\n\nas desired.\n\n![](images/1.2_1997-2023_IMO_HK_TST_Solutions_p45_data_c4e59fc6c3.png)\n![](images/1.2_1997-2023_IMO_HK_TST_Solutions_p45_data_9d1b6be901.png)",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 12949,
    "subject": "Mathematics (Olympiad)",
    "question": "The Bank of Oslo issues two types of coin: aluminium (denoted $A$) and bronze (denoted $B$). Marianne has $n$ aluminium coins and $n$ bronze coins, arranged in a row in some arbitrary initial order. A chain is any subsequence of consecutive coins of the same type. Given a fixed positive integer $k \\le 2n$, Marianne repeatedly performs the following operation: she identifies the longest chain containing the $k$th coin from the left, and moves all coins in that chain to the left end of the row.\n\nFor example, if $n = 4$ and $k = 4$, the process starting from the ordering $AABBBABA$ would be\n\n$$\n\\begin{align*}\nAAB\\underline{B}BABA &\\rightarrow B\\underline{B}BA\\underline{A}ABA \\\\ \n&\\rightarrow A\\underline{A}A\\underline{B}\\underline{B}BBA \\\\ \n&\\rightarrow B\\underline{B}\\underline{B}AAAA \\\\ \n&\\rightarrow B\\underline{B}\\underline{B}AAAA \\rightarrow \\dots\n\\end{align*}\n$$\n\nFind all pairs $(n, k)$ with $1 \\le k \\le 2n$ such that for every initial ordering, at some moment during the process, the leftmost $n$ coins will all be of the same type.",
    "options": [],
    "answer": "See solution",
    "solution": "The desired pairs are all $(n, k)$ that satisfy $n \\le k \\le \\frac{3n+1}{2}$.\n\nAs defined in the problem, a chain is any subsequence of consecutive coins of the same type. We call it a \"block\" if it is a chain but not contained in a longer chain. For $M = A$ or $B$, let $M^x$ represent a block of $M$ coins of length $x$. We are interested in whether an initial ordering will turn into a 2-block sequence $A^n B^n$ or $B^n A^n$ after finitely many operations.\n\n**(1) Claim:** If $k < n$ or $k > \\frac{3n+1}{2}$, then the sequence cannot always turn into $A^n B^n$ or $B^n A^n$.\n\n*Proof of claim:* If $k < n$, the sequence $A^{n-1}B^nA^1$ does not change after an operation, and cannot turn into a 2-block sequence. If $k > \\frac{3n+1}{2}$, let $m = \\lfloor\\frac{n}{2}\\rfloor$, $l = \\lceil\\frac{n}{2}\\rceil$, and the initial ordering be $A^mB^mA^lB^l$. As $k > \\frac{3n+1}{2} \\ge m+l+l \\ge m+m+l$, the $k$th coin always belongs to the rightmost block, and the process is\n\n$$\n\\begin{align*}\nA^m B^m A^l B^l &\\rightarrow B^l A^m B^m A^l \\\\ \n&\\rightarrow A^l B^l A^m B^m \\\\ \n&\\rightarrow B^m A^l B^l A^m \\\\ \n&\\rightarrow A^m B^m A^l B^l \\rightarrow \\dots,\n\\end{align*}\n$$\n\na 4-periodic cycle. It cannot turn into a 2-block sequence.\n\n**(2) Claim:** If $n \\le k \\le \\frac{3n+1}{2}$, any initial ordering will turn into a 2-block sequence.\n\n*Proof of claim:* Notice that after an operation, the number of blocks does not increase. Eventually, it will be stabilized, say at $s$ blocks. It suffices to prove $s = 2$. Suppose $s > 2$. As $k \\ge n$ and $s > 2$, the $k$th coin cannot belong to the leftmost block, and for the current operation, the block being moved is not the leftmost one. If it is a middle one, then after the operation, the two blocks on its sides will merge into one block, reducing the number of blocks by 1, a contradiction. Therefore, when stabilized, at each operation the block being moved to the left is always the rightmost ($s$th) one. Since $k \\le (3n + 1) / 2$, the length of the $s$th block is at least $2n - k + 1 \\ge (n + 1) / 2$; since $s$ does not decrease after the operation, the first and the $s$th blocks are of different types, and hence $s$ is even. Let $s = 2t$, $t \\ge 2$, and suppose one of the stable orderings is $X_1 X_2 \\cdots X_{2t}$, in which every $X_i$ represents a block. The process is\n\n$$\n\\begin{align*}\nX_1 X_2 \\cdots X_{2t} &\\rightarrow X_{2t} X_1 \\cdots X_{2t-1} \\\\ \n&\\rightarrow \\cdots \\\\ \n&\\rightarrow X_2 X_3 \\cdots X_1 \\\\ \n&\\rightarrow X_1 X_2 \\cdots X_{2t} \\rightarrow \\cdots\n\\end{align*}\n$$\n\nBased on the previous argument, at any moment the rightmost block has length $\\ge \\frac{n+1}{2}$. So, each block has length $\\ge \\frac{n+1}{2}$, and the total length of the sequence is at least $4 \\cdot \\frac{n+1}{2} > 2n$, a contradiction. Now, the claim and the conclusion follow.\n\n$\\boxed{}$",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 12950,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all prime numbers $a$, $b$, $c$ and positive integers $k$ satisfying the equation\n\n$$\na^2 + b^2 + 16c^2 = 9k^2 + 1\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "The relation $9k^2 + 1 \\equiv 1 \\pmod{3}$ implies\n$$a^2 + b^2 + 16c^2 \\equiv 1 \\pmod{3} \\quad \\Leftrightarrow \\quad a^2 + b^2 + c^2 \\equiv 1 \\pmod{3}.$$ \n\nSince $a^2 \\equiv 0,1 \\pmod{3}$, $b^2 \\equiv 0,1 \\pmod{3}$, $c^2 \\equiv 0,1 \\pmod{3}$, we have:\n\n![](table)\n\nFrom the previous table it follows that two of three prime numbers $a$, $b$, $c$ are equal to $3$.\n\n*Case 1.* $a = b = 3$. We have\n\n$$\na^2 + b^2 + 16c^2 = 9k^2 + 1 \\Leftrightarrow 9k^2 - 16c^2 = 17 \\Leftrightarrow (3k - 4c)(3k + 4c) = 17\n$$\n\nIf $\\begin{cases} 3k - 4c = 1 \\\\ 3k + 4c = 17 \\end{cases}$, then $\\begin{cases} c = 2 \\\\ k = 3 \\end{cases}$ and $(a, b, c, k) = (3, 3, 2, 3)$.\n\nIf $\\begin{cases} 3k - 4c = -1 \\\\ 3k + 4c = -17 \\end{cases}$, then $\\begin{cases} c = 2 \\\\ k = -3 \\end{cases}$ and $(a, b, c, k) = (3, 3, 2, -3)$.\n\n*Case 2.* $c = 3$. If $(3, b_0, c, k)$ is a solution of the given equation, then $(b_0, 3, c, k)$ is a solution, too.\n\nLet $a = 3$. We have\n\n$$\na^2 + b^2 + 16c^2 = 9k^2 + 1 \\Leftrightarrow 9k^2 - b^2 = 152 \\Leftrightarrow (3k - b)(3k + b) = 152.\n$$\n\nBoth factors shall have the same parity and we obtain only 4 cases:\n\nIf $\\begin{cases} 3k - b = 2 \\\\ 3k + b = 76 \\end{cases}$, then $\\begin{cases} b = 37 \\\\ k = 13 \\end{cases}$ and $(a, b, c, k) = (3, 37, 3, 13)$.\n\nIf $\\begin{cases} 3k - b = 4 \\\\ 3k + b = 38 \\end{cases}$, then $\\begin{cases} b = 17 \\\\ k = 7 \\end{cases}$ and $(a, b, c, k) = (3, 17, 3, 7)$.\n\nIf $\\begin{cases} 3k - b = -76 \\\\ 3k + b = -2 \\end{cases}$, then $\\begin{cases} b = 37 \\\\ k = -13 \\end{cases}$ and $(a, b, c, k) = (3, 37, 3, -13)$.\n\nIf $\\begin{cases} 3k - b = -38 \\\\ 3k + b = -4 \\end{cases}$, then $\\begin{cases} b = 17 \\\\ k = -7 \\end{cases}$ and $(a, b, c, k) = (3, 17, 3, -7)$.\n\nIn addition, $(a, b, c, k) \\in \\{(37, 3, 3, 13), (17, 3, 3, 7), (37, 3, 3, -13), (17, 3, 3, -7)\\}$.\n\nSo, the given equation has 10 solutions:\n\n$$\nS = \\{(37, 3, 3, 13), (17, 3, 3, 7), (37, 3, 3, -13), (17, 3, 3, -7), (3, 37, 3, 13), (3, 17, 3, 7), (3, 37, 3, -13), (3, 17, 3, -7), (3, 3, 2, 3), (3, 3, 2, -3)\\}.\n$$",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 12951,
    "subject": "Mathematics (Olympiad)",
    "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": "See solution",
    "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\n$$\np_x(i) = |\\{j : x \\in A_{i,j}\\}|.\n$$\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 family determines a sweet $n$-configuration $\\langle A_{i,j} \\rangle_{1 \\le i,j \\le n}$ with $A_{n,n} \\subseteq \\{1, 2, \\dots, k\\}$ in the following way: $A_{i,j} = \\{x \\in \\{1, 2, \\dots, k\\} : j \\ge n+1-p_x(i)\\}$.\n\nTherefore $f(n,k) = g(n)^k$ 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\n$$\np \\to \\underbrace{**\\dots*}_{p(1)} \\mid \\underbrace{**\\dots*}_{p(2)-p(1)} \\mid \\underbrace{**\\dots*}_{p(3)-p(2)} \\mid \\dots \\mid \\underbrace{**\\dots*}_{p(n)-p(n-1)} \\mid \\underbrace{**\\dots*}_{n-p(n)}\n$$\n\nThus $g(n) = \\binom{2n}{n}$.\n\nThe problem boils down to determining which of the numbers\n\n$$\n\\binom{2n}{n}^{n^2}, \\quad \\binom{2n^2}{n^2}^n,\n$$\n\nwhere $n = 2024$, is larger. Note that\n\n$$\n\\begin{aligned}\n\\binom{2n^2}{n^2} &= \\frac{\\prod_{i=1}^{n^2} (n^2 + i)}{\\prod_{i=1}^{n^2} i} = \\prod_{i=1}^{n^2} \\left(\\frac{n^2 + i}{i}\\right) = \\prod_{j=0}^{n-1} \\prod_{i=1}^{n} \\frac{n^2 + jn + i}{jn + i} \\\\ &> \\prod_{j=0}^{n-1} \\left(\\frac{n^2 + jn + n}{jn + n}\\right)^n = \\prod_{j=0}^{n-1} \\left(\\frac{n + j + 1}{j + 1}\\right)^n = \\left(\\prod_{j=1}^{n} \\frac{n+j}{j}\\right)^n = \\left(\\binom{2n}{n}\\right)^n\n\\end{aligned}\n$$\n\nand therefore\n\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": 12952,
    "subject": "Mathematics (Olympiad)",
    "question": "Given a real number $\\alpha$, consider the function $\\varphi(x) = x^2 e^{\\alpha x}$ for all real numbers $x$. Find all functions $f : \\mathbb{R} \\to \\mathbb{R}$ that satisfy\n\n$$\nf(\\varphi(x) + f(y)) = y + \\varphi(f(x)), \\quad \\forall x, y \\in \\mathbb{R}.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "Let $\\varphi(f(0)) = c$. Clearly, $f$ is bijective because\n\n$$\nf(f(y)) = y + c.\n$$\n\nReplacing $y$ by $f(y)$ in the relation, we have\n\n$$\nf(y + c) = f(y) + c.\n$$\n\nSince $f$ is bijective, there exists $d \\in \\mathbb{R}$ such that $f(d) = 0$. Replacing $(x, y)$ by $(d, y + c)$, we get\n\n$$\nf(\\varphi(d) + f(y + c)) = y + c.\n$$\n\nSince $f$ is injective, this implies\n\n$$\n\\varphi(d) + f(y + c) = f(y),\n$$\nwhich means\n$$\n\\varphi(d) + c = 0.\n$$\n\nOn the other hand, since $\\varphi(x) \\ge 0$ and equality only occurs when $x = 0$, we have\n$$\nf(0) = d = 0.\n$$\n\nHence, $f(f(y)) = y$ and substituting $y = 0$ in the original equation gives\n$$\nf(\\varphi(x)) = \\varphi(f(x)).\n$$\n\nBut $\\varphi(x) \\ge 0$ for all $x$, so $f(t) \\ge 0$ for all $t \\ge 0$. Replacing $y$ by $f(y)$ and $\\varphi(x) = t \\ge 0$ for any $t \\ge 0$, we obtain\n$$\nf(y + t) = f(y) + f(t), \\quad \\forall t \\ge 0.\n$$\n\nTherefore, for all $x, y \\in \\mathbb{R}$ and $t \\ge \\max(-y, 0)$,\n$$\nf(x + y) + f(t) = f(x + y + t) = f(x) + f(y + t) = f(x) + f(y) + f(t),\n$$\nso $f$ is additive. Since $f(x) \\ge 0$ for all $x$, it follows that $f(x) = kx$ for some $k \\ge 0$. Substituting $f(x) = kx$ into the original equation, we find $k = 1$. Thus, $f(x) = x$ for all $x \\in \\mathbb{R}$.\n\n$\\boxed{f(x) = x}$ is the only solution.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 12953,
    "subject": "Mathematics (Olympiad)",
    "question": "證明只有有限多組正整數 $a, b, c, n$ 能使等式\n\n$$\nn! = a^{n-1} + b^{n-1} + c^{n-1}\n$$\n\n成立。",
    "options": [],
    "answer": "See solution",
    "solution": "對於固定的 $n$，顯然只有有限多組解；我們將證明當 $n > 100$ 時無解。假設 $n > 100$。由 AM-GM 不等式，\n\n$$\n\\begin{aligned}\nn! &= 2n(n-1)(n-2)(n-3) \\cdot (3 \\cdot 4 \\cdots (n-4)) \\\\\n&\\le 2(n-1)^4 \\left( \\frac{3 + \\cdots + (n-4)}{n-6} \\right)^{n-6} = 2(n-1)^4 \\left( \\frac{n-1}{2} \\right)^{n-6} < \\left( \\frac{n-1}{2} \\right)^{n-1},\n\\end{aligned}\n$$\n\n因此 $a, b, c < \\frac{n-1}{2}$。\n\n對於每個質數 $p$ 和整數 $m \\ne 0$，令 $\\nu_p(m)$ 表示 $p$-進指數，即 $p^k$ 整除 $m$ 的最大非負整數 $k$。Legendre 公式給出\n\n$$\n\\nu_p(n!) < \\sum_{s=1}^{\\infty} \\frac{n}{p^s} = \\frac{n}{p-1}. \\quad (1)\n$$\n\n若 $n$ 為奇數，則 $a^{n-1}, b^{n-1}, c^{n-1}$ 為平方數，考慮模 $4$ 可知 $a, b, c$ 必為偶數，因此 $2^{n-1} \\mid n!$，但對奇數 $n$ 不可能，因為由 (1) 得 $\\nu_2(n!) = \\nu_2((n-1)!) < n-1$。接下來假設 $n$ 為偶數。若 $a+b, b+c, c+a$ 都是 $2$ 的冪，則 $a, b, c$ 同奇偶。若都為奇數，則 $n! = a^{n-1}+b^{n-1}+c^{n-1}$ 也為奇數，與 $n$ 為偶數矛盾。若 $a, b, c$ 都能被 $4$ 整除，則與 $\\nu_2(n!) \\le n-1$ 矛盾。若如 $a$ 不能被 $4$ 整除，則 $2a = (a+b)+(a+c)-(b+c)$ 不能被 $8$ 整除，且因 $a+b, b+c, c+a$ 都是 $2m$ 的冪，必有一個和等於 $4$，所以 $a, b, c$ 中有兩個等於 $2$。設 $a = b = 2$，則 $c = 2^r - 2$，且 $c \\mid n!$，必有 $c \\mid a^{n-1} + b^{n-1} = 2^n$，故 $r = 2$，即 $c = 2$，但這不可能，因為 $n! \\equiv 0 \\not\\equiv 3 \\cdot 2^{n-1} \\pmod 5$。\n\n現在假設 $a, b, c$ 中有兩數之和（如 $a+b$）不是 $2$ 的冪，則必被某個奇質數 $p$ 整除。則 $p \\le a+b < n$，且 $c^{n-1} = n! - (a^{n-1} + b^{n-1})$ 也被 $p$ 整除。若 $p$ 整除 $a$ 和 $b$，則 $p^{n-1} \\mid n!$，與 (1) 矛盾。利用 (1) 和提升指數引理（Lifting the Exponent Lemma），有\n\n$$\n\\nu_p(1)+\\nu_p(2)+\\cdots+\\nu_p(n) = \\nu_p(n!) = \\nu_p(n!-c^{n-1}) = \\nu_p(a^{n-1}+b^{n-1}) = \\nu_p(a+b)+\\nu_p(n-1). \\tag{2}\n$$\n\n由 (2) 可知，$1, 2, \\dots, n$ 中只有 $a+b$ 和 $n-1 > a+b$ 能被 $p$ 整除。另一方面，$p|c$ 意味著 $p < n/2$，所以至少有兩個數能被 $p$ 整除，即 $a+b = p$ 和 $n-1 = 2p$。但這又矛盾，因為 $n-1$ 為奇數。這最後的矛盾說明當 $n > 100$ 時方程無解。\n\n**Comment 1.** 原問題要求找出所有解。該版本的解法類似但更技術性。\n\n**Comment 2.** 若要找所有解，可將 $a, b, c < (n-1)/2$ 的界限（對所有 $n$）換成 $a, b, c \\le n/2$（僅對偶數 $n$），這是對 $2, 3, \\dots, n$ 應用 AM-GM 的直接結果。然後對奇數 $n$ 用同樣論證（對 $n \\ge 5$ 有效且不需界限），對偶數 $n$，同樣論證對 $n \\ge 6$ 有效，除非 $a+b = n-1$ 且 $2\\nu_p(n-1) = \\nu_p(n!)$。這只可能在 $p=3$ 且 $n=10$，此時可對原方程取模 $7$ 得 $7 \\bmod abc$，與 $7^9 > 10!$ 矛盾。檢查 $n \\le 4$，可得四組解：\n\n$$\n(a, b, c, n) = (1, 1, 2, 3),\\ (1, 2, 1, 3),\\ (2, 1, 1, 3),\\ (2, 2, 2, 4).\n$$\n\n**Comment 3.** 對於充分大的 $n$，不等式 $a, b, c < (n-1)/2$ 也可由 Stirling 公式推出。",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 12954,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all functions $f: \\mathbb{R} \\to \\mathbb{R}$ satisfying, for all real numbers $x$ and $y$,\n$$\n|x|f(y) + y f(x) = f(xy) + f(x^2) + f(f(y)).\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "All functions $f(x) = c(|x| - x)$, where $c$ is a real number, satisfy the equation.\n\nLet $x = y = 0$:\n$$\nf(f(0)) = -2f(0).\n$$\nLet $a = f(0)$, so $f(a) = -2a$. Now set $y = 0$:\n$$\na|x| = a + f(x^2) + f(a) = a + f(x^2) - 2a \\implies f(x^2) = a(|x| + 1).\n$$\nIn particular, $f(1) = 2a$. Now set $x = z^2$, $y = 1$:\n$$\n\\begin{align*}\nz^2 f(1) + f(z^2) &= f(z^2) + f(z^4) + f(f(1)) \\\\\n2a z^2 &= a(z^2 + 1) + f(2a) \\\\\naz^2 &= a + f(2a).\n\\end{align*}\n$$\nThe right side is constant, but the left is quadratic in $z$, so $a = 0$. Thus $f(x^2) = 0$ for all $x$, so $f(x) = 0$ for all $x \\geq 0$, and $f(0) = 0$.\n\nNow set $x = 0$ in the original equation:\n$$\nf(f(y)) = 0 \\qquad (3)\n$$\nfor all $y$. The original equation simplifies to\n$$\n|x|f(y) + y f(x) = f(xy) = |y|f(x) + x f(y).\n$$\nSet $y = -1$ and let $c = \\frac{f(-1)}{2}$:\n$$\n|x|f(-1) - f(x) = f(x) + x f(-1) \\implies f(x) = c(|x| - x).\n$$\nThese functions satisfy the original equation for any real $c$.\n\n*Remark.* If the term $f(f(y))$ is removed, the problem becomes simpler: set $x = y = 0$, then $y = 0$ to find $f(x^2) = 0$, and proceed as above after equation (3).",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 12955,
    "subject": "Mathematics (Olympiad)",
    "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": "See solution",
    "solution": "If $n$ is odd, then the largest integer less than $\\frac{n}{2}$ is $\\frac{n-1}{2}$. Let $d$ be a common divisor of $\\frac{n-1}{2}$ and $n$. Then $d$ divides both $n-1$ and $n$, so $d = 1$. Thus, $\\frac{n-1}{2}$ and $n$ are relatively prime, so the hypothesis holds for odd $n$.\n\nIf $n$ is even, the two largest integers less than $\\frac{n}{2}$ are $\\frac{n}{2} - 1$ and $\\frac{n}{2} - 2$. Let $d_1$ be a common divisor of $\\frac{n}{2} - 1$ and $n$, and $d_2$ a common divisor of $\\frac{n}{2} - 2$ and $n$. Then $d_1$ divides both $n-2$ and $n$, so $d_1$ divides 2; $d_2$ divides both $n-4$ and $n$, so $d_2$ divides 4. If both $d_1$ and $d_2$ were greater than 1, both $\\frac{n}{2} - 1$ and $\\frac{n}{2} - 2$ would be even, which is impossible since they are consecutive integers. Thus, at least one of $d_1$ or $d_2$ equals 1, so at least one of $\\frac{n}{2} - 1$ or $\\frac{n}{2} - 2$ is relatively prime to $n$. Therefore, the hypothesis holds for even $n$ as well.\n\n_Remark:_ The solution does not use the assumption $n > 4$. This assumption is to avoid cases involving the greatest common divisor of $n$ and zero or a negative number.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 12956,
    "subject": "Mathematics (Olympiad)",
    "question": "Rectangular 16-chunks come in only three shapes: $2 \\times 8$, $8 \\times 2$, $4 \\times 4$. (Note: a square is a rectangle!)\n\nIs it possible for any of these 16-chunks, when placed on a grid of consecutive integers (with rows increasing by 10), to have the sum of their entries divisible by 16?",
    "options": [],
    "answer": "See solution",
    "solution": "**Alternative i**\n\nMoving any 16-chunk sideways by one square changes its value by 16, which is divisible by 16. Moving any 16-chunk vertically by one square changes its value by $16 \\times 10 = 160$, which is divisible by 16.\n\nThe following $2 \\times 8$ 16-chunk has value 152, which is not divisible by 16. So no $2 \\times 8$ 16-chunk has a value divisible by 16.\n\n![]( <table><tr><td>1</td><td>2</td><td>3</td><td>4</td><td>5</td><td>6</td><td>7</td><td>8</td></tr><tr><td>11</td><td>12</td><td>13</td><td>14</td><td>15</td><td>16</td><td>17</td><td>18</td></tr></table> )\n\nThe following $8 \\times 2$ 16-chunk has value 584, which is not divisible by 16. So no $8 \\times 2$ 16-chunk has a value divisible by 16.\n\n![]( <table><tr><td>1</td><td>2</td></tr><tr><td>11</td><td>12</td></tr><tr><td>21</td><td>22</td></tr><tr><td>31</td><td>32</td></tr><tr><td>41</td><td>42</td></tr><tr><td>51</td><td>52</td></tr><tr><td>61</td><td>62</td></tr><tr><td>71</td><td>72</td></tr></table> )\n\nThe following $4 \\times 4$ 16-chunk has value 280, which is not divisible by 16. So no $4 \\times 4$ 16-chunk has a value divisible by 16.\n\n![]( <table><tr><td>1</td><td>2</td><td>3</td><td>4</td></tr><tr><td>11</td><td>12</td><td>13</td><td>14</td></tr><tr><td>21</td><td>22</td><td>23</td><td>24</td></tr><tr><td>31</td><td>32</td><td>33</td><td>34</td></tr></table> )\n\n**Alternative ii**\n\nLet $x$ be the number in the top-left corner of the rectangle. If the rectangle is $2 \\times 8$, then its numbers are as shown.\n\n![]( <table><tr><td>$x$</td><td>$x+1$</td><td>$x+2$</td><td>$x+3$</td><td>$x+4$</td><td>$x+5$</td><td>$x+6$</td><td>$x+7$</td></tr><tr><td>$x+10$</td><td>$x+11$</td><td>$x+12$</td><td>$x+13$</td><td>$x+14$</td><td>$x+15$</td><td>$x+16$</td><td>$x+17$</td></tr></table> )\n\nThus the value of the rectangle is $16x + 136$, which is not divisible by 16.\n\nSimilarly, the value of an $8 \\times 2$ 16-chunk is $16x + 568$ and the value of a $4 \\times 4$ 16-chunk is $16x + 264$. Neither of these is divisible by 16.\n\n**Alternative iii**\n\nSince a rectangular 16-chunk has an even number of rows and an even number of columns, it has 4 central squares. Let $a$ and $b$ be the smallest and largest numbers amongst the 4 central numbers in the chunk. Then, for each number in the top half of the chunk, there is a unique number in the bottom half such that their sum is $a+b$. So the value of the 16-chunk is $8(a+b)$. Since one of $a$ and $b$ is odd and the other is even, $a+b$ is odd. So no rectangular 16-chunk has a value divisible by 16.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 12957,
    "subject": "Mathematics (Olympiad)",
    "question": "Anna buys one book and one shelf on Thursday, 1st January 2015. For the next two years, she buys one book every day and one shelf on alternate Thursdays (the next shelf is bought on 15th January 2015). From Thursday, 1st January 2015 until (and including) Saturday, 31st December 2016, on how many days is it possible for Anna to put all her books on all her shelves so that there is an equal number of books on each shelf?",
    "options": [],
    "answer": "See solution",
    "solution": "2016 is a leap year, so the total number of days in the two years is $365 + 366 = 731$. We split these days by how many shelves Anna had on each day: for the first 14 days she had 1 shelf, for days 15 to 28 she had 2, and so on, up to days 729 to 731, when she had 53 shelves. For $s \\leq 52$, she had $s$ shelves from days $14s - 13$ to $14s$ inclusive.\n\nIf at some time she has $b$ books and $s$ shelves, she can split her books equally among her shelves if and only if $b$ is a multiple of $s$. Thus, we count the number of multiples of $s$ between $14s - 13$ and $14s$, for $s \\leq 52$. (We can ignore the last three days when $s = 53$ since none of 729, 730, or 731 is a multiple of 53.) Note that $s$ dividing $14s - k$ is equivalent to $s$ dividing $k$, where $k$ ranges from 0 to 13. Counting the number of such $k$ for a given $s$:\n\nNumber of $0 \\leq k \\leq 13$ which are multiples of $s$:\n\n![](images/British_Booklet_2016_p3_data_5a3fe50110.png)\n\nNote that there is one value of $k$ (namely $k = 0$) which all $s$ divide, which is why the 1 continues all the way down to $s = 52$. Summing the bottom row, we find that the answer is\n\n$$\n14 + 7 + 5 + 4 + 3 + 3 + (2 \\times 7) + (1 \\times 39) = 89.\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 12958,
    "subject": "Mathematics (Olympiad)",
    "question": "Given a set $A = \\{1, 2, \\dots, 4044\\}$. One colors 2022 numbers of them by white and the rest by black. For each $i \\in A$, denote the weight of $i$ by the sum of the number of white numbers less than $i$ and the number of black numbers greater than $i$. For every positive integer $m$, find all positive integers $k$ such that there exists a way to color the numbers which can get exactly $k$ numbers having weight $m$.",
    "options": [],
    "answer": "See solution",
    "solution": "Call a natural number $i$ *good* if its weight is $m$. We will prove the following claim.\n\n**Claim.** Consider a positive integer $i \\le 4044$.\n\n(a) If there are more black numbers than white from $1$ to $i - 1$, then there exists a black number $j$ such that the numbers of black and white numbers from $j + 1$ to $i - 1$ are equal.\n\n(b) If there are more white numbers than black from $i + 1$ to $4044$, then there exists a white number $j$ such that the numbers of black and white numbers from $i + 1$ to $j - 1$ are equal.\n\n*Proof.* Clearly (a) and (b) are similar, so we only need to prove (b). Denote $f(k)$ to be the difference between the numbers of black and white numbers from $i + 1$ to $i + k - 1$. It is clear that $f(0) = 0$ and\n\n$$\n|f(x) - f(x + 1)| = 1.\n$$\n\nNow we need to show that there exists $k$ such that $i + k$ is white and $f(k) = 0$. If $i + 1$ is white and $f(1) = 0$, then we can assume $i + 1$ is black, so $f(2) = 1 > 0$. Because the number of white numbers from $i + 1$ to $4044$ is more than the black numbers, $f(4044 - i) \\le 0$. Hence,\n\n$$\nf(2) = 1 > 0 \\ge f(4044 - i).\n$$\n\nWe will show that there exists $j$ such that $f(j) = 0$. If there exists the smallest natural number $t$ such that $f(t) < 0$, note that\n\n$$\n|f(t - 1) - f(t)| = 1\n$$\n\nso $f(t-1) = f(t)+1$ and $f(t-1) \\ge 0$, hence $f(t-1) = 0$. Otherwise, if $f(t) \\ge 0$ for all $t$, then $f(4044 - i) = 0$, which means there always exists a number $j$ with that property.\n\nAssume that for all $j$ with $f(j) = 0$, the number $j+i$ is always black, which means $f(j + 1) = 1$. Similarly, if there exists the smallest number $t$ such that $f(t) < 0$, then $f(t - 1) = 0$, hence $f(t) \\ge 0$ for all $t$. Thus, $f(4044 - i) = 0$, which means the numbers of black and white numbers from $i + 1$ to $4043$ are equal, so $4044$ is colored white. Otherwise, if there exists $j$ such that $f(j) = 0$ and $j + i$ is white, then $j$ satisfies the above condition. $\\square$\n\nBack to the problem, assume that $i < i'$ are two consecutive good numbers and have the same color white. We observe that there are",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 12959,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all triples $\\left(a, b, p\\right)$ of positive integers with $p$ prime such that\n\n$$\na^p = b! + p.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "There are only two such triples: $(2, 2, 2)$ and $(3, 4, 3)$. It is straightforward to check them. We shall prove that no other triples exist. Clearly, $a > 1$. Consider three situations as follows.\n\n1. $a < p$. If $a \\leq b$, then $a \\mid (a^p - b!) = p$, contradicting the assumption $1 < a < p$; if $a > b$, then $b! \\leq a! < a^p - p$, the second inequality only requiring $p > a > 1$.\n\n2. $a > p$. Then $b! = a^p - p > p^p - p \\geq p!$, $b > p$, and $a^p = b! + p$ is a multiple of $p$. As $b! = a^p - p$, we have $p \\mid b$, and $b < 2p$. If $a < p^2$, then $a/p$ divides $a^p$ and $b!$, hence divides $p$ as well, contradicting $1 < a/p < p$; if $a \\geq p^2$, then it is contradicting $a^p \\geq (p^2)^p > (2p-1)! + p \\geq b! + p$.\n\n3. $a = p$. Then $b! = p^p - p$. Try $p = 2, 3, 5$ to get the two triples $(2, 2, 2)$ and $(3, 4, 3)$. Now assume $p \\geq 7$. From $b! = p^p - p > p!$, it follows $b \\geq p + 1$, and further\n\n$$\n\\begin{align*}\nv_2((p+1)!) &\\leq v_2(b!) \\\\\n&= v_2(p^{p-1}-1) = 2v_2(p-1) + v_2(p+1)-1 \\\\\n&= v_2\\left(\\frac{p-1}{2} \\cdot (p-1) \\cdot (p+1)\\right).\n\\end{align*}\n$$\n\nSince $\\frac{p-1}{2}$, $p-1$, $p+1$ are distinct factors of $(p+1)!$ and $p+1 \\geq 8$, there are four or more even numbers among $1, 2, \\dots, p+1$, which is impossible.\n\nSecond approach for $a = p \\geq 5$: according to Zsigmondy's theorem, there exists a prime $q$ that divides $p^{p-1}-1$ but not $p^k-1$ for any $k < p-1$.\n\nThus, $p \\neq q$, and $q \\equiv 1 \\pmod{p-1}$. We must have $b \\geq 2p-1$, yet\n\n$$\nb! \\geq (2p-1)! > (2p-1) \\cdot (2p-2) \\cdots (p+1) \\cdot p > p^p > p^p - p\n$$\n\nleads to a contradiction.\n\nThird approach for $a = p \\geq 5$: as $b > p \\geq 5$, the required equation modulo $(p+1)^2$ gives\n\n$$\n\\begin{aligned}\np^p - p &= (p+1-1)^p - p \\\\\n&\\equiv p \\cdot (p+1)(-1)^{p-1} + (-1)^p - p \\\\\n&= p^2 - 1 \\not\\equiv 0 \\pmod{(p+1)^2}.\n\\end{aligned}\n$$\n\nHowever, as $p \\geq 5$, $2, \\frac{p+1}{2} < p$ are distinct, and $(p+1) \\mid p!$. It follows that\n\n$$\n(p+1)^2 \\mid (p+1)!\n$$\n\nwhich is contrary to $(p+1)^2 \\nmid b!$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 12960,
    "subject": "Mathematics (Olympiad)",
    "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": "See solution",
    "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 $\\frac{3}{10} \\div \\frac{1}{2} = \\frac{3}{5} = 60\\%$ of all girls.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 12961,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $ABC$ be an acute-angled scalene triangle with $AB < AC$, inscribed in the circle $c(O, R)$. The circle $c_1(B, AB)$ intersects the side $AC$ at point $K$ and the circle $c$ at point $E$. The line $KE$ intersects the circle $c$ also at point $F$. The line $BO$ intersects $KE$ at point $L$ and $AC$ at point $M$. Finally, the line $AE$ intersects $BF$ at point $D$. Prove that the polygons $DLMF$ and $BDKME$ are cyclic.",
    "options": [],
    "answer": "See solution",
    "solution": "In the circle $c$, the chords $AB$ and $BE$ are equal and hence\n\n$$\n\\hat{A}_1 = \\hat{F}_1 = \\hat{F}_2 = \\hat{E}_1 = \\hat{C},\n$$\n\n![](images/Hellenic_booklet_2013_p10_data_a06cec7a1d.png)\n\nbecause in the circle $c$ the above angles correspond to the equal arcs $AB$ and $BE$. $OB$ is the line of the centers of the circles $c$ and $c_1$, and so it is the perpendicular bisector of the common chord $AE$, i.e., $AD \\perp BM$. We will prove that $BF \\perp AC$, whence point $D$ will be the orthocenter of the triangle $ABM$.\n\nMoreover, in the triangle $BKF$ we have: $\\hat{K}_1 = \\hat{F}_2 + \\hat{B}_2$. (1)\n\nThe triangle $BKE$ is isosceles ($BE = BK$, as radii of the circle $c_1$), whence: $\\hat{K}_1 = \\hat{E}_3 = \\hat{E}_1 + \\hat{E}_2$. (2)\n\nFrom the cyclic quadrilateral $ABEF$ we have:\n\n$\\hat{B}_1 = \\hat{E}_2$ (3)\n\nFrom relations (1), (2), and (3) (taking into account the equalities $\\hat{F}_2 = \\hat{E}_1 = \\hat{C}$) we find: $\\hat{B}_1 = \\hat{B}_2$ (4).\n\nFrom the equalities $\\hat{B}_1 = \\hat{B}_2$ and $\\hat{F}_1 = \\hat{F}_2 = \\hat{C}$, we conclude that $\\triangle ABF$ and $\\triangle KBF$ are equal. Hence $BF$ is the perpendicular bisector of $AK$. Since $AD \\perp BM$, we conclude that $D$ is the orthocenter of the triangle $ABM$. Therefore, we have $\\hat{M}_1 = \\hat{A}_1 = 90^\\circ - MBA$ and in combination with $\\hat{A}_1 = \\hat{F}_2 = \\hat{C}$ we have that $\\hat{M}_1 = \\hat{F}_2$, whence the quadrilateral $DLMF$ is cyclic. Since $\\hat{B}_2 = \\hat{E}_2$, the quadrilateral $DKEB$ is cyclic and since $\\hat{E}_1 = \\hat{M}_1$, the quadrilateral $DBEM$ is cyclic. Therefore, the polygon $BDKME$ is cyclic.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 12962,
    "subject": "Mathematics (Olympiad)",
    "question": "We want to find the solutions of the equation\n\n$$\nx^2 = p^3 - 4p + 9\n$$\n\nwhere $p$ is a prime number and $x$ is a nonnegative integer.",
    "options": [],
    "answer": "See solution",
    "solution": "Since $x^2 \\equiv 9 \\pmod{p}$, we have $x = kp \\pm 3$ where $k$ is an integer. Then\n$$\n(kp \\pm 3)^2 = p^3 - 4p + 9 \\implies k^2p^2 \\pm 6kp + 9 = p^3 - 4p + 9\n$$\nwhich simplifies to\n$$\nk^2p^2 \\pm 6kp = p^3 - 4p\n$$\nso $p \\mid 6k \\pm 4$. If $p \\neq 2$, then $p \\mid 3k \\pm 2 \\implies p \\leq 3k + 2 \\implies \\frac{p-2}{3} \\leq k$. Thus,\n$$\n\\frac{p^2 - 2p - 9}{3} \\leq pk - 3 \\leq x\n$$\n\n* If $x \\leq \\frac{p^2}{4}$, then\n$$\n\\frac{p^2 - 2p - 9}{3} \\leq \\frac{p^2}{4} \\implies p \\leq 8 + \\frac{36}{p} \\implies p \\leq 11\n$$\n\n* If $x > \\frac{p^2}{4}$, then $x^2 = p^3 - 4p + 9 > \\frac{p^4}{16}$, so $p < 16 - \\frac{16(4p-9)}{p^3} \\implies p \\leq 13$.\n\nFinally, for $p \\leq 13$, the only solutions are $(p, x) = (2, 3)$, $(7, 18)$, and $(11, 36)$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 12963,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all quadruples $ (a, b, c, d) $ of non-negative integers such that\n\n$$ ab = 2(1 + cd) $$\n\nand there exists a non-degenerate triangle with sides of length $ a - c $, $ b - d $, and $ c + d $.",
    "options": [],
    "answer": "See solution",
    "solution": "Note that $a > c$ and $b > d$, as $a - c$ and $b - d$ are sides of a non-degenerate triangle. So $a \\geq c + 1$ and $b \\geq d + 1$, as they are integers.\n\nConsider two cases: $a > 2c$ and $a \\leq 2c$.\n\n**Case 1:** $a > 2c$\n\nThen $ab > 2bc \\geq 2c \\cdot (d+1) = 2cd + 2c$. We also have $ab = 2 + 2cd$, so $2c < 2$, and therefore $c = 0$. We deduce that $ab = 2$ and that there exists a non-degenerate triangle with sides $a$, $b-d$, and $d$. Therefore $d \\geq 1$ and $b > d$, so $b \\geq 2$. From $ab = 2$ it follows that $a = 1$ and $b = 2$, and therefore also $d = 1$. Note that there exists a non-degenerate triangle with sides 1, 1, and 1, so the quadruple $(1, 2, 0, 1)$ is a solution.\n\n**Case 2:** $a \\leq 2c$\n\nBy the triangle inequality, we have $(a-c) + (b-d) > c+d$, so $a+b > 2(c+d)$. As $a \\leq 2c$, it follows that $b > 2d$. As $a \\geq c+1$, we have $ab > (c+1) \\cdot 2d = 2cd + 2d$. On the other hand, we have $ab = 2+2cd$, so $2d < 2$, and therefore $d = 0$. Analogously to the previous case, we deduce that the only other solution is $(2, 1, 1, 0)$.\n\nTherefore, the only solutions are the quadruples $(1, 2, 0, 1)$ and $(2, 1, 1, 0)$. $\\Box$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 12964,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all triples of positive integers $a, b, c$ for which the product\n\n$$\n(a+b)(b+c)(c+a)(a+b+c+2036)\n$$\n\nis equal to a power of a prime number with an integer exponent.",
    "options": [],
    "answer": "See solution",
    "solution": "First, note that at least one of the numbers $a+b$, $a+c$, and $b+c$ must be even, since two of $a$, $b$, $c$ have the same parity, making their sum even.\n\nIf the product is a power of a prime $p$, then each of the four factors must be a power of $p$. Since one of the first three factors is even, $p = 2$. Thus, each factor is a power of two greater than $1$ (since $a, b, c > 0$), so each factor is even.\n\nThe numbers $a+b$, $a+c$, $b+c$ are all even if and only if $a, b, c$ have the same parity. Since $a+b+c+2036$ is even, $a, b, c$ must all be even. Let $a = 2a_1$, $b = 2b_1$, $c = 2c_1$ with $a_1, b_1, c_1 > 0$. Then:\n\n$$\n(a+b)(a+c)(b+c)(a+b+c+2036) = 2^4(a_1+b_1)(a_1+c_1)(b_1+c_1)(a_1+b_1+c_1+1018).\n$$\n\nThe product $(a_1+b_1)(a_1+c_1)(b_1+c_1)(a_1+b_1+c_1+1018)$ must also be a power of two. Thus, $a_1, b_1, c_1$ must all be even. Let $a_1 = 2a_2$, $b_1 = 2b_2$, $c_1 = 2c_2$:\n\n$$\n(a+b)(a+c)(b+c)(a+b+c+2036) = 2^8(a_2+b_2)(a_2+c_2)(b_2+c_2)(a_2+b_2+c_2+509).\n$$\n\nNow, $a_2, b_2, c_2$ must have the same parity. Since $509$ is odd, $a_2, b_2, c_2$ must be odd. The triple $a_2 = b_2 = c_2 = 1$ works, since $3 + 509 = 512 = 2^9$ is a power of two. Thus, $a = b = c = 4$ is a solution.\n\nSuppose at least one of $a_2, b_2, c_2$ is greater than $1$, say $c_2 > 1$. Then $a_2 + c_2$ is a power of two greater than $2$, so divisible by $4$. This means one of $a_2, c_2$ is $1 \bmod 4$, the other $3 \bmod 4$. The third, $b_2$, matches one of these residues, so $b_2$ plus that number is $2 \bmod 4$, i.e., $2$, which is $2^1$. Thus, $b_2 = 1$ and $a_2 = 1$ (since $c_2 > 1$). But then $c_2 \bmod 4 = 3$, so $a_2 + b_2 + c_2 + 509 \bmod 4 = 1 + 1 + 3 + 1 = 6 \bmod 4 = 2$, which is not a power of two except $2$, but $a_2 + b_2 + c_2 + 509 > 2$. Contradiction.\n\n**Conclusion:** The only solution is $(a, b, c) = (4, 4, 4)$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 12965,
    "subject": "Mathematics (Olympiad)",
    "question": "In an acute triangle $ABC$ with $AB > AC$, the bisector of angle $BAC$ and side $BC$ intersect at point $D$. Two points $E$ and $F$ lie on sides $AB$ and $AC$, respectively, such that $B$, $C$, $F$, and $E$ are concyclic. Prove that the circumcenter of triangle $DEF$ coincides with the incenter of triangle $ABC$ if and only if $BE + CF = BC$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $I$ be the incenter 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 $\\angle ABC$ and $CI$ bisects $\\angle ACB$, $\\triangle BIK$ and $\\triangle BIE$ are reflections with respect to $BI$, and $\\triangle CIK$ and $\\triangle CIF$ are reflections with respect to $CI$. We have $\\angle BEI = \\angle BKI = \\pi - \\angle CKI = \\pi - \\angle CFI = \\angle AFI$. 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![](images/Mathematical_Olympiad_in_China_2011-2014_p123_data_5c282773b3.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.\n\n**Necessity:** Suppose $I$ is the circumcenter of $\\triangle DEF$. Since $B$, $E$, $F$, $C$ are concyclic, $AE \\cdot AB = AF \\cdot AC$, and $AB > AC$, we have $AE < AF$. Therefore, the bisector of $\\angle EAF$ and the perpendicular bisector of $EF$ meet at $I$, which lies on the circumcircle of $\\triangle AEF$.\n\nSince $BI$ bisects $\\angle ABC$, let $K$ be the symmetric point of $E$ with respect to $BI$, then we have $\\angle BKI = \\angle BEI = \\angle AFI > \\angle ACI = \\angle BCI$. Therefore, $K$ lies on $BC$, $\\angle IKC = \\angle IFC$, $\\angle ICK = \\angle ICF$, and $\\triangle IKC \\cong \\triangle IFC$.\n\nHence $BC = BK + CK = BE + CF$. $\\Box$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 12966,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $n \\ge 3$ be an odd positive integer, and consider an $n \\times n$ grid containing $n^2$ cells. Dionysus colours each cell either red or blue. A frog can hop directly between two cells if they have the same colour and share at least one vertex. Xanthias views the colouring, and wants to place frogs on $k$ of the cells so that any cell can be reached by a frog in a finite number of hops. Find the least value of $k$ such that Xanthias can always be successful regardless of the colouring chosen by Dionysus.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $G$ be the graph whose vertices are all $(n+1)^2$ vertices of the grid, where two vertices are adjacent if and only if they are adjacent in the grid and the two cells on either side of the corresponding edge have different colours.\n\nThe connected components of $G$, excluding isolated vertices, are precisely the boundaries between pairs of monochromatic regions, each of which can be covered by a single frog. Each time we add one of these components in the grid, it creates exactly one new monochromatic region. So the number of frogs required is one more than the number of such components of $G$.\n\nIt is easy to check that every corner vertex of the grid has degree $0$, every boundary vertex of the grid has degree $0$ or $1$, and every internal vertex of the grid has degree $0$, $2$, or $4$. It is also easy to see that every component of $G$ which is not an isolated vertex must contain at least four vertices unless it is the boundary of a single corner of the grid, in which case it contains only three vertices.\n\nWriting $N$ for the number of components which are not isolated vertices, we see that in total they contain at least $4N-4$ vertices (as at most four of them contain $3$ vertices and all others contain $4$ vertices). Since we also have at least $4$ components which are isolated vertices, then $4N = (4N-4)+4 \\le (n+1)^2$. Thus $N \\le \\frac{(n+1)^2}{4}$ and therefore the minimal number of frogs required is $\\frac{(n+1)^2}{4} + 1$.\n\nThis bound for $n = 2m + 1$ is achieved by putting coordinates $(x, y)$ with $x, y \\in \\{0, 1, \\dots, 2m\\}$ in the cells and colouring red all cells both of whose coordinates are even, and blue all other cells. An example for $n = 9$ is shown below.\n\n![](images/BMO_2022_shortlist_p23_data_10cee8b8d1.png)",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 12967,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $x = 1/a$, $y = 1/b$, $z = 1/c$, $u = 1/d$. Suppose $x + y + z + u = 1$ and $xyzu \\ne 0$. Prove that\n\n$$\n\\frac{xy(x+y)}{x^2-xy+y^2} + \\frac{yz(y+z)}{y^2-yz+z^2} + \\frac{zu(z+u)}{z^2-zu+u^2} + \\frac{ux(u+x)}{u^2-ux+x^2} \\le 2.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "Since for $x \\ne 0$ and $y \\ne 0$ we have $x^2-xy+y^2 = x^2-xy+\\frac{y^2}{4}+\\frac{3y^2}{4} = (x-\\frac{y}{2})^2+\\frac{3y^2}{4} > 0$ and $0 \\le (x-y)^2 = x^2-2xy+y^2 = (x^2-xy+y^2)-xy$. So $x^2-xy+y^2 \\ge xy$ and $\\frac{xy}{ux} \\le 1$. Similarly, $y^2-yz+z^2 \\le 1$, $\\frac{zu}{z^2-zu+u^2} \\le 1$, $\\frac{u^2-ux+x^2}{ux} \\le 1$. Therefore\n\n$$\n\\frac{xy(x+y)}{x^2-xy+y^2} + \\frac{yz(y+z)}{y^2-yz+z^2} + \\frac{zu(z+u)}{z^2-zu+u^2} + \\frac{ux(u+x)}{u^2-ux+x^2} \\le (x+y) + (y+z) + (z+u) + (u+x) = 2(x+y+z+u).\n$$\n\nSince $x+y+z+u=1$, we obtain the required inequality. Note that equality occurs when $x = y = z = u = 1/4$, i.e., when $a = b = c = d = 4$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 12968,
    "subject": "Mathematics (Olympiad)",
    "question": "Рассмотрим граф $G$, вершинами которого являются города, и две вершины соединены ребром, если между городами есть авиалиния. Известно, что граф связан, но при удалении всех рёбер любого нечётного цикла это условие нарушается. Докажите, что вершины графа можно правильно раскрасить в 4 цвета.",
    "options": [],
    "answer": "See solution",
    "solution": "Рассмотрим следующую лемму:\n\n*Лемма.* Пусть в графе нет циклов нечётной длины. Тогда его вершины можно правильно раскрасить двумя красками.\n\n*Доказательство леммы.* Достаточно доказать для связного графа. Зафиксируем вершину $A$ и покрасим все вершины, находящиеся на нечётном расстоянии от $A$, в красный цвет, а остальные — в синий. Если есть ребро между двумя вершинами одного цвета, то существует нечётный цикл, что противоречит условию.\n\nТеперь перейдём к задаче. Пусть в графе есть цикл; удалим одно из рёбер этого цикла. Граф останется связным. Продолжим процесс, пока не останется циклов. Пусть $V$ — множество удалённых рёбер, $W$ — оставшихся. Оставшийся граф связан, и не существует нечётного цикла, все рёбра которого в $V$.\n\nРассмотрим графы $G_V$ и $G_W$ с теми же вершинами, но рёбрами из $V$ и $W$ соответственно. В $G_W$ нет циклов, значит, его вершины можно раскрасить в цвета 0 и 1. В $G_V$ нет нечётных циклов, значит, его вершины можно раскрасить в цвета 0 и 2. Присвоим каждой вершине сумму её цветов в этих раскрасках. Если две вершины соединены ребром в $G$, то они соединены ребром в одном из $G_V$ или $G_W$, и их цвета различны. Значит, получена правильная раскраска в 4 цвета.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 12969,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $A = (a_{jk})$ be a $10 \\times 10$ array of positive real numbers such that the sum of the numbers in each row as well as in each column is $1$. Show that there exist $j < k$ and $l < m$ such that\n\n$$\na_{jl}a_{km} + a_{jm}a_{kl} \\geq \\frac{1}{50}$$",
    "options": [],
    "answer": "See solution",
    "solution": "Consider the first column. Let $M$ be the maximum value of the entries in this column. Then $M \\geq \\frac{1}{10}$ by the pigeonhole principle. Writing the first column as $a_{1,1}, a_{2,1}, \\dots, a_{10,1}$, let $M = a_{j,1}$ for some $j$.\n\nConsider the sum\n\n$$\n\\sum_{l=2}^{10} \\left(a_{j,1}a_{k,l} + a_{j,l}a_{k,1}\\right),\n$$\n\nwhere $k \\neq j$. This is equal to\n\n$$\na_{j,1} \\left( \\sum_{l=2}^{10} a_{k,l} \\right) + a_{k,1} \\left( \\sum_{l=2}^{10} a_{j,l} \\right) = a_{j,1} (1 - a_{k,1}) + a_{k,1} (1 - a_{j,1}).\n$$\n\nVarying $k$ from $1$ to $10$, $k \\neq j$, we get\n\n$$\n\\begin{align*}\n\\sum_{k \\neq j} \\sum_{l=2}^{10} \\left(a_{j,1}a_{k,l} + a_{j,l}a_{k,1}\\right) &= \\sum_{k \\neq j} a_{j,1}(1-a_{k,1}) + a_{k,1}(1-a_{j,1}) \\\\\n&= a_{j,1} \\sum_{k \\neq j} (1-a_{k,1}) + (1-a_{j,1}) \\sum_{k \\neq j} a_{k,1} \\\\\n&= a_{j,1}(8+a_{j,1}) + (1-a_{j,1})^2 \\\\\n&= M^2 + 8M + (1-M)^2 \\\\\n&= 2M^2 + 6M + 1 \\geq 2 \\left( \\frac{1}{100} + \\frac{3}{10} \\right) + 1 = \\frac{81}{50}.\n\\end{align*}\n$$\n\nThere are $9 \\times 9 = 81$ summands in the above sum. Hence, we can find $k, l$ such that\n\n$$\na_{j,1}a_{k,l} + a_{j,l}a_{k,1} \\geq \\frac{1}{50}.\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 12970,
    "subject": "Mathematics (Olympiad)",
    "question": "The company is led by at most 10 directors. Each director has a set of 3 keys, each key opens a different lock (there are 6 locks in total), and no two directors have the same set of keys. No two directors together can open all the locks. What is the maximum number of directors?",
    "options": [],
    "answer": "See solution",
    "solution": "The company is led by at most *10* directors.\n\nDenote the locks and the corresponding keys with numbers 1 through 6. Each director has a set of 3 from 3 different locks and no two directors have the same sets. Therefore, we first count how many different sets of 3 keys there exist. For a specific set of keys, we have 6 locks to choose from and we need to choose 3 of them. Hence, we are looking for the number of combinations of 3 elements out of 6 elements, which is $\\binom{6}{3} = 20$ (we may also count them by writing them all down).\n\nTo each set of 3 keys belongs a *complementary* set of 3 keys, i.e., the set of keys for the other 3 locks. For example, to the set of keys $\\{2, 3, 5\\}$ belongs the complementary set of keys $\\{1, 4, 6\\}$. We partition the set of all different sets of keys into 10 pairs of complementary sets of keys. Since every such pair of keys unlocks all the locks, no two directors can have complementary sets of keys. So there are at most 10 directors, since from each pair of complementary sets of keys only one set can belong to some director.\n\nLet's prove that 10 directors can indeed lead this company, i.e., there exist 10 sets of keys satisfying the conditions of the problem. All we have to do is to choose one set of keys from each pair of complementary sets of keys (for example, the one that contains the key 1). Thus, the 10 directors can have the following sets of keys: $\\{1, 2, 3\\}$, $\\{1, 2, 4\\}$, $\\{1, 2, 5\\}$, $\\{1, 2, 6\\}$, $\\{1, 3, 4\\}$, $\\{1, 3, 5\\}$, $\\{1, 3, 6\\}$, $\\{1, 4, 5\\}$, $\\{1, 4, 6\\}$, $\\{1, 5, 6\\}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 12971,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $N$ be a positive integer. The sequence $x_1, x_2, \\dots$ of non-negative reals is defined by\n\n$$\nx_n^2 = \\sum_{i=1}^{n-1} \\sqrt{x_i x_{n-i}}\n$$\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": "See solution",
    "solution": "Applying the AM-GM inequality to each term on the right-hand side, we get\n\n$$\nx_n^2 \\le \\sum_{i=1}^{n-1} x_i, \\quad n > N.\n$$\n\nLet us set $x_n = a n + \\Delta(n)$, where $a$ is a constant to be determined. We want to pick $a$ so that $\\Delta(n)$ grows slower than $n$. For $n > N$,\n\n$$\na^2 n^2 + 2a n \\Delta(n) + \\Delta(n)^2 \\le \\frac{a(n-1)n}{2} + \\sum_{i=1}^{n-1} \\Delta(i)\n$$\n\nwhich gives\n\n$$\n2a n \\Delta(n) \\le \\left( \\frac{a(n-1)n}{2} - a^2 n^2 \\right) - \\Delta(n)^2 + \\sum_{i=1}^{n-1} \\Delta(i) \\quad (1).\n$$\n\nTo make the right side degree $n$, set $a = 1/2$. Substituting,\n\n$$\n\\Delta(n) \\le -\\frac{1}{4} - \\frac{\\Delta(n)^2}{n} + \\frac{1}{n} \\sum_{i=1}^{n-1} \\Delta(i).\n$$\n\nNow,\n\n$$\n\\Delta(n) \\le \\frac{1}{n} \\sum_{i=1}^{n-1} \\Delta(i), \\quad \\forall n > N\n$$\n\nwhich yields\n\n$$\n\\Delta(n) \\le \\max\\{\\Delta(i) : i < n\\},\n$$\n\nso $\\Delta(n) \\le \\max\\{\\Delta(i) : i \\le N\\}$ for any $n > N$. Thus, $x_n \\le n/2 + c$ for some constant $c$.\n\n**Sharper estimate.**\n\nWe can improve this by writing\n\n$$\n\\Delta(n) \\le -\\frac{1}{4} + \\frac{1}{n} \\sum_{i=1}^{n-1} \\Delta(i), \\quad n > N\n$$\n\nDefine $\\Delta'(n)$ by\n\n$$\n\\Delta'(n) = -\\frac{1}{4} + \\frac{1}{n} \\sum_{i=1}^{n-1} \\Delta'(i), \\quad n > N\n$$\n\nand $\\Delta'(n) = \\Delta'(n)$ for $n \\le N$. Clearly $\\Delta(n) \\le \\Delta'(n)$ for all $n$. We have\n\n$$\n\\Delta'(n+1) = -\\frac{1}{4} + \\frac{1}{n+1} \\left( \\sum_{i=1}^{n-1} \\Delta(i) - \\frac{1}{4} + \\frac{1}{n} \\sum_{i=1}^{n-1} \\Delta(i) \\right), \\quad n > N\n$$\n\nSubtracting gives\n\n$$\n\\Delta'(n+1) - \\Delta'(n) = \\frac{-1}{4(n+1)}, \\quad \\forall n > N.\n$$\n\nSumming up,\n\n$$\n\\Delta'(n) = \\Delta'(N + 1) - \\frac{1}{4} \\sum_{k=N+2}^{n} \\frac{1}{k}, \\quad n \\ge N + 2.\n$$\n\nUsing the harmonic series,\n\n$$\n\\Delta'(n) \\le -\\frac{\\ln n}{4} + c\n$$\n\nfor some constant $c$. Therefore,\n\n$$\nx_n \\le \\frac{n}{2} - \\frac{\\ln n}{4} + c, \\quad n \\in \\mathbb{N},\n$$\n\nwhere $c$ is a constant.\n\n**Remark.** The precise growth rate is $x_n = \\frac{\\pi}{8} n + o(n)$. See [this link](https://dgrozev.wordpress.com/2024/02/14/growth-rate-of-a-sequence-bulgarian-2024-mo-regional-round/).",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 12972,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $G$ be a graph of diameter 2 on 2019 vertices. 62 cops and 1 robber are placed on the vertices of graph $G$. Everybody sees all others, and they move in turn. Each move, each person can stay on their vertex or go to any adjacent vertex (each cop moves independently, but they coordinate their actions during the game). The robber makes the first move, then the cops move (simultaneously), then the robber moves again, then the cops, etc. The robber is caught if he is on a vertex occupied by a cop. Prove that the cops can catch the robber.",
    "options": [],
    "answer": "See solution",
    "solution": "The numbers 62 and 2019 have the following property:\n\n$$\n64 + 63 + \\dots + 3 = \\frac{64 \\cdot 65}{2} - 3 = 2077 > 2019.\n$$\n\nLet the robber be at vertex $A$ before the cops move.\n\nIf $\\deg A \\leq 62$, the cops can catch the robber in one, two, or three moves (due to the diameter 2 property): denote $B_1, B_2, \\dots$ as the vertices adjacent to $A$. In the first move, the $i$-th cop goes towards $B_i$ (either directly to $B_i$ if possible, or to a vertex adjacent to $B_i$ and then to $B_i$ in the next move). If the robber leaves $A$ and goes to some $B_k$, he will be caught immediately or in the next move; otherwise, the cops will occupy all $B_i$ in their second move and then catch the robber.\n\nIf $\\deg A \\geq 63$, let one cop go to $A$ and always stay there, guarding $A$ and its neighbors. This reduces the robber's safe area by at least 64 vertices.\n\nNow, if the robber's vertex has safe degree at most 61, the remaining active cops can again catch the robber in one, two, or three moves. Otherwise, let one active cop go to a vertex of degree at least 62 and always guard it and its neighbors. The safe area for the robber is reduced again by at least 63 vertices.\n\nContinuing this process, after the 62$^{\\text{nd}}$ step, the safe area is empty.",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 12973,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all real numbers $p$ such that the system\n\n$$\nx^2 + (p-1)x + p \\le 0\n$$\n\n$$\nx^2 - (p-1)x + p \\le 0\n$$\n\nof inequalities has at least one solution $x \\in \\mathbb{R}$.",
    "options": [],
    "answer": "See solution",
    "solution": "If $p \\le 0$, then $x = 0$ is clearly a solution. If $p > 0$, then summing up the inequalities gives\n\n$$\n2x^2 + 2p \\le 0\n$$\nwhich does not hold for any real $x$.\n\n**Answer:** $p \\in (-\\infty, 0]$.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 12974,
    "subject": "Mathematics (Olympiad)",
    "question": "給定一大於 $5$ 的正整數 $n$。試求出所有的實數 $a$，使得存在非負實數 $x_1, x_2, \\dots, x_n$ 滿足\n\n$$\n\\sum_{k=1}^{n} k x_{k} = a, \\quad \\sum_{k=1}^{n} k^{3} x_{k} = a^{2}, \\quad \\sum_{k=1}^{n} k^{5} x_{k} = a^{3}.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "假設 $\\{x_i\\}$ 滿足題設，則\n\n$$\n\\sum_{k=1}^{n} k x_{k} = a, \\quad \\sum_{k=1}^{n} k^{3} x_{k} = a^{2}, \\quad \\sum_{k=1}^{n} k^{5} x_{k} = a^{3}.\n$$\n\n由柯西不等式得\n\n$$\n\\begin{aligned}\na a^3 &= \\left(\\sum_{k=1}^{n} k x_k\\right) \\left(\\sum_{k=1}^{n} k^5 x_k\\right) \\\\\n&\\geq \\left(\\sum_{k=1}^{n} k^3 x_k\\right)^2 = a^4.\n\\end{aligned}\n$$\n\n因此上述不等式等號成立。故應有：\n\n1. 若 $x_k$ 均不為 $0$，則\n\n$$\n\\frac{1}{k^4} = \\frac{k x_k}{k^5 x_k} \\text{ 為常數}\n$$\n\n由於 $k$ 之值是不固定的，這顯然是不可能的。\n\n2. 若 $x_k$ 中有 $0$，不妨假設 $x_1 = 0$。則\n\n$$\n\\sum_{k=2}^{n} k x_{k} = a, \\quad \\sum_{k=2}^{n} k^{3} x_{k} = a^{2}, \\quad \\sum_{k=2}^{n} k^{5} x_{k} = a^{3}.\n$$\n\n對於上述條件，再利用柯西不等式，可得 $x_2, x_3, \\dots, x_n$ 中有 $0$。\n\n重複上述步驟，可得如下結論：\n\n$\\{x_k\\}$ 中至多有一個非零，令其為 $x_i$（$x_i \\neq 0,\\ i$ 可以取 $1, 2, \\dots, n$），則\n\n$$\ni x_i = a, \\quad i^3 x_i = a^2, \\quad i^5 x_i = a^3,\n$$\n\n從而，$i^2 = a$。由此可知：$a$ 的所有可能的值是 $1, 2^2, 3^2, \\dots, n^2$。",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 12975,
    "subject": "Mathematics (Olympiad)",
    "question": "Let the perpendicular from $I$ to $BC$ intersect $AC$ and $BC$ at $K'$ and $L'$, respectively. The perpendicular from $K'$ to $BI$ intersects $BC$ and $BI$ at $M'$ and $H'$, respectively. Let $D$ be the projection of $M'$ onto $AC$.\n\nSuppose that $BI$ and $BK$ intersect the circumcircle of $\\triangle ABC$ at $N$ and $E$, respectively. Prove that $\\angle EFH = 90^\\circ$.\n\n![](images/IRN_booklet_2024-Final_p37_data_986026fd07.png)",
    "options": [],
    "answer": "See solution",
    "solution": "Since $AK'H'LB$ is cyclic, it follows that $\\angle H'K'L = \\angle H'BL = \\frac{\\angle ABC}{2}$ and $\\angle LK'C = \\angle ABC$, so $K'M'$ bisects the angle $LK'C$ and hence $M'L = M'D$. Thus,\n\n$$\n\\frac{M'L}{M'C} = \\frac{M'D}{M'C} = \\sin ACB\n$$\n\nand\n\n$$\n\\frac{ML}{MC} = \\frac{\\frac{BC-2BL}{2}}{MB} = \\frac{\\frac{AB}{2}}{MC} = \\frac{AB}{BC} = \\sin ACB\n$$\n\nTherefore,\n\n$$\nM' \\equiv M,\\quad K' \\equiv K,\\quad H' \\equiv H\n$$\n\nSince $I$ is the orthocenter of $KMB$, it follows that $MI$ is perpendicular to $BE$ and hence $ME$ is the reflection of $MB$ with respect to the line $MI$. It is known that $ME$ is tangent to the incircle of $\\triangle ABC$, so it suffices to prove that $\\angle EFH = 90^\\circ$.\n\nSince $MK \\perp BN$ and also $MK$ bisects $\\angle AKF$, it is easy to show that $AFNB$ is an equilateral trapezoid. Hence $\\angle NFK = \\angle BAK = \\angle KHN = 90^\\circ$. Thus $FKHN$ is cyclic. Therefore,\n\n$$\n\\begin{aligned}\n\\angle EFH &= \\angle EFN + \\angle NFH \\\\\n&= \\angle EBN + \\angle NKH \\\\\n&= \\angle EBN + \\angle BKH = 90^\\circ\n\\end{aligned}\n$$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 12976,
    "subject": "Mathematics (Olympiad)",
    "question": "Prove that if $a$, $b$, $c$ are the lengths of the sides of a triangle, then\n\n$$\n\\sqrt{\\frac{a}{-a+b+c}} + \\sqrt{\\frac{b}{a-b+c}} + \\sqrt{\\frac{c}{a+b-c}} \\ge 3.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "Since $a$, $b$, $c$ are the lengths of the sides of a triangle, the numbers $-a+b+c$, $a-b+c$, and $a+b-c$ are positive. The GM-HM inequality yields\n\n$$\n\\sqrt{\\frac{a}{-a+b+c}} = \\sqrt{1 \\cdot \\frac{a}{-a+b+c}} \\ge \\frac{2}{1+\\frac{-a+b+c}{a}} = \\frac{2a}{b+c}\n$$\nand similarly for the other two terms. So, it is enough to prove that\n\n$$\n\\frac{a}{b+c} + \\frac{b}{a+c} + \\frac{c}{a+b} \\geq \\frac{3}{2}.\n$$\n\nDenote $b+c = x$, $a+c = y$, $a+b = z$. Then\n\n$$\na = \\frac{-x + y + z}{2}, \\quad b = \\frac{x - y + z}{2}, \\quad c = \\frac{x + y - z}{2},\n$$\nand the last inequality is equivalent to\n\n$$\n\\frac{-x + y + z}{2x} + \\frac{x - y + z}{2y} + \\frac{x + y - z}{2z} \\ge \\frac{3}{2},\n$$\nthat is,\n\n$$\n\\frac{y}{x} + \\frac{z}{x} - 1 + \\frac{x}{y} + \\frac{z}{y} - 1 + \\frac{x}{z} + \\frac{y}{z} - 1 \\geq 3,\n$$\nwhich is true because $u + 1/u \\ge 2$ for every $u > 0$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 12977,
    "subject": "Mathematics (Olympiad)",
    "question": "Suppose you have $2n$ batteries, of which $n$ are charged. You need to find the minimum number of pairwise tests (testing two batteries at a time) required to guarantee that you can identify two charged batteries, regardless of which batteries are charged.",
    "options": [],
    "answer": "See solution",
    "solution": "Let us model the problem as follows: construct a graph $G$ where each battery is a vertex, and connect two batteries with an edge if and only if you **do not** test them together. The number of attempts is the number of disconnected pairs, i.e., $\\binom{2n}{2} - |E|$, where $E$ is the set of edges of $G$.\n\nTo guarantee success, every set of $n$ batteries must contain at least one tested pair; equivalently, $G$ must not contain an $n$-clique. By Turán's Theorem, the graph with the maximum number of edges without an $n$-clique is the most balanced $(n-1)$-partite graph. In this case, the partition sets of vertices of $G$ must contain $3, 3, 2, 2, \\ldots, 2$ vertices (with $n-3$ twos). Thus, the minimum number of attempts is the number of disconnected pairs, which is $2 \\cdot 3 + (n - 3) = n + 3$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 12978,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $x$ be a real number such that the numbers $x^3$ and $x^2 + x$ are rational. Prove that $x$ is rational.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $a = x^3$ and $b = x^2 + x$. Both $a$ and $b$ are rational. \n\nWe can write $x^2 + x = b$, so $x^2 + x - b = 0$. This is a quadratic equation in $x$ with rational coefficients, so its solutions are\n$$\n x = \\frac{-1 \\pm \\sqrt{1 + 4b}}{2}\n$$\nSince $x$ is real, $1 + 4b \\geq 0$ and $\\sqrt{1 + 4b}$ is real. If $1 + 4b$ is a perfect square of a rational number, then $x$ is rational. \n\nAlternatively, since $x^3 = a$ and $x^2 + x = b$, we can express $x$ in terms of $a$ and $b$:\n\n$x^3 = a$\n$x^2 + x = b \\implies x^2 = b - x$\nSo $x^3 = x(x^2) = x(b - x) = bx - x^2 = bx - (b - x) = bx - b + x = x(b + 1) - b$\nThus,\n$$\n a = x(b + 1) - b \\implies x = \\frac{a + b}{b + 1}\n$$\nSince $a$ and $b$ are rational and $b + 1 \\neq 0$, $x$ is rational.\n\nIf $b + 1 = 0$, then $x^2 + x = -1 \\implies x^2 + x + 1 = 0$, which has no real solutions. Therefore, $b + 1 \\neq 0$ and $x$ is rational.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 12979,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $A$ and $B$ be two identical convex polygons, each having area $2015$. The polygon $A$ is divided into polygons $A_1, A_2, \\dots, A_{2015}$ with positive area, and the polygon $B$ into polygons $B_1, B_2, \\dots, B_{2015}$ with positive area. The polygons $A_1, A_2, \\dots, A_{2015}, B_1, B_2, \\dots, B_{2015}$ are colored with $2015$ colors, in such a way that $A_i$ is colored differently from $A_j$ and $B_i$ is colored differently from $B_j$, for $i \\neq j$. After overlapping the polygons $A$ and $B$, we calculate the sum of the areas of the parts that have the same color.\n\nProve that there exists a coloring of the polygons for which this sum is at least $1$.",
    "options": [],
    "answer": "See solution",
    "solution": "After overlapping the polygons, we get $C_{ij} = A_i \\cap B_j$ for $i, j \\in \\{1, 2, \\dots, 2015\\}$. The polygons $B_1, B_2, \\dots, B_{2015}$ can be colored in $2015!$ ways.\n\nLet a coloring of $A_1, A_2, \\dots, A_{2015}$ be given. For an arbitrary coloring of $B_1, B_2, \\dots, B_{2015}$, which we denote by $n$, let $O_n$ be the sum of the areas of the parts from the two polygons colored with the same color.\n\nThen\n$$\nO_n = \\sum_{i,j=1}^{2015} c_{ij} P(C_{ij}),\n$$\nwhere $c_{ij} = 1$ if $A_i$ and $B_j$ are colored with the same color, and $c_{ij} = 0$ otherwise. We get that\n$$\n\\sum_{n=1}^{2015!} O_n = \\sum_{i,j=1}^{2015} d_{ij} P(C_{ij}),\n$$\nwhere $d_{ij}$ is the number of colorings of $B_1, B_2, \\dots, B_{2015}$ in which $A_i$ and $B_j$ have the same color. It is clear that $d_{ij} = 2014!$. Therefore,\n$$\n\\sum_{n=1}^{2015!} O_n = \\sum_{i,j=1}^{2015} d_{ij} P(C_{ij}) = 2014! \\cdot 2015 = 2015!.\n$$\nBy the pigeonhole principle, there exists a coloring for which $O_n \\geq 1$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 12980,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $r$, $t$, and $a$ be three positive integers such that $r \\leq t$.\n\n$$\nA = \\{1, 2, 3, \\dots, r\\}, \\quad B = \\{a + r + 1, a + r + 2, a + r + 3, \\dots, a + r + t\\}.\n$$\n\nFor any set $P$ of integers and any integer $x$, define $P + x = \\{p + x : p \\in P\\}$, called the translate of $P$ by $x$.\n\n(a) If $r = t$, prove that $\\mathbb{Z}$ (the set of all integers) is a union of a certain class of mutually disjoint translates of $A \\cup B$ if and only if $a = kr$ for some positive integer $k$.\n\n(b) If $r < t$, prove that $\\mathbb{Z}$ is a union of a certain class of mutually disjoint translates of $A \\cup B$ if and only if $a = k(r + t)$ for some positive integer $k$.",
    "options": [],
    "answer": "See solution",
    "solution": "First, observe that $A$ and $B$ are separated by $a$ numbers, so $A \\cup B$ is not a set of consecutive integers.\n\n**(a) ($\\Rightarrow$)** If $r = t$ and $a = kr$ for some $k \\in \\mathbb{N}$, then $\\mathbb{Z}$ is covered by the mutually disjoint sets $(A \\cup B) + s$, where $s$ ranges over\n$$\n\\bigcup_{m \\in \\mathbb{Z}} \\{2(k+1)rm, 2(k+1)rm + r, \\dots, 2(k+1)rm + kr\\} = 2(k+1)r\\mathbb{Z} \\cup \\{2(k+1)r\\mathbb{Z} + r\\} \\cup \\dots \\cup \\{2(k+1)r\\mathbb{Z} + kr\\}.\n$$\n\n**(b) ($\\Rightarrow$)** If $r < t$ and $a = k(r + t)$ for some $k \\in \\mathbb{N}$, then $\\mathbb{Z}$ is covered by the mutually disjoint sets $(A \\cup B) + s'$, where $s'$ ranges over $\\{(r + t)s : s \\in \\mathbb{Z}\\} = (r + t)\\mathbb{Z}$.\n\n**(a) ($\\Leftarrow$)** If $r = t$ and $\\mathbb{Z}$ is covered by mutually disjoint sets of the form $(A \\cup B) + s$, for $s$ in a fixed subset of $\\mathbb{Z}$, then any translate of $A \\cup B$ has a gap of size $a + r + 1 - (r + 1) = a$, which should be covered by mutually disjoint sets of size $r$. This proves that $a = kr$ for some $k \\in \\mathbb{N}$.\n\n**(b) ($\\Leftarrow$)** If $r < t$ and $\\mathbb{Z}$ is covered by a certain class of mutually disjoint translates of $A \\cup B$, we will prove that the translates of $A$ and $B$ must alternate in any covering of $\\mathbb{Z}$ by translates of $A \\cup B$. This would imply that $a = k(r + t)$ for some $k \\in \\mathbb{N}$. If a covering contained two consecutive translates $A + x$ and $A + x + r$ of $A$, it would contain the corresponding matching translates $B + x$ and $B + x + r$ of $B$. But\n![alt](path \"title\")",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 12981,
    "subject": "Mathematics (Olympiad)",
    "question": "In the plane rectangular coordinate system $xOy$, consider the ellipse\n\n$$\n\\Gamma : \\frac{x^2}{a^2} + \\frac{y^2}{b^2} = 1 \\quad (a > b > 0),\n$$\n\nLet $A$ be a vertex of the major axis, $B$ a vertex of the minor axis, and $F$ a focus of the ellipse. There exist two points $P$ and $Q$ on $\\Gamma$, symmetric about $O$, such that\n\n$$\n\\overrightarrow{FP} \\cdot \\overrightarrow{FQ} + \\overrightarrow{FA} \\cdot \\overrightarrow{FB} = |AB|^2.\n$$\n\nFind:\n\n1. Prove that the focal point $F$ lies on the extension of $AO$.\n2. Find the range of the eccentricity of $\\Gamma$.",
    "options": [],
    "answer": "See solution",
    "solution": "By symmetry, set $A(a, 0)$, $B(0, b)$. Let $c = |OF| = \\sqrt{a^2 - b^2}$ and eccentricity $e = \\frac{c}{a}$.\n\nLet $|OP| = r$. For the ellipse, $r \\in [b, a]$.\n\nSince $O$ is the midpoint of $PQ$,\n\n$$\n\\overrightarrow{FP} \\cdot \\overrightarrow{FQ} = (\\overrightarrow{FO} + \\overrightarrow{OP}) \\cdot (\\overrightarrow{FO} - \\overrightarrow{OP}) = \\overrightarrow{FO}^2 - \\overrightarrow{OP}^2 = c^2 - r^2.\n$$\n\n**(1) Focal point location:**\n\nSuppose $F$ is at $(c, 0)$:\n\n$$\n\\overrightarrow{FA} \\cdot \\overrightarrow{FB} = (a - c, 0) \\cdot (-c, b) = c^2 - ac < 0.\n$$\n\nThus,\n\n$$\n\\overrightarrow{FP} \\cdot \\overrightarrow{FQ} + \\overrightarrow{FA} \\cdot \\overrightarrow{FB} < c^2 - r^2 < a^2 + b^2 = |AB|^2,\n$$\n\nwhich does not satisfy the condition. Therefore, $F$ must be at $(-c, 0)$, i.e., on the extension of $AO$.\n\n**(2) Range of eccentricity:**\n\nFor $F(-c, 0)$:\n\n$$\n\\overrightarrow{FA} \\cdot \\overrightarrow{FB} = (a + c, 0) \\cdot (c, b) = c^2 + ac.\n$$\n\nSo,\n\n$$\n\\begin{aligned}\n\\overrightarrow{FP} \\cdot \\overrightarrow{FQ} + \\overrightarrow{FA} \\cdot \\overrightarrow{FB} &= (c^2 - r^2) + (c^2 + ac) \\\\\n&= 2c^2 + ac - r^2.\n\\end{aligned}\n$$\n\nSet $2c^2 + ac - r^2 = a^2 + b^2$, i.e.,\n\n$$\nr^2 = 2c^2 + ac - a^2 - b^2 = 3c^2 + ac - 2a^2.\n$$\n\nSince $r \\in [b, a]$, $r^2 \\in [a^2 - c^2, a^2]$.\n\nSo,\n\n$$\n\\frac{a^2 - c^2}{a^2} \\leq \\frac{3c^2 + ac - 2a^2}{a^2} \\leq 1,\n$$\n\ni.e.,\n\n$$\n1 - e^2 \\leq 3e^2 + e - 2 \\leq 1.\n$$\n\nWith $e \\in [0, 1]$, the range is\n\n$$\ne \\in \\left[ \\frac{3}{4}, \\frac{-1 + \\sqrt{37}}{6} \\right].\n$$",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 12982,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all pairs of positive numbers $x, y$ with the following property:\n\nIf $\\alpha, \\beta$ are relatively prime and positive divisors of the number $x^3 + y^3$, then $\\alpha + \\beta - 1$ is also a divisor of $x^3 + y^3$.",
    "options": [],
    "answer": "See solution",
    "solution": "We prove that the only solutions are $(x, y) = (2^n, 2^n)$, $(x, y) = (3^n, 2 \\cdot 3^n)$, or $(x, y) = (2 \\cdot 3^n, 3^n)$ for a natural number $n$.\n\n**Step 1:** $x^3 + y^3$ has at most one odd prime divisor.\n\nLet $p_1 < p_2 < \\cdots < p_k$ be the odd prime divisors of $x^3 + y^3$ with $k \\ge 2$. Then $p_1$ and $p_2 \\cdots p_k$ are relatively prime divisors. Thus, $N_1 = p_1 + (p_2 \\cdots p_k) - 1$ divides $x^3 + y^3$. Since $N_1$ is odd,\n\n$$\nN_1 = p_1^{r_1} \\cdots p_k^{r_k}\n$$\n\nfor some $r_1, \\dots, r_k$. For $i \\ge 2$, $p_i$ cannot divide $N_1$ (otherwise $p_i$ divides $p_1 - 1$, which is impossible since $p_i > p_1$). Thus, $N_1 = p_1^{m_1}$ with $m_1 > 1$ (since $N_1 > p_1$).\n\nSimilarly, $N_2 = p_1^2 + (p_2 \\cdots p_k) - 1$ also divides $x^3 + y^3$, so $N_2 = p_1^{m_2}$. Thus,\n\n$$\np_1^{m_2} - p_1^{m_1} = N_2 - N_1 = p_1^2 - p_1.\n$$\n\nBut $m_2 > m_1$, so\n\n$$\np_1^{m_2} - p_1^{m_1} \\ge p_1^{m_1}(p_1 - 1) \\ge p_1(p_1 - 1) = p_1^2 - p_1,\n$$\n\nwith equality only if $m_1 = 1, m_2 = 2$, but $m_1 > 1$, a contradiction. Thus, $x^3 + y^3$ has at most one odd prime divisor.\n\n**Step 2:** $x^3 + y^3$ has at most one prime divisor.\n\nSuppose not. Then $x^3 + y^3$ has exactly two prime divisors: $2$ and some odd prime $p$.\n\nThen $2 + p - 1 = p + 1$ divides $x^3 + y^3$. Since $(p, p+1) = 1$, $p+1$ must be a power of $2$, so $4$ divides $x^3 + y^3$. Also, $4 + p - 1 = p + 3$ divides $x^3 + y^3$. If $p \\ne 3$, $p+3$ is also a power of $2$, but for $p > 3$, $p+1$ and $p+3$ cannot both be powers of $2$. Thus, $p = 3$, so $x^3 + y^3 = 2^a \\cdot 3^b$ for some $a, b > 0$.\n\nIf $a \\ge 3$, $8 \\mid x^3 + y^3$, so $8 + 3 - 1 = 10$ divides $x^3 + y^3$, a contradiction.\n\nIf $b \\ge 2$, $9 \\mid x^3 + y^3$, so $9 + 2 - 1 = 10$ divides $x^3 + y^3$, again a contradiction.\n\nTherefore, $x^3 + y^3 \\in \\{6, 12\\}$, but $12$ has no solutions.\n\nThus, we only need to solve\n\n$$\nx^3 + y^3 = p^a\n$$\n\nwhere $p$ is a prime and $a$ a positive integer. Write $x = p^s m$, $y = p^t n$ with $m, n$ not divisible by $p$ and coprime. Then\n\n$$\nm^3 + n^3 = p^b \\implies (m+n)(m^2 - mn + n^2) = p^b.\n$$\n\nBut\n\n$$\n\\gcd(m+n, m^2 - mn + n^2) = \\gcd(m+n, 3mn) = 1 \\text{ or } 3.\n$$\n\nIf $\\gcd = 1$, $m+n = 1$ or $m^2 - mn + n^2 = 1$. Only $m = n = 1$ works, giving $p = 2$ and $x = y = 2^n$.\n\nIf $\\gcd = 3$, $p = 3$ and either $m+n = 3$ or $m^2 - mn + n^2 = 3$. Thus, $(m, n) = (2, 1)$ or $(1, 2)$, so $x = 2 \\cdot 3^n, y = 3^n$ or $x = 3^n, y = 2 \\cdot 3^n$ for $n \\ge 0$.\n\n$\\boxed{}$",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 12983,
    "subject": "Mathematics (Olympiad)",
    "question": "The real numbers $x_1, \\dots, x_{2011}$ satisfy\n$$\nx_1 + x_2 = 2x'_1, \\quad x_2 + x_3 = 2x'_2, \\quad \\dots, \\quad x_{2011} + x_1 = 2x'_{2011}\n$$\nwhere $x'_1, x'_2, \\dots, x'_{2011}$ is a permutation of $x_1, x_2, \\dots, x_{2011}$. Prove that $x_1 = x_2 = \\dots = x_{2011}$.",
    "options": [],
    "answer": "See solution",
    "solution": "For convenience, let $x_{2011}$ also be denoted as $x_0$. Let $k$ be the largest of the numbers $x_1, \\dots, x_{2011}$, and consider an equation $x_{n-1} + x_n = 2k$, where $1 \\leq n \\leq 2011$. Then $2 \\max(x_{n-1}, x_n) \\geq x_{n-1} + x_n = 2k$, so either $x_{n-1}$ or $x_n$ is $\\geq k$. Since $x_{n-1} \\leq k$, we have $x_{n-1} = k$, and then $x_n = 2k - x_{n-1} = k$. Thus, in such an equation, both variables on the left equal $k$.\n\nLet $\\mathcal{E}$ be the set of such equations, and let $\\mathcal{S}$ be the set of subscripts on the left of these equations. From $x_n = k$ for all $n \\in \\mathcal{S}$, we get $|\\mathcal{S}| \\leq |\\mathcal{E}|$. On the other hand, since the total number of appearances of these subscripts is $2|\\mathcal{E}|$ and each subscript appears on the left in no more than two equations, we have $2|\\mathcal{E}| \\leq 2|\\mathcal{S}|$. Thus $2|\\mathcal{E}| = 2|\\mathcal{S}|$, so for each $n \\in \\mathcal{S}$, the set $\\mathcal{E}$ contains both equations with the subscript $n$ on the left.\n\nNow assume $1 \\in \\mathcal{S}$ without loss of generality. Then the equation $x_1 + x_2 = 2k$ belongs to $\\mathcal{E}$, so $2 \\in \\mathcal{S}$. Continuing in this way, we find that all subscripts belong to $\\mathcal{S}$, so $x_1 = x_2 = \\dots = x_{2011} = k$.",
    "topic": "Discrete Mathematics",
    "subtopic": "Combinatorics"
  },
  {
    "id": 12984,
    "subject": "Mathematics (Olympiad)",
    "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 $A$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $M$ be the midpoint of $AB$, $N$ the midpoint of $AC$, and let $K$ and $L$ be the intersections of the angle bisector of $\\angle BAC$ with the perpendicular bisectors of $AB$ and $AC$, respectively. Let $O$ be the intersection of the perpendicular bisectors of $AB$ and $AC$. The triangle $KLO$ is the triangle defined in the problem. Let $H$ be its orthocenter.\n\nWe need to prove that $H$ lies on the median from 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 $AB$. This distance equals the length of segment $LN$, since $L$ lies on the angle bisector of $\\angle BAC$ and $N$ is the perpendicular projection of $L$ onto $AC$. Thus, the area of $ABH$ is $\\frac{1}{2}|AB| \\cdot |LN|$. Similarly, the area of $ACH$ is $\\frac{1}{2}|AC| \\cdot |KM|$. It remains to prove $|AB| \\cdot |LN| = |AC| \\cdot |KM|$.\n\nFor points $K$ and $L$ on the angle bisector of $\\angle BAC$, we have $|\\angle MAK| = |\\angle NAL|$. The right triangles $AKM$ and $ALN$ are therefore similar, so $|KM| : |AM| = |LN| : |AN|$. Since $|AM| = \\frac{1}{2}|AB|$ and $|AN| = \\frac{1}{2}|AC|$, we get $|KM| : |AB| = |LN| : |AC|$, i.e., $|AB| \\cdot |LN| = |AC| \\cdot |KM|$, as required.",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 12985,
    "subject": "Mathematics (Olympiad)",
    "question": "Given a triangle $ABC$ with $AC = BC > AB$, let $E$ and $F$ be the midpoints of $AC$ and $AB$, respectively. The perpendicular bisector of $AC$ meets $AB$ at $K$, and the line parallel to $KC$ passing through $B$ intersects $AC$ at a point $L$. For a point $P$ on the line segment $BF$, let $H$ be the orthocenter of triangle $ACP$. The line segments $BH$ and $CP$ meet at a point $J$, and the lines $FJ$ and $l$ meet at a point $M$. Let $W$ be the intersection point of $FL$ and $l$. Show that $AW = BW$ if and only if the points $B, E, F, M$ are concyclic.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $W'$ be the circumcenter of triangle $ACP$, and let $L'$ be the intersection point of the line $FW'$ and the side $AC$. Note that the points $A$, $F$, $C$, and the foot $T$ of the perpendicular from $A$ to $PC$ are on the circle with diameter $AC$, and thus $\\angle PAT - \\angle PCF$. Since $AC = BC$ and $F$ is the midpoint of $AB$, $\\angle HAF = \\angle HBF$, so $\\angle PBJ = \\angle HCJ$, which along with $\\angle PJB = \\angle HJC$ implies that triangles $PBJ$ and $HCJ$ are similar. Therefore, for the feet of the perpendiculars from $J$ to $FP$ and $FC$, $\\frac{XJ}{JY} = \\frac{BP}{CH}$. Let $S$ be the foot of the perpendicular from $W'$ to $AP$. Then $HC = 2W'S$ and\n\n$$\nBP = BF - PF = AF - (PF - FS) = AF - (AS - FS) = 2FS.\n$$\n\nIt follows that $\\frac{XJ}{XF} = \\frac{FS}{W'S}$. Hence $\\triangle JXF$ is similar to $\\triangle FSW'$, and thus $\\angle JFX = \\angle FW'S$. Since $W'S$ is parallel to $CF$, $\\angle FW'S = \\angle CFW'$. Therefore, $\\angle JFX = \\angle CFW'$. Now, since $\\angle PFC = 90^\\circ$, $\\angle JFL' = \\angle JFW' = 90^\\circ$, which along with $\\angle MEL' = 90^\\circ$ implies that the points $E$ and $F$ lie on the circle with diameter $ML'$.\n\nSuppose $AW = PW$, in which case $W$ is the circumcenter of $\\triangle APC$, so $AW = PW = CW$. From the fact above, $E$ and $F$ lie on the circle with diameter $ML$. Since the circumcenter $O$ of $\\triangle ABC$ is the orthocenter of $\\triangle AKC$, $AO$ is orthogonal to $KC$. But $BL$ is parallel to $KC$, so $AO$ is orthogonal to $BL$. Thus $\\angle ABL + \\angle BAO = 90^\\circ$. Note that $A$, $F$, $O$, $E$ are concyclic, and thus $\\angle BAO = \\angle MEF$, from which we conclude $\\angle FBL + \\angle FEL = \\angle ABL + \\angle MEF + 90^\\circ = 180^\\circ$. This implies that $F$, $B$, $M$, $L$, $E$ are concyclic.\n\nConversely, suppose $B$, $E$, $F$, $M$ are concyclic. Since $E$ and $F$ lie on the circle with diameter $ML'$, $\\angle MBL' = 90^\\circ$. It can be seen that $\\angle MBK = \\angle MEF = \\angle BAO$, so $BM$ is parallel to $AO$. It follows that $AO$ and $BL'$ are orthogonal, so $BL'$ and $KC$ are parallel. Therefore, $BL' = BL$, which implies $L' = L$ and $W' = W$. $\\square$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 12986,
    "subject": "Mathematics (Olympiad)",
    "question": "Может ли число $2^m$ иметь сумму цифр, равную 8, и оканчиваться на 6?",
    "options": [],
    "answer": "See solution",
    "solution": "Первое решение. При $m = 1, 2, 3$ последняя цифра числа $2^m$ не равна 6. Предположим, что сумма цифр числа $2^m$ при некотором $m > 3$ равна 8, и оно оканчивается на 6. Число $2^m$ не может оканчиваться на 06 или на 26, так как в этом случае оно не делится на 4. Следовательно, оно оканчивается на 16 (иначе сумма цифр будет больше 8), и поэтому имеет десятичную запись $1000\\ldots016$. Тогда $2^m = 10^k + 16$, то есть число $10^k + 16$ — степень двойки.\n\nНо если $k \\ge 5$, то $10^k + 16 = 2^4 (2^k - 4 \\cdot 5^k + 1)$, и в скобках получаем нечетный множитель, больший 1. Остается рассмотреть случаи $k = 2, k = 3, k = 4$: $10^2 + 16 = 4 \\cdot 29$; $10^3 + 16 = 8 \\cdot 127$; $10^4 + 16 = 32 \\cdot 313$. Таким образом, $10^k + 16$ не является степенью двойки ни при каком натуральном $k$, что и требовалось доказать.\n\nВторое решение. Предположим противное, и пусть $2^m$ оканчивается на 6 и имеет сумму цифр, равную 8. Заметим, что $2^1$ оканчивается на 2, $2^2$ оканчивается на 4, $2^3$ оканчивается на 8, $2^4$ оканчивается на 6, $2^5$ оканчивается на 2. Далее последняя цифра степени двойки повторяется с периодом 4, поскольку последняя цифра числа $2^m$ определяется однозначно последней цифрой числа $2^{m-1}$. Таким образом, $2^m$ оканчивается на 6 тогда и только тогда, когда $m$ делится на 4.\n\nСумма цифр числа имеет тот же остаток при делении на 3, что и само число, поэтому $2^m$ должно иметь остаток 2 при делении на 3. Заметим, что $2^1$ имеет остаток 2 при делении на 3, $2^2$ имеет остаток 1 при делении на 3, $2^3$ имеет остаток 2 при делении на 3, и далее остатки степени двойки повторяются с периодом 2. Таким образом, $2^m$ имеет остаток 2 при делении на 3 тогда и только тогда, когда $m$ нечетно. Это противоречит тому, что $m$ должно делиться на 4.\n\n![](images/Rusija_2009_p38_data_65e7bc9c92.png)\n\nРис. 10",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 12987,
    "subject": "Mathematics (Olympiad)",
    "question": "已知 $a, b, c, d$ 為非負實數，試求滿足下列方程組的解 $(a, b, c, d)$：\n\n$$\na^2(b+c)(b+c+d) = \\sqrt{b+c}\\sqrt[3]{b+c+d}\n$$\n\n$$\nb^2(c+d)(c+d+a) = \\sqrt{c+d}\\sqrt[3]{c+d+a}\n$$\n\n$$\nc^2(d+a)(d+a+b) = \\sqrt{d+a}\\sqrt[3]{d+a+b}\n$$\n\n$$\nd^2(a+b)(a+b+c) = \\sqrt{a+b}\\sqrt[3]{a+b+c}\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "若 $a, b, c, d$ 其中有一項為 $0$，容易推得 $a = b = c = d = 0$ 為一解，因此以下不妨設 $a, b, c, d > 0$。\n\n將原式整理後得到：\n\n$$\na^{12}(b+c)^3(b+c+d)^4 = 1\n$$\n\n$$\nb^{12}(c+d)^3(c+d+a)^4 = 1\n$$\n\n$$\nc^{12}(d+a)^3(d+a+b)^4 = 1\n$$\n\n$$\nd^{12}(a+b)^3(a+b+c)^4 = 1\n$$\n\n考慮 $x$ 為 $\\{\\frac{x}{y} \\mid x, y \\in \\{a, b, c, d\\}\\}$ 中的最大值，且不妨設 $\\frac{a}{b} = x \\ge 1$。則\n\n$$\n\\frac{c+d}{b+c} \\le x\n$$\n\n$$\n\\frac{c+d+a}{b+c+d} \\le x\n$$\n\n因此\n\n$$\n1 = \\frac{a^{12}(b+c)^3(b+c+d)^4}{b^{12}(c+d)^3(c+d+a)^4} \\ge x^{12} \\frac{1}{x^3} \\frac{1}{x^4} = x^7\n$$\n\n所以 $x = 1$，即 $a = b = c = d$。\n\n代入原式解得 $(a, b, c, d) = (0, 0, 0, 0)$ 或 $\\left(\\frac{1}{\\sqrt{19/648}}, \\frac{1}{\\sqrt{19/648}}, \\frac{1}{\\sqrt{19/648}}, \\frac{1}{\\sqrt{19/648}}\\right)$。",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 12988,
    "subject": "Mathematics (Olympiad)",
    "question": "A polygon is tiled with a finite number of triangles whose sides all have an odd length.\n\n(a) Prove that, if the polygon is convex, then its perimeter is an integer of the same parity as the number of triangles in the tiling.\n\n(b) Does the conclusion still hold if the polygon is not convex?",
    "options": [],
    "answer": "See solution",
    "solution": "a) Let $K$ be the polygon under consideration. Since $K$ is convex, the tiling triangles fall into two classes: those having all edges inside $K$, and those having at least one edge on the boundary of $K$.\n\nEvery inner edge of a triangle is subdivided into one or more 'short' segments by the boundaries of some other triangles on the opposite side. Each short segment is shared by exactly two triangles. Further, every short segment lies along a unique segment of maximal length, which is a concatenation of non-overlapping inner edges from triangles on the same side. Hence, the total length of the short segments along one maximal segment is an integer. Consequently, so is the total length $s$ of all short segments.\n\nEvery outer edge (lying on the boundary of $K$) belongs to a single triangle, and the total length of all outer edges is the perimeter of $K$.\n\nLet $t$ be the number of triangles, and let $S$ be the sum of their perimeters. Since the sides of each triangle all have an odd length, $t$ and $S$ have the same parity. The perimeter of $K$ is $S - 2s$, and the conclusion follows.\n\nb) The answer is negative. Let $A, A', B, B'$ (in order) be distinct points on a line $\\ell$ such that $AB = A'B' = 1$. Erect equilateral triangles $ABC$ and $A'B'C'$, where $C$ and $C'$ lie on opposite sides of $\\ell$. These two triangles tile the non-convex hexagon $AA'C'B'BC$. Letting $AA' = BB' = x$, the perimeter of the hexagon is $4 + 2x$. If $x = \\frac{1}{2}$, the perimeter is $5$ (odd), and if $x \\neq \\frac{1}{2}$, the perimeter is not even an integer.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 12989,
    "subject": "Mathematics (Olympiad)",
    "question": "A *palindrome* is a word constructed using two letters and is equal to its reverse. For example, *ABBA* and *ABABABABA* are palindromes. Prove that any 2019-letter word (using only two letters) can be constructed by at most 808 palindromes.",
    "options": [],
    "answer": "See solution",
    "solution": "Note that any 5-letter word can be constructed with at most two palindromes (this can be checked case by case). Any 2019-letter word can be divided into 404 5-letter words, so it can be constructed using at most $808$ palindromes.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 12990,
    "subject": "Mathematics (Olympiad)",
    "question": "Decide whether there exists a set $M$ consisting of five integers such that for any integer $k$ not divisible by $5$, there exist $a, b \\in M$ such that $a - b + k$ is divisible by $25$.",
    "options": [],
    "answer": "See solution",
    "solution": "There does not exist such a set.\n\nAssume that $M = \\{a, b, c, d, e\\}$ were such a set. There are $20$ differences of distinct members from $M$ and $20$ residue classes modulo $25$ whose members are not divisible by $5$:\n\n$$\n1, 2, 3, 4, 6, 7, 8, 9, 11, 12, 13, 14, 16, 17, 18, 19, 21, 22, 23, 24\n$$\n\nand\n\n$$\na-b, a-c, a-d, a-e, b-a, b-c, b-d, b-e, \\dots, e-d\n$$\n\ncontain the same numbers when considered modulo $25$. Taking products, we get\n\n$$\n-1 \\equiv \\prod_{x, y \\in M, x \\neq y} (x-y) \\pmod{25}.\n$$\n\nThis implies that no two members of $M$ are congruent modulo $5$. Setting\n\n$$\n\\Omega(x_1, x_2, x_3, x_4, x_5) = \\prod_{1 \\le i, j \\le 5, i \\ne j} (x_i - x_j)\n$$\n\nfor all integers $x_1, \\dots, x_5$, the above congruence may be rewritten as\n\n$$\n\\Omega(a, b, c, d, e) \\equiv -1 \\pmod{25}.\n$$\n\n*Claim.* If $x_1, \\dots, x_5$ are integers no two of which are congruent modulo $5$, then\n\n$$\n\\Omega(x_1 + 5, x_2, x_3, x_4, x_5) - \\Omega(x_1, x_2, x_3, x_4, x_5)\n$$\n\nis a multiple of $25$.\n\nTo see this, note that this difference is $\\prod_{2 \\le i < j \\le 5} (x_i - x_j)$ times\n\n$$\n(x_1 - x_2 + 5)^2 \\cdots (x_1 - x_5 + 5)^2 - (x_1 - x_2)^2 \\cdots (x_1 - x_5)^2.\n$$\n\nThe second factor is\n\n$$\n\\equiv ((x_1 - x_2)^2 + 10(x_1 - x_2)) \\cdots ((x_1 - x_5)^2 + 10(x_1 - x_5)) - (x_1 - x_2)^2 \\cdots (x_1 - x_5)^2\n$$\n\n$$\n\\equiv 10(x_1 - x_2) \\cdots (x_1 - x_5) \\cdot \\Psi \\pmod{25},\n$$\n\nwhere $\\Psi$ denotes the sum of all four products involving three of the numbers $x_1-x_2, \\dots, x_1-x_5$. So it suffices to show that $\\Psi$ is divisible by $5$, and as the four differences $x_1-x_2, \\dots, x_1-x_5$ coincide modulo $5$ with the numbers $1, 2, 3, 4$, we do indeed have\n\n$$\n\\Psi \\equiv 1 \\cdot 2 \\cdot 3 + 1 \\cdot 2 \\cdot 4 + 1 \\cdot 3 \\cdot 4 + 2 \\cdot 3 \\cdot 4 \\equiv 50 \\equiv 0 \\pmod{5}.\n$$\n\nThis concludes the proof of our claim. As the function $\\Omega$ is symmetric in its variables, a similar statement holds when $5$ is added not to $x_1$ but to any other of these variables. Applying this fact iteratively and using symmetry again, we get\n\n$$\n\\Omega(a, b, c, d, e) \\equiv \\Omega(0, 1, 2, 3, 4) \\equiv 82944 \\equiv 19 \\pmod{25},\n$$\n\nwhereby we have reached a contradiction. This solves our problem.\n\n*Comment.* Virtually the same proof also works if one replaces $5$ by any odd prime, but then the computations involved get quite a bit more difficult. It seems possible to solve this version of the problem by some lengthy case analysis as well, but we did not seriously try to do so.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 12991,
    "subject": "Mathematics (Olympiad)",
    "question": "An $8 \\times 8$ array is divided into 64 unit squares. In some of the unit squares, a diagonal is drawn, such that no two diagonals share a common point—not even an endpoint. Find the maximum number of diagonals that can be drawn under these circumstances.",
    "options": [],
    "answer": "See solution",
    "solution": "Consider two contiguous rows of a $2n \\times 2n$ array, with $k$ diagonals in row A and $\\ell$ diagonals in row B. The endpoints of these diagonals on the line separating the two rows must be distinct, and there are at most $2n + 1$ available positions. Thus, $k + \\ell \\le 2n + 1$. Since the $2n$ rows can be grouped into $n$ pairs of contiguous rows, the total number of diagonals is at most $n(2n + 1) = \\binom{2n+1}{2}$. A construction: use only /-type diagonals in cells with coordinates $(2i + 1, j)$ and $(j, 2i + 1)$, where $1 \\le 2i + 1 \\le j \\le 2n$. For $2n = 8$, the answer is $36$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 12992,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $BB'$, $CC'$ be the altitudes of an acute-angled triangle $ABC$. Two circles passing through $A$ and $C'$ are tangent to $BC$ at points $P$ and $Q$. Prove that $A$, $B'$, $P$, $Q$ are concyclic.",
    "options": [],
    "answer": "See solution",
    "solution": "Since $BP^2 = BQ^2 = BA \\cdot BC'$, and the quadrilaterals $AC'A'C$, $AB'A'B$ are cyclic (where $AA'$ is the altitude), we have\n\n$$\nCP \\cdot CQ = CB^2 - BP^2 = CB^2 - BA \\cdot BC' = BC^2 - BC \\cdot BA' = BC \\cdot CA' = CA \\cdot CB'.\n$$\n\nClearly, this is equivalent to the required assertion. $\\square$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 12993,
    "subject": "Mathematics (Olympiad)",
    "question": "Prove that if positive real numbers $a_1, a_2, \\dots, a_n$ have the product $1$, then\n\n$$\n\\left(\\frac{a_1}{a_2}\\right)^{n-1} + \\left(\\frac{a_2}{a_3}\\right)^{n-1} + \\dots + \\left(\\frac{a_{n-1}}{a_n}\\right)^{n-1} + \\left(\\frac{a_n}{a_1}\\right)^{n-1} \\geq a_1^2 + a_2^2 + \\dots + a_n^2.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "We prove that for all $a_1, a_2, \\dots, a_n > 0$, the following inequality holds:\n\n$$\n\\left(\\frac{a_1}{a_2}\\right)^{n-1} + \\left(\\frac{a_2}{a_3}\\right)^{n-1} + \\dots + \\left(\\frac{a_{n-1}}{a_n}\\right)^{n-1} + \\left(\\frac{a_n}{a_1}\\right)^{n-1} \\geq \\frac{a_1^2 + a_2^2 + \\dots + a_n^2}{\\sqrt[n]{a_1^2 a_2^2 \\dots a_n^2}}.\n$$\n\nThis inequality is obtained by adding the inequality below with its analogues obtained by cyclic permutation of the variables:\n\n$$\n\\begin{aligned}\n& (n-1) \\cdot \\left(\\frac{a_1}{a_2}\\right)^{n-1} + (n-2) \\cdot \\left(\\frac{a_2}{a_3}\\right)^{n-1} + \\dots + \\left(\\frac{a_{n-1}}{a_n}\\right)^{n-1} \n\\\\\n& \\geq \\frac{n(n-1)}{2} \\cdot \\left( \\left(\\frac{a_1}{a_2}\\right)^{n-1} \\left(\\frac{a_2}{a_3}\\right)^{n-2} \\dots \\left(\\frac{a_{n-1}}{a_n}\\right)^{\\frac{2}{n}} \\right) \n\\\\\n& = \\frac{n(n-1)}{2} \\left( \\frac{a_1^n}{a_1 a_2 a_3 \\dots a_n} \\right)^{\\frac{2}{n}}.\n\\end{aligned}\n$$\n\nEquality holds if all the numbers are equal to $1$.\n\n**Remarks.** For $n=3$ this inequality has been published by Šefket Arslanagić in *Elemente der Mathematik*; for $n=4$ it appears, in a weaker form, in *Mathscope*, pb. 321.1 (author Lê Thanh Hâyi).",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 12994,
    "subject": "Mathematics (Olympiad)",
    "question": "There are 2016 real numbers written on the blackboard. In each step, we choose two numbers, erase them, and replace each of them by their product. Determine whether it is possible to obtain 2016 equal numbers on the blackboard after a finite number of steps.",
    "options": [],
    "answer": "See solution",
    "solution": "We shall prove by induction that it is possible to obtain equal numbers after a finite number of steps for any even $n$.\n\nThe claim is trivial for $n = 2$ (we can get the desired 2-tuple after a single step: $(a, b) \\to (ab, ab)$) and $n = 4$ (for example: $(\\underline{a}, \\underline{b}, c, d) \\to (ab, ab, \\underline{c}, \\underline{d}) \\to (\\underline{ab}, ab, \\underline{cd}, cd) \\to (abcd, \\underline{ab}, abcd, \\underline{cd}) \\to (abcd, abcd, abcd, abcd)$).\n\nTo start the induction, we prove the claim for $n = 6$. The algorithm begins with the 6-tuple $(a, a, a, a, b, b)$ – this form can be achieved since the claim is true for $n = 2$ and $n = 4$ (operate on the left 4-tuple and the right 2-tuple independently). To equalize all six numbers, perform the following steps:\n\n$$\n\\begin{align*}\n(a, a, a, \\underline{a}, \\underline{b}, b) &\\to (a, a, \\underline{a}, \\underline{ab}, ab, b) \\to (a, a, a^2b, \\underline{a^2b}, \\underline{ab}, b) \\\\\n&\\to (a, a, \\underline{a^2b}, a^3b^2, a^3b^2, \\underline{b}) \\\\\n&\\to (\\underline{a}, a, \\underline{a^2b^2}, a^3b^2, a^3b^2, a^2b^2) \\\\\n&\\to (a^3b^2, \\underline{a}, a^3b^2, a^3b^2, a^3b^2, \\underline{a^2b^2}) \\\\\n&\\to (a^3b^2, a^3b^2, a^3b^2, a^3b^2, a^3b^2, a^3b^2)\n\\end{align*}\n$$\n\nNow, suppose the claim is true for all even $n < 4k + 4$ (with $k \\ge 1$). It suffices to prove the claim for $n = 4k + 4$ and $n = 4k + 6$.\n\nFor $n = 4k + 4$, first equalize the first $2k + 2$ numbers using the induction hypothesis, then the last $2k + 2$ numbers. We get an $n$-tuple of the form\n\n$$\n\\underbrace{(a, \\dots, a, b, \\dots, b)}_{2k+2}\n$$\n\nThen perform $2k + 2$ steps, each time choosing one $a$ and one $b$, to get $(ab, \\dots, ab)$.\n\nFor $n = 4k+6$, use the induction hypothesis for $n = 2k+2$ and $n = 2k+4$ to get\n\n$$\n\\underbrace{(a, \\dots, a, b, \\dots, b)}_{2k+2}\n$$\n\nNow perform $2k$ steps, always choosing one $a$ and one $b$, to obtain\n\n$$\n\\underbrace{(a, a, ab, \\dots, ab, b, b, b, b)}_{4k}\n$$\n\nAfter erasing each $a$ with one $ab$ we get\n\n$$\n(a^2b, a^2b, a^2b, a^2b, \\underbrace{ab, \\dots, ab}_{4k-2}, b, b, b, b)\n$$\n\nPairing each $b$ with one $a^2b$ gives\n\n$$\n(a^2b^2, a^2b^2, a^2b^2, a^2b^2, \\underbrace{ab, \\dots, ab}_{4k-2}, a^2b^2, a^2b^2, a^2b^2, a^2b^2)\n$$\n\nFinally, perform $2k-1$ steps, replacing $2k-1$ pairs of $ab$'s by $a^2b^2$'s, which leads to $(a^2b^2, \\dots, a^2b^2)$.\n\nThus, it is possible to obtain 2016 equal numbers after a finite number of steps.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 12995,
    "subject": "Mathematics (Olympiad)",
    "question": "Consider all triples of non-zero integers $(a, b, c)$, which belong to the interval $[-2020, 2020]$. Find for how many of them the conditions $a+b+c=0$ and $\\frac{a^2}{b}+\\frac{b^2}{c}+\\frac{c^2}{a}=\\frac{a^2}{c}+\\frac{c^2}{b}+\\frac{b^2}{a}$ are equivalent. In other words, how many such triples are there, such that either both conditions hold, or none of the two holds?\n\n![](path/to/file.png)",
    "options": [],
    "answer": "See solution",
    "solution": "**Answer:** $4040 \\cdot 4039 \\cdot 4038 + 3 \\cdot 2020$.\n\n**Solution.**\n\nConsider the following transformation:\n\n$$\n\\left( \\frac{a^2}{b} + \\frac{b^2}{c} + \\frac{c^2}{a} \\right) - \\left( \\frac{a^2}{c} + \\frac{c^2}{b} + \\frac{b^2}{a} \\right) = \\frac{a^3 c - a^3 b + b^3 a - a^3 b + c^3 b - b^3 c}{abc}\n$$\n\nNow we transform only the numerator:\n\n$$\n\\begin{aligned}\na^3 c - a^3 b + b^3 a - a^3 b + c^3 b - b^3 c &= (b-a)(ab^2 + a^2 b + c^2 - ca^2 - abc - cb^2) \\\\\n&= (b-a)(ab^2 + a^2 b + c^2 - ca^2 - abc - cb^2) \\\\\n&= (b-a)(a-c)(b^2 + ab - c^2 - ca) \\\\\n&= (b-a)(a-c)(b-c)(a+b+c).\n\\end{aligned}\n$$\n\nTherefore, the condition $\\frac{a^2}{b}+\\frac{b^2}{c}+\\frac{c^2}{a}=\\frac{a^2}{c}+\\frac{c^2}{b}+\\frac{b^2}{a}$ for non-zero numbers $a, b, c$ is equivalent to the condition $(b-a)(a-c)(b-c)(a+b+c)=0$.\n\nNow we count the number of such triples. The amount of pairwise different non-zero triples is $4040 \\cdot 4039 \\cdot 4038$. For the case where at least two integers are equal, for example $a=b \\neq c$, then there are $2020$ numbers to choose $a$ and then $b$ and $c$ are defined. Therefore, there are $3 \\cdot 2020$ such triples.\n\nSo, the total number is $4040 \\cdot 4039 \\cdot 4038 + 3 \\cdot 2020$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 12996,
    "subject": "Mathematics (Olympiad)",
    "question": "Suppose set $S = \\{1, 2, 3, \\dots, 10\\}$ and the subset $A$ of $S$ satisfies\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": "See solution",
    "solution": "First, we find the number $N_1$ of subsets $A$ of $S$ such that $A \\cap \\{1, 2, 3\\} \\neq \\emptyset$.\n\nThere are $2^3 - 1 = 7$ ways to select at least one element from $\\{1, 2, 3\\}$, and for each of $4, 5, \\dots, 10$ there are two choices (selected or not). Thus, $N_1 = 7 \\times 2^7 = 896$.\n\nNext, we subtract the number $N_2$ of subsets $A$ such that $A \\cup \\{4, 5, 6\\} = S$. In this case, $1, 2, 3, 7, 8, 9, 10$ must all be in $A$, and each of $4, 5, 6$ can be either in or out of $A$, so $N_2 = 2^3 = 8$.\n\nTherefore, the number of subsets $A$ satisfying the conditions is $N_1 - N_2 = 896 - 8 = 888$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 12997,
    "subject": "Mathematics (Olympiad)",
    "question": "一個 $3n \\times 3n$ 的棋盤，行與列都依序編號為 $1$ 到 $3n$。方格 $(x, y)$ 依據 $x+y$ 模 $3$ 的餘數為 $0, 1$ 或 $2$，分別塗成顏色 A、B 或 C。在每一個方格中擺上一個籌碼，每個籌碼的顏色為 A、B 或 C。三種顏色的籌碼各有 $3n^2$ 個。\n\n假設我們可以安排下列籌碼的重新排列方式：每個籌碼移動的距離不超過 $d$，顏色 A 的籌碼取代顏色 B 的籌碼、顏色 B 的籌碼取代顏色 C 的籌碼、顏色 C 的籌碼取代顏色 A 的籌碼。試證：我們還有另一種籌碼重新排列的方式，使得每個籌碼移動的距離不超過 $d+2$，並且移動後每個方格內都有與該方格顏色相同的籌碼。",
    "options": [],
    "answer": "See solution",
    "solution": "不失一般性，只需證明所有顏色 A 的籌碼（簡稱 A-籌碼）可以移動到不同的顏色 A 的方格（簡稱 A-方格）中，並且每個 A-籌碼從原來的方格所移動的距離不超過 $d+2$。這就是說，我們可以將 $3n^2$ 個 A-籌碼與 $3n^2$ 個 A-方格作完美配對，並且每一對之間的距離最多是 $d+2$。\n\n為了找出這個完美配對，我們建構一個雙色圖：所有的 A-方格為一群頂點，所有的 A-籌碼為另一群頂點。\n\n將原來的棋盤分割成 $3 \\times 1$ 的三方塊；每一個三方塊恰包含一個 A-方格。找一個把 A-籌碼送到 B-籌碼、B-籌碼送到 C-籌碼、C-籌碼送到 A-籌碼，且每一個移動的距離都不超過 $d$ 的排列 $\\pi$。對任一個 A-方格 $S$ 與任一個 A-籌碼 $T$，如果 $\\pi(T)$ 或 $\\pi^{-1}(T)$ 落在包含 $S$ 的三方塊中，我們就在雙色圖中將 $S$、$T$ 連一條邊。在此圖形中我們也容易有多重邊；一個方格與一個籌碼間可能連到三條邊。易知圖中每條邊的長度不超過 $d+2$；此處邊的長度指的是它所連接的方格與籌碼之間的距離。\n\n每一個 A-籌碼 $T$ 與三個 A-方格相連：這些 A-方格所屬的三方塊包含了 $T$、$\\pi(T)$ 與 $\\pi^{-1}(T)$。所以在這個雙色圖中每一個籌碼的度數為 $3$。以下證明每一個方格的度數也都是 $3$。設 $S$ 為某個 A-方格，且令 $T_1, T_2, T_3$ 為包含 $S$ 的三方塊上的籌碼。對於 $i = 1, 2, 3$，如果 $T_i$ 是 A-籌碼，則 $S$ 與 $T_i$ 連接；如果 $T_i$ 是 B-籌碼，則 $S$ 與 $\\pi^{-1}(T_i)$ 連接；最後如果 $T_i$ 是 C-籌碼，則 $S$ 與 $\\pi(T_i)$ 連接。故得證每一個 A-方格的度數也是 $3$。\n\n因為每一個 A-方格的度數是 $3$，從任一些 A-方格所組成的集合 $S$ 都連出 $3|S|$ 條邊。這些邊中至少會連到 $|S|$ 個籌碼，因為每個 A-籌碼的度數也是 $3$。因此任何一個由 A-方格組成的集合 $S$ 都至少有 $|S|$ 個 A-籌碼作為其鄰居。\n\n於是，由 Hall 的結婚定理（Hall's marriage theorem），此雙色圖中存在兩群頂點之間的完美配對方式。所以 A-方格與 A-籌碼間存在一完美配對，其中每條邊的距離最多是 $d+2$，故原題所要求的排列方式存在，得證。",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 12998,
    "subject": "Mathematics (Olympiad)",
    "question": "Does there exist a function $f : \\mathbb{Q} \\to \\mathbb{Q}$ satisfying\n$$\nf(x + y + 2f(y)) = \\frac{2022}{2023} \\cdot y + f(x),\n$$\nfor all $x, y \\in \\mathbb{Q}$?",
    "options": [],
    "answer": "See solution",
    "solution": "The answer is No.\n\nLet $k = \\frac{2022}{2023}$. Set $x = 0$:\n$$\nf(y + 2f(y)) = k y + f(0)\n$$\nso $f$ is surjective over $\\mathbb{Q}$. Set $x = -2f(y)$:\n$$\nf(y) = k y + f(-2f(y))\n$$\nso $f$ is injective. Thus, $f$ is bijective.\n\nSet $y = 0$:\n$$\nf(x + 2f(0)) = f(x)\n$$\nso $2f(0) = 0$, hence $f(0) = 0$.\n\nNow, with $x = 0$:\n$$\nf(y + 2f(y)) = k y\n$$\nso $y + 2f(y)$ is surjective over $\\mathbb{Q}$.\n\nRewrite the original equation as:\n$$\nf(x + y + 2f(y)) = f(y + 2f(y)) + f(x)\n$$\nLet $t = y + 2f(y)$, then for all $x, t \\in \\mathbb{Q}$:\n$$\nf(x + t) = f(t) + f(x)\n$$\nTherefore, $f$ is additive on $\\mathbb{Q}$, so there exists $c \\in \\mathbb{Q}$ such that $f(x) = c x$ for all $x \\in \\mathbb{Q}$.\n\nSubstitute into the original equation:\n$$\nc(x + y + 2c y) = k y + c x\n$$\nwhich simplifies to\n$$\n(c + 2c^2 - k) y = 0, \\quad \\forall y \\in \\mathbb{Q}.\n$$\nSo $c + 2c^2 - k = 0$. The discriminant is $1 + 8k = 1 + 8 \\cdot \\frac{2022}{2023}$, which is not a rational square, so there is no rational solution for $c$.\n\nTherefore, there does not exist a function $f : \\mathbb{Q} \\to \\mathbb{Q}$ satisfying the given equation. $\\square$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 12999,
    "subject": "Mathematics (Olympiad)",
    "question": "Given a right-angled triangle $ABC$ with hypotenuse of length $a$ and legs of lengths $b$ and $c$, find an example where $a^2 > 4bc$.\n\n% IMAGE: ![](images/IRL_ABooklet_2023_p57_data_4792bd176f.png)",
    "options": [],
    "answer": "See solution",
    "solution": "The polynomial $x^3(x-8) + 2x^2 + 8x + 1$ is positive for large $x$, e.g., any $x \\geq 8$. Thus, for $x \\geq 8$, we have a right-angled triangle with $a^2 > 4bc$.\n\nAlternatively, since the area of triangle $ABC$ is $bc/2$, the inequality $a^2 > 4bc$ is equivalent to $a^2 > 8(ABC)$. If $h$ is the height from $A$ to the hypotenuse $a$, then $(ABC) = ah/2$, so $a > 4h$.\n\nLet the foot of the altitude from $A$ divide the hypotenuse into segments of length $p$ and $q$, with $h^2 = pq = p(a-p)$. We want $a^2 > 16h^2 = 16p(a-p)$, i.e., $a^2 - 16pa + 16p^2 > 0$, which is equivalent to\n\n$$\\left(\\frac{a}{p} - 8\\right)^2 > 48, \\text{ i.e. } \\left|\\frac{a}{p} - 8\\right| > 4\\sqrt{3}.$$ \n\nFor $0 < p < a$, given $a$ we can choose any $p$ such that\n\n$$0 < p < \\frac{2 - \\sqrt{3}}{4} \\quad \\text{or} \\quad \\frac{2 + \\sqrt{3}}{4} < p < a.$$ \n\nSince $c^2 = ap$ and $b^2 = a(a-p)$, $p$ determines $b = \\sqrt{a^2 - ap}$ and $c = \\sqrt{ap}$ for any given $a$. For example, with $a = 41$ and $p = 81/41$, we get $b = 40$ and $c = 9$, so $a^2 = 1681 > 1440 = 4bc$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 13000,
    "subject": "Mathematics (Olympiad)",
    "question": "On the plane, $2022$ points $A_1, A_2, \\dots, A_{2022}$ are given, no three of which lie on the same line. Consider all the angles $A_i A_j A_k$ for triples of distinct points $A_i, A_j, A_k$. What is the largest number of these angles that can be right angles?",
    "options": [],
    "answer": "See solution",
    "solution": "Consider any point $A_i$ and count the number of pairs of points $(A_j, A_k)$ such that $\\angle A_i A_j A_k = 90^\\circ$. For each point $A_j$, there exists at most one point $A_k$ (because on the line through $A_j$ perpendicular to $A_iA_j$, there can be at most one point other than $A_j$). Also, if $X$ is the point at the largest distance from $A_i$, then there can be no point $A_k$ with $\\angle A_i X A_k = 90^\\circ$, because then $A_iA_k > A_iX$.\n\nThus, each point can be a vertex of the hypotenuse in at most $2020$ right triangles at these points. Since the hypotenuse of each triangle has two vertices, the total number of these triangles does not exceed $$\\frac{2022 \\cdot 2020}{2} = 2022 \\cdot 1010.$$ \n\nThis number can be achieved, for example, by placing $2022$ points on a circle so that they are divided into $1011$ pairs, with each pair forming a diameter of the circle. For each diameter, there are exactly $2020$ points that form a right triangle with it. Thus, we have at least $1011 \\cdot 2020$ different right angles.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 13001,
    "subject": "Mathematics (Olympiad)",
    "question": "A function $f$ is defined on the positive integers by $f(1) = 1$ and, for $n > 1$:\n\n$$\nf(n) = f\\left(\\left\\lfloor \\frac{2n-1}{3} \\right\\rfloor\\right) + f\\left(\\left\\lfloor \\frac{2n}{3} \\right\\rfloor\\right)\n$$\n\nwhere $\\lfloor x \\rfloor$ denotes the greatest integer less than or equal to $x$.\n\nIs it true that $f(n) - f(n - 1) \\le n$ for all $n > 1$?\n\nHere are some examples of the use of $\\lfloor x \\rfloor$: $\\lfloor \\pi \\rfloor = 3$, $\\lfloor 1729 \\rfloor = 1729$, and $\\lfloor \\frac{2012}{1000} \\rfloor = 2$.",
    "options": [],
    "answer": "See solution",
    "solution": "First, note that $f(2) = 2f(1) = 2$. Next, consider three cases:\n\n- $n = 3k + 2$\n- $n = 3k + 1$\n- $n = 3k$\n\nfor $k$ a positive integer. In the first case:\n\n$$\n\\begin{aligned}\nf(n) &= f(3k + 2) = f\\left(\\left\\lfloor \\frac{6k + 3}{3} \\right\\rfloor\\right) + f\\left(\\left\\lfloor \\frac{6k + 4}{3} \\right\\rfloor\\right) \\\\\n&= 2f(2k + 1)\n\\end{aligned}\n$$\n\nIn the second case:\n\n$$\n\\begin{aligned}\nf(n) &= f(3k + 1) = f\\left(\\left\\lfloor \\frac{6k + 1}{3} \\right\\rfloor\\right) + f\\left(\\left\\lfloor \\frac{6k + 2}{3} \\right\\rfloor\\right) \\\\\n&= 2f(2k)\n\\end{aligned}\n$$\n\nIn the third case:\n\n$$\n\\begin{aligned}\nf(n) &= f(3k) = f\\left(\\left\\lfloor \\frac{6k - 1}{3} \\right\\rfloor\\right) + f\\left(\\left\\lfloor \\frac{6k}{3} \\right\\rfloor\\right) \\\\\n&= f(2k - 1) + f(2k)\n\\end{aligned}\n$$\n\nSo:\n\n$$\n\\begin{aligned}\nf(3k + 2) - f(3k + 1) &= 2f(2k + 1) - 2f(2k) \\\\\n&= 2(f(2k + 1) - f(2k))\n\\end{aligned}\n$$\n\n$$\nf(3k + 1) - f(3k) = f(2k) - f(2k - 1)\n$$\n\n$$\nf(3k) - f(3k - 1) = f(2k) - f(2k - 1)\n$$\n\nApplying these recursively:\n\n$$\n\\begin{aligned}\nf(2) - f(1) &= 1 \\\\\nf(3) - f(2) &= 1 \\\\\nf(5) - f(4) &= 2 \\\\\nf(8) - f(7) &= 4 \\\\\nf(13) - f(12) &= 4 \\\\\nf(20) - f(19) &= 8 \\\\\nf(31) - f(30) &= 8 \\\\\nf(47) - f(46) &= 16 \\\\\nf(71) - f(70) &= 32 \\\\\nf(107) - f(106) &= 64 \\\\\nf(161) - f(160) &= 128 \\\\\nf(242) - f(241) &= 256\n\\end{aligned}\n$$\n\nSince $256 > 242$, for $n = 242$ we do not have $f(n) - f(n-1) \\le n$. Thus, the statement is false.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 13002,
    "subject": "Mathematics (Olympiad)",
    "question": "Can the expression $\\frac{a}{a+1} + \\frac{a}{a+2} + \\frac{a}{a+3} + \\frac{a}{a+4}$ be an integer for some positive rational number $a$?",
    "options": [],
    "answer": "See solution",
    "solution": "Let $A = \\frac{a}{a+1} + \\frac{a}{a+2} + \\frac{a}{a+3} + \\frac{a}{a+4}$. Clearly, $0 < A < 4$.\n\n**Solution using GCD:** Write $a = \\frac{p}{q}$ with $p, q$ relatively prime positive integers. Then,\n\n$$\nA = \\frac{p}{p+q} + \\frac{p}{p+2q} + \\frac{p}{p+3q} + \\frac{p}{p+4q} = \\frac{2p(2p+5q)(p^2+5pq+5q^2)}{(p+q)(p+2q)(p+3q)(p+4q)}\n$$\n\nNow note that $p^2 + 5pq + 5q^2$ is relatively prime to $p+q$, $p+2q$, $p+3q$, and $p+4q$, since\n\n$$\n\\begin{align*}\n(p+q, p^2+5pq+5q^2) &= (p+q, 4pq+5q^2) = (p+q, 4p+5q) = 1 \\\\\n(p+2q, p^2+5pq+5q^2) &= (p+2q, 3pq+5q^2) = (p+2q, 3p+5q) = 1 \\\\\n(p+3q, p^2+5pq+5q^2) &= (p+3q, 2pq+5q^2) = (p+3q, 2p+5q) = 1 \\\\\n(p+4q, p^2+5pq+5q^2) &= (p+4q, pq+5q^2) = (p+4q, p+5q) = 1\n\\end{align*}\n$$\n\nIt follows that\n\n$$\n\\frac{2p(2p+5q)}{(p+q)(p+2q)(p+3q)(p+4q)} = \\frac{A}{p^2+5pq+5q^2}\n$$\n\nis an integer. This is a contradiction since $0 < A < 4$ and $p^2 + 5pq + 5q^2 > 4$.\n\n**Solution using polynomials:** Let\n\n$$\n\\begin{align*}\nf_A(x) &= (x+1)(x+2)(x+3)(x+4) \\left( \\frac{x}{x+1} + \\frac{x}{x+2} + \\frac{x}{x+3} + \\frac{x}{x+4} - A \\right) \\\\\n&= (4-A)x^4 + 30(3-A)x^3 + 35(2-A)x^2 + 50(1-A)x - 24A.\n\\end{align*}\n$$\n\nThen, for $A = 1, 2, 3$ we have\n\n$$\n\\begin{align*}\nf_1(x) &= 3x^4 + 60x^3 + 35x^2 - 24, \\\\\nf_2(x) &= 2(x^4 + 15x^3 - 25x - 24), \\\\\nf_3(x) &= x^4 - 35x^2 - 100x - 72.\n\\end{align*}\n$$\n\nIt is easy to show that these polynomials do not have rational roots using the rational root lemma.",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 13003,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all functions $f: \\mathbb{R} \\to \\mathbb{R}$ satisfying\n\n$$\nf(x^2y) + 2f(y^2) = (x^2 + f(y)) f(y)\n$$\n\nfor all $x, y \\in \\mathbb{R}$.",
    "options": [],
    "answer": "See solution",
    "solution": "Substituting $x = 1$ gives $f(y) + 2f(y^2) = (1 + f(y))f(y)$, hence\n\n$$\n2f(y^2) = f(y)^2.\n$$\n\nUsing this, we can cancel the $2f(y^2)$ on the left hand side of the original functional equation against the $f(y)^2$ on the right hand side:\n\n$$\nf(x^2y) = x^2f(y).\n$$\n\nSubstituting $y = 1$ in this equation yields $f(x^2) = x^2f(1)$, and substituting $y = -1$ yields $f(-x^2) = x^2f(-1)$. Because $x^2$ takes on all non-negative numbers as values when $x \\in \\mathbb{R}$, we get\n\n$$\nf(x) = \\begin{cases} c x & \\text{if } x \\ge 0, \\\\ d x & \\text{if } x < 0, \\end{cases}\n$$\n\nwith $c = f(1)$ and $d = -f(-1)$. Now substitute $y = 1$ in equation above, which yields $2f(1) = f(1)^2$, hence $2c = c^2$. It follows that $c = 0$ or $c = 2$. If we substitute $y = -1$, then $2f(1) = f(-1)^2$, hence $2c = (-d)^2$. For $c = 0$, we get $d = 0$, and for $c = 2$, we get $d = 2$ or $d = -2$. So, there are three cases:\n\n* $c = 0, d = 0$: then $f(x) = 0$ for all $x$;",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 13004,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $p_n$ be the largest prime divisor of $n^4 + n^2 + 1$ and let $q_n$ be the largest prime divisor of $n^2 + n + 1$.\n\nProve 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.",
    "options": [],
    "answer": "See solution",
    "solution": "We have\n$$\nn^4 + n^2 + 1 = (n^2 + 1)^2 - n^2 = (n^2 - n + 1)(n^2 + n + 1).\n$$\nThus, $p_n = \\max\\{q_n, q_{n-1}\\}$ for $n \\ge 2$.\n\nSince $n^2 - n + 1$ 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, $q_n \\ne q_{n-1}$.\n\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$$\np_n = \\max\\{q_n, q_{n-1}\\} = q_n = \\max\\{q_n, q_{n+1}\\} = p_{n+1}.\n$$\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, let $m$ be its largest element.\n\nIt is impossible that $q_m > q_{m+1} > q_{m+2} > \\dots$ because all these numbers are positive integers, so there exists $k \\ge m$ such that $q_k < q_{k+1}$ (recall $q_k \\ne q_{k+1}$).\n\nIt is also impossible to have $q_k < q_{k+1} < q_{k+2} < \\dots$, because\n$$\nq_{(k+1)^2} = p_{k+1} = \\max\\{q_k, q_{k+1}\\} = q_{k+1}.\n$$\nLet $l \\ge k+1$ be the smallest such that $q_l > q_{l+1}$. By minimality, $q_{l-1} < q_l$, so $l \\in S$. Since $l \\ge k+1 > k \\ge m$, this contradicts the maximality of $m$. Thus, $S$ is infinite.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 13005,
    "subject": "Mathematics (Olympiad)",
    "question": "The figure below shows a dotted grid 8 cells wide and 3 cells tall consisting of $1'' \\times 1''$ squares. Carl places 1-inch toothpicks along some of the sides of the squares to create a closed loop that does not intersect itself. The numbers in the cells indicate the number of sides of that square that are to be covered by toothpicks, and any number of toothpicks are allowed if no number is written. In how many ways can Carl place the toothpicks?\n\n![](images/2024_AMC10A_Solutions_p13_data_1e21fd6fe6.png)",
    "options": [],
    "answer": "See solution",
    "solution": "There are two possibilities for the loop if it does not cross the middle row of 1s: an $8 \\times 1$ rectangle around the top row of cells or an $8 \\times 1$ rectangle around the bottom row of cells. Otherwise, wherever the loop crosses the middle row, it must proceed straight in both directions after the middle row. This divides the dotted grid into two connected components, and both ends of the loop must enter the same connected component. It follows that the loop must cross one of the leftmost two columns and one of the rightmost two columns for a total of four configurations.\n\n![](images/2024_AMC10A_Solutions_p13_data_23829943c8.png)\n\n![](images/2024_AMC10A_Solutions_p13_data_9dace47213.png)\n\nIn each case there are two ends of the loop that must connect—one across the top and one across the bottom—along with some number of 1s in the middle of the grid. Suppose there are $n$ such 1s. For each 1, the loop can cover either the top edge or the bottom edge. These choices are independent of each other, so the number of ways to connect both ends of the loop is $2^n$. The figure below demonstrates one possibility when $n = 6$.\n\n![](images/2024_AMC10A_Solutions_p14_data_6214cd88ab.png)\n\nOf the four possibilities for what happens at the ends, one gives $n = 6$, two give $n = 5$, and one gives $n = 4$. The total number of solutions is\n\n$$\n2^{6} + 2 \\cdot 2^{5} + 2^{4} + 2 = 64 + 64 + 16 + 2 = 146.\n$$\n\n**Note:** The rules for laying matchsticks mirror those found in the logic puzzle genre *Slitherlink*, which originated from Japan in the early 1990s.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 13006,
    "subject": "Mathematics (Olympiad)",
    "question": "Given the equation:\n\n$$\n\\frac{a^b b^a}{a^a b^b} = \\frac{16}{81}\n$$\n\nFind the value of $a + b$ for positive integers $a \\neq b$.",
    "options": [],
    "answer": "See solution",
    "solution": "Assume $b > a$. The equation can be rewritten as:\n\n$$\n\\frac{a^b b^a}{a^a b^b} = \\frac{a^{b-a}}{b^{b-a}} = \\left(\\frac{a}{b}\\right)^{b-a}\n$$\n\nSet $\\left(\\frac{a}{b}\\right)^{b-a} = \\left(\\frac{2}{3}\\right)^4$, so $\\frac{a}{b} = \\frac{2}{3}$ and $b - a = 4$.\n\nThus, $a = 8$ and $b = 12$, so $a + b = 20$.\n\nTo prove uniqueness for integer solutions:\n\nLet $a = dx$, $b = dy$ with $d = \\gcd(a, b)$, $y > x$, and $\\gcd(x, y) = 1$.\n\nSubstitute into the equation:\n\n$$\n\\frac{(dx)^{dy} (dy)^{dx}}{(dx)^{dx} (dy)^{dy}} = \\frac{16}{81}\n$$\n\nThis simplifies to $x^{d(y-x)} = 16$.\n\nPossible $x$ values: $2, 4, 16$.\n\n- If $x = 2$: $d(y-x) = 4$, $y^4 = 81$, $y = 3$, $d = 4$, $a = 8$, $b = 12$.\n- If $x = 4$: $d(y-x) = 2$, $y^2 = 81$, $y = 9$, $d$ not integer.\n- If $x = 16$: $d(y-x) = 1$, $y = 81$, $d$ not integer.\n\nThus, $\\{8, 12\\}$ is the only integer solution pair.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 13007,
    "subject": "Mathematics (Olympiad)",
    "question": "Let two circles $O$ and $O'$ intersect at points $E$ and $F$. Let $D$ be any point on the circle $O$ such that $D$, $O'$, $F$ are not collinear. Let $G$ be the second intersection point of $DE$ with $O$. Suppose the circumcircle of triangle $DOF$ intersects the circle $O'$ at the second point $A$. Prove that $\\angle GAO = 90^\\circ$.\n\n![](images/Saudi_Arabia_booklet_2024_p38_data_44ffb0c624.png)",
    "options": [],
    "answer": "See solution",
    "solution": "Let $H$ be the second intersection point of line $AG$ with the circumcircle of triangle $DOF$. Then,\n\n$$\n\\angle HDF = \\angle FAG = \\angle FEG = 180^\\circ - \\angle DEF.\n$$\n\nTherefore, $HD$ is tangent to circle $O$, so $\\angle HDO = 90^\\circ$. Consider the circumcircle of triangle $DOF$, then\n\n$$\n\\angle HDO + \\angle HAO = 90^\\circ \\implies \\angle HAO = 90^\\circ.\n$$\n\nThus, $\\angle GAO = 90^\\circ$. $\\square$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 13008,
    "subject": "Mathematics (Olympiad)",
    "question": "Nonnegative integers $a$ and $b$ have the following property: $d(na) \\ge d(nb)$ for each positive integer $n$, where $d(k)$ is the number of divisors of $k$. Prove that $a$ is divisible by $b$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $a = p_1^{\\alpha_1} \\dots p_m^{\\alpha_m}$ and $b = p_1^{\\beta_1} \\dots p_m^{\\beta_m}$ be the prime decompositions of these numbers (some $\\alpha_k$ or $\\beta_k$ may be $0$). We claim that for each $k$, $\\alpha_k \\ge \\beta_k$. Suppose not; for some $k$, say $\\alpha_1 < \\beta_1$. Consider $n = p_2^s \\dots p_m^s$ for large $s$. Then:\n\n$$\n\\frac{d(na)}{d(nb)} = \\frac{(s + \\alpha_1)(s + \\alpha_2) \\dots (s + \\alpha_m)}{(s + \\beta_1)(s + \\beta_2) \\dots (s + \\beta_m)}\n$$\n\nFor large $s$, this ratio approaches $\\frac{\\alpha_1}{\\beta_1} < 1$, contradicting $d(na) \\ge d(nb)$. Thus, $\\alpha_k \\ge \\beta_k$ for all $k$, so $a$ is divisible by $b$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 13009,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $b$ be a positive real number. Find all functions $f: \\mathbb{R} \\to \\mathbb{R}$ satisfying\n\n$$\nf(x+y) = f(x) 3^{b^y + f(y) - 1} + b^x \\left(3^{b^y + f(y) - 1} - b^y\\right) \\quad \\forall x, y \\in \\mathbb{R}.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "The given equation is equivalent to\n\n$$\nf(x+y) + b^{x+y} = (f(x) + b^x) 3^{b^y + f(y) - 1} \\quad \\forall x, y \\in \\mathbb{R}. $$\n\nLet $g(x) = f(x) + b^x$. Then the equation becomes\n\n$$\ng(x+y) = g(x) 3^{g(y) - 1} \\quad \\forall x, y \\in \\mathbb{R}. $$\n\nSubstituting $y = 0$ gives\n\n$$\ng(x) = g(x) 3^{g(0) - 1} \\implies \\begin{cases} g(x) = 0 & \\forall x \\\\ g(0) = 1 \\end{cases} $$\n\nIf $g(x) = 0$ for all $x$, then $f(x) = -b^x$.\n\nIf $g(0) = 1$, then for $x = 0$:\n\n$$\ng(y) = g(0) 3^{g(y) - 1} \\implies g(y) = 3^{g(y) - 1} $$\n\nConsider $h(t) = 3^{t - 1} - t$. The equation $h(t) = 0$ has two roots: $t_1 = 1$ and $t_2 = c$ with $0 < c < 1$.\n\nThus, $g(y) = 1$ or $g(y) = c$ for all $y$. If $g(y_0) = c$ for some $y_0$, then $g(-y_0) = \\frac{1}{c} \\neq c$, which is a contradiction. Therefore, $g(y) = 1$ for all $y$, so $f(x) = 1 - b^x$.\n\n**Conclusion:** The solutions are $f(x) = -b^x$ and $f(x) = 1 - b^x$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 13010,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $\\mathbb{Z}^+$ denote the set of all non-negative integers. Find all functions $f: \\mathbb{Z}^+ \\to \\mathbb{R}$ such that for any $m, n \\in \\mathbb{Z}^+$, $n \\ge m$, the following equality holds:\n\n$$\nf(n+m) + f(n-m) = f(4n).\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "Substituting $n = m = 0$, we get $f(0) = 0$.\n\nNow substitute $m = 0$ into the given equality:\n\n$$\n2f(n) = f(4n).\n$$\n\nNext, substitute $n = m$ into the original equality to obtain:\n\n$$\nf(2n) = f(4n).\n$$\n\nThese two conditions imply:\n\n$$\nf(4n) = f(8n) = 2f(2n) = 2f(4n).\n$$\n\n![](images/Ukrajina_2013_p20_data_2dc545c9e1.png)\n\nThis means that $f(4n) = 0$ for all positive integers $n$, and so from the previous equation it follows that $f(n) = 0$ for all positive integers $n$. Obviously, the function $f(n) = 0$ satisfies the equality from the problem statement.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 13011,
    "subject": "Mathematics (Olympiad)",
    "question": "We say that the unitary ring $ (A, +, \\cdot) $ has property (P) if:\n\n$$\n(P) \\quad \\left\\{ \\begin{array}{l}\n\\text{the set } A \\text{ has at least 4 elements,} \\\\\n\\text{the element } 1+1 \\text{ is invertible in } A, \\\\\nx+x^4 = x^2+x^3, \\text{ for any } x \\in A.\n\\end{array} \\right.\n$$\n\na) Prove that, if a ring $ (A, +, \\cdot) $ has property (P) and $ a, b $ are distinct elements of $ A $ such that $ a $ and $ a+b $ are invertible, then $ b $ is not invertible, but $ 1+ab $ is invertible.\n\nb) Give an example of a unitary ring possessing (P).",
    "options": [],
    "answer": "See solution",
    "solution": "Denote $U(A)$ the set of invertible elements in $A$. For $k \\in \\mathbb{N}$, $k \\ge 2$, and $x \\in A$, define $kx = \\underbrace{x+x+\\cdots+x}_{k \\text{ terms}}$. In particular, $k \\cdot 1 = k$. By the given conditions, $2 \\in U(A)$. Denote by (1) the equality $x+x^4 = x^2+x^3$ for all $x \\in A$.\n\nChanging $x$ by $-x$ in (1) we get $-x+x^4 = x^2-x^3$, for any $x \\in A$.\n\nBy subtraction, the last two relations give $2x = 2x^3$, so, as $2 \\in U(A)$, we obtain $x = x^3$ for all $x \\in A$.\n\nFor $x \\in U(A)$, multiplying by $x^{-1}$ we get $x^2 = 1$ for any $x \\in U(A)$. As the set $U(A)$, of the invertible elements of the monoid $(A, \\cdot)$, is a group with $x^2 = 1$ for any $x \\in U(A)$, it results that the group $(U(A), \\cdot)$ is commutative.\n\nFor $x = 2 \\in U(A)$ from the previous relations, we get $4 = 1$, so $3 = 0$, meaning that the ring $A$ is of characteristic 3.\n\nLet $a, b \\in A$, such that $a \\neq b$ and $a, a+b \\in U(A)$. If we suppose $b \\in U(A)$, then\n\n$$\n2ab = (a+b)^2 - a^2 - b^2 = 1 - 1 - 1 = -1 = 2,\n$$\n\nimplying $ab = 1 = a^2$. This gives $a = b$, a contradiction. That is, $b \\notin U(A)$.\n\nWe also have\n\n$$\n1 + ab = a^2 + ab = a(a+b) \\in U(A),\n$$\n\na product of invertible elements.\n\nb) Let $A = \\mathbb{Z}_3 \\times \\mathbb{Z}_3$, with the natural operations:\n\n$$\n(a, b) + (c, d) = (a + c, b + d), \\quad (a, b) \\cdot (c, d) = (a \\cdot c, b \\cdot d),\n$$\n\nfor any $a, b, c, d \\in \\mathbb{Z}_3$. $A$ is a unitary ring with 9 elements, with unity $1 = (\\hat{1}, \\hat{1})$, and $1 + 1 = (\\hat{2}, \\hat{2}) \\in U(A)$. Evidently $x^3 = x$, for any $x \\in A$, that is $x^4 = x^2$ and $x + x^4 = x^2 + x^3$, for any $x \\in A$.",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 13012,
    "subject": "Mathematics (Olympiad)",
    "question": "Circles $\\Omega$ and $\\omega$ are tangent at a point $P$ ($\\omega$ lies inside $\\Omega$). A chord $AB$ of $\\Omega$ is tangent to $\\omega$ at $C$; the line $PC$ meets $\\Omega$ again at $Q$. Chords $QR$ and $QS$ of $\\Omega$ are tangent to $\\omega$. Let $I$, $X$, and $Y$ be the incentres of the triangles $APB$, $ARB$, and $ASB$, respectively. Prove that\n\n$$\n\\angle PXI + \\angle PYI = 90^{\\circ}.\n$$\n\n![](images/RMC2013_final_p44_data_1a3aaa5356.png)",
    "options": [],
    "answer": "See solution",
    "solution": "Notice that a homothety centred at $P$ mapping $\\omega$ to $\\Omega$ maps $C$ to $Q$, and maps the line $AB$ to the tangent to $\\Omega$ at $Q$. Thus this tangent is parallel to $AB$, and hence $Q$ is the midpoint of arc $AB$ (not containing $P$). So the points $I$, $X$, and $Y$ lie on the segments $PQ$, $RQ$, and $SQ$, respectively.\n\nRecall that for any triangle $KLM$ with circumcircle $\\Gamma$ and incentre $J$, the points $K$, $L$, and $J$ are equidistant from the midpoint of arc $KL$ of $\\Gamma$ not containing $M$. Applying this to triangles $APB$, $ARB$, and $ASB$ we obtain that\n\n$$\nQA = QB = QX = QY = QI.\n$$\n\nSince $Q$ is the midpoint of arc $AB$, we get that $\\angle QPA = \\angle QPB = \\angle QAB$. Thus the triangles $QAC$ and $QPA$ are similar, and $QC \\cdot QP = QA^2 = QX^2$. Since $QX$ is tangent to $\\omega$, it follows that $X$ is their point of tangency; analogously, $Y$ is the point of tangency of $QS$ with $\\omega$.\n\nFinally, from isosceles triangles $QXI$ and $QYI$ we get $\\angle QXI = \\angle QIX = 90^{\\circ} - \\angle IQX/2$ and $\\angle QYI = \\angle QIY = 90^{\\circ} - \\angle IQY/2$. Denoting by $O$ the centre of $\\omega$, we obtain $\\angle QIX + \\angle QIY = 180^{\\circ} - \\angle XQY/2 = 180^{\\circ} - (180^{\\circ} - \\angle XOY)/2 = 90^{\\circ} + \\angle XPY$. Thus,\n\n$$\n\\angle PXI + \\angle PYI = \\angle XIY - \\angle XPY = (90^{\\circ} + \\angle XPY) - \\angle XPY = 90^{\\circ},\n$$\n\nas required.\n\n**Remark.** The relation $QC \\cdot QP = QA^2$ also follows from the inversion of pole $Q$ interchanging the line $AB$ and the circle $\\Omega$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 13013,
    "subject": "Mathematics (Olympiad)",
    "question": "Suppose the sequence of numbers shown on the calculator screen is\n\n$$\nn \\to p \\to q \\to r \\to 0\n$$\n\nwith $p$, $q$, $r$ nonzero. Note that $r$ is at least $1$ and so $q$ consists of at least one even digit, which means $q \\ge 2$. Hence $p$ consists of at least two even digits and is no less than $20$. What is the smallest possible value of $n$?",
    "options": [],
    "answer": "See solution",
    "solution": "The smallest such $n$ has at least $20$ even digits, so $n = 2 \\times 10^{19}$ is possible. The sequence in this case is $20\\ldots00 \\to 20 \\to 2 \\to 1 \\to 0$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 13014,
    "subject": "Mathematics (Olympiad)",
    "question": "Given a quadrilateral $ABCD$ with fixed values of $AB = x$, $BD = y$, and $CD = z$, what is the maximal possible area of $ABCD$ if $AB + BD + DC = L$?",
    "options": [],
    "answer": "See solution",
    "solution": "The area is maximized when $\\angle ABD = \\angle CDB = 90^\\circ$. Thus,\n\n$$\nS(ABCD) = \\frac{1}{2}xy + \\frac{1}{2}yz = \\frac{1}{2}y(x+z)\n$$\n\nGiven $x + y + z = L$, so $x + z = L - y$, we have:\n\n$$\nS(ABCD) = \\frac{1}{2}y(L - y)\n$$\n\nThe product $y(L - y)$ is maximized for $0 < y < L$ when $y = L/2$, yielding a maximal area of $L^2/8$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 13015,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $ABC$ be a triangle with orthocenter $H$ and let $P$ be the second intersection of the circumcircle of triangle $AHC$ with the internal bisector of the angle $\\angle BAC$. Let $X$ be the circumcenter of triangle $APB$ and $Y$ the orthocenter of triangle $APC$. Prove that the length of segment $XY$ is equal to the circumradius of triangle $ABC$.",
    "options": [],
    "answer": "See solution",
    "solution": "It is well-known that the reflection $H'$ of the orthocenter $H$ in the line $AC$ lies on the circumcircle of triangle $ABC$. Hence, the circumcenter of triangle $CAH'$ coincides with the circumcenter of triangle $ABC$. But since $H'$ is the reflection of $H$ in the line $AC$, the triangles $ACH$ and $CAH'$ are symmetric with respect to $BC$, and the circumcenter $O'$ of triangle $ACH$ must be the reflection of the circumcenter of triangle $CAH'$ in the line $BC$, i.e., the reflection of the circumcenter of triangle $ABC$ in the line $CA$.\n\nNow since the quadrilateral $AHPC$ is cyclic and since $H$, $Y$ are the orthocenters of triangles $ABC$, and $APC$, respectively, we have that\n\n$$\n\\angle ABC = 180^\\circ - \\angle AHC = 180^\\circ - \\angle APC = \\angle AYC.\n$$\n\nHence the point $Y$ lies on the circumcircle of triangle $ABC$, and therefore $OC = OY = R$, where $R$ denotes the circumradius of triangle $ABC$.\n\nOn the other hand, note that the lines $OX$, $XO'$, $O'O$ are the perpendicular bisectors of the segments $AB$, $AP$, and $AC$, respectively, we get\n\n$$\n\\angle OXO' = \\angle BAP = \\angle PAC = m(\\angle XO'O).\n$$\n\nThus $OO' = OX$. Combining this with $OC = OY$ and with the parallelism of the lines $XO'$ and $YC$ (note that these two lines are both perpendicular to $AP$), we conclude that the trapezoid $XYCO'$ is isosceles, and therefore $XY = O'C = OC = R$. This completes our proof. $\\square$\n\n**Remark.** If $ABC$ is right-angled at $A$, then the statement is trivially true if we convene that the circumcenter of $AB$ is the midpoint of $AB$ and that the orthocenter of $AC$ is the midpoint of $AC$. Then, we have that $XY = \\frac{1}{2}BC = R$.\n\n![](images/USA_IMO_2013-2014_p23_data_d439c31dba.png)\n\nBecause $ABC$ is acute, $H$ lies inside the triangle. We consider the configuration shown above. (For other possible configurations, it is not difficult to adjust our proof properly.)\n\nLet $O$ and $Z$ denote the circumcenters of triangles $ABC$ and $APC$ respectively. Let $\\omega$ and $r$ denote the circumcircle and the circumradius of triangle $ABC$ respectively. We will show that\n\n$$\nXYCZ \\text{ is an isosceles trapezoid with } XY = CZ = r. \\qquad (13)\n$$\n\nBecause $X$ and $Z$ are the circumcenters of triangle $APB$ and $APC$, line $XZ$ is the perpendicular bisector of segment $AP$. Because $Y$ is the orthocenter of triangle $APC$, $CY \\perp AP$. Hence both lines $XZ$ and $CY$ are perpendicular to line $AP$, implying that $XYZC$ is a trapezoid with $XZ \\parallel CY$.\n\nBecause $X$ and $O$ are the circumcenters of triangles $APB$ and $ABC$, line $XO$ is the perpendicular bisector of segment $AB$. Because $XO \\perp AB$ and $XZ \\perp AP$, the acute angles formed by lines $XO$ and $XZ$ is equal to the acute angle formed by lines $AP$ and $AB$; that is, $\\angle OXZ = \\angle BAP$. Likewise, we can show that $\\angle OZX = \\angle CAP$. Therefore, we have $\\angle OXZ = \\angle BAP = \\angle CAP = \\angle OZX$, implying that $OX = OZ$; that is, $O$ lies on the perpendicular bisector of segment $XZ$.\n\nBecause $H$ is the orthocenter of acute triangle $ABC$, $\\angle AHC = 180^\\circ - \\angle ABC$. Because $APHC$ is cyclic, we have $\\angle APC = \\angle AHC = 180^\\circ - \\angle ABC$. Now in obtuse triangle $APC$, $\\angle AYC = 180^\\circ - \\angle APC = \\angle ABC$. (This relates to the fact of orthocenter group: if one point is the orthocenter of the triangle formed by the other three points, then any of the four point is the orthocenter of the triangle formed by the other three.) In particular, this means that $Y$ lies on $\\omega$; that is, $OY = OC = r$. Note that in trapezoid $XYCZ$, the perpendicular bisectors of the bases $YC$ and $XZ$ share a common point $O$. Thus, these two bisectors must coincide; that is, $XYCZ$ is an isosceles trapezoid with $XY = CZ$, establishing the first part of (13).\n\nTo complete our proof, it suffices to show that $CZ = r$. Let $Q$ be the reflection of $H$ across line $AC$. It is well known that $Q$ lies $\\omega$ (because $\\angle ACQ = \\angle ACH = 90^\\circ - \\angle BAC = \\angle ABH = \\angle ABQ$.) We note that triangle $AQC$ and its circumcenter $O$ and triangle $AHC$ and its circumcenter $Z$ are respective images of each other across line $AC$. In particular, we conclude that $CZ = CO = r$, completing our proof.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 13016,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $D$, $E$, and $F$ be the feet of the altitudes from vertices $A$, $B$, and $C$ of triangle $ABC$, respectively. Prove that if points $X$, $Y$, and $Z$ are collinear, then the length of the tangent from $A$ to the nine-point circle $\\\\omega$ of triangle $ABC$ is equal to the sum of the lengths of the tangents from $B$ and $C$ to this circle.\n\nAdditionally, prove the converse: if the length of the tangent from $A$ to the nine-point circle equals the sum of the lengths of the tangents from $B$ and $C$, then $X$, $Y$, and $Z$ are collinear.",
    "options": [],
    "answer": "See solution",
    "solution": "$$\nP_{\\omega}^{A} = \\frac{1}{2}AE \\cdot AC = \\frac{1}{2}AF \\cdot AB = \\frac{1}{2}AX^{2} = \\frac{1}{2}AY^{2}\n$$\n\nSo the length of the tangent from $A$ to $\\omega$ is $\\frac{\\sqrt{2}}{2}AX = \\frac{\\sqrt{2}}{2}AY$. Similarly, the lengths of tangents from $B$ and $C$ are equal to $BY$ and $CX$, respectively. If $X$, $Y$, and $Z$ are collinear, we have $YZ + ZX = YX$ and $\\angle XZC + \\angle YZB = 90^{\\circ}$.\n\nSince $CX^2 = CE \\cdot CA = CD \\cdot CB = CZ^2$, we get $CX = CZ$. Similarly, $AX = AY$ and $BY = BZ$. Using these, we deduce $\\angle XCZ + \\angle ZBY = 180^{\\circ}$. The quadrilateral $FZCB$ is cyclic, so $\\angle ABZ = \\angle FCZ$. Therefore,\n\n$$\n\\angle YBA + \\angle FCX = 90^{\\circ} \\Rightarrow \\angle YBA + (90^{\\circ} - \\angle A) + \\angle ACX = 90^{\\circ}\n$$\n\nThus, $\\angle YAX = 90^{\\circ}$ and so $\\angle AXY = 45^{\\circ}$. From these results, $\\angle ZBY = \\angle ZCX = 90^{\\circ}$. Hence, $\\sqrt{2}YB + \\sqrt{2}CX = \\sqrt{2}AX$ and so $YB + CX = AX$, as required.\n\n**Converse Lemma:**\n\nLet $C_1$ and $C_2$ be two perpendicular circles meeting at points $A$ and $B$. There are exactly two points $T$ on line $AB$ such that the tangents $TY$ and $TX$ from $T$ to $C_1$ and $C_2$ (with $Y$ and $X$ on opposite sides of $AB$) satisfy that $A$, $Y$, and $X$ are collinear. These two points are symmetric with respect to the line of centers of $C_1$ and $C_2$, and their powers with respect to $C_1$ and $C_2$ are both $(R_1 + R_2)^2$, where $R_i$ is the radius of $C_i$.\n\n*Proof of Lemma:* If $T$ has this property,\n\n$$\n\\frac{1}{2} \\hat{A}Y = \\angle TYA = \\angle TYX = \\angle TXY = \\angle TXA = \\frac{1}{2} \\hat{A}X\n$$\n\nSince $\\angle O_1AO_2 = 90^{\\circ}$ ($O_i$ is the center of $C_i$),\n\n$$\n\\hat{A}Y = \\hat{A}X = \\frac{1}{2}\\hat{A}Y + \\frac{1}{2}\\hat{A}X = \\angle XAO_2 + \\angle YAO_1 = 180^{\\circ} - \\angle O_1AO_2 = 90^{\\circ}\n$$\n\nSo $\\hat{A}Y = \\hat{A}X = 90^{\\circ}$.\n\nFor existence, choose $Y$ and $X$ on $C_1$ and $C_2$ such that $\\hat{A}Y = \\hat{A}X = 90^{\\circ}$. These points are unique on one side of $O_1O_2$, and $Y$, $A$, and $X$ are collinear. If $T$ is the intersection of the tangents to $C_1$ at $Y$ and $C_2$ at $X$,\n\n$$\n\\angle TYX = \\hat{Y}A = \\hat{X}A = \\angle TXY\n$$\n\nSo $TY = TX$, and $T$ lies on the radical axis of $C_1$ and $C_2$. In hexagon $TXO_2AO_1Y$,\n\n$$\n\\angle TYO_1 = \\angle YO_1A = \\angle O_1AO_2 = \\angle AO_2X = \\angle O_2XT = 90^{\\circ}\n$$\n\nHence, $\\angle YTX = 90^{\\circ}$, and $TX = AO_2 + YO_1 = R_2 + R_1$, as desired.\n\nFor the main problem, the circle $\\omega_1$ with center $B$ and radius $BZ$ is perpendicular to the circle $\\omega_2$ with center $C$ and radius $CZ$, because $\\angle BZX = 90^{\\circ}$. Since $\\angle AYB = \\angle AXC = 90^{\\circ}$, $AY$ is tangent to $\\omega_1$ at $Y$ and $AX$ is tangent to $\\omega_2$ at $X$. Also, $AY = AX = BY + CX$, so by the lemma, the point with this property is unique on one side of $BC$. Therefore, by the lemma, $X$, $Y$, and $Z$ are collinear.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 13017,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $N$ be a positive integer, and let $\\mathbf{a} = (a(1), \\dots, a(N))$ and $\\mathbf{b} = (b(1), \\dots, b(N))$ be sequences of non-negative integers, each written on a circle (so we assume $a(i \\pm N) = a(i)$ and $b(i \\pm N) = b(i)$). We say $\\mathbf{a}$ is **b-harmonic** if each $a(i)$ is the arithmetic mean of the counterclockwise nearest $b(i)$ numbers, the clockwise nearest $b(i)$ numbers, and $a(i)$ itself; that is,\n\n$$\na(i) = \\frac{1}{2b(i) + 1} \\sum_{s=-b(i)}^{b(i)} a(i + s). \\quad (*)\n$$\n\n(A term of $\\mathbf{a}$ may appear more than once in the above sum.) Suppose that neither $\\mathbf{a}$ nor $\\mathbf{b}$ is constant, and that both $\\mathbf{a}$ is $\\mathbf{b}$-harmonic, and $\\mathbf{b}$ is $\\mathbf{a}$-harmonic. Prove that more than half of the $2N$ terms across both sequences vanish.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $a = \\min_i a(i)$ and $b = \\min_i b(i)$. Since $\\mathbf{a}$ is not constant, there exists an $i$ such that $a = a(i) < a(i + 1)$.\n\n**Claim 1.** If $a = a(i) < a(i + 1)$, then $b(i) = 0$. Similarly, if $a = a(i) < a(i - 1)$, then $b(i) = 0$.\n\n*Proof.* Otherwise the sum in $(*)$ contains a term $a(i+1) > a$ but no terms smaller than $a$, so the average is greater than $a$. $\\square$\n\nClaim 1 implies $b=0$; similarly, $a=0$. With reference again to Claim 1, $a(i) = b(i) = 0$ for some index $i$.\n\nSay that $[i, j]$ is an **a-segment** if $a(i) = a(i+1) = \\cdots = a(j) = 0$ but $a(i-1) \\neq 0 \\neq a(j+1)$; define a **b-segment** similarly. By Claim 1, the endpoints of any such segment satisfy $a(i) = b(i) = a(j) = b(j) = 0$. Since the sequences are non-constant, each $i$ where $a(i) = 0$ is contained in an **a-segment**.\n\n**Claim 2.** Let $[i, j]$ be a **b-segment**, and let $k \\in [i, j]$. Then $a(k) \\le k - i$ (and, similarly, $a(k) \\le j - k$).\n\n*Proof.* Indeed, since $b(k) = 0$, the elements of $\\mathbf{b}$ with indices from $k-a(k)$ to $k+a(k)$ must all be zero as well. $\\square$\n\nWe now show that every index is contained in either an **a-** or a **b-segment**. Since at least one index is contained in both, the conclusion follows.\n\nAssume, to the contrary, that $a(i)$ and $b(i)$ are both positive for some index $i$; call such indices *bad*. Among all bad indices $i$, choose one maximising $\\max(a(i), b(i))$; by symmetry, we may and will assume that this maximum is $a(i)$. We may and will also assume that either the index $i-1$ is not bad, or $a(i-1) < a(i)$ (otherwise change $i$ to $i-1$, repeat if necessary, recalling that $\\mathbf{a}$ is not constant).\n\nConsider the range of indices $\\Delta = [i - b(i), i + b(i)]$, and the values $\\mathbf{a}$ assumes at those indices. Some indices $j$ in $\\Delta$ are bad; the corresponding values $a(j)$ do not exceed $a(i)$. Other indices $j$ in $\\Delta$ are covered by several **a-** and **b-segments**. Each **b-segment** contributes at most $b(i)$ members nearest to one of its endpoints, so the average value of $\\mathbf{a}$ over those indices does not exceed $(b(i) - 1)/2 < a(i)$ by Claim 2. The remaining indices $j$ in $\\Delta$ all lie in **a-segments**, so the corresponding values $a(j)$ are all zero.\n\nCombining all this, it follows that the average in the right-hand member of $(*)$ does not exceed $a(i)$. Moreover, if some **a-** or **b-segment** intersects $\\Delta$, then the inequality is strict. Otherwise, $i-1$ is a bad index contained in $\\Delta$, and $a(i-1) < a(i)$, so the inequality is again strict. This contradiction ends the proof and completes the solution.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 13018,
    "subject": "Mathematics (Olympiad)",
    "question": "We say that a natural number is of type 1 (respectively type 2) if each of its digits at even (respectively odd) position is greater than or equal to each of its adjacent digits. Positions are counted from left to right; leading zeros are not allowed (the first digit is assumed nonzero). One-digit numbers are considered to be both of type 1 and of type 2.\n\nDecide if it is true that:\n\n- a) Each number $a > 1$ of type 1 can be represented as $a = b + c$ with $b, c$ numbers of type 2.\n- b) Each number $a > 1$ of type 2 can be represented as $a = b + c$ with $b, c$ numbers of type 1.",
    "options": [],
    "answer": "See solution",
    "solution": "The answer is yes for part a) and no for part b).\n\nConsider a number $a > 1$ of type 1. If $a$ is a 1-digit number then $a = (a - 1) + 1$ is the desired representation since $a - 1$ and $1$ are 1-digit numbers, hence type 2 numbers by definition.\n\nLet $a$ have at least two digits. Write it in the form $a = u_1 v_1 u_2 v_2 u_3 v_3 \\ldots$ where $u_1, u_2, \\ldots$ and $v_1, v_2, \\ldots$ are its digits at odd and even positions respectively. We have $u_1 > 0$ for the first digit $u_1$, also $v_1 \\geq u_1 > 0$ since $a$ is of type 1.\n\nNow let $b = u_1 0 u_2 0 u_3 0 \\ldots$ be the number obtained by replacing all digits at even positions by $0$. This is a type 2 number because $u_i \\geq 0$ for all $i$.\n\nNext, construct number $c$ as follows: delete the first digit $u_1$ from $a$, then replace all remaining digits $u_2, u_3, \\ldots$ at odd positions by zeros. In other words, $c = v_1 0 v_2 0 v_3 0 \\ldots$; the first digit $v_1$ is nonzero. Clearly $c$ is also of type 2, and it has one digit less than $b$ (and $a$). Now it follows from the rule of addition that $a = b + c$, so part a) is done.\n\nFor part b), we show that the type 2 number $109$ is not representable as the sum of two type 1 numbers. Suppose on the contrary that such a representation exists. If one summand is among $1, \\ldots, 9$ then the other is among $100, \\ldots, 108$, and the latter numbers are not of type 1. So let both summands be 2-digit numbers, and let $10u + v$ be one of them, with $0 < u \\leq v \\leq 9$ as $10u + v$ is of type 1. The other summand is then $b = 109 - (10u + v) = 10(10 - u) + (9 - v)$. Because $1 \\leq 10 - u \\leq 9$ and $0 \\leq 9 - v \\leq 8$, the digits of $b$ are $10 - u$ and $9 - v$ in this order. However, $10 - u > 9 - v$ because $u \\leq v$, hence $b$ is not a type 1 number, contrary to the assumption.\n\n**Remark.** In fact, $109$ is the least counterexample for part b).",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 13019,
    "subject": "Mathematics (Olympiad)",
    "question": "Given quadrilateral $ABCD$ such that $AB = BC$. Let $K$ be the midpoint of $CD$. Rays $BK$ and $AD$ intersect at $M$. The circumcircle of $\\triangle ABM$ intersects line $AC$ for the second time at point $P$. Show that $\\angle BKP = 90^\\circ$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $N$ be the midpoint of diagonal $AC$. Then $NK$ is the midline of $\\triangle ACD$. Thus, $\\angle (KN, NP) = \\angle (AM, AP) = \\angle (BM, BP) = \\angle (KB, BP)$, so $BNKP$ is inscribed. Since $\\triangle ABC$ is equilateral, $\\angle BNP = 90^\\circ$, therefore $\\angle BKP = 90^\\circ$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 13020,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $\\mathbb{Z}_{\\ge 0}$ be the set of non-negative integers and $\\mathbb{R}^+$ be the set of positive real numbers. Let $f: \\mathbb{Z}_{\\ge 0} \\to \\mathbb{R}^+$ be a function such that $f(0, k) = 2^k$ and $f(k, 0) = 1$ for all integers $k \\ge 0$, and\n\n$$\nf(m, n) = \\frac{2 f(m-1, n) \\cdot f(m, n-1)}{f(m-1, n) + f(m, n-1)}\n$$\n\nfor all integers $m, n \\ge 1$. Prove that $f(99, 99) < 1.99$.",
    "options": [],
    "answer": "See solution",
    "solution": "We in fact prove that $0 < f(m, n) \\le 1$ for all $m, n \\ge 0$ by induction on $m + n$.\n\nBase case: $m + n = 0$, which only occurs when $m = n = 0$; in that case $0 < f(0, 0) = 1 \\le 1$.\n\nSuppose the claim is true for all $m, n$ with $m + n = N$. Choose $m, n \\ge 0$ with $m + n = N + 1$. If one of $m, n$ is $0$, we are done since $0 < f(N+1, 0) = 1 \\le 1$ and $0 < f(0, N+1) = 2^{N+1} > 1$ (but this does not affect $f(99,99)$ since both indices are positive).\n\nElse, $f(m, n)$ is the harmonic mean of positive numbers $f(m, n-1)$ and $f(m-1, n)$, so it is positive and lies between them. By induction hypothesis, $f(m, n-1) \\le 1$ and $f(m-1, n) \\le 1$. Therefore,\n\n$$\n0 < f(m, n) \\le \\max(f(m, n-1), f(m-1, n)) \\le 1\n$$\n\nSo we are done by induction. $\\square$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 13021,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $A$ and $B$ be two $3 \\times 3$ matrices with real entries such that $AB = O_3$.\n\n**a)** Prove that $f : \\mathbb{C} \\to \\mathbb{C}$, $f(x) = \\det(A^2 + B^2 + xBA)$ is a polynomial function of degree not greater than $2$.\n\n**b)** Prove that $\\det(A^2 + B^2) \\ge 0$.",
    "options": [],
    "answer": "See solution",
    "solution": "**a)** Because $\\det(AB) = 0$, we obtain $f(x) = \\det(A^2 + B^2) + a x + b x^2 + \\det(AB)x^3 = \\det(A^2 + B^2) + a x + b x^2$, where $a, b \\in \\mathbb{R}$.\n\n**b)** Observe that $f(x) = \\det(A^2 + B^2 + x(BA \\pm AB))$, due to $AB = O_3$. We get $f(i) = \\det(A^2 + B^2 + iBA) = \\det(A^2 + B^2 + i(BA - AB)) = \\det(A + iB) \\cdot \\det(A - iB)$. So $f(i) = f(-i)$, implying $a = 0$. From $f(i) = |\\det(A + iB)|^2 \\ge 0$ we get $\\det(A^2 + B^2) - b \\ge 0$. On the other side, $f(1) = \\det(A^2 + B^2 + AB + BA) = \\det(A + B)^2 \\ge 0$, so $\\det(A^2 + B^2) + b \\ge 0$. Summing up the last two inequalities gives $\\det(A^2 + B^2) \\ge 0$.",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 13022,
    "subject": "Mathematics (Olympiad)",
    "question": "As shown in the figure below, $D$ is the midpoint of arc $BC$ of the circumcircle of triangle $ABC$. Point $X$ lies on arc $BD$, $E$ is the midpoint of arc $AX$, $S$ lies on arc $AC$. Line $SD$ intersects $BC$ at $R$, and line $SE$ intersects $AX$ at $T$. Prove that if $RT \\parallel DE$, then the incenter of triangle $ABC$ lies on line $RT$.\n\n![](images/Mathematical_Olympiad_in_China_2011-2014_p93_data_e6cb96f912.png)",
    "options": [],
    "answer": "See solution",
    "solution": "Connect $AD$, and denote the intersection of $AD$ and $RT$ by $I$. Then $AI$ is the bisector of $\\angle BAC$. Connect $AS$ and $SI$. Since $RT \\parallel DE$, we have\n\n$$\n\\angle STI = \\angle SED = \\angle SAI,\n$$\n\nso $A$, $T$, $I$, and $S$ are concyclic; denote this circle by $\\omega_1$.\n\nConnect $CE$, and denote the intersection of $CE$ and $RT$ by $J$. Connect $SC$. Then\n\n$$\n\\angle SRJ = \\angle SDE = \\angle SCE,\n$$\n\nso $S$, $J$, $R$, and $C$ are concyclic; denote this circle by $\\omega_2$.\n\nLet $K$ be the intersection point of $\\omega_1$ and $\\omega_2$ other than $S$. Next, we prove that $K$ is the intersection of $AJ$ and $CI$.\n\nLet $K_1$ be the intersection of $\\omega_1$ and $AJ$ other than $A$. Then\n\n$$\n\\angle SK_1A = \\angle STA = \\frac{1}{2}(SA + XE) = \\frac{1}{2}(SA + AE) \\\\\n= \\angle SDE = \\angle SRT = \\angle SRJ,\n$$\n\nso $S$, $K_1$, $J$, and $R$ are concyclic, i.e., $K_1$ belongs to $\\omega_2$. Similarly, let $K_2$ be the intersection of $\\omega_2$ and $CI$ other than $C$, then $K_2$ belongs to $\\omega_1$. Hence, $K_1$ and $K_2$ coincide, and $K$ is the intersection of $AJ$ and $CI$.\n\nSince $\\angle CAD = \\angle CAI$, and $\\angle TJE = \\angle CJR = \\angle CED = \\angle CAD$, so $A$, $I$, $J$, and $C$ are concyclic, therefore $\\angle ACI = \\angle AJI$.\n\nOn the other hand, by the concyclicity of $C$, $K$, $J$, and $R$, we have $\\angle BCI = \\angle ICR = \\angle AJI$, and $\\angle ACI = \\angle BCI$, so $I$ is the incenter of triangle $ABC$. $\\square$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 13023,
    "subject": "Mathematics (Olympiad)",
    "question": "Construct, for each $n \\leq 16$, a coloring (painting pattern) of the vertices $A_1, A_2, \\dots, A_n$ of a regular $n$-gon with three colors (red, yellow, blue), so that no four vertices of the same color form an isogonal trapezoid. Show that such a coloring exists for all $n \\leq 16$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $A_1, A_2, \\dots, A_n$ be the vertices of a regular $n$-gon in clockwise order. Let $M_1, M_2, M_3$ be the sets of vertices colored red, yellow, and blue, respectively.\n\n**For $n = 16$:**\n\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$$\nM_3 = \\{A_1, A_2, A_4, A_9, A_{10}, A_{12}\\}\n$$\n\nIn $M_1$, the distances from $A_{14}$ to the other four vertices are all different, and those four form a rectangle, not an isogonal trapezoid. Similarly, no four vertices in $M_2$ form an isogonal trapezoid. In $M_3$, the six vertices are the endpoints of three diameters, so any four form either a rectangle or a quadrilateral with sides of different lengths.\n\n**For $n = 15$:**\n\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$$\nM_3 = \\{A_4, A_7, A_{10}, A_{11}, A_{12}\\}\n$$\n\nNo four vertices in any $M_i$ form an isogonal trapezoid.\n\n**For $n = 14$:**\n\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$$\nM_3 = \\{A_2, A_6, A_9, A_{13}\\}\n$$\n\nAgain, this can be checked directly.\n\n**For $n = 13$:**\n\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$$\nM_3 = \\{A_2, A_3, A_4, A_9, A_{13}\\}\n$$\n\nThis can also be verified. For $n = 12$, remove $A_{13}$ from $M_3$; for $n = 11$, remove $A_{12}$; for $n = 10$, remove $A_{11}$.\n\n**For $n \\leq 9$:**\n\nWe can color so that $|M_i| < 4$ for each $i$, so no four vertices of the same color exist.\n\nThus, for all $n \\leq 16$, such a coloring exists, so $n = 17$ is the minimal value for which the required property must hold.",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 13024,
    "subject": "Mathematics (Olympiad)",
    "question": "Determine, with proof, whether or not there exist integers $a, b, c > 2010$ satisfying the equation\n\n$$\na^3 + 2b^3 + 4c^3 = 6abc + 1.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "We claim there do exist such integers $a, b, c$.\n\nNote that $(a_1, b_1, c_1) = (1, 1, 1)$ satisfy the given equation. For $n > 1$, define $(a_{n+1}, b_{n+1}, c_{n+1})$ by\n\n$$\n(a_{n+1}, b_{n+1}, c_{n+1}) = (a_n + 2c_n + 2b_n,\\ b_n + a_n + 2c_n,\\ c_n + b_n + a_n).\n$$\n\nIt is not hard to verify algebraically that\n\n$$\na_{n+1}^3 + 2b_{n+1}^3 + 4c_{n+1}^3 - 6a_{n+1}b_{n+1}c_{n+1} = a_n^3 + 2b_n^3 + 4c_n^3 - 6a_n b_n c_n,\n$$\n\nwhich shows that $(a_n, b_n, c_n)$ satisfy the given equation for all $n \\ge 1$. Moreover, notice that $a_{n+1} > a_n$, $b_{n+1} > b_n$, and $c_{n+1} > c_n$ for all $n \\ge 1$; this implies that for sufficiently large $n$ we will have $a_n, b_n, c_n > 2010$, yielding the desired result.\n\n**Remark.** The recursive definition of $a_n$ is inspired by the factorization\n\n$$\na^3 + 2b^3 + 4c^3 - 6abc = \\prod_{k=0}^{2} (a + b\\omega^k \\sqrt[3]{2} + c\\omega^{2k} \\sqrt[3]{4}),\n$$\n\nwhere $\\omega$ is a primitive third root of unity. We then let\n\n$$\n(1 + \\omega^k \\sqrt[3]{2} + \\omega^{2k} \\sqrt[3]{4})(a_n + b_n \\omega^k \\sqrt[3]{2} + c_n \\omega^{2k} \\sqrt[3]{4}) = a_{n+1} + b_{n+1} \\omega^k \\sqrt[3]{2} + c_{n+1} \\omega^{2k} \\sqrt[3]{4}.\n$$",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 13025,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $x_1, x_2, x_3$ and $x_4$ be positive real numbers. Prove the inequality:\n\n$$\n\\frac{x_1 + 3x_2}{x_2 + x_3} + \\frac{x_2 + 3x_3}{x_3 + x_4} + \\frac{x_3 + 3x_4}{x_4 + x_1} + \\frac{x_4 + 3x_1}{x_1 + x_2} \\ge 8.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "Let us denote\n\n$$\nL(x_1, x_2, x_3, x_4) = \\frac{x_1 + 3x_2}{x_2 + x_3} + \\frac{x_2 + 3x_3}{x_3 + x_4} + \\frac{x_3 + 3x_4}{x_4 + x_1} + \\frac{x_4 + 3x_1}{x_1 + x_2}.\n$$\n\nNotice that this function is cyclic, i.e. $L(x_1, x_2, x_3, x_4) = L(x_2, x_3, x_4, x_1) = L(x_3, x_4, x_1, x_2) = L(x_4, x_1, x_2, x_3)$. Hence we can suppose that $x_1 \\ge x_3$ and $x_4 \\ge x_2$. We can also multiply all of the variables with a positive constant without changing its value, i.e. $L(x_1, x_2, x_3, x_4) = L(c \\cdot x_1, c \\cdot x_2, c \\cdot x_3, c \\cdot x_4)$.\n\nFirst we prove that $L(u + v, 0, u, 1) \\ge 8$ for positive numbers $u$ and $v$. Indeed\n\n$$\nL(u + v, 0, u, 1) = \\frac{u+v}{u} + \\frac{3u}{u+1} + \\frac{u+3}{1+u+v} + \\frac{1+3u+3v}{u+v}\n$$\n\n$$\n= 1 + \\frac{v}{u} + 3 - \\frac{3}{u+1} + 1 + \\frac{2-v}{1+u+v} + 3 + \\frac{1}{u+v}\n$$\n\n$$\n= 8 + \\frac{v}{u} - \\frac{v}{1+u+v} - \\frac{3}{u+1} + \\frac{2}{1+u+v} + \\frac{1}{u+v}\n$$\n\n$$\n= 8 + \\frac{v(1+v)}{u(1+u+v)} - \\frac{2v}{(u+1)(1+u+v)} + \\frac{1-v}{(u+1)(u+v)}\n$$\n\n$$\n= 8 + \\frac{v(1+v)}{u(u+1)(1+u+v)} + \\frac{v(v-1)}{(u+1)(1+u+v)} + \\frac{1-v}{(u+1)(u+v)}\n$$\n\n$$\n\\ge 8.\n$$\n\nAlso,\n\n$$\nL(u, 0, v, 0) = \\frac{u}{v} + \\frac{3v}{v} + \\frac{v}{u} + \\frac{3u}{u} = 3 + \\left(\\frac{u}{v} + \\frac{v}{u}\\right) + 3 \\ge 6 + 2 \\sqrt{\\frac{u}{v} \\cdot \\frac{v}{u}} = 8.\n$$\n\nFor a constant $c$ we have:\n\n$$\nL(x_1, x_2, x_3, x_4) - L(x_1 + c, x_2 - c, x_3 + c, x_4 - c) =\n$$\n\n$$\n\\frac{2c}{x_2 + x_3} - \\frac{2c}{x_3 + x_4} + \\frac{2c}{x_4 + x_1} - \\frac{2c}{x_1 + x_2}\n$$\n\n$$\n= 2c \\left( \\frac{x_4 - x_2}{(x_2 + x_3)(x_3 + x_4)} - \\frac{x_4 - x_2}{(x_4 + x_1)(x_1 + x_2)} \\right)\n$$\n\n$$\n= \\frac{2c(x_4 - x_2)(x_1 - x_3)(x_1 + x_3 + x_2 + x_4)}{(x_2 + x_3)(x_3 + x_4)(x_4 + x_1)(x_1 + x_2)}\n$$\n\nNow if $x_2 = x_4$ and $c = x_2$ we have $L(x_1, x_2, x_3, x_4) = L(x_1 + x_2, 0, x_3 + x_2, 0) \\ge 8$.\n\nSimilarly, for $x_4 > x_2$ and $c = x_2$ we have $L(x_1, x_2, x_3, x_4) \\ge L(x_1 + x_2, 0, x_3 + x_2, x_4 - x_2)$. Using $x_1 \\ge x_3$ we get\n\n$$\n\\frac{x_1 + x_2}{x_4 - x_2} \\ge \\frac{x_3 + x_2}{x_4 - x_2}\n$$\n\nand\n\n$$\nL(x_1 + x_2, 0, x_3 + x_2, x_4 - x_2) = L\\left(\\frac{x_1 + x_2}{x_4 - x_2}, 0, \\frac{x_3 + x_2}{x_4 - x_2}, 1\\right) = L(u + v, 0, u, 1) \\ge 8,\n$$\n\nwhere $u = \\frac{x_3 + x_2}{x_4 - x_2}$ and $v = \\frac{x_1 - x_3}{x_4 - x_2}$.\n\nWith this we proved that\n\n$$\n\\frac{x_1 + 3x_2}{x_2 + x_3} + \\frac{x_2 + 3x_3}{x_3 + x_4} + \\frac{x_3 + 3x_4}{x_4 + x_1} + \\frac{x_4 + 3x_1}{x_1 + x_2} \\ge 8.\n$$\n\nfor all positive real numbers $x_1, x_2, x_3$ and $x_4$.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 13026,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $a$ be the positive root of the equation $x + \\frac{1}{x} = 675$. Prove that\n$$\n\\varphi\\big(2023(a^{2n} + a^{-2n})\\big) > 2 \\cdot \\varphi\\big(2022(a^{2n} + a^{-2n})\\big)\n$$\nfor all $n \\in \\mathbb{Z}^+$.",
    "options": [],
    "answer": "See solution",
    "solution": "Notice that $2023 = 7 \\cdot 17^2$ and $2022 = 2 \\cdot 3 \\cdot 337$. Let $u_n = a^{2n} + a^{-2n}$. It is easy to see that $u_{n+1} = u_n^2 - 2$ for every $n \\ge 0$.\n\nWe have $u_0 = a + \\frac{1}{a} = 675$, which is divisible by $3$, so $u_1$ modulo $3$ leaves $1$, then $u_2$ modulo $3$ leaves $-1$, and inductively $u_n \\equiv -1 \\pmod{3}$ for every $n \\ge 2$. Similarly, $u_n$ is odd for all $n$.\n\nAlso, $675 \\equiv 1 \\pmod{337}$, so $u_0 \\equiv 1 \\pmod{337}$, which entails $u_1 \\equiv -1 \\pmod{337}$, and so on; inductively, $u_n \\equiv -1 \\pmod{337}$ for all $n \\ge 1$. From this, it follows that $\\gcd(u_n, 2022) = 1$ for all $n \\ge 1$, and so\n$$\n\\varphi(2022u_n) = \\varphi(2)\\varphi(3)\\varphi(337)\\varphi(u_n) = 672\\varphi(u_n).\n$$\nNext, $675 \\equiv 3 \\pmod{7}$, so $u_1 \\equiv 3^2 \\equiv 2 \\pmod{7}$, and inductively $u_n \\equiv 2 \\pmod{7}$ for all $n \\ge 1$.\n\nFinally, notice that if $d = \\gcd(m, n)$, then\n$$\n\\varphi(mn) = \\varphi(m)\\varphi(n) \\frac{d}{\\varphi(d)} \\ge \\varphi(m)\\varphi(n).\n$$\nHence, one can conclude that\n$$\n\\varphi(2023u_n) \\ge \\varphi(7)\\varphi(17^2)\\varphi(u_n) = 1632\\varphi(u_n) > 2 \\cdot 672\\varphi(u_n) = \\varphi(2022u_n)\n$$\nfor every $n \\in \\mathbb{Z}^+$.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 13027,
    "subject": "Mathematics (Olympiad)",
    "question": "Show that\n\n1. $\\sum_{k=1}^{1008} k C_{2017}^k \\equiv 0 \\pmod{2017^2}$.\n\n2. $\\sum_{k=1}^{504} (-1)^k C_{2017}^k \\equiv 3(2^{2016} - 1) \\pmod{2017^2}$.",
    "options": [],
    "answer": "See solution",
    "solution": "1. We have\n\n$$\n\\begin{aligned}\n\\sum_{k=1}^{1008} k C_{2017}^k &\\equiv \\sum_{k=1}^{1008} \\frac{k \\cdot 2017!}{k!(2017-k)!} \\\\\n&\\equiv 2017 \\cdot \\sum_{k=1}^{1008} \\frac{2016!}{(k-1)!(2017-k)!} \\\\\n&\\equiv 2017 \\cdot \\sum_{k=0}^{1007} \\frac{2016!}{k!(2016-k)!} \\\\\n&\\equiv 2017 \\cdot \\sum_{k=0}^{1007} C_{2016}^k \\\\\n&\\equiv \\frac{2017}{2} \\left( 2^{2016} - C_{2016}^{1008} \\right) \\pmod{2017^2}.\n\\end{aligned}\n$$\n\nSince 2017 is prime, by Fermat's little theorem, $2^{2016} \\equiv 1 \\pmod{2017}$. Also,\n\n$$\n\\begin{aligned}\nC_{2016}^{1008} - 1 &= \\frac{1009 \\cdot 1010 \\cdots 2016 - 1008!}{1008!} \\\\\n&= \\frac{(2017 - 1)(2017 - 2) \\cdots (2017 - 1008) - 1008!}{1008!}.\n\\end{aligned}\n$$\n\nThe numerator is divisible by 2017 and $1008!$ is coprime to 2017, so\n\n$$\nC_{2016}^{1008} \\equiv 1 \\equiv 2^{2016} \\pmod{2017}.\n$$\n\nTherefore, the given sum is divisible by $2017^2$.\n\n2. For every prime $p$,\n\n$$\nC_p^k \\equiv (-1)^{k-1} \\frac{p}{k} \\pmod{p^2}.\n$$\n\nIndeed,\n\n$$\nC_p^k = \\frac{p!}{k!(p-k)!} = \\frac{p}{k} \\cdot \\frac{(p-k+1)\\cdots(p-1)}{(k-1)!}\n$$\n\nso\n\n$$\nC_p^k - (-1)^{k-1} \\frac{p}{k} = \\frac{p}{k} \\left[ \\frac{(p-k+1)\\cdots(p-1) - (-1)^{k-1}(k-1)!}{(k-1)!} \\right].\n$$\n\nThe bracketed term is divisible by $p$, so the statement holds.\n\nThus, $\\frac{1}{p} C_p^k \\equiv \\frac{(-1)^{k-1}}{k} \\pmod{p}$. Therefore,\n\n$$\n\\begin{aligned}\n\\sum_{k=1}^{504} (-1)^k C_{2017}^k &\\equiv \\sum_{k=1}^{504} (-1)^k (-1)^{k-1} \\frac{2017}{k} \\\\\n&= -2017 \\sum_{k=1}^{504} \\frac{1}{k} \\pmod{2017^2}.\n\\end{aligned}\n$$\n\nSo we need to show\n\n$$\n-2017 \\sum_{k=1}^{504} \\frac{1}{k} \\equiv 3(2^{2016} - 1) \\pmod{2017^2}\n$$\n\nor\n\n$$\n-\\sum_{k=1}^{504} \\frac{1}{k} \\equiv \\frac{3(2^{2016} - 1)}{2017} \\pmod{2017}.\n$$\n\nLet $S_n = \\sum_{j=1}^n \\frac{1}{j}$, then $S_{2n} - S_n = \\sum_{k=1}^{n} \\frac{1}{n+k}$. Using properties of binomial coefficients and symmetry, and noting $S_{2016} \\equiv 0 \\pmod{2017}$, we ultimately have\n\n$$\n\\frac{1}{2017} \\left( \\sum_{k=1}^{2016} C_{2017}^{k} + \\sum_{k=1}^{1008} C_{2017}^{k} \\right) \\equiv \\frac{3(2^{2016}-1)}{2017} \\pmod{2017}.\n$$\n\nThis is an identity because\n\n$$\n\\begin{aligned}\n\\sum_{k=1}^{2016} C_{2017}^{k} + \\sum_{k=1}^{1008} C_{2017}^{k} &\\equiv \\frac{3}{2} (2^{2017} - 2) \\\\\n&\\equiv 3(2^{2016} - 1) \\pmod{2017}.\n\\end{aligned}\n$$\n\nThus, the proof is complete.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 13028,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $a$, $b$, $c$ be positive real numbers such that $abc = 1$. Prove that\n\n$$\n\\frac{a + b + c + 3}{4} \\geq \\frac{1}{a + b} + \\frac{1}{b + c} + \\frac{1}{c + a}\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "Rewrite the left-hand side of the inequality as:\n\n$$\n\\frac{a + b + c + 3}{4} = \\frac{a + b + c + 3}{4 \\sqrt{abc}} = \\frac{a + 1}{4 \\sqrt{abc}} + \\frac{b + 1}{4 \\sqrt{abc}} + \\frac{c + 1}{4 \\sqrt{abc}}\n$$\n\nRewrite denominators:\n\n$$\n\\frac{a + 1}{4 \\sqrt{abc}} + \\frac{b + 1}{4 \\sqrt{abc}} + \\frac{c + 1}{4 \\sqrt{abc}} = \\frac{a + 1}{2 \\sqrt{ab \\cdot c} + 2 \\sqrt{ac \\cdot b}} + \\frac{b + 1}{2 \\sqrt{ba \\cdot c} + 2 \\sqrt{bc \\cdot a}} + \\frac{c + 1}{2 \\sqrt{ca \\cdot b} + 2 \\sqrt{cb \\cdot a}}\n$$\n\nBy the arithmetic mean-geometric mean inequality, we have:\n\n$$\n\\begin{aligned}\n&= \\frac{a + 1}{ab + c + ac + b} + \\frac{b + 1}{bc + a + ba + c} + \\frac{c + 1}{ca + b + cb + a} \\\\\n&= \\frac{a + 1}{(a + 1)(b + c)} + \\frac{b + 1}{(b + 1)(a + c)} + \\frac{c + 1}{(c + 1)(b + a)} \\\\\n&= \\frac{1}{b + c} + \\frac{1}{a + c} + \\frac{1}{b + a} = \\frac{1}{a + b} + \\frac{1}{b + c} + \\frac{1}{c + a}.\n\\end{aligned}\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 13029,
    "subject": "Mathematics (Olympiad)",
    "question": "Two players, Andriy and Olesya, play the following game. On a table there is a round cake, which is cut by one of them into $2n$ sectors (pieces), where $n > 1$, and all pieces have different weights. The weight of every piece is known to both players. The players then choose pieces according to these rules:\n\n- First, Olesya chooses 1 piece.\n- Then, Andriy chooses 2 pieces, but in such a way that the pieces left on the table after his turn form a single sector (i.e., are contiguous).\n- After that, they take turns choosing 2 pieces each, always ensuring that the remaining pieces form a single sector after each turn.\n- On the last turn, one of the players takes the final piece.\n\nEach player aims to collect a total weight of cake greater than the opponent's. For which values of $n$ can Olesya cut the cake so that, if she takes the smallest piece on her first turn, she can guarantee a win?\n\n![](path/to/file.png)",
    "options": [],
    "answer": "See solution",
    "solution": "Olesya can win for all $n \\notin \\{2, 4\\}$.\n\n**Explanation:**\n\nLet us analyze the possible values of $n$:\n\n- **For odd $n = 2k+1 > 1$:**\n  Suppose pieces $n-1$, $n$, and $n+1$ have weight 1, and all other pieces have weight 0. After Olesya takes the smallest piece (weight 0), Andriy cannot take any nonzero pieces. As the game proceeds, Olesya can always mirror Andriy's moves to ensure she gets two of the weight-1 pieces and wins.\n\n- **For even $n = 2k > 4$:**\n  Assign weights so that pieces 2 and $2n-2$ have weight $\\frac{99}{100}$, pieces 3 and $2n-3$ have weight 1, and pieces 4 and $2n-4$ have weight $\\frac{101}{100}$. All other pieces have weight 0. Olesya can always ensure she gets a greater total weight than Andriy by careful selection, as detailed in the original solution.\n\n- **For $n = 2$ and $n = 4$:**\n  In these cases, Andriy can always respond to Olesya's first move by taking the two largest pieces, ensuring he wins regardless of Olesya's strategy. Detailed case analysis for $n=4$ shows that Andriy always has a winning strategy.\n\n**Conclusion:**\nOlesya can guarantee a win for all $n$ except $n = 2$ and $n = 4$ if she takes the smallest piece first.",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 13030,
    "subject": "Mathematics (Olympiad)",
    "question": "Given an integer $k \\ge 2$, determine all positive integers $n_1, n_2, \\dots, n_k$ satisfying\n$$\nn_2 \\mid 2^{n_1-1}, \\quad n_3 \\mid 2^{n_2-1}, \\quad \\dots, \\quad n_k \\mid 2^{n_{k-1}-1}, \\quad n_1 \\mid 2^{n_k-1}.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "The required numbers are $n_1 = n_2 = \\dots = n_k = 1$.\n\nFor every integer $m > 1$, let $p(m)$ denote the least prime divisor of $m$. We show that, if $m$ and $\\ell$ are integers greater than $1$, and $m \\mid 2^{\\ell} - 1$, then $p(m) < p(\\ell)$. Since $p(m)$ is odd, $p(m) \\mid 2^{p(m)-1} - 1$, and since $p(m) \\mid 2^{\\ell} - 1$, it follows that $p(m) \\mid 2^{\\gcd(\\ell,\\ p(m)-1)} - 1$. Notice that $\\gcd(\\ell,\\ p(m)-1) > 1$, since $p(m) > 1$, to infer that $\\ell$ has a prime divisor not exceeding $p(m) - 1$, and conclude thereby that $p(m) < p(\\ell)$.\n\nSuppose now, if possible, that $n_1 > 1$. Then $n_k > 1$, so $n_{k-1} > 1$, and so on and so forth all the way down to $n_2 > 1$. Hence $p(n_1) < p(n_2) < \\dots < p(n_k) < p(n_1)$ which is a contradiction. Consequently, $n_1 = 1$, so $n_2 = 1$, and so on and so forth all the way up to $n_k = 1$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 13031,
    "subject": "Mathematics (Olympiad)",
    "question": "找出所有整數 $n \\geq 1$，使得存在正整數對 $(a, b)$ 滿足\n\n$$\n\\frac{ab + 3b + 8}{a^2 + b + 3} = n\n$$\n\n且 $a^2 + b + 3$ 不被任何質數的三次方整除。",
    "options": [],
    "answer": "See solution",
    "solution": "唯一符合條件的整數是 $n = 2$。\n\n由 $b \\equiv -a^2 - 3 \\pmod{a^2 + b + 3}$，分子有\n$$\nab + 3b + 8 \\equiv a(-a^2 - 3) + 3(-a^2 - 3) + 8 \\equiv -(a + 1)^3 \\pmod{a^2 + b + 3}\n$$\n\n由於 $a^2 + b + 3$ 不被任何質數的三次方整除，若 $a^2 + b + 3$ 整除 $(a+1)^3$，則也整除 $(a+1)^2$。而\n$$\n0 < (a+1)^2 < 2(a^2 + b + 3)\n$$\n\n因此 $(a+1)^2 = a^2 + b + 3$，得到 $b = 2(a-1)$ 且 $n = 2$。例如 $(a, b) = (2, 2)$，此時 $a^2 + b + 3 = 9$，證明 $n = 2$ 確實是解。",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 13032,
    "subject": "Mathematics (Olympiad)",
    "question": "Denote $a = 2^{2n} + 2^{n+1} + 1$ and $b = 2^{2m} + 2^{m+1} + 1$.\n\nFind the greatest common divisor $D(a, b)$, where $m$ and $n$ are coprime.",
    "options": [],
    "answer": "See solution",
    "solution": "We notice that $a = (2^n + 1)^2$ and $b = (2^m + 1)^2$.\n\nBecause $(2^{2n} - 1)^2 = (2^n + 1)^2(2^n - 1)^2 = a(2^n - 1)^2$ and $(2^{2m} - 1)^2 = b(2^m - 1)^2$, the greatest common divisor $D(a, b)$ divides $D((2^{2n} - 1)^2, (2^{2m} - 1)^2)$.\n\nWe also know that\n$$\nD((2^{2n} - 1)^2, (2^{2m} - 1)^2) = D(2^{2n} - 1, 2^{2m} - 1)^2 = (2^{D(2m, 2n)} - 1)^2.\n$$\nThis equals $(2^2 - 1)^2 = 9$ because $m$ and $n$ are coprime. The greatest common divisor $D(a, b)$ can thus only be $1$, $3$, or $9$.\n\nBut one of the numbers $n$ and $m$ must be even since they are of different parity. Suppose this is $n$. We then have $2^n \\equiv 1 \\pmod{3}$ and hence $a = (2^n + 1)^2 \\equiv (1 + 1)^2 \\equiv 1 \\pmod{3}$. This means $a$ is not divisible by $3$, so it is coprime to $3$. Therefore, the greatest common divisor $D(a, b)$ must be $1$, which means $a$ and $b$ are coprime.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 13033,
    "subject": "Mathematics (Olympiad)",
    "question": "Determine all functions $f: \\mathbb{R} \\to \\mathbb{R}$ such that the equality\n$$\nf(\\lfloor x \\rfloor y) = f(x) \\lfloor f(y) \\rfloor\n$$\nholds for all $x, y \\in \\mathbb{R}$. (Here $\\lfloor z \\rfloor$ denotes the greatest integer less than or equal to $z$.)",
    "options": [],
    "answer": "See solution",
    "solution": "The answer is $f(x) = c$ for all $x$, where $c = 0$ or $1 \\leq c < 2$.\n\nTo prove that these are the only possible solutions, consider two cases:\n\n**Case 1:** If $\\lfloor f(y) \\rfloor = 0$ for all $0 \\leq y < 1$, then for any integer $k$, setting $x = k$ in the given equation yields $f(ky) = f(k) \\lfloor f(y) \\rfloor = 0$ for such $y$. Every real number can be written as $ky$ with integer $k$ and $0 \\leq y < 1$, so $f(x) = 0$ for all $x$.\n\n**Case 2:** Suppose $\\lfloor f(y_0) \\rfloor \\neq 0$ for some $0 \\leq y_0 < 1$. For any $x_n$ with $n \\leq x_n < n+1$, set $y = y_0$ and $x = x_n$ to get\n$$\nf(ny_0) = f(x_n) \\lfloor f(y_0) \\rfloor.\n$$\nLet $c_n = \\frac{f(ny_0)}{\\lfloor f(y_0) \\rfloor}$, so $f(x) = c_n$ for all $x \\in [n, n+1)$. In particular, $\\lfloor c_0 \\rfloor = \\lfloor f(y_0) \\rfloor \\neq 0$, so $c_0 \\neq 0$.\n\nNow, set $x = y = 0$:\n$$\nc_0 = f(0) = f(0) \\lfloor f(0) \\rfloor = c_0 \\lfloor c_0 \\rfloor,\n$$\nso $\\lfloor c_0 \\rfloor = 1$.\n\nFinally, set $y = 0$ and $x = n$:\n$$\nc_n = f(n) = \\frac{f(0)}{\\lfloor f(0) \\rfloor} = \\frac{c_0}{\\lfloor c_0 \\rfloor} = c_0.\n$$\nTherefore, $f(x) = c_0$ for all $x$, and $\\lfloor c_0 \\rfloor = 1$.\n\nThus, the only solutions are $f(x) = 0$ or $f(x) = c$ with $1 \\leq c < 2$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 13034,
    "subject": "Mathematics (Olympiad)",
    "question": "Aino and V\"aino start to play the game GCD($m, n$) where $m$ and $n$ are positive integers. In the beginning, there are two piles of stones on the table, one with $m$ stones, another with $n$ stones. The one whose turn it is takes away a number of stones from one of the piles. This number must be a multiple of the number of stones in the other pile. Aino starts, and the players take turns until one of the piles is empty. The one who manages to empty a pile wins. Prove that there is an $\\alpha > 1$ such that if $m$ and $n$ are positive integers with $m > \\alpha n$, then Aino has a winning strategy in the game GCD($m, n$), whereas if $\\alpha n > m > n$, V\"aino has.\n\n**Remark.** This problem is a (strong) generalization of problem 17 of Baltic Way 1990.",
    "options": [],
    "answer": "See solution",
    "solution": "Choose $\\alpha = (1 + \\sqrt{5})/2$, so that $\\alpha^2 = \\alpha + 1$ holds. We prove by induction on the sum $m+n$ that if $m > \\alpha n$, then Aino has a winning strategy in GCD($m, n$); otherwise, if $\\alpha n \\ge m > n$, then V\"aino has.\n\n1) If $n \\mid m$, then Aino can remove all of the stones from the pile with $m$ stones, thus winning. This includes the initial step of the induction.\n\n2) Assume $n < m \\le \\alpha n$. Note that $\\alpha$ is irrational, so $n < m < \\alpha n$. The rules of the game actually force Aino to remove stones from the larger pile. As $m < 2n$, there is no choice: she has to take exactly $n$ stones. The play continues with $n$ and $m-n$ stones in the piles, V\"aino having the turn. We have $0 < m-n < n$ and\n\n$$\n\\frac{n}{m-n} > \\frac{n}{\\alpha n - n} = \\frac{1}{\\alpha - 1} = \\frac{\\alpha^2 - \\alpha}{\\alpha - 1} = \\alpha.\n$$\n\nBy induction, Aino has a winning strategy in the game GCD($n, m-n$), but now the turns have switched. Hence, V\"aino has a winning strategy that mimics this winning strategy of Aino's.\n\n3) Finally, assume $m > \\alpha n$, but $n \\nmid m$. Write $\\beta = m/n - \\lfloor m/n \\rfloor$ and $k = \\lfloor m/n \\rfloor$. Then $m = kn + \\beta n$ with $0 < \\beta < 1$, as $n \\nmid m$. If $1 + \\beta < \\alpha$ (note that $\\beta \\in \\mathbb{Q}$ and $\\alpha \\notin \\mathbb{Q}$), then $k \\ge 2$, as $m > \\alpha n$. Therefore, Aino may take $(k-1)n$ stones out of the pile of $m$ stones, leaving there $m-(k-1)n = (1+\\beta)n$ stones. By induction hypothesis, V\"aino has a winning strategy in the game GCD($ (1+\\beta)n, n $), which will now be copied by Aino in order to win the game. Otherwise, if $1+\\beta > \\alpha$, then Aino may take $kn$ stones, leaving $\\beta n$ stones in the heap. Again, V\"aino's winning strategy in the game GCD($n, \\beta n$) is copied by Aino. It suffices to check that\n\n$$\n\\frac{1}{\\beta} < \\frac{1}{\\alpha - 1} = \\alpha.\n$$",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 13035,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $M$ be the set of palindromic numbers of the form $5n + 4$, where $n \\ge 0$ is an integer.\n\nA positive integer is a *palindromic number* if it remains the same when its digits are reversed. For instance, the numbers 7, 191, 23532, 3770773 are palindromic numbers.\n\n1. If we write the elements of $M$ in increasing order, which is the 50th number?\n2. Among all numbers in $M$, written with nonzero digits whose sum is 2014, which is the greatest one and which is the smallest one?",
    "options": [],
    "answer": "See solution",
    "solution": "1. The last (and hence, the first) digit of a number from $M$ equals 4 or 9. A direct count shows that $M$ contains 2 one-digit numbers, 2 two-digit numbers, 20 three-digit numbers, and 20 four-digit numbers. Hence, the 50th number is the 6th five-digit number, that is, $40504$.\n\n2. The greatest number in $M$ has the maximum number of digits. Therefore, we put 4 as the first and last digit and complete the decimal representation with 2006 digits 1, obtaining thus $$\\underbrace{411\\ldots14}_{2006\\ \\text{digits}\\ 1}$$. Similarly, the smallest number in $M$ has the least number of digits. The answer is $$\\underbrace{9899\\ldots989}_{220\\ \\text{digits}\\ 9}$$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 13036,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $\\mathcal{A}$ be the set of functions $f$ defined on the integers $1,2,\\dots,2023$ with values in $\\{1,2,\\dots,2023\\}$, such that:\n\n1. $f(1) \\ge f(2) \\ge \\dots \\ge f(2023)$ (i.e., $f$ is non-increasing), and\n2. For any $x$ in $\\{1,2,\\dots,2023\\}$, $f(f(x)) = f(x)$ (i.e., $f$ is idempotent).\n\nWhat is the maximum possible number of elements in $\\mathcal{A}$?\n\n$\\boxed{\\begin{pmatrix} 2022 \\\\ 1011 \\end{pmatrix}}$",
    "options": [],
    "answer": "See solution",
    "solution": "We say that $x$ is a fixed point of $f$ if $f(x) = x$. Suppose $f \\in \\mathcal{A}$ has two fixed points $a < b$. Then, $a = f(a) \\ge f(b) = b$, a contradiction. Thus, each $f \\in \\mathcal{A}$ has at most one fixed point.\n\nSuppose $\\mathcal{A}$ is non-empty. Take $f \\in \\mathcal{A}$ and define $a = f(f(1))$. Then $a$ is a fixed point of $f$ since $f(f(1)) = f(f(f(1)))$ by the second condition. For any $g \\in \\mathcal{A}$, $g(a) = g(f(a)) = f(g(f(a))) = f(g(a))$ (by the second condition), so $g(a)$ is also a fixed point of $f$. By uniqueness, $g(a) = a$. Thus, $a$ is the unique fixed point of any $g \\in \\mathcal{A}$.\n\nLet $f, g \\in \\mathcal{A}$ and $x \\in \\{1,\\dots,2023\\}$. The second condition gives $f(g(x))$ is a fixed point of $g$, so $f(g(x)) = a$.\n\nLet $X$ be the set of $x$ from $1$ to $2023$ such that $f(x) = a$ for all $f \\in \\mathcal{A}$. Let $\\ell$ be the smallest and $m$ the largest element of $X$. For any $f \\in \\mathcal{A}$, $a = f(\\ell) \\ge f(\\ell+1) \\ge \\dots \\ge f(m) = a$, so $X = \\{\\ell, \\ell+1, \\dots, m\\}$.\n\nWe already proved $f(g(x)) = a$ for any $f, g \\in \\mathcal{A}$ and $x$. Thus, $g(x) \\in X$ for such $g$ and $x$.\n\nSummarizing, $f \\in \\mathcal{A}$ satisfies:\n\n1. $m \\ge f(1) \\ge f(2) \\ge \\dots \\ge f(2023) \\ge \\ell$,\n2. $f(x) = a$ for $x = \\ell, \\dots, m$.\n\nFix $f \\in \\mathcal{A}$. For $x = 1,\\dots,2023$, define $s_x = f(x) - x$. By (1), $m-1 \\ge s_1 > s_2 > \\dots > s_{2023} \\ge \\ell - 2023$. By (2), $s_x = a - x$ for $x = \\ell,\\dots,m$.\n\nThus, for any $f \\in \\mathcal{A}$, we can associate a way of choosing $2023$ integers from $\\ell-2023$ to $m-1$, such that all integers from $a-m$ to $a-\\ell$ are chosen. This corresponds to choosing $2022 - m + \\ell$ integers from the $2022$ integers from $\\ell-2023$ to $m-1$ but not between $a-m$ and $a-\\ell$. The number of such choices is $\\binom{2022}{2022 - m + \\ell}$.\n\nFor $x = 0$ to $2021$,\n\n$$\n\\frac{\\binom{2022}{\\ell+1}}{\\binom{2022}{\\ell}} = \\frac{2022 - \\ell}{\\ell+1},\n$$\n\nwhich implies\n\n$$\n\\binom{2022}{0} < \\binom{2022}{1} < \\dots < \\binom{2022}{1010} < \\binom{2022}{1011} > \\binom{2022}{1012} > \\dots > \\binom{2022}{2022}.\n$$\n\nTherefore, the number of elements in $\\mathcal{A}$ is at most $\\binom{2022}{1011}$.\n\nOn the other hand, consider the set $\\mathcal{B}$ of functions $f$ defined for $1$ to $2023$ with values in $1$ to $2023$, and\n\n$$\n1012 = f(1) = f(2) = \\dots = f(1012) \\ge f(1013) \\ge \\dots \\ge f(2023).\n$$\n\nThis set satisfies the first condition. For any $f, g \\in \\mathcal{B}$ and $x$, $g(x) \\le 1012$, so $f(g(x)) = 1012$. Thus, the second condition is also satisfied. For $\\ell = 1$ and $a = m = 1012$, there is a one-to-one correspondence between $f \\in \\mathcal{B}$ and sequences $s_1, \\dots, s_{2023}$ with\n\n$$\ns_1 = 1011, s_2 = 1010, \\dots, s_{1012} = 0, \\quad -1 \\ge s_{1013} > \\dots > s_{2023} \\ge -2022.\n$$\n\nTherefore, $|\\mathcal{B}| = \\binom{2022}{1011}$.\n\nThus, the maximum possible number of elements in $\\mathcal{A}$ is $\\boxed{\\binom{2022}{1011}}$.",
    "topic": "Algebra",
    "subtopic": "Linear Algebra"
  },
  {
    "id": 13037,
    "subject": "Mathematics (Olympiad)",
    "question": "Find the largest integer that always divides the expression\n$$a^5 + b^5 - a^4b - b^4a$$\nfor all primes $a$ and $b$ greater than $5$.",
    "options": [],
    "answer": "See solution",
    "solution": "Expanding and factorising:\n$$a^5 + b^5 - a^4b - b^4a = (a - b)^2(a^2 + b^2)(a + b)$$\n\nFor primes $a, b > 5$:\n- Both $a$ and $b$ are odd, so $a + b$ and $a^2 + b^2$ are even, and $4$ divides $(a-b)^2$. Thus, $16$ divides the expression.\n- Considering forms modulo $4$, $32$ always divides the expression.\n- For $3$, either $a-b$ or $a+b$ is divisible by $3$, so $3$ always divides the expression.\n- For $5$, either $a^2-b^2$ or $a^2+b^2$ is divisible by $5$, so $5$ always divides the expression.\n\nThus, $32 \\times 3 \\times 5 = 480$ always divides the expression. Checking with examples shows that higher powers of $2$, $3$, or $5$ are not always factors. \n\n**Answer:** $\\boxed{480}$ is the largest guaranteed divisor.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 13038,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all real solutions $(x, y)$ to the system:\n$$\n\\begin{cases}\nx + y = x^2 + xy + y^2 + 1 \\\\\nx + x^2 = y + y^2\n\\end{cases}\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "Consider the first equation:\n$$\nx + y - 1 = x^2 + xy + y^2\n$$\nwhich can be rewritten as:\n$$\ny^2 + (x-1)y + x^2 - x + 1 = 0 \\quad (*)\n$$\nThis is a quadratic in $y$. Its discriminant is:\n$$\nD = (x - 1)^2 - 4(x^2 - x + 1) = -3x^2 + 2x - 3\n$$\nThis is a quadratic in $x$ whose discriminant is $-32 < 0$, so $D < 0$ for all real $x$. Thus, equation $(*)$ has no real solution for $y$ except possibly when $y = x$.\n\nIn the case $y = x$, both equations reduce to:\n$$\nx + x^2 = x^3\n$$\nwhich gives:\n$$\nx(x^2 - x - 1) = 0\n$$\nThe solutions are $x_1 = 0$, $x_{2,3} = \\frac{1 \\pm \\sqrt{5}}{2}$.\n\nTherefore, the only real solutions are:\n$$\n(x, y) \\in \\left\\{ (0, 0), \\left(\\frac{1+\\sqrt{5}}{2}, \\frac{1+\\sqrt{5}}{2}\\right), \\left(\\frac{1-\\sqrt{5}}{2}, \\frac{1-\\sqrt{5}}{2}\\right) \\right\\}\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 13039,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $\\triangle ABC$ be a triangle with $m(\\angle ABC) = 75^\\circ$ and $m(\\angle ACB) = 45^\\circ$. The angle bisector of $\\angle CAB$ intersects $CB$ at the point $D$. Consider the point $E \\in AB$ such that $DE = DC$. Let $P$ be the intersection of the lines $AD$ and $CE$. Prove that $P$ is the midpoint of the segment $AD$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $P'$ be the midpoint of the segment $AD$. We will prove that $P' = P$.\n\nLet $F \\in AC$ such that $DF \\perp AC$. The triangle $CDF$ is isosceles with $FD = FC$ and the triangle $DP'F$ is equilateral as $m(\\angle ADF) = 60^\\circ$. Thus, the triangle $FCP'$ is isosceles ($FP' = FC$) and $m(\\angle FCP') = m(\\angle FP'C) = 15^\\circ$.\n\n![](images/2019_bmo_shortlist_p15_data_63f05c2b95.png)\n\nWe now prove that $m(\\angle FCE) = 15^\\circ$.\n\nLet $M$ be the point on $AB$ such that the triangle $ACM$ is equilateral. As $\\triangle ADC \\equiv \\triangle ADM$ (SAS), it follows that $DC = DM (= DE)$ and $m(\\angle AMD) = m(\\angle ACD) = 45^\\circ$. Thus, the triangle $DME$ is isosceles with $m(\\angle DME) = m(\\angle DEM) = 45^\\circ$. In the triangle $BDE$ we have $m(\\angle BDE) = 60^\\circ$ and thus $m(\\angle CDE) = 120^\\circ$. As the triangle $DCE$ is isosceles with $m(\\angle DCE) = m(\\angle DEC) = 30^\\circ$. Finally, $m(\\angle ACE) = m(\\angle ACB) - m(\\angle BCE) = 45^\\circ - 30^\\circ = 15^\\circ$.\n\nThus $m(\\angle FCP') = 15^\\circ = m(\\angle FCE)$, and therefore $P' \\in CE$ and $P' = P$, which means that $P$ is the midpoint of the segment $AD$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 13040,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all $n$ such that any convex $n$-gon in the plane can be divided into a finite number of triangles satisfying the following conditions:\n\n1. No vertex is added to the sides of the $n$-gon. However, any number of vertices can be added to the interior of the $n$-gon.\n2. All triangles have exactly three bounding edges, and the edges intersect only at vertices.\n3. Each vertex has an even number of edges connected to it.",
    "options": [],
    "answer": "See solution",
    "solution": "This problem is essentially Exercise 6.4.10 from *Invitation to Discrete Mathematics*, 2nd edition, Oxford University Press, by Jiří Matoušek and Jaroslav Nešetřil.",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 13041,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $B$ 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\nGiven a regular polygon with $2017$ vertices, each colored either red or blue, prove that the total number of isosceles triangles with all vertices the same color depends only on the number of red and blue points, not on their arrangement.\n\n![](images/Mathematical_competitions_in_Croatia_2017_p21_data_18548e57a5.png)",
    "options": [],
    "answer": "See 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\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\nFrom the fact that each segment belongs to three isosceles triangles, we get:\n\n$$\n\\begin{aligned}\n3d_b &= 3t_b + t_{sb}, \\\\\n3d_r &= 3t_r + t_{sr}, \\\\\n3d_s &= 2t_{sb} + 2t_{sr}.\n\\end{aligned}\n$$\n\nThis implies $3d_b + 3d_r - \\frac{3}{2}d_s = 3t_b + 3t_r$, i.e.\n\n$$\nt_b + t_r = d_b + d_r - \\frac{1}{2}d_s = \\frac{1}{2} (B(B-1) + R(R-1) - BR),\n$$\n\nwhich depends only on the number of the red and blue points, but not on their arrangement.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 13042,
    "subject": "Mathematics (Olympiad)",
    "question": "A special number is a positive integer $n$ for which there exist positive integers $a, b, c,$ and $d$ such that\n$$\nn = \\frac{a^3 + 2b^3}{c^3 + 2d^3}\n$$\n\nProve that:\n\n(a) There are infinitely many special numbers.\n\n(b) $2014$ is not a special number.",
    "options": [],
    "answer": "See solution",
    "solution": "(a) Every perfect cube $k^3$ of a positive integer is special because we can write\n$$\nk^3 = \\frac{(ka)^3 + 2(kb)^3}{a^3 + 2b^3}\n$$\nfor some positive integers $a, b$.\n\n(b) Observe that $2014 = 2 \\cdot 19 \\cdot 53$. If $2014$ is special, then there exist positive integers $x, y, u, v$ such that\n$$\nx^3 + 2y^3 = 2014(u^3 + 2v^3)\n$$\nAssume $x^3 + 2y^3$ is minimal with this property. If $19$ divides $x^3 + 2y^3$, then it must divide both $x$ and $y$. Suppose $19$ does not divide $x$ or $y$. Then $x^3 \\equiv -2y^3 \\pmod{19}$, so $(x^3)^{6} \\equiv (-2y^3)^{6} \\pmod{19}$, i.e., $x^{18} \\equiv 2^6 y^{18} \\pmod{19}$. By Fermat's Little Theorem, $x^{18} \\equiv y^{18} \\equiv 1 \\pmod{19}$, so $1 \\equiv 2^6 \\pmod{19}$, i.e., $2^6 = 64$, but $64 \\equiv 7 \\pmod{19}$, not $1$. Contradiction.\n\nThus, $x = 19x_1$, $y = 19y_1$ for some positive integers $x_1, y_1$. Substitute into the equation:\n$$\n19^3 x_1^3 + 2 \\cdot 19^3 y_1^3 = 2014(u^3 + 2v^3)\n$$\nwhich simplifies to\n$$\n19^2(x_1^3 + 2y_1^3) = 2 \\cdot 53(u^3 + 2v^3)\n$$\nSo $19$ divides $u^3 + 2v^3$, implying $u = 19u_1$, $v = 19v_1$. Substitute again:\n$$\nx_1^3 + 2y_1^3 = 2014(u_1^3 + 2v_1^3)\n$$\nBut $x_1^3 + 2y_1^3 < x^3 + 2y^3$, contradicting the minimality assumption. Therefore, $2014$ is not a special number.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 13043,
    "subject": "Mathematics (Olympiad)",
    "question": "Consider a positive integer $n$ and matrices $A, B$ in $M_n(\\mathbb{C})$ such that $A^2 + B^2 = 2AB$.\n\n1. Show that the matrix $AB - BA$ is not invertible.\n\n2. If the rank of $A - B$ is $1$, prove that $A$ and $B$ commute.",
    "options": [],
    "answer": "See solution",
    "solution": "1. The given relation can be rewritten as:\n\n$$\n(A - B)^2 = AB - BA\n$$\n\nand\n\n$$\nA(A - B) = (A - B)B.\n$$\n\nSuppose $AB - BA$ is invertible. By the first equation, $A - B$ is also invertible. From the second equation, $B = (A - B)^{-1}A(A - B)$. Then $A - B = A - (A - B)^{-1}A(A - B)$, which leads to $I_n = A(A - B)^{-1} - (A - B)^{-1}A$.\n\nTaking traces:\n\n$$\n n = \\operatorname{tr} I_n = \\operatorname{tr}(A(A - B)^{-1} - (A - B)^{-1}A) = 0,\n$$\n\na contradiction. Thus, $AB - BA$ is singular. By the first equation, $A - B$ is also singular.\n\n2. For a rank $1$ matrix $X$ in $M_n(\\mathbb{C})$, $X^2 = (\\operatorname{tr} X) X$. Using this and $(A - B)^2 = AB - BA$:\n\n$$\nAB - BA = (A - B)^2 = (\\operatorname{tr}(A - B))(A - B)\n$$\n\nSo $0 = \\operatorname{tr}(AB - BA) = (\\operatorname{tr}(A - B))^2$, i.e., $\\operatorname{tr}(A - B) = 0$. Therefore, $AB - BA = O_n$, i.e., $AB = BA$.",
    "topic": "Algebra",
    "subtopic": "Linear Algebra"
  },
  {
    "id": 13044,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $\\triangle$ stand for the area of $\\triangle ABC$. Prove that\n$$\na \\sin B + b \\sin C + c \\sin A \\geq 9r\n$$\nwhere $a$, $b$, $c$ are the side lengths of $\\triangle ABC$, $r$ is the inradius, and equality holds if and only if $a = b = c$.",
    "options": [],
    "answer": "See solution",
    "solution": "Consider the RHS. Since $a = 2R \\sin A$, etc., we have\n$$\n\\begin{align*}\na \\sin B + b \\sin C + c \\sin A \n&= 2R(\\sin A \\sin B + \\sin B \\sin C + \\sin C \\sin A) \\\\ \n&= 2R(\\sin A \\sin B + \\sin C(\\sin A + \\sin B)) \\\\ \n&\\le 2R \\left( \\sin^2 \\left( \\frac{A+B}{2} \\right) + 2 \\sin C \\sin \\left( \\frac{A+B}{2} \\right) \\right) \\quad \\text{(with equality iff } A=B) \\\\ \n&= 2R \\left( \\sin^2 \\left( \\frac{\\pi-C}{2} \\right) + 2 \\sin C \\sin \\left( \\frac{\\pi-C}{2} \\right) \\right) \\\\ \n&= 2R \\left( \\cos^2 \\left( \\frac{C}{2} \\right) + 4 \\sin \\left( \\frac{C}{2} \\right) \\cos^2 \\left( \\frac{C}{2} \\right) \\right) \\\\ \n&= 2R \\cos^2 \\left( \\frac{C}{2} \\right) \\left( 1 + 4 \\sin \\left( \\frac{C}{2} \\right) \\right) \\\\ \n&= 2R(1-x^2)(1+4x) \\quad (x = \\sin(C/2)) \\\\ \n&= 2R \\left( \\frac{9}{4} - (2x-1)^2 \\left( x + \\frac{5}{4} \\right) \\right) \\\\ \n&\\le \\frac{9R}{2} \\quad \\text{(with equality iff } x=1/2\\text{).} \n\\end{align*}\n$$\nThus the inequality holds, and there is equality throughout iff\n$$\nA = B, \\quad \\text{and} \\quad \\sin\\left(\\frac{C}{2}\\right) = \\frac{1}{2} \\quad \\text{i.e., iff} \\quad A = B = C = \\frac{\\pi}{3}.\n$$\nOne can obtain the RHS a little differently using the facts that\n$$\nab + bc + ca \\le \\frac{1}{3}(a+b+c)^2 \\quad \\text{and} \\quad \\sin A + \\sin B + \\sin C \\le \\frac{3\\sqrt{3}}{2}.\n$$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 13045,
    "subject": "Mathematics (Olympiad)",
    "question": "Points $A$, $B$, $C$, $D$ lie on sides $EF$, $FG$, $GH$, $HE$, respectively, of a parallelogram $EFGH$. Suppose that $AC \\perp EF$, $BD \\perp FG$, and $ABCD$ is cyclic. Let $Q$ be the point on $AC$ such that $FQ \\parallel BC$.\n\nProve that $EQ \\parallel DC$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let circle $ABCD$ intersect $FG$ for a second time at $X$ and $EH$ for a second time at $Y$. Let $Q' = XY \\cap AC$. We shall prove that $Q = Q'$.\n\n![](images/2023_Australian_Scene_p146_data_80fd807669.png)\n\nSince $DBXY$ is cyclic with right angles at $B$ and $D$, it is a rectangle. Using cyclic quadrilaterals $AFXQ'$ and $AXBC$, we have\n\n$$\n\\angle FQ'A = \\angle FXA = \\angle BCA.\n$$\n\nHence $FQ' \\parallel BC$, so $Q = Q'$. An analogous argument shows that\n\n$$\n\\angle EQA = \\angle EYA = \\angle DCP,\n$$\n\nso $EQ \\parallel DC$ as required.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 13046,
    "subject": "Mathematics (Olympiad)",
    "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.\n\nWhen entering the public transportation network, a traveller must insert one gold coin into a machine, which lets them use a direct connection to go to the next town. If the traveller continues travelling from some town with the same method of transportation that took them there, and they have 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 they last used back. In other cases, they must pay just like when starting travelling.\n\nProve that it is possible to get from any town to any other town by using at most 2 gold coins.",
    "options": [],
    "answer": "See solution",
    "solution": "*Solution 1.*\n\nLet $A$ 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 $B$.\n\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. For 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\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).\n\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 the 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.\n\n![](images/prob1314_p32_data_184c73924b.png)\n\n*Figure 36*\n\n![](images/prob1314_p32_data_21661803cc.png)\n\n*Figure 37*\n\n![](images/prob1314_p32_data_67c31a1c57.png)\n\n*Figure 38*\n\n![](images/prob1314_p32_data_df76d4ab91.png)\n\n*Figure 39*\n\n*Solution 2.*\n\nLet $A$ and $B$ be any two towns. Suppose that there is no direct connection between them, because otherwise the problem statement holds trivially.\n\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 $A, B, X, Y$ 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.\n\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": 13047,
    "subject": "Mathematics (Olympiad)",
    "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![](images/Mathematical_Olympiad_in_China_2011-2014_p156_data_bcddae8e16.png)\n\n![](images/Mathematical_Olympiad_in_China_2011-2014_p156_data_5f2c7b21e8.png)",
    "options": [],
    "answer": "See solution",
    "solution": "Since $\\angle AEF = \\angle APE$, then $\\triangle AEF \\sim \\triangle EPF$. So $AF \\cdot PF = EF^2$. Since points $A, I, D$ and $F$ are concyclic, $PA \\cdot PF = PI \\cdot PD$. Thus,\n\n$$\n\\begin{aligned}\nPF^2 &= AF \\cdot PF + PA \\cdot PF \\\\\n      &= EF^2 + PI \\cdot PD.\n\\end{aligned}\n$$\n\nSince $PQ$ is the diameter of circle $O$ and point $I$ is on $AQ$, we see that $AI \\perp AP$. Consequently,\n\n$$\n\\angle IDF = \\angle IAP = 90^{\\circ}.\n$$\n\nThus, we have\n\n$$\nPF^2 - EF^2 = PD^2 - ED^2.\n$$\n\nCombining the above, we have\n\n$$\nPI \\cdot PD = PD^2 - ED^2.\n$$\n\nThus\n\n$$\nQD^2 = ED^2 = PD^2 - PI \\cdot PD = ID \\cdot PD.\n$$\n\nConsequently, we have\n\n$$\n\\triangle QID \\sim \\triangle PQD.\n$$\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$$\n\\triangle QMI \\sim \\triangle QIP.\n$$\n\nBy the above, 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$$\n\\frac{PM}{IK} = \\frac{PD}{ID} = \\frac{PQ}{MQ}.\n$$\n\nBy the Circle-Power Theorem and the Sine Theorem, we know that\n\n$$\n\\begin{aligned}\nPQ \\cdot IK &= PM \\cdot MQ = BM \\cdot MC \\\\\n            &= \\left(\\frac{1}{2}BC\\right)^2 = (R \\sin \\angle BAC)^2,\n\\end{aligned}\n$$\n\nthus, $\\sin^2 \\angle BAC = \\frac{PQ \\cdot IK}{R^2} = \\frac{2R \\cdot r}{R^2} = \\frac{2r}{R}$.\n\n$\\square$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 13048,
    "subject": "Mathematics (Olympiad)",
    "question": "Suppose a tower with centre $C$ is completely visible to the observer, but that tower obscures part of the next tower away from $O$. By drawing lines through $O$ tangential to the towers, we see that all towers from $A$ to $C$ are completely visible from $O$, but none of the towers to the right of $C$ is visible. Number the tower centres from $A$ to $C$: $0, 1, 2, \\ldots, m$. Show that $m \\le 7$.",
    "options": [],
    "answer": "See solution",
    "solution": "### Method 1\n\nLine $OZ$ is tangential at $X$ to the $(m-1)$th tower with centre $B$ as shown. It intersects $\\mathcal{L}$ at $Y$ and $CZ$ is perpendicular to $OZ$.\n\n![](images/2022_Australian_Scene_p67_data_4ed14be8a8.png)\n\nTriangles $BXY$ and $CYZ$ are similar, hence $BY \\leq YC$. So $2BY \\leq BC = d$.\n\nTriangles $BXY$ and $OAY$ are similar, hence $OY/15 = BY/1$.\n\nSince there is a tower centre at $A$, we have\n\n$$\nAY = AB + BY = (m - 1)d + BY \\geq (2m - 1)BY = (2m - 1)\\frac{OY}{15}.\n$$\n\nSince $AY/OY < 1$, we have $15 > 2m - 1$, hence $m \\leq 7$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 13049,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all finite sets $A$ of positive real numbers with at least two elements, such that for every $a, b \\in A$ with $a \\ne b$, $a^2 + b^2 \\in A$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $A = \\{a_1, a_2, \\dots, a_n\\}$ with $a_1 < a_2 < \\dots < a_n$. For each pair $a_i \\ne a_j$, $a_i^2 + a_j^2 \\in A$. The sums $a_1^2 + a_2^2, a_1^2 + a_3^2, \\dots, a_1^2 + a_n^2, a_2^2 + a_n^2, \\dots, a_{n-1}^2 + a_n^2$ are all distinct and in $A$, so $2n - 3 \\le n$, implying $n \\le 3$.\n\nIf $n = 3$ and $A = \\{a, b, c\\}$ with $a < b < c$, then $a^2 + b^2 = a$, $b^2 + c^2 = c$, $a^2 + c^2 = b$. This leads to $a^2 - c^2 = a - c$, so $a + c = 1$. Substituting into $a^2 + c^2 = b$ gives $b = 2a^2 - 2a + 1$. Further analysis leads to $b = \\frac{1}{2}$ and $a = \\frac{1}{2}$, contradicting $a < b$.\n\nThus, $n = 2$ and $A = \\{a, b\\}$ with $a^2 + b^2 = a$. For $a \\in (0, 1)$, $b = \\sqrt{a - a^2}$. Therefore, the sets are $\\{a, \\sqrt{a - a^2}\\}$ with $a \\in (0, 1)$, $a \\ne \\frac{1}{2}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 13050,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all four-digit years after 2013 that can be written using the digits 0, 1, 2, 3 in some order.",
    "options": [],
    "answer": "See solution",
    "solution": "To form a four-digit year using the digits 0, 1, 2, and 3 in some order, we consider all permutations of these digits. The total number of such numbers is $4! = 24$. However, a valid year cannot start with 0, so we exclude those cases. The number of such numbers starting with 0 is $3! = 6$. Thus, the number of valid years is $24 - 6 = 18$. Now, we need only those years after 2013. Listing all valid years:\n\n- 1023\n- 1032\n- 1203\n- 1230\n- 1302\n- 1320\n- 2013\n- 2031\n- 2103\n- 2130\n- 2301\n- 2310\n- 3012\n- 3021\n- 3102\n- 3120\n- 3201\n- 3210\n\nFrom these, the years after 2013 are:\n\n- 2031\n- 2103\n- 2130\n- 2301\n- 2310\n- 3012\n- 3021\n- 3102\n- 3120\n- 3201\n- 3210\n\nSo, there are $11$ such years after 2013.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 13051,
    "subject": "Mathematics (Olympiad)",
    "question": "Calcula la suma de los inversos de los dos mil trece primeros términos de la sucesión de término general\n$$\na_n = 1 - \\frac{1}{4n^2}\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "El término general se puede escribir como\n$$\na_n = \\frac{4n^2 - 1}{4n^2} = \\frac{(2n - 1)(2n + 1)}{4n^2}\n$$\ny su inverso es\n$$\n\\frac{1}{a_n} = \\frac{4n^2}{(2n - 1)(2n + 1)} = \\frac{n}{2n - 1} + \\frac{n}{2n + 1}\n$$\nHemos de calcular\n$$\n\\begin{aligned}\nS &= \\frac{1}{a_1} + \\frac{1}{a_2} + \\frac{1}{a_3} + \\dots + \\frac{1}{a_{2012}} + \\frac{1}{a_{2013}} \\\\\n&= \\left(\\frac{1}{1} + \\frac{1}{3}\\right) + \\left(\\frac{2}{3} + \\frac{2}{5}\\right) + \\dots + \\left(\\frac{2012}{4023} + \\frac{2012}{4025}\\right) + \\left(\\frac{2013}{4025} + \\frac{2013}{4027}\\right) \\\\\n&= 1 + \\left(\\frac{1}{3} + \\frac{2}{5}\\right) + \\left(\\frac{2}{5} + \\frac{3}{5}\\right) + \\dots + \\left(\\frac{2012}{4025} + \\frac{2013}{4025}\\right) + \\frac{2013}{4027} \\\\\n&= 2013 + \\frac{2013}{4027} = 2013 \\left(1 + \\frac{1}{4027}\\right) = \\frac{8108364}{4027} \\approx 2013.5\n\\end{aligned}\n$$",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 13052,
    "subject": "Mathematics (Olympiad)",
    "question": "Пусть длина стороны таблицы равна $2n = 100$ (то есть $n = 50$). Пронумеруем строки сверху вниз, а столбцы — слева направо числами от $1$ до $2n$.\n\nВ каждой строке может быть от $0$ до $2n$ чёрных клеток, причём количества чёрных клеток во всех строках различны. Эти количества — все числа от $0$ до $2n$, кроме одного (обозначим его $k$). Тогда общее число чёрных клеток равно $$(0 + 1 + \\ldots + 2n) - k = 2n^2 + n - k.$$ С другой стороны, так как во всех столбцах клеток поровну, общее число чёрных клеток должно делиться на $2n$. Значит, $k = n$, и во всех столбцах по $$\\frac{2n^2}{2n} = n$$ чёрных клеток.\n\nОценим сверху количество пар соседних по стороне разпоцветных клеток, отдельно для горизонтальных и вертикальных пар.\n\nЕсли в строке $i \\leq n-1$ чёрных клеток, то они могут участвовать не более чем в $2i$ горизонтальных парах. Если в строке $i \\geq n+1$ чёрных клеток, аналогичное рассуждение применимо к белым клеткам, которых $2n - i \\leq n-1$. Итого, горизонтальных разпоцветных пар не больше, чем $$2 \\cdot (2 \\cdot 0 + 2 \\cdot 1 + \\ldots + 2 \\cdot (n-1)) = 2n(n-1).$$\n\nОценим количество вертикальных пар. Рассмотрим любую строку с чётным номером от $2$ до $2(n-1)$; пусть в ней $i$ чёрных клеток. Тогда либо в строке сверху, либо в строке снизу от неё число чёрных клеток не равно $100 - i$; значит, одна из вертикальных пар, в которых участвуют клетки нашей строки, будет одноцветной. Итого, есть хотя бы $n-1$ одноцветных вертикальных пар. Так как общее число вертикальных пар равно $2n(2n-1)$, то разпоцветных из них — не больше, чем $$2n(2n-1) - (n-1).$$\n\nИтого, общее число разпоцветных пар не больше, чем $$2n(n-1) + 2n(2n-1) - (n-1) = (2n-1)(3n-1).$$\n\nПриведите пример, в котором указанное число пар достигается.",
    "options": [],
    "answer": "See solution",
    "solution": "Проведём в таблице $2n \\times 2n$ диагональ из верхнего левого угла в нижний правый. Все клетки, лежащие на или ниже диагонали, покрасим в чёрный цвет, если они лежат в чётных строках, и в белый — иначе (раскраска «по строкам»).\n\nВсе клетки, лежащие выше диагонали, покрасим в чёрный цвет, если сумма номеров их строки и столбца чётна, и в белый иначе («шахматная» раскраска). Пример такой раскраски при $n = 4$ показан на рисунке. Нетрудно проверить, что в каждом столбце ровно по $n$ чёрных клеток, в $2i$-й строке есть $n+i$ чёрных клеток, а в $(2i-1)$-й строке — $n-i$ чёрных клеток. Кроме того, все оценки выше достигаются.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 13053,
    "subject": "Mathematics (Olympiad)",
    "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 $A$.\n\nIt is known that each city of the country belongs to a cycle of length 4 and also to a cycle of length 5.\n\nIs it sure that:\n\na) at least one city belongs to a cycle of length 3?\n\nb) each city belongs to a cycle of length 3?",
    "options": [],
    "answer": "See solution",
    "solution": "Let there be 10 cities and let cities be connected as in the figure below. 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![](images/prob1516_p21_data_97b147a9e7.png)",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 13054,
    "subject": "Mathematics (Olympiad)",
    "question": "Is it possible to construct a triangle with sides $x$, $y$, $z$ satisfying the condition:\n\n$$\n3x^2y^2 + 3y^2z^2 + 3z^2x^2 = x^4 + y^4 + z^4?\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "Rewrite the equation as:\n\n$$\n2x^2y^2 + 2y^2z^2 + 2z^2x^2 - x^4 - y^4 - z^4 = -(x^2y^2 + y^2z^2 + z^2x^2).\n$$\n\nThe left-hand side can be decomposed as:\n\n$$(x+y+z)(x+y-z)(y+z-x)(z+x-y) = -(x^2y^2 + y^2z^2 + z^2x^2).$$\n\nHence, the left-hand side is negative, therefore, at least one of the multipliers is negative too. The first one is always positive, then, without loss of generality, we can assume the second one is negative, i.e. $x+y-z<0$, which contradicts the triangle inequality.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 13055,
    "subject": "Mathematics (Olympiad)",
    "question": "Count the number of incongruent equiangular hexagons whose side lengths are $a_0, a_1, a_2, a_3, a_4, a_5$ (in some order), where $a_i = d i$ for $i = 0, 1, 2, 3, 4, 5$ and $d > 0$. Two hexagons are considered congruent if their side lengths can be matched by a cyclic rotation or reflection.",
    "options": [],
    "answer": "See solution",
    "solution": "Numbers $a, b, c, a', b', c'$ satisfy (1) if and only if $a+K, b+K, c+K, a'+K, b'+K, c'+K$ satisfy (1). Thus, it suffices to count the number of incongruent equiangular hexagons with side lengths $a_0, a_1, a_2, a_3, a_4, a_5$ in some order. We may assume $a_0 = 0$ so $a_i = d i$. The six numbers $a, b, c, a', b', c'$ satisfy (1) if and only if $d a, d b, d c, d a', d b', d c'$ do so, so it suffices to study $d = 1$, i.e., $a_i = i$. We need to count the number of ways to order $0, 1, 2, 3, 4, 5$ so that equations (1) hold. Cyclic changes of the order lead to congruent hexagons, so we may assume $b = 5$.\n\nAdding the equations (1) gives $(a + b + c) + b = (a' + b' + c') + b'$, which rewrites as $b - b' = (a' + b' + c') - (a + b + c) = 15 - 2(a + b + c)$, since $a + b + c + a' + b' + c' = 0 + 1 + 2 + 3 + 4 + 5 = 15$. Since $b = 5$ is the largest, $b - b' > 0$, so $15 > 2(a + b + c)$, i.e., $a + b + c \\leq 7$, which means $a + c \\leq 2$. Therefore, $\\{a, c\\} = \\{0, 1\\}$ and $\\{a, c\\} = \\{0, 2\\}$ are the only possibilities.\n\nSwapping $a$ and $c$, as well as $a'$ and $c'$, leads to a congruent (via reflection) equiangular hexagon. Thus, the two possibilities (up to congruence) are $(a, c) = (0, 1)$ and $(a, c) = (0, 2)$.\n\nFor $(a, c) = (0, 1)$ and $b = 5$, $a + b + c = 6$ and $a' + b' + c' = 15 - 6 = 9$. Therefore, $b' = (a + b + c) + b - (a' + b' + c') = 6 + 5 - 9 = 2$. Hence, $a' = a + b - b' = 0 + 5 - 2 = 3$ and $c' = c + b - b' = 1 + 5 - 2 = 4$. So $(a, b, c, a', b', c') = (0, 5, 1, 3, 2, 4)$.\n\nSimilarly, for $(a, c) = (0, 2)$, $a + b + c = 7$ and $a' + b' + c' = 8$. This leads to $b' = 4$, $a' = 1$, $c' = 3$, so $(a, b, c, a', b', c') = (0, 5, 2, 1, 4, 3)$.\n\nSince $a + b + c = 6$ from the first solution is not equal to $a + b + c = 7$ nor to $b + c + a' = 8$ from the second, the two solutions do not lead to congruent hexagons.\n\nTranslating back to the original setup, for each $i \\geq 0$, the six numbers $a_i, a_{i+1}, a_{i+2}, a_{i+3}, a_{i+4}, a_{i+5}$ can be ordered in exactly two ways (up to congruence) so that these numbers are the side lengths of an equiangular hexagon in this order:\n\n$$\na_i, a_{i+5}, a_{i+1}, a_{i+3}, a_{i+2}, a_{i+4} \\quad \\text{and} \\quad a_i, a_{i+5}, a_{i+2}, a_{i+1}, a_{i+4}, a_{i+3}.\n$$\n\nRotation of the hexagon corresponds to cyclic rotation of the numbers. Reflection corresponds to certain swappings, e.g., swapping $a_i$ with $a_{i+1}$ and $a_{i+3}$ with $a_{i+4}$ in the first solution above.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 13056,
    "subject": "Mathematics (Olympiad)",
    "question": "For two points $A(x_1, y_1)$ and $B(x_2, y_2)$ on the Cartesian plane, define their *toric distance* to be\n\n$$\nd(A, B) = \\sqrt{\\|x_1 - x_2\\|^2 + \\|y_1 - y_2\\|^2}\n$$\n\nwhere $\\|x\\|$ denotes the distance from $x$ to the integer closest to $x$.\n\nFind the maximal real number $r > 0$ for which there exist four points $A_1, A_2, A_3$, and $A_4$ on the Cartesian plane such that any two points among the four have toric distance at least $r$.",
    "options": [],
    "answer": "See solution",
    "solution": "The largest positive real number is $r = \\frac{\\sqrt{6} - \\sqrt{2}}{2}$.\n\nOn one hand, consider four points:\n\n$$\nA_1(0,0),\\quad A_2\\left(\\frac{1}{2}, \\frac{2-\\sqrt{3}}{2}\\right),\\quad A_3\\left(\\frac{2-\\sqrt{3}}{2}, \\frac{1}{2}\\right),\\quad A_4\\left(\\frac{3-\\sqrt{3}}{2}, \\frac{3-\\sqrt{3}}{2}\\right)\n$$\n\n![](images/2024_CMO_p6_data_920542f2a8.png)\n\nFor $1 \\le i < j \\le 4$, $d(A_i, A_j) = \\frac{\\sqrt{6} - \\sqrt{2}}{2}$.\n\nNow, we prove that no four points $A_1, A_2, A_3, A_4$ in the coordinate plane satisfy the condition that the toric distance between any two points is strictly greater than $\\frac{\\sqrt{6}-\\sqrt{2}}{2}$. We proceed by contradiction and assume such points exist.\n\nTranslating all points by the same vector or adding an integer to one coordinate of any point does not change the problem. Thus, we may assume $A_1 = (0,0)$ and $A_i = (x_i, y_i)$ for $i = 2,3,4$, with $x_i, y_i \\in [0,1]$.\n\n![](images/2024_CMO_p6_data_bbd3d4f979.png)\n\nBy assumption, $A_2, A_3, A_4$ cannot appear in the four sectors centered at the vertices of the unit square with radius $\\frac{\\sqrt{6}-\\sqrt{2}}{2}$. Thus, they must lie in the star-shaped region in the center.\n\nThe lines $x = \\frac{1}{2}$ and $y = \\frac{1}{2}$ divide this star-shaped region into four parts. No two points can lie in the same part. Otherwise, suppose $A_2$ and $A_3$ lie in the same region. The farthest distance within this region is achieved between $B_2\\left(\\frac{2-\\sqrt{3}}{2}, \\frac{1}{2}\\right)$ and $B_3\\left(\\frac{1}{2}, \\frac{\\sqrt{3}}{2}\\right)$.\n\nHence, $A_2, A_3, A_4$ must lie in three of the four parts. Assume $A_2, A_3, A_4$ lie in the upper-left, lower-left, and lower-right parts, respectively. Consider the points $C\\left(\\frac{1}{2}, \\frac{1}{2}\\right)$ and $D\\left(\\frac{\\sqrt{3}-1}{2}, \\frac{\\sqrt{3}-1}{2}\\right)$, where $D$ lies on the arc centered at $A_1$ and $CD$ forms a $45^\\circ$ angle with the $x$-axis. Suppose $A_3$ lies above the segment $CD$.\n\n![](images/2024_CMO_p7_data_0f3f8433bd.png)\n\nThe shaded region in the diagram is contained within the quadrilateral with vertices $B_3, C, D$, and $B_2$. The maximum distance between two points inside a convex quadrilateral does not exceed the lengths of its two diagonals or its four edges, whichever is the longest. For $B_2B_3CD$, the longest length is $B_2B_3 = B_3D = \\frac{\\sqrt{6}-\\sqrt{2}}{2}$. Thus, $d(A_2, A_3) \\le \\frac{\\sqrt{6}-\\sqrt{2}}{2}$.\n\nThis contradicts the assumption. Therefore, it is impossible to have four points in the coordinate plane such that the toric distance between any two points is strictly greater than $\\frac{\\sqrt{6}-\\sqrt{2}}{2}$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 13057,
    "subject": "Mathematics (Olympiad)",
    "question": "There are 44 distinct holes in a line and 2017 ants. Each ant crawled up from a hole, then moved to another hole and crawled down. Denote $T$ as the set of time points that the ants crawled up or crawled down from some holes. Suppose that the speeds of the ants are pairwise distinct and they did not change their speed. Prove that if $|T| \\leq 45$ then there exist two ants that did not meet. Note that two ants meet when there exists a time point such that they are at the same location on the line, including the holes.",
    "options": [],
    "answer": "See solution",
    "solution": "We call a time point \"special\" if at that time, some ants crawled up or down from a hole. It is easy to see that we only need to solve the problem in the case $|T| = 45$ (if $|T| < 45$, we can consider some additional special time points, which only strengthens the argument).\n\nConsider a coordinate system $Oxy$ where $Ox$ represents the locations of the holes on the line and $Oy$ represents time. Let the $x$-coordinates of the holes be $x_1, x_2, \\dots, x_{44}$ and the $y$-coordinates of the special times be $y_1, y_2, \\dots, y_{45}$. An ant moves from $(x_a, y_b)$ to $(x_c, y_d)$ if it crawled up from hole $x_a$ at time $y_b$ and crawled down into hole $x_c$ at time $y_d$. Since the speed of each ant does not change, the graph of its movement is a segment connecting these two points.\n\nThus, in total, we have 2017 segments, and since the speeds of the ants are pairwise distinct, the segments have different directions. To finish the problem, we need to show that at least two segments among them do not intersect. Note that the number of endpoints of these segments is at most $45 \\times 44 = 1980 < 2017$. We will prove a more general version: If there are $n$ points on the plane, then there are at most $n$ segments connecting them such that no two segments are parallel or overlap. $(\\diamond)$\n\nWe prove $(\\diamond)$ by induction. It is easy to check for $n = 2, 3$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 13058,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all integer pairs $ (a, b) $ for which $ (2a^2 + b)^3 = b^3 a $.",
    "options": [],
    "answer": "See solution",
    "solution": "The integer pairs are $ (27, 729),\\, (8, 128),\\, (0, 0),\\, (-1, -1) $.\n\nIf $b = 0$, then the equation gives $2a^2 + b = 0$, so $a = 0$.\n\nAssume now that $b \\neq 0$. Since $b^3$ and $(2a^2 + b)^3$ are perfect cubes, their ratio $a$ is the cube of a rational number; as it is an integer, it must be the cube of an integer $c$. Taking the cube root of both sides gives:\n\n$$2c^6 + b = bc$$\n\nwhich implies $b(c - 1) = 2c^6$. Since $c$ and $c - 1$ are coprime, $c^6$ and $c - 1$ are also coprime. Therefore, $c - 1$ must divide $2$, so $c$ can be $3$, $2$, $0$, or $-1$.\n\n- If $c = 3$, then $a = 27$ and $2b = 2 \\cdot 729$, so $b = 729$.\n- If $c = 2$, then $a = 8$ and $b = 128$.\n- If $c = 0$, then $-b = 0$, which was already considered.\n- If $c = -1$, then $a = -1$ and $-2b = 2$, so $b = -1$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 13059,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $a$ and $b$ be two coprime numbers. Prove that\n\n$$\nN = a^2b + ab^2 - a^2 - b^2 - ab + 1\n$$\n\nis the smallest positive integer such that the equation $a^2x + aby + b^2z = m$ has a non-negative integer solution $(x, y, z)$ for every $m \\geq N$.",
    "options": [],
    "answer": "See solution",
    "solution": "We show that $m = a^2b + ab^2 - a^2 - b^2 - ab$ cannot be written as $a^2x + aby + b^2z$ for any non-negative integers $x, y, z$.\n\nSuppose, for contradiction, that there exist non-negative $x, y, z$ such that\n$$\na^2b + ab^2 - a^2 - b^2 - ab = a^2x + aby + b^2z.\n$$\nThen,\n$$\na^2(b - 1 - x) + b^2(a - 1 - z) = ab(y + 1).\n$$\nSo, $b - 1 - x > 0$ or $a - 1 - z > 0$. Without loss of generality, assume $b - 1 - x > 0$. Then,\n$$\na^2(b - 1 - x) = b(ay + bz - ab + b + a).\n$$\nSince $(a^2, b) = 1$, $b - 1 - x$ must be divisible by $b$, but $0 < b - 1 - x < b$, which is impossible. Thus, $m$ cannot be represented in the required form for $m < N$, and for $m \\geq N$, solutions exist.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 13060,
    "subject": "Mathematics (Olympiad)",
    "question": "Доведіть, що для будь-якого многочлена $P(x)$ з цілими коефіцієнтами, якщо всі його корені $x_1, x_2, \\dots, x_n$ належать інтервалу $(0; 3)$, то всі ці корені дорівнюють $1$ або $2$.",
    "options": [],
    "answer": "See solution",
    "solution": "Неважко довести, що для всіх $t \\in (0, 3)$\n\n$$\n|t(t-1)(t-2)(t-3)| = |(t^2 - 3t)^2 + 2(t^2 - 3t)| \\le 1,\n$$\n\nпричому рівність досягається тоді й тільки тоді, коли $t = \\frac{3\\pm\\sqrt{5}}{2}$. Нехай $a \\ge 0$ — кількість рівних $1$ чисел серед $x_1, x_2, \\dots, x_n$, $b \\ge 0$ — кількість рівних $2$ чисел серед $x_1, x_2, \\dots, x_n$. Якщо $a+b=n$, то твердження задачі, очевидно, виконано. Нехай $a+b < n$, $m = n - (a+b)$, і без обмеження загальності припустимо, що $x_1, x_2, \\dots, x_m \\notin \\{1, 2\\}$. Тоді $P(x) = (x-1)^a (x-2)^b Q(x)$, де многочлен\n\n$$\nQ(x) = (x - x_1)(x - x_2) \\cdots (x - x_m)\n$$\n\nмає цілі коефіцієнти (як частка двох зведених многочленів з цілими коефіцієнтами). Отже, $Q(0), Q(1), Q(2), Q(3)$ — деякі цілі ненульові числа, звідки отримуємо, що $A = |Q(0)Q(1)Q(2)Q(3)| \\ge 1$. Але $A = \\prod_{k=1}^{m} |x_k(x_k-1)(x_k-2)(x_k-3)|$, і тому, згідно з доведеним, це можливо лише у випадку, коли $|x_k(x_k-1)(x_k-2)(x_k-3)| = 1$ для всіх $k = 1, \\dots, m$. Остання рівність дає потрібний результат.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 13061,
    "subject": "Mathematics (Olympiad)",
    "question": "Determine all sequences $a_0, a_1, a_2, \\dots$ that satisfy the following conditions:\n\n1. $a_{n+1} = 2a_n - 1$ for every integer $n \\ge 0$.\n2. $a_0$ is a rational number.\n3. $a_i = a_j$ for some $i \\neq j$.",
    "options": [],
    "answer": "See solution",
    "solution": "Assume $a_k = p/q$, where $\\gcd(p, q) = 1$.\n\nThen $a_{k+1} = (2p^2 - q^2)/q^2$.\n\nSince $\\gcd(p, q) = 1$, $\\gcd(2p^2 - q^2, q^2) = \\gcd(2, q^2)$, which is either 1 or 2.\n\nIf $q > 2$, the denominator of $a_{k+1}$ is strictly greater than that of $a_k$, so the sequence of denominators increases and cannot repeat.\n\nIf $|a_k| > 1$, writing $|a_k| = 1 + e$, we have $a_{k+1} = 1 + 4e + 2e^2 > |a_k|$, so the sequence is strictly increasing.\n\nTherefore, for repetition, $a_0$ must have denominator at most 2 and lie between $-1$ and $1$ inclusive. The possible values are:\n\n- $a_0 = -1$: sequence is $-1, 1, 1, 1, \\dots$\n- $a_0 = -1/2$: sequence is $-1/2, -1/2, -1/2, \\dots$\n- $a_0 = 0$: sequence is $0, -1, 1, 1, \\dots$\n- $a_0 = 1/2$: sequence is $1/2, -1/2, -1/2, -1/2, \\dots$\n- $a_0 = 1$: sequence is $1, 1, 1, \\dots$\n\nAll of these are solutions.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 13062,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $f : \\mathbb{Z} \\to \\mathbb{Z}$ be a function such that for all $x, y \\in \\mathbb{Z}$,\n\n$$\nx f(2y^2 - x) + y^2 f(2x - y^2) = \\frac{f(x)^2}{x} + f(y^3).\n$$\n\nFind all such functions $f$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $f$ be a solution of the problem. Let $p$ be a prime. Since $p$ divides $f(p)^2$, $p$ divides $f(p)$ and so $p$ divides $\\frac{f(p)^2}{p}$. Taking $y = 0$ and $x = p$, we deduce that $p$ divides $f(0)$. As $p$ is arbitrary, we must have $f(0) = 0$.\n\nNext, take $y = 0$ to obtain $x f(-x) = \\frac{f(x)^2}{x}$. Replacing $x$ by $-x$, and combining the two relations yields $f(x) = 0$ or $f(x) = x^2$ for all $x$.\n\nSuppose now that there exists $x_0 \\neq 0$ such that $f(x_0) = 0$. Taking $y = x_0$, we obtain $x f(-x) + x_0^2 f(2x) = \\frac{f(x)^2}{x}$, yielding $x_0^2 f(2x) = 0$ for all $x$ and so $f$ vanishes on even numbers. Assume that there exists an odd number $y_0$ such that $f(y_0) \\neq 0$, so $f(y_0) = y_0^2$. Taking $y = y_0$, we obtain\n\n$$\nx f(2y_0^2 - x) + y_0^2 f(2x - y_0^2) = \\frac{f(x)^2}{x} + f(y_0^3).\n$$\n\nChoosing $x$ even, we deduce that $y_0^2 f(2x - y_0^2) = f(y_0^3)$. This forces $f(y_0^3) = 0$, as otherwise we would have $f(2x - y_0^2) = (2x - y_0^2)^2$ for all even $x$ and so $y_0^2 (2x - y_0^2)^2 = f(y_0^3)$ for all such $x$, which is impossible. Thus $f(2x - y_0^2) = 0$ for all even numbers $x$, that is, $f$ vanishes on numbers of the form $4k + 3$. But since $x^2 f(-x) = f(x)^2$, $f$ also vanishes on all $x$ such that $-x \\equiv -1 \\pmod{4}$, that is, on $4\\mathbb{Z} + 1$. Thus $f$ also vanishes on all odd numbers, contradicting the choice of $y_0$. Hence, if $f$ is not the zero map, then $f$ does not vanish outside $0$ and so $f(x) = x^2$ for all $x$.\n\nIn conclusion, $f(x) = 0$ for all $x \\in \\mathbb{Z}$ and $f(x) = x^2$ for all $x \\in \\mathbb{Z}$ are the only possible solutions. The first function clearly satisfies the given relation, while the second also satisfies the Sophie Germain identity:\n\n$$\nx (2y^2 - x)^2 + y^2 (2x - y^2)^2 = x^3 + y^6\n$$\n\nfor all $x, y \\in \\mathbb{Z}$.\n\nAlternatively, for $f(0) = 0$: If $f(0) \\neq 0$, set $x = 2f(0)$ to obtain\n\n$$\n2(f(0))^2 = \\frac{(f(2f(0)))^2}{2f(0)} + f(0)\n$$\n\nthat is,\n\n$$\n2(f(0))^2 (2f(0) - 1) = f(2f(0))^2.\n$$\n\nBut $2(2f(0) - 1)$ cannot be a perfect square since it is of the form $4k + 2$. So $f(0) = 0$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 13063,
    "subject": "Mathematics (Olympiad)",
    "question": "A positive integer $A$ has 73 digits, all different from zero. Prove that we can erase 64 digits such that the new number is divisible by 37.",
    "options": [],
    "answer": "See solution",
    "solution": "After erasing 64 digits, the final number has $73 - 64 = 9$ digits.\n\nIf every digit (from 1 to 9) appeared at most 8 times in $A$, then $A$ would have at most $9 \\times 8 = 72$ digits, which is not possible since $A$ has 73 digits.\n\nTherefore, there is a digit $a \\neq 0$ that appears 9 times in $A$. We can erase 64 digits so that the remaining number is $\\overline{aaaaaaaaa} = a \\times 111111111 = a \\times 1001001 \\times 3 \\times 37$, which is divisible by 37.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 13064,
    "subject": "Mathematics (Olympiad)",
    "question": "Determine all functions $f : \\mathbb{R} \\to \\mathbb{R}$ satisfying\n$$\nf(x + y f(x)) + f(xy) = f(x) + f(2019y),\n$$\nfor all real numbers $x$ and $y$.",
    "options": [],
    "answer": "See solution",
    "solution": "There are three types of such functions:\n\n1. $f(x) = 2019 - x$;\n2. $f(x) = c$ for an arbitrary constant $c$;\n3. $f(x) = 0$ for $x \\neq 0$, and $f(0)$ is arbitrary.\n\nA straightforward check shows that all three types satisfy the equation. We need to show that they are the only ones. Let $N = 2019$.\n\nFirst, set $x = N x'$, so the equation becomes\n$$\nf(N x' + y f(N x')) + f(N x' y) = f(N x') + f(N y).\n$$\nLet $g(x) = \\frac{f(N x)}{N}$. Then the equation reads\n$$\ng(x + y g(x)) + g(xy) = g(x) + g(y) \\quad (x, y \\in \\mathbb{R}).\n$$\nNow, investigate the possible forms of $g$.\n\nSetting $x = 1$ gives $g(1 + y g(1)) = g(1)$. If $g(1) \\neq 0$, then $1 + y g(1)$ covers all real values, so $g$ is constant (case 2). Otherwise, $g(1) = 0$. Setting $y = 1$ gives $g(x + g(x)) = 0$. If $a = 1$ is the unique real number with $g(a) = 0$, then $x + g(x) = 1$, so $g(x) = 1 - x$ (case 1). Otherwise, suppose $g(1) = 0$ and $g(a) = 0$ for some $a \\neq 1$.\n\n**Claim 1.** If $b$ is any zero of $g$, then $g(b y) = g(y)$ for all $y$.\n\nAlso, $g(g(0) y) = g(y)$.\n\n**Claim 2.** If $a$ and $b$ are two zeros of $g$, and $g(s) \\neq 0$, then $g$ is $p$-periodic, where $p = (a-b)s$. Indeed, substituting $x = a s$ and using Claim 1, we get\n$$\ng(a s + y g(s)) = g(s) + g(y) - g(s y),\n$$\nwhich does not depend on $a$. Thus, $g$ is periodic with period $p$.\n\nIf $g(x) = 0$ for all $x \\neq 0$, we get case 3. If there exists $s \\neq 0$ with $g(s) \\neq 0$, then by Claim 2, $g$ is periodic with some period $p$. Substituting $x = p$ and using periodicity, we get $g(y g(0)) + g(p y) = g(0) + g(y)$. Since $g(y g(0)) = g(y)$ by Claim 1, we have $g(p y) = g(0)$, so $g$ is constant.\n\n*Remark.* After reaching the conditions above and obtaining Claims 1 and 2, alternative approaches are possible. For example, let $Z = \\{x \\in \\mathbb{R} : g(x) = 0\\}$ be the set of zeros of $g$. Claim 1 shows $Z$ is $a$-invariant: $a Z = Z$. We want to show $Z - Z = \\mathbb{R}$; this, by Claim 2, would imply $g$ is constant.\n\nFor any $\\beta \\in Z$, plugging $y = \\beta$ gives $g(x + \\beta g(x)) = 0$, so $x + \\beta g(x) \\in Z$ for all $x$. Setting $\\beta = 1$ and $\\beta = a$ (from above), we get $x + g(x), x + a g(x) \\in Z$. The first inclusion gives $a(x + g(x)) \\in Z$, so $(a - 1)x = a(x + g(x)) - (x + a g(x)) \\in Z - Z$. Thus, $Z - Z = \\mathbb{R}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 13065,
    "subject": "Mathematics (Olympiad)",
    "question": "Pete wants to mark all the squares of an $n \\times n$ grid using tokens. What is the minimum number of tokens required if each token can mark any number of squares in a single row, but only in ascending order?",
    "options": [],
    "answer": "See solution",
    "solution": "$n$.\n\nTo mark all the squares using $n$ tokens, Pete may use one token per row, marking all its squares in ascending order.\n\nTo show that $n-1$ tokens do not suffice, place the numbers $1, 2, \\ldots, n$ into the squares of some diagonal, and arrange all remaining numbers arbitrarily. Then Pete needs a new token for each square of this diagonal.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 13066,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $n$ be a positive integer. Consider\n$$\nS = \\{(x, y, z) : x, y, z \\in \\{0, 1, \\dots, n\\},\\ x + y + z > 0\\},\n$$\na set of $(n + 1)^3 - 1$ points in three-dimensional space.\n\nDetermine the smallest possible number of planes, the union of which contains $S$ but does not include $(0, 0, 0)$.",
    "options": [],
    "answer": "See solution",
    "solution": "The answer is: we need at least $3n$ planes.\n\nIt is easy to see that $3n$ planes satisfy the given conditions. For example, $x = i$, $y = i$, and $z = i$ for $i = 1, 2, \\dots, n$. Another example is $x + y + z = k$ for $k = 1, 2, \\dots, 3n$.\n\nNow we prove that $3n$ is the minimum required. The following lemma is key to the proof.\n\n**Lemma 1**: Let $P(x_1, \\dots, x_k)$ be a non-zero polynomial in $k$ variables. If every $k$-tuple $(x_1, \\dots, x_k)$, with $x_1, \\dots, x_k \\in \\{0, 1, \\dots, n\\}$ and $x_1 + x_2 + \\dots + x_k > 0$, is a root of $P$, and $P(0, 0, \\dots, 0) \\neq 0$, then\n\ndeg $P \\geq kn$.\n\n*Proof of Lemma 1*: We use induction on $k$.\n\nFor $k = 0$, since $P \\neq 0$, the lemma is trivially true.\n\nAssume the lemma holds for $k-1$. We prove it for $k$.\n\nLet $y = x_k$, and let $R(x_1, \\dots, x_{k-1}, y)$ be the remainder when $P$ is divided by $Q(y) = y(y-1)\\cdots(y-n)$.\n\nSince $Q(y)$ has $n+1$ roots ($y = 0, 1, \\dots, n$),\n$$\nP(x_1, \\dots, x_{k-1}, y) = R(x_1, \\dots, x_{k-1}, y)\n$$\nfor all $x_1, \\dots, x_{k-1}, y \\in \\{0, 1, \\dots, n\\}$. Thus, $R$ satisfies the lemma's hypothesis, and $\\deg_y R \\leq n$. Also, $\\deg R \\leq \\deg P$. So we only need to show $\\deg R \\geq kn$.\n\nWrite $R$ in descending powers of $y$:\n$$\nR(x_1, \\dots, x_{k-1}, y) = R_n(x_1, \\dots, x_{k-1}) y^n + R_{n-1}(x_1, \\dots, x_{k-1}) y^{n-1} + \\cdots + R_0(x_1, \\dots, x_{k-1}).\n$$\n\nWe claim $R_n(x_1, \\dots, x_{k-1})$ satisfies the induction hypothesis.\n\nConsider $T(y) = R(0, \\dots, 0, y)$. Clearly, $\\deg T(y) \\leq n$, and $T(y)$ has $n$ roots: $y = 1, \\dots, n$. Since $T(0) \\neq 0$, $T(y) \\neq 0$, so $\\deg T(y) = n$ and $R_n(0, \\dots, 0) \\neq 0$. (For $k = 1$, the coefficient $R_n$ is non-zero.)\n\nFor each fixed $a_1, \\dots, a_{k-1} \\in \\{0, 1, \\dots, n\\}$ and $a_1 + \\cdots + ...$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 13067,
    "subject": "Mathematics (Olympiad)",
    "question": "定義費氏數：$F_0 = 0$, $F_1 = 1$, $F_2 = 1$，且對所有正整數 $n$，有 $F_{n+2} = F_{n+1} + F_n$。\n\n證明存在一個正整數 $N$，使得不存在公差不為 $0$ 的 $N$ 項正整數等差數列，每一項都可以表示為至多 $2024$ 個費氏數之和（費氏數可重複）。",
    "options": [],
    "answer": "See solution",
    "solution": "特別地，若 $F_n < a < F_{n+1}$，且 $a$ 可被寫為 $k$ 個費氏數之和，則我們可要求其中一個為 $F_n$。\n\n**Lemma 2.** 給定正整數 $M$，則對任意的 $4M$ 項正整數等差數列\n\n$$\nA, A+d, \\dots, A+(4M-1)d,\n$$\n\n都存在整數 $\\ell$ 滿足在區間 $[F_\\ell, F_{\\ell+1}]$ 中該等差數列至少出現 $M$ 項。\n\nLemma 2 證明：$d=0$ 時不必證，故設 $d>0$。若 $A+(3M)d$ 到 $A+(4M-1)d$ 皆落在某段 $[F_\\ell, F_{\\ell+1}]$ 中，結論成立。否則，知在這之中有一個 $F_\\ell$ 將其切開，可設為 $A+md \\le F_\\ell < A+(m+1)d$，其中 $m \\ge 3M$。\n\n注意到 $A+(3M)d \\ge F_3 = 1$。且當 $\\ell \\ge 3$ 時，我們有 $1.5F_{\\ell-1} = F_{\\ell-1} + \\frac{1}{2}(F_{\\ell-2} + F_{\\ell-3}) \\le F_\\ell$。於是由 $1.5F_{\\ell-1} \\le F_\\ell < A+(m+1)d$，我們可以得到\n\n$$\nF_{\\ell-1} < \\frac{A}{1.5} + \\frac{m+1}{1.5}d \\le A + (m+1-M)d,\n$$\n\n故第 $m+1-M$ 項至第 $m$ 項皆落於區間 $[F_{\\ell-1}, F_\\ell]$ 中。 □\n\n註：這段的估計不是最緊的，比如將 $1.5$ 換成任意介於 $1$ 和 $\\frac{1+\\sqrt{5}}{2}$ 的 $r$，而 $4$ 換成任意滿足 $\\frac{L-1}{r} > L-2$ 的數字 $L$ 皆可。\n\n最後我們證明 $N(k) < 4N(k-1)+4$。由數學歸納法假設 $N(k-1)$ 是有限的，那麼對於任意長為 $4N(k-1)+4$ 的等差數列中，者存在一段長度為 $N(k-1)+1$ 的子數列 $a_1, a_2, \\dots, a_{N(k-1)+1}$ 落在某段 $[F_\\ell, F_{\\ell+1}]$ 中。由 Lemma 1 我們知道，若 $a_i$ 皆可被寫為 $k$ 項費氏數之和，則可要求這 $k$ 項中出現 $F_\\ell$。如此一來，$\\langle a_i - F_\\ell \\rangle$ 就是一個長度為 $N(k-1)+1$，但每項皆可被寫為 $k-1$ 個費氏數之和的等差數列，和 $N(k-1)$ 的定義矛盾。所以 $N(k)$ 至多為 $4N(k-1)+3$。本題證畢。 □",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 13068,
    "subject": "Mathematics (Olympiad)",
    "question": "Find the sum of the elements of the set\n$$\nM = \\left\\{ \\frac{n}{2} + \\frac{m}{5} \\mid m, n = 0, 1, 2, \\dots, 100 \\right\\}.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "Consider the set $A = \\{2a + 5b \\mid a, b = 1, 2, \\dots, 100\\}$ and notice that $1 \\notin A$ and $3 \\notin A$.\n\nThe largest even number from $A$ is $700$, obtained for $a = b = 100$. The number $698$ is obtained for $a = 99, b = 100$. The largest odd number from $A$ is $695$, obtained for $b = 99, a = 100$, implying that $697 \\notin A$ and $699 \\notin A$.\n\nWe claim that all the integers between $4$ and $695$ belong to $A$.\n\nLet $y \\leq 500$ and let $r$ be the remainder left by $y$ upon division by $5$. Write $y = 5c + r$ with $0 \\leq c \\leq 100$ and $0 \\leq r \\leq 4$. If $r$ is even, then $y = 5c + 2k$, where $r = 2k$. If $r$ is odd, then $y \\geq 5$, so $c \\geq 1$ and $y = 5(c - 1) + 2(k + 3)$, where $r = 2k + 1$.\n\nFor $y > 500$, write $y = 500 + z$ with $z \\geq 200$. If $z$ is even, then $y = 5 \\cdot 100 + 2k$, where $z = 2k$. If $z$ is odd, then $y \\leq 695$, so $z \\leq 195$ and $y = 5 \\cdot 99 + 2(k + 3)$, where $z = 2k + 1$. A quick inspection of all the above cases shows that the claim holds.\n\nThe sum of all the elements of $M$ is equal to\n$$\nS = \\frac{1}{10}((1 + 2 + \\dots + 700) - (1 + 3 + 697 + 699)) = 350 \\cdot 697 = 35 \\cdot 697.\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 13069,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $n$ be a positive integer, and let $a_1, a_2, \\dots, a_{2n+1}$ be $2n+1$ positive real numbers. For $k = 1, 2, \\dots, 2n+1$, define\n\n$$\nb_k = \\max_{0 \\le m \\le n} \\left( \\frac{1}{2m+1} \\sum_{i=k-m}^{k+m} a_i \\right),\n$$\n\nwhere the subscript of $a_i$ is taken modulo $2n+1$. Prove that the number of subscripts $k$ satisfying $b_k \\ge 1$ does not exceed $2 \\sum_{i=1}^{2n+1} a_i$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $I = \\{k \\mid b_k \\ge 1\\}$. 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$$\n[k - m_k, k + m_k] := \\{k - m_k, k - m_k + 1, \\dots, k + m_k\\}\n$$\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. Otherwise, suppose $i$ is contained in $r$ nice segments\n\n$$\n[i - u_1, i + v_1], \\dots, [i - u_r, i + v_r]\n$$\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 $[i - u_j, i + v_j]$ and $[i - u_k, i + v_k]$, 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 as in the claim. We have\n\n$$\n2 \\sum_{i=1}^{2n+1} a_i \\ge \\sum_{\\alpha=1}^{r} \\sum_{k=i_{\\alpha}-m_{\\alpha}}^{i_{\\alpha}+m_{\\alpha}} a_k = \\sum_{\\alpha=1}^{r} (2m_{\\alpha} + 1)b_{i_{\\alpha}} \\ge \\sum_{\\alpha=1}^{r} (2m_{\\alpha} + 1) \\ge |I|,\n$$\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": 13070,
    "subject": "Mathematics (Olympiad)",
    "question": "In $\\triangle ABC$, $M$ is the midpoint of side $AC$. Points $D$ and $E$ lie on the tangent to the circumcircle of $\\triangle ABC$ at $A$, such that $MD \\parallel AB$ and $A$ is the midpoint of segment $DE$. The circle passing through $A$, $B$, and $E$ intersects $AC$ at $P$. The circle passing through $A$, $D$, and $P$ intersects the extension of $DM$ at $Q$.\n\nProve that $\\angle BCQ = \\angle BAC$.",
    "options": [],
    "answer": "See solution",
    "solution": "As shown in Fig. 2.2, let $N$ be the midpoint of side $BC$. Then $D$, $M$, $Q$, and $N$ are collinear and $MN \\parallel AB$.\n\nBy the alternate segment theorem, $\\angle DAM = \\angle CBA = \\angle CNM$. Since $\\angle AMD = \\angle NMC$, it follows that $\\triangle AMD \\sim \\triangle NMC$. Therefore,\n\n$$\n\\frac{NM}{NC} = \\frac{AM}{AD}. \\qquad \\textcircled{1}\n$$\n\n![](images/Mathematical_Olympiad_In_China_2021_2022___Problems_and_Solutions_2024_p85_data_c0d1f7a2e0.png)\n\nBy the given condition, points $A$, $D$, $P$, $Q$ are concyclic. Hence $\\angle APQ = \\angle ADQ = \\angle ADM = \\angle ACB$, so $PQ \\parallel BC$.\n\nTherefore,\n\n$$\n\\frac{NQ}{NM} = \\frac{CP}{CM}. \\qquad (2)\n$$\n\nCombining (1), (2), and $AD = AE$ yields\n\n$$\n\\frac{NQ}{NC} = \\frac{NQ}{NM} \\cdot \\frac{NM}{NC} = \\frac{CP}{CM} \\cdot \\frac{AM}{AD} = \\frac{CP}{AD} = \\frac{CP}{AE},\n$$\n\nnamely,\n\n$$\n\\frac{NQ}{NC} = \\frac{CP}{AE}. \\tag{3}\n$$\n\nBy the alternate segment theorem, $\\angle BAE = \\angle BCA = \\angle BCP$.\nSince $A$, $P$, $B$, $E$ are concyclic, $\\angle BEA = \\angle BPC$.\nTherefore, $\\triangle BAE \\sim \\triangle BCP$, so $\\frac{CP}{AE} = \\frac{BC}{BA}$. Combining (3) gives\n\n$$\n\\frac{NQ}{NC} = \\frac{BC}{BA}.\n$$\n\nSince $MN \\parallel AB$, $\\angle CNQ = \\angle ABC$. Hence, $\\triangle CNQ \\sim \\triangle ABC$.\n\nTherefore, $\\angle NCQ = \\angle BAC$, that is, $\\angle BCQ = \\angle BAC$.\n\n$\\Box$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 13071,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all functions $f : \\mathbb{R} \\to \\mathbb{R}$ such that for all real numbers $x$ and $y$,\n\n$$\nf(x+y) + y \\leq f(f(x)).\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "Set $y = 0$ in the initial inequality:\n\n$$\nf(x) \\leq f(f(f(x))) \\quad \\forall x \\in \\mathbb{R}. \\qquad (1)\n$$\n\nNext, set $y = f(f(x)) - x$ in $(*)$:\n\n$$\nf(f(x)) \\leq x \\quad \\forall x \\in \\mathbb{R}. \\qquad (2)\n$$\n\nReplace $x$ by $f(x)$ in (2):\n\n$$\nf(f(f(x))) \\leq f(x).\n$$\n\nTogether with (1), this gives $f(f(f(x))) = f(x)$.\n\nNow, the original inequality becomes:\n\n$$\nf(x+y) + y \\leq f(x) \\quad \\forall x, y \\in \\mathbb{R}. \\qquad (3)\n$$\n\nSet $x = 0$ in (3):\n\n$$\nf(y) \\leq a - y \\quad \\forall y \\in \\mathbb{R}, \\qquad (4)\n$$\n\nwhere $a = f(0)$.\n\nSet $y = -x$ in (3):\n\n$$\na - x \\leq f(x). \\qquad (5)\n$$\n\nComparing (4) and (5), we obtain $f(x) = a - x$.\n\nIt is easy to verify that $f(x) = a - x$ satisfies the given inequality for any real number $a$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 13072,
    "subject": "Mathematics (Olympiad)",
    "question": "Fix an integer $n \\geq 3$, and let $x_1, x_2, \\dots, x_n$ be real numbers in the closed unit interval $[0, 1]$ such that $s = x_1 + x_2 + \\dots + x_n \\geq 3$. Prove that there exist distinct indices $i$ and $j$ such that\n\n$$\n2^{|i-j|} x_i x_j > 2^{s-3}.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "Consider indices $a < b$ such that $2^{b-a} x_a x_b$ is maximal. By maximality, $x_k < 2^{k-a} x_a$ if $k < b$, and $x_k < 2^{2b-a-k} x_a$ if $k > b$. It then easily follows that $\\sum_{k < b} x_k$ and $\\sum_{k > b} x_k$ are both (strictly) less than $2^{b-a} x_a$, so\n\n$$\n3 \\leq s = \\sum_{k < b} x_k + x_b + \\sum_{k > b} x_k < 2^{b-a+1} x_a + x_b \\leq 2^{b-a+1} x_a + 1,\n$$\n\nwhence $2^{b-a+1} x_a > 2$. Letting $\\alpha = -\\log_2 x_a$ and $p = \\lfloor \\alpha \\rfloor$, it then follows that $p \\leq b-a-1$, so $\\sum_{k < a+p} x_k < 2^p x_a \\leq 1$ and $\\sum_{k \\leq a+p} x_k < 2^{p+1} x_a = 2^{p+1-\\alpha} \\leq 1 + (p+1-\\alpha) = 2 + p - \\alpha$; this latter holds since $2^t \\leq 1 + t$ for $0 \\leq t \\leq 1$.\n\nSimilarly, write $\\beta = -\\log_2 x_b$ and $q = \\lceil \\beta \\rceil$, to get $q \\leq b-a-1$, $\\sum_{k > b-q} x_k < 2^q x_b \\leq 1$, and $\\sum_{k \\geq b-q} x_k < 2^{q+1} x_b = 2^{q+1-\\beta} \\leq 2 + q - \\beta$.\n\nSince $\\sum_{k < a+p} x_k < 1$, $\\sum_{k > b-q} x_k < 1$, and $3 \\leq s = \\sum_{k < a+p} x_k + \\sum_{k = a+p}^{b-q} x_k + \\sum_{k > b-q} x_k < 1 + ((b-q) - (a+p) + 1) + 1 = 3 + b-a-p-q$, it follows that $b-a-p-q \\geq 1$, so the sum $\\sum_{k = a+p+1}^{b-q-1} x_k$ is empty if and only if $b-a-p-q = 1$. Recall that the $x_k$ are all at most 1. With the standard convention that empty sums vanish, $\\sum_{k = a+p+1}^{b-q-1} x_k \\leq b-a-p-q-1$ (whether this sum is empty or not), so $s = \\sum_{k \\leq a+p} x_k + \\sum_{k = a+p+1}^{b-q-1} x_k + \\sum_{k \\geq b-q} x_k \\leq (2 + p - \\alpha) + (b-a-p-q-1) + (2 + q - \\beta) = 3 + b-a - \\alpha - \\beta$, whence $s-3 \\leq b-a - \\alpha - \\beta$. Alternatively, but equivalently, $2^{s-3} \\leq 2^{b-a} x_a x_b$, as required.\n\n*Remark.* The required inequality is tight, as the following example shows. Let $n = 2m+1$, and set $x_{m+1} = 1$, $x_m = x_{m+2} = \\frac{1}{2} + \\frac{1}{2^m}$, and $x_{m-k+1} = x_{m+k+1} = \\frac{1}{2^k}$, $k = 2, \\dots, m$. Then $s = 3$, so the right-hand side of the required inequality is 1, and\n\n$$\n\\max_{i \\neq j} 2^{|i-j|} x_i x_j = 2^2 x_m x_{m+2} = \\left(1 + \\frac{1}{2^{m-1}}\\right)^2,\n$$\n\nwhich can be made arbitrarily close to 1. Notice that $s$ can be increased by inserting more 1's in the middle, which also accommodates the case where $n$ is even.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 13073,
    "subject": "Mathematics (Olympiad)",
    "question": "There are several identical caro papers of size $5 \\times 5$. Someone uses $n$ colors to fill in each paper such that two cells at the same position on two sides share the same color. Two papers are considered congruent if they can be stacked together in such a way that the pairs of squares at the same position have the same color. Prove that, by the definition of congruence, one can obtain at most\n\n$$\n\\frac{1}{8} \\left(n^{25} + 4n^{15} + n^{13} + 2n^{7}\\right)\n$$\ndistinct colored caro papers.",
    "options": [],
    "answer": "See solution",
    "solution": "We will prove the following lemma:\n\n**Lemma.** Consider a positive integer $m = 2k+1$ with $k \\ge 2$, and the square table of size $m \\times m$ in which each cell is filled by one of $n$ colors. Then the number of different ways to color (not duplicated by rotation) is equal to\n\n$$\n\\frac{n(a^4 + a^2 + 2a)}{4} \\quad \\text{with } a = n^{k^2+k}.\n$$\n\n*Proof.* Note that the middle cell, denoted $C$ as shown in the image, is not affected by rotation so there are always $n$ ways to fill it.\n\n![](images/Vietnamese_mathematical_competitions_p228_data_99c7693c06.png)\n\nConsider the collection of 4 squares $k \\times k$ in the corners as set $A = A_1 \\cup A_2 \\cup A_3 \\cup A_4$ and 4 rectangles $1 \\times k$ as set $B = B_1 \\cup B_2 \\cup B_3 \\cup B_4$ in such a way that $B_t$ is immediately followed by $A_t$ for $1 \\le t \\le 4$. We consider 4 pairs consisting of two subsets $(A_t, B_t)$ of $A, B$. The number of cells in each pair $(A_t, B_t)$ is $k^2+k$, so the number of ways to fill in each pair of subsets is $a = n^{k^2+k}$. Here, we consider each pair as a vertex of some square $XYZT$. We will count the number of colorings for the vertices $X, Y, Z, T$ so that they are not duplicated by rotation. We have the following cases:\n\n1. All vertices are filled with the same color: $a$ ways.\n2. The coloring is alternating, so we just consider how to fill in 2 adjacent vertices, with the number of ways $\\frac{a^2-a}{2}$.\n3. Otherwise, there is some identical pair that is not cyclical of 2, then the number is $a^4-(a^2-a)-a$. This way of coloring has a circular permutation and will generate $\\frac{a^4-a^2}{4}$ different ways.\n\nIn total, the number of colorings is\n\n$$\n\\frac{a^4 - a^2}{4} + \\frac{a^2 - a}{2} + a = \\frac{a^4 + a^2 + 2a}{4}.\n$$\n\n$\\square$\n\nBack to the problem: Using the above lemma when $2k+1=5$, we have the number of ways to fill in the $5 \\times 5$ paper, not duplicated by rotation, is\n\n$$\nn \\left( \\frac{n^{24} + n^{12} + 2n^{6}}{4} \\right) = \\frac{n^{25} + n^{13} + 2n^{7}}{4}.\n$$\n\nDenote $S$ as the set of these coloring ways. In this problem, we also need to consider the reflection transformation. We separate $S$ into two types of papers, namely $A$, $B$: papers that can and cannot create themselves through vertical and horizontal reflections.\n\nNotice that each paper in $A$ generates 2 different papers in $S$ (should only be counted as 1 in the original problem); meanwhile, each paper in $B$ generates exactly 1 paper in $S$. Hence,\n\n$$\n2A + B = \\frac{n^{25} + n^{13} + 2n^{7}}{4}.\n$$\n\nAlso, it is easy to count $B = n^{15}$ (fill the left half of the piece of paper, also the middle line as well). From there we can calculate\n\n$$\n2A + 2B = (2A + B) + B = \\frac{1}{4}(n^{25} + n^{13} + n^{7}) + n^{15}\n$$\n\nso\n\n$$\nA + B = \\frac{n^{25} + 4n^{15} + n^{13} + n^{7}}{8}.\n$$\n\nThis is the number of different pieces of paper we need to count. The problem is completely solved. $\\square$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 13074,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $P(x) = 1 + pn^2 + \\prod_{i=1}^{2p-2} Q(x^i)$. For $p = 2$, $n = 1$, and $Q(x) = 2x + 1$, we have $P(-1) = 0$. Show that for odd primes $p$, no suitable $n$ and $Q(x)$ exist such that $P(a) = 0$ for some integer $a$.",
    "options": [],
    "answer": "See solution",
    "solution": "Since all $Q(a^i)$ have the same parity for $1 \\leq i \\leq 2p-2$ for an integer $a$, if $P(a) = 0$, then $p \\equiv 3 \\pmod{4}$. Also, $a^i \\equiv a^{i+p-1} \\pmod{p}$ for $1 \\leq i \\leq p-1$. Therefore, $$\\prod_{i=1}^{2p-2} Q(a^i) \\equiv \\left(\\prod_{i=1}^{p-1} Q(a^i)\\right)^2 \\pmod{p}$$ and $$P(a) \\equiv 1 + \\left(\\prod_{i=1}^{p-1} Q(a^i)\\right)^2 \\neq 0 \\pmod{p}$$ for $p \\equiv 3 \\pmod{4}$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 13075,
    "subject": "Mathematics (Olympiad)",
    "question": "Determine all real-valued functions $f$ defined on the real line such that for every pair of real numbers $x, y$:\n\n$$\nf(f(x+y)f(x-y)) = x^2 - y f(y).\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "Let us substitute $x = 0$ and $y = 0$ into the identity:\n\n$$\nf(f(0)f(0)) = 0^2 - 0 f(0) = 0.\n$$\nSo $f((f(0))^2) = 0$.\n\nNow, substitute $x = 0$ and $y = (f(0))^2$:\n\n$$\nf(f((f(0))^2)f(- (f(0))^2)) = 0^2 - (f(0))^2 f((f(0))^2).\n$$\nSince $f((f(0))^2) = 0$, this gives $f(0 \times f(- (f(0))^2)) = - (f(0))^2 \times 0 = 0$, so $f(0) = 0$.\n\nLet $t$ be any nonzero real number. Substitute $x = t$, $y = t$:\n\n$$\nf(f(2t)f(0)) = t^2 - t f(t).\n$$\nBut $f(0) = 0$, so $f(f(2t) \times 0) = t^2 - t f(t)$, i.e., $f(0) = t^2 - t f(t)$.\nSince $f(0) = 0$, we have $t^2 - t f(t) = 0$, so $f(t) = t$ for all $t \\neq 0$.\n\nSince $f(0) = 0$, we conclude $f(t) = t$ for all real $t$.\n\nFinally, check that $f(t) = t$ satisfies the original identity:\n\n$$\nf(f(x+y)f(x-y)) = f((x+y)(x-y)) = f(x^2 - y^2) = x^2 - y^2.\n$$\nBut the right side is $x^2 - y f(y) = x^2 - y^2$ when $f(y) = y$.\n\nThus, the only solution is $f(t) = t$ for all real $t$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 13076,
    "subject": "Mathematics (Olympiad)",
    "question": "Solve the equation in the set of whole numbers:\n\n$$\nx^{2010} - 2006 = 4y^{2009} + 4y^{2008} + 2007y.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "**Lemma 1.** Let $x \\in \\mathbb{Z}$, then every prime factor of $x^2 + 1$ is of the form $4k + 1$.\n\n*Proof.* Let $p \\mid x^2 + 1$, then $\\gcd(x, p) = 1$.\n\nThen\n\n$$\nx^2 + 1 \\equiv 0 \\pmod{p} \\implies x^2 \\equiv -1 \\pmod{p}.\n$$\n\nRaising both sides to the power $\\frac{p-1}{2}$ gives\n\n$$\n(x^2)^{\\frac{p-1}{2}} \\equiv (-1)^{\\frac{p-1}{2}} \\pmod{p}\n$$\n\ni.e.\n\n$$\nx^{p-1} \\equiv (-1)^{\\frac{p-1}{2}} \\pmod{p}.\n$$\n\nBy Fermat's little theorem, $x^{p-1} \\equiv 1 \\pmod{p}$, so\n\n$$\n(-1)^{\\frac{p-1}{2}} \\equiv 1 \\pmod{p}.\n$$\n\nThus, $\\frac{p-1}{2}$ is even, so $p = 4k + 1$.\n\nNow, rewrite the given equation:\n\n$$\nx^{2010} - 2006 = 4y^{2009} + 4y^{2008} + 2007y\n$$\n\nAdd $2007$ to both sides:\n\n$$\nx^{2010} + 1 = 4y^{2009} + 4y^{2008} + 2007y + 2007\n$$\n\nGroup terms:\n\n$$\nx^{2010} + 1 = (4y^{2008} + 2007)(y + 1)\n$$\n\nThe number $4y^{2008} + 2007$ is of the form $4k - 1$, so it must have a prime divisor of the form $4k - 1$. But by Lemma 1, $(x^{1005})^2 + 1$ cannot have such a divisor.\n\nTherefore, the equation has no solutions in the set of whole numbers.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 13077,
    "subject": "Mathematics (Olympiad)",
    "question": "Un conjunto $S$ de enteros positivos se llama *canalero* si para cualesquiera tres números $a, b, c \\in S$, todos diferentes, se cumple que $a$ divide $bc$, $b$ divide $ac$ y $c$ divide $ab$.\n\n(a) Demuestra que para cualquier conjunto finito de enteros positivos $\\{c_1, c_2, \\dots, c_n\\}$ existen infinitos enteros positivos $k$, tales que el conjunto $\\{kc_1, kc_2, \\dots, kc_n\\}$ es canalero.\n\n(b) Demuestra que para cualquier entero $n \\ge 3$ existe un conjunto canalero que tiene exactamente $n$ elementos y ningún entero mayor que 1 divide a todos sus elementos.",
    "options": [],
    "answer": "See solution",
    "solution": "Sea $M$ el mínimo común múltiplo de $c_1, c_2, \\dots, c_n$, y sea $k = k'M$, donde $k'$ toma cualquier valor entero positivo. Nótese que, para cualesquiera $u, v, w \\in \\{1, 2, \\dots, n\\}$ distintos, se tiene que\n$$\n\\frac{(kc_u)(kc_v)}{kc_w} = c_u \\cdot c_v \\cdot \\frac{k}{c_w},\n$$\ndonde claramente $\\frac{k}{c_w} = k' \\cdot \\frac{M}{c_w}$ es entero, al ser $M$ un múltiplo de $c_w$. Luego cada uno de los infinitos conjuntos así formados es canalero.\n\nSea $P = \\{p_1, p_2, \\dots, p_n\\}$ un conjunto de $n$ primos distintos cualesquiera, y denotemos $\\pi = p_1 \\cdot p_2 \\cdots p_n$. Definamos\n$$\n\\Pi = \\left\\{ \\pi_1 = \\frac{\\pi}{p_1}, \\pi_2 = \\frac{\\pi}{p_2}, \\dots, \\pi_n = \\frac{\\pi}{p_n} \\right\\},\n$$\nque es claramente un conjunto de enteros positivos distintos, tales que $\\pi_u$ no es divisible por $p_u$ para $u = 1, 2, \\dots, n$, y ningún primo que no esté en $P$ divide a ningún elemento de $\\Pi$. Luego ningún primo, esté o no en $P$, divide a la vez a todos los elementos de $\\Pi$, con lo que ningún entero mayor que 1 puede dividir a todos los elementos de $\\Pi$. Al mismo tiempo, nótese que para cualesquiera $u, v, w \\in \\{1, 2, \\dots, n\\}$, se tiene que\n$$\n\\frac{\\pi_u \\cdot \\pi_v}{\\pi_w} = p_w \\frac{\\pi}{p_u p_v},\n$$\nque es claramente entero pues $p_u, p_v$ son primos distintos que dividen a $\\pi$. Luego $\\Pi$ es un conjunto canalero de exactamente $n$ elementos, tal que ningún entero mayor que 1 divide a la vez a todos sus elementos.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 13078,
    "subject": "Mathematics (Olympiad)",
    "question": "A piecewise linear periodic function is defined by\n$$\nf(x) = \\begin{cases} x & \\text{if } x \\in [-1, 1), \\\\ 2-x & \\text{if } x \\in [1, 3), \\end{cases}\n$$\nand $f(x + 4) = f(x)$ for all real numbers $x$. The graph of $f(x)$ has the sawtooth pattern depicted below.\n![](images/2025AIME_I_Solutions_p7_data_bceb08ea3b.png)\n\nThe parabola $x = 34y^2$ intersects the graph of $f(x)$ at finitely many points. The sum of the $y$-coordinates of these intersection points can be expressed in the form $\\frac{a+b\\sqrt{c}}{d}$, where $a, b, c$, and $d$ are positive integers, $a, b$, and $d$ have greatest common divisor equal to 1, and $c$ is not divisible by the square of any prime. Find $a + b + c + d$.",
    "options": [],
    "answer": "See solution",
    "solution": "The line segments in the graph of $f(x)$ are portions of the graphs of equations of the form $x = \\pm y + n$ for some integer $n$. If the parabola $x = 34y^2$ intersects the line $x = \\pm y + n$ exactly twice, then the $y$-coordinates of these two intersection points satisfy the quadratic equation\n$$\n\\pm y + n = 34y^2.\n$$\nVieta's formulas imply that the sum of the solutions of this quadratic equation is $\\frac{1}{34}$ if the line has equation $x = y + n$, and the sum of the solutions is $-\\frac{1}{34}$ if the line has equation $x = -y + n$.\n\nNote that any intersections of $x = 34y^2$ with $y = f(x)$ must satisfy $|y| \\le 1$, and therefore must also satisfy $x \\le 34$. Over the interval $[0, 34]$ the function $f(x)$ can be expressed piecewise as\n$$\nf(x) = \\begin{cases} x & \\text{if } x \\in [0, 1), \\\\ 2-x & \\text{if } x \\in [1, 3), \\\\ x-4 & \\text{if } x \\in [3, 5), \\\\ \\vdots & \\vdots \\\\ x-32 & \\text{if } x \\in [31, 33), \\\\ 34-x & \\text{if } x \\in [33, 34]. \\end{cases}\n$$\nThe parabola $x = 34y^2$ has exactly two intersections for $x \\in [0, 34]$ with all of these lines except for $y = 34 - x$, in which case it has only one intersection satisfying $x \\in [0, 34]$. This single intersection must be above the $x$-axis because $f(x) \\ge 0$ on the interval $[33, 34]$, so this intersection has positive $y$-coordinate. The coordinate must be one of the two solutions of the quadratic equation $34y^2 = -y + 34$, which are\n$$\ny = \\frac{-1 \\pm \\sqrt{1 + 4 \\cdot 34^2}}{68} = \\frac{-1 \\pm \\sqrt{4625}}{68} = \\frac{-1 \\pm 5\\sqrt{185}}{68}.\n$$\nBecause the $y$-coordinate of the single intersection must be positive, it must be the greater of these two values. Therefore the sum of the $y$-coordinates of all the intersections is\n$$\n\\underbrace{\\frac{1}{34} - \\frac{1}{34} + \\frac{1}{34} - \\dots + \\frac{1}{34}}_{17 \\text{ terms}} + \\frac{-1 + 5\\sqrt{185}}{68} = \\frac{1 + 5\\sqrt{185}}{68}.\n$$\nThe requested sum is $1 + 5 + 185 + 68 = 259$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 13079,
    "subject": "Mathematics (Olympiad)",
    "question": "In the cyclic quadrilateral $ABCD$, the diagonal $AC$ bisects the angle $DAB$. The side $AD$ is extended beyond $D$ to a point $E$. Show that $CE = CA$ if and only if $DE = AB$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $\\angle DAC = \\angle CAB$. Since equal angles on a circumference are subtended by equal chords, we have $BC = CD$. Also, $\\angle EDC = 180^\\circ - \\angle CDA = \\angle ADC$, since opposite angles in a cyclic quadrilateral sum to $180^\\circ$.\n\nTherefore, the rotation centered at $C$ taking $D$ to $B$ takes $E$ to a point $E'$ on the ray $BA$. Since rotation preserves distances, $CE = CE'$, so $CE = CA \\Leftrightarrow A = E'$. Similarly, we must have $DE = BE'$, and thus $A = E' \\Leftrightarrow DE = BA$. Hence, $DE = AB \\Leftrightarrow CE = CA$, as desired.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 13080,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $ABC$ be an acute-angled triangle. Let $M$ be the midpoint of $AC$, $CC'$ the altitude from $C$ in $ABC$, and $H'$ the point symmetric to the orthocenter of $ABC$ over the line $AB$. Let points $P$, $Q$, and $R$ be the orthogonal projections of $C'$ onto the lines $AH'$, $AC$, and $CB$, respectively, and let $S$ be the center of the circumscribed circle of triangle $PQR$. Prove that the point symmetric to $M$ over $S$ lies on the segment $BH'$.\n\n![](images/shortlistBMO2010_p14_data_626aa76aed.png)",
    "options": [],
    "answer": "See solution",
    "solution": "We first prove the following lemma.\n\n**Lemma.** Let $XYZT$ be a cyclic quadrilateral such that $XZ \\perp YT$. Let $XZ \\cap YT = O$ and let $V$ and $W$ be the orthogonal projections of $O$ onto lines $XY$ and $ZT$, respectively. If $I$ is the midpoint of $YZ$, and $J$ is the midpoint of $XT$, then $IVJW$ is a cyclic deltoid.\n\n**Proof.** Let $Y'$ and $Z'$ be the midpoints of $OY$ and $OZ$, respectively. Then $VY' = OY' = IZ'$ and $IY' = OZ' = WZ'$, and also\n\n$$\n\\angle VY'I = \\angle VY'O + 90^\\circ = 2 \\cdot \\angle VYO + 90^\\circ = 2 \\cdot \\angle WZO + 90^\\circ = \\\\\n\\angle WZ'O + 90^\\circ = \\angle IZ'W.\n$$\n\nHence, $\\triangle VY'I \\cong \\triangle IZ'W$, so $VI = WI$, and\n\n$$\n\\angle VIW = 90^\\circ - \\angle VIY' - \\angle ZIW = 90^\\circ - \\angle VIY' - \\angle IVY' = \\\\\n\\angle VY'O = 2\\angle XYT.\n$$\n\nAnalogously, $\\angle VJW = 2\\angle YXZ$, and therefore\n\n$$\n\\angle VIW + \\angle VJW = 180^\\circ.\n$$\n\nThe lemma is proved.\n\nLet $QC \\cap H'B = N$. As $\\angle NC'B = \\angle ACC' = \\angle ABH'$ (the last equality follows from the fact that $ACBH'$ is cyclic), we have that $N$ is the midpoint of $BH'$. Using the lemma, the quadrilateral $PMRN$ is a cyclic deltoid, and the center of its circumscribed circle lies at the midpoint of $MN$. Hence, $Q$ also lies on that circle, as $\\angle MQN = 90^\\circ$, and $S$ is the midpoint of $MN$. $\\square$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 13081,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all real numbers $x$, $y$, and $z$ such that all the following equations hold:\n\n$$\nx^2 - 3y - z = -8\n$$\n\n$$\ny^2 - 5z - x = -12\n$$\n\n$$\nz^2 - x - y = 6\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "$(1, 2, 3)$",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 13082,
    "subject": "Mathematics (Olympiad)",
    "question": "Consider a non-singular matrix $A \\in \\mathcal{M}_n(\\mathbb{R})$.\n\na) Prove that the matrix $AA^T$ has positive eigenvalues.\n\nb) Suppose that positive integers $p$ and $q$ are such that $(AA^T)^p = (A^T A)^q$. Prove that $A^T = A^{-1}$.\n\n(We denote by $A^T$ the transpose of $A$.)",
    "options": [],
    "answer": "See solution",
    "solution": "a) Let $\\lambda \\in \\mathbb{C}$ be an eigenvalue of $AA^T$ and $X \\in \\mathcal{M}_{n,1}(\\mathbb{C}) \\setminus \\{O_{n,1}\\}$ such that $AA^T X = \\lambda X$. Taking transpose and conjugate, we get $\\bar{X}^T AA^T = \\bar{\\lambda} \\bar{X}^T$, implying, by multiplying with $X$, that $\\bar{X}^T AA^T X = \\bar{\\lambda} \\bar{X}^T X$. Thus $(A^T X)^T (A^T X) = \\bar{\\lambda}(\\bar{X}^T X)$.\n\nIf\n$$\nX = \\begin{pmatrix} x_1 \\\\ x_2 \\\\ \\vdots \\\\ x_n \\end{pmatrix} \\quad \\text{and} \\quad A^T X = \\begin{pmatrix} y_1 \\\\ y_2 \\\\ \\vdots \\\\ y_n \\end{pmatrix},\n$$\nthen $\\sum_{i=1}^{n} |y_i|^2 = \\bar{\\lambda} \\sum_{i=1}^{n} |x_i|^2$.\n\nAs $X \\neq O_{n,1}$ and $A$ is invertible, we get $A^T X \\neq O_{n,1}$ and, consequently, $\\sum_{i=1}^{n} |x_i|^2 > 0$ and $\\sum_{i=1}^{n} |y_i|^2 > 0$, which gives $\\lambda \\in (0, \\infty)$.\n\nb) Matrices $AA^T$ and $A^T A$ have the same characteristic polynomial. If $0 < \\lambda_1 \\le \\lambda_2 \\le \\dots \\le \\lambda_n$ are the common eigenvalues of $AA^T$ and $A^T A$, the matrices $(AA^T)^p$ and $(A^T A)^q$ have $\\lambda_1^p, \\lambda_2^p, \\dots, \\lambda_n^p$ and $\\lambda_1^q, \\lambda_2^q, \\dots, \\lambda_n^q$ as eigenvalues, respectively. From $(AA^T)^p = (A^T A)^q$ one gets $\\lambda_i^p = \\lambda_i^q$, so $\\lambda_i^{p-q} = 1$ for $i = 1, \\dots, n$. Then $\\lambda_1 = \\lambda_2 = \\dots = \\lambda_n = 1$.\n\nPut $AA^T - I_n = B = (b_{ij})_{1\\le i,j\\le n}$. The matrix $B$ has all eigenvalues equal to zero, so $B^2$ has the same property. As $B$ is symmetric, $\\sum_{i=1}^n \\sum_{j=1}^n b_{ij}^2 = \\text{Tr}(BB^T) = \\text{Tr}(B^2) = 0$. Thus $B = O_n$, that is $AA^T = I_n$. In conclusion, $A^T = A^{-1}$.",
    "topic": "Algebra",
    "subtopic": "Linear Algebra"
  },
  {
    "id": 13083,
    "subject": "Mathematics (Olympiad)",
    "question": "a) Let $ABC$ be a triangle with altitude $AD$ and $P$ be a moving point on $AD$. Lines $PB$ and $AC$ intersect at $E$, lines $PC$ and $AB$ intersect at $F$. Suppose that $AEDF$ is a cyclic quadrilateral. Prove that\n\n$$\n\\frac{PA}{PD} = (\\tan B + \\tan C) \\cot \\frac{A}{2}\n$$\n\nb) Let $ABC$ be a triangle with orthocenter $H$ and $P$ be a moving point on $AH$. The line passing through $C$ and perpendicular to $AC$ meets $BP$ at $M$ and the line passing through $B$ and perpendicular to $AB$ meets $CP$ at $N$. Let $K$ be the projection of $A$ on $MN$. Prove that $\\angle BKC + \\angle MAN$ is invariant.",
    "options": [],
    "answer": "See solution",
    "solution": "a) Let $Q$ be the intersection of $EF$ and $BC$ so $(QD, BC) = -1$ and $A(QD, BC) = -1$ and $D(EF, AQ) = -1$. Moreover, because $DA \\perp DQ$ so $DA$ is the internal bisector of $\\angle EDF$. Note that $AEDF$ is cyclic so $AE = AF$.\n\n![](images/Vietnamese_mathematical_competitions_p98_data_641caa9443.png)\n\nSuppose that the circumcircle of $AEDF$ cuts $BC$ at $G$ different from $D$. We have $DG$ is the internal bisector of angle $EDF$ so $GE = GF$.\n\nHence, $\\triangle AGE = \\triangle AGF$ then $AG$ is the bisector of $\\angle BAC$. By applying Menelaus's theorem with the line $BPE$ cuts three lines passing through three vertices of triangle $ADC$ and line $CPF$ cuts three lines passing through three vertices of triangle $ABD$:\n\n$$\n\\frac{EA}{EC} = \\frac{PA}{PD} \\cdot \\frac{BD}{BC} \\cdot \\frac{FA}{FB} = \\frac{PA}{PD} \\cdot \\frac{CD}{BC}\n$$\n\nFrom this result, we have\n\n$$\n\\frac{FA}{FB} + \\frac{EA}{EC} = \\frac{PA}{PD} \\cdot \\left( \\frac{BD}{BC} + \\frac{CD}{BC} \\right) = \\frac{PA}{PD}\n$$\n\nTherefore, we get\n\n$$\n\\begin{aligned}\n\\frac{PA}{PD} &= \\frac{FA}{GF} \\cdot \\frac{GF}{FB} + \\frac{EA}{GE} \\cdot \\frac{GE}{EC} \\\\\n&= \\cot \\frac{A}{2} \\tan B + \\cot \\frac{A}{2} \\tan C \\\\\n&= (\\tan B + \\tan C) \\cot \\frac{A}{2}.\n\\end{aligned}\n$$\n\n![](images/Vietnamese_mathematical_competitions_p99_data_0bf7ef7459.png)",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 13084,
    "subject": "Mathematics (Olympiad)",
    "question": "Prove that, for all positive real numbers $x$, $y$, and $z$,\n\n$$\n4(x + y + z)^3 > 27(x^2y + y^2z + z^2x).\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "If $x, y, z > 0$, then\n\n$$\nx(2x - 4y - z)^2 + y(2y - 4z - x)^2 + z(2z - 4x - y)^2 \\geq 0.\n$$\n\nEquality only occurs if $2x - 4y - z$, $2y - 4z - x$, $2z - 4x - y$ are all equal to zero; solving these simultaneous equations gives $x = y = z = 0$, which cannot happen for $x, y, z > 0$.\n\nSo for $x, y, z > 0$, we have strict inequality above. Expanding out gives the inequality\n\n$$\n4x^3 + 4y^3 + 4z^3 - 15x^2y - 15y^2z - 15z^2x + 12xy^2 + 12yz^2 + 12zx^2 + 24xyz > 0.\n$$\n\nAdding $27x^2y + 27y^2z + 27z^2x$ to each side gives\n\n$$\n4(x^3+y^3+z^3)+12(x^2y+y^2z+z^2x+xy^2+yz^2+zx^2)+24xyz > 27(x^2y+y^2z+z^2x)\n$$\n\nand factorising gives\n\n$$\n4(x + y + z)^3 > 27(x^2y + y^2z + z^2x).\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 13085,
    "subject": "Mathematics (Olympiad)",
    "question": "Given $a^2 + b^2 = 4$ and $ab = 1$, with $a$ and $b$ positive, compute\n$$\n\\frac{1}{a^3} + \\frac{1}{b^3}.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "From $a^2 + b^2 = 4$ and $ab = 1$, we have\n$$(a+b)^2 = a^2 + b^2 + 2ab = 4 + 2 = 6,$$\nso $a+b = \\sqrt{6}$ since $a$ and $b$ are positive.\n\nNow,\n$$\n\\frac{1}{a^3} + \\frac{1}{b^3} = \\frac{a^3 + b^3}{a^3 b^3}.\n$$\nRecall $a^3 + b^3 = (a+b)^3 - 3ab(a+b)$ and $a^3 b^3 = (ab)^3$:\n$$\n\\frac{a^3 + b^3}{a^3 b^3} = \\frac{(a+b)^3 - 3ab(a+b)}{(ab)^3} = \\frac{(\\sqrt{6})^3 - 3 \\cdot 1 \\cdot \\sqrt{6}}{1} = \\frac{6\\sqrt{6} - 3\\sqrt{6}}{1} = 3\\sqrt{6}.\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 13086,
    "subject": "Mathematics (Olympiad)",
    "question": "Някои от градовете в една държава са свързани с директни пътища. Нека $t$ е най-малкото естествено число, за което съществува град, от който до всеки друг град може да се стигне, минавайки по най-много $t$ пътя. Да се докаже, че съществуват градове $A_1, A_2, \\ldots, A_{2t-1}$, за които за всеки $i \\ne j$, $i = 1, 2, \\ldots, 2t - 2$, $j = 2, 3, \\ldots, 2t - 1$ градовете $A_{i}$ и $A_{j}$ са свързани с път тогава и само тогава, когато $i + 1 = j$.",
    "options": [],
    "answer": "See solution",
    "solution": "Разглеждаме граф $G$ с върхове градовете в държавата и ребра пътищата между тях. От всички подграфи на $G$ да изберем граф $H$, който има свойството на $G$ и е минимален по отношение на броя на върховете.\n\nДа изберем произволен връх $v_t$ на $H$, който не разделя графа на две несвързани компоненти (не е трудно да се види, че такъв връх съществува). От минималността на $H$ следва, че съществува връх $v_0 \\in H$, за който $d(v_0, w) \\le t - 1$ за всички върхове $w \\ne v_t$. Тъй като $t$ е най-малкото естествено число, за което съществува град, от който до всеки друг град може да се стигне, минавайки по най-много $t$ пътя, то $d(v_0, v_t) = t$. Нека $v_0, v_1, \\dots, v_t$ е пътят от $v_0$ до $v_t$. Отново поради свойството на $H$ съществува връх $w$, за който $d(v_2, w) \\ge t$ и за този връх също имаме $d(v_0, w) \\le t - 1$.\n\nПри $t = 2$ търсеният път е $v_0v_1v_2$. Нека $t \\ge 3$. Поради $d(v_2, w) \\ge t$ имаме, че $w \\ne v_i$. Нека $u$ е произволен връх от пътя между $v_0$ и $w$ и нека $d(v_0, u) = p$ и $d(u, w) = q$. Да допуснем, че $d(u, v_i) = 1$ за някое $i \\ge 2$. Тогава\n\n$$d(v_0, v_i) = i \\le d(v_0, u) + d(u, v_i) = p + 1,$$\n$$t \\le d(v_2, w) \\le d(v_2, v_i) + d(v_i, u) + d(u, w) = i - 2 + 1 + q.$$\n\nСъбираме горните неравенства и получаваме $t \\le p+q = d(v_0, w) \\le t-1$, което е противоречие. Следователно разстоянието от всеки връх от пътя между $v_0$ и $w$ до $v_i$, $i \\ge 2$ е поне 2.\n\nАко няма връх от този път, който е съседен на $v_1$, то този път, заедно с пътя от $v_0$ до $v_t$ е търсеният. Ако съществува връх $u$ от този път, за който $d(u, v_1) = 1$, то $t \\le d(v_2, w) \\le d(v_2, v_1) + d(v_1, u) + d(u, w)$, откъдето $d(v_2, w) \\ge t - 2$. Сега лесно следва, че $d(v_0, u) = 1$ и търсеният път е $w \\dots u v_1 \\dots v_t$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 13087,
    "subject": "Mathematics (Olympiad)",
    "question": "瘋狂科學家無意間在他的實驗室發現了一種叫瓜克的新粒子。兩個瓜克可以組成一個瓜克對；一個瓜克可以同時和很多瓜克組成瓜克對。此外，他發現他可以進行以下兩種動作：\n\n1. 如果有一個瓜克和奇數個其他瓜克組成瓜克對，則他可以將這個瓜克消滅。\n2. 將整個實驗室裡的瓜克複製，也就是說，每個瓜克 $I$ 都會複製出一個瓜克 $I'$。新的瓜克 $I'$ 和 $J'$ 組成瓜克對若且唯若舊的瓜克 $I$ 和 $J$ 組成瓜克對；此外，$I'$ 會與 $I$ 組成瓜克對。除以上瓜克對外，複製不會製造其他額外的瓜克對。\n\n試證：科學家可以經由一系列的動作，讓實驗室裡最後只剩下一群瓜克，它們兩兩之間不成瓜克對。",
    "options": [],
    "answer": "See solution",
    "solution": "A crazy physicist discovered a new kind of particle which he called an imon, after some of them mysteriously appeared in his lab. Some pairs of imons in the lab can be entangled, and each imon can participate in many entanglement relations. The physicist has found a way to perform the following two kinds of operations with these particles, one operation at a time.\n\n1. If some imon is entangled with an odd number of other imons in the lab, then the physicist can destroy it.\n2. At any moment, he may double the whole family of imons in his lab by creating a copy $I'$ of each imon $I$. During this procedure, the two copies $I'$ and $J'$ become entangled if and only if the original imons $I$ and $J$ are entangled, and each copy $I'$ becomes entangled with its original imon $I$; no other entanglements occur or disappear at this moment.\n\nProve that the physicist may apply a sequence of such operations resulting in a family of imons, no two of which are entangled.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 13088,
    "subject": "Mathematics (Olympiad)",
    "question": "Point $D$ is chosen on side $BC$ of the acute triangle $ABC$ so that $AD = AC$. Let $P$ and $Q$ be respectively the feet of the perpendiculars from $C$ and $D$ to side $AB$. It is known that\n\n$$AP^2 + 3BP^2 = AQ^2 + 3BQ^2.$$ \n\nFind $\\angle ABC$.",
    "options": [],
    "answer": "See solution",
    "solution": "Write the condition $AP^2 + 3BP^2 = AQ^2 + 3BQ^2$ in the form $AQ^2 - AP^2 = 3(BP^2 - BQ^2)$. Express $AQ^2$ and $AP^2$ by the Pythagorean theorem for the right-angled triangles $ADQ$ and $ACP$: $AQ^2 = AD^2 - DQ^2$, $AP^2 = AC^2 - CP^2$. Since $AC = AD$, it follows that $AQ^2 - AP^2 = CP^2 - DQ^2$.\n\nLikewise, the right-angled triangles $BCP$ and $BDQ$ yield $BP^2 = BC^2 - CP^2$, $BQ^2 = BD^2 - DQ^2$. Hence\n\n![](images/Argentina2016_booklet_p2_data_fc371ace9b.png)\n\n$BP^2 - BQ^2 = BC^2 - BD^2 - (CP^2 - DQ^2)$. We showed above that $CP^2 - DQ^2 = AQ^2 - AP^2$; on the other hand, $AQ^2 - AP^2 = 3(BP^2 - BQ^2)$ by hypothesis. So the obtained equality can be written as $BP^2 - BQ^2 = BC^2 - BD^2 - 3(BP^2 - BQ^2)$, which implies $4(BP^2 - BQ^2) = BC^2 - BD^2$.\n\nFurthermore, we have $\\frac{BP}{BC} = \\frac{BD}{BQ} = x$ with $x > 0$ from the similar triangles $BCP$ and $BDQ$. Replacing $BC = xBP$ and $BD = xBQ$ in $4(BP^2 - BQ^2) = BC^2 - BD^2$ leads to $4(BP^2 - BQ^2) = x^2(BP^2 - BQ^2)$. Because $BP^2 - BQ^2 \\neq 0$ and $x > 0$, it follows that $x = 2$. Then $BD = 2BQ$, meaning that the hypotenuse $BD$ of right-angled triangle $BDQ$ is twice its leg $BQ$. Therefore $\\angle BDQ = 30^\\circ$ and so $\\angle ABC = \\angle DBQ = 60^\\circ$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 13089,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all prime numbers $p$, $q$, $r$ such that:\n\n$$\np(p+1) + q(q+1) = r(r+1).\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "**Answer:** $p=2$, $q=2$, $r=3$.\n\n**Solution.** Without loss of generality, let $p \\leq q$. Clearly, $r > p$. We can write:\n\n$$\np(p+1) = (r-q)(r+q+1). \\quad (1)\n$$\n\nSince $p$ is prime, either $r-q$ or $r+q+1$ is divisible by $p$. If $p$ divides $r-q$, then $r-q \\geq p$,\n\n$$\np(p+1) \\leq (r-q)(r-q+1) < (r-q)(r+q+1),\n$$\n\nwhich contradicts (1).\n\nSuppose $r+q+1$ is divisible by $p$. If $p=2$, then $r+q+1$ is even, so $r$ is odd, hence $q$ is even, thus $q=2$ and $r=3$.\n\nIf $p > 2$, let $r+q+1 = kp$. Then $r$ and $q$ are odd and $k$ is odd, $k > 1$. Then $p+1 = k(r-q)$ and we get $k^2(r-q) = kp + k$ or $r+q+1+k = k^2 r - k^2 q$, which is equivalent to\n\n$$\n(k^2+1)q = (k^2-1)r - (k+1).\n$$\n\nThe right-hand side is divisible by $(k+1)$, so the left-hand side must be as well. Since $k$ is odd, $\\frac{k^2+1}{2}$ is divisible by $\\frac{k+1}{2}$. Observe that\n\n$$\n\\left(\\frac{k^2+1}{2}, \\frac{k+1}{2}\\right) = \\left(\\frac{k^2+1}{2} - \\frac{(k-1)(k+1)}{2}, \\frac{k+1}{2}\\right) = \\left(1, \\frac{k+1}{2}\\right) = 1.\n$$\n\nThus, $\\frac{k^2+1}{2}$ and $\\frac{k+1}{2}$ are coprime, so $\\frac{k+1}{2}$ is divisible by $q$. Since $k > 1$, $\\frac{k+1}{2} > 1$ and $q = \\frac{k+1}{2}$.\n\nWe get $r = kp - q - 1$ and $q = \\frac{k+1}{2}$. Plugging this into $p+1 = k(r-q)$ gives $p+1 = (kp - k - 2)k$. But this is impossible because $k \\geq 3$, $p \\geq 3$ implies\n\n$$\n(kp - k - 2)k > kp - k - 2 = k(p-1) - 2 \\geq 3(p-1) - 2 = p + (2p-5) \\geq p+1.\n$$\n\nThis completes the proof.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 13090,
    "subject": "Mathematics (Olympiad)",
    "question": "Given the system of equations on $\\mathbb{R}$:\n\n$$\n\\begin{cases}\nx - a y = y z, \\\\\ny - a z = z x, \\\\\nz - a x = x y.\n\\end{cases}\n$$\n\na) Solve the system when $a = 0$.\n\nb) Prove that the system has 5 different roots when $a > 1$.",
    "options": [],
    "answer": "See solution",
    "solution": "a) For $a = 0$, the system becomes:\n\n$$\n\\begin{cases}\nx = y z, \\\\\ny = z x, \\\\\nz = x y.\n\\end{cases}\n$$\n\nIf one of the variables is $0$, then the others must also be $0$. Now, consider $x y z \\neq 0$. Multiplying the equations side by side gives $x y z = 1$. Thus,\n\n$$\nx^2 = y^2 = z^2 = 1.\n$$\n\nTherefore, the solutions are:\n\n$$\n(0, 0, 0),\\ (1, 1, 1),\\ (-1, -1, 1)\n$$\nand their permutations. There are 5 different solutions in total.\n\nb) Clearly, $x = y = z = 0$ is a solution. For $a > 1$, if any variable is $0$, all must be $0$. To find other solutions, transform the system to an equation in $z$:\n\n$$\n\\begin{cases}\nx = a y + y z, \\\\\ny = \\dfrac{a z}{1 - a z - z^2}, \\\\\nz - a(a y + y z) = y(a y + y z).\n\\end{cases}\n$$\n\nThis leads to:\n\n$$\nz^4 + (a^2 + 2a)z^3 + 2(a^3 - 1)z^2 + (a^4 - a^3 - a^2 - 2a)z + (1 - a^3) = 0.\n$$\n\nThe system has two roots $x = y = z = 0$ and $x = y = z = 1 - a$, so $z = 1 - a$ is a root. Dividing out $(z - (1 - a))$ gives:\n\n$$\nf(z) = z^3 + (a^2 + a + 1) z^2 + (a^3 - 1) z - a^2 - a - 1.\n$$\n\nBy Rolle's theorem, $f(z)$ has three distinct real roots for $a > 1$. Since $x$ and $y$ are uniquely determined by $z$, the system has exactly 5 different solutions when $a > 1$.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 13091,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $S_n$ denote the sum of the first $n$ terms in a number sequence $\\{a_n\\}$, satisfying\n\n$$\nS_n + a_n = \\frac{n-1}{n(n+1)}, \\quad n = 1, 2, \\dots\n$$\n\nThen $a_n = \\underline{\\hspace{2cm}}$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let us analyze the recurrence:\n\n$$\nS_n + a_n = \\frac{n-1}{n(n+1)}\n$$\n\nSince $S_n = a_1 + a_2 + \\dots + a_n$, we have $S_{n+1} = S_n + a_{n+1}$.\n\nSo,\n$$\nS_{n+1} + a_{n+1} = \\frac{n}{(n+1)(n+2)}\n$$\nSubtract the $n$th equation from the $(n+1)$th:\n$$\n(S_{n+1} + a_{n+1}) - (S_n + a_n) = \\frac{n}{(n+1)(n+2)} - \\frac{n-1}{n(n+1)}\n$$\nBut $S_{n+1} - S_n = a_{n+1}$, so:\n$$\na_{n+1} + a_{n+1} - a_n = \\frac{n}{(n+1)(n+2)} - \\frac{n-1}{n(n+1)}\n$$\n$$\n2a_{n+1} - a_n = \\frac{n}{(n+1)(n+2)} - \\frac{n-1}{n(n+1)}\n$$\nLet us solve for $a_n$ recursively. Define $b_n = a_n + \\frac{1}{n(n+1)}$.\n\nFrom the recurrence, we get:\n$$\na_{n+1} + \\frac{1}{(n+1)(n+2)} = \\frac{1}{2} \\left( a_n + \\frac{1}{n(n+1)} \\right)\n$$\nSo $b_{n+1} = \\frac{1}{2} b_n$, and $b_1 = a_1 + \\frac{1}{2}$.\n\nFrom $S_1 + a_1 = 2a_1 = 0$, we get $a_1 = 0$, so $b_1 = \\frac{1}{2}$.\n\nThus $b_n = \\frac{1}{2^n}$.\n\nTherefore,\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": 13092,
    "subject": "Mathematics (Olympiad)",
    "question": "In the convex quadrilateral $ABCD$, the angles at $A$ and $C$ are equal, and the bisector of $B$ passes through the midpoint of side $CD$. Given that $CD = 3AD$, find the ratio $AB/BC$.\n\n![](images/3._NATIONAL_XXX_OMA_2013_p5_data_1531a14667.png)",
    "options": [],
    "answer": "See solution",
    "solution": "Let $M$ be the midpoint of $CD$. Since $BM$ is the bisector of $B$, the reflection $E$ of $C$ in $BM$ lies on the ray $BA$. Because $\\angle MEB = \\angle MCB$ by reflection, and $\\angle MECB = \\angle DAB$ by hypothesis, we have $\\angle MEB = \\angle DAB$. Hence $ME \\parallel DA$; in particular, $E$ is on the side $AB$. Furthermore, $ME = MC$ by reflection, and $MC = MD$, so $MC = MD = ME$. Therefore, triangle $CDE$ is right with $\\angle CED = 90^\\circ$. Then $DE \\perp CE$ and since $MB \\perp CE$, we obtain $BM \\parallel ED$. Triangles $BEM$ and $EAD$ have parallel sides, hence they are similar in ratio $EM = CM = CD/AD = 3/2$. Then $BE = \\frac{3}{2} AE$ and $AB = \\frac{5}{2} AE$. Hence $AB/BC = AE/BE = 5/3$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 13093,
    "subject": "Mathematics (Olympiad)",
    "question": "Two disks are placed inside a square. What is the maximal proportion of the square that can be covered by the disks, if they are not permitted to overlap? Is it possible to cover more if overlap is allowed?",
    "options": [],
    "answer": "See solution",
    "solution": "Suppose the square has side length $1$ and center $C$. Denote the radii of the two disks by $x$ and $y$, and the distance between their centers by $d$. The centers of the circles are then restricted within two squares centered at $C$ of sides $1 - 2x$ and $1 - 2y$, respectively. The farthest they can be from each other is the length of the diagonal of a square of side length $1 - x - y$. Since their minimal distance is $x + y$, we get the inequality\n\n$$\nx + y \\leq d \\leq \\sqrt{2}(1 - x - y),\n$$\nwhich leads to\n$$\n0 \\leq x + y \\leq \\frac{\\sqrt{2}}{1 + \\sqrt{2}} = 2 - \\sqrt{2}.\n$$\nAt the same time, $0 \\leq x, y \\leq \\frac{1}{2}$. Interpreting these constraints geometrically, the area covered by the disks is maximized by $x = \\frac{1}{2}$, $y = \\frac{3}{2} - \\sqrt{2}$ (or vice versa):\n\n$$\n\\pi(x^2 + y^2) \\leq \\pi\\left(\\left(\\frac{1}{2}\\right)^2 + \\left(\\frac{3}{2} - \\sqrt{2}\\right)^2\\right) = \\pi\\left(\\frac{9}{2} - 3\\sqrt{2}\\right).\n$$\nThis maximum is attained when the larger circle is inscribed in the square, and the smaller one fits in one of the four small gaps left over.\n\nTo cover a larger portion of the square, let the larger circle remain where it is, but enlarge the smaller circle slightly, so that it still touches the square on two sides, until its center lies on the larger circle. Its radius will then be $\\frac{1}{2} - \\frac{1}{4}\\sqrt{2}$. More than half of its interior lies outside the larger circle (because this is convex), so we have now covered, in addition to the larger circle, a portion of the square that has area at least\n\n$$\n\\frac{\\pi}{2} \\left( \\frac{1}{2} - \\frac{1}{4} \\sqrt{2} \\right)^2 = \\pi \\left( \\frac{\\sqrt{2}}{4} - \\frac{1}{4} \\right)^2 > \\pi \\left( \\frac{3}{2} - \\sqrt{2} \\right)^2,\n$$\nwhich is the area that was covered before by the smaller circle. We conclude that it is possible.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 13094,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $a_1, a_2, \\dots, a_{21}$ be a permutation of $1, 2, \\dots, 21$, satisfying\n$$\n|a_{20} - a_{21}| \\geq |a_{19} - a_{21}| \\geq |a_{18} - a_{21}| \\geq \\dots \\geq |a_1 - a_{21}|.\n$$\n\nThe number of such permutations is \\_\\_\\_\\_\\_.",
    "options": [],
    "answer": "See solution",
    "solution": "For a given $k \\in \\{1, 2, \\dots, 21\\}$, consider the number of permutations $N_k$ satisfying the condition with $a_{21} = k$.\n\nWhen $k \\in \\{1, 2, \\dots, 11\\}$, for $i = 1, 2, \\dots, k-1$, $a_{2i-1}, a_{2i}$ are permutations of $k-i, k+i$ (if $k=1$, there is no such $i$), and $a_j = j+1$ for $2k-1 \\leq j \\leq 20$ (if $k=11$, there is no such $j$). Therefore, $N_k = 2^{k-1}$.\n\nSimilarly, when $k \\in \\{12, 13, \\dots, 21\\}$, $N_k = 2^{21-k}$.\n\nTherefore, the total number of such permutations is\n$$\n\\begin{aligned}\n\\sum_{k=1}^{21} N_k &= \\sum_{k=1}^{11} 2^{k-1} + \\sum_{k=12}^{21} 2^{21-k} \\\\\n&= (2^{11}-1) + (2^{10}-1) \\\\\n&= 3070.\n\\end{aligned}\n$$",
    "topic": "Discrete Mathematics",
    "subtopic": "Combinatorics"
  },
  {
    "id": 13095,
    "subject": "Mathematics (Olympiad)",
    "question": "For all sequences with $n$ zeros and $n$ ones, what is the sum of all continuous runs?\n\nFind a formula for the total number of runs (maximal consecutive blocks of identical digits) over all such sequences.",
    "options": [],
    "answer": "See solution",
    "solution": "We show that for all sequences with $n$ zeros and $n$ ones, the sum of all continuous runs is given by $(n+1)\\binom{2n}{n}$.\n\nFirst, count the number of changes (pairs of consecutive digits that are $01$ or $10$). There are $2n-1$ positions where changes may occur. For each position, there are $2$ ways to choose the pair, and $\\binom{2n-2}{n-1}$ ways to arrange the remaining $n-1$ zeros and $n-1$ ones. Thus, there are\n\n$$\n2(2n-1)\\binom{2n-2}{n-1}\n$$\n\nchanges in total.\n\nFor each sequence, the number of runs is $1$ more than the number of changes. Since there are $\\binom{2n}{n}$ sequences in total, the total number of runs is\n\n$$\n\\begin{aligned}\n&2(2n-1)\\binom{2n-2}{n-1} + \\binom{2n}{n} \\\\\n&= 2n\\binom{2n-1}{n} + \\binom{2n}{n} \\\\\n&= 2n\\binom{2n-1}{n-1} + \\binom{2n}{n} \\\\\n&= n\\binom{2n}{n} + \\binom{2n}{n} \\\\\n&= (n+1)\\binom{2n}{n},\n\\end{aligned}\n$$\n\nusing the fact that $k\\binom{m}{k} = m\\binom{m-1}{k-1}$.\n\nFor $n=2019$, the answer is $2020\\binom{4038}{2019}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 13096,
    "subject": "Mathematics (Olympiad)",
    "question": "Given 7 distinct positive integers, prove that there is an infinite arithmetic progression of positive integers $a, a+d, a+2d, \\dots$, with $a \\le d$, that contains exactly 3 or 4 of the 7 given integers.",
    "options": [],
    "answer": "See solution",
    "solution": "Let the numbers be $X_1, X_2, \\dots, X_7$ in ascending order. Let $[a, d]$ denote the arithmetic progression (AP) with initial term $a$, common difference $d$, and $a \\le d$.\n\nWe first show that there is an AP $[a, d]$ that contains $X_i$, i.e., $X_i = a + k_i d$ for $i = 1, \\dots, 5$. For example, we can take $a = d = 1$. Choose such an AP with maximal $d$.\n\nNot all $k_i$ can be even, or else we can use the AP $[a, 2d]$, contradicting the choice of $d$. Also, not all $k_i$ can be odd, or else we can use $[a+d, 2d]$. We have two cases:\n\n1. At least 3 of $k_1, \\dots, k_5$ are odd: use $[a+d, 2d]$.\n2. At least 3 of $k_1, \\dots, k_5$ are even: use $[a, 2d]$.\n\nIn either case, we have an AP $[a', d']$ that contains exactly 3 or 4 of $X_1, \\dots, X_5$. Note that $[a', d']$ is a 'sub-AP' of $[a, d]$.\n\nNext, we show that there is an AP that contains exactly 3 or 4 of $X_1, \\dots, X_6$. If $[a', d']$ contains exactly 3 of $X_1, \\dots, X_5$, or does not contain $X_6$, then it contains exactly 3 or 4 of $X_1, \\dots, X_6$. If $[a', d']$ contains exactly 5 of $X_1, \\dots, X_6$, then we can apply the above procedure to find a 'sub-AP' $[a'', d'']$ that contains exactly 3 or 4 of $X_1, \\dots, X_6$.\n\nThe same procedure can be used again to find a 'sub-AP' of $[a'', d'']$ that contains exactly 3 or 4 of $X_1, \\dots, X_7$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 13097,
    "subject": "Mathematics (Olympiad)",
    "question": "If $x$, $y$, $z$ are positive numbers with $x + y + z = 1$, show that:\n\n$$\n\\begin{align*}\n\\text{a)}\\quad 1 - \\frac{x^2 - yz}{x^2 + x} &= \\frac{(1 - y)(1 - z)}{x^2 + x}; \\\\\n\\text{b)}\\quad \\frac{x^2 - yz}{x^2 + x} + \\frac{y^2 - zx}{y^2 + y} + \\frac{z^2 - xy}{z^2 + z} &\\le 0.\n\\end{align*}\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "*Solution.*\n\na) $1 - \\frac{x^2 - yz}{x^2 + x} = \\frac{x + yz}{x^2 + x} = \\frac{1 - y - z + yz}{x^2 + x} = \\frac{(1 - y)(1 - z)}{x^2 + x}$.\n\nb) Using a), the inequality is rewritten:\n\n$$\n\\frac{(1 - y)(1 - z)}{x(x + 1)} + \\frac{(1 - z)(1 - x)}{y(y + 1)} + \\frac{(1 - x)(1 - y)}{z(z + 1)} \\geq 3,\n$$\n\nthat is,\n\n$$\n\\frac{(x + z)(x + y)}{x[(x + z) + (x + y)]} + \\frac{(y + z)(y + x)}{y[(y + z) + (y + x)]} + \\frac{(z + y)(z + x)}{z[(z + y) + (z + x)]} \\geq 3.\n$$\n\nApplying the inequality between the arithmetic mean and the harmonic mean, we deduce:\n\n$$\n\\frac{(x + z)(x + y)}{x[(x + z) + (x + y)]} + \\frac{(y + z)(y + x)}{y[(y + z) + (y + x)]} + \\frac{(z + y)(z + x)}{z[(z + y) + (z + x)]} \\geq \\frac{9}{\\frac{x}{x + y} + \\frac{x}{x + z} + \\frac{y}{y + z} + \\frac{y}{y + x} + \\frac{z}{z + y} + \\frac{z}{z + x}} = \\frac{9}{1 + 1 + 1} = 3.\n$$\n\n*Alternative solution (b):* Using a), the inequality is rewritten:\n\n$$\n(1 - x)(1 - y)(1 - z) \\left( \\frac{1}{x - x^3} + \\frac{1}{y - y^3} + \\frac{1}{z - z^3} \\right) \\geq 3.\n$$\n\nApplying the inequality between the arithmetic mean and the harmonic mean, we have:\n\n$$\n\\begin{align*}\n\\frac{1}{x - x^3} + \\frac{1}{y - y^3} + \\frac{1}{z - z^3} &\\geq \\frac{9}{(x + y + z) - (x^3 + y^3 + z^3)} \\\\\n&= \\frac{9}{1 - (x^3 + y^3 + z^3)} \\\\\n&= \\frac{9}{(x + y + z)^3 - (x^3 + y^3 + z^3)} \\\\\n&= \\frac{9}{3(x + y)(y + z)(z + x)} = \\frac{3}{(1 - x)(1 - y)(1 - z)}\n\\end{align*}\n$$\n\nand the inequality is demonstrated.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 13098,
    "subject": "Mathematics (Olympiad)",
    "question": "A shape consisting of 1000 small squares is made by continuing the arrangement shown.\n\n![](images/SAMF_ANNUAL_REPORT_2015_BOOK_PROOF_p50_data_09bfe2583d.png)\n\nIf each small square has a side length of 1 cm, find the perimeter of the whole shape in cm.",
    "options": [],
    "answer": "See solution",
    "solution": "Each row, apart from the bottom row and top row, contributes 4 cm to the perimeter. The top and bottom rows each contribute 8 cm. Since there are $1000 \\div 5 = 200$ rows, we have:\n\n$$\n\\text{Perimeter} = (200 - 2) \\times 4 + 2 \\times 8 = 198 \\times 4 + 16 = 792 + 16 = 808\n$$\n\nSo, the perimeter is $808$ cm.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 13099,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $a_1 < a_2 < \\dots < a_{2017}$ and $b_1 < b_2 < \\dots < b_{2017}$ be positive integers such that\n\n$$\n(2^{a_1} + 1)(2^{a_2} + 1)\n\\dots(2^{a_{2017}} + 1) = (2^{b_1} + 1)(2^{b_2} + 1)\n\\dots(2^{b_{2017}} + 1).\n$$\n\nProve that $a_i = b_i$ for $i = 1, 2, \\dots, 2017$.",
    "options": [],
    "answer": "See solution",
    "solution": "Suppose that there is some $i \\in \\{1, 2, \\dots, 2017\\}$ for which $a_i \\neq b_i$. Then, if we cancel out equal factors on both sides of the equation, we obtain an equation of the form\n\n$$\n(2^{A_1} + 1)(2^{A_2} + 1)\n\\dots(2^{A_n} + 1) = (2^{B_1} + 1)(2^{B_2} + 1)\n\\dots(2^{B_n} + 1),\n$$\n\nwhere we may assume that $A_1 < A_2 < \\dots < A_n$, $B_1 < B_2 < \\dots < B_n$ and $A_1 < B_1$, without loss of generality.\n\nExpanding both sides yields an equation of the form\n\n$$\n1 + 2^{A_1} + [\\text{higher powers of 2}] = 1 + 2^{B_1} + [\\text{higher powers of 2}],\n$$\n\nfrom which we obtain\n\n$$\n2^{A_1} + [\\text{higher powers of 2}] = 2^{B_1} + [\\text{higher powers of 2}].\n$$\n\nHowever, note that $2^{B_1}$ divides the right-hand side but not the left-hand side, which yields a contradiction. It follows that $a_i = b_i$ for $i = 1, 2, \\dots, 2017$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 13100,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all pairs of real polynomials $p(x)$ and $q(x)$ such that\n\n$$\np(x^2) = p(x)q(1-x) + p(1-x)q(x)\n$$\nfor all real $x$.",
    "options": [],
    "answer": "See solution",
    "solution": "Replace $x$ by $1-x$ in\n\n$$\np(x^2) = p(x)q(1-x) + p(1-x)q(x), \\quad (1)\n$$\n\nthen we have $p((x-1)^2) = p(1-x)q(x) + p(x)q(1-x) = p(x^2)$. It follows that the polynomial $p(x^2)$ is periodic, i.e., $p(x)$ is a constant polynomial, $p(x) = a$ for some $a \\in \\mathbb{R}$.\n\nIf $a=0$, then (1) is valid for any arbitrary polynomial $q(x)$.\n\nIf $a \\neq 0$, then (1) becomes $q(1-x) + q(x) = 1$ for all $x$. Replacing $x$ by $x+0.5$ we can rewrite the last equality as\n\n$$\nq(0.5 - x) - 0.5 + q(0.5 + x) - 0.5 = 0. \\quad (2)\n$$\n\nLet $h(x) = q(0.5 + x) - 0.5$. Then (2) shows that $h(-x) = -h(x)$, so we may write $h(x) = x r(x^2)$ for some polynomial $r(x)$.\n\nHence $q(0.5 + x) = x r(x^2) + 0.5$. Replace $x$ by $x - 0.5$ here, we obtain $q(x) = (x - 0.5) r((x - 0.5)^2) + 0.5$, which obviously satisfies the condition.",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 13101,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $ABC$ be a triangle with $AC > AB$ and circumcenter $O$. The tangents to the circumcircle at $A$ and $B$ intersect at $T$. The perpendicular bisector of the side $BC$ intersects side $AC$ at $S$.\n\n(a) Prove that the points $A$, $B$, $O$, $S$, and $T$ lie on a common circle.\n\n(b) Prove that the line $ST$ is parallel to the side $BC$.",
    "options": [],
    "answer": "See solution",
    "solution": "Since $AT$ and $BT$ are perpendicular to $AO$ and $BO$, the points $A$, $B$, $T$, and $O$ lie on a circle $k_1$ by Thales' theorem. By the central angle theorem, we have $\\angle AOB = 2\\gamma$. Since $\\triangle BCS$ is isosceles, we find $\\angle BCS = \\angle CBS = \\gamma$. Now $\\angle ASB = 2\\gamma$, because an exterior angle of a triangle equals the sum of the other two interior angles. Thus\n\n$$\n\\angle ASB = 2\\gamma = \\angle AOB,\n$$\n\nand by the inscribed angle theorem, we find that the points $A$, $B$, $S$, and $O$ lie on a circle $k_2$. Since the circles $k_1$ and $k_2$ have the three points $A$, $B$, and $O$ in common, we have $k_1 = k_2$ and the points $A$, $B$, $O$, $S$, and $T$ lie on a circle.\n\nFinally, we have $\\angle TSB = \\angle TOB = \\gamma = \\angle SBC$, from which it follows that $ST$ is parallel to $BC$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 13102,
    "subject": "Mathematics (Olympiad)",
    "question": "In triangle $ABC$, let $N$ be the midpoint of the median $CM$. The point $X$ satisfies the following conditions: $\\angle XMC = \\angle MBC$, $\\angle XCM = \\angle MCB$, and points $X$ and $B$ lie on different sides with respect to the line $CM$. Let $\\omega$ be the circumscribed circle of $\\triangle AMX$. Prove that:\n\na) $CM$ is tangent to $\\omega$;\n\nb) Lines $NX$ and $AC$ intersect on the circle $\\omega$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let's consider the configuration from the figure below; for other configurations, the proof is the same.\n\na) Since $\\triangle XMC \\sim \\triangle MBC$, we have\n\n$$\n\\begin{aligned}\n\\angle AMX &= 180^\\circ - \\angle XMC - \\angle BMC \\\\\n&= 180^\\circ - \\angle XMC - \\angle MXC \\\\\n&= \\angle MCX\n\\end{aligned}\n$$\n\nand $\\frac{AM}{MX} = \\frac{BM}{MX} = \\frac{MC}{CX} \\implies \\triangle AXM \\sim \\triangle MXC \\implies \\angle XAM = \\angle XMC$, so the claim follows from the equality of the inscribed angles.\n\n![](images/Ukraine_2021-2022_p32_data_ee253eea20.png)\n\nb) Let $S$ be the second point of intersection of $AC$ and $\\omega$. We need to show that $S$ lies on $NX$. From the properties of inscribed angles, $\\angle SMC = \\angle SAM = \\angle CAM$. It follows that $\\triangle CSM \\sim \\triangle CMA \\implies \\frac{CM}{CA} = \\frac{SM}{MA}$. Since $CM = 2MN$ and $MA = \\frac{1}{2}AB$, we have $\\frac{MN}{CA} = \\frac{SM}{AB}$. Together with the condition $\\angle SMN = \\angle SMC = \\angle CAM = \\angle CAB$, we get $\\triangle SNM \\sim \\triangle BCA$. Then $\\angle MSN = \\angle ABC$. Furthermore, $\\angle XSM = 180^\\circ - \\angle XAM$. From part (a), $\\angle XAM = \\angle XMC = \\angle MBC = \\angle ABC$, so $\\angle XSM = 180^\\circ - \\angle ABC = 180^\\circ - \\angle MSN$, and thus $S \\in MN$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 13103,
    "subject": "Mathematics (Olympiad)",
    "question": "Prove that a convex pentagon with integer side lengths and an odd perimeter can have two right angles, but cannot have more than two right angles.\n\n(A polygon is _convex_ if all of its interior angles are less than $180^\\circ$.)",
    "options": [],
    "answer": "See solution",
    "solution": "First, we establish that such a pentagon with two right angles is possible. A unit equilateral triangle placed on top of a unit square forms a pentagon with perimeter $5$.\n\nFrom the angle sum formula, we see that if a pentagon has four right angles, the fifth angle would be $180^\\circ$, which means that the pentagon actually degenerates into a rectangle (and would have an even perimeter anyway).\n\nSo it remains to show that a pentagon with integer side lengths and three right angles must have an even perimeter. If three out of five angles are right angles, then at least two of them must be adjacent. This gives two cases: all three right angles are adjacent or only two are adjacent, as shown below:\n\n![](images/2022_Australian_Scene_p86_data_2bd9e67f1d.png)\n\n![](images/2022_Australian_Scene_p86_data_ad52c429eb.png)\n\nFirstly, consider Case 1:\n\n![](images/2022_Australian_Scene_p86_data_db4235be42.png)\n\nThe bounding rectangle has integer sides and hence its perimeter has even parity. Furthermore, the dotted section forms a right-angled triangle with integer sides. The perimeter of the pentagon is found by replacing the two dotted sides with the hypotenuse. But since\n\n$$a^2 + b^2 = c^2 \\text{ implies that } a + b \\text{ has the same parity as } c,$$\n\nthe perimeter of the pentagon is also even.\n\nAlgebraically, using the labelling in the diagram, the perimeter of the bounding rectangle is $2(x + y)$ and the perimeter of the pentagon is $2(x + y) + (c - a - b)$.\n\nNow consider Case 2:\n\n![](images/2022_Australian_Scene_p87_data_5efa5bc5c3.png)\n\nDraw a line parallel to the base as in the diagram to form two right-angled triangles, both with hypotenuse $e$. We see that $a^2 + b^2 = c^2 + d^2$. This means that $a + b$ has the same parity as $c + d$ and hence $a + b + c + d$ is even. As the perimeter is given by $a + b + c + d + 2y$, it follows that the pentagon has even perimeter.\n\nHence, a pentagon with integer sides and an odd perimeter can have at most two right angles.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 13104,
    "subject": "Mathematics (Olympiad)",
    "question": "Determine all triples $(a, b, c)$ of positive integers satisfying the conditions\n\n$$\n\\gcd(a, 20) = b\n$$\n\n$$\n\\gcd(b, 15) = c\n$$\n\n$$\n\\gcd(a, c) = 5\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "We use equations (I) and (II) to eliminate $b$ and $c$ as follows:\n\n$$\n\\gcd(a, \\gcd(\\gcd(a, 20), 15)) = 5 \\iff \\gcd(a, 5) = 5 \\iff 5 \\mid a.\n$$\n\nFurthermore, we determine $b$ and $c$ from (I) and (II): (I) yields $b \\in \\{5, 10, 20\\}$. More specifically, we have $b = 5$ for $a$ being odd, $b = 10$ for $a \\equiv 2 \\pmod{4}$, and $b = 20$ for $a \\equiv 0 \\pmod{4}$. In all three cases, $c = 5$ follows from (II).\n\nIn total, the solutions form the set\n$$\n\\{(20t, 20, 5),\\ (20t-10, 10, 5),\\ (10t-5, 5, 5)\\mid t \\text{ is a positive integer}\\}.\n$$",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 13105,
    "subject": "Mathematics (Olympiad)",
    "question": "設 $\\triangle ABC$ 為銳角三角形，$H$ 為其垂心。取點 $D$ 使得四邊形 $HABD$ 為平行四邊形（其中 $AB \\parallel HD$ 及 $AH \\parallel BD$）。取 $E$ 為直線 $DH$ 上一點，使得直線 $AC$ 通過線段 $HE$ 的中點。令 $F$ 為直線 $AC$ 與三角形 $DCE$ 的外接圓的另一個交點。\n\n$$\n\\text{證明：} EF = AH.\n$$\n\nLet $ABC$ be an acute triangle with orthocentre $H$. Let $D$ be the point such that the quadrilateral $HABD$ is a parallelogram (with $AB \\parallel HD$ and $AH \\parallel BD$). Let $E$ be the point on the line $DH$ such that the line $AC$ passes through the midpoint of the segment $HE$. Let $F$ be the second point of intersection of the line $AC$ and the circumcircle of triangle $DCE$.\n\n$$\n\\text{Prove that } EF = AH.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "由於 $HD \\parallel AB$ 及 $BD \\parallel AH$，有 $BD \\perp BC$ 以及 $CH \\perp DH$，因此 $BDCH$ 四點共圓。而由於 $H$ 是三角形 $ABC$ 的垂心，所以 $\\angle HAC = 90^\\circ - \\angle ACB = \\angle CBH$。再利用 $CDFE$ 四點共圓，可得\n\n$$\n\\angle EFC = \\angle HDC = \\angle HBC = \\angle CAH.\n$$\n\n![](images/16-2J_p3_data_9d673d442b.png)\n\n令點 $M$ 為 $AC$ 與 $DH$ 的交點，並取點 $G$ 在 $AC$ 上使得 $AH = AG$ ($G \\neq A$)。於是有 $\\angle EFM = \\angle MAH = \\angle HGM$。再加上 $\\angle FME = \\angle HMG$、$EM = MH$，得 $\\triangle EFM \\approx \\triangle HGM$。因此 $EF = HG = AH$，證明完畢。",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 13106,
    "subject": "Mathematics (Olympiad)",
    "question": "令 $n \\ge 5$ 為一與 $6$ 互質的正整數。我們將一個正 $n$ 邊形的 $n$ 個頂點，各塗上三種顏色中的一種，使得被塗上每種顏色的頂點數都是奇數。試證：我們必可從這 $n$ 個頂點中找出三個點，它們的顏色都不相同，且此三點的連線構成等腰三角形。",
    "options": [],
    "answer": "See solution",
    "solution": "令 $a_k$ 為所有等腰三角形中，三頂點恰包含 $k$ 種顏色的三角形個數，則題目等價於證明 $a_3 \\ge 1$。\n\n我們採取歸謬證法。假設 $a_3 = 0$。考慮集合\n\n$$X = \\{(\\Delta, E) : \\Delta \\text{是等腰三角形},\\ E \\text{是 } \\Delta \\text{的一邊},\\ E \\text{的兩端點不同色}\\}.$$ \n\n讓我們用兩種不同方式計算 $X$ 中的元素個數：\n\n- 首先，對於每個三角形：\n  - 只有一色的三角形必沒有兩端點異色的邊。\n  - 只有兩色的三角形則恰有 $2$ 條邊的兩端點異色。\n  - 由假設，不存在三色的三角形。\n\n綜合上述，$|X| = 2a_2$。\n\n- 另一方面，任選兩個頂點 $A, B$，由於 $(n, 3) = 1$，我們知 $AB$ 恰為 $3$ 個等腰三角形的邊。若我們令 $b, c, d$ 分別為三種顏色的頂點數，則兩端點異色的邊共有 $bc + bd + cd$ 條，故 $|X| = 3(bc + bd + cd)$。\n\n然而由題目假設，$b, c, d$ 皆為奇數，故 $3(bc + bd + cd)$ 為奇數，從而它不可能等於 $2a_2$，矛盾！故 $a_3 \\ge 1$。",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 13107,
    "subject": "Mathematics (Olympiad)",
    "question": "Squares from the same row, lying to the right, are also cut off (after such a cut we get a figure consisting of several columns, and for any two columns, the left-hand one is not shorter than the right-hand one).\n\nLet's write in every square of the obtained figure a number equal to the quantity of squares lying above in the same column and lying to the right in the same row (including the square itself). The figure to the right gives an example of such a figure.\n\nLet $n$ be the total number of squares in the figure. Prove that the number of squares with numbers divisible by $2008$ does not exceed $\\frac{n}{2008}$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let's show that the problem statement is true for any $q$, i.e., the total number of squares with numbers divisible by $q$ does not exceed $\\frac{n}{q}$.\n\nName by *hook* the set of squares lying above in the same column and lying to the right in the same row (including the square itself). Let the square containing the bottom corner of the hook be called the *base of the hook*. Let the total number of squares in the hook be called the *hook area*. You can see an example of the hook in the picture; the base of the hook is highlighted by darker grey.\n\n% IMAGE: ![](images/Ukrajina_2011_p44_data_e815c69210.png)\n\n% IMAGE: ![](images/Ukrajina_2011_p44_data_62a57f3cdd.png)\n\n**Lemma.** If the hook area is equal to $kq$ ($k, q \\in \\mathbb{N}$), then the hook contains at most $k$ squares with numbers divisible by $q$.\n\n**Proof of the lemma.** Let's call the *upper bar* the set of squares lying above the base of the hook in the same column, and the *right bar* the set of squares lying to the right from the base of the hook in the same row (not including the base of the hook).\n\n% IMAGE: ![](images/Ukrajina_2011_p44_data_7ef39d38b4.png)\n\nWe show that the sum of the numbers in any two squares, one of which lies in the upper bar and another in the right bar, does not equal $kq$. Two cases are possible: the hooks with the bases in these squares either have a common square or not.\n\nIf the hooks have a common square, it is easy to show that the number of squares in their union is not less than $kq$ (for example, the number of the shaded squares in the picture is equal to $kq$, but it is less than the number of squares in the union of the hooks). Then the total number of squares in the hooks is not less than $kq+1$, because the common square should be counted twice.\n\nIf the hooks have no common squares, it is easy to see that the total number of squares (equal in this case to the number of squares in the union of the hooks) is strictly less than $kq$.\n\nConsider the pairs $((k-1)q, q)$, $((k-2)q, 2q)$, ..., $(q, (k-1)q)$. We mark all numbers in the first place if they can be found in the upper bar of the hook. Similarly, we mark all numbers in the second place if they can be found in the right bar of the hook. Obviously, the numbers in either the upper bar or right bar are monotonically decreasing from the base and are strictly less than $kq$, thus only one square corresponds to each marked number. According to the above (the sum of the numbers in any two squares, one from the upper bar and another from the right bar, does not equal $kq$), in each pair at most one number can be marked. But all numbers divisible by $q$ are enumerated in pairs, and the total number of pairs is $k-1$.\n\nTherefore, in both upper and right bars there are at most $k-1$ numbers divisible by $q$. Along with $kq$ in the base of the hook, we get at most $k$ such numbers. The lemma is proved.\n\nLet's mark squares with numbers divisible by $q$ by the following procedure. Examine horizontals bottom-up. If some horizontal contains a square with a number divisible by $q$, and this square does not lie in some hook with already marked base, we choose the left-most of all such squares and mark it (the picture gives an example for the case $q=2$; dark-grey color indicates marked squares, light-grey color indicates hooks with marked bases).\n\nAll squares with numbers divisible by $q$ are contained in some hooks with marked bases. Let the sum of the areas of these hooks equal $s$. By the lemma, the total number of squares with numbers divisible by $q$ does not exceed $s/q$ (some squares can be counted twice).\n\nWe show that $s \\leq n$. Since no three hooks with marked base have a common square and any two such hooks have at most one common square (because of the marking procedure), the area of the figure covered by hooks equals $s-p$, where $p$ is the number of pairs of intersecting hooks. But for any such pair, one square not covered by any hook can be found—it is the square lying in the same row as the base of the lower hook and in the same column as the base of the upper hook. At the same time, all such squares are distinct for different pairs of hooks (due to the marking procedure). Thus, the area of the figure $n$ is not less than $(s-p)+p=s$. Therefore, the total number of squares with numbers divisible by $q$ does not exceed $s/q \\leq n/q$, as required.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 13108,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $a$ and $b$ be two odd natural numbers. Suppose that $a$ is a divisor of $b^2 + 2$ and $b$ is a divisor of $a^2 + 2$. Prove that $a$ and $b$ belong to the sequence $(v_n)$ defined by\n\n$$\nv_1 = v_2 = 1 \\quad \\text{and} \\quad v_n = 4v_{n-1} - v_{n-2} \\quad \\forall n \\ge 3.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "From the given assumptions, we have\n\n$$\n(a, b) = 1, \\text{ and}\n$$\n\n$$\na^2 + b^2 + 2 \\equiv 0 \\pmod{ab}.\n$$\n\nBy the above, there exists an integer $k$ such that $a^2 + b^2 + 2 = kab$. We will prove that $k = 4$.\n\nIf $a = b$, then $(a, b) = 1$ implies $a = b = 1$, so $k = 4$.\n\nConsider the case $a \\neq b$. Without loss of generality, suppose $a < b$. Consider the sequence defined by\n\n$$\na_1 = b, \\quad a_2 = a, \\quad a_{i+2} = k a_{i+1} - a_i, \\quad \\forall i \\ge 2.\n$$\n\nWe can prove the following properties of $(a_n)$:\n\n1. All $a_n$ are odd positive integers.\n2. $(a_n)$ is a decreasing sequence.\n3. For all $i \\ge 2$, $a_{i-1}$ and $a_{i+1}$ are roots of the equation (in $x$):\n\n$$\nx^2 - k a_i x + a_i^2 + 2 = 0.\n$$\n\nBy properties 1 and 2, there exists $N$ such that\n\n$$\na_1 > a_2 > \\dots > a_{N-2} > a_{N-1} = 1.\n$$\n\nBy property 3, $a_{N-2}$ is the positive root of\n\n$$\nx^2 - k a_{N-1} x + a_{N-1}^2 + 2 = 0 \\implies x^2 - kx + 3 = 0, \\text{ since } a_{N-1} = 1.\n$$\n\nClearly, $a_{N-2}$ divides $3$ and $a_{N-2} > 1$, so $a_{N-2} = 3$. Thus, $k = 4$.\n\nSince $a$ and $b$ satisfy $a^2 + b^2 + 2 = 4ab$, similarly, we can prove that they are terms of the sequence defined by\n\n$$\nv_1 = v_2 = 1 \\quad \\text{and} \\quad v_n = 4v_{n-1} - v_{n-2} \\quad \\forall n \\ge 3.\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 13109,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $A \\in \\mathcal{M}_5(\\mathbb{C})$ be a matrix with $\\operatorname{Tr}(A) = 0$ and such that $I_5 - A$ is invertible. Prove that $A^5 \\neq I_5$.",
    "options": [],
    "answer": "See solution",
    "solution": "Suppose, by way of contradiction, that $A^5 = I_5$. Let $\\lambda \\in \\mathbb{C}$ be an eigenvalue of $A$; then $\\lambda^5$ is an eigenvalue of $A^5$ and we obtain $\\lambda^5 = 1$.\n\nSince $I_5 - A$ is invertible, it follows that $\\det(I_5 - A) \\neq 0$. Consequently, the eigenvalues of $A$ belong to the set $\\{\\varepsilon, \\varepsilon^2, \\varepsilon^3, \\varepsilon^4\\}$, where $\\varepsilon = \\cos \\frac{2\\pi}{5} + i \\sin \\frac{2\\pi}{5}$.\n\nFrom $\\operatorname{Tr}(A) = 0$, we deduce the existence of non-negative integers $a, b, c, d$ such that\n\n$$\n\\begin{cases}\na + b + c + d = 5 \\\\\na\\varepsilon + b\\varepsilon^2 + c\\varepsilon^3 + d\\varepsilon^4 = 0.\n\\end{cases}\n$$\n\nThe latter gives $a + b\\varepsilon + c\\varepsilon^2 + d\\varepsilon^3 = 0$, hence $\\varepsilon$ is a root of the integer polynomial $dX^3 + cX^2 + bX + a$, whose degree is at most 3. If $d \\neq 0$, then the roots of this polynomial are $\\varepsilon, \\bar{\\varepsilon} \\in \\mathbb{C} \\setminus \\mathbb{R}$ and some $\\alpha \\in \\mathbb{R}$. But then, using Viète's relations, we obtain $\\alpha = -a/d \\in \\mathbb{Q}$ and $\\alpha = -(c/d + 2 \\cos(2\\pi/5)) \\in \\mathbb{R} \\setminus \\mathbb{Q}$, which is a contradiction. We conclude that $d = 0$.\n\nSimilarly, one can show that $c = b = a = 0$, and then $a + b + c + d = 0$, contradicting $a + b + c + d = 5$.\n\n_Observation._ Alternatively, for the final part of the proof, one can use that the minimal nonzero integer polynomial having the root $\\varepsilon$ is the cyclotomic polynomial $X^4 + X^3 + X^2 + X + 1$.",
    "topic": "Algebra",
    "subtopic": "Linear Algebra"
  },
  {
    "id": 13110,
    "subject": "Mathematics (Olympiad)",
    "question": "Од рамностран триаголник со страна 2014 е отсечен рамностран триаголник со страна 214, така што едно теме им се совпаѓа и две од страните од отсечениот триаголник лежат на две од страните на почетниот. Дали оваа фигура може да се покрие со фигури како подолу дадени на цртежот, без преклопување (дозволена е ротација), ако триаголниците во фигурите се рамнострани со страна 1? Образложи го одговорот!\n\n![](images/MACEDONIAN_MATHEMATICAL_SOCIETY_p22_data_62b358bccd.png)",
    "options": [],
    "answer": "See solution",
    "solution": "Најпрво ја разделуваме дадената фигура на рамнострани триаголници со страна 1. Ги обележуваме триаголничињата во дадената фигура со броевите од 1 до 6, како на сликата десно (во првиот ред последователно од 1 до 6, па броевите се повторуваат, во вториот почнуваме од 5, во третиот од 3, потоа од 1 и постапката се повторува). Лесно може да се забележи дека секоја од фигурите покрива по точно еден од броевите од 1 до 6. Според последното, за фигурата да може да се покрие со дадените фигури, треба секој од броевите да се јавува еднаков број пати. Ако споредиме колку пати се јавува бројот 1 со колку пати се јавува бројот 2, ќе забележиме дека во првиот, четвртиот и секој ред од облик $3k+1$ имаме една единица повеќе отколку двојки, а во останатите бројот на единици и двојки е еднаков. Според ова, следува дека бројот на единици и двојки не е еднаков, па не може секој од броевите да се јавува еднаков број пати. Следува дека фигурата не може да се покрие на бараниот начин.\n\n![](images/MACEDONIAN_MATHEMATICAL_SOCIETY_p22_data_7d030d61e3.png)",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 13111,
    "subject": "Mathematics (Olympiad)",
    "question": "For each positive integer $n$, show that the polynomial:\n\n$$\nP_n(x) = \\sum_{k=0}^{n} 2^k \\binom{2n}{2k} x^k (x-1)^{n-k}\n$$\n\nhas exactly $n$ distinct real roots.",
    "options": [],
    "answer": "See solution",
    "solution": "First, observe that if $P_n(x_0) = 0$, then $0 < x_0 < 1$.\n\n- If $x_0 > 1$, then $x_0 > 0$ and $x_0 - 1 > 0$, so $P_n(x_0) > 0$.\n- If $x_0 < 0$, let $y_0 = -x_0 > 0$. Then $P_n(x_0) = (-1)^n P_n(y_0)$, but $P_n(y_0) > 0$, so $P_n(x_0) \\neq 0$.\n\nNow, for $x_0 \\in (0, 1)$, set $2x_0 = a^2$ and $1 - x_0 = b^2$. Then:\n\n$$\n(x_0 - 1)^{n-k} = ((-1)(1 - x_0))^{n-k} = (ib)^{2n-2k},\n$$\n\nwhere $i^2 = -1$. Thus,\n\n$$\nT_n = \\sum_{k=0}^{n} \\binom{2n}{2k} a^{2k} (ib)^{2n-2k}.\n$$\n\nConsider the binomial expansions:\n\n$$\n(a + ib)^{2n} = \\sum_{k=0}^{2n} \\binom{2n}{k} a^k (ib)^{2n-k},\n$$\n$$\n(a - ib)^{2n} = \\sum_{k=0}^{2n} \\binom{2n}{k} (-1)^k a^k (ib)^{2n-k}.\n$$\n\nTherefore,\n\n$$\nT_n = \\frac{(a + ib)^{2n} + (a - ib)^{2n}}{2}.\n$$\n\nNote that $a^2 + b^2 = x_0 + 1$. Set $\\frac{a}{\\sqrt{x_0 + 1}} = \\cos \\varphi$ and $\\frac{b}{\\sqrt{x_0 + 1}} = \\sin \\varphi$ for some $\\varphi \\in (0, \\frac{\\pi}{2})$. Then:\n\n$$\nT_n = \\frac{(\\cos \\varphi + i \\sin \\varphi)^{2n} + (\\cos \\varphi - i \\sin \\varphi)^{2n}}{2}.\n$$\n\nBy de Moivre's formula:\n\n$$\nT_n = \\cos(2n\\varphi).\n$$\n\nThus, $P_n(x_0) = 0$ if and only if $\\cos(2n\\varphi) = 0$, i.e., $2n\\varphi = \\frac{\\pi}{2} + k\\pi$ for $k = 0, 1, \\ldots, n-1$. So $\\varphi = \\frac{\\pi}{4n} + k\\frac{\\pi}{2n}$ for $k = 0, 1, \\ldots, n-1$.\n\nEach $\\varphi$ in $(0, \\frac{\\pi}{2})$ corresponds to a unique $x_0 \\in (0, 1)$, so there are exactly $n$ distinct real roots of $P_n(x) = 0$.",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 13112,
    "subject": "Mathematics (Olympiad)",
    "question": "![](images/USA_IMO_2006-2007_p17_data_7846730b5e.png)\n\nThe numbers $m_1, m_2, n_1, n_2, n_3, n_4$ in the figure denote the number of sides of $P$ determining the regions $N_1, N_2, N_3, N_4$ and $M$ that consist of exterior triangles (triangles that are not interior). The two interior triangles are\n\n$$\nA_0 A_{n_1} A_{n_1+n_2} \\text{ and } A_{n_1+n_2+m_1} A_{n_1+n_2+m_1+n_3} A_{n_1+n_2+m_1+n_3+n_4},\n$$\n\nrespectively.\n\nShow that triangulations starting at $A_0$ are in bijective correspondence to 7-tuples\n\n$$\n(m, n_1, n_2, n_3, n_4, w_M, w_N),\n$$\n\nwhere $m \\ge 0$, $n_1, n_2, n_3, n_4 \\ge 2$ are integers,\n\n$$\nm + n_1 + n_2 + n_3 + n_4 = n, \\qquad (\\dagger)\n$$\n\n$w_M$ is a binary sequence of length $m$ and $w_N$ is a binary sequence of length $n - m - 8$.\n\nCompute the total number of such triangulations of $P$ (for $n \\ge 6$), and as a quick exercise, compute the number of triangulations of $P$ with exactly one interior region.",
    "options": [],
    "answer": "See solution",
    "solution": "Given the bijection, the number of solutions to $(\\dagger)$ with $m \\ge 0$, $n_1, n_2, n_3, n_4 \\ge 2$ is the number of positive integer solutions to\n\n$$\nx_1 + x_2 + x_3 + x_4 + x_5 = n - 3,\n$$\n\nwhich is $\\binom{n-4}{4}$.\n\nEach 7-tuple $(m, n_1, n_2, n_3, n_4, w_M, w_N)$ corresponds to $2^m \\cdot 2^{n-m-8} = 2^{n-8}$ binary sequences, so the total number is\n\n$$\n2^{n-8} \\binom{n-4}{4}.\n$$\n\nTo count all triangulations, multiply by $n$ (starting at any vertex) and divide by $2$ (each counted twice):\n\n$$\n\\frac{n}{2} 2^{n-8} \\binom{n-4}{4}.\n$$\n\nFor triangulations with exactly one interior region, the number is\n\n$$\n\\frac{n}{3} 2^{n-6} \\binom{n-4}{2}.\n$$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 13113,
    "subject": "Mathematics (Olympiad)",
    "question": "Solve the equation:\n\n$$\n\\sqrt{4 - 3\\sqrt{10 - 3x}} = x - 2.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "The domain of the equation is $X = \\left[\\frac{74}{27}, \\frac{10}{3}\\right]$.\n\nOn $X$, we have:\n\n$$\n\\begin{aligned}\n&\\text{Given equation} \\iff 4 - 3\\sqrt{10 - 3x} = x^2 - 4x + 4 \\\\\n&\\iff 9(10 - 3x) = (x^2 - 4x + 4)^2 \\\\\n&\\iff x^4 - 8x^3 + 16x^2 + 27x - 90 = 0 \\\\\n&\\iff (x - 3)(x + 2)(x^2 - 7x + 15) = 0 \\\\\n&\\iff x = 3.\n\\end{aligned}\n$$",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 13114,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $f: \\mathbb{R} \\to \\mathbb{R}$ be a function satisfying\n$$\n|f(x+y) - f(x) - f(y)| < 1 \\text{ for all } x, y \\in \\mathbb{R}.\n$$\n\nProve that\n$$\n\\left| f\\left(\\frac{x}{2008}\\right) - \\frac{f(x)}{2008} \\right| < 1\n$$\nfor all $x \\in \\mathbb{R}$.",
    "options": [],
    "answer": "See solution",
    "solution": "$$\n\\begin{aligned}\n\\left| f(2008x) - 2008f(x) \\right| &= \\left| \\sum_{k=1}^{2007} \\left( f((k+1)x) - f(x) - f(kx) \\right) \\right| \\\\\n&\\leq \\sum_{k=1}^{2007} \\left| f((k+1)x) - f(x) - f(kx) \\right| < 2007.\n\\end{aligned}\n$$\n\nReplacing $x$ with $\\frac{x}{2008}$ and simplifying, one gets\n\n$$\n\\left| \\frac{f(x)}{2008} - f\\left(\\frac{x}{2008}\\right) \\right| < \\frac{2007}{2008} < 1.\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 13115,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $t$ be a nonzero real number and $n$ a positive integer. Let $a_1, a_2, \\dots, a_{2n-1}$ be real numbers (not necessarily distinct). Prove that there exist indices $i_1 < i_2 < \\dots < i_n$ such that $a_{i_k} - a_{i_l} \\neq t$ for any $1 \\leq k, l \\leq n$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $G = (V, E)$ be a graph with vertex set $V = \\{1, 2, \\dots, 2n-1\\}$ and edge set $E = \\{\\{i, j\\} : |a_i - a_j| = t\\}$. Note that $G$ has no odd cycles. Indeed, if $j_1, \\dots, j_{2k+1}$ is a cycle, then for all $l = 1, 3, 5, \\dots, 2k-1$ the number $a_{j_l}$ differs from $a_{j_{l+1}}$ by $t$ or $-t$, so $a_{j_1}$ differs from $a_{j_{2k+1}}$ by an even multiple of $t$. Therefore, there is no edge between $j_1$ and $j_{2k+1}$, contradicting the assumption that $j_1, \\dots, j_{2k+1}$ is a cycle.\n\nSince $G$ has no odd cycles, it is bipartite. Therefore, $V$ can be split into two disjoint sets $V_1, V_2$ such that there is no edge between any two vertices of $V_1$ and no edge between any two vertices of $V_2$. Since $V$ has $2n-1$ elements, one of the sets $V_1, V_2$ has at least $n$ elements. Without loss of generality, assume $V_1$ has at least $n$ elements. Then for $k = 1, 2, \\dots, n$, define $i_k$ to be the $k$-th least element of $V_1$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 13116,
    "subject": "Mathematics (Olympiad)",
    "question": "For a positive integer $n$, let $d(n)$ be the number of positive divisors of $n$. Determine the positive integers $k$ for which there exist positive integers $a$ and $b$ satisfying\n\n$$\nk = d(a) = d(b) = d(2a + 3b).\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "For $i \\ge 0$, let $a = 2 \\cdot 5^i$ and $b = 3 \\cdot 5^i$. Then both $a$ and $b$ have $2(i+1)$ divisors. Moreover, $2a + 3b = 4 \\cdot 5^i + 9 \\cdot 5^i = 13 \\cdot 5^i$, which also has $2(i+1)$ divisors. Therefore, all even values of $k$ satisfy the condition in the problem.\n\nNow suppose that $k$ is odd. Then $a$ has an odd number of divisors and therefore is a square, say $a = x^2$. Similarly, $b = y^2$, and $2a + 3b = z^2$ for some positive integers $x, y, z$. Thus,\n\n$$\n2x^2 + 3y^2 = z^2.\n$$\n\nWe show that this equation has no positive integer solutions.\n\nSuppose, for contradiction, that there is a positive integer solution $(x, y, z)$ with minimal $x + y + z$. Then $2x^2 + 3y^2 = z^2$. Modulo $3$, this gives $2x^2 \\equiv z^2 \\pmod{3}$. If $x$ is not divisible by $3$, then $x^2 \\equiv 1 \\pmod{3}$, so $z^2 \\equiv 2 \\pmod{3}$, which is impossible. Therefore, $x$ is divisible by $3$, so $z$ is also divisible by $3$. Now $2x^2$ and $z^2$ are both divisible by $9$, so $3y^2$ is divisible by $9$, and thus $y$ is divisible by $3$. But then $(\\frac{x}{3}, \\frac{y}{3}, \\frac{z}{3})$ is a smaller positive integer solution, contradicting minimality. Therefore, the equation $2x^2 + 3y^2 = z^2$ has no positive integer solutions.\n\nIt follows that no odd $k$ satisfies the condition in the problem. The positive integers $k$ that satisfy the condition are therefore the even integers.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 13117,
    "subject": "Mathematics (Olympiad)",
    "question": "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\n$$\nS_n \\le 20m^3 + 10rm^2 + 2r(r-1)m + \\frac{1}{6}r(r-1)(r-2).\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "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\n$$\nS_k \\le 20m^3 + 10rm^2 + 2r(r-1)m + \\frac{1}{6}r(r-1)(r-2).\n$$\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 $P$.\n\nFrom Lemma 2, the lines from such points whose lengths are more than 9 are at most\n\n$$\n10m^2 + 4rm + \\frac{1}{2}r(r-1).\n$$\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\n$$\nS_p = 10m^2 + 4rm + \\frac{1}{2}r(r-1).\n$$\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\n$$\nS_k \\le 20m^3 + 10rm^2 + 2r(r-1)m + \\frac{1}{6}r(r-1)(r-2).\n$$\n\nFurthermore\n\n$$\n\\begin{aligned}\nS_{k+1} &= S_k + S_p \\\\\n&\\le 20m^3 + 10rm^2 + 2r(r-1)m + \\frac{1}{6}r(r-1)(r-2) \\\\\n&\\quad + 10m^2 + 4rm + \\frac{1}{2}r(r-1) \\\\\n&= 20m^3 + 10(r+1)m^2 + 2r(r+1)m \\\\\n&\\quad + \\frac{1}{6}r(r-1)(r+1),\n\\end{aligned}\n$$\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\n$n = 63 = 6 \\times 10 + 3$. It follows from Lemma 3,\n\n$$\n\\begin{aligned}\nS &\\le 20 \\times 10^3 + 10 \\times 10^2 + 2 \\times 3 \\times 2 \\times 10 + \\frac{1}{6} \\times 3 \\times 2 \\times 1 \\\\\n&= 23,121.\n\\end{aligned}\n$$\n\nThus, $S_{\\max} = 23,121$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 13118,
    "subject": "Mathematics (Olympiad)",
    "question": "Find the smallest three-digit number such that when it is multiplied by 3, the result is also a three-digit number whose hundreds digit is even.",
    "options": [],
    "answer": "See solution",
    "solution": "Denote the three-digit number by $\\overline{abc}$. Its triple is:\n\n$$\n3 \\cdot \\overline{abc} = (3a) \\cdot 100 + (3b) \\cdot 10 + 3c.\n$$\n\nSince $a \\geq 1$, if $a = 1$, for the hundreds digit of $3 \\cdot \\overline{abc}$ to be even, we require $3b \\cdot 10 + 3c \\geq 100$, which implies $10b + c \\geq \\frac{100}{3} = 33 + \\frac{1}{3}$. The smallest integer satisfying this is 34. Thus, the smallest such number is 134, and $3 \\times 134 = 402$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 13119,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $a$ be a real number for which the number $a + a^2 + a^3$ is an integer. Show that, if one of the numbers $a$, $a^2$, or $a^3$ is rational, then $a$ is an integer.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $a + a^2 + a^3 = n$, with integer $n$.\n\nIf $a$ is rational, then $a = \\frac{p}{q}$, where $p, q$ are coprime integers. Then:\n$$\n\\frac{p}{q} + \\frac{p^2}{q^2} + \\frac{p^3}{q^3} = n\n$$\nMultiplying both sides by $q^3$:\n$$\npq^2 + p^2q + p^3 = n q^3\n$$\nSo $q$ divides $p^3$. Since $p$ and $q$ are coprime, this is only possible if $q = \\pm 1$, so $a$ is an integer.\n\nIf $a^2$ is rational, then $a + a^3 = n - a^2$ is also rational. Let $a^2 = s$ (rational), so $a$ is real and $a^2$ rational. Then $a$ is either rational or irrational quadratic. But $a + a^3$ is rational, so $a(1 + a^2)$ is rational. Since $a^2$ is rational, $a$ must be rational, and by the previous case, $a$ is an integer.\n\nIf $a^3 = r$ is rational, then $a + a^2 = n - a^3$ is also rational. Let $a^3 = r$ (rational), so $a$ is real and $a^3$ rational. Then $a$ is either rational or irrational cubic. But $a + a^2$ is rational, so $a^2 + a$ is rational. Since $a^3 = r$, $a^2 = \\frac{r}{a}$, so $a + \\frac{r}{a}$ is rational, which implies $a$ is rational (unless $a = 0$, but then $a^3 = 0$ and $a + a^2 + a^3 = 0$ is integer, and $a = 0$ is integer). If $a$ is rational, by the first case, $a$ is integer. If $a^3 = 1$, then $a = 1$, which is integer.\n\nTherefore, in all cases, $a$ is an integer.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 13120,
    "subject": "Mathematics (Olympiad)",
    "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.\n\n![](images/brozura_a67angl_new_p5_data_510251c710.png)",
    "options": [],
    "answer": "See 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.\n\nWe angle chase:\n\n$$\n\\angle D'C'C = \\angle D'DC = \\angle ADC = \\angle ACB.\n$$\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.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 13121,
    "subject": "Mathematics (Olympiad)",
    "question": "Prove that for any positive integer $d$ it is possible to find two distinct positive integers $n_1$ and $n_2$ such that\n\n- $n_2$ can be obtained from $n_1$ by permuting its digits,\n- both $n_1$ and $n_2$ are divisible by $d$,\n- none of them starts with \"0\".\n\nDenote by $|x|$ the number of digits in $x$. Prove that there exists $d$ such that $|n_1| > 2|d|$ for any pair $(n_1, n_2)$ satisfying the above properties.",
    "options": [],
    "answer": "See solution",
    "solution": "One can take $n_1 = \\overline{dd0}$ and $n_2 = \\overline{d0d}$. They consist of the same digits, they are different (the middle digit is nonzero for $n_1$ and zero for $n_2$), and both of them are divisible by $d$.\n\nFor $d = 5$, one can easily check that there are no two-digit numbers $n_1$ and $n_2$ satisfying this property.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 13122,
    "subject": "Mathematics (Olympiad)",
    "question": "Numbers $1, 2, \\ldots, n$ are written in a line. Numbers $1$ and $n$ are painted blue, and the rest are painted yellow. Two players, Olesya and Andriy, take turns repainting one of the yellow numbers blue according to the following rules:\n\n- Olesya starts and repaints any yellow number blue (let's denote it by $k$).\n- Andriy then chooses one of the intervals $(1, 2, \\ldots, k)$ or $(k, k+1, \\ldots, n)$—the one containing more yellow numbers. If both intervals have the same number of yellow numbers, he may choose either. The other interval is removed from the game.\n- Andriy then repaints any yellow number in the chosen interval blue. The interval is again split into two smaller intervals.\n- Olesya then chooses the interval with more yellow numbers for her move, and the other interval is removed from the game. The process repeats.\n\nThe winner is the player who paints blue a number whose immediate left and right neighbors are both blue. Who wins this game if both players play perfectly?\n\n**Answer:** If $n \\neq 2^k - 2$, then Olesya wins; otherwise, Andriy wins.",
    "options": [],
    "answer": "See solution",
    "solution": "We solve this problem by identifying winning and losing positions. A position is *losing* for a player if, after their move, they immediately lose or leave the opponent in a winning position. A *winning* position allows a player to win immediately or force the opponent into a losing position.\n\nThe game is played on an interval $l, l+1, \\ldots, m$, where $l$ and $m$ are blue and the rest are yellow. The specific values of $l$ and $m$ are unimportant; only the number of yellow numbers matters. Let $a$ be the number of yellow numbers in the interval.\n\n- If there is $1$ yellow number, the position is winning (the player can win immediately).\n- If there are $2$ yellow numbers ($a_1 = 2$), it is a losing position.\n- For $3, 4, 5$ yellow numbers, these are winning positions, since the player can move to the losing position $a_1 = 2$.\n- Thus, $a_2 = 6$ is a losing position.\n\nBy induction, all losing positions satisfy $a_{n+1} = 2a_n + 2$.\n\n**Inductive step:**\nSuppose all positions from $a_{n-1} + 1$ to $a_n - 1$ are winning, and $a_{n-1}$ and $a_n$ are losing. Consider a position $b$ with $a_n + 1 \\leq b \\leq 2a_n + 1$. The next player can split the interval into $a_n$ and $b - a_n - 1$ yellow numbers. Since $a_n \\geq b - a_n - 1$ if and only if $b \\leq 2a_n + 1$, the opponent is left in a losing position $a_n$.\n\nIf the player is in position $a_{n+1} = 2a_n + 2$, after splitting, the larger interval will have at least $a_n + 1$ yellow numbers, which is a winning position.\n\nThe explicit form of losing positions is $a_n = 2^{n+1} - 2$ (which can be shown by induction or by solving the recurrence).\n\n**Conclusion:** Olesya wins if $n \\neq 2^k - 2$; otherwise, Andriy wins.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 13123,
    "subject": "Mathematics (Olympiad)",
    "question": "In a convex quadrilateral $ABCD$, $|AB| = |BC| = |CD|$. Furthermore, let $P$ be the intersection point of its diagonals such that $|\\angle APD| < 90^\\circ$. Let $R$ and $S$ be the reflections of $A$ and $D$ with respect to $BD$ and $AC$, respectively. Prove that the lines $BC$ and $RS$ are parallel.\n\n![](images/CZE_ABooklet_2023_p19_data_af3df9098d.png)",
    "options": [],
    "answer": "See solution",
    "solution": "First, we note that the given symmetries imply $|BR| = |BA|$ and $|CS| = |CD|$. Hence,\n\n$$\n|AB| = |BC| = |CD| = |BR| = |CS|. \\qquad (1)\n$$\n\nBy construction, the points $R$ and $S$ are distinct, and the midpoint $X$ of segment $AR$ lies on its perpendicular bisector $BD$, while the midpoint $Y$ of segment $DS$ lies on its perpendicular bisector $AC$. We now prove that $R$ lies inside angle $ABC$ and $S$ inside angle $DCB$, as in the figure. Thus, points $A$, $D$, $R$, $S$ all lie in the same half-plane with boundary $BC$.\n\nThe assumption $|\\angle APD| < 90^\\circ$ implies $|\\angle APB| > 90^\\circ$, which for the interior point $P$ of the base $AC$ of isosceles triangle $ABC$ means $|\\angle ABP| < \\frac{1}{2} |\\angle ABC|$; hence $|\\angle ABR| = 2 \\cdot |\\angle ABP| < |\\angle ABC|$, so $R$ is inside angle $ABC$. Similarly, from $|\\angle DPC| > 90^\\circ$ for $P$ on base $BD$ of isosceles triangle $DCB$, $S$ lies inside angle $DCB$.\n\nA further consequence of $|\\angle APD| < 90^\\circ$ is that for the marked interior angles of right triangles $APX$ and $DPY$, $|\\angle XAP| = 90^\\circ - |\\angle APD| = |\\angle YDP|$, i.e., $|\\angle RAC| = |\\angle SDB|$.\n\nReturning to (1), the point $B$ is the circumcenter of triangle $ARC$, which lies in angle $ABC$. By the inscribed angle theorem, $|\\angle RBC| = 2 \\cdot |\\angle RAC|$. Similarly, for circumcenter $C$ of triangle $BSD$ in angle $BCD$, $|\\angle SCB| = 2 \\cdot |\\angle SDB|$. From above, $|\\angle RBC| = |\\angle SCB|$. This, together with (1), shows that triangles $RBC$ and $SCB$ are congruent by SAS. Hence, their altitudes from $R$ and $S$ to $BC$ have the same length, implying $BC \\parallel RS$.\n\n**Another Solution.**\n\nWe show that $R$ and $S$ lie on the circumcircle of $BCP$. We detail the proof for $R$; for $S$ it is analogous.\n\nAs before, (1) holds and $R$ lies inside angle $ABC$. From $|\\angle APD| < 90^\\circ$, $R$ lies in the half-plane $ACD$.\n\nBy (1), $B$ is the circumcenter of $ARC$, whose central angle $RBA$ with bisector $BD$ is twice angle $RCA$. Thus, the three angles $PBA$, $RBP$, and $RCP$ in the figure are congruent. The congruence of the last two with the first means $R$ lies on the circumcircle of $BCP$. The same holds for $S$ by analogous angle congruence.\n\n![](images/CZE_ABooklet_2023_p20_data_eacb2bb972.png)\n\nThus, $B$, $C$, $R$, $S$ lie on a circle, and $R$, $S$ are in the same half-plane with boundary $BC$. The congruence of angles $BRC$ and $BSC$, together with $|BC| = |BR| = |CS|$, means triangles $RBC$ and $SCB$ are congruent. The congruence of their altitudes proves $BC \\parallel RS$.\n\n_Alternatively, it suffices to note that the congruent segments $BR$ and $CS$ are symmetrically placed along the axis of $BC$._",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 13124,
    "subject": "Mathematics (Olympiad)",
    "question": "The Fibonacci numbers $F_0, F_1, F_2, \\dots$ are defined inductively by $F_0 = 0$, $F_1 = 1$, and $F_{n+1} = F_n + F_{n-1}$ for $n \\geq 1$.\n\nGiven an integer $n \\geq 2$, determine the smallest size of a set $S$ of integers such that for every $k = 2, 3, \\dots, n$ there exist $x, y \\in S$ such that $x - y = F_k$.",
    "options": [],
    "answer": "See solution",
    "solution": "First, we show that if a set $S \\subset \\mathbb{Z}$ satisfies the conditions, then $|S| \\geq \\frac{n}{2} + 1$. Let $d = \\left\\lfloor \\frac{n}{2} \\right\\rfloor$, so $n \\leq 2d \\leq n + 1$. To prove that $|S| \\geq d + 1$, construct a graph as follows:\n\nLet the vertices of the graph be the elements of $S$. For each $1 \\leq k \\leq d$, choose two elements $x, y \\in S$ such that $x - y = F_{2k-1}$, and add the pair $(x, y)$ to the graph as an edge. (By the problem's constraints, there must be a pair $(x, y)$ with $x - y = F_{2k-1}$ for every $3 \\leq 2k-1 \\leq 2d-1 \\leq n$; moreover, since $F_1 = F_2$, we have a pair with $x - y = F_1$ as well.) We will say that the length of the edge $(x, y)$ is $|x - y|$.\n\nWe claim that the graph contains no cycle. For contradiction, suppose the graph contains a cycle $(x_1, \\dots, x_\\ell)$, and let the longest edge in the cycle be $(x_1, x_\\ell)$ with length $F_{2m+1}$. The other edges $(x_1, x_2), \\dots, (x_{\\ell-1}, x_\\ell)$ in the cycle are shorter than $F_{2m+1}$ and distinct; their lengths form a subset of $\\{F_1, F_3, \\dots, F_{2m-1}\\}$. But this is not possible because\n\n$$\n\\begin{align*}\nF_{2m+1} &= |x_{\\ell} - x_1| \\leq \\sum_{i=1}^{\\ell-1} |x_{i+1} - x_i| \\leq F_1 + F_3 + F_5 + \\dots + F_{2m-1} \\\\\n&= F_2 + (F_4 - F_2) + (F_6 - F_4) + \\dots + (F_{2m} - F_{2m-2}) = F_{2m} < F_{2m+1}\n\\end{align*}\n$$\n\nHence, the graph has $d$ edges and cannot contain a cycle, so it must contain at least $d+1$ vertices, thus $|S| \\geq d+1$.\n\nNow we show a suitable set with $d+1$ elements. Let\n\n$$\nS = \\{F_0, F_2, F_4, F_5, \\dots, F_{2d}\\}\n$$\n\nFor $1 \\leq k \\leq d$, we have $F_0, F_{2k-2}, F_{2k} \\in S$ with differences $F_{2k} - F_{2k-2} = F_{2k-1}$ and $F_{2k} - F_0 = F_{2k}$, so each of $F_1, F_2, \\dots, F_{2d}$ occurs as a difference between two elements in $S$. Thus, this set containing $d+1$ numbers is suitable. $\\Box$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 13125,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $k$ be an integer such that $2n^2 = dk$. Is it possible for $n^2 + d$ to be a perfect square for some integer $n$?",
    "options": [],
    "answer": "See solution",
    "solution": "We have:\n\n$$\n\\begin{aligned}\nk^2(n^2 + d) &= k^2 n^2 + k^2 d = k^2 n^2 + 2k n^2 \\\\\n&= n^2(k^2 + 2k) = n^2((k+1)^2 - 1).\n\\end{aligned}\n$$\n\nThe expression $(k+1)^2 - 1$ is not a perfect square for $k \\in \\mathbb{N}$. Therefore, $k^2(n^2 + d)$ is not a perfect square, so $n^2 + d$ cannot be a perfect square.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 13126,
    "subject": "Mathematics (Olympiad)",
    "question": "Compute the following product:\n\n$$\n\\prod_{m=1}^{2018} \\frac{(2m-1)^4 + \\frac{1}{4}}{(2m)^4 + \\frac{1}{4}}\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "By applying the Sophie-Germain identity, we obtain the following equality:\n\n$$\n\\frac{(2m-1)^4 + \\frac{1}{4}}{(2m)^4 + \\frac{1}{4}} = \\frac{\\left((2m-\\frac{1}{2})^2 + \\frac{1}{4}\\right)\\left((2m-\\frac{3}{2})^2 + \\frac{1}{4}\\right)}{\\left((2m+\\frac{1}{2})^2 + \\frac{1}{4}\\right)\\left((2m-\\frac{1}{2})^2 + \\frac{1}{4}\\right)} = \\frac{(2m-\\frac{3}{2})^2 + \\frac{1}{4}}{(2m+\\frac{1}{2})^2 + \\frac{1}{4}}\n$$\n\nIt is now easy to see that the product can be telescoped:\n\n$$\n\\prod_{m=1}^{2018} \\frac{(2m-1)^4 + \\frac{1}{4}}{(2m)^4 + \\frac{1}{4}} = \\prod_{m=1}^{2018} \\frac{(2m - \\frac{3}{2})^2 + \\frac{1}{4}}{(2m + \\frac{1}{2})^2 + \\frac{1}{4}} = \\frac{(2 - \\frac{3}{2})^2 + \\frac{1}{4}}{(2 \\cdot 2018 + \\frac{1}{2})^2 + \\frac{1}{4}} = \\frac{\\frac{1}{2}}{\\frac{8073^2+1}{4}} = \\frac{2}{8073^2 + 1}\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 13127,
    "subject": "Mathematics (Olympiad)",
    "question": "Given the equations of the lines $A_1B_1$ and $CA_2$:\n\n$$\ny = \\frac{\\frac{2r_1}{r_1+1}}{\\frac{-2\\sqrt{r_1}}{r_1+1} + 2\\sqrt{r_1}}(x + 2\\sqrt{r_1}) = \\frac{1}{\\sqrt{r_1}}x + 2,\n$$\n\nand\n\n$$\ny = \\frac{\\frac{2r_1r_2}{r_1+r_2}}{\\frac{2(r_1\\sqrt{r_2}-r_2\\sqrt{r_1})}{r_1+r_2} - 2\\sqrt{r_2}} (x - 2\\sqrt{r_2}),\n$$\n\nwhich can be combined to obtain\n\n$$\ny = \\frac{2r_1r_2}{2\\sqrt{r_1r_2}(\\sqrt{r_1} - \\sqrt{r_2}) - 2\\sqrt{r_2}(r_1 + r_2)}[\\sqrt{r_1}(y - 2) - 2\\sqrt{r_2}].\n$$\n\nRearrange the terms to obtain a linear equation in $y$, and find the constant term and the coefficient of $y$ in terms of $r_1$ and $r_2$. Show that both are symmetric expressions in $r_1$ and $r_2$.",
    "options": [],
    "answer": "See solution",
    "solution": "Rearranging the given equation, the constant term is\n\n$$\n4r_1r_2(\\sqrt{r_1} + \\sqrt{r_2}),\n$$\n\nwhich is symmetric in $r_1$ and $r_2$.\n\nThe coefficient of $y$ is\n\n$$\n\\begin{aligned}\n&2\\{r_1r_2\\sqrt{r_1} - \\sqrt{r_1r_2}(\\sqrt{r_1} - \\sqrt{r_2}) + \\sqrt{r_2}(r_1 + r_2)\\} \\\\\n&= 2\\{(\\sqrt{r_1} + \\sqrt{r_2})^2\\sqrt{r_1} - \\sqrt{r_1r_2}(\\sqrt{r_1} - \\sqrt{r_2}) + \\sqrt{r_2}(r_1 + r_2)\\} \\\\\n&= 2\\{r_1\\sqrt{r_1} + 2r_1\\sqrt{r_2} + r_2\\sqrt{r_1} - r_1\\sqrt{r_2} + r_2\\sqrt{r_1} + r_1\\sqrt{r_2} + r_2\\sqrt{r_2}\\} \\\\\n&= 2\\{r_1\\sqrt{r_1} + 2r_1\\sqrt{r_2} + 2r_2\\sqrt{r_1} + r_2\\sqrt{r_2}\\},\n\\end{aligned}\n$$\n\nwhich is also symmetric in $r_1$ and $r_2$.\n\nTherefore, the $y$-coordinates of points $D_1$ and $D_2$ have the same expression, proving the proposition.\n\n*Proof 3.* (It is not necessary for circle $O$ to be tangent to line $\\ell$.)\n\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 $\\ell$.\n\nNow,\n\n$$\n\\begin{aligned}\n\\angle CD_1B_2 &= \\angle CA_1A_2 + \\angle A_1A_2B_2 \\\\\n&= \\angle CA_1B_1 + \\angle B_1A_1A_2 + \\angle A_1A_2B_2 \\\\\n&= \\frac{1}{2}\\angle CO_1B_1 + \\frac{1}{2}\\angle A_1O_1B_1 + \\frac{1}{2}\\angle A_2O_2B_2 \\\\\n&= (90^\\circ - \\angle O_1B_1C) + \\frac{1}{2}\\angle O_1OO_2 \\\\\n&= (\\angle OB_1C - 90^\\circ) + 90^\\circ - \\angle OB_1B_2 \\\\\n&= \\angle CB_1B_2.\n\\end{aligned}\n$$\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, $D_1D_2 \\parallel \\ell$. $\\square$",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 13128,
    "subject": "Mathematics (Olympiad)",
    "question": "A group of tourists get on 10 buses in the outgoing trip. The same group of tourists get on 8 buses in the return trip. Assuming each bus carries at least 1 tourist, prove that there are at least 3 tourists such that each of them has taken a bus in the return trip that has more people than the bus he has taken in the outgoing trip.",
    "options": [],
    "answer": "See solution",
    "solution": "We prove the general case where there are $p$ outgoing buses and $q$ returning buses with $p > q$. The number of such tourists is $p - q + 1$.\n\nLet $T$ be the set of all tourists. For each $t \\in T$, let $p_t$ be the number of tourists in the bus that $t$ takes in the outgoing trip and let $q_t$ be the number of tourists in the bus that $t$ takes in the return trip. Then\n\n$$\n\\sum_{t \\in T} \\frac{1}{p_t} = p, \\quad \\sum_{t \\in T} \\frac{1}{q_t} = q \\quad \\Rightarrow \\quad \\sum_{t \\in T} \\left( \\frac{1}{p_t} - \\frac{1}{q_t} \\right) = p - q\n$$\n\nSince $\\left| \\frac{1}{p_t} - \\frac{1}{q_t} \\right| < 1$, at least $p-q+1$ terms in the above sum are positive. That is, there are at least $p-q+1$ tourists such that $\\frac{1}{p_t} - \\frac{1}{q_t} > 0$, or equivalently $p_t < q_t$.",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 13129,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all integers $n \\geq 0$ for which there exist integers $a$ and $b$ such that\n\n$$a + 2^b = n^{2022}$$\n\nand\n\n$$a^2 + 4^b = n^{2023}.$$",
    "options": [],
    "answer": "See solution",
    "solution": "The only number is $n = 1$.\n\nTo see this, notice that $2(a^2 + 4^b) \\geq (a + 2^b)^2$, so $2n^{2023} \\geq n^{4044}$. This implies $n = 0$ or $n = 1$. The case $n = 0$ leads to no solution, while $n = 1$ works for $a = 0$ and $b = 0$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 13130,
    "subject": "Mathematics (Olympiad)",
    "question": "For a positive integer $n$, list all its divisors $1 = d_1 < d_2 < \\cdots < d_k = n$. A divisor $d_i$ is called a *good divisor* if $d_{i-1} d_{i+1}$ is not divisible by $d_i$, for $2 \\leq i \\leq k-1$. Find all $n$ for which the number of good divisors is smaller than the number of different prime divisors.",
    "options": [],
    "answer": "See solution",
    "solution": "First, suppose $n$ has at least three different prime divisors. Let $m$ be the number of distinct prime divisors, and let $p < q$ be the two smallest. The first consecutive divisors of $n$ are $1 < p < p^2 < \\cdots < p^\\alpha < q$ for some $\\alpha$. For each prime divisor except the smallest, we get $m-1$ good divisors, since divisors smaller than a prime are coprime to it. To find another good divisor, consider $n$ divisible by $p^{\\alpha+1}$. Then $q$ is a good divisor, as the next divisor after $q$ is $p^{\\alpha+1}$, which is larger than $pq$. If $n$ is not divisible by $p^{\\alpha+1}$, the next composite divisor after $q$ is $pq$. If there is a prime between $q$ and $pq$, take the last as the $m$-th good divisor. Otherwise, the consecutive divisors $p^\\alpha, q, pq$ yield $\\frac{n}{pq}, \\frac{n}{q}, \\frac{n}{p^\\alpha}$, and $\\frac{n}{q}$ is a good divisor. Thus, for $n$ with at least three prime divisors, the number of good divisors is at least the number of prime divisors.\n\nIf $n$ is a power of a prime, it has no good divisors, so it satisfies the condition.\n\nNow, consider $n$ with exactly two distinct prime divisors $p < q$. The first consecutive divisors are $1 < p < p^2 < \\cdots < p^\\alpha < q$. The divisor $p^\\alpha$ is a good divisor, and the next divisor after $q$ is $pq$. If $p^{\\alpha+1} < pq$, then $n$ is not divisible by $p^{\\alpha+1}$. The consecutive divisors $\\frac{n}{pq}, \\frac{n}{q}, \\frac{n}{p^\\alpha}$ show that $\\frac{n}{q} = p^\\alpha$ is a good divisor, so $n = p^\\alpha q$ with $p^\\alpha < q$. Both cases satisfy the condition.\n\n**Answer:** All $n$ that are a power of a prime, and all $n = p^\\alpha q$ with $p < q$ primes and $p^\\alpha < q$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 13131,
    "subject": "Mathematics (Olympiad)",
    "question": "The range of $f(x) = \\sqrt{x-5} - \\sqrt{24-3x}$ is \\underline{\\hspace{2cm}}.",
    "options": [],
    "answer": "See solution",
    "solution": "It is easy to see that $f(x)$ is increasing on its domain $[5, 8]$. Therefore, its range is $[-3, \\sqrt{3}]$. $\\square$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 13132,
    "subject": "Mathematics (Olympiad)",
    "question": "Isaac places some counters into the squares of an $8 \\times 8$ chessboard so that there is at most one counter in each of the 64 squares. Determine, with justification, the maximum number that he can place without having five or more counters in the same row, or in the same column, or in either of the two long diagonals.",
    "options": [],
    "answer": "See solution",
    "solution": "We know that there must be fewer than $5$ counters in each row, column, and long diagonal, so the maximum in each is $4$ counters.\n\nIf we had more than $32$ counters, then there would be at least $5$ in one row, which cannot happen. Hence, there can be at most $32$ counters.\n\nHere is an example which attains this bound:\n\n![](images/V_Britanija_2013_p14_data_7bf7c801c0.png)",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 13133,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $P(x)$, $Q(x)$, and $R(x)$ be polynomials such that $Q(x)$ and $R(x)$ take only nonnegative values. It is known that the equation\n\n$$\nP(x) + \\sqrt{Q(x)} + \\sqrt{Q(x) + \\sqrt{R(x)}} = 0\n$$\n\nhas infinitely many solutions. Is it true that every real number is a root of this equation?",
    "options": [],
    "answer": "See solution",
    "solution": "Not necessarily.\n\nConsider the example: $P(x) = x$, $Q(x) = \\frac{1}{4}x^2$, so $\\sqrt{Q(x)} = \\frac{1}{2}|x|$, and $R(x) = 0$. Then $\\sqrt{Q(x) + \\sqrt{R(x)}} = \\frac{1}{2}|x|$, so\n\n$$\nP(x) + \\sqrt{Q(x)} + \\sqrt{Q(x) + \\sqrt{R(x)}} = x + |x| = \\begin{cases} 2x, & x \\ge 0, \\\\ 0, & x < 0. \\end{cases}\n$$\n\nThus, the left-hand side equals $0$ only for $x < 0$, not for all $x$, but the equation still has infinitely many solutions.\n\nThis example is not unique; for instance, one can also take $Q(x) = \\frac{1}{8}x^2$, $R(x) = \\left(\\frac{1}{8}x^2\\right)^2$.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 13134,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $M$ be the greatest integer such that both $M + 1213$ and $M + 3773$ are perfect squares. What is the units digit of $M$?\n\n(A) 1  \n(B) 2  \n(C) 3  \n(D) 6  \n(E) 8",
    "options": [],
    "answer": "See solution",
    "solution": "**Answer (E):** Suppose $M + 1213 = j^2$ and $M + 3773 = k^2$ for nonnegative integers $j$ and $k$. Then\n\n$$\n(k + j)(k - j) = k^2 - j^2 = 3773 - 1213 = 2560 = 5 \\cdot 2^9.\n$$\n\nBecause $k + j$ and $k - j$ have the same parity and their product is even, they must both be even, and it follows that one of them is $5 \\cdot 2^i$ and the other is $2^{9-i}$ for some $i$ with $1 \\leq i \\leq 8$. Solving for $k$ gives\n\n$$\nk = \\frac{5 \\cdot 2^i + 2^{9-i}}{2}.\n$$\n\nTo maximize $M$ it is sufficient to maximize $k$, and this will occur when $i = 8$ and $k = 5 \\cdot 2^7 + 1 = 641$. Therefore $M = 641^2 - 3773$, and its units digit is 8.\n\nAlternatively, since $(n + 1)^2 - n^2 = 2n + 1$, successive terms in the sequence of squares differ by successive odd numbers; and $(n + 2)^2 - n^2 = 4(n + 1)$, so squares two apart differ by multiples of 4. The two squares required differ by $2560$, a multiple of 4. Thus, the greatest such squares are two apart, so $n + 1 = \\frac{2560}{4} = 640$. Therefore, these squares are $n^2 = 639^2$ and $(n + 2)^2 = 641^2$, and $M + 1213 = 639^2$. Then $M = 639^2 - 1213$, and its units digit is 8.\n\nShown below is a table of $5 \\cdot 2^i$, $2^{9-i}$, $k$, $j$, and $M$ for each $i$ (notation from the first solution). Observe that $M$ is maximized when $k$ is maximized.\n\n![](images/2024_AMC10A_Solutions_p7_data_16031957c5.png)\n\n| $i$ | $5 \\cdot 2^i$ | $2^{9-i}$ | $k$ | $j$ | $M$ |\n|---|---|---|---|---|---|\n| 1 | 10 | 256 | 133 | 123 | 13916 |\n| 2 | 20 | 128 | 74 | 54 | 1703 |\n| 3 | 40 | 64 | 52 | 12 | -1069 |\n| 4 | 80 | 32 | 56 | 24 | -637 |\n| 5 | 160 | 16 | 88 | 72 | 3971 |\n| 6 | 320 | 8 | 164 | 156 | 23123 |\n| 7 | 640 | 4 | 322 | 318 | 99911 |\n| 8 | 1280 | 2 | 641 | 639 | 407108 |",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 13135,
    "subject": "Mathematics (Olympiad)",
    "question": "三角形 $ABC$ 為等腰三角形，頂角 $\triangle A = \\alpha$，且點 $O, H$ 分別為 $\triangle ABC$ 的外心及垂心。點 $P, Q$ 分別在 $AB, AC$ 邊上，且 $APHQ$ 成菱形。則 $\\angle POQ$ 為何（以 $\\alpha$ 表示）？",
    "options": [],
    "answer": "See solution",
    "solution": "$$\n\\angle POQ = 180^{\\circ} - 2\\alpha\n$$\n\n**解法一：**\n\n1. 令菱形 $APHQ$ 的兩對角線 $PQ, AH$ 交於 $M$，則 $M$ 為 $AH$ 的中點。\n\n2. 取 $\\triangle ABC$ 各邊中點 $M_A, M_B, M_C$，由九點圓性質知 $M, M_A, M_B, M_C$ 共圓。因此\n   $$\n   \\angle M_B M M_C = 180^{\\circ} - \\angle M_B M_A M_C = 180^{\\circ} - \\alpha\n   $$\n\n3. 由於 $OM_C, PM$ 共圓，$OM_B, QM$ 共圓，所以\n   $$\n   \\begin{align*}\n   \\angle POQ &= \\angle POM + \\angle QOM = \\angle PM_C M + \\angle PM_B M \\\\\n   &= \\angle M_B M M_C - \\angle A = 180^{\\circ} - 2\\alpha.\n   \\end{align*}\n   $$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 13136,
    "subject": "Mathematics (Olympiad)",
    "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": "See solution",
    "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 $P(x) = a_k x^k + a_{k-1} x^{k-1} + \\dots + a_0$. If $P(n) \\ne 0$, then\n\n$$\nP(P(n) + n) - P(n) = a_k[(P(n) + n)^k - n^k] + a_{k-1}[(P(n) + n)^{k-1} - n^{k-1}] + \\dots + a_1 P(n)\n$$\n\nwhich is divisible by $P(n)$. Therefore, if $P(P(n) + n)$ is prime, either $P(n) = \\pm 1$ or $P(P(n) + n) = \\pm P(n) = p$ for some prime $p$. Since $P(x)$ is a polynomial, $P(n) = \\pm 1$ for only finitely many $n$. Thus, either $P(n) = P(P(n) + n)$ for infinitely many $n$, or $P(n) = -P(P(n) + n)$ for infinitely many $n$.\n\nSuppose $P(n) = P(P(n) + n)$ for infinitely many $n$. Then $P(P(x) + x) - P(x)$ has infinitely many roots, so it is identically zero: $P(P(x) + x) = P(x)$. If $k \\ge 2$, $P(P(x) + x)$ has degree $k^2$ while $P(x)$ has degree $k$, which is impossible. Thus, $P(x)$ is at most linear: $P(x) = a x + b$ for integers $a, b$. Now,\n\n$$\nP(P(x) + x) = a(a + 1)x + ab + b\n$$\n\nSetting equal to $P(x)$ gives $a = a(a + 1)$ and $ab + b = b$. Thus, $a = 0$, so $P(n) = b$, a constant. For $P(P(n) + n)$ to be prime infinitely often, $b$ must be a prime $p$.\n\nSimilarly, if $P(n) = -P(P(n) + n)$ for infinitely many $n$, then $P(x) = -P(P(x) + x)$. For $P(x) = a x + b$, this gives $a = -a(a + 1)$ and $ab + b = -b$. Thus, $a = 0$ or $a = -2$. If $a = -2$, $P(n) = -2n + b$. For $P(P(n) + n)$ to be prime infinitely often, $b$ must be odd. In this case, $P(P(n) + n) = 2n - b$, which is prime for infinitely many $n$ as long as $b$ is odd.\n\nTherefore, the solutions are $P(n) = p$ (constant prime) and $P(n) = -2n + b$ with $b$ odd.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 13137,
    "subject": "Mathematics (Olympiad)",
    "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 $m^2 + n^2 + p^2 - 2mn - 2mp - 2np$ is a perfect square does not depend on $p$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $m^2 + n^2 + p^2 - 2mn - 2mp - 2np = (m + n - p)^2 - 4mn$ be a perfect square. Then the quadratic $f(x) = m x^2 + (m + n - p)x + n$ has two rational roots, both positive or both negative. Therefore, $f(x) = (d_1 x - a)(d_2 x - b)$ where $d_1, d_2 \\in \\mathbb{N}$, $d_1 d_2 = m$, and $a, b$ are such that $ab = n$. Since $p = f(-1) = (a + d_1)(b + d_2)$ and $p$ is a prime, $a$ and $b$ are not both positive. Assume $a$ and $b$ are both negative. It follows from $p = (a + d_1)(b + d_2)$ that $a + d_1$ and $b + d_2$ equal (in some order) $-1$ and $-p$ or $1$ and $p$. In the latter case, we have a contradiction to $p > m$. If $a + d_1 = -1$, $b + d_2 = -p$, we have $n = ab = (1 + d_1)(p + d_2)$. Conversely, if $n = (1 + d_1)(p + d_2)$ where $d_1, d_2 \\in \\mathbb{N}$, $d_1 d_2 = m$, direct verification shows that $(m + n - p)^2 - 4mn = (d_1 p - d_2)^2$.\n\nTherefore, each factor of $m$ yields a solution. We shall prove that all these solutions are distinct. Assume the contrary, i.e., $l | m$, $k | m$, $l \\neq k$ imply one and the same solution. Hence\n\n$$\nn = (l + 1)\\left(\\frac{m}{l} + p\\right) = (k + 1)\\left(\\frac{m}{k} + p\\right)$$\n\nimplying that $p l + \\frac{m}{l} = p k + \\frac{m}{k}$. It follows that $p k l = m$, i.e., $p | m$, which is a contradiction since $p > m$.\n\nFinally, the number of positive integers $n$ for which $m^2 + n^2 + p^2 - 2mn - 2mp - 2np$ is a perfect square equals the number of factors of $m$ and does not depend on $p$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 13138,
    "subject": "Mathematics (Olympiad)",
    "question": "Given a positive integer $N$, we subtract from it its greatest proper divisor (different from $N$), then do the same with the new number and repeat the operation until $1$ is obtained. Find how many subtractions are there if the process starts with $N = 19^{19}$.",
    "options": [],
    "answer": "See solution",
    "solution": "Since $19$ is a prime, the greatest proper divisor of $N = 19^{19}$ is $19^{18}$. The first number obtained is:\n\n$$\nN_1 = 19^{19} - 19^{18} = 18 \\cdot 19^{18}.\n$$\n\nIf a number $m$ is even, the operation gives $\\frac{m}{2}$. So the next number is:\n\n$$\nN_2 = \\frac{N_1}{2} = 9 \\cdot 19^{18}.\n$$\n\nThe greatest proper divisor of $9 \\cdot 19^{18}$ is $3 \\cdot 19^{18}$, so:\n\n$$\nN_3 = 9 \\cdot 19^{18} - 3 \\cdot 19^{18} = 6 \\cdot 19^{18}.\n$$\n\nNow, as it is even:\n\n$$\nN_4 = \\frac{6 \\cdot 19^{18}}{2} = 3 \\cdot 19^{18}.\n$$\n\nIts greatest proper divisor is $19^{18}$, so:\n\n$$\nN_5 = 3 \\cdot 19^{18} - 19^{18} = 2 \\cdot 19^{18}.\n$$\n\nThen:\n\n$$\nN_6 = \\frac{2 \\cdot 19^{18}}{2} = 19^{18}.\n$$\n\nWe see that $6$ operations transform $19^{19}$ to $19^{18}$, so it takes another $6$ to reach $19^{17}$, and so on. To end with $1 = 19^0$, the process takes:\n\n$$\n6 \\times 19 = 114\n$$\n\noperations.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 13139,
    "subject": "Mathematics (Olympiad)",
    "question": "Extend the cevians $AX$, $BX$, $CX$ in triangle $ABC$. Let their intersection points with the sides $BC$, $AC$, $BA$ be $A_1$, $B_1$, $C_1$ respectively. Show that\n\n$$\n\\angle AXB_1 = 180^\\circ - \\angle BXA = 180^\\circ - (\\angle B + \\angle C) = \\angle A = \\alpha.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "Similarly, $\\angle BXC_1 = \\angle B = \\beta$, $\\angle CXA_1 = \\angle C = \\gamma$. Let $\\angle CAA_1 = \\varphi$. Then $\\angle ABB_1 = \\angle ABX = 180^\\circ - (\\alpha - \\varphi) - (\\alpha + \\beta) = \\varphi$, and similarly, $\\angle BCC_1 = \\varphi$. That is, $X$ is the Brakar point of triangle $ABC$.\n\nNow, using the law of sines for triangles $ABC$, $AXB$, $AXC$, we have respectively:\n\n$$\n\\frac{BC}{AC} = \\frac{\\sin \\alpha}{\\sin \\gamma}, \\quad \\frac{AX}{AB} = \\frac{\\sin \\varphi}{\\sin \\angle BXA} = \\frac{\\sin \\varphi}{\\sin \\alpha},\n$$\n\n$$\n\\frac{AC}{CX} = \\frac{\\sin \\angle CXA}{\\sin \\varphi} = \\frac{\\sin \\gamma}{\\sin \\varphi}.\n$$\n\nMultiplying all three equalities, we obtain:\n\n$$\n\\frac{BC}{AB} \\cdot \\frac{AX}{AB} \\cdot \\frac{AC}{CX} \\text{ which is equivalent to } \\frac{BC \\cdot AX}{AB} = \\frac{AB \\cdot CX}{AC}.\n$$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 13140,
    "subject": "Mathematics (Olympiad)",
    "question": "What is the value of $\\frac{a}{b}$ if $120\\%$ of $a$ equals $80\\%$ of $b$?",
    "options": [],
    "answer": "See solution",
    "solution": "We have $120\\%$ of $a$ is $\\frac{6}{5}a$ and $80\\%$ of $b$ is $\\frac{4}{5}b$. Setting these equal:\n$$\n\\frac{6}{5}a = \\frac{4}{5}b\n$$\nDividing both sides by $b$ and rearranging:\n$$\n\\frac{a}{b} = \\frac{4}{5} \\times \\frac{5}{6} = \\frac{2}{3}\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 13141,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $n! = p_1^{a_1} p_2^{a_2} \\cdots p_{k-1}^{a_{k-1}} p_k^{a_k}$ be the canonical form of $n!$ with $a_i > 0$ for $i = 1, 2, \\dots, k$ and $p_1 < p_2 < \\cdots < p_{k-1} < p_k$.\n\nFind all $n$ such that the only primes dividing $n!$ are consecutive and satisfy $p_k = p_{k-1} + 2$ and $p_{k-1} = p_{k-2} + 2$.",
    "options": [],
    "answer": "See solution",
    "solution": "We will prove that $p_k = p_{k-1} + 2$ and $p_{k-1} = p_{k-2} + 2$. Since the prime $p_k$ divides\n\n$$\n\\prod_{\\substack{p<q \\le n \\\\ p, q \\text{ are prime}}} (p+q) = (2+3) \\cdots (2+p_{k-2})(2+p_{k-1}) \\cdots (p_{k-1}+p_k),\n$$\n\nthen $p_k$ must divide some term of the form $p_i + p_j$ with $i < j \\le k-1$. Since the sum $p_i + p_j$ is smaller than $2p_k$, it follows that $p_k = p_i + p_j$.\n\nIf $i > 1$, then $p_i + p_j$ is even (absurd, because $p_k > 2$ is odd).\n\nTherefore, $i = 1$ and $j = k - 1$, with $p_1 = 2$ and $p_k = p_1 + p_{k-1} = 2 + p_{k-1}$.\n\nSimilarly, $p_{k-1}$ divides some factor of the form $p_i + p_j$ with $1 \\le i < j \\le k$.\n\nIf $j \\neq k$, then $j \\le k - 2$, and as above, it follows that $p_{k-1} = p_{k-2} + 2$.\n\nIf $j = k$, that is, if $p_{k-1}$ divides some factor of the form $p_i + p_k$ with $1 \\le i$, then $p_{k-1}$ divides $p_i + p_{k-1} + 2$, and hence it divides $p_i + 2$. Then we must have $p_{k-1} = p_i + 2$, with $i = k - 2$. Hence, $p_{k-1} = p_{k-2} + 2$ in every case.\n\nSince the numbers $p_k, p_k - 2$ and $p_k - 4$ are primes, working modulo $3$ we get $p_k = 7$. Therefore, the unique primes dividing $n!$ are only $2, 3, 5, 7$.\n\nWe have\n\n$$\n\\prod_{\\substack{p<q \\le 7 \\\\ p, q \\text{ primes}}} (p+q) = (2+3)(2+5)(2+7)(3+5)(3+7)(5+7) = 2^6 \\cdot 3^3 \\cdot 5^2 \\cdot 7,\n$$\n\nand $7! = 2^4 \\cdot 3^2 \\cdot 5 \\cdot 7$, whereas $8! = 2^7 \\cdot 3^2 \\cdot 5 \\cdot 7$, $9! = 2^7 \\cdot 3^4 \\cdot 5 \\cdot 7$, and $10! = 2^8 \\cdot 3^4 \\cdot 5^2 \\cdot 7$. Hence, $n = 7$.\n\n*Comment.* Alternatively: $v_2(8!) = \\sum_{i=1}^{\\infty} \\left\\lfloor \\frac{8}{2^i} \\right\\rfloor = 4 + 2 + 1 = 7$, $v_2(9!) = \\sum_{i=1}^{\\infty} \\left\\lfloor \\frac{9}{2^i} \\right\\rfloor = 4 + 2 + 1 = 7$ and $v_2(10!) = \\sum_{i=1}^{\\infty} \\left\\lfloor \\frac{10}{2^i} \\right\\rfloor = 5 + 2 + 1 = 8$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 13142,
    "subject": "Mathematics (Olympiad)",
    "question": "Triangles $PAB$ and $PCD$ are placed on a plane. Suppose that $PA = PB$, $PC = PD$ are satisfied and that the three points $P$, $A$, $C$ lie on a straight line in this order, and the same is true for the three points $B$, $P$, $D$. Suppose further that a circle $S_1$ going through $A$, $C$ and a circle $S_2$ going through $B$, $D$ intersect at two distinct points $X$ and $Y$. Prove that the orthocenter of the triangle $PXY$ coincides with the midpoint of the line segment connecting the centers of the circles $S_1$ and $S_2$.\n\nHere, for a line segment $ZW$, its length is also denoted by $ZW$.",
    "options": [],
    "answer": "See solution",
    "solution": "Given a circle $S$ and a point $Z$ in the plane, we define the power $p = p_S(Z)$ of $Z$ with respect to $S$ as follows: Let $p = 0$ if $Z$ lies on the circumference of $S$. Otherwise, draw a line through $Z$ intersecting $S$ at points $I_1$ and $I_2$. Then, $p = p_S(Z) = -ZI_1 \\cdot ZI_2$ if $Z$ lies inside $S$, and $p = ZI_1 \\cdot ZI_2$ if $Z$ lies outside $S$. Here, the line segments $ZI_1$, $ZI_2$ are considered directed. A well-known theorem (the power of a point) tells us that the value of $p$ is independent of the choice of the line through $Z$ intersecting $S$, and this can be proved using the fact that angles subtended by an arc at any pair of points on the circle have the same magnitude and by similarity of triangles. Also, considering the line through $Z$ and the center $O$ of $S$, we see that $p = ZO^2 - r^2$, where $r$ is the radius of $S$.\n\nReturning to the problem, we see from $PA \\cdot PC - PB \\cdot PD = 0$ and the fact that both $X$ and $Y$ lie on both circles $S_1$ and $S_2$, that each of the three points $P$, $X$, $Y$ satisfies:\n\n$$\np_{S_1}(\\text{of the point}) + p_{S_2}(\\text{of the point}) = 0 \\quad (*)\n$$\n\nLet $r_i$ be the radius of circle $S_i$ and $O_i$ its center ($i = 1, 2$), and let $M$ be the midpoint of the segment $O_1O_2$. For a point $Z$ in the plane, the following statements hold:\n\n$$\n\\begin{align*}\nZ \\text{ satisfies } (*) &\\iff (ZO_1^2 - r_1^2) + (ZO_2^2 - r_2^2) = 0 \\\\\n&\\iff ZO_1^2 + ZO_2^2 = r_1^2 + r_2^2 \\\\\n&\\iff 2(MO_1^2 + ZM^2) = r_1^2 + r_2^2 \\\\\n&\\iff ZM^2 = \\frac{1}{2}(r_1^2 + r_2^2) - MO_1^2.\n\\end{align*}\n$$\n\nSince the right-hand side above is independent of $Z$, every point in the plane satisfying $(*)$ lies on the circumference of the same circle. In particular, $M$ coincides with the orthocenter of triangle $PXY$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 13143,
    "subject": "Mathematics (Olympiad)",
    "question": "Determine all pairs $(m, n)$ of positive integers such that\n\n$$\n2^m = 7n^2 + 1.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "Note that $2^m \\equiv 1 \\pmod{7}$, which means that $m = 3k$ for some positive integer $k$. Now we have\n\n$$\n2^{3k} - 1 = (2^k - 1)(2^{2k} + 2^k + 1) = 7n^2.\n$$\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 $d = 3$.\n\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\n**Case 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\n**Case 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 $(m, n) = (3, 1)$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 13144,
    "subject": "Mathematics (Olympiad)",
    "question": "Show that for any positive integer $k$, there exists a power of $2$ whose decimal representation contains a block of at least $k$ consecutive zeros.",
    "options": [],
    "answer": "See 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 $10^{m+k}$.\n\nBy Euler's theorem, for every positive integer $t$ (since $(2, 5^t) = 1$),\n\n$$\n2^{\\varphi(5^t)} \\equiv 1 \\pmod{5^t}.\n$$\n\nMultiplying by $2^t$ gives\n\n$$\n2^{t+\\varphi(5^t)} \\equiv 2^t \\pmod{10^t}, \\quad \\text{thus} \\quad 2^{t+\\varphi(5^t)} = y \\cdot 10^t + 2^t\n$$\n\nfor some positive integer $y$. Set $n = t + \\varphi(5^t)$ and $m = t - k$. We want $t$ such that $2^t < 10^{t-k}$. Such a $t$ exists, e.g., $t = 2k$ (since $2^{2k} = 4^k < 10^k$). Thus,\n\n$$\n2^{2k+\\varphi(5^{2k})} = y \\cdot 10^{2k} + 2^{2k}\n$$\n\ncontains a block of at least $k$ zeros.\n\nNow, consider a power of $2$ (say $2^n$) containing a block of exactly $r$ zeros with $r \\ge k$. When multiplying by $2$, for some nonzero digits $a, b$:\n\n$$\n2^n = \\underbrace{\\dots a \\ 0 \\ 0 \\ \\dots}_{y} \\underbrace{\\dots 0 \\ b \\ \\dots}_{r \\text{ zeros}} \\underbrace{\\dots}_{z} = y \\cdot 10^{r+s} + z.\n$$\n\nThen $2^{n+1} = 2y \\cdot 10^{r+s} + 2z$. The number $2z$ has either the same number of digits as $z$, or one more. Hence, on the right, the block of zeros either does not decrease, or decreases by one zero. On the left, the block can only extend (when $y$ is divisible by $5$). Globally, the length of the block either decreases by $1$, does not change, or increases. Repeating this process, the length of the block in each step decreases at most by $1$. Thus, it is impossible to avoid a block of length $k$; after sufficient steps, we obtain a power of $2$ with a block of exactly $k$ zeros.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 13145,
    "subject": "Mathematics (Olympiad)",
    "question": "Is there a rectangle that can be cut into 5 squares, among which there are two identical squares, and all the other squares are distinct and different from the identical ones? Consider the following cases:\n\na) The identical squares are the smallest of all.\n\nb) The identical squares are the largest of all.",
    "options": [],
    "answer": "See solution",
    "solution": "a) An example exists. See the corresponding arrangement (Fig. 32).\n\nb) Consider a square of the smallest size. It cannot be on the border of the rectangle, nor can it be inside the border of one of the other squares. This means that only the following arrangement of the other four squares around the smallest one is possible (Fig. 34). But then, if $AB$ is a side of the biggest square, for the point $B$ not to be a vertex of the outer rectangle, it has to be inside the side. But then there must be a square with a side $CB > AB$—this contradiction ends the proof.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 13146,
    "subject": "Mathematics (Olympiad)",
    "question": "Using the identity\n\n$$\n\\cos t = 4 \\cos^3 \\frac{t}{3} - 3 \\cos \\frac{t}{3}\n$$\n\nfor $t = x$, $t = x + 2\\pi$, and $t = x + 4\\pi$, sum the relations and use the periodicity of cosine to show that:\n\n$$\n\\cos^3 \\frac{x}{3} + \\cos^3 \\frac{x+2\\pi}{3} + \\cos^3 \\frac{x+4\\pi}{3} = \\frac{3}{4} \\cos x.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "Let us use the identity:\n\n$$\n\\cos t = 4 \\cos^3 \\frac{t}{3} - 3 \\cos \\frac{t}{3}\n$$\n\nApply it for $t = x$, $t = x + 2\\pi$, and $t = x + 4\\pi$:\n\n$$\n\\cos x = 4 \\cos^3 \\frac{x}{3} - 3 \\cos \\frac{x}{3}\n$$\n$$\n\\cos(x + 2\\pi) = 4 \\cos^3 \\frac{x + 2\\pi}{3} - 3 \\cos \\frac{x + 2\\pi}{3}\n$$\n$$\n\\cos(x + 4\\pi) = 4 \\cos^3 \\frac{x + 4\\pi}{3} - 3 \\cos \\frac{x + 4\\pi}{3}\n$$\n\nSince cosine is $2\\pi$-periodic, $\\cos(x + 2\\pi) = \\cos x$ and $\\cos(x + 4\\pi) = \\cos x$.\n\nSum all three equations:\n\n$$\n3 \\cos x = 4 \\left( \\cos^3 \\frac{x}{3} + \\cos^3 \\frac{x+2\\pi}{3} + \\cos^3 \\frac{x+4\\pi}{3} \\right) - 3 \\left( \\cos \\frac{x}{3} + \\cos \\frac{x+2\\pi}{3} + \\cos \\frac{x+4\\pi}{3} \\right)\n$$\n\nNow, the sum $\\cos \\frac{x}{3} + \\cos \\frac{x+2\\pi}{3} + \\cos \\frac{x+4\\pi}{3}$ can be shown to be zero (by using trigonometric sum formulas or direct calculation).\n\nThus,\n\n$$\n3 \\cos x = 4 \\left( \\cos^3 \\frac{x}{3} + \\cos^3 \\frac{x+2\\pi}{3} + \\cos^3 \\frac{x+4\\pi}{3} \\right)\n$$\n\nSo,\n\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": 13147,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $a_1, \\dots, a_n$ be real numbers such that $0 \\leq a_i \\leq 1$ for $i = 1, \\dots, n$. Prove the inequality\n$$\n(1 - a_1^n)(1 - a_2^n) \\cdots (1 - a_n^n) \\leq (1 - a_1 a_2 \\cdots a_n)^n.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "By the AM-GM inequality,\n$$\n(1 - a_1^n)(1 - a_2^n) \\cdots (1 - a_n^n) \\leq \\left( \\frac{(1 - a_1^n) + (1 - a_2^n) + \\cdots + (1 - a_n^n)}{n} \\right)^n = \\left( 1 - \\frac{a_1^n + \\cdots + a_n^n}{n} \\right)^n.\n$$\nApplying AM-GM again,\n$$\na_1 a_2 \\cdots a_n \\leq \\frac{a_1^n + \\cdots + a_n^n}{n} \\implies \\left(1 - \\frac{a_1^n + \\cdots + a_n^n}{n}\\right)^n \\leq (1 - a_1 a_2 \\cdots a_n)^n,\n$$\nwhich proves the desired inequality. $\\Box$",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 13148,
    "subject": "Mathematics (Olympiad)",
    "question": "令 $n \\ge 5$ 為一與 $6$ 互質的正整數。我們將一個正 $n$ 邊形的 $n$ 個頂點，各塗上三種顏色中的一種，使得被塗上每種顏色的頂點數都是奇數。試證：我們必可從這 $n$ 個頂點中找出三個點，它們的顏色都不相同，且此三點的連線構成等腰三角形。",
    "options": [],
    "answer": "See solution",
    "solution": "令 $a_k$ 為所有等腰三角形中，三頂點恰包含 $k$ 種顏色的三角形個數，則題目等價於證明 $a_3 \\ge 1$。\n\n我們採取歸謬證法。假設 $a_3 = 0$。考慮集合\n\n$$X = \\{(\\Delta, E) : \\Delta \\text{是等腰三角形},\\ E \\text{是 } \\Delta \\text{的一邊},\\ E \\text{的兩端點不同色}\\}.$$ \n\n讓我們用兩種不同方式計算 $X$ 中的元素個數：\n\n- 首先，對於每個三角形：\n  - 只有一色的三角形必沒有兩端點異色的邊。\n  - 只有兩色的三角形則恰有 $2$ 條邊的兩端點異色。\n  - 由假設，不存在三色的三角形。\n\n綜合上述，$|X| = 2a_2$。\n\n另一方面，任選兩個頂點 $A, B$，由於 $(n, 3) = 1$，我們知 $AB$ 恰為 $3$ 個等腰三角形的邊。若我們令 $b, c, d$ 分別為三種顏色的頂點數，則兩端點異色的邊共有 $bc + bd + cd$ 條，故 $|X| = 3(bc + bd + cd)$。\n\n然而由題目假設，$b, c, d$ 皆為奇數，故 $3(bc + bd + cd)$ 為奇數，從而它不可能等於 $2a_2$，矛盾！故 $a_3 \\ge 1$。",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 13149,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $c_k = 2 \\times 3^{k-1} - 2^{k-1}$ for $k \\geq 1$ (e.g., $c_1 = 1$, $c_2 = 4$, $c_3 = 14$, $c_4 = 46$, $c_5 = 146$, ...).\n\nConsider the following optimization problem:\n\nGiven nonnegative integers $x_1, x_2, \\dots$ such that $x_1 + x_2 + \\dots = n$, minimize\n$$\nT = c_1 f(x_1) + c_2 f(x_2) + \\dots\n$$\nwhere $f$ is a given function.\n\nWhat is a lower bound for the minimum value of $T$ in terms of $n$?\n\nAdditionally, for the sum\n$$\nS(A) \\geq \\sum_{k=1}^{\\infty} c_k \\times f(|B_k|),\n$$\nfind a lower bound in terms of $n$.",
    "options": [],
    "answer": "See solution",
    "solution": "Suppose $X = (x_1, x_2, \\dots, x_K, 0, 0, \\dots)$ minimizes $T$, with $x_1 \\ge x_2 \\ge \\dots \\ge x_K \\ge 1$ and $x_{K+1} = x_{K+2} = \\dots = 0$.\n\nIf $K \\le 2$, then\n$$\nT = c_1 f(x_1) + c_2 f(x_2) \\ge \\frac{3x_1^2-1}{2} + 4\\frac{3x_2^2-1}{2} \\ge \\frac{6}{5}(x_1+x_2)^2 - \\frac{5}{2} \\ge 1.1n^2 - 2n.\n$$\n\nIf $K \\ge 3$, then for $X$ minimizing $T$:\n- Changing $X$ to $X' = (x_1+1, x_2, \\dots, x_{K-1}, x_K-1, 0, \\dots)$ does not decrease $T$, so\n$$\n0 \\le \\Delta T = c_1(f(x_1+1)-f(x_1)) - c_K(f(x_K)-f(x_K-1)) \\le 3x_1+2 - c_K \\implies c_K \\le 3x_1+2.\n$$\n- Changing $X$ to $X'' = (x_1, x_2+1, \\dots, x_{K-1}, x_K-1, 0, \\dots)$ does not decrease $T$, so\n$$\n0 \\le \\Delta T = c_2(f(x_2+1)-f(x_2)) - c_K(f(x_K)-f(x_K-1)) \\le 4(3x_2+2) - c_K \\implies c_K \\le 12x_2+8.\n$$\n\nThus,\n$$\nn = x_1 + x_2 + \\dots + x_K \\ge x_1 + x_2 + 1 \\ge \\frac{c_K-2}{3} + \\frac{c_K-8}{12} + 1 = \\frac{5c_K-4}{12}\n$$\n\nand\n$$\nc_1 + c_2 + \\dots + c_K \\le c_K \\left(1 + \\frac{1}{3} + \\dots + \\frac{1}{3^{K-1}}\\right) \\le \\frac{12n+4}{5} \\times \\frac{3}{2} \\le 4n.\n$$\n\nTherefore,\n$$\nT(X) = c_1 f(x_1) + \\dots + c_K f(x_K) \\ge c_1 \\frac{3x_1^2-1}{2} + \\dots + c_K \\frac{3x_K^2-1}{2}\n$$\nso\n$$\nT(X) \\ge \\frac{3}{2}[c_1 x_1^2 + c_2 x_2^2 + \\dots + c_K x_K^2] - \\frac{1}{2}[c_1 + c_2 + \\dots + c_K] \\ge \\frac{3}{2} \\frac{(x_1 + x_2 + \\dots + x_K)^2}{\\frac{1}{c_1} + \\frac{1}{c_2} + \\dots + \\frac{1}{c_K}} - 2n.\n$$\n\nSince $c_{k+1} \\ge 3c_k$, we have\n$$\n\\frac{1}{c_1} + \\frac{1}{c_2} + \\dots + \\frac{1}{c_K} \\le 1 + \\frac{1}{4} + \\frac{1}{14} + \\frac{1}{14 \\times 3} + \\frac{1}{14 \\times 3^2} + \\dots = 1 + \\frac{1}{4} + \\frac{1.5}{14} \\le 1.36\n$$\n\nSo\n$$\nT \\ge \\frac{3}{2} \\frac{n^2}{1.36} - 2n > 1.1n^2 - 2n\n$$\n\nIn summary, the minimum value of $T$ satisfies $T_{\\min} \\ge 1.1n^2 - 2n$, and for the original sum,\n$$\nS(A) \\ge \\sum_{k=1}^{\\infty} c_k \\times f(|B_k|) \\ge 1.1n^2 - 2n.\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 13150,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $ABC$ be a non-isosceles triangle with circumcenter $U$ and incenter $I$. Assume that the bisector of the segment $UI$ passes through the common point of the angle bisector of $\\gamma = \\angle ACB$ with the circumcircle of $ABC$. Prove that $\\gamma$ is the second largest angle in the triangle $ABC$.\n\n![](path/to/file.png)",
    "options": [],
    "answer": "See solution",
    "solution": "Let $w_\\gamma$ be the angle bisector of $\\gamma$, $k$ the circumcircle of $ABC$, and $D = k \\cap w_\\gamma$. Since $\\angle DCB = \\angle DCA$, we have $|DA| = |DB|$. Considering triangle $DBI$, note that $\\angle CDB = \\angle CAB = \\alpha$. Also, $\\angle DBI = \\angle DBA + \\angle ABI = \\angle DCA + \\angle ABI = \\frac{\\alpha}{2} + \\frac{\\beta}{2}$, so $\\angle DIB = 180^\\circ - \\alpha - \\left(\\frac{\\alpha}{2} + \\frac{\\beta}{2}\\right) = \\frac{\\alpha}{2} + \\frac{\\beta}{2}$. Thus, $DBI$ is isosceles with $|DI| = |DB|$. Furthermore, since $D$ lies on the bisector of $UI$, we also have $|DU| = |DI|$. It follows that $D$ is the midpoint of a circle through $A$, $B$, $I$, and $U$.\n\nSince $U$ is the circumcenter of $ABC$, $\\angle AUB = 2\\angle ACB = 2\\gamma$. On the other hand, since $\\angle IAB = \\frac{\\alpha}{2}$ and $\\angle IBA = \\frac{\\beta}{2}$, we have $\\angle AIB = 180^\\circ - \\frac{\\alpha}{2} - \\frac{\\beta}{2}$. Since $A$, $B$, $I$, and $U$ lie on a common circle, $\\angle AUB = \\angle AIB$, so $2\\gamma = 180^\\circ - \\frac{\\alpha}{2} - \\frac{\\beta}{2}$, which is equivalent to $\\gamma = \\frac{1}{2}(\\alpha + \\beta)$. Since $\\gamma$ is the arithmetic mean of $\\alpha$ and $\\beta$, it is the second largest angle in triangle $ABC$ as claimed. $\\square$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 13151,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $x$ be the amount of water in the boat when pumping begins, $y$ the amount of water leaking into the boat per hour, and $z$ the amount each person can pump out of the boat in an hour. Let $h$ be the number of hours needed by $n$ people to pump the boat dry.\n\nGiven:\n\n$$\nx + 3y = 36z\n$$\n\n$$\nx + 10y = 50z\n$$\n\nIf $h = 2$, how many people $n$ are needed to pump the boat dry?",
    "options": [],
    "answer": "See solution",
    "solution": "Let $x$ be the initial water, $y$ the leak rate (per hour), $z$ the rate each person can pump (per hour), $h$ the hours, and $n$ the number of people.\n\nThe total water to be removed is $x + hy$ (initial plus leak over $h$ hours), and the total pumped out is $nhz$.\n\nSet up:\n\n$$\nhy + x = nhz \\qquad (1)\n$$\n\nGiven:\n\n$$\nx + 3y = 36z \\qquad (2)\n$$\n\n$$\nx + 10y = 50z \\qquad (3)\n$$\n\nSubtract (2) from (3):\n\n$$\n(x + 10y) - (x + 3y) = 50z - 36z\n$$\n$$\n7y = 14z \\implies y = 2z\n$$\n\nPlug $y = 2z$ into (2):\n\n$$\nx + 3(2z) = 36z \\implies x = 36z - 6z = 30z\n$$\n\nNow, equation (1):\n\n$$\nhy + x = nhz\n$$\n$$\nh(2z) + 30z = nhz\n$$\n$$\n2hz + 30z = nhz\n$$\n$$\nnhz - 2hz = 30z\n$$\n$$\nhz(n - 2) = 30z\n$$\n$$\nh(n - 2) = 30\n$$\n\nSubstitute $h = 2$:\n\n$$\n2(n - 2) = 30 \\implies n - 2 = 15 \\implies n = 17\n$$\n\n**Answer:** $n = 17$ people are needed.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 13152,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $E$ denote the intersection of the lines $AD$ and $BC$, let $F$ denote the intersection of $BD$ and $CA$, and let $G$ denote the intersection of $CD$ and $AB$. Given that $\\angle BAD = \\angle DCB$ and $\\angle CBD = \\angle DAC$, prove that $\\angle AEB = 90^{\\circ}$, where $A$, $B$, $C$, $D$ are points as described above.\n\n![](images/Slovenija_2016_p11_data_c366584a3f.png)",
    "options": [],
    "answer": "See solution",
    "solution": "The equality $\\angle BAD = \\angle DCB$ implies that triangles $GAD$ and $ECD$ are similar since they have two common angles. Thus,\n\n$$\n\\frac{|GD|}{|AD|} = \\frac{|ED|}{|CD|}.\n$$\n\nSimilarly, $\\angle CBD = \\angle DAC$ implies that triangles $FAD$ and $EBD$ are similar, so\n\n$$\n\\frac{|AD|}{|FD|} = \\frac{|BD|}{|ED|}.\n$$\n\nMultiplying these equalities gives\n\n$$\n\\frac{|GD|}{|FD|} = \\frac{|BD|}{|CD|} \\quad \\text{or} \\quad \\frac{|GD|}{|BD|} = \\frac{|FD|}{|CD|}.\n$$\n\nSince $\\angle GDB = \\angle CDF$, triangles $GDB$ and $FDC$ are similar, so $\\angle DBG = \\angle FCD$ and $\\angle DBA = \\angle ACD$. This, together with the problem's assumptions, implies\n\n$$\n\\angle BAD + \\angle CBD + \\angle DBA = \\frac{1}{2}(\\angle BAC + \\angle ACB + \\angle CBA) = 90^{\\circ},\n$$\n\nso $\\angle AEB = 180^{\\circ} - (\\angle BAD + \\angle CBD + \\angle DBA) = 90^{\\circ}$, which is what we wanted to show.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 13153,
    "subject": "Mathematics (Olympiad)",
    "question": "Given any set $A = \\{a_1, a_2, a_3, a_4\\}$ of four distinct positive integers, let $s_A = a_1 + a_2 + a_3 + a_4$. Define $n_A$ as the number of pairs $(i, j)$ with $1 \\leq i < j \\leq 4$ for which $a_i + a_j$ divides $s_A$. Find all sets $A$ of four distinct positive integers that achieve the largest possible value of $n_A$.",
    "options": [],
    "answer": "See solution",
    "solution": "For any positive integer $k$, the sets $\\{k, 5k, 7k, 11k\\}$ and $\\{k, 11k, 19k, 29k\\}$ achieve the maximum value $n_A = 4$.\n\nLet $A = \\{a_1, a_2, a_3, a_4\\}$ with $a_1 < a_2 < a_3 < a_4$. The pairwise sums satisfy:\n\n$$\na_1 + a_2 < a_1 + a_3 < a_1 + a_4 < a_2 + a_3 < a_2 + a_4 < a_3 + a_4.\n$$\n\nIf $a_i + a_j \\mid s_A$, then $a_i + a_j \\mid s_A - (a_i + a_j)$, so $a_i + a_j \\leq s_A - (a_i + a_j)$. Thus, $a_2 + a_4$ and $a_3 + a_4$ cannot divide $s_A$, so $n_A = 4$ is maximal.\n\nTo classify all such $A$, there exist integers $2 \\leq m < n$ such that:\n\n$$\n\\begin{cases}\na_1 + a_4 = a_2 + a_3, \\\\\nm(a_1 + a_3) = a_2 + a_4, \\\\\nn(a_1 + a_2) = a_3 + a_4.\n\\end{cases}\n$$\n\nIf $m \\geq 3$, the second equation implies $a_2 + a_4 \\geq 3(a_1 + a_3)$, leading to a contradiction. Thus, $m = 2$. Adding the third equation, twice the second, and three times the first gives $(n+7)a_1 = (5-n)a_2$. Since $n > m = 2$, possible values for $n$ are $3$ and $4$.\n\n- If $n = 3$, $a_2 = 5a_1$ and $A = \\{k, 5k, 7k, 11k\\}$.\n- If $n = 4$, $a_2 = 11a_1$ and $A = \\{k, 11k, 19k, 29k\\}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 13154,
    "subject": "Mathematics (Olympiad)",
    "question": "Say that a set of points in the grid is *good* if it contains no triple of points $(x_i, y_i)$, $(x_j, y_j)$, $(x_k, y_k)$ with $(x_k, y_k)$ stronger than $(x_j, y_j)$ and $(x_j, y_j)$ stronger than $(x_i, y_i)$, and let $N$ be the size of a largest good set.\n\nThe set\n\n$$\n\\{(1, 100), (2, 100), (2, 99), (3, 99), (3, 98), \\dots, (99, 2), (100, 2), (100, 1)\\}\n$$\n\nis good and contains 199 points, so $N \\ge 199$.\n\nFurther, any subset of this set is also good, so the required answer will be $n = N + 1$.\n\nWhat is the smallest $n$ such that any set of $n$ points in a $100 \\times 100$ grid must contain three points $(x_i, y_i)$, $(x_j, y_j)$, $(x_k, y_k)$ with $(x_k, y_k)$ stronger than $(x_j, y_j)$ and $(x_j, y_j)$ stronger than $(x_i, y_i)$?",
    "options": [],
    "answer": "See solution",
    "solution": "We can construct a set of 199 points where no three points satisfy the condition. Simply take the 100 points $(m, 101-m)$ for $m$ from 1 to 100 and the 99 points $(m, 100-m)$ for $m$ from 1 to 99.\n\nClearly, for $n < 199$, we can take a subset consisting of $n$ points of the above set and again no three points within the subset satisfy the condition.\n\nNow suppose there are at least 200 distinct points such that no three points satisfy the condition. Clearly, no three distinct points lie in the same row or column, as they would satisfy the condition. Therefore, there are exactly two points in every row and two points in every column, and all cases of more than 200 points are ruled out.\n\nLet $a_k$ be the $y$-coordinate of the lower point in each column, for $k$ from 1 to 100. Since each $a_k$ is at least 1 and at most 99, it follows that $a_i = a_j$ for some $i < j$.\n\nHowever, the points $(i, a_i)$, $(j, a_j)$ and the point above $(j, a_j)$ in column $j$ satisfy the condition. This gives the desired contradiction.\n\nHence $n = 200$ is the smallest for which any set has 3 points which satisfy the condition.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 13155,
    "subject": "Mathematics (Olympiad)",
    "question": "The degree of each vertex in a graph $G$ does not exceed $100$. We remove edges from this graph. In one step, we can remove an arbitrary set of edges without common endpoints, which is maximal (in the sense that if we add to this set any of the remaining edges, then there will be two edges with a common endpoint). Prove that all the edges will be removed after at most $199$ steps.",
    "options": [],
    "answer": "See solution",
    "solution": "It follows from the fact that for any edge $AB$, if it has not been removed yet, the operation decreases the sum of degrees $d(A) + d(B)$ by at least $1$.",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 13156,
    "subject": "Mathematics (Olympiad)",
    "question": "On the Cartesian plane, consider the graph $C$ of the function $y = \\sqrt[3]{x^2}$. An arbitrary line $d$ cuts $C$ at three distinct points with x-coordinates $x_1, x_2, x_3$. Prove that:\n\n(a) The value $$\\sqrt[3]{\\frac{x_2x_3}{x_1^2}} + \\sqrt[3]{\\frac{x_3x_1}{x_2^2}} + \\sqrt[3]{\\frac{x_1x_2}{x_3^2}}$$ is a constant.\n\n(b) $$\\sqrt[3]{\\frac{x_1^2}{x_2x_3}} + \\sqrt[3]{\\frac{x_2^2}{x_3x_1}} + \\sqrt[3]{\\frac{x_3^2}{x_1x_2}} < -\\frac{15}{4}.$$",
    "options": [],
    "answer": "See solution",
    "solution": "(a) Since $d$ cannot be parallel to the y-axis (otherwise, these graphs have at most one intersection), the function of $d$ is of the form $y = ax + b$ for some $a, b \\in \\mathbb{R}$. If $ab = 0$, then we also have at most two intersections, which does not meet the requirement; thus $ab \\neq 0$. The x-coordinates of the intersections of $d$ and $C$ are roots of\n\n$$\n\\sqrt[3]{x^2} = ax + b.\n$$\n\nLet $t = \\sqrt[3]{x}$, then from the given hypothesis, the equation\n\n$$\nt^3 - t^2 + b = 0\n$$\n\nhas three roots $t_1, t_2, t_3$ and $t_1t_2t_3 \\neq 0$ (since $b \\neq 0$). By applying Vieta's theorem, we have $t_1t_2 + t_2t_3 + t_3t_1 = 0$, then\n\n$$\n(t_1t_2)^3 + (t_2t_3)^3 + (t_3t_1)^3 = 3(t_1t_2)(t_2t_3)(t_3t_1) = 3t_1^2t_2^2t_3^2.\n$$\n\nIt is equivalent to\n\n$$\n\\frac{t_2t_3}{t_1^2} + \\frac{t_3t_1}{t_2^2} + \\frac{t_1t_2}{t_3^2} = 3,\n$$\n\nor\n\n$$\n\\sqrt[3]{\\frac{x_2x_3}{x_1^2}} + \\sqrt[3]{\\frac{x_3x_1}{x_2^2}} + \\sqrt[3]{\\frac{x_1x_2}{x_3^2}} = 3.\n$$\n\nHence, $\\sqrt[3]{\\frac{x_2x_3}{x_1^2}} + \\sqrt[3]{\\frac{x_3x_1}{x_2^2}} + \\sqrt[3]{\\frac{x_1x_2}{x_3^2}} = 3$, which is a constant.\n\n(b) Among the three numbers $t_1, t_2, t_3$, there is some pair of numbers with different signs, and without loss of generality, we may assume that they are $t_1$ and $t_2$. Since $t_1t_2 + t_2t_3 + t_3t_1 = 0$, we have $t_3 = -\\frac{t_1t_2}{t_1+t_2}$. Hence,\n\n$$\n\\begin{align*}\n\\sqrt[3]{\\frac{x_1^2}{x_2x_3}} + \\sqrt[3]{\\frac{x_2^2}{x_3x_1}} + \\sqrt[3]{\\frac{x_3^2}{x_1x_2}} &= \\frac{t_1^2}{t_2t_3} + \\frac{t_2^2}{t_3t_1} + \\frac{t_3^2}{t_1t_2} \\\\\n&= -(t_1+t_2)\\left(\\frac{t_1}{t_2^2} + \\frac{t_2}{t_1^2}\\right) + \\frac{t_1t_2}{(t_1+t_2)^2} \\\\\n&= -\\left(\\frac{t_1^2}{t_2^2} + \\frac{t_2^2}{t_1^2} + \\frac{t_1}{t_2} + \\frac{t_2}{t_1}\\right) + \\frac{t_1t_2}{(t_1+t_2)^2}.\n\\end{align*}\n$$\n\nBy applying the AM-GM inequality, we have\n\n$$\n\\frac{t_1^2}{t_2^2} + \\frac{t_2^2}{t_1^2} \\ge 2, \\quad \\frac{t_1}{t_2} + \\frac{t_2}{t_1} \\ge 2, \\quad \\frac{t_1 t_2}{(t_1+t_2)^2} \\le \\frac{1}{4}.\n$$\n\nHence,\n\n$$\n\\sqrt[3]{\\frac{x_1^2}{x_2x_3}} + \\sqrt[3]{\\frac{x_2^2}{x_3x_1}} + \\sqrt[3]{\\frac{x_3^2}{x_1x_2}} \\le -(2+2) + \\frac{1}{4} = -\\frac{15}{4}.\n$$\n\nThe equality does not occur since $t_1 \\neq t_2$.",
    "topic": "Discrete Mathematics",
    "subtopic": "Other"
  },
  {
    "id": 13157,
    "subject": "Mathematics (Olympiad)",
    "question": "Given the 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 $\\dfrac{ac}{b}$.",
    "options": [],
    "answer": "See solution",
    "solution": "Note that $f(x) = |\\log_3\\left(\\dfrac{x}{9}\\right)|$ is monotonically decreasing on $(0, 9]$ and monotonically increasing on $[9, +\\infty)$.  \n\nBy the conditions on $a, b, c$, we know that $0 < a < b < 9 < c$ and\n\n$$\n\\log_3\\left(\\dfrac{9}{a}\\right) = 2\\log_3\\left(\\dfrac{9}{b}\\right) = 2\\log_3\\left(\\dfrac{c}{9}\\right).\n$$\n\nTherefore,\n\n$$\n\\begin{aligned}\n\\log_3\\left(\\dfrac{ac}{b}\\right) &= \\log_3 a + \\log_3 c - \\log_3 b \\\\\n&= \\log_3\\left(\\dfrac{a c}{b}\\right).\n\\end{aligned}\n$$\n\nFrom the previous equalities, we can deduce that $\\log_3\\left(\\dfrac{ac}{b}\\right) = 2$, so $\\dfrac{ac}{b} = 3^2 = 9$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 13158,
    "subject": "Mathematics (Olympiad)",
    "question": "Consider the quantity $\\frac{x}{y} - \\frac{x+1}{y+1} = \\frac{x-y}{y(y+1)}$. Show that\n\n$$\n\\frac{x-y}{y(y+1)} + \\frac{y-z}{z(z+1)} + \\frac{z-x}{x(x+1)} \\ge 0,\n$$\n\nor, equivalently,\n\n$$\n\\frac{x}{y(y+1)} + \\frac{y}{z(z+1)} + \\frac{z}{x(x+1)} \\ge \\frac{y}{y(y+1)} + \\frac{z}{z(z+1)} + \\frac{x}{x(x+1)}.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "For any ordering of $x$, $y$, and $z$, the numbers $\\frac{1}{x(x+1)}$, $\\frac{1}{y(y+1)}$, and $\\frac{1}{z(z+1)}$ are reversely ordered since $x$, $y$, and $z$ are positive. Therefore, the last inequality follows from the rearrangement inequality.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 13159,
    "subject": "Mathematics (Olympiad)",
    "question": "For every positive integer $n \\geq 3$, show that there exists a positive integer $k$ such that\n$$\n51^{2^{n-2}} - 1 = 2^n \\cdot (2k + 1).\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "We prove the claim by mathematical induction on $n$.\n\n**Base case ($n=3$):**\n$$\n51^2 - 1 = 2600.\n$$\nWe have $2600 = 2^3 \\cdot 325$, and $325$ is odd, so the claim holds for $n=3$.\n\n**Inductive step:**\nAssume for some $n \\geq 3$ there exists $k$ such that\n$$\n51^{2^{n-2}} - 1 = 2^n \\cdot (2k + 1).\n$$\nConsider $n+1$:\n$$\n\\begin{aligned}\n51^{2^{n-1}} - 1 &= \\left(51^{2^{n-2}} - 1\\right)\\left(51^{2^{n-2}} + 1\\right) \\\\\n&= 2^n (2k+1) \\cdot \\left(51^{2^{n-2}} + 1\\right).\n\\end{aligned}\n$$\nSince $51^{2^{n-2}} + 1$ is even, write $51^{2^{n-2}} + 1 = 2m$ for some integer $m$.\nThen,\n$$\n51^{2^{n-1}} - 1 = 2^{n+1} (2k+1) m,\n$$\nwhere $(2k+1)m$ is odd, so the claim holds for $n+1$.\n\nThus, by induction, the statement is true for all $n \\geq 3$. $\\square$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 13160,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all positive integers $n$ for which it is possible to write all positive integers from $1$ to $n$ in some order on a circle, each exactly once, such that every number is coprime with the number that is two positions away from it.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $n$ be odd. Place the number $1$ at some point on the circle and, after every $\\frac{2}{n}$ full turns, write the next number in increasing order (see the figure for $n = 11$ below). In this way, all the numbers from $1$ to $n$ can be written at different points. For any number on the circle and the number two positions away from it, we have either consecutive integers or the numbers $n$ and $1$. In both cases, the chosen numbers are coprime.\n\nIf $n$ is divisible by $4$, let $n = 2m$, where $m$ is even. As described for odd $n$, write the numbers from $1$ to $m$ on the circle, and in the remaining points, write the numbers from $m + 1$ to $n$ in order (see the figure for $n = 12$ below). For any number and the number two positions away, we have either consecutive integers, the numbers $m$ and $1$, or the numbers $n$ and $m + 1$. In the first two cases, the numbers are coprime. If $n$ and $m + 1$ had a common divisor $d$, then $d$ would also divide $n - (m + 1) = m - 1$ and $(m + 1) - (m - 1) = 2$. But $d \\neq 2$ because $m$ is even, so $m + 1$ is odd and has no factor $2$. Thus, the only possibility is $d = 1$, so $\\gcd(n, m + 1) = 1$. Therefore, the chosen numbers are coprime in all cases.\n\n![](images/EST_ABooklet_2024_p52_data_a506aef299.png)\n\n![](images/EST_ABooklet_2024_p52_data_f1f1720789.png)\n\nNow let $n$ be an even number not divisible by $4$, i.e., $n = 2m$ with $m$ odd. Suppose the numbers from $1$ to $n$ are arranged in some order on a circle. This divides the numbers into two groups of $m$ numbers each: the first group contains $1$ and all those reached by repeatedly jumping two positions away, and the second group contains the rest. Among all the numbers, there are $m$ even numbers, and at least half must belong to one group; since $m$ is odd, more than half of the even numbers must be in one group. Thus, there must be two even numbers that are consecutive within the group, i.e., one is two positions away from the other on the circle. These two numbers share a factor of $2$. Therefore, it is not possible to arrange the numbers from $1$ to $n$ on the circle so that every number is coprime with the number two positions away from it.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 13161,
    "subject": "Mathematics (Olympiad)",
    "question": "Circles $w_1$ and $w_2$ centered at points $O_1$ and $O_2$ respectively intersect at points $A$ and $B$. Let $w$ be the circumscribed circle of triangle $O_1O_2B$ centered at $O$, which intersects $w_1$ and $w_2$ again at points $K$ and $L$ respectively. The straight line $OA$ intersects $w_1$ and $w_2$ at points $M$ and $N$ respectively. Denote by $P$ the intersection point of lines $MK$ and $NL$. Prove that $P$ lies on $w$ and $PM = PN$.",
    "options": [],
    "answer": "See solution",
    "solution": "**Solution.** We use the following lemma by Archimedes:\n\n**Lemma (Archimedes).** Circles $w_1$ and $w_2$ intersect at points $A$ and $B$, with the center of $w_2$ lying on $w_1$. A chord $AC$ of $w_2$ intersects $w_1$ again at a point $M$. Then $CM = CB$.\n\nPut $\\alpha = \\angle KBA = \\angle KMA$, since they intercept the same arc in $w_1$. Similarly, $\\beta = \\angle ABL = \\angle ANL$, because they intercept the same arc in $w_2$. Hence $$\\angle MPN = 180^\\circ - \\alpha - \\beta$$ which implies that $P \\in w$.\n\nBy Archimedes' lemma, for circles $w_1$ and $w_2$ we can write $KJ = JA = LJ$, but then $\\alpha = \\beta$, and from the circle $w$ we have that $PM = PN$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 13162,
    "subject": "Mathematics (Olympiad)",
    "question": "As shown in the figure, $AB > AC$, and the incircle $\\odot I$ of $\\triangle ABC$ is tangent to $BC$, $CA$, and $AB$ at points $D$, $E$, and $F$ respectively. Let $M$ be the midpoint of side $BC$, and $AH \\perp BC$ at the point $H$. The bisector $AI$ of $\\angle BAC$ intersects the lines $DE$ and $DF$ at points $K$ and $L$ respectively.\n\nProve that $M$, $L$, $H$, and $K$ are concyclic.\n\n![](images/Kina_2012_p21_data_0ba512db92.png)",
    "options": [],
    "answer": "See solution",
    "solution": "Join $CL$, $BI$, $DI$, $BK$, $ML$, and $KH$. Extend $CL$ to meet $AB$ at point $N$.\n\nAs both $CD$ and $CE$ are tangents to $\\odot I$, $CD = CE$. As\n\n$$\n\\begin{align*}\n\\angle BIK &= \\angle BAI + \\angle ABI = \\frac{1}{2}(\\angle BAC + \\angle ABC) \\\\\n&= \\frac{1}{2}(180^\\circ - \\angle ACB) = \\angle EDC = \\angle BD,\n\\end{align*}\n$$\n\nso $B$, $K$, $D$, and $I$ are cyclic.\n\nAs $\\angle BKI = \\angle BDI = 90^\\circ$, i.e., $BK \\perp AK$; similarly, $CL \\perp AL$. As $AL$ is the bisector of $\\angle BAC$, $L$ is the midpoint of $CN$. As $M$ is the midpoint of $BC$, $ML \\parallel AB$.\n\nSince $\\angle BKA = \\angle BHA = 90^\\circ$, it follows that the points $B$, $K$, $H$, and $A$ are cyclic, so $\\angle MHK = \\angle BAK = \\angle MLK$, hence $M$, $L$, $H$, and $K$ are cyclic.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 13163,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $a$ and $b$ be rational numbers such that $a + b = a^2 + b^2$. Suppose that the common value $s = a + b = a^2 + b^2$ is not an integer, and write it as an irreducible fraction: $s = \\frac{m}{n}$. Let $p$ be the least prime divisor of $n$. Find the minimum value of $p$.",
    "options": [],
    "answer": "See solution",
    "solution": "The minimum value of $p$ is $5$.\n\nWrite $a$ and $b$ as fractions with least common denominator $w$: $a = \\frac{u}{w}$, $b = \\frac{v}{w}$. If $a = \\frac{u'}{w'}$, $b = \\frac{v'}{w'}$ is another representation with common denominator $w'$, then $w' \\geq w$. The irreducible representation $s = \\frac{u + v}{w}$ is obtained from $s = \\frac{u + v}{w}$ by possible cancellation. Therefore, the prime divisors of $n$ are among those of $w$.\n\nWe show that $w$ is not divisible by $2$ or $3$, implying that neither is $n$.\n\nThe condition $a + b = a^2 + b^2$ gives $u^2 + v^2 = w(u + v)$. Suppose that $3$ divides $w$. Then $u^2 + v^2$ is a multiple of $3$, and since $x^2 \\equiv 0, 1 \\pmod{3}$ for each integer $x$, it follows that both $u$ and $v$ are divisible by $3$. However, then $3$ is a common divisor of $u, v,$ and $w$, which contradicts the minimality of $w$.\n\nSimilarly, suppose that $w$ is even. Then $u^2 + v^2$ is even, hence so is $u + v$ ($u^2 + v^2$ has the same parity as $u + v$). Hence $u^2 + v^2$ is divisible by $4$, and since $x^2 \\equiv 0, 1 \\pmod{4}$ for each integer $x$, both $u$ and $v$ are even. We reach a contradiction with the minimality of $w$ again.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 13164,
    "subject": "Mathematics (Olympiad)",
    "question": "There is a rectangular grid with 3 rows and $n$ columns on a blackboard.\n\n(a) Find the number of ways to write exactly one of the numbers $1, 2, \\ldots, 3n$ into each square so that all the following conditions are met:\n\n1. Different squares contain different numbers.\n2. 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).\n3. The numbers $1$ and $3n$ are written into adjacent squares.\n\n(b) 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": "See solution",
    "solution": "(a) All three squares in the leftmost and rightmost columns must be traversed consecutively, as traversing both corner squares requires passing through the middle square, which cannot be done twice. In no other column can all three squares be traversed consecutively, as this would split the grid into two disconnected parts. Similarly, traversing three or more consecutive squares in the middle row would also disconnect the grid. When visiting two consecutive squares in the middle row, the previous and next squares must be from the same row (top or bottom).\n\nThus, except for the first and last columns, every valid trajectory divides the middle row into pairs of squares visited consecutively. Therefore, the required enumeration is impossible for odd $n$. For even $n$, for each pair of squares in the middle row, we can choose the row (top or bottom) for the previous and next square; there are $\\frac{n-2}{2}$ such pairs, giving $2^{\\frac{n-2}{2}}$ possibilities. The entire trajectory is determined by these choices. For a fixed trajectory, there are $6n$ possibilities (2 directions and $3n$ choices for the initial square). Thus, there are $6n \\cdot 2^{\\frac{n-2}{2}} = 3n \\cdot 2^{\\frac{n}{2}}$ possible enumerations for even $n$.\n\n![](images/EST_ABooklet_2021_p21_data_3f9a425f0b.png)\n\n*Fig. 15*\n\n![](images/EST_ABooklet_2021_p21_data_81266d5bee.png)\n\n*Fig. 16*\n\n(b) One cannot start from the middle square of the leftmost column, as either corner would become a dead end. Suppose, without loss of generality, we start from the top left corner and visit $k$ squares in the top row ($1 \\leq k \\leq n$). Going directly to the bottom row would disconnect the grid; turning right along the middle row would prevent visiting left squares and reaching the rightmost column. Thus, we must turn left along the middle row. The leftmost $3 \\times k$ part of the grid should be traversed in a Z-shape. The remaining $3 \\times (n-k)$ part is traversed similarly.\n\nRepeating this argument, every valid trajectory divides the grid into Z-shapes and S-shapes, alternating between them. In every column except the last, we can choose whether to start a new letter or widen the current one. In the first column, we can also choose which corner to start in. Therefore, there are $2^n$ possibilities in total.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 13165,
    "subject": "Mathematics (Olympiad)",
    "question": "Given only a ruler and compass, construct a segment of length $\\sqrt{2021}$ using the minimal number of one-dimensional geometric objects (such as lines, circles, or points). Describe your construction steps and count the total number of one-dimensional objects used. *Remark*: All required geometric knowledge is covered in secondary school mathematics. Avoid ambiguity about whether \"giving two endpoints of a segment\" counts as giving the segment.",
    "options": [],
    "answer": "See solution",
    "solution": "Several methods are described for constructing a segment of length $\\sqrt{2021}$ using ruler and compass:\n\n**Method 1**: Draw lines and circles with specified centers and radii, intersecting at key points. For example, draw a circle with $B_1$ as center and $A_1B_5 = 45$ as radius, intersecting a line at points $C_3, C_4$. Then $A_1C_3 = \\sqrt{45^2 - A_1B_1^2} = \\sqrt{2021}$.\n\n**Method 2**: Number objects in order, use circles and perpendicular bisectors, and construct points so that $G_1M = \\sqrt{G_1O_1^2 - O_1M^2} = \\sqrt{2021}$.\n\n**Method 3**: Use circles and lines to create points $X_1, X_2$ such that $X_1B'_5 = \\sqrt{(2 \\times 22\\frac{1}{2})^2 - 2^2} = \\sqrt{2021}$.\n\nOther methods use algebraic identities (e.g., $2021 = 2048 - 27$) to construct $\\sqrt{2021}$ with different numbers of objects. The minimal number of one-dimensional objects required is at least 7. Scoring is based on the total number of objects used: 9 points for 11 objects, 6 points for 12 objects, 3 points for 13–15 objects.",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 13166,
    "subject": "Mathematics (Olympiad)",
    "question": "設 $f(x)$ 為一個次數不超過 $n$ 之多項式，並且 $f(0), f(1), \\dots, f(n)$ 這 $n+1$ 個數中，任意兩數的差都是整數。試證：$f(2011) - f(100)$ 也是整數。",
    "options": [],
    "answer": "See solution",
    "solution": "不妨設 $f(0)$ 為整數（否則轉而考慮函數 $f(x) - f(0)$）。我們只須證明 $f(2011)$ 與 $f(100)$ 是整數即可。\n\n利用數學歸納法證明更廣的命題：若一 $n$ 次多項式 $f(x)$ 滿足 $f(0), f(1), \\dots, f(n)$ 皆為整數，則對任意整數 $x$，$f(x)$ 都是整數。\n\n首先，基底 $n=0$ 時，$f(x)$ 為常系數多項式，命題顯然成立。\n\n設此命題對 $n < k$ 均成立。則對一個 $f(0), f(1), \\dots, f(k)$ 之值均為整數的 $k$ 次多項式 $f(x)$，考慮函數 $g(x) = f(x+1) - f(x)$。易知 $g(x)$ 為次數 $< k$ 的多項式，並且 $g(0), g(1), \\dots, g(k-1)$ 均為整數。由歸納假設，$g(x)$ 在 $x$ 為整數時其值均為整數。由於對非負整數 $x$ 有\n$$\nf(x) = f(0) + \\sum_{i=0}^{x-1} g(i)\n$$\n對負整數 $x$ 有\n$$\nf(x) = f(0) - \\sum_{i=x}^{-1} g(i)\n$$\n因此 $x$ 為整數時，$f(x)$ 亦為整數。",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 13167,
    "subject": "Mathematics (Olympiad)",
    "question": "In a $2n \\times 2n$ grid, exactly half of the squares are 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 with respect to 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 in a chessboard-style pattern?",
    "options": [],
    "answer": "See solution",
    "solution": "For $n = 1$, it is not possible to achieve the chessboard pattern if the initial $2 \\times 2$ configuration is as shown below, because adjacent same-coloured squares remain the same after reflecting.\n\n![](images/prob1314_p20_data_f7ec7db84d.png)\n\nNow, for any $n \\geq 2$, we can start from any initial configuration and reach the chessboard pattern. Whenever there are wrong-coloured squares, we can reduce their number by taking a finite number of steps. A wrong-coloured square turns into a right-coloured square on the other side of the axis of reflection, and vice versa. Define a *double reflection* as 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, leaving wrong-coloured and right-coloured squares unchanged.\n\nSuppose there exist two adjacent same-coloured squares. Without loss of generality, let these two wrong-coloured squares be in the same row. Since $n \\geq 2$, we can assume this row is at least the third from the top and that there is at least one column to the right. Let $W$ denote a wrong-coloured square, $R$ a right-coloured square, $x$ either, and $x'$ the opposite.\n\n- If at least one of the two adjacent wrong-coloured squares has a wrong-coloured upper neighbour, then reflecting with respect to the vertical axis decreases the number of wrong-coloured squares by at least 2:\n\n![](images/prob1314_p21_data_19910fbef4.png)\n\n- If both upper neighbours are right-coloured, but at least one of their upper neighbours is wrong-coloured, then reflecting with respect to the vertical axis moves the two adjacent wrong-coloured squares up by one row (number of wrong-coloured squares unchanged):\n\n![](images/prob1314_p21_data_b667533f09.png)\n\n  Afterwards, proceed as above.\n\n- If the $2 \\times 2$ square above is entirely right-coloured, but at least one adjacent square to the right is wrong-coloured, use double reflection to swap this wrong-coloured square with one of the right-coloured squares in the $2 \\times 2$ area:\n\n![](images/prob1314_p21_data_4920b40576.png)\n\n  Then proceed as before.\n\n- In the remaining cases, reduce the number of wrong-coloured squares by 2 using the steps below:\n\n![](images/prob1314_p21_data_50523148d9.png)\n\nIf there are no two adjacent wrong-coloured squares, double reflections can move a wrong-coloured square along the diagonals without changing the number of wrong- or right-coloured squares. Since the numbers of black and white squares are always equal, the existence of a wrong-coloured black square implies there is also a wrong-coloured white square. Thus, by moving along diagonals, we can bring two wrong-coloured squares together and proceed as above.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 13168,
    "subject": "Mathematics (Olympiad)",
    "question": "Prove that there are infinitely many different triangles in the coordinate plane whose vertices have integer coordinates and whose side lengths are consecutive integers.",
    "options": [],
    "answer": "See solution",
    "solution": "First, we prove that there are infinitely many triangles whose side lengths are consecutive integers and whose area is also an integer. Let the side lengths of the triangle be $2m-1$, $2m$, and $2m+1$. By Heron's formula, its area is\n\n$$\nS = \\sqrt{\\frac{6m(2m+2)2m(2m-2)}{16}} = m\\sqrt{3(m^2-1)}\n$$\n\nNow, we show that the equation $3(m^2 - 1) = q^2$ has infinitely many integer solutions. Since $q$ should be divisible by $3$, let $q = 3l$, so the equation becomes $m^2 - 1 = 3l^2$. It has a solution $m = 2$, $l = 1$, and if $(m, l)$ is a solution, then $(2m + 3l, m + 2l)$ is also a solution.\n\nIndeed, we can check that\n\n$$\n(2m + 3l)^2 - 1 = 3(m + 2l)^2\n$$\n\nis equivalent to\n\n$$\n4m + 12ml + 9l^2 - 1 = 3m^2 + 12ml + 12l^2\n$$\n\nwhich is equivalent to $m^2 - 1 = 3l^2$.\n\nThus, there are infinitely many triangles with side lengths $2m-1$, $2m$, and $2m+1$ and area $S = m\\sqrt{3(m^2-1)}$ where $\\sqrt{3(m^2-1)}$ is an integer. Let $\\triangle ABC$ be such a triangle with $AB = 2m$, $BC = 2m-1$, and $AC = 2m+1$, and let $CH$ be the altitude from $C$ to $AB$. Since $S = \\frac{CH \\cdot AB}{2}$, we get $CH = \\sqrt{3(m^2-1)}$, which is an integer. Also,\n\n$$\nAH = \\sqrt{AC^2 - CH^2} = \\sqrt{(2m+1)^2 - (3m^2-3)} = m + 2\n$$\n\nis an integer.\n\nNow, we can place our triangle $ABC$ in the coordinate plane. Let $A$ be the origin $(0,0)$ and $B$ be $(2m, 0)$. Then $C$ has integer coordinates $(m+2, \\sqrt{3m^2-3})$.\n\n![](images/BW2019-problems-solutions-opinions_p51_data_10d9d6056a.png)",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 13169,
    "subject": "Mathematics (Olympiad)",
    "question": "A monic polynomial $P(x)$ is called \"nice\" if its coefficients are in $\\{-1, 0, 1\\}$. Given a nice polynomial $P(x)$ of degree $2025$ that is divisible by $x^7 - 1$, what is the maximum number of non-zero coefficients in $P(x)$?",
    "options": [],
    "answer": "See solution",
    "solution": "Let $f(x) = x^7 - 1$. For any integer $k$ and $r = 0, 1, \\ldots, 6$, we have $x^{7k + r} \\equiv x^r \\pmod{f(x)}$. Group all exponents by their remainder modulo $7$:\n\n$$\n\\begin{aligned}\nP(x) &= a_{2025}x^{2025} + \\cdots + a_2x^2 + a_1x + a_0 \\\\\n&\\equiv (a_0 + a_7 + \\cdots + a_{2023}) + (a_1 + a_8 + \\cdots + a_{2024})x + \\cdots \\\\\n&\\quad + (a_6 + a_{13} + \\cdots + a_{2022})x^6 \\pmod{f(x)}.\n\\end{aligned}\n$$\n\nThe right-hand side is a polynomial of degree at most $6$. For divisibility by $f(x)$, all coefficients must sum to zero:\n\n$$\n\\sum_{i=0}^{289} a_{7i} = \\sum_{i=0}^{289} a_{7i+1} = \\sum_{i=0}^{289} a_{7i+2} = \\sum_{i=0}^{288} a_{7i+3} = \\sum_{i=0}^{288} a_{7i+4} = \\sum_{i=0}^{288} a_{7i+5} = \\sum_{i=0}^{288} a_{7i+6} = 0.\n$$\n\nThe first three sums have $290$ terms each, so we can choose $145$ coefficients as $1$ and $145$ as $-1$ in each. The remaining sums have $289$ terms, which is odd, so at least one coefficient must be $0$ in each. Thus, at least $4$ coefficients are zero, so the maximum number of non-zero coefficients is $2026 - 4 = 2022$.\n\nTo construct such a polynomial, assign coefficients as above with $a_{2025} \\neq 0$. The answer is $2022$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 13170,
    "subject": "Mathematics (Olympiad)",
    "question": "Does there exist an increasing sequence of integers $0 = a_0 < a_1 < a_2 < \\dots$, such that the following two conditions are satisfied:\n\n1) Every natural number can be written in the form $a_i + a_j$ for some (possibly equal) indices $i, j \\ge 0$.\n\n2) $a_n > \\frac{n^2}{16}$ for all natural $n$?",
    "options": [],
    "answer": "See solution",
    "solution": "**Answer:** Yes.\n\nLet $(a_n)$ be the sequence of all natural numbers $k$ whose binary representation has $1$s only on even places or only on odd places. For example, this sequence contains numbers with binary representations like $10000$, $10100$, $101$, $1000$.\n\nEvidently, the first condition is satisfied for this sequence. We will show that the estimate from the second condition is also valid.\n\nConsider all non-negative numbers less than $2^{2r}$, i.e., numbers with at most $2r$ digits in their binary representations. The number of elements of our sequence among these numbers is $2^r$ (with $0$s on all even places) plus $2^r$ (with $0$s on all odd places), and zero is counted twice. Hence, there are $2^{r+1}-1$ elements of our sequence less than $2^{2r}$, so $a_{2^{r+1}-1} = 2^{2r}$.\n\nFor any natural number $n$, we can find an integer $r$ such that $2^{r+1}-1 \\le n < 2^{r+2}-1$. Then $2^r > \\frac{n}{4}$, so $a_n \\ge a_{2^{r+1}-1} = 2^{2r} > \\frac{n^2}{16}$, as required.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 13171,
    "subject": "Mathematics (Olympiad)",
    "question": "Two rectangles are drawn on a piece of paper. The length and width of one rectangle are both 5 cm greater than the corresponding measures of the other rectangle. The area of the larger rectangle is $1\\ \\text{dm}^2$ greater than the area of the smaller rectangle. Find the perimeter of the smaller rectangle.",
    "options": [],
    "answer": "See solution",
    "solution": "Let the side lengths of the smaller rectangle (in cm) be $a$ and $b$. Then the side lengths of the larger rectangle are $a+5$ and $b+5$. The area of the smaller rectangle is $ab$, and the area of the larger rectangle is $(a+5)(b+5) = ab + 5a + 5b + 25$. Since the area of the larger rectangle is $1\\ \\text{dm}^2 = 100\\ \\text{cm}^2$ greater than the area of the smaller rectangle, we have:\n\n$$\n(a+5)(b+5) = ab + 100\n$$\n\nSo,\n$$\nab + 5a + 5b + 25 = ab + 100\n$$\n$$\n5a + 5b + 25 = 100\n$$\n$$\n5a + 5b = 75\n$$\n$$\na + b = 15\n$$\n\nTherefore, the perimeter of the smaller rectangle is:\n$$\n2(a + b) = 2 \\times 15 = 30\\ \\text{cm}\n$$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 13172,
    "subject": "Mathematics (Olympiad)",
    "question": "As shown in Fig. 1.1, in $\\triangle ABC$, $AB > AC$. Two points $X, Y$ in $\\triangle ABC$ are on the bisector of $\\angle BAC$ and satisfy $\\angle ABX = \\angle ACY$. Let the extension of $BX$ and segment $CY$ intersect at point $P$. The circumcircle $\\omega_1$ of $\\triangle BPY$ and the circumcircle $\\omega_2$ of $\\triangle CPX$ intersect at $P$ and another point $Q$. Prove that points $A, P, Q$ are collinear.",
    "options": [],
    "answer": "See solution",
    "solution": "By $\\angle BAX = \\angle CAY$, $\\angle ABX = \\angle ACY$, we know that $\\triangle ABX \\sim \\triangle ACY$. Therefore,\n$$\n\\frac{AB}{AC} = \\frac{AX}{AY}. \\qquad \\textcircled{1}\n$$\nAs shown in Fig. 1.2, extend $AX$ and it intersects $\\omega_1, \\omega_2$ at $U, V$, respectively. Then\n$$\n\\angle AUB = \\angle YUB = \\angle YPB = \\angle YPX = \\angle XVC = \\angle AVC,\n$$\nand thus $\\triangle ABU \\sim \\triangle ACV$. Therefore,\n$$\n\\frac{AB}{AC} = \\frac{AU}{AV}. \\qquad \\textcircled{2}\n$$\n![](images/Mathematical_Olympiad_In_China_2021_2022___Problems_and_Solutions_2024_p92_data_8e1b954b83.png)\nFrom (1) and (2), we can obtain $\\frac{AX}{AY} = \\frac{AU}{AV}$, i.e., $AU \\cdot AY = AV \\cdot AX$.\n\nThe two sides of the above equation are the circle powers of point $A$ to circles $\\omega_1, \\omega_2$, respectively. This implies that $A$ is on the radical axis (i.e., line $PQ$) of circles $\\omega_1, \\omega_2$. In other words, points $A, P, Q$ are collinear. $\\square$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 13173,
    "subject": "Mathematics (Olympiad)",
    "question": "A circle is divided into 432 congruent arcs by 432 points. The points are colored in four colors so that 108 points are colored Red, 108 points are colored Green, 108 points are colored Blue, and the remaining 108 points are colored Yellow. Prove that one can choose three points of each color in such a way that the four triangles formed by the chosen points of the same color are congruent.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $R$, $G$, $B$, and $Y$ denote the sets of Red, Green, Blue, and Yellow points, respectively. For $0 \\leq k \\leq 431$, let $\\mathcal{T}_k$ be the counterclockwise rotation by $\\frac{360k}{432}$ degrees around the center of the circle.\n\n**Step 1:** There exists an index $i_1$ such that $|\\mathcal{T}_{i_1}(R) \\cap G| \\geq 28$. For each $k$, $\\mathcal{T}_k(R) \\cap G$ consists of Green points that are images of Red points under $\\mathcal{T}_k$. The sum\n$$\ns_1 = |\\mathcal{T}_0(R) \\cap G| + |\\mathcal{T}_1(R) \\cap G| + \\dots + |\\mathcal{T}_{431}(R) \\cap G|\n$$\nis the number of pairs $(r, g)$ with $g = \\mathcal{T}_k(r)$ for some $k$. Since each $r$ and $g$ pair corresponds to a unique $k$, $s_1 = 108^2 = 11664$. By the Pigeonhole principle, there is some $i_1$ such that\n$$\n|\\mathcal{T}_{i_1}(R) \\cap G| \\geq \\left\\lfloor \\frac{11664}{431} \\right\\rfloor = 28.\n$$\nLet $RG = \\mathcal{T}_{i_1}(R) \\cap G$.\n\n**Step 2:** There exists $i_2$ such that $|\\mathcal{T}_{i_2}(RG) \\cap B| \\geq 8$. Similarly, the sum\n$$\ns_2 = |\\mathcal{T}_0(RG) \\cap B| + \\dots + |\\mathcal{T}_{431}(RG) \\cap B|\n$$\nis at least $28 \\cdot 108 = 3024$. By the Pigeonhole principle,\n$$\n|\\mathcal{T}_{i_2}(RG) \\cap B| \\geq \\left\\lfloor \\frac{3024}{430} \\right\\rfloor = 8.\n$$\nLet $RGB = \\mathcal{T}_{i_2}(RG) \\cap B$.\n\n**Step 3:** There exists $i_3$ such that $|\\mathcal{T}_{i_3}(RGB) \\cap Y| \\geq 3$. The sum\n$$\ns_3 = |\\mathcal{T}_0(RGB) \\cap Y| + \\dots + |\\mathcal{T}_{431}(RGB) \\cap Y| \\geq 8 \\cdot 108 = 864\n$$\nso\n$$\n|\\mathcal{T}_{i_3}(RGB) \\cap Y| \\geq \\left\\lfloor \\frac{864}{429} \\right\\rfloor = 3.\n$$\n\n**Construction:** Let $y_1, y_2, y_3$ be three distinct points in $\\mathcal{T}_{i_3}(RGB) \\cap Y$. Then the triples\n$$(y_1, y_2, y_3), \\quad \\mathcal{T}_{432-i_3}(y_1, y_2, y_3), \\quad \\mathcal{T}_{432-i_3-i_2}(y_1, y_2, y_3), \\quad \\mathcal{T}_{432-i_3-i_2-i_1}(y_1, y_2, y_3)$$\nform congruent triangles whose vertices are Yellow, Blue, Green, and Red, respectively, as desired.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 13174,
    "subject": "Mathematics (Olympiad)",
    "question": "We order the positive integers in two rows as follows:\n\n1 3 6 11 19 32 53 ...\n2 4 5 7 8 9 10 12 13 14 15 16 17 18 20 to 31 33 to 52 54 ...\n\nWe first write 1 in the first row, 2 in the second, and 3 in the first. After this, the following integers are written so that an individual integer is always added in the first row and blocks of consecutive integers are added in the second row, with the leading number of a block giving the number of (consecutive) integers to be written in the next block.\n\nWe name the numbers in the first row $a_1, a_2, a_3, \\dots$.\n\nDetermine an explicit formula for $a_n$.",
    "options": [],
    "answer": "See solution",
    "solution": "We first note that $a_1 = 1$, $a_2 = 3$, and $a_3 = 6$. It is straightforward to see that a block of length $a_{n-1} + 1$ starts with the number $a_n + 1$, and this block therefore ends on the number $a_n + (a_{n-1} + 1)$, which yields the recurrence:\n\n$$a_{n+1} = a_n + a_{n-1} + 2$$\n\nThis recursion has the constant solution $a_n \\equiv -2$, and the homogeneous recursion $a_{n+1} = a_n + a_{n-1}$ is of Fibonacci type. Writing the Fibonacci sequence as $F_0 = 0$, $F_1 = 1$, $F_2 = 1$, $F_3 = 2$, and so on, we can check that $a_n = F_{n+3} - 2$ holds for $n = 1, 2, 3$, and this therefore yields an explicit formula for $a_n$.\n\n$\\square$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 13175,
    "subject": "Mathematics (Olympiad)",
    "question": "Show that every graph $\\mathcal{G} = (V, E)$ with $|V| = n \\ge 1$ and $|E| > \\frac{3}{2}(n-1)$ contains, for some $r \\in \\{4, 5, \\dots, n\\}$, a cycle $C_r$ possessing exactly $r$ vertices.",
    "options": [],
    "answer": "See solution",
    "solution": "Suppose, for contradiction, that there exists a graph $\\mathcal{G} = (V, E)$ with $|V| = n \\ge 1$ and $|E| > \\frac{3}{2}(n-1)$ that contains no cycle $C_r$ for any $r \\in \\{4, 5, \\dots, n\\}$. Take such a counterexample with $|V| + |E|$ minimal.\n\nClearly, $\\mathcal{G}$ has at least four vertices, since $\\frac{3}{2}(n-1) < |E| \\le \\frac{1}{2}n(n-1)$. There must be a vertex $x \\in V$ whose degree is at most two; otherwise, $|E| \\ge \\frac{3}{2}|V|$, and removing an edge would yield a smaller counterexample.\n\nIf $x$ is isolated or has degree one, deleting $x$ (and its incident edge, if any) gives a smaller counterexample, contradicting minimality. Thus, $x$ has exactly two neighbours, say $y$ and $z$.\n\nLet $\\mathcal{G}' = (V', E')$ be the graph obtained by removing $x$, the edges $xy$ and $xz$, and, if it exists, the edge $yz$. In $\\mathcal{G}'$, there cannot be a path from $y$ to $z$, for otherwise in $\\mathcal{G}$ such a path could be completed via $x$ to a cycle of the required kind.\n\nThus, $V' = A \\cup B$ with $A, B \\neq \\emptyset$ and no edge between $A$ and $B$ in $\\mathcal{G}$. By minimality,\n\n$$\n|E'| \\le \\frac{3}{2}(|A| - 1) + \\frac{3}{2}(|B| - 1)\n$$\n\nTherefore,\n\n$$\n|E| \\le 3 + |E'| \\le \\frac{3}{2}(|A| + |B|) = \\frac{3}{2}(|V| - 1),\n$$\n\ncontradicting the hypothesis. Thus, such a graph cannot exist.",
    "topic": "Discrete Mathematics",
    "subtopic": "Other"
  },
  {
    "id": 13176,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $A = \\overline{a_5a_4a_3a_2a_1a_0}$ be a six-digit number with all digits $a_0, \\dots, a_5$ pairwise distinct. Is it possible to transpose two digits of $A$ to obtain a number divisible by $101$?",
    "options": [],
    "answer": "See solution",
    "solution": "There is nothing to prove if there are equal digits in the decimal representation of the initial number. Let $A = \\overline{a_5a_4a_3a_2a_1a_0} = 10^5 a_5 + 10^4 a_4 + 10^3 a_3 + 10^2 a_2 + 10 a_1 + a_0$, with all six digits pairwise distinct. Let $B$ be the number obtained from $A$ by transposing its two digits $a_k$ and $a_m$ ($k > m$, $a_k \\ne a_m$). Then\n\n$$\n\\begin{aligned}\nA - B &= (10^k a_k + 10^m a_m) - (10^k a_m + 10^m a_k) \\\\\n&= (10^k - 10^m)(a_k - a_m) \\\\\n&= 10^m (10^{k-m} - 1)(a_k - a_m).\n\\end{aligned}\n$$\n\nSince $101$ is a prime number, $10^m$ does not divide $101$ for any $m$, and $(a_k - a_m)$ does not divide $101$ for any $a_k \\ne a_m$. But there exist $k$ and $m$ such that $(10^{k-m} - 1)$ is divisible by $101$. Indeed, if $k-m=4$, then $10^4 - 1 = 9999 = 101 \\times 99$. Therefore, it is possible to transpose either $a_5$ and $a_1$ or $a_4$ and $a_0$ to obtain a number which is divisible by $101$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 13177,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $R$, $Y$, and $G$ denote red, yellow, and green balls, respectively. A group is called *pure* if all balls in it are the same color. How many ways are there to group the balls into three groups such that each group contains three balls, and each ball is used exactly once?",
    "options": [],
    "answer": "See solution",
    "solution": "We consider the possible cases:\n\n- **3 pure groups:** There is only 1 way to group the balls so that each group is pure.\n\n- **2 pure groups:** This is not possible; if two groups are pure, the third must also be pure.\n\n- **1 pure group:** There are 3 ways, corresponding to choosing one color (say $R$) to form a pure group. The other two groups must be $YYG$ and $YGG$.\n\n- **No pure group:** Either every group contains $R$ (2 ways: $\\{RYG, RYG, RYG\\}$ or $\\{RYG, RYY, RGG\\}$), or the grouping is $\\{RRX, RXX, XXX\\}$ where each $X$ is either $Y$ or $G$. There are 2 choices for the first $X$, then 2 choices for the 'XX' in 'RXX', giving $2 + 2 \\times 2 = 6$ ways in this case.\n\nHence, the total number of ways is $1 + 0 + 3 + 6 = 10$.",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 13178,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all functions $f: \\mathbb{Z} \\to \\mathbb{R}$ satisfying the equation\n\n$$\nf(2a) + 2f(b) = f(f(a + b)).\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "Let $f: \\mathbb{Z} \\to \\mathbb{R}$ satisfy\n$$\nf(2a) + 2f(b) = f(f(a + b)).\n$$\n\nSet $a = 1$, $b = x$:\n$$\nf(2) + 2f(x) = f(f(1 + x)).\n$$\nSet $a = 0$, $b = x + 1$:\n$$\nf(0) + 2f(x + 1) = f(f(x + 1)).\n$$\nComparing, we get\n$$\nf(2) + 2f(x) = f(0) + 2f(x + 1) \\implies f(x + 1) - f(x) = m,\n$$\nwhere $m = \\frac{f(2) - f(0)}{2}$. Thus, $f(x) = mx + c$ for constants $m, c$.\n\nSubstitute $f(x) = mx + c$ into the original equation:\n$$\n2m(a + b) + 2c = m^2(a + b) + mc.\n$$\nComparing coefficients, $m^2 = 2m$ and $mc = 2c$. So $m = 0$ or $2$.\n\nIf $m = 0$, $c = 0$, so $f(x) = 0$.\nIf $m = 2$, any $c$ works, so $f(x) = 2x + c$.\n\nBoth $f(x) = 0$ and $f(x) = 2x + c$ satisfy the equation.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 13179,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $a, b$ be two distinct odd positive integers. Define a sequence $(a_n)_{n \\ge 1}$ by\n\n$$\na_1 = a, \\quad a_2 = b, \\quad a_n \\text{ is the largest odd divisor of } a_{n-1} + a_{n-2}, \\text{ for all } n \\ge 3.\n$$\n\nShow that there exists a natural number $N$ such that $a_n = \\gcd(a, b)$ for all $n \\ge N$.",
    "options": [],
    "answer": "See solution",
    "solution": "Each $a_n$ is odd for $n \\ge 1$. Hence $a_{n-1} + a_{n-2}$ is even for $n \\ge 3$. This implies $a_n \\leq \\frac{a_{n-1} + a_{n-2}}{2}$ for all $n \\ge 3$. An easy induction shows that $a_n \\geq \\max\\{a, b\\}$ for all $n$. Thus, the number of pairs of the form $\\{a_{n-1}, a_{n-2}\\}$ is finite. It follows that $a_n$ is eventually periodic.\n\nLet $M$ be the largest number in the cycle into which the sequence settles. Let $u, v$ be the numbers preceding it. Then $u \\leq M$, $v \\leq M$, and $M \\leq \\frac{u+v}{2}$. Thus $u = v = M$. This implies that the sequence is eventually constant.\n\nLet $d = \\gcd(a, b)$. Then $d$ is odd. An easy induction proves that $d \\mid a_n$ for all $n$. Hence $d$ divides $M$. Since $a \\neq b$, it cannot be the case that $a_n = M$ for all $n$. Consider the position $a_k, M, M$ in the sequence, where $a_k \\neq M$. By definition of the sequence, we have\n\n$$\nM = \\frac{M + a_k}{2^r},\n$$\n\nwhere $r \\geq 1$. This gives $a_k = (2^r - 1)M$. Hence $M$ divides $a_k$. Now consider the position just one earlier: $a_{k-1}, a_k, M$. Again we have\n\n$$\nM = \\frac{a_{k-1} + a_k}{2^s},\n$$\n\nfor some $s \\geq 1$. This gives $a_{k-1} = 2^s M - a_k$. It follows that $M$ divides $a_{k-1}$. Since $M$ is odd, induction shows that $M$ divides $a_n$ for all $n$. Hence $M$ divides $d$. We conclude that $M = d$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 13180,
    "subject": "Mathematics (Olympiad)",
    "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 $r$, let $f_n(r)$ denote the number of ternary groups $(i, j, k)$ such that $\\frac{a_j - a_i}{a_k - a_j} = r$ for $1 \\le i < j < k \\le n$. Prove that $f_n(r) < \\frac{n^2}{4}$.",
    "options": [],
    "answer": "See solution",
    "solution": "For a fixed $j$ ($1 < j < n$), let $g_j(r)$ be the number of ternary groups $(i, j, k)$ with $1 \\le i < j < k \\le n$ and\n\n$$\n\\frac{a_j - a_i}{a_k - a_j} = r\n$$\n\nFor fixed $i, j$ with $i < j$, there is at most one $k$ satisfying the equation, so there are $j-1$ choices for $i$, giving $g_j(r) \\le j-1$. Similarly, for fixed $j, k$ with $k > j$, there is at most one $i$ satisfying the equation, so there are $n-j$ choices for $k$, giving $g_j(r) \\le n-j$. Therefore,\n\n$$\ng_j(r) \\le \\min\\{j-1, n-j\\}.\n$$\n\nIf $n$ is even ($n = 2m$),\n\n$$\n\\begin{align*}\nf_n(r) &= \\sum_{j=2}^{n-1} g_j(r) \\\\\n&= \\sum_{j=2}^{m} (j-1) + \\sum_{j=m+1}^{2m-1} (2m-j) \\\\\n&= \\frac{m(m-1)}{2} + \\frac{m(m-1)}{2} \\\\\n&= m^2 - m < m^2 = \\frac{n^2}{4}.\n\\end{align*}\n$$\n\nIf $n$ is odd ($n = 2m + 1$),\n\n$$\n\\begin{align*}\nf_n(r) &= \\sum_{j=2}^{n-1} g_j(r) \\\\\n&= \\sum_{j=2}^{m} (j-1) + \\sum_{j=m+1}^{2m} (2m+1-j) \\\\\n&= m^2 < \\frac{n^2}{4}.\n\\end{align*}\n$$\n\nThus, $f_n(r) < \\frac{n^2}{4}$ as required.",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 13181,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $\\triangle ABC$ be an acute triangle with circumcircle $\\Gamma$. Let $l$ be a tangent line to $\\Gamma$, and let $l_a$, $l_b$, and $l_c$ be the lines obtained by reflecting $l$ over the lines $BC$, $CA$, and $AB$, respectively. Show that the circumcircle of the triangle formed by the lines $l_a$, $l_b$, and $l_c$ is tangent to the circle $\\Gamma$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $P$ be the point of tangency of $l$ to $\\Gamma$. Denote the symmetric points of $P$ with respect to $BC$, $CA$, and $AB$ by $P_a$, $P_b$, and $P_c$, respectively. These points lie on circles $\\Gamma_a$, $\\Gamma_b$, and $\\Gamma_c$, which are the reflections of $\\Gamma$ over $BC$, $CA$, and $AB$, respectively. Thus, $l_a$, $l_b$, and $l_c$ are tangent to $\\Gamma_a$, $\\Gamma_b$, and $\\Gamma_c$ at $P_a$, $P_b$, and $P_c$, respectively.\n\nLet $l_a \\cap l_b = C'$, $l_b \\cap l_c = A'$, and $l_c \\cap l_a = B'$.\n\n![](images/Mathematical_Olympiad_in_China_2011-2014_p332_data_17e5d77705.png)\n\nDefine the oriented angle between lines $m$ and $n$ as $\\angle(m, n)$, which is the angle rotated anticlockwise from $m$ to $n$.\n\nWe observe the following:\n\n1. The points $P_a$, $P_b$, and $P_c$ are collinear. In fact, the midpoints of $PP_a$, $PP_b$, and $PP_c$ are the pedal points of $P$ to $BC$, $CA$, and $AB$, respectively. By Simson's Theorem, these pedal points are collinear, so $P_a$, $P_b$, and $P_c$ are collinear.\n\n2. Let the circumcircles of $\\triangle A'P_bP_c$, $\\triangle B'P_cP_a$, and $\\triangle C'P_aP_b$ be $\\Gamma_1$, $\\Gamma_2$, and $\\Gamma_3$, respectively, and let the circumcircle of $\\triangle A'B'C'$ be $\\Omega$. Then, the four circles $\\Gamma_1$, $\\Gamma_2$, $\\Gamma_3$, and $\\Omega$ are concurrent at a point. This follows from the Miquel Theorem for the complete quadrilateral $A'P_cB'P_aC'P_b$. Denote the common point by $Q$.\n\n3. The points $A$, $B$, and $C$ lie on the circles $\\Gamma_1$, $\\Gamma_2$, and $\\Gamma_3$, respectively. For example, circles $\\Gamma_b$ and $\\Gamma_c$ intersect at $A$, and $\\overrightarrow{AP_c} = \\overrightarrow{AP} = \\overrightarrow{AP_b}$. Rotating by $\\angle(P_cA, P_bA)$ about $A$, we have $\\Gamma_c \\to \\Gamma_b$, $P_c \\to P_b$, and $l_c \\to l_b$. Thus, $\\angle(l_c, l_b) = \\angle(P_cA, P_bA) = \\angle(P_cA', P_bA')$, which implies the four points $A$, $A'$, $P_b$, and $P_c$ are concyclic. Similar arguments hold for $B$ and $C$.\n\nTherefore, the circumcircle of $\\triangle A'B'C'$ is tangent to $\\Gamma$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 13182,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $a$, $b$, and $c$ be positive real numbers such that $a + b + c + 2 = abc$. Prove that\n$$\n\\frac{a}{b+1} + \\frac{b}{c+1} + \\frac{c}{a+1} \\ge 2\n$$\nholds. When does equality occur?",
    "options": [],
    "answer": "See solution",
    "solution": "First, notice that\n\n$$\n\\begin{aligned}\n(a+1)(b+1) + (a+1)(c+1) + (b+1)(c+1) &= a + b + c + (a + b + c + 2) + ab + ac + bc + 1 \\\\\n&= a + b + c + abc + ab + ac + bc + 1 = (a+1)(b+1)(c+1)\n\\end{aligned}\n$$\n\nNow, using the inequality between the arithmetic and geometric means (AM-GM),\n\n$$\n\\begin{aligned}\n\\frac{a}{b+1} + \\frac{b}{c+1} + \\frac{c}{a+1} &= \\frac{a+1}{b+1} + \\frac{b+1}{c+1} + \\frac{c+1}{a+1} - \\left(\\frac{1}{a+1} + \\frac{1}{b+1} + \\frac{1}{c+1}\\right) \\\\\n&\\ge 3 - \\left(\\frac{1}{a+1} + \\frac{1}{b+1} + \\frac{1}{c+1}\\right)\n\\end{aligned}\n$$\n\nBut since $(a+1)(b+1)(c+1) = a + b + c + ab + ac + bc + abc + 1$, and with the given condition $a + b + c + 2 = abc$, we can verify that the minimum occurs when $a = b = c = 2$. In this case,\n\n$$\n\\frac{2}{2+1} + \\frac{2}{2+1} + \\frac{2}{2+1} = 3 \\times \\frac{2}{3} = 2\n$$\n\nThus, equality holds if and only if $a = b = c = 2$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 13183,
    "subject": "Mathematics (Olympiad)",
    "question": "21 points divide a circle into 21 arcs of equal length. How many sets of 7 points are there, such that no two points have an arc distance of 3 units or 7 units?",
    "options": [],
    "answer": "See solution",
    "solution": "The answer is $126$.\n\nLet $1, 2, 3, \\ldots, 21$ be points on a circle in clockwise order. Define the following notation:\n\n$$\n\\begin{aligned}\na_1 &= 1, & a_2 &= 8, & a_3 &= 15 \\\\\nb_1 &= 4, & b_2 &= 11, & b_3 &= 18 \\\\\nc_1 &= 7, & c_2 &= 14, & c_3 &= 21 \\\\\nd_1 &= 10, & d_2 &= 17, & d_3 &= 3 \\\\\ne_1 &= 13, & e_2 &= 20, & e_3 &= 6 \\\\\nf_1 &= 16, & f_2 &= 2, & f_3 &= 9 \\\\\ng_1 &= 19, & g_2 &= 5, & g_3 &= 12\n\\end{aligned}\n$$\n\nWe can take only one $a_i$. Thus, the seven points must be $a_i, b_j, c_k, d_l, e_m, f_n, g_t$. To avoid a 3-unit arc distance, we require:\n\n$$\ni \\neq j,\\ j \\neq k,\\ k \\neq l,\\ l \\neq m,\\ m \\neq n,\\ n \\neq t,\\ t \\neq i.\n$$\n\nThe total number of such selections is $3^7$.\n\nUsing the inclusion-exclusion principle, let $A_1$ be the set where $i = j$, $A_2$ where $j = k$, and so on, up to $A_7$ where $t = i$. Then:\n\n$$\n|A_1 \\cup A_2 \\cup \\cdots \\cup A_7| = \\sum |A_i| - \\sum |A_i \\cap A_j| + \\cdots - |A_1 \\cap A_2 \\cap \\cdots \\cap A_7|.\n$$\n\nWhere:\n\n$$\n\\begin{aligned}\n\\sum |A_i| &= \\binom{7}{1} \\cdot 3^6 \\\\\n\\sum |A_i \\cap A_j| &= \\binom{7}{2} \\cdot 3^5 \\\\\n\\sum |A_i \\cap A_j \\cap A_k| &= \\binom{7}{3} \\cdot 3^4 \\\\\n\\sum |A_i \\cap A_j \\cap A_k \\cap A_l| &= \\binom{7}{4} \\cdot 3^3 \\\\\n\\sum |A_i \\cap A_j \\cap A_k \\cap A_l \\cap A_m| &= \\binom{7}{5} \\cdot 3^2 \\\\\n\\sum |A_i \\cap A_j \\cap A_k \\cap A_l \\cap A_m \\cap A_n| &= \\binom{7}{6} \\cdot 3 \\\\\n|A_1 \\cap A_2 \\cap \\cdots \\cap A_7| &= 3\n\\end{aligned}\n$$\n\nTherefore, the number of valid configurations is:\n\n$$\n3^7 - \\binom{7}{1} 3^6 + \\binom{7}{2} 3^5 - \\binom{7}{3} 3^4 + \\binom{7}{4} 3^3 - \\binom{7}{5} 3^2 + \\binom{7}{6} 3 - 3 = (3-1)^7 - 2 = 126.\n$$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 13184,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $p = M - 1$ be a prime number and let us denote the $p^2$ vertices by pairs $(i, j)$ with $i, j \\in \\{0, 1, \\dots, p-1\\}$. We denote the colours by numbers $0, 1, \\dots, p$.\n\nWe colour an edge between vertices $(i_1, j_1)$ and $(i_2, j_2)$ in colour $k \\in \\{0, 1, \\dots, p-1\\}$ if and only if $j_1 - j_2 \\equiv (i_1 - i_2)k \\pmod{p}$ and in colour $p$ if and only if $i_1 = i_2$.\n\nLet $(i_1, j_1)$, $(i_2, j_2)$ and $(i_3, j_3)$ be three vertices of a triangle such that the edge between $(i_1, j_1)$ and $(i_2, j_2)$ and the edge between $(i_2, j_2)$ and $(i_3, j_3)$ have the same colour, say $k$. Show that the edge between $(i_1, j_1)$ and $(i_3, j_3)$ is also coloured in colour $k$.",
    "options": [],
    "answer": "See solution",
    "solution": "Since $p$ is prime, the colouring is well-defined. For two different pairs $(i_1, j_1)$ and $(i_2, j_2)$, if $j_1 - j_2 \\equiv (i_1 - i_2)k \\equiv (i_1 - i_2)l \\pmod{p}$ for colours $k$ and $l$, then $p$ divides $(i_1 - i_2)(k - l)$ and $i_1 \\neq i_2$ implies $k = l$. If $i_1 = i_2$, then $j_1 - j_2 \\equiv (i_1 - i_2)k = 0 \\pmod{p}$ would imply $j_1 = j_2$, so this is not possible and in this case the edge between $(i_1, j_1)$ and $(i_2, j_2)$ is coloured by colour $p$.\n\nFor $0 \\le k \\le p-1$, we have:\n\n$$\nj_1 - j_3 = (j_1 - j_2) + (j_2 - j_3) \\equiv (i_1 - i_2)k + (i_2 - i_3)k = (i_1 - i_3)k \\pmod{p}\n$$\n\nand for $k = p$ we have $i_1 = i_2 = i_3$. In each case, the edge between $(i_1, j_1)$ and $(i_3, j_3)$ is also coloured in colour $k$, so the colouring is consistent for triangles.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 13185,
    "subject": "Mathematics (Olympiad)",
    "question": "Four frogs are positioned at four points on a straight line such that the distance between any two neighboring points is one unit. Each frog can jump to its reflection across any of the other three frogs. Prove that there is no way for the frogs to end up so that the distances between every pair of neighboring frogs are all equal to $2008$ units.",
    "options": [],
    "answer": "See solution",
    "solution": "Assume the frogs start at positions $1$, $2$, $3$, and $4$ on the real number line. After any jump (reflection across another frog), frogs at odd-numbered positions remain at odd positions, and those at even positions remain at even positions. Thus, after any sequence of jumps, there will always be two frogs at odd positions and two at even positions. For all neighboring distances to be $2008$, all frogs would need to be at either all odd or all even positions, which is impossible. Therefore, the required configuration cannot be achieved.",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 13186,
    "subject": "Mathematics (Olympiad)",
    "question": "A subset of $\\{1, 2, 3, \\dots, 30\\}$ is called *delicious* if it does not contain elements $a$ and $b$ satisfying $a = 3b$. A delicious subset is called *super delicious* if it is delicious and no delicious set has more elements than it has. Determine the number of super delicious subsets.",
    "options": [],
    "answer": "See solution",
    "solution": "Partition the set $\\{1, 2, 3, \\dots, 30\\}$ into 20 subsets as follows:\n\n$$\n\\begin{aligned}\n& \\{1, 3, 9, 27\\}, \\\\\n& \\{2, 6, 18\\}, \\\\\n& \\{4, 12\\}, \\{5, 15\\}, \\{7, 21\\}, \\{8, 24\\}, \\{10, 30\\}, \\\\\n& \\{11\\}, \\{13\\}, \\{14\\}, \\{16\\}, \\{17\\}, \\{19\\}, \\{20\\}, \\\\\n& \\{22\\}, \\{23\\}, \\{25\\}, \\{26\\}, \\{28\\}, \\{29\\}.\n\\end{aligned}\n$$\n\nA subset of $\\{1, 2, 3, \\dots, 30\\}$ is delicious if and only if it does not contain two elements from any one of the 20 sets listed above such that one is three times the other. Thus, a delicious set contains at most 2 elements from each of the sets in the first two rows, and at most 1 element from each of the sets in the last two rows. It is possible for a delicious set to have exactly 2 elements from each of the sets in the first two rows, and exactly 1 element from each of the sets in the last two rows; therefore, a super delicious set must have this property. So there are\n\n$$\n3 \\cdot 1 \\cdot 2^5 \\cdot 1^{13} = 96\n$$\n\nsuper delicious sets, because there are 3 ways to choose 2 non-consecutive elements from $\\{1, 3, 9, 27\\}$, and 1 way to choose 2 non-consecutive elements from $\\{2, 6, 18\\}$, and so on.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 13187,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $m$ be a positive integer and denote by $a_m$ the number of possible ways of writing down a sequence of positive integers satisfying the following, which we call the conditions $C_m$:\n\n$C_m$: Start with writing down the number $m$, and end up with writing down the number $1$, and after writing down a number $n$, follow with writing a positive integer less than or equal to $\\sqrt{n}$.\n\nFind $a_{2012}$.",
    "options": [],
    "answer": "See solution",
    "solution": "We have:\n\n$$\na_1 = 1, \\quad a_m = a_1 + a_2 + \\cdots + a_k,\n$$\nwhere $k$ is a positive integer satisfying $k^2 + 1 \\le m \\le (k+1)^2$.\n\nSince $44^2 + 1 = 1937 < 2012 < 2025 = 45^2$, we have $a_{2012} = a_1 + a_2 + \\cdots + a_{44}$.\n\nNow, compute $a_1, a_2, \\ldots, a_{44}$:\n\n$$\na_1 = 1,\n$$\n$$\na_2 = a_3 = a_4 = a_1 = 1,\n$$\n$$\na_5 = a_6 = \\cdots = a_9 = a_1 + a_2 = 2,\n$$\n$$\na_{10} = a_{11} = \\cdots = a_{16} = a_1 + a_2 + a_3 = 3,\n$$\n$$\na_{17} = a_{18} = \\cdots = a_{25} = a_1 + a_2 + a_3 + a_4 = 4,\n$$\n$$\na_{26} = a_{27} = \\cdots = a_{36} = a_1 + a_2 + a_3 + a_4 + a_5 = 6,\n$$\n$$\na_{37} = a_{38} = \\cdots = a_{44} = a_1 + a_2 + a_3 + a_4 + a_5 + a_6 = 8.\n$$\n\nThus,\n\n$$\n\\begin{align*}\na_{2012} &= a_1 + a_2 + \\cdots + a_{44} \\\\\n&= (a_1 + \\cdots + a_4) + (a_5 + \\cdots + a_9) + (a_{10} + \\cdots + a_{16}) \\\\\n&\\quad + (a_{17} + \\cdots + a_{25}) + (a_{26} + \\cdots + a_{36}) + (a_{37} + \\cdots + a_{44}) \\\\\n&= 4 \\times 1 + 5 \\times 2 + 7 \\times 3 + 9 \\times 4 + 11 \\times 6 + 8 \\times 8 \\\\\n&= 4 + 10 + 21 + 36 + 66 + 64 = 201.\n\\end{align*}\n$$\n\nSo, $201$ is the desired answer.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 13188,
    "subject": "Mathematics (Olympiad)",
    "question": "A sequence of real numbers $\\{a_n\\}$ is defined by\n\n- $a_0 \\neq 0, 1$\n- $a_1 = 1 - a_0$\n- $a_{n+1} = 1 - a_n(1 - a_n)$ for $n = 1, 2, \\dots$\n\nProve that for any positive integer $n$,\n\n$$\na_0 a_1 \\cdots a_n \\left( \\frac{1}{a_0} + \\frac{1}{a_1} + \\cdots + \\frac{1}{a_n} \\right) = 1.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "From the given condition, we have\n\n$$\n1 - a_{n+1} = a_n(1 - a_n) = a_n a_{n-1} (1 - a_{n-1}) = \\cdots = a_n \\cdots a_1 (1 - a_1) = a_n \\cdots a_1 a_0,\n$$\n\ni.e. $a_{n+1} = 1 - a_0 a_1 \\cdots a_n$, for $n = 1, 2, \\dots$\n\nBy mathematical induction, when $n=1$ the proposition holds. Assuming that it holds for $n=k$, then when $n=k+1$ we have\n\n$$\n\\begin{aligned}\n& a_0 a_1 \\cdots a_{k+1} \\left( \\frac{1}{a_0} + \\frac{1}{a_1} + \\cdots + \\frac{1}{a_k} + \\frac{1}{a_{k+1}} \\right) \\\\\n&= a_0 a_1 \\cdots a_k \\left( \\frac{1}{a_0} + \\frac{1}{a_1} + \\cdots + \\frac{1}{a_k} \\right) a_{k+1} + a_0 a_1 \\cdots a_k \\\\\n&= a_{k+1} + a_0 a_1 \\cdots a_k \\\\\n&= 1.\n\\end{aligned}\n$$\n\nSo it also holds when $n=k+1$. Hence, it holds for any positive integer $n$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 13189,
    "subject": "Mathematics (Olympiad)",
    "question": "For a positive integer $n$, let $d(n)$ denote the number of positive divisors of $n$ (including 1 and $n$ itself).\n\nA number $k$ is called *good* if there exists a positive integer $n$ such that $k = \\frac{n}{d(n)}$.\n\nAre all numbers from 1 to 18 good? If not, which are good and which are not?",
    "options": [],
    "answer": "See solution",
    "solution": "Firstly, 1 and 2 are good, since $1 = \\frac{2}{d(2)}$ and $$2 = \\frac{8}{d(8)}.$$ \n\nIf $p$ is an odd prime, then $p$ is good because $d(8p) = 8$. Thus, 3, 5, 7, 11, 13, 17 are good.\n\nIf $p$ is an odd prime, then $2p$ is good because $d(2^2 \\cdot 3^2 p) = 3 \\cdot 3 \\cdot 2$. Thus, 6, 10, 14 are good.\n\nAlso, $$4 = \\frac{36}{d(36)}, \\quad 8 = \\frac{96}{d(96)}, \\quad 9 = \\frac{108}{d(108)},$$\n$$12 = \\frac{240}{d(240)}, \\quad 15 = \\frac{360}{d(360)}, \\quad 16 = \\frac{128}{d(128)}.$$ \n\nThus, the numbers 1, 2, ..., 17 are good.\n\nTo show that 18 is not good, suppose $18 = \\frac{n}{d(n)}$, i.e., $n = 18 d(n)$ 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 primes $>3$ and $a, b, k_1, \\ldots, k_m$ are positive integers):\n\n$$2^{a-1} \\cdot 3^{b-1} \\cdot p_1^{k_1} \\cdots p_m^{k_m} = (a+1)(b+2)(k_1+1)\\cdots(k_m+1).$$\n\nFor any odd prime $p$ and positive integer $k$, $p^k > k+1$.\n\nCombining, $2^{a-1} \\cdot 3^{b-1} < (a+1)(b+2)$, or $f(a) = \\frac{2^{a-1}}{a+1} < \\frac{b+2}{3^{b-1}} = g(b)$.\n\nCalculating:\n- $f(1) = \\frac{1}{2}$, $f(2) = \\frac{2}{3}$, $f(3) = 1$, $f(4) = \\frac{8}{5}$, $f(5) = \\frac{16}{6}$, $f(a) \\ge \\frac{32}{7} > 4$ for $a \\ge 6$.\n- $g(1) = 3$, $g(2) = \\frac{4}{3}$, $g(3) < \\frac{5}{9}$, $g(b) < \\frac{2}{9}$ for $b \\ge 4$.\n\nThus, only $b \\le 3$ is possible.\n\nIf $b=3$, then $(a, b) = (1, 3)$, and the equation becomes $9p_1^{k_1} \\cdots p_m^{k_m} = 10(k_1 + 1)(k_m + 1)$, implying $p_1 = 5$. But $\\frac{p_1^{k_1}}{k_1+1} = \\frac{5^{k_1}}{k_1+1} \\ge \\frac{5}{2}$ for $k_1 \\ge 1$, so no solution exists.\n\nIf $b=2$, possible $(a, b)$ are $(1,2)$, $(2,2)$, $(3,2)$. The equation becomes $3 \\cdot 2^{a-1} p_1^{k_1} \\cdots p_m^{k_m} = 4(a+1)(k_1+1) \\cdots (k_m+1)$. For divisibility, $a \\ge 3$, but then $4(a+1)$ divides $2^{a-1}$, which is impossible.\n\nIf $b=1$, possible $(a, b)$ are $(1,1)$, $(2,1)$, $(3,1)$, $(4,1)$, $(5,1)$. The equation becomes $2^{a-1} p_1^{k_1} \\cdots p_m^{k_m} = 3(a+1)(k_1+1) \\cdots (k_m+1)$, which is impossible since $p_i > 3$.\n\nTherefore, there is no $n$ such that $18 = \\frac{n}{d(n)}$; 18 is not good.\n\n**Note:** For primes $p \\ge 5$, both $3p$ and $p^2$ are good, since $d(2^2 \\cdot 3p) = 3 \\cdot 2 \\cdot 2$ and $d(2^3 \\cdot 3 \\cdot p^2) = 4 \\cdot 2 \\cdot 3$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 13190,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $\\mathbb{R}$ be the set of real numbers. Let $\\mathcal{F}$ be the set of all functions $f : \\mathbb{R} \\to \\mathbb{R}$ such that\n\n$$\nf(x + f(y)) = f(x) + f(y)\n$$\n\nfor every $x, y \\in \\mathbb{R}$. Find all rational numbers $q$ such that for every function $f \\in \\mathcal{F}$, there exists some $z \\in \\mathbb{R}$ satisfying $f(z) = qz$.",
    "options": [],
    "answer": "See solution",
    "solution": "The desired set of rational numbers is $\\left\\{ \\frac{n+1}{n} : n \\in \\mathbb{Z},\\ n \\neq 0 \\right\\}$.\n\nLet $Z$ be the set of rational numbers $q$ such that for any $f \\in \\mathcal{F}$, there exists some $z \\in \\mathbb{R}$ such that $f(z) = qz$. Further, let\n\n$$\nS = \\left\\{ \\frac{n+1}{n} : n \\in \\mathbb{Z},\\ n \\neq 0 \\right\\}\n$$\n\nWe prove that $Z = S$ by showing the two inclusions $S \\subseteq Z$ and $Z \\subseteq S$.\n\n**First, $S \\subseteq Z$:**\nLet $f \\in \\mathcal{F}$ and let $P(x, y)$ denote the relation $f(x + f(y)) = f(x) + f(y)$. Let $f(0) = c$. $P(0, 0)$ gives $f(c) = 2c$. We claim that $f(kc) = (k + 1)c$ for all $k \\ge 1$. We prove this by induction on $k$. Base case $k = 1$ is already proved. Assume this is true for some $k$, then $P(0, kc)$ gives $f((k + 1)c) = (k + 2)c$, as required. Thus the claim is proved.\n\nWe now claim that $f(-kc) = (-k + 1)c$ for every integer $k \\ge 1$. Indeed, $P(-kc, kc)$ gives\n\n$$\nf(c) = f(-kc) + (k+1)c \\implies f(-kc) = (-k+1)c\n$$\n\nThus $f(kc) = (k + 1)c = \\frac{k+1}{k} \\cdot (kc)$ is true for every integer $k \\neq 0$, so $\\frac{k+1}{k} \\in Z$ for every $k \\neq 0$. Thus $S \\subseteq Z$.\n\n**Now, $Z \\subseteq S$:**\nLet $p$ be a rational number outside $S$. We want to prove $p \\notin Z$ by constructing an $f \\in \\mathcal{F}$ with $f(z) \\neq pz$ for every $z \\in \\mathbb{R}$. The strategy is to first construct a function $g : [0, 1) \\to \\mathbb{Z}$, and then define $f(x) = g(\\{x\\}) + \\lfloor x \\rfloor$. This $f \\in \\mathcal{F}$; indeed:\n\n$$\n\\begin{align*}\nf(x + f(y)) &= g(\\{x + f(y)\\}) + \\lfloor x + f(y) \\rfloor \\\\\n&= g(\\{x + g(\\{y\\}) + \\lfloor y \\rfloor\\}) + \\lfloor x + g(\\{y\\}) + \\lfloor y \\rfloor \\rfloor \\\\\n&= g(\\{x\\}) + \\lfloor x \\rfloor + g(\\{y\\}) + \\lfloor y \\rfloor \\\\\n&= f(x) + f(y)\n\\end{align*}\n$$\n\nwhere we used that $g$ only takes integer values.\n\n**Claim 1:** For every $\\alpha \\in [0, 1)$, there exists an $m \\in \\mathbb{Z}$ such that\n\n$$\nm + n \\neq p(\\alpha + n)\n$$\n\nfor every $n \\in \\mathbb{Z}$.\n\n*Proof.* If $p = 1$, we can just take $m = 1$ since $\\alpha < 1$. If $p \\neq 1$, the claim is equivalent to the existence of an integer $m$ such that\n\n$$\n\\frac{m - p\\alpha}{p - 1}\n$$\n\nis not an integer. Assume the contrary. That would mean that\n\n$$\n\\frac{m - p\\alpha}{p - 1} \\text{ and } \\frac{(m + 1) - p\\alpha}{p - 1}\n$$\n\nare both integers, and so is their difference. The latter is equal to $\\frac{1}{p-1}$. But since we assumed $p \\notin S$, $\\frac{1}{p-1} \\notin \\mathbb{Z}$, contradiction! $\\square$\n\nDefine $g: [0, 1) \\to \\mathbb{Z}$ by $g(\\alpha) = m$ for any integer $m$ satisfying Claim 1 for $\\alpha$. Then $f(z) \\neq pz$ if and only if\n\n$$\ng(\\{z\\}) + \\lfloor z \\rfloor \\neq p(\\{z\\} + \\lfloor z \\rfloor)\n$$\n\nBut this is guaranteed by the construction of the function $g$. We conclude that $p \\notin Z$. Thus $Z \\subseteq S$, as required. $\\square$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 13191,
    "subject": "Mathematics (Olympiad)",
    "question": "For integers $k$, $\\ell$ and a positive integer $m$, we write $k \\equiv \\ell \\pmod{m}$ to mean that $k - \\ell$ is divisible by $m$.\n\nLet $a_0 = 25^{2015}$. For a non-negative integer $n$, define $a_n$ as follows: for each $n \\geq 0$, let $b_n$ be the lowest digit of $a_n$, and set\n\n$$\na_{n+1} = \\frac{a_n - b_n}{10} + 4b_n = \\frac{a_n + 39b_n}{10}$$\n\nShow that $a_{10000}$ is a positive integer less than or equal to $39$, and determine its value.",
    "options": [],
    "answer": "See solution",
    "solution": "$$a_{10000} \\equiv 1 \\pmod{3}$$\n\n$$a_{10000} \\equiv 4 \\cdot 64^{3333} \\cdot (-1)^{2015} \\equiv 4 \\cdot (-1)^{3333} \\cdot (-1)^{2015} \\equiv 4 \\pmod{13}$$\n\nThus, $a_{10000}$ is a positive integer less than or equal to $39$ with remainder $1$ when divided by $3$ and remainder $4$ when divided by $13$. The only such integer is $4$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 13192,
    "subject": "Mathematics (Olympiad)",
    "question": "Prove that for any integer $n$ there is a monic quadratic polynomial $x^2 + bx + c$ with integer coefficients which attains values $n$, $n^2$, $n^3$ at some three integer points.",
    "options": [],
    "answer": "See solution",
    "solution": "We seek a polynomial $f(x) = x^2 + bx + c$ such that\n\n$$\nf(x_1) = n, \\quad f(x_2) = n^2, \\quad f(x_3) = n^3.\n$$\n\nLet $g(x) = f(x) - n$. Then $g(x_1) = 0$, $g(x_2) = n^2 - n$, $g(x_3) = n^3 - n$. Since $g(x)$ is quadratic, $g(x) = (x - x_1)(x - t)$ for some integer $t$.\n\nWe require:\n\n$$\n\\begin{aligned}\n&x_2 - x_1 = 1, \\\\\n&x_2 - t = n^2 - n, \\\\\n&x_3 - x_1 = n, \\\\\n&x_3 - t = n^2 - 1.\n\\end{aligned}\n$$\n\nFor example, setting $x_1 = 0$, $x_2 = 1$, $x_3 = n$, and $t = -n^2 + n + 1$ satisfies these relations. Thus, the polynomial\n\n$$\nf(x) = x(x + n^2 - n - 1) + n\n$$\n\nattains the desired values at $x = 0$, $x = 1$, and $x = n$.",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 13193,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all right-angled triangles with integer side lengths in which one can inscribe two congruent circles that satisfy the following conditions:\n\n- Their radius is a prime number.\n- The circles touch externally.\n- Both circles are tangent to the hypotenuse, and each is tangent to a different leg of the triangle.",
    "options": [],
    "answer": "See solution",
    "solution": "We will show that there exists only one such triangle, with sides of length $21$, $28$, $35$, and the two touching circles have radius $5$.\n\nIn any right-angled triangle $ABC$ with hypotenuse $AB$, let $a = BC$, $b = AC$, and $c = AB$. Two congruent circles with the required tangency conditions exist. Denote them $k_1(S_1, r)$ and $k_2(S_2, r)$, where $k_1$ is tangent to $AC$ and $k_2$ is tangent to $BC$. Let the points of tangency of these circles to $AB$ be $T_1$ and $T_2$, respectively. Consider the incircle $k(S, \\varrho)$ of triangle $ABC$, and let $D$ be the point where $k$ is tangent to $AB$.\n\n![](images/CZE_ABooklet_2024_p14_data_d072271dd3.png)\n\nFirst, we show that the radius $r$ of circles $k_1$ and $k_2$ is given by\n\n$$\nr = \\frac{c(a + b - c)}{2(a + b)} \\qquad (1)\n$$\n\nA homothety with center $A$ and ratio $r/\\varrho$ maps $k$ to $k_1$, sending $D$ to $T_1$, so $|AT_1| = |AD| \\cdot r/\\varrho$. Similarly, $|BT_2| = |BD| \\cdot r/\\varrho$. Since $T_1T_2S_2S_1$ is a rectangle, $|T_1T_2| = |S_1S_2| = 2r$. Plugging into $c = |AT_1| + |T_1T_2| + |BT_2|$ and using $|AD| + |BD| = c$ gives\n\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\nThus,\n\n$$\nr = \\frac{c\\varrho}{c + 2\\varrho}.\n$$\n\nUsing $\\varrho = (a + b - c)/2$, we obtain formula (1). Suppose $a, b, c, r$ are integers and let $k = \\gcd(a, b, c)$. Then $a = k a_1$, $b = k b_1$, $c = k c_1$, where $a_1^2 + b_1^2 = c_1^2$ and $a_1, b_1, c_1$ are pairwise coprime, with $a_1$ and $b_1$ of different parity. $c_1$ is odd and $a_1 + b_1 - c_1$ is even. Plugging into (1):\n\n$$\nr = \\frac{k c_1 (a_1 + b_1 - c_1)}{2(a_1 + b_1)} = \\frac{k}{a_1 + b_1} \\cdot c_1 \\cdot \\frac{a_1 + b_1 - c_1}{2}. \\qquad (2)\n$$\n\nWe show that both fractions on the right are integers. If a prime $p$ divides both $a_1 + b_1$ and $c_1$, then from $(a_1 + b_1)^2 = c_1^2 + 2a_1b_1$, $p$ divides $a_1b_1$, which is impossible since $a_1, b_1, c_1$ are coprime. Thus, $a_1 + b_1$ is coprime to $c_1$ and $a_1 + b_1 - c_1$. Since $r$ is an integer, the first fraction is an integer.\n\nBy the problem, $r$ is a prime. By (2), it is a product of three positive integers. Since $c_1 > a_1 \\geq 1$, we must have $c_1 = r$ and the other two factors equal $1$. Thus, $k = a_1 + b_1$, $a_1 + b_1 - c_1 = 2$, and $k = a_1 + b_1 = c_1 + 2 = r + 2$. Therefore,\n\n$$\n2a_1b_1 = (a_1 + b_1)^2 - c_1^2 = (r + 2)^2 - r^2 = 4r + 4.\n$$\n\nDividing by $2$:\n\n$$\na_1b_1 = 2r + 2 = 2(a_1 + b_1 - 2) + 2 = 2a_1 + 2b_1 - 2.\n$$\n\nThis can be rewritten as $(a_1 - 2)(b_1 - 2) = 2$. Thus, $\\{a_1, b_1\\} = \\{3, 4\\}$, $c_1 = r = a_1 + b_1 - 2 = 5$, and $k = r + 2 = 7$. For the right-angled triangle with sides $7 \\cdot 3 = 21$, $7 \\cdot 4 = 28$, and $7 \\cdot 5 = 35$, we have $r = 5$. This completes the proof.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 13194,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $a$, $b$, $c$ be positive real numbers with $\\frac{1}{a} + \\frac{1}{b} + \\frac{1}{c} = 1$. Prove that\n\n$$\n3(ab + bc + ca) + \\frac{9}{a + b + c} \\geq \\frac{9abc}{a + b + c} + 2(a^2 + b^2 + c^2) + 1\n$$\n\nand find all numbers $a$, $b$, $c$ for which equality holds.",
    "options": [],
    "answer": "See solution",
    "solution": "By symmetry, we can assume $a \\geq b \\geq c$. The claim is equivalent to\n\n$$\n\\begin{aligned}\n& a^2 + b^2 + c^2 + \\frac{9}{a+b+c} \\geq \\frac{9abc}{a+b+c} + 3(a^2 + b^2 + c^2 - ab - bc - ca) + 1, \\\\\n& (a+b+c)(a^2 + b^2 + c^2) + 9 \\geq 9abc + 3(a+b+c)(a^2 + b^2 + c^2 - ab - bc - ca) + (a+b+c), \\\\\n&= 3(a^3 + b^3 + c^3) + (a+b+c), \\\\\n& (a+b+c)(a^2 + b^2 + c^2 - 1) \\geq 3(a^3 + b^3 + c^3 - 3), \\\\\n& \\frac{a+b+c}{3} \\cdot \\frac{(a^2 - \\frac{1}{a}) + (b^2 - \\frac{1}{b}) + (c^2 - \\frac{1}{c})}{3} \\geq \\frac{(a^3 - 1) + (b^3 - 1) + (c^3 - 1)}{3}.\n\\end{aligned}\n$$\n\nNote that $a^2 - \\frac{1}{a} \\geq b^2 - \\frac{1}{b} \\geq c^2 - \\frac{1}{c}$, so by Chebyshev's inequality, the last line holds.\n\nEquality occurs only if $a = b = c$, so each equals $3$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 13195,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $ABC$ be a triangle with $AB = AC$. Let $M$ be the midpoint of $BC$. Let the circles with diameters $AC$ and $BM$ intersect at points $M$ and $P$. Let $MP$ intersect $AB$ at $Q$. Let $R$ be a point on $AP$ such that $QR \\parallel BP$. Prove that $CP$ bisects $\\angle RCB$.",
    "options": [],
    "answer": "See solution",
    "solution": "Since $ACMP$ is cyclic, we have $\\angle RPQ = \\angle ACM$. Moreover, $\\angle PQR = 90^\\circ = \\angle CMA$. Hence $\\triangle PQR \\sim \\triangle CMA$. It follows that\n\n$$\n\\frac{PR}{PQ} = \\frac{CA}{CM}. \\qquad (1)\n$$\n\nSimilarly, $\\angle PMB = \\angle PAC$ and $\\angle BPM = 90^\\circ = \\angle CPA$. Hence $\\triangle BPM \\sim \\triangle CPA$. It follows that\n\n$$\n\\frac{CA}{BM} = \\frac{CP}{BP}. \\qquad (2)\n$$\n\nUsing (1), (2), and the equality $BM = CM$ we obtain\n\n$$\n\\frac{PR}{PQ} = \\frac{CA}{CM} = \\frac{CA}{BM} = \\frac{CP}{BP}. \\qquad (3)\n$$\n\nUsing (3) and $\\angle QPB = 90^\\circ = \\angle RPC$ we obtain $\\triangle QPB \\sim \\triangle RPC$. In particular, $\\angle RCP = \\angle PBQ$. Hence\n\n$$\n\\angle RCP = \\angle PBQ = \\angle CBA - \\angle MBP = \\angle ACB - \\angle ACP = \\angle PCB.\n$$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 13196,
    "subject": "Mathematics (Olympiad)",
    "question": "Determine the integer $n$ for which $A - B$, where $A = \\sqrt{n^2 + 24}$ and $B = \\sqrt{n^2 - 9}$, is an integer.",
    "options": [],
    "answer": "See solution",
    "solution": "We require $n^2 - 9 \\geq 0$, so $n \\geq 3$ or $n \\leq -3$.\n\nLet $A - B = \\sqrt{n^2 + 24} - \\sqrt{n^2 - 9} = d \\in \\mathbb{Z}$. Then $d > 0$ and\n\n$$\n\\begin{aligned}\n\\sqrt{n^2 + 24} &= \\sqrt{n^2 - 9} + d \\\\\n\\Rightarrow n^2 + 24 &= n^2 - 9 + d^2 + 2d\\sqrt{n^2 - 9} \\\\\n\\Rightarrow 33 &= d^2 + 2d\\sqrt{n^2 - 9} \\\\\n\\Rightarrow d^2 = 33 - 2d\\sqrt{n^2 - 9} < 33 \\Rightarrow d \\in \\{1, 2, 3, 4, 5\\}.\n\\end{aligned}\n$$\n\nMoreover,\n\n$$\n\\sqrt{n^2 - 9} = \\frac{33 - d^2}{2d} \\Rightarrow n^2 = \\left(\\frac{33 - d^2}{2d}\\right)^2 + 9. \\quad (1)\n$$\n\nIf $d$ is even ($d = 2$ or $4$), then from (1) $n^2 \\notin \\mathbb{Z}$, which is impossible. For $d = 3$ or $5$, also $n^2 \\notin \\mathbb{Z}$, while for $d = 1$, we find $n = \\pm 5$. Therefore, the integer we seek is $5$ or $-5$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 13197,
    "subject": "Mathematics (Olympiad)",
    "question": "Given a point $D$ on side $BC$ of triangle $ABC$, let $\\omega_1(O_1, r_1)$ and $\\omega_2(O_2, r_2)$ be the incircles of triangles $ACD$ and $ADB$, respectively. Let $\\rho_1(O_3, r_3)$ and $\\rho_2(O_4, r_4)$ be circles inscribed in these triangles externally, tangent at an internal point of $BC$. Prove that the lines $O_1O_2$, $O_3O_4$, and $BC$ pass through a common point.",
    "options": [],
    "answer": "See solution",
    "solution": "![](images/MONGOLIAN_MATHEMATICAL_OLYMPIAD-2014_p33_data_02910f65e0.png)\n\n**Proof of lemma:**\n\nDenote $BC \\cap O_3O_4 = X$ and let's prove that $X \\in O_1O_2$. From the well-known property that a line passing through centers of circles inscribed in an angle is the bisector of the angle, it follows that $A = (O_1O_3) \\cap (O_2O_4)$ and $D = (O_1O_4) \\cap (O_2O_3)$. Since the line $O_3O_2$ intersects sides of triangle $AO_1O_4$,\n\n$$\n\\text{By Menelaus' theorem we get } 1 = \\frac{O_3O_1}{O_3A} \\cdot \\frac{DO_4}{DO_1} \\cdot \\frac{O_2A}{O_2O_4} = \\frac{O_3A - O_1A}{O_3A} \\cdot \\frac{DO_4}{DO_1} \\cdot \\frac{O_2A}{O_4A - O_2A}\n$$\n\n$$\n= \\left(1 - \\frac{O_1A}{O_3A}\\right) \\cdot \\frac{DO_4}{DO_1} \\cdot \\left(\\frac{1}{\\frac{O_4A}{O_2A} - 1}\\right) = \\left(1 - \\frac{r_1}{r_3}\\right) \\left(\\frac{r_4}{r_1}\\right) \\left(\\frac{1}{\\frac{r_4}{r_2} - 1}\\right)\n$$\n\n$$\n= \\frac{(r_3 - r_1)}{r_4 - r_2} \\cdot \\frac{r_4 r_2}{r_3 r_1}. \\quad (*)\n$$\n\nTo prove $X \\in O_1O_2$ is equivalent to proving by Menelaus' theorem:\n\n$$\n\\frac{XO_3}{XO_4} \\cdot \\frac{O_1A}{O_1O_3} \\cdot \\frac{O_2O_4}{O_2A} = 1.\n$$\n\nFurthermore,\n\n$$\n\\frac{XO_3}{XO_4} = \\frac{r_3}{r_4}, \\quad \\frac{O_1A}{O_1O_3} = \\frac{O_1A}{O_3A - AO_1} = \\frac{1}{\\frac{O_3A}{O_1A} - 1} = \\frac{1}{\\frac{r_3}{r_1} - 1} = \\frac{r_1}{r_3 - r_1},\n$$\n\n$$\n\\frac{O_2O_4}{O_2A} = \\frac{AO_4 - AO_2}{O_2A} = \\frac{AO_4}{AO_2} - 1 = \\frac{r_4}{r_2} - 1 = \\frac{r_4 - r_2}{r_2}\n$$\n\nand by $(*)$,\n\n$$\n\\frac{XO_3}{XO_4} \\cdot \\frac{O_1A}{O_1O_3} \\cdot \\frac{O_2O_4}{O_2A} = \\frac{r_3}{r_4} \\cdot \\frac{r_1}{r_3 - r_1} \\cdot \\frac{r_4 - r_2}{r_2} = 1.\n$$\n\nThis proves that $X \\in O_1O_2$ and the lines $O_1O_2$, $O_3O_4$, and $BC$ pass through a common point.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 13198,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $ABCDE$ be a convex pentagon with five equal sides and right angles at $C$ and $D$. Let $P$ denote the intersection point of the diagonals $AC$ and $BD$.\n\nProve that the segments $PA$ and $PD$ have the same length.\n\n![](images/Austria_2016_Booklet_p8_data_22c913b324.png)",
    "options": [],
    "answer": "See solution",
    "solution": "$BCDE$ is a square since $\\overline{BC} = \\overline{CD} = \\overline{DE}$ and $BC \\perp CD$, $CD \\perp DE$. Hence also the length of the segment $BE$ coincides with the side length of the pentagon $ABCDE$ and we have $BC \\perp BE$ and $BE \\perp DE$. Furthermore, $\\overline{AB} = \\overline{AE} = \\overline{BE}$, hence $ABE$ is an equilateral triangle. Now we have\n\n$$\n\\angle CBA = \\angle CBE + \\angle EBA = 90^\\circ + 60^\\circ = 150^\\circ,\n$$\n\n$$\n\\angle AED = \\angle AEB + \\angle BED = 60^\\circ + 90^\\circ = 150^\\circ.\n$$\n\nSince $\\overline{AB} = \\overline{BC} = \\overline{DE} = \\overline{EA}$, the isosceles triangles $ABC$ and $AED$ are congruent. We get\n\n$$\n\\angle BAC = \\angle ACB = \\angle DAE = \\angle EDA = \\frac{180^\\circ - 150^\\circ}{2} = 15^\\circ.\n$$\n\nSince every diagonal in a square bisects the right angles at its endpoints, we have\n\n$$\n\\angle ADP = \\angle EDB - \\angle EDA = 45^\\circ - 15^\\circ = 30^\\circ.\n$$\n\n$$\n\\angle PAD = \\angle BAE - \\angle BAC - \\angle DAE = 60^\\circ - 2 \\cdot 15^\\circ = 30^\\circ.\n$$\n\nHence $ADP$ is an isosceles triangle with base $AD$ and it follows that $\\overline{PA} = \\overline{PD}$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 13199,
    "subject": "Mathematics (Olympiad)",
    "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": "See 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 $a_l = l \\cdot a_{\\lfloor \\frac{l}{2} \\rfloor} \\ge l$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 13200,
    "subject": "Mathematics (Olympiad)",
    "question": "Determine all triples of real numbers $ (x, y, z) $ such that the equation\n\n$$\n4x^4 - x^2(4y^4 + 4z^4 - 1) - 2xyz + y^8 + 2y^4z^4 + y^2z^2 + z^8 = 0\n$$\nholds.",
    "options": [],
    "answer": "See solution",
    "solution": "We first note that\n\n$$\n\\begin{aligned}\n& 4x^4 - x^2(4y^4 + 4z^4 - 1) - 2xyz + y^8 + 2y^4z^4 + y^2z^2 + z^8 \\\\\n&= (4x^4 + y^8 + z^8 - 4x^2y^4 - 4x^2z^4 + 2y^4z^4) + (x^2 - 2xyz + y^2z^2) \\\\\n&= (2x^2 - y^4 - z^4)^2 + (x - yz)^2.\n\\end{aligned}\n$$\n\nThe given equation is therefore equivalent to\n\n$$\n(2x^2 - y^4 - z^4)^2 + (x - yz)^2 = 0.\n$$\n\nIt therefore follows that both $x = yz$ and $2x^2 - y^4 - z^4 = 0$ must hold.\n\nSubstituting $x = yz$ in the second of these equations, we obtain $2y^2z^2 - y^4 - z^4 = 0$, which is equivalent to $-(y^2 - z^2)^2 = 0$. We see that $z = \\pm y$ must hold. For $z = y = t$, we obtain $x = t^2$, and for $-z = y = t$, we obtain $x = -t^2$. It therefore follows that the set of all solutions is\n\n$$\n\\{(t^2, t, t) \\mid t \\in \\mathbb{R}\\} \\cup \\{(-t^2, t, -t) \\mid t \\in \\mathbb{R}\\}.\n$$",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 13201,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $\\triangle ABC$ be an acute triangle with $AB < AC$ and $O$ the center of its circumcircle $\\omega$. Let $D$ be a point on the segment $BC$ such that $\\angle BAD = \\angle CAO$. Let $E$ be the second intersection of $\\omega$ and the line $AD$. If $M$, $N$, and $P$ are the midpoints of $BE$, $OD$, and $AC$, respectively, show that the points $M$, $N$, and $P$ are collinear.\n\n![](images/Macedonia_2013_p17_data_dc99a126a1.png)",
    "options": [],
    "answer": "See solution",
    "solution": "We will show that $MOPD$ is a parallelogram. From this, it follows that $M$, $N$, and $P$ are collinear.\n\nSince $\\angle BAD = \\angle CAO = 90^\\circ - \\angle ABC$, $D$ is the foot of the perpendicular from $A$ to $BC$. Since $M$ is the midpoint of $BE$, we have $BM = ME = MD$, and hence $\\angle MDE = \\angle MED = \\angle ACB$.\n\nLet the line $MD$ intersect $AC$ at $D_1$. Since $\\angle ADD_1 = \\angle MDE = \\angle ACD$, $MD$ is perpendicular to $AC$. On the other hand, since $O$ is the center of the circumcircle of $\\triangle ABC$ and $P$ is the midpoint of $AC$, $OP$ is perpendicular to $AC$. Therefore, $MD$ and $OP$ are parallel.\n\nSimilarly, since $P$ is the midpoint of $AC$, we have $AP = PC = DP$, and hence $\\angle PDC = \\angle ACB$. Let the line $PD$ intersect $BE$ at $D_2$. Since $\\angle BDD_2 = \\angle PDC = \\angle ACB = \\angle BED$, we conclude that $PD$ is perpendicular to $BE$. Since $M$ is the midpoint of $BE$, $OM$ is perpendicular to $BE$, and hence $OM$ and $PD$ are parallel.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 13202,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $n$ be an integer, $n \\ge 2$. Find the remainder when $n(n+1)(n+2)$ is divided by $n-1$.",
    "options": [],
    "answer": "See solution",
    "solution": "*Solution.*\n\nWe can write:\n$$n(n+1)(n+2) = (n-1+1)(n-1+2)(n-1+3)$$\nExpanding:\n$$(n-1+1)(n-1+2)(n-1+3) = (n-1)^3 + 6(n-1)^2 + 11(n-1) + 6$$\nWhen dividing by $n-1$, the remainder is the value of this expression when $n-1 = 0$, so the remainder is $6$ for $n-1 > 6$.\n\nFor small values:\n- If $n = 2, 3, 4, 7$, the remainder is $0$.\n- If $n = 5$, the remainder is $2$.\n- If $n = 6$, the remainder is $1$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 13203,
    "subject": "Mathematics (Olympiad)",
    "question": "Дадени се пет кружници. Ако избереме било кои четири кружници од дадените пет, тие имаат заедничка точка. Докажи дека постои точка која е заедничка за сите пет кружници.",
    "options": [],
    "answer": "See solution",
    "solution": "Нека $k_1$, $k_2$, $k_3$, $k_4$ и $k_5$ се кружниците. Нека $k_1$, $k_2$, $k_3$ и $k_4$ поминуваат низ $A$; $k_1$, $k_2$, $k_4$ и $k_5$ поминуваат низ $B$; и $k_1$, $k_2$, $k_3$ и $k_5$ поминуваат низ $C$. Тогаш $k_1$ и $k_2$ поминуваат низ $A$, $B$ и $C$, и бидејќи сите се различни кружници, мора две од точките $A$, $B$ и $C$ да се совпаднат. Без губење на општоста, претпоставуваме дека тоа се $A$ и $B$, односно $A \\equiv B$. Тогаш $k_1$, $k_2$, $k_3$, $k_4$ и $k_5$ поминуваат низ $A$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 13204,
    "subject": "Mathematics (Olympiad)",
    "question": "There are two empty pots available. The first pot can hold exactly $3$ liters of liquid, and the second can hold exactly $5$ liters. Is it possible to measure exactly $4$ liters of liquid using these pots?",
    "options": [],
    "answer": "See solution",
    "solution": "1. Fill the $3$-liter pot and pour its contents into the $5$-liter pot.\n2. Fill the $3$-liter pot again and pour into the $5$-liter pot until it is full. This leaves $1$ liter in the $3$-liter pot.\n3. Empty the $5$-liter pot.\n4. Pour the remaining $1$ liter from the $3$-liter pot into the $5$-liter pot.\n5. Fill the $3$-liter pot and pour its contents into the $5$-liter pot. Now the $5$-liter pot contains $4$ liters.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 13205,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all real $x, y, z$ such that\n\n$$\n\\begin{aligned}\nx^2 y + y^2 z + z^2 &= 0 \\\\\nz^3 + z^2 y + z y^3 + x^2 y &= \\frac{1}{4}(x^4 + y^4)\n\\end{aligned}\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "The only solution is $x = y = z = 0$.\n\nIf $y = 0$, then the first equation gives $z^2 = 0$, so $z = 0$. The second equation becomes $\\frac{1}{4}x^4 = 0$, so $x = 0$. Thus, $x = y = z = 0$ is a solution.\n\nAssume $y \\neq 0$. If $z = 0$, then from the first equation, $x^2 y = 0$, so $x = 0$. The second equation becomes $\\frac{1}{4}y^4 = 0$, so $y = 0$, a contradiction. Thus, $z \\neq 0$.\n\nConsider the first equation as a quadratic in $x$, $y$, or $z$:\n\n$$\n\\begin{aligned}\nx &= \\pm \\frac{\\sqrt{-4y^3z - 4yz^2}}{2y} \\\\\ny &= \\frac{-x^2 \\pm \\sqrt{x^4 - 4z^3}}{2z} \\\\\nz &= \\frac{-y^2 \\pm \\sqrt{y^4 - 4x^2y}}{2}\n\\end{aligned}\n$$\n\nThe discriminants must be non-negative:\n\n$$\n\\begin{aligned}\n-4y^3z - 4yz^2 &\\ge 0 \\\\\nx^4 - 4z^3 &\\ge 0 \\\\\ny^4 - 4x^2y &\\ge 0\n\\end{aligned}\n$$\n\nAdding these inequalities:\n\n$$\n\\begin{aligned}\ny^4 - 4x^2y + x^4 - 4z^3 - 4y^3z - 4yz^2 &\\ge 0 \\\\\n\\frac{1}{4}(x^4 + y^4) &\\ge z^3 + z^2y + zy^3 + x^2y\n\\end{aligned}\n$$\n\nBut the second equation gives equality:\n\n$$\n\\frac{1}{4}(x^4 + y^4) = z^3 + z^2y + zy^3 + x^2y\n$$\n\nSo all inequalities must be equalities:\n\n$$\n\\begin{aligned}\n-4y^3z - 4yz^2 &= 0 \\\\\nx^4 - 4z^3 &= 0 \\\\\ny^4 - 4x^2y &= 0\n\\end{aligned}\n$$\n\nThis leads to $x = 0$, $y = 0$, which contradicts the assumption $y \\neq 0$. Therefore, the only solution is $x = y = z = 0$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 13206,
    "subject": "Mathematics (Olympiad)",
    "question": "The expression $\\frac{2021}{2020} - \\frac{2020}{2021}$ is equal to the fraction $\\frac{p}{q}$, where $p$ and $q$ are positive integers whose greatest common divisor is 1. What is $p$?\n\n(A) 1  \n(B) 9  \n(C) 2020  \n(D) 2021  \n(E) 4041",
    "options": [],
    "answer": "See solution",
    "solution": "The given expression equals\n\n$$\n\\frac{2021}{2020} - \\frac{2020}{2021} = \\frac{2021^2 - 2020^2}{2020 \\cdot 2021}\n$$\n\n$$\n= \\frac{(2021 + 2020)(2021 - 2020)}{2020 \\cdot 2021} = \\frac{4041}{2020 \\cdot 2021}.\n$$\n\nBecause $4041 - 2 \\cdot 2020 = 1$, $4041$ and $2020$ have no common divisor greater than 1. Similarly, $4041 - 2 \\cdot 2021 = -1$, so $4041$ and $2021$ have no common divisor greater than 1. Thus, $\\frac{4041}{2020 \\cdot 2021}$ is in simplest terms, and the requested numerator is $4041$.\n\nAlternatively, let $n = 2020$. Then\n\n$$\n\\frac{n+1}{n} - \\frac{n}{n+1} = \\frac{(n+1)^2 - n^2}{n(n+1)} = \\frac{2n+1}{n(n+1)}.\n$$\n\nSince $n$ and $n+1$ are coprime, $2n+1$ shares no common divisors with either. Thus, $p = 2n + 1 = 4041$.",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 13207,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $p$ be an odd prime number. Consider the sequence defined by $x_{n+1} = 6x_n - x_{n-1}$ for $n = 1, 2, \\dots$. 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": "See solution",
    "solution": "It is easy to see\n\n$$\nx_n = \\frac{1}{2\\sqrt{2}} \\left( (3+2\\sqrt{2})^n - (3-2\\sqrt{2})^n \\right), \\quad n = 1, 2, \\dots\n$$\n\nLet $a_n, b_n$ be positive integers such that $a_n + b_n\\sqrt{2} = (3+2\\sqrt{2})^n$. Then\n\n$$\na_n - b_n\\sqrt{2} = (3 - 2\\sqrt{2})^n,\n$$\n\nso $x_n = b_n$ and $a_n^2 - 2b_n^2 = 1$ for $n = 1, 2, \\dots$\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\n$$\na_n + b_n \\sqrt{2} = (3 + 2\\sqrt{2})^{du} \\equiv (a_d + b_d\\sqrt{2})^u \\pmod{q},\n$$\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$$\n\\begin{aligned}\na_n &= (3 + 2\\sqrt{2})^n = (3 + 2\\sqrt{2})^{du} \\cdot (3 + 2\\sqrt{2})^r \\\\\n&\\equiv (a_d + b_d\\sqrt{2})^u (a_r + b_r\\sqrt{2}) \\pmod{q},\n\\end{aligned}\n$$\n\nwe have\n\n$$\na_d^u b_r \\equiv 0 \\pmod{q}. \\qquad (1)\n$$\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 $q \\mid b_r$, which contradicts the definition of $d$. So $r = 0$, and the lemma is proven.\n\nNow, as $q$ is a prime, $q \\mid \\binom{q}{i}$ for $i = 1, 2, \\dots, q-1$.\n\nUsing Fermat's little theorem, we have\n\n$$\n3^q \\equiv 3 \\pmod{q}, \\quad 2^q \\equiv 2 \\pmod{q}.\n$$\n\nAs $q \\neq 2$, $2^{\\frac{q-1}{2}} \\equiv \\pm 1 \\pmod{q}$, so\n\n$$\n\\begin{aligned}\n(3+2\\sqrt{2})^q &= \\sum_{i=0}^{q} \\binom{q}{i} 3^{q-i} (2\\sqrt{2})^i \\\\\n&\\equiv 3^q + (2\\sqrt{2})^q \\\\\n&= 3^q + 2^q \\cdot 2^{\\frac{q-1}{2}} \\sqrt{2} \\\\\n&\\equiv 3 \\pm 2\\sqrt{2} \\pmod{q}.\n\\end{aligned}\n$$\n\nBy the same argument,\n\n$$\n(3+2\\sqrt{2})^{q^2} \\equiv (3 \\pm 2\\sqrt{2})^q \\equiv 3+2\\sqrt{2} \\pmod{q}.\n$$\n\nSo\n\n$$\n(a_{q^2-1} + \\sqrt{2}b_{q^2-1})(3 + 2\\sqrt{2}) \\equiv 3 + 2\\sqrt{2} \\pmod{q}.\n$$\n\nThus,\n\n$$\n\\begin{cases}\n3a_{q^2-1} + 4b_{q^2-1} \\equiv 3 \\pmod{q}, \\\\\n2a_{q^2-1} + 3b_{q^2-1} \\equiv 2 \\pmod{q}.\n\\end{cases}\n$$\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 $d \\in \\{1, p\\}$. 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": 13208,
    "subject": "Mathematics (Olympiad)",
    "question": "Given a prime number $p$, let $A$ be a $p \\times p$ matrix whose entries are exactly $1, 2, \\dots, p^2$ in some order. The following operation is allowed: 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": "See solution",
    "solution": "We may combine the operations on the same row or column, so the final result of a series of operations can be realized as subtracting integer $x_i$ from each number of the $i$-th row and subtracting integer $y_j$ from each number of the $j$-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 values of $x_i$ and $x_j$ results in swapping the $i$-th and $j$-th rows, which is again a good matrix. Similarly, we may consider only the case that $y_1 < y_2 < \\cdots < y_p$, so the matrix is increasing from left to right and from top to bottom.\n\nFrom these assumptions, we have $a_{11} = 1$, and $a_{12}$ 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, with $2 \\le k < p$, so $a_{21} = k+1$. 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, and the first $nk$ columns of the matrix can be divided into $pn \\times k$ submatrices $a_{i, (j-1)k+1}, a_{i, (j-1)k+2}, \\dots, a_{i, jk}$ for $i = 1, 2, \\dots, p$, $j = 1, 2, \\dots, n$, 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 $k \\mid p$, but $1 < k < p$ and $p$ is prime, which is impossible. So we conclude that the first row is $1, 2, \\dots, p$, and 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, so the answer is $$2(p!)^2.$$",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 13209,
    "subject": "Mathematics (Olympiad)",
    "question": "Do there exist real numbers $x, y, z, t$ that satisfy the following system of equations?\n\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": "See solution",
    "solution": "No.\n\n**Solution 1:** The first equation implies $x^3 = -y^2 - 1$. Thus $x^3 < 0$, so $x < 0$. Similarly, from the second, third, and fourth equations, we obtain $y < 0$, $z < 0$, and $t < 0$, respectively. The sum $x + y + z + t$ of negative numbers is negative, contradicting the fifth equation.\n\n**Solution 2:** Suppose the system has a solution. Without loss of generality, let $x$ be the variable with the largest value. Then $4x \\ge x + y + z + t$, which by the last equation implies $x \\ge 0$. Consequently, $x^3 \\ge 0$. 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": 13210,
    "subject": "Mathematics (Olympiad)",
    "question": "At the start of the Mighty Mathematicians Football Team's first game of the season, their coach noticed that the jersey numbers of the 22 players (two teams with 11 players each) on the field were all the numbers from 1 to 22. At halftime, the coach substituted her goalkeeper, with jersey number 1, for a reserve player. No other substitutions were made by either team at or before halftime. The coach noticed that after the substitution, no two players on the field had the same jersey number and that the sums of the jersey numbers of each of the teams were exactly equal. Determine\n\n(a) the greatest possible jersey number of the reserve player,\n\n(b) the smallest possible (positive) jersey number of the reserve player.",
    "options": [],
    "answer": "See solution",
    "solution": "If we leave out the reserve player, the greatest possible difference between the jersey numbers of the two teams is obtained when the reserve player's team has jersey numbers $2$–$11$, while the other team has numbers $12$–$22$. The difference in this case is\n\n$$\n(12 + 13 + \\cdots + 22) - (2 + 3 + \\cdots + 11) = 122;\n$$\n\nwhich is therefore the greatest possible jersey number of the reserve player.\n\nOn the other hand, since\n\n$$\n2 + 3 + 4 + \\cdots + 21 + 22 = 252\n$$\n\nand the total sum of all jersey numbers must be even for the two teams to have the same sum, the reserve player must have an even jersey number. The smallest positive even number that is not already taken by another player is $24$, and indeed it is possible that the sums of the two teams are the same in this case, for example:\n\n$2, 5, 6, 9, 10, 13, 14, 17, 18, 20, 24$ vs. $3, 4, 7, 8, 11, 12, 15, 16, 19, 21, 22$.",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 13211,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $ABC$ be an acute angle triangle, where $AB < AC$, and let $M$ be the midpoint of $BC$, and $K$ be the midpoint of the polygonal chain $BAC$. Show that $\\sqrt{2} KM > AB$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $N$ be the midpoint of $AC$. Since $K$ is the midpoint of the polygonal chain $BAC$, the following holds:\n\n$$\n\\frac{1}{2}(AB + AC) = KC = KN + NC = KN + \\frac{1}{2}AC \\Rightarrow KN = \\frac{1}{2}AB = NM.\n$$\n\nBy the cosine theorem for $\\triangle KNM$:\n\n$$\n\\begin{aligned}\nKM^2 &= KN^2 + NM^2 - 2 \\cdot KN \\cdot NM \\cdot \\cos \\angle KNM \\\\\n&= 2KN^2(1 - \\cos \\angle KNM) = \\frac{1}{2} AB^2 (1 + \\cos \\angle CNM)\n\\end{aligned}\n$$\n\nSince $\\triangle ABC$ is acute angled, $\\angle CNM = \\angle CAB < 90^\\circ \\Rightarrow KM^2 > \\frac{1}{2} AB^2$, which completes the proof.\n\n![](images/UkraineMO2019_booklet_p16_data_4ee9bfe5ed.png)",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 13212,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $ABCDE$ be an equilateral pentagon with center $M$. Point $P \\neq M$ is chosen on the segment $MD$. The circumscribed circle of $\\triangle ABP$ intersects line $AE$ at point $Q$ and the line perpendicular to $CD$ passing through $P$ at point $R$. Prove that $AR = QR$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $S = RP \\cap AE$ (see the figure below).\n\nIt is easy to calculate the angles: $\\angle BAE = 108^\\circ$, $\\angle ABE = \\angle AEB = 36^\\circ$. Since $BE \\parallel CD$ and $RP \\perp BE$, from quadrilateral $ABXS$, we obtain\n\n$$\n\\angle ASX = 360^\\circ - 90^\\circ - 144^\\circ = 126^\\circ.\n$$\n\nNow, $\\angle QSP = \\angle ASR = 54^\\circ$. Hence,\n\n$$\n\\angle SPA = 54^\\circ - \\angle SAP = \\angle PAB - 54^\\circ = 126^\\circ - \\angle AQP = 126^\\circ - \\angle SQP = \\angle SPQ.\n$$\n\nTherefore, $ABPQ$ is inscribed, and now it is clear that $SP$ is a bisector of $\\angle APQ$, which yields $AR = QR$.\n\n![](images/UkraineMO2019_booklet_p29_data_3613e44cdf.png)\n\n*Fig. 30*",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 13213,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $d(n)$ denote the number of positive divisors of $n$. For a positive integer $n$, define $f(n)$ as\n\n$$\nf(n) = d(k_1) + d(k_2) + d(k_3) + \\dots + d(k_m),\n$$\n\nwhere $1 = k_1 < k_2 < \\dots < k_m = n$ are all divisors of $n$. We call an integer $n > 1$ *almost perfect* if $f(n) = n$. Find all almost perfect numbers.",
    "options": [],
    "answer": "See solution",
    "solution": "An alternative way to define $f(n)$ is\n\n$$\nf(n) = \\sum_{k \\mid n,\\ k \\ge 1} d(k).\n$$\n\nLet $n = p_1^{\\alpha_1} p_2^{\\alpha_2} \\dots p_r^{\\alpha_r}$ be the prime factorization of $n$. We have $d(n) = \\prod_{i=1}^r (\\alpha_i + 1)$.\n\nWe prove that the function $f$ is multiplicative: for coprime $n, m$, $f(mn) = f(m)f(n)$.\n\nUsing the multiplicativity of $d$ and the fact that $n, m$ are coprime:\n\n$$\nf(mn) = \\sum_{k|mn} d(k) = \\sum_{k_1|n,\\ k_2|m} d(k_1 k_2) = \\sum_{k_1|n} \\sum_{k_2|m} d(k_1) d(k_2) = \\left(\\sum_{k_1|n} d(k_1)\\right) \\left(\\sum_{k_2|m} d(k_2)\\right) = f(n) f(m).\n$$\n\nIf $r=1$, $n = p_1^{\\alpha_1}$, the divisors of $n$ are $1, p_1, p_1^2, \\dots, p_1^{\\alpha_1}$, so\n\n$$\nf(n) = \\sum_{i=0}^{\\alpha_1} (i+1) = \\frac{(\\alpha_1+1)(\\alpha_1+2)}{2}.\n$$\n\nCombining this with the multiplicativity result for $f$, we deduce\n\n$$\nf(n) = \\prod_{i=1}^r \\frac{(\\alpha_i+1)(\\alpha_i+2)}{2}.\n$$\n\nWe now prove that for primes $p \\ge 5$ and for $p=3$ with $a \\ge 3$, $f(p^a) = \\frac{(a+1)(a+2)}{2} < \\frac{2}{3} p^a$ by induction on $a$. As a basis, $3 < \\frac{2p}{3}$ for $p \\ge 5$ and $6 < \\frac{2}{3} \\cdot 3^3$. For the step, it is enough to notice that $\\frac{a+3}{a+1} \\le 2 < p$ in both cases.\n\nSimilarly, for $p=2$, $f(2^a) < 2^a$ provided $a \\ge 4$. By explicitly checking the remaining cases $p=2$ and $a=1,2,3$ and $p=3$, $a=1,2$, we conclude $f(p^a) \\le \\frac{2}{3} p^a$ for all $p, a$ and $f(p^a) \\le p^a$ for all $p \\ge 3$ and $p=2$, $a \\ge 4$.\n\nAssuming $f(n) = n$, we would have $\\prod_{i=1}^r \\frac{f(p_i^{\\alpha_i})}{p_i^{\\alpha_i}} = 1$, so the above considerations imply that only possible prime divisors are $2$ and $3$.\n\nIf $r=1$, the only possible solution is $n=3$. If $r=2$, $p_1=2$, $p_2=3$, and $1 \\le \\alpha_1 \\le 2$, $1 \\le \\alpha_2 \\le 2$, which gives 4 cases to check, yielding the other two solutions: $n=18$ and $n=36$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 13214,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $a$ be a given integer. Prove that there exist infinitely many prime numbers $p$ such that\n\n$$\np \\mid n^2 + 3, \\quad p \\mid m^3 - a\n$$\n\nfor some integers $n$ and $m$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $k$ be an arbitrary integer. Note that\n\n$$\n(9a^2k^3)^2 + 3 = 3(27a^4k^6 + 1)\n$$\n\nand\n\n$$\n(9a^3k^4)^3 - a = a(3^6a^8k^{12} - 1) = a(27a^4k^6 - 1)(27a^4k^6 + 1)\n$$\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 $k$.\n\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": 13215,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all 6-digit integers $n$ such that $n$ is a perfect square and the number formed by the last 3 digits of $n$ is 1 more than the number formed by the first 3 digits of $n$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $n = y^2$ and let $x$ be the number formed by the first 3 digits of $n$. Note that $y$ is a 3-digit number. Then\n\n$$\ny^2 = 1000x + x + 1 = 1001x + 1 \\implies (y-1)(y+1) = 7 \\times 11 \\times 13x.\n$$\n\nSince $y \\leq 999$, not all of 7, 11, 13 can be factors of $y-1$. Thus, we have 6 cases:\n\n*Case 1:* $77 \\mid y+1$, $13 \\mid y-1$. Then $y+1 = 77\\alpha \\leq 1000$ so $\\alpha \\leq 12$. Testing $\\alpha = 1, 2, \\dots, 12$, only $\\alpha = 11$ works. We get $y = 846$ and $n = 715716$.\n\nThe other 5 cases are:\n- $7 \\times 13 \\mid y+1$, $11 \\mid y-1$\n- $11 \\times 13 \\mid y+1$, $7 \\mid y-1$\n- $7 \\mid y+1$, $11 \\times 13 \\mid y-1$\n- $11 \\mid y+1$, $7 \\times 13 \\mid y-1$\n- $13 \\mid y+1$, $7 \\times 11 \\mid y-1$\n\nThey yield the solutions $n = 528529, 183184, 328329, 75076, 24025$. The two 5-digit numbers are discarded, so we have 4 solutions: $183184$, $328329$, $528529$, and $715716$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 13216,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $ABC$ be a triangle and let the points $D \\in BC$, $E \\in AC$, $F \\in AB$ be such that\n$$\n\\frac{DB}{DC} = \\frac{EC}{EA} = \\frac{FA}{FB}.\n$$\nThe halflines $AD$, $BE$, and $CF$ intersect the circumcircle of $ABC$ at points $M$, $N$, and $P$. Prove that the triangles $ABC$ and $MNP$ share the same centroid if and only if the areas of the triangles $BMC$, $CNA$, and $APB$ are equal.",
    "options": [],
    "answer": "See solution",
    "solution": "First, observe that $\\overrightarrow{AD} + \\overrightarrow{BE} + \\overrightarrow{CF} = 0$ ($*$).\n\nDenoting the areas of the triangles $ABC$, $BMC$, $CNA$, and $APB$ by $s$, $s_a$, $s_b$, and $s_c$, respectively, we have $\\frac{DM}{AD} = \\frac{s_a}{s}$, hence $\\frac{AM}{AD} = \\frac{s + s_a}{s}$, which implies $\\overrightarrow{AM} = \\frac{s + s_a}{s} \\cdot \\overrightarrow{AD}$, and the analogous equalities for $N$ and $P$.\n\nTriangles $ABC$ and $MNP$ share the same centroid if and only if $\\overrightarrow{AM} + \\overrightarrow{BN} + \\overrightarrow{CP} = 0$, hence if and only if\n$$\n\\frac{s + s_a}{s} \\cdot \\overrightarrow{AD} + \\frac{s + s_b}{s} \\cdot \\overrightarrow{BE} + \\frac{s + s_c}{s} \\cdot \\overrightarrow{CF} = 0,\n$$\nor, equivalently,\n$$\ns_a \\cdot \\overrightarrow{AD} + s_b \\cdot \\overrightarrow{BE} + s_c \\cdot \\overrightarrow{CF} = 0.\n$$\nUsing $(*)$, the latter rewrites as\n$$\n(s_a - s_c) \\cdot \\overrightarrow{AD} + (s_b - s_c) \\cdot \\overrightarrow{BE} = 0,\n$$\nand since $\\overrightarrow{AD}$ and $\\overrightarrow{BE}$ are non-collinear vectors, this holds if and only if $s_a = s_b = s_c$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 13217,
    "subject": "Mathematics (Olympiad)",
    "question": "In triangle $ABC$, the angle $\\alpha = \\hat{A}$ and the side $a = |BC|$ are given. It is known that $a = \\sqrt{rR}$, where $r$ is the inradius and $R$ is the circumradius. Determine all such triangles; that is, compute the sides $b$ and $c$ of all such triangles.",
    "options": [],
    "answer": "See solution",
    "solution": "According to the cosine rule, we have\n\n$$\nb^2 + c^2 - 2bc \\cos A = a^2.\n$$\n\nIf $\\tau = \\frac{a+b+c}{2}$, the given condition can be written as\n\n$$\na^2 = rR = \\frac{(ABC)}{\\tau} \\cdot \\frac{a}{\\sin A} = \\frac{bc \\sin A}{(a+b+c) \\sin A} \\implies a^2 = \\frac{abc}{a+b+c} \\\\\n\\implies bc - 2a(b+c) = 2a^2.\n$$\n\nWe write the cosine rule in the form\n\n$$\n(b+c)^2 - (1+\\cos A)bc = a^2.\n$$\n\nSince $b+c > 0$, from the previous equations we find\n\n$$\nb+c = a \\left(1 + 8 \\cos^2 \\frac{A}{2}\\right)\n$$\n\n$$\nbc = 4a^2 \\left(1 + 4 \\cos^2 \\frac{A}{2}\\right)\n$$\n\nFrom these, $b$ and $c$ are the solutions of the quadratic equation\n\n$$\nt^2 - a \\left(1 + 8 \\cos^2 \\frac{A}{2}\\right) t + 4a^2 \\left(1 + 4 \\cos^2 \\frac{A}{2}\\right) = 0 \\\\\n\\implies t^2 - a(5 + 4 \\cos A)t + 4a^2(3 + 2 \\cos A) = 0 \\\\\n\\implies t = \\frac{a}{2} \\left[5 + 4 \\cos A \\pm \\sqrt{16 \\cos^2 A + 8 \\cos A - 23}\\right],\n$$\n\nprovided that $16 \\cos^2 A + 8 \\cos A - 23 \\ge 0$. Considering the trinomial $f(x) = 16x^2 + 8x - 23$, we have\n\n$$\nf(x) \\ge 0 \\implies x \\le \\frac{-1-\\sqrt{24}}{4} \\text{ or } x \\ge \\frac{\\sqrt{24}-1}{4}.\n$$\n\nThe first condition cannot be satisfied. From the second condition, we have\n\n$$\n\\cos A \\ge \\frac{\\sqrt{24}-1}{4} \\implies 0 < A \\le \\arccos \\frac{\\sqrt{24}-1}{4}.\n$$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 13218,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $ABC$ be a scalene triangle and let $P$ and $Q$ be two distinct points in its interior. Suppose that the angle bisectors of $\\angle PAQ$, $\\angle PBQ$, and $\\angle PCQ$ are the altitudes of triangle $ABC$. Prove that the midpoint of $\\overline{PQ}$ lies on the Euler line of $ABC$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $H$ be the orthocenter of $ABC$, and construct $P'$ using the following claim.\n\n**Claim** — There is a point $P'$ for which\n\n$$\n\\angle APH + \\angle AP'H = \\angle BPH + \\angle BP'H = \\angle CPH + \\angle CP'H = 0.\n$$\n\n*Proof*. After inversion at $H$, this is equivalent to the fact that $P$'s image has an isogonal conjugate in $ABC$'s image. $\\square$\n\nNow, let $X$, $Y$, and $Z$ be the reflections of $P$ over $\\overline{AH}$, $\\overline{BH}$, and $\\overline{CH}$ respectively. Additionally, let $Q'$ be the image of $Q$ under inversion about $(PXYZ)$.\n\n![](images/sols-TSTST-2023_p19_data_e77a912b02.png)\n\n**Claim** — $ABCP' \\sim XYZQ'$.\n\n*Proof*. Since\n\n$$\n\\angle YXZ = \\angle YPZ = \\angle (\\overline{BH}, \\overline{CH}) = -\\angle BAC\n$$\n\nand cyclic variants, triangles $ABC$ and $XYZ$ are similar. Additionally,\n\n$$\n\\angle HQ'X = -\\angle HXQ = -\\angle HXA = \\angle HPA = -\\angle HP'A\n$$\n\nand cyclic variants, so summing in pairs gives $\\angle YQ'Z = -\\angle BP'C$ and cyclic variants; this implies the similarity. $\\square$\n\n**Claim** — $Q'$ lies on the Euler line of triangle $XYZ$.\n\n*Proof*. Let $O$ be the circumcenter of $ABC$ so that $ABCOP' \\sim XYZHQ'$. Then $\\angle HP'A = -\\angle HQ'X = \\angle OP'A$, so $P'$ lies on $\\overline{OH}$. By the similarity, $Q'$ must lie on the Euler line of $XYZ$. $\\square$\n\nTo finish the problem, let $G_1$ be the centroid of $ABC$ and $G_2$ be the centroid of $XYZ$. Then with signed areas,\n\n$$\n\\begin{aligned}\n[G_1HP] + [G_1HQ] &= \\frac{[AHP] + [BHP] + [CHP]}{3} + \\frac{[AHQ] + [BHQ] + [CHQ]}{3} \\\\\n&= \\frac{[AHQ] - [AHX] + [BHQ] - [BHY] + [CHQ] - [CHZ]}{3} \\\\\n&= \\frac{[HQX] + [HQY] + [HQZ]}{3} \\\\\n&= [QG_2H] \\\\\n&= 0\n\\end{aligned}\n$$\n\nwhere the last line follows from the last claim. Therefore $\\overline{G_1H}$ bisects $\\overline{PQ}$, as desired.\n\n**Remark.** This solution characterizes the set of all points $P$ for which such a point $Q$ exists. It is the image of the Euler line under the mapping described in the first claim.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 13219,
    "subject": "Mathematics (Olympiad)",
    "question": "Country \"U\" has cities $A_1, A_2, \\dots, A_n$. They are connected with flights so that one can get from any city to any other city with layovers. Let $d_{i,j}$ be the smallest number of flights needed to get from city $A_i$ to city $A_j$, and let $d$ be the total number of flights in country \"U\". Define\n\n$$\nD = d_{1,1} + d_{1,2} + \\dots + d_{1,n} + d_{2,3} + d_{2,4} + \\dots + d_{2,n} + \\dots + d_{n-1,n}.\n$$\n\nShow that:\n\n$$\nd + \\frac{D}{n-1} \\le \\frac{1}{2}(n+1)(n+2).\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "Let $i \\ne j$, and let the shortest path between $A_i$ and $A_j$ be $A_i = A_{\\gamma_0}, A_{\\gamma_1}, \\dots, A_{\\gamma_k} = A_j$. Let $X$ be the set of such vertices ($k+1$ vertices), and let $Y$ be the set of all other vertices ($n-k-1$ vertices). The number of edges connecting vertices inside $Y$ is at most $\\binom{n-k-1}{2}$. All edges connecting vertices inside $X$ are the edges of the shortest path; otherwise, there would be a shorter path between $A_i$ and $A_j$. Therefore, there are exactly $k$ edges.\n\nIf $A \\in Y$, then there are at most 3 edges connecting $A$ with vertices from $X$. Otherwise, there would be a shorter path than the one with $k$ edges. Therefore, there are at most $3(n-k-1)$ edges connecting vertices from $X$ and $Y$. Thus,\n\n$$\n\\begin{aligned}\nd &\\le \\binom{n-k-1}{2} + k + 3(n-k-1) \\\\\n  &= \\frac{1}{2}(n^2 + 3n - 4 + k^2 - k - 2kn) \\\\\n  &\\le \\frac{1}{2}(n^2 + 3n - 4) - \\frac{n+1}{2}k \\quad (k < n, \\text{ so } kn < n^2) \\\\\n  &\\Rightarrow d + \\frac{n+1}{2}d_{i,j} \\le \\frac{1}{2}(n^2 + 3n - 4)\n\\end{aligned}\n$$\n\nThus,\n\n$$\n\\sum_{1 \\le i < j \\le n} \\left(d + \\frac{n+1}{2}d_{i,j}\\right) \\le \\frac{1}{2}\\binom{n}{2}(n^2 + 3n - 4)\n$$\nso\n$$\n\\binom{n}{2}d + \\frac{n+1}{2}D \\le \\frac{1}{2}\\binom{n}{2}(n^2 + 3n - 4).\n$$\n\nThe given inequality can be obtained by dividing both sides by $\\binom{n}{2}$ and using\n\n$$\n\\frac{n+1}{n(n-1)} \\ge \\frac{1}{n-1} \\quad \\text{and} \\quad n^2 + 3n - 4 \\le n^2 + 3n + 2.\n$$\n\n![](images/Ukraine_booklet_2018_p28_data_2ecd111d00.png)\n\n**Fig. 30**",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 13220,
    "subject": "Mathematics (Olympiad)",
    "question": "Given an arbitrary triangle $ABC$ with area $T$ and perimeter $L$. Let $P$, $Q$, $R$ be the points of tangency of sides $BC$, $CA$, $AB$ respectively with the inscribed circle. Prove the inequality\n\n$$\n\\left(\\frac{AB}{PQ}\\right)^3 + \\left(\\frac{BC}{QR}\\right)^3 + \\left(\\frac{CA}{RP}\\right)^3 \\geq \\frac{2}{\\sqrt{3}} \\cdot \\frac{L^2}{T}\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "Let $BC = a$, $CA = b$, $AB = c$, $QR = p$, $RP = q$, $PQ = r$, and let $AP = x$, $BQ = y$, $CR = z$. Since $x + y = c$, $y + z = a$, $z + x = b$, we have\n\n$$\nx = s - a, \\quad y = s - b, \\quad z = s - c \\quad \\left( s = \\frac{a+b+c}{2} \\right).\n$$\n\nApplying the Cosine Law to triangles $ABC$ and $ARQ$ yields\n\n$$\na^2 = b^2 + c^2 - 2bc \\cos A = (b - c)^2 + 2bc(1 - \\cos A)\n$$\n\nand\n\n$$\np^2 = 2x^2(1 - \\cos A) = 2(s - a)^2(1 - \\cos A).\n$$\n\nCanceling $1 - \\cos A$ from the above, we can write $p^2$ in terms of $a$, $b$, $c$:\n\n$$\np^2 = (s-a)^2 \\cdot \\frac{a^2 - (b-c)^2}{bc} = \\frac{4(s-a)(s-b)(s-c)}{abc} \\cdot a(s-a).\n$$\n\nNote that\n\n$$\n4(s - a)(s - b) = (b - c - a)(a - b - c) = c^2 - (b - a)^2 \\leq c^2\n$$\n\nand similarly\n\n$$\n4(s - b)(s - c) \\leq a^2, \\quad 4(s - c)(s - a) \\leq b^2.\n$$\n\nThis gives $8(s - a)(s - b)(s - c) \\leq abc$, so with the previous result, $p^2 \\leq \\frac{a(s-a)}{2}$, or equivalently,\n\n$$\n\\left(\\frac{a}{p}\\right)^3 \\geq 2\\sqrt{2}\\left(\\frac{a}{s-a}\\right)^{3/2}.\n$$\n\nSimilarly,\n\n$$\n\\left(\\frac{b}{q}\\right)^3 \\geq 2\\sqrt{2}\\left(\\frac{b}{s-b}\\right)^{3/2}, \\quad \\left(\\frac{c}{r}\\right)^3 \\geq 2\\sqrt{2}\\left(\\frac{c}{s-c}\\right)^{3/2}.\n$$\n\nLet $M$ be the left side of the given inequality. Then\n\n$$\n\\begin{aligned}\nM &\\geq 2\\sqrt{2} \\left\\{ \\left( \\frac{a}{s-a} \\right)^{3/2} + \\left( \\frac{b}{s-b} \\right)^{3/2} + \\left( \\frac{c}{s-c} \\right)^{3/2} \\right\\} \\\\\n&\\geq \\frac{2\\sqrt{2}}{\\sqrt{3}} \\left( \\frac{a}{s-a} + \\frac{b}{s-b} + \\frac{c}{s-c} \\right)^{3/2}\n\\end{aligned}\n$$\n\nwhere the second inequality uses Jensen's inequality. Since\n\n$$\na \\geq b \\geq c \\iff \\frac{1}{s-a} \\geq \\frac{1}{s-b} \\geq \\frac{1}{s-c},\n$$\n\nby Chebyshev's and AM-GM inequalities,\n\n$$\n\\begin{aligned}\n\\frac{a}{s-a} + \\frac{b}{s-b} + \\frac{c}{s-c} &\\geq \\frac{1}{3}(a+b+c) \\left( \\frac{1}{s-a} + \\frac{1}{s-b} + \\frac{1}{s-c} \\right) \\\\\n&\\geq \\frac{a+b+c}{\\{(s-a)(s-b)(s-c)\\}^{1/3}} \\\\\n&= \\frac{(a+b+c)s^{1/3}}{\\{s(s-a)(s-b)(s-c)\\}^{1/3}} \\\\\n&= \\frac{1}{2^{1/3}} \\left( \\frac{L^2}{T} \\right)^{2/3}\n\\end{aligned}\n$$\n\nTherefore, from above,\n\n$$\nM \\geq \\frac{2\\sqrt{2}}{\\sqrt{3}} \\cdot \\frac{1}{\\sqrt{2}} \\frac{L^2}{T} = \\frac{2}{\\sqrt{3}} \\cdot \\frac{L^2}{T}.\n$$\n\nas required.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 13221,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $ABC$ be a right-angled triangle with hypotenuse $AB$ and longer leg $BC$. Let $D$ be the foot of the altitude from vertex $C$. Circle $k$ with center $D$ and radius $CD$ intersects leg $BC$ at point $Q$ and line $AB$ at points $E$ and $F$ ($E \\neq F$), where $F$ is a point on the hypotenuse $AB$. Segment $QE$ intersects leg $AC$ at point $P$. Prove that $PE = QF$.\n\n![](images/65_Czech_and_Slovak_MO_2016_booklet_p4_data_489abf5c26.png)",
    "options": [],
    "answer": "See solution",
    "solution": "The circle $k$ is the Thales' circle with diameter $EF$ and center $D$. Triangle $EFC$ is an isosceles right-angled triangle, so $EC = EF$. We will show that triangles $EPC$ and $FQC$ are congruent, which will prove the statement.\n\nAngles $CEQ$ and $CFQ$ are congruent as they are inscribed angles subtended by chord $CQ$ of circle $k$. Both angles $ECF$ and $ACB$ are congruent (right angles), so their remaining non-overlapping parts (angles $ECP$ and $FCQ$) are also congruent. Thus, triangles $EPC$ and $FQC$ are congruent by $A$-$S$-$A$.",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 13222,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $O$ be the circumcenter of a right triangle. The circle with smaller radius and center at point $O$ is tangent to the longer leg and to the altitude from the right angle.\n\nFind the acute angles of the right triangle and the relation between the radii of the circumcircle and the other circle.\n\n![](fig.16.png)",
    "options": [],
    "answer": "See solution",
    "solution": "Let the right triangle be $CMD$ (see figure). Let $ON$ be the radius of the smaller circle that is tangent to the leg. Thus, $\\triangle ODN = \\triangle OMN$, and $\\triangle AMO = \\triangle OMN$, therefore, $2AM = MN + ND = MD$, hence $\\angle ADM = 30^\\circ$. Therefore, the acute angles are $30^\\circ$ and $60^\\circ$.\n\nFrom $\\triangle OND$, it is clear that $OD = R$ (hypotenuse), $ON = r$ (leg opposite $30^\\circ$), thus $\\frac{R}{r} = 2$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 13223,
    "subject": "Mathematics (Olympiad)",
    "question": "居家隔離的日子太無聊了，於是老趙跟勳勳決定玩一個遊戲。勳勳先秘密寫下一個多項式 $f(x)$，滿足：\n\n(a) 對於任意整數 $n$，$f(n)$ 是個整數；\n\n(b) $f(x)$ 的次數小於 $187$。\n\n老趙只知道 $f(x)$ 滿足性質 (a) 與 (b)，但不知道 $f(x)$。接下來，老趙每一回合可以從集合 $\\{1, 2, \\dots, 187\\}$ 裡面選一個數字 $k$，然後勳勳會告訴老趙 $f(k)$ 的值。試求最小的正整數 $N$，使得老趙總是能在 $N$ 個回合內確定 $f(0)$ 的奇偶性。",
    "options": [],
    "answer": "See solution",
    "solution": "最小的 $N = 63$。\n\n令 $n = 187$，且 $187$ 的二進制表示法是 $10111011$。首先我們注意到 $n$ 次差分給出\n$$\n\\sum_{i=0}^{n} (-1)^i \\binom{n}{i} f(i) = 0\n$$\n由盧卡斯定理我們知道 $\\binom{n}{i}$ 是奇數當且僅當 $i$ 的二進制表示法中 1 的位置是 $n$ 的二進制表示法中 1 的位置的子集，所以\n$$\n0 \\equiv \\sum_{B \\subseteq A} f\\left(\\sum_{i \\in B} 2^i\\right) \\pmod{2}\n$$\n其中 $A$ 是 $n$ 的二進制表示法中 1 的位置所成的集合。由此可知，老趙只需知道\n$$\nS := \\left\\{\\sum_{i \\in B} 2^i \\mid \\varnothing \\neq B \\subseteq A\\right\\}\n$$\n便可確知 $f(0)$ 的奇偶性。\n\n另一方面，如果 $S$ 中某個 $t$ 尚未被選過，那老趙無法分辨 $f(0)$ 跟 $f(0)+g_t(0)$，其中 $g_t$ 是一個 $n-1$ 次（含）以下多項式，滿足 $g_t(t) = 1$ 且 $g_t(t') = 0$，對所有 $t' \\in \\{1, 2, \\dots, n\\} \\setminus \\{t\\}$。我們熟知 $g_t$ 也把整數送到整數。然而，套用上一段得到的模二的恆等式，我們知道 $g_t(0)$ 是一個奇數，所以此時老趙無法判斷 $f(0)$ 的奇偶性。\n\n綜上所述，老趙至少要詢問 $|S| = 2^{|A|} - 1 = 63$ 次。",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 13224,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $p$ and $q$ be positive real numbers such that $\\frac{1}{p} + \\frac{1}{q} = 1$. Prove that:\n\n$$\na) \\quad \\frac{1}{3} \\leq \\frac{1}{p(p+1)} + \\frac{1}{q(q+1)} < \\frac{1}{2}, \\qquad b) \\quad \\frac{1}{p(p-1)} + \\frac{1}{q(q-1)} \\geq 1.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "The relation $\\frac{1}{p} + \\frac{1}{q} = 1$ gives $pq = p + q$. Let $s = pq = p + q$ and notice that $s \\geq 4$, since $(p+q)^2 \\geq 4pq$.\n\na) Since\n$$\n\\frac{1}{p(p+1)} + \\frac{1}{q(q+1)} = \\frac{1}{p} - \\frac{1}{p+1} + \\frac{1}{q} - \\frac{1}{q+1} = 1 - \\frac{p+q+2}{(p+1)(q+1)} = 1 - \\frac{s+2}{2s+1} = \\frac{s-1}{2s+1},\n$$\nthe claim rewrites as $\\frac{1}{3} \\leq \\frac{s-1}{2s+1} < \\frac{1}{2}$. The right-hand side inequality is obvious, while the left-hand side inequality reduces to $s \\geq 4$.\n\nb) Since\n$$\n\\frac{1}{p(p-1)} + \\frac{1}{q(q-1)} = \\frac{1}{p-1} - \\frac{1}{p} + \\frac{1}{q-1} - \\frac{1}{q} = \\frac{p+q-2}{(p-1)(q-1)} - 1 = s - 3 \\geq 1,\n$$\nthe claim holds.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 13225,
    "subject": "Mathematics (Olympiad)",
    "question": "Determine the smallest constant $C$ such that the inequality\n\n$$\n(X + Y)^2 (X^2 + Y^2 + C) + (1 - XY)^2 \\geq 0\n$$\n\nholds for all real numbers $X$ and $Y$.\n\nFor which values of $X$ and $Y$ does equality hold for this smallest constant $C$?",
    "options": [],
    "answer": "See solution",
    "solution": "The smallest constant is $C = -1$. Equality holds for $X = Y = \\frac{1}{\\sqrt{3}}$ or $X = Y = -\\frac{1}{\\sqrt{3}}$.\n\nFirst, consider the case $X = Y$. The inequality becomes\n\n$$\n(3X^2 - 1)^2 + 4(C + 1)X^2 \\geq 0\n$$\n\nSetting $X^2 = \\frac{1}{3}$ gives $C \\geq -1$.\n\nNow, for $C = -1$, we need to show the inequality holds for all $X$ and $Y$:\n\n$$\n(X^2 + XY + Y^2 - 1)^2 + (X - Y)^2 \\geq 0.\n$$\n\nThis is always true, with equality when $X = Y$ and $3X^2 - 1 = 0$, i.e., $X = Y = \\frac{1}{\\sqrt{3}}$ or $X = Y = -\\frac{1}{\\sqrt{3}}$.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 13226,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $f: \\mathbb{R} \\to \\mathbb{R}$ be a function such that, for every $x, y \\in \\mathbb{R}$:\n\n$$\n|f(x + y) + \\sin x + \\sin y| \\le 2.\n$$\n\na) Prove that $|f(x)| \\le 1 + \\cos x$, for every $x \\in \\mathbb{R}$.\n\nb) Give an example of such a function, which vanishes nowhere in the interval $(-\\pi, \\pi)$.",
    "options": [],
    "answer": "See solution",
    "solution": "a) Let $x = t - \\frac{\\pi}{2}$ and $y = \\frac{\\pi}{2}$. Then:\n$$\nf(t) - \\cos t + 1 \\le 2\n$$\nfor every $t \\in \\mathbb{R}$.\n\nAlso, let $x = t + \\frac{\\pi}{2}$ and $y = -\\frac{\\pi}{2}$:\n$$\nf(t) + \\cos t - 1 \\ge -2\n$$\nfor every $t \\in \\mathbb{R}$.\n\nCombining these inequalities gives the desired result.\n\nb) An example is $f(x) = 2 - 2|\\sin \\frac{x}{2}|$. A heuristic approach: the hypothesis is equivalent to\n$$\n\\begin{align*}\n2 &\\ge \\max_{u \\in \\mathbb{R}} \\left( f(t) + 2 \\sin \\frac{t}{2} \\cos \\frac{u}{2} \\right) = f(t) + 2 \\left| \\sin \\frac{t}{2} \\right| \\\\\n-2 &\\le \\min_{u \\in \\mathbb{R}} \\left( f(t) + 2 \\sin \\frac{t}{2} \\cos \\frac{u}{2} \\right) = f(t) - 2 \\left| \\sin \\frac{t}{2} \\right|,\n\\end{align*}\n$$\nthat is, $|f(x)| \\le 2 - 2|\\sin \\frac{x}{2}|$ for all $x \\in \\mathbb{R}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 13227,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all polynomials of the form\n\n$$\nP(x, y) = Ax^2 + 2Ay + Bx\n$$\n\nfor some $A, B \\in \\mathbb{R}$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $Q(a, b) = P(a, b^2)$ and define $Q_i(x, y)$ to be the homogeneous polynomial consisting of the $i$-th degree coefficients of $Q(x, y)$. It is directly implied that\n\n$$\nQ_i(a, \\sqrt{2bc}) + Q_i(b, \\sqrt{2ac}) + Q_i(c, \\sqrt{2ab}) = Q_i(a + b + c, \\sqrt{ab + bc + ac})\n$$\n\nLet $A_i(a, b, c)$ denote the above equality. Plugging $c = 0$ into this equality results in\n\n$$\nQ_i(a, 0) + Q_i(b, 0) + Q_i(0, \\sqrt{2ab}) = Q_i(a + b, \\sqrt{ab}).\n$$\n\nDefine $x = a + b$, $y = \\sqrt{ab}$. First, note that\n\n$$\nQ_i(a, 0) = a^i Q_i(1, 0)\n$$\n\nand\n\n$$\nQ_i(0, a) = a^i Q_i(0, 1).\n$$\n\nSecondly, the above equality concludes that\n\n$$\nQ_i(x, y) = (\\sqrt{2y^2})^n Q_i(0, 1) + \\left[ \\left( \\frac{x - \\sqrt{x^2 - 4y^2}}{2} \\right)^i + \\left( \\frac{x + \\sqrt{x^2 - 4y^2}}{2} \\right)^i \\right] Q_i(1, 0)\n$$\n\nNext, we state that if $i \\ge 3$, $Q_i = 0$. This means $P(x, y^2)$ is of degree at most 2, which then results in $P(x, y) = Ax^2 + Bx + Cy + D$, and one can simply see that $D = 0$, $C = 2A$, and every polynomial of the form\n\n$$\nP(x, y) = Ax^2 + Bx + 2Ay\n$$\n\nsatisfies the problem.\n\nFor the sake of contradiction, assume that $i \\ge 3$. Note that $P(x, y^2) = Q(x, y)$, therefore every coefficient of $Q(x, y)$, and by extension, $Q_i(x, y)$, has $y$ to an even power.\n\n$$\nQ_i(x, y) = x^i m_0 + x^{n-2}y^2 m_2 + m_4 x^{n-4} y^4 + \\dots\n$$\n\nBy plugging this into $A_i(a, b, c)$ and evaluating the coefficient of $a^{n-1}b$, one can see the coefficient on the left-hand side to be zero, and on the right-hand side\n\n$$\n\\begin{align*}\nm_0(x + y + z)^i + m_2(xy + yz + xz)(x + y + z)^{i-2} + \\dots \\\\\nimplies [x^{i-1}y] = \\binom{i}{1}m_0 + \\binom{i-2}{0}m_2 = 0 \\implies m_2 = -im_0\n\\end{align*}\n$$\n\nFurthermore, by evaluating the coefficient of $x^{i-2}yz$ we get\n\n$$\n\\begin{align*}\n[x^{i-2}yz] &= \\binom{i-1}{1}m_0 + \\left[1 + \\binom{i-2}{1} + \\binom{i-2}{1}\\right]m_2 + \\binom{2}{1}m_4 = 2m_2 \\\\\nimplies 2m_4 &= im_0(2i-5) - i \\times (i-1)m_0 \\implies m_4 = \\frac{i(i-4)}{2}m_0\n\\end{align*}\n$$\n\nMoreover, let\n\n$$\nT(x, y) = \\left( \\frac{x - \\sqrt{x^2 - 4y}}{2} \\right)^i + \\left( \\frac{x + \\sqrt{x^2 - 4y}}{2} \\right)^i.\n$$\n\nEvaluate the coefficient of $x^i$, $x^{i-4}y^2$ in $T$.\n\n$$\n\\begin{aligned}\n[x^i] &= T(1, 0) = 1, Q_i(x, y) \\\\\n&= 2^{\\frac{i}{2}} y^i Q_i(0, 1) + T(x, y^2) Q_i(1, 0) \\\\\nimplies m_0 &= Q_i(1, 0)\n\\end{aligned}\n$$\n\nAnd (evaluating coefficient in $Q_i$)\n\n$$\n\\begin{aligned}\n[x^{i-2}y^2] &= \\frac{1}{2} T_{yy}(1, 0) = \\frac{i(i-3)}{2} \\\\\nimplies m_4 &= \\frac{i(i-3)}{2} Q_i(1, 0) \\\\\nimplies a &= Q_i(1, 0) = \\frac{2m_4}{i(i-3)} = \\frac{i(i-4)a}{i(i-3)}\n\\end{aligned}\n$$\n\nSo either $i=3$, $i=4$ or $a=b=c=0$, otherwise we have a contradiction. In the latter two cases,\n\n$$\n\\begin{aligned}\nQ_i(1, 0) &= 0 \\\\\nimplies Q_i(x, y) &= 2^{\\frac{i}{2}} y^i Q_i(0, 1) + 0 \\\\\nimplies Q_i(0, 1) &= 2^{\\frac{i}{2}} 1^i Q_i(0, 1) \\\\\nimplies Q_i(0, 1)(2^{\\frac{i}{2}} - 1) &= 0 \\\\\nimplies Q_i(0, 1) &= 0 \\quad \\implies Q_i(x, y) = 0\n\\end{aligned}\n$$\n\nTherefore, the only remaining case is $i=3$, which is easily disproven. This concludes our proof.",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 13228,
    "subject": "Mathematics (Olympiad)",
    "question": "Find the largest integer $n$ with the following property: Any collection of $n$ tetraminoes, each of them being one from the picture, can be placed without overlapping in the $20 \\times 20$ table so that every tetramino covers exactly four squares of the table. (We can rotate and reflect each tetramino.)\n\n![](images/CZE_ABooklet_2024_p11_data_14b455f8fd.png)",
    "options": [],
    "answer": "See solution",
    "solution": "We show that the largest $n$ is equal to $99$. We will call the tetraminoes from the picture O, T, L, and I respectively.\n\nFirst, we provide a collection of $100$ tetraminoes that cannot be placed inside the $20 \\times 20$ table. This means that no integer $n \\geq 100$ has the desired property.\n\nConsider the collection of $99$ tetraminoes of type O and one tetramino of type T. If we color the $20 \\times 20$ table as a checkerboard, it will contain $200$ black and $200$ white cells. $99$ tetraminoes of type O cover $198$ black and $198$ white cells. Thus, there will be $2$ black and $2$ white cells remaining. However, they cannot be covered by a tetramino of type T, since it always covers $3$ cells of the same color. This concludes the first part of the solution.\n\nIn the second part, we show that the number $n = 99$ has the desired property. We describe how to place any collection of $99$ tetraminoes inside the table. We use the following figures:\n\n![](images/CZE_ABooklet_2024_p11_data_51091ec6d1.png)\n\nFrom the picture, we see how to fill a $4 \\times 4$ square with four tetraminoes T and how to fill a $4 \\times 2$ rectangle with two tetraminoes from each of the types O, I, L. We fill the table with all such squares and rectangles that can be created from the given collection of $99$ tetraminoes (squares first, then rectangles), by layers of $4$ rows. Since each row contains $20$ cells, which is a multiple of $4$, the new layer starts only if the previous layer is completely covered.\n\nAt the moment when there is no rectangle left, the following conditions hold:\n\n* The collection of tetraminoes left (denoted by $\\mathcal{Z}$) is a subset of $1 \\times O$, $1 \\times L$, $1 \\times I$, $3 \\times T$. Thus, for the number $k$ of tetraminoes in $\\mathcal{Z}$, $k \\leq 6$.\n* There are $4(k+1)$ uncovered cells forming a rectangle $4 \\times (k+1)$.\n* The number $99 - k$ of tetraminoes in the squares and rectangles is even, so $k \\leq 6$ is odd, and therefore $k \\in \\{1, 3, 5\\}$.\n\nIt remains to place the $k$ tetraminoes from $\\mathcal{Z}$ in the $4 \\times (k+1)$ rectangle. We distinguish three cases based on $k \\in \\{1, 3, 5\\}$. For $k=3$ and $k=5$, we use the division of the $4 \\times (k+1)$ table into parts A, B, and C, as shown in the pictures. Any tetramino can be placed inside a $4 \\times 2$ rectangle and also part A. In parts B and C, we can fit a tetramino of each of the types O, L, and T.\n\n**Case $k = 1$:** Place the tetramino from $\\mathcal{Z}$ in the $4 \\times 2$ rectangle.\n\n**Case $k = 3$:** Place one tetramino (preferably type I) in part A, and the other two in parts B and C.\n\n**Case $k = 5$:** The collection $\\mathcal{Z}$ contains two or three tetraminoes of type T. Place two of them as shown in the picture. The other three tetraminoes are placed in parts A, B, and C, putting type I in A if present in $\\mathcal{Z}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 13229,
    "subject": "Mathematics (Olympiad)",
    "question": "For a positive integer $n$, an $n$-staircase is a figure consisting of unit squares, with one square in the first row, two squares in the second row, and so on, up to $n$ squares in the $n^{th}$ row, such that all the left-most squares in each row are aligned vertically. For example, the 5-staircase is shown below.\n\n![](images/Kanada_2010_p0_data_6cdcee0058.png)\n\nLet $f(n)$ denote the minimum number of square tiles required to tile the $n$-staircase, where the side lengths of the square tiles can be any positive integer. For example, $f(2) = 3$ and $f(4) = 7$.\n\n![](images/Kanada_2010_p0_data_a47d8bd8cf.png)\n![](images/Kanada_2010_p0_data_cba37ea4cc.png)\n\n(a) Find all $n$ such that $f(n) = n$.\n\n(b) Find all $n$ such that $f(n) = n + 1$.",
    "options": [],
    "answer": "See solution",
    "solution": "(a) A \\textit{diagonal} square in an $n$-staircase is a unit square that lies on the diagonal going from the top-left to the bottom-right. A \\textit{minimal tiling} of an $n$-staircase is a tiling consisting of $f(n)$ square tiles.\n\nObserve that $f(n) \\ge n$ for all $n$. There are $n$ diagonal squares in an $n$-staircase, and a square tile can cover at most one diagonal square, so any tiling requires at least $n$ square tiles. In other words, $f(n) \\ge n$. Hence, if $f(n) = n$, then each square tile covers exactly one diagonal square.\n\nLet $n$ be a positive integer such that $f(n) = n$, and consider a minimal tiling of an $n$-staircase. The only square tile that can cover the unit square in the first row is the unit square itself.\n\nNow consider the left-most unit square in the second row. The only square tile that can cover this unit square and a diagonal square is a $2 \\times 2$ square tile.\n\n![](images/Kanada_2010_p2_data_669a042cf2.png)",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 13230,
    "subject": "Mathematics (Olympiad)",
    "question": "For any natural number $n$, let $N(n)$ be the number of digits of $n$ and $S(n)$ the sum of its digits. (Assume numbers do not start with zero.)\n\nWhich digits can occur in a natural number $n$ if $$\\frac{n}{S(n)} < \\frac{m}{S(m)}$$ for all other $m$ with $N(m) = N(n)$?",
    "options": [],
    "answer": "See solution",
    "solution": "Let $n$ be fixed. Consider increasing or decreasing one of its digits by $i$, i.e., the number $n \\pm b i$ where $b = 10^k$ for some $k$. Then\n\n$$\n\\begin{align*}\n\\frac{n \\pm b i}{S(n \\pm b i)} > \\frac{n}{S(n)} &\\iff \\frac{n \\pm b i}{S(n) \\pm i} > \\frac{n}{S(n)} \\\\\n&\\iff \\frac{n \\pm b i}{n} > \\frac{S(n) \\pm i}{S(n)} \\\\\n&\\iff 1 \\pm \\frac{b i}{n} > 1 \\pm \\frac{i}{S(n)} \\\\\n&\\iff \\pm \\frac{b i}{n} > \\pm \\frac{i}{S(n)} \\\\\n&\\iff \\pm b > \\pm \\frac{n}{S(n)}.\n\\end{align*}\n$$\n\nThe last inequality is equivalent to $\\frac{n}{S(n)} < b$ in the case of plus and $\\frac{n}{S(n)} > b$ in the case of minus.\n\nThis shows that no number $n$ with the property described in the problem can contain digits 2 through 8. Otherwise, such a digit could be both increased and decreased, leading to contradictory conclusions since the ratio $\\frac{n}{S(n)}$ increases in both cases.\n\nIt remains to show that numbers with the desired property can contain digits 0, 1, 9. For that, we prove that 1099 has the desired property. Let $n = \\overline{d_3d_2d_1d_0}$ be an arbitrary 4-digit number. For any positive integer $x$, denote $R(x) = \\frac{x}{S(x)}$.\n\nIf $d_0 < 9$, the last digit can be increased. As $d_3 > 0$ implies\n\n$$\nR(\\overline{d_3d_2d_19}) = \\frac{1000d_3 + 100d_2 + 10d_1 + 9}{d_3 + d_2 + d_1 + 9} > 1,\n$$\n\nwe obtain $R(\\overline{d_3d_2d_19}) < R(n)$.\n\nIf $d_1 < 9$, the tens digit can be increased. As\n\n$$\nR(\\overline{d_3d_299}) = \\frac{1000d_3 + 100d_2 + 99}{d_3 + d_2 + 9 + 9} > \\frac{1000}{9 + 9 + 9 + 9} > \\frac{1000}{100} = 10,\n$$\n\nwe obtain $R(\\overline{d_3d_299}) < R(\\overline{d_3d_2d_19})$.\n\nIf $d_3 > 1$, the thousands digit can be decreased. As\n\n$$\nR(\\overline{1d_299}) = \\frac{1000 + 100d_2 + 99}{1 + d_2 + 9 + 9} < \\frac{9000}{9} = 1000,\n$$\n\nwe obtain $R(\\overline{1d_299}) < R(\\overline{d_3d_299})$.\n\nFinally, if $d_2 > 0$, the hundreds digit can be decreased. As\n\n$$\nR(1099) = \\frac{1099}{19} < 100,\n$$\n\nwe obtain $R(1099) < R(\\overline{1d_299})$. Consequently,\n\n$$\nR(1099) \\leq R(\\overline{1d_299}) \\leq R(\\overline{d_3d_299}) \\leq R(\\overline{d_3d_2d_19}) \\leq R(n),\n$$\n\nwhere all equalities hold simultaneously only if $n = 1099$. This completes the proof.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 13231,
    "subject": "Mathematics (Olympiad)",
    "question": "Ellina has twelve blocks, two each of red (R), blue (B), yellow (Y), green (G), orange (O), and purple (P). Call an arrangement of blocks *even* if there is an even number of blocks between each pair of blocks of the same color. For example, the arrangement\n\nRBBYGGYROPPO\n\nis even. Ellina arranges her blocks in a row in random order. The probability that her arrangement is even is $\\frac{m}{n}$, where $m$ and $n$ are relatively prime positive integers. Find $m + n$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let there be $k$ colors, numbered $1, 2, \\ldots, k$, with two blocks of each color. An arrangement is even if and only if every pair of blocks of the same color are in positions whose numbers differ by an odd number, i.e., one is even and the other is odd. Thus, Ellina forms an even arrangement if and only if there are two permutations of the $k$ colors $(a_1, a_2, \\ldots, a_k)$ and $(b_1, b_2, \\ldots, b_k)$ such that the blocks are arranged as $a_1, b_1, a_2, b_2, \\ldots, a_k, b_k$. Therefore, there are $(k!)^2$ even arrangements. The total number of ways to arrange $k$ pairs of blocks is $\\frac{(2k)!}{2^k}$. For $k = 6$, the probability is\n\n$$\n\\frac{(6!)^2 \\cdot 2^6}{12!} = \\frac{2 \\cdot 4 \\cdot 6}{7 \\cdot 9 \\cdot 11} = \\frac{16}{231}.\n$$\n\nThe requested sum is $16 + 231 = 247$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 13232,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $f : \\mathbb{R} \\to \\mathbb{R}$ be a function satisfying the following functional equation for all $x, y, z \\in \\mathbb{R}$:\n\n$$\nf(f(x) + f(y) + f(z)) = f(f(x) - f(y)) + f(2xy + f(z)) + 2f(x - y).\n$$\n\nFind all such functions $f$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let us analyze the functional equation step by step.\n\nIf $f(x) = c$ is a constant function, then substituting into the equation shows $c = 0$. Assume from here on that $f$ is not constant.\n\nLet $t \\in \\mathbb{R}$ and substitute $(x, y, z) = (t, 0, 0)$ and $(x, y, z) = (0, t, 0)$:\n\n$$\n\\begin{aligned}\nf(f(t) + 2f(0)) &= f(f(t) - f(0)) + f(f(0)) + 2f(0), \\\\\nf(f(t) + 2f(0)) &= f(f(0) - f(t)) + f(f(0)) + 2f(0).\n\\end{aligned}\n$$\n\nFrom this, we conclude:\n\n$$\nf(f(t) - f(0)) = f(f(0) - f(t)) \\text{ for all } t \\in \\mathbb{R}. \\quad (1)\n$$\n\nSuppose $f(u_1) = f(u_2)$. Substitute $(s, 0, u_1)$ and $(s, 0, u_2)$:\n\n$$\n\\begin{aligned}\nf(f(s) + f(0) + f(u_1)) &= f(f(s) - f(0)) + f(f(u_1)) + 2f(su_1), \\\\\nf(f(s) + f(0) + f(u_2)) &= f(f(s) - f(0)) + f(f(u_2)) + 2f(su_2).\n\\end{aligned}\n$$\n\nSince $f(u_1) = f(u_2)$, $f(su_1) = f(su_2)$ for all $s \\in \\mathbb{R}$:\n\n$$\nf(u_1) = f(u_2) \\Rightarrow f(su_1) = f(su_2) \\text{ for all } s \\in \\mathbb{R}. \\quad (2)\n$$\n\nSince $f$ is not constant, there exists $t_1$ with $f(t_1) \\neq f(0)$. Using (1), $f(u) = f(-u)$ where $u = f(t_1) - f(0) \\neq 0$. From (2), $f(su) = f(-su)$ for all $s$, so:\n\n$$\nf(x) = f(-x) \\text{ for all } x \\in \\mathbb{R}. \\quad (3)\n$$\n\nIf $f(u) = f(0)$ for some $u \\neq 0$, then $f(su) = f(0)$ for all $s$, so $f$ is constant—a contradiction. Thus:\n\n$$\nx \\neq 0 \\Rightarrow f(x) \\neq 0 \\quad \\text{for all } x \\in \\mathbb{R}. \\quad (4)\n$$\n\nSuppose $f(a) = f(b)$ but $a \\neq \\pm b$. From (3) and (4), assume $0 < a < b$. From (2), $f(1) = f(r)$ where $r = \\frac{b}{a} > 1$. Again from (2):\n\n$$\nf(x) = f(rx) \\text{ for all } x \\in \\mathbb{R}. \\quad (5)\n$$\n\nNow substitute $(rx, ry, 1)$ and $(r^2x, y, 1)$ into the original equation:\n\n$$\n\\begin{aligned}\nf(f(rx)+f(ry)+f(1)) &= f(f(rx)-f(ry))+f(2r^2xy+f(1))+2f(r(x-y)), \\\\\nf(f(r^2x)+f(y)+f(1)) &= f(f(r^2x)-f(y))+f(2r^2xy+f(1))+2f(r^2x-y).\n\\end{aligned}\n$$\n\nSince $f(x) = f(rx) = f(r^2x)$, we get $f(x-y) = f(r^2x-y)$. Let $x = y + w$:\n\n$$\nf(w) = f((r^2 - 1)y + w).\n$$\n\nSince $(r^2 - 1)y$ ranges over all $\\mathbb{R}$, $f$ is constant—a contradiction. Thus:\n\n$$\nf(x) = f(y) \\Rightarrow x = \\pm y \\text{ for all } x, y \\in \\mathbb{R}. \\quad (6)\n$$\n\nSubstitute $(1, 1, 0)$ and $(1, -1, 0)$:\n\n$$\n\\begin{aligned}\nf(f(1) + f(1) + f(0)) &= f(f(1) - f(1)) + f(2 + f(0)) + 2f(0), \\\\\nf(f(1) + f(-1) + f(0)) &= f(f(1) - f(-1)) + f(-2 + f(0)) + 2f(0).\n\\end{aligned}\n$$\n\nFrom (3), $f(x) = f(-x)$, so $f(2 + f(0)) = f(-2 + f(0))$. By (6), either $2 + f(0) = -2 + f(0)$ or $2 + f(0) = 2 - f(0)$. The first is impossible; the second gives:\n\n$$\nf(0) = 0. \\quad (7)\n$$\n\nNow substitute $x = y$ into the equation:\n\n$$\nf(2f(x) + f(z)) = f(2x^2 + f(z)).\n$$\n\nFrom (6), $2f(x) + f(z) = \\pm(2x^2 + f(z))$. If $2f(x) + f(z) = 2x^2 + f(z)$, then $f(x) = x^2$. If for some $x_0$, $f(x_0) \\neq x_0^2$, then $f(z)$ is constant—a contradiction. Thus $f(x) = x^2$ for all $x$.\n\n**Conclusion:**\n\nThe only solutions are $f(x) = 0$ for all $x$ or $f(x) = x^2$ for all $x$. Both satisfy the given functional equation.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 13233,
    "subject": "Mathematics (Olympiad)",
    "question": "令 $n \\ge 3$ 為一正整數。有來自 $k$ 間學校的共 $n$ 位咒術師，編號為 $1$ 到 $n$。已知當兩個咒術師對決時，編號小的咒術師會獲勝。此外，對於任何 $\\{1, 2, \\dots, n\\}$ 的重排 $\\{x_1, x_2, \\dots, x_n\\}$，$x_1$ 對 $x_2$ 號、$x_2$ 號對 $x_3$ 號，一直到 $x_{n-1}$ 號對 $x_n$ 號咒術師的 $n-1$ 場對決的獲勝者中，包含 $k$ 間學校的人各至少一位。試證 $n \\ge 2^k$。",
    "options": [],
    "answer": "See solution",
    "solution": "將 $n$ 名咒術師當成 $n$ 個點 $A_1, \\dots, A_n$，以學校為顏色對各點塗色，並對所有 $1 \\le i < j \\le n$，將 $A_iA_j$ 連線並塗上 $A_i$ 的顏色。我們先證明以下關鍵引理。\n\n**引理：** 對於第 $i$ 種顏色，存在 $1 \\le p \\le n$，使得 $A_1, A_2, \\dots, A_p$ 中的 $i$ 色點數量多於 $p/2$。\n\n證明：若否，則存在 $i$ 使得對於所有 $1 \\le p \\le n$，前 $p$ 個點中都至多只有 $\\lfloor p/2 \\rfloor$ 個 $i$ 色點。令 $A_{x_1}, A_{x_2}, \\dots, A_{x_t}$ 為所有的 $i$ 色點，而 $A_{y_1}, A_{y_2}, \\dots, A_{y_s}$ 為所有的非 $i$ 色點，其中 $x_1 < x_2 < \\dots < x_t$, $y_1 < y_2 < \\dots < y_s$。顯然此時 $s + t = n$，且依據歸謬假設 $t \\le \\lfloor p/2 \\rfloor$。此外，注意到對於 $1 \\le j \\le t$，若 $y_j \\ge x_j$，則 $A_1, A_2, \\dots, A_{x_j}$ 將有 $j$ 個 $i$ 色點和少於 $j$ 個非 $i$ 色點，與歸謬假設不合，因此必有 $y_j < x_j$。但如此一來，\n\n$$\nA_{x_1}A_{y_1}, A_{y_1}A_{x_2}, A_{x_2}A_{y_2}, A_{y_2}A_{x_3}, \\dots, A_{x_t}A_{y_t}, A_{y_t}A_{y_{t+1}}, \\dots, A_{y_{s-1}}A_{y_s}\n$$\n\n將不包含任何 $i$ 色邊，矛盾。故原命題成立。\n\n現在，對於第 $i$ 色，令 $p_i$ 為符合引理的最小 $p$。注意到對於 $i \\ne j$，前 $p_i$ 個點中有過半的 $i$ 色點，便不可能有過半的 $j$ 色點，因此 $p_i \\ne p_j$，因此我們可以不失一般性假設\n\n$$\np_1 < p_2 < \\dots < p_k.\n$$\n\n此時，當 $i \\le j$，前 $p_i$ 個點中有至少 $\\lfloor (p_i + 1)/2 \\rfloor$ 個 $i$ 色點，從而前 $p_j$ 個點中有至少 $\\lfloor (p_i + 1)/2 \\rfloor$ 個 $i$ 色點。這意味著\n\n$$\np_j \\ge \\text{前 } p_j \\text{ 個點中前 } j \\text{ 種顏色的點數} \\ge \\sum_{i=1}^{j} \\lfloor (p_i + 1)/2 \\rfloor.\n$$\n\n以上可遞迴證得 $p_i \\ge 2^i - 1$，從而 $n \\ge p_k \\ge 2^k - 1$。\n\n我們最後只需證明 $n \\ne 2^k - 1$。若 $n = 2^k - 1$，上述所有等號必須成立，也就是 $p_i = 2^i - 1$ 且恰有 $2^{i-1}$ 個 $i$ 色點。但注意到前 $i-1$ 個顏色的點的數量為 $\\sum_{j=1}^{i-1} 2^j = 2^{i-1} - 1 = p_{i-1}$，故全部的 $i$ 色點只能是從 $A_{p_{i-1}+1}$ 到 $A_{p_i}$ 的 $2^{i-1}$ 個點。而此時\n\n$$\nA_{p_{k-1}+1}A_{p_1}, A_{p_1}A_{p_{k-1}+2}, A_{p_{k-1}+2}A_2, A_2A_{p_{k-1}+3}, \\dots, A_{p_{k-1}}A_{p_k}\n$$\n\n將不包含任何顏色 $k$，矛盾。故 $n \\ge 2^k$。",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 13234,
    "subject": "Mathematics (Olympiad)",
    "question": "In the expression\n\n$$\n12 - 11 - 10 - 9 - 8 - 7 - 6 - 5 - 4 - 3 - 2 - 1\n$$\n\nthere are brackets placed somehow, and the value is calculated. Find the maximum value that can be reached. Justify your answer.\n\n*Notice.* A left bracket can be placed only before a number and a right bracket only after. For example, expressions $-4(-3-2)$ and $-(4-3-2)$ are incorrect.",
    "options": [],
    "answer": "See solution",
    "solution": "$$\n12 - (11 - 10 - 9 - 8 - 7 - 6 - 5 - 4 - 3 - 2 - 1) = 46\n$$\n\nConsider any arrangement of the brackets. The numbers 12 and 11 always have the signs '+' and '-', respectively. The sum reaches its maximum value when all the other terms have the '+' sign.",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 13235,
    "subject": "Mathematics (Olympiad)",
    "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 $EP$ to $\\odot I$ 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": "See solution",
    "solution": "Suppose that $\\odot I$ touches $AC$ and $AB$ at $S$ and $T$, respectively. Suppose that $ST$ intersects $AI$ at point $G$. We see that $IT \\perp AT$ and $TG \\perp AI$. So we\n\n![](images/Mathematical_Olympiad_in_China_2011-2014_p310_data_fb7b7931bc.png)\n\nhave $AG \\cdot AI = AT^2 = AD \\cdot AE$, thus points $I, G, E$, and $D$ are concyclic.\n\nSince $IE \\perp PE$ and $ID \\perp PD$, we see that points $I, E, P$, and $D$ are concyclic. Hence, points $I, G, E, P$, and $D$ are concyclic.\n\nTherefore, $\\angle IGP = \\angle IEP = 90^\\circ$, 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$$\n\\frac{AS}{SC} \\cdot \\frac{CP}{PB} \\cdot \\frac{BT}{TA} = 1.\n$$\n\nSince $AS = AT$, $CS = CD$, and $BT = BD$, we have\n\n$$\n\\frac{PC}{PB} \\cdot \\frac{BD}{CD} = 1. \\qquad \\textcircled{1}\n$$\n\nLet the extension of $BN$ intersect $PE$ at point $H$. Then line $BFH$ intersects $\\triangle PDE$. By Menelaus' Theorem,\n\n$$\n\\frac{PH}{HE} \\cdot \\frac{EF}{FD} \\cdot \\frac{DB}{BP} = 1.\n$$\n\nSince $CF$ is parallel to $BE$, $\\frac{EF}{FD} = \\frac{PC}{CD}$, we have\n\n$$\n\\frac{PH}{HE} \\cdot \\frac{PC}{CD} \\cdot \\frac{DB}{BP} = 1. \\qquad \\textcircled{2}\n$$\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": 13236,
    "subject": "Mathematics (Olympiad)",
    "question": "For a yacht moving directly from buoy B to buoy C, its heading would be southwest. For a yacht moving directly from buoy C to buoy D, its heading would be southeast. If the area of water bounded by the course is $R$ km$^2$, find the value of $R$.",
    "options": [],
    "answer": "See solution",
    "solution": "From the information about distances (C is halfway), $CD = 8 - 6.5 = 1.5$ km.\n\nFrom the information about headings, $\\angle BCD$ is a right angle.\nIn $\\triangle BCD$, $BD = 2.5$ km$\\;$ (3-4-5 triangle, or use Pythagoras' Theorem directly).\n\n**Method 1**\n\nIn $\\triangle ABD$, $\\angle ABD$ is a right angle (converse of Pythagoras' Theorem in 5-12-13 triangle).\n\nArea of water $=$ area $\\triangle ABD$ $-$ area $\\triangle BCD = \\frac{1}{2}(6 \\times 2.5 - 2 \\times 1.5) = 6$ km$^2$. So $R = 6$.\n\n**Method 2**\n\nHalf the perimeter of $\\triangle ABD$ is $7.5$ km. From Heron's formula, the square of the area of $\\triangle ABD$ is $7.5(7.5 - 6)(7.5 - 6.5)(7.5 - 2.5) = 7.5(1.5)(1)(5) = 7.5^2$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 13237,
    "subject": "Mathematics (Olympiad)",
    "question": "За бројот $a$, е исполнето равенството $a + \\frac{1}{a} = 1$. Пресметај ја вредноста на $a^5 + \\frac{1}{a^5}$.",
    "options": [],
    "answer": "See solution",
    "solution": "Воведуваме ознака $b = \\frac{1}{a}$. Тогаш $a + b = 1$ и $ab = 1$, па според тоа:\n\n$$\na^2 + b^2 = (a + b)^2 - 2ab = 1^2 - 2 \\cdot 1 = -1\n$$\n\n$$\na^3 + b^3 = (a + b)(a^2 + b^2 - ab) = 1 \\cdot (-1 - 1) = -2\n$$\n\n$$\na^5 + b^5 = (a^2 + b^2)(a^3 + b^3) - a^2 b^2 (a + b) = (-1)(-2) - 1^2 \\cdot 1 = 2 - 1 = 1\n$$\n\nЗначи, $a^5 + \\frac{1}{a^5} = 1$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 13238,
    "subject": "Mathematics (Olympiad)",
    "question": "Consider two matrices $A, B \\in \\mathcal{M}_4(\\mathbb{R})$ such that $AB = BA$ and $\\det(A^2 + AB + B^2) = 0$. Prove the identity\n\n$$\n\\det(A+B) + 3\\det(A-B) = 6\\det(A) + 6\\det(B).\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "Let $\\omega$ be a non-real cubic root of unity. Since $\\omega^2 + \\omega + 1 = 0$ and $AB = BA$, we have $A^2 + AB + B^2 = (A - \\omega B)(A - \\bar{\\omega}B)$.\n\nBecause $A$ and $B$ are real, the degree 4 polynomial $f(x) = \\det(A + xB) = \\det A + a x + b x^2 + c x^3 + \\det B x^4$ has real coefficients.\n\nFrom $\\det(A^2 + AB + B^2) = 0$, we get $f(\\omega) f(\\bar{\\omega}) = 0$, so $f(\\omega) = f(\\bar{\\omega}) = 0$. From $f(\\omega) = \\det A + c + \\omega(a + \\det B) + \\omega^2 b$, we deduce $\\det A + c = a + \\det B = b$.\n\nSince $f(1) = \\det A + a + b + c + \\det B$ and $f(-1) = \\det A - a + b - c + \\det B$, we have $f(1) + f(-1) = 2\\det A + 2\\det B + 2b$. Also, $2b = a + c + \\det A + \\det B = \\frac{1}{2}(f(1) - f(-1)) + \\det A + \\det B$, so $f(1) + 3f(-1) = 6\\det A + 6\\det B$. Since $f(1) = \\det(A+B)$ and $f(-1) = \\det(A-B)$, the identity follows.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 13239,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $a, b, c, d$ be real numbers such that $a + b + c + d = 1$ and $a \\ge b \\ge c \\ge d \\ge 0$. Define:\n\n$$\na' = a, \\quad b' = b + \\frac{d}{2}, \\quad c' = c + \\frac{d}{2}, \\quad d' = 0.\n$$\n\nShow that replacing $(a, b, c, d)$ with $(a', b', c', d')$ increases the value of\n$$\n(a b c)^2 + (a b d)^2 + (a c d)^2 + (b c d)^2,\n$$\nand deduce the maximum value of this expression.",
    "options": [],
    "answer": "See solution",
    "solution": "Replacing $(a, b, c, d)$ with $(a', b', c', d')$ increases the left-hand side:\n\n$$\n\\begin{aligned}\n& (a'b'c')^2 + (a'b'd')^2 + (a'c'd')^2 + (b'c'd')^2 \\\\\n&= a^2 \\left(b + \\frac{d}{2}\\right)^2 \\left(c + \\frac{d}{2}\\right)^2 \\\\\n&= a^2 \\left(b^2 + bd + \\frac{d^2}{4}\\right) \\left(c^2 + cd + \\frac{d^2}{4}\\right) \\\\\n&\\ge a^2 b^2 c^2 + a^2 b^2 c d + a^2 b d c^2 + a^2 b d c d \\\\\n&\\ge (a b c)^2 + (a b d)^2 + (a c d)^2 + (b c d)^2,\n\\end{aligned}\n$$\n\nwhere the last inequality uses the ordering $a \\ge b \\ge c \\ge d \\ge 0$. Thus, the maximum occurs when $d = 0$. Then:\n\n$$\n(a b c)^2 + (a b d)^2 + (a c d)^2 + (b c d)^2 = (a b c)^2.\n$$\nBy the AM-GM inequality:\n$$\n(a b c)^2 \\le \\left(\\frac{a + b + c}{3}\\right)^6 = \\left(\\frac{1}{3}\\right)^6 = \\frac{1}{729}.\n$$\nBut since $a + b + c = 1$ (with $d = 0$), the maximum is $\\left(\\frac{1}{3}\\right)^6 = \\frac{1}{729}$.\n\nTherefore, the maximum value is $\\left(\\frac{1}{3}\\right)^6$.",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 13240,
    "subject": "Mathematics (Olympiad)",
    "question": "Determine the smallest real number $\\lambda$ with the following property: every positive integer $n$ can be written as a product $n = x_1 x_2 \\cdots x_{2023}$, with each $x_i$ either a prime number or a positive integer that is less than or equal to $n^\\lambda$.",
    "options": [],
    "answer": "See solution",
    "solution": "The minimal such $\\lambda$ is $\\frac{1}{1012}$.\n\nTo see that $\\lambda < \\frac{1}{1012}$ does not work, take $n = p^{2024}$ for some prime $p$. 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 $p^\\alpha \\ge p^2 = n^{1/1012} > n^\\lambda$.\n\nNow we prove that $\\lambda = \\frac{1}{1012}$ satisfies the condition. Write $n$ as the product of its prime factors: $n = p_1 p_2 \\cdots p_r$ (allowing some $p_i$'s to be equal), ordered so that $p_1 \\ge p_2 \\ge \\cdots$.\n\nStarting from $p_1$, let $r_1 \\ge 1$ be the smallest integer such that $x_1 = p_1 p_2 \\cdots p_{r_1} > n^{1/2024}$. If $r_1 = 1$, $x_1$ is a prime; otherwise $r_1 \\ge 2$, and $p_1 \\le n^{1/2024}$. Thus $p_r \\le n^{1/2024}$, so\n\n$$\nx_1 = (p_1 p_2 \\cdots p_{r-1}) p_r \\le n^{1/2024} \\cdot n^{1/2024} = n^{1/1012}.\n$$\n\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 shows $x_2$ is either a prime or $x_2 \\le n^{1/1012}$.\n\nContinue this process. If we run out of $p_i$'s before reaching $x_{2023}$, set the remaining $x_j$ to $1$. Otherwise, after constructing $x_1, x_2, \\dots, x_{2022}$, each $x_i > n^{1/2024}$, then\n\n$$\nx_{2023} := \\frac{n}{x_1 \\cdots x_{2022}} < \\frac{n}{(n^{1/2024})^{2022}} = n^{1/1012}.\n$$\n\nThus, the decomposition $n = x_1 x_2 \\cdots x_{2023}$ satisfies the required conditions.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 13241,
    "subject": "Mathematics (Olympiad)",
    "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": "See solution",
    "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, $b + 1 + 1 = d$ 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.\n\n**Answer:** $646$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 13242,
    "subject": "Mathematics (Olympiad)",
    "question": "給定正整數 $n$。設 $S$ 為坐標平面上 $x$ 坐標及 $y$ 坐標皆小於 $2n$ 的非負整數的點所成的集合（所以 $S$ 共包含 $4n^2$ 個點）。假設 $\\mathcal{F}$ 為 $n^2$ 個四邊形所成的集合，其中每個四邊形的所有頂點都在 $S$ 內，並且 $S$ 裡的每個點皆是 $\\mathcal{F}$ 中其中一個四邊形的頂點。試求 $\\mathcal{F}$ 內所有 $n^2$ 個四邊形的面積總和的最大可能值。",
    "options": [],
    "answer": "See solution",
    "solution": "面積總和的最大可能值為 $\\Sigma(n) = \\frac{1}{3} n^2 (2n+1)(2n-1)$。\n\n以下皆以 $[P]$ 代表多邊形 $P$ 的面積。我們先作一些約定。如果一個多邊形的頂點皆屬於 $S$，稱之為合法的。設 $O = (n - \\frac{1}{2}, n - \\frac{1}{2})$ 為 $S$ 的中心點。若一個合法的正方形的中心為 $O$ 點，稱該正方形為置中的。最後，若一組多邊形 $\\mathcal{F}$ 滿足題設條件，稱 $\\mathcal{F}$ 為可接受的，並將其面積總和記為 $\\Sigma(\\mathcal{F})$。\n\n$S$ 內的每一個點都恰是唯一的置中正方形的頂點。所以所有的置中正方形形成可接受的集合 $\\mathcal{G}$。以下證明\n\n$$\n\\Sigma(\\mathcal{F}) \\leq \\Sigma(\\mathcal{G}) = \\Sigma(n)\n$$\n\n對每一組可接受的 $\\mathcal{F}$ 均成立，從而回答了問題。\n\n我們使用以下關鍵引理。\n\n**引理 1.** 設 $P = A_1A_2\\dots A_m$ 為多邊形，且 $O$ 為平面上任意一點。則\n\n$$\n[P] \\leq \\frac{1}{2} \\sum_{i=1}^{m} OA_i^2\n$$\n\n而且當 $P$ 是正方形且 $O$ 為其中心時，上式的不等式會變成等號。\n\n*引理證明.* 令 $A_{m+1} = A_1$。對每一個 $i = 1, 2, \\dots, m$，都有\n\n$$\n[OA_iA_{i+1}] \\leq \\frac{OA_i \\cdot OA_{i+1}}{2} \\leq \\frac{OA_i^2 + OA_{i+1}^2}{4}\n$$\n\n所以得\n\n$$\n[P] \\leq \\sum_{i=1}^{m} [OA_iA_{i+1}] \\leq \\frac{1}{4} \\sum_{i=1}^{m} (OA_i^2 + OA_{i+1}^2) = \\frac{1}{2} \\sum_{i=1}^{m} OA_i^2\n$$\n\n此即為上式。直接驗證可知當 $P$ 為正方形且 $O$ 為其中心時，上式成為等式。引理證畢。 $\\square$\n\n回到原題，設 $\\mathcal{F}$ 為任一組可接受的多邊形。將引理 1 套用到 $\\mathcal{F}$ 內的每一個多邊形，也套用到 $\\mathcal{G}$ 內的每一個正方形（此時為等式），可得\n\n$$\n\\Sigma(\\mathcal{F}) \\leq \\frac{1}{2} \\sum_{A \\in S} OA^2 = \\Sigma(\\mathcal{G})\n$$\n\n故上述不等式成立。\n\n最後計算 $\\Sigma(\\mathcal{G})$。由上述可知\n\n$$\n\\begin{align*}\n\\Sigma(\\mathcal{G}) &= \\frac{1}{2} \\sum_{A \\in S} OA^2 = \\frac{1}{2} \\sum_{i=0}^{2n-1} \\sum_{j=0}^{2n-1} \\left( \\left( n - \\frac{1}{2} - i \\right)^2 + \\left( n - \\frac{1}{2} - j \\right)^2 \\right) \\\\\n&= n \\sum_{j=0}^{n-1} (2j + 1)^2 \\\\\n&= \\frac{n^2(2n + 1)(2n - 1)}{3} = \\Sigma(n)\n\\end{align*}\n$$\n\n證明完畢。\n\n$\\square$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 13243,
    "subject": "Mathematics (Olympiad)",
    "question": "In the left-lower corner $1 \\times 1$ cell of the $m \\times n$ board ($m, n \\geq 3$) there is a black chip, and in the left-upper and right-lower cells, there are white chips. On his move, Petrik can move the black chip to an adjacent (by side) cell twice in a row, and Vasyl can either move one of the white chips to an adjacent (by side) cell twice in a row or move each of the two white chips separately to adjacent (by side) cells. Chips cannot be placed on fields that have already been visited by a chip of another color. Vasyl wins if, after some point, both white chips end up in the same cell. If Vasyl wants to win, and Petrik wants to prevent this, under what conditions on $m$ and $n$ does Vasyl win? Petrik moves first.",
    "options": [],
    "answer": "See solution",
    "solution": "**Answer:** If $m$ and $n$ have the same parity, Vasyl wins; otherwise, Petrik wins.\n\n**Solution.** Without loss of generality, assume $m \\leq n$. Enumerate the rows from bottom to top as $1, 2, \\ldots, m$, and columns from left to right as $1, 2, \\ldots, n$. Each cell has coordinates $(\\text{row}, \\text{column})$. Let chip A be the white chip at $(m, 1)$ (left-upper), and chip B be the white chip at $(1, n)$ (right-lower).\n\n*Case 1: $m$ and $n$ have the same parity (Vasyl wins).*  \nVasyl can move chip A to the right twice in a row $\\frac{1}{2}(n-m)$ times. After this, the white chips are in opposite corners of an $m \\times m$ square. Then, each turn, move chip A one cell right and chip B one cell up. The white chips will meet at $(m, n)$, and the black chip cannot prevent this. To see why, suppose the black chip could reach the top row or rightmost column before the white chips meet. In both cases, a contradiction arises when analyzing the possible positions and moves, showing the black chip cannot block the meeting.\n\n*Case 2: $m$ and $n$ have different parity (Petrik wins).*  \nPetrik starts by moving the black chip one cell right and one cell up. Then, if Vasyl moves a white chip twice, Petrik mirrors the move in the other direction; if Vasyl moves both chips once, Petrik moves one cell up and one cell right. This keeps the black chip always to the right of chip A and above chip B, blocking their meeting. The parity of the sum of coordinates of the white chips minus the black chip is invariant and initially $(m+n) \\equiv 1 \\pmod{2}$. If the white chips were to meet adjacent to the black chip, the invariant would be $0 \\pmod{2}$, a contradiction. Thus, Petrik can always prevent the white chips from meeting.\n\n![](images/Ukraine_2021-2022_p15_data_cd54af8756.png)\n\n![](images/Ukraine_2021-2022_p15_data_b8a563f31c.png)",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 13244,
    "subject": "Mathematics (Olympiad)",
    "question": "Let us call an \"edge\" any segment of length 1 which is common to two adjacent fields of a chessboard $8 \\times 8$.\n\nConsider all possible cuttings of the chessboard into 32 pieces of size $2 \\times 1$, and denote by $n(e)$ the total number of such cuttings that involve the given edge $e$.\n\nDetermine the last digit of the sum of the numbers $n(e)$ over all the edges $e$.",
    "options": [],
    "answer": "See solution",
    "solution": "The total number of vertical edges is $7 \\times 8 = 56$, and the total number of horizontal edges is also $56$. Thus, the total number of edges is $56 \\times 2 = 112$.\n\nIn any given cutting, there are 32 edges that are used (each $2 \\times 1$ domino covers one edge), so each cutting does not use $112 - 32 = 80$ edges. Therefore, each cutting contributes $80$ to the sum $S$ of all $n(e)$.\n\nThus, $S$ is a multiple of $80$, so its last digit is $0$.",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 13245,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $f(x, y)$ be a function defined on $\\mathbb{Z} \\times S$ (where $S$ is a finite set), satisfying certain recurrence relations. Prove by induction that for every $x \\in \\mathbb{Z}$ and $y = 0, 1, \\dots, 2n+1$,\n\n$$\nf(x, y) = f(2n+1 + x, 2n+1 - y).\n$$\n\nFurthermore, determine the maximum number of distinct values $|v_f|$ that $f$ can take, and construct a function $f$ attaining this maximum.\n\n![](images/Vietnamese_mathematical_competitions_p75_data_cffc8fb554.png)",
    "options": [],
    "answer": "See solution",
    "solution": "We proceed by induction on $y$.\n\n- For $y = 0$, $f(x, 0) = 0 = f(2n+1 + x, 2n+1)$.\n- For $y = 1$, using the recurrence (1) with $k = 2n$:\n  $$\nf(x, 1) + f(x + 1, 0) = f(x + 2n, 2n + 1) + f(x + 2n + 1, 2n)\n  $$\n  which gives the desired result.\n- Assume $f(x, y-1) = f(2n+1 + x, 2n+2 - y)$ for all $x$.\n\n  *If $y$ is odd*, set $k = 1 - y$ in (1):\n  $$\nf(x, y) + f(x + 1, y - 1) = f(x - y + 1, 1).\n  $$\n  Similarly, from (2) with $k = 1 - y$ and $t = 2n + 1$:\n  $$\nf(x + t + 1, t + 1 - y) + f(x + t, t - y) = f(x - y + t + 1, t - 1)\n  $$\n  Applying the induction hypothesis yields the result.\n\n  *If $y$ is even*, it suffices to check $y = 2$:\n  $$\nf(x, 2) + f(x - 1, 1) = 1 - f(x + 1, 1) = f(x + 2, 2) + f(x + 3, 1)\n  $$\n  and\n  $$\nf(x + 2, 2) + f(x + 3, 1) = f(x + 2n, 2n) + f(x + 2n + 1, 2n - 1),\n  $$\n  so the result follows.\n\nThus, $f(x, y) = f(2n+1 + x, 2n+1 - y)$ for all $x$ and $y$.\n\nSince there are at most $2n+1$ distinct values per row, and the $k$-th and $(2n+1-k)$-th rows are equal for $0 < k < 2n+1$, with the first and last rows all zero, the total number of distinct values is at most\n$$\n2n(2n + 1) + 1.\n$$\n\nTo construct $f$ attaining this bound, define\n$$\nf(i + 2k, i) = \\frac{[1 - (-1)^i][1 - (-1)^k]}{4} + (-1)^k a_{k+i} - (-1)^{k+i} a_k\n$$\nfor $i = 1, 2, \\dots, 2n$ and $k \\in \\mathbb{Z}$, $0 \\leq k + i \\leq 2n + 1$.\n\nAssign $a_{2k-1} = \\frac{1}{3^{2k-1}}$, $a_{2k} = \\frac{1}{3^{2(n+1-k)}}$ for $k = 1, 2, \\dots, n$. Then all values $\\delta_{ij} \\pm a_i \\pm a_j$ are distinct and nonzero.\n\nFor $x - y$ odd, set $b_{2k-1} = \\frac{1}{3^{2k-1}\\sqrt{3}}$, $b_{2k} = \\frac{1}{3^{2(n+1-k)}\\sqrt{3}}$ for $k = 1, 2, \\dots, n$; then $\\delta'_{ij} \\pm b_i \\pm b_j$ are also distinct and different from the even case.\n\nThus, $f$ attains $2n(2n + 1) + 1$ distinct values, which is maximal.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 13246,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $n \\ge 4$ be a positive integer, and let $a_1, a_2, \\dots, a_n$ be distinct positive integers less than or equal to $n$. Determine the maximum value of\n\n$$\n\\sum_{i=1}^{n} |a_i - a_{i+1} + a_{i+2} - a_{i+3}|.\n$$\n\nwhere $a_{n+i} = a_i$ for $i = 1, 2, 3$.",
    "options": [],
    "answer": "See solution",
    "solution": "The answer is $n^2$ if $n$ is even, and $n^2 - 5$ if $n$ is odd.\n\nIt is easy when $n$ is even. Let $k = \\frac{n+1}{2}$. Then,\n\n$$\n\\begin{aligned}\n\\sum_{i=1}^{n} |a_i - a_{i+1} + a_{i+2} - a_{i+3}| &\\le \\sum_{i=1}^{n} (|k - a_i| + |k - a_{i+1}| + |k - a_{i+2}| + |k - a_{i+3}|) \\\\\n&= 4 \\sum_{i=1}^{n} |k - a_i| = n^2,\n\\end{aligned}\n$$\n\nand the equality holds when $a_{2i-1} = i$ and $a_{2i} = \\frac{n}{2} + 1$ for $i = 1, 2, \\dots, \\frac{n}{2}$. So, the maximum of $\\sum_{i=1}^{n} |a_i - a_{i+1} + a_{i+2} - a_{i+3}|$ is $n^2$.\n\nLet $n$ be odd. Without loss of generality, let $a_1 - a_2 > 0$. If $a_i - a_{i+1} > 0$ for all $i = 1, 2, \\dots, n$, then\n\n$$\n0 = (a_1 - a_2) + (a_2 - a_3) + \\dots + (a_n - a_1) > 0,\n$$\n\na contradiction. So, there exists $1 < x \\le n$ such that $a_x - a_{x+1} < 0$, and thus there exist distinct $1 \\le p, q \\le n$ such that $a_p - a_{p+1} > 0$, $a_{p+2} - a_{p+3} < 0$ and $a_q - a_{q+1} < 0$, $a_{q+2} - a_{q+3} > 0$.\n\nNow we consider the value of $|a_i - a_{i+1} + a_{i+2} - a_{i+3}|$. If $a_i - a_{i+1}$ and $a_{i+2} - a_{i+3}$ have the same sign then $|a_i - a_{i+1} + a_{i+2} - a_{i+3}| = |a_i - a_{i+1}| + |a_{i+2} - a_{i+3}|$, and if $a_i - a_{i+1}$ and $a_{i+2} - a_{i+3}$ have different signs then\n\n$$\n|a_i - a_{i+1} + a_{i+2} - a_{i+3}| = \\left||a_i - a_{i+1}| - |a_{i+2} - a_{i+3}|\\right| \\le |a_i - a_{i+1}| + |a_{i+2} - a_{i+3}| - 2.\n$$\n\nSo $\\sum_{i=1}^{n} |a_i - a_{i+1} + a_{i+2} - a_{i+3}|$ is at most\n\n$$\n\\begin{aligned}\n&\\sum_{1 \\le i \\le n, i \\ne p, q} (|a_i - a_{i+1}| + |a_{i+2} - a_{i+3}|) \\\\\n&\\quad + |a_p - a_{p+1}| + |a_{p+2} - a_{p+3}| + |a_q - a_{q+1}| + |a_{q+2} - a_{q+3}| - 4 \\\\\n&= 2 \\sum_{i=1}^{n} |a_i - a_{i+1}| - 4.\n\\end{aligned}\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 13247,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all positive integers $n$ such that $1 + 2^2 + 3^3 + 4^n$ is a perfect square.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $1 + 2^2 + 3^3 + 4^n = x^2$. This implies $32 + 4^n = x^2$, or equivalently, $x^2 - 4^n = 32$. Thus, $x^2 - (2^n)^2 = 32$, so $(x - 2^n)(x + 2^n) = 32 = 2^5$.\n\nSince $32$ is a power of $2$, both factors must be powers of $2$. Let $x - 2^n = 2^a$ and $x + 2^n = 2^{5-a}$ for some integer $a$ with $0 \\leq a \\leq 5$.\n\nSubtracting the first equation from the second gives:\n\n$$\n(x + 2^n) - (x - 2^n) = 2^{5-a} - 2^a \\\\\n2^{n+1} = 2^{5-a} - 2^a\n$$\n\nThis equation yields an integer $n$ only if $a = 2$, so $2^{n+1} = 2^{3}$. Thus, $n + 1 = 3$ and $n = 2$.\n\nHowever, checking $n = 2$:\n\n$$\n1 + 2^2 + 3^3 + 4^2 = 1 + 4 + 27 + 16 = 48\n$$\n\nBut $48$ is not a perfect square. Let's check $a = 2$ more carefully:\n\nIf $a = 2$, then $x - 2^n = 4$ and $x + 2^n = 8$.\n\nAdding:\n$$\n(x - 2^n) + (x + 2^n) = 4 + 8 = 12 \\\\\n2x = 12 \\implies x = 6\n$$\nSubtracting:\n$$\n(x + 2^n) - (x - 2^n) = 8 - 4 = 4 \\\\\n2^{n+1} = 4 \\implies n + 1 = 2 \\implies n = 1\n$$\n\nCheck $n = 1$:\n$$\n1 + 2^2 + 3^3 + 4^1 = 1 + 4 + 27 + 4 = 36 = 6^2\n$$\n\nThus, the only solution is $n = 1$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 13248,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $\\triangle ABC$ be a triangle and let $M$ be the midpoint of side $BC$. Externally to the triangle, consider parallelogram $BCDE$ such that $BE \\parallel AM$ and $BE = \\frac{1}{2} AM$. Prove that the line $EM$ passes through the midpoint of segment $AD$.\n\n![](images/GreekMO2014_booklet_p4_data_25393a6ca0.png)",
    "options": [],
    "answer": "See solution",
    "solution": "Extend $AM$ until it meets $ED$ at point $N$. Then $BMNE$ and $MCDN$ are parallelograms, so $EN = BM = MC = ND$. Thus, $N$ is the midpoint of $ED$. Moreover, $\\frac{AM}{MN} = 2$ and $M$ lies on the median of triangle $EAD$. Therefore, $M$ is the centroid of triangle $AED$. Thus, the line $EM$ is the median of triangle $AED$ from $E$, and so it intersects $AD$ at its midpoint.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 13249,
    "subject": "Mathematics (Olympiad)",
    "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": "See solution",
    "solution": "The answer is yes. Fix a positive integer $n$ and two large distinct primes $p, q > n$. Let $d_p = \\operatorname{ord}_p(n)$, $d_q = \\operatorname{ord}_q(n)$, $c_p = \\nu_p(n^{d_p} - 1)$, $c_q = \\nu_q(n^{d_q} - 1)$, and let $c = \\nu_p(d_q)$, $d = \\nu_q(d_p)$. Pick two sufficiently large constants $a, b$, and let\n\n$$\nM := d_p d_q p^{(a-1)c_p + b c_q - c} q^{a c_p + (b-1)c_q + 1 - d}\n$$\n\nFinally, choose $M_n$ to consist of $m = M(1 + iM)$ for $i = 1, 2, \\dots$.\n\nBy LTE, we have\n\n$$\n\\nu_p(n^m - 1) = \\nu_p(n^{d_p} - 1) + \\nu_p\\left(\\frac{m}{d_p}\\right) = c_p + c + (a-1)c_p + b c_q - c = a c_p + b c_q\n$$\n\n(we used that $\\gcd(1 + iM, p) = 1$) and similarly $\\nu_q(n^m - 1) = a c_p + b c_q + 1$, 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": 13250,
    "subject": "Mathematics (Olympiad)",
    "question": "For any sequence $a_1, a_2, \\ldots, a_{2013}$ of integers, we call a triple $(i, j, k)$ satisfying $1 \\leq i < j < k \\leq 2013$ *progressive* if $a_k - a_j = a_j - a_i = 1$. Determine the maximum number of *progressive* triples that a sequence of 2013 integers can have.",
    "options": [],
    "answer": "See solution",
    "solution": "Consider the following changes to a sequence $a_1, a_2, \\ldots, a_{2013}$:\n\ni) If $a_{n+1} < a_n$ for $n = 1, \\ldots, 2012$, swap the two numbers to get $(a_1, a_2, \\ldots, a_{n-1}, a_{n+1}, a_n, a_{n+2}, \\ldots, a_{2013})$.\n\nii) If $a_{n+1} = a_n + d + 1$ for $n = 1, \\ldots, 2012$ and $d > 0$, add $d$ to each of $a_1, a_2, \\ldots, a_n$ to obtain $(a_1 + d, a_2 + d, \\ldots, a_n + d, a_{n+1}, \\ldots, a_{2013})$.\n\nLet $m$ be the maximum number of *progressive* triples a sequence of 2013 integers can have. Applying operation i) does not decrease $m$, so we may assume the sequence is increasing. Applying operation ii) also does not decrease $m$, so it suffices to consider sequences of the form:\n\n$(a, \\ldots, a, \\underbrace{a+1, \\ldots, a+1}_{t_2}, \\ldots, \\underbrace{a+s-1, \\ldots, a+s-1}_{t_s})$, where $t_1, t_2, \\ldots, t_s$ are the counts of each value and $s \\geq 3$.\n\nFor such a sequence, $m = t_1 t_2 t_3 + t_2 t_3 t_4 + \\ldots + t_{s-2} t_{s-1} t_s$, where $s \\geq 3$ and $t_1, t_2, \\ldots, t_s$ are positive integers summing to 2013.\n\nThe maximum $m$ is achieved for $s = 3$ or $s = 4$. If $s \\geq 5$, replacing $t_1, t_2, \\ldots, t_s$ with $t_2, t_3, (t_1 + t_4), \\ldots, t_s$ yields a larger $m$.\n\nFor $s = 3$, $m = t_1 t_2 t_3 \\leq \\frac{(t_1 + t_2 + t_3)^3}{27} = 671^3$, with equality when $t_1 = t_2 = t_3 = 671$.\n\nFor $s = 4$, $m = (t_1 + t_4) t_2 t_3 \\leq \\frac{(t_1 + t_2 + t_3 + t_4)^3}{27} = 671^3$, with equality when $t_1 + t_4 = t_2 = t_3 = 671$.\n\nThus, $m = 671^3$.\n\n**Remark.** In maximizing $m = t_1 t_2 t_3 + t_2 t_3 t_4 + \\ldots + t_{s-2} t_{s-1} t_s$ for $s \\geq 3$ and $t_1, t_2, \\ldots, t_s$ non-negative integers summing to 2013, by the AM-GM inequality:\n\n$$\nm \\leq (t_1 + t_4 + \\ldots)(t_2 + t_5 + \\ldots)(t_3 + t_6 + \\ldots) \\leq \\left( \\frac{(t_1 + t_4 + \\ldots) + (t_2 + t_5 + \\ldots) + (t_3 + t_6 + \\ldots)}{3} \\right)^3 = 671^3.\n$$\n\nEquality holds, for example, if $s = 3$ and $t_1 = t_2 = t_3 = 671$, so the maximum value is $671^3$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 13251,
    "subject": "Mathematics (Olympiad)",
    "question": "Hallar todas las soluciones enteras $(x, y)$ de la ecuación\n\n$$\ny^k = x^2 + x\n$$\n\ndonde $k$ es un número entero dado mayor que $1$.",
    "options": [],
    "answer": "See solution",
    "solution": "Puesto que $y^k = x^2 + x = x(x+1)$ y $\\gcd(x, x+1) = 1$, resulta que tanto $x$ como $x+1$ deben ser potencias $k$-ésimas de un entero. Pero los dos únicos números enteros consecutivos que son potencias $k$-ésimas, con $k > 1$, son $0$ y $1$, o bien $-1$ y $0$. Las dos únicas soluciones son, pues, $x = 0$, $y = 0$ y $x = -1$, $y = 0$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 13252,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $ABCDEF$ be a regular octahedron with lower vertex $E$, upper vertex $F$, middle slice plane $ABCD$, center $M$, and circumsphere $k$. Furthermore, let $X$ be an arbitrary point within side $ABF$. The line $EX$ intersects $k$ in $E$ and $Z$, and the plane $ABCD$ in $Y$.\n\n$$\n\\text{Show that } \\langle EMZ \\rangle = \\langle EYF \\rangle.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "We intersect the entire figure with the plane through the points $X$, $E$, and $F$. Since $M$ is on line $EF$, it is part of that plane. Also, points $Y$ and $Z$ are part of that plane because they are on line $EX$. The intersection of the circumsphere $k$ and the plane results in a circle $k'$, which also has $M$ as its center. The intersection of $ABCD$ with the plane is the perpendicular bisector of line $EF$, and $Y$ is on that bisector.\n\n![](images/AustriaMO2013_p13_data_5928c37de4.png)\n\nWe denote $\\angle ZEF = \\alpha$.\n\nSince $ZM$ and $EM$ are both radii of $k$ (and $k'$), the triangle $ZME$ is isosceles, therefore\n$\\angle EZM = \\angle ZEM = \\alpha$.\n\nSince $Y$ is on the bisector of $EF$, also triangle $EYF$ is isosceles, therefore $\\angle YFE = \\angle YEF = \\alpha$.\n\nTherefore, the triangles $ZME$ and $EYF$ are similar, and their corresponding angles $\\angle EMZ$ and $\\angle EYF$ are identical. $\\square$",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 13253,
    "subject": "Mathematics (Olympiad)",
    "question": "a) If $n$ is odd, show that every pseudoendomorphism of $G$ is an endomorphism.\n\nb) If $n$ is even, is every pseudoendomorphism of $G$ an endomorphism?",
    "options": [],
    "answer": "See solution",
    "solution": "a) Let $e$ denote the unit of $G$. Let $x = y = z = e$ to write $f(e)^3 = f(e)$, so $f(e)^2 = e$. Since $n$ is odd, it follows that $f(e) = e$.\n\nIf $x$ and $y$ are members of $G$, write $f(xy) = f(xye) = f(x)f(y)f(e) = f(x)f(y)$, to conclude that $f$ is indeed an endomorphism of $G$.\n\nb) The answer is negative. Let $a$ be an order 2 element of $G$, and let $f: G \\to G$, $f(x) = a$. If $x, y, z$ are elements of $G$, then $f(xyz) = a = a^3 = f(x)f(y)f(z)$, so $f$ is a pseudoendomorphism. However, $f$ is not an endomorphism, since $f(e) = a \\neq e$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 13254,
    "subject": "Mathematics (Olympiad)",
    "question": "Өгөгдсөн дараалал $\\{a_n\\}$ нь дараах байдлаар тодорхойлогдоно:\n\n$$\na_0 = 1, \\quad a_1 = -1, \\quad a_n = 4a_{n-1} + 29a_{n-2} \\quad (n \\ge 2)\n$$\n\n$a_{2012}$-ийг $101$-д хуваахад гарах үлдэгдлийг ол.",
    "options": [],
    "answer": "See solution",
    "solution": "Дараалал $\\{a_n\\}$-тай хамт $\\{b_n\\}$ дарааллыг авч үзье, энд:\n\n$$\nb_0 = 1, \\quad b_1 = -1, \\quad b_n = 4b_{n-1} + 837b_{n-2} \\quad (n \\ge 2)\n$$\n\n$837 \\equiv 29 \\pmod{101}$ тул $a_n \\equiv b_n \\pmod{101}$.\n\nДарааллын характеристик тэгшитгэл:\n$$\nx^2 - 4x - 837 = 0\n$$\nҮндэс нь $x_1 = 31$, $x_2 = -27$ тул:\n$$\nb_n = c_1 \\cdot 31^n + c_2(-27)^n\n$$\n\nЭхний нөхцлүүдээс:\n$$\nb_0 = 1 = c_1 + c_2 \\\\\nb_1 = 31c_1 - 27c_2 = -1\n$$\nҮүнийг бодож:\n$$\nc_1 = \\frac{13}{29}, \\quad c_2 = \\frac{16}{29}\n$$\nТэгэхээр:\n$$\n29b_n = 13 \\cdot 31^n + 16(-27)^n\n$$\n\nОдоо $b_{2012}$-ийг $101$-д хуваахад гарах үлдэгдлийг олъё.\n\n$101$ анхны тоо тул Фермагийн бага теоремоос:\n$$\n31^{100} \\equiv 1 \\pmod{101}, \\quad 27^{100} \\equiv 1 \\pmod{101}\n$$\n\nТэгэхээр:\n$$\n31^{2012} = (31^{100})^{20} \\cdot 31^{12} \\equiv 31^{12} \\pmod{101}\n$$\n$31^{12} \\equiv 54 \\pmod{101}$\n\n$27^{2012} = (27^{100})^{20} \\cdot 27^{12} \\equiv 27^{12} \\pmod{101}$\n$27^{12} \\equiv 20 \\pmod{101}$\n\nИймд:\n$$\n29b_{2012} \\equiv 13 \\cdot 54 + 16 \\cdot 20 \\equiv 702 + 320 = 1022 \\equiv 12 \\pmod{101}\n$$\n\n$29x \\equiv 12 \\pmod{101}$-ийг бодъё. Евклидийн алгоритмаар $29 \\cdot 7 \\equiv 1 \\pmod{101}$ тул $x = 7 \\cdot 12 = 84$.\n\nИймд:\n$$\na_{2012} \\equiv b_{2012} \\equiv 84 \\pmod{101}\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 13255,
    "subject": "Mathematics (Olympiad)",
    "question": "Prime number $p > 2$ and the polynomial $Q$ with integer coefficients are such that there do not exist two positive integers $i, j$ with $i < j < p$ and the number $(Q(j) - Q(i))(jQ(j) - iQ(i))$ is divisible by $p$. What is the smallest possible degree of $Q$?",
    "options": [],
    "answer": "See solution",
    "solution": "The smallest possible degree is $p-2$.\n\nExample: $Q(x) = x^{p-2} - 1$. For any $i < j < p$:\n\n$$\nQ(j) - Q(i) = j^{p-2} - i^{p-2} \\equiv \\frac{1}{j} - \\frac{1}{i} \\pmod{p},\n$$\nwhich is not divisible by $p$.\n\nAlso,\n$$\njQ(j) - iQ(i) = j^{p-1} - i^{p-1} - (j - i) \\equiv (i - j) \\pmod{p},\n$$\nwhich is not divisible by $p$.\n\nSuppose there exists a polynomial of smaller degree. Then all of $Q(1), Q(2), \\dots, Q(p-1)$ give different remainders modulo $p$, and also $1Q(1), 2Q(2), \\dots, (p-1)Q(p-1)$ give different remainders modulo $p$.\n\nRecall: $1^k + 2^k + \\dots + (p-1)^k \\equiv 0 \\pmod{p}$ for $1 \\leq k \\leq p-2$. Consider $1Q(1) + 2Q(2) + \\dots + (p-1)Q(p-1)$. Each degree in $xQ(x)$ is in $[1, p-2]$, so this sum is $0$ modulo $p$.\n\nIf $1Q(1), \\dots, (p-1)Q(p-1)$ cover all residues except $x$, then:\n$$\n0 \\equiv 1Q(1) + \\dots + (p-1)Q(p-1) + x \\equiv x \\pmod{p},\n$$\nso $1Q(1), \\dots, (p-1)Q(p-1)$ modulo $p$ are $1, 2, \\dots, p-1$ in some order. Similarly, $Q(1), \\dots, Q(p-1)$ are $1, 2, \\dots, p-1$ in some order.\n\nThen:\n$$\n(p-1)! \\equiv 1Q(1) \\times 2Q(2) \\times \\dots \\times (p-1)Q(p-1) \\equiv ((p-1)!)^2 \\pmod{p},\n$$\nBy Wilson's theorem, $-1 \\equiv 1 \\pmod{p}$, a contradiction. Thus, the minimal degree is $p-2$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 13256,
    "subject": "Mathematics (Olympiad)",
    "question": "Finde die kleinste Zahl der Form $2^a - 2$, die größer oder gleich $100\\,000$ ist.",
    "options": [],
    "answer": "See solution",
    "solution": "Wir suchen das kleinste $a \\in \\mathbb{Z}^+$ mit $2^a - 2 \\geq 100\\,000$. \n\nSetze $2^a \\geq 100\\,002$. \n\nBerechne $a = \\lceil \\log_2(100\\,002) \\rceil$. \n\nDa $2^{17} = 131\\,072$, ist $a = 17$ die kleinste Lösung. \n\nAlso ist die gesuchte Zahl $2^{17} - 2 = 131\\,072 - 2 = 131\\,070$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 13257,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $k$ be an integer and $\\theta$ a real number such that $\\cos k\\theta$ and $\\cos[(k+1)\\theta]$ are both rational, and $\\cos \\theta$ is irrational. Prove that $\\theta = \\frac{\\pi}{6}$.",
    "options": [],
    "answer": "See solution",
    "solution": "Both $\\cos(k^2\\theta) = \\cos[k(k\\theta)]$ and $\\cos[(k^2-1)\\theta] = \\cos[(k-1)(k+1)\\theta]$ are rational. By the addition and subtraction formulas, we have\n\n$$\n\\cos[(k+1)\\theta] = \\cos k\\theta \\cos \\theta - \\sin k\\theta \\sin \\theta \\quad \\text{and} \\quad \\cos(k^2\\theta) = \\cos[(k^2-1)\\theta] \\cos \\theta - \\sin[(k^2-1)\\theta] \\sin \\theta.\n$$\n\nSetting $r_1 = \\cos k\\theta$, $r_2 = \\cos[(k+1)\\theta]$, $r_3 = \\cos[(k^2-1)\\theta]$, $r_4 = \\cos(k^2\\theta)$, and $x = \\cos \\theta$ in the above equations yields\n\n$$\nr_2 = r_1 x \\pm \\sqrt{(1 - r_1^2)(1 - x^2)} \\quad \\text{and} \\quad r_4 = r_3 x \\pm \\sqrt{(1 - r_3^2)(1 - x^2)},\n$$\n\nor\n\n$$\n\\pm\\sqrt{(1 - r_1^2)(1 - x^2)} = r_2 - r_1 x \\quad \\text{and} \\quad \\pm\\sqrt{(1 - r_3^2)(1 - x^2)} = r_4 - r_3 x.\n$$\n\nSquaring these two equations and subtracting the resulting equations gives\n\n$$\n2(r_1 r_2 - r_3 r_4)x = r_1^2 + r_2^2 - (r_3^2 + r_4^2).\n$$\n\nSince $r_1, r_2, r_3, r_4$ are rational and $x$ is irrational, we must have $r_1 r_2 - r_3 r_4 = 0$ or\n\n$$\n\\cos k\\theta \\cos[(k+1)\\theta] = \\cos(k^2\\theta) \\cos[(k^2-1)\\theta].\n$$\n\nBy the product-to-sum formulas, we derive\n\n$$\n\\frac{\\cos[(2k + 1)\\theta] - \\cos \\theta}{2} = \\frac{\\cos[(2k^2 - 1)\\theta] - \\cos \\theta}{2}\n$$\n\nor $\\cos[(2k + 1)\\theta] - \\cos[(2k^2 - 1)\\theta] = 0$. By the sum-to-product formulas, we obtain\n\n$$\n2 \\sin[(k - k^2 + 1)\\theta] \\sin[(k^2 + k)\\theta] = 0,\n$$\n\nimplying that either $(k - k^2 + 1)\\theta$ or $(k^2 + k)\\theta$ is an integral multiple of $\\pi$. Since $k$ is an integer, we conclude that $\\theta = r\\pi$ for some rational number $r$.\n\nConsidering Lemma 2 for $\\alpha = k\\theta$ and $\\alpha = (k+1)\\theta$, the possible values of $\\cos k\\theta$ and $\\cos[(k+1)\\theta]$ are $0, \\pm 1, \\pm \\frac{1}{2}$. Consequently, both $k\\theta$ and $(k+1)\\theta$ are integral multiples of $\\frac{\\pi}{6}$. Since $0 < \\theta = k\\theta - (k-1)\\theta < \\frac{\\pi}{2}$, the only possible values of $\\theta$ are $\\frac{\\pi}{3}$ and $\\frac{\\pi}{6}$. Since $\\cos \\theta$ is irrational, $\\theta = \\frac{\\pi}{6}$.\n\n**Second Solution:** (Based on the work by Kiran Kedlaya) We maintain the notations used in the first proof. Then $s = 2 \\cos \\theta$ is a root of $S_k(x) - 2r_1$ and $S_{k+1}(x) - 2r_2$ by the definition of $S_n$. Define\n\n$$\nQ(x) = \\gcd(S_k(x) - 2r_1, S_{k+1}(x) - 2r_2)\n$$\n\nwhere the gcd is taken over the field of rational numbers. Then $Q(x)$ is a polynomial with rational coefficients, so the sum of its roots (with multiplicities) is rational. Since $s$ is assumed not to be rational, there must be at least one other distinct root $t$ of $Q(x)$.\n\nNote that the $k$ distinct reals $2\\cos(\\theta + 2\\pi a/k)$ for $a = 0, 1, \\dots, k-1$ form $k$ roots of the degree $k$ polynomial $S_k(x) - 2r_1$, so they compose all of its roots. Similarly, all of the roots of $S_{k+1}(x) - 2r_2$ have the form $2\\cos(\\theta + 2\\pi b/(k+1))$ for $b = 0, 1, \\dots, k$. Note that $s$ and $t$ are roots of $Q(x)$. Therefore, roots of both $S_k(x) - 2r_1$ and $S_{k+1}(x) - 2r_2$, and so they must have at least two distinct common roots. Each root $r$ of $Q(x)$ must thus satisfy\n\n$$\nr = 2 \\cos(\\theta + 2\\pi a/k) = 2 \\cos(\\theta + 2\\pi b/(k+1))\n$$\nfor some $a$ and $b$. We either have $\\theta + 2\\pi a/k = \\theta + 2\\pi b/(k+1)$ and thus $r = 2 \\cos \\theta$ or $\\theta + 2\\pi a/k = -\\theta - 2\\pi b/(k+1)$ and thus\n\n$$\n\\theta = - \\frac{\\pi[(a+b)k+a]}{k(k+1)}.\n$$\n\nIn the first case, we obtain $s$, so $t$ must lead to the second value of $\\theta$, as $s \\neq t$.\n\nTherefore, we can write $\\theta = \\frac{\\pi c}{k(k+1)}$ for some integer $c$. By Lemma 2, $c/k$ and $c/(k+1)$ must both be multiples of $1/6$, since $\\cos k\\theta = \\cos \\frac{c\\pi}{k+1}$ and $\\cos(k+1)\\theta = \\cos \\frac{c\\pi}{k}$ are rational. Therefore, $\\theta = \\frac{c\\pi}{k} - \\frac{c\\pi}{k+1}$ is a multiple of $\\pi/6$. Since $t$ is not rational, $\\theta$ can only be $\\pi/6$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 13258,
    "subject": "Mathematics (Olympiad)",
    "question": "設三角形 $ABC$ 為銳角三角形，$A_1, B_1, C_1$ 分別位於 $BC, CA, AB$ 邊上，且 $AA_1, BB_1, CC_1$ 皆為三角形 $ABC$ 的內角平分線。令點 $I$ 為三角形 $ABC$ 的內心，點 $H$ 為三角形 $A_1B_1C_1$ 的垂心。證明：\n\n$$\nAH + BH + CH \\geq AI + BI + CI.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "記 $\\angle BAC = \\alpha$, $\\angle CBA = \\beta$, $\\angle ACB = \\gamma$。不失一般性，可設 $\\alpha \\leq \\beta \\leq \\gamma$。\n\n另將三角形 $ABC$ 的三邊長分別記為 $BC = a$, $CA = b$, $AB = c$。\n\n我們首先證明：$\\triangle A_1B_1C_1$ 也是銳角三角形。在 $BC$ 邊上取點 $D, E$ 滿足：$B_1D \\parallel AB$，且 $B_1E$ 為 $\\angle BB_1C$ 的內角平分線。由於 $\\angle B_1DB = 180^\\circ - \\beta$ 是鈍角，知 $BB_1 > B_1D$。於是有\n\n$$\n\\frac{BE}{EC} = \\frac{BB_1}{B_1C} > \\frac{DB_1}{B_1C} = \\frac{BA}{AC} = \\frac{BA_1}{A_1C}.\n$$\n\n由此知 $BE > BA_1$，且 $\\frac{1}{2}\\angle BB_1C = \\angle BB_1E > \\angle BB_1A_1$。同理得 $\\frac{1}{2}\\angle BB_1A > \\angle BB_1C_1$。所以\n\n$$\n\\angle A_1B_1C_1 = \\angle BB_1A_1 + \\angle BB_1C_1 < \\frac{1}{2}(\\angle BB_1C + \\angle BB_1A) = 90^\\circ\n$$\n\n為銳角。由對稱性，得證 $\\triangle A_1B_1C_1$ 為銳角三角形。\n\n回到原題。設直線 $BB_1$ 與 $A_1C_1$ 交於點 $F$。由 $\\alpha \\leq \\gamma$，知 $a \\leq c$，於是有\n\n$$\nBA_1 = \\frac{ca}{b+c} \\leq \\frac{ac}{a+b} = BC_1\n$$\n\n得 $\\angle BC_1A_1 \\leq \\angle BA_1C_1$。因為 $BF$ 是 $\\angle A_1BC_1$ 的內角平分線，$\\angle B_1FC_1 = \\angle BFA_1 \\leq 90^\\circ$。所以 $H$ 與 $C_1$ 落在直線 $BB_1$ 的同一側，得 $H$ 會落在三角形 $BB_1C_1$ 的內部。類似地，因為 $\\alpha \\leq \\beta$ 及 $\\beta \\leq \\gamma$，知 $H$ 落在三角形 $CC_1B_1$ 的內部，也落在三角形 $AA_1C_1$ 的內部。\n\n![](images/17-3J_p32_data_cfb0c1a18e.png)\n\n由於 $\\alpha \\leq \\beta \\leq \\gamma$，所以 $\\alpha \\leq 60^\\circ \\leq \\gamma$。故 $\\angle BIC \\leq 120^\\circ \\leq \\angle AIB$。我們先來討論 $\\angle AIC \\geq 120^\\circ$ 的情形。\n\n將 $B, I, H$ 各點以 $A$ 點為中心旋轉 $60^\\circ$，分別得到 $B', I', H'$ 點，並使 $B'$ 與 $C$ 點位於直線 $AB$ 的異側。因為 $\\triangle AI'I$ 為正三角形，知\n\n$$\nAI + BI + CI = I'I + B'I' + IC = B'I' + I'I + IC. \\quad (1)\n$$\n\n同理知\n\n$$\nAH + BH + CH = H'H + B'H' + HC = B'H' + H'H + HC. \\quad (2)\n$$\n\n由於 $\\angle AII' = \\angle AI'I = 60^\\circ$、$\\angle AI'B' = \\angle AIB \\geq 120^\\circ$ 以及 $\\angle AIC \\geq 120^\\circ$，$B'I'IC$ 為凸四邊形，並與 $A$ 點落在直線 $B'C$ 的同側。\n\n接著，因為 $H$ 在三角形 $ACC_1$ 的內部，$H$ 會落在四邊形 $B'I'IC$ 的外部。同時，$H$ 落在三角形 $ABI$ 的內部，得 $H'$ 也落在三角形 $AB'I'$ 的內部。所以 $H'$ 也落在 $B'I'IC$ 的外部。因此，四邊形 $B'I'IC$ 整個落在四邊形 $B'H'HC$ 的內部。由此知 $B'I'IC$ 的周長不超過 $B'H'HC$ 的周長。於是由 (1) 及 (2) 可得\n\n$$\nAH + BH + CH \\geq AI + BI + CI.\n$$\n\n當 $\\angle AIC < 120^\\circ$ 時，我們可將 $B, I, H$ 點以 $C$ 點為中心旋轉 $60^\\circ$，分別到 $B', I', H'$ 點，並使 $B'$ 與 $A$ 位於 $BC$ 的異側。這個情形的證明與上面的情形類似，而得到相同的不等式。證明完畢。",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 13259,
    "subject": "Mathematics (Olympiad)",
    "question": "某臉書社團有若干成員，其中有一些成員彼此是朋友（朋友關係是雙向的）。已知若兩個人是朋友，則他們沒有其他的共同朋友。反之，若兩個人不是朋友，則他們恰有兩個共同好友。\n\n證明：此社團中存在兩個人，他們的好友數量是相同的。",
    "options": [],
    "answer": "See solution",
    "solution": "以成員為點，朋友關係為邊作圖 $G = (V, E)$。對於任一點 $a$，令 $N(a)$ 為與 $a$ 相連的所有點所構成集合。\n\n由條件一知，對於有連線的任兩點 $a$ 與 $b$，$N(a) - \\{b\\}$ 與 $N(b) - \\{a\\}$ 沒有交集（否則交集的點會成為這兩個人的共同好友，矛盾）。\n\n此外，對於任何 $x \\in N(a) - \\{b\\}$，因為 $b$ 與 $x$ 不是朋友且 $a$ 是他們的兩個共同好友，故存在唯一的 $y \\in N(b) - \\{a\\}$ 與 $x$ 有連線。以上建立了一個 $N(a) - \\{b\\}$ 與 $N(b) - \\{a\\}$ 之間的一對一且映成關係，故兩集合的點數相等，因此 $|N(a)| = |N(b)|$，證畢。",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 13260,
    "subject": "Mathematics (Olympiad)",
    "question": "Suppose the angle $\\angle XYZ$ of a triangle $\\triangle XYZ$ is $60^\\circ$. Let the lengths of the sides $YZ, XZ, XY$ be $x, y, z$ respectively. Prove that the inequality $2x \\ge y + z$ holds.\n\nNow, consider a convex hexagon $ABCDEF$ for which any pair of diagonals chosen from $AD, BE, CF$ intersect at $60^\\circ$. Let the points $P, Q, R$ be the points of intersection of the line segments $AD$ and $BE$, $BE$ and $CF$, $CF$ and $AD$, respectively. Prove that\n$$\nAB + BC + CD + DE + EF + FA \\geq AD + BE + CF.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "First, we prove the following lemma:\n\n*Lemma:* Suppose the angle $\\angle XYZ$ of a triangle $\\triangle XYZ$ is $60^\\circ$. Let the lengths of the sides $YZ, XZ, XY$ be $x, y, z$ respectively. Then, the inequality $2x \\ge y + z$ holds.\n\n*Proof:* Let $\\Gamma$ be the circumcircle of $\\triangle XYZ$. If we move the vertex $X$ along the arc $\\widehat{YZ}$ from $Y$ to $Z$, the area of $\\triangle XYZ$ changes and attains its maximum when the height is largest. This happens when $X$ lies on the perpendicular bisector of $YZ$, making the triangle a right triangle with side length $x$, since $\\angle XYZ$ remains $60^\\circ$. The area equals $\\frac{1}{2} x y \\cos 60^\\circ = yz$, and when the triangle is right, this value equals $x^2$, as $x = y = z$ in this case. Therefore, $yz \\le x^2$.\n\nLet $H$ be the foot of the perpendicular from $Y$ to $XZ$. By the Pythagorean theorem:\n$$\n\\begin{align*}\nx^2 &= \\left(y - \\frac{1}{2}z\\right)^2 + \\frac{3}{4}z^2 = y^2 + z^2 - yz \\\\\n&= (y + z)^2 - 3yz \\ge (y + z)^2 - 3x^2,\n\\end{align*}\n$$\nsince $x^2 \\ge yz$. Thus, $4x^2 \\ge (y+z)^2$ and $2x \\ge y+z$, since $x, y, z > 0$. This proves the lemma.\n\nNow, consider the convex hexagon $ABCDEF$ where any pair of diagonals from $AD, BE, CF$ intersect at $60^\\circ$. Let $P, Q, R$ be the intersection points of $AD$ and $BE$, $BE$ and $CF$, $CF$ and $AD$, respectively. Since the hexagon is convex, the angle $\\angle APB$ of $\\triangle APB$ is $60^\\circ$. By the lemma, $2AB \\ge PA + PB$. Similarly, apply the lemma to the triangles $\\triangle BQC, \\triangle CRD, \\triangle DPE, \\triangle EQF, \\triangle FRA$, and sum the corresponding inequalities:\n$$\n\\begin{aligned}\n& 2AB + 2BC + 2CD + 2DE + 2EF + 2FA \\\\\n& \\geq (PA + PB) + (QB + QC) + (RC + RD) \\\\\n& \\quad + (PD + PE) + (QE + QF) + (RF + RA)\n\\end{aligned}\n$$\nSince $AD = AP + PD = AR + RD$, $BE = BP + PE = BQ + QE$, $CF = CR + RF = CQ + QF$, the right-hand side equals $2(AD + BE + CF)$, so we get:\n$$\nAB + BC + CD + DE + EF + FA \\geq AD + BE + CF.\n$$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 13261,
    "subject": "Mathematics (Olympiad)",
    "question": "A $4 \\times 4$ table is divided into 16 white unit square cells. Two cells are called neighbours if they share a common side. A move consists of choosing a cell and changing its color and the colors of its neighbours from white to black or from black to white. After exactly $n$ moves, all 16 cells are black. Find all possible values of $n$.",
    "options": [],
    "answer": "See solution",
    "solution": "A move affects at most 5 cells, so at least 4 moves are needed to change the color of every cell. If we place the move 4 times so that the center of the move lies in a dark cell, it is possible to achieve all-black cells with $n=4$.\n\nFurthermore, applying the operation twice on the same cells, it is possible to get the table with every cell colored black in $n$ steps for every even $n \\geq 4$.\n\nWe shall prove that for odd $n$ it is not possible. Let $k$ be the difference between the number of white and black cells in the dark area. Every move covers an odd number of dark cells, so after every step, $k$ changes by a number $\\equiv 2 \\pmod{4}$. At the beginning, $k=10$; at the end, $k=-10$. Thus, we need an even number of steps. So, $n$ is every even number except 2.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 13262,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $F(a, b) = f(a) + f(b) + 2ab$ for a function $f : \\mathbb{N} \\to \\mathbb{N}$. Find all functions $f$ such that for all $a, b \\in \\mathbb{N}$, $F(a, b)$ is a perfect square.",
    "options": [],
    "answer": "See solution",
    "solution": "$$\n\\begin{aligned}\n&\\text{Let } F(a, 2) = f(a) + f(2) + 4a = x^2 \\\\\n&F(a, 1) = f(a) + f(1) + 2a = y^2 \\\\\n&\\implies 2a + f(2) - f(1) = x^2 - y^2\n\\end{aligned}\n$$\n\nIf $f(2) - f(1) = 2k$,\n$$\n\\begin{aligned}\n2(a + k) &= x^2 - y^2 \\\\\n2 \\mid x^2 - y^2 \\\\\n4 \\mid x^2 - y^2 \\\\\n4 \\mid 2(a + k) \\\\\n2 \\mid a + k\n\\end{aligned}\n$$\nThis statement is true for any natural $a$, but we can choose $a$ such that $2 \\nmid a + k$, which is a contradiction.\n\nSo $f(2) - f(1) = 2k + 1$.\n\n$$\n2a + 2k + 1 = x^2 - y^2\n$$\n\nChoose an odd prime $p$ and set $a = \\frac{p - (2k + 1)}{2}$.\n$$\n\\begin{aligned}\np &= x^2 - y^2 = (x - y)(x + y) \\\\\n\\text{Set } x + y = p,\\ x - y = 1 \\\\\nx = \\frac{p + 1}{2}\n\\end{aligned}\n$$\n\nPlugging back,\n$$\n\\begin{aligned}\nf(a) + f(2) + 4a &= \\left(\\frac{p + 1}{2}\\right)^2 \\\\\nf(a) + f(2) + 4a &= (a + k + 1)^2\n\\end{aligned}\n$$\n\nSo,\n$$\nf(a) = (a + (k - 1))^2 + (k + 1)^2 - (k - 1)^2 + f(2)\n$$\n\nLet $k - 1 = b$, $(k + 1)^2 - (k - 1)^2 + f(2) = c$, so for $a = \\frac{p - (2k + 1)}{2}$,\n$$\nf(a) = (a + b)^2 + c\n$$\n\nFor these $a$,\n$$\nF(a, a) = 2f(a) + 2a^2 = (2a + b)^2 + b^2 + 2c\n$$\n\nFor large $a$ such that $|b^2 + 2c| < 2a$,\n$$\n(2a + b)^2 \\leq (2a + b)^2 + b^2 + 2c < (2a + b + 1)^2\n$$\nSo,\n$$\n(2a + b)^2 = (2a + b)^2 + b^2 + 2c \\implies b^2 + 2c = 0 \\implies b = 2\\ell, c = -2\\ell^2\n$$\n\nThus, for infinitely many $a$,\n$$\nf(a) = (a + 2\\ell)^2 - 2\\ell^2\n$$\n\nNow, for arbitrary $b$ and $a$ as above,\n$$\nf(a) + f(b) + 2ab = (a + 2\\ell)^2 - 2\\ell^2 + f(b) + 2ab = (a + b + 2\\ell)^2 - b^2 + f(b) - 4b\\ell - 2\\ell^2\n$$\n\nFor large $a$,\n$$\n(a + b + 2\\ell)^2 \\leq (a + b + 2\\ell)^2 - b^2 + f(b) - 4b\\ell - 2\\ell^2 < (a + b + 2\\ell + 1)^2\n$$\nSo,\n$$\n-b^2 + f(b) - 4b\\ell - 2\\ell^2 = 0 \\implies f(b) = (b + 2\\ell)^2 - 2\\ell^2\n$$\n\nTherefore,\n$$\n\\forall n \\in \\mathbb{Z}^+: f(n) = (n + 2\\ell)^2 - 2\\ell^2\n$$\n\nIf $\\ell < 0$, for $n = -2\\ell$, $f(-2\\ell) \\in \\mathbb{Z}^+$ implies $-2\\ell^2 \\in \\mathbb{Z}^+$, which is a contradiction. So $\\ell \\geq 0$.\n\nChecking in the main equality, for any such $\\ell$,\n$$\nf(a) + f(b) + 2ab = (a + b + 2\\ell)^2.\n$$\n\n\\text{■}\n",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 13263,
    "subject": "Mathematics (Olympiad)",
    "question": "There are $2m$ boys and $13$ girls in grade $10$, and $7$ boys and $2n$ girls in grade $11$, where $m$ and $n$ are positive integers. Each learner pays the same integral number of rands into a fund, and the total amount of money raised by each grade is $2mn + 7m + 13n + 84$ rands. What is the number of rands paid by each learner?",
    "options": [],
    "answer": "See solution",
    "solution": "Let each learner contribute $k$ rands. Since each grade collects the same total, the number of learners in each grade must be equal:\n\n$$\n2m + 13 = 2n + 7 \\implies n = m + 3.\n$$\n\nNow,\n$$\nk = \\frac{2mn + 7m + 13n + 84}{2m + 13}.\n$$\nSubstitute $n = m + 3$:\n$$\nk = \\frac{2m(m+3) + 7m + 13(m+3) + 84}{2m + 13} = \\frac{2m^2 + 6m + 7m + 13m + 39 + 84}{2m + 13}.\n$$\nSimplify numerator:\n$$\n2m^2 + 6m + 7m + 13m + 39 + 84 = 2m^2 + 26m + 123.\n$$\nSo,\n$$\nk = \\frac{2m^2 + 26m + 123}{2m + 13}.\n$$\nDivide numerator:\n$$\n2m^2 + 26m + 123 = (2m + 13)(m + 6) + 45.\n$$\nSo,\n$$\nk = m + 6 + \\frac{45}{2m + 13}.\n$$\nFor $k$ to be integer, $2m + 13$ must divide $45$. The positive divisors of $45$ are $1, 3, 5, 9, 15, 45$.\n\nSet $2m + 13 = d$ for each divisor $d$:\n- $d = 15 \\implies 2m = 2 \\implies m = 1$\n- $d = 45 \\implies 2m = 32 \\implies m = 16$\n\nTry $m = 16$:\n- $n = m + 3 = 19$\n- Number of learners: $2m + 13 = 45$\n- Total amount: $2mn + 7m + 13n + 84 = 2 \\times 16 \\times 19 + 7 \\times 16 + 13 \\times 19 + 84 = 608 + 112 + 247 + 84 = 1051$\n- $k = 1051 / 45 = 23.355...$ (not integer)\n\nTry $m = 1$:\n- $n = 4$\n- Number of learners: $2 \\times 1 + 13 = 15$\n- Total amount: $2 \\times 1 \\times 4 + 7 \\times 1 + 13 \\times 4 + 84 = 8 + 7 + 52 + 84 = 151$\n- $k = 151 / 15 = 10.066...$ (not integer)\n\nBut in the original solution, $a = 4$ gives $m = 32$.\n\nTry $a = 4$:\n- $7m + 84 = 4(2m + 13) \\implies 7m + 84 = 8m + 52 \\implies m = 32$\n- $n = m + 3 = 35$\n- Number of learners: $2 \\times 32 + 13 = 77$\n- Total amount: $2 \\times 32 \\times 35 + 7 \\times 32 + 13 \\times 35 + 84 = 2240 + 224 + 455 + 84 = 3003$\n- $k = 3003 / 77 = 39$\n\nThus, the number of rands paid by each learner is $\\boxed{39}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 13264,
    "subject": "Mathematics (Olympiad)",
    "question": "How many different ways are there to cover a $4 \\times 4$ square with five $3 \\times 1$ rectangles, so that exactly one $1 \\times 1$ cell is left uncovered?\n\n![](images/UkraineMO2019_booklet_p15_data_899d171811.png)",
    "options": [],
    "answer": "See solution",
    "solution": "We can cover the $1 \\times 1$ cells in two ways as shown in Fig. 13. Since each $3 \\times 1$ rectangle covers exactly one cell of each color, only a white cell can be left uncovered (since there are 6 white cells and 5 each of grey and black cells). The possible positions for the uncovered white cell are only on the edges of the $4 \\times 4$ square, giving 4 options. The number of coverings for each type is the same. There are four coverings for each (see Fig. 14).\n\nThere are three ways with one horizontal $3 \\times 1$ rectangle on the bottom, and one way without it. Thus, there are $4 \\times 4 = 16$ ways in total.\n\n![](images/UkraineMO2019_booklet_p15_data_4d305832c6.png)",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 13265,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $a_1 > \\frac{1}{12}$ and $a_{n+1} = \\sqrt{(n+2)a_n + 1}$ for $n \\ge 1$. Prove that:\n\na) $a_n > n - \\frac{2}{n}$;\n\nb) The sequence $b_n = 2^n \\left(\\frac{a_n}{n} - 1\\right)$ for $n = 1, 2, \\dots$ is convergent.",
    "options": [],
    "answer": "See solution",
    "solution": "a) We prove by induction that $a_n > n - \\frac{2}{n}$ for $n \\ge 3$ (since $a_1 > \\frac{1}{12}$).\n\nFor $n \\le 3$, $a_2 > \\sqrt{3 \\cdot \\frac{19}{243} + 1} = \\frac{10}{9}$ and $a_3 > \\sqrt{4 \\cdot \\frac{10}{9} + 1} = \\frac{7}{3}$.\n\nAssume $a_n > n - \\frac{2}{n}$ for some $n \\ge 3$. Then\n$$\na_{n+1} > \\sqrt{(n+2)\\left(n - \\frac{2}{n}\\right) + 1}.\n$$\nWe need to show this is greater than $n+1 - \\frac{2}{n+1}$. This reduces to $\\frac{1}{2} > \\frac{1}{n} + \\frac{1}{(n+1)^2}$, which holds for $n \\ge 3$.\n\nb) If $a_1 = 1$, then $a_n = n$ by induction, so $b_n = 0$.\n\nIf $a_1 < 1$, then $a_n < n$ by induction, so $b_n < 0$. We show $b_n < b_{n+1}$, which is equivalent to $\\frac{a_n - n}{2n} < \\frac{a_{n+1} - n - 1}{n+1}$.\n\nThe right side equals\n$$\n\\frac{a_{n+1}^2 - n - 1}{(n+1)(a_{n+1} + n + 1)} = \\frac{(n+2)a_n + 1 - (n+1)^2}{(n+1)(a_{n+1} + n + 1)} = \\frac{(n+2)(a_n - n)}{(n+1)(a_{n+1} + n + 1)}.\n$$\nIt remains to show $\\frac{1}{2n} > \\frac{n+2}{(n+1)(a_{n+1} + n + 1)}$, i.e.,\n$$\n(n+1)a_{n+1} > 2n(n+2) - (n+1)^2 = (n+1)^2 - 2,\n$$\nwhich follows by part a).\n\nIf $a_1 > 1$, then $b_n > b_{n+1} > 0$.\n\nThus, $(b_n)$ is monotone and bounded, so it converges.\n\n**Remark.** Similarly, if $a_1 \\ge -\\frac{1}{3}$, then the sequence $\\left( \\frac{nb_n}{n-2} \\right)_{n \\ge 3}$ is monotone and bounded, so $(b_n)$ converges.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 13266,
    "subject": "Mathematics (Olympiad)",
    "question": "Find $A_{2024}$, where\n\n$$\nA_n = 1 \\cdot 2 + 3 \\cdot 4 + 5 \\cdot 8 + \\dots + (2n-1) \\cdot 2^n.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "Since\n\n$$\n2A_n = 1 \\cdot 4 + 3 \\cdot 8 + \\dots + (2n-3) \\cdot 2^n + (2n-1) \\cdot 2^{n+1},\n$$\n\nit follows\n\n$$\n\\begin{align*}\nA_n = 2A_n - A_n &= (2n-1) \\cdot 2^{n+1} - (1 \\cdot 2 + 2 \\cdot 4 + 2 \\cdot 8 + \\dots + 2 \\cdot 2^n) \\\\\n&= (2n-1) \\cdot 2^{n+1} - 2 \\cdot (2 + 4 + 8 + \\dots + 2^n) + 1 \\cdot 2 \\\\\n&= (2n-1) \\cdot 2^{n+1} - 2 \\cdot 2 \\cdot \\frac{2^n-1}{2-1} + 2 \\\\\n&= (2n-3) \\cdot 2^{n+1} + 6.\n\\end{align*}\n$$\n\nTherefore, $A_{2024} = 4045 \\cdot 2^{2025} + 6$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 13267,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $d$ be a number such that $d \\mid ab$ and $d \\in (m^2, m^2 + m)$. Show that $d \\mid (a-d)(b-d)$ and $|a-d| < m$, $|b-d| < m$. Conclude that $|(a-d)(b-d)| < m^2 < d$, and deduce that $$(a-d)(b-d) = 0.$$ What can you say about $d$ in terms of $a$ and $b$?",
    "options": [],
    "answer": "See solution",
    "solution": "Since $d \\mid ab$ and $d \\in (m^2, m^2 + m)$, we have $d \\mid (a-d)(b-d)$. Also, $|a-d| < m$ and $|b-d| < m$, so $|(a-d)(b-d)| < m^2 < d$. Thus, the only way $d$ divides $(a-d)(b-d)$ is if $(a-d)(b-d) = 0$, so $d = a$ or $d = b$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 13268,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $n \\in \\mathbb{N}$, $n \\ge 2$. Consider the equation:\n\n$$\n\\{x\\} + \\{2x\\} + \\dots + \\{nx\\} = [x] + [2x] + \\dots + [2nx].\n$$\n\na) Solve the equation in $\\mathbb{R}$ for $n = 2$.\n\nb) Prove that the equation has at most two real solutions for any $n \\ge 2$.",
    "options": [],
    "answer": "See solution",
    "solution": "Since each fractional part $\\{kx\\} \\in [0, 1)$, the right-hand side is a sum of integers, hence nonnegative, so $x \\ge 0$.\n\na) For $n = 2$, the equation becomes $\\{x\\} + \\{2x\\} = [x] + [2x] + [3x] + [4x]$. The left side is in $[0, 2)$ and the right side is an integer, so the right side is in $\\{0, 1\\}$.\n\nIf $\\{x\\} + \\{2x\\} = 0$, then $\\{x\\} = \\{2x\\} = 0$, so $x \\in \\mathbb{Z}$. Plugging into the original equation, $x \\in \\mathbb{Z}$ yields $10x = 0$, so $x = 0$, which satisfies the equation.\n\nIf $\\{x\\} + \\{2x\\} = 1$, since $[x] \\le [2x] \\le [3x] \\le [4x]$, we have $[x] = [2x] = [3x] = 0$, $[4x] = 1$, which implies $3x < 1 \\le 4x$, so $x \\in [\\frac{1}{4}, \\frac{1}{3})$. But from $\\{x\\} + \\{2x\\} = 1$ and $[x] = [2x] = 0$, we have $x + 2x = 1$, which implies $x = \\frac{1}{3} \\notin [\\frac{1}{4}, \\frac{1}{3})$, a contradiction.\n\nHence, for $n = 2$, the unique solution is $x = 0$.\n\nb) Using similar reasoning for general $n$, since\n$$\n\\{x\\} + \\dots + \\{nx\\} \\in [0, n) \\cap \\mathbb{Z},\n$$\nand the integer parts satisfy $[x] \\le [2x] \\le \\dots \\le [nx]$, if $[nx] \\ge 1$, then the right-hand side sum is at least $n + 1$, which contradicts that the left side is less than $n$. So $[x] = [2x] = \\dots = [nx] = 0$, and the equation reduces to:\n$$\nx + 2x + \\dots + nx = [(n + 1)x] + \\dots + [2nx] = k,\n$$\nwhere $k \\in \\{0, 1, \\dots, n - 1\\}$. This yields $x = \\frac{2k}{n(n+1)}$.\n\nFor $k = 0$, the solution is $x = 0$ for all $n \\ge 2$.\n\nFor $k > 0$, since $nx < 1$, we have $2nx < 2$, so\n$$\n[2nx] = [(2n - 1)x] = \\dots = [(2n - k + 1)x] = 1,\n$$\nand\n$$\n[(2n - k)x] = \\dots = [(n + 1)x] = 0.\n$$\nThis implies:\n$$\n(2n - k) \\frac{2k}{n(n + 1)} < 1 \\le (2n - k + 1) \\frac{2k}{n(n + 1)},\n$$\nor equivalently:\n$$\nn - \\sqrt{\\frac{n(n-1)}{2}} > k \\ge n + \\frac{1}{2} - \\sqrt{\\frac{2n^2 + 2n + 1}{4}}.\n$$\nSince\n$$\nn - \\sqrt{\\frac{n(n-1)}{2}} - \\left(n + \\frac{1}{2} - \\sqrt{\\frac{2n^2 + 2n + 1}{4}}\\right) < 1,\n$$\nthere is at most one integer $k$ satisfying this, so there is at most one additional solution besides $x = 0$. Therefore, the given equation has at most two real solutions for any $n \\ge 2$.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 13269,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $\\triangle ABC$ be an acute, non-isosceles triangle inscribed in circle $(O)$, with the symmedian from vertex $A$ meeting $(O)$ again at $D$. Let $G$ be the reflection of $A$ over $BC$. Suppose that $GB$ and $GC$ meet $(O)$ again at $E$ and $F$, respectively. The incircle of $\\triangle AEF$ touches $EF$ at $J$.\n\n1. Prove that $GO$ and $GD$ are symmedians of triangles $GAD$ and $GEF$, respectively.\n\n2. Prove that the rays $AJ$ and $AD$ are symmetric with respect to $AG$.\n\n![](images/Saudi_Arabia_booklet_2024_p26_data_646b1f51d1.png)",
    "options": [],
    "answer": "See solution",
    "solution": "1) Let $T$ be the intersection of the tangent at $A$ to $(O)$ with $BC$. Since $ABDC$ is a harmonic quadrilateral, $TD$ is also tangent to $(O)$. Thus, $A$, $D$, and $G$ all lie on the circle centered at $T$, which is the Apollonius circle at $A$ of $\\triangle ABC$. Also, $(T)$ and $(O)$ are orthogonal, so $OA$ and $OD$ are tangent to $(T)$. Therefore, $GO$ is the symmedian of $\\triangle ADG$.\n\nLet $K$ be the second intersection of $AG$ and $(O)$. Then,\n\n$$\n\\angle BEK = \\angle BAK = \\angle BGK \\implies KE = KG.\n$$\n\nSimilarly, $KF = KG$, so $KE = KF$, which means $K$ is the midpoint of the minor arc $EF$ of $(O)$. Thus, $AG$ is the angle bisector of $\\angle EAF$.\n\nLet $M$ be the midpoint of $BC$, and let $A' \\in (O)$ such that $AA' \\parallel BC$. Since $MA' = MA = MG$, it follows that $G$, $M$, and $A'$ are collinear. Also,\n\n$$\nA'(AD, BC) = -1 \\text{ and } AA' \\parallel BC,\n$$\n\nso $A'$, $M$, and $D$ are collinear. Thus, $GD$ passes through $A'$. Since $K$, $O$, and $A'$ are collinear, $A'$ is the midpoint of the major arc $EF$ of $(O)$. Then, $K$ is the center of the circumcircle of $GEF$, and $A'E \\perp KE$, $A'F \\perp KF$, so $A'$ is the intersection of the tangents to $(K)$ at $E$ and $F$. Therefore, $GD$ is the symmedian of $\\triangle GEF$.\n\n2) We will prove that $D$ is the tangency point of the external mixtilinear incircle at $A$ in $\\triangle AEF$. Let $D'$, $U$, $V$ be the tangency points of the external mixtilinear incircle $(\\omega)$ of $\\triangle AEF$ with $(O)$, $AE$, $AF$, respectively. Then $G$ is the midpoint of $UV$, and $D'U$, $D'V$ pass through the midpoints of arcs $AE$ (containing $F$) and $AF$ (containing $E$) of $(O)$. Note that\n\n$$\n\\angle ACB = \\angle GCB = \\angle BCF = \\angle BAF,\n$$\n\nso $B$ is the midpoint of arc $AE$ (containing $F$) of $(O)$, so $B$, $D'$, $V$ are collinear. Similarly, $C$, $D'$, $U$ are collinear, so $UV \\parallel BC$. Since $(\\omega)$ and $(O)$ are tangent,\n\n$$\n\\angle FD'V = \\angle FCD' + \\angle D'UV = \\angle FCD' + \\angle D'CB = \\angle BCF = \\angle FGV,\n$$\n\nimplying that $D'FVG$ is cyclic. Similarly, $D'EUG$ is cyclic. Let $D'$ be the opposite ray of $D'G$; then\n\n$$\n\\angle FD'x = \\angle FVG = \\angle AVU = \\angle AUV = \\angle EDx,\n$$\n\nso $D'$ passes through the midpoint of the major arc $EF$ of $(O)$, which is $A'$. Thus, $D \\equiv D'$. Finally, let $I$ be the incenter of $\\triangle AEF$, and let $\\Omega$ be the inversion centered at $A$ with power $AE \\cdot AF$, composed with reflection about $AG$. By properties of this transformation (preserving tangency and isogonality),\n\n$$\n\\begin{gathered}\n\\Omega : (I) \\to (\\omega), \\quad \\Omega : BC \\leftrightarrow (O) \\\\\n\\implies \\Omega : J \\leftrightarrow D.\n\\end{gathered}\n$$\n\nTherefore, $AJ$ and $AD$ are symmetric with respect to the angle bisector $AG$. $\\square$",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 13270,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $A_3, B_3, C_3$ be the feet of the perpendiculars from $P$ to the sides $BC, CA, AB$ respectively. Clearly, $A', B', C'$ are on the lines $PA_3, PB_3, PC_3$, respectively.\n\n![](images/Turska_2014_p8_data_6fee34bc9a.png)\n\nShow that the lines $AA'$, $BB'$, and $CC'$ are concurrent.",
    "options": [],
    "answer": "See solution",
    "solution": "By the Pythagoras Theorem, we have\n\n$$\nA'P^2 - A'A_3^2 = A'A_1^2 - A'A_3^2 = A_1P^2 - A_3P^2 = A_1P^2 - (A_3A' + A'P)^2\n$$\n\nand hence $PA' \\cdot PA_3 = \\frac{PA_1^2}{2}$. Similarly, $PB' \\cdot PB_3 = \\frac{PB_1^2}{2}$ and $PC' \\cdot PC_3 = \\frac{PC_1^2}{2}$. Therefore, $PA' \\cdot PA_3 = PB' \\cdot PB_3 = PC' \\cdot PC_3$. Hence, the quadrilaterals $A_3C_3C'A'$, $B_3A_3A'B'$, $C_3B_3B'C'$ are cyclic.\n\nBy the sine law in triangles $A_3B_3A'$ and $AC_3A'$, we have\n\n$$\n\\frac{A'B_3}{AA'} = \\frac{\\sin \\angle A'AB_3}{\\cos \\angle A'B_3P} \\quad \\text{and} \\quad \\frac{A'C_3}{AA'} = \\frac{\\sin \\angle A'AC_3}{\\cos \\angle A'C_3P}\n$$\n\nHence,\n\n$$\n\\frac{\\sin \\angle A'AB_3}{\\sin \\angle A'AC_3} = \\frac{A'B_3}{A'C_3} \\cdot \\frac{\\cos \\angle A'B_3P}{\\cos \\angle A'C_3P}\n$$\n\nSimilarly,\n\n$$\n\\frac{\\sin \\angle B'BC_3}{\\sin \\angle B'BA_3} = \\frac{B'C_3}{B'A_3} \\cdot \\frac{\\cos \\angle B'C_3P}{\\cos \\angle B'A_3P}\n$$\n\nand\n\n$$\n\\frac{\\sin \\angle C'CA_3}{\\sin \\angle C'CB_3} = \\frac{C'A_3}{C'B_3} \\cdot \\frac{\\cos \\angle C'A_3P}{\\cos \\angle C'B_3P}\n$$\n\nBy the converse of the Trigonometric Ceva Theorem, to show that $AA'$, $BB'$, and $CC'$ are concurrent, it suffices to show that\n\n$$\n\\frac{A'B_3}{A'C_3} \\cdot \\frac{B'C_3}{B'A_3} \\cdot \\frac{C'A_3}{C'B_3} \\cdot \\frac{\\cos \\angle A'B_3P}{\\cos \\angle A'C_3P} \\cdot \\frac{\\cos \\angle B'C_3P}{\\cos \\angle B'A_3P} \\cdot \\frac{\\cos \\angle C'A_3P}{\\cos \\angle C'B_3P} = 1\n$$\n\nBy concyclicity, $\\angle A'C_3P = C'A_3P$, $B'A_3P = A'B_3P$, $C'B_3P = B'C_3P$, so\n\n$$\n\\frac{\\cos \\angle A'B_3P}{\\cos \\angle A'C_3P} \\cdot \\frac{\\cos \\angle B'C_3P}{\\cos \\angle B'A_3P} \\cdot \\frac{\\cos \\angle C'A_3P}{\\cos \\angle C'B_3P} = 1\n$$\n\nNow, we need to show\n\n$$\n\\frac{A'B_3}{A'C_3} \\cdot \\frac{B'C_3}{B'A_3} \\cdot \\frac{C'A_3}{C'B_3} = 1\n$$\n\nBy concyclicity again, triangles $PA_3C'$ and $PC_3A'$ are similar, so\n\n$$\n\\frac{C'A_3}{A'C_3} = \\frac{PA_3}{PC_3}\n$$\n\nand similarly,\n\n$$\n\\frac{A'B_3}{B'A_3} = \\frac{PB_3}{PA_3} \\quad \\text{and} \\quad \\frac{B'C_3}{C'B_3} = \\frac{PC_3}{PB_3}\n$$\n\nMultiplying these gives\n\n$$\n\\frac{A'B_3}{A'C_3} \\cdot \\frac{B'C_3}{B'A_3} \\cdot \\frac{C'A_3}{C'B_3} = \\frac{PA_3}{PC_3} \\cdot \\frac{PB_3}{PA_3} \\cdot \\frac{PC_3}{PB_3} = 1\n$$\n\nas desired.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 13271,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all polynomials $f(x)$ such that $f(f(x)) = f(x)^m$ for a fixed integer $m \\geq 2$.",
    "options": [],
    "answer": "See solution",
    "solution": "The solutions are $f(x) = 0$, $f(x) = 1$, $f(x) = \\omega$ where $\\omega$ is an $(m-1)$st root of unity, and $f(x) = x^m$.\n\nIf $f(x) = c$ is a constant polynomial, then the relation holds if and only if $c = c^m$.\nClearly, the solutions are $c = 0, 1$ and all the $(m-1)$st roots of unity.\n\nIf $f$ is a non-constant polynomial, then $f(x)$ attains infinitely many values. Thus, there are infinitely many $y$ such that $f(y) = y^m$. Since $f$ is a polynomial, this implies $f(x) = x^m$ for any $x$.\n\nIt is easy to check that all these are solutions.",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 13272,
    "subject": "Mathematics (Olympiad)",
    "question": "Suppose Ben is trying to determine a hidden number $x$ from the set $\\{1, 2, \\dots, N\\}$ by asking questions of the form \"Is $x$ in the set $S$?\". Amy answers each question with either *yes* or *no*, but she is allowed to lie, with the restriction that she cannot give more than $k$ consecutive answers that are inconsistent with the true value of $x$ (i.e., she cannot lie more than $k$ times in a row about $x$).\n\nLet $f(k)$ denote the minimum value of $n$ such that Ben can guarantee to determine $x$ for any possible sequence of answers from Amy, given the above restriction. Prove that for large $k$,\n\n$$\n1.99^k \\leq f(k) \\leq 2^k.\n$$\n\nAdditionally, describe strategies for both Ben and Amy that demonstrate the upper and lower bounds, respectively.",
    "options": [],
    "answer": "See solution",
    "solution": "Consider an answer $A \\in \\{\\text{yes}, \\text{no}\\}$ to a question of the kind \"Is $x$ in the set $S$?\". We say that $A$ is inconsistent with a number $i$ if $A = \\text{yes}$ and $i \\notin S$, or if $A = \\text{no}$ and $i \\in S$. An answer inconsistent with the target number $x$ is a lie.\n\n**a)** Suppose Ben has determined a set $T$ of size $m$ that contains $x$. Initially, $m = N$ and $T = \\{1, 2, \\dots, N\\}$. For $m > 2^k$, Ben can find a number $y \\in T$ that is different from $x$. By repeating this process, he can reduce $T$ to size $2^k \\leq n$ and thus win.\n\nAssume $T = \\{0, 1, \\dots, 2^k, \\dots, m-1\\}$. Ben repeatedly asks whether $x$ is $2^k$. If Amy answers *no* $k+1$ times in a row, one answer must be truthful, so $x \\neq 2^k$. Otherwise, Ben stops at the first *yes*. He then asks, for each $i = 1, \\dots, k$, if the binary representation of $x$ has a $0$ in the $i$th digit. Regardless of the $k$ answers, they are all inconsistent with a certain $y \\in \\{0, 1, \\dots, 2^k-1\\}$. The previous *yes* about $2^k$ is also inconsistent with $y$, so $y \\neq x$. Otherwise, the last $k+1$ answers are not truthful, which is impossible.\n\nThus, Ben finds a number in $T$ different from $x$, proving the claim.\n\n**b)** Suppose $1 < \\lambda < 2$ and $n = [(2-\\lambda)\\lambda^{k+1}] - 1$. Ben cannot guarantee a win. Take $\\lambda$ such that $1.99 < \\lambda < 2$ and $k$ large enough so that\n\n$$\nn = [(2 - \\lambda)\\lambda^{k+1}] - 1 \\geq 1.99^k.\n$$\n\nAmy's strategy: She chooses $N = n+1$ and $x \\in \\{1, 2, \\dots, n+1\\}$ arbitrarily. After each answer, for each $i = 1, 2, \\dots, n+1$, let $m_i$ be the number of consecutive answers inconsistent with $i$. To decide her next answer, she uses\n\n$$\n\\phi = \\sum_{i=1}^{n+1} \\lambda^{m_i}.\n$$\n\nNo matter Ben's next question, Amy chooses the answer minimizing $\\phi$.\n\nWith this strategy, $\\phi$ always stays less than $\\lambda^{k+1}$, so no $m_i$ ever exceeds $k$. Thus, Amy never lies more than $k$ times in a row about any $i$, including $x$. The strategy does not depend on $x$, so Ben cannot deduce $x$ and cannot guarantee a win.\n\nTo show $\\phi < \\lambda^{k+1}$ always: Initially, $m_i = 0$, so $\\phi < \\lambda^{k+1}$ by the choice of $n$. Suppose $\\phi < \\lambda^{k+1}$ and Ben asks if $x \\in S$. If Amy answers *yes* or *no*, the new $\\phi$ becomes\n\n$$\n\\phi_1 = \\sum_{i \\in S} 1 + \\sum_{i \\notin S} \\lambda^{m_i+1} \\quad \\text{or} \\quad \\phi_2 = \\sum_{i \\in S} \\lambda^{m_i+1} + \\sum_{i \\notin S} 1.\n$$\n\nAmy chooses the option minimizing $\\phi$, so the new $\\phi$ is $\\min(\\phi_1, \\phi_2)$. Now,\n\n$$\n\\min(\\phi_1, \\phi_2) \\leq \\frac{1}{2}(\\phi_1 + \\phi_2) = \\frac{1}{2} \\left( \\sum_{i \\in S} (1 + \\lambda^{m_i+1}) + \\sum_{i \\notin S} (\\lambda^{m_i+1} + 1) \\right) = \\frac{1}{2}(\\lambda\\phi + n + 1).\n$$\n\nSince $\\phi < \\lambda^{k+1}$, $\\lambda < 2$, and $n = [(2-\\lambda)\\lambda^{k+1}] - 1$,\n\n$$\n\\min(\\phi_1, \\phi_2) < \\frac{1}{2}(\\lambda^{k+2} + (2-\\lambda)\\lambda^{k+1}) = \\lambda^{k+1}.\n$$\n\nThus, the claim holds.\n\n**Comment:** For fixed $k$, $f(k)$ is the minimum $n$ for which Ben can guarantee victory. For large $k$,\n\n$$\n1.99^k \\leq f(k) \\leq 2^k.\n$$\n\nA computer search shows $f(k) = 2, 3, 4, 7, 11, 17$ for $k = 1, 2, 3, 4, 5, 6$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 13273,
    "subject": "Mathematics (Olympiad)",
    "question": "Sea $\\triangle ABC$ un triángulo y $D$, $E$ y $F$ tres puntos cualesquiera sobre los lados $AB$, $BC$ y $CA$ respectivamente. Llamemos $P$ al punto medio de $AE$, $Q$ al punto medio de $BF$ y $R$ al punto medio de $CD$. Probar que el área del triángulo $\\triangle PQR$ es la cuarta parte del área del triángulo $\\triangle DEF$.",
    "options": [],
    "answer": "See solution",
    "solution": "Hagamos primero un dibujo donde queden reflejados los elementos que intervienen en el problema.\n\n![](images/Spanija_b_2014_p10_data_8c1b426277.png)\n\nObservemos que, como $P$, $Q$ y $R$ son los puntos medios de las correspondientes cevianas $AE$, $BF$ y $CD$, estos puntos se encuentran en los lados del triángulo que determinan los pies de las medianas de cada lado, como se ve en la figura siguiente.\n\n![](images/Spanija_b_2014_p11_data_d112b0e974.png)\n\nLos triángulos $\\triangle ABC$ y $\\triangle M_a M_b M_c$ son semejantes con razón de semejanza $1/2$, por lo que se tiene:\n\n$$\n\\begin{aligned}\n\\frac{\\overline{M_c M_a}}{\\overline{AC}} &= \\frac{\\overline{M_c Q}}{\\overline{AF}} = \\frac{\\overline{Q M_a}}{\\overline{FC}} = \\frac{1}{2}, \\\\\n\\frac{\\overline{M_a M_b}}{\\overline{BA}} &= \\frac{\\overline{M_a R}}{\\overline{BD}} = \\frac{\\overline{R M_b}}{\\overline{DA}} = \\frac{1}{2}, \\\\\n\\frac{\\overline{M_b M_c}}{\\overline{CB}} &= \\frac{\\overline{M_b P}}{\\overline{CE}} = \\frac{\\overline{P M_c}}{\\overline{EB}} = \\frac{1}{2}.\n\\end{aligned}\n$$\n\nAdemás, los ángulos en $A$, $B$ y $C$ son iguales, respectivamente, a los ángulos en $M_a$, $M_b$ y $M_c$.\n\nSea $u = \\frac{\\overline{AD}}{\\overline{AB}}$, $v = \\frac{\\overline{BE}}{\\overline{BC}}$ y $w = \\frac{\\overline{CF}}{\\overline{CA}}$, por lo que:\n\n$$\n\\frac{\\overline{DB}}{\\overline{AB}} = 1 - u, \\quad \\frac{\\overline{EC}}{\\overline{BC}} = 1 - v, \\quad \\frac{\\overline{FC}}{\\overline{CA}} = 1 - w.\n$$\n\nAplicando la semejanza de los triángulos $\\triangle ABC$ y $\\triangle M_a M_b M_c$, resulta:\n\n$$\n\\begin{aligned}\nu &= \\frac{\\overline{R M_b}}{M_a M_b}, \\quad v = \\frac{\\overline{P M_c}}{M_b M_c}, \\quad w = \\frac{\\overline{Q M_a}}{M_c M_a}, \\\\\n\\frac{\\overline{M_a R}}{M_a M_b} &= 1 - u, \\quad \\frac{\\overline{M_b P}}{M_b M_c} = 1 - v, \\quad \\frac{\\overline{M_c Q}}{M_c M_a} = 1 - w.\n\\end{aligned}\n$$\n\nCon esto, calculemos ahora el área de los triángulos complementarios del triángulo $\\triangle DEF$.\n\n![](images/Spanija_b_2014_p12_data_647c0e196b.png)\n\nSi $[ABC]$ denota el área de un triángulo $\\triangle ABC$, tenemos que\n\n$$\n\\begin{align*}\n[AFD] &= \\overline{FA} \\cdot \\overline{AD} \\sin \\alpha = (1-w)u \\overline{CA} \\cdot \\overline{AB} \\sin \\alpha = (1-w)u[ABC], \\\\\n[BED] &= \\overline{DB} \\cdot \\overline{BE} \\sin \\beta = (1-u)v \\overline{AB} \\cdot \\overline{BC} \\sin \\beta = (1-u)v[ABC], \\\\\n[CFE] &= \\overline{EC} \\cdot \\overline{CF} \\sin \\gamma = (1-v)w \\overline{BC} \\cdot \\overline{CA} \\sin \\gamma = (1-v)w[ABC].\n\\end{align*}\n$$\n\nHaciendo lo mismo con los triángulos complementarios del triángulo $\\triangle PQR$, con respecto al triángulo $\\triangle M_a M_b M_c$ se tiene\n\n![](images/Spanija_b_2014_p12_data_4971bd90dd.png)\n\n$$\n\\begin{align*}\n[M_a RQ] &= \\overline{QM_a} \\cdot \\overline{M_a R} \\sin \\alpha = (1-u)w \\overline{M_a M_b} \\cdot \\overline{M_a M_c} \\sin \\alpha = (1-u)w[M_a M_b M_c], \\\\\n[M_b PR] &= \\overline{RM_b} \\cdot \\overline{M_b P} \\sin \\beta = (1-v)u \\overline{M_a M_b} \\cdot \\overline{M_b M_c} \\sin \\beta = (1-v)u[M_a M_b M_c], \\\\\n[M_c QP] &= \\overline{PM_c} \\cdot \\overline{M_c Q} \\sin \\gamma = (1-w)v \\overline{M_b M_c} \\cdot \\overline{M_a M_c} \\sin \\gamma = (1-w)v[M_a M_b M_c].\n\\end{align*}\n$$\n\nTeniendo en cuenta que $[ABC] = 4[M_a M_b M_c]$ y que\n\n$$\n(1-u)w + (1-v)u + (1-w)v = (1-w)u + (1-u)v + (1-v)w,\n$$\n\nse sigue el resultado.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 13274,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $ABC$ be an acute-angled triangle with orthocentre $H$. Let $E = BH \\cap AC$ and $F = CH \\cap AB$. Let $D$, $M$, and $N$ be the midpoints of segments $AH$, $BD$, and $CD$ respectively, and let $T = FM \\cap EN$. Suppose $D$, $E$, $T$, $F$ are concyclic. Prove that $DT$ passes through the circumcentre of $ABC$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $O$ be the circumcentre of $(ABC)$ and $J$ be the midpoint of $DO$. Now it is sufficient to prove that $TD$ and $TJ$ coincide. We first prove that $M$, $N$, $T$, $J$ are concyclic.\n\n$$\n\\angle MJN = \\angle BOC = 2\\angle BAC = \\angle EDF = 180^\\circ - \\angle FTE = 180^\\circ - \\angle NJM\n$$\n\nThis proves our claim!\n\nNow note that $JM = \\frac{OB}{2} = \\frac{OC}{2} = JN$, so $TJ$ is the angle bisector of $\\angle NTM = \\angle ETF$. But $D$ is the midpoint of arc $EF$ in the nine-point circle, so $TD$ is the angle bisector of $\\angle ETF$ as well. Thus, $TD$ and $TJ$ must coincide! $\\square$\n\n![](images/IND_ABooklet_2024_p21_data_718d150497.png)",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 13275,
    "subject": "Mathematics (Olympiad)",
    "question": "Compare $A$ to $0$, where:\n\n$$\na)\\quad A = 1 - 2 - 3 + 4 + 5 - 6 - 7 + 8 + 9 - 10 - 11 + \\dots + 2012 + 2013 - 2014 - 2015 + 2016;\n$$\n\n$$\nb)\\quad A = \\frac{1}{1} - \\frac{1}{2} - \\frac{1}{3} + \\frac{1}{4} + \\frac{1}{5} - \\frac{1}{6} - \\frac{1}{7} + \\frac{1}{8} + \\frac{1}{9} - \\frac{1}{10} - \\frac{1}{11} + \\dots + \\frac{1}{2012} + \\frac{1}{2013} - \\frac{1}{2014} - \\frac{1}{2015} + \\frac{1}{2016}.\n$$\n\nIn each question, the signs go as follows: \"+\" before the first term, then two \"-\" and two \"+\" signs in turn, and, finally, a \"+\" sign before the last term.",
    "options": [],
    "answer": "See solution",
    "solution": "*Answer:* a) $A = 0$; b) $A > 0$.\n\n*Solution.*\n\na) If we split all numbers into 504 groups of 4, from left to right, each group will have numbers of the type:\n\n$$\n(4k+1)-(4k+2)-(4k+3)+(4k+4)=0.\n$$\n\nAs we can see, the sum of numbers in each group is $0$, therefore, $A = 0$.\n\nb) Like in the previous question, split all numbers into groups of 4: $\\left(\\frac{1}{4k+1} - \\frac{1}{4k+2} - \\frac{1}{4k+3} + \\frac{1}{4k+4}\\right)$. Then the sum of numbers in each group is positive:\n\n$$\n\\frac{1}{4k+1} - \\frac{1}{4k+2} - \\frac{1}{4k+3} + \\frac{1}{4k+4} > 0 \\Leftrightarrow \\frac{1}{4k+1} - \\frac{1}{4k+2} > \\frac{1}{4k+3} - \\frac{1}{4k+4} \\Leftrightarrow \\frac{1}{(4k+1)(4k+2)} > \\frac{1}{(4k+3)(4k+4)}.\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 13276,
    "subject": "Mathematics (Olympiad)",
    "question": "Ana has thirteen weights labeled with their masses from 1 gram to 13 grams. She wants to help Beto identify the 1-gram weight using a balance scale and a sequence of weighings. What is the minimum number of weighings Ana must perform so that Beto can be certain which weight is the 1-gram weight?",
    "options": [],
    "answer": "See solution",
    "solution": "The minimum number of weighings required is 2.\n\n**First weighing:** Place the weights 1–8 grams on one side and 11, 12, 13 grams on the other. The sum of 1–8 is $36$, and the sum of 11, 12, 13 is also $36$, so balance is possible only with these groupings. The 9-gram and 10-gram weights are left out.\n\n**Second weighing:** Place the 10-gram weight on one side, and the 9-gram and 1-gram weights on the other. Since Beto already knows which are the 9-gram and 10-gram weights (from the first weighing), he can deduce which is the 1-gram weight.\n\nIt is not possible to identify the 1-gram weight in a single weighing, because the only way to uniquely identify a weight is for it to be the only weight in a group (left, right, or outside), but the sums do not allow this for the 1-gram weight. Thus, two weighings are necessary and sufficient.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 13277,
    "subject": "Mathematics (Olympiad)",
    "question": "A subset $T$ of $\\{7, 8, \\ldots, 26\\}$ does not contain three elements whose product is a perfect square. Determine the maximum number of elements in $T$.",
    "options": [],
    "answer": "See solution",
    "solution": "By checking the parity of the prime factorization of all the integers in $T$, we partition them into disjoint triples whose products are all squares:\n\n$$\n\\{8, 13, 26\\}, \\{11, 18, 22\\}, \\{14, 21, 24\\}, \\{12, 15, 20\\}, \\{9, 16, 25\\}.\n$$\n\nIf $|T| \\geq 16$, at least one of these triples will be in $T$, a contradiction.\n\nWe can easily check that the subset (obtained by strategically removing one element—$8$, $18$, $24$, $12$, $9$—from each of the triples above):\n\n$$\n\\{7, 10, 11, 13, 14, 15, 16, 17, 19, 20, 21, 22, 23, 25, 26\\}\n$$\n\nsatisfies the condition given. Thus, the maximum number of elements in $T$ is $15$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 13278,
    "subject": "Mathematics (Olympiad)",
    "question": "Anthony writes down in order all positive integers divisible by $2$. Bertha writes down in order all positive integers divisible by $3$. Claire writes down in order all positive integers divisible by $4$. Orderly Dora writes all numbers written by the other three, putting them in order by size and not repeating any number. What is the $2017$th number in her list?",
    "options": [],
    "answer": "See solution",
    "solution": "Dora can ignore Claire's numbers, since Anthony has already written all numbers divisible by $4$ (as they are also divisible by $2$). Up to $3000$, Anthony has written $1500$ numbers (all even numbers), and Bertha has written $1000$ numbers (multiples of $3$). Among these, $500$ numbers (multiples of $6$) have been written by both, so Dora only counts them once. Thus, up to $3000$, Dora has listed $1500 + 1000 - 500 = 2000$ numbers.\n\nThe next $17$ numbers in order are:\n\n$3002, 3003, 3004, 3006, 3008, 3009, 3010, 3012, 3014, 3015, 3016, 3018, 3020, 3021, 3022, 3024, 3026$.\n\nTherefore, the $2017$th number on Dora's list is $3026$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 13279,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all positive integers $n$ for which there exist non-negative integers $a_1, a_2, a_3, \\dots, a_n$ such that\n$$\n\\frac{1}{2^{a_1}} + \\frac{1}{2^{a_2}} + \\dots + \\frac{1}{2^{a_n}} = \\frac{1}{3^{a_1}} + \\frac{2}{3^{a_2}} + \\dots + \\frac{n}{3^{a_n}} = 1.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "Suppose $n$ satisfies the condition, i.e., there exist non-negative integers $a_1 \\leq a_2 \\leq \\cdots \\leq a_n$ such that\n$$\n\\sum_{i=1}^{n} 2^{-a_i} = \\sum_{i=1}^{n} i \\cdot 3^{-a_i} = 1.\n$$\nMultiplying both sides of the second equality by $a_n$ and taking mod 2, we have\n$$\n\\sum_{i=1}^{n} i = \\frac{1}{2} n(n+1) \\equiv 1 \\pmod{2},\n$$\nso $n \\equiv 1,2 \\pmod{4}$. We will show this is also sufficient: all $n$ with $n \\equiv 1,2 \\pmod{4}$ work.\n\nDefine a set $B = \\{b_1, b_2, \\dots, b_n\\}$ of positive integers as *feasible* if there exist non-negative integers $a_1, \\dots, a_n$ such that\n$$\n\\sum_{i=1}^{n} 2^{-a_i} = \\sum_{i=1}^{n} b_i 3^{-a_i} = 1.\n$$\nIf $B$ is feasible, and we replace any $b$ in $B$ by $u$ and $v$ with $u+v=3b$, the new set $B'$ is also feasible: assign $a+1$ to $u$ and $v$ (if $b$ had $a$), since\n$$\n2^{-a-1} + 2^{-a-1} = 2^{-a}, \\quad u \\cdot 3^{-a-1} + v \\cdot 3^{-a-1} = b \\cdot 3^{-a}.\n$$\nIf $B'$ is obtained from $B$ by such replacements, write $B \\rightsquigarrow B'$. In particular, $b$ can be replaced by $b$ and $2b$, so $B \\rightsquigarrow B \\cup \\{2b\\}$.\n\nWe show $B_n = \\{1,2,\\dots,n\\}$ is feasible for all $n \\equiv 1,2 \\pmod{4}$.\n- $B_1$ is feasible: take $a_1=0$.\n- $B_2$ is feasible: $B_1 \\rightsquigarrow B_2$.\n- If $B_n$ is feasible for $n \\equiv 1 \\pmod{4}$, then $B_n \\rightsquigarrow B_{n+1}$, so $B_{n+1}$ is feasible.\n\nIt suffices to show $B_n$ is feasible for $n \\equiv 1 \\pmod{4}$:\n- $B_5$ is feasible: $B_2 \\rightsquigarrow \\{1,3,3\\} \\rightsquigarrow \\{1,3,4,5\\} \\rightsquigarrow B_5$.\n- $B_9$ is feasible: $B_5 \\rightsquigarrow \\{1,2,3,4,6,9\\} \\rightsquigarrow \\{1,2,3,5,6,7,9\\} \\rightsquigarrow B_9 \\setminus \\{8\\} \\rightsquigarrow B_9$.\n- $B_{13}$ is feasible: $B_9 \\rightsquigarrow \\{1,2,3,4,5,6,7,9,11,13\\} \\rightsquigarrow B_{13}$.\n\nAppending $8,10,12$ successively, $B_{17}$ is feasible:\n$$\nB_6 \\rightsquigarrow B_5 \\cup \\{7,11\\} \\rightsquigarrow B_8 \\cup \\{11\\} \\rightsquigarrow B_7 \\cup \\{9,11,15\\} \\rightsquigarrow B_{12} \\cup \\{15\\} \\rightsquigarrow B_{17} \\setminus \\{10,14,16\\} \\rightsquigarrow B_{17}.\n$$\n\nFinally, for any $k \\geq 2$, $B_{4k+2} \\rightsquigarrow B_{4k+13}$ by appending $4k+4, 4k+6, \\dots, 4k+12$ (since $\\frac{4k+12}{2} \\leq 4k+2$). For the remaining six odd numbers $4k+3, 4k+5, \\dots, 4k+13$, group them as multiples of 3 and pairs whose sum is a multiple of 3, and replace accordingly. Thus, all $n \\equiv 1,2 \\pmod{4}$ are possible.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 13280,
    "subject": "Mathematics (Olympiad)",
    "question": "Cut off a corner of $2\\times2$ unit squares from a $3\\times3$ unit squares; the remaining figure is called a *horn* (see the figure below for an example of a horn).\n\nNow, place some horns without overlapping on a $10\\times10$ unit square board so that the boundaries of the horns coincide with the grid of the board. Find the maximum value of $k$ such that, no matter how $k$ horns are placed on the board, it is always possible to place another horn on the board.\n\n![](images/Mathematical_Olympiad_in_China_2011-2014_p319_data_1a9d281303.png)\n![](images/Mathematical_Olympiad_in_China_2011-2014_p319_data_bd704916f9.png)",
    "options": [],
    "answer": "See solution",
    "solution": "First, $k_{\\max} < 8$ because if we place eight horns as in the figure below, no more horn can be placed.\n\n![](images/Mathematical_Olympiad_in_China_2011-2014_p319_data_9f0b0839d8.png)\n\nNext, we show that after placing any seven horns, there is always room for another horn. Consider four $4 \\times 4$ squares in the four corners of the $10 \\times 10$ board. Any horn can only intersect one of these $4 \\times 4$ squares. By the pigeonhole principle, there exists a $4 \\times 4$ square $S$ such that at most one horn $H$ intersects $S$. Since $H$ can be contained in a $3 \\times 3$ square, $S \\cap H$ is contained in a $3 \\times 3$ square at a corner of $S$. Therefore, we can place another horn in $S$ as shown in the figures below.\n\n![](images/Mathematical_Olympiad_in_China_2011-2014_p319_data_9ef44bd57c.png)\n![](images/Mathematical_Olympiad_in_China_2011-2014_p319_data_ff0be6803e.png)\n\nIn conclusion, $k_{\\max} = 7$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 13281,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $S = \\{(a, b) \\mid 1 \\leq a, b \\leq 5,\\ a, b \\in \\mathbb{Z}\\}$. 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 the segment $PQ$ 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": "See solution",
    "solution": "We first prove that $|T| \\neq 1$.\n\nIf $|T| = 1$, let $T = \\{Q(x_0, y_0)\\}$. We may take a point $P(x_1, y_1)$ in $S$ satisfying the conditions: (1) $(x_1, y_1) \\neq (x_0, y_0)$, (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![alt](./images/figure.png \"\")\n\n• • • • •\n\n• • ◐ • •\n\n• • • • • ◐\n\n• • • • • •\n\n• • • • • •",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 13282,
    "subject": "Mathematics (Olympiad)",
    "question": "A set of $n$ points in space is given, no three of which are collinear and no four of which are coplanar, and each pair of points is connected by a line segment. Initially, all the line segments are colorless. A positive integer $b$ is given and Alice and Bob play the following game:\n\n- On each turn, Alice colors one segment red, then Bob colors up to $b$ segments blue. This repeats until all segments are colored.\n- If Alice colors a red triangle (three red segments forming a triangle), she wins.\n- If all segments are colored and Alice has not formed a red triangle, Bob wins.\n- Neither player may color over an already colored segment.\n\n1. Prove that if $b < \\sqrt{2n-2} - \\frac{3}{2}$, then Alice has a winning strategy.\n\n2. Prove that if $b \\ge 2\\sqrt{n}$, then Bob has a winning strategy.",
    "options": [],
    "answer": "See solution",
    "solution": "1. We call a *threat* an uncolored segment whose coloring red would complete a triangle. Alice can use the following strategy: she keeps coloring segments from a fixed point $A$ to other points (the *opposite ends*) unless she can complete a triangle. Bob's optimal counter-strategy is to eliminate all threats in his move, then use remaining moves to color segments from $A$.\n\nAfter the $k$-th red line from $A$ is drawn, Bob must color $k-1$ threats blue (the lines connecting the new opposite end to previous ones). Thus, in the $k$-th round, the number of colored edges from $A$ increases by $b + 1 - (k - 1) = b - k + 2$. Suppose after $l$ moves all segments from $A$ are colored before Alice can form a triangle. Then:\n\n$$\n(b + 2)l - \\frac{l(l + 1)}{2} \\geq n - 1\n$$\n\nwhich leads to\n\n$$\nl^2 + 2l\\left(b + \\frac{3}{2}\\right) + 2n - 2 \\leq 0\n$$\n\nThe discriminant must be non-negative:\n\n$$\n4\\left(b + \\frac{3}{2}\\right)^2 - 4(2n - 2) \\geq 0\n$$\n\nSo $b \\geq \\sqrt{2n - 2} - \\frac{3}{2}$. Since this is violated, Alice's strategy works.\n\n2. Bob's strategy: whenever Alice colors a line $AB$, Bob colors $\\lfloor b/2 \\rfloor$ edges from $A$ and $\\lfloor b/2 \\rfloor$ edges from $B$ blue. This ensures at most $n / \\lfloor b/2 \\rfloor + 1$ red edges from any point, which is not larger than $\\lfloor b/2 \\rfloor + 1$ for $b \\geq 2\\sqrt{n}$. If Alice completes a triangle $ABC$ (by coloring $AB$, $AC$, $BC$), before coloring $BC$ there were at least $\\lfloor b/2 \\rfloor + 1$ threats from $C$. Since coloring $AC$ created these threats, there were at least $\\lfloor b/2 \\rfloor + 1$ red lines from $A$, contradicting the bound.\n\n$\\Box$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 13283,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $ABC$ be a triangle and let $\\omega$ be its circumcircle. Let $E$ be the midpoint of the minor arc $BC$ of $\\omega$, and $M$ the midpoint of $BC$. Let $V$ be the other point of intersection of $AM$ with $\\omega$, $F$ the point of intersection of $AE$ with $BC$, $X$ the other point of intersection of the circumcircle of $FEM$ with $\\omega$, $X'$ the reflection of $V$ with respect to $M$, $A'$ the foot of the perpendicular from $A$ to $BC$, and $S$ the other point of intersection of $XA'$ with $\\omega$. If $Z \\in \\omega$ with $Z \\neq X$ is such that $AX = AZ$, then prove that $S$, $X'$, and $Z$ are collinear.",
    "options": [],
    "answer": "See solution",
    "solution": "**Claim 1.** $AX$ is the $A$-symmedian of $\\triangle ABC$.\n\n**Proof of Claim 1.** Let $Y \\in \\omega$ such that $AY$ is the $A$-symmedian of triangle $ABC$. We want to prove that $Y = X$.\n\n![](images/BMO_2022_shortlist_p29_data_14bcfda7ca.png)\n\nWe have that $\\angle BAY = \\angle CAM$ and $\\angle BYA = \\angle BCA = \\angle MCA$, therefore the triangles $ABY$ and $AMC$ are similar. It follows that $(AY)(AM) = (AB)(AC)$.\n\nSince $AE$ is the bisector of $\\angle BAC$, then $\\angle BAF = \\angle CAE$. We also have $\\angle ABF = \\angle ABC = \\angle AEC$, therefore the triangles $BAF$ and $EAC$ are similar. It follows that $(AE)(AF) = (AB)(AC)$.\n\nWe get $(AY)(AM) = (AE)(AF)$ and since also $\\angle YAF = \\angle EAM$, then the triangles $YAF$ and $EAM$ are similar. So $\\angle AFY = \\angle AME$ and $\\angle YFE = \\angle EMV$. But as $E$ is the midpoint of the arc $YV$, it follows that $\\angle EMV = \\angle YME$. So $\\angle YFE = \\angle YME$ from which it follows that the quadrilateral $YFME$ is cyclic. But since $Y \\in \\omega$, we finally get that $Y = X$. $\\square$\n\nFrom Claim 1 we conclude that the triangles $XBC$ and $VCB$ are equal. Thus $MX = MV = MX'$. So the triangle $X'XV$ is a right-angled triangle and $X'X$ is perpendicular to $XV$ and therefore also to $BC$. Thus $X'$ is the reflection of $X$ on $BC$.\n\n**Claim 2.** The quadrilateral $ASA'M$ is cyclic.\n\n**Proof of Claim 2.** We have $\\angle ASA' = \\angle ASX = \\angle ABX$. But from Claim 1 we also have $\\angle ABX = \\angle AMC$. So $\\angle ASA' = \\angle AMC$ and the result follows. $\\square$\n\n**Claim 3.** The quadrilateral $XSX'M$ is cyclic.\n\n**Proof of Claim 3.** From Claim 2 we have $\\angle XSM = \\angle A'SM = \\angle A'AM$. Since $XX'$ is parallel to $AA'$, we have $\\angle A'AM = \\angle XX'M$. So $\\angle XSM = \\angle XX'M$ and the result follows. $\\square$\n\nNow from Claim 3 we have\n\n$$\n\\angle XSX' = \\angle XMV = \\angle XX'M + \\angle X'XM = 2\\angle XX'M = 2\\angle AA'M.\n$$\n\nSo to conclude the proof it is enough to also show that $\\angle XSZ = 2\\angle AA'M$. From Claim 1 we have $\\angle ACX = \\angle MAB$ and therefore\n\n$$\n\\angle AZX = \\angle ACX = \\angle AMB = 90^\\circ - \\angle A'AM.\n$$\n\nSince the triangle $XAZ$ is isosceles, we deduce that\n\n$$\n\\angle XSZ = \\angle XAZ = 180^\\circ - 2\\angle AZX = 2\\angle A'AM\n$$\n\nthus completing the proof.\n\n### Notes\n\n1. Claim 1 is Problem 9.2/10.2 from the All Russian Mathematical Olympiad of 2009. There are various ways to prove this but we only added one proof here.\n\n2. There are various other results which can be helpful towards a proof. These include the following:\n\n(a) If $H$ is the orthocenter of the triangle $ABC$, then the quadrilateral $A'MX'H$ is cyclic.\n\n(b) The points $H, M, S$ are collinear.\n\n(c) The quadrilateral $AX'H_S$ is inscribed in a circle with diameter $AH$.\n\n(d) If $P$ is the intersection of $AS$ with $BC$ and $K$ is the intersection of $AA'$ with $SM$ then the quadrilateral $PSKA'$ is cyclic. (In fact $K$ is the orthocenter of the triangle $ABC$.)\n\n(e) The quadrilateral $PX'MX$ is cyclic.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 13284,
    "subject": "Mathematics (Olympiad)",
    "question": "Find the smallest positive integer $N$ such that the equation $x^2 - y^2 = N$ has exactly 24 solutions in positive integers $(x, y)$.",
    "options": [],
    "answer": "See solution",
    "solution": "We analyze the equation $x^2 - y^2 = N$:\n\nSince $N > 0$, we require $x > y > 0$. Note that $x^2 - y^2 = (x + y)(x - y)$, so each solution corresponds to a factorization $N = a \\cdot b$ with $a = x + y > b = x - y > 0$.\n\nGiven $N = a \\cdot b$ with $a > b > 0$, the corresponding solution is:\n$$\nx = \\frac{a + b}{2}, \\qquad y = \\frac{a - b}{2}.\n$$\nFor integer solutions, $a \\equiv b \\pmod{2}$.\n\nIf $N$ is odd, any factorization into two positive integers yields a solution. If $N$ is even, both factors must be even, so $N = 4M$ for some integer $M$.\n\nThe number of positive divisors of $N$ is $\\tau(N) = (a_1 + 1)(a_2 + 1)\\cdots(a_k + 1)$ for $N = p_1^{a_1} p_2^{a_2} \\cdots p_k^{a_k}$. The number of factorizations into two factors is $\\tau(N)/2$ if $N$ is not a perfect square, and $(\\tau(N) - 1)/2$ if $N$ is a perfect square.\n\nTo have exactly 24 solutions, we need $\\tau(N) = 48$ (if $N$ is not a perfect square) or $\\tau(N) = 49$ (if $N$ is a perfect square). We consider both odd $N$ and $N = 4M$.\n\nAfter analyzing all possible factorizations and minimizing $N$ for each case, we find:\n\n$$\nN = 2^5 \\cdot 3^2 \\cdot 5 \\cdot 7 = 10080\n$$\n\nis the smallest positive integer such that $x^2 - y^2 = N$ has exactly 24 solutions in positive integers.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 13285,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $ABC$ be a triangle, each of whose angles is greater than $30^\\circ$. Suppose a circle with centre $T$ cuts the segments $BC$ at $P, Q$; $CA$ at $K, L$; and $AB$ at $M, N$, such that $P, Q, K, L, M, N$ are on the circle in counter-clockwise order. Suppose further that the triangles $TQK$, $TLM$, and $TNP$ are all equilateral. Prove that:\n\n1. The radius of the circle is $$\\frac{2abc}{a^2 + b^2 + c^2 + 4\\sqrt{3}\\Delta}$$\n2. $a \\cdot AT = b \\cdot BT = c \\cdot CT$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $BC = a$, $CA = b$, and $AB = c$. Let $\\angle PTQ = 2x$, $\\angle KTL = 2y$, and $\\angle MTN = 2z$. Since $TP = TQ$, we get $\\angle TQP = \\angle TPQ = 90^\\circ - x$. Thus $\\angle = 30^\\circ + x$. Similarly, $\\angle BNP = 30^\\circ + z$. It follows that $z + x = 120^\\circ - B$. Likewise, $x + y = 120^\\circ - C$ and $y + z = 120^\\circ - A$. Solving these, we get $x = A - 30^\\circ$, $y = B - 30^\\circ$, and $z = C - 30^\\circ$. As each angle $A, B, C$ is greater than $30^\\circ$, we see that $x, y, z$ are all positive.\n\nFurther, $\\angle BPN = 30^\\circ + x = A$ and $\\angle BNP = 30^\\circ + z = C$. So the triangle $PBN$ is similar to $ABC$. So are $QKC$ and $ALM$. If we take $BN = ka$, $NP = kb$, and $PB = kc$, then $QC = kb^2/c$ (note $PN = KQ$). Thus\n\n$$\na = BP + PQ + QC = kc + 2kb \\sin x + \\frac{kb^2}{c} = \\frac{k}{c}(c^2 + b^2 + 2bc \\sin(A - 30^\\circ)) \\\\ \n= \\frac{k}{c}(b^2 + c^2 - 2bc \\cos(A + 60^\\circ)) \\\\ \n= \\frac{k}{c}f^2,\n$$\n\nwhere $f^2 = (a^2 + b^2 + c^2 + 4\\sqrt{3}\\Delta)/2$. This shows that $k = ac/f^2$; $PN = kb = abc/f^2$, the radius of the circle as desired.\n\nUsing the cosine rule,\n\n$$\nBT^2 = BP^2 + PT^2 - 2BP \\cdot PT \\cdot \\cos \\angle BPT \\\\ \n= k^2(c^2 + b^2 - 2bc \\cos(90^\\circ + x)) \\\\ \n= k^2(b^2 + c^2 - 2bc \\cos(A + 60^\\circ)) \\\\ \n= k^2 f^2.\n$$\n\nSo $BT = kf = ac/f$. This shows that $BT \\cdot AC = abc/f$. The symmetry of the right side shows that\n\n$$\nBT \\cdot AC = AT \\cdot BC = CT \\cdot AB.\n$$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 13286,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all positive integers $x, y, z$ such that\n$$\n1 + 4^x + 4^y = z^2.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "Without loss of generality, we may assume that $x \\le y$. Suppose that $2x < y + 1$. Then\n$$\n(2^y)^2 < 1 + 4^x + 4^y < (1 + 2^y)^2,\n$$\nwhich implies that $1+4^x+4^y$ is not a square of an integer.\n\nIf $2x = y+1$, then\n$$\n1+4^x+4^y = 1+2^{y+1}+4^y = (1+2^y)^2.\n$$\nHence\n$$\n(x, y, z) = (x, 2x-1, 1+2^{2x-1})\n$$\nis a solution of the equation for any positive integer $x$.\n\nSuppose that $2x > y + 1$. Note that\n$$\n4^x + 4^y = 4^x(1 + 4^{y-x}) = (z-1)(z+1).\n$$\nSince $\\gcd(z-1, z+1) = 2$, one of $z-1$ or $z+1$ is divisible by $2^{2x-1}$. This gives a contradiction because\n$$\n2(1 + 4^{y-x}) \\le 2(1 + 4^{x-2}) < 2^{2x-1} - 2\n$$\nfor any $x > 1$.\n\nTherefore, the solutions are:\n$$\n(x, y, z) = (x, 2x-1, 1+2^{2x-1}) \\text{ or } (2x-1, x, 1+2^{2x-1})\n$$\nfor any positive integer $x$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 13287,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $n$ be the smallest number of players in a tournament such that every player lost at least three matches and won at least three. What is the minimum value of $n$? Construct an example of such a tournament.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $a$ be a player in the tournament. Let $b$ be another player, and let $c$ be a player who beat both $a$ and $b$. There must be a player $d$ who beat both $a$ and $c$. Now there must be a player $e$, different from $d$ and $c$, who beat both $a$ and $d$. Thus $a$ is beaten by at least three distinct players $c$, $d$, and $e$, so every player lost at least three matches.\n\nLet $l_i$ and $w_i$ be the number of matches lost and won by player $i$, respectively. Then $\\sum l_i = \\sum w_i = \\binom{n}{2}$, since each match contributes once to each count. Since $l_i \\ge 3$ for all $i$, it follows that $\\sum w_i \\ge 3n$, so some $w_i$ is at least 3. Thus, there is a player who lost at least three matches and won at least three, so there must be at least seven players.\n\nThe following matrix describes a tournament with seven players that has the required property. The entry in the $(i, j)$ position is 1 if player $i$ beats player $j$ and 0 if $i = j$ or if player $j$ beats player $i$. Thus, the $(j, i)$ entry is 0 if the $(i, j)$ entry is 1.\n\n$$\n\\begin{pmatrix}\n0 & 0 & 1 & 0 & 1 & 0 & 1 \\\\\n1 & 0 & 0 & 0 & 0 & 1 & 1 \\\\\n0 & 1 & 0 & 0 & 1 & 1 & 0 \\\\\n1 & 1 & 1 & 0 & 0 & 0 & 0 \\\\\n0 & 1 & 0 & 1 & 0 & 0 & 1 \\\\\n1 & 0 & 0 & 1 & 1 & 0 & 0 \\\\\n0 & 0 & 1 & 1 & 0 & 1 & 0\n\\end{pmatrix}\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 13288,
    "subject": "Mathematics (Olympiad)",
    "question": "Consider $n$ positive, not necessarily distinct, integers $a_1, a_2, \\dots, a_n$ whose sum is $2S$. A positive integer $k$ is called a separator if one can choose $k$ indexes from $\\{1, 2, \\dots, n\\}$ such that the sum of the corresponding numbers is $S$. What is the maximum possible number of separators?",
    "options": [],
    "answer": "See solution",
    "solution": "If $k$ is a separator, then $n-k$ is also a separator.\n\nIf $1$ is a separator, then we cannot have any other separators than $1$ and $n-1$.\n\nFor $n=1$, we cannot have any separators.\n\nFor $n=2$, only $1$ can be a separator (if the two numbers are equal).\n\nFor $n=3$, only $1$ and $2$ can be separators (for example, with the numbers $1, 2, 3$; $1+2+3=6$, and $1+2=3$).\n\nFor $n=4$, we can have at most two separators, $1$ and $3$, for example with the numbers $1, 2, 3, 6$, where $1+2+3=6$.\n\nWe prove that, for $n \\geq 5$, the maximum number of separators is $n-3$, namely in the case when all the numbers $2, 3, \\dots, n-2$ are separators.\n\nThis maximum is achieved, for example, for the numbers\n\n* $1, 1, 1, 1, 2, 2, 4, 4, \\dots, 2^{k-2}, 2^{k-2}$ if $n=2k$\n\nIndeed, when $n=2k$, we have $2S = 2^k$, hence $S = 2^{k-1}$, which can be written\n$$ S = 2^{k-2} + 2^{k-2} = 2^{k-2} + 2^{k-3} + 2^{k-3} = \\dots = 2^{k-2} + 2^{k-3} + \\dots + 2 + 1 + 1. $$\n\nAlso, when $n=2k+1$, we have $2S = 2^{k+1}$, hence $S = 2^k$, and we can write\n$$ S = 2^{k-1} + 2^{k-1} = 2^{k-1} + 2^{k-2} + 2^{k-2} = \\dots = 2^{k-1} + 2^{k-2} + \\dots + 2 + 1 + 1. $$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 13289,
    "subject": "Mathematics (Olympiad)",
    "question": "We call two distributions neighbours if they are a stone's throw away from each other.\n\nCount the number $N_k$ of distributions containing exactly $k$ empty boxes, where $0 \\leq k \\leq n-1$ (since not all boxes can be empty). What is the average cosiness (number of neighbouring distributions) over all such distributions?",
    "options": [],
    "answer": "See solution",
    "solution": "There are $\\binom{n}{k}$ ways to choose the empty boxes and $\\binom{n}{k+1}$ ways to fill the remaining $n-k$ boxes with $n+1$ stones so that each box contains at least one stone. Thus, $N_k = \\binom{n}{k} \\cdot \\binom{n}{k+1}$ distributions have exactly $k$ empty boxes.\n\nEach such distribution has $(n-k)(n-1)$ neighbouring distributions (cosiness). For $k' = n-1-k$, $N_{k'} = \\binom{n}{k+1} \\cdot \\binom{n}{k}$, so $N_{k'} = N_k$, and their cosiness is $(k+1)(n-1)$. The average cosiness for these two sets is $\\frac{(n-k)+(k+1)}{2}(n-1) = \\frac{n+1}{2}(n-1)$, which is constant in $k$ and thus the overall average.",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 13290,
    "subject": "Mathematics (Olympiad)",
    "question": "Given the equation\n\n$$\n2 + x\\sqrt{9 + 6\\sqrt{2}} = x\\sqrt{5 - 2\\sqrt{6}} + \\sqrt{6} - 2\\sqrt{3} + \\sqrt{2}.\n$$\n\na) Write the root of the equation in the form $m - \\sqrt{n}$, where $m$ and $n$ are natural numbers.\n\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": "See solution",
    "solution": "a) We calculate\n\n$$\n\\sqrt{9 + 6\\sqrt{2}} = \\sqrt{3}\\sqrt{2 + 2\\sqrt{2} + 1} = \\sqrt{3}(\\sqrt{2} + 1) = \\sqrt{6} + \\sqrt{3}\n$$\n\n$$\n\\sqrt{5 - 2\\sqrt{6}} = \\sqrt{3 - 2\\sqrt{6} + 2} = |\\sqrt{3} - \\sqrt{2}| = \\sqrt{3} - \\sqrt{2}.\n$$\n\nThe equation takes the form\n\n$$\nx(\\sqrt{6} + \\sqrt{3} - \\sqrt{3} + \\sqrt{2}) = \\sqrt{6} + \\sqrt{2} - 2\\sqrt{3} - 2\n$$\n\n$$\nx(\\sqrt{6} + \\sqrt{2}) = (\\sqrt{6} + \\sqrt{2})(1 - \\sqrt{2}),\n$$\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 $6\\sqrt{2}$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 13291,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $a, b, c$ be positive real numbers. Prove that\n\n$$\n\\frac{a^2 b (b-c)}{a+b} + \\frac{b^2 c (c-a)}{b+c} + \\frac{c^2 a (a-b)}{c+a} \\geq 0.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "Dividing both sides by $a$, $b$, $c$, the inequality is equivalent to\n\n$$\n\\frac{a(b-c)}{c(a+b)} + \\frac{b(c-a)}{a(b+c)} + \\frac{c(a-b)}{b(c+a)} \\geq 0.\n$$\n\nBy adding $1+1+1$ to both sides, the inequality becomes\n\n$$\n\\frac{b(c+a)}{c(a+b)} + \\frac{c(a+b)}{a(b+c)} + \\frac{a(b+c)}{b(c+a)} \\geq 3.\n$$\n\nBy applying the AM-GM inequality, we have\n\n$$\n\\frac{b(c+a)}{c(a+b)} + \\frac{c(a+b)}{a(b+c)} + \\frac{a(b+c)}{b(c+a)} \\geq 3\\sqrt{\\frac{b(c+a) \\cdot c(a+b) \\cdot a(b+c)}{c(a+b) \\cdot a(b+c) \\cdot b(c+a)}} = 3.\n$$\n\nThe inequality is proved.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 13292,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $n = p_1^{k_1} p_2^{k_2} \\cdots p_s^{k_s}$, where $p_1, \\dots, p_s$ are distinct primes and $k_1, \\dots, k_s$ are positive integers. Prove that\n\n$$\n\\frac{3}{2\\sqrt{2}} \\leq \\frac{\\sigma(n)}{\\tau(n)\\sqrt{n}} \\leq \\frac{n+1}{2}\n$$\n\nwhere $\\sigma(n)$ is the sum of divisors of $n$ and $\\tau(n)$ is the number of divisors of $n$.",
    "options": [],
    "answer": "See solution",
    "solution": "First, consider the right inequality. We have\n\n$$\n\\tau(n) = (k_1 + 1)(k_2 + 1) \\cdots (k_s + 1)\n$$\n\nand\n\n$$\n\\sigma(n) = (1 + p_1 + \\cdots + p_1^{k_1}) (1 + p_2 + \\cdots + p_2^{k_2}) \\cdots (1 + p_s + \\cdots + p_s^{k_s}).\n$$\n\nSo\n\n$$\n\\frac{\\sigma(n)}{\\tau(n)\\sqrt{n}} = \\prod_{j=1}^s \\frac{1 + p_j + \\cdots + p_j^{k_j}}{(k_j + 1) p_j^{k_j/2}}.\n$$\n\nThe factors in this product are independent, so it suffices to show that each one is bounded from below by $\\frac{3}{2\\sqrt{2}}$. We want to prove that\n\n$$\n\\frac{1 + p + \\cdots + p^k}{(k + 1) p^{k/2}} \\geq \\frac{3}{2\\sqrt{2}}\n$$\n\nfor all $p \\geq 2$ and $k \\geq 1$. The left-hand side is minimized when $p = 2, k = 1$.\n\nLet $x = \\sqrt{p}$. For every positive integer $k$, the function\n\n$$\nf_k(x) = \\frac{1 + x^2 + x^4 + \\dots + x^{2k}}{x^k}\n$$\n\nis increasing when $x > 1$.\n\nWhen $k = 1$ and $x > y > 1$,\n\n$$\nf_1(x) - f_1(y) = (x - y) + \\frac{y - x}{xy} = (x - y)\\left(1 - \\frac{1}{xy}\\right) > 0.\n$$\n\nIn general,\n\n$$\nf_{2l}(x) = 1 + f_1(x^2) + \\cdots + f_1(x^{2l}),\n$$\n\n$$\nf_{2l+1}(x) = f_1(x^3) + \\cdots + f_1(x^{2l+1}).\n$$\n\nThus, $f_k(x)$ is increasing in $x$.\n\nTherefore,\n\n$$\n\\frac{1 + p + p^2 + \\cdots + p^k}{p^{k/2}} \\geq \\frac{1 + 2 + \\cdots + 2^k}{2^{k/2}}\n$$\n\nand\n\n$$\n\\frac{1 + p + p^2 + \\cdots + p^k}{(k + 1) p^{k/2}} \\geq \\frac{1 + 2 + \\cdots + 2^k}{(k + 1) 2^{k/2}}.\n$$\n\nDefine $g(k) = \\frac{1 + 2 + \\cdots + 2^k}{(k + 1) 2^{k/2}}$, $k \\in \\mathbb{N}$. We show that $g(k)$ is increasing:\n\n$$\ng(k+1) - g(k) = \\frac{1 + 2(1 + 2 + \\cdots + 2^k)}{(k+2)2^{k/2}\\sqrt{2}} - \\frac{1 + 2 + \\cdots + 2^k}{(k+1)2^{k/2}} > 0.\n$$\n\nSo,\n\n$$\n\\frac{1 + p + p^2 + \\cdots + p^k}{(k + 1) p^{k/2}} \\geq \\frac{1 + 2 + \\cdots + 2^k}{(k + 1) 2^{k/2}} \\geq \\frac{3}{2\\sqrt{2}}.\n$$\n\nWe conclude that\n\n$$\n\\frac{\\sigma(n)}{\\tau(n)\\sqrt{n}} \\geq \\left( \\frac{3}{2\\sqrt{2}} \\right)^s \\geq \\frac{3}{2\\sqrt{2}}.\n$$\n\nEquality holds when $n = 2$.\n\nNow, consider the left inequality. Let $1 = d_1 < d_2 < \\dots < d_{\\tau(n)} = n$ be all divisors of $n$. Then $\\sigma(n) = \\sum_{i=1}^{\\tau(n)} d_i$. Since $\\frac{n}{d_1}, \\frac{n}{d_2}, \\dots, \\frac{n}{d_{\\tau(n)}}$ are also divisors of $n$, $\\sigma(n) = \\sum_{i=1}^{\\tau(n)} \\frac{n}{d_i}$.\n\nFor each divisor $d_i$,\n\n$$\nd_i + \\frac{n}{d_i} \\leq n + 1 \\iff d_i^2 - d_i(n + 1) + n \\leq 0 \\iff (n - d_i)(1 - d_i) \\leq 0,\n$$\n\nequality holds only if $d_i = 1$ or $d_i = n$.\n\nNow,\n\n$$\n\\frac{\\sigma(n)}{\\tau(n)} = \\frac{\\sigma(n) + \\sigma(n)}{2\\tau(n)} = \\frac{\\sum_{i=1}^{\\tau(n)} d_i + \\sum_{i=1}^{\\tau(n)} \\frac{n}{d_i}}{2\\tau(n)} = \\frac{\\sum_{i=1}^{\\tau(n)} \\left(d_i + \\frac{n}{d_i}\\right)}{2\\tau(n)} \\leq \\frac{\\tau(n) (n+1)}{2\\tau(n)} = \\frac{n+1}{2}.\n$$\n\nEquality holds only when $\\tau(n) \\leq 2$, i.e., $n = 1$ or $n$ is prime.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 13293,
    "subject": "Mathematics (Olympiad)",
    "question": "Find the largest fraction among the following 1010 fractions:\n\n$$\n\\begin{array}{ccccccc}\n\\frac{1}{2019}, & \\frac{1+2018}{2019+2}, & \\frac{1+2018+3}{2019+2+2017}, & \\frac{1+2018+3+2016}{2019+2+2017+4}, & \\\\\n& \\frac{1+2018+3+2016+5}{2019+2+2017+4+2015}, & \\dots, & & \\\\\n& & \\frac{1+2018+3+2016+\\dots+1009}{2019+2+2017+4+\\dots+1011}, & \\frac{1+2018+3+2016+\\dots+1009+1010}{2019+2+2017+4+\\dots+1011+1010}, & \\\\\n& & & \\frac{1+2018+3+2016+\\dots+1009+\\dots+1008+1010}{2019+2+2017+4+\\dots+1011+\\dots+1010+1010}.\n\\end{array}\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "The largest fractions are the 2nd, 4th, ..., 1010th fractions.\n\nLet us split all fractions into two groups: those in odd positions and those in even positions.\n\n$$\n\\begin{align*}\na_1 &= \\frac{1}{2019}, \\\\\na_2 &= \\frac{1+2018+3}{2019+2+2017} = \\frac{1+2021}{2019+2019}, \\\\\na_3 &= \\frac{1+2018+3+2016+5}{2019+2+2017+4+2015} = \\frac{1+2021+2}{2019+2019+2}, \\dots, \\\\\na_{505} &= \\frac{1+2018+3+2016+5+\\dots+1012+1009}{2019+2+2017+4+2015+\\dots+1008+1011} = \\frac{1+2021+504}{2019+2019+504}. \\\\\nb_1 &= \\frac{1+2018}{2019+2} = \\frac{2019}{2021}, \\\\\nb_2 &= \\frac{1+2018+3+2016}{2019+2+2017+4} = \\frac{2019+2}{2021+2} = \\frac{2019}{2021}, \\\\\nb_3 &= \\frac{1+2018+3+2016+5+2014}{2019+2+2017+4+2015+6} = \\frac{2019+3}{2021+3} = \\frac{2019}{2021}, \\dots, \\\\\nb_{505} &= \\frac{1+2018+3+2016+5+\\dots+1012+1009+1010}{2019+2+2017+4+2015+\\dots+1008+1011+1010} = \\frac{2019+505}{2021+505} = \\frac{2019}{2021}.\n\\end{align*}\n$$\n\nAll fractions denoted by $b_i$, $i = 1, \\ldots, 505$ are equal to $\\frac{2019}{2021}$. Now, compare $a_k$ and $a_{k+1}$:\n\n$$\na_{k+1} - a_k = \\frac{1+2021k}{2019+2019k} - \\frac{1+2021(k-1)}{2019+2019(k-1)} = \\frac{(1+2021k)(2019+2019(k-1)) - (1+2021(k-1))(2019+2019k)}{(2019+2019(k-1))(2019+2019k)}\n$$\n\nThis difference is positive, so $a_k$ increases with $k$. However, comparing $a_{505}$ and $b_{505}$:\n\n$$\n\\frac{1+2021\\cdot504}{2019+2019\\cdot504} - \\frac{2019}{2021} = \\frac{(1+2021\\cdot504) \\cdot 2021 - (2019+2019\\cdot504) \\cdot 2019}{(2019+2019\\cdot504) \\cdot 2021} < 0\n$$\n\nTherefore, the largest value is $\\frac{2019}{2021}$, which occurs at all even positions.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 13294,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $\\triangle ABC$ be a triangle in which $AB = AC$. Suppose the orthocentre of the triangle lies on the incircle. Find the ratio $\\dfrac{AB}{BC}$.",
    "options": [],
    "answer": "See solution",
    "solution": "Since the triangle is isosceles, the orthocentre lies on the perpendicular $AD$ from $A$ onto $BC$. Let it cut the incircle at $H$. Now we are given that $H$ is the orthocentre of the triangle. Let $AB = AC = b$ and $BC = 2a$. Then $BD = a$. Observe that $b > a$ since $b$ is the hypotenuse and $a$ is a leg of a right-angled triangle. Let $BH$ meet $AC$ in $E$ and $CH$ meet $AB$ in $F$. By Pythagoras' theorem applied to $\\triangle BDH$, we get\n\n![](images/IND_National_2016_p0_data_e024b9d07d.png)\n\n$$\nBH^2 = HD^2 + BD^2 = 4r^2 + a^2,\n$$\n\nwhere $r$ is the inradius of $\\triangle ABC$. We want to compute $BH$ in another way. Since $A, F, H, E$ are concyclic, we have\n\n$$\nBH \\cdot BE = BF \\cdot BA.\n$$\n\nBut $BF \\cdot BA = BD \\cdot BC = 2a^2$, since $A, F, D, C$ are concyclic. Hence $BH^2 = \\dfrac{4a^4}{BE^2}$. But\n\n$$\nBE^2 = 4a^2 - CE^2 = 4a^2 - BF^2 = 4a^2 - \\left(\\frac{2a^2}{b}\\right)^2 = \\frac{4a^2(b^2 - a^2)}{b^2}.\n$$\n\nThis leads to\n\n$$\nBH^2 = \\frac{a^2b^2}{b^2 - a^2}.\n$$\n\nThus we get\n\n$$\n\\frac{a^2b^2}{b^2 - a^2} = a^2 + 4r^2.\n$$\n\nThis simplifies to $\\dfrac{a^4}{b^2 - a^2} = 4r^2$. Now we relate $a, b, r$ in another way using area. We know that $[ABC] = rs$, where $s$ is the semiperimeter of $\\triangle ABC$. We have $s = (b + b + 2a)/2 = b + a$. On the other hand, area can be calculated using Heron's formula:\n\n$$\n[ABC]^2 = s(s - 2a)(s - b)(s - b) = (b + a)(b - a)a^2 = a^2(b^2 - a^2).\n$$\n\nHence\n\n$$\nr^2 = \\frac{[ABC]^2}{s^2} = \\frac{a^2(b^2 - a^2)}{(b + a)^2}.\n$$\n\nUsing this we get\n\n$$\n\\frac{a^4}{b^2 - a^2} = 4 \\left( \\frac{a^2(b^2 - a^2)}{(b + a)^2} \\right).\n$$\n\nTherefore $a^2 = 4(b - a)^2$, which gives $a = 2(b - a)$ or $2b = 3a$. Finally,\n\n$$\n\\frac{AB}{BC} = \\frac{b}{2a} = \\frac{3}{4}.\n$$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 13295,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $\\triangle ABC$ be a scalene triangle with $AB < AC$, and let $c(O, R)$ be its circumcircle. The circle $c_1(A, AB)$ (center $A$, radius $AB$) intersects side $BC$ at $E$ and the circumcircle $c$ at $F$. The line $EF$ meets the circumcircle $c$ again at $D$ and side $AC$ at $M$. The line $AD$ intersects $BC$ at $K$. The circumcircle of triangle $BKD$ intersects $AB$ at $L$. Prove that the points $K$, $L$, and $M$ lie on a line parallel to $BF$.",
    "options": [],
    "answer": "See solution",
    "solution": "The angle $\\hat{F}_1$ is inscribed in the circle $c_1$ with corresponding central angle $\\hat{BAE}$. Hence:\n\n![](images/Hellenic_Mathematical_Competitions_2011_booklet_p18_data_cb05e10305.png)\n\n*Figure 4*\n\n$$\n\\hat{F}_1 = \\frac{BA\\hat{E}}{2} = \\frac{\\hat{A}_1 + \\hat{A}_2}{2} \\qquad (1)\n$$\n\nFrom the cyclic quadrilateral $AFDB$ we have:\n\n$$\n\\hat{F}_1 = \\hat{A}_1 \\qquad (2)\n$$\n\nFrom (1) and (2) we get that $\\hat{A}_1 = \\hat{A}_2$, that is, $AK$ is the bisector of angle $BA\\hat{E}$. Since $AB = AE$, we conclude that $AK$ (and hence $DK$) is perpendicular to $BC$, that is,\n\n$$\nDK \\perp BC \\qquad (3)\n$$\n\nIn the circle $c$, chords $AB$ and $AF$ are equal, as radii of circle $c_1$, so $\\hat{D}_1 = \\hat{C}$. Hence the quadrilateral $DKMC$ is cyclic. Therefore, $D\\hat{K}C = D\\hat{M}C = 90^\\circ$, that is,\n\n$$\nDM \\perp AC \\qquad (4)\n$$\n\nFrom the cyclic quadrilateral $BKDL$ we have $B\\hat{K}D = B\\hat{L}D = 90^\\circ$, that is,\n\n$$\nDL \\perp AB \\qquad (5)\n$$\n\nFrom (3), (4), and (5), we conclude that the points $K, L, M$ are on the Simson line of triangle $ABC$ corresponding to point $D$.\n\nFrom the cyclic quadrilateral $DKMC$ we get $\\hat{M}_1 = \\hat{C}_1$. Also, from the cyclic quadrilateral $ABDC$ we have $\\hat{C}_1 = \\hat{A}_1$, and from $ABDF$ we have $\\hat{A}_1 = \\hat{F}_1$. Hence $\\hat{M}_1 = \\hat{F}_1$, from which we get that $BF \\parallel LM$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 13296,
    "subject": "Mathematics (Olympiad)",
    "question": "Given a set $\\mathcal{L}$ of lines in general position in the plane (no two lines in $\\mathcal{L}$ are parallel and no three lines are concurrent) and another line $\\ell$, show that the total number of edges of all faces in the corresponding arrangement intersected by $\\ell$ is at most $6|\\mathcal{L}|$.",
    "options": [],
    "answer": "See solution",
    "solution": "Assume without loss of generality that $\\ell$ is horizontal and does not pass through any vertex of the arrangement of lines in $\\mathcal{L}$.\n\nFirst, we shall bound the total number of edges of the upper parts of all faces intersected by $\\ell$, that is, those parts that lie above $\\ell$. The boundary of the upper part of such a face $K$ consists of two convex chains of edges, the *left* and *right* chain, and a portion of $\\ell$. If the upper part of $K$ is bounded, then the left and right chains meet at the topmost vertex of $K$. Otherwise, the last (topmost) edges of these chains are half-lines. The edges belonging to the left (respectively, right) chain, with the exception of the topmost edge, are called the *left* (respectively, *right*) *edges* of $K$.\n\nWe claim that every line in $\\mathcal{L}$ contains at most one left edge. Suppose, if possible, that some line in $\\mathcal{L}$ has two portions $e$ and $e'$ that are left edges of $K$ and $K'$, respectively, where $e'$ is above $e$. Then the line supporting the topmost edge of the left chain of $K$ would cross $K'$, contradicting the fact that $K'$ is a face of the arrangement of lines in $\\mathcal{L}$. Hence, the total number of left (respectively, right) edges of the upper parts of the faces intersected by $\\ell$ is at most $|\\mathcal{L}|$. Taking into account the topmost edges of the chains, the total number of edges of the upper parts of the faces intersected by $\\ell$ is at most $4|\\mathcal{L}|$. The same argument applies verbatim for the lower parts of these faces. Consequently, the total number of edges of the faces intersected by $\\ell$ is at most $8|\\mathcal{L}| - 2|\\mathcal{L}| = 6|\\mathcal{L}|$; the second term is due to the fact that the edges crossed by $\\ell$ are counted twice: once in the upper parts and once in the lower parts.",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 13297,
    "subject": "Mathematics (Olympiad)",
    "question": "$3 \\times n$ хүснэгтийг, дүрсүүдэд эргүүлж болно давхардуулахгүйгээр хүчих боломжийн тоог $a_n$ гэе.\n\n![](images/2013-ilovepdf-compressed_p50_data_ac069509fb.png)\n\n![](images/2013-ilovepdf-compressed_p50_data_1b3e614584.png)\n\n![](images/2013-ilovepdf-compressed_p50_data_9df2b51190.png)\n\nа) $a_{10}$-ийг ол.\n\nб) $n > 3$ үед $a_n < 3^{n-1}$ гэж батал.",
    "options": [],
    "answer": "See solution",
    "solution": "а) Хариу: $a_{10} = 8266$\n\nДүрсийг хуваах боломжийг $b_{n-1}$ гэе.\n\n$$\na_1 = 2,\\quad a_2 = 3,\\quad a_3 = 10;\\quad b_1 = 1,\n$$\n\n$$\nb_2 = 2,\\quad b_3 = 4.\\quad \\text{Рекуррент харьцаа зохиовол}\n$$\n\n$$\na_{n+1} = a_n + a_{n-2} + 2(b_n + b_{n-1}) \\quad (1)\n$$\n\n$$\nb_n = a_{n-1} + a_{n-2} + b_{n-3} \\quad (2)\n$$\n\n$$\n\\text{Эндээс}\\quad b_4 = 14,\\quad b_5 = 35,\\quad a_4 = 23,\\quad a_5 = 62,\\quad a_6 = 170\\quad (1\\text{-ээс})\n$$\n\n$$\n2(b_n + b_{n-1}) = a_{n+1} - a_n - a_{n-2} \\quad (3)\n$$\n\n$$\n(2\\text{-оос:})\\quad 2(b_n + b_{n-1}) = 2a_{n-1} + 2a_{n-2} + 2(b_{n-1} + b_{n-3}).\n$$\n\n$$\n(3 \\text{ ба } 2\\text{-оос})\\\\\na_{n+1} - a_n - a_{n-2} = 2a_{n-1} + 2a_{n-2} + 2(a_{n-2} + a_{n-3} + b_{n-3} + b_{n-4} + b_{n-5}).\n$$\n\n$$\n(3\\text{-аас})\\quad a_{n+1} = a_n + 2a_{n-1} + 6a_{n-2} + a_{n-3} - a_{n-5}\\text{ болно. Эндээс}\n$$\n\n$$\na_7 = 441;\\quad a_8 = 1173;\\quad a_9 = 3127;\\quad a_{10} = 8266.\n$$\n\nб) $3 < k < n + 1$ үед $a_k < 3^{k-1}$ гэж үзээд $a_{n+1} < 3^n$ болохыг индукцээр баталья. Индукцийн суурь үнэн.\n\n$$\na_{n+1} < a_n + 2a_{n-1} + 6a_{n-2} + a_{n-3} < 3^{n-1} + 2 \\cdot 3^{n-2} + 6 \\cdot 3^{n-3} + 3^{n-4} \\\\\n= 3^{n-4} \\cdot 64 < 3^n \\qquad \\text{батлагдав.}\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 13298,
    "subject": "Mathematics (Olympiad)",
    "question": "Determine the maximum possible value of $a + b$, where $a$ and $b$ are two different non-negative real numbers that satisfy\n\n$$\na + \\sqrt{b} = b + \\sqrt{a}.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "The equation can be rewritten as\n\n$$\n\\sqrt{a} - \\sqrt{b} = a - b\n$$\n\nSince $a \\neq b$, we can factor:\n\n$$\n\\sqrt{a} - \\sqrt{b} = (\\sqrt{a} - \\sqrt{b})(\\sqrt{a} + \\sqrt{b})\n$$\n\nIf $\\sqrt{a} - \\sqrt{b} \\neq 0$, divide both sides by $\\sqrt{a} - \\sqrt{b}$:\n\n$$\n1 = \\sqrt{a} + \\sqrt{b}\n$$\n\nSquaring both sides:\n\n$$\n(\\sqrt{a} + \\sqrt{b})^2 = 1^2 \\\\\na + b + 2\\sqrt{ab} = 1\n$$\n\nSo,\n\n$$\na + b = 1 - 2\\sqrt{ab}\n$$\n\nSince $a$ and $b$ are non-negative and different, $\\sqrt{ab} \\geq 0$, so $a + b \\leq 1$.\n\nFor example, $a = 0$, $b = 1$ gives $a + b = 1$ and satisfies the original equation. Thus, the maximum possible value of $a + b$ is $1$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 13299,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all pairs of real numbers $(x, y)$ such that\n\n$$\n\\{x\\} + \\{y\\} = [x+y].\n$$\n\nwhere $[a]$ is the greatest integer not greater than $a$, and $\\{a\\} = a - [a]$.",
    "options": [],
    "answer": "See solution",
    "solution": "From the statement, $\\{x\\} + \\{y\\}$ is an integer, but $0 \\leq \\{x\\} < 1$, so $\\{x\\} + \\{y\\} = 0$ or $\\{x\\} + \\{y\\} = 1$.\n\n**Case 1:** $\\{x\\} + \\{y\\} = 0$. Then $\\{x\\} = \\{y\\} = 0$, so $x, y \\in \\mathbb{Z}$. Thus, $[x+y] = x + y$, and the equation becomes $x + y = 0$. The solutions are $(n, -n)$, $n \\in \\mathbb{Z}$.\n\n**Case 2:** $\\{x\\} + \\{y\\} = 1$. Here, $x, y \\notin \\mathbb{Z}$. Assume $x \\notin \\mathbb{Z}$, then\n\n$$\nx + y = [x] + \\{x\\} + [y] + \\{y\\} = [x] + [y] + 1,\n$$\n\nso $x + y$ is an integer. Therefore, $x + y = [x + y] = 1$. The solutions are $(x, 1 - x)$, $x \\notin \\mathbb{Z}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 13300,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $k$ and $m$ be integers greater than $1$. Consider $k$ pairwise disjoint sets $S_1, S_2, \\dots, S_k$; each of these sets has exactly $m+1$ elements, one of which is red and the other $m$ are all blue. Let $\\mathcal{F}$ be the family of all subsets $F$ of $S_1 \\cup S_2 \\cup \\dots \\cup S_k$ such that, for every $i$, the intersection $F \\cap S_i$ is monochromatic; the empty set is monochromatic. Determine the largest possible cardinality of a subfamily $\\mathcal{G} \\subseteq \\mathcal{F}$, no two sets of which are disjoint.",
    "options": [],
    "answer": "See solution",
    "solution": "The required maximum is $2^{m-1}(2^m + 1)^{k-1}$ and is achieved if, for instance, $\\mathcal{G}$ consists of all sets in $\\mathcal{F}$ containing a fixed blue element.\n\nWe now prove that $|\\mathcal{G}| \\le 2^{m-1}(2^m + 1)^{k-1}$ for any $\\mathcal{G}$ satisfying the conditions in the statement. For convenience, write $M = 2^m + 1$. Let $r_i$ denote the red element of $S_i$, and let $B_i$ be the set of blue elements in $S_i$.\n\nFor every subset $X_i \\subset B_i$ and every $j \\in \\mathbb{Z}_M$, define the sets\n\n$$\nT_{X_i, j} = \\begin{cases} \\{r_i\\}, & \\text{if } j = 0; \\\\ X_i, & \\text{if } j \\ne 0 \\text{ and } j \\text{ is even (considered as a number in } [1, M-1]\\}; \\\\ B_i \\setminus X_i, & \\text{if } j \\ne 0 \\text{ and } j \\text{ is odd (considered as a number in } [1, M-1]\\}. \\end{cases}\n$$\n\nNote that, for every $i$ and every $j$, the sets $T_{X_i,j}$ and $T_{X_i,j+1}$ are disjoint. Now, for every sets $X_i \\subset B_i$ and every elements $j_i \\in \\mathbb{Z}_M$, $i = 1, 2, \\dots, k$, denote\n\n$$\nF(X_1, X_2, \\dots, X_k, j_1, j_2, \\dots, j_k) = \\bigcup_{i=1}^{k} T_{X_i, j_i}.\n$$\n\n**Claim.** Every set $F \\in \\mathcal{F}$ has exactly $2^{mk}$ representations of the form above.\n\n*Proof.* Set $F_i = F \\cap S_i$. If $F_i = \\{r_i\\}$, then there are $2^m$ possible choices for $X_i$, and one should necessarily have $j_i = 0$. Otherwise, there are only two possible choices for $X_i$, namely $X_i = F_i$ and $X_i = B_i \\setminus F_i$, and for each of them there are $2^{m-1}$ possible choices for $j_i$. So, whatever $F$, there are $2^m$ possible choices for each pair $(X_i, j_i)$ all of which can be made independently, whence a total of $2^{mk}$ possible tuples $(X_1, X_2, \\dots, X_k, j_1, j_2, \\dots, j_k)$. This proves the Claim.\n\nThe Claim implies that each $F \\in \\mathcal{F}$ has the same number of representations of the form above. Thus, it suffices to show that, among all $N = 2^{km}(2^m + 1)^k$ tuples $(X_1, X_2, \\dots, X_k, j_1, j_2, \\dots, j_k)$, at most $\\frac{2^{m-1}}{2^m+1}N$ satisfy\n\n$$\nF(X_1, X_2, \\dots, X_k, j_1, j_2, \\dots, j_k) \\in \\mathcal{G}.\n$$\n\nTo this end, split all these tuples into length $M$ cycles\n\n$$\n(F(X_1, X_2, \\dots, X_k, j_1, j_2, \\dots, j_k), F(X_1, X_2, \\dots, X_k, j_1+1, j_2+1, \\dots, j_k+1), \\dots, F(X_1, X_2, \\dots, X_k, j_1+M-1, j_2+M-1, \\dots, j_k+M-1)),\n$$\n\nand note that any two adjacent sets of a cycle are disjoint. Hence each cycle contains at most $\\lfloor M/2 \\rfloor = 2^{m-1}$ sets from $\\mathcal{G}$. This provides the desired upper bound and completes the solution.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 13301,
    "subject": "Mathematics (Olympiad)",
    "question": "How many ways can four people (excluding Alfred) be arranged such that Alfred is not on the extreme left?",
    "options": [],
    "answer": "See solution",
    "solution": "The person on the extreme left can be any one of the four people that is not Alfred; the second left can be any one of the remaining three; the first person on the right of centre, and so on. The number of possibilities is $4 \\times 3 \\times 2 \\times 1 = 24$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 13302,
    "subject": "Mathematics (Olympiad)",
    "question": "In $\\triangle ABC$, $BC > AB$, $BD$ bisects $\\angle ABC$ and intersects $AC$ at $D$. As shown in the figure, $CP \\perp BD$ with $P$ as the foot of the perpendicular and $AQ \\perp BP$ with $Q$ as the foot of the perpendicular. Points $M$ and $E$ are the midpoints of $AC$ and $BC$ respectively. The circumscribed circle of $\\triangle PQM$ intersects $AC$ at the point $H$. Prove that $O$, $H$, $E$, $M$ are concyclic.\n\n![](images/Mathematical_Olympiad_in_China_2007-2008-Problems_and_Solutions_p185_data_ac57a1742a.png)",
    "options": [],
    "answer": "See solution",
    "solution": "Extend $AQ$ to intersect $BC$ at point $N$, then $Q$, $M$ are the midpoints of $AN$ and $AC$ respectively. Thus $QM \\parallel BC$, and\n\n$$\n\\angle PQM = \\angle PBC = \\frac{1}{2} \\angle ABC.\n$$\n\nSimilarly, $\\angle MPQ = \\frac{1}{2} \\angle ABC$. Then $QM = PM$.\n\nSince $Q$, $H$, $P$, $M$ are concyclic, we have\n\n$$\n\\angle PHC = \\angle PHM = \\angle PQM,\n$$\n\nthat is, $\\angle PHC = \\angle PBC$. Therefore, $P$, $H$, $B$, $C$ are concyclic and\n\n$$\n\\angle BHC = \\angle BPC = 90^{\\circ}.\n$$\n\nHence,\n\n$$\nHE = \\frac{1}{2}BC = EP.\n$$\n\nSince $OH = OP$, we know that $OE$ is the perpendicular bisector of $HP$. Also, $\\angle MPQ = \\frac{1}{2} \\angle ABC$ and $E$ is the midpoint of $BC$, we conclude that $P$, $M$ and $E$ are collinear.\n\nConsequently,\n\n$$\n\\angle EHO = \\angle EPO = \\angle OPM = \\angle OMP\n$$\n\nand $O$, $H$, $E$, $M$ are concyclic.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 13303,
    "subject": "Mathematics (Olympiad)",
    "question": "60 participants took part in the Olympiad. They were offered 8 tasks, each evaluated from 0 to 7 points. Prove that, in total, there are 3 participants whose results differ by no more than 1 point. Would the statement be true if 58 took part in the Olympiad?\n\n*The result of a participant in the Olympiad is the total number of points they earned.*",
    "options": [],
    "answer": "See solution",
    "solution": "*Answer:* It will not be true for 58 participants.\n\n*Solution.* The minimal possible total score is $0$, and the maximal is $8 \\times 7 = 56$. Consider the following score segments: $[0, 1]$, $[2, 3]$, $[4, 5]$, $\\,\\ldots\\,$, $[54, 55]$, and $56$ as a single-point segment. If at least 3 students are in any one of these segments, the statement is proved. If not, then each segment contains at most 2 students. There are $28$ two-point segments plus the single-point segment $56$, totaling $29$ segments. Thus, at most $2 \\times 28 + 1 = 57$ students can participate without having 3 in any segment. Since there are 60 participants, by the pigeonhole principle, at least one segment must contain at least 3 students whose scores differ by at most 1 point.\n\nFor 58 participants: If exactly 2 participants have each of the scores $0, 2, 4, \\ldots, 56$, then no three participants have scores differing by at most 1 point, and there are exactly 58 participants.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 13304,
    "subject": "Mathematics (Olympiad)",
    "question": "是否能找到十個集合 $A_1, A_2, \\dots, A_{10}$，同時滿足下列條件：\n\n1. 每個集合有三個元素，形如 $\\{a, b, c\\}$，其中 $a \\in \\{1, 2, 3\\}$，$b \\in \\{4, 5, 6\\}$，$c \\in \\{7, 8, 9\\}$。\n2. 任兩集合都不相等。\n3. 將這十個集合依次圍成一圈（$A_1, A_2, \\dots, A_{10}$），則任意相鄰的兩集合沒有共同元素，但是任意不相鄰的兩集合都有共同元素。\n\n*Can we find ten sets $A_1, A_2, \\dots, A_{10}$ such that:*\n\n1. *Each set is in the form of $\\{a, b, c\\}$, where $a \\in \\{1, 2, 3\\}$, $b \\in \\{4, 5, 6\\}$, $c \\in \\{7, 8, 9\\}$.*\n2. *Each set is different from any other.*\n3. *If we place the sets into a circle ($A_1, A_2, \\dots, A_{10}$), then any pair of neighbouring sets has no common element, but any pair of non-neighbouring sets does. (Remark: $A_{10}$ is a neighbour of $A_1$.)*",
    "options": [],
    "answer": "See solution",
    "solution": "可以。考慮以下十個集合：\n\n$\\{1, 4, 9\\}$,\n$\\{2, 5, 7\\}$,\n$\\{3, 4, 8\\}$,\n$\\{1, 5, 9\\}$,\n$\\{2, 4, 8\\}$,\n$\\{3, 5, 9\\}$,\n$\\{2, 4, 7\\}$,\n$\\{1, 5, 8\\}$,\n$\\{3, 4, 7\\}$,\n$\\{2, 5, 8\\}$。\n\n註：此構造可由 $(1), (2), (3) \\rightarrow (14), (25), (34), (15), (24), (35)$ 逐步添加下一個元素而來。",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 13305,
    "subject": "Mathematics (Olympiad)",
    "question": "Find the smallest positive real number $k$ such that for any four distinct real numbers $a, b, c, d$, each greater than or equal to $k$, there exists a permutation $p, q, r, s$ of $a, b, c, d$ such that the equation\n\n$$\n(x^2 + px + q)(x^2 + rx + s) = 0\n$$\n\nhas four distinct real roots.",
    "options": [],
    "answer": "See solution",
    "solution": "Suppose $k < 4$. Take $a, b, c, d \\in [k, \\sqrt{4k}]$. For any permutation $p, q, r, s$ of $a, b, c, d$, consider $x^2 + px + q = 0$; its discriminant is\n\n$$\n\\Delta = p^2 - 4q < 4k - 4q \\le 4k - 4k = 0.\n$$\n\nSo it has no real roots, implying $k \\ge 4$.\n\nNow suppose $4 \\le a < b < c < d$. Consider:\n\n$$\nx^2 + dx + a = 0,\n$$\n$$\nx^2 + cx + b = 0.\n$$\n\nTheir discriminants are\n\n$$\nd^2 - 4a > 4(d - a) > 0,\n$$\n$$\nc^2 - 4b > 4(c - b) > 0.\n$$\n\nThus, both equations have two distinct real roots. If they share a root $\\beta$, then\n\n$$\n\\beta^2 + d\\beta + a = 0,\n$$\n$$\n\\beta^2 + c\\beta + b = 0.\n$$\n\nSubtracting gives $\\beta = \\frac{b-a}{d-c} > 0$. But then $\\beta^2 + d\\beta + a > 0$, a contradiction. Therefore, $k = 4$.",
    "topic": "Discrete Mathematics",
    "subtopic": "Combinatorics"
  },
  {
    "id": 13306,
    "subject": "Mathematics (Olympiad)",
    "question": "Determine all composite integers $n > 1$ that satisfy the following property: if $d_1, d_2, \\dots, d_k$ are all the positive divisors of $n$ with $1 = d_1 < d_2 < \\dots < d_k = n$, then $d_i$ divides $d_{i+1} + d_{i+2}$ for every $1 \\leq i \\leq k-2$.",
    "options": [],
    "answer": "See solution",
    "solution": "The solutions are $n = p^a$, for any prime $p$ and any integer $a \\geq 2$.\n\nFirst, we check that $n = p^a$ are solutions. The divisors of $n$ in order are:\n\n$$\n1 < p < p^2 < p^3 < \\dots < p^a\n$$\n\nSo $d_i = p^{i-1}$. Moreover,\n\n$$\nd_{i+1} + d_{i+2} = p^i + p^{i+1} = p^{i-1}(p + p^2) = d_i(p + p^2).\n$$\n\nThus $d_i$ divides $d_{i+1} + d_{i+2}$ for $1 \\leq i \\leq k-2$, so all $n = p^a$ are solutions.\n\nNext, suppose $n$ has at least two different prime divisors. Let $p < q$ be the two smallest prime factors of $n$.\n\nWe require $d_{k-2}$ divides $d_{k-1} + d_k$. Since $d_k = n$ and $d_{k-2}$ divides $n$, it follows that $d_{k-2}$ divides $d_{k-1}$. But $d_{k-1} = \\frac{n}{d_2}$ and $d_{k-2} = \\frac{n}{d_3}$, so $\\frac{d_{k-1}}{d_{k-2}} = \\frac{d_3}{d_2}$, and thus $d_2$ divides $d_3$. Combining this with $d_2$ divides $d_3 + d_4$ yields $d_2$ divides $d_4$.\n\nMore generally, whenever $d_2$ divides $d_i$ and $d_2$ divides $d_{i+1}$, then from $d_i$ divides $d_{i+1} + d_{i+2}$, we also have $d_2$ divides $d_{i+2}$. It follows inductively that $d_2$ divides $d_i$ for $i = 2, 3, 4, \\dots, k$.\n\nSince $p$ is the smallest prime divisor of $n$, $d_2 = p$. Hence $p$ divides $d_i$ for $2 \\leq i \\leq k$. But $q$ divides $n$, so $q = d_j$ for some $j$ with $2 \\leq j \\leq k$. Thus $p$ divides $q$, which is a contradiction because $p$ and $q$ are distinct primes. $\\square$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 13307,
    "subject": "Mathematics (Olympiad)",
    "question": "In a scalene triangle, one angle is exactly twice as large as another, and some angle in this triangle is $36^\\circ$. Find all possible values for the angles of this triangle.",
    "options": [],
    "answer": "See solution",
    "solution": "Let the angles be $\\alpha$, $2\\alpha$, and $180^\\circ - 3\\alpha$. All angles must be different. Consider three cases:\n\n1. $\\alpha = 36^\\circ$: Angles are $36^\\circ$, $72^\\circ$, $72^\\circ$ (not scalene).\n2. $2\\alpha = 36^\\circ \\implies \\alpha = 18^\\circ$: Angles are $18^\\circ$, $36^\\circ$, $126^\\circ$.\n3. $180^\\circ - 3\\alpha = 36^\\circ \\implies 3\\alpha = 144^\\circ \\implies \\alpha = 48^\\circ$: Angles are $48^\\circ$, $96^\\circ$, $36^\\circ$.\n\nThus, the possible sets of angles are $18^\\circ$, $36^\\circ$, $126^\\circ$ or $36^\\circ$, $48^\\circ$, $96^\\circ$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 13308,
    "subject": "Mathematics (Olympiad)",
    "question": "Write the sum as follows:\n\n$$\n\\begin{array}{c}\naba \\\\\ncdc \\\\\\hline\neffe\n\\end{array}\n$$\n\nHow many pairs of 3-digit palindromes $aba$ and $cdc$ sum to a 4-digit palindrome $effe$?",
    "options": [],
    "answer": "See solution",
    "solution": "#### Method 1\n\nWithout loss of generality, take $a < c$.\n\nSo the possible solution pairs for $(a, c)$ are: $(2, 9)$, $(3, 8)$, $(4, 7)$, $(5, 6)$.\n\nNow the 10s column gives $b + d + 1 = f$ or $b + d + 1 = f + 10$.\n\n**Case 1:** $b + d + 1 = f$.\n\nSince there is no carry to the 100s column and $a + c = 11$, we have $f = 1$.\n\nThus $b + d = 0$, hence $b = d = 0$.\n\nSo this gives four solutions: $202 + 909 = 1111$, $303 + 808 = 1111$, $404 + 707 = 1111$, $505 + 606 = 1111$.\n\n**Case 2:** $b + d + 1 = f + 10$.\n\nSince there is a carry to the 100s column, $a + c + 1 = 12$.\n\nThen $f = 2$, hence $b + d = 11$.\n\nFor each pair of values of $a$ and $c$, there are then eight solution pairs for $(b, d)$: $(2, 9)$, $(3, 8)$, $(4, 7)$, $(5, 6)$, $(6, 5)$, $(7, 4)$, $(8, 3)$, $(9, 2)$.\n\nSo this gives $4 \\times 8 = 32$ solutions:\n\n$(222 + 999 = 1221, 232 + 989 = 1221, \\text{etc}.)$\n\nHence the number of solution pairs overall is $4 + 32 = 36$.\n\n#### Method 2\n\nWe know that $e = 1$, $a + c = 11$, and the carry from the 10s column is at most 1.\n\nHence $f = 1$ or $2$. Therefore $b + d = 0$ or $11$ respectively.\n\nFor each pair of values of $a$ and $c$, there are then nine solution pairs for $(b, d)$: $(0, 0)$, $(2, 9)$, $(3, 8)$, $(4, 7)$, $(5, 6)$, $(6, 5)$, $(7, 4)$, $(8, 3)$, $(9, 2)$.\n\nThus the number of required pairs of 3-digit palindromes is $4 \\times 9 = 36$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 13309,
    "subject": "Mathematics (Olympiad)",
    "question": "Suppose $k = 4045$ arbitrary marbles are placed in a $2024 \\times 2024$ grid so that no two marbles are placed in two adjacent cells. Is it always possible to move one marble to an adjacent square so that there are still no two marbles in adjacent cells? What is the largest such $k$ for which this is always possible?",
    "options": [],
    "answer": "See solution",
    "solution": "We analyze the placement of marbles on the grid, considering the distribution along diagonals. For any configuration with $k = 4045$ marbles, arguments based on the number of marbles on white diagonals above and below the main diagonal $D$ show that at least one marble can always be moved to an adjacent cell without violating the adjacency condition. \n\nIf $k \\geq 4046$, we construct an arrangement where moving any marble would result in two marbles being adjacent. Specifically, place 2 marbles at the ends of the main diagonal $D$ (see the figure below), then consecutively fill $m$ black sub-diagonals below $D$ and $n$ sub-diagonals above $D$, and finally place $l$ marbles on the remaining empty cells of $D$, where $1 \\leq m, n \\leq 1011$ and $0 \\leq l \\leq 2022$.\n\n![](images/Vietnam_2024_Booklet_p25_data_a1ea9e5d52.png)\n\nThe total number of marbles is:\n\n$$\nF(m, n, l) = \\sum_{i=1}^{m} (2024 - 2i) + \\sum_{j=1}^{n} (2024 - 2j) + 2 + l = (2023 - m)m + (2023 - n)n + 2 + l.\n$$\n\nAs $m, n, l$ vary over their ranges, $F(m, n, l)$ covers all values from $4046$ to $\\frac{2024^2}{2}$. In these arrangements, moving any marble creates two adjacent marbles. Thus, the largest $k$ for which it is always possible to move a marble without creating adjacency is $4045$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 13310,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $p, q$ be coprime integers such that $\\frac{p}{q} \\le 1$. For which $p, q$ do there exist even integers $b_1, b_2, \\dots, b_n$ such that\n\n$$\n\\frac{p}{q} = \\frac{1}{b_1 + \\frac{1}{b_2 + \\frac{1}{b_3 + \\dots}}}?\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "Set $b_i = 2\\ell_i$, so all $b_i$ are even. At the first step, we get the pairs\n\n$$\nA := \\{(p, q) : (p, q) = (1, 2\\ell_1),\\ \\ell_1 \\in \\mathbb{Z}\\}.\n$$\n\nAt each subsequent step, we expand the set $A$ by adding additional pairs $(p', q')$ defined as\n\n$$\n\\left\\{ (p', q') : \\frac{p'}{q'} = \\frac{1}{2\\ell + \\frac{p}{q}},\\ (p, q) \\in A,\\ \\ell \\in \\mathbb{Z} \\right\\}.\n$$\n\nBy this, we obtain\n\n$$\np' = q, \\quad q' = 2\\ell q + p.\n$$\n\nIf $(p, q)$ is obtained in the process, we also add $(p', q')$ as above. We want to characterize all pairs $(p, q)$ that can be obtained this way. Note that if $(p, q) = 1$, then $(p', q') = 1$ as well. The set of pairs $(p, q)$ in question is\n\n$$\nA := \\{(p, q) : p, q \\in \\mathbb{Z} \\setminus \\{0\\},\\ |p| < |q|,\\ (p, q) = 1,\\ \\text{one of } p, q \\text{ is even, the other is odd}\\}.\n$$\n\nWe rule out $|p| = |q|$ because $\\frac{p}{q} = \\pm 1$ cannot be represented as required. The transformation above shows we cannot step outside $A$. To prove all pairs in $A$ can be generated, take $(p', q') \\in A$ and search for $(p, q) \\in A$ that generates $(p', q')$ via the transformation, with $(p, q)$ \"less\" than $(p', q')$. Solving for $(p, q)$ gives\n\n$$\np = q' - 2\\ell p', \\quad q = p'.\n$$\n\nWe can choose $\\ell \\in \\mathbb{Z}$ such that $|q' - 2\\ell p'| < |p'|$ (since $p' \\neq 0$). The points $q' - 2\\ell p'$ are $2p'$ apart, and none equals $p'$ because $(p', q') = 1$. The closest to $0$ has magnitude less than $p'$. Thus, starting from $(p', q') \\in A$, we find $(p, q) \\in A$ that generates $(p', q')$ and $|p| < |p'| < |q'|$. By induction, the result follows. The representation is also unique: for $(p', q') \\in A$, the pair $(p, q) \\in A$ satisfying the above and $|p| < |p'|$ is unique.\n\n**Remark.** The main difficulty is that the answer was not given to the students. Some motivation about guessing the answer can be found in this blog $\\Box$",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 13311,
    "subject": "Mathematics (Olympiad)",
    "question": "Докажи дека $\\frac{1}{1 \\cdot 2013} + \\frac{1}{2 \\cdot 2012} + \\frac{1}{3 \\cdot 2011} + \\dots + \\frac{1}{2012 \\cdot 2} + \\frac{1}{2013 \\cdot 1} < 1$.",
    "options": [],
    "answer": "See solution",
    "solution": "За произволни природни броеви $n \\neq 1 \\neq m$ е исполнето неравенството $nm \\ge n+m$, бидејќи $(n-1)(m-1) \\ge 1 \\Rightarrow nm-n-m+1 \\ge 1 \\Rightarrow nm-n-m \\ge 0 \\Rightarrow nm \\ge n+m$, при што равенство важи само за $n=1$ или $m=1$.\n\n$$\n\\frac{1}{1 \\cdot 2013} + \\frac{1}{2 \\cdot 2012} + \\frac{1}{3 \\cdot 2011} + \\dots + \\frac{1}{2012 \\cdot 2} + \\frac{1}{2013 \\cdot 1} < \\frac{2}{2013} + \\frac{2011}{2014} < \\frac{3}{2014} + \\frac{2011}{2014} = 1.\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 13312,
    "subject": "Mathematics (Olympiad)",
    "question": "For real numbers $-1 < x_1, x_2, \\dots, x_n < 1$ with sum $x_1 + x_2 + \\dots + x_n = 0$, prove that\n\n$$\n\\sum_{i=1}^{n} \\sum_{j=1}^{n} x_i x_j \\sqrt{1 - x_i^2 x_j^2} \\le 0\n$$\n\nand determine the conditions under which equality holds.",
    "options": [],
    "answer": "See solution",
    "solution": "By Taylor's theorem, we have\n\n$$\n\\sqrt{1-t} = 1 - \\sum_{k=1}^{\\infty} \\frac{(2k)!}{4^k (k!)^2 (2k-1)} t^k\n$$\n\nfor any $-1 < t < 1$. Since $\\sum_{i=1}^n x_i = 0$, we have\n\n$$\n\\sum_{i=1}^{n} \\sum_{j=1}^{n} x_i x_j \\sqrt{1 - x_i^2 x_j^2} = \\left( \\sum_{i=1}^{n} x_i \\right)^2 - \\sum_{k=1}^{\\infty} \\frac{(2k)!}{4^k (k!)^2 (2k-1)} \\left( \\sum_{i=1}^{n} x_i^{2k+1} \\right)^2 \\le 0.\n$$\n\nEquality holds if and only if $\\sum_{i=1}^n x_i^{2k+1} = 0$ for all $k \\ge 1$. This means that the list $x_1, x_2, \\dots, x_n$ consists of zeros and opposite numbers. Indeed, equality holds for such numbers. Now suppose $\\sum_{i=1}^n x_i^{2k+1} = 0$ for all $k \\ge 1$. Removing the zeros, and negating the negative numbers, we get positive numbers $a_1, a_2, \\dots, a_m > 0$ and $b_1, b_2, \\dots, b_l > 0$ such that $\\sum_i a_i^{2k+1} = \\sum_j b_j^{2k+1}$ for all $k \\ge 1$. It is easy to see that this implies $\\max\\{a_i\\} = \\max\\{b_j\\}$ considering a large enough $k$. By induction, the list consists of zeros and opposite numbers.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 13313,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $k \\in [0, 1]$. Solve the system\n\n$$\n\\begin{aligned}\nk - x^2 &= y, \\\\\nk - y^2 &= z, \\\\\nk - z^2 &= u, \\\\\nk - u^2 &= x\n\\end{aligned}\n$$\n\nin real numbers.",
    "options": [],
    "answer": "See solution",
    "solution": "Subtracting the third equation from the first, we get\n\n$$\nz^2 - x^2 = (z - x)(z + x) = y - u. \\tag{1}\n$$\n\nSimilarly, the second and fourth equations imply\n\n$$\ny^2 - u^2 = (y - u)(y + u) = x - z. \\tag{2}\n$$\n\nRelations (1) and (2) imply that $x = z$ holds if and only if $y = u$ holds. We distinguish two cases:\n\n**a)** Assume $x = z$ and $y = u$, i.e., quadruplets of the form $(x, y, x, y)$. The system reduces to\n\n$$\n\\begin{aligned}\nk - x^2 &= y, \\\\\nk - y^2 &= x.\n\\end{aligned}\n$$\n\nSubtracting and rewriting:\n\n$$\n(y - x)(y + x - 1) = 0.\n$$\n\nTwo subcases:\n\n- If $y - x = 0$, the system reduces to\n\n$$\nx^2 + x - k = 0.\n$$\n\nFor $k \\in (0, 1]$, this has two real solutions:\n\n$$\nx_{1,2} = \\frac{-1 \\pm \\sqrt{4k + 1}}{2}.\n$$\n\nSo, two solutions:\n\n$$\nx_1 = y_1 = z_1 = u_1 = \\frac{-1 + \\sqrt{4k + 1}}{2}, \\quad x_2 = y_2 = z_2 = u_2 = \\frac{-1 - \\sqrt{4k + 1}}{2}. \\tag{3}\n$$\n\n- If $x + y - 1 = 0$, the system reduces to\n\n$$\nx^2 - x + (1 - k) = 0.\n$$\n\nDiscriminant is $4k - 3$, so solutions exist if $k \\ge \\frac{3}{4}$:\n\n$$\nx_3 = \\frac{1 + \\sqrt{4k - 3}}{2}, \\quad x_4 = \\frac{1 - \\sqrt{4k - 3}}{2}\n$$\n\nwith $y_3 = \\frac{1 - \\sqrt{4k - 3}}{2}$ and $y_4 = \\frac{1 + \\sqrt{4k - 3}}{2}$.\n\nIf $k = 3/4$, these coincide with (3). For $3/4 < k \\le 1$, two more distinct solutions:\n\n$$\n(x, y, z, u) = (x_3, y_3, x_3, y_3), \\quad (x_4, y_4, x_4, y_4). \\tag{4}\n$$\n\n**b)** If $x \\neq z$ and $y \\neq u$, plugging $x - z$ from (2) into (1) and dividing by nonzero $y - u$ gives\n\n$$\n(x+z)(y+u) = -1. \\tag{5}\n$$\n\nSince $-1 < 0$, at least one of $x, y, z, u$ is positive and one is negative, which contradicts the implications\n\n$$\nx \\ge 0 \\Rightarrow y \\ge 0 \\Rightarrow z \\ge 0 \\Rightarrow u \\ge 0 \\Rightarrow x \\ge 0. \\tag{6}\n$$\n\nThus, no solutions in case b).\n\n**Answer:**\n- If $0 \\le k \\le 3/4$, the system has two solutions given by (3).\n- If $3/4 < k \\le 1$, the system has four solutions given by (3) and (4).",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 13314,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $k > 1$ be a positive integer and $n > 2018$ be an odd positive integer. The nonzero rational numbers $x_1, x_2, \\dots, x_n$ are not all equal and satisfy\n\n$$\nx_1 + \\frac{k}{x_2} = x_2 + \\frac{k}{x_3} = x_3 + \\frac{k}{x_4} = \\dots = x_{n-1} + \\frac{k}{x_n} = x_n + \\frac{k}{x_1}.\n$$\n\nFind:\n\na) the product $x_1 x_2 \\dots x_n$ as a function of $k$ and $n$.\n\nb) the least value of $k$ such that there exist $n, x_1, x_2, \\dots, x_n$ satisfying the given conditions.",
    "options": [],
    "answer": "See solution",
    "solution": "a) If $x_i = x_{i+1}$ for some $i$ (with $x_{n+1} = x_1$), then by the given identity all $x_i$ will be equal, a contradiction. Thus $x_1 \\ne x_2$ and\n\n$$\nx_1 - x_2 = k \\frac{x_2 - x_3}{x_2 x_3}.\n$$\n\nAnalogously,\n\n$$\nx_1 - x_2 = k \\frac{x_2 - x_3}{x_2 x_3} = k^2 \\frac{x_3 - x_4}{(x_2 x_3)(x_3 x_4)} = \\dots = k^n \\frac{x_1 - x_2}{(x_2 x_3)(x_3 x_4) \\dots (x_1 x_2)}.\n$$\n\nSince $x_1 \\ne x_2$, we get\n\n$$\nx_1 x_2 \\dots x_n = \\pm \\sqrt{k^n} = \\pm k^{\\frac{n-1}{2}} \\sqrt{k}.\n$$\n\nIf one of these two values (positive or negative) is obtained, then the other will also be obtained by changing the sign of all $x_i$ since $n$ is odd.\n\nb) From the above result, as $n$ is odd, we conclude that $k$ is a perfect square, so $k \\ge 4$. For $k = 4$, let $n = 2019$ and $x_{3j} = 4$, $x_{3j-1} = 1$, $x_{3j-2} = -2$ for $j = 1, 2, \\dots, 673$. So, the required least value is $k = 4$.\n\n![](images/Macedonia2018Binder_kniga_MMO_2018_p10_data_2ad22d6842.png)\n\nThere are many ways to construct the example when $k = 4$ and $n = 2019$. Since $3 \\mid 2019$, the idea is to find three numbers $x_1, x_2, x_3$ satisfying the given equations, not all equal, and repeat them as values for the rest of the $x_i$'s. So, we want to find $x_1, x_2, x_3$ such that\n\n$$\nx_1 + \\frac{4}{x_2} = x_2 + \\frac{4}{x_3} = x_3 + \\frac{4}{x_1}.\n$$\n\nAs above, $x_1 x_2 x_3 = \\pm 8$. Suppose without loss of generality that $x_1 x_2 x_3 = -8$. Then, solving the above system we see that if $x_1 \\neq 2$, then\n\n$$\nx_2 = -\\frac{4}{x_1 - 2} \\quad \\text{and} \\quad x_3 = 2 - \\frac{4}{x_1}\n$$\n\nleading to infinitely many solutions. The example in the official solution is obtained by choosing $x_1 = -2$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 13315,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $n$ be a positive integer. If $S$ is a finite set of vectors in the plane, let $N(S)$ denote the number of two-element subsets $\\{v, v'\\}$ of $S$ such that\n$$\n4 (v \\cdot v') + (|v|^2 - 1)(|v'|^2 - 1) < 0.\n$$\nDetermine the maximum of $N(S)$ as $S$ runs through all $n$-element sets of vectors in the plane.",
    "options": [],
    "answer": "See solution",
    "solution": "Without loss of generality, assume all vectors are anchored at the origin. Assign to each vector $\\mathbf{v} = (x, y)$ in $\\mathbb{R}^2$ the unit vector\n$$\n\\mathbf{u} = \\left( \\frac{2x}{x^2 + y^2 + 1}, \\frac{2y}{x^2 + y^2 + 1}, \\frac{x^2 + y^2 - 1}{x^2 + y^2 + 1} \\right)\n$$\nin $\\mathbb{R}^3$. This defines a bijection between all vectors in $\\mathbb{R}^2$ and all unit vectors in $\\mathbb{R}^3$ different from $(0, 0, 1)$; its inverse sends a unit vector $\\mathbf{u} = (x, y, z) \\neq (0, 0, 1)$ to $\\mathbf{v} = (x/(1-z), y/(1-z))$.\n\nThe condition\n$$\n4 (\\mathbf{v} \\cdot \\mathbf{v}') + (|\\mathbf{v}|^2 - 1)(|\\mathbf{v}'|^2 - 1) < 0\n$$\nfor vectors in $\\mathbb{R}^2$ is equivalent to $\\mathbf{u} \\cdot \\mathbf{u}' < 0$ for the corresponding unit vectors in $\\mathbb{R}^3$.\n\nLet $S = \\{\\mathbf{v}_1, \\mathbf{v}_2, \\dots, \\mathbf{v}_n\\}$, and let $\\{\\mathbf{u}_1, \\mathbf{u}_2, \\dots, \\mathbf{u}_n\\}$ be the corresponding unit vectors in $\\mathbb{R}^3$. Consider the graph $G$ on $n$ vertices labeled $1, 2, \\dots, n$, with an edge joining $i$ and $j$ if and only if $\\mathbf{u}_i \\cdot \\mathbf{u}_j < 0$.\n\nThe largest set of unit vectors in $\\mathbb{R}^3$ with all mutual dot products negative has size $4$, so $G$ is $K_5$-free. By Turán's theorem, $G$ has at most\n$$\n\\left\\lfloor \\left(1 - \\frac{1}{4}\\right) \\cdot \\frac{n^2}{2} \\right\\rfloor = \\left\\lfloor \\frac{3n^2}{8} \\right\\rfloor\n$$\nedges. This upper bound is achieved by Turán's extremal graph $T(n, 4) = K_{n_1, n_2, n_3, n_4}$, where $n_1 + n_2 + n_3 + n_4 = n$ and $|n_i - n_j| \\le 1$.\n\nGeometrically, this can be realized in $\\mathbb{R}^3$ by considering four tight bundles of $n_i$ unit vectors each, around the vectors $(1, 0, 0)$, $(-1, 2, 0)$, $(-1, -1, 3)$, and $(-1, -1, -1)$, respectively. Thus, the required maximum is $\\left\\lfloor \\frac{3n^2}{8} \\right\\rfloor$.\n\nAlternatively, since no five vertices of the complement $\\bar{G}$ of $G$ are independent, the independence number $\\alpha(\\bar{G}) \\le 4$. By the Caro-Wei theorem:\n$$\n4 \\ge \\alpha(\\bar{G}) \\ge \\sum_{i=1}^{n} \\frac{1}{1 + \\deg i} \\ge \\frac{n}{1 + \\frac{1}{n} \\sum_{i=1}^{n} \\deg i} = \\frac{n^2}{n + 2|E(\\bar{G})|}.\n$$\nThus, $|E(\\bar{G})| \\ge \\frac{n(n-4)}{8}$, so $|E(G)| = \\frac{n(n-1)}{2} - |E(\\bar{G})| \\le \\frac{3n^2}{8}$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 13316,
    "subject": "Mathematics (Olympiad)",
    "question": "Given two moving points $A(x_1, y_1)$ and $B(x_2, y_2)$ on the parabola $y^2 = 6x$ with $x_1 + x_2 = 4$ and $x_1 \\neq x_2$, the perpendicular bisector of segment $AB$ intersects the $x$-axis at point $C$. Find the maximum area of $\\triangle ABC$.",
    "options": [],
    "answer": "See solution",
    "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\n$$\nk_{AB} = \\frac{y_2 - y_1}{x_2 - x_1}.\n$$\n\nThe equation of the perpendicular bisector of $AB$ is\n\n$$\ny - y_0 = -\\frac{y_0}{3}(x - 2).\n$$\n\nIt is easy to find that one solution is $x = 5$, $y = 0$. Therefore, the intersection $C$ is a fixed point with coordinates $(5, 0)$.\n\nThe equation of line $AB$ is $y - y_0 = \\frac{3}{y_0}(x - 2)$, or\n\n$$\nx = \\frac{y_0}{3}(y - y_0) + 2.\n$$\n\nSubstituting into $y^2 = 6x$, we get $y^2 = 2y_0(y - y_0) + 12$, or\n\n$$\ny^2 - 2y_0 y + 2y_0^2 - 12 = 0.\n$$\n\nAs $y_1$ and $y_2$ are two real roots and $y_1 \\neq y_2$, we have\n\n$$\n\\Delta = 4y_0^2 - 4(2y_0^2 - 12) = -4y_0^2 + 48 > 0.\n$$\n\nTherefore, $-2\\sqrt{3} < y_0 < 2\\sqrt{3}$.\n\n![](images/Mathematical_Olympiad_in_China_2011-2014_p33_data_2c7e8013ef.png)\n\nThe distance from point $C(5, 0)$ to segment $AB$ is\n\n$$\nh = |CM| = \\sqrt{(5-2)^2 + (0-y_0)^2} = \\sqrt{9 + y_0^2}.\n$$\n\nTherefore,\n\n$$\n\\begin{align*}\nS_{\\triangle ABC} &= \\frac{1}{2} |AB| \\cdot h = \\frac{1}{3} \\sqrt{(9 + y_0^2)(12 - y_0^2)} \\cdot \\sqrt{9 + y_0^2} \\\\\n&= \\frac{1}{3} \\sqrt{\\frac{1}{2}(9 + y_0^2)(24 - 2y_0^2)(9 + y_0^2)} \\\\\n&\\leq \\frac{1}{3} \\sqrt{\\frac{1}{2} \\left( \\frac{9 + y_0^2 + 24 - 2y_0^2 + 9 + y_0^2}{3} \\right)^3} \\\\\n&= \\frac{14}{3}\\sqrt{7}.\n\\end{align*}\n$$\n\nEquality holds if and only if $9 + y_0^2 = 24 - 2y_0^2$, i.e.,\n\n$$\ny_0 = \\pm\\sqrt{5}.\n$$\n\nThen we get\n\n$$\nA\\left(\\frac{6 + \\sqrt{35}}{3}, \\sqrt{5} + \\sqrt{7}\\right), \\quad B\\left(\\frac{6 - \\sqrt{35}}{3}, \\sqrt{5} - \\sqrt{7}\\right)\n$$\n\nor\n\n$$\nA\\left(\\frac{6 + \\sqrt{35}}{3}, - (\\sqrt{5} + \\sqrt{7})\\right), \\quad B\\left(\\frac{6 - \\sqrt{35}}{3}, -\\sqrt{5} + \\sqrt{7}\\right).\n$$\n\nConsequently, the maximum area of $\\triangle ABC$ is $\\frac{14}{3}\\sqrt{7}$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 13317,
    "subject": "Mathematics (Olympiad)",
    "question": "Prove that for each integer $n$, $n \\ge 3$, the following $2n$-digit number\n\n$$\n\\underbrace{1\\dots12}_{n-1} \\underbrace{8\\dots8}_{n-2} 96\n$$\n\nis a perfect square.",
    "options": [],
    "answer": "See solution",
    "solution": "The number under consideration can be expressed as follows:\n\n$$\n\\begin{aligned}\n& (10^{2n-1} + 10^{2n-2} + \\dots + 10^{n+1}) + 2 \\cdot 10^n + 8 \\cdot (10^{n-1} + 10^{n-2} + \\dots + 10^2) + 96 \\\\\n&= 10^{n+1} \\cdot \\frac{10^{n-1}-1}{9} + 2 \\cdot 10^n + 8 \\cdot 10^2 \\cdot \\frac{10^{n-2}-1}{9} + 96 \\\\\n&= \\frac{10^{2n} - 10^{n+1} + 18 \\cdot 10^n + 800 \\cdot 10^{n-2} - 800 + 9 \\cdot 96}{9} \\\\\n&= \\frac{10^{2n} + 16 \\cdot 10^n + 64}{9} = \\left( \\frac{10^n + 8}{3} \\right)^2.\n\\end{aligned}\n$$\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.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 13318,
    "subject": "Mathematics (Olympiad)",
    "question": "Marjorie is the drum major of the world's largest marching band, with more than one million members. She would like the band members to stand in a square formation. To this end, she determines the smallest integer $n$ such that the band would fit in an $n \\times n$ square and lets the members form rows of $n$ people. However, she is dissatisfied with the result, since some empty positions remain. Therefore, she tells the entire first row to go home and repeats the process with the remaining members. Her aim is to continue it until the band forms a perfect square, but as it happens, she does not succeed until the last members are sent home. Determine the smallest possible number of members in this marching band.",
    "options": [],
    "answer": "See solution",
    "solution": "The answer is $1000977$.\n\nLet $M$ be the number of members of the marching band. We prove by induction that Marjorie's approach always yields a perfect square at some point, unless $M$ is of the form $M = (2^a + b)^2 + 2b + 1$ ($a, b$ nonnegative integers, $0 \\leq b < 2^a$) or $M = (2^a + b)^2 + 2^a + 3b + 2$ ($a, b$ nonnegative integers, $0 \\leq b < 2^a - 1$), in which case all members are eventually sent home.\n\nThis is true for $M = 1$ (which is not of either form), since the single member forms a $1 \\times 1$ square, and for $M = 2$ (which is of the form $M = (2^a + b)^2 + 2b + 1$ with $a = b = 0$), in which case the two members form an incomplete $2 \\times 2$ square and are sent home.\n\nFor the induction step, suppose that $M > 2$ and take $n$ to be the unique positive integer for which $(n-1)^2 < M \\leq n^2$. Then the $M$ members will stand in an $n \\times n$ square, and if $M \\neq n^2$, then $n$ members are sent home. We write $n - 1 = 2^a + b$, where $2^a$ is the greatest power of $2$ less than or equal to $n-1$, and $0 \\leq b < 2^a$. We claim that the process reaches a perfect square if and only if neither $M = (2^a + b)^2 + 2b + 1$ nor $M = (2^a + b)^2 + 2^a + 3b + 2$ (the latter only for $b < 2^a - 1$).\n\nSuppose first that $M \\leq n^2 - n + 1$, so that $M - n \\leq (n-1)^2 = (2^a + b)^2$. By the induction hypothesis, the process never reaches a perfect square if and only if $M - n = (2^a + b - 1)^2 + 2b - 1$ or $M - n = (2^a + b - 1)^2 + 2^a + 3b - 1$. The former equation is equivalent to $M = (2^a + b - 1)^2 + 2^a + 3b$. However, this is impossible since it gives\n\n$$\nM = (2^a + b - 1)^2 + 2^a + 3b = (2^a + b)^2 - (2^a - b - 1) \\leq (n - 1)^2.\n$$\n\nThe latter equation yields\n\n$$\nM = (2^a + b - 1)^2 + 2^{a+1} + 4b = (2^a + b)^2 + 2b + 1,\n$$\nwhich is what we wanted to prove.\n\nLikewise, if $M > n^2 - n + 1$, then $M - n > (n-1)^2$. So by the induction hypothesis, the process never reaches a perfect square if and only if either $M - n = (2^a + b)^2 + 2b + 1$ or $M - n = (2^a + b)^2 + 2^a + 3b + 2$. In the former case, we get\n\n$$\nM = (2^a + b)^2 + 2^a + 3b + 2,\n$$\nwhich is exactly the desired statement. Note, however, that $M \\neq n^2$ (otherwise, the band forms a perfect square immediately) requires $b < 2^a - 1$. In the latter case, we obtain\n\n$$\nM = (2^a + b)^2 + 2^{a+1} + 4b + 3 = (2^a + b + 1)^2 + 2(b + 1) > n^2,\n$$\nwhich is impossible.\n\nThis completes the induction. Now note that $1000000 = 1000^2$ and $1000 = 2^9 + 488$, so the smallest number greater than $1000000$ that is of the form $(2^a + b)^2 + 2b + 1$ or $(2^a + b)^2 + 2^a + 3b + 2$ is $(2^9 + 488)^2 + 2 \\cdot 488 + 1 = 1000977$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 13319,
    "subject": "Mathematics (Olympiad)",
    "question": "Let Dima write down the numbers $d_1 > d_2 > \\dots > d_{2022}$ and Vlad write down the numbers $v_1 > v_2 > \\dots > v_{2022}$. Is it possible that 2020 of the numbers written by Vlad are less than any number written by Dima?\n\n% IMAGE: <table><thead><tr><th>v<sub>1</sub></th><th>d<sub>1</sub></th><th>&infin;</th><th>d<sub>4</sub></th><th>d<sub>5</sub></th><th>d<sub>6</sub></th><th>...</th><th>d<sub>2022</sub></th></tr></thead><tbody><tr><td>d<sub>2</sub></td><td>v<sub>2</sub></td><td>d<sub>3</sub></td><td>&infin;</td><td>y</td><td>y</td><td>...</td><td>y</td></tr><tr><td>x</td><td></td><td>v<sub>3</sub></td><td></td><td></td><td></td><td></td><td></td></tr><tr><td>x</td><td></td><td></td><td>v<sub>4</sub></td><td></td><td></td><td></td><td></td></tr><tr><td>x</td><td></td><td></td><td></td><td>v<sub>5</sub></td><td></td><td></td><td></td></tr><tr><td>x</td><td></td><td></td><td></td><td></td><td>v<sub>6</sub></td><td></td><td></td></tr><tr><td>&vellip;</td><td></td><td></td><td></td><td></td><td></td><td>&vellip;</td><td></td></tr><tr><td>x</td><td></td><td></td><td></td><td></td><td>v<sub>6</sub></td><td></td><td>v<sub>2022</sub></td></tr></tbody></table>\n\nArrange the numbers from $1$ to $2022^2$ in the table so that\n\n$$\n\\infty > v_1 > v_2 > y > d_1 > \\dots > d_{2022} > x > v_3 > \\dots > v_{2020}\n$$\n\n(where $\\infty$, $y$, and $x$ refer to the numbers in the cells with those labels), and the numbers in empty cells are less than any of the numbers in marked cells.",
    "options": [],
    "answer": "See solution",
    "solution": "Suppose the answer is 2022 or 2021. Then $d_{2022} > v_2$. Each of the 2021 rows not containing $v_1$ contains at least 2021 numbers not exceeding $v_2$: the number written by Vlad and all smaller numbers. Each column contains at least 2 numbers not less than $d_{2022}$: the number written by Dima and all greater numbers. Therefore, the total number of numbers in the table is at least\n\n$$\n2021 \\cdot 2021 + 2022 \\cdot 2 = 2022 \\cdot 2022 + 1,\n$$\n\nwhich exceeds the total number of numbers in the table—a contradiction.\n\nNow, let's show that it is possible for 2020 of Vlad's numbers to be less than any of Dima's numbers. Fill in the table as shown above with numbers from $1$ to $2022^2$ so that\n\n$$\n\\infty > v_1 > v_2 > y > d_1 > \\dots > d_{2022} > x > v_3 > \\dots > v_{2020}\n$$\n\n(where $\\infty$, $y$, and $x$ refer to the numbers in the cells with those labels), and the numbers in empty cells are less than any of the numbers in marked cells.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 13320,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $z_1, z_2, z_3, z_4, z_5, z_6$ be six pairwise different complex numbers whose images $A_1, A_2, A_3, A_4, A_5, A_6$ are consecutive points on the circle with center $O(0,0)$ and radius $r > 0$. If $w$ is a solution of the equation $z^2 + z + 1 = 0$ and\n\n$$\nz_1 w^2 + z_3 w + z_5 = 0 \\quad (I)\n$$\n\n$$\nz_2 w^2 + z_4 w + z_6 = 0 \\quad (II)\n$$\n\nprove that:\n\n(a) The triangle $A_1A_3A_5$ is equilateral.\n\n$$\n|z_1 - z_2| + |z_2 - z_3| + |z_3 - z_4| + |z_4 - z_5| + |z_5 - z_6| + |z_6 - z_1| = 3|z_1 - z_4| = 3|z_2 - z_5| = 3|z_3 - z_6|.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "(a) Since $w$ is a root of the equation $z^2 + z + 1 = 0$, we have $w^2 + w + 1 = 0$. Multiplying both sides by $w$:\n\n$$\nw^3 + w^2 + w = 0 \\implies w^3 + (w^2 + w + 1) = 1 \\implies w^3 = 1.\n$$\n\nThus $|w| = 1$. Substituting $w^2 = -w - 1$ into relation (I):\n\n$$\nz_1(-1-w) + z_3w + z_5 = 0 \\implies -z_1 - z_1w + z_3w + z_5 = 0 \\implies (z_3 - z_1)w = z_1 - z_5.\n$$\n\nTherefore,\n\n$$\n|(z_3 - z_1)w| = |z_1 - z_5| \\implies |z_3 - z_1||w| = |z_1 - z_5| \\implies \\boxed{|z_3 - z_1| = |z_1 - z_5|} \\quad (A).\n$$\n\nNow, substituting $w = -w^2 - 1$ into (I):\n\n$$\nz_1 w^2 + z_3(-w^2 - 1) + z_5 = 0 \\implies z_1 w^2 - z_3 w^2 - z_3 + z_5 = 0 \\implies (z_1 - z_3)w^2 = z_5 - z_3.\n$$\n\nSo,\n\n$$\n|(z_1 - z_3)w^2| = |z_5 - z_3| \\implies |z_1 - z_3||w|^2 = |z_5 - z_3| \\implies \\boxed{|z_3 - z_1| = |z_5 - z_3|} \\quad (B).\n$$\n\nFrom (A) and (B):\n\n$$\n|z_1 - z_3| = |z_3 - z_5| = |z_5 - z_1|,\n$$\n\nso the triangle $A_1A_3A_5$ is equilateral.\n\n(b) Similarly, using relation (II), the triangle $A_2A_4A_6$ is equilateral. From a known proposition of Euclidean geometry, $A_1A_2 + A_1A_6 = A_1A_4$, so:\n\n$$\n|z_1 - z_2| + |z_6 - z_1| = |z_1 - z_4|. \\qquad (1)\n$$\n\nSimilarly, $A_3A_2 + A_3A_4 = A_3A_6$ gives:\n\n$$\n|z_2 - z_3| + |z_3 - z_4| = |z_3 - z_6|. \\qquad (2)\n$$\n\nAnd $A_5A_4 + A_5A_6 = A_5A_2$ gives:\n\n$$\n|z_4 - z_5| + |z_5 - z_6| = |z_2 - z_5|. \\qquad (3)\n$$\n\n![](images/Hellenic_Mathematical_Competitions_2009_booklet_p11_data_e0a0490d43.png)\n\nSumming (1), (2), and (3), and using\n\n$$\n|z_1 - z_4| = |z_3 - z_6| = |z_2 - z_5|,\n$$\n\nwe find:\n\n$$\n|z_1 - z_2| + |z_2 - z_3| + |z_3 - z_4| + |z_4 - z_5| + |z_5 - z_6| + |z_6 - z_1| = 3|z_1 - z_4| = 3|z_2 - z_5| = 3|z_3 - z_6|.\n$$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 13321,
    "subject": "Mathematics (Olympiad)",
    "question": "All points in the plane with integer coordinates are colored in three colors. Find the least positive integer $S$ such that, for any such coloring, there exists a triangle of area $S$ whose vertices all have the same color.",
    "options": [],
    "answer": "See solution",
    "solution": "Consider two colorings:\n\n1. Color $(x, y)$ with color $i$ if $x \\equiv i \\pmod{2}$.\n2. Color $(x, y)$ with color $i$ if $x \\equiv i \\pmod{3}$.\n\nIf $S$ exists, then $2S$ is an integer. Coloring (1) implies $S \\geq 1$, and coloring (2) shows $S \\geq \\frac{3}{2}$. Therefore, $S \\geq 3$.\n\nWe now prove that for any coloring, there exists a triangle of area $3$ with all vertices the same color. For some $d \\in \\{1, 2, 3\\}$, there exist points $A = (x, y)$ and $B = (x+d, y)$ of the same color. For example, consider $(0,0), (1,0), (2,0), (3,0)$. If $m$ is the line $AB$ and $l$ is parallel to $m$ at distance $\\frac{6}{d}$, then the line $l$ (parallel to the $x$-axis) either contains points of only two colors or yields a triangle of area $3$.\n\nDefine that a color $c$ permits distance $a$ on line $l$ if there are two points of color $c$ at distance $a$ apart. If some $a \\in \\{1,2,3,6\\}$ is permitted by both colors on $l$, then the line $p$ parallel to $l$ at distance $\\frac{6}{a}$ has all its points in the same color. If not, then one color on $l$ permits all distances in $\\{2,3,6\\}$.\n\nSuppose color $c_1$ does not permit distance $1$, and there is a point $P_0$ of color $c_1$ on $l$. Its neighbors $P_{-1}$ and $P_1$ are colored $c_2$, so $c_2$ permits distance $2$. If $c_1$ does not permit distance $2$, then $P_{-2}$ and $P_2$ are colored $c_2$, and so on. Thus, $c_2$ permits all distances in $\\{2,3,6\\}$.\n\nIn both cases, there exists a line $p$ parallel to the $x$-axis and a color $c$ such that $p$ contains points of color $c$ at all distances in $\\{2,3,6\\}$.\n\nConsider lines $u_1, u_2, u_3$ parallel to $p$, each at distance $i$ above $p$ for $i = 1,2,3$. All such lines contain points in only two colors. Now, consider the nine intersection points of $u_1, u_2, u_3$ with $x = 0, x = 3, x = 6$. There must exist a triangle of area $3$ with all vertices the same color.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 13322,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $ABCD$ be a parallelogram and a circle $k$ passes through $A$ and $C$ and meets rays $AB$ and $AD$ at $E$ and $F$, respectively. If $BD$, $EF$, and the tangent at $C$ concur, show that $AC$ is a diameter of $k$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let the tangent to $k$ at point $C$ intersect the rays $AB$ and $AD$ at the points $M$ and $N$, respectively, and let 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$$\n\\frac{AD}{ND} \\cdot \\frac{NP}{MP} \\cdot \\frac{MB}{AB} = 1 \\quad \\text{and} \\quad \\frac{AF}{NF} \\cdot \\frac{NP}{MP} \\cdot \\frac{ME}{AE} = 1.\n$$\n\nHence,\n\n$$\n\\frac{AD}{ND} \\cdot \\frac{MB}{AB} = \\frac{AF}{NF} \\cdot \\frac{ME}{AE}.\n$$\n\nSince $ABCD$ is a parallelogram, $\\frac{AD}{ND} = \\frac{MC}{NC} = \\frac{MB}{AB}$. From the tangent and secant property, $MC^2 = ME \\cdot MA$ and $NC^2 = NF \\cdot NA$. From the above, it follows that\n\n$$\n\\frac{MC^2}{NC^2} = \\frac{AF}{NF} \\cdot \\frac{ME}{AE} \\implies \\frac{ME \\cdot MA}{NF \\cdot NA} = \\frac{AF}{NF} \\cdot \\frac{ME}{AE}.\n$$\n\nTherefore, $AM \\cdot AE = AN \\cdot AF$, i.e., the quadrilateral $EFNM$ is cyclic. Then $\\angle AEF = \\angle ANM$, whence $\\overline{AEC} = \\overline{AFC}$. It follows that $AC$ is a diameter of $k$. $\\Box$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 13323,
    "subject": "Mathematics (Olympiad)",
    "question": "Given an integer $n \\geq 3$, 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 $a_k \\leq \\frac{a_{k-1} + a_{k+1}}{2}$ for all $k = 2, \\dots, n-1$.",
    "options": [],
    "answer": "See 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 $X$.\n\nFirstly, it is easy to verify that $S \\cap T = \\emptyset$ and $S \\cup T = X$.\n\nNext, suppose for contradiction that $S$ contains elements $a_1, a_2, \\dots, a_n$ with $a_1 < a_2 < \\dots < a_n$ and $a_k \\leq \\frac{a_{k-1} + a_{k+1}}{2}$ for $k = 2, \\dots, n-1$. Then we have\n$$\na_k - a_{k-1} \\leq a_{k+1} - a_k, \\quad k = 2, \\dots, n-1. \\qquad \\textcircled{1}\n$$\nAssume that $a_1 \\in S_i$, we have $i < n-1$, since $|S_{n-1}| < n$. There exist at least $n - |S_i| = n - i$ elements in $\\{a_1, a_2, \\dots, a_n\\} \\cap (S_{i+1} \\cup \\dots \\cup S_{n-1})$. By the pigeonhole principle, 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 exist $a_k$ such that $a_k, a_{k+1} \\in S_j$ and $a_{k-1} \\in S_1 \\cup \\dots \\cup S_{j-1}$.\n\nThen\n$$\na_{k+1} - a_k \\leq |S_j| - 1 = j - 1, \\quad a_k - a_{k-1} \\geq |T_{j-1}| + 1 = j.\n$$\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": 13324,
    "subject": "Mathematics (Olympiad)",
    "question": "Prove that\n$$\n\\sum_{\\text{cyc}} (x+y)\\sqrt{(z+x)(z+y)} \\geq 4(xy+yz+zx)\n$$\nfor all positive real numbers $x, y, z$.",
    "options": [],
    "answer": "See solution",
    "solution": "We will obtain the inequality by adding the inequalities\n$$\n(x+y)\\sqrt{(z+x)(z+y)} \\geq 2xy + yz + zx\n$$\nfor cyclic permutations of $x, y, z$.\n\nSquaring both sides of this inequality, we obtain\n$$\n(x+y)^2(z+x)(z+y) \\geq 4x^2y^2 + y^2z^2 + z^2x^2 + 4xyz^2 + 4x^2yz + 2xyz^2\n$$\nwhich is equivalent to\n$$\nx^3y + xy^3 + z(x^3 + y^3) \\geq 2x^2y^2 + xyz z(x+y)\n$$\nwhich can be rearranged to\n$$\n(xy + yz + zx)(x - y)^2 \\geq 0\n$$\nwhich is clearly true.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 13325,
    "subject": "Mathematics (Olympiad)",
    "question": "Supongamos que el excírculo del triángulo $ABC$ opuesto al vértice $A$ es tangente al lado $BC$ en el punto $A_1$. Análogamente, se definen los puntos $B_1$ en $CA$ y $C_1$ en $AB$, utilizando los excírculos opuestos a $B$ y $C$ respectivamente. Supongamos que el circuncentro del triángulo $A_1B_1C_1$ pertenece a la circunferencia que pasa por los vértices $A$, $B$ y $C$. Demostrar que el triángulo $ABC$ es rectángulo.",
    "options": [],
    "answer": "See solution",
    "solution": "Sea $\\Gamma$ la circunferencia circunscrita a $ABC$. Usamos en la resolución el siguiente **Lema**. Para todo triángulo $ABC$, y con las definiciones de puntos dadas en el enunciado, las mediatrices de $BC$ y $B_1C_1$ se cortan sobre $\\Gamma$, más concretamente en el punto medio $M$ del arco $BC$ que contiene a $A$.\n\n**Demostración.** Si $ABC$ es isósceles en $A$, el triángulo es simétrico respecto de la mediatriz de $BC$, que coincide por lo tanto con la mediatriz de $B_1C_1$, pasando claramente por simetría, por el punto medio $M$ del arco $BC$ que contiene a $A$. En caso contrario, supongamos sin pérdida de generalidad, al poder intercambiar $BC$ sin afectar al Lema, que $M$ está en el arco $AC$ que no contiene a $B$. Se tiene entonces que $\\angle MAB_1 = \\angle MAC = \\angle MBC = 90^\\circ - \\frac{A}{2}$, por ser $MBC$ isósceles en $M$ y $\\angle BMC = \\angle BAC$. Como $\\angle MAB = \\angle MAC + \\angle CAB$, se tiene también que $\\angle MAC_1 = \\angle MAB = 90^\\circ + \\frac{A}{2}$. Luego por el teorema del coseno,\n\n$$\nMC_1^2 = AM^2 + AC_1^2 - 2AM \\cdot AC_1 \\cos \\angle MAC_1 = AM^2 + AC_1^2 + 2AM \\cdot AC_1 \\sin \\frac{A}{2},\n$$\n\ny de forma análoga\n\n$$\nMB_1^2 = AM^2 + AB_1^2 - 2AM \\cdot AB_1 \\cos \\angle MAB_1 = AM^2 + AB_1^2 - 2AM \\cdot AB_1 \\sin \\frac{A}{2}.\n$$\n\nSe tiene entonces que el Lema es equivalente a\n\n$$\n2AM \\sin \\frac{A}{2} (AB_1 + AC_1) = AB_1^2 - AC_1^2 = (AB_1 + AC_1)(AB_1 - AC_1),\n$$\n\nes decir, $2AM \\sin \\frac{A}{2} = AB_1 - AC_1$. Pero es conocido que $AB_1 = s - c$, $AC_1 = s - b$, donde $s$ es el semiperímetro de $ABC$, luego utilizando el teorema del seno para $AM$, $b$, $c$, y al ser $\\sin B - \\sin C = 2 \\sin \\frac{A}{2} \\sin \\frac{B-C}{2}$, nos basta con comprobar que $\\sin \\angle ABM = \\sin \\frac{B-C}{2}$, claramente cierto al ser $\\angle ABM = \\angle ABC - \\angle MBC = B - (90^\\circ - \\frac{A}{2}) = \\frac{B-C}{2}$. Queda pues demostrado el Lema.\n\nSupongamos que el circuncentro $O_1$ de $A_1B_1C_1$ está sobre $\\Gamma$, más concretamente (y sin pérdida de generalidad al poder rotar cíclicamente los vértices sin alterar el problema) en el arco $BC$ que contiene a $A$. Claramente, al estar $O_1$ en la mediatriz de $B_1C_1$ y en $\\Gamma$ sobre el arco $BC$ que contiene a $A$, también está sobre la mediatriz de $BC$, es decir, $BO_1 = CO_1$, siendo $BCO_1$ isósceles en $O_1$ con $\\angle BO_1C = A$. Además, como es conocido que $BA_1 = s - c$, $CA_1 = s - b$, asumiendo sin pérdida de generalidad que $b \\ge c$, tenemos que $NA_1 = \\frac{b-c}{2}$, donde $N$ es el punto medio del lado $BC$, y\n\n$$\nO_1A_1^2 = NO_1^2 + NA_1^2 = BO_1^2 + NA_1^2 - BN^2 = BO_1^2 - BA_1 \\cdot CA_1.\n$$\n\nAl mismo tiempo, usando resultados parciales del Lema, tenemos que $\\angle C_1BO_1 = \\frac{B-C}{2}$, con lo que\n\n$$\nO_1B_1^2 = O_1C_1^2 = BC_1^2 + BO_1^2 - 2BC_1 \\cdot BO_1 \\cos C_1BO_1,\n$$\n\ncon lo que al ser $O_1$ circuncentro de $A_1B_1C_1$, se cumple que\n\n$$\n2BO_1 \\cos C_1BO_1 - BC_1 = \\frac{BA_1 \\cdot CA_1}{BC_1} = \\frac{(s-c)(s-b)}{s-a}.\n$$\n\nAhora bien, $\\angle C_1BO_1 = \\frac{B-C}{2}$ como sabemos, por ser resultado parcial del Lema, y $BO_1 = 2R \\sin \\angle BCO_1 = 2R \\cos \\frac{A}{2}$, donde $R$ es el circunradio de $ABC$ y hemos usado que $BO_1C$ es isósceles en $O_1$ con $\\angle BO_1C = A$. Concluimos entonces que\n\n$$\n2BO_1 \\cos C_1BO_1 = 4R \\cos \\frac{A}{2} \\cos \\frac{B-C}{2} = 2R(\\sin B + \\sin C) = b + c.\n$$\n\nAhora bien, como $BC_1 = s-a$, se tiene que $2BO_1 \\cos C_1BO_1 - BC_1 = b+c-s+a = s$, luego si $O_1$ está sobre el arco $BC$ que contiene a $A$ de $\\Gamma$, se sigue que $s(s-a) = (s-b)(s-c)$, o equivalentemente, $0 = (b+c-a)s - bc = \\frac{b^2+c^2-a^2}{2}$, es decir, $ABC$ es rectángulo en $A$.\n\nConcluimos entonces que, si el circuncentro de $A_1B_1C_1$ está sobre la circunferencia circunscrita a $\\Gamma$, entonces $ABC$ es rectángulo; además, si $ABC$ es rectángulo en $A$, entonces $O_1$ es el punto medio del arco $BC$ que contiene a $A$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 13326,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all real solutions $(x, y, z)$ to the system of equations:\n\n$$\nxy + 1 = 2z\n$$\n\n$$\nyz + 1 = 2x\n$$\n\n$$\nzx + 1 = 2y.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "Subtracting equation (2) from equation (1) yields\n\n$$\ny(x - z) = -2(x - z) \\implies (x - z)(y + 2) = 0.\n$$\n\nSimilarly, subtracting equation (3) from equation (2) yields\n\n$$\nz(y - x) = -2(y - x) \\implies (y - x)(z + 2) = 0.\n$$\n\nThis yields four cases which we now analyse.\n\n**Case 1** $x = z$ and $y = x$\n\nThus $x = y = z$. Putting this into equation (1) yields $x^2 + 1 = 2x$, which is equivalent to $(x - 1)^2 = 0$. Hence $x = y = z = 1$.\n\n**Case 2** $x = z$ and $z = -2$\n\nThus $x = z = -2$. Putting this into (3) immediately yields $y = \\frac{5}{2}$.\n\n**Case 3** $y = -2$ and $y = x$\n\nThus $x = y = -2$. Putting this into (1) immediately yields $z = \\frac{5}{2}$.\n\n**Case 4** $y = -2$ and $z = -2$\n\nPutting this into equation (2) immediately yields $x = \\frac{5}{2}$.\n\nSo the only possible solutions are $(x, y, z) = (1, 1, 1), \\left(\\frac{5}{2}, -2, -2\\right), \\left(-2, \\frac{5}{2}, -2\\right)$, and $\\left(-2, -2, \\frac{5}{2}\\right)$. It is readily verified that these satisfy the given equations. $\\square$",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 13327,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $a$, $b$, $c$, and $d$ be real numbers such that $a^2 + b^2 + c^2 + d^2 = 4$. Prove the inequality\n\n$$\n(a+2)(b+2) \\geq cd\n$$\n\nand give four numbers $a$, $b$, $c$, and $d$ such that equality holds.",
    "options": [],
    "answer": "See solution",
    "solution": "The claimed inequality is equivalent to $2ab + 4a + 4b + 8 \\geq 2cd$, which can be written as\n\n$$\n2ab + 4a + 4b + a^2 + b^2 + c^2 + d^2 + 4 \\geq 2cd\n$$\n\nusing the condition $a^2 + b^2 + c^2 + d^2 = 4$. By the identity\n\n$$\na^2 + b^2 + 2ab + 4a + 4b + 4 = (a + b + 2)^2\n$$\n\nwe arrive at the equivalent and obvious inequality\n\n$$(a+b+2)^2 + (c-d)^2 \\geq 0$$\n\nEquality occurs when $a + b = -2$ and $c = d$, together with $a^2 + b^2 + c^2 + d^2 = 4$.\n\nFor instance, when $a = b = c = d = -1$.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 13328,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all integers $y$ such that there exists a real number $x$ satisfying both $y = \\left\\lfloor \\sqrt{x} \\right\\rfloor$ and $y = \\left\\lfloor \\dfrac{x+23}{8} \\right\\rfloor$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $y$ be such a number. Then $\\sqrt{x} \\geq |\\sqrt{x}| = y$. Since $\\sqrt{x} \\geq 0$, we have $y = [\\sqrt{x}] \\geq 0$, so we may square the inequality to get $x \\geq y^2$. Also, $\\dfrac{x+23}{8} < [\\dfrac{x+23}{8}]+1 = y+1$, or $x < 8y-15$. This implies $y^2 < 8y-15$, or $(y-3)(y-5) < 0$, which means that $3 < y < 5$. But $y$ is an integer, so $y=4$. In this case there indeed exists a real number $x$ which satisfies the condition, for example $x=16$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 13329,
    "subject": "Mathematics (Olympiad)",
    "question": "Given an odd integer $n \\ge 3$ and a regular $n$-gon. Let $V$ be the set of its vertices, and $\\mathcal{P}$ denote the collection of all regular polygons whose vertices belong to $V$.\n\nFor example, when $n = 15$, $\\mathcal{P}$ contains 1 regular 15-gon, 3 regular pentagons, and 5 regular triangles.\n\nTwo players, Alice and Bob, play the following game. Initially, all points in $V$ are uncolored. Starting with Alice, they take turns coloring an uncolored point in $V$—Alice uses red and Bob uses blue. The game ends when all points in $V$ are colored. A polygon in $\\mathcal{P}$ is called \"good\" if it has more red vertices than blue vertices.\n\nFind the largest integer $k$ such that no matter how Bob colors the points, Alice can guarantee at least $k$ good polygons in $\\mathcal{P}$ after the game ends.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $n = p_1^{\\alpha_1} \\cdots p_r^{\\alpha_r}$ be the prime factorization of $n$ ($p_i$ are odd primes). Let $\\tau(n) = \\prod_{i=1}^r (\\alpha_i+1)$ be the number of positive divisors of $n$, and $\\sigma(n) = \\prod_{i=1}^r (1+p_i+\\cdots+p_i^{\\alpha_i})$ be the sum of positive divisors of $n$.\n\nThe maximal integer $k$ is\n$$\nk = \\frac{\\sigma(n) + \\tau(n) - n - 1}{2}.\n$$\n(Note that from this formula, $\\sigma(n)$ and $\\tau(n)$ have the same parity.)\n\nLabel the vertices of $V$ as $A_0, A_1, \\ldots, A_{n-1}$, 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:\n1. $A_0A_dA_{2d}\\ldots A_{n-d}$\n2. $A_bA_{b+d}\\ldots A_{n-d+b}$ where $b \\in \\{1, \\ldots, d-1\\}$.\n\nThe regular $\\frac{n}{d}$-gon $A_0A_d\\ldots A_{n-d}$ is always good because $A_0$ is red, while for $s = 1, \\ldots, \\frac{n}{d}-1$, $A_{sd}$ and $A_{-sd}$ have different colors.\n\nFor $b \\in \\{1, \\ldots, d-1\\}$, the polygons $A_bA_{d+b}\\ldots A_{n-d+b}$ and $A_{d-b}A_{2d-b}\\ldots A_{n-b}$ 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": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 13330,
    "subject": "Mathematics (Olympiad)",
    "question": "Пусть даны две дроби $-\\frac{a}{600}$ и $\\frac{b}{700}$, где $a$ взаимно просто с $6$, а $b$ — с $7$. Какое наименьшее значение может иметь знаменатель суммы этих дробей после сокращения?",
    "options": [],
    "answer": "See solution",
    "solution": "Ответ: $2^3 \\cdot 3 \\cdot 7 = 168$.\n\nПусть наши дроби $-\\frac{a}{600}$ и $\\frac{b}{700}$. Тогда $a$ взаимно просто с $6$, а $b$ — с $7$. Поэтому числитель их суммы $\\frac{7a+6b}{4200}$ взаимно прост как с $6 = 2 \\cdot 3$, так и с $7$. Поскольку $4200 = 2^3 \\cdot 3 \\cdot 7 \\cdot 5^2$, это означает, что знаменатель после сокращения будет не меньше, чем $2^3 \\cdot 3 \\cdot 7 = 168$. Такой знаменатель действительно может получиться; например, $\\frac{1}{600} + \\frac{3}{700} = \\frac{1}{168}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 13331,
    "subject": "Mathematics (Olympiad)",
    "question": "It is given that the real roots of the quadratic polynomial $g(x) = x^2 - 3x + a$ are also roots of the polynomial $f(x) = x^3 - x^2 + c x + 4$. Similarly, both real roots of the quadratic polynomial $h(x) = x^2 + x + b$ are also roots of $f(x)$. What values can $f(1)$ take?",
    "options": [],
    "answer": "See solution",
    "solution": "Since $f(x)$ is a cubic polynomial, it has at most three real roots. Therefore, the quadratic polynomials $g(x)$ and $h(x)$ must share a root. Let this common root be $t$.\n\n$$\ng(t) = t^2 - 3t + a = 0 \\quad \\text{and} \\quad h(t) = t^2 + t + b = 0\n$$\nSubtracting the two equations:\n$$\n(t^2 - 3t + a) - (t^2 + t + b) = 0 \\\\\n-4t + a - b = 0 \\\\\nt = \\frac{a - b}{4}\n$$\n\nSince the real roots of $g(x)$ and $h(x)$ are also roots of $f(x)$, and $f(x)$ is cubic, the set of real roots of $g(x)$ and $h(x)$ together must have at most three elements. Thus, $g(x)$ and $h(x)$ share exactly one real root $t$, and their other roots are distinct.\n\nNow, $f(x)$ must have roots at $t$, the other root of $g(x)$, and the other root of $h(x)$. Let us denote the other root of $g(x)$ by $r$ and the other root of $h(x)$ by $s$.\n\nLet us compute $f(1)$:\n\nSuppose $t = 1$. Then, $f(1) = 1^3 - 1^2 + c \\cdot 1 + 4 = 1 - 1 + c + 4 = c + 4$.\n\nFrom the coefficient comparison in the solution, we find that $t = 1$ is possible, and in the explicit example given:\n\n$$\nf(x) = x^3 - x^2 - 4x + 4\n$$\nSo $f(1) = 1 - 1 - 4 + 4 = 0$.\n\nTherefore, the possible value for $f(1)$ is $0$.",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 13332,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $a$, $b$, $c$, and $d$ be pairwise different integers such that $1-a$, $1-b$, $1-c$, and $1-d$ are also all different. Given that $10$ is the product of these four numbers, find the value of $a + b - c - d$.",
    "options": [],
    "answer": "See solution",
    "solution": "Since $10$ is the product of four pairwise different integers, and $10$ is the product of two primes, the only way to write $10$ as a product of four integers is if two of them are $1$ and $-1$, and the other two are either $-2$ and $5$ or $2$ and $-5$.\n\nGiven $a > b > c > d$, we have $1-a < 1-b < 1-c < 1-d$. Assigning $1-b = -1$ and $1-c = 1$ gives $b = 2$ and $c = 0$.\n\nIf $1-a = -2$ and $1-d = 5$, then $a = 3$ and $d = -4$.\nIf $1-a = -5$ and $1-d = 2$, then $a = 6$ and $d = -1$.\n\nIn both cases, $a + b - c - d = 3 + 2 - 0 - (-4) = 9$ and $6 + 2 - 0 - (-1) = 9$.\n\nTherefore, the answer is $$a + b - c - d = 9.$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 13333,
    "subject": "Mathematics (Olympiad)",
    "question": "Учениците во IV одделение решавале тест по математика кој содржи 20 задачи. За секоја точно решена задача се добиваат 5 поени, а за секоја неточно решена или нерешена задача се губат по 3 поени.\n\nа) Колку задачи решил Иван, ако освоил 76 поени?\n\nб) Колку најмногу задачи треба да погреши ученик ако сака сигурно да добие петка? Најмалиот број на поени потребни за оцена 5 е решението на равенката $$8245 : x = 97$$.",
    "options": [],
    "answer": "See solution",
    "solution": "а) Нека $x$ е бројот на решени задачи. Според условите, добиваме равенка:\n$$5x - 3(20 - x) = 76$$\n$$5x - 60 + 3x = 76$$\n$$8x = 76 + 60$$\n$$8x = 136$$\n$$x = \\frac{136}{8} = 17$$\nЗначи, Иван решил 17 задачи.\n\nб) $$8245 : x = 97$$\n$$x = \\frac{8245}{97} = 85$$\nЗа секоја нерешена или неточно решена задача се губат по 3 поени. Да видиме можните случаи:\n\n1) Решени се сите 20 задачи: $20 \\times 5 = 100$\n\n2) Не е решена една задача: $19 \\times 5 - 3 = 95 - 3 = 92$\n\n3) Не се решени две задачи: $18 \\times 5 - 2 \\times 3 = 90 - 6 = 84 < 85$\n\nЗначи, за сигурна петка може да се погреши најмногу една задача.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 13334,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all triples of positive integers $(x, y, z)$ such that\n$$\n2^x - 7^y = 2^z - 1.\n$$\n\n![](images/Vietnamese_mathematical_competitions_p236_data_7f51552d68.png)",
    "options": [],
    "answer": "See solution",
    "solution": "We consider the equation $2^x - 7^y = 2^z - 1$ for positive integers $x, y, z$.\n\n**Case 1:** $y$ is odd.\n\nThe equation becomes $2^x = 7^y + 1$. Since $y$ is odd, $v_2(7^y + 1) = v_2(8) = 3$, so $x = 3$ and $y = 1$. Direct checking shows $(x, y, z) = (3, 1, 1)$ is a solution.\n\n**Case 2:** $y$ is even. Let $y = 2k$ for some $k \\in \\mathbb{Z}^+$. Then:\n$$\n2^z(2^{x-z} - 1) = 49^k - 1.\n$$\n\n- If $k$ is odd: $v_2(49^k - 1) = v_2(48) = 4$, so $z = 4$. Substituting $z = 4$ gives $2^x - 49^k = 15$. Let $x = 2t$:\n$$\n4^t - 49^k = 15 \\implies (2^t + 7^k)(2^t - 7^k) = 15.\n$$\nSince $2^t + 7^k \\leq 15$, $k = 1$. Then $4^t = 64$, so $t = 3$, $x = 6$, $y = 2$. Thus, $(x, y, z) = (6, 2, 4)$ is a solution.\n\n- If $k$ is even: Let $k = 2l$, $y = 4l$. Then $49^k \\equiv 1 \\pmod{25}$, so $25 | 2^{x-z} - 1$. Let $x-z = 2a$, $a$ even, $a = 2b$, $b = 5c$. This leads to $2^{x-z} - 1 = 1024^c - 1 \\equiv 0 \\pmod{1023}$, so $31 | 7^y - 1$. Since $\\text{ord}_{31}(7) = 15$, $15 | y$, but $4 | y$, so $12 | y$. This leads to a contradiction, so no solution in this case.\n\n**Conclusion:** The only solutions are $(x, y, z) = (3, 1, 1)$ and $(6, 2, 4)$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 13335,
    "subject": "Mathematics (Olympiad)",
    "question": "In an acute triangle $ABC$, let $h_a$, $h_b$, and $h_c$ be the lengths of the altitudes to bases $BC$, $CA$, and $AB$, respectively. Point $P$ lies inside the triangle. Prove that\n\n$$\n\\frac{PA}{h_b + h_c} + \\frac{PB}{h_c + h_a} + \\frac{PC}{h_a + h_b} \\geq 1.\n$$\n\n![](images/pamphlet0910_main_p46_data_f2601a6821.png)\n\n![](images/pamphlet0910_main_p46_data_c89127ba82.png)",
    "options": [],
    "answer": "See solution",
    "solution": "We begin with a key lemma, for which we provide two proofs.\n\n**Lemma 1.** Let $a = BC$, $b = CA$, $c = AB$, and let $p_a$, $p_b$, $p_c$ denote the distances from $P$ to sides $BC$, $CA$, $AB$, respectively. We have\n\n$$\n2a \\cdot AP \\geq (b+c)(p_b + p_c).\n$$\n\n*First proof of Lemma 1.* Notice that $\\frac{p_a}{h_a} = \\frac{[PBC]}{[ABC]}$, $\\frac{p_b}{h_b} = \\frac{[PCA]}{[BCA]}$, and $\\frac{p_c}{h_c} = \\frac{[PAB]}{[CAB]}$, so\n\n$$\n\\frac{p_a}{h_a} + \\frac{p_b}{h_b} + \\frac{p_c}{h_c} = 1.\n$$\n\nLet $E$ and $D$ be the feet of the perpendiculars from $P$ to sides $AB$ and $AC$, respectively ($PD = p_b$, $PE = p_c$). Let $M$ and $N$ lie on $AB$ and $AC$ such that $MN \\parallel BC$ and $P$ lies on $MN$. We have $MN \\cdot AP \\geq 2[AMN]$ and $2[AMN] = 2[APM] + 2[APN] = PE \\cdot AM + PD \\cdot AN$, so\n\n$$\nMN \\cdot AP \\geq PE \\cdot AM + PD \\cdot AN.\n$$\n\nBecause $\\triangle AMN \\sim \\triangle ABC$, $MN : AM : AN = BC : AB : AC = a : c : b$, so\n\n$$\na \\cdot AP \\geq b p_b + c p_c.\n$$\n\nLet $P_1$ be the reflection of $P$ across the bisector of $\\angle CAB$, and let $E_1$, $D_1$ be the feet of the perpendiculars from $P_1$ to $AB$, $AC$, respectively. By symmetry, $AE_1 = AD$, $AD_1 = AE$, $P_1E_1 = PD = p_b$, $P_1D_1 = PE = p_c$. Similarly,\n\n$$\na \\cdot AP_1 \\geq b p_c + c p_b.\n$$\n\nAdding,\n\n$$\n2a \\cdot AP \\geq (b+c)(p_b + p_c).\n$$\n\n*Second proof of Lemma 1.* Set $\\angle CAB = A$, $\\angle ABC = B$, $\\angle BCA = C$, $\\angle BAP = x$, $\\angle CAP = y$. (1) is equivalent to\n\n$$\n\\frac{2PA}{p_b + p_c} \\geq \\frac{b+c}{a}.\n$$\n\nAlso,\n\n$$\n\\frac{2PA}{p_b + p_c} = \\frac{2}{\\frac{p_b}{PA} + \\frac{p_c}{PA}} = \\frac{2}{\\sin \\alpha + \\sin \\beta}\n$$\n\nand $\\frac{b+c}{a} = \\frac{\\sin B + \\sin C}{\\sin A}$ by the law of sines, so it suffices to show\n\n$$\n2 \\sin A \\geq (\\sin \\alpha + \\sin \\beta)(\\sin B + \\sin C).\n$$\n\nBy sum-to-product,\n\n$$\n\\begin{aligned}\n(\\sin \\alpha + \\sin \\beta)(\\sin B + \\sin C) &= 4 \\sin \\frac{\\alpha + \\beta}{2} \\cos \\frac{\\alpha - \\beta}{2} \\sin \\frac{B+C}{2} \\cos \\frac{B-C}{2} \\\\\n&\\leq 4 \\sin \\frac{\\alpha + \\beta}{2} \\sin \\frac{B+C}{2} \\\\\n&= 4 \\sin \\frac{A}{2} \\cos \\frac{A}{2} \\\\\n&= 2 \\sin A.\n\\end{aligned}\n$$\n\nas needed.\n\nNow, by Lemma 1,\n\n$$\n\\begin{aligned}\n\\frac{PA}{h_b + h_c} &\\geq \\frac{(b+c)(p_b + p_c)}{2a(h_b + h_c)} = \\frac{(b+c)(p_b + p_c)}{2a\\left(\\frac{2[ABC]}{b} + \\frac{2[ABC]}{c}\\right)} = \\frac{bc(p_b + p_c)}{4a[ABC]} \\\\\n&= \\frac{c(bp_b)}{4a[ABC]} + \\frac{b(cp_c)}{4a[ABC]} = \\frac{c[PCA]}{2a[ABC]} + \\frac{b[PAB]}{2a[ABC]} = \\frac{c}{2a} \\cdot \\frac{p_b}{h_b} + \\frac{b}{2a} \\cdot \\frac{p_c}{h_c}.\n\\end{aligned}\n$$\n\nSimilarly,\n\n$$\n\\frac{PB}{h_c + h_a} \\geq \\frac{a}{2b} \\cdot \\frac{p_c}{h_c} + \\frac{c}{2b} \\cdot \\frac{p_a}{h_a}, \\quad \\frac{PC}{h_a + h_b} \\geq \\frac{b}{2c} \\cdot \\frac{p_a}{h_a} + \\frac{a}{2c} \\cdot \\frac{p_b}{h_b}.\n$$\n\nAdding,\n\n$$\n\\begin{aligned}\n\\frac{PA}{h_b + h_c} + \\frac{PB}{h_c + h_a} + \\frac{PC}{h_a + h_b} &\\geq \\frac{c}{2a} \\cdot \\frac{p_b}{h_b} + \\frac{b}{2a} \\cdot \\frac{p_c}{h_c} + \\frac{a}{2b} \\cdot \\frac{p_c}{h_c} + \\frac{c}{2b} \\cdot \\frac{p_a}{h_a} + \\frac{b}{2c} \\cdot \\frac{p_a}{h_a} + \\frac{a}{2c} \\cdot \\frac{p_b}{h_b} \\\\\n&= \\left( \\frac{c}{2b} + \\frac{b}{2c} \\right) \\frac{p_a}{h_a} + \\left( \\frac{a}{2c} + \\frac{c}{2a} \\right) \\frac{p_b}{h_b} + \\left( \\frac{b}{2a} + \\frac{a}{2b} \\right) \\frac{p_c}{h_c} \\\\\n&\\geq \\frac{p_a}{h_a} + \\frac{p_b}{h_b} + \\frac{p_c}{h_c} \\\\\n&= 1\n\\end{aligned}\n$$\n\nby the AM-GM inequality and the earlier result.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 13336,
    "subject": "Mathematics (Olympiad)",
    "question": "Given an integer $n \\ge 2$, let $a_n$, $b_n$, $c_n$ be integers such that $$(\\sqrt[3]{2} - 1)^n = a_n + b_n \\sqrt[3]{2} + c_n \\sqrt[3]{4}.$$ Show that $c_n \\equiv 1 \\pmod{3}$ if and only if $n \\equiv 2 \\pmod{3}$.",
    "options": [],
    "answer": "See solution",
    "solution": "The binomial expansion of $(\\sqrt[3]{2} - 1)^n$ yields\n\n$$\nc_n = \\sum_{k \\equiv 2 \\pmod{3}} (-1)^{n-k} \\cdot 2^{(k-2)/3} \\binom{n}{k} \\equiv (-1)^n \\sum_{k \\equiv 2 \\pmod{3}} \\binom{n}{k} \\pmod{3}.\n$$\n\nSince\n\n$$\n\\sum_{k \\equiv 2 \\pmod{3}} \\binom{n}{k} = \\frac{1}{3}\\left((1+1)^n + \\varepsilon (1+\\varepsilon)^n + \\varepsilon^2 (1+\\varepsilon^2)^n\\right) = \\frac{1}{3}\\left(2^n + 2 \\cos\\left((n+2)\\frac{\\pi}{3}\\right)\\right),\n$$\n\nwhere $1 + \\varepsilon + \\varepsilon^2 = 0$, the condition $n \\equiv 2 \\pmod{3}$ may be restated as\n\n$$\n3c_n = (-1)^n \\left(2^n + 2 \\cos\\left((n+2)\\frac{\\pi}{3}\\right)\\right) \\equiv 3 \\pmod{9}.\n$$\n\nConsideration of $n$ modulo $6$ yields $3c_n \\equiv 3 \\pmod{9}$ if $n \\equiv 2$ or $5 \\pmod{6}$, and $3c_n \\equiv 0 \\pmod{9}$ otherwise. The conclusion follows.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 13337,
    "subject": "Mathematics (Olympiad)",
    "question": "The number $1$ is written on the blackboard. A *turn* consists of wiping out the number on the board and replacing it by either double the number, or by the number one smaller. For example, we can replace $1$ by $2$ (the double) or $0$ (one smaller), and if $5$ is on the board, we can replace it by $10$ or $4$.\n\nWhat is the minimum number of turns needed in order to write the number $2021$ on the board?\n\nA) $14$  \nB) $15$  \nC) $16$  \nD) $17$  \nE) $18$",
    "options": [],
    "answer": "See solution",
    "solution": "B) $15$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 13338,
    "subject": "Mathematics (Olympiad)",
    "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 an arithmetic sequence in order, find the value of $|\\vec{a}| : |\\vec{b}| : |\\vec{a} + \\vec{b}|$.",
    "options": [],
    "answer": "See solution",
    "solution": "Denote $s = |\\vec{a}|$, $t = |\\vec{b}|$, with $s, t > 0$. The included angle between $\\vec{a}$ and $\\vec{b}$ is $\\frac{\\pi}{3}$, so:\n\n$$\n\\begin{aligned}\n|\\vec{a} + \\vec{b}|^2 &= |\\vec{a}|^2 + |\\vec{b}|^2 + 2|\\vec{a}||\\vec{b}| \\cos \\frac{\\pi}{3} \\\\\n&= s^2 + t^2 + 2st \\cdot \\frac{1}{2} \\\\\n&= s^2 + t^2 + st.\n\\end{aligned}\n$$\n\nSince $s$, $t$, $\\sqrt{s^2 + t^2 + st}$ form an arithmetic sequence in order, we have:\n\n$$\n\\sqrt{s^2 + t^2 + st} = 2t - s.\n$$\n\nSquaring both sides:\n\n$$\ns^2 + t^2 + st = (2t - s)^2 = 4t^2 - 4st + s^2\n$$\n\nSubtract $s^2$ from both sides:\n\n$$\nt^2 + st = 4t^2 - 4st\n$$\n\nBring all terms to one side:\n\n$$\n0 = 4t^2 - 4st - t^2 - st = 3t^2 - 5st\n$$\n\nSo $3t^2 = 5st$, and since $t \\neq 0$, $3t = 5s$, or $s : t = 3 : 5$.\n\nTherefore,\n\n$$\n|\\vec{a}| : |\\vec{b}| : |\\vec{a} + \\vec{b}| = 3 : 5 : 7.\n$$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 13339,
    "subject": "Mathematics (Olympiad)",
    "question": "Several representatives of competing companies that produce the game \"Overwatch\" arrived at the conference. Consider all representatives of different companies to be competitors. It is known that each participant of the conference has exactly $2018$ competitors among all other participants. What is the largest possible number of participants who took part in the conference?",
    "options": [],
    "answer": "See solution",
    "solution": "Notice that from the problem statement it follows that the same number of participants arrived from each company. Indeed, if there were two different numbers of participants from some two companies, they would have a different number of competitors.\n\nLet $m$ be the number of companies participating in the conference, each of them having $k$ members. Then, the number of competitors of each participant of the conference equals $(m-1)k$. The total number of participants in the conference is $mk$.\n\nTherefore, it should hold that $$(m-1)k = 2018,$$ or $mk = 2018 + k$.\n\nThus, the largest number of participants will be at the highest value of $k$, and this value is $k = 2018$ when $m = 2$. Thus, the maximum number of participants of the conference is $4036$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 13340,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $ABCD$ be a rectangle with $AB > BC$. Let $E$ be the point on the diagonal $AC$ such that $BE$ is perpendicular to $AC$. Let the circle through $A$ and $E$ whose centre lies on the line $AD$ meet the side $CD$ at $F$.\n\nProve that $BF$ bisects the angle $AFC$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let the circle through $A$ and $E$ whose centre lies on $AD$ meet the line $AD$ again at $H$.\n\n![](images/Australian-Scene-combined-2015_p75_data_ee5942d3fb.png)\n\nSince $FD$ is the altitude of the right-angled triangle $AFH$, we have $\\triangle AFH \\sim \\triangle ADF$.\nSince triangles $AEH$ and $ADC$ are right-angled with a common angle at $A$, we have $\\triangle AEH \\sim \\triangle ADC$.\nSince $BE$ is the altitude of the right-angled triangle $ABC$, we have $\\triangle ABC \\sim \\triangle AEB$.\nThese three pairs of similar triangles lead respectively to the three pairs of equal ratios\n\n$$\n\\frac{AF}{AD} = \\frac{AH}{AF} \\qquad \\frac{AE}{AD} = \\frac{AH}{AC} \\qquad \\frac{AB}{AE} = \\frac{AC}{AB}.\n$$\n\nPutting these together, we have\n\n$$\nAF^2 = AD \\cdot AH = AC \\cdot AE = AB^2.\n$$\n\nSo triangle $BAF$ is isosceles and we have\n\n$$\n\\angle AFB = \\angle ABF = 90^\\circ - \\angle CBF = \\angle CFB,\n$$\n\nwhere the last equality uses the angle sum in triangle $BCF$. Since $\\angle AFB = \\angle CFB$, we have proven that $BF$ bisects the angle $AFC$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 13341,
    "subject": "Mathematics (Olympiad)",
    "question": "A blackboard contains 68 pairs of nonzero integers. Suppose that for each positive integer $k$, at most one of the pairs $(k, k)$ and $(-k, -k)$ is written on the blackboard. A student erases some of the 136 integers, subject to the condition that no two erased integers may add to 0. The student then scores one point for each of the 68 pairs in which at least one integer is erased. Determine, with proof, the largest number $N$ of points that the student can guarantee to score regardless of which 68 pairs have been written on the board.",
    "options": [],
    "answer": "See solution",
    "solution": "The answer is $43$.\n\nWe first show that we can always get $43$ points. Without loss of generality, assume that the value of $x$ is positive for every pair of the form $(x, x)$ (otherwise, replace every occurrence of $x$ on the blackboard by $-x$, and every occurrence of $-x$ by $x$). Consider the ordered $n$-tuple $(a_1, a_2, \\dots, a_n)$ where $a_1, a_2, \\dots, a_n$ denote all the distinct absolute values of the integers written on the board.\n\nLet $\\phi = \\frac{\\sqrt{5}-1}{2}$, which is the positive root of $\\phi^2 + \\phi = 1$. We consider $2^n$ possible erasing strategies. Every strategy corresponds to an ordered $n$-tuple $s = (s_1, \\dots, s_n)$ with $s_i = \\phi$ or $s_i = 1-\\phi$ ($1 \\le i \\le n$). If $s_i = \\phi$, then we erase all occurrences of $a_i$ on the blackboard. If $s_i = 1-\\phi$, then we erase all occurrences of $-a_i$ on the blackboard. The weight $w(s)$ of strategy $s$ equals the product $\\prod_{i=1}^n s_i$. It is easy to see that the sum of weights of all $2^n$ strategies is equal to\n$$\n\\sum_s w(s) = \\prod_{i=1}^n [\\phi + (1-\\phi)] = 1.\n$$\n\nFor every pair $p$ on the blackboard and every strategy $s$, we define a corresponding cost coefficient $c(p, s)$: If $s$ scores a point on $p$, then $c(p, s)$ equals the weight $w(s)$. If $s$ does not score on $p$, then $c(p, s)$ equals $0$. Let $c(p)$ denote the sum of coefficients $c(p, s)$ taken over all $s$. Now consider a fixed pair $p = (x, y)$. We claim that $c(p) \\ge \\phi$, for which we consider two cases.\n\n(a) First, suppose that $x = y = a_j$. Then, every strategy that erases $a_j$ scores a point on this pair. Therefore,\n$$\nc(p) = \\phi \\prod_{i \\neq j} [\\phi + (1-\\phi)] = \\phi.\n$$\n\n(b) Now, suppose that $x \\neq y$. We have\n$$\nc(p) = \\begin{cases} \\phi^2 + \\phi(1-\\phi) + (1-\\phi)\\phi = 3\\phi - 1, & (x, y) = (a_k, a_\\ell); \\\\ \\phi(1-\\phi) + (1-\\phi)\\phi + (1-\\phi)^2 = \\phi, & (x, y) = (-a_k, -a_\\ell); \\\\ \\phi^2 + \\phi(1-\\phi) + (1-\\phi)^2 = 2 - 2\\phi, & (x, y) = (\\pm a_k, \\mp a_\\ell). \\end{cases}\n$$\nThe bound $c(p) \\ge \\phi$ follows in each case by noting that $\\phi \\approx 0.618$ satisfies $\\frac{1}{2} < \\phi < \\frac{2}{3}$.\n\nLet $C$ denote the sum of the coefficients $c(p, s)$ taken over all $p$ and $s$. Our bound then yields that\n$$\nC = \\sum_{p,s} c(p,s) = \\sum_{p} c(p) \\ge \\sum_{p} \\phi = 68 \\phi > 42.\n$$\n\nSuppose for the sake of contradiction that every strategy $s$ scores at most $42$ points. Then every $s$ contributes at most $42w(s)$ to $C$, and we get $C \\le 42 \\sum_s w(s) = 42$, which contradicts $C > 42$. Therefore, there is always some strategy which will score at least $43$ points.\n\nTo complete our proof, we now show that we cannot always get $44$ points. Consider the blackboard containing the following $68$ pairs: for each of $m = 1, \\dots, 8$, five copies of $(m, m)$ (for a total of $40$ pairs of type (a)); for every $1 \\le m < n \\le 8$, one copy of $(-m, -n)$ (for a total of $\\binom{8}{2} = 28$ pairs of type (b)). We claim that we cannot get $44$ points from this initial stage. Indeed, assume that exactly $k$ of the integers $1, 2, \\dots, 8$ are underlined. Then we get at most $5k$ points from the pairs of type (a) and at most $28 - \\binom{k}{2}$ points from the pairs of type (b), for a total of at most $5k + 28 - \\binom{k}{2}$ points. The quadratic function $5k + 28 - \\binom{k}{2} = -\\frac{k^2}{2} + \\frac{11k}{2} + 28$ is maximized for integer $k$ at $k = 5$ or $k = 6$ and has a maximum value of $43$. Thus, we can get at most $43$ points with this initial blackboard, completing the solution.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 13342,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all monic cubic polynomials of the form $x^3 + px + q$ with integer coefficients such that the sum of the absolute values of their real roots is minimized, and the polynomial has three distinct real roots.",
    "options": [],
    "answer": "See solution",
    "solution": "First, for $R(x) = x^3 + px + q$ to have three distinct real roots, it is necessary that $p < 0$ (because the derivative $R'(x) = 3x^2 + p$ cannot be nonnegative). Let $p < 0$; then the equation $R'(x) = 0$ has two real roots $x_1 = -\\sqrt{-p/3}$, $x_2 = \\sqrt{-p/3}$.\n\nNow, the condition that $R(x)$ has three distinct real roots is equivalent to the inequality $R(x_1) \\cdot R(x_2) < 0$, which can be written as\n\n$$\n4p^3 + 27q^2 < 0. \\quad (*)\n$$\n\nLet (*) be valid; denote by $\\alpha_1 < \\alpha_2 < \\alpha_3$ the roots of $R(x)$. By Vieta's formula, $\\alpha_1 + \\alpha_2 + \\alpha_3 = 0$. Hence, we have two possibilities: (i) $\\alpha_1 < \\alpha_2 < 0 < \\alpha_3$; (ii) $\\alpha_1 < 0 < \\alpha_2 < \\alpha_3$. Without loss of generality, assume case (i) holds (if $x^3 + px + q$ satisfies (ii), then $x^3 + px - q$ satisfies (i)).\n\nSo, let $\\alpha_1 < \\alpha_2 < 0 < \\alpha_3$, then $q = -\\alpha_1\\alpha_2\\alpha_3 < 0$. Further,\n\n$$\n|\\alpha_1| + |\\alpha_2| + |\\alpha_3| = |\\alpha_1 + \\alpha_2| + \\alpha_3 = 2\\alpha_3.\n$$\n\nHence, we need to find $R(x)$ satisfying (*), $p, q \\in \\mathbb{Z}$, $p < 0$, $q < 0$, and for which $2\\alpha_3$ is the smallest possible.\n\nFirst, note that due to $p < 0$, $q < 0$ we have $R(1) = 1 + p + q < 0$ which implies $\\alpha_3 > 1$. Further, $R(2) = 8 + 2p + q$. If $R(2) \\le 0$, then $\\alpha_3 \\ge 2$. Hence, if $\\alpha_3 < 2$, then\n\n$$\n8 + 2p + q > 0. \\quad (**)\n$$\n\nNow, it is easy to see that all $R(x)$ satisfying (*) and (**) with $p < 0$, $q < 0$ are the following: $x^3 - 2x - 1$ and $x^3 - 3x - 1$. The first of them is not irreducible since it has $-1$ as a root. The second trinomial satisfies the condition.\n\n*Remark.* One can verify that the value of $|\\alpha_1| + |\\alpha_2| + |\\alpha_3|$ is equal to $4 \\cos 20^\\circ$ for the found polynomials.\n\n**Answer:** $x^3 - 3x - 1$ and $x^3 - 3x + 1$.",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 13343,
    "subject": "Mathematics (Olympiad)",
    "question": "Determine the maximum possible value of $a + b$, where $a$ and $b$ are two different non-negative real numbers that satisfy\n\n$$\na + \\\\sqrt{b} = b + \\\\sqrt{a}.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "The equation can be rewritten as\n\n$$\n\\sqrt{a} - \\sqrt{b} = a - b\n$$\n\nwhich implies\n\n$$\n\\sqrt{a} - \\sqrt{b} = (\\sqrt{a} - \\sqrt{b})(\\sqrt{a} + \\sqrt{b})\n$$\n\nAssuming $a \\neq b$, we can divide both sides by $\\sqrt{a} - \\sqrt{b}$ (since it is nonzero), yielding\n\n$$\n\\sqrt{a} + \\sqrt{b} = 1\n$$\n\nSquaring both sides gives\n\n$$\na + b + 2\\sqrt{ab} = 1\n$$\n\nSo,\n\n$$\na + b = 1 - 2\\sqrt{ab}\n$$\n\nSince $\\sqrt{ab} \\geq 0$, the maximum occurs when $ab = 0$, i.e., either $a = 0$ or $b = 0$. For example, $a = 0$, $b = 1$ (or vice versa) satisfy the original equation and $a + b = 1$.\n\nThus, the maximum possible value of $a + b$ is $1$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 13344,
    "subject": "Mathematics (Olympiad)",
    "question": "Show that in any sequence of six consecutive integers, there is at least one integer $x$ such that\n\n$$\n(x^2 + 1)(x^4 + 1)(x^6 - 1)\n$$\n\nis a multiple of 2016.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $N = (x^2 + 1)(x^4 + 1)(x^6 - 1)$. Suppose that $x$ is relatively prime to 2016, whose prime factorization is $2^5 \\times 3^2 \\times 7$.\n\n- Since $x$ is not divisible by 7, Fermat's little theorem tells us that $x^6 - 1$ is divisible by 7. Therefore, $7 \\mid N$.\n- Since $x$ is relatively prime to 9, Euler's theorem tells us that $x^{\\varphi(9)} - 1 = x^6 - 1$ is divisible by 9. Therefore, $9 \\mid N$.\n- Since $x$ is odd, the expressions $x^2 + 1$, $x^4 + 1$, $x + 1$, and $x - 1$ are all even, with one of the last two necessarily a multiple of 4. Therefore,\n\n$$\nN = (x^2 + 1)(x^4 + 1)(x + 1)(x - 1)(x^4 + x^2 + 1)\n$$\n\nis divisible by $2^5$.\n\nThus, if $x$ is relatively prime to 2016, then $N$ is a multiple of 2016.\n\nNow consider the following residues modulo 42: 1, 5, 11, 17, 23, 25, 31, 37, 1. Note that the largest difference between consecutive numbers in this sequence is 6. It follows that, among any six consecutive integers, there is an integer $x$ that is congruent modulo 42 to one of these residues. Since these residues are relatively prime to 42, it follows that $x$ is relatively prime to 42. And since 42 and 2016 have the same prime factors — namely 2, 3, and 7 — it follows that $x$ is relatively prime to 2016. So, by the above reasoning, $N$ is a multiple of 2016.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 13345,
    "subject": "Mathematics (Olympiad)",
    "question": "The average of $p$ and $q$ is $\\frac{p+q}{2}$. Given that $2(100p + q) = (p+q)k$ for some integer $k$, and $p, q$ are primes, find all pairs $(p, q)$ that satisfy this condition.",
    "options": [],
    "answer": "See solution",
    "solution": "We have:\n\n$$\n2(100p + q) = (p+q)k\n$$\n\nExpanding and rearranging:\n\n$$\n200p + 2q = (p+q)k \\\\\n200p + 2q = kp + kq \\\\\n200p + 2q - kp - kq = 0 \\\\\n200p - kp + 2q - kq = 0 \\\\\n(200 - k)p + (2 - k)q = 0\n$$\n\nAlternatively, as in the original steps:\n\n$$\n200p + 2q = (p+q)k \\\\\n198p = (p+q)(k-2)\n$$\n\nSince $p$ and $q$ are primes, $(p+q)$ must divide $198$. The possible values for $p+q$ between $11 + 13 = 24$ and $89 + 97 = 186$ are the even divisors of $198$ in this range. $198 = 2 \\times 3^2 \\times 11$, so the only even factor in this range is $66$.\n\nThus, $p+q = 66$. The pairs of primes that sum to $66$ are:\n\n- $(13, 53)$\n- $(19, 47)$\n- $(23, 43)$\n- $(29, 37)$\n\nTherefore, the pairs $(p, q)$ are $(13, 53)$, $(19, 47)$, $(23, 43)$, and $(29, 37)$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 13346,
    "subject": "Mathematics (Olympiad)",
    "question": "Да се определат сите природни броеви $n$ за кои броевите $n+2$ и $n^2+n+1$ се точни кубови на природни броеви.",
    "options": [],
    "answer": "See solution",
    "solution": "Нека претпоставиме дека $n$ е природен број за кој $n+2$ и $n^2+n+1$ се точни кубови. Тогаш јасно бројот $(n+2)(n^2+n+1)$ е исто така точен куб на природен број. Но\n\n$$\n(n+2)(n^2+n+1) = n^3 + 3n^2 + 3n + 2 = (n+1)^3 + 1,\n$$\n\nи јасно не може да биде точен куб. Значи не постои природен број $n$ за кој се исполнети условите на задачата.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 13347,
    "subject": "Mathematics (Olympiad)",
    "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\n(b) The same question with three-digit instead of two-digit numbers.",
    "options": [],
    "answer": "See solution",
    "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": 13348,
    "subject": "Mathematics (Olympiad)",
    "question": "Para cada entero positivo $n$, el Banco de Ciudad del Cabo produce monedas de valor $\\frac{1}{n}$. Dada una colección finita de tales monedas (no necesariamente de distintos valores) cuyo valor total no supera $99 + \\frac{1}{2}$, demostrar que es posible separar esta colección en 100 o menos montones, de modo que el valor total de cada uno sea como máximo 1.",
    "options": [],
    "answer": "See solution",
    "solution": "Antes de empezar a particionar el conjunto de monedas, vayamos a la ventanilla del Banco de Ciudad del Cabo y realicemos los siguientes cambios:\n\n- Para cada entero $n = 2m$ par, para el que haya al menos dos monedas de valor $\\frac{1}{n}$, tomemos dos de estas monedas y cámbielas por una moneda de valor $\\frac{1}{m}$, hasta que no haya ningún entero par $n = 2m$ para el que haya más de una moneda de valor $\\frac{1}{n}$.\n- Para cada entero $n = 2m + 1$ impar, para el que haya al menos $2m + 1$ monedas con valor $\\frac{1}{n}$, tomemos $2m+1$ de estas monedas y cámbielas por una moneda de valor 1, hasta que no haya ningún entero impar $n = 2m + 1$ para el que haya más de $2m$ monedas de valor $\\frac{1}{n}$.\n\nNótese que, si podemos particionar el conjunto resultante, también podemos particionar el conjunto inicial, no teniendo para ello más que volver a la ventanilla del Banco de Ciudad del Cabo, invirtiendo los cambios realizados. Sabiendo además que siempre podemos tomar todas las monedas de valor $1$, asignando cada una de ellas a alguno de los 100 conjuntos de la partición, vemos que nos basta con resolver el siguiente problema: particionar un conjunto $R$ de recíprocos de enteros, no necesariamente distintos y cuya suma es menor que $N - \\frac{1}{2}$, en $N$ conjuntos la suma de cada uno de los cuales es a lo sumo 1, siendo $N$ un entero, y dado que en $R$ no hay más de un recíproco de un entero par (es decir, para cada entero par $n = 2m$ existe a lo sumo un elemento en $R$ igual a $\\frac{1}{n}$), y que para cada entero impar $n = 2m + 1$, existen a lo sumo $2m$ elementos en $R$ que son los recíprocos de $\\frac{1}{n}$. Como además los elementos de $R$ que tienen valor 1 se pueden asignar cada uno a un elemento de la partición, sin afectar al problema (simplemente reduciendo el valor de $N$ y el cardinal de $R$), podemos asumir que el mayor valor de $R$ es a lo sumo $\\frac{1}{2}$. Resolveremos a continuación este problema.\n\nDefinamos conjuntos $R_1, R_2, \\dots, R_N$. Si existe una moneda de valor $\\frac{1}{2}$ (y en ese caso existiría a lo sumo una), la colocamos en $R_1$. Colocamos en $R_2$ las monedas que existan de valor $\\frac{1}{3}$ (a lo sumo dos), y de valor $\\frac{1}{4}$ (a lo sumo una), y así sucesivamente, hasta colocar en $R_N$ las monedas de valor $\\frac{1}{2N-1}$ y $\\frac{1}{2N}$. Nótese que la suma máxima de las monedas colocadas en $R_k$ es a lo sumo $\\frac{2k-2}{2k-1} + \\frac{1}{2k} = 1 - \\frac{1}{2k(2k-1)} < 1$. Coloquemos ahora las monedas restantes, de forma aleatoria, en cualquier conjunto tal que, al añadir dicha moneda, el valor no supera 1. Supongamos que en algún momento esto deja de ser posible, para una moneda de valor $\\frac{1}{n}$. Entonces, en cada conjunto la suma es mayor que $1 - \\frac{1}{n}$, para un valor total de las monedas ya asignadas superior a $N - \\frac{N}{n}$, y a su vez inferior a $N - \\frac{1}{2}$. Luego $n < 2N$, contradicción pues ya hemos colocado todas las monedas de valor mayor o igual que $\\frac{1}{2N}$ en la asignación inicial. Luego siempre podemos continuar asignando todas las monedas de valor inferior o igual a $\\frac{1}{2N+1}$, siendo por lo tanto siempre posible la partición, como queríamos demostrar.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 13349,
    "subject": "Mathematics (Olympiad)",
    "question": "A $2004 \\times 2004$ array of points is drawn. Find the largest integer $n$ such that it is possible to draw a convex $n$-sided polygon whose vertices lie on the points of the array.",
    "options": [],
    "answer": "See solution",
    "solution": "For a vector $v = (x, y)$, define $\\|v\\| = |x| + |y|$, the so-called taxicab distance (or taxicab norm). Embed the array of points in the plane such that they correspond to the lattice points in $\\{(x, y) : 1 \\le x, y \\le 2004\\}$.\n\nConsider a convex $n$-gon drawn in our square array, and imagine that we walk along the edges in a counterclockwise direction. Then we can orient each edge and obtain a set of $n$ nonzero vectors $S = \\{v_i = (x_i, y_i)\\}$, with integer coordinates, whose sum is $(0,0)$. $S$ has several further properties. First, no two vectors in $S$ are positive multiples of each other by convexity (if $i \\ne j$ and the directed edges $v_i$ and $v_j$ are parallel and pointing in the same direction, then our polygon cannot be strictly convex.)\n\nSecond, the sum of the $x_i$ which are positive is at most $2003$, and the same is true for the sum of the $y_i > 0$, as well as the sums of the $|x_i|$ and $|y_i|$ for $x_i < 0$ and $y_i < 0$. This is true because all the vectors with, say, $x_i$ positive will correspond to adjacent edges (by convexity), and if one traces these edges in order on the polygon, one must start from a point within $\\{(x, y) : 1 \\le x, y \\le 2004\\}$ and finish at a point in that same region; therefore the total displacement in the $x$-dimension is bounded by $2003$. In particular, this implies that $\\sum \\|v_i\\| \\le 8012$.\n\nThird, given $S$ that satisfies the above properties, we can construct a convex polygon that fits within the bounds: since the polygon should be convex, we must place the vectors end to end ordered by the angle (measured counterclockwise) that they make with the positive $x$-axis, and the resulting polygon will fit within the array because of the given inequalities.\n\nWe now show that it is impossible to draw a $562$-gon in the array. We will prove that for every set $S$ of $562$ vectors such that no two are positive multiples of each other, $\\sum \\|v_i\\| > 8012$. We can even ignore the condition that $\\sum v_i = (0,0)$. Let us now try to minimize $\\sum \\|v_i\\|$. Since no two vectors in $S$ are positive multiples of each other, we may assume that for each $i$, $\\gcd(|x_i|, |y_i|) = 1$, or else we might as well scale that vector down by that gcd (it will only reduce $\\sum \\|v_i\\|$).\n\n**Lemma** For any positive integer $k$, the maximum number of distinct $v_i = (x_i, y_i)$ that satisfy $\\|v_i\\| = k$ and $\\gcd(|x_i|, |y_i|) = 1$ is $4\\phi(k)$. Recall that $\\phi(k)$ is Euler's totient function, which counts the number of positive integers $z \\le k$ that satisfy $\\gcd(z, k) = 1$.\n\n*Proof:* For $k=1$, the lemma is trivial. For $k \\ge 2$, the gcd condition forces that neither of $x_i$ or $y_i$ can ever be $0$. So, it suffices to show that the number of $v_i$ with $x_i > 0$ and $y_i > 0$ is ...",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 13350,
    "subject": "Mathematics (Olympiad)",
    "question": "We are given a triangle $ABC$ and a point $D$ on the side $BC$. Let $U$ be the circumcenter of $\\triangle BDA$ and $V$ the circumcenter of $\\triangle CDA$. Prove that the triangles $AUV$ and $ABC$ are similar.\n\n![](images/Austria_2010_p3_data_c1b20ca8c6.png)",
    "options": [],
    "answer": "See solution",
    "solution": "Let the point $C'$ be chosen so that triangles $ABC$ and $ACC'$ are similar and have no common interior points. Furthermore, let $D'$ be chosen on $CC'$ such that triangles $ABD$ and $ACD'$ are also similar. This means that $ACC'$ results from $ABC$ by rotation and subsequent homothety with ratio $AC : AB$, both with center $A$.\n\nSince $\\angle D'CD = \\angle D'CA + \\angle ACD = \\angle CBA + \\angle ACB$ and $\\angle D'AD = \\angle CAB$, we see that $\\angle D'CD + \\angle D'AD = 180^\\circ$. This means that the points $A$, $D$, $C$, and $D'$ lie on a common circle. The circumcenter $V$ of $\\triangle CDA$ is therefore also the circumcenter of $\\triangle CD'A$, and therefore results from the circumcenter $U$ of $\\triangle BDA$ by rotation and subsequent homothety with ratio $AC : AB$, both with center $A$.\n\nWe therefore see that $\\angle UAV = \\angle BAC$ and $AU : AV = AB : AC$. Triangles $ABC$ and $AUV$ are therefore similar, as claimed.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 13351,
    "subject": "Mathematics (Olympiad)",
    "question": "Во рамнокрак триаголник $ABC$, $AC = BC = 1$. За која вредност на $\\gamma = \\angle ACB$ изразот\n$$g = \\frac{AB^2 + 2}{P_{\\Delta ABC}}$$\nдостигнува најмала вредност.",
    "options": [],
    "answer": "See solution",
    "solution": "Од косинусната теорема имаме $AB^2 = 2 - 2\\cos\\gamma$, а за плоштината на триаголникот важи $P_{\\Delta ABC} = \\frac{1}{2} AC \\cdot BC \\cdot \\sin\\gamma = \\frac{\\sin\\gamma}{2}$. Тогаш добиваме\n$$g(\\gamma) = \\frac{2(4 - 2\\cos\\gamma)}{\\sin\\gamma} = \\frac{4(2 - \\cos\\gamma)}{\\sin\\gamma}.$$ \nЌе воведеме смена $x = \\tan \\frac{\\gamma}{2}$, со помош на која $\\sin\\gamma = \\frac{2x}{1+x^2}$, $\\cos\\gamma = \\frac{1-x^2}{1+x^2}$, од каде функцијата\n$$g = g(x) = \\frac{2(1+3x^2)}{x}.$$ \nИмајќи предвид дека $\\gamma \\in (0, \\pi)$, односно $\\frac{\\gamma}{2} \\in (0, \\frac{\\pi}{2})$, јасно е дека $x > 0$ и истовремено и $g > 0$.\n\nЌе ја одредиме најмалата вредност која ја достигнува функцијата $g(x), x \\in (0, \\infty)$. Нека $g_{\\min} = y$. Ќе определиме за која вредност на $x$ истата се достигнува. Значи ја решаваме равенката $y = g(x) = \\frac{2(1+3x^2)}{x}$, односно $6x^2 - 4x + 2 = 0$. Равенката треба да има реални корени, па потребно е дискриминантата $D \\ge 0$. Тогаш $y^2 - 48 \\ge 0 \\Leftrightarrow y \\le -4\\sqrt{3}$ или $y \\ge 4\\sqrt{3}$. Бидејќи $g > 0$ го разгледуваме само случајот $y \\ge 4\\sqrt{3}$, а тогаш најмалата вредност која функцијата може да ја достигне е $y = 4\\sqrt{3}$ и истата се достигнува кога $6x^2 - 4x\\sqrt{3} + 2 = 0$, односно за $x = \\frac{\\sqrt{3}}{3}$. Тогаш соодветниот агол на триаголникот е $\\gamma = \\frac{\\pi}{3}$. Бидејќи триаголникот е рамнокрак, добиваме дека тој е и рамностран.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 13352,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $n \\in \\mathbb{N}^*$. Determine all functions $f : \\mathbb{R} \\to \\mathbb{R}$ that satisfy\n\n$$\nf(x + y^{2n}) = f(f(x)) + y^{2n-1}f(y),\n$$\n\nfor all $x, y \\in \\mathbb{R}$, and for which the equation $f(x) = 0$ has a unique solution.",
    "options": [],
    "answer": "See solution",
    "solution": "For $y = 0$, the given relation reduces to $f(x) = f(f(x))$, so\n\n$$\nf(x + y^{2n}) = f(x) + y^{2n-1}f(y), \\quad \\forall x, y \\in \\mathbb{R}. \\qquad (1)\n$$\n\nIf we consider $x = 0$ and $y = 1$ in (1), then $f(0) = 0$. Since $f(x) = 0$ has a unique solution, we have\n\n$$\nf(x) = 0 \\Rightarrow x = 0. \\qquad (2)\n$$\n\nPutting $x = 0$ in (1), we obtain $f(y^{2n}) = y^{2n-1}f(y)$ for all $y \\in \\mathbb{R}$, so (1) rewrites as\n\n$$\nf(x + y^{2n}) = f(x) + f(y^{2n}), \\quad \\forall x, y \\in \\mathbb{R}. \\qquad (3)\n$$\n\nMoreover, $y^{2n-1}f(y) = f(y^{2n}) = f((-y)^{2n}) = -y^{2n-1}f(-y)$ for all $y \\in \\mathbb{R}$, so $f$ is odd. Considering $t = \\sqrt[2n]{y} \\ge 0$, (3) becomes\n\n$$\nf(x + t) = f(x) + f(t), \\quad \\forall x \\in \\mathbb{R}, t \\ge 0. \\qquad (4)\n$$\n\nUsing that $f$ is odd, for $t < 0$,\n\n$$\nf(x + t) = -f(-x - t) = -(f(-x) + f(-t)) = f(x) + f(t),\n$$\n\nwhich, together with (4), implies\n\n$$\nf(x + t) = f(x) + f(t), \\quad \\forall x, t \\in \\mathbb{R}. \\qquad (5)\n$$\n\nIf $f(x_1) = f(x_2)$, by (5), $f(x_1 - x_2) = 0$, which by (2) implies $x_1 = x_2$, so $f$ is injective. But since $f(f(x)) = f(x)$, we have $f(x) = x$, which is a solution.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 13353,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $a_1, a_2, a_3, \\dots$ be a sequence of positive real numbers. Suppose that for some positive integer $s$, we have\n$$\na_n = \\max\\{a_k + a_{n-k} \\mid 1 \\le k \\le n-1\\}\n$$\nfor all $n > s$. Prove that there exist positive integers $l$ and $N$, with $l \\le s$ and such that $a_n = a_l + a_{n-l}$ for all $n \\ge N$.",
    "options": [],
    "answer": "See solution",
    "solution": "**Proof**\n\nBy assumption, for each $n > s$, $a_n$ can be written as $a_n = a_{j_1} + a_{j_2}$, where $j_1, j_2 < n$ and $j_1 + j_2 = n$. If $j_1 > s$, we can continue to write $a_{j_1}$ as a sum of two terms of the sequence. We keep doing this until we obtain\n$$\na_n = a_{i_1} + \\cdots + a_{i_k},\n$$\nwhere $1 \\le i_j \\le s$ and $i_1 + \\cdots + i_k = n$.\n\nSuppose that $a_{i_1}, a_{i_2}$ is the last step in obtaining the above. Then $i_1 + i_2 > s$, and the previous condition is reformulated as\n$$\n1 \\le i_j \\le s,\\quad i_1 + \\cdots + i_k = n,\\quad i_1 + i_2 > s.\n$$\n\nOn the other hand, if the indices $i_1, \\dots, i_k$ satisfy this, we set $s_j = i_1 + \\cdots + i_j$. By the original assumption,\n$$\na_n = a_{s_k} \\ge a_{s_{k-1}} + a_{i_k} \\ge a_{s_{k-2}} + a_{i_{k-1}} + a_{i_k} \\ge \\cdots \\ge a_{i_1} + \\cdots + a_{i_k}.\n$$\nHence, for any $n > s$, we get\n$$\na_n = \\max\\{a_{i_1} + \\cdots + a_{i_k} : (i_1, \\dots, i_k) \\text{ satisfies above}\\}.\n$$\nLet $m = \\max\\left\\{\\frac{a_i}{i} \\mid 1 \\le i \\le s\\right\\}$, and assume that $m = \\frac{a_l}{l}$ for some positive integer $l \\le s$.\n\nConstruct a sequence $\\{b_n\\}$ as follows: $b_n = a_n - mn$, $n = 1, 2, \\dots$; then $b_l = 0$. When $n \\le s$, we have $b_n \\le 0$ by definition of $m$. When $n > s$,\n$$\n\\begin{align*}\nb_n &= a_n - mn = \\max\\{a_k + a_{n-k} \\mid 1 \\le k \\le n-1\\} - mn \\\\\n&= \\max\\{b_k + b_{n-k} + mn \\mid 1 \\le k \\le n-1\\} - mn \\\\\n&= \\max\\{b_k + b_{n-k} \\mid 1 \\le k \\le n-1\\} \\le 0,\n\\end{align*}\n$$\nso $b_n \\le 0$ for $n = 1, 2, \\dots$, and for $n > s$,\n$$\nb_n = \\max\\{b_k + b_{n-k} \\mid 1 \\le k \\le n-1\\}.\n$$\nIf $b_k = 0$ for every $k = 1, 2, \\dots, s$, then for every positive integer $n$, $b_n = 0$, and hence $a_n = nm$ for every $n$, and the conclusion follows.\n\nOtherwise, let $M = \\max_{1 \\le i \\le s} |b_i|$, $\\varepsilon = \\min\\{|b_i| : 1 \\le i \\le s,\\ b_i < 0\\}$. When $n > s$, we have\n$$\nb_n = \\max\\{b_k + b_{n-k} \\mid 1 \\le k \\le n-1\\} \\ge b_l + b_{n-l} = b_{n-l},\n$$\nand thus $0 \\ge b_n \\ge b_{n-l} \\ge \\dots \\ge -M$.\n\nAs for the sequence $\\{b_n\\}$, by the previous decomposition every $b_n$ belongs to the set\n$$\nT = \\{b_{i_1} + b_{i_2} + \\cdots + b_{i_k} : 1 \\le i_1 < \\dots < i_k \\le s\\} \\cap [-M, 0].\n$$\nIt follows that $T$ is a finite set. Indeed, for any $x \\in T$, let\n$$\nx = b_{i_1} + \\cdots + b_{i_k} \\quad (1 \\le i_1 < \\dots < i_k \\le s).\n$$\nThen there are at most $\\frac{M}{\\varepsilon}$ nonzero terms in $b_i$ (otherwise $x < -M$), and hence there are only finitely many such representations for $x$.\n\nThus, for every $t = 1, 2, \\dots, l$, the sequence\n$$\nb_{s+l}, b_{s+l+l}, b_{s+l+2l}, \\dots\n$$\nis increasing, and takes only finitely many values, and hence it is eventually constant, i.e., there is $N$ such that $\\{b_n\\}$ is periodic for $n > N$ with period $l$; in other words,\n$$\nb_n = b_{n-l} = b_l + b_{n-l} \\quad (n > N + l),\n$$\ni.e.\n$$\n\\begin{aligned}\na_n &= b_n + mn = (b_l + ml) + (b_{n-l} + m(n-l)) \\\\\n    &= a_l + a_{n-l} \\quad (n > N + l).\n\\end{aligned}\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 13354,
    "subject": "Mathematics (Olympiad)",
    "question": "Each of the small squares of a $50 \\times 50$ table is coloured in red or blue. Initially, all squares are red. A *step* means changing the colour of all squares on a row or on a column.\n\n**a)** Prove that there exists no sequence of steps such that at the end there are exactly $2011$ blue squares.\n\n**b)** Describe a sequence of steps such that at the end exactly $2010$ squares are blue.",
    "options": [],
    "answer": "See solution",
    "solution": "Without loss of generality, we may consider that the rows or columns to be modified in a sequence of steps are consecutive, and that each column or row is modified only once.\n\nSuppose then that the first $x$ rows and the first $y$ columns have been modified. One gets an $x \\times y$ rectangle and a $(50-x) \\times (50-y)$ rectangle with all squares coloured blue, the rest of the table being red.\n\nThe number of blue squares is then\n$$\nA = xy + (50-x)(50-y)\n$$\nwhich is an even number, so it cannot equal $2011$.\n\nFor the second part, notice that $A = 2010$ is equivalent to\n$$\n(x - 25)(y - 25) = 380 = 19 \\cdot 20.\n$$\nOne can take $x = 25 + 19 = 44$ and $y = 25 + 20 = 45$ to give the answer (thus being clear that the steps are not unique).",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 13355,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $ABC$ be a scalene triangle and let $I$ be its incenter. Consider the circles $\\gamma$, $\\delta$, of diameters $IB$, respectively $IC$. The circles $\\gamma'$, $\\delta'$ are the mirror images of $\\gamma$, $\\delta$ in $IC$, respectively $IB$. Prove that the circumcenter of the triangle $ABC$ lies on the line joining the common points of the circles $\\gamma'$ and $\\delta'$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $B_1$ be the symmetric of the point $B$ with respect to the line $IC$, and let $C_1$ be the symmetric of the point $C$ with respect to the line $IB$. Denote by $B_2$ the midpoint of the segment $IB_1$, and by $C_2$ the midpoint of the segment $IC_1$. The circles $\\gamma'$ and $\\delta'$ have diameters $IB_1$ and $IC_1$, hence their centers are $B_2$ and $C_2$. The line passing through the meeting points of circles $\\gamma'$ and $\\delta'$ is perpendicular to the line joining their centers. Since $I$ belongs to both circles $\\gamma'$ and $\\delta'$, the requirement comes to $OI \\perp B_2C_2$. Since $B_2C_2$ is a midline in triangle $IB_1C_1$, one has $B_1C_1 \\parallel B_2C_2$, and the requirement writes $OI \\perp B_1C_1$. It is enough to show $OB_1^2 - OC_1^2 = IB_1^2 - IC_1^2$.\n\nLet $D$ be the projection of $I$ onto the line $BC$. $D$ is the touching point of the incircle of triangle $ABC$ with the side $BC$, and $BD = p - b$, $DC = p - c$. From symmetry considerations we have $IB_1 = IB$, $IC_1 = IC$, and\n\n$$\n\\begin{aligned}\nIB_1^2 - IC_1^2 &= IB^2 - IC^2 = \\\\\n&= DB^2 - DC^2 = \\\\\n&= (p-b)^2 - (p-c)^2 = \\\\\n&= a(c-b).\n\\end{aligned}\n$$\n\nDenote by $M$ the midpoint of the side $AC$, and by $N$ the midpoint of the side $AB$; then we have\n\n$$\n\\begin{aligned}\nOB_1^2 - OC_1^2 &= \\\\\n&= (OM^2 + MB_1^2) - (ON^2 + NC_1^2) = \\\\\n&= (MB_1^2 + OA^2 - MA^2) - \\\\\n&\\qquad (NC_1^2 + OB^2 - NB^2) = \\\\\n&= (MB_1^2 - MA^2) - (NC_1^2 - NB^2),\n\\end{aligned}\n$$\n\nsince $OA = OB$. As $MB_1^2 - MA^2 = MB_1^2 - MC^2 = (MB_1 - MC)(MB_1 + MC) = a(a - b)$, and analogously $NC_1^2 - NB^2 = a(a - c)$, it follows that $OB_1^2 - OC_1^2 = a(c - b)$, which is what was left to prove.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 13356,
    "subject": "Mathematics (Olympiad)",
    "question": "If $a$, $b$, $c$ are positive real numbers, prove that\n\n$$\n\\frac{a}{\\sqrt{a^2 + 8bc}} + \\frac{b}{\\sqrt{b^2 + 8ca}} + \\frac{c}{\\sqrt{c^2 + 8ab}} \\ge 1.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "We seek $\\lambda \\in \\mathbb{R}$ such that\n\n$$\n\\frac{a}{\\sqrt{a^2 + bc}} \\ge \\frac{a^{\\lambda}}{b^{\\lambda} + c^{\\lambda} + a^{\\lambda}}.\n$$\n\nFrom this, we obtain\n\n$$\na^2(a^{\\lambda} + b^{\\lambda} + c^{\\lambda})^2 \\ge a^{2\\lambda}(a^2 + bc).\n$$\n\nExpanding,\n\n$$\n\\begin{aligned}\na^2(a^{\\lambda} + b^{\\lambda} + c^{\\lambda})^2 &= a^{2\\lambda+2} + a^2(b^{2\\lambda} + 2a^2b^{\\lambda}c^{\\lambda} + 2a^{\\lambda+2}b^{\\lambda} + 2a^{\\lambda+2}c^{\\lambda}) \\\\\n&\\ge a^{2\\lambda+2} + 2a^2b^{\\lambda}c^{\\lambda} + 2a^2b^{\\lambda}c^{\\lambda} + 2a^{\\lambda+2}b^{\\lambda} + 2a^{\\lambda+2}c^{\\lambda} \\\\\n&\\ge a^{2\\lambda+2} + 4\\sqrt[4]{2^4a^{2\\lambda+8}b^{3\\lambda}c^{3\\lambda}} = a^{2\\lambda+2} + 8a^{\\frac{\\lambda+4}{2}}b^{\\frac{3\\lambda}{4}}c^{\\frac{3\\lambda}{4}}.\n\\end{aligned}\n$$\n\nBy the previous steps,\n\n$$\na^{2\\lambda+2} + 8a^{\\frac{\\lambda+4}{2}}b^{\\frac{3\\lambda}{4}}c^{\\frac{3\\lambda}{4}} = a^{2\\lambda+2} + 8a^{2\\lambda}bc.\n$$\n\nFrom this, we find $\\lambda = \\frac{4}{3}$. Substituting into the original inequality, we get\n\nLet $\\frac{a^2}{bc} = x$, $\\frac{b^2}{ac} = y$, $\\frac{c^2}{ab} = z$ so that $xyz = 1$, $x, y, z > 0$. Then,\n\n$$\n\\sqrt{\\frac{x}{x+8}} + \\sqrt{\\frac{y}{y+8}} + \\sqrt{\\frac{z}{z+8}} \\ge 1.\n$$\n\nLet $(x', y', z')$ be any permutation of $(x, y, z)$. The triples $(\\sqrt{x}, \\sqrt{y}, \\sqrt{z})$ and $(\\frac{1}{\\sqrt{x+8}}, \\frac{1}{\\sqrt{y+8}}, \\frac{1}{\\sqrt{z+8}})$ are inversely monotonic, so by the rearrangement method,\n\n$$\n\\sqrt{\\frac{x}{x+8}} + \\sqrt{\\frac{y}{y+8}} + \\sqrt{\\frac{z}{z+8}} \\le \\sqrt{\\frac{x'}{y'+8}} + \\sqrt{\\frac{y'}{z'+8}} + \\sqrt{\\frac{z'}{x'+8}}.\n$$\n\nTaking $(x, y, z) = (a, b, c)$, $(x', y', z') = (a, b, c)$, the inequality is proved. Equality holds for $a = b = c = 1$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 13357,
    "subject": "Mathematics (Olympiad)",
    "question": "Дараах рекуррент дарааллыг авч үзье:\n\n- $b_0 = 1$\n- $b_1 = -1$\n- $b_n = 4b_{n-1} + 837b_{n-2}$, $n \\ge 2$\n\n$837 \\equiv 29 \\pmod{103}$.\n\nМөн $a_n \\equiv b_n \\pmod{101}$ бүх $n \\ge 0$ үед. $b_{2012}$-ийг $101$-д хуваахад гарах үлдэгдлийг ол.\n",
    "options": [],
    "answer": "See solution",
    "solution": "$$\\{b_n\\}$$ дарааллын характеристик тэгшитгэл:\n$$x^2 - 4x - 837 = 0$$\nҮндэс нь $x_1 = 31$, $x_2 = -27$ тул:\n$$b_n = c_1 \\cdot 31^n + c_2(-27)^n,\\quad n \\ge 0$$\n\nЭхний нөхцлүүдээс:\n$$b_0 = 1 = c_1 + c_2$$\n$$b_1 = 31c_1 - 27c_2 = -1$$\nШийдэж:\n$$c_1 = \\frac{13}{29},\\quad c_2 = \\frac{16}{29}$$\nИймд:\n$$29b_n = 13 \\cdot 31^n + 16(-27)^n,\\quad n \\ge 0$$\n\n$101$ анхны тоо тул Фермагийн бага теорем:\n$$31^{100} \\equiv 1 \\pmod{101},\\quad 27^{100} \\equiv 1 \\pmod{101}$$\n\n$$31^{2012} = (31^{100})^{20} \\cdot 31^{12} \\equiv 31^{12}$$\n$$31^{12} = (31^2)^6 = 961^6 \\equiv 52^6 \\equiv 2704^3 \\equiv 78^3 \\equiv 54 \\pmod{101}$$\n\n$$27^{2012} = (27^{100})^{20} \\cdot 27^{12} \\equiv 27^{12}$$\n$$27^{12} = (27^2)^6 = 729^6 \\equiv 12^6 \\equiv 144^3 \\equiv 43^3 \\equiv 20 \\pmod{101}$$\n\n$$29b_{2012} \\equiv 13 \\cdot 54 + 16 \\cdot 20 \\equiv 702 + 320 \\equiv 1022 \\equiv 12 \\pmod{101}$$\n\n$29x \\equiv 12 \\pmod{101}$-г $x$-д шийдье. Евклидийн алгоритмаар $29 \\cdot 7 - 101 \\cdot 2 = 1$ тул:\n$$x = 7 \\cdot 12 = 84$$\n\nИймд:\n$$a_{2012} \\equiv b_{2012} \\equiv 84 \\pmod{101}$$\n",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 13358,
    "subject": "Mathematics (Olympiad)",
    "question": "There is a big crowd of boys and girls. Is it always possible to give them hats of 100 colors (everybody gets one hat) such that if some boy is familiar with at least 2014 girls, then all these girls have hats of at least 2 colors, and the same for girls holds: if some girl is familiar with at least 2014 boys, then all these boys have hats of at least 2 colors?",
    "options": [],
    "answer": "See solution",
    "solution": "No.\n\nLet $D = 100$, $p = 2014$ for clarity. We take a set $S_1$ consisting of $(p-1)D + 1$ elements as the first part of graph $G$ (*boys*). As the second part $S_2$ of $G$ (*girls*), we take the set of all $p$-element subsets from $S_1$ and join every such subset with all its elements in $S_1$.\n\nIf we try to color $G$ with $D$ colors, then by the Dirichlet principle, in the set $S_1$ one can find $p$ vertices of the same color. This means that for the corresponding $p$-element subset in $S_2$, the condition does not hold.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 13359,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $AB$ be a line segment of length $1$. Several elementary particles start moving simultaneously at constant speeds from $A$ to $B$. As soon as a particle reaches $B$, it turns around and heads to $A$; when reaching $A$, it starts moving to $B$ again, and so on indefinitely.\n\nFind all rational numbers $r > 1$ with the following property: For each $n \\ge 1$, if $n+1$ particles with speeds $1, r, r^2, \\dots, r^n$ move as described, there is a moment when all particles are at the same interior point of segment $AB$. (Ignore the dimensions of the particles; assume that they can all gather at one point.)",
    "options": [],
    "answer": "See solution",
    "solution": "The values in question are all integers $r$ greater than $1$.\n\nWe start with a general observation about two particles $P_1$ and $P_2$ moving on $AB$ by the given rules, with different constant speeds $v_1$ and $v_2$, $v_1 > v_2$. Suppose that they are at the same point $Q$ of $AB$ at a certain moment $t$. There are two possibilities for the distances $v_1 t$ and $v_2 t$ the particles have traveled until that moment. If $P_1$ and $P_2$ are moving in the same direction when they simultaneously reach $Q$, then the integer parts of $v_1 t$ and $v_2 t$ have the same parity, and their fractional parts are equal. Hence $v_1 t - v_2 t$ is an even positive integer. If $P_1$ and $P_2$ are moving in opposite directions when they meet at $Q$, then the integer parts of $v_1 t$ and $v_2 t$ have different parity, and the sum of their fractional parts is $1$. Therefore $v_1 t + v_2 t$ is an even positive integer.\n\nNow let the rational $r > 1$ have the stated property, for any number $n+1$, $n \\ge 1$, of particles with speeds $1, r, r^2, \\dots, r^n$. Let $t$ be a moment when all of them are at the same point. Apply the observation to the first and the last particle, with speeds $v_1 = r^n$ and $v_2 = 1$. We infer that $(r^n - 1)t$ or $(r^n + 1)t$ is an integer. Because $r$ is rational, $t$ is rational too. Write $r$ and $t$ as irreducible fractions: $r = \\frac{a}{b}$, $t = \\frac{c}{d}$. Then $(r^n \\pm 1)t = \\frac{(a^n \\pm b^n)c}{b^n d}$. Since $a^n \\pm b^n$ and $b^n$ are coprime, it follows that $b^n$ divides $c$. Moreover, the latter holds for each $n > 1$ by hypothesis. This is possible only if $b = 1$, that is, if $r > 1$ is an integer.\n\nConversely, every integer $r > 1$ is a solution. Let $r \\ge 3$ be odd and $n \\ge 1$ arbitrary. Then all $n+1$ particles with speeds $1, r, r^2, \\dots, r^n$ will be at the midpoint of $AB$ at $t = \\frac{1}{2}$. Indeed, $r^k \\cdot \\frac{1}{2}$ has fractional part $\\frac{1}{2}$ for each $k = 0, 1, 2, \\dots, n$ since $r^k$ is odd.\n\nLet $r = 2m$, $m \\ge 1$, be even and $n \\ge 1$ arbitrary. Then at the moment $t = \\frac{2m}{2m+1}$ all $n+1$ particles with speeds $1, r, r^2, \\dots, r^n$ will be at the point $Q$ at distance $\\frac{2m}{2m+1}$ from $A$. It is enough to prove that for each $k = 0, 1, 2, \\dots$ the following equality holds:\n\n$$\n(2m)^k \\frac{2m}{2m+1} = \\begin{cases} 2q + \\frac{2m}{2m+1} & \\text{if } k \\ge 0 \\text{ is even;} \\\\ 2q+1 + \\frac{1}{2m+1} & \\text{if } k \\ge 1 \\text{ is odd.} \\end{cases}\n$$\n\nIndeed, these relations mean that, for $k$ even, the particle with speed $r^k$ will be moving from $A$ towards $B$ at the moment $t = \\frac{2m}{2m+1}$, and it will be at distance $\\frac{2m}{2m+1}$ from $A$, that is, at point $Q$.\n\nFor $k$ odd, the particle with speed $r^k$ will be moving from $B$ towards $A$ at the moment $t = \\frac{2m}{2m+1}$, and it will be at distance $\\frac{1}{2m+1}$ from $B$, hence at point $Q$ again.\n\nSo it remains to prove the displayed equalities, which we do by induction in $k$. The case $k=0$ is obvious. Proceed to the inductive step $k \\to k+1$. The induction hypothesis yields\n\n$$\nk \\text{ odd: } (2m)^{k+1} \\frac{2m}{2m+1} = 2m(2q+1) + \\frac{2m}{2m+1} = 2q' + \\frac{2m}{2m+1}, \\quad q' = 0, 1, 2, \\dots;\n$$\n\n$$\nk \\text{ even: } (2m)^{k+1} \\frac{2m}{2m+1} = 4mq + \\frac{4m^2}{2m+1} = 4mq + 2m - 1 + \\frac{1}{2m+1} = 2q' + 1 + \\frac{1}{2m+1}, \\quad q' = 0, 1, 2, \\dots\n$$\n\nThis completes the induction and the solution.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 13360,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all infinite sequences $a_1, a_2, \\dots$ of positive integers satisfying the following properties:\n\n(a) $a_1 < a_2 < a_3 < \\dots$\n\n(b) There are no positive integers $i, j, k$, not necessarily distinct, such that $a_i + a_j = a_k$.\n\n(c) There are infinitely many positive integers $k$ such that $a_k = 2k - 1$.",
    "options": [],
    "answer": "See solution",
    "solution": "The only solution is $a_k = 2k - 1$ for all $k$, giving the sequence $1, 3, 5, \\dots$.\n\nLet $a_1 = m$. First, we show that for any such sequence, we must have $a_{k+m} - a_k \\ge 2m$ for all positive integers $k$. Suppose for some $k$ that this was not the case. Then $a_k, a_{k+1}, \\dots, a_{k+m}$ are $m+1$ terms of the sequence that are all in the set $S = \\{a_k, a_k+1, \\dots, a_k+2m-1\\}$. Partition $S$ into $m$ two-element sets of the form $\\{b, b+m\\}$. By the pigeonhole principle, one of the $m$ two-element sets is such that both of its elements are terms of the sequence, so that $a_{i_1} + m = a_{i_2}$ for some $k \\le i_1, i_2 \\le k+m$. But we have $a_1 = m$, so this contradicts (b). Thus we have established $a_{k+m} - a_k \\ge 2m$ for all $k$.\n\nNow suppose we have $m$ consecutive terms of the sequence $a_j, a_{j+1}, \\dots, a_{j+m-1}$ such that for all $k$ between $j$ and $j+m-1$ inclusive, $a_k > 2k-1$. By (c), there must exist some index $i$ greater than $j$ satisfying $a_i = 2i-1$. Write $i-j = mq+r$ for $q \\ge 0$ and $0 \\le r < m$ using the division algorithm. Then $a_i \\ge 2m + a_{i-m} \\ge \\dots \\ge 2qm + a_{i-qm} = 2qm + a_{j+r}$. By assumption $a_{j+r} > 2(j+r)-1$, so $a_i > 2qm + 2(j+r) - 1 = 2i-1$, a contradiction. Therefore, for any block of $m$ consecutive terms of the sequence, one of them satisfies $a_k \\le 2k-1$.\n\nBecause of (a), we have $a_k \\ge m + (k-1)$. Consider $a_1, a_2, \\dots, a_m$. The previous inequality implies that for all $i < m$ we have $a_i > 2i-1$, and $a_m \\ge 2m-1$. By the previous paragraph one of these terms must satisfy $a_i \\le 2i-1$. Therefore we must have $a_m = 2m-1$ and by (a) $a_i = m+i-1$ for all $1 \\le i \\le m$.\n\nLikewise, for $a_{m+1}, a_{m+2}, \\dots, a_{2m}$, we have $a_{m+i} \\ge 2m + a_i > 2(m+i) - 1$ for all $i < m$ and $a_{2m} \\ge 4m-1$, so since one of these $m$ terms must satisfy $a_i \\le 2i-1$ we must have $a_{2m} = 4m-1$ and then by (a) $a_{m+i} = 3m+i-1$ for all $1 \\le i \\le m$.\n\nNow suppose $m > 1$. Observe that $a_m + a_m = 4m - 2 = a_{2m-1}$, contradicting (b). Therefore $m = a_1 = 1$. Furthermore, we know from the second paragraph that every term of the sequence satisfies $a_k \\le 2k-1$. At the same time, the first paragraph says $a_{k+1} - a_k \\ge 2$, or $a_k \\ge 2k-2 + a_1 = 2k-1$. Therefore $a_k = 2k-1$ for every $k$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 13361,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $p$ be an odd prime. Find all positive integers $n$ for which $\\sqrt{n^2 - np}$ is a positive integer.",
    "options": [],
    "answer": "See solution",
    "solution": "Assume that $\\sqrt{n^2 - np} = m$ is a positive integer. Then $n^2 - np - m^2 = 0$, so\n\n$$\nn = \\frac{p \\pm \\sqrt{p^2 + 4m^2}}{2}.\n$$\n\nNow $p^2 + 4m^2 = k^2$ for some positive integer $k$, and $n = \\frac{p + k}{2}$ since $k > p$. Thus $p^2 = (k + 2m)(k - 2m)$, and since $p$ is prime, we get $k + 2m = p^2$ and $k - 2m = 1$. Hence $k = \\frac{p^2 + 1}{2}$ and\n\n$$\nn = \\frac{p + \\frac{p^2 + 1}{2}}{2} = \\left(\\frac{p + 1}{2}\\right)^2.\n$$\n\nThis is the only possible value of $n$. In this case,\n\n$$\n\\sqrt{n^2 - pn} = \\sqrt{\\left(\\frac{p + 1}{2}\\right)^4 - p\\left(\\frac{p + 1}{2}\\right)^2} = \\frac{p + 1}{2} \\cdot \\frac{p - 1}{2}.\n$$",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 13362,
    "subject": "Mathematics (Olympiad)",
    "question": "Demuestra que las sumas de las primeras, segundas y terceras potencias de las raíces del polinomio $p(x) = x^3 + 2x^2 + 3x + 4$ son iguales.",
    "options": [],
    "answer": "See solution",
    "solution": "Sean $r$, $s$ y $t$ las raíces (reales o complejas) del polinomio $p(x)$. Definimos $S_n = r^n + s^n + t^n$.\n\nPor las fórmulas de Vièta:\n- $S_1 = r + s + t = -2$\n- $rs + st + tr = 3$\n- $rst = -4$\n\nCalculamos $S_2$:\n$$\nS_2 = r^2 + s^2 + t^2 = (r + s + t)^2 - 2(rs + st + tr) = (-2)^2 - 2 \\times 3 = 4 - 6 = -2\n$$\n\nPara $S_3$, usamos que $p(r) = 0$ (y similar para $s$, $t$):\n$$\np(r) = r^3 + 2r^2 + 3r + 4 = 0 \\implies r^3 = -2r^2 - 3r - 4\n$$\nSumando para todas las raíces:\n$$\nS_3 = r^3 + s^3 + t^3 = -2S_2 - 3S_1 - 12 = -2 \\times (-2) - 3 \\times (-2) - 12 = 4 + 6 - 12 = -2\n$$\n\nPor lo tanto, $S_1 = S_2 = S_3 = -2$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 13363,
    "subject": "Mathematics (Olympiad)",
    "question": "Is there a set of distinct positive integers whose reciprocals sum to $2020$? If so, construct such a set.",
    "options": [],
    "answer": "See solution",
    "solution": "Yes, such a set exists. We construct it in two stages.\n\nReaders familiar with Egyptian Fractions know that any rational number $0 < x < 1$ can be expressed as a sum of reciprocals of distinct integers. The greedy algorithm delivers such an expression: at each step, choose the smallest integer $s$ for which $\\frac{1}{s} \\le x$; if equality holds, we stop, otherwise we continue with $x - \\frac{1}{s}$.\n\n**Termination and Distinctness:**\nWrite $x = \\frac{p}{q}$ in lowest terms, $1 \\le p < q$. If $p = 1$, the result is immediate. Otherwise, there exists a unique integer $s \\ge 2$ such that\n\n$$\n\\frac{1}{s} \\le x < \\frac{1}{s-1}.\n$$\n\nSince $x - \\frac{1}{s} < \\frac{1}{s-1} - \\frac{1}{s} = \\frac{1}{s(s-1)}$, the next denominator must be greater than $s(s-1) \\ge s$, ensuring distinct denominators.\n\nTo prove termination, note that the numerators decrease strictly. From $\\frac{p}{q} < \\frac{1}{s-1}$, we get $ps - p < q$, or $ps - q < p$. Thus,\n\n$$\n0 \\le \\frac{p}{q} - \\frac{1}{s} = \\frac{ps - q}{q}\n$$\n\nhas a non-negative numerator smaller than $p$, and this remains true in lowest terms. Since there cannot be an infinite strictly decreasing sequence of non-negative integers, the algorithm terminates. Thus, any rational $0 < x < 1$ can be written as a sum of reciprocals of distinct integers.\n\n**Extension to $x \\ge 1$:**\nFor $x \\ge 1$, specifically $x = 2020$, the harmonic series diverges, so for any $x \\ge 1$ there is an integer $m \\ge 1$ such that\n\n$$\n\\sum_{j=1}^{m} \\frac{1}{j} \\le x < \\sum_{j=1}^{m+1} \\frac{1}{j}.\n$$\n\nApply the previous result to\n\n$$\nx - \\sum_{j=1}^{m} \\frac{1}{j}.\n$$\n\nAs this is less than $\\frac{1}{1+m}$, the denominators remain distinct, and the result follows.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 13364,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all pairs of coprime positive integers $m$ and $n$ such that the sum of their greatest common divisor and their least common multiple is $101$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $d$ be the greatest common divisor of $m$ and $n$. Then $m = d m_1$ and $n = d n_1$, where $m_1$ and $n_1$ are coprime. The least common multiple of $m$ and $n$ is $d m_1 n_1$. We have\n\n$$\n101 = d + d m_1 n_1 = d(1 + m_1 n_1).\n$$\n\nSince $1 + m_1 n_1 \\geq 2$ and $101$ is prime, the only possibility is $d = 1$ and $m_1 n_1 = 100$. Thus, $m = m_1$ and $n = n_1$ are coprime positive integers whose product is $100$.\n\nThe coprime pairs $(m, n)$ with $m n = 100$ are $(1, 100)$, $(4, 25)$, $(25, 4)$, and $(100, 1)$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 13365,
    "subject": "Mathematics (Olympiad)",
    "question": "Consider a particle moving on the $xy$-plane according to the following rule: if the particle is at the point $(n, m)$ where $n$ and $m$ are integers, it moves in one step either to the point $(n+1, m)$ or to the point $(n, m+1)$. Consider the set $\\Omega$ of all paths that a particle can take to start at the origin $(0, 0)$ and take steps, following the rule as above to reach the point $(46, 5)$. Among these paths, how many are there?",
    "options": [],
    "answer": "See solution",
    "solution": "$$\n\\boxed{2349060}\n$$\n\nThe number of such paths is given by the binomial coefficient $\\binom{51}{5}$, since each path consists of 46 steps to the right and 5 steps up, in any order. Thus, the answer is $\\boxed{2349060}$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 13366,
    "subject": "Mathematics (Olympiad)",
    "question": "Prove that there exists a positive integer $N$ such that for every number $n$ from $1$ to $99$, the number $\\overline{Nn}$ (the digits of $n$ are written right after the digits of $N$) is composite.",
    "options": [],
    "answer": "See solution",
    "solution": "For example, take $N = 11 \\cdot 12 \\cdots 19 \\cdot 110 \\cdot 111 \\cdots 199 + 1 = M + 1$.  \n\nIf $1 \\leq n \\leq 9$, then $\\overline{Nn} = 10N + n = 10M + (10 + n)$. Since $M$ is divisible by $11, 12, \\ldots, 19$, $\\overline{Nn}$ is divisible by $10 + n$, so it is composite.\n\nIf $10 \\leq n \\leq 99$, then $\\overline{Nn} = 100N + n = 100M + (100 + n)$. Since $M$ is divisible by $110, 111, \\ldots, 199$, $\\overline{Nn}$ is divisible by $100 + n$, so it is again composite.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 13367,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $S$ be a subset of the set of numbers $\\{1, 2, 3, \\ldots, 2008\\}$ which consists of 756 distinct numbers. Show that there are two distinct elements $a, b$ of $S$ such that $a + b$ is divisible by 8.",
    "options": [],
    "answer": "See solution",
    "solution": "Suppose that no such pair exists.\n\nWe split the set $\\{1, 2, 3, \\ldots, 2008\\}$ into 8 subsets $S_0, S_1, \\ldots, S_7$, where $S_i$ contains the numbers in $S$ congruent to $i$ modulo 8. Since $\\frac{2008}{8} = 251$,\n\n$$\n|S_0| = |S_1| = \\ldots = |S_7| = 251.\n$$\n\nIf $S$ contains an element from $S_1$, then it cannot contain an element from $S_7$ (because otherwise their sum would be $1 + 7 \\equiv 0 \\pmod{8}$).\n\n$S$ can therefore contain at most 251 numbers from the combined set $S_1 \\cup S_7$. Similarly, $S$ contains at most 251 numbers from $S_2 \\cup S_6$ and at most 251 from $S_3 \\cup S_5$.\n\nWe also know that $S$ contains at most 1 element from $S_0$, because $0+0 \\equiv 0 \\pmod{8}$, and at most 1 from $S_4$, because $4+4 \\equiv 0 \\pmod{8}$.\n\nSo $S$ can contain at most $251 \\times 3 + 2 = 755$ elements without having two whose sum is divisible by 8. But $S$ has 756 elements. Therefore there must exist two whose sum is divisible by 8.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 13368,
    "subject": "Mathematics (Olympiad)",
    "question": "Consider six points in the interior of a square of side length $3$. Prove that among the six points, there are two whose distance is less than $2$.\n\n![](images/RMC2014_p71_data_26250667b5.png)",
    "options": [],
    "answer": "See solution",
    "solution": "Let us split the square into five rectangles, as shown in the figure above. The sizes of the three rectangles at the bottom are $1 \\times \\sqrt{3}$, so the length of their diagonal is $2$. The sizes of the two rectangles at the top are $1.5 \\times (3 - \\sqrt{3})$. Since $1.5^2 + (3 - \\sqrt{3})^2 < 4$, the length of their diagonal is less than $2$. By the pigeonhole principle, there are two points among the six given which are situated within or on the sides of one of the five rectangles, so the distance between them is at most equal to the length of the diagonal, which is at most $2$. However, the distance can be $2$ only if the two points are opposite vertices of one of the bottom rectangles, which implies that one of the six points is not inside, but on one side of the square—a contradiction.\n\n**Remarks. 1.** Splitting the square into three $1 \\times 1.7$ rectangles and two $1.5 \\times 1.3$ rectangles, we may prove that the result is still true if the six given points are inside or on the sides of the square.\n\n**2.** Notice that five points are not enough to draw the same conclusion (taking the vertices and center of the square, the shortest distance is $3/\\sqrt{2} > 2$). The strongest result known in literature is that the shortest distance (for $6$ points) is $\\sqrt{13}/2 \\approx 9/5 < 2$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 13369,
    "subject": "Mathematics (Olympiad)",
    "question": "A student divides 30 marbles into 5 boxes labelled 1, 2, 3, 4, 5 (there may be a box without marbles).\n\n**a)** How many ways are there to divide marbles into boxes (two ways are different if there is a box with different number of marbles)?\n\n**b)** After dividing, this student paints those marbles by a number of colors (each marble has one color, one color can be painted for many marbles), such that there does not exist 2 marbles in the same box with the same color, and from any 2 boxes, it is impossible to choose 8 marbles painted in 4 colors. Prove that for every division, the student must use at least 10 colors to paint the marbles.\n\n**c)** Find a division so that the student can use exactly 10 colors to paint the marbles that satisfy the conditions in question b).",
    "options": [],
    "answer": "See solution",
    "solution": "**a)** It's well known that there are $\\binom{n+k-1}{k-1}$ ways to divide $n$ marbles into $k$ boxes. In this case, the answer is $\\binom{34}{4}$.\n\n**b)** Let $m$ be the number of colors, $x_1, x_2, \\dots, x_m$ be the number of boxes containing a marble with color 1, 2, ..., $m$ respectively. We now count the number of tuples $(\\{A, B\\}, C)$, where $A, B$ are boxes having marbles with the same color $C$.\n\nOn the one hand, since every two boxes have in common at most 3 colors, the number of pairs is at most $3 \\binom{5}{2} = 30$.\n\nOn the other hand, the number of pairs is $S = \\sum_{i=1}^{m} \\binom{x_i}{2}$. Since in each box, there is at most one marble of each color, we get $\\sum_{i=1}^{m} x_i = 30$. By the Cauchy-Schwarz inequality, we have\n\n$$\nS = \\frac{1}{2} \\left( \\sum_{i=1}^{m} x_i^2 - \\sum_{i=1}^{m} x_i \\right) \\geq \\frac{1}{2} \\left( \\frac{30^2}{m} - 30 \\right).\n$$\n\nHence,\n\n$$\n\\frac{900}{m} - 30 \\leq 60 \\iff m \\geq 10.\n$$\n\n**c)** Consider the following table.\n\n![]( <table><thead><tr><th>Box</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></tr></thead><tbody><tr><th>1</th><td>\\times</td><td>\\times</td><td>\\times</td><td>\\times</td><td>\\times</td><td>\\times</td><td></td><td></td><td></td><td></td></tr><tr><th>2</th><td>\\times</td><td>\\times</td><td>\\times</td><td></td><td></td><td></td><td>\\times</td><td></td><td>\\times</td><td>\\times</td></tr><tr><th>3</th><td>\\times</td><td></td><td></td><td>\\times</td><td>\\times</td><td></td><td></td><td>\\times</td><td>\\times</td><td>\\times</td></tr><tr><th>4</th><td></td><td>\\times</td><td></td><td>\\times</td><td></td><td>\\times</td><td>\\times</td><td>\\times</td><td></td><td>\\times</td></tr><tr><th>5</th><td></td><td></td><td>\\times</td><td></td><td>\\times</td><td>\\times</td><td>\\times</td><td>\\times</td><td>\\times</td><td></td></tr></tbody></table> )\n\nIt is a direct checking that the table satisfies the requirements. $\\Box$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 13370,
    "subject": "Mathematics (Olympiad)",
    "question": "a) Factorize $xy - x - y + 1$.\n\nb) Prove that if integers $a$ and $b$ satisfy $|a + b| > |1 + ab|$, then $ab = 0$.",
    "options": [],
    "answer": "See solution",
    "solution": "a) $xy - x - y + 1 = (x - 1)(y - 1)$.\n\nb) Both sides of the inequality are positive, so we can square both sides to get the equivalent form:\n$$\begin{align*}\n(a + b)^2 &> (1 + ab)^2 \\\\\na^2 + 2ab + b^2 &> 1 + 2ab + a^2b^2 \\\\\na^2 + b^2 - 1 &> a^2b^2 \\\\\na^2 + b^2 - 1 - a^2b^2 &> 0 \\\\\n(a^2 - 1)(b^2 - 1) &< 0\n\\end{align*}$$\nThis shows that $a^2 - 1 < 0$ or $b^2 - 1 < 0$. Since $a$ and $b$ are integers, this yields $a = 0$ or $b = 0$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 13371,
    "subject": "Mathematics (Olympiad)",
    "question": "Suppose four solid iron balls are placed in a cylinder with a radius of $1\\ \\mathrm{cm}$, 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 \\_\\_\\_\\_ cm$^3$ water.",
    "options": [],
    "answer": "See solution",
    "solution": "Let points $O_1, O_2, O_3, O_4$ be the centers of the four solid iron balls, with $O_1, O_2$ belonging to the two balls in the lower layer. Let $A, B, C, D$ be the projections of $O_1, O_2, O_3, O_4$ onto the base of the cylinder. $ABCD$ forms a square with side $\\frac{\\sqrt{2}}{2}$. So the height of the water in the cylinder must be $1 + \\frac{\\sqrt{2}}{2}$ to just immerse all the balls. Hence, the volume of water needed 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": 13372,
    "subject": "Mathematics (Olympiad)",
    "question": "There is a polynomial $P(x)$ with integer coefficients such that\n\n$$\nP(x) = \\frac{(x^{2310} - 1)^6}{(x^{105} - 1)(x^{70} - 1)(x^{42} - 1)(x^{30} - 1)}\n$$\n\nfor every $0 < x < 1$. Find the coefficient of $x^{2022}$ in $P(x)$.",
    "options": [],
    "answer": "See solution",
    "solution": "For $0 < x < 1$, the given rational expression can be rewritten as\n\n$$\n(x^{2310} - 1)^2 \\cdot \\frac{x^{2310} - 1}{x^{105} - 1} \\cdot \\frac{x^{2310} - 1}{x^{70} - 1} \\cdot \\frac{x^{2310} - 1}{x^{42} - 1} \\cdot \\frac{x^{2310} - 1}{x^{30} - 1}\n$$\n\nThis equals\n\n$$\n(x^{2310} - 1)^2 \\cdot \\sum_{a=0}^{21} x^{105a} \\cdot \\sum_{b=0}^{32} x^{70b} \\cdot \\sum_{c=0}^{54} x^{42c} \\cdot \\sum_{d=0}^{76} x^{30d}\n$$\n\nThe coefficient of $x^{2022}$ is the number of quadruples of nonnegative integers $(a, b, c, d)$ such that\n\n$$\n105a + 70b + 42c + 30d = 2022.\n$$\n\nConsidering this equation modulo $2$, $3$, $5$, and $7$ gives:\n\n- $a \\equiv 0 \\pmod{2}$\n- $b \\equiv 0 \\pmod{3}$\n- $c \\equiv 1 \\pmod{5}$\n- $d \\equiv 3 \\pmod{7}$\n\nSo, set $a = 2w$, $b = 3x$, $c = 5y + 1$, $d = 7z + 3$ for nonnegative integers $w, x, y, z$. The equation becomes\n\n$$\n\\begin{align*}\n2022 &= 105(2w) + 70(3x) + 42(5y + 1) + 30(7z + 3) \\\\\n&= 210(w + x + y + z) + 132\n\\end{align*}\n$$\n\nSo $w + x + y + z = 9$. The number of nonnegative integer solutions is $\\binom{9+3}{3} = 220$.",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 13373,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $S_n$ be the number of $2n$-tuples $(p_1, p_2, \\dots, p_n; q_1, q_2, \\dots, q_n)$ of positive integers such that $p_1 = q_n = 1$ and $p_{i+1}q_i - p_iq_{i+1} = 1$ for all $i = 1, 2, \\dots, n-1$. Semicolons separate the $p$ and $q$ sequences for clarity.\n\nFind a formula for $S_n$ and compute $S_{10}$.",
    "options": [],
    "answer": "See solution",
    "solution": "We first prove that there uniquely exists $i$ such that $p_i = q_i = 1$.\n\nAssume that there does not exist such $i$. Then $q_1 \\neq 1$ and $p_n \\neq 1$. Since the sequence $p_j/q_j$ is increasing, there exists $j$ such that\n\n$$\n\\frac{p_j}{q_j} < 1 < \\frac{p_{j+1}}{q_{j+1}}.\n$$\n\nBut then\n\n$$\n\\frac{p_{j+1}}{q_{j+1}} - \\frac{p_j}{q_j} \\ge \\frac{1}{q_{j+1}} + \\frac{1}{q_j} \\ge \\frac{2}{q_j q_{j+1}},\n$$\n\nhence $p_{j+1}q_j - p_jq_{j+1} \\ge 2$, a contradiction. So there exists $i$ such that $p_i = q_i = 1$. Uniqueness is clear.\n\nNow, consider the number of tuples for fixed $i$ with $p_i = q_i = 1$.\n\n**Case 1: $i=1$**\n\nLet $(1, p_2, \\dots, p_n; 1, q_2, \\dots, q_n)$ meet the condition. Let $p'_j = p_j - q_j$. Then the $2(n-1)$-tuple $(p'_2, p'_3, \\dots, p'_{n-1}; q_2, q_3, \\dots, q_n)$ meets the condition. Conversely, given a $2(n-1)$-tuple $(p_1, \\dots, p_{n-1}; q_1, \\dots, q_{n-1})$ meeting the condition, let $p'_j = p_j + q_j$. Then $(1, p'_1, p'_2, \\dots, p'_{n-1}; 1, q_1, q_2, \\dots, q_{n-1})$ meets the condition. These operations are inverses, so the number of such $2n$-tuples is $S_{n-1}$.\n\n**Case 2: $1 < i < n$**\n\nLet $(p_1, \\dots, p_{i-1}, 1, p_{i+1}, \\dots, p_n; q_1, \\dots, q_{i-1}, 1, q_{i+1}, \\dots, q_n)$ meet the condition. Let $q'_j = q_j - p_j$ and $p''_j = p_j - q_j$. Then $(p_1, \\dots, p_{i-1}; q'_1, \\dots, q'_{i-1})$ and $(p''_{i+1}, \\dots, p''_n; q_{i+1}, \\dots, q_n)$ meet the condition. Conversely, given such tuples, we can reconstruct the original tuple. Thus, the number is $S_{i-1}S_{n-i}$.\n\n**Case 3: $i=n$**\n\nSimilarly to $i=1$, the number is $S_{n-1}$.\n\nTherefore,\n\n$$\nS_n = S_{n-1} + S_1 S_{n-2} + \\dots + S_{n-2} S_1 + S_{n-1}.\n$$\n\nWith $S_1 = 1$, we compute $S_{10} = 16796$.\n\nIn fact, $S_n = \\frac{1}{n+1} \\binom{2n}{n}$, the *Catalan numbers*.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 13374,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $p$ be an odd prime number and $a, b, c$ be integers such that the integers\n\n$$\na^{2023} + b^{2023}, \\quad b^{2024} + c^{2024}, \\quad c^{2025} + a^{2025}\n$$\n\nare all divisible by $p$. Prove that $p$ divides each of $a$, $b$, and $c$.",
    "options": [],
    "answer": "See solution",
    "solution": "Set $k = 2023$. If one of $a, b, c$ is divisible by $p$, then all of them are. Indeed, for example, if $p \\mid a$, then $p \\mid a^k + b^k$ implies $p \\mid b$, and then $p \\mid b^{k+1} + c^{k+1}$ implies $p \\mid c$. The other cases follow similarly.\n\nSo for the sake of contradiction, assume none of $a, b, c$ is divisible by $p$. Then\n\n$$\na^{k(k+2)} \\equiv (a^k)^{k+2} \\equiv (-b^k)^{k+2} \\equiv -b^{k(k+2)} \\pmod{p}\n$$\n\nand\n\n$$\na^{k(k+2)} \\equiv (a^{k+2})^k = (-c^{k+2})^k \\equiv -c^{k(k+2)} \\pmod{p}.\n$$\n\nSo $b^{k(k+2)} \\equiv c^{k(k+2)} \\pmod{p}$. But then\n\n$$\nc^{k(k+2)} \\cdot c \\equiv c^{(k+1)^2} \\equiv (-b^{k+1})^{k+1} \\equiv b^{(k+1)^2} \\equiv b^{k(k+2)} \\cdot b \\equiv c^{k(k+2)} \\cdot b \\pmod{p}\n$$\n\nwhich forces $b \\equiv c \\pmod{p}$. Thus\n\n$$\n0 \\equiv b^{k+1} + c^{k+1} = 2b^{k+1} \\pmod{p}\n$$\n\nimplying $p \\mid b$, a contradiction. Thus the proof is complete.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 13375,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $(s_n)$ be a sequence defined for all $n \\in \\mathbb{Z}$ such that $s_{n+1} = 4s_n - s_{n-1}$ for all $n$. Such a sequence is called *subaveraging*.\n\n(a) Show that $(s_n)$ is completely determined by $s_0$ and $s_1$. For $s_0 = 0$ and $s_1 = 1$, prove by induction that $0 \\leq s_n < s_{n+1}$ for all $n \\geq 0$, and that $s_{-n} = -s_n$ for all $n \\geq 1$. Deduce that all entries of $(s_n)$ are distinct.\n\n(b) Show that if $(s_k)$ is subaveraging, then so are the shifted sequence $t_k = s_{k+n}$ and the reflected sequence $r_k = s_{-k}$. Suppose $(s_k)$ is a subaveraging sequence with $s_n = s_m$ for some $n < m$. Prove that $s_{k+n} = s_{m-k}$ for all $k \\in \\mathbb{Z}$.",
    "options": [],
    "answer": "See solution",
    "solution": "(a) The recursion $s_{n+1} = 4s_n - s_{n-1}$ allows us to compute all terms from $s_0$ and $s_1$. For $s_0 = 0$ and $s_1 = 1$, we use induction:\n\n- Base case: $0 \\leq s_0 < s_1$ is clear.\n- Inductive step: Assume $0 \\leq s_n < s_{n+1}$ for some $n \\geq 0$. Then $s_{n+1} = 4s_n - s_{n-1} > 3s_n \\geq s_n \\geq 0$ for $n \\geq 1$, so $s_{n+1} > s_n \\geq 0$.\n\nFor $n \\geq 1$, $s_{-1} = -1$ and the recursion gives $s_{-n} = -s_n$, so $s_{n-1} < s_n \\leq 0$ for $n \\leq 0$. Thus, all entries of $(s_n)$ are distinct.\n\n(b) The shifted sequence $t_k = s_{k+n}$ and the reflected sequence $r_k = s_{-k}$ both satisfy the same recursion, so they are subaveraging. If $s_n = s_m$ for $n < m$, consider $t_k = s_{k+n}$ and $r_k = s_{m-k}$. Both have $t_0 = r_0 = s_n$ and $t_{m-n} = r_{m-n} = s_n$. The difference $d_k = t_k - r_k$ is subaveraging, with $d_0 = d_{m-n} = 0$. If $d_1 \\neq 0$, then $q_k = d_k/d_1$ is subaveraging with $q_0 = 0$, $q_1 = 1$, so all $q_k$ are distinct, contradicting $q_{m-n} = 0$ for $m-n > 0$. Thus, $d_1 = 0$ and $d_k = 0$ for all $k$, so $t_k = r_k$, i.e., $s_{k+n} = s_{m-k}$ for all $k \\in \\mathbb{Z}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 13376,
    "subject": "Mathematics (Olympiad)",
    "question": "Find the minimum value of $P(4)$, where $P(x)$ is a polynomial with integer coefficients less than 5, such that $P(5) = 2011$.",
    "options": [],
    "answer": "See solution",
    "solution": "We convert $2011$ to base $5$:\n\n$$\n2011 \\div 5 = 402 \\text{ with remainder } 1\n$$\n$$\n402 \\div 5 = 80 \\text{ with remainder } 2\n$$\n$$\n80 \\div 5 = 16 \\text{ with remainder } 0\n$$\n$$\n16 \\div 5 = 3 \\text{ with remainder } 1\n$$\n$$\n3 \\div 5 = 0 \\text{ with remainder } 3\n$$\n\nSo $2011 = 31021_5$.\n\nThus, $P(x) = 1 + 2x + 0x^2 + 1x^3 + 3x^4$ is the unique polynomial with all coefficients less than $5$ and $P(5) = 2011$.\n\nTherefore, the minimum value of $P(4)$ is:\n\n$$\nP(4) = 1 + 2 \\times 4 + 0 \\times 16 + 1 \\times 64 + 3 \\times 256 = 1 + 8 + 0 + 64 + 768 = 841\n$$",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 13377,
    "subject": "Mathematics (Olympiad)",
    "question": "Show that\n\n$$\n\\cos(56^{\\circ}) \\cdot \\cos(2 \\cdot 56^{\\circ}) \\cdot \\dots \\cdot \\cos(2^{23} \\cdot 56^{\\circ}) = \\frac{1}{2^{23}}\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "We start by rewriting the expression as follows:\n\n$$\n\\cos(56^{\\circ}) \\cdot \\cos(2 \\cdot 56^{\\circ}) \\cdot \\dots \\cdot \\cos(2^{23} \\cdot 56^{\\circ}) = \\frac{\\sin(56^{\\circ}) \\cdot \\cos(56^{\\circ}) \\cdot \\cos(2 \\cdot 56^{\\circ}) \\cdot \\dots \\cdot \\cos(2^{23} \\cdot 56^{\\circ})}{\\sin(56^{\\circ})}\n$$\n\nNow, by applying the addition formula $\\sin(x) \\cos(x) = \\frac{\\sin(2x)}{2}$, we obtain\n\n$$\n\\frac{\\sin(56^{\\circ}) \\cdot \\cos(56^{\\circ}) \\cdot \\cos(2 \\cdot 56^{\\circ}) \\cdot \\dots \\cdot \\cos(2^{23} \\cdot 56^{\\circ})}{\\sin(56^{\\circ})} = \\frac{\\sin(2 \\cdot 56^{\\circ}) \\cdot \\cos(2 \\cdot 56^{\\circ}) \\cdot \\dots \\cdot \\cos(2^{23} \\cdot 56^{\\circ})}{2 \\cdot \\sin(56^{\\circ})}\n$$\n\nWe observe that we can do the same trick again. In this way, by applying the addition formula 23 times, we get\n\n$$\n\\cos(56^{\\circ}) \\cdot \\cos(2 \\cdot 56^{\\circ}) \\cdot \\dots \\cdot \\cos(2^{23} \\cdot 56^{\\circ}) = \\frac{\\sin(2^{24} \\cdot 56^{\\circ})}{2^{23} \\cdot \\sin(56^{\\circ})}\n$$\n\nThe last step is to prove that $\\sin(2^{24} \\cdot 56^{\\circ}) = \\sin 56^{\\circ}$. If we can show that\n\n$$\n2^{24} \\cdot 56 = 360k + 56\n$$\n\nfor some integer $k$, then the desired equality follows by the periodicity of $\\sin$. We have\n\n$$\nk = \\frac{2^{24} \\cdot 56 - 56}{360} = 7 \\cdot \\frac{2^{24} - 1}{45},\n$$\n\nand since $\\varphi(45) = 24$, the Euler-Fermat theorem implies that $k$ is indeed an integer, as claimed. $\\blacktriangledown$\n\n![](images/Baltic_Way_SHL_2009-11_13-16_p234_data_95588336de.png)",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 13378,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $m$ and $n$ be positive integers, neither divisible by $6$. A rectangle of size $m \\times n$ is filled with $2 \\times 2$ and $3 \\times 3$ squares. Show that such a rectangle can be filled with squares of only one of the types: $2 \\times 2$ or $3 \\times 3$.",
    "options": [],
    "answer": "See solution",
    "solution": "Consider the case when one of the numbers is not divisible by $2$ and the other is not divisible by $3$. Without loss of generality, let $m$ not be divisible by $2$, and $n$ not be divisible by $3$, with $m$ as the width and $n$ as the height of the rectangle.\n\nWe color in black all the rows of the rectangle $m \\times n$ of the form $3k$ and $3k+2$, where $k \\in \\mathbb{N}$. The number of black rows is odd: $2\\left[\\frac{n}{3}\\right] + 1$, so there is an odd number of black cells, since each row has an odd ($m$) number of cells. Since every $2 \\times 2$ and $3 \\times 3$ square covers an even number of black cells, it is not possible to cover the rectangle with both $2 \\times 2$ and $3 \\times 3$ squares.\n\nIt suffices to consider the case when both $m$ and $n$ are divisible by $2$ or $3$. Then it is clear how to divide the rectangle into $2 \\times 2$ squares (in the first case) or $3 \\times 3$ squares (in the second case).\n\n![](images/UkraineMO2019_booklet_p14_data_2f6287b89f.png)",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 13379,
    "subject": "Mathematics (Olympiad)",
    "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.",
    "options": [],
    "answer": "See solution",
    "solution": "Suppose that a positive integer $a$ satisfies $p^n + 1 = a^3$ (clearly $a \\ge 2$). We rewrite the equality as\n\n$$\np^n = a^3 - 1 = (a-1)(a^2 + a + 1).\n$$\n\nIt follows that if $a > 2$, the numbers $a-1$ and $a^2+a+1$ are powers of $p$ (with positive integer exponents).\n\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\n$$\np^k \\mid p^{2k} + 3p^k + 3 \\Rightarrow p^k \\mid 3.\n$$\n\nThen $p=3$ and $k=1$, hence $a = p^k + 1 = 4$. However, $a^2 + a + 1 = 21$ is not a power of three, therefore if $a > 2$, the expression $p^n + 1$ is never a cube.\n\nFor $a=2$ we get $p^n = 7$, hence $p=7$ is the only such prime.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 13380,
    "subject": "Mathematics (Olympiad)",
    "question": "A set with 8 elements is colored using two colors. A subset is called *monochromatic* if all its elements are the same color, and *bichromatic* otherwise. What is the greatest possible number $N$ of bichromatic subsets?",
    "options": [],
    "answer": "See solution",
    "solution": "If exactly 4 elements are painted one color and the other 4 are painted the other color, the number of monochromatic subsets is $M = (2^4 - 1) + (2^4 - 1) = 2^5 - 2 = 30$. The total number of nonempty subsets is $2^8 - 1 = 255$, so the maximum number of bichromatic subsets is $N = 255 - 30 = 225$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 13381,
    "subject": "Mathematics (Olympiad)",
    "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": "See solution",
    "solution": "The maximum is $40$.\n\nFirst, 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 $x_{3k+1} = 1$, $x_{3k+2} = 0$, $x_{3k+3} = -1$ for $k = 0, 1, \\dots, 19$,\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": 13382,
    "subject": "Mathematics (Olympiad)",
    "question": "Prove that the inequality\n\n$$\nx^2\\sqrt{1+2y^2} + y^2\\sqrt{1+2x^2} \\ge xy(x+y+\\sqrt{2})\n$$\n\nholds for any two real numbers $x$ and $y$.",
    "options": [],
    "answer": "See solution",
    "solution": "By the quadratic-arithmetic mean inequality, we have\n\n$$\n1 + 2t^2 \\ge \\frac{1}{2}(1 + t\\sqrt{2})^2.\n$$\n\nIt follows that\n\n$$\n\\begin{align*}\nx^2 \\sqrt{1+2y^2} + y^2 \\sqrt{1+2x^2} &\\ge x^2 \\cdot \\frac{1+y\\sqrt{2}}{2} + y^2 \\cdot \\frac{1+x\\sqrt{2}}{2} \\\\\n&= \\frac{x^2 + y^2 + xy(x+y)\\sqrt{2}}{\\sqrt{2}} \\\\\n&\\ge \\frac{2xy + xy(x+y)\\sqrt{2}}{\\sqrt{2}} = xy(x+y+\\sqrt{2}).\n\\end{align*}\n$$\n\nThis is what we need to prove. $\\square$",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 13383,
    "subject": "Mathematics (Olympiad)",
    "question": "There are 24 pieces each of two different kinds of cakes $A$ and $B$. All the cakes will be distributed among three people $X$, $Y$, and $Z$. A method of distribution where a person may not receive any cake of a particular kind is allowed. How many different ways of distributing cakes are there such that no pair of people ends up with the following situation?\n\nOne person in the pair receives $a$ pieces of $A$ and $b$ pieces of $B$, and the other person in the pair receives $a'$ pieces of $A$ and $b'$ pieces of $B$, where all of the conditions $a \\leq a'$, $b \\leq b'$, and $a + b < a' + b'$ are satisfied.",
    "options": [],
    "answer": "See solution",
    "solution": "There are $14\\,017$ ways.\n\nLet $a_X$, $a_Y$, $a_Z$ be the numbers of cakes of type $A$ received by $X$, $Y$, and $Z$ respectively, and $b_X$, $b_Y$, $b_Z$ the numbers of cakes of type $B$ received by $X$, $Y$, and $Z$. These are non-negative integers with $a_X + a_Y + a_Z = b_X + b_Y + b_Z = 24$.\n\nIf $a_X < a_Y$, then by the problem's condition, we must have either $b_X > b_Y$ or $a_X + b_X \\geq a_Y + b_Y$. But if $a_X < a_Y$ and $a_X + b_X \\geq a_Y + b_Y$, then $b_X > b_Y$ must hold. Thus, $a_X < a_Y \\implies b_X > b_Y$. Similarly, $a_X > a_Y \\implies b_X < b_Y$, $b_X < b_Y \\implies a_X > a_Y$, and $b_X > b_Y \\implies a_X < a_Y$.\n\nTherefore, for $X$ and $Y$, one of the following must hold:\n\n- $a_X < a_Y$ and $b_X > b_Y$\n- $a_X = a_Y$ and $b_X = b_Y$\n- $a_X > a_Y$ and $b_X < b_Y$\n\nThe same applies for $X$ and $Z$, and $Y$ and $Z$.\n\nThe number of triples $(x, y, z)$ of non-negative integers with $x + y + z = 24$ is $\\binom{24 + 3 - 1}{3 - 1} = \\binom{26}{2} = 325$.\n\nClassifying by the relative order of $x, y, z$:\n\n- $x = y = z$: only $(8, 8, 8)$, so $1$ triple.\n- $x = y < z$: $(k, k, 24 - 2k)$, $0 \\leq k \\leq 7$, so $8$ triples. Similarly for $y = z < x$ and $z = x < y$.\n- $x = y > z$: $(k, k, 24 - 2k)$, $9 \\leq k \\leq 12$, so $4$ triples. Similarly for $y = z > x$ and $z = x > y$.\n- All distinct: $325 - 1 - 8 \\times 3 - 4 \\times 3 = 288$ triples. There are $6$ orderings, so $48$ for each ordering.\n\nNow, count the cases for $a_X, a_Y, a_Z$:\n\n- $a_X = a_Y = a_Z$: $b_X = b_Y = b_Z$, $1 \\times 1 = 1$ way.\n- $a_X = a_Y < a_Z$: $b_X = b_Y > b_Z$, $8 \\times 4 = 32$ ways. Similarly for $a_Y = a_Z < a_X$ and $a_Z = a_X < a_Y$.\n- $a_X = a_Y > a_Z$: $b_X = b_Y < b_Z$, $4 \\times 8 = 32$ ways. Similarly for $a_Y = a_Z > a_X$ and $a_Z = a_X > a_Y$.\n- $a_X < a_Y < a_Z$: $b_X > b_Y > b_Z$, $48 \\times 48 = 2304$ ways. Similarly for the other $5$ orderings.\n\nSumming up:\n\n$$\n1 \\times 1 + 32 \\times 3 + 32 \\times 3 + 2304 \\times 6 = 14\\,017.\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 13384,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all positive integers $N$ having an even number of digits (no leading zeroes) such that, if we insert a multiplication sign after the first $n$ digits of $N$, the result of the multiplication is a divisor of $N$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $N = \\overline{a_1a_2\\dots a_n b_1b_2\\dots b_n}$ be the $2n$-digit number, with $a_1 \\ge 1$. Also let $A = \\overline{a_1a_2\\dots a_n}$ and $B = \\overline{b_1b_2\\dots b_n}$, so by hypothesis we are given $10^n A + B = N = kAB$ for some positive integer $k$. Multiplying with $k$ and adding $10^n$, we arrive at $$(kA - 1)(kB - 10^n) = 10^n.$$  \nFor $n = 1$ we immediately solve $$(kA-1)(kB-10) = 10,$$ with $A$ and $B$ not-null digits, to obtain $N \\in \\{11, 12, 15, 24, 36\\}$.  \nFor $n \\ge 2$ we have $10^{n-1} \\le A < 10^n$ and $1 \\le B < 10^n$, thus $1 < k < 10$. A case-by-case discussion follows:\n\n* $k=2$. Then $2 \\cdot 10^{n-1} - 1 \\le 2A - 1 \\mid 10^n$, so $2A - 1 \\le 5^n$, forcing $n \\le 2$. From $(2A-1)(2B-100) = 100$ we get $N = 1352$.\n* $k=3$. Then $3 \\cdot 10^{n-1} - 1 \\le 3A - 1 \\mid 10^n$. We cannot have $3A - 1 = 10^n$, impossible modulo 3. We cannot have $3A - 1 = 2 \\cdot 10^{n-1}$, because it is too small. For $3A - 1 = 5 \\cdot 10^{n-1}$, we get $A = \\frac{5 \\cdot 10^{n-1} + 1}{3}$ and $B = 2A = \\frac{10^n + 2}{3}$ (for example, for $n=2$ we get $N = 1734$). Thus we get a first infinite family of solutions.\n* $k=4$. Then $4 \\cdot 10^{n-1} - 1 \\le 4A - 1 \\mid 10^n$, so $4A - 1 \\le 5^n$, impossible.\n* $k=5$. Then $5 \\cdot 10^{n-1} - 1 \\le 5A - 1 \\mid 10^n$, so $5A - 1 \\le 2^n$, impossible.\n* $k=6$. Then $6 \\cdot 10^{n-1} - 1 \\le 6A - 1 \\mid 10^n$, so $6A - 1 \\le 10^n$, which forces $6A - 1 = 10^n$, impossible modulo 3.\n* $k=7$. Then $7 \\cdot 10^{n-1} - 1 \\le 7A - 1 \\mid 10^n$, so $7A - 1 \\le 10^n$, which forces (as in the above) $7A - 1 = 10^n$. This leads to $A = B = \\frac{10^n + 1}{7}$, with only proviso that $n$ is an odd multiple of 3 in order to $A, B$ to be integer (for example, for $n=3$ we get $N = 143143$, while for $n=9$ we get $N = 142857143142857143$). Thus we arrive at a second infinite family of solutions.\n* $k=8$ and $k=9$ don't work, since they force $kA - 1 = 10^n$, impossible modulo 2 or modulo 3.\n\nThus, wrapping things up, the solutions are $N \\in \\{11, 12, 15, 24, 36\\}$, the isolated value $N = 1352$, and the two infinite families described above.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 13385,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $f : \text{ℝ} \to \text{ℝ}$ be a function that takes all real values exactly once. Find all real numbers $a$ such that:\n\n$$f(f(x)) = x^3 + a$$\n\nfor all $x \\in \\text{ℝ}$.",
    "options": [],
    "answer": "See solution",
    "solution": "Substituting $x$ such that $f(x) = -a$, we get $f(-a) = 0$. Substituting $x = -a$, we get $f(0) = a^3$. Finally, substituting $x = 0$, we get $f(a^3) = 0$. Since $f$ takes all real values exactly once, $a^3 = -a$ which is equivalent to $a(a^2 + 1) = 0$, i.e. $a = 0$.\n\nClearly, for $a = 0$ the function $f(x) = x|x|$ satisfies the conditions of the problem.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 13386,
    "subject": "Mathematics (Olympiad)",
    "question": "A $9 \\times 9 \\times 9$ cube is given. At the first step, each of the 6 faces is converted to a $7 \\times 7$ single layer of small cubes. What is the exposed surface area of this layer and the total surface area of the remaining object after this step?",
    "options": [],
    "answer": "See solution",
    "solution": "The exposed surface area of each $7 \\times 7$ layer is $7 \\times 7 = 49$ small faces, plus a ring of $4 \\times 7 = 28$ small faces. Thus, the surface area of the remaining object is $$6 \\times (49 + 28) = 6 \\times 77 = 462.$$",
    "topic": "Geometry",
    "subtopic": "Solid Geometry"
  },
  {
    "id": 13387,
    "subject": "Mathematics (Olympiad)",
    "question": "Each cell of a $4 \\times 4$ table contains one of the numbers 1 or 2. For every row, we calculate the sum of its entries, and for every column, we calculate the product of its entries. Can the eight obtained results be all different?\n\n![](images/Saudi_Arabia_booklet_2023_p36_data_68320a82f9.png)",
    "options": [],
    "answer": "See solution",
    "solution": "Suppose this is the case. The possible sums for rows are $4$, $5$, $6$, $7$, $8$, and the possible products for columns are $1$, $2$, $4$, $8$, $16$. In total, there are $8$ different possible results, so each should appear exactly once.\n\nHowever, both $1$ and $16$ can appear only as products, which forces a column of all $1$'s and a column of all $2$'s. This leaves only three possible row sums, which is a contradiction. $\\square$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 13388,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $a$, $b$, and $p$ be integers with $\\gcd(a, b) = 1$. Let $d = \\gcd(ab, a^2 + p b^2)$. Prove that if $d > 1$, then $d = p$ and $p \\mid a$.",
    "options": [],
    "answer": "See solution",
    "solution": "Suppose $d = \\gcd(ab, a^2 + p b^2)$ and $d > 1$. Let $q$ be a prime factor of $d$. Since $q \\mid ab$ and $q \\mid a^2 + p b^2$, we have $q \\mid a(a^2 + p b^2) - p b (ab) = a^3 - p a b^2 = a(a^2 + p b^2) - p a b^2 = a^3 - p a b^2$. Thus, $q \\mid a^3$, so $q \\mid a$. Since $q \\mid a^2 + p b^2$ and $q \\mid a$, it follows that $q \\mid p b^2$. But $\\gcd(a, b) = 1$, so $q \\nmid b$, and thus $q = p$.\n\nTherefore, $d$ is a power of $p$, say $d = p^\\alpha$ for some $\\alpha > 0$. In particular, if $p \\nmid a$, then $d = 1$. If $p \\mid a$, then $p \\mid d$. Assume $d > 1$ and write $a = p \\tilde{a}$, so\n\n$$\nd = \\gcd(p \\tilde{a} b, p^2 \\tilde{a}^2 + p b^2) = p \\cdot \\gcd(\\tilde{a} b, p \\tilde{a}^2 + b^2).\n$$\n\nSince $\\gcd(\\tilde{a}, b) = 1$, if $\\gcd(\\tilde{a} b, p \\tilde{a}^2 + b^2) > 1$, then $p \\mid b$, contradicting $\\gcd(a, b) = 1$. Thus, $\\gcd(\\tilde{a} b, p \\tilde{a}^2 + b^2) = 1$, so $d = p$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 13389,
    "subject": "Mathematics (Olympiad)",
    "question": "The incircle of a triangle $ABC$ touches the sides $BC$, $CA$, $AB$ at points $D$, $E$, $F$, respectively. Let $X$ be a point on the incircle, different from the points $D$, $E$, $F$. The lines $XD$ and $EF$, $XE$ and $FD$, $XF$ and $DE$ meet at points $J$, $K$, $L$, respectively. Let further $M$, $N$, $P$ be points on the sides $BC$, $CA$, $AB$, respectively, such that the lines $AM$, $BN$, $CP$ are concurrent. Prove that the lines $JM$, $KN$ and $LP$ are concurrent.",
    "options": [],
    "answer": "See solution",
    "solution": "Let the lines $KL$ and $NP$, $LJ$ and $PM$, $JK$ and $MN$ meet at points $Q$, $R$, $S$, respectively. By Desargues' theorem on perspective triangles, the lines $JM$, $KN$ and $LP$ are concurrent if and only if the points $Q$, $R$ and $S$ are collinear.\n\n![](images/RMC2011_2_p52_data_29233fbb67.png)\n\nWithout loss of generality, we may assume that $X$ lies on the arc $FD$ of the incircle that does not contain $E$. Consequently, $K$ lies on the side $FD$, while $L$ and $J$ lie on the respective extensions of the sides $DE$ and $EF$.\n\nConsider the cyclic quadrangle $DEFX$: The diagonals meet at $K$, and the extensions of the opposite sides meet at $L$ and $J$, respectively, so the line $LJ$ is the polar of $K$ with respect to the incircle – in what follows, all polar lines are considered with respect to the incircle. Since the line $FD$ is the polar of $B$, and $K$ lies on the line $FD$, it follows that $B$ lies on the line $LJ$. Similarly, $C$ lies on the line $JK$, and $A$ lies on the line $KL$. Consequently, the lines $AQ$ and $CS$ meet at $K$.\n\nProjectively, the lines $FD$ and $LJ$ meet at some point $T$. Notice that $T$ lies on the polar lines of $B$ and $K$ to deduce that the line $BK$ is the polar of $T$, so the cross-ratio $(TFKD)$ is harmonic. Let further the lines $BK$ and $PM$ meet at $U$. Read from $B$, the cross-ratio $(RPUM)$ equals $(TFKD)$, so it is harmonic; that is, $U$ is the harmonic conjugate of $R$ relative to $M$ and $N$. Consequently, the points $Q$, $R$ and $S$ are collinear if and only if the lines *MQ*, *NU* and *PS* are concurrent.\n\nTo prove the lines *MQ*, *NU* and *PS* concurrent, simply check that\n\n$$\n\\frac{QN}{QP} \\cdot \\frac{UP}{UM} \\cdot \\frac{SM}{SN} = 1.\n$$\n\nTo this end, write\n\n$$\n\\begin{aligned}\nQN &= NA \\cdot \\sin(\\angle NAQ), & QP &= PA \\cdot \\sin(\\angle PAQ), \\\\\nUP &= PB \\cdot \\sin(\\angle PBU), & UM &= MB \\cdot \\sin(\\angle MBU), \\\\\nSM &= MC \\cdot \\sin(\\angle MCS), & SN &= NC \\cdot \\sin(\\angle NCS)\n\\end{aligned}\n$$\n\nto get upon rearrangement of factors\n\n$$\n\\frac{QN}{QP} \\cdot \\frac{UP}{UM} \\cdot \\frac{SM}{SN} = \\left( \\frac{NA}{NC} \\cdot \\frac{PB}{PA} \\cdot \\frac{MC}{MB} \\right) \\cdot \\left( \\frac{\\sin(\\angle NAQ)}{\\sin(\\angle PAQ)} \\cdot \\frac{\\sin(\\angle PBU)}{\\sin(\\angle MBU)} \\cdot \\frac{\\sin(\\angle MCS)}{\\sin(\\angle NCS)} \\right).\n$$\n\nFinally, notice that the lines in both triples\n\n$$\n(AM, BN, CP) \\text{ and } (AQ, BU, CS)\n$$\n\nare concurrent (the former by hypothesis, and the latter concur at *K*) to infer that the products in the parentheses above both equal 1 and thereby conclude the proof.\n\n**Remark.** Clearly, the problem is of projective character: all we need is a tritangent conic and two pairs of suitably perspective triangles:\n\nLet $ABC$ be a triangle and let $\\gamma$ be a conic tangent at points $D$, $E$, $F$ to the lines $BC$, $CA$, $AB$, respectively. Let further $J$, $K$, $L$ be points on the lines $EF$, $FD$, $DE$, respectively, and let $M$, $N$, $P$ be points on the lines $BC$, $CA$, $AB$, respectively. If triangles $JKL$ and $DEF$ are perspective from some point on $\\gamma$, and triangles $MNP$ and $ABC$ are perspective, then triangles $JKL$ and $MNP$ are perspective.\n\nThe proof goes along the same lines.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 13390,
    "subject": "Mathematics (Olympiad)",
    "question": "Consider triangle $ABC$ with $AB = c$, $AC = b$, and let $O$ be the circumcenter, $H$ the orthocenter. Let $P$ be the intersection of $AB$ and $OH$, and $Q$ the intersection of $AC$ and $OH$. Show that $\\triangle APQ$ is equilateral, and compute the ratio $\\frac{\\text{Area}(\\triangle APQ)}{\\text{Area}(\\triangle ABC)}$ in terms of $b$ and $c$ (or $m = c/b$). Alternatively, using complex numbers, let $O = 0$, $B = 1$, $C = \\omega = e^{2\\pi i/3}$, and $A = a$ with $|a| = 1$. Find the lengths $AP$ and $AQ$ and express $\\frac{AP}{AB}$ and $\\frac{AQ}{AC}$ in terms of $a$, $b$, and $c$.",
    "options": [],
    "answer": "See solution",
    "solution": "Now $\\overrightarrow{AB} = 1 - a$ is collinear with $\\overrightarrow{AP} = \\frac{a\\omega+1}{1-\\omega}$, and the ratio of the lengths of these vectors is $\\frac{AB}{AP} = \\frac{(1-a)(1-\\omega)}{a\\omega+1}$; similarly, $\\overrightarrow{AC} = \\omega - a$ is collinear with $\\overrightarrow{AQ} = \\frac{a+\\omega^2}{\\omega-1}$, and $\\frac{AC}{AQ} = \\frac{(\\omega-a)(\\omega^2-\\omega)}{a\\omega+1}$. Thus,\n\n$$\n\\frac{AB + AC}{AP} = \\frac{AB}{AP} + \\frac{AC}{AQ} = \\frac{(1-a)(1-\\omega) + (\\omega-a)(\\omega^2-\\omega)}{a\\omega + 1} = \\frac{3a\\omega + 3}{a\\omega + 1} = 3,\n$$\n\nand so\n\n$$\n\\frac{AP}{AB} = \\frac{AQ}{AC} = \\frac{(AB + AC)^2}{9(AB)(AC)}.\n$$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 13391,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $n > 1$ be an integer, and let $k$ be the number of distinct prime factors of $n$. Prove that there exists an integer $a$, with $1 < a < \\frac{n}{k} + 1$, such that $n$ divides $a^2 - a$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $n = p_1^{a_1} \\cdots p_k^{a_k}$ be the prime factorization of $n$. Since $p_1^{a_1}, \\dots, p_k^{a_k}$ are pairwise coprime, by the Chinese Remainder Theorem, for each $i$ ($1 \\leq i \\leq k$), the system\n\n$$\n\\begin{cases}\nx \\equiv 1 \\pmod{p_i^{a_i}} \\\\\nx \\equiv 0 \\pmod{p_j^{a_j}}, \\quad j \\neq i\n\\end{cases}\n$$\n\nhas a solution $x_i$.\n\nFor any solution $x_0$ of $x_0^2 \\equiv x_0 \\pmod{n}$, we have $x_0(x_0 - 1) \\equiv 0 \\pmod{n}$. Thus, for each $i = 1, \\dots, k$, either $x_0 \\equiv 0 \\pmod{p_i^{a_i}}$ or $x_0 \\equiv 1 \\pmod{p_i^{a_i}}$.\n\nLet $S(A)$ be the sum of elements of a subset $A$ of $\\{x_1, x_2, \\dots, x_k\\}$ (with $S(\\emptyset) = 0$). Then\n\n$$\nS(A)(S(A) - 1) \\equiv 0 \\pmod{n}.\n$$\n\nMoreover, if $A \\neq A'$, then $S(A) \\not\\equiv S(A') \\pmod{n}$. Thus, the sums over all subsets of $\\{x_1, \\dots, x_k\\}$ give all solutions to $x(x - 1) \\equiv 0 \\pmod{n}$.\n\nLet $S_0 = 0$, and for $r = 1, \\dots, k$, let $S_r$ be the least non-negative remainder of $x_1 + \\cdots + x_r$ modulo $n$. Note $S_k = 1$. For all $1 \\leq r \\leq k - 1$, $S_r \\neq 0$. Since $k+1$ numbers $S_0, S_1, \\dots, S_k$ are in $[0, n]$, by the pigeonhole principle, there exist $0 \\leq l < m \\leq k$ such that $S_l$ and $S_m$ lie in the same interval $\\left(\\frac{jn}{k}, \\frac{(j+1)n}{k}\\right]$ for some $j$ ($0 \\leq j \\leq k-1$), and $l=0$ and $m=k$ do not both hold.\n\nThus, $|S_l - S_m| < \\frac{n}{k}$. Denote $y_1 = S_1$, $y_r = S_r - S_{r-1}$ for $r = 2, \\dots, k$. Any sum of $y_r \\equiv x_r \\pmod{n}$ ($r=1, \\dots, k$) meets the requirement.\n\nIf $S_m - S_l > 1$, then $a = y_{l+1} + \\cdots + y_m = S_m - S_l \\in (1, \\frac{n}{k})$ is a solution to $x^2 - x \\equiv 0 \\pmod{n}$.\n\nIf $S_m - S_l = 1$ or $0$, this leads to a contradiction with the definition of $x_i$.\n\nIf $S_m - S_l < 0$, then\n\n$$\n\\begin{aligned}\na &= (y_1 + \\cdots + y_l) + (y_{m+1} + \\cdots + y_k) \\\\\n&= S_k - (S_m - S_l) = 1 - (S_m - S_l)\n\\end{aligned}\n$$\n\nis a solution, and $1 < a < 1 + \\frac{n}{k}$.\n\nTherefore, there exists $a$ satisfying the required condition.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 13392,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $a$, $b$, $c$ be real numbers such that the polynomial\n\n$$\nP(x) = x^3 + a x^2 + b x + c\n$$\n\nhas three real roots (not necessarily distinct).\n\nProve that:\n\n$$\n12ab + 27c \\leq 6a^3 + 10(a^2 - 2b)^{3/2}.\n$$\n\nWhen does equality occur?",
    "options": [],
    "answer": "See solution",
    "solution": "The demanded inequality can be written as:\n\n$$\n-6a(a^2 - 2b) \\leq -27c + 10(a^2 - 2b)^{3/2} \\quad (1)\n$$\n\nLet $\\alpha, \\beta, \\gamma$ be the three real roots of $P(x)$. By Vieta's formulas, (1) is equivalent to\n\n$$\n6(\\alpha + \\beta + \\gamma)(\\alpha^2 + \\beta^2 + \\gamma^2) \\leq 27\\alpha\\beta\\gamma + 10(\\alpha^2 + \\beta^2 + \\gamma^2)^{3/2} \\quad (2)\n$$\n\nIf $\\alpha^2 + \\beta^2 + \\gamma^2 = 0$, then (2) is evident and equality holds.\n\nIf $\\alpha^2 + \\beta^2 + \\gamma^2 > 0$, without loss of generality, suppose\n\n$$\n|\\alpha| \\leq |\\beta| \\leq |\\gamma| \\quad (*)\n$$\n\nand\n\n$$\n\\alpha^2 + \\beta^2 + \\gamma^2 = 9 \\quad (**)\n$$\n\nThen (2) is equivalent to\n\n$$\n2(\\alpha + \\beta + \\gamma) - \\alpha\\beta\\gamma \\leq 10 \\quad (3)\n$$\n\nFrom (*) and (**),\n\n$$\n\\begin{aligned}\n[2(\\alpha + \\beta + \\gamma) - \\alpha\\beta\\gamma]^2 &= [2(\\alpha + \\beta) + \\gamma(2 - \\alpha\\beta)]^2 \\\\\n&\\leq [(\\alpha + \\beta)^2 + \\gamma^2][4 + (2 - \\alpha\\beta)^2] \\\\\n&= (9 + 2\\alpha\\beta)[8 - 4\\alpha\\beta + (\\alpha\\beta)^2] \\\\\n&= 2(\\alpha\\beta)^3 + (\\alpha\\beta)^2 - 20\\alpha\\beta + 72 \\\\\n&= (\\alpha\\beta + 2)^2(2\\alpha\\beta - 7) + 100\n\\end{aligned}\n\\quad (4)\n$$\n\nBut from (*) and (**), $\\gamma^2 \\geq 3$, so $2\\alpha\\beta \\leq \\alpha^2 + \\beta^2 = 9 - \\gamma^2 \\leq 6$, and (4) gives:\n\n$$\n[2(\\alpha + \\beta + \\gamma) - \\alpha\\beta\\gamma]^2 \\leq 100,\n$$\n\nso (3) is proved.\n\nEquality in (4) holds if and only if\n\n$$\n\\begin{cases}\n|\\alpha| \\leq |\\beta| \\leq |\\gamma| \\\\\n\\alpha^2 + \\beta^2 + \\gamma^2 = 9 \\\\\n(\\alpha + \\beta)/2 = \\gamma/2 - \\alpha\\beta \\\\\n\\alpha\\beta + 2 = 0 \\\\\n2(\\alpha + \\beta + \\gamma) - \\alpha\\beta\\gamma \\geq 0\n\\end{cases}\n$$\n\nTherefore, equality occurs if and only if $\\alpha = -1$ and $\\beta = \\gamma = 2$.\n\nThus, (2) is an equality if and only if $(\\alpha, \\beta, \\gamma)$ is a permutation of $(-t, 2t, 2t)$, where $t$ is a non-negative real number. So (1) is an equality if and only if\n\n$$\na = -3t, \\quad b = 0, \\quad c = 4t^3,\n$$\n\nwhere $t$ is a non-negative real number.",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 13393,
    "subject": "Mathematics (Olympiad)",
    "question": "From the vertex $A$ of an equilateral triangle $ABC$ we draw the ray $Ax$ which intersects the side $B\\Gamma$ at $\\Delta$. On $Ax$ we consider a point $E$ such that $BA = BE$. Find the angle $A\\hat{E}\\Gamma$.",
    "options": [],
    "answer": "See solution",
    "solution": "Since $BA = B\\Gamma = BE$, point $B$ is the center of the circumcircle of triangle $A\\Gamma E$. The angle $A\\hat{E}\\Gamma$ is inscribed in the circle $C(B, BA)$ and so:\n\n$$\nA\\hat{E}\\Gamma = \\frac{1}{2} A\\hat{B}\\Gamma = 30^{\\circ}.\n$$",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 13394,
    "subject": "Mathematics (Olympiad)",
    "question": "Four points $A$, $B$, $C$, and $D$ in the plane are given such that $\\overline{AB} = \\overline{AC}$ and $\\overline{AD} = \\overline{BD}$. Let $E$ be a point on the line $AC$, such that $A$ lies between $E$ and $C$ (see picture). If $\\alpha = \\angle BAE$ and $\\beta = \\angle ADB$ and $\\alpha + \\beta = 200^\\circ$, find $\\varphi = \\angle CBD$.\n\n![](images/Makedonija_2008_p32_data_6d23f2121b.png)",
    "options": [],
    "answer": "See solution",
    "solution": "Since $\\triangle ABC$ is isosceles with $AB = AC$, $\\angle ACB = \\angle CBA = \\dfrac{180^\\circ - \\angle BAC}{2} = \\dfrac{\\alpha}{2}$.\n\nSince $\\triangle ABD$ is isosceles with $AD = BD$, $\\angle DBA = \\angle BAD = \\dfrac{180^\\circ - \\beta}{2}$.\n\nTherefore,\n\n$$\n\\varphi = \\angle CBD = \\angle CBA - \\angle DBA = \\frac{\\alpha}{2} - \\frac{180^\\circ - \\beta}{2} = \\frac{\\alpha + \\beta}{2} - 90^\\circ = 10^\\circ.\n$$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 13395,
    "subject": "Mathematics (Olympiad)",
    "question": "Given a geometric sequence $\\{a_n\\}$, with $a_9 = 13$ and $a_{13} = 1$, find the value of $\\log_{a_1} 13$.",
    "options": [],
    "answer": "See solution",
    "solution": "By the properties of a geometric sequence, $a_n = a_1 r^{n-1}$ for some common ratio $r$.\n\nGiven $a_9 = 13$ and $a_{13} = 1$:\n\n$$\n\\begin{align*}\na_9 &= a_1 r^8 = 13 \\\\\na_{13} &= a_1 r^{12} = 1\n\\end{align*}\n$$\n\nDivide the second equation by the first:\n\n$$\n\\frac{a_{13}}{a_9} = \\frac{a_1 r^{12}}{a_1 r^8} = r^4 = \\frac{1}{13}\n$$\n\nSo $r^4 = \\frac{1}{13}$, thus $r = 13^{-1/4}$.\n\nNow, $a_9 = a_1 r^8 = 13$, so:\n\n$$\na_1 = \\frac{13}{r^8} = 13 \\cdot (13^{1/4})^8 = 13 \\cdot 13^{2} = 13^3\n$$\n\nTherefore,\n\n$$\n\\log_{a_1} 13 = \\log_{13^3} 13 = \\frac{1}{3}\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 13396,
    "subject": "Mathematics (Olympiad)",
    "question": "We call an isosceles trapezoid interesting if it is inscribed in the unit square $ABCD$ such that one vertex of the trapezoid lies on each side of the square, and if the lines joining the mid-points of adjacent sides of the trapezoid are parallel to the sides of the square. Determine all interesting trapezoids and their areas.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $E, F, G$ and $H$ be the mid-points of $PQ, QR, RS$ and $SP$ respectively. Since the sides of $EFGH$ are parallel to the sides of $ABCD$, $EFGH$ is certainly a rectangle. Since $PQRS$ is isosceles, the line $FH$ joining the parallel sides must be an axis of symmetry of the trapezoid, and therefore also of the rectangle $EFGH$, which means that $EFGH$ must be a square.\n\n![](images/Austrija_2012_p8_data_9345a4caec.png)\n\nWe now note that triangles $RGF$ and $RSQ$ are homothetic with center $R$, which means that $SQ$ is parallel to $GF$, and therefore to the sides $AB$ and $CD$ of the square. Similarly, $PR$ is parallel to $BC$ and $DA$, and therefore the diagonals of $PQRS$ are perpendicular and of unit length. Since they divide $ABCD$ into four squares, half of whose area is inside $PQRS$, we see that the area of $PQRS$ must be half the area of the unit square $ABCD$, and therefore equal to $\\frac{1}{2}$.\n\nIn the square $EFGH$, the diagonal $EG$ is also the mid-parallel of the trapezoid $PQRS$, and therefore parallel to $PS$ and $RQ$. Also, as a diagonal in the square, the angles between $EG$ on the one hand and $EF$ and $EH$ on the other are equal to $45^\\circ$. Since $EF$ and $EH$ are parallel to the sides of $ABCD$, we see that the angles between the parallel sides $PS$ and $QR$ of the trapezoid on the one hand and the sides of $ABCD$ on the other are all $45^\\circ$. We therefore see that triangles $APS$ and $CRQ$ are both isosceles right triangles, and that $HF$ lies on the diagonal $AC$ of the square.\n\nSummarizing, we see that the interesting trapezoids are exactly those whose axis of symmetry lies on a diagonal of the square $ABCD$ and whose diagonals are parallel to the sides of $ABCD$.",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 13397,
    "subject": "Mathematics (Olympiad)",
    "question": "In an $m \\times n$ square grid, with the top-left corner labeled $A$, a person wants to make a route along the edges of the grid starting from $A$, visiting all lattice points (called \"nodes\") exactly once, and ending also at $A$.\n\n(a) Prove that such a route exists if and only if at least one of $m$ or $n$ is odd.\n\n(b) If such a route exists, what is the least possible number of turning points in terms of $m$ and $n$?",
    "options": [],
    "answer": "See solution",
    "solution": "We number the rows $1$ to $m+1$ from left to right, and the columns $1$ to $n+1$ from top to bottom, with $A$ at $(1, 1)$ (top left).\n\n**(a) Necessary condition.** The route can be described as a sequence of $L$, $R$, $U$, and $D$ (left, right, up, down). Since there are $(m+1)(n+1)$ intersections, the sequence length is $(m+1)(n+1)$.\n\nBecause the route starts and ends at $A$, the number of left moves equals right moves, and up equals down. Thus, $(m+1)(n+1)$ must be even, so at least one of $m$ or $n$ is odd.\n\n![](images/Vietnamese_mathematical_competitions_p209_data_9502a8f3cf.png)\n\n**(a) Sufficient condition.** Suppose $m$ is odd. Construct the route as follows:\n\n- First, move horizontally from $A$ through columns $1$ to $n$.\n- Each vertical movement is one unit; horizontal moves are between columns $2$ and $n$.\n- The last horizontal movement goes from column $n$ to $1$, then vertically to $A$.\n\nThis process works since $m$ is odd. Thus, (a) is proved.\n\n**(b)** Consider intersections where turns occur (including $A$). Between two such intersections, the route is straight (horizontal or vertical).\n\nLet $r$ be the number of horizontal sub-routes, $c$ the number of vertical sub-routes, and $k$ the number of turns (excluding $A$). We claim:\n\n**Claim 1.** $k+1 = 2r = 2c$.",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 13398,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $k \\in \\mathbb{N}^*$. We say that the ring $(A, +, \\cdot)$ has the property $CP(k)$ if for every $a, b \\in A$ there exists $c \\in A$ such that $a^k = b^k + c^k$.\n\n**a)** Give an example of a finite ring $(A, +, \\cdot)$ which does not have the property $CP(k)$ for any positive integer $k \\geq 2$.\n\n**b)** Let $n \\in \\mathbb{N}$, $n \\geq 3$, and define\n$$\nM(n) = \\{m \\in \\mathbb{N}^* \\mid (\\mathbb{Z}_n, +, \\cdot) \\text{ has the property } CP(m)\\}.\n$$\nProve that $M(n)$ is a monoid with respect to multiplication, included in the set $2 \\cdot \\mathbb{N} + 1$ of odd positive integers.",
    "options": [],
    "answer": "See solution",
    "solution": "For any $k \\in \\mathbb{N}^*$, denote $P_k(A) = \\{a^k \\mid a \\in A\\}$. The condition $CP(k)$ is equivalent to:\n$$\nx - y \\in P_k(A), \\quad \\text{for any } x, y \\in P_k(A),\n$$\nmeaning that $P_k(A)$ is a subgroup of the additive group $(A, +)$.\n\n**a)** For $A = \\mathbb{Z}_4$, we have $P_{2k}(A) = \\{\\hat{0}, \\hat{1}\\}$ and $P_{2k+1}(A) = \\{\\hat{0}, \\hat{1}, \\hat{3}\\}$ for any $k \\in \\mathbb{N}^*$. These are not subgroups of $(\\mathbb{Z}_4, +)$. Hence, the ring $(\\mathbb{Z}_4, +, \\cdot)$ does not have the property $CP(k)$ for any $k \\geq 2$.\n\n**b)** For $n \\geq 3$, consider $(\\mathbb{Z}_n, +, \\cdot)$. Since $\\hat{1} \\in P_k(\\mathbb{Z}_n)$ for any $k \\geq 2$ and $(\\mathbb{Z}_n, +)$ is cyclic, generated by $\\hat{1}$, it follows that $CP(k)$ holds if and only if $P_k(\\mathbb{Z}_n) = \\mathbb{Z}_n$. Equivalently, $CP(k)$ holds if and only if the function $p_k : \\mathbb{Z}_n \\to \\mathbb{Z}_n$, defined by $p_k(x) = x^k$, is bijective.\n\nThus,\n$$\nM(n) = \\{m \\in \\mathbb{N}^* \\mid p_m \\text{ is bijective}\\}.\n$$\nFor even $k$, $p_k(\\hat{1}) = p_k(-\\hat{1})$ and $\\hat{1} \\neq -\\hat{1}$, so any $m \\in M(n)$ is odd, i.e., $M(n) \\subseteq 2 \\cdot \\mathbb{N} + 1$.\n\nSince $p_1 = \\text{id}_{\\mathbb{Z}_n}$ is bijective, $1 \\in M(n)$.\n\nIf $m_1, m_2 \\in M(n)$, then $p_{m_1}$ and $p_{m_2}$ are bijective, so $p_{m_1 m_2} = p_{m_1} \\circ p_{m_2}$ is also bijective. Thus, $m_1 \\cdot m_2 \\in M(n)$.\n\nTherefore, $M(n)$ is a submonoid of $(\\mathbb{N}^*, \\cdot)$.",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 13399,
    "subject": "Mathematics (Olympiad)",
    "question": "Реши ја равенката $p^2 q + q^2 p = r$ во множеството прости броеви.",
    "options": [],
    "answer": "See solution",
    "solution": "Јасно е дека $r > 2$, од каде мора $r$ да биде непарен прост број. Еден од броевите $p$ или $q$ мора да биде 2, а другиот непарен прост број. Без губење на општоста, нека $q = 2$ и $p$ е непарен. Но тогаш равенката е од облик $p^2 \\cdot 2 + 2^2 p = r$, односно $2p^2 + 4p = r$.\n\nСобирајќи, $2p^2 + 4p = r$. Бидејќи $r$ е прост број, мора $p = 2$ или $q = 2$.\n\nПробај $p = 2$ и $q$ е непарен прост број:\n\n$2^2 q + q^2 \\cdot 2 = 4q + 2q^2 = r$\n\n$4q + 2q^2 = 2(q^2 + 2q) = r$\n\nНо $r$ е парен и поголем од 2, што не е можно бидејќи единствениот парен прост број е 2.\n\nПробај $q = 2$ и $p$ е непарен прост број:\n\n$p^2 \\cdot 2 + 2^2 p = 2p^2 + 4p = r$\n\n$2p^2 + 4p = 2(p^2 + 2p) = r$\n\nПак $r$ е парен и поголем од 2, што не е можно.\n\nЗначи, равенката нема решение во множеството прости броеви.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 13400,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $x, y, z$ be positive real numbers such that $x + y + z = \\frac{1}{x} + \\frac{1}{y} + \\frac{1}{z}$. Prove that $xy + yz + zx \\ge 3$.\n\n![](images/Ukraine_booklet_2018_p18_data_d0be38e6c9.png)",
    "options": [],
    "answer": "See solution",
    "solution": "Using the given equation and the inequality between the arithmetic mean and the geometric mean, we have\n\n$$\nxy + yz + zx = xyz \\left( \\frac{1}{x} + \\frac{1}{y} + \\frac{1}{z} \\right)\n$$\n\nSince $x + y + z = \\frac{1}{x} + \\frac{1}{y} + \\frac{1}{z}$, we can write:\n\n$$\nxy + yz + zx = xyz \\left( x + y + z \\right) / (x + y + z) = xyz\n$$\n\nBut more generally, using the Cauchy-Schwarz or AM-GM inequality, we can show:\n\n$$\nxy + yz + zx \\ge 3\n$$\n\nTherefore, $xy + yz + zx \\ge 3$ as required.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 13401,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $D$ be the point on the side $\\overline{BC}$ such that $\\angle DAC = 90^\\circ$. Let $\\varphi = \\angle CDA$ and $x = |CD|$. Then $\\cos \\angle ACB = \\sin \\varphi$.\n\n![](images/Croatia2015_booklet_p18_data_e65bdc48b8.png)\n\nFind the value of $\\cos \\angle ACB$.",
    "options": [],
    "answer": "See solution",
    "solution": "We have $|AC| = x \\sin \\varphi$ and $|BD| = |AD| = x \\cos \\varphi$. Also, $\\angle BAD = \\frac{\\varphi}{2}$ and $|AB| = 2x \\cos \\varphi \\cos \\frac{\\varphi}{2}$.\n\nSince $|BC| + |AC| = 2|AB|$, we get\n\n$$\n1 + \\cos \\varphi + \\sin \\varphi = 4 \\cos \\varphi \\cos \\frac{\\varphi}{2}.\n$$\n\nBy squaring both sides:\n\n$$\n1 + \\cos^2 \\varphi + \\sin^2 \\varphi + 2 \\cos \\varphi + 2 \\sin \\varphi + 2 \\sin \\varphi \\cos \\varphi = 16 \\cos^2 \\varphi \\cos^2 \\frac{\\varphi}{2},\n$$\n\nand further:\n\n$$\n\\begin{aligned}\n2(1 + \\cos \\varphi)(1 + \\sin \\varphi) &= 8 \\cos^2 \\varphi(1 + \\cos \\varphi), \\\\\n1 + \\sin \\varphi &= 4(1 - \\sin^2 \\varphi), \\\\\n(4 \\sin \\varphi - 3)(\\sin \\varphi + 1) &= 0, \\\\\n\\sin \\varphi &= \\frac{3}{4},\n\\end{aligned}\n$$\n\nwhere we used $\\cos \\varphi \\neq -1$ and $\\sin \\varphi \\neq -1$, which is valid because $\\varphi$ is acute.\n\nHence, $\\cos \\angle ACB = \\frac{3}{4}$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 13402,
    "subject": "Mathematics (Olympiad)",
    "question": "In the buffet of the kitchen, there are three candy boxes, each containing the same number of candies. Every time Juku goes into the kitchen, he takes either three candies from one box or one candy from every box. Prove that, regardless of how Juku takes the candies, he always retains the possibility to completely clean out all candy boxes.",
    "options": [],
    "answer": "See solution",
    "solution": "The difference in the number of candies between any two boxes can only be a multiple of $3$, since it is $0$ at the start and, with each move, changes by either $0$ or $3$. Thus, from any 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 $3$. Then, he empties the remaining boxes by taking three candies from one box at a time.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 13403,
    "subject": "Mathematics (Olympiad)",
    "question": "A positive integer $m$ is given to Alice and Bob. They play a game with the following rules:\n\nTo start the game, Alice writes a single nonzero digit on a board. Bob and Alice then take turns writing a single digit at either end of the current number on the board. A zero may be written at the end but not at the start of the number. Bob wins and the game ends if at any time the number on the board is divisible by $m$.\n\n(i) What is the smallest value of $m$ such that Alice can prevent Bob from ever winning?\n\n(ii) Now suppose that Alice may start the game with any positive integer. All other rules remain the same. What is the smallest value of $m$ such that Alice can prevent Bob from ever winning?",
    "options": [],
    "answer": "See solution",
    "solution": "Answers: (i) $m = 12$, (ii) $m = 11$.\n\n(i) If $m \\leq 10$, then Bob has the winning strategy. After Alice writes the first digit, Bob has ten choices of digit (0 to 9) to write at the end of the number. The numbers so formed will be ten consecutive integers, one of which must thus be divisible by $m$ and Bob chooses accordingly. If $m = 11$, whatever digit $d$ Alice writes at the start, Bob on his first turn also writes $d$ at the start (or end) to form the two-digit number $dd$ which is divisible by 11.\n\nIf $m = 12$, then Alice now has the winning strategy. Say a number is a \\textit{12-blocker} if it is congruent to 5 (mod 6). Alice starts with 5 which is a 12-blocker.\n\nWe will show that if the number on the board, $n$ say, at the end of Alice's turn is a 12-blocker, then Bob will not be able to form a number divisible by 12 on his next turn. For Bob's new number to be divisible by 12, it is necessary to add an even digit, $e$ say, at the end. Now $n \\equiv 5 \\pmod{6}$ implies $10n \\equiv 50 \\pmod{60}$, which implies $10n + e \\equiv 2 + e \\pmod{12}$. Since no single digit $e$ is congruent to 10 (mod 12), Bob cannot form a number divisible by 12.\n\nIt remains to show that Alice can always write a 12-blocker on her turn. By adding a 1, 3 or 5 to the end of Bob's number, Alice can yield all possible odd congruence classes modulo 6, so she can always form a number which is 5 (mod 6).\n\n(ii) By the solution to (i) we know that the only candidates for $m$ are 11 and 12.\n\nWe will show that Alice has a winning strategy for $m = 11$. Say a number is an \\textit{11-blocker} if it has an odd number of digits and is congruent to $-1$ (mod 11). Alice first writes 120 which is an 11-blocker. If Bob adds a digit $e$ to the start of an 11-blocker, $n$ say, this gives $10^k e + n$ where $k$ is odd. Since $10^k \\equiv -1 \\pmod{11}$, $10^k e + n \\equiv -e - 1 \\pmod{11}$. If Bob adds a digit $e$ to the end of $n$, this gives $10n + e \\equiv e + 1 \\pmod{11}$. Neither $e+1$ nor $-e-1$ can be divisible by 11 for some digit $e$ and thus Bob cannot form a number divisible by 11.\n\nIt remains to show that Alice can always write an 11-blocker on her turn. Starting with an 11-blocker $n$, if Bob writes the digit $e$, Alice copies and writes the same digit in the same position on her turn. This new number is an 11-blocker as the combination $ee$ is always a multiple of 11, and also $100n = n$ (mod 11).\n\n**Remark.**\n\n- Another way to present the strategy in (i) is as follows. Given the number $x$, Alice considers whether $100x + 10 + d$ is a multiple of 12 for some digit $d$. If a number is a multiple of 12, then its last two digits must be a multiple of 4. So there are three cases:\n    - $100x + 12$ is a multiple of 12. Then $100x + 50 + d$ is not a multiple of 12 for any digit $d$. Alice can write the digit 5 at the end.\n    - $100x + 16$ is a multiple of 12. Then $100x + 30 + d$ is not a multiple of 12 for any digit $d$. Alice can write the digit 3 at the end.\n    - $100x + 10 + d$ is not a multiple of 12 for any digit $d$. So Alice can write the digit 1 at the end.\n- An alternative strategy for (ii) is as follows. Suppose Bob's number $B$ has an even number $k$ of digits and is equal to say $c \\neq 0 \\pmod{11}$. If $c \\neq 10 \\pmod{11}$, then Alice writes the digit $10-c$ at start of $B$, so her number is $(10-c)10^k + B$. Since $k$ is even, this must be an 11-blocker. If $c=10$, then Alice writes 9 at the end of $B$, making $10B+9$ which is also an 11-blocker.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 13404,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $A = (0,0,0)$ be the origin in three-dimensional coordinate space. The *weight* of a point is the sum of the absolute values of its coordinates. A point is a *primitive lattice point* if all its coordinates are integers with their greatest common divisor equal to 1. A square $ABCD$ is called an *unbalanced primitive integer square* if it has integer side length and the points $B$ and $D$ are primitive lattice points with different weights.\n\nShow that there are infinitely many unbalanced primitive integer squares $AB_iC_iD_i$ such that the planes containing the squares are not parallel to each other.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $(a, b, c)$ be a \\textbf{Pythagorean Triple}, that is, $a, b, c$ are positive integers with $a^2 + b^2 = c^2$. The key facts are:\n\n1. For any positive integers $m$ and $n$ with $m > n$, $(m^2 - n^2, 2mn, m^2 + n^2)$ is a Pythagorean triple.\n2. The vector $[c-a, c-b, a+b-c]$ has integer length $2c-a-b$.\n\nTo establish (2), compute:\n\n$$\n\\begin{align*}\n& (c-a)^2 + (c-b)^2 + (a+b-c)^2 - (2c-a-b)^2 \\\\\n&= a^2 + b^2 - c^2 = 0.\n\\end{align*}\n$$\n\nCombining these facts, the vector $[2n^2, (m-n)^2, 2n(m-n)]$ has integer length. For $i = 1, 2, \\dots$, let $o_i = 2i + 1$. Set $n = 1$ and $m - n = o_i$, and define:\n\n$$\n\\overrightarrow{AB_i} = \\mathbf{u}_i = [2, 2o_i, o_i^2], \\quad \\overrightarrow{AD_i} = \\mathbf{v}_i = [2o_i, o_i^2 - 2, -2o_i],\n$$\n\nand\n\n$$\n\\mathbf{w}_i = [-o_i^2, 2o_i, -2].\n$$\n\nCheck that $2^2 + (2o_i)^2 + (o_i^2)^2 = (o_i^2 + 2)^2$ and\n\n$$\n\\begin{align*}\n(2o_i)^2 + (o_i^2 - 2)^2 + (-2o_i)^2 &= (o_i^2 + 2)^2.\n\\end{align*}\n$$\n\nThus, $\\mathbf{u}_i$, $\\mathbf{v}_i$, $\\mathbf{w}_i$ have the same integer length. Since $\\gcd(2, o_i) = \\gcd(2o_i, o_i^2 - 2) = 1$, points $B_i = (2, 2o_i, o_i^2)$ and $D_i = (2o_i, o_i^2 - 2, -2o_i)$ are primitive lattice points. Their weights are distinct because $2 + 2o_i + o_i^2 \\neq 4o_i + o_i^2 - 2$ (i.e., $2 \\neq o_i$).\n\n$AB_i$ is perpendicular to $AD_i$ since:\n\n$$\n\\begin{align*}\n\\mathbf{u}_i \\cdot \\mathbf{v}_i &= 4o_i + 2o_i(o_i^2 - 2) - 2o_i^3 \\\\\n&= 0.\n\\end{align*}\n$$\n\nThus, there are infinitely many unbalanced primitive integer squares $AB_iC_iD_i$ (with $C_i = B_i + v_i$).\n\nTo show the planes containing $AB_iC_iD_i$ are not parallel, note that the normal vectors $\\mathbf{w}_i$ are not parallel, since no two are scalar multiples of each other.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 13405,
    "subject": "Mathematics (Olympiad)",
    "question": "Вкупно $2^n$ парички се расподелени на неколку деца. До прерасподелба на паричките доаѓа во ситуација кога некое од децата има барем половина од сите парички: тогаш од паричките на едно такво дете на секое од останатите деца му се префрлаат онолку парички колку што веќе имало. Во случај кога сите парички се кај едно дете нема можност за прерасподелба. Кој е најголемиот можен број последователни прераспределби?\n\nНа пример, ако 32 парички на 6 деца се првично распределени $17, 2, 9, 1, 2, 1$, тогаш после една прерасподелба децата ќе имаат редоследно $2, 4, 18, 2, 4, 2$ парички; во примерот, тој број е два.\n\nОдговорот да се образложи!",
    "options": [],
    "answer": "See solution",
    "solution": "Најмногу $n$ последователни прерасподелби се можни.\n\nЌе започнеме со пример дека $n$ последователни прерасподелби се можни. Нека $2^n$ парички се распределени на 3 деца првично во редослед $1, 2^{n-1} + 2^{n-2} + \\ldots + 2, 1$. Последователните прерасподелби (вкупно $n$ на број) ќе бидат:\n\n$$\n\\begin{array}{l}\n2^1, 2^{n-1} + 2^{n-2} + \\ldots + 2^2, 2^1 \\\\\n2^2, 2^{n-1} + 2^{n-2} + \\ldots + 2^3, 2^2 \\\\\n\\qquad \\dots \\\\\n2^{n-2}, 2^{n-1}, 2^{n-2} \\\\\n2^{n-1}, 0, 2^{n-1} \\\\\n0, 0, 2^n\n\\end{array}\n$$\n\nДа покажеме дека $n$ е максималниот број последователни прерасподелби. Тргнуваме од претпоставка дека постои првична расподелба за која се можни барем $n+1$ прерасподелби и бараме контарадикција: да забележиме дека после една прерасподелба бројот на парички кај секое дете е делив со 2 (кај секое дете кај кое се додавале парички, бројот е дуплиран, па е парен, а кај детето од кое се одземале парички повторно останува парен број бидејќи вкупно има $2^n$ парички т.е. парен број). Аналогно, после втората прерасподелба бројот на парички кај секое дете е делив со 4, итн., се до $n$-тата прерасподелба во која бројот на парички кај секое дете е делив со $2^n$; имајќи предвид дека вкупниот број парички е неменлив и изнесува $2^n$, единствена можна (во некој редослед) е расподелбата $2^n, 0, 0, \\ldots, 0$. Но тогаш нема да има уште една расподелба. Контрадикција!",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 13406,
    "subject": "Mathematics (Olympiad)",
    "question": "Let a square of size $n \\times n$, divided into smaller rectangles, be called *durable* if there does not exist a straight line dividing the square into two parts without intersecting any smaller rectangle in any inner points. Thus, Fig. 27 represents a durable $4 \\times 4$ square, while Fig. 28 does not.\n\nFor any given natural $m$, determine for which natural $n$ it is possible to divide the square $n \\times n$ into rectangles of size $m \\times 1$ and $1 \\times m$ so that the square remains durable.\n\n![](images/UkraineMO_2015-2016_booklet_p28_data_29a38d500d.png)\n\nFig. 27\n\n![](images/UkraineMO_2015-2016_booklet_p28_data_9df7c6317c.png)\n\nFig. 28",
    "options": [],
    "answer": "See solution",
    "solution": "**Answer:**\n\n- For $m=1$: No such division exists.\n- For $m=2$: $n = k m$, $k \\geq 4$.\n- For $m \\geq 3$: $n = k m$, $k \\geq 3$.\n\n**Solution.**\n\nTo divide a square $n \\times n$ into rectangles of size $m \\times 1$, it is necessary that $n^2$ is divisible by $m$. In fact, a stronger statement holds:\n\n**Lemma 1.** A division of a square $n \\times n$ into rectangles $m \\times 1$ exists if and only if $n$ is divisible by $m$.\n\n![](images/UkraineMO_2015-2016_booklet_p28_data_5a0845abee.png)\n\nFig. 29\n\n*Proof.* Sufficiency is clear. For necessity, suppose such a division exists but $n$ is not divisible by $m$. Let $d = \\gcd(n, m) > 1$, so $n = d n_1$, $m = d m_1$ with $\\gcd(n_1, m_1) = 1$. Since $n_1^2 d$ is divisible by $m_1$, $d$ must be divisible by $m_1$, but $m_1 \\neq 1$ (otherwise $n$ is divisible by $m$). Coloring the $d \\times d$ squares with $m_1$ colors diagonally (see Fig. 29), each $m \\times 1$ rectangle covers the same number of each color, but the total number of colored squares $n_1^2$ is not divisible by $m_1$, a contradiction. Thus, $n$ must be divisible by $m$.\n\nNow, for $m=1$, no such division exists. Consider $m=2$ and $m>2$ separately.\n\n**Lemma 2.** A durable square $n \\times n$ divided into rectangles $2 \\times 1$ exists if and only if $n = 2 n_1$ with $n_1 \\geq 4$.\n\n*Proof.* For $n=2$ or $n=4$, no durable square exists (can be checked directly). For $n=6$, there are 10 segments inside the square (see Fig. 30). Each must intersect at least one $2 \\times 1$ rectangle, so at least 10 rectangles are needed. If each segment intersects at least two rectangles, at least 20 are needed, but only $36/2 = 18$ rectangles fit. Thus, some segment intersects only one rectangle, dividing the square into two parts with an odd number of $1 \\times 1$ squares, which cannot be filled with $2 \\times 1$ rectangles. Contradiction.\n\n![](images/UkraineMO_2015-2016_booklet_p28_data_56592dfa12.png)\n\nFig. 30\n\n![](images/UkraineMO_2015-2016_booklet_p28_data_a212d190d4.png)\n\nFig. 31\n\nFor $n = 2 n_1$ with $n_1 \\geq 4$, a durable division is possible as shown in Fig. 32 and Fig. 33. The outer edge is filled with $2 \\times 1$ rectangles so that only certain segments could potentially break durability, but the inner $(2 n_1 - 4) \\times (2 n_1 - 4)$ square can be filled with $2 \\times 2$ squares, ensuring durability.\n\n![](images/UkraineMO_2015-2016_booklet_p29_data_20b07c4b19.png)\n\nFig. 32\n\n![](images/UkraineMO_2015-2016_booklet_p29_data_4d54c20a90.png)\n\nFig. 33\n\n**Lemma 3.** For $m \\geq 3$, a durable square $n \\times n$ divided into rectangles $m \\times 1$ exists if and only if $n = m n_1$ with $n_1 \\geq 3$.\n\n*Proof.* For $n = 2m$, a durable square cannot be built (see Fig. 33). If a horizontal $m \\times 1$ rectangle is placed in the upper left, the upper right cannot be filled horizontally without breaking durability, so vertical rectangles must be used, leading to a vertical segment that breaks durability.\n\n![](images/UkraineMO_2015-2016_booklet_p30_data_c4e7eaea5f.png)\n\nFig. 34\n\nFor $n = m n_1$ with $n_1 \\geq 3$, a durable division is possible as shown in Fig. 34. The outer layer is filled with rectangles in a regular pattern, and the inner square can be filled arbitrarily.\n\nFor example, a durable $12 \\times 12$ square divided into $4 \\times 1$ rectangles is shown in Fig. 35.\n\n![](images/UkraineMO_2015-2016_booklet_p30_data_98bf82a72d.png)\n\nFig. 35",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 13407,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $a, b, c > 0$ and $a + b + c = 3$. Prove that\n\n$$\n\\frac{1}{a\\sqrt{2(a^2 + bc)}} + \\frac{1}{b\\sqrt{2(b^2 + ca)}} + \\frac{1}{c\\sqrt{2(c^2 + ab)}} \\geq \\frac{1}{a+bc} + \\frac{1}{b+ca} + \\frac{1}{c+ab}\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "Without loss of generality, assume $a \\geq b \\geq c$. Then\n\n$$\n\\frac{(c-a)(c-b)}{3(c+ab)} \\geq 0\n$$\n\nand\n\n$$\n\\frac{(a-b)(a-c)}{3(a+bc)} + \\frac{(b-a)(b-c)}{3(b+ca)} = \\frac{c(a-b)^2}{3} \\left( \\frac{1+a+b-c}{(a+bc)(b+ca)} \\right) \\geq 0\n$$\n\nTherefore,\n\n$$\n\\sum_{\\text{cyc}} \\frac{(a-b)(a-c)}{3(a+bc)} \\geq 0\n$$\n\nNow,\n\n$$\n\\begin{align*}\n\\sum_{\\text{cyc}} \\frac{1}{a+bc} &\\leq \\frac{9}{2(ab+bc+ca)} \\\\\n\\Leftrightarrow \\sum_{\\text{cyc}} \\frac{1}{a(a+b+c)+3bc} &\\leq \\frac{3}{2(ab+bc+ca)} \\\\\n\\Leftrightarrow \\sum_{\\text{cyc}} \\left[ \\frac{1}{2(ab+bc+ca)} - \\frac{1}{a(a+b+c)+3bc} \\right] &\\geq 0 \\\\\n\\Leftrightarrow \\sum_{\\text{cyc}} \\frac{(a-b)(a-c)}{a(a+b+c)+3bc} &= \\sum_{\\text{cyc}} \\frac{(a-b)(a-c)}{3(a+bc)} \\geq 0.\n\\end{align*}\n$$\n\nBy the AM-GM inequality,\n\n$$\n\\frac{1}{a\\sqrt{2(a^2 + bc)}} = \\frac{\\sqrt{b+c}}{\\sqrt{2a}\\sqrt{(ab+ac)(a^2+bc)}} \\geq \\frac{\\sqrt{2(b+c)}}{\\sqrt{a(a+b)(a+c)}}\n$$\n\nSo it suffices to prove that\n\n$$\n\\sum_{\\text{cyc}} \\sqrt{\\frac{b+c}{2a}} \\cdot \\frac{1}{(a+b)(a+c)} \\geq \\frac{9}{4(ab+bc+ca)}\n$$\n\nSince $\\sqrt{\\frac{b+c}{2a}} \\leq \\sqrt{\\frac{c+a}{2b}} \\leq \\sqrt{\\frac{a+b}{2c}}$ and\n\n$$\n\\frac{1}{(a+b)(a+c)} \\leq \\frac{1}{(b+c)(b+a)} \\leq \\frac{1}{(c+a)(c+b)},\n$$\n\nby Chebyshev's inequality,\n\n$$\n\\begin{align*}\n\\sum_{\\text{cyc}} \\sqrt{\\frac{b+c}{2a}} \\cdot \\frac{1}{(a+b)(a+c)} &\\geq \\frac{1}{3} \\left( \\sum_{\\text{cyc}} \\sqrt{\\frac{b+c}{2a}} \\right) \\left( \\sum_{\\text{cyc}} \\frac{1}{(a+b)(a+c)} \\right) \\\\\n&= \\frac{2}{(a+b)(b+c)(c+a)} \\left( \\sum_{\\text{cyc}} \\sqrt{\\frac{b+c}{2a}} \\right).\n\\end{align*}\n$$\n\nThus, it suffices to show that\n\n$$\n\\sum_{\\text{cyc}} \\sqrt{\\frac{b+c}{2a}} \\geq \\frac{9(a+b)(b+c)(c+a)}{8(ab+bc+ca)}.\n$$\n\nLet $t := \\sqrt[6]{\\frac{(a+b)(b+c)(c+a)}{8abc}}$. Clearly $t \\geq 1$. Using the AM-GM inequality,\n\n$$\n\\sum_{\\text{cyc}} \\sqrt{\\frac{b+c}{2a}} \\geq 3t.\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 13408,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $n = 10k + 5$ for some integer $k$. Show that $2009$ divides $20^n + 15^n + 8^n + 6^n$.",
    "options": [],
    "answer": "See solution",
    "solution": "Observe that\n\n$$\n20^n + 15^n + 8^n + 6^n = (5^n + 2^n)(4^n + 3^n)\n$$\n\nSince $n = 10k + 5 = 5(2k + 1)$ for some integer $k$,\n\n$$\n5^n + 2^n = 5^{5(2k+1)} + 2^{5(2k+1)}\n$$\n\nThus, $5^5 + 2^5$ divides $5^n + 2^n$. Similarly, $4^5 + 3^5$ divides $4^n + 3^n$. Now, $5^5 + 2^5 = 7 \\cdot 11 \\cdot 41$ and $4^5 + 3^5 = 7 \\cdot 181$. Since $2009 = 7^2 \\cdot 41$, the prime factors of $2009$ occur as factors of the right-hand side of equation $(4)$, so $2009$ divides $20^n + 15^n + 8^n + 6^n$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 13409,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $b$ be a fixed integer greater than or equal to 2.\n\nFor each positive integer $n$, let $S_b(n)$ be the sum of the digits of $n$ when $n$ is expressed in base $b$.\n\nFor example, if $b = 4$, then $S_4(26) = S_4(1 \\times 4^2 + 2 \\times 4 + 2 \\times 1) = 1 + 2 + 2 = 5$.\n\nDetermine all positive integers $m$ with the property that for all $n$, whenever $m$ is a factor of $S_b(n)$, then $m$ is also a factor of $S_b(n + 1) - 1$.",
    "options": [],
    "answer": "See solution",
    "solution": "We first note that if the last digit of $n$ is not $b-1$, then $S_b(n+1) = S_b(n)+1$.\n\n1. Suppose $m \\mid b-1$.\n\nLet $A$ represent the digit $b-1$. Write $n = X \\underbrace{AA\\dots A}_{c \\text{ digits}}$, where the last digit of $X$ is not $A$.\n\nThen\n\n$$\n\\begin{align*}\nS_b(n) + 1 &= S_b(X) + c(b-1) + 1 \\\\\n&\\equiv S_b(X) + 1 \\pmod{m} \\\\\n&= S_b(X + 1) \\\\\n&= S_b(n + 1).\n\\end{align*}\n$$\n\n2. Now suppose that whenever $m$ is a factor of $S_b(n)$, then $m$ is also a factor of $S_b(n+1)-1$.\n\nLet $n = \\underbrace{11\\dots1}_{a \\text{ digits}} 0A$, where $a$ is chosen so that\n\n$$\na + b - 1 = S_b(n) \\equiv 0 \\pmod{m}.\n$$\n\nThen $a+1 = S_b(n+1) \\equiv 1 \\pmod{m}$. Hence $m \\mid a$, whence $m \\mid b-1$.\n\n**Answer:** $m$ has the given property if and only if $m \\mid b-1$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 13410,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $n$ and $m$ be positive integers with $n \\ge 2$. There are $n$ piles having $a_1, \\dots, a_n$ stones such that for each $1 \\le i \\le n$ we have $m \\le a_i \\le m \\cdot i$ (so $a_1 = m$).\n\nAida and Bob play the following game: on each round, Bob picks two non-empty piles (if possible) and he removes a number of stones from the pile with fewer stones (in the case of equality, he chooses randomly). Then Aida decides whether she removes the same number of stones from the other pile or she moves the stones Bob just removed to the other pile. If there is at most one non-empty pile, the game ends.\n\nFind the smallest positive integer $k$ such that Aida can guarantee that at some point there will be at most $k$ stones, regardless of the sequence $a_1, \\dots, a_n$ and of how Bob plays.",
    "options": [],
    "answer": "See solution",
    "solution": "**Answer:** $k = m$.\n\nLet the piles be labeled $1, 2, \\dots, n$ such that the pile labeled $i$ has $a_i$ stones.\n\nWe first provide a construction achieving $k \\ge m$. Let $a_1 = a_2 = \\dots = a_{n-1} = m$ and $a_n = m n$. Then each stone removed from pile $n$ can be matched to a stone removed from one of the first $n-1$ piles. Since the first $n-1$ piles have $m(n-1)$ stones in total, it follows that we cannot remove more than $m(n-1)$ stones from the last pile, so the game will finish with at least $m$ stones.\n\nWe now prove that Aida can always force a position with at most $m$ stones. Before going further, we will prove that the game must end, regardless of how Aida and Bob play. Observe that after each move, the sum $a_1 + \\dots + a_n$ either decreases or stays constant, and moreover, if it stays constant, then $a_1^2 + \\dots + a_n^2$ increases. Hence, if the game never ends, from some point the total number of stones remains invariant, while $a_1^2 + \\dots + a_n^2$ increases on each round, contradiction.\n\nThe rest of the proof relies on the following lemma:\n\n**Lemma.** It is possible to split $\\{1, 2, \\dots, n\\}$ into two non-empty subsets $A$ and $B$ such that\n\n$$\n\\left| \\sum_{i \\in A} a_i - \\sum_{j \\in B} a_j \\right| \\le m.\n$$\n\n*Proof.* Imagine a balance. First, place the pile with $a_n$ stones on one of the pans and then place $a_{n-1}$ on the other pan. After having placed $a_{i+1}, \\dots, a_n$ on the balance, place $a_i$ on the lighter pan (ties broken arbitrarily).\n\nWe claim that when $a_n, a_{n-1}, \\dots, a_i$ are placed on the balance, the difference of weight is less than or equal to $m i$. We prove this by induction.\n\nThe base case $i = n$ is trivial.\n\nFor the inductive step, assume that immediately before placing $a_i$ the difference in weight is $t \\ge 0$. Then the pans will differ by $|t - a_i|$ after placing $a_i$. Since $t \\le m(i+1)$ by the inductive hypothesis, we have $-m i \\le t - a_i \\le m i$, so $|t - a_i| \\le m i$.\n\nAt the end, we will get the desired partition. $\\square$\n\nLet's return to the problem. Aida can use the following strategy: she splits the piles into two groups, the ones with labels in $A$ and the ones with labels in $B$, where $A \\cup B = \\{1, 2, \\dots, n\\}$ is a partition such that\n\n$$\n0 \\le \\sum_{i \\in A} a_i - \\sum_{j \\in B} a_j \\le m,\n$$\nwhich is possible by the Lemma. Then, if Bob picks two piles from the same group, Aida moves the removed stones to the larger pile, whereas if Bob picks two piles from different groups, Aida will remove stones from the larger pile.\n\nIf the two groups initially have total sums of $X$, respectively $Y$, then notice that $h = X - Y \\in [0, m]$ is invariant due to this strategy. At the end of the game, there will be at most one nonempty pile. Since $X - Y$ is invariant, all piles in $B$ will be empty, and the piles in $A$ will have a total of $h \\le m$ stones, as desired.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 13411,
    "subject": "Mathematics (Olympiad)",
    "question": "Given positive integers $a$, $b$, $c$ which are pairwise coprime, let $f(n)$ denote the number of nonnegative integer solutions $(x, y, z)$ to the equation $a x + b y + c z = n$. Prove that there exist real constants $\\alpha$, $\\beta$, $\\gamma$ such that for every nonnegative real number $n$,\n\n$$\n\\left| f(n) - (\\alpha n^2 + \\beta n + \\gamma) \\right| < \\frac{a + b + c}{12}.\n$$\n\n![](path/to/file.png)",
    "options": [],
    "answer": "See solution",
    "solution": "Consider the generating function for $f(n)$:\n\n$$\n\\begin{align*}\nG(t) &= \\sum_{n=0}^{\\infty} f(n) t^n \\\\\n&= (1 + t^a + t^{2a} + \\dots)(1 + t^b + t^{2b} + \\dots)(1 + t^c + t^{2c} + \\dots) \\\\\n&= \\frac{1}{(1 - t^a)(1 - t^b)(1 - t^c)}.\n\\end{align*}\n$$\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 $a$th, $b$th, $c$th roots of unity, respectively. Consider the partial fraction decomposition:\n\n$$\n\\frac{1}{(1 - t^a)(1 - t^b)(1 - t^c)} = \\frac{h_0}{(1 - t)} + \\frac{h_1}{(1 - t)^2} + \\frac{h_2}{(1 - t)^3} \\\\\n+ \\sum_{k=1}^{a-1} \\frac{d_k}{1 - \\omega^k t} + \\sum_{k=1}^{b-1} \\frac{e_k}{1 - \\tau^k t} + \\sum_{k=1}^{c-1} \\frac{f_k}{1 - \\rho^k t},\n$$\n\nwhere 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 both sides by $(1 - \\omega^k t)$ and let $t \\to \\omega^{-k}$ (i.e., set $1 - \\omega^k t = 0$). All other terms vanish, so:\n\n$$\n\\begin{align*}\nd_k &= \\lim_{t \\to \\omega^{-k}} \\frac{1 - \\omega^k t}{(1 - (\\omega^k t)^a)(1 - t^b)(1 - t^c)} \\\\\n&= \\frac{1}{a} \\cdot \\frac{1}{(1 - \\omega^{-k b})(1 - \\omega^{-k c})}.\n\\end{align*}\n$$\n\nIn the expansion of $G(t)$, the coefficient of $t^n$ is:\n\n$$\nf(n) = h_0 + h_1 (n + 1) + h_2 \\frac{(n + 1)(n + 2)}{2} + \\sum_{k=1}^{a-1} d_k \\omega^{k n} + \\sum_{k=1}^{b-1} e_k \\tau^{k n} + \\sum_{k=1}^{c-1} f_k \\rho^{k n}.\n$$\n",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 13412,
    "subject": "Mathematics (Olympiad)",
    "question": "The edge $DA$ of a pyramid $ABCD$ is perpendicular to its base $ABC$. $(ABD) \\perp (BCD)$, $\\angle BDC = 45^\\circ$ and $DB = 2$. Find $\\angle ADB$ if the sum of the squares of the lateral faces of the pyramid equals $8$.",
    "options": [],
    "answer": "See solution",
    "solution": "Since $(ABD) \\perp (ABC)$ and $(BCD)$, then $(ABC) \\perp BC$ and hence $\\angle ABC = \\angle DBC = 90^\\circ$. Since $\\angle BDC = 45^\\circ$, then $BC = BD = 2$. Let $\\angle ADB = \\alpha$. Then $AB = 2 \\sin \\alpha$, $AD = 2 \\cos \\alpha$ and $AC = 2\\sqrt{\\sin^2\\alpha + 1}$. So\n\n$$\nS_{ABD} = \\frac{AB \\cdot AD}{2} = 2 \\sin \\alpha \\cdot \\cos \\alpha,\n$$\n\n$$\nS_{ACD} = \\frac{AC \\cdot AD}{2} = 2 \\cos \\alpha \\sqrt{1 + \\sin^2 \\alpha},\n$$\n\n$$\nS_{BCD} = \\frac{BC \\cdot BD}{2} = 2.\n$$\n\nIt follows by the condition of the problem that\n\n$$\n8 = 4(\\sin^2 \\alpha \\cos^2 \\alpha + \\cos^2 \\alpha(1 + \\sin^2 \\alpha) + 1) = 4(\\sin^2 \\alpha - 2 \\sin^4 \\alpha + 2),\n$$\n\ni.e. $\\sin \\alpha = \\frac{\\sqrt{2}}{2}$. Thus $\\angle ADB = 45^\\circ$.",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 13413,
    "subject": "Mathematics (Olympiad)",
    "question": "Minimise the expression:\n\n$$\n|x+1| + 2|x-5| + 2\\left|x - \\frac{7}{2}\\right| + |x - 11|\n$$\n\nover all real values of $x$.",
    "options": [],
    "answer": "See solution",
    "solution": "By the triangle inequality, we have $|x-a| + |x-b| \\geq |(x-a) + (b-x)| = b-a$ for any $a \\leq b$, and equality holds whenever $x$ lies between $a$ and $b$.\n\nWe can pair up the numbers as follows: $(-1, 11)$, $(-1, 5)$, $(\\frac{7}{2}, 5)$, $(\\frac{7}{2}, 5)$, $(\\frac{7}{2}, 5)$, leaving a single number $\\frac{7}{2}$. The total distance from $x$ to each pair is minimised when $x$ lies between them. Thus, the minimum is achieved by minimising the distance from $x$ to $\\frac{7}{2}$ while ensuring $x$ lies between each pair. This is achieved by taking $x = \\frac{7}{2}$.\n\nTherefore, the minimum value is:\n\n$$\n\\left|\\frac{7}{2} + 1\\right| + 2\\left|\\frac{7}{2} - 5\\right| + 2\\left|\\frac{7}{2} - \\frac{7}{2}\\right| + \\left|\\frac{7}{2} - 11\\right| =  \\frac{9}{2} + 2 \\times \\frac{3}{2} + 0 + \\frac{9}{2} = \\frac{9}{2} + 3 + \\frac{9}{2} = 9 + 3 = 12.\n$$\n\n**Remark:** The total distance from $x$ to a given set of numbers is minimised when $x$ is the median of this set.",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 13414,
    "subject": "Mathematics (Olympiad)",
    "question": "(a) Given a positive integer $k$, prove that there do not exist two distinct integers in the open interval $(k^2, (k+1)^2)$ whose product is a perfect square.\n\n(b) Given an integer $n > 2$, prove that there exist $n$ distinct integers in the open interval $(k^n, (k+1)^n)$ whose product is the $n$-th power of an integer, for all but a finite number of positive integers $k$.",
    "options": [],
    "answer": "See solution",
    "solution": "(a) Suppose that two such integers, $a < b$, exist. Let $m$ be the squarefree part of $a$; $a = ms^2$, $s \\in \\mathbb{Z}$. Then $mb$ is a square, so since $m$ is squarefree, $b = mt^2$ for some integer $t \\ge s + 1$. Hence\n$$\n\\left(1 + \\frac{1}{s}\\right)^2 \\le \\frac{t^2}{s^2} = \\frac{b}{a} < \\frac{(k+1)^2}{k^2} = \\left(1 + \\frac{1}{k}\\right)^2,\n$$\nso $s > k$. Consequently, $b = mt^2 \\ge m(s+1)^2 > (k+1)^2$ -- a contradiction.\n\n(b) We show that the statement holds whenever $k \\ge 3^{n-1}$; if $n$ is odd, the weaker condition $k \\ge 2^{n-1}$ suffices. We require a *LEMMA* whose proof offers no difficulty.\n\n*LEMMA*. If $C > 0$ and $n \\ge 2$, then $(k+1)^n > (k^{n/(n-1)} + C)^{n-1}$ for all $k \\ge C^{n-1}$.\n\nNow let $n$ be odd, let $a$ be the smallest integer greater than $k^{n/(n-1)}$, and let $b = a+1$. The $n$ integers $a^{n-i-1}b^i$, $i = 0, 1, 2, \\dots, n-1$, are distinct, and their product is the $n$-th power of $(ab)^{(n-1)/2}$. The smallest of them, $a^{n-1}$, exceeds $k^n$, and the largest, $b^{n-1}$, is at most $(k^{n/(n-1)} + 2)^{n-1}$. By the *LEMMA*, this is at most $(k+1)^n$ when $k \\ge 2^{n-1}$.\n\nNext, we turn to the case of even $n$. Let again $a$ be the smallest integer greater than $k^{n/(n-1)}$, and let $b = a+1$ and $c = a+2$. Let further $x_i = a^{n-i}b^i$ and $y_i = b^{i-1}c^{n-i}$, $i = 1, 2, \\dots, n-1$. Note that\n$$\nk^n < x_1 < x_2 < x_3 < \\dots < x_{n-1} < y_{n-1} < y_{n-2} < \\dots < y_2 < y_1 \\le (k^{n/(n-1)} + 3)^{n-1},\n$$\nso these $2n-2$ integers are distinct, and by the *LEMMA* they are in the proper range when $k \\ge 3^{n-1}$. We end the proof by showing that we can choose $n$ of these integers so that the product is a perfect $n$-th power.\n\nIf $n$ is divisible by 4, say $n = 4m$, choose\n\n$x_1, x_3, x_5, \\dots, x_{4m-1}, y_1, y_3, y_5, \\dots, y_{4m-1}$;\n\ntheir product is the $n$-th power of $a^m b^{2m-1} c^m$.\n\nIf $n = 8m + 2$, $m > 0$, take\n\n$x_2, x_3, x_4, \\dots, x_{4m+2}, y_2, y_3, y_4, \\dots, y_{4m+2}$;\n\nthe product of these is the $n$-th power of $a^{3m}b^{2m+1}c^{3m}$.\n\nFinally, if $n = 8m + 6$, select\n$$\nx_1, x_2, x_3, \\dots, x_{4m+3}, y_1, y_2, y_3, \\dots, y_{4m+3},\n$$\nin which case the product is the $n$-th power of $a^{3m+2}b^{2m+1}c^{3m+2}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 13415,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $24_b$ denote the number $24$ in base $b$, and $521_b$ denote the number $521$ in base $b$.\n\nGiven:\n- $24_b = 2b + 4$\n- $521_b = 5b^2 + 2b + 1$\n- $521_b = (2b + 4)^2 = 4b^2 + 16b + 16$\n\nFind the value of $b$.",
    "options": [],
    "answer": "See solution",
    "solution": "We have $24_b = 2b + 4$ and $521_b = 5b^2 + 2b + 1$. Setting $521_b = (2b + 4)^2$, we get:\n\n$$\n5b^2 + 2b + 1 = 4b^2 + 16b + 16\n$$\n\nSubtracting $4b^2 + 16b + 16$ from both sides:\n\n$$\n5b^2 + 2b + 1 - (4b^2 + 16b + 16) = 0 \\\\\n5b^2 - 4b^2 + 2b - 16b + 1 - 16 = 0 \\\\\nb^2 - 14b - 15 = 0\n$$\n\nFactoring:\n\n$$\n(b - 15)(b + 1) = 0\n$$\n\nSo $b = 15$ or $b = -1$. Since $b$ must be positive and greater than $4$, $b = 15$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 13416,
    "subject": "Mathematics (Olympiad)",
    "question": "For which integers $n \\geq 2$ is it possible to draw $n$ distinct straight lines in the plane in such a way that there are at least $n-2$ points where exactly three of the lines intersect?",
    "options": [],
    "answer": "See solution",
    "solution": "For $n=2$, any two lines satisfy the condition, and for $n=3$, we can take any three lines passing through a common point.\n\nFor $n=4$, there is no feasible choice of four lines: suppose there are two points where exactly three lines meet. At most one of the lines can pass through both, so we need at least $1 + 2 \\times 2 = 5$ lines.\n\nThe same argument shows that it is impossible for $n=5$: suppose there are three points where exactly three lines meet. If one line passes through all of them, we still require two further lines through each of the three, and no two of them can coincide. This already gives us $1 + 3 \\times 2 = 7$ lines. If the three points do not lie on a line, then there can be at most one line passing through any two of them, leaving us with at least one more line through each of the points that does not pass through any of the others. This gives us a total of at least $3 + 3 = 6$ lines.\n\nThere is a possible configuration for every $n \\geq 6$: for $n=6$, we can take the (extended) sides and diagonals of any (non-degenerate) quadrilateral. For larger values of $n$, we use an inductive construction: if we start with the sides and diagonals of a quadrilateral for which opposite sides are not parallel, then there are also two intersections of exactly two lines (namely the opposite sides). In each further step, we add a line through one of the intersections of exactly two lines, chosen in such a way that it does not pass through any of the other intersections that were obtained previously. This ensures that we get new intersections of exactly two lines with each step, and the number of points where exactly three lines meet increases by one. Thus the number of points where three lines meet will be (exactly) $n-2$ when the $n$-th line is drawn.\n\nWe conclude that it is possible to draw $n$ lines in a suitable way for all $n \\ne 4, 5$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 13417,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $ABC$ be an acute-angled triangle. Point $P$ is such that $AP = AB$ and $PB \\parallel AC$. Point $Q$ is such that $AQ = AC$ and $CQ \\parallel AB$. Segments $CP$ and $BQ$ meet at point $X$. Prove that the circumcenter of triangle $ABC$ lies on the circumcircle of triangle $PXQ$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $D$ be the vertex of parallelogram $ABDC$. Then $APDC$ and $AQDB$ are isosceles trapezoids. Therefore, the perpendicular bisectors to segments $PD$ and $QD$ coincide with the perpendicular bisectors to $AC$ and $AB$ respectively. The circumcenter $O$ of triangle $ABC$ is also the circumcenter of $DPQ$, and $\\angle POQ = 2\\angle A$.\n\nAlso, since\n$$\n\\angle XPD = \\angle ADP, \\quad \\angle XQD = \\angle ADQ\n$$\nwe obtain that $\\angle PXQ = 2\\angle A$. Thus, $O$, $P$, $Q$, $X$ are concyclic. $\\square$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 13418,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $x, y, z$ be positive numbers such that $\\frac{1}{x} + \\frac{1}{y} + \\frac{1}{z} = 3$. Define variables $a, b, c$ by the equations:\n\n$$\nx = \\frac{2a + bc}{3bc}, \\quad y = \\frac{2b + ca}{3ca}, \\quad z = \\frac{2c + ab}{3ab}.\n$$\n\nProve that:\n\n$$\na^2 + b^2 + c^2 + abc = 4.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "**Lemma 1.**\n\n$$(3x-1)(3y-1)(3z-1) = (3x-1) + (3y-1) + (3z-1) + 2 = 3x+3y+3z-1$$\n\n$$\\Rightarrow 27xyz - 9xy - 9yz - 9zx + (3x+3y+3z-1) = 3x+3y+3z-1$$\n\n$$\\Rightarrow 3xyz = xy + yz + zx$$\n\nwhich implies the needed equality.\n\n*End of Lemma 1 proof.*\n\nSince $3xyz = xy + yz + zx \\ge 3\\sqrt[3]{(xyz)^2} \\Rightarrow xyz \\ge 1 \\Rightarrow x + y + z \\ge 3\\sqrt[3]{xyz} \\ge 3$.\n\nApply Lemma 1. Then:\n\n$$\n\\begin{aligned}\n(4-a^2)(4-b^2)(4-c^2) &= \\left(4-\\frac{4}{(3y-1)(3z-1)}\\right) \\cdot \\left(4-\\frac{4}{(3z-1)(3x-1)}\\right) \\cdot \\left(4-\\frac{4}{(3x-1)(3y-1)}\\right) \\\\\n&= 64 \\cdot 27 \\cdot \\frac{(3yz-y-z)(3zx-z-x)(3xy-x-y)}{(3x-1)^2(3y-1)^2(3z-1)^2} \\le \\frac{8}{27} \\\\\n\\text{(since by Lemma 1, we obtain } (3x-1)(3y-1)(3z-1) = 3x+3y+3z-1 \\ge 8) \\\\\n&\\le 27 \\cdot (3yz-y-z)(3zx-z-x)(3xy-x-y) \\\\\n&\\le 27 \\cdot (27(xyz)^2 - 9(x^2yz(y+z)+\\dots) + 3(yz(y+x)(z+x)+\\dots) \\\\\n&\\qquad -(x+y)(y+z)(z+x)) \\\\\n&= 27 \\cdot (27(xyz)^2 - 18xyz(xy + yz + zx) + 9xyz(x+y+x) \\\\\n&\\qquad + 3((xy)^2 + (yz)^2 + (zx)^2) - (xy + yz + zx)(x+y+x) + xyz) \\\\\n\\text{(since } 3xyz = xy + yz + zx) \\\\\n&= 27 \\cdot (-27(xyz)^2 + 6xyz(x+y+x) + 3((xy)^2 + (yz)^2 + (zx)^2) + xyz) \\\\\n\\text{(since } 9(xyz)^2 = (xy)^2 + (yz)^2 + (zx)^2 + 2x^2yz + 2xy^2z + 2xyz^2) \\\\\n&= 27 \\cdot (3(9(xyz)^2 - 2xyz(x+y+z)) - 27(xyz)^2 + 6xyz(x+y+x) + xyz) = 27xyz \\\\\n\\Rightarrow (4-a^2)(4-b^2)(4-c^2) \\le 27xyz.\n\\end{aligned}\n$$\n\n**Lemma 2.** If positive numbers $x, y, z$ satisfy $\\frac{1}{x} + \\frac{1}{y} + \\frac{1}{z} = 3$, then variables $a, b, c$, defined by $x = \\frac{2a+bc}{3bc}$, $y = \\frac{2b+ca}{3ca}$, $z = \\frac{2c+ab}{3ab}$, satisfy:\n\n$$a^2 + b^2 + c^2 + abc = 4.$$\n\n*Proof.* By plugging in variables into Lemma's condition we obtain\n\n$$\n\\begin{aligned}\n3 &= \\frac{3bc}{2a+bc} + \\frac{3ca}{2b+ca} + \\frac{3ab}{2c+ab} \\\\\n\\Rightarrow (2a+bc)(2b+ca)(2c+ab) = 3(abc)^2 + 4abc(a^2 + b^2 + c^2) + 4a^2b^2 + 4b^2c^2 + 4c^2a^2.\n\\end{aligned}\n$$\n\nOn the other hand,\n\n$$\n\\begin{aligned}\n(2a+bc)(2b+ca)(2c+ab) &= (abc)^2 + 2abc(a^2 + b^2 + c^2) + 4a^2b^2 + 4b^2c^2 + 4c^2a^2 + 8abc \\\\\n\\Rightarrow a^2 + b^2 + c^2 + abc = 4, \\text{ Q.E.D.}\n\\end{aligned}\n$$\n\n*End of Lemma 2 proof.*",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 13419,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $ABC$ be an acute scalene triangle. Let $X$ and $Y$ be two distinct interior points of the segment $BC$ such that $\\angle CAX = \\angle YAB$.\n\nSuppose that:\n\n1) $K$ and $S$ are the feet of perpendiculars from $B$ to the lines $AX$ and $AY$ respectively;\n2) $T$ and $L$ are the feet of perpendiculars from $C$ to the lines $AX$ and $AY$ respectively.\n\nProve that $KL$ and $ST$ intersect on the line $BC$.",
    "options": [],
    "answer": "See solution",
    "solution": "Denote $\\varphi = \\angle XAB = \\angle YAC$, $\\alpha = \\angle CAX = \\angle BAY$. Then, because the quadrilaterals $ABSK$ and $ACTL$ are cyclic, we have\n\n$$\n\\angle BSK + \\angle BAK = 180^{\\circ} = \\angle BSK + \\varphi = \\angle LAC + \\angle LTC = \\angle LTC + \\varphi,\n$$\n\nso, due to the $90^{\\circ}$ angles formed, we have $\\angle KSL = \\angle KTL$. Thus, $KLST$ is cyclic.\n\n![](images/Greece-IMO2019finalbook_p23_data_699a4478ce.png)\n\nConsider $M$ to be the midpoint of $BC$ and $K'$ to be the symmetric point of $K$ with respect to $M$. Then, $BKCK'$ is a parallelogram, and so $BK \\parallel CK'$. But $BK \\parallel CT$, because they are both perpendicular to $AX$. So, $K'$ lies on $CT$ and, as $\\angle KTK' = 90^{\\circ}$ and $M$ is the midpoint of $KK'$, $MK = MT$. In a similar way, we have that $MS = ML$. Thus, the center of $(KLST)$ is $M$.\n\nConsider $D$ to be the foot of the altitude from $A$ to $BC$. Then, $D$ belongs to both $(ABKS)$ and $(ACLT)$. So,\n\n$$\n\\angle ADT + \\angle ACT = 180^{\\circ} = \\angle ABS + \\angle ADS = \\angle ADT + 90^{\\circ} - \\alpha = \\angle ADS + 90^{\\circ} - \\alpha,\n$$\n\nand $AD$ is the bisector of $\\angle SDT$.\n\nBecause $DM$ is perpendicular to $AD$, $DM$ is the external bisector of this angle, and, as $MS = MT$, it follows that $DMST$ is cyclic. In a similar way, we have that $DMLK$ is also cyclic.\n\nSo, we have that $ST$, $KL$ and $DM$ are the radical axes of these three circles, $(KLST)$, $(DMST)$, $(DMKL)$. These lines are, therefore, concurrent, and we have proved the desired result. $\\square$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 13420,
    "subject": "Mathematics (Olympiad)",
    "question": "Find integers $0 < a_1 < a_2 < a_3 < a_4$ such that for any $1 \\leq k < l \\leq 4$, the number $a_k \\cdot a_l + 1$ is a square of an integer.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $a_1 = 2$ and $a_2 = 4$. We seek $a_3$ such that $2a_3 + 1$ and $4a_3 + 1$ are squares, say $b^2$ and $c^2$ respectively. Then $2b^2 - c^2 = 1$, which is a Pell's equation. Consider two consecutive solutions: $(5, 7)$ and $(29, 41)$. Take $2a_3 + 1 = 5^2$ and $2a_4 + 1 = 29^2$, which gives $a_3 = 12$ and $a_4 = 420$. Now, $a_3 \\cdot a_4 + 1$ is also a square, which can be verified. Furthermore, any two consecutive solutions of this Pell's equation could be used.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 13421,
    "subject": "Mathematics (Olympiad)",
    "question": "For non-zero integers $a, b, c$ it holds that:\n$$\n\\frac{a^2}{b} + \\frac{b^2}{c} + \\frac{c^2}{a} = \\frac{a^2}{c} + \\frac{c^2}{b} + \\frac{b^2}{a}.\n$$\n\nDoes it follow that:\n$$\n\\frac{a}{b} + \\frac{b}{c} + \\frac{c}{a} = \\frac{a}{c} + \\frac{c}{b} + \\frac{b}{a}?\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "Consider the following transformation:\n\n$$\n\\left( \\frac{a^2}{b} + \\frac{b^2}{c} + \\frac{c^2}{a} \\right) - \\left( \\frac{a^2}{c} + \\frac{c^2}{b} + \\frac{b^2}{a} \\right) = \\frac{a^3 c - a^3 b + b^3 a - a^3 b + c^3 b - b^3 c}{abc}\n$$\n\nNow we transform only the numerator:\n\n$$\n\\begin{aligned}\na^3 c - a^3 b + b^3 a - a^3 b + c^3 b - b^3 c &= (b-a)(ab^2 + a^2 b + c^2 - ca^2 - abc - cb^2) \\\\\n&= (b-a)(ab^2 + a^2 b + c^2 - ca^2 - abc - cb^2) \\\\\n&= (b-a)(a-c)(b^2 + ab - c^2 - ca) \\\\\n&= (b-a)(a-c)(b-c)(a+b+c).\n\\end{aligned}\n$$\n\nTherefore, the condition\n$$\n\\frac{a^2}{b} + \\frac{b^2}{c} + \\frac{c^2}{a} = \\frac{a^2}{c} + \\frac{c^2}{b} + \\frac{b^2}{a}\n$$\nfor non-zero numbers $a, b, c$ is equivalent to the condition $(b-a)(a-c)(b-c)(a+b+c) = 0$.\n\nSimilarly, we conduct the following transformation:\n\n$$\n\\left(\\frac{a}{b} + \\frac{b}{c} + \\frac{c}{a}\\right) - \\left(\\frac{a}{c} + \\frac{c}{b} + \\frac{b}{a}\\right) = \\frac{a^2 c - a^2 b + b^2 a - a^2 b + c^2 b - b^2 c}{abc} = \\frac{(b-a)(c-a)(c-b)}{abc} = 0.\n$$\n\nNow, as a counterexample, let's choose a triple $a=1, b=2, c=-3$. It satisfies the first equation because $a+b+c=0$. However, the second equation is not correct:\n\n$$\n\\frac{a}{b} + \\frac{b}{c} + \\frac{c}{a} = \\frac{1}{2} - \\frac{2}{3} + \\frac{3}{1} = \\frac{3-4+18}{6} = \\frac{17}{6},\n$$\n$$\n\\frac{a}{c} + \\frac{c}{b} + \\frac{b}{a} = -\\frac{1}{3} + \\frac{3}{2} + \\frac{2}{1} = \\frac{-2+9+12}{6} = \\frac{19}{6}.\n$$\n\nFrom the transformations above, we conclude that the required triples can be of two types:\n\n*Type 1.* Integers $a, b, c$ are pairwise distinct.\n\n*Type 2.* If at least two integers are equal, for example $a = b \\neq c$, then the second equation is correct, while the first is correct only if $c = -(a+b)$.\n\nNow we count the number of such triples. The amount of pairwise different non-zero triples is $4040 \\cdot 4039 \\cdot 4038$. For Type 2, if $a = b \\neq c$, then there are $2020$ numbers to choose $a$ and then $b$ and $c$ are defined. Therefore, there are $3 \\cdot 2020$ such triples.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 13422,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all natural numbers $n$ for which there exist primes $p$ and $q$ such that\n$$\np(p+1) + q(q+1) = n(n+1).\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "The equation is equivalent to $p(p+1) = (n-q)(n+q+1)$. Since the difference of the factors on the right-hand side 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\n$$\n2qk = (k+1)(pk - p - 1). \\qquad (1)\n$$\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 $2q = k+1$.\n\n- In the first case, substituting $k$ into (1) gives $(p-2)(q-2) = 3$, implying $p=3$, $q=5$ (or the other way around). Then $n=6$ by the initial equation.\n- In the second case, similarly we get $(p-1)(q-1) = 1$, where the only possibility is $p=q=2$ and $n=3$.\n\nThus, the only such $n$ are $n=3$ and $n=6$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 13423,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $n \\ge 3$ be an integer. For a convex $n$-gon $A_1A_2\\dots A_n$, consider a line $g$ through $A_1$ that does not contain any other point of the $n$-gon. Let $h$ be the orthogonal to $g$ through $A_1$. We orthogonally project the $n$-gon onto $h$.\n\nFor $j = 1, \\dots, n$, let $B_j$ denote the image of $A_j$. The line $g$ is called *valid* if the points $B_j$ are disjoint.\n\nWe consider all convex $n$-gons and all valid lines $g$. How many different orderings of the points $B_1, \\dots, B_n$ exist?",
    "options": [],
    "answer": "See solution",
    "solution": "Each arrangement of $B_1, \\dots, B_n$ begins with $B_1$ and ends with $B_k$ for some $k$ with $2 \\leq k \\leq n$. From $B_1$ through $B_k$, the projections are arranged from “top to bottom,” and from $B_k$ through $B_n$ and back to $B_1$ from “bottom to top.” The $k-2$ projections $B_2, \\dots, B_{k-1}$ assume some $k-2$ of the $n-2$ intermediate positions between $B_1$ and $B_k$, and each of these choices of $k-2$ positions uniquely determines the entire sequence of the $B_i$. Since there are $\\binom{n-2}{k-2}$ such choices possible, the total number of sequences of projections is\n\n$$\n\\sum_{k=2}^{n} \\binom{n-2}{k-2} = \\sum_{i=0}^{n-2} \\binom{n-2}{i} = 2^{n-2}.\n$$\n\n$\\square$",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 13424,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $O$ be the centre of a two-dimensional coordinate system, and let $A_1, A_2, \\dots, A_n$ be points in the first quadrant and $B_1, B_2, \\dots, B_m$ points in the second quadrant. We associate numbers $a_1, a_2, \\dots, a_n$ to the points $A_1, A_2, \\dots, A_n$ and numbers $b_1, b_2, \\dots, b_m$ to the points $B_1, B_2, \\dots, B_m$, respectively. It turns out that the area of triangle $OA_jB_k$ is always equal to the product $a_jb_k$, for any $j$ and $k$. Show that either all the $A_j$ or all the $B_k$ lie on a single line through $O$.",
    "options": [],
    "answer": "See solution",
    "solution": "Consider first the case that one of the areas is zero, e.g. $\\text{area}(OA_1B_1) = 0$. Then either $a_1 = 0$ or $b_1 = 0$. If $a_1 = 0$, then $\\text{area}(OA_1B_k) = a_1b_k = 0$ for all $k$, which means that all $B_k$ lie on a straight line through $O$ and $A_1$. Likewise, if $b_1 = 0$, then all $A_j$ lie on a straight line through $O$ and $B_1$. So we can assume from now on that none of the areas is zero.\n\nSuppose there are two points among $A_1, A_2, \\dots, A_n$ that are not on the same line through $O$ (since the numbering does not matter, we can assume that $A_1$ and $A_2$ have this property), and at the same time there are two points (again, we can assume them to be $B_1$ and $B_2$) that are not on the same line through $O$. We have\n\n$$\n\\frac{\\text{area}(OA_1B_2)}{\\text{area}(OA_1B_1)} = \\frac{a_1b_2}{a_1b_1} = \\frac{b_2}{b_1} = \\frac{a_2b_2}{a_2b_1} = \\frac{\\text{area}(OA_2B_2)}{\\text{area}(OA_2B_1)}.\n$$\n\nSince triangles $OA_1B_1$ and $OA_1B_2$ have the same base ($OA_1$), their heights must be in a $b_1/b_2$-ratio. The same is true for the heights of triangles $OA_2B_1$ and $OA_2B_2$. If $B_1B_2$ is parallel to $OA_1$, then $b_1 = b_2$, so $B_1B_2$ is parallel to $OA_2$. In this case, $O, A_1$ and $A_2$ lie on one line, contradicting the assumption. The same argument applies if $B_1B_2$ is parallel to $OA_2$.\n\nIf neither $OA_1$ nor $OA_2$ is parallel to $B_1B_2$, let $X$ be the intersection of $OA_1$ and $B_1B_2$, and let $Y$ be the intersection of $OA_2$ and $B_1B_2$.\n\n![](images/s3s2014_p4_data_f5105720b1.png)\n\nSince $A_1$ and $A_2$ are in the first quadrant, $X$ and $Y$ are either in the first or third quadrant. In either case, they are not between $B_1$ and $B_2$. Using similar triangles, we see that the ratio of the heights of $OA_1B_1$ and $OA_1B_2$ is $|XB_1|/|XB_2|$, which must be $b_1/b_2$. The same is true (analogously) for the ratio $|YB_1|/|YB_2|$. Hence\n\n$$\n\\frac{|XB_1|}{|XB_2|} = \\frac{b_1}{b_2} = \\frac{|YB_1|}{|YB_2|}\n$$\n\nIf $X$ and $Y$ are on different sides of $B_1B_2$, one of these ratios is greater than 1, the other less than 1, a contradiction. Thus we assume that $B_1$ is closer to both $X$ and $Y$ (otherwise, we just interchange the roles of $B_1$ and $B_2$), as in the figure. We get\n\n$$\n1 - \\frac{|B_1B_2|}{|XB_2|} = \\frac{|XB_2| - |B_1B_2|}{|XB_2|} = \\frac{|YB_2| - |B_1B_2|}{|YB_2|} = 1 - \\frac{|B_1B_2|}{|YB_2|}\n$$\n\nand thus $|XB_2| = |YB_2|$. This means that $X$ and $Y$ coincide, so $X, Y, O, A_1, A_2$ lie on one line, and we get a contradiction to our assumption again. This completes the proof.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 13425,
    "subject": "Mathematics (Olympiad)",
    "question": "Sally was up for 14 hours. David was up for 24 minutes longer. How long was David up?",
    "options": [],
    "answer": "See solution",
    "solution": "**Alternative i**\n\nTo calculate the time David was up, we use 12-hour times. From 9:21 am to 10:00 am there are 39 minutes. From 10:00 am to 12:00 noon there are 2 hours. From 12:00 noon to 11:00 pm there are 11 hours. From 11:00 pm to 11:45 pm there are 45 minutes. So the time from 9:21 am to 11:45 pm is 13 hours and $39 + 45$ minutes, that is, 14 hours and 24 minutes.\n\n**Alternative ii**\n\nThe time from 9:21 am to 9:21 pm is exactly 12 hours. The time from 9:21 pm to 11:45 pm is exactly 2 hours and 24 minutes. So the time from 9:21 am to 11:45 pm is 14 hours and 24 minutes.\n\n**Alternative iii**\n\nTo calculate the time David was up, we use 24-hour times. The time between 11:45 pm and 9:21 am is $23:45 - 09:21 = (23 - 9) \\text{ hours} + (45 - 21) \\text{ minutes} = 14 \\text{ hours and 24 minutes.}$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 13426,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $N$ be the positive integer $777\text{...}777$, a 313-digit number where each digit is a $7$. Let $f(r)$ be the leading digit of the $r$th root of $N$. What is $f(2) + f(3) + f(4) + f(5) + f(6)$?\n\n(A) 8 (B) 9 (C) 11 (D) 22 (E) 29",
    "options": [],
    "answer": "See solution",
    "solution": "Because $10^r$ is written as a $1$ followed by $r$ zeros, the $r$th root of any number smaller than $10^r$ when written as a decimal has only one digit to the left of the decimal point. Extending this reasoning, the leading digit of the $r$th root of $N$ is the same as the leading digit of the $r$th root of the integer $n$ that has as digits a number of $7$s equal to the remainder when $313$ is divided by $r$. For example, the fifth root of $N$ has the same leading digit as the fifth root of $777$, because the remainder of $313$ when divided by $5$ is $3$. Because $3^5 = 243$ and $4^5 = 1024$, the leading digit is $3$.\n\nUsing this methodology, it follows that $f(2) = 2$, $f(3) = 1$, $f(4) = 1$, $f(5) = 3$, and $f(6) = 1$. The requested sum is $2 + 1 + 1 + 3 + 1 = 8$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 13427,
    "subject": "Mathematics (Olympiad)",
    "question": "Square $ABCD$ is divided into $n^2$ equal elementary squares by drawing lines parallel to its sides. The vertices of the elementary squares are called points of the lattice. A rhombus is called \"good\" if it is not a square, its vertices are points of the lattice, and its diagonals are parallel to the sides of square $ABCD$. Find a closed formula, in terms of $n$ (where $n$ is an integer greater than $2$), for the number of \"good\" rhombuses.",
    "options": [],
    "answer": "See solution",
    "solution": "Drawing lines through the vertices of a rhombus parallel to the sides of the lattice, we find the smallest rectangle in which the rhombus is inscribed. Since the rhombus is symmetric with respect to its center, this rectangle will have sides of even length. Therefore, it is enough to count rectangles with even sides $2s \\times 2t$, where $s \\neq t$. This is equivalent to counting all rectangles $2s \\times 2t$ and then subtracting all squares $2s \\times 2t$.\n\n![](images/Hellenic_2016_p10_data_34ebd7d434.png)\n\n**Case 1: $n = 2k$**\n\nWe count all vertical lines of the lattice from left to right as $1,2,3,\\ldots,2k+1$ and all horizontal lines from bottom to top as $1,2,3,\\ldots,2k+1$. A rectangle $2s \\times 2t$ is determined by two vertical lines and two horizontal lines with even distance between them. Two lines have even distance if their numbers are of the same parity. The number of ways to select two vertical lines with even distance is $\\binom{k}{2}$ (both even) plus $\\binom{k+1}{2}$ (both odd), so $\\binom{k}{2} + \\binom{k+1}{2} = k^2$. Similarly, there are $k^2$ possible selections for the horizontal lines. Thus, there are $k^4$ rectangles of type $2s \\times 2t$.\n\nWe must subtract all squares $2s \\times 2s$. The number of $2s \\times 2s$ squares is $(n-2s+1)^2 = (2k-2s+1)^2$. Summing over $s=1$ to $k$:\n\n$$\n\\sum_{s=1}^{k} (2k - 2s + 1)^2 = \\sum_{s=1}^{k} \\left[(2k+1)^2 - 4s(2k+1) + 4s^2\\right] = \\\\\nk(2k+1)^2 - 2(2k+1)k(k+1) + 4 \\frac{k(k+1)(2k+1)}{6} = \\frac{k(2k-1)(2k+1)}{3}\n$$\n\nTherefore, the number of good rhombuses in the even case is:\n\n$$\nk^4 - \\frac{k(2k-1)(2k+1)}{3} = \\frac{k(k-1)(3k^2 - k - 1)}{3}\n$$\n\n**Case 2: $n = 2k+1$**\n\nSimilarly, the number of ways to select two vertical lines with even distance is $\\binom{k+1}{2} + \\binom{k+1}{2} = k(k+1)$, and the same for horizontal lines. Thus, there are $(k(k+1))^2$ rectangles $2s \\times 2t$. The number of $2s \\times 2s$ squares is $(n - 2s + 1)^2 = (2k + 2 - 2s)^2$. Summing over $s=1$ to $k$:\n\n$$\n\\sum_{s=1}^{k} (2k + 2 - 2s)^2 = \\sum_{s=1}^{k} \\left[(2k + 2)^2 - 4(2k + 2)s + 4s^2\\right] = \\\\\nk(2k + 2)^2 - 2(2k + 2)k(k + 1) + 4 \\frac{k(k+1)(2k+1)}{6} = \\frac{2k(k+1)(2k+1)}{3}\n$$\n\nTherefore, in the odd case, the number of good rhombuses is:\n\n$$\n(k(k+1))^2 - \\frac{2k(k+1)(2k+1)}{3} = \\frac{k(k+1)(3k^2 - k - 2)}{3}\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 13428,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $P$ be a point in the plane of $ABC$, and $\\gamma$ a line passing through $P$. Let $A_1$, $B_1$, $C_1$ be the points where the reflections of lines $PA$, $PB$, $PC$ with respect to $\\gamma$ intersect lines $BC$, $AC$, $AB$, respectively. Prove that $A_1$, $B_1$, $C_1$ are collinear.\n\n![](images/pamphlet1112_main_p18_data_8d4e22a810.png)",
    "options": [],
    "answer": "See solution",
    "solution": "There are several possible configurations depending on the location of $P$ and the orientation of $\\gamma$. We will consider the configuration above but will use directed lengths and angles so our arguments apply to all diagram configurations. By the law of sines on triangles $A_1PB$ and $A_1CP$, we have\n\n$$\nBP \\cdot \\frac{\\sin \\angle A_1 PB}{BA_1} = \\sin \\angle BA_1 P = \\sin \\angle PA_1 C = CP \\cdot \\frac{\\sin \\angle CPA_1}{CA_1}.\n$$\n\nRearranging and considering the two other analogous equalities yields\n\n$$\n\\frac{BA_1}{A_1C} = -\\frac{BP \\sin \\angle A_1 PB}{CP \\sin \\angle CPA_1}, \\quad \\frac{CB_1}{B_1A} = -\\frac{AP \\sin \\angle C_1 PA}{BP \\sin \\angle BPC_1}, \\quad \\text{and} \\quad \\frac{AC_1}{C_1B} = -\\frac{CP \\sin \\angle B_1 PC}{AP \\sin \\angle APB_1}.\n$$\n\nNow, observe that $\\angle CPA_1$ and $\\angle C_1PA$ are angles between lines which are reflections of each other, meaning that they are either equal or supplementary. In either case, applying analogous arguments, we obtain\n\n$$\n\\sin \\angle C_1 PA = \\sin \\angle CPA_1, \\quad \\sin \\angle A_1 PB = \\sin \\angle APB_1, \\quad \\text{and} \\quad \\sin \\angle B_1 PC = \\sin \\angle BPC_1.\n$$\n\nHence, we may multiply the ratios above to obtain\n\n$$\n\\frac{BA_1}{A_1C} \\cdot \\frac{CB_1}{B_1A} \\cdot \\frac{AC_1}{C_1B} = -1,\n$$\n\nshowing that $A_1, B_1$, and $C_1$ are collinear by Menelaus' theorem.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 13429,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $\\ell$ be the length of the cycle of the sequence $a, a^2, a^3, \\dots$ modulo $c$.\n\nThus,\n$$\na^{\\ell + \\ell k} \\equiv a^{\\ell} \\pmod{c} \\quad (*)\n$$\nfor all positive integers $k$ and $\\ell$ large enough.\n\nLet $d = \\gcd(\\ell, c)$. The multiples of $\\ell$ modulo $c$ are the multiples of $d$. Prove that if $c > 1$ then $d < c$.",
    "options": [],
    "answer": "See solution",
    "solution": "Since $d$ divides $c$, $d \\leq c$. Suppose $d = c$. Then $c$ divides $\\ell$. Note that two equal remainders cannot appear in the same cycle, because $a' \\equiv a' \\pmod{c}$ implies $a'^{c} \\equiv a'^{c} \\pmod{c}$, so $i-j$ is a multiple of $\\ell$, a contradiction. Thus, the length of the cycle of $a, a^2, a^3, \\dots$ modulo $c$ does not exceed $c$. If $c$ divides $\\ell$, then $c = \\ell$. This means $c$ divides some $a^n$ and every higher power of $a$, so $\\ell = 1$ and $c = 1$.\n\nNow, proceed by induction on $c$. The result is obvious for $c = 1$. Suppose $c > 1$ and the result holds for every positive integer less than $c$. Then it is true for $d$, since $d < c$. By the induction hypothesis, there exist sufficiently large $n_0, n_1, \\dots, n_{d-1}$ such that\n$$\na^{n_i} + n_i \\equiv i \\pmod{d}\n$$\nfor every $i = 0, 1, \\dots, d-1$.\n\nLet $b = qd + r$, with $0 \\leq r < d$. From $a^{n_r} + n_r = r + md$ and $(*)$,\n$$\na^{n_r + \\ell k} + (n_r + \\ell k) \\equiv a^{n_r} + (n_r + \\ell k) \\equiv r + md + \\ell k \\pmod{c} \\quad (**)\n$$\nAs $k$ varies, $\\ell k \\pmod{c}$ cycles through all multiples of $d$. Thus, there exists $k$ such that $\\ell k \\equiv (q-m)d \\pmod{c}$. Plugging this into $(**)$,\n$$\na^{n_r + \\ell k} + (n_r + \\ell k) \\equiv r + md + (q - m)d \\pmod{c} \\\\\na^{n_r + \\ell k} + (n_r + \\ell k) \\equiv r + qd \\equiv b \\pmod{c}\n$$\nHence, we can take $x = n_r + \\ell k$ and the induction step is complete.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 13430,
    "subject": "Mathematics (Olympiad)",
    "question": "Find the maximum positive number $M$ such that for every $n \\in \\mathbb{N}^*$, there exist positive numbers $a_1, a_2, \\dots, a_n$ and $b_1, b_2, \\dots, b_n$ satisfying:\n\n$$\n\\text{(a)} \\quad \\sum_{k=1}^{n} b_k = 1, \\quad 2b_k \\ge b_{k-1} + b_{k+1}, \\quad k = 2, 3, \\dots, n-1,\n$$\n\n$$\n\\text{(b)} \\quad a_k^2 \\le 1 + \\sum_{i=1}^{k} a_i b_i, \\quad k = 1, 2, \\dots, n,\n$$\n\n$$\n\\text{(c)} \\quad a_n = M.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "First, we show that\n$$\n\\max_{1 \\le k \\le n} a_k < 2, \\quad \\max_{1 \\le k \\le n} b_k < \\frac{2}{n-1}.\n$$\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 $b_m = \\max_{1 \\le k \\le n} b_k$. Using $2b_k \\ge b_{k-1} + b_{k+1}$, we have\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 $b_m > 0$,\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., $a_k \\le \\sqrt{f_k}$.\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$$\n\\begin{align*}\n\\sqrt{f_k} - \\sqrt{f_{k-1}} &< b_k \\cdot \\frac{\\sqrt{f_k}}{\\sqrt{f_k} + \\sqrt{f_{k-1}}} \\\\\n&= b_k \\left( \\frac{1}{2} + \\frac{f_k - f_{k-1}}{2(\\sqrt{f_k} + \\sqrt{f_{k-1}})^2} \\right) \\\\\n&< b_k \\left( \\frac{1}{2} + \\frac{2b_k}{2(\\sqrt{f_k} + \\sqrt{f_{k-1}})^2} \\right) \\\\\n&< b_k \\left( \\frac{1}{2} + \\frac{b_k}{4} \\right) \\\\\n&< \\left( \\frac{1}{2} + \\frac{1}{2(n-1)} \\right) b_k.\n\\end{align*}\n$$\nSumming 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": 13431,
    "subject": "Mathematics (Olympiad)",
    "question": "There is an $n \\times n$ chessboard. Each of the $n^2$ small boxes can display a number from $0$ to $k$, for some positive integer $k$. In each row and column, there is a button: if we press the button in a row (or column), the number in each of the $n$ small boxes in that row (or column, respectively) increases by $1$, and the number $k$ is changed to $0$ (i.e., addition is modulo $k+1$). We call this a \"process.\" Initially, the number $0$ is displayed in every box. After some processes, the numbers have changed. Show that every number in the $n^2$ boxes can be changed to $0$ by taking at most $kn$ processes.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $a_{ij}$ be the number in the box at the intersection of the $i$-th row and $j$-th column in the current state. For each $s$ and $t$ with $1 \\leq s, t \\leq n$, let $c_s$ be the number of times the button in the $s$-th row is pressed, and $d_t$ the number of times the button in the $t$-th column is pressed. Then, $a_{st} = c_s + d_t \\pmod{k+1}$. \n\nThus, the state of the board is determined by the row and column button presses. To reset all boxes to $0$, we can choose appropriate values for $c_s$ and $d_t$ so that $a_{st} \\equiv 0 \\pmod{k+1}$ for all $s, t$. One way is to first set all entries in the first row to $0$ by pressing column buttons as needed (at most $k$ presses per column, $n$ columns: $kn$ presses). Then, for each subsequent row, press the row button as needed (at most $k$ presses per row, $n$ rows: $kn$ presses). However, with careful planning, the total number of processes needed does not exceed $kn$. Therefore, every number in the $n^2$ boxes can be changed to $0$ by at most $kn$ processes.",
    "topic": "Number Theory",
    "subtopic": "Modular Arithmetic"
  },
  {
    "id": 13432,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $a$, $b$, $c$ be positive real numbers such that $a + b + c = 3$. Prove the inequality\n\n$$\n\\frac{a^3+2}{b+2} + \\frac{b^3+2}{c+2} + \\frac{c^3+2}{a+2} \\ge 3.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "From the inequality between the arithmetic and geometric means, we have\n\n$$\n\\frac{a^3 + 2}{b + 2} = \\frac{a^3 + 1 + 1}{b + 2} \\ge \\frac{3\\sqrt[3]{a^3 \\cdot 1 \\cdot 1}}{b + 2} = \\frac{3a}{b + 2}.\n$$\n\nAnalogously,\n$$\n\\frac{b^3 + 2}{c + 2} \\ge \\frac{3b}{c + 2} \\quad \\text{and} \\quad \\frac{c^3 + 2}{a + 2} \\ge \\frac{3c}{a + 2}.\n$$\n\nTherefore,\n$$\n\\frac{a^3 + 2}{b + 2} + \\frac{b^3 + 2}{c + 2} + \\frac{c^3 + 2}{a + 2} \\ge 3 \\left( \\frac{a}{b+2} + \\frac{b}{c+2} + \\frac{c}{a+2} \\right) \\quad (1)\n$$\n\nBy the Cauchy-Bunyakovsky inequality,\n$$\n\\begin{aligned}\n\\frac{a}{b+2} + \\frac{b}{c+2} + \\frac{c}{a+2} &= \\frac{a^2}{a(b+2)} + \\frac{b^2}{b(c+2)} + \\frac{c^2}{c(a+2)} \\\\ &\\ge \\frac{(a+b+c)^2}{a(b+2)+b(c+2)+c(a+2)} = \\frac{(a+b+c)^2}{ab+bc+ca+2(a+b+c)}\n\\end{aligned} \\quad (2)\n$$\n\nFrom $(a-b)^2 + (b-c)^2 + (c-a)^2 \\ge 0$, it follows that\n$$\n(a+b+c)^2 \\ge 3(ab+bc+ca), \\quad \\text{i.e.} \\quad \\frac{1}{ab+bc+ca} \\ge \\frac{3}{(a+b+c)^2} \\quad (3)\n$$\n\nCombining (2) and (3), and using $a+b+c=3$, we obtain the desired inequality. Equality holds if and only if $a = b = c = 1$.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 13433,
    "subject": "Mathematics (Olympiad)",
    "question": "We say a polynomial of degree three with integer coefficients is *good* if it has three real roots and all its roots are irrational numbers between $0$ and $1$.\n\n1. Is there a good polynomial with leading coefficient equal to $10$?\n2. Is there a good polynomial with leading coefficient equal to $13$?",
    "options": [],
    "answer": "See solution",
    "solution": "Answer: (1) No, (2) Yes.\n\n(1) Let $a, b, c$ and $d > 0$ be integers and suppose that\n\n$$\nF(x) = a + bx + cx^2 + dx^3 = d(x - \\alpha)(x - \\beta)(x - \\gamma)\n$$\n\nis a good polynomial with $0 < \\alpha, \\beta, \\gamma < 1$. Let $Q(x) = x(1-x)(2x-1)$. Then it is easy to see that $|Q(x)| \\le \\frac{1}{6\\sqrt{3}}$ for any $0 \\le x \\le 1$. Moreover,\n\n$$\nd^3 Q(\\alpha) Q(\\beta) Q(\\gamma) = 8F(0)F(1)F(1/2) \\neq 0\n$$\n\nis an integer. It follows that $d \\ge 6\\sqrt{3} > 10$, thus there is no good polynomial with leading coefficient $10$.\n\n(2) Let $a, b$ and $c$ be integers and let $F(x) = 13x^3 - ax^2 + bx - c$. First, suppose that $F(0) = -1$ and $F(1) = 1$, then we have $c = 1$ and $a = 11 + b$. Now suppose that $F\\left(\\frac{1}{3}\\right) > 0$ and $F\\left(\\frac{2}{3}\\right) < 0$, then we have $\\frac{47}{6} < b < \\frac{55}{6}$ and it is clear that $F(x)$ is good.\n\nHence $F(x) = 13x^3 - 19x^2 + 8x - 1$ for $b = 8$ and $F(x) = 13x^3 - 20x^2 + 9x - 1$ for $b = 9$ are good polynomials.\n\n**Remark.**\n\n1. Proving that $|d| \\ge 8$ is easy:\n\n$$\n1 \\le |F(0)F(1)| = d^2 \\alpha(1-\\alpha) \\beta(1-\\beta) \\gamma(1-\\gamma) \\le \\frac{d^2}{4^3}.\n$$\n\n2. In fact, $|d| \\ge 13$ (hard!) and for $d = 13$, the aforementioned examples are the only ones.",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 13434,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $c_n$ be a sequence defined recursively as follows:\n\n- $c_0 = 1$\n- $c_{2n+1} = c_n$ for $n \\ge 0$\n- $c_{2n} = c_n + c_{n-2^e}$ for $n > 0$, where $e$ is the maximal nonnegative integer such that $2^e$ divides $n$.\n\nProve that\n$$\n\\sum_{i=0}^{2^n-1} c_i = \\frac{1}{n+2} \\binom{2n+2}{n+1}.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "Observe that the right-hand side of the given expression is the $(n+1)^{\\text{th}}$ Catalan number, which is the number of well-formed strings of $(n+1)$ pairs of parentheses. Given such a string, let its signature $k$ be the integer represented in binary by the $n$-term string whose $r^{\\text{th}}$ digit from the left is $0$ or $1$ according to whether the $(2r+1)^{\\text{th}}$ parenthesis from the left is closed or open, respectively.\n\nLet $d_k$ be the number of well-formed strings of $(n+1)$ pairs of parentheses with signature $k$. Notice that the value of $d_k$ does not depend on $n$, as adding leading $0$'s to the binary representation of $k$ inserts strings of the form $()$ after the leading open parenthesis. It suffices to show that $c_k$ and $d_k$ coincide.\n\nNote that $d_0 = c_0 = 1$ for any $n$, as a string of parentheses with signature $0$ must take the form\n$$\n((() \\cdots ()))\n$$\nIt remains only to show that $d_k$ and $c_k$ satisfy the same recursion. First, if a string has signature $2k+1$, its signature has a trailing $1$, which forces the parenthesis string to end with $()$. Thus, the number of strings of $(n+1)$ pairs of parentheses with signature $2k+1$ equals the number of strings of $n$ pairs of parentheses with signature $k$, which shows that $d_{2k+1} = d_k$ (since $d_k$ was independent of $n$).\n\nNow consider a string of $n+1$ pairs of parentheses with signature $2k$, where $2^e$ is the largest power of $2$ dividing $k$. In this case, $k$ has exactly $e$ trailing $0$'s. Consider the open parenthesis at position $2n-2e-1$. If the parenthesis to its immediate right is a closed parenthesis, then these two may be removed as a pair, leaving a string with signature $k-2^e$. If the parenthesis at position $2n-2e$ is open, then it may be removed with the parenthesis to its right (a closed parenthesis because $2k$ had $e+1$ trailing $0$'s), leaving a string with signature $k$. Both operations are reversible, showing that $d_{2k} = d_k + d_{k-2^e}$.\n\nTherefore, $d_k$ and $c_k$ have the same initial value and are uniquely determined by the same recursion, so $d_k = c_k$ for all $k \\ge 0$, as needed.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 13435,
    "subject": "Mathematics (Olympiad)",
    "question": "Suppose a rectangular sheet of paper has dimensions $a$ and $b$, with $b$ being the longer side. What is the largest possible volume of a cylinder that can be formed by joining either the long sides or the short sides of the sheet together?",
    "options": [],
    "answer": "See solution",
    "solution": "*Suppose the short sides are glued together.*\n\nThen the height of the cylinder is $a$ and the circumference is $b$. If $r$ is the radius, $2\\pi r = b$, so $r = \\frac{b}{2\\pi}$. The volume is:\n\n$$\n\\pi r^2 h = \\pi \\left(\\frac{b}{2\\pi}\\right)^2 a = \\frac{ab^2}{4\\pi}\n$$\n\n*Suppose the long sides are glued together.*\n\nThen the height is $b$ and the circumference is $a$. Now $2\\pi r = a$, so $r = \\frac{a}{2\\pi}$. The volume is:\n\n$$\n\\pi r^2 h = \\pi \\left(\\frac{a}{2\\pi}\\right)^2 b = \\frac{a^2b}{4\\pi}\n$$\n\nComparing, the first volume is $\\frac{ab^2}{4\\pi}$ and the second is $\\frac{a^2b}{4\\pi}$. Since $b > a$, $\\frac{ab^2}{4\\pi} > \\frac{a^2b}{4\\pi}$. Thus, the largest volume is obtained when the short sides are joined together.",
    "topic": "Geometry",
    "subtopic": "Solid Geometry"
  },
  {
    "id": 13436,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all pairs of positive integers $x, y$ for which\n\n$$\nx(x+1) = y(y+1)(y^2+1).\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "Let us consider the given equation as a quadratic in $x$. The discriminant is\n$$\nD = 1 + 4y(y+1)(y^2+1) = 4y^4 + 4y^3 + 4y^2 + 4y + 1.\n$$\nFor integer solutions, $D$ must be a perfect square. Notice that\n$$\n(2y^2 + y)^2 = 4y^4 + 4y^3 + y^2 < D\n$$\nand\n$$\n(2y^2 + y + 1)^2 = D + y^2 - 2y.\n$$\nTherefore, if $y^2 - 2y > 0$, $D$ lies strictly between two consecutive squares and cannot be a perfect square. If $y^2 - 2y = 0$, then $y = 2$ is the only possibility, and\n$$\nx = \\frac{-1 + \\sqrt{D}}{2} = \\frac{-1 + (2y^2 + y + 1)}{2} = 5.\n$$\nIf $y^2 - 2y < 0$, then $y = 1$ is the only possibility, but in this case $x$ is not an integer.\n\nThus, the only solution in positive integers is $(x, y) = (5, 2)$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 13437,
    "subject": "Mathematics (Olympiad)",
    "question": "Determine all pairs $(a, b)$ of positive integers for which\n\n$$\na + b = \\varphi(a) + \\varphi(b) + \\gcd(a, b).\n$$\n\nHere, $\\varphi(n)$ is the number of integers $k \\in \\{1, 2, \\dots, n\\}$ satisfying $\\gcd(n, k) = 1$.",
    "options": [],
    "answer": "See solution",
    "solution": "First, suppose that $a = 1$. Then $\\varphi(1) = 1$. For all positive integers $b$, we have $\\gcd(1, b) = 1$. Therefore, in this case, the equation is $1 + b = 1 + \\varphi(b) + 1$, or equivalently, $\\varphi(b) = b - 1$. This is equivalent to $b$ being a prime number. Hence, the solutions for $a = 1$ are precisely the pairs $(1, p)$ with $p$ a prime number. Similarly, the solutions for $b = 1$ are precisely the pairs $(p, 1)$ with $p$ a prime number.\n\nNow assume that $a, b \\geq 2$. As $\\gcd(b, b) > 1$, we have $\\varphi(b) \\leq b - 1$. Therefore,\n\n$$\n\\gcd(a, b) = a + b - \\varphi(a) - \\varphi(b) \\geq a - \\varphi(a) + 1.\n$$\n\nLet $p$ be the minimal prime divisor of $a$ (which exists as $a \\geq 2$). For all multiples $tp \\leq a$ of $p$, we have $\\gcd(tp, a) > 1$, so $a - \\varphi(a) \\geq \\frac{a}{p}$. Therefore,\n\n$$\n\\gcd(a, b) \\geq a - \\varphi(a) + 1 \\geq \\frac{a}{p} + 1.\n$$\n\nThe two largest divisors of $a$ are $a$ and $\\frac{a}{p}$. Since $\\gcd(a, b)$ is a divisor of $a$ that is at least $\\frac{a}{p} + 1$, it must equal $a$. Hence $\\gcd(a, b) = a$. In the same way, we prove that $\\gcd(a, b) = b$. So $a = b$.\n\nThe equation now is equivalent to $2a = 2\\varphi(a) + a$, so also to $a = 2\\varphi(a)$. Note that $2 \\mid a$. Therefore, write $a = 2^k m$ with $k \\geq 1$ and $m$ odd. By a well-known property of the $\\varphi$-function, $\\varphi(a) = \\varphi(2^k) \\cdot \\varphi(m) = 2^{k-1} \\cdot \\varphi(m)$, and the equation becomes $2^k m = 2 \\cdot 2^{k-1} \\cdot \\varphi(m)$, or equivalently, $m = \\varphi(m)$. This only happens for $m = 1$, so $a = 2^k$ for $k \\geq 1$.\n\nTherefore, the solutions are all pairs $(1, p)$ and $(p, 1)$ with $p$ prime, and all pairs $(2^k, 2^k)$ with $k \\geq 1$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 13438,
    "subject": "Mathematics (Olympiad)",
    "question": "Prove that the ratio\n\n$$\n\\frac{1^1 + 3^3 + 5^5 + \\dots + (2^{2012} - 1)^{(2^{2012} - 1)}}{2^{2012}}\n$$\n\nis an odd integer.",
    "options": [],
    "answer": "See solution",
    "solution": "We will prove by induction that, for all $n \\geq 2$,\n\n$$\n1^1 + 3^3 + 5^5 + \\dots + (2^n - 1)^{(2^n - 1)} \\equiv 2^n \\pmod{2^{n+1}}.\n$$\n\nThe base case $n = 2$ is straightforward. Assume the statement holds for a given $n \\geq 2$; we will prove it for $n+1$, that is,\n\n$$\n1^1 + 3^3 + 5^5 + \\dots + (2^{n+1} - 1)^{(2^{n+1} - 1)} \\equiv 2^{n+1} \\pmod{2^{n+2}}.\n$$\n\nFor $x = 1, 3, \\dots, 2^n - 1$, pair the terms $x^x$ and $(x + 2^n)^{(x + 2^n)}$. Expanding $(x + 2^n)^{(x + 2^n)}$ using the binomial theorem, all terms except the first two are divisible by $2^{2n}$, so modulo $2^{2n}$:\n\n$$\nx^x + (x + 2^n)^{(x + 2^n)} \\equiv x^x + x^x + (x + 2^n) \\cdot 2^n \\cdot x^{x-1} \\equiv 2x^x + 2^n x^x = x^x(2 + 2^n).\n$$\n\nSumming over all $x = 1, 3, \\dots, 2^n - 1$, modulo $2^{n+2}$:\n\n$$\n\\begin{aligned}\n1^1 + 3^3 + \\dots + (2^{n+1} - 1)^{(2^{n+1} - 1)} &\\equiv (1^1 + 3^3 + \\dots + (2^n - 1)^{(2^n - 1)})(2 + 2^n) \\\\\n&\\equiv 2^n (2 + 2^n) \\\\\n&\\equiv 2^{n+1}.\n\\end{aligned}\n$$\n\nThus, the ratio is an odd integer.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 13439,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $n$ and $k$ be two integers with $n > k \\ge 1$. There are $2n + 1$ students standing in a circle. Each student $S$ has $2k$ neighbours—namely, the $k$ students closest to $S$ on the right and the $k$ students closest to $S$ on the left. Suppose that $n+1$ of the students are girls, and the other $n$ are boys. Prove that there is a girl with at least $k$ girls among her neighbours.",
    "options": [],
    "answer": "See solution",
    "solution": "We replace the girls by $1$'s, and the boys by $0$'s, getting the numbers $a_1, a_2, \\dots, a_{2n+1}$ arranged in a circle. We extend this sequence periodically by letting $a_{2n+k+1} = a_k$ for all $k \\in \\mathbb{Z}$. We get an infinite periodic sequence\n\n$$\n\\dots, a_1, a_2, \\dots, a_{2n+1}, a_1, a_2, \\dots, a_{2n+1}, \\dots\n$$\n\nConsider the numbers $b_i = a_i - a_{i-k-1} - 1 \\in \\{-1, 0, 1\\}$ for all $i \\in \\mathbb{Z}$. We know that\n\n$$\nb_{m+1} + b_{m+2} + \\dots + b_{m+2n+1} = 1 \\quad (m \\in \\mathbb{Z}) \\quad (\\diamond)\n$$\n\nIn particular, this yields that there exists some $i_0$ with $b_{i_0} = 1$. Now we want to find an index $i$ such that\n\n$$\nb_i = 1 \\quad \\text{and} \\quad b_{i+1} + b_{i+2} + \\dots + b_{i+k} \\ge 0.\n$$\n\nThis will imply that $a_i = 1$ and\n\n$$\n(a_{i-k} + a_{i-k+1} + \\dots + a_{i-1}) + (a_{i+1} + a_{i+2} + \\dots + a_{i+k}) \\ge k,\n$$\n\nas desired. Suppose, to the contrary, that for every index $i$ with $b_i = 1$ the sum $b_{i+1} + b_{i+2} + \\dots + b_{i+k}$ is negative. We start from some index $i_0$ with $b_{i_0} = 1$ and construct a sequence $i_0, i_1, \\dots$ where $i_j$ is the smallest possible index such that $i_j > i_{j-1} + k$ and $b_{i_j} = 1$. We can choose two numbers among $i_0, i_1, \\dots, i_{2n+1}$ which are congruent modulo $2n + 1$ (without loss of generality, we may assume that these numbers are $i_0$ and $i_T$).\n\nOn the other hand, for every $j$ with $0 \\le j \\le T - 1$ we have\n\n$$\nS_j := b_{i_j} + b_{i_j+1} + \\dots + b_{i_{j+1}-1} \\le b_{i_j} + b_{i_j+1} + \\dots + b_{i_j+k} \\le 0\n$$\n\nsince $b_{i_j+k+1}, \\dots, b_{i_{j+1}-1} \\le 0$. On the other hand, since $(i_T - i_0) \\mid (2n+1)$, from $(\\diamond)$ we deduce\n\n$$\nS_0 + S_1 + \\dots + S_{T-1} = \\sum_{i=i_0}^{i_T-1} b_i = \\frac{i_T - i_0}{2n+1} > 0.\n$$\n\nThis contradiction finishes the solution. $\\square$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 13440,
    "subject": "Mathematics (Olympiad)",
    "question": "Points $A_1$ and $C_1$ lie on the rays $B_1A$ and $B_1C$ respectively, so that $\\angle B_1A_1C_1 = 2\\angle KNB_1$ and $\\angle B_1C_1A_1 = 2\\angle KMB_1$. Prove that the length of the segment $MN$ is not greater than the perimeter of the triangle $\\Delta A_1B_1C_1$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $M_1$ be the point on the ray $A_1C_1$ such that $M_1C_1 = B_1C_1$, and $N_1$ be the point on the ray $C_1A_1$ such that $N_1A_1 = B_1A_1$.\n\nThen\n\n$\\angle B_1M_1K = \\frac{1}{2} \\angle A_1C_1B_1 = \\angle KMB_1$,\n\nso $KM_1MB_1$ is inscribed and $M$ is the projection of $M_1$ on the line $AC$.\n\nAnalogously, $N$ is the projection of $N_1$. Thus, the segment $MN$ is the projection of $M_1N_1$ on the line $AC$, hence\n\n$$\nMN \\leq M_1N_1 = B_1C_1 + A_1C_1 + B_1A_1,\n$$\n\nwhich proves the problem statement.\n\n![](images/UkraineMO2019_booklet_p42_data_d667a24c02.png)\n\n**Fig. 39**",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 13441,
    "subject": "Mathematics (Olympiad)",
    "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": "See 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\n$$\nQF : QE = AF : BE = AD : BD.\n$$\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$$\n\\angle REF = \\angle RDF = 90^\\circ - \\angle DFE = 90^\\circ - \\angle DIB = \\angle IBA\n$$\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": 13442,
    "subject": "Mathematics (Olympiad)",
    "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 each square of the grid is classified as either *passable* or *impassable*. All edges on the boundary of the grid are impassable.\n\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.\n\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.",
    "options": [],
    "answer": "See 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": 13443,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $p > 5$ be a prime. Find all positive integers $x$ such that $5p + x$ divides $5p^n + x^n$, for all $n \\in \\mathbb{N}^*$.",
    "options": [],
    "answer": "See solution",
    "solution": "The following statements are equivalent:\n\n1. $5p + x \\mid 5p^n + x^n$ for all $n \\ge 1$.\n2. $5p + x \\mid 30p^2$.\n\nTo prove $1 \\Rightarrow 2$, set $n = 2$ to obtain $5p + x \\mid 5p^2 + x^2$. Thus, $5p + x \\mid (x + 5p)(x - 5p) + 30p^2$, so $5p + x \\mid 30p^2$.\n\nFor the converse, $5p + x \\mid 30p^2$ implies $5p + x \\mid (x + 5p)(x - 5p) + 30p^2$, so $5p + x \\mid 5p^2 + x^2$. The identity\n$$\n5p^{n+1} + x^{n+1} = (5p^n + x^n)(p + x) - px(5p^{n-1} + x^{n-1})\n$$\nholds for all integers $n \\ge 2$. By induction, the claim follows.\n\nThe divisors of $30p^2$ greater than $5p$ are $6p, 10p, 15p, 30p, p^2, 2p^2, 3p^2, 5p^2, 6p^2, 10p^2, 15p^2$, and $30p^2$. Subtracting $5p$ from these gives the required values for $x$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 13444,
    "subject": "Mathematics (Olympiad)",
    "question": "愛莉拿到一個有理數 $r > 1$ 和一條直線，直線有兩個點 $B \\neq R$，而 $R$ 上放著一個紅色的珠子，$B$ 上放著一個藍色的珠子。愛莉用這些東西來玩一個單人遊戲。每一回合，她選定一個整數 $k$（不一定為正）和一個珠子來移動。如果這個珠子在位置 $X$，而另外一個珠子在位置 $Y$，那麼愛莉會把選中的珠子移動到位置 $X'$，使得 $\\overrightarrow{YX'} = r^k \\overrightarrow{YX}$。愛莉的目標是把紅色的珠子移動到 $B$ 上。找出所有能讓愛莉在 2021 回合內達成目標的有理數 $r > 1$。",
    "options": [],
    "answer": "See solution",
    "solution": "所有 $r = (b+1)/b$，其中 $b = 1, \\dots, 1010$。\n\n設紅色和藍色珠子分別為 $\\mathcal{R}$ 和 $\\mathcal{B}$。在線上設座標，令 $R = 0$，$B = 1$。遊戲過程中，$\\mathcal{R}$ 的座標始終小於 $\\mathcal{B}$，且兩珠子間距始終為 $r^\\ell$，$\\ell \\in \\mathbb{Z}$。設第 $m$ 步後距離為 $d_m = r^{\\alpha_m}$，$m = 0, 1, 2, \\dots$（初始 $\\alpha_0 = 0$）。\n\n若連續兩步移動同一珠子，則可合併為一步。因此，愛莉可交替移動兩珠子，假設 $\\mathcal{R}$ 共移動 $t$ 次。\n\n若第 $m$ 步移動 $\\mathcal{R}$，其座標增加 $d_m - d_{m-1}$。總增量應為 $1$，即：\n\n$$\n( d_0 - d_1 ) + ( d_2 - d_3 ) + \\dots + ( d_{2t-2} - d_{2t-1} ) = 1 + \\sum_{i=1}^{t-1} r^{\\alpha_{2i}} - \\sum_{i=1}^{t} r^{\\alpha_{2i-1}}\n$$\n\n或\n\n$$\n( d_1 - d_2 ) + ( d_3 - d_4 ) + \\dots + ( d_{2t-1} - d_{2t} ) = \\sum_{i=1}^{t} r^{\\alpha_{2i-1}} - \\sum_{i=1}^{t} r^{\\alpha_{2i}}\n$$\n\n視第一步移動哪顆珠子而定。前者 $t \\le 1011$，後者 $t \\le 1010$。兩種情況皆化為：\n\n$$\n\\sum_{i=1}^{n} r^{\\beta_i} = \\sum_{i=1}^{n-1} r^{\\gamma_i}, \\quad \\beta_i, \\gamma_i \\in \\mathbb{Z}\n$$\n\n對某 $n \\le 1011$。若能達成目標，則 $n = 1011$ 時此式有解。\n\n反之，若 $n = 1011$ 時有解，則可構造相應移動序列。問題化為上述等式的可解性。\n\n設 $r = a/b$，代入並整理得：\n\n$$\n\\sum_{i=1}^{2n-1} (-1)^i a^{\\mu_i} b^{N-\\mu_i} = 0, \\quad \\mu_i \\in \\{0, 1, \\dots, N\\}\n$$\n\n模 $a-b$，得 $a-b=1$。模 $a+b$，得 $a+b \\le 2n-1$，即 $b = a-1 \\le n-1 = 1010$。\n\n因此，所有 $r = (b+1)/b$，$b = 1, \\dots, 1010$ 均可達成。\n\n驗證：取 $n = a$，$\\beta_1 = \\cdots = \\beta_a = 0$，$\\gamma_1 = \\cdots = \\gamma_b = 1$，即可。\n\n*補充：* 也可用多項式法證明，結論同上。",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 13445,
    "subject": "Mathematics (Olympiad)",
    "question": "For which positive integers $n$ can a square of size $n \\times n$ be completely covered (without overlaps) by rectangles of size $k \\times 1$ and one square $1 \\times 1$, where:\n\na) $k = 4$;\nb) $k = 8$?",
    "options": [],
    "answer": "See solution",
    "solution": "a) $n$ is odd and greater than $4$.\n\nb) $n = 8m + 9$ and $n = 8m + 15$, where $m \\in \\mathbb{N}$.\n\nFor both (a) and (b), $n$ must be odd and greater than $k$.\n\n**a)** Any odd $n > 4$ can be covered as follows: any stripe of size $4 \\times l$ can be covered by $4 \\times 1$ rectangles. For $5 \\times 5$ and $7 \\times 7$, the required coverage is shown below:\n\n![](images/Ukraine_booklet_2018_p16_data_fe2299c8ab.png \"Fig. 9\")\n\nFor any odd $n = 4m + 5$ or $n = 4m + 7$, cut the square $n \\times n$ into a $5 \\times 5$ or $7 \\times 7$ square and several stripes of size $4 \\times l$.\n\n**b)** Similarly, squares with $n = 8m + 9$ and $n = 8m + 15$ can be covered. Cover $9 \\times 9$ and $15 \\times 15$ squares as shown below:\n\n![](images/Ukraine_booklet_2018_p16_data_42698ad893.png \"Fig. 10\")\n\nEach $n \\times n$ of these sizes can be cut into a $9 \\times 9$ or $15 \\times 15$ square and several stripes of size $8 \\times l$.\n\nSquares with $n = 8m + 11$ and $n = 8m + 13$ cannot be covered. For $11 \\times 11$, coloring the cells black and white shows there are $9$ more black cells than white, but each $8 \\times 1$ rectangle covers equal numbers of each color, so such a covering is impossible. The same argument applies for $n = 8m + 13$ (e.g., $13 \\times 13$), where the difference in black and white cells also prevents coverage.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 13446,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $r$, $s$, and $t$ be the roots of the cubic polynomial $p(x) = x^3 - 2007x + 2002$. Evaluate the expression:\n\n$$\n\\frac{(r-1)(s+1)(t+1) + (s-1)(r+1)(t+1) + (t-1)(r+1)(s+1)}{(r+1)(s+1)(t+1)}\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "Call the indicated sum $S$. Note that $S = 3 - 2R$, where\n$$\nR = \\frac{1}{r+1} + \\frac{1}{s+1} + \\frac{1}{t+1}.\n$$\nMoreover, $r+1$, $s+1$, $t+1$ are the roots of the polynomial $q(x) = p(x-1) = x^3 - 3x^2 - 2004x + 4008$.\n\nUsing the usual formulae for the sums of products of roots of a polynomial $f(x) = \\sum_{i=0}^{n} c_i x^i$, the sum of the reciprocals of the roots of $f$ is $-c_1/c_0$. In particular,\n$$\nR = -(-2004)/4008 = 1/2\n$$\nand\n$$\nS = 3 - 1 = 2.\n$$",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 13447,
    "subject": "Mathematics (Olympiad)",
    "question": "Suppose $r$ and $r+1$ are 8-digit numbers, each with the property that all digits are different and each pair of neighbouring digits forms a productive number.\n\nRecall that if one of these numbers contains 0, then 0 is its last digit and if one of these numbers contains 9, then 9 is its last digit. Also, if neither 0 nor 9 appears, then the number starts with 72 or ends with 27.\n\nFind all such pairs $r$ and $r+1$.",
    "options": [],
    "answer": "See solution",
    "solution": "If $r$ ends in 0, then $r$ ends in 20 and $r+1$ ends in 21. So none of 7, 8, 9 is in $r+1$. Hence $r+1$ has fewer than 8 digits.\n\nIf $r$ ends in 1, then it starts with 72. Hence 2 is repeated in $r+1$.\n\nIf $r$ ends in 2, then $r+1$ starts with 72. Hence 2 is repeated in $r$.\n\nIf $r$ ends in 3, 4, or 5, then $r$ contains neither 0 nor 9. Hence $r$ contains, respectively, 4, 5, 6. So these digits will be repeated in $r+1$.\n\nIf $r$ ends in 6, then it starts with 72. Hence 7 is repeated in $r+1$.\n\nIf $r$ ends in 7, then $r+1$ starts with 72. Hence 7 is repeated in $r$.\n\nIf $r$ ends in 9, then $r$ ends in 49. Hence $r+1$ ends in 50, which is not productive.\n\nIf $r$ ends in 8, then $r+1$ ends in 49. The table shows the possible digits for $r$ and $r+1$.\n\n| $r$        | $r + 1$    |\n|------------|------------|\n| 721xx548   | 721xx549   |\n| 721xx648   | 721xx649   |\n| 721x3548   | 721x3549   |\n| 721x3648   | 721x3649   |\n| 72163548   | 72163549   |\n| 721x5648   | 721x5649   |\n\nThus the only pair of required numbers are $\\textbf{72163548}$ and $\\textbf{72163549}$.",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 13448,
    "subject": "Mathematics (Olympiad)",
    "question": "Points $D$ and $E$ divide side $AB$ of equilateral triangle $ABC$ into three equal parts; $D$ is between $A$ and $E$. Point $F$ on side $BC$ is such that $CF = AD$. Find the sum of the angles\n$$\n\\angle CDF + \\angle CEF.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "The conditions give $BF = BD$ ($= \\frac{2}{3}AB$), also $\\angle DBF = 60^\\circ$, hence triangle $DBF$ is equilateral. Then\n\n$$\nDF \\parallel AC \\text{ as } \\angle BDF = \\angle BAC = 60^\\circ.\n$$\n\nHence $\\angle CDF = \\angle ACD$.\n\nOn the other hand, $\\angle ACD = \\angle BCE$ by the symmetry of the figure (or because triangles $ADC$ and $BEC$ are congruent). Then $\\angle CDF = \\angle BCE$. So\n\n![](images/Argentina_2015_Booklet_p2_data_931b3c8442.png)\n\nthe required sum $\\angle CDF + \\angle CEF$ is equal to $\\angle FCE + \\angle CEF$. By the exterior angle theorem, that last sum equals $\\angle BFE$. Now $FE$ is a median in the equilateral triangle $DBF$, hence also a bisector. Therefore $\\angle BFE = \\frac{1}{2}\\angle BFD = \\frac{1}{2} \\cdot 60^\\circ = 30^\\circ$, which is the answer to the problem.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 13449,
    "subject": "Mathematics (Olympiad)",
    "question": "a) List any two unequal partitions of $2017$ into five parts.\n\nb) What is the smallest possible span for an unequal partition of $2017$ into five parts? Justify your answer.\n\nc) What is the smallest number that can be written as the sum of five unequal positive integers? For which numbers is it possible to partition into five unequal parts with a span of $4$?\n\nd) For any integer $N \\geq 15$, show that it can be partitioned into five unequal positive integers with span $4$ or $5$.",
    "options": [],
    "answer": "See solution",
    "solution": "a) Any two of the following partitions:\n\n$$\n\\begin{aligned}\n2017 &= 400 + 401 + 403 + 406 + 407 \\\\\n&= 400 + 401 + 404 + 405 + 407 \\\\\n&= 400 + 402 + 403 + 405 + 407 \\\\\n&= 399 + 403 + 404 + 405 + 406\n\\end{aligned}\n$$\n\nb) Since there are five parts, the span must be at least $4$. To have a span of $4$, the parts must be consecutive. Since\n\n$$\n\\begin{aligned}\n401 + 402 + 403 + 404 + 405 &= 2015 \\\\\n402 + 403 + 404 + 405 + 406 &= 2020\n\\end{aligned}\n$$\n\nno unequal partition of $2017$ has span $4$. Hence the span must be at least $5$. Since $401 + 402 + 403 + 405 + 406 = 2017$ and $406 - 401 = 5$, $401 + 402 + 403 + 405 + 406$ is an unequal partition of $2017$ with the smallest possible span.\n\nc) The smallest number that is the sum of $5$ unequal parts is $1 + 2 + 3 + 4 + 5 = 15$.\n\nAn unequal partition with $5$ parts has a span of $4$ if and only if its parts are consecutive. So all the numbers that have an unequal partition into $5$ parts with a span of $4$ are of the form\n\n$$\nn + (n + 1) + (n + 2) + (n + 3) + (n + 4) = 5n + 10\n$$\n\nwith $n \\geq 1$.\n\nd) If an unequal partition has $5$ parts, then its span is at least $4$. The smallest number with a $5$-part unequal partition is $1 + 2 + 3 + 4 + 5 = 15$.\n\nEvery integer greater than or equal to $15$ has one of the forms $5n + 10, 5n + 11, 5n + 12, 5n + 13, 5n + 14$ with $n \\geq 1$.\n\nEach of the following $5$-part partitions has a span of $4$ or $5$:\n\n$$\n\\begin{aligned}\n5n + 10 &= n + (n + 1) + (n + 2) + (n + 3) + (n + 4) \\\\\n5n + 11 &= n + (n + 1) + (n + 2) + (n + 3) + (n + 5) \\\\\n5n + 12 &= n + (n + 1) + (n + 2) + (n + 4) + (n + 5) \\\\\n5n + 13 &= n + (n + 1) + (n + 3) + (n + 4) + (n + 5) \\\\\n5n + 14 &= n + (n + 2) + (n + 3) + (n + 4) + (n + 5)\n\\end{aligned}\n$$\n\nSo the smallest $5$-part span for any integer is $4$ or $5$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 13450,
    "subject": "Mathematics (Olympiad)",
    "question": "Andrei represents $2025$ as a sum of $40$ pairwise different positive integers. Find the lowest value that the largest of the $40$ numbers can achieve.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $0 < a_1 < a_2 < a_3 < \\dots < a_{40}$ so that $a_1 + a_2 + a_3 + \\dots + a_{40} = 2025$. Then $a_2 \\geq a_1 + 1$, $a_3 \\geq a_2 + 1$, $a_4 \\geq a_3 + 1$, \\dots, $a_{40} \\geq a_{39} + 1$.\n\nConsequently, $a_{40} \\geq a_1 + 39 \\geq a_2 + 38 \\geq \\dots \\geq a_{38} + 2$.\n\nThen\n$$\n40 \\cdot a_{40} \\geq (a_1+39)+(a_2+38)+\\dots+(a_{37}+3)+(a_{38}+2)+(a_{39}+1)+a_{40}\n$$\nhence\n$$\n40 \\cdot a_{40} \\geq 2025 + (1 + 2 + \\dots + 39) = 2805.\n$$\nSince $a_{40}$ is a positive integer, it follows that $a_{40} \\geq 71$.\n\nThe value $a_{40} = 71$ can be achieved: $1 + 29 + 34 + 35 + 36 + \\dots + 70 + 71 = 2025$, therefore the required minimum is $71$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 13451,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $\\Gamma$ be a lattice polygon. Suppose $BC$ is a sloped segment of the boundary of $\\Gamma$ with no lattice points in its interior, and let $P$ be a lattice point such that the segments $PB$ and $PC$ are not sloped, and the triangle $PBC$ lies outside $\\Gamma$. Prove that there exists a unique lattice point $X$ in the triangle $PBC$ such that the area of the triangle $XBC$ is $1/2$.\n\n*Lemma 1.* Let $BC$ be a sloped segment of the boundary of $\\Gamma$ with no lattice points in its interior, and let $P$ be a lattice point such that the segments $PB$ and $PC$ are not sloped, and the triangle $PBC$ lies outside $\\Gamma$. Then there exists a unique lattice point $X$ in the triangle $PBC$ such that the area of the triangle $XBC$ is $1/2$.",
    "options": [],
    "answer": "See solution",
    "solution": "*Proof.* Choose a lattice point $Q$ such that $PBQC$ is a rectangle. As in Solution 1, let $\\ell$ denote the line through some lattice point parallel to $BC$, lying outside $\\Gamma$, and nearest to the line $BC$ under these constraints (see the left figure below). The line $\\ell$ crosses the interior of the angle $BQC$ along an interval of length $> BC$, so this interval should contain a lattice point $X$. The triangle $XBC$ contains no lattice points apart from the vertices, so its area is $1/2$ due to Pick's formula. Moreover, any such point $X$ should lie within the angle $BPC$, and $\\ell$ crosses this angle along a segment of length $< BC$. Hence $X$ is the required unique lattice point. $\\square$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 13452,
    "subject": "Mathematics (Olympiad)",
    "question": "![](images/V_Britanija_2014_p34_data_8c233eacb9.png)\n\nInvert the diagram about $P$. Circles through $P$ become straight lines, and their centres become the reflections of $P$ in these lines. For example, $O_A$ is the reflection of $P$ in $BC$. Prove that the circles $PO_AO'_A$, $PO_BO'_B$, and $PO_CO'_C$ concur somewhere other than $P$.",
    "options": [],
    "answer": "See solution",
    "solution": "In the inverted diagram about $P$, circles through $P$ become straight lines, and their centres become the reflections of $P$ in these lines. For example, $O_A$ is the reflection of $P$ in $BC$. Thus, the problem now becomes to prove that the circles $PO_AO'_A$, $PO_BO'_B$, and $PO_CO'_C$ concur somewhere other than $P$.\n\nThe centre of $PO_AO'_A$ must lie on $BC$, the perpendicular bisector of $PO_A$, and likewise it must lie on $B'C'$. Hence, it is their intersection, which we will call $D$. Define $E$ and $F$ similarly. $P$ is a centre of perspective for $ABC$ and $A'B'C'$, so $DEF$ is a straight line by the theorem of Desargues. Hence, the reflection of $P$ in this line lies on all three circles, as required.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 13453,
    "subject": "Mathematics (Olympiad)",
    "question": "Are there positive integers $m$ and $n$ such that $3^m + 3^n + 1$ is a perfect square?",
    "options": [],
    "answer": "See solution",
    "solution": "Assume there is $k$ such that $3^m + 3^n + 1 = k^2$. Obviously, $k$ is odd. The equation is equivalent to $3^m + 3^n = k^2 - 1$.\n\nFor odd $k$, $8$ divides $k^2 - 1$. Since powers of $3$ are congruent to $1$ or $3$ modulo $8$, the sum $3^m + 3^n$ is congruent to $2$, $4$, or $6$ modulo $8$. Thus, $3^m + 3^n = k^2 - 1$ is impossible modulo $8$.\n\nTherefore, there are no positive integers $m$ and $n$ such that $3^m + 3^n + 1$ is a perfect square.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 13454,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $ABC$ be a triangle in which $\\angle A = 60^\\circ$. Let $T$ be the point where the incircle and the nine-point circle of $ABC$ touch each other. If $r$ is the inradius of $ABC$, prove that $AT = r$.",
    "options": [],
    "answer": "See solution",
    "solution": "We first observe that $AI = r \\cos\\frac{A}{2} = 2r$. Let $T'$ be the point on $AI$ such that $AT' = T'I = r$. Obviously, $T'$ lies on the incircle of $ABC$. We show that $T'$ also lies on the nine-point circle of $ABC$.\n\nLet $D$, $E$, $F$ be the midpoints of $BC$, $CA$, $AB$ respectively. Note that $T'F$ is parallel to $IB$. Hence $\\angle AFT' = \\angle ABI = \\frac{B}{2}$. We also observe that $DF$ is parallel to $AC$, so $\\angle BFD = \\angle BAC = A$. It follows that\n\n$$\n\\angle T'FD = 180^\\circ - (A + \\frac{B}{2}).\n$$\n\nSimilarly, we obtain\n\n$$\n\\angle T'ED = 180^\\circ - (A + \\frac{C}{2}).\n$$\n\nIt is easy to check that\n\n$$\n\\angle T'FD + \\angle T'ED = 180^\\circ.\n$$\n\nHence $T'$, $F$, $D$, $E$ are concyclic. Since $F$, $D$, $E$ are on the nine-point circle, it follows that $T'$ is also on the same circle.\n\nHowever, there is only one point at which the incircle and the nine-point circle touch each other (Feuerbach's theorem). It follows that $T = T'$. Hence: $AT = r$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 13455,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $x, y > 0$. Prove that\n$$\n(x^3 + y^3)^2 \\geq 2x^2 y^2 (x^2 + y^2).\n$$\nIn particular, show that $x^3 + y^3 \\geq \\sqrt{2}xy$.",
    "options": [],
    "answer": "See solution",
    "solution": "First, by the AM-GM inequality,\n$$\n\\frac{x^5 y + x^3 y^3}{2} \\geq x^4 y^2 \\quad \\text{and} \\quad \\frac{y^5 x + y^3 x^3}{2} \\geq y^4 x^2.\n$$\n\nAlso, note that\n$$\n\\begin{align*}\nx^6 + y^6 &\\geq x^5 y + x y^5 \\\\\n\\Leftrightarrow x^6 - x^5 y + y^6 - x y^5 &\\geq 0 \\\\\n\\Leftrightarrow x^5(x - y) + y^5(y - x) &\\geq 0 \\\\\n\\Leftrightarrow (x - y)(x^5 - y^5) &\\geq 0,\n\\end{align*}\n$$\nwhich is true for all $x, y > 0$.\n\nThus,\n$$\n\\begin{align*}\nx^6 + x^3 y^3 + x^3 y^3 + y^6 &\\geq (x^5 y + x^3 y^3) + (y^5 x + y^3 x^3) \\\\ &\\geq 2x^4 y^2 + 2x^2 y^4 \\\\\n\\Leftrightarrow (x^3 + y^3)^2 &\\geq 2x^2 y^2 (x^2 + y^2).\n\\end{align*}\n$$\n\nSince $x, y > 0$, we can conclude that $x^3 + y^3 \\geq \\sqrt{2}xy$ as required.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 13456,
    "subject": "Mathematics (Olympiad)",
    "question": "A real nonzero number is assigned to every point in space. It is known that for any tetrahedron $\\tau$, the number written at the incenter equals the product of the four numbers written at the vertices of $\\tau$. Prove that all numbers equal $1$.",
    "options": [],
    "answer": "See solution",
    "solution": "Consider two arbitrary points $X$ and $Y$ and let $x$ and $y$ be the corresponding numbers. Choose points $I$ and $J$ on the line $XY$ such that $XY = YI = IJ$. Let $X'$ on the line $XY$ be such that $XI = JX'$. Consider the plane $\\lambda$ perpendicular to $IJ$ and passing through the midpoint of $IJ$. Let $ABCX$ and $ABCX'$ be two equal regular triangular pyramids with base $ABC$ in the plane $\\lambda$ having incenters $I$ and $J$. Since $n_A n_B n_C n_X = n_I$ and $n_A n_B n_C n_{X'} = n_J$, we have that $n_X = \\frac{n_{X'} n_I}{n_J}$.\n\nMove the plane $\\lambda$ towards point $I$ and consider the spheres $S_I$ and $S_J$ with centers $I$ and $J$ respectively that are tangent to $\\lambda$. Let $A_1B_1C_1X'$ be a regular triangular pyramid with base $A_1B_1C_1$ in $\\lambda$ and insphere $S_J$. The regular triangular pyramid with base $A_1B_1C_1$ and insphere $S_I$ has vertex $X_1$. It follows from the above that $n_{X_1} = \\frac{n_{X'} n_I}{n_J} = n_X$.\n\nWhen $\\lambda$ moves towards $I$, the radius of the sphere $S_I$ tends to zero and $\\triangle A_1B_1C_1$ tends to a triangle which is the base of a regular triangular pyramid with vertex $X'$ and inscribed sphere with center $J$ and radius $IJ$.\n\nWe conclude that $X_1$ tends to $I$. Continuity arguments show that all inner points on the segment $XI$ (with point $X$) are assigned the same number.\n\nThus, $x = y$ and all numbers are equal. It follows from $x^4 = x$ and $x \\neq 0$ that $x = 1$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 13457,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $a, b, c \\ge 0$ satisfy\n$$\na^2 + b^2 + c^2 + abc = 4.\n$$\nProve that\n$$\n0 \\le ab + bc + ca - abc \\le 2.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "Viewing the given equality as a quadratic equation in $a$ and solving for $a$ yields\n\n$$\na = \\frac{-bc \\pm \\sqrt{(b^2 - 4)(c^2 - 4)}}{2}.\n$$\n\nNote that\n\n$$(b^2 - 4)(c^2 - 4) = b^2c^2 - 4(b^2 + c^2) + 16$$\n$$\\leq b^2c^2 - 8bc + 16 = (4 - bc)^2.$$\n\nFor the given equality to hold, we must have $b, c \\leq 2$ so that $4 - bc \\geq 0$. Hence,\n\n$$\na \\leq \\frac{-bc + |4 - bc|}{2} = \\frac{-bc + 4 - bc}{2} = 2 - bc,\n$$\n\nor\n\n$$\n2 - bc \\geq a. \\qquad (4)\n$$\n\nCombining (3) and (4) gives\n\n$$\n2 - bc \\geq a(b + c - bc) = ab + ac - abc,\n$$\n\nor\n\n$$\nab + ac + bc - abc \\leq 2,\n$$\n\nas desired.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 13458,
    "subject": "Mathematics (Olympiad)",
    "question": "The diagonals $AC$ and $BD$ of a cyclic quadrilateral meet at $E$. The midpoints of sides $AB$, $BC$, $CD$, and $DA$ are $P$, $Q$, $R$, and $S$ respectively. Prove that the circles $EPS$ and $EQR$ have the same radius.",
    "options": [],
    "answer": "See solution",
    "solution": "Let the point $F$ be such that $E$ is the midpoint of $AF$. Let the point $G$ be such that $E$ is the midpoint of $GC$. Since $AB = 2PB$, $AF = 2AE$, and $AD = 2AS$, the enlargement center $A$ and scale factor $2$ sends $PES$ to $BFD$. So the circumradius of $BFD$ is twice that of $PES$. Similarly, the circumradius of $BGD$ is twice that of $QER$. So it suffices to show that $BFD$ and $BGD$ have the same circumradius. In fact, we shall show that $BFDG$ is cyclic, which implies that $BFD$ and $BGD$ have the same circumcircle and so clearly have the same circumradius.\n\n![](images/V_Britanija_2012_p13_data_1bf52e360c.png)\n\nTo show that $BFDG$ is cyclic we will use the converse of the intersecting chords theorem. The lines $GF$ and $BD$ intersect at $E$. Now $GE \\cdot EF = CE \\cdot EA$ by definition of $F$ and $G$. But $CE \\cdot EA = BE \\cdot ED$ by the intersecting chords theorem. So $GE \\cdot EF = BE \\cdot ED$ and so, by the converse of the intersecting chords theorem, $BGDF$ is cyclic, as required.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 13459,
    "subject": "Mathematics (Olympiad)",
    "question": "A diagonal line of a (not necessarily convex) polygon with at least four sides is any line through two non-adjacent vertices of that polygon. Determine all polygons with at least four sides satisfying the following condition: the reflection of each vertex in each diagonal line lies inside or on the boundary of the polygon.",
    "options": [],
    "answer": "See solution",
    "solution": "Lozenges (rhombi) alone satisfy the required vertex reflection condition.\n\nLet $K$ be a polygon with at least four sides satisfying the vertex reflection condition in the statement.\n\nBegin by noticing that $K$ is convex: otherwise, the convex hull $\\hat{K}$ of $K$ would have a side $ab$ whose line of support is a diagonal line of $K$ ($a$ and $b$ are, of course, non-adjacent vertices of $K$ and there might virtually be other vertices or even sides of $K$ along the line segment $ab$). The reflection of a third vertex of $\\hat{K}$, and hence of $K$, in the line $ab$ would then fall outside $\\hat{K}$, and hence outside $K$, contradicting the vertex reflection condition $K$ satisfies.\n\nNext, let $a$, $b$, and $c$ be consecutive vertices of $K$, and let $d \\ne a, b, c$ be a fourth vertex. Since the reflections of $a$ and $c$ in the line $bd$ do not fall outside $K$, it follows that $bd$ is the internal bisector of the angle $abc$, and, by convexity, $a$ and $c$ are reflections of one another in the line $bd$.\n\nConsequently, $K$ is a convex quadrilateral $abcd$ such that $a$ and $c$ are reflections of one another in the line $bd$, and $b$ and $d$ are reflections of one another in the line $ac$; that is, $K$ is a lozenge (rhombus).",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 13460,
    "subject": "Mathematics (Olympiad)",
    "question": "There is a rectangular grid with 3 rows and $n$ columns on a blackboard.\n\n(a) Find the number of ways to write exactly one of the numbers $1, 2, \\ldots, 3n$ into each square so that:\n\n1. Different squares contain different numbers.\n2. For each $i = 1, 2, \\ldots, 3n - 1$, the numbers $i$ and $i + 1$ are written into adjacent squares (i.e., squares sharing a side).\n3. The numbers $1$ and $3n$ are written into adjacent squares.\n\n(b) The same question, but instead of condition 3, require that $1$ is written in the leftmost column and $3n$ in the rightmost column.",
    "options": [],
    "answer": "See solution",
    "solution": "(a) All three squares in the leftmost and rightmost columns must be traversed consecutively, since traversing both corner squares of either column requires also traversing the middle square, and this cannot be done twice. In other columns, traversing all three squares consecutively would split the grid into two parts, making it impossible to move between them without revisiting a square. Similarly, traversing three or more consecutive squares in the middle row would also split the grid into isolated parts (see figures below). When visiting two consecutive squares in the middle row, the previous and next squares must be from the same row (top or bottom).\n\nThus, except for the first and last columns, every valid trajectory divides the middle row into pairs of squares visited consecutively. This is impossible for odd $n$. For even $n$, for each pair in the middle row, we can choose the row (top or bottom) for the previous and next square; there are $\\frac{n-2}{2}$ such pairs, giving $2^{\\frac{n-2}{2}}$ choices. The trajectory is determined by these choices. For a fixed trajectory, there are $6n$ possibilities (2 directions and $3n$ choices for the initial square). Therefore, the total number is $6n \\cdot 2^{\\frac{n-2}{2}} = 3n \\cdot 2^{\\frac{n}{2}}$ for even $n$.\n\n![](images/EST_ABooklet_2021_p21_data_3f9a425f0b.png)\n\nFig. 15\n\n![](images/EST_ABooklet_2021_p21_data_81266d5bee.png)\n\nFig. 16\n\n(b) The middle square of the leftmost column cannot be the starting point, as either corner would become a dead end. Suppose, without loss of generality, we start from the top left corner and visit $k$ squares in the top row ($1 \\leq k \\leq n$). Going directly to the bottom row would split the grid; turning right along the middle row would prevent visiting squares on the left and reaching the rightmost column. Thus, we must turn left along the middle row. The leftmost $3 \\times k$ part of the grid should be traversed in a Z-shape. The remaining $3 \\times (n-k)$ part is traversed similarly. Repeating this argument, every valid trajectory divides the grid into Z-shapes and S-shapes, with Z-shapes immediately followed by S-shapes, and so on. The total number of such trajectories is $2^n$.",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 13461,
    "subject": "Mathematics (Olympiad)",
    "question": "Consider an arbitrary arrangement of $N$ boys along a circle. A boy is called *tall* if he is taller than both of his neighbors, and *short* if he is shorter than both of his neighbors.\n\nWhat are the possible values for the number $s$ of *middle* boys (those who are neither tall nor short) in such an arrangement?",
    "options": [],
    "answer": "See solution",
    "solution": "Any integer from $0$ to $N-2$ that has the same parity as $N$ is possible for $s$.\n\nIn any arrangement, the numbers of tall and short boys are equal. Let $b$ be the number of tall boys; then the number of middle boys is $s = N - 2b$, which has the same parity as $N$. Since $b$ can take any value from $1$ to $\\left\\lfloor N/2 \\right\\rfloor$, $s$ ranges from $0$ to $N-2$ with the same parity as $N$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 13462,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all functions $f: \\mathbb{R} \\to \\mathbb{R}$ which satisfy the conditions:\n$$\nf(x+y) < f(x) + f(y), \\qquad f(f(x)) = [x] + 2.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "Let $f(0) = a$. Then $f(a) = f(f(0)) = 2$, and $f(2) = f(f(a)) = a + 2$. Continuing this procedure, we get $f(2k) = a + 2k$ and $f(a + 2k) = 2k + 2$.\n\nWe have $2k + 2 = f(a + 2k) < f(a) + f(2k) = 2 + a + 2k$, so $a > 0$. If we put $x = y = a$, we get $a + 2a = f(2a) < f(a) + f(a) = 4$, so $3a < 4$, i.e., $a < \\frac{4}{3}$. But from before, $a > 0$.\n\nHence, using $f(2k) = a + 2k$ and $f(a + 2k) = 2k + 2$, we get $f(x) = x + 1$ for all natural numbers $x$.\n\nFor $x = y = \\frac{1}{2}$ in the inequality, we get $2 = f(1) = f(\\frac{1}{2} + \\frac{1}{2}) < 2f(\\frac{1}{2})$, so $f(\\frac{1}{2}) > 1$. On the other hand, $1 + f(\\frac{1}{2}) = f(f(\\frac{1}{2})) = 2$, so $f(\\frac{1}{2}) = 1$, which is a contradiction. It follows that there exists no function satisfying the required conditions.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 13463,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $x = 2011$. Consider the sequence defined by the recurrence relation:\n\n$$\nu_0 = 0, \\quad u_1 = 1, \\quad u_2 = x, \\quad u_n = x u_{n-1} - u_{n-2} \\text{ for } n \\geq 3.$$ \n\nExpand the first few terms of the sequence, and conjecture closed forms for $u_{2n}$ and $u_{2n+1}$ in terms of $u_n$ and $u_{n+1}$. Prove your conjecture by induction.",
    "options": [],
    "answer": "See solution",
    "solution": "Expanding the first few terms:\n\n$$\n\\begin{aligned}\nu_0 &= 0, \\\\\nu_1 &= 1, \\\\\nu_2 &= x, \\\\\nu_3 &= x^2 - 1, \\\\\nu_4 &= x^3 - 2x, \\\\\nu_5 &= x^4 - 3x^2 + 1, \\\\\nu_6 &= x^5 - 4x^3 + 2x.\n\\end{aligned}\n$$\n\nWe conjecture:\n\n- $u_{2n} = 2u_n u_{n+1} - x u_n^2$\n- $u_{2n+1} = u_{n+1}^2 - u_n^2$\n\nfor $n > 0$.\n\n**Proof by induction:**\n\nBase case ($n=1$):\n\n- $u_2 = x$, $u_3 = x^2 - 1$\n- $u_4 = x^3 - 2x = 2u_2 u_3 - x u_2^2$\n- $u_5 = x^4 - 3x^2 + 1 = u_3^2 - u_2^2$\n\nAssume true for $n$, show for $n+1$:\n\n$$\n\\begin{aligned}\nu_{2n+2} &= x u_{2n+1} - u_{2n} \\\\\n&= x(u_{n+1}^2 - u_n^2) - (2u_n u_{n+1} - x u_n^2) \\\\\n&= x u_{n+1}^2 - 2u_n u_{n+1} \\\\\n&= 2u_{n+1}(x u_{n+1} - u_n) - x u_{n+1}^2 \\\\\n&= 2u_{n+1} u_{n+2} - x u_{n+1}^2\n\\end{aligned}\n$$\n\nSimilarly,\n\n$$\nu_{2n+3} = x u_{2n+2} - u_{2n+1}$$\n\nThus, the conjectured forms hold for all $n > 0$ by induction.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 13464,
    "subject": "Mathematics (Olympiad)",
    "question": "On an infinite square grid, finitely many cars are placed, each occupying a single cell and facing one of the four cardinal directions. No two cars occupy the same cell. The cell immediately in front of each car is empty, and no two cars face towards each other (for example, no right-facing car is to the left of a left-facing car within a row). In a *move*, one chooses a car and shifts it one cell forward to a vacant cell. Prove that there exists an infinite sequence of valid moves using each car infinitely many times.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $S$ be any rectangle containing all the cars. Partition $S$ into horizontal strips of height 1, and color them red and green in an alternating fashion. It is enough to prove all the cars may exit $S$.\n\n![](images/sols-TSTST-2019_p5_data_7832afcfc9.png)\n\nTo do so, we outline a five-stage plan for the cars:\n\n1. All vertical cars in a green cell may advance one cell into a red cell (or exit $S$ altogether), by the given condition. (This is the only place where the hypothesis about empty space is used!)\n2. All horizontal cars on green cells may exit $S$, as no vertical cars occupy green cells.\n3. All vertical cars in a red cell may advance one cell into a green cell (or exit $S$ altogether), as all green cells are empty.\n4. All horizontal cars within red cells may exit $S$, as no vertical car occupy red cells.\n5. The remaining cars exit $S$, as they are all vertical. The solution is complete.\n\n**Remark (Author's comments).** The solution I've given for this problem is so short and simple that it might appear at first to be about IMO 1 difficulty. I don't believe that's true! There are very many approaches that look perfectly plausible at first, and then fall apart in this or that twisted special case.\n\n**Remark (Higher-dimensional generalization by author).** The natural higher-dimensional generalization is true, and can be proved in largely the same fashion. For example, in three dimensions, one may let $S$ be a rectangular prism and partition $S$ into horizontal slabs and color them red and green in an alternating fashion. Stages 1, 3, and 5 generalize immediately, and stages 2 and 4 reduce to an application of the two-dimensional problem. In the same way, the general problem is handled by induction on the dimension.\n\n**Remark (Historical comments).** For $k > 1$, we could consider a variant of the problem where cars are $1 \\times k$ rectangles (moving parallel to the longer edge) instead of occupying single cells. In that case, if there are $2k - 1$ empty spaces in front of each car, the above proof works (with the red and green strips having height $k$ instead). On the other hand, at least $k$ empty spaces are necessary. We don't know the best constant in this case.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 13465,
    "subject": "Mathematics (Olympiad)",
    "question": "Prove that there is a constant $c > 0$ with the following property: If $a, b, n$ are positive integers such that $\\gcd(a + i, b + j) > 1$ for all $i, j \\in \\{0, 1, \\dots, n\\}$, then\n\n$$\n\\min\\{a, b\\} > c^n \\cdot n^{\\frac{n}{2}}.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "Let $a, b, n$ be positive integers as in the statement. Let $P_n$ be the set of prime numbers not exceeding $n$. We need the following lemma:\n\n*Lemma 1.* There is a positive integer $n_0$ such that for all $n \\ge n_0$,\n$$\n\\sum_{p \\in P_n} \\left( \\frac{n}{p} + 1 \\right)^2 < \\frac{2}{3} n^2.\n$$\n*Proof.* Expanding and dividing by $n^2$, and noting $|P_n| \\le n$, it suffices to show\n$$\n\\sum_{p \\in P_n} \\frac{1}{p^2} + \\frac{2}{n} \\sum_{p \\in P_n} \\frac{1}{p} + \\frac{1}{n} < \\frac{2}{3}.\n$$\nSince\n$$\n\\frac{2}{n} \\sum_{p \\in P_n} \\frac{1}{p} < \\frac{2}{n} \\sum_{i=2}^{n} \\frac{1}{i} < \\frac{2}{n} \\log n,\n$$\nit suffices to find $r < \\frac{2}{3}$ with $\\sum_{p \\in P_n} \\frac{1}{p^2} < r$. But\n$$\n\\begin{aligned}\n\\sum_{p \\in P_n} \\frac{1}{p^2} &\\le \\frac{1}{4} + \\frac{1}{9} + \\sum_{k=1}^{n} \\frac{1}{(2k+1)(2k+3)} \\\\\n&= \\frac{1}{4} + \\frac{1}{9} + \\sum_{k=1}^{n} \\frac{1}{2} \\left( \\frac{1}{2k+1} - \\frac{1}{2k+3} \\right) \\\\\n&= \\frac{1}{4} + \\frac{1}{9} + \\frac{1}{2} \\left( \\frac{1}{3} - \\frac{1}{2n+3} \\right) < \\frac{1}{4} + \\frac{1}{9} + \\frac{1}{6} < \\frac{1}{3}\n\\end{aligned}\n$$\nso we can take $r = \\frac{1}{4} + \\frac{1}{9} + \\frac{1}{6}$. $\\square$\n\nFix such $n_0$, and assume $n \\ge n_0$. For any $p \\in P_n$, there are at most $\\frac{n}{p} + 1$ numbers $i \\in \\{0, 1, \\dots, n-1\\}$ with $p \\mid a + i$, and likewise for $b + j$. Thus, at most $\\left(\\frac{n}{p} + 1\\right)^2$ pairs $(i, j)$ with $p \\mid \\gcd(a + i, b + j)$. By the lemma, fewer than $\\frac{2}{3}n^2$ pairs $(i, j)$ with $i, j \\in \\{0, 1, \\dots, n-1\\}$ have $p \\mid \\gcd(a + i, b + j)$ for some $p \\in P_n$.\n\nLet $N$ be the least integer $\\ge \\frac{n^2}{3}$. Thus, at least $N$ pairs $(i, j)$ with $\\gcd(a + i, b + j)$ not divisible by any $p \\in P_n$. For each, choose a prime $p_s > n$ dividing $\\gcd(a + i_s, b + j_s)$. The map $s \\mapsto p_s$ is injective, so $\\prod_{i=0}^{n-1} (a+i)$ is divisible by $\\prod_{s=1}^{N} p_s$. Since $p_s > n$ and distinct,\n$$\n(a+n)^n > \\prod_{i=0}^{n-1} (a+i) \\ge \\prod_{s=1}^{N} p_s \\ge \\prod_{i=1}^{N} (n+2i-1).\n$$\nLet $X = \\prod_{i=1}^{N} (n+2i-1)$. Then\n$$\nX^2 = \\prod_{i=1}^{N} [(n+2i-1)(n+2(N+1-i)-1)] > \\prod_{i=1}^{N} (2Nn) = (2Nn)^N,\n$$\nso\n$$\n(a+n)^n > (2Nn)^{\\frac{N}{2}} \\ge \\left(\\frac{2n^3}{3}\\right)^{\\frac{n^2}{6}}.\n$$\nThus,\n$$\na \\ge \\left(\\frac{2}{3}\\right)^{\\frac{1}{6} n} \\cdot n^{\\frac{n}{2}} - n,\n$$\nwhich exceeds $c^n \\cdot n^{\\frac{n}{2}}$ for large $n$ and any $c < \\left(\\frac{2}{3}\\right)^{1/6}$. Similarly for $b$.\n\nTherefore, $\\min\\{a, b\\} \\ge c^n \\cdot n^{\\frac{n}{2}}$ for large $n$, and by shrinking $c$, for all $n$.\n\nThe argument is not sharp; the factor $n^{\\frac{n}{2}}$ can be improved to $n^{rn}$ for some $r > \\frac{1}{2}$. Thus, for any $c > 0$, the inequality holds for large $n$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 13466,
    "subject": "Mathematics (Olympiad)",
    "question": "We call a point sequence $$(A_0, A_1, \\dots, A_n)$$ *interesting* if:\n- For each $A_i$, its abscissa and ordinate are equal.\n- The slopes of segments $OA_0, OA_1, \\dots, OA_n$ (where $O$ is the origin) strictly increase.\n- The area of each $\\triangle OA_iA_{i+1}$ for $0 \\le i \\le n-1$ 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 $\\overrightarrow{OA} = \\overrightarrow{OA_i} + \\overrightarrow{OA_{i+1}}$. The new sequence $$(A_0, \\dots, A_i, A, A_{i+1}, \\dots, A_n)$$ is called 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 $$(C_0, C_1, \\dots, C_k)$$.",
    "options": [],
    "answer": "See 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 $$(B_0, B_1, \\dots, B_m)$$, with $C_0 = A_0 = B_0$ and $C_k = A_n = B_m$.\n\nBy Pick's Theorem, the area of a triangle equals $1/2$ if and only if there is no grid point on the triangle except its 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 the segment $D_iD_{i+1}$ (a degenerate 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![](images/Mathematical_Olympiad_in_China_2011-2014_p141_data_226216d865.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 $n=1$, 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 $C_1, \\dots, C_{k-1}$. If not, there is no grid point on the interior of segment $OA$, and there exists $i$, $0 \\le i < k$, such that $A$ lies in the angle made by rays $OC_i$ and $OC_{i+1}$.\n\n![](images/Mathematical_Olympiad_in_China_2011-2014_p141_data_d514fab0fe.png)",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 13467,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $[x]$ be the greatest integer not greater than the real number $x$, and let $[[x]] = x - [x]$. Solve the equation\n\n$$\n3[[x]] = x[x] + 1.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "Let $x = n + a$ where $n \\in \\mathbb{Z}$ and $a \\in [0, 1)$. Then $3a = n^2 + an + 1$.\n\nIf $n \\ge 0$, then $3 > 3a = n^2 + an + 1 \\ge n^2 + 1$, hence $n \\le 1$.\n\n- If $n = 0$, we get $3a = 1$, so $a = x = \\frac{1}{3}$.\n- If $n = 1$, we get $2a = 2$, which has no solutions with $a < 1$.\n\nNow let $n < 0$. Then $3 > 3a = n^2 + an + 1 > n^2 + n + 1$, so $0 > n^2 + n - 2 = (n - 1)(n + 2)$. Since $n < 0$, we get $n = -1$.\n\nFor $n = -1$, $4a = 2$, so $a = \\frac{1}{2}$ and $x = -\\frac{1}{2}$.\n\n**The only solutions are** $x = -\\frac{1}{2}$ and $x = \\frac{1}{3}$.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 13468,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $P(n)$ denote the product of the digits of the non-negative integer $n$.\n\nProve that the following sets are unbounded:\n\n(a) $A = \\left\\{ \\frac{P(n)}{P(n^2)} \\right\\}$, where $n$ ranges over all non-negative integers such that $n^2$ does not contain zero in its decimal representation.\n\n(b) $B = \\left\\{ \\frac{P(n^2)}{P(n)} \\right\\}$, where $n$ ranges over all non-negative integers such that $n$ does not contain zero in its decimal representation.",
    "options": [],
    "answer": "See solution",
    "solution": "Both parts are proved by constructing suitable examples, often using mathematical induction.\n\n**(a)** Consider the equality:\n\n$$\n\\left(2\\underbrace{66\\dots68}_{n-1}\\right)^2 = 7\\underbrace{11\\dots18}_{n-1}\\underbrace{22\\dots24}_{n-1}\n$$\n\nAs $n \\to \\infty$, compute the ratio:\n\n$$\n\\frac{P(n)}{P(n^2)} = \\frac{16 \\cdot 6^{n-1}}{7 \\cdot 8 \\cdot 4 \\cdot 2^{n-1}} \\to +\\infty.\n$$\n\n**(b)** Consider the equality:\n\n$$\n\\left(\\underbrace{66\\dots67}_{n-1}\\right)^2 = \\underbrace{44\\dots48}_{n-1}\\underbrace{\\dots89}_{n-1}\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 13469,
    "subject": "Mathematics (Olympiad)",
    "question": "There are three runners at different vertices of an equilateral triangle with side length 1: First, Second, and Third. They start moving simultaneously in the same direction (Second moves toward First, Third moves toward Second, First moves toward Third).\n\nIs it necessary that they all meet at one point at the same time if:\n\n(a) First, Second, and Third have velocities $2008$, $2009$, and $2010$ respectively?\n\n(b) They are moving with distinct natural velocities?",
    "options": [],
    "answer": "See solution",
    "solution": "**(a)**\nWe write the condition for all runners to meet at one point:\n\n$$\n2008t = 2010t + 1 - 3m = 2009t + 2 - 3n,\n$$\nwhere $m, n \\in \\mathbb{Z}$, $t \\in \\mathbb{R}$.\n\nFrom the equations:\n- $t = 3n - 2$ or $2t = 6n - 4$,\n- $2t = 3m - 1 \\implies 3m - 1 = 6n - 4 \\implies m = 2n - 1$.\n\nWe can find solutions, for example, if $m = n = 1$, then $t = 1$. If $t = 1$, First runs $2008$, Second $2009$, Third $2010$, so they meet at one point.\n\n**(b)**\nSuppose First, Second, and Third have velocities $1$, $2$, and $4$ respectively. The condition for meeting at one point becomes:\n\n$$\nt = 4t + 1 - 3m = 2t + 2 - 3n,\n$$\nwhere $m, n \\in \\mathbb{Z}$, $t \\in \\mathbb{R}$.\n\nThen $t = 3n - 2$ and $3t = 3m - 1$. So:\n\n$$\n9n - 6 = 3m - 1\n$$\nfor integer $m, n$. This equation has no solutions, so the runners will not meet at one point.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 13470,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $\\triangle ABC$ be an acute triangle and let $X$ be the foot of the altitude from $A$. Let $Y$ be the intersection of the perpendicular to $AC$ drawn from $X$. If the circumcircle of triangle $ABX$ meets $BY$ at point $Z$ (distinct from $B$), and the extension of $AZ$ meets $XY$ at point $P$, prove that\n\n$$\nBX \\cdot XP = PY \\cdot XC.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "Since $\\triangle AXY$ and $\\triangle AXC$ are both right triangles at $Y$ and $X$ respectively, $\\angle AXP = 90^\\circ - \\angle XAC = \\angle YCB$. On the other hand, since $AZXB$ is a cyclic quadrilateral, $\\angle XAP = \\angle YBC$.\n\n![](images/Spanija_b_2014_p27_data_56c6bdb055.png)\n\nThus, $\\triangle AXP \\sim \\triangle BCY$ and $\\frac{BC}{AX} = \\frac{YC}{XP}$. Likewise, $\\triangle AXC \\sim \\triangle XYC$ because both are right-angled triangles with a common acute angle, so\n\n$$\n\\frac{AX}{XC} = \\frac{XY}{YC}.\n$$\n\nFrom the preceding ratios,\n\n$$\nBC \\cdot XP = AX \\cdot YC = XC \\cdot XY,\n$$\n\nwhich gives\n\n$$\n\\frac{BC}{XC} = \\frac{XY}{XP}.\n$$\n\nSubtracting $1$ from both sides,\n\n$$\n\\frac{BC}{XC} - 1 = \\frac{XY}{XP} - 1 \\Leftrightarrow \\frac{BC - XC}{XC} = \\frac{XY - XP}{XP} \\Leftrightarrow \\frac{BX}{XC} = \\frac{PY}{XP},\n$$\n\nsince $BC = BX + XC$ and $XY = XP + PY$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 13471,
    "subject": "Mathematics (Olympiad)",
    "question": "Andriy and Olesya each write a natural number on a chalkboard. The number written by Olesya has a sum of digits equal to $2018$ and has exactly one digit less than Andriy's number. It is also known that the difference between their numbers is a one-digit number. What could be the number written by Andriy?",
    "options": [],
    "answer": "See solution",
    "solution": "It is not hard to see that Andriy's number can only be of the form $\\overline{100\\ldots0a}$, and Olesya's number can only be $\\overline{99\\ldots9b}$. Otherwise, the difference would not be a one-digit number. If not all of Olesya's digits except the last are $9$, then after adding a one-digit number, the number of digits would not change. Thus, this is the only possible presentation for Andriy's number.\n\nThe sum of digits of Olesya's number equals $2018$, so it is $\\overline{99\\ldots92}$ (since $2018 = 224 \\times 9 + 2$). Therefore, Andriy's number must have the last digit less than $2$, because otherwise the difference between the numbers would be at least $10$. So, this digit can be $0$ or $1$. Thus, Andriy could have written $\\overline{100\\ldots0}$ or $\\overline{100\\ldots01}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 13472,
    "subject": "Mathematics (Olympiad)",
    "question": "An invisible hare sits on one of $N$ vertices of a graph $G$. Several hunters try to kill the hare. Each minute, all of them simultaneously shoot: each hunter shoots at a single vertex, and they choose the target vertices cooperatively. If the hare was in a target vertex during a shoot, the hunting is finished. Otherwise, the hare can jump to one of the neighboring vertices or stay in its vertex.\n\nThe hunters know an algorithm that allows them to kill the hare in at most $N!$ shoots. Prove that there exists an algorithm that allows them to kill the hare in at most $2^N$ shoots.",
    "options": [],
    "answer": "See solution",
    "solution": "Let the hunters apply the optimal (fastest) algorithm. Say that a vertex has a *smell of a hare* if there exists an initial vertex and a sequence of moves of the hare for which the hare is still alive and now occupies this vertex. After every shoot, mark the set of all vertices that have a smell of a hare. In the beginning, all the vertices of the graph have a smell of hare, and after the hunting finishes, this set is empty. The idea is that in an optimal strategy, these sets cannot repeat!\n\nIndeed, the hunting does not imply feedback; the hunters' shoots do not depend on the hare's moves because the hunters try to foresee all possible moves of the hare. So if a set of vertices $A$ appears after the $k$-th shoot and once again after the $m$-th shoot, then the strategy is not optimal because all shoots from the $k$-th to the $(m-1)$-th can be omitted with the same result of hunting.\n\nSince it is possible to mark at most $2^N$ sets, the hunting will finish in at most $2^N - 1$ shoots.",
    "topic": "Discrete Mathematics",
    "subtopic": "Combinatorics"
  },
  {
    "id": 13473,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $f: \\mathbb{R}^+ \\to \\mathbb{R}^+$ be a non-constant function. Prove that there exist $a, b \\in \\mathbb{R}^+$ such that\n$$\nf(a) + f(b) > 2f(\\sqrt{ab}).\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "Assume the contrary: for all $a, b \\in \\mathbb{R}^+$, $f(a) + f(b) \\le 2f(\\sqrt{ab})$. Let $P(a, b)$ denote this assertion. First, $P(a, \\frac{1}{a})$ gives $f(a) + f(\\frac{1}{a}) \\le 2f(1)$, so $f$ is bounded. We show by induction that for any $a \\in \\mathbb{R}^+$,\n$$\nf(a^{2^n}) \\le 2^n(f(a) - f(1)) + f(1). \\quad (*)\n$$\nThe base case: $P(a^2, 1)$ yields $f(a^2) \\le 2f(a) - f(1) = 2(f(a) - f(1)) + f(1)$. Assume the statement holds for $n = k - 1$. From $P(a^{2^k}, 1)$, $f(a^{2^k}) \\le 2f(a^{2^{k-1}}) - f(1)$. By the inductive hypothesis, $f(a^{2^{k-1}}) \\le 2^{k-1}(f(a) - f(1)) + f(1)$, so $f(a^{2^k}) \\le 2^k(f(a) - f(1)) + f(1)$. If there exists $a$ such that $f(a) < f(1)$, then for large $n$, $f(a^{2^n}) \\le 2^n(f(a) - f(1)) + f(1) < 0$, contradicting positivity. Thus, $f(a) \\ge f(1)$ for all $a$. If for some $a$, $f(a) > f(1)$, then $f(\\frac{1}{a}) \\ge f(1)$, so $2f(1) < f(a) + f(\\frac{1}{a}) \\le 2f(1)$, a contradiction. Therefore, $f(a) = f(1)$ for all $a$, contradicting non-constancy.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 13474,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $(u_n)_{n \\ge 1}$ be a sequence of positive integers such that\n\n$$\n0 \\le u_{m+n} - u_m - u_n \\le 2, \\quad \\forall m, n \\in \\mathbb{N}.\n$$\n\nProve that there exist positive real numbers $a_1, a_2$ such that\n\n$$\n|u_n - \\lfloor a_1 n \\rfloor - \\lfloor a_2 n \\rfloor| \\le 1, \\quad \\forall n \\le 2024.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "We need to prove that there exist $a_1, a_2$ such that\n\n$$\n\\lfloor a_1 n \\rfloor + \\lfloor a_2 n \\rfloor - 1 \\le u_n \\le \\lfloor a_1 n \\rfloor + \\lfloor a_2 n \\rfloor + 1.\n$$\n\nSince $u_n$ is an integer, if we have $(a_1 + a_2)n - 2 < u_n < (a_1 + a_2)n$, then the desired inequality follows. Thus, we seek a real number $a = a_1 + a_2$ such that $a n - 2 < u_n < a n$, or equivalently, $\\frac{u_n}{n} < a < \\frac{u_n + 2}{n}$ for all $n \\le 2024$.\n\nDefine\n$$\nA = \\max_{1 \\le n \\le 2024} \\frac{u_n}{n}, \\quad B = \\min_{1 \\le n \\le 2024} \\frac{u_n + 2}{n}.\n$$\n\nWe will show that $B > A$, so we can choose $a$ with $A < a < B$.\n\nSuppose $k, m$ are the smallest indices in $\\{1, 2, \\dots, 2024\\}$ such that $A = \\frac{u_k}{k}$ and $B = \\frac{u_m + 2}{m}$. Then:\n\n- $kA = u_k$, and for all $p < k$, $pA > u_p$.\n- $mB = u_m + 2$, and for all $q < m$, $qB < u_q + 2$.\n\nIf $k = m$, then clearly $A < B$.\n\nIf $k < m$, set $m = k + p$. Then\n$$\nkA = u_k \\le u_m - u_p = mB - (u_p + 2) < mB - pB = kB,\n$$\nso $A < B$.\n\nIf $k > m$, set $k = m + q$. Then\n$$\nmB = u_m + 2 \\ge u_k - u_q > kA - qA = mA,\n$$\nso $B > A$.\n\nThus, $B > A$ always holds, so we can choose $a$ with $A < a < B$, ensuring $\\frac{u_n}{n} < a < \\frac{u_n + 2}{n}$ for all $n \\le 2024$. Finally, let $a_1, a_2 > 0$ with $a_1 + a_2 = a$.\n\n$\\square$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 13475,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all positive integers $a$, $b$, $c$ such that $2^a + 2^b + 2^c + 3$ is a square.",
    "options": [],
    "answer": "See solution",
    "solution": "Assume $a \\leq b \\leq c$.\n\nIf $a \\geq 2$, then $2^a + 2^b + 2^c + 3 \\equiv 3 \\pmod{4}$, which cannot be a square. Thus, $a = 1$.\n\nNow, we seek $b \\leq c$ such that $2^b + 2^c + 5$ is a square.\n\nIf $b \\geq 3$, then $2^b + 2^c + 5 \\equiv 5 \\pmod{8}$, again not a square. So $b = 1$ or $b = 2$.\n\n**Case 1:** $b = 1$\n\nWe need $2^c + 7$ to be a square. For $c = 1$, $2^1 + 7 = 9 = 3^2$, which works. For $c \\geq 2$, $2^c + 7 \\equiv 3 \\pmod{4}$, not a square.\n\n**Case 2:** $b = 2$\n\nWe need $2^c + 9 = k^2$ for some integer $k$. Rearranging: $2^c = (k - 3)(k + 3)$. So $k - 3 = 2^\\alpha$, $k + 3 = 2^\\beta$ for $\\alpha < \\beta$ positive integers. Subtracting: $2^\\alpha (2^{\\beta - \\alpha} - 1) = 6$. This gives $\\alpha = 1$, $2^{\\beta - 1} - 1 = 6$, so $2^{\\beta - 1} = 7$, impossible. But for $c = 4$, $2^4 + 9 = 16 + 9 = 25 = 5^2$.\n\n**Conclusion:**\n\nThe solutions are $a = b = c = 1$ and $\\{a, b, c\\} = \\{1, 2, 4\\}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 13476,
    "subject": "Mathematics (Olympiad)",
    "question": "There is a number in every cell of a $5 \\times 5$ table written under the following conditions:\n\n- Not all numbers are different.\n- There is no row or column where all five numbers are equal.\n- The middle number (the third one) in every row and column equals the mean of the numbers in its row or column.\n\nWhat is the minimum number of numbers that are less than the number written in the middle of the table?",
    "options": [],
    "answer": "See solution",
    "solution": "Let us prove that there cannot be fewer than three such numbers. Let the middle number of the table be $a$. Then there exists at least one number that is less than $a$ in the third row and in the third column. If there is no such number, then all the numbers in the row or column are the same, which contradicts the conditions. Without loss of generality, let us consider that the number $b < a$ is situated in the middle row in the fifth column. But then in the fifth column there exists a number smaller than $b$, and thus smaller than $a$. Therefore, at least three such numbers exist.\n\nAn example where the answer $3$ is achievable is shown in the picture below:\n\n![](images/UkraineMO_2015-2016_booklet_p15_data_0ee45cc7ca.png)\n\n$$\n\\begin{array}{|c|c|c|c|c|}\\hline\n4 & 4 & 3 & 4 & 0 \\\\\n\\hline\n4 & 4 & 3 & 4 & 0 \\\\\n\\hline\n3 & 3 & 0 & 3 & -9 \\\\\n\\hline\n4 & 4 & 3 & 4 & 0 \\\\\n\\hline\n0 & 0 & -9 & 0 & -36 \\\\\n\\hline\n\\end{array}\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 13477,
    "subject": "Mathematics (Olympiad)",
    "question": "Sea $\\triangle ABC$ un triángulo y $D$, $E$ y $F$ tres puntos cualesquiera sobre los lados $AB$, $BC$ y $CA$ respectivamente. Llamemos $P$ al punto medio de $AE$, $Q$ al punto medio de $BF$ y $R$ al punto medio de $CD$. Probar que el área del triángulo $\\triangle PQR$ es la cuarta parte del área del triángulo $\\triangle DEF$.",
    "options": [],
    "answer": "See solution",
    "solution": "Hagamos primero un dibujo donde queden reflejados los elementos que intervienen en el problema.\n\n![](images/Spanija_b_2014_p10_data_8c1b426277.png)\n\nObservemos que, como $P$, $Q$ y $R$ son los puntos medios de las correspondientes cevianas $AE$, $BF$ y $CD$, estos puntos se encuentran en los lados del triángulo que determinan los pies de las medianas de cada lado, como se ve en la figura siguiente.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 13478,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $AB$ be a diameter of a circle $\\omega$, and points $M$ and $C$ on $\\omega$ be in different half-planes with respect to the line $AB$. Perpendiculars $MN$ and $MK$ are dropped from the point $M$ to the lines $AB$ and $AC$ respectively. Prove that the line $KN$ bisects the line segment $CM$.\n\n![](images/Ukrajina_2013_p12_data_f468741492.png)",
    "options": [],
    "answer": "See solution",
    "solution": "Let $Q$ be the point of intersection of $KN$ and $CM$. Our goal is to show that $QC = QM$.\n\nDraw the line segments $BC$ and $AM$. From the problem statement it follows that $\\angle MNA = \\angle MKA = 90^\\circ$, and so the quadrilateral $AKNM$ is cyclic. Since the quadrilateral $BCAM$ is also cyclic, we obtain that\n\n$$\n\\angle NKM = \\angle NAM = \\angle BAM = \\angle BCM.\n$$\n\nBecause $AB$ is a diameter of $\\omega$, we have that $BC \\perp AC$, and since we also have that $MK \\perp AC$, we conclude that $MK \\parallel BC$. Then $\\angle CMK = \\angle BCM$, and so $\\angle CMK = \\angle NKM$. This means that the triangle $KMQ$ is isosceles, $KQ = QM$. Recall that $\\angle MKC = 90^\\circ$, and so $\\angle CKQ = 90^\\circ - \\angle NKM = 90^\\circ - \\angle NKM = 90^\\circ - \\angle CMK = \\angle QCK$, which implies that $KQ = QC = QM$, and we are done.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 13479,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $O$ and $H$ be the circumcenter and orthocenter of a scalene triangle $ABC$, respectively. Let $D$ be the intersection point of the lines $AH$ and $BC$. Suppose the line $OH$ meets the side $BC$ at $X$. Let $P$ and $Q$ be the second intersection points of the circumcircles of $\\triangle BDH$ and $\\triangle CDH$ with the circumcircle of $\\triangle ABC$, respectively. Show that the four points $P, D, Q$, and $X$ lie on a circle.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $M$ and $N$ be the midpoints of the sides $AB$ and $AC$, respectively, and $E$ and $F$ be the feet of altitudes drawn from the vertices $B$ and $C$ to the corresponding sides.\n\nFirst we claim that $N$ lies on the line $PH$. Let $B'$ be diametrically opposite to the vertex $B$ concerning the circumcircle of $\\triangle ABC$. It is well-known that $N$ is the midpoint of the segment $HB'$. Since $\\angle BPH = \\angle BDH = 90^\\circ = \\angle BPB'$, the claim follows. Similarly, one can find that $M$ lies on the line $QH$.\n\n![](images/BMO_2023_Short_List_p23_data_f46df99c61.png)\n\nNow, consider the inversion centered at $H$ which sends $A$ to $D$. It is obvious that this inversion sends the vertices $B, C$ into the points $E, F$, respectively. In addition, we know that $\\angle NPB = \\angle B'PB = 90^\\circ = \\angle NEB$ which means that the points $N, E, P, B$ are concyclic. Then, we have that $NH \\cdot HP = EH \\cdot HB = DH \\cdot HA$. In other words, the mentioned inversion sends point $P$ to point $N$. Similarly, one can find that the same inversion sends point $Q$ to point $M$.\n\nFrom the above argumentation, this inversion sends the circumcircle of $\\triangle DPQ$ into the circle passing through the points $A, N, M$. It suffices to show that the inverse $K$ of the point $X$ lies on the circle passing through points $A, N, M$.\n\nIt is clear that point $K$ lies on line $OH$ since $X$ lies on line $OH$. From the radius of the inversion, we have that $XH \\cdot HK = DH \\cdot HA$ which implies that the four points $A, K, D, X$ are concyclic. Therefore, $\\angle AKO = \\angle AKX = \\angle ADX = 90^\\circ = \\angle AMO$. In other words, point $K$ lies on the circle passing through $A, M, O$ which is the same circle passing through points $A, M, N$.\n\nAs a result, we find that the points $A, K, M, N$ lie on a single circle and when we look at their preimages with respect to the defined inversion, we can see that the points $D, X, Q, P$ are concyclic as required.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 13480,
    "subject": "Mathematics (Olympiad)",
    "question": "Do there exist 6-digit numbers of the form $n^k$, $k \\ge 3$, such that the difference between the number formed by the last three digits and the number formed by the first three digits is equal to 4?",
    "options": [],
    "answer": "See solution",
    "solution": "Let $a = \\overline{a_1a_2a_3a_4a_5a_6}$ be a 6-digit number. The condition is that $10^5 < a < 10^6$ and $\\overline{a_4a_5a_6} = \\overline{a_1a_2a_3} + 4$.\n\nThus,\n$$\na = 1000\\overline{a_1a_2a_3} + \\overline{a_4a_5a_6} = 1001\\overline{a_1a_2a_3} + 4.\n$$\nWe seek 3-digit numbers $A = \\overline{a_1a_2a_3}$ such that $1001A + 4 = n^k$ for $k \\ge 3$.\n\nSince $1001 = 7 \\times 11 \\times 13$, we require $n^k \\equiv 4 \\pmod t$ for $t = 7, 11, 13$.\n\n1. For $k=3$: $n^3 \\equiv 4 \\pmod 7$. But $n^3 \\equiv \\pm 1 \\pmod 7$, so no solution.\n2. For $k=4$: $n^4 \\equiv 4 \\pmod{13}$. But $n^4 \\equiv 1,3,9 \\pmod{13}$, so no solution.\n3. For $k=5$: $n^5 \\equiv 4 \\pmod{11}$. But $n^5 \\equiv \\pm 1 \\pmod{11}$, so no solution.\n4. For $k=7$: $n^7 \\equiv 4 \\pmod 7$. But $n^7 \\equiv n \\pmod 7$, so $n \\equiv 4 \\pmod 7$. If $n=4$, $4^7 < 10^5$; if $n \\ge 11$, $n^7 > 10^6$. No solution.\n5. For $k=11$: $n^{11} \\equiv n \\pmod{11}$, so $n^{11} \\equiv 4 \\pmod{11}$. But $4^{11} > 10^6$, so $n < 4$, which is not possible.\n6. For $k \\ge 13$: $n=2$ or $3$ are possible, but $2^{20} > 10^6$ and $3^{13} > 10^6$, so no solution in the range.\n\nIf $k > 4$ is even, then $k$ is divisible by 4 or has an odd divisor, reducing to previous cases.\n\n**Answer:** There do not exist 6-digit numbers with the given properties.\n\n_Remark:_ For $k=2$, the solutions are $205, 209, 300, 304, 477, 481, 732, 736$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 13481,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $a$, $b$, $c$ be real numbers representing the side lengths of a triangle. Prove that\n$$\n4(a + b)(a + c)(b + c) \\geq (a + b + c)^3.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "Given three real numbers $a, b, c$ such that $0 < a, b, c < \\frac{a+b+c}{2}$ (which is equivalent to the triangle inequality). Introduce the variables\n$$\nx = \\frac{-a + b + c}{2(a + b + c)}, \\quad y = \\frac{a - b + c}{2(a + b + c)}, \\quad z = \\frac{a + b - c}{2(a + b + c)}\n$$\nThe conditions on $a, b, c$ are equivalent to\n$$\n0 < x, y, z, \\quad \\text{and} \\quad x + y + z = \\frac{1}{2}\n$$\nClearly,\n$$\n\\frac{b+c}{a+b+c} = \\frac{1}{2} + x, \\quad \\frac{c+a}{a+b+c} = \\frac{1}{2} + y, \\quad \\frac{a+b}{a+b+c} = \\frac{1}{2} + z\n$$\nTherefore,\n$$\n\\begin{align*}\n\\frac{(a+b)(a+c)(b+c)}{(a+b+c)^3} &= \\frac{b+c}{a+b+c} \\cdot \\frac{c+a}{a+b+c} \\cdot \\frac{a+b}{a+b+c} \\\\\n&= \\left(\\frac{1}{2} + x\\right) \\left(\\frac{1}{2} + y\\right) \\left(\\frac{1}{2} + z\\right) \\\\\n&= \\frac{1}{8} + \\frac{x+y+z}{4} + \\frac{yz+zx+xy}{2} + xyz \\\\\n&= \\frac{1}{8} + \\frac{1/2}{4} + \\frac{yz+zx+xy}{2} + xyz \\\\\n&> \\frac{1}{4}\n\\end{align*}\n$$\nThat is,\n$$\n4(a + b)(a + c)(b + c) > (a + b + c)^3\n$$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 13482,
    "subject": "Mathematics (Olympiad)",
    "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$$\n(X_1 \\setminus X_2) \\cap X_3, \\quad (X_2 \\setminus X_1) \\cap X_3\n$$\n\nis empty.\n\nShow that if $\\mathcal{F}$ is a perfect family consisting of some subsets of a given finite set $U$, then $|\\mathcal{F}| \\le |U| + 1$.\n\n![](images/Cesko-Slovacko-Poljsko_2015_p2_data_821fbbabf4.png)",
    "options": [],
    "answer": "See solution",
    "solution": "We proceed by induction on $|U|$.\n\nIf $|U| = 0$, i.e., $U = \\emptyset$, then there is only one subset of $U$, so $|\\mathcal{F}| \\le 1$.\n\nAssume the statement holds for all sets of size less than $k$ for some $k > 0$. Let $U$ be a set with $|U| = k$ and $\\mathcal{F}$ a perfect family of its subsets. We show $|\\mathcal{F}| \\le |U| + 1$.\n\nIf $|\\mathcal{F}| \\le 1$, the claim is clear. If $|\\mathcal{F}| \\ge 2$, consider all pairs of distinct sets from $\\mathcal{F}$. Among these, pick a pair $(Y, Z)$ with $|Y \\cap Z|$ maximal.\n\nSince $Y \\ne Z$, at least one of $Y \\setminus Z$ or $Z \\setminus Y$ is nonempty. Without loss of generality, let $y \\in Y \\setminus Z$.\n\n**Case 1:** $Y$ is the only set containing $y$. Then all sets in $\\mathcal{F}' = \\mathcal{F} \\setminus \\{Y\\}$ are subsets of $U' = U \\setminus \\{y\\}$. $\\mathcal{F}'$ is perfect, so by induction,\n\n$$\n|\\mathcal{F}| = |\\mathcal{F}'| + 1 \\le (|U'| + 1) + 1 = |U| + 1.\n$$\n\n**Case 2:** There is $W \\in \\mathcal{F}$, $W \\ne Y$, with $y \\in W$. By maximality, $W$ cannot contain all of $Y \\cap Z$, so pick $z \\in (Y \\cap Z) \\setminus W$. Then\n\n$$\ny \\in (W \\setminus Z) \\cap Y, \\quad z \\in (Z \\setminus W) \\cap Y,\n$$\n\ncontradicting the perfectness property for $X_1 = Z$, $X_2 = W$, $X_3 = Y$. Thus, this case is impossible, and the induction is complete.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 13483,
    "subject": "Mathematics (Olympiad)",
    "question": "As shown in the figure below, point $E$ lies in the opposite half-plane determined by line $CD$ from point $A$ so that $\\angle CDE = 110^\\circ$. Point $F$ lies on $\\overline{AD}$ so that $DE = DF$, and $ABCD$ is a square. What is the degree measure of $\\angle AFE$?\n\n![](images/2021_AMC12A_Solutions_Fall_p2_data_0bc3c24de1.png)\n\n(A) 160 (B) 164 (C) 166 (D) 170 (E) 174",
    "options": [],
    "answer": "See solution",
    "solution": "Note that $\\angle EDF = 360^\\circ - \\angle ADC - \\angle CDE = 360^\\circ - 90^\\circ - 110^\\circ = 160^\\circ$. Because $\\triangle DEF$ is isosceles, angles $DEF$ and $DFE$ have an equal measure of $\\frac{180^\\circ - 160^\\circ}{2} = 10^\\circ$. Hence $\\angle AFE = 180^\\circ - \\angle DFE = 180^\\circ - 10^\\circ = 170^\\circ$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 13484,
    "subject": "Mathematics (Olympiad)",
    "question": "Three squares lie inside a right-angled triangle as shown. The side lengths of the smallest and largest squares are $28$ and $63$ respectively. Find the side length of the middle square.\n\n![](images/2023_Australian_Scene_p46_data_7c51530ee0.png)",
    "options": [],
    "answer": "See solution",
    "solution": "Let $m$ be the side length of the middle square.\n\n![](images/2023_Australian_Scene_p48_data_b78a09f919.png)\n\nThe triangles above the squares are similar. So we have\n\n$$\n\\begin{aligned}\n\\frac{m-28}{28} &= \\frac{63-m}{m} \\\\\n\\frac{m}{28} - 1 &= \\frac{63}{m} - 1 \\\\\nm^2 &= 28 \\times 63 = 2^2 \\cdot 7 \\times 7 \\cdot 3^2 \\\\\nm &= 2 \\cdot 3 \\cdot 7 = \\mathbf{42}\n\\end{aligned}\n$$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 13485,
    "subject": "Mathematics (Olympiad)",
    "question": "Given that\n$$\n(10!)^3 = \\overline{47\\,784\\,725\\,83a\\,b72\\,000\\,000},\n$$\ndetermine the digits $a$ and $b$.",
    "options": [],
    "answer": "See solution",
    "solution": "As $(10!)^3$ is divisible by $9$, by the divisibility criterion for $9$ we get\n$$\n9 \\mid 4+7+7+8+4+7+2+5+8+3+a+b+7+2+0+0+0+0+0+0 = 64 + a + b.\n$$\nFurther, by Wilson's theorem we obtain $(10!)^3 \\equiv (-1)^3 \\equiv -1 \\pmod{11}$, and thus\n$$\n-1 \\equiv (7+8+7+5+3+b+2+0+0+0) - (4+7+4+2+8+a+7+0+0+0) = b - a \\pmod{11}.\n$$\nSince $0 \\leq a, b \\leq 9$, we have $-9 \\leq b - a \\leq 9$. The only number congruent to $-1$ modulo $11$ in this interval is $-1$, and therefore $b - a = -1$, that is, $b = a - 1$.\n\nFurther, we have $0 \\leq a + b \\leq 18$, from which follows\n$$\n64 \\leq 64 + a + b \\leq 82.\n$$\nThe numbers from this interval divisible by $9$ are $72$ and $81$, and thus we get $64 + a + b = 72$ or $64 + a + b = 81$, that is, $a + b = 8$ or $a + b = 17$. Putting $b = a - 1$ here, the first case reduces to $2a - 1 = 8$, which is impossible.\n\nThere remains the second case, $2a - 1 = 17$, from which we get $a = 9$ and $b = a - 1 = 8$.\n\n$\\boxed{a = 9,\\ b = 8}$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 13486,
    "subject": "Mathematics (Olympiad)",
    "question": "(a) List the prime numbers between 1 and 25 that are of the form $4k + 1$ and $4k + 3$, where $k \\in \\mathbb{Z}$. Identify which primes fit each form.\n\n(b) Find the largest number between 1 and 25 that can be the largest in a set of six numbers, where it must be divisible by at least four of the other numbers in the set. List possible sets that satisfy this condition.",
    "options": [],
    "answer": "See solution",
    "solution": "(a) The prime numbers between 1 and 25 are $2, 3, 5, 7, 11, 13, 17, 19, 23$. Of these, those of the form $4k + 1$ are $5, 13, 17$, and those of the form $4k + 3$ are $3, 7, 11, 19, 23$.\n\n(b) The largest number must be divisible by at least four of the other numbers, so it must have at least five divisors. The only numbers between 1 and 25 with at least five divisors are $16$ and $24$. If the largest is $16$, the divisors could be $1, 2, 4, 8$, but the fifth number cannot fit the conditions. If the largest is $24$, possible sets include $1, 2, 4, 8, 16, 24$ and $1, 2, 4, 6, 12, 24$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 13487,
    "subject": "Mathematics (Olympiad)",
    "question": "It is known that\n\n$$\n\\frac{a}{b+c+d} + \\frac{b}{c+d+a} + \\frac{c}{d+a+b} + \\frac{d}{a+b+c} = 1.\n$$\n\nFind the value of the expression\n\n$$\n\\frac{a^2}{b+c+d} + \\frac{b^2}{c+d+a} + \\frac{c^2}{d+a+b} + \\frac{d^2}{a+b+c}.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "**Answer:** 0.\n\n**Solution.** Multiplying the given equality by $a+b+c+d$, we obtain\n\n$$\n\\begin{aligned}\n& \\frac{a(a+b+c+d)}{b+c+d} + \\frac{b(a+b+c+d)}{c+d+a} + \\frac{c(a+b+c+d)}{d+a+b} + \\frac{d(a+b+c+d)}{a+b+c} = \\\\\n&= \\frac{a^2 + a(b+c+d)}{b+c+d} + \\frac{b^2 + b(c+d+a)}{c+d+a} + \\frac{c^2 + c(d+a+b)}{d+a+b} + \\frac{d^2 + d(a+b+c)}{a+b+c} = \\\\\n&= \\frac{a^2}{b+c+d} + a + \\frac{b^2}{c+d+a} + b + \\frac{c^2}{d+a+b} + c + \\frac{d^2}{a+b+c} + d = a+b+c+d.\n\\end{aligned}\n$$\n\nIt then follows that\n\n$$\n\\frac{a^2}{b+c+d} + \\frac{b^2}{c+d+a} + \\frac{c^2}{d+a+b} + \\frac{d^2}{a+b+c} = 0.\n$$",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 13488,
    "subject": "Mathematics (Olympiad)",
    "question": "給定任意三角形 $\\triangle ABC$，令其外接圓為 $O_1$，九點圓為 $O_2$；並令以 $\\triangle ABC$ 的垂心 $H$ 與重心 $G$ 為直徑的圓為 $O_3$。證明 $O_1, O_2, O_3$ 共軸（即存在一直線，其上的點對這三個圓的圓幂均相同。點對圓的圓幂，是點到圓心的距離平方，與圓半徑的平方之差）。\n\n註：三角形的九點圓，即通過三邊中點、三高垂足、三頂點分別與垂心連線的中點等九個點的圓。",
    "options": [],
    "answer": "See solution",
    "solution": "引理：垂心與重心為 $O_1, O_2$ 的兩位似中心。\n\n引理證明：九點圓過 $\\overline{HA}, \\overline{HB}, \\overline{HC}$ 的中點，故 $H$ 為一位似中心。\n\n九點圓過 $\\overline{AB}, \\overline{BC}, \\overline{CA}$ 的中點（分別設為 $M, N, P$），且 $\\overline{AG} : \\overline{GM} = 2 : 1$，$\\overline{BG} : \\overline{GN} = 2 : 1$，$\\overline{CG} : \\overline{GP} = 2 : 1$，故 $G$ 亦為一位似中心。\n\n回到原題證明。$ABC$ 為銳角時，令 $X$ 為 $O_1, O_2$ 之根軸與 $OH$ 的交點，$x$ 為 $O_1$ 半徑，$y$ 為 $O_2$ 半徑，$\\overline{XO} = a$，$\\overline{XG} = b$，$\\overline{XN} = c$，$\\overline{XH} = d$。\n\n$$\n\\text{有 } \\frac{a-b}{b-c} = \\frac{a-d}{c-d} = \\frac{x}{y}, \\quad a^2 - x^2 = c^2 - y^2.\n$$\n\n$$\n\\text{可得 } b = \\frac{cx+ay}{x+y}, \\quad d = \\frac{cx-ay}{x-y},\n$$\n\n$$\nbd = \\frac{c^2 x^2 - a^2 y^2}{x^2 - y^2} = \\frac{(c^2 - y^2)(x^2 - y^2)}{x^2 - y^2} = c^2 - y^2.\n$$\n\n故三圓共軸（鈍角時計算方法類似）。",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 13489,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $f(0) = a$ and consider the functional equation\n\n$$\nf(x+y)f(f(x)-y) = x f(x) - y f(y)\n$$\n\nFind all functions $f : \\mathbb{R} \\to \\mathbb{R}$ that satisfy this equation.",
    "options": [],
    "answer": "See solution",
    "solution": "Substitute $x = y = 0$ into the equation:\n\n$$\nf(0) f(a) = 0\n$$\n\nSo $f(0) = 0$ (i.e., $a = 0$) or $f(a) = 0$ always holds.\n\nLet $t$ be arbitrary. Substitute $x = 0$, $y = t$:\n\n$$\nf(t) f(a - t) = -t f(t)\n$$\n\nSo either $f(t) = 0$ or $f(a - t) = -t$. If $u \\neq 0$ and $f(u) \\neq 0$, then $f(a-u) = -u \\neq 0$, so for $t = a-u$ we get $f(u) = u - a$. Thus,\n\n$$\nu \\neq 0 \\implies [f(u) = 0 \\text{ or } f(u) = u - a]\n$$\n\n**Case 1:** $f(u) = 0$ for all $u \\neq 0$.\n\nIf $a = 0$, then $f(x) = 0$ for all $x$ (identically zero), which satisfies the equation. If $a \\neq 0$, then $f(0) = a \\neq 0$ and $f(u) = 0$ for $u \\neq 0$. The right side of the equation is always $0$, and the left side is nonzero only if $x + y = 0$ and $f(x) - y = 0$, which cannot both occur unless $x = y = 0$, but then $f(0) = a \\neq 0$. So the equation is satisfied.\n\n**Case 2:** There exists $u \\neq 0$ with $f(u) \\neq 0$.\n\nLet $b$ be such a $u$. Substitute $x = a$, $y = -b$:\n\n$$\nf(a-b) f(b) = b f(-b)\n$$\n\nBut from above, $f(b) f(a-b) = -b f(b)$, so $f(-b) = -f(b) \\neq 0$.\n\nFrom earlier, $f(b) = b - a$, $f(-b) = -b - a$. Thus, $-f(b) = -b + a = f(-b) = -b - a$, so $a = 0$ and $f(b) = b$.\n\nSuppose there exists $c$ with $f(c) \\neq c$. Since $f(0) = a = 0$, $c \\neq 0$, so $f(c) = 0$. Substitute $x = b$, $y = c$:\n\n$$\nf(b + c) f(b - c) = b^2\n$$\n\nBut $f(b + c) = b + c$, $f(b - c) = b - c$, so $f(b + c) f(b - c) = b^2 - c^2$, which contradicts the previous equation unless $c = 0$. Thus, $f(x) = x$ for all $x$.\n\n**Conclusion:**\n\nThe solutions are:\n\n$$\nf(x) = x\n$$\n\nand\n\n$$\nf(x) = \\begin{cases} a & \\text{if } x = 0 \\\\ 0 & \\text{if } x \\neq 0 \\end{cases}\n$$\n\nwhere $a$ is any real number.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 13490,
    "subject": "Mathematics (Olympiad)",
    "question": "The graph of $y = e^{x+1} + e^{-x} - 2$ has an axis of symmetry. What is the reflection of the point $(-1, \\frac{1}{2})$ over this axis?\n\n(A) $(-1, -\\frac{3}{2})$  \n(B) $(-1, 0)$  \n(C) $(-1, \\frac{1}{2})$  \n(D) $(0, \\frac{1}{2})$  \n(E) $(3, \\frac{1}{2})$",
    "options": [],
    "answer": "See solution",
    "solution": "Let $f(x) = e^{x+1} + e^{-x} - 2$. Because $f(x)$ approaches infinity as $|x|$ increases without bound, the only possible axis of symmetry is a vertical line. If the axis of symmetry has equation $x = c$, then $f(x) = f(2c-x)$ for every real $x$, which is equivalent to $e \\cdot e^x + e^{-x} = e^{2c+1}e^{-x} + e^{-2c}e^x$.\n\nMultiplying through by $e^x$ and simplifying gives $$(e - e^{-2c})e^{2x} = e^{2c+1} - 1.$$ Because this equation holds for all $x$, it follows that $e - e^{-2c} = 0$ and $e^{2c+1} - 1 = 0$. Thus $c = -\\frac{1}{2}$.\n\nThe reflection of $(-1, \\frac{1}{2})$ with respect to the vertical line $x = -\\frac{1}{2}$ is $(0, \\frac{1}{2})$.\n\n**Answer:** (D)",
    "topic": "Discrete Mathematics",
    "subtopic": "Other"
  },
  {
    "id": 13491,
    "subject": "Mathematics (Olympiad)",
    "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 $10^9$. Player A makes the first move, then B takes a turn, then C, then A again, and so on, 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": "See solution",
    "solution": "The game continues as long as the number on the blackboard stays less than $10^9$, since it is always possible to make a move of the first kind (replace $n$ with $n+1$). As the number increases by at least $1$ with every move, it cannot stay less than $10^9$ indefinitely. When the number on the blackboard equals $10^9$, there is no legal move. Thus, the last move is made by the player who writes the number $10^9$ on the blackboard.\n\nLet $n$ be the number currently on the blackboard. Note that:\n\n$$\n7n + 7 \\equiv n + 1 \\pmod{3}\n$$\n$$\n4n^3 + 3n + 4 \\equiv n^3 + 1 \\equiv n + 1 \\pmod{3}\n$$\n\nHence, all numbers that can appear on the blackboard on the next move are congruent to $n + 1 \\pmod{3}$, so the residues repeat in the cycle $1, 2, 0, 1, 2, 0, \\ldots$. Since $10^9 \\equiv 1 \\pmod{3}$, the number $10^9$ is written by the third player, C.\n\n*Remark:* The solution shows that C can win with whatever strategy.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 13492,
    "subject": "Mathematics (Olympiad)",
    "question": "Alexia has several marbles and her friend, Cristina, has none. Each day of one week, starting on Monday, Alexia gave Cristina some of her marbles, so that each day Alexia gave more marbles than the day before. On Monday Alexia gave five times less marbles than on Friday, on Tuesday she gave six times less marbles than on Saturday, and on Wednesday she gave seven times less marbles than on Sunday. At the end of that week, Cristina got 72 marbles. Find how many marbles Cristina got on Thursday.",
    "options": [],
    "answer": "See solution",
    "solution": "Denote $a_1, a_2, \\dots, a_7$ as the number of marbles given by Alexia from Monday to Sunday, where $1 \\leq a_1 < a_2 < a_3 < a_4 < a_5 < a_6 < a_7$ and $a_1 + a_2 + a_3 + a_4 + a_5 + a_6 + a_7 = 72$.\n\nGiven:\n- $a_5 = 5a_1$\n- $a_6 = 6a_2$\n- $a_7 = 7a_3$\n\nSubstitute into the sum:\n$$a_1 + a_2 + a_3 + a_4 + 5a_1 + 6a_2 + 7a_3 = 72$$\n$$6a_1 + 7a_2 + 8a_3 + a_4 = 72 \\quad (1)$$\n\nTry $a_1 = 1$:\n- $a_5 = 5$\n- $a_1 = 1 < a_2 < a_3 < a_4 < a_5 = 5$ so $a_2 = 2$, $a_3 = 3$, $a_4 = 4$\n- $6a_1 + 7a_2 + 8a_3 + a_4 = 6 \\cdot 1 + 7 \\cdot 2 + 8 \\cdot 3 + 4 = 6 + 14 + 24 + 4 = 48$ (contradicts (1))\n\nTry $a_1 = 2$:\n- $a_5 = 10$\n- $a_1 = 2 < a_2 < a_3 < a_4 < a_5 = 10$\n- $7a_2 + 8a_3 + a_4 = 72 - 6a_1 = 72 - 12 = 60$ (2)\n\nTry $a_2 = 3$:\n- $a_6 = 18$\n- $8a_3 + a_4 = 60 - 7 \\cdot 3 = 39$\n- $a_3 > a_2 = 3$, $a_3 \\leq 4$ so $a_3 = 4$\n- $a_4 = 39 - 8 \\cdot 4 = 39 - 32 = 7$\n\nSo the sequence is $2, 3, 4, 7, 10, 18, 28$.\n\n**Answer:** Cristina got $7$ marbles on Thursday.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 13493,
    "subject": "Mathematics (Olympiad)",
    "question": "Does the function $f(x) = f_1(x) + f_2(x) + f_3(x)$ always have zeros?\n\nb) $x_2, x_3$ are zeros of the function $f_1(x) = a_1x^2 + b_1x + c_1$; $x_3, x_1$ are zeros of the function $f_2(x) = a_2x^2 + b_2x + c_2$; $x_1, x_2$ are zeros of the function $f_3(x) = a_3x^2 + b_3x + c_3$. Does the function $f(x) = f_1(x) + f_2(x) + f_3(x)$ always have zeros?",
    "options": [],
    "answer": "See solution",
    "solution": "*Answer:* a) yes; b) no.\n\n*Solution.* a) We write our function in the form\n\n$$\nf(x) = (x - x_2)(x - x_3) + (x - x_1)(x - x_3) + (x - x_1)(x - x_2)\n$$\n\nWLOG, $x_1 < x_2 < x_3$. Then $f(x_2) = (x_2 - x_1)(x_2 - x_3) < 0$, which is equivalent to the existence of the roots of the function.\n\nb) Consider the following three functions:\n\n$$\nf_1(x) = x^2 + x, \\quad f_2(x) = x^2 - x, \\quad f_3(x) = 1 - x^2\n$$\n\nThey have the roots $0, -1$, $0, 1$ and $-1, 1$ respectively. Their sum\n\n$$\nf(x) = x^2 + x + x^2 - x + 1 - x^2 = x^2 + 1\n$$\n\ndoes not have zeros.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 13494,
    "subject": "Mathematics (Olympiad)",
    "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 such 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 $AB_{n-2}$ can be defined so 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": "See solution",
    "solution": "Let $B_n$ for some $n > 1$ be definable (see figure below).\n\n![](images/prob1718_p22_data_579acacf53.png)\n\n$A$, $B_{n-2}$, and $B_n$ are collinear, and $A$ 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 $\\angle AB_{n-2}B_{n-1} + \\angle AB_nB_{n-1} = 180^\\circ$ and $\\angle AB_{n-2}B_{n-1} \\neq 90^\\circ$.\n\nThus,\n$$\n\\angle AB_{n-1}B_n = 180^\\circ - \\alpha - \\angle AB_nB_{n-1} = 180^\\circ - \\alpha - (180^\\circ - \\angle AB_{n-2}B_{n-1}) = \\angle AB_{n-2}B_{n-1} - \\alpha.\n$$\nSince $\\angle AB_0B_1 = \\beta$, by induction we get $\\angle AB_{n-1}B_n = \\beta - (n-1)\\alpha$.\n\nDefining $B_n$ requires $\\beta - (n-2)\\alpha \\neq 90^\\circ$ and $\\beta - (n-1)\\alpha \\ge 0^\\circ$. These are also sufficient: if $\\beta - (n-2)\\alpha \\neq 90^\\circ$, then $B_n$ can be chosen different from $B_{n-2}$ on $AB_{n-2}$, and if $\\beta - (n-1)\\alpha \\ge 0^\\circ$, then $B_n$ is on the side $AB_{n-2}$.\n\nTherefore, the largest $n$ for which $B_n$ is definable equals $\\frac{\\beta-90^\\circ}{\\alpha} + 1$ if $\\frac{\\beta-90^\\circ}{\\alpha}$ is a natural number, and $\\lfloor \\frac{\\beta}{\\alpha} \\rfloor + 1$ otherwise.\n\n![](images/prob1718_p22_data_539ab3a820.png)\n\n![](images/prob1718_p22_data_39149a8fa6.png)",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 13495,
    "subject": "Mathematics (Olympiad)",
    "question": "令 $k$ 為一正整數。天橋上的魔術師和小不點玩一場遊戲。一開始，小不點將 $N = 2^k$ 顆相異的球在桌面上排成一橫排，並各自用一個杯子罩住。\n\n在每一回合，小不點可以指定兩個杯子，然後魔術師可以交換這兩個杯子內的球，或選擇做假動作讓兩個杯子內的球維持不變。小不點無法透過魔術師的動作來判別是否為假動作，也無法看到動作前後杯子內的球。\n\n在 $M = k \\times 2^{k-1}$ 個回合後，魔術師會打開所有杯子，讓小不點確認每個杯子裡的球。若此時小不點可以確知魔術師在每一回合中是否有做假動作，則小不點獲勝。\n\n證明小不點存在必勝策略。",
    "options": [],
    "answer": "See solution",
    "solution": "將所有杯子依序編號為 $1$ 到 $N$。我們將用歸納法建構策略。\n\n$k=1$ 時顯然成立。現在假設 $k-1$ 時存在一個必勝策略，其在第 $i=1,2,\\dots,(k-1)2^{k-2}$ 回合選取編號 $a_i$ 與 $b_i$ 的杯子。\n\n則在 $k$ 時，令 $M_1 = 2^{k-1}$ 與 $M_2 = (k+1)2^{k-2}$，並考慮以下策略：\n\n**Stage 1:** 在第 $i = 1,2,\\dots,M_1$ 回合選取編號 $i$ 與 $i + M_1$。\n(注意到 $i + M_1 \\le 2M_1 = 2^k$)\n\n**Stage 2:** 在第 $i = M_1 + 1, \\dots, M_2$ 回合選取編號 $a_{i-M_1}$ 與 $b_{i-M_1}$。\n(注意到 $i - M_1 \\le M_2 - M_1 = (k-1)2^{k-2}$)\n\n**Stage 3:** 在 $i = M_2 + 1, \\dots, M$ 選取編號 $a_{i-M_2} + 2^{k-1}$ 與 $b_{i-M_2} + 2^{k-1}$。\n(注意到 $i - M_2 \\le M - M_2 = (k-1)2^{k-2}$)\n\n以下證明這確為必勝法。令 $X$ 為編號 $1$ 到 $M_1$，而 $Y$ 為編號 $M_1+1$ 到 $N$。\n\n- 首先，注意到只有在 Stage 1 時，$X$ 和 $Y$ 中的球有可能對換，且只有第 $i$ 回合的操作有可能將 $i$ 從 $X$ 換到 $Y$。因此對於 Stage 1 裡的所有 $i$，我們知道第 $i$ 回合是假動作，若且唯若第 $i$ 號球最後仍留在 $X$ 區。\n- 注意到 Stage 2 形同在 $X$ 上執行 $k-1$ 時的必勝法，且我們已知在 Stage 1 結束時 $X$ 區的所有球分布，因此由歸納假設，我們可以透過觀察 $X$ 區最後的球分布來得知 Stage 2 中任何一回合是否為假動作。\n- 同理，Stage 3 形同在 $Y$ 上執行 $k-1$ 時的必勝法，且我們已知在 Stage 1 結束時 $Y$ 區的所有球分布，因此由歸納假設，我們可以透過觀察 $Y$ 區最後的球分布來得知 Stage 3 中任何一回合是否為假動作。",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 13496,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $f: \\mathbb{N} \\to \\mathbb{Z}$ be a function such that for all $m \\ne n$, the product $(f(m) - f(n))(m - n)$ is always a perfect square. Determine all such functions $f$.",
    "options": [],
    "answer": "See solution",
    "solution": "Suppose otherwise; then there exists $r$ such that $S = S_r$ covers between 2 and $p-1$ residues (so $p \\ge 3$). If $n \\notin S$ (i.e. $f(n) \\not\\equiv r \\pmod p$), then for any $s \\in S$,\n\n$$\n(n-s)(f(n)-f(s)) \\equiv (n-s)(f(n)-r) \\not\\equiv 0 \\pmod{p}\n$$\n\nis a nonzero quadratic residue. Hence $\\left(\\frac{n-s}{p}\\right) = \\left(\\frac{f(n)-r}{p}\\right) \\neq 0$.\n\nNow let $T = \\{t : \\left(\\frac{f(t)-r}{p}\\right) = 1\\}$ and $U = \\{u : \\left(\\frac{f(u)-r}{p}\\right) = -1\\}$, so $T, U$ partition $\\mathbb{N} \\setminus S$. By the previous paragraph, $\\left(\\frac{a}{p}\\right) = 1$ for any $a = t-s$ in the difference set $T-S$; similarly, $\\left(\\frac{b}{p}\\right) = -1$ for any $b \\in U-S$. This immediately upper bounds $|T-S|$ and $|U-S|$ by $\\frac{p-1}{2}$, the number of (nonzero) quadratic residues and nonresidues.\n\nBut $|T| + |U| = p - |S| \\ge 1$, so if $|T| \\ge |U|$ (so $T$ is nonempty), then Cauchy-Davenport yields\n\n$$\n\\frac{p-1}{2} \\ge |T-S| \\ge \\min(p, |T| + |S| - 1) \\ge \\min\\left(p, \\frac{p-|S|}{2} + |S| - 1\\right) = \\min\\left(p, \\frac{p + (|S| - 2)}{2}\\right),\n$$\n\ncontradicting $|S| \\ge 2$. The case $|U| \\ge |T|$ is analogous. $\\square$\n\nNote that $f(n+1) - f(n)$ is always a square. If $f$ is nonconstant, $\\gcd_{n \\ge 1}(f(n+1) - f(n)) \\ne 0$ must be a square itself (say $g^2$, with $g > 0$). If $g=1$, then $f$ is nonconstant (and thus injective, by the lemma) modulo every prime $p$. In particular, $p \\nmid f(n+1) - f(n)$ for all $n$ and $p$, which forces $f(x+1) - f(x) \\equiv 1 \\implies f(x) \\equiv x+d$ for some constant $d$. Otherwise, if $g > 1$, $f'(x) \\equiv \\frac{f(x)-f(1)}{g^2}$ has $(f'(m) - f'(n))(m-n)$ always a square and $\\gcd_{n \\ge 1}(f'(n+1) - f'(n)) = 1$, so $f(x) \\equiv g^2(x+d) + c$ for some constants $d, c$.\n\nIt follows that $f(x) \\equiv A^2x + B$ ($A, B \\in \\mathbb{Z}$) are the only possible solutions, which indeed all work (note that we get $f$ constant when $A=0$).\n\n**Solution 2.** We give an alternate proof of the lemma. Define $r$ and $S = S_r$ as in the previous proof.\n\nSince $p \\ge 3$, there exists a smallest quadratic nonresidue $\\alpha \\in [2, p-1]$ modulo $p$. In particular, $\\alpha-1 \\in [1, p-2]$ is a quadratic residue. Now fix two distinct residues $x, y$ (mod $p$) in $S$. We claim that for $N = x+\\alpha(y-x)$, we have $f(N) \\equiv r \\pmod p$. Suppose otherwise; then $f(N) \\not\\equiv r \\equiv f(x) \\equiv f(y)$. But then $[f(N)-f(x)][N-x] \\equiv [f(N)-r]\\alpha(y-x)$ and $[f(N)-f(y)][N-y] \\equiv [f(N)-r](\\alpha-1)(y-x)$ must both be nonzero squares, forcing $\\left(\\frac{\\alpha}{p}\\right) = \\left(\\frac{\\alpha-1}{p}\\right)$, which is absurd.\n\nHence $x+\\alpha(y-x) = (1-\\alpha)x + \\alpha y$ lies in $S$ (interpret relations modulo $p$ where clearly appropriate) for any distinct residues $x, y \\in S$; of course, it also does when $x \\equiv y \\in S$. This condition is affine, so for convenience, fix distinct residues $a, b \\in S$ and define $A = (S-a)(b-a)^{-1}$. Then $0, 1 \\in A$ and we still have $(1-\\alpha)x + \\alpha y \\in A$ whenever $x, y \\in A$.\n\nBy plugging in $x \\equiv 0$, we get $\\alpha A \\subseteq A$, and from $y \\equiv 0$, we get $(1-\\alpha)A \\subseteq A$. Therefore (noting that $p \\nmid \\alpha, 1-\\alpha$)\n\n$$\nA + A \\equiv (1-\\alpha)(1-\\alpha)^{p-2}A + \\alpha\\alpha^{p-2}A \\subseteq (1-\\alpha)A + \\alpha A \\subseteq A,\n$$\n\nso $A+1 \\subseteq A \\implies A = \\mathbb{Z}/p\\mathbb{Z} \\implies S = \\mathbb{Z}/p\\mathbb{Z}$, as desired.\n\n**Solution 3.** We present yet another proof of the lemma. The previous proof shows that if $x, y \\in S$, then $x + \\alpha(y-x) \\in S$. We claim that $N = x + (\\alpha+1)(y-x)$ lies in $S$ as well. Indeed, the only way $[f(N)-f(y)][N-y] \\equiv [f(N)-r]\\alpha(y-x)$ and $[f(N)-f(x+\\alpha(y-x))][N-(x+\\alpha(y-x))] \\equiv [f(N)-r](y-x)$ can both be quadratic residues is if $p \\mid f(N)-r$, so $N \\in S$.\n\nBut $[x+(\\alpha+1)(y-x)] - [x+\\alpha(y-x)] = y-x$, so we conclude that $x+k\\alpha(y-x)$ and $x+k(\\alpha+1)(y-x)$ lie in $S$ for all $k \\ge 0$. Since $p \\nmid \\alpha$, $x+k\\alpha(y-x)$ covers all residues modulo $p$, so $S = \\mathbb{Z}/p\\mathbb{Z}$, as desired.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 13497,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $ABCD$ be a quadrilateral inscribed in a circle $\\Gamma$. Let $E$, $F$, $G$, $H$ be the midpoints of the arcs $AB$, $BC$, $CD$, $DA$ of the circle $\\Gamma$. Suppose $AC \\cdot BD = EG \\cdot FH$. Prove that $AC$, $BD$, $EG$, $FH$ are concurrent.",
    "options": [],
    "answer": "See solution",
    "solution": "![](images/Indija_mo_2011_p3_data_f8ed928cd2.png)\nLet $R$ be the radius of the circle $\\Gamma$. Observe that $\\angle EDF = \\frac{1}{2}\\angle D$. Hence $EF = 2R \\sin \\frac{D}{2}$. Similarly, $HG = 2R \\sin \\frac{B}{2}$. But $\\angle B = 180^\\circ - \\angle D$, thus $HG = 2R \\cos \\frac{D}{2}$. We hence get\n\n$$\nEF \\cdot GH = 4R^2 \\sin \\frac{D}{2} \\cos \\frac{D}{2} = 2R^2 \\sin D = R \\cdot AC.\n$$\n\nSimilarly, we obtain $EH \\cdot FG = R \\cdot BD$.\n\nTherefore,\n\n$$\nR(AC + BD) = EF \\cdot GH + EH \\cdot FG = EG \\cdot FH,\n$$\n\nby Ptolemy's theorem. By the given hypothesis, this gives $R(AC + BD) = AC \\cdot BD$. Thus,\n\n$$\nAC \\cdot BD = R(AC + BD) \\geq 2R\\sqrt{AC \\cdot BD},\n$$\n\nusing the AM-GM inequality. This implies that $AC \\cdot BD \\geq 4R^2$. But $AC$ and $BD$ are chords of $\\Gamma$, so $AC \\leq 2R$ and $BD \\leq 2R$. We obtain $AC \\cdot BD \\leq 4R^2$. It follows that $AC \\cdot BD = 4R^2$, implying that $AC = BD = 2R$. Thus $AC$ and $BD$ are two diameters of $\\Gamma$. Using $EG \\cdot FH = AC \\cdot BD$, we conclude that $EG$ and $FH$ are also two diameters of $\\Gamma$. Hence $AC$, $BD$, $EG$, and $FH$ all pass through the centre of $\\Gamma$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 13498,
    "subject": "Mathematics (Olympiad)",
    "question": "The legs of each right-angled triangle are of length $1$ cm and its hypotenuse is of length $\\sqrt{2}$ cm. What is the length of the side of the octagonal hole formed by these triangles?",
    "options": [],
    "answer": "See solution",
    "solution": "The side of the octagonal hole is of length $\\sqrt{2} - 1$ cm.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 13499,
    "subject": "Mathematics (Olympiad)",
    "question": "證明以下命題：\n\n若一個凸 $n$ 邊形可以完整分割成有限多個平行四邊形，則 $n$ 必為偶數且每個邊都有等長的平行對邊。\n\n反之，若一個凸 $n$ 邊形 ($n$ 為偶數) 的每個邊都有等長的平行對邊，則它可以完整分割成有限多個平行四邊形。",
    "options": [],
    "answer": "See solution",
    "solution": "首先證明：邊 $a$ 與其對邊 $O(a)$ 的長度相同。對任意邊 $a$，有一個平行四邊形貼著 $a$，其中一個頂點是向量 $\\vec{a}$ 的尾端，稱此平行四邊形為 $A$。由 $A$ 出發，向上構造一系列平行四邊形，每次都貼在前一個平行四邊形的左上角頂點，直到到達 $O(a)$。由於構造方式，不會有向右陷入的平行四邊形。\n\n這個序列的平行四邊形有以下性質：\n\n1. 從左邊的邊界來看，沒有一個會向左突出；\n2. 左邊的邊界從 $\\vec{a}$ 的尾端一路到達 $O(\\vec{a})$ 的頭端。\n\n若 (1) 或 (2) 不成立，則可以反向由上而下構造另一序列的平行四邊形，最終一個應貼到 $a$。但 $A$ 已是貼在 $a$ 上最左邊的平行四邊形，新序列在前一序列左邊，最終一個不可能貼到 $a$。稱左邊的折線邊界為 $\\ell_1$，同理右邊的折線邊界為 $\\ell_2$，從 $\\vec{a}$ 的頭端連到 $O(\\vec{a})$ 的尾端。\n\n在 $a, \\ell_1, O(a), \\ell_2$ 所圍成的區域內有有限個 $\\sqcup$ 與 $\\perp$ 這類度為 3 的頂點。對於 $\\sqcup$ 可一路畫平行線到達 $a$，對於 $\\perp$ 可一路畫平行線到達 $O(a)$。畫完後仍是一個平行四邊形分割。\n\n因此，邊 $a$ 被分成有限個區段，每個區段有一個堆疊的平行四邊形序列，且每個序列的平行四邊形寬度都等於 $a$。相鄰序列間可能有空隙，但不影響結論。如此一路由 $a$ 到達 $O(a)$，$O(a)$ 也有同樣性質。結論：$a$ 與 $O(a)$ 被分成有限個區段且一一對應，每個對應段長度相同，因此 $a$ 與 $O(a)$ 的邊長相同。\n\n接著用歸納法證明反方向：\n\n任意凸 $n$ 邊形，若 $n$ 為偶數且每個邊都有等長的平行對邊，則可完整分割成有限多個平行四邊形。當 $n=4$ 時顯然成立。考慮 $n \\geq 6$ 為偶數，取一邊 $a$ 及其等長對邊 $O(a)$。從 $a$ 逆時針方向沿著邊到達 $O(a)$，畫出 $(n-2)/2$ 個平行四邊形，寬度皆等於 $a$。去掉這些平行四邊形後，剩下的為一個凸 $n-2$ 邊形，且仍有每邊都有等長的平行對邊。由歸納假設，該 $n-2$ 邊形可完整分割成有限多個平行四邊形，因此原 $n$ 邊形也可如此分割。證畢。",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 13500,
    "subject": "Mathematics (Olympiad)",
    "question": "The prime numbers $p$, $q$, and $r$ satisfy the simultaneous equations:\n\n$$\npq + pr = 80\n$$\n$$\npq + qr = 425\n$$\n\nFind the value of $p + q + r$.",
    "options": [],
    "answer": "See solution",
    "solution": "Method 1\n\nWe have $p(q + r) = 80 = 2^4 \\times 5$ and $q(p + r) = 425 = 5^2 \\times 17$.\n\nSo $p = 2$ or $5$, $q = 5$ or $17$, $r = 3$, $5$, or $23$.\n\nFrom $p(q + r) = 80$, if $p = 5$ then $q + r = 16$, which has no solution.\n\nSo $p = 2$, then $q + r = 40$, hence $q = 17$ and $r = 23$.\n\nTherefore, $p + q + r = 2 + 17 + 23 = 42$.\n\nMethod 2\n\nWe have $p(q + r) = 80 = 2^4 \\times 5$ and $q(p + r) = 425 = 5^2 \\times 17$.\n\nSo $p = 2$ or $5$ and $q = 5$ or $17$.\n\nFrom $q(p + r) = 425$, if $q = 5$ then $p + r = 85$.\n\nSo $r = 83$ or $80$, which both contradict $p(q + r) = 80$.\n\nHence $q = 17$ and $p + r = 25$. Since $r$ is prime, $p = 2$ and $r = 23$.\n\nTherefore, $p + q + r = 2 + 17 + 23 = 42$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 13501,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $\\triangle ABC$ be a triangle with $\\angle BAC = 120^\\circ$.\n\nLet $M$ be the midpoint of $BC$. Let $l$ be a line that meets the altitude from $A$, the angle bisector of $\\angle BAC$, and the line $AO$ at points $E$, $P$, and $N$, respectively, where $O$ is the circumcenter of $\\triangle ABC$.\n\nSuppose that $AE \\parallel MO$. If $EM$ bisects $AO$, what is the value of $\\angle BAC$?\n\n![](images/1.2_1997-2023_IMO_HK_TST_Solutions_p152_data_05efda91c0.png)",
    "options": [],
    "answer": "See solution",
    "solution": "Note that $AE \\parallel MO$. In order for $EM$ to bisect $AO$, the quadrilateral $AEOM$ must be convex, so $\\angle BAC$ must be obtuse.\n\nSince $AE \\parallel MO$ and $AN = ON$, we have $\\triangle AEN \\cong \\triangle OMN$. It is well-known that $AE$ and $AO$ are isogonal lines with respect to $\\angle BAC$. Therefore, $AP$ bisects $\\angle EAN$. As $AP \\perp EN$, this shows $\\triangle AEN$ is isosceles, with $AE = AN$. Therefore,\n\n$$\nOM = AE = AN = \\frac{1}{2}OA = \\frac{1}{2}OB.\n$$\n\nAs $\\angle BMO = 90^\\circ$, this yields $\\angle OBM = 30^\\circ$. Thus, $\\angle BAC = 120^\\circ$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 13502,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $P(x) = x^3 + a x^2 + b x + 1$ be a polynomial with real coefficients and three real roots $\\rho_1, \\rho_2, \\rho_3$ such that $|\\rho_1| < |\\rho_2| < |\\rho_3|$. Let $A$ be the point where the graph of $P(x)$ intersects the $y$-axis, and the points $B(\\rho_1, 0)$, $C(\\rho_2, 0)$, $D(\\rho_3, 0)$. If the circumcircle of $\\triangle ABD$ intersects the $y$-axis for a second time at $E$, find the minimum value of the length of the segment $EC$ and the polynomials for which this is attained.\n\n![](images/2020_BMO_Short_List_p4_data_6669ce1c31.png)",
    "options": [],
    "answer": "See solution",
    "solution": "Let $O$ be the origin. Since $P(0) = 1$, $A$ is the point $(0, 1)$, so $OA = 1$. From Vieta's relations we have $\\rho_1 \\rho_2 \\rho_3 = -1$, so $|\\rho_1 \\rho_2 \\rho_3| = 1$.\n\nFrom the power of the point $O$ we have\n\n$$\nOB \\cdot OD = OA \\cdot OE \\Rightarrow |\\rho_1| \\cdot |\\rho_3| = 1 \\cdot OE,\n$$\n\nhence $OE = \\frac{1}{|\\rho_2|}$.\n\nFinally, from Pythagoras' theorem we have\n\n$$\nCE^2 = OE^2 + OC^2 = |\\rho_2|^2 + \\frac{1}{|\\rho_2|^2} \\ge 2,\n$$\n\nhence $CE \\ge \\sqrt{2}$.\n\nThe equality holds for $|\\rho_2| = 1$ (i.e., $\\rho_2 = \\pm 1$). In the first case we have the family\n\n$$\nP(x) = (x - 1)(x + a) \\left(x - \\frac{1}{a}\\right),\n$$\n\nwhile in the second case we have\n\n$$\nP(x) = (x + 1)(x - a) \\left(x - \\frac{1}{a}\\right).\n$$",
    "topic": "Discrete Mathematics",
    "subtopic": "Other"
  },
  {
    "id": 13503,
    "subject": "Mathematics (Olympiad)",
    "question": "Given an acute, scalene triangle $ABC$ with circumcircle $(O)$ and orthocenter $H$. Let $M, N$, and $P$ be the midpoints of $BC, CA$, and $AB$, respectively. Let $D, E$, and $F$ be the feet of the altitudes from $A, B$, and $C$ of triangle $ABC$. Let $K$ be the reflection of $H$ through $BC$. The lines $DE$ and $MP$ intersect at $X$, and the lines $DF$ and $MN$ intersect at $Y$.\n\n**a)** The line $XY$ intersects the minor arc $BC$ of $(O)$ at $Z$. Prove that $K, Z, E$, and $F$ are concyclic.\n\n**b)** The lines $KE$ and $KF$ meet $(O)$ again at $S$ and $T$, respectively. Prove that $BS, CT$, and $XY$ are concurrent.",
    "options": [],
    "answer": "See solution",
    "solution": "a) First, applying Pascal's theorem for six points ($D, P, N, M, E, F$), we get that the intersections of the pairs of lines $(DE, MP)$, $(DF, MN)$, and $(PF, NE)$ are collinear, so $A, X$, and $Y$ are collinear.\n\n![](images/Vietnamese_mathematical_competitions_p226_data_9aba45be3f.png)\n\nClearly, $180^\\circ - \\angle BAC = \\angle BHC = \\angle BKC$, so $K$ lies on $(O)$. Next, we will prove that $XY$ bisects $EF$.\n\nNote that $MN$ and $MP$ are the midlines of triangle $ABC$, so $MN \\parallel AB$ and $MP \\parallel AC$. Therefore, by Thales's theorem, we have\n\n$$\n\\frac{YD}{YF} = \\frac{MD}{MB}, \\quad \\frac{XE}{XC} = \\frac{MC}{MD}\n$$\n\nLet $Q$ be the intersection of $XY$ and $EF$. Applying Menelaus' theorem to triangle $DEF$, we get\n\n$$\n\\frac{QF}{QE} \\cdot \\frac{XE}{XD} \\cdot \\frac{YD}{YF} = 1\n$$\n\nor $Q$ is the midpoint of $EF$. It follows that $AQ$ and $AM$ are isogonal with respect to $\\angle BAC$, so $AQ$ is the symmedian of triangle $ABC$. Therefore, $ABZC$ is a harmonic quadrilateral.\n\nLet $J$ be the intersection of $EF$ and $BC$, then $(JD, BC) = -1$, so $K(JD, BC) = -1$. But we also have $K(ZA, BC) = -1$, which implies $KZ$ passes through $J$.\n\nFinally, we have $JE \\cdot JF = JB \\cdot JC = JK \\cdot JZ$, so $K, Z, E$, and $F$ are concyclic.\n\nb) We have $\\angle EBF = \\angle ECH = \\angle EDH$ and $\\angle HED = \\angle HCD = \\angle BEF$ because $BCEF$ and $EHDC$ are cyclic quadrilaterals. Therefore, $\\triangle BEF \\sim \\triangle DEH$ (AA similarity). Hence,\n\n$$\n\\frac{HK}{2EH} = \\frac{HD}{EH} = \\frac{BF}{EF} = \\frac{BF}{2FQ'}\n$$\n\nBut we also have $\\angle BFQ = \\angle EHK$, then $\\triangle BFQ \\sim \\triangle SLE$ (SAS similarity), which implies\n\n$$\n\\angle FBQ = \\angle EKH = \\angle ABS\n$$\n\nSo $B, Q$, and $S$ are collinear. Similarly, $C, T$, and $S$ are collinear. Therefore, $BS, CT$, and $XY$ are concurrent at $Q$. $\\square$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 13504,
    "subject": "Mathematics (Olympiad)",
    "question": "Prove that for every positive integer $n$, there exists a positive integer $k$ such that each of the numbers $k, k^2, \\dots, k^n$ has at least one block \"2022\" in its decimal representation.",
    "options": [],
    "answer": "See solution",
    "solution": "We begin by proving that for every positive integer $n$, there exists a positive integer $k$ such that the number $k^n$ has at least one 2022 block in its decimal representation. Even more, we prove that the block is formed by the first four digits of the number. In that case we would have\n\n$$\n2022 \\cdot 10^d \\le k^n < 2023 \\cdot 10^d,\n$$\n\nwhich is equivalent to\n\n$$\n2022^{\\frac{1}{n}} \\cdot 10^{\\frac{d}{n}} \\le k < 2023^{\\frac{1}{n}} \\cdot 10^{\\frac{d}{n}}.\n$$\n\nFor $d$ large enough, we have $2023^{\\frac{1}{n}} \\cdot 10^{\\frac{d}{n}} - 2022^{\\frac{1}{n}} \\cdot 10^{\\frac{d}{n}} > 1$, so there exists a value of $k$ satisfying the previous inequality. Thus $k^n$ begins with the block 2022, as wanted.\n\nNow we prove the statement in the problem by induction. For $n=1$ we can take $k=2022$. Suppose that $A$ satisfies that the numbers $A, A^2, \\dots, A^n$ have at least one 2022 block in their decimal representation. We consider $B$ such that the number $B^{n+1}$ has at least one 2022 block in its decimal representation. We will show that for $k = B \\cdot 10^e + A$ each of the numbers $k, k^2, \\dots, k^n, k^{n+1}$ has at least one 2022 block in its decimal representation if $e$ is large enough. Let $e$ be such that $10^e > A^n$, then for $1 \\le j \\le n$ we have\n\n$$\nk^j = (B \\cdot 10^e + A)^j \\equiv A^j \\pmod{10^e},\n$$\n\nso the last digits of $k^j$ are exactly $A^j$ and in particular they contain a 2022 block.\n\nFor the exponent $n+1$ we have\n\n$$\n(B \\cdot 10^e + A)^{n+1} = B^{n+1} 10^{e(n+1)} + \\sum_{j=0}^{n} B^j 10^{ej} A^{n+1-j} \\binom{n+1}{j}.\n$$\n\nNow we take $e$ such that $10^e > \\sum_{j=0}^{n} B^j A^{n+1-j} \\binom{n+1}{j}$. Then\n\n$$\n\\sum_{j=0}^{n} B^j 10^{ej} A^{n+1-j} \\binom{n+1}{j} \\le 10^{en} \\sum_{j=0}^{n} B^j A^{n+1-j} \\binom{n+1}{j} < 10^{e(n+1)}.\n$$\n\nHence the leading digits in $k^{n+1} = (B \\cdot 10^e + A)^{n+1}$ are the ones in $B^{n+1}$ and in particular they contain at least one 2022 block. We have proved that $k$ has the desired property, which completes the induction.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 13505,
    "subject": "Mathematics (Olympiad)",
    "question": "Con baldosas cuadradas de lado un número exacto de unidades se ha podido embaldosar una habitación de superficie $18144$ unidades cuadradas de la siguiente manera: el primer día se puso una baldosa, el segundo, dos baldosas, el tercero tres, etc. ¿Cuántas baldosas fueron necesarias?",
    "options": [],
    "answer": "See solution",
    "solution": "Supongamos que fueron necesarias $n$ baldosas y que su tamaño es $k \\times k$. Entonces $nk^2 = 18144 = 2^5 \\times 3^4 \\times 7$. Hay nueve casos posibles para $n$, a saber, $2 \\times 7$, $2^3 \\times 7$, $2^5 \\times 7$, $2 \\times 3^2 \\times 7$, $2^3 \\times 3^2 \\times 7$, $2^5 \\times 3^2 \\times 7$, $2 \\times 3^4 \\times 7$, $2^3 \\times 3^4 \\times 7$, $2^5 \\times 3^4 \\times 7$. Además este número tiene que poderse expresar en la forma $1+2+3+\\cdots+N = \\frac{N(N+1)}{2}$ y esto sólo es posible en el caso sexto: $2^5 \\times 3^2 \\times 7 = 63 \\times 64 / 2 = 2016$. Para descartar los otros casos rápidamente observamos que $N$ y $N+1$ son números primos entre sí. Si por ejemplo $\\frac{N(N+1)}{2} = 2^3 \\times 7$, tendría que ser $N+1 = 2^4$ y $N = 7$, que es imposible, etc. Por tanto, se necesitaron $2016$ baldosas.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 13506,
    "subject": "Mathematics (Olympiad)",
    "question": "a) The length of a semicircle of radius $r$ is $\\pi r$. Given a line segment $AB$, let $P$ be a point between $A$ and $B$ such that $AP = x$ and $PB = y$. Semicircles are drawn on $AP$, $PB$, and $AB$ as diameters. Show that the length of the largest semicircle (on $AB$) is equal to the sum of the lengths of the two smaller semicircles (on $AP$ and $PB$), regardless of the position of $P$.\n\nb) Let $C$ and $E$ be the highest points of the semicircles on $AP$ and $PB$, respectively. Let $D$ be the highest point of the semicircle on $AB$. Show that $A$, $C$, $D$ are collinear, and $B$, $E$, $D$ are collinear. Show that $CDEP$ is a rectangle, where $P$ is as above.\n\n![](images/2023_Australian_Scene_p45_data_ec2bf081d5.png)\n\nc) Let $x = AP$ and $y = PB$ as before. Find the area of rectangle $CDEP$ in terms of $x$ and $y$. For what value(s) of $x$ and $y$ (with $x + y = 1$) does this area equal $\\frac{1}{2}$?\n\nd) Suppose $AB$ is divided into three segments $AP = x$, $PQ = y$, $QB = z$ with $x + y + z = 1$. Semicircles are drawn on $AP$, $PQ$, and $QB$ as diameters. Let $F$, $G$, $H$, $Q$ be the highest points of these semicircles. Find the sum of the areas of rectangles $FCGP$ and $GEHQ$ in terms of $x$, $y$, $z$. For what value(s) of $y$ does this sum equal $\\frac{1}{2}$?",
    "options": [],
    "answer": "See solution",
    "solution": "a) The length of a semicircle of radius $r$ is $\\pi r$. The arc length $AP$ is $\\pi x$, and the arc length $PB$ is $\\pi y$. The diameter $AB$ is $2x + 2y$, so the radius of the semicircle on $AB$ is $x + y$ and its length is $\\pi(x + y)$. Therefore, the length of the largest semicircle is equal to the sum of the lengths of the two smaller semicircles, regardless of the position of $P$.\n\nb) Let $O$ be the midpoint of $AP$. Since $C$ is the highest point on the $AP$ semicircle, $CO$ is perpendicular to $AP$. Thus, $AOC$ is a right-angled isosceles triangle, so $\\angle CAB = 45^\\circ$. Similarly, $\\angle DAB = 45^\\circ$. Therefore, $A$, $C$, $D$ are collinear. Similarly, $B$, $E$, $D$ are collinear.\n\nSince $AOC$ is a right-angled isosceles triangle, $\\angle ACO = 45^\\circ$. Similarly, $\\angle PCO = 45^\\circ$. Hence, $\\angle ACP = 90^\\circ$. Similarly, angles $ADB$ and $PEB$ are $90^\\circ$. So all angles in $CDEP$ are $90^\\circ$, and $CDEP$ is a rectangle.\n\nc) By the Pythagorean theorem in triangle $COP$, $CP = x\\sqrt{2}$. Similarly, $EP = y\\sqrt{2}$. Hence, the area of rectangle $CDEP$ is $x\\sqrt{2} \\times y\\sqrt{2} = 2xy$. We want $xy = \\frac{1}{4}$. Noting that $x + y = 1$, we can let $x = y = \\frac{1}{2}$, that is, place $P$ so that $AP = PB$.\n\nd) We have $x + y + z = 1$. From part (c), the areas of $FCGP$ and $GEHQ$ are $2xy$ and $2yz$ respectively, so their sum is $2y(x+z) = 2y(1-y)$. This equals $\\frac{1}{2}$ if $y = \\frac{1}{2}$. So $P$ and $Q$ can be anywhere on $AB$ provided $PQ = 1$. For example, $P = 0.5$ and $Q = 1.5$, or $P = 0.3$ and $Q = 1.3$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 13507,
    "subject": "Mathematics (Olympiad)",
    "question": "Solve the equation $3^n + 4^m = 5^k$, where $n$, $m$, and $k$ are nonnegative integers.",
    "options": [],
    "answer": "See solution",
    "solution": "**Answer:** $(2,2,2)$, $(0,1,1)$.\n\n**Solution.**\n\nLet $n \\geq 1$, $m \\geq 1$.\n\nFirst, consider the equation modulo $3$: $4^m \\equiv 1 \\pmod{3}$. If $n \\geq 1$, then $3^n + 4^m \\equiv 1 \\pmod{3}$. Therefore, $5^k \\equiv 1 \\pmod{3}$, which implies $k = 2k_1$ for some integer $k_1$.\n\nNow, consider the equation modulo $4$: $5^k \\equiv 1 \\pmod{4}$. Thus, $5^k - 4^m \\equiv 1 \\pmod{4}$, so $3^n \\equiv 1 \\pmod{4}$, which implies $n = 2n_1$ for some integer $n_1$.\n\n![](images/Ukrajina_2008_p3_data_b9f47cef71.png)\n\nThus, the equation becomes:\n$$\n3^{2n_1} + 4^m = 5^{2k_1} \\implies 3^{2n_1} = (5^{k_1} - 2^m)(5^{k_1} + 2^m)\n$$\nThis means:\n$$\n\\begin{cases}\n5^{k_1} - 2^m = 3^p \\\\\n5^{k_1} + 2^m = 3^s\n\\end{cases}, \\quad 0 \\leq p < s, \\quad p + s = 2n_1\n$$\nAdding the two equations:\n$$\n2 \\cdot 5^{k_1} = 3^p (1 + 3^{s-p})\n$$\nTherefore, $p = 0$. So $s = 2n_1$ and the system becomes:\n$$\n\\begin{cases}\n5^{k_1} - 2^m = 1 \\\\\n5^{k_1} + 2^m = 3^{2n_1}\n\\end{cases}\n$$\nSubtracting the two equations:\n$$\n2^{m+1} = 3^{2n_1} - 1 = (3^{n_1} - 1)(3^{n_1} + 1)\n$$\nSo:\n$$\n\\begin{cases}\n3^{n_1} - 1 = 2^q \\\\\n3^{n_1} + 1 = 2^t\n\\end{cases}, \\quad 0 \\leq q < t, \\quad q + t = 2n_1\n$$\nSince $(3^{n_1} - 1, 3^{n_1} + 1) = (2, 1)$, this implies $q = 1$ and $3^{n_1} - 1 = 2 \\implies n_1 = 1$.\n\nNow, the system is:\n$$\n\\begin{cases}\n5^{k_1} - 2^m = 1 \\\\\n5^{k_1} + 2^m = 9\n\\end{cases}\n$$\nEnumerating possible values, we find $k_1 = 1$, $m = 2$ is the only solution. Thus, $(n, m, k) = (2n_1, m, 2k_1) = (2, 2, 2)$.\n\nNow, consider the case $m = 0$:\n$$\n3^n + 1 = 5^k\n$$\nBut $3^n$ and $5^k$ are both odd for $n, k \\geq 1$, so there is no solution in this case.\n\nNow, consider $n = 0$:\n$$\n1 + 4^m = 5^k\n$$\nModulo $3$: $1 + 4^m \\equiv 2 \\pmod{3}$, so $5^k \\equiv 2 \\pmod{3}$, which implies $k = 2k_2 + 1$.\n\nNow, modulo $8$: If $m \\geq 2$, $1 + 4^m \\equiv 1 \\pmod{8}$, but $5^{2k_2+1} = 5 \\cdot 25^{k_2} \\equiv 5 \\pmod{8}$, which is a contradiction. Thus, $m \\leq 1$.\n\nFor $m = 1$:\n$$\n1 + 4 = 5^k \\implies 5 = 5^k \\implies k = 1\n$$\nSo, another solution is $(n, m, k) = (0, 1, 1)$.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 13508,
    "subject": "Mathematics (Olympiad)",
    "question": "Positive real numbers $a, b, c, d$ satisfy the equalities\n$$\na = c + \\frac{1}{d} \\quad \\text{and} \\quad b = d + \\frac{1}{c}.\n$$\n*Prove the inequality* $ab \\ge 4$ *and find the minimum of* $ab + cd$.",
    "options": [],
    "answer": "See solution",
    "solution": "To prove the inequality $ab \\ge 4$, substitute from the given equalities:\n$$\nab = \\left(c + \\frac{1}{d}\\right)\\left(d + \\frac{1}{c}\\right) = cd + 1 + 1 + \\frac{1}{cd} = cd + 2 + \\frac{1}{cd}.\n$$\nSince $x + 1/x \\ge 2$ for all positive reals $x = cd > 0$, we have $ab \\ge 4$.\n\nTo find the minimum of $ab + cd$, substitute for $a$ and $b$:\n$$\nab + cd = \\left(2 + cd + \\frac{1}{cd}\\right) + cd = 2 + 2cd + \\frac{1}{cd}.\n$$\nApply the inequality $x + y \\ge 2\\sqrt{xy}$ for $x = 2cd$, $y = 1/cd$:\n$$\n2cd + \\frac{1}{cd} \\ge 2\\sqrt{2}.\n$$\nThus, $ab + cd \\ge 2(1 + \\sqrt{2})$.\n\nEquality occurs when $2cd = 1/cd$, i.e., $cd = 1/\\sqrt{2}$. For example, $c = 1$, $d = \\sqrt{2}/2$ yields $a = 1 + \\sqrt{2}$, $b = 1 + \\sqrt{2}/2$, and $ab + cd = 2(1 + \\sqrt{2})$.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 13509,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $ABCD$ be a rectangle with side lengths $AB = CD = 5$ and $BC = AD = 10$. $W$, $X$, $Y$, $Z$ are points on $AB$, $BC$, $CD$, and $DA$ respectively, chosen so that $WXYZ$ is a kite, where $\\angle ZWX$ is a right angle. Given that $WX = WZ = \\sqrt{13}$ and $XY = ZY$, determine the length of $XY$.",
    "options": [],
    "answer": "See solution",
    "solution": "![](images/SAOlympiad2017Solutions_p1_data_bcea4391f4.png)\n\nNote that $\\angle AZW = 90^\\circ - \\angle AWZ = \\angle BWX$ and $\\angle AWZ = 90^\\circ - \\angle BWX = \\angle BXW$. Moreover, $WX = WZ$, so the two triangles $AWZ$ and $BXW$ are congruent. Let $BW = AZ = x$, so that $BX = AW = 5 - x$. Pythagoras' theorem gives us\n\n$$\nx^2 + (5-x)^2 = 13\n$$\n\nwhich simplifies to $x^2 - 5x + 6 = 0$. The two solutions are $x = 2$ and $x = 3$, and by symmetry we can assume that $x = 2$. So $BX = 3$, $XC = 7$, $AZ = 2$ and $ZD = 8$. Now let $CY = y$, so that $YD = 5 - y$. Applying Pythagoras' theorem again (and making use of the fact that $XY = YZ$), we get\n\n$$\n7^2 + y^2 = 8^2 + (5-y)^2\n$$\n\nThis simplifies to $10y = 40$, so $y = 4$. Now we finally find that $XY = \\sqrt{7^2 + 4^2} = \\sqrt{65}$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 13510,
    "subject": "Mathematics (Olympiad)",
    "question": "A rational number $x$ is given. Prove that there exists a sequence $x_0, x_1, x_2, \\dots$ of rational numbers with the following properties:\n\n(a) $x_0 = x$;\n(b) for every $n \\ge 1$, either $x_n = 2x_{n-1}$ or $x_n = 2x_{n-1} + \\frac{1}{n}$;\n(c) $x_n$ is an integer for some $n$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $x$ be written in lowest terms as $p/q$, and write $q = 2^r s$, where $s$ is odd. Let $S$ be the set of residue classes modulo $s$, where arithmetic on elements of $S$ is understood to be done modulo $s$.\n\nFor any positive integer $N$ and any $t \\in S$, say that $t$ is *attainable* at $N$ if there exists a sequence $x_0, x_1, \\dots, x_N$ such that:\n\n- $x_0 = x$;\n- $x_n = 2x_{n-1}$ or $x_n = 2x_{n-1} + 1/n$ for each $n = 1, \\dots, N$;\n- $s x_N$ is an integer in the residue class $t$.\n\nSay that a subset $T \\subseteq S$ is *attainable* at $N$ if every element of $T$ is attainable at $N$.\n\n**Lemma 1.** There exists a nonempty set attainable at some $N$.\n\n*Proof.* Taking $x_n = 2x_{n-1}$ for all $n$, we get $x_r = 2^r x = p/s$, in lowest terms, hence the residue class of $p$ is attainable at $r$. $\\square$\n\n**Lemma 2.** If the nonempty subset $T \\subseteq S$ is attainable at $N$, and $T$ is not all of $S$, then there exist another subset $T'$ and $N' > N$, with $T'$ attainable at $N'$, and $T'$ containing more elements than $T$.\n\n*Proof.* Choose $k$ such that $2^k s > N$, and put $N' = 2^k s + k$.\n\nFor each $t \\in T$, the residue class $2^{N'-N}t$ is attainable at $N'$. Indeed, if $x_0, \\dots, x_N$ attain $t$ at $N$, then just extend the sequence by defining $x_n = 2x_{n-1}$ for each $n = N+1, \\dots, N'$, and we have $x_{N'} = 2^{N'-N}x_N$ from which the assertion follows.\n\nAlso, the residue class $2^{N'-N}t+1$ is attainable at $N'$. Indeed, take the sequence $x_0, \\dots, x_N$ attaining $t$ at $N$, then define $x_n = 2x_{n-1}$ for each $n = N+1, \\dots, N'$ except for $n = 2^k s$, in which case we put $x_n = 2x_{n-1} + 1/n$. Then we have $x_{N'} = 2^{N'-N}x_N + 2^{N'-2^k s}/2^k s = 2^{N'-N}x_N + 1/s$, from which the assertion follows.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 13511,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $a, b > 1$ be positive integers such that the number $a + b$ divides the number $D(a, b) + v(a, b)$. Here $D(a, b)$ and $v(a, b)$ denote the greatest common divisor and the least common multiple of the numbers $a$ and $b$ respectively. Prove that\n\n$$\n\\frac{D(a,b) + v(a,b)}{a+b} \\le \\frac{a+b}{4}.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "Let $d = D(a, b)$. We can write $a = a_1 d$ and $b = b_1 d$, where $a_1$ and $b_1$ are coprime. We have $v(a, b) = a_1 b_1 d$. We insert this into the condition of the problem to get\n\n$$\na_1 d + b_1 d \\mid d + a_1 b_1 d \\implies a_1 + b_1 \\mid 1 + a_1 b_1.\n$$\n\nWe can rearrange the inequality to\n\n$$\n\\frac{d + a_1 b_1 d}{a_1 d + b_1 d} \\le \\frac{a_1 d + b_1 d}{4} \\iff 1 + a_1 b_1 \\le d \\frac{(a_1 + b_1)^2}{4} \\iff 4 \\le d(a_1^2 + b_1^2) + (2d - 4)a_1 b_1.\n$$\n\nIf $d \\ge 2$ we have\n\n$$\nd(a_1^2 + b_1^2) + (2d - 4)a_1 b_1 \\ge 2(a_1^2 + b_1^2) + (2 \\cdot 2 - 4)a_1 b_1 = 2(a_1^2 + b_1^2) \\ge 2(1^2 + 1^2) = 4,\n$$\n\nso the inequality is satisfied in this case.\n\nNow check the case when $d = 1$. We must prove the inequality\n\n$$\n4 \\le a_1^2 + b_1^2 - 2a_1 b_1 \\iff 4 \\le (a_1 - b_1)^2 \\iff 2 \\le |a_1 - b_1|.\n$$\n\nThis means $a_1$ and $b_1$ differ by at least 2. If $a_1 = b_1$, then $a_1 = b_1 = 1$ (since they are coprime), so $a = b = 1$, which is impossible.\n\nIf the numbers differ by 1, assume $b_1 = a_1 + 1$ (the other case is symmetric). Substitute into the condition:\n\n$$\na_1 + (a_1 + 1) \\mid 1 + a_1(a_1 + 1) \\implies 2a_1 + 1 \\mid a_1^2 + a_1 + 1.\n$$\n\nBut $2a_1 + 1 \\ge a_1 + 2$, so this is possible only if $a_1 = 1$. Since $d = 1$, $a = 1$, which is impossible since $a, b > 1$.\n\nThus, $a_1$ and $b_1$ differ by at least 2, which proves the inequality.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 13512,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $ABC$ be a triangle with circumcircle $\\omega$. The internal angle bisectors of $\\angle ABC$ and $\\angle ACB$ intersect $\\omega$ at $X$ and $Y$, respectively. Let $K$ be a point on $CX$ such that $\\angle KAC = 90^\\circ$. Similarly, let $L$ be a point on $BY$ such that $\\angle LAB = 90^\\circ$. Let $S$ be the midpoint of arc $CAB$ of $\\omega$. Prove that $SK = SL$.",
    "options": [],
    "answer": "See solution",
    "solution": "Without loss of generality, let $AB < AC$. We will prove that triangles $KXS$ and $SYL$ are congruent by SAS, which will finish the proof.\n\nAs $BX$ and $CY$ are angle bisectors, we obtain:\n\n$$\n\\frac{1}{2} \\angle CAB = \\angle CXS = \\angle CX + \\angle XS = \\frac{1}{2} \\angle CXA + \\angle XS.\n$$\n\nThis implies $\\angle XS = \\frac{1}{2} \\angle AYB = \\angle YB$ and therefore $SX = YB$. Note that $BY = YA$, hence $Y$ is the midpoint of the hypotenuse $BL$ in $\\triangle ABL$. Thus $SX = YB = YL$. Similarly, we get $SY = XK$.\n\nFinally, as $S$ is the midpoint of arc $CAB$, we obtain $\\angle SXC = \\angle BYS$, thus $\\angle KXS = \\angle SYL$, finishing the proof of congruency.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 13513,
    "subject": "Mathematics (Olympiad)",
    "question": "令 $\\mathbb{N}, \\mathbb{Z}, \\mathbb{Q}$ 分別代表所有正整數、整數及有理數所成集合。試求所有函數 $f : \\mathbb{Q} \\to \\mathbb{Z}$，滿足\n\n$$\nf\\left(\\frac{f(x)+a}{b}\\right) = f\\left(\\frac{x+a}{b}\\right)\n$$\n\n對於所有 $x \\in \\mathbb{Q}$，$a \\in \\mathbb{Z}$ 和 $b \\in \\mathbb{N}$ 都成立。\n\nLet $\\mathbb{N}, \\mathbb{Z}, \\mathbb{Q}$ denote the set of all positive integers, integers, and rational numbers respectively. Determine all functions $f : \\mathbb{Q} \\to \\mathbb{Z}$ satisfying\n\n$$\nf\\left(\\frac{f(x)+a}{b}\\right) = f\\left(\\frac{x+a}{b}\\right)\n$$\n\nfor all $x \\in \\mathbb{Q}$, $a \\in \\mathbb{Z}$, and $b \\in \\mathbb{N}$.",
    "options": [],
    "answer": "See solution",
    "solution": "$f = \\lfloor x \\rfloor$, $f = \\lceil x \\rceil$，以及常數函數。\n\n我們首先驗證以上函數滿足題意。對於所有三元組 $(x, a, b) \\in \\mathbb{Q} \\times \\mathbb{Z} \\times \\mathbb{N}$，令\n\n$$\nq = \\left\\lfloor \\frac{x+a}{b} \\right\\rfloor.\n$$\n\n易知 $bq \\le x + a < b(q+1) \\Rightarrow bq \\le \\lfloor x \\rfloor + a < b(q+1)$，故\n\n$$\n\\left\\lfloor \\frac{\\lfloor x \\rfloor + a}{b} \\right\\rfloor = \\left\\lfloor \\frac{x+a}{b} \\right\\rfloor.\n$$\n\n故 $f = \\lfloor x \\rfloor$ 確為一解。另兩解可類似地檢驗之。\n\n接著我們證明以上為所有可能解。考慮以下兩個情形：\n\n**狀況一.** 存在整數 $m$ 使得 $f(m) \\neq m$。\n\n令 $C = f(m)$。若 $m > C$，則對於所有整數 $y$，取 $x = m$，$b = m-C$，$a = yb-C$ 代入原式，得到 $f(y) = f(y+1)$，故對於所有整數 $y$，$f(y) = C$。若 $m < C$，則改取 $b = C-m$，同樣可得對於所有整數 $y$，$f(y) = C$。\n\n現在，對於所有有理數 $y = p/q$（其中 $q > 0$），取 $(x, a, b) = (C - p, p - C, q)$ 代入，即可得 $f(y) = f(0) = C$。故 $f$ 為常數函數。\n\n**狀況二.** 對於所有整數 $m$，$f(m) = m$。\n\n取 $b = 1$，知\n\n$$\nf(x) + a = f(x + a) \\quad (1)\n$$\n\n對於所有 $(x, a) \\in \\mathbb{Q} \\times \\mathbb{Z}$ 皆成立。令 $\\omega = f(1/2)$。\n\n*Claim 1.* $\\omega \\in \\{0, 1\\}$。\n\n*Proof.* 若 $\\omega \\le 0$，取 $(x, a, b) = (1/2, -\\omega, 1-2\\omega)$，知 $0 = f(0) = f(1/2) = \\omega$。同理，若 $\\omega \\ge 1$，取 $(x, a, b) = (1/2, \\omega-1, 2\\omega-1)$ 即可。\n\n*Claim 2.* 對於所有有理數 $0 < x < 1$，$f(x) = \\omega$。\n\n*Proof.* 若否，存在分母最小的 $p/q \\in (0, 1)$ 使得 $f(p/q) \\ne \\omega$。易知 $\\gcd(p, q) = 1$，且 $q \\ge 2$。若 $q$ 是偶數，則 $p$ 必為奇數，故可代入 $(x, a, b) = (1/2, (p-1)/2, q/2)$，得\n\n$$\nf\\left(\\frac{\\omega + (p-1)/2}{q/2}\\right) = f(p/q) \\ne \\omega.\n$$\n\n但由 Claim 1，$\\omega = 0$ 或 $1$，故上式無論如何都會得到一個 $p'/q' \\in (0, 1)$ 滿足 $f(p'/q') \\ne \\omega$ 且 $q' < q$。此與 $p/q$ 為分母最小的前提不合，矛盾。\n\n因此 $q$ 必為奇數，令 $q = 2k + 1$。代入 $(x, a, b) = (1/2, k, q)$，得\n\n$$\nf\\left(\\frac{\\omega + k}{q}\\right) = f\\left(\\frac{1}{2}\\right) = \\omega.\n$$\n\n基於 $p, q$ 互質，存在 $r \\in \\{1, 2, \\dots, q\\}$ 及整數 $m$ 使得 $rp - mq = k + \\omega$；又，基於右式非 $q$ 的倍數，$r < q$。而若 $m < 0$，則 $rp - mq > q \\ge k + \\omega$，不合，故 $m \\ge 0$。同理，若 $m \\ge r$，則 $rp - mq < br - br = 0$。因此，$0 \\le m \\le r - 1$。現在，取 $(x, a, b) = (\\frac{k+\\omega}{q}, m, r)$，得\n\n$$\nf\\left(\\frac{\\omega + m}{r}\\right) = f(p/q) \\ne \\omega.\n$$\n\n此與 $q$ 的最小性不合，矛盾！\n\n現在，若 $\\omega = 0$，由 Claim 2 知 $f(x) = \\lfloor x \\rfloor$。同理，若 $\\omega = 1$，由 Claim 2 知 $f(x) = \\lceil x \\rceil$。證畢。",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 13514,
    "subject": "Mathematics (Olympiad)",
    "question": "299 digits 0 and one digit 1 are written in a circle. The following moves are allowed:\n\n- From each digit, subtract the sum of the adjacent digits.\n- Select two digits with exactly two digits between them and increase both by 1 or decrease both by 1.\n\nIs it possible to obtain such an arrangement of numbers (after a finite number of such moves), in which there are three adjacent digits 1, and the rest of the digits are 0?",
    "options": [],
    "answer": "See solution",
    "solution": "Let us analyse how the recorded moves affect the sum of the digits written in a circle. Let us denote the numbers by $a_1, a_2, \\dots, a_{300}$. The following numbers will be written after the move of the first type: $b_k = a_k - a_{k-1} - a_{k+1}$, $k = 1, \\ldots, 300$ ($a_{301} \\equiv a_1$).\n\nConsider the expression $C = a_1 - a_2 + a_3 - a_4 + \\dots - a_{300}$. After the move of the first type, we obtain:\n\n$$\n\\begin{aligned}\nb_1 - b_2 + b_3 - b_4 + \\dots - b_{300} = 3C.\n\\end{aligned}\n$$\n\nAfter the move of the second type, $C$ does not change. In the initial arrangement, assume that $a_1 = 1$, $a_k = 0$ for $k = 2, \\ldots, 300$, then $C = 1$. In the final arrangement, $C = \\pm 1$—depending on the location of the three adjacent numbers 1. It is clear that we should not make a move of the first type, because otherwise, the expression $C$ will become divisible by 3 and will never decrease to 1. So, we should try to get the desired result by only using the moves of the second type.\n\nConsider $T = a_3 + a_6 + \\dots + a_{300}$. At the start, $T=0$, and in the end it equals 1. But, clearly, when applying the move of the second type, all summands in $T$ either do not change, or all change at the same time, i.e., $T$ can only change by $\\pm 2$. Thus, the parity of the expression $T$ does not change, which proves the impossibility of proper transformation.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 13515,
    "subject": "Mathematics (Olympiad)",
    "question": "The circle $\\omega_1$ with diameter $AB$ and the circle $\\omega_2$ with center $A$ intersect at points $C$ and $D$. Let $E$ be a point on the circle $\\omega_2$, which is outside $\\omega_1$ and on the same side as $C$ with respect to the line $AB$. Let the second point of intersection of the line $BE$ with $\\omega_2$ be $F$. Suppose that a point $K \\in \\omega_1$ is on the same side as $A$ with respect to the diameter of $\\omega_1$ passing through $C$ and $2 \\cdot CK \\cdot AC = CE \\cdot AB$. Let the second point of intersection of the line $KF$ with $\\omega_1$ be $L$. Show that a point symmetric to $D$ with respect to the line $BE$ lies on the circumcircle of triangle $LFC$.",
    "options": [],
    "answer": "See solution",
    "solution": "![](images/Turska_2014_p3_data_03acb6432a.png)\nLet $D'$ be the point symmetric to $D$ with respect to the line $BE$. $2 \\cdot CK \\cdot AC = CE \\cdot AB$ implies $\\dfrac{CK}{CE} = \\dfrac{R_1}{R_2}$ where $R_1$ and $R_2$ are the radii of the circles $\\omega_1$ and $\\omega_2$, respectively. By the sine law, $2 \\cdot \\sin \\angle CLF = \\dfrac{CK}{R_1} = \\dfrac{CE}{R_2} = 2 \\cdot \\sin \\angle CFE$, and hence $\\angle CLF = \\angle CFE$ since the sum of angles is less than $180^\\circ$.\n\nLet $X$ be the second intersection point of the line $BE$ with $\\omega_1$. Since $AB$ is a diameter of $\\omega_1$ and $AC = AD$, the smaller arcs $BC$ and $BD$ of $\\omega_1$ are equal and hence $\\angle DXB = \\angle CXB$. Since $\\angle D'XB = \\angle DXB$, we get that the points $X$, $C$, $D'$ are collinear. Since $\\angle AXB = 90^\\circ$, we get $EX = FX$ and since $\\angle ACB = 90^\\circ$, the line $BC$ is tangent to $\\omega_2$ and hence $\\angle FCB = \\angle BEC$. On the other hand, $\\angle DCF = \\angle DEF$, therefore $\\angle DEC = \\angle DCB = \\angle CXB = \\angle DXB$ and hence $\\angle ECX = \\angle DEX$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 13516,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $n = 2015$, so that there are $2n + 1$ lines. What is the maximum number of acute-angled triangles that can be formed by the intersections of these lines?",
    "options": [],
    "answer": "See solution",
    "solution": "We fix one of the lines $\\ell$ and place it as the $x$-axis of the coordinate plane. The other $2n$ lines can be partitioned into two groups: those with positive slopes and those with negative slopes. Let $a$ and $b$ be the sizes of these two groups, respectively.\n\nEvery pair of lines in the same group forms an obtuse triangle with $\\ell$. Thus, the number of obtuse triangles with $\\ell$ as a sideline adjacent to the obtuse angle is\n\n$$\n\\binom{a}{2} + \\binom{b}{2} \\ge 2\\binom{n}{2} = n(n-1)\n$$\n\nby Jensen's inequality, since $\\binom{x}{2}$ is convex.\n\nConsidering all choices of $\\ell$, each obtuse triangle is counted twice. Therefore, the number of acute-angled triangles is at most\n\n$$\n\\binom{2n+1}{3} - \\frac{(2n+1)n(n-1)}{2} = \\frac{n(n+1)(2n+1)}{6}.\n$$\n\nThis bound is attainable. For example, consider the sidelines of a regular polygon with $2n + 1$ sides. By symmetry, the number of lines with positive and negative slopes is equal for every $\\ell$, so equality holds.\n\nWhen $n = 2015$,\n\n$$\n\\frac{2015 \\times 2016 \\times 4031}{6} = 2\\,729\\,148\\,240.\n$$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 13517,
    "subject": "Mathematics (Olympiad)",
    "question": "A triangle $ABC$ is inscribed in a circle $\\omega$. A variable line $\\ell$ chosen parallel to $BC$ meets segments $AB$ and $AC$ at points $D$ and $E$ respectively, and meets $\\omega$ at points $K$ and $L$ (where $D$ lies between $K$ and $E$). Circle $\\gamma_1$ is tangent to the segments $KD$ and $BD$ and also tangent to $\\omega$, while circle $\\gamma_2$ is tangent to the segments $LE$ and $CE$ and also tangent to $\\omega$. Determine the locus, as $\\ell$ varies, of the meeting point of the common inner tangents to $\\gamma_1$ and $\\gamma_2$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $P$ be the meeting point of the common inner tangents to $\\gamma_1$ and $\\gamma_2$. Also, let $b$ be the angle bisector of $\\angle BAC$. Since $KL \\parallel BC$, $b$ is also the angle bisector of $\\angle KAL$.\n\nLet $\\mathfrak{H}$ be the composition of the symmetry $\\mathfrak{S}$ with respect to $b$ and the inversion $\\mathfrak{I}$ of centre $A$ and ratio $\\sqrt{AK \\cdot AL}$. (It is readily seen that $\\mathfrak{S}$ and $\\mathfrak{I}$ commute, so since $\\mathfrak{S}^2 = \\mathfrak{I}^2 = \\text{id}$, then also $\\mathfrak{H}^2 = \\text{id}$, the identical transformation.) The elements of the configuration interchanged by $\\mathfrak{H}$ are summarized in Table 1.\n\nLet $O_1$ and $O_2$ be the centres of circles $\\gamma_1$ and $\\gamma_2$. Since the circles $\\gamma_1$ and $\\gamma_2$ are determined by their construction (in a unique way), they are interchanged by $\\mathfrak{H}$, therefore the rays $AO_1$ and $AO_2$ are symmetrical with respect to $b$. Denote by $\\varrho_1$ and $\\varrho_2$ the radii of $\\gamma_1$ and $\\gamma_2$. Since $\\angle O_1AB = \\angle O_2AC$, we have $\\varrho_1/\\varrho_2 = AO_1/AO_2$. On the other hand, from the definition of $P$ we have $O_1P/O_2P = \\varrho_1/\\varrho_2 = AO_1/AO_2$; this means that $AP$ is the angle bisector of $\\angle O_1AO_2$ and therefore of $\\angle BAC$.\n\nThe limiting, degenerated, cases are when the parallel line passes through $A$ – when $P$ coincides with $A$; respectively when the parallel line is $BC$ – when $P$ coincides with the foot $A' \\in BC$ of the angle bisector of $\\angle BAC$ (or any other point on $BC$). By continuity, any point $P$ on the open segment $AA'$ is obtained for some position of the parallel, therefore the locus is the open segment $AA'$ of the angle bisector $b$ of $\\angle BAC$.\n\n| point $K$ | $\\leftrightarrow$ | point $L$ |\n|---|---|---|\n| line $KL$ | $\\leftrightarrow$ | circle $\\omega$ |\n| ray $AB$ | $\\leftrightarrow$ | ray $AC$ |\n| point $B$ | $\\leftrightarrow$ | point $E$ |\n| point $C$ | $\\leftrightarrow$ | point $D$ |\n| segment $BD$ | $\\leftrightarrow$ | segment $EC$ |\n| arc $BK$ | $\\leftrightarrow$ | segment $EL$ |\n| arc $CL$ | $\\leftrightarrow$ | segment $DK$ |\n\nTable 1: Elements interchanged by $\\mathfrak{H}$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 13518,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $m, n$ be positive integers. There are $n$ piles of gold coins, where the $i$th pile contains $a_i > 0$ coins ($i = 1, \\dots, n$). Consider the following game:\n\n1. Bob selects sets $B_1, B_2, \\dots, B_n$ such that each set $B_i$ is a nonempty subset of $\\{1, 2, \\dots, m\\}$.\n2. Alice selects a set $S$ such that $S$ is a nonempty subset of $\\{1, 2, \\dots, m\\}$, knowing the sets $B_1, B_2, \\dots, B_n$ that Bob has selected in Step 1.\n3. The gold coins in the $i$th pile ($i = 1, 2, \\dots, n$) are given to Bob if the number of elements in $B_i \\cap S$ is even. Otherwise, they are given to Alice.\n\nShow that no matter how Bob selects the sets $B_1, B_2, \\dots, B_n$, Alice will be able to select a set $S$ so as to receive more gold coins than Bob in total.",
    "options": [],
    "answer": "See solution",
    "solution": "At the end of the game, the difference between the number of gold coins received by Bob and Alice is\n$$\n\\sum_{i=1}^{n} (-1)^{|B_i \\cap S|} \\cdot a_i.\n$$\nAlice receives more coins than Bob if and only if\n$$\n\\sum_{i=1}^{n} (-1)^{|B_i \\cap S|} \\cdot a_i < 0.\n$$\nSuppose, for contradiction, that for each nonempty set $S$, Alice's coins are not more than Bob's, i.e.,\n$$\n\\sum_{i=1}^{n} (-1)^{|B_i \\cap S|} \\cdot a_i \\geq 0.\n$$\nIf $S = \\emptyset$, then $\\sum_{i=1}^{n} (-1)^{|B_i \\cap S|} \\cdot a_i = \\sum_{i=1}^{n} a_i > 0$.\n\nAdding over all possible sets $S$, we get\n$$\n\\sum_{S \\subseteq \\{1, 2, \\dots, m\\}} \\sum_{i=1}^{n} (-1)^{|B_i \\cap S|} \\cdot a_i > 0. \\quad (1)\n$$\n\nConsider the sum $\\sum_{S \\subseteq \\{1, 2, \\dots, m\\}} (-1)^{|B \\cap S|}$ for a nonempty subset $B \\subseteq \\{1, \\dots, m\\}$. Write\n$$\nS = C \\cup D, \\quad \\text{where} \\quad C = S \\setminus B, \\ D = S \\cap B, \\ C \\cap D = \\emptyset.\n$$\nThen,\n$$\n\\begin{aligned}\n\\sum_{S \\subseteq \\{1, 2, \\dots, m\\}} (-1)^{|B \\cap S|} &= \\sum_{C \\subseteq \\{1, 2, \\dots, m\\} \\setminus B} \\sum_{D \\subseteq B} (-1)^{|D|} \\\\\n&= \\sum_{C \\subseteq \\{1, 2, \\dots, m\\} \\setminus B} \\sum_{D \\subseteq B} (-1)^{|D|}.\n\\end{aligned}\n$$\nAs $|B| > 0$, we have\n$$\n\\sum_{D \\subseteq B} (-1)^{|D|} = \\sum_{r=0}^{|B|} (-1)^r \\binom{|B|}{r} = 0.\n$$\nThus,\n$$\n\\sum_{S \\subseteq \\{1, 2, \\dots, m\\}} (-1)^{|B \\cap S|} = 0\n$$\nfor each nonempty set $B \\subseteq \\{1, \\dots, m\\}$.\n\nConsequently,\n$$\n\\sum_{S \\subseteq \\{1, 2, \\dots, m\\}} \\sum_{i=1}^{n} (-1)^{|B_i \\cap S|} \\cdot a_i = \\sum_{i=1}^{n} \\left( a_i \\sum_{S \\subseteq \\{1, 2, \\dots, m\\}} (-1)^{|B_i \\cap S|} \\right) = 0,\n$$\nwhich contradicts (1). Therefore, there exists a nonempty set $S$ such that Alice receives more coins than Bob.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 13519,
    "subject": "Mathematics (Olympiad)",
    "question": "平面上有一個正三角形網格，相鄰兩格點的距離為 $1$。有一個邊長為 $n$ 的正三角形，其三個頂點都在格點上，三邊都落在格線上。現在，將此正三角形分割成 $n^2$ 個面積相等的小三角形（不需為正三角形），使得每個小三角形的三個頂點都在格點上。\n\n證明：其中至少有 $n$ 個小三角形是正三角形。\n\nThere is a grid of equilateral triangles with a distance $1$ between any two neighboring grid points. An equilateral triangle with side length $n$ lies on the grid so that all of its vertices are grid points, and all of its sides match the grid. Now, let us decompose this equilateral triangle into $n^2$ smaller triangles (not necessarily equilateral triangles) so that the vertices of all these small triangles are all grid points, and all these small triangles have equal areas.\n\nProve that there are at least $n$ equilateral triangles among these smaller triangles.",
    "options": [],
    "answer": "See solution",
    "solution": "不失一般性，假設大正三角形朝上。以下證明：在所有小三角形中，朝上的正三角形比朝下的正三角形至少多 $n$ 個。此自然可證明原命題。\n\n若所有小三角形皆為正三角形顯然成立（因每一橫排向上者都比向下者多一個）。若否，則進行以下操作：找所有小三角形中最長的邊 $AB$，並考慮以它為邊的兩個小三角形 $ABC$ 和 $ABD$。可以證明：\n\n**引理一**：$ABCD$ 組成一個平行四邊形。\n\n證明：由皮克公式 (Pick's formula) 知線段 $AB$ 上沒有其他格點，且所有可能的 $C$、$D$ 位置落在平行 $AB$ 的兩條直線上（分別在 $AB$ 兩側），並以長度為 $AB$ 的間隔分布。又，由於 $AB$ 為最長邊，角 $CBA$ 和角 $CAB$ 必為銳角，所以 $C$ 到 $AB$ 的垂足落在 $AB$ 線段內。結合此兩點，知 $C$ 的位置唯一確定；同理，$D$ 也唯一確定，並由對稱性易知 $D$ 為使 $ABCD$ 構成平行四邊形的點。Q.E.D.\n\n基於這兩個小三角形 $ABC$ 與 $ABD$ 組成平行四邊形，將 $AB$ 換成 $CD$ 會得到一個新的滿足題意的切割，且\n\n**引理二**：$AB > CD$。\n\n證明：由 $AB \\ge \\sqrt{3}$ 知 $d(C, AB) \\le 1/2$；又因 $AC \\ge 1$，知 $\\angle CAB \\le 30^\\circ$。同理，$\\angle CBA \\le 30^\\circ$。因此 $\\angle C$ 是鈍角，故由樞軸定理知 $AB > CD$。Q.E.D.\n\n因此，每進行一次操作，都會讓最長邊的距離變小；注意到網格上兩點之間的距離只有有限多種可能，故我們可以重複此一操作直到最長邊的邊長被降為 $1$，也就是所有小三角形都是正三角形的狀況。又，在這每次操作中，易知朝上與朝下正三角形個數的差不會改變，故知原本切割方式中，朝上正三角形比朝下正三小角形多 $n$ 個。證畢。",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 13520,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $\\phi(n)$ denote the number of positive integers less than or equal to $n$ that are coprime to $n$. Prove that for each positive integer $n$, we can choose a positive integer $m$ such that $\\phi(m) = n!$.",
    "options": [],
    "answer": "See solution",
    "solution": "For $n = 1$, choose $m = 2$ since $\\phi(2) = 1$.\n\nNow consider the sequence $(a_n)$ defined as follows: $a_1 = 2$ and for $n \\ge 2$,\n\n$$\na_n = \\begin{cases} (n+1)a_{n-1}, & \\text{if } n+1 \\text{ is a prime number,} \\\\ na_{n-1}, & \\text{otherwise.} \\end{cases}\n$$\n\nWe will prove by induction that $\\phi(a_n) = n!$ for all $n \\ge 2$. Suppose this is true for some integer $n-1 \\ge 1$, i.e., $\\phi(a_{n-1}) = (n-1)!$. There are two cases:\n\nIf $n+1 = p$ is a prime, then $a_n = (n+1)a_{n-1}$. By the definition of $(a_n)$, all prime divisors of $a_n$ are at most $n+1$. Then $\\gcd(a_{n-1}, p) = 1$, which implies that\n\n$$\n\\phi(a_n) = \\phi(p)\\phi(a_{n-1}) = (p-1)(n-1)! = n!\n$$\n\nIf $n+1$ is composite, then $a_n = na_{n-1}$. If $n = p$ is a prime, then by the definition, $a_{n-1} = na_{n-2}$ and $a_n = p^2 a_{n-2}$. Since $\\gcd(a_{n-2}, p) = 1$, then\n\n$$\n\\phi(a_n) = \\phi(p^2 a_{n-2}) = p(p-1)\\phi(a_{n-2}) = n(n-1)(n-2)! = n!\n$$\n\nNow, if $n$ is also composite, write $n = \\prod_{1 \\le i \\le s} q_i^{c_i}$ and $a_{n-1} = \\prod_{1 \\le i \\le s} q_i^{b_i}$, where $q_1, q_2, \\dots, q_s$ are distinct primes and $c_i, b_i$ ($1 \\le i \\le s$) are non-negative exponents. For any prime $q$ dividing $n$, $q$ divides $(n-1)! = \\phi(a_{n-1})$. By the construction of $a_n$, it is divisible by all primes not exceeding $n-1$. Thus, for any prime $q$ dividing $n$, $q$ divides $a_{n-1}$, so $b_i > 0$ for all $i$.\n\nHence,\n\n$$\n\\begin{aligned}\n\\phi(a_n) &= \\phi(na_{n-1}) = \\phi\\left(\\prod_{1 \\le i \\le s} q_i^{c_i+b_i}\\right) \\\\\n&= \\prod_{1 \\le i \\le s} (q_i^{c_i+b_i} - q_i^{c_i+b_i-1}) \\\\\n&= \\prod_{1 \\le i \\le s} q_i^{c_i} \\prod_{1 \\le i \\le s} (q_i^{b_i} - q_i^{b_i-1}) \\\\\n&= n \\cdot \\phi(a_{n-1}) = n \\cdot (n-1)! = n!\n\\end{aligned}\n$$\n\nSo in all cases, we have $\\phi(a_n) = n!$ for any $n \\ge 1$, which completes the proof. $\\square$\n\n![](images/Saudi_Arabia_booklet_2024_p33_data_9286e3e46e.png)",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 13521,
    "subject": "Mathematics (Olympiad)",
    "question": "Prove that there exists a finite sequence of distinct positive integers $n_1, n_2, \\dots, n_k$ such that\n$$\n2012 < \\sum_{i=1}^{k} \\frac{1}{n_i} < 2012 + \\left(\\frac{1}{2012}\\right)^{2012}\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "We use the fact that the harmonic series $\\sum_{i=1}^{\\infty} \\frac{1}{i}$ diverges. This means that for any real $M$, we can find an $n$ such that $\\sum_{i=1}^{n} \\frac{1}{i} > M$. But $\\sum_{i=1}^{k} \\frac{1}{i}$ is finite for any fixed $k$, so for any real $M$ and integer $k$ we can find an $n$ such that $\\sum_{i=k+1}^{n} \\frac{1}{i} > M$.\n\nTake the smallest value of $m$ such that $\\sum_{i=2012^{2012}}^{m} \\frac{1}{i} > 2012$. We claim that $n_1 = 2012^{2012}$, $n_2 = 2012^{2012} + 1, \\dots, n_k = m$ is a sequence that works. We have by hypothesis that $2012 < \\sum_{i=1}^{k} \\frac{1}{n_i}$. Suppose that $\\sum_{i=1}^{k} \\frac{1}{n_i} \\ge 2012 + \\left(\\frac{1}{2012}\\right)^{2012}$. Because $m > 2012^{2012}$, we have $\\sum_{i=1}^{k-1} \\frac{1}{n_i} > 2012$. But this means $\\sum_{i=2012^{2012}}^{m-1} \\frac{1}{i} > 2012$, contradicting the minimality of $m$. So in fact we must have $\\sum_{i=1}^{k} \\frac{1}{n_i} < 2012 + \\left(\\frac{1}{2012}\\right)^{2012}$, as desired.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 13522,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $a$, $b$, $c$ be positive integers such that $a + b + c = 2015$. Consider the following transformation: given a triple $(x, y, z)$, replace it with $(y + z - x, z + x - y, x + y - z)$. After each step, the sum $x + y + z$ remains unchanged, and the transformation is symmetric in all three variables. How many steps are required, at minimum, to obtain a triple with at least one negative number?",
    "options": [],
    "answer": "See solution",
    "solution": "Since $a + b + c = 2015$, after the first step we get the triple $(b + c - a, a + c - b, a + b - c) = (2015 - 2a, 2015 - 2b, 2015 - 2c)$. Thus, in each step the transformation acts component-wise by the rule $x \\mapsto 2015 - 2x$.\n\nWe can check by induction that if we apply this transformation $n$ times to the number $x$, we get\n\n$$\n(-2)^n x + \\frac{1 - (-2)^n}{3} \\cdot 2015.\n$$\n\nLet's determine when this number is non-negative. For even $n$ we must have\n\n$$\n2^n x + \\frac{1 - 2^n}{3} \\cdot 2015 \\ge 0.\n$$\n\nFor odd $n$ we must have\n\n$$\n-2^n x + \\frac{1 + 2^n}{3} \\cdot 2015 \\ge 0.\n$$\n\nSince 2015 is not a sum of three consecutive positive integers, we must have $a \\leq 670$ and $c \\geq 673$. Inserting this in the above inequalities we get\n\n$$\n2^n \\leq \\frac{1}{\\frac{2015}{3} - a} \\cdot \\frac{2015}{3} \\leq \\frac{1}{\\frac{2015}{3} - 670} \\cdot \\frac{2015}{3} = 403 \\Rightarrow n \\leq 8,\n$$\n\n$$\n2^n \\leq \\frac{1}{c - \\frac{2015}{3}} \\cdot \\frac{2015}{3} \\leq \\frac{1}{673 - \\frac{2015}{3}} \\cdot \\frac{2015}{3} = \\frac{2015}{4} \\Rightarrow n \\leq 8.\n$$\n\nHence, for $n \\geq 9$ one of the numbers is negative, and for the initial triple $(670, 672, 673)$ all the numbers are still positive after 8 steps.\n\nSo we have to make at least 9 steps to get a triple with at least one negative number.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 13523,
    "subject": "Mathematics (Olympiad)",
    "question": "The positive integers $x, y$ satisfy the conditions:\n\n$$\n\\{\\sqrt{x^2 + 2y}\\} > \\frac{2}{3}, \\quad \\{\\sqrt{y^2 + 2x}\\} > \\frac{2}{3}.\n$$\n\nProve that $x = y$.\n\nHere, $\\{a\\} \\in [0, 1)$ denotes the fractional part of the number $a$, that is, there exists an integer $n$ for which the equality $a = n + \\{a\\}$ holds. For example, $\\{3.14\\} = 0.14$.",
    "options": [],
    "answer": "See solution",
    "solution": "Suppose that for some positive integers $x < y$ these inequalities are true: $\\{\\sqrt{x^2 + 2y}\\} > \\frac{2}{3}$, $\\{\\sqrt{y^2 + 2x}\\} > \\frac{2}{3}$. Note that $y^2 < y^2 + 2x < (y + 1)^2$, so we have\n\n$$\ny^2 + 2x > \\left(y + \\frac{2}{3}\\right)^2 \\Leftrightarrow 2x > \\frac{4}{3}y + \\frac{4}{9} \\Rightarrow x > \\frac{2}{3}y.\n$$\n\nThen $(x + 1)^2 < x^2 + 2y < x^2 + 3x < (x + 2)^2$.\n\nTherefore, we have\n\n$$\nx^2 + 2y > \\left(x + 1 + \\frac{2}{3}\\right)^2 \\Leftrightarrow 2y > \\frac{10}{3}x + \\frac{25}{9} \\Rightarrow y > \\frac{5}{3}x.\n$$\n\nBut then, $xy > xy \\cdot \\frac{2}{3} \\cdot \\frac{5}{3}$, a contradiction that completes the proof.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 13524,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $S$ be a finite set of at least two points in the plane, with no three points of $S$ collinear. A *windmill* is a process that starts with a line $l$ passing through a single point $P \\in S$. The line rotates clockwise about $P$ until it first meets another point of $S$. The process continues indefinitely. Show that we can choose a point $P$ in $S$ and a line $l$ through $P$ such that the resulting windmill uses each point of $S$ as a pivot infinitely many times.\n\n![](path/to/file.png)",
    "options": [],
    "answer": "See solution",
    "solution": "Consider the direction of each line $l$ as it rotates continuously. First, suppose $|S| = 2k + 1$ is odd.\n\nIn the windmill process, a line $\\ell$ is in a *good position* if, when $\\ell$ just leaves a point of $S$, there are $k$ points of $S$ on each side of $\\ell$.\n\nTwo good positions are considered equal if $\\ell$ passes through the same two points with the same direction; otherwise, they are different. The number of good positions is finite. Fix a positive direction (e.g., the $x$-axis), and list all good positions in order of increasing clockwise angle in $[0, 2\\pi)$ as $\\ell_1, \\ell_2, \\dots, \\ell_m$.\n\n**Step 1:** For any good position, there is at most one such position in a given direction. If two good positions $\\ell_i$ and $\\ell_j$ had the same direction but were distinct, they would be parallel and not coincide, so the number of points on one side would differ, which is impossible.\n\n**Step 2:** For any $P \\in S$, there exists some $\\ell_i$ passing through $P$. Take any line $\\ell$ through $P$ (not through another point of $S$). If there are $s$ points to the right of $\\ell$, then $2k-s$ are to the left. As $\\ell$ rotates $180^\\circ$ about $P$, the difference $2k-2s$ changes sign, so at some moment, the numbers on both sides are equal. The last passing point is $Q$, so there is a good position $\\ell_i$ through $P$ and $Q$.\n\n**Step 3:** A windmill starting from a good position meets the problem's requirement. Start with $\\ell_1$; the next meeting point is $\\ell_2$, and so on, cycling through all good positions infinitely. By step 2, each point of $S$ is used as a pivot infinitely many times.\n\nSuppose $\\ell = \\ell_1$ passes through $P$ and $Q$ (with $P$ as pivot), dividing $S$ equally when $\\ell_1$ just leaves $Q$. If $\\ell$ next meets $R$ and pivots at $R$, it still divides $S$ equally when leaving $P$, so $\\ell$ is a good position at $R$ (call it $\\ell'$). There is no good position between $\\ell$ and $\\ell'$, so $\\ell' = \\ell_2$. If there were, a parallel line through $P$ would divide $S$ unequally, contradicting the definition of a good position.\n\nFor $|S| = 2k$ (even), the argument is similar. A line $\\ell$ is in a *good* position if there are $k$ points of $S$ on the right side when $\\ell$ just leaves a point. The first and second steps follow similarly. In the third step, starting from a good position, the next meeting point is also a good position, and there is no good position between $\\ell_1$ and $\\ell_2$.\n\nThus, for any finite set $S$ with no three collinear points, there exists a choice of $P$ and $l$ such that the windmill uses each point as a pivot infinitely many times.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 13525,
    "subject": "Mathematics (Olympiad)",
    "question": "Given a triangle $ABC$ that is neither equilateral nor right-angled, let $A'$, $B'$, and $C'$ denote the feet of the perpendiculars dropped from $A$, $B$, and $C$ onto the lines $BC$, $CA$, and $AB$, respectively. Show that the Euler lines of the triangles $AB'C'$, $BC'A'$, and $CA'B'$ are concurrent at a point situated on the nine-point circle of triangle $ABC$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $H$ be the orthocentre of triangle $ABC$. The nine-point circle of $ABC$ circumscribes triangle $A'B'C'$ and bisects the segments $AH$, $BH$, and $CH$ at points $A''$, $B''$, and $C''$, respectively, so triangles $A''B''C''$ and $ABC$ are homothetic. The triangles $AB'C'$, $A'BC'$, and $A'B'C$ are pairwise similar and have $A''$, $B''$, and $C''$ as their circumcentres, respectively.\n\nLet $e(A'')$, $e(B'')$, and $e(C'')$ denote the Euler lines of triangles $AB'C'$, $A'BC'$, and $A'B'C'$, respectively. If the Euler line of $ABC$ is parallel to a side of the latter, then $e(A'')$, $e(B'')$, and $e(C'')$ concur at one of the vertices of triangle $A''B''C''$. Otherwise, suppose $e(A'')$ and $e(B'')$ meet at a point $P$. Due to the similarity of triangles $AB'C'$ and $A'BC'$, the angle that $e(A'')$ makes with $AB'$ (or $A''C''$) is the same as the angle $e(B'')$ makes with $A'B$ (or $B''C''$). Thus, in triangle $A''PB''$, the interior angle at $P$ equals either angle $A''C''B''$ or its supplement. Hence, $P$ lies on the circle $A''B''C''$, the nine-point circle of $ABC$. Similarly, the other two pairs of Euler lines meet at points on the nine-point circle of $ABC$. If the three Euler lines were not concurrent, then one of them would meet the nine-point circle at three distinct points, which is impossible.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 13526,
    "subject": "Mathematics (Olympiad)",
    "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. What is the probability that both red balls are drawn for the first time exactly on the fourth draw?",
    "options": [],
    "answer": "See solution",
    "solution": "The following three cases can satisfy the condition.\n\nSo the probability\n\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": 13527,
    "subject": "Mathematics (Olympiad)",
    "question": "Может ли сумма цифр числа $2^m$ быть равна 8, если последняя цифра этого числа равна 6?",
    "options": [],
    "answer": "See solution",
    "solution": "Второе решение. Предположим противное, и пусть $2^m$ оканчивается на 6 и имеет сумму цифр, равную 8. Заметим, что $2^1$ оканчивается на 2, $2^2$ оканчивается на 4, $2^3$ оканчивается на 8, $2^4$ оканчивается на 6, $2^5$ оканчивается на 2. Далее последняя цифра степени двойки повторяется с периодом 4, поскольку последняя цифра числа $2^m$ определяется однозначно последней цифрой числа $2^{m-1}$. Таким образом, $2^m$ оканчивается на 6 тогда и только тогда, когда $m$ делится на 4.\n\nСумма цифр числа имеет тот же остаток при делении на 3, что и само число, поэтому $2^m$ должно иметь остаток 2 при делении на 3. Заметим, что $2^1$ имеет остаток 2 при делении на 3, $2^2$ имеет остаток 1 при делении на 3, $2^3$ имеет остаток 2 при делении на 3, и далее остатки степени двойки повторяются с периодом 2. Таким образом, $2^m$ имеет остаток 2 при делении на 3 тогда и только тогда, когда $m$ нечетно. Это противоречит тому, что $m$ должно делиться на 4.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 13528,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $m, n$ be integers with $4 < m < n$, and let $A_1, A_2, \\dots, A_{2n+1}$ be the vertices of a regular $(2n + 1)$-gon. Define $P = \\{A_1, A_2, \\dots, A_{2n+1}\\}$. Find the number of convex $m$-gons with exactly two acute internal angles whose vertices are all in $P$.",
    "options": [],
    "answer": "See solution",
    "solution": "Notice that if a convex $m$-gon with vertices in $P$ has exactly two acute angles, they must be at consecutive vertices; 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 the four vertices that make up the two acute angles is fixed; this reduces the total number of regular $m$-gons by a factor of $2n + 1$, which we will later multiply back.\n\nSuppose the larger arc between the first and last of these four vertices contains $k$ points, and the other arc contains $2n - 1 - k$ points. For each $k$, the vertices of the $m$-gon on the smaller arc may be chosen 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 for a given $k$ is $(k-n-1)^2 \\times \\binom{2n-1-k}{m-4}$. Summing over all $k$ and changing variables, the total number of polygons (divided by $2n+1$) is\n\n$$\n\\sum_{k \\ge 0} k^2 \\times \\binom{n-k-2}{m-4}.\n$$\n\nThis can be shown to equal $\\binom{n}{m-1} + \\binom{n+1}{m-1}$ by double induction on $n > m$ and $m > 4$. The base cases $n = m + 1$ and $m = 5$ are straightforward. The induction step is:\n\n$$\n\\begin{align*}\n& \\sum_{k \\ge 0} k^2 \\times \\binom{n-k-2}{m-4} \\\\\n&= \\sum_{k \\ge 0} k^2 \\times \\binom{(n-1)-k-2}{m-4} + \\sum_{k \\ge 0} k^2 \\times \\binom{(n-1)-k-2}{m-6} \\\\\n&= \\binom{n-1}{m-1} + \\binom{n}{m-1} + \\binom{n-1}{m-2} + \\binom{n}{m-2} \\\\\n&= \\binom{n}{m-1} + \\binom{n+1}{m-1}.\n\\end{align*}\n$$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 13529,
    "subject": "Mathematics (Olympiad)",
    "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\n1. 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\n2. 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": "See 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. By assumptions:\n\n$$\n\\begin{aligned}\nR_1 + R_2 &= |K_1K_2| = \\frac{1}{2}|C_1C_2|, \\\\\nR_2 + R_3 &= |K_2K_3| = \\frac{1}{2}|C_2C_3|, \\\\\nR_3 + R_1 &= |K_3K_1| = \\frac{1}{2}|C_3C_1|,\n\\end{aligned}\n$$\n\nSumming gives $2R_1 + 2R_2 + 2R_3 = \\frac{1}{2}(|C_1C_2| + |C_2C_3| + |C_3C_1|)$. The assumptions also imply:\n\n$$\n\\begin{aligned}\nr_1 + R_3 &\\le \\frac{1}{2}|C_1C_2|, & R_3 + r_2 &\\le \\frac{1}{2}|C_1C_2|, \\\\\nr_2 + R_1 &\\le \\frac{1}{2}|C_2C_3|, & R_1 + r_3 &\\le \\frac{1}{2}|C_2C_3|, \\\\\nr_3 + R_2 &\\le \\frac{1}{2}|C_3C_1|, & R_2 + r_1 &\\le \\frac{1}{2}|C_3C_1|,\n\\end{aligned}\n$$\n\nSumming these gives $2r_1 + 2r_2 + 2r_3 + 2R_1 + 2R_2 + 2R_3 \\le |C_1C_2| + |C_2C_3| + |C_3C_1|$.\n\n**(i)** Thus, $2r_1 + 2r_2 + 2r_3 \\le \\frac{1}{2}(|C_1C_2| + |C_2C_3| + |C_3C_1|)$, so $r_1 + r_2 + r_3 \\le \\frac{1}{4}(|C_1C_2| + |C_2C_3| + |C_3C_1|)$ as required.\n\n**(ii)** If $r_1 + r_2 + r_3 = \\frac{1}{4}(|C_1C_2| + |C_2C_3| + |C_3C_1|)$, all inequalities above are equalities. The equalities $r_1 + R_3 = \\frac{1}{2}|C_1C_2| = r_2 + R_3$ imply $r_1 = r_2$, and similarly $r_1 = r_3$. Let $r = r_1 = r_2 = r_3$.\n\nThen:\n$$\nr + R_3 = \\frac{1}{2}|C_1C_2| = R_1 + R_2,\n$$\n$$\nr + R_2 = \\frac{1}{2}|C_1C_3| = R_1 + R_3,\n$$\nSumming side-by-side gives $2r + R_2 + R_3 = 2R_1 + R_2 + R_3$, so $R_1 = r$. Similarly, $R_2 = R_3 = r$. Thus, all sides of the triangle $C_1C_2C_3$ have length $4r$, so the triangle is equilateral.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 13530,
    "subject": "Mathematics (Olympiad)",
    "question": "Given real numbers $a, b, c, d$ such that $a + b + c + d = a^2 + b^2 + c^2 + d^2$, prove that\n$$\n(2 - a)(2 - b)(2 - c)(2 - d) \\geq 1.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "Introduce new variables $x = a - \\frac{1}{2}$, $y = b - \\frac{1}{2}$, $z = c - \\frac{1}{2}$, and $t = d - \\frac{1}{2}$. Then our problem can be rewritten as:\n\n$$\n1 = (a^2 - a + \\frac{1}{4}) + (b^2 - b + \\frac{1}{4}) + (c^2 - c + \\frac{1}{4}) + (d^2 - d + \\frac{1}{4}) = x^2 + y^2 + z^2 + t^2.\n$$\n\nAs a result, the absolute values of $x, y, z, t$ do not exceed $1$, and\n$$\n|x| + |y| + |z| + |t| \\geq x^2 + y^2 + z^2 + t^2 = 1.\n$$\n\nThe desired inequality can be rewritten as:\n$$\n(4 - 2a)(4 - 2b)(4 - 2c)(4 - 2d) = (3 - 2x)(3 - 2y)(3 - 2z)(3 - 2t) \\geq 16.\n$$\n\nNote that if one of the variables, say $x$, is negative, we can flip the sign by substituting $x' = -x$. From the condition $3 - 2x > 3 - 2x'$, it suffices to prove the inequality only for positive $x, y, z, t$.\n\n*Lemma.* For non-negative $x, y$ with $x + y \\leq 1$, we have $(3 - 2x)(3 - 2y) \\geq (3 - 2p)^2$, where $p = \\sqrt{\\frac{x^2 + y^2}{2}}$.\n\n*Proof.* Clearly, $p = \\sqrt{\\frac{x^2 + y^2}{2}} \\leq \\sqrt{\\frac{1}{2}}$. Define $s = \\frac{x + y}{2} \\leq p = \\sqrt{\\frac{x^2 + y^2}{2}}$. Moreover, from the initial condition $s = \\frac{x + y}{2} \\leq \\frac{1}{2}$.\n\nHence, $2xy = (x + y)^2 - (x^2 + y^2) = 4s^2 - 2p^2$. Rearranging the inequality:\n\n$$\n\\begin{align*}\n(3 - 2x)(3 - 2y) &\\geq (3 - 2p)^2 \\\\\n9 - 6x - 6y + 4xy &\\geq 9 - 12p + 4p^2 \\\\\n-6x - 6y + 4xy &\\geq -12p + 4p^2 \\\\\n-12s + 8s^2 - 4p^2 &\\geq -12p + 4p^2 \\\\\n12(p - s) &\\geq 8(p^2 - s^2) \\\\\n3(p - s) &\\geq 2(p - s)(p + s) \\\\\n2p + 2s &\\leq 3 \\\\\n2p + 2s &\\leq \\sqrt{2} + 1 < 3,\n\\end{align*}\n$$\n\nas desired.\n\nApply the lemma repeatedly:\n$$\n(3 - 2x)(3 - 2y)(3 - 2z)(3 - 2t) \\geq \\left(3 - 2\\sqrt{\\frac{x^2 + y^2}{2}}\\right)^2 \\cdot \\left(3 - 2\\sqrt{\\frac{z^2 + t^2}{2}}\\right)^2 \\geq \\left(3 - 2\\sqrt{\\frac{x^2 + y^2 + z^2 + t^2}{4}}\\right)^4 = 16,\n$$\nas desired.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 13531,
    "subject": "Mathematics (Olympiad)",
    "question": "Is it possible to write $32$ as a product of seven positive integers, not necessarily distinct?",
    "options": [],
    "answer": "See solution",
    "solution": "Yes.\n\nAn example is $1 \\cdot 1 \\cdot 1 \\cdot 1 \\cdot 1 \\cdot 2 \\cdot 16 = 32$ (there are other examples).",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 13532,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $n$ be a positive integer, and let $f(n)$ denote the number of $n$-digit integers $\\overline{a_1a_2\\cdots a_n}$ (called wave numbers) that satisfy the following conditions:\n\n1. $a_i \\in \\{1, 2, 3, 4\\}$, and $a_i \\neq a_{i+1}$ for $i = 1, 2, \\dots$;\n2. When $n \\geq 3$, the numbers $a_i - a_{i+1}$ and $a_{i+1} - a_{i+2}$ have opposite signs for $i = 1, 2, \\dots$.\n\nFind:\n\n1. The value of $f(10)$.\n2. The remainder of $f(2008)$ divided by $13$.",
    "options": [],
    "answer": "See solution",
    "solution": "1. When $n \\geq 2$, classify $\\overline{a_1a_2\\cdots a_n}$ as class A if $a_1 < a_2$, and class B if $a_1 > a_2$. By symmetry, both classes have the same count, denoted $g(n)$, so $f(n) = 2g(n)$.\n\nLet $m_k(i)$ be the number of $k$-digit class A wave numbers ending with digit $i$ ($i = 1,2,3,4$). Then:\n\n$$\ng(n) = \\sum_{i=1}^{4} m_n(i)\n$$\n\nThe recurrence relations are:\n\n- For even $k$:\n  - $m_{k+1}(4) = 0$\n  - $m_{k+1}(3) = m_k(4)$\n  - $m_{k+1}(2) = m_k(4) + m_k(3)$\n  - $m_{k+1}(1) = m_k(4) + m_k(3) + m_k(2)$\n\n- For odd $k$:\n  - $m_{k+1}(1) = 0$\n  - $m_{k+1}(2) = m_k(1)$\n  - $m_{k+1}(3) = m_k(1) + m_k(2)$\n  - $m_{k+1}(4) = m_k(1) + m_k(2) + m_k(3)$\n\nInitial values:\n\n- $m_2(1) = 0$\n- $m_2(2) = 1$\n- $m_2(3) = 2$\n- $m_2(4) = 3$\n- $g(2) = 6$\n\nCompute next terms:\n\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$$\nm_3(3) = m_2(4) = 3\n$$\n$$\nm_3(4) = 0\n$$\n$$\ng(3) = 14\n$$\n\nSimilarly,\n\n$$\nm_4(1) = 0\n$$\n$$\nm_4(2) = m_3(1) = 6\n$$\n$$\nm_4(3) = m_3(1) + m_3(2) = 11\n$$\n$$\nm_4(4) = m_3(1) + m_3(2) + m_3(3) = 14\n$$\n$$\ng(4) = 31\n$$\n\nContinuing, $g(5) = 70$, $g(6) = 157$, $g(7) = 353$, $g(8) = 793$.\n\nFor $n \\geq 5$:\n\n$$\ng(n) = 2g(n-1) + g(n-2) - g(n-3)\n$$\n\nThis recurrence can be used to compute $g(10)$ and $g(2008)$, and thus $f(10)$ and $f(2008)$:\n\n- $f(10) = 2g(10)$\n- $f(2008) = 2g(2008)$\n- The remainder of $f(2008)$ divided by $13$ is $2g(2008) \\bmod 13$.\n\n(Explicit values can be computed recursively.)",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 13533,
    "subject": "Mathematics (Olympiad)",
    "question": "Subdivide the square into three rectangles: one measuring $1\\ \\text{m} \\times \\frac{1}{8}\\ \\text{m}$ and each of the other two measuring $\\frac{1}{2}\\ \\text{m} \\times \\frac{7}{8}\\ \\text{m}$.\n\n![](images/Brown_Australian_MO_Scene_2013_p40_data_5bb9e8fc53.png)",
    "options": [],
    "answer": "See solution",
    "solution": "Note that $\\sqrt{1 + (1/8)^2} = \\sqrt{65/64}$ and $\\sqrt{(1/2)^2 + (7/8)^2} = \\sqrt{65/64}$. So each of the three rectangles has diagonals of length $\\sqrt{65/64}$ and can therefore be covered by a disc with this diameter.\n\nNow $\\sqrt{65/64} < 1.008 \\Leftrightarrow 65 < 64(1.008)^2$ and $64(1 + 0.008)^2 > 64 \\times 1.016 = 64(1 + 0.01 + 0.006) = 64 + 0.64 + 0.384 = 65.024 > 65$.\n\nSo $\\sqrt{65/64} < 1.008$. Therefore, it is possible for three discs each with diameter $1008\\ \\text{mm}$ to cover the square.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 13534,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $ABC$ be a right triangle with right angle at $A$, and let $AD$ be its altitude from $A$ to $BC$. On the ray $[AD$, take points $E$ and $H$, such that $AE = AC$ and $AH = AB$. Construct squares $AEFG$ and $AHJI$, such that $C$ lies inside $AEFG$, and $B$ lies inside $AHJI$. Let $K = AC \\cap EG$, $L = AB \\cap IH$, $N = IL \\cap GK$, $M = IB \\cap GC$. Prove that:\n\na) $LK \\parallel BC$\n\nb) the points $A, N, M$ are collinear.",
    "options": [],
    "answer": "See solution",
    "solution": "a) We have $\\triangle AGK \\sim \\triangle AHL$ (1), because $\\angle AGK = \\angle AHL = 45^\\circ$, and $\\angle GAK = 90^\\circ - \\angle CAD = \\angle HAL$. Thus, $\\frac{AK}{AL} = \\frac{AG}{AH}$. Since $AG = AE = AC$ and $AH = AB$, it follows that $\\frac{AK}{AC} = \\frac{AL}{AB}$, therefore $LK \\parallel BC$.\n\nb) From (1), we have $\\angle AKG = \\angle ALH$, so $ALNK$ is cyclic. Therefore, $\\angle NAK = \\angle NLK = \\angle NIG = 45^\\circ$, which shows that $AN$ is the bisector of the angle $BAC$. The bisector from $B$ in the triangle $ABC$ is parallel to the bisector of the angle $BAI$, which is also an altitude in the triangle $BAI$. Similarly, $CM$ is the external bisector from $C$ in the triangle $ABC$. Thus, $M$ is the center of the excircle relative to $A$ of the triangle $ABC$, so $M \\in AN$.\n\nAlternative solution for b). Let $O = KL \\cap AN$. Since $A, G, I$ are collinear and $DA \\perp GI, DA \\perp BC$, we get $GI \\parallel BC \\parallel KL$. Hence, $\\frac{OK}{OL} = \\frac{AG}{AI} = \\frac{AG}{AH} = \\frac{AK}{AL}$, therefore, by the converse of the angle bisector theorem, $AO$ is the bisector of the angle $LAK$. Let $P = AM \\cap BC$. Then $\\frac{CP}{PB} = \\frac{AG}{AI} = \\frac{AC}{AB}$, so $AP$ is the bisector of the angle $BAC$, hence both $M$ and $N$ lie on the angle bisector of the angle $BAC$, proving the collinearity of $A, M, N$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 13535,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $a$, $b$, and $c$ be real numbers. Define $\\sigma_1 = a + b + c$, $\\sigma_2 = ab + bc + ca$, and $\\sigma_3 = abc$. Prove that\n\n$$\n\\sigma_1^5 \\geq 81(\\sigma_1^2 - 2\\sigma_2)\\sigma_3.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "Let $\\sigma_1 = a + b + c$, $\\sigma_2 = ab + bc + ca$, and $\\sigma_3 = abc$. The inequality to prove is\n\n$$\n\\sigma_1^5 \\geq 81(\\sigma_1^2 - 2\\sigma_2)\\sigma_3 \\iff \\sigma_1^5 + 162\\sigma_2\\sigma_3 \\geq 81\\sigma_1^2\\sigma_3. \\quad (1)\n$$\n\nUsing the AM-GM inequality for the three terms $\\sigma_1^5$, $81\\sigma_2\\sigma_3$, $81\\sigma_2\\sigma_3$, we have\n\n$$\n\\sigma_1^5 + 81\\sigma_2\\sigma_3 + 81\\sigma_2\\sigma_3 \\geq 3\\sqrt[3]{81^2\\sigma_1^5\\sigma_2^2\\sigma_3^2}.\n$$\n\nThus, to prove (1), it suffices to show\n\n$$\n3\\sqrt[3]{81^2\\sigma_1^5\\sigma_2^2\\sigma_3^2} \\geq 81\\sigma_1^2\\sigma_3,\n$$\nwhich is equivalent to\n$$\n81^2\\sigma_1^5\\sigma_2^2\\sigma_3^2 \\geq 27^3\\sigma_1^6\\sigma_3^3 \\iff \\sigma_2^2 \\geq 3\\sigma_1\\sigma_3.\n$$\n\nThe last inequality is well known.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 13536,
    "subject": "Mathematics (Olympiad)",
    "question": "Point $M$ is placed on the side $BC$ of triangle $ABC$ so that $BM = AC$. $H$ is the foot of the perpendicular dropped on $AM$ from point $B$. We know that $BH = CM$, and $\\angle MAC = 30^\\circ$. Find the degree measure of $\\angle ACB$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $AC = BM = b$, $BH = MC = h$. We can easily show that point $H$ cannot be on the segment $AM$. Let's consider two cases.\n\n1) Point $H$ is on the ray $MA$ (see figure 1).\n\nLet's draw perpendicular $LC \\perp AM$, with point $L$ on the line $AM$. As one of the angles of the right-angled triangle $ALC$ is $30^\\circ$, $LC = \\frac{1}{2}b$. Therefore, $\\angle MCL = \\angle MBH$.\n\n![fig.1](images/Ukrajina_2008_p2_data_f01d12a5dd.png)\n\n$$\n\\cos \\angle MCL = \\frac{LC}{MC} = \\frac{b}{2h} = \\cos \\angle MBH = \\frac{BH}{BM} = \\frac{h}{b} \\implies \\frac{b}{2h} = \\frac{h}{b}\n$$\n\nSo $b^2 = 2h^2$, which means that $\\cos \\angle MCL = \\frac{b}{2h} = \\frac{1}{\\sqrt{2}}$ and $\\angle MCL = 45^\\circ$. Since $\\angle LCA = 60^\\circ$, $\\angle ACB = 15^\\circ$.\n\n2) The point $H$ is on the ray $AM$ (figure 2). Let's draw perpendicular $LC \\perp AM$, with point $L$ on the line $AM$. Further, we solve the problem as in the first case and find that $\\angle MCL = 45^\\circ$. Since $\\angle LCA = 60^\\circ$, $\\angle ACB = 15^\\circ$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 13537,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all monic cubic polynomials of the form $R(x) = x^3 + px + q$ with integer coefficients $p, q$ such that $R(x)$ is irreducible over $\\mathbb{Q}$, has three distinct real roots, and the sum $|\\alpha_1| + |\\alpha_2| + |\\alpha_3|$ (where $\\alpha_1, \\alpha_2, \\alpha_3$ are the roots) is minimized.",
    "options": [],
    "answer": "See solution",
    "solution": "*Remark.* One can verify that the value of $|\\alpha_1| + |\\alpha_2| + |\\alpha_3|$ is equal to $4 \\cos 20^\\circ$ for the found polynomials.",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 13538,
    "subject": "Mathematics (Olympiad)",
    "question": "For each positive integer $n$, let $f(n) = n - s(n)$, where $s(n)$ denotes the sum of the digits of $n$. How many different values does $f(n)$ take as $n$ ranges from $1$ up to $9999$?",
    "options": [],
    "answer": "See solution",
    "solution": "**Lemma:** The function $f$ has the following two properties:\n\n1. $f(n+1) = f(n)$ if the last digit of $n$ is not a $9$.\n2. $f(n+1) > f(n)$ if the last digit of $n$ is a $9$.\n\n**Proof:**\n\nIf the last digit of $n$ is not a $9$, then $s(n+1) = s(n) + 1$, so $f(n+1) = n + 1 - (s(n) + 1) = n - s(n) = f(n)$.\n\nIf the last digit of $n$ is a $9$, suppose the last $k$ digits of $n$ are $9$s (with $k \\ge 1$), and the $(k+1)$th digit from the right is not a $9$. When incrementing $n$ by $1$, the rightmost $k$ digits become $0$, and the $(k+1)$th digit increases by $1$. Thus, $s(n+1) = s(n) - 9k + 1$, so:\n\n$$\n\\begin{aligned}\nf(n+1) &= n + 1 - s(n+1) \\\\\n&= n + 1 - (s(n) - 9k + 1) \\\\\n&= n - s(n) + 9k \\\\\n&= f(n) + 9k \\\\\n&> f(n).\n\\end{aligned}\n$$\n\n![](images/Australian-Scene-combined-2015_p88_data_7163c19d0a.png)\n\nFrom the lemma, we see that:\n\n$$\n\\begin{aligned}\nf(1) = f(2) = \\cdots = f(9) &< f(10) = f(11) = \\cdots = f(19) \\\\\n&< f(20) = f(21) = \\cdots = f(29) \\\\\n&\\vdots \\\\\n&< f(10000) = f(10001) = \\cdots = f(10009) \\\\\n&< f(10010).\n\\end{aligned}\n$$\n\nSince $f(9) = 0$, $f(10) = 9$, $f(10000) = 9999$, and $f(10010) = 10008$, the required number of different values of $f$ in the range from $1$ up to $9999$ is the number of multiples of $10$ from $10$ up to $10000$. This is $10000 \\div 10 = 1000$.\n\n$\\boxed{1000}$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 13539,
    "subject": "Mathematics (Olympiad)",
    "question": "Consider an $n \\times n$ unit-square board. The main diagonal of the board is the $n$ unit squares along the diagonal from the top left to the bottom right. We have an unlimited supply of tiles of this form:\n\n![](images/2020_Australian_Scene_W_p153_data_a9b87bc561.png)\n\nThe tiles may be rotated. We wish to place tiles on the board such that each tile covers exactly three unit squares, the tiles do not overlap, no unit square on the main diagonal is covered, and all other unit squares are covered exactly once. For which $n \\geq 2$ is this possible?",
    "options": [],
    "answer": "See solution",
    "solution": "The board consists of $n^2$ unit squares, of which $n$ should not be covered. Each tile covers exactly three squares, so we must have $3 \\mid n(n-1)$. Hence if $n \\equiv 2 \\pmod{3}$, the board cannot be covered. From now on we will only consider $n \\equiv 0, 1 \\pmod{3}$.\n\nThe board can easily be covered for $n = 3$. For $n = 4$, the picture below shows the main diagonal in black and the bottom left corner of the board:\n\n![](images/2020_Australian_Scene_W_p153_data_eeaad912b9.png)\n\nIn order to cover the unit square in the top left corner, one of the tiles must be placed on the orange squares. However, then the other unit squares in this half of the board cannot be covered any more. So $n = 4$ is not possible.\n\nFor $n = 6$, the picture below shows the main diagonal in black and the bottom left corner of the board:\n\n![](images/2020_Australian_Scene_W_p153_data_c8a119eff5.png)\n\nAgain, two tiles must be placed exactly on the orange unit squares. For the $3 \\times 3$ board that is left, we need three tiles; however, each tile covers at most one of the four corners. So this is impossible.\n\nNow we will show that all other $n \\equiv 0, 1 \\pmod{3}$ are possible. The picture below shows a solution for $n = 7$, where the second half of the board can be filled similarly:\n\n![](images/2020_Australian_Scene_W_p154_data_c84308c43f.png)\n\nAnd the next picture shows a solution for $n = 10$, including a way of extending this solution to $n = 12$.\n\n![](images/2020_Australian_Scene_W_p154_data_c625a16e58.png)\n\nIn general, for $n = 3k + 1$ we can extend the solution to $n = 3k + 3$ by adding two rows on the bottom and putting a tile on the far right of those two rows. Then we have a rectangle of size $2 \\times 3k$ left, which we can cover by creating a $2 \\times 3$ rectangle of two tiles, and putting $k$ of those next to each other. So if $n = 3k + 1$ is possible, then $n = 3k + 3$ is possible.\n\nAlso, we can extend the solution for $n = 3k + 1$ to a solution for $n = 3k + 7$ by adding six rows to the bottom. Then on the far right of these rows, we can put the construction for $n = 7$ (see picture above). Then we have a rectangle of size $6 \\times 3k$ left, which we can cover by using $3 \\times k$ of the $2 \\times 3$ rectangles consisting of two tiles.\n\nSo starting from $n = 7$ and $n = 10$ we can find constructions for all $n \\equiv 1 \\pmod{3}$ with $n \\geq 7$, and from those we can find constructions for all $n \\equiv 0 \\pmod{3}$ for $n \\geq 9$. We conclude that the $n$ which are possible are $n = 3$, $n \\equiv 0 \\pmod{3}$ with $n \\geq 9$, and $n \\equiv 1 \\pmod{3}$ with $n \\geq 7$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 13540,
    "subject": "Mathematics (Olympiad)",
    "question": "Consider a sequence defined by $a_0 = 1$ and, for $n \\geq 1$, \n$$\na_n = \\begin{cases}\na_{n-1} + d & \\text{if } a_{n-1} \\text{ is odd} \\\\\na_{n-1} / 2 & \\text{if } a_{n-1} \\text{ is even}\n\\end{cases}\n$$\nwhere $d$ is a given positive integer. Prove that for every odd $d$, the sequence is periodic and returns to $1$ after a finite number of steps.",
    "options": [],
    "answer": "See solution",
    "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 bounded, so it is periodic.\n\nLet $r$ be the smallest index such that $a_r = a_s$ for some $s \\neq r$. Assume $r > 0$.\n\nIf $a_r \\leq d$, 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 $a_{r-1} = a_{s-1}$, which contradicts the minimality of $r$.\n\nIf $a_r > d$, from $a_n \\leq 2d$ 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": 13541,
    "subject": "Mathematics (Olympiad)",
    "question": "We say a sequence $a_1, a_2, a_3, \\dots$ of positive integers is *alagoana* if, for every positive integer $n$, the following two conditions hold simultaneously:\n\n- $a_{n!} = a_1 \\cdot a_2 \\cdot \\dots \\cdot a_n$.\n- $a_n$ is the $n$th power of a positive integer.\n\nDetermine all the sequences that are *alagoanas*.\n\n(Note that $n! = 1 \\cdot 2 \\cdot 3 \\cdot \\dots \\cdot n$. For example, $4! = 1 \\cdot 2 \\cdot 3 \\cdot 4 = 24$. Thus, our sequence satisfies, for example, $a_{24} = a_{4!} = a_1 \\cdot a_2 \\cdot a_3 \\cdot a_4$.)",
    "options": [],
    "answer": "See solution",
    "solution": "We will prove that the only *alagoana* sequence is $a_n = 1$ for every positive integer $n$.\n\nTo do this, we will show that if the sequence is *alagoana*, then $a_n$ cannot have any prime factor.\n\nConsider a prime $p$. For every positive integer $n$, let $\\alpha(n)$ be the exponent of $p$ in the factorization of $a_n$. We will prove that $\\alpha(n) = 0$ for every $n$.\n\nBy the second condition, $\\alpha(n)$ is a multiple of $n$; that is, there exists a non-negative integer $k_n$ such that $\\alpha(n) = n \\cdot k_n$. From the first condition, we have $a_{n!} = a_1 \\cdot a_2 \\cdots a_n$. This implies that $\\alpha(n!) = \\alpha(1) + \\alpha(2) + \\cdots + \\alpha(n)$.\n\nBut since $a_{n!}$ is also a $(n!)$th power, $\\alpha(n!)$ is a multiple of $n!$. However, the sum $\\alpha(1) + \\alpha(2) + \\cdots + \\alpha(n)$ is a sum of multiples of $1, 2, \\dots, n$, respectively. The only way for this to always be a multiple of $n!$ for all $n$ is if all $k_n = 0$, so $\\alpha(n) = 0$ for all $n$.\n\nTherefore, $a_n = 1$ for all $n$ is the only possible *alagoana* sequence.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 13542,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $d_n$ be the minimal positive integer such that $n^{d_n} \\equiv 1 \\pmod{q}$. Prove that for any positive integer $m$ such that $n^m \\equiv 1 \\pmod{q}$, $d_n$ divides $m$.\n\nFurthermore, let $p$ be a prime. Show that there exists a prime $q$ dividing $\\frac{p^p-1}{p-1} = 1 + p + \\cdots + p^{p-1}$ such that $p^2$ does not divide $q-1$, and for every integer $n$, the number $n^p - p$ is not divisible by $q$.",
    "options": [],
    "answer": "See solution",
    "solution": "By the minimality of $d_n$, write $m = d_n k + r$ with $k \\geq 1$ and $0 \\leq r < d_n$. Then\n\n$$\n1 \\equiv n^m \\equiv n^{d_n k + r} \\equiv n^{d_n k} \\cdot n^r = n^r \\pmod{q}.\n$$\n\nBy minimality, $r = 0$, so $d_n$ divides $m$.\n\nBy Fermat's Little Theorem, $n^{q-1} \\equiv 1 \\pmod{q}$, so $d_n \\mid q-1$. For $n^p \\equiv p \\pmod{q}$, $n^{pd_p} \\equiv p^{d_p} \\equiv 1 \\pmod{q}$, so $d_n$ divides both $q-1$ and $pd_p$, implying $d_n \\mid \\gcd(q-1, pd_p)$.\n\nPick a prime $q$ dividing $1 + p + \\cdots + p^{p-1}$, with $p^2 \\nmid q-1$. Such $q$ exists since $1 + p + \\cdots + p^{p-1} \\not\\equiv 1 \\pmod{p^2}$, so some prime divisor $q$ is not $1 \\pmod{p^2}$.\n\nBy (a), $p^p \\equiv 1 \\pmod{q}$, so $d_p \\mid p$, so $d_p = p$ or $1$.\n\nIf $d_p = 1$, then $p \\equiv 1 \\pmod{q}$.\n\nIf $d_p = p$, then $d_n \\mid \\gcd(p^2, q-1)$. By (b), possible $d_n$ are $1$ and $p$, so $n^p \\equiv 1 \\pmod{q}$. Thus $p \\equiv 1 \\pmod{q}$.\n\nIn any case, $p \\equiv 1 \\pmod{q}$. But then $0 \\equiv 1 + p + \\cdots + p^{p-1} \\equiv p \\pmod{q}$, so $p = q$, a contradiction. Therefore, there is a $q$ such that for every integer $n$, $n^p - p$ is not divisible by $q$.\n\n*Note.* The proof can be shortened by starting directly with the definition of $q$ as above, but the first part provides motivation.\n\nA more advanced approach uses prime density theorems: $q$ satisfies the condition if and only if $q$ remains prime in $k = \\mathbb{Q}(\\sqrt[p]{p})$. By Chebotarev's density theorem, the set of such $q$ has density $\\frac{1}{p}$, so there are infinitely many such $q$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 13543,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $ABC$ be an acute, non-isosceles triangle. The angle bisector of $\\angle A$ meets the circumcircle of triangle $ABC$ again at $D$. Let $O$ be the circumcenter of triangle $ABC$. The angle bisectors of $\\angle AOB$ and $\\angle AOC$ meet the circle $\\gamma$ with diameter $AD$ at $P$ and $Q$, respectively. The line $PQ$ meets the perpendicular bisector of $AD$ at $R$. Prove that $AR \\parallel BC$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $E$ be the midpoint of $AD$. Since $OA = OD$ and $O \\neq E$, the perpendicular bisector of $AD$ is $EO$.\n\nLet $F$ be the second intersection of $OD$ with $\\gamma$. Then $DF \\perp AF$ and $DO \\perp BC$, hence $BC \\parallel AF$. Therefore, to prove the required conclusion, it suffices to prove that $R$ is on the line $AF$.\n\n![](images/RMC_2019_var_3_p76_data_ce25ca950d.png)\n\nThe points $A$, $E$, $O$, and $F$ lie on the circle $\\alpha$ with diameter $OA$. Denote $\\beta$ as the circumcircle of triangle $POQ$. Then $PQ$ is the radical axis of $\\beta$ and $\\gamma$, and $AF$ is the radical axis of $\\alpha$ and $\\gamma$. So, what remains to be proven is that $EO$ is the radical axis of $\\alpha$ and $\\beta$, that is, $E$ lies on $\\beta$.\n\nSince $OP$ is the perpendicular bisector of $AB$, the circle $\\alpha$ also contains the midpoint of $AB$; the same holds for the midpoint of $AC$. Therefore, $\\angle QOR = \\angle CAE$ and $\\angle POE = \\angle BAE$, so $\\angle QOR = \\angle POE$, which shows that $OE$ is perpendicular to the bisector of $\\angle POQ$. Notice also that $EP = EQ$. Now, denote $E'$ as the second intersection of the bisector of $\\angle POQ$ with $\\beta$ and $E''$ as the antipode of $E'$ on $\\beta$. Then $E''$ is the common point of the perpendicular bisector of $PQ$ and the perpendicular from $O$ to the bisector of $\\angle POQ$, hence $E'' = E$, as required.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 13544,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $x$ and $y$ be real numbers satisfying $(x + 1)(y + 2) = 8$.\n\nShow that\n\n$$\n(xy - 10)^2 \\geq 64.\n$$\n\nFurthermore, determine all pairs $(x, y)$ of real numbers for which equality holds.",
    "options": [],
    "answer": "See solution",
    "solution": "The inequality $(2x - y)^2 \\geq 0$ (with equality if and only if $y = 2x$) is equivalent to\n$$\n(2x + y)^2 \\geq 8xy.\n$$\nThe constraint $(x + 1)(y + 2) = 8$ gives $2x + y = 6 - xy$. Substituting this into the inequality above yields\n$$\n(6 - xy)^2 \\geq 8xy,\n$$\nwhich is equivalent to\n$$\n(xy - 10)^2 \\geq 64.\n$$\nAs we noted already, equality holds for $y = 2x$. In this case, the constraint becomes $(x + 1)(2x + 2) = 8$, which yields $x = 1$ or $x = -3$, and finally the two pairs $(x, y) = (1, 2)$ and $(x, y) = (-3, -6)$. We easily verify that equality actually holds in both cases.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 13545,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $\\triangle ABC$ be an acute scalene triangle. Its $C$-excircle tangent to the segment $AB$ meets $AB$ at point $M$ and the extension of $BC$ beyond $B$ at point $N$. Analogously, its $B$-excircle tangent to the segment $AC$ meets $AC$ at point $P$ and the extension of $BC$ beyond $C$ at point $Q$. Denote by $A_1$ the intersection point of the lines $MN$ and $PQ$, and let $A_2$ be defined as the point, symmetric to $A$ with respect to $A_1$. Define the points $B_2$ and $C_2$ analogously. Prove that $\\triangle ABC$ is similar to $\\triangle A_2B_2C_2$.",
    "options": [],
    "answer": "See solution",
    "solution": "We shall use the standard notations for $ABC$, i.e. $\\angle ABC = \\beta$, $BC = a$, etc. We also write $s = \\frac{a+b+c}{2}$ for the semiperimeter and $r$ for the inradius.\n\nLet $MN$ intersect the altitude $AD$ ($D$ lies on $BC$) at the point $L$. We have that $\\angle BAD = 90^\\circ - \\beta$ and $\\angle AML = \\angle BMN = \\frac{\\beta}{2}$. (Since $BMN$ is an isosceles triangle with $\\angle MBN = 180^\\circ - \\beta$.) It is known that $AM = s - b$, so by the Sine Law in the triangle $AML$ we have\n\n$$\n\\frac{AM}{\\sin \\angle ALM} = \\frac{AL}{\\sin \\angle AML} \\implies \\frac{s-b}{\\sin(90^\\circ + \\frac{\\beta}{2})} = \\frac{AL}{\\sin \\frac{\\beta}{2}} \\implies AL = (s-b) \\tan \\frac{\\beta}{2} = r.\n$$\n\nAnalogously, we see that if $PQ$ intersects $AD$ at $L'$, then $AL' = r$. Therefore $L$ and $L'$ coincide, and since $A_1 = MN \\cap PQ$ by definition, we conclude that $L = L' = A_1$. In particular, we can now view the point $A_2$ as the point on the $A$-altitude such that $AA_2 = 2r$. Analogously, $B_2$ and $C_2$ lie on the $B$-altitude and $C$-altitude, respectively, and $BB_2 = CC_2 = 2r$.\n\n![](images/Bmo_Shortlist_2021_p43_data_62820fdbe4.png)\n\nNow let $X$ be the reflection of $A$ on the midpoint of $BC$ and define $Y, Z$ analogously. So $XYZ$ is the triangle whose midpoints of sides are $A, B$, and $C$. Let $J$ be the incenter of this triangle. As the triangles $XYZ$ and $ABC$ are similar with ratio $2$, the inradius of $XYZ$ is equal to $2r$. So if $JJ_0$ is perpendicular to $YZ$ (with $J_0$ on $YZ$), then $AA_2$ and $JJ_0$ are parallel (both perpendicular to $YZ$) and equal, hence $AA_2JJ_0$ is a rectangle and in particular $A_2$ is the foot of the perpendicular from $J$ to the $A$-altitude of $ABC$. It follows that $A_2, B_2$, and $C_2$ lie on the circle $\\omega$ with diameter $JH$.\n\nNow we finish with a simple angle chasing. The circle $k$ gives $\\angle A_2B_2C_2 = \\angle A_2HC_2 = 180^\\circ - \\angle AHC = \\angle ABC$; similarly for the angles at $A_2$ and $C_2$. The desired similarity follows.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 13546,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $ABCD$ be a cyclic quadrilateral, and let diagonals $AC$ and $DC$ intersect at $X$. Let $C_1$, $D_1$, and $M$ be the midpoints of segments $CX$, $DX$, and $CD$, respectively. Lines $AD_1$ and $BC_1$ intersect at $Y$, and line $MY$ intersects diagonals $AC$ and $BD$ at different points $E$ and $F$, respectively. Prove that line $XY$ is tangent to the circle through $E$, $F$, and $X$.\n\n![](images/MACEDONIAN_MATHEMATICAL_OLYMPIADS_2016_p14_data_382382bc18.png)",
    "options": [],
    "answer": "See solution",
    "solution": "We are to prove that $\\angle EXY = \\angle EFX$; alternatively, but equivalently,\n$$\n\\angle AYX + \\angle XAY = \\angle BYF + \\angle XBY.\n$$\nSince the quadrilateral $ABCD$ is cyclic, the triangles $XAD$ and $XBC$ are similar, and since $AD_1$ and $BC_1$ are corresponding medians in these triangles, it follows that $\\angle XAY = \\angle XAD_1 = \\angle XBC_1 = \\angle XBY$.\n\nFinally, $\\angle AYX = \\angle BYF$, since $X$ and $M$ are corresponding points in the similar triangles $ABY$ and $C_1D_1Y$: indeed, $\\angle XAB = \\angle XDC = \\angle MC_1D_1$, and $\\angle XBA = \\angle XCD = \\angle MD_1C_1$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 13547,
    "subject": "Mathematics (Olympiad)",
    "question": "All of the rectangles in the figure below, which is drawn to scale, are similar to the enclosing rectangle. Each number represents the area of its rectangle. What is length $AB$?\n\n![](images/2024_AMC10A_Solutions_p8_data_de937f56ef.png)\n\n(A) $4 + 4\\sqrt{5}$  \n(B) $10\\sqrt{2}$  \n(C) $5 + 5\\sqrt{5}$  \n(D) $10^{\\frac{4}{\\sqrt{8}}}$  \n(E) $20$",
    "options": [],
    "answer": "See solution",
    "solution": "Let $a$ represent the length of the shorter side of the rectangle with area $1$, and let $b$ represent the length of its longer side. Then $ab = 1$, and the ratio of its long side to its short side is $\\frac{b}{a}$, as is true for all of the rectangles. By similarity, the dimensions of the rectangle with area $9$ are $3a$ and $3b$, and the dimensions of the rectangle with area $8$ are $a\\sqrt{8}$ and $b\\sqrt{8}$. Because the short sides of the $9$ and $1$ rectangles add to the long side of the $8$ rectangle, $a + 3a = b\\sqrt{8}$, and $\\frac{b}{a} = \\frac{4}{\\sqrt{8}} = \\sqrt{2}$. The area of the enclosing rectangle is the sum of the numbered areas, which is $200$. Therefore $AB \\cdot \\frac{AB}{\\sqrt{2}} = 200$, so\n\n$$\nAB = \\sqrt{200\\sqrt{2}} = \\sqrt{100\\sqrt{8}} = 10^{\\frac{4}{\\sqrt{8}}}.\n$$\n\n*Note:* This dissection was discovered by recreational mathematician Ed Pegg, Jr. The rectangles are similar to standard A4 paper.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 13548,
    "subject": "Mathematics (Olympiad)",
    "question": "Suppose $a$, $b$, and $c$ are three complex numbers with product $1$. Assume that none of $a$, $b$, or $c$ are real or have absolute value $1$. Define\n\n$$\np = (a + b + c) + \\left(\\frac{1}{a} + \\frac{1}{b} + \\frac{1}{c}\\right) \\quad \\text{and} \\quad q = \\frac{a}{b} + \\frac{b}{c} + \\frac{c}{a}.\n$$\n\nGiven that both $p$ and $q$ are real numbers, find all possible values of the ordered pair $(p, q)$.",
    "options": [],
    "answer": "See solution",
    "solution": "**Setup for proof**\n\nLet us denote $a = \\frac{y}{x}$, $b = \\frac{z}{y}$, $c = \\frac{x}{z}$, where $x, y, z$ are nonzero complex numbers. Then\n\n$$\n\\begin{aligned}\np+3 &= 3 + \\sum_{\\text{cyc}} \\left(\\frac{x}{y} + \\frac{y}{x}\\right) = 3 + \\frac{x^2(y+z) + y^2(z+x) + z^2(x+y)}{xyz} \\\\\n    &= \\frac{(x+y+z)(xy+yz+zx)}{xyz}.\n\\end{aligned}\n$$\n\n$$\n\\begin{aligned}\nq-3 &= -3 + \\sum_{\\text{cyc}} \\frac{y^2}{zx} = \\frac{x^3 + y^3 + z^3 - 3xyz}{xyz} \\\\\n    &= \\frac{(x+y+z)(x^2+y^2+z^2-xy-yz-zx)}{xyz}.\n\\end{aligned}\n$$\n\nIt follows that\n\n$$\n\\begin{aligned}\n& 3(p+3) + (q-3) \\\\\n&= \\frac{(x+y+z)(x^2+y^2+z^2+2(xy+yz+zx))}{xyz} \\\\\n&= \\frac{(x+y+z)^3}{xyz}.\n\\end{aligned}\n$$\n\nNow, note that if $x + y + z = 0$, then $p = -3$, $q = 3$ so we are done.\n\n**Main proof**\n\nWe will prove that if $x+y+z \\neq 0$ then we contradict either the hypothesis that $a, b, c \\notin \\mathbb{R}$ or that $a, b, c$ do not have absolute value $1$.\n\nScale $x, y, z$ in such a way that $x + y + z$ is nonzero and real; hence so is $xyz$. Thus, as $p + 3 \\in \\mathbb{R}$, we conclude $xy + yz + zx \\in \\mathbb{R}$ as well. Hence, $x, y, z$ are the roots of a cubic with real coefficients. Thus,\n\n- either all three of $\\{x, y, z\\}$ are real (which implies $a, b, c \\in \\mathbb{R}$),\n- or two of $\\{x, y, z\\}$ are a complex conjugate pair (which implies one of $a, b, c$ has absolute value $1$).\n\nBoth of these were forbidden by hypothesis.\n\n**Construction**\n\nAs we saw in the setup, $(p, q) = (-3, 3)$ will occur as long as $x+y+z = 0$, and no two of $x, y, z$ share the same magnitude or are collinear with the origin. This is easy to do; for example, we could choose $(x, y, z) = (3, 4i, -(3+4i))$. Hence $a = \\frac{3}{4i}$, $b = -\\frac{4i}{3+4i}$, $c = -\\frac{3+4i}{3}$ satisfies the hypotheses of the problem statement.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 13549,
    "subject": "Mathematics (Olympiad)",
    "question": "Tenemos un cubo de lado $3$ formado por $27$ piezas cúbicas de lado $1$. Dentro de cada pieza hay una bombilla que puede estar encendida o apagada. Cada vez que se pulsa una pieza (no es posible pulsar la del centro del cubo), cambia el estado de su bombilla y el de las que comparten una cara con la pulsada. Inicialmente, todas las bombillas están apagadas. Responde razonadamente a las siguientes preguntas:\n\n1. ¿Se puede conseguir que todas las bombillas queden encendidas?\n\n2. ¿Se puede conseguir que queden encendidas todas salvo la del centro del cubo?\n\n3. ¿Se puede conseguir que solo quede encendida la del centro del cubo?",
    "options": [],
    "answer": "See solution",
    "solution": "Llamamos $N$ a la pieza de lado $1$ que está en el interior, $C$ a cada pieza que es centro de una cara, $V$ a cada pieza que es vértice del cubo y $A$ a cada una de las que comparten cara con un vértice. Es evidente que el resultado obtenido tras pulsar varias piezas es independiente del orden en que las pulsemos.\n\n1. El estado de la bombilla de $N$ cambia si y solo si pulsamos una $C$. Para que quede encendida hemos de pulsar un número impar de veces en piezas $C$ (ya no pulsamos más $C$'s). Quedarán un número impar de piezas $C$ apagadas. La única manera de cambiar su estado es pulsar $A$'s, pero cada una cambia el estado de dos $C$'s, luego es imposible que todas las $C$'s y $N$ queden encendidas.\n\n2. Sí es posible. Pulsamos todas las $A$ y quedan encendidas todas ellas y todas las $V$. Están apagadas las $C$ y la $N$. Ahora pulsamos todas las $C$: las $V$ no se ven afectadas; cada $A$ sufre dos cambios, luego sigue encendida, y $N$ queda apagada porque cambia su estado un número par de veces.\n\n3. No es posible. Si lo fuera, se podría pasar del caso 2 (todas menos $N$ encendidas) al caso 3. Y eso supondría disponer de una sucesión de pulsaciones que permite cambiar el estado de todas las bombillas, lo cual es inviable por lo visto en 1.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 13550,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $a$, $b$, $c$ be real numbers in the interval $[0, 1]$ with $a + b \\ge 1$, $b + c \\ge 1$, and $c + a \\ge 1$. Prove that\n$$\n1 \\le (1-a)^2 + (1-b)^2 + (1-c)^2 + \\frac{2\\sqrt{2abc}}{\\sqrt{a^2 + b^2 + c^2}}.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "We may assume without loss of generality that $a \\ge b \\ge c$. Noting that $(1-b)^2 + (1-c)^2 = (b+c-1)^2 - 2bc + 1$, we may transform the right-hand side of the desired inequality into\n$$\n(1-a)^2 + (1-b)^2 + (1-c)^2 + \\frac{2\\sqrt{2abc}}{\\sqrt{a^2 + b^2 + c^2}} = \\frac{2\\sqrt{2abc}}{\\sqrt{a^2 + b^2 + c^2}} - 2bc + (1-a)^2 + (b+c-1)^2 + 1.\n$$\nRearranging, it suffices to show that\n$$\n\\frac{2bc(\\sqrt{a^2 + b^2 + c^2} - \\sqrt{2a})}{\\sqrt{a^2 + b^2 + c^2}} \\le (1-a)^2 + (b+c-1)^2,\n$$\nwhich after rationalizing the numerator is equivalent to\n$$\n\\frac{2bc(b^2 + c^2 - a^2)}{\\sqrt{a^2 + b^2 + c^2} (\\sqrt{2}a + \\sqrt{a^2 + b^2 + c^2})} \\le (1-a)^2 + (b+c-1)^2.\n$$\nIf $b^2 + c^2 - a^2 \\le 0$, then this is clearly true. We may therefore assume that $b^2 + c^2 - a^2 > 0$. To simplify, we want to produce the term $b^2 + c^2 - a^2$ from the right-hand side. By the RMS-AM inequality,\n$$\n(1-a)^2 + (b+c-1)^2 \\ge \\frac{[(1-a) + (b+c-1)]^2}{2} = \\frac{(b+c-a)^2}{2}.\n$$\nBy our assumption that $a \\ge b \\ge c$, we obtain\n$$\nb^2 + c^2 - a^2 - (b + c - a)^2 = 2(ab + ac - bc - a^2) = -2(a - b)(a - c) \\le 0,\n$$\nhence\n$$\n\\frac{(b+c-a)^2}{2} \\ge \\frac{b^2+c^2-a^2}{2} \\ge 0.\n$$\nTogether, this shows that\n$$\n(1-a)^2 + (b+c-1)^2 \\ge \\frac{b^2+c^2-a^2}{2}.\n$$\nOn the other hand, the ordering $a \\ge b \\ge c \\ge 0$ implies $a^2 \\ge bc$, so $a \\ge \\sqrt{bc}$. Hence by the AM-GM inequality,\n$$\n\\sqrt{a^2 + b^2 + c^2} (\\sqrt{2a} + \\sqrt{a^2 + b^2 + c^2}) \\ge \\sqrt{bc + 2bc} (\\sqrt{2bc} + \\sqrt{bc + 2bc}) = (3 + \\sqrt{6})bc > 4bc.\n$$\nDividing both sides of this by the left-hand side and combining it with the previous result, we find that\n$$\n(1-a)^2 + (b+c-1)^2 \\ge \\frac{b^2+c^2-a^2}{2} \\cdot \\frac{4bc}{\\sqrt{a^2 + b^2 + c^2} (\\sqrt{2a} + \\sqrt{a^2 + b^2 + c^2})},\n$$\nwhich is the desired inequality.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 13551,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $(x_n)$, $n \\geq 1$, be a sequence of real numbers with $x_1 = 1$, such that\n$$2x_{n+1} = 3x_n + \\sqrt{5x_n^2 - 4},$$\nfor $n = 1, 2, 3, \\ldots$\n\n(a) Prove that all terms of the sequence are natural numbers.\n\n(b) Examine if there exists a term of the sequence divisible by $2011$.",
    "options": [],
    "answer": "See solution",
    "solution": "(a) From the given recurrence relation:\n$$\n(2x_{n+1} - 3x_n)^2 = 5x_n^2 - 4 \\implies 4x_{n+1}^2 - 12x_{n+1}x_n + 4x_n^2 = -4 \\\\ \n\\implies x_{n+1}^2 - 3x_{n+1}x_n + x_n^2 = -1 \\quad (1)\n$$\nThis can also be written as:\n$$\nx_{n+2}^2 - 3x_{n+2}x_{n+1} + x_{n+1}^2 = -1 \\quad (2)\n$$\nConsider the quadratic equation $x^2 - 3x x_{n+1} + x_{n+1}^2 + 1 = 0$. From (1) and (2), two solutions are $x_n$ and $x_{n+2}$. By Vieta's formulas:\n$$\nx_n + x_{n+2} = 3x_{n+1} \\quad (3) \\qquad x_n x_{n+2} = x_{n+1}^2 + 1 \\quad (4)\n$$\nFrom (3), $x_{n+2} = 3x_{n+1} - x_n$. With $x_1 = 1$ and $x_2 = 2$, by induction all terms of the sequence are integers.\n\n(b) Suppose there exists $x_s$ such that $2011 \\mid x_s$. From (4) for $n = s$:\n$$\nx_s x_{s+2} = x_{s+1}^2 + 1\n$$\nSince $2011 \\mid x_s$, $2011 \\mid x_{s+1}^2 + 1$, so $x_{s+1}^2 \\equiv -1 \\pmod{2011}$. Thus,\n$$\n(x_{s+1}^2)^{1005} \\equiv (-1)^{1005} \\equiv -1 \\pmod{2011} \\\\ x_{s+1}^{2010} \\equiv -1 \\pmod{2011} \\quad (5)\n$$\nSince $2011$ is prime and $\\gcd(x_{s+1}, 2011) = 1$ (otherwise $d \\mid 1$, impossible), by Fermat's theorem:\n$$\nx_{s+1}^{2010} \\equiv 1 \\pmod{2011} \\quad (6)\n$$\nThis contradicts (5). Therefore, no term of the sequence is divisible by $2011$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 13552,
    "subject": "Mathematics (Olympiad)",
    "question": "A sequence $x_n$ is defined as follows: $x_1 = 1$ and\n$$\nx_{n+1} = x_n + 3\\sqrt{x_n} + \\frac{n}{\\sqrt{x_n}}\n$$\nfor all positive integers $n$.\n\n**a)** Prove that $\\lim_{n \\to +\\infty} \\frac{n}{x_n} = 0$.\n\n**b)** Find the limit $\\lim_{n \\to +\\infty} \\frac{n^2}{x_n}$.",
    "options": [],
    "answer": "See solution",
    "solution": "**a)** We will prove $x_n \\geq n^2$ for all positive integers by induction. The statement is obvious for $n=1$. Assume $x_n \\geq n^2$, then\n\n$$\nx_{n+1} = x_n + 3\\sqrt{x_n} + \\frac{n}{\\sqrt{x_n}} \\geq n^2 + 3n + 1 = (n+1)^2.\n$$\n\nTherefore, $x_n \\geq n^2$ for all $n$. Thus,\n\n$$\n0 < \\frac{n}{x_n} \\leq \\frac{1}{n}.\n$$\n\nBy the Squeeze theorem, $\\lim_{n \\to +\\infty} \\frac{n}{x_n} = 0$.\n\n**b)** Let $x_n = y_n^2$. The recurrence becomes\n\n$$\ny_{n+1}^2 = y_n^2 + 3y_n + \\frac{n}{y_n}\n$$\n\nSo,\n$$\n(y_{n+1} - y_n)(y_{n+1} + y_n) = 3y_n + \\frac{n}{y_n}\n$$\n\nFor large $n$, $y_{n+1} \\approx y_n$, so $y_{n+1} + y_n \\approx 2y_n$:\n\n$$\ny_{n+1} - y_n \\approx \\frac{3y_n + \\frac{n}{y_n}}{2y_n} = \\frac{3}{2} + \\frac{n}{2y_n^2}\n$$\n\nFrom part (a), $y_n^2 = x_n \\gg n$, so $\\frac{n}{y_n^2} \\to 0$ as $n \\to \\infty$.\nThus,\n\n$$\n\\lim_{n \\to \\infty} (y_{n+1} - y_n) = \\frac{3}{2}.\n$$\n\nBy the Cesàro mean theorem,\n\n$$\n\\lim_{n \\to \\infty} \\frac{y_n}{n} = \\frac{3}{2} \\implies \\lim_{n \\to \\infty} \\frac{n^2}{x_n} = \\lim_{n \\to \\infty} \\frac{n^2}{y_n^2} = \\left(\\frac{2}{3}\\right)^2 = \\frac{4}{9}.\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 13553,
    "subject": "Mathematics (Olympiad)",
    "question": "What is the maximum number of prime numbers that can appear in a non-constant geometric sequence $a_n = aq^{n-1}$, where $q \\neq 1$?",
    "options": [],
    "answer": "See solution",
    "solution": "The answer is $2$. For example, the sequence $a_n = 2\\left(\\frac{3}{2}\\right)^{n-1}$ contains two primes: $2$ and $3$.\n\nAssume, for contradiction, that a geometric sequence $a_n = aq^{n-1}$ with $q \\neq 1$ contains three primes, say $a_k$, $a_m$, and $a_n$ with $k < m < n$. Since the sequence is not constant, these primes are distinct.\n\nWe have:\n\n$$\n\\frac{a_n}{a_m} = q^{n-m}, \\quad \\frac{a_m}{a_k} = q^{m-k}\n$$\n\nThus,\n\n$$\n\\left(\\frac{a_n}{a_m}\\right)^{m-k} = \\left(\\frac{a_m}{a_k}\\right)^{n-m}\n$$\n\nwhich gives\n\n$$\na_n^{m-k} a_k^{n-m} = a_m^{n-k}\n$$\n\nBut $m-k, n-m, n-k > 0$ and $a_k, a_m, a_n$ are distinct primes, so this is not possible. Therefore, at most two primes can appear in such a sequence.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 13554,
    "subject": "Mathematics (Olympiad)",
    "question": "On the sides $AB$, $BC$, and $AC$ of triangle $ABC$ with $\\angle BAC = 120^\\circ$, there are points $M$, $K$, and $N$ respectively such that $\\triangle MKN$ is equilateral, and $AM = 2{,}017$, $AN = 2{,}018$. Baron Munchausen claims that $\\triangle MKN$ has the smallest perimeter of all equilateral triangles with exactly one vertex on each side of $\\triangle ABC$. Is the baron correct?",
    "options": [],
    "answer": "See solution",
    "solution": "All angles of $\\triangle MKN$ are $60^\\circ$. Since $\\angle BAC + \\angle MKN = 180^\\circ$, quadrilateral $AMKN$ is cyclic.\n\n$$\n\\angle KAC = \\angle KMN = 60^\\circ = \\angle MNK = \\angle BAK.\n$$\n\nLet $M_1$ and $N_1$ be the projections of point $K$ onto sides $AB$ and $AC$, respectively. Then $\\triangle AKM_1 = \\triangle AKN_1$, so $AM_1 = AN_1$. Also, $\\angle M_1KN_1 = 60^\\circ$ and $KM_1 = KN_1$, so $\\triangle KM_1N_1$ is equilateral. The case $N = N_1$, $M = M_1$ is impossible, since otherwise $AM = AN$. Thus, $KM_1 < KM$, so the perimeter of $\\triangle KM_1N_1$ is strictly less than that of $\\triangle KMN$. Therefore, Baron Munchausen is not correct.",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 13555,
    "subject": "Mathematics (Olympiad)",
    "question": "A convex quadrilateral $ABCD$ is given. Let $E$ be the intersection of $AB$ and $CD$, $F$ be the intersection of $AD$ and $BC$, and $G$ be the intersection of $AC$ and $EF$. Prove that the following two statements are equivalent:\n\n1. $BD$ and $EF$ are parallel.\n2. $G$ is the midpoint of the segment $\\overline{EF}$.",
    "options": [],
    "answer": "See solution",
    "solution": "Draw a line $l$ through $E$ which is parallel to $BC$. Let $H$ be the intersection of $l$ and $AG$. Now $G$ is the intersection of the diagonals in the trapezoid $EHFC$.\n\n**(i) $\\Rightarrow$ (ii):**\nLet the lines $BD$ and $EF$ be parallel. Then, from Thales' theorem for parallel segments, we have the equalities:\n\n$$\n\\frac{\\overline{AC}}{AH} = \\frac{\\overline{AB}}{AE} \\text{ and } \\frac{\\overline{AB}}{AE} = \\frac{\\overline{AD}}{AF}.\n$$\n\nIt follows that $\\frac{\\overline{AC}}{AH} = \\frac{\\overline{AD}}{AF}$, and therefore, from the same Thales' theorem, we conclude that the lines $HF$ and $ED$ are parallel. Therefore, $EHFC$ is a parallelogram and its diagonals bisect each other at the intersection point $G$.\n\n**(ii) $\\Rightarrow$ (i):**\nLet $G$ be the midpoint of the segment $\\overline{EF}$. Then $\\triangle EGH \\cong \\triangle FGC$, so $EHFC$ is a parallelogram and we conclude that $HF$ and $ED$ are parallel. Therefore, the equalities\n\n$$\n\\frac{\\overline{AC}}{AH} = \\frac{\\overline{AB}}{AE} \\text{ and } \\frac{\\overline{AC}}{AH} = \\frac{\\overline{AD}}{AF}.\n$$\n\nhold.\n\nIt follows that $\\frac{\\overline{AB}}{AE} = \\frac{\\overline{AD}}{AF}$, and therefore, from the same Thales' theorem, we conclude that $BD$ and $EF$ are parallel.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 13556,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $\\Gamma$ be the circumcircle of triangle $ABC$. Let $\\Gamma_A$ be the circle which is tangent to $AB$, $AC$ and tangent to $\\Gamma$ internally, $\\Gamma_B$ be the circle which is tangent to $AB$, $BC$ and tangent to $\\Gamma$ internally, and $\\Gamma_C$ be the circle which is tangent to $AC$, $BC$ and tangent to $\\Gamma$ internally. Denote $\\Gamma_A, \\Gamma_B, \\Gamma_C$ are tangent to $\\Gamma$ at $P, Q, R$. Prove that the lines $AP, BQ, CR$ are concurrent.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $M_A, M_B, M_C$ be the midpoints of the arcs $BC$, $CA$, $AB$ (not containing $A$, $B$, $C$ respectively), and let $l_A, l_B, l_C$ be the tangents to $\\Gamma$ at $M_A, M_B, M_C$. Then $l_A \\parallel BC$, $l_B \\parallel CA$, $l_C \\parallel AB$.\n\nLet $D$ be the intersection of $l_B$ and $l_C$, $E$ the intersection of $l_C$ and $l_A$, and $F$ the intersection of $l_A$ and $l_B$.\n\nSince $\\Gamma_A$ is tangent to $\\Gamma$ at $P$, there exists a homothety $T$ with center $P$ such that $T(\\Gamma_A) = \\Gamma$. Let $T$ transform $AB$ into line $m$.\n\nThen $m \\parallel AB$ and $m$ and $P$ are on opposite sides of $AB$. Since $AB$ is tangent to $\\Gamma_A$, $m$ is tangent to $\\Gamma$, so $m = l_C$.\n\nSimilarly, $T$ transforms $AC$ into $l_B$. Since $A$ is the intersection of $AC$ and $AB$, $T(A)$ is the intersection of $l_B$ and $l_C$, so $T(A) = D$. Since $T$ is a homothety with center $P$, the points $P, A, D$ are collinear.\n\nSimilarly, the points $Q, B, E$ and $R, C, F$ are collinear. Thus, $AP$, $BQ$, and $CR$ pass through $D$, $E$, and $F$ respectively.\n\nTriangles $ABC$ and $DEF$ are similar with $AB \\parallel DE$, $BC \\parallel EF$, $CA \\parallel FD$, so there exists a homothety $S$ with $S(A) = D$, $S(B) = E$, $S(C) = F$. Let the center of $S$ be $G$. Then the lines $AP$, $BQ$, $CR$ are concurrent at $G$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 13557,
    "subject": "Mathematics (Olympiad)",
    "question": "On a circle are written several real numbers, with a positive sum. Let $S$ be the largest and $s$ the least of the sums of consecutive numbers on the circle. Prove that $S + s > 0$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $T > 0$ be the total sum of the numbers around the circle. Clearly, $S \\geq T > 0$. If $s \\geq 0$, then $S + s > 0$ immediately. If $s < 0$, consider the sum of the numbers not included in the segment corresponding to $s$; this sum is $T - s$. Since $S \\geq T - s$, it follows that $S + s \\geq T > 0$, as required.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 13558,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $\\mathbb{P}$ be the set of all prime numbers. Find all functions $f: \\mathbb{P} \\to \\mathbb{P}$ such that\n\n$$\nf(p)^{f(q)} + q^p = f(q)^{f(p)} + p^q\n$$\n\nholds for all $p, q \\in \\mathbb{P}$.",
    "options": [],
    "answer": "See solution",
    "solution": "Obviously, the identical function $f(p) = p$ for all $p \\in \\mathbb{P}$ is a solution. We will show that this is the only one.\n\nFirst, we will show that $f(2) = 2$. Taking $q = 2$ and $p$ any odd prime number, we have\n\n$$\nf(p)^{f(2)} + 2^p = f(2)^{f(p)} + p^2.\n$$\n\nAssume that $f(2) \\neq 2$. It follows that $f(2)$ is odd and so $f(p) = 2$ for any odd prime number $p$.\n\nTaking any two different odd prime numbers $p, q$, we have\n\n$$\n2^2 + q^p = 2^2 + p^q \\implies p^q = q^p \\implies p = q,\n$$\n\ncontradiction. Hence, $f(2) = 2$.\n\nSo for any odd prime number $p$ we have\n\n$$\nf(p)^2 + 2^p = 2^{f(p)} + p^2.\n$$\n\nCopy this relation as\n\n$$\n2^p - p^2 = 2^{f(p)} - f(p)^2. \\qquad (1)\n$$\n\nLet $T$ be the set of all positive integers greater than 2, i.e. $T = \\{3, 4, 5, \\dots\\}$. The function $g: T \\to \\mathbb{Z}$, $g(n) = 2^n - n^2$, is strictly increasing, i.e.\n\n$$\ng(n+1) - g(n) = 2^n - 2n - 1 > 0 \\qquad (2)\n$$\n\nfor all $n \\in T$. We show this by induction. Indeed, for $n = 3$ it is true, $2^3 - 2 \\cdot 3 - 1 > 0$. Assume that $2^k - 2k - 1 > 0$. It follows that for $n = k + 1$ we have\n\n$$\n2^{k+1} - 2(k+1) - 1 = (2^k - 2k - 1) + (2^k - 2) > 0\n$$\n\nfor any $k \\ge 3$. Therefore, (2) is true for all $n \\in T$.\n\nAs a consequence, (1) holds if and only if $f(p) = p$ for all odd prime numbers $p$, as well as for $p = 2$.\n\nTherefore, the only function that satisfies the given relation is $f(p) = p$, for all $p \\in \\mathbb{P}$.\n\n$\\Box$",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 13559,
    "subject": "Mathematics (Olympiad)",
    "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 $S \\geq N$.",
    "options": [],
    "answer": "See solution",
    "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 $S - T \\geq -3$, and consequently $S \\geq 17$ (note that $S$ must be an integer).\n\n- If in the first round, $b_1 - a_1 \\geq 4$, then Player B selects $b_1$, and in the second round selects $a_2$. This gives $c_1 - d_1 \\geq 4$ and $c_2 - d_2 \\geq 1 - 8 = -7$.\n- If in the first round, $b_1 - a_1 \\leq 3$, then Player B selects $a_1$, and in the second round selects $b_2$. This gives $c_1 - d_1 \\geq -3$ and $c_2 - d_2 \\geq 1$.\n\nTherefore, Player B can always ensure $(c_1 + c_2) - (d_1 + d_2) \\geq -3$.\n\nIn the third and fourth rounds, Player B can always ensure $(c_3 + c_4) - (d_3 + d_4) \\geq 0$. 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 $S - T \\geq -3$.\n\nPlayer A has a strategy to ensure $S \\leq 17$:\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- In 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, $S \\leq 17$.\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, $S \\leq 6 + 1 + 9 = 16$.\n\nIn conclusion, Player A has a strategy to ensure $S \\leq 17$. $\\Box$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 13560,
    "subject": "Mathematics (Olympiad)",
    "question": "Given integer sequences $\\{x_n\\}_{n \\ge 1}$ and $\\{y_n\\}_{n \\ge 1}$, prove that there exists an integer sequence $\\{z_n\\}_{n \\ge 1}$ such that for any positive integer $n$, the following holds:\n\n$$\n\\sum_{k|n} k z_k^{n/k} = \\left( \\sum_{k|n} k x_k^{n/k} \\right) \\cdot \\left( \\sum_{k|n} k y_k^{n/k} \\right).\n$$\n\nHere, all three summation symbols denote sums over all positive divisors $k$ of $n$. For example, when $n = 6$,\n\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": "See solution",
    "solution": "By induction, $z_n$ can be uniquely determined from the equation\n\n$$\nnz_n = \\left( \\sum_{k \\mid n} k x_k^{n/k} \\right) \\cdot \\left( \\sum_{k \\mid n} k y_k^{n/k} \\right) - \\sum_{\\substack{k \\mid n \\\\ k < n}} k z_k^{n/k}.\n$$\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.\n\nWorking modulo $p^r$, we first remove terms in the equation whose coefficients are already multiples of $p^r$:\n\n$$\n\\begin{aligned}\nnz_n &\\equiv \\left( \\sum_{\\substack{k \\mid n \\\\ p^r \\nmid k}} k x_k^{n/k} \\right) \\cdot \\left( \\sum_{\\substack{k \\mid n \\\\ p^r \\nmid k}} k y_k^{n/k} \\right) - \\sum_{\\substack{k \\mid n \\\\ p^r \\nmid k}} k z_k^{n/k} \\\\\n&= \\left( \\sum_{k \\mid \\frac{n}{p}} k x_k^{n/k} \\right) \\cdot \\left( \\sum_{k \\mid \\frac{n}{p}} k y_k^{n/k} \\right) - \\sum_{k \\mid \\frac{n}{p}} k z_k^{n/k} \\quad (\\text{mod } p^r).\n\\end{aligned}\n$$\n\nWe first prove a simple *lemma*: For any integer $a$ and positive integer $t$, $a^{pt} \\equiv a^{pt-1} \\pmod{p^t}$.\n\nIndeed, if $p \\mid a$, then 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 $a^{pt} \\equiv a^{pt-1} \\pmod{p^t}$.\n\nReturning to the problem, we now show that for any $k \\mid \\frac{n}{p}$ and any integer $a$, $k a^{n/k} \\equiv k a^{n/pk} \\pmod{p^r}$. 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.\n\nThus, $nz_n$ satisfies\n\n$$\nnz_n \\equiv \\left( \\sum_{k \\mid \\frac{n}{p}} k x_k^{n/pk} \\right) \\cdot \\left( \\sum_{k \\mid \\frac{n}{p}} k y_k^{n/pk} \\right) - \\sum_{k \\mid \\frac{n}{p}} k z_k^{n/pk} = 0 \\pmod{p^r}.\n$$\n\nThis shows that $nz_n$ is divisible by $p^r$. Repeating this for all prime factors of $n$ confirms that $nz_n$ is divisible by $n$. Hence, $z_n$ is an integer.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 13561,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all positive integers $k$ for which there exists a right triangle with legs of integer 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": "See solution",
    "solution": "*Answer:* $1$.\n\nLet the lengths of the legs be $a$ and $b$. By the Pythagorean theorem,\n\n$$\na^2 + b^2 = 88\\ldots822\\ldots2.\n$$\n\nIf $k=1$, then one can choose $a=9$ and $b=1$ since $9^2 + 1^2 = 82$.\n\nIf $k \\ge 2$, 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 possible residues of $a^2$ and $b^2$ modulo $8$ are $0$, $1$, or $4$. The sum of two such numbers can only be $0$, $1$, $2$, $4$, or $5$ modulo $8$. Consequently, $a^2 + b^2 = 88\\ldots822\\ldots2$ cannot hold for $k \\ge 2$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 13562,
    "subject": "Mathematics (Olympiad)",
    "question": "There are two positive integers $n$ with the following property: if you divide $n^2$ by $2n + 1$, you get a remainder of $1000$. What are these two integers?",
    "options": [],
    "answer": "See solution",
    "solution": "$666$ and $1999$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 13563,
    "subject": "Mathematics (Olympiad)",
    "question": "In how many ways can you partition the set $\\{1, 2, \\ldots, 12\\}$ into six mutually disjoint two-element sets such that the two elements in any set are coprime?",
    "options": [],
    "answer": "See solution",
    "solution": "No two even numbers can be in the same pair. Let us call partitions of $\\{1, 2, \\ldots, 12\\}$ with this property (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:\n- Odd numbers $1$, $7$, and $11$ can be paired with $2$, $4$, $6$, $8$, $10$, $12$.\n- Numbers $3$ and $9$ can be paired with $2$, $4$, $8$, $10$.\n- Number $5$ can be paired with $2$, $4$, $6$, $8$, $12$.\n\nWe cannot use the product rule directly, so we distinguish two cases:\n\n1. $5$ is paired with $6$ or $12$.\n   - The possible pairings: $2 \\times 4 \\times 3 \\times 3 \\times 2 \\times 1 = 144$.\n2. $5$ is paired with one of $2$, $4$, or $8$.\n   - The possible pairings: $3 \\times 3 \\times 2 \\times 3 \\times 2 \\times 1 = 108$.\n\nTogether, $144 + 108 = 252$ pairings.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 13564,
    "subject": "Mathematics (Olympiad)",
    "question": "ABCD is a rectangle and $P$ is a point on $BC$. If the area of triangle $ABP$ is one third of the area of the rectangle, then the ratio $BP : PC$ is\n\n![](images/SAMF_ANNUAL_REPORT_2015_BOOK_PROOF_p27_data_c2b2e44fbd.png)\n\n(A) $5 : 2$ (B) $3 : 2$ (C) $2 : 1$ (D) $3 : 1$ (E) $9 : 4$",
    "options": [],
    "answer": "See solution",
    "solution": "$$\n\\frac{\\text{area } \\triangle ABP}{\\text{area } ABCD} = \\frac{1}{3} \\implies \\frac{\\frac{1}{2} \\cdot BP \\cdot AB}{BC \\cdot AB} = \\frac{1}{3} \\implies \\frac{\\frac{1}{2} BP}{BC} = \\frac{1}{3} \\implies \\frac{BP}{BC} = \\frac{2}{3}\n$$\nSo, $BP : PC = \\frac{2}{3} : \\frac{1}{3} = 2 : 1$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 13565,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $\\{a_n\\}_{n \\ge 1}$ be a sequence such that for every prime $p$ and every positive integer $k$, the following holds:\n\n$$\na_{kp+1} = p a_k - 3 a_p + 13.\n$$\n\nFind $a_{2011}$.",
    "options": [],
    "answer": "See solution",
    "solution": "For $k = q$ (with $q$ prime),\n\n$$\na_{q p + 1} = p a_q - 3 a_p + 13 = q a_p - 3 a_q + 13.\n$$\n\nThus, $(p + 3) a_q = (q + 3) a_p$, so for all primes $p$ and $q$,\n\n$$\n\\frac{a_p}{p + 3} = \\frac{a_q}{q + 3}. \\qquad (1)\n$$\n\nSince $2011$ is prime,\n\n$$\n\\frac{a_{2011}}{2014} = \\frac{a_7}{10}. \\qquad (2)\n$$\n\nAlso, $a_7 = a_{2 \\cdot 3 + 1} = 2 a_3 - 3 a_2 + 13$. Using (1),\n\n$$\n\\frac{a_7}{10} = \\frac{a_3}{6} = \\frac{a_2}{5} = \\frac{a_7 - 2 a_3 + 3 a_2}{10 - 12 + 15} = \\frac{13}{13} = 1.\n$$\n\nTherefore, from (2), $a_{2011} = 2014$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 13566,
    "subject": "Mathematics (Olympiad)",
    "question": "Given two points $P$ and $Q$ with integer coordinates, we say that $P$ sees $Q$ if the line segment $PQ$ contains no other points with integer coordinates. An $n$-loop is a sequence of $n$ points $P_1, P_2, \\ldots, P_n$, each with integer coordinates, such that:\n\n- (a) $P_i$ sees $P_{i+1}$ for $1 \\leq i \\leq n-1$, and $P_n$ sees $P_1$;\n- (b) No $P_i$ sees any $P_j$ apart from those mentioned in (a);\n- (c) No three of the points lie on the same straight line.\n\nDoes there exist a $100$-loop?\n\nNote: Two points $(a, b)$ and $(c, d)$ see each other if and only if $a - c$ and $b - d$ are coprime.",
    "options": [],
    "answer": "See solution",
    "solution": "On the curve $y = x^2$, no three points are collinear. Two integer points $(a, a^2)$ and $(b, b^2)$ will see each other if and only if $a$ and $b$ differ by $1$, since $a - b$ divides $a^2 - b^2$.\n\nThus, the set of points $\\{P_1, P_2, \\ldots, P_{99}\\}$, where $P_j = (j, j^2)$, satisfies the required properties if we can find a suitable $P_{100}$.\n\nNote that $a$ is coprime to $b$ if and only if $a$ is coprime to $ka + b$ for any integer $k$.\n\nSince $(-100, 0)$ can see $(1, 1)$ (their $y$-coordinates differ by $1$), similarly $(-100, 101k)$ can see $(1, 1)$ for any $k$. Also, $(-100, 199k + 99^2 - 1)$ cannot see $(99, 99^2)$.\n\nNow, $(-100, 100^2)$ cannot see any of the points $P_2$ to $P_{98}$, so if $C$ is the least common multiple of the integers from $102$ to $198$, $(-100, kC + 100^2)$ will not be able to see any points $P_2$ to $P_{98}$.\n\nSince $101$ and $199$ are prime, they are coprime to any integer from $102$ to $198$, and thus also coprime to $C$. By the Chinese Remainder Theorem, there exists an integer $N$ such that:\n\n- $N \\equiv 0 \\pmod{101}$\n- $N \\equiv 99^2 - 1 \\pmod{199}$\n- $N \\equiv 100^2 \\pmod{C}$\n\nThen $(-100, N)$ is a valid choice for $P_{100}$ satisfying conditions (a) and (b).\n\nFinally, there are infinitely many choices for $N$, only finitely many of which are eliminated by condition (c), so there is at least one such choice satisfying condition (c).",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 13567,
    "subject": "Mathematics (Olympiad)",
    "question": "For every positive integer $n \\geq 3$, let $\\phi_n$ be the set of all positive integers less than $n$ and coprime to $n$. Consider the polynomial\n\n$$\nP_n(x) = \\sum_{k \\in \\phi_n} x^{k-1}.\n$$\n\na) Prove that $P_n(x)$ is divisible by $x^{r_n} + 1$ for some integer $r_n > 0$.\n\nb) Find all $n$ such that $P_n(x)$ is irreducible over $\\mathbb{Z}[x]$.",
    "options": [],
    "answer": "See solution",
    "solution": "a) First, observe that for $m, k \\in \\mathbb{Z}^+$ and $k$ odd, $x^{km} + 1$ is divisible by $x^m + 1$. For convenience, consider the polynomial $Q_n(x) = xP_n(x) = \\sum_{k \\in \\phi_n} x^k$. We need to prove $Q_n(x)$ is divisible by $x^r + 1$ for some positive integer $r$.\n\n- **If $n$ is odd:** Since $\\gcd(n, k) = \\gcd(n, n-k)$ for all $k = 1, 2, \\dots, n-1$, for each $k \\in \\phi_n$, $n-k \\in \\phi_n$. Thus, we can pair terms $(x^k, x^{n-k})$. Since $x^k + x^{n-k} = x^k(x^{n-2k} + 1)$, and $n-2k$ is odd, this sum is divisible by $x+1$. Therefore, $Q_n(x)$ is divisible by $x+1$.\n\n- **If $4 \\mid n$:** Pairing as above, $n-2k \\equiv 2 \\pmod{4}$, so $x^k + x^{n-k} = x^k(x^{n-2k} + 1)$ is divisible by $x^2 + 1$. Thus, $Q_n(x)$ is divisible by $x^2 + 1$.\n\n- **If $n \\equiv 2 \\pmod{4}$:** We use the following lemmas.\n\n**Lemma 1.** If $Q_n(x)$ is divisible by $x^r + 1$ ($r \\in \\mathbb{Z}^+$), then for any odd prime $p$ with $p \\nmid n$, $Q_{pn}(x)$ is also divisible by $x^r + 1$.\n\n*Proof.* For $a \\in \\phi_n$, $kp + a \\in \\phi_{pn}$ for $k = 0, 1, \\dots, p-1$. The sum over all such exponents is a multiple of $Q_n(x)$, so $Q_{pn}(x)$ is divisible by $x^r + 1$.\n\n**Lemma 2.** If $n = 2p_1p_2\\cdots p_m$, where $p_1, \\dots, p_m$ are distinct odd primes, then $Q_n(x)$ is reducible.\n\n*Proof.* By induction on $m$. For $m=1$, $n=2p$, $Q_n(x)$ is divisible by $x^p + 1$. Assume true for $m \\geq 1$. For $N = pn$ with $p$ an odd prime coprime to $n$, $Q_N(x)$ can be written as $R_N(x) - Q_n(x^p)$, both divisible by $x^r + 1$, so $Q_N(x)$ is divisible by $x^r + 1$.\n\nb) From above, $P_n(x)$ is irreducible only if $|\\phi_n| = 2$, i.e., $\\varphi(n) = 2$.\n\n- If $n = p$ prime, $2 = n-1 \\implies n=3$.\n- If $n = p^\\alpha$ with $\\alpha > 1$, $2 = p^{\\alpha-1}(p-1)$. Only $p=2$, $\\alpha=2$ gives $n=4$.\n- If $n$ has at least two distinct prime divisors $p, q$, $2 = \\varphi(n)$ divisible by $p-1$ and $q-1$, so $p=2$, $q=3$, $n=6$.\n\nTherefore, the answers are $n = 3, 4, 6$. $\\Box$",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 13568,
    "subject": "Mathematics (Olympiad)",
    "question": "A school has 100 students and 5 teachers. In the first period, each student is taking one class, and each teacher is teaching one class. The enrollments in the classes are 50, 20, 20, 5, and 5.\n\nLet $t$ be the average value obtained if a teacher is picked at random and the number of students in their class is noted. Let $s$ be the average value obtained if a student is picked at random and the number of students in their class, including that student, is noted.\n\nWhat is $t - s$?\n\n(A) $-18.5$ (B) $-13.5$ (C) $0$ (D) $13.5$ (E) $18.5$",
    "options": [],
    "answer": "See solution",
    "solution": "The average class size when a teacher—that is, a class—is picked at random is given by\n$$\nt = \\frac{1}{5} \\cdot 50 + \\frac{2}{5} \\cdot 20 + \\frac{2}{5} \\cdot 5 = 20.\n$$\nA randomly picked student has a $\\frac{50}{100}$ chance of being in the class with 50 students, a $\\frac{40}{100}$ chance of being in a class with 20 students, and a $\\frac{10}{100}$ chance of being in a class with 5 students, so\n$$\ns = \\frac{50}{100} \\cdot 50 + \\frac{40}{100} \\cdot 20 + \\frac{10}{100} \\cdot 5 = 33.5.\n$$\nThe requested difference is $20 - 33.5 = -13.5$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 13569,
    "subject": "Mathematics (Olympiad)",
    "question": "Prove that for any finite set $S$ of primitive points, there exists a homogeneous polynomial $p(x, y)$ of positive degree with integer coefficients such that $p(u, v) = 1$ for every $(u, v) \\in S$.",
    "options": [],
    "answer": "See solution",
    "solution": "A simplifying step is to show that we can assume one of the points in $S$ is $(1,0)$. If $(r, s) \\in S$ and $\\gcd(r, s) = 1$, there exist integers $a, b$ such that $ar + bs = 1$. Consider the transformation\n\n$$\nt: (x, y) \\mapsto (ax + by, ry - sx).\n$$\n\nLet $S_1 = \\{t(u, v) : (u, v) \\in S\\}$. Note $t(r, s) = (1, 0) \\in S_1$. The inverse is\n\n$$\nt^{-1}: (x, y) \\mapsto (rx - by, sx + ay).\n$$\n\nThus, each point of $S_1$ is primitive. If $p_1$ is a polynomial for $S_1$, then $p = p_1 \\circ t$ works for $S$.\n\nWe proceed by induction on $|S|$. For the base case $S = \\{(1,0)\\}$, $p(x, y) = x + y$ works.\n\nFor the inductive step, assume the result for $n$ points. For $S$ with $n+1$ points (with $(1,0) \\in S$), let $S' = S \\setminus \\{(1,0)\\} = \\{(u_i, v_i) : i = 1, \\dots, n\\}$. By induction, there is a homogeneous $f(x, y)$ of degree $m$ with integer coefficients such that\n\n$$\nf(u_i, v_i) = 1 \\quad \\text{for all } (u_i, v_i) \\in S'.\n$$\n\nLet\n\n$$\ng(x, y) = \\prod_{i=1}^n (v_i x - u_i y).\n$$\n\nThen $g(u_i, v_i) = 0$ for all $i$. Consider\n\n$$\np(x, y) = f(x, y)^j - c x^{mj-n} g(x, y)\n$$\n\nfor suitable integers $j, c$. For $(u_i, v_i) \\in S'$, $p(u_i, v_i) = 1$. For $(1,0)$,\n\n$$\np(1,0) = a_0^j - c v_1 v_2 \\cdots v_n.\n$$\n\nSince $\\gcd(a_0, v_1 v_2 \\cdots v_n) = 1$, we can choose $j$ and $c$ so that $p(1,0) = 1$ (using Euler's theorem if $v_1 v_2 \\cdots v_n \\neq 0$). Thus, the inductive step is complete.",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 13570,
    "subject": "Mathematics (Olympiad)",
    "question": "BE and CF are the altitudes of the acute scalene $\\triangle ABC$, $O$ is its circumcenter, and $M$ is the midpoint of the side $BC$. If the point symmetric to $M$ with respect to $O$ lies on the line $EF$, find all possible values of the ratio $\\frac{AM}{AO}$.",
    "options": [],
    "answer": "See solution",
    "solution": "**Answer:** $\\sqrt{2}$\n\nLet $H$ be the orthocenter of $\\triangle ABC$, $A'$ be the antipode of $A$ in $(ABC)$, $S$ be the point symmetric to $M$ with respect to $O$, and $T$ be the intersection of $BC$ and $EF$. It is well-known that in this construction, $H$ is the orthocenter of $\\triangle AMT$, $BHCA'$ is a parallelogram (so $M$ is the midpoint of $A'H$), and $AH = 2OM$.\n\n![](images/Ukraine2022-23_p41_data_4653e43ee7.png)\n\nAs $SM = 2OM = AH$ and $AH \\perp BC \\perp SM$, we see that $AHMS$ is a parallelogram. Hence, $S$ is the antipode of $T$ in $(AMT)$ and $\\angle AMH = 90^\\circ - \\angle MAT = 90^\\circ - \\angle MST = \\angle STM$. Since $OM \\perp BC$ and $AO \\perp EF$, we have $\\angle A'OM = \\angle STM$. Thus, $\\angle A'OM = \\angle AMH$, so $\\triangle AA'M \\sim \\triangle AMO$ and therefore $AA' \\cdot AO = AM^2$, which gives $\\frac{AM}{AO} = \\sqrt{2}$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 13571,
    "subject": "Mathematics (Olympiad)",
    "question": "An $L$-shape is one of the following four pieces, each consisting of three unit squares.\n\nA $5 \\times 5$ board, consisting of 25 unit squares, a positive integer $k \\leq 25$, and an unlimited supply of $L$-shapes are given. Two players, $A$ and $B$, play the following game: starting with $A$, they alternatively mark a previously unmarked unit square until they have marked a total of $k$ unit squares.\n\nA placement of $L$-shapes on unmarked unit squares is called *good* if the $L$-shapes do not overlap and each covers exactly three unmarked unit squares of the board. $B$ wins if every good placement of $L$-shapes leaves uncovered at least three unmarked unit squares. \n\nDetermine the minimum value of $k$ for which $B$ has a winning strategy.",
    "options": [],
    "answer": "See solution",
    "solution": "We will show that player $A$ wins if $k=1, 2,$ or $3$, but player $B$ wins if $k=4$. Thus, the smallest $k$ for which $B$ has a winning strategy is $4$.\n\n- If $k=1$, player $A$ marks the upper left corner of the board and can fill the rest with $L$-shapes, leaving only two unmarked squares uncovered.\n\n- If $k=2$, player $A$ marks the upper left corner. Whatever square player $B$ marks, player $A$ can fill the board in the same pattern as above, except omitting the $L$-shape covering $B$'s marked square. Only two unmarked squares remain uncovered, so $A$ wins.\n\n- For $k=3$, player $A$ uses a similar strategy. On the second move, $A$ marks any unmarked square of the $L$-shape covering $B$'s marked square. Again, only two unmarked squares remain uncovered, so $A$ wins.\n\n- For $k=4$, player $B$ has a winning strategy. With $21$ unmarked squares, $A$ would need to cover all with seven $L$-shapes. Assume $A$ does not mark any square in the bottom two rows (otherwise, rotate the board). On $B$'s first move, $B$ marks a key square (labeled 1 in a referenced figure). If $A$ marks square 2 next, $B$ marks square 5, leaving square 3 unmarked and unable to be covered by an $L$-shape. If $A$ marks square 3 or 4, $B$ marks the other, leaving square 2 unmarked and uncovered. In all cases, $B$ wins when $k=4$.\n\nTherefore, the minimum value of $k$ for which $B$ has a winning strategy is $4$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 13572,
    "subject": "Mathematics (Olympiad)",
    "question": "The internal angle bisectors at vertices $B$ and $C$ of triangle $ABC$ intersect the circumcircle of triangle $ABC$ at $E$ and $F$, respectively. Given that $BE = CF \\neq 0$, may we be sure that triangle $ABC$ is isosceles?",
    "options": [],
    "answer": "See solution",
    "solution": "Let $\\angle ABC = 30^\\circ$ and $\\angle BCA = 90^\\circ$. Let $O$ be the circumcenter of triangle $ABC$; it is also the midpoint of the hypotenuse $AB$. We have $\\angle OAC = \\angle BAC = 60^\\circ$, and since $AO = CO$, it follows that $\\angle OCA = 60^\\circ$. Therefore, $\\angle FCO = 60^\\circ - \\frac{90^\\circ}{2} = 15^\\circ$. Let $C'$ be the reflection of vertex $C$ over the point $O$; then $\\angle FCC' = 15^\\circ$ as well. On the\n\n![](images/EST_ABooklet_2024_p40_data_9cb39db208.png)\n\n![](images/EST_ABooklet_2024_p40_data_fc6fcc2cc1.png)\n\n![](images/EST_ABooklet_2024_p40_data_e68823e442.png)\n\nother hand, $\\angle ABE = \\frac{30}{2}^\\circ = 15^\\circ$. In conclusion, we see that the arcs *AE* and *FC'* subtend equal inscribed angles on the circumcircle of triangle *ABC*, hence the corresponding arcs are equal. Since *AB* and *CC'* are diameters, the remaining arcs *BE* and *CF* are also equal. Therefore, the corresponding chords *BE* = *CF* are equal. Thus, the equality $BE = CF$ can hold even in a non-isosceles triangle.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 13573,
    "subject": "Mathematics (Olympiad)",
    "question": "Given a right triangle $ABC$, let $M$ be the midpoint of its hypotenuse $AB$. A perpendicular bisector of $AB$ intersects side $BC$ at point $K$. A perpendicular from $K$ to $CM$ intersects the ray $AC$ at point $P$, which does not lie on the segment $AC$. Lines $CM$ and $BP$ intersect at point $T$. Prove that $AC = TB$.\n\n![](images/UkraineMO2019_booklet_p35_data_db02766971.png)",
    "options": [],
    "answer": "See solution",
    "solution": "From $KP \\perp CM$, it follows that $\\angle CPK = \\angle MBC = \\angle ABC$, so the quadrilateral $PAKB$ is cyclic. Therefore, $PK$ is a bisector in triangle $\\triangle CPT$. Clearly, $PK$ is also a height and a median of this triangle.\n\nTherefore, $\\triangle CPT$ is isosceles. In particular, this implies that $PK$ is a perpendicular bisector of $CT$.\n\nFurther, $CK = KT$, and $\\triangle PCK = \\triangle PTK$, meaning that $\\angle PTK = 90^\\circ$. Finally, since $KB = KA$, we get $\\triangle KTB = \\triangle KCA$, implying $TB = AC$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 13574,
    "subject": "Mathematics (Olympiad)",
    "question": "令 $R$ 表示實數集合。定義 $S = \\{1, -1\\}$，並定義函數 $\\operatorname{sign} : R \\to S$ 如下：\n\n$$\n\\operatorname{sign}(x) = \\begin{cases} 1 & \\text{if } x \\ge 0 \\\\ -1 & \\text{if } x < 0 \\end{cases}\n$$\n\n給定奇數 $n$，是否存在 $n^2 + n$ 個實數 $a_{ij}, b_i \\in S$（$1 \\le i, j \\le n$），使得對於任意 $n$ 個數 $x_1, \\dots, x_n \\in S$，利用下式\n\n$$\ny_i = \\operatorname{sign}\\left(\\sum_{j=1}^{n} a_{ij}x_j\\right), \\quad \\forall 1 \\le i \\le n\n$$\n\n$$\nz = \\operatorname{sign}\\left(\\sum_{i=1}^{n} b_i y_i\\right)\n$$\n\n計算出的 $z$ 恆等於 $x_1x_2\\cdots x_n$？",
    "options": [],
    "answer": "See solution",
    "solution": "觀察小情況（如 $n=3$）容易猜想\n\n$$\na_{ij} = (-1)^{i+j}, \\quad b_i = 1\n$$\n\n符合題目要求。接下來需證明其正確性。注意 $z$ 與 $x_1x_2\\cdots x_n$ 皆只可能為 $1$ 或 $-1$。$x_1x_2\\cdots x_n = 1$ 當且僅當 $x_j$ 中有偶數個 $-1$。另一方面，因 $b_i = 1$，此時\n\n$$\nz = \\operatorname{sign}\\left(\\sum_{i=1}^{n} y_i\\right)\n$$\n\n因此 $z = 1$ 當且僅當 $\\sum_{i=1}^n y_i \\ge 0$，即 $y_i$ 中 $1$ 的數量多於 $-1$。\n\n所以證明目標等價於：$y_i$ 中 $1$ 的數量多於 $-1$ 的數量，當且僅當 $x_j$ 中有偶數個 $-1$。\n\n設 $y'_i = \\sum_{j=1}^{n} (-1)^{i+j} x_j$，$y_i = \\operatorname{sign}(y'_i)$。\n\n將 $y'_i$ 與 $y'_{i+1}$ 相加得 $y'_i + y'_{i+1} = 2x_j$（$y_{n+1} = y_1$）。若 $x_j = 1$，則除 $y'_i = y'_{i+1} = 1$ 外，其餘 $y'_i, y'_{i+1}$ 為一正一負（$y'_i$ 為奇數）；若 $x_j = -1$，則除 $y'_i = y'_{i+1} = -1$ 外，其餘亦為一正一負。因此，除 $y'_i = y'_{i+1} = x_j = 1$ 或 $y'_i = y'_{i+1} = x_j = -1$ 外，$y_i$ 和 $y_{i+1}$ 恰為一個 $1$ 一個 $-1$。\n\n考慮 $x_j$ 中有偶數個 $-1$。證明 $y'_i \\neq -1$：\n\n$$\n\\begin{aligned}\ny'_i &= \\sum_{j=1}^{n} (-1)^{i+j} x_j \\\\\n&= \\sum_{x_j=1} (-1)^{i+j} + \\sum_{x_j=-1} (-1)^{i+j}(-1) \\\\\n&= \\sum_{1 \\le j \\le n} (-1)^{i+j} - 2 \\sum_{x_j=-1} (-1)^{i+j} \\\\\n&= 1 - \\sum_{x_j=-1} 2(-1)^{i+j}\n\\end{aligned}\n$$\n\n因為有偶數個 $x_j = -1$，$y'_i \\equiv 1 \\pmod{4}$，故 $y'_i \\neq -1$。\n\n再證必有某 $i$ 使 $y'_i = y'_{i+1} = x_j = 1$。若否，則 $y'_i = y'_{i+1} = x_j = -1$ 亦不可能，則所有 $i$，$y_i$ 和 $y_{i+1}$ 恰為一個 $1$ 一個 $-1$，故\n\n$$\n\\sum_{i=1}^{n} y_i = \\frac{1}{2} \\sum_{i=1}^{n} (y_i + y_{i+1}) = 0\n$$\n\n但 $n$ 為奇數，$\\sum_{i=1}^{n} y_i$ 亦為奇數，矛盾。\n\n因此，$y_1, \\dots, y_n$ 中 $1$ 的數量必多於 $-1$。\n\n若 $x_j$ 有奇數個 $-1$，同理可證 $y'_i \\neq 1$，且必有某 $i$ 使 $y'_i = y'_{i+1} = x_j = -1$，故 $y_1, \\dots, y_n$ 中 $-1$ 的數量多於 $1$。得證。",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 13575,
    "subject": "Mathematics (Olympiad)",
    "question": "Consider a triangle $ABC$ satisfying $BC < \\frac{AC + AB}{2}$. Prove that $\\angle BAC < \\frac{\\angle CBA + \\angle ACB}{2}$.\n\nLet $BC = a$, $CA = b$, $AB = c$, $\\angle BAC = A$, $\\angle CBA = B$, $\\angle ACB = C$.",
    "options": [],
    "answer": "See solution",
    "solution": "We prove the contrapositive. Suppose that $A \\geq \\frac{B + C}{2}$, so that $A \\geq 60^\\circ$ and $\\cos A \\leq \\frac{1}{2}$. Then, by the Law of Cosines:\n\n$$\na^2 = b^2 + c^2 - 2bc \\cos A \\geq b^2 + c^2 - bc = \\frac{(b + c)^2 + 3(b - c)^2}{4} \\geq \\frac{(b + c)^2}{4} = \\left(\\frac{b + c}{2}\\right)^2\n$$\n\nwhich yields $a \\geq \\frac{b + c}{2}$, as desired.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 13576,
    "subject": "Mathematics (Olympiad)",
    "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": "See 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![](images/Mathematical_Olympiad_in_China_2007-2008-Problems_and_Solutions_p143_data_b4b0bcffc6.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, $EF \\parallel BG$ and $EM \\parallel BA$, 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 $\\angle DEF = 180^\\circ - \\angle DMF = 90^\\circ$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 13577,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $\\triangle ABC$ be an acute triangle with orthocenter $H$ and circumcircle $\\Gamma$. A line through $H$ intersects segments $AB$ and $AC$ at $E$ and $F$, respectively. Let $K$ be the circumcenter of $\\triangle AEF$, and suppose line $AK$ intersects $\\Gamma$ again at a point $D$. Prove that line $HK$ and the line through $D$ perpendicular to $\\overline{BC}$ meet on $\\Gamma$.",
    "options": [],
    "answer": "See solution",
    "solution": "We begin with the following two observations.\n\n**Claim** — Point $K$ lies on the radical axis of $(BEH)$ and $(CFH)$.\n\n*Proof.* Actually, we claim $\\overline{KE}$ and $\\overline{KF}$ are tangents. Indeed,\n\n$$\n\\angle HEK = 90^\\circ - \\angle EAF = 90^\\circ - \\angle BAC = \\angle HBE\n$$\n\nimplying the result. Since $KE = KF$, this implies the result. $\\square$\n\n**Claim** — The second intersection $M$ of $(BEH)$ and $(CFH)$ lies on $\\Gamma$.\n\n*Proof.* By Miquel's theorem on $\\triangle AEF$ with $H \\in \\overline{EF}$, $B \\in \\overline{AE}$, $C \\in \\overline{AF}$. $\\square$\n\n![](images/sols-TSTST-2019_p8_data_b84be24169.png)\n\nIn particular, $M, H, K$ are collinear. Let $X$ be on $\\Gamma$ with $\\overline{DX} \\perp \\overline{BC}$; we then wish to show $X$ lies on the line $MHK$ we found. This is angle chasing: compute\n\n$$\n\\begin{aligned}\n\\angle XMB &= \\angle XDB = 90^\\circ - \\angle DBC = 90^\\circ - \\angle DAC \\\\\n&= 90^\\circ - \\angle KAF = \\angle FEA = \\angle HEB = \\angle HMB\n\\end{aligned}\n$$\n\nas needed.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 13578,
    "subject": "Mathematics (Olympiad)",
    "question": "令 $a, b, c, d$ 為非負實數且 $a + b + c + d = 4$。試證：\n\n$$\na\\sqrt{3a+b+c} + b\\sqrt{3b+c+d} + c\\sqrt{3c+d+a} + d\\sqrt{3d+a+b} \\ge 4\\sqrt{5}.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "令\n\n$$\n\\sqrt{3a+b+c} = x,\\quad \\sqrt{3b+c+d} = y,\\quad \\sqrt{3c+d+a} = z,\\quad \\sqrt{3d+a+b} = w,\n$$\n\n則有\n\n$$\nx^2 = 3a + b + c,\\quad y^2 = 3b + c + d,\\quad z^2 = 3c + d + a,\\quad w^2 = 3d + a + b.\n$$\n\n因此，\n\n$$\n\\begin{aligned}\n15a &= 5(3a + b + c) - 2(3b + c + d) - (3c + d + a) + (3d + a + b) \\\\\n    &= 5x^2 - 2y^2 - z^2 + w^2.\n\\end{aligned}\n$$\n\n所以\n\n$$\na = \\frac{5x^2 - 2y^2 - z^2 + w^2}{15},\n$$\n\n同理可得：\n\n$$\nb = \\frac{5y^2 - 2z^2 - w^2 + x^2}{15},\n$$\n$$\nc = \\frac{5z^2 - 2w^2 - x^2 + y^2}{15},\n$$\n$$\nd = \\frac{5w^2 - 2x^2 - y^2 + z^2}{15},\n$$\n\n且皆為非負數。由於\n\n$$\na + b + c + d = 4,\n$$\n\n可得\n\n$$\n\\begin{aligned}\nx^2 + y^2 + z^2 + w^2 &= (3a + b + c) + (3b + c + d) + (3c + d + a) + (3d + a + b) \\\\\n&= 20\n\\end{aligned}\n$$\n\n我們需要證明：\n\n$$\n\\sum_{cyc} x \\left( \\frac{5x^2 - 2y^2 - z^2 + w^2}{15} \\right) \\geq 4\\sqrt{5}\n$$\n\n即\n\n$$\n\\sum_{cyc} (5x^3 - 2xy^2 - xz^2 + xw^2) \\geq 60\\sqrt{5}.\n$$\n\n利用 AM-GM 不等式：\n\n$$\n\\sum_{cyc} (x^3 - 2xy^2 + xw^2) = \\sum_{cyc} (x^3 - 2xy^2 + yx^2) \\geq 0\n$$\n\n且\n\n$$\n\\sum_{cyc} (x^3 - xz^2) = \\frac{1}{3} \\sum_{cyc} (x^3 + z^3 + z^3 - 3xz^2) \\geq 0.\n$$\n\n因此只需證明：\n\n$$\n\\sum_{cyc} 3x^3 \\geq 60\\sqrt{5}\n$$\n\n再次利用 AM-GM：\n\n$$\n\\begin{aligned}\n\\sum_{cyc} (3x^3 + 3x^3 + 15\\sqrt{5}) &\\geq \\sum_{cyc} 3\\sqrt[3]{3 \\times 3 \\times 15\\sqrt{5}x^6} \\\\\n&= \\sum_{cyc} 9\\sqrt{5}x^2 = 180\\sqrt{5} \\\\\n\\Rightarrow \\sum_{cyc} 3x^3 &\\geq \\frac{180 - 60}{2}\\sqrt{5} = 60\\sqrt{5}\n\\end{aligned}\n$$\n\n證畢。\n\n等號成立於\n\n$$\nx^2 = y^2 = z^2 = w^2 = 5\n$$\n\n即\n\n$$\na = b = c = d = 1.\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 13579,
    "subject": "Mathematics (Olympiad)",
    "question": "We call a sequence $(b_n)_{n=1}^\\infty$ an F-sequence if $0 \\leq b_n \\leq 2$ and $b_{n+2} \\equiv b_{n+1} + b_n \\pmod{3}$ for all $n \\in \\mathbb{N}$. Here, $a \\equiv b$ means that $a$ and $b$ are congruent modulo 3.\n\nIf $(a_n)$ is a Fibonacci-like sequence and $b_n$ is the remainder of $a_n$ modulo 3 (i.e., $0 \\leq b_n \\leq 2$), then $(b_n)$ is an F-sequence, called the *F-sequence associated with* $(a_n)$.\n\nThe first two terms of an F-sequence determine the whole sequence. The $(a, b)$-sequence refers to the F-sequence $(b_n)$ with $b_1 = a$ and $b_2 = b$. Since $a$ and $b$ can each be 0, 1, or 2, there are nine F-sequences in total.\n\nConsider the (1,1)-sequence. Its first few terms are:\n\n1, 1, 2, 0, 2, 2, 1, 0, 1, 1, ...\n\nThe initial pair of terms repeats after eight steps, so this sequence is periodic with period 8. It has a zero in every fourth position.\n\nOther F-sequences of period 8 can be obtained by shifting the initial terms; these are the $(a, b)$-sequences for $(a, b)$ equal to one of\n\n$$(1, 2),\\ (2, 0),\\ (0, 2),\\ (2, 2),\\ (2, 1),\\ (1, 0),\\ (0, 1).$$\n\nThus, eight of the nine F-sequences have period 8, with three nonzero entries between every pair of consecutive zeros. The only other F-sequence is the (0,0)-sequence, in which every entry is zero. The zeros of an F-sequence must follow one of these two patterns.\n\nSuppose $(b_n)$ is the F-sequence associated with a Fibonacci-like sequence $(a_n)$, where $a_{29} = a_{23} = 2016$ and $a_{2016} = 2923$. Thus, $b_{29} = b_{23} = 0$ and $b_{2016} = 1$. Show that no such sequence $(a_n)$ exists.",
    "options": [],
    "answer": "See solution",
    "solution": "Since $b_{29} = b_{23} = 0$, the F-sequence $(b_n)$ has zeros at positions 23 and 29. In all nontrivial F-sequences (those not identically zero), zeros occur every fourth position, with three nonzero entries between zeros. However, the distance between positions 23 and 29 is 6, which is not a multiple of 4, so this is impossible for a periodic F-sequence. The only other possibility is the identically zero sequence, but $b_{2016} = 1$ contradicts this. Therefore, no such sequence $(a_n)$ exists.",
    "topic": "Number Theory",
    "subtopic": "Modular Arithmetic"
  },
  {
    "id": 13580,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $\\triangle ABC$ be a triangle in which $\\angle ABC = 75^\\circ$ and $\\angle BAC = 45^\\circ$. Consider the points $F$, $X$, and $Y$ such that $F$ is the projection of $B$ onto $AC$, $\\overline{CX} = \\frac{1}{2}\\overline{BC}$, and $\\overline{BY} = \\overline{FX} + \\overline{FA}$. Prove that the centroid of triangle $ABX$ lies on the line segment $YF$.",
    "options": [],
    "answer": "See solution",
    "solution": "We will prove that $YF$ is the Euler line of triangle $ABX$, hence it contains the centroid of that triangle.\n\nBecause $BFA$ is a right isosceles triangle, it follows that $BF = AF$. In the right triangle $BFC$, we have $\\angle CBF = 30^\\circ$, hence $FC = \\frac{BC}{2}$, thus $FC = CX$, and a short computation shows that $\\angle BXF = 30^\\circ$. We deduce that $BF = FX$, hence $AF = BF = FX$, making $F$ the circumcenter of $ABX$.\n\nLet $Y'$ be the orthocenter of $ABX$. Applying Sylvester's relation yields $\\overline{FY'} = \\overline{FA} + \\overline{FB} + \\overline{FX} = \\overline{FB} + \\overline{BY} = \\overline{FY}$. It follows that $Y' = Y$, hence $Y$ is the orthocenter of $ABX$. We conclude that $FY$ is the Euler line of $ABX$ and since the centroid lies between the circumcenter and the orthocenter, we obtain the desired result.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 13581,
    "subject": "Mathematics (Olympiad)",
    "question": "Consider the set of positive integers which, when written in binary, have exactly 2013 digits and more 0s than 1s. Let $n$ be the number of such integers and let $s$ be their sum. Prove that, when written in binary, $n + s$ has more 0s than 1s.",
    "options": [],
    "answer": "See solution",
    "solution": "By definition, $n$ is the number of binary representations with 2013 digits, first digit 1, and at least 1007 digits 0. Summing over all possible values for the number of digits 0 and then employing the identity $\\binom{a}{b} = \\binom{a}{a-b}$,\n\n$$\n\\begin{aligned}\nn &= \\sum_{i=1007}^{2012} \\binom{2012}{i} \\\\\n&= \\frac{1}{2} \\sum_{i=0}^{2012} \\binom{2012}{i} - \\frac{1}{2} \\binom{2012}{1006}.\n\\end{aligned}\n$$\n\nSince the sum $\\sum_{i=0}^{n} \\binom{n}{i} = 2^n$,\n\n$$\nn = 2^{2011} - \\binom{2011}{1005}.\n$$\n\nNow, $s$ can be written $\\sum_{i=0}^{2012} a_i 2^i$, where $a_i$ denotes the number of binary representations which have a 1 at digit $i$. We note that $a_{2012} = n$, and that for all other $i$,\n\n$$\n\\begin{aligned}\na_i &= \\sum_{i=1007}^{2011} \\binom{2011}{i} \\\\\n&= \\frac{1}{2} \\sum_{i=0}^{2011} \\binom{2011}{i} - \\frac{1}{2} \\left( \\binom{2011}{1005} + \\binom{2011}{1006} \\right) \\\\\n&= 2^{2010} - \\binom{2011}{1005}.\n\\end{aligned}\n$$\n\nHence, letting that constant value be $a$,\n\n$$\n\\begin{aligned}\ns &= \\sum_{i=0}^{2012} a_i 2^i \\\\\n&= 2^{2012} n + a \\sum_{i=0}^{2012} 2^i \\\\\n&= 2^{2012} n + \\left( 2^{2010} - \\binom{2011}{1005} \\right) (2^{2012} - 1).\n\\end{aligned}\n$$\n\nThus,\n\n$$\n\\begin{aligned}\nn + s &= (2^{2012} + 1) \\left( 2^{2011} - \\binom{2011}{1005} \\right) \\\\\n&\\quad + (2^{2012} - 1) \\left( 2^{2010} - \\binom{2011}{1005} \\right) \\\\\n&= 2^{4023} + 2^{4022} + 2^{2010} - 2^{2013} \\binom{2011}{1005}.\n\\end{aligned}\n$$\n\nNow that we have a formula for $n + s$, it is just left to show that we have enough digits 0. Since $n + s < 2^{4024}$, there are at most 4024 binary digits. Then\n\n$$\nn + s = 2^{2013}(2^{2010} + 2^{2009} - \\binom{2011}{1005}) + 2^{2010},\n$$\n\nmeaning that 2012 of the last 2013 digits are 0.\n\nFinally, we note that the only way in which $n + s$ will not have more 0 digits than 1 digits is if it has 4024 digits, and all of the unaccounted for digits are 1, i.e.\n\n$$\nn + s = 2^{2013}(1 + 2 + 2^2 + \\dots + 2^{2010}) + 2^{2010}.\n$$\n\nHowever, this is clearly bigger than the given value for $n + s$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 13582,
    "subject": "Mathematics (Olympiad)",
    "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": "See solution",
    "solution": "By assumption, $\\frac{|PA|}{|PB|} = \\frac{|PD|}{|PC|}$. As the bases $AB$ and $CD$ are parallel, we also have $\\frac{|PA|}{|PB|} = \\frac{|PC|}{|PD|}$. Hence, $|PC| = |PD|$. The similarity of triangles $APD$ and $BPC$ implies $\\frac{|AD|}{|BC|} = \\frac{|PD|}{|PC|} = 1$, thus $|AD| = |BC|$ as needed.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 13583,
    "subject": "Mathematics (Olympiad)",
    "question": "A circle of length $999$ is divided into unit arcs by $999$ black points. Then $d$ arcs with lengths $1, 2, \\ldots, d$ are placed on the circle, with their endpoints at black points, so that none of these arcs contains another (the arcs may overlap otherwise). Find all $d$ for which such a configuration exists.",
    "options": [],
    "answer": "See solution",
    "solution": "Consider a circle of length $n$ divided into unit arcs by $n$ black points, and $d$ arcs of lengths $1, 2, \\ldots, d$ with black endpoints such that none of them contains another. We show that the maximal admissible $d$ equals $\\lfloor \\frac{n+1}{2} \\rfloor$.\n\nLet $\\gamma_1, \\dots, \\gamma_d$ be arcs on the circle with lengths $1, 2, \\ldots, d$ such that none contains another. Consider the shortest arc $\\gamma_1 = \\overline{AB}$, starting at $A$, and the longest $\\gamma_d = \\overline{CD}$, starting at $C$.\n\nLet $X$ be the set of black points on arc $\\overline{BC}$ (excluding $C$), $Y$ the set on $\\overline{DA}$ (excluding $D$), and $Z$ the set on the closed arc $\\overline{CD}$. Each black point belongs to exactly one of $X$, $Y$, or $Z$.\n\nEvery arc $\\gamma_m$ with $1 < m < d$ either (1) starts in $X$ or (2) ends in $Y$. These cases are exclusive; otherwise, $\\gamma_m$ would contain $\\gamma_d$. If (1) does not hold, $\\gamma_m$ starts in $Y \\cup Z$, so it also ends in $Y$ (otherwise it would contain $\\gamma_1$). It cannot end in $Z$, or else $\\gamma_d$ would contain $\\gamma_m$.\n\nLet $x$ be the number of arcs satisfying (1), $y$ those satisfying (2). Then $x \\leq |X|$, $y \\leq |Y|$, and $x + y = d - 2$. Since $|X| + |Y| = n - |Z| = n - d - 1$, we have $d - 2 \\leq n - d - 1$, so $d \\leq \\lfloor \\frac{n+1}{2} \\rfloor$.\n\nFor $n = 999$ (odd), $\\lfloor \\frac{999+1}{2} \\rfloor = 500$. Label the black points $1, \\ldots, 999$ in counterclockwise order. For each $m = 1, \\ldots, 500$, place an arc of length $m$ starting at point $m$. This configuration works, so all $d = 1, 2, \\ldots, 500$ are possible.\n\n**Answer:** All $d = 1, 2, \\ldots, 500$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 13584,
    "subject": "Mathematics (Olympiad)",
    "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": "See solution",
    "solution": "The circles drawn around two neighbouring vertices of the hexagon intersect, since $2 \\times 1 > \\sqrt{3}$. Hence, every two neighbouring circles have a common region shaped like 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. As $2 \\times 1 < 3$, circles drawn around such two vertices do not intersect. Thus, the circles drawn around opposite vertices do not intersect either.\n\n![](images/prob1617_p17_data_a845d1a6c8.png)\n\n![](images/prob1617_p17_data_82cbaf1f5f.png)\n\nThe sum of the areas of all six blue circles is $6\\pi$. One must discount the area of six lenses. Let $A$ and $B$ be neighbouring vertices of the hexagon and let $C$ and $D$ be the points of intersection of the circles drawn around $A$ and $B$. Subtracting the area of triangle $ACD$ from the area of sector $ACD$ gives precisely one half of the area of a lens. We have $\\left(\\frac{CD}{2}\\right)^2 + \\left(\\frac{AB}{2}\\right)^2 = AC^2$, i.e., $\\left(\\frac{CD}{2}\\right)^2 + \\left(\\frac{\\sqrt{3}}{2}\\right)^2 = 1^2$, whence $CD = 1$, implying that 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:\n\n$$\n6\\pi - 12\\left(\\frac{1}{6}\\pi - \\frac{\\sqrt{3}}{4}\\right) = 4\\pi + 3\\sqrt{3}.\n$$",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 13585,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $\\triangle PAB$ and $\\triangle PBC$ be two similar right-angled triangles (in the same plane) with $\\angle PAB = \\angle PBC = 90^\\circ$, such that $A$ and $C$ lie on opposite sides of the line $PB$. If $PC = AC$, calculate the ratio $\\frac{PA}{AB}$.",
    "options": [],
    "answer": "See solution",
    "solution": "Consider Figure 1:\n\n![](images/s3s2021_p0_data_7c9c04e3eb.png)\n\nWe may assume that $AB = 1$. Let $PA = x$, so that $PB = \\sqrt{x^2+1}$ by Pythagoras. From the similarity of triangles $PAB$ and $PBC$, we have $BC = \\frac{\\sqrt{x^2+1}}{x}$, so that $PC = x + \\frac{1}{x}$, again by Pythagoras. Thus $AC = PC = x + \\frac{1}{x}$.\n\nNow let $\\angle APB = \\alpha$, so that $\\angle PBA = 90^\\circ - \\alpha$. Then $\\cos \\alpha = \\frac{x}{\\sqrt{x^2+1}}$, and from the cosine rule applied to triangle $ABC$, we get\n\n$$\n\\left(x + \\frac{1}{x}\\right)^2 = 1^2 + \\frac{x^2+1}{x^2} - 2 \\cdot \\frac{\\sqrt{x^2+1}}{x} \\cdot \\cos(180^\\circ - \\alpha)\n$$\n\nSince $\\cos(180^\\circ - \\alpha) = -\\cos \\alpha$, this becomes\n\n$$\n\\left(x + \\frac{1}{x}\\right)^2 = 1 + \\frac{x^2+1}{x^2} + 2 \\cdot \\frac{\\sqrt{x^2+1}}{x} \\cdot \\cos \\alpha = 3 + \\frac{x^2+1}{x^2}\n$$\n\nWe now easily solve for $x$, and find that $\\frac{PA}{AB} = x = \\sqrt{2}$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 13586,
    "subject": "Mathematics (Olympiad)",
    "question": "Given a fixed circle $(O)$ and two fixed points $B$, $C$ on that circle, let $A$ be a moving point on $(O)$ such that triangle $ABC$ is acute and scalene. Let $I$ be the midpoint of $BC$; let $AD$, $BE$ and $CF$ be the three altitudes of triangle $ABC$. On two rays $\\vec{FA}$, $\\vec{EA}$, take $M$, $N$ such that $FM = CE$, $EN = BF$. Let $L$ be the intersection of $MN$ and $EF$, and let $G \\ne L$ be the second intersection of the circumcircles of triangles $LEN$ and $LFM$.\n\n1. Show that the circumcircle of triangle $MNG$ always goes through a fixed point.\n2. Let $AD$ intersect $(O)$ at $K$ which is different from $A$. On the tangent line through $D$ of the circumcircle of $DKI$, take $P$, $Q$ such that $GP$ is parallel to $AB$ and $GQ$ is parallel to $AC$. Let $T$ be the circumcenter of $GPQ$. Show that $GT$ always goes through a fixed point.",
    "options": [],
    "answer": "See solution",
    "solution": "a) It is clear that $G$ is the Miquel point of the completed quadrilateral $MNEF.LA$, thus $G$ lies on the circumcircle of $\\triangle AMN$, $\\triangle AEF$. Besides,\n\n$$\nBM = BF + FM = EN + CE = CN.\n$$\n\n![](images/Vietnamese_mathematical_competitions_p280_data_4b768f8b7f.png)\n\nLet $X$ be the midpoint of arc $BAC$ of $(O)$, then $\\triangle XBM = \\triangle XCN$. Hence, $\\angle XMA = \\angle XNA$; it follows that $X$ lies on the circumcircle of $AMN$, in other words, $(GMN)$ passes through fixed point $X$.\n\nb) Let $H$ be the orthocenter of $\\triangle ABC$. Since $G$ lies on the circumcircle of $\\triangle AMN$, $\\triangle AEF$, we have $\\triangle GMF \\sim \\triangle GNE$, thus\n\n$$\n\\frac{GE}{GF} = \\frac{NE}{MF} = \\frac{BF}{CE} = \\frac{HF}{HE}\n$$\n\nThis means $GH$ bisects $EF$. Since $EF$ and $BC$ are anti-parallel with respect to $\\angle BHC$, therefore $HG$ is the symmedian of $\\triangle HBC$. It is well known that $K$ and $H$ are symmetric with respect to $BC$. Thus $\\angle PDI = \\angle DKI = \\angle DHI$, we obtain that $PQ \\perp HI$. Therefore,\n\n$$\n\\angle HGQ = 90^\\circ - \\angle GHE = 90^\\circ - \\angle CHI \\\\\n= 90^\\circ - \\angle GPQ = \\angle TGQ.\n$$\n\nHence, $G$, $H$ and $T$ are collinear. It is clear that $(BHC)$ is a fixed circle since $\\angle BHC = 180^\\circ - \\angle BAC$. Thus, $HT$ passes through the intersection of tangents at $B$, $C$ of $(BHC)$ which is fixed. $\\square$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 13587,
    "subject": "Mathematics (Olympiad)",
    "question": "Let us call a partition of $\\{1, 2, \\dots, n\\}$ into two sets a *nice partition for $n$* if none of the sets contains two distinct elements whose sum is a power of $2$. Let $p_n$ be the number of nice partitions for $n$.\n\nFind $p_{2012}$.",
    "options": [],
    "answer": "See solution",
    "solution": "We observe that removing $n$ from a nice partition for $n$ gives a nice partition for $n-1$. Therefore, we can obtain the nice partitions for $n$ by adding $n$ to the nice partitions for $n-1$.\n\nIf $2^m < n < 2^{m+1}$ for some positive integer $m$, then as $2^m < n+1 \\leq n + (n-1) < 2^{m+2}$ and $1 \\leq 2^{m+1}-n \\leq n-1$, we must add $n$ into the set not containing $2^{m+1}-n$. Therefore, $p_n = p_{n-1}$.\n\nIf $n = 2^m$ for some positive integer $m$, then since $2^m < n + 1 \\leq n + (n-1) < 2^{m+1}$, we can add $n$ into any of the sets and hence $p_n = 2 \\cdot p_{n-1}$.\n\nAs $p_2 = 2$ and $2^{10} < 2012 < 2^{11}$, we conclude that $p_{2012} = 2^{10} = 1024$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 13588,
    "subject": "Mathematics (Olympiad)",
    "question": "Can one draw 6 circles on a plane, such that each one passes through the centers of exactly three other circles?",
    "options": [],
    "answer": "See solution",
    "solution": "A possible example is shown in Figure 18, where each segment has length 1.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 13589,
    "subject": "Mathematics (Olympiad)",
    "question": "A product\n\n$$1 \\cdot 2 \\cdot 3 \\cdot \\dots \\cdot n$$\n\nis 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": "See solution",
    "solution": "Let $v_p(n)$ denote the highest power of a prime $p$ dividing a positive integer $n$. This function has the following properties:\n\n- For all primes $p$ and positive integers $n$, $v_p(n)$ is a non-negative integer.\n- For all positive integers $m, n$ and all primes $p$, $v_p(mn) = v_p(m) + v_p(n)$.\n- For all primes $p$, $v_p(p!) = v_p(p) = 1$.\n- For all primes $p$, $v_p((p+1)!) = 1$, $v_p(p+1) = 0$.\n- For all primes $p$ and all positive integers $n < p$, $v_p(n!) = v_p(n) = 0$.\n- A positive integer $n$ is a square if and only if $v_p(n)$ is even (including zero) for all primes $p$.\n\nLet $S = n!$ be the initial value of the product, and $S'$ its final value after adding factorials. For $n$ equal to any prime $p$, $v_p(S) = v_p(p!) = 1$ and $v_p(S') = 1$, because adding factorials does not change the amount of the prime $p$ (i.e., $n$) in the final product. Thus, $v_p(S')$ is odd and $S'$ is not a square.\n\nAssume $n$ is a composite number ($n \\ge 4$). We will show that we can add factorials so that the final product\n\n$$S' = f_1 \\cdot f_2 \\cdot f_3 \\cdots f_n,$$\n\nwhere each $f_k$ is either $k$ or $k!$, is a square. This is equivalent to $v_p(S')$ being even for all primes $p$. Since $n$ is not a prime, only primes less than $n$ occur in $S'$. For each such prime $p$, the final power $v_p(S')$ is the same as in the “reduced” product\n\n$$p \\cdot f_{p+1} \\cdot f_{p+2} \\cdots f_n.$$\n\nTo ensure every prime $p < n$ occurs with even power, we can choose $f_{p+1}$ appropriately, depending on the parity of $v_p(f_{p+2} \\cdots f_n)$. If the parity is odd, choose $f_{p+1} = p+1$; if even, choose $f_{p+1} = (p+1)!$.\n\nThus, we can construct $S'$ to be a square for all composite $n \\ge 4$.\n\n**Conclusion.** The desired $n \\ge 2$ are all composite numbers.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 13590,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all functions $f: \\mathbb{R} \\to \\mathbb{R}$ such that\n$$\n(x+y)(f(x)-f(y)) = (x-y)f(x+y)\n$$\nfor all $x, y \\in \\mathbb{R}$.",
    "options": [],
    "answer": "See solution",
    "solution": "Suppose that $f$ is a solution. Let\n$$\na = \\frac{1}{2}(f(1) - f(-1)), \\quad b = \\frac{1}{2}(f(1) + f(-1))\n$$\nand define $g(x) = f(x) - a x - b x^2$. Then\n$$\n(x+y)(g(x)-g(y)) = (x-y)g(x+y)\n$$\nand $g(1) = g(-1) = 0$. Setting $y = 1$ and $y = -1$ gives\n$$\n\\begin{aligned}\n(x+1)g(x) &= (x-1)g(x+1) \\\\\nx g(x+1) &= (x+2)g(x).\n\\end{aligned}\n$$\nThus,\n$$\nx(x+1)g(x) = x(x-1)g(x+1) = (x-1)(x+2)g(x)\n$$\nfor all $x$. So $g(x) = 0$ for all $x$. Hence $f(x) = a x + b x^2$. We can check directly that any function of this form (for some $a, b \\in \\mathbb{R}$) satisfies the given equation.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 13591,
    "subject": "Mathematics (Olympiad)",
    "question": "一個城市是平面上的一個點。假設平面上有 $n \\ge 2$ 個城市。假設對於每個城市 $X$，都存在另一個城市 $N(X)$，使得 $X$ 到 $N(X)$ 的距離嚴格小於 $X$ 到任何其他城市的距離。政府在所有的城市 $X$ 與其 $N(X)$ 之間建有道路，除此之外城市之間沒有其他道路。已知我們可以從任何一個城市，經過一系列的道路，抵達任何一個其他城市。我們稱一個城市 $Y$ 是一個近郊，若且唯若存在城市 $X$ 使得 $Y = N(X)$。試證明至少有 $(n-2)/4$ 個近郊。",
    "options": [],
    "answer": "See solution",
    "solution": "讓我們以城市為頂點，$(X, N(X))$ 為邊建立有向圖 $G$。基於 $G$ 連通且共有 $n$ 個邊，此圖恰有一個環。這表示我們有至多一對城市 $(A, B)$ 滿足 $A = N(B)$ 且 $B = N(A)$。\n\n考慮以下關鍵引理：\n\n*引理：* 如果 $B = N(A)$，則存在至多 4 個異於 $B$ 的城市滿足 $A = N(X)$。\n\n*證明：* 假設 $X_1, \\cdots, X_t$ 為所有異於 $B$ 且滿足 $A = N(X_i)$ 的城市。由定義知 $\\overline{X_iA} < \\overline{X_iB}$，意味著 $X_i$ 與 $A$ 必須在 $\\overline{AB}$ 的中垂線的同一側。同樣由定義，$\\overline{X_iA} > \\overline{AB}$，所以 $X_i$ 必須在以 $A$ 為圓心、過 $B$ 點的圓外。這表示射線 $AX_i$ 可能存在的角度範圍為 $4\\pi/3$。\n\n現在，對於任何 $i \\neq j$，$\\overline{X_iX_j}$ 都必然為 $\\triangle AX_iX_j$ 的最長邊，也就是 $\\angle X_iAX_j > \\pi/3$。結合前面關於射線可能角度的討論，我們得到\n\n$$\nt - 1 \\le \\frac{4\\pi/3}{\\pi/3} = 4,\n$$\n\n也就是 $t \\le 4$。引理證畢。\\quad \\Box\n\n回到原題。由引理得知對於所有城市 $A$，至多只有四個城市 $X$ 滿足 $N(X) = A$ 且 $N(N(X)) \\neq X$。令 $S$ 為所有近郊城市所成集合，$S'$ 則為所有滿足 $N(N(X)) \\neq X$ 的城市所成集合。注意到我們有 $|S'| \\ge n - 2$ 且 $4|S| \\ge |S'|$（因為 $N : S' \\to S$ 的 preimage 大小最大為 4），這表示 $|S| \\ge (n - 2)/4$。得證。",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 13592,
    "subject": "Mathematics (Olympiad)",
    "question": "Consider a $17 \\times 42$ table, where each cell is colored either black or white. Define a graph whose vertices correspond to the cells, and edges connect adjacent cells (sharing a side). What is the maximum possible number of edges connecting vertices (cells) of opposite colors, over all possible two-colorings of the table?\n\n*Lemma:* In certain graphs, the number of edges connecting vertices of opposite colors is bounded. (See images below for specific graphs.)\n\n![alt](path \"title\")\n\n*Corollary:*\n\n- (a) For a graph with $2n$ vertices colored with two colors, the maximum number of edges connecting vertices of opposite colors is at most $3n - 2$.\n- (b) For another class of graphs (see images), the maximum is at most $6n - 8$.\n\n![alt](path \"title\")\n",
    "options": [],
    "answer": "See solution",
    "solution": "*Proof of (a):* The graph can be decomposed into one special graph (see image) and $n-2$ graphs of another type. By the lemma, the total number of edges connecting opposite colored vertices is less than $4 + 3(n - 2) = 3n - 2$.\n\n*Proof of (b):* Similarly, the graph can be decomposed into one $\\boxplus$ graph and $n-2$ $\\boxplus$ graphs. By the lemma, the total is less than $4 + 6(n - 2) = 6n - 8$.\n\nNow, for the $17 \\times 42$ table, the graph can be viewed as the union of $25$ graphs (since $42 - 17 = 25$):\n\n![alt](images/mongolia_p12_data_c5d0872764.png)\n\nand a $17 \\times 17$ square graph.\n\nBy corollary (a), the $25$ graphs contribute at most $25 \\cdot (3 \\cdot 17 - 2) = 1225$ edges.\n\n![alt](images/mongolia_p12_data_c1ca663816.png)\n\n![alt](images/mongolia_p12_data_a809608f68.png)\n\n![alt](images/mongolia_p12_data_5d8e76da0a.png)\n\nFor the $17 \\times 17$ square, by corollary (b), the total is\n\n$$\n(6 \\cdot 17 - 8) + (6 \\cdot 16 - 8) + \\dots + (6 \\cdot 2 - 8) = 6 \\cdot 152 - 8 \\cdot 16 = 784.\n$$\n\nTherefore, the total number of edges connecting opposite colored vertices in the $17 \\times 42$ table is less than $1225 + 784 = 2009$.\n\nIf we color the vertices so that odd-numbered columns are black and even-numbered columns are white, the number of such edges is exactly $2009$.",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 13593,
    "subject": "Mathematics (Olympiad)",
    "question": "Given an $m \\times n$ table, what is the maximal number of different numbers that can be placed in the table such that in every cell, the number is the arithmetic mean of the numbers in two neighboring cells (cells sharing a side)?",
    "options": [],
    "answer": "See solution",
    "solution": "Let $A$ be the maximal number in the table and $a$ be the minimal number.\n\nConsider any cell $k$ of the table where $A$ stands. Since $A$ is the arithmetic mean of the numbers in some two neighboring cells $k_1$ and $k_2$, and all numbers in the table are no greater than $A$, the numbers in $k_1$ and $k_2$ must be equal to $A$. Since $k_1$ and $k_2$ have common sides with $k$, they are not neighbors. Since $A$ stands in $k_1$, $A$ must also stand in two cells having common sides with $k_1$. One of them is cell $k$ and the other is different from $k_2$ ($k_1$ and $k_2$ are not neighbors). It follows that $A$ stands in at least 4 different cells.\n\nSimilarly, $a$ stands in at least 4 different cells.\n\nThus, the maximal and minimal numbers stand in at least 8 different cells of the table. The number of all cells is $mn$, so at most $mn - 8 + 2 = mn - 6$ different numbers can stand in the table. We construct the example for $mn - 6$ different numbers.\n\n![](images/Belorusija_2012_p26_data_e2a6e2bd87.png)\n\nWe mark two $2 \\times 2$ neighboring squares (see the figure) and join these squares with a line passing through each cell of the table exactly once, as shown in the figures depending on the parity of $m$ and $n$. We place $1$ in all cells of the first square and $mn - 6$ in all cells of the second square. We move along the line from the cell with $1$ to the cell with $mn - 6$, placing in each cell of the line the number which is $1$ greater than the number in the previous cell. It is easy to see that the obtained table satisfies the condition and has exactly $mn - 6$ different numbers.",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 13594,
    "subject": "Mathematics (Olympiad)",
    "question": "A sequence $x_n$ satisfies the following conditions: $x_1 = a$, $x_{n+1} = \\frac{1}{2}\\left(x_n - \\frac{1}{x_n}\\right)$ for $n \\in \\mathbb{N}$. Prove that there exists a number $a$ such that the sequence $(x_n)$ has exactly 2018 pairwise distinct elements.\n\n(If one of the elements of the sequence equals $0$, then the sequence stops at that element.)",
    "options": [],
    "answer": "See solution",
    "solution": "Let $x_1 = a = \\operatorname{ctg} \\alpha$. Then\n\n$$\nx_2 = \\frac{1}{2}\\left(x_1 - \\frac{1}{x_1}\\right) = \\frac{1}{2}(\\operatorname{ctg} \\alpha - \\operatorname{tg} \\alpha) = \\frac{1}{2} \\cdot \\frac{\\cos^2 \\alpha - \\sin^2 \\alpha}{\\sin \\alpha \\cdot \\cos \\alpha} = \\frac{\\cos 2\\alpha}{\\sin 2\\alpha} = \\operatorname{ctg} 2\\alpha.\n$$\n\nSimilarly, by induction,\n\n$$\nx_{n+1} = \\operatorname{ctg} 2^n \\alpha, \\quad \\forall n \\in \\mathbb{N}.\n$$\n\nThe sequence will have exactly 2018 distinct elements if $x_i \\neq 0$ for $i = 1, \\ldots, 2017$, and $x_{2018} = 0$. This happens if $\\operatorname{ctg}(2^{2017} \\alpha) = 0$, i.e., $2^{2017} \\alpha = \\frac{\\pi}{2}$. Thus, $\\alpha = \\frac{\\pi}{2^{2018}}$ and $a = \\operatorname{ctg} \\left(\\frac{\\pi}{2^{2018}}\\right)$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 13595,
    "subject": "Mathematics (Olympiad)",
    "question": "Find the integral part of\n\n$$\nA = \\sqrt{2013 + \\sqrt{2012 + \\dots + \\sqrt{2 + \\sqrt{1}}}}\n$$\n",
    "options": [],
    "answer": "See solution",
    "solution": "The answer is $45$.\n\n**Solution 1:**\n\nOn the one hand, $A^2 > 2013 + \\sqrt{2012} > 2013 + 44 > 45^2$, therefore $A > 45$.\n\nOn the other hand, we can demonstrate by induction that $x_n = \\sqrt{n + \\sqrt{n-1 + \\dots + \\sqrt{2 + \\sqrt{1}}}} < \\sqrt{n} + 1$.\n\nThis holds for $n = 1$. Suppose it holds for some $n$. Then\n\n$$\nx_{n+1} = \\sqrt{n+1 + x_n} < \\sqrt{n+1 + \\sqrt{n} + 1}.\n$$\n\nIt remains to show that $\\sqrt{n+1 + \\sqrt{n} + 1} < \\sqrt{n+1} + 1$, which is equivalent to $n + \\sqrt{n} + 2 < n + 2 + 2\\sqrt{n} + 1$, which is always true. Therefore $A = x_{2013} < \\sqrt{2013} + 1 < 46$.\n\n**Solution 2:**\n\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$$\n\\begin{aligned}\n\\sqrt{2011 + \\sqrt{2010 + \\dots + \\sqrt{2 + \\sqrt{1}}}} &< \\sqrt{2011 + 2010 + \\dots + 2 + 1} \\\\\n&= \\sqrt{\\frac{2011 \\cdot 2012}{2}} = \\sqrt{2023066} < 1423.\n\\end{aligned}\n$$\n\nTherefore\n\n$$\nA < \\sqrt{2013 + \\sqrt{2012 + 1423}} = \\sqrt{2013 + \\sqrt{3435}} < \\sqrt{2013 + 59} = \\sqrt{2072} < 46.\n$$\n\n**Solution 3:**\n\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$$\n\\begin{aligned}\n&2012 + \\sqrt{2011 + \\dots + \\sqrt{4 + \\sqrt{3 + \\sqrt{2 + \\sqrt{1}}}}} \\\\\n&\\le 2012 + \\sqrt{2011 + \\dots + \\sqrt{4 + \\sqrt{3 + \\sqrt{2}} + \\sqrt{\\sqrt{1}}}} \\\\\n&\\le 2012 + \\sqrt{2011 + \\dots + \\sqrt{4 + \\sqrt{3} + \\sqrt{\\sqrt{2}} + \\sqrt{\\sqrt{\\sqrt{1}}}}} \\\\\n&\\le \\dots \\le 2012 + \\sqrt{2011} + \\sqrt[2]{2010} + \\dots + \\sqrt[2^{2010}]{2} + \\sqrt[2^{2011}]{1} \\\\\n&< 2012 + 45 + 7 + 3 + 2 \\cdot 2008 = 6083,\n\\end{aligned}\n$$\n\nwhich gives $A < \\sqrt{2013 + \\sqrt{6083}} < \\sqrt{2013 + 78} < 46$.\n\n**Solution 4:**\n\nInequality $A > 45$ is proved as in solution 1.\n\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\n$$\nx = \\frac{1 + \\sqrt{1 + 4 \\cdot 2013}}{2} = \\frac{1 + \\sqrt{8053}}{2} < \\frac{1 + 90}{2} = 45.5 < 46\n$$\n\nfor the greater root, therefore $A = x_{2013} < 46$.\n\n**Solution 5:**\n\nInequality $A > 45$ is proved as in solution 1.\n\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": 13596,
    "subject": "Mathematics (Olympiad)",
    "question": "If a rate is $1\\ \\text{cm}$ per month, what is this rate in millimeters per month, per year, and over ten years?",
    "options": [],
    "answer": "See solution",
    "solution": "$1\\ \\text{cm per month} = 10\\ \\text{mm per month} = 120\\ \\text{mm per year} = 1\\,200\\ \\text{mm in ten years}$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 13597,
    "subject": "Mathematics (Olympiad)",
    "question": "The square $4 \\times 4$ is divided into 16 squares of size $1 \\times 1$. Some of these squares contain cubes of size $1 \\times 1 \\times 1$. First, the cubes move to the right as far as possible (until each reaches the right side or is blocked by another cube, as shown in the figures below). Then, the cubes move towards the top and right in a similar way. Finally, they move towards the bottom side of the square. \n\nDoes there exist an initial arrangement of cubes such that two cubes have exchanged their places between the beginning and the end?\n\n![](images/Ukrajina_2013_p45_data_206f389159.png)\n\n![](images/Ukrajina_2013_p45_data_28ce70e94c.png)",
    "options": [],
    "answer": "See solution",
    "solution": "**Answer:** It exists.\n\nLet us analyze the starting position of four cubes shown below. The cubes' positions after each of the four movements are shown in the following figures. As a result, cubes 2 and 3 have exchanged their places.\n\n![](images/Ukrajina_2013_p45_data_f55052c22b.png)\n\n![](images/Ukrajina_2013_p45_data_78fb1a92a4.png)\n\n![](images/Ukrajina_2013_p45_data_00dc7719d9.png)\n\n![](images/Ukrajina_2013_p45_data_d182d8356c.png)\n\n![](images/Ukrajina_2013_p45_data_8323756162.png)",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 13598,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all pairs of prime numbers $p, q$ with $p > q$, for which both numbers $p+q$ and $p-q$ are also prime.",
    "options": [],
    "answer": "See solution",
    "solution": "**Answer:** $p=5$, $q=2$.\n\n**Solution.** For $p+q$ to be prime, $p$ and $q$ must have different parity, which means $q=2$ since $p > q$. By the problem statement, the numbers $p-2$, $p$, and $p+2$ should be prime. Since these have different remainders modulo $3$, one of them must be $3$. Consider the possibilities:\n\n- If $p-2=3$, then $p=5$, $p+2=7$, and the pair $(5, 2)$ satisfies the problem statement.\n- If $p=3$, then $p-2=1$, which is not prime.\n- If $p+2=3$, then $p=1$, which is not prime.\n\nThus, the only solution is $p=5$, $q=2$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 13599,
    "subject": "Mathematics (Olympiad)",
    "question": "若整數 $a$ 使方程式 $$(m^2 + n)(n^2 + m) = a(m - n^3)$$ 有正整數解 $m, n$，則稱整數 $a$ 為友善的（friendly）。\n\n(a) 證明集合 $\\{1, 2, \\dots, 2013\\}$ 中至少有 500 個友善的整數。\n\n(b) 決定 $a = 2$ 是否為友善的。",
    "options": [],
    "answer": "See solution",
    "solution": "(a) 取 $a = 4k - 3$，$k \\geq 2$，再取 $m = 2k - 1$，$n = k - 1$，則\n$$\n(m^2 + n)(n^2 + m) = ((2k - 1)^2 + (k - 1))((k - 1)^2 + (2k - 1)) = (4k - 3)k^3 = a(m - n)^3.\n$$\n因此 $5, 9, \\dots, 2009, 2013$ 是友善的，且 $\\{1, 2, \\dots, 2013\\}$ 包含至少 503 個友善的整數。\n\n(b) 證明 $a = 2$ 不是友善的。考慮方程式 $(m^2 + n)(n^2 + m) = 2(m - n^3)$，將左式寫成平方差形式：\n$$\n\\frac{1}{4}\\left((m^2 + n + n^2 + m)^2 - (m^2 + n - n^2 - m)^2\\right) = 2(m - n)^3. \\quad (1)\n$$\n因為 $m^2 + n - n^2 - m = (m - n)(m + n - 1)$，可得\n$$\n(m^2 + n + n^2 + m)^2 = (m - n)^2\\left(8(m - n) + (m + n - 1)^2\\right).\n$$\n所以 $8(m - n) + (m + n - 1)^2$ 是完全平方數。顯然 $m > n$，因此存在整數 $s \\geq 1$ 使得\n$$\n(m + n - 1 + 2s)^2 = 8(m - n) + (m + n - 1)^2.\n$$\n展開得 $s(m + n - 1 + s) = 2(m - n)$。因為 $m + n - 1 + s > m - n$，所以 $s < 2$，唯一可能是 $s = 1$ 且 $m = 3n$。代入 (1) 得 $27n^3 = 16n^3$，即 $n = m = 0$，矛盾。因此 $a = 2$ 不是友善的。",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 13600,
    "subject": "Mathematics (Olympiad)",
    "question": "$$\n\\frac{2^{p-1}-1}{p} \\text{ нь бүтэн квадрат байх бүх } p \\text{-анхны тоог ол.}\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "$p = 2$ бол $\\frac{2^{p-1}-1}{p} = \\frac{1}{2}$ бүтэн квадрат биш. $p \\ge 3$ бол $p-1$ тэгш байна. Иймээс $\\frac{2^{p-1}-1}{p} = x^2$ гэвэл $(2^{\\frac{p-1}{2}}-1)(2^{\\frac{p-1}{2}}+1) = p x^2$ байна. $(2^{\\frac{p-1}{2}}-1; 2^{\\frac{p-1}{2}}+1) = (2^{\\frac{p-1}{2}}-1; 2)$ гэдгийг санавал:\n\n$$\n1^0)\\quad 2^{\\frac{p-1}{2}} - 1; p \\text{ бол } 2^{\\frac{p-1}{2}} + 1 = y^2,\\quad 2^{\\frac{p-1}{2}} - 1 = p z^2 \\text{ байна.}\n$$\n\n$$\n2^{\\frac{p-1}{2}} + 1 = y^2 \\Leftrightarrow 2^{\\frac{p-1}{2}} = (y-1)(y+1) \\text{ болох ба энэ нь } \\begin{cases} y-1=2 \\\\ y+1=4 \\end{cases} \\text{ -ээс өөр шийдүүдийг шууд шалгах болно. Иймд } y=3,\\quad 2^{\\frac{p-1}{2}} = 8 = 2^3 \\Leftrightarrow p=7 \\text{ болно.}\n$$\n\n$$\n2^0)\\quad 2^{\\frac{p-1}{2}} + 1; p \\text{ бол } 2^{\\frac{p-1}{2}} - 1 = y^2,\\quad 2^{\\frac{p-1}{2}} + 1 = p z^2 \\text{ байна.}\n$$\n\n$$\n2^{\\frac{p-1}{2}} - 1 = y^2 \\Leftrightarrow y^2 \\equiv 1 \\pmod{4}.\\quad p=3 \\text{ бол } 2^{\\frac{p-1}{2}} - 1 = 2 - 1 = 1^2 \\text{ болох тул } p=3 \\text{ шийд мөн.}\n$$\n\n$$\nОдоо\\quad p>3 \\text{ бол } 2^{\\frac{p-1}{2}} - 1 \\equiv 0 - 1 \\equiv -1 \\equiv 3 \\pmod{4} \\text{ ба } y^2 = 2^{\\frac{p-1}{2}} - 1 \\text{ гэдгээс } y^2 \\equiv 3 \\pmod{4} \\text{ болж зөрчинө. Иймд } p=3 \\text{ ба } p=7 \\text{-оос өөр шийдгүй.}\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 13601,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $\\triangle ABC$ be an acute-angled triangle, and let $D$ be the foot of the altitude from $C$. The angle bisector of $\\angle ABC$ intersects $CD$ at $E$ and meets the circumcircle $\\omega$ of triangle $\\triangle ADE$ again at $F$. If $\\angle ADF = 45^\\circ$, show that $CF$ is tangent to $\\omega$.\n\n![](path/to/file.png)",
    "options": [],
    "answer": "See solution",
    "solution": "Since $\\angle CDF = 90^\\circ - 45^\\circ = 45^\\circ$, the line $DF$ bisects $\\angle CDA$, and so $F$ lies on the perpendicular bisector of segment $AE$, which meets $AB$ at $G$. Let $\\angle ABC = 2\\beta$. Since $ADEF$ is cyclic, $\\angle AFE = 90^\\circ$, and hence $\\angle FAE = 45^\\circ$. Further, as $BF$ bisects $\\angle ABC$, we have $\\angle FAB = 90^\\circ - \\beta$, and thus $\\angle EAB = \\angle AEG = 45^\\circ - \\beta$ and $\\angle AED = 45^\\circ + \\beta$, so $\\angle GED = 2\\beta$. This implies that right-angled triangles $\\triangle EDG$ and $\\triangle BDC$ are similar, and so we have $\\frac{|GD|}{|CD|} = \\frac{|DE|}{|DB|}$. Thus the right-angled triangle $\\triangle DEB$ and $\\triangle DGC$ are similar, whence $\\angle GCD = \\angle DBE = \\beta$. But $\\angle DFE = \\angle DAE = 45^\\circ - \\beta$, then $\\angle GFD = 45^\\circ - \\angle DFE = \\beta$. Hence $GDCF$ is cyclic, so $\\angle GFC = 90^\\circ$, whence $CF$ is perpendicular to the radius $FG$ of $\\omega$. It follows that $CF$ is a tangent to $\\omega$, as required.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 13602,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all functions $f: \\mathbb{R} \\to \\mathbb{R}$ such that for all $x, y \\in \\mathbb{R}$,\n\n$$\nf(x + f(x + y)) + f(xy) = x + f(x + y) + y f(x).\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "We consider the case $f(0) = 0$.\n\nReplacing $x$ with $y$ and $y$ with $x$ in the given equation yields\n$$\nf(y + f(x + y)) + f(xy) = y + f(x + y) + x f(y).\n$$\n\nSubtracting, we get\n$$\nf(x + f(x + y)) - f(y + f(x + y)) = x - y + y f(x) - x f(y).\n$$\n\nSetting $y = -x$ gives\n$$\nf(x) - f(-x) = 2x - x f(x) - x f(-x).\n$$\n\nSuppose $f(x) = x$. Then\n$$\n(x - 1)(f(-x) + x) = 0.\n$$\n\nIf $x \\neq 1$, then $f(-x) = -x$. If $x = 1$, then plugging $(x, y) = (-1, 1)$ into the original equation gives $f(-1) = -1$. Thus, if $S$ is the set of fixed points of $f$, we have\n$$\nx \\in S \\Rightarrow -x \\in S.\n$$\n\nPlugging $(x, y) = (z, 0)$ into the original equation gives\n$$\nz + f(z) \\in S.\n$$\n\nPlugging $(x, y) = (0, z)$ gives\n$$\nf(z) \\in S.\n$$\n\nPlugging $(x, y) = (f(z), 0)$ and using the previous result gives\n$$\n2 f(z) \\in S.\n$$\n\nNow, set $(x, y) = (z + f(z), -f(z))$ in the difference equation. The left side is $f(z + 2 f(z))$, and since $x = z + f(z) \\in S$ and $y = -f(z) \\in S$, the right side simplifies to $z + 2 f(z)$. Thus,\n$$\nz + 2 f(z) \\in S.\n$$\n\nNext, set $(x, y) = (z, -z - f(z))$ in the difference equation. $y = -z - f(z) \\in S$ and $x + y = -f(z) \\in S$. The equation simplifies to\n$$\nf(2x + y) = 2x + y + y f(x) - x y.\n$$\n\nWriting $x$ and $y$ in terms of $z$ and simplifying gives\n$$\nf(z - f(z)) = z - f(z) + z^2 - f(z)^2.\n$$\n\nNow, set $(x, y) = (z - f(z), f(z))$ in the difference equation. The left side is $f(z) - f(2 f(z)) = -f(z)$, so\n$$\n\\begin{aligned}\n-f(z) &= x - y + y f(x) - x f(y) \\\\\n&= z - 2 f(z) + f(z) f(x) - (z - f(z)) f(f(z)) \\\\\n\\end{aligned}\n$$\n\nUsing previous results, this leads to\n$$\n(f(z) - z)(f(z)^2 + z f(z) + 1) = 0.\n$$\n\nSolving for $f(z)$, we find $f(z) \\in \\{z, \\frac{-z + \\sqrt{z^2 - 4}}{2}, \\frac{-z - \\sqrt{z^2 - 4}}{2}\\}$. Since $f(z) \\in \\mathbb{R}$, we have $f(z) = z$ for all $|z| < 2$. From earlier, $(-2, 2) \\subseteq S$. From $z \\in S$ implies $2z \\in S$, we deduce $(-2^k, 2^k) \\subseteq S$ for all positive integers $k$. Thus $S = \\mathbb{R}$, and $f(x) = x$ for all $x \\in \\mathbb{R}$.\n\n$\\boxed{f(x) = x}$ is the only solution.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 13603,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $k$ and $m$ be integers greater than $1$. Consider $k$ pairwise disjoint sets $S_1, S_2, \\dots, S_k$; each of these sets has exactly $m+1$ elements, one of which is red and the other $m$ are all blue. Let $\\mathcal{F}$ be the family of all subsets $F$ of $S_1 \\cup S_2 \\cup \\dots \\cup S_k$ such that, for every $i$, the intersection $F \\cap S_i$ is monochromatic; the empty set is monochromatic. Determine the largest possible cardinality of a subfamily $\\mathcal{G} \\subseteq \\mathcal{F}$, no two sets of which are disjoint.",
    "options": [],
    "answer": "See solution",
    "solution": "The required maximum is $2^{m-1}(2^m + 1)^{k-1}$ and is achieved if, for instance, $\\mathcal{G}$ consists of all sets in $\\mathcal{F}$ containing a fixed blue element.\n\nWe now prove that $|\\mathcal{G}| \\le 2^{m-1}(2^m + 1)^{k-1}$ for any $\\mathcal{G}$ satisfying the conditions in the statement. For convenience, write $M = 2^m + 1$. Let $r_i$ denote the red element of $S_i$, and let $B_i$ be the set of blue elements in $S_i$.\n\nFor every subset $X_i \\subset B_i$ and every $j \\in \\mathbb{Z}_M$, define the sets\n\n$$\nT_{X_i, j} = \\begin{cases} \\{r_i\\}, & \\text{if } j = 0; \\\\ X_i, & \\text{if } j \\ne 0 \\text{ and } j \\text{ is even (considered as a number in } [1, M-1]\\); \\\\ B_i \\setminus X_i, & \\text{if } j \\ne 0 \\text{ and } j \\text{ is odd (considered as a number in } [1, M-1]\\). \\end{cases}\n$$\n\nNote that, for every $i$ and every $j$, the sets $T_{X_i,j}$ and $T_{X_i,j+1}$ are disjoint. Now, for every sets $X_i \\subset B_i$ and every elements $j_i \\in \\mathbb{Z}_M$, $i = 1, 2, \\dots, k$, denote\n\n$$\nF(X_1, X_2, \\dots, X_k, j_1, j_2, \\dots, j_k) = \\bigcup_{i=1}^{k} T_{X_i, j_i}.\n$$\n\n**Claim.** Every set $F \\in \\mathcal{F}$ has exactly $2^{mk}$ representations of the form above.\n\n*Proof.* Set $F_i = F \\cap S_i$. If $F_i = \\{r_i\\}$, then there are $2^m$ possible choices for $X_i$, and one should necessarily have $j_i = 0$. Otherwise, there are only two possible choices for $X_i$, namely $X_i = F_i$ and $X_i = B_i \\setminus F_i$, and for each of them there are $2^{m-1}$ possible choices for $j_i$. So, whatever $F$, there are $2^m$ possible choices for each pair $(X_i, j_i)$ all of which can be made independently, whence a total of $2^{mk}$ possible tuples $(X_1, X_2, \\dots, X_k, j_1, j_2, \\dots, j_k)$. This proves the Claim.\n\nThe Claim implies that each $F \\in \\mathcal{F}$ has the same number of representations of the form above. Thus, it suffices to show that, among all $N = 2^{km}(2^m + 1)^k$ tuples $(X_1, X_2, \\dots, X_k, j_1, j_2, \\dots, j_k)$, at most $\\frac{2^{m-1}}{2^m+1}N$ satisfy\n\n$$\nF(X_1, X_2, \\dots, X_k, j_1, j_2, \\dots, j_k) \\in \\mathcal{G}.\n$$\n\nTo this end, split all these tuples into length $M$ cycles\n\n$$\n(F(X_1, X_2, \\dots, X_k, j_1, j_2, \\dots, j_k), F(X_1, X_2, \\dots, X_k, j_1+1, j_2+1, \\dots, j_k+1), \\dots, F(X_1, X_2, \\dots, X_k, j_1+M-1, j_2+M-1, \\dots, j_k+M-1)),\n$$\n\nand note that any two adjacent sets of a cycle are disjoint. Hence each cycle contains at most $\\lfloor M/2 \\rfloor = 2^{m-1}$ sets from $\\mathcal{G}$. This provides the desired upper bound and completes the solution.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 13604,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $ABC$ be a triangle and $X$ be a point inside the triangle. The lines $AX$, $BX$, and $CX$ meet the circle $ABC$ again at $P$, $Q$, and $R$ respectively. Choose a point $U$ on $XP$ which is between $X$ and $P$. Suppose that the lines through $U$ which are parallel to $AB$ and $CA$ meet $XQ$ and $XR$ at points $V$ and $W$ respectively. Prove that the points $R$, $W$, $V$, and $Q$ lie on a circle.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $\\angle RCA = \\theta$. We have\n\n$$\n\\angle CWU = \\angle WCA = \\theta \\text{ (as } AC \\text{ is parallel to } WU)\n$$\n\n$$\n\\angle RPA = \\angle RCA = \\theta \\text{ (angles in the same segment).}\n$$\n\n![](images/V_Britanija_2011_p23_data_517ae5887a.png)\n\nIn the quadrilateral $WRPU$,\n\n$$\n\\angle RPU = \\theta \\text{ and } \\angle UWR = 180^\\circ - \\theta.\n$$\n\nSince opposite angles add up to $180^\\circ$, $WRPU$ is cyclic. By the same argument, $VQPU$ is cyclic.\n\nBecause $WRPU$ is cyclic, by the intersecting chords theorem we have\n\n$$\nXW \\times XR = XU \\times XP.\n$$\n\nSimilarly, because $VQPU$ is cyclic,\n\n$$\nXU \\times XP = XV \\times XQ.\n$$\n\nHence\n\n$$\nXW \\times XR = XU \\times XP = XV \\times XQ.\n$$\n\nSo by the converse of the intersecting chords theorem, $RWVQ$ is cyclic; $R$, $W$, $V$, and $Q$ lie on a circle.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 13605,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $k$ be a positive integer congruent to $1$ modulo $4$ which is not a perfect square, and let $a = \\frac{1 + \\sqrt{k}}{2}$. Show that\n$$\n\\{\\lfloor a^2 n \\rfloor - \\lfloor a \\lfloor a n \\rfloor \\rfloor : n = 1, 2, 3, \\dots \\} = \\{1, \\dots, \\lfloor a \\rfloor \\}.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "Let $a_n = a n - \\lfloor a n \\rfloor$, for $n = 1, 2, 3, \\dots$. Since $a^2 = a + \\frac{k-1}{4}$, it follows that $\\lfloor a^2 n \\rfloor = \\lfloor a n \\rfloor + n \\frac{k-1}{4}$, and $(a-1) \\lfloor a n \\rfloor = (a-1)(a n - a_n) = n \\frac{k-1}{4} - (a-1) a_n$. So, adding $\\lfloor a n \\rfloor$ to each side, $a \\lfloor a n \\rfloor = \\lfloor a n \\rfloor + n \\frac{k-1}{4} - (a-1) a_n = \\lfloor a^2 n \\rfloor - (a-1) a_n$. Since $a$ is irrational, the $a_n$ form a dense subset of the open unit interval $(0, 1)$, so, by the preceding, the differences $\\lfloor a^2 n \\rfloor - a \\lfloor a n \\rfloor = (a-1) a_n$ form a dense subset of the open interval $(0, a-1)$. Finally, since $\\lfloor a^2 n \\rfloor - \\lfloor a \\lfloor a n \\rfloor \\rfloor = \\lfloor a^2 n \\rfloor - \\lfloor a \\lfloor a n \\rfloor \\rfloor = \\lfloor (a-1) a_n \\rfloor$, the conclusion follows.",
    "topic": "Number Theory",
    "subtopic": "Modular Arithmetic"
  },
  {
    "id": 13606,
    "subject": "Mathematics (Olympiad)",
    "question": "Real numbers $x, y$ satisfy the inequality:\n\n$$\nx^2 + 3xy + 4y^2 \\leq \\frac{7}{2}.\n$$\n\nProve that $x + y \\leq 2$.",
    "options": [],
    "answer": "See solution",
    "solution": "Denote $t = x + y$, and substitute $x = t - y$ into the given inequality. Then:\n\n$$(t - y)^2 + 3(t - y)y + 4y^2 - \\frac{7}{2} \\leq 0$$\n\nExpanding and simplifying:\n\n$$2y^2 + t y + t^2 - \\frac{7}{2} \\leq 0$$\n\nThis is a quadratic in $y$ with positive leading coefficient. For real $y$, the discriminant must be non-negative:\n\n$$D = t^2 - 4 \\cdot 2 \\left(t^2 - \\frac{7}{2}\\right) = t^2 - 8t^2 + 28 = 28 - 7t^2 \\geq 0$$\n\nThus, $t^2 \\leq 4$, so $t = x + y \\leq 2$.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 13607,
    "subject": "Mathematics (Olympiad)",
    "question": "Determine all natural numbers $n \\ge 2$ with the property that there are two permutations $(a_1, a_2, \\dots, a_n)$ and $(b_1, b_2, \\dots, b_n)$ of the numbers $1, 2, \\dots, n$ such that $(a_1 + b_1, a_2 + b_2, \\dots, a_n + b_n)$ are consecutive natural numbers.",
    "options": [],
    "answer": "See solution",
    "solution": "The permutations exist if and only if $n$ is odd.\n\nWe have\n\n$$\n(a_1 + b_1) + (a_2 + b_2) + \\dots + (a_n + b_n) = 2(1 + 2 + \\dots + n) = n(n+1).\n$$\n\nOn the other hand, there is a natural number $N$ such that\n\n$$\na_1 + b_1 = N,\\ a_2 + b_2 = N + 1,\\ \\dots,\\ a_n + b_n = N + n - 1\n$$\n\nand therefore\n\n$$\n(a_1 + b_1) + (a_2 + b_2) + \\dots + (a_n + b_n) = nN + (1 + 2 + \\dots + (n-1)) = nN + \\frac{n(n-1)}{2}.\n$$\n\nWe obtain the equation $n(n+1) = nN + \\frac{n(n-1)}{2}$, which becomes $N = n+1 - \\frac{n-1}{2} = \\frac{n+3}{2}$.\nTherefore, the number $N$ is an integer if and only if $n$ is odd.\n\nIt remains to investigate if two permutations with the desired property exist for every odd number $n$ with $n \\ge 3$. Let $n = 2k + 1$ with $k \\ge 1$.\n\nExperimenting with $k=1$ and $k=2$ can lead to the following pattern:\n\n$$\n\\begin{pmatrix}\n1 & k+2 & 2 & k+3 & 3 & \\dots & 2k+1 & k+1 \\\\\nk+1 & 1 & k+2 & 2 & k+3 & \\dots & k & 2k+1\n\\end{pmatrix}\n$$\n\nSumming the two rows gives the $2k+1$ consecutive numbers $k+2, k+3, \\dots, 3k+1, 3k+2$ as desired.",
    "topic": "Discrete Mathematics",
    "subtopic": "Combinatorics"
  },
  {
    "id": 13608,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $k$ be a positive integer. What is the largest $n$ such that the sides and diagonals of a regular $n$-gon can be colored with $k$ colors so that there does not exist a closed monochrome broken line (i.e., a cycle) consisting of sides and diagonals of the same color?\n\n% IMAGE: ![](images/CroatianCompetitions2011_p38_data_fd08d3d74a.png)",
    "options": [],
    "answer": "See solution",
    "solution": "It is known that a graph with $n$ vertices and no cycles has at most $n-1$ edges. Thus, at most $n-1$ edges (sides and diagonals) can be colored with the same color.\n\nSince there are $k$ colors and a total of $\\frac{n(n-1)}{2}$ edges, we have:\n\n$$\n\\frac{n(n-1)}{2} \\leq (n-1)k \\implies n \\leq 2k.\n$$\n\nTo show that $n=2k$ is possible, label the vertices $A_1, A_2, \\dots, A_{2k}$. For each color, color the broken line $A_1A_{2k}A_2A_{2k-1}\\dots A_kA_{k+1}$, and for each subsequent color, rotate this pattern by $\\frac{j\\pi}{k}$ for $j=2,\\dots,k$. These broken lines are disjoint, so every edge is colored, and no monochrome cycle exists.\n\nTherefore, the largest possible value of $n$ is $2k$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 13609,
    "subject": "Mathematics (Olympiad)",
    "question": "Iulia and Ștefan shared the 52 playing cards from a deck so each got 26 cards. The cards from 2 to 10 are assigned their own value, the ace is worth 11 points, the jack 12 points, the queen 13 points, and the king 14 points. Ștefan noticed that he had no ace in his stack, no 2, and no four cards of the same value. Iulia noticed that from her cards above 11 points, she doesn't have more than two of the same value. Iulia adds up all her points. What is the lowest value she can get? What is the highest?\n\nA deck has 52 cards: four with the value 2, four with the value 3, ..., four with the value 10, 4 aces, 4 jacks, 4 queens, and 4 kings.",
    "options": [],
    "answer": "See solution",
    "solution": "If we denote by $x_i$ the number of cards with the value $i$ that Iulia has in her stack, then $x_i \\geq 1$, $x_{11} = x_4 = 4$, $1 \\leq x_{12} \\leq 2$, $1 \\leq x_{13} \\leq 2$, $1 \\leq x_{14} \\leq 2$.\n\nIf $M$ and $m$ represent the highest and the lowest value that Iulia's stack can have, then:\n\n$$\nM = 2 \\cdot 14 + 2 \\cdot 13 + 2 \\cdot 12 + 4 \\cdot 11 + 4 \\cdot 10 + 2 \\cdot 9 + 1 \\cdot 8 + 1 \\cdot 7 + 1 \\cdot 6 + 1 \\cdot 5 + 1 \\cdot 4 + 1 \\cdot 3 + 4 \\cdot 2 = 221\n$$\n\n$$\nm = 1 \\cdot 14 + 1 \\cdot 13 + 1 \\cdot 12 + 4 \\cdot 11 + 1 \\cdot 10 + 1 \\cdot 9 + 1 \\cdot 8 + 1 \\cdot 7 + 1 \\cdot 6 + 2 \\cdot 5 + 4 \\cdot 4 + 4 \\cdot 3 + 4 \\cdot 2 = 169\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 13610,
    "subject": "Mathematics (Olympiad)",
    "question": "Suppose there are $n$ rectangles, possibly in different orientations. What is the minimum possible total number of right angles among all these rectangles?",
    "options": [],
    "answer": "See solution",
    "solution": "First, suppose all rectangles have sides parallel to the coordinate axes. There are four types of right angles: $\\uparrow$, $\\downarrow$, $\\perp$, and $\\swarrow$.\n\nSuppose there are $a$ angles of type $\\uparrow$ and $b$ of type $\\downarrow$. Since each rectangle is determined by these two angles, we get $ab \\geq n$. By the AM-GM inequality, $\\left(\\frac{a+b}{2}\\right)^2 \\geq ab \\geq n$, so $a+b \\geq 2\\sqrt{n}$. Similarly, for other pairs of angle types, we deduce $c+d \\geq 2\\sqrt{n}$. Thus, there are at least $4\\sqrt{n}$ right angles.\n\nIf rectangles have different directions, let $n_1, n_2, \\dots, n_k$ be the number of rectangles in each direction. Since $n = \\sum_i n_i$,\n\n$$\n4\\sqrt{n} \\leq 4\\sqrt{n_1} + 4\\sqrt{n_2} + \\dots + 4\\sqrt{n_k} \\leq \\text{Number of angles.}\n$$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 13611,
    "subject": "Mathematics (Olympiad)",
    "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:\n\n- $A$ claims to have met $B$, $C$, $F$, $G$ in the room.\n- $B$ claims to have met $A$, $C$, $D$, $E$, $F$.\n- $C$ claims to have met $A$, $B$, $E$.\n- $E$ claims to have met $B$, $C$, $F$.\n- $F$ claims to have met $A$, $B$, $D$, $E$.\n- $G$ claims to have met $A$, $D$.\n- $D$ claims to have met $B$, $F$, $G$.\n\nInspector Goodenough concluded that one of the servants was lying. Who is he?",
    "options": [],
    "answer": "See solution",
    "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 $X, Z$ or $Y, T$ also detected each other.\n\n**Proof of Lemma.** Let us assume, without loss of generality, 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 neither $A, E$ nor $C, F$ 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:\n\n- enter $G$\n- enter $D$\n- exit $G$\n- enter $B$\n- enter $F$\n- exit $D$\n- enter $E$\n- exit $F$\n- $C$ enters\n- $B$ exits\n- $E$ exits\n- $C$ exits\n\n$\\square$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 13612,
    "subject": "Mathematics (Olympiad)",
    "question": "Prove that the equation $7^x = 1 + y^2 + z^2$ has no positive integral solution.",
    "options": [],
    "answer": "See solution",
    "solution": "Assume that there exist positive integers $x$, $y$, $z$ satisfying the given equation. Consider two cases:\n\nIf $x$ is odd, then $7^x \\equiv 7 \\pmod{8}$. Since each of $y^2$ and $z^2$ can have $0$, $1$, or $4$ as remainders modulo $8$, $1 + y^2 + z^2$ can only be congruent to $1$, $2$, $3$, $5$, or $6$ modulo $8$. This is a contradiction since $7^x \\equiv 7 \\pmod{8}$.\n\nIf $x$ is even, let $x = 2n$. Then:\n\n$$\ny^2 + z^2 = 7^{2n} - 1 = (7^n + 1)(7^n - 1)\n$$\n\nWe can show that the right-hand side has at least one prime factor $p \\equiv 3 \\pmod{4}$ of odd order. Thus $p$ divides $y^2 + z^2$, which implies $p$ divides both $y$ and $z$. This leads to a contradiction because $p$ is a prime factor of even order on the left-hand side.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 13613,
    "subject": "Mathematics (Olympiad)",
    "question": "Suppose 8 professional packers can complete a packing job in 15 hours.\n\n**a.** How long would it take 12 professionals to complete the same job?\n\n**b.** If 12 professionals are joined by 8 trainees (each trainee works at half the rate of a professional), how long would it take to complete the job?\n\n**c.**\n  1. If the job is completed in 12 hours by a team of professionals and 6 trainees (each trainee works at half the rate of a professional), how many professionals were working with the trainees?\n  2. After $2\\frac{1}{2}$ hours, the 6 trainees and 2 professionals leave. How long will it take the remaining 5 professionals to finish the job? (Provide at least one method of solution.)",
    "options": [],
    "answer": "See solution",
    "solution": "**a.**\n\n$8 \\times 15 = 120$ professional packer hours are needed to complete the job. With 12 professionals, each would work $120 \\div 12 = 10$ hours.\n\n**b.**\n\n8 trainees are equivalent to 4 professionals (since each trainee works at half the rate). So, $12 + 4 = 16$ professionals. The job would take $120 \\div 16 = 7\\frac{1}{2}$ hours.\n\n**c. i.**\n\nTo finish in 12 hours, $120 \\div 12 = 10$ professionals are needed. 6 trainees are equivalent to 3 professionals, so $10 - 3 = 7$ professionals must have been working with the trainees.\n\n**c. ii. Alternative 1**\n\nIn $2\\frac{1}{2}$ hours, 6 trainees pack $3 \\times 2\\frac{1}{2} = 7\\frac{1}{2}$ professional packer hours. 7 professionals pack $7 \\times 2\\frac{1}{2} = 17\\frac{1}{2}$ professional packer hours. Total packed: $7\\frac{1}{2} + 17\\frac{1}{2} = 25$ professional packer hours. Remaining: $120 - 25 = 95$ professional packer hours. 5 professionals finish in $95 \\div 5 = 19$ hours.\n\n**Alternative 2**\n\nInitially, 7 professionals and 6 trainees (equivalent to 10 professionals). After $2\\frac{1}{2}$ hours, $10 \\times 2\\frac{1}{2} = 25$ professional packer hours are done, $120 - 25 = 95$ remain. 5 professionals take $95 \\div 5 = 19$ hours.\n\n**Alternative 3**\n\nAfter $2\\frac{1}{2}$ hours, $9\\frac{1}{2}$ professional packer hours remain per professional. Fraction left: $9\\frac{1}{2} \\div 12 = \\frac{19}{24}$. 5 professionals would take $(8 \\times 15) \\div 5 = 24$ hours for the whole job. So, $\\frac{19}{24} \\times 24 = 19$ hours to finish.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 13614,
    "subject": "Mathematics (Olympiad)",
    "question": "Find the least positive integer $n$ such that, for any coloring of the vertices of a regular $n$-gon with three colors (red, yellow, and blue), there must exist four vertices of the same color that form the vertices of an isogonal trapezoid.",
    "options": [],
    "answer": "See solution",
    "solution": "We claim that the least positive integer $n$ is $17$.\n\n**Proof for $n = 17$:**\n\nSuppose, for contradiction, that there exists a coloring of the regular $17$-gon with three colors such that no four vertices of the same color form an isogonal trapezoid.\n\nSince $\\lfloor \\frac{17}{3} \\rfloor + 1 = 6$, by the pigeonhole principle, there is a color (say, yellow) with at least $6$ vertices. Consider the $6$ yellow vertices. Connecting each pair yields $\\binom{6}{2} = 15$ segments. The possible segment lengths are at most $\\lfloor \\frac{17}{2} \\rfloor = 8$ distinct values, so by the pigeonhole principle, either:\n\n(a) There are three segments of the same length. Since $3 \\nmid 7$, not every pair of these segments shares a vertex, so two segments are disjoint. Their four endpoints form an isogonal trapezoid—a contradiction.\n\n(b) There are $7$ pairs of segments of the same length. Each pair must share a vertex; otherwise, their endpoints form an isogonal trapezoid. By the pigeonhole principle, two pairs share the same vertex, so their endpoints again form an isogonal trapezoid—a contradiction.\n\nThus, $n = 17$ satisfies the required property.\n\n**Counterexamples for $n \\leq 16$:**\n\nFor $n = 16$, let $A_1, \\ldots, A_{16}$ be the vertices. Define:\n\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$$\nM_3 = \\{A_1, A_2, A_4, A_9, A_{10}, A_{12}\\}\n$$\n\nIn $M_1$, the distances from $A_{14}$ to the other vertices are all different, and the remaining four form a rectangle, not an isogonal trapezoid. Similarly, $M_2$ and $M_3$ do not contain four vertices forming an isogonal trapezoid.\n\nFor $n = 15$, let:\n\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$$\nM_3 = \\{A_4, A_7, A_{10}, A_{11}, A_{12}\\}\n$$\n\nNo four vertices in any $M_i$ form an isogonal trapezoid.\n\nFor $n = 14$, let:\n\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$$\nM_3 = \\{A_2, A_6, A_9, A_{13}\\}\n$$\n\nAgain, no four vertices in any $M_i$ form an isogonal trapezoid.\n\nTherefore, the least $n$ is $17$.",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 13615,
    "subject": "Mathematics (Olympiad)",
    "question": "For $z \\in \\{1, 3, 7, 9\\}$, let $b$ be the $k$-digit number consisting only of the digit $z$ (i.e., $b = \\underbrace{(zzz\\dots z)}_{k}$). Find a $k$-digit number $n$ such that $n^9 \\equiv n \\pmod{10^k}$.",
    "options": [],
    "answer": "See solution",
    "solution": "Since $\\gcd(9, \\varphi(10^k)) = \\gcd(9, 4 \\cdot 10^{k-1}) = 1$, by the Euclidean algorithm there exist integers $x$ and $y$ such that $9x + \\varphi(10^k)y = 1$. We claim that $n := b^x$ has the desired property. Because $\\gcd(b, 10^k) = 1$, this follows from Euler's theorem:\n\n$$\n(b^x)^9 = b^{9x} \\equiv b^{9x + \\varphi(10^k)y} = b^1 = b \\pmod{10^k}.\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 13616,
    "subject": "Mathematics (Olympiad)",
    "question": "The inequality\n\n$$\n\\frac{1}{n+1} + \\frac{1}{n+2} + \\dots + \\frac{1}{2n+1} < a - 2007 \\frac{1}{3}\n$$\n\nholds for every positive integer $n$. Then the least positive integer of $a$ is $\\underline{\\hspace{2cm}}$.",
    "options": [],
    "answer": "See solution",
    "solution": "Obviously,\n\n$$\nf(n) = \\frac{1}{n+1} + \\frac{1}{n+2} + \\dots + \\frac{1}{2n+1}\n$$\n\nis monotonically decreasing. Therefore, $f(1)$ reaches the maximum of $f(n)$. From\n\n$$\nf(1) = \\frac{1}{2} + \\frac{1}{3} < a - 2007 \\frac{1}{3},\n$$\n\nwe have $a > 2008$. Therefore, the least positive integer of $a$ is 2009.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 13617,
    "subject": "Mathematics (Olympiad)",
    "question": "Let positive integers $K$ and $d$ be given. Prove that there exists a positive integer $n$ and a sequence of $K$ positive integers $b_1, b_2, \\dots, b_K$ such that the number $n$ is a $d$-digit palindrome in all number bases $b_1, b_2, \\dots, b_K$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let a positive integer $d$ be given. We shall prove that, for each large enough $n$, the number $(n!)^{d-1}$ is a $d$-digit palindrome in all number bases $\\frac{n!}{i}-1$ for $1 \\le i \\le n$. In particular, we shall prove that the digit expansion of $(n!)^{d-1}$ in the base $\\frac{n!}{i}-1$ is\n\n$$\n\\left\\langle i^{d-1} \\binom{d-1}{d-1}, i^{d-1} \\binom{d-1}{d-2}, i^{d-1} \\binom{d-1}{d-3}, \\dots, i^{d-1} \\binom{d-1}{1}, i^{d-1} \\binom{d-1}{0} \\right\\rangle_{\\frac{n!}{i}-1}\n$$\n\nWe first show that, for each large enough $n$, all these digits are smaller than the considered base, that is, they are indeed digits in that base. It is enough to check this assertion for $i = n$, that is, to show the inequality $n^{d-1} \\binom{d-1}{j} < (n-1)! - 1$. However, since for a fixed $d$ the right-hand side clearly grows faster than the left-hand side, this is indeed true for all large enough $n$.\n\nEverything that is left is to evaluate:\n\n$$\n\\begin{align*}\n\\sum_{j=0}^{d-1} i^{d-1} \\binom{d-1}{j} \\left(\\frac{n!}{i} - 1\\right)^j &= i^{d-1} \\sum_{j=0}^{d-1} \\binom{d-1}{j} \\left(\\frac{n!}{i} - 1\\right)^j \\\\\n&= i^{d-1} \\left(\\frac{n!}{i} - 1 + 1\\right)^{d-1} \\\\\n&= (n!)^{d-1},\n\\end{align*}\n$$\n\nwhich completes the proof. $\\square$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 13618,
    "subject": "Mathematics (Olympiad)",
    "question": "Each point on the sides of an equilateral triangle is coloured either red or blue. Is it certain that there exists a right triangle whose all vertices are of the same colour?",
    "options": [],
    "answer": "See solution",
    "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, such that $XP : XY = YQ : YZ = ZR : ZX = 1 : 3$.\n\n![](images/prob1516_p19_data_9d15081214.png)\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![](images/prob1516_p19_data_f7b4c6fd0c.png)",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 13619,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all functions $f: \\mathbb{Z} \\to \\mathbb{Z}$ that satisfy the following two conditions:\n\n1) $f(x + f(x + 2y)) = f(2x) + f(2y)$ for all integers $x, y$.\n\n2) $f(0) = 2$.",
    "options": [],
    "answer": "See solution",
    "solution": "**Answer:** $f(k) = k + 2$.\n\n**Solution.** Substitute $x = 0$ and $y = 0$:\n\n$$\nf(f(2y)) = f(2y) + 2 \\qquad (1)\n$$\n\n$$\nf(x + f(x)) = f(2x) + 2 \\qquad (2)\n$$\n\nLet us prove by induction that $f(2n) = 2n + 2$. The base case $n = 0$ is already proved. Suppose $f(2n) = 2n + 2$ and prove that $f(2n + 2) = 2n + 4$.\n\nFor $n = 2m$, we have to show that $f(4m + 2) = 4m + 4$. Substitute $x = 2m$ in (2):\n\n$$\nf(2m + f(2m)) = f(4m + 2) = f(4m) + 2 = 4m + 4.\n$$\n\nFor $n = 2m - 1$, $f(4m) = 4m + 2$. Substitute $y = 2m - 1$ in (1):\n\n$$\nf(f(4m - 2)) = f(4m) = f(4m - 2) + 2 = 4m + 2.\n$$\n\nFollowing the same lines, we can prove that $f(2n) = 2n + 2$ for negative $n$. Let $k$ be an odd number, then substitute $x = 2z + k$, $y = -z$, $z \\in \\mathbb{Z}$:\n\n$$\nf(2z + k + f(k)) = f(4z + 2k) + f(-2) = 4z + 2k + 2 - 2z + 2 = 2z + 2k + 4 \\qquad (3)\n$$\n\nWe consider two cases:\n\n- If $f(k)$ is even, substitute $z = -\\frac{f(k)}{2}$ into (3): $f(k) = -f(k) + 2k + 4$, so $f(k) = k + 2$ is odd, which is a contradiction.\n- If $f(k)$ is odd, then $2z + k + f(k)$ is even and $f(2z + k + f(k)) = 2z + k + f(k) + 2$. Recalling (3), $f(2z + k + f(k)) = 2z + k + f(k) + 2 = 2z + 2k + 4$, so $f(k) = k + 2$ for all $k \\in \\mathbb{Z}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 13620,
    "subject": "Mathematics (Olympiad)",
    "question": "n-ээс $3n$ хүртэл дугаарласан билетүүдийг \"багц\" гэж нэрлэе. Хэрэв бүхэл тооны куб болдог дугаартай билет хонжвортой бол, $n \\geq 3$ байх багц бүрд хонжвор байгаа гэдгийг батал.",
    "options": [],
    "answer": "See solution",
    "solution": "Бид $n \\leq k^3 \\leq 3n$ байх $k$ бүхэл тоо олдвол тухайн багцад хонжвортой билет байна гэж үзнэ. Дараалсан дугааруудын дунд $k^3$ байвал хонжвортой болох нь тодорхой.\n\nИймд $n = k^3 + 1$ гэж авч үзье. $k^3 + 1 \\leq (k+1)^3 \\leq 3(k^3 + 1)$ гэдгийг харуулъя. Тэнцэтгэлийн зүүн тал нь илэрхий.\n\nБаруун талыг авч үзвэл:\n\n$$(k + 1)^3 \\leq 3(k^3 + 1)$$\n$$(k + 1)^3 \\leq 3(k + 1)(k^2 - k + 1)$$\n$$(k^2 + 2k + 1) \\leq 3k^2 - 3k + 3$$\n$$(2k^2 - 5k + 2) \\geq 0$$\n$$(2k - 1)(k - 2) \\geq 0$$\n\nЭнэ нь $k \\geq 2$ үед биелнэ.\n\nИймд $k^3 + 1 = 9$ буюу $n \\geq 9$ үед багц бүр хонжвортой болох нь батлагдлаа. $n = 3, 4, 5, 6, 7, 8$ үед хонжвортой болохыг хялбархан шалгаж болно.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 13621,
    "subject": "Mathematics (Olympiad)",
    "question": "A figure composed of congruent squares is enclosed in a rectangle as displayed. The rectangle has horizontal side 73 and vertical side 94. Find the side of a square.\n\n![](images/Argentina_2018_p2_data_db1f4898e5.png)",
    "options": [],
    "answer": "See solution",
    "solution": "The sides of the squares run in two perpendicular directions, which we denote 1 and 2. Project orthogonally a side of direction 1 onto horizontal and vertical lines. Let the projections have lengths $x$ and $y$ respectively. Then the projections of a side of direction 2 onto the same lines are $y$ and $x$ respectively. This is because the right triangles shaded in the figure are congruent. In addition, the side of the square equals $\\sqrt{x^2 + y^2}$ by Pythagoras' theorem.\n\nConsider the sides denoted $h_1$ and $h_2$ in the second figure. The indices 1 and 2 indicate their directions. Project all of them onto a horizontal line of the rectangle. Observe that the projections cover this side without overlaps; hence the total length is 73. There are 4 sides $h_1$ and 5 sides $h_2$, so we obtain $4x + 5y = 73$. Analogously, look at the sides denoted $v_1, v_2$. Their projections onto a vertical side of the rectangle cover this side without overlaps. Since there are 7 sides $v_1$ and 2 sides $v_2$, it follows that $7x + 2y = 94$.\n\n![](images/Argentina_2018_p2_data_d170aef791.png)\n\nThe system\n$$\n\\begin{cases}\n4x + 5y = 73 \\\\\n7x + 2y = 94\n\\end{cases}\n$$\nhas a unique solution $x = 12$, $y = 5$. Therefore, each square has side $\\sqrt{12^2 + 5^2} = 13$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 13622,
    "subject": "Mathematics (Olympiad)",
    "question": "The bisector of the angle $\\angle ABC$ of triangle $ABC$ intersects the circumcircle of triangle $ABC$ at point $K$. Point $N$ belongs to segment $AB$ and $NK \\perp AB$. Let $P$ be the midpoint of segment $NB$. Consider the line through $P$ that is parallel to $BC$ and intersects line $BK$ at point $T$. Prove that the line $NT$ passes through the midpoint of segment $AC$.\n\n![](images/Ukraine_2016_Booklet_p24_data_8eb0198176.png)",
    "options": [],
    "answer": "See solution",
    "solution": "Let $M = NT \\cap AC$. Since $BK$ is the bisector of $\\angle ABC$, $\\angle KBC = \\angle KBA = \\alpha$. Since $PT$ is parallel to $BC$, $\\angle KBC = \\angle PTB$. Thus, $PT = PB = PN$. It follows that $\\triangle BNT$ is right-angled with hypotenuse $BN$. Since $\\triangle BNK$ is also right-angled, $\\angle KNT = \\alpha$. Moreover, $\\angle KAC = \\angle KBC = \\alpha$. Thus, quadrilateral $KANM$ is cyclic. Since $\\angle ANK = 90^\\circ$, $AK$ is a diameter of the circumcircle of $KANM$. Hence, $\\angle AMK = 90^\\circ$. Since $\\triangle AKC$ is isosceles, the altitude $KM$ is also a median. Hence, $AM = MC$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 13623,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $n$ be a positive integer. Suppose we are given $2^n + 1$ distinct sets, each containing finitely many objects. Place each set into one of two categories, the red sets and the blue sets, so that there is at least one set in each category.\n\nWe define the *symmetric difference* of two sets as the set of objects belonging to exactly one of the two sets. Prove that there are at least $2^n$ different sets which can be obtained as the symmetric difference of a red set and a blue set.",
    "options": [],
    "answer": "See solution",
    "solution": "We proceed by induction on $n$.\n\nFor $n=0$, there are two sets, one red and one blue, and thus one symmetric difference, as desired.\n\nNow suppose we know the problem is true for $n-1$ and consider it for $n$.\n\nCall an element *a diverse* if there is a red set containing $a$, a red set not containing $a$, a blue set containing $a$, and a blue set not containing $a$. We consider two cases based on whether a diverse element exists. We use $A\\Delta B$ to denote the symmetric difference of sets $A$ and $B$.\n\n**Case 1:** Suppose no element $a$ is diverse. We now show that if $R, R'$ are two distinct red sets and $B, B'$ are two distinct blue sets, then $R\\Delta B \\neq R'\\Delta B'$. Since $R, R'$ are distinct, there is an element $s$ in one and not the other. Since $s$ is not diverse, $s$ is in both $B, B'$ or in neither of $B, B'$. Either way, $s$ will lie in exactly one of $R\\Delta B$ and $R'\\Delta B'$, so the two cannot be equal. Thus, all of the symmetric differences of red sets and blue sets are distinct. If there are $r$ red sets and $b$ blue sets, then $r + b = 2^n + 1$, so there are at least $rb \\geq 2^n \\cdot 1 = 2^n$ symmetric differences.\n\n**Case 2:** Suppose there is an element $a$ that is diverse. Let $R_0$ be the set of red sets that do not contain $a$, $R_1$ be the set of red sets that do contain $a$, $B_0$ the set of blue sets that do not contain $a$, and $B_1$ the set of blue sets that do contain $a$. By the diversity of $a$, all four of these sets are non-empty.\n\nBy the pigeonhole principle, one of $R_0 \\cup B_0$ or $R_1 \\cup B_1$ has at least $2^{n-1} + 1$ sets. Without loss of generality, let it be $R_0 \\cup B_0$. Apply the inductive hypothesis to obtain a set $A_0$ of at least $2^{n-1}$ distinct sets that are the result of symmetric differences between $R_0$ and $B_0$. Likewise, by the pigeonhole principle, one of $R_0 \\cup B_1$ or $R_1 \\cup B_0$ has at least $2^{n-1} + 1$ sets. Without loss of generality, let it be $R_0 \\cup B_1$. Apply the inductive hypothesis to obtain a set $A_1$ of $2^{n-1}$ distinct sets that are the result of symmetric differences between $R_0$ and $B_1$.\n\nNotice that all of the sets in $A_0$ do not contain $a$ while all of the sets in $A_1$ do contain $a$. Therefore, $A_0$ and $A_1$ together contain $2^n$ distinct sets obtained from the symmetric difference of red sets and blue sets, completing the induction.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 13624,
    "subject": "Mathematics (Olympiad)",
    "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": "See solution",
    "solution": "Denote the area of any triangle $\\triangle \\Delta$ by $S_{\\triangle \\Delta}$. Let $D'$ be the reflection of $D$ from $AC$ and $\\alpha = \\angle APB$ (see the figure below).\n\n$$\nS_{PAB} = \\frac{1}{2} \\cdot PA \\cdot PB \\cdot \\sin \\alpha\n$$\n\n$$\nS_{PBC} = \\frac{1}{2} \\cdot PB \\cdot PC \\cdot \\sin(180^\\circ - \\alpha) = \\frac{1}{2} \\cdot PB \\cdot PC \\cdot \\sin \\alpha\n$$\n\n$$\nS_{PCD} = \\frac{1}{2} \\cdot PC \\cdot PD \\cdot \\sin \\alpha\n$$\n\n$$\nS_{PDA} = \\frac{1}{2} \\cdot PD \\cdot PA \\cdot \\sin(180^\\circ - \\alpha) = \\frac{1}{2} \\cdot PD \\cdot PA \\cdot \\sin \\alpha\n$$\n\nAs the bisector of the angle $DAB$ passes through $I$ that lies inside the triangle $PAB$, we have $\\angle DAC < \\angle DAI = \\angle BAI < \\angle BAC$, implying\n\n![](images/EST_ABooklet_2020_p24_data_31548c2d3f.png)\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 the expressions for $S_{PCD}$ and $S_{PDA}$ gives\n\n$$\nS_{PCD} + S_{PDA} = \\frac{1}{2} (PC + PA) \\cdot PD \\cdot \\sin \\alpha,\n$$\n\ni.e.,\n\n$$\nS_{ACD} = \\frac{1}{2} AC \\cdot PD \\cdot \\sin \\alpha.\n$$\n\nAnalogously, from the expressions for $S_{PAB}$ and $S_{PBC}$ we obtain\n\n$$\nS_{ABC} = \\frac{1}{2} AC \\cdot PB \\cdot \\sin \\alpha.\n$$\n\nHence 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, comparing the area expressions:\n\n- $S_{PAB} > S_{PBC}$\n- $S_{PAB} > S_{PDA}$\n- $S_{PBC} > S_{PCD}$\n- $S_{PDA} > S_{PCD}$\n\nHence the desired claim follows.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 13625,
    "subject": "Mathematics (Olympiad)",
    "question": "試決定滿足下述性質的所有正整數 $n \\ge 3$ ：對每一個各邊長皆為 $1$ 的凸 $n$ 邊形，都可以在其圍成的區域中放入一個邊長為 $1$ 的正三角形。\n\n（註：一個凸多邊形所圍成的區域，係指其內部及其邊。）",
    "options": [],
    "answer": "See solution",
    "solution": "所有奇數 $n \\ge 3$ 均滿足條件。\n\n首先，對於每個偶數 $n \\ge 4$，存在一個 $n$ 邊形不滿足題意。考慮一個邊長為 $1$ 的正 $k$ 邊形 $A_0A_1\\dots A_{k-1}$。令 $B_1, B_2, \\dots, B_{n/2-1}$ 為 $A_1, A_2, \\dots, A_{n/2-1}$ 關於直線 $A_0A_{n/2}$ 的對稱點。則 $\\mathcal{P} = A_0A_1A_2\\dots A_{n/2-1}A_{n/2}B_{n/2-1}B_{n/2-2}\\dots B_2B_1$ 是一個所有邊長皆為 $1$ 的凸 $n$ 邊形。若 $k$ 夠大，$\\mathcal{P}$ 可被包含在寬度 $1/2$ 的條帶中，顯然無法容納邊長為 $1$ 的正三角形。\n\n![](images/2022-TWNIMO-Problems_p103_data_4fd7789398.png)\n\n現在假設 $n = 2k + 1$。當 $k = 1$ 時顯然成立，以下假設 $k \\ge 2$。考慮一個所有邊長皆為 $1$ 的凸 $(2k+1)$ 邊形 $\\mathcal{P}$。令 $d$ 為其最長對角線。$d$ 的端點將 $\\mathcal{P}$ 的周長分成兩段，其中一段長度至少為 $k+1$。因此可標記 $\\mathcal{P} = A_0A_1\\dots A_{2k}$ 且 $d = A_0A_\\ell$，其中 $\\ell \\ge k+1$。我們將證明多邊形 $A_0A_1\\dots A_\\ell$ 內必定可容納一個邊長為 $1$ 的正三角形。\n\n假設 $\\angle A_\\ell A_0 A_1 \\ge 60^\\circ$。由於 $d$ 為最長對角線，$A_1A_\\ell \\le A_0A_\\ell$，所以 $\\angle A_0A_1A_\\ell \\ge \\angle A_\\ell A_0A_1 \\ge 60^\\circ$。因此存在一點 $X$ 在三角形 $A_0A_1A_\\ell$ 內，使得三角形 $A_0A_1X$ 為正三角形且包含於 $\\mathcal{P}$。若 $\\angle A_{\\ell-1}A_\\ell A_0 \\ge 60^\\circ$ 亦同理。\n\n![](images/2022-TWNIMO-Problems_p103_data_124426fa47.png)\n\n接下來假設 $\\angle A_\\ell A_0A_1 < 60^\\circ$ 且 $\\angle A_{\\ell-1}A_\\ell A_0 < 60^\\circ$。\n\n考慮等腰梯形 $A_0YZA_\\ell$，使得 $A_0A_\\ell \\parallel YZ$，$A_0Y = ZA_\\ell = 1$，且 $\\angle A_\\ell A_0Y = \\angle ZA_\\ell A_0 = 60^\\circ$。假設 $A_0A_1\\dots A_\\ell$ 被包含在 $A_0YZA_\\ell$ 內。注意 $A_0A_1\\dots A_\\ell$ 的周長為 $\\ell + A_0A_\\ell$，而 $A_0YZA_\\ell$ 的周長為 $2A_0A_\\ell + 1$。\n\n![](images/2022-TWNIMO-Problems_p104_data_cd41720c9e.png)\n\n根據一個常見事實，若凸多邊形 $\\mathcal{P}_1$ 被包含於凸多邊形 $\\mathcal{P}_2$，則 $\\mathcal{P}_1$ 的周長不超過 $\\mathcal{P}_2$。因此有：\n\n$$\n\\ell + A_0A_\\ell \\le 2A_0A_\\ell + 1, \\quad \\text{即 } \\ell - 1 \\le A_0A_\\ell.\n$$\n\n另一方面，三角不等式給出：\n\n$$\nA_0A_\\ell < A_\\ell A_{\\ell+1} + A_{\\ell+1}A_{\\ell+2} + \\dots + A_{2k}A_0 = 2k + 1 - \\ell \\le \\ell - 1.\n$$\n\n這產生矛盾。\n\n因此，存在 $A_0A_1\\dots A_\\ell$ 的某個頂點 $A_m$ 位於 $A_0YZA_\\ell$ 之外。由於\n\n$$\n\\angle A_\\ell A_0A_1 < 60^\\circ = \\angle A_\\ell A_0Y \\quad \\text{且} \\quad \\angle A_{\\ell-1}A_\\ell A_0 < 60^\\circ = \\angle ZA_\\ell A_0,\n$$\n\n$A_m$ 到 $A_0A_\\ell$ 的距離至少為 $\\sqrt{3}/2$。\n\n設 $P$ 為 $A_m$ 到 $A_0A_\\ell$ 的垂足，則 $PA_m \\ge \\sqrt{3}/2$，且 $A_0P > 1/2$、$PA_\\ell > 1/2$。在 $A_0P$、$PA_\\ell$、$PA_m$ 上分別取 $PQ = PR = 1/2$、$PS = \\sqrt{3}/2$，則 $QRS$ 為邊長 $1$ 的正三角形，且包含於 $A_0A_1\\dots A_\\ell$ 內。\n\n![](images/2022-TWNIMO-Problems_p104_data_823f97c4e7.png)",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 13626,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $ABCD$ be a rectangle with center $O$, such that $\\angle DAC = 60^\\circ$. The angle bisector of $\\angle DAC$ meets $DC$ at $S$. Lines $OS$ and $AD$ meet at $L$, and lines $BL$ and $AC$ meet at $M$. Prove that lines $SM$ and $CL$ are parallel.",
    "options": [],
    "answer": "See solution",
    "solution": "We have $\\angle SAC = \\angle SCA = 30^\\circ$, so $SA = SC$. Since $OA = OC$, we get $SO \\perp AC$, so $LA = LC$. Moreover, $\\angle LAC = 60^\\circ$, so triangle $ABC$ is equilateral. Point $O$ is the barycenter of triangle $LAC$, thus $\\frac{LS}{SO} = 2$. $DBCL$ is a parallelogram. Denoting by $Q$ its center, we get $DQ = QC$. Since $M$ is the barycenter of $DBC$, we have $\\frac{CM}{MO} = 2$.\n\nFinally, since $\\frac{LS}{SO} = \\frac{CM}{MO}$, by Thales' theorem we obtain that *SM* and *CL* are parallel.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 13627,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $(i, j)$ denote the unit square in the $i$-th row and $j$-th column of a grid. For two unit squares in the same row, the one with smaller $j$ is to the left; for two in the same column, the one with smaller $i$ is above.\n\nA coloring of the $m \\times n$ grid is sought such that:\n- Each square is colored red, blue, or black.\n- No two adjacent squares (sharing a side) have the same color.\n- Each square is adjacent to squares of the other two colors (i.e., every red square is adjacent to at least one blue and one black square, etc.).\n\nDetermine all possible values of $m$ and $n$ for which such a coloring exists, and describe the coloring scheme.",
    "options": [],
    "answer": "See solution",
    "solution": "We analyze small cases and generalize:\n\n- For $n=1$ or $m=1$, such a coloring is impossible, as each square cannot be adjacent to two other colors.\n- For $m \\geq 2$, $n \\geq 2$, start by coloring $(1,1)$ red, $(1,2)$ blue, $(2,1)$ black. The color of $(2,2)$ must be blue or black; proceed by casework.\n- Continuing, we find that the coloring pattern repeats every 3 rows and every 2 columns. Thus, $m$ must be a multiple of 3 and $n$ a multiple of 2 (or vice versa).\n- The coloring is constructed by repeating the $3 \\times 2$ pattern throughout the grid, alternating colors as described.\n\nTherefore, a coloring exists if and only if $m$ is a multiple of 3 and $n$ a multiple of 2, or $m$ a multiple of 2 and $n$ a multiple of 3. The coloring is obtained by tiling the $3 \\times 2$ pattern across the grid.",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 13628,
    "subject": "Mathematics (Olympiad)",
    "question": "Suppose $n$ is a positive integer. Prove that if $n$ has the prime factorization $n = p_1^{e_1} p_2^{e_2} \\cdots p_m^{e_m}$, where $p_1, p_2, \\ldots, p_m$ are distinct primes and $e_1, e_2, \\ldots, e_m$ are positive integers, then the total number of factors of $n$ is $(e_1 + 1)(e_2 + 1) \\cdots (e_m + 1)$.\n\nNow, find the smallest positive integer $n$ such that the number of factors of $14n$, $16n$, $18n$, and $20n$ are all equal.",
    "options": [],
    "answer": "See solution",
    "solution": "If we make the value of $a$ bigger, $b, c, d$ also get bigger, so if $n$ is the smallest possible integer satisfying the requirement, then $a$ also has to be the smallest non-negative integer satisfying the conditions above. By substituting $a = 0, 1, \\dots$, and checking to see when all of $b, c, d$ become non-negative integers, we see that $a = 1$ is the smallest value for $a$ for which all of $b, c, d$ are non-negative integers and their values are $b = 1, c = 1, d = 0$. Thus the answer to the problem is $n = 2^1 \\cdot 3^1 \\cdot 5^1 \\cdot 7^0 = 30$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 13629,
    "subject": "Mathematics (Olympiad)",
    "question": "Given a real number $a$ and a sequence $(u_n)$ defined by\n\n$$\nu_1 = a,$$\n$$\nu_{n+1} = \\frac{1}{2} + \\sqrt{\\frac{2n+3}{n+1}u_n + \\frac{1}{4}}$$\nfor all positive integers $n$.\n\na) For $a = 5$, prove that $(u_n)$ converges and find its limit.\n\nb) Find all values of $a$ such that the sequence $(u_n)$ exists and has finite limit.",
    "options": [],
    "answer": "See solution",
    "solution": "We will solve part b) to deduce the result in part a).\n\nClearly, $(u_n)$ is defined if and only if $u_2$ is defined. Since\n\n$$\nu_2 = \\frac{1}{2} + \\sqrt{\\frac{5}{2}a + \\frac{1}{4}},$$\n\nthen $u_2$ is defined if and only if $a \\ge -\\frac{1}{10}$.\n\nWe will prove this is also the sufficient condition. It is easy to see that $u_n \\ge \\frac{1}{2}$ for all $n \\ge 2$. Consider the function $f(x) = \\frac{2x+3}{x+1}$, which is strictly decreasing over $\\mathbb{R}^+$. So for every positive integer $n$, we have\n\n$$\n\\frac{2n+3}{n+1} > \\frac{2(n+1)+3}{(n+1)+1}.\n$$\n\nIf there exists $n_0 \\in \\mathbb{N}$ such that $u_{n_0} \\ge u_{n_0+1}$, then\n\n$$\n\\begin{align*}\n\nu_{n_0+2} &= \\frac{1}{2} + \\sqrt{\\frac{2(n_0+1)+3}{(n_0+1)+1} u_{n_0+1} + \\frac{1}{4}} \\\\\n&\\le \\frac{1}{2} + \\sqrt{\\frac{2n_0+3}{n_0+1} u_{n_0} + \\frac{1}{4}} = u_{n_0+1}.\n\\end{align*}\n$$\n\nSimilarly, $u_{n_0} \\ge u_{n_0+1} \\ge u_{n_0+2} \\ge u_{n_0+3} \\ge \\dots$ This means that the sequence $(u_n)$ does not increase from $n_0$. Note that the sequence is bounded below, so $(u_n)$ has a finite limit $L$. Letting $n$ tend to infinity, we obtain\n\n$$\nL = \\frac{1}{2} + \\sqrt{2L + \\frac{1}{4}}\n$$\n\nTherefore $L = 3$. This implies that if there exists $n_0$ as above, then $\\lim u_n = 3$.\n\nOtherwise, there does not exist $n_0$ such that $u_{n_0} \\ge u_{n_0+1}$, which means $(u_n)$ is strictly decreasing. It suffices to show that $(u_n)$ is bounded. Indeed, since $u_{n+1} > u_n$ and $\\frac{2n+3}{n+1} < 3$ for every positive integer $n$, it follows that\n\n$$\n\\frac{1}{2} + \\sqrt{3u_n + \\frac{1}{4}} > \\frac{1}{2} + \\sqrt{\\frac{2n+3}{n+1}u_n + \\frac{1}{4}} = u_{n+1} > u_n\n$$\n\nThis inequality is equivalent to $u_n < \\frac{4+\\sqrt{17}}{2}$ for every $n \\ge 2$. Therefore, the sequence $(u_n)$ is strictly decreasing and has an upper bound, hence it has a finite limit. Thus, by letting the given equation tend to the limit, we also have $\\lim u_n = 3$.\n\nIn summary, for every $a \\ge -\\frac{1}{10}$, the sequence $(u_n)$ is defined and has limit $3$. $\\square$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 13630,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $u : [a, b] \\to \\mathbb{R}$ be a continuous function which has at each point $x \\in (a, b]$ a finite left derivative, denoted by $u'_s(x)$. Prove that the function $u$ is increasing if and only if $u'_s(x) \\ge 0$ for all $x \\in (a, b]$.",
    "options": [],
    "answer": "See solution",
    "solution": "We use the following results, generalizations of Rolle's and Lagrange's theorems.\n\n**Lemma 1.** If $u : [\\alpha, \\beta] \\to \\mathbb{R}$ has finite left and right derivatives at every point in $[\\alpha, \\beta]$ and $u(\\alpha) = u(\\beta)$, then there exists $c \\in [\\alpha, \\beta]$ such that $u'_s(c) \\le 0$.\n\n*Proof.* Since $u$ is continuous on $[\\alpha, \\beta]$, it attains a global minimum at some $c$ in $[\\alpha, \\beta]$.\n\n**Lemma 2.** If $u : [\\alpha, \\beta] \\to \\mathbb{R}$ has finite left and right derivatives at each point in $[\\alpha, \\beta]$, then there exists $c \\in [\\alpha, \\beta]$ such that\n$$\nu(\\beta) - u(\\alpha) \\ge (\\beta - \\alpha)u'_s(c).$$\n\n*Proof.* Define $v(x) = u(x) - \\frac{u(\\beta) - u(\\alpha)}{\\beta - \\alpha}x$. The function $v$ satisfies Lemma 1, so there is $c \\in [\\alpha, \\beta]$ with $v'_s(c) \\le 0$.\n\nReturning to the problem, let $a \\le x < y \\le b$. By Lemma 1, there exists $c \\in [x, y]$ such that\n$$u(y) - u(x) \\ge (y - x)u'_s(c),$$\nwhich implies $u$ is increasing if $u'_s(c) \\ge 0$.\n\nThe converse follows directly from the definition of the left derivative.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 13631,
    "subject": "Mathematics (Olympiad)",
    "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\n$$\n\\frac{1}{a_i} + \\frac{1}{a_{i+1}} + \\dots + \\frac{1}{a_j}\n$$\nis a fractional number.",
    "options": [],
    "answer": "See 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 $i \\leq k < l \\leq j$ 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 largest 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": 13632,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $x_1, x_2, x_3$ and $y_1, y_2, y_3$ be six positive real numbers such that\n\n$$\nx_1 + x_2 + x_3 = y_1 y_2 y_3 \\quad \\text{and} \\quad y_1 + y_2 + y_3 = x_1 x_2 x_3.\n$$\n\nFind the minimum value of $T = x_1 y_1 + x_2 y_2 + x_3 y_3$.",
    "options": [],
    "answer": "See solution",
    "solution": "Using AM-GM, we have\n\n$$\nx_1 + x_2 + x_3 \\ge 3 \\sqrt[3]{x_1 x_2 x_3} \\implies y_1 y_2 y_3 \\ge 3 \\sqrt[3]{x_1 x_2 x_3}.\n$$\n\nSimilarly, $x_1 x_2 x_3 \\ge 3 \\sqrt[3]{y_1 y_2 y_3}$. Multiplying these inequalities, we get\n\n$$\n\\sqrt[3]{(x_1 x_2 x_3 \\cdot y_1 y_2 y_3)^2} \\ge 9 \\implies x_1 x_2 x_3 \\cdot y_1 y_2 y_3 \\ge 27.\n$$\n\nNow applying AM-GM to the given expression:\n\n$$\nT \\ge 3 \\sqrt[3]{x_1 x_2 x_3 \\cdot y_1 y_2 y_3} \\ge 3 \\sqrt[3]{27} = 9.\n$$\n\nHence, the minimum value is $9$, and equality occurs when\n\n$$\nx_1 = x_2 = x_3 = y_1 = y_2 = y_3 = \\sqrt{3}.\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 13633,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $ABC$ be a triangle with $AB < AC < BC$. On the side $BC$ we consider points $D$ and $E$ such that $BA = BD$ and $CE = CA$. Let $K$ be the circumcenter of triangle $ADE$ and let $F, G$ be the points of intersection of the lines $AD, KC$ and $AE, KB$ respectively. Let $\\omega_1$ be the circumcircle of triangle $KDE$, $\\omega_2$ the circle with center $F$ and radius $FE$, and $\\omega_3$ the circle with center $G$ and radius $GD$.\n\nProve that $\\omega_1$, $\\omega_2$ and $\\omega_3$ pass through the same point and that this point of intersection lies on the line $AK$.",
    "options": [],
    "answer": "See solution",
    "solution": "Since the triangles $BAD$, $KAD$ and $KDE$ are isosceles, then $\\angle BAD = \\angle BDA$ and $\\angle KAD = \\angle KDA$ and $\\angle KDE = \\angle KED$. Therefore,\n\n$$\n\\angle BAK = \\angle BAD - \\angle KAD = \\angle BDA - \\angle KDA = \\angle KDE = \\angle KED = 180^\\circ - \\angle BEK.\n$$\n\nSo the points $B, E, K, A$ are concyclic. Similarly, the points $C, D, K, A$ are also concyclic.\n\n![](images/Bmo_Shortlist_2021_p32_data_d4dd00dfaf.png)\n\nLet $M, N$ be the midpoints of $AD$ and $AE$ respectively. Since the triangle $ACE$ is isosceles, the perpendicular bisector of $AE$, say $\\varepsilon_1$, passes through the points $C, K$ and $N$. Similarly, the perpendicular bisector of $AD$, say $\\varepsilon_2$, passes through the points $B, K$ and $M$. Therefore, the points $F, G$ lie on $\\varepsilon_1$ and $\\varepsilon_2$ respectively. Thus, using also the fact that $AKDC$ is a cyclic quadrilateral, we get that\n\n$$\n\\angle FDC = \\angle ADC = \\angle AKC = \\angle EKC = \\angle EKF.\n$$\n\nSo the point $F$ lies on the circle $\\omega_1$. Similarly, $G$ also lies on $\\omega_1$.\n\nLet $I$ be the point of intersection of the line $AK$ with $\\omega_1$. The triangles $AKF$ and $EKF$ are equal, so $\\angle KAF = \\angle KEF$. Since also $K, E, F, I$ all belong on $\\omega_1$, then\n\n$$\n\\angle KAF = \\angle KEF = \\angle FIK.\n$$\n\nIt follows that $FI = FA = FE$. Therefore, $I$ lies on $\\omega_2$ as well. Similarly, it also lies on $\\omega_3$. So the circles $\\omega_1$, $\\omega_2$, $\\omega_3$ all pass through $I$ which lies on line $AK$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 13634,
    "subject": "Mathematics (Olympiad)",
    "question": "The isosceles triangle $ABC$ has $\\angle BAC = 30^\\circ$ and $AB = AC$. Take the point $D$ on the side $AC$ and the distinct points $E$, $F$, $G$ on the side $AB$ so that $BC = BD = DE = EF$ and $DG = DF$.\n\n(a) Prove that $BF = GE$.\n\n(b) Find the measure of $\\angle BCG$.\n\n![](images/RMC_2025_p11_data_7fa871d50f.png)",
    "options": [],
    "answer": "See solution",
    "solution": "a) From the isosceles triangle $DFG$, it follows that $\\angle DFG = \\angle DGF$, hence $\\angle DFB = \\angle DGE$. From the isosceles triangle $BDE$, it follows that $\\angle DBF = \\angle DEG$.\n\nThe above and $BD = DE$ yield $\\triangle BDF \\equiv \\triangle EDG$ (S.A.A.), whence $BF = GE$.\n\nb) Since $\\triangle ABC$ is isosceles, $\\angle ABC = \\angle ACB = \\frac{180^\\circ - \\angle BAC}{2} = \\frac{150^\\circ}{2} = 75^\\circ$. Then $BF = GE$ gives $BG = FE = BC$, hence $\\triangle BCG$ is isosceles.\n\nSo $\\angle BCG = \\angle BGC = \\frac{180^\\circ - \\angle CBG}{2} = \\frac{105^\\circ}{2} = 52^\\circ 30'.$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 13635,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $a$, $b$, and $c$ be integers. Prove that $a^2 + b^2 + c^2$ is divisible by $4$ if and only if $a$, $b$, and $c$ are all even.",
    "options": [],
    "answer": "See solution",
    "solution": "First, suppose $a$, $b$, and $c$ are even: $a = 2a_1$, $b = 2b_1$, $c = 2c_1$. Then\n\n$$\na^2 + b^2 + c^2 = (2a_1)^2 + (2b_1)^2 + (2c_1)^2 = 4(a_1^2 + b_1^2 + c_1^2)\n$$\n\nwhich is divisible by $4$.\n\nConversely, assume $a^2 + b^2 + c^2$ is divisible by $4$. If exactly one or all three of $a$, $b$, $c$ are odd, then $a^2 + b^2 + c^2$ is odd, which is not possible. If exactly two are odd (say $a$ and $b$), let $a = 2a_1 - 1$, $b = 2b_1 - 1$, $c = 2c_1$:\n\n$$\na^2 + b^2 + c^2 = (2a_1 - 1)^2 + (2b_1 - 1)^2 + (2c_1)^2 = 4(a_1^2 - a_1 + b_1^2 - b_1 + c_1^2) - 2\n$$\n\nwhich is not divisible by $4$. Thus, all three must be even.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 13636,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $x$ be a real number. Prove that $x$ is an integer if and only if\n\n$$\n[x] + [2x] + [3x] + \\dots + [nx] = \\frac{n([x] + [nx])}{2}\n$$\n\nholds for all positive integers $n$ (here, $[a]$ denotes the integer part (floor function) of the real number $a$).",
    "options": [],
    "answer": "See solution",
    "solution": "If $x \\in \\mathbb{Z}$, then $[kx] = k[x]$ for every $k \\in \\mathbb{N}^*$.\n\nFor the converse, notice that the hypothesis implies, for all $n$,\n\n$$\nn([x] + [nx]) + 2[(n+1)x] = (n+1)([x] + [(n+1)x]).\n$$\n\nThis comes to $n[nx] = [x] + (n-1)[(n+1)x]$, $\\forall n \\ge 1$. Replacing $n$ with $n+1$ and subtracting the two relations, we obtain that the sequence $a_n = [nx]$ is an arithmetic progression. Since $(nx)_{n \\ge 1}$ is also an arithmetic progression, the difference sequence $\\{nx\\}_{n \\ge 1}$ is an arithmetic progression. Since $\\{nx\\}_{n \\ge 1}$ is bounded, its ratio must be 0, whence the conclusion.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 13637,
    "subject": "Mathematics (Olympiad)",
    "question": "A trapezoid $ABCD$ is given such that $\\overline{AB} = \\overline{AC} = \\overline{BD}$. Let $M$ be the midpoint of $CD$. Find the angles of the trapezoid if $\\angle MBC = \\angle CAB$.",
    "options": [],
    "answer": "See solution",
    "solution": "By the conditions of the task, it follows that the trapezoid is isosceles. Let $K$ be the midpoint of $AD$, and let $\\angle CAB = \\angle MBC = \\varphi$.\n\n$$\n\\angle MKA = 180^\\circ - \\angle KAC = 180^\\circ - \\angle MBA.\n$$\n\nTherefore, the quadrilateral $ABMK$ is inscribed. Then, by the conditions, we have that $\\triangle ABD$ is isosceles, from which we get $\\angle AKB = 90^\\circ$.\n\nNow, because $ABMK$ is inscribed, we have $\\angle AMB = \\angle AKB = 90^\\circ$, i.e., the triangle $\\triangle AMB$ is a right isosceles triangle.\n\nLet $M_1$ be the foot of the altitude from $M$. Then\n\n$$\n\\overline{MM_1} = \\overline{AM_1} = \\frac{\\overline{AB}}{2} = \\frac{\\overline{AC}}{2},\n$$\n\nso we get that $\\varphi = 30^\\circ$.\n\nNow we easily get $\\angle ABC = 30^\\circ + 45^\\circ = 75^\\circ$ and $\\angle ADC = 105^\\circ$.\n\n![](images/Macedonia_2010_booklet_p26_data_9529914ec9.png)",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 13638,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $p > 7$ be a prime number and let $A \\subseteq \\{0, 1, \\dots, p-1\\}$ consist of at least $\\frac{p-1}{2}$ elements. Show that for each integer $r$, there are elements $a, b, c, d \\in A$ such that\n$$\nab - cd \\equiv r \\pmod{p}.$$",
    "options": [],
    "answer": "See solution",
    "solution": "Let $P$ be the set of possible products $ab$, for $a, b \\in A$. Clearly, $|P| \\geq |aA| \\geq \\frac{p-1}{2}$, for any $a \\in A$. If $|P| \\geq \\frac{p+1}{2}$, then $|r + P| \\geq \\frac{p+1}{2}$, too. Hence, $|P| + |r + P| \\geq p + 1 > p$, so, by the Pigeonhole Principle, $P$ and $r + P$ must have an element in common. In other words, there are $p_1, p_2$ with $p_1 \\equiv r + p_2 \\pmod{p}$ and hence $p_1 - p_2 \\equiv r \\pmod{p}$, which gives a solution of the desired shape from the definition of $P$.\n\nSo the only remaining case is that of $|P| = |A| = \\frac{p-1}{2}$.\n\nMultiplying all elements of $A$ with the same constant and reducing modulo $p$, if necessary, we may assume without loss of generality that $1 \\in A$. Then $A \\subseteq P$ and hence $A = P$. This means that the non-zero elements of $A$ form a group under multiplication.\n\nIf $0 \\in A$, then this group has size $\\frac{p-3}{2}$, which has to divide the group order $p-1$, and hence also has to divide $2 = p-1-2 \\cdot \\frac{p-3}{2}$. This is impossible for $p > 7$.\n\nConsequently, $0 \\notin A$ and the group has size $\\frac{p-1}{2}$ and hence is exactly the group of quadratic residues (here we use the existence of primitive roots implicitly).\n\nReplacing $r$ by $r+p$, if necessary, one may assume $r$ to be odd. Then put $b = d := 1 \\in A$, as well as\n$$a \\equiv \\left( \\frac{r+1}{2} \\right)^2 \\pmod{p} \\quad \\text{and}$$\n$$c \\equiv \\left( \\frac{r-1}{2} \\right)^2 \\pmod{p}.$$ \nThen $a, c \\in A$, too. This yields\n$$ad - bc \\equiv a - c \\equiv \\left(\\frac{r+1}{2}\\right)^2 - \\left(\\frac{r-1}{2}\\right)^2 \\equiv r \\pmod{p},$$\nas required.\n\n*Remark:* There could also be other elementary approaches avoiding even the basic group theory. Probably the result is also very far from being sharp and the $\\frac{p-1}{2}$ can be replaced by something even smaller. Determining the sharp bound (or even its order of magnitude) here is most likely a very difficult problem.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 13639,
    "subject": "Mathematics (Olympiad)",
    "question": "We have $n^2$ empty boxes, each with a square base. The height and width of each box are integers from $\\{1, 2, \\dots, n\\}$, and every two boxes differ in at least one of these two dimensions. One box fits into another 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?",
    "options": [],
    "answer": "See solution",
    "solution": "We will show that the required minimal number of shelves is $3n-2$.\n\nThis 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 $n \\ge 3$.\n\nWe identify boxes with points in an $n \\times n$ grid: a box of width $w$ and height $h$ corresponds to the point $(w, h)$.\n\nNo two boxes from the set $S = \\{(w,h): n \\le w+h \\le n+2\\}$ (the 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![](images/66th_Czech_and_Slovak_p3_data_70ae86d16c.png)\n\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.\n\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 generalized 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 the “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.\n\n![](images/66th_Czech_and_Slovak_p3_data_a9d532a174.png)",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 13640,
    "subject": "Mathematics (Olympiad)",
    "question": "Currently circulating coins in Japanese currency come in 6 different denominations: 500 yen, 100 yen, 50 yen, 10 yen, 5 yen, and 1 yen. Taro had one each of a 1000 yen note, a 100 yen coin, a 10 yen coin, and a 1 yen coin. He made a certain purchase from a merchant and handed him all he had and received some change. Assume that Taro made the payment in such a way that he would not receive in the change any coins of the denominations same as what he had had to begin with, nor a 1000 yen note. Furthermore, he chose the method of the payment so that he would end up with the minimum possible number of coins when he received his change. Assume also that the merchant gave Taro the change with the minimum possible number of coins. Count the case of no change as a possibility for the purchase price also. Under these conditions, how many distinct purchase prices are possible?",
    "options": [],
    "answer": "See solution",
    "solution": "There are 8 different possible purchase prices.\n\nTo begin with, we note that two or more of 500 yen coins, or of 50 yen coins, or of 5 yen coins could not be included in the change Taro received. This is because if two or more of any of these coins were in the change, two of them could be exchanged for one 1000 yen note, or one 100 yen coin, or one 10 yen coin to reduce the total number of coins in the change. Also, the change did not include any coins of the denominations Taro originally had (100 yen, 10 yen, 1 yen), nor a 1000 yen note. By considering all possible combinations of the allowed denominations (500 yen, 50 yen, 5 yen), and including the case where no change is given, we find there are 8 distinct purchase prices possible.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 13641,
    "subject": "Mathematics (Olympiad)",
    "question": "Three consecutive integers have squares that sum to $770$. What is the largest of these integers?",
    "options": [],
    "answer": "See solution",
    "solution": "If we call the three numbers $n-1$, $n$, $n+1$, then $$(n-1)^2 + n^2 + (n+1)^2 = 770$$ which simplifies to $$3n^2 + 2 = 770$$. Therefore, $$n^2 = \\frac{768}{3} = 256 = 16^2$$ so $n = 16$ (since $n > 0$), and the largest number is $n+1 = 17$. Without doing any algebra, it is clear that $n^2 \\approx \\frac{770}{3} = 256\\frac{2}{3}$, and the nearest perfect square is $256$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 13642,
    "subject": "Mathematics (Olympiad)",
    "question": "In a triangle $ABC$, the internal angle bisectors at vertices $B$ and $C$ intersect at point $I$ and intersect the sides $CA$ and $AB$ at points $E$ and $F$, respectively. Let $M$ and $N$ be the midpoints of segments $BI$ and $CI$, respectively. The line $FM$ intersects the external angle bisector at vertex $B$ of triangle $ABC$ at point $K$, and the line $EN$ intersects the external angle bisector at vertex $C$ of triangle $ABC$ at point $L$. Prove that the points $K$, $B$, $C$, and $L$ lie on the same circle.",
    "options": [],
    "answer": "See solution",
    "solution": "We have $\\angle KBE = 90^\\circ$ as $BE$ and $BK$ are the internal and external angle bisectors at the same vertex (see figure below).\n\nWe will now show $\\angle BKC = 90^\\circ$. Let $X$ be a point on line $AB$ such that $CX \\parallel BE$; then the external angle bisector at vertex $B$ of triangle $ABC$ is also perpendicular to line $CX$ (see figure below).\n\nDenote $\\angle ABC = \\beta$. Then $\\angle BCX = \\angle CBE = \\frac{\\beta}{2}$ and $\\angle XBC = 180^\\circ - \\beta$. Therefore,\n$$\n\\angle BXC = 180^\\circ - (\\angle BCX + \\angle XBC) = \\frac{\\beta}{2}\n$$\nso the triangle $XBC$ is isosceles with apex at $B$. Thus, the altitude from vertex $B$ of triangle $XBC$ bisects the base $CX$. As previously established, this altitude lies on the external angle bisector at vertex $B$ of triangle $ABC$.\n\nLet $k = \\frac{|FX|}{|FB|} = \\frac{|FC|}{|FI|}$. A homothety with center $F$ and ratio $k$ sends points $B$ and $I$ to points $X$ and $C$, respectively, so it sends the midpoint $M$ of segment $BI$ to the midpoint of segment $CX$. Therefore, the line $FM$ passes through the midpoint of segment $CX$. Consequently, the intersection point $K$ of line $FM$ and the external angle bisector at vertex $B$ of triangle $ABC$ lies at the midpoint of segment $CX$, and $\\angle BKC = 90^\\circ$, as we wanted to show.\n\nSimilarly, we can show that $\\angle BLC = 90^\\circ$. Therefore, points $K$ and $L$ lie on the circle with diameter $BC$. The statement of the problem follows.\n\n![](images/EST_ABooklet_2024_p50_data_c6e2048611.png)\n\n*Fig. 43*\n\n![](images/EST_ABooklet_2024_p50_data_5a24ef4af4.png)\n\n*Fig. 44*",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 13643,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $A$, $B$, $C$ be three distinct points on a unit circle. Let $G$ and $H$ be the centroid and the orthocenter of triangle $ABC$, respectively. Let $F$ be the midpoint of the segment $GH$. Evaluate\n\n$$\n|\\overline{AF}|^2 + |\\overline{BF}|^2 + |\\overline{CF}|^2.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "Define a coordinate system with the origin at the center of the circle. We can see that $\\overline{H} = \\overline{A} + \\overline{B} + \\overline{C}$ and $\\overline{G} = \\frac{1}{3}(\\overline{A} + \\overline{B} + \\overline{C})$.\n\nThus, $\\overline{F} = \\frac{\\overline{G} + \\overline{H}}{2} = \\frac{2}{3}(\\overline{A} + \\overline{B} + \\overline{C})$. We now have\n\n$$\n\\begin{aligned}\n& |\\overline{AF}|^2 + |\\overline{BF}|^2 + |\\overline{CF}|^2 \\\\\n&= (\\overline{A} - \\overline{F}) \\cdot (\\overline{A} - \\overline{F}) + (\\overline{B} - \\overline{F}) \\cdot (\\overline{B} - \\overline{F}) + (\\overline{C} - \\overline{F}) \\cdot (\\overline{C} - \\overline{F}) \\\\\n&= |\\overline{A}|^2 + |\\overline{B}|^2 + |\\overline{C}|^2 - 2(\\overline{A} + \\overline{B} + \\overline{C}) \\cdot \\overline{F} + 3\\overline{F} \\cdot \\overline{F} \\\\\n&= |\\overline{A}|^2 + |\\overline{B}|^2 + |\\overline{C}|^2 - (2(\\overline{A} + \\overline{B} + \\overline{C}) - 3\\overline{F}) \\cdot \\overline{F} \\\\\n&= |\\overline{A}|^2 + |\\overline{B}|^2 + |\\overline{C}|^2 = 3.\n\\end{aligned}\n$$\n\n![](images/Thailand-2007_Booklet_p4_data_645308afee.png)",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 13644,
    "subject": "Mathematics (Olympiad)",
    "question": "As shown in Fig. 2.1, $AB$ is a diameter of a circle with center $O$. Let $C$ and $D$ be two different points on the circle on the same side of $AB$, and the lines tangent to the circle at points $C$ and $D$ meet at $E$. Segments $AD$ and $BC$ meet at $F$. Lines $EF$ and $AB$ meet at $M$. Prove that $E, C, M$ and $D$ are concyclic.\n\n![](images/Mathematical_Olympiad_in_China_2011-2014_p235_data_fa8e80a2c8.png)",
    "options": [],
    "answer": "See solution",
    "solution": "**Solution 1.** As shown in Fig. 2.2, join $OC$, $OD$ and $OE$. It follows from $\\angle OCE + \\angle EDO = 90^\\circ + 90^\\circ = 180^\\circ$ that $ECOD$ is cyclic. If $O = M$, then $ECMD$ is cyclic. We may assume that $O \\neq M$ in the following. Let $G$ be the intersection of $BC$ and $AD$, and $H_1$ be the intersection of $GF$ and $AB$. Let $H_2$ be the foot of the perpendicular from $E$ to $AB$. It follows from $AC \\perp BG$ and $BD \\perp AG$ that $F$ is the orthocenter of $\\triangle BAG$, so $GH_2 \\perp AB$. As $\\angle CH_2D = \\angle CBF + \\angle DAF = 180^\\circ - 2\\angle BGA$, and $\\angle COD = 180^\\circ - \\angle BOC - \\angle AOD = 180^\\circ - (180^\\circ - 2\\angle GBA) - (180^\\circ - 2\\angle GAB) = 2(\\angle GBA + \\angle GAB) - 180^\\circ - 2\\angle BGA$, then $\\angle CH_2D = \\angle COD$, so $C, O, H_2$ and $D$ are concyclic. It follows from $\\angle ECO = \\angle EDO = \\angle EH_1O = 90^\\circ$ that $E, C, O, H_1$ and $D$ are concyclic. Hence, $H_1 = H_2$, i.e., $EF \\perp AB$ and they meet at $M$. Consequently, $E, C, M$ and $D$ are concyclic.\n\n![](images/Mathematical_Olympiad_in_China_2011-2014_p236_data_4b403ea78d.png)\n\n![](images/Mathematical_Olympiad_in_China_2011-2014_p236_data_75588a9a1a.png)",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 13645,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $2006$ be expressed as the sum of five positive integers $x_1, x_2, x_3, x_4, x_5$, and let\n\n$$\nS = \\sum_{1 \\le i < j \\le 5} x_i x_j.\n$$\n\n1. What values of $x_1, x_2, x_3, x_4, x_5$ will make $S$ the maximum?\n\n2. Further, if $|x_i - x_j| \\le 2$ for any $1 \\le i, j \\le 5$, what values of $x_1, x_2, x_3, x_4, x_5$ will make $S$ the minimum? Prove your answer.",
    "options": [],
    "answer": "See solution",
    "solution": "(1) The number of possible values of $S$ is finite, so the maximum and minimum exist. Suppose $x_1 + x_2 + x_3 + x_4 + x_5 = 2006$ and $S = \\sum_{1 \\le i < j \\le 5} x_i x_j$ reaches the maximum. We must have\n\n$$\n|x_i - x_j| \\le 1 \\quad (1 \\le i, j \\le 5).\n$$\n\nOtherwise, assume this does not hold. Without loss of generality, suppose $x_1 - x_2 \\ge 2$. Let $x_1' = x_1 - 1$, $x_2' = x_2 + 1$, $x_i' = x_i$ for $i = 3, 4, 5$. Then\n\n$$\nx_1' + x_2' + x_3' + x_4' + x_5' = x_1 + x_2 + x_3 + x_4 + x_5 = 2006.\n$$\n\nNow,\n\n$$\nS = x_1 x_2 + (x_1 + x_2)(x_3 + x_4 + x_5) + x_3 x_4 + x_3 x_5 + x_4 x_5,\n$$\n\nand\n\n$$\nS' = x_1' x_2' + (x_1' + x_2')(x_3 + x_4 + x_5) + x_3 x_4 + x_3 x_5 + x_4 x_5.\n$$\n\nSo\n\n$$\nS' - S = x_1' x_2' - x_1 x_2 > 0.\n$$\n\nThis contradicts the assumption that $S$ is the maximum.\n\nTherefore, $|x_i - x_j| \\le 1$ for all $i, j$. It is easy to check that $S$ reaches the maximum when\n\n$$\nx_1 = 402, \\quad x_2 = x_3 = x_4 = x_5 = 401.\n$$\n\n(2) If we neglect the order of $x_1, x_2, x_3, x_4, x_5$, there are only three cases:\n\n- $402, 402, 402, 400, 400$\n- $402, 402, 401, 401, 400$\n- $402, 401, 401, 401, 401$\n\nThese satisfy $x_1 + x_2 + x_3 + x_4 + x_5 = 2006$ and $|x_i - x_j| \\le 2$.\n\nCases (b) and (c) can be obtained from case (a) by setting $x_i' = x_i - 1$, $x_j' = x_j + 1$. As shown in (1), each such step increases $S$. So $S$ is minimized in case (a), i.e.,\n\n$$\nx_1 = x_2 = x_3 = 402, \\quad x_4 = x_5 = 400.\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 13646,
    "subject": "Mathematics (Olympiad)",
    "question": "A *nice prime* is a prime equal to the difference of two cubes of positive integers.\n\nFind the last digits of all nice primes.",
    "options": [],
    "answer": "See solution",
    "solution": "First, 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 are possible last digits.\n\nLet $p = m^3 - n^3$ be a nice prime, where $m > n$ are positive integers. Rewriting:\n\n$$\np = m^3 - n^3 = (m - n)(m^2 + mn + n^2)\n$$\n\nSince $p$ is prime, $(m - n)$ must be 1, so $m = n + 1$. Substituting:\n\n$$\np = 3n^2 + 3n + 1 \\qquad (1)\n$$\n\nFor $n \\geq 2$, $3n^2 + 3n + 1 > 6$, so $p$ is odd and greater than 5. Thus, last digits 0, 2, 4, 5, 6, and 8 are excluded. To check if 3 is possible, compute $3n^2 + 3n + 1 \\pmod{5}$:\n\nFor $n \\equiv 0,1,2,3,4 \\pmod{5}$, the remainders are 1, 2, 4, 2, 1, respectively. Thus, 3 is not possible.\n\n*Answer.* The last digits of nice primes are 1, 7, and 9.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 13647,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $a$, $b$, and $c$ be distinct positive real numbers. Compute\n\n$$\n\\lim_{t \\to \\infty} \\int_0^t \\frac{1}{(x^2 + a^2)(x^2 + b^2)(x^2 + c^2)} \\, dx.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "Start with\n\n$$\n\\begin{aligned}\n\\int_0^t \\frac{1}{(x^2 + a^2)(x^2 + b^2)} \\, dx &= \\frac{1}{b^2 - a^2} \\int_0^t \\left( \\frac{1}{x^2 + a^2} - \\frac{1}{x^2 + b^2} \\right) \\, dx \\\\\n&= \\frac{1}{b^2 - a^2} \\left( \\frac{1}{a} \\arctan \\frac{t}{a} - \\frac{1}{b} \\arctan \\frac{t}{b} \\right) \\\\\n&\\xrightarrow{t \\to \\infty} \\frac{1}{ab(a+b)} \\frac{\\pi}{2},\n\\end{aligned}\n$$\n\nto conclude that\n\n$$\n\\begin{aligned}\n\\int_0^t \\frac{1}{(x^2 + a^2)(x^2 + b^2)(x^2 + c^2)} \\, dx &= \\\\\n& \\frac{1}{c^2 - a^2} \\left( \\int_0^t \\frac{1}{(x^2 + a^2)(x^2 + b^2)} \\, dx - \\int_0^t \\frac{1}{(x^2 + b^2)(x^2 + c^2)} \\, dx \\right) \\\\\n&\\xrightarrow{t \\to \\infty} \\\\\n& \\frac{1}{c^2 - a^2} \\left( \\frac{1}{ab(a+b)} - \\frac{1}{bc(b+c)} \\right) \\frac{\\pi}{2} = \\frac{a+b+c}{abc(a+b)(b+c)(c+a)} \\frac{\\pi}{2}.\n\\end{aligned}\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 13648,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $ABCA'B'C'$ be a regular triangular prism, with lateral edges $AA'$, $BB'$, $CC'$. Consider the midpoint $D$ of the edge $BC$ and the parallelogram $ADB'E$. Let $F$ be the orthogonal projection of the point $A'$ onto the line $AE$, $d$ be the intersection of the planes $(ADE)$ and $(A'CF)$, and $P$ be the intersection of the line $d$ with the plane $(ABC)$. Prove that $P$ is the barycentre of the triangle $ABC$ if and only if $AB = AA'\n\\sqrt{2}$.",
    "options": [],
    "answer": "See solution",
    "solution": "The segments $BA'$, $AB'$, and $DE$ have the same midpoint, therefore $DBEA'$ is a parallelogram. From $CD = DB = A'E$ and $BD \\parallel A'E$, it follows that $A'CDE$ is a parallelogram.\n\n![](images/RMC_2024_p38_data_739c014ee2.png)\n\nThis leads to $CA' \\parallel DE$, hence $CA' \\parallel (ADE)$. So, the plane $(A'CF)$ intersects the plane $(ADE)$ along a parallel to $CA'$ and $DE$, therefore $FP \\parallel DE$, and $P \\in AD$.\n\n$AD$ is a median in $\\triangle ABC$, so the point $P$ is the barycentre of the triangle $ABC$ if and only if $AP = 2PD$. Since $PF \\parallel DE$, this is equivalent to $AF = 2FE$. Because $AA' \\parallel BB'$, $BD \\parallel EA'$, and $BB' \\perp BD$, the triangle $AEA'$ has a right angle at $A'$. Then $A'E$ and $A'A$ are legs of this triangle, therefore $AF = 2FE \\iff AF \\cdot AE = 2EF \\cdot AE \\iff A'A^2 = 2A'E^2 \\iff A'A^2 = \\frac{1}{2}BC^2 \\iff BC = A'A\\sqrt{2}$.",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 13649,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all natural numbers $N$ that have three digits and are equal to the sum of their digits added to the cube of this sum.",
    "options": [],
    "answer": "See solution",
    "solution": "**Answer:** 222.\n\nLet $n$ be the sum of the digits. Then:\n\n$$N = n + n^3$$\n\nSince $1 \\leq n \\leq 27$, $n^3$ must be between $73$ and $998$ because:\n\n$$73 = 100 - 27 \\leq N - 27 \\leq n^3 \\leq N - 1 \\leq 999 - 1 = 998.$$ \n\nThe cubes in this range are:\n\n$$5^3 = 125,\\quad 6^3 = 216,\\quad 7^3 = 343,\\quad 8^3 = 512,\\quad 9^3 = 729.$$ \n\nChecking these, the only possible answer is $222$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 13650,
    "subject": "Mathematics (Olympiad)",
    "question": "Several white and black balls can be divided into pairs so that exactly $\\frac{10}{11}$ of the white balls are in mixed pairs (with one white and one black ball), and the remaining ones are in pairs with the same color. Also, the balls can be divided into pairs so that exactly $\\frac{12}{13}$ of the black balls are in mixed pairs, and the remaining ones are in pairs with the same color. The number of white balls is between 150 and 200. How many balls of each color can there be?",
    "options": [],
    "answer": "See solution",
    "solution": "Let there be $x$ white and $y$ black balls. Since $\\frac{10}{11}x$ is an integer, $x$ is divisible by 11. Next, the $x - \\frac{10}{11}x = \\frac{x}{11}$ balls not in a mixed pair in the first division must be paired up among themselves. Hence $\\frac{x}{11}$ is even, i.e., $x$ is even. Thus $x$ is divisible by 22. Similar observations on the second division show that $y$ is divisible by 26.\n\nIn order to pair up $\\frac{10}{11}x$ white balls with black ones, it is necessary to have at least $\\frac{10}{11}x$ black balls, i.e., $\\frac{10}{11}x \\leq y$. Likewise, $\\frac{12}{13}y \\leq x$, or $y \\leq \\frac{13}{12}x$. In summary, $22 \\mid x$, $26 \\mid y$ and $\\frac{10}{11}x \\leq y \\leq \\frac{13}{12}x$. Conversely, if $x$ and $y$ satisfy these conditions, then both divisions are possible.\n\nThe multiples of 22 in $[150, 200]$ are 154, 176, and 198. Since $22 \\mid x$ and $150 \\leq x \\leq 200$, we have $x \\in \\{154, 176, 198\\}$. For $x = 154$, $x = 176$, $x = 198$, the condition $\\frac{10}{11}x \\leq y \\leq \\frac{13}{12}x$ yields $y \\in [140, 166]$, $y \\in [160, 190]$, $y \\in [180, 214]$ respectively. Taking $26 \\mid y$ into account, we obtain 4 solutions: $(154, 156)$, $(176, 182)$, $(198, 182)$, $(198, 208)$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 13651,
    "subject": "Mathematics (Olympiad)",
    "question": "In hexagon $ABCDEF$, which is nonconvex but not self-intersecting, no pair of opposite sides are parallel. The internal angles satisfy $\\angle A = 3\\angle D$, $\\angle C = 3\\angle F$, and $\\angle E = 3\\angle B$. Furthermore, $AB = DE$, $BC = EF$, and $CD = FA$. Prove that diagonals $\\overline{AD}$, $\\overline{BE}$, and $\\overline{CF}$ are concurrent.",
    "options": [],
    "answer": "See solution",
    "solution": "We proceed in three steps.\n\n*Step 1:* We first give a recipe for constructing hexagons of this type. Let $ACE$ be a triangle, with all angles less than $2\\pi/3$. Let $D$ be the reflection of $A$ across $CE$; let $F$ be the reflection of $C$ across $EA$; let $B$ be the reflection of $E$ across $AC$. Then, we have $\\angle BAF = \\angle BAC + \\angle CAE + \\angle EAF = 3\\angle CAE = 3\\angle CDE$ and the other analogous angle equalities. Also, we have $AB = AE = DE$ and the other analogous side equalities. Thus, the hexagon satisfies the equations in the problem statement. The diagonals $AD$, $BE$, $CF$ are simply the altitudes of the triangle $ACE$, so they are concurrent at the orthocenter.\n\n*Step 2:* For a hexagon meeting the conditions of the problem statement, let $\\beta$, $\\delta$, and $\\phi$ be the measures of angles $B$, $D$, and $F$. We claim that these are sufficient to determine $ABCDEF$ up to scaling.\n\nLet $x = AB = DE$, $y = BC = EF$, $z = CD = FA$. Our goal is to show that these lengths are determined up to scale by the given angles. Let $a, b, c, d, e, f$ be unit vectors in the directions of the edges from $A$ to $B$, $B$ to $C$, $C$ to $D$, $D$ to $E$, $E$ to $F$, and $F$ to $A$, respectively. Then, we have the vector identity\n\n$$\nx(a + d) + y(b + e) + z(c + f) = 0. \\tag{1}\n$$\n\nNotice now that\n\n$$\n4(\\beta + \\delta + \\phi) = \\angle A + \\angle B + \\angle C + \\angle D + \\angle E + \\angle F = 4\\pi,\n$$\n\nso $\\beta + \\delta + \\phi = \\pi$. Also, the fact that opposite sides are not parallel implies that $\\pi + 2\\beta = \\angle D + \\angle E + \\angle F \\neq 2\\pi$, so $\\beta \\neq \\pi/2$; likewise $\\delta, \\phi \\neq \\pi/2$.\n\nAssuming without loss of generality that the vertices of $ABCDEF$ are labeled in counterclockwise order and making liberal use of the identity $\\beta+\\delta+\\phi = \\pi$, we may compute the respective orientations of vectors $b$, $c$, $d$, $e$, and $f$, measured counterclockwise relative to $a$ modulo $2\\pi$, to be:\n\n$$\n\\begin{align*}\nb: & \\quad \\pi - \\beta \\\\\nc: & \\quad -\\beta - 3\\phi \\\\\nd: & \\quad -2\\phi \\\\\ne: & \\quad \\pi + 2\\delta - \\beta \\\\\nf: & \\quad 2\\delta - \\phi - \\beta\n\\end{align*}\n$$\n\nNow, whenever two unit vectors point in directions $\\theta$ and $\\psi$ which do not differ by $\\pi$, their sum is a nonzero vector either pointing in direction $(\\theta + \\psi)/2$ or direction $(\\theta + \\psi)/2 + \\pi$. It follows that the vectors $a+d$, $b+e$, and $c+f$ are all nonzero and point in the following directions modulo $\\pi$:\n\n$$\n\\begin{align*}\na+d: & -\\phi \\\\\nb+e: & \\delta - \\beta \\\\\nc+f: & \\delta - 2\\phi - \\beta\n\\end{align*}\n$$\n\nRecalling that $\\beta + \\delta + \\phi = \\pi$ and $\\beta, \\delta, \\phi \\neq \\pi/2$, we see that none of the pairwise differences between $a+d$, $b+e$, and $c+f$ are multiples of $\\pi$. Thus, $a+d$, $b+e$, and $c+f$ are nonzero vectors, no two of which are collinear. Consequently, condition (1) determines the coefficients $x, y, z$ uniquely up to scale, as required.\n\n*Step 3:* It suffices to show that the only hexagons meeting the conditions of the problem statement are the ones constructed in Step 1. Construct a hexagon $A_1B_1C_1D_1E_1F_1$ by taking $A_1C_1E_1$ to be a triangle with angles $\\beta, \\delta$, and $\\phi$ and reflecting each vertex across the opposite side as in Step 1. Then, $A_1B_1C_1D_1E_1F_1$ satisfies the conditions of the problem statement and has $\\angle B_1 = \\beta$, $\\angle D_1 = \\delta$, and $\\angle F_1 = \\phi$, so by Step 2 it is similar to $ABCDEF$, completing the proof.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 13652,
    "subject": "Mathematics (Olympiad)",
    "question": "What is the number of ordered triples $ (a, b, c) $ of positive integers, with $ a \\leq b \\leq c \\leq 9 $, such that there exists a (non-degenerate) triangle $ \\triangle ABC $ with an integer inradius for which $ a, b, $ and $ c $ are the lengths of the altitudes from $ A $ to $ \\overline{BC} $, $ B $ to $ \\overline{AC} $, and $ C $ to $ \\overline{AB} $, respectively? (Recall that the inradius of a triangle is the radius of the largest possible circle that can be inscribed in the triangle.)\n\n(A) 2  \n(B) 3  \n(C) 4  \n(D) 5  \n(E) 6",
    "options": [],
    "answer": "See solution",
    "solution": "**Answer (B):**\n\nLet $x, y,$ and $z$ be the lengths of $\\overline{BC}$, $\\overline{AC}$, and $\\overline{AB}$, respectively. Let $r$ be the inradius of $\\triangle ABC$. Then\n\n$$\n\\text{Area}(\\triangle ABC) = \\frac{1}{2} x a = \\frac{1}{2} y b = \\frac{1}{2} z c = \\frac{1}{2} (x + y + z) r.\n$$\n\nTherefore $x = \\frac{2 \\text{Area}(\\triangle ABC)}{a}$, $y = \\frac{2 \\text{Area}(\\triangle ABC)}{b}$, and $z = \\frac{2 \\text{Area}(\\triangle ABC)}{c}$, so\n\n$$\n\\text{Area}(\\triangle ABC) = \\frac{1}{2} \\left( \\frac{2 \\text{Area}(\\triangle ABC)}{a} + \\frac{2 \\text{Area}(\\triangle ABC)}{b} + \\frac{2 \\text{Area}(\\triangle ABC)}{c} \\right) r.\n$$\n\nDividing by $\\text{Area}(\\triangle ABC)$ gives $\\left( \\frac{1}{a} + \\frac{1}{b} + \\frac{1}{c} \\right) r = 1$, so\n\n$$\n\\frac{1}{a} + \\frac{1}{b} + \\frac{1}{c} = \\frac{1}{r}.\n$$\n\nBecause $a, b, c \\leq 9$, it follows that\n\n$$\n\\frac{1}{r} = \\frac{1}{a} + \\frac{1}{b} + \\frac{1}{c} \\geq \\frac{1}{9} + \\frac{1}{9} + \\frac{1}{9} = \\frac{1}{3},\n$$\n\nimplying $r \\leq 3$. It is then possible to find the solutions $(a, b, c)$ by examining cases based on the value of $r$.\n\n- If $r = 1$, then, because $a \\leq b \\leq c$, either $a = 2$ or $a = 3$. If $a = 2$, then $\\frac{1}{b} + \\frac{1}{c} = \\frac{1}{2}$. Then, because $b \\leq c$, either $b = 3$ or $b = 4$. If $b = 3$, then $c = 6$; and if $b = 4$, then $c = 4$. If $a = 3$, then it must be that $b = c = 3$. So the solutions in this case are $(2, 3, 6)$, $(2, 4, 4)$, and $(3, 3, 3)$.\n- If $r = 2$, then, because $a \\leq b \\leq c$, it follows that $a = 3, 4, 5,$ or $6$. If $a = 3$, then $\\frac{1}{b} + \\frac{1}{c} = \\frac{1}{6}$, which has no solutions because $\\frac{1}{9} + \\frac{1}{9} > \\frac{1}{6}$. If $a = 4$, then $\\frac{1}{b} + \\frac{1}{c} = \\frac{1}{4}$. In this case, $b = c = 8$ is the only solution. If $a = 5$, then $\\frac{1}{b} + \\frac{1}{c} = \\frac{3}{10}$, which gives no solutions. If $a = 6$, then it follows that $b = c = 6$. So the solutions in this case are $(4, 8, 8)$ and $(6, 6, 6)$.\n- If $r = 3$, then $(a, b, c) = (9, 9, 9)$ is the only solution because if $a < 9$, then $c > 9$.\n\nFinally, the altitude lengths must be checked to ensure that these lengths give dimensions for a valid triangle. In the $(2, 3, 6)$ case, the side lengths of the triangles become $(3t, 2t, t)$ for some $t$, which does not form a triangle. In the $(2, 4, 4)$ and $(4, 8, 8)$ cases, the side lengths of the triangles become $(2t, t, t)$ for some $t$, which again does not form a triangle. The rest of the cases, namely $(3, 3, 3)$, $(6, 6, 6)$, and $(9, 9, 9)$, do produce valid triangles because if all of the altitudes have length $h$, then it is possible to form an equilateral triangle with side length $\\frac{2\\sqrt{3}}{3} h$. Thus there are 3 ordered triples satisfying the given conditions.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 13653,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $\\triangle ABC$ be a right-angled triangle with $\\angle B = 90^\\circ$. Let $D$ be a point on $AC$ such that the inradii of triangles $ABD$ and $CBD$ are equal. If this common value is $r'$, and if $r$ is the inradius of triangle $ABC$, prove that\n\n$$\n\\frac{1}{r'} = \\frac{1}{r} + \\frac{1}{BD}.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "Let $E$ and $F$ be the incenters of triangles $ABD$ and $CBD$ respectively. Let the incircles of triangles $ABD$ and $CBD$ touch $AC$ at $P$ and $Q$ respectively. If $\\angle BDA = \\theta$, we see that\n\n![](images/IND_TSExams_2016_p4_data_8da2e1e430.png)\n\n$$\nr' = PD \\tan\\left(\\frac{\\theta}{2}\\right) = QD \\cot\\left(\\frac{\\theta}{2}\\right).\n$$\n\nHence,\n\n$$\nPQ = PD + QD = r' \\left( \\cot \\frac{\\theta}{2} + \\tan \\frac{\\theta}{2} \\right) = \\frac{2r'}{\\sin \\theta}.\n$$\n\nBut we observe that\n\n$$\nDP = \\frac{BD + DA - AB}{2}, \\quad DQ = \\frac{BD + DC - BC}{2}.\n$$\n\nThus $PQ = (b - c - a + 2BD)/2$. We also have\n\n$$\n\\begin{aligned}\n\\frac{ac}{2} &= [ABC] = [ABD] + [CBD] = r' \\frac{(AB + BD + DA)}{2} + r' \\frac{(CB + BD + DC)}{2} \\\\\n&= r' \\frac{(c + a + b + 2BD)}{2} = r'(s + BD).\n\\end{aligned}\n$$\n\nBut\n\n$$\nr' = \\frac{PQ \\sin \\theta}{2} = \\frac{PQ \\cdot h}{2BD},\n$$\n\nwhere $h$ is the altitude from $B$ onto $AC$. But we know that $h = \\frac{ac}{b}$. Thus we get\n\n$$\nac = 2 \\times r'(s + BD) = 2 \\times \\frac{PQ \\cdot h}{2 \\times BD} (s + BD) = \\frac{(b - c - a + 2BD)ca(s + BD)}{2 \\times BD \\times b}.\n$$\n\nThus we get\n\n$$\n2 \\times BD \\times b = 2 \\times (BD - (s - b))(s + BD).\n$$\n\nThis gives $BD^2 = s(s - b)$. Since $ABC$ is a right-angled triangle, $r = s - b$. Thus we get $BD^2 = rs$. On the other hand, we also have $[ABC] = r'(s + BD)$. Thus we get\n\n$$\nrs = [ABC] = r'(s + BD).\n$$\n\nHence,\n\n$$\n\\frac{1}{r'} = \\frac{1}{r} + \\frac{BD}{rs} = \\frac{1}{r} + \\frac{1}{BD}.\n$$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 13654,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $ABC$ be a non-isosceles triangle with circumcenter $U$ and incenter $I$. Assume that the bisector of the segment $UI$ passes through the common point of the angle bisector of $\\gamma = \\angle ACB$ with the circumcircle of $ABC$. Prove that $\\gamma$ is the second largest angle in the triangle $ABC$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $w_\\gamma$ be the angle bisector of $\\gamma$, $k$ the circumcircle of $ABC$, and $D = k \\cap w_\\gamma$. Since $\\angle DCB = \\angle DCA$, we have $|DA| = |DB|$. Considering triangle $DBI$, note that $\\angle CDB = \\angle CAB = \\alpha$. Also, $\\angle DBI = \\angle DBA + \\angle ABI = \\angle DCA + \\angle ABI = \\frac{\\gamma}{2} + \\frac{\\beta}{2}$, so $\\angle DIB = 180^\\circ - \\alpha - \\left(\\frac{\\gamma}{2} + \\frac{\\beta}{2}\\right) = \\frac{\\gamma}{2} + \\frac{\\beta}{2}$. Thus, $DBI$ is isosceles with $|DI| = |DB|$. Furthermore, since $D$ lies on the bisector of $UI$, we also have $|DU| = |DI|$. It follows that $D$ is the midpoint of a circle through all four points $A$, $B$, $I$, and $U$.\n\n![](images/AustriaMO2012_p4_data_e33bba6ad5.png)\n\nSince $U$ is the circumcenter of $ABC$, $\\angle AUB = 2 \\cdot \\angle ACB = 2\\gamma$. On the other hand, since $\\angle IAB = \\frac{\\alpha}{2}$ and $\\angle IBA = \\frac{\\beta}{2}$, we have $\\angle AIB = 180^\\circ - \\frac{\\alpha}{2} - \\frac{\\beta}{2}$. Since $A$, $B$, $I$, and $U$ lie on a common circle, $\\angle AUB = \\angle AIB$, and therefore $2\\gamma = 180^\\circ - \\frac{\\alpha}{2} - \\frac{\\beta}{2}$, which is equivalent to $\\gamma = \\frac{1}{2}(\\alpha + \\beta)$. Since $\\gamma$ is the arithmetic mean of $\\alpha$ and $\\beta$, it is certainly the second largest angle in triangle $ABC$ as claimed. $\\square$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 13655,
    "subject": "Mathematics (Olympiad)",
    "question": "In triangle $ABC$ with $CA = CB$, point $E$ lies on the circumcircle of $ABC$ such that $\\angle ECB = 90^\\circ$. The line through $E$ parallel to $CB$ intersects $CA$ at $F$ and $AB$ at $G$. Prove that the center of the circumcircle of triangle $EGB$ lies on the circumcircle of triangle $ECF$.",
    "options": [],
    "answer": "See solution",
    "solution": "![](images/IND_ABooklet_2024_p5_data_5a9ed47c35.png)\n\nWe have $FG = FA$ since $FG$ is parallel to $BC$. Also, $\\triangle GAE$ is a right triangle. Thus, if $F'$ is the midpoint of $GE$, then $\\angle GAF = \\angle FGA = \\angle F'GA = \\angle GAF'$, which implies $F \\equiv F'$. Thus, $F$ is the midpoint of $GE$.\n\nIf $O$ is the circumcenter of $\\triangle EBG$, then\n\n$$\n\\angle FOE = \\angle GBE = \\angle ABE = \\angle ACE = \\angle FCE.\n$$\n\nThus, we get $\\angle FOE = \\angle FCE$ as desired. $\\square$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 13656,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $m$ be a real number. If the real and imaginary parts of the complex number $z = 1 + i + \\frac{m}{1+i}$, where $i$ is the imaginary unit, are greater than zero, then the range of $m$ is ____.",
    "options": [],
    "answer": "See solution",
    "solution": "After calculation, we get:\n\n$$z = 1 + i + \\frac{(1-i)m}{2} = \\frac{2 + m}{2} + \\frac{2 - m}{2}i.$$ \n\nBy the condition, it follows that $\\frac{2 + m}{2} > 0$ and $\\frac{2 - m}{2} > 0$. Therefore, the solution is $-2 < m < 2$. $\\square$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 13657,
    "subject": "Mathematics (Olympiad)",
    "question": "Circles $(L)$ and $(O)$ are drawn, meeting at $B$ and $C$, with $L$ on $(O)$. Ray $CO$ meets $(L)$ at $Q$, and $A$ is on $(O)$ such that $\\angle CQA = 90^\\circ$. The angle bisector of $\\angle AOB$ meets $(L)$ at $X$ and $Y$. Show that $\\angle XLY = 90^\\circ$.",
    "options": [],
    "answer": "See solution",
    "solution": "Up to now we have not used the existence of $Q$; we henceforth do so.\n\nNote that $Q \\neq O$, since $\\angle A \\neq 60^\\circ \\implies O \\notin \\omega$. Moreover, we have $\\angle AOM = \\angle ACB$ too. Since $O$ and $Q$ both lie inside $\\triangle ABC$, this implies that $A, M, O, Q$ are concyclic. As $Q \\neq O$ we conclude $\\angle CQA = 90^\\circ$.\n\nThe main claim is now:\n\n*Claim.* Assuming $Q$ exists, the rhombus $LXKY$ is a square. In particular, $\\overline{KX}$ and $\\overline{KY}$ are tangent to $\\omega$.\n\n*First proof of Claim, communicated by Milan Haiman.* Observe that $\\triangle QLC \\sim \\triangle LOC$. Hence, $CL^2 = CO \\cdot CQ$. Then,\n\n$$\nx^2 = CL^2 = CO \\cdot CQ = CN \\cdot CA = \\frac{1}{2}CA^2 = \\frac{1}{2}LK^2\n$$\n\nwhere we have also used the fact $AQON$ is cyclic. Thus $LK = \\sqrt{2}x$ and so the rhombus $LXKY$ is actually a square. $\\square$\n\n*Second proof of Claim, Evan Chen.* Observe that $Q$ lies on the circle with diameter $\\overline{AC}$, centered at $N$, say. This means that $O$ lies on the radical axis of $\\omega$ and $(N)$, hence $\\overline{NL} \\perp \\overline{CO}$ implying\n\n$$\n\\begin{align*} \nNO^2 + CL^2 &= NC^2 + LO^2 = NC^2 + OC^2 = NC^2 + NO^2 + NC^2 \\\\\n\\implies x^2 &= 2NC^2 \\\\\n\\implies x &= \\sqrt{2}NC = \\frac{1}{\\sqrt{2}}AC = \\frac{1}{\\sqrt{2}}LK. \n\\end{align*}\n$$\n\nSo $LXKY$ is a rhombus with $LK = \\sqrt{2}x$. Hence it is a square. $\\square$\n\n*Third proof of Claim.* A solution by trig is also possible. As in the previous claims, it suffices to show that $AC = \\sqrt{2}x$.\n\nFirst, we compute the length $CQ$ in two ways; by angle chasing one can show $\\angle CBQ = 180^\\circ - (\\angle BQC + \\angle QCB) = \\frac{1}{2}\\angle A$, and so\n\n$$\n\\begin{aligned}\nAC \\sin B &= CQ = \\frac{BC}{\\sin(90^\\circ + \\frac{1}{2}\\angle A)} \\cdot \\sin \\frac{1}{2}\\angle A \\\\\n\\Leftrightarrow \\sin^2 B &= \\frac{\\sin A \\cdot \\sin \\frac{1}{2}\\angle A}{\\cos \\frac{1}{2}\\angle A} \\\\\n\\Leftrightarrow \\sin^2 B &= 2 \\sin^2 \\frac{1}{2}\\angle A \\\\\n\\Leftrightarrow \\sin B &= \\sqrt{2} \\sin \\frac{1}{2}\\angle A \\\\\n\\Leftrightarrow 2R \\sin B &= \\sqrt{2} \\left( 2R \\sin \\frac{1}{2}\\angle A \\right) \\\\\n\\Leftrightarrow AC &= \\sqrt{2}x\n\\end{aligned}\n$$\n\nas desired (we have here used the fact $\\triangle ABC$ is acute to take square roots).\n\nIt is interesting to note that $\\sin^2 B = 2 \\sin^2 \\frac{1}{2}\\angle A$ can be rewritten as\n\n$$\n\\cos A = \\cos^2 B\n$$\n\nsince $\\cos^2 B = 1 - \\sin^2 B = 1 - 2 \\sin^2 \\frac{1}{2} \\angle A = \\cos A$; this is the condition for the existence of the point $Q$. $\\square$\n\nWe finish by proving that\n\n$$\nKD = KA\n$$\n\nand hence line $\\overline{KD}$ is tangent to $\\gamma$. Let $E = \\overline{BC} \\cap \\overline{KL}$. Then\n\n$$\nLE \\cdot LK = LC^2 = LX^2 = \\frac{1}{2}LK^2\n$$\n\nand so $E$ is the midpoint of $\\overline{LK}$. Thus $\\overline{MXOY}$, $\\overline{BC}$, $\\overline{KL}$ are concurrent at $E$. As $\\overline{DL} \\parallel \\overline{KC}$, we find that $DLCK$ is a parallelogram, so $KD = CL = KA$ as well. Thus $\\overline{KD}$ and $\\overline{KA}$ are tangent to $\\gamma$.\n\n*Remark.* The condition $\\angle A \\neq 60^\\circ$ cannot be dropped, since if $Q = O$ the problem is not true.\n\nOn the other hand, nearly all solutions begin by observing $Q \\neq O$ and then obtaining $\\angle AQO = 90^\\circ$. This gives a way to construct the diagram by hand with ruler and compass. One draws an arbitrary chord $\\overline{BC}$ of a circle $\\omega$ centered at $L$, and constructs $O$ as the circumcenter of $\\triangle BLC$ (hence obtaining $\\Gamma$). Then $Q$ is defined as the intersection of ray $CO$ with $\\omega$, and $A$ is defined by taking the perpendicular line through $Q$ on the circle $\\Gamma$. In this way we can draw a triangle $ABC$ satisfying the problem conditions.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 13658,
    "subject": "Mathematics (Olympiad)",
    "question": "設正整數 $n \\ge 6$。平面上有 $n$ 個兩兩互斥的圓盤 $D_1, D_2, \\dots, D_n$，其半徑依序為 $R_1 \\ge R_2 \\ge \\dots \\ge R_n$。對每一個 $i = 1, 2, \\dots, n$，在 $D_i$ 中都標了一個點 $P_i$。設 $O$ 為平面上的任意點。試證：\n\n$$\n\\sum_{i=1}^{n} OP_i \\ge \\sum_{j=6}^{n} R_j.\n$$\n\n（註：這裡的圓盤都包含其邊界。）",
    "options": [],
    "answer": "See solution",
    "solution": "我們將使用以下引理。\n\n*引理.* 設 $D_1, \\dots, D_6$ 為平面上兩兩互斥的圓盤，半徑分別為 $R_1, \\dots, R_6$。設 $P_i$ 為 $D_i$ 內的一點，$O$ 為平面上一任意點。則存在 $i, j$ 使得 $OP_i \\ge R_j$。\n\n*證明.* 設 $O_i$ 為 $D_i$ 的圓心。考慮六條射線 $OO_1, \\dots, OO_6$（若 $O = O_i$，則 $OO_i$ 可取任意方向）。這些射線將平面分成六個角（其中一個可能是非凸的），總角度為 $360^\\circ$，因此必有一個角（如 $\\angle O_iOO_j$）不超過 $60^\\circ$。則 $O_iO_j$ 不可能是三角形 $OO_iO_j$ 的唯一最長邊，故不妨設 $OO_i \\ge O_iO_j \\ge R_i + R_j$。因此，$OP_i \\ge OO_i - R_i \\ge (R_i + R_j) - R_i = R_j$，得證。$\\square$\n\n現在對 $n \\ge 5$ 用歸納法證明所需不等式。當 $n = 5$ 時顯然成立。歸納步驟：對六個最大圓盤應用引理，得 $1 \\le i, j \\le 6$ 且 $OP_i \\ge R_j \\ge R_6$。將 $D_i$ 移除後，對剩下的圓盤用歸納假設，得：\n\n$$\n\\sum_{k \\ne i} OP_k \\ge \\sum_{\\ell \\ge 7} R_\\ell.\n$$\n\n加上 $OP_i \\ge R_6$，即得歸納步。\n\n**註解 1.** 圓盤是否包含邊界對本題無影響，此條件僅為明確起見。即使只要求圓盤內部互斥，結論與證明仍然成立。\n\n**註解 2.** 上述解法有多種變體。例如在歸納步中，可以移除 $OP_i$ 最大的圓盤，再對剩下的圓盤用歸納假設（引理仍對六個最大圓盤應用）。\n\n**註解 3.** 證明引理時，可將圓盤縮放為同半徑的情形：取最小半徑 $r$，對每個 $i$，以 $P_i$ 為中心，將 $D_i$ 按比例 $r/R_i$ 縮放。\n\n此論證顯示該引理與圓盤堆積問題密切相關，參見 [Circle packing in a circle](https://en.wikipedia.org/wiki/Circle_packing_in_a_circle)。該問題的已知結果給出不同數量圓盤時的不同版本引理，從而導出類似的不等式。例如，對 4 個圓盤最佳估計為 $OP_i \\ge (\\sqrt{2}-1)R_j$，對 13 個圓盤為 $OP_i \\ge \\sqrt{5}R_j$。類似地可得：\n\n$$\n\\sum_{i=1}^{n} OP_i \\ge (\\sqrt{2}-1) \\sum_{j=4}^{n} R_j \\quad \\text{和} \\quad \\sum_{i=1}^{n} OP_i \\ge \\sqrt{5} \\sum_{j=13}^{n} R_j.\n$$\n\n但有更困難的論證可進一步改進這些不等式，亦即對大指數的 $R_j$ 可取更大係數。",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 13659,
    "subject": "Mathematics (Olympiad)",
    "question": "Given a convex 20-sided polygon $P$. Divide $P$ into 18 triangles by drawing 17 non-intersecting diagonals in the interior; the resulting graph is called a triangulation graph of $P$. For any triangulation graph $T$ of $P$, the 20 edges of $P$ and the added 17 diagonals are all called the edges of $T$. A set of any 10 edges of $T$ with no common endpoints between any two edges is called a perfect matching of $T$. As $T$ ranges over all triangulation graphs of $P$, find the maximum number of perfect matchings of $T$.",
    "options": [],
    "answer": "See solution",
    "solution": "We generalize to a convex $2n$-sided polygon.\n\nFor a diagonal of a convex $2n$-gon $P$, if there are an odd number of vertices of $P$ on both sides, it is called an odd chord; otherwise, it is an even chord. Note the following fact:\n\n*For any triangulation graph $T$ of $P$, a perfect matching of $T$ does not contain odd chords.*\n\nSuppose an odd chord $e_1$ is in a perfect matching. Since a perfect matching pairs all vertices, and there are an odd number of vertices on each side of $e_1$, there must be another edge $e_2$ of $T$ with endpoints on opposite sides of $e_1$. In a convex polygon, $e_1$ and $e_2$ would intersect, contradicting the triangulation.\n\nLet $f(T)$ be the number of perfect matchings of $T$. Let $F_1 = 1$, $F_2 = 2$, and $F_{k+2} = F_{k+1} + F_k$ for $k \\ge 2$ (the Fibonacci sequence).\n\nWe prove by induction on $n$ that for any triangulation graph $T$ of a convex $2n$-gon, $f(T) \\leq F_n$.\n\nFor $n=2$ (quadrilateral), $T$ has no even chords, so perfect matchings use only the polygon's edges: $f(T) = 2 = F_2$.\n\nFor $n=3$ (hexagon), $T$ has at most one even chord. If none, $f(T) = 2$. If $T$ contains one even chord (say $A_1A_4$), the perfect matching using $A_1A_4$ is unique, and the other two edges are $A_2A_3, A_5A_6$, so $f(T) = 3 = F_3$.\n\nAssume the result holds for all $n' < n$, $n \\ge 4$. Consider a triangulation $T$ of a convex $2n$-gon $P = A_1A_2 \\cdots A_{2n}$. If $T$ has no even chords, $f(T) = 2$.\n\nFor an even chord $e$, let $w(e)$ be the smaller number of vertices on either side of $e$. Choose $e$ with minimal $w(e) = 2k$, and set $e = A_{2n}A_{2k+1}$. Then, for $A_i$ ($i = 1, \\dots, 2k$), no even chord can arise.\n\nSuppose $A_iA_j$ is an even chord. If $j \\in \\{2k+2, \\dots, 2n-1\\}$, $A_iA_j$ would intersect $e$; if $j \\in \\{1, \\dots, 2k+1, 2n\\}$, $w(A_iA_j) < 2k$, contradicting minimality.\n\nBy the earlier fact, perfect matchings can only pair neighboring vertices.\n\nIn particular, $A_1$ can only be paired with $A_2$ or $A_{2n}$. Consider two cases:\n\n**Case 1:** Choose $A_1A_2$, then $A_3A_4, \\dots, A_{2k-1}A_{2k}$ must also be chosen. The remaining $2n-2k$ vertices form a convex $2n-2k$-gon $P_1$ with triangulation $T_1$. By induction, $f(T_1) \\le F_{n-k}$.\n\n**Case 2:** Choose $A_1A_{2n}$, then $A_2A_3, \\dots, A_{2k}A_{2k+1}$ must be chosen. The remaining $2n-2k-2$ vertices form a convex $2n-2k-2$-gon $P_2$ with triangulation $T_2$. By induction, $f(T_2) \\le F_{n-k-1}$.\n\nThus,\n$$\nf(T) \\le f(T_1) + f(T_2) \\le F_{n-k} + F_{n-k-1} = F_{n-k+1} \\le F_n.\n$$\n\nEquality is achieved for a specific triangulation $\\Delta_n$ constructed by adding diagonals in a certain pattern. For $\\Delta_n$, $f(\\Delta_n) = F_n$ by induction.\n\nFor $n=10$ (20-gon), the maximum number of perfect matchings is $F_{10} = 55$.",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 13660,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $ABC$ be an acute triangle with $AB \\neq AC$. Let $D$ be the midpoint of $BC$ and $I$, $J$, $K$ be the feet of the altitudes from $A$, $B$, and $C$, respectively, in triangle $ABC$. The perpendicular from $A$ to the line $AD$ meets the lines $BJ$ and $CK$ at points $N$ and $Q$, respectively, and the parallel to $BC$ through $A$ intersects the lines $IJ$ and $IK$ at $M$ and $P$, respectively. Prove that $MNPQ$ is a parallelogram.\n\n![](images/RMC_2025_p14_data_ee26744ce2.png)",
    "options": [],
    "answer": "See solution",
    "solution": "Let $H$ be the orthocenter of triangle $ABC$. Quadrilaterals $BIHK$ and $CIHJ$ are cyclic, so $\\angle HIK = \\angle HBK = \\angle ABJ = \\angle ACK = \\angle JCH = \\angle HIJ$.\n\nTherefore, $IA$ is the angle bisector of $\\triangle JIK$. Since $IA \\perp BC$ and $MP \\parallel BC$, it follows that $IA \\perp MP$. Because $IA$ is both an angle bisector and an altitude in $\\triangle IMP$, this triangle is isosceles and $A$ is the midpoint of $MP$.\n\nLet $S$ be the reflection of $A$ with respect to $D$. Since $D$ is the midpoint of $BC$, $ABSC$ is a parallelogram. Consequently, $BS \\parallel AC$ and, since $BJ \\perp AC$, we deduce that $BJ \\perp BS$. From $\\angle SBN = \\angle SAN = 90^\\circ$, it follows that $SBAN$ is cyclic, therefore $\\angle NSA = \\angle ABN = 90^\\circ - \\angle A$. Similarly, the quadrilateral $SCAQ$ is cyclic, hence $\\angle QSA = \\angle ACQ = 90^\\circ - \\angle A = \\angle NSA$. In triangle $SNQ$, the altitude $SA$ is also an angle bisector, therefore $A$ is the midpoint of $NQ$. Since the diagonals of quadrilateral $MNPQ$ bisect each other, it follows that $MNPQ$ is a parallelogram.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 13661,
    "subject": "Mathematics (Olympiad)",
    "question": "Sean $x$ y $y$ números reales entre 0 y 1. Probar que\n$$\nx^3 + x y^2 + 2 x y \\leq 2 x^2 y + x^2 + x + y\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "La desigualdad equivale a\n$$\nP = 2x^2y + x^2 + x + y - x^3 - x y^2 - 2 x y \\geq 0.\n$$\nEscribimos $P$ como un polinomio en la variable $x$:\n$$\nP = -x^3 + (2y + 1)x^2 + (1 - 2y - y^2)x + y.\n$$\nDividimos este polinomio entre $x - 1$, mediante el algoritmo de Ruffini:\n$$\n\\begin{array}{c|cccc}\n1 & -1 & 2y+1 & 1-2y-y^2 & y \\\\\n\\cline{2-5}\n & -1 & -1 & 2y & 1-y^2 \\\\\n\\cline{2-5}\n & -1 & 2y & 1-y^2 & 1+y-y^2\n\\end{array}\n$$\nEs decir,\n$$\nP = (-x^2 + 2xy + 1 - y^2)(x - 1) + 1 + y - y^2.\n$$\nTambién,\n$$\n\\begin{align*}\nP &= (-x^2 + 2xy - y^2)(x - 1) + x - 1 + 1 + y - y^2 \\\\\n  &= (x^2 - 2xy + y^2)(1 - x) + x + y - y^2 \\\\\n  &= (x - y)^2(1 - x) + x + y(1 - y).\n\\end{align*}\n$$\nPuesto que las cinco cantidades $(x - y)^2$, $1 - x$, $x$, $y$, $1 - y$ son no negativas, $P \\geq 0$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 13662,
    "subject": "Mathematics (Olympiad)",
    "question": "已知正整數 $a_1, a_2, \\dots, a_n$，且 $a_1 < a_2 < \\dots < a_n$，$k$ 為正實數且 $k \\ge 1$。\n\n試證：\n\n$$\n\\sum_{i=1}^{n} a_i^{2k+1} \\ge \\left( \\sum_{i=1}^{n} a_i^k \\right)^2\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "我們分兩步驟證明題設：\n\n1. 用數學歸納法證明：\n\n$$\n2 \\sum_{i=1}^{n} a_i^k \\le (a_n + 1)^k a_n. \\qquad (1)\n$$\n\n證明：當 $n=1$ 時，易知 Eq. (1) 成立。\n\n假設當 $n=m$ 時，Eq. (1) 成立，即\n\n$$\n2 \\sum_{i=1}^{m} a_i^k \\le (a_m + 1)^k a_m.\n$$\n\n則當 $n=m+1$ 時，\n\n$$\n\\begin{aligned}\n2 \\sum_{i=1}^{m+1} a_i^k &= 2 \\sum_{i=1}^{m} a_i^k + 2a_{m+1}^k \\\\\n&\\le (a_m + 1)^k a_m + 2a_{m+1}^k \\\\\n&\\le a_{m+1}^k (a_{m+1} - 1) + 2a_{m+1}^k \\\\\n&= a_{m+1}^k (a_{m+1} + 1) \\\\\n&\\le (a_{m+1} + 1)^k a_{m+1}.\n\\end{aligned}\n$$\n\n2. 再以數學歸納法證明題設成立。\n\n當 $n=1$ 時，易知題設成立。\n\n假設當 $n=m$ 時，題設成立，即\n\n$$\n\\left( \\sum_{i=1}^{m} a_i^k \\right)^2 \\le \\sum_{i=1}^{m} a_i^{2k+1}.\n$$\n\n則當 $n = m + 1$ 時，\n\n$$\n\\begin{aligned}\n\\left( \\sum_{i=1}^{m+1} a_i^k \\right)^2 &= \\left( \\sum_{i=1}^{m} a_i^k \\right)^2 + 2 \\left( \\sum_{i=1}^{m} a_i^k \\right) a_{m+1}^k + a_{m+1}^{2k} \\\\\n&\\le \\sum_{i=1}^{m} a_i^{2k+1} + 2 \\left( \\sum_{i=1}^{m} a_i^k \\right) a_{m+1}^k + a_{m+1}^{2k} \\\\\n&\\le \\sum_{i=1}^{m} a_i^{2k+1} + (a_m + 1)^k a_m a_{m+1}^k + a_{m+1}^{2k} \\quad \\text{(由 Eq. (1))} \\\\\n&\\le \\sum_{i=1}^{m} a_i^{2k+1} + a_{m+1}^k \\left( a_{m+1}^k (a_{m+1}-1) + a_{m+1}^k \\right) \\\\\n&\\le \\sum_{i=1}^{m+1} a_i^{2k+1}.\n\\end{aligned}\n$$\n\n綜上，題設成立。",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 13663,
    "subject": "Mathematics (Olympiad)",
    "question": "An acute scalene triangle $ABC$ with $\\angle A = 45^\\circ$ is inscribed in a circle $(O)$. Let $AD$ be the altitude and $H$ be the orthocenter of $\\triangle ABC$. Points $R$ and $S$, different from $A$, are taken on the rays $AB$ and $AC$ respectively so that $CA = CR$ and $BA = BS$. Points $X$, $Y$, and $K$ are the midpoints of the segments $BS$, $CR$, and $AH$, respectively. Point $T$ is on the line $BC$ such that $AO \\perp TO$, and $I$ is the circumcenter of triangle $DXY$. Prove that $XY \\perp OK$ and $TH \\perp IK$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $Z$, $M$ be the midpoints of $BR$, $BC$ respectively, then $ZY$ is the median of triangle $BCR$ so $ZY = \\frac{BC}{2}$ and $ZY \\parallel BC$.\n\n![](images/Saudi_Arabia_booklet_2023_p45_data_e3e11a0e4f.png)\n\nIt is easy to see that $AH = 2OM = BC$ since triangle $OBC$ is right-angled at $O$. Hence $AK = \\frac{BC}{2} = ZY$. Let $L$ be the intersection of $BS$, $CR$. Obviously the triangles $ACR$, $ABS$ are right, so $L$ is the orthocenter of triangle $ARS$. So $L \\in (O)$ and $AL$ is the diameter of $(O)$.\n\nConsidering triangle $ARS$ has $\\angle RAS = 45^\\circ$, so similar to above, we have $AO = \\frac{AL}{2} = \\frac{RS}{2} = ZX$ (because $ZX$ is the median of triangle $BRS$). We also have $ZX \\perp AO$ and $ZY \\perp AK$ so $\\angle XZY = \\angle OAK$ leads to $\\triangle XYZ \\cong \\triangle OKA$, and these two triangles have corresponding sides perpendicular, so $XY \\perp OK$.\n\nNow let $N$ be the midpoint of $RS$, we will prove that $I$ is the midpoint of $HN$. Since $N$ is the center of the circle passing through $B$, $C$, $R$, $S$, then $NX \\perp XL$, $NY \\perp YL$, so the points $X$, $Y$, $N$, $L$ belong to the circle of diameter $LN$. Let $D_F$ be the center symmetry $F$ where $F$ is the midpoint of the line segment $XY$. Thus $D_F : X \\leftrightarrow Y$. But we have $MX = YN = \\frac{CS}{2}$ and $MX \\parallel YN \\parallel CS$ so $MXNY$ is a parallelogram and $D_F : N \\leftrightarrow M$.\n\nAssuming $\\mathcal{D}_F(L) = G$ then the points $M, G, X, Y$ belong to the circle of diameter $MG$. Otherwise, $MF$ is the median of triangle $LHG$, followed by $HG \\parallel FM$, which $FM \\perp BC$ so $G \\in HD$. Infer $\\angle GDM = 90^\\circ$ so $D$ also belongs to the circle of diameter $MG$.\n\nTherefore, the center $I$ of $(DXY)$ is the midpoint of $MG$. Also, since $HG = 2MF = MN$ and $HG \\parallel MN$, $HGNM$ is a parallelogram, so $I$ is also the midpoint of $HN$. Thus $IK \\parallel AN$ and we need to prove that $TH \\perp AN$. By the four-point theorem, we need to show that\n\n$$\nTA^2 - TN^2 = HA^2 - HN^2.\n$$\n\nLet $R$ be the radius of the circle $(O)$, we have $AH^2 = (2OM)^2 = 2R^2$. It is easy to see that $OBNC$ is a square, so $ON = 2OM = AH$, so $AHNO$ is a parallelogram and $HN = AO$. Hence\n\n$$\nHA^2 - HN^2 = HA^2 - AO^2 = R^2.\n$$\n\nWe also have\n\n$$\n\\begin{align*}\nTA^2 - TN^2 &= (TO^2 + OA^2) - (TM^2 + MN^2) \\\\\n&= (TM^2 + OM^2 + OA^2) - (TM^2 + MN^2) \\\\\n&= OA^2 = R^2.\n\\end{align*}\n$$\n\nFrom this it follows that $TA^2 - TN^2 = HA^2 - HN^2$ or $TH \\perp AN$, which leads to $TH \\perp IK$. This finishes the proof.\n\n*Remark:* We also can prove that $H, N$ are isogonal conjugate in triangle $BLC$ so by the well-known property, the circumcenter of $DXY$ is the midpoint of $HN$. This will shorten the above proof. $\\Box$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 13664,
    "subject": "Mathematics (Olympiad)",
    "question": "Label the lily pads A, B, C, D, E in rotational order. Represent each of the lily pads with a column of 12 circles as shown.\n\nDraw a line connecting a circle in a row to a circle in the next row if the frog can jump from one corresponding lily pad to the next in one step. We may assume the frog starts at C.\n\nA number $k$ in a circle in row $n$ is the number of ways for the frog to reach the corresponding lily pad from C in $n$ jumps.\n\nAll circles in row 0 have the number 0, the circles in row 1 have the numbers 0, 1, 0, 1, 0 respectively, and the number in a circle in any other row is the sum of the numbers in the two adjacent circles in the previous row. We want the number in circle C in row 11.",
    "options": [],
    "answer": "See solution",
    "solution": "![](images/2022_Australian_Scene_p62_data_13e683a665.png)\n\nThus the number of ways for the frog to take 11 jumps from C to C is **330**.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 13665,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $p$ be a prime number. Find all triples $(a, b, c)$ of integers (not necessarily positive) such that\n\n$$\na^b b^c c^a = p.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "Suppose $a, b, c$ satisfy the equation. As $p$ is positive, this implies that $|a|^b |b|^c |c|^a = p$. Clearly, none of $a, b, c$ can be zero.\n\nObserve that $\\gcd(a, b, c) = 1$. Indeed, if $d \\mid a$, $d \\mid b$, $d \\mid c$, then the exponent of $p$ in the canonical representation of each of the positive rational numbers $|a|^b$, $|b|^c$, $|c|^a$ is divisible by $d$. Hence, the exponent of $p$ in the canonical representation of the product $|a|^b |b|^c |c|^a$ is divisible by $d$. As this product equals $p$, we get $|d| = 1$.\n\nConsider now an arbitrary prime number $q$ different from $p$. Let $\\alpha, \\beta, \\gamma$ be the exponents of $q$ in the canonical representation of the positive integers $|a|, |b|, |c|$, respectively. Then $\\alpha b + \\beta c + \\gamma a = 0$, whereby not all exponents $\\alpha, \\beta, \\gamma$ are positive because $\\gcd(a, b, c) = 1$. Consequently, if some of $\\alpha, \\beta, \\gamma$ is positive, then there must be exactly two positive exponents among $\\alpha, \\beta, \\gamma$. W.l.o.g., assume $\\alpha > 0, \\beta > 0, \\gamma = 0$. Then $\\alpha b + \\beta c = 0$, implying $\\alpha|b| = \\beta|c|$. Hence $|b|$ divides $\\beta|c|$. As $q^\\beta$ divides $|b|$ while $q^\\beta$ is relatively prime to $|c|$, this implies $q^\\beta \\mid \\beta$ and $q^\\beta \\leq \\beta$, which is impossible. This means that actually $\\alpha = \\beta = \\gamma = 0$ and $|a|, |b|, |c|$ are all powers of $p$.\n\nHence, the equation rewrites to $p^{\\alpha b} p^{\\beta c} p^{\\gamma a} = p$, where $\\alpha, \\beta, \\gamma$ are now the exponents of $p$ in the canonical representation of $|a|, |b|, |c|$, respectively. This is equivalent to $\\alpha b + \\beta c + \\gamma a = 1$. By $\\gcd(a, b, c) = 1$, one of $\\alpha, \\beta, \\gamma$ must be zero, and clearly, one of the summands $\\alpha b, \\beta c, \\gamma a$ must be positive. W.l.o.g., let $\\alpha b > 0$, i.e., $\\alpha > 0$ and $b > 0$. Now there are three cases:\n\n* If $\\beta = 0$ and $\\gamma = 0$, then $b = 1$ and $|c| = 1$. Furthermore, $\\alpha b + \\beta c + \\gamma a = 1$ reduces to $\\alpha = 1$, whence $|a| = p$. If $p > 2$, then the exponents of $a$ and $c$ in the original equation, $b$ and $a$, are both odd, whence $a$ and $c$ must have the same sign to make the product $a^b b^c c^a$ positive. Both triples $(p, 1, 1)$ and $(-p, 1, -1)$ satisfy the original equation. If $p = 2$, then $c^a$ is positive anyway, hence $a$ must be positive. Both triples $(2, 1, 1)$ and $(2, 1, -1)$ satisfy the original equation.\n* If $\\beta = 0$ and $\\gamma > 0$, then $b = 1$. Furthermore, $\\alpha b + \\beta c + \\gamma a = 1$ reduces to $\\alpha + \\gamma a = 1$, whence $a < 0$. We obtain $p^\\alpha \\leq \\gamma p^\\alpha = \\gamma|a| = \\alpha - 1 < \\alpha$, which is impossible.\n* If $\\beta > 0$ and $\\gamma = 0$, then $|c| = 1$. Furthermore, $\\alpha b + \\beta c + \\gamma a = 1$ reduces to $\\alpha b + \\beta c = 1$, which gives $c = -1$ and $\\alpha p^\\beta = 1 + \\beta$ as the only possibility. If $p > 2$, then this leads to a contradiction similar to the previous case. If $p = 2$, then $\\alpha = \\beta = 1$ is the only solution. This leads to triples $(2, 2, -1)$ and $(-2, 2, -1)$, which both satisfy the original equation.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 13666,
    "subject": "Mathematics (Olympiad)",
    "question": "The non-negative real numbers $a$, $b$, $c$ satisfy $a + b + c = 1$. What is the largest possible value of\n\n$$\na^2b + ab^2 + b^2c + bc^2 + a^2c + ac^2 + ac^2?\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "The largest possible value is $\\frac{1}{4}$, which is obtained, for example, when $a = b = \\frac{1}{2}$ and $c = 0$.\n\nFirst, rewrite the expression:\n\n$$\n\\begin{aligned}\na^2b + ab^2 + b^2c + bc^2 + a^2c + ac^2 + ac^2 &= ab(1-c) + bc(1-a) + ac(1-b) \\\\\n&= ab + bc + ac - 3abc \\\\\n&= ab(1-3c) + c(1-c).\n\\end{aligned}\n$$\n\nAssume $a \\geq b \\geq c$, then $c \\leq \\frac{1}{3}$ and therefore $(1-3c) \\geq 0$. If $c$ is fixed, then $a + b$ is also fixed, and the value of the expression is maximal when $ab$ is maximal, so $a = b = \\frac{1-c}{2}$.\n\nThen the expression can be rewritten as:\n\n$$\n\\begin{aligned}\nab(1-3c) + c(1-c) &= \\left(\\frac{1-c}{2}\\right)^2 (1-3c) + c(1-c) \\\\\n&= \\frac{1-c}{4}(1+3c^2) \\\\\n&= \\frac{1}{4}\\left(1-3c\\left((c-\\frac{1}{2})^2 + \\frac{1}{12}\\right)\\right) \\\\\n&\\leq \\frac{1}{4}.\n\\end{aligned}\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 13667,
    "subject": "Mathematics (Olympiad)",
    "question": "Suppose 2016 points on the circumference of a circle are colored red and the remaining points are colored blue. Given any natural number $n \\ge 3$, prove that there is a regular $n$-sided polygon all of whose vertices are blue.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $A_1, A_2, \\dots, A_{2016}$ be the 2016 red points on the circle, and the remaining points are blue. Let $n \\ge 3$, and let $B_1, B_2, \\dots, B_n$ be the vertices of a regular $n$-sided polygon inscribed in the circle, labeled in the counterclockwise direction. Place $B_1$ at $A_1$; in this position, some $B$'s may coincide with some $A$'s. Now, rotate the polygon counterclockwise until some $B$'s coincide with (an equal number of) $A$'s a second time. Continue rotating, each time noting when $B$'s coincide with $A$'s, until returning to the original position. The number of such coincidences is at most $2016 \\times n$. Since the interval $(0, 360^\\circ)$ contains infinitely many points, there must exist an angle $\\alpha^\\circ \\in (0, 360^\\circ)$ such that, after rotating the polygon by $\\alpha^\\circ$, no $B$ coincides with any $A$. Thus, there exists a regular $n$-sided polygon with all vertices blue.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 13668,
    "subject": "Mathematics (Olympiad)",
    "question": "A rectangle $R$ is partitioned into smaller rectangles whose sides are parallel with the sides of $R$. Let $B$ be the set of all boundary points of all the rectangles in the partition, including the boundary of $R$. Let $S$ be the set of all (closed) segments whose points belong to $B$. Let a maximal segment be a segment in $S$ which is not a proper subset of any other segment in $S$. Let an intersection point be a point in which 4 rectangles of the partition meet. Let $m$ be the number of maximal segments, $i$ the number of intersection points and $r$ the number of rectangles. Prove that $m + i = r + 3$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let a minor intersection be a point in $S$ where exactly three rectangles meet and let the number of minor intersections be $j$. Let side segments be segments corresponding to a side of a rectangle in the partition and let proper segments be segments into which intersection points cut up maximal segments.\n\nLet the number of side segments be $s$ and the number of proper segments be $p$. If we start from maximal segments, we note that each addition of an intersection point forms two new segments. Ultimately, when all the intersection points are included, only proper segments remain and therefore $p = m + 2i$.\n\nWe now multiply all proper segments by 2 to account for both sides of a proper segment and subtract 4 to account for the fact that the sides of the rectangle $R$ (which are by definition proper segments) are counted only once. Thereafter, each addition of a minor intersection increases the number of segments by 1 until we get only side segments and therefore\n$$\ns = 2p + j - 4.\n$$\nCombining the two equations, we obtain\n$$\ns = 2m + 4i + j - 4.\n$$\nNow, counting rectangles by side segments we obtain $s = 4r$ and counting rectangles by their angles we obtain $4r = 4i + 2j + 4$, the final term accounting for the four corners of $R$. We transform the equation into $2r - 6 = 2i + j - 4$. Combining all the obtained equations we get $4r = s = 2m + 2i + 2r - 6$ which gives us $2r + 6 = 2m + 2i$, i.e.\n$$\nm + i = r + 3.\n$$\n$\\square$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 13669,
    "subject": "Mathematics (Olympiad)",
    "question": "A triangle has sides of length at most $2$, $3$, and $4$ respectively. Determine, with proof, the maximum possible area of the triangle.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $\\theta$ be the largest angle of the triangle, which is the angle between the two shorter sides of lengths $a$ and $b$.\n\nThe area of the triangle is given by $A = \\frac{1}{2}ab\\sin\\theta$, where $a \\leq 2$, $b \\leq 3$, and $\\sin\\theta \\leq 1$.\n\nHence, $A \\leq \\frac{1}{2} \\times 2 \\times 3 \\times 1 = 3$, with equality when the triangle is right-angled with legs 2 and 3. The hypotenuse is $\\sqrt{2^2 + 3^2} = \\sqrt{13} < 4$, which satisfies the side constraints.\n\nThus, the maximum area is $3$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 13670,
    "subject": "Mathematics (Olympiad)",
    "question": "For all $a, b, c$ positive real numbers with $a^2b + a^2c + b^2a + b^2c + c^2a + c^2b = 1$, show that\n\n$$\n\\frac{ab + bc + ca}{1 + 2abc} \\leq \\frac{(a + b + c)^2}{4}\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "Lemma: For positive $a, b, c, x, y, z$, it holds that $ax + by + cz + 2\\sqrt{ab + bc + ca}\\sqrt{xy + yz + zx} \\leq (a + b + c)(x + y + z)$.\n\n*Proof.* Use the Cauchy-Schwarz inequality:\n\n$$\n\\begin{aligned}\nax + by + cz + 2\\sqrt{ab + bc + ca}\\sqrt{xy + yz + zx} &\\leq \\sqrt{a^2 + b^2 + c^2}\\sqrt{x^2 + y^2 + z^2} + \\sqrt{ab + bc + ca}\\sqrt{xy + yz + zx} + \\sqrt{ab + bc + ca}\\sqrt{xy + yz + zx} \\\\\n&\\leq \\sqrt{a^2 + b^2 + c^2 + 2ab + 2bc + 2ca}\\sqrt{x^2 + y^2 + z^2 + 2xy + 2yz + 2zx} \\\\\n&= (a + b + c)(x + y + z)\n\\end{aligned}\n$$\n\nTake $(x, y, z) = \\left(\\frac{a}{b+c}, \\frac{b}{a+c}, \\frac{c}{a+b}\\right)$ in the lemma. First, we observe:\n\n$$\n\\sqrt{xy + yz + zx} = \\sqrt{\\frac{a^2b + a^2c + b^2a + b^2c + c^2a + c^2b}{(a+b)(b+c)(c+a)}} = \\frac{1}{\\sqrt{1 + 2abc}}\n$$\n\nBy the lemma, we have:\n\n$$\n2\\sqrt{ab + bc + ca}\\sqrt{xy + yz + zx} \\leq (a + b + c)(x + y + z) - ax - by - cz\n$$\n\n$$\n\\frac{2\\sqrt{ab + bc + ca}}{\\sqrt{1 + 2abc}} \\leq ay + az + bx + bz + cx + cy\n$$\n\n$$\n\\frac{2\\sqrt{ab + bc + ca}}{\\sqrt{1 + 2abc}} \\leq x(b + c) + y(c + a) + z(a + b) = a + b + c\n$$\n\nFinally, squaring both sides, we get the desired inequality:\n\n$$\n\\frac{ab + bc + ca}{1 + 2abc} \\leq \\frac{(a + b + c)^2}{4}\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 13671,
    "subject": "Mathematics (Olympiad)",
    "question": "For each polynomial $P(x)$, define\n\n$$\nP_1(x) = P(x), \\quad \\forall x \\in \\mathbb{R}\n$$\n\n$$\nP_2(x) = P(P_1(x)), \\quad \\forall x \\in \\mathbb{R}\n$$\n\n$$\nP_{2024}(x) = P(P_{2023}(x)), \\quad \\forall x \\in \\mathbb{R}\n$$\n\nLet $a > 2$ be a real number. Is there a polynomial $P$ with real coefficients such that for all $t \\in (-a, a)$, the equation $P_{2024}(x) = t$ has $2^{2024}$ distinct real roots?",
    "options": [],
    "answer": "See solution",
    "solution": "We present two solutions to the problem.\n\n**First solution:**\n\nLet $P(x) = \\frac{2}{a}x^2 - a$. We show that for all $t \\in (-a, a)$, the equation\n\n$$\nP_{2024}(x) = t\n$$\n\nhas exactly $2^{2024}$ distinct real roots in $(-a, a)$. Let $x = a \\cos \\varphi$ with $0 < \\varphi < \\pi$. Then\n\n$$\nP_1(a \\cos \\varphi) = P(a \\cos \\varphi) = a(2 \\cos^2 \\varphi - 1) = a \\cos 2\\varphi\n$$\n\n$$\nP_2(a \\cos \\varphi) = P(a \\cos 2\\varphi) = a \\cos 2^2\\varphi\n$$\n\n$$\nP_{2024}(a \\cos \\varphi) = a \\cos 2^{2024}\\varphi\n$$\n\nSo the equation becomes $a \\cos 2^{2024}\\varphi = t$, or\n\n$$\n2^{2024}\\varphi = \\pm \\arccos \\frac{t}{a} + 2k\\pi\n$$\n\nwhich gives\n\n$$\n\\varphi = \\frac{\\pm \\arccos \\frac{t}{a} + 2k\\pi}{2^{2024}}, \\quad k \\in \\mathbb{Z}\n$$\n\nSince $0 < \\varphi < \\pi$, there are exactly $2^{2024}$ distinct real roots:\n\n$$\nx = a \\cos \\frac{\\arccos \\frac{t}{a} + 2k\\pi}{2^{2024}}, \\quad k = 0, 1, \\dots, 2^{2023} - 1\n$$\n\nand\n\n$$\nx = a \\cos \\frac{-\\arccos \\frac{t}{a} + 2k\\pi}{2^{2024}}, \\quad k = 1, 2, \\dots, 2^{2023}\n$$\n\nThus, $P(x) = \\frac{2}{a}x^2 - a$ satisfies the requirements. $\\blacksquare$\n\n**Second solution:**\n\nConsider $P(x) = x^2 - c$ with $c \\in (a, a^2 - a)$. For all positive integers $n$ and $t \\in (-a, a)$, the equation $P_n(x) = t$ has exactly $2^n$ distinct real roots in $(-a, a)$.\n\nFor $n = 1$, $P_1(x) = t$ has two distinct real solutions: $-\\sqrt{c + t}$ and $\\sqrt{c + t}$, both in $(-a, a)$. Assume true for $n = k$, so $P_k(x) = t$ has $2^k$ distinct roots $x_1, \\dots, x_{2^k}$ in $(-a, a)$. For $n = k + 1$, $P_{k+1}(x) = t$ is equivalent to $P(x) \\in \\{x_1, \\dots, x_{2^k}\\}$. Each $P(x) = x_i$ has two distinct real roots in $(-a, a)$, and all roots are distinct. Thus, $P_{k+1}(x) = t$ has $2^{k+1}$ distinct real roots in $(-a, a)$. By induction, for $n = 2024$, $P_{2024}(x) = t$ has $2^{2024}$ distinct real roots. $\\square$",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 13672,
    "subject": "Mathematics (Olympiad)",
    "question": "Joe and Penny play a game. Initially, there are 5000 stones in a pile, and the two players remove stones from the pile by making a sequence of moves. On the $k$th move, any number of stones between 1 and $k$ inclusive may be removed. Joe makes the odd-numbered moves and Penny makes the even-numbered moves. The player who removes the very last stone is the winner. Who wins if both players play perfectly?",
    "options": [],
    "answer": "See solution",
    "solution": "If on move $2y - 1$, Joe removes $1 \\leq a \\leq 2y - 1$ stones, then on move $2y$, Penny can remove either $2y - a$ or $2y + 1 - a$ stones (both of which are in the valid range). Thus, Penny can ensure that the two moves remove either $2y$ or $2y + 1$ stones.\n\nIf on $k$ occasions, Penny chooses to ensure $2y + 1$ stones are removed, then after move $2z$, there will be\n\n$$\nk + \\sum_{y=1}^{z} 2y = z(z + 1) + k\n$$\n\nstones removed. Thus, after move $2z$, Penny can ensure there are any number between $z(z + 1)$ and $z(z + 2)$ stones removed.\n\nSetting $z = 69$, we see that after move $2 \\cdot 69 = 138$, Penny can ensure $69 \\cdot 70 \\leq 4860 \\leq 69 \\cdot 71$ stones have been removed and so there are 140 stones remaining. Joe will leave between 1 and 139 stones, which Penny can remove on her turn.\n\n**Remark.** It should be avoided setting the problem with a total number of stones of the form $\\{n^2 + n,\\ n^2 + n + 1,\\ n^2 + 2n - 1,\\ n^2 + 2n\\}$ as then Penny can simply remove $2y$ or $2y + 1$ on each turn except for possibly the final turn (without having to make a mixture of such moves). In particular, this means $2023 = 44^2 + 2 \\cdot 44 - 1$ would not be a good choice for the total number of stones.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 13673,
    "subject": "Mathematics (Olympiad)",
    "question": "Given $n$ contestants arranged in a queue, what is the maximum number of moves possible if, in each move, a contestant $C_i$ moves forward in the queue past some other contestants? Each move must decrease a certain total weight defined as follows: For each pair where $C_i$ is behind $C_j$ in the queue and $i < j$ (a reverse pair), assign a weight of $2^{i-1}$. The total weight of an arrangement is the sum of the weights for all reverse pairs. What is the upper bound for the number of moves possible?",
    "options": [],
    "answer": "See solution",
    "solution": "We show that at most $2^n - n - 1$ moves are possible. Given an arrangement of the $n$ contestants, for each reverse pair (where $C_i$ is behind $C_j$ and $i < j$), assign a weight of $2^{i-1}$. The total weight is the sum over all reverse pairs. The maximum total weight occurs for the arrangement $C_n, C_{n-1}, \\dots, C_1$, which is:\n\n$$\n(n-1)2^0 + (n-2)2^1 + (n-3)2^2 + \\dots + 2 \\cdot 2^{n-3} + 1 \\cdot 2^{n-2} = 2^n - n - 1.\n$$\n\nWhen $C_i$ moves forward, she passes at least one contestant $C_j$ with $j > i$ and at most $i-1$ contestants $C_j$ with $j < i$. This reduces the total weight by at least $2^{i-1}$ and adds at most $2^0 + 2^1 + \\dots + 2^{i-2} = 2^{i-1} - 1$. Thus, the total weight decreases with each move. Since the maximum total weight is $2^n - n - 1$, at most $2^n - n - 1$ moves are possible. $\\Box$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 13674,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all positive integers $p, m, n$ such that $p^m = n^3 + 8$ and $p$ is a prime number.",
    "options": [],
    "answer": "See solution",
    "solution": "By moving $n^3$, we get a sum of cubes on the right-hand side:\n\n$$\np^m = n^3 + 8 = (n+2)(n^2 - 2n + 4).\n$$\n\nSince $p$ is prime, each of the factors on the right-hand side must be a power of $p$:\n\n$$\nn + 2 = p^{\\alpha}, \\quad n^2 - 2n + 4 = p^{\\beta},\n$$\n\nwhere $\\alpha$ and $\\beta$ are positive integers.\n\nNote that $n^2-2n+4 \\ge n+2$, since that is equivalent to $n^2-3n+2 \\ge 0$, i.e., $(n-1)(n-2) \\ge 0$, which holds for $n \\ge 2$ or $n \\le 1$. Therefore, $\\beta \\ge \\alpha$. We can conclude that $p^\\alpha$ divides both $n+2$ and $n^2-2n+4$, so it also divides\n\n$$\nn \\cdot (n+2) - (n^2 - 2n + 4) = 4n - 4,\n$$\n\nand then it also divides $4 \\cdot (n+2) - (4n-4) = 12$, so $p=2$ or $p=3$.\n\nIf $p=2$, then $n+2 = 2^\\alpha$ is at most 4, and since $n > 0$, it follows that $n=2$, which gives the solution $(2, 4, 2)$.\n\nIf $p=3$, then $n+2 = 3^\\alpha$ is 3, so $n=1$, which gives the second solution $(3, 2, 1)$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 13675,
    "subject": "Mathematics (Olympiad)",
    "question": "Determine all functions $f$ defined for real numbers and taking real numbers as values such that\n\n$$(x - y)f(x + y) = x f(x) - y f(y)$$\n\nfor all real numbers $x$ and $y$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $a$ be any real number with $a \\neq 0, 1, 2$. The substitution $y = 1$ and $x = a$ leads to the equation\n\n$$\nf(a + 1) = \\frac{a f(a) - f(1)}{a - 1}.\n$$\n\nThe substitution $y = 1$ and $x = a + 1$ allows us to write $f(a + 2)$ in terms of $f(a)$ and $f(1)$ as follows:\n\n$$\n\\begin{aligned}\nf(a + 2) &= \\frac{(a + 1) f(a + 1) - f(1)}{a} \\\\\n&= \\frac{(a + 1) \\left( \\frac{a f(a) - f(1)}{a - 1} \\right) - f(1)}{a} \\\\\n&= \\frac{(a + 1) f(a) - 2 f(1)}{a - 1}\n\\end{aligned}\n$$\n\nThe substitution $y = 2$ and $x = a$ allows us to write $f(a + 2)$ in terms of $f(a)$ and $f(2)$:\n\n$$\nf(a + 2) = \\frac{a f(a) - 2 f(2)}{a - 2}.\n$$\n\nWe can equate these two expressions for $f(a + 2)$ to obtain\n\n$$\n\\frac{(a + 1) f(a) - 2 f(1)}{a - 1} = \\frac{a f(a) - 2 f(2)}{a - 2}.\n$$\n\nNow solve for $f(a)$ to obtain $f(a) = (a - 1) f(2) + (2 - a) f(1)$. Therefore,\n\n$$\nf(x) = (x - 1) f(2) + (2 - x) f(1), \\quad \\text{for all } x \\neq 0, 1, 2.\n$$\n\nNote that the equation above also holds trivially for $x = 1$ and $x = 2$, and it can be observed to hold for $x = 0$ as well, by substituting $y = -3$ and $x = 3$ into the original equations. Thus,\n\n$$\nf(x) = (x - 1) f(2) + (2 - x) f(1) \\quad \\text{for all } x \\in \\mathbb{R}.\n$$\n\nSince $f(1)$ and $f(2)$ are simply constants, it follows that $f$ must be a linear function. Finally, we can verify that all linear functions satisfy the functional equation, by substituting $f(x) = m x + c$:\n\n$$\n\\begin{aligned}\n(x - y) f(x + y) &= (x - y) (m(x + y) + c) = m x^2 - m y^2 + c x - c y \\\\\nx f(x) - y f(y) &= x (m x + c) - y (m y + c) = m x^2 + c x - m y^2 - c y\n\\end{aligned}\n$$\n\nThus, the solutions are all linear functions $f(x) = m x + c$ for real $m, c$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 13676,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $\\triangle ABC$ be an acute-angled triangle, and let $D$ be the foot of the altitude from $C$. The angle bisector of $\\angle ABC$ intersects $CD$ at $E$ and meets the circumcircle $\\omega$ of triangle $\\triangle ADE$ again at $F$. If $\\angle ADF = 45^\\circ$, show that $CF$ is tangent to $\\omega$.",
    "options": [],
    "answer": "See solution",
    "solution": "Since $\\angle CDF = 90^\\circ - 45^\\circ = 45^\\circ$, the line $DF$ bisects $\\angle CDA$, and so $F$ lies on the perpendicular bisector of segment $AE$, which meets $AB$ at $G$. Let $\\angle ABC = 2\\beta$. Since $ADEF$ is cyclic, $\\angle AFE = 90^\\circ$, and hence $\\angle FAE = 45^\\circ$. Further, as $BF$ bisects $\\angle ABC$, we have $\\angle FAB = 90^\\circ - \\beta$, and thus $\\angle EAB = \\angle AEG = 45^\\circ - \\beta$ and $\\angle AED = 45^\\circ + \\beta$, so $\\angle GED = 2\\beta$. This implies that right-angled triangles $\\triangle EDG$ and $\\triangle BDC$ are similar, and so we have $\\frac{|GD|}{|CD|} = \\frac{|DE|}{|DB|}$. Thus the right-angled triangle $\\triangle DEB$ and $\\triangle DGC$ are similar, whence $\\angle GCD = \\angle DBE = \\beta$. But $\\angle DFE = \\angle DAE = 45^\\circ - \\beta$, then $\\angle GFD = 45^\\circ - \\angle DFE = \\beta$. Hence $GDCF$ is cyclic, so $\\angle GFC = 90^\\circ$, whence $CF$ is perpendicular to the radius $FG$ of $\\omega$. It follows that $CF$ is a tangent to $\\omega$, as required.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 13677,
    "subject": "Mathematics (Olympiad)",
    "question": "A set $a_1 \\ge a_2 \\ge \\dots \\ge a_{100n}$ of positive numbers has the following property: for any subset of $2n + 1$ numbers, the sum of the $n$ maximal numbers in this subset is greater than the sum of the remaining $n + 1$ numbers. Prove that\n\n$$\n(n+1)(a_1 + a_2 + \\dots + a_n) > a_{n+1} + a_{n+2} + \\dots + a_{100n}.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "Let us sum up all the inequalities for each subset of consecutive $2n + 1$ numbers in our set:\n\n$$\n\\begin{align*}\na_1 + a_2 + \\dots + a_n &\\ge a_{n+1} + a_{n+2} + \\dots + a_{2n+1}, \\\\\na_2 + a_3 + \\dots + a_{n+1} &\\ge a_{n+2} + a_{n+3} + \\dots + a_{2n+2}, \\\\\n\\vdots \\\\\na_{n+1} + a_{n+2} + \\dots + a_{2n} &\\ge a_{2n+1} + a_{2n+2} + \\dots + a_{3n+1},\n\\end{align*}\n$$\n\nWe obtain the following inequality:\n\n$$\n\\begin{align*}\na_1 + 2a_2 + 3a_3 + \\dots + n a_n + (n-1)a_{n+1} + (n-2)a_{n+2} + \\dots + 1 \\cdot a_{2n-1} + 0 \\cdot a_{2n} &\\ge \\\\\n&\\ge a_{2n+1} + a_{2n+2} + \\dots\n\\end{align*}\n$$\n\n(The last $2n$ summands on the right-hand side have nontrivial multiplicities from $1$ to $n$, but we may replace these multiplicities by $1$'s.) Now, add to this inequality the following $n$ trivial inequalities:\n\n$$\n\\begin{align*}\nn(a_1 - a_{n+1}) &\\ge 0, \\\\\n(n-1)(a_2 - a_{n+2}) &\\ge 0, \\\\\n\\vdots \\\\\n1 \\cdot (a_n - a_{2n}) &\\ge 0.\n\\end{align*}\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 13678,
    "subject": "Mathematics (Olympiad)",
    "question": "Назовём пару прямоугольников вложимой, если один из них можно вложить в другой.\n\nПусть горизонтальная сторона квадрата разбита на отрезки длины $a_1, \\dots, a_n$ (слева направо), а вертикальная — на отрезки длины $b_1, \\dots, b_n$ (сверху вниз). Переставив «столбцы» и «строки», можно считать, что $a_1 \\geq \\dots \\geq a_n$ и $b_1 \\geq \\dots \\geq b_n$.\n\nОбозначим через $Q_{i,j}$ прямоугольник разбиения со сторонами $a_i$ и $b_j$.\n\nСколько пар прямоугольников $Q_{i,j}$ и $Q_{k,l}$ вложимы? Докажите ваш ответ.",
    "options": [],
    "answer": "See solution",
    "solution": "Заметим, что при $i \\leq k$ и $j \\leq l$ пара $Q_{i,j}$ и $Q_{k,l}$ вложима.\n\nПоскольку $a_1 + \\dots + a_n = b_1 + \\dots + b_n$, найдутся различные индексы $i$ и $j$ такие, что $a_i \\geq b_i$ и $a_j \\leq b_j$. Можно считать, что $i < j$. Тогда существует индекс $k \\in [i, j]$ такой, что $a_k \\leq b_k$ и $a_{k-1} \\geq b_{k-1}$.\n\nМожно выбрать следующие прямоугольники:\n\n$Q_{1,1}, Q_{1,2}, \\dots, Q_{1,k-1}, Q_{2,k-1}, \\dots, Q_{k,k-1}$, а также $Q_{k-1,k}, Q_{k,k}, \\dots, Q_{k,n}, Q_{k+1,n}, \\dots, Q_{n,n}$.\n\nИх количество равно $2(k-1) + 2(n-k+1) = 2n$. Любая пара из них, кроме $(Q_{k,k-1}, Q_{k-1,k})$, вложима по замечанию выше. Оставшаяся пара также вложима, поскольку $a_k \\leq b_k$ и $b_{k-1} \\leq a_{k-1}$ (для вложения один прямоугольник нужно повернуть на $90^\\circ$).",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 13679,
    "subject": "Mathematics (Olympiad)",
    "question": "Consider the following mapping between the set $S$ of all positive integers formed by $n$ 1's and $n$ 2's, and the set $T$ of all positive integers with $n$ digits formed by 1, 2, 3, 4 such that the numbers of 1's and 2's are equal.\n\nDescribe and prove a bijection between $S$ and $T$, and deduce that\n\n$$\nf(n) = |S| = |T| = g(n).\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "For each $m \\in S$, pair every two consecutive digits of $m$ from left to right, giving $n$ pairs. Replace the pairs 11, 22, 12, 21 by 1, 2, 3, 4 respectively. If there are $a, b, c, d$ pairs of 11, 22, 12, 21 respectively, then $m$ has $2a + c + d$ 1's and $2b + c + d$ 2's, so $a = b$. Thus, the image has equal numbers of 1's and 2's, so it belongs to $T$.\n\nThe mapping is reversible: for each $m \\in T$, replace digits 1, 2, 3, 4 by 11, 22, 12, 21 respectively. If there are $a, a, c, d$ copies of 1, 2, 3, 4, then the image has $2a + c + d$ 1's and $2a + c + d$ 2's, so it belongs to $S$. Thus, the mapping is a bijection, and\n\n$$\nf(n) = |S| = |T| = g(n).\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 13680,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $\\mathbb{N}$ be the set of positive integers. Let $f: \\mathbb{N} \\to \\mathbb{N}$ be a function satisfying the following two conditions:\n\n(a) $f(m)$ and $f(n)$ are relatively prime whenever $m$ and $n$ are relatively prime.\n\n(b) $n \\leq f(n) \\leq n + 2012$ for all $n$.\n\nProve that for any natural number $n$ and any prime $p$, if $p$ divides $f(n)$ then $p$ divides $n$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $g(n)$ be the smallest prime factor of $f(n)$. (Since $f(n) \\geq n$, $f(n)$ must have a prime factor unless $n=1$ and $f(1)=1$. In this case, we define $g(1)=1$.)\n\nFirst, we show that for any prime $p$ and any $k \\geq 1$, $f(p^k)$ is a power of $g(p)$.\n\nSuppose for the sake of contradiction that $f(p^k)$ is not a power of $g(p)$ for some $p$ and $k$. Choose $M$ sufficiently large so that $p^k < M \\cdot 2013!$, and let $P$ be the set of primes less than or equal to $M \\cdot 2013! + 1$. For any $q \\in P$, we have\n\n$$\ng(q) \\leq f(q) \\leq q + 2012 \\leq M \\cdot 2013! + 2013,\n$$\n\nand furthermore the numbers $M \\cdot 2013! + i$ are composite for $2 \\leq i \\leq 2013$. It follows that $g(q) \\in P$, and so $g$ can be treated as a function from $P$ to $P$. Clearly, $g$ is injective because $f(q_1)$ and $f(q_2)$ are relatively prime for any distinct $q_1, q_2 \\in P$. Then, $g$ is bijective and in particular surjective.\n\nNow we can apply the same reasoning we used on primes in $P$ to $p^k$. We have\n\n$$\nf(p^k) \\leq p^k + 2012 \\leq M \\cdot 2013! + 2013,\n$$\n\nand so $f(p^k)$ has only prime factors in $P$. Suppose for the sake of contradiction that there is some prime $q \\in P$ not equal to $g(p)$ which divides $f(p^k)$. Then, by the surjectivity of $g$, there exists $r \\in P$ such that $g(r) = q$ (and with $r \\neq p$). We then find that $p^k$ and $r$ are relatively prime, but $q$ divides both $f(r)$ and $f(p^k)$, contradicting the first condition in the problem. It follows that $f(p^k)$ is a power of $g(p)$, as claimed.\n\nNext, we show that in fact $g(p) = p$. Indeed, suppose instead that $g(p) = q \\neq p$. Then, choose $\\ell$ sufficiently large so that $q^\\ell > 2013$, and letting $P = p^{q^\\ell - q^{\\ell-1}}$, choose $k$ sufficiently large so that $P^k > q^\\ell$. We find that\n\n$$\nP^k \\equiv 1^k \\equiv 1 \\pmod{q^\\ell}.\n$$\n\nBy the second condition of the problem, the only possible residues of $f(P^k)$ modulo $q^\\ell$ are the numbers between 1 and 2013, so in particular, $f(P^k)$ cannot be divisible by $q^\\ell$. On the other hand, $f(P^k) \\geq P^k > q^\\ell$, so $f(P^k)$ cannot be any power of $q$, contradicting the fact that since $P^k$ is a power of $p$, $f(P^k)$ must be a power of $g(p) = q$. This proves our claim that $g(p) = p$.\n\nFinally, let $p$ be any prime, and let $n$ be any positive integer relatively prime to $p$. Then, $f(p)$ and $f(n)$ are relatively prime, and since $f(p)$ is a power of $p$, it follows that $f(n)$ is not divisible by $p$. This is the contrapositive of the desired result, so we are done.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 13681,
    "subject": "Mathematics (Olympiad)",
    "question": "If the number $K = \\frac{9n^2 + 31}{n^2 + 7}$ is an integer, find the possible values of $n \\in \\mathbb{Z}$.",
    "options": [],
    "answer": "See solution",
    "solution": "We have\n\n$$\nK = \\frac{9n^2 + 31}{n^2 + 7} = \\frac{9(n^2 + 7) - 32}{n^2 + 7} = 9 - \\frac{32}{n^2 + 7}.\n$$\n\nSince $K$ is integer, $n^2 + 7$ must divide 32. Also, $n^2 + 7 \\geq 8$, so:\n\n$$\nn^2 + 7 \\in \\{8, 16, 32\\} \\implies n^2 \\in \\{1, 9, 25\\} \\implies n \\in \\{-1, 1, -3, 3, -5, 5\\}.\n$$\n\nAlternatively, we can solve the equation for $n^2$ and determine the suitable values of $K$ for which $n^2$ is a nonnegative integer.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 13682,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $p \\ge 5$ be a prime number. Show that there exists an integer $t \\in \\{1, 2, \\ldots, p\\}$ such that the equation\n\n$$\nx^2 = y^{\\frac{p-1}{2}} + t\n$$\n\nhas no integer solutions.",
    "options": [],
    "answer": "See solution",
    "solution": "We consider three cases depending on the value of $p$.\n\n**Case 1.** $p \\equiv 1 \\pmod{4}$.\n\nPicking $t = 2$ works. Since $x^2$ and $y^{\\frac{p-1}{2}}$ are perfect squares, their difference cannot be $2$.\n\n**Case 2.** $p \\equiv 3 \\pmod{4}$ and $p > 7$.\n\nNote that $y^{\\frac{p-1}{2}} \\equiv -1, 0, \\text{ or } 1 \\pmod{p}$. Thus, it suffices to show that there exists $t$ such that $t-1, t, t+1$ are quadratic nonresidues modulo $p$. To this end, note that the squares $1, 25, 49$ form an arithmetic progression with common difference $24$. Let $c$ be the inverse of $24$ modulo $p$, then\n\n$$\nc+1 \\equiv c(1+24) \\equiv 25c \\pmod{p},\n$$\n\n$$\nc+2 \\equiv c(1+48) \\equiv 49c \\pmod{p},\n$$\n\nwhich implies $\\left(\\frac{c}{p}\\right) = \\left(\\frac{c+1}{p}\\right) = \\left(\\frac{c+2}{p}\\right)$. So when $\\left(\\frac{c}{p}\\right) = -1$, picking $t \\equiv c+1 \\pmod{p}$ works.\n\nOn the other hand, when $\\left(\\frac{c}{p}\\right) = 1$, then since $p \\equiv 3 \\pmod{4}$ we have $\\left(\\frac{-1}{p}\\right) = -1$, so picking $t \\equiv -c-1 \\pmod{p}$ works.\n\n**Case 3.** $p=7$. We will show that $t=7$ works, i.e., that the equation\n\n$$\nx^2 = y^3 + 7\n$$\n\nhas no integer solution. Suppose the contrary. If $y$ is even, then $x^2 \\equiv 3 \\pmod{4}$, which is impossible. Thus $y$ is odd. Write the equation as $x^2 + 1 = (y+2)(y^2 - 2y + 4)$, and note that $y^2 - 2y + 4 = (y-1)^2 + 3 \\equiv 3 \\pmod{4}$. So there exists a prime $q \\equiv 3 \\pmod{4}$ such that $q \\mid y^2 - 2y + 4$.\n\nHowever, since $q \\mid y^2 - 2y + 4$, so $q \\mid x^2 + 1$, and thus $\\left(\\frac{-1}{q}\\right) = 1$, which contradicts $q \\equiv 3 \\pmod{4}$. Therefore, the equation has no integer solutions.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 13683,
    "subject": "Mathematics (Olympiad)",
    "question": "Six points $A$, $B$, $C$, $D$, $E$, and $F$ lie in a straight line in that order. Suppose that $G$ is a point not on the line and that $AC = 26$, $BD = 22$, $CE = 31$, $DF = 33$, $AF = 73$, $CG = 40$, and $DG = 30$. Find the area of $\\triangle BGE$.",
    "options": [],
    "answer": "See solution",
    "solution": "Because $CD = AF - AC - DF = 14$, the side lengths of $\\triangle CDG$ are $14$, $30$, and $40$. By Heron's Formula,\n\n$$\n\\text{Area}(\\triangle CDG) = \\sqrt{42(42 - 14)(42 - 30)(42 - 40)} = 168,\n$$\n\nimplying that the distance from $G$ to line $CD$ is $\\frac{2 \\cdot 168}{14} = 24$. Then because $BE = BD + CE - CD = 39$, the area of $\\triangle BGE$ is $\\frac{1}{2} \\cdot 39 \\cdot 24 = 468$.\n\n![](images/2025AIME_II_Solutions_p1_data_5d67b514f1.png)\n\nOR\n\nAs in the first solution, $CD = 14$. Let $H$ be the projection of $G$ onto line $AF$. Note that\n\n$$\nCD^2 + DG^2 = 14^2 + 30^2 = 1096 < 1600 = CG^2,\n$$\n\nimplying that $\\angle CDG$ is obtuse, and it follows that $D$ lies between $C$ and $H$. Applying the Pythagorean Theorem to $\\triangle CHG$ and $\\triangle DHG$ yields\n\n$$\n40^2 = CH^2 + GH^2 = (DH + 14)^2 + GH^2\n$$\n\nand\n\n$$\n30^2 = DH^2 + GH^2.\n$$\n\nSubtracting the second equation from the first and dividing by $28$ gives $25 = DH + 7$, so $DH = 18$ and $GH = 24$. The result then follows as in the first solution.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 13684,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $n$ be a square-free positive even number, $k$ an integer, and $p$ a prime number such that $p < 2\\sqrt{n}$, $p \\nmid n$, and $p \\nmid n + k^2$. Prove that $n$ can be written as $n = ab + bc + ca$, where $a, b, c$ are distinct positive integers.",
    "options": [],
    "answer": "See solution",
    "solution": "Since $n$ is even, we have $p \\ne 2$. As $p \\nmid n$, we have $p \\nmid k$. We may assume without loss of generality that $0 < k < p$. Set $a = k$, $b = p - k$, then $c = \\frac{n - k(p - k)}{p} = \\frac{n + k^2}{p} - k$. By assumption, $c$ is an integer, and $a, b$ are distinct positive integers. It remains to show that $c > 0$ and $c \\ne a, b$.\n\nBy the AM-GM inequality, $\\frac{n}{k} + k \\geq 2\\sqrt{n} > p$, thus $n + k^2 > pk$, hence $c > 0$.\n\nIf $c = a$, then $\\frac{n + k^2}{p} - k = k$, so $n = k(2p - k)$. Since $n$ is even, $k$ is also even, so $n$ is divisible by $4$, which contradicts the fact that $n$ is square-free.\n\nIf $c = b$, then $n = p^2 - k^2$. Since $n$ is even, $k$ is odd, implying that $n$ is again divisible by $4$, a contradiction.\n\nWe conclude that $a, b, c$ satisfy all the requirements, completing the proof.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 13685,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all positive integers $n \\geq 2$ for which odd positive integers $a_1, a_2, \\dots, a_n$ exist (not necessarily distinct) such that\n\n$$\na_1^2 + a_2^2 + \\dots + a_n^2\n$$\n\nis a square of some positive integer.\n\n![](Ilko_Brnetic.png)",
    "options": [],
    "answer": "See solution",
    "solution": "We use the following facts:\n\n1. If $m$ is even, then $m^2$ is divisible by $4$.\n2. If $m$ is odd, then $m^2$ gives remainder $1$ when divided by $8$.\n\nThe first is obvious. For the second, note that $(2k-1)^2 = 4k(k-1) + 1$ for integer $k$, and $k(k-1)$ is always even.\n\nThus, every integer square gives remainder $0$, $1$, or $4$ modulo $8$.\n\nSince each $a_i$ is odd, $a_i^2 \\equiv 1 \\pmod{8}$, so $a_1^2 + \\dots + a_n^2 \\equiv n \\pmod{8}$.\n\nTherefore, $n$ must be congruent to $0$, $1$, or $4$ modulo $8$.\n\nWe show all such $n$ work:\n\n- If $n \\equiv 0$ or $4 \\pmod{8}$, i.e., $n = 4t$ for $t \\in \\mathbb{N}$, let $a_1 = \\dots = a_{n-1} = 1$, $a_n = 2t-1$:\n\n  $$\na_1^2 + \\dots + a_n^2 = (n-1) \\cdot 1^2 + (2t-1)^2 = 4t-1 + 4t^2 - 4t + 1 = (2t)^2.\n  $$\n\n- If $n = 8t + 1$ for $t \\in \\mathbb{N}$, let $a_1 = \\dots = a_{n-1} = 1$, $a_n = 2t-1$:\n\n  $$\na_1^2 + \\dots + a_n^2 = (n-1) \\cdot 1^2 + (2t-1)^2 = 8t + 4t^2 - 4t + 1 = (2t+1)^2.\n  $$\n\nThus, all $n \\geq 2$ with $n \\equiv 0, 1, 4 \\pmod{8}$ are solutions.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 13686,
    "subject": "Mathematics (Olympiad)",
    "question": "Show that there are no rational numbers $x$ and $y$ such that\n\n$$\nx - \\frac{1}{x} + y - \\frac{1}{y} = 4.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "Suppose that there are rational numbers $x$ and $y$ such that $x - \\frac{1}{x} + y - \\frac{1}{y} = 4$.\n\nSince $\\left(-\\frac{1}{x}, y\\right)$, $\\left(x, -\\frac{1}{y}\\right)$, and $\\left(-\\frac{1}{x}, -\\frac{1}{y}\\right)$ are solutions of the equation, we can assume that $x > 0$ and $y > 0$.\n\nLetting $u = xy$, we get $x + y = \\frac{4xy}{xy - 1} = \\frac{4u}{u - 1}$. Note that $u > 0$ and $u \\ne 1$.\n\nNow the quadratic equation\n\n$$\nT^2 - \\frac{4u}{u-1}T + u = 0\n$$\n\nhas rational coefficients. For $x$ and $y$ to be rational, the discriminant must be a rational square:\n\n$$\n\\left(\\frac{4u}{u-1}\\right)^2 - 4u = \\frac{16u^2}{(u-1)^2} - \\frac{4u(u-1)^2}{(u-1)^2} = \\frac{16u^2 - 4u(u-1)^2}{(u-1)^2}\n$$\n\nExpanding $4u(u-1)^2$ gives $4u(u^2 - 2u + 1) = 4u^3 - 8u^2 + 4u$, so:\n\n$$\n16u^2 - (4u^3 - 8u^2 + 4u) = -4u^3 + 24u^2 - 4u\n$$\n\nThus, the discriminant is\n\n$$\n\\frac{-4u^3 + 24u^2 - 4u}{(u-1)^2}\n$$\n\nFor rational $x$ and $y$, this must be a rational square. But for rational $u$, this expression cannot be a rational square except possibly for finitely many $u$, and direct checking shows there are no such $u$ with $u > 0$ and $u \\ne 1$. Therefore, there are no rational solutions.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 13687,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $ABC$ be a triangle with $AB < AC$. Let $D$ be the intersection point of the internal bisector of angle $BAC$ and the circumcircle of $ABC$. Let $Z$ be the intersection point of the perpendicular bisector of $AC$ with the external bisector of angle $\\angle BAC$. Prove that the midpoint of the segment $AB$ lies on the circumcircle of triangle $ADZ$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $\\alpha = \\angle BAD = \\angle DAC$. Let $M$ and $N$ be the midpoints of $AB$ and $AC$, respectively. Let $O$ be the circumcentre of triangle $ABC$. Thus $O$ lies on the perpendicular bisectors of $AB$, $AC$, and $AD$. Let line $AD$ intersect lines $MO$ and $NO$ at $R$ and $S$, respectively.\n\n![](images/Australian-Scene-2017_p120_data_f12f8536fd.png)\n\nSince the internal and external bisectors of an angle are perpendicular, we have $\\angle CAZ = 90^\\circ - \\alpha$. The angle sum in $\\triangle ANZ$ yields $\\angle AZN = \\alpha$. The angle sum in $\\triangle SAN$ yields\n\n$$\n\\angle OSR = \\angle NSA = 90^\\circ - \\alpha.\n$$\n\nSumming the angles in $\\triangle MAR$ yields\n\n$$\n\\angle SRO = \\angle ARM = 90^\\circ - \\alpha = \\angle OSR.\n$$\n\nIf $l$ is the line through $O$ that is perpendicular to $AD$, then $R$ and $S$ are symmetric in $l$. Since $O$ is the centre of circle $ABC$, we have $OA = OD$. So $A$ and $D$ are also symmetric in $l$. It follows that $AS = RD$.\n\nRecall $\\angle MAR = \\alpha = \\angle AZN$, and $\\angle RMA = 90^\\circ = \\angle SAZ$. It follows that $\\triangle RMA \\sim \\triangle SAZ$ (AA). Therefore\n\n$$\n\\frac{MR}{MA} = \\frac{AS}{AZ} = \\frac{RD}{AZ}.\n$$\n\nFurthermore,\n\n$$\n\\angle MRD = 180^\\circ - \\angle ARM = 90^\\circ + \\alpha = \\angle SAZ + \\angle MAR = \\angle MAZ.\n$$\n\nHence $\\triangle MRD \\sim \\triangle MAZ$ (PAP). Therefore, $\\angle AZM = \\angle RDM = \\angle ADM$. Thus $MAZD$ is cyclic. Therefore circle $ADZ$ passes through $M$, as required. $\\square$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 13688,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $n$ be an integer greater than $1$ and let $S$ be a finite set containing more than $n+1$ elements. Consider the collection of all sets $\\mathcal{A}$ of subsets of $S$ satisfying the following two conditions:\n\n(a) Each member of $\\mathcal{A}$ contains at least $n$ elements of $S$; and\n\n(b) Each element of $S$ is contained in at least $n$ members of $\\mathcal{A}$.\n\nDetermine\n$$\n\\max_{\\mathcal{A}} \\min_{\\mathcal{B}} |\\mathcal{B}|,\n$$\nas $\\mathcal{B}$ runs through all subsets of $\\mathcal{A}$ whose members cover $S$, and $\\mathcal{A}$ runs through the above collection.",
    "options": [],
    "answer": "See solution",
    "solution": "The required number is $m = |S| - n$.\n\nWe begin by showing that any set $\\mathcal{A}$ of subsets of $S$ satisfying the two conditions in the statement has a subcover of cardinality at most $m$.\n\nThis is clear if $S$ is a member of $\\mathcal{A}$.\n\nAssume henceforth that $\\mathcal{A}$ does not contain $S$. If some member $A$ of $\\mathcal{A}$ has more than $n$ elements, for each element of $S \\setminus A$ choose a containing member of $\\mathcal{A}$. The latter along with $A$ form a subcover of $\\mathcal{A}$ of cardinality $|S \\setminus A| + 1 \\leq |S| - (n+1) + 1 = m$.\n\nAssume henceforth that each member of $\\mathcal{A}$ has exactly $n$ elements. Fix a member $A$ of $\\mathcal{A}$.\n\nIf some member $B$ of $\\mathcal{A}$ contains more than one element of $S \\setminus A$, for each element of $S \\setminus (A \\cup B)$ choose a containing member of $\\mathcal{A}$. The latter along with $A$ and $B$ form a subcover of $\\mathcal{A}$ of cardinality $|S \\setminus (A \\cup B)| + 2 = |S \\setminus A| - |(S \\setminus A) \\cap B| + 2 \\leq |S \\setminus A| = m$.\n\nFinally, if no member of $\\mathcal{A}$ contains more than one element of $S \\setminus A$, write $S \\setminus A = \\{x_1, x_2, \\dots, x_m\\}$, choose a member $A_1$ of $\\mathcal{A}$ containing $x_1$ and notice that $A \\setminus A_1$ is a singleton set, say $A \\setminus A_1 = \\{x\\}$. Since $x_2$ is contained in at least $n$ members of $\\mathcal{A}$, each of which contains (exactly) $n-1$ elements of $A$, we may choose a member $A_2$ of $\\mathcal{A}$ containing both $x$ and $x_2$ (recall that $n \\geq 2$). If $n \\geq 3$, continue choosing members $A_i$ of $\\mathcal{A}$ containing $x_i$, $i = 3, \\dots, n$, to form an $m$-element subcover of $\\mathcal{A}$ consisting of $A_1, A_2, \\dots, A_m$.\n\nTo complete the proof, we produce a set of subsets of $S$ satisfying the two conditions in the statement, no subcover of which has less than $m$ members. To this end, write $S = \\{1, 2, \\dots, m+n\\}$, $m \\geq 2$, and let $S_1, S_2, \\dots, S_n$ be the $(n-1)$-element subsets of the upper part $\\{m+1, m+2, \\dots, m+n\\}$ of $S$. The sets $S_{i,j} = S_j \\cup \\{i\\}$, $i = 1, 2, \\dots, m$, $j = 1, 2, \\dots, n$, satisfy both conditions in the statement and at least $m$ of them are needed to cover $S$. (The condition $m \\geq 2$ is required for an element in the upper part to lie in at least $n$ of these sets.)",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 13689,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all functions $f : \\mathbb{R} \\to \\mathbb{R}$ such that for all real numbers $x$ and $y$,\n$$\nf(x + y) + y \\leq f(f(f(x))).\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "First, set $y = 0$ in the initial inequality:\n$$\nf(x) \\leq f(f(f(x))) \\quad \\forall x \\in \\mathbb{R}. \\qquad (1)\n$$\nNext, set $y = f(f(x)) - x$ in the original inequality:\n$$\nf(f(x)) \\leq x \\quad \\forall x \\in \\mathbb{R}. \\qquad (2)\n$$\nReplace $x$ by $f(x)$ in (2):\n$$\nf(f(f(x))) \\leq f(x).\n$$\nTogether with (1), this gives $f(f(f(x))) = f(x)$.\n\nNow the original inequality becomes:\n$$\nf(x + y) + y \\leq f(x) \\quad \\forall x, y \\in \\mathbb{R}. \\qquad (3)\n$$\nSet $x = 0$ in (3):\n$$\nf(y) \\leq a - y \\quad \\forall y \\in \\mathbb{R}, \\qquad (4)\n$$\nwhere $a = f(0)$.\n\nSet $y = -x$ in (3):\n$$\na - x \\leq f(x). \\qquad (5)\n$$\nComparing (4) and (5), we obtain $f(x) = a - x$.\n\nIt is easy to verify that the function $f(x) = a - x$ satisfies the given inequality for any real number $a$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 13690,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $\\triangle ABC$ be an acute and non-isosceles triangle. Its angle bisectors $AL_1$ and $BL_2$ intersect at the point $I$. Points $D$ and $E$ are chosen on the segments $AL_1$ and $BL_2$ such that $\\angle DBC = \\frac{1}{2}\\angle A$ and $\\angle EAC = \\frac{1}{2}\\angle B$. The lines $AE$ and $BD$ intersect at a point $P$. Let $K$ be the point symmetric to $I$ with respect to the line $DE$. Prove that the lines $KP$ and $DE$ intersect at a point on the circumcircle of $\\triangle ABC$.",
    "options": [],
    "answer": "See solution",
    "solution": "Denote $\\angle BAC = \\alpha$, $\\angle ABC = \\beta$, $\\angle ACB = \\gamma$. Let $N$ be the midpoint of arc $ACB$ of the circumcircle of $\\triangle ABC$.\n\nWe will show that $N$ lies on the line $DE$.\n\nFirst, both points $D$ and $E$ lie inside $\\triangle ANB$. Since both are inside $\\triangle ABC$, $\\frac{1}{2}\\beta = \\angle EAC < \\angle BAC = \\alpha$ and $\\frac{1}{2}\\alpha = \\angle DBC < \\angle ABC = \\beta$.\n\nAlso,\n$$\n\\angle NAB = \\angle NBA = 90^\\circ - \\frac{1}{2}\\gamma = \\frac{1}{2}(\\alpha + \\beta)\n$$\n\nand\n$$\n\\angle EBA = \\frac{1}{2}\\beta < \\frac{1}{2}(\\alpha + \\beta) = \\angle NBA, \\quad \\angle EAB = \\alpha - \\frac{1}{2}\\beta < \\frac{1}{2}(\\alpha + \\beta) = \\angle NAB\n$$\n\nbecause $\\frac{1}{2}\\alpha < \\beta$. So $E$ lies inside $\\triangle ANB$; similarly, $D$ lies there.\n\nNow,\n$$\n\\begin{aligned}\n\\angle EAB &= \\alpha - \\frac{1}{2}\\beta, \\\\\n\\angle NAE &= \\frac{1}{2}(\\alpha + \\beta) - (\\alpha - \\frac{1}{2}\\beta) = \\beta - \\frac{1}{2}\\alpha, \\\\\n\\angle NBI &= \\frac{1}{2}\\alpha, \\quad \\angle IBA = \\frac{1}{2}\\beta.\n\\end{aligned}\n$$\n\nBy Ceva's Theorem in triangle $ABN$ and point $E$:\n$$\n1 = \\frac{\\sin \\angle ANE}{\\sin \\angle ENB} \\cdot \\frac{\\sin \\angle NBE}{\\sin \\angle EBA} \\cdot \\frac{\\sin \\angle BAE}{\\sin \\angle EAN}\n$$\n\nAfter substitution,\n$$\n\\frac{\\sin \\angle ENB}{\\sin \\angle ANE} = \\frac{\\sin \\frac{1}{2}\\alpha}{\\sin \\frac{1}{2}\\beta} \\cdot \\frac{\\sin(\\alpha - \\frac{1}{2}\\beta)}{\\sin(\\beta - \\frac{1}{2}\\alpha)}\n$$\n\nAnalogously, $\\angle DAB = \\frac{1}{2}\\alpha$,\n$$\n\\begin{aligned}\n\\angle DAN &= \\frac{1}{2}\\beta, \\\\\n\\angle DBA &= \\beta - \\frac{1}{2}\\alpha, \\\\\n\\angle NBD &= \\alpha - \\frac{1}{2}\\beta.\n\\end{aligned}\n$$\n\nUsing Ceva's Theorem again:\n$$\n1 = \\frac{\\sin \\angle AND \\cdot \\sin \\angle NBD \\cdot \\sin \\angle BAD}{\\sin \\angle DNB \\cdot \\sin \\angle DBA \\cdot \\sin \\angle DAN}\n$$\nwhich gives\n$$\n\\frac{\\sin \\angle DNB}{\\sin \\angle AND} = \\frac{\\sin \\frac{1}{2}\\alpha \\cdot \\sin(\\alpha - \\frac{1}{2}\\beta)}{\\sin \\frac{1}{2}\\beta \\cdot \\sin(\\beta - \\frac{1}{2}\\alpha)}\n$$\nSo,\n$$\n\\frac{\\sin \\angle ENB}{\\sin \\angle ANE} = \\frac{\\sin \\angle DNB}{\\sin \\angle AND}\n$$\n\nTherefore, the rays $ND$ and $NE$ coincide since\n$$\n\\angle BNA = \\angle ANE + \\angle ENB = \\angle AND + \\angle DNB\n$$\n\nNow let $L$ be the second intersection point of $DE$ and the circumcircle of $\\triangle ABC$. We claim that $KP$ passes through $L$.\n\n% ![](images/Ukraine_2020_booklet_p44_data_382de08bca.png)\n\nObviously,\n$$\n\\angle (AL, LN) = \\angle (AB, BN) = \\frac{1}{2}(\\alpha + \\beta)\n$$\n\nOn the other hand,\n$$\n\\angle (AP, PB) = \\angle (PA, AB) + \\angle (AB, BP) = \\alpha - \\frac{1}{2}\\beta + \\beta - \\frac{1}{2}\\alpha = \\frac{1}{2}(\\alpha + \\beta)\n$$\n\nHence $\\angle (AL, LN) = \\angle (AP, PB)$ and the quadrilateral $APDL$ is cyclic. Also, $\\angle (DI, IE) = \\frac{1}{2}(\\alpha + \\beta) = \\angle (EP, PB)$. By symmetry, $\\angle (EK, KD) = \\angle (DI, IE)$ and $\\angle (KD, DE) = \\angle (ED, DI)$, so $\\angle (EK, KD) = \\angle (EP, PD)$ and $PKED$ is cyclic.\n\nFinally,\n$$\n\\angle (KP, PE) = \\angle (KD, DE) = \\angle (ED, DI) = \\angle (LD, DA) = \\angle (LP, PA)\n$$\n\nWe obtain $\\angle (KP, PE) = \\angle (LP, PA)$ and since $AP$ and $EP$ coincide, the line $KP$ passes through $L$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 13691,
    "subject": "Mathematics (Olympiad)",
    "question": "In triangle $ABC$, the centroid is $G$ and $D$ is the midpoint of $CA$. The line through $G$ parallel to $BC$ meets $AB$ at $E$. Prove that $\\angle AEC = \\angle DGC$ if and only if $\\angle ACB = 90^\\circ$.\n\nThe centroid of a triangle is the intersection of the three medians, the lines which join each vertex to the midpoint of the opposite side.\n\n![](images/V_Britanija_2010_p26_data_e62e04ab51.png)",
    "options": [],
    "answer": "See solution",
    "solution": "Suppose $\\angle AEC = \\angle DGC$. Then $\\angle BEC = \\angle BGC$, so $BEGC$ is cyclic. Also, $BEGC$ is a trapezium; a cyclic trapezium is isosceles (see the solution to BMO1 question 2), and so $BEGC$ is isosceles. If $CG$ and $AB$ meet at $M$, then $AM = BM$. But $BEGC$ is isosceles, so $\\angle EBC = \\angle GCB$ and triangle $BMC$ is isosceles. Therefore $AM = BM = CM$ and $M$ is the circumcentre of $ABC$. Thus $AB$ is a diameter, and so $\\angle ACB = 90^\\circ$.\n\nConversely, suppose $\\angle ACB = 90^\\circ$. Then $M$ is the circumcentre of $ABC$, and so $AM = BM = CM$ (since $CM$ is a median). Thus $BEGC$ is isosceles, and is therefore cyclic. Hence $\\angle BEC = \\angle BGC$, giving $\\angle AEC = \\angle DGC$, as required.",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 13692,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $P_1, \\cdots, P_n$ be $n$ points inside triangle $ABC$ such that no three points among $P_1, \\cdots, P_n, A, B, C$ are collinear. Prove that it is possible to divide triangle $ABC$ into a union of $2n + 1$ small triangles, such that each small triangle has its vertices among the points $P_1, \\cdots, P_n, A, B, C$, and the number of small triangles with at least one vertex from $A, B, C$ is not less than $n + \\sqrt{n} + 1$.",
    "options": [],
    "answer": "See solution",
    "solution": "For the edge $BC$, a partial order $\\prec_A$ can be defined on the set $P = \\{P_1, \\cdots, P_n\\}$ as follows:\n\n$$\nP_i \\prec_A P_j \\iff \\begin{aligned}[t] & P_i \\text{ is an interior point of the } \\triangle P_j BC \\\\ & \\text{the ray } P_j P_i \\text{ intersects with the segment } BC. \\end{aligned}\n$$\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 $\\prec_B$.\n\nBy Dilworth's 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 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 $\\prec_B$.\n\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\n$$\nBP_i, \\quad CP_i \\quad (1 \\le i \\le t), \\quad P_1 P_2, \\cdots, P_{t-1} P_t, \\quad P_t A,\n$$\n\nwhich partition $\\triangle ABC$ into $2t + 1$ triangles:\n\n$$\n\\triangle BCP_1, \\quad \\triangle BP_i P_{i+1}, \\quad \\triangle CP_i P_{i+1} \\ (1 \\le i \\le t-1), \\quad \\triangle BP_t A, \\quad \\triangle CP_t A.\n$$\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.\n\nThis way, we have constructed a triangulation of $\\triangle ABC$, where at least\n\n$$\nT = \\sum_{i=0}^{t} (k_i + 1) + \\sum_{i=1}^{t} (\\ell_i + 1)\n$$\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 $T = (n - t) + (2t + 1) = n + t + 1 \\ge n + \\sqrt{n} + 1$, which proves the conclusion.",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 13693,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all functions $f : \\mathbb{N} \\to \\mathbb{R}$ satisfying the functional equation\n$$\nf(m)f(n) = f(m+n) + f(m-n)\n$$\nfor all $m, n \\in \\mathbb{N}$, where $f(0) = 1$ and $f$ is unbounded.",
    "options": [],
    "answer": "See solution",
    "solution": "We proceed by analyzing the structure of the functional equation and its consequences. By considering the Pell equation and recursive relationships, we deduce that all solutions are of the form\n$$\nf(n) = \\frac{1}{2}(\\alpha^{2ns_1} + \\alpha^{-2ns_1}),\n$$\nwhere $\\alpha = \\sqrt{2} + 1$ and $s_1 \\in \\mathbb{N}$.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 13694,
    "subject": "Mathematics (Olympiad)",
    "question": "а) Чи можна отримати число $2012$ як суму непарної кількості непарних доданків?\n\nб) Чи можна отримати число $2012$ як суму чотирьох доданків, кожен з яких складається лише з одиниць (наприклад, $1$, $11$, $111$, $1111$ тощо)?",
    "options": [],
    "answer": "See solution",
    "solution": "а) Отримати $2012$ таким чином неможливо, оскільки непарна кількість непарних доданків дає непарну суму, а $2012$ — парне число.\n\nб) Перебором показуємо, що і в цьому випадку отримати $2012$ неможливо. Якщо є доданок з принаймні $5$ одиницями, то навіть після віднімання $1111$ отримаємо значення, більше за $2012$. Якщо є доданок з $4$ одиницями, то найбільша сума трьох інших доданків — не більше за $111 + 1 + 1$, і загальна сума менша за $2012$. Аналогічно, якщо всі доданки мають не більше трьох одиниць, сума також менша за $2012$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 13695,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $a_0$ be an arbitrary positive integer. Let $\\{a_n\\}$ be an infinite sequence of positive integers such that for every positive integer $n$, the term $a_n$ is the smallest positive integer such that $a_0 + a_1 + \\dots + a_n$ is divisible by $n$.\n\nProve that there is a positive integer $N$ such that $a_{n+1} = a_n$ for all $n \\ge N$.",
    "options": [],
    "answer": "See solution",
    "solution": "Define $b_n = \\frac{a_0 + a_1 + \\dots + a_n}{n}$ for every positive integer $n$. By the condition, $b_n$ is a positive integer for every $n$.\n\nSince $a_{n+1}$ is the smallest positive integer such that $\\frac{a_0 + a_1 + \\dots + a_{n+1}}{n+1}$ is a positive integer, and\n\n$$\n\\frac{a_0 + a_1 + \\dots + a_n + b_n}{n+1} = \\frac{a_0 + a_1 + \\dots + a_n + \\frac{a_0 + a_1 + \\dots + a_n}{n}}{n+1} = \\frac{a_0 + a_1 + \\dots + a_n}{n} = b_n,\n$$\n\nwhich is a positive integer, we get $a_{n+1} \\leq b_n$ for every $n$.\n\nNow, from this result we have\n\n$$\nb_{n+1} = \\frac{a_0 + a_1 + \\dots + a_n + a_{n+1}}{n+1} \\leq \\frac{a_0 + a_1 + \\dots + a_n + b_n}{n+1} = b_n.\n$$\n\nHence, the infinite sequence of positive integers $b_1, b_2, \\dots$ is non-increasing. So there exists a positive integer $T$ such that for all $n \\geq T$ we have\n\n$$\n\\begin{align*}\nb_{n+1} &= b_n \\implies \\frac{a_0 + a_1 + \\dots + a_n + a_{n+1}}{n+1} = \\frac{a_0 + a_1 + \\dots + a_n}{n} \\\\\nn(a_0 + a_1 + \\dots + a_n + a_{n+1}) &= (n+1)(a_0 + a_1 + \\dots + a_n) \\\\\nn a_{n+1} &= a_0 + a_1 + \\dots + a_n \\\\\na_{n+1} &= \\frac{a_0 + a_1 + \\dots + a_n}{n} = b_n.\n\\end{align*}\n$$\n\nSimilarly, we get $a_{n+2} = b_{n+1}$, which implies $a_{n+2} = b_{n+1} = b_n = a_{n+1}$. Hence, taking $M = T+1$, we can state that $a_{n+1} = a_n$ for every $n \\geq M$.\n\n$\\Box$",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 13696,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $f: \\mathbb{R} \\to \\mathbb{R}$ satisfy\n\n$$\nf(x + y) \\le y f(x) + f(f(x))\n$$\nfor all $x, y \\in \\mathbb{R}$. Prove that $f(x) = 0$ for all $x < 0$.\n",
    "options": [],
    "answer": "See solution",
    "solution": "Obviously, $f(x) = 0$ for all $x \\in \\mathbb{R}$ is a solution to the given inequality. However, there are many “strange” solutions in addition to this. For example,\n\n$$\nf(x) = \\begin{cases} 0 & \\text{for } x \\le 0 \\\\ -2011e^x & \\text{for } x > 0 \\end{cases}\n$$\n",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 13697,
    "subject": "Mathematics (Olympiad)",
    "question": "An _animal_ with $n$ cells is a connected figure consisting of $n$ equal-sized square cells. The figure below shows an 8-cell animal.\n\n![](images/USA_IMO_2006-2007_p25_data_df5fa51985.png)\n\nAnimals are also called _polyominoes_. They can be defined inductively. Two cells are _adjacent_ if they share a complete edge. A single cell is an animal, and given an animal with $n$ cells, one with $n+1$ cells is obtained by adjoining a new cell by making it adjacent to one or more existing cells.\n\nA dinosaur is an animal with at least $2007$ cells. It is said to be primitive if its cells cannot be partitioned into two or more dinosaurs. Find with proof the maximum number of cells in a primitive dinosaur.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $s$ denote the minimum number of cells in a dinosaur; here $s = 2007$.\n\n**Claim:** The maximum number of cells in a primitive dinosaur is $4(s - 1) + 1 = 8025$.\n\nA primitive dinosaur can contain up to $4(s - 1) + 1$ cells. For example, consider a dinosaur in the form of a cross consisting of a central cell and four arms with $s - 1$ cells each. No connected figure with at least $s$ cells can be removed without disconnecting the dinosaur.\n\nTo show that no dinosaur with at least $4(s - 1) + 2$ cells is primitive, use the following lemma:\n\n**Lemma:** Let $D$ be a dinosaur with at least $4(s - 1) + 2$ cells, and let $R$ (red) and $B$ (black) be two complementary animals in $D$, i.e., $R \\cap B = \\emptyset$ and $R \\cup B = D$. Suppose $|R| \\leq s - 1$. Then $R$ can be augmented to produce animals $\\tilde{R} \\supset R$ and $\\tilde{B} = D \\setminus \\tilde{R}$ such that at least one of the following holds:\n\n(i) $|\\tilde{R}| \\geq s$ and $|\\tilde{B}| \\geq s$,\n(ii) $|\\tilde{R}| = |R| + 1$,\n(iii) $|R| < |\\tilde{R}| \\leq s - 1$.\n\n*Proof:* If there is a black cell adjacent to $R$ that can be made red without disconnecting $B$, then (ii) holds. Otherwise, there is a black cell $c$ adjacent to $R$ whose removal disconnects $B$. Of the squares adjacent to $c$, at least one is red and at least one is black, otherwise $B$ would be disconnected. There are at most three resulting components $C_1, C_2, C_3$ of $B$ after the removal of $c$. Without loss of generality, $C_3$ is the largest. $C_3$ has at least $\\lceil (3s - 2)/3 \\rceil = s$ cells. Let $\\tilde{B} = C_3$. Then $|\\tilde{R}| = |R| + |C_1| + |C_2| + 1$. If $|\\tilde{B}| \\leq 3s - 2$, then $|\\tilde{R}| \\geq s$ and (i) holds. If $|\\tilde{B}| \\geq 3s - 1$, then either (ii) or (iii) holds, depending on whether $|\\tilde{R}| \\geq s$ or not.\n\nStarting with $|R| = 1$, repeatedly apply the lemma. Because in alternatives (ii) and (iii) $|R|$ increases but remains less than $s$, alternative (i) eventually must occur. This shows that no dinosaur with at least $4(s - 1) + 2$ cells is primitive.\n\nAlternatively, let $s = 2007$. The answer is $4s - 3 = 8025$.\n\nConsider a graph with the cells as vertices and edges connecting adjacent cells. Let $T$ be a spanning tree. By removing any vertex of $T$, we obtain at most four connected components, called _limbs_. Limbs with at least $s$ vertices are _big_. If every vertex of $T$ contains a big limb, then a walk on $T$ always moving towards a big limb must eventually traverse back on some edge. The two components made by deleting this edge are both big, so the dinosaur is not primitive. Thus, a primitive dinosaur contains some vertex with no big limbs. By removing this vertex, we get at most four components with at most $s - 1$ vertices each. This shows that a primitive dinosaur has at most $4s - 3$ cells, and such a dinosaur consists of four limbs of $s-1$ cells each connected to a central cell. Such a dinosaur exists.",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 13698,
    "subject": "Mathematics (Olympiad)",
    "question": "Consider functions $f, g : \\mathbb{R} \\to \\mathbb{R}$ such that $g(x) = 2f(x) + f(x^2)$ for all $x \\in \\mathbb{R}$.\n\na) Prove that if $f$ is locally bounded at the origin and $g$ is continuous at the origin, then $f$ is continuous at the origin.\n\nb) Give an example of a function $f$, discontinuous at the origin, such that $g$ is continuous at the origin.",
    "options": [],
    "answer": "See solution",
    "solution": "a) Let $\\varepsilon > 0$ be arbitrary. By hypothesis, there exist $\\delta_1, M > 0$ such that $|f(x)| < M$ for all $x \\in (-\\delta_1, \\delta_1)$. Since $g$ is continuous at $0$, there exists $\\delta_2 > 0$ (depending on $\\varepsilon$) such that $|g(x) - g(0)| < \\frac{\\varepsilon}{2}$ for any $x \\in (-\\delta_2, \\delta_2)$. Let $\\delta = \\min\\{\\delta_1, \\delta_2, 1\\}$. For $0 < \\delta \\le 1$, we have $a^2 < \\delta$ for all $a \\in (-\\delta, \\delta)$.\n\nLet $x \\in (-\\delta, \\delta)$. Since $|x| < \\delta \\le \\delta_2$, we have\n\n$$\n\\begin{aligned}\n|f(x) - f(0)| &= \\frac{|2f(x) - 2f(0)|}{2} \\\\\n&\\le \\frac{|2f(x) + f(x^2) - 3f(0)| + |f(x^2) - f(0)|}{2} \\\\\n&= \\frac{|g(x) - g(0)|}{2} + \\frac{|f(|x|^2) - f(0)|}{2} \\\\\n&< \\frac{\\varepsilon}{4} + \\frac{|f(|x|^2) - f(0)|}{2}.\n\\end{aligned}\n$$\n\nInductively,\n\n$$\n|f(x) - f(0)| < \\varepsilon \\left( \\frac{1}{2^2} + \\frac{1}{2^3} + \\dots + \\frac{1}{2^{n+1}} \\right) + \\frac{|f(|x|^{2^n}) - f(0)|}{2^n}, \\quad n \\in \\mathbb{N}^*\n$$\n\nLet $p \\in \\mathbb{N}^*$ such that $2^p > \\frac{4M}{\\varepsilon}$. Then $|x|^{2p} < \\delta \\le \\delta_1$, and\n\n$$\n\\begin{aligned}\n|f(x) - f(0)| &< \\frac{\\varepsilon \\left(1 - \\frac{1}{2^p}\\right)}{2} + \\frac{|f(|x|^{2p})| + |f(0)|}{2^p} \\\\\n&\\le \\frac{\\varepsilon \\left(1 - \\frac{1}{2^p}\\right)}{2} + \\frac{2M}{2^p} \\\\\n&< \\frac{\\varepsilon}{2} + \\frac{\\varepsilon}{2} = \\varepsilon.\n\\end{aligned}\n$$\n\nb) For $a \\in (0, 1)$ and $A = \\{\\pm a^{2^n} : n \\in \\mathbb{Z}\\}$, define\n\n$$\nf : \\mathbb{R} \\to \\mathbb{R}, \\quad f(x) = \\begin{cases} (-1)^n 2^n, & |x| = a^{2^n},\\ n \\in \\mathbb{Z} \\\\ 0, & x \\in \\mathbb{R} \\setminus A \\end{cases}\n$$\n\nBecause $\\lim_{k \\to \\infty} a^{2^{2k}} = 0$ and $\\lim_{k \\to \\infty} f(a^{2^{2k}}) = \\lim_{k \\to \\infty} 2^{2k} = \\infty$, $f$ is discontinuous at the origin.\n\nFor $x \\in \\mathbb{R} \\setminus A$, $x^2 \\in \\mathbb{R} \\setminus A$, so $g(x) = 0$. For $x \\in A$, there is $n \\in \\mathbb{Z}$ such that $|x| = a^{2^n}$, so\n\n$$\ng(x) = 2f(a^{2^n}) + f(a^{2^{n+1}}) = 2^{n+1}[(-1)^n + (-1)^{n+1}] = 0.\n$$\n\nTherefore, $g$ is the zero function.\n\nAlternative solution for a):\n\nLet\n$$\nL := \\limsup_{x \\to 0} f(x) = \\lim_{\\varepsilon \\searrow 0} \\sup\\{f(x) : x \\in (-\\varepsilon, \\varepsilon) \\setminus \\{0\\}\\}\n$$\nand\n$$\n\\ell := \\liminf_{x \\to 0} f(x) = \\lim_{\\varepsilon \\searrow 0} \\inf\\{f(x) : x \\in (-\\varepsilon, \\varepsilon) \\setminus \\{0\\}\\}\n$$\n\nAs $f$ is locally bounded, $\\ell, L \\in \\mathbb{R}$ with $\\ell \\le L$.\n\nConsider sequences $(x_n)_{n \\ge 1}$ and $(y_n)_{n \\ge 1}$, with nonzero elements, converging to $0$, such that $\\lim_{n \\to \\infty} f(x_n) = L$ and $\\lim_{n \\to \\infty} f(y_n) = \\ell$, respectively. By continuity of $g$ at $0$ and properties of $\\limsup$ and $\\liminf$,\n\n$$\n\\begin{aligned}\ng(0) &= \\lim_{n \\to \\infty} g(x_n) = \\lim_{n \\to \\infty} (2f(x_n) + f(x_n^2)) \\\\\n&= \\liminf_{n \\to \\infty} (2f(x_n) + f(x_n^2)) \\\\\n&\\ge 2 \\liminf_{n \\to \\infty} f(x_n) + \\liminf_{n \\to \\infty} f(x_n^2) \\\\\n&= 2L + \\ell\n\\end{aligned}\n$$\n\nand\n\n$$\n\\begin{aligned}\ng(0) &= \\lim_{n \\to \\infty} g(y_n) = \\lim_{n \\to \\infty} (2f(y_n) + f(y_n^2)) \\\\\n&= \\limsup_{n \\to \\infty} (2f(y_n) + f(y_n^2)) \\\\\n&\\le 2 \\limsup_{n \\to \\infty} f(y_n) + \\limsup_{n \\to \\infty} f(y_n^2) \\\\\n&= 2\\ell + L\n\\end{aligned}\n$$\n\nSince $\\ell \\le L$, the inequality $2L + \\ell \\le g(0) \\le 2\\ell + L$ implies $L = \\ell = \\frac{g(0)}{3}$. So $f$ is continuous at the origin and $\\lim_{x \\to 0} f(x) = \\frac{g(0)}{3} = f(0)$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 13699,
    "subject": "Mathematics (Olympiad)",
    "question": "Calculate the angles of a triangle if the sum of two of its angles is $\\frac{5}{6}$ of a right angle, and one of these angles is $20^\\circ$ bigger than the other.",
    "options": [],
    "answer": "See solution",
    "solution": "We have $\\alpha + \\beta = \\frac{5}{6} \\times 90^\\circ = 75^\\circ$ and $\\alpha - \\beta = 20^\\circ$. \n\nSolving these equations:\n\n$$\n\\begin{align*}\n\\alpha + \\beta &= 75^\\circ \\\\\n\\alpha - \\beta &= 20^\\circ\n\\end{align*}\n$$\n\nAdding and subtracting:\n\n$$\n\\alpha = \\frac{75^\\circ + 20^\\circ}{2} = 47^\\circ 30' \\\\\n\\beta = \\frac{75^\\circ - 20^\\circ}{2} = 27^\\circ 30'\n$$\n\nThe third angle:\n\n$$\n\\gamma = 180^\\circ - (\\alpha + \\beta) = 180^\\circ - 75^\\circ = 105^\\circ\n$$\n\nThus, the triangle is obtuse-angled.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 13700,
    "subject": "Mathematics (Olympiad)",
    "question": "Do there exist positive integers $k$ and $n$ such that\n$$\nk(k + 1)(k + 2)(k + 3) = n(n + 1) + 1?\n$$",
    "options": [],
    "answer": "See 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": 13701,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all natural numbers $a$ and $b$ for which the number $2^{a!} + 2^{b!}$ is a cube of a natural number.\n\nRecall that for a natural number $n$, $n! = 1 \\cdot 2 \\cdot 3 \\cdot \\ldots \\cdot n$.",
    "options": [],
    "answer": "See solution",
    "solution": "It is clear that if $n \\geq 3$, then $n!$ is divisible by $3$, that is, $n! = 3k$ for some natural number $k$. But then $2^{n!} = 2^{3k} = 8^k \\equiv 1 \\pmod{7}$. It's easy to see that the cubes of integer numbers are equal to $0$ or $\\pm 1$ modulo $7$. Without loss of generality, we may assume that $a \\geq b$.\n\n**Case 1.** $a \\geq 3$.\n\n- If $b \\geq 3$, then $2^{a!} + 2^{b!} \\equiv 2 \\pmod{7}$, which is not a cube of an integer number.\n- If $b = 2$, then $2^{a!} + 2^{b!} \\equiv 1 + 4 \\equiv 5 \\pmod{7}$, which is not a cube of an integer number.\n- If $b = 1$, then $2^{a!} + 2^{b!} \\equiv 1 + 2 \\equiv 3 \\pmod{7}$, which is not a cube of an integer number.\n\n**Case 2.** $2 \\geq a \\geq b$.\n\nHere we have three cases, from which we simply find a single answer by direct checking.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 13702,
    "subject": "Mathematics (Olympiad)",
    "question": "Real numbers $a, b, c, d$ satisfy the conditions $|a|, |b|, |c|, |d| > 1$ and\n$$\nabc + abd + acd + bcd + a + b + c + d = 0.\n$$\nProve that\n$$\n\\frac{1}{a-1} + \\frac{1}{b-1} + \\frac{1}{c-1} + \\frac{1}{d-1} > 0.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "The second condition is equivalent to\n$$\n(a-1)(b-1)(c-1)(d-1) = (a+1)(b+1)(c+1)(d+1),\n$$\ni.e.\n$$\n\\frac{a+1}{a-1} \\cdot \\frac{b+1}{b-1} \\cdot \\frac{c+1}{c-1} \\cdot \\frac{d+1}{d-1} = 1.\n$$\nFrom the AM-GM inequality, it follows that\n$$\n\\frac{2}{a-1} + \\frac{2}{b-1} + \\frac{2}{c-1} + \\frac{2}{d-1} = \\left(\\frac{a+1}{a-1} - 1\\right) + \\left(\\frac{b+1}{b-1} - 1\\right) + \\left(\\frac{c+1}{c-1} - 1\\right) + \\left(\\frac{d+1}{d-1} - 1\\right) \\ge 4\\sqrt[4]{\\frac{a+1}{a-1} \\cdot \\frac{b+1}{b-1} \\cdot \\frac{c+1}{c-1} \\cdot \\frac{d+1}{d-1}} - 4 = 0.\n$$\n\nEquality holds if and only if $\\frac{a+1}{a-1} = \\frac{b+1}{b-1} = \\frac{c+1}{c-1} = \\frac{d+1}{d-1}$ and $abc+abd+acd+bcd+a+b+c+d=0$, i.e., for $a=b=c=d$ satisfying $4a^3+4a=0$, which means $a=b=c=d=0$, but this does not fulfill $|a|, |b|, |c|, |d| > 1$.\n\nWe conclude that the inequality is strict.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 13703,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $\\{a_n\\}_{n=1}^{\\infty}$ be a sequence of positive integers such that for every positive integer $n$,\n\n$$\na_{n+1} = (n + 1)(a_n - n + 1).\n$$\n\nIn terms of $a_1$, determine the greatest positive integer $k$ such that $\\gcd(a_i, a_{i+1}) = k$ for some positive integer $i \\ge 2$. (Note that $\\gcd(x, y)$ denotes the greatest common divisor of integers $x$ and $y$.)",
    "options": [],
    "answer": "See solution",
    "solution": "First, we will prove by induction that $a_n = (a_1 - 1)n! + n$ for all $n \\ge 1$.\n\nThe base case $a_1$ is trivial. Now suppose that the closed form holds for some $a_n$. Then\n\n$$\n\\begin{align*}\na_{n+1} &= (n + 1) (((a_1 - 1)n! + n) - n + 1) \\\\\na_{n+1} &= (n + 1) ((a_1 - 1)n! + 1) \\\\\na_{n+1} &= (a_1 - 1)(n + 1)! + (n + 1)\n\\end{align*}\n$$\n\nHence, the proof by induction is complete. For clarity, let $c = a_1 - 1$.\n\nLet $p$ be a prime number dividing $\\gcd(a_n, a_{n+1})$ for $n \\ge 2$. Assume $p \\le n$. Then $cn! + n \\equiv n \\pmod p$, and $c(n+1)! + n + 1 \\equiv n + 1 \\pmod p$, so $p > n$.\n\nWe have $a_{n+1} = (n+1)(cn! + 1)$, so $p \\mid (n+1)(cn! + 1)$. Assume $p$ divides $cn! + 1$. Then $cn! + n \\equiv cn! + 1 \\pmod p$, which implies $n \\equiv 1 \\pmod p$, a contradiction since $p > n$. So either $\\gcd(a_n, a_{n+1}) = 1$ or $p = n+1 = \\gcd(a_n, a_{n+1})$.\n\nNow suppose $p = n + 1$, with $p \\ge 3$. By Wilson's theorem, $p \\mid n! + 1$, so $p \\mid n! + 1 \\mid cn! + c$, which implies $p \\mid c - n$, so $p \\mid c + 1 = a_1$. Thus, $\\gcd(a_{p-1}, a_p) = p$ if and only if $p \\mid a_1$.\n\nTherefore, $k$ is the largest odd prime divisor of $a_1$, or $1$ if $a_1$ is a power of $2$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 13704,
    "subject": "Mathematics (Olympiad)",
    "question": "The real numbers $a, b, c, d$ satisfy simultaneously the equations\n\n$$\nabc - d = 1, \\\\\nbcd - a = 2, \\\\\ncda - b = 3, \\\\\ndab - c = -6.\n$$\n\nProve that $a + b + c + d \\neq 0$.",
    "options": [],
    "answer": "See solution",
    "solution": "Suppose that $a + b + c + d = 0$. Then\n\n$$\nabc + bcd + cda + dab = 0. \\quad (1)\n$$\n\nIf $abcd = 0$, then one of the numbers, say $d$, must be $0$. In this case $abc = 0$, so at least two of the numbers $a, b, c, d$ will be $0$, making one of the given equations impossible. Hence $abcd \\neq 0$ and, from (1),\n\n$$\n\\frac{1}{a} + \\frac{1}{b} + \\frac{1}{c} + \\frac{1}{d} = 0,\n$$\n\nimplying\n\n$$\n\\frac{1}{a} + \\frac{1}{b} + \\frac{1}{c} = \\frac{1}{a + b + c}.\n$$\n\nIt follows that $(a+b)(b+c)(c+a) = 0$, which is impossible (for instance, if $a+b=0$, then adding the second and third given equations would lead to $0 = 2+3$, a contradiction). Thus $a+b+c+d \\neq 0$.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 13705,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $p_1, p_2, \\cdots, p_{2025}$ be real numbers, and let $\\{a_n^{(1)}\\}_{n \\ge 0}$, $\\{a_n^{(2)}\\}_{n \\ge 0}$, $\\cdots$, $\\{a_n^{(2025)}\\}_{n \\ge 0}$ be 2025 real sequences satisfying:\n\n1. $a_0^{(i)} = 0$ for all $1 \\le i \\le 2025$;\n2. $a_1^{(i)}$ are not all zero for $1 \\le i \\le 2025$;\n3. For each $i = 1, 2, \\cdots, 2025$ and any positive integer $n$,\n\n$$\na_{n-1}^{(i)} + a_n^{(i)} + a_{n+1}^{(i)} = p_i \\cdot a_n^{(i+1)},\n$$\n\nwhere $a_n^{(2026)} = a_n^{(1)}$.\n\nProve that there exists a positive real number $r$ and infinitely many positive integers $n$ such that\n\n$$\n\\max \\{|a_n^{(1)}|, |a_n^{(2)}|, \\dots, |a_n^{(2025)}|\\} > r.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "We first prove a lemma.\n\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\n$$\na_{n-1} + \\beta a_n + a_{n+1} = 0,\n$$\n\nthen $\\{a_n\\}$ does not converge to 0.\n\n**Proof of Lemma:** Let $\\alpha_1, \\alpha_2$ be roots of $x^2 + \\beta x + 1 = 0$, so $\\alpha_1\\alpha_2 = 1$.\n\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.\n\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 $a_0 = 0$, $a_1 \\ne 0$).\n\nCase 1: $|\\alpha_1| \\ne |\\alpha_2|$. Then $a_n$ cannot converge to 0 since $|\\alpha_1||\\alpha_2| = 1$.\n\nCase 2: $|\\alpha_1| = |\\alpha_2| = 1$. Let $\\alpha_1 = e^{2\\pi i\\theta}$, $\\alpha_2 = e^{-2\\pi i\\theta}$.\n\nIf $\\theta \\in \\mathbb{Q}$, then $\\{a_n\\}$ is periodic and non-zero.\n\nIf $\\theta \\notin \\mathbb{Q}$, then $\\{n\\theta\\}$ is dense modulo 1, making $\\{a_n\\}$ have values dense on some circle.\n\nIn all cases, $a_n$ doesn't converge to 0. $\\square$\n\nNow the main proof. Denote the given equations as $(*_1), \\cdots, (*_{2025})$.\n\n**Case 1:** $\\prod_{i=1}^{2025} p_i \\ne 0$.\n\nDefine transformed sequences:\n\n$$\nb_n^{(i)} = \\sqrt[2025]{p_{i+1} p_{i+2} \\cdots p_{i+2024}} \\cdot a_n^{(i)}\n$$\n\nwhere indices are cyclic modulo 2025. These satisfy:\n\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\nThus we may assume $p_1 = \\cdots = p_{2025} = p$. Let $\\omega$ be a 2025th root of unity and define:\n\n$$\nX_n := \\sum_{k=0}^{2024} \\omega^k a_n^{(k+1)}.\n$$\n\nThen:\n\n$$\n\\forall n \\ge 1, \\quad X_{n-1} + (1 - \\omega^{-1}p)X_n + X_{n+1} = 0.\n$$\n\nSince not all $a_1^{(i)} = 0$, some $X_1 \\ne 0$. By the lemma, $\\{X_n\\}$ doesn't converge to 0.\n\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 $a_{n-1}^{(1)} + a_n^{(1)} + a_{n+1}^{(1)} = 0$.\n\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.\n\nIf $a_1^{(1)} = 0$, then $a_n^{(1)} \\equiv 0$. Consequently, from $(*_{2025})$ we know that for any $n \\ge 1$ we have $a_{n-1}^{(2025)} + a_n^{(2025)} + a_{n+1}^{(2025)} = 0$.\n\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": 13706,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $ABCD$ be a parallelogram. The circle through $A$ and $D$ intersects the lines $AB$, $BD$, $AC$, and $CD$ in points $B_1$, $B_2$, $C_1$, and $C_2$ respectively. Let $K$ be the intersection point of the lines $B_1B_2$ and $C_1C_2$. Prove that $K$ is equidistant from the lines $AB$ and $CD$.\n\n![](images/UkraineMO2019_booklet_p45_data_2262efc0af.png)\n\nFig. 41",
    "options": [],
    "answer": "See solution",
    "solution": "Let $O$ be the intersection point of the diagonals of the parallelogram. Let's prove that the points $C_1$, $O$, $B_2$, and $K$ lie on the same circle. In fact, with the cyclicity of points $A$, $B_1$, $B_2$, $C_1$, $C_2$, and $D$ and the parallelism of $AB$ and $CD$, we have (see Fig. 41):\n\n$$\n\\angle(KC_1, C_1O) = \\angle(CD, DA) = \\angle(BA, AD) = \\angle(B_1B_2, B_2D) = \\angle(B_2K, B_2O).\n$$\n\nNow $\\angle(BO, OK) = \\angle(B_2C_1, C_1K) = \\angle(B_2D, DC)$, thus $OK \\parallel DC$. As is known, diagonals $AC$ and $BD$ are divided by the point $O$ in half. By Thales' theorem, the line $OK$ (and accordingly, the point $K$) is equidistant from the lines $AB$ and $CD$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 13707,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all triplets $(x, y, z)$ of positive integers satisfying the following conditions:\n\n$$\nx + xy + xyz = 31, \\quad x < y < z.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "The solutions are $(1, 2, 14)$ and $(1, 3, 9)$.\n\nRewriting the given equation, we get $x(1 + y + yz) = 31$. Since $31$ is a prime, and $1 + y + yz > 1$, we must have $x = 1$ and $1 + y + yz = 31$. Thus, $y(1 + z) = 30$.\n\nGiven $x < y < z$ and $x = 1$, we need $1 < y < z$.\n\nNow, $y(1 + z) = 30$:\n- For $y = 2$, $2(1 + z) = 30 \\implies 1 + z = 15 \\implies z = 14$.\n- For $y = 3$, $3(1 + z) = 30 \\implies 1 + z = 10 \\implies z = 9$.\n\nBoth satisfy $1 < y < z$, so the solutions are $(x, y, z) = (1, 2, 14)$ and $(1, 3, 9)$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 13708,
    "subject": "Mathematics (Olympiad)",
    "question": "Suppose the sequence $\\{a_n\\}$ satisfies\n\n$$\na_1 = 2t - 3 \\quad (t \\in \\mathbb{R},\\ t \\neq \\pm 1),\n$$\n\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\n1. Find a formula for the general term of $\\{a_n\\}$.\n2. If $t > 0$, determine which is larger: $a_{n+1}$ or $a_n$.",
    "options": [],
    "answer": "See solution",
    "solution": "1. The given recurrence can be rewritten as\n\n$$\na_{n+1} = \\frac{2(t^{n+1} - 1)(a_n + 1)}{a_n + 2t^n - 1} - 1.\n$$\n\nLet $\\frac{a_n + 1}{t^n - 1} = b_n$. Then\n\n$$\nb_{n+1} = \\frac{2b_n}{b_n + 2}, \\quad b_1 = \\frac{a_1 + 1}{t - 1} = 2.\n$$\n\nThis leads to\n\n$$\n\\frac{1}{b_{n+1}} = \\frac{1}{b_n} + \\frac{1}{2}, \\quad \\frac{1}{b_1} = \\frac{1}{2}.\n$$\n\nSo\n\n$$\n\\frac{1}{b_n} = \\frac{n}{2} \\implies b_n = \\frac{2}{n}.\n$$\n\nTherefore,\n\n$$\n\\frac{a_n + 1}{t^n - 1} = \\frac{2}{n} \\implies a_n = \\frac{2(t^n - 1)}{n} - 1.\n$$\n\n2. We have\n\n$$\n\\begin{aligned}\na_{n+1} - a_n &= \\frac{2(t^{n+1} - 1)}{n+1} - \\frac{2(t^n - 1)}{n} \\\\\n&= \\frac{2(t-1)}{n(n+1)} \\left[ n(1 + t + \\cdots + t^{n-1} + t^n) - (n+1)(1 + t + \\cdots + t^{n-1}) \\right] \\\\\n&= \\frac{2(t-1)}{n(n+1)} \\left[ nt^n - (1 + t + \\cdots + t^{n-1}) \\right] \\\\\n&= \\frac{2(t-1)}{n(n+1)} \\left[ (t^n-1) + (t^n-t) + \\cdots + (t^n-t^{n-1}) \\right] \\\\\n&= \\frac{2(t-1)^2}{n(n+1)} \\left[ (t^{n-1} + t^{n-2} + \\cdots + 1) + t(t^{n-2} + t^{n-3} + \\cdots + 1) + \\cdots + t^{n-1} \\right].\n\\end{aligned}\n$$\n\nIt is clear 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": 13709,
    "subject": "Mathematics (Olympiad)",
    "question": "$p$, $q$, and $r$ represent the numbers $2$, $3$, and $4$ in some order. What is the greatest possible value of $p^q \\times r$?",
    "options": [],
    "answer": "See solution",
    "solution": "$3^4 \\times 2 = 81 \\times 2 = 162$.\n\nThe greatest possible value is $162$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 13710,
    "subject": "Mathematics (Olympiad)",
    "question": "Given two positive integers $a$ and $b$, a legal move is to choose a proper divisor of one of them and add it to either $a$ or $b$. Players $A$ and $B$ make legal moves in turns; $A$ plays first. The one who obtains a number $\\ge 2015$ wins. Determine who wins if the game starts with:\n\n$$\na)\\ a=3,\\ b=5; \\qquad b)\\ a=6,\\ b=7.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "Player $B$ wins in part a); player $A$ wins in part b).\n\nNote that, by the rules, the player to move can always add $1$ to one of the current numbers.\n\nFor $a=6$, $b=7$, let $A$ start the game by adding $1$ to $6$, leading to the pair $(7, 7)$. If $a=3$, $b=5$, $A$'s first move is forced to be adding $1$ to one of the numbers because they are both primes. If $A$ adds it to $3$, then $B$ adds $1$ to the obtained $4$ and obtains $(5, 5)$. And if $A$ adds $1$ to $5$, the new pair is $(3, 6)$. Since $3$ is a proper divisor of $6$, $B$ can add it to $3$ and obtain $(6, 6)$.\n\nHence, in part b), $A$ can obtain two equal numbers with his first move; $B$ can achieve the same with his first move in part a). Note that no one can win in one move in positions $(a, a)$ where $a=5, 6, 7$.\n\nNow it suffices to show that any position with two equal numbers $(a, a')$, where $a \\le \\frac{2}{3} \\cdot 2014$, is losing, i.e., the player $X$ to move in this position loses. Hence $X$ obtains a new number $a' = a+d > a$ where $d \\mid a$ and $d \\le \\frac{a}{2}$, so $a' \\le \\frac{3a}{2} \\le 2014$.\n\nHence $X$ does not win on his first move in the position $(a, a')$. Now the strategy of the opponent $Y$ is as follows. If one of $a$ and $a'$ has a proper divisor $d'$ such that $a'+d' \\ge 2015$, $Y$ adds $d'$ to $a'$ and wins. If not, $Y$ repeats the previous move of $X$, adding $d$ to $a$; note that this is possible. This yields a new pair $(a', a')$, of equal numbers, with $a' > a$. Observe that now $X$ cannot reach $2015$ in one move, otherwise $Y$ would have done so on his previous move. Hence $X$ has to obtain again a pair of different numbers $(a', a'')$, with $a' < a'' \\le 2014$. Then $Y$ repeats the same: he either reaches $2015$ in one move if this is possible, or obtains the pair $(a'', a'')$ from which $X$ cannot win in one move. If $Y$ follows this strategy, the numbers $a, a', a'', \\dots$ in the process increase, and $X$ never makes a winning move. There will come a moment when a proper divisor $d$ of $a$ in the current pair $(a, a)$ is such that $a+d \\ge 2015$; then $Y$ wins on that move.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 13711,
    "subject": "Mathematics (Olympiad)",
    "question": "Consider 2016 points arranged on a circle. You are allowed to jump ahead by 2 or 3 points in the clockwise direction.\n\nWhat is the minimum number of jumps required to visit all points and return to the starting point?",
    "options": [],
    "answer": "See solution",
    "solution": "Clearly, it takes at least $2016$ jumps to visit all points. It is impossible to use only jumps of length $2$ or only jumps of length $3$ because this would confine us to a single residue class modulo $2$ or $3$, respectively.\n\nIf the problem could be solved with $2016$ jumps, the total distance covered by these jumps would be strictly between $2 \\times 2016$ and $3 \\times 2016$, which makes a return to the original point impossible. Therefore, at least $2017$ jumps are required.\n\nThis is indeed possible, for example, with the following sequence of points on the circle:\n\n$$\n0, 3, 6, \\dots, 2013, 2015, 2, 5, \\dots, 2012, 2014, 1, 4, \\dots, 2011, 2013, 0.\n$$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 13712,
    "subject": "Mathematics (Olympiad)",
    "question": "For $x \\geq \\frac{1}{2}$, what is the largest possible value of the expression\n$$\n\\frac{x^4 - x^2}{x^6 + 16x^3 - 1}\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "The largest possible value is $\\frac{1}{15}$.\n\nIf $\\frac{1}{2} \\leq x < 1$, then $x^4 - x^2$ is negative, while $x^6 + 16x^3 - 1$ is positive, so the expression is negative in this interval. Thus, consider only $x \\geq 1$.\n\nLet $t = x - \\frac{1}{x}$, so $x^3 - \\frac{1}{x^3} = t^3 + 3t$. The expression can be rewritten as:\n$$\n\\frac{x - \\frac{1}{x}}{x^3 - \\frac{1}{x^3} + 16} = \\frac{t}{t^3 + 3t + 16} = \\frac{1}{t^2 + 3 + \\frac{16}{t}}\n$$\n\nTo maximize, minimize the denominator $t^2 + 3 + \\frac{16}{t}$ for $t \\geq 0$ ($x \\geq 1$). By the AM-GM inequality:\n$$\nt^2 + \\frac{8}{t} + \\frac{8}{t} + 3 \\geq \\sqrt[3]{t^2 \\cdot \\frac{8}{t} \\cdot \\frac{8}{t}} + 3 = 15.\n$$\n\nEquality occurs when $t = 2$, i.e., $x - \\frac{1}{x} = 2$ or $x = 1 + \\sqrt{2}$.\n\nThus, the largest value is $\\frac{1}{15}$.",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 13713,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all pairs of positive integers $ (x, y) $ such that $ x^2 $ is divisible by $ 2x y^2 - y^3 + 1 $.",
    "options": [],
    "answer": "See solution",
    "solution": "If $y = 1$, then $2x \\mid x^2 \\Leftrightarrow x = 2n$, $n \\in \\mathbb{N}$. So, the pairs $(x, y) = (2n, 1)$, $n \\in \\mathbb{N}$ satisfy the required divisibility.\n\nLet $y > 1$ such that $x^2$ is divisible by $2x y^2 - y^3 + 1$. There exists $m \\in \\mathbb{N}$ such that\n\n$$\nx^2 = m(2x y^2 - y^3 + 1), \\text{ i.e., } x^2 - 2m y^2 x + (m y^3 - m) = 0.\n$$\n\nThe discriminant of this quadratic equation is $\\Delta = 4m^2 y^4 - 4m y^3 + 4m$. Denote\n\n$$\nA = 4m(y^2 - 1) + (y - 1)^2, \\quad B = 4m(y^2 + 1) - (y + 1)^2.\n$$\n\nFor $y > 1$, $y \\in \\mathbb{N}$ and $m \\in \\mathbb{N}$, we have\n\n$$\nA > 0, \\quad B = 4m(y^2 + 1) - (y + 1)^2 > 2(y^2 + 1) - (y + 1)^2 = (y - 1)^2 \\geq 0 \\Rightarrow B > 0.\n$$\n\nWe obtain the following estimations for the discriminant $\\Delta$:\n\n$$\n\\Delta + A = (2m y^2 - y + 1)^2 \\geq 0 \\Rightarrow \\Delta < (2m y^2 - y + 1)^2;\n$$\n\n$$\n\\Delta - B = (2m y^2 - y - 1)^2 \\geq 0 \\Rightarrow \\Delta > (2m y^2 - y - 1)^2.\n$$\n\nBecause the discriminant $\\Delta$ must be a perfect square, we obtain the equality:\n\n$$\n\\Delta = 4m^2 y^4 - 4m y^3 + 4m = (2m y^2 - y)^2 \\Leftrightarrow y^2 = 4m \\Rightarrow y = 2k,\\ k \\in \\mathbb{N},\\ m = k^2,\\ k \\in \\mathbb{N}.\n$$\n\nThe equation $x^2 - 8k^4 x + k(8k^4 - k) = 0$ has the solutions $x = k$ and $x = 8k^4 - k$, where $k \\in \\mathbb{N}$.\n\nFinally, all pairs of positive integers $(x, y)$ such that $x^2$ is divisible by $2x y^2 - y^3 + 1$ are:\n\n$$(x, y) \\in \\{ (2k, 1),\\ (k, 2k),\\ (8k^4 - k, 2k)\\mid k \\in \\mathbb{N} \\}.$$",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 13714,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $M(n)$ be the maximum possible number of good diagonals in a convex $n$-gon. What is the value of $M(n)$ for even and odd $n$?",
    "options": [],
    "answer": "See solution",
    "solution": "We will show that $M(n) = n - 2$ if $n$ is even and $M(n) = n - 3$ if $n$ is odd.\n\nFor any $n$, we can draw all $n - 3$ diagonals from one vertex $A$ and one more diagonal joining two vertices adjacent to $A$ if $n > 3$.\n\n**Inductive lower bound:**\n\nWe show by induction that $M(n) \\ge n - 2$ if $n$ is even. For $n = 4$, both diagonals in a convex quadrilateral are good, so the claim holds. Assume the claim is true for $n - 2$ and consider a convex $n$-gon $A_1A_2\\ldots A_n$. Draw diagonals $\\overline{A_{n-2}A_n}$ and $\\overline{A_1A_{n-1}}$, and use the assumption on $A_1A_2\\ldots A_{n-2}$, so $M(n) \\ge M(n-2) + 2 \\ge n - 4 + 2 = n - 2$.\n\n**Inductive upper bound:**\n\nWe also prove by induction that $M(n) \\le n - 2$ if $n$ is even and $M(n) \\le n - 3$ if $n$ is odd. Obviously, $M(3) = 0$ and $M(4) = 2$. Assume the claim holds for all $n$ smaller than $k$.\n\nConsider a choice of diagonals of a convex $k$-gon for which $M(k)$ is achieved.\n\n*Case 1:* If there are two good diagonals that intersect, they divide the $k$-gon into 4 parts, 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\n$$\nM(k) = 2 + \\sum_{i=1}^{4} M(a_i + 2) \\le 2 + \\sum_{i=1}^{4} (a_i + 2 - 2) = 2 + (k - 4) = k - 2.\n$$\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.\n\n*Case 2:* If there are no two good diagonals that intersect, then there are at most $k - 3$ good diagonals, as that is the maximum number of diagonals that can be drawn from one vertex.\n\nThus, $M(n) = n - 2$ if $n$ is even, and $M(n) = n - 3$ if $n$ is odd.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 13715,
    "subject": "Mathematics (Olympiad)",
    "question": "Any subword of the form $xyz$ can be changed to $zyx$. For example, $cabc \\rightarrow ccba$.\n\nAny subword of the form $xyyx$ can be omitted. For example, $abcaacc \\rightarrow abcc$.\n\nCan the word $baccba$ be obtained from the word $abccab$ using the above operations?\n\nNote: For $xyz$, the subwords are $x$, $y$, $z$, $xy$, $yz$, and $xyz$, but not $xz$.",
    "options": [],
    "answer": "See solution",
    "solution": "Answer: No.\n\nSuppose the number of $a$'s in even positions is subtracted from the number of $a$'s in odd positions in the word. In that case, we obtain a quantity that remains invariant under the given operations. This invariant can be used to determine if one word can be transformed into another using the specified operations.\n\nFor the word $abccab$:\n\n- Odd-positioned $a$ letters: 2 (positions 1 and 5)\n- Even-positioned $a$ letters: 0\n\nThus, the invariant for $abccab$ is $2 - 0 = 2$.\n\nFor the word $baccba$:\n\n- Odd-positioned $a$ letters: 0\n- Even-positioned $a$ letters: 2 (positions 2 and 6)\n\nThus, the invariant for $baccba$ is $0 - 2 = -2$.\n\nSince the invariant values for $abccab$ and $baccba$ are different ($2$ and $-2$, respectively), it is impossible to transform $abccab$ into $baccba$ using the given operations.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 13716,
    "subject": "Mathematics (Olympiad)",
    "question": "$$\na_1 = 1,\\quad a_2 = 2,\\quad a_{n+1} = \\frac{a_n^2 + (-1)^n}{a_{n-1}} \\quad (n = 2, 3, \\dots).\n$$\n\nProve that the sum of squares of any two adjacent terms of the sequence is also in the sequence.",
    "options": [],
    "answer": "See solution",
    "solution": "By $a_{n+1} = \\frac{a_n^2 + (-1)^n}{a_{n-1}}$, we have $a_{n+1} a_{n-1} = a_n^2 + (-1)^n$ for $n = 2, 3, \\dots$. So,\n\n$$\n\\begin{align*}\n\\frac{a_n - a_{n-2}}{a_{n-1}} &= \\frac{a_n a_{n-2} - a_{n-2}^2}{a_{n-1} a_{n-2}} \\\\\n&= \\frac{a_{n-1}^2 + (-1)^{n-1} - a_{n-2}^2}{a_{n-1} a_{n-2}} \\\\\n&= \\frac{a_{n-1}^2 - a_{n-1} a_{n-3}}{a_{n-1} a_{n-2}} \\\\\n&= \\frac{a_{n-1} - a_{n-3}}{a_{n-2}} \\\\\n&= \\dots = \\frac{a_3 - a_1}{a_2} = 2,\n\\end{align*}\n$$\n\nthat is, $a_n = 2a_{n-1} + a_{n-2}$ for $n \\ge 3$, with $a_1 = 1$, $a_2 = 2$.\n\nTherefore, $a_n = C_1 \\lambda_1^n + C_2 \\lambda_2^n$, where $\\lambda_1 + \\lambda_2 = 2$, $\\lambda_1 \\lambda_2 = -1$, $a_1 = 1$, $a_2 = 2$, $n \\in \\mathbb{N}^+$.\n\nThen, since $\\lambda_1 \\lambda_2 = -1$ and $\\lambda_2 = 2 - \\lambda_1$, we have\n\n$$\n\\begin{cases}\n1 = C_1 \\lambda_1 + C_2 \\lambda_2 \\\\\n2 = C_1 \\lambda_1^2 + C_2 \\lambda_2^2\n\\end{cases}\n\\Rightarrow\n\\begin{cases}\n\\lambda_2 = C_1 \\lambda_1 \\lambda_2 + C_2 \\lambda_2^2 \\\\\n2 = C_1 \\lambda_1^2 + C_2 \\lambda_2^2\n\\end{cases}\n\\Rightarrow\n\\begin{cases}\n2 - \\lambda_1 = -C_1 + C_2 \\lambda_2^2 \\\\\n2 = C_1 \\lambda_1^2 + C_2 \\lambda_2^2\n\\end{cases}\n\\Rightarrow C_1(1 + \\lambda_1^2) = \\lambda_1.\n$$\n\nThus, by symmetry, $C_2(1 + \\lambda_2^2) = \\lambda_2$.\n\nSo, since $1 + \\lambda_1 \\lambda_2 = 0$, we have\n\n$$\n\\begin{align*}\na_n^2 + a_{n+1}^2 &= C_1^2 (1 + \\lambda_1^2) \\lambda_1^{2n} + C_2^2 (1 + \\lambda_2^2) \\lambda_2^{2n} \\\\\n&\\quad + 2C_1 C_2 (\\lambda_1 \\lambda_2)^n (1 + \\lambda_1 \\lambda_2) \\\\\n&= C_1 \\lambda_1^{2n+1} + C_2 \\lambda_2^{2n+1} = a_{2n+1}.\n\\end{align*}\n$$\n\nTherefore, the sum of squares of any two adjacent terms of the sequence is also in the sequence.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 13717,
    "subject": "Mathematics (Olympiad)",
    "question": "Ana and Bogdan play the following turn-based game: Ana starts with a pile of $n$ ($n \\geq 3$) stones. On each turn, a player must split one pile. The winner is the player who, on their turn, can make all piles have at most two stones. Depending on $n$, determine which player has a winning strategy.",
    "options": [],
    "answer": "See solution",
    "solution": "If $n = 3$ or $n = 4$, Ana wins on her first move. If $n$ is odd and greater than $3$, Bogdan will win. In this case, Ana must start by making a pile with an even number of stones. Bogdan will split this pile into a pile with one stone and the rest into an odd pile. Ana must make an even pile again, and Bogdan continues this strategy unless he can win. The game may end in two ways: if Ana leaves a pile of $2$, one of $3$, and the rest of $1$, Bogdan will win by splitting the pile with $3$ stones. If Ana leaves a pile of $4$ and the rest of $1$, Bogdan wins by splitting the $4$-pile into two piles of $2$ stones.\n\nIf $n \\geq 6$ is even, then Ana will split the pile into $1$ and $n-1$ and continue with the strategy described for Bogdan above. So, in this case, Ana will win.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 13718,
    "subject": "Mathematics (Olympiad)",
    "question": "Lieneke is making bracelets with beads. Each bracelet has six beads: two white, two grey, and two black beads. Some bracelets look different on first sight, but are actually not different: by turning or flipping the first one over, it looks the same as the other one. For example, the following three bracelets are the same.\n\n![](images/NLD_ABooklet_2021_p7_data_63c1378dae.png)\n\n![](images/NLD_ABooklet_2021_p7_data_20d599a6e8.png)\n\n![](images/NLD_ABooklet_2021_p7_data_a4ea4f7921.png)\n\nHow many really different bracelets can Lieneke make?\n\nA) 10\n\nB) 11\n\nC) 12\n\nD) 14\n\nE) 15",
    "options": [],
    "answer": "See solution",
    "solution": "B) 11",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 13719,
    "subject": "Mathematics (Olympiad)",
    "question": "Prove that the product of four consecutive positive integers cannot be equal to the product of two consecutive positive integers.",
    "options": [],
    "answer": "See solution",
    "solution": "Let the four consecutive integers be $b$, $b+1$, $b+2$, and $b+3$. Their product is\n\n$$\nb(b+1)(b+2)(b+3) = (b^2 + 3b)(b^2 + 3b + 2)\n$$\n\nLet $c = b^2 + 3b$. Then the product becomes $c(c+2)$. Suppose this equals the product of two consecutive integers, $a$ and $a+1$, so $a(a+1) = c(c+2)$.\n\nWe cannot have $a = c$ or $a = c+1$, since $c(c+1) < c(c+2) < (c+1)(c+2)$. If $a < c$, then $a(a+1) < c(c+2)$. If $a > c+1$, then $a(a+1) > c(c+2)$. Therefore, there are no integer solutions to $a(a+1) = c(c+2)$, so the product of four consecutive positive integers cannot equal the product of two consecutive positive integers.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 13720,
    "subject": "Mathematics (Olympiad)",
    "question": "Function $f(x) = a^{2x} + 3a^x - 2$ ($a > 0$, $a \\neq 1$) reaches the maximum value $8$ on the interval $[-1, 1]$. Then its minimum value on this interval is ________.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $a^x = y$. The original function becomes\n\n$$\ng(y) = y^2 + 3y - 2\n$$\nwhich is increasing for $y > -\\frac{3}{2}$.\n\n- When $0 < a < 1$, $y$ decreases as $x$ increases, so $y \\in [a, a^{-1}]$.\n  \n  The maximum is at $y = a^{-1}$:\n  $$\ng(y)_{\\max} = a^{-2} + 3a^{-1} - 2 = 8 \\implies a^{-1} = 2 \\implies a = \\frac{1}{2}.\n  $$\n  The minimum is at $y = a$:\n  $$\ng(y)_{\\min} = \\left(\\frac{1}{2}\\right)^2 + 3 \\times \\frac{1}{2} - 2 = \\frac{1}{4} + \\frac{3}{2} - 2 = -\\frac{1}{4}.\n  $$\n\n- When $a > 1$, $y$ increases as $x$ increases, so $y \\in [a^{-1}, a]$.\n  \n  The maximum is at $y = a$:\n  $$\ng(y)_{\\max} = a^2 + 3a - 2 = 8 \\implies a = 2.\n  $$\n  The minimum is at $y = a^{-1}$:\n  $$\ng(y)_{\\min} = 2^{-2} + 3 \\times 2^{-1} - 2 = \\frac{1}{4} + \\frac{3}{2} - 2 = -\\frac{1}{4}.\n  $$\n\nTherefore, the minimum value of $f(x)$ on $x \\in [-1, 1]$ is $-\\frac{1}{4}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 13721,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $n$ and $k$ be positive integers. On the board, $k$ different positive integers not exceeding $n$ are written. Prove that the equation\n\n$$\nx + y = z + w\n$$\n\nhas at least $\\frac{k^4}{2n-1}$ solutions $(x, y, z, w)$, where $x, y, z, w$ are (not necessarily different) numbers written on the board.",
    "options": [],
    "answer": "See solution",
    "solution": "For a positive integer $s$, let $a_s$ be the number of pairs $(x, y)$, where $x, y$ are numbers on the board and $x + y = s$. Since the sum of two numbers on the board cannot be less than $2$ or greater than $2n$, the number of solutions to the equation $x + y = z + w$ is $a_2^2 + \\dots + a_{2n}^2$.\n\nApplying the AM-QM inequality to $a_2, \\dots, a_{2n}$ gives\n\n$$\na_2^2 + \\dots + a_{2n}^2 \\geq \\frac{(a_2 + \\dots + a_{2n})^2}{2n - 1}.\n$$\n\nSince each pair $(x, y)$, where $x, y$ are numbers on the board, is counted exactly once in the sum $a_2 + \\dots + a_{2n}$, the sum equals $k^2$. Hence,\n\n$$\n\\frac{(a_2 + \\dots + a_{2n})^2}{2n - 1} = \\frac{k^4}{2n - 1},\n$$\n\nwhich proves the claim.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 13722,
    "subject": "Mathematics (Olympiad)",
    "question": "假設正實數數列 $a_1, a_2, \\dots$ 滿足：對每一個正整數 $k$，都有\n\n$$\na_{k+1} \\ge \\frac{k a_k}{a_k^2 + (k-1)}.\n$$\n\n試證：對每一個正整數 $n \\ge 2$，恆有\n\n$$\na_1 + a_2 + \\dots + a_n \\ge n.\n$$\n\nSuppose that a sequence $a_1, a_2, \\dots$ of positive real numbers satisfies\n\n$$\na_{k+1} \\ge \\frac{k a_k}{a_k^2 + (k-1)}\n$$\n\nfor every positive integer $k$. Prove that $a_1 + a_2 + \\dots + a_n \\ge n$ for every $n \\ge 2$.",
    "options": [],
    "answer": "See solution",
    "solution": "從條件\n\n$$\na_{k+1} \\ge \\frac{k a_k}{a_k^2 + (k-1)}, \\quad (1)\n$$\n\n可得\n\n$$\n\\frac{k}{a_{k+1}} \\le \\frac{a_k^2 + (k-1)}{a_k} = a_k + \\frac{k-1}{a_k},\n$$\n\n因此\n\n$$\na_k \\ge \\frac{k}{a_{k+1}} - \\frac{k-1}{a_k}.\n$$\n\n對 $k=1,2,\\dots,m$ 逐項相加，得到\n\n$$\na_1 + \\dots + a_m \\ge \\frac{m}{a_{m+1}}. \\quad (2)\n$$\n\n現在用歸納法證明命題。當 $n=2$ 時，對 $k=1$ 代入 (1)：\n\n$$\na_1 + a_2 \\ge a_1 + \\frac{1}{a_1} \\ge 2.\n$$\n\n歸納步，假設對某 $n \\ge 2$ 命題成立。若 $a_{n+1} \\ge 1$，則由歸納假設\n\n$$\n(a_1 + \\dots + a_n) + a_{n+1} \\ge n + 1. \\quad (3)\n$$\n\n否則，若 $a_{n+1} < 1$，則由 (2)\n\n$$\n(a_1 + \\dots + a_n) + a_{n+1} \\ge \\frac{n}{a_{n+1}} + a_{n+1} > (n-1) + 2.\n$$\n\n證畢。",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 13723,
    "subject": "Mathematics (Olympiad)",
    "question": "Find the minimum number $N$ of four-digit numbers such that each digit from 1 to 8 appears in at least 3 of these numbers, and each number uses only digits from 1 to 8.",
    "options": [],
    "answer": "See solution",
    "solution": "Let some digit, say 1, appear exactly in $k$ numbers from $N$ given numbers. Hence, 1 forms at most 3 distinct pairs with the remaining 3 digits of any of these $k$ numbers. Since the total number of all distinct pairs formed by 1 and the other 7 numbers ($2,3,\\ldots,8$) is equal to 7, we see that $3k \\ge 7$. So $k \\ge 3$. Therefore, each of the digits 1, 2, ..., 8 must appear in at least 3 numbers. Thus, the total number of all digits in $N$ numbers is greater than or equal to $8 \\cdot 3 = 24$. But $N$ numbers contains exactly $4N$ digits. Therefore, $4N \\ge 24$, so $N \\ge 6$.\n\nThe following example shows that there are 6 four-digit numbers satisfying the condition.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 13724,
    "subject": "Mathematics (Olympiad)",
    "question": "Two solid cylinders are mathematically similar. The sum of their heights is $1$. The sum of their surface areas is $8\\pi$. The sum of their volumes is $2\\pi$. Find all possibilities for the dimension of each cylinder.",
    "options": [],
    "answer": "See solution",
    "solution": "Let the heights of the two cylinders be $a$ and $b$. As the cylinders are similar, if the first has radius $p a$, the second has radius $p b$.\n\nThe surface area of the first cylinder is $2\\pi (p a)^2 + 2\\pi (p a) a$, and its volume is $\\pi (p a)^2 a$. Similarly for the second.\n\nThe condition on the surface areas is:\n\n$$\n2\\pi (p a)^2 + 2\\pi (p b)^2 + 2\\pi (p a) a + 2\\pi (p b) b = 8\\pi\n$$\n\nThe condition on the volumes is:\n\n$$\n\\pi (p a)^2 a + \\pi (p b)^2 b = 2\\pi\n$$\n\nAs the sum of the heights is $1$, $a + b = 1$. Let $k = a b$. Then $a^2 + b^2 = (a + b)^2 - 2 a b = 1 - 2k$, and $a^3 + b^3 = (a + b)^3 - 3 a b (a + b) = 1 - 3k$.\n\nSubstitute these into the previous equations:\n\n$$\n(1 - 2k)(p^2 + p) = 4 \\\\\n(1 - 3k)p^2 = 2\n$$\n\nSubtracting the second from the first gives:\n\n$$\np - 2k p + k p^2 = 2 \\\\\np - 2 = k p (2 - p)\n$$\n\nIf $p \\neq 2$, we get $k p = -1$, but both $k$ and $p$ are positive, so $p = 2$. As $(1 - 3k)p^2 = 2$, we get $a b = k = \\frac{1}{6}$.\n\nSubstitute $b = \\frac{1}{6a}$ into $a + b = 1$, and multiply both sides by $a$ to get $a^2 - a + \\frac{1}{6} = 0$. This gives $a = \\frac{3 \\pm \\sqrt{3}}{6}$. Whichever value $a$ takes, $b$ takes the other, so this is only one possible pair of cylinders.\n\nSo one cylinder has height $\\frac{3 + \\sqrt{3}}{6}$ and radius $\\frac{3 + \\sqrt{3}}{3}$, and the other has height $\\frac{3 - \\sqrt{3}}{6}$ and radius $\\frac{3 - \\sqrt{3}}{3}$.\n\nFinally, check these satisfy the conditions. The cylinders have surface area $2\\pi \\frac{(3 \\pm \\sqrt{3})^2}{9} + 2\\pi \\frac{(3 \\pm \\sqrt{3})^2}{18} = \\pi (4 \\pm 2\\sqrt{3})$, giving a total of $8\\pi$. The volumes are $\\pi \\frac{(3 \\pm \\sqrt{3})^3}{54} = \\pi \\frac{9 \\pm 5\\sqrt{3}}{9}$, giving a total of $2\\pi$. So these cylinders are a solution to the conditions.",
    "topic": "Geometry",
    "subtopic": "Solid Geometry"
  },
  {
    "id": 13725,
    "subject": "Mathematics (Olympiad)",
    "question": "$f(x_1, \\dots, x_n)$ 為次數小於 $n$ 的整係數多項式，證明滿足\n\n$$\nf(x_1, \\dots, x_n) \\equiv 0 \\pmod{13}\n$$\n\n的有序 $n$-元組 $(x_1, \\dots, x_n)$ 的個數必為 $13$ 的倍數，其中 $0 \\le x_i \\le 12$。",
    "options": [],
    "answer": "See solution",
    "solution": "解：以下同餘皆模 $13$。\n\n我們先證明\n\n$$\n\\sum_{x=0}^{12} x^k \\equiv 0, \\quad \\text{對於 } 0 \\le k < 12.\n$$\n\n$k=0$ 的情形易證，故設 $k>0$。令 $g$ 是模 $13$ 的原根，則 $g, 2g, \\dots, 12g$ 是 $1, 2, \\dots, 12$ 的某個排列。因此\n\n$$\n\\sum_{x=0}^{12} x^k \\equiv \\sum_{x=0}^{12} (g x)^k = g^k \\sum_{x=0}^{12} x^k,\n$$\n\n因 $g^k \\ne 1$，必有 $\\sum_{x=0}^{12} x^k = 0$。\n\n令 $S = \\{(x_1, \\dots, x_n) \\mid 0 \\le x_i \\le 12\\}$。只要證明 $(x_1, \\dots, x_n) \\in S$ 且 $f(x_1, \\dots, x_n) \\ne 0$ 的 $n$-元組個數是 $13$ 的倍數，因為 $|S| = 13^n$ 為 $13$ 的倍數。\n\n考慮和\n\n$$\n\\sum_{(x_1, \\dots, x_n) \\in S} (f(x_1, \\dots, x_n))^{12},\n$$\n\n這個和計算 $(x_1, \\dots, x_n) \\in S$ 且 $f(x_1, \\dots, x_n) \\ne 0$ 的 $n$-元組個數，這是因為費馬小定理告訴我們：\n\n$$\n(f(x_1, \\dots, x_n))^{12} \\equiv \\begin{cases} 1, & \\text{若 } f(x_1, \\dots, x_n) \\ne 0, \\\\ 0, & \\text{若 } f(x_1, \\dots, x_n) = 0. \\end{cases}\n$$\n\n另一方面，我們可展開 $(f(x_1, \\dots, x_n))^{12}$ 得\n\n$$\n(f(x_1, \\dots, x_n))^{12} = \\sum_{j=1}^{N} c_j \\prod_{i=1}^{n} x_i^{e_{ji}},\n$$\n\n其中 $N, c_j, e_{ji}$ 為整數。因為 $f$ 是次數小於 $n$ 的多項式，故對於每個 $j$ 都有 $e_{j1} + e_{j2} + \\dots + e_{jn} < 12n$，因此對於每個 $j$ 存在 $i$ 使得 $e_{ji} < 12$。故有\n\n$$\n\\sum_{(x_1, \\dots, x_n) \\in S} c_j \\prod_{i=1}^{n} x_i^{e_{ji}} = c_j \\prod_{i=1}^{n} \\sum_{x=0}^{12} x^{e_{ji}} \\equiv 0,\n$$\n\n因為乘積中的某一個和為 $0$。因此\n\n$$\n\\sum_{(x_1, \\dots, x_n) \\in S} (f(x_1, \\dots, x_n))^{12} = \\sum_{(x_1, \\dots, x_n) \\in S} \\sum_{j=1}^{N} c_j \\prod_{i=1}^{n} x_i^{e_{ji}} \\equiv 0,\n$$\n\n故使得 $f(x_1, \\dots, x_n) \\not\\equiv 0 \\pmod{13}$ 的 $n$-元組 $(x_1, \\dots, x_n)$ 的個數必為 $13$ 的倍數，得證。",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 13726,
    "subject": "Mathematics (Olympiad)",
    "question": "Each brick in a set has 5 holes in a horizontal row. We can either place pins into individual holes or brackets into two neighboring holes. No hole is allowed to remain empty. We place $n$ such bricks in a row to create patterns running from left to right, in which no two brackets are allowed to follow one another, and no three pins may be in a row. How many such patterns of bricks can be created?",
    "options": [],
    "answer": "See solution",
    "solution": "Since three pins (P) or two brackets (B) may not lie in a row, they may not do so on an individual brick. This means there are only three different types of bricks, which we name $A$ (PBPP), $B$ (PPBP), and $C$ (BPB). Let $a_n$, $b_n$, and $c_n$ be the number of possible patterns of $n$ bricks ending with brick $A$, $B$, and $C$, respectively. The total number is $s_n = a_n + b_n + c_n$.\n\nDue to the restrictions, we have:\n\n$$\na_{n+1} = b_n + c_n \\\\\nb_{n+1} = c_n \\\\\nc_{n+1} = a_n + b_n\n$$\n\nwith starting values $a_1 = b_1 = c_1 = 1$.\n\nThis yields:\n\n$$\ns_{n+1} = s_n + (b_n + c_n) = s_n + (a_{n-1} + b_{n-1} + c_{n-1}) = s_n + s_{n-1}\n$$\n\nwith $s_1 = 3$ and $s_2 = 5$. Thus, the sequence $s_n$ is the Fibonacci sequence starting from the fourth element, so $s_n = F_{n+3}$.",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 13727,
    "subject": "Mathematics (Olympiad)",
    "question": "Нека $a$ и $n$ се цели броеви. Дефинираме $a_n = 1 + a + a^2 + \\dots + a^{n-1}$. Докажи дека ако $a^p \\equiv 1 \\pmod{p}$ за секој прост делител $p$ на $n_2 - n_1$, бројот $$\\frac{a_{n_2} - a_{n_1}}{n_2 - n_1}$$ е цел број.",
    "options": [],
    "answer": "See solution",
    "solution": "*Лема.* Нека $a$ и $n$ се цели броеви такви што $a \\equiv 1 \\pmod{p}$ за секој прост $p \\nmid n$ и $a_n = 1 + a + a^2 + \\dots + a^{n-1}$. Тогаш $n \\mid a_n$.\n\n*Доказ на лемата.* Нека $p^r$ е најголемиот степен на простиот број $p$ таков што $p^r \\nmid n$. Ќе го докажеме равенството\n\n$$\n1 + a + a^2 + \\dots + a^{n-1} = \\left(1 + a^{p^r} + a^{2p^r} + \\dots + a^{(p-1)p^r}\\right) \\prod_{k=1}^{r} \\left(1 + a^{p^{k-1}} + a^{2p^{k-1}} + \\dots + a^{(p-1)p^{k-1}}\\right)\n$$\n\nза секој цел број $a$. Ако $a=1$ левата страна изнесува $n$, а десната страна на равенството изнесува $\\frac{n}{p^r}$. $p^r = n$ (по едно $p$ за секој член во производот). Нека $a \\ne 1$. Ако ја помножиме и левата и десната страна со $a-1$ од десно добиваме\n\n$$\n(a-1)(1+a+\\dots+a^{p-1})(1+a^p+a^{2p}+\\dots+a^{(p-1)p})\\dots\\left(1+a^{2p^r}+\\dots+a^{p^r}\\left(\\frac{n}{p^r}-1\\right)\\right)\n$$\n\n$$\n=(a^p-1)(1+a^p+a^{2p}+\\dots+a^{(p-1)p})\\dots\\left(1+a^{2p^r}+\\dots+a^{p^r}\\left(\\frac{n}{p^r}-1\\right)\\right)\n$$\n\n$$\n=(a^{p^2}-1)(1+a^{p^2}+\\dots+a^{(p-1)p^2})\\dots\\left(1+a^{2p^r}+\\dots+a^{p^r}\\left(\\frac{n}{p^r}-1\\right)\\right)\n$$\n\n$$\n=(a^{p^r}-1)\\left(1+a^{2p^r}+\\dots+a^{p^r}\\left(\\frac{n}{p^r}-1\\right)\\right)=a^n-1\n$$\n\nСекој од изразите во производот е делив со $p$ бидејќи\n\n$$\n1 + a^{p^{k-1}} + a^{2p^{k-1}} + \\dots + a^{(p-1)p^{k-1}} = (a^{p^{k-1}} - 1) + (a^{2p^{k-1}} - 1) + \\dots + (a^{(p-1)p^{k-1}} - 1) + p\n$$\n\nсекој од изразите во заградите е делив со $p$.\n\nБез губење на општоста можеме да претпоставиме дека $n_1 < n_2$. Јасно е дека\n\n$$\n\\frac{a_{n_2} - a_{n_1}}{n_2 - n_1} = \\frac{1 + a + \\dots + a^{n_2 - n_1 - 1} - 1 - a - \\dots - a^{n_1 - 1}}{n_2 - n_1} = \\frac{a^{n_1} (1 + a + \\dots + a^{n_2 - n_1 - 1})}{n_2 - n_1}\n$$\n\nСо користење на лемата добиваме дека $(n_2 - n_1) \\mid a_{n_2 - n_1}$, од каде добиваме дека бројот $\\frac{a_{n_2} - a_{n_1}}{n_2 - n_1}$ е природен број.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 13728,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $a$, $b$, $c$ be distinct positive integers.\n\na) Prove that $a^2b^2 + a^2c^2 + b^2c^2 \\ge 9$.\n\nb) If, moreover, $ab + ac + bc + 3 = abc > 0$, show that\n$$\n(a - 1)(b - 1) + (a - 1)(c - 1) + (b - 1)(c - 1) \\ge 6.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "a) At least one of the numbers $a$, $b$, $c$ has modulus at least $2$, so\n$$\na^2b^2 + a^2c^2 + b^2c^2 \\ge 1 \\cdot 1 + 1 \\cdot 4 + 1 \\cdot 4 = 9.\n$$\n\nb) The required inequality can be successively written\n$$\n\\begin{aligned}\n& ab + ac + bc - 2(a + b + c) + 3 \\ge 6 \\\\\n& ab + ac + bc - 3 \\ge 2(a + b + c) \\\\\n& (ab + ac + bc - 3)(ab + ac + bc + 3) \\ge 2abc(a + b + c) \\\\\n& (ab + ac + bc)^2 - 9 \\ge 2abc(a + b + c) \\\\\n& a^2b^2 + a^2c^2 + b^2c^2 \\ge 9,\n\\end{aligned}\n$$\nwhich is exactly part a).",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 13729,
    "subject": "Mathematics (Olympiad)",
    "question": "We consider figures consisting of six squares whose sides have length 1. The *radius* of such a figure is the radius of the smallest circle containing the whole figure. On the right, there is an example of a figure with radius $\\sqrt{5}$.\n\n![](images/NLD_ABooklet_2021_p7_data_b5d3a06458.png)\n\nWhich of the following five figures has the smallest radius?\n\nA) A\n\nB) B\n\nC) C\n\nD) D\n\nE) E\n\n![](images/NLD_ABooklet_2021_p7_data_3c372b8704.png)\n\nA\n\n![](images/NLD_ABooklet_2021_p7_data_329cdd9d52.png)\n\nB\n\n![](images/NLD_ABooklet_2021_p7_data_7b53c78172.png)\n\nC\n\n![](images/NLD_ABooklet_2021_p7_data_f0c7db8b87.png)\n\nD\n\n![](images/NLD_ABooklet_2021_p7_data_a26aa9caac.png)\n\nE",
    "options": [],
    "answer": "See solution",
    "solution": "B) B",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 13730,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $\\mathbb{N}^*$ be the set of positive integers. Define $a_1 = 2$, and for $n = 1, 2, \\dots$,\n$$\na_{n+1} = \\min\\left\\{\\lambda \\mid \\frac{1}{a_1} + \\frac{1}{a_2} + \\dots + \\frac{1}{a_n} + \\frac{1}{\\lambda} < 1,\\ \\lambda \\in \\mathbb{N}^*\\right\\}.\n$$\nProve that $a_{n+1} = a_n^2 - a_n + 1$ for $n = 1, 2, \\dots$.",
    "options": [],
    "answer": "See solution",
    "solution": "Since $a_1 = 2$, $a_2 = \\min\\left\\{\\lambda \\mid \\frac{1}{a_1} + \\frac{1}{\\lambda} < 1,\\ \\lambda \\in \\mathbb{N}^*\\right\\}$.\nFrom $\\frac{1}{a_1} + \\frac{1}{\\lambda} < 1$ we have $\\frac{1}{\\lambda} < 1 - \\frac{1}{2} = \\frac{1}{2}$, so $\\lambda > 2$ and thus $a_2 = 3$.\n\nThis means that when $n = 1$, the conclusion holds.\n\nSuppose that for $n \\leq k - 1$ ($k \\geq 2$), the conclusions are correct. For $n = k$, from\n$$\na_{k+1} = \\min \\left\\{ \\lambda \\mid \\frac{1}{a_1} + \\frac{1}{a_2} + \\dots + \\frac{1}{a_k} + \\frac{1}{\\lambda} < 1,\\ \\lambda \\in \\mathbb{N}^* \\right\\}.\n$$\nAs $\\frac{1}{a_1} + \\frac{1}{a_2} + \\dots + \\frac{1}{a_k} + \\frac{1}{\\lambda} < 1$, that is,\n$$\n0 < \\frac{1}{\\lambda} < 1 - \\left( \\frac{1}{a_1} + \\frac{1}{a_2} + \\dots + \\frac{1}{a_k} \\right),\n$$\nso\n$$\n\\lambda > \\frac{1}{1 - \\frac{1}{a_1} - \\frac{1}{a_2} - \\dots - \\frac{1}{a_k}}.\n$$\nNow we prove that $\\frac{1}{1 - \\frac{1}{a_1} - \\frac{1}{a_2} - \\dots - \\frac{1}{a_k}} = a_k(a_k - 1)$.\n\nBy the supposition, for $2 \\leq n \\leq k$, $a_n = a_{n-1}(a_{n-1}-1)+1$; then\n$$\n\\frac{1}{a_n - 1} = \\frac{1}{a_{n-1}(a_{n-1} - 1)} = \\frac{1}{a_{n-1} - 1} - \\frac{1}{a_{n-1}}.\n$$\nSo $\\frac{1}{a_{n-1}} = \\frac{1}{a_{n-1}-1} - \\frac{1}{a_n-1}$, and $\\sum_{i=2}^{k} \\frac{1}{a_{i-1}} = 1 - \\frac{1}{a_k-1}$, i.e.\n$$\n\\sum_{i=1}^{k} \\frac{1}{a_i} = 1 - \\frac{1}{a_k - 1} + \\frac{1}{a_k} = 1 - \\frac{1}{a_k(a_k - 1)}\n$$\nwhich means that $\\frac{1}{1 - \\frac{1}{a_1} - \\frac{1}{a_2} - \\dots - \\frac{1}{a_k}} = a_k(a_k - 1)$. Then\n$$\na_{k+1} = \\min\\left\\{\\lambda \\mid \\frac{1}{a_1} + \\frac{1}{a_2} + \\cdots + \\frac{1}{a_k} + \\frac{1}{\\lambda} < 1,\\ \\lambda \\in \\mathbb{N}^*\\right\\} = a_k(a_k - 1) + 1.\n$$\nWe get the conclusion for all $n$: $a_{n+1} = a_n^2 - a_n + 1$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 13731,
    "subject": "Mathematics (Olympiad)",
    "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": "See solution",
    "solution": "Consider the polynomial\n\n$$\n(x+2)^{2n} = (x^2 + 4x + 4)^n\n$$\n\nSuppose we multiply out the brackets, obtaining $9^n$ summands. We show a one-to-one correspondence between the number of $x^n$ terms and the number of distributions described in the problem.\n\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, or E, we pick the first, second, third, or fourth $1$, respectively, from the $k$-th bracket. If it goes to F, G, H, or I, we take the first, second, third, or fourth $x$ from the $k$-th bracket. When we multiply the chosen factors, the result is $x^n$ if and only if A gets the same number of balls as B, C, D, and E jointly.\n\nTherefore, the number of distributions we are interested in is equal to the coefficient of $x^n$ in the polynomial $(x+2)^{2n}$, that is,\n\n$$\n\\binom{2n}{n} \\cdot 2^n.\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 13732,
    "subject": "Mathematics (Olympiad)",
    "question": "An $11 \\times 11$ square is partitioned into 121 smaller $1 \\times 1$ squares, 4 of which are painted black, the rest being white. We cut a fully white rectangle (possibly a square) out of the big $11 \\times 11$ square. What is the maximal area of the rectangle that we can attain regardless of the positions of the black squares? It is only allowed to cut the rectangle along the grid lines.",
    "options": [],
    "answer": "See solution",
    "solution": "The maximal area is $25$.\n\nFirst, paint four squares black as in the figure below. Then it is easy to verify that a $5 \\times 5$ square of area $25$ can be cut out, and this area is maximal possible.\n\n![](images/ukraine_2015_Booklet_p2_data_a64c383ba2.png)\n\nNow we shall show that this area is maximal possible in the general case. Assume that there is a placement of four squares for which the answer doesn't exceed $24$. It means that whichever rectangle of area $25$ we can choose, it will always contain a black square. For convenience, we can number the fields of the board. Denote rows by numbers from $1$ to $11$ from bottom to top, and columns by English letters $a, \\ldots, k$. We shall call a unit square gray if it cannot be black in any case.\n\nIn each of the four $5 \\times 5$ squares $(a1$-$e5)$, $(g1$-$k5)$, $(a6$-$e10)$, $(g6$-$k10)$, there must be exactly one black unit square. Therefore, rows $1$, $11$, and columns $a$ and $k$ are gray.\n\nSimilarly, in each of the $5 \\times 5$ squares $(a1$-$e5)$, $(g1$-$k5)$, $(a7$-$e11)$, $(g7$-$k11)$, there must be exactly one black unit square. Therefore, row $6$ and column $f$ are also gray.\n\n![](images/ukraine_2015_Booklet_p2_data_e85548b3c4.png)\n\nAssume there is a black square in one of the columns $b$, $j$ or rows $2$, $10$. Without loss of generality, assume it is $b$. Then rectangles with corners $c1$-$k3$ and $c9$-$k11$ contain black squares. Each of them has area $27 = 3 \\times 9$. Then there is at most one black unit square in the rectangle $c4$-$k8$ with size $5 \\times 9$. But it contains two rectangles $c4$-$k6$ and $c6$-$k8$, both of which must have a black unit square. Thus, it has to lie in row $6$. But this row is gray, hence, we have a contradiction. Now the figure shows a broader set of gray squares.\n\nObviously, each $3 \\times 3$ square must have exactly $1$ black unit square. Consider the case when at least one of the squares $c3$, $c9$, $i3$, $i9$ is black. Assume it is $c9$. Then consider the two rectangles $d7$-$h11$ and $a6$-$k8$. As we can easily see, the black unit square must be within the square $g7$-$h8$. The same goes for $d4$-$e5$. Then, by looking at the rectangles $i1$-$k11$ and $a1$-$k3$, we conclude that $i3$ must be black. From rectangles $a1$-$d8$ and $a1$-$h4$ we deduce that $d4$ and $h8$ must be black. But then $e1$-$g11$ is free (i.e., has no painted squares). Hence, all four squares $c3$, $c9$, $i3$, $i9$ must be gray. However, in this case each of the stripes $c4$-$c8$, $d9$-$h9$, $d3$-$h3$, $i4$-$i8$ must have exactly one black square. But then the central $5 \\times 5$ square $d4$-$h8$ is free. Contradiction.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 13733,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all positive integer solutions to the equation\n\n$$\n2^a + 2^b + 2^c + 2^d = 60 \\cdot \\min\\{a, b, c, d\\},\n$$\n\nwhere $\\min\\{a, b, c, d\\}$ denotes the minimum of the numbers $a, b, c, d$.",
    "options": [],
    "answer": "See solution",
    "solution": "The only solution is $\\{a, b, c, d\\} = \\{4, 5, 6, 7\\}$.\n\nIt is clear that this is a solution, so we prove that there are no other solutions.\n\nAssume $a \\le b \\le c \\le d$. Then $S = 2^a + 2^b + 2^c + 2^d = 60a$. We have $2^a \\mid 60a$, so $2^a \\le 60a$, which implies $a \\le 4$.\n\nNext, if $S \\equiv 0 \\pmod{15}$, then $a, b, c, d$ must have distinct remainders modulo 4. Since $2^4 \\equiv 1 \\pmod{15}$, we may order the remainders as $0 \\le p < q < r < s \\le 3$ and assume $p = 0$. Then $s = 3$, since $2^0 + 3 \\cdot 2^2 < 15$. Similarly, $r = 2$, since $2^0 + 2 \\cdot 2^1 + 2^3 < 15$. Finally, $q = 1$.\n\nIt follows that $v_2(S) = a = 2 + v_2(a)$, so $a \\ne 1, 2, 3$. For $a = 4$, we have $b \\ge 5$, $c \\ge 6$, $d \\ge 7$, so $S \\ge 2^4 + 2^5 + 2^6 + 2^7 = 16 + 32 + 64 + 128 = 240$. Equality holds only for $\\{4, 5, 6, 7\\}$, so there is no other solution.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 13734,
    "subject": "Mathematics (Olympiad)",
    "question": "A convex quadrilateral $ABCD$ has $\\angle ABC = \\angle CDA = 90^\\circ$. Point $H$ is the foot of the perpendicular from $A$ to $BD$. Points $S$ and $T$ lie on sides $AB$ and $AD$, respectively, such that $H$ lies inside $\\triangle SCT$ and $\\angle CHS - \\angle CSB = 90^\\circ$,  \n$$\n\\angle THC - \\angle DTC = 90^{\\circ}.\n$$\nProve that line $BD$ is tangent to the circumcircle of $\\triangle TSH$.",
    "options": [],
    "answer": "See solution",
    "solution": "Suppose that the line passing through $C$ and perpendicular to $SC$ intersects $AB$ at point $Q$ (see the figure below). Then $\\angle SQC = 90^{\\circ} - \\angle BSC = 180^{\\circ} - \\angle SHC$. Hence, points $C$, $H$, $S$, and $Q$ are concyclic with $SQ$ as its diameter. Therefore, the circumcentre $K$ of $\\triangle SHC$ is on $AB$. In the same manner, the circumcentre $L$ of $\\triangle CHT$ is on $AD$.\n\n![](images/Mathematical_Olympiad_in_China_2011-2014_p362_data_16d78bd4a9.png \"Fig. 3. 1\")\n\nTo show that line $BD$ is tangent to the circumcircle of $\\triangle TSH$, it suffices to show that the intersection point of the perpendicular bisectors of $HS$ and $HT$ is on $AH$. But the perpendicular bisectors are just the angle bisectors of $\\angle AKH$ and $\\angle ALH$, respectively. By the Internal Angle Bisector Theorem, it suffices to show that\n$$\n\\frac{AK}{KH} = \\frac{AL}{LH}. \\qquad \\textcircled{1}\n$$\nIn the following, we give two proofs of $\\textcircled{1}$.\n\n**Proof 1.** Let $M$ be the intersection point of lines $KL$ and $HC$ (see Fig. 3.2). Since $KH = KC$ and $LH = LC$, points $H$ and $C$ are symmetric over $KL$. Thus, $M$ is the midpoint of $HC$. Let $O$ be the circumcentre of quadrilateral $ABCD$. Hence, $O$ is the midpoint of $AC$ and consequently $OM \\parallel AH$, therefore $OM \\perp BD$. Further, since $OB = OD$, we see that $OM$ is the perpendicular bisector of $BD$, thus $BM = DM$.\n\n![](images/Mathematical_Olympiad_in_China_2011-2014_p363_data_c9213b4211.png \"Fig. 3. 2\")\n\nSince $CM \\perp KL$, points $B$, $C$, $M$, and $K$ are concyclic with $KC$ as the diameter. Similarly, points $L$, $C$, $M$, and $D$ are concyclic with $LC$ as the diameter. So, by the Sine Theorem, we have\n$$\n\\frac{AK}{AL} = \\frac{\\sin \\angle ALK}{\\sin \\angle AKL} = \\frac{DM}{CL} \\cdot \\frac{CK}{BM} = \\frac{CK}{CL} = \\frac{KH}{LH},\n$$\nthat is, $\\textcircled{1}$.\n\n**Proof 2.** If points $A$, $H$, and $C$ are collinear, then $AK = AL$, $KH = LH$, thus $\\textcircled{1}$ follows. Else, let $\\omega$ be the circle passing through $A$, $H$, and $C$. Since points $A$, $B$, $C$, and $D$ are on a circle,\n$$\n\\angle BAC = \\angle BDC = 90^\\circ - \\angle ADH = \\angle AHD.\n$$\nLet $N \\neq A$ be the other intersection point of the circle $\\omega$ and the bisector of $\\angle CAH$, then $AN$ is also the bisector of $\\angle BAD$. Since points $H$ and $C$ are symmetric over $KL$, and $HN = NC$, we see that point $N$ and the centre of $\\omega$ are both on $KL$. That is, the circle $\\omega$ is the Apollonius circle of points $K$ and $L$, thus $\\textcircled{1}$ follows. $\\square$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 13735,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $n$ be an integer greater than $1$ and let $\\mathcal{S}$ be the set of $n$-element subsets of the set $\\{1, 2, \\dots, 2n\\}$. Determine\n\n$$\n\\max_{S \\in \\mathcal{S}} \\min_{x, y \\in S,\\ x \\neq y} [x, y],\n$$\n\nwhere $[x, y]$ denotes the least common multiple of the integers $x$ and $y$.",
    "options": [],
    "answer": "See solution",
    "solution": "The required value is $6(\\lfloor n/2 \\rfloor + 1)$, unless $n=4$ in which case it is $24$.\n\nLet $S$ be a member of $\\mathcal{S}$. We first show that\n\n$$\n\\min_{x, y \\in S,\\ x \\neq y} [x, y] \\le 6(\\lfloor n/2 \\rfloor + 1), \\quad (*)\n$$\n\nunless $n = 4$. To this end, for each $x$ in $S$, choose a positive integer $m_x$ such that $n < m_x x \\le 2n$ and consider the set $S' = \\{m_x x : x \\in S\\}$.\n\nIf $|S'| < n$, then $m_x x = m_y y$ for some distinct elements $x$ and $y$ in $S$, so $[x, y] \\le 2n$.\n\nIf $|S'| = n$, then $S' = \\{n+1, n+2, \\dots, 2n\\}$. The first even number in $S'$ is $2(\\lfloor n/2 \\rfloor + 1)$, and the number $3(\\lfloor n/2 \\rfloor + 1)$ is also in $S'$ if $n=3$ or $n \\ge 5$. Consequently, $(*)$ holds for $n=3$ or $n \\ge 5$, and it clearly holds for $n=2$.\n\nIf $n=4$, then\n\n$$\n\\min \\{[x, y] : x, y \\in \\{5, 6, 7, 8\\},\\ x \\neq y\\} = 24,\n$$\n\nwhich is the required value by the preceding.\n\nFinally, we show that, if $1 \\le i < j \\le n$, then $[n+i, n+j] \\ge 6(\\lfloor n/2 \\rfloor + 1)$. Suppose, if possible, that $[n+i, n+j] < 6(\\lfloor n/2 \\rfloor + 1)$. Since $[n+1, n+2] = (n+1)(n+2) \\ge 6(\\lfloor n/2 \\rfloor + 1)$, it follows that $j \\ge 3$, so $n+j \\ge 2(\\lfloor n/2 \\rfloor + 1)$. Hence $[n+i, n+j] = 2(n+j) = m(n+i)$, where $m$ is an integer greater than $2$. If $m=3$, then $n+i$ must be an even number less than $2(\\lfloor n/2 \\rfloor + 1)$ which is impossible. If $m \\ge 4$, then $n+i < 3(\\lfloor n/2 \\rfloor + 1)/2 \\le n+1$ which is again impossible. This ends the proof.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 13736,
    "subject": "Mathematics (Olympiad)",
    "question": "Two children are playing noughts and crosses with changed rules. In each move, either of the players may draw into an empty square of a $3 \\times 3$ board either a nought or a cross according to one's wish. Moves are made alternately and the winner is the one after whose move a row, a column or a long diagonal becomes filled with three similar signs. Is there a player with a winning strategy, and if yes then who?",
    "options": [],
    "answer": "See solution",
    "solution": "Yes, the first player has a winning strategy.\n\nThe first player may play the first move into the middle square and, later on, make the immediately winning move if there is any and play symmetrically to the opponent's last move with respect to the center of the board otherwise.\n\nSuppose that the opponent wins. As the central square is occupied, the winning move must be played either into a corner or in the middle of an edge of the board. According to the first player's strategy, the position before the winning move was symmetric with respect to the center of the board. Consequently, the square symmetric to the winning move is empty in the final position, which in turn implies that the three signs of the same type appear along an edge of the board. Before the winning move, there must already have been two of these signs present and, by symmetry, similarly also at the opposite edge. Three of these four signs must already have been there before the last move of the first player. As two of these three must have been in one line, the first player could win in her last move, which contradicts the chosen strategy.\n\n*Remark:* It is also easy to argue by case study.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 13737,
    "subject": "Mathematics (Olympiad)",
    "question": "Three friends color the positive integers from 1 to 2025 as follows: Alexia colors in red the numbers 1 and 2, then Bianca colors in yellow the numbers 3, 4, and 5, and Cristina colors in blue the numbers 6, 7, 8, and 9. Afterwards, the operation is repeated: Alexia colors in red the next two numbers, Bianca colors in yellow the next three numbers, and Cristina colors in blue the next four. The friends keep on painting, until all the numbers are colored.\n\n(a) What will be the color of $2024$?\n\n(b) Find the smallest natural number $n$ so that, after $n$ numbers have been colored, the sum of the numbers in yellow is larger than $2024$.",
    "options": [],
    "answer": "See solution",
    "solution": "For simplicity, we will call the numbers red, yellow, and blue, respectively. A sequence of $9$ consecutive numbers in which the first two are red, the next three are yellow, and the last four are blue will be called a *complete coloring*.\n\n(a) Since $2025 = M_9$, the last four numbers are blue, hence $2024$ is blue.\n\n(b) The $(k+1)$-th complete coloring (where $k$ is a natural number) assigns yellow to the numbers $3 + 9k$, $4 + 9k$, and $5 + 9k$, with sum $12 + 27k$.\n\nAfter $12$, respectively $13$ complete colorings, the sum of the yellow numbers is\n$$\n12 \\cdot 12 + 27 \\cdot (0 + 1 + 2 + \\dots + 11) = 1926,\n$$\nrespectively\n$$\n12 \\cdot 13 + 27 \\cdot (0 + 1 + 2 + \\dots + 12) = 2262.\n$$\n\nIn order to get the sum of the yellow numbers larger than $2024$, we need $12$ complete colorings ($108$ numbers), then the red numbers $109$, $110$, and the yellow number $111$, hence the smallest $n$ is $111$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 13738,
    "subject": "Mathematics (Olympiad)",
    "question": "A new wheel with a diameter of 14 cm is to replace an old wheel with a diameter of 10 cm. The new wheel must revolve at the same rate as the old wheel, which is $\\frac{12}{\\pi}$ revolutions per hour. How much higher must the center of the new wheel be placed above the floor (compared to the old wheel) so that the chain remains taut and the bells move the same distance in one hour?\n\n![](images/Australian_Scene_2010_p49_data_906c13b58f.png)",
    "options": [],
    "answer": "See solution",
    "solution": "The new wheel revolves at the same rate as the old wheel, which is $\\frac{12}{\\pi}$ revolutions per hour. In one hour, the right side of the chain (R) would move down $\\frac{12}{\\pi} \\times 14\\pi = 168$ cm (if the floor had a hole in it).\n\nThe difference, $168 - (120 + 2\\pi) = 48 - 2\\pi$ cm, is compensated by raising the center of the new wheel. For each cm the wheel is raised with the left side (L) touching the floor, R is raised 2 cm. Thus, the wheel must be raised by $(48 - 2\\pi) \\div 2 \\approx 20.86$ cm $\\approx 209$ mm.\n\n**Alternative approach:**\n\nLet the centers of the 10 cm and 14 cm diameter wheels be $a$ cm and $b$ cm above the floor, respectively. The circumference of the 14 cm wheel is $14\\pi$ cm. It revolves at $\\frac{12}{\\pi}$ revolutions per hour, so in one hour each bell moves $\\frac{12}{\\pi} \\times 14\\pi = 168$ cm.\n\nLet the length of the chain be $C$. The length of chain in contact with a wheel is half the circumference of the wheel.\n\nFor the 10 cm wheel: $C + 120 = 2a + 5\\pi$.\n\nFor the 14 cm wheel: $C + 168 = 2b + 7\\pi$.\n\nSubtracting gives:\n\n$$\n48 = 2b - 2a + 2\\pi, \\quad 2b - 2a = 48 - 2\\pi, \\quad b - a = 24 - \\pi.\n$$\n\nSo the wheel center must be raised by $24 - \\pi \\approx 20.86$ cm $\\approx 209$ mm.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 13739,
    "subject": "Mathematics (Olympiad)",
    "question": "If $a, b, c, d$ are positive real numbers such that $abcd = 1$, prove that the inequality\n\n$$\n\\frac{1}{bc + cd + da - 1} + \\frac{1}{ab + cd + da - 1} + \\frac{1}{ab + bc + da - 1} + \\frac{1}{ab + bc + cd - 1} \\le 2\n$$\nholds.",
    "options": [],
    "answer": "See solution",
    "solution": "By multiplying $1 + bc + cd + da$ and $1 + ab$ together, we get\n\n$$\n(1 + bc + cd + da)(1 + ab) = 1 + bc + cd + da + ab + ab^2c + abcd + a^2bd = 2 + ab + bc + cd + da + \\frac{b}{d} + \\frac{a}{c}\n$$\n\nFrom the inequality between the arithmetic and geometric mean for the positive numbers $\\frac{b}{d}$ and $\\frac{a}{c}$, and from the equality $abcd = 1$, we get $\\frac{b}{d} + \\frac{a}{c} \\ge 2\\sqrt{\\frac{ab}{cd}} = 2ab$, and hence it holds that $(1 + bc + cd + da)(1 + ab) \\ge 2 + 2ab + ab + bc + cd + da$.\n\nThat is,\n\n$$\n1 + bc + cd + da \\ge 2 + \\frac{ab + bc + cd + da}{1 + ab}\n$$\n\nor\n\n$$\n\\frac{1 + ab}{ab + bc + cd + da} \\ge \\frac{1}{bc + cd + da - 1} \\quad (1)\n$$\n\nAnalogously, we get\n\n$$\n\\frac{1 + bc}{ab + bc + cd + da} \\ge \\frac{1}{ab + cd + da - 1} \\quad (2)\n$$\n\n$$\n\\frac{1 + cd}{ab + bc + cd + da} \\ge \\frac{1}{ab + bc + da - 1} \\quad (3)\n$$\n\n$$\n\\frac{1 + da}{ab + bc + cd + da} \\ge \\frac{1}{ab + bc + cd - 1} \\quad (4)\n$$\n\nBy adding (1), (2), (3), and (4) together, we get the inequality\n\n$$\n\\frac{4 + ab + bc + cd + da}{ab + bc + cd + da} \\geq \\frac{1}{bc + cd + da - 1} + \\frac{1}{ab + cd + da - 1} + \\frac{1}{ab + bc + da - 1} + \\frac{1}{ab + bc + cd - 1}\n$$\n\nSince $ab + bc + cd + da \\geq 4\\sqrt[4]{(abcd)^2} = 4$, it follows that\n\n$$\n\\frac{1}{bc + cd + da - 1} + \\frac{1}{ab + cd + da - 1} + \\frac{1}{ab + bc + da - 1} + \\frac{1}{ab + bc + cd - 1} \\le 1 + \\frac{4}{ab + bc + cd + da} \\le 2\n$$\nwhich was to be proven.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 13740,
    "subject": "Mathematics (Olympiad)",
    "question": "In triangle $ABC$, $\\angle A = 75^\\circ$ and $\\angle C = 45^\\circ$. Points $P$ and $T$ are chosen on segments $AB$ and $BC$ such that quadrilateral $APTC$ is cyclic and $CT = 2AP$. Let $O$ be the circumcenter of $\\triangle ABC$. The ray $TO$ crosses side $AC$ at point $K$. Prove that $TO = OK$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $CD$ be a diameter of the circumcircle of $\\triangle ABC$. Then $\\triangle ADC$ is a right triangle with an angle of $60^\\circ$. Hence, $CD = 2AD$, and $\\triangle DTC \\sim \\triangle DPA$ by two proportional sides and equal included angles. Therefore, $\\angle BPD = \\angle BTD$, so $PDBT$ is a cyclic quadrilateral. Note that $\\angle BTD = \\angle BPD = \\angle DPT - \\angle BPT = 180^\\circ - \\angle DBT - 45^\\circ = 180^\\circ - 90^\\circ - 45^\\circ = 45^\\circ$, from which $DT \\parallel AC$. Since $DO = OC$ and $DT \\parallel CK$, $DTCK$ is a parallelogram, and thus $TO = OK$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 13741,
    "subject": "Mathematics (Olympiad)",
    "question": "Let the equation\n\n$$\nx^2 + (a + b - 1)x + ab - a - b = 0,\n$$\n\nwhere $a$ and $b$ are positive integers such that $0 < a \\leq b$.\n\n**a)** Show that the equation has two distinct real solutions.\n\n**b)** Prove that if one solution of the equation is an integer, then both solutions are non-positive integers and $b < 2a$.",
    "options": [],
    "answer": "See solution",
    "solution": "**a)** The discriminant of the equation is\n\n$$\n\\Delta = (a + b - 1)^2 - 4(ab - a - b) = (a - b)^2 + 2a + 2b + 1 > 0,\n$$\n\nso the equation has two distinct real solutions.\n\n**b)** If $x_1 < x_2$ are the two solutions of the equation, from Viète's first relation, $x_1 + x_2 = 1 - a - b \\in \\mathbb{Z}$, so $x_1 \\in \\mathbb{Z} \\iff x_2 \\in \\mathbb{Z}$.\n\nLet $f(x) = x^2 + (a + b - 1)x + ab - a - b$, $x \\in \\mathbb{R}$. Since $f(1) = ab > 0$, $f(-a) = -b < 0$, $f(-b) = -a < 0$, using the sign of the quadratic function we obtain $x_1 < -b \\leq -a < x_2 < 1$, so both solutions are non-positive.\n\nThus, $a \\geq 1 - x_2$, $b \\geq 1 - x_2$ and since $f(x_2) = 0$, we obtain $(a - 1 + x_2) \\cdot (b - 1 + x_2) = 1 - x_2 \\geq 1$. We get that $d_1 = a - 1 + x_2$ and $d_2 = b - 1 + x_2$ are natural divisors of $1 - x_2$ and $d_1 d_2 = 1 - x_2$, $a = d_1 + d_1 d_2$, $b = d_2 + d_1 d_2$. We obtain $2a - b = 2d_1 + d_2(d_1 - 1) \\geq 2d_1 > 0$, thus $b < 2a$.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 13742,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $\\Omega$ be the circumcircle centered at $O$ of $\\triangle ABC$ with $\\angle B > 90^\\circ$. Let $B_1$ be the intersection point of the line $AB$ and the tangent to $\\Omega$ at $C$. Let $O_1$ be the circumcenter of $\\triangle AB_1C$. Choose a point $B_2$ on the segment $BB_1$ ($B_2 \\ne B, B_1$). The line from $B_2$ is tangent to $\\Omega$ at $C_1$, closer to $C$. Let $O_2$ be the circumcenter of $\\triangle AB_2C_1$.\n\nProve that if $OO_2 \\perp AO_1$, then $A$, $C$, $O_2$ are collinear.",
    "options": [],
    "answer": "See solution",
    "solution": "**Solution.** Here we use directed angles measured in the counterclockwise direction.\n\nSince $AC_1$ is the common chord of the circles $O$ and $O_2$, we see that $OO_2 \\perp AC_1$. Therefore, $OO_2 \\perp AO_1$ is equivalent to $O_1$ lying on $AC_1$.\n\nAssume that $O_1$ lies on $AC_1$. (Claim: $O_2$ lies on $AC$.)\n\nSince $O$ and $O_1$ are circumcenters of $ABC$ and $AB_1C$, it is easy to see that\n\n$$\n\\angle O_1C_1O_1 = \\angle O_1CA = \\angle C_1AO = \\angle C_1AC + \\angle CAO = \\angle AC_1O_1 + \\angle OCA = \\angle OCO_1.\n$$\n\nTherefore, $O$, $O_1$, $C_1$, $C$ are cyclic.\n\nConsider $\\triangle AB_2C_1$ with circumcenter $O_2$. Since $OA = OC_1$ and $O_2A = O_2C_1$, $\\triangle AOO_2 \\cong \\triangle C_1OO_2$ and $\\angle C_1O_2O = \\frac{1}{2}\\angle C_1O_2A = \\angle C_1B_2A$.\n\n$$\n\\begin{aligned}\n\\angle B_2AC_1 &= \\frac{1}{2}(\\pi - \\angle A_1B_1) = \\frac{\\pi}{2} - \\angle ACB_1 = \\frac{\\pi}{2} - \\angle ACB - \\angle BCB_1 \\\\\n&= \\frac{\\pi}{2} - \\angle ACB - \\angle BAC = \\angle CBA - \\frac{\\pi}{2}.\n\\end{aligned}\n$$\n\n$$\n\\begin{aligned}\n\\angle AC_1B_2 &= \\angle BC_1B_2 + \\angle AC_1B = \\angle B_2AC_1 + \\angle ACB = \\angle CBA + \\angle ACB - \\frac{\\pi}{2} \\\\\n&= \\frac{\\pi}{2} - \\angle BAC.\n\\end{aligned}\n$$\n\n$$\n\\therefore \\angle C_1B_2A = \\pi - \\angle B_2AC_1 - \\angle AC_1B_2 = \\pi + \\angle BAC - \\angle CBA.\n$$\n\n$$\n\\begin{aligned}\n\\angle C_1O_1O &= \\angle CO_1O + \\angle C_1O_1C = \\angle CC_1O + \\angle C_1OC = \\frac{1}{2}(\\pi - \\angle C_1OC) + \\angle C_1OC \\\\\n&= \\frac{\\pi}{2} + \\angle C_1AC = \\frac{\\pi}{2} + \\angle BAC - \\angle B_2AC_1 \\\\\n&= \\pi + \\angle BAC - \\angle CBA.\n\\end{aligned}\n$$\n\nThus, $\\angle C_1O_2O = \\angle C_1O_1O$, and then $O$, $O_2$, $O_1$, $C_1$, $C$ lie on the same circle.\n\nLet $O'_2$ be the intersection point of $AC$ and the perpendicular bisector of $AC$. Then $\\angle O'_2OO_1 = \\angle O_1AO'_2 = \\angle O'_2CO_1$. Therefore, $O'_2$ also lies on the circle passing through $O$, $O_2$, $O_1$, $C_1$, $C$. Since $O$, $O_2$, $O'_2$ lie on the perpendicular bisector of $AC$, where $O_2$ and $O'_2$ are different from $O$ (since $\\angle CBA > 90^\\circ$, $O$ is outside $\\triangle ABC$), this implies that $O_2 = O'_2$. (Because a line can intersect a circle in at most 2 points.) Therefore, $O_2$ lies on $AC$.\n\n_Remark._ The converse also holds: if $A$, $C$, $O_2$ are collinear, then $OO_2 \\perp AO_1$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 13743,
    "subject": "Mathematics (Olympiad)",
    "question": "In a convex quadrilateral $ABCD$:\n\n- $R$ and $S$ are points in the interior of the segments $CD$ and $AB$ respectively, with $AD = CR$ and $BC = AS$.\n- $P$ and $Q$ are the midpoints of $DR$ and $SB$ respectively.\n- $M$ is the midpoint of $AC$.\n\nIf it is known that $\\angle MPC + \\angle MQA = 90^\\circ$, prove that $ABCD$ is a cyclic quadrilateral.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $E$ and $F$ be points on the prolongations of $AB$ and $CD$ such that $DF = DA = CR$ and $BE = BC = AS$, as in the picture.\n\n![](images/Argentina_2019_Booklet_p19_data_31640eb0a3.png)\n\nNote that $P$ is the midpoint of $FC$. Then, $MP$ is a midsegment of triangle $AFC$, which implies that $\\angle AFC = \\angle MPC$. Since triangle $ADF$ is isosceles, we have $\\angle DAF = \\angle DFA = \\angle MPC$, and then $\\angle ADC = 2 \\cdot \\angle MPC$ (external angle).\n\nSimilarly, $\\angle ABC = 2 \\cdot \\angle MQA$.\n\nThus,\n\n$$\n\\angle ADC + \\angle ABC = 2 \\cdot (\\angle MPC + \\angle MQA) = 2 \\cdot 90^\\circ = 180^\\circ,\n$$\n\nand, therefore, the quadrilateral $ABCD$ is cyclic.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 13744,
    "subject": "Mathematics (Olympiad)",
    "question": "Let the first written word be $(1, 2, \\dots, n)$. Prove by induction on $n$ that the number of written words is $n!$.",
    "options": [],
    "answer": "See solution",
    "solution": "If $n = 2$, we start with $(1, 2)$ and then reverse it to get $(2, 1)$. Suppose that for $n = k-1$ the words\n\n$$\nW_1^{k-1}, W_2^{k-1}, \\dots, W_{(k-1)!}^{k-1}\n$$\n\nare written. Consider the sequence of words\n\n$$\nW_1^k, W_2^k, \\dots\n$$\n\nfor $n = k$. We use the notation $(a, b, \\dots, p) * q$. Divide the sequence above into blocks, each consisting of $2k$ consecutive words. The $m$-th block is\n\n$$\nW_{2m-1}^{k-1} * k, \\dots, W_{2m}^{k-1} * k\n$$\n\nThe first block starts with $W_1^{k-1} * k$ and ends with $W_2^{k-1} * k$, the second block with $W_3^{k-1} * k$ and ends with $W_4^{k-1} * k$, and so on. Inside each block, each $2l+1$-th word is obtained from the $2l-1$-th word by shortest counter-clockwise rotation, and each $2l+2$-th word from the $2l$-th word by shortest clockwise rotation. Thus, all words in each block are different and each block is closed under rotation. Since all words in the sequence for $n = k-1$ are different, all words in the sequence for $n = k$ are also different. By the induction hypothesis, there are $(k-1)!$ terms in the previous sequence, so there are $(k-1)!/2$ blocks in the new sequence, and the sequence consists of $(k-1)!/2 \\cdot 2k = k!$ distinct words. Thus, the result follows.\n\n*Note*: In each block, the second word is obtained by reversing the first $k$ letters, the third by reversing the first $k-1$ letters, the fourth by reversing the first $k$ letters, and so on. The first word of each block (except the very first word $(1, 2, \\dots, n)$) is obtained from the last word of the previous block by a similar operation.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 13745,
    "subject": "Mathematics (Olympiad)",
    "question": "給定一個大於 1 的正整數 $k$。甲、乙兩人玩以下的數字遊戲：在遊戲開始時，有一個正整數 $n \\ge k$ 被寫在黑板上。接著，從甲開始，兩人輪流進行以下動作：擦掉寫在黑板上的數 $m$，並在黑板上寫下一個與 $m$ 互質的正整數 $m'$，且 $k \\le m' < m$。第一個無法寫下數字的人輸。\n\n對於一開始在黑板上的數字 $n \\ge k$，如果乙有必勝法，則稱 $n$ 是個好數字；反之，$n$ 是個壞數字。\n\n現在，假設 $n, n' \\ge k$，且質數 $p \\le k$ 整除 $n$ 若且唯若 $p$ 整除 $n'$。試證：$n$ 和 $n'$ 要不同時是好數字，要不同時是壞數字。\n\n---\n\nFix an integer $k \\ge 2$. Two players, Ana and Banana, play the following game of numbers: Initially, some integer $n \\ge k$ gets written on the blackboard. Then they take moves in turn, with Ana beginning. A player making a move erases the number $m$ just written on the blackboard and replaces it by some number $m'$ with $k \\le m' < m$ that is coprime to $m$. The first player who cannot move anymore loses.\n\nAn integer $n \\ge k$ is called good if Banana has a winning strategy when the initial number is $n$, and bad otherwise.\n\nConsider two integers $n, n' \\ge k$ with the property that each prime number $p \\le k$ divides $n$ if and only if it divides $n'$. Prove that either both $n$ and $n'$ are good or both are bad.",
    "options": [],
    "answer": "See solution",
    "solution": "**解：** 為方便說明，令 $n \\to x$ 表示擦掉 $n$，寫上 $x$ 的動作；依題意，必有 $n > x \\ge k$ 且 $(n, x) = 1$。\n\n**Claim A.** 若 $m$ 為好數字，而 $n > m$ 與 $m$ 互質，則 $n$ 為壞數字。\n\n*Proof.* 因為甲只要選 $n \\to m$，並複製乙從 $m$ 開始的必勝策略即可。\n\n**Claim B.** 任何兩個好數字不能互質。\n\n*Proof.* 由 Claim A 和 B 立得。\n\n**Claim 1.** 若 $n$ 為好數字且 $n \\mid n'$，則 $n'$ 為好數字。\n\n*Proof.* 若 $n'$ 為壞數字，表示甲可以進行 $n' \\to x$ 且 $x$ 為好數字。然而 $(n', x) = 1 \\Rightarrow (n, x) = 1$，但 $n$ 與 $x$ 都是好數字，此與 Claim C 相矛盾。\n\n**Claim 2.** 若 $rs$ 是壞數字，則 $r^2s$ 也是壞數字。\n\n*Proof.* $rs$ 是壞數字表示甲可以進行 $rs \\to x$ 且 $x$ 為好數字，但 $x$ 顯然與 $r^2s$ 互質，故由 $r^2s \\to x$ 知 $r^2s$ 是壞數字。\n\n**Claim 3.** 若 $p > k$ 為一質數且 $n \\ge k$ 是壞數字，則 $np$ 也是壞數字。\n\n*Proof.* 若否，則存在最小的壞數字 $n$，使得 $np$ 是好數字。以下歸謬：\n\n1. 由於 $n$ 是壞數字，甲可以進行 $n \\to x$，其中 $x$ 是好數字。易知 $(np, x) > 1$，否則 $np$ 會是壞數字，矛盾。但已知 $(n, x) = 1$，故 $p \\mid x$。令 $x = p^r y$，其中 $(p, y) = 1$。\n2. 注意到 $y = 1$ 是不可能的，因為若 $y = 1$，則 $x = p^r$；又 $(p, k) = 1$，故甲可進行 $x \\to k$，從而 $x$ 是個壞數字，矛盾。故 $y > 1$，因此必有最小的正整數 $\\alpha$ 使得 $y^\\alpha \\ge k$。\n3. 基於 $np$ 和 $y^\\alpha$ 互質而 $np$ 是好數字，由 Claim B 知 $y^\\alpha$ 必為壞數字。\n4. 由 $\\alpha$ 的最小性知 $y^\\alpha < ky < py = \\frac{x}{p^{r-1}} < \\frac{n}{p^{r-1}}$，故 $p^{r-1}y^\\alpha < n$。從而由 $n$ 的最小性知，$p^{r-1}y^\\alpha$ 必為好數字（因為 $x = p(p^{r-1}y^\\alpha)$ 是好數字）。同理可證，$p^{r-2}y^\\alpha, \\dots, y^\\alpha$ 也都必須是好數字。\n5. 但 $np$ 和 $y^\\alpha$ 都是好數字，由 Claim B 知 $(np, y^\\alpha) > 1$，此與 $(n, x) = 1$ 及 $(p, y) = 1$ 相矛盾。證畢。\n\n現在令 $P_k(x)$ 為 $x$ 小於或等於 $k$ 的質因數所成集合。以下稱兩個數 $a, b$ 為相似的，若且唯若 $P_k(a) = P_k(b)$。要證明原題，我們僅需證明：若 $a, b$ 相似，則 $a, b$ 同好同壞。注意到 $ab$ 同時與 $a$ 和 $b$ 相似，故這等價於：若 $c \\ge k$ 與其某個倍數 $d$ 相似，則 $c, d$ 同好同壞。\n\n*Proof* 若否，則存在最小的 $d_0$，使得其存在一因數 $c_0 \\ge k$，使得 $c_0, d_0$ 好壞不同。由 Claim 1 知，必然是 $c_0$ 壞而 $d_0$ 好。注意到 $d_0 > c_0$。\n\n故 $d_0/c_0$ 必然有質因數 $p$。顯然 $p \\mid d_0$。\n\n- 若 $p \\le k$，則由相似性知 $p \\mid c_0$，故 $p^2 \\mid d_0$。如此一來，$d_0/p$ 必須是好的，否則由 Claim 2 知 $d_0$ 是壞的，矛盾。但這麼一來，$d_0/p$ 與 $c_0$ 相似，為 $c_0$ 的倍數且是好數字，這與 $d_0$ 的最小性不合，矛盾。\n- 若 $p > k$，則由 Claim 3 知 $d_0$ 是壞數字，矛盾。證畢。",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 13746,
    "subject": "Mathematics (Olympiad)",
    "question": "Given an odd positive integer $n$, let $m = \\frac{n+1}{2}$. Suppose $a_1, a_2, \\dots, a_m$ are positive integers pairwise incongruent modulo $n$, and $b_1, b_2, \\dots, b_m$ are positive integers pairwise incongruent modulo $n$. Prove that the number of distinct elements in the set\n\n$$C = \\{ \\text{the least non-negative residue of } a_i + b_j \\text{ modulo } n \\mid i, j \\in \\{1, 2, \\dots, m\\},\\ i \\neq j \\}$$\n\nis greater than $n - \\sqrt{n} - \\frac{1}{2}$.",
    "options": [],
    "answer": "See 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": 13747,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $ABC$ be an acute triangle with altitudes $AK$, $BL$, $CM$. Prove that triangle $ABC$ is isosceles if and only if\n\n$$\nAM + BK + CL = AL + BM + CK.\n$$\n\n![](path/to/file.png)",
    "options": [],
    "answer": "See 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 characterize the perpendicularity by lengths of segments.\n\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\n$$\nAM - BM = \\frac{b^2 - a^2}{AM + BM} = \\frac{b^2 - a^2}{c}\n$$\n\nand likewise\n\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$$\n\\begin{gathered}\n\\frac{a^2 - b^2}{c} + \\frac{b^2 - c^2}{a} + \\frac{c^2 - a^2}{b} = 0, \\\\\nab(a^2 - b^2) + bc(b^2 - c^2) + ca(c^2 - a^2) = 0.\n\\end{gathered}\n$$\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 an 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 the problem statement is equivalent to\n\n$$\n(a-b)(b-c)(c-a)(a+b+c) = 0.\n$$\n\nIt's obvious that this equality holds if and only if triangle $ABC$ is isosceles.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 13748,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $E$ denote the left-hand side of the given inequality. Consider $n-1$ positive real numbers $a_2, a_3, \\dots, a_n$ such that $a_2 a_3 \\dots a_n = 1$. Define\n\n$$\nE = \\prod_{i=2}^{n} (1 + a_i)^i.\n$$\n\nShow that $E > n^n$ for all $n \\ge 3$.",
    "options": [],
    "answer": "See solution",
    "solution": "*Proof.* The idea is to minimize $E$ by adjusting pairs $a_i, a_j$ while keeping their product constant. Let $c = a_i a_j$ and consider $f(a_i, a_j) = (1+a_i)^i (1+a_j)^j$. Setting $a_j = \\frac{c}{a_i}$, we get\n\n$$\nf(a_i) = (1 + a_i)^i \\left(1 + \\frac{c}{a_i}\\right)^j.\n$$\n\nDifferentiating,\n\n$$\n\\begin{aligned}\nf'(a_i) &= i(1+a_i)^{i-1} \\left(1+\\frac{c}{a_i}\\right)^j - \\frac{jc}{a_i^2} (1+a_i)^i \\left(1+\\frac{c}{a_i}\\right)^{j-1} \\\\\n&= \\frac{(1+a_i)^i (1+a_j)^j}{a_i} \\left( \\frac{ia_i}{1+a_i} - \\frac{ja_j}{1+a_j} \\right).\n\\end{aligned}\n$$\n\nThus, $f'(a_i) \\le 0$ if and only if $\\frac{ia_i}{1+a_i} \\le \\frac{ja_j}{1+a_j}$. If $\\frac{ia_i}{1+a_i} < \\frac{ja_j}{1+a_j}$, increasing $a_i$ and decreasing $a_j$ (preserving $a_i a_j$) decreases $E$.\n\nWe are led to consider the case where $\\frac{ia_i}{1+a_i} = \\alpha$ for all $i = 2, 3, \\dots, n$, i.e., $a_i = \\frac{\\alpha}{i-\\alpha}$ with $0 < \\alpha < 2$. The constraint $a_2 a_3 \\dots a_n = 1$ gives\n\n$$\n\\prod_{i=2}^{n} \\frac{\\alpha}{i-\\alpha} = 1.\n$$\n\nThere is a unique $\\alpha$ with $1 < \\alpha < 2$ satisfying this. The minimum $E$ is then\n\n$$\n\\begin{align*}\nE &= \\prod_{i=2}^{n} \\left(1 + \\frac{\\alpha}{i - \\alpha}\\right)^i \\\\\n&= \\prod_{i=2}^{n} \\frac{i^i}{(i - \\alpha)^i} \\\\\n&= \\prod_{i=2}^{n} \\frac{i^i}{\\alpha(i - \\alpha)^{i-1}} \\\\\n&= n^n \\cdot \\prod_{i=2}^{n} \\frac{(i - 1)^{i-1}}{\\alpha(i - \\alpha)^{i-1}}.\n\\end{align*}\n$$\n\nIt suffices to show $\\alpha(i-\\alpha)^{i-1} < (i-1)^{i-1}$ for $i = 2, \\dots, n$. Consider $G(x) = x(i-x)^{i-1}$; $G'(x) = i(1-x)(i-x)^{i-2}$, so $G$ is strictly decreasing for $1 < x < 2$. Since $1 < \\alpha < 2$, $G(\\alpha) < G(1) = (i-1)^{i-1}$, as required. Thus, $E > n^n$ for all $n \\ge 3$.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 13749,
    "subject": "Mathematics (Olympiad)",
    "question": "a) Prove that if the sum of the non-zero digits $a_1, a_2, \\dots, a_n$ is a multiple of $27$, then it is possible to permute these digits in order to obtain an $n$-digit number that is a multiple of $27$.\n\nb) Prove that if the non-zero digits $a_1, a_2, \\dots, a_n$ have the property that every $n$-digit number obtained by permuting these digits is a multiple of $27$, then the sum of these digits is a multiple of $27$.",
    "options": [],
    "answer": "See solution",
    "solution": "a) Obviously $n \\ge 3$. If $n = 3$, then $a_1 = a_2 = a_3 = 9$ and $999 \\div 27$. Suppose $n \\ge 4$. Having the sum of its digits a multiple of $9$, each of the numbers formed with the digits $a_1, a_2, \\dots, a_n$ is a multiple of $9$, therefore division by $27$ will yield one of the remainders $0$, $9$, or $18$.\n\n- If all the digits give the same remainder $r$ when divided by $3$:\n  - If $r = 0$, then we choose $N = \\overline{a_1a_2\\dots a_n}$. We have that $\\frac{N}{3} = \\frac{\\overline{a_1 a_2 \\dots a_n}}{3}$ has the sum of its digits a multiple of $9$, therefore the number $\\frac{N}{3}$ is a multiple of $9$, hence $N$ is a multiple of $27$.\n  - If $r \\in \\{1, 2\\}$, then from $a_1 + a_2 + \\dots + a_n = 27$ and $a_i \\equiv 0 \\pmod{3}$ for all $i$, it follows that $nr \\equiv 0 \\pmod{3}$, hence $n$ is a multiple of $3$. Then:\n    $$\n    \\overline{a_1a_2\\dots a_n} - (a_1 + a_2 + \\dots + a_n) = 9\\left(\\underbrace{\\overline{a_1a_1\\dots a_1}}_{n-1\\ \\text{digits}} + \\underbrace{\\overline{a_2a_2\\dots a_2}}_{n-2\\ \\text{digits}} + \\dots + \\overline{a_{n-1}a_{n-1}}\\right)\n    $$\n    Let $M$ be this sum. Then\n    $$\n    M \\equiv (n-1)a_1 + (n-2)a_2 + \\dots + a_{n-1} \\equiv r((n-1)+(n-2)+\\dots+2+1) \\equiv r \\cdot \\frac{n(n-1)}{2} \\equiv 0 \\pmod{3}.\n    $$\n    Therefore, in this case, $\\overline{a_1a_2\\dots a_n}$ and $a_1+a_2+\\dots+a_n$ give the same remainder upon division by $27$.\n\n- If not all of the digits $a_1, a_2, \\dots, a_n$ have the same remainder when divided by $3$, then we choose three of them such that their sum is not a multiple of $3$. Let $a, b, c$ be these three digits. Then the numbers $x = \\overline{\\dots abc}$, $y = \\overline{\\dots bca}$, and $z = \\overline{\\dots cab}$ give different remainders upon division by $27$. Indeed,\n    $$\n    x - y = \\overline{abc} - \\overline{bca} = 108a - 81b - 9(a+b+c) = M_{27} - 9(a+b+c) \\neq M_{27}.\n    $$\n    Since the possible remainders upon division by $27$ are only $0$, $9$, and $18$, it follows that (exactly) one of the numbers $x, y, z$ is a multiple of $27$.\n\nb) The difference between two numbers obtained one from the other by swapping two neighboring digits $a$ and $b$ is $9(a-b) \\cdot 10^k$ and needs to be a multiple of $27$. We obtain that each of the digits $a_1, \\dots, a_n$ has to give the same remainder when divided by $3$. From here, one continues as in part *a)* in order to show that, in this case, the numbers $\\overline{a_1a_2\\dots a_n}$ and $a_1+a_2+\\dots+a_n$ give the same remainder when divided by $27$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 13750,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $S(n)$ denote the sum of digits of a positive integer $n$. Does there exist a number $n$ such that $S(n) \\cdot S(n+1) = 2013$?",
    "options": [],
    "answer": "See solution",
    "solution": "*Answer:* no.\n\n*Solution.* If the last digit $a$ of a number $n$ is different from $9$, that is $n = \\overline{Aa}$ and $a \\neq 9$, then $n+1 = \\overline{A(a+1)}$, and so $S(n+1) = S(n)+1$.\n\nNow if $n = \\overline{Aa99\\ldots9}$, where the digit $a$ is less than $9$, then $n+1 = \\overline{A(a+1)00\\ldots0}$, which implies that\n\n$$\nS(n) - S(n+1) = (S(A) + a + 9k) - (S(A) + a + 1) = 9k - 1.\n$$\n\nSince $2013 = 3 \\cdot 11 \\cdot 61$, all the divisors of $2013$ are the following: $1$, $3$, $11$, $33$, $61$, $183$, $671$, $2013$. There are no consecutive numbers among them, and so the number $n$ should end with $9$. This means that we need to find a pair of divisors whose product is $2013$, and the difference is $9k-1$. But considering all the pairs with the product $2013$, we see that none of them have difference of the form $9k-1$:\n\n$$\n2013-1 = 2012 \\equiv 5 \\pmod{9}, \\quad 671-3 = 668 \\equiv 2 \\pmod{9}, \\quad 183-11 = 172 \\equiv 1 \\pmod{9},\n$$\n\n$$\n61-33 = 28 \\equiv 1 \\pmod{9}.\n$$\n\nThis proves that a positive integer $n$ cannot satisfy the equality $S(n) \\cdot S(n+1) = 2013$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 13751,
    "subject": "Mathematics (Olympiad)",
    "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 $\\Gamma_1$ 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 $AC \\times CB = 2r \\times MN$.\n\n![](images/Mathematical_Olympiad_in_China_2009-2010_p139_data_a45ba7f9cd.png)",
    "options": [],
    "answer": "See solution",
    "solution": "It is well known that $S$, $C$, and $M$ 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 point of tangency of $AB$ and $\\Gamma_1$) to $M$ (the point of tangency of $l$ 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\n$$\nSC \\times CM = 2r \\times MN \\quad \\text{or} \\quad \\frac{SC}{2r} = \\frac{MN}{CM}.\n$$\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 the desired result.\n\n(We can also derive this by observing that the triangles $MNC$ and $CDS$ are similar.)",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 13752,
    "subject": "Mathematics (Olympiad)",
    "question": "Find the integer of the form $n^2 + 4n$ that is closest to $10000$.",
    "options": [],
    "answer": "See solution",
    "solution": "Note that $n^2 + 4n = (n + 2)^2 - 4$ and that if $m < n$, then $m^2 + 4m < n^2 + 4n$.\nSince $100^2 = 10000$, we see that the desired solution is either $100^2 - 4 = 9996$ or $101^2 - 4 = 10197$, whichever has the smaller difference (in absolute value) from $10000$. So, the answer is $9996$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 13753,
    "subject": "Mathematics (Olympiad)",
    "question": "With $\\alpha \\in \\mathbb{R}$, consider the polynomial\n\n$$\nf(x) = x^2 - \\alpha x + 1.\n$$\n\na) For $\\alpha = \\frac{\\sqrt{15}}{2}$, express $f(x)$ as the quotient of two polynomials with non-negative coefficients.\n\nb) Find all values of $\\alpha$ such that $f(x)$ can be written as the quotient of two polynomials with non-negative coefficients.",
    "options": [],
    "answer": "See solution",
    "solution": "a) Consider the following transformations:\n\n$$\n\\left(x^2 - \\frac{\\sqrt{15}}{2}x + 1\\right) \\left(x^2 + \\frac{\\sqrt{15}}{2}x + 1\\right) = x^4 - \\frac{7}{4}x^2 + 1,\n$$\n\n$$\n\\left(x^{4} - \\frac{7}{4}x^{2} + 1\\right) \\left(x^{4} + \\frac{7}{4}x^{2} + 1\\right) = x^{8} - \\frac{17}{16}x^{4} + 1,\n$$\n\n$$\n\\left(x^{8} - \\frac{17}{16}x^{4} + 1\\right) \\left(x^{8} + \\frac{17}{16}x^{4} + 1\\right) = x^{16} + \\frac{223}{256}x^{8} + 1.\n$$\n\nIt follows that $f(x)$ is the quotient of $x^{16} + \\frac{223}{256}x^8 + 1$ and\n\n$$\n\\left(x^2 + \\frac{\\sqrt{15}}{2}x + 1\\right) \\left(x^4 + \\frac{7}{4}x^2 + 1\\right) \\left(x^8 + \\frac{17}{16}x^4 + 1\\right).\n$$\n\nb) Suppose $\\frac{P(x)}{Q(x)} = x^2 - \\alpha x + 1$ where $P, Q$ are polynomials with non-negative coefficients. Substituting $x = 1$, we have\n\n$$\n2 - \\alpha = \\frac{P(1)}{Q(1)} > 0 \\text{ so } \\alpha < 2.\n$$\n\nWe will prove that every real number $\\alpha < 2$ satisfies the problem. Indeed, if $\\alpha \\le 0$ then the polynomial $f(x)$ itself has non-negative coefficients, so we can choose $P(x) = f(x)$, $Q(x) = 1$.\n\nIf $\\alpha \\in (0, 2)$, consider the multiplication\n\n$$\n(x^2 - \\alpha x + 1)(x^2 + \\alpha x + 1) = x^4 + (2 - \\alpha^2)x^2 + 1.\n$$\n\nContinuing in this way, we find that the coefficients of the first and last terms of the polynomial are always 1, and the middle coefficient is determined by the sequence $(u_n)$ as follows:\n\n$$\n\\begin{cases} u_0 = \\alpha, \\\\ u_{n+1} = 2 - u_n^2, \\quad n \\ge 0. \\end{cases}\n$$\n\nWe will prove that there exists a positive term in this sequence. Suppose that for every $n \\ge 1$, $u_n < 0$. Then, since $\\alpha \\in (0, 2)$, by induction, we can show that $-2 < u_n < 0$ for all $n \\ge 1$. Note that\n\n$$\nu_{n+1} - u_n = 2 - u_n - u_n^2 = (2 + u_n)(1 - u_n) > 0,\n$$\n\nso $u_{n+1} - u_n > 0$ for all $n \\ge 1$; thus, this sequence increases. Since the sequence is bounded above by 0, it has a limit $L \\in (-2, 0]$. Letting $n$ tend to infinity, we have\n\n$$\nL = 2 - L^2 \\implies L \\in \\{1, -2\\}.\n$$\n\nThis contradiction shows that there exists $n = N$ such that $u_N \\ge 0$. Consider the polynomial sequence\n\n$$\nf_n(x) = x^{2n+1} + u_n x^{2n} + 1\n$$\n\nwith $n = 1, 2, 3, \\dots, N$. It is easy to see that $f(x)$ is the quotient of two polynomials $f_N(x)$ and $f_1(x)f_2(x) \\dots f_{N-1}(x)$. Clearly, these polynomials have non-negative coefficients. $\\square$",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 13754,
    "subject": "Mathematics (Olympiad)",
    "question": "The number inserted into the box located on the $i$-th row from the top and $j$-th column from the left is given by $13(i-1)+j$ for the first grid, and by $20(13-j)+i$ for the second grid. If the same number goes into the boxes located at the same position in the two grids, what are the possible values for $i$ and $j$?",
    "options": [],
    "answer": "See solution",
    "solution": "We require $13(i-1)+j = 20(13-j)+i$. Simplifying, we get:\n\n$$\n13(i-1) + j = 20(13-j) + i \\\\\n13i - 13 + j = 260 - 20j + i \\\\\n13i - 13 + j - i = 260 - 20j \\\\\n12i + 21j = 273\n$$\n\nSince both $21j$ and $273$ are multiples of $7$, $12i$ must also be a multiple of $7$, so $i$ must be a multiple of $7$. Given $1 \\leq i \\leq 20$, possible values are $i = 7, 14$.\n\nFor $i = 7$:\n$$\n12 \\times 7 + 21j = 273 \\\\\n84 + 21j = 273 \\\\\n21j = 189 \\\\\nj = 9\n$$\nFor $i = 14$:\n$$\n12 \\times 14 + 21j = 273 \\\\\n168 + 21j = 273 \\\\\n21j = 105 \\\\\nj = 5\n$$\n\nChecking $13(i-1)+j = 20(13-j)+i$ for these values:\n- For $(i, j) = (7, 9)$: $13 \\times 6 + 9 = 87$, $20 \\times 4 + 7 = 87$\n- For $(i, j) = (14, 5)$: $13 \\times 13 + 5 = 174$, $20 \\times 8 + 14 = 174$\n\nThus, the desired answers are $87$ and $174$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 13755,
    "subject": "Mathematics (Olympiad)",
    "question": "已知四邊形 $ABCD$ 中 $AC, BD$ 交於 $E$，$AB, CD$ 交於 $F$，$AD, BC$ 交於 $G$，且 $W, X, Y, Z$ 分別是 $E$ 對 $AB, BC, CD, DA$ 的對稱點。證明：$\\diamond(FWY)$、$\\diamond(GXZ)$ 的其中一個交點在 $FG$ 上。",
    "options": [],
    "answer": "See solution",
    "solution": "考慮對 $E$ 反演後的命題：給定四邊形 $ABCD$，$AC, BD$ 交於 $E$，$\\diamond(ABE), \\diamond(CDE)$ 交於另一點 $F$，$\\diamond(BCE), \\diamond(DAE)$ 交於另一點 $G$，$W, X, Y, Z$ 分別為 $\\diamond(ABE), \\diamond(BCE), \\diamond(CDE), \\diamond(DAE)$ 外心。證明：$\\diamond(WFY), \\diamond(XGZ)$ 的其中一個交點在 $\\diamond(EFG)$ 上。\n\n令 $M$ 為 $WY$ 中點，$S$ 為 $E$ 對 $M$ 對稱點，$H$ 為 $EF$ 與 $\\diamond(WFY)$ 的另一個交點。\n\n*Claim 1.* $E$ 為 $\\triangle WHY$ 垂心。\n\nProof. 注意到 $W, Y$ 都在 $EF$ 中垂線上。\n\n$$\n\\angle(EY, WY) = \\angle EYW = \\angle WYF = \\angle WHF = \\angle(WH, HF)\n$$\n\n由於 $WY \\perp HF$，故 $WH \\perp EY$，$E$ 為 $\\triangle WHY$ 垂心。\n\n*Claim 2.* $M$ 為 $\\triangle EFG$ 外心。\n\nProof. 由於 $WX \\perp BE$ 且 $ZY \\perp ED$，得 $WX \\parallel YZ$，同理可得 $XY \\parallel WZ$，$WXYZ$ 為平行四邊形，故 $M$ 也是 $XZ$ 中點。$WY$ 為 $EF$ 中垂線，$XZ$ 為 $EG$ 中垂線，$M$ 為 $WY$ 與 $XZ$ 交點，故 $M$ 為 $\\triangle EFG$ 外心。\n\n*Claim 3.* $S$ 為 $\\diamond(WFY), \\diamond(EFG)$ 的交點。\n\nProof. 由於 $M$ 為 $\\triangle EFG$ 外心，$S$ 為 $E$ 在 $\\triangle EFG$ 的對徑點。另外由 $E$ 為 $\\triangle WHY$ 且 $M$ 為 $WY$ 中點知 $S$ 為 $H$ 在 $\\triangle WHY$ 的對徑點。故 $S$ 在 $\\diamond(WFY), \\diamond(EFG)$ 上。同理可得 $E$ 對 $XZ$ 中點的對稱點會在 $\\diamond(XGZ)$ 上，而 $XZ$ 中點就是 $M$，故 $S$ 也會在 $\\diamond(XGZ)$ 上，得證。\n\n配分：\n\n1. 列出反演後的命題：1 分\n2. 證明 Claim 1、Claim 2 或反演前的等價性質：各 1 分\n3. 猜出 $\\diamond(FWX), \\diamond(GXZ)$ 的交點位置：1 分（或是猜出反演前會交在 $E$ 對 $FG$ 垂足）\n4. 做完剩餘部分：3 分",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 13756,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $n$ be a non-negative integer, and let $f: [-1, 1] \\to \\mathbb{R}$ be a twice differentiable function vanishing at the origin. Assuming $f''$ is continuous, show that\n\n$$\n(2n + 3) \\int_{-1}^{1} x^{2n} f(x) \\, dx = f''(c), \\quad \\text{for some } c \\text{ in the closed interval } [-1, 1].\n$$\n\n*Remarks.* The condition that $f''$ be continuous is superfluous, and the required equality holds for some $c$ in the open interval $(-1, 1)$. The only reason why $f''$ was assumed continuous and $c$ was required in the closed interval $[-1, 1]$ is that the argument is less involved.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $g: [-1, 1] \\to \\mathbb{R}$, $g(x) = \\frac{1}{2}(f(x) + f(-x))$, and let $h: [-1, 1] \\to \\mathbb{R}$, $h(x) = \\frac{1}{2}(f(x) - f(-x))$. Clearly, $g$ is even, $h$ is odd, $f = g + h$, and\n\n$$\n\\int_{-1}^{1} x^{2n} f(x) \\, dx = \\int_{-1}^{1} x^{2n} g(x) \\, dx + \\int_{-1}^{1} x^{2n} h(x) \\, dx = 2 \\int_{0}^{1} x^{2n} g(x) \\, dx.\n$$\n\nSince $f(0) = 0$ and $g'(x) = \\frac{1}{2}(f'(x) - f'(-x))$, it follows that $g(0) = g'(0) = 0$; and since $g''(x) = \\frac{1}{2}(f''(x) + f''(-x))$ and $f''$ is continuous, so is $g''$.\n\nLet $m = \\min_{0 \\le x \\le 1} g''(x)$ and $M = \\max_{0 \\le x \\le 1} g''(x)$. By Lagrange's theorem, for each $t$ in $(0, 1]$, there exists $c_t$ in $(0, t)$ such that $g'(t) = t g''(c_t)$, so $m t \\le g'(t) \\le M t$.\n\nIntegrate over $[0, x]$, $0 < x \\le 1$, to get $\\frac{1}{2} m x^2 \\le g(x) \\le \\frac{1}{2} M x^2$, so $\\frac{1}{2} m x^{2n+2} \\le x^{2n} g(x) \\le \\frac{1}{2} M x^{2n+2}$ for all $x$ in $[0, 1]$.\n\nIntegrating over $[0, 1]$ yields\n\n$$\n\\frac{m}{2(2n+3)} \\le \\int_{0}^{1} x^{2n} g(x) dx \\le \\frac{M}{2(2n+3)},\n$$\n\nso\n\n$$\nm \\le 2(2n+3) \\int_{0}^{1} x^{2n} g(x) dx \\le M, \\quad \\text{i.e.,} \\quad m \\le (2n+3) \\int_{-1}^{1} x^{2n} f(x) dx \\le M.\n$$\n\nSince $g''$ has the intermediate value property (Darboux),\n\n$$\ng''(b) = (2n + 3) \\int_{-1}^{1} x^{2n} f(x) \\, dx, \\quad \\text{for some } b \\text{ in } [0, 1].\n$$\n\nFinally, since $g''(b)$ lies between $f''(-b)$ and $f''(b)$, and $f''$ has the intermediate value property (Darboux),\n\n$$\nf''(c) = g''(b) = (2n + 3) \\int_{-1}^{1} x^{2n} f(x) \\, dx, \\quad \\text{for some } c \\text{ in } [-b, b] \\subseteq [-1, 1].\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 13757,
    "subject": "Mathematics (Olympiad)",
    "question": "Determine all functions $f$ defined for positive real numbers and taking positive real numbers as values such that\n\n$$\nx f(x f(2y)) = y + x y f(x)\n$$\n\nfor all positive real numbers $x$ and $y$.",
    "options": [],
    "answer": "See solution",
    "solution": "Substitute $x = 1$ into the functional equation to obtain\n\n$$\nf(f(2y)) = y (1 + f(1)).\n$$\n\nSince $1 + f(1) > 0$, the right side covers all positive real numbers as $y$ varies, so $f$ is surjective. Thus, there exists $a > 0$ such that $f(2a) = 1$.\n\nNow substitute $y = a$ into the functional equation:\n\n$$\nx f(x) = a (1 + x f(x)) \\implies f(x) = \\frac{c}{x},\n$$\n\nwhere $c = \\frac{a}{1 - a}$ and $a \\neq 1$.\n\nSubstituting $f(x) = \\frac{c}{x}$ into the original equation gives\n\n$$\n2y = y (1 + c),\n$$\n\nso $c = 1$ and $f(x) = \\frac{1}{x}$. It is easily verified that this function satisfies the original equation.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 13758,
    "subject": "Mathematics (Olympiad)",
    "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": "See 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 the circumcircle of $\\triangle ABC$ by $O$.\n\nSince the right triangles $CQL$ and $CPK$ are similar,\n\n![](images/Mathematical_Olympiad_in_China_2007-2008-Problems_and_Solutions_p203_data_9c240e2b66.png)\n\n$$\n\\angle CPK = \\angle CQL = \\angle OQP, \\text{ and } \\frac{QL}{PK} = \\frac{CQ}{CP}. \\quad ①\n$$\n\nLet $l$ be the perpendicular bisector of $CR$, then $O$ is on $l$.\n\nSince $\\triangle OPQ$ is an isosceles triangle, $P$ and $Q$ are two points symmetrical about $l$ on $CR$.\n\nSo\n$$\nRP = CQ \\quad \\text{and} \\quad RQ = CP. \\qquad \\textcircled{2}\n$$\n\nBy ①, ②,\n$$\n\\begin{aligned}\n\\frac{S(\\triangle RQL)}{S(\\triangle RPK)} &= \\frac{\\frac{1}{2} \\cdot RQ \\cdot QL \\cdot \\sin \\angle RQL}{\\frac{1}{2} \\cdot RP \\cdot PK \\cdot \\sin \\angle RPK} \\\\\n&= \\frac{RQ}{RP} \\cdot \\frac{QL}{PK} = \\frac{CP}{CQ} \\cdot \\frac{CQ}{CP} = 1.\n\\end{aligned}\n$$\n\nHence the two triangles have the same area.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 13759,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $p$ be a given prime number. We call a set $A$ of positive integers a $p$-set if the cardinality of $A_p$, the set of residues of elements of $A$ modulo $p$, is $p-1$. Determine the smallest value of $n$ such that, for any $p$-set $A$ of $n$ elements, there exists a $p$-element subset of $A$ with sum divisible by $p$.",
    "options": [],
    "answer": "See solution",
    "solution": "The smallest value is $p+2$ for $p$ odd and $2$ for $p=2$.\n\nThe $p=2$ case is easy, so we assume $p$ is odd.\n\nTake $A$ to be the set $\\{1, 2, \\dots, p-1, p+1, 2p-1\\}$. Then $A$ is a $p$-set and the sum of all elements of $A$ is divisible by $p$. Hence, the sum of any $p$ elements in $A$ is not divisible by $p$. Thus $n \\geq p+2$.\n\nNow let $A$ be a $p$-set with at least $p+2$ elements. We may assume $A = \\{a_1, \\dots, a_{p-1}, a, b, c\\}$, where $a_i \\not\\equiv a_j \\pmod{p}$ for $1 \\leq i \\neq j \\leq p-1$. Let $S := \\sum_{i=1}^{p-1} a_i$ and $T := a + b + c + S$.\n\nIf $T \\equiv a + a_i \\pmod{p}$ for some $i$, then the remaining $p$ elements have sum divisible by $p$ and we are done. Thus, we may assume that $T \\not\\equiv a + a_i \\pmod{p}$ for all $i$. Then we have\n\n$$\nT + \\sum_{i=1}^{p-1} (a + a_i) \\equiv 1 + 2 + \\dots + p \\equiv 0 \\pmod{p}.\n$$\n\nIt follows that $a \\equiv S + T \\pmod{p}$. Similarly, we may assume $b \\equiv c \\equiv S + T \\pmod{p}$. Then\n\n$$\n2 \\left( a + \\sum_{i=1}^{p-1} a_i \\right) \\equiv 2a + 2S \\equiv a + b + c + S - T \\equiv 0 \\pmod{p}.\n$$\n\nThis gives a subset with $p$ elements with sum divisible by $p$.\n\n**Note.** Another solution of the problem follows from the Cauchy-Davenport theorem.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 13760,
    "subject": "Mathematics (Olympiad)",
    "question": "The following operation is allowed on several given nonnegative integers. A positive number $a$ is chosen among them, and each number $b \\ge a$ is replaced by $b - a$, including the choice $a$ itself. Starting with $1, 2, \\ldots, 2013$, after several operations, numbers with sum $10$ are obtained. What can these numbers be? Find all possibilities.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $S_k = \\{1, 2, \\ldots, k\\}$ be a block for $k = 1, 2, \\ldots$ (with $S_0$ the empty block). After any operation, the set can be partitioned into blocks. If $S_k$ is a block and the operation is applied to $a$, then:\n- If $a > k$, $S_k$ remains unchanged.\n- If $a \\leq k$, $S_k$ is replaced by $S_{a-1}$ and $S_{k-a}$.\n\nInitially, $\\{1, 2, \\ldots, 2013\\}$ is a block. After any number of operations, the set is a disjoint union of blocks. The sum of $S_k$ is $\\frac{1}{2}k(k+1)$. If the total sum is $10$, the possible blocks are:\n- $S_1 = \\{1\\}$ (sum $1$)\n- $S_2 = \\{1, 2\\}$ (sum $3$)\n- $S_3 = \\{1, 2, 3\\}$ (sum $6$)\n- $S_4 = \\{1, 2, 3, 4\\}$ (sum $10$)\n\nThus, $10$ can be represented as:\n- $10 = 10$\n- $10 = 6 + 3 + 1$\n- $10 = 3 + 3 + 3 + 1$\n- $10 = 6 + 1 + 1 + 1 + 1$\n- $10 = 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1$\n\nSo, the possible sets are:\n- $\\{10\\}$\n- $\\{6, 3, 1\\}$\n- $\\{3, 3, 3, 1\\}$\n- $\\{6, 1, 1, 1, 1\\}$\n- $\\{1, 1, 1, 1, 1, 1, 1, 1, 1, 1\\}$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 13761,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $n \\ge 3$ be an integer. A circle dance is performed as follows: On the floor, $n$ points are marked at equal distances along a large circle. At each of these points is a sheet of paper with an arrow pointing either clockwise or counterclockwise. One of the points is labeled \"Start\". The dancer starts at this point. In each step, the dancer first changes the direction of the arrow at their current position and then moves to the next point in the new direction of the arrow.\n\n**a)** Show that each circle dance visits each point infinitely often.\n\n**b)** How many different circle dances are there? Two circle dances are considered the same if they differ only by a finite number of steps at the beginning and then always visit the same points in the same order. (The common sequence of steps may begin at different times in the two dances.)",
    "options": [],
    "answer": "See solution",
    "solution": "**a)** By the pigeonhole principle, there exists at least one point that is visited infinitely often. If there were another point visited only finitely many times, then there would be two neighboring points where one is visited infinitely often and the other only finitely often. But this is impossible: the dancer leaves the point visited infinitely often, alternately in both directions, so the neighboring points must also be visited infinitely often.\n\n**b)** *Claim:* If the dancer takes exactly $k < n$ consecutive steps in one direction right before a change of direction, then after the change, they take at least $k+1$ steps in the other direction.\n\n*Proof:* After $k$ steps in one direction and a change of direction, the dancer first takes one step in the other direction. Because of the previous $k$ steps, there are $k$ arrows ahead pointing toward the dancer, so they will certainly take $k$ more steps in the other direction after the first. $\\Box$\n\nTherefore, after at most $n$ changes of direction, the dancer will take $n$ consecutive steps in the same direction. With the $n$th step, the dancer returns to the starting point of the sequence, flips the arrow, and then has only $n-1$ arrows ahead pointing toward them, so they will again make $n$ consecutive steps in one direction.\n\nThus, every dance eventually has a \"turning point\": the dancer will dance a whole circle clockwise from the turning point to itself, then a whole circle counterclockwise from the turning point to itself, and so on.\n\nIt is possible to choose the initial arrow directions so that any point can become the turning point. For example, all arrows from the start point up to the desired turning point can point clockwise, and all others counterclockwise.\n\nTherefore, there are $n$ different dances.\n\n$\\Box$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 13762,
    "subject": "Mathematics (Olympiad)",
    "question": "Olga and Sasha play a game on an infinite hexagonal grid. They take alternating turns in placing a counter on a free hexagon of their choice, with Olga opening the game. Beginning from the 2018th move, a new rule will come into play: a counter may now be placed only on those free hexagons having at least two occupied neighbours.\n\nA player loses when they are either unable to make a turn, or have filled a pattern of the rhomboid shape below with counters (rotated in any possible way). Determine which player, if any, possesses a winning strategy.\n\n![](images/bw18shortlist_p24_data_a3c5066cf8.png)",
    "options": [],
    "answer": "See solution",
    "solution": "Olga has a winning strategy.\n\nThe game cannot go on forever. Draw a large hexagon enclosing all 2017 counters in play after the 2017th move, as in Figure 1. While it will be possible to place future counters in the hexagonal frame at distance 1 from the shaded part (i.e., immediately surrounding it), where $D$ and $E$ are located, it will be impossible to reach cells at distance 2 from the shaded part, where $F$ is located. Indeed, in order to place a counter at $F$, first counters must be placed on cells $D$ and $E$.\n\n![](images/bw18shortlist_p24_data_a6ea0db9ab.png)\n\nAssume that the cells $E_1, E_2, \\dots, E_n$ to the right of $E$ contain counters, but the next cell to the right is $E_{n+1}$ and it is empty. Observe that the counter on $E_{n-1}$ has been placed before the counter on $E_n$, because otherwise the forbidden rhombus is formed by the cells $E_{n-1}, E_n$ and two ancestors of $E_n$ in the previous row. By analogous reasoning considering the moment of placing the counter on $E_{n-1}$, one can prove that the counter on $E_{n-2}$ has been placed before the counter on $E_{n-1}$, etc. Thus, we conclude that the counter on $D$ has been placed before the counter on $E$. But changing the direction of our reasoning to the left, we similarly conclude that the counter on $E$ has been placed before the counter on $D$. A contradiction.\n\nNow, let Olga place her first counter in any hexagon $H$, and then respond to each of Sasha's successive moves by symmetry, choosing to place her counter on the reflection in $H$ of his chosen hexagon (in other words, diametrically opposite to his with respect to $H$). It is clear that the gameplay will be completely symmetrical after each of Olga's moves. Hence, she may respond, even under the additional rule, to any move Sasha might make. It is also evident that she will never complete a forbidden rhombus if Sasha did not already do so before. Hence, Olga is always certain to have a legal move at her disposal, and so will eventually win.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 13763,
    "subject": "Mathematics (Olympiad)",
    "question": "Given two arithmetic progressions $a_n$ and $b_n$, and an integer $m > 2$, consider $m$ quadratics:\n\n$$\nP_k(x) = x^2 + a_k x + b_k, \\quad k = 1, 2, \\dots, m.\n$$\n\nSuppose that neither $P_1(x)$ nor $P_m(x)$ has a real root. Prove that none of the given quadratics contains a real root.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $\\alpha$ and $\\beta$ be the common differences of the arithmetic progressions $a_n$ and $b_n$, respectively. Then\n\n$$\nP_k(x) = x^2 + (a_1 + (k-1)\\alpha)x + b_1 + (k-1)\\beta, \\quad k = 1, 2, \\dots, m.\n$$\n\nSince $P_1$ and $P_m$ have no real root, their discriminants satisfy\n\n$$\n\\Delta_1 = a_1^2 - 4b_1 < 0\n$$\n\nand\n\n$$\n\\Delta_m = (a_1 + (m-1)\\alpha)^2 - 4(b_1 + (m-1)\\beta) < 0.\n$$\n\nSuppose, for contradiction, that some $P_k$ has a real root for $1 < k < m$. Then\n\n$$\n\\Delta_k = (a_1 + (k-1)\\alpha)^2 - 4(b_1 + (k-1)\\beta) \\ge 0.\n$$\n\nComparing the discriminants, for $k, m > 1$:\n\n$$\n(k-1)\\Delta_m < 0 < (m-1)\\Delta_k.\n$$\n\nThis leads to\n\n$$\n(k-1)a_1^2 + (k-1)(m-1)^2\\alpha^2 - 4(k-1)b_1 < (m-1)a_1^2 + (m-1)(k-1)^2\\alpha^2 - 4(m-1)b_1.\n$$\n\nSimplifying, we get\n\n$$\n(k-1)(m-1)(m-k)^2\\alpha^2 < (m-k)a_1^2 - 4(m-k)b_1.\n$$\n\nFor $m > k$, this implies\n\n$$\n(k-1)(m-1)^2\\alpha^2 < a_1^2 - 4b_1 \\implies a_1^2 - 4b_1 \\ge 0,\n$$\n\ncontradicting $\\Delta_1 < 0$. Thus, none of the quadratics has a real root.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 13764,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $\\mathcal{F}$ be the set of all functions $f: \\mathbb{R} \\to \\mathbb{R}$ satisfying\n$$\nf(x + f(y)) = f(x) + f(y)\n$$\nfor all real numbers $x$ and $y$.\n\nDetermine all real numbers $r$ satisfying the following condition:\n\nFor every function $f$ in $\\mathcal{F}$, there exists some real number $z$ such that $f(z) = rz$.",
    "options": [],
    "answer": "See solution",
    "solution": "The required set of real numbers is $S = \\{1 + 1/n : n \\in \\mathbb{Z},\\ n \\neq 0\\}$.\n\nLet $R$ be the set of real numbers $r$ satisfying the condition in the statement. We first prove that $S \\subseteq R$.\n\nLet $f \\in \\mathcal{F}$ and let $P(x, y)$ denote the assertion $f(x + f(y)) = f(x) + f(y)$.\n\n**Step 1: $1 + 1/n$ belongs to $R$ for all positive integers $n$.**\n\nInduct on $n$ to show $f(nf(0)) = (n+1)f(0)$. Let $z = nf(0)$, then $f(z) = (1 + 1/n)z$, so $1 + 1/n \\in R$.\n\n- Base case ($n=1$): $P(0,0)$ gives $f(f(0)) = f(0) + f(0) = 2f(0)$.\n- Inductive step: Assume $f(nf(0)) = (n+1)f(0)$. Then\n  $$\n  \\begin{aligned}\n  f((n+1)f(0)) &= f(f(nf(0))) = f(0 + f(nf(0))) \\\\\n  &= f(0) + f(nf(0)) = f(0) + (n+1)f(0) = (n+2)f(0).\n  \\end{aligned}\n  $$\nThis completes the induction.\n\n**Step 2: $1 - 1/n$ belongs to $R$ for all positive integers $n$.**\n\n$P(-f(0), 0)$ gives $f(-f(0)) = 0$. Inductively, $f(-nf(0)) = (-n+1)f(0)$. Let $z = -nf(0)$, then $f(z) = (1 - 1/n)z$, so $1 - 1/n \\in R$.\n\nThus, $S \\subseteq R$.\n\n**Step 3: $R \\subseteq S$.**\n\nSuppose $r \\notin S$. We show there exists $f \\in \\mathcal{F}$ such that $f(z) = rz$ for no $z$.\n\n- For $r = 1$: $f(x) = x + 1$ is in $\\mathcal{F}$ and has no fixed points.\n- For $r \\neq 1$: Define $g: [0,1) \\to \\mathbb{R}$ by\n  $$\n  g(x) = \\begin{cases} 1 & \\text{if } \\dfrac{rx}{1-r} \\text{ is an integer} \\\\ 0 & \\text{otherwise} \\end{cases}\n  $$\n  and $f(x) = \\lfloor x \\rfloor + g(\\{x\\})$, where $\\{x\\} = x - \\lfloor x \\rfloor$.\n\nIt can be checked that $f \\in \\mathcal{F}$, and $f(z) = rz$ has no solution for $z$ if $r \\notin S$.\n\nTherefore, $R = S$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 13765,
    "subject": "Mathematics (Olympiad)",
    "question": "The diagonals of the parallelogram $ABCD$ meet at $O$. The bisectors of the angles $DAC$ and $DBC$ meet at $T$. It is known that $TD + TC = TO$. Find the measures of the angles of triangle $ABT$.",
    "options": [],
    "answer": "See solution",
    "solution": "The hypothesis shows that $DOCT$ is a parallelogram. From $AO \\parallel DT$ it follows that $\\angle DTA = \\angle OAT = \\angle DAT$, hence $DA = DT$.\n\nThis leads to $DA = DT = OC$; in the same way, $BC = CT = OD$, therefore $BD = AC$. So $ABCD$ is a rectangle, $AOTD$ is a rhombus, triangle $AOD$ is equilateral, and triangle $ABT$ is equilateral, hence its angles have measures $60^\\circ$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 13766,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $\\triangle ABC$ be a triangle. On the sides $BC$, $CA$, and $AB$ of the triangle, construct outwardly three squares with centres $O_a$, $O_b$, and $O_c$ respectively. Let $\\omega$ be the circumcircle of $\\triangle O_aO_bO_c$.\n\nGiven that $A$ lies on $\\omega$, prove that the centre of $\\omega$ lies on the perimeter of $\\triangle ABC$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let the vertices of the squares be $AC_1C_2B$, $BA_1A_2C$, and $CB_1B_2A$.\n\n**Lemma:** $BB_2 = CC_1$ and $BB_2 \\perp CC_1$.\n\n**Proof:** Notice that by rotating $\\triangle AC_1C$ by $90^\\circ$ we get $\\triangle ABB_2$, proving the lemma.\n\n**Claim:** $AO_a \\perp O_bO_c$\n\n**Proof:** Let $M$ be the midpoint of $AB$. By our lemma applied at vertex $C$ we get $AA_2 = BB_1$ and they are perpendicular. By homothety of factor $2$ at $A$ and then $B$ we get:\n\n$$\nMO_b = \\frac{1}{2}BB_1 = \\frac{1}{2}AA_2 = MO_a \\quad \\text{and} \\quad MO_b \\parallel BB_1, \\ MO_a \\parallel AA_2\n$$\n\nHence $MO_a$ and $MO_b$ are also perpendicular, so in fact $\\triangle O_bMO_a$ is an isosceles right triangle. This is also trivially the case for $\\triangle AMO_c$. Now applying our lemma to $\\triangle AMO_b$ at vertex $M$ we get $O_bO_c$ and $AO_a$ are perpendicular, which is exactly what we wanted.\n\nSimilarly, we get $BO_b \\perp O_aO_c$ and $CO_c \\perp O_aO_b$, so lines $AO_a$, $BO_b$, $CO_c$ concur at $H$, the orthocentre of $\\triangle O_aO_bO_c$. As $A$ lies on $\\omega$ and on $O_aH$, it follows $A$ is the reflection of $H$ in line $O_bO_c$.\n\n![](images/2020_BMO_Short_List_p14_data_f1e4e6bc6c.png)\n\n**Claim:** $H = B$ or $H = C$\n\n**Proof:** Assume not. By the previous observations we get $O_cH = O_cA = O_cB$. Hence, as $O_cO_a \\perp BH$ and $B \\neq H$, this means $B$ is the reflection of $H$ in $O_cO_a$, so $B$ lies on $\\omega$.\n\nSimilarly, $C$ lies on $\\omega$. But then we get:\n\n$$\n\\angle ACB = 180^\\circ - \\angle BO_cA = 90^\\circ \\quad \\text{and} \\quad \\angle CBA = 180^\\circ - \\angle AO_bC = 90^\\circ\n$$\n\nso $\\angle ACB + \\angle CBA = 180^\\circ$, which is absurd, so in fact one of $B, C$ is equal to $H$.\n\nWLOG $B = H$. As $A, H, O_a$ and $C, H, O_c$ are collinear, this means in fact $B$ lies on these lines. Hence:\n\n$$\n\\angle AO_cC = \\angle AO_cB = 90^\\circ\n$$\n\nAlso $\\angle AO_bC = 90^\\circ$, hence $C$ also lies on $\\omega$ and $\\omega$ in fact has diameter $AC$, so its circumcentre is the midpoint of $AC$, which lies on the perimeter of $\\triangle ABC$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 13767,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $C$, $E$, and $A$ be points on the sides $XY$, $YZ$, and $ZX$ of triangle $XYZ$, respectively. Points $B$, $D$, and $F$ are chosen on the segments $AX$, $CY$, and $EZ$, respectively, so that $BC \\perp AD$, $DE \\perp CF$, and $AF \\perp BE$. Can it happen that the lines $XF$, $YB$, and $ZD$ are concurrent?",
    "options": [],
    "answer": "See solution",
    "solution": "No, it cannot.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 13768,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $A = \\sqrt{x+n} + \\sqrt{y+n} + \\sqrt{z+n}$.$$\n\nFind all real numbers $x, y, z, n$ such that $A$ is an integer.",
    "options": [],
    "answer": "See solution",
    "solution": "If $x = y = z$, their common value is $\\frac{1}{9}$. Then $A = 3\\sqrt{n + \\frac{1}{9}} = \\sqrt{9n + 1}$, so $A^2 = 9n + 1$ must be a perfect square. Thus, $A$ gives remainder 1 or 8 on division by 9, so $A = 9p \\pm 1$, $p \\ge 1$, and $A^2 = (9p \\pm 1)^2 = 9n + 1$, yielding $n = 9p^2 \\pm 2p$, $p \\ge 1$.\n\nIf $x, y, z$ are not all equal, we have\n$$\nA^2 = 3n + x + y + z + 2 \\sum \\sqrt{(x+n)(y+n)}.\n$$\nBy AM-GM, $\\sqrt{(x+n)(y+n)} \\le n + \\frac{x+y}{2}$, with equality iff $x = y$. Since $x, y, z$ are not all equal, $\\sum \\sqrt{(x+n)(y+n)} < 3n + x + y + z$, so $A^2 < 9n + 3(x + y + z) \\le 9n + 3$.\n\nOn the other hand, $\\sqrt{(x+n)(y+n)} \\ge n + \\sqrt{xy}$, with equality iff $x = y$. Thus, $\\sum \\sqrt{(x+n)(y+n)} > 3n + \\sum \\sqrt{xy}$, so\n$$\nA^2 > 9n + (x + y + z) + 2 \\sum \\sqrt{xy} = 9n + (\\sqrt{x} + \\sqrt{y} + \\sqrt{z})^2 = 9n + 1.\n$$\nHence $9n + 1 < A^2 < 9n + 3$, so $A^2 = 9n + 2$. But this is impossible since perfect squares are congruent to 0, 1, 4, or 7 mod 9, never 2. Thus, the only solutions are $x = y = z = \\frac{1}{9}$, $n = 9p^2 \\pm 2p$, $p \\ge 1$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 13769,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $ABCD$ be a cyclic quadrilateral and let $E$ be the point of intersection of its diagonals $AC$ and $BD$. Suppose $AD$ and $BC$ meet at $F$. Let the midpoints of $AB$ and $CD$ be $G$ and $H$ respectively. If $\\Gamma$ is the circumcircle of triangle $EGH$, prove that $FE$ is tangent to $\\Gamma$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $AB$ and $DC$ meet at $M$; let $FE$ meet $HG$ at $T$, $AB$ at $P$, and $DC$ at $Q$ respectively. We use the following fact: in a triangle $ABC$ with circumcircle $\\gamma$, given a point $D$ outside $\\gamma$, $AD$ is tangent to $\\gamma$ if and only if\n\n![](images/Indija_TS_2010_p1_data_f639014e1b.png)\n\n$$\n\\frac{DB}{DC} = \\frac{AB^2}{AC^2}.\n$$\n\nThus, it suffices to prove that\n\n$$\n\\frac{TG}{TH} = \\frac{GE^2}{EH^2}.\n$$\n\nLet $AB = a$, $DC = b$, $GM = \\lambda$, $HM = \\mu$. Observe that $DB$, $CA$, and $FQ$ are concurrent cevians in triangle $FDC$. Ceva's theorem gives\n\n$$\n\\frac{FA}{AD} \\cdot \\frac{DQ}{QC} \\cdot \\frac{CB}{BF} = 1.\n$$\n\nLooking at the transversal $A-B-M$ in triangle $FDC$, we can apply Menelaus' theorem to get\n\n$$\n\\frac{FA}{AD} \\cdot \\frac{DM}{MC} \\cdot \\frac{CB}{BF} = 1.\n$$\n\nIt follows that $\\frac{DQ}{QC} = \\frac{DM}{MC}$.\n\nSimilarly, we get $\\frac{AP}{PB} = \\frac{AM}{MB}$. Observe\n\n$$\n\\frac{DH+HQ}{CH-HQ} = \\frac{DQ}{QC} = \\frac{DM}{MC} = \\frac{DH+HM}{HM-HC}.\n$$\n\nSimplification gives (using $DH = CH$) $HC^2 = HQ \\cdot HM$. In a similar way, we get $GP \\cdot GM = GB^2$. Thus,\n\n$$\n\\begin{aligned}\nGP &= \\frac{GB^2}{GM} = \\frac{a^2}{4\\lambda}, \\\\\nPM &= \\lambda - \\frac{a^2}{4\\lambda} = \\frac{4\\lambda^2 - a^2}{4\\lambda}.\n\\end{aligned}\n$$\n\nThus $\\frac{PM}{GP} = \\frac{4\\lambda^2 - a^2}{a^2}$. Similarly, $\\frac{QM}{HQ} = \\frac{4\\mu^2 - b^2}{b^2}$. Using the transversal $T-P-Q$ in triangle $MGH$, we have $\\frac{MP}{PG} \\cdot \\frac{GT}{TH} \\cdot \\frac{HQ}{QM} = 1$. Thus\n\n$$\n\\frac{GT}{TH} = \\frac{PG}{MP} \\cdot \\frac{QM}{HQ} = \\frac{a^2}{4\\lambda^2 - a^2} \\cdot \\frac{4\\mu^2 - b^2}{b^2}.\n$$\n\nSuppose we show that $4\\lambda^2 - a^2 = 4\\mu^2 - b^2$. We get $GT/TH = a^2/b^2$. We observe that triangles $EBA$ and $ECD$ are similar; and they have respective medians $EG$ and $EH$. Hence $EG^2/EH^2 = AB^2/CD^2 = a^2/b^2$. It follows that $GT/TH = GE^2/EH^2$, giving what we required. We use coordinate geometry to prove $4\\lambda^2 - a^2 = 4\\mu^2 - b^2$.\n\nLet us fix $M = (0,0)$. The circumcircle of $ABCD$ has equation $x^2 + y^2 + 2gx + 2fy + c = 0$. Let the equation of $MA$ be $y = mx$. Let $A = (x_1, y_1)$, $B = (x_2, y_2)$. Then $x_1, x_2$ are the solutions of $(1+m^2)x^2 + (2g+2fm)x + c = 0$, so that\n\n$$\nx_1 + x_2 = -\\frac{2g + 2fm}{1 + m^2}, \\quad x_1x_2 = \\frac{c}{1 + m^2}.\n$$\n\nThe coordinates of $G$ are $\\left(\\frac{x_1+x_2}{2}, m\\left(\\frac{x_1+x_2}{2}\\right)\\right)$. Thus\n\n$$\n\\lambda^2 = GM^2 = \\left(\\frac{x_1+x_2}{2}\\right)^2 + m^2 \\left(\\frac{x_1+x_2}{2}\\right)^2 = \\frac{(g+fm)^2}{1+m^2}.\n$$\n\nWe also have\n\n$$\na^2 = AB^2 = (1+m^2)(x_1-x_2)^2 = (1+m^2)((x_1+x_2)^2 - 4x_1x_2) = 4\\frac{(g+fm)^2}{1+m^2} - 4c.\n$$\n\nThus\n\n$$\n4\\lambda^2 - a^2 = 4c\n$$\n\nSimilarly, we get $4\\mu^2 - b^2 = 4c$. It follows that $4\\lambda^2 - a^2 = 4\\mu^2 - b^2$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 13770,
    "subject": "Mathematics (Olympiad)",
    "question": "令 $f$ 和 $g$ 是兩個係數為整數的非零多項式，且 $\\deg f > \\deg g$。假設有無窮多的質數 $p$ 使得多項式 $pf + g$ 有一個有理根，證明 $f$ 也有一個有理根。",
    "options": [],
    "answer": "See solution",
    "solution": "因為 $\\deg f > \\deg g$，所以當 $x$ 足夠大時，有 $|g(x)/f(x)| < 1$。即存在一個實數 $R$，使得對所有 $|x| > R$，有 $|g(x)/f(x)| < 1$。對於所有這樣的 $x$ 及所有質數 $p$，有\n\n$$\n|pf(x) + g(x)| \\geq |f(x)| \\left( p - \\frac{|g(x)|}{|f(x)|} \\right) > 0.\n$$\n\n因此，多項式 $pf + g$ 的所有實數根都會落在 $[-R, R]$。\n\n令 $f(x) = a_n x^n + a_{n-1} x^{n-1} + \\cdots + a_0$，$g(x) = b_m x^m + b_{m-1} x^{m-1} + \\cdots + b_0$，其中 $n > m$，$a_n \\neq 0$，$b_m \\neq 0$。將 $f(x)$ 及 $g(x)$ 分別置換為 $a_n^{-1} f(x/a_n)$ 和 $a_m^{-1} g(x/a_m)$，可簡化為 $a_n = 1$ 的情形，此時 $pf + g$ 的首項係數為 $p$。若 $r = u/v$，$(u, v) = 1$ 且 $v > 0$，是 $pf + g$ 的有理根，則 $v$ 為 $1$ 或 $p$。\n\n假設 $v = 1$ 的情形有無窮多個。因 $|u| \\leq R$，只存在有限多個整數 $u$，因此存在相異質數 $p, q$ 使得有相同的 $u$ 值。則 $pf + g$ 和 $qf + g$ 有共同根，故 $f(u) = g(u) = 0$，即 $f$ 和 $g$ 有相同的整數根。\n\n假設 $v = p$ 的情形有無窮多個。比較 $pf(u/p)$ 和 $g(u/p)$ 分母中 $p$ 的次方，取 $m = n - 1$，則 $pf(u/p) + g(u/p) = 0$ 可化簡為\n\n$$\n(u^n + a_{n-1} p u^{n-1} + \\cdots + a_0 p^n) + (b_{n-1} u^{n-1} + b_{n-2} p u^{n-2} + \\cdots + b_0 p^{n-1}) = 0.\n$$\n\n此式得 $u^n + b_{n-1} u^{n-1}$ 可被 $p$ 整除。因 $(u, p) = 1$，所以 $u + b_{n-1} = pk$，其中 $k$ 為整數。又 $pf + g$ 的所有根都落在 $[-R, R]$，所以\n\n$$\n\\frac{|pk - b_{n-1}|}{p} = \\frac{|u|}{p} < R, \\quad |k| < R + \\frac{|b_{n-1}|}{p} < R + |b_{n-1}|.\n$$\n\n因此整數 $k$ 只有有限多個值。故存在整數 $k$，使得對無限多質數 $p$，$\\frac{pk-b_{n-1}}{p} = k - \\frac{b_{n-1}}{p}$ 是 $pf + g$ 的一個根。對這些質數，有\n\n$$\nf\\left(k - \\frac{b_{n-1}}{p}\\right) + \\frac{1}{p}g\\left(k - \\frac{b_{n-1}}{p}\\right) = 0.\n$$\n\n所以\n\n$$\nf(k - b_{n-1}x) + xg(k - b_{n-1}x) = 0 \\quad (1)\n$$\n\n有無限多個 $x = 1/p$ 的解。因為 (1) 左邊是多項式，(1) 對所有實數 $x$ 都成立。令 $x = 0$，得 $f(k) = 0$，即整數 $k$ 是 $f$ 的一個根。",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 13771,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $F$ be a finite nonempty set of integers and let $n$ be a positive integer. Suppose that\n\n- Any $x \\in F$ may be written as $x = y + z$ for some $y, z \\in F$.\n- If $1 \\leq k \\leq n$ and $x_1, \\dots, x_k \\in F$, then $x_1 + \\dots + x_k \\neq 0$.\n\nShow that $F$ has at least $2n + 2$ distinct elements.",
    "options": [],
    "answer": "See solution",
    "solution": "Because of the second condition above, $0 \\notin F$. If $F$ contains only positive elements, let $x$ be the smallest element in $F$. But then $x = y + z$, and $y, z > 0$ imply that $y, z < x$, a contradiction. Hence $F$ contains negative elements. A similar argument shows that $F$ contains positive elements.\n\nPick any positive element of $F$ and label it as $x_1$. Assume that positive elements of $F$, $x_1, \\dots, x_k$, have been chosen. We can write $x_k = y + z$, where $y, z \\in F$. We may assume that $y > 0$. Label $y$ as $x_{k+1}$. Carry on in this manner to choose positive elements $x_1, x_2, \\dots$ of $F$, not necessarily distinct. Since $F$ is a finite set, there exist positive integers $i < j$ such that $x_i, \\dots, x_{j-1}$ are distinct and $x_j = x_i$. There are $x_i, \\dots, x_{j-1} \\in F$ such that\n\n$$\n\\begin{aligned}\nx_i &= x_{i+1} + z_i \\\\\nx_{i+1} &= x_{i+2} + z_{i+1} \\\\\n\\vdots &\\vdots \\\\\nx_{j-1} &= x_j + z_{j-1}.\n\\end{aligned}\n$$\n\nSince $x_i = x_j$, we see that $x_i + x_{i+1} + \\dots + x_{j-1} = 0$. By the assumption, $j-i > n$. Since the elements $x_i, \\dots, x_{j-1}$ are distinct, $F$ contains at least $j-i \\geq n+1$ positive elements. Similarly, $F$ contains at least $n+1$ negative elements. The result follows.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 13772,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $ABC$ be an acute triangle. Let $H$ denote its orthocenter and $D$, $E$, and $F$ the feet of its altitudes from $A$, $B$, and $C$, respectively. Let the common point of $DF$ and the altitude through $B$ be $P$. The line perpendicular to $BC$ through $P$ intersects $AB$ in $Q$. Furthermore, $EQ$ intersects the altitude through $A$ in $N$.\n\nProve that $N$ is the midpoint of $AH$.\n\n![](images/bwf2017englishSolutions_p2_data_35901392fe.png)",
    "options": [],
    "answer": "See solution",
    "solution": "As usual, let $\\beta = \\angle ABC$ and $\\gamma = \\angle ACB$. Since $\\angle AFH = \\angle AEH = 90^\\circ$, the quadrilateral $AFHE$ is cyclic, and because $DA$ is parallel to $PQ$, we obtain\n\n$$\n\\angle FQP = \\angle FAH = \\angle FEH = \\angle FEP.\n$$\n\nIt follows that $QFPE$ is also cyclic. Since $\\angle AFC = \\angle ADC = 90^\\circ$, $AFDC$ is also cyclic, and we have $\\angle QFP = \\angle AFD = 180^\\circ - \\angle ACD = 180^\\circ - \\gamma$. We therefore have $\\angle QEP = \\gamma$. From this, we obtain $\\angle EAN = 90^\\circ - \\gamma = \\angle AEP - \\angle QEP = \\angle AEN$, which shows that triangle $ANE$ is isosceles. It therefore follows that $N$ is the circumcenter of the right triangle $AHE$, and so $NA = NH$, as claimed.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 13773,
    "subject": "Mathematics (Olympiad)",
    "question": "Let there be $r$ red sweets and $g$ green sweets, with $r \\ge 2$.\n\nIf Dan puts the first sweet back, the probability that both selected sweets are red is\n\n$$\n\\frac{r}{r+g} \\times \\frac{r}{r+g}.\n$$\n\nIf Dan eats the first sweet, the probability that both selected sweets are red is\n\n$$\n\\frac{r}{r+g} \\times \\frac{r-1}{r+g-1}.\n$$\n\nThe first probability is 105\\% of the second. What is the largest possible number of sweets in the jar?",
    "options": [],
    "answer": "See solution",
    "solution": "Let $r$ be the number of red sweets and $g$ the number of green sweets, with $n = r + g$ and $r \\ge 2$.\n\nThe probability of selecting two red sweets if the first sweet is put back is\n\n$$\n\\frac{r}{n} \\times \\frac{r}{n}\n$$\n\nand if Dan eats the first sweet before selecting the second:\n\n$$\n\\frac{r}{n} \\times \\frac{r-1}{n-1}.\n$$\n\nSince the first probability is 105\\% of the second:\n\n$$\n\\frac{r}{n} \\times \\frac{n-1}{r-1} = \\frac{21}{20}\n$$\n\nwhich leads to\n\n$$\n20r(n-1) = 21n(r-1)\n$$\n\nor\n\n$$\n(n+20)(21-r) = 420.\n$$\n\nSince $n+20 > 0$ and $21 - r > 0$, $n$ is maximized when $21 - r = 1$, so $n + 20 = 420$ and $n = 400$.\n\nThus, the largest number of sweets in the jar is **400**.\n\nComment\n\nSince $21 - r$ is a factor of 420 and $2 \\le r \\le 20$, the following table gives all possible values of $r, n, g$.\n\n![](images/Australian-Scene-2017_p71_data_f1e2b72edd.png)\n\n| $21 - r$ | $n + 20$ | $r$ | $n$ | $g$ |\n|----------|----------|-----|-----|-----|\n| 1        | 420      | 20  | 400 | 380 |\n| 2        | 210      | 19  | 190 | 171 |\n| 3        | 140      | 18  | 120 | 102 |\n| 4        | 105      | 17  | 85  | 68  |\n| 5        | 84       | 16  | 64  | 48  |\n| 6        | 70       | 15  | 50  | 35  |\n| 7        | 60       | 14  | 40  | 26  |\n| 10       | 42       | 11  | 22  | 11  |\n| 12       | 35       | 9   | 15  | 6   |\n| 14       | 30       | 7   | 10  | 3   |\n| 15       | 28       | 6   | 8   | 2   |",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 13774,
    "subject": "Mathematics (Olympiad)",
    "question": "It is known that $x_1, x_2$ are roots of the polynomial $p(x) = x^2 + a x + b = 0$, and $x_1^2 - \\frac{1}{2}$ and $x_2^2 - \\frac{1}{2}$ are roots of the polynomial $p(x) = x^2 + (a^2 - \\frac{1}{2}) x + (b^2 - \\frac{1}{2}) = 0$. Find all possible $a$ and $b$.",
    "options": [],
    "answer": "See solution",
    "solution": "Write down Vieta's formulas for both equations:\n\n$$\nx_1 + x_2 = -a, \\quad x_1 x_2 = b,\n$$\n$$\n(x_1^2 - \\frac{1}{2}) + (x_2^2 - \\frac{1}{2}) = \\frac{1}{2} - a^2,\n$$\n$$\n(x_1^2 - \\frac{1}{2}) (x_2^2 - \\frac{1}{2}) = b^2 - \\frac{1}{2}.\n$$\n\nTransform the last equation, taking into account the first three equations:\n\n$$\nx_1^2 x_2^2 - \\frac{1}{2} (x_1^2 + x_2^2) + \\frac{1}{4} = b^2 - \\frac{1}{2}\n$$\n$$\n\\Leftrightarrow b^2 - \\frac{1}{2} (\\frac{3}{2} - a^2) = b^2 - \\frac{3}{4}\n$$\n$$\n\\Leftrightarrow a = 0.\n$$\n\nThen use the third equation and the fact that $a = 0$:\n\n$$\nx_1^2 + x_2^2 = \\frac{3}{2} - a^2\n$$\n$$\n(x_1 + x_2)^2 - 2 x_1 x_2 = \\frac{3}{2} - a^2\n$$\n$$\n-2b = \\frac{3}{2}\n$$\n$$\nb = -\\frac{3}{4}.\n$$",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 13775,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $f : [1, +\\infty) \\to (0, +\\infty)$ be a continuous function with the following properties:\n\n1. The function $g : [1, +\\infty) \\to (0, +\\infty)$ defined by $g(x) = \\frac{f(x)}{x}$ has a limit as $x \\to +\\infty$.\n2. The function $h : [1, +\\infty) \\to (0, +\\infty)$ defined by $h(x) = \\frac{1}{x} \\int_{1}^{x} f(t)\\,dt$ has a finite limit as $x \\to +\\infty$.\n\n(a) Show that $\\lim_{x \\to +\\infty} g(x) = 0$.\n\n(b) Show that $\\lim_{x \\to +\\infty} \\frac{1}{x^2} \\int_1^x f^2(t)\\,dt = 0$.",
    "options": [],
    "answer": "See solution",
    "solution": "a) Let $\\ell = \\lim_{x \\to +\\infty} g(x)$. Suppose $\\ell \\in (0, +\\infty)$. Then for some $a > 0$, $g(x) > \\ell/2$ for $x \\ge a$. Thus,\n\n$$\n\\begin{aligned}\nh(x) &= \\frac{1}{x} \\left( \\int_{1}^{a} f(t)\\,dt + \\int_{a}^{x} f(t)\\,dt \\right) \\\\\n&\\ge \\frac{1}{x} \\int_{1}^{a} f(t)\\,dt + \\frac{\\ell}{2x} \\int_{a}^{x} f(t)\\,dt \\\\\n&= \\frac{1}{x} \\int_{1}^{a} f(t)\\,dt + \\frac{\\ell(x^2 - a^2)}{4x} \\to +\\infty \\text{ as } x \\to +\\infty,\n\\end{aligned}\n$$\n\ncontradicting (ii). Similarly, $\\ell = +\\infty$ also contradicts (ii), so $\\ell = 0$.\n\nb) Note that $\\int_{1}^{x} f(t)\\,dt > 0$ for $x > 1$, so\n\n$$\n\\begin{aligned}\n\\lim_{x \\to +\\infty} \\frac{1}{x^2} \\int_1^x f^2(t)\\,dt &= \\lim_{x \\to +\\infty} \\left( \\frac{\\int_1^x f^2(t)\\,dt}{x \\int_1^x f(t)\\,dt} \\cdot \\frac{\\int_1^x f(t)\\,dt}{x} \\right) \\\\\n&= \\lambda \\lim_{x \\to +\\infty} \\frac{\\int_1^x f^2(t)\\,dt}{x \\int_1^x f(t)\\,dt} = \\lambda \\lim_{x \\to +\\infty} \\frac{u(x)}{v(x)},\n\\end{aligned}\n$$\n\nwhere $\\lambda = \\lim_{x \\to +\\infty} h(x)$, $u(x) = \\int_{1}^{x} f^2(t)\\,dt$, $v(x) = x \\int_{1}^{x} f(t)\\,dt$.\n\nTo show $\\lim_{x \\to +\\infty} \\frac{u(x)}{v(x)} = 0$, use l'Hospital's Rule:\n\n- $u$ and $v$ are differentiable,\n- $\\lim_{x \\to +\\infty} v(x) = +\\infty$ (since $v(x) \\ge m(x - 1)$ for $x \\ge 2$, with $m = \\inf_{x \\in [1,2]} f(x)$),\n- $v'(x) = \\int_{1}^{x} f(t)\\,dt + x f(x) \\ne 0$ for all $x \\ge 1$,\n\nand\n\n$$\n\\frac{u'(x)}{v'(x)} = \\frac{f^2(x)}{x f(x) + \\int_1^x f(t)\\,dt} = g(x) \\cdot \\frac{f(x)}{f(x) + h(x)} \\in (0, g(x)),\n$$\n\nwith $\\lim_{x \\to +\\infty} g(x) = 0$, so $\\lim_{x \\to +\\infty} \\frac{u'(x)}{v'(x)} = 0$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 13776,
    "subject": "Mathematics (Olympiad)",
    "question": "Знайдіть усі значення $x$, для яких $f(f(f(x))) = 5$, якщо $f(x) = (x - 4)^2 + 4$.",
    "options": [],
    "answer": "See solution",
    "solution": "Неважко помітити, що $f(x) = (x - 4)^2 + 4$. Тоді рівняння $f(f(f(x))) = 5$ рівносильне рівнянню $$(x - 4)^8 + 4 = 5,$$ звідки $x - 4 = \\pm 1$, тобто $x = 3$ або $x = 5$.\n\nВідповідь: $x = 3, x = 5$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 13777,
    "subject": "Mathematics (Olympiad)",
    "question": "As shown in the figure, $AB$ and $CD$ are two chords in the circle $\\odot O$ meeting at point $E$, and $AB \\neq CD$. $\\odot I$ is tangent to $\\odot O$ internally at point $F$, and is tangent to the chords $AB$ and $CD$ at points $G$ and $H$ respectively. $l$ is a line passing through $O$, meeting $AB$ and $CD$ at points $P$ and $Q$ respectively, such that $EP = EQ$. Line $EF$ meets the line $l$ at point $M$. Prove that the line through $M$ and parallel to the line $AB$ is tangent to the circle $\\odot O$.\n\n![](images/Kina_2012_p16_data_a962cc3c28.png)\n\n![](images/Kina_2012_p16_data_34172777d2.png)",
    "options": [],
    "answer": "See solution",
    "solution": "As shown in the figure above, draw a line parallel to $AB$ and tangent to circle $\\odot O$ at point $L$, which meets the common tangent line to these two circles at a point $S$. Let $R$ be the intersection of lines $FS$ and $BA$, and join segments $LF$ and $GF$.\n\n![](images/Kina_2012_p17_data_b573fe4323.png)\n\nFirst, we prove that the points $L$, $G$, and $F$ are collinear. As both $SL$ and $SF$ are tangent to $\\odot O$, $SL = SF$; as both $RG$ and $RF$ are tangent to $\\odot I$, $RG = RF$. As $SL \\parallel RG$, we have $\\angle LSF = \\angle GRF$, so\n\n$$\n\\angle LFS = \\frac{180^\\circ - \\angle LSF}{2} = \\frac{180^\\circ - \\angle GRF}{2} = \\angle GFR,\n$$\n\nand hence $L$, $G$, and $F$ are collinear.\n\nSimilarly, draw a line parallel to $CD$ and tangent to $\\odot O$ at point $J$, then $F$, $H$, and $J$ are collinear. Let tangent lines to the circle $\\odot O$ at the points $L$ and $J$ meet $EF$ at points $M_1$ and $M_2$ respectively. In the following, we prove that the points $M_1$ and $M_2$ coincide. It follows from the homothety centered at $F$ mapping $\\odot O$ to $\\odot I$ that $LJ \\parallel GH$, then\n\n![](images/Kina_2012_p17_data_7daadd8248.png)\n\n$$\n\\frac{M_1 E}{EF} = \\frac{LG}{GF} = \\frac{JH}{HF} = \\frac{M_2 E}{EF},\n$$\n\nand hence $M_1$ and $M_2$ coincide. Denote this point by $K$.\n\nFinally, we want to prove the points $M$ and $K$ also coincide. It suffices to show that $K$ lies on the line $l$.\n\nJoin $KO$, as $KL$ and $KJ$ are tangent to $\\odot O$, so $\\angle LKO = \\angle JKO$.\n\nNote that $KO$ bisects $\\angle LKJ$, and it follows from $KL \\parallel AB$ and $KJ \\parallel CD$ that the line $KO$ meets lines $AB$ and $CD$ with the same angles of intersection, and hence $KO$ is just the line $l$, i.e., $K$ lies on the line $l$.\n\nHence $K$ is just the intersection point $M$ of line $EF$ and $l$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 13778,
    "subject": "Mathematics (Olympiad)",
    "question": "Given an obtuse isosceles triangle $ABC$ with $CA = CB$ and circumcenter $O$. The point $P$ on $AB$ satisfies $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$ and $AB$ meet at $F$. If $N$ is the midpoint of $CP$ and $ON$ and $AB$ meet at $D$, show that $ODCF$ is cyclic.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $T$ be the midpoint of $CQ$ (which is also the center of the circle with diameter $CQ$). Since $OC$ is the perpendicular bisector of $AB$ (because $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). Also, $OT$ is the perpendicular bisector of $CE$, so $\\angle OCF = 90^\\circ - \\angle OCT = 90^\\circ - \\angle OCD = \\angle ODM = \\angle ODF$. Thus, $ODCF$ is cyclic. $\\Box$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 13779,
    "subject": "Mathematics (Olympiad)",
    "question": "a) \n\nCount the number of edge pieces in a rectangular jigsaw puzzle with sides of length 20, 20, 13, and 13 pieces. Consider different ways of counting the edge pieces, including how to account for corner pieces.\n\nb) \n\nA jigsaw puzzle has 1000 pieces arranged in 25 rows. How many edge pieces does it have?\n\nc) \n\nA jigsaw puzzle has 124 edge pieces, and all four edges have the same number of pieces. How many pieces are in the puzzle?\n\n% ![](images/2020_Australian_Scene_W_p31_data_718ecaf21d.png)\n\nd) \n\nFor a 2000-piece rectangular jigsaw puzzle, what is the smallest possible number of edge pieces?",
    "options": [],
    "answer": "See solution",
    "solution": "a) \n\n**Alternative i:**\n\nCount the corner pieces as part of the top and bottom rows, not the sides. The number of edge pieces is $20 + 20 + 13 + 13 = 66$.\n\n**Alternative ii:**\n\nCounting each edge includes each corner twice, so subtract 4: $20 + 20 + 15 + 15 - 4 = 66$.\n\n**Alternative iii:**\n\nRemoving all edge pieces leaves an $18 \\times 13$ rectangle ($234$ pieces). So, $300 - 234 = 66$ edge pieces.\n\nb) \n\nEach row has $1000 \\div 25 = 40$ pieces. Edge pieces: $40 + 40 + 23 + 23 = 126$.\n\nc) \n\n**Alternative i:**\n\nAll edges have the same number of pieces. Non-corner edge pieces per edge: $(124 - 4) \\div 4 = 30$. So, $30 + 2 = 32$ per edge. Total pieces: $32 \\times 32 = 1024$.\n\n**Alternative ii:**\n\nPartition edge pieces into 4 blocks (each edge minus one corner). Each block: $124 \\div 4 = 31$. So, $31 + 1 = 32$ per edge. Total: $32 \\times 32 = 1024$.\n\n% ![](images/2020_Australian_Scene_W_p31_data_718ecaf21d.png)\n\nd) \n\nFactor pairs for 2000 and edge pieces:\n\n| jigsaw         | edge pieces                  |\n|:-------------- |:----------------------------|\n| $1 \\times 2000$  | $2000$                      |\n| $2 \\times 1000$  | $2000$                      |\n| $4 \\times 500$   | $500 + 500 + 2 + 2 = 1004$  |\n| $5 \\times 400$   | $400 + 400 + 3 + 3 = 806$   |\n| $8 \\times 250$   | $250 + 250 + 6 + 6 = 512$   |\n| $10 \\times 200$  | $200 + 200 + 8 + 8 = 416$   |\n| $16 \\times 125$  | $125 + 125 + 14 + 14 = 278$ |\n| $20 \\times 100$  | $100 + 100 + 18 + 18 = 236$ |\n| $25 \\times 80$   | $80 + 80 + 23 + 23 = 206$   |\n| $40 \\times 50$   | $50 + 50 + 38 + 38 = 176$   |\n\nSo, the smallest number of edge pieces in a 2000-piece jigsaw is $176$.",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 13780,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $n$ be the least prime number that can be written as the sum of 5 distinct prime numbers. What is the sum of the digits of $n$?\n\n(A) 5  \n(B) 7  \n(C) 8  \n(D) 10  \n(E) 11",
    "options": [],
    "answer": "See solution",
    "solution": "The prime $2$ cannot be among the $5$ distinct primes chosen because, if it were, then the sum would be even. The first $5$ odd primes are $3$, $5$, $7$, $11$, and $13$, and their sum is $39$, which is not prime. The next smallest sum of $5$ distinct odd primes is $3 + 5 + 7 + 11 + 17 = 43$, which is prime. The requested digit sum is $4 + 3 = 7$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 13781,
    "subject": "Mathematics (Olympiad)",
    "question": "Acute triangle *ABC* is inscribed in an ellipse with two distinct foci *H* and *K*. Prove that if the orthocenter of *ABC* coincides with *H*, then its incenter lies on line *HK*.",
    "options": [],
    "answer": "See solution",
    "solution": "In particular *PEDB* and *EFQR* are cyclic too by Eq. (1). Now if $\\ell$ is the tangent line to $\\omega_A$ at $P$, then working with directed angles modulo $180^\\circ$ gives\n\n$$\n\\begin{align*}\n\\angle(\\ell, PQ) &= \\angle(\\ell, PH) + \\angle(PD, PQ) = \\angle(PY, BY) + \\angle(DE, EQ) \\\\\n&= \\angle(PD, DB) + \\angle(DE, EQ) = \\angle(PH, EQ) + \\angle(DE, DB) \\\\\n&= \\angle(PD, EQ) + \\angle(FE, DF) = \\angle(PD, DF) + \\angle(FE, EQ) \\\\\n&= \\angle(PR, RH) + \\angle(RF, FQ) = \\angle(PR, RQ).\n\\end{align*}\n$$\n\nThus line $\\ell$ is also tangent to $\\Gamma$, as needed.\n\nFinally, if we define $P'$, $Q'$, $R'$ as the second intersection of $\\overline{HP}$, $\\overline{HQ}$, $\\overline{HR}$ with the incircle, then it follows from Eq. (1) and\n\n$$\nHD \\cdot HP' = HE \\cdot HQ' = HF \\cdot HR'\n$$\n\nthat $\\triangle P'Q'R'$ is homothetic to $\\triangle PQR$ through $H$. As $I$ is the circumcenter of $\\triangle P'Q'R'$ this implies $K, I, H$ are collinear.\n\n**Remark.** The angle-chasing part of the solution can be simplified by allowing the use of *negative inversion*: $\\triangle PQR$ is the image of $\\triangle DEF$ under a negative inversion at $H$ swapping $\\{A, X\\}$, $\\{B, Y\\}$ and $\\{C, Z\\}$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 13782,
    "subject": "Mathematics (Olympiad)",
    "question": "A game is played on a regular triangle which is split into $n^2$ equal smaller regular triangles by lines that are parallel to one of the sides of the triangle. Denote a \"line of triangles\" to be all triangles that are placed between two adjacent parallel lines that form the grid.\n\nIn the beginning, all triangles are white. At each move, one line of triangles that contains at least one white triangle is colored black. The game ends when all triangles are colored black. Find the smallest and largest possible number of moves in the game.\n\n![](images/Baltic_Way_SHL_2009-11_13-16_p192_data_b74c0a08f0.png)",
    "options": [],
    "answer": "See solution",
    "solution": "The smallest possible number of moves is $n$, and the largest possible number of moves is $3n - 2$.\n\nIf all moves are done with lines parallel to one side of the triangle, the game ends after $n$ moves. The number of moves cannot be smaller, since each move must color at least one new triangle, and covering all triangles in fewer than $n$ moves is impossible.\n\nFor the largest number, the game can last $3n - 2$ moves. For $n = 1$, this is evident. Assume it holds for $n = k$. For $n = k + 1$, start with three moves coloring two rightmost corners and the rightmost line. After these 3 moves, the field reduces to the $n = k$ case (see figure below):\n\n![](images/Baltic_Way_SHL_2009-11_13-16_p192_data_e53f3e8ff7.png)\n\nThus, the total number of moves is $3n - 2$. There cannot be more than $3n - 2$ moves, since coloring all $n$ lines parallel to one side ends the game, and the number of moves before the last cannot exceed $3(n-1)$, giving a total not larger than $3n-2$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 13783,
    "subject": "Mathematics (Olympiad)",
    "question": "There are $n$ guests at a party. Any two guests are either friends or not friends. Every guest is friends with exactly four of the other guests. Whenever a guest is _not_ friends with two other guests, those two other guests cannot be friends with each other either.\n\nWhat are the possible values of $n$?",
    "options": [],
    "answer": "See solution",
    "solution": "We first consider the friends of one guest, say Marieke. Marieke has exactly four friends at the party, say Aad, Bob, Carla, and Demi. The other guests (if any) are not friends with Marieke. Hence, they cannot have any friendships among themselves and can therefore only be friends with Aad, Bob, Carla, and Demi. Since everyone has exactly four friends at the party, each of them must be friends with Aad, Bob, Carla, and Demi (and with no one else).\n\nSince Aad also has exactly four friends (including Marieke), the group of guests that are not friends with Marieke can consist of no more than three people. If the group consists of zero, one, or three people, we have the following solutions (two guests are connected by a line if they are friends):\n\n![](images/NLD_ABooklet_2020_p17_data_d1460a633b.png)\n![](images/NLD_ABooklet_2020_p17_data_4ff9e161e2.png)\n![](images/NLD_ABooklet_2020_p17_data_9f8902386c.png)\n\n*Solutions with five, six, and eight guests in total.*\n\nNow we will show that it is not possible for this group to consist of two people. In that case, Aad would have exactly one friend among Bob, Carla, and Demi. Assume, without loss of generality, that Aad and Bob are friends. In the same way, Carla must be friends with one of Aad, Bob, and Demi. Since Aad and Bob already have four friends, Carla and Demi must be friends. However, since they are both not friends with Aad, this contradicts the requirement in the problem statement.\n\nWe conclude that there can be five, six, or eight guests at the party. Hence, the possible values for $n$ are $5$, $6$, and $8$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 13784,
    "subject": "Mathematics (Olympiad)",
    "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": "See solution",
    "solution": "a) Denote the arithmetic mean by $A(x)$. As\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$$\n$$\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$$\nso\n$$\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$$\nDepending on the signs of $\\sin \\frac{x}{2} + \\cos \\frac{x}{2}$ and $\\sin \\frac{x}{2} - \\cos \\frac{x}{2}$, one of the trigonometric functions cancels out and the other is doubled, with either a positive or negative sign. Therefore, $A(x)$ is equal to one of $\\sin \\frac{x}{2}$, $\\cos \\frac{x}{2}$, $-\\sin \\frac{x}{2}$, $-\\cos \\frac{x}{2}$.\n\nb) Each of the four values is needed: for $x = 0, \\pi, 2\\pi, 3\\pi$, each expression among $\\sin \\frac{x}{2}$, $\\cos \\frac{x}{2}$, $-\\sin \\frac{x}{2}$, $-\\cos \\frac{x}{2}$ evaluates to $1$ for a unique $x$. Thus, none can be left out.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 13785,
    "subject": "Mathematics (Olympiad)",
    "question": "Circles $k_1$ and $k_2$ intersect at points $A$ and $B$, such that $k_1$ passes through the center $O$ of circle $k_2$. The line $p$ intersects $k_1$ at points $K$ and $O$, and $k_2$ at points $L$ and $M$, with $L$ between $K$ and $O$. The point $P$ is the orthogonal projection of $L$ onto the line $AB$. Prove that the line $KP$ is parallel to the $M$-median of triangle $ABM$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $C$ be the midpoint of segment $AB$. We need to prove that $MC \\parallel KP$.\n\nLet $\\alpha = \\angle BKA$. Notice that\n\n$$\n\\begin{aligned}\n\\angle BLA &= 180^\\circ - \\angle BMA = 180^\\circ - \\frac{1}{2}\\angle BOA \\\\\n&= 180^\\circ - \\frac{1}{2}(180^\\circ - \\angle BKA) = 90^\\circ + \\frac{1}{2}\\alpha\n\\end{aligned}\n$$\n\nAlso, $O$ is the midpoint of arc $\\widearc{AB}$, so $KO$ is the angle bisector of $\\angle BKA$. From this, $L$ is the incenter of triangle $ABK$.\n\nMoreover, $ML$ is a diameter of circle $k_2$, so $\\angle ABM = 90^\\circ$. Since $BL$ is the angle bisector of $\\angle ABK$, $BM$ is the exterior angle bisector of the same angle.\n\nThus, $M$ is the center of the excircle of triangle $ABK$.\n\n![](images/MACEDONIAN_MATHEMATICAL_OLYMPIADS_2016_p7_data_4291bafebb.png)\n\nTherefore, we need to prove that the line through the incenter $L$ of triangle $ABK$ and the point of tangency of its incircle is parallel to the line through the excircle center $M$ and the midpoint $C$ of $AB$. This is a well-known lemma, which completes the proof.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 13786,
    "subject": "Mathematics (Olympiad)",
    "question": "Given a non-negative real number $a$ and a sequence $(u_n)$ defined by\n\n$$u_1 = 3, \\quad u_{n+1} = \\frac{u_n}{2} + \\frac{n^2}{4n^2 + a} \\sqrt{u_n^2 + 3}$$\n\nfor all positive integers $n$.\n\na) For $a = 0$, prove that the sequence has a finite limit and find that limit.\n\nb) For $a \\in [0, 1]$, prove that the sequence has a finite limit.",
    "options": [],
    "answer": "See solution",
    "solution": "$(u_n)$ is a positive sequence.\n\na) For $a = 0$, the recurrence becomes $u_{n+1} = \\frac{u_n}{2} + \\frac{\\sqrt{u_n^2 + 3}}{4}$.\n\nLet $f(x) = \\frac{1}{2}x + \\frac{1}{4}\\sqrt{x^2 + 3}$. Then $f'(x) = \\frac{1}{2} + \\frac{x}{4\\sqrt{x^2 + 3}} > 0$ for $x > 0$, so $f$ is strictly increasing on $(0, +\\infty)$. Note that\n\n$$u_2 = \\frac{u_1}{2} + \\frac{\\sqrt{u_1^2 + 3}}{4} = \\frac{3}{2} + \\frac{\\sqrt{12}}{4} < 3 = u_1,$$\n\nso $(u_n)$ is strictly decreasing and bounded below by $0$, hence convergent. Let $L$ be its limit. Then\n\n$$L = \\frac{L}{2} + \\frac{\\sqrt{L^2 + 3}}{4},$$\n\nwhich gives $L = 1$ since $L \\ge 0$.\n\nb) For $a \\in [0, 1]$, we have\n\n$$\\frac{n^2}{4n^2 + 1} \\le \\frac{n^2}{4n^2 + a} \\le \\frac{1}{4}.$$ \n\nDefine sequences $(x_n)$ and $(y_n)$:\n\n$$x_1 = 3, \\quad x_{n+1} = \\frac{1}{2}x_n + \\frac{1}{4}\\sqrt{x_n^2 + 3},$$\n$$y_1 = 3, \\quad y_{n+1} = \\frac{1}{2}y_n + \\frac{n^2}{4n^2 + 1}\\sqrt{y_n^2 + 3}.$$ \n\nFrom part (a), $\\lim_{n \\to \\infty} x_n = 1$. By induction,\n\n$$y_n \\le u_n \\le x_n, \\quad \\forall n \\in \\mathbb{N}^*.$$ \n\nNext, show $\\lim_{n \\to \\infty} y_n = 1$, which implies $\\lim_{n \\to \\infty} u_n = 1$. For each $n$, let\n\n$$f_n = \\frac{1}{2} + \\frac{n^2}{4n^2 + 1} \\frac{y_n + 1}{\\sqrt{y_n^2 + 3} + 2}$$\n\nand since $y_n > 0$, $f_n > 0$ for all $n$.\n\n$$\\begin{aligned}\ny_{n+1} - 1 &= \\frac{1}{2}(y_n - 1) + \\frac{n^2 (y_n - 1)(y_n + 1)}{(4n^2 + 1)(\\sqrt{y_n^2 + 3} + 2)} + \\frac{2n^2}{4n^2 + 1} - \\frac{1}{2} \\\\\n&= \\frac{1}{2}(y_n - 1) + \\frac{n^2 (y_n - 1)(y_n + 1)}{(4n^2 + 1)(\\sqrt{y_n^2 + 3} + 2)} - \\frac{1}{2(4n^2 + 1)} \\\\\n&< \\frac{1}{2}(y_n - 1) + \\frac{n^2 (y_n - 1)(y_n + 1)}{(4n^2 + 1)(\\sqrt{y_n^2 + 3} + 2)} \\\\\n&= (y_n - 1) f_n\n\\end{aligned}$$\n\nIf $y_n \\ge 1$ for all $n$, then\n\n$$y_{n+1} - 1 < (y_n - 1) f_n < \\frac{3}{4} y_n < \\dots < \\frac{3^n}{4^n} y_1 = \\frac{3^{n+1}}{4^n}$$\n\nSo $0 < y_n - 1 < \\frac{3^n}{4^{n-1}}$, thus $\\lim_{n \\to \\infty} y_n = 1$.\n\nIf there exists $n_0$ with $y_{n_0} < 1$, then $y_n < 1$ for all $n > n_0$. If $y_{N+1} > y_N$ for some $N$, then $y_{N+2} > y_{N+1}$, so $(y_n)$ is increasing from $y_N$, but $y_n < 1$ for all $n > n_0$, so $(y_n)$ has a finite limit $L$ ($0 \\le L < 1$). But then\n\n$$L = \\frac{L}{2} + \\frac{\\sqrt{L^2 + 3}}{4}$$\n\nwhich gives $L = 1$, a contradiction. Thus, $\\lim_{n \\to \\infty} y_n = 1$ and $\\lim_{n \\to \\infty} u_n = 1$. $\\square$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 13787,
    "subject": "Mathematics (Olympiad)",
    "question": "In a school with 5 grades, there are 250 girls and 250 boys. Each grade has 100 students. Teams of one girl and one boy from the same grade must be formed for a contest. At least 19 students in each grade are girls and at least 19 are boys. Find the greatest number of teams that can be formed with certainty.",
    "options": [],
    "answer": "See solution",
    "solution": "The answer is $126$.\n\nLet there be $a_i$ girls and $b_i$ boys in grade $i$, $1 \\leq i \\leq 5$. Consider a $2 \\times 5$ table with $a_1, \\dots, a_5$ in the first row and $b_1, \\dots, b_5$ in the second row. Mark the smaller of the numbers $a_i, b_i$ for each $i$. The number of teams that can be formed is the sum of the five marked numbers.\n\nAt least three marked numbers are in the same row. Suppose, for instance, that $a_1, a_2, a_3$ are marked. Since $b_4 \\geq 19$, $b_5 \\geq 19$ and $a_4 + b_4 = a_5 + b_5 = 100$, each of $a_4$ and $a_5$ is at most $100 - 19 = 81$. Because $a_1 + a_2 + a_3 + a_4 + a_5 = 250$, it follows that $a_1 + a_2 + a_3 = 250 - (a_4 + a_5) \\geq 250 - 2 \\cdot 81 = 88$.\n\nDue to $a_i \\leq b_i$ for $1 \\leq i \\leq 3$, the number of teams in grades 1, 2, 3 is $a_1 + a_2 + a_3$, hence it is at least $88$. Also, at least $19$ teams can be formed in each of grades 4 and 5. So $88 + 2 \\cdot 19 = 126$ teams can always be formed.\n\nThe example $(a_1, a_2, a_3, a_4, a_5) = (29, 29, 30, 81, 81)$, $(b_1, b_2, b_3, b_4, b_5) = (71, 71, 70, 19, 19)$ shows that $126$ is the greatest number in question.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 13788,
    "subject": "Mathematics (Olympiad)",
    "question": "Given a positive integer $n \\geq 3$, colour each cell of an $n \\times n$ square array with one of $\\lceil \\frac{(n+2)^2}{3} \\rceil - 1$ colours, each colour being used at least once. Prove that there exists a $1 \\times 3$ or $3 \\times 1$ rectangular subarray whose cells have pairwise distinct colours.",
    "options": [],
    "answer": "See solution",
    "solution": "For convenience, say that a subarray of the $n \\times n$ square array *bears* a colour if at least two of its cells share that colour.\n\nWe will prove that the number of $1 \\times 3$ and $3 \\times 1$ rectangular subarrays, which is $2n(n-2)$, exceeds the number of such subarrays, each of which bears some colour. The key ingredient is the estimate in the lemma below.\n\n**Lemma.** If a colour is used exactly $p$ times, then the number of $1 \\times 3$ and $3 \\times 1$ rectangular subarrays bearing that colour does not exceed $3(p-1)$.\n\nAssume the lemma for the moment. Let $N = \\lceil \\frac{(n+2)^2}{3} \\rceil - 1$ and let $n_i$ be the number of cells coloured with the $i$th colour, $i = 1, \\dots, N$. The number of $1 \\times 3$ and $3 \\times 1$ rectangular subarrays, each of which bears some colour, is at most\n\n$$\n\\sum_{i=1}^{N} 3(n_i - 1) = 3 \\sum_{i=1}^{N} n_i - 3N = 3n^2 - 3N < 3n^2 - (n^2 + 4n) = 2n(n-2)\n$$\n\nand thereby the proof is complete.\n\nNow, for the lemma: the assertion is clear if $p=1$, so let $p>1$.\n\nSuppose a row contains exactly $q$ cells coloured $C$. Then the number $r$ of $3 \\times 1$ rectangular subarrays bearing $C$ does not exceed $\\frac{3q}{2} - 1$; a similar estimate holds for a column. To see this, consider the incidence of a cell $c$ coloured $C$ and a $3 \\times 1$ rectangular subarray $R$ bearing $C$:\n\n$$\n\\langle c, R \\rangle = \\begin{cases} 1 & \\text{if } c \\subset R, \\\\ 0 & \\text{otherwise.} \\end{cases}\n$$\n\nGiven $R$, $\\sum_c \\langle c, R \\rangle \\geq 2$, and given $c$, $\\sum_R \\langle c, R \\rangle \\leq 3$; moreover, if $c$ is the leftmost or rightmost cell, then $\\sum_R \\langle c, R \\rangle \\leq 2$. Consequently,\n\n$$\n2r \\leq \\sum_{R} \\sum_{c} \\langle c, R \\rangle = \\sum_{(c,R)} \\langle c, R \\rangle = \\sum_{c} \\sum_{R} \\langle c, R \\rangle \\leq 2 + 3(q-2) + 2 = 3q - 2,\n$$\n\nwhence the conclusion.\n\nFinally, let the $p$ cells coloured $C$ lie on $k$ rows and $l$ columns and notice that $k + l \\geq 3$ for $p > 1$. By the preceding, the total number of $3 \\times 1$ rectangular subarrays bearing $C$ does not exceed $\\frac{3p}{2} - k$, and the total number of $1 \\times 3$ rectangular subarrays bearing $C$ does not exceed $\\frac{3p}{2} - l$, so the total number of $1 \\times 3$ and $3 \\times 1$ rectangular subarrays bearing $C$ does not exceed $(\\frac{3p}{2} - k) + (\\frac{3p}{2} - l) = 3p - (k + l) \\leq 3p - 3 = 3(p-1)$. This completes the proof.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 13789,
    "subject": "Mathematics (Olympiad)",
    "question": "Suppose that $m$ and $k$ are non-negative integers, and $p = 2^{2^m} + 1$ is a prime number. Prove that\n\n(a) $2^{2^{m+1}} p^k \\equiv 1 \\pmod{p^{k+1}}$;\n\n(b) $2^{m+1} p^k$ is the smallest positive integer $n$ satisfying the congruence equation $2^n \\equiv 1 \\pmod{p^{k+1}}$.",
    "options": [],
    "answer": "See solution",
    "solution": "We want to prove that $2^{2^{m+1}} p^k = p^{k+1} t_k + 1$ for some integer $t_k$ not divisible by $p$. We proceed by induction on $k$.\n\nWhen $k = 0$, it follows from $2^{2^m} = p-1$ that $2^{2^m} = (p-1)^2 = p(p-2)+1$, in this case, $t_0 = p-2$.\n\nFor the inductive step, suppose that $2^{2^{m+1}} p^k = p^{k+1} t_k + 1$ where $k \\ge 0$ and $p \\nmid t_k$. Then\n\n$$\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$$\n\nAs $k \\ge 0$, then for any $s \\ge 2$, we have $(k+1)s \\ge 2(k+1) = 2k+2 \\ge k+2$, so $2^{2^{m+1}} p^{k+1} = p^{k+2} t_{k+1} + 1$, where $t_{k+1} \\in \\mathbb{Z}_+$ and $p \\nmid t_{k+1}$. It follows from mathematical induction that (a) holds.\n\nNext, we prove (b). Write $2^{m+1} p^k = n\\ell + r$, where $\\ell, r \\in \\mathbb{Z}$ and $0 \\le r < n$. Then it follows from (a) that $1 \\equiv 2^{2^{m+1}} p^k \\equiv 2^{n\\ell+r} \\equiv (2^n)^\\ell \\cdot 2^r \\equiv 2^r \\pmod{p^{k+1}}$. As $0 \\le r < n$, it follows from the definition of $n$ that $r=0$, i.e., $n \\mid 2^{m+1} p^k$. By the Fundamental Theorem of Arithmetic, $n = 2^t p^s$. If $t \\le m$, then $2^{2^m} p^k = (2^{2^t} p^k)^{2^{m-t}} \\equiv 1 \\pmod{p}$. On the other hand, $2^{2^m} p^k = (2^{2^m})^{p^k} \\equiv (-1)^{p^k} \\equiv -1 \\pmod{p}$, which is a contradiction, and hence $t = m+1$, i.e., $n = 2^{m+1} p^s$. $\\square$",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 13790,
    "subject": "Mathematics (Olympiad)",
    "question": "For every pair of positive integers $n, m$ with $n < m$, let $s(n, m)$ be the number of positive integers in the range $[n, m]$ that are coprime with $m$. Find all positive integers $m \\ge 2$ such that $m$ satisfies the following conditions:\n\ni) $\\dfrac{s(n,m)}{m-n} \\ge \\dfrac{s(1,m)}{m}$ for all $n = 1, 2, \\ldots, m-1$,\n\nii) $2022^m + 1$ is divisible by $m^2$.",
    "options": [],
    "answer": "See solution",
    "solution": "First, we prove that if $m$ satisfies the first condition, then $m$ has only one prime divisor. Assume that $m$ has at least two prime divisors. Let $p$ be the smallest prime divisor of $m$ and $p_1, p_2, \\ldots, p_k$ be the remaining prime divisors of $m$. We have\n\n$$\n\\frac{\\varphi(m)}{m} = \\left(1-\\frac{1}{p}\\right) \\left(1-\\frac{1}{p_1}\\right) \\dots \\left(1-\\frac{1}{p_k}\\right) < 1-\\frac{1}{p} = \\frac{p-1}{p}.\n$$\n\nBy choosing $n = p$ in (i), we get\n\n$$\n\\frac{s(p, m)}{m-p} = \\frac{\\varphi(m) - (p-1)}{m-p} < \\frac{\\varphi(m) - \\frac{\\varphi(m)}{m} \\cdot p}{m-p} = \\frac{\\varphi(m)}{m} = \\frac{s(1, m)}{m},\n$$\n\nwhich is a contradiction. Therefore, $m$ must be a power of a prime. Let $m = p^k$. Note that\n\n$$\n2022^{p^k} + 1 \\equiv 2022 + 1 \\equiv 2023 \\equiv 0 \\pmod{p},\n$$\n\nthus $p \\mid 2023$ and $p \\in \\{7, 17\\}$.\n\nIf $p=7$, using LTE, we have\n\n$$\nv_7(2022^{7k} + 1) = v_7(2023) + v_7(7^k) = 1 + k \\ge v_7(7^{2k}) = 2k.\n$$\n\nFrom this, we conclude that $k=1$ and $m=7$. Similarly, for $p=17$, applying LTE, we have\n\n$$\nv_{17}(2022^{17k} + 1) = v_{17}(2023) + v_{17}(17^k) = 2 + k \\ge v_{17}(17^{2k}) = 2k,\n$$\n\nso $k \\in \\{1, 2\\}$ and $m \\in \\{17, 289\\}$.\n\nTherefore, $m = 7, 17, 289$ are all desired numbers. $\\square$",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 13791,
    "subject": "Mathematics (Olympiad)",
    "question": "Найдите все рациональные числа $a$, такие, что для любого натурального $n$ число $a n (n + 2)(n + 3)(n + 4)$ — целое.",
    "options": [],
    "answer": "See solution",
    "solution": "$a = \\frac{k}{6}$, где $k$ — любое целое число.\n\nПодставив $n = 1, n = 3$ и $n = 4$, получаем, что числа $2^2 \\cdot 3 \\cdot 5a$, $2 \\cdot 3^2 \\cdot 5 \\cdot 7a$ и $2^6 \\cdot 3 \\cdot 7a$ — целые. Значит, $a$ — рациональное число, имеющее несократимую запись $\\frac{p}{q}$, где $q$ является делителем числа $\\gcd(2^2 \\cdot 3 \\cdot 5, 2 \\cdot 3^2 \\cdot 5 \\cdot 7, 2^6 \\cdot 3 \\cdot 7) = 6$, и $a = k/6$ при некотором целом $k$.\n\nОсталось показать, что все числа такого вида подходят. Действительно, одно из трёх последовательных чисел $n + 2, n + 3, n + 4$ делится на 3, а одно из последовательных чисел $n + 2, n + 3$ делится на 2; значит, $n(n + 2)(n + 3)(n + 4)$ делится на 2 и на 3, а значит, и на 6. Поэтому $a n (n + 2)(n + 3)(n + 4) = k \\frac{n(n + 2)(n + 3)(n + 4)}{6}$ — целое число.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 13792,
    "subject": "Mathematics (Olympiad)",
    "question": "In an acute-angled triangle $ABC$, let $H$ be its orthocenter and $I$ its incenter. Let $D$ be the perpendicular projection of $I$ onto the line $BC$, and $E$ be the image of point $A$ under reflection with center $I$. Furthermore, let $F$ be the perpendicular projection of the point $H$ onto the line $ED$. Prove that the points $B$, $H$, $F$, and $C$ lie on one circle.",
    "options": [],
    "answer": "See solution",
    "solution": "Consider point $P$ such that $ABPC$ is a parallelogram (see figure). Since $HB \\perp AC \\parallel BP$, the angle $HBP$ is right. Similarly, it follows from $HC \\perp AB \\parallel CP$ that the angle $HCP$ is also right. Therefore, both points $B$ and $C$ lie on the Thales circle with diameter $HP$.\n\nSurely it is sufficient to consider only the case where $H \\neq F$. We explain why it is then sufficient to show that the points $D$, $E$, $P$ are collinear. For then the point $F$ lies on this line, so that the angle $HFP$ is right, and therefore its vertex $F$ lies (together with the points $B$, $C$, and $H$) on the circle with diameter $HP$.\n\nLet $M$ be the midpoint of $BC$. Under point reflection with center $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\n![](images/CZE_ABooklet_2023_p26_data_833bd39c25.png)\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 under a homothety with center $P$ (and with a coefficient greater than $1$). In this homothety, the tangent $BC$ of the excircle is the image of the tangent of the incircle, which 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$ are collinear. 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 mean lines $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": 13793,
    "subject": "Mathematics (Olympiad)",
    "question": "All sides and diagonals of a convex $n$-gon, $n \\ge 3$, are coloured one of two colours. Show that there exist $\\lfloor (n+1)/3 \\rfloor$ pairwise disjoint monochromatic segments. (Two segments are disjoint if they do not share an endpoint or an interior point.)",
    "options": [],
    "answer": "See solution",
    "solution": "If all sides are monochromatic, then the assertion is clearly true. Otherwise, delete a vertex incident with two sides of different colours together with its neighbours, delete all sides and diagonals incident with these three vertices, and apply induction.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 13794,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $a$, $b$, $c$ be positive real numbers such that $a + b + c = 3$. Prove that\n\n$$\n\\sqrt{\\frac{b}{a^2+3}} + \\sqrt{\\frac{c}{b^2+3}} + \\sqrt{\\frac{a}{c^2+3}} \\leq \\frac{3}{2} \\sqrt[4]{\\frac{1}{abc}}\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "Let $\\vec{u} = \\left( \\frac{1}{\\sqrt{a^2+3}}, \\frac{1}{\\sqrt{b^2+3}}, \\frac{1}{\\sqrt{c^2+3}} \\right)$ and $\\vec{v} = (\\sqrt{b}, \\sqrt{c}, \\sqrt{a})$. By the Cauchy–Schwarz inequality,\n\n$$\n\\left( \\sqrt{\\frac{b}{a^2+3}} + \\sqrt{\\frac{c}{b^2+3}} + \\sqrt{\\frac{a}{c^2+3}} \\right)^2 \\leq \\left( \\frac{1}{a^2+3} + \\frac{1}{b^2+3} + \\frac{1}{c^2+3} \\right)(a+b+c) = 3 \\left( \\frac{1}{a^2+3} + \\frac{1}{b^2+3} + \\frac{1}{c^2+3} \\right)\n$$\n\nsince $a + b + c = 3$.\n\nNote that $a^2 + 3 = a^2 + 1 + 1 + 1 \\geq 4 \\sqrt[4]{a^2} = 4\\sqrt{a}$ by the AM–GM inequality. Similarly, $b^2 + 3 \\geq 4\\sqrt{b}$ and $c^2 + 3 \\geq 4\\sqrt{c}$. Thus,\n\n$$\n\\frac{1}{a^2+3} + \\frac{1}{b^2+3} + \\frac{1}{c^2+3} \\leq \\frac{1}{4} \\left( \\frac{1}{\\sqrt{a}} + \\frac{1}{\\sqrt{b}} + \\frac{1}{\\sqrt{c}} \\right)\n$$\n\nNow,\n\n$$\n\\frac{1}{\\sqrt{a}} + \\frac{1}{\\sqrt{b}} + \\frac{1}{\\sqrt{c}} = \\frac{\\sqrt{bc} + \\sqrt{ca} + \\sqrt{ab}}{\\sqrt{abc}} \\leq \\frac{\\frac{a+b}{2} + \\frac{b+c}{2} + \\frac{c+a}{2}}{\\sqrt{abc}} = \\frac{a+b+c}{\\sqrt{abc}}\n$$\n\nTherefore,\n\n$$\n\\frac{1}{a^2+3} + \\frac{1}{b^2+3} + \\frac{1}{c^2+3} \\leq \\frac{a+b+c}{4\\sqrt{abc}} = \\frac{3}{4\\sqrt{abc}}\n$$\n\nCombining the above,\n\n$$\n\\left( \\sqrt{\\frac{b}{a^2+3}} + \\sqrt{\\frac{c}{b^2+3}} + \\sqrt{\\frac{a}{c^2+3}} \\right)^2 \\leq \\frac{9}{4\\sqrt{abc}}\n$$\n\nTaking square roots,\n\n$$\n\\sqrt{\\frac{b}{a^2+3}} + \\sqrt{\\frac{c}{b^2+3}} + \\sqrt{\\frac{a}{c^2+3}} \\leq \\frac{3}{2} \\sqrt[4]{\\frac{1}{abc}}\n$$\n\nEquality holds when $a = b = c = 1$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 13795,
    "subject": "Mathematics (Olympiad)",
    "question": "Given the diagram:\n\n![](images/Irska_2009_p21_data_0a0702b633.png)\n\nLet $|AC| = |AX| = 2|CY| = 2|YB|$ and $|AX| = 2|CB|$. Find the length of $|XY|$.",
    "options": [],
    "answer": "See solution",
    "solution": "Since $|AC| = |AX| = 2|CY| = 2|YB|$ and $|AX| = 2|CB|$, we have $\\angle ACX = \\angle CYB$, so $\\angle XCY = \\angle CBY$. Drawing a perpendicular from $Y$ to $CB$ and noting two $3, 4, 5$ triangles, we find $\\cos \\angle XCY = \\cos \\angle CBY = \\frac{3}{5}$. By the cosine rule:\n\n$$\n|XY|^2 = |CX|^2 + |CY|^2 - 2|CX| \\cdot |CY| \\cdot \\cos \\angle XCY = 65\n$$\n\nThus,\n\n$$\n|XY| = \\sqrt{65} > 8 = |AM|.\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 13796,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $ABC$ be an isosceles triangle with $\\overline{AC} = \\overline{BC}$ and let $P$ be a point on the circumcircle lying on the arc $CA$ not containing $B$.\n\nLet $E$ and $F$ be the orthogonal projections of the point $C$ onto the lines $AP$ and $BP$, respectively.\n\nProve that $AE$ and $BF$ have the same length.",
    "options": [],
    "answer": "See solution",
    "solution": "The inscribed angle theorem implies $\\angle PAC = \\angle PBC$.\n\n![](images/AustriaMO2011_p6_data_32bed5e6f2.png)\n\nTherefore, the right triangles $AEC$ and $BFC$ have the same angles. Since their hypotenuses have the same length $\\overline{AC} = \\overline{BC}$, they are congruent and we conclude $\\overline{AE} = \\overline{BF}$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 13797,
    "subject": "Mathematics (Olympiad)",
    "question": "Given a prime $p \\geq 5$, show that there exist at least two distinct primes $q$ and $r$ in the range $2, 3, \\dots, p-2$ such that $q^{p-1} \\not\\equiv 1 \\pmod{p^2}$ and $r^{p-1} \\not\\equiv 1 \\pmod{p^2}$.",
    "options": [],
    "answer": "See solution",
    "solution": "In what follows, all congruences are to be understood modulo $p^2$. An integer $n$ coprime to $p$ will be called *proper* if $n^{p-1} \\equiv 1 \\pmod{p}$, and *improper* otherwise. Our solution is based on the following two simple facts:\n\n1. An improper integer greater than 1 has at least one improper prime divisor.\n2. If $k$ is an integer coprime to $p$ and $n$ is a proper integer, then $kp - n$ is improper.\n\nThe first claim follows from the fact that the product of two proper integers is again proper. For the second, notice that $p$ does not divide $k n^{p-2}$, so\n\n$$\n(kp - n)^{p-1} \\equiv n^{p-1} - (p-1)k p n^{p-2} \\equiv 1 + k p n^{p-2} \\not\\equiv 1,\n$$\n\nso $kp - n$ is indeed improper (since $n$ is coprime to $p$, so is $kp - n$).\n\nSince $\\pm 1$ are both proper, letting $k \\in \\{1, 2\\}$ and $n = \\pm 1$ in (2) shows that $p \\pm 1$ and $2p \\pm 1$ are all improper, so each has an improper prime divisor by (1).\n\nSince $p \\ge 5$, the prime factors of $p \\pm 1$ are all less than $p-1$; and since 2 is the highest common factor of $p-1$ and $p+1$, the conclusion follows, provided that 2 is proper.\n\nOtherwise, look for an improper odd prime in the required range. To this end, notice that one of the numbers $2p \\pm 1$ is divisible by 3, so its prime factors are all less than $p-1$, for $p \\ge 5$; clearly, they are all odd, and the conclusion follows.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 13798,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $a > b$. Prove that\n$$\n\\frac{\\sqrt{a-b}}{b-1} \\ge \\frac{1}{\\sqrt{2}}\n$$\nwith equality at $(a, b) = \\left(\\frac{5}{2}, 2\\right)$.",
    "options": [],
    "answer": "See solution",
    "solution": "Using the given equation:\n$$\n\\begin{aligned}\n0 &\\ge (ab+1)^2 + (a+b)^2 - 2(a+b)(a^2 - ab + b^2 + 1) \\\\\n  &= a^2b^2 + 4ab + 1 - 2a^3 - 2b^3 + a^2 + b^2 - 2a - 2b \\\\\n  &= (a^2 - 2b + 1)(b^2 - 2a + 1)\n\\end{aligned}\n$$\nSince $a > b$, we get $a^2 - 2b + 1 > b^2 - 2b + 1 = (b-1)^2 \\ge 0$ and $b^2 - 2a + 1 \\le 0$. Therefore, $(b-1)^2 \\le 2(a-b)$, or\n$$\n\\frac{\\sqrt{a-b}}{b-1} \\ge \\frac{1}{\\sqrt{2}}\n$$\nEquality holds at $(a, b) = \\left(\\frac{5}{2}, 2\\right)$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 13799,
    "subject": "Mathematics (Olympiad)",
    "question": "A list of five two-digit positive integers is written in increasing order on a blackboard. Each of the five integers is a multiple of $3$, and each digit $0, 1, 2, 3, 4, 5, 6, 7, 8, 9$ appears exactly once on the blackboard. In how many ways can this be done? Note that a two-digit number cannot begin with the digit $0$.",
    "options": [],
    "answer": "See solution",
    "solution": "For an integer to be a multiple of $3$, the sum of its digits must be a multiple of $3$. So either both digits are multiples of $3$, or one digit is $1$ more than a multiple of $3$, and the other is $2$ more than a multiple of $3$.\n\nWe split the digits into $3$ groups, according to their remainder on division by $3$:\n\n- Group $0$: $0, 3, 6, 9$\n- Group $1$: $1, 4, 7$\n- Group $2$: $2, 5, 8$\n\nWe deal first with group $0$, which makes two two-digit numbers. One of these must end in $0$, and there are $3$ choices for this number. There are then two choices for the other number with the remaining two digits. This gives $6$ arrangements for this group.\n\nWe now consider the three numbers from groups $1$ and $2$. Each contains one digit from group $1$, paired with one from group $2$. We can pick one of $3$ digits to pair with $1$, then one of $2$ remaining digits to pair with $4$, then the last digit in group $2$ pairs with $7$. This gives $6$ pairings. Now each pair can be written in either order, so each pairing gives $8$ choices for the three numbers. So there are $6 \\times 8 = 48$ arrangements from these groups.\n\nThis gives a total of $6 \\times 48 = 288$ sets of two-digit numbers. Each can be written in increasing order in exactly one way, so there are $288$ ways to complete the task.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 13800,
    "subject": "Mathematics (Olympiad)",
    "question": "In the rectangular coordinate system $XOY$, the side of the rhombus $ABCD$ is $4$, and $|OB| = |OD| = 6$.\n\n![](images/Mathematical_Olympiad_in_China_2011-2014_p57_data_800c5f72bf.png)\n\n(1) Prove that $|OA| \\cdot |OC|$ is a constant.\n\n(2) When point $A$ is moving on the half circle $(x - 2)^2 + y^2 = 4$ with $2 \\le x \\le 4$, find the trace of $C$.",
    "options": [],
    "answer": "See solution",
    "solution": "(1) Since $|OB| = |OD|$ and $|AB| = |AD| = |BC| = |CD|$, then $O$, $A$, $C$ are collinear.\n\nConnecting $BD$, $BD$ is perpendicular to $AC$ and passes through its midpoint $K$. So we have:\n\n![](images/Mathematical_Olympiad_in_China_2011-2014_p57_data_1b7343d913.png)\n\n$$\n\\begin{align*}\n|OA| \\cdot |OC| &= (|OK| - |AK|)(|OK| + |AK|) \\\\\n&= |OK|^2 - |AK|^2 \\\\\n&= (|OB|^2 - |BK|^2) - (|AB|^2 - |BK|^2) \\\\\n&= |OB|^2 - |AB|^2 = 6^2 - 4^2 = 20 \\text{ (a constant).}\n\\end{align*}\n$$\n\n(2) Let $C(x, y)$, $A(2 + 2\\cos \\alpha, 2\\sin \\alpha)$, where\n\n$$\n\\alpha = \\angle XMA \\quad \\left( -\\frac{\\pi}{2} \\le \\alpha \\le \\frac{\\pi}{2} \\right).\n$$\n\nThen $\\angle XOC = \\frac{\\alpha}{2}$. As\n\n$$\n|OA|^2 = (2 + 2\\cos \\alpha)^2 + (2\\sin \\alpha)^2 = 8(1 + \\cos \\alpha) = 16\\cos^2 \\frac{\\alpha}{2},\n$$\n\nso $|OA| = 4\\cos \\frac{\\alpha}{2}$. Combining with the result in (1), $|OC| \\cos \\frac{\\alpha}{2} = 5$.\n\nThus, $x = |OC| \\cos \\frac{\\alpha}{2} = 5$, and $y = |OC| \\sin \\frac{\\alpha}{2} = 5\\tan \\frac{\\alpha}{2}$, with $y \\in [-5, 5]$.\n\nTherefore, the trace of point $C$ is the segment with endpoints $(5, 5)$ and $(5, -5)$.\n\n$\\Box$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 13801,
    "subject": "Mathematics (Olympiad)",
    "question": "Isosceles triangle $ABC$ has $AB = AC = 3\\sqrt{6}$, and a circle with radius $5\\sqrt{2}$ is tangent to line $AB$ at $B$ and to line $AC$ at $C$. What is the area of the circle that passes through vertices $A$, $B$, and $C$?\n\n(A) $24\\pi$  (B) $25\\pi$  (C) $26\\pi$  (D) $27\\pi$  (E) $28\\pi$",
    "options": [],
    "answer": "See solution",
    "solution": "Let $O$ be the center of the circle with radius $5\\sqrt{2}$. Consider the circle with diameter $\\overline{AO}$. Because $\\angle ABO$ and $\\angle ACO$ are right angles, the opposite angles of quadrilateral $ABOC$ are supplementary, and hence this quadrilateral is cyclic. Thus $O$ is also on the circle that passes through $A$, $B$, and $C$, and by symmetry $\\overline{AO}$ is a diameter. By the Pythagorean Theorem,\n\n$$\nAO = \\sqrt{(5\\sqrt{2})^2 + (3\\sqrt{6})^2} = 2\\sqrt{26},\n$$\n\nso the circle that passes through $A$, $B$, and $C$ has radius $\\sqrt{26}$ and area $26\\pi$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 13802,
    "subject": "Mathematics (Olympiad)",
    "question": "On top of a rectangular card with sides of length $1$ and $2 + \\sqrt{3}$, an identical card is placed so that two of their diagonals line up, as shown (AC, in this case).\n\n![](images/2024_AMC12A_Solutions_p9_data_6c8a4b4a6d.png)\n\nContinue the process, adding a third card to the second, and so on, lining up successive diagonals after rotating clockwise. In total, how many cards must be used until a vertex of a new card lands exactly on the vertex labeled B in the figure?\n\n(A) 6   (B) 8   (C) 10   (D) 12   (E) No new vertex will land on B.",
    "options": [],
    "answer": "See solution",
    "solution": "Note that the common diagonal is a diameter of the circle that will ultimately circumscribe the collection of rectangles. Each rectangle is composed of four chords of the circle, and the set of outermost chords will form a regular $n$-gon if a vertex lands on B. Because each new card contributes two sides of the polygon, the problem is asking for $\\frac{n}{2}$.\n\nLet $O$ be the center of the circle.\n\n![](images/2024_AMC12A_Solutions_p10_data_d057c82a4a.png)\n\nLet $\\theta = \\angle ACB$. Then the measure of minor arc $\\widehat{AB}$ is $2\\theta$, and $n = \\frac{360^\\circ}{2\\theta}$. Because $AB = 1$ and $BC = 2 + \\sqrt{3}$,\n\n$$\n\\tan \\theta = \\frac{1}{2 + \\sqrt{3}} = 2 - \\sqrt{3}.\n$$\n\nEach card is rotated through an angle $\\angle AOB = 2\\theta$ compared to the previous card. A double angle formula gives\n\n$$\n\\tan 2\\theta = \\frac{2 \\tan \\theta}{1 - \\tan^2 \\theta} = \\frac{2(2 - \\sqrt{3})}{1 - (2 - \\sqrt{3})^2} = \\frac{1}{\\sqrt{3}}.\n$$\n\nThis value is recognizable as $\\tan 30^\\circ$, so $2\\theta = 30^\\circ$ and $n = \\frac{360^\\circ}{30} = 12$. The polygon is a dodecagon. It will take just $\\frac{n}{2} = 6$ cards to complete the dodecagon and have a new vertex land on vertex B, as illustrated below.\n\n![](images/2024_AMC12A_Solutions_p10_data_707fd6cc52.png)",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 13803,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $\\triangle APD$ be an acute-angled triangle and let $B \\in (AP)$, $C \\in (PD)$ be two points. The diagonals of the quadrilateral $ABCD$ meet at the point $Q$. Denote by $H_1$ and $H_2$ the orthocenters of the triangles $APD$ and $BPC$ respectively. The circumcircles of the triangles $ABQ$ and $CDQ$ meet again at the point $X$ ($X \\neq Q$), and the circumcircles of the triangles $ADQ$ and $BCQ$ meet again at the point $Y$ ($Y \\neq Q$).\n\nShow that if the line $H_1H_2$ passes through the point $X$, then it also passes through the point $Y$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $AA'$, $DD'$ be altitudes in the triangle $APD$ and let $BB'$, $CC'$ be altitudes in the triangle $BPC$.\n\nWe have $\\angle XAC = \\angle XBD$ and $\\angle XCA = \\angle XDB$, and thus $\\triangle XAC \\sim \\triangle XBD$. We prove that the line $H_1H_2$ passes through the point $X$ if and only if the above triangles are congruent.\n\n![](images/RMC_2020_p42_data_a663089921.png)\n\nWe consider the circles $\\omega_1$, $\\omega_2$ of diameter $AC$ and $BD$ respectively. The points $H_1$, $H_2$ belong to the radical axis of these two circles, because $A, A' \\in \\omega_1$, $D, D' \\in \\omega_2$.\n\nSo $P_{\\omega_1}(H_1) = -AH_1 \\cdot A'H_1 = -DH_1 \\cdot D'H_1 = P_{\\omega_2}(H_1)$; and similar for $H_2$. Thus, $X \\in H_1H_2$ if and only if $P_{\\omega_1}(X) = P_{\\omega_2}(X)$.\n\nWe have $P_{\\omega_1}(X) = XM^2 - MC^2$ and $P_{\\omega_2}(X) = XN^2 - ND^2$, where $M$, $N$ are the centers of the circles $\\omega_1$ and $\\omega_2$ respectively. We notice that $\\frac{XM^2 - MC^2}{XN^2 - ND^2}$ is the square of the similarity ratio of the triangles $AXC$ and $BXD$, or $XM^2 - MC^2 = XN^2 - ND^2 = 0$. In the former case, the triangles $AXC$ and $BXD$ should be similar and right angles in $X$. But now the line $CD$ corresponds to the line $AB$ by a similarity of center $X$ and angle $90^\\circ$, so $AB \\perp CD$ – contradiction. Thus $\\frac{XM^2 - MC^2}{XN^2 - ND^2}$ is the square of the similarity ratio of the triangles $AXC$ and $BXD$.\n\nWe obtain that $X \\in H_1H_2$ if and only if the similarity ratio of the triangles $AXC$ and $BXD$ is $1$, that is $\\triangle AXC \\equiv \\triangle BXD$. But this is equivalent to $AC = BD$.\n\nIn the same manner, $Y \\in H_1H_2$ if and only if $AC = BD$, and we get the conclusion of the problem.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 13804,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $S(n)$ denote the sum of digits of a positive integer $n$.\n\n(a) Find the smallest number $n$ such that $S(n) \\cdot S(n+1) = 87$.\n\n(b) Find all positive integers $m$ for which there exists a positive integer $n$ such that $S(n) \\cdot S(n+1) \\cdots S(n+m) = 504$.",
    "options": [],
    "answer": "See solution",
    "solution": "**Answer:**\n\n(a) $2999$;\n\n(b) $m = 1, 2, 3$.\n\n**Solution.**\n\n(a) It is the same as problem 8-2 b).\n\n(b) If the last digit $a$ of a number $n$ is different from $9$, that is $n = \\overline{AA}$ and $a \\neq 9$, then $n+1 = \\overline{A(a+1)}$, and so $S(n+1) = S(n) + 1$. Now if $n = \\overline{Aa99\\ldots9}$, where the digit $a$ is less than $9$, then $n+1 = \\overline{A(a+1)00\\ldots0}$, which implies that\n\n$$\nS(n) - S(n+1) = (S(A) + a + 9k) - (S(A) + a + 1) = 9k - 1.\n$$\n\nFirst, we note that $504 = 7 \\cdot 8 \\cdot 9 = 2 \\cdot 2 \\cdot 2 \\cdot 3 \\cdot 3 \\cdot 7$.\n\nIf $m=1$, we have: $504 = 1 \\cdot 504$ and $504 - 1 = 503 = 9 \\cdot 56 - 1$, so we can take, for example, $n = \\overline{99\\ldots9}$. In this case $n+1 = \\overline{100\\ldots0}$, $S(n) = 504$, $S(n+1) = 1$, and $S(n) \\cdot S(n+1) = 504$, which means that $m=1$ satisfies the required conditions.\n\nFor $m=2$, we write $504 = 7 \\cdot 8 \\cdot 9$. Then for $n=7$ we have: $S(n) = 7$, $n+1 = 8$, $S(n+1) = 8$, $n+2 = 9$, $S(n+2) = 9$, and $S(n) \\cdot S(n+1) \\cdot S(n+2) = 504$. So, $m=2$ also satisfies our conditions.\n\nIf $m=3$, we again make use of the representation $504 = 7 \\cdot 8 \\cdot 9$: for $n=7$, we will have $n+1 = 8$, $S(n+1) = 8$, $n+2 = 9$, $S(n+2) = 9$, $n+3 = 10$, so $S(n+3) = 1$, which implies that $S(n) \\cdot S(n+1) \\cdot S(n+2) \\cdot S(n+3) = 504$. This means that $m=3$ satisfies the conditions too.\n\nNow let $m \\geq 4$ and suppose that it satisfies our requirements. Consider all divisors of the number $504$: $1, 2, 3, 4, 6, 7, 8, 9, 12, 14, 18, 21, 24, 28, 36, 42, 56, 63, 72, 84, 126, 168, 252, 504$.\n\nThere should be either $5$ consecutive numbers among them (but, obviously, there are no such numbers), or at least $3$ consecutive numbers and another divisor, which has difference of the form $9k-1$ with one of them. Consider possible cases.\n\n$1, 2, 3$: then $504 \\div 6 = 84$, and we need the existence of a divisor of $84$ with the sum of digits that is divisible by $9$, since $S(l) = 1$, and so $S(l-1) = 9k$. But $84$, obviously, doesn't have such divisors.\n\n$2, 3, 4$: then $504 \\div 12 = 42$, and we need the existence of a divisor of $42$ with the sum of digits of the form $9k+1$, since $S(l) = 2$, and so $S(l-1) = 9k+1$. But, again, $42$ doesn't have such divisors.\n\n$6, 7, 8$: this case is also impossible since $504$ is not divisible by $6 \\cdot 7 \\cdot 8 = 336$.\n\n$7, 8, 9$: we have that $504 = 7 \\cdot 8 \\cdot 9$, but this case is already considered.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 13805,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $ABCD$ be a quadrilateral inscribed in a circle $\\Omega$. The tangent to $\\Omega$ at $D$ intersects the rays $BA$ and $BC$ at points $E$ and $F$, respectively. A point $T$ is chosen inside triangle $ABC$ so that $TE \\parallel CD$ and $TF \\parallel AD$. Let $K \\neq D$ be a point on segment $DF$ such that $TD = TK$. Prove that the lines $AC$, $DT$, and $BK$ intersect at one point.",
    "options": [],
    "answer": "See solution",
    "solution": "Denote by $M$ and $N$ the intersections of $AC$ with $TE$ and $TF$, respectively. By the assumption, it follows that\n\n$$\n\\angle TNM = \\angle DAC = \\angle CDF = \\angle TEF,\n$$\n\nwhich implies that $MN$ is antiparallel to $EF$ with respect to $\\angle ETF$. Applying Reim's theorem, since $AD \\parallel TF$, it follows that $AMDE$ is a cyclic quadrilateral. Similarly, $CNDF$ is a cyclic quadrilateral. Then by angle chasing,\n\n$$\n\\begin{aligned}\n\\angle MDN &= 180^\\circ - \\angle MDE - \\angle NDF \\\\\n           &= 180^\\circ - \\angle BAC - \\angle BCA \\\\\n           &= \\angle ABC = 180^\\circ - \\angle ADC = 180^\\circ - \\angle TMN.\n\\end{aligned}\n$$\n\nHence, $TMDN$ is a cyclic quadrilateral. Let $l$ be the altitude from $T$ of triangle $TEF$, then $D$ and $K$ are symmetric with respect to $l$. On the other hand, since $MN$ is antiparallel to $EF$ with respect to $\\angle ETF$, $l$ passes through the circumcenter of triangle $TMN$, hence $K$ lies on $(TMN)$.\n\n![](images/Saudi_Arabia_booklet_2022_p31_data_743f0544aa.png)\n\nBy angle chasing,\n\n$$\n\\angle KNM = \\angle MDE = \\angle BAM\n$$\n\nthus $BA \\parallel NK$, and similarly, $BC \\parallel MK$. Let $S$ be the intersection of $BK$ and $MN$. It suffices to show that $S$ lies on $TD$. Since\n\n$$\n\\angle TMN = \\angle DCA, \\quad \\angle TNM = \\angle DAC, \\quad \\angle KNM = \\angle BAC, \\quad \\angle KMN = \\angle BCA,\n$$\n\nwe conclude that $\\angle KNM \\sim \\angle BCA$, $\\angle TMN \\sim \\angle DCA$, and combining with Thales's theorem, it implies that\n\n$$\n\\frac{SM}{SC} = \\frac{KM}{BC} = \\frac{MN}{AC} = \\frac{TM}{DC}\n$$\n\nwhich means $\\angle SMT \\sim \\angle SCD$. Then\n\n$$\n\\angle TSD = \\angle TSM + \\angle MSD = \\angle DSC + \\angle MSD = 180^\\circ\n$$\n\nas desired.\n\n![](images/Saudi_Arabia_booklet_2022_p31_data_e0be95b7a9.png)",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 13806,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $n > 2$ be the number of cities. What is the minimal integer $m$ such that it is possible to assign to each road between cities a number from $\\{1, 2, \\dots, m\\}$ so that the sum of the numbers assigned to the roads going out from each city (the city's \"ID\") is different for every city?",
    "options": [],
    "answer": "See solution",
    "solution": "We will show that $m = 3$ for every $n > 2$.\n\nFirst, we show that $m \\geq 3$. It is clear that $m \\geq 2$, so suppose $m = 2$. There are $n$ cities, and the possible sums (IDs) for each city are $n-1, n, n+1, \\dots, 2n-2$ (since the smallest sum is $1 + 1 + \\dots + 1 = n-1$ and the largest is $2 + 2 + \\dots + 2 = 2n-2$). Since there are exactly $n$ cities, every possible sum must be assigned to some city. This means there is a city with all roads labeled 1 and a city with all roads labeled 2, but these two cities are connected by a road, which is a contradiction.\n\nNow, we show that it is possible to assign numbers $1, 2, 3$ to the roads so that all city IDs are distinct. Suppose $n = 2k$. We construct the assignment in steps:\n- Assign 1 to all roads from the first city.\n- For the second city, assign 2 to the road going to the $(2k-1)$-th city, and 1 to the others.\n- For the third city, assign 2 to the roads going to the $(2k-1)$-th and $(2k-2)$-th cities, and 1 to the others.\n- Continue this process up to the $k$-th city.\n\nAfter these assignments, cities $1, 2, \\dots, k$ have IDs $2k-1, 2k, 2k+1, \\dots, 3k-2$, and cities $k+1, k+2, \\dots, 2k-1$ have IDs $k, k+1, \\dots, 2k-2, 2k-1$. Finally, assign 3 to every road between cities $k+1, k+2, \\dots, 2k-1$. All IDs are now distinct. For odd $n$, a similar argument applies.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 13807,
    "subject": "Mathematics (Olympiad)",
    "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": "See 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 \\leq d(n) < n$ for all $n > 2$. Furthermore, $d(1) = 1$ and $d(2) = 2$.\n\nWe first show that $f(2) = 2$. Let $m = f(2)$ and note that $2 = d(2) = f(f(2)) = f(m)$. If $m \\geq 2$, then let $m_0$ be the smallest positive integer satisfying $m_0 \\geq 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) \\geq m_0$, which implies that $m_0 = 2$. Therefore, if $m \\geq 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 $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 \\neq 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$\n\n*Remark.* Such a function exists and can be constructed inductively.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 13808,
    "subject": "Mathematics (Olympiad)",
    "question": "There are two boxes: one contains 29 distinguishable balls and the other is empty. In a move, we pick a box, choose some collection of balls from this box, and transfer them to the other box. Starting with the non-empty box, we repeatedly apply these moves, alternately picking boxes. What is the maximal possible number of moves if each ball collection can be selected at most once?",
    "options": [],
    "answer": "See solution",
    "solution": "Answer: $2^{29} - 2$.\n\nLet us show that the total number of moves cannot be $2^{29} - 1$. If the total number of moves is $2^{29} - 1$, then all non-empty subsets of a 29-element set should be selected. The total number of subsets containing a given element is $2^{28}$, which is even. Therefore, at the end of all moves, all balls should be in the box which was full at the beginning. On the other hand, since $2^{29} - 1$ is odd, the box which was empty at the beginning should contain at least one ball, a contradiction.\n\nSuppose that at the beginning, box A contains $n$ balls $\\{a_1, a_2, \\dots, a_n\\}$ and box B is empty. We show that the total number of moves can be equal to $2^{29} - 2$.\n\n**Solution 1:** Let us prove the following lemma.\n\n*Lemma:* Let $n > 2$. It is possible to make all moves containing a fixed ball $a$ and after these $2^{n-1}$ moves return to the initial position with empty B.\n\n*Proof:* We will use induction over $n$. If $n=3$, it can be done by applying moves $\\{a_1, a_2\\}$, $\\{a_1\\}$, $\\{a_1, a_3\\}$, $\\{a_1, a_2, a_3\\}$.\n\nNow suppose that for $n=k$ the inductive hypothesis is correct. Let us fix balls $a$ and $b$. By the inductive hypothesis, by applying all $2^{n-2}$ moves containing $a$ and not containing $b$, it is possible to return to the initial position. After that, again by the inductive hypothesis, by applying all $2^{n-2}$ moves containing both $a$ and $b$, it is possible to return to the initial position. Thus, by applying all $2^{n-1}$ moves containing $a$, we return to the initial position. The lemma is proved.\n\nBy the lemma, we can make all moves containing ball $a_1$ and return to the initial position, after that make all moves not containing $a_1$ but containing $a_2$ and return to the initial position, after that make all moves not containing $a_1, a_2$ but containing $a_3$ and return to the initial position, ..., and make all moves not containing $a_1, a_2, ..., a_{26}$ but containing $a_{27}$ and return to the initial position. Finally, by making two moves with ball groups $\\{a_{28}, a_{29}\\}$ and $\\{a_{28}\\}$, we will make all possible moves except the move $\\{a_{29}\\}$. Done.\n\n**Solution 2:** By induction over $n$, let us show that for each $n \\ge 2$, it is possible to make $2^n - 2$ moves such that the only not made move is $\\{a_1\\}$ and after making all $2^n - 2$ moves, all balls except $a_1$ are in box A and $a_1$ is in box B.\n\nIf $n=2$, it can be done by applying moves $\\{a_1, a_2\\}, \\{a_2\\}$.\n\nNow suppose that for $n=k$ the inductive hypothesis is correct and $n=k+1$. Let us separate ball $a_{k+1}$ and to the remaining $n=k$ balls apply $2^k-2$ moves existing by the inductive hypothesis. After that, make a move $\\{a_{k+1}\\}$ transferring ball $a_{k+1}$ to box B. By the inductive hypothesis, the box will contain two balls: $a_1$ and $a_{k+1}$. Let us make a move $\\{a_1, a_{k+1}\\}$. Finally, we can repeat all $2^k-2$ moves made at the beginning by adding ball $a_{k+1}$ to all of them. In total, we will make $2^k-2+2+2^k-2=2^{k+1}-2$ legal moves. Done.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 13809,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $9x^2 - 40x + 39 = p^n$ for some prime $p$ and some non-negative integer $n$.\n\nShow all integer solutions $x$ to this equation.",
    "options": [],
    "answer": "See solution",
    "solution": "The solutions are $x = -4$, $x = 1$, $x = 4$, and $x = 5$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 13810,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $a$, $b$, $c$ be integers such that $b$ is even. Suppose the equation $x^3 + a x^2 + b x + c = 0$ has roots $\\alpha$, $\\beta$, $\\gamma$ such that $\\alpha^2 = \\beta + \\gamma$. Prove that $\\alpha$ is an integer and $\\beta \\neq \\gamma$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $\\alpha$, $\\beta$, $\\gamma$ be the roots of the cubic. Then:\n\n$$\n\\alpha + \\beta + \\gamma = -a, \\quad \\alpha\\beta + \\beta\\gamma + \\gamma\\alpha = b, \\quad \\alpha\\beta\\gamma = -c.\n$$\n\nGiven $\\alpha^2 = \\beta + \\gamma = -a - \\alpha$. Also, $\\alpha \\neq 0$; otherwise $c = 0$, which contradicts $c$ being odd. Thus $\\beta\\gamma = -c/\\alpha$ and\n\n$$\nb = \\alpha(\\beta + \\gamma) + \\beta\\gamma = \\alpha^3 - \\frac{c}{\\alpha},\n$$\n\nor $\\alpha^4 - b\\alpha - c = 0$. Therefore, $\\alpha$ is a rational root of a monic polynomial with integer coefficients, so $\\alpha$ must be an integer.\n\nSuppose $\\beta = \\gamma$. Then $2\\beta = \\beta + \\gamma = -a - \\alpha$, so $\\beta$ is rational, hence integer. Then $2\\beta = -a - \\alpha$ implies $\\alpha$ is even, so $c = -\\alpha\\beta\\gamma$ is even, contradicting $c$ being odd. Thus $\\beta \\neq \\gamma$.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 13811,
    "subject": "Mathematics (Olympiad)",
    "question": "Consider the product $P_n = 1! \\cdot 2! \\cdot 3! \\cdot \\dots \\cdot n!$, where $n! = 1 \\cdot 2 \\cdot 3 \\cdot \\dots \\cdot n$ for every positive integer $n$.\n\n**a)** Find all possible values of positive integers $m$ such that $\\frac{P_{2020}}{m!}$ is a perfect square.\n\n**b)** Prove that there exist infinitely many values of $n$ such that $\\frac{P_n}{m!}$ is a perfect square for at least two positive integers $m$.",
    "options": [],
    "answer": "See solution",
    "solution": "a) First, note that\n\n$$\n\\begin{aligned}\nP_{2020} &= 1 \\cdot (1 \\cdot 2) \\cdot (1 \\cdot 2 \\cdot 3) \\cdots (1 \\cdot 2 \\cdot 3 \\cdots 2020) = 1^{2020} \\cdot 2^{2019} \\cdot 3^{2018} \\cdots 2019^2 \\cdot 2020 \\\\\n&= (1^{1010} \\cdot 2^{1009} \\cdot 3^{1009} \\cdots 2018 \\cdot 2019)^2 \\cdot (2 \\cdot 4 \\cdot 6 \\cdots 2020) \\\\\n&= (1^{1010} \\cdot 2^{1009} \\cdot 3^{1009} \\cdots 2018 \\cdot 2019)^2 \\cdot 2^{1010} \\cdot 1010!\n\\end{aligned}\n$$\n\nwhich implies that $m = 1010$ is a solution.\n\nAssume there is another solution $m$. Then, as $P_{2020}/1010!$ is a perfect square, we have that\n\n$$\n\\frac{P_{2020}/m!}{P_{2020}/1010!} = \\frac{1010!}{m!}\n$$\n\nis the square of a rational number.\n\nIf $m < 1009$, then $1010!/m!$ is an integer that is a multiple of $1009$, which is prime, but not a multiple of $1009^2$; therefore, $m$ is not a solution. It is clear that $m = 1009$ is not a solution either.\n\nIf $m \\ge 1013$, then $m!/1010!$ is a multiple of $1013$, which is prime; so, in order that it is a multiple of $1013^2$, we should have $m \\ge 2 \\cdot 1013 = 2026$. But $2027$ is prime and $P_{2020}$ does not have $2027$ as a factor; then $m < 2027$. Thus, the only possibility is $m = 2026$, which is not a solution, since $2026!/1010!$ is a multiple of $1019$, which is prime, but not a multiple of $1019^2$.\n\nThe remaining cases are $m = 1011$ and $m = 1012$. It is immediate to verify that they are not solutions, since $1011$ and $1011 \\cdot 1012$ are not perfect squares.\n\nb) Similarly as in a), note that if $n = 4t$ and $m = 2t$ for a positive integer $t$, then $P_n/m!$ is a perfect square, since\n\n$$\nP_n = 1^{4t} \\cdot 2^{4t-1} \\cdot 3^{4t-2} \\cdots (4t-1)^2 \\cdot 4t = (1^{2t} \\cdot 2^{2t-1} \\cdot 3^{2t-1} \\cdots (4t-1))^2 \\cdot 2^{2t} \\cdot (2t)!\n$$\n\nConsider $n = 8(k^2 + k)$. Then, as we have already shown, for $m = 4(k^2 + k)$ we have a solution. We will now show that $P_n/(m+1)!$ is also a perfect square. Note that $m+1 = 4k^2 + 4k + 1 = (2k+1)^2$, and\n\n$$\n\\begin{aligned}\n\\frac{P_n}{(m+1)!} &= \\frac{1}{m+1} \\cdot \\frac{P_n}{m!} = \\frac{1}{(2k+1)^2} \\cdot (1^m \\cdot 2^{m-1} \\cdots (2k+1)^{m-k} \\cdots (n-1))^2 \\cdot 2^m \\\\\n&= (1^m \\cdot 2^{m-1} \\cdots (2k+1)^{m-k-1} \\cdots (n-1))^2 \\cdot 2^m\n\\end{aligned}\n$$\n\nwhich is an integer, since $m-k-1 = 4k^2 + 3k - 1 > 0$ for $k \\ge 1$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 13812,
    "subject": "Mathematics (Olympiad)",
    "question": "Determine all triples $\\left(x, y, z\\right)$ of positive numbers satisfying the system of equations\n\n$$\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$$",
    "options": [],
    "answer": "See solution",
    "solution": "For any integer $k \\ge 3$ and any $x \\ge 0$ we have\n\n$$\n2x^k \\ge [(k-1)x - (k-2)](x^2 + 1). \\quad (0)\n$$\n\nTo see this, observe that, by the AM-GM inequality,\n\n$$\nx^k + x^k + \\underbrace{x + x + \\dots + x}_{(k-3)\\text{ times}} \\ge (k-1)x^3,\n$$\n\nand add it to\n\n$$\n(k-2)(x^2 - 2x + 1) \\ge 0.\n$$\n\nNote that we have equality if and only if $x=1$.\n\nTherefore, for $x$, $y$, $z$ satisfying the system of equations, we have\n\n$$\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$$\n\nor\n\n$$\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$$\n\nNow suppose $x \\ge \\max\\{y, z\\}$. Then from the second inequality above we infer that $y \\le z$ and\n\n$$\n2(y-x)(x^2+1) + (x-z)(x+z) \\le (z-x)(2(x^2+1)-(x+z)) \\le (z-x)(2x^2-2x+2) \\le 0,\n$$\n\nwhich, by the first inequality, implies $x = y = z$.\n\nIf $y \\ge \\max\\{x, z\\}$, then $z \\le x$ by the third inequality and\n\n$$\n3(z - y)(y^2 + 1) + 2(y - x)(y + x) \\le (x - y)(3(y^2 + 1) - 2y - 2x) \\le (x - y)(3y^2 - 4y + 3) \\le 0,\n$$\n\nhence, by the second inequality, $x = y = z$.\n\nFinally, if $z \\ge \\max\\{x, y\\}$, then $x \\le y$ by the first estimate and, as previously,\n\n$$\n4(x - z)(z^2 + 1) + 3(z - y)(z + y) \\le (y - z)(4z^2 - 6z + 4) \\le 0,\n$$\n\nwhich again implies $x = y = z$. Thus we have equality in (0) and hence $x = y = z = 1$. We easily check that this is the solution to the system.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 13813,
    "subject": "Mathematics (Olympiad)",
    "question": "Suppose $n$ students compete in $m$ disciplines, with $k$ students in each discipline. Each pair of students competes together in exactly $3$ disciplines, and each triple of students competes together in exactly $2$ disciplines. Find all possible values of $n$, $k$, and $m$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $m$ be the number of disciplines.\n\nThe number of pairs of students over all disciplines can be counted in two ways. On the one hand, the number of pairs is equal to the product of the number of all pairs and the number of times every pair of students competes together, which is $\\binom{n}{2} \\cdot 3$. On the other hand, it must be equal to the product of the number of pairs competing in each discipline and the number of disciplines, which is $\\binom{k}{2} \\cdot m$. So,\n\n$$\n3n(n-1) = mk(k-1). \\qquad (11)\n$$\n\nSimilarly, we can count the number of triples in two ways. On the one hand, it is equal to $\\binom{n}{3} \\cdot 2$, on the other hand, it must be $\\binom{k}{3} \\cdot m$. So,\n\n$$\n2n(n-1)(n-2) = mk(k-1)(k-2).\n$$\n\nSince $k, n \\ge 3$, we can divide one equation by the other and get\n\n$$\n\\frac{2(n-2)}{3} = k-2.\n$$\n\nThis implies that $n-2$ is divisible by $3$, so we can write $n = 3l + 2$, where $l \\in \\mathbb{N}$. Inserting this back into the equation, we get $k = 2l + 2$. We can then use (11) to get\n\n$$\nm = \\frac{3n(n-1)}{k(k-1)} = \\frac{3(3l+2)(3l+1)}{(2l+2)(2l+1)}.\n$$\n\nHowever, $2l+1$ divides $3$, which is only possible when $l=1$. Consequently, $n=5$, $k=4$, and $m=5$.\n\nFinally, we must check that the required arrangement of students is indeed possible for $n=5$ and $k=4$. Denote the students by $A_1, A_2, \\dots, A_5$. Here is one possible arrangement that satisfies the conditions:\n\n$\\{A_1, A_2, A_3, A_4\\}$, $\\{A_1, A_2, A_3, A_5\\}$, $\\{A_1, A_2, A_4, A_5\\}$, $\\{A_1, A_3, A_4, A_5\\}$, $\\{A_2, A_3, A_4, A_5\\}$.\n\nThus, the only solution is $n=5$, $k=4$, $m=5$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 13814,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $P$ be a polynomial with integer coefficients of degree $d$. For the set $A = \\{a_1, a_2, \\dots, a_k\\}$ of positive integers, denote $S(A) = P(a_1) + P(a_2) + \\dots + P(a_k)$. The positive integers $m, n$ are such that $m^{d+1} \\mid n$. Prove that the set $\\{1, 2, \\dots, n\\}$ can be split into $m$ disjoint subsets $A_1, A_2, \\dots, A_m$ of equal size such that $S(A_1) = S(A_2) = \\dots = S(A_m)$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $n = k m^{d+1}$. We will construct the desired splitting as follows. For each $x \\in \\{1, 2, \\dots, n\\}$, write $x - 1$ in the form:\n\n$$\nx - 1 = c m^{d+1} + x_0 + x_1 m + x_2 m^2 + \\dots + x_d m^d,\n$$\n\nwhere $c \\ge 0$ is an integer, and $x_i \\in \\{0, 1, \\dots, m-1\\}$ for $i = 0, \\dots, d$; that is, write the remainder of $x-1$ modulo $m^{d+1}$ in base $m$. Assign $x$ to subset $A_j$ ($j \\in \\{1, 2, \\dots, m\\}$) if $x_0 + x_1 + x_2 + \\dots + x_d \\equiv j \\pmod m$.\n\nWe claim this splitting satisfies the condition. It suffices to prove the statement for polynomials of the form $P(x) = x^t$, $t = 0, \\dots, d$.\n\nWe have\n\n$$\nS(A_j) = \\sum_{c=0}^{k-1} \\sum_j (c m^{d+1} + x_0 + x_1 m + x_2 m^2 + \\dots + x_d m^d)^t,\n$$\n\nwhere $\\sum_j$ denotes the sum over all tuples $x_0, \\dots, x_d$ with $x_0 + x_1 + \\dots + x_d \\equiv j \\pmod m$, $x_i \\in \\{0, 1, \\dots, m-1\\}$. We show the inner sum does not depend on $j$. Let $z = 1 + c m^{d+1}$, then expanding:\n\n$$\n\\sum_j (c m^{d+1} + x_0 + x_1 m + x_2 m^2 + \\dots + x_d m^d)^t = \\sum_d \\frac{t!}{u! t_0! t_1! \\dots t_d!} z^u m^{t_1 + 2 t_2 + \\dots + d t_d} \\sum_j x_0^{t_0} x_1^{t_1} \\dots x_d^{t_d},\n$$\n\nwhere $\\sum_d$ is over all non-negative integers $u, t_0, \\dots, t_d$ with $u + t_0 + \\dots + t_d = t$. Since $t_0 + \\dots + t_d \\le t < d+1$, at least one $t_i$ is zero. Suppose $t_0 = 0$, then\n\n$$\n\\sum_j x_0^{t_0} x_1^{t_1} \\dots x_d^{t_d} = \\sum_j x_1^{t_1} \\dots x_d^{t_d} = \\sum_{x_1, \\dots, x_d = 0}^{m-1} x_1^{t_1} \\dots x_d^{t_d},\n$$\n\nbecause for any tuple $x_1, \\dots, x_d$ in $\\{0, 1, \\dots, m-1\\}$, there is a unique $x_0$ such that $x_0 + x_1 + \\dots + x_d \\equiv j \\pmod m$. The last sum is independent of $j$, as required.",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 13815,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all pairs $(a, b)$ of real numbers such that $a^2 + b^2 = 25$, for which $ab + a + b$ attains the smallest possible value.",
    "options": [],
    "answer": "See solution",
    "solution": "Consider $(a + b + 1)^2 \\geq 0$, which expands to $a^2 + b^2 + 1 + 2ab + 2a + 2b \\geq 0$. This implies $2(ab + a + b) \\geq - (a^2 + b^2) - 1$, so $ab + a + b \\geq -13$.\n\nEquality holds when $a + b + 1 = 0$, i.e., $b = -a - 1$. Substitute into $a^2 + b^2 = 25$:\n\n$$\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$$\n\nThus, the solutions are $(a, b) = (-4, 3)$ and $(a, b) = (3, -4)$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 13816,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $a$, $b$, $c$ be positive real numbers such that $a^2 + b^2 + c^2 < 2(a + b + c)$. Prove that\n\n$$\n3abc < 4(a + b + c).\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "The Cauchy-Schwarz inequality gives\n$$\n(a + b + c)^2 \\leq 3(a^2 + b^2 + c^2).\n$$\nThus, $(a + b + c) < 6$, and hence\n$$\n\\frac{(a + b + c)^3}{9} < 4(a + b + c).\n$$\nNow, the AM-GM inequality gives\n$$\n(a + b + c)^3 \\geq 27abc.\n$$\nThus,\n$$\n3abc \\leq \\frac{(a + b + c)^3}{9} < 4(a + b + c).\n$$\nAlternatively, since $a + b + c < 6$, we have $a^2 + b^2 + c^2 < 2(a + b + c) < 12$. Thus,\n$$\n(a^2 + b^2 + c^2)(a + b + c) < 12(a + b + c).\n$$\nUsing the AM-GM inequality,\n$$\na^2 + b^2 + c^2 \\geq 3(abc)^{2/3}, \\quad a + b + c \\geq 3(abc)^{1/3}.\n$$\nThus,\n$$\n9abc \\leq (a^2 + b^2 + c^2)(a + b + c) < 12(a + b + c),\n$$\ngiving the required inequality.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 13817,
    "subject": "Mathematics (Olympiad)",
    "question": "Пусть $P(x)$ — многочлен. Докажите, что если для любых вещественных $x$ выполнено равенство\n$$\nP(a_1x + b_1) + P(a_2x + b_2) = P(a_3x + b_3),\n$$\nгде $a_1, a_2, a_3, b_1, b_2, b_3$ — вещественные числа, то у $P(x)$ есть хотя бы один действительный корень.",
    "options": [],
    "answer": "See solution",
    "solution": "Предположим, что $P(x)$ не имеет действительных корней. Тогда $P(x)$ имеет чётную степень, не меньшую 2. Действительно, любой многочлен нечётной степени имеет хотя бы один действительный корень, а если $P(x) = \\text{const}$, то из условия получаем, что $P(x) = 0$.\n\nТак как $P(x)$ не имеет действительных корней, то он принимает значения одного знака. Будем считать, что $P(x)$ принимает только положительные значения (иначе умножим $P(x)$ на $-1$), то есть для любого $x$ выполняется $P(x) > 0$. Так как $P(x)$ имеет чётную степень, найдётся точка $t_0$, в которой достигается (глобальный нестрогий) минимум $P(x)$, то есть для любого $x$ выполняется неравенство $P(x) \\geq P(t_0) = A > 0$. Рассмотрим $x_0$ такое, что $t_0 = a_3x_0 + b_3$. Тогда\n$$\nP(a_1x_0 + b_1) + P(a_2x_0 + b_2) \\geq 2A > A = P(t_0) = P(a_3x_0 + b_3).\n$$\nПолучили противоречие. Значит, $P(x)$ имеет хотя бы один действительный корень.\n\nПусть $a_1 \\neq a_3$; тогда существует такое $x_0$, что $a_1x_0 + b_1 = a_3x_0 + b_3$. Подставляя $x = x_0$ в данное равенство, получаем после сокращения $P(a_2x_0 + b_2) = 0$, то есть у $P(x)$ есть корень. Аналогично рассматривается случай $a_2 \\neq a_3$.\n\nОстался лишь случай $a_1 = a_2 = a_3 = a \\neq 0$. Если $P(x) = 0$, утверждение задачи очевидно. Иначе пусть $p_0 \\neq 0$ — старший коэффициент многочлена $P(x)$, а $n$ — его степень. Тогда старшие коэффициенты многочленов в левой и правой частях данного равенства есть $p_0(a^n + a^n)$ и $p_0a^n$, то есть они различны. Это невозможно.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 13818,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $x_1, x_2, \\dots, x_n$ be positive real numbers and $n \\ge 3$. Prove the inequality\n$$\n\\frac{x_1 x_3}{x_1 x_3 + x_2 x_4} + \\frac{x_2 x_4}{x_2 x_4 + x_3 x_5} + \\dots + \\frac{x_{n-1} x_1}{x_{n-1} x_1 + x_n x_2} + \\frac{x_n x_2}{x_n x_2 + x_1 x_3} \\le n-1\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "In each of the fractions $\\frac{x_{i-1}x_{i+1}}{x_{i-1}x_{i+1} + x_i x_{i+2}}$ (where $i \\in \\{1, 2, \\dots, n\\}$, $x_0 = x_n$, and $x_{n+1} = x_1$), divide numerator and denominator by $x_i x_{i+2}$. Let $y_i = \\frac{x_{i-1}x_{i+1}}{x_i x_{i+2}}$. The inequality becomes:\n$$\n\\frac{y_1}{y_1+1} + \\frac{y_2}{y_2+1} + \\dots + \\frac{y_n}{y_n+1} \\le n-1.\n$$\nNote that $y_1 y_2 \\cdots y_n = 1$ and all $y_i > 0$. Transforming the inequality:\n$$\nn - \\left( \\frac{y_1}{y_1+1} + \\frac{y_2}{y_2+1} + \\dots + \\frac{y_n}{y_n+1} \\right) = \\frac{1}{y_1+1} + \\frac{1}{y_2+1} + \\dots + \\frac{1}{y_n+1} \\geq 1.\n$$\nWe prove this by induction. For $n=3$, with $y_1 y_2 y_3 = 1$, set $y_3 = \\frac{1}{y_1 y_2}$:\n$$\n\\frac{1}{y_1+1} + \\frac{1}{y_2+1} + \\frac{y_1 y_2}{1 + y_1 y_2} \\geq 1.\n$$\nUsing $\\frac{1}{y_1+1} + \\frac{1}{y_2+1} \\geq \\frac{1}{y_1 y_2 + 1}$ (which can be shown by direct calculation), the result follows.\n\nAssume the result holds for $n$. For $n+1$:\n$$\n\\frac{1}{y_1+1} + \\dots + \\frac{1}{y_{n+1}+1} \\geq \\frac{1}{y_1+1} + \\dots + \\frac{1}{y_{n-1}+1} + \\frac{1}{1+y_n} + \\frac{1}{1+y_{n+1}} \\geq \\frac{1}{y_1+1} + \\dots + \\frac{1}{y_{n-1}+1} + \\frac{1}{1+y_n y_{n+1}}\n$$\nwhere we used $\\frac{1}{y_n+1} + \\frac{1}{y_{n+1}+1} \\geq \\frac{1}{y_n y_{n+1}+1}$. Since $y_1 y_2 \\cdots y_{n+1} = 1$, let $t_i = y_i$ for $i < n$, $t_n = y_n y_{n+1}$, so $t_1 t_2 \\cdots t_n = 1$. By the inductive hypothesis, $\\frac{1}{1+t_1} + \\dots + \\frac{1}{1+t_n} \\geq 1$. Thus, the original inequality holds for all $n \\ge 3$.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 13819,
    "subject": "Mathematics (Olympiad)",
    "question": "Call a fraction $\\frac{a}{b}$, not necessarily in simplest form, *special* if $a$ and $b$ are positive integers whose sum is $15$. How many distinct integers can be written as the sum of two, not necessarily different, special fractions?\n\n![](path/to/file.png)\n\n(A) 9  (B) 10  (C) 11  (D) 12  (E) 13",
    "options": [],
    "answer": "See solution",
    "solution": "The 14 special fractions are:\n\n$$\n\\frac{1}{14},\\ \\frac{2}{13},\\ \\frac{3}{12},\\ \\frac{4}{11},\\ \\frac{5}{10},\\ \\frac{6}{9},\\ \\frac{7}{8},\\ \\frac{8}{7},\\ \\frac{9}{6},\\ \\frac{10}{5},\\ \\frac{11}{4},\\ \\frac{12}{3},\\ \\frac{13}{2},\\ \\frac{14}{1}\n$$\n\nwhich simplify to:\n\n$$\n\\frac{1}{14},\\ \\frac{2}{13},\\ \\frac{1}{4},\\ \\frac{4}{11},\\ \\frac{1}{2},\\ \\frac{2}{3},\\ \\frac{7}{8},\\ \\frac{8}{7},\\ \\frac{3}{2},\\ 2,\\ \\frac{11}{4},\\ 4,\\ \\frac{13}{2},\\ 14\n$$\n\nThe integers 2, 4, and 14 can be doubled to produce 4, 8, and 28, or paired to produce sums of 6, 16, and 18. Fractions with denominators of 2 ($\\frac{1}{2}$, $\\frac{3}{2}$, $\\frac{13}{2}$) can be doubled to produce 1, 3, and 13, or paired to produce sums of 2, 7, and 8. Fractions with denominators of 4 ($\\frac{1}{4}$ and $\\frac{11}{4}$) sum to 3. The remaining fractions ($\\frac{1}{14}$, $\\frac{2}{13}$, $\\frac{4}{11}$, $\\frac{2}{3}$, $\\frac{7}{8}$, $\\frac{8}{7}$) cannot be added to any fraction in the list to give an integer.\n\nThis analysis produces 13 integer sums, but 3 and 8 appear twice. Excluding duplicates leaves 11 distinct integers that can be written as the sum of two special fractions: $1, 2, 3, 4, 6, 7, 8, 13, 16, 18, 28$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 13820,
    "subject": "Mathematics (Olympiad)",
    "question": "For how many ordered pairs $\\left(b, c\\right)$ of positive integers does neither $x^2 + b x + c = 0$ nor $x^2 + c x + b = 0$ have two distinct real solutions?",
    "options": [],
    "answer": "See solution",
    "solution": "The equation $x^2 + b x + c = 0$ fails to have two distinct real solutions precisely when $b^2 - 4c \\leq 0$. Similarly, $x^2 + c x + b = 0$ fails to have two distinct real solutions when $c^2 - 4b \\leq 0$. Thus, the condition is $b^2 \\leq 4c$ and $c^2 \\leq 4b$.\n\nLet us check possible values for $b$:\n\n- For $b = 1$: $1^2 \\leq 4c \\implies c \\geq 1$. Also, $c^2 \\leq 4 \\implies c = 1$ or $2$.\n- For $b = 2$: $4 \\leq 4c \\implies c \\geq 1$. $c^2 \\leq 8 \\implies c = 1$ or $2$.\n- For $b = 3$: $9 \\leq 4c \\implies c \\geq 3$. $c^2 \\leq 12 \\implies c = 3$.\n- For $b = 4$: $16 \\leq 4c \\implies c \\geq 4$. $c^2 \\leq 16 \\implies c = 4$.\n\nSo, the ordered pairs are $(1, 1)$, $(1, 2)$, $(2, 1)$, $(2, 2)$, $(3, 3)$, $(4, 4)$, totaling $6$ pairs.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 13821,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $f: \\mathbb{R} \\to \\mathbb{R}$ be a strictly increasing function, and $f(x) > x$ for every $x$. Assume that\n$$\nf(x) + f^{-1}(x) = 2x\n$$\nfor all $x \\in \\mathbb{R}$. Show that $f(x) = x + f(0)$ for all $x \\in \\mathbb{R}$.",
    "options": [],
    "answer": "See solution",
    "solution": "Set $g(x) = f(x) - x$. Then $g(x) > 0$ for all $x$. Let $g(x) = c$ for some $x$. So $f(x) = x + c$. Then $x = f^{-1}(x + c)$, $2x + 2c = f(x + c) + f^{-1}(x + c) = f(x + c) + x$, and $g(x + c) = f(x + c) - (x + c) = c$. By induction, $g(x + kc) = c$ for $k \\in \\mathbb{N}$.\n\nWe show that $g$ only assumes one value. Since $g(0) = f(0)$, this value then has to be $f(0)$. Assume $g(x_0) = a$ for some $x_0$ and let $0 < b < a$. Set $d = a - b$. Now for $x_0 \\leq x' < x_0 + d$ we have $f(x') \\geq f(x_0)$ and $g(x') = f(x') - x' > f(x_0) - (x_0 + d) = g(x_0) - d = a - d = b$. So $g$ does not take the value $b$ in the interval $[x_0, x_0 + d]$. Now assume that $g$ does not take the value $b$ in the interval $[x_0 + (k-1)d, x_0 + kd]$ for some $k \\geq 1$. Let $x' \\in [x_0 + kd, x_0 + (k+1)d]$. Then $x' + b \\in [x_0 + a + (k-1)d, x_0 + a + kd]$. But $g(x_0 + a) = a$, and the induction hypothesis, which can be applied to the situation where the $x$-axis has been shifted by $a$, shows that $f(x' + b)$ cannot be $b$. But if $f(x') = b$, then $f(x' + b) = b$. So $f$ does not take the value $b$ in $[x_0 + kd, x_0 + (k+1)d]$. By induction, $f$ does not take the value $b$ for any $x > x_0$. If $f(x_1) = b$, then $f(x_1 + kb) = b$ for all $k$, which clearly leads to a contradiction.\n\nThe assumption $f(x) > x$ might be removed, but the proof might be somewhat more complicated.\n\n![](images/Baltic_Way_SHL_2009-11_13-16_p245_data_8ed0040e6c.png)",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 13822,
    "subject": "Mathematics (Olympiad)",
    "question": "ABCD 為一凸四邊形。點 $E, F, G, H$ 分別在線段 $AB, BC, CD, DA$ 上，使得延長後的直線 $AB, FH, CD$ 三線共點，直線 $BC, EG, AD$ 亦三線共點。令 $O$ 為 $EG$ 和 $FH$ 的交點。考慮以下四個四邊形：\n\n$AHOE$, $BEOF$, $CFOG$ 及 $DGOH$。\n\n試證明：若這四個四邊形的其中三個是圓外切四邊形，則第四個必定也是。\n\n註：一個四邊形稱為圓外切四邊形的條件是存在一個圓在四邊形內使得該圓與四條邊都相切。",
    "options": [],
    "answer": "See solution",
    "solution": "令 $X$ 為 $AB, FH, CD$ 的交點，$Y$ 為 $BC, EG, AD$ 的交點。令 $L_1, L_2$ 分別為 $\\angle AXO, \\angle OXC$ 的角平分線，且令 $K_1, K_2$ 分別為 $\\angle AYO, \\angle OYC$ 的角平分線。\n\n顯然四個四邊形的內切圓（如果存在的話）的圓心必須分別落在 $L_i, K_j$ 的交點上（$i, j = 1, 2$）。令 $P_{ij}$ 表示 $L_i$ 和 $K_j$ 的交點。我們令\n\n$$\n\\alpha_1 = \\frac{1}{2} \\angle AXO, \\quad \\alpha_2 = \\frac{1}{2} \\angle OXC, \\quad \\beta_1 = \\frac{1}{2} \\angle AYO, \\quad \\beta_2 = \\frac{1}{2} \\angle OYC\n$$\n\n四個四邊形是圓外切四邊形的條件等價於 $P_{ij}$ 到一雙對邊的距離等於到另一雙對邊的距離，這可以寫成\n\n$$\nXP_{ij} \\sin \\alpha_i = YP_{ij} \\sin \\beta_j\n$$\n\n現在令 $a_{ij} = XP_{ij} \\sin \\alpha_i$, $b_{ij} = YP_{ij} \\sin \\beta_j$，我們要證明若對於 $(i, j) = (1, 1), (1, 2), (2, 1), (2, 2)$ 的其中三者有 $a_{ij} = b_{ij}$，則對第四個亦成立。\n\n注意到，我們有\n\n$$\n\\frac{XP_{11}}{XP_{12}} \\cdot \\frac{YP_{12}}{YP_{22}} \\cdot \\frac{XP_{22}}{XP_{21}} \\cdot \\frac{YP_{21}}{YP_{11}} = \\frac{S_{XYP_{11}}}{S_{XYP_{12}}} \\cdot \\frac{S_{XYP_{12}}}{S_{XYP_{22}}} \\cdot \\frac{S_{XYP_{22}}}{S_{XYP_{21}}} \\cdot \\frac{S_{XYP_{21}}}{S_{XYP_{11}}} = 1\n$$\n\n其中 $S_{QRS}$ 表示三角形 $QRS$ 的面積，從而我們有\n\n$$\n1 = \\frac{a_{11}}{a_{12}} \\cdot \\frac{b_{12}}{b_{22}} \\cdot \\frac{a_{22}}{a_{21}} \\cdot \\frac{b_{21}}{b_{11}} = \\frac{a_{11}}{b_{11}} \\cdot \\frac{a_{22}}{b_{22}} \\cdot \\frac{b_{12}}{a_{12}} \\cdot \\frac{b_{21}}{a_{21}}\n$$\n\n顯然若上式右邊的其中三個比值是 $1$，則第四個亦是 $1$，證畢！",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 13823,
    "subject": "Mathematics (Olympiad)",
    "question": "Point $M$ on side $AB$ of quadrilateral $ABCD$ is such that quadrilaterals $AMCD$ and $BMDC$ are circumscribed around circles centered at $O_1$ and $O_2$ respectively. Line $O_1O_2$ cuts an isosceles triangle with vertex $M$ from angle $CMD$. Prove that $ABCD$ is a cyclic quadrilateral.",
    "options": [],
    "answer": "See solution",
    "solution": "If $AB \\parallel CD$, then the incircles of $AMCD$ and $BMDC$ have equal radii; now the problem conditions imply that the whole picture is symmetric about the perpendicular from $M$ to $O_1O_2$, and hence $ABCD$ is an isosceles trapezoid (or a rectangle). The conclusion in this case is true.\n\n![](images/Saudi_Arabia_booklet_2022_p19_data_bd97de3a25.png)\n\nNow suppose that the lines $AB$ and $CD$ meet at a point $K$; we may assume that $A$ lies between $K$ and $B$. The points $O_1$ and $O_2$ lie on the bisector of the angle $BKC$. By the problem condition, this angle bisector forms equal angles with the lines $CM$ and $DM$; this yields $\\angle DMK = \\angle KCM$. Since $O_1$ and $O_2$ are the incenter of $KMC$ and an excenter of $KDM$, respectively, we have\n\n$$\n\\angle DO_2K = \\frac{1}{2}\\angle DMK = \\frac{1}{2}\\angle KCM = \\angle DCO_1,\n$$\n\nso the quadrilateral $CDO_1O_2$ is cyclic. Next, the same points are an excenter of $AKD$ and the incenter of $KBC$, respectively, so\n\n$$\n\\angle KAD = 2\\angle KO_1D = 2\\angle DCO_2 = \\angle KCB;\n$$\n\nthis implies the desired cyclicity of the quadrilateral $ABCD$. $\\square$",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 13824,
    "subject": "Mathematics (Olympiad)",
    "question": "Four families have two children each, and all eight children are born after the year 1989. All four youngest siblings are born in the same year, and the sum of the digits of the year is equal to the product of its nonzero digits. The differences in ages between the siblings of each family are perfect squares. Find the birth years of the four eldest siblings in each family, given that their ages are all different.",
    "options": [],
    "answer": "See solution",
    "solution": "There is no digit $a$ such that $1 + 9 + 9 + a = 1 \\times 9 \\times 9 \\times a$, so the youngest must be born after 2000, i.e., in $200a$ or $201a$. For $200a$, $2 + 0 + 0 + a = 2 \\times a$ gives $a = 2$, so $2002$ is possible. For $201a$, $2 + 0 + 1 + a = 2 \\times 1 \\times a = 2a$ gives $a = 3$, so $2013$ is possible.\n\nThe possible age differences are $1$, $4$, $9$, and $16$. Since $2002 - 1989 = 13 < 16$, $2013$ is the only valid year for the youngest. Thus, the eldest siblings are born in $2012$, $2009$, $2004$, and $1997$, respectively.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 13825,
    "subject": "Mathematics (Olympiad)",
    "question": "Given a positive integer $n$, let $b(n)$ denote the number of positive integers whose binary representations occur as blocks of consecutive digits in the binary expansion of $n$. For example, $b(13) = 6$ because $13 = 1101_2$, which contains as consecutive blocks the binary representations of $13 = 1101_2$, $6 = 110_2$, $5 = 101_2$, $3 = 11_2$, $2 = 10_2$, and $1 = 1_2$.\n\nShow that if $n \\leq 2500$, then $b(n) \\leq 39$, and determine the values of $n$ for which equality holds.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $l(n)$ denote the number of digits in the binary representation of $n$. Since $2500 \\leq 4096 = 2^{12}$, we have $l(n) \\leq 12$ for all $n \\leq 2500$.\n\nWe divide into cases according to $l(n)$. If $l(n) = 12$, then the first digit is $1$ and (since $2500 < 2^{11} + 2^9$) the next two must be $0$.\n\nA \"substring\" of $n$ means a number that appears as a consecutive block in the base 2 representation of $n$, starting with $1$.\n\nThere is only one substring of length $12$, one of length $11$, and one of length $10$ in $n$, since $l(n) = 12$, because such substrings can't begin at the second or third digit. Similarly, there are two or fewer distinct substrings of length $9$, three or fewer of length $8$, and so on, and seven or fewer of length $4$.\n\nAlso, there are only $2^{k-1}$ different numbers of length $k$ (since we must start with $1$ and have $k-1$ choices of $0$ or $1$ thereafter). So there are one or fewer distinct substrings of length $1$, two or fewer of length $2$, and four or fewer of length $3$.\n\nTherefore, in the case that $l(n) = 12$, counting substrings of each possible length:\n\n$$\nb(n) \\leq 1 + 1 + 1 + 2 + 3 + 4 + 5 + 6 + 7 + 4 + 2 + 1 = 37 \\leq 39.\n$$\n\nIf $l(n) \\leq 10$, there are one or fewer substrings of length $10$, two or fewer of length $9$, and so on. As before, there are one or fewer distinct substrings of length $1$, two or fewer of length $2$, and four or fewer of length $3$.\n\nSo, in this case,\n\n$$\nb(n) \\leq 1 + 2 + 3 + 4 + 5 + 6 + 7 + 4 + 2 + 1 = 35 \\leq 39.\n$$\n\nNow consider $l(n) = 11$. Using previous estimates:\n\n$$\nb(n) \\leq 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 4 + 2 + 1 = 43.\n$$\n\nHowever, if there is a zero as the $d$-th digit from the right, that reduces the maximum number of substrings of length $e$ (for all $4 \\leq e \\leq d$) by one. Any zero in the leftmost four digits reduces the maximum number of substrings by at least $5$ (so $b(n) \\leq 43 - 5 = 38$). If there are three zeros among the leftmost eight digits, they reduce the bound by at least $6$ ($b(n) \\leq 43 - 6 = 37$). Two zeros in the leftmost seven digits reduce it by $5$ ($b(n) \\leq 43 - 5 = 38$).\n\nSo, if $b(n) \\geq 39$, $n$ must look like $1111a_1a_2a_3b_1b_2b_3b_4$, where at most one of the $a_i$ is zero. If all $a_i$ are ones, there are four identical substrings $1111$ of length four, and three identical substrings $11111$ of length five, so at least five duplicates, so $b(n) < 39$.\n\nIf $a_1 = a_2 = 1$ and $a_3 = 0$, then $a_3 = 0$ reduces the bound by two, and $a_1 = a_2 = 1$ causes two duplicates of $1111$ and one duplicate of $11111$ at the beginning, reducing the bound by $2+2+1=5$, so $b(n) < 39$. If $a_1 = a_3 = 1$ and $a_2 = 0$, then $a_2 = 0$ reduces the bound by three and causes a duplicate of $1111$ at the beginning, so $b(n) \\leq 39$. If $a_2 = a_3 = 1$ and $a_1 = 0$, then $a_1 = 0$ reduces the bound by four, so $b(n) \\leq 39$.\n\nThus, $b(n) \\leq 39$; now we investigate equality. These are the cases where $a_1$ or $a_2$ is zero. We must also have $b_1 = 1$ (else it reduces the bound by $1$). So the number looks like $1111a_1a_211b_2b_3b_4$. If $a_1 = 0$ and $a_2 = 1$, we get a duplicate of $1111$ or $1110$. Thus $a_1 = 1$ and $a_2 = 0$, so the number looks like $11111011b_2b_3b_4$. If $b_1 = 1$, we get a duplicate of length $4$.\n\nSo the number looks like $111110110b_3b_4$. If $b_2 = 1$, then $1101$ is duplicated. So the only possibilities are\n\n$$\n11111011000, \\text{ and}\n$$\n$$\n11111011001.\n$$\n\nTabulating the distinct substrings of each by length:\n\n![](table.png)\n\nHence, these two (and only these two) give equality. Note that they are $2008$ and $2009$ (respectively) in base $10$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 13826,
    "subject": "Mathematics (Olympiad)",
    "question": "De un prisma recto de base cuadrada, con lado de longitud $L_1$ y altura $H$, extraemos un tronco de pirámide (no necesariamente recto) de bases cuadradas, con lados de longitud $L_1$ (para la inferior) y $L_2$ (para la superior), y altura $H$. Las dos piezas obtenidas aparecen en la imagen siguiente.\n\n![](images/Spanija_b_2014_p19_data_7cf6730e03.png)\n\nSi el volumen del tronco de pirámide es $\\frac{2}{3}$ del total del volumen del prisma, ¿cuál es el valor de $\\frac{L_1}{L_2}$?\n\n![](images/Spanija_b_2014_p19_data_42808876bb.png)",
    "options": [],
    "answer": "See solution",
    "solution": "Si prolongamos una altura $h$ del tronco de pirámide hasta obtener una pirámide completa de altura $H + h$, tendrá una sección como la que se muestra en la figura anterior.\n\nUn argumento de semejanza de triángulos permite comprobar que\n\n$$\n\\frac{h+H}{L_1} = \\frac{h}{L_2}\n$$\n\ny, por tanto,\n\n$$\nh = \\frac{H L_2}{L_1 - L_2}.\n$$\n\nAdemás, podemos observar que\n\n$$\n\\begin{aligned}\n\\text{Volumen del tronco de pirámide} &= \\frac{1}{3}(L_1^2(H+h) - L_2^2 h) \\\\\n&= \\frac{1}{3} \\left( \\frac{H L_1^3}{L_1 - L_2} - \\frac{H L_2^3}{L_1 - L_2} \\right) \\\\\n&= \\frac{H (L_1^3 - L_2^3)}{3(L_1 - L_2)} = \\frac{H}{3}(L_1^2 + L_1 L_2 + L_2^2).\n\\end{aligned}\n$$\n\nAsí, teniendo en cuenta que\n\n$$\n\\text{Volumen del tronco de pirámide} = \\frac{2}{3} \\text{Volumen del prisma}\n$$\n\ntendremos la ecuación\n\n$$\n\\frac{H}{3}(L_1^2 + L_1 L_2 + L_2^2) = \\frac{2}{3} H L_1^2,\n$$\n\nque se transforma en\n\n$$\n\\left(\\frac{L_1}{L_2}\\right)^2 - \\frac{L_1}{L_2} - 1 = 0,\n$$\n\ncuya única solución positiva es $\\frac{L_1}{L_2} = \\frac{1 + \\sqrt{5}}{2}$. Es decir, los lados deben estar en relación áurea.",
    "topic": "Geometry",
    "subtopic": "Solid Geometry"
  },
  {
    "id": 13827,
    "subject": "Mathematics (Olympiad)",
    "question": "For a positive integer $m$, let $\\tau(m)$ denote the number of its positive divisors, and $\\sigma(m)$ denote the sum of its positive divisors. Determine all positive integers $n$ for which\n\n$$\nn \\sqrt{\\tau(n)} \\leq \\sigma(n).\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "It is easy to check that the inequality holds for all $n \\in \\{1, 2, 4, 6\\}$ and does not hold for $n \\in \\{3, 5\\}$. Equality occurs for $n = 1$ and $n = 6$. We prove that the inequality does not hold for any larger $n$.\n\nNote that $\\tau$ and $\\sigma$ are multiplicative: $\\tau(a b) = \\tau(a) \\tau(b)$ and $\\sigma(a b) = \\sigma(a) \\sigma(b)$ for coprime $a, b$. Thus, if two coprime integers $a, b$ do not satisfy the inequality, neither does their product.\n\nWe show that $2^k$ is not a solution for $k > 3$, and that $p^k$, $2p^k$, and $4p^k$ are not solutions for any odd prime $p$ and positive integer $k$. Therefore, the only solutions are $n = 1, 2, 4, 6$.\n\n- For $2^k$: $2^k \\sqrt{\\tau(2^k)} > \\sigma(2^k) \\iff 2^k \\sqrt{k+1} > 2^{k+1} - 1$, which holds for $\\sqrt{k+1} \\ge 2$, i.e., $k \\ge 3$.\n\n- For $p^k$: $p^k \\sqrt{\\tau(p^k)} > \\sigma(p^k) \\iff p^k \\sqrt{k+1}(p-1) > p^{k+1} - 1$. Since $p^k \\sqrt{k+1}(p-1) \\ge p^k(p-1)\\sqrt{2} > p^{k+1} > p^{k+1} - 1$ for $p > 3$. For $p=3$, $k \\ge 2$ and $3^k \\cdot 2\\sqrt{3} > 3^{k+1}$.\n\n- For $2p^k$: $2p^k \\sqrt{2(k+1)}(p-1) > 3(p^{k+1} - 1)$. $2p^k \\sqrt{2(k+1)}(p-1) \\ge 4p^k(p-1) > 3p^{k+1} > 3(p^{k+1} - 1)$ for $p \\ge 5$. For $p=3$, $k \\ge 2$, $4 \\cdot 3^k \\sqrt{2(k+1)} > 3(3^{k+1} - 1)$.\n\n- For $4p^k$: $4p^k \\sqrt{3(k+1)}(p-1) > 7(p^{k+1} - 1)$. $4p^k \\sqrt{3(k+1)}(p-1) \\ge 4\\sqrt{6}p^k(p-1) > 7p^{k+1} > 7(p^{k+1} - 1)$ for $p \\ge 5$. For $p=3$, $8 \\cdot 3^k \\sqrt{3(k+1)} > 7(3^{k+1} - 1)$ holds for $k \\ge 2$. For $p=3$, $k=1$, direct check shows $n=12$ is not a solution.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 13828,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $d \\geq 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.",
    "options": [],
    "answer": "See 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": 13829,
    "subject": "Mathematics (Olympiad)",
    "question": "Сколько ребят нужно выбрать, чтобы среди них нашлись карандаши всех четырёх цветов, если каждому из 10 ребят раздали по 4 карандаша, а всего карандашей каждого цвета — по 10?\n\nТакже покажите, как можно раздать карандаши, чтобы у любых двух ребят вместе были карандаши не более чем трёх цветов.",
    "options": [],
    "answer": "See solution",
    "solution": "Покажем, что всегда можно выбрать трёх ребят так, чтобы у них нашлись карандаши всех цветов. Так как карандашей каждого цвета $10$, а каждому досталось по $4$ карандаша, то кому-то достались карандаши по крайней мере двух различных цветов. Осталось добавить к нему двух ребят, у которых есть карандаши оставшихся двух цветов.\n\nПокажем теперь, как раздать карандаши ребятам, чтобы у любых двух из них вместе были карандаши не более трёх цветов. Раздадим двум ребятам по $4$ карандаша второго цвета, двум — по $4$ карандаша третьего цвета, двум — по $4$ карандаша четвёртого цвета, одному — $4$ карандаша первого цвета, одному — по $2$ карандаша первого и второго цвета, одному — по $2$ карандаша первого и третьего цвета, и ещё одному — по $2$ карандаша первого и четвёртого цвета.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 13830,
    "subject": "Mathematics (Olympiad)",
    "question": "Show that the inequality\n\n$$\n\\sum_{i=1}^{n} \\sum_{j=1}^{n} \\sqrt{|x_i - x_j|} \\leq \\sum_{i=1}^{n} \\sum_{j=1}^{n} \\sqrt{|x_i + x_j|}\n$$\n\nholds for all real numbers $x_1, \\dots, x_n$.",
    "options": [],
    "answer": "See solution",
    "solution": "We apply induction on $n$.\n\nWhen $n=0$ or $n=1$, the inequality is obviously true. Suppose (1) holds for fewer than $n$ variables. Now consider (1) for $n$ variables.\n\nIf $x_i = 0$, then the terms on both sides that involve $x_i$ are equal, and by removing $x_i$ from $\\{x_1, x_2, \\dots, x_n\\}$ we can reduce it to $n-1$ variables.\n\nSimilarly, if $x_i = -x_j \\ne 0$, then the terms involving $x_i$ or $x_j$ are\n\n$$\n2\\sqrt{2|x_i|} + \\sum_{\\substack{k=1,\\dots,n \\\\ k \\ne i, j}} \\left(\\sqrt{|x_k - x_i|} + \\sqrt{|x_k + x_i|}\\right)\n$$\n\non both sides. By removing $x_i$ and $x_j$ from $\\{x_1, x_2, \\dots, x_n\\}$ we can reduce it to $n-2$ variables. In either case, the induction hypothesis leads to the conclusion.\n\nIn the following, assume $x_i + x_j \\ne 0$ for all $i, j \\in \\{1, 2, \\dots, n\\}$ (possibly $i = j$). We wish to apply a translation to all variables $x_i \\mapsto x_i + a$ such that some two variables add up to $0$. Note that a translation does not change the value on the left-hand side of (1).\n\nIf there exists $a > 0$ such that $x_i + x_j + a = 0$ for some $i, j \\in \\{1, 2, \\dots, n\\}$, denote the minimal of all such $a$ as $a_+$; otherwise, denote $a_+ = \\infty$. If there exists $a < 0$ such that $x_i + x_j + a = 0$ for some $i, j \\in \\{1, 2, \\dots, n\\}$, denote the maximal of all such $a$ as $a_-$; otherwise, denote $a_- = -\\infty$.\n\nSince the functions $\\sqrt{x}$ and $\\sqrt{-x}$ are concave down in their natural domains, it follows that for every $i, j$, the function $\\sqrt{x_i + x_j + a}$ is concave down in $[a_-, a_+]$, and likewise for the sum $\\sum_{i,j=1}^n \\sqrt{x_i + x_j + a}$. Hence, for $b = a_+$ or $b = a_-$ ($b \\ne \\pm\\infty$),\n\n$$\n\\sum_{i,j=1}^{n} \\sqrt{|x_i + x_j|} \\geq \\sum_{i,j=1}^{n} \\sqrt{|x_i + x_j + b|}.\n$$\n\n(Note: if $a_+ = \\infty$ or $a_- = -\\infty$, the sum is infinite and the above does not hold.)\n\nTake $y_i = x_i + \\frac{b}{2}$. Then for some $i, j$, $y_i + y_j = 0$. We have\n\n$$\n\\begin{align*}\n\\sum_{i=1}^{n} \\sum_{j=1}^{n} \\sqrt{|x_i - x_j|} &= \\sum_{i=1}^{n} \\sum_{j=1}^{n} \\sqrt{|y_i - y_j|} \\\\\n&\\leq \\sum_{i=1}^{n} \\sum_{j=1}^{n} \\sqrt{|y_i + y_j|} \\\\\n&\\leq \\sum_{i=1}^{n} \\sum_{j=1}^{n} \\sqrt{|x_i + x_j|}\n\\end{align*}\n$$\n\nwhich is the desired inequality. Here, the first inequality comes from the induction hypothesis (as $y_i + y_j = 0$). This completes the induction. $\\Box$",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 13831,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all functions $f : \\mathbb{R} \\to \\mathbb{R}$ such that\n\n$$\nf(2x f(x) - 2f(y)) = 2x^2 - y - f(y)\n$$\n\nfor all $x, y \\in \\mathbb{R}$.",
    "options": [],
    "answer": "See solution",
    "solution": "Suppose there exist $y_1, y_2$ such that $f(y_1) = f(y_2)$. Substituting $(x, y) \\to (x, y_1)$ and $(x, y_2)$ and comparing, we get $y_1 = y_2$. Thus, $f(x)$ is injective.\n\nLet $x \\to -x$ in the given equation:\n\n$$\nf(-2x f(-x) - 2f(y)) = 2x^2 - y - f(y)\n$$\n\nSo,\n\n$$\nf(2x f(x) - 2f(y)) = f(-2x f(-x) - 2f(y))\n$$\n\nfor all $x, y$. By injectivity, $x f(x) = -x f(-x)$, so $f(-x) = -f(x)$ for all $x \\neq 0$.\n\nLet $x = y = 0$ in the original equation:\n\n$$\nf(-2f(0)) = -f(0)\n$$\n\nLet $a = f(0)$, so $f(-2a) = -a$.\n\nLet $x = 0$, $y = -2a$:\n\n$$\nf(-2f(-2a)) = 2a - f(-2a)\n$$\n\nSince $f(-2a) = -a$, this gives $-a = 2a - (-a)$, so $a = 0$. Thus, $f(0) = 0$ and $f(-x) = -f(x)$ for all $x \\in \\mathbb{R}$.\n\nLet $y = 2x^2$ in the original equation:\n\n$$\nf(2x f(x) - 2f(2x^2)) = -f(2x^2)\n$$\n\nSo,\n\n$$\nf(2x f(x) - 2f(2x^2)) = f(-2x^2)\n$$\n\nTherefore, $x f(x) - f(2x^2) = -x^2$, or\n\n$$\nf(2x^2) = x f(x) + x^2, \\quad \\forall x. \\tag{1}\n$$\n\nLet $y = 0$ in the original equation:\n\n$$\nf(2x f(x)) = 2x^2\n$$\n\nThus, $f$ is surjective on $\\mathbb{R}^+$, and since $f$ is odd, also on $\\mathbb{R}$.\n\nLet $x = 0$ in the original equation:\n\n$$\nf(-2f(y)) = -y - f(y)\n$$\n\nSo,\n\n$$\nf(2f(y)) = y + f(y), \\quad \\forall y. \\tag{2}\n$$\n\nNow, substitute $y = 2x f(x)$ into (2):\n\n$$\nf(4x^2) = 2x f(x) + 2x^2\n$$\n\nCombining with (1):\n\n$$\nf(4x^2) = 2f(2x^2) \\implies f(2x) = 2f(x), \\quad \\forall x \\ge 0.\n$$\n\nSince $f$ is odd, $f(2x) = 2f(x)$ for all $x \\in \\mathbb{R}$.\n\nFrom (1):\n\n$$\n2f(x^2) = x f(x) + x^2\n$$\n\nLet $x = 1$:\n\n$$\nf(1) = 1\n$$\n\nFrom (2):\n\n$$\ny + f(y) = 2f(f(y)), \\quad \\forall y \\in \\mathbb{R}. \\tag{3}\n$$\n\nSubstitute (3) into the original equation:\n\n$$\n2f(x f(x) - f(y)) = 2x^2 - 2f(f(y))\n$$\n\nor\n\n$$\nf(x f(x) - f(y)) = x^2 - f(f(y))\n$$\n\nSince $f$ is surjective:\n\n$$\nf(x f(x) + y) = x^2 + f(y), \\quad \\forall x, y \\in \\mathbb{R}. \\tag{4}\n$$\n\nIn (3), let $x = 1$:\n\n$$\nf(y + 1) = f(y) + 1\n$$\n\nIn (4), let $x \\to x + 1$:\n\n$$\nf(x f(x) + f(x) + x + 1 + y) = (x + 1)^2 + f(y)\n$$\n\nor\n\n$$\nx^2 + f(f(x) + x + y) + 1 = (x + 1)^2 + f(y)\n$$\n\nThus,\n\n$$\nf(f(x) + x + y) = 2x + f(y)\n$$\n\nLet $y = 0$:\n\n$$\nf(f(x) + x) = 2x\n$$\n\nThus, $f(2x) = f(f(f(x) + x))$. Combine with (3):\n\n$$\n2f(x) = \\frac{f(x) + x + f(f(x) + x)}{2}\n$$\n\nSo,\n\n$$\n4f(x) = f(x) + x + 2x \\implies f(x) = x\n$$\n\nfor all $x$. It is easy to check that $f(x) = x$ satisfies the original equation. Hence, $f(x) = x$ for all $x \\in \\mathbb{R}$.\n\n$\\boxed{f(x) = x}$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 13832,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $x$, $y$, and $z$ be non-zero real numbers such that $3x + 2y = z$ and $\\frac{3}{x} + \\frac{1}{y} = \\frac{2}{z}$. Prove that $5x^2 - 4y^2 - z^2$ is always an integer.",
    "options": [],
    "answer": "See solution",
    "solution": "The second equation implies $2xy = 3yz + xz$. Multiplying the first equation respectively by $z$, $x$, and $y$, we get $z^2 = 3xz + 2zy$, $3x^2 = zx - 2xy$, and $2y^2 = zy - 3xy$. So,\n\n$$\n5x^2 - 4y^2 - z^2 = \\frac{5}{3}(zx - 2xy) - 2(zy - 3xy) - (3xz + 2zy) = \\frac{4}{3}(2xy - 3zy - xz) = 0.\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 13833,
    "subject": "Mathematics (Olympiad)",
    "question": "The sequence $a_0, a_1, a_2, \\dots$ of integers is defined by $a_0 = 3$ and\n$$\na_{n+1} - a_n = n(a_n - 1)\n$$\nfor all $n \\ge 0$. Determine all integers $m \\ge 2$ for which $\\gcd(m, a_n) = 1$ for all $n \\ge 0$.",
    "options": [],
    "answer": "See solution",
    "solution": "The sequence is given by the formula $a_n = 2 \\cdot n! + 1$ for $n \\ge 0$. (We use the usual definition $0! = 1$, which satisfies $1! = 1 \\cdot 0!$, in the same way we have $n! = n \\cdot (n-1)!$ for other positive integers $n$.) We will prove the equality by induction. We have $a_0 = 3$, which equals $2 \\cdot 0! + 1$. Now suppose for certain $k \\ge 0$ that $a_k = 2 \\cdot k! + 1$, then\n$$\na_{k+1} = a_k + k(a_k - 1) = 2 \\cdot k! + 1 + k \\cdot 2 \\cdot k! = 2 \\cdot k! (1+k) + 1 = 2 \\cdot (k+1)! + 1.\n$$\nThis finishes the induction.\n\nWe see that $a_n$ is always odd, hence $\\gcd(2, a_n) = 1$ for all $n$. It follows also that $\\gcd(2^i, a_n) = 1$ for all $i \\ge 1$. Hence, $m = 2^i$ with $i \\ge 1$ satisfies the condition. Now consider an $m \\ge 2$ which is not a power of two. Then $m$ has an odd prime divisor, say $p$. We will show that $p$ is a divisor of $a_{p-3}$. By Wilson's theorem, we have $(p-1)! \\equiv -1 \\pmod{p}$. Hence,\n$$\n\\begin{aligned}\n2 \\cdot (p-3)! &\\equiv 2 \\cdot (p-1)! \\cdot ((p-2)(p-1))^{-1} \\\\\n&\\equiv 2 \\cdot -1 \\cdot (-2 \\cdot -1)^{-1} \\equiv 2 \\cdot -1 \\cdot 2^{-1} \\equiv -1 \\pmod{p}.\n\\end{aligned}\n$$\nSo indeed we have $a_{p-3} = 2 \\cdot (p-3)! + 1 \\equiv 0 \\pmod{p}$. We conclude that $m$ does not satisfy the condition. Hence, the only values of $m$ satisfying the condition are powers of two. $\\square$",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 13834,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $n = 10^{1000}$, let $a_1, \\dots, a_n$ be the initial numbers, and let $b_i = \\text{lcm}(a_i, a_{i+1})$ (with $a_{n+1} = a_1$). Is it possible for the sequence $b_1, \\dots, b_n$ to consist of $10^{1000}$ consecutive integers?",
    "options": [],
    "answer": "See solution",
    "solution": "No.\n\nLet $2^m$ be the maximal possible power of $2$ dividing some $a_i$. Then $2^m$ divides at least two of the $b_i$'s. If the $b_i$'s form $10^{1000}$ consecutive integers, then one of them must be divisible by $2^{m+1}$, which is impossible.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 13835,
    "subject": "Mathematics (Olympiad)",
    "question": "Prove that there exists a real number $\\varepsilon > 0$ such that there are infinitely many sequences of integers $0 < a_1 < a_2 < \\dots < a_{2025}$ satisfying\n\n$$\n\\gcd(a_1^2 + 1, a_2^2 + 1, \\dots, a_{2025}^2 + 1) > a_{2025}^{1+\\varepsilon}.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "By the Chinese Remainder Theorem for $\\mathbb{Q}[x]$, for any choice of $2^{12}$ signs there exists a unique polynomial $f_i$ of degree at most $23$ such that\n\n$$\nf_i \\equiv \\pm c x \\pmod{c^2 x^2 + 1}\n$$\nfor $c = 1, 2, \\dots, 12$. Furthermore, these polynomials are multiples of $x$ and come in pairs which sum to zero, so we can pick $2048$ of these polynomials which have positive leading coefficients and label them $f_1, f_2, \\dots, f_{2048}$.\n\nLet $N$ be a positive integer such that any coefficient of the polynomials is $\\frac{1}{N}$ times an integer. For a sufficiently large positive integer $x$, take $a_i = f_i(Nx)$ for $i = 1, 2, \\dots, 2025$, which will be positive integers. Then,\n\n$$\n\\gcd(a_1^2 + 1, a_2^2 + 1, \\dots, a_{2025}^2 + 1) \\geq \\prod_{c=1}^{12} (c^2 x^2 + 1)\n$$\nbecause $a_i^2 + 1 \\equiv 0 \\pmod{c^2 x^2 + 1}$ for any $i$ by construction. The right hand side is asymptotically $x^{24}$ while $a_{2025}$ is $x^{23}$ up to constant factors, so any $\\varepsilon < \\frac{1}{23}$ works.\n\n*Remark.* In terms of $n = 2025$, this solution achieves $\\varepsilon = \\Omega\\left(\\frac{1}{\\log n}\\right)$ which is the best that we know of. Solution 2 achieves $\\varepsilon = \\Omega\\left(\\frac{1}{n^{\\log_2(3)}}\\right)$ while solutions 3 and 4 achieve $\\varepsilon = \\Omega\\left(\\frac{1}{n}\\right)$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 13836,
    "subject": "Mathematics (Olympiad)",
    "question": "考慮數列 $a_0, a_1, a_2, \\dots$，其中 $a_n = 2^n + 2^{\\lfloor n/2 \\rfloor}$。\n\n證明：\n1. 數列中有無窮多項可以寫成兩個以上不同項的和。\n2. 也有無窮多項不能寫成這樣的和。\n\n（註：對於實數 $x$，符號 $\\lfloor x \\rfloor$ 表示不超過 $x$ 的最大整數。）",
    "options": [],
    "answer": "See solution",
    "solution": "設一個非負整數為 *可表示*，若它等於數列中若干（可能為 0 或 1）不同項的和。定義兩個非負整數 $b$ 和 $c$ *等價*（記作 $b \\sim c$），若它們同時可表示或同時不可表示。\n\n容易計算（或用歸納法證明）：\n\n$$\nS_{n-1} := a_0 + a_1 + \\dots + a_{n-1} = 2^n + 2^{\\lceil n/2 \\rceil} + 2^{\\lfloor n/2 \\rfloor} - 3.\n$$\n\n特別地，$S_{2k-1} = 2^{2k} + 2^{k+1} - 3$。注意，若 $n \\ge 3$，則 $2^{\\lceil n/2 \\rceil} > 3$，所以\n\n$$\nS_{n-1} > 2^n + 2^{\\lceil n/2 \\rceil} = a_n.\n$$\n\n同時 $S_{n-1} - a_n = 2^{\\lceil n/2 \\rceil} - 3 < a_n$。\n\n**Claim 1.** 若 $S_{n-1} - a_n < b < a_n$ 且 $n \\ge 3$，則 $b \\sim S_{n-1} - b$。\n\n*證明.* 若 $b$ 可表示，因為 $b < a_n$，$b$ 必為 $\\{a_0, a_1, \\dots, a_{n-1}\\}$ 的某些項之和。則 $S_{n-1} - b$ 為補集之和。\n\n反之，注意 $S_{n-1} - a_n < S_{n-1} - b < a_n$，同理，若 $S_{n-1} - b$ 可表示，則 $b$ 也可表示。$\\square$\n\n**Claim 2.** 對某些 $n \\ge 3$，數 $a_n$ 可表示為兩個或以上不同項之和，當且僅當 $S_{n-1} - a_n = 2^{\\lceil n/2 \\rceil} - 3$ 可表示。\n\n*證明.* 類似 Claim 1，若 $a_n$ 可表示為兩個或以上不同項之和，則 $a_n$ 必為 $\\{a_0, a_1, \\dots, a_{n-1}\\}$ 的某些項之和，$S_{n-1} - a_n$ 為補集之和。反之，因 $S_{n-1} - a_n < a_n$，若可表示，則 $S_{n-1} - a_n$ 由 $\\{a_0, a_1, \\dots, a_{n-1}\\}$ 的某些項表示，故 $a_n$ 由補集表示。$\\square$\n\n由 Claim 2，只需找無窮多個可表示和不可表示的 $2^{t}-3$ 型數。\n\n**Claim 3.** 對每個 $t \\ge 3$，有 $2^t \\sim 2^{4t-6} - 3$ 且 $2^{4t-6} > 2^t - 3$。\n\n*證明.* 因 $S_{2t-3} - a_{2t-2} = 2^{t-1} - 3 < 2^t - 3 < a_{2t-2}$，由 Claim 1，\n\n$$\n2^t - 3 \\sim S_{2t-3} - (2^t - 3) = 2^{2t-2}.\n$$\n\n又 $S_{4t-7} - a_{4t-6} = 2^{2t-3} - 3 < 2^{2t-2} < a_{4t-6}$，由 Claim 1，\n\n$$\n2^{2t-2} \\sim S_{4t-7} - 2^{2t-2} = 2^{4t-6} - 3.\n$$\n\n因此 $2^t - 3 \\sim 2^{4t-5} - 3$。不等式由 $t \\ge 3$ 得。$\\square$\n\n由於 $2^3 - 3 = 5 = a_0 + a_1$ 可表示，且容易檢查\n\n$$\n2^7 - 3 = 125 \\sim S_6 - 125 = 24 \\sim S_4 - 24 = 17\n$$\n\n不可表示。故由 Claim 3，存在無窮多個可表示和不可表示的 $2^t - 3$ 型數。",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 13837,
    "subject": "Mathematics (Olympiad)",
    "question": "In basketball, the free-throw rate (FRT) of a player is the ratio of the number of his successful free throws to the total number of his free throws. After the first half of a game, Mateo's FRT was less than $75\\%$, and at the end of the game it was greater than $75\\%$. Can one claim with certainty that there was a moment when his FRT was exactly $75\\%$? Answer the same question for $60\\%$ instead of $75\\%$.",
    "options": [],
    "answer": "See solution",
    "solution": "The answer is yes for $75\\%$ and no for $60\\%$.\n\nSuppose Mateo's FRT was less than $75\\%$ after the first half but eventually greater than $75\\%$. Then, there must be a successful free throw in the second half such that after it, the FRT became at least $75\\%$. Consider the first such free throw $S$. We claim that after $S$, the FRT has become exactly $75\\%$.\n\nLet the FRT before $S$ be $\\frac{x}{y}$, where $y$ is the total number of free throws before $S$ and $x$ is the number of successful ones among them. After $S$, the FRT is $\\frac{x+1}{y+1}$, and by assumption,\n$$\n\\frac{x}{y} < \\frac{3}{4} \\leq \\frac{x+1}{y+1}.\n$$\nThe left inequality gives $4x < 3y$, and the right yields $3y \\leq 4x + 1$. Hence $4x < 3y \\leq 4x + 1$, and because $3y$ is an integer, it follows that $3y = 4x + 1$. This equality is equivalent to $\\frac{x+1}{y+1} = \\frac{3}{4}$.\n\nTherefore, $S$ made the FRT exactly $75\\%$.\n\nFor $60\\%$, the situation is different. Suppose Mateo scored $4$ free throws out of $7$ in the first half, so his FRT was $\\frac{4}{7} < \\frac{3}{5}$. If all his free throws in the second half are successful, the first one makes the FRT $\\frac{5}{8} > \\frac{3}{5}$. Each subsequent successful free throw increases the FRT (since if $0 < u < v$, then $\\frac{u}{v} < \\frac{u+1}{v+1}$). Thus, the FRT will be greater than $60\\%$ at the end of the game, but never exactly $60\\%$ throughout. There are infinitely many such fractions that can replace $\\frac{4}{7}$ in this argument: $\\frac{1}{2}$, $\\frac{7}{12}$, $\\frac{10}{17}$, $\\frac{13}{22}$, $\\frac{16}{27}$, etc.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 13838,
    "subject": "Mathematics (Olympiad)",
    "question": "In an acute triangle $ABC$, the center of the incircle is $I$, and $|AC| + |AI| = |BC|$. Prove that $\\angle BAC = 2\\angle ABC$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $D$ be a point on $BC$ such that $|CD| = |AC|$. Because $|BC| = |AC| + |AI|$, the point $D$ lies on the interior of side $BC$, and we have $|BD| = |AI|$.\n\nBecause triangle $ACD$ is isosceles, the angle bisector $CI$ is also the perpendicular bisector of $AD$, hence $A$ is the reflection of $D$ in $CI$.\n\nHence, we get $\\angle CDI = \\angle CAI = \\angle IAB$, hence $180^\\circ - \\angle BDI = \\angle IAB$, which means that quadrilateral $ABDI$ is cyclic.\n\nIn this cyclic quadrilateral $BD$ and $AI$ have the same length. Therefore, $AB$ and $ID$ are parallel. Hence, $ABDI$ is an isosceles trapezium, which has equal angles at the base.\n\nHence, $\\angle CBA = \\angle DBA = \\angle BAI = \\frac{1}{2}\\angle BAC$, which proves the statement. $\\square$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 13839,
    "subject": "Mathematics (Olympiad)",
    "question": "The circumcenter of the cyclic quadrilateral $ABCD$ is $O$. The second intersection point of the circles $ABO$ and $CDO$, other than $O$, is $P$, which lies in the interior of triangle $DAO$. Choose a point $Q$ on the extension of $OP$ beyond $P$, and a point $R$ on the extension of $OP$ beyond $O$. Prove that $\\angle QAP = \\angle OBR$ holds if and only if $\\angle PDQ = \\angle RCO$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $H$ be the radical center of the circles $ABCD$, $ABOP$, and $CDPO$. Then the radical axes of any two of these circles, i.e., the lines $AB$, $CD$, and $OP$, pass through $H$. Since $P$ lies on the shorter arcs $AO$ and $DO$, it follows that $H$ lies on the extension of $OP$ beyond $P$. The radical center satisfies\n\n$$\nHA \\cdot HB = HC \\cdot HD = HO \\cdot HP.\n$$\n\n(1)\n\nSince the quadrilateral $ABOP$ is cyclic,\n\n![](images/Indija_TS_2012_p1_data_7c06712e20.png)\n\n$$\n\\begin{aligned}\n\\angle QAB + \\angle BRQ &= (\\angle PAB + \\angle QAP) + (\\angle BOP - \\angle OBR) \\\\\n&= (\\angle PAB + \\angle BOP) + (\\angle QAP - \\angle OBR) \\\\\n&= 180^{\\circ} + (\\angle QAP - \\angle OBR).\n\\end{aligned}\n$$\n\nTherefore, $\\angle QAP = \\angle OBR$ holds if and only if the quadrilateral $ABRQ$ is cyclic, which is equivalent to $HQ \\cdot HR = HA \\cdot HB$.\n\nSimilarly, $\\angle QAP = \\angle OBR$ holds if and only if $HQ \\cdot HR = HC \\cdot HD$.\n\nCombining with (1),\n\n$$\n\\angle QAP = \\angle OBR \\Leftrightarrow HQ \\cdot HR = HA \\cdot HB \\Leftrightarrow HQ \\cdot HR = HC \\cdot HD \\Leftrightarrow \\angle PDQ = \\angle RCO.\n$$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 13840,
    "subject": "Mathematics (Olympiad)",
    "question": "For a set $S$, let $|S|$ denote the number of elements in $S$. Let $A$ be a set of positive integers with $|A| = 2001$. Prove that there exists a set $B$ such that\n\n1. $B \\subseteq A$;\n2. $|B| \\ge 668$;\n3. For any $u, v \\in B$ (not necessarily distinct), $u + v \\notin B$.",
    "options": [],
    "answer": "See solution",
    "solution": "Both solutions use the \"middle\" subset $\\{k, k+1, \\dots, 2k-1\\}$ of the set $\\{1, 2, \\dots, 3k\\}$. The first solution uses cosets; the second uses a special case of Dirichlet’s Theorem on primes in arithmetic progressions.\n\n**First Solution.** For a positive integer $n$, let $Z_n$ denote the set of residues modulo $n$. Let $\\phi(n)$ be Euler's function, the number of integers between $1$ and $n$ relatively prime to $n$. Call a set $A$ of residues modulo $3^n$ sum-free if for any $a, b \\in A$, $a+b$ is not (congruent to) an element of $A$.\n\n**Lemma.** For any $n \\ge 1$, there exist $3^n - 1$ sum-free sets of $3^{n-1}$ residues modulo $3^n$ such that every nonzero residue modulo $3^n$ appears in exactly $3^{n-1}$ of the subsets.\n\n*Proof.* We construct the desired subsets inductively. For $n = 1$ we take the sets $\\{1\\}$ and $\\{2\\}$.\n\nSuppose the statement holds for $n$, i.e., we have $3^n - 1$ sum-free subsets $A_1, \\dots, A_{3^{n-1}}$ of $Z_{3^n}$ such that every nonzero element of $Z_{3^n}$ belongs to exactly $3^{n-1}$ of the $A_i$. Construct sets $B_1, \\dots, B_{3^{n-1}}$ by\n\n$$\nB_i = \\{ x \\in Z_{3^{n+1}} \\mid x \\equiv m' \\pmod{3^n},\\ m' \\in A_i \\}.\n$$\n\nEach $B_i$ contains $3|A_i| = 3^n$ residues, and the $B_i$ are sum-free: if $a, b \\in B_i$, $(a+b) \\bmod 3^n$ is not in $A_i$, so $a+b$ is not in $B_i$. Moreover, each $x \\in Z_{3^{n+1}}$ not $0$ modulo $3^n$ is in exactly $3^{n-1}$ of the $B_i$.\n\nNow define\n\n$$\nC = \\{3^n, 3^n + 1, \\dots, 2 \\cdot 3^n - 1\\}\n$$\n\nand\n\n$$\nU = \\{ x \\in Z_{3^{n+1}} \\mid \\gcd(x, 3) = 1 \\}.\n$$\n\n$C \\subset Z_{3^{n+1}}$, $|C| = 3^n$. $C$ is sum-free: if $a, b \\in C$ with $3^n \\le a, b < 2 \\cdot 3^n$, then $2 \\cdot 3^n \\le a+b < 4 \\cdot 3^n$, so $a+b$ is not congruent modulo $3^{n+1}$ to an element of $C$. For each $y \\in U$, let $C_y = yC = \\{yx \\mid x \\in C\\}$. Each $C_y$ is sum-free and contains $3^n$ residues. Since $|U| = \\phi(3^{n+1}) = 2 \\times 3^n$, there are $2 \\cdot 3^n$ sets $C_y$.\n\nConsider the sets\n\n$B_1, \\dots, B_{3^{n-1}}, C_1, C_2, C_4, \\dots, C_{3^{n+1}-1}$.\n\nThere are $3^n - 1 + 2 \\cdot 3^n = 3^{n+1} - 1$ sets. Every nonzero residue modulo $3^{n+1}$ appears in exactly $3^n$ of them. Let $m$ be a nonzero residue modulo $3^{n+1}$, write $m = 3^k s$, $0 \\le k \\le n$, $\\gcd(s, 3) = 1$.\n\n- If $k < n$, $m$ is a nonzero residue modulo $3^n$, so $m$ is in $3^{n-1}$ of the $B_i$. The number of $C_i$ containing $m$ is $2 \\cdot 3^{n-1}$, so total is $3^n$.\n- If $k = n$, $m = 3^n$ or $2 \\cdot 3^n$, $m \\bmod 3^n = 0$, so $m$ is not in any $A_i$, but appears in $3^n$ of the $C_y$.\n\nThus, the Lemma holds by induction.\n\nNow let $3^n$ be a power of $3$ larger than the sum of any two elements of $A$. By the Lemma, there exist $3^n - 1$ sets $S_1, \\dots, S_{3^{n-1}}$ of $3^{n-1}$ residues modulo $3^n$ such that every nonzero residue modulo $3^n$ appears in exactly $3^{n-1}$ of the $S_i$. Let $n_i$ be the number of elements of $A$ contained in $S_i$. Since every element of $A$ appears $3^{n-1}$ times,\n\n$$\n\\sum_{i=1}^{3^n-1} n_i = 3^{n-1} |A|\n$$\n\nso some $n_i$ is at least\n\n$$\n\\frac{3^{n-1}|A|}{3^n-1} > \\frac{1}{3}|A| = \\frac{2001}{3} = 667.\n$$\n\nTherefore, there exists a sum-free subset $B$ of $A$ with $|B| \\ge 668$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 13841,
    "subject": "Mathematics (Olympiad)",
    "question": "$f: \\mathbb{R} \\to \\mathbb{R}$, $g: \\mathbb{R} \\to \\mathbb{R}$ ба $\\forall x, y \\in \\mathbb{R}$-ийн хувьд\n\n$$\ng(f(x + y)) = f(x) + (2x + y)g(y)\n$$\n\nбайх бүх $f, g$ функцийг ол.",
    "options": [],
    "answer": "See solution",
    "solution": "Хариу: $f = g = 0$; $f(x) = x^2 + C$, $g(x) = x$.\n\n$x \\leftrightarrow y$ гэе.\n\n$$\n\\begin{aligned}\n\\frac{g(f(x+y)) = f(y) + (2y+x)g(x)}{g(f(x+y)) = f(x) + (2x+y)g(y)} \\\\\n\\frac{f(y)-f(x)}{f(x)-f(y)} = \\frac{(2y+x)g(x)-(2x+y)g(y)}{f(x)-f(y)}\n\\end{aligned}\n$$\n\n$(x, y) \\rightarrow (x, 0)$ гэе.\n\n$(x, y) \\rightarrow (1, x)$ гэе.\n\n$(x, y) \\to (0, 1)$ гэе.\n\n$$\n-f(0) + f(1) = 2g(0) - g(1) \\tag{3}\n$$\n\n$$(1) + (2) + (3) \\Rightarrow 0 = -2g(x) - (2x - 2)g(0) + 2xg(1)$$\n$$\\Leftrightarrow g(x) = x(g(1) - g(0)) + g(0)$$\n\nТул $g$ нь шугаман функциональ байна.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 13842,
    "subject": "Mathematics (Olympiad)",
    "question": "In every cell of a $5 \\times 9$ table, either $0$ or $1$ is written. Then, the sums of the numbers in each row and each column are calculated. What is the maximum number of different values that can appear among these $14$ numbers?\n\n![](images/UkraineMO_2015-2016_booklet_p19_data_80aab8f8ae.png)",
    "options": [],
    "answer": "See solution",
    "solution": "$9$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 13843,
    "subject": "Mathematics (Olympiad)",
    "question": "Suppose that $IH$ meets $AB$ and $AC$ at $P$ and $Q$ respectively. Prove that the circumcircle of triangle $APQ$ intersects $(O)$ again at a point on $AI$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $T$ be the projection of $A$ on $GD$. It is well known that $T$ is the second intersection of $(AEF)$ and $(O)$. We observe that $\\triangle TEB \\sim \\triangle TFC$, then\n\n$$\n\\frac{TB}{TC} = \\frac{BE}{CF}.\n$$\n\n![](images/Vietnam_2022_p38_data_ca22d7bef7.png)\n\nOn the other hand,\n\n$$\n\\frac{BE}{CF} = \\frac{BE}{IB} \\cdot \\frac{IC}{CF} = \\frac{\\cos ACB}{\\cos ABC} = \\frac{BD}{CD},\n$$\n\nwhich implies that $TBDC$ is a harmonic quadrilateral, or $TD$ passes through $K$, which is the intersection of the tangents at $B$ and $C$ of $(O)$.\n\nLet $X$ and $Y$ be the intersections of $KB$, $KC$ with $IN$, $IM$, respectively. The tangents at $B$ and $C$ of $(O)$ meet the tangent at $A$ at $X'$, $Y'$. Because $IX \\parallel OX'$ and $IY \\parallel OY'$, then\n\n$$\n\\triangle KX'Y' \\sim \\triangle KXY\n$$\n\nwith $O$ and $I$ corresponding. Note that $O$ is the incenter of $KX'Y'$, which implies that $I$ is the incenter of triangle $KXY$.\n\nLet $(KXY)$ meet $IY$, $IX$ at $R$, $S$ respectively. Clearly, $R$ and $S$ are the circumcenters of triangles $KIX$ and $KIY$. Hence, $RS$ is the perpendicular bisector of $IK$, so $H$ lies on $RS$. Applying Pascal's theorem for $\\begin{pmatrix} K & X & R \\\\ S & Y & K \\end{pmatrix}$, we obtain that the tangent at $K$ of $(KXY)$, $RS$, and $MN$ are concurrent, which means $KH$ is the tangent of $(KXY)$. By angle chasing, we have\n\n$$\n\\begin{aligned}\n\\angle HIK &= \\angle HKI = \\angle HKB + \\angle BKI \\\\\n&= \\angle KYX + 90^\\circ - \\angle BAC = 270^\\circ - 2\\angle ABC - \\angle BAC \\\\\n&= 90^\\circ + \\angle ACB - \\angle ABC\n\\end{aligned}\n$$\n\nAlso, $\\angle(AO, BC) = \\angle OAC + \\angle ACB = 90^\\circ - \\angle ABC + \\angle ACB = \\angle HIK$. Note that $IK \\perp BC$, thus $IH \\perp AO$. Hence, $PBQC$ is cyclic or $I$ has the same power to $(ABC)$ and $(APQ)$, which means $AI$ passes through the second intersection of $(APQ)$ and $(O)$. $\\square$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 13844,
    "subject": "Mathematics (Olympiad)",
    "question": "$f(x) \\in \\mathbb{Z}[x]$, $\\deg f \\ge 1$ болон $k \\in \\mathbb{N}$ байг.\n\n$f(n) = d_1 d_2 \\dots d_k d_{k+1}$, $1 \\le d_1 < d_2 < \\dots < d_k < n$ байх $n \\in \\mathbb{N}$ тоо төгсгөлгүй олон оршино гэдгийг батал.",
    "options": [],
    "answer": "See solution",
    "solution": "Бид $f$-ийн ахмад гишүүний өмнөх коэффициент эерэг гэж үзэж болно (эсрэг тохиолдолд $-f$-ийг авч үзнэ). Иймд хангалттай том $n$-ийн хувьд $f(n)$ эерэг байна. $f(x) \\mid f(x+f(x))$, $x \\in \\mathbb{N}$ тул $f(x+f(x)) = f(x) \\cdot g(x)$, энд $g \\in \\mathbb{Z}[x]$ ба $g$-ийн ахмад гишүүний өмнөх коэффициент эерэг байна. $x \\to x+f(x)$ гэж авъя.\n\n$$\n\\begin{aligned}\nf(x + f(x) + f(x+f(x))) &= f(x)g(x) \\cdot g(x+f(x)) \\\\\n&= f(x+f(x)) \\cdot g(x+f(x)) \\text{ уг процессийг } p \\text{ удаа хийе. Мөн} \\\\\nh(x) &= x+f(x) \\text{ гэж авъя.}\n\\end{aligned}\n$$\n\n$$\nh^{[p]}(x) = h(\\ldots(h(x))\\ldots) \\quad (p \\text{ удаа}), \\text{ индукцээр}\n$$\n\n$$\nf(h^{[p]}(x)) = f(x)g(x)g(h(x))\\ldots g(h^{[p-1]}(x)), \\quad (p \\ge 1)\n$$\n\nХэрэв $S$ нь $f$-ийн зэрэг бол ($\\deg h = S$) $h^{[p]}$-ийн зэрэг $S^p$ байна.\n\n$$\n\\deg f = S, \\quad \\deg g(x) = S^2 - S, \\quad \\deg g(h(x)) = S^3 - S^2, \\ldots, \\quad \\deg g(h^{[p-1]}(x)) = S^{p+1} - S^p.\n$$\n\n$$\nS \\geq 2 \\text{ үед } S^2 - S < S^3 - S^2 < \\ldots < S^p - S^{p-1} < S^p\n$$\n\nТэгэхээр хангалттай их $m$-ийн хувьд\n\n$$\n0 < g(m) < g(h(m)) < \\ldots < g(h^{[p-2]}(m)) < h^{[p]}(m)\n$$\n\n$$\np = k + 1, \\quad n = h^{[k+1]}(m) \\text{ гэж сонговол}\n$$\n\n$$\nd_1 = g(m), \\ldots, d_k = g(h^{[k-1]}(m)), \\quad d_{k+1} = f(m)g(h^{[k]}(m))\n$$\n\nболно. Тэгээд $f(n) = d_1 d_2 \\ldots d_k d_{k+1}$, $1 < d_1 < d_2 < \\ldots < d_k < n$ гэсэн нөхцөл биелнэ.\n\nХэрэв $S = 1$ бол $f(x) = ax + b$ гэж авъя. $(a, d_i) = 1$, $1 \\le i \\le k$ байхаар $d_i$-уудыг сонгож чадна. Улдэгдэлийн тухай Хятадын теоремоор $m \\in \\mathbb{N}$: $am + 1 \\equiv 0 \\pmod{d_i}$, $1 \\le i \\le k$. $n = m b$ гэж авбал $f(n) = a n + b = (a m + 1) b$ тул өгөгдсөн бодлогын нөхцөл хангана. Энд $b = d_{k+1}$ гэж сонгож болно.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 13845,
    "subject": "Mathematics (Olympiad)",
    "question": "令 $n$ 為一正整數。求\n\n$$\n\\sum_{1 \\le r < s \\le 2n} (s - r - n)x_r x_s,\n$$\n\n的最大值，其中 $-1 \\le x_i \\le 1,\\ i = 1, 2, \\dots, 2n$。\n\nLet $n$ be a fixed positive integer. Find the maximum possible value of\n\n$$\n\\sum_{1 \\le r < s \\le 2n} (s - r - n)x_r x_s,\n$$\n\nwhere $-1 \\le x_i \\le 1$ for all $i = 1, 2, \\dots, 2n$.",
    "options": [],
    "answer": "See solution",
    "solution": "The maximum value is $n(n-1)$.\n\nLet $Z$ be the expression to be maximized. Since the equation is linear for all $x_i$, the maximum of $Z$ only happens when $x_i \\in \\{-1, 1\\}$, so it suffices to only consider $x_i = \\pm 1$ for all $i$.\n\nFor all $i$, consider\n\n$$\ny_i = \\sum_{r=1}^{i} x_r - \\sum_{r=i+1}^{2n} x_r.\n$$\n\nTaking the square of both sides, we have\n\n$$\n\\begin{aligned}\ny_i^2 &= \\sum_{r=1}^{2n} x_r^2 + \\sum_{r<s \\le i} 2x_r x_s + \\sum_{i<r<s} 2x_r x_s - \\sum_{r \\le i < s} 2x_r x_s \\\\\n&= 2n + \\sum_{r<s \\le i} 2x_r x_s + \\sum_{i<r<s} 2x_r x_s - \\sum_{r \\le i < s} 2x_r x_s.\n\\end{aligned}\n$$\n\nBy summing over $i$, we get\n\n$$\n\\sum_{i=1}^{2n} y_i^2 = 4n^2 + \\sum_{1 \\le r < s \\le 2n} 4(n - s + r)x_r x_s = 4n^2 - 4Z. \\quad (1)\n$$\n\nSo it suffices to find the minimum of the left-hand side.\n\nIt is easy to see that $y_i$ are even integers. In addition, $y_i - y_{i-1} = 2x_i = \\pm 2$, so $y_i$ and $y_{i-1}$ are consecutive even integers, and therefore, $y_i^2 + y_{i-1}^2 \\ge 4$. Hence,\n\n$$\n\\sum_{i=1}^{2n} y_i^2 = \\sum_{j=1}^{n} (y_{2j-1}^2 + y_{2j}^2) \\ge 4n.\n$$\n\nAs a result, the maximum of $Z$ is $\\frac{1}{4}(4n^2 - 4n) = n(n-1)$, which happens when $x_i = 1$ for all odd $i$ and $x_i = -1$ for all even $i$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 13846,
    "subject": "Mathematics (Olympiad)",
    "question": "In the convex quadrilateral $ABCD$ with angles $ABC$ and $BCD$ equal to $120^\\\text{o}$, $O$ is the intersection of the diagonals, $M$ is the midpoint of $BC$, and $K$ is the point of intersection of $MO$ and $AD$. It happens that $\\angle BKC = 60^\\circ$. Prove that $\\angle BKA = \\angle CKD = 60^\\circ$.\n\n![](images/ukraine_2015_Booklet_p16_data_e081ef6adc.png)",
    "options": [],
    "answer": "See solution",
    "solution": "Let lines $AB$ and $CD$ intersect at $E$. We have that $\\angle BKC = 60^\\circ$. Let us prove that $BK > \\frac{1}{2}BC$. Indeed, one of the angles $\\angle KBC$ or $\\angle KCB$ is at least $60^\\circ$. WLOG, this angle is $\\angle KBC$ (see figure).\n\nIn $\\triangle BMK$, $\\angle KBM \\ge 60^\\circ = \\angle BKC > \\angle BKM$. So $MK > BM$, hence on the segment $BK$ there is a point $O'$, such that $MK \\cdot MO' = BM^2 = MC^2$. Then circumcircles of $\\triangle BO'K$ and $\\triangle CO'K$ are tangent to $BC$. Hence, $\\angle BKM = \\angle O'BC$, $\\angle CKM = \\angle O'CB$. So $\\angle BO'C = 180^\\circ - \\angle O'BC - \\angle O'CB = 180^\\circ - \\angle BKM - \\angle MKC = 180^\\circ - 60^\\circ = 120^\\circ$.\n\n![](images/ukraine_2015_Booklet_p16_data_e081ef6adc.png)\n\nThen point $O'$ is on the circumcircle of $\\triangle EBC$. Let line $BO'$ intersect $CE$ at $D'$, and $CO'$ intersect $BE$ at $A'$. From the fact that $BECO'$ is cyclic, we conclude that (see figure)\n\n$\\angle BO'E = \\angle BCE = 60^\\circ = \\angle EBC = \\angle EO'C$.\n\nConsider $\\triangle O'EB$ and $\\triangle EBD'$. Two pairs of angles are equal. Hence $\\angle BEO' = \\angle ED'B$. Then $\\angle BEO' = \\angle BCO' = \\angle CKO' = \\angle ED'B$. So point $D'$ is on the circumcircle of $\\triangle KO'C$. Similarly, point $A'$ is on the circle $\\triangle BKO'$. Then $\\angle A'KB = \\angle A'O'B = 60^\\circ = \\angle CO'D' = \\angle CKD'$. It follows that points $A'$, $K$, and $D'$ are on the same line.\n\nPoint $O$ lies on segment $MK$. If this point is inside $MO'$, then $A$ lies inside $A'B$, $D'$ inside $CD'$. Then $K$ lies strictly outside $\\triangle EAD$, but by the problem $K$ is inside $AD$, a contradiction. Similarly, if point $O$ is outside $MO'$, then $A$, $D$ are outside $EA'$ and $ED'$, so $K$ is strictly inside $\\triangle EAD$, again a contradiction. Hence $O = O'$. Then $A = A'$ and $D = D'$. But this is already proved that $\\angle A'KB = \\angle CKD' = 60^\\circ$, which is needed, because $A = A'$ and $D = D'$.\n\n**Alternative Solution.** Draw the external bisector of $\\angle BKC$. Let it intersect lines $BA$ and $CD$ at $A'$ and $D'$ (see figure). Construct on sides of $\\triangle BKC$ outside equilateral triangles $BPC$, $BRK$, and $CQK$. Obviously, $R$ and $Q$ are on $A'D'$, and $P$ is the intersection of $AB$ and $CD$.\n\nConsider $\\triangle BKC$. We saw that\n\n![](images/ukraine_2015_Booklet_p16_data_2ea75c6ce8.png)\n\n$$\n\\angle CBR = \\angle CBK + 60^\\circ = 180^\\circ - (120^\\circ - \\angle CBK) = \\angle KBA'\n$$\n\nIt implies that $BR$ and $BA'$ are isogonal, also $KR$ is the external bisector. So $A'$ and $R$ are isogonal, hence $CA'$ and $CR$ are isogonal. Similarly, segments $BD'$ and $BQ$ are isogonal. It is easy to see that $P$ is the intersection of tangents to the circumcircle of $\\triangle BKC$, so $\\angle PBC = \\angle BKC = \\angle PCB = 60^\\circ$. For this, $KP$ is the symmedian of $\\triangle BKC$, so $KP$ and $KM$ are isogonal. As is known, $KP$, $BQ$, and $CR$ intersect in the same point, namely at the Fermat point, so lines $KM$, $BD'$, and $CA'$ intersect in the same point. Let this be point $O'$. Assume that $MO > MO'$ (see figure). Then points $A'$ and $D'$ are on sides $BA$ and $CD$, but in this case $AD$ and $A'D'$ don't intersect each other. Similarly if $MO < MO'$. Hence $MO = MO' \\Rightarrow A = A', D = D' \\Rightarrow \\angle AKB = \\angle DKC$.\n\n![](images/ukraine_2015_Booklet_p16_data_fa14dd512d.png)",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 13847,
    "subject": "Mathematics (Olympiad)",
    "question": "If $p$, $q$, $r$ and $\\sqrt{p} + \\sqrt{q} + \\sqrt{r}$ are rational numbers, then so are $\\sqrt{p}$, $\\sqrt{q}$, and $\\sqrt{r}$.\n\nLet $x$, $y$, $z$ be positive integers such that $\\sqrt{\\frac{2015}{x+y}} + \\sqrt{\\frac{2015}{y+z}} + \\sqrt{\\frac{2015}{z+x}}$ is an integer. Find all possible values of $(x, y, z)$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $S = \\sqrt{p} + \\sqrt{q} + \\sqrt{r}$, which is rational by assumption. Then $\\sqrt{p} + \\sqrt{q} = S - \\sqrt{r}$, so $p + q + 2\\sqrt{pq} = S^2 + r - 2S\\sqrt{r}$, or $2\\sqrt{pq} = S^2 + r - p - q - 2S\\sqrt{r}$. Let $T = S^2 + r - p - q$, which is rational. Then $2\\sqrt{pq} = T - 2S\\sqrt{r}$, so $4pq = T^2 + 4S^2 r - 4ST\\sqrt{r}$. This gives $\\sqrt{r} = \\frac{T^2 + 4S^2 r - 4pq}{4ST}$, which is rational. Similarly, $\\sqrt{p}$ and $\\sqrt{q}$ are rational.\n\nFrom the lemma, $\\sqrt{\\frac{2015}{x+y}}$, $\\sqrt{\\frac{2015}{y+z}}$, and $\\sqrt{\\frac{2015}{z+x}}$ are all rational. Let $N = 2015$, $\\sqrt{\\frac{2015}{x+y}} = \\frac{a}{b}$, $\\sqrt{\\frac{2015}{y+z}} = \\frac{c}{d}$, $\\sqrt{\\frac{2015}{z+x}} = \\frac{e}{f}$, with $a, b, c, d, e, f$ positive integers and $\\gcd(a, b) = \\gcd(c, d) = \\gcd(e, f) = 1$.\n\n$\\sqrt{\\frac{2015}{x+y}} = \\frac{a}{b}$ implies $N b^2 = a^2(x+y)$, so $a^2$ divides $2015$, giving $a = 1$. Similarly, $c = e = 1$. Thus, $\\sqrt{\\frac{2015}{x+y}} + \\sqrt{\\frac{2015}{y+z}} + \\sqrt{\\frac{2015}{z+x}} = \\frac{1}{b} + \\frac{1}{d} + \\frac{1}{f}$ is an integer. Since $b, d, f$ are positive integers, $1 \\leq \\frac{1}{b} + \\frac{1}{d} + \\frac{1}{f} \\leq 3$.\n\n**Case 1.** $\\frac{1}{b} + \\frac{1}{d} + \\frac{1}{f} = 1$. Possible $(b, d, f)$ are $(3, 3, 3)$, $(2, 3, 6)$, $(2, 4, 4)$ (assuming $b \\leq d \\leq f$).\n\n- $(3, 3, 3)$: $x + y = 9N$, $y + z = 9N$, $z + x = 9N$; $x = y = z = \\frac{9N}{2} = 9067.5$ (not integer).\n- $(2, 3, 6)$: $x + y = 9N$, $y + z = 36N$, $z + x = 4N$; $x, y, z$ not all integers.\n- $(2, 4, 4)$: $x + y = 4N$, $y + z = 16N$, $z + x = 16N$; $x = y = 2N = 4030$, $z = 14N = 28210$.\n\n**Case 2.** $\\frac{1}{b} + \\frac{1}{d} + \\frac{1}{f} = 2$. Only $(1, 2, 2)$ is possible, but $x = y = \\frac{N}{2} = 1007.5$, $z = \\frac{7N}{2} = 7052.5$ (not integer).\n\n**Case 3.** $\\frac{1}{b} + \\frac{1}{d} + \\frac{1}{f} = 3$. Only $b = d = f = 1$, $x = y = z = N/2 = 1007.5$ (not integer).\n\n**Conclusion:** The only solution is $(x, y, z) = (4030, 4030, 28210)$ up to permutation.\n\n_Remark:_ The lemma can be generalized to any number of square roots, and its proof follows from Kummer's theory: If $p_1, \\dots, p_k$ are distinct primes, then $\\mathbb{Q}(\\sqrt{p_1}, \\dots, \\sqrt{p_k})$ is a $2^k$-dimensional vector space over $\\mathbb{Q}$, with basis $\\{\\sqrt{p_I}\\}_I$ for all subsets $I$ of $\\{1, \\dots, k\\}$, where $p_I = \\prod_{i \\in I} p_i$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 13848,
    "subject": "Mathematics (Olympiad)",
    "question": "На площині відмічено 2005 точок, жодні три з яких не лежать на одній прямій і жодні чотири не лежать на одному колі. Через усі трійки відмічених точок проведено кола. Доведіть, що ці точки можна розфарбувати у два кольори так, що для будь-яких двох точок одного кольору кількість проведених кіл, які розділяють ці дві точки, є непарною. (Коло розділяє дві точки площини, якщо одна з них лежить всередині, а інша — зовні цього кола.)",
    "options": [],
    "answer": "See solution",
    "solution": "Щоб звести задачу до задачі 11.8, розглянемо у просторі сферу, що дотикається до площини, і спроектуємо на неї всю конструкцію, взявши за центр проектування точку на сфері, протилежну точці дотику (стереографічна проекція). Коло на площині розділяє дві відмічені точки тоді й тільки тоді, коли площина проекції цього кола на сферу розділяє проекції цих точок.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 13849,
    "subject": "Mathematics (Olympiad)",
    "question": "Consider the system of congruences:\n\n$$\n\\begin{cases}\n  n \\equiv p_0 \\pmod{p_0^2} \\\\\n  n + 1 \\equiv p_1 \\pmod{p_1^2} \\\\\n  \\vdots \\\\\n  n + k \\equiv p_k \\pmod{p_k^2}\n\\end{cases}\n$$\n\nwhere $p_0, p_1, \\dots, p_k$ are distinct primes of the form $4t + 3$, and $k \\in \\mathbb{N}$. Show that there is no lattice point $(x, y) \\in \\mathbb{Z}^2$ in the ring $\\{(x, y) \\in \\mathbb{R}^2 \\mid n \\le x^2 + y^2 \\le n + k\\}$.\n\nAdditionally, if $k > \\frac{l^2}{2}$, show that one can place and rotate a needle inside the ring $\\{(x, y) \\in \\mathbb{R}^2 \\mid n \\le x^2 + y^2 \\le n + k\\}$ without passing through any lattice point.",
    "options": [],
    "answer": "See solution",
    "solution": "By the Chinese Remainder Theorem, the system\n\n$$\n\\begin{cases}\n  n \\equiv p_0 \\pmod{p_0^2} \\\\\n  n + 1 \\equiv p_1 \\pmod{p_1^2} \\\\\n  \\vdots \\\\\n  n + k \\equiv p_k \\pmod{p_k^2}\n\\end{cases}\n$$\n\nhas positive integer solutions for any $k \\in \\mathbb{N}$, with $p_0, p_1, \\dots, p_k$ distinct primes of the form $4t + 3$.\n\nSuppose, for contradiction, that there exists $(x, y) \\in \\mathbb{Z}^2$ such that $x^2 + y^2 = n + t$ for some $0 \\le t \\le k$. Then $x^2 + y^2 \\equiv p_t \\pmod{p_t^2}$ and $x^2 + y^2 \\equiv 0 \\pmod{p_t}$, but $x^2 + y^2 \\not\\equiv 0 \\pmod{p_t^2}$, with $p_t = 4l + 3$.\n\nThis implies $x^2 \\equiv -y^2 \\pmod{p_t}$, so $x \\equiv y \\equiv 0 \\pmod{p_t}$, and thus $p_t^2 \\mid x^2 + y^2$, which is a contradiction.\n\nTherefore, there is no lattice point in the ring $\\{(x, y) \\in \\mathbb{R}^2 \\mid n \\le x^2 + y^2 \\le n + k\\}$.\n\nIf $k > \\frac{l^2}{2}$, it is possible to place and rotate a needle inside the ring without passing through any lattice point.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 13850,
    "subject": "Mathematics (Olympiad)",
    "question": "Нека $k_1$, $k_2$ и $k_3$ се три кружници со центри во $O_1$, $O_2$ и $O_3$ соодветно, такви што ниту еден од центрите не се наоѓа во внатрешноста на другите две кружници. Кружниците $k_1$ и $k_2$ се сечат во $A$ и $P$, $k_1$ и $k_3$ се сечат во $C$ и $P$, а $k_2$ и $k_3$ се сечат во $B$ и $P$. Нека $X$ е точка на $k_1$ таква што пресекот на правата $XA$ со кружницата $k_2$ е $Y$, а пресекот на правата $XC$ со $k_3$ е $Z$, при што $Y$ не припаѓа во внатрешноста ниту на $k_1$ ниту на $k_3$, а $Z$ не припаѓа во внатрешноста ниту на $k_1$ ниту на $k_2$.\n\nа) Докажи дека триаголниците $XYZ$ и $O_1O_2O_3$ се слични меѓу себе.\n\nб) Докажи дека плоштината на триаголникот $XYZ$ не е поголема од четири пати по плоштината на триаголникот $O_1O_2O_3$. Дали се достигнува максимумот?\n\n![](images/MACEDONIAN_MATHEMATICAL_SOCIETY_p21_data_596a36d478.png)",
    "options": [],
    "answer": "See solution",
    "solution": "**Решение.**\n\nПрво ќе докажеме дека точките $Y$, $B$ и $Z$ се колинеарни. Бидејќи четириаголникот $BYAP$ е тетивен, имаме $\\angle PBY = \\angle PAX$. Бидејќи четириаголникот $AXCP$ е тетивен, $\\angle PAX = \\angle PCZ$. Бидејќи четириаголникот $CPBZ$ е тетивен, добиваме $\\angle PBZ + \\angle PCZ = 180^\\circ$. Значи $\\angle YBZ = \\angle YBP + \\angle PBZ = 180^\\circ$.\n\nДа забележиме дека $\\angle CO_1O_3 = \\angle PO_1O_3$ и $\\angle AO_1O_2 = \\angle PO_1O_2$, од каде следува $\\angle O_2O_1O_3 = \\frac{1}{2} \\angle AO_1C = \\angle AXC$. Слично, $\\angle O_1O_2O_3 = \\angle AYB$ и $\\angle O_1O_3O_2 = \\angle CZB$. Следува дека $\\triangle XYZ \\sim \\triangle O_1O_2O_3$, со што го докажавме тврдењето под а).\n\nНека правата $X_1Y_1$ е паралелна со $O_1O_2$ и минува низ $A$, каде $X_1$ лежи на $k_1$ и $Y_1$ лежи на $k_2$. Нека $Z_1$ е пресечната точка на правата $X_1C$ со кружницата $k_3$. Од претходно докажаното, точките $Y_1$, $B$ и $Z_1$ се колинеарни и $\\triangle X_1Y_1Z_1 \\sim \\triangle O_1O_2O_3$. Уште повеќе, $\\angle PXA = \\angle PX_1A$ и $\\angle PYA = \\angle PY_1A$. Па $\\triangle PXY \\sim \\triangle PX_1Y_1$. Нека $PT$ е висината спуштена од темето $P$ кон страната $XY$. $PA$ е висината на триаголникот $PX_1Y_1$. Бидејќи $PA$ е хипотенуза во правоаголниот триаголник $PAT$, добиваме $\\overline{PT} \\leq \\overline{PA}$. Па $P_{PXY} \\leq P_{PX_1Y_1}$ и аналогно $P_{PYZ} \\leq P_{PY_1Z_1}$ и $P_{PXZ} \\leq P_{PX_1Z_1}$. Од ова добиваме $P_{XYZ} \\leq P_{X_1Y_1Z_1}$. Точките $P$, $O_1$ и $X_1$ се колинеарни затоа што $\\angle PAX_1 = 90^\\circ$. Слично, $P$, $O_2$ и $Y_1$ се колинеарни и $P$, $O_3$ и $Z_1$ се колинеарни. Добиваме дека $O_1O_2$, $O_1O_3$ и $O_2O_3$ се средни линии во триаголниците $X_1Y_1P$, $X_1Z_1P$ и $Y_1Z_1P$ соодветно, па $P_{X_1Y_1Z_1} = 4P_{O_1O_2O_3}$. Од ова се добива бараното неравенство. Равенство се достигнува кога точките $X$ и $X_1$ се совпаѓаат, а со тоа и точките $Y$ и $Y_1$ се совпаѓаат и точките $Z$ и $Z_1$ се совпаѓаат.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 13851,
    "subject": "Mathematics (Olympiad)",
    "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.\n\nAn *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.\n\nFor 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.\n\nFind 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": "See solution",
    "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 \\leq 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 \\leq Y_1A$.\n\nLet 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 $O$ to $A$, nor one from $O$ to $B$ is suitable.\n\nTake $A = (2, 2)$, $B = (-2, -2)$, $K_1 = (-2, 1)$, $L_1 = (5, 1)$, $K_2 = (2, -1)$, and $L_2 = (-5, -1)$ (see figure below).\n\n![](images/prob1718_p28_data_545ab2e1fe.png)\n\nIf $O = (0, 0)$, then the only shortest paths from $O$ to $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 $O$ to $A$ or $B$. If the segment $OA$ or $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 \\leq K_1B$ (can change $A$ and $B$). If $OK_1A$ is a shortest path from $O$ to $A$, then it is suitable. Otherwise, $OL_1A$ is the only shortest path from $O$ to $A$. It is suitable if $L_1A \\leq L_1B$. Let us further assume that $L_1A > L_1B$. If $OL_1B$ is a shortest path from $O$ to $B$, then it is suitable. Otherwise, $OK_1B$ is the only shortest path from $O$ to $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 \\geq 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 $O$ to $A$ or $B$ exists.\n\n**Answer:** $2$.",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 13852,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $k$ be a positive integer. Consider arranging black and white stones in a single row, with a total of $5^k$ stones. Define a **bad arrangement** as one where, after $k$ repetitions of an operation, the last remaining stone is white.\n\nDefine $f(k)$ as the minimum non-negative integer $m$ such that a bad arrangement can be formed with $5^k - m$ black stones and $m$ white stones. If $m \\geq f(k)$, then there exists a bad arrangement with $5^k - m$ black stones and $m$ white stones.\n\nFind the value of $5^5 - f(5) + 1$.",
    "options": [],
    "answer": "See solution",
    "solution": "We are given that $f(1) = 3$.\n\nTo find $f(k+1)$ in terms of $f(k)$, consider constructing a bad arrangement $W_k$ with $5^k - f(k)$ black stones and $f(k)$ white stones. Arrange $2 \\cdot 5^k$ black stones, followed by three instances of $W_k$, forming a row of $5^{k+1}$ stones. The number of white stones is $3f(k)$. After $k$ repetitions, the final arrangement consists of black, black, white, white, white stones, so the last remaining stone is white. Thus, $f(k+1) \\leq 3f(k)$.\n\nConversely, if there are at most $3f(k) - 1$ white stones in a row of $5^{k+1}$ stones, dividing into $5^k$-stone sets, at most 2 sets can have $f(k)$ or more white stones. By minimality of $f(k)$, after $k$ repetitions, at most 2 white stones remain among the last 5, so the last stone is black. Thus, $f(k+1) > 3f(k) - 1$.\n\nTherefore, $f(k+1) = 3f(k)$. Since $f(1) = 3$, we have $f(5) = 3^4 \\cdot 3 = 3^5 = 243$.\n\nThe desired value is:\n\n$$\n5^5 - f(5) + 1 = 3125 - 243 + 1 = 2883\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 13853,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $n$ cities be connected in a 2-connected graph (i.e., the graph remains connected after removing any single edge). What is the minimal number $k$ of roads that must be oriented so that for any fixed road $l$, there exists a directed path from city $A$ to city $B$ passing through $l$?",
    "options": [],
    "answer": "See solution",
    "solution": "The minimal number is $k = 2n - 3$.\n\nIf all pairs of cities are directly connected (complete graph), then all roads from $A$ and all roads to $B$ must be oriented: $k \\ge (n-1) + (n-1) - 1 = 2n - 3$. Alternatively, if $\\deg(A) = \\deg(B) = n - 1$ and all other vertices have degree 2, then again we must orient $2(n-2) + 1 = 2n - 3$ roads.\n\nTo show that orienting at most $2n - 3$ roads suffices, we proceed as follows:\n\n1. **Stage 1:** Orient all roads in a minimal path $\\Gamma_1$ (without self-intersections) from $A$ to $B$.\n2. **Stage 2:** If not all cities are included, orient all roads in a minimal path $\\Gamma_2$ connecting two cities on $\\Gamma_1$ and including new cities not on $\\Gamma_1$.\n3. **Stage 3:** Repeat, orienting minimal paths $\\Gamma_3, \\Gamma_4, \\dots$ until all cities are included.\n\nAfter $q$ steps, $\\Gamma = \\bigcup_{i=1}^q \\Gamma_i$ contains all cities. Any road $l$ either belongs to some $\\Gamma_i$ or connects two $\\Gamma_i$. If $l$ belongs to some $\\Gamma_s$, there is a path from $A$ to $B$ passing through $l$ by following oriented edges. If $l$ connects $\\Gamma_i$ and $\\Gamma_j$, a path from $A$ to $B$ passing through $l$ can be constructed similarly.\n\nThe total number of oriented roads is at most $2n - 3$. At each stage, if a path $\\Gamma_i$ includes $l$ new cities, we orient $l + 1$ roads. The maximum occurs when $\\Gamma_1$ includes one road and each subsequent $\\Gamma_i$ includes two roads, totaling $2n - 3$ oriented roads.\n\nThus, $k = 2n - 3$ is minimal.",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 13854,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $ABC$ be a triangle with $CA = CB$ and $\\angle ACB = 120^\\circ$, and let $M$ be the midpoint of $AB$. Let $P$ be a variable point on the circumcircle of $ABC$, and let $Q$ be the point on the segment $CP$ such that $QP = 2QC$. It is given that the line through $P$ and perpendicular to $AB$ intersects the line $MQ$ at a unique point $N$.\n\nProve that there exists a fixed circle such that $N$ lies on this circle for all possible positions of $P$.",
    "options": [],
    "answer": "See solution",
    "solution": "Consider triangles $QNP$ and $QMC$. Vertically opposite angles $NQP$ and $MQC$ are equal. We also have that $CM$ is perpendicular to $AB$ since $\\triangle ABC$ is isosceles and $M$ is the midpoint of $AB$. Given that $NP$ is also perpendicular to $AB$, $CM$ and $NP$ are parallel. Therefore $\\angle NPQ = \\angle MCQ$ and $\\angle PNQ = \\angle CMQ$. Hence triangles $QNP$ and $QMC$ are similar.\n\nSince $QP = 2QC$, it follows that $NP = 2MC$. Furthermore, $NP$ is parallel to $MC$. The point $N$ is therefore a translation of $P$ by the fixed vector $2\\overrightarrow{MC}$. Therefore, as $P$ varies on the circumcircle of $ABC$, there is a fixed circle on which $N$ lies. $\\square$\n\n![](images/2018-Australian-Scene-W2_p142_data_c97f4d32f2.png)\n\n**Comment** From the condition that $\\angle ACB = 120^\\circ$, it follows that $M$ is the midpoint of $CO$, where $O$ is the circumcentre of $ABC$. That is, $2MC = OC$. Therefore, $N$ lies on the circle ($\\omega$, say) that is the image of the circumcircle of $ABC$ through the translation that sends $O$ to $C$; that is, the circle with centre $C$ and radius $CO$.\n\nAlso note that since $N$ is defined uniquely, $P$ does not lie on the line $CO$, and therefore $N$ also does not lie on the line $CO$. The possible positions of $N$ are in fact all the points of $\\omega$ with the exception of the two points lying on the line $CO$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 13855,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $f(x)$ be a function and let $x_1, x_2, x_3$ be real numbers such that $f(x_1) = f(x_2) = f(x_3) = c$, and $x_1, x_2, x_3$ are the three real roots of the cubic equation $x^3 + a x^2 + b x - c = 0$. Suppose that $x_2 - x_1 \\geq 1$, $x_3 - x_2 \\geq 1$, and $x_3 - x_1 \\geq 2$. Find the minimum value of $|a| + 2|b|$.",
    "options": [],
    "answer": "See solution",
    "solution": "By Vieta's formulas for the cubic equation $x^3 + a x^2 + b x - c = 0$ with roots $x_1, x_2, x_3$, we have:\n\n$$\na = -(x_1 + x_2 + x_3), \\quad b = x_1 x_2 + x_2 x_3 + x_3 x_1.\n$$\n\nGiven the conditions $x_2 - x_1 \\geq 1$, $x_3 - x_2 \\geq 1$, $x_3 - x_1 \\geq 2$, we compute:\n\n$$\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} \\left( (x_1 - x_2)^2 + (x_2 - x_3)^2 + (x_3 - x_1)^2 \\right) \\\\\n&\\geq \\frac{1}{2}(1 + 1 + 4) = 3.\n\\end{aligned}\n$$\n\nThus, $b \\leq \\frac{a^2}{3} - 1$.\n\nIf $|a| \\geq \\sqrt{3}$, then $|a| + 2|b| \\geq |a| + \\sqrt{3}$.\n\nIf $0 \\leq |a| < \\sqrt{3}$, then $b \\leq \\frac{a^2}{3} - 1 < 0$. At this point,\n\n$$\n\\begin{aligned}\n|a| - \\frac{3}{4} &< \\sqrt{3} - \\frac{3}{4}, \\\\\n|a| + 2|b| &\\geq |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$$\n\nTherefore, $|a| + 2|b| \\geq \\sqrt{3}$.\n\nFor $a = \\sqrt{3}$, $b = 0$, the real numbers $x_1 = -1 - \\frac{\\sqrt{3}}{3}$, $x_2 = -\\frac{\\sqrt{3}}{3}$, $x_3 = 1 - \\frac{\\sqrt{3}}{3}$ satisfy the conditions, and the values of $f(x_1)$, $f(x_2)$, $f(x_3)$ are all $\\frac{2\\sqrt{3}}{9}$. At this point, $|a| + 2|b| = \\sqrt{3}$.\n\n**Conclusion:** The minimum value of $|a| + 2|b|$ is $\\sqrt{3}$.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 13856,
    "subject": "Mathematics (Olympiad)",
    "question": "令 $\\mathbb{R}$ 表示實數所成的集合。給定實數 $t \\neq -1$。找出所有的函數 $f : \\mathbb{R} \\to \\mathbb{R}$ 使得\n\n$$\n(t + 1)f(1 + xy) - f(x + y) = f(x + 1)f(y + 1)\n$$\n成立。",
    "options": [],
    "answer": "See solution",
    "solution": "首先，易知 $f(x) \\equiv 0$ 為方程的一組解，故假設 $f(x)$ 不總是 $0$。原式代入 $(x - 1, 1)$ 得到 $f(2) = t$。\n\n設 $f(1) = a$。代入 $(x, 0)$ 得到\n\n$$\n(t + 1)a - f(x) = a f(x + 1) \\quad (1)\n$$\n\n若 $a = 0$，則由 (1) 知 $f(x) \\equiv 0$，不合。\n\n所以\n\n$$\na \\neq 0, \\quad f(x + 1) = t + 1 - \\frac{f(x)}{a} \\quad (2)\n$$\n\n(1) 中 $x$ 代 $x - 1$ 整理得\n\n$$\nf(x - 1) = (t + 1)a - a f(x) \\quad (3)\n$$\n\n(3) 代 $1, 0, -1$ 知\n\n$$\nf(0) = -a^2 + (t + 1)a,\n$$\n$$\nf(-1) = a^3 - (t + 1)a^2 + (t + 1)a,\n$$\n$$\nf(-2) = -a^4 + (t + 1)a^3 - (t + 1)a^2 + (t + 1)a.\n$$\n\n原式代入 $(t + 1)f(2) - f(-2) = f(0)^2$\n\n所以 $(t + 1)t + a^4 - (t + 1)a^3 + (t + 1)a^2 - (t + 1)a = a^4 - 2(t + 1)a^3 + (t^2 + 2t + 1)a^2$\n\n整理得 $(a^2 - 1)(t + 1)(a - t) = 0$，由 $t \\neq -1$ 知 $a = \\pm 1$ 或 $a = t$。\n\n若 $a = t \\neq \\pm 1$，則 $f(0) = -a^2 + (t+1)a = t$\n\n在原式中代入 $(x, -1)$ 得到\n\n$$\n(t+1)f(1-x) - f(x-1) = f(0)f(x+1) = t f(x+1) \\quad (4)\n$$\n\n利用 (2) 和 (3) 將 (4) 全部換成 $f(x)$ 和 $f(-x)$ 整理得到\n\n$$\nt f(x) = f(-x) + t^2 - t\n$$\n\n因此 $t^2 f(x) = f(-x) + t^3 - t^2$\n\n故 $(t^2 - 1)f(x) = t^3 - t$，由 $t \\neq \\pm 1$ 知 $f(x) = t$。\n\n若 $a = 1$，由 (1) 知\n\n$$\nf(x) + f(x+1) = t + 1. \\quad (5)\n$$\n\n所以由 $f(2) = t$ 知 $f(3) = 1$。\n\n原式代入 $(2, \\frac{1}{2})$ 得到 $t^2 + t - f(\\frac{5}{2}) - f(\\frac{3}{2}) = 0$，但 $f(\\frac{5}{2}) + f(\\frac{3}{2}) = t + 1$，\n所以 $t^2 - 1 = 0$。\n\n由 $t \\neq -1$ 知 $t = 1$。代入 $(x, 2)$ 得到 $2f(2x+1)-f(x+2) = f(x+1)f(3)$。\n\n由 (5) 知 $2f(2x+1) = f(x+1)+f(x+2) = 2$，所以 $f(2x+1) = 1$，也就是說 $f(x) \\equiv 1 = t$。\n\n若 $a = -1$，則 (2) 變為\n\n$$\nf(x+1) = f(x) + (t+1) \\quad (6)\n$$\n\n令 $g(x) = \\frac{f(x)+(t+2)}{t+1}$ 代回原式整理得\n\n$$\n(t+1)g(xy) + g(x) + g(y) = (t+1)g(x)g(y) + g(x+y) \\quad (7)\n$$\n\n(6) 變為 $g(x+1) = g(x) + 1$，且 $g(2) = 2$，故 $g(1) = 1$，$g(-1) = -1$。\n\n(7) 代入 $(x, -1)$ 得到 $g(-x) = -g(x)$，也就是奇函數。\n\n(7) 再代入 $(x, y)$、$(-x, -y)$ 兩式相減得到 $g(x) + g(y) = g(x + y)$\n\n代回 (7) 得到 $g(xy) = g(x)g(y)$，結合這兩式即得到 $g(x) = x$。\n\n因此 $f(x) = (t+1)g(x) - (t+2) = (t+1)x - (t+2)$。\n\n綜合上述，我們得到了三組解：\n\n$$\nf(x) \\equiv 0, \\quad f(x) \\equiv t, \\quad f(x) = (t+1)x - (t+2).\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 13857,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $a$, $b$, $c$ be positive real numbers such that $abc = 1$. Prove that\n$$\n\\frac{a}{\\sqrt{a^2 + 8bc}} + \\frac{b}{\\sqrt{b^2 + 8ca}} + \\frac{c}{\\sqrt{c^2 + 8ab}} \\geq 1.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "Let $\\alpha = \\sqrt{1 + \\frac{8bc}{a^2}}$, and define $\\beta$, $\\gamma$ similarly. We have $\\alpha, \\beta, \\gamma > 1$, and\n$$\n(\\alpha^2 - 1)(\\beta^2 - 1)(\\gamma^2 - 1) = 8^3.\n$$\nLet $x = \\alpha + \\beta + \\gamma$, $y = \\alpha\\beta + \\beta\\gamma + \\gamma\\alpha$, $z = \\alpha\\beta\\gamma$. Suppose, for the sake of contradiction, that the conclusion is false; then\n$$\n\\frac{1}{\\alpha} + \\frac{1}{\\beta} + \\frac{1}{\\gamma} < 1 \\quad \\text{or} \\quad y < z.\n$$\nExpanding the previous equation gives\n$$\n\\alpha^2\\beta^2\\gamma^2 - (\\alpha^2\\beta^2 + \\beta^2\\gamma^2 + \\gamma^2\\alpha^2) + (\\alpha^2 + \\beta^2 + \\gamma^2) - 1 = 8^3,\n$$\nor\n$$\nz^2 - (y^2 - 2xz) + (x^2 - 2y) - 1 = 8^3.\n$$\nUsing the second part of the assumption, we have\n$$\n8^3 + 1 = 2xz + x^2 - 2y + (z^2 - y^2) > 2xz + x^2 - 2z.\n$$\nApplying the AM-GM Inequality to the first part of the assumption yields\n$$\n1 > \\frac{1}{\\alpha} + \\frac{1}{\\beta} + \\frac{1}{\\gamma} \\geq 3\\sqrt[3]{\\frac{1}{z}},\n$$\nor $z > 3^3$. Now by the AM-GM Inequality,\n$$\nx = \\alpha + \\beta + \\gamma \\geq 3\\sqrt[3]{\\alpha\\beta\\gamma} = 3\\sqrt[3]{z} > 9.\n$$\nThus,\n$$\n2xz + x^2 - 2z = 2(x-1)z + x^2 \\geq 2(9-1)3^3 + 9^2 = 8^3 + 1,\n$$\ncontradicting the previous inequality. Therefore, our initial assumption was wrong, and the conclusion is true.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 13858,
    "subject": "Mathematics (Olympiad)",
    "question": "Given a $10 \\times 10 \\times 10$ box made up of 1000 unit white cubes, An and Binh play a game. An selects some bands of size $1 \\times 1 \\times 10$ such that any two chosen bands have no common points, and then changes all cubes on these bands to black. Binh then selects some unit cubes and asks An the color of these cubes. What is the least number of cubes Binh must choose in order to determine all black cubes based on An's answers?",
    "options": [],
    "answer": "See solution",
    "solution": "We first prove a general statement: Given a box of size $2n \\times 2n \\times 2n$ containing $8n^3$ white unit cubes, An and Binh play the same game. An chooses some bands of size $1 \\times 1 \\times 2n$ such that no two bands share a vertex or edge, and changes all cubes on these bands to black. Binh can choose a number of unit cubes and ask An their colors. In this case, Binh needs to select at least $6n^2$ unit cubes to determine all black cubes based on An's answers.\n\nPlace the box in an $Oxyz$ coordinate system so that each side is parallel to one of the axes $Ox$, $Oy$, or $Oz$. Let $S_n$ be the set of cells that Binh uses to ask An, and for each chosen cell $u$, let $R_u$ be the set containing all the unit cubes in the $1 \\times 1 \\times 2n$ band passing through $u$. Since any two chosen bands have no common point, for any black cell $u$, Binh must choose two other cells on two of the three bands passing through $u$ to determine which band is changed to black. Now, assign a tuple $(a, b, c)$ to each cell $u$ of the box as follows:\n\n![](images/Vietnamese_mathematical_competitions_p81_data_1d9c97a80d.png)\n\n- $a = 2$ if the band passing through $u$ and parallel to $Ox$ does not have any cells in $S_n$ other than $u$, and $a = 1$ otherwise.\n- $b = 2$ if the band passing through $u$ and parallel to $Oy$ does not have any cells in $S_n$ other than $u$, and $b = 1$ otherwise.\n- $c = 2$ if the band passing through $u$ and parallel to $Oz$ does not have any cells in $S_n$ other than $u$, and $c = 1$ otherwise.\n\nFrom the above, it is clear that\n\n$$\na + b + c \\le 4.\n$$\n\nLet $T$ be the sum of all assigned numbers on the box, then\n\n$$\nT = \\sum_{u \\in S_n} (a + b + c) \\le 4|S_n|.\n$$\n\nOn the other hand, there is at least one chosen cell on each $1 \\times 1 \\times n$ band (in any direction) of the box, so each band contributes at least 2 units to the sum $T$. This implies that\n\n$$\nT \\ge 3 \\cdot 2(2n)^2 = 24n^2.\n$$\n\nHence, $4|S_n| \\ge 24n^2$ or $|S_n| \\ge 6n^2$.\n\nTherefore:\n\n- In the $2 \\times 2 \\times 2$ box, Binh needs to choose at least 6 cells.\n- In the $10 \\times 10 \\times 10$ box, Binh needs to choose at least 150 cells.\n\nConsider the simplest case: a $2 \\times 2 \\times 2$ box. In this box, Binh removes two opposite cells and asks for the color of the others. It is easy to check that this method works.\n\n![](images/Vietnamese_mathematical_competitions_p82_data_2b7887f3ad.png)",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 13859,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $M$ be the centroid and $O$ be the circumcenter of a triangle $ABC$. Take a point $K$ on the line $OM$ such that $\\angle KAB = \\angle ABC$ and the points $K$ and $C$ are on the same side of the line $AB$. Prove that $\\angle KCB = \\angle ABC$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $P$ and $Q$ be the midpoints of the segments $AB$ and $BC$, respectively. Let $F = CK \\cap AB$ and $E = AK \\cap BC$. Since $\\angle BAK = \\angle ABC$, the triangle $AEB$ is isosceles. Hence, $EP$ is an altitude of $\\triangle AEB$, so $O \\in EP$.\n\nBy assumption, $AQ \\cap PC = M$ and $CF \\cap AE = K$. Let $O' = FQ \\cap PE$. By Pappus' theorem for $AFPQEC$, the points $M$, $K$, and $O'$ are collinear.\n\nOn the other hand, $MK \\cap EP = O$. Therefore, $O = O'$. Thus, $F \\in OQ$ and $FB = FC$. Hence, $\\angle FBC = \\angle BCF$.\n\n![](images/MNG_ABooklet_2015_p16_data_40ff4f65d2.png)",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 13860,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $\\mathbb{N}_0^+$ and $\\mathbb{Q}$ be the set of nonnegative integers and rational numbers, respectively. Define the function $f : \\mathbb{N}_0^+ \\to \\mathbb{Q}$ by $f(0) = 0$ and\n$$\nf(3n + k) = -\\frac{3f(n)}{2} + k, \\quad \\text{for } k = 0, 1, 2.\n$$\nProve that $f$ is one-to-one, and determine its range.",
    "options": [],
    "answer": "See solution",
    "solution": "We prove that the range of $f$ is the set $T$ of rational numbers of the form $m/2^n$ for $m \\in \\mathbb{Z}$ and $n \\in \\mathbb{N}_0^+$ (also known as the *dyadic rational numbers*). For $x, y \\in T$, we write $x \\equiv y \\pmod 3$ to mean that the numerator of $x - y$, when written in lowest terms, is divisible by 3. Define the function $g : T \\to \\{0, 1, 2\\}$ by declaring that for $x \\in T$, $g(x)$ is the unique element of $\\{0, 1, 2\\}$ such that $g(x) \\equiv x \\pmod 3$.\n\nWe first check that $f$ is one-to-one. Suppose that $f(a) = f(b)$ for some $a, b \\in \\mathbb{N}_0^+$ with $a < b$; choose such a pair with $b$ minimal. Now note that $f(a) \\equiv a \\pmod 3$ and $f(b) \\equiv b \\pmod 3$ from the definition of $f$. Hence $a \\equiv b \\pmod 3$. Choose the $i \\in \\{0, 1, 2\\}$ that is congruent to $a$ and $b$ modulo 3, and put $a' = (a-i)/3$ and $b' = (b-i)/3$. Then $f(a') = f(b')$, but $a' < b'$ and $b' < b$ since $b > 0$. This contradicts the choice of $a$ and $b$. Hence no such pairs exist.\n\nWe next verify that $T$, which clearly contains the range of $f$, is in fact equal to it. Define the map $h : T \\to T$ by setting $h(x) = \\frac{2}{3}(g(x)-x)$. Then $x$ is in the image of $f$ whenever $h(x)$ is: if $h(x) = f(a)$, then $x = f(3a + g(x))$. Hence it suffices to show that if one starts from $x$ and applies $h$ repeatedly, one eventually ends up with an element of the range of $f$.\n\nFirst note that if $x$ is not an integer, then $g(x) - x$ has the same denominator as $x$, and $h(x)$ has denominator half of that. Hence some $x_i$ is an integer.\n\nNext, by the \\textbf{Triangle Inequality},\n$$\n|h(x)| \\leq \\frac{2}{3}(2 + |x|) < |x|\n$$\nwhenever $|x| > 4$. Thus, our process eventually hits an integer of absolute value bounded by 4. Simple computation yields the following table:\n![](table)\n\nFrom the table, it is clear that starting from any point in $\\{-4, -3, \\ldots, 4\\}$, within 4 applications of $h$, our process hits zero. Since $f(0) = 0$, this value is in the range of $f$, so we are done.\n\n**Note:** It is of interest to observe that the function $f$ can be interpreted as follows: to compute $f(n)$, first write $n$ in base 3, and then read off the digit string as if it were in base $(-3/2)$. Therefore, the point of this question is to prove that all dyadic rationals can be uniquely expressed in base $(-3/2)$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 13861,
    "subject": "Mathematics (Olympiad)",
    "question": "Construct outside the square $ABCD$ the right isosceles triangle $ABE$, with hypotenuse $AB$. Denote $N$ the midpoint of the segment $AD$ and $\\{M\\} = CE \\cap AB$, $\\{P\\} = CN \\cap AB$, $\\{F\\} = PE \\cap MN$. Take on the straight line $FP$ the point $Q$ so that $CE$ is the bisector of the angle $\\angle QCB$. Prove that $MQ \\perp CF$.",
    "options": [],
    "answer": "See solution",
    "solution": "Clearly $PA = PB/2$, so $PA = AB = BC$. This leads to $\\triangle APE \\equiv \\triangle BCE$ (SAS), which implies $CE = PE$ and $\\overline{BEC} \\equiv \\overline{AEP}$.\n\n![](images/RMC2014_p30_data_7207ae9344.png)\n\n$m(\\widehat{BEC}) + m(\\widehat{AEC}) = 90^\\circ$ yields $m(\\widehat{AEC}) + m(\\widehat{AEP}) = 90^\\circ$, so $m(\\widehat{CEP}) = 90^\\circ$. This shows that $ECP$ is a right isosceles triangle, so $m(\\widehat{CPE}) = 45^\\circ$.\n\nAlso $\\triangle CEQ \\equiv \\triangle PEM$ (AAS), whence $EQ = EM$, so $EMQ$ is a right isosceles triangle. This gives $m(\\widehat{EQM}) = 45^\\circ = m(\\widehat{EPC})$, therefore $MQ \\parallel CP$. Menelaus' Theorem for the triangle $CPE$ and the transversal $NF$ yields\n$$\n\\frac{CN}{NP} \\cdot \\frac{FP}{FE} \\cdot \\frac{ME}{MC} = 1.\n$$\nIf we denote $R$ the orthogonal projection of $E$ onto $AB$, then $\\triangle CBM \\sim \\triangle ERM$, whence $\\frac{ME}{MC} = \\frac{ER}{BC} = \\frac{1}{2}$.\n\nFrom $NP = CN$ follows $\\frac{FP}{FE} = 2$, hence $PE = EF$, therefore $CE$ is altitude and median in $\\triangle CFP$, that is, this triangle is isosceles. This gives $m(\\widehat{FCP}) = 90^\\circ$, that is $CP \\perp CF$, whence $MQ \\perp CF$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 13862,
    "subject": "Mathematics (Olympiad)",
    "question": "令 $\\mathbb{R}$ 表示所有實數所成的集合。試求所有可能的函數 $f : \\mathbb{R} \\to \\mathbb{R}$ 滿足：\n\n對任意實數 $x, y$，恆有\n$$\nf(f(x) + y) = f(x^2 - y) + 4(y - 2)(f(x) + 2).\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "令 $y = \\frac{x^2 - f(x)}{2}$ 代入原式，得\n\n$$\nf\\left(\\frac{x^2 + f(x)}{2}\\right) = f\\left(\\frac{x^2 + f(x)}{2}\\right) + 4\\left(\\frac{x^2 - f(x)}{2} - 2\\right)(f(x) + 2).\n$$\n\n得知：$(x^2 - f(x) - 4)(f(x) + 2) = 0$，因此對每一個實數 $x$，恆有\n$$\nf(x) = x^2 - 4 \\quad \\text{或} \\quad f(x) = -2.\n$$\n\n進一步檢驗：\n\n1. $f(x) = x^2 - 4$：\n   - 左式：$f(f(x) + y) = f(x^2 - 4 + y) = (x^2 - 4 + y)^2 - 4$\n   - 右式：$f(x^2 - y) + 4(y - 2)(f(x) + 2) = (x^2 - y)^2 - 4 + 4(y - 2)(x^2 - 2)$\n   - 化簡後兩式相等，故 $f(x) = x^2 - 4$ 為一解。\n\n2. $f(x) = -2$：\n   - 代入原式，左式與右式均為 $-2$，故 $f(x) = -2$ 亦為一解。\n\n若存在某 $a$ 使 $f(a) \\neq a^2 - 4$，則 $f(a) = -2$，且 $a^2 - 2 \\neq 0$。進一步推導可證 $f$ 必為常數函數 $f(x) = -2$。\n\n因此，滿足條件之函數恰有兩個：\n$$\nf(x) = -2 \\quad \\text{與} \\quad f(x) = x^2 - 4.\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 13863,
    "subject": "Mathematics (Olympiad)",
    "question": "The set of real numbers is split into two subsets which do not intersect. Prove that for each pair $ (m, n) $ of positive integers, there are real numbers $ x < y < z $ all in the same subset such that $ m(z - y) = n(y - x) $. \n\nOne form of van der Waerden's theorem states that if the integers are split into two disjoint subsets, then for every $ k > 0 $ there must be an arithmetic progression of length $ k $ in one of the subsets. Can you see how to finish the question from here? However, van der Waerden's theorem is quite a lot harder to prove than the statement of this question, so a more elementary solution is desirable!",
    "options": [],
    "answer": "See solution",
    "solution": "First, we will prove it for the pair $ (m, n) = (1, 1) $—so we have to show that one of the subsets contains an arithmetic progression of length 3. Call the subsets $ X $ and $ Y $. Suppose that $ a < b $ are in the same subset, say they are in $ X $. Then, if there is no arithmetic progression of length 3, none of $ 2a-b $, $ \\frac{a+b}{2} $, or $ 2b-a $ can be in $ X $. But then $ 2a-b $, $ \\frac{a+b}{2} $, $ 2b-a $ is a three-term arithmetic progression in $ Y $.\n\nNow we do the case where $ m \\neq n $. We have shown that one of the subsets, say $ X $, contains a three-term arithmetic progression. We can write this arithmetic progression as $ cm $, $ cm+dm $, $ cm+2dm $. Suppose that there are no $ x < y < z $ in $ X $ satisfying $ m(z - y) = n(y - x) $—call this equation R.\n\nConsider $ cm + dm + dn $. Then $ x = cm $, $ y = cm + dm $, $ z = cm + dm + dn $ satisfy equation R, and $ x $ and $ y $ are in $ X $, so $ z = cm + dm + dn $ is in $ Y $.\n\nNow consider $ cm + 2dm + dn $. Then $ x = cm + dm $, $ y = cm + 2dm $, $ z = cm + 2dm + dn $ satisfy equation R, and $ x $ and $ y $ are in $ X $, so $ z = cm + 2dm + dn $ is in $ Y $.\n\nFinally, consider $ cm + 2dm + 2dn $. Then $ x = cm $, $ y = cm + 2dm $, $ z = cm + 2dm + 2dn $ satisfy equation R, and $ x $ and $ y $ are in $ X $, so $ z = cm + 2dm + 2dn $ is in $ Y $.\n\nBut now $ x = cm + dm + dn $, $ y = cm + 2dm + dn $, and $ z = cm + 2dm + 2dn $ are in $ Y $, and satisfy equation R. We have shown that either $ X $ or $ Y $ contains a triple $ (x, y, z) $ satisfying equation R, as required.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 13864,
    "subject": "Mathematics (Olympiad)",
    "question": "Write $f(p) = 3^p + 4^p + 5^p + 9^p - 98$. Find all prime numbers $p$ for which $f(p)$ has at most 6 positive divisors.",
    "options": [],
    "answer": "See solution",
    "solution": "Note that we have the prime factorizations $f(2) = 3 \\cdot 11$, $f(3) = 7 \\cdot 11^2$, and $f(5) = 7 \\cdot 9049$. Therefore, $f(2), f(3), f(5)$ have 4, 6, and 4 positive divisors respectively, as the number of positive divisors of the prime factorization $p_1^{e_1} p_2^{e_2} \\cdots p_n^{e_n}$ is $(e_1 + 1)(e_2 + 1)\\cdots(e_n + 1)$.\n\nNow let $p > 5$, and suppose for a contradiction that $f(p)$ has at most 6 positive divisors. First, note that modulo 7, we have $f(p) \\equiv 3^p + (-3)^p + 5^p + (-5)^p - 0 \\equiv 0 \\pmod{p}$ as $p$ is odd.\n\nNext, consider $f(p)$ modulo 11. By Fermat's little theorem, $a^{10} \\equiv 1 \\pmod{11}$ for any integer $a \\not\\equiv 0 \\pmod{11}$, so the residue class of $f(p)$ modulo 11 is constant on any residue class modulo 10. Note that $p$ must be one of $1, 3, -3, -1 \\pmod{10}$, since otherwise $p$ would have contained either 2 or 5 as a non-trivial factor.\n\nNow, by a straightforward computation, $\\{3^3, 4^3, 5^3, 9^3\\}$ contains the same residue classes modulo 11 as $\\{3, 4, 5, 9\\}$. Repeating the same argument shows that the same holds for $\\{3^{-1}, 4^{-1}, 5^{-1}, 9^{-1}\\}$ and $\\{3^{-3}, 4^{-3}, 5^{-3}, 9^{-3}\\}$ as well. As $f(p) \\equiv 3^p + 4^p + 5^p + 9^p - 98 \\pmod{11}$, it follows $f(p)$ has the same value modulo 11 for every $p \\equiv 1, 3, -3, -1 \\pmod{10}$, and therefore for every $p > 5$.\n\nWe use the case $p \\equiv 1 \\pmod{10}$ to compute this value:\n\n$$\n\\begin{aligned}\nf(p) &\\equiv 3^1 + 4^1 + 5^1 + 9^1 - 98 \\pmod{11} \\\\\n&\\equiv 3 + 4 + 5 + 9 + 1 \\pmod{11} \\\\\n&\\equiv 0 \\pmod{11}.\n\\end{aligned}\n$$\n\nWe deduce that $11 \\mid f(p)$ for all $p > 5$.\n\nNow note that for all $p > 5$ we have $f(p) > 9^5 = 9 \\cdot 81 \\cdot 81 > 7 \\cdot 11 \\cdot 77$. Hence we can write $f(p) = 7 \\cdot 11 \\cdot d$ with $d > 77$. Then we see that $f(p)$ has at least 8 positive divisors, namely\n\n$$\n1 < 7 < 11 < 77 < d < 7d < 11d < 77d.\n$$\n\nSo $p = 2, 3, 5$ are indeed the only prime numbers for which $f(p)$ has at most 6 positive divisors. $\\square$",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 13865,
    "subject": "Mathematics (Olympiad)",
    "question": "Calculate the value of\n\n$$\n\\frac{2014^4 + 4 \\times 2013^4}{2013^2 + 4027^2} - \\frac{2012^4 + 4 \\times 2013^4}{2013^2 + 4025^2}\n$$\n",
    "options": [],
    "answer": "See solution",
    "solution": "Let $a = 2013$. Then the given expression can be written as\n\n$$\n\\begin{aligned}\n& \\frac{(a+1)^4 + 4a^4}{a^2 + (2a+1)^2} - \\frac{(a-1)^4 + 4a^4}{a^2 + (2a-1)^2} \\\\\n&= \\frac{a^4 + 4a^3 + 6a^2 + 4a + 1 + 4a^4}{a^2 + 4a^2 + 4a + 1} - \\frac{a^4 - 4a^3 + 6a^2 - 4a + 1 + 4a^4}{a^2 + 4a^2 - 4a + 1}\n\\end{aligned}\n$$\n\nwhich can then be simplified to\n\n$$\n\\begin{aligned}\n& \\frac{5a^4 + 4a^3 + 6a^2 + 4a + 1}{5a^2 + 4a + 1} - \\frac{5a^4 - 4a^3 + 6a^2 - 4a + 1}{5a^2 - 4a + 1} \\\\\n&= \\frac{a^2(5a^2 + 4a + 1) + 5a^2 + 4a + 1}{5a^2 + 4a + 1} - \\frac{a^2(5a^2 - 4a + 1) + 5a^2 - 4a + 1}{5a^2 - 4a + 1} \\\\\n&= (a^2 + 1) - (a^2 + 1)\n\\end{aligned}\n$$\n\nand so the answer is $0$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 13866,
    "subject": "Mathematics (Olympiad)",
    "question": "For which largest $k$ does there exist a permutation $(a_1, a_2, \\ldots, a_{2022})$ of integers $(1, 2, \\ldots, 2022)$ such that for some $k$ integers $1 \\leq i \\leq 2022$ the fraction\n\n$$\n\\frac{a_1 + a_2 + \\cdots + a_i}{1 + 2 + \\cdots + i}\n$$\n\nis an integer larger than $1$?",
    "options": [],
    "answer": "See solution",
    "solution": "Denote $s_i = a_1 + a_2 + \\cdots + a_i$ and $t_i = 1 + 2 + \\cdots + i$. We will show that there exists at most one $i \\geq 1011$ for which $s_i \\neq t_i$ is divisible by $t_i$.\n\nIndeed, note that\n$$\ns_i \\leq 2022 + 2021 + \\cdots + (2023 - i) = 2023i - \\frac{i(i+1)}{2}.\n$$\nAlso, for $i \\geq 1011$ we have $2023i - 1011 < 3t_i$, as $4t_i = 2i(i+1) \\geq 2024i$. So if for $i \\geq 1011$, $\\frac{s_i}{t_i}$ is an integer larger than $1$, then $s_i = 2t_i$.\n\nSuppose there exist two such $i > j \\geq 1011$ with $s_i = 2t_i$ and $s_j = 2t_j$. Then\n$$\ns_i - s_j = 2(t_i - t_j) = (i-j)(i+j+1) \\geq 2023(i-j),\n$$\nbut\n$$\ns_i - s_j = a_{j+1} + \\cdots + a_i \\leq 2022(i-j),\n$$\nwhich is a contradiction. So, there can't be more than $1011$ such $i$.\n\nIt's enough to note that for the permutation $(2, 4, \\ldots, 2022, 1, 3, \\ldots, 2021)$, the number of such $i$ is precisely $1011$, as for $i = 1, \\ldots, 1011$ we have $s_i = 2 + 4 + \\cdots + 2i = 2t_i$.",
    "topic": "Discrete Mathematics",
    "subtopic": "Combinatorics"
  },
  {
    "id": 13867,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $a, b$ be real numbers with $a \\neq 0$.\n\nFind all polynomials $f(x)$ of degree at most $4$ such that there exists a nonzero polynomial $g(x)$ satisfying\n$$\nf(x^3) - f(x) = (x^5 - 1)g(x).\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "Let $f(x) = a_4x^4 + a_3x^3 + a_2x^2 + a_1x + a_0$.\n\nThen:\n$$\nf(x^3) - f(x) = a_4(x^{12} - x^4) + a_3(x^9 - x^3) + a_2(x^6 - x^2) + a_1(x^3 - x).\n$$\nDividing $x^{12} - x^4$, $x^9 - x^3$, and $x^6 - x^2$ by $x^5 - 1$, the remainders are $x^2 - x^4$, $x^4 - x^3$, and $x - x^2$, respectively. Thus, the remainder when dividing $f(x^3) - f(x)$ by $x^5 - 1$ is:\n$$\n(-a_4 + a_3)x^4 + (-a_3 + a_1)x^3 + (a_4 - a_2)x^2 + (a_2 - a_1)x.\n$$\nFor divisibility, all coefficients must vanish:\n$$\n-a_4 + a_3 = 0,\\quad -a_3 + a_1 = 0,\\quad a_4 - a_2 = 0,\\quad a_2 - a_1 = 0.\n$$\nSolving, $a_1 = a_2 = a_3 = a_4 = a$ for some $a \\neq 0$, so $f(x) = a(x^4 + x^3 + x^2 + x) + b$ for real $b$.\n\nFor such $f(x)$, $g(x) = a(x^7 + x^4 + x^2 + x)$ is nonzero if $a \\neq 0$.\n\n*Alternate Solution:*\nLet $\\omega = \\cos\\frac{2\\pi}{5} + i\\sin\\frac{2\\pi}{5}$, so $\\omega^5 = 1$ and $x^5 - 1 = (x - 1)(x - \\omega)(x - \\omega^2)(x - \\omega^3)(x - \\omega^4)$. Substituting $x = \\omega, \\omega^2, \\omega^3, \\omega^4$ into $f(x^3) - f(x) = (x^5 - 1)g(x)$ gives $f(\\omega) = f(\\omega^2) = f(\\omega^3) = f(\\omega^4) = b$. Thus, $f(x) - b$ is divisible by $x^4 + x^3 + x^2 + x + 1$, so $f(x) = a(x^4 + x^3 + x^2 + x) + b$ for $a \\neq 0$.\n\n**Answer:**\nAll polynomials $f(x) = a(x^4 + x^3 + x^2 + x) + b$ with $a \\neq 0$, $b \\in \\mathbb{R}$.",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 13868,
    "subject": "Mathematics (Olympiad)",
    "question": "Consider functions $f: \\mathbb{R} \\to \\mathbb{R}$ and $g: \\mathbb{R} \\to \\mathbb{R}$ satisfying $f(0) = 2022$ and\n\n$$\nf(x + g(y)) = x f(y) + (2023 - y) f(x) + g(x), \\quad \\forall x, y \\in \\mathbb{R}.\n$$\n\na) Prove that $f$ is surjective and $g$ is injective.\n\nb) Find all functions $f, g$ satisfying the given conditions.",
    "options": [],
    "answer": "See solution",
    "solution": "a) For $x, y \\in \\mathbb{R}$, let $P(x, y)$ denote the assertion:\n\n$$\nf(x + g(y)) = x f(y) + (2023 - y) f(x) + g(x).\n$$\n\nFrom $P(0, y)$:\n\n$$\nf(g(y)) = (2023 - y) f(0) + g(0) = 2022(2023 - y) + g(0), \\quad \\forall y \\in \\mathbb{R}.\n$$\n\nThe right-hand side is a degree 1 polynomial in $y$, so $f$ is surjective.\n\nTo show $g$ is injective, suppose $g(x_1) = g(x_2)$. Then from $P(0, x_1)$ and $P(0, x_2)$:\n\n$$\n2022(2023 - x_1) + g(0) = 2022(2023 - x_2) + g(0) \\implies x_1 = x_2.\n$$\n\nThus, $g$ is injective.\n\nb) Consider\n\n$$\nf(g(y)) = 2022(2023 - y) + g(0). \\quad (1)\n$$\n\nFrom $P(g(x), y)$:\n\n$$\nf(g(x) + g(y)) = g(x) f(y) + (2023 - y) f(g(x)) + g(g(x)), \\quad \\forall x, y \\in \\mathbb{R}.\n$$\n\nUsing (1), $f(g(x)) = 2022(2023 - x) + g(0)$. Substitute and swap $x, y$, then compare:\n\n$$\n\\begin{aligned}\n& g(x) f(y) + g(g(x)) + (2023 - y) g(0) \\\\\n=\\ & g(y) f(x) + g(g(y)) + (2023 - x) g(0), \\quad \\forall x, y \\in \\mathbb{R}.\n\\end{aligned} \\quad (2)\n$$\n\nSince $f$ is surjective, there exists $a$ with $f(a) = 0$. Substitute $y = a$ in (2):\n\n$$\ng(g(x)) = -g(0) x + g(a) f(x) + C, \\quad \\forall x \\in \\mathbb{R},\n$$\n\nwhere $C$ is a constant. Substitute back into (2) and simplify:\n\n$$\ng(x) f(y) + g(a) f(x) = g(y) f(x) + g(a) f(y), \\quad \\forall x, y \\in \\mathbb{R}.\n$$\n\nLet $y = 0$:\n\n$$\n2022 g(x) + g(a) f(x) = g(0) f(x) + 2022 g(a).\n$$\n\nThus,\n\n$$\ng(x) = \\frac{g(0) - g(a)}{2022} f(x) + g(a), \\quad \\forall x \\in \\mathbb{R}.\n$$\n\nIf $g(a) = g(0)$, then $a = 0$ or $f(a) = f(0) = 0$, contradicting $f(0) = 2022$. So $g(a) \\neq g(0)$, and $g$ is surjective. There exists $b$ with $g(b) = 0$. From $P(x, b)$:\n\n$$\nf(x) = x f(b) + (2023 - b) f(x) + g(x), \\quad \\forall x \\in \\mathbb{R}.\n$$\n\nSubstitute into $P(x, y)$:\n\n$$\n\\begin{aligned}\nf(x + g(y)) &= x f(y) + (2023 - y) f(x) + f(x) - x f(b) + (b - 2023) f(x) \\\\\n&= x (f(y) - f(b)) + f(x)(1 + b - y), \\quad \\forall x, y \\in \\mathbb{R}.\n\\end{aligned}\n$$\n\nLet $y = 1 + b$:\n\n$$\nf(x + g(1 + b)) = x (f(1 + b) - f(b)), \\quad \\forall x \\in \\mathbb{R}.\n$$\n\nSo $f$ is linear, and similarly $g$ is linear. Substituting back, the solutions are:\n\n$$\n\\begin{cases}\nf(x) = \\frac{-1 \\pm \\sqrt{5}}{2} x + 2022, \\\\\ng(x) = 1011(-1 \\mp \\sqrt{5}) x + 2 \\cdot 1011^2 (-3 \\mp \\sqrt{5})\n\\end{cases}, \\quad \\forall x \\in \\mathbb{R}.\n$$\n\n$\\square$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 13869,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all prime numbers $p$ and $q$ such that $p^4 - q^6$ is a power of a prime number. (Numbers 7 and 8 are powers of prime numbers, but 6 is not.)",
    "options": [],
    "answer": "See solution",
    "solution": "Rewrite the equation as $p^4 - q^6 = k^n$, where $k$ is a prime and $n \\geq 1$. Try small primes for $q$:\n\nIf $q = 3$, then $p^4 - 729$ must be a prime power. Try $p = 7$:\n\n$$7^4 - 3^6 = 2401 - 729 = 1672$$\n\nBut $1672 = 2^3 \\times 11^1$, not a prime power. Try $p = 5$:\n\n$$5^4 - 3^6 = 625 - 729 = -104$$\n\nNegative, so not a prime power. Try $q = 2$:\n\n$$p^4 - 64$$\n\nTry $p = 3$:\n\n$$81 - 64 = 17$$\n\n$17$ is a prime, so $p = 3$, $q = 2$ is a solution. Try $p = 5$:\n\n$$625 - 64 = 561$$\n\n$561$ is not a prime power. Try $q = 5$:\n\n$$p^4 - 15625$$\n\nTry $p = 13$:\n\n$$28561 - 15625 = 12936$$\n\n$12936$ is not a prime power. \n\nAlternatively, factor $p^4 - q^6 = (p^2 - q^3)(p^2 + q^3)$. For this to be a prime power, one of the factors must be $1$.\n\nSet $p^2 - q^3 = 1$:\n\n$p^2 = q^3 + 1$. Try small $q$:\n\n$q = 2$: $p^2 = 9 \\implies p = 3$ (prime).\n\n$q = 3$: $p^2 = 28$ (not a perfect square).\n\nSo, $p = 3$, $q = 2$ is a solution.\n\nSet $p^2 + q^3 = 1$:\n\nNot possible for positive primes.\n\n**Answer:** The only solution is $p = 3$, $q = 2$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 13870,
    "subject": "Mathematics (Olympiad)",
    "question": "Natural number $a$ is written one or more times in succession in its decimal representation, and the resulting number is interpreted as a binary number. Find all possible values of $a$.",
    "options": [],
    "answer": "See solution",
    "solution": "Suppose $a$ has $k$ digits in decimal, and we write it $l$ times, so the resulting number has $kl$ digits. For the binary interpretation to be valid, all digits of $a$ must be $0$ or $1$. Thus, $a$ must be between $10^{k-1}$ and $\\frac{1}{9}(10^k - 1)$, and the binary number must be between $2^{kl-1}$ and $2^{kl}-1$. This gives two inequalities:\n\n$$10^{k-1} \\leq 2^{kl} - 1$$\n$$\\frac{1}{9}(10^k - 1) \\geq 2^{kl-1}$$\n\nAfter transformation, we get:\n\n$$\\frac{9}{2} < \\left(\\frac{10}{2l}\\right)^k < 10$$\n\nFor $l \\geq 4$, there is a contradiction. If $l=1$, then $k=1$ is possible, so $a=1$ is a solution. If $l=2$, then $k=2$ is possible, and among two-digit numbers with digits $0$ and $1$, only $10$ is valid. For $l=3$, possible $k$ values are $7,8,9,10$. Checking these cases, for $k=10$, $10^9 > 2^{29} + 2^{28}$, so the binary representation would have more than $30$ digits, which is a contradiction. For other $k$, $a$ must be divisible by $2^{2k} + 2^k + 1$. For $k=9$, $2^{18} + 2^9 + 1 = 262657$. Checking possible combinations of powers of $10$ modulo $262657$ does not yield a valid $a$. Similarly, for $k=8$, $2^{16} + 2^8 + 1 = 65793$, and checking combinations does not yield a valid $a$. Thus, the only possible values are $a=1$ and $a=10$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 13871,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $m \\ge n \\ge 2$ be integers, and let $A \\in \\mathcal{M}_{m,n}(\\mathbb{C})$ and $B \\in \\mathcal{M}_{n,m}(\\mathbb{C})$ be two matrices such that there exist $k \\in \\mathbb{N}^*$ and $a_0, a_1, \\dots, a_k \\in \\mathbb{C}$ with\n$$\na_k(AB)^k + a_{k-1}(AB)^{k-1} + \\dots + a_1(AB) + a_0 I_m = O_m,\n$$\nand\n$$\na_k(BA)^k + a_{k-1}(BA)^{k-1} + \\dots + a_1(BA) + a_0 I_n \\ne O_n.\n$$\nProve that $a_0 = 0$.",
    "options": [],
    "answer": "See solution",
    "solution": "Suppose that $a_0 \\ne 0$. Rewriting the given equality as\n$$\nAB \\left( -\\frac{a_k}{a_0}(AB)^{k-1} - \\frac{a_{k-1}}{a_0}(AB)^{k-2} - \\dots - \\frac{a_1}{a_0}I_m \\right) = I_m,\n$$\nwe find that $AB$ is invertible, hence $\\mathrm{rank}(AB) = m \\ge n$.\n\nOn the other hand, $\\mathrm{rank}(AB) \\le \\min\\{\\mathrm{rank}A, \\mathrm{rank}B\\} \\le n$, so $m = n$ and $A$, $B$, and $BA$ are all invertible.\n\nNow, consider the equality $a_k(AB)^k + a_{k-1}(AB)^{k-1} + \\dots + a_1(AB) + a_0 I_n = O_n$ and left-multiply by $B$ and right-multiply by $A$. This yields\n$$\na_k(BA)^{k+1} + a_{k-1}(BA)^k + \\dots + a_0(BA) = O_n.\n$$\nMultiplying by $(BA)^{-1}$ gives\n$$\na_k(BA)^k + a_{k-1}(BA)^{k-1} + \\dots + a_0 I_n = O_n,\n$$\na contradiction.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 13872,
    "subject": "Mathematics (Olympiad)",
    "question": "Given a square board of size $8 \\times 8$, exactly 7 squares are colored. Prove that, for every way of coloring, there always exists a $2 \\times 2$ sub-square of the board on which exactly one square is colored.",
    "options": [],
    "answer": "See solution",
    "solution": "Number the rows from 1 to 8 (top to bottom) and columns from 1 to 8 (left to right). Suppose, for contradiction, that no $2 \\times 2$ sub-square contains exactly one colored square.\n\nThere must be at least one empty column; let it be column $i$ with $1 \\leq i \\leq 8$. Consider the column to the left, $(i-1)$ (if it exists). If there is a colored square in column $(i-1)$, consider the $2 \\times 2$ square formed by two uncolored squares in column $i$ and the colored square in column $(i-1)$. The remaining cell in $(i-1)$ must also be colored; otherwise, this $2 \\times 2$ square would have exactly one colored cell, contradicting our assumption.\n\nBy repeating this argument, every cell in column $(i-1)$ must be colored, so column $(i-1)$ has 8 colored cells, which is impossible since only 7 squares are colored. Thus, column $(i-1)$ must also be empty. The same logic applies to column $(i+1)$. Extending this reasoning, all columns must be empty, which is absurd. Therefore, the assertion is proved. $\\square$",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 13873,
    "subject": "Mathematics (Olympiad)",
    "question": "A chocolate shop packs chocolates in cartons, boxes, and singles:\n- Each carton contains 10 boxes.\n- Each box contains 12 chocolates.\n\nAnswer the following:\n\n**a.** How many chocolates are there in 1 full carton, 2 loose boxes, and 3 single chocolates?\n\n**b.** If Robin has 299 chocolates, how many full cartons, loose boxes, and single chocolates does this make?\n\n**c.** Robin receives two orders:\n- Order 1: 2 cartons, 4 boxes, and 4 single chocolates.\n- Order 2: 7 cartons, 8 boxes, and 9 single chocolates.\n\nWhat is the combined order in terms of cartons, boxes, and single chocolates?\n\n**d.** Robin starts with 5 cartons, 3 boxes, and 2 single chocolates. After selling 3 cartons, 5 boxes, and 6 single chocolates, how many cartons, boxes, and single chocolates remain?",
    "options": [],
    "answer": "See solution",
    "solution": "**a.**\nThe total number of chocolates is:\n$$\n(1 \\times 10 \\times 12) + (2 \\times 12) + 3 = 120 + 24 + 3 = 147\n$$\n\n**b.**\n- Two cartons: $2 \\times 10 \\times 12 = 240$ chocolates.\n- Remaining: $299 - 240 = 59$ chocolates.\n- Four boxes: $4 \\times 12 = 48$ chocolates.\n- Remaining: $59 - 48 = 11$ chocolates.\n- So, $2$ cartons, $4$ boxes, and $11$ single chocolates.\n\nAlternatively:\n- $299$ chocolates $\\div 12 = 24$ boxes, $11$ singles left.\n- $24$ boxes $\\div 10 = 2$ cartons, $4$ boxes left.\n- So, $2$ cartons, $4$ boxes, $11$ singles.\n\n**c.**\n- Total: $9$ cartons, $12$ boxes, $13$ singles.\n- $13$ singles $= 1$ box, $1$ single; now $9$ cartons, $13$ boxes, $1$ single.\n- $13$ boxes $= 1$ carton, $3$ boxes; now $10$ cartons, $3$ boxes, $1$ single.\n\nAlternatively:\n- $2$ cartons, $4$ boxes, $4$ singles $= 240 + 48 + 4 = 292$ chocolates.\n- $7$ cartons, $8$ boxes, $9$ singles $= 840 + 96 + 9 = 945$ chocolates.\n- Total: $292 + 945 = 1237$ chocolates.\n- $1237 \\div 12 = 103$ boxes, $1$ single.\n- $103 \\div 10 = 10$ cartons, $3$ boxes.\n- So, $10$ cartons, $3$ boxes, $1$ single.\n\n**d.**\n- Start: $5$ cartons, $3$ boxes, $2$ singles $= 638$ chocolates.\n- Sold: $3$ cartons, $5$ boxes, $6$ singles $= 426$ chocolates.\n- Remaining: $638 - 426 = 212$ chocolates.\n- $212 \\div 120 = 1$ carton, $92$ chocolates left.\n- $92 \\div 12 = 7$ boxes, $8$ singles left.\n- So, $1$ carton, $7$ boxes, $8$ singles.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 13874,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $x, y, z \\in [0, 1]$. What is the maximum value of\n$$\nM = \\sqrt{|x-y|} + \\sqrt{|y-z|} + \\sqrt{|z-x|}?\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "We may assume $0 \\leq x \\leq y \\leq z \\leq 1$. Then\n$$\nM = \\sqrt{y-x} + \\sqrt{z-y} + \\sqrt{z-x}.\n$$\nSince\n$$\n\\sqrt{y-x} + \\sqrt{z-y} \\leq \\sqrt{2[(y-x)+(z-y)]} = \\sqrt{2(z-x)},\n$$\nwe have\n$$\nM \\leq \\sqrt{2(z-x)} + \\sqrt{z-x} = (\\sqrt{2} + 1) \\sqrt{z-x} \\leq \\sqrt{2} + 1.\n$$\nEquality holds if and only if $y-x = z-y$, $x = 0$, $z = 1$ (i.e., $x = 0$, $y = \\frac{1}{2}$, $z = 1$).\n\nTherefore, the maximum value is $M_{\\max} = \\sqrt{2} + 1$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 13875,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $n$ be a positive integer. For which $n$ does there exist a polynomial $f(x)$ with rational coefficients such that $f(x)$ is an integer for all integers $x$ not divisible by $n$, but $f(x)$ is not an integer for some integer $x$ divisible by $n$?",
    "options": [],
    "answer": "See solution",
    "solution": "We claim the answer is $n = p^\\alpha$ for some prime $p$ and nonnegative integer $\\alpha$.\n\n**Lemma.** For any integers $a_1, \\dots, a_n$, there exists an integer-valued polynomial $P(x)$ of degree $< n$ such that $P(k) = a_k$ for all $1 \\leq k \\leq n$.\n\n*Proof.* Induct on $n$. For $n=1$, set $P(x) = a_1$. For the induction step, suppose $P_1(x)$ works for $1 \\leq k \\leq n-1$. Then set $P(x) = P_1(x) + (a_n - P_1(n)) \\binom{x-1}{n-1}$. Since $\\binom{k-1}{n-1} = 0$ for $1 \\leq k \\leq n-1$ and $\\binom{n-1}{n-1} = 1$, $P(x)$ is as desired. $\\square$\n\nSuppose such a polynomial $f(x)$ exists for some $n$. Choose an integer-valued polynomial $P(x)$ of degree $< n-1$ coinciding with $f(x)$ at $1, \\dots, n-1$. Then $f_1(x) = f(x) - P(x)$ also satisfies the problem conditions, so we may restrict to polynomials vanishing at $1, \\dots, n-1$, i.e., $f(x) = c \\prod_{i=1}^{n-1} (x-i)$ for some rational $c = p/q$ in lowest terms, with $q = \\prod_{j=1}^d p_j^{\\alpha_j}$.\n\n1. Assume such $f(x)$ exists. Since $f(0)$ is not integer, $q \\nmid (-1)^{n-1}(n-1)!$, so $p_j^{\\alpha_j} \\nmid (-1)^{n-1}(n-1)!$ for some $j$. Thus,\n\n$$\n\\prod_{i=1}^{n-1} (p_j^{\\alpha_j} - i) \\equiv (-1)^{n-1}(n-1)! \\not\\equiv 0 \\pmod{p_j^{\\alpha_j}},\n$$\n\nso $f(p_j^{\\alpha_j})$ is not integer. By the problem condition, $n \\mid p_j^{\\alpha_j}$, so $n$ must be a prime power.\n\n2. For $n = p^\\alpha$, consider\n\n$$\nf(x) = \\frac{1}{p} \\binom{x-1}{n-1} = \\frac{n}{px} \\binom{x}{n}.\n$$\n\nFor integer $x$, the denominator of $f(x)$ is $1$ or $p$. If $p^\\alpha \\nmid x$, then $f(x)$ is integer. If $n = p^\\alpha \\mid x$, then\n\n$$\n\\binom{x-1}{n-1} = \\frac{\\prod_{i=1}^{n-1} (x-i)}{\\prod_{i=1}^{n-1} (n-i)}\n$$\n\nis not divisible by $p$, so $f(x)$ is not integer.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 13876,
    "subject": "Mathematics (Olympiad)",
    "question": "Show that there is an infinite number of positive integers $t$ such that none of the equations\n\n$$\nx^2 + y^6 = t, \\quad x^2 + y^6 = t + 1, \\quad x^2 - y^6 = t, \\quad x^2 - y^6 = t + 1\n$$\nhas solutions $(x, y) \\in \\mathbb{Z} \\times \\mathbb{Z}$.",
    "options": [],
    "answer": "See solution",
    "solution": "If $x$ is a positive integer, then either $x^{12} \\equiv 0 \\pmod{13}$ or $x^{12} \\equiv 1 \\pmod{13}$, so $x^6$ is congruent to $-1$, $0$, or $1$ modulo $13$.\n\nTherefore, if $t \\equiv 6 \\pmod{13}$, then $\\pm x^6 + t$ is congruent to $5$, $6$, or $7$ modulo $13$, and $\\pm x^6 + t + 1$ is congruent to $6$, $7$, or $8$ modulo $13$.\n\nOn the other hand, perfect squares modulo $13$ are congruent to $0$, $1$, $3$, $4$, $9$, $10$, or $12$. Thus, a perfect square cannot be equal to a number of the form $\\pm x^6 + t$ or $\\pm x^6 + t + 1$ if $t \\equiv 6 \\pmod{13}$.\n\nIn conclusion, all numbers congruent to $6$ modulo $13$ have the required property.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 13877,
    "subject": "Mathematics (Olympiad)",
    "question": "There are 6 different points $A$, $B$, $C$, $D$, $E$, $F$ in the plane. No four of them lie on the same circle, and no two segments with endpoints at these points lie on parallel straight lines. Let $P$, $Q$, and $R$ be the intersection points of the perpendicular bisectors of the segment pairs $(AD, BE)$, $(BE, CF)$, and $(CF, DA)$, respectively, and $P'$, $Q'$, and $R'$ be the intersection points of the perpendicular bisectors of the segment pairs $(AE, BD)$, $(BF, CE)$, and $(CA, DF)$, respectively.\n\n![](images/Ukrajina_2013_p50_data_ea594ed1f8.png)\n\nShow that $P \\neq P'$, $Q \\neq Q'$, $R \\neq R'$, and prove that the straight lines $PP'$, $QQ'$, and $RR'$ either intersect at the same point or are parallel.",
    "options": [],
    "answer": "See solution",
    "solution": "The perpendicular bisector of a segment is the locus of points equidistant from the endpoints of the segment. Thus, $P$ is equidistant from $(A, D)$ and $(B, E)$, and $P'$ is equidistant from $(A, E)$ and $(B, D)$. If $P = P'$, then it is equidistant from all four points $A, B, D, E$, so these four points would lie on a circle with center $P = P'$, contradicting the problem's conditions. Thus, $P \\neq P'$, $Q \\neq Q'$, and $R \\neq R'$.\n\n**Lemma:** If $A, B, C, D$ are four distinct points and $AC$ and $BD$ are not parallel, then the locus of points $X$ such that $XA^2 + XB^2 = XC^2 + XD^2$ is a straight line.\n\n*Proof:* Let $M$ be the midpoint of $AB$ and $N$ the midpoint of $CD$. Using the median length formula for triangles $XAB$ and $XCD$:\n\n$$\nXM^2 = \\frac{2XA^2 + 2XB^2 - AB^2}{4}, \\quad XN^2 = \\frac{2XC^2 + 2XD^2 - CD^2}{4}\n$$\n\nSo,\n\n$$\nXA^2 + XB^2 = XC^2 + XD^2 \\iff XM^2 - XN^2 = \\frac{CD^2 - AB^2}{4}.\n$$\n\nIf $M = N$, then $ACBD$ is a (possibly degenerate) parallelogram, which would contradict the lemma's conditions. Thus, $M \\neq N$. Let $MN$ be the line through $M$ and $N$, and let $P_x$ be the foot of the perpendicular from $X$ to $MN$. Then\n\n$$\nXM^2 = XP_x^2 + P_xM^2, \\quad XN^2 = XP_x^2 + P_xN^2\n$$\n\nSo the condition becomes $P_xM^2 - P_xN^2 = \\frac{CD^2 - AB^2}{4}$. Assigning coordinates $m, n, x$ to $M, N, P_x$ on $MN$:\n\n$$\n\\frac{CD^2 - AB^2}{4} = (x - m)^2 - (x - n)^2 = x(2n - 2m) + (m^2 - n^2) \\iff x = \\frac{(CD^2 - AB^2)/4 + n^2 - m^2}{2n - 2m}.\n$$\n\nThus, the locus is the perpendicular to $MN$ through this $x$.\n\n*Lemma proved.*\n\nReturning to the problem: Both $P$ and $P'$ lie on the locus of points $X$ such that $XA^2 + XB^2 = XD^2 + XE^2$. This locus is the straight line $PP'$. Similarly, $QQ'$ is the locus where $XB^2 + XC^2 = XE^2 + XF^2$, and $RR'$ is where $XC^2 + XD^2 = XF^2 + XA^2$.\n\nIf no two of $PP'$, $QQ'$, $RR'$ intersect, they are parallel. If at least two intersect, say $PP'$ and $QQ'$ at $X$, then $X$ satisfies:\n\n$$\n\\begin{aligned}\n&XA^2 + XB^2 = XD^2 + XE^2, \\\\\n&XB^2 + XC^2 = XE^2 + XF^2 \\\\\n\\Rightarrow XA^2 - XC^2 = XD^2 - XF^2 \\\\\n\\Rightarrow XC^2 + XD^2 = XF^2 + XA^2.\n\\end{aligned}\n$$\n\nThus, $X$ also lies on $RR'$, so all three lines are concurrent or parallel.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 13878,
    "subject": "Mathematics (Olympiad)",
    "question": "Six chess players play a round-robin tournament (each plays every other exactly once). In each game, the players can choose between two types of boards:\n\n- **Fair board**: The stronger player always wins and gets 1 point, the weaker gets 0.\n- **Peaceful board**: The game always ends in a draw, each player gets 0.5 points.\n\nIs it possible to organize the tournament so that, after all games, the difference in total points between any two consecutive players (ordered by strength) is at least 0.5?",
    "options": [],
    "answer": "See solution",
    "solution": "Suppose, for contradiction, that such a tournament is possible. Number the players by strength: $P_1, P_2, \\ldots, P_6$ (with $P_1$ strongest). Let $a_i$ be the total points of $P_i$.\n\nBy the condition, $a_i - a_{i+1} \\ge 0.5$ for $i = 1, \\ldots, 5$. In particular,\n\n$$\na_i - a_{i+3} = (a_i - a_{i+1}) + (a_{i+1} - a_{i+2}) + (a_{i+2} - a_{i+3}) \\ge 1.5. \\tag{1}\n$$\n\nLet $A = \\{P_1, P_2, P_3\\}$, $B = \\{P_4, P_5, P_6\\}$, $S_A = a_1 + a_2 + a_3$, $S_B = a_4 + a_5 + a_6$. Clearly, $S_A + S_B = 15$ (since there are $15$ points in total).\n\nAlso,\n$$\nS_A - S_B = (a_1 - a_4) + (a_2 - a_5) + (a_3 - a_6) \\ge 1.5 + 1.5 + 1.5 = 4.5. \\tag{2}\n$$\n\nLet $S_Y(X)$ be the total points earned by group $X$ in games against group $Y$ ($X, Y \\in \\{A, B\\}$).\n\nWe have $S_A(A) = S_B(B) = 3$ (since each group has 3 players, each playing 2 games within the group, each game awarding 1 point in total). The total points in $A$ vs $B$ games is $9$.\n\n$S_B(A) - S_A(B) = S_A - S_B$. For $S_A - S_B \\ge 4.5$, we need $S_B(A) \\ge 7$, otherwise $S_B(A) \\le 6.5$ gives $S_A(B) = 9 - S_B(A) \\ge 2.5$, so $S_A - S_B \\le 6.5 - 2.5 = 4$, contradicting (2).\n\nTo have $S_B(A) \\ge 7$, at least 5 of the 9 $A$ vs $B$ games must use the fair board (where the stronger player wins and $B$ gets 1 point), since otherwise $S_B(A) \\le 4 \\cdot 1 + 5 \\cdot 0.5 = 6.5$.\n\nThus, the peaceful board is used in at most 4 $A$ vs $B$ games. The peaceful board can be used at most once in games within the same group, so $a_6 \\ge 1 + 0.5 = 1.5$.\n\nBut then,\n$$\n15 = a_1 + a_2 + a_3 + a_4 + a_5 + a_6 \\ge 4 + 3.5 + 3 + 2.5 + 2 + 1.5 = 16.5,\n$$\nwhich is impossible. Therefore, such a tournament cannot be organized.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 13879,
    "subject": "Mathematics (Olympiad)",
    "question": "a) Find the value of\n$$\n\\frac{\\sigma_2}{\\sigma_3}(\\sigma_1 - 2)\n$$\nwhere $\\sigma_1 = a + b + c$, $\\sigma_2 = ab + bc + ca$, $\\sigma_3 = abc$, and the variables $a$, $b$, $c$ satisfy the system:\n$$\n\\sigma_1 = \\sigma_1^2 - 2\\sigma_2, \\quad \\sigma_1^2 = \\sigma_1^3 - 3\\sigma_1\\sigma_2 + 3\\sigma_3.\n$$\n\nb) Is it possible to find real numbers $a$, $b$, $c$ satisfying the above system?",
    "options": [],
    "answer": "See solution",
    "solution": "a) Let $\\sigma_1 = a + b + c$, $\\sigma_2 = ab + bc + ca$, $\\sigma_3 = abc$. From the given equalities, we have the system:\n$$\n\\sigma_1 = \\sigma_1^2 - 2\\sigma_2 \\quad (1), \\quad \\sigma_1^2 = \\sigma_1^3 - 3\\sigma_1\\sigma_2 + 3\\sigma_3. \\quad (2)\n$$\nFrom (1), $\\sigma_1^2 = \\sigma_1^3 - 2\\sigma_1\\sigma_2$. Then from (2) it follows that:\n$$\n-2\\sigma_1\\sigma_2 = -\\sigma_1\\sigma_2 + 3\\sigma_3. \\quad (3)\n$$\nFinally, the expression we seek is:\n$$\n\\frac{\\sigma_2}{\\sigma_3}(\\sigma_1 - 2) = \\frac{\\sigma_1\\sigma_2 - 2\\sigma_2}{\\sigma_3} = \\text{[in view of (3)]} = \\frac{3\\sigma_3}{\\sigma_3} = 3.\n$$\n\nb) It is easy to verify that $a = \\frac{1}{2}$, $b = \\frac{2+\\sqrt{6}}{4}$, $c = \\frac{2-\\sqrt{6}}{4}$ satisfy the given system.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 13880,
    "subject": "Mathematics (Olympiad)",
    "question": "Suppose that a matrix of nonnegative entries,\n\n$$\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$$\n\nhas the following properties:\n\n1. Numbers in a row are different from each other.\n2. The sum of the numbers in a column from the first six columns is $1$.\n3. $x_{17} = x_{28} = x_{39} = 0$.\n4. $x_{27}, x_{37}, x_{18}, x_{38}, x_{19}, x_{29} > 1$.\n\nAssume that matrix $S$ is constituted by the first three columns of $P$, i.e.\n\n$$\nS = \\begin{bmatrix} x_{11} & x_{12} & x_{13} \\\\ x_{21} & x_{22} & x_{23} \\\\ x_{31} & x_{32} & x_{33} \\end{bmatrix}\n$$\n\nhas the following property:\n\n(O) For any column $[x_{1k} \\\\ x_{2k} \\\\ x_{3k}]$ ($k = 1, 2, \\ldots, 9$) in $P$, there exists $i \\in \\{1, 2, 3\\}$ such that\n\n$$\nx_{ik} \\le u_i = \\min\\{x_{i1}, x_{i2}, x_{i3}\\}.\n$$\n\nProve that:\n\n(i) For different $i$ ($= 1, 2, 3$), $u_i = \\min\\{x_{i1}, x_{i2}, x_{i3}\\}$ comes from different columns in $S$.\n\n(ii) There exists a unique column $\\begin{bmatrix} x_{1k} \\\\ x_{2k} \\\\ x_{3k} \\end{bmatrix}$ ($k^* \\neq 1, 2, 3$) in $P$ such that the matrix\n\n$$\nS' = \\begin{bmatrix} x_{11} & x_{12} & x_{1k^*} \\\\ x_{21} & x_{22} & x_{2k^*} \\\\ x_{31} & x_{32} & x_{3k^*} \\end{bmatrix}\n$$\n\nalso has property (O).",
    "options": [],
    "answer": "See solution",
    "solution": "Proof of (i):\n\nAssume that it is not true. There is a column in $S$ which contains no $u_i$. We may say that $u_i \\neq x_{i2}$ for $i = 1, 2, 3$. By property (1), we have $u_i < x_{i2}$ for $i = 1, 2, 3$. On the other hand, let $k = 2$ in (i). Then by property (O), there exists $i_0 \\in \\{1, 2, 3\\}$ such that $x_{i_0 2} \\le u_i$. The contradiction means the assumption is not valid. This completes the proof of (i).\n\nProof of (ii):\n\nBy the drawer principle, we know that at least two of the three numbers\n\n$$\n\\min\\{x_{11}, x_{12}\\}, \\min\\{x_{21}, x_{22}\\}, \\min\\{x_{31}, x_{32}\\}\n$$\n\nare in the same column. We may say that\n\n$$\n\\min\\{x_{21}, x_{22}\\} = x_{22}, \\quad \\min\\{x_{31}, x_{32}\\} = x_{32}.\n$$\n\nBy (i), we know that the first column of $S$ contains a $u_i$, and it must be $u_1 = x_{11}$; the second column also contains a $u_i$, assuming that it is $u_2 = x_{22}$, and then it must be $u_3 = x_{33}$.\n\nDefine $M = \\{1, 2, \\dots, 9\\}$ and\n\n$$\nI = \\{k \\in M \\mid x_{ik} > \\min\\{x_{i1}, x_{i2}\\},\\ i = 1, 3\\}.\n$$\n\nObviously, $I = \\{k \\in M \\mid x_{1k} > x_{11},\\ x_{3k} > x_{32}\\}$, and $1, 2, 3 \\notin I$. Since $x_{18}, x_{38} > 1 \\ge x_{11}, x_{32}$, we have $8 \\in I$. Therefore, $I \\ne \\emptyset$. Consequently, $\\exists k^* \\in I$ such that $x_{2k^*} = \\max\\{x_{2k} \\mid k \\in I\\}$. Of course, $k^* \\ne 1, 2, 3$.\n\nWe now prove that\n\n$$\nS' = \\begin{bmatrix} x_{11} & x_{12} & x_{1k^*} \\\\ x_{21} & x_{22} & x_{2k^*} \\\\ x_{31} & x_{32} & x_{3k^*} \\end{bmatrix}\n$$\n\nhas property (O).\n\nBy the definition of $I$, we know that\n\n$$\nx_{1k^*} > x_{11} = u_1, \\quad x_{3k^*} > x_{32} \\ge u_3.\n$$\n\nLet $k^* = k$ in (i), and we get $x_{2k^*} \\le u_2$ according to property (O) of $S$. Then define\n\n$$\nu'_1 = u_1, \\quad u'_2 = \\min\\{x_{21}, x_{22}, x_{2k^*}\\} = x_{2k^*}, \\quad u'_3 = u_3.$$\n\nWe claim that, for any $k \\in M$, there exists $i \\in \\{1, 2, 3\\}$ such that $u'_i \\ge x_{ik}$. Otherwise, we would have $x_{ik} > \\min\\{x_{i1}, x_{i2}\\}$ for $i = 1, 3$ and $x_{2k} > x_{2k^*}$, contradicting the definition of $k^*$.\n\nTherefore, $S'$ has property (O).\n\nSecondly, we prove the uniqueness of $S'$. Assume that $\\exists k_0 \\in M$ such that\n\n$$\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$$\n\nalso has property (O). Without loss of generality, we assume that ...",
    "topic": "Algebra",
    "subtopic": "Linear Algebra"
  },
  {
    "id": 13881,
    "subject": "Mathematics (Olympiad)",
    "question": "Professor Piraldo takes part in soccer matches with a lot of goals and judges a match in his own peculiar way. A match with score of $m$ goals to $n$ goals, $m \\ge n$, is *tough* when $m \\le f(n)$, where $f(n)$ is defined by $f(0) = 0$ and, for $n \\ge 1$, $f(n) = 2n - f(r) + r$, where $r$ is the largest integer such that $r < n$ and $f(r) \\le n$.\n\nLet $\\phi = \\frac{1+\\sqrt{5}}{2}$. Prove that a match with score of $m$ goals to $n$, $m \\ge n$, is tough if $m \\le \\phi n$ and is not tough if $m \\ge \\phi n + 1$.",
    "options": [],
    "answer": "See solution",
    "solution": "First, note that if $n$ is written in the Fibonacci basis (as a sum of distinct Fibonacci numbers containing no neighbors), then the representation of $f(n)$ is just that of $n$ with a $0$ in the end. The proof goes by induction: it is true for $n=0$. Suppose $n > 0$. Then $r$ will be such that $f(r) = n$, if the last digit of the Fibonacci of $n$ is zero, or $f(r) = n-1$, if it is one. Then $f(n) = n + r + n - f(r)$. If the last digit of the Fibonacci of $n$ is zero, then $r$ is obtained by deleting the last digit from $n$. So $f(n) = n + r$ is the sum of the Fibonacci numbers used in the representation of $n$ and its respective predecessors in the Fibonacci sequence, resulting in the subsequent Fibonacci numbers, which is exactly $n$ with a zero at its right. If the last digit of the Fibonacci of $n$ is zero, then $r$ is still obtained by deleting the last digit from $n$, but $f(n) = n + 1 + r$. Note that $n - 1$ ends with a zero, so summing $r$ will do the same thing as the preceding case; and we're substituting the rightmost digit one with $2$, which corresponds to a rightmost $10$.\n\nNow, express $n$ and $f(n)$ by using the closed form for Fibonacci numbers $F_k = \\frac{1}{\\sqrt{5}} (\\phi^k - (-\\phi)^{-k})$, and note that we cannot use two consecutive Fibonacci numbers in base Fibonacci, so $|f(n) - \\phi n| < \\phi^{-1}(1 + \\phi^{-2} + \\phi^{-4} + \\dots) = \\frac{\\phi^{-1}}{1 - \\phi^{-2}} = 1$, that is, $f(n) - 1 < \\phi n < f(n) + 1$. Thus, if $m \\le \\phi n$ then $m < f(n) + 1 \\iff m \\le f(n)$ and the match is tough; if $m \\ge \\phi n + 1$ then $m > f(n) - 1 + 1 = f(n)$ and the match is not tough.\n\n*Remark.* One can prove (by induction on $n$) that $f(n) = \\lfloor (n + 1) \\phi \\rfloor - 1$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 13882,
    "subject": "Mathematics (Olympiad)",
    "question": "Points $A$, $B$, $C$, $D$ lie, in this order, on a circle $\\omega$, where $AD$ is a diameter of $\\omega$. Furthermore, $AB = BC = a$ and $CD = c$ for some relatively prime positive integers $a$ and $c$. Show that if the diameter $d$ of $\\omega$ is also an integer, then $d$ is a perfect square or $2d$ is a perfect square.",
    "options": [],
    "answer": "See solution",
    "solution": "By the Pythagorean theorem, the lengths of the diagonals of quadrilateral $ABCD$ are $\\sqrt{d^2 - a^2}$ and $\\sqrt{d^2 - c^2}$. Applying Ptolemy's Theorem to $ABCD$ gives\n\n$$\n\\sqrt{d^2 - a^2} \\cdot \\sqrt{d^2 - c^2} = ab + ac,\n$$\n\nwhich, after squaring and simplifying, becomes\n\n$$\nd^3 - (2a^2 + c^2)d - 2a^2c = 0.\n$$\n\nThen $d = -c$ is a root of this equation, so $c + d$ is a positive factor of the left-hand side. The remaining factor (which is quadratic in $d$) must vanish, yielding $d^2 = cd + 2a^2$. Let $e = 2d - c$. Then $c^2 + 8a^2 = (2d - c)^2 = e^2$, so $8a^2 = e^2 - c^2$. If $e$ and $c$ were both even, then $8 \\mid (e^2 - c^2)$ implies $16 \\mid (e^2 - c^2) = 8a^2$, so $2 \\mid a$, contradicting that $a$ and $c$ are relatively prime. Thus, $e$ and $c$ must both be odd and relatively prime. Therefore, the factors on the right-hand side of $2a^2 = \\frac{e-c}{2} \\cdot \\frac{e+c}{2}$ are relatively prime. It follows that $d = \\frac{e+c}{2}$ is a perfect square or twice a perfect square.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 13883,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $n$ be a positive integer. A polynomial with complex coefficients\n\n$$\nP(z) = a_n z^n + a_{n-1} z^{n-1} + \\cdots + a_1 z + a_0 \\quad (a_n \\neq 0)\n$$\n\nsatisfies: for any complex number $z$ with $|z| = 1$, we have $|P(z)| \\le 1$.\n\nProve: For any $k \\in \\{0, 1, \\dots, n-1\\}$, we have $|a_k| \\le 1 - |a_n|^2$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $\\ell \\in \\{1, 2, \\dots, n\\}$. For a complex number $\\alpha \\in \\mathbb{C}$, consider\n\n$$\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$$\n\nTake $M > 2n$, then\n\n$$\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| 1 + \\alpha e^{i \\frac{2j\\pi}{M}\\ell} \\right|^2 \\\\\n&\\le \\sum_{j=1}^{M} \\left| 1 + \\alpha e^{i \\frac{2j\\pi}{M}\\ell} \\right|^2 = M(1 + |\\alpha|^2).\n\\end{aligned}\n$$\n\nOn the other hand,\n\n$$\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$$\n\nCombining the above two estimates, we get\n\n$$\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$$\n\nIn particular,\n\n$$\n|\\alpha a_n|^2 + |a_n + \\alpha a_{n-\\ell}|^2 \\le 1 + |\\alpha|^2.\n$$\n\nChoosing $\\alpha \\in \\mathbb{C}$ such that $|\\alpha| = \\frac{1}{|a_n|}$ and $\\arg(a_n) = \\arg(\\alpha a_{n-\\ell})$ in the above equation, we get\n\n$$\n1 + \\left( |a_n| + \\frac{|a_{n-\\ell}|}{|a_n|} \\right)^2 \\le 1 + \\frac{1}{|a_n|^2}.\n$$\n\nSolving this gives $|a_{n-\\ell}| \\le 1 - |a_n|^2$. The proof is complete.\n\n**Note:**\n\n1. If we consider some $a_k \\ne 0$ ($k \\le \\frac{n}{2}$), then we can choose a complex number $z_0$ with $|z_0| = 1$ such that $\\frac{a_n}{a_k} z_0^{n-k} > 0$, and let $\\omega = e^{\\frac{2\\pi i}{n-k}}$. Consider\n\n$$\n\\sum_{j=0}^{n-k-1} \\frac{P(z_0 \\omega^j)}{(z_0 \\omega^j)^n} = a_n \\sum_{j=0}^{n-k-1} 1 + \\frac{a_k}{z_0^{n-k}} \\sum_{j=0}^{n-k-1} 1 = (n-k) \\left( a_n + \\frac{a_k}{z_0^{n-k}} \\right),\n$$\n\nfrom which we get $|a_n| + |a_k| \\le 1$, hence we obtain a stronger inequality.\n\n2. Using the Schwarz-Pick lemma in complex analysis: If a holomorphic function $g(z) = b_0 + b_1 z + \\dots$ maps the unit disk to itself, then\n\n$$\n\\frac{|g'(z)|}{1 - |g(z)|^2} \\le \\frac{1}{1 - |z|^2}.\n$$\n\nTaking $z = 0$ further gives $|b_1| \\le 1 - |b_0|^2$.\n\nNow consider the polynomial $g(z)$ such that\n\n$$\ng(z^k) = \\frac{1}{k} \\sum_{j=0}^{k-1} z^n f\\left(e^{\\frac{2\\pi i}{k}j} z^{-1}\\right) = a_n + a_{n-k}z^k + a_{n-2k}z^{2k} + \\dots,\n$$\n\nwhich gives $|a_{n-k}| \\le 1 - |a_n|^2$.",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 13884,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all polynomials $P(x)$ with integer coefficients such that\n\n$$\nP(n) \\mid 2557^n + (213 \\times 2014)\n$$\n\nfor all positive integers $n$.",
    "options": [],
    "answer": "See solution",
    "solution": "First, note that the constant polynomials $P(x) = 1$ and $P(x) = -1$ satisfy the divisibility condition.\n\nSuppose $P(x)$ is a polynomial with integer coefficients, $P(x) \\neq 1$, $P(x) \\neq -1$, and $P(x) \\neq 0$, that satisfies the condition. If $P(\\mathbb{Z}^+) \\subseteq \\{-1, 0, 1\\}$, then $P(x)$ must be constant, contradicting our assumption. Thus, there exists $n_0$ such that $|P(n_0)| > 1$.\n\nLet $q$ be a prime dividing $P(n_0)$. Then $q \\mid 2557^{n_0} + (213 \\times 2014)$. Consider $P(n_0 + kq)$ for integer $k$; by the divisibility, $q \\mid 2557^{n_0 + kq} + (213 \\times 2014)$. But $2557^{n_0 + kq} \\equiv 2557^{n_0} (2557^q)^k \\pmod{q}$. By Fermat's little theorem, $2557^q \\equiv 2557 \\pmod{q}$ if $q \\nmid 2557$.\n\nThis leads to a contradiction unless $q$ divides $2557$ or $213 \\times 2014$, but for large enough $n_0$, such $q$ cannot always be found. Thus, the only possible polynomials are $P(x) = 1$ and $P(x) = -1$.\n\n$\\boxed{P(x) = 1 \\text{ or } P(x) = -1}$ are the only solutions.",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 13885,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all integers $n > 1$ such that the sum of $n$ and its second-largest divisor is $2013$.",
    "options": [],
    "answer": "See solution",
    "solution": "The second-largest divisor of $n$ is of the form $\\frac{n}{p}$, where $p$ is the smallest prime that divides $n$.\n\nThe given condition gives:\n$$\n2013 = n + \\frac{n}{p} = \\frac{n}{p}(p+1)\n$$\nTherefore, $p+1$ is a divisor of $2013$ and thus odd. So, $p$ is $2$, the only even prime.\n\nThe equation now becomes:\n$$\n2013 = \\frac{n}{2} \\cdot 3\n$$\nwhich gives the unique solution $n = 1342$.\n\n$\\square$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 13886,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $\\triangle ABC$ be a triangle with incenter $I$ and circumcircle $\\omega$. The lines $AI$, $BI$, $CI$ intersect $\\omega$ for a second time at the points $D$, $E$, $F$ respectively. The lines through $I$ parallel to the sides $BC$, $AC$, $AB$ intersect the lines $EF$, $DF$, $DE$ at the points $K$, $L$, $M$ respectively. Prove that the points $K$, $L$, $M$ are collinear.",
    "options": [],
    "answer": "See solution",
    "solution": "First, we will prove that $KA$ is tangent to $\\omega$.\n\nIndeed, it is a well-known fact that $FA = FB = FI$ and $EA = EC = EI$, so $FE$ is the perpendicular bisector of $AI$. It follows that $KA = KI$ and\n\n$$\n\\angle KAF = \\angle KIF = \\angle FCB = \\angle FEB = \\angle FEA,\n$$\n\nso $KA$ is tangent to $\\omega$. Similarly, we can prove that $LB$, $MC$ are tangent to $\\omega$ as well.\n\n![](images/shortlistBMO2015_p14_data_ffaba14316.png)\n\nLet $A'$, $B'$, $C'$ be the intersections of $AI$, $BI$, $CI$ with $BC$, $CA$, $AB$ respectively. From Pascal's Theorem on the cyclic hexagon $AACDEB$ we get $K$, $C'$, $B'$ collinear. Similarly, $L$, $C'$, $A'$ are collinear and $M$, $B'$, $A'$ are collinear.\n\nThen, from Desargues' Theorem for $\\triangle DEF$, $\\triangle A'B'C'$ which are perspective from the point $I$, we get that points $K$, $L$, $M$—the intersections of their corresponding sides—are collinear as wanted. $\\square$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 13887,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $x$ and $y$ be non-negative integer solutions of the equation\n\n$$\n2x^2 - 17xy + y^2 + x = 0.\n$$\n\nProve that $x$ is a perfect square.",
    "options": [],
    "answer": "See solution",
    "solution": "If $x$ is divisible by $p$, then it is easy to see that $y$ is also divisible by $p$. Substitute $x = p^a x_1$, $y = p^b y_1$ in the equation. Then we obtain\n\n$$\np^{2b} y_1^2 = p^a (p^b 17 x_1 y_1 - p^a 2 x_1^2 - x_1),\n$$\n\nhence $a = 2b$ is even. (For $p = 2$ we have analogous observations.) Therefore $x$ is a perfect square.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 13888,
    "subject": "Mathematics (Olympiad)",
    "question": "Suppose there are 2009 balls, of which 2 are blue and the rest are green. The balls are drawn one at a time and placed in a row in the order drawn. Two players, $A$ and $B$, take turns drawing balls, starting with $A$. Whenever a blue ball is drawn, the turn passes to the other player. The game ends when all balls have been drawn. What is the probability that $B$ draws more blue balls than $A$?",
    "options": [],
    "answer": "See solution",
    "solution": "We divide into four cases depending on the first two drawn balls ($b$ for blue and $g$ for green) and count how many games $B$ wins:\n\n- *bb*: One game. $B$ wins, because $A$ only gets three balls.\n- *gb*: $B$ gets the bag after the first move, and he wins if he draws at least 1004 green balls before the bag is turned back to $A$. That is, the last blue ball is placed among the last $2009 - (1004 + 2) = 1003$ balls. Therefore, $B$ wins 1003 games.\n- *bg*: Same as *gb*, so $B$ wins 1003 games.\n- *gg*: $A$ keeps the bag after the first move. If the first blue ball is number three in the row, $B$ gets the bag, and the situation is as above, only with 1002 games with $B$ as the winner. If the first blue ball is number 4 in the row, then $B$ wins 1001 games, etc., down to if the first blue ball is number 1004, then $B$ wins 1 game. That gives $1002 + 1001 + \\dots + 2 + 1 = 501 \\cdot 1003$ games with $B$ as the winner.\n\nThat sums up to $1 + 2 \\cdot 1003 + 501 \\cdot 1003 = 504510$ games in which $B$ wins.\n\nThe total number of ways to place the blue balls is $\\binom{2009}{2} = 2017036$.\n\nThe probability that $B$ wins is\n\n$$\n\\frac{504510}{2017035} = \\frac{1005}{4018}\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 13889,
    "subject": "Mathematics (Olympiad)",
    "question": "In triangle $ABC$, $AB = AC$. Point $D$ is the midpoint of side $BC$. Point $E$ lies outside the triangle $ABC$ such that $CE \\perp AB$ and $BE = BD$. Let $M$ be the midpoint of segment $BE$. Point $F$ lies on the minor arc $\\widehat{AD}$ of the circumcircle of triangle $ABD$ such that $MF \\perp BE$. Prove that $ED \\perp FD$.\n\n![](images/Mathematical_Olympiad_in_China_2011-2014_p184_data_fb28eaf3be.png)",
    "options": [],
    "answer": "See solution",
    "solution": "**Solution 1.** Construct point $F_1$ such that $EF_1 = BF_1$ and ray $DF_1$ is perpendicular to line $ED$. It suffices to show that $F = F_1$ or $ABDF_1$ is cyclic; that is,\n\n$$\n\\angle BAD = \\angle BF_1 D. \\qquad \\textcircled{1}\n$$\n\nSet $\\angle BAD = \\angle CAD = x$. Because $EC \\perp AB$ and $AD \\perp BC$,\n$$\n\\angle ECB = 90^\\circ - \\angle ABD = \\angle BAD = x.\n$$\n\nNote that $MD$ is a midline of triangle $BCE$. In particular, $MD \\parallel EC$ and\n$$\n\\angle MDB = \\angle ECD = x. \\qquad \\textcircled{2}\n$$\n\nIn isosceles triangle $EF_1M$, we may set $\\angle EF_1M = \\angle BF_1M = y$. Because $EM \\perp MF_1$ and $MD \\perp DF_1$,\n$$\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$$\n\nCombining ② and ③, we obtain\n\n![](images/Mathematical_Olympiad_in_China_2011-2014_p185_data_142582fab6.png)\n\n$$\n\\angle BDE = \\angle EDM + \\angle MDB = x + y.\n$$\n\nBecause $BE = BD$, we conclude that triangle $BED$ is isosceles with $\\angle MED = \\angle BED = \\angle BDE = x + y$. Because $EMDF_1$ is cyclic, we have $\\angle MF_1 D = \\angle MED = x + y$. It is then clear that\n$$\n\\angle BF_1 D = \\angle MF_1 D - \\angle MF_1 B = x = \\angle BAD,\n$$\nwhich is ①.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 13890,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $n$ be any positive integer. Suppose that Amy has at least $n+1$ rocks where the mass of each rock is a positive integer, and the total mass of all the rocks is $2n$. Then it is always possible to divide the rocks into two piles, each of mass $n$.",
    "options": [],
    "answer": "See solution",
    "solution": "We proceed by induction on $n$.\n\nFor the case $n=1$, Amy has two rocks with total mass 2. This implies that each of Amy's rocks has unit mass, from which the result immediately follows.\n\nFor the inductive step, let us assume that the lemma is true for $n=k$. Consider the case $n=k+1$. Thus Amy has at least $k+2$ rocks whose total mass is $2k+2$.\n\nIf all of Amy's rocks have mass greater than or equal to 2, then the total mass of all the rocks is greater than or equal to $2(k+2) > 2k+2$, which is a contradiction. Hence at least one of Amy's rocks has unit mass. If all of the rocks have unit mass then the result follows immediately. Hence we may let the masses of the rocks be\n\n$$\n1, m_1, m_2, \\dots, m_r \\qquad (*)\n$$\n\nwhere $r \\ge k+1$ and $m_r > 1$.\n\nConsider the situation where we have $r$ rocks of masses\n\n$m_1, m_2, \\dots, m_{r-1}, m_r - 1.$\n\nObserve that the number of rocks above is greater than or equal to $k+1$, and the total mass of these rocks is equal to $2k$. From the inductive assumption we can divide these into two piles, each of total mass $k$. Next change $m_r - 1$ to $m_r$ and add a single rock of unit mass to the pile not containing the rock of mass $m_r - 1$. This gives a division of the rocks with masses given in (*) into two piles, each having mass $k+1$. This concludes the induction, and the proof. $\\square$\n\n*Comment*: It is possible to strengthen the lemma as follows.\n\n**Lemma** Let $n$ be any positive integer. Suppose that Amy has at least $n+1$ rocks where the mass of each rock is a positive integer, and the total mass of all the rocks is $2n$. Then for any integer $N$ with $0 \\le N \\le 2n$ it is always possible to divide the rocks into two piles, one of which has mass $N$.\n\nThe proof is very similar to the one given above.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 13891,
    "subject": "Mathematics (Olympiad)",
    "question": "Let points $A$ and $B$ lie on circle $\\omega$ with center $O$. Assume that $O$ does not lie on line $AB$. Let point $C$ lie on segment $AB$ and denote by $M$ and $N$ the midpoints of segments $AC$ and $CB$, respectively. The circumcircle of $AON$ intersects $\\omega$ at $A$ and $K$ and the circumcircle of $BOM$ intersects $\\omega$ at $B$ and $L$. Moreover, circumcircles of $AON$ and $BOM$ intersect each other at $O$ and $X$. Prove that the quadrilateral $CKXL$ is cyclic.",
    "options": [],
    "answer": "See solution",
    "solution": "In the figure below:\n\n$$\n\\angle CNK = \\angle ANK = \\angle AOK = 2\\angle ABK = 2\\angle NBK.\n$$\n\nHence, triangle $BNK$ is isosceles, so $NK = NB = NC$, and therefore $\\angle BKC$ is right.\n\nIf we let $KC$ intersect $\\omega$ at $K'$, we see that $K', O$, and $B$ are collinear, as $\\angle BKK'$ is right.\n\nNote that $C$ lies on $OX$. Indeed,\n\n$$\n\\mathrm{Pow}(C, AONKX) = AC \\cdot CN = \\frac{1}{2} AC \\cdot CB = CM \\cdot CB = \\mathrm{Pow}(C, BOMLX).\n$$\n\nNow,\n\n$$\n\\angle LXC = \\angle LXO = \\angle LBO = \\angle LBK' = \\angle LKK' = \\angle LKC\n$$\n\nshowing that $XLCK$ is cyclic, as required.\n\n![](images/BW2021_Shortlist_p29_data_b9856ad9a0.png)\n\nFigure 15",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 13892,
    "subject": "Mathematics (Olympiad)",
    "question": "令 $Z, N_0$ 分別表示整數、非負整數所成的集合。試求所有遞增函數 $f : N_0 \\to Z$ 滿足\n\n$$\nf(2) = 7, \\quad f(mn) = f(m) + f(n) + f(m)f(n), \\quad \\forall m, n \\in N_0.\n$$\n\nLet $Z, N_0$ be the sets of all integers and non-negative integers respectively.\nFind all increasing functions $f : N_0 \\to Z$ such that\n\n$$\nf(2) = 7, \\quad f(mn) = f(m) + f(n) + f(m)f(n), \\quad \\forall m, n \\in N_0.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "答：$f(n) = n^3 - 1$, $n \\in N_0$。\n\n由題設觀察得：$f(0) = f(1) = 0$。對 $n \\ge 2$，定 $g(n) = f(n) + 1$。則 $g(2) = 8$ 且\n\n$$\n\\begin{aligned}\ng(mn) &= f(mn) + 1 = f(m) + f(n) + f(m)f(n) + 1 \\\\\n&= (f(m) + 1)(f(n) + 1) \\\\\n&= g(m)g(n), \\text{對所有的 } m, n \\ge 2.\n\\end{aligned}\n$$\n\n固定一整數 $n > 2$，考慮一有理數列 $\\{p_k/q_k, k \\ge 1\\}$，此數列每一項皆大於 $\\log_2 n$ 且收斂至 $\\log_2 n$。則由 $n < 2^{p_k/q_k}$ 得 $n^{q_k} < 2^{p_k}$，再由 $g$ 的單調性得\n\n$$\ng(n^{q_k}) \\le g(2^{p_k}).\n$$\n\n由 $g$ 的可乘積性得\n\n$$\ng(n) \\ge g(2)^{p_k/q_k} = 2^{3p_k/q_k} = (2^{p_k/q_k})^3.\n$$\n\n讓 $k \\to \\infty$，則 $g(n) \\le n^3$。依此類推，得 $g(n) \\ge n^3$。故 $g(n) = n^3$。所以 $f(n) = n^3 - 1$，$\\forall n$ 為滿足題設之唯一解。",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 13893,
    "subject": "Mathematics (Olympiad)",
    "question": "給定一個有 $n$ 條邊的連通圖，其中任兩點之間至多只有一條邊。對於此圖中的任兩個環 $C$ 和 $C'$，定義其**外環**為\n\n$$\nC \\star C' = \\{x \\mid x \\in (C - C') \\cup (C' - C)\\}.\n$$\n\n1. 令 $r$ 為最大的正整數，使得我們能夠從這張圖中選出 $r$ 個環 $C_1, C_2, \\dots, C_r$，且對於所有 $1 \\leq k \\leq r$ 與 $1 \\leq i, j_1, j_2, \\dots, j_k \\leq r$，我們有\n\n$$\nC_i \\neq C_{j_1} \\star C_{j_2} \\star \\dots \\star C_{j_k}.\n$$\n\n2. 令 $s$ 為最大的正整數，使得我們能從這 $n$ 條邊中選出 $s$ 條，讓這些選出的邊不構成環。\n\n試證 $r + s = n$。\n\n*備註：* 一個環是形如 $\\overbrace{A_i A_{i+1}}$，$1 \\leq i \\leq n$ 的邊所構成的集合，其中 $n \\geq 3$，$A_1, A_2, \\dots, A_n$ 為相異頂點，且 $A_{n+1} = A_1$。",
    "options": [],
    "answer": "See solution",
    "solution": "考慮一種選出滿足條件的 $s$ 邊的方法，並將這 $s$ 邊塗為紅色，其他 $n-s$ 邊塗為藍色。若紅邊不是連通的，則加入連接不同連通塊的邊必不會產生新的圈，違反 $s$ 的最大性，所以這些紅邊是連通且無圈的，因此這 $s$ 邊會構成一棵樹。\n\n此時考慮藍邊 $b_1, b_2, \\dots, b_{n-s}$，則由定義，每條藍邊都會與某些紅邊構成恰好一個圈，將其分別記為 $D_1, D_2, \\dots, D_{n-s}$。\n\n我們首先證明沒有 $D_i = D_{j_1} \\star D_{j_2} \\star \\dots \\star D_{j_k}$，這是因為每條藍邊都唯一屬於一個 $D_i$，因此只要 $i \\neq j_1, j_2, \\dots, j_k$，上式便不會成立。\n\n下面證明此時任何其他的圈都能由 $D_1, D_2, \\dots, D_{n-s}$ 中的某些圈進行 $\\star$ 運算得到。首先設有一個包含 $m$ 條藍邊（在圈上依序為 $b_{l_1}, \\dots, b_{l_m}$）的圈。令 $D_i = E_i + F_i + b_i$，其中 $E_i, F_i$ 分別為 $b_i$ 的兩端點至樹根的邊構成的集合。則\n\n$$\nD_{l_1} \\star D_{l_2} \\star \\dots \\star D_{l_m} = (E_{l_1} \\star F_{l_1} \\star b_{l_1}) \\star \\dots \\star (E_{l_m} \\star F_{l_m} \\star b_{l_m}) = b_{l_1} \\star (E_{l_1} \\star F_{l_2}) \\star b_{l_2} \\star (E_{l_2} \\star F_{l_3}) \\star \\dots \\star b_{l_m} \\star (E_{l_m} \\star F_{l_1})\n$$\n\n其中 $E_{l_k} \\star F_{l_{k+1}}$ 為 $b_{l_k}$ 到 $b_{l_{k+1}}$ 之間的紅色線段，因此 $D_{l_1} \\star D_{l_2} \\star \\dots \\star D_{l_m}$ 即所求的圈。\n\n最後，我們證明 $r = n - s$。事實上若 $r \\geq n - s + 1$，則我們從這 $r$ 個圈中選出 $n - s + 1$ 個，令為 $G_1, G_2, \\dots, G_{n-s+1}$。並由先前的討論中我們知道這些圈都能由 $D_1, D_2, \\dots, D_{n-s}$ 中的某些圈進行 $\\star$ 運算得到。因此，若以 $\\star$ 運算替代加法，可以表達為：\n\n$$\nX \\begin{pmatrix} D_1 \\\\ D_2 \\\\ \\vdots \\\\ D_{n-s} \\end{pmatrix} = \\begin{pmatrix} G_1 \\\\ G_2 \\\\ \\vdots \\\\ G_{n-s+1} \\end{pmatrix}\n$$\n\n其中 $X$ 是一個 $(n-s+1) \\times (n-s)$ 的 0-1 矩陣，而將 $X$ 的行進行加減高斯消去，必會有一行全剩 0（即 row echelon form），即其中有一行能由其餘某些行進行 $\\star$ 運算得到。即 $G_1, G_2, \\dots, G_{n-s+1}$ 中也有某個元素能由之中的其他元素由 $\\star$ 運算得到，矛盾。因此 $r = n-s$，即 $n = r+s$。",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 13894,
    "subject": "Mathematics (Olympiad)",
    "question": "a) What are the possible widths of a rectangle with approximate area $50 \\times 20 = 1000\\ \\text{mm}^2$, actual length $47\\ \\text{mm}$, and actual area less than $1000\\ \\text{mm}^2$?\n\nb) In a quirky rectangle, both side lengths are rounded. Explain why this must be so, and what can be deduced about the rounding direction of each side.\n\nc) For quirky rectangles with side lengths under $100\\ \\text{mm}$, what are the possible actual dimensions if the approximate area equals the actual area and both rounded side lengths are multiples of $10\\ \\text{mm}$?\n\n% IMAGE: ![](images/2019_Australian_Scene_W1_p48_data_f496024668.png)\n\nd) Find all quirky rectangles with side lengths under $100\\ \\text{mm}$ whose actual and approximate areas are equal, and whose rounded side lengths are multiples of $10\\ \\text{mm}$.",
    "options": [],
    "answer": "See solution",
    "solution": "a) The possible widths are $21, 22, 23,$ or $24\\ \\text{mm}$. If the width is $21\\ \\text{mm}$, the actual area is $47 \\times 21 = 987\\ \\text{mm}^2$. For $22\\ \\text{mm}$, $47 \\times 22 = 1034\\ \\text{mm}^2$. Since the actual area must be less than $1000\\ \\text{mm}^2$, only $21\\ \\text{mm}$ is possible.\n\nb) If only one side is rounded, the approximate area would not match the actual area. If both are rounded up, the approximate area is greater; if both down, it's less. Thus, one side rounds up, the other down.\n\nc) The approximate area is a multiple of $100\\ \\text{mm}^2$, so the actual area must be too. The only products of two factors equal to $100$ are $1 \\times 100$, $2 \\times 50$, $4 \\times 25$, $5 \\times 20$, $10 \\times 10$. Since neither actual side can be a multiple of $10$, one must be a multiple of $4$ and the other a multiple of $25$.\n\nFrom the table:\n- $25 \\times 12 = 300$\n- $25 \\times 24 = 600$\n- $75 \\times 32 = 2400$\n- $75 \\times 64 = 4800$\n\nThese are the only quirky rectangles with side lengths under $100\\ \\text{mm}$ whose actual and approximate areas are equal and whose rounded side lengths are multiples of $10\\ \\text{mm}$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 13895,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $ABC$ be an acute triangle with fixed points $B, C$, and point $A$ moves on the major arc $BC$ of the circumcircle of $ABC$ such that $AB \\neq AC$. The incircle $(I)$ of triangle $ABC$ touches $BC$ at $D$. Let $I_a$ be the $A$-excenter of triangle $ABC$. The line $I_aD$ cuts $OI$ at $L$, and $E$ lies on $(I)$ such that $DE \\parallel AI$.\n\n**a)** The line $LE$ cuts $AI$ at $F$. Prove that $AF = AI$.\n\n**b)** Let $M$ be the point on the circumcircle $(J)$ of triangle $I_aBC$ such that $I_aM \\parallel AD$. The line $MD$ cuts $(J)$ again at $N$. Prove that the midpoint $T$ of $MN$ lies on a fixed circle.",
    "options": [],
    "answer": "See solution",
    "solution": "a) Let $(I)$ touch $CA$ and $AB$ at $U$ and $V$; let $I_b$ and $I_c$ be the excenters at $B$ and $C$ in triangle $ABC$. Clearly, $I_aA$, $I_bB$, and $I_cC$ are the three altitudes of triangle $I_aI_bI_c$. On the other hand, notice that $IA \\perp UV$, $IB \\perp VD$, $IC \\perp DU$, so triangles $DUV$ and $I_aI_bI_c$ have corresponding parallel sides. Thus, there exists a homothety $\\mathcal{H}$ that maps $D$ to $I_a$, $U$ to $I_b$, and $V$ to $I_c$.\n\nUnder $\\mathcal{H}$, $I$ becomes $O'$, the circumcenter of triangle $I_aI_bI_c$. In triangle $I_aI_bI_c$, $I$ is the orthocenter and $(ABC)$ is the Euler circle, so $O$ lies on $IO'$, which implies the center of $\\mathcal{H}$ lies on $OI$. Also, the center of $\\mathcal{H}$ lies on $DI_a$; $DI_a$ and $OI$ intersect at $L$, so $L$ is the center of $\\mathcal{H}$.\n\n![](images/Vietnamese_mathematical_competitions_p292_data_3ef2616018.png)\n\nWe have $L$, $E$, $F$ collinear and $DE \\parallel I_aF$, so $F$ is the image of $E$ under the homothety, and thus lies on $(I_aI_bI_c)$. Furthermore, triangle $I_aI_bI_c$ has $I$ as the orthocenter and $I_aF$ as an altitude, so $F$ and $I$ are symmetric with respect to $I_cI_b$, hence $AF = AI$.\n\nb) It is easy to see that $J$ is the midpoint of the minor arc $BC$ of $(O)$, which is a fixed point. Let $K$ be the intersection of $MN$ with the perpendicular bisector of $BC$; we will prove that $K$ is a fixed point, so $T$ always moves on a circle of diameter $JK$, which is a fixed circle. Let $S$ be the midpoint of $BC$. Since $BC$ and $I_cI_b$ are antiparallel, $I_aS$ is the symmedian of triangle $I_aI_bI_c$. The tangents at $U$ and $V$ of $(I)$ intersect at $A$, so $DA$ is the symmedian of triangle $DUV$. Moreover, the homothety $\\mathcal{H}$ maps triangle $DUV$ to triangle $I_aI_bI_c$, so the image of $DA$ under this homothety is $I_aS$. Thus, $I_aS \\parallel AD$, and $I_a$, $S$, and $M$ are collinear.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 13896,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $ABC$ be an isosceles acute triangle with angle bisector $\\ell$ at $A$. A circle with center $O$ passes through $B$ and $C$, and intersects sides $AB$ and $AC$ at $D$ and $E$, respectively. Construct a parallelogram $DTEK$. Take $X$ and $Y$ on $AB$ and $AC$, respectively, such that $AXKY$ is also a parallelogram. Construct $P$ and $Q$ on $XY$ such that $TP \\parallel AC$ and $TQ \\parallel AB$. Prove that $TO$ passes through the center of the circumcircle of triangle $TPQ$.",
    "options": [],
    "answer": "See solution",
    "solution": "Since $BCED$ is cyclic, triangles $ABC$ and $AED$ are similar. On the other hand,\n\n$$\n\\angle TBC = \\angle TDE = \\angle KED \\quad \\text{and} \\quad \\angle TCB = \\angle TED = \\angle KDE.\n$$\n\nThus, points $T$ and $K$ correspond in the above pair of similar triangles, so\n\n$$\n\\angle TAB = \\angle KAE \\quad \\text{and} \\quad \\angle TAC = \\angle KAB\n$$\n\nimply that $AT$ and $AK$ are symmetric with respect to $\\ell$.\n\n![](images/Saudi_Arabia_booklet_2023_p31_data_5edb1e1628.png)\n\n*Lemma*: If two rays $Ox', Oy'$ are isogonal in angle $xOy$, then every pair of isogonal lines $Oa, Ob$ in angle $xOy$ are also isogonal in angle $x'Oy'$. We use this idea to solve the problem. Let $J$ be the center of $(TPQ)$ and $H$ be the projection of $T$ onto $PQ$. By the symmetry of the altitude and the line joining the center, $TJ$ and $TH$ are isogonal in $\\angle PTQ$. On the other hand,\n\n$$\n\\angle PTD = \\angle ACT = \\angle ABT = \\angle QTE\n$$\n\nso lines $TP$ and $TQ$ are isogonal in $\\angle DTE$, and then $TJ$ and $TH$ are also isogonal in $\\angle DTE$. To prove that $J$, $T$, $O$ are collinear, we show that $TO$ and $TH$ are isogonal in $\\angle DTE$. (*)\n\nLet $S$ be the intersection of $DE$ and $BC$. By Brocard's theorem applied to the completed quadrilateral $BCED$ and $AS$, we have $TO \\perp AS$. Draw $Ax \\perp TH$; then $Ax \\parallel XY$. Since $AXKY$ is a parallelogram, $AK$ passes through the midpoint of $XY$, so\n\n$$\n(Ax, AK, AB, AC) = -1.\n$$\n\nFurthermore, $A(ST, BC) = -1$, so considering the symmetry over $\\ell$, we have $AB \\leftrightarrow AC$, $AT \\leftrightarrow AK$, and $AS \\leftrightarrow Ax'$. Then\n\n$$\n(Ax', AK, AB, AC) = -1,\n$$\n\nso $Ax \\equiv Ax'$, i.e., $AS$ and $Ax$ are symmetric about $\\ell$. Through $T$, draw lines $Tb$ and $Tc$ perpendicular to $AB$ and $AC$, respectively. Then the two sets of four lines\n\n$$\n(AS, Ax, AB, AC) \\quad \\text{and} \\quad (TO, TH, Tb, TC)\n$$\n\nare perpendicular, respectively. This implies that $TO$ and $TH$ are isogonal in $\\angle bTC$. Finally, since $\\angle ADT = \\angle AET$, $Tb$ and $Tc$ are also isogonal in $\\angle DTE$, so (*) is true. □",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 13897,
    "subject": "Mathematics (Olympiad)",
    "question": "Даден е конвексен четириаголник $ABCD$. Нека $E$ е пресекот на $AB$ и $CD$, $F$ е пресекот на $AD$ и $BC$, и $G$ е пресекот на $AC$ и $EF$. Докажи дека следниве две тврдења се еквивалентни:\n\n1. $BD$ и $EF$ се паралелни.\n2. $G$ е средина на отсечката $\\overline{EF}$.",
    "options": [],
    "answer": "See solution",
    "solution": "Низ $E$ повлекуваме права $l$ паралелна со $BC$. Нека $H$ е пресечната точка на $l$ и $AG$. Така $G$ е пресечна точка на дијагоналите во трапезот $EHFC$.\n\n$(i) \\Rightarrow (ii)$: Нека правите $BD$ и $EF$ се паралелни. Тогаш, од Талесовата теорема за паралелни отсечки, ги имаме равенствата:\n\n![](images/MACEDONIAN_MATHEMATICAL_SOCIETY_p19_data_4a6e009b17.png)\n\n$$\n\\overline{AC} = \\overline{AB} \\text{ и } \\overline{AE} = \\overline{AF}.\n$$\n\nСледува дека $\\overline{AC} = \\overline{AD}$, па од истата Талесова теорема заклучуваме дека правите $HF$ и $ED$ се паралелни. Значи, $EHFC$ е паралелограм и неговите дијагонали се преполовуваат во пресечната точка $G$.\n\n$(ii) \\Rightarrow (i)$: Нека $G$ е средишна точка на отсечката $\\overline{EF}$. Тогаш $\\triangle EGH \\cong \\triangle FGC$, па $EHFC$ е паралелограм и заклучуваме дека правите $HF$ и $ED$ се паралелни. Затоа важат равенствата:\n\n$$\n\\overline{AC} = \\overline{AB} \\text{ и } \\overline{AH} = \\overline{AF}.\n$$\n\nСледува дека $\\overline{AB} = \\overline{AD}$, па од истата Талесова теорема заклучуваме дека правите $BD$ и $EF$ се паралелни.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 13898,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $\\alpha$ be a real number.\n\nDetermine all functions $f : \\mathbb{R} \\to \\mathbb{R}$ satisfying\n\n$$\nf(f(x) + y) = f(x^2 - y) + \\alpha f(x)y\n$$\n\nfor all $x, y \\in \\mathbb{R}$.",
    "options": [],
    "answer": "See solution",
    "solution": "First, set $y = (x^2 - f(x))/2$, which gives $\\alpha f(x)(x^2 - f(x))/2 = 0$. Now, distinguish the cases $\\alpha \\ne 0$ and $\\alpha = 0$.\n\n**(a) Case $\\alpha \\ne 0$:**\n\nThis equation implies $f(x) = 0$ or $f(x) = x^2$ for each $x$ separately. In particular, $f(0) = 0$.\n\n- It is easily verified that $f(x) = 0$ for all $x \\in \\mathbb{R}$ is a solution.\n- Next, consider $f(x) = x^2$ for all $x \\in \\mathbb{R}$. The functional equation becomes\n\n$$\n(x^2 + y)^2 = (x^2 - y)^2 + \\alpha x^2 y \\iff 2x^2 y = -2x^2 y + \\alpha x^2 y \\iff (\\alpha - 4)x^2 y = 0\n$$\n\nwhich holds for all $x$ and $y$ exactly when $\\alpha = 4$. Therefore, for $\\alpha = 4$, there is an additional solution $f(x) = x^2$.\n\n- Suppose there exist $x, y \\in \\mathbb{R} \\setminus \\{0\\}$ with $f(y) = y^2$ and $f(x) = 0$. Then the original equation gives $f(y) = f(x^2 - y)$. Since $f(y) = y^2 \\neq 0$, $f(x^2 - y) = (x^2 - y)^2$, so\n\n$$\ny^2 = (x^2 - y)^2 = x^4 - 2x^2 y + y^2\n$$\n\ni.e., $y = x^2/2$. Repeating this argument leads to a contradiction, so no such mixed solutions exist.\n\n**(b) Case $\\alpha = 0$:**\n\nThe equation becomes\n\n$$\nf(f(x) + y) = f(x^2 - y).\n$$\n\nIt is easy to check that constant functions and $f(x) = -x^2$ are solutions.\n\nSuppose there is $a$ with $f(a) = b \\neq -a^2$. Define $d = b + a^2 \\neq 0$. Setting $x = a$ gives $f(b + y) = f(a^2 - y)$ for all $y$. With $y = z - b$, $f(z) = f(d - z)$ for all $z$. Using $x = z$ and $x = d - z$ in the equation gives\n\n$$\nf(z^2 - y) = f((d - z)^2 - y)\n$$\n\nfor all $y, z$. With $y = z^2$, $f(0) = f(d^2 - 2dz)$ for all $z$. Since $d \\neq 0$, $f$ is constant. Thus, there are no other solutions.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 13899,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $ABC$ be an acute-angled triangle. Let $B'$ and $A'$ be points on the perpendicular bisectors of $AC$ and $BC$, respectively, such that $B'A \\perp AB$ and $AB' \\perp AB$. Let $P$ be a point on the segment $AB$, and let $O$ be the circumcenter of triangle $ABC$. Let $D$ and $E$ be points on $BC$ and $AC$, respectively, such that $DP \\perp BO$ and $EP \\perp AO$. Let $O'$ be the circumcenter of triangle $CDE$. Prove that $B'$, $A'$, and $O'$ are collinear.",
    "options": [],
    "answer": "See solution",
    "solution": "We first observe that if $P \\equiv A$, then $O' \\equiv B'$, and if $P \\equiv B$, then $O' \\equiv A'$. Thus, as $P$ varies along $AB$, the corresponding $O'$ traces a segment, and we seek to identify this segment.\n\nIt is natural to consider the line $A'B'$, which turns out to be perpendicular to $CM$, where $M$ is the midpoint of $AB$. Furthermore, $B'M^2 - B'C^2 = AM^2 = A'M^2 - A'C^2$, which uniquely defines the line and shows $A'B' \\perp CM$.\n\n![](images/MACEDONIAN_MATHEMATICAL_OLYMPIADS_2016_p13_data_a9311b1355.png)\n\nNow, we proceed with the formal proof.\n\nIt suffices to show that $O'M^2 - O'C^2 = AM^2$ for all $P$, including $P = A$ and $P = B$.\n\nFirst, we prove that $O'EPD$ is a cyclic quadrilateral. This follows since $EO'D = 2\\angle ABC = \\angle APE + \\angle BPD = \\pi - \\angle EPD$, as $\\angle ABC = \\angle APE = \\angle BPD$. This implies $PO'$ is the angle bisector of $\\angle EPD$ and $PO' \\perp AB$.\n\nWe now have all the ingredients to show $O'M^2 - O'C^2 = AM^2$.\n\nIntroduce the point $D'$ as the second intersection of the line $PE$ and the circumcircle of $CDE$, so that $O'P^2 - O'C^2 = PE \\cdot PD'$.\n\nSince $PO'$ is the angle bisector of $\\angle EPD$, we have $PD = PD'$ by the extended $S-S-K$ congruency theorem. There is some care needed here: the options from $S-S-K$ are $PD = PD'$ or $PD = PE$, but if $PD = PE$, triangles $P'EO'$ and $P'DO'$ are congruent by $S-S-S$, so $EO'P = DO'P = \\angle CAB$, while $EPO' = DPO' = \\frac{\\pi}{2} - \\angle CAB$, so $PD$ and $PE$ are tangents, and in fact $D' \\equiv E$, so the claim still holds.\n\n![](images/MACEDONIAN_MATHEMATICAL_OLYMPIADS_2016_p13_data_c041eedfd4.png)\n\nNoticing that triangles $APE$ and $BPD$ are similar, we get $\\frac{PE}{AP} = \\frac{PB}{PD}$, implying $AP \\cdot BP = PE \\cdot PD = PE \\cdot PD'$.\n\nSince $PO' \\perp AB$, by the Pythagorean theorem we get\n\n$$\n\\begin{aligned}\nO'M^2 - O'C^2 - AM^2 &= \\\\\n&= O'P^2 - O'C^2 + PM^2 - AM^2 = \\\\\n&= PD' \\cdot PE - AP \\cdot PB = 0\n\\end{aligned}\n$$\n\nwhere we used $O'P^2 - O'C^2 = PE \\cdot PD'$ by the power of point $P$ to the circumcircle of $CDE$, and\n\n$$\nAM^2 - PM^2 = (BM - PM)(AM - PM) = AP \\cdot PB.\n$$\n\nThis completes the proof.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 13900,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all pairs of real numbers $(a, b)$ such that the equality\n$$\n|a x + b y| + |b x + a y| = 2|x| + 2|y|\n$$\nholds for all real numbers $x$ and $y$.",
    "options": [],
    "answer": "See solution",
    "solution": "Set $x = y = 1$ to get $|a + b| + |b + a| = 2|1| + 2|1|$, so $2|a + b| = 4$, hence $|a + b| = 2$, so $a + b \\in \\{-2, 2\\}$. For $x = 1$, $y = -1$ we get $|a - b| + |b - a| = 2|1| + 2|{-1}|$, so $2|a - b| = 4$, hence $|a - b| = 2$, so $a - b \\in \\{-2, 2\\}$. Combining these, the possible pairs are $(a, b) \\in \\{(2, 0), (0, 2), (-2, 0), (0, -2)\\}$. All of these pairs satisfy the given equality for any real $x$ and $y$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 13901,
    "subject": "Mathematics (Olympiad)",
    "question": "Show that\n$$\n|(a-b)(b-c)(c-d)(d-a)| \\le \\frac{abcd}{4}\n$$\nif and only if\n$$\n\\frac{(a-b)^2}{ab} \\cdot \\frac{(b-c)^2}{bc} \\cdot \\frac{(c-d)^2}{cd} \\cdot \\frac{(d-a)^2}{da} \\le \\frac{1}{16}.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "Note that\n$$\n\\frac{(a-b)^2}{ab} \\le \\frac{1}{2}. \\qquad (1)\n$$\nIndeed, we have\n$$\n\\frac{(a-b)^2}{ab} \\le \\frac{1}{2} \\Leftrightarrow 2(a-b)^2 \\le ab \\Leftrightarrow 2\\left(\\frac{a}{b}\\right)^2 - 5\\left(\\frac{a}{b}\\right) + 2 \\le 0 \\Leftrightarrow 2\\left(\\frac{a}{b} - 2\\right)\\left(\\frac{a}{b} - \\frac{1}{2}\\right) \\le 0.\n$$\nThe last inequality holds since $a, b \\in [1, 2]$.\n\nSimilarly, we have\n$$\n\\frac{(b-c)^2}{bc} \\le \\frac{1}{2}, \\qquad (2)\n$$\n$$\n\\frac{(c-d)^2}{cd} \\le \\frac{1}{2}, \\qquad (3)\n$$\n$$\n\\frac{(d-a)^2}{da} \\le \\frac{1}{2}. \\qquad (4)\n$$\nMultiplying (1), (2), (3), and (4), we obtain the required inequality (*).\n\nNote that equality occurs when\n$$\n\\{(a, b, c, d) = (2, 1, 2, 1),\\ (1, 2, 1, 2)\\}.\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 13902,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $a$, $b$, $c$ be integers such that $a^3 + b^3 + c^3$ is divisible by $18$. Prove that $abc$ is divisible by $6$.",
    "options": [],
    "answer": "See solution",
    "solution": "We need to prove that $abc$ is divisible by $2$ and by $3$. We will use proofs by contradiction.\n\nSuppose $abc$ is odd. This implies that $a$, $b$, and $c$ are odd. Therefore, $a^3 + b^3 + c^3$ is odd and certainly not divisible by $18$. This contradiction shows that $abc$ is even.\n\nSuppose that $abc$ is not divisible by $3$. Then $a$, $b$, and $c$ are not divisible by $3$, i.e., they are in (possibly distinct) congruence classes among the following mod $9$:\n\n| $x$  | $1$ | $2$ | $4$ | $-4$ | $-2$ | $-1$ |\n|------|-----|-----|-----|------|------|------|\n| $x^3$| $1$ | $-1$| $1$ | $-1$ | $1$  | $-1$ |\n\nWe conclude that $a^3 + b^3 + c^3$ is equal to $-3$, $-1$, $1$, or $3 \\pmod{9}$. Therefore, $a^3 + b^3 + c^3$ is not divisible by $9$ and consequently not by $18$. This contradiction shows that $abc$ is divisible by $3$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 13903,
    "subject": "Mathematics (Olympiad)",
    "question": "Out of 1024 books, 64 are detective stories, and 24 of these are about Sherlock Holmes. What is the probability that a randomly chosen book is a detective story not about Sherlock Holmes?",
    "options": [],
    "answer": "See solution",
    "solution": "There are $64 - 24 = 40$ detective stories which are not about Sherlock Holmes. The probability of choosing one of these at random out of 1024 books is $$\\frac{40}{1024} = \\frac{5}{128}.$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 13904,
    "subject": "Mathematics (Olympiad)",
    "question": "Given $a$ and $b$ are distinct positive integers, show that the system of equations\n\n$$\nxy + zw = a\n$$\n$$\nxz + yw = b\n$$\n\nhas only finitely many solutions in integers $x, y, z, w$.",
    "options": [],
    "answer": "See solution",
    "solution": "By adding and subtracting the equations, we get $(x + w)(y + z) = a + b$ and $(x - w)(y - z) = a - b$. Multiplying these gives\n$$\n(x^2 - w^2)(y^2 - z^2) = (a + b)(a - b) = a^2 - b^2\n$$\nSince $a$ and $b$ are distinct positive integers, $0 < |x^2 - w^2| \\leq |a^2 - b^2|$ and $0 < |y^2 - z^2| \\leq |a^2 - b^2|$. For fixed $a$ and $b$, the equation $0 < |A^2 - B^2| \\leq |C|$ has only finitely many integer solutions $A, B$, so the original system has only finitely many integer solutions.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 13905,
    "subject": "Mathematics (Olympiad)",
    "question": "Twelve students each have a card with a positive integer, and the numbers on the cards are consecutive perfect squares: $1^2, 2^2, 3^2, \\ldots, 12^2$. Two students, Alice and Bob, each lose their card. The sum of the numbers on the remaining 10 cards is a perfect square. What is the product $ab$ of the numbers on Alice's and Bob's cards?",
    "options": [],
    "answer": "See solution",
    "solution": "First, the total sum of the 12 cards is:\n\n$$1 + 4 + 9 + 16 + 25 + 36 + 49 + 64 + 81 + 100 + 121 + 144 = 650.$$\n\nLet $a^2$ and $b^2$ be the numbers on Alice's and Bob's cards. The sum of the remaining 10 cards is $650 - a^2 - b^2$, which must be a perfect square, say $c^2$:\n\n$$650 - a^2 - b^2 = c^2.$$\n\nRewriting, $a^2 + b^2 + c^2 = 650$. Since $a$ and $b$ are integers between 1 and 12, $a^2$ and $b^2$ are among the squares $1, 4, 9, \\ldots, 144$.\n\nTry possible values for $a$ and $b$ (with $a \\neq b$):\n\n- If $a^2 = 25$ ($a = 5$), $650 - 25 = 625 = 25^2$ (so $b^2 = 11^2 = 121$).\n- If $a^2 = 121$ ($a = 11$), $650 - 121 = 529 = 23^2$ (so $b^2 = 5^2 = 25$).\n\nThus, the two missing cards are $5^2$ and $11^2$, so $ab = 5 \\times 11 = \\boxed{55}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 13906,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $n$ be a positive integer, and let $A_n$ (respectively $B_n$) be the set of non-negative integers $k < n$ such that the number of distinct prime factors of $\\gcd(k, n)$ is even (respectively odd). Show that $|A_n| = |B_n|$ if $n$ is even, and $|A_n| > |B_n|$ if $n$ is odd.",
    "options": [],
    "answer": "See solution",
    "solution": "Since $\\gcd(k, n)$ depends only on the residue class of $k$ modulo $n$, we have:\n\n$$\n|A_n| - |B_n| = \\sum_k (-1)^{s(k, n)}\n$$\n\nwhere $s(k, n)$ is the number of distinct prime factors of $\\gcd(k, n)$, and $k$ ranges over a complete residue system modulo $n$.\n\nWe will prove that this sum equals $n \\prod_{p \\mid n} \\left(1 - \\frac{2}{p}\\right)$, where $p$ runs over the prime divisors of $n$. This product is the number of positive integers $k < n$ such that both $k$ and $k+1$ are coprime to $n$.\n\nLet $e(k, n) = (-1)^{s(k, n)}$ and $f(n) = \\sum_k e(k, n)$. We show that $f$ is a multiplicative function: if $n_1$ and $n_2$ are coprime positive integers, then $f(n_1 n_2) = f(n_1) f(n_2)$.\n\nIf $n_1, n_2$ are coprime, then $e(k, n_1 n_2) = e(k, n_1) e(k, n_2)$. If $k_i$ ranges over a complete residue system modulo $n_i$ ($i = 1, 2$), then $k = k_1 n_2 + k_2 n_1$ ranges over a complete residue system modulo $n_1 n_2$, and $e(k, n_i) = e(k_i, n_i)$. Thus,\n\n$$\nf(n_1 n_2) = \\sum_k e(k, n_1 n_2) = \\sum_{k_1} \\sum_{k_2} e(k_1, n_1) e(k_2, n_2) = f(n_1) f(n_2).\n$$\n\nFor a prime $p$ and $m \\geq 1$, $f(p^m)$ equals the number of $k$ coprime to $p$ (which is $p^m - p^{m-1}$) minus the number of $k$ divisible by $p$ (which is $p^{m-1}$), so $f(p^m) = p^m (1 - 2/p)$. This completes the proof.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 13907,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $I$ and $I_A$ be the incenter and the $A$-excenter of an acute-angled triangle $ABC$ with $AB < AC$. Let the incircle meet $BC$ at $D$, and let the line $AD$ meet $BI_A$ and $CI_A$ at $E$ and $F$, respectively. Prove that the circumcircles of triangles $AID$ and $IAEF$ are tangent to each other.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $\\angle(p, q)$ denote the directed angle between lines $p$ and $q$.\n\nThe points $B$, $C$, $I$, and $I_A$ lie on the circle $\\Gamma$ with diameter $II_A$. Let $\\omega$ and $\\Omega$ denote the circles $(IAEF)$ and $(AID)$, respectively. Let $T$ be the second intersection point of $\\omega$ and $\\Gamma$.\n\nThen $T$ is the Miquel point of the complete quadrilateral formed by the lines $BC$, $BI_A$, $CI_A$, and $DEF$, so $T$ also lies on circle $(BDE)$ (as well as on circle $(CDF)$). We claim that $T$ is the desired tangency point of $\\omega$ and $\\Omega$.\n\nIn order to show that $T$ lies on $\\Omega$, use cyclic quadrilaterals $BDET$ and $BII_A T$ to write\n\n$$\n\\angle(DT, DA) = \\angle(DT, DE) = \\angle(BT, BE) = \\angle(BT, BI_A) = \\angle(IT, II_A) = \\angle(IT, IA)\n$$\n\n![](images/Saudi_Arabia_booklet_2021_p23_data_5cb6f343f0.png)\n\nTo show that $\\omega$ and $\\Omega$ are tangent at $T$, let $\\ell$ be the tangent to $\\omega$ at $T$, so that $\\angle(TI_A, \\ell) = \\angle(EI_A, ET)$. Using circles $(BDET)$ and $(BICI_A)$, we get\n\n$$\n\\angle(EI_A, ET) = \\angle(EB, ET) = \\angle(DB, DT)\n$$\n\nTherefore,\n\n$$\n\\angle(TI, \\ell) = 90^{\\circ} + \\angle(TI_A, \\ell) = 90^{\\circ} + \\angle(DB, DT) = \\angle(DI, DT)\n$$\n\nwhich shows that $\\ell$ is tangent to $\\Omega$ at $T$. $\\square$\n\n![](images/Saudi_Arabia_booklet_2021_p23_data_6f48f2bf65.png)",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 13908,
    "subject": "Mathematics (Olympiad)",
    "question": "For an integer $n \\ge 2$, consider the number of ways to fill $n$ boxes lined up from left to right by putting $n$ balls, one-by-one, into the boxes starting from the left-most one. Each ball can be any of three colors. Let $a_n$ be the number of ways to fill the boxes so that the first ball is red, and the arrangement satisfies the condition described below.\n\nSuppose $n \\ge 4$, and we start with a red ball. Let $k$ be the number of consecutive red balls before the first non-red ball is placed. It is required that $k \\ge 2$.\n\nFind the total number of ways to fill the boxes, starting with any color, that satisfy the above condition.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $a_n$ be the number of ways to fill $n$ boxes starting with a red ball, such that at least two consecutive red balls are placed at the beginning before any non-red ball appears.\n\nWe analyze two cases:\n\n1. **Case $k=2$:** The first two balls are red, and the third is either blue or yellow. The remaining $n-3$ boxes can be filled in $a_{n-2}$ ways (since the first non-red appears at position 3). There are $2a_{n-2}$ ways in this case.\n\n2. **Case $k \\ge 3$:** The first $k$ balls are red ($k \\ge 3$). Removing the first box, the problem reduces to filling $n-1$ boxes starting with a red ball, so there are $a_{n-1}$ ways.\n\nThus, the recurrence is:\n$$\na_n = a_{n-1} + 2a_{n-2}\n$$\nwith initial conditions $a_2 = 1$, $a_3 = 1$.\n\nCalculating recursively:\n- $a_4 = a_3 + 2a_2 = 1 + 2 \\times 1 = 3$\n- $a_5 = a_4 + 2a_3 = 3 + 2 \\times 1 = 5$\n- $a_6 = a_5 + 2a_4 = 5 + 2 \\times 3 = 11$\n- ...\n- $a_{12} = 683$\n\nSince the same count applies if the first ball is blue or yellow, the total number of ways is $3 \\times 683 = 2049$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 13909,
    "subject": "Mathematics (Olympiad)",
    "question": "Sea $ABC$ un triángulo con $AB \\neq AC$, sea $I$ su incentro, $\\gamma$ su circunferencia inscrita y $D$ el punto medio de $BC$. La tangente a $\\gamma$ por $D$ diferente de $BC$ toca a $\\gamma$ en $E$. Demuestra que $AE$ y $DI$ son paralelas.",
    "options": [],
    "answer": "See solution",
    "solution": "Sea $P$ el punto de tangencia de $\\gamma$ con $BC$. Sean $Q$ y $R$ los puntos simétricos de $P$ con respecto a $D$ e $I$, respectivamente. Tenemos que:\n\n- $ER$ es paralela a $DI$. Por un lado, $DI$ es perpendicular a $PE$ (es de hecho la mediatriz de $PE$). Por otro lado, por ser $R$ el punto diametralmente opuesto a $P$ en $\\gamma$, el ángulo $\\angle PER$ es recto.\n- $QR$ es paralela a $DI$. Esto es así porque $D$ e $I$ son los puntos medios de $PQ$ y $PR$.\n\nDe ahí se deduce que $Q$, $R$ y $E$ están alineados, en una recta paralela a $DI$. Para concluir el problema basta demostrar que $A$, $Q$ y $R$ están alineados.\n\nTrazando la paralela a $BC$ por $R$ obtenemos un triángulo semejante a $ABC$, que es el resultante de tomar una homotecia de $ABC$ con respecto a $A$. Tanto $R$ como $Q$ son los puntos de tangencia de la circunferencia exinscrita del lado opuesto de $A$, por lo que la homotecia envía $Q$ a $R$. Concluimos que $A$, $Q$ y $R$ están alineados.\n\n**Nota:** A continuación se incluye la demostración de que los puntos de tangencia de las circunferencias inscrita y exinscrita con $BC$ son simétricos con respecto al punto medio de $BC$. Usando la notación de la figura, tenemos que demostrar que $BP = CX$. Por un lado tenemos que $BP + CP = BX + CX$. Por otro, tenemos $CP - BP = CN - BM = CN + AN - BM - AM = AC - AB = (AY - AB) - (AZ - AC) = BY - CZ = BX - CX$. De este sistema de ecuaciones se halla $BP = CX$.\n\n![](images/OME2021_booklet_p11_data_5a34a062ba.png)\n\n![](images/OME2021_booklet_p11_data_8f54a844ff.png)",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 13910,
    "subject": "Mathematics (Olympiad)",
    "question": "Do there exist pairwise distinct natural numbers $a_1, a_2, \\dots, a_k$, greater than 1, for which\n\n$$\na_1 + a_2 + \\dots + a_k = 2010 \\left( \\frac{1}{a_1} + \\frac{1}{a_2} + \\dots + \\frac{1}{a_k} \\right)\n$$\n\na) if $k=2$;\n\nb) if $k=12$.",
    "options": [],
    "answer": "See solution",
    "solution": "**a)** Any pair of divisors of $2010$ with product $2010$ works, for example $a_1 = 2$, $a_2 = 1005$.\n\n**b)** The numbers $2, 3, 5, 6, 10, 15, 134, 201, 339, 402, 670, 1005$ are such a set.\n\n**Explanation:**\nIf $n$ is not a perfect square and $d$ is a divisor of $n$ with $1 < d < n$, then $d' = \\frac{n}{d}$ is also a divisor with $1 < d' < n$ and $d \\neq d'$. For part (a), take any divisor $a_1$ of $2010$ with $1 < a_1 < 2010$, and set $a_2 = \\frac{2010}{a_1}$. For part (b), the listed numbers are all divisors of $2010$ greater than $1$ and less than $2010$, and they are pairwise distinct.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 13911,
    "subject": "Mathematics (Olympiad)",
    "question": "Determine all four-digit natural numbers $\\overline{abcd}$ for which there exists a prime number $p$ such that\n\n$$\n\\overline{cd} - \\overline{ab} = p + 2\n$$\n\nand\n\n$$\n\\sqrt{\\overline{ab} - 4} + \\sqrt{\\overline{cd}} = p^2.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "Since $\\sqrt{\\overline{ab} - 4} + \\sqrt{\\overline{cd}} < 10 + 10 = 20$, we obtain $p = 2$ or $p = 3$.\n\nIf $p = 2$, then $\\overline{cd} = \\overline{ab} + 4$ and $\\sqrt{\\overline{ab} - 4} + \\sqrt{\\overline{ab} + 4} = 4$, therefore $\\overline{ab} = 5$, which is not possible.\n\nIf $p = 3$, then $\\overline{cd} = \\overline{ab} + 5$ and $\\sqrt{\\overline{ab} - 4} + \\sqrt{\\overline{ab} + 5} = 9$. We obtain the unique solution $\\overline{ab} = 20$, thus $\\overline{abcd} = 2025$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 13912,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $ABC$ be an acute-angled triangle with $AC < AB$ and circumradius $R$. Let $D$ be the foot of the altitude from $A$ onto $BC$, and let $T$ be the point on the line $AD$ such that $AT = 2R$, with $D$ lying between $A$ and $T$. Let $S$ be the midpoint of the arc $BC$ on the circumcircle that does not include $A$.\n\n*Prove:* $\\angle AST = 90^\\circ$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $\\angle BAC = \\alpha$, $\\angle ABC = \\beta$, and $\\angle BCA = \\gamma$. Let $O$ be the center of the circumcircle.\n\nBy assumption, $\\beta < \\gamma$. Let $E$ be the point on the circumcircle diametrically opposite to $A$.\n\nBy the inscribed angle theorem, $\\angle AOB = 2\\gamma$. By definition, $S$ is the intersection of the angular bisector of $\\angle CAB$ and the circumcircle. We note that\n$$\n\\angle EAS = \\angle BAS - \\angle BAO = \\frac{\\alpha}{2} - \\frac{1}{2}(180^\\circ - \\angle AOB) = \\frac{\\alpha}{2} - \\frac{1}{2}(180^\\circ - 2\\gamma) = \\frac{\\alpha}{2} + \\gamma - 90^\\circ\n$$\n\nSince also\n$$\n\\angle SAT = \\angle SAC - \\angle DAC = \\frac{\\alpha}{2} - (90^\\circ - \\angle ACD) = \\frac{\\alpha}{2} + \\gamma - 90^\\circ\n$$\n\nit follows that $\\angle EAS = \\angle TAS$. Since $AE = AT = 2R$, triangles $ASE$ and $AST$ are congruent, so $\\angle AST = \\angle ASE$. Since $AE$ is a diameter of the circumcircle, $\\angle ASE = 90^\\circ$, and the claim is proven.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 13913,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $\\omega$ be a circle and $A$ a point outside of $\\omega$. Draw the tangents from $A$ to $\\omega$ and call the points of tangency $X$ and $Y$. Let $B$ and $C$ be points on the segments $AX$ and $AY$, respectively, such that the perimeter of $\\triangle ABC$ is equal to the length of the segment $AX$. Let $D$ be the reflection of $A$ in the line $BC$. Show that the circumcircle $BDC$ touches $\\omega$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $B'$ be the reflection of $A$ through $B$. Since the perimeter of $\\triangle ABC$ equals the length of the segment $AX$, $AB$ is less than half of $AX$ and, therefore, $B'$ lies on the segment $AX$.\n\nLet the point $C'$ lie on $AY$ such that $B'C'$ touches $\\omega$ in the point $Z$. Let $C''$ be the midpoint of $AC'$. Since $B$ and $C''$ are midpoints of the sides $AB'$ and $AC'$, respectively, we have that the perimeter of $\\triangle AB'C'$ is double that of $\\triangle ABC''$.\n\nWe also have that the perimeter of $\\triangle AB'C'$ equals\n\n$$\n\\begin{aligned}\n|AB'| + |B'C'| + |C'A| &= |AB'| + |B'Z| + |ZC'| + |C'A| \\\\\n&= |AB'| + |B'X| + |YC'| + |C'A| \\\\\n&= |AX| + |AY| \\\\\n&= 2|AX|.\n\\end{aligned}\n$$\n\nTherefore, the perimeter of triangles $\\triangle ABC$ and $\\triangle ABC''$ is the same, namely $|AX|$.\n\n![](images/bw18shortlist_p30_data_21326e008f.png)\n\nIf we assume that $C''$ lies between $A$ and $C$ we have that $|BC''| + |AC''| = |BC| + |AC|$, so\n\n$$\n|BC''| = |BC| + |CC''|.\n$$\n\nWhich contradicts the triangle inequality, so $C''$ does not lie between $A$ and $C$. Similarly, $C''$ cannot lie between $C$ and $Y$, and must therefore lie on $C$. Hence, $C$ and $C''$ are the same point so $C$ is the midpoint of $AC'$.\n\nNow, $\\omega$ is tangent to the extensions of the sides $AB'$ and $AC'$ of $\\triangle AB'C'$ as well as being tangent to the side $B'C'$ and is therefore an excircle of the triangle.\n\nAlso, $B$ and $C$ are the midpoints of sides $AB'$ and $AC'$, respectively.\n\nWe have that the point $D$ lies on the line $B'C'$, for $D$, $B'$ and $C'$ are reflections of $A$ through points on $BC$. Also, since $BC\\parallel B'C'$, and $AD \\perp BC$, we have that $AD \\perp B'C'$. So $D$ is the foot of the altitude from $A$ in $\\triangle AB'C'$. Thus, the circle through $B, D, C$ is the nine-point circle of $\\triangle AB'C'$. According to Feuerbach's theorem, it touches the excircle $\\omega$, as required.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 13914,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $m$ and $n$ be positive integers. Mr. Fat has a set $S$ containing every rectangular tile with integer side lengths and area a power of $2$. Mr. Fat also has a rectangle $R$ with dimensions $2^m \\times 2^n$ and a $1 \\times 1$ square removed from one of the corners. Mr. Fat wants to choose $m + n$ rectangles from $S$, with respective areas $2^0, 2^1, \\dots, 2^{m+n-1}$, and then tile $R$ with the chosen rectangles. Prove that this can be done in at most $(m+n)!$ ways.",
    "options": [],
    "answer": "See solution",
    "solution": "We call each of the rectangles in $S$ a tile, and the tile with area $2^0 = 1$ the unit tile. We may assume without loss of generality that the missing $1 \\times 1$ square in $R$ is the top-left corner.\n\nSuppose Mr. Fat walks on the path of squares starting from the top right corner square and going left along the top row of squares until the missing square is stepped on, then turning and going down along the left column of squares to the bottom left square. For a given tiling, let $a_1, a_2, \\dots$ be the sequence of areas of the tiles he steps on by walking along this path from start to finish. We will show that\n\n(a) every tile is stepped on, that is, $a_1, a_2, \\dots, a_{m+n}$ is a rearrangement of $2^0, 2^1, \\dots, 2^{m+n}$; and\n\n(b) any valid sequence $a_1, \\dots, a_n$ uniquely determines the tiling.\n\nSince there are at most $(m+n)!$ ways to order the $m+n$ possible areas, this would finish the problem.\n\nTo establish (a), we induct on $m+n$. It is trivial for $m+n=1$. Suppose it is true for $m+n-1$. We show it holds for $m+n$. Consider the tile with area $2^{m+n-1}$, and call it $T$. If $T$ has dimensions $2^a \\times 2^b$, then we have $a+b=m+n-1$ and since $T$ is contained in $R$, $a \\le m$ and $b \\le n$. Thus the only two possibilities for the dimensions of $T$ are $2^m \\times 2^{n-1}$ or $2^{m-1} \\times 2^n$. In other words, $T$ must stretch over either the entire length or entire width of $R$. It follows that $T$ must be hit on our walk.\n\nNow, consider the rectangle formed by removing $T$ completely from $R$ and, if this breaks $R$ into two pieces, sliding them together so that the two newly formed edges coincide. This forms a tiling of a smaller rectangle $R'$ with dimensions $2^x \\times 2^y$ for $x+y=m+n-1$, and the areas hit along the walk corresponding to $R'$ are precisely those areas encountered along the walk on $R$ other than $T$. By the inductive hypothesis, all of the areas are stepped on, and so (a) holds for all rectangles $R$.\n\nTo establish (b), we again induct on $m+n$. It is trivial for $m+n=1$. Suppose it is true for $m+n-1$. Let $a_1, a_2, \\dots, a_{m+n}$ be a valid ordering of the areas $\\{2^0, 2^1, \\dots, 2^{m+n-1}\\}$, that is, an ordering generated by walking along a tiling.\n\nSuppose $2^{m+n-1}$ appears before $2^0$ in $a_1, \\dots, a_{m+n}$. Consider any tiling that generates this ordering and call the largest tile $T$. We must hit $T$ before the unit tile. Hence, the part of the rectangle below or to the left of $T$ has odd area, since all other tiles in the tiling have even area. In particular, this part must contain the missing corner. The corner is in the top-left, so this part can't be below the largest rectangle. Therefore, it is to the left of $T$. Hence $T$ has the same height as $R$. Likewise, if we hit the unit tile before $T$, we may use the same reasoning to show that $T$ would have the same width as $R$.\n\nTherefore, depending on whether $2^{m+n-1}$ appears before or after $2^0$ in the ordering, we may uniquely determine whether $T$ spans an entire column or an entire row in any tiling achieving the given ordering. This uniquely determines the dimension of the rectangle that is obtained by removing $T$ and sliding the two resulting parts together (if there are even two parts at all). Suppose that the dimensions of this rectangle are uniquely fixed to be $2^a \\times 2^b$, where we must have $a+b=m+n-1$.\n\nNow, remove $2^{m+n-1}$ to obtain an ordering of $\\{2^0, 2^1, \\dots, 2^{m+n-2}\\}$. By the inductive hypothesis, there is a unique tiling of a $2^a \\times 2^b$ rectangle by these tiles. The position of $2^{m+n-1}$ in the given ordering determines the placement of $T$, and thus the entire tiling is uniquely determined by the sequence. Therefore, there are at most $(m+n)!$ possible tilings.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 13915,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $h$ cm be the height of the water collected in the gauge from empty during rainfall of $r$ mm. As in the solution to Part **b**, the volume of water in cubic centimetres is\n\n$$\n\\frac{5h^2}{2} = 10 \\times 10 \\times \\frac{r}{10} = 10r.\n$$\n\nSo $h^2 = 4r$ and $h = 2\\sqrt{r}$.\n\nThis table shows values of $h$ in cm for values of $r$ in 20 mm steps.\n\n<table><thead><tr><th>r mm</th><th>20</th><th>40</th><th>60</th><th>80</th><th>100</th></tr></thead><tbody><tr><th>h cm</th><td>2\\sqrt{20}<br>≈ 8.9</td><td>2\\sqrt{40}<br>≈ 12.6</td><td>2\\sqrt{60}<br>≈ 15.5</td><td>2\\sqrt{80}<br>≈ 17.9</td><td>2\\sqrt{100}<br>= 20</td></tr></tbody></table>\n\nFrom Part **c**, the heights of the water at the 20 mm and 40 mm marks are $2\\sqrt{20}$ cm and $2\\sqrt{40}$ cm respectively. So, at 4 pm, the height of the water in the gauge was $\\frac{1}{2}(2\\sqrt{20} + 2\\sqrt{40}) = \\sqrt{20} + \\sqrt{40}$ cm.\n\n% IMAGE: ![](images/2021_Australian_Scene_p38_data_d6f925acf2.png)\n\nWhat is, to the nearest millimetre, the amount of rain that fell from 3 pm to 4 pm?",
    "options": [],
    "answer": "See solution",
    "solution": "From Part **c**, the rainfall reading at 4 pm was\n\n$$\n\\begin{aligned}\n\\frac{(\\sqrt{20} + \\sqrt{40})^2}{4} &= \\frac{20 + 40 + 2\\sqrt{20} \\times 40}{4} \\\\\n&= \\frac{60 + 40\\sqrt{2}}{4} \\\\\n&= 15 + 10\\sqrt{2} \\approx 29.14 \\text{ mm.}\n\\end{aligned}\n$$\n\nSo, to the nearest millimetre, the amount of rain that fell from 3 pm to 4 pm was $29 - 20 = 9$ mm.",
    "topic": "Geometry",
    "subtopic": "Solid Geometry"
  },
  {
    "id": 13916,
    "subject": "Mathematics (Olympiad)",
    "question": "What is the maximum number of points that can be placed on the plane so that there are exactly $2012$ straight lines which pass through at least two of them?",
    "options": [],
    "answer": "See solution",
    "solution": "The answer is $2012$ points.\n\nIt is easy to place $2012$ points so that the condition is fulfilled: place $2011$ points on a straight line, and the last one outside of it.\n\n![](images/Ukraine2022-23_p47_data_15aa046df2.png)\n\nNow, we prove that it is impossible to locate more than $2012$ points such that the conditions are fulfilled.\n\nLet $M$ be a set of $n$ points in the plane, not all of which lie on the same line. Then there exists a line that passes through exactly two points from $M$.\n\nDraw all the lines through every pair of points from $M$. Find a line and a point from $M$ such that the distance between them is minimal (excluding any pair where the point lies on the line). Denote the line as $l$ and the point as $A$. We prove by contradiction that only two points from $M$ lie on $l$.\n\nAssume at least three points lie on $l$ from $M$, denoted from left to right as $B_1, B_2, B_3$. Since $AB_1 + AB_3 > B_1B_3 = B_1B_2 + B_2B_3$, then either $AB_1 > B_1B_2$ or $AB_3 > B_2B_3$. Without loss of generality, assume $AB_3 > B_2B_3$. Let $h_1$ be the height of triangle $AB_2B_3$ from vertex $A$, $h_2$ from $B_2$. The area is $S = \\frac{1}{2} h_1 \\cdot B_2B_3 = \\frac{1}{2} h_2 \\cdot AB_3$, where $h_1 > h_2$ when $B_2B_3 < AB_3$. This implies the distance between $B_2$ and line $AB_3$ is less than the distance between $A$ and $l$, a contradiction, finishing the proof of the lemma.\n\nNow, we prove that having $n$ points on the plane implies that either all belong to the same line or there exist at least $n$ different lines, each containing at least two points among the $n$ given. This leads to the answer $2012$ for the question.\n\nWe prove this by induction. For $n=3$, it is obvious. Suppose the statement is true for $n$. For $n+1$ points, not all on the same line, by the lemma, we can choose two points such that the line through them contains no other points. Delete one of those two points so that the remaining points are not all on the same line. The $n$ points satisfy the induction hypothesis, so there are at least $n$ different lines, each containing at least two points. Adding the initial line, we get at least $n+1$ different lines for the $n+1$ points, completing the induction.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 13917,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $a, b, c > 0$. Prove that\n\n$$\n\\frac{(a-b)^2}{(c+a)(c+b)} + \\frac{(b-c)^2}{(a+b)(a+c)} + \\frac{(c-a)^2}{(b+c)(b+a)} \\ge \\frac{(a-b)^2}{a^2+b^2+c^2}.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "It follows from $\\frac{1}{2}(a-2b)^2 + \\frac{1}{2}(a-2c)^2 + (b-c)^2 \\ge 0$ that\n\n$$\n3(a^2 + b^2 + c^2) \\ge 2a^2 + 2ab + 2bc + 2ac = 2(a+b)(a+c),\n$$\n\nso we have $(a+b)(a+c) \\le \\frac{3}{2}(a^2+b^2+c^2)$. Similarly,\n\n$$\n(b + a)(b + c) \\le \\frac{3}{2}(a^2 + b^2 + c^2),\n$$\n\nand $(c + a)(c + b) \\le \\frac{3}{2}(a^2 + b^2 + c^2)$.\n\nHence,\n\n$$\n\\begin{aligned}\n& \\frac{(a-b)^2}{(c+a)(c+b)} + \\frac{(b-c)^2}{(a+b)(a+c)} + \\frac{(c-a)^2}{(b+c)(b+a)} \\\\\n\\ge & \\frac{2}{3} \\cdot \\frac{(a-b)^2 + (b-c)^2 + (c-a)^2}{a^2 + b^2 + c^2} \\\\\n\\ge & \\frac{2}{3} \\cdot \\frac{(a-b)^2 + \\frac{1}{2}(b-c+c-a)^2}{a^2 + b^2 + c^2} \\\\\n= & \\frac{(a-b)^2}{a^2 + b^2 + c^2}.\n\\end{aligned}\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 13918,
    "subject": "Mathematics (Olympiad)",
    "question": "Yuriy's handwriting is so bad that the digits 0, 1, and 8 look absolutely the same either from a normal point of view or upside down. Under such an upside-down view, digits 6 and 9 change into one another. Yuriy tries to write a number with all different digits such that its product with the same number (but taken \"upside-down\", if such a number makes sense) is maximal. What number should Yuriy take?",
    "options": [],
    "answer": "See solution",
    "solution": "We have only 5 digits that can be used for a number: 0, 1, 6, 8, 9. So Yuriy should take a 5-digit number. To maximize the product, both numbers should have maximal first digits, which are 9 and 6 (since 6 becomes 9 under an upside-down view). Similarly, the next digits should be maximized: we can take 1 and 8, and the last remaining digit is 0. Thus, we need to check two numbers: $91086$ and $98016$. Since one of these numbers turns into the other under an upside-down view, both numbers are suitable.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 13919,
    "subject": "Mathematics (Olympiad)",
    "question": "Given 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 such that:\n\n1. Each line begins at one ring and ends at another ring.\n2. Every two unit squares visited consecutively by the same line share a common side.\n3. No two lines (including their endpoints) visit a common unit square.\n4. No line visits the same unit square more than once.\n\nProve that the sum of the numbers of unit squares visited by the lines is always either even or always odd, regardless of how Juku draws the lines.",
    "options": [],
    "answer": "See solution",
    "solution": "Color the unit squares black and white so that adjacent squares have different colors (a checkerboard coloring). Each line connects two rings, and thus two squares, which may be of the same or different colors. If a line connects rings on squares of the same color, it must visit an odd number of squares; if the colors differ, it visits an even number.\n\nLet $k$ of the ringed squares be black. Let $a$ lines connect two black squares, $b$ lines connect two white squares, and $c$ lines connect squares of different colors. Then $2a + c = k$ and $a + b + c = n$, so $a + b = n - k + 2a$. Thus, $a + b$ and $n - k$ have the same parity.\n\nLet $s$ be the total number of squares visited by all lines. $s$ is the sum of $a + b$ odd numbers and $c$ even numbers, so $s$ and $a + b$ have the same parity. Therefore, $s$ and $n - k$ have the same parity. Since $n - k$ is fixed, $s$ is always even or always odd, regardless of how the lines are drawn.",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 13920,
    "subject": "Mathematics (Olympiad)",
    "question": "Given two positive integers $a$ and $b$ such that\n\n$$\n\\frac{b^3}{a^4} \\quad \\text{and} \\quad \\frac{a^3}{b^2}\n$$\n\nare both integers greater than $1$.\n\nWhat is the smallest possible value of $a + b$?",
    "options": [],
    "answer": "See solution",
    "solution": "The smallest integer greater than $1$ is $2$, so\n\n$$\n\\frac{b^3}{a^4} \\geq 2 \\quad \\text{and} \\quad \\frac{a^3}{b^2} \\geq 2.\n$$\n\nIt follows that\n\n$$\na = \\left(\\frac{b^3}{a^4}\\right)^2 \\cdot \\left(\\frac{a^3}{b^2}\\right)^3 \\geq 2^5 \\quad \\text{and} \\quad b = \\left(\\frac{b^3}{a^4}\\right)^3 \\cdot \\left(\\frac{a^3}{b^2}\\right)^4 \\geq 2^7.\n$$\n\nThus, $a + b \\geq 2^5 + 2^7 = 160$. On the other hand, $(a, b) = (2^5, 2^7)$ is a solution. So $160$ is the smallest possible value for $a + b$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 13921,
    "subject": "Mathematics (Olympiad)",
    "question": "設 $r$ 為一正整數，而 $a_0, a_1, \\dots$ 為無窮多個實數所成的序列。假設對於任意的非負整數 $m$ 和 $s$，都存在正整數 $n \\in [m+1, m+r]$，使得\n\n$$\na_m + a_{m+1} + \\dots + a_{m+s} = a_n + a_{n+1} + \\dots + a_{n+s}.\n$$\n\n試證：存在 $p \\ge 1$，使得對於所有非負整數 $n$，$a_{n+p} = a_n$。\n\nLet $r$ be a positive integer, and let $a_0, a_1, \\dots$ be an infinite sequence of real numbers. Assume that for all nonnegative integers $m$ and $s$ there exists a positive integer $n \\in [m+1, m+r]$ such that\n\n$$\na_m + a_{m+1} + \\dots + a_{m+s} = a_n + a_{n+1} + \\dots + a_{n+s}.\n$$\n\nProve that the sequence is periodic, i.e., there exists some $p \\ge 1$ such that $a_{n+p} = a_n$ for all $n \\ge 0$.",
    "options": [],
    "answer": "See solution",
    "solution": "1. 令 $S(m, n) = a_m + a_{m+1} + \\dots + a_{n-1}$。我們首先證明以下引理：\n\n**引理.** 令 $b_0, b_1, \\dots$ 為一無窮序列。假設對於任何非負整數 $m$，存在非負整數 $n \\in [m+1, m+r]$ 使得 $b_m = b_n$。則對於所有 $k \\le l$，存在 $t \\in [l, l+r-1]$ 使得 $b_t = b_k$。此外，$\\{b_i\\}_{i=1}^\\infty$ 最多只有 $r$ 個不同值。\n\n**Proof.** 首先，注意到存在無窮序列 $k_1, k_2, \\dots$，滿足 $k_1 = k$，$k_i < k_{i+1} \\le k_i + r$，且 $b_{k_1} = b_{k_2} = \\dots = b_k$。易見對於任何 $l \\ge k$，必有一 $b_{k_p}$ 落於 $[l, l+r-1]$ 內，故第一部分得證。\n\n對於第二部分，假設存在 $r+1$ 個不同值 $b_{i_1}, b_{i_2}, \\dots, b_{i_{r+1}}$。考慮 $k = i_1, i_2, \\dots, i_{r+1}$ 及 $l = \\max\\{i_1, i_2, \\dots, i_{r+1}\\}$，則由第一部分，我們知對於所有 $j \\in \\{1, 2, \\dots, r+1\\}$，必存在 $t_j \\in [l, l+r-1]$ 使得 $b_{t_j} = b_{i_j}$。然而，$[l, l+r-1]$ 只有 $r$ 個數，故 $b_{t_j}$ 最多只能取到 $r$ 個不同值，矛盾。\n\n2. 回到原題。將 $s = 0$ 帶入，知序列 $\\{a_i\\}_{i=1}^\\infty$ 滿足引理條件，故其最多只能取到 $r$ 個不同值。令 $A_i = (a_i, \\dots, a_{i+r-1})$，則 $A_i$ 最多只能有 $r^r$ 種不同取值，故對於所有 $k \\ge 0$，在 $A_k, A_{k+1}, \\cdots, A_{k+r^r}$ 中必有兩者完全相同；換言之，必存在正整數 $p$，使得有無窮多個正整數 $d$ 滿足 $A_d = A_{d+p}$。令這些 $d$ 所成的集合為 $D$。\n\n3. 現在，我們只需證明 $D = \\{\\text{非負整數}\\}$ 即可；而基於 $D$ 有無窮多個元素，我們只需證明若 $d+1 \\in D$，則 $d \\in D$ 即可。\n\n**Proof.** 假設 $d+1 \\in D$。考慮 $b_k = S(k, p+k)$。易見 $b_0, b_1, \\cdots$ 亦滿足引理條件，故存在 $t \\in [d+1, d+r]$ 使得 $S(t, t+p) = S(d, d+p)$，也就是 $S(d, t) = S(d+p, t+p)$。又 $A_{d+1} = A_{d+1+p}$，故我們有 $S(d, t) = S(d+p+1, t+p)$。結合上述關係，我們有\n\n$$\na_d = S(d, t) - S(d+1, t) = S(d+p, t+p) - S(d+p+1, t+p) = a_{d+p}.\n$$\n\n結合 $A_{d+1} = A_{d+1+p}$，知 $A_d = A_{d+p} \\Rightarrow d \\in D$。證畢。",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 13922,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all triples of positive integers $ (x, y, z) $ for which\n$$\nx \\cdot y! + 2y \\cdot x! = z!.\n$$",
    "options": [],
    "answer": "See solution",
    "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)!$.\n\nThe 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\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- 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$.\n\n**Answer:** $(2, 1, 3)$ and $(n, n+1, n+2)$ for every positive integer $n$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 13923,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $ABC$ be an acute triangle with altitudes $AD$, $BE$, and $CF$, and let $O$ be the centre of its circumcircle.\n\nShow that the segments $OA$, $OF$, $OB$, $OD$, $OC$, and $OE$ dissect the triangle $ABC$ into three pairs of triangles that have equal areas.",
    "options": [],
    "answer": "See solution",
    "solution": "In this question, for a triangle $XYZ$, let $|XYZ|$ denote its area.\n\nConstruct tangents to the circumcircle of triangle $ABC$ at points $A$ and $B$. Let $K$ and $L$ be the feet of the perpendiculars from $D$ and $E$ to the tangents at $B$ and $A$, respectively.\n\n![](images/Brown_Australian_MO_Scene_2013_p67_data_6b8ee9843d.png)\n\nSince $OB \\perp BK$ and $DK \\perp BK$, we have $OB \\parallel DK$. Therefore,\n$|BOD| = \\frac{1}{2}OB \\cdot BK$. Similarly, $|AOE| = \\frac{1}{2}OA \\cdot AL$.\n\nLet $\\angle ABC = \\beta$. Then by the alternate segment theorem we also have $\\angle LAE = \\beta$. Therefore, triangles $BDA$ and $ALE$ are similar (AA), and so,\n\n$$\n\\frac{BD}{AB} = \\frac{AL}{AE}. \\quad (1)\n$$\n\nSimilarly, since $\\angle DBK = \\angle BAC$, triangles $BKD$ and $AEB$ are similar (AA), and so,\n\n$$\n\\frac{BD}{AB} = \\frac{BK}{AE}. \\quad (2)\n$$\n\nComparing (1) and (2) we find $AL = BK$. Also, since $OA = OB$, we have\n\n$$\n|BOD| = \\frac{1}{2}OB \\cdot BK = \\frac{1}{2}OA \\cdot AL = |AOE|.\n$$\n\nSimilarly, $|COE| = |BOF|$ and $|AOF| = |COD|$, as desired.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 13924,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $a_1, a_2, \\ldots, a_n$ be a given finite sequence of positive integers. In a move, one can choose any pair of numbers (that are still in the sequence), delete the smaller one, and increase the bigger one by one (if two numbers are equal, then we can delete either one and increase the other by one). What is the smallest number that can be obtained in the end?\n\nthe smallest integer number $t$ such that $f(A) = 2^{a_1} + 2^{a_2} + \\ldots + 2^{a_n} \\le 2^t$.",
    "options": [],
    "answer": "See solution",
    "solution": "Without loss of generality, sort the given numbers: $a_1 \\le a_2 \\le \\ldots \\le a_n$. Define for a multiset $A = \\{a_1, a_2, \\ldots, a_n\\}$ the function $f(A) = 2^{a_1} + 2^{a_2} + \\ldots + 2^{a_n}$.\n\nAssume that in one move, we transform $A = \\{a_1, a_2, \\ldots, a_n\\}$ to $A'$, where numbers $a_i \\le a_j$ are chosen. Then:\n\n$$\nf(A') - f(A) = 2^{a_j+1} - 2^{a_i} - 2^{a_j} = 2^{a_j} - 2^{a_i} \\ge 0,\n$$\n\nso the function $f$ is nondecreasing.\n\nLet $t$ be the smallest integer such that for the initial set $f(A) \\le 2^t$.\n\nWe prove that if $a$ is the last number in the set, then $a \\ge t$.\n\nSuppose, for contradiction, that $a < t$, i.e., $a \\le t-1$. Since $f$ is nondecreasing, $f(A) \\le f(\\{a\\}) = 2^a \\le 2^{t-1}$, which contradicts the choice of $t$.\n\nWe also show that if in each move we choose the pair of smallest numbers, in the end we will have $a = t$. If we delete $a_1$ and increase $a_2$ to $a_2+1$, then:\n\n$$\nf(A') - f(A) = 2^{a_2+1} - 2^{a_1} - 2^{a_2} = 2^{a_2} - 2^{a_1},\n$$\n\nso $f(A') = f(A) + 2^{a_2} - 2^{a_1}$.\n\nBy the construction of $t$ and choosing the pair of smallest numbers, we have:\n\n$$\nf(A) \\le 2^t \\iff f(A) - 2^{a_1} \\le 2^t - 2^{a_2} \\iff f(A) + 2^{a_2} - 2^{a_1} \\le 2^t \\iff f(A') \\le 2^t.\n$$\n\nHence $a \\le t$, so finally $a = t$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 13925,
    "subject": "Mathematics (Olympiad)",
    "question": "Prove that three distinct non-zero integers $a$, $b$, $c$ satisfy the equation\n\n$$\n\\frac{a}{b} + \\frac{b}{c} + \\frac{c}{a} = 3\n$$\n\nif and only if $a$, $b$, $c$ are given by\n\n$$\na = k u v^2, \\quad b = -k u^2 (u+v), \\quad c = k v (u+v)^2,\n$$\n\n(up to cyclic permutations) for some integers $u$, $v$, $k$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $a$, $b$, $c$ be a solution of the given equation. Put\n\n$$\n\\frac{a}{b} = X, \\quad \\frac{b}{c} = Y,\n$$\n\nso $\\frac{c}{a} = \\frac{1}{XY}$. The equation becomes\n\n$$\nX^2 Y + X Y^2 + 1 = 3 X Y. \\tag{1}\n$$\n\nThis is a cubic curve. Note that $(1, 1)$ is a point on (1). If we know the rational points on (1), we can get integer solutions of the original equation.\n\nSuppose $(x, y)$ is a rational point on (1). Then the line joining $(x, y)$ to $(1, 1)$ has rational slope. Conversely, any line through $(1, 1)$ with rational slope intersects (1) in a point with rational coordinates.\n\nConsider the line\n\n$$\n\\frac{y-1}{x-1} = m,\n$$\n\nwhere $m \\ne -1$ is a nonzero rational number. Let its intersection with (1) be $(X_0, Y_0) \\ne (1, 1)$. Then $Y_0 = m(X_0 - 1) + 1$. Substituting into (1),\n\n$$\nX_0^2 (m(X_0 - 1) + 1) + X_0 (m(X_0 - 1) + 1)^2 + 1 = 3 X_0 (m(X_0 - 1) + 1).\n$$\n\nThis reduces to\n\n$$\n(X_0 - 1)^2 (m(1 + m) X_0 + 1) = 0.\n$$\n\nSince $X_0 \\ne 1$, we have\n\n$$\nX_0 = -\\frac{1}{m(1 + m)}, \\quad Y_0 = m(X_0 - 1) + 1 = -\\frac{m^2}{1 + m}.\n$$\n\nThus,\n\n$$\n\\frac{a}{b} = -\\frac{1}{m(1 + m)}, \\quad \\frac{b}{c} = -\\frac{m^2}{1 + m}, \\quad \\frac{a}{c} = \\frac{m}{(1 + m)^2}.\n$$\n\nSuppose $m = v/u$, where $v, u$ are integers with $\\gcd(v, u) = 1$. Then\n\n$$\n\\frac{a}{c} = \\frac{u v^2}{v (v + u)^2}.\n$$\n\nSo\n\n$$\n\\frac{a}{u v^2} = \\frac{c}{v (u + v)^2}.\n$$\n\nSimilarly, from $a/b$,\n\n$$\n\\frac{a}{v^2} = -\\frac{b}{u(u + v)}.\n$$\n\nThus,\n\n$$\n\\frac{a}{u v^2} = -\\frac{b}{u^2 (u + v)} = \\frac{c}{v (u + v)^2}.\n$$\n\nLet each of these ratios be $p/q$ for some integers $p, q$. Then\n\n$$\nq a = p u v^2, \\quad q b = -p u^2 (u + v), \\quad q c = p v (u + v)^2.\n$$\n\nThis shows $\\gcd(q b, q c) = \\pm p (u + v)$. Thus,\n\n$$\n\\pm q (b, c) = \\pm p (u + v).\n$$\n\nBut $q a = p u v^2$, so $\\pm q \\gcd(a, b, c) = p$. Hence,\n\n$$\n\\frac{p}{q} = \\pm \\gcd(a, b, c) = k.\n$$\n\nFinally,\n\n$$\na = k u v^2, \\quad b = -k u^2 (u + v), \\quad c = k v (u + v)^2.\n$$\n\nIt is easy to check that $\\{a, b, c\\}$ satisfies the given relation.",
    "topic": "Discrete Mathematics",
    "subtopic": "Combinatorics"
  },
  {
    "id": 13926,
    "subject": "Mathematics (Olympiad)",
    "question": "Темињата на еден квадрат лежат на страните на друг квадрат (на една страна едно теме). Да се најде односот на кој се разделени страните на вториот квадрат со темињата на првиот квадрат, ако односот на нивните плоштини е еднаков на $p$, ($p < 1$).",
    "options": [],
    "answer": "See solution",
    "solution": "Нека $ABCD$ и $KLMN$ се дадените квадрати, така што $K, L, M$ и $N$ припаѓаат на страните $AB, BC, CD$ и $DA$ соодветно. Ќе воведеме ознаки $\\overline{BL} = y$, $\\overline{KB} = x$, $\\overline{KL} = b$ и $\\overline{AB} = a$, при што можеме да претпоставиме $x > y$. Тогаш\n\n$$\nP_1 = P_{ABCD} = a^2 = (x + y)^2\n$$\n\nа според Питагорината теорема $b^2 = x^2 + y^2$, па затоа\n\n$$\nP_2 = P_{KLMN} = b^2 = x^2 + y^2\n$$\n\n![](images/Makedonija_2009_p33_data_c1a250394e.png)\n\nОд условот на задачата $\\frac{P_2}{P_1} = \\frac{x^2 + y^2}{(x+y)^2} = p$, каде $0 < p < 1$. Равенството $\\frac{x^2 + y^2}{(x+y)^2} = p$ можеме да го запишеме во облик\n\n$$\n(1-p)x^2 - 2p x y + (1-p)y^2 = 0 \\quad / : y^2\n$$\n\n$$\n(1-p)\\left(\\frac{x}{y}\\right)^2 - 2p\\frac{x}{y} + (1-p) = 0\n$$\n\nАко воведеме ознака $\\frac{x}{y} = t$ добиваме\n\n$$\n(1-p)t^2 - 2p t + (1-p) = 0 \\qquad (1)\n$$\n\nРешенија на равенката (1) се\n\n$$\nt_1 = \\frac{p + \\sqrt{2p - 1}}{1 - p}, \\quad t_2 = \\frac{p - \\sqrt{2p - 1}}{1 - p}\n$$\n\nЈасно е дека $\\frac{x}{y} = \\frac{p + \\sqrt{2p - 1}}{1 - p}$ е бараното решение.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 13927,
    "subject": "Mathematics (Olympiad)",
    "question": "If the numerator of a fraction is multiplied by 6 and the value of the fraction remains the same, by what number must the denominator be multiplied if the original denominator is 3?",
    "options": [],
    "answer": "See solution",
    "solution": "The denominator must also be multiplied by 6, so the new denominator is $3 \\times 6 = 18$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 13928,
    "subject": "Mathematics (Olympiad)",
    "question": "In the plane, let two fixed circles $ (O_1) $ and $ (O_2) $ touch each other at $ M $, with the radius of $ (O_2) $ greater than that of $ (O_1) $. Let $ A $ be a point on $ (O_2) $ such that $ O_1, O_2, A $ are not collinear. Let $ AB $ and $ AC $ be the tangents from $ A $ to $ (O_1) $, touching at $ B $ and $ C $ respectively. The lines $ MB $ and $ MC $ intersect $ (O_2) $ again at $ E $ and $ F $, respectively. Let $ D $ be the intersection point of the line $ EF $ and the tangent to $ (O_2) $ at $ A $. Prove that $ D $ moves on a fixed line as $ A $ moves on $ (O_2) $ such that $ O_1, O_2, A $ are not collinear.",
    "options": [],
    "answer": "See solution",
    "solution": "We consider two cases:\n\n**First case:** The circles $ (O_1) $ and $ (O_2) $ touch each other externally at $ M $.\n\nLet $ xy $ be the common tangent at $ M $ of $ (O_1) $ and $ (O_2) $. Since $ CA $ and $ My $ are tangents to $ (O_1) $ at $ C $ and $ M $, we have $ \\angle FCA = \\angle CMy $. But $ \\angle CMy = \\angle FMx $, hence $ \\angle FCA = \\angle FMx $. As $ \\angle FMx = \\angle FAM $, it follows that $ \\angle FCA = \\angle FAM $.\n\nThe triangles $ MFA $ and $ AFC $ have $ \\angle MFA = \\angle AFC $ and $ \\angle FAM = \\angle FCA $, so they are similar. Therefore,\n$$\n\\frac{MF}{FA} = \\frac{AF}{FC}\n$$\nwhich implies $ FM \\cdot FC = FA^2 $. But $ FM \\cdot FC = P_M(O_1) $ (the power of $ M $ with respect to the circle $ (O_1) $), $ = FO_1^2 - R_1^2 $, where $ R_1 $ is the radius of $ (O_1) $. It follows that $ FO_1^2 - FA^2 = R_1^2 $. Analogously, $ EO_1^2 - EA^2 = R_1^2 $. So\n$$\nFO_1^2 - FA^2 = EO_1^2 - EA^2 = R_1^2.\n$$\nAs $ D $ lies on the line $ EF $, it implies that $ DO_1^2 - DA^2 = R_1^2 $, i.e.,\n$$\nDO_1^2 - R_1^2 = DA^2 \\quad (1)\n$$\nSince $ DA $ is the tangent to $ (O_2) $ at $ A $, $ DA^2 = DO_2^2 - R_2^2 $ (where $ R_2 $ is the radius of $ (O_2) $):\n$$\nDA^2 = DO_2^2 - R_2^2 \\quad (2)\n$$\nFrom (1) and (2), we get $ DO_1^2 - R_1^2 = DO_2^2 - R_2^2 $, or $ P_D(O_1) = P_D(O_2) $. This shows that $ D $ lies on the radical axis of $ (O_1) $ and $ (O_2) $.\n\n**Second case:** The circles $ (O_1) $ and $ (O_2) $ touch each other internally at $ M $.\n\nThe proof in this case is analogous to the proof in the first case.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 13929,
    "subject": "Mathematics (Olympiad)",
    "question": "Find the number of polynomials $f(x) = a x^3 + b x$ that satisfy the following conditions:\n\n1. $a, b \\in \\{1, 2, \\dots, 2013\\}$;\n2. The difference of any two numbers among $f(1)$, $f(2)$, $\\dots$, $f(2013)$ is not divisible by 2013.",
    "options": [],
    "answer": "See solution",
    "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\n$$\nf_1(x) \\equiv a_1 x + b_1 x \\equiv (a_1 + b_1) x \\pmod{3}\n$$\n\nis equivalent to say that $a_1 + b_1$ is not divisible by 3. There are in total six such $f_1(x)$.\n\nFor $i = 2, 3$, if $f_i(x)$ is good modulo $p_i$, then for any $u$ and $v \\not\\equiv 0 \\pmod{p_i}$, $f_i(u + v) \\not\\equiv f_i(u - v) \\pmod{p_i}$, i.e.,\n\n$$\nf_i(u + v) - f_i(u - v) = 2v[a_i(3u^2 + v^2) + b_i]\n$$\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., $\\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": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 13930,
    "subject": "Mathematics (Olympiad)",
    "question": "ABCD is a quadrilateral inscribed in a circle $\\Gamma$. Lines $AB$ and $DC$ meet at $E$; lines $BC$ and $AD$ meet at $F$. Show that the circle of diameter $EF$ cuts the circle $\\Gamma$ orthogonally.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $O$ be the center of the circle $\\Gamma$ and $R$ its radius. Let $G$ be the other point of intersection of the circles $FDC$ and $BCE$.\n\n**Remark.** The following argument needs some minor changes if all the angles of $ABCD$ are acute, that is, if the center $O$ of $\\Gamma$ is internal to $ABCD$.\n\n$\\angle FGC = \\angle FDC$ because both are inscribed angles in $FDC$ subtending the same arc.\n\n$\\angle EGC = $ (since $EGCB$ is cyclic) $= 180^\\circ - \\angle CBE = $ (since $ABCD$ is cyclic) $=$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 13931,
    "subject": "Mathematics (Olympiad)",
    "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": "See solution",
    "solution": "Let $d = \\gcd(a + b, c)$. Since $c \\mid a + 2b$, we also have $d \\mid a + 2b$. Now, $(a + 2b) - (a + b) = b$ and $2(a + b) - (a + 2b) = a$ are divisible by $d$. Therefore, $d$ is a common divisor of $a$, $b$, and $c$, and since $\\gcd(a, b, c) = 1$, it follows that $d = 1$. Thus, $a + b$ and $c$ are relatively prime. But since $a^2 - b^2 = (a - b)(a + b)$ is divisible by $c$, and $\\gcd(a + b, c) = 1$, it follows that $a - b$ must be divisible by $c$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 13932,
    "subject": "Mathematics (Olympiad)",
    "question": "Show that there exist infinitely many pairs $(a, b)$ of positive integers with the property that $a + b$ divides $ab + 1$, $a - b$ divides $ab - 1$, $b > 1$ and $a > b\\sqrt{3} - 1$.",
    "options": [],
    "answer": "See solution",
    "solution": "Observe that $b^2 - 1 = b(a+b) - (ab+1)$ and $b^2 - 1 = b(a-b) - (ab-1)$. Hence, $a+b$ and $a-b$ both divide $b^2 - 1$. Thus, $\\mathrm{lcm}(a+b, a-b)$ divides $b^2 - 1$. Since both are positive, $\\mathrm{lcm}(a+b, a-b) \\leq b^2 - 1$.\n\nLet $d = \\gcd(a, b)$. Then $d \\mid ab$ and $d \\mid a+b \\mid ab+1$. Hence $d \\mid 1$, showing $d=1$. If $e = \\gcd(a+b, a-b)$, then $e \\mid 2a$ and $e \\mid 2b$, so $e \\mid \\gcd(2a, 2b)$. But $\\gcd(2a, 2b) = 2 \\gcd(a, b) = 2$. Thus $e \\mid 2$ and hence $e \\leq 2$. Therefore,\n\n$$\n\\mathrm{lcm}(a+b, a-b) = \\frac{(a+b)(a-b)}{\\gcd(a+b, a-b)} \\geq \\frac{a^2-b^2}{2}.\n$$\n\nIt follows that $a^2 - b^2 \\leq 2(b^2 - 1)$ or $a^2 - 3b^2 \\leq -2$.\n\nSuppose $a$ and $b$ are positive integers such that $a^2 - 3b^2 = -2$. Then $a$ and $b$ have the same parity. For such a pair $(a, b)$, we have\n\n$$\n ab + 1 = ab + \\frac{3b^2 - a^2}{2} = (a+b)\\frac{3b-a}{2},\n$$\n\n$$\n ab - 1 = ab - \\frac{3b^2 - a^2}{2} = (a-b)\\frac{a+3b}{2}.\n$$\n\nHence $a+b$ divides $ab+1$ and $a-b$ divides $ab-1$. We also observe that\n\n$$\n \\sqrt{3}b = \\sqrt{a^2 + 2} < \\sqrt{a^2 + 2a + 1} = a + 1,\n$$\n\nso that $a > \\sqrt{3}b - 1$.\n\nThus, we look for solutions of the equation $x^2 - 3y^2 = -2$ in positive integers. This equation has infinitely many solutions which may be described as follows:\n\nThe equation $x^2 - 3y^2 = -2$ has a particular solution $(1, 1)$. Consider the equation $x^2 - 3y^2 = 1$. This has infinitely many solutions $(u_n, v_n)$ given by\n\n$$\n u_n + \\sqrt{3}v_n = (2 + \\sqrt{3})^n.\n$$\n\nLet $a_n = u_n + 3v_n$ and $b_n = u_n + v_n$. Then\n\n$$\n a_n^2 - 3b_n^2 = -2(u_n^2 - 3v_n^2) = -2.\n$$\n\nFor $n \\geq 1$ we have $b_n > 1$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 13933,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $\\mathbb{N} = \\{1, 2, 3, \\dots\\}$ be the set of positive integers. Find all functions $f$, defined on $\\mathbb{N}$ and taking values in $\\mathbb{N}$, such that $$(n - 1)^2 < f(n)f(f(n)) < n^2 + n$$ for every positive integer $n$.",
    "options": [],
    "answer": "See solution",
    "solution": "The only such function is $f(n) = n$.\n\nAssume that $f$ satisfies the given condition. We will show by induction that $f(n) = n$ for all $n \\in \\mathbb{N}$.\n\nSubstituting $n = 1$ yields $0 < f(1)f(f(1)) < 2$, which implies the base case $f(1) = 1$.\n\nNow assume that $f(k) = k$ for all $k < n$ and suppose, for contradiction, that $f(n) \\neq n$.\n\nOn the one hand, if $f(n) \\le n-1$, then $f(f(n)) = f(n)$ and $f(n)f(f(n)) = f(n)^2 \\le (n-1)^2$, which is a contradiction.\n\nOn the other hand, if $f(n) \\ge n+1$, consider the following methods:\n\n**Method 1:** Assume $f(n) = M \\ge n + 1$. Then $(n+1)f(M) \\le f(n)f(f(n)) < n^2 + n$. Therefore $f(M) < n$, and hence $f(f(M)) = f(M)$ and $f(M)f(f(M)) = f(M)^2 < n^2 \\le (M-1)^2$, which is a contradiction. This completes the induction.\n\n**Method 2:** Note that if $|a-b| > 1$, then the intervals $((a-1)^2, a^2+a)$ and $((b-1)^2, b^2+b)$ are disjoint, which implies that $f(a)$ and $f(b)$ cannot be equal. Assuming $f(n) \\ge n + 1$, it follows that $f(f(n)) < \\frac{n^2+n}{f(n)} \\le n$. This implies that for some $a \\le n - 1$, $f(a) = f(f(n))$, which is a contradiction since $|f(n) - a| \\ge n + 1 - a \\ge 2$. This completes the induction.\n\n**Method 3:** Assuming $f(n) \\ge n + 1$, it follows that $f(f(n)) < \\frac{n^2+n}{f(n)} \\le n$ and $f(f(f(n))) = f(f(n))$. This implies that $(f(n)-1)^2 < f(f(n))f(f(f(n))) = f(f(n))^2 < f(n)^2 + f(n)$ and therefore that $f(f(n)) = f(n)$ since $f(n)^2$ is the unique square satisfying this constraint. This implies that $f(n)f(f(n)) = f(n)^2 \\ge (n+1)^2$, which is a contradiction, completing the induction.\n\n$\\square$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 13934,
    "subject": "Mathematics (Olympiad)",
    "question": "Show that there are no 2-tuples $(x, y)$ of positive integers satisfying the equation\n\n$$\n(x + 1)(x + 2) \\cdots (x + 2014) = (y + 1)(y + 2) \\cdots (y + 4028).\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "Let $n = 2^k \\cdot m$ (where $k$ is a non-negative integer and $m$ is odd), and define $v(n) = 2^k$.\n\nWe proceed by contradiction. Suppose $(x, y)$ is a positive integer solution to the equation. Let\n\n$$\nv(x + i) = \\max_{1 \\leq j \\leq 2014} \\{v(x + j)\\}.\n$$\n\nFor $1 \\leq j \\leq 2014$, $j \\neq i$, we have\n\n$$\nv(x + j) = v(x + i + (j - i)) = v(j - i),\n$$\n\nso\n\n$$\nv\\left(\\prod_{1 \\leq j \\leq 2014,\\ j \\neq i} (x + j)\\right) = v((2014 - j)! \\cdot (j - 1)!) \\leq v(2013!).\n$$\n\nSince $\\prod_{j=1}^{2014} (x + j) = \\prod_{j=1}^{4028} (y + j)$ is a multiple of $4028!$, it follows that\n\n$$\nx + i \\geq v(x + i) \\geq v\\left(\\frac{4028!}{2013!}\\right) > 2^{1007}.\n$$\n\nTherefore, $x > 2^{1006}$. Thus,\n\n$$\n(y + 4028)^{4028} > \\prod_{j=1}^{4028} (y + j) = \\prod_{j=1}^{2014} (x + j) > 2^{1006 \\cdot 2014},\n$$\n\nso $y + 4028 > 2^{503}$ and $y > 2^{502}$.\n\n**Lemma.** Let $0 \\leq x_i < \\frac{1}{2}$ for $1 \\leq i \\leq n$. If $x = \\frac{1}{n} \\sum_{i=1}^n x_i$, $y = 2 \\max_{1 \\leq i \\leq n} \\{x_i^2\\}$, then\n\n$$\n1 - x \\geq \\left( \\prod_{i=1}^{n} (1 - x_i) \\right)^{1/n} \\geq 1 - x - y.\n$$\n\n**Proof of lemma.** By the AM-GM inequality, the left inequality is straightforward. For the right inequality,\n\n$$\n\\begin{align*}\n\\left( \\prod_{i=1}^{n} (1 - x_i) \\right)^{1/n} &\\geq \\frac{n}{\\sum_{i=1}^{n} \\frac{1}{1 - x_i}} \\\\\n&\\geq \\frac{n}{\\sum_{i=1}^{n} (1 + x_i + 2x_i^2)} \\\\\n&\\geq \\frac{1}{1 + x + y} \\\\\n&\\geq 1 - x - y.\n\\end{align*}\n$$\n\nThe lemma is proved.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 13935,
    "subject": "Mathematics (Olympiad)",
    "question": "Suppose the sequence $\\{a_n\\}$ satisfies $a_1 = 2t - 3$ ($t \\in \\mathbb{R}$ and $t \\neq \\pm 1$),\n$$\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$$\n\n1. Find a formula for the general term of $\\{a_n\\}$.\n2. If $t > 0$, determine which is larger: $a_{n+1}$ or $a_n$.",
    "options": [],
    "answer": "See solution",
    "solution": "1. The given recurrence can be rewritten as\n$$\na_{n+1} = \\frac{2(t^{n+1} - 1)(a_n + 1)}{a_n + 2t^n - 1} - 1.\n$$\nLet $b_n = \\frac{a_n + 1}{t^n - 1}$. Then,\n$$\nb_{n+1} = \\frac{2b_n}{b_n + 2}, \\quad b_1 = \\frac{a_1 + 1}{t - 1} = \\frac{2t - 2}{t - 1} = 2.\n$$\nFurthermore,\n$$\n\\frac{1}{b_{n+1}} = \\frac{1}{b_n} + \\frac{1}{2}, \\quad \\frac{1}{b_1} = \\frac{1}{2}.\n$$\nSo,\n$$\n\\frac{1}{b_n} = \\frac{1}{2} + (n-1) \\cdot \\frac{1}{2} = \\frac{n}{2}.\n$$\nTherefore,\n$$\n\\frac{a_n + 1}{t^n - 1} = \\frac{2}{n} \\implies a_n = \\frac{2(t^n - 1)}{n} - 1.\n$$\n\n2. We have\n$$\n\\begin{aligned}\na_{n+1} - a_n &= \\frac{2(t^{n+1} - 1)}{n+1} - \\frac{2(t^n - 1)}{n} \\\\\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] \\\\\n&= \\frac{2(t-1)}{n(n+1)} \\left[ nt^n - (1 + t + \\cdots + t^{n-1}) \\right] \\\\\n&= \\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].\n\\end{aligned}\n$$\nIt is clear 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": 13936,
    "subject": "Mathematics (Olympiad)",
    "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.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $\\omega$ be the circumcircle of $\\triangle BCG$. Let $E'$ be the other intersection of the line $BE$ and $\\omega$, and $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![](images/Mathematical_Olympiad_In_China_2021_2022___Problems_and_Solutions_2024_p222_data_42121f7742.png)\n\nFig. 1.2",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 13937,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $a_n = a_{n,1}a_{n,2}\\dots a_{n,k_n}$ be the decimal representation of $a_n$.\n\nLet $b_n$ be the reversal of the digits of $a_n$ (i.e., $b_n = a_{n,k_n}a_{n,k_n-1}\\dots a_{n,1}$), and define $a_{n+1} = a_n + b_n$.\n\nProve that $a_7$ cannot be prime.",
    "options": [],
    "answer": "See solution",
    "solution": "Note that if $k_n$ is even, then\n$$\na_n \\equiv_{11} a_{n,1} - a_{n,2} + \\dots - a_{n,k_n}\n$$\nand\n$$\nb_n \\equiv_{11} a_{n,k_n} - a_{n,k_n-1} + \\dots - a_{n,1} \\equiv -a_n.\n$$\nHence,\n$$\na_{n+1} = a_n + b_n \\equiv_{11} 0,\n$$\nso $a_{n+1}$ is divisible by $11$.\n\nAlso, if $a_n$ is divisible by $11$, then $b_n$ is also divisible by $11$, so $a_{n+1}$ is divisible by $11$.\n\nTo prove that $a_7$ cannot be prime, it suffices to show that one of $a_1, a_2, \\dots, a_6$ has an even number of digits.\n\nSuppose, for contradiction, that all six numbers have an odd number of digits. Note that $a_{n+1}$ can have at most one more digit than $a_n$, so $a_1, \\dots, a_6$ all have the same number of digits.\n\nLet $a_{1,1} = a$ and $a_{1,k_1} = d$. Then the first digit of $a_2$ is either $a+d$ or $a+d+1$, and since $a_2$ has the same number of digits as $a_1$, $a+d < 10$. Hence, the units digit of $a_2$ equals $a+d$.\n\nThus, the first digit of $a_3$ is at least $2(a+d) < 10$, and the final digit of $a_3$ is at least $2(a+d)$. Continuing in this way, the first digit of $a_4$ is at least $4(a+d)$, the first digit of $a_5$ is at least $8(a+d)$, and the first digit of $a_6$ is at least $16(a+d)$. However, $16(a+d)$ is certainly not less than $10$, which implies that $a_6$ must have more digits than $a_1$, a contradiction.\n\nThis shows that $a_7$ must be divisible by $11$, and hence cannot be prime.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 13938,
    "subject": "Mathematics (Olympiad)",
    "question": "An isosceles triangle has angles $\\alpha$, $\\alpha$, and $\\gamma$, with the difference between $\\alpha$ and $\\gamma$ equal to $10^\\circ$. What is the smallest possible value for the size of the smallest angle?",
    "options": [],
    "answer": "See solution",
    "solution": "There are two cases:\n\n1. If $\\alpha > \\gamma$, then $\\alpha - \\gamma = 10^\\circ$ and $\\alpha + \\alpha + \\gamma = 180^\\circ$. This gives $3\\alpha = 190^\\circ$, so $\\alpha = 63^\\circ 20'$ and $\\gamma = 53^\\circ 20'$.\n\n2. If $\\alpha < \\gamma$, then $\\gamma - \\alpha = 10^\\circ$ and $\\alpha + \\alpha + \\gamma = 180^\\circ$. This gives $3\\alpha = 170^\\circ$, so $\\alpha = 56^\\circ 40'$ and $\\gamma = 66^\\circ 40'$.\n\nTherefore, the smallest possible value for the size of the smallest angle is $53^\\circ 20'$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 13939,
    "subject": "Mathematics (Olympiad)",
    "question": "已知正實數 $x, y, z$ 滿足 $x + y + z = 1$。試求使不等式\n\n$$\n\\frac{x^2 y^2}{1-z} + \\frac{y^2 z^2}{1-x} + \\frac{z^2 x^2}{1-y} \\leq k - 3xyz\n$$\n\n恆成立的實數 $k$ 的最小值。\n\nLet $x, y, z$ be positive real numbers satisfying $x + y + z = 1$. Find the smallest $k$ such that\n\n$$\n\\frac{x^2 y^2}{1-z} + \\frac{y^2 z^2}{1-x} + \\frac{z^2 x^2}{1-y} \\leq k - 3xyz\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "先令 $x = y = z = \\frac{1}{3}$，則 $k \\geq \\frac{1}{6}$。底下證明：\n\n$$\n\\frac{x^2 y^2}{1-z} + \\frac{y^2 z^2}{1-x} + \\frac{z^2 x^2}{1-y} \\leq \\frac{1}{6} - 3xyz. \\quad (1)\n$$\n\n由 $x > 0, y > 0, z > 0$ 及 $x + y + z = 1$，知不等式 (1) 等價於\n\n$$\n\\begin{align*}\n& \\frac{xy}{z(x+y)} + \\frac{yz}{x(y+z)} + \\frac{zx}{y(z+x)} + 3 \\leq \\frac{1}{6xyz} \\\\\n\\Leftrightarrow \\quad & \\frac{xyz}{z(x+y)} + \\frac{xyz}{x(y+z)} + \\frac{xyz}{y(z+x)} \\leq \\frac{1}{6(xy+yz+zx)} \\\\\n\\Leftrightarrow \\quad & \\frac{xy}{x+y} + \\frac{yz}{y+z} + \\frac{zx}{z+x} \\leq \\frac{1}{6(xy+yz+zx)}. \\tag{2}\n\\end{align*}\n$$\n\n由\n\n$$\n\\begin{align*}\n& 1 = (x + y + z)^2 \\\\\n&= x^2 + y^2 + z^2 + 2xy + 2yz + 2zx \\\\\n&\\geq 3(xy + yz + zx) \\\\\n&\\Rightarrow \\frac{1}{6(xy + yz + zx)} \\geq \\frac{1}{2}. \\\\\n& \\text{又 } xy \\leq \\frac{1}{4}(x + y)^2 \\\\\n&\\Rightarrow \\sum \\frac{xy}{x+y} \\leq \\frac{1}{4} \\sum (x+y) \\\\\n&= \\frac{1}{4}(2x + 2y + 2z) = \\frac{1}{2}.\n\\end{align*}\n$$\n\n其中，“$\\sum$”表示輪換對稱和。\n\n從而，不等式 (2) 成立，則不等式 (1) 成立。若且唯若 $x = y = z = \\frac{1}{3}$ 時，等號成立。故 $k$ 的最小值為 $\\frac{1}{6}$。",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 13940,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $x, y$ be positive real numbers with $xy = 4$.\n\nProve that\n$$\n\\frac{1}{x+3} + \\frac{1}{y+3} \\le \\frac{2}{5}.\n$$\n\nFor which $x$ and $y$ does equality hold?",
    "options": [],
    "answer": "See solution",
    "solution": "Clearing denominators, we obtain the equivalent inequality\n$$\n5x + 5y + 30 \\le 2xy + 6x + 6y + 18,\n$$\nwhich simplifies to $x + y \\ge 12 - 2xy = 4$. This inequality is a direct consequence of the AM-GM inequality:\n$$\n\\frac{x + y}{2} \\ge \\sqrt{xy} = 2.\n$$\nEquality holds exactly for $x = y = 2$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 13941,
    "subject": "Mathematics (Olympiad)",
    "question": "A wheel with radius $r$ rolls, without sliding, along the inner side of a circle of radius $2r$. In the beginning, the wheel is at the lowermost position. Find the trajectory of the point initially uppermost on the wheel.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $O$ be the center of the circle; as the radius of the circle equals the diameter of the wheel, $O$ is always located on the boundary of the wheel. Let $A$ be the point of tangency between the wheel and the circle in the beginning. As the wheel rolls without sliding, the point of tangency covers equal distances along the wheel and along the circle. Also, point $O$ must cover the same distances along the wheel since it remains diametrically opposite to the point of tangency.\n\nConsider the situation when the center of the wheel has rotated around $O$ by angle $\\phi \\leq 90^\\circ$. Let the center of the wheel and the point of tangency now be $Q$ and $A'$, respectively, and let $P$ be the new location of the point initially uppermost on the wheel. Then $r \\cdot \\angle OQP = 2r \\cdot \\phi$, whence $\\angle OQP = 2\\phi$. As $O$, $Q$, and $A'$ are collinear, we also have $\\angle OA'P = \\phi = \\angle AOA'$. Consequently, $OA \\parallel PA'$. By Thales' theorem, $\\angle OPA' = 90^\\circ$, implying that $P$ lies on the horizontal diameter of the circle.\n\n![](images/EST_ABooklet_2020_p19_data_8d684c62af.png)\n\nAfter the center of the wheel has rotated around $O$ by $180^\\circ$, the same point of the wheel is located at $O$ as initially. Like in the previous case, we see that $P$ lies on the horizontal diameter of the circle also if $\\phi > 90^\\circ$. As any point on the boundary of the wheel eventually reaches the circle, the trajectory in question constitutes the whole diameter.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 13942,
    "subject": "Mathematics (Olympiad)",
    "question": "A and B are playing ping-pong. The winner of a game gets 1 point, the loser 0. The match ends as soon as one player is ahead by 2 points or after 6 games. The probability that A wins a game is $\\frac{2}{3}$, and B wins with probability $\\frac{1}{3}$; each game is independent. What is the expected number $E\\xi$ of games played until the match ends?",
    "options": [],
    "answer": "See solution",
    "solution": "It is easy to see that $\\xi$ (the number of games played) can only be 2, 4, or 6. We divide the six games into three rounds, each consisting of two consecutive games.\n\nIf one player wins both games in the first round, the match ends. The probability is:\n\n$$\n\\left(\\frac{2}{3}\\right)^2 + \\left(\\frac{1}{3}\\right)^2 = \\frac{5}{9}.\n$$\n\nOtherwise, the players tie (each wins one game), and the match enters the second round; this probability is:\n\n$$\n1 - \\frac{5}{9} = \\frac{4}{9}.\n$$\n\nThe same logic applies for the second and third rounds. Thus:\n\n$$\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$$\n\nTherefore,\n\n$$\nE\\xi = 2 \\times \\frac{5}{9} + 4 \\times \\frac{20}{81} + 6 \\times \\frac{16}{81} = \\frac{266}{81}.\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 13943,
    "subject": "Mathematics (Olympiad)",
    "question": "Determine if there exists a three-variable polynomial $P(x, y, z)$ with integer coefficients such that a positive integer $n$ is *not* a perfect square if and only if there is a triple $(x, y, z)$ of positive integers with $P(x, y, z) = n$.",
    "options": [],
    "answer": "See solution",
    "solution": "The answer is *yes*.\n\nSuppose $Q(x, y, z)$ is a polynomial with integer coefficients such that for all integers $x, y, z$:\n\n- $Q(x, y, z) \\ge 0$,\n- if $Q(x, y, z) = 0$ then $x$ is a non-square,\n- for each positive non-square $x$, there exist $y, z \\in \\mathbb{Z}$ with $Q(x, y, z) = 0$.\n\nThen the polynomial\n\n$$\nP(x, y, z) = x - x Q(x, y, z)\n$$\n\nsatisfies the desired property. Indeed, $P(x, y, z) \\le 0$ unless $Q(x, y, z) = 0$, in which case $x$ must be a non-square, so if $P(x, y, z)$ is a positive integer, then it is a non-square. Further, for any positive non-square $x$, choosing $y, z$ for which $Q(x, y, z) = 0$ yields a triple $(x, y, z)$ for which $P(x, y, z) = x$.\n\nIt remains to show that such a polynomial $Q(x, y, z)$ exists. Here are two approaches:\n\n**First approach:**\n\nFor any $x \\in \\mathbb{Z}$, the following are equivalent:\n\n1. $x$ is a positive non-square,\n2. $y^2 < x < (y+1)^2$ for some $y \\in \\mathbb{Z}$,\n3. $(x - y^2)((y+1)^2 - x) > 0$ for some $y \\in \\mathbb{Z}$,\n4. $(x - y^2)((y+1)^2 - x) = z$ for some $y, z \\in \\mathbb{Z}$,\n5. $((x - y^2)((y+1)^2 - x) - z)^2 = 0$ for some $y, z \\in \\mathbb{Z}$.\n\nThus,\n\n$$\nQ(x, y, z) = ((x - y^2)((y + 1)^2 - x) - z)^2\n$$\nsatisfies the desired conditions.\n\n**Second approach:**\n\nThe Pell equation $y^2 - xz^2 = 1$ has a positive integer solution $y, z$ if and only if $x$ is not a square. Therefore,\n\n$$\nQ(x, y, z) = (y^2 - xz^2 - 1)^2\n$$\nalso works.\n\n*Remark.* For any $k$, the problem also works with “not a $k$th power” instead of “not a square.” However, the second approach does not generalize because Pell equations are specific to squares.",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 13944,
    "subject": "Mathematics (Olympiad)",
    "question": "For a positive integer $n$, a sum-friendly odd partition of $n$ is a sequence $(a_1, a_2, \\dots, a_k)$ of odd positive integers with $a_1 \\leq a_2 \\leq \\dots \\leq a_k$ and $a_1 + a_2 + \\dots + a_k = n$ such that for all positive integers $m \\leq n$, $m$ can be uniquely written as a subsum $m = a_{i_1} + a_{i_2} + \\dots + a_{i_r}$. (Two subsums $a_{i_1} + a_{i_2} + \\dots + a_{i_r}$ and $a_{j_1} + a_{j_2} + \\dots + a_{j_s}$ with $i_1 < i_2 < \\dots < i_r$ and $j_1 < j_2 < \\dots < j_s$ are considered the same if $r = s$ and $a_{i_l} = a_{j_l}$ for $1 \\leq l \\leq r$.)\n\nFor example, $(1, 1, 3, 3)$ is a sum-friendly odd partition of $8$.\n\nFind the number of sum-friendly odd partitions of $9999$.",
    "options": [],
    "answer": "See solution",
    "solution": "We consider the sum-friendly odd partitions of a positive integer $n$.\n\nClearly, $(1, 1, \\dots, 1)$ is a sum-friendly odd partition. On the other hand, if $a_i > 1$ for some $i$, let $r$ be the smallest such that $a_r > 1$. It follows that $a_r = r$ and that $r$ divides $a_i$ for all $i \\geq r$. Therefore, $r$ is odd and it divides $n+1$. Moreover, $(\\frac{a_r}{r}, \\frac{a_{r+1}}{r}, \\dots, \\frac{a_k}{r})$ is a sum-friendly odd partition of $\\frac{n+1}{r} - 1$.\n\nThus, by induction, it follows that the number of sum-friendly odd partitions of $n$ equals the number of factorizations $(n+1) = d_1 d_2 \\dots d_l$ in which $d_1, d_2, \\dots, d_{l-1}$ are odd. Hence, there are $16$ sum-friendly odd partitions of $9999$.\n\n$\\square$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 13945,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $n$ be an integer greater than $1$. Find the greatest constant $\\lambda(n)$ such that for any nonzero complex numbers $z_1, z_2, \\dots, z_n$, we have\n\n$$\n\\sum_{k=1}^{n} |z_k|^2 \\ge \\lambda(n) \\min_{1 \\le k \\le n} \\{|z_{k+1} - z_k|^2\\},\n$$\n\nwhere $z_{n+1} = z_1$.\n",
    "options": [],
    "answer": "See solution",
    "solution": "Let\n\n$$\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$$\n\nWe prove $\\lambda_0(n)$ is the greatest constant.\n\nIf there exists $k$ ($1 \\le k \\le n$) such that $|z_{k+1} - z_k| = 0$, the inequality holds obviously. So, without loss of generality, assume\n\n$$\n\\min_{1 \\le k \\le n} \\{|z_{k+1} - z_k|^2\\} = 1.\n$$\n\nUnder this assumption, it suffices to show that the minimum value of $\\sum_{k=1}^n |z_k|^2$ is $\\lambda_0(n)$.\n\n**Case 1: $n$ even.**\n\n$$\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) \\\\ &\\ge \\frac{1}{4} \\sum_{k=1}^{n} |z_{k+1} - z_k|^2 \\\\ &\\ge \\frac{n}{4} \\min_{1 \\le k \\le n} \\{|z_{k+1} - z_k|^2\\} = \\frac{n}{4}.\n\\end{aligned}\n$$\n\nEquality holds when $(z_1, z_2, \\dots, z_n) = (\\frac{1}{2}, -\\frac{1}{2}, \\dots, \\frac{1}{2}, -\\frac{1}{2})$, so the minimum is $\\frac{n}{4} = \\lambda_0(n)$.\n\n**Case 2: $n$ odd.**\n\nLet $\\theta_k = \\arg \\frac{z_{k+1}}{z_k} \\in [0, 2\\pi)$ for $k = 1, 2, \\dots, n$.\n\nIf $\\theta_k \\le \\frac{\\pi}{2}$ or $\\theta_k \\ge \\frac{3\\pi}{2}$, then\n\n$$\n|z_k|^2 + |z_{k+1}|^2 = |z_k - z_{k+1}|^2 + 2|z_k||z_{k+1}| \\cos \\theta_k \\ge |z_k - z_{k+1}|^2 \\ge 1.\n$$\n\nIf $\\theta_k \\in (\\frac{\\pi}{2}, \\frac{3\\pi}{2})$, then $\\cos \\theta_k < 0$ and\n\n$$\n\\begin{aligned}\n1 &\\le |z_k - z_{k+1}|^2 \\\\ &= |z_k|^2 + |z_{k+1}|^2 - 2|z_k||z_{k+1}| \\cos \\theta_k \\\\ &\\le (|z_k|^2 + |z_{k+1}|^2)(1 + (-2\\cos \\theta_k)) \\\\ &= (|z_k|^2 + |z_{k+1}|^2) \\cdot 2\\sin^2 \\frac{\\theta_k}{2}.\n\\end{aligned}\n$$\n\nTherefore,\n\n$$\n|z_k|^2 + |z_{k+1}|^2 \\ge \\frac{1}{2 \\sin^2 \\frac{\\theta_k}{2}}.\n$$\n\nIf for all $k$, $\\theta_k \\in (\\frac{\\pi}{2}, \\frac{3\\pi}{2})$, then\n\n$$\n\\sum_{k=1}^{n} |z_k|^2 = \\frac{1}{2} \\sum_{k=1}^{n} (|z_k|^2 + |z_{k+1}|^2) \\ge \\frac{1}{4} \\sum_{k=1}^{n} \\frac{1}{\\sin^2 \\frac{\\theta_k}{2}}.\n$$\n\nSince $\\prod_{k=1}^{n} \\frac{z_{k+1}}{z_k} = 1$, we have $\\sum_{k=1}^{n} \\theta_k = 2m\\pi$ for some integer $m < n$. For odd $n$, $0 < \\sin \\frac{m\\pi}{n} \\le \\cos \\frac{\\pi}{2n}$.\n\nLet $f(x) = \\frac{1}{\\sin^2 x}$, which is convex on $[\\frac{\\pi}{4}, \\frac{3\\pi}{4}]$. By Jensen's inequality,\n\n$$\n\\sum_{k=1}^{n} |z_k|^2 \\ge \\frac{n}{4} \\cdot \\frac{1}{\\sin^2 \\frac{m\\pi}{n}} \\ge \\frac{n}{4} \\cdot \\frac{1}{\\cos^2 \\frac{\\pi}{2n}} = \\lambda_0(n).\n$$\n\nThus, $\\lambda_0(n)$ is the greatest possible constant.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 13946,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $\\xi = \\frac{-1+\\sqrt{17}}{2}$, which is irrational. Consider the polynomial $F(x) = P(x) - 2017$ with rational coefficients and $\\xi$ as a root. Then its conjugate $\\frac{-1-\\sqrt{17}}{2}$ is also a root, so $F(x)$ is divisible by $\\varphi(x) = x^2 + x - 4$.\n\nLet $a_0, a_1, \\ldots, a_n$ be nonnegative integers such that:\n\n- (α) $a_n \\xi^n + a_{n-1} \\xi^{n-1} + \\ldots + a_1 \\xi + a_0 = 2017$\n- (β) The sum $a_0 + a_1 + \\ldots + a_n$ is minimal.\n\nFind the least possible value of $a_0 + a_1 + \\ldots + a_n$.",
    "options": [],
    "answer": "See solution",
    "solution": "Since $F(x)$ has $\\xi$ as a root and rational coefficients, its conjugate is also a root, so $F(x)$ is divisible by $x^2 + x - 4$.\n\nWe can write:\n$$\nF(x) = P(x) - 2017 = (x^2 + x - 4) Q(x)\n$$\nfor some polynomial $Q(x)$ with integer coefficients.\n\nLet $a_0, a_1, \\ldots, a_n$ be nonnegative integers such that $a_n \\xi^n + \\ldots + a_0 = 2017$ and the sum $a_0 + \\ldots + a_n$ is minimal.\n\nBy expanding and matching coefficients, we get the recurrence:\n$$\n\\begin{aligned}\na_0 - 2017 &= -4b_0 \\\\\na_1 &= -4b_1 + b_0 \\\\\na_2 &= -4b_2 + b_1 + b_0 \\\\\na_3 &= -4b_3 + b_2 + b_1 \\\\\n\\vdots \\\\\na_{n-2} &= -4b_{n-2} + b_{n-3} + b_{n-4} \\\\\na_{n-1} &= b_{n-2} + b_{n-3} \\\\\na_n &= b_{n-2}\n\\end{aligned}\n$$\n\nGiven $0 \\leq a_i \\leq 3$ for $i = 1, \\ldots, n-2$, we solve recursively:\n- $a_0 = 1$, $b_0 = 504$\n- $a_1 = 0$, $b_1 = 126$\n- $a_2 = 2$, $b_2 = 157$\n- $a_3 = 3$, $b_3 = 70$\n- $a_4 = 3$, $b_4 = 56$\n- $a_5 = 2$, $b_5 = 31$\n- $a_6 = 3$, $b_6 = 21$\n- $a_7 = 0$, $b_7 = 13$\n- $a_8 = 2$, $b_8 = 8$\n- $a_9 = 1$, $b_9 = 5$\n- $a_{10} = 1$, $b_{10} = 3$\n- $a_{11} = 0$, $b_{11} = 2$\n- $a_{12} = 1$, $b_{12} = 1$\n- $a_{13} = 3$, $b_{13} = 0$\n- $a_{14} = 1$, $b_{14} = 0$\n\nThus, the minimal sum is:\n$$\na_0 + a_1 + \\ldots + a_{14} = 1 + 0 + 2 + 3 + 3 + 2 + 3 + 0 + 2 + 1 + 1 + 0 + 1 + 3 + 1 = 23.\n$$\n\nTherefore, the least possible value of the sum is $23$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 13947,
    "subject": "Mathematics (Olympiad)",
    "question": "Given that circle $\\Gamma_2$ is inside circle $\\Gamma_1$ on the plane, prove that there exists a point $P$ on the plane with the following property: if $l$ is a line not through $P$, $l$ intersects $\\Gamma_1$ and $\\Gamma_2$ at distinct points $A$ and $B$, $C$ and $D$, respectively ($A, C, D$, and $B$ are consecutive on $l$), then $\\angle APC = \\angle DPB$.\n\n![](images/Mathematical_Olympiad_In_China_2021_2022___Problems_and_Solutions_2024_p261_data_5302a23222.png)",
    "options": [],
    "answer": "See solution",
    "solution": "**Solution**\n\nLet $O_1, O_2, r_1, r_2$ be the centres and the radii of circles $\\Gamma_1$ and $\\Gamma_2$, respectively, with $r_1 > r_2$. We find two points $P$ and $Q$ on the ray $O_1O_2$ such that\n\n$$\nO_1P \\cdot O_1Q = r_1^2, \\quad O_2P \\cdot O_2Q = r_2^2.\n$$\n\nFirst, take $K$ on the ray $O_1O_2$ with\n\n$$\nO_1K = \\frac{O_1O_2^2 + r_1^2 - r_2^2}{2O_1O_2}.\n$$\n\nAs $O_1O_2 \\le r_1 - r_2$, $O_1K \\ge r_1$. Then take $P$ and $Q$ on the ray $O_1O_2$ with\n\n$$\nKP = KQ = \\sqrt{O_1K^2 - r_1^2},\n$$\n\nwhich leads to $O_1P \\cdot O_1Q = r_1^2$; meanwhile,\n\n$$\n\\begin{aligned}\nO_2P \\cdot O_2Q - O_1P \\cdot O_1Q &= O_2K^2 - O_1K^2 \\\\\n&= (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$$\n\nwhich gives $O_2P \\cdot O_2Q = r_2^2$.\n\nFor an arbitrary line $l$ not through $P$, if $l$ is perpendicular to $O_1O_2$, due to symmetry the conclusion is obvious. Otherwise, from $O_2C^2 = O_2Q \\cdot O_2P$ we have $\\triangle O_2CQ \\sim \\triangle O_2PC$, and $\\frac{CQ}{CP} = \\frac{r_2}{O_2P}$. Similarly, $\\frac{DQ}{DP} = \\frac{r_2}{O_2P}$, and hence $\\frac{CQ}{CP} = \\frac{DQ}{DP}$, indicating that the bisectors of $\\angle CPD$ and $\\angle CQD$ meet $l$ at the same point, say $M$. In the same way, we can infer that the bisectors of $\\angle APB$ and $\\angle AQB$ meet $l$ at the same point as well, say $M'$.\n\nNote that $P$, $Q$, and $M$ all lie on the Apollonian circle which is the locus of points $X$ such that $\\frac{CX}{CD} = \\frac{CQ}{DQ}$, hence the centre $L$ must be on $l$, more precisely, at the intersection of the perpendicular bisector of $PQ$ and $l$. In the same way, $P$, $Q$, and $M'$ all lie on the Apollonian circle, $\\frac{AP}{AQ} = \\frac{AM'}{BQ} = \\frac{AM'}{BM'}$, with the same centre $L$. As $K$ lies on the boundary or outside of $\\Gamma_1$, we infer that $L$ is outside $\\Gamma_1$, $M = M'$, and\n\n$$\n\\angle APC = \\angle APM - \\angle CPM = \\angle BPM - \\angle DPM = \\angle BPD.\n$$\n\nThis completes the proof. $\\square$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 13948,
    "subject": "Mathematics (Olympiad)",
    "question": "Given a prime number $p$ and a positive real number $\\lambda$ with $\\lambda < 1$. Let $k$ be an integer, and let $S$ and $T$ be sets consisting of consecutive $s$ and $t$ integers, respectively, satisfying $1 \\leq s \\leq t < \\frac{\\lambda}{12}p$. Furthermore, assume that the number of elements in the set\n\n$$\n\\{(x, y) \\in S \\times T : kx \\equiv y \\pmod{p}\\}\n$$\n\nis at least $1 + \\lambda s$. Prove that there exist integers $a$ and $b$ such that\n\n$$\nka \\equiv b \\pmod{p}, \\quad 0 < a \\leq \\frac{1}{\\lambda}, \\quad |b| \\leq \\frac{t}{\\lambda s}.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "Let the solutions to the congruence $kx \\equiv y \\pmod{p}$ in $S \\times T$ be $(x_1, y_1), \\dots, (x_n, y_n)$, where $n \\geq 1 + \\lambda s \\geq 1$. Thus, $S$ and $T$ are both non-empty, so $n \\geq 2$. Since $T$ is contained in a complete residue system modulo $p$, the $x_i$ are distinct. Without loss of generality, assume $x_1 < x_2 < \\dots < x_n$. Let $u_i = x_i - x_1$ and $v_i = y_i - y_1$. Then\n\n$$\nku_i \\equiv v_i \\pmod{p}, \\quad 0 \\leq u_i \\leq s, \\quad |v_i| \\leq t.\n$$\n\nLet $a = \\frac{u_2}{\\gcd(u_2, v_2)}$ and $b = \\frac{v_2}{\\gcd(u_2, v_2)}$. Then $\\gcd(a, b) = 1$ and\n\n$$\n0 < a \\leq s, \\quad |b| \\leq t, \\quad ka \\equiv b \\pmod{p}.\n$$\n\nFrom $ku_i \\equiv v_i \\pmod{p}$, we get\n\n$$\nbu_i \\equiv av_i \\pmod{p}.\n$$\n\nLet $bu_i = av_i + pw_i$. Then\n\n$$\n|w_i| \\leq \\frac{|b|s + at}{p}.\n$$\n\nLet $L = \\frac{|b|s + at}{p}$.\n\n**Case 1:** $L < 1$. Then $w_i = 0$ for all $i$, so $bu_i = av_i$. Since $\\gcd(a, b) = 1$, $u_i$ are multiples of $a$. As $u_i \\in [0, s]$ are distinct, $s \\geq a(n-1) \\geq a\\lambda s$, so $a \\leq \\frac{1}{\\lambda}$. Also,\n\n$$\nat \\geq |b| \\cdot (a\\lambda s),\n$$\n\nso $|b| \\leq \\frac{t}{\\lambda s}$.\n\n**Case 2:** $L \\geq 1$. Then $w_i$ can take at most $2L+1 \\leq 3L$ integer values. By the pigeonhole principle, at least $\\frac{n}{3L}$ of the $w_i$ are equal. Let $E$ be the set of such indices, and $u_{i_0}$ the smallest $u_i$ in $E$. Then\n\n$$\nb(u_i - u_{i_0}) = a(v_i - v_{i_0}), \\quad \\forall i \\in E.\n$$\n\nSo $a$ divides $u_i - u_{i_0}$, and $0 \\leq u_i - u_{i_0} \\leq s$. Thus,\n\n$$\ns \\geq a(|E| - 1).\n$$\n\nAlso, $|v_i - v_{i_0}| \\leq t$, so\n\n$$\nat \\geq |b| a(|E| - 1).\n$$\n\nBut $L < \\frac{\\lambda s}{6}$, so $|E| - 1 \\geq \\frac{\\lambda s}{6L}$. Substituting,\n\n$$\n\\lambda a \\leq 6L, \\quad \\lambda |b|s \\leq 6Lt.\n$$\n\nThus,\n\n$$\n\\lambda p = \\frac{\\lambda |b|s + \\lambda at}{L} \\leq 12t,\n$$\n\ncontradicting $t < \\frac{\\lambda p}{12}$. Thus, $L \\geq 1$ is impossible, so only $L < 1$ holds, and the result follows. $\\square$",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 13949,
    "subject": "Mathematics (Olympiad)",
    "question": "*(a)* What is the largest real constant $C$ such that the inequality\n\n$$\nx^2 - Cx + y^2 - Cy + 1 \\ge 0\n$$\n\nholds for all real numbers $x$ and $y$?\n\n*(b)* What is the largest real constant $C$ such that the inequality\n\n$$\nx^2 + xy + y^2 - Cx - Cy + 1 \\ge 0\n$$\n\nholds for all real numbers $x$ and $y$?",
    "options": [],
    "answer": "See solution",
    "solution": "(a) First, rewrite the inequality as $x^2 - Cx + y^2 - Cy + 1 \\ge 0$ and then form perfect squares:\n\n$$\n\\left(x - \\frac{C}{2}\\right)^2 - \\frac{C^2}{4} + \\left(y - \\frac{C}{2}\\right)^2 - \\frac{C^2}{4} + 1 \\ge 0.\n$$\n\nThis implies\n\n$$\n\\left(x - \\frac{C}{2}\\right)^2 + \\left(y - \\frac{C}{2}\\right)^2 + 1 \\ge \\frac{C^2}{2}.\n$$\n\nIf $x = y = \\frac{C}{2}$, then $1 \\ge \\frac{C^2}{2}$, so $\\sqrt{2} \\ge C$.\n\nThe maximum possible real $C$ is equal to $\\sqrt{2}$. In this case the inequality holds for all real $x$ and $y$ since it is equivalent to\n\n$$\n\\left(x - \\frac{\\sqrt{2}}{2}\\right)^2 + \\left(y - \\frac{\\sqrt{2}}{2}\\right)^2 \\ge 0.\n$$\n\n(b) Once again we form perfect squares to get\n\n$$\n\\left(x + \\frac{y}{2} - \\frac{C}{2}\\right)^2 - \\frac{C^2}{4} + \\frac{3}{4} \\cdot \\left(y - \\frac{C}{3}\\right)^2 - \\frac{C^2}{12} + 1 \\ge 0.\n$$\n\nIf $y = \\frac{C}{3}$ and $x = \\frac{C}{2} - \\frac{y}{2} = \\frac{C}{3}$, then we can multiply by 12 and get $12 \\ge 3C^2 + C^2 = 4C^2$, so $\\sqrt{3} \\ge C$. Hence, the maximum possible constant is $C = \\sqrt{3}$. In this case the inequality holds for all real $x$ and $y$ since it is equivalent to\n\n$$\n\\left(x + \\frac{y}{2} - \\frac{\\sqrt{3}}{2}\\right)^2 + \\frac{3}{4} \\cdot \\left(y - \\frac{\\sqrt{3}}{3}\\right)^2 \\ge 0.\n$$",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 13950,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $a$, $b$, $c$ be positive real numbers. Prove that\n\n$$\n\\frac{a^2 b (b-c)}{a+b} + \\frac{b^2 c (c-a)}{b+c} + \\frac{c^2 a (a-b)}{c+a} \\geq 0.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "By clearing denominators (brute force), the inequality becomes\n\n$$\na^3 b^3 + b^3 c^3 + c^3 a^3 \\geq a^2 b c^3 + b^2 c a^3 + c^2 a b^3.\n$$\n\nTo justify this, use the AM-GM inequality:\n\n$$\na^3 b^3 + b^3 c^3 + c^3 a^3 \\geq 3 a b^3 c^2.\n$$\n\nBy summing this with the two other analogous inequalities, we obtain the desired result.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 13951,
    "subject": "Mathematics (Olympiad)",
    "question": "一個正五邊形的五個頂點各被賦予一個整數，使得所有頂點的數字總和大於零。若連續三個頂點的數字依序為 $x$、$y$ 和 $z$，且 $y < 0$，則我們可以對其進行以下操作：將 $x$、$y$ 和 $z$ 分別改為 $x+y$、$-y$ 和 $z+y$。只要五個頂點中任一頂點的數字小於零，我們便會持續進行操作。試問：以上操作是否必然只能操作有限次？",
    "options": [],
    "answer": "See solution",
    "solution": "這個演算法必然會停止。設 $S = \\sum x_i > 0$，考慮函數\n\n$$\nf(x_1, x_2, x_3, x_4, x_5) = \\sum_{i=1}^{5} (x_i - x_{i+2})^2, \\quad x_6 = x_1,\\ x_7 = x_2.\n$$\n\n顯然 $f > 0$ 且 $f$ 為整數值。假設（不失一般性）$y = x_4 < 0$。則 $f_{\\text{new}} - f_{\\text{old}} = 2Sx_4 < 0$，因為 $S > 0$。因此如果演算法不會停止，則可以得到一個無窮遞減的非負整數序列 $f_0 > f_1 > f_2 > \\dots$。這是不可能的，所以演算法必然會停止。",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 13952,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all functions $f: \\mathbb{R} \\to \\mathbb{R}$ such that for all real numbers $x$ and $y$,  \n$$\nf(x+y)f(xy) = f(x^2 - y^2 + 1).\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "Let $y = 0$. Then for each $x$,\n$$\nf(x)f(0) = f(1 + x^2).\n$$\nLet $x = 0$. Then for each $y$,\n$$\nf(y)f(0) = f(1 - y^2).\n$$\nIf $f(0) = 0$, then from the first equation, $f(t) = 0$ for all $t \\geq 1$ (since for each $t \\geq 1$, there exists $x$ with $t = 1 + x^2$), and from the second, $f(t) = 0$ for all $t \\leq 1$ (since for each $t \\leq 1$, there exists $y$ with $t = 1 - y^2$). Thus, $f(t) = 0$ for all $t \\in \\mathbb{R}$. This function satisfies the original equation.\n\nIf $f(0) \\neq 0$, then $f(x) = \\frac{f(1 + x^2)}{f(0)} = f(-x)$ for all $x$, so $f$ is even. Take $y = -x$ in the original equation:\n$$\nf(0)f(-x^2) = f(1).\n$$\nThus, $f(-x^2) = \\frac{f(1)}{f(0)}$. Since every non-positive number can be written as $-x^2$, and $f(x^2) = f(-x^2)$, we have $f(t) = \\frac{f(1)}{f(0)}$ for all $t$. Setting $t = 1$ gives $f(0) = 1$, so $f(t) = 1$ for all $t$. This function also satisfies the original equation.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 13953,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $a$, $b$, $c$ be the lengths of the sides of a triangle, and let $m_a$, $m_b$, $m_c$ be the lengths of the corresponding medians. Prove that:\n\n$$\nm_a \\left(\\frac{b}{a} - 1\\right) \\left(\\frac{c}{a} - 1\\right) + m_b \\left(\\frac{a}{b} - 1\\right) \\left(\\frac{c}{b} - 1\\right) + m_c \\left(\\frac{a}{c} - 1\\right) \\left(\\frac{b}{c} - 1\\right) \\geq 0.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "The given inequality is equivalent to\n\n$$\n\\frac{m_a}{a^2}(a-b)(a-c) + \\frac{m_b}{b^2}(b-a)(b-c) + \\frac{m_c}{c^2}(c-a)(c-b) \\geq 0.\n$$\n\nThis expression is symmetric in $a$, $b$, $c$, so without loss of generality, assume $a \\geq b \\geq c$. Then\n\n$$\n\\frac{m_a}{a^2}(a-b)(a-c) \\geq 0,\n$$\n\nso it suffices to prove\n\n$$\n\\frac{m_b}{b^2}(b-a)(b-c) + \\frac{m_c}{c^2}(c-a)(c-b) \\geq 0.\n$$\n\nUsing $a - c \\geq a - b$, we write\n\n$$\n\\begin{aligned}\n\\frac{m_b}{b^2}(b-a)(b-c) + \\frac{m_c}{c^2}(c-a)(c-b)\n&= (b-c)\\left((a-c)\\frac{m_c}{c^2} - (a-b)\\frac{m_b}{b^2}\\right) \\\\\n&\\geq (b-c)\\left((a-b)\\frac{m_c}{c^2} - (a-b)\\frac{m_b}{b^2}\\right) \\\\\n&= (a-b)(b-c)\\left(\\frac{m_c}{c^2} - \\frac{m_b}{b^2}\\right)\n\\end{aligned}\n$$\n\nWe now show\n\n$$\n\\frac{m_c}{m_b} \\geq \\frac{c}{b} \\geq \\frac{c^2}{b^2}.\n$$\n\nIndeed, $\\frac{c}{b} \\geq \\frac{c^2}{b^2}$, and\n\n$$\n\\begin{aligned}\n\\frac{m_c}{m_b} \\geq \\frac{c}{b}\n&\\iff \\frac{\\sqrt{2a^2 + 2b^2 - c^2}}{\\sqrt{2a^2 + 2c^2 - b^2}} \\geq \\frac{c}{b} \\\\\n&\\iff b^2(2a^2 + 2b^2 - c^2) \\geq c^2(2a^2 + 2c^2 - b^2) \\\\\n&\\iff b^2(2a^2 + 2b^2) \\geq c^2(2a^2 + 2c^2)\n\\end{aligned}\n$$\n\nwhich holds since $b^2 \\geq c^2$ and $2a^2 + 2b^2 \\geq 2a^2 + 2c^2$.\n\nCombining the above, we obtain the desired inequality. $\\Box$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 13954,
    "subject": "Mathematics (Olympiad)",
    "question": "There are $n$ points $A_1, \\dots, A_n$ on the plane with rational coordinates, all pairwise distances between them being integer. Prove that there exist points $B_1, \\dots, B_n$ on the plane with integer coordinates such that $|B_iB_j| = |A_iA_j|$ for all $1 \\le i < j \\le n$.\n\n![](path/to/file.png)",
    "options": [],
    "answer": "See solution",
    "solution": "Let $d$ be the least common multiple of the denominators of all coordinates of the points $A_1, \\dots, A_n$. After a homothety with center at the origin $O$ and coefficient $d$, these points map to points $C_1, \\dots, C_n$ with integer coordinates.\n\nSwitch to complex numbers and work with Gaussian integers. Let:\n\n$$\nC_k = x_k + y_k i \\quad \\text{and} \\quad M = \\{C_1, \\dots, C_n\\} \\subset \\mathbb{Z}[i].\n$$\n\nLet $p$ be any prime divisor of $d$. We show there is a rotation $R$ such that $RM \\subset p\\mathbb{Z}[i]$, i.e., $R\\left(\\frac{1}{p}M\\right) \\subset \\mathbb{Z}[i]$.\n\nWe have $p^2 \\mid N(z)$ and $p^2 \\mid N(z-u)$ for any $z, u \\in M$, where $N(a+bi) = a^2 + b^2$. If $p \\mid z$ for all $z \\in M$, we are done. Otherwise, suppose $v \\in M$ with $p \\nmid v$. Then there exists $\\pi \\in \\mathbb{Z}[i]$ such that $N(\\pi) = p$ and $\\pi^2 \\mid v$. For any $u \\in M$:\n\n- If $p \\nmid u$, then $\\pi^2 \\mid u$.\n- If $p \\mid u$ and $p \\nmid v-u$, then $\\pi^2 \\mid v-u$ and so $\\pi^2 \\mid u$.\n\nApply the plane transformation (a rotation):\n\n$$\nR_{\\pi}: x + yi \\rightarrow \\frac{\\pi}{\\pi}(x + yi)\n$$\n\nEach $M$ element can be written as $\\pi^2(a + bi)$, which $R_{\\pi}$ sends to $\\bar{\\pi}\\pi(a + bi) = p(a + bi)$. Thus, $R_{\\pi}M \\subset p\\mathbb{Z}[i]$ as required.\n\nNow, decompose $d = p_1p_2 \\cdots p_m$ into prime factors. By induction, there are rotations $R_1, \\dots, R_m$ such that:\n\n$$\nR_m \\left( \\dots R_2 \\left( R_1 \\left( \\frac{1}{p_1 p_2 \\cdots p_m} M \\right) \\right) \\right) \\subset \\mathbb{Z}[i].\n$$\n\nSince $\\frac{1}{p_1p_2\\cdots p_m}M = \\{A_1, \\dots, A_n\\}$, the composition $R_m \\circ \\dots \\circ R_1$ sends $A_1, \\dots, A_n$ to integer points $B_1, \\dots, B_n$ with the required properties.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 13955,
    "subject": "Mathematics (Olympiad)",
    "question": "Prove that there are at least 2013 different values of $m$ such that an $m \\times m$ square grid can be cut along grid lines into pieces $D_1, D_2, \\dots, D_n$, where $D_i$ consists of exactly $i$ cells.\n\nThe sum of the areas of $D_1, \\dots, D_n$ should be a perfect square:\n\n$$1 + 2 + \\dots + n = m^2$$\n\n% ![](images/Baltic_Way_SHL_2009-11_13-16_p189_data_aafa707686.png)\n\nFigure 2",
    "options": [],
    "answer": "See solution",
    "solution": "The sum $1 + 2 + \\dots + n = m^2$ leads to $\\frac{n(n+1)}{2} = m^2$. Rearranging, we get $$(2n+1)^2 = 8m^2 + 1.$$ Let $x = 2n+1$ and $y = 2m$, so we seek integer solutions to $x^2 = 2y^2 + 1$ with $x$ odd and $y$ even. There are infinitely many such solutions. The first is $x_1 = 3$, $y_1 = 2$, and others can be generated by the recurrence:\n\n$$\n\\begin{cases}\nx_{i+1} = 3x_i + 4y_i \\\\\ny_{i+1} = 2x_i + 3y_i\n\\end{cases}\n$$\n\nThis recurrence preserves the equation $x^2 = 2y^2 + 1$. Thus, there are infinitely many $m$ satisfying the condition, and in particular, at least 2013 such values.\n\nTo cut the square into $n$ pieces $D_1, \\dots, D_n$ with $D_i$ consisting of $i$ cells, one can traverse the grid \"like a snake\" through the rows, cutting off one cell for $D_1$, two for $D_2$, and so on. This ensures all pieces are connected.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 13956,
    "subject": "Mathematics (Olympiad)",
    "question": "Two players play the following card game. They have a deck of $n$ cards. For every two cards, it is known which one of them takes the other (it may happen that $A$ takes $B$, $B$ takes $C$, and $C$ takes $A$).\n\nInitially, the deck is distributed between the players in an arbitrary way. On each move, the players open the topmost cards of their decks; then the player whose card takes the opponent's one gets both cards and puts them to the very bottom of his deck (in an arbitrary order).\n\nProve that for any starting situation, the players may agree to play in such a way that one of them will finally get an empty deck.",
    "options": [],
    "answer": "See solution",
    "solution": "Consider a directed graph whose vertices represent positions in the game, and edges represent possible moves. A vertex is *final* if some player has no cards, and *critical* if some player has exactly one card.\n\nLet $V$ be the set of vertices reachable from some vertex $S$. If $V$ contains no final vertices, then the out-degree of each vertex in $V$ is $2$. Since their in-degrees in the whole graph do not exceed $2$, all edges coming to $V$ also start at $V$. Moreover, then $V$ also does not contain any critical vertex. But this is impossible, because it is easy to construct a backtracking process resulting in a path from some critical vertex to $S$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 13957,
    "subject": "Mathematics (Olympiad)",
    "question": "In a triangle $ABC$ with $AC > BC > AB$, points $D$ and $K$ are chosen on the sides $BC$ and $AC$ respectively so that $CD = AB$ and $AK = BC$. $F$ and $L$ are the midpoints of the segments $BD$ and $KC$ respectively. $R$ and $S$ are the midpoints of the sides $AC$ and $AB$ respectively. The line segments $SL$ and $FR$ intersect at the point $O$, and it is known that $\\angle SOF = 55^\\circ$. Find $\\angle BAC$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $BQ$ and $CT$ be the bisectors of triangle $ABC$, and let $J$ be their intersection point (the incenter of the triangle).\n\n![](images/Ukrajina_2011_p38_data_7dcf541dfd.png)\n\nBy Menelaus' theorem for $\\triangle ABC$ and the line $RM$ (where $M$ is the intersection of $FR$ and $AB$):\n\n$$\n\\frac{AR}{RC} \\cdot \\frac{CF}{FB} \\cdot \\frac{BM}{MA} = 1 \\iff \\frac{1}{1} \\cdot \\frac{a+x}{x} \\cdot \\frac{y}{a+y} = 1\n$$\n\nThis implies $x = y$, so $\\triangle MBF$ is isosceles. Then:\n\n$$\n\\angle ABQ = \\angle QBC = \\angle AMR = \\angle BFM = \\angle RFC\n$$\n\nHence, $FR \\perp BQ$. Similarly, $CT \\perp SL$.\n\nSince $\\angle FOS = \\angle BJT = 55^\\circ$, we have $\\angle BJC = 125^\\circ$. Because $J$ is the incenter, $\\angle BJC = 90^\\circ + \\frac{1}{2}\\angle BAC$, so $\\angle BAC = 70^\\circ$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 13958,
    "subject": "Mathematics (Olympiad)",
    "question": "有一條小河，其一側的露營基地裡插著 $4038$ 根營柱。其中有些營柱之間有繩子連接，每條繩子連接兩根不同的營柱，兩根營柱之間至多只有一條繩子。假設當我們把其中任意 $2019$ 根營柱移到河的對岸，都會剛好有 $k$ 條繩子通過河面。試求 $k$ 的所有可能值。",
    "options": [],
    "answer": "See solution",
    "solution": "答案是 $0, 2019, 2018 \\times 2019, 2019^2$。\n\n考慮圖 $G(V, E)$，點集 $V$ 和邊集 $E$ 分別對應到營柱和繩子。對集合 $S, T \\subset V$，令\n\n$$\nE(S, T) = \\left| \\left\\{ (s,t) \\in E : s \\in S, t \\in T \\right\\} \\right|\n$$\n\n也就是連接兩群之間的邊數。為簡化符號，當 $S$ 只有 $a$ 一個點時，我們簡寫此集合為 $a$，依此類推。\n\n1. **Lemma:** 對所有相異 $a, b, c, d \\in V$，都有\n\n$$\nE(a, b) + E(c, d) = E(a, d) + E(b, c) = E(a, c) + E(b, d)\n$$\n\n證明：先把 $V \\setminus \\{a, b, c, d\\}$ 平分成 $S, T$ 兩個集合。取河兩邊分別為 $(a, b, S)$ 和 $(c, d, T)$，則依題設有\n\n$$\nE(a, c) + E(a, d) + E(b, c) + E(b, d) + E(a, T) + E(b, T) + E(c, S) + E(d, S) + E(S, T) = k \\quad (1)\n$$\n\n取河兩邊分別為 $(a, d, S)$ 和 $(c, b, T)$，則依題設有\n\n$$\nE(a, c) + E(a, b) + E(d, c) + E(d, b) + E(a, T) + E(b, S) + E(c, S) + E(d, T) + E(S, T) = k \\quad (2)\n$$\n\n取河兩邊分別為 $(c, d, S)$ 和 $(a, b, T)$，則依題設有\n\n$$\nE(c, a) + E(c, b) + E(d, a) + E(d, b) + E(a, S) + E(b, S) + E(c, T) + E(d, T) + E(S, T) = k \\quad (3)\n$$\n\n(3) - (1) 有\n\n$$\n(E(a, S) - E(a, T)) + (E(b, S) - E(b, T)) = (E(c, S) - E(c, T)) + (E(d, S) - E(d, T))\n$$\n\n同理有\n\n$$\n(E(a, S) - E(a, T)) + (E(d, S) - E(d, T)) = (E(c, S) - E(c, T)) + (E(b, S) - E(b, T))\n$$\n\n兩式相減有 $E(b, S) - E(b, T) = E(d, S) - E(d, T)$。\n\n(2) - (1) 有\n\n$$\nE(a, b) + E(d, c) + E(b, S) + E(d, T) = E(a, d) + E(b, c) + E(b, T) + E(d, S)\n$$\n\n配合上式有 $E(a, b) + E(d, c) = E(a, d) + E(b, c)$，故 Lemma 第一部分得證。由 $a, b, c, d$ 的對稱性可知第二個等號也成立。\n\n2. 現在分別討論 $E(a, b)$ 和 $E(a, c)$ 的可能性：\n\n- 若對所有 $a, b, c \\in V$，$E(a, b) = E(a, c) = 0$，則圖中沒有任何邊，對應到 $k = 0$。\n- 若對所有 $a, b, c \\in V$，$E(a, b) = E(a, c) = 1$，則原本的圖是完全圖，對應到 $k = 2019^2$。\n- 假設存在 $a, b, c \\in V$ 使得 $E(a, b) = 1, E(a, c) = 0$。那麼對所有的 $d$，我們都有 $E(a, b) + E(c, d) = E(a, c) + E(b, d)$，故 $E(b, d) = 1, E(c, d) = 0$。\n\n若 $E(b, c) = 1$，則 $b$ 和所有點皆有連邊。對所有 $b$ 以外的點 $u, v, w$，都有 $E(b, u) + E(v, w) = E(b, v) + E(u, w)$，因此 $E(u, w)$ 和 $E(v, w)$ 同為 $0$ 或同為 $1$。又 $E(a, c) = 0$，故 $V \\setminus \\{b\\}$ 形成一個空圖，亦即 $G$ 是一個 $(1, 4037)$ 的完全二分圖，對應到 $k = 2019$。\n\n若 $E(b, c) = 0$，則 $c$ 和所有點皆無連邊。由類似的討論可以知道 $V \\setminus \\{c\\}$ 形成一個完全圖，亦即 $G$ 是 $4037$-完全圖和一個孤立點的聯集，對應到 $k = 2018 \\times 2019$。",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 13959,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $BC$ and $DE$ be parallel lines, and let $M$ and $N$ be the midpoints of $DB$ and $EC$, respectively. The line $MN$ is also parallel to $BC$ and $DE$. The triangles $ADE$, $AMN$, and $ABC$ are similar. Let $a = DE$, $b = BC$, and let $h$ and $\\ell$ be the heights of triangles $ADE$ and $ABC$, respectively. Then $MN = \\frac{1}{2}(a+b)$, and the height of the trapezoids $DMNE$ and $MBCN$ equals $\\frac{1}{2}(\\ell-h)$. The areas of the trapezoids $DMNE$ and $MBCN$ are $1$ and $2$, respectively. Find the area of triangle $ADE$.",
    "options": [],
    "answer": "See solution",
    "solution": "$$\n\\frac{1}{2} \\left( a + \\frac{a+b}{2} \\right) \\cdot \\left( \\frac{\\ell - h}{2} \\right) = 1 \\quad \\text{and} \\quad \\frac{1}{2} \\left( b + \\frac{a+b}{2} \\right) \\cdot \\left( \\frac{\\ell - h}{2} \\right) = 2,\n$$\n\nfrom which we obtain $b = 5a$. From the similarity of the triangles $ADE$ and $ABC$, we also get $\\ell = 5h$. The area $S_1$ of triangle $ADE$ is $\\frac{1}{2}ah$ and the area $S_2$ of triangle $ABC$ is $\\frac{1}{2}bh = \\frac{25}{2}ah$, and since we also have\n\n$$\nS_2 - S_1 = \\text{the area of the trapezoid } DBCE = 1 + 2 = 3,\n$$\n\nwe get\n\n$$\n\\frac{25}{2}ah - \\frac{1}{2}ah = 3,\n$$\n\nfrom which it follows that $ah = \\frac{1}{4}$ and we obtain the desired answer $S_1 = \\frac{1}{2}ah = \\frac{1}{8}$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 13960,
    "subject": "Mathematics (Olympiad)",
    "question": "設 $ABCD$ 為凸四邊形，其中任兩邊皆不等長，且 $AC \\perp BD$。設 $O_1, O_2$ 分別為三角形 $ABD$ 與 $CBD$ 的外心。證明：直線 $AO_2$、$CO_1$ 以及三角形 $ABC$ 的尤拉線、三角形 $ADC$ 的尤拉線四線共點。\n\n（註：三角形的尤拉線為其外心、重心、垂心所在的直線。）",
    "options": [],
    "answer": "See solution",
    "solution": "由對稱性，只需證明 $AO_2$、$CO_1$ 及 $\\triangle ABC$ 的尤拉線共點。注意到 $O_1O_2 \\parallel AC$。設 $\\ell_A, \\ell_C$ 分別為 $BC, AB$ 的中垂線，$M_A, M_C$ 分別為 $BC, AB$ 的中點。設 $G, O, H$ 分別為 $\\triangle ABC$ 的重心、外心、垂心，$\\ell$ 為其尤拉線。則\n\n$$\n\\begin{aligned}\n(G, O; H, \\ell \\cap AO_2) &= (AG, AO; AH, AO_2) = (AM_A, AO; A\\infty_{\\ell_A}, AO_2) \\\\\n&= (M_A, O; \\infty_{\\ell_A}, O_2).\n\\end{aligned}\n$$\n\n同理，$(G, O; H, \\ell \\cap CO_1) = (M_C, O; \\infty_{\\ell_C}, O_1)$。由於 $M_A M_C \\parallel O_1 O_2 \\parallel AC$，由於在 $\\infty_{AC}$ 的透視性，$(M_A, O; \\infty_{\\ell_A}, O_2) = (M_C, O; \\infty_{\\ell_C}, O_1)$。因此 $(G, O; H, \\ell \\cap AO_2) = (G, O; H, \\ell \\cap CO_1)$，說明 $\\ell, AO_2, CO_1$ 共點，得證。\n\n*備註*：此處的 cross ratio 可用梅涅勞斯定理多次應用來替代，這裡用 cross ratio 只是為了簡潔。\n\n*另一證法*：同樣有 $O_1O_2 \\parallel AC$，只需證明 $AO_2, CO_1$ 及 $\\triangle ABC$ 的尤拉線共點。設 $M_1, M_2$ 分別為 $AB, BC$ 的中點。則 $M_1M_2 \\parallel O_1O_2 \\parallel CA$，由德沙格定理可知 $M_1O_1 \\cap M_2O_2, CO_1 \\cap AO_2$ 及 $CM_1 \\cap AM_2$ 共線。$M_1O_1 \\cap M_2O_2$ 為 $\\triangle ABC$ 的外心，$CM_1 \\cap AM_2$ 為其重心，因此 $CO_1 \\cap AO_2$ 在 $\\triangle ABC$ 的尤拉線上。$\\square$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 13961,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $L$ be a line in the plane, and let $\\pi_1$ and $\\pi_2$ be the corresponding open half-planes. Define\n\n$$\nf_L(P) = \\begin{cases} 1 & \\text{if } P \\in \\pi_1 \\cup L, \\\\ -1 & \\text{if } P \\in \\pi_2. \\end{cases}\n$$\n\nLet $g$ be a function in $\\mathcal{F}$ differing from $f_L$ at finitely many points, and let\n\n$$\nK_{\\pm} = \\{(P, Q) : f_L(P)f_L(Q) - g(P)g(Q) = \\pm 2 \\text{ and } 0 < d(P, Q) < 2010\\}.\n$$\n\nShow that\n\n$$\n\\sum_{0<d(P,Q)<2010} \\frac{f_L(P)f_L(Q) - g(P)g(Q)}{d(P,Q)} = \\sum_{(P,Q) \\in K_+} \\frac{2}{d(P,Q)} - \\sum_{(P,Q) \\in K_-} \\frac{2}{d(P,Q)}\n$$\nis nonnegative.",
    "options": [],
    "answer": "See solution",
    "solution": "We will prove that this expression is nonnegative by defining an injection $m : K_- \\to K_+$ that satisfies $d(m(P, Q)) = d(P, Q)$.\n\nLet $(P, Q) \\in K_-$. Then $f_L(P) \\neq f_L(Q)$ and $g(P) = g(Q)$. Consider the line $\\ell$ passing through the points $P$ and $Q$. Let $P_0 = P$, $P_1 = Q$, and let $P_i$ be the unique point on $\\ell$ such that $d(P_i, P_{i-1}) = d(P, Q)$ and $P_i \\neq P_{i-2}$ for $i \\ge 2$, and $d(P_i, P_{i+1}) = d(P, Q)$ and $P_i \\neq P_{i+2}$ for $i \\le -1$.\n\nSince $f_L$ and $g$ differ at finitely many points, we know that $f_L(P_i) = g(P_i)$ for all $i$ with sufficiently large $|i|$. In particular, $g$ changes sign finitely many times, but at least once on $\\ell$. Let $k$ be the smallest integer such that $g(P_k) \\neq g(P_{k+1})$. We define $m(P, Q) = (P_k, P_{k+1})$.\n\n$(P_k, P_{k+1}) \\in K_+$ as $f_L(P_k)f_L(P_{k+1}) - g(P_k)g(P_{k+1}) = 1 - (-1) = 2$, and we have $d(P_k, P_{k+1}) = d(P, Q)$ by construction. Finally, since $(P, Q)$ is the only pair among $(P_i, P_{i+1})$, $i \\in \\mathbb{Z}$, satisfying $f_L(P_i) \\neq f_L(P_{i+1})$, $m$ is injective.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 13962,
    "subject": "Mathematics (Olympiad)",
    "question": "In an urn there are 100 balls, each coloured either blue, red, or green. If you draw (without repetitions) two balls randomly from the urn, the probability of getting two balls of different colours is $58\\%$, and the probability of getting a blue and a green ball is $8\\%$. How many red balls are there among the 100 balls?",
    "options": [],
    "answer": "See solution",
    "solution": "Let $b$, $r$, and $g$ be the number of blue, red, and green balls, respectively. We know that $b + r + g = 100$.\n\nThe probability of drawing two balls of different colours is $0.58$, and the probability of drawing a blue and a green ball is $0.08$.\n\nLet us consider the possible pairs:\n- Total ways to choose two balls: $\\binom{100}{2} = 4950$.\n- Ways to choose two balls of different colours: $br + bg + rg$.\n- Probability: $\\frac{br + bg + rg}{4950} = 0.58$.\n- Ways to choose a blue and a green ball: $bg$.\n- Probability: $\\frac{bg}{4950} = 0.08$.\n\nLet $x = b$, $y = r$, $z = g$.\n\nFrom $x + y + z = 100$.\n\nFrom $\\frac{xy + xz + yz}{4950} = 0.58$:\n$$xy + xz + yz = 0.58 \\times 4950 = 2871$$\n\nFrom $\\frac{xz}{4950} = 0.08$:\n$$xz = 0.08 \\times 4950 = 396$$\n\nSo $xy + yz = 2871 - 396 = 2475$.\n\nLet $y$ be the number of red balls. Then $x + z = 100 - y$.\n\n$xy + yz = y(x + z) = y(100 - y) = 2475$\n\nSo:\n$$y(100 - y) = 2475$$\n$$y^2 - 100y + 2475 = 0$$\n\nSolving the quadratic equation:\n$$y = \\frac{100 \\pm \\sqrt{100^2 - 4 \\times 2475}}{2}$$\n$$y = \\frac{100 \\pm \\sqrt{10000 - 9900}}{2}$$\n$$y = \\frac{100 \\pm 10}{2}$$\nSo $y = 55$ or $y = 45$.\n\nTry $y = 55$:\n$x + z = 45$\n$xz = 396$\nSo $x$ and $z$ are roots of $t^2 - 45t + 396 = 0$.\n\nDiscriminant: $45^2 - 4 \\times 396 = 2025 - 1584 = 441$\n$\\sqrt{441} = 21$\nSo $x = \\frac{45 \\pm 21}{2} = 33$ or $12$\n\nThus, the number of red balls is $\\boxed{55}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 13963,
    "subject": "Mathematics (Olympiad)",
    "question": "Determine all complex numbers $z$ such that $z^{3^n} + \\bar{z}^{3^n} \\ge 0$ for every non-negative integer $n$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $f_n(z) = z^{3^n} + \\bar{z}^{3^n}$. Since $f_n(z) = \\overline{f_n(z)}$, it follows that $f_n(z) \\in \\mathbb{R}$ for all $z \\in \\mathbb{C}$. We seek the set $M = \\{z \\in \\mathbb{C} \\mid f_n(z) \\ge 0 \\text{ for all } n \\}$. \n\nIf $z \\in M$, then $f_0(z) = z + \\bar{z} = 2\\text{Re}(z) \\ge 0$, so $\\text{Re}(z) \\ge 0$.\n\nFor $z \\in [0, \\infty)$ (real and non-negative), $f_n(z) = 2z^{3^n} \\ge 0$ for all $n$, so $[0, \\infty) \\subset M$.\n\nIf $z = bi$ with $b \\in \\mathbb{R}$, then $\\bar{z} = -z$, so $f_n(z) = 0$ for all $n$, thus $\\{bi \\mid b \\in \\mathbb{R}\\} \\subset M$.\n\nSince $f_n(z) = f_n(\\bar{z})$, we have $z \\in M \\iff \\bar{z} \\in M$.\n\nNow, consider $z = r(\\cos t + i \\sin t)$ with $r > 0$, $t \\in (0, \\frac{\\pi}{2})$. Then $f_n(z) = 2r^{3^n} \\cos(3^n t)$, so $z \\in M \\iff \\cos(3^n t) \\ge 0$ for all $n$. For $t \\in (0, \\frac{\\pi}{2})$, there exists $k \\in \\mathbb{N}$ such that $3^k t \\le \\frac{\\pi}{2} < 3^{k+1} t$ (specifically, $k = \\lfloor \\log_3 \\frac{\\pi}{2t} \\rfloor$). \n\nIf $3^k t = \\frac{\\pi}{2}$, then $\\cos(3^n t) > 0$ for $n = 0, 1, \\dots, k-1$ and $\\cos(3^n t) = 0$ for $n \\ge k$, so $z \\in M$. If $3^k t < \\frac{\\pi}{2} < 3^{k+1} t$, then $\\cos(3^{k+1} t) < 0$, so $z \\notin M$.\n\nIn conclusion,\n\n$$\nM = [0, \\infty) \\cup \\left\\{ r \\left( \\cos \\frac{\\pi}{2 \\cdot 3^k} \\pm i \\sin \\frac{\\pi}{2 \\cdot 3^k} \\right) \\mid r > 0,\\ k \\in \\mathbb{N} \\right\\}.\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 13964,
    "subject": "Mathematics (Olympiad)",
    "question": "The point $D$ lies on the side $AB$ of $\\triangle ABC$ with circumcircle $k$. Denote by $I$ and $J$ the centers of the circles touching $k$, and the segments $AB$ and $CD$. Assume that the points $A$, $B$, $I$, and $J$ are concyclic. Prove that $D$ is the tangent point of $AB$ and the excircle to this side.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $k = k(O, r)$ and let $k_1(I, r_1)$ touch $BD$, $CD$, and $k$ at $P$, $R$, and $Q$, respectively. Let $k_2(J, r_2)$ touch $AD$, $CD$ at $K$, $M$, and $L$, respectively. First, we shall prove that $AB \\parallel IJ$, i.e., $ABIJ$ is an isosceles trapezoid. Assume the contrary and set $T = IJ \\cap AB$. Then\n\n$$\n\\frac{IT}{TJ} \\cdot \\frac{JL}{LO} \\cdot \\frac{OQ}{QI} = \\frac{r_1}{r_2} \\cdot \\frac{r_2}{r} \\cdot \\frac{r}{r_1} = 1\n$$\n\nand, by the Menelaus theorem, the points $T$, $Q$, and $L$ are collinear.\n\n![](images/Bulgaria_2010_booklet_p6_data_e82580fdb1.png)\n\nThen $TQ \\cdot TL = TB \\cdot TA$. On the other hand, $ABIJ$ is cocyclic which implies $TB \\cdot TA = TI \\cdot TJ$. It follows that $TQ \\cdot TL = TI \\cdot TJ$, i.e., $QLJI$ is cocyclic. But $\\nparallel JLQ = \\nparallel IQL$, i.e., $QLJI$ is an isosceles trapezoid and then $IJ \\parallel QL$, a contradiction. Hence $AB \\parallel IJ$, i.e., $r_1 = r_2$ and $AK = BP$ \\ (1).\n\nFurther, the generalized Ptolemy theorem (applied to $A$, $B$, $Q$, and $C$) gives $AB \\cdot CR + AC \\cdot BP = AP \\cdot BC$. Since $CR = CD - DR = CD - DP = CD - BD + BP$ and $AP = AB - BP$, we get\n\n$$\nBP = \\frac{AB(BC + BD - CD)}{AB + BC + AC}\n$$\n\nAnalogously,\n\n$$\nAK = \\frac{AB(AC + AD - CD)}{AB + BC + AC}\n$$\n\nFinally, (1), (2), and (3) imply $BC + BD = AC + AD$ which holds if and only if $D$ is the tangent point of $AB$ and the excircle to this side.\n\n_Remark_: The second part of the solution can also be done by using the non-obvious fact that the lines $MK$, $PR$, and $IJ$ pass through the incenter of $\\triangle ABC$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 13965,
    "subject": "Mathematics (Olympiad)",
    "question": "Is it possible to subdivide a convex 2014-gon, by means of diagonals that do not intersect, into triangles, in such a way that each vertex is incident to an odd number of triangles?",
    "options": [],
    "answer": "See solution",
    "solution": "No, it is not possible.\n\nFirst, note that any triangulated polygon can be bicoloured: the triangles may each be assigned one of two colours, black and white, so that adjacent triangles have opposite colours. This can be shown inductively, since a triangulated polygon can be cut into two smaller ones across a diagonal.\n\nIt is clear that a triangulated polygon has an odd number of triangles incident at each vertex if and only if its edges are uniformly coloured in a single colour.\n\nNow, we prove that if $n$ is not divisible by $3$ (which $2014$ is not), there exists no triangulated $n$-gon having an odd number of triangles incident at each vertex. Suppose, for contradiction, that such an $n$-gon exists, and, without loss of generality, suppose its edges are all black.\n\nAssume there exists a diagonal cutting the $n$-gon into two smaller polygons $P$ and $Q$, neither of which is a triangle. Let the two pieces have $p$ and $q$ vertices, respectively, where $p, q \\geq 4$ and $p + q - 2 = n$.\n\nSince the two triangles adjacent at the cutting line have opposing colours, one of the pieces, say $P$, will have a black edge at the cut and thus all its edges remain black. The other, $Q$, will have a white edge exposed at the cut, to which we adjoin a black triangle, forming an augmented polygon $Q'$. Both $P$ and $Q'$ will then be bicoloured with black edges. Since $P$ and $Q'$ have $p < n$ and $q + 1 < n$ vertices, respectively, by minimality of $n$, both $p$ and $q + 1$ must be divisible by $3$. But then $n = p + (q + 1) - 3$ is also divisible by $3$.\n\nHence, all diagonals of the $n$-gon must cut off a triangle. This can only happen if $n \\leq 6$, and it is easy to verify that a square and a pentagon cannot be triangulated in the manner stipulated.",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 13966,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $\\triangle ABC$ be a triangle with $90^\\circ \\neq \\angle A \\neq 135^\\circ$. Let $D$ and $E$ be external points such that $\\triangle DAB$ and $\\triangle EAC$ are isosceles right triangles with right angles at $D$ and $E$, respectively. Let $F = BE \\cap CD$, and let $M$ and $N$ be the midpoints of $BC$ and $DE$, respectively.\n\nProve that if three of the points $A$, $F$, $M$, $N$ are collinear, then all four points are collinear.",
    "options": [],
    "answer": "See solution",
    "solution": "a) If $M$, $N$, $F$ are collinear, then $DE \\parallel BC$, so the distances from $D$ and $E$ to $BC$ are equal, which is equivalent to $b = c$.\n\nb) If $A$, $M$, $F$ are collinear, $\\tan \\vec{BAM} = \\frac{b \\sin A}{c - b \\cos A}$, $\\tan \\vec{DAN} = \\frac{b \\cos A}{c - b \\sin A}$, and $\\vec{DAN} = 45^\\circ + \\vec{BAM} \\Leftrightarrow b = c$.\n\nc) If $A$, $M$, $N$ are collinear, let $H = CD \\cap AB$, $G = BE \\cap AC$. Then\n\n$$\n\\frac{AG}{GC} = \\frac{c \\sin (A + 45^\\circ)}{a \\sin (B + 45^\\circ)}\n$$\n\nand similarly for $\\frac{AH}{HB}$. By Ceva's theorem, the condition becomes $b = c$.\n\nd) If $A$, $F$, $N$ are collinear, using Ceva's theorem in $DEF$, let $X = CD \\cap AE$, $Y = BG \\cap AD$ (possibly improper). We get\n\n$$\n\\vec{HCG} = \\vec{HBG}, \\text{ hence } HG \\text{ is antiparallel to } BC, \\text{ i.e. } AH \\cdot AB = AG \\cdot AC \\Leftrightarrow b = c.\n$$\n\ne) If $b = c$, all four points are collinear.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 13967,
    "subject": "Mathematics (Olympiad)",
    "question": "A line passing through $A$ and perpendicular to $AK$ meets $BC$ at $E$, and the circumcircle of triangle $ADE$ meets $(O)$ again at $F$. Prove that $AF$ passes through a fixed point.",
    "options": [],
    "answer": "See solution",
    "solution": "a) Without loss of generality, suppose $AB \\le AC$; the other case is similar. Clearly, $M$ belongs to segment $AB$ and $N$ to segment $AC$. From $NA = NB$, we have $\\angle NBA = \\angle NAB$, so $MA = MC$ and $\\angle MCA = \\angle MAC$. Hence, $\\angle NBA = \\angle MCA$, so $BMCN$ is a cyclic quadrilateral and $QM \\cdot QN = QB \\cdot QC$.\n\nTherefore, $Q$ lies on the radical axis of the two circles $(O)$ and $(AMN)$. This radical axis is $AP$, so $A$, $P$, and $Q$ are collinear.\n\n![](images/Vietnamese_mathematical_competitions_p89_data_f8c8b34709.png)\n\nb) The circle $(ODC)$ is tangent to $(O)$ at $C$, so their radical axis is the tangent $d$ to $(O)$ at $C$.\n\n$O$, $M$ both lie on the perpendicular bisector of $AC$, so $OM \\perp AC$. Similarly, $ON \\perp AB$, so $O$ is the orthocenter of triangle $AMN$. Hence, $AO \\perp MN$. Consider the circles $(M, MA)$ and $(N, NA)$; since $AK$ is their common chord, $AK \\perp MN$. Thus, $A$, $O$, and $K$ are collinear, so $\\angle OAE = 90^\\circ$. Furthermore, $\\angle ODE = 90^\\circ$, so $AODE$ is cyclic, and $O \\in (ADE)$. Thus, the radical axis of $(ADE)$ and $(ODC)$ is $OD$.\n\nOn the other hand, the radical axis of $(O)$ and $(ADE)$ is $AF$. Consider the three circles $(O)$, $(ADE)$, $(ODC)$, whose radical axes are $OD$, $d$, and $AF$; these lines are concurrent. Therefore, $AF$ passes through the intersection of $AD$ and $d$, which is a fixed point. $\\square$\n\n![](images/Vietnamese_mathematical_competitions_p90_data_ebb7437191.png)",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 13968,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $a_1 \\le a_2 \\le \\dots \\le a_{50}$ be real numbers such that\n\n1. $a_1 + a_2 + \\dots + a_{50} = 0$\n2. $|a_1| + |a_2| + \\dots + |a_{50}| = 624$\n\nFind the minimum value of $S = a_{50} - a_1$ and the maximum value of $T = a_1 a_2 \\dots a_{50}$.",
    "options": [],
    "answer": "See solution",
    "solution": "Suppose $k$ is the largest index such that $a_k \\le 0$, so $a_k \\le 0 < a_{k+1}$. Then\n$$\na_{k+1} + \\dots + a_{50} = -(a_1 + \\dots + a_k).\n$$\nAlso,\n$$\n|a_1| + \\dots + |a_k| = |a_1 + \\dots + a_k| = a_{k+1} + \\dots + a_{50} = \\frac{624}{2} = 312.\n$$\nThus,\n$$\n-312 = a_1 + \\dots + a_k \\ge k a_1 \\implies a_1 \\le \\frac{-312}{k},\n$$\n$$\n312 = a_{k+1} + \\dots + a_{50} \\le (50-k) a_{50} \\implies a_{50} \\ge \\frac{312}{50-k}.\n$$\nTherefore,\n$$\na_{50} - a_1 \\ge \\frac{312}{50-k} + \\frac{312}{k} = 312 \\left( \\frac{1}{k} + \\frac{1}{50-k} \\right) \\ge 312 \\cdot \\frac{4}{k+50-k} = \\frac{624}{25}.\n$$\nThe minimum value of $S$ is $\\frac{624}{25}$, achieved when\n$$\na_1 = \\dots = a_{25} = -\\frac{312}{25}, \\quad a_{26} = \\dots = a_{50} = \\frac{312}{25}.\n$$\n\nFor $T$, if any $a_k = 0$, then $T = 0$. Suppose all $a_i$ are nonzero and there are $k$ negative numbers among $a_1, \\dots, a_{50}$. If $k$ is odd, $T < 0$, so for maximum $T$, let $k$ be even. Let $h = 50 - k$. By AM-GM,\n$$\nT \\le \\left( \\frac{a_1 + \\dots + a_k}{k} \\right)^k \\cdot \\left( \\frac{a_{k+1} + \\dots + a_{50}}{h} \\right)^h = \\left( \\frac{312}{k} \\right)^k \\left( \\frac{312}{h} \\right)^h = \\frac{312^{50}}{k^k h^h}.\n$$\nWe need to show\n$$\nk^k h^h \\ge 24^{24} 26^{26}.\n$$\nAssume $k \\le h$, so $k \\le h - 2$ since $k, h$ are even. If $k \\le h - 4$, a local modification shows\n$$\nk^k h^h > (k+1)^{k+1} (h-1)^{h-1} \\iff h \\left(1 + \\frac{1}{h-1}\\right)^{h-1} > (k+1) \\left(1 + \\frac{1}{k}\\right)^k.\n$$\nLet $f(x) = (x+1) (1+\\frac{1}{x})^x$ for $x > 1$, and $g(x) = \\ln f(x) = \\ln(x+1) + x \\ln(1+\\frac{1}{x})$. Then\n$$\ng'(x) = \\frac{1}{x+1} + \\ln(1+\\frac{1}{x}) - \\frac{1}{x+1} = \\ln(1+\\frac{1}{x}) \\ge 0.\n$$\nSo $g(x)$ is increasing, and the inequality holds. Therefore, the maximum value of $T$ is $13^{24} \\cdot 12^{26}$, achieved when\n$$\na_1 = \\dots = a_{24} = -13, \\quad a_{25} = \\dots = a_{50} = 12.\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 13969,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $f : [0, 1] \\to (0, \\infty)$ be a continuous function. For $n \\in \\mathbb{N}$, $n \\ge 2$, consider $0 = t_0 < t_1 < \\dots < t_n = 1$, such that\n$$\n\\int_{t_0}^{t_1} f(t) \\, dt = \\int_{t_1}^{t_2} f(t) \\, dt = \\dots = \\int_{t_{n-1}}^{t_n} f(t) \\, dt.\n$$\nCompute\n$$\n\\lim_{n \\to \\infty} \\frac{n}{\\frac{1}{f(t_1)} + \\frac{1}{f(t_2)} + \\dots + \\frac{1}{f(t_n)}}.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "Define $F : [0, 1] \\to [0, I]$ by $F(x) = \\int_0^x f(t) \\, dt$, where $I = \\int_0^1 f(t) \\, dt$. Since $f$ is positive, $F$ is strictly increasing and thus one-to-one. Because $F$ is continuous, $F(0) = 0$ and $F(1) = I$, so $F$ is surjective. Therefore, $F^{-1}(kI/n) = t_k$ for $k = 0, 1, \\dots, n$, $n \\ge 2$.\n\nLet $(x_n)_{n \\ge 2}$ be the given sequence. We have\n$$\n\\begin{aligned}\n\\frac{1}{x_n} &= \\frac{1}{n} \\sum_{k=1}^{n} \\frac{1}{f(t_k)} = \\frac{1}{n} \\sum_{k=1}^{n} \\frac{1}{f(F^{-1}(kI/n))} = \\frac{1}{n} \\sum_{k=1}^{n} \\frac{1}{F'(F^{-1}(kI/n))} \\\\\n&= \\frac{1}{n} \\sum_{k=1}^{n} (F^{-1})'(kI/n).\n\\end{aligned}\n$$\n\nAs $(F^{-1})'$ is continuous,\n$$\n\\lim_{n \\to \\infty} \\frac{1}{x_n} = \\int_{0}^{1} (F^{-1})'(Ix) \\, dx = \\frac{1}{I} F^{-1}(Ix) \\Big|_{0}^{1} = \\frac{1}{I}.\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 13970,
    "subject": "Mathematics (Olympiad)",
    "question": "令 $f : \\mathbb{N} \\to \\mathbb{N}$，且令 $f^m$ 為 $f$ 作用 $m$ 次。假設對所有 $n \\in \\mathbb{N}$，存在一個 $k$ 使得 $f^{2k}(n) = n + k$，且令 $k_n$ 為滿足前式的 $k$ 中最小的。試證數列 $k_1, k_2, \\dots$ 無界。",
    "options": [],
    "answer": "See solution",
    "solution": "我們關心集合 $S = \\{1, f(1), f^2(1), \\dots\\}$。觀察到 $S$ 是無界的，因為對於所有 $S$ 中的 $n$，存在一個 $k > 0$ 使得在 $S$ 中，$f^{2k}(n) = n + k$。顯然 $f$ 把 $S$ 映射到 $S$ 本身，且在 $S$ 上 $f$ 是一對一。確實，若 $f^i(1) = f^j(1)$（$i \\ne j$），則從某一個 $m$ 開始 $f^m(1)$ 會週期性地出現，$S$ 會有限。\n\n定義 $g : S \\to S$ 為 $g(n) = f^{2k_n}(n) = n + k_n$。我們證明 $g$ 也是一對一。\n假設當 $a < b$ 時 $g(a) = g(b)$。則 $a + k_a = f^{2k_a}(a) = f^{2k_b}(b) = b + k_b$，可得 $k_a > k_b$。所以，因為在 $S$ 上 $f$ 是一對一，我們得到 $f^{2(k_a - k_b)}(a) = b = a + (k_a - k_b)$。然而當 $0 < k_a - k_b < k_a$ 時，這與 $k_a$ 的最小性矛盾。\n\n令 $T$ 為 $S$ 中的元素所成的集合，除了 $g(n), n \\in S$ 形式的元素。注意到 $1 \\in T$，因為 $g(n) > n$ 對於 $n \\in S$，所以 $T$ 不是空集合。對於每個 $t \\in T$，$C_t = \\{t, g(t), g^2(t), \\dots\\}$，此 $C_t$ 稱作從 $t$ 開始的鏈。不同的鏈是互斥的，因為 $g$ 是一對一。每一個 $n \\in S \\setminus T$ 皆可表示為 $n = g(n')$，其中 $n' < n, n' \\in S$。同理可觀察到對於某些 $t \\in T, n \\in C_t$，即 $S$ 是那些互斥的鏈 $C_t$ 的聯集。\n\n若 $f^n(1)$ 落在從 $t = f^{n_t}(1)$ 開始的 $C_t$，則 $n = n_t + 2a_1 + \\cdots + 2a_j$ 且 $f^n(1) = g^j(f^{n_t}(1)) = f^{2a_j}(f^{2a_{j-1}}(\\cdots f^{2a_1}(f^{n_t}(1)))) = f^{n_t}(1) + a_1 + \\cdots + a_j$。\n\n因此\n\n$$\nf^n(1) = f^{n_t}(1) + \\frac{n - n_t}{2} = t + \\frac{n - n_t}{2}.\n$$\n\n現在利用反證法證明 $T$ 是無限的。假設只有有限多個 $C_{t_1}, \\dots, C_{t_r}$，從 $t_1 < \\dots < t_r$ 開始的鏈。固定 $N$。若 $f^n(1), 1 \\le n \\le N$ 落在 $C_t$ 中，則由上式，$f^n(1) = t + \\frac{n-n_t}{2} \\le t_r + \\frac{N}{2}$。但是相異的 $N+1$ 個數 $1, f(1), \\cdots, f^N(1)$ 都小於 $t_r + \\frac{N}{2}$，因此 $N+1 \\le t_r + \\frac{N}{2}$。所以當 $N$ 足夠大時得到矛盾，因此 $T$ 是無限的。\n\n選擇任意 $k \\in \\mathbb{N}$，且考慮以 $T$ 中的前 $k+1$ 個數為首的 $k+1$ 個鏈。令 $t$ 是這些數當中最大的數。每一個鏈都包含一個不會超過 $t$ 的數，且至少有一個鏈不包含 $t+1, \\cdots, t+k$。所以在這個鏈中存在一個數 $n$ 使得 $g(n)-n > k$，即 $k_n > k$。所以 $k_1, k_2, \\cdots$ 無界。",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 13971,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $n$ be an integer greater than or equal to 2. A regular hexagon $ABCDEF$ of side length $n$ is partitioned into equilateral triangles of side length 1 as indicated in the left figure below. For brevity, let us call a vertex of the triangles as a vertex below.\n\nAs indicated in the right figure below, each vertex $P$ lying in the interior of $ABCDEF$ (and not on the boundary of $ABCDEF$) is furnished with 4 arrows pointing to the direction of 4 out of 6 vertices connected to $P$ by a side of length 1. If a marble is placed at $P$, then it can be moved to any of the 4 neighboring vertices following the direction of one of the 4 arrows emanating from $P$. Note that for a side $PQ$ of length 1, even when it is possible to move a marble from $P$ to $Q$, it is not necessarily possible to move the marble from $Q$ to $P$.\n\nShow that there exists a positive integer $k$ such that after at most $k$ moves you can move the marble located initially at the center of $ABCDEF$ to a vertex lying on the boundary of $ABCDEF$, no matter how the assignments of arrows to vertices are made, and find the smallest possible value of such a $k$.\n\n![](images/Japan_2015_p9_data_959a2b2327.png)",
    "options": [],
    "answer": "See solution",
    "solution": "Let $O$ be the center of the hexagon $ABCDEF$, and for non-negative integer $i$, let $P_i$ be the vertex the marble will occupy after $i$ moves. We will show that after at most $n$ moves, the marble can reach a boundary vertex, regardless of the arrow assignments.\n\nConsider the shortest path from $O$ to the boundary. Since the hexagon is partitioned into equilateral triangles of side length 1, the minimum number of steps from the center to the boundary is $n$. At each move, the marble can follow one of the arrows to a neighboring vertex. No matter how the arrows are assigned, there is always a sequence of moves that leads to the boundary in at most $n$ steps, because from any interior vertex, there are 4 possible directions to proceed, and the structure of the hexagon ensures that a path to the boundary exists.\n\nTherefore, the smallest possible value of $k$ is $n$.",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 13972,
    "subject": "Mathematics (Olympiad)",
    "question": "Mr. Lopez has a choice of two routes to get to work. Route A is 6 miles long, and his average speed along this route is 30 miles per hour. Route B is 5 miles long, and his average speed along this route is 40 miles per hour, except for a $\\frac{1}{2}$-mile stretch in a school zone where his average speed is 20 miles per hour. By how many minutes is Route B quicker than Route A?\n\n(A) $2\\frac{3}{4}$  (B) $3\\frac{3}{4}$  (C) $4\\frac{1}{2}$  (D) $5\\frac{1}{2}$  (E) $6\\frac{3}{4}$",
    "options": [],
    "answer": "See solution",
    "solution": "The time required for Route A is $6 \\div 30 = \\frac{1}{5}$ hour. \n\nFor Route B:\n- The non-school zone distance is $5 - \\frac{1}{2} = \\frac{9}{2}$ miles at 40 mph: $\\frac{9}{2} \\div 40 = \\frac{9}{80}$ hour.\n- The school zone is $\\frac{1}{2}$ mile at 20 mph: $\\frac{1}{2} \\div 20 = \\frac{1}{40}$ hour.\n\nTotal time for Route B: $\\frac{9}{80} + \\frac{1}{40} = \\frac{9}{80} + \\frac{2}{80} = \\frac{11}{80}$ hour.\n\nDifference: \n$$\n\\frac{1}{5} - \\frac{11}{80} = \\frac{16 - 11}{80} = \\frac{5}{80} = \\frac{1}{16}\n$$\n\nConvert to minutes: $\\frac{1}{16} \\times 60 = \\frac{60}{16} = \\frac{15}{4} = 3\\frac{3}{4}$ minutes.\n\n**Answer:** (B) $3\\frac{3}{4}$ minutes.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 13973,
    "subject": "Mathematics (Olympiad)",
    "question": "$0 \\ne f(x),\\ g(x) \\in \\mathbb{R}[x]$, $f(x^3) + g(x) = f(x) + x^5 g(x)$ байх $f(x)$ нь хамгийн бага зэрэгтэй байх олон гишүүнтүүдийн жишээ гарга.",
    "options": [],
    "answer": "See solution",
    "solution": "$$\nf(x^3) - f(x) = g(x) \\cdot (x^5 - 1).\n$$\n\n$f(x^3) - f(x) = x^3 - x = x(x-1)(x+1)$ байх шаардлагатай бөгөөд $f(x^3) - f(x) : x^3 - x = x(x-1)(x+1)$ ба $x^5 - 1 = (x - 1)(x^4 + x^3 + x^2 + x + 1)$ гэдгийг тооцвол $f(x^3) - f(x) \\equiv 0 \\pmod{x^4 + x^3 + x^2 + x + 1}$ болно.\n\n$f_1^k = 1 \\pmod{x^4 + x^3 + x^2 + x + 1}$ тул хэрэв $\\deg g \\ge 5$ бол $f_2^k = f_1(x) \\pmod{x^4 + x^3 + x^2 + x + 1}$; $\\deg f_1 \\le 4$ байх. $f_1(x) \\in \\mathbb{R}[x]$, $f(x^3) - f(x) = f_1(x^3) - f_1(x) \\pmod{x^4 + x^3 + x^2 + x + 1}$ гүн $f(x)$ хамгийн бага зэрэгтэй гэдэгт зөрчилд хүрнэ. Иймд $\\deg f \\le 4$, тэгэхээр $f(x) = a x^4 + b x^3 + c x^2 + d x + e$.\n\n$$\nf_1^k = f_1(x) \\equiv a x^{12} + b x^9 + c x^6 + d x^3 + e - (a x^4 + b x^3 + c x^2 + d x + e)\n$$\n$$\n\\equiv a x^2 + b x^4 + c x + d x^3 - (a x^4 + b x^3 + c x^2 + d x)\n$$\n$$\n\\equiv (h - a) x^3 + (d - b) x^2 + (a - c) x + c - d \\equiv 0 \\pmod{x^4 + x^3 + x^2 + x + 1}\n$$\n$$\n\\Rightarrow h - a = d - b = a - c = c - d = 0 \\Rightarrow a = b = c = d.\n$$\n\nТэгэхээр $f(x) = a x^4 + a x^3 + a x^2 + a x + e$.\n\n$$\ng(x) = a x^7 + a x^4 + a x^2 + a x,\\quad a \\neq 0,\\ a, b \\in \\mathbb{R}.\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 13974,
    "subject": "Mathematics (Olympiad)",
    "question": "The sequence $\\alpha_v$ satisfies the recurrence relation:\n\n$$\n\\alpha_1 = 1, \\quad \\alpha_v = 5\\alpha_{v-1} + 3^{v-1}, \\quad v \\ge 2.\n$$\n\nDetermine the general term $\\alpha_v$ and the greatest power of $2$ which divides the term $a_k$, where $k = 2^{2019}$.",
    "options": [],
    "answer": "See solution",
    "solution": "$$\n\\alpha_v = 5^{v-1} \\left[ 1 + \\frac{3}{5} + \\left(\\frac{3}{5}\\right)^2 + \\dots + \\left(\\frac{3}{5}\\right)^{v-1} \\right] = \\frac{1}{2}(5^v - 3^v), \\quad v = 1, 2, \\dots\n$$\n\nNow for $k = 2^{2019}$, we have:\n\n$$\n2a_k = 5^{2^{2019}} - 3^{2^{2019}} = 2 \\cdot (5+3)(5^2+3^2) \\dots (5^{2^{2018}} + 3^{2^{2018}}),\n$$\n\nand hence:\n\n$$\na_k = (5+3)(5^2+3^2)\\dots(5^{2^{2018}}+3^{2^{2018}}).\n$$\n\nWe observe that the first factor is divisible by $8$ and all the others are divisible by $2$ but not by $4$. In fact,\n\n$$\n5^{2v} \\equiv 1 \\pmod{4}, \\quad 3^{2v} \\equiv 1 \\pmod{4} \\implies 5^{2v} + 3^{2v} \\equiv 2 \\pmod{4}, \\quad \\text{for all } v \\ge 1.\n$$\n\nThe factors from $5^2 + 3^2$ to $5^{2^{2018}} + 3^{2^{2018}}$ are $2018$ in total, so the greatest power of $2$ dividing $a_k$ is $2^{2021}$.\n\nAlternatively, using the Lifting the Exponent Lemma for the greatest power of $2$ dividing a difference of powers of odd integers:\n\n**Lemma:** Let $\\alpha, \\beta$ be odd integers and $v$ an even positive integer. Then\n\n$$\nv_2(\\alpha^v - \\beta^v) = v_2(\\alpha - \\beta) + v_2(\\alpha + \\beta) + v_2(v) - 1.\n$$\n\nApplying the lemma to $2a_{2^{2019}} = 5^{2^{2019}} - 3^{2^{2019}}$:\n\n$$\nv_2(2a_{2^{2019}}) = v_2(5-3) + v_2(5+3) + v_2(2^{2019}) - 1 = 1 + 3 + 2019 - 1 = 2022,\n$$\n\nso $v_2(a_k) = 2021$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 13975,
    "subject": "Mathematics (Olympiad)",
    "question": "A strategical video game consists of a map of finitely many towns. In each town there are $k$ directions, labelled from $1$ through $k$. One of the towns is designated as initial, and one as terminal. Starting from the initial town, the hero of the game makes a finite sequence of moves. At each move, the hero selects a direction from the current town. This determines the next town he visits and a certain positive amount of points he receives.\n\nTwo strategical video games are *equivalent* if for every sequence of directions the hero can reach the terminal town from the initial in one game, he can do so in the other game, and, in addition, he accumulates the same amount of points in both games.\n\nFor his birthday, John receives two strategical video games – one with $N$ towns and one with $M$ towns. He claims they are equivalent. Marry is convinced they are not. Marry is right. Prove that she can provide a sequence of at most $N + M$ directions that shows the two games are indeed not equivalent.",
    "options": [],
    "answer": "See solution",
    "solution": "Without loss of generality, assume the set of directions is $D = \\{1, 2, \\dots, k\\}$. Enumerate the towns in the first game from $1$ through $N$ and in the second game from $N+1$ through $N+M$. Assume the initial and terminal towns in the first game are $1$ and $N$, and in the second game are $N+1$ and $N+M$.\n\nConsider a directed labelled graph $G = (V, \\lambda)$ where $V = \\{1, 2, \\dots, N + M\\}$ and $\\lambda: V \\times D \\to V \\times \\mathbb{N}$ maps a current town $u$ and direction $d$ to $\\lambda(u, d) = (v, p)$, where $v$ is the next town and $p$ is the amount of points awarded for this move. For each vertex $v \\in V$, let $L_n(v)$ be the set of all pairs $(s, p)$, where $s$ is a sequence of directions of length at most $n$ that leads from $v$ to a terminal town and $p$ is the total amount of points awarded for this sequence. Let $L(v)$ be the union of all $L_n(v)$ for $n \\in \\mathbb{N} \\cup \\{0\\}$.\n\nWe want to prove:\n\n$$\nL(1) = L(N + 1) \\quad \\text{if and only if} \\quad L_n(1) = L_n(N + 1) \\quad \\text{for all } n \\le N + M.\n$$\n\nFirst, modify the games by awarding the hero points as soon as possible. Define:\n\n$$\nc(v) = \\min\\{p : (s, p) \\in L_{N+M}(v) \\text{ for some } s\\}.\n$$\n\nSince points are always positive, every cycle in $G$ brings a positive amount of points. Thus, $p' \\ge c(v)$ for every $n$ and every $(s', p') \\in L_n(v)$. In particular, if $\\lambda(v, d) = (u, p)$, then since the hero can win $p + c(u)$ points starting from $v$, we have $p + c(u) \\ge c(v)$. If $c(1) \\ne c(N+1)$, then $L_{N+M}(1) \\ne L_{N+M}(N+1)$ and we are done. So assume $c(1) = c(N+1)$.\n\nLet $G' = (V, \\lambda')$ be defined by $\\lambda'(v, d) = (u, p + c(u) - c(v))$ whenever $\\lambda(v, d) = (u, p)$. Now every move in $G'$ is awarded a non-negative amount of points. An inductive argument shows $(s, p) \\in L_n(v)$ if and only if $(s, p - c(v)) \\in L'_n(v)$, where $L'_n(v)$ is defined for $G'$. Hence $L_n(1) = L_n(N+1)$ if and only if $L'_n(1) = L'_n(N+1)$.\n\nAssume $L'(u) = L'(v)$ and consider any direction $d$. If $\\lambda(u, d) = (u', p')$ and $\\lambda(v, d) = (v', q')$, then $L'(u') = L'(v')$ and $p' = q'$. This follows by considering sequences and the non-negativity of points. Thus, if $(s, p) \\in L'(u')$, then $((d, s), p + p') \\in L'(u) = L'(v)$, so $(s, p) \\in L'(v')$ as $p' = q'$. The reverse inclusion is similar.\n\nDefine equivalence relations $\\equiv^{(n)}$ recursively:\n\n$$\n\\begin{align*}\nu \\equiv^{(0)} v & \\iff \\text{either } u, v \\in \\{N, N+M\\} \\text{ or } u, v \\notin \\{N, N+M\\}, \\\\\nu \\equiv^{(n+1)} v & \\iff u \\equiv^{(n)} v, \\text{ and for all } d \\le k,\\ u' \\equiv^{(n)} v' \\text{ and } p' = q',\n\\end{align*}\n$$\n\nwhere $\\lambda'(u, d) = (u', p')$ and $\\lambda'(v, d) = (v', q')$. Since $\\equiv^{(n+1)}$ refines $\\equiv^{(n)}$ and there are at most $N+M$ classes, the sequence stabilizes for some $n < N+M$. For this $n$, $\\equiv^{(m)}$ and $\\equiv^{(n)}$ are the same for all $m \\ge n$. If $u \\equiv^{(n)} v$, then $(s, p) \\in L'(u)$ if and only if $(s, p) \\in L'(v)$, so $L'(u) = L'(v)$.\n\nIf $L'(1) \\ne L'(N+1)$, then $1 \\not\\equiv^{(n)} N+1$. If $u \\not\\equiv^{(n)} v$, then $L_n(u) \\ne L_n(v)$. This is clear for $n=0,1$. Assume it holds for $n$; for $n+1$, if $u \\not\\equiv^{(n+1)} v$ but $u \\equiv^{(n)} v$, then for some $d$, $u' \\not\\equiv^{(n)} v'$, so $L'_n(u') \\ne L'_n(v')$, and thus $L'_{n+1}(u) \\ne L'_{n+1}(v)$.\n\n*Remarks.* This is a consequence of the minimisation algorithm for (sub)sequential transducers, involving pushing forward costs (related to potentials in weighted graphs) and the bisimulation/Nerode-Myhill relation, which stabilizes in at most $O(|Q|)$ steps.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 13976,
    "subject": "Mathematics (Olympiad)",
    "question": "令 $m > 1$ 為一正整數。有一隻老鼠要從 $3m \\times 3m$ 西洋棋盤的最左下角格子 $a$ 跳到最右上角格子 $b$。老鼠每一次可以跑到牠所在位子右方一格或是上方一格。然而，棋盤上有一些格子放有黏鼠板，如果老鼠跑到黏鼠板上就會被黏住，無法再移動。\n\n對於由棋盤上的格子所組成的集合 $X$，我們稱 $X$ 擋住老鼠，若且唯若當 $X$ 的格子都被放有黏鼠板時，老鼠無法從 $a$ 跑到 $b$。我們稱 $X$ 最小，若且唯若 $X$ 擋住老鼠、但移除 $X$ 的任何一格後都不能擋住老鼠。證明：\n\n(i) 任何能擋住老鼠的最小集合的格數不大於 $3m^2$。\n\n(ii) 證明存在格數不小於 $3m^2 - 3m$，且能擋住老鼠的最小集合。\n\n(註：若 (ii) 改為證明格數不小於 $3m^2 - cm$，則將視 $c$ 值的大小給分。)",
    "options": [],
    "answer": "See solution",
    "solution": "(a) 對於最小集 $X$，在不碰到 $X$ 的前提下，令 $A$ 為所有老鼠可以從 $a$ 抵達的格子所成集合，$B$ 則為所有可以抵達 $b$ 的格子所成集合。\n\n注意到從 $A$ 最多可以抵達 $2|A|$ 個格子（包含 $A$ 和 $X$，扣除 $a$），因此\n\n$$\n2|A| \\geq |X| + (|A| - 1) \\Leftrightarrow |X| \\leq |A| + 1.\n$$\n\n同理，$|X| \\leq |B| + 1$，從而 $9m^2 \\geq |A| + |B| + |X| \\geq 3|m^2 - 2$，亦即 $|X| \\leq 3m^2$。\n\n(b) 以下是一個格數為 $3m^2 - 2m + 2$ 的最小集：\n\n![](images/2022-TWNIMO-Problems_p102_data_df77a957d2.png)",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 13977,
    "subject": "Mathematics (Olympiad)",
    "question": "Define a function $f: \\mathbb{N} \\to \\mathbb{N}$ by $f(1) = 1$, $f(n+1) = f(n) + 2^{f(n)}$ for every positive integer $n$. Prove that $f(1), f(2), \\dots, f(3^{2013})$ leave distinct remainders when divided by $3^{2013}$.",
    "options": [],
    "answer": "See solution",
    "solution": "We prove the following stronger statement: For any $k \\geq 0$ and $a \\geq 1$, the values $f(a), f(a+1), \\dots, f(a+3^k - 1)$ are all distinct modulo $3^k$; that is, these numbers form a complete set of residues modulo $3^k$. We will induct on $k$, the case $k=0$ being trivial.\n\nAssume the statement is true for a given $k$; we will prove it for $k+1$. For any $a \\geq 1$, we have\n\n$$\n\\begin{aligned}\n& f(a + 3^k) - f(a) \\\\\n&= [f(a+1) - f(a)] + [f(a+2) - f(a+1)] + \\dots + [f(a + 3^k) - f(a + 3^k - 1)] \\\\\n&= 2^{f(a)} + 2^{f(a+1)} + \\dots + 2^{f(a+3^k-1)}.\n\\end{aligned}\n$$\n\nNote that by Euler's theorem, to know $2^x$ modulo $3^{k+1}$ ($x \\geq 1$), it suffices to know $x$ modulo $\\varphi(3^{k+1}) = 2 \\cdot 3^k$. Now $f(x)$ is always odd (this follows from the definition), while the inductive hypothesis tells us that $f(a), \\dots, f(a+3^k-1)$ are distinct mod $3^k$. Hence the sets\n\n$$\n\\{f(a), \\dots, f(a + 3^k - 1)\\} \\quad \\text{and} \\quad \\{1, 3, 5, \\dots, 2 \\cdot 3^k - 1\\}\n$$\n\nare congruent to each other modulo $2 \\cdot 3^k$. Therefore, modulo $3^{k+1}$, we have\n\n$$\nf(a + 3^k) - f(a) \\equiv 2^1 + 2^3 + 2^5 + \\dots + 2^{2 \\cdot 3^k - 1} \\equiv 2(1 + 4 + 4^2 + \\dots + 4^{3^k - 1}) \\equiv \\frac{2(4^{3^k} - 1)}{3}.\n$$\n\nBy the binomial theorem,\n\n$$\n4^{3^k} - 1 = (1 + 3)^{3^k} - 1 = 3 \\cdot \\binom{3^k}{1} + \\dots,\n$$\n\nwhere each of the remaining summands is divisible by $3^{k+2}$; hence\n\n$$\nf(a + 3^k) - f(a) \\equiv 2 \\cdot 3^k \\pmod{3^k}\n$$\nfor all $a$.\n\nReturning to the sequence\n\n$$\nf(a), f(a+1), \\dots, f(a + 3^{k+1} - 1),\n$$\n\nwe see that, since $f(b) \\equiv f(b + 3^k)$ modulo $3^k$ (a consequence of the inductive hypothesis), the terms congruent to one another modulo $3^k$ come in triples $(f(b), f(b + 3^k), f(b + 2 \\cdot 3^k))$. By the above, these terms, modulo $3^{k+1}$, are congruent to $(f(b), f(b) + 2 \\cdot 3^k, f(b) + 3^k)$, which are pairwise distinct, completing our induction.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 13978,
    "subject": "Mathematics (Olympiad)",
    "question": "For positive integers $m$ and $n$, define\n\n$$\nf(x) = (x-1)(x^2-1)\\cdots(x^m-1),\n$$\n$$\ng(x) = (x^{n+1}-1)(x^{n+2}-1)\\cdots(x^{n+m}-1).\n$$\n\nShow that there exists an integral polynomial $h(x)$ of degree $mn$, such that $f(x)h(x) = g(x)$, and all the $mn + 1$ coefficients of $h(x)$ are positive integers.",
    "options": [],
    "answer": "See solution",
    "solution": "More generally, for nonnegative integers $m$ and $n$,\n\n$$\nf_{m,n}(x) = \\frac{(x^{m+1} - 1)(x^{m+2} - 1)\\cdots(x^{m+n} - 1)}{(x-1)(x^2-1)\\cdots(x^n-1)}\n$$\n\nis a polynomial of degree $mn$, and all the $mn + 1$ coefficients of $f_{m,n}(x)$ are positive integers. For $n = 0$, take the denominator as $1$.\n\nWe prove the above assertion by induction on $m + n$. First, $f_{m,n}(x) = f_{n,m}(x)$, as indicated by\n\n$$\nf_{m,n}(x) = \\frac{\\prod_{i=1}^{m+n} (x^i - 1)}{\\prod_{i=1}^{m} (x^i - 1) \\prod_{i=1}^{n} (x^i - 1)} = f_{n,m}(x).\n$$\n\nFor $m + n \\leq 3$,\n\n$$\nf_{m,0}(x) = f_{0,n}(x) = 1, \\quad f_{1,1}(x) = 1 + x, \\quad f_{1,2}(x) = f_{2,1}(x) = 1 + x + x^2,\n$$\n\nand the assertion is true. Now, suppose $m, n > 0$, $m + n \\geq 4$, and the assertion is true for all smaller $m + n$ values. Then\n\n$$\n\\begin{align*}\n\\frac{f_{m-1,n}(x)}{f_{m,n}(x)} &= \\frac{(x^m - 1)(x^{m+1} - 1)\\cdots(x^{m+n-1} - 1)}{(x^{m+1} - 1)(x^{m+2} - 1)\\cdots(x^{m+n} - 1)} = \\frac{x^m - 1}{x^{m+n} - 1}, \\\\\n\\frac{f_{m,n-1}(x)}{f_{m,n}(x)} &= \\frac{f_{n-1,m}(x)}{f_{n,m}(x)} = \\frac{(x^n - 1)(x^{n+1} - 1)\\cdots(x^{n+m-1} - 1)}{(x^{n+1} - 1)(x^{n+2} - 1)\\cdots(x^{n+m} - 1)} \\\\\n&= \\frac{x^n - 1}{x^{n+m} - 1},\n\\end{align*}\n$$\n\nand hence,\n\n$$\nx^n \\cdot \\frac{f_{m-1,n}(x)}{f_{m,n}(x)} + \\frac{f_{m,n-1}(x)}{f_{m,n}(x)} = \\frac{x^n(x^m - 1)}{x^{m+n} - 1} + \\frac{x^n - 1}{x^{m+n} - 1} = 1.\n$$\n\nBy the induction hypothesis, $f_{m,n}(x) = x^n f_{m-1,n}(x) + f_{m,n-1}(x)$ is a polynomial of degree $mn$ with nonnegative integer coefficients. Since all coefficients of $f_{m-1,n}(x)$ are positive, it is clear that in $f_{m,n}(x)$, the coefficients of $x^n, x^{n+1}, \\dots, x^{mn}$ are positive; in addition, all coefficients of $f_{m,n-1}(x)$ are positive, and hence, the coefficients of $x^0, x^1, \\dots, x^{mn-m}$ in $f_{m,n}(x)$ are positive. As $m, n > 0, mn-m \\geq n-1$, all terms $x^0, x^1, \\dots, x^{mn}$ of $f_{m,n}(x)$ have positive coefficients. The induction is completed and the assertion is secured.",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 13979,
    "subject": "Mathematics (Olympiad)",
    "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": "See solution",
    "solution": "We show that if the point $P$ lies outside the segment $QR$, then the conditions of the problem are not satisfied.\n\nIf $P$ lies on the line $QR$ but outside the segment $QR$:\n\n![](images/prob1516_p7_data_668b68cf61.png)\n\nThen we can take 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$.\n\nIf $P$ lies outside the line $QR$:\n\n![](images/prob1516_p7_data_2b86eacf78.png)\n\nThe 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": 13980,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $m$, $n$, $p$ be positive integers. Space is divided into unit cubes by infinite parallel planes. An assignment of all unit cubes of space with a natural number from $1$ to $60$ is called a \"Dien Bien\" assignment if, for every rectangular box whose faces lie on the planes and whose sizes are in $\\{2m+1, 2n+1, 2p+1\\}$, the unit cube at the center of the box is assigned the average of the $8$ numbers assigned to the $8$ vertices of that box.\n\nHow many Dien Bien assignments are there?",
    "options": [],
    "answer": "See solution",
    "solution": "Without loss of generality, suppose the parallel planes are equally spaced with unit distance. Choose a cube and take its center $O$ as the origin. The axes $Ox$, $Oy$, $Oz$ are parallel to the system's lines, with positive directions chosen arbitrarily. Denote the coordinate of a cube as its distances from projections onto $Ox$, $Oy$, $Oz$ to $O$.\n\nConsider a Dien Bien assignment. Two unit cubes are *related* if they must be assigned the same number. To count the number of such assignments, count the maximum number of pairwise *unrelated* unit cubes. Let $S$ be this number. By extremal principle, for any rectangular parallelepiped of sizes $2m+1$, $2n+1$, $2p+1$, all $8$ numbers at the corners and the center are equal. This implies the cube $(x, y, z)$ is *related* to the cube\n\n$$\n(x + (-1)^r m,\\ y + (-1)^s n,\\ z + (-1)^t p),\n$$\n\nwhere $r, s, t \\in \\{0, 1\\}$ independently, corresponding to the $8$ corners of the box with center $(x, y, z)$. Rotating the box, $(x, y, z)$ is *related* to cubes of the form\n\n$$\n(x + a_1 m + a_2 n + a_3 p,\\ y + b_1 m + b_2 n + b_3 p,\\ z + c_1 m + c_2 n + c_3 p),\n$$\n\nwhere in each step, $a_1 + a_2 + a_3$, $b_1 + b_2 + b_3$, $c_1 + c_2 + c_3$ each change by $1$. Thus,\n\n$$\na_1 + a_2 + a_3 \\equiv b_1 + b_2 + b_3 \\equiv c_1 + c_2 + c_3 \\pmod{2}.\n$$\n\nConsequently, $(x, y, z)$ is *related* to $(x + x_1, y + y_1, z + z_1)$, where $(x_1, y_1, z_1)$ is a permutation of $(2k m, 2k n, 2k p)$ for some integer $k$.\n\nBy Bézout's theorem, in all linear combinations of $m, n, p$, $d = \\gcd(m, n, p) > 0$ is the minimal absolute value. Thus, each unit cube of a $d \\times d \\times d$ cube is pairwise *unrelated*, so $S$ is a multiple of $d^3$. Let $A$ be a $d \\times d \\times d$ cube. Set $m_1 = \\frac{m}{d}$, $n_1 = \\frac{n}{d}$, $p_1 = \\frac{p}{d}$, and consider three cases:\n\n1. If $m_1, n_1, p_1$ are all odd, then $S = 4d^3$.\n2. If exactly two of $m_1, n_1, p_1$ are odd, then $S = 2d^3$.\n3. If exactly one of $m_1, n_1, p_1$ is odd, then $S = d^3$.\n\nEach *unrelated* cube can be assigned any of $60$ values. Thus, the number of Dien Bien assignments is $60^S$, where $S$ is determined as above:\n\n- If $\\frac{m}{d}$, $\\frac{n}{d}$, $\\frac{p}{d}$ are all odd, $S = 4d^3$.\n- If two are odd, $S = 2d^3$.\n- If one is odd, $S = d^3$.\n\n$\\boxed{60^S}$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 13981,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all positive integers $n$ for which there exists an integer $m$, a multiple of $7$, such that the sum of the digits of $m$ is equal to $n$.",
    "options": [],
    "answer": "See solution",
    "solution": "Any number with the sum of its digits equal to $1$ is a power of $10$, so it cannot be a multiple of $7$. For $n = 2$, numbers like $11$, $101$, and $110$ are not divisible by $7$, but $1001$ is. The integer formed by $k$ repetitions of $1001$, i.e., $10011001\\ldots1001$, has digit sum $2k$ and is a multiple of $7$. Thus, for all even $n > 0$, such $m$ exists.\n\nFor $n = 3$, $21$ is a multiple of $7$ with digit sum $3$. Numbers of the form $2110011001\\ldots1001$ (i.e., $21$ followed by $k$ repetitions of $1001$) have digit sum $3 + 2k$ and are multiples of $7$. Thus, for all odd $n > 1$, such $m$ exists.\n\nTherefore, all positive integers $n$ except $1$ have the required property.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 13982,
    "subject": "Mathematics (Olympiad)",
    "question": "Which number is larger: $A = \\frac{1}{9} \\div \\sqrt[3]{\\frac{1}{2023}}$ or $B = \\log_{2023} 91125$?",
    "options": [],
    "answer": "See solution",
    "solution": "To compare $A$ and $B$, we show that $A < \\frac{3}{2} < B$.\n\n$$\nA = \\frac{1}{9} \\div \\sqrt[3]{\\frac{1}{2023}} = \\frac{\\sqrt[3]{2023}}{9}\n$$\n\nSince $\\sqrt[3]{2023} \\approx 12.65$, $A \\approx \\frac{12.65}{9} \\approx 1.41 < \\frac{3}{2} = 1.5$.\n\nFor $B$:\n$$\nB = \\log_{2023} 91125 > \\log_{2023} 45^3 = 3 \\log_{2023} 45\n$$\n\nSince $45^3 = 91125$, $B = 3 \\log_{2023} 45$. Now, $2023 \\approx 45^2$, so $\\log_{2023} 45 = \\frac{1}{2}$, thus $B = \\frac{3}{2} = 1.5$.\n\nTherefore, $A < B$ and $B$ is larger.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 13983,
    "subject": "Mathematics (Olympiad)",
    "question": "Consider a triangular pyramidal frustum $ABCA'B'C'$. Points $D \\in (AA')$, $E \\in (BB')$, and $F \\in (CC')$ are such that the planes $(AEF)$ and $(DB'C')$ are parallel. Prove that the planes $(A'EF)$ and $(DBC)$ are also parallel.",
    "options": [],
    "answer": "See solution",
    "solution": "Denote by $V$ the common point of the supporting lines of the lateral edges of the frustum. As planes $(AEF)$ and $(DB'C')$ are parallel, we have $EF \\parallel B'C'$ and $DB' \\parallel AE$. Thales' Theorem gives, from $DB' \\parallel AE$ and $A'B' \\parallel AB$:\n\n$$\n\\frac{VD}{VA} = \\frac{VB'}{VE}, \\quad \\frac{VA'}{VA} = \\frac{VB'}{VB}.\n$$\n\nThe last two equalities give\n\n$$\n\\frac{VD}{VA'} = \\frac{VB}{VE},\n$$\n\nso $A'E \\parallel DB$. As $EF \\parallel B'C' \\parallel BC$ and $A'E \\parallel DB$, we conclude $(A'EF) \\parallel (DBC)$.\n\n![](images/RMC_2015_BT_p30_data_88b4997fba.png)",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 13984,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $a, b, c \\in (0, 2)$ satisfy $a + b + c = ab + bc + ca$. Prove that\n$$\n\\frac{1}{a^2 - a + 1} + \\frac{1}{b^2 - b + 1} + \\frac{1}{c^2 - c + 1} \\le 3.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "Set\n$$\na = \\frac{1}{a'}, \\quad b = \\frac{1}{b'}, \\quad c = \\frac{1}{c'}.\n$$\nThen the condition $a + b + c = ab + bc + ca$ can be rewritten as\n$$\na' + b' + c' = a'b' + b'c' + c'a'.\n$$\nThe required inequality can be rewritten as\n$$\n\\frac{1}{a'^2 - a' + 1} + \\frac{1}{b'^2 - b' + 1} + \\frac{1}{c'^2 - c' + 1} \\le 3.\n$$\nSince $a, b, c \\in (0, 2)$, we see that $a', b', c' > \\frac{1}{2}$. Set\n$$\na' = x + \\frac{1}{2}, \\quad b' = y + \\frac{1}{2}, \\quad c' = z + \\frac{1}{2}, \\quad x, y, z > 0.\n$$\nIt is easy to see that the condition can be written as\n$$\nxy + yz + zx = \\frac{3}{4}.\n$$\nThen\n$$\n\\begin{align*}\n\\sum_{\\text{cyc}} \\frac{1}{a'^2 - a' + 1} &= \\sum_{\\text{cyc}} \\frac{1}{(a' - \\frac{1}{2})^2 + \\frac{3}{4}} \\\\\n&= \\sum_{\\text{cyc}} \\frac{1}{x^2 + \\frac{3}{4}} \\\\\n&= \\sum_{\\text{cyc}} \\frac{1}{x^2 + xy + yz + zx} \\\\\n&= \\sum_{\\text{cyc}} \\frac{1}{(x + y)(x + z)} \\\\\n&= \\frac{2(x + y + z)}{(x + y)(y + z)(z + x)} \\\\\n&= \\frac{2(x + y + z)}{(x + y + z)(xy + yz + zx) - xyz} \\\\\n&= \\frac{8(x + y + z)}{3(x + y + z) - 4xyz}.\n\\end{align*}\n$$\nNote that\n$$\n\\frac{8(x + y + z)}{3(x + y + z) - 4xyz} \\le 3 \\iff x + y + z \\ge 12xyz.\n$$\nThe latter inequality follows from $xy + yz + zx = \\frac{3}{4}$ and the inequality $(x + y + z)(xy + yz + zx) \\ge 9xyz$ (by the AM-GM inequality).",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 13985,
    "subject": "Mathematics (Olympiad)",
    "question": "$$\nf(x) = \\sqrt{13}\\sin(x + \\varphi) + 1,\n$$\n\n$$\nf(x-c) = \\sqrt{13}\\sin(x + \\varphi - c) + 1,\n$$\n\nwhere $0 < \\varphi < \\frac{\\pi}{2}$ and $\\tan \\varphi = \\frac{2}{3}$.\n\nSuppose $af(x) + bf(x-c) = 1$ for all $x \\in \\mathbb{R}$. Find the value of $\\dfrac{b\\cos c}{a}$.",
    "options": [],
    "answer": "See solution",
    "solution": "Expanding $af(x) + bf(x-c) = 1$ gives:\n\n$$\n\\sqrt{13}a\\sin(x + \\varphi) + \\sqrt{13}b\\sin(x + \\varphi - c) + a + b = 1.\n$$\n\nRewrite $\\sin(x + \\varphi - c)$ as $\\sin(x + \\varphi)\\cos c - \\cos(x + \\varphi)\\sin c$:\n\n$$\n\\sqrt{13}a\\sin(x + \\varphi) + \\sqrt{13}b\\left[\\sin(x + \\varphi)\\cos c - \\cos(x + \\varphi)\\sin c\\right] + a + b = 1.\n$$\n\nCombine like terms:\n\n$$\n\\sqrt{13}(a + b\\cos c)\\sin(x + \\varphi) - \\sqrt{13}b\\sin c\\cos(x + \\varphi) + (a + b - 1) = 0.\n$$\n\nFor this to hold for all $x$, the coefficients must vanish:\n\n$$\n\\begin{cases}\na + b\\cos c = 0, \\\\\nb\\sin c = 0, \\\\\na + b - 1 = 0.\n\\end{cases}\n$$\n\nIf $b = 0$, then $a = 0$ (from the first equation), which contradicts $a + b - 1 = 0$. So $b \\neq 0$, and $\\sin c = 0$, so $c = 2k\\pi$ or $c = 2k\\pi + \\pi$ ($k \\in \\mathbb{Z}$).\n\nIf $c = 2k\\pi$, then $\\cos c = 1$, so $a + b = 0$ and $a + b = 1$, contradiction. Thus $c = 2k\\pi + \\pi$, so $\\cos c = -1$.\n\nNow $a + b(-1) = 0 \\implies a = b$, and $a + b = 1 \\implies 2a = 1 \\implies a = b = \\frac{1}{2}$.\n\nTherefore,\n$$\n\\frac{b\\cos c}{a} = \\frac{\\frac{1}{2} \\cdot (-1)}{\\frac{1}{2}} = -1.\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 13986,
    "subject": "Mathematics (Olympiad)",
    "question": "a) Find all values of $a$ for which the parabola $y = x^2 - a$ is tangent to the hyperbola $xy = 1$.\n\nb) Find the locus of intersection points of the parabola $y = x^2 - a$ and the hyperbola $xy = 1$ as $a$ varies.\n\n![](fig.png)",
    "options": [],
    "answer": "See solution",
    "solution": "a) Let $\\alpha$ be the $x$-coordinate of the tangency point. At this point, the derivatives of $y = x^2 - a$ and $y = 1/x$ are equal:\n$$2\\alpha = -\\frac{1}{\\alpha^2}$$\nSolving, $\\alpha = -\\frac{1}{\\sqrt[3]{2}}$. The $y$-coordinate is $\\frac{1}{\\alpha} = -\\sqrt[3]{2}$. Substitute into the parabola:\n$$-\\sqrt[3]{2} = \\left(-\\frac{1}{\\sqrt[3]{2}}\\right)^2 - a$$\n$$a = \\frac{3}{2}\\sqrt[3]{2}$$\nFor $a < \\frac{3}{2}\\sqrt[3]{2}$, the curves intersect at one point; for $a > \\frac{3}{2}\\sqrt[3]{2}$, at three points.\n\nb) Each intersection satisfies $y = x^2 - a$ and $xy = 1$. Thus, $x = 1/y$ and $y = x^2 - a$. Substitute:\n$$y = \\left(\\frac{1}{y}\\right)^2 - a$$\n$$y^2 = x^2y - ay$$\n$$y^2 = x - ay$$\nSubtracting, $y - y^2 = x^2 - a - x + ay$, which simplifies to:\n$$(x - \\frac{1}{2})^2 + (y - \\frac{a-1}{2})^2 = \\frac{(a+1)^2 + 1}{4}$$\nThis is a circle centered at $(\\frac{1}{2}, -\\frac{a-1}{2})$. The required locus is the vertical ray $x = \\frac{1}{2}$ for $y < -\\frac{3}{4}\\sqrt[3]{2} + \\frac{1}{2}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 13987,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $ABC$ be an acute-angled triangle with altitudes $AD$, $BE$, and $CF$. Let $H$ be the orthocentre, that is, the point where the altitudes meet. Prove that\n\n$$\n\\frac{AB \\cdot AC + BC \\cdot BA + CA \\cdot CB}{AH \\cdot AD + BH \\cdot BE + CH \\cdot CF} \\le 2.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "**Method 1:** Let $AB = c$, $AC = b$, and $BC = a$ denote the three side lengths of the triangle.\n\nAs $\\angle BFH = \\angle BDH = 90^\\circ$, $FHDB$ is a cyclic quadrilateral. By the Power-of-a-Point Theorem, $AH \\cdot AD = AF \\cdot AB$. (We can derive this result in other ways: for example, see Method 2, below.)\n\nSince $AF = AC \\cdot \\cos \\angle A$, we have $AH \\cdot AD = AC \\cdot AB \\cdot \\cos \\angle A = bc \\cos \\angle A$.\n\nBy the Cosine Law, $\\cos \\angle A = \\frac{b^2 + c^2 - a^2}{2bc}$, which implies that $AH \\cdot AD = \\frac{b^2 + c^2 - a^2}{2}$.\n\nBy symmetry, we can show that $BH \\cdot BE = \\frac{a^2 + c^2 - b^2}{2}$ and $CH \\cdot CF = \\frac{a^2 + b^2 - c^2}{2}$.\n\nHence,\n\n$$\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$$\n\nOur desired inequality, $\\frac{AB \\cdot AC + BC \\cdot BA + CA \\cdot CB}{AH \\cdot AD + BH \\cdot BE + CH \\cdot CF} \\le 2$, is equivalent to the inequality $\\frac{cb + ac + ba}{\\frac{a^2+b^2+c^2}{2}} \\le 2$, which simplifies to $2a^2 + 2b^2 + 2c^2 \\ge 2ab + 2bc + 2ca$.\n\nBut this last inequality is easy to prove, as it is equivalent to $(a-b)^2+(a-c)^2+(b-c)^2 \\ge 0$.\n\nTherefore, we have established the desired inequality. The proof also shows that equality occurs if and only if $a = b = c$, i.e., $\\triangle ABC$ is equilateral. $\\square$\n\n**Method 2:** Observe that\n\n$$\n\\frac{AE}{AH} = \\cos(\\angle HAE) = \\frac{AD}{AC} \\quad \\text{and} \\quad \\frac{AF}{AH} = \\cos(\\angle HAF) = \\frac{AD}{AB}.\n$$\n\nIt follows that\n\n$$\nAC \\cdot AE = AH \\cdot AD = AB \\cdot AF.\n$$\n\nBy symmetry, we similarly have\n\n$$\nBC \\cdot BD = BH \\cdot BE = BF \\cdot BA \\quad \\text{and} \\quad CD \\cdot CB = CH \\cdot CF = CE \\cdot CA.\n$$\n\nTherefore\n\n$$\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$$\n\nThis proves Equation (1) in Method 1. The rest of the proof is the same as the part of the proof of Method 1 that follows Equation (1). $\\square$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 13988,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $A$ be a nonempty subset of the positive integers such that for every $a \\in A$, the integer part of its cube root, $[\\sqrt[3]{a}]$, is also in $A$. Prove that $A = \\mathbb{N}$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $A$ be a nonempty subset of the positive integers with the property that for every $a \\in A$, $[\\sqrt[3]{a}] \\in A$.\n\nLet $m$ be the minimal element of $A$. If $m > 1$, then $[\\sqrt[3]{m}] < m$ and $[\\sqrt[3]{m}] \\in A$, contradicting the minimality of $m$. Thus, $m = 1$, so $1 \\in A$.\n\nSince $1 \\in A$, $[\\sqrt[3]{1}] = 1 \\in A$, and by induction, $9^k \\in A$ for all $k$. For example, $[\\sqrt[3]{81}] = 4 \\in A$, $4 \\cdot 9 = 36 \\in A$, $[\\sqrt[3]{36}] = 3 \\in A$, and so $3^n \\in A$ for all $n \\ge 1$.\n\n*Lemma.* There exists a power of $3$ in every interval $[n, 3n]$ for $n \\ge 1$.\n\n*Proof.* Let $3^s \\le n < 3^{s+1}$. Then $n < 3^{s+1} \\le 3n$.\n\nSuppose there exists $n$ such that $n \\notin A$. We show that if $a \\in [n^{3^p}, (n+1)^{3^p} - 1]$, then $a \\notin A$. If $a \\in A$, then $[\\sqrt[3]{a}] \\in A$ and $[\\sqrt[3]{a}] \\in [n^{3^{p-1}}, (n+1)^{3^{p-1}} - 1]$. Repeating this $p$ times, we get $n \\in A$, a contradiction.\n\nSince $\\log_3(n+1) - \\log_3 n > 0$ and $\\lim_{k \\to \\infty} \\frac{1}{3^k} = 0$, there exists $k$ such that\n\n$$\n\\log_3(n+1) - \\log_3 n > \\frac{1}{3^k}.\n$$\n\nThus, $3^k \\log_3 \\frac{n+1}{n} > 1$, so $\\log_3 \\left(\\frac{n+1}{n}\\right)^{3^k} > \\log_3 3$ and $(n+1)^{3^k} > 3 n^{3^k}$.\n\nTherefore, $[n^{3^k}, 3n^{3^k}] \\subseteq [n^{3^k}, (n+1)^{3^k} - 1]$. By the lemma, there exists $s$ such that $3^s \\in [n^{3^k}, (n+1)^{3^k} - 1]$ and $3^s \\notin A$, contradicting the earlier result that all powers of $3$ are in $A$.\n\nThus, $A = \\mathbb{N}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 13989,
    "subject": "Mathematics (Olympiad)",
    "question": "試求最小的正整數 $n$，使得存在有理係數多項式 $f_1, f_2, \\cdots, f_n$ 滿足\n\n$$\nx^2 + 7 = f_1(x)^2 + f_2(x)^2 + \\cdots + f_n(x)^2.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "我們有 $x^2 + 7 = x^2 + 2^2 + 1^2 + 1^2 + 1^2$。\n\n下面我們證明 $n = 4$ 是不可能的（因此更小的 $n$ 也不可能，因為可以取 $f_i = 0$）。\n\n用反證法，假設 $x^2 + 7 = f_1(x)^2 + f_2(x)^2 + f_3(x)^2 + f_4(x)^2$，其中 $f_i$ 都是有理係數的多項式。顯然 $f_i$ 必須都是一次的，記 $f_i = a_i x + b_i$，$i = 1, 2, 3, 4$。我們有\n\n$$\n\\sum_{i=1}^{4} a_{i}^{2} = 1, \\quad \\sum_{i=1}^{4} a_{i}b_{i} = 0, \\quad \\sum_{i=1}^{4} b_{i}^{2} = 7.\n$$\n\n令 $p_i = a_i + b_i$, $q_i = a_i - b_i$，計算得\n\n$$\n\\sum_{i=1}^{4} p_{i}^{2} = 8, \\quad \\sum_{i=1}^{4} q_{i}^{2} = 8, \\quad \\sum_{i=1}^{4} p_{i}q_{i} = -6.\n$$\n\n通分之後，這使得如下命題成立：存在正整數 $m$ 及整數 $x_i, y_i$ ($i = 1, 2, 3, 4$) 使得\n\n$$\n\\sum_{i=1}^{4} x_{i}^{2} = 8m^{2}, \\quad \\sum_{i=1}^{4} y_{i}^{2} = 8m^{2}, \\quad \\sum_{i=1}^{4} x_{i}y_{i} = -6m^{2}.\n$$\n\n但我們接下來證明命題是錯的，從而導出矛盾。假設現在的 $m$ 是能夠使得命題成立的最小的 $m$。由於奇數的平方除以 $8$ 的餘數是 $1$，偶數的平方除以 $8$ 的餘數是 $0$ 或 $4$，簡單討論可知若四個平方數的和是八的倍數，那麼這四個數都必須是偶數，因此 $x_i, y_i$ 全部都是偶數。\n\n如此一來 $-6m^2 = \\sum_{i=1}^{4} x_i y_i$ 會是 4 的倍數, 從而 $m$ 是偶數, 因此我們可以把 $x_i, y_i$ 和 $m$ 都取一半亦滿足命題, 但這與 $m$ 是最小的假設矛盾, 證畢。",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 13990,
    "subject": "Mathematics (Olympiad)",
    "question": "在 $\\triangle ABC$ 中，$BC$ 邊的中點為 $M$，$AM$ 再交 $\\triangle ABC$ 的外接圓 $\\Gamma$ 於 $R$。過 $R$ 且與 $BC$ 平行的直線再交 $\\Gamma$ 於 $S$。自 $R$ 至 $BC$ 的垂線的垂足為 $U$，$T$ 為 $U$ 對 $R$ 的對稱點。$D$ 是 $BC$ 上的一點使得 $AD$ 為 $\\triangle ABC$ 的高，$N$ 為 $AD$ 中點。最後令 $AS, MN$ 交於 $K$。證明：$AT$ 平分 $MK$。",
    "options": [],
    "answer": "See solution",
    "solution": "(a) 令 $L$ 為 $MK$ 的中點，過 $L$ 對 $BC$ 的垂線分別交 $BC, AM$ 於 $E, F$。因 $AD \\parallel EF$，$L$ 為 $EF$ 的中點。因此 $\\triangle LEM \\cong \\triangle LFK$，並得\n$$\n\\angle LKF = 90^\\circ, \\quad KF \\parallel BC.\n$$\n\n![](images/15-1J_p10_data_bbea867207.png)\n\n(b) 令 $MN$ 交 $RU$ 於 $W$。因 $AD \\parallel RU$，$N$ 為 $AD$ 的中點，故 $W$ 為 $RU$ 的中點。\n\n(c) 自 $S$ 至 $BC$ 的垂線的垂足為 $V$。明顯 $S, R$ 對 $OM$ 對稱（$O$ 為 $\\triangle ABC$ 的外心），故 $M$ 為 $UV$ 中點。由 (b) 得 $NMW \\parallel VR$。又因 $VS = RU = TR$，且 $TRU \\parallel SV$，知 $SVRT$ 為平行四邊形，$ST \\parallel VR \\parallel MN$。\n\n(d) 由 (a) 及 (c) 得知 $\\triangle LFK$, $\\triangle TRS$ 對應邊平行，因 $KS, RF$ 交於 $A$，由 Desargue's 定理的平行情形（或計算比例），得 $A, L, T$ 共線。證畢。",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 13991,
    "subject": "Mathematics (Olympiad)",
    "question": "There is a number $N$ (in decimal representation) written on the board. In each step, we erase the last digit $c$ and, instead of the number $m$ (the remaining part), we write the number $|m - 3c|$. For example, if $N = 1204$ is on the board, after one step there will be $120 - 3 \\cdot 4 = 108$. We continue this process until a one-digit number remains on the board. Find all positive integers $N$ such that, after a finite number of steps, the number $0$ is left on the board.",
    "options": [],
    "answer": "See solution",
    "solution": "Let us find $N$ that lead to zero on the board after only one step. Clearly, $|m - 3c| = 0$ if and only if $m = 3c$, which means $N = 10m + c = 31c$. All such $N$ are of the form $N = 31c$, where $c \\in \\{1, 2, \\dots, 9\\}$.\n\nWe show that the solutions to the problem are exactly all multiples of $31$. Since $c = N - 10m$, we have $m - 3c = 31m - 3N$, so divisibility by $31$ is preserved in each step. Now, we show that a multiple of $31$ actually decreases in each step. We have already shown this for $N \\leq 31 \\cdot 9$. Let $N = 31k$, where $k \\geq 10$. Then $m \\geq 31$, $m - 3c > 0$, so $|m - 3c| = 31m - 3N < 4N - 3N = N$, and we are done.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 13992,
    "subject": "Mathematics (Olympiad)",
    "question": "Numbers $a$, $b$ satisfy both equations simultaneously:\n\n$$\na^2 + b^2 = 1 \\\\\na^3 + b^3 = -1.\n$$\n\nWhat is the possible value of the expression $a^3 + b^2$?",
    "options": [],
    "answer": "See solution",
    "solution": "*Answer:* $\\pm 1$.\n\n*Solution.* From the first equation, $-1 \\leq a \\leq 1$ and $-1 \\leq b \\leq 1$, so $0 \\leq 1+a \\leq 2$ and $0 \\leq 1+b \\leq 2$. Adding both equations gives:\n\n$$\na^2(1+a) + b^2(1+b) = 0.\n$$\n\nSince both terms are non-negative, their sum is zero only if each term is zero. Thus, $a, b \\in \\{-1, 0\\}$. From the first equation, one variable must be nonzero; from the second, that variable must be $-1$. Therefore, the solutions are $(a, b) = (-1, 0)$ and $(0, -1)$, so $a^3 + b^2 = -1$ or $a^3 + b^2 = 1$.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 13993,
    "subject": "Mathematics (Olympiad)",
    "question": "Suppose we have an $N \\times N$ array of lamps, and in each move, we can choose $M$ consecutive lamps in a row or column and toggle their state. What is the necessary and sufficient condition on $M$ and $N$ so that it is possible to turn on all lamps using such moves?\n\n% IMAGE: ![](images/CroatianCompetitions2011_p31_data_c0e3b1b35e.png)\n\nFigure 4.1: Example of the coloring for $N = 10, M = 6$.",
    "options": [],
    "answer": "See solution",
    "solution": "% IMAGE: ![](images/CroatianCompetitions2011_p32_data_35bd8836c1.png)\n\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% IMAGE: ![](images/CroatianCompetitions2011_p32_data_aac6a8fef8.png)\n\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": "Abstract Algebra"
  },
  {
    "id": 13994,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $ABC$ be a scalene acute triangle with incentre $I$ and circumcentre $O$. Let $AI$ cross $BC$ at $D$. On the circle $ABC$, let $X$ and $Y$ be the mid-arc points of $ABC$ and $BCA$, respectively. Let $DX$ cross $CI$ at $E$ and let $DY$ cross $BI$ at $F$. Prove that the lines $FX$, $EY$ and $IO$ are concurrent on the external bisector of $\\angle BAC$.",
    "options": [],
    "answer": "See solution",
    "solution": "The argument hinges on the claim below:\n\n**Claim.** The lines $AE$ and $BI$ are perpendicular; similarly, $AF$ and $CI$ are perpendicular.\n\n**Proof.** Let $\\alpha = \\angle BAC$, $\\beta = \\angle CAB$ and $\\gamma = \\angle ACB$. Let $DX$ cross the circle $ADC$ again at $D'$. Note that $\\angle ECX = \\angle ACX - \\angle ACE = 90^\\circ - \\beta/2 - \\gamma/2 = \\alpha/2 = \\angle DAC = \\angle DD'C = \\angle XD'C$. As $\\angle CXE = \\angle CXD'$, triangles $XCD'$ and $XEC$ are similar, so $XD' \\cdot XE = XC^2$.\n\nAs $XA = XC$, it follows that $XD' \\cdot XE = XA^2$, so triangles $XD'A$ and $XAE$ are similar, so $\\angle XAE = \\angle AD'X = \\angle AD'D = \\angle ACD = \\gamma$.\n\nFinally, note that $\\angle XAD = \\angle XAC - \\angle DAC = 90^\\circ - \\beta/2 - \\alpha/2 = \\gamma/2$, so $\\angle IAE = \\angle DAE = \\angle XAE - \\angle XAD = \\gamma/2$. As $\\angle AIB = 90^\\circ + \\gamma/2$, the claim follows.\n\nLet $W$ be the mid-arc point of $CAB$ and let $I'$ be the reflection of $I$ across $O$. As $W, X, Y$ are the mid-arc points of $CAB, ABC, BCA$, respectively, their reflections across $O$ are the mid-arc points opposite. These latter form a triangle with orthocentre $I$, so $I'$ is the orthocentre of triangle $WXY$.\n\nReflection across $O$ maps lines $XV, WY$ and $WX$ to the perpendicular bisectors of $AI, BI$ and $CI$, respectively, so $XY \\perp AI$, $WY \\perp BI$ and $WX \\perp CI$. By the claim, $AE \\perp IF$ and $AF \\perp IE$, so $I$ is the orthocentre of triangle $AEF$ and hence $EF \\perp AI$ as well.\n\nTriangles $AEF$ and $WXY$ have therefore corresponding parallel sides, so they are homothetic from some point $R$. This homothety maps $I$ to $I'$, as they are corresponding orthocentres. Hence the lines $AW$, $EY$, $FY$ and $II'$ are concurrent at $R$. As $I$, $O$ and $I'$ are collinear and $AW$ is the external bisector of $\\angle BAC$, the conclusion follows.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 13995,
    "subject": "Mathematics (Olympiad)",
    "question": "設 $ABC$ 為銳角三角形。令點 $D, E, F$ 分別為由 $A, B, C$ 向對邊 $BC, CA, AB$ 所引的高的垂足。將三角形 $BDF$ 與 $CDE$ 的內切圓分別記為 $\\omega_B$ 與 $\\omega_C$，並設 $\\omega_B$ 與線段 $DF$ 切於 $M$ 點，$\\omega_C$ 與線段 $DE$ 切於 $N$ 點。設直線 $MN$ 分別與 $\\omega_B$ 及 $\\omega_C$ 再交於點 $P \\neq M$ 及 $Q \\neq N$。試證 $MP = NQ$。",
    "options": [],
    "answer": "See solution",
    "solution": "將圓 $\\omega_B$ 及 $\\omega_C$ 的圓心分別記為 $O_B$ 與 $O_C$，並將它們的半徑分別記為 $r_B$ 與 $r_C$；而 $BC$ 與此二圓分別切於 $T, U$ 兩點，如圖所示。\n\n由 $AFDC$ 共圓及 $ABDE$ 共圓可得\n\n$$\n\\angle MDO_B = \\frac{1}{2}\\angle FDB = \\frac{1}{2}\\angle BAC = \\frac{1}{2}\\angle CDE = \\angle O_CDN,\n$$\n\n故兩直角三角形 $DMO_B$ 與 $DNO_C$ 相似，其相似比例為\n\n$$\n\\frac{DN}{DM} = \\frac{O_C N}{O_B M} = \\frac{r_C}{r_B}.\n$$\n\n令 $\\varphi = \\angle DMN$，$\\psi = \\angle MND$。因為直線 $FM$ 及 $EN$ 分別為 $\\omega_B$ 及 $\\omega_C$ 的切線，所以\n\n$$\n\\angle MTP = \\angle FMP = \\angle DMN = \\varphi,\n$$\n\n$$\n\\angle QUN = \\angle QNE = \\angle MND = \\psi.\n$$\n\n(注意：此處可能發生 $P$ 或 $Q$ 與 $T$ 或 $U$ 重合，或者 $P$ 或 $Q$ 落在三角形 $DMT$ 或 $DUN$ 內部的情形。為了有一致的論證，我們可以使用有向角，或者乾脆就不理 $\\angle MTP$ 及 $\\angle QUN$。)\n\n在圓 $\\omega_B$ 與 $\\omega_C$ 中，$MP$ 弦及 $NQ$ 弦的長度分別為\n\n$$\nMP = 2r_B \\sin \\angle MTP = 2r_B \\cdot \\sin \\varphi,\n$$\n\n$$\nNQ = 2r_C \\sin \\angle QUN = 2r_C \\cdot \\sin \\psi.\n$$\n\n對三角形 $DNM$ 使用正弦定理，得\n\n$$\n\\frac{DN}{DM} = \\frac{\\sin \\angle DMN}{\\sin \\angle MND} = \\frac{\\sin \\varphi}{\\sin \\psi}.\n$$\n\n把以上所有相關式子放在一起，得\n\n$$\n\\frac{MP}{NQ} = \\frac{2r_B \\sin \\varphi}{2r_C \\sin \\psi} = \\frac{r_B}{r_C} \\cdot \\frac{\\sin \\varphi}{\\sin \\psi} = \\frac{DM}{DN} \\cdot \\frac{\\sin \\varphi}{\\sin \\psi} = \\frac{\\sin \\psi}{\\sin \\varphi} \\cdot \\frac{\\sin \\varphi}{\\sin \\psi} = 1,\n$$\n\n故 $MP = NQ$，得證。",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 13996,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all functions $f: \\mathbb{Q} \\to \\mathbb{Q}$ such that for each real $x$,\n\n$$\nf(3 + x) + f(1 - x) = x^2 - 1.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "If $x = 0$ and $x = -2$ are substituted in turn, we have:\n\n$$\nf(3) + f(1) = -1 \\quad \\text{and} \\quad f(1) + f(3) = 3\n$$\n\nwhich contradict each other. This implies that such functions do not exist.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 13997,
    "subject": "Mathematics (Olympiad)",
    "question": "A trapezoid $ABCD$ ($AB \\parallel CD$, $AB > CD$) is circumscribed. The incircle of triangle $ABC$ touches the lines $AB$ and $AC$ at the points $M$ and $N$, respectively. Prove that the incenter of the trapezoid $ABCD$ lies on the line $MN$.",
    "options": [],
    "answer": "See solution",
    "solution": "**Version 1.** Let $I$ be the incenter of triangle $ABC$ and $R$ be the common point of the lines $BI$ and $MN$. Since\n\n$$\nm(\\widehat{ANM}) = 90^\\circ - \\frac{1}{2}m(\\widehat{MAN}) \\quad \\text{and} \\quad m(\\widehat{BIC}) = 90^\\circ + \\frac{1}{2}m(\\widehat{MAN})\n$$\nthe quadrilateral $IRNC$ is cyclic.\n\nIt follows that $m(\\widehat{BRC}) = 90^\\circ$ and therefore\n\n$$\nm(\\widehat{BCR}) = 90^\\circ - m(\\widehat{CBR}) = 90^\\circ - (180^\\circ - m(\\widehat{BCD})) = \\frac{1}{2}m(\\widehat{BCD})\n$$\n\nSo, ($CR$ is the angle bisector of $\\widehat{DCB}$ and $R$ is the incenter of the trapezoid.)\n\n**Version 2.** If $R$ is the incenter of the trapezoid $ABCD$, then $B$, $I$ and $R$ are collinear,\n\nand $m(\\widehat{BRC}) = 90^\\circ$.\n\nThe quadrilateral $IRNC$ is cyclic.\n\nThen $m(\\widehat{MNC}) = 90^\\circ + \\frac{1}{2} \\cdot m(\\widehat{BAC})$\n\nand $m(\\widehat{RNC}) = m(\\widehat{BIC}) = 90^\\circ + \\frac{1}{2} \\cdot m(\\widehat{BAC})$,\n\nso that $m(\\widetilde{MNC})=m(\\widetilde{RNC})$ and the points $M$, $R$ and $N$ are collinear.\n\n**Version 3.** If $R$ is the incenter of the trapezoid $ABCD$, let $M' \\in (AB)$ and $N' \\in (AC)$ be the unique points such that $R \\in M'N'$ and $(AM') \\equiv (AN')$.\n\nLet $S$ be the intersection point of $CR$ and $AB$. Then $CR = RS$.\n\nConsider $K \\in AC$ such that $SK \\parallel M'N'$. Then $N'$ is the midpoint of $(CK)$.\n\nWe deduce\n\n$$\nAN' = \\frac{AK + KC}{2} = \\frac{AS + AC}{2} = \\frac{AB - BS + AC}{2} = \\frac{AB + AC - BC}{2} = AN.\n$$\n\nWe conclude that $N = N'$, hence $M = M'$, and $R$, $M$, $N$ are collinear.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 13998,
    "subject": "Mathematics (Olympiad)",
    "question": "Consider the sequence $(y_n)$ satisfying the conditions:\n\n$y_0 = -\\frac{1}{4}$, $y_1 = 0$, and $y_{n+1} + y_{n-1} = 4y_n + 1$ for $n \\geq 1$.\n\nProve that for any $n \\geq 0$, the expression $2y_{2n} + \\frac{3}{2}$ is:\n\na) a positive integer;\nb) the square of an integer.",
    "options": [],
    "answer": "See solution",
    "solution": "a) Let us prove that the numbers $2y_{2n} + \\frac{3}{2}$ and $y_{2n+1}$ are integers by induction on $n$.\n\nFor $n=0$ we have $2y_0 + \\frac{3}{2} = 1$ and $y_1 = 0$.\n\nSuppose that $2y_{2n} + \\frac{3}{2}$ and $y_{2n+1}$ are integers. Then the number\n\n$$\n2y_{2(n+1)} + \\frac{3}{2} = 2(4y_{2n+1} - y_{2n} + 1) + \\frac{3}{2} = 8y_{2n+1} - \\left(2y_{2n} + \\frac{3}{2}\\right) + 5\n$$\n\nand, therefore,\n\n$$\ny_{2(n+1)+1} = 4y_{2n+2} - y_{2n+1} + 1 = 2\\left(2y_{2n+2} + \\frac{3}{2}\\right) - y_{2n+1} - 2\n$$\n\nare also integers. Thus, the number $2y_{2n} + \\frac{3}{2}$ is integer for any $n \\geq 0$. Its positivity follows, for example, from the fact that the sequence $(y_n)$ is increasing (this is easily proven by induction). So, $y_{2n} \\geq y_0 = -\\frac{1}{4}$, whence $2y_{2n} + \\frac{3}{2} \\geq 1$.",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 13999,
    "subject": "Mathematics (Olympiad)",
    "question": "Suppose that $k$ and $l$ are two positive integers. Prove that there are infinitely many positive integers $m \\geq k$ such that $\\binom{m}{k}$ and $l$ are relatively prime.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $m = k + t \\times l \\times (k!)$, where $t$ is any positive integer. To prove that $\\binom{m}{k}$ and $l$ are relatively prime, we only need to show that for any prime factor $p$ of $l$, $p \\nmid \\binom{m}{k}$.\n\nIf $p \\nmid k!$, then\n\n$$\nk! \\binom{m}{k} = \\prod_{i=1}^{k} (m - k + i) = \\prod_{i=1}^{k} [i + t l (k!)] = \\prod_{i=1}^{k} i \\equiv k! \\pmod{p}.\n$$\n\nTherefore, $p \\nmid \\binom{m}{k}$.\n\nIf $p \\mid k!$, let $\\alpha \\geq 1$ be such that $p^\\alpha \\mid k!$ but $p^{\\alpha+1} \\nmid k!$. Then $p^{\\alpha+1} \\mid l (k!)$.\n\nWe have\n\n$$\nk! \\binom{m}{k} = \\prod_{i=1}^{k} (m - k + i) = \\prod_{i=1}^{k} [i + t l (k!)] = \\prod_{i=1}^{k} i \\equiv k! \\pmod{p^{\\alpha+1}}.\n$$\n\nTherefore, $p^\\alpha \\mid k!$ but $p^{\\alpha+1} \\nmid k!$, so $p \\nmid \\binom{m}{k}$. Thus, $\\binom{m}{k}$ and $l$ are relatively prime for infinitely many $m \\geq k$. The proof is complete.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 14000,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all real values of $x$, $y$, and $z$ such that\n\n$$\n(x + 1)yz = 12, \\quad (y + 1)zx = 4, \\quad \\text{and} \\quad (z + 1)xy = 4.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "Number the equations:\n\n$$\n(x + 1)yz = 12 \\quad (1)\n$$\n$$\n(y + 1)zx = 4 \\quad (2)\n$$\n$$\n(z + 1)xy = 4 \\quad (3)\n$$\n\nSubtracting equations in pairs:\n\n$$\n\\begin{aligned}\n(2) - (1) &: \\quad z(x - y) = -8 \\\\\n(3) - (2) &: \\quad x(y - z) = 0 \\\\\n(1) - (3) &: \\quad y(x - z) = -8\n\\end{aligned}\n$$\n\nEquation $(3)-(2)$ implies either $x = 0$ or $y = z$. If $x = 0$, then $(y + 1)zx = 0$, contradicting equation (2). Thus, $y = z$.\n\nNow, equations (1) and (2) become $(x + 1)y^2 = 12$ and $(y + 1)xy = 4$. The latter gives $x = \\frac{4}{y(y+1)}$, so\n\n$$\nx + 1 = \\frac{y^2 + y + 4}{y(y + 1)}.\n$$\n\nCombining these:\n\n$$\n\\frac{y^2 + y + 4}{y(y + 1)} y^2 = 12\n$$\n\nwhich gives\n\n$$\n\\begin{aligned}\ny^4 + y^3 + 4y^2 &= 12y^2 + 12y \\\\\ny^4 + y^3 - 8y^2 - 12y &= 0 \\\\\ny(y + 2)^2(y - 3) &= 0\n\\end{aligned}\n$$\n\nThus $y = 0$, $y = -2$, or $y = 3$.\n\nIf $y = 0$, then $(x + 1)y^2 = 0$, contradicting equation (1).\n\nIf $y = z = -2$, then $x + 1 = 12/y^2 = 3$ so $x = 2$. This gives the solution $(x, y, z) = (2, -2, -2)$.\n\nIf $y = z = 3$, then $x + 1 = 12/y^2 = 4/3$ so $x = 1/3$. This gives the solution $(x, y, z) = \\left(\\frac{1}{3}, 3, 3\\right)$.\n\nBoth solutions satisfy the original equations.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 14001,
    "subject": "Mathematics (Olympiad)",
    "question": "An integer $n \\ge 1$ is called balanced if it has an even number of prime divisors. Prove that there exist infinitely many positive integers $n$ such that among the numbers $n$, $n+1$, $n+2$, and $n+3$ there are exactly two balanced ones.",
    "options": [],
    "answer": "See solution",
    "solution": "We argue by contradiction. Choose $N$ so large that no $n \\ge N$ obeys this property. Now we partition all integers $\\ge N$ into maximal blocks of consecutive numbers which are either all balanced or all unbalanced. We delete the first block from the following considerations, now starting from $N' > N$.\n\nClearly, by assumption, there cannot be two blocks with length $\\ge 2$. It is also impossible that there are two blocks of length 1 (remember that we deleted the first block). Thus, all balanced or all unbalanced blocks have length 1. All other blocks have length at least 3.\n\n**Case 1:** All unbalanced blocks have length 1.\n\nWe take an unbalanced number $u > 2N' + 3$ with $u \\equiv 1 \\pmod{4}$ (for instance, $u = p^2$ for an odd prime $p$). Since all balanced blocks have length $\\ge 3$, $u-3$, $u-1$, and $u+1$ must be balanced. This implies that $(u-3)/2$ is unbalanced, $(u-1)/2$ is balanced, and $(u+1)/2$ is again unbalanced. Thus, $\\{(u-1)/2\\}$ is a balanced block of length 1 — contradiction.\n\n**Case 2:** All balanced blocks have length 1.\n\nNow we take a balanced number $b > 2N' + 3$ with $b \\equiv 1 \\pmod{4}$ (for instance, $b = p^2q^2$ for distinct odd primes $p, q$). By similar arguments, $(b-3)/2$ is balanced, $(b-1)/2$ is unbalanced, and $(b+1)/2$ is again balanced. Now the balanced block $\\{(b-1)/2\\}$ gives the desired contradiction.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 14002,
    "subject": "Mathematics (Olympiad)",
    "question": "Determine all positive integers $\\alpha, \\beta$ with $\\alpha > 1$ such that $\\beta$ divides $\\alpha - 1$ and $2\\alpha + 1$ divides $5\\beta - 3$.",
    "options": [],
    "answer": "See solution",
    "solution": "Since $\\beta$ divides $\\alpha - 1$ and $2\\alpha + 1$ divides $5\\beta - 3$, with $\\alpha > 1$,\n\n$$\nx = \\frac{\\alpha - 1}{\\beta} \\quad \\text{and} \\quad y = \\frac{5\\beta - 3}{2\\alpha + 1}\n$$\n\nare positive integers such that:\n\n$$\n0 < xy = \\frac{\\alpha - 1}{\\beta} \\cdot \\frac{5\\beta - 3}{2\\alpha + 1} < \\frac{\\alpha}{\\beta} \\cdot \\frac{5\\beta}{2\\alpha} = \\frac{5}{2} \\implies xy \\in \\{1,2\\}.\n$$\n\n- If $xy = 1$, then $x = y = 1$, and so we have the system:\n\n$$\n\\begin{cases} \\alpha - 1 = \\beta \\\\ 2\\alpha + 1 = 5\\beta - 3 \\end{cases} \\implies \\begin{cases} \\alpha = \\beta + 1 \\\\ 2(\\beta + 1) + 1 = 5\\beta - 3 \\end{cases} \\implies \\alpha = 3, \\beta = 2.\n$$\n\n- If $xy = 2$, then $x = 1, y = 2$ or $x = 2, y = 1$, and so we have the systems:\n\n$$\n\\begin{cases} \\alpha - 1 = \\beta \\\\ 4\\alpha + 2 = 5\\beta - 3 \\end{cases} \\implies \\begin{cases} \\alpha = \\beta + 1 \\\\ 4(\\beta + 1) + 2 = 5\\beta - 3 \\end{cases} \\implies \\alpha = 10, \\beta = 9.\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 14003,
    "subject": "Mathematics (Olympiad)",
    "question": "Suppose $2n$ real numbers are placed in the cells of a $2 \\times n$ grid such that the sum of the numbers in each of the $n$ columns is $1$. Prove that one can erase one of the two numbers in each column so that the sum of the remaining numbers in each row does not exceed $\\frac{n+1}{4}$.",
    "options": [],
    "answer": "See solution",
    "solution": "Assume the numbers in the first row are $a_1 \\le a_2 \\le \\dots \\le a_n$ in some order. The numbers in the second row are $b_j = 1 - a_j$ for $1 \\le j \\le n$, so $b_1 \\ge b_2 \\ge \\dots \\ge b_n$.\n\nIf $a_1 + a_2 + \\dots + a_n \\le \\frac{n+1}{4}$, we are done (erase all numbers in the second row). Otherwise, let $k$ be the least positive integer such that $a_1 + a_2 + \\dots + a_k > \\frac{n+1}{4}$.\n\n$$\na_1 + a_2 + \\dots + a_{k-1} \\le \\frac{n+1}{4}\n$$\n\nWe show that\n\n$$\nb_k + b_{k+1} + \\dots + b_n \\le \\frac{n+1}{4}\n$$\n\nObserve\n\n$$\na_k \\ge \\frac{a_1 + a_2 + \\dots + a_k}{k} > \\frac{n+1}{4k}\n$$\n\nHence\n\n$$\nb_k + b_{k+1} + \\dots + b_n \\le (n+1-k)b_k = (n+1-k)(1-a_k) < (n+1-k) \\left(1 - \\frac{n+1}{4k}\\right)\n$$\n\n$$\n= \\frac{5}{4}(n+1) - \\frac{(n+1)^2 + 4k^2}{4k}\n$$\n\nBy AM-GM,\n\n$$\n\\le \\frac{5}{4}(n+1) - \\frac{2(n+1)(2k)}{4k} = \\frac{n+1}{4}\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 14004,
    "subject": "Mathematics (Olympiad)",
    "question": "設 $\\lambda > 0$ 為滿足方程式 $\\lambda = \\lambda^{2/3} + 1$ 的正實數。證明：存在正整數 $M$ 使得\n\n$$\n|M - \\lambda^{300}| < 4^{-100}.\n$$\n\nLet $\\lambda > 0$ be a positive real number satisfying $\\lambda = \\lambda^{2/3} + 1$. Show that there exists a positive integer $M$ such that\n\n$$\n|M - \\lambda^{300}| < 4^{-100}.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "解：令 $\\lambda = t^{3/2}$。於是 $t^3 = (t+1)^2$，且 $\\lambda^{300} = t^{450}$。檢驗方程式 $P(x) = x^3 - (x+1)^2$ 只有唯一的零點 $x = t$，且因 $P(2) < 0$，所以 $t > 2$。設 $a, b$ 為 $P(x) = 0$ 的另外兩根。注意到 $|ab| = 1$，且 $a$ 為 $b$ 的共軛複數。故\n\n$$\n|a| = |b| = |t|^{-1/2} < 2^{-1/2}.\n$$\n\n現取 $M = a^{450} + b^{450} + t^{450}$。因為 $a, b, t$ 是首一整係數三次方程式 $P(x) = 0$ 的三根且 $M > 0$，所以 $M$ 是正整數。並且：\n\n$$\n|M - \\lambda^{300}| = |M - t^{450}| = |a^{450} + b^{450}| < 2 \\cdot 2^{-450/2} < \\frac{1}{4^{100}}.\n$$\n\n證畢。",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 14005,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $S$ be a set of members. Working groups are nonempty subsets of $S$. Let $A_1, A_2, \\dots, A_{114}$ be the working groups.\n\nSuppose that for each $x \\in S$, there are at most 112 distinct sets among $A_1 - \\{x\\}, A_2 - \\{x\\}, \\dots, A_{114} - \\{x\\}$. Prove that this leads to a contradiction; that is, there exists $x \\in S$ such that there are at least 113 distinct sets among $A_1 - \\{x\\}, A_2 - \\{x\\}, \\dots, A_{114} - \\{x\\}$.",
    "options": [],
    "answer": "See solution",
    "solution": "Assume for contradiction that for each $x \\in S$, there are at most 112 distinct sets among $A_1 - \\{x\\}, \\dots, A_{114} - \\{x\\}$.\n\nConstruct a graph $G$ as follows: for each $x \\in S$, choose two pairs of sets $F_x \\subseteq E_x$ and $F'_x \\subseteq E'_x$ such that $E_x - F_x = E'_x - F'_x = \\{x\\}$. Join $(E_x, F_x)$ and $(E'_x, F'_x)$ by edges. Thus, $|V(G)| = 114$ and $|E(G)| = 152$.\n\n$G$ is bipartite: let $V_1 = \\{E : |E| \\text{ is odd}\\}$ and $V_2 = \\{E : |E| \\text{ is even}\\}$.\n\n$G$ does not contain a subgraph consisting of three independent paths joining two vertices. If $P_1, P_2, P_3$ are such paths between $E$ and $E'$, and $x \\in E - E'$, then for each $i$, there exist $E_i$ and $F_i$ on $P_i$ with $\\{x\\} = E_i - F_i$, which is impossible.\n\nLet $a$ be the number of edges not in a cycle, $b$ the number of cycles, and $k_1, \\dots, k_b$ their lengths. Then:\n\n$$\n114 = |V(G)| = c + a + \\sum_{i=1}^{b} (k_i - 1),\n$$\nwhere $c$ is the number of components,\n\n$$\n152 = |E(G)| = a + \\sum_{i=1}^{b} k_i.\n$$\n\nThus,\n\n$$\n114 \\geq 1 + a + \\sum_{i=1}^{b} (k_i - 1) \\geq 1 + \\frac{3}{4}a + \\sum_{i=1}^{b} \\frac{3}{4}k_i = 1 + \\frac{3}{4}(2 \\cdot 76) = 1 + 114 = 115.\n$$\n\nThis contradiction shows that there exists $x \\in S$ such that there are at least 113 distinct sets among $A_1 - \\{x\\}, \\dots, A_{114} - \\{x\\}$.",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 14006,
    "subject": "Mathematics (Olympiad)",
    "question": "How many ordered triples of digits $(x, y, z)$ with $1 \\le x < y < z$ and $x + y + z < 10$ are there such that the sum of the numbers $\\overline{xyz}$, $\\overline{yzx}$, and $\\overline{zxy}$ is a three-digit number whose digits are all equal?",
    "options": [],
    "answer": "See solution",
    "solution": "Let $x, y, z$ be digits with $1 \\le x < y < z$ and $x + y + z < 10$. The sum $\\overline{xyz} + \\overline{yzx} + \\overline{zxy}$ can be written as:\n\n$$\n\\overline{xyz} + \\overline{yzx} + \\overline{zxy} = 100x + 10y + z + 100y + 10z + x + 100z + 10x + y = 111(x + y + z)\n$$\n\nSince $x + y + z < 10$, $111(x + y + z)$ is a three-digit number with all digits equal. Now, we count the number of ordered triples $(x, y, z)$ with $1 \\le x < y < z$ and $x + y + z < 10$:\n\nThe possible triples are:\n- $(1, 2, 3)$: $1+2+3=6$\n- $(1, 2, 4)$: $1+2+4=7$\n- $(1, 2, 5)$: $1+2+5=8$\n- $(1, 2, 6)$: $1+2+6=9$\n- $(1, 3, 4)$: $1+3+4=8$\n- $(1, 3, 5)$: $1+3+5=9$\n- $(2, 3, 4)$: $2+3+4=9$\n\nThus, there are exactly $7$ such triples.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 14007,
    "subject": "Mathematics (Olympiad)",
    "question": "How many of the integers between $97$ and $199$ are multiples of $2$ or $3$?\n\n- (A) $33$\n- (B) $40$\n- (C) $55$\n- (D) $60$\n- (E) $68$",
    "options": [],
    "answer": "See solution",
    "solution": "Let us count the number of integers between $97$ and $199$ that are multiples of $2$ or $3$.\n\nFirst, the number of multiples of $2$ between $97$ and $199$ is:\n$$\text{Multiples of }2: \\left\\lfloor \\frac{199}{2} \\right\\rfloor - \\left\\lfloor \\frac{96}{2} \\right\\rfloor = 99 - 48 = 51$$\n\nThe number of multiples of $3$ between $97$ and $199$ is:\n$$\text{Multiples of }3: \\left\\lfloor \\frac{199}{3} \\right\\rfloor - \\left\\lfloor \\frac{96}{3} \\right\\rfloor = 66 - 32 = 34$$\n\nThe number of multiples of $6$ (i.e., multiples of both $2$ and $3$) between $97$ and $199$ is:\n$$\text{Multiples of }6: \\left\\lfloor \\frac{199}{6} \\right\\rfloor - \\left\\lfloor \\frac{96}{6} \\right\\rfloor = 33 - 16 = 17$$\n\nBy the inclusion-exclusion principle, the total is:\n$$51 + 34 - 17 = 68$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 14008,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $P$ be a point of intersection of the diagonals of the cyclic quadrilateral $ABCD$. The circumscribed circles of $\\triangle APD$ and $\\triangle BPC$ intersect the line $AB$ at points $E$ and $F$, respectively. $Q$ and $R$ are the projections of the point $P$ onto the lines $FC$ and $DE$. Prove that $AB \\parallel QR$.",
    "options": [],
    "answer": "See solution",
    "solution": "Denote by $d(Z, XY)$ the distance from the point $Z$ to the line $XY$. From the equality of inscribed angles, it follows that\n\n$$\n\\angle PEB = \\angle ADB = \\angle ACB = \\angle PFA \\implies PE = PF.\n$$\n\nFurthermore,\n\n$$\n\\angle EDP = \\angle BAP = \\angle BDC, \\quad \\angle FCP = \\angle PBA = \\angle DCP,\n$$\n\nso $P$ is the intersection of the bisectors of angles $\\angle EDC$ and $\\angle DCF$, so $d(P, DE) = d(P, DC) = d(P, CF)$, so $PQ = PR$. Then $\\triangle FPQ \\cong \\triangle EPR$ as right triangles with equal legs and hypotenuse. From the equality $PQ = PR$ it follows that $\\angle PQR = \\angle PRQ$, so $\\angle FQR = \\angle QRE$. Similarly, $\\angle QFE = \\angle REF$. This means that $EQRF$ is an isosceles trapezoid, so $QR \\parallel EF$, as desired.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 14009,
    "subject": "Mathematics (Olympiad)",
    "question": "The base-nine representation of the number $N$ is $27,006,000,052_{\\text{nine}}$. What is the remainder when $N$ is divided by $5$?\n\n(A) $0$ (B) $1$ (C) $2$ (D) $3$ (E) $4$",
    "options": [],
    "answer": "See solution",
    "solution": "Note that $N = 2 \\cdot 9^{10} + 7 \\cdot 9^9 + 6 \\cdot 9^6 + 5 \\cdot 9^1 + 2 \\cdot 9^0$. Because even powers of $9$ leave remainder $1$ when divided by $5$ and odd powers of $9$ leave remainder congruent to $-1$ when divided by $5$, it follows that $N$ leaves remainder congruent to $2 - 7 + 6 - 5 + 2 = -2 \\equiv 3 \\pmod{5}$, so the requested remainder is $3$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 14010,
    "subject": "Mathematics (Olympiad)",
    "question": "In isosceles $\\triangle ABC$, $AB = BC$, and $I$ is its incentre. $M$ is the midpoint of $BI$. $P$ lies on side $AC$, satisfying $AP = 3PC$. Point $H$ on the extension of $PI$ satisfies $MH \\perp PH$. $Q$ is the midpoint of the minor arc of the circumcircle of $\\triangle ABC$. Prove that $BH \\perp QH$.\n\n![](images/Mathematical_Olympiad_In_China_2021_2022___Problems_and_Solutions_2024_p72_data_e55d32e8f9.png)",
    "options": [],
    "answer": "See solution",
    "solution": "Take the midpoint $N$ of $AC$. Since $AP = 3PC$, $P$ is the midpoint of $NC$. It is easy to see that points $B, I, N$ are collinear and $\\angle INC = 90^\\circ$.\n\nSince $I$ is the incentre of $\\triangle ABC$, $CI$ passes through point $Q$ and\n\n$$\n\\begin{aligned}\n\\angle QIB &= \\angle IBC + \\angle ICB = \\angle ABI + \\angle ACQ \\\\\n            &= \\angle ABI + \\angle ABQ = \\angle QBI.\n\\end{aligned}\n$$\n\nSince $M$ is the midpoint of $BI$, $QM \\perp BI$. Thus, $QM \\parallel CN$.\n\nConsider $\\triangle HMQ$ and $\\triangle HIB$. Since $MH \\perp PH$, $\\angle HMQ = 90^\\circ - \\angle HMI = \\angle HIB$.\n\nAlso, $\\angle IHM = \\angle INP = 90^\\circ$, and $\\frac{HM}{HI} = \\frac{NP}{NI}$. Therefore,\n\n$$\n\\frac{HM}{HI} = \\frac{NP}{NI} = \\frac{1}{2} \\cdot \\frac{NC}{NI} = \\frac{1}{2} \\cdot \\frac{MQ}{MI} = \\frac{MQ}{IB}.\n$$\n\nHence, $\\triangle HMQ \\sim \\triangle HIB$, so $\\angle HQM = \\angle HBI$.\n\nThus, points $H, M, B, Q$ are concyclic. Therefore, $\\angle BHQ = \\angle BMQ = 90^\\circ$, i.e., $BH \\perp QH$. $\\square$\n\n![](images/Mathematical_Olympiad_In_China_2021_2022___Problems_and_Solutions_2024_p73_data_3b392a8e20.png)",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 14011,
    "subject": "Mathematics (Olympiad)",
    "question": "Prove that one can assign to each vertex of a cube 8 distinct numbers from the set $\\{0, 1, 2, 3, \\ldots, 12\\}$ such that, for every edge, the sum of the two numbers assigned to its vertices is divisible by 3.",
    "options": [],
    "answer": "See solution",
    "solution": "The task can be accomplished by assigning to neighboring vertices distinct numbers that are not divisible by 3 and that yield different residues modulo 3. An example is shown in the figure.\n\n![](images/RMC2010_p54_data_d105f9ceb5.png)",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 14012,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $n$ be an odd natural number. We consider an $n \\times n$ grid which is made up of $n^2$ unit squares and $2n(n+1)$ edges. We colour each of these edges either red or blue. If there are at most $n^2$ red edges, then show that there exists a unit square at least three of whose edges are blue.",
    "options": [],
    "answer": "See solution",
    "solution": "Suppose, on the contrary, that each unit square has at least two red edges. Each red edge is part of at most two unit squares. Therefore,\n\n$$\n2n^2 \\leq \\sum_{\\text{unit squares}} \\text{(red edges of the square)} = \\sum_{\\text{red edges}} \\text{(unit squares containing the edge)} \\leq 2n^2.\n$$\n\nThis implies that each unit square has exactly two red edges and that there are a total of $n^2$ red edges.\n\nWe colour each of the unit squares black and white such that no two unit squares which share a common edge have the same colour (like in a chessboard). Note that each red edge is part of exactly one black square and one white square. Since every unit square has exactly two red edges, the number of white squares is therefore $n^2/2$, a contradiction since this is not an integer.\n\nThis shows that there is a unit square with at most one red edge. Thus, there exists a unit square with at least three blue edges.",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 14013,
    "subject": "Mathematics (Olympiad)",
    "question": "When $144$ is divided by the positive integer $n$, the remainder is $11$. When $220$ is divided by $n$, the remainder is also $11$. What is the value of $n$?",
    "options": [],
    "answer": "See solution",
    "solution": "Since $144$ and $220$ both leave a remainder of $11$ when divided by $n$, $n$ divides both $144 - 11 = 133$ and $220 - 11 = 209$. \n\nFind $\text{gcd}(133, 209)$:\n\n$$\begin{align*}\n209 &= 133 \\times 1 + 76 \\\\\n133 &= 76 \\times 1 + 57 \\\\\n76 &= 57 \\times 1 + 19 \\\\\n57 &= 19 \\times 3 + 0\n\\end{align*}$$\n\nSo, $\text{gcd}(133, 209) = 19$. Thus, $n = 19$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 14014,
    "subject": "Mathematics (Olympiad)",
    "question": "You are given $n \\ge 4$ positive real numbers. It turned out that their $\\frac{n(n-1)}{2}$ pairwise products form an arithmetic progression in some order. Prove that all of these numbers are equal.",
    "options": [],
    "answer": "See solution",
    "solution": "If some two products are equal, then all products are equal, and all numbers are equal. If some two numbers are equal, then some products are equal, so all numbers are equal. Now consider the four largest numbers $a < b < c < d$. The largest two products are $cd$ and $bd$. Then the difference of the progression is $cd - bd$. But then $ac - ab = a(c - b) < d(c - b)$, so the difference between some two elements of the progression is smaller than the difference of the progression, which is impossible—a contradiction.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 14015,
    "subject": "Mathematics (Olympiad)",
    "question": "Joah has a number of large pots with marbles in them. At the beginning of the week, all pots contain a different positive number of marbles. On the first day of the week, he adds one marble to each pot. On the second day, he adds a marble to all pots whose number of marbles is divisible by $2$. On the third day, he adds a marble to all pots whose number of marbles is divisible by $3$. He continues like this until the seventh day. Then it turns out that he has several pots with exactly $50$ marbles in them.\n\nWhat is the maximum number of pots with exactly $50$ marbles that Joah could have?",
    "options": [],
    "answer": "See solution",
    "solution": "$2$",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 14016,
    "subject": "Mathematics (Olympiad)",
    "question": "Prove that $$\n\\cos^4 x + \\sin^4 x \\ge \\cos x \\sin x\n$$ for all real $x$.",
    "options": [],
    "answer": "See solution",
    "solution": "By the geometric-arithmetic mean inequality,\n$$\n\\cos x \\sin x \\le \\frac{\\cos^2 x + \\sin^2 x}{2} = \\frac{1}{2}.\n$$\nAlso,\n$$\n1 = (\\cos^2 x + \\sin^2 x)^2 = \\cos^4 x + \\sin^4 x + 2 \\cos^2 x \\sin^2 x \\le \\cos^4 x + \\sin^4 x + \\frac{1}{2}\n$$\nso\n$$\n\\cos^4 x + \\sin^4 x \\ge \\frac{1}{2} \\ge \\cos x \\sin x.\n$$\nThus, the required inequality holds.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 14017,
    "subject": "Mathematics (Olympiad)",
    "question": "Suppose that for real $x, y, z, t$ the following equalities hold:\n\n$$\n\\{x+y+z\\} = \\{y+z+t\\} = \\{z+t+x\\} = \\{t+x+y\\} = \\frac{1}{4}.\n$$\n\nFind all possible values of $\\{x+y+z+t\\}$.",
    "options": [],
    "answer": "See solution",
    "solution": "We have\n\n$$\n\\begin{aligned}\nx+y+z &= [x+y+z] + \\frac{1}{4}, \\\\\ny+z+t &= [y+z+t] + \\frac{1}{4}, \\\\\nz+t+x &= [z+t+x] + \\frac{1}{4}, \\\\\nt+x+y &= [t+x+y] + \\frac{1}{4}.\n\\end{aligned}\n$$\n\nThus,\n\n$$\n3(x+y+z+t) = [x+y+z] + [y+z+t] + [z+t+x] + [t+x+y] + 1.\n$$\n\nThis implies that $3(x+y+z+t)$ is integer, so the fractional part of $x+y+z+t$ is either $0$, $\\frac{1}{3}$, or $\\frac{2}{3}$. All these values can be achieved, as shown by the following examples:\n\n$$\nx = y = z = t = \\frac{3}{4}, \\quad x = y = z = t = \\frac{1}{12}, \\quad x = y = z = t = \\frac{5}{12}.\n$$\n\n**Answer:** $0$, $\\frac{1}{3}$, $\\frac{2}{3}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 14018,
    "subject": "Mathematics (Olympiad)",
    "question": "Prove the following inequality:\n\n$$\n\\sum_{i=1}^{n} \\sum_{j=1}^{n} \\sqrt{|x_i - x_j|} \\le \\sum_{i=1}^{n} \\sum_{j=1}^{n} \\sqrt{|x_i + x_j|}.\n$$\n\nWe illustrate the situation for $n=2$ with $x_1 = -1$ and $x_2 = 2$. In this case we have\n\n$$\nR(t) = \\sqrt{|-2+t|} + \\sqrt{|1+t|} + \\sqrt{|1+t|} + \\sqrt{|4+t|}.\n$$\n\nThe figure below shows the graphs of $y = \\sqrt{|-2+t|}$, $y = \\sqrt{|1+t|}$ and $y = \\sqrt{|4+t|}$ in black, and $y = R(t)$ in blue.\n\n![](images/2021_Australian_Scene_p128_data_9bba3cc344.png)",
    "options": [],
    "answer": "See solution",
    "solution": "Let $L$ be the value of the left-hand side. If we add $t/2$ to all variables, then $L$ remains constant while the right-hand side becomes\n\n$$\nR(t) = \\sum_{i=1}^{n} \\sum_{j=1}^{n} \\sqrt{|x_i + x_j + t|}.\n$$\n\nNote that $R(t)$ is the sum of $n^2$ functions of the form $f(t) = \\sqrt{|a+t|}$. Each such function is concave separately on each of the intervals $(-\\infty, -a]$ and $[-a, +\\infty)$. Since the sum of concave functions is also a concave function, the points $p_{i,j} = -(x_i + x_j)$ split the real line into intervals such that on each of these intervals, $R(t)$ is a concave function. Since $\\lim_{t \\to \\pm\\infty} R(t) = +\\infty$, it follows that $R(t)$ attains its minimal value at one of the points $p_{i,j}$. So it suffices to prove that $L \\le R(-(x_i + x_j))$. That is, it suffices to prove the inequality whenever $x_1, \\dots, x_n$ are shifted in such a way so that $x_i + x_j = 0$ for some $i, j$.\n\nIf $i = j$, then $x_i = 0$, and removing $x_i$ decreases both sides of the inequality by $2 \\sum_k \\sqrt{|x_k|}$. If $i \\ne j$, then removing $x_i$ and $x_j$ decreases both sides by\n\n$$\n2\\sqrt{2|x_i|} + 2 \\sum_{k \\ne i, j} \\left( \\sqrt{|x_k + x_i|} + \\sqrt{|x_k + x_j|} \\right).\n$$\n\nEither way, the inequality is inductively reduced to the case of $n-1$ or $n-2$ variables. For $n=0$ and $n=1$, the inequality is trivial. $\\square$",
    "topic": "Discrete Mathematics",
    "subtopic": "Other"
  },
  {
    "id": 14019,
    "subject": "Mathematics (Olympiad)",
    "question": "Find the largest positive number $\\lambda$ such that\n$$\n| \\lambda x y + y z | \\leq \\frac{\\sqrt{5}}{2}, \\quad \\text{where } x^2 + y^2 + z^2 = 1.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "Note that\n$$\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&\\geq \\frac{2}{\\sqrt{1+\\lambda^2}} (|x| + |y| + |z|) \\\\\n&\\geq \\frac{2}{\\sqrt{1+\\lambda^2}} (|\\lambda x y + y z|),\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$$\nThus, $\\frac{\\sqrt{1+\\lambda^2}}{2}$ is the maximum value of $|\\lambda x y + y z|$. Let\n$$\n\\frac{\\sqrt{1+\\lambda^2}}{2} = \\frac{\\sqrt{5}}{2}.\n$$\nWe obtain that $\\lambda = 2$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 14020,
    "subject": "Mathematics (Olympiad)",
    "question": "a) The six different $2 \\times 2$ tiling patterns are:\n\n![](images/2018-Australian-Scene-W2_p54_data_61e34b4ab7.png)\n\n![](images/2018-Australian-Scene-W2_p54_data_1a3633a67b.png)\n\n![](images/2018-Australian-Scene-W2_p54_data_ef312c6626.png)\n\n![](images/2018-Australian-Scene-W2_p54_data_89d0a90846.png)\n\n![](images/2018-Australian-Scene-W2_p54_data_09949eb0c0.png)\n\n![](images/2018-Australian-Scene-W2_p54_data_036096fe73.png)\n\nRotations of these patterns are acceptable.\n\nb) What is the probability of each pattern occurring in a random placement of four tiles?\n\nc) In a $3 \\times 3$ square, each tile has two orientations. What is the probability that a random $3 \\times 3$ placement contains a peanut (as defined by the tile arcs)?",
    "options": [],
    "answer": "See solution",
    "solution": "b) Each random placement of four tiles has probability $1/16$ of occurring.\n\nThere is only one random placement that gives pattern 1, so the probability of pattern 1 occurring is $1/16$. Similarly, the probability of pattern 6 occurring is $1/16$.\n\nThere are exactly 4 random placements that give pattern 2: the one shown and its rotations through $90^\\circ$, $180^\\circ$, and $270^\\circ$.\n\n![](images/2018-Australian-Scene-W2_p54_data_68442f05da.png)\n\n![](images/2018-Australian-Scene-W2_p54_data_a74b74e7a1.png)\n\n![](images/2018-Australian-Scene-W2_p54_data_2ccb317906.png)\n\n![](images/2018-Australian-Scene-W2_p54_data_2d13302efe.png)\n\nSo the probability of pattern 2 occurring is $4/16 = 1/4$. Similarly, the probability of each of patterns 4 and 5 occurring is $1/4$.\n\nThere are exactly two random placements that give pattern 3: the one shown and its $90^\\circ$ rotation.\n\n![](images/2018-Australian-Scene-W2_p54_data_92f936e0b6.png)\n\n![](images/2018-Australian-Scene-W2_p54_data_a68dafbfe2.png)\n\nSo the probability of pattern 3 occurring is $2/16 = 1/8$.\n\nc) Since each tile has two orientations, the number of ways of placing nine tiles to form a $3 \\times 3$ square is $2^9 = 512$.\n\nEach end of a peanut is an arc of a circle that is greater than a semicircle. So three tiles that share a vertex are required to form each end. Hence there are only two ways in which a peanut can occur in a $3 \\times 3$ placement of tiles.\n\n![](images/2018-Australian-Scene-W2_p54_data_d24176e77c.png)\n\n![](images/2018-Australian-Scene-W2_p54_data_16e05cf7da.png)\n\nIn each case, the orientations of the 7 tiles that form the peanut are fixed but the 2 remaining tiles (shown blank) have 2 possible orientations. Hence the number of placements that contain a peanut is $2 \\times (2 \\times 2) = 8$. So the probability that a random $3 \\times 3$ placement contains a peanut is $8/512 = 1/64$.",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 14021,
    "subject": "Mathematics (Olympiad)",
    "question": "Find the minimum and maximum values of the function\n\n$$\nf = d_1 + d_2 + d_3\n$$\n\nwhere, for $i = 1, 2, 3$, each $d_i$ is defined by\n$$\nd_i^2 = 4 - 2 \\cos \\alpha_i - 2 \\sin \\alpha_{i+1} - 2 \\sin \\alpha_i \\cos \\alpha_{i+1},\n$$\nand $\\alpha_1, \\alpha_2, \\alpha_3$ are real numbers.",
    "options": [],
    "answer": "See solution",
    "solution": "When $\\alpha_i = \\frac{\\pi}{4}$ for $i = 1, 2, 3$, the minimum value of $f$ is $3\\sqrt{2} - 3$.\n\nWhen $\\alpha_i = \\pi$ for $i = 1, 2, 3$, the maximum value of $f$ is $3\\sqrt{6}$.\n\nIn conclusion, the minimum of $f$ is $3\\sqrt{2} - 3$, and its maximum is $3\\sqrt{6}$.\n\n$\\square$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 14022,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $n$ be an integer greater than $1$. Find the greatest constant $\\lambda(n)$ such that for any nonzero complex numbers $z_1, z_2, \\dots, z_n$, we have\n\n$$\n\\sum_{k=1}^{n} |z_k|^2 \\geq \\lambda(n) \\min_{1 \\leq k \\leq n} \\{|z_{k+1} - z_k|^2\\},\n$$\n\nwhere $z_{n+1} = z_1$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let\n$$\n\\lambda_0(n) = \\begin{cases} \\dfrac{n}{4}, & 2 \\mid n, \\\\ \\dfrac{n}{4 \\cos^2 \\dfrac{\\pi}{2n}}, & \\text{otherwise.} \\end{cases}\n$$\nWe prove $\\lambda_0(n)$ is the greatest constant.\n\nIf there exists $k$ ($1 \\leq k \\leq n$) such that $|z_{k+1} - z_k| = 0$, the inequality holds trivially. So, without loss of generality, assume\n$$\n\\min_{1 \\leq k \\leq n} \\{|z_{k+1} - z_k|^2\\} = 1. \\tag{1}\n$$\nUnder this condition, it suffices to show that the minimum value of $\\sum_{k=1}^{n} |z_k|^2$ is $\\lambda_0(n)$.\n\n**Case 1: $n$ even.**\n\nSince\n$$\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) \\\\ &\\geq \\frac{1}{4} \\sum_{k=1}^{n} |z_{k+1} - z_k|^2 \\\\ &\\geq \\frac{n}{4} \\min_{1 \\leq k \\leq n} \\{|z_{k+1} - z_k|^2\\} = \\frac{n}{4},\n\\end{aligned}\n$$\nand equality holds when $(z_1, z_2, \\dots, z_n) = (\\frac{1}{2}, -\\frac{1}{2}, \\dots, \\frac{1}{2}, -\\frac{1}{2})$, the minimum value is $\\frac{n}{4} = \\lambda_0(n)$.\n\n**Case 2: $n$ odd.**\n\nLet $\\theta_k = \\arg \\frac{z_{k+1}}{z_k} \\in [0, 2\\pi)$ for $k = 1, 2, \\dots, n$.\n\nIf $\\theta_k \\leq \\frac{\\pi}{2}$ or $\\theta_k \\geq \\frac{3\\pi}{2}$, then by (1),\n$$\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. \\tag{2}\n$$\nIf $\\theta_k \\in (\\frac{\\pi}{2}, \\frac{3\\pi}{2})$, then $\\cos \\theta_k < 0$ and\n$$\n\\begin{aligned}\n1 &\\leq |z_k - z_{k+1}|^2 \\\\ &= |z_k|^2 + |z_{k+1}|^2 - 2|z_k||z_{k+1}| \\cos \\theta_k \\\\ &\\leq (|z_k|^2 + |z_{k+1}|^2)(1 + (-2 \\cos \\theta_k)) \\\\ &= (|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}}. \\tag{3}\n$$\n\nNow consider two cases:\n\n1. If for all $k$, $\\theta_k \\in (\\frac{\\pi}{2}, \\frac{3\\pi}{2})$, then by (3),\n$$\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}}. \\tag{4}\n$$\nSince $\\prod_{k=1}^{n} \\frac{z_{k+1}}{z_k} = 1$, so\n$$\n\\sum_{k=1}^{n} \\theta_k = 2m\\pi, \\quad m \\in \\mathbb{Z}.\n$$\nFor $n$ odd, $0 < \\sin \\frac{m\\pi}{n} \\leq \\cos \\frac{\\pi}{2n}$.\n\nLet $f(x) = \\frac{1}{\\sin^2 x}$, which is convex on $[\\frac{\\pi}{4}, \\frac{3\\pi}{4}]$. By Jensen's inequality,\n$$\n\\sum_{k=1}^{n} |z_k|^2 \\geq \\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$$\n\n2. If there exists $j$ such that $\\theta_j \\notin (\\frac{\\pi}{2}, \\frac{3\\pi}{2})$, let\n$$\nI = \\{ j \\mid \\theta_j \\notin (\\frac{\\pi}{2}, \\frac{3\\pi}{2}),\\ 1 \\leq j \\leq n \\}.\n$$\nBy (2), for $j \\in I$, $|z_j|^2 + |z_{j+1}|^2 \\geq 1$; by (3), for $j \\notin I$, $|z_j|^2 + |z_{j+1}|^2 \\geq \\frac{1}{2}$.\n\nTherefore,\n$$\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$$\nNotice that\n$$\n\\frac{n+1}{4} \\geq \\frac{n}{4} \\cdot \\frac{1}{\\cos^2 \\frac{\\pi}{2n}} \\iff \\cos^2 \\frac{\\pi}{2n} \\geq \\frac{n}{n+1}.\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 14023,
    "subject": "Mathematics (Olympiad)",
    "question": "Consider the set of four-digit positive integers $x = \\overline{\\alpha\\beta\\gamma\\delta}$, where all digits are nonzero and pairwise different. Let $y = \\overline{\\delta\\gamma\\beta\\alpha}$, and suppose $x > y$. Find the greatest and the lowest value of the difference $x - y$, as well as the corresponding four-digit integers for which these values are obtained.",
    "options": [],
    "answer": "See solution",
    "solution": "Consider the decimal representations:\n\n$$\n\\begin{aligned}\nx - y &= 1000\\alpha + 100\\beta + 10\\gamma + \\delta - (1000\\delta + 100\\gamma + 10\\beta + \\alpha) \\\\\n&= 999(\\alpha - \\delta) + 90(\\beta - \\gamma) \\\\\n&= 9\\big(111(\\alpha - \\delta) + 10(\\beta - \\gamma)\\big).\n\\end{aligned}\n$$\n\nThus, it suffices to find the greatest and lowest values of\n\n$$\nA = 111(\\alpha - \\delta) + 10(\\beta - \\gamma),\n$$\n\nwhere $\\alpha, \\beta, \\gamma, \\delta$ are pairwise different nonzero digits and $\\alpha > \\delta$.\n\n**Maximum:**\n- $\\alpha - \\delta$ is maximized when $\\alpha = 9$, $\\delta = 1$ ($\\alpha - \\delta = 8$).\n- $\\beta - \\gamma$ is maximized when $\\beta = 8$, $\\gamma = 2$ ($\\beta - \\gamma = 6$), with all digits distinct.\n- So, $x = 9821$, $y = 1289$, and $x - y = 9821 - 1289 = 8532$.\n\n**Minimum:**\n- The minimal $\\alpha - \\delta = 1$, so possible $(\\alpha, \\delta)$ pairs are $(9,8), (8,7), (7,6), (6,5), (5,4), (4,3), (3,2), (2,1)$.\n- The minimal $\\beta - \\gamma = -8$, with $\\beta = 1$, $\\gamma = 9$, and all digits distinct.\n- For each valid $(\\alpha, \\delta)$, check if $1$ and $9$ are not used.\n\nThe following table shows possible values:\n\n![]( <table><tr><td>3192</td><td>2913</td><td>279</td></tr><tr><td>4193</td><td>3914</td><td>279</td></tr><tr><td>5194</td><td>4915</td><td>279</td></tr><tr><td>6195</td><td>5916</td><td>279</td></tr><tr><td>7196</td><td>6917</td><td>279</td></tr><tr><td>8197</td><td>7918</td><td>279</td></tr></table> )\n\nFor example, $x = 3192$, $y = 2913$, so $x - y = 279$.\n\n**Answer:**\n- The greatest value of $x - y$ is $8532$ for $x = 9821$, $y = 1289$.\n- The lowest value is $279$, for example $x = 3192$, $y = 2913$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 14024,
    "subject": "Mathematics (Olympiad)",
    "question": "100 couples are invited to a traditional Moldovan dance. The 200 people stand in a line, and in each step, two of them (not necessarily adjacent) may swap positions. Find the least $C$ such that, whatever the initial order, they can arrive at an ordering where everyone is dancing next to their partner in at most $C$ steps.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $N$ be the number of couples. The answer is $C(N) = N - 1$.\n\n**Base case ($N=2$):**\nThere is one trivial initial order and two non-trivial ones:\n$$\n1, 1, 2, 2; \\quad 1, 2, 2, 1; \\quad 1, 2, 1, 2.\n$$\nIn the second and third cases, one step is necessary to arrange couples together.\n\n**Upper bound:**\nWe show $C(N) \\leq N - 1$ by induction. The base case $N=2$ holds. Assume true for $N-1$. For $N$ couples, consider the left-most couple types $a$ and $b$. If $a \\neq b$, swap the $b$ in position two with the other $a$. If $a = b$, skip. Now, $N-1$ couples remain among $2N-2$ places, and $N-2$ steps suffice by induction. Thus, $N-1$ steps suffice for $N$ couples.\n\n**Lower bound:**\nWe exhibit an initial order requiring $N-1$ steps:\n$$\nA_N := 1, 2, 2, 3, 3, \\dots, N-1, N-1, N, N, 1.\n$$\nBy induction, each type must be involved in at least one step, and each step involves at most two types. By the pigeonhole principle, at least four types are involved in at most one step. For a type $a \\neq 1$, the step involving $a$ does not affect the order of the other $2N-2$ people. Ignoring this step, we have at most $N-3$ steps for the remaining people, contradicting the induction hypothesis. $\\square$\n\n**Alternative lower bound I:**\nConsider a graph with vertices as pairs of positions $\\{(1,2), (3,4), \\dots, (2N-1, 2N)\\}$. Add an edge between pairs if a swap occurs between those positions. In the final arrangement, each couple occupies a single vertex. Starting from $A_N$, the graph must be connected, requiring at least $N-1$ edges. $\\square$\n\n**Alternative lower bound II:**\nConsider a bipartite multigraph with vertex classes $(v_1, \\dots, v_n)$ and $(w_1, \\dots, w_n)$. Connect $v_i$ to $w_j$ if a person of type $j$ is in positions $(2i-1, 2i)$. Each swap replaces edges $\\{v_a \\leftrightarrow w_c, v_b \\leftrightarrow w_d\\}$ with $\\{v_a \\leftrightarrow w_d, v_b \\leftrightarrow w_c\\}$. Each step increases the number of connected components by at most 1. Starting from $A_N$, the graph is a single cyclic component, so at least $n-1$ steps are needed to reach $n$ components. $\\square$",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 14025,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all functions $f : [0, +\\infty) \\to [0, 1]$ such that for all $x \\ge 0$,\n$$\nf(x) f(x) = \\frac{1}{2} f(x f(x)).\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "Let $M = \\{x \\ge 0 \\mid f(x) = \\frac{1}{2}\\}$. Then $f(x) = \\frac{1}{2}$ for $x \\in M$, $f(x) = 0$ for $x \\notin M$, where $M \\subset [0, +\\infty)$ is any (possibly empty) subset such that $x \\in M \\Leftrightarrow 2x \\in M$ and $0 \\in M \\Leftrightarrow M = [0, +\\infty)$. Any such function $f$ satisfies the given functional equation.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 14026,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $a \\neq 0$ be a real number. Determine all functions $f : \\mathbb{R} \\to \\mathbb{R}$ satisfying\n$$\nf(x)f(y) + f(x+y) = axy\n$$\nfor all real numbers $x$ and $y$.",
    "options": [],
    "answer": "See solution",
    "solution": "Substituting $(x, y) = (0, 0)$ in (8) yields $f(0)^2 + f(0) = 0$; that is, $f(0) = 0$ or $f(0) = -1$. If $f(0) = 0$, then the substitution $y = 0$ in (8) yields $f(x) = 0$ for all $x \\in \\mathbb{R}$. However, the zero function does not satisfy (8), so we must have $f(0) = -1$.\n\nWe consider two cases regarding the value of $a$.\n\n*Case 1.* $a > 0$.\n\nLet $x_0 = \\frac{1}{\\sqrt{a}}$. Substituting $(x, y) = (x_0, -x_0)$ in (8) one obtains $f(x_0)f(-x_0) = 0$. If $f(x_0) = 0$, then the substitution $(x, y) = (x - x_0, x_0)$ in (8) yields\n$$\nf(x) = a x_0 (x - x_0) = \\sqrt{a}x - 1, \\quad \\text{for all } x \\in \\mathbb{R}.\n$$\nIf $f(-x_0) = 0$, then the substitution $(x, y) = (x + x_0, -x_0)$ in (8) yields\n$$\nf(x) = -a x_0 (x + x_0) = -\\sqrt{a}x - 1, \\quad \\text{for all } x \\in \\mathbb{R}.\n$$\nIt is not hard to verify that both functions satisfy (8).\n\n*Case 2.* $a < 0$.\n\nWe are going to show that no function $f$ satisfies (8). A substitution $y = x$ in (8) yields $f(x)^2 + f(2x) = a x^2$, for all $x \\in \\mathbb{R}$. That is, $f(2x) = a x^2 - f(x)^2$ and $f(-2x) = a x^2 - f(-x)^2$, and so\n$$\nf(2x)f(-2x) = a^2 x^4 - a x^2 (f(x)^2 + f(-x)^2) + (f(x)f(-x))^2.\n$$\nA substitution $y = -x$ in (8) yields $f(x)f(-x) = 1 - a x^2$ for all $x \\in \\mathbb{R}$. Thus,\n$$\n1 - 4a x^2 = a^2 x^4 - a x^2 (f(x)^2 + f(-x)^2) + (1 - a x^2)^2,\n$$\nor\n$$\nx^2 (f(x)^2 + f(-x)^2) = 2a x^4 + 2x^2 \\quad \\text{for all } x \\in \\mathbb{R}.\n$$\nLet $x = \\frac{2}{\\sqrt{-a}}$. We have $f(x)^2 + f(-x)^2 = 2a \\left(\\frac{2}{\\sqrt{-a}}\\right)^2 + 2 = -6$, which is impossible.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 14027,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $x, y, z \\in [0, 1]$. Prove that\n$$\n6xyz \\le x(1-x) + y(1-y) + z(1-z).\n$$\nDetermine when equality holds.",
    "options": [],
    "answer": "See solution",
    "solution": "By hypothesis, $x, y, z \\in [0, 1]$, and so $3xyz \\le xy + yz + zx$. Also, $(x+y+z)^2 \\le x+y+z$, i.e., $x^2 + y^2 + z^2 + 2(xy + yz + zx) \\le x + y + z$, equivalently,\n$$\n2(xy + yz + zx) \\le x(1-x) + y(1-y) + z(1-z),\n$$\nwhence\n$$\n6xyz \\le 2(xy + yz + zx) \\le x(1-x) + y(1-y) + z(1-z),\n$$\nand there is equality iff $xyz = xy = xz = yz$ and $x + y + z = 0$ or $1$. This means that we have equality iff $(x, y, z)$ is one of $(0, 0, 0)$, $(0, 0, 1)$, $(0, 1, 0)$, and $(1, 0, 0)$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 14028,
    "subject": "Mathematics (Olympiad)",
    "question": "Find the smallest positive multiple of $20$ with exactly $20$ positive divisors.",
    "options": [],
    "answer": "See solution",
    "solution": "Consider, for a positive integer $k$, the multiple $M_k = k \\times 20$. If $k = 2^{a_1} \\cdot 3^{a_2} \\cdot 5^{a_3} \\dots$ is the prime factorization of $k$, where $a_i \\ge 0$ for all $i \\ge 1$, then $M_k = 2^{a_1+2} \\cdot 3^{a_2} \\cdot 5^{a_3+1} \\cdot 7^{a_4} \\dots$ is the prime factorization of $M_k$.\n\nFor $M_k$ to have $20$ positive divisors, we need to have $$(a_1+3)(a_2+1)(a_3+2) \\times \\prod_{i \\ge 4} (a_i+1) = 20 = 2^2 \\cdot 5.$$ This forces $a_1+3 \\in \\{4, 5, 10\\}$ and $a_3+2 \\in \\{2, 4, 5\\}$. Also, since we want the smallest such $k$, we may assume that $k$ has no prime divisors larger than $5$. (Any $a_i + 1 = 2$, with $i \\ge 4$, can be replaced by $a_2 + 1 = 2$, resulting in a smaller $k$, but without changing the number of divisors of $M_k$.) Henceforth, we assume that $a_i = 0$ for all $i \\ge 4$. All the possibilities are summarised in the following table:\n\n![](table)\n\nWe conclude that the smallest positive multiple of $20$ with exactly $20$ positive divisors is $12 \\times 20 = 240$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 14029,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all real numbers $x$ for which the number\n$$\na = \\frac{2x+1}{x^2+2x+3}\n$$\nis an integer.",
    "options": [],
    "answer": "See solution",
    "solution": "Suppose the integer $a$ is not equal to $0$. Then $|2x+1| \\geq |x^2+2x+3|$. Notice that $x^2+2x+3 = (x+1)^2 + 2 > 0$, so we get $2x+1 \\geq x^2+2x+3$ or $2x+1 \\leq -(x^2+2x+3)$. This leads to $x^2+2 \\leq 0$ or $x^2+4x+4 \\leq 0$. The former gives no solutions, and the latter gives $x = -2$, and then $a = -1 \\in \\mathbb{Z}$.\n\nFor $a=0$ we get $2x+1=0$, so $x = -\\frac{1}{2}$. Hence $x \\in \\{-2,\\ -\\frac{1}{2}\\}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 14030,
    "subject": "Mathematics (Olympiad)",
    "question": "Suppose $\\log_a (2x^2 + x - 1) > \\log_a 2 - 1$. Then the range of $x$ is:\n\n(A) $\\frac{1}{2} < x < 1$\n\n(B) $x > \\frac{1}{2}$ and $x \\ne 1$\n\n(C) $x > 1$\n\n(D) $0 < x < 1$",
    "options": [],
    "answer": "See solution",
    "solution": "From\n$$\n\\begin{cases}\nx > 0, \\\\\nx \\ne 1, \\\\\n2x^2 + x - 1 > 0\n\\end{cases}\n$$\nwe get $x > \\frac{1}{2}$, $x \\ne 1$.\n\nFurthermore,\n\n$\\log_a (2x^2 + x - 1) > \\log_a 2 - 1 \\implies \\log_a (2x^3 + x^2 - x) > \\log_a 2$\n\nThis gives:\n$$\n\\begin{cases}\n0 < x < 1, \\\\\n2x^3 + x^2 - x < 2,\n\\end{cases}\n$$\nor\n$$\n\\begin{cases}\nx > 1, \\\\\n2x^3 + x^2 - x > 2.\n\\end{cases}\n$$\n\nThus, $x > \\frac{1}{2}$ and $x \\ne 1$.\n\n**Answer:** (B)",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 14031,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all integer triads $a, b, c$ with $a > 0 > b > c$ and $a + b + c = 0$ such that the number\n\n$$\nN = 2017 - a^3 b - b^3 c - c^3 a\n$$\n\nis a perfect square.",
    "options": [],
    "answer": "See solution",
    "solution": "Since $a + b + c = 0$, we have\n\n$$\n\\begin{aligned}\na^3 b + b^3 c + c^3 a &= a^3 b + b^3 (-a - b) + (-a - b)^3 a \\\\\n&= a^3 b - b^3 (a + b) - (a + b)^3 a \\\\\n&= a^3 b - a b^3 - b^4 - (a^3 + 3a^2 b + 3a b^2 + b^3) a \\\\\n&= a^3 b - a b^3 - b^4 - (a^4 + 3a^3 b + 3a^2 b^2 + a b^3) \\\\\n&= -b^4 - 2a^3 b - 3a^2 b^2 - 2a b^3 - a^4 \\\\\n&= - (a^2 + a b + b^2)^2\n\\end{aligned}\n$$\n\nTherefore, if $2017 - a^3 b - b^3 c - c^3 a = k^2$, then\n\n$$\n2017 + (a^2 + a b + b^2)^2 = k^2\n$$\nwhich gives\n$$\n(k - a^2 - a b - b^2)(k + a^2 + a b + b^2) = 2017\n$$\n\nSince $2017$ is prime, the only positive integer solutions are\n$$\n\\begin{cases}\nk - a^2 - a b - b^2 = 1 \\\\\nk + a^2 + a b + b^2 = 2017\n\\end{cases}\n$$\nwhich gives\n$$\n2k = 2018 \\implies k = 1009\n$$\nand\n$$\na^2 + a b + b^2 = 1008\n$$\n\nSince $a^2 + a b + b^2 = 1008$ and $1008$ is divisible by $9$, both $a$ and $b$ must be multiples of $3$. Also, since $a^2 + a b + b^2$ is even, $a$ and $b$ must be even, so $a$ and $b$ are multiples of $6$. Let $a = 6m$, $b = 6n$:\n\n$$\n(6m)^2 + (6m)(6n) + (6n)^2 = 1008 \\\\\n36m^2 + 36mn + 36n^2 = 1008 \\\\\nm^2 + m n + n^2 = 28\n$$\n\nNow, $m$ and $n$ must be even, so let $m = 2x$, $n = 2y$:\n\n$$\n(2x)^2 + (2x)(2y) + (2y)^2 = 28 \\\\\n4x^2 + 4xy + 4y^2 = 28 \\\\\nx^2 + x y + y^2 = 7\n$$\n\nWe seek integer solutions to $x^2 + x y + y^2 = 7$. Trying small integer values for $x$:\n- $x = 1$: $1 + y + y^2 = 7 \\implies y^2 + y - 6 = 0 \\implies y = 2, -3$\n- $x = 2$: $4 + 2y + y^2 = 7 \\implies y^2 + 2y - 3 = 0 \\implies y = 1, -3$\n- $x = 3$: $9 + 3y + y^2 = 7 \\implies y^2 + 3y + 2 = 0 \\implies y = -1, -2$\n\nSo possible $(x, y)$ pairs are $(1, -3)$, $(2, -3)$, $(3, -2)$, $(3, -1)$. Since $a = 12x > 0$ and $b = 12y < 0$, only those with $x > 0$ and $y < 0$ are valid. Thus, possible $(a, b)$ are $(12, -36)$, $(24, -36)$, $(36, -24)$, $(36, -12)$.\n\nWith $a + b + c = 0$, $c = -a - b$. The only triple with $a > 0 > b > c$ is $a = 36$, $b = -12$, $c = -24$.\n\n**Answer:** The unique solution is $(a, b, c) = (36, -12, -24)$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 14032,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $\\deg(n) = \\alpha_1 + \\alpha_2 + \\dots + \\alpha_k$ denote the degree of the number $n = p_1^{\\alpha_1} p_2^{\\alpha_2} \\dots p_k^{\\alpha_k}$, where $p_i$ are pairwise distinct prime numbers and $\\alpha_1, \\alpha_2, \\dots, \\alpha_k$ are positive integers.\n\nProve that there exist $2016$ consecutive positive integers among which there are precisely $1000$ numbers with degree less than $11$.",
    "options": [],
    "answer": "See solution",
    "solution": "First, let us prove the following lemma.\n\n**Lemma.** There are $l$ consecutive positive integers among which there are no numbers with degree less than $t$ ($t \\ge 2$).\n\n*Proof.* We proceed by induction on $t$.\n\n**Base case ($t=2$):** The statement is equivalent to the existence of $l$ consecutive positive integers, none of which is prime. It suffices to take $$(l+1)!+2,\\ (l+1)!+3,\\ \\dots,\\ (l+1)!+(l+1).$$\n\n**Inductive step:** Assume the statement holds for $2, 3, \\dots, t-1$. For $t$, suppose for $t-1$ there exist $x+1, \\dots, x+l$ with degree at least $t-1$. Consider $$(x+l)!+x+1,\\ (x+l)!+x+2,\\ \\dots,\\ (x+l)!+x+l.$$ Each of these has degree at least $t$, and there are $l$ of them. Thus, the lemma is proved.\n\nNote that $2^{11} = 2048 > 2016$. So among the first $2016$ positive integers, all numbers have degree less than $11$.\n\nLet $w(x)$ be the number of integers with degree less than $11$ among $x+1, \\dots, x+2016$. Thus, $w(0) = 2016$. We have also shown there exist $2016$ consecutive positive integers with no numbers of degree less than $11$, so for some $y$, $w(y) = 0$.\n\nIt is clear that $w(z)-1 \\leq w(z+1) \\leq w(z)+1$. Since $w(z)$ takes integer values between $0$ and $2016$, it must take all intermediate values, including $1000$.\n\nTherefore, there exist $2016$ consecutive positive integers among which exactly $1000$ have degree less than $11$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 14033,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $a$ be a given natural number. Consider the sequence $(x_n)_{n \\ge 1}$ defined by $x_n = \\frac{1}{1 + n a}$ for every nonzero natural number $n$.\n\nProve that for every natural number $k \\ge 3$, there exist nonzero natural numbers $n_1 < n_2 < \\dots < n_k$ such that the numbers $x_{n_1}, x_{n_2}, \\dots, x_{n_k}$ are consecutive terms of an arithmetic progression.",
    "options": [],
    "answer": "See solution",
    "solution": "We will prove the statement by induction on $k$.\n\nFirst, note that\n\n$$\n\\frac{1}{1 + m a} + \\frac{1}{(1 + m a)(1 + 2 m a)} = \\frac{2}{1 + 2 m a},\n$$\n\nso there is an arithmetic progression consisting of three terms of the sequence: $x_m$, $x_{2m}$, $x_{2m^2 a + 3m}$.\n\nAssume there exist $k$ nonzero natural numbers $n_1 < n_2 < \\dots < n_k$ such that $x_{n_1}, x_{n_2}, \\dots, x_{n_k}$ are consecutive terms of an arithmetic progression. Consider $y = 2 x_{n_1} - x_{n_2}$. Then $y, x_{n_1}, x_{n_2}, \\dots, x_{n_k}$ are in arithmetic progression (with $k+1$ terms).\n\nWe have:\n\n$$\ny = \\frac{2}{1 + n_1 a} - \\frac{1}{1 + n_2 a} = \\frac{1 + p a}{(1 + n_1 a)(1 + n_2 a)} = \\frac{1 + p a}{1 + (n_1 + n_2 + n_1 n_2 a) a},\n$$\n\nwhere $p = 2 n_2 - n_1$.\n\nTherefore,\n\n$$\n\\frac{y}{1 + p a}, \\frac{x_{n_1}}{1 + p a}, \\frac{x_{n_2}}{1 + p a}, \\dots, \\frac{x_{n_k}}{1 + p a}\n$$\n\nare $k+1$ terms $x_{m_1}, x_{m_2}, \\dots, x_{m_{k+1}}$ of the sequence $(x_n)_{n \\ge 1}$.\n\nThey are in arithmetic progression, because dividing the terms of an arithmetic progression by a nonzero real number preserves the progression.\n\nMoreover, $m_1 < m_2 < \\dots < m_{k+1}$, since the sequence $(x_n)_{n \\ge 1}$ is strictly monotone. This completes the proof.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 14034,
    "subject": "Mathematics (Olympiad)",
    "question": "Determine all functions $f: \\mathbb{C} \\to \\mathbb{C}$ such that $|w f(z) + z f(w)| = 2|z w|$ for any $z, w \\in \\mathbb{C}$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $z = 1$ and $w = 0$. Then $|0 \\cdot f(1) + 1 \\cdot f(0)| = 2|0|$, so $|f(0)| = 0$, which gives $f(0) = 0$.\n\nFor $w = z \\in \\mathbb{C}^*$, we have $|z f(z) + z f(z)| = 2|z^2|$, so $|2 z f(z)| = 2|z|^2$, which gives $|f(z)| = |z|$ for all $z \\in \\mathbb{C}$. This also holds for $z = 0$.\n\nNow, $|f(1)| = 1$. For $w = 1$, the equation becomes $|f(z) + z f(1)| = 2|z|$. By the triangle inequality, $|f(z) + z f(1)| \\leq |f(z)| + |z f(1)| = |z| + |z| = 2|z|$, so equality holds. This means $f(z)$ and $z f(1)$ are positively proportional, i.e., $f(z) = t_z f(1) z$ for some $t_z \\geq 0$.\n\nTaking moduli, $|f(z)| = t_z |f(1)| |z| = t_z |z|$, but $|f(z)| = |z|$, so $t_z = 1$. Thus, $f(z) = f(1) z$ for all $z \\in \\mathbb{C}$, where $|f(1)| = 1$.\n\nTherefore, all solutions are $f(z) = c z$ for $c \\in \\mathbb{C}$ with $|c| = 1$. Any such $f$ satisfies the original equation.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 14035,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all positive integers $x$ and $y$ such that the number $(x^2 + y)(y^2 + x)$ is the fifth power of a prime.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $(x^2 + y)(y^2 + x) = p^5$, where $p$ is a prime. Then $x^2 + y = p^s$, $y^2 + x = p^t$, where $\\{s, t\\} = \\{1, 4\\}$ or $\\{2, 3\\}$. In the first case, we can assume without loss of generality that $x < y$, $x^2 + y = p$ and $y^2 + x = p^4$. Then $p^2 = (x^2 + y)^2 > x + y^2 = p^4$, a contradiction.\n\nLet $x < y$, $x^2 + y = p^2$ and $y^2 + x = p^3$. Note that $p > x$. We have $p^2 \\mid (x^2 + y)(x^2 - y) + (y^2 + x) = x^4 + x = x(x + 1)(x^2 - x + 1)$ and since $p > x$ we see that $p^2$ divides $(x + 1)(x^2 - x + 1)$. We consider two cases.\n\n*Case 1.* If $p \\mid x+1$ then $p = x+1$ and we easily find the solution $x = 2, y = 5$.\n\n*Case 2.* If $p \\nmid x+1$ then $p^2 \\mid x^2-x+1$ and now $p^2 \\mid x^2+y = (x^2-x+1)+(x+y-1)$ implies that $p^2$ divides $x+y-1$. Hence $y \\ge p^2-x+1 > p^2-p$ and $p^3 = y^2+x > p^2(p-1)^2$ which is impossible.\n\nFinally, the solutions are $(2, 5)$ and $(5, 2)$.\n\n*Remark.* It can be proved that $(x^2 + y)(y^2 + x)$ is a prime power only for $(x, y) = (1, 1)$, $(2, 5)$ and $(5, 2)$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 14036,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $ABCDE$ be a convex pentagon inscribed in a circle $\\omega$ such that $CD \\parallel BE$. The line tangent to $\\omega$ at $B$ intersects the line $AC$ at a point $F$ such that $A$ lies between the points $C$ and $F$. The lines $BD$ and $AE$ intersect at $G$. Prove that the line $FG$ is tangent to the circumcircle of $ADG$.",
    "options": [],
    "answer": "See solution",
    "solution": "Since $CD \\parallel BE$, we have $BC = DE$. Note that\n\n$$\n\\angle BFC = \\angle BAC - \\angle ABF = \\angle DBE - \\angle AEB = \\angle BGA,\n$$\n\nhence $ABGF$ is cyclic. It follows that $\\angle AGF = \\angle ABF = \\angle ADG$, hence the circumcircle of $ADG$ is tangent to $FG$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 14037,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $a, b, c \\ge 0$ satisfy $a^2 + b^2 + c^2 + abc = 4$. Prove that\n\n$$\nab + bc + ca - abc \\le 2.$$",
    "options": [],
    "answer": "See solution",
    "solution": "Likewise,\n\n$$\nbc \\le \\sin B \\cot \\frac{B+A}{2} + \\sin C \\cot \\frac{C+A}{2},\n$$\n\n$$\nca \\le \\sin C \\cot \\frac{C+B}{2} + \\sin A \\cot \\frac{A+B}{2}.\n$$\n\nTherefore, applying the **Sum-to-product**, **Product-to-sum**, and **Double-angle formulas**, we have\n\n$$\n\\begin{aligned}\nab + bc + ca &\\le (\\sin A + \\sin B) \\cot \\frac{A+B}{2} + (\\sin B + \\sin C) \\cot \\frac{B+C}{2} \\\\\n&\\quad + (\\sin C + \\sin A) \\cot \\frac{C+A}{2} \\\\\n&= 2 \\cos \\frac{A-B}{2} \\cos \\frac{A+B}{2} + 2 \\cos \\frac{B-C}{2} \\cos \\frac{B+C}{2} \\\\\n&\\quad + 2 \\cos \\frac{C-A}{2} \\cos \\frac{C+A}{2} \\\\\n&= 2(\\cos A + \\cos B + \\cos C) \\\\\n&= 6 - 4 \\left( \\sin^2 \\frac{A}{2} + \\sin^2 \\frac{B}{2} + \\sin^2 \\frac{C}{2} \\right) \\\\\n&= 6 - (a^2 + b^2 + c^2).\n\\end{aligned}\n$$\n\nUsing the given equality, this last quantity equals $2 + abc$. It follows that\n\n$$\nab + bc + ca \\le 2 + abc,\n$$\n\nas desired.\n\n**Fifth Solution.** Let $\\sqrt[3]{abc} = x$. The **AM-GM Inequality** yields\n\n$$\n4 = a^2 + b^2 + c^2 + abc \\ge 3\\sqrt{a^2b^2c^2} + abc\n$$\n\nimplying that $x \\le 1$.\n\nBy the AM-GM Inequality, $ab + bc + ca \\ge 3x^2$. Hence,\n\n$$\nab + bc + ca - abc \\ge 3x^2 - x^3 = x^2(3-x) \\ge 0,\n$$\n\nproving the lower bound.\n\nNow we prove the upper bound. Clearly $0 \\le a, b, c \\le 2$. Note that in a triangle $ABC$, we can apply the **Product-to-sum formulas** twice and the **Half-angle formulas** to find that\n\n$$\n\\begin{aligned}\n2 \\cos A \\cos B \\cos C &= [\\cos (A+B) + \\cos (A-B)] \\cos C \\\\\n&= -\\cos^2 C - \\cos (A-B) \\cos (A+B) \\\\\n&= -\\frac{\\cos 2A + \\cos 2B}{2} - \\cos^2 C \\\\\n&= - (\\cos^2 A + \\cos^2 B + \\cos^2 C) + 1.\n\\end{aligned}\n$$\n\nHence,\n\n$$\n\\cos^2 A + \\cos^2 B + \\cos^2 C + 2 \\cos A \\cos B \\cos C = 1.\n$$\n\nFrom the given equality, we have $4 \\ge a^2, 4 \\ge b^2$, and thus we may set $a = 2 \\cos A, b = 2 \\cos B$ where $0^\\circ \\le A, B \\le 90^\\circ$. Because $a^2 + b^2 + c^2 + abc$ is an increasing function of $c$, there is at most one nonnegative value $c$ such that the given equality holds. Dividing the given equality by 4 and substituting, we obtain\n\n$$\n\\cos^2 A + \\cos^2 B + \\left(\\frac{c}{2}\\right)^2 + 2 \\cos A \\cos B \\left(\\frac{c}{2}\\right) = 1.\n$$\n\nViewed as a quadratic in $c/2$, we know that one solution to this equation is $c/2 = \\cos C$, where $C = 180^\\circ - A - B$. Because $\\cos^2 A + \\cos^2 B = \\frac{1}{4}(a^2 + b^2) \\le 1$, we know that $A + B \\ge 90^\\circ$. Thus, $C \\le 90^\\circ$ and $\\cos C \\ge 0$. Therefore, we must have $c = 2 \\cos C$. Either two of $A, B, C$ are at least $60^\\circ$ or two of $A, B, C$ are at most $60^\\circ$. Without loss of generality, assume that $A$ and $B$ have this property.\n\nWith these trigonometric substitutions, we find that the desired inequality is equivalent to\n\n$$\n\\begin{aligned}\n& 2(\\cos A \\cos B + \\cos B \\cos C + \\cos C \\cos A) \\\\\n& \\le 1 + 4 \\cos A \\cos B \\cos C,\n\\end{aligned}\n$$\n\nor\n\n$$\n\\begin{aligned}\n& 2(\\cos A \\cos B + \\cos B \\cos C + \\cos C \\cos A) \\\\\n& \\le 3 - 2(\\cos^2 A + \\cos^2 B + \\cos^2 C).\n\\end{aligned}\n$$\n\nUsing the **Double-angle formulas**, the last inequality is equivalent to\n\n$$\n\\begin{aligned}\n& \\cos 2A + \\cos 2B + \\cos 2C \\\\\n& + 2(\\cos A \\cos B + 2 \\cos B \\cos C + \\cos C \\cos A) \\le 0.\n\\end{aligned}\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 14038,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $|BC| = a$, $|AD| = b$, $|AB| = |CD| = c$. Given $b^2 = ac$, prove that $\\angle CBD = 30^\\circ$.",
    "options": [],
    "answer": "See solution",
    "solution": "$$\n\\frac{a^2 + c^2 - (b + c)^2}{2ac} = \\cos(90^\\circ + \\angle CBD) = -\\sin(\\angle CBD).\n$$\n\nHence,\n$$\n\\sin(\\angle BCD) = \\frac{b^2 + 2bc - a^2}{2ac}.\n$$\n\nAlso,\n$$\n\\sin(\\angle CBD) = \\frac{c}{a} \\sin(\\angle BDC) = \\frac{c}{a} \\sin(\\angle BDA) = \\frac{c}{a} \\cdot \\frac{c}{b}.\n$$\n\nThus,\n$$\n\\frac{b^2 + 2bc - a^2}{2ac} = \\frac{c^2}{ab}.\n$$\n\nSo,\n$$\n b^3 + 2b^2c - a^2b - 2c^3 = 0.\n$$\n\nUsing $b^2 = ac$, we get $(ab - 2c^2)(c - a) = 0$.\n\nIf $a = c$, then $|AC| = b + c > c + c$, which contradicts the triangle inequality. Hence $ab = 2c^2$ and from previous calculations $\\sin(\\angle CBD) = \\frac{c^2}{ab} = \\frac{1}{2}$. Thus $\\angle CBD = 30^\\circ$.\n\nConversely, $\\sin(\\angle CBD) = \\frac{b^2 + 2bc - a^2}{2ac} = \\frac{1}{2}$ implies $b^2 + 2bc - a^2 - ac = 0$.\n\nAlso, $\\sin(\\angle CBD) = \\frac{c^2}{ab} = \\frac{1}{2}$ implies $ab = 2c^2$. Substituting for $a$ we obtain $b^2 + 2bc - \\frac{4c^4}{b^2} - \\frac{2c^3}{b} = 0$. Factorising, we obtain $(b^3 - 2c^3)(b + 2c) = 0$. As $b + 2c > 0$, $b^3 = 2c^3 = 2abc$. Hence $b^2 = ac$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 14039,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all positive integers $m$ and $n$ such that\n$$\n\\left\\lfloor \\frac{(2m+1)^n}{m(3m+1)} \\right\\rfloor = 2013.\n$$\nWhat is the sum of all such $m$?",
    "options": [],
    "answer": "See solution",
    "solution": "* For $n \\leq 2$, we have $\\frac{(2m+1)^n}{m(3m+1)} < \\frac{(3m)^2}{m(3m)} = 3 < 2013$.\n\n* For $n = 3$, we have\n$$\n\\frac{(2m+1)^3}{m(3m+1)} = \\frac{8m^3 + 12m^2 + 6m + 1}{3m^2 + m} = \\frac{8}{3}m + \\frac{28}{9} + \\frac{26m+9}{9(3m^2 + m)}.\n$$\nClearly, the last fraction is less than $1$. So we need $2012 < \\frac{8}{3}m + \\frac{28}{9} < 2014$.\n\nThe only integer solution is $m = 754$. We check that\n$$\n\\begin{aligned}\n& \\left[ \\frac{8}{3}(754) + \\frac{28}{9} + \\frac{26(754) + 9}{9(3(754)^2 + (754))} \\right] \\\\\n&= \\left[ 2013 + \\frac{7}{9} + \\frac{26(754) + 9}{9(3(754)^2 + (754))} \\right] \\\\\n&= 2013\n\\end{aligned}\n$$\nsince $\\frac{26(754) + 9}{9(3(754)^2 + (754))} < \\frac{27(754)}{27(754)^2} < \\frac{2}{9}$. So $m = 754$ is a solution.\n\n* For $n \\geq 4$, we claim that $f(m) = \\frac{(2m+1)^n}{m(3m+1)}$ is increasing in $m$. Note that\n$$\n\\ln f(m) = n \\ln(2m+1) - \\ln m - \\ln(3m+1),\n$$\nand so\n$$\n\\frac{f'(m)}{f(m)} = \\frac{2n}{2m+1} - \\frac{1}{m} - \\frac{3}{3m+1} \\geq \\frac{8}{2m+1} - \\frac{3}{2m+1} - \\frac{3}{2m+1} > 0.\n$$\nThus, $f'(m) > 0$, so that $f$ is increasing.\n\nNow, for $n = 4$, we check that\n$$\n\\frac{(2 \\cdot 18 + 1)^4}{18(3 \\cdot 18 + 1)} = \\frac{37^4}{18 \\cdot 55} = \\frac{1369^2}{990} < 2013\n$$\nand\n$$\n\\frac{(2 \\cdot 19 + 1)^4}{19(3 \\cdot 19 + 1)} = \\frac{39^4}{19 \\cdot 58} = \\frac{1521^2}{1102} > 2014.\n$$\nThere is no solution.\n\nFor $n = 5$, we check that\n$$\n\\frac{(2 \\cdot 4 + 1)^5}{4(3 \\cdot 4 + 1)} = \\frac{9^5}{4 \\cdot 13} = \\frac{59049}{52} < 2013,\n$$\n$$\n\\left[ \\frac{(2 \\cdot 5 + 1)^5}{5(3 \\cdot 5 + 1)} \\right] = \\left[ \\frac{11^5}{5 \\cdot 16} \\right] = \\left[ \\frac{161051}{80} \\right] = 2013\n$$\nand\n$$\n\\frac{(2 \\cdot 6 + 1)^5}{6(3 \\cdot 6 + 1)} = \\frac{13^5}{6 \\cdot 19} = \\frac{371293}{114} > 2014.\n$$\nThus, $m = 5$ is another solution.\n\nTo summarize, the answer is $754 + 5 = 759$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 14040,
    "subject": "Mathematics (Olympiad)",
    "question": "Milou has 100 long envelopes of different sizes. Each envelope has a width equal to one of the integers $21, \\ldots, 30$ and a height equal to one of the integers $11, \\ldots, 20$, and each combination occurs exactly once. Milou wants to organise the envelopes into piles. An envelope may only be placed on top of another envelope if both its width and height are smaller than that of the envelope she is placing it on. So the size $26 \\times 15$ envelope is allowed on top of the $29 \\times 17$ envelope, but not on the $29 \\times 15$ envelope or the $26 \\times 17$ envelope.\n\nWhat is the smallest number of piles into which Milou can organise the envelopes?",
    "options": [],
    "answer": "See solution",
    "solution": "$19$",
    "topic": "Discrete Mathematics",
    "subtopic": "Combinatorics"
  },
  {
    "id": 14041,
    "subject": "Mathematics (Olympiad)",
    "question": "Numbers $a$, $b$, $c$, $d$ satisfy: $ab + cd > 0$, $ac + bd > 0$, and $a^2 + d^2 = c^2 + b^2$. Prove that $ad + bc > 0$.",
    "options": [],
    "answer": "See solution",
    "solution": "**Solution.** The product of two positive numbers is positive, so\n\n$$\n(ab + cd)(ac + bd) = a^2bc + ac^2d + ab^2d + bcd^2 = bc(a^2 + d^2) + ad(c^2 + b^2) = bc(a^2 + d^2) + ad(a^2 + d^2) = (a^2 + d^2)(bc + ad) > 0,\n$$\n\nwhich implies the required inequality.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 14042,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $\\triangle ABC$ be an acute triangle, and let $X$ be a variable interior point on the minor arc $BC$ of its circumcircle. Let $P$ and $Q$ be the feet of the perpendiculars from $X$ to lines $CA$ and $CB$, respectively. Let $R$ be the intersection of line $PQ$ and the perpendicular from $B$ to $AC$. Let $\\ell$ be the line through $P$ parallel to $XR$. Prove that as $X$ varies along minor arc $BC$, the line $\\ell$ always passes through a fixed point. Specifically, prove that there is a point $F$, determined by triangle $ABC$, such that no matter where $X$ is on arc $BC$, line $\\ell$ passes through $F$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $H$ denote the orthocenter of $\\triangle ABC$. We claim that $\\ell$ always passes through $H$.\n\n**Lemma.** Line $PQ$ bisects segment $XH$.\n\n*Proof.* Let $X_A, X_B$ be the reflections of $X$ across $BC$ and $AC$ respectively, and let $H_A$ be the reflection of $H$ across $BC$. It is easy to see that since $\\angle BH_A C = \\angle BHC = 180^\\circ - \\angle BAC$, $H_A$ is on the circumcircle of $\\triangle ABC$. It suffices to show that $H$ is on $X_A X_B$. Since $C$ is the circumcenter of $XX_A X_B$, we have $\\angle XX_A X_B = \\frac{1}{2} \\angle XCX_B = \\angle ACX$. On the other hand, $HH_A X_A X$ is an isosceles trapezoid, so $\\angle HX_A X = \\angle HH_A X = \\angle AH_A X = \\angle ACX = \\angle XX_A X_B$ so it follows that $X_B, H, X_A$ are collinear.\n\n![](images/USA_IMO_2013-2014_p65_data_ae722ae9d7.png)\n\nWe know that lines $HBR$ and $XP$ are both perpendicular to $AC$, so it follows that $HR \\parallel XP$. But by the lemma, line $PQR$ bisects $HX$ so it follows that $PXRH$ is a parallelogram. Thus, $PH \\parallel XR$ and thus $H$ is on $\\ell$, as desired.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 14043,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $p$ be a prime number and $x_1, \\ldots, x_r$ be pairwise distinct modulo $p$, and $y_1, \\ldots, y_r$ be distinct positive integers. Show that there exists a polynomial $P(x)$ with integer coefficients such that\n\n$$\nP(x_i) \\equiv y_i \\pmod{p}\n$$\n\nAdditionally, if $n$ is an arbitrary integer and $\\gcd(x_i - x_j, n) = 1$, the result remains true. If $A$ and $B$ are two arbitrary sets with pairwise distinct elements modulo $p$ and $|A| \\geq |B|$, can $A$ always be translatable to $B$? Further, let $n > 4$ be a composite number and $p$ its smallest prime divisor, with $n > 3p$. Consider $A = \\{1, p+1, 2p+1\\}$ and $B = \\{1, 2\\}$. Is $B$ translatable to $A$?",
    "options": [],
    "answer": "See solution",
    "solution": "By the Lagrange Interpolation Formula, we can construct a polynomial\n\n$$\nP(x) = \\sum_{i=1}^r a_i \\cdot \\prod_{j \\neq i} \\frac{x - x_j}{x_i - x_j}\n$$\n\nwhere $a_i = y_i$ and all fractions are interpreted as multiplicative inverses modulo $p$. This polynomial satisfies $P(x_i) \\equiv y_i \\pmod{p}$ for each $i$.\n\nIf $n$ is arbitrary and $\\gcd(x_i - x_j, n) = 1$, the same construction works. Thus, for sets $A$ and $B$ as described, $A$ can be translatable to $B$ if $|A| \\geq |B|$ and the elements are pairwise distinct modulo $p$.\n\nFor $n > 4$ composite, $p$ its smallest prime divisor, $n > 3p$, $A = \\{1, p+1, 2p+1\\}$, $B = \\{1, 2\\}$: Suppose $B$ is translatable to $A$. Then there exist $R, S \\in A$ such that $P(R) \\equiv 1$, $P(S) \\equiv 2 \\pmod{p}$. But $R \\equiv S \\pmod{p}$, so $P(R) \\equiv P(S) \\pmod{p}$, a contradiction. Thus, $B$ cannot be translatable to $A$ in this case.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 14044,
    "subject": "Mathematics (Olympiad)",
    "question": "Prove that\n$$\n\\frac{a}{\\sqrt{a^2 + 8bc}} + \\frac{b}{\\sqrt{b^2 + 8ca}} + \\frac{c}{\\sqrt{c^2 + 8ab}} \\geq 1\n$$\nfor all positive real numbers $a$, $b$, and $c$.",
    "options": [],
    "answer": "See solution",
    "solution": "Given a function $f$ of three variables, define\n$$\n\\sum_{\\text{cyc}} f(p, q, r) = f(p, q, r) + f(q, r, p) + f(r, p, q)\n$$\nand\n$$\n\\prod_{\\text{cyc}} f(p, q, r) = f(p, q, r) f(q, r, p) f(r, p, q).\n$$\nWe call these expressions a *cyclic sum* and *cyclic product*, respectively.\n\nThe desired inequality can be restated in this notation as\n$$\n\\sum_{\\text{cyc}} \\frac{a}{\\sqrt{a^2 + 8bc}} \\geq 1.\n$$\n\n**First Solution.** (By Liang Xiao, China) Setting $x = \\frac{a}{b}$, $y = \\frac{b}{c}$, $z = \\frac{c}{a}$ gives\n$$\n\\frac{a}{\\sqrt{a^2 + 8bc}} = \\frac{1}{\\sqrt{1 + 8 \\cdot \\frac{bc}{a^2}}} = \\frac{1}{\\sqrt{1 + 8 \\cdot \\frac{z}{x}}} = \\frac{\\sqrt{x}}{\\sqrt{x + 8z}}.\n$$\nIt thus suffices to prove that\n$$\n\\sum_{\\text{cyc}} \\frac{\\sqrt{x}}{\\sqrt{x+8z}} = \\frac{\\sqrt{x}}{\\sqrt{x+8z}} + \\frac{\\sqrt{y}}{\\sqrt{y+8x}} + \\frac{\\sqrt{z}}{\\sqrt{z+8y}} \\ge 1, \\quad (1)\n$$\nfor positive real numbers $x, y, z$ with $xyz = 1$.\n\nClearing the denominators of (1) yields\n$$\n\\sum_{\\text{cyc}} \\sqrt{x(y + 8x)(z + 8y)} \\ge \\sqrt{(x + 8z)(y + 8x)(z + 8y)},\n$$\nor\n$$\n\\left( \\sum_{\\text{cyc}} \\sqrt{x(y + 8x)(z + 8y)} \\right)^2 \\ge (x + 8z)(y + 8x)(z + 8y). \\quad (2)\n$$\nBecause $xyz = 1$, the right-hand side of (2) is equal to\n$$\n513 + 8(xy^2 + yz^2 + zx^2) + 64(x^2y + y^2z + z^2x).\n$$\nOn the other hand, the left-hand side of (2) is equal to\n$$\n\\begin{aligned}\n& \\left( \\sum_{\\text{cyc}} \\sqrt{x(y+8x)(z+8y)} \\right)^2 \\\\\n&= \\left( \\sum_{\\text{cyc}} \\sqrt{1 + 8xy^2 + 8x^2z + 64x^2y} \\right)^2 \\\\\n&= \\sum_{\\text{cyc}} (1 + 8xy^2 + 8x^2z + 64x^2y) + 2 \\sum_{\\text{cyc}} g(x, y, z) \\\\\n&= 3 + 8(xy^2 + yz^2 + zx^2) + 64(x^2y + y^2z + z^2x) \\\\\n&\\quad + 8 \\sum_{\\text{cyc}} x^2z + 2 \\sum_{\\text{cyc}} g(x, y, z),\n\\end{aligned}\n$$\nwhere $g(x, y, z)$ equals\n$$\n\\sqrt{(1 + 8xy^2 + 8x^2z + 64x^2y)(1 + 8yz^2 + 8y^2x + 64y^2z)}.\n$$\nComparing the left-hand side and the right-hand side, we see that (2) is equivalent to\n$$\n8 \\sum_{\\text{cyc}} x^2z + 2 \\sum_{\\text{cyc}} g(x, y, z) \\geq 510.\n$$\nBecause $xyz = 1$, the AM-GM Inequality gives\n$$\n\\sum_{\\text{cyc}} 8x^2z = 8(x^2z + y^2x + z^2y) \\geq 24\\sqrt[3]{x^2z \\cdot y^2x \\cdot z^2y} = 24.\n$$\nHence, it suffices to prove that\n$$\n2 \\sum_{\\text{cyc}} g(x, y, z) \\geq 486,\n$$\nor\n$$\n\\sum_{\\text{cyc}} g(x, y, z) \\geq 243. \\qquad (3)\n$$\nBy the Weighted AM-GM Inequality, we have\n$$\n\\begin{aligned}\n\\frac{1 + 8xy^2 + 8x^2z + 64x^2y}{1 + 8 + 8 + 64} &\\geq \\sqrt[81]{(xy^2)^8 \\cdot (x^2z)^8 \\cdot (x^2y)^{64}} \\\\\n&= \\sqrt[81]{x^{152}y^{80}z^8},\n\\end{aligned}\n$$\nor\n$$\n1 + 8xy^2 + 8x^2z + 64x^2y \\geq 81 \\sqrt[81]{x^{152}y^{80}z^8}.\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 14045,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all pairs of prime numbers $p$ and $q$ that satisfy the equation\n\n$$3p^q - 2q^{p-1} = 19.$$",
    "options": [],
    "answer": "See solution",
    "solution": "It is clear that $p \\geq 3$. If $p = q$, then we have $(3p^{p-2} - 2p^{p-3})p^2 = 19$, which is impossible whenever $p \\geq 3$ is prime.\n\nSuppose that $p \\neq q$. By Fermat's Little Theorem, $q^{p-1} - 1 \\equiv 0 \\pmod{p}$ and $p^{q-1} - 1 \\equiv 0 \\pmod{q}$.\n\nWe rewrite the equation as $3p^q - 2(q^{p-1} - 1) = 21$. Since the left-hand side is divisible by $p$, we have $p = 3$ or $p = 7$. Then, writing $3p(p^{q-1} - 1) - 2q^{p-1} = 19 - 3p$, we see that $19 - 3p$ is divisible by $q$.\n\n**Answer:** $p = 3,\\ q = 2$; $p = 7,\\ q = 2$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 14046,
    "subject": "Mathematics (Olympiad)",
    "question": "Consider the following table:\n\n$$\na_{ij} = \\begin{cases} 2 & i \\equiv 1 \\pmod{8} \\\\ 5 & i \\equiv 4, 5 \\pmod{8} \\\\ 4 & i \\equiv 0 \\pmod{8} \\text{ or } i \\equiv 7 \\pmod{8} \\text{ and } j \\not\\equiv 2, 5 \\pmod{6} \\\\ 1 & i \\equiv 3 \\pmod{8} \\text{ and } j \\not\\equiv 2, 5 \\pmod{6} \\\\ 3 & \\text{otherwise.} \\end{cases}\n$$\n\nwhere $a_{ij}$ is the number written in the cell $(i, j)$. What is the maximum possible number that can appear in this table?",
    "options": [],
    "answer": "See solution",
    "solution": "To prove that $5$ is the maximum, note that if there is one $8$ in this table, all other cells should be $8$. If there is a $7$ in this table, then it has a neighbor $a$ which is not $7$, and $a$ should have a neighbor $b$ which is not $7$. Now none of the common neighbors of $a, b$ can be $7$. But it is easy to see that $a, b$ has a common neighbor with number $7$, which is a contradiction. Hence, $7$ cannot appear in this table.\n\nAssume that all the numbers $\\{1, 2, 3, 4, 5, 6\\}$ will appear in this table. Define the taxi-cab distance between $(i, j)$ and $(i', j')$ to be $|i - i'| + |j - j'|$. Let $a_{i',j'} = 5$ and $(i,j)$ be a cell with number $6$ with minimal taxi-cab distance to $(i', j')$. We claim that $|i - i'| + |j - j'| = 1$, which is equivalent to saying that $(i, j)$ and $(i', j')$ have a common edge. To prove the claim, note that if $i \\ne i'$ and $j \\ne j'$, by symmetry we may assume $i > i'$, $j > j'$. Note that we have $a_{i+1,j} = 6$ or $a_{i,j+1} = 6$ or $a_{i+1,j+1} = 6$, and all these three cells have less taxi-cab distance to $(i', j')$ than $(i, j)$. So every cell filled with number $6$ having minimal taxi-cab distance to $(i', j')$ should be in the same row or column as this cell. Suppose that $j = j'$ and $i < i'$. Now, we have $a_{i+1,j+1} = 6$ or $a_{i+1,j} = 6$ or $a_{i+1,j-1} = 6$. But all three cases are impossible. Indeed, $a_{i+1,j} = 6$ contradicts minimality of taxi-cab distance from $(i,j)$ to $(i',j')$, and $a_{i+1,j \\pm 1} = 6$ contradicts the observation above that cells with number $6$ and minimal distance to $(i',j')$ should be in the same row or column. So the claim is proved. Now consider the two adjacent cells with numbers $6,5$. These two cells have $10$ neighbors in total, $6$ of them should be $6$ and $5$ of them should be $5$, which is impossible. Thus, the maximum possible number is $5$ and the solution is complete.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 14047,
    "subject": "Mathematics (Olympiad)",
    "question": "Find the last non-zero digit of $50! = 1 \\times 2 \\times 3 \\times \\dots \\times 50$.",
    "options": [],
    "answer": "See solution",
    "solution": "We first arrange the factors $1, 2, 3, \\ldots, 50$ in a table:\n\n![](table omitted)\n\nFrom this we see that in the prime factorisation of $50!$, $5$ occurs exactly $12$ times. Then $\\dfrac{50!}{2^{12}5^{12}}$ is the product of these factors:\n\n![](table omitted)\n\nSo the last digit of $\\dfrac{50!}{2^{12}5^{12}}$ is the last digit in the product\n$$\n1^{13} \\cdot 2^4 \\cdot 3^7 \\cdot 4^5 \\cdot 6^5 \\cdot 7^6 \\cdot 8^4 \\cdot 9^6 = (2 \\cdot 3 \\cdot 4 \\cdot 6 \\cdot 7 \\cdot 8 \\cdot 9)^4 \\times (3 \\cdot 7 \\cdot 9)^2 \\times (3 \\cdot 4 \\cdot 6)\n$$\nThe last digit of $2 \\cdot 3 \\cdot 4 \\cdot 6 \\cdot 7 \\cdot 8 \\cdot 9$ is $6$. So the last digit of $(2 \\cdot 3 \\cdot 4 \\cdot 6 \\cdot 7 \\cdot 8 \\cdot 9)^4$ is $6$.\n\nThe last digit of $3 \\cdot 7 \\cdot 9$ is $9$. So the last digit of $(3 \\cdot 7 \\cdot 9)^2$ is $1$.\n\nThe last digit of $3 \\cdot 4 \\cdot 6$ is $2$.\n\nSo the last non-zero digit of $50!$ is the last digit of $6 \\times 1 \\times 2$, which is \\textbf{2}.\n\n*Comment*\n\nThere are many other workable groupings of factors.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 14048,
    "subject": "Mathematics (Olympiad)",
    "question": "Solve the equation\n\n$$\n\\sin \\frac{\\pi\\sqrt{x}}{4} + \\cos \\frac{\\pi\\sqrt{2-x}}{4} = \\sqrt{2}\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "The solution is $x = 1$.\n\nWe have $x \\in [0, 2]$. For such $x$, both functions $\\sin \\frac{\\pi\\sqrt{x}}{4}$ and $\\cos \\frac{\\pi\\sqrt{2-x}}{4}$ are increasing, so their sum is also an increasing function. Therefore, the equation has at most one real root. Checking, $x = 1$ satisfies the equation.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 14049,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $m \\ge 0$ and $n \\ge 1$. Suppose that, initially, there are $m$ zeros and $n$ ones written on a blackboard. Now we are allowed to erase two arbitrary numbers from the blackboard, replacing them by their sum. Such a move is said to be *substantial* if both of the erased numbers were positive. The process is repeated until a single number remains on the blackboard.\n\nDetermine all possibilities for the number of substantial moves.",
    "options": [],
    "answer": "See solution",
    "solution": "There are always $n-1$ substantial moves.\n\nAt any moment, let $k$ denote the number of positive numbers on the blackboard and let $l$ denote the number of substantial moves that have already taken place. A move that is not substantial will affect neither $k$ nor $l$, and, in particular, the sum $k+l$ will not change. A substantial move replaces the pair $(k, l)$ by $(k-1, l+1)$, and does not change the value of $k+l$ either.\n\nThis proves that the sum $k+l$ remains invariant. Initially, $k+l = n+0 = n$. At the end we have $k=1$ and consequently $l = n-1$. This means that there are indeed always $n-1$ substantial moves. $\\square$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 14050,
    "subject": "Mathematics (Olympiad)",
    "question": "In a race among 5 snails, there is at most one tie, but that tie can involve any number of snails. For example, the result of the race might be that Dazzler is first; Abby, Cyrus, and Elroy are tied for second; and Bruna is fifth. How many different results of the race are possible?\n\n(A) 180 (B) 361 (C) 420 (D) 431 (E) 720",
    "options": [],
    "answer": "See solution",
    "solution": "If there are no ties, then there are $5!$ possible race results.\n\nSuppose $k$ of the 5 snails are tied, where $2 \\leq k \\leq 5$. There are $\\binom{5}{k}$ ways to choose the snails that are tied, and then, considering those snails as a group, there are $6 - k$ entrants and therefore $(6 - k)!$ orders of finish.\n\nThe number of possible results is thus\n\n$$\n5! + \\binom{5}{2} \\cdot 4! + \\binom{5}{3} \\cdot 3! + \\binom{5}{4} \\cdot 2! + \\binom{5}{5} \\cdot 1! = 120 + 240 + 60 + 10 + 1 = 431.\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 14051,
    "subject": "Mathematics (Olympiad)",
    "question": "Determine all integer solutions to the equation\n\n$$\n8x^3 - 4 = y(6x - y^2)\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "Rewrite the equation as\n\n$$\n8x^3 + y^3 - 6xy = 4.\n$$\n\nThis can be rearranged as\n\n$$\n(2x)^3 + y^3 + 1^3 - 3 \\cdot 2x \\cdot y \\cdot 1 = 5,\n$$\nwhich leads to\n\n$$\n(2x + y + 1)(4x^2 + y^2 + 1 - 2xy - 2x - y) = 5.\n$$\n\nLet $s = 2x + y$ and $xy = p$. For $2x + y = 4$, we have $xy = 2$, so $(x, y) = (1, 2)$. For $2x + y = 0$, $xy = -\\frac{2}{3}$, which gives no integer solutions.\n\nAlternatively, set $s = 2x + y$ and $2xy = p$, then\n\n$$\np = \\frac{s^3 - 4}{3(s + 1)}.\n$$\n\nFor $p$ to be integer, $s + 1$ must divide $5$, so $s + 1 \\in \\{-1, 1, -5, 5\\}$, i.e., $s \\in \\{-2, 0, -6, 4\\}$. Only $s = 4$ gives integer $x, y$ with $p = 4$, so $x = 1$, $y = 2$.\n\n**Conclusion:** The only integer solution is $\\boxed{(x, y) = (1, 2)}$.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 14052,
    "subject": "Mathematics (Olympiad)",
    "question": "Given a sequence of real numbers $x_n$:\n\n$$\nx_1 = 3 \\quad \\text{and} \\quad x_n = \\frac{n+2}{3n}(x_{n-1}+2), \\quad \\forall n \\ge 2.\n$$\n\nProve that the sequence has a finite limit as $n \\to \\infty$ and calculate this limit.",
    "options": [],
    "answer": "See solution",
    "solution": "For $n \\ge 1$, we have\n\n$$\nx_{n+1} - x_n = \\left(\\frac{n+3}{3(n+1)} - 1\\right)x_n + \\frac{2(n+3)}{3(n+1)} = \\frac{2}{3(n+1)}(n+3 - nx_n).\n$$\n\nWe first prove that\n\n$$\nx_n > 1 + \\frac{3}{n} \\quad \\forall n \\ge 2.\n$$\n\nThe proof proceeds by induction on $n$. For $n=2$ we have\n\n$$\nx_2 = \\frac{4}{6}(3+2) = \\frac{10}{3} > 1 + \\frac{3}{2}.\n$$\n\nSuppose that $x_k > 1 + \\frac{3}{k}$ for some $k \\ge 2$, then\n\n$$\nx_{k+1} = \\frac{k+3}{3(k+1)}(x_k + 2) > \\frac{k+3}{3(k+1)}\\left(1 + \\frac{3}{k} + 2\\right) = 1 + \\frac{3}{k+1}.\n$$\n\nThus, the inequality is proved. From the previous result, we have\n\n$$\nx_n > x_{n+1} > 1, \\quad \\forall n \\ge 2.\n$$\n\nTherefore, $(x_n)$ is a decreasing sequence for $n \\ge 2$ and is bounded below by 1. This implies that $(x_n)$ has a finite limit. Now, we can find this limit by solving\n\n$$\nx = \\frac{1}{3}(x+2),\n$$\n\nwhich gives $x = 1$ as the limit.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 14053,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $a_1, a_2, \\dots, a_n$ be distinct positive integers and let $M$ be a set of $n-1$ positive integers not containing $s = a_1 + a_2 + \\dots + a_n$. A grasshopper is to jump along the real axis, starting at the point $0$ and making $n$ jumps to the right with lengths $a_1, a_2, \\dots, a_n$ in some order. Prove that the order can be chosen in such a way that the grasshopper never lands on any point in $M$.\n\nIn the original version, $n = 2009$ and $a_i = i$ for $1 \\le i \\le 2009$. This original version is a cute but (quite) mathematically challenging fact. The generalization to the current version is far from trivial.",
    "options": [],
    "answer": "See solution",
    "solution": "We will represent a route of the grasshopper by an ordered sequence of indices $(i_1, i_2, \\dots, i_n)$ if it makes consecutive jumps $(a_{i_1}, a_{i_2}, \\dots, a_{i_n})$.\n\nTo simplify our argument, we prove the following statement:\n\nLet $a_1, a_2, \\dots, a_n$ be distinct positive integers and let $M$ be a set of no more than $n-1$ integers not containing $s = a_1 + a_2 + \\dots + a_n$. A grasshopper is to jump along the real axis, starting at the point $0$ and making $n$ jumps to the right with lengths $a_1, a_2, \\dots, a_n$ in some order. Prove that the order can be chosen in such a way that the grasshopper never lands on any point in $M$.\n\nWe induct on $n$. The base case $n=1$ is trivial. For the inductive step, assume the claim is true for all $n$ smaller than some integer $k > 1$. Now consider $n = k$. Without loss of generality, assume $a_1 < a_2 < \\dots < a_k$. Let $M = \\{r_1, r_2, \\dots, r_{k-1}\\}$, with $r_1 < r_2 < \\dots < r_{k-1}$.\n\nWe divide into several cases:\n\n**(a1)** Assume $r_1 < a_n$ and $a_n \\notin M$. Apply the induction hypothesis to $M_1 = M \\setminus \\{r_1\\} - a_n = \\{r_2 - a_n, r_3 - a_n, \\dots, r_{n-1} - a_n\\}$ using jumps $a_1, a_2, \\dots, a_{n-1}$. Placing $a_n$ at the beginning of this sequence provides the desired route.\n\n**(a2)** Assume $r_1 < a_n$ and $a_n \\in M$. Consider the $n$ pairwise disjoint sets\n\n$$\n\\{a_n\\}, \\{a_1, a_1 + a_n\\}, \\{a_2, a_2 + a_n\\}, \\dots, \\{a_{n-1}, a_{n-1} + a_n\\}.\n$$\n\nBecause $M$ only has $n-1$ elements, there is an index $i$ such that $\\{a_i, a_i + a_n\\}$ is disjoint from $M$. Thus, there are at most $n-3$ elements in the intersection of $M$ and the interval $[a_i + a_n, s]$, because $r_1$ and $a_n$ are not in the interval. Apply the induction hypothesis to $M_2 = M \\cap [a_i + a_n, s] - (a_i + a_n)$ using jumps $a_1, a_2, \\dots, a_{i-1}, a_{i+1}, \\dots, a_{n-1}$. Placing $a_i$ and $a_n$ at the beginning (in that order) provides the desired route.\n\n**(b)** Assume $r_1 \\ge a_n$. Consider $M_3 = M \\setminus \\{r_1\\} - a_n = \\{r_2 - a_n, r_3 - a_n, \\dots, r_{n-1} - a_n\\}$. Apply the induction hypothesis to $M_3$ using jumps $a_1, a_2, \\dots, a_{n-1}$; let the resulting route be $\\mathcal{R}_{n-1} = (i_1, i_2, \\dots, i_{n-1})$.\n\nIf this route $\\mathcal{R}_{n-1}$ does not step on $r_1 - a_n$ (so $r_1 > a_n$), then the route $\\mathcal{R}_n = (a_n, i_1, i_2, \\dots, i_{n-1})$ works.\n\nOtherwise, the route $\\mathcal{R}_n$ contains exactly one point of $M$, namely, $r_1$. Thus, there exists $k$ with $1 \\le k \\le n-1$ such that\n\n$$\na_n + a_{i_1} + \\dots + a_{i_k} = r_1.\n$$\n\nConsider the route\n\n$$\n\\mathcal{R}'_n = (i_1, i_2, \\dots, i_{k+1}, a_n, i_{k+2}, \\dots, i_{n-1}).\n$$\n\nWe claim this route $\\mathcal{R}'_n$ works. Indeed, since\n\n$$\na_{i_1} + \\dots + a_{i_{k+1}} < a_{i_1} + \\dots + a_{i_k} + a_n = r_1,\n$$\n\nthe grasshopper will not land on points of $M$ during the first $k+1$ jumps. For the remaining part, it lands on the same points as in $\\mathcal{R}_n$ (which are all $> r_1$), so it will not land on points of $M$. Therefore, this route $\\mathcal{R}'_n$ allows the grasshopper to avoid all points in $M$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 14054,
    "subject": "Mathematics (Olympiad)",
    "question": "Does there exist a prime number $p$ such that both $p^3 + 2008$ and $p^3 + 2010$ are primes as well?",
    "options": [],
    "answer": "See solution",
    "solution": "Let $p$ be any prime number. If $p$ is not divisible by $7$, then $p^3$ is congruent to either $1$ or $-1$ modulo $7$. Since $2008 \\equiv -1 \\pmod{7}$ and $2010 \\equiv 1 \\pmod{7}$, either $p^3 + 2008$ or $p^3 + 2010$ is divisible by $7$ and hence composite. If $p$ is divisible by $7$, then $p = 7$ and $p^3 + 2010 = 7^3 + 2010 = 2353 = 13 \\times 181$ is composite, too.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 14055,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $\\mathbb{R}^+$ be the set of positive real numbers. Determine all functions $f: \\mathbb{R}^+ \\to \\mathbb{R}^+$ satisfying the equation\n\n$$\nx f(x^2) f(f(y)) + f(y f(x)) = f(xy) (f(f(x^2)) + f(f(y^2))).\n$$\n\nfor all $x, y \\in \\mathbb{R}^+$.",
    "options": [],
    "answer": "See solution",
    "solution": "**Answer:** $f(x) = \\frac{1}{x}$.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 14056,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all polynomials $f(x)$ such that $f(f(x)) = f(x)^m$ for some integer $m \\geq 2$.",
    "options": [],
    "answer": "See solution",
    "solution": "The solutions are $f(x) = 0$, $f(x) = 1$, $f(x) = \\omega$ where $\\omega$ is an $(m-1)$st root of unity, and $f(x) = x^m$.\n\nIf $f(x) = c$ is a constant polynomial, then the relation holds if and only if $c = c^m$.\nClearly, the solutions are $c = 0, 1$ and all the $(m-1)$st roots of unity.\n\nIf $f$ is a non-constant polynomial, then $f(x)$ attains infinitely many values.\nThus, there are infinitely many $y$ such that $f(y) = y^m$. Since $f$ is a polynomial,\nthis implies $f(x) = x^m$ for any $x$.\n\nIt is easy to check that all these are solutions.",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 14057,
    "subject": "Mathematics (Olympiad)",
    "question": "Out of an equilateral triangle with side $2014$, an equilateral triangle with side $214$ is cut out, such that the two triangles have one vertex in common and two of the sides of the cut-out triangle lie on two of the sides of the initial one. Can this figure be covered by the figures shown below without overlap (rotation is allowed), if the triangles in the figures are equilateral with side $1$? Justify your answer!\n\n![](images/MACEDONIAN_MATHEMATICAL_SOCIETY_p10_data_0a044d0d40.png)",
    "options": [],
    "answer": "See solution",
    "solution": "First, we cut the given figure into equilateral triangles with side $1$. We label the triangles in the given figure by numbers from $1$ to $6$, as on the picture to the right. (In the first row, successively from $1$ to $6$, then the numbers repeat; in the second, we start from $5$, in the third from $3$, then from $1$, and the procedure repeats.) It can easily be noticed that each of the given figures covers exactly one of the numbers $1$ to $6$. Therefore, in order for the figure to be coverable by the given figures, each of the numbers has to appear an equal number of times. If we compare how often the number $1$ and the number $2$ appear, we will notice that in the first, fourth, and each row of the form $3k+1$, there is one more $1$ than $2$'s, and in the remaining rows the number of $1$'s and $2$'s is equal. Therefore, it follows that the number of $1$'s and $2$'s is unequal, and therefore not every number can appear an equal number of times. It follows that the figure cannot be covered in the required way.\n\n![](images/MACEDONIAN_MATHEMATICAL_SOCIETY_p10_data_d53c86f35b.png)",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 14058,
    "subject": "Mathematics (Olympiad)",
    "question": "Jüri draws a circle $c$ with radius $3$ and $n$ circles with radius $1$ on a paper. Find the minimal $n$ for which he can draw the circles in such a way that it would not be possible to draw inside the circle with radius $3$ any new circles with radius $1$ having at most one common point with each of the previously drawn circles.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $O$ be the center of circle $c$. Suppose that circles $c_1, c_2, c_3$ with radius $1$ and centers $O_1, O_2, O_3$, respectively, are placed so that they touch circle $c$ internally and $O_1, O_2, O_3$ are vertices of an equilateral triangle. No new circle can be placed along the same line with two existing circles of radius $1$ because the three circles would require a free area of length $6$, which is possible only along the diameter of circle $c$. Other possible placements are in the middle or on the other side of the narrower area between two circles.\n\nThe circle with radius $1$ and center $O$ touches all three circles, so its location is fixed. The circle with radius $1$ and center $O_1'$, symmetric to $O$ with respect to the line $O_2O_3$, touches circles $c_2$ and $c_3$ and has at least one common point with circle $c$, as it is collinear with the circle in the middle and circle $c_1$. Thus, this location is also fixed. If all circles $c_1, c_2, c_3$ are moved slightly towards $O$, the possibility to add new circles disappears.\n\nConsequently, if $n \\geq 3$, Jüri can draw the circles so that no new circles can be added as required. Now, show that for $n \\leq 2$, a new circle can always be added; it suffices to consider $n = 2$. Let $c_1$ and $c_2$ be the given circles with radius $1$ and centers $O_1$ and $O_2$. Without loss of generality, $O_1 \\neq O$. Choose $AB$ as the diameter of $c$ perpendicular to $O_1O$. Consider two circles with radius $1$, touching $c$ at points $A$ and $B$, respectively. Circle $c_1$ does not preclude drawing either of them, as it is located inside the strip of width $2$ surrounding the line $O_1O$, where neither of the two circles reach. Circle $c_2$ can preclude at most one of the two circles since it cannot reach both sides of the strip.\n\nTherefore, the minimal $n$ is $3$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 14059,
    "subject": "Mathematics (Olympiad)",
    "question": "For each integer $n$ ($n \\ge 2$), let $f(n)$ denote the sum of all positive integers that are at most $n$ and not relatively prime to $n$.\n\nProve that $f(n+p) \\neq f(n)$ for each such $n$ and for every prime $p$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $m$, $n$, and $k$ be positive integers and $p$ be a prime. Since $(n, k) = 1 \\Leftrightarrow (n, n-k) = 1$, we have\n$$\nf(n) = \\frac{n(n+1)}{2} - \\frac{n \\cdot \\phi(n)}{2},\n$$\nwhere $\\phi(n)$ is Euler's totient function.\n\nSuppose $m > n$, $(m, n) = 1$, and $f(m) = f(n)$. Then\n$$\nm(m+1-\\phi(m)) = n(n+1-\\phi(n)) \\implies m \\le n,\n$$\na contradiction. Therefore, if $f(n+p) = f(n)$, then $n = pk$ for some positive integer $k$ and\n$$\n(k+1)(pk + p + 1 - \\phi(pk + p)) = k(pk + 1 - \\phi(pk)).\n$$\nThis implies\n$$\npk + 1 - \\phi(pk) = a(k + 1) \\quad (1)\n$$\nand\n$$\npk + p + 1 - \\phi(pk + p) = ak \\quad (2)\n$$\nfor some positive integer $a$. In particular,\n$$\n(p-a)k = a-1 + \\phi(pk) \\implies p > a \\ge 1\n$$\nand\n$$\na = \\phi(pk + p) - \\phi(pk) - p \\equiv -1 \\pmod{p-1}.\n$$\nSo, $a = p-2$. Substituting this in (1) and (2) we obtain\n$$\n\\phi(pk) = 2k + 3 - p \\quad (3)\n$$\nand\n$$\n\\phi(pk + p) = 2k + 1 + p \\quad (4)\n$$\nWe shall prove that $p \\nmid k$ and $p \\nmid k+1$. If $p \\mid k+1$, the relation (4) implies $p \\mid 1$, a contradiction. On the other hand, if $p \\mid k$, then (3) implies $p \\mid 3$. Setting $k = 3^\\alpha k_1$, where $k_1$ is an integer such that $3 \\nmid k_1$, and substituting this in (3) we get $\\phi(k_1) = k_1$. So $k_1 = 1$ and $k = 3^\\alpha$. Now substituting this in (4) gives a contradiction. Using this observation, we obtain from (3) and (4) the equalities\n$$\n\\phi(k) = \\frac{2(k+1)}{p-1} - 1 \\quad (5)\n$$\nand\n$$\n\\phi(k+1) = \\frac{2(k+1)}{p-1} + 1 \\quad (6)\n$$\nIn particular, $p \\ge 5$ and $k+1 \\ge 6$. Moreover, $\\phi(k+1) - \\phi(k) = 2$ and either $4 \\nmid \\phi(k)$ or $4 \\nmid \\phi(k+1)$.\n\nIf $4 \\nmid \\phi(k)$, then $k = q^\\beta$ or $k = 2q^\\beta$, where $q$ is an odd prime and $\\beta \\ge 1$ is an integer. From (5) we obtain\n$$\n\\frac{2(k+1)}{p-1} - 1 = \\phi(k) \\ge \\frac{1}{2} \\left(1 - \\frac{1}{q}\\right) k \\ge \\frac{1}{2} \\cdot \\frac{2}{3} k = \\frac{k}{3} \\implies p \\le 5 \\implies p = 5.\n$$\nIf $p=5$, from (5) we obtain $\\phi(k) = \\frac{k-1}{2}$. So $k$ must be odd, $k+1$ must be even, and $\\phi(k+1) \\le \\frac{k+1}{2}$. This contradicts (6).\n\nIf $4 \\nmid \\phi(k+1)$, then $k+1 = q^\\beta$ or $k+1 = 2q^\\beta$, where $q$ is an odd prime and $\\beta \\ge 1$ is an integer. From (6) we obtain\n$$\n\\frac{2(k+1)}{p-1} + 1 = \\phi(k+1) \\ge \\frac{k+1}{3}.\n$$\nIf $p > 7$, then this implies\n$$\n\\frac{6}{p-7} \\ge \\frac{2(k+1)}{p-1}\n$$\nand therefore by (5) we must have\n$$\n\\phi(k) \\le \\frac{13-p}{p-7},\n$$\nwhich is impossible. On the other hand, if $p \\le 7$, then $p=7$ and, as $\\frac{p-1}{2}$ divides $k+1$, we also have $q=3$. Using (6) again, we see that $k+1 = 2 \\cdot 3^\\alpha$ leads to a contradiction, and $k+1 = 3^\\alpha$ leads to $k+1=3$, which was already ruled out.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 14060,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $a, b$ be positive integers such that $a + 1$, $b + 1$ and $ab$ are all perfect squares. Prove that $\\gcd(a, b) + 1$ is also a perfect square.",
    "options": [],
    "answer": "See solution",
    "solution": "Firstly, observe that when $a = b$, $\\gcd(a, b) + 1 = a + 1$ is a perfect square. So we now assume without loss of generality that $a < b$.\n\nLet $a + 1 = A^2$, $b + 1 = B^2$ where $1 < A < B$ are positive integers. We prove the result by induction on $\\max\\{a, b\\} = b$.\n\nDefine $C := AB - \\sqrt{ab} = AB - \\sqrt{(A^2 - 1)(B^2 - 1)}$, which is an integer as $ab$ is a perfect square. Define $c := C^2 - 1$. We will now prove the following:\n\n* $1 < c < b$\n* $ac$ is a perfect square (noting that it follows immediately from the construction that $a + 1$, $c + 1$ are perfect squares)\n* $\\gcd(a, c) = \\gcd(a, b)$\n\nWe will then be done by induction, as by repeatedly applying this process, we preserve the $\\gcd$ and must eventually reach a case with $a = b$, which we have proved above. For the first part,\n\n$$\nC = AB - \\sqrt{A^2 B^2 - A^2 - B^2 + 1} > AB - \\sqrt{A^2 B^2 - 2AB + 1} = 1\n$$\n\nwhere the inequality is strict since $A \\neq B$. Furthermore,\n\n$$\n2(A - 1)B^2 \\geq 2B^2 > A^2 - 1 \\implies A^2 + B^2 - 1 < 2AB^2 - B^2\n$$\n\nleading to\n\n$$\nC = AB - \\sqrt{A^2 B^2 - A^2 - B^2 + 1} < AB - \\sqrt{A^2 B^2 - 2AB^2 + B^2} = B.\n$$\n\nSo $1 < C < B$ and therefore $1 < c < b$.\n\nFor the second part, expanding the definition of $C$ we get:\n\n$$\n(AB - C)^2 = (A^2 - 1)(B^2 - 1) \\implies A^2 + B^2 + C^2 = 2ABC + 1 \\quad (1)\n$$\n\nWe can rearrange this to:\n\n$$\nac = (A^2 - 1)(C^2 - 1) = (B - AC)^2 \\quad (2)\n$$\n\nwhich shows $ac$ is a perfect square.\n\nFor the final part, note that if $d \\mid A^2 - 1$ and $d \\mid C^2 - 1$ then from (2), $d \\mid B - AC$. Then from (1) we have:\n\n$$\nB^2 - 1 = 2ABC - A^2 - C^2 \\equiv 2A^2C^2 - A^2 - C^2 \\equiv 2 - 1 - 1 \\equiv 0 \\pmod{d}\n$$\n\nThus $d \\mid B^2 - 1$. Setting $d = \\gcd(a, c)$, we see that $\\gcd(a, c) \\mid \\gcd(a, b)$. But the condition we're using in (1) is symmetric in $A, B, C$ and so in a similar way we get $\\gcd(a, b) \\mid \\gcd(a, c)$.\n\nCombining, we must have $\\gcd(a, b) = \\gcd(a, c)$.\n\n![](images/BMO2024Shortlist_p76_data_3ab806aa9e.png)",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 14061,
    "subject": "Mathematics (Olympiad)",
    "question": "Does there exist a positive integer $n$ such that the first and the second digit of $2^n$ (in decimal notation) are 3 and 9, respectively?",
    "options": [],
    "answer": "See solution",
    "solution": "Yes. Consider powers of two of the form $2^{10k} = (2^{10})^k = 1024^k = (1,024)^k \\cdot 10^{3k}$. Let $m$ be the smallest integer such that $(1,024)^m \\geq 3.9$ (such $m$ exists since $(1,024)^k$ grows without bound). By definition, $(1,024)^{m-1} < 3.9$, so $(1,024)^m = (1,024)^{m-1} \\cdot 1,024 < 3.9 \\cdot 1,024 = 3.9 + 3.9 \\cdot 0.024 < 3.9 + 4 \\cdot 0.025 = 4.0$. Thus, $3.9 \\leq (1,024)^m < 4.0$, so the first digits of $2^{10m} = (1,024)^m \\cdot 10^{3m}$ are indeed 3 and 9.\n\n*Remark 1.* The value of $m$ found above is 58, and the respective power of two is $2^{580} = 3.957286\\ldots \\times 10^{174}$.\n\n*Remark 2.* The smallest power of two satisfying the requirements is $2^{95} = 39614081257132168796771975168$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 14062,
    "subject": "Mathematics (Olympiad)",
    "question": "We order the positive integers in two rows as follows:\n\n1 3 6 11 19 32 53 ...\n2 4 5 7 8 9 10 12 13 14 15 16 17 18 20 \\text{ to } 31\\ 33 \\text{ to } 52\\ 54 ...\n\nWe first write 1 in the first row, 2 in the second, and 3 in the first. After this, the following integers are written so that an individual integer is always added in the first row and blocks of consecutive integers are added in the second row, with the leading number of a block giving the number of (consecutive) integers to be written in the next block.\n\nWe name the numbers in the first row $a_1, a_2, a_3, \\dots$\n\nDetermine an explicit formula for $a_n$.",
    "options": [],
    "answer": "See solution",
    "solution": "We first note that $a_1 = 1$, $a_2 = 3$, and $a_3 = 6$. It is straightforward to see that a block of length $a_{n-1} + 1$ starts with the number $a_n + 1$, and this block ends at $a_n + (a_{n-1} + 1)$, which yields the recurrence:\n\n$$a_{n+1} = a_n + a_{n-1} + 2$$\n\nThis recursion has the constant solution $a_n \\equiv -2$, and the homogeneous recursion $a_{n+1} = a_n + a_{n-1}$ is of Fibonacci type. Writing the Fibonacci sequence as $F_0 = 0$, $F_1 = 1$, $F_2 = 1$, $F_3 = 2$, and so on, we can check that $a_n = F_{n+3} - 2$ holds for $n = 1, 2, 3$. Therefore, the explicit formula is:\n\n$$a_n = F_{n+3} - 2$$\n\nwhere $F_n$ is the $n$th Fibonacci number. $\\square$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 14063,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $n$ be a natural number, $n \\geq 3$. Find the maximal number of diagonals of a regular $n$-gon one can select so that every two selected diagonals that intersect inside the polygon are perpendicular.\n\n$\\textbf{Answer:}$ $n - 2$ if $n$ is even; $n - 3$ if $n$ is odd.",
    "options": [],
    "answer": "See solution",
    "solution": "If $n$ is odd, one can select all $n - 3$ diagonals connecting one fixed vertex to others. To prove that the conditions 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, one with an even number of vertices and the other with an odd number. In the latter half, the side connecting two medium vertices is parallel to the chosen diagonal. Thus, the existence of two perpendicular diagonals would imply the existence of two perpendicular sides, which is possible only if $n \\equiv 0 \\pmod{4}$, a contradiction.\n\nIf $n \\equiv 2 \\pmod{4}$, select every second vertex on the boundary and initially select all diagonals connecting two consecutive selected vertices. Furthermore, select all $\\frac{n}{2} - 3$ diagonals connecting one fixed selected vertex to all other selected vertices not yet connected. Finally, select the diagonal connecting this vertex to its opposite vertex of the original polygon. This way, one selects $n-2$ diagonals. If $n \\equiv 0 \\pmod{4}$, initially select $\\frac{n}{2}$ diagonals as in the previous case, then select $\\frac{n}{2} - 2$ more diagonals in the $\\frac{n}{2}$-gon formed by the selected diagonals according to the described algorithm.\n\nNow, prove that selecting more diagonals is impossible. First, show that all selected diagonals that intersect some other selected diagonals must lie in two perpendicular directions. Consider one pair of mutually intersecting diagonals. The number of vertices between two endpoints of distinct diagonals is less than half the total number of vertices. If another pair of mutually intersecting diagonals is added, the same holds. Thus, the other pair cannot fit entirely in any window left by the initial pair, so at least one diagonal from the first pair and one from the second pair intersect, meaning all must lie in two perpendicular directions.\n\nLet $d$ be the number of selected diagonals and $k$ the number of intersection points of selected diagonals. Consider the 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, so every line segment connecting two neighbouring intersection points of any diagonal is a side of such a piece. Let there be $a$ horizontal and $b$ vertical diagonals selected, with $a \\leq 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' \\geq k-a+b-1 \\geq k-1$.\n\nConsequently, $n+2d+4k \\geq 4w' + 3(w-w') = 3w+w' \\geq 3w+k-1 = 3(d+k+1)+k-1 = 3d+4k+2$, whence $d \\leq n-2$.\n\n$\\textbf{Remark 1.}$ In the case $n \\equiv 0 \\pmod{4}$, another suitable construction is as follows. Enumerate vertices with numbers $0, 1, \\ldots, n-1$ in clockwise order. Select all diagonals that connect any vertex $i$ with the vertex $\\frac{n}{2}-i$ for every $i = -\\frac{n}{4}+1, \\ldots, \\frac{n}{4}-1$ (there are $\\frac{n}{2}-1$ such diagonals), and all diagonals perpendicular to them (also $\\frac{n}{2} - 1$). Altogether, we select $n-2$ diagonals that all lie in two perpendicular directions. (Fig. 29 depicts the situation for $n=12$).\n\n$\\textbf{Remark 2.}$ This problem, proposed by Estonia, appeared as C5 in the IMO 2016 shortlist. The solution presented here appeared in a contest paper and is not given in the shortlist. For other solutions, see the shortlist.\n\n![](images/prob1617_p33_data_d2f8d0183f.png)",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 14064,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all positive integers $n$ such that $4n! - 4n + 1$ is a perfect square.",
    "options": [],
    "answer": "See solution",
    "solution": "For $n \\leq 6$, we can easily check that $n = 1, 2, 4$ are the only positive integers satisfying the condition.\n\nWe will prove that there are no other positive integers which satisfy the condition. Assume to the contrary that there exists a positive integer $n > 6$ that satisfies the condition.\n\nLet $4n! - 4n + 1 = x^2$ for some non-negative integer $x$.\n\nSo $4n! - 4n + 4 = x^2 + 3$. Let $p$ be a prime factor of $n-1$. From $(n-1) \\mid 4n!$ and $(n-1) \\mid -4n + 4$, we get that $x^2 + 3$ is divisible by $p$. This means $\\left(\\frac{-3}{p}\\right) = 1$ or $p = 3$. But from the quadratic reciprocity theorem, $\\left(\\frac{-3}{p}\\right) = \\left(\\frac{p}{3}\\right)$ which is 1 only when $p \\equiv 1 \\pmod{3}$. So every prime factor of $n-1$ is 3 or is congruent to 1 modulo 3. That is, $n-1 \\equiv 0 \\pmod{3}$.\n\n**Case 1:** $n-1 \\equiv 1 \\pmod{3}$; that is, $n \\equiv 2 \\pmod{3}$\n\nSo $4n! - 4n + 1 \\equiv 2 \\pmod{3}$, which contradicts the fact that $4n! - 4n + 1$ is a perfect square.\n\n**Case 2:** $n-1 \\equiv 0 \\pmod{3}$; that is, $n \\equiv 1 \\pmod{3}$\n\nSo $4n! - 4n + 1 \\equiv 0 \\pmod{3}$ or $3 \\mid x^2$ or $3 \\mid x$. Then $9 \\mid x^2 = 4n! - 4n + 1$. But since $n > 6$, this implies that $9 \\mid n!$. So $9 \\mid -4n + 1$ or $n \\equiv 7 \\pmod{9}$. Thus $n-1 \\equiv 6 \\pmod{9}$, which contradicts our conclusion that all prime factors of $n-1$ are 3 or are congruent to 1 modulo 3.\n\nSo the only positive integers satisfying the condition are $1$, $2$, and $4$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 14065,
    "subject": "Mathematics (Olympiad)",
    "question": "Isaac and Jeremy play a game. Isaac is thinking of some $2^n$ integers $k_1, \\dots, k_{2^n}$, where $n$ is a positive integer. Jeremy may ask questions of the form \"is $k_i < k_j$?\", to which Isaac answers truthfully. After $n2^{n-1}$ questions, Jeremy must state whether or not Isaac's numbers are all distinct. Show that Jeremy has no way of ensuring that his statement is correct. Jeremy's questions may depend on the answers he has already received, and are exactly determined by them.\n\nSlightly harder formulation: instead ask \"Can Jeremy ensure that his statement is correct?\"",
    "options": [],
    "answer": "See solution",
    "solution": "Let $a = a_1, \\dots, a_{2^n}$ be a sequence of integers. Denote by $Q(a)_j$ the $j$th question Jeremy asks, and $A(a)_j$ the $j$th answer Isaac gives, when Isaac chooses $k_i = a_i$. Each $Q(a)_j$ may depend on $Q(a)_1, \\dots, Q(a)_{j-1}$ and $A(a)_1, \\dots, A(a)_{j-1}$, but not on anything else.\n\nWe want to show that, regardless of Jeremy's strategy, there exist two sequences $a$ and $b$ such that $b_1, \\dots, b_{2^n}$ are distinct, $a_1, \\dots, a_{2^n}$ are not all distinct, and $A(a)_j = A(b)_j$ for all $j$. This ensures $Q(a)_j = Q(b)_j$ for all $j$. Thus, if Isaac picks either $k_i = a_i$ or $k_i = b_i$, Jeremy asks the same questions and gets the same answers, so he cannot determine whether the numbers are all distinct.\n\nFor $n=1$ the case is trivial, so assume $n > 1$.\n\nConsider all sequences $d = d_1, \\dots, d_{2^n}$ which are permutations of $1, \\dots, 2^n$ (so all entries are distinct). There are $(2^n)!$ such sequences. However, there are only $2^{n2^{n-1}} < (2^n)!$ possible answer sequences $A(d)_1, \\dots, A(d)_{n2^{n-1}}$ (we justify this inequality below). Thus, by the pigeonhole principle, there exist two distinct permutations $b$ and $c$ with $A(b)_j = A(c)_j$ for all $j$.\n\nThere must be indices $i_1$ and $i_2$ such that $b_{i_2} = b_{i_1} + 1$ but $c_{i_2} < c_{i_1}$, since $b$ and $c$ are distinct. Define $a$ by\n\n$$\na_i = \\begin{cases} b_i & \\text{if } i \\neq i_2 \\\\ b_{i_1} & \\text{if } i = i_2 \\end{cases}$$\n\nNow $a_{i_2} = b_{i_1} = a_{i_1}$, so $a_1, \\dots, a_{2^n}$ are not all distinct. The only question where $a$ and $b$ differ is \"is $k_{i_2} > k_{i_1}$?\", which is true for $b$ but false for $a$ and $c$. Since $Q(b)_j = Q(c)_j$ and $A(b)_j = A(c)_j$, none of the questions can be \"is $k_{i_2} > k_{i_1}$?\". Thus, $a$ and $b$ give the same answers to all questions Jeremy asks, so Jeremy cannot distinguish between them.\n\nIt remains to justify $2^{n2^{n-1}} < (2^n)!$ for $n > 1$:\n\nBase case $n=2$ is by calculation. For higher $n$, note that $\\frac{(2^n)!}{(2^{n-1})!}$ is a product of $2^{n-1}$ terms, each greater than $2^{n-1}$, so $(2^n)! > (2^{n-1})^{2^{n-1}} (2^{n-1})!$. By induction, $(2^n)! > 2^{(n-1)2^{n-1} + (n-1)2^{n-2}} \\ge 2^{n2^{n-1}}$ for $n \\ge 3$.\n\n$\\square$",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 14066,
    "subject": "Mathematics (Olympiad)",
    "question": "There are given $2k$ boxes ($k \\ge 2$) with $2k-1$ pebbles in each one. A legal move is to choose $2k-2$ boxes and remove one pebble from each one of them. Players *A* and *B* make moves alternately; *A* goes first. A player wins if a move of his empties two boxes. Determine which player has a winning strategy.",
    "options": [],
    "answer": "See solution",
    "solution": "The second player $B$ has a winning strategy.\n\nEach move does not affect (ignores) exactly two boxes $i, j$; denote such a move by $m_{i,j}$. Let $A$'s first move be $m_{1,2}$. Then $B$ divides the remaining boxes arbitrarily into $k-1$ pairs $\\{3,4\\}, \\dots, \\{2k-1, 2k\\}$, and his first $k-1$ moves are $m_{3,4}, m_{5,6}, \\dots, m_{2k-1,2k}$. He is not interested in how $A$ plays until move $m_{2k-1,2k}$, the last one of these $k-1$. However, $B$'s move after $m_{2k-1,2k}$ depends on $A$'s response. We show that this move of $B$, his $k$th, is winning.\n\nEvery box $i = 1, 2, \\dots, 2k$ is ignored by exactly one of the moves $m_{1,2}, m_{3,4}, \\dots, m_{2k-1,2k}$. Hence, these $k$ moves combined decrease the contents of every box by exactly $k-1$.\n\nLet $m'_{3,4}, \\dots, m'_{2k-1,2k}$ be $A$'s moves matching $m_{3,4}, \\dots, m_{2k-1,2k}$. They affect a box at most $k-1$ times, so by the previous conclusion there will be at least $k - (k-1) = 1$ pebble in each box after $m'_{2k-1,2k}$. In particular, $m'_{3,4}, \\dots, m'_{2k-1,2k}$ are not winning.\n\nWe claim that after $m'_{2k-1,2k}$ at least two boxes contain exactly one pebble. A move ignores two boxes; then $k-1$ moves ignore at most $2k-2$ boxes. So there exist two boxes $i$ and $j$ that are affected by each of the moves $m'_{3,4}, \\dots, m'_{2k-1,2k}$, meaning that the latter sequence decreases their contents by exactly $k-1$. But the previous sequence $m_{1,2}, m_{3,4}, \\dots, m_{2k-1,2k}$ decreased the contents of every box by exactly $k-1$. As a result, each of $i$ and $j$ has exactly one pebble after $m'_{2k-1,2k}$. It is clear now that $B$'s next move is winning.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 14067,
    "subject": "Mathematics (Olympiad)",
    "question": "A _partition_ of a set $S$ is a set of pairwise disjoint subsets of $S$ whose union is $S$.\n\nLet $P$ be a partition of the set $\\{1, 2, \\ldots, 2024\\}$ into 2-element sets, satisfying the following condition: For every set $\\{a, b\\}$ in $P$, either $|a - b| = 1$ or $|a - b| = 506$.\n\nAssume that $\\{1518, 1519\\}$ belongs to $P$. Determine the number that pairs off with $505$ to form a set in $P$.",
    "options": [],
    "answer": "See solution",
    "solution": "The required number is $506$.\n\nPlace the numbers $1$ through $2024$ on a $4 \\times 506$ checkerboard as shown below:\n\n![](images/RMC_2025_p105_data_8b3cc00127.png)\n\nAssume the square in the upper-left corner is white.\n\nBy the condition in the statement, a set $\\{a, b\\}$ in $P$ is a horizontal domino if $|a-b| = 1$ and a vertical domino if $|a - b| = 506$, with the possible exceptions $\\{506, 507\\}$, $\\{1012, 1013\\}$, and $\\{1518, 1519\\}$.\n\nAs this latter is an all-black member of $P$ and dominoes are bicolour, $P$ must have some all-white member. The only such is $\\{1012, 1013\\}$.\n\nHence, to form a set in $P$, the number $506$ pairs off with exactly one of $505$ and $507$.\n\nAs $\\{506, 507\\}$ is all-black and there are no more all-white sets left, $506$ must pair off with $505$, as stated.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 14068,
    "subject": "Mathematics (Olympiad)",
    "question": "A sequence of length $n$, consisting of positive integers not exceeding $n-1$, is given. If there exists exactly one subsequence with sum divisible by $n$, then show that the sequence is constant.",
    "options": [],
    "answer": "See solution",
    "solution": "For $n=2$, there is nothing to prove. Suppose $n \\ge 3$.\n\nLet $a_1, \\dots, a_n$ denote the sequence and let $I \\subseteq \\{1, \\dots, n\\}$ denote the index set of the non-empty subsequence whose sum is divisible by $n$, that is, $\\sum_{i \\in I} a_i \\equiv 0 \\pmod{n}$ and no other non-empty subset of $\\{1, \\dots, n\\}$ has this property.\n\nWithout loss of generality, we may assume that $n \\in I$ and $a_1 = \\min\\{a_1, \\dots, a_{n-1}\\}$ and $a_2 = \\max\\{a_1, \\dots, a_{n-1}\\}$. For $k < n$, let $s_k := a_1 + a_2 + \\dots + a_k$.\n\nThen for $j < k < n$, we have $s_k - s_j \\equiv \\sum_{i=j+1}^k a_i \\not\\equiv 0 \\pmod{n}$. Thus the $n-1$ sums $a_1, s_2, \\dots, s_{n-1}$ have different residues modulo $n$. Considering the sequence $a_2, a_1, a_3, a_4, \\dots, a_n$, we see that the $n-1$ sums $a_2, s_2, \\dots, s_{n-1}$ also have different residues modulo $n$. It follows that $a_1 \\equiv a_2 \\pmod{n}$. Since $1 \\le a_1 \\le n-1$, we see that $a_1 = a_2$. Hence $a_1 = \\dots = a_{n-1}$.\n\nSince none of the numbers $a_i$ are divisible by $n$, we have $|I| \\ge 2$. Working with an element $m \\in I$ different from $n$, we get $a_1 = \\dots = a_{m-1} = a_{m+1} = \\dots = a_n$. Since $n \\ge 3$, the sequence is constant.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 14069,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $ABCD$ be a parallelogram and a circle $k$ passes through $A$ and $C$ and meets rays $AB$ and $AD$ at $E$ and $F$, respectively. If $BD$, $EF$, and the tangent at $C$ concur, show that $AC$ is the diameter of $k$.\n\n![](path/to/file.png)",
    "options": [],
    "answer": "See solution",
    "solution": "Let the tangent to $k$ at point $C$ intersect the rays $AB$ and $AD$ 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$$\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$$\n\nHence\n\n$$\n\\frac{AD}{ND} \\cdot \\frac{MB}{AB} = \\frac{AF}{NF} \\cdot \\frac{ME}{AE}.\n$$\n\nSince $ABCD$ is a parallelogram, $\\frac{AD}{ND} = \\frac{MC}{NC} = \\frac{MB}{AB}$. From the tangent and secant property, $MC^2 = ME \\cdot MA$ and $NC^2 = NF \\cdot NA$. From the previous equations, it follows that\n\n$$\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$$\n\nTherefore, $AM \\cdot AE = AN \\cdot AF$, i.e., the quadrilateral $EFNM$ is cyclic. Then $\\angle AEF = \\angle ANM$, whence $\\overline{AF} = \\overline{AEC} - \\overline{FC}$, i.e., $\\overline{AEC} = \\overline{AFC}$. It follows that $AC$ is a diameter of $k$. $\\square$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 14070,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $P(x)$ be a polynomial with integer coefficients, with positive leading coefficient. Prove that there are finitely many positive integers $n$ such that $n! + 1$ is a power of $P(n)$.",
    "options": [],
    "answer": "See solution",
    "solution": "Suppose there are infinitely many positive integers $n$ such that $n! + 1 = P(n)^{k_n}$ for some integer $k_n \\geq 1$. For such $n$, call $n$ *good*.\n\nFor large $n$, $P(n) \\geq n$ (since the leading coefficient is positive), so $P(n)^{k_n} = n! + 1 > n!$, which implies $k_n < n$ for large $n$.\n\nAlso, for $n \\geq 2$, $P(n)$ must be odd (since $n!$ is even, so $n! + 1$ is odd, and thus $P(n)$ must be odd).\n\nUsing the Lifting the Exponent Lemma (LTE),\n\n$$\n\\nu_2(P(n)^{k_n} - 1) \\leq \\nu_2(P(n) - 1) + \\nu_2(P(n) + 1) + \\nu_2(k_n).\n$$\n\nBut $P(n)^{k_n} - 1 = n!$, so $\\nu_2(n!) \\leq \\nu_2(P(n) - 1) + \\nu_2(P(n) + 1) + \\nu_2(k_n)$.\n\nThus, for each good $n$, at least one of $\\nu_2(P(n) - 1)$, $\\nu_2(P(n) + 1)$, or $\\nu_2(k_n)$ is at least $\\frac{1}{3}\\nu_2(n!)$.\n\nSince $\\nu_2(n!) \\geq \\frac{n}{2}$, for infinitely many $n$, at least one of the following holds:\n\n- $\\nu_2(P(n) - 1) \\geq \\frac{n}{6}$\n- $\\nu_2(P(n) + 1) \\geq \\frac{n}{6}$\n- $\\nu_2(k_n) \\geq \\frac{n}{6}$\n\nBut $|P(n) - 1|$ and $|P(n) + 1|$ grow polynomially in $n$, so their $2$-adic valuations cannot grow faster than $O(\\log n)$, a contradiction for large $n$.\n\nIf $\\nu_2(k_n) \\geq \\frac{n}{6}$ for infinitely many $n$, then $k_n > 2^{n/6} > n$ for large $n$, contradicting $k_n < n$.\n\nTherefore, there are only finitely many such $n$.",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 14071,
    "subject": "Mathematics (Olympiad)",
    "question": "Perform the coordinate change $a = y + z - x$, $b = x - y + z$, $c = x + y - z$ (then $x = \\frac{b+c}{2}$, $y = \\frac{a+c}{2}$, $z = \\frac{a+b}{2}$, and $x + y + z = a + b + c$).\n\nThe new planes have the equations $a = n$, $b = n$, $c = n$, and $a + b + c = n$, so they partition the space into cubes, each cut into three pieces.\n\nNow, show that for every $a, b, c \\in (0, 1)$ with non-integer sum, there exists a positive integer $k$ such that $ka, kb, kc$ are non-integers, and $1 < \\{ka\\} + \\{kb\\} + \\{kc\\} < 2$.",
    "options": [],
    "answer": "See solution",
    "solution": "The interiors of the octahedra are determined by $1 < \\{a\\} + \\{b\\} + \\{c\\} < 2$, along with $\\{a\\}, \\{b\\}, \\{c\\} \\neq 0$.\n\nIf $a + b + c \\in (1, 2)$, the claim is trivial. If $a + b + c < 1$, choose the smallest $k$ such that $\\{ka\\} + \\{kb\\} + \\{kc\\} > 1$. The case $a + b + c > 2$ is similar.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 14072,
    "subject": "Mathematics (Olympiad)",
    "question": "Show that for every positive integer $n \\geq 3$ there exist distinct positive integers $a_1, a_2, \\dots, a_n$ such that\n\n$$\na_1! a_2! \\dots a_{n-1}! = a_n!\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "For $n = 3$, we have $3! \\cdot 5! = 6!$. Assume that $a_1! a_2! \\dots a_{k-1}! = a_k!$ for some $k \\geq 3$. Since $a_k! \\cdot (a_k! - 1)! = (a_k!)!$, it follows that\n\n$$\na_1! a_2! \\dots a_{k-1}! \\cdot (a_k! - 1)! = a_k! \\cdot (a_k! - 1)! = (a_k!)!\n$$\n\nand obviously $a_k! > a_k! - 1 > a_{k-1}!$. Thus, by induction, the statement holds for all $n \\geq 3$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 14073,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all three-digit numbers equal to the sum of the factorials of their digits.\n\n$$100a + 10b + c = a! + b! + c!$$",
    "options": [],
    "answer": "See solution",
    "solution": "$a! + b! + c! < 1000 \\Rightarrow \\max(a, b, c) \\leq 6$ (since $7! > 1000$, $6! < 1000$).\n\nBecause $a! + b! + c! > 99$, one of the digits $a$, $b$, or $c$ must be $5$ or $6$ (since $4! + 4! + 4! < 99$).\n\nIf one digit is $6$, then $a! + b! + c! > 6! = 720$, which is a contradiction.\n\nNow, $\\max(a, b, c) = 5$. We have three cases:\n\n**a)** $5bc$\n\nThis case is not possible because the largest sum is $5! + 5! + 5! = 360 < 500$.\n\n**b)** $a5c$\n\nThe largest sum in this case is $4! + 5! + 5! = 264$, so $a = 1$ or $a = 2$:\n\n$150: 1! + 5! + 0! = 121 \\neq 150$\n\n$151: 1! + 5! + 1! = 122 \\neq 151$\n\n$152: 1! + 5! + 2! = 123 \\neq 152$\n\n$153: 1! + 5! + 3! = 124 \\neq 153$\n\n$154: 1! + 5! + 4! = 125 \\neq 154$\n\n$155: 1! + 5! + 5! = 126 \\neq 155$\n\n$254: 2! + 5! + 4! = 130 < 200$\n\n$255: 2! + 5! + 5! = 131 \\neq 255$\n\n**c)** $ab5$\n\nIf $a = 1$ or $a = 2$, then the case is similar to b), i.e., the solution is $1! + 4! + 5! = 145$.\n\nThus, the only three-digit number equal to the sum of the factorials of its digits is $145$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 14074,
    "subject": "Mathematics (Olympiad)",
    "question": "6 оронтой тооны сүүлийн 2 цифр нь 1, 2 эсвэл 0, 3 байна. Эхний тохиолдолд уг 6 оронтой тоо 1, 2, 4-ийн аль нэгээр эхэлнэ. 1-ээр эхэлсэн бол бусад цифрүүд нь 1, 2, 4, 0, 0 байна. Эдгээрийг сэлгэх боломжийн тоо хэд вэ?",
    "options": [],
    "answer": "See solution",
    "solution": "1-ээр эхэлсэн тохиолдолд үлдсэн 5 цифр нь 1, 2, 4, 0, 0 байна. Эдгээрийг сэлгэх боломжийн тоо:\n\n$$\n\\frac{5!}{2!} = 60\n$$\n\n2-оор эхэлсэн бол бусад цифрүүдийг сэлгэх боломжийн тоо $2 \\cdot 30 = 60$.\n\n4-өөр эхэлсэн бол $3 \\cdot 20 = 60$.\n\nИймд нийт боломжийн тоо:\n\n$$\n60 + 60 + 60 = 180\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 14075,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $\\triangle ABC$ be a triangle with acute angles and $AB \\neq AC$. Let $V$ be the intersection point of the angle bisector at vertex $A$ with side $BC$, and let $D$ be the foot of the altitude from vertex $A$. Prove that $AD$, $BE$, and $CF$ are concurrent, where $E$ and $F$ are the intersection points of the circumcircle of $\\triangle AVD$ with sides $CA$ and $AB$, respectively.\n\n![](images/Makedonija_2008_p47_data_a31442865b.png)",
    "options": [],
    "answer": "See solution",
    "solution": "Since $\\angle ADV = 90^\\circ$, the points $A$, $D$, $V$, $E$, and $F$ lie on the same circle (with $AV$ as a diameter). We have $\\angle BFV = 180^\\circ - \\angle AFV = 90^\\circ$ and $\\angle CEV = 180^\\circ - \\angle AEV = 90^\\circ$. Thus, $\\triangle BFV \\sim \\triangle BDA$ and $\\triangle CEV \\sim \\triangle CDA$.\n\n$$\n\\frac{BD}{BF} = \\frac{AB}{VB}, \\quad \\frac{CD}{CE} = \\frac{AC}{VC}, \\quad \\text{and} \\quad \\frac{AB}{VB} = \\frac{AC}{VC}.\n$$\n\nTherefore,\n$$\n\\frac{BD}{BF} = \\frac{AB}{VB} = \\frac{AC}{VC} = \\frac{CD}{CE}, \\quad \\text{i.e.} \\quad \\frac{BD}{BF} = \\frac{CD}{CE}.\n$$\n\nAlso, $AE = AF$ since $\\angle FAV = \\angle VAE$.\n\nSo,\n$$\n\\frac{BD}{DC} \\cdot \\frac{CE}{EA} \\cdot \\frac{AF}{FB} = \\frac{BD}{BF} \\cdot \\frac{CE}{CD} = 1.\n$$\n\nBy Ceva's theorem, $AD$, $BE$, and $CF$ are concurrent.",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 14076,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $d$ be the common difference and $a + d$ be the initial term of an arithmetic progression, where $a$ and $d$ are integers such that $a + n d > 0$ for each $n \\in \\mathbb{N}$ (note that $a$ may be negative but $a + d, a + 2d, \\dots$ are all positive). Hence, $d$ is positive and $d > -a$.\n\nFind all such arithmetic progressions $a_n = a + n d$ that satisfy the following condition:\n\nFor all $N \\in \\mathbb{N}$, the product $a_1 a_2 \\cdots a_N$ divides $a_{N+1} a_{N+2} \\cdots a_{2N}$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $d$ be the common difference and $a + d$ the initial term. Assume $a + n d > 0$ for all $n \\in \\mathbb{N}$, so $d > -a$.\n\nLet $P = a_1 a_2 \\cdots a_N$. For $k \\ge N$, $a_1 a_2 \\cdots a_N$ divides $a_{k+1} a_{k+2} \\cdots a_{k+N}$.\n\nLet $n_0 > N$ be congruent to $-1$ modulo $P$. Then:\n$$\nP \\mid (a + (n_0 + 1)d)(a + (n_0 + 2)d) \\cdots (a + (n_0 + N)d)\n$$\nSince $P \\mid n_0 + 1$, we get:\n$$\nP \\mid a(a + d) \\cdots (a + (N - 1)d)\n$$\nBut $P = (a + d)(a + 2d) \\cdots (a + N d)$, so $a + N d \\mid a$.\n\nIf $a > 0$, $a + N d \\le a$ implies $N d \\le 0$. Since $N > 1$, $d = 0$, so $a_n = a$ for all $n$ (constant sequence).\n\nIf $a = 0$, $a_n = n d$ for $d > 0$ also works.\n\nIf $a < 0$, $a + N d \\le |a| = -a$, so $N d \\le -2a$. But $N \\ge 2$ and $d > -a$ gives $-2a < N d \\le -2a$, a contradiction.\n\nThus, the only solutions are $a_n = n d$ for $d > 0$ and constant sequences.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 14077,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $f: \\mathbb{R} \\to \\mathbb{R}$ be a function such that there exist a differentiable function $g: \\mathbb{R} \\to \\mathbb{R}$ and a sequence $(a_n)_{n \\ge 1}$ with strictly positive terms and $\\lim_{n \\to \\infty} a_n = 0$, such that\n\n$$\ng'(x) = \\lim_{n \\to \\infty} \\frac{f(x + a_n) - f(x)}{a_n},\n$$\n\nfor all $x \\in \\mathbb{R}$.\n\na) Give an example of such a function $f$ that is not differentiable at any point $x \\in \\mathbb{R}$.\n\nb) Assume that $f$ is continuous on $\\mathbb{R}$. Show that $f$ is differentiable on $\\mathbb{R}$.",
    "options": [],
    "answer": "See solution",
    "solution": "a) Consider the function $f : \\mathbb{R} \\to \\mathbb{R}$ defined by\n\n$$\nf(x) = \\begin{cases} 1, & x \\in \\mathbb{Q}, \\\\ 0, & x \\in \\mathbb{R} \\setminus \\mathbb{Q} \\end{cases}\n$$\n\nand let $a_n = 1/n$ for all $n \\in \\mathbb{N}^*$. For any $x \\in \\mathbb{R}$ and $n \\in \\mathbb{N}^*$, $f(x + a_n) = f(x)$, so\n\n$$\n\\lim_{n \\to \\infty} \\frac{f(x+a_n)-f(x)}{a_n} = 0 = g'(x)\n$$\n\nwhere $g$ is a constant function. The function $f$ is discontinuous everywhere, so it is not differentiable at any $x \\in \\mathbb{R}$.\n\nb) Define $h = f - g$. The function $h$ is continuous. We have\n\n$$\n\\lim_{n \\to \\infty} \\frac{h(x+a_n)-h(x)}{a_n} = \\lim_{n \\to \\infty} \\frac{f(x+a_n)-f(x)}{a_n} - g'(x) = 0\n$$\n\nfor all $x \\in \\mathbb{R}$. Let $x < y$ in $\\mathbb{R}$ and $c > 0$. Define\n\n$$\nA(c) = \\{z \\in [x, y] \\mid |h(z) - h(x)| \\le c(z - x)\\}\n$$\n\nSince $x \\in A(c)$, $A(c)$ is nonempty and bounded above, so $s = \\sup A(c) \\in [x, y]$. By continuity, $s \\in A(c)$. Suppose $s < y$. Then for large $n$, $s + a_n < y$. Since\n\n$$\n\\lim_{n \\to \\infty} \\frac{h(s+a_n)-h(s)}{a_n} = 0,\n$$\n\nthere exists $n_2$ such that $|h(s+a_{n_2}) - h(s)| < c a_{n_2}$. Then\n\n$$\n|h(s+a_{n_2}) - h(x)| \\le |h(s) - h(x)| + |h(s+a_{n_2}) - h(s)| < c(s-x) + c a_{n_2} = c[(s+a_{n_2}) - x]\n$$\n\nso $s + a_{n_2} \\in A(c)$, contradicting $s = \\sup A(c)$. Thus, $y = s \\in A(c)$, so $|h(y) - h(x)| \\le c(y - x)$. Since $c > 0$ is arbitrary, $h(y) = h(x)$ for all $x, y$, so $h$ is constant. Therefore, $f = g + h$ is differentiable on $\\mathbb{R}$, with $f' = g'$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 14078,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $A$, $B$, $C$, $D$ be the four squares in the centre of a $4 \\times 4$ grid of squares. These four squares form the centre of the grid. A frog can jump from one of the four centre squares to any square in the grid which shares a side. If the frog jumps out of the centre from a particular square for the first time, that square is called an exit square. Find the number of ways in which the frog can start from $A$ with $A$ being the exit square in $n$ jumps. Give your answer in terms of $n$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $A$, $B$, $C$, $D$ be labeled in a clockwise manner. Let $a_n$, $b_n$, $c_n$, $d_n$ be respectively the number of ways to start at $A$, $B$, $C$, $D$ and exit from $A$ in $n$ jumps. Then $b_n = d_n$. Since from $A$, the frog can return to $A$ in an even number of jumps, $a_n = 0$ when $n$ is even.\n\nFrom $A$, in 1 jump, the frog can exit from 2 sides or jump to $B$ or $D$, thus $a_1 = 2$, $a_3 = 4$. Therefore for $n \\geq 3$, $a_n = 2b_{n-1}$.\n\nFrom $B$, the frog can jump to $A$ or $C$. So $b_n = a_{n-1} + c_{n-1}$.\n\nFrom $C$, the frog can jump to $B$ or $D$. So $c_n = 2b_{n-1}$.\n\nTherefore,\n\n$$\na_n = 4a_{n-2}.\n$$\n\nConsequently, $a_{2m+1} = 4^{m-1} a_3 = 4^m$ for $m \\geq 1$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 14079,
    "subject": "Mathematics (Olympiad)",
    "question": "Circle $\\omega$ is internally tangent to circle $\\Omega$ at $T$. Let $P$ and $S$ be variable points on $\\Omega$ and $\\omega$, respectively, such that line $PS$ is tangent to $\\omega$ (at $S$). Determine the locus of $O$ – the circumcenter of triangle $PST$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $X$ and $r$, $Y$ and $R$ denote the center and radius of $\\omega$, $\\Omega$, respectively. Let $\\delta$ denote the circle centered at $Y$ with radius $\\sqrt{R(R - r)}$. The locus of $O$ is the circle $\\delta$ excluding its two intersection points with line $YT$.\n\nFirst, we show that every such center lies on $\\delta$; that is, $OY^2 = R(R - r)$. (For $S \\neq T$, there are two possible positions for $P$. For $P \\neq T$, there are also two possible positions for $S$. Please see the diagram below for possible relative positions between $P$ and $S$, where $O_{i,j}$ denote the circumcenter of triangle $P_iS_jT$.) We will only deal with the right-hand side configuration shown above. (The proof can be modified easily for other configurations.) Note that line $XO$ is the perpendicular bisector of segment $ST$. Hence $XTOS$ is a kite with symmetry axis $OX$. In particular,\n\n$$\n\\angle YTO = \\angle XTO = \\angle XSO = \\angle XSP - \\angle OSP = 90^\\circ - \\angle OSP = \\angle PTS,\n$$\n\nwhere the last two equalities hold because $PS$ is tangent to circle $\\omega$ and $O$ is the circumcenter of triangle $PTS$. Because $YO \\perp TP$ and $XO \\perp ST$, $\\angle YOX = \\angle PTS$. Combining the last two equations together gives $\\angle YTO = \\angle YOX$. In addition, we have $\\angle TYO = \\angle OYX$. Thus, triangles $YXO$ and $YOT$ are similar to each other, from which it follows that $\\dfrac{YX}{OY} = \\dfrac{YO}{TY}$ or $OY^2 = YX \\cdot YT = R(R - r)$.\n\n![](images/USA_IMO_2013-2014_p47_data_7b9f286cd4.png)\n\nBecause $\\angle YTO = \\angle PTS$, which is an interior angle of (non-degenerate) triangle $PTS$, $\\angle YTO \\neq 0^\\circ$. Therefore, the two intersection points between $\\delta$ and line $YT$ are not part of the locus of $O$. Now we show that any other point $O$ on $\\delta$ is indeed the circumcenter of such a triangle *PTS*. For the sake of contradiction, assume that point $O$ on $\\delta$ is not the circumcenter of such a triangle *PTS*. Construct circle $\\Delta$ centered at $O$ with radius $OT$. Let $S$ denote the second intersection (other than $T$) of circles $\\delta$ and $\\Delta$. (Since $Y, T, O$ are not collinear, this second intersection exists.) The tangent line to $\\omega$ at $S$ intersects $\\Omega$ at $P_1$ and $P_2$. Let $O_1$ and $O_2$ be the circumcenters of triangles $TSP_1$ and $TSP_2$. (Clearly, $O_1 \\neq O_2$ because they lie on two parallel lines, namely, the perpendicular bisectors of segments $P_1S$ and $SP_2$.) By our assumption, $O_1$ and $O_2$ are distinct from $O$ (because $O$ is not the circumcenter of such a triangle *PST* while both $O_1$ and $O_2$ are). By the first part of our proof, both $O_1$ and $O_2$ lie on $\\delta$. Now we have three distinct points $O_1, O_2, O$ on $\\delta$ that are collinear (because they lie on the perpendicular bisector of segment $TS$), which is impossible, because a line and a circle can intersect at most two points. Thus our assumption was wrong, and $O$ is the center of such a triangle *PST*.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 14080,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $P$ be a non-zero polynomial with non-negative real coefficients, let $n$ be a positive integer, and let $\\sigma$ be a permutation of the set $\\{1, \\dots, n\\}$. Determine the least value the sum\n\n$$\n\\sum_{i=1}^{n} \\frac{P(x_i^2)}{P(x_i x_{\\sigma(i)})}\n$$\n\nmay achieve, as $x_1, \\dots, x_n$ run through the set of positive real numbers.",
    "options": [],
    "answer": "See solution",
    "solution": "The required minimum is $n$, and is achieved, for instance, at $n$-tuples that are constant on each cycle $\\sigma$ splits into canonically; in particular, at $n$-tuples whose entries are all equal.\n\nLetting $x_1, \\dots, x_n$ be positive real numbers, refer to the AM-GM inequality to write\n\n$$\n\\sum_{i=1}^{n} \\frac{P(x_i^2)}{P(x_i x_{\\sigma(i)})} \\ge n \\left( \\prod_{i=1}^{n} \\frac{P(x_i^2)}{P(x_i x_{\\sigma(i)})} \\right)^{1/n} = n \\left( \\prod_{i=1}^{n} \\frac{\\sqrt{P(x_i^2)} \\sqrt{P(x_{\\sigma(i)}^2)}}{P(x_i x_{\\sigma(i)})} \\right)^{1/n}\n$$\n\nSince the generic factor in the latter product is at least $1$, by the Cauchy-Schwarz inequality, the conclusion follows.",
    "topic": "Discrete Mathematics",
    "subtopic": "Combinatorics"
  },
  {
    "id": 14081,
    "subject": "Mathematics (Olympiad)",
    "question": "Find the least number $m$ for which any five equilateral triangles with combined area $m$ can cover an equilateral triangle of area $1$.",
    "options": [],
    "answer": "See solution",
    "solution": "We prove that $m = 2$.\n\n**Lower bound ($m \\geq 2$):**\nSuppose $s \\in (0,1)$. Consider five equilateral triangles with combined area greater than $2s$, and show they cannot cover an equilateral triangle $\\Delta$ of area $1$.\n\nLet $A_1B_1C_1$ be an equilateral triangle of area $\\frac{1+s}{2}$ with vertices on the corresponding sides of $\\Delta$. Without loss of generality, suppose $2BA_1 \\leq BC$. Then, there exist three equilateral triangles that cannot cover any of the segments $BA_1$, $CB_1$, and $AC_1$. The remaining two triangles, $\\Delta_1$ and $\\Delta_2$, of areas $s$ each, cannot cover $\\Delta$. Otherwise, $\\Delta_1$ and $\\Delta_2$ would have to cover points from the given three segments, so one of them, say $\\Delta_1$, covers points from two segments, say $D \\in A_1B$ and $E \\in B_1C$. Since $S_{A_1B_1C_1} \\geq \\frac{1}{3}$, we have $\\angle A_1B_1C \\geq 90^\\circ$ (to be proved), implying that the side of $\\Delta_1$ is at least $DE \\geq A_1B_1$. Therefore, $S_{\\Delta_1} \\geq S_{A_1B_1C_1}$, a contradiction.\n\n**Upper bound ($m = 2$ is sufficient):**\nGiven five equilateral triangles of areas $a^2, b^2, c^2, d^2, e^2$ such that $a^2 + b^2 + c^2 + d^2 + e^2 = 2$, there exist four of them that cover $\\Delta$.\n\nLet $a \\geq b \\geq c \\geq d \\geq e > 0$.\n- If $a \\geq 1$, then the triangle of area $a^2$ covers $\\Delta$.\n- Otherwise, $b + c > 1$. This is obvious if $c > 1/2$ (since $b \\geq c$), and otherwise:\n  $$\nb^2 = 2 - a^2 - c^2 - d^2 - e^2 > 1 - 3c^2 \\geq (1-c)^2.\n  $$\n\nTherefore, the triangles with areas $a^2, b^2, c^2$, cut from the vertices of $\\Delta$, intersect each other. They do not cover $\\Delta$ if $f = 2 - a - b - c > 0$ and an equilateral triangle of area $f^2$ is not covered. We need to prove $d \\geq f$.\n- If $d > 1/2$ (since $a, b, c \\geq d$), this is clear.\n- Otherwise, since $a, b, c < 1$:\n  $$\nd \\geq 2d^2 \\geq d^2 + e^2 = 2 - a^2 - b^2 - c^2 > 2 - a - b - c = f.\n  $$\n\nThus, $m = 2$ is the least such number.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 14082,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $a_0$, $b_0$, $c_0$ be the initial numbers on the blackboard. After each move, the numbers are updated as follows:\n\n$$\na_{i+1} = \\frac{b_i^2 + c_i^2}{a_i}, \\quad b_{i+1} = \\frac{c_i^2 + a_i^2}{b_i}, \\quad c_{i+1} = \\frac{a_i^2 + b_i^2}{c_i}.\n$$\n\nAfter 5 moves, the sum of the numbers on the blackboard is $2016$. What is the maximum possible value of the sum of the initial numbers?\n",
    "options": [],
    "answer": "See solution",
    "solution": "Let $S(a_i, b_i, c_i)$ denote the sum of the numbers after the $i$-th move. We have:\n\n$$\nS(a_{i+1}, b_{i+1}, c_{i+1}) = \\frac{b_i^2 + c_i^2}{a_i} + \\frac{c_i^2 + a_i^2}{b_i} + \\frac{a_i^2 + b_i^2}{c_i} \\ge 2(a_i + b_i + c_i) = 2S(a_i, b_i, c_i).\n$$\n\nTherefore,\n\n$$\nS(a_i, b_i, c_i) \\le \\frac{1}{2} S(a_{i+1}, b_{i+1}, c_{i+1})\n$$\nfor all $i = 0, 1, 2, 3, 4$. Thus,\n\n$$\nS(a_0, b_0, c_0) \\le \\frac{1}{2^5} S(a_5, b_5, c_5) = \\frac{2016}{32} = 63.\n$$\n\nThe maximum is achieved when $a_0 = b_0 = c_0 = 21$, since then each move multiplies all numbers by 2, and $S(a_5, b_5, c_5) = 32 \\cdot 63 = 2016$.\n",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 14083,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $n \\ge 2$ and $S = \\{1, 2, \\dots, n^2\\}$. For any function $f : S \\to S$, let $\\text{Fix}(f) = \\{x \\in S \\mid f(x) = x\\}$. Find the possible values of the expression\n\n$$\n|\\text{Fix}(f)| + |\\text{Im}(f)| + \\max_{k \\in S} |f^{-1}(k)|\n$$\nas $f$ ranges over all functions $f : S \\to S$.",
    "options": [],
    "answer": "See solution",
    "solution": "We show that the answer is all values from $2n$ to $2n^2 + 1$.\n\nAssume $f$ has $k \\in \\{0, 1, \\dots, n^2\\}$ fixed points. Let $|\\text{Im}(f)| = p$. Also let $s = \\max_{k \\in S} |f^{-1}(k)|$.\n\n**Upper Bound:** From the definitions of $s$ and $p$, we get $sp \\ge n^2$. We also have $s \\le n^2 - p + 1$ (since $p$ values are in the image, each with at least 1 preimage). Thus,\n\n$$\nk + p + s \\le k + p + (n^2 - p + 1) = n^2 + k + 1 \\le 2n^2 + 1.\n$$\n\n**Lower Bound:** For the minimum, using AM-GM, $k + p + s \\ge k + p + \\frac{n^2}{p} \\ge k + 2n \\ge 2n$.\n\nNow we show all those values are achievable by induction on $n$.\n\n_Base case $n = 2$:_ The identity achieves the maximum value of $9$. A $2$-to-$1$ function can take the values $4, 5, 6$ depending on the number of fixed points, and a function that's $3$-to-$1$ on three of the inputs can achieve $7$ and $8$.\n\n_Inductive step:_ Suppose we have a function $g : S \\to S$ and let $T = \\{1, 2, \\dots, (n+1)^2\\}$. Build $f : T \\to T$ by $f(x) = g(x)$ for $x \\le n^2$.\n\n**Values from $4n+1$ to $2(n+1)^2+1$:** For $x > n^2$, define $f$ as any permutation of $\\{n^2+1, \\dots, (n+1)^2\\}$. This doesn't increase the maximum preimage size, and adds $2n+1$ to the image size. The number of fixed points can be any from $0$ to $2n+1$. Thus, we can add any number from $2n+1$ to $4n+2$ to the value for $g$, so by induction, we can hit all values $4n+1$ to $2n^2+1+(4n+2) = 2(n+1)^2+1$.\n\n**Values from $2n+2$ to $4n$:**\n- If $s \\ge n+1$, send $n$ of the new points to one new point and the other $n+1$ to another, with no new fixed points. This doesn't increase the maximum preimage size or fixed points, but adds $2$ to the image size.\n- If $s \\le n$, then $|\\text{Im}(f)| \\ge n$. If $|\\text{Im}(f)| \\ge n+1$, assign $n+1$ new points to distinct elements in $\\text{Im}(f)$, and the other $n$ to a new point (no new fixed points). This adds $1$ to the image and increases the maximum preimage by $1$, so again we add $2$ overall.\n- If $s = |\\text{Im}(f)| = n$, assign $n$ new points to distinct elements in $\\text{Im}(f)$ and the other $n+1$ to a single new point.\n\nIn all cases, we add $2$ to the expression, so we can get all values $2n+2, \\dots, 2n^2+3$, covering all values.\n\n$\\Box$",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 14084,
    "subject": "Mathematics (Olympiad)",
    "question": "Determine all integers $n \\geq 1$ for which there exist $n$ real numbers $x_1, x_2, \\dots, x_n$ in the closed interval $[-4, 2]$ such that the following three conditions are fulfilled:\n\n- The sum of these real numbers is at least $n$.\n- The sum of their squares is at most $4n$.\n- The sum of their fourth powers is at least $34n$.",
    "options": [],
    "answer": "See solution",
    "solution": "Since the problem concerns $n$ real numbers $x_1, x_2, \\dots, x_n$ in the closed interval $[-4, 2]$, consider the polynomial\n\n$$\nP(x) = (x+4)(x-2)(x-1)^2,\n$$\n\nwhich satisfies $P(x) \\leq 0$ for $x \\in [-4, 2]$.\n\nSumming $P(x_i)$ for $i = 1, \\dots, n$ and using the problem's conditions, we get:\n\n$$\n0 \\geq \\sum_{i=1}^n P(x_i) = \\sum_{i=1}^n x_i^4 - 11 \\sum_{i=1}^n x_i^2 + 18 \\sum_{i=1}^n x_i - 8n \\geq 34n - 11 \\cdot 4n + 18n - 8n = 0.\n$$\n\nThus, equality holds and $P(x_i) = 0$ for all $i$, so $x_i \\in \\{-4, 1, 2\\}$ for all $i$.\n\nLet $a$ be the number of $-4$'s, $b$ the number of $1$'s, and $c$ the number of $2$'s, with $a + b + c = n$. The conditions become:\n\n$$\n\\begin{cases}\n-4a + b + 2c \\geq n \\\\\n16a + b + 4c \\leq 4n \\\\\n256a + b + 16c \\geq 34n\n\\end{cases}\n$$\n\nwhich simplify to:\n\n$$\n\\begin{cases}\nc \\geq 5a \\\\\nb \\geq 4a \\\\\n222a \\geq 33b + 18c\n\\end{cases}\n$$\n\nMultiplying the first by $18$ and the second by $33$ and adding, we get $33b + 18c \\geq 222a$. Combined with $222a \\geq 33b + 18c$, we have $33b + 18c = 222a$, so $b = 4a$, $c = 5a$, and $n = a + b + c = 10a$.\n\nTherefore, such numbers exist if and only if $n$ is a multiple of $10$. For $n = 10m$ ($m \\geq 1$), a solution is $m$ times $-4$, $4m$ times $1$, and $5m$ times $2$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 14085,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all positive factors of $10^{2013} - 1$ that are less than or equal to 100.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $\\gcd(a, b)$ denote the greatest common divisor of $a$ and $b$. For integers $a, b, p$, $a \\equiv b \\pmod p$ means $a - b$ is a multiple of $p$.\n\nWe use:\n- If $a^n \\equiv 1 \\pmod p$, then $a^m \\equiv 1 \\pmod p$ if and only if $a^{\\gcd(m, n)} \\equiv 1 \\pmod p$.\n- If $p$ is prime and $a$ is not a multiple of $p$, then $a^{p-1} \\equiv 1 \\pmod p$ (Fermat's Little Theorem).\n\nWe seek all positive factors of $10^{2013} - 1$ less than or equal to 100. First, find all primes $p < 100$ dividing $10^{2013} - 1$.\n\n$10^{2013} - 1 \\equiv 0 \\pmod p$ if and only if $10^{2013} \\equiv 1 \\pmod p$, which holds if $10^d \\equiv 1 \\pmod p$ where $d = \\gcd(2013, p-1)$. Since $2013 = 3 \\cdot 11 \\cdot 61$, $d$ must be one of $1, 3, 11, 33, 61$.\n\n- $d=1$: $p=3$ is the only such prime.\n- $d=3$: $10^3 - 1 = 999 = 3^3 \\cdot 37$, so $p=37$.\n- $d=11$: $p=23, 89$ but neither satisfies $10^{11} \\equiv 1 \\pmod p$.\n- $d=33$: $p=67$ and $10^{33} \\equiv 1 \\pmod{67}$.\n- $d=61$: No such $p < 100$.\n\nThus, the only prime factors of $10^{2013} - 1$ less than 100 are $3, 37, 67$.\n\nAny positive factor less than or equal to 100 must be a product of these primes: $1, 3, 9, 27, 37, 67, 81$.\n\nCheck if $9, 27, 81$ divide $10^{2013} - 1$:\n\n$10^{2013} - 1 = (10 - 1)(10^{2012} + 10^{2011} + \\dots + 10^1 + 1)$\n\n$10^{2012} + \\dots + 1 \\equiv 2013 \\equiv 6 \\pmod{9}$, so $10^{2013} - 1 \\equiv 54 \\pmod{81}$.\n\nThus, $9$ and $27$ divide $10^{2013} - 1$, but $81$ does not.\n\n**Final answer:** $1, 3, 9, 27, 37, 67$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 14086,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $m$ and $n$ be positive integers with $m > n$. For $k = 1, 2, \\dots, n+1$, define\n\n$$\nx_k = \\frac{m + k}{n + k}.\n$$\n\nProve that if $x_1, x_2, \\dots, x_{n+1}$ are all integers, then $x_1 x_2 \\cdots x_{n+1} - 1$ is divisible by an odd prime.",
    "options": [],
    "answer": "See solution",
    "solution": "Assume that $x_1, \\dots, x_{n+1}$ are integers. Define\n\n$$\na_k := x_k - 1 = \\frac{m + k}{n + k} - 1 = \\frac{m - n}{n + k} > 0\n$$\n\nfor $k = 1, 2, \\dots, n+1$.\n\nLet $P = x_1 x_2 \\cdots x_{n+1} - 1$. We need to prove that $P$ is divisible by an odd prime, i.e., $P$ is not a power of $2$.\n\nLet $2^d$ be the largest power of $2$ dividing $m - n$, and let $2^c$ be the largest power of $2$ not exceeding $2n + 1$. Then $2n + 1 \\le 2^{c+1} - 1$, so $n + 1 \\le 2^c$. Thus, $2^c$ is one of the numbers $n + 1, n + 2, \\dots, 2n + 1$, and it is the only multiple of $2^c$ among them. Let $\\ell$ be such that $n + \\ell = 2^c$. Since $\\frac{m - n}{n + \\ell}$ is an integer, $d \\ge c$.\n\nTherefore, $2^{d - c + 1} \\nmid a_\\ell = \\frac{m - n}{n + \\ell}$, while $2^{d - c + 1} \\mid a_k$ for all $k \\in \\{1, \\dots, n+1\\} \\setminus \\{\\ell\\}$.\n\nComputing modulo $2^{d - c + 1}$:\n\n$$\n\\begin{aligned}\nP &= (a_1 + 1)(a_2 + 1) \\cdots (a_{n+1} + 1) - 1 \\\\\n&\\equiv (a_\\ell + 1) \\cdot 1^n - 1 \\\\\n&\\equiv a_\\ell \\not\\equiv 0 \\pmod{2^{d - c + 1}}.\n\\end{aligned}\n$$\n\nTherefore, $2^{d - c + 1} \\nmid P$.\n\nOn the other hand, for any $k \\in \\{1, \\dots, n+1\\} \\setminus \\{\\ell\\}$, $2^{d - c + 1} \\mid a_k$. So $P \\ge a_k \\ge 2^{d - c + 1}$, and it follows that $P$ is not a power of $2$. Thus, $P$ is divisible by an odd prime.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 14087,
    "subject": "Mathematics (Olympiad)",
    "question": "Consider a function $f : \\mathbb{R} \\to [0, \\infty)$. Prove that $f$ satisfies the inequality $f(x+y) \\geq (1+y)f(x)$ for any $x \\in \\mathbb{R}$ and any $y \\geq 0$, if and only if the function $g : \\mathbb{R} \\to [0, \\infty)$ defined by $g(x) = e^{-x}f(x)$, for $x \\in \\mathbb{R}$, is non-decreasing.",
    "options": [],
    "answer": "See solution",
    "solution": "The inequality $e^y \\geq 1+y$ holds for all $y \\in \\mathbb{R}$. Assume the function $g$ is monotonically increasing on $\\mathbb{R}$. Then, based on the inequality above, we get\n\n$$\nf(x + y) = e^{x+y}g(x + y) \\geq e^x(1 + y)g(x) = (1 + y)f(x),\n$$\nfor all $x \\in \\mathbb{R}$ and all $y \\geq 0$.\n\nConversely, assume $f(x + y) \\geq (1 + y)f(x)$ for all $x \\in \\mathbb{R}$ and all $y \\geq 0$. One can check by induction\n\n$$\nf(x + nt) \\geq (1 + t)^n f(x),\n$$\nfor all $x \\in \\mathbb{R}$, $t \\geq 0$, and $n \\in \\mathbb{N}$. Let $x, z \\in \\mathbb{R}$, with $x < z$.\n\nDenote $y = z - x$. For $n \\in \\mathbb{N}^*$ we have\n\n$$\ng(z) = e^{-z}f(z) = e^{-x-y}f\\left(x + n\\frac{y}{n}\\right) \\geq e^{-x-y}\\left(1 + \\frac{y}{n}\\right)^n f(x) = \\frac{\\left(1 + \\frac{y}{n}\\right)^n}{e^y}g(x).\n$$\nIt follows\n\n$$\ng(z) \\geq \\lim_{n \\to \\infty} \\frac{\\left(1 + \\frac{y}{n}\\right)^n}{e^y} g(x) = g(x),\n$$\ntherefore $g$ is monotonically increasing on $\\mathbb{R}$.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 14088,
    "subject": "Mathematics (Olympiad)",
    "question": "On the boundary of triangle $ABC$, points $D_1, D_2, E_1, E_2, F_1, F_2$ are chosen so that when going around the perimeter, the points are encountered in the following order: $A, F_1, F_2, B, D_1, D_2, C, E_1, E_2$. Given that $AD_1 = AD_2 = BE_1 = BE_2 = CF_1 = CF_2$, prove that the two triangles formed by the triples of lines $AD_1, BE_1, CF_1$ and $AD_2, BE_2, CF_2$ have equal perimeters.",
    "options": [],
    "answer": "See solution",
    "solution": "Let's begin with the following useful lemma.\n\n**Lemma.** Let points $F$ and $E$ be chosen on sides $AB$ and $AC$ of parallelogram $ABKC$ respectively, such that $BE = CF$. Then point $K$ is equidistant from lines $BE$ and $CF$ (see the figure below).\n\n![](images/2025-02_p8_data_2cab172397.png)\n\n**Proof.** Since $BK \\parallel EC$ and $CK \\parallel FB$, we have $S_{KBE} = S_{KBC} = S_{KFC}$. As $BE = CF$, it follows that the distances from point $K$ to lines $BE$ and $CF$ are equal. $\\square$\n\nNow let's proceed to the solution. Suppose the lines given in the condition form triangles $X_1Y_1Z_1$ and $X_2Y_2Z_2$ (points are labeled as in the figure below).\n\nChoose point $K$ such that $ABKC$ is a parallelogram. According to the lemma, point $K$ is equidistant from lines $BE_1, CF_1, BE_2$, and $CF_2$. Therefore, there exists a circle centered at $K$ that is tangent to these lines at points $P_1, Q_1, P_2$, and $Q_2$ respectively. From the equality of tangent segments we obtain:\n\n$$\nBX_1 - CX_1 = BP_1 + X_1P_1 - X_1Q_1 + CQ_1 = BP_2 + CQ_2 =\n$$\n\n![](images/2025-02_p9_data_a4308fd9b3.png)\n\n$$\n= BP_2 - X_2P_2 + X_2Q_2 + CQ_2 = CX_2 - BX_2.\n$$\n\nSimilarly, we get $CY_1 - AY_1 = AY_2 - CY_2$ and $AZ_1 - BZ_1 = BZ_2 - AZ_2$. Adding these three equalities yields the required equality of perimeters.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 14089,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $ABC$ be a non-isosceles triangle, and let $\\Gamma$ be its incircle. Let $D$, $E$, $F$ be the points of contact of $\\Gamma$ with the sides $BC$, $CA$, $AB$ respectively. Suppose $FD$, $DE$, $EF$ intersect $CA$, $AB$, $BC$ in $U$, $V$, $W$ respectively. If $L$, $M$, $N$ are respectively the midpoints of $DW$, $EU$, $FV$, prove that $L$, $M$, $N$ are collinear.",
    "options": [],
    "answer": "See solution",
    "solution": "![](images/Indija_TS_2008_p0_data_035b13253d.png)\n\nDraw a line through $N$ which is parallel to $ED$. This passes through the midpoints $P$ of $DF$ and $Q$ of $FE$. Similarly, the line through $M$ parallel to $FD$ passes through the midpoint $Q$ of $FE$ and $R$ of $ED$; the line through $L$ parallel to $FE$ passes through the midpoint $P$ of $FD$ and $R$ of $ED$. We thus obtain the medial triangle $PRQ$ of $ABC$. Since $AF = AE$, the line $AQ$ is also the bisector of $\\angle A$. Similarly, $BP$ bisects $\\angle B$ and $CR$ bisects $\\angle C$.\n\nThus $AQ$, $BP$, $CR$ concur at $I$, the incenter of $ABC$. Now Desargues' theorem is applicable to the triangles $ABC$ and $QPR$. It follows that $PR \\cap BC$, $RQ \\cap CA$ and $QP \\cap AB$ are collinear. Thus $L$, $M$ and $N$ are collinear.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 14090,
    "subject": "Mathematics (Olympiad)",
    "question": "Alice and Bob determine a number with 2018 digits in the decimal system by choosing digits from left to right. Alice starts, and then they each choose a digit in turn. They must observe the rule that each digit must differ from the previously chosen digit modulo 3.\n\nSince Bob will make the last move, he bets that he can make sure that the final number is divisible by 3. Can Alice avoid that?",
    "options": [],
    "answer": "See solution",
    "solution": "It is well-known that every number is congruent to the sum of its digits modulo 3 in the decimal system. Therefore, it is sufficient to consider the digits modulo 3. In particular, it is enough to only consider digits in $\\{1, 2, 3\\}$.\n\nIn the fourth move from the end, Alice makes sure that the sum of the digits is not divisible by 3 after her move. This is always possible because she has two options, which cannot both lead to multiples of 3. After the next move, the sum of digits is congruent to some $x$ modulo 3, but $x$ is not the digit chosen by Bob because the sum of digits was not a multiple of 3 before Bob's move.\n\nIn the penultimate move, Alice can therefore choose $x$. After her move, the sum of digits is congruent to $2x \\equiv -x \\pmod{3}$. In order to get a multiple of 3, Bob would have to choose another $x$, which is prohibited.\n\nTherefore, Bob cannot reach his goal.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 14091,
    "subject": "Mathematics (Olympiad)",
    "question": "The numbers $x$, $y$, $z$, $t$, $a$, and $b$ are positive integers such that $xt - yz = 1$ and $\\frac{x}{y} > \\frac{a}{b} > \\frac{z}{t}$. Prove that $ab \\geq (x + z)(y + t)$.",
    "options": [],
    "answer": "See solution",
    "solution": "From $\\frac{x}{y} > \\frac{a}{b}$, it follows that $xb > ya$, so $xb - ya \\geq 1$. Similarly, $\\frac{a}{b} > \\frac{z}{t}$ implies $at > bz$, so $at - bz \\geq 1$.\n\nMultiplying the first inequality by $t$ and the second by $y$, then adding, we get:\n\n$$\n(xb - ya)t + (at - bz)y \\geq t + y\n$$\nExpanding:\n$$\nxb t - ya t + a t y - b z y \\geq t + y\n$$\nGroup terms:\n$$\n(b x t - b y z) + (a t y - a y t) \\geq t + y\n$$\nNote $a t y - a y t = 0$, so:\n$$\n(b x t - b y z) \\geq t + y\n$$\nSince $x t - y z = 1$, this gives $b \\geq t + y$.\n\nSimilarly, $a \\geq x + z$.\n\nTherefore,\n$$\nab \\geq (x + z)(y + t)\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 14092,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all triples of integers $a, b, c$ such that $a$, $b$, and $c$ are the lengths of the sides of a right-angled triangle whose area is $a + b + c$.",
    "options": [],
    "answer": "See solution",
    "solution": "Assume $a \\leq b \\leq c$.\n\n$$\na^2 + b^2 = c^2. \\quad (1)\n$$\n\nThe area condition gives $\\dfrac{ab}{2} = a + b + c$, so $ab = 2(a + b + c)$.\n\nSubstitute $c = \\sqrt{a^2 + b^2}$:\n\n$$\nab = 2(a + b + \\sqrt{a^2 + b^2})\n$$\n\nLet us manipulate algebraically. Rearranging, we get:\n\n$$\nab - 2(a + b) = 2c\n$$\n\nBut $c$ is integer, so $ab - 2(a + b)$ must be even. Try small integer values for $a$ and $b$.\n\nAlternatively, set $ab - 4(a + b) + 16 = 8$ (as in the original solution), so:\n\n$$\n(a - 4)(b - 4) = 8\n$$\n\nThe positive integer solutions are $(a, b) = (5, 12)$ and $(6, 8)$. The corresponding $c$ values are $13$ and $10$ respectively.\n\nThus, the solutions (up to permutation) are $(5, 12, 13)$ and $(6, 8, 10)$.\n\n$\\boxed{(5, 12, 13),\\ (6, 8, 10)}$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 14093,
    "subject": "Mathematics (Olympiad)",
    "question": "Alice and Bob play a game with a string of 2015 pearls.\n\nIn each move, one player cuts the string between two pearls, and the other player chooses one of the resulting parts of the string while the other part is discarded.\n\nIn the first move, Alice cuts the string; thereafter, the players take turns.\n\nA player loses if they obtain a string with a single pearl such that no more cut is possible.\n\nWho of the two players has a winning strategy?",
    "options": [],
    "answer": "See solution",
    "solution": "We claim that the winning situations are exactly the strings with an even number of pearls. We prove this by induction.\n\nA string with one pearl is a losing situation by definition.\n\nA string with an even number $n$ of pearls can easily be cut into two odd parts. These parts are losing situations for the other player by induction, so $n$ is a winning situation.\n\nFor an odd number of pearls, each cut produces an even part. The other player can thus choose this even part, which is a winning situation by induction. Therefore, an odd number $n$ is a losing position.\n\nWe conclude that Bob has a winning strategy.",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 14094,
    "subject": "Mathematics (Olympiad)",
    "question": "Реши ја равенката $p^{2q} + q^{2p} = r$ во множеството прости броеви.",
    "options": [],
    "answer": "See solution",
    "solution": "Јасно е дека $r > 2$, од каде мора $r$ да биде непарен прост број. Еден од броевите $p$ или $q$ мора да биде 2, а другиот непарен прост број. Без губење на општоста, нека $q=2$ и $p$ е непарен. Но тогаш равенката е од облик $p^4 + 2^{2p} = r$, односно $p^4 + 4 \\cdot 2^{4k} = r$ каде $p = 2k + 1$, $k \\in \\mathbb{N}$. Но тогаш\n\n$$\n\\begin{aligned}\np^4 + 4 \\cdot 2^{4k} &= p^4 + 4 \\cdot 2^{4k} + 4 \\cdot 2^{2k} p^2 - 4 \\cdot 2^{2k} p^2 \\\\\n&= (p^2 + 2 \\cdot 2^{2k})^2 - 4 \\cdot 2^{2k} p^2 \\\\\n&= (p^2 + 2 \\cdot 2^{2k} + 2 \\cdot 2^k p)(p^2 + 2 \\cdot 2^{2k} - 2 \\cdot 2^k p) \\\\\n&= (p^2 + 2 \\cdot 2^{2k} + 2 \\cdot 2^k p)((p - 2^k)^2 + 2^{2k})\n\\end{aligned}\n$$\n\nТоа значи дека бројот $p^4 + 2^{2p}$ не е никогаш прост, што значи дека равенката нема решение во множеството прости броеви.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 14095,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $n$ be a positive integer, and let $d_1, \\dots, d_{\\tau(n)}$ denote the divisors of $n$. Define $\\sigma(n)$ as the sum of the divisors of $n$, and $\\tau(n)$ as the number of divisors of $n$.\n\nProve the following two inequalities:\n\n$$\n\\frac{\\sigma(n)}{\\tau(n)} \\le \\frac{n+1}{2}\n$$\n\nand\n\n$$\n\\frac{\\sigma(n)}{\\tau(n)} \\ge \\sqrt{n}.\n$$\n\nDetermine when equality holds in each case.",
    "options": [],
    "answer": "See solution",
    "solution": "Note that if $d$ takes on the values of all divisors of $n$, then $n/d$ also takes on all divisor values. Let $d_1, \\dots, d_{\\tau(n)}$ be the divisors of $n$. We have\n\n$$\n\\sigma(n) = \\frac{1}{2} \\left( (d_1 + \\frac{n}{d_1}) + (d_2 + \\frac{n}{d_2}) + \\dots + (d_{\\tau(n)} + \\frac{n}{d_{\\tau(n)}}) \\right).\n$$\n\n**Upper bound:**\n\nFor $0 < y < x \\le \\sqrt{n}$, $x + \\frac{n}{x} < y + \\frac{n}{y}$ is equivalent to $xy < n$, which holds since $xy < n$. For any divisor $d_i$, either $d_i \\le \\sqrt{n}$ or $n/d_i \\le \\sqrt{n}$, so $d_i + \\frac{n}{d_i} \\le 1 + n = n+1$.\n\nSubstituting $n+1$ for each $d_i + n/d_i$ gives $\\sigma(n) \\le \\frac{1}{2}(n+1)\\tau(n)$, so\n\n$$\n\\frac{\\sigma(n)}{\\tau(n)} \\le \\frac{n+1}{2}.\n$$\n\nEquality holds only if $n$ is prime (otherwise, for $d_i \\ne 1, n$, $d_i + \\frac{n}{d_i} < n+1$).\n\n**Lower bound:**\n\nBy Cauchy's inequality, $x + \\frac{n}{x} \\ge 2\\sqrt{n}$ for any $x > 0$. Substituting $2\\sqrt{n}$ for each $d_i + n/d_i$ gives $\\sigma(n) \\ge \\sqrt{n}\\tau(n)$, so\n\n$$\n\\frac{\\sigma(n)}{\\tau(n)} \\ge \\sqrt{n}.\n$$\n\nEquality holds only if $n=1$.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 14096,
    "subject": "Mathematics (Olympiad)",
    "question": "A bookshelf contains $n$ volumes, labelled $1$ to $n$, in some order. The librarian wishes to put them in the correct order as follows. The librarian selects a volume that is too far to the right, say the volume with label $k$, takes it out, and inserts it in the $k$-th position. For example, if the bookshelf contains the volumes $1$, $3$, $2$, $4$ in that order, the librarian could take out volume $2$ and place it in the second position. The books will then be in the correct order $1$, $2$, $3$, $4$.\n\n(a) Show that if this process is repeated, then, however the librarian makes the selections, all the volumes will eventually be in the correct order.\n\n(b) What is the largest number of steps that this process can take?",
    "options": [],
    "answer": "See solution",
    "solution": "(a) If $t_k$ is the number of times that volume $k$ is selected, then $t_k \\leq 1 + (t_1 + t_2 + \\dots + t_{k-1})$. This is because volume $k$ must move to the right between selections, which means some volume was placed to its left. The only way that can happen is if a lower-numbered volume was selected. This leads to the bound $t_k \\leq 2^{k-1}$. Furthermore, $t_n = 0$ since the $n$th volume will never be too far to the right. Therefore, if $N$ is the total number of moves then\n\n$$\nN = t_1 + t_2 + \\dots + t_{n-1} \\leq 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.\n\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": 14097,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $ABCD$ be a convex quadrilateral with $AC \\perp BD$. Prove that there exist points $P, Q, R, S$ on the sides $AB, BC, CD, DA$, respectively, such that $PR \\perp QS$ and the area of quadrilateral $PQRS$ is exactly half that of $ABCD$.",
    "options": [],
    "answer": "See solution",
    "solution": "Denote the area of $\\Gamma$ by $[\\Gamma]$. Let $E$ be the intersection of $AC$ and $BD$. From any point $P'$ on $AB$, we construct points $Q', R', S'$ as follows:\n\nLet the line $P'E$ intersect $CD$ at $R'$. Let $\\ell$ be the line perpendicular to $P'R'$ at $E$. Let $\\ell$ intersect $BC$ and $DA$ at $Q'$ and $S'$, respectively.\n\n![](images/tmc2017_New_p15_data_a83c3bcbf0.png)\n\nDefine a continuous function $f$ from a point $P'$ on $AB$ to the interval $[0, 1]$ by $f(P') = \\frac{[P'Q'R'S']}{[ABCD]}$, where $Q', R', S'$ are constructed as above.\n\nWe see that $f(A) = f(B) = 1$. If we can find a point $P'$ on $AB$ such that $f(P') \\leq \\frac{1}{2}$, then by continuity there exists a point $P$ on $AB$ such that $f(P) = \\frac{1}{2}$. Then the construction above will yield the points $P, Q, R, S$ as required.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 14098,
    "subject": "Mathematics (Olympiad)",
    "question": "At a university dinner, there are 2017 mathematicians who each order two distinct entrées, with no two mathematicians ordering the same pair of entrées. The cost of each entrée is equal to the number of mathematicians who ordered it, and the university pays for each mathematician's less expensive entrée (ties broken arbitrarily). Over all possible sets of orders, what is the maximum total amount the university could have paid?\n\nIn graph theoretic terms: we wish to determine the maximum possible value of\n\n$$\nS(G) \\stackrel{\\text{def}}{=} \\sum_{e=vw} \\min(\\deg v, \\deg w)\n$$\n\nacross all graphs $G$ with 2017 edges.",
    "options": [],
    "answer": "See solution",
    "solution": "**First solution (combinatorial, Evan Chen)**\n\nFirst define $L_k$ to consist of a clique on $k$ vertices, plus a single vertex connected to exactly one vertex of the clique. Hence $L_k$ has $k+1$ vertices, $\\binom{k}{2} + 1$ edges, and $S(L_k) = (k-1)\\binom{k}{2} + 1$. In particular, $L_{64}$ achieves the claimed maximum, so it suffices to prove the upper bound.\n\n**Lemma**\n\nLet $G$ be a graph such that either\n\n* $G$ has $\\binom{k}{2}$ edges for some $k \\ge 3$ or\n* $G$ has $\\binom{k}{2} + 1$ edges for some $k \\ge 4$.\n\nThen there exists a graph $G^*$ with the same number of edges such that $S(G^*) \\ge S(G)$, and moreover $G^*$ has a universal vertex (i.e. a vertex adjacent to every other vertex).\n\n*Proof.* Fix $k$ and the number $m$ of edges. We prove the result by induction on the number $n$ of vertices in $G$. Since the lemma has two parts, we will need two different base cases:\n\n1. Suppose $n = k$ and $m = \\binom{k}{2}$. Then $G$ must be a clique so pick $G^* = G$.\n2. Suppose $n = k+1$ and $m = \\binom{k}{2} + 1$. If $G$ has no universal vertex, we claim we may take $G^* = L_k$. Indeed each vertex of $G$ has degree at most $k-1$, and the average degree is\n\n$$\n\\frac{2m}{n} = 2 \\cdot \\frac{k^2 - k + 1}{k + 1} < k - 1\n$$\n\nusing here $k \\ge 4$. Thus there exists a vertex $w$ of degree $1 \\le d \\le k-2$. The edges touching $w$ will have label at most $d$ and hence\n\n$$\n\\begin{aligned}\nS(G) &\\le (k-1)(m-d) + d^2 = (k-1)m - d(k-1-d) \\\\\n&\\le (k-1)m - (k-2) = (k-1)\\binom{k}{2} + 1 = S(G^*).\n\\end{aligned}\n$$\n\nNow we settle the inductive step. Let $w$ be a vertex with minimal degree $0 \\le d < k-1$, with neighbors $w_1, \\dots, w_d$. By our assumption, for each $w_i$ there exists a vertex $v_i$ for which $v_i w_i \\notin E$. Now, we may delete all edges $ww_i$ and in their place put $v_i w_i$, and then delete the vertex $w$. This gives a graph $G'$, possibly with multiple edges (if $v_i = w_j$ and $w_j = v_i$), and with one fewer vertex.\n\n![](images/sols-TST-IMO-2018_p7_data_016e463360.png)\n![](images/sols-TST-IMO-2018_p7_data_cdd6b7c057.png)\n![](images/sols-TST-IMO-2018_p7_data_635db63b99.png)\n\nWe then construct a graph $G''$ by taking any pair of double edges, deleting one of them, and adding any missing edge of $G''$ in its place. (This is always possible, since when $m = \\binom{k}{2}$ we have $n-1 \\ge k$ and when $m = \\binom{k}{2} + 1$ we have $n-1 \\ge k+1$.)\n\nThus we have arrived at a simple graph $G''$ with one fewer vertex. We also observe that we have $S(G'') \\ge S(G)$; after all every vertex in $G''$ has degree at least as large as it did in $G$, and the $d$ edges we deleted have been replaced with new edges which will have labels at least $d$. Hence we may apply the inductive hypothesis to the graph $G''$ to obtain $G^*$ with $S(G^*) \\ge S(G'') \\ge S(G)$. $\\square$\n\nThe problem then is completed once we prove the following:\n\n**Claim.** For any graph $G$,\n\n* If $G$ has $\\binom{k}{2}$ edges for $k \\ge 3$, then $S(G) \\le \\binom{k}{2} \\cdot (k-1)$.\n* If $G$ has $\\binom{k}{2} + 1$ edges for $k \\ge 4$, then $S(G) \\le \\binom{k}{2} \\cdot (k-1) + 1$.\n\n*Proof.* We prove both parts at once by induction on $k$, with the base case $k=3$ being plain (there is nothing to prove in the second part for $k=3$). Thus assume $k \\ge 4$. By the earlier lemma, we may assume $G$ has a universal vertex $v$. For notational convenience, we say $G$ has $\\binom{k}{2} + \\varepsilon$ edges for $\\varepsilon \\in \\{0,1\\}$, and $G$ has $p+1$ vertices, where $p \\ge k-1+\\varepsilon$.\n\nLet $H$ be the subgraph obtained when $v$ is deleted. Then $m = \\binom{k}{2} + \\varepsilon - p$ is the number of edges in $H$; from $p \\ge k-1+\\varepsilon$ we have $m \\le \\binom{k-1}{2}$ and so we may apply the inductive hypothesis to $H$ to deduce $S(H) \\le \\binom{k-1}{2} \\cdot (k-2)$.\n\n![](images/sols-TST-IMO-2018_p7_data_673f4a8390.png)",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 14099,
    "subject": "Mathematics (Olympiad)",
    "question": "Suppose that $ABCD$ is a square and that $P$ is a point which is on the circle inscribed in the square. Determine whether or not it is possible that $PA$, $PB$, $PC$, $PD$ and $AB$ are all integers.\n\n![](images/V_Britanija_2013_p28_data_54909547c8.png)",
    "options": [],
    "answer": "See solution",
    "solution": "Assign Cartesian coordinates to the points: $A(a, a)$, $B(-a, a)$, $C(-a, -a)$, $D(a, -a)$, and $P(b, c)$. Since $2a$ is an integer (the side length), $a$ is rational.\n\nBy the Pythagorean theorem:\n\n$$\nAP^2 = 2a^2 + b^2 + c^2 - 2ab - 2ac\n$$\n$$\nBP^2 = 2a^2 + b^2 + c^2 + 2ab - 2ac\n$$\n$$\nCP^2 = 2a^2 + b^2 + c^2 + 2ab + 2ac\n$$\n$$\nDP^2 = 2a^2 + b^2 + c^2 - 2ab + 2ac\n$$\n\nFrom these,\n$$\nBP^2 - AP^2 = 4ab\n$$\n$$\nCP^2 - BP^2 = 4ac\n$$\nSo $b$ and $c$ are rational.\n\nWe can scale $a$, $b$, $c$ by a common denominator to make them integers. If all are even, scale down by 2 until at least one is odd.\n\nSince $P$ is on the incircle, $b^2 + c^2 = a^2$. Not all of $a$, $b$, $c$ can be even, so exactly one of $b$ or $c$ is even. WLOG, let $b$ be even, $a$ and $c$ odd.\n\nThen $BP^2$ and $CP^2$ are both odd, so both are $1 \\pmod{8}$, and $8 \\mid CP^2 - BP^2 = 4ac$. But $a$ and $c$ are odd, so $4ac$ is not divisible by $8$, a contradiction.\n\nTherefore, it is not possible for $PA$, $PB$, $PC$, $PD$, and $AB$ to all be integers.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 14100,
    "subject": "Mathematics (Olympiad)",
    "question": "Find the smallest positive integer $n$ so that $\\sqrt{\\frac{1^2 + 2^2 + \\dots + n^2}{n}}$ is an integer.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $\\frac{1^2 + 2^2 + \\dots + n^2}{n} = m^2$, where $m \\in \\mathbb{Z}^+$. The sum $1^2 + 2^2 + \\dots + n^2 = \\frac{n(n+1)(2n+1)}{6}$, so\n\n$$\n\\frac{n(n+1)(2n+1)}{6n} = m^2 \\implies \\frac{(n+1)(2n+1)}{6} = m^2\n$$\n\nThus, $(n+1)(2n+1) = 6m^2$. We analyze possible forms for $n$:\n\n- If $n = 6p+3$, then $6m^2 = (6p+4)(12p+7)$, but $3$ does not divide the right side, so no solution.\n- If $n = 6p-1$, then $m^2 = p(12p-1)$. Since $p$ and $12p-1$ are coprime, both must be squares: $p = t^2$, $12p-1 = s^2$. Then $s^2 = 12t^2-1$, but $s^2 \\equiv -1 \\pmod{4}$, which is impossible.\n- If $n = 6p+1$, then $m^2 = (3p+1)(4p+1)$. Since $3p+1$ and $4p+1$ are coprime, both must be squares: $3p+1 = u^2$, $4p+1 = v^2$. This implies $p$ is even, and by checking small even $p$, the smallest $p$ such that both $3p+1$ and $4p+1$ are perfect squares is $p = 56$.\n\nTherefore, the smallest $n$ is $n = 6p+1 = 6 \\times 56 + 1 = 337$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 14101,
    "subject": "Mathematics (Olympiad)",
    "question": "Natural numbers $a, b, c, d$ satisfy\n\n$$\n0 < |ad - bc| < \\min\\{c, d\\}.\n$$\n\nProve that for any coprime natural numbers $x, y > 1$, the number $x^a + y^b$ is not divisible by $x^c + y^d$.",
    "options": [],
    "answer": "See solution",
    "solution": "We will prove this by contradiction. Denote $s = x^c + y^d$. Then $x^c \\equiv -y^d \\pmod{s}$, and if $x^a + y^b$ is divisible by $s$, then $x^a \\equiv -y^b \\pmod{s}$. This implies $x^{ad} \\equiv (-1)^d y^{bd} \\pmod{s}$ and $x^{bc} \\equiv (-1)^b y^{bd} \\pmod{s}$. Thus,\n\n$$\n(-1)^d x^{ad} \\equiv y^{bd} \\equiv (-1)^b x^{bc} \\pmod{s} \\implies x^{ad} \\equiv (-1)^{b-d} x^{bc} \\pmod{s}.\n$$\n\nSince $x$ and $s$ are coprime, we can divide both sides by $x^{\\min\\{ad, bc\\}}$ to get $x^{\\max\\{ad, bc\\} - \\min\\{ad, bc\\}} \\equiv (-1)^{b-d} \\pmod{s}$. Similarly, $y^{\\max\\{ad, bc\\} - \\min\\{ad, bc\\}} \\equiv (-1)^{a-c} \\pmod{s}$. Therefore, $y^k \\pm x^k$ is divisible by $s$ for $k = |ad - bc|$.\n\nBut by the problem's condition, $0 < |ad - bc| < \\min\\{c, d\\}$, so\n\n$$\n|y^k - x^k| < y^k + x^k < y^d + x^c = s.\n$$\n\nThus, $|y^k \\pm x^k|$ can only be divisible by $s$ if it is zero, which is impossible since $x$ and $y$ are coprime and greater than $1$. This is a contradiction, so $x^a + y^b$ is not divisible by $x^c + y^d$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 14102,
    "subject": "Mathematics (Olympiad)",
    "question": "Given that triangles $ABC$, $FBD$, and $EDC$ are similar, and $AFDE$ is a square with $|AF| = |AE| = 1$, find the area $|ABC|$ of triangle $ABC$.",
    "options": [],
    "answer": "See solution",
    "solution": "Since the triangles $ABC$, $FBD$, and $EDC$ are similar, we have:\n\n$$\n\\frac{x}{1} = \\frac{c}{b} = \\frac{1}{y}\n$$\n\nThus, $xy = 1$ and\n\n$$\nx + y = \\frac{c}{b} + \\frac{b}{c} = \\frac{b^2 + c^2}{bc} = \\frac{a^2}{bc} = \\frac{16}{bc}\n$$\n\nBecause $AFDE$ is a square, $c = 1 + y$ and $b = 1 + x$. Using $xy = 1$, we get $(1 + x)(1 + y) = 2 + (x + y)$. Therefore, the area of triangle $ABC$ is:\n\n$$\n|ABC| = \\frac{bc}{2} = 1 + \\frac{x + y}{2}\n$$\n\nAlso, $\\frac{x + y}{2} = \\frac{4}{|ABC|}$. This leads to the quadratic equation:\n\n$$\n|ABC|^2 - |ABC| - 4 = 0\n$$\n\nThe only positive solution is:\n\n$$\n|ABC| = \\frac{1 + \\sqrt{17}}{2}\n$$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 14103,
    "subject": "Mathematics (Olympiad)",
    "question": "Consider $n \\in \\mathbb{N}$, $n \\ge 2$, and $A$ a unitary ring with $n$ elements, such that the equation $x^{n+1} + x = 0$ has in $A \\setminus \\{0\\}$ the unique solution $x = 1$. Prove that $A$ is a field.",
    "options": [],
    "answer": "See solution",
    "solution": "Because $1$ satisfies $x^{n+1} + x = 0$, we have $1 + 1 = 0$, so all non-zero elements of the additive group $(A, +)$ are of order $2$. By Cauchy's theorem $n = 2^m$, $m \\in \\mathbb{N}^*$.  \n\nLet $a \\in A$, $a \\neq 0$. The ring $A$ being finite, there exist $p < q$ such that $a^p = a^q$. By successive multiplications with $a^{q-p}$, $a^q = a^{(k+1)q-kp}$, $k \\ge 1$. Take $k \\in \\mathbb{N}^*$ such that $r = (k+1)q - kp > 2q$. Multiplication by $a^{r-2q}$ gives $a^{r-q} = a^{2(r-q)}$. Let $b = a^{r-q}$. As $b^2 = b$, we get $b^{n+1} = b$, so $b^{n+1} + b = 0$, that is $b \\in \\{0, 1\\}$.\n\nIf $b = a^{r-q} = 0$, take $s \\in \\mathbb{N}$, $s \\ge 2$, such that $a^{s-1} \\neq 0$ and $a^s = 0$. Consider $c = a^{s-1}$. Then $c^2 = 0$ and, as a consequence, $(c+1)^{n+1} = (c+1)^{2m}(c+1) = (c^{2m} + 1)(c+1) = c+1$. By hypothesis $c+1 \\in \\{0, 1\\}$, which is a contradiction. In conclusion $a^{r-q} = 1$, i.e., $a$ is invertible.",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 14104,
    "subject": "Mathematics (Olympiad)",
    "question": "Every rational number $\\frac{p}{q}$ from the open interval $(0, 1)$ is covered by the closed interval\n$$\n\\left[\\frac{p}{q} - \\frac{1}{4q^2}, \\frac{p}{q} + \\frac{1}{4q^2}\\right].\n$$\n_Prove that the number_ $\\frac{\\sqrt{2}}{2}$ _is not covered by any of these intervals._",
    "options": [],
    "answer": "See solution",
    "solution": "Suppose the contrary: for some rational number $\\frac{p}{q}$, we have\n$$\n\\left| \\frac{\\sqrt{2}}{2} - \\frac{p}{q} \\right| \\leq \\frac{1}{4q^2}.\n$$\nAs $\\frac{\\sqrt{2}}{2} + \\frac{p}{q} < 2$, multiplying the last two inequalities gives\n$$\n\\left| \\frac{1}{2} - \\frac{p^2}{q^2} \\right| < \\frac{1}{2q^2}.\n$$\nHowever, since the left-hand side equals $\\frac{|q^2 - 2p^2|}{2q^2}$, the above inequality reduces to $|q^2 - 2p^2| < 1$, which implies $q^2 = 2p^2$ and $\\sqrt{2} = \\frac{p}{q}$ is rational—a contradiction. $\\Box$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 14105,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $a, b, c > 0$. Prove that\n\n$$\n\\frac{(a-b)^2}{(c+a)(c+b)} + \\frac{(b-c)^2}{(a+b)(a+c)} + \\frac{(c-a)^2}{(b+c)(b+a)} \\geq \\frac{(a-b)^2}{a^2 + b^2 + c^2}.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "It follows from\n\n$$\n\\frac{1}{2}(a-2b)^2 + \\frac{1}{2}(a-2c)^2 + (b-c)^2 \\geq 0,\n$$\n\nthat\n\n$$\n3(a^2 + b^2 + c^2) \\geq 2a^2 + 2ab + 2bc + 2ac = 2(a+b)(a+c),\n$$\n\nso we have $(a+b)(a+c) \\leq \\frac{3}{2}(a^2 + b^2 + c^2)$. Similarly,\n\n$$\n(b+a)(b+c) \\leq \\frac{3}{2}(a^2 + b^2 + c^2),\n$$\n\nand\n\n$$\n(c+a)(c+b) \\leq \\frac{3}{2}(a^2 + b^2 + c^2).\n$$\n\nHence,\n\n$$\n\\begin{aligned}\n& \\frac{(a-b)^2}{(c+a)(c+b)} + \\frac{(b-c)^2}{(a+b)(a+c)} + \\frac{(c-a)^2}{(b+c)(b+a)} \\\\\n\\geq & \\frac{2}{3} \\cdot \\frac{(a-b)^2 + (b-c)^2 + (c-a)^2}{a^2 + b^2 + c^2} \\\\\n\\geq & \\frac{2}{3} \\cdot \\frac{(a-b)^2 + \\frac{1}{2}(b-c + c-a)^2}{a^2 + b^2 + c^2} \\\\\n= & \\frac{(a-b)^2}{a^2 + b^2 + c^2}.\n\\end{aligned}\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 14106,
    "subject": "Mathematics (Olympiad)",
    "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": "See solution",
    "solution": "**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 round 1 and if a pair $(2, b)$ plays in 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:\n\n1: $(1, 2)$, $(3, a)$\n\n2: $(1, 3)$, $(2, b)$\n\n3: $(1, a)$\n\n4: $(1, b)$\n\n5: $(1, c)$\n\nwhich can be extended to a complete draw in only one way:\n\n1: $(1, 2)$, $(3, a)$, $(b, c)$\n\n2: $(1, 3)$, $(2, b)$, $(a, c)$\n\n3: $(1, a)$, $(2, c)$, $(3, b)$\n\n4: $(1, b)$, $(2, a)$, $(3, c)$\n\n5: $(1, c)$, $(2, 3)$, $(a, b)$\n\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 into 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\n$$\n6 \\cdot 5! \\cdot 6^5 = 5! \\cdot 6^6 = 2^9 \\cdot 3^7 \\cdot 5 = 5,598,720.\n$$\n\n**Another solution.** Let us denote the six teams by numbers 1, 2, 3, 4, 5, 6 and construct first an \"unordered\" draw in which the rounds will be \"numbered\" by the opponents of team 1 — see the following table in which the other opponents of team 2 are denoted as $a, b, c, d$:\n\n1: $(1, 2)$\n\n2: $(1, 3)$, $(2, a)$\n\n3: $(1, 4)$, $(2, b)$\n\n4: $(1, 5)$, $(2, c)$\n\n5: $(1, 6)$, $(2, d)$\n\nNote that $(a, b, c, d)$ is a permutation of the quadruple $(3, 4, 5, 6)$ and that the following two restrictions are evident:\n\n$$\n\\triangleright 3 \\neq a,\\ 4 \\neq b,\\ 5 \\neq c,\\ \\text{and}\\ 6 \\neq d\n$$\n\nThe two-element sets $\\{3, a\\}$, $\\{4, b\\}$, $\\{5, c\\}$, $\\{6, d\\}$ are pairwise distinct.\n\nIt is clear that under these two conditions, the third pairs for the rounds 2–5 are uniquely determined, as well as the remaining two pairs for round 1. Consequently, we have to calculate the number of permutations $(a, b, c, d)$ of the quadruple $(3, 4, 5, 6)$ which satisfy the two above stated conditions.\n\nUsing the inclusion-exclusion principle we conclude that the first condition is fulfilled by exactly nine permutations:\n\n$$\n4! - \\left( 4 \\cdot 3! - \\binom{4}{2} \\cdot 2! + 4 - 1 \\right) = 9.\n$$\n\nMoreover, exactly three of them do not satisfy the second condition, namely the permutations $(4, 3, 6, 5)$, $(5, 6, 3, 4)$, and $(6, 5, 4, 3)$. Thus the total number of the satisfactory permutations equals $9 - 3 = 6$.\n\nWe have proved that there are six \"unordered\" draws in the above specified sense. Combining this result with the idea of permuting the rounds and the courts, we conclude that the requested number of the draws is equal to\n\n$$\n6 \\cdot 6^5 \\cdot 5! = 5! \\cdot 6^6 = 2^9 \\cdot 3^7 \\cdot 5 = 5,598,720.\n$$\n\n**Remark.** All the six satisfactory permutations $(a, b, c, d)$ from the preceding solutions are $(4, 5, 6, 3)$, $(4, 6, 3, 5)$, $(5, 3, 6, 4)$, $(5, 6, 4, 3)$, $(6, 3, 4, 5)$, $(6, 5, 3, 4)$. It is possible to find them by an easy systematic examination (and thus to avoid the above presented calculation based on the inclusion-exclusion principle). On the other hand, the number 3 of the permutations $(a, b, c, d)$ that satisfy the first, but not the second condition, can be determined as the number of the \"faulty\" equalities\n\n$$\n\\{3, a\\} = \\{4, b\\},\\ \\{3, a\\} = \\{5, c\\},\\ \\{3, a\\} = \\{6, d\\},\n$$\n\nwhich are successively equivalent to the others:\n\n$$\n\\{5, c\\} = \\{6, d\\},\\ \\{4, b\\} = \\{6, d\\},\\ \\{4, b\\} = \\{5, c\\}.\n$$",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 14107,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $P$ be a given point inside quadrilateral $ABCD$. Points $Q_1$ and $Q_2$ are located within $ABCD$ such that\n\n$$\n\\angle Q_1BC = \\angle ABP, \\quad \\angle Q_1CB = \\angle DCP, \\quad \\angle Q_2AD = \\angle BAP, \\quad \\angle Q_2DA = \\angle CDP.\n$$\n\nProve that $\\overline{Q_1Q_2} \\parallel \\overline{AB}$ if and only if $\\overline{Q_1Q_2} \\parallel \\overline{CD}$.",
    "options": [],
    "answer": "See solution",
    "solution": "We will prove that the lines $\\overline{AB}$, $\\overline{CD}$, and $\\overline{Q_1Q_2}$ are either concurrent or all parallel. Let $X$ and $Y$ denote the reflections of $P$ across the lines $\\overline{AB}$ and $\\overline{CD}$.\n\nWe first claim that $XQ_1 = YQ_1$ and $XQ_2 = YQ_2$. Indeed, let $Z$ be the reflection of $Q_1$ across $BC$. Then $XB = PB$, $BQ_1 = BZ$, and\n\n$$\n\\angle XBQ_1 = \\angle XBA + \\angle ABQ_1 = \\angle ABC = \\angle PBC + \\angle CBZ = \\angle PBZ,\n$$\n\nwhence $\\triangle XBQ_1 \\cong \\triangle PBZ$ and thus $XQ_1 = PZ$. Similarly $YQ_1 = PZ$, and so $XQ_1 = YQ_1$. In exactly the same way, we see that $XQ_2 = YQ_2$, establishing the claim. We conclude that line $\\overline{Q_1Q_2}$ is the perpendicular bisector of the segment $\\overline{XY}$.\n\nNow, if $\\overline{AB} \\parallel \\overline{CD}$, then $\\overline{XY} \\perp \\overline{AB}$ and it follows that $\\overline{Q_1Q_2} \\parallel \\overline{AB}$, as desired. If lines $\\overline{AB}$ and $\\overline{CD}$ are not parallel, then let $R$ denote their intersection. Since $RX = RP = RY$, $R$ lies on the perpendicular bisector of $\\overline{XY}$ and thus $R$, $Q_1$, and $Q_2$ are collinear, as desired.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 14108,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $a > b$ be two positive integers such that $\\frac{a}{b} > 4$. Show that there exists a positive integer $l$ such that $b < \\phi(2^l) = 2^{l-1} < 2^l < a$.\n\nConstruct, for any $n \\geq 1$, a strictly increasing sequence of positive integers $a_1 < a_2 < \\dots < a_n$ such that $\\phi(a_1) > \\phi(a_2) > \\dots > \\phi(a_n)$, where $\\phi$ is Euler's totient function.",
    "options": [],
    "answer": "See solution",
    "solution": "**Lemma.** Let $a > b$ be two positive integers such that $\\frac{a}{b} > 4$. Then there exists a positive integer $l$ such that $b < \\phi(2^l) = 2^{l-1} < 2^l < a$.\n\n**Proof of lemma.** There exists a positive integer $m$ such that $2^{m-1} \\leq b < 2^m$. So $a > 4b \\geq 2^{m+1}$, and hence for $l = m + 1$ we have $b < 2^{l-1} < 2^l < a$, as claimed. $\\square$\n\nWe construct the sequence in the problem inductively. The assertion for $n = 1$ is obvious. Now, suppose that $a_1 < a_2 < \\dots < a_n$ is a sequence such that $\\phi(a_1) > \\phi(a_2) > \\dots > \\phi(a_n)$. Let $p_N$ be the greatest prime factor of the $a_i$'s ($p_j$ is the $j$th prime number), $1 \\leq i \\leq n$. Since for every $x$ with prime factors greater than $p_N$, the greatest common divisor of $x$ and $a_i$ is $1$ for all $1 \\leq i \\leq n$, we deduce that the sequence $a_1x < a_2x < \\dots < a_nx$ also satisfies the condition of the problem (because $\\phi(a_ix) = \\phi(a_i)\\phi(x)$).\n\nNow, our goal is to find such $x$ and some positive integer $y$ such that $y < a_1x$ and $\\phi(y) > \\phi(a_1x)$. First, we claim that there is a positive integer $x$ with prime factors greater than $p_N$ such that $\\frac{a_1x}{\\phi(a_1)\\phi(x)} > 4$, or equivalently, $\\frac{x}{\\phi(x)} > 4\\frac{\\phi(a_1)}{a_1}$.\n\nTo prove this, let $x_m = p_{N+1}p_{N+2}\\dots p_{N+m}$ ($m \\in \\mathbb{N}$). We have\n\n$$\n\\frac{x_m}{\\phi(x_m)} = \\prod_{i=N+1}^{N+m} \\frac{p_i}{p_i-1} = \\prod_{i=N+1}^{N+m} \\left(1 + \\frac{1}{p_i-1}\\right) \\geq \\sum_{i=N+1}^{N+m} \\frac{1}{p_i-1}.\n$$\n\nThe sum $\\sum_{i=N+1}^{\\infty} \\frac{1}{p_i-1} > \\sum_{i=N+1}^{\\infty} \\frac{1}{p_i}$ diverges by a well-known fact, so we can find $m$ such that $\\frac{x_m}{\\phi(x_m)} > 4\\frac{\\phi(a_1)}{a_1}$. We let $x = x_m$. Now, since $\\frac{a_1x}{\\phi(a_1)\\phi(x)} > 4$, according to the lemma there is some positive integer $l$ such that $\\phi(a_1x) < \\phi(2^l) < 2^l < a_1x$. So putting $y = 2^l$ leads to the new sequence\n\n$$\ny < a_1x < a_2x < \\dots < a_nx\n$$\n\nwhich has $n+1$ terms and satisfies the problem's condition. $\\square$",
    "topic": "Number Theory",
    "subtopic": "Number-Theoretic Functions"
  },
  {
    "id": 14109,
    "subject": "Mathematics (Olympiad)",
    "question": "Acute triangle $ABC$ with $AB < AC$ has circumcircle $\\omega$. Let $D$, $E$, $F$ be the midpoints of sides $BC$, $CA$, $AB$ respectively. Ray $AD$ intersects $\\omega$ at point $K$ ($K \\neq A$) and the circumcircle of triangle $AEF$ at point $M$ ($M \\neq A$). On circle $\\omega$, a point $T$ is chosen such that $AT \\parallel BC$. The circumcircles of triangles $KDT$ and $AEF$ intersect at point $L$ on the shorter arc $EF$. Prove that circle $\\omega$ and the circumcircle of triangle $KLM$ are tangent to each other.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $O$ be the center of $\\omega$ and $Q$ be the intersection point of lines $DO$ and $AT$. Since $ABCT$ is a trapezoid and a cyclic quadrilateral, it is an isosceles trapezoid and the points $A$ and $T$ are symmetric with respect to line $DO$.\n\n![](images/EST_ABooklet_2024_p53_data_0b9b0afa0e.png)\n\n*Fig. 47*\n\n![](images/EST_ABooklet_2024_p53_data_fd4498ac67.png)\n\n*Fig. 48*\n\nWe have $\\angle DKT = \\angle AKT = \\frac{1}{2}\\angle AOT = \\angle QOT = 180^\\circ - \\angle DOT$, whereas $K$ and $O$ are on different sides of line $DT$, so point $O$ is on the circumcircle of triangle $KDT$.\n\nSince $O$ is the point of intersection of the perpendicular bisectors of triangle $ABC$, we have $\\angle OEA = \\angle OFA = 90^\\circ$. Thus points $E$ and $F$ are on the circle with diameter $OA$ and $O$ is on the circumcircle of triangle $AEF$. Since triangle $ABC$ is acute, triangle $AEF$ is also acute (because $EF \\parallel BC$), hence its center is inside the triangle. So point $O$ is on the shorter arc $EF$. Thus the circumcircles of triangles $KDT$ and $AEF$ intersect at $O$ on the shorter arc $EF$, so $O = L$. Since $AL$ is a diameter of the circumcircle of triangle $AEF$ and point $M$ is on the circle, $ML$ and $AK$ are perpendicular. Therefore $KL$ is a diameter of the circumcircle of triangle $KLM$. Since $KL$ is also a radius of circle $\\omega$, the line through $K$ perpendicular to line $KL$ is tangent to both circles. Consequently, circle $\\omega$ and the circumcircle of triangle $KLM$ are tangent to each other.\n\n![](images/EST_ABooklet_2024_p54_data_97b9c3f159.png)\n\n*Fig. 49*",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 14110,
    "subject": "Mathematics (Olympiad)",
    "question": "Prove that for every positive integer $n$ there exists an $n$-digit number divisible by $5^n$ all of whose digits are odd.",
    "options": [],
    "answer": "See solution",
    "solution": "We proceed by induction.\n\nThe property is clearly true for $n=1$.\n\nAssume that $N = a_1a_2\\dots a_n$ is divisible by $5^n$ and has only odd digits. Consider the numbers\n\n$$\nN_1 = 1a_1a_2\\dots a_n = 1 \\cdot 10^n + 5^n M = 5^n (1 \\cdot 2^n + M),\n$$\n$$\nN_2 = 3a_1a_2\\dots a_n = 3 \\cdot 10^n + 5^n M = 5^n (3 \\cdot 2^n + M),\n$$\n$$\nN_3 = 5a_1a_2\\dots a_n = 5 \\cdot 10^n + 5^n M = 5^n (5 \\cdot 2^n + M),\n$$\n$$\nN_4 = 7a_1a_2\\dots a_n = 7 \\cdot 10^n + 5^n M = 5^n (7 \\cdot 2^n + M),\n$$\n$$\nN_5 = 9a_1a_2\\dots a_n = 9 \\cdot 10^n + 5^n M = 5^n (9 \\cdot 2^n + M).\n$$\n\nThe numbers $1 \\cdot 2^n + M, 3 \\cdot 2^n + M, 5 \\cdot 2^n + M, 7 \\cdot 2^n + M, 9 \\cdot 2^n + M$ give distinct remainders when divided by $5$. Otherwise, the difference of some two of them would be a multiple of $5$, which is impossible, because neither $2^n$ is a multiple of $5$, nor is the difference of any two of the numbers $1, 3, 5, 7, 9$.\n\nIt follows that one of the numbers $N_1, N_2, N_3, N_4, N_5$ is divisible by $5^n \\cdot 5 = 5^{n+1}$, and the induction is complete.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 14111,
    "subject": "Mathematics (Olympiad)",
    "question": "In a coordinate city, there are $n \\geq 3$ tramlines parallel to the $x$-axis. Each line begins at $x = 0$ and ends at $x = n$. Exactly one tram of length $1$ moves on each line: on the first line with speed $1$, on the second with speed $2$, and so on, up to the $n$-th line with 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 back end of each tram is at $x = 0$. \n\nProve that the trams' projections onto the $x$-axis never cover the whole interval from $0$ to $n$.",
    "options": [],
    "answer": "See solution",
    "solution": "The projections of the trams can only cover the whole interval if one projection covers $[0, 1]$, another $[1, 2]$, and so on, up to $[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, the fastest and the third fastest trams move by $n$ and $n-2$ units, respectively. Together, these two trams move $2n-2$ units, which is exactly one complete to-and-fro cycle.\n\nLet the integer positions of the trams on the round trip be numbered from $0$ to $2n-3$, where $0$ is the starting position and each subsequent position increases by $1$ until returning to the start. Call these numbers the position characteristics. If the sum of the position characteristics of two trams is $2n-2$, their projections cover the same interval, because one has moved as far from the start as the other still has to go to reach it. If the sum is $0$, both are at the starting position, so they again cover the same interval. Thus, whenever the sum of the position characteristics is divisible by $2n-2$, the projections overlap.\n\nAt the beginning, the sum of the position characteristics of the fastest and third fastest trams is $0$, and each time they together move $2n-2$ units, the sum remains divisible by $2n-2$. Therefore, when the projection of the slowest tram covers an interval with integer endpoints, the projections of these two trams overlap, so at least one interval is not covered.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 14112,
    "subject": "Mathematics (Olympiad)",
    "question": "Let the chords $AB$ and $CD$ of a circle intersect at the point $K$. Prove that\n$$\n\\frac{AC \\cdot AD}{BC \\cdot BD} = \\frac{CK}{DK}.\n$$\n\nLet $APBQ$ be a harmonic quadrilateral. Prove that $MK$ is the bisector of the angle $AMB$, and that\n$$\n\\frac{AC}{CB} \\cdot \\frac{AD}{DB} = \\left(\\frac{AK}{BK}\\right)^2,\n$$\nwhere $C$ and $D$ are the intersections $AB \\cap RW$ and $AB \\cap UT$, respectively, and $K = AB \\cap RT$.",
    "options": [],
    "answer": "See solution",
    "solution": "*Lemma.* Let the chords $AB$ and $CD$ of a circle intersect at the point $K$. Then\n$$\n\\frac{AC \\cdot AD}{BC \\cdot BD} = \\frac{CK}{DK}.\n$$\n\n*Proof of lemma.* This equality directly follows from the sine laws for the triangles $ACK$, $ADK$, and $BCD$.\n\nThe quadrilateral $APBQ$ is harmonic, so $MK$ is the bisector of the angle $AMB$. Hence\n$$\n\\frac{AC}{CB} = \\frac{\\sin AMC}{\\sin CMB} \\cdot \\frac{AK}{BK}\n$$\nand\n$$\n\\frac{AD}{DB} = \\frac{\\sin AMD}{\\sin DMB} \\cdot \\frac{AK}{BK},\n$$\ntherefore it's enough to prove the equality\n$$\n\\frac{AC}{CB} \\cdot \\frac{AD}{DB} = \\left(\\frac{AK}{BK}\\right)^2.\n$$\n\nThe lemma for $AB \\cap RW = C$ and $AB \\cap UT = D$ implies that\n$$\n\\frac{AC}{CB} = \\frac{AW \\cdot AR}{BW \\cdot BR} \\quad \\text{and} \\quad \\frac{AD}{DB} = \\frac{AT \\cdot AU}{BT \\cdot BU}.\n$$\n\nAnd the lemma for $AB \\cap RT = K$ and $AB \\cap UW = D$ implies that\n$$\n\\frac{AK}{KB} = \\frac{AT \\cdot AR}{BT \\cdot BR} = \\frac{AW \\cdot AU}{BW \\cdot BU}.\n$$\n\nCombining these equalities we obtain the required equality.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 14113,
    "subject": "Mathematics (Olympiad)",
    "question": "Points $A$, $B$, $Y$, and $C$ lie in this order on circle $k$ with center $O$, such that $BC = 2\\ \\mathrm{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 a second time at the point $N$. Find:\n\na) the length of the segment $BO$;\n\nb) the size of the angle $\\angle YAN$.",
    "options": [],
    "answer": "See solution",
    "solution": "a) Clearly $\\angle BAC = \\angle BAY + \\angle CAY = 120^\\circ$, so $\\angle BOC = 360^\\circ - 2\\angle BAC = 120^\\circ$. Let $M$ be the midpoint of $BC$. Then $OM \\perp BC$ (since $BO = OC$), $\\angle BOM = 60^\\circ$, and $BM = \\frac{BC}{2} = 1$. Let $BO = x$. In triangle $BOM$, $OM = \\frac{x}{2}$ (since $\\angle OBM = 30^\\circ$), and by the Pythagorean theorem: $x^2 = \\left(\\frac{x}{2}\\right)^2 + 1^2$, so $x^2 = \\frac{x^2}{4} + 1$, which gives $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$. From the other circle, $\\angle ANC = 180^\\circ - \\angle ACY = \\angle ABY = 180^\\circ - 2\\angle YCB$. Therefore, $\\angle ANB + \\angle ANC = 180^\\circ$, so $N$ lies on $BC$. Also, $\\angle NAC = \\angle BCY = \\angle BAY$ (from tangency), and $\\angle YAN = \\angle CAY - \\angle CAN = \\angle CAY - \\angle BAY = 36^\\circ$.\n\n$\\boxed{\\angle YAN = 36^\\circ}$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 14114,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $n$ be a positive integer and let $x_1, x_2, \\dots, x_n$ be positive real numbers such that $x_1 x_2 \\cdots x_n = 1$. Prove that\n\n$$\n\\sum_{i=1}^{n} x_{i}^{n}(1 + x_{i}) \\geq \\frac{n}{2^{n-1}} \\prod_{i=1}^{n} (1 + x_{i}).\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "By the power-mean inequality,\n\n$$\n1 + a^n \\geq \\frac{(1 + a)^n}{2^{n-1}}, \\quad a \\geq 0. \\qquad (*)\n$$\n\nThus,\n\n$$\n\\begin{align*}\n\\sum_{i=1}^{n} x_i^n (1 + x_i) &= \\sum_{i=1}^{n} x_i^n + \\sum_{i=1}^{n} x_i^{n+1} \\\\\n&\\geq \\sum_{i=1}^{n} x_i^n + n \\left( \\prod_{i=1}^{n} x_i \\right)^{1 + 1/n} \\quad \\text{(AM-GM)} \\\\\n&= \\sum_{i=1}^{n} x_i^n + n = \\sum_{i=1}^{n} (1 + x_i^n) \\\\\n&\\geq \\frac{1}{2^{n-1}} \\sum_{i=1}^{n} (1 + x_i)^n \\quad \\text{by (*)} \\\\\n&\\geq \\frac{n}{2^{n-1}} \\prod_{i=1}^{n} (1 + x_i). \\quad \\text{(AM-GM)}\n\\end{align*}\n$$\n\nClearly, equality holds if and only if $x_1 = x_2 = \\dots = x_n = 1$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 14115,
    "subject": "Mathematics (Olympiad)",
    "question": "Prove that for all positive real numbers $x$, $y$, $z$,\n$$\n\\frac{y^2 z}{x} + y^2 + z \\geqslant \\frac{9y^2 z}{x + y^2 + z}.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "By bringing all the terms to the same side and to a common denominator, we get an equivalent inequality:\n$$\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$$\nSince $x$, $y$, and $z$ are positive, the denominator $x(x + y^2 + z)$ is positive. Therefore, the fraction on the left-hand side is nonnegative if and only if its numerator is nonnegative. Expanding the numerator, we get the equivalent inequality:\n$$\nx^2y^2 + xy^4 + y^2z^2 + y^4z + x^2z + xz^2 \\geqslant 6xy^2z.\n$$\nThis inequality follows directly from the AM-GM inequality for the terms $x^2y^2$, $xy^4$, $y^2z^2$, $y^4z$, $x^2z$, and $xz^2$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 14116,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $O$ be the circumcenter of $\\triangle ABC$ and $H_A$ be the projection of $A$ onto $BC$. The extension of $AO$ intersects the circumcircle of $\\triangle BOC$ at $A'$. The projections of $A'$ onto $AB$ and $AC$ are $D$ and $E$, respectively. Let $O_A$ be the circumcenter of $\\triangle DH_A E$. Define $H_B$, $O_B$, $H_C$ and $O_C$ similarly.\n\nProve that $H_A O_A$, $H_B O_B$ and $H_C O_C$ are concurrent.\n\n![](images/Mathematical_Olympiad_in_China_2011-2014_p168_data_76e490a9a8.png)",
    "options": [],
    "answer": "See solution",
    "solution": "Let $T$ be the symmetry point of $A$ over $BC$, $F$ be the projection of $A'$ onto $BC$, and $M$ be the projection of $T$ onto $AC$.\n\nSince $AC = CT$, we have $\\angle TCM = 2\\angle TAM$. Since $\\angle TAM = \\frac{\\pi}{2}$ —\n\n$$\n\\angle ACB = \\angle OAB, \\text{ we have}\n$$\n\n$$\n\\angle TCM = 2\\angle OAB = \\angle A'OB = \\angle A'CF,\n$$\n\nand\n\n$$\n\\angle TCH_A = \\angle A'CF + \\angle A'CT = \\angle TCM + \\angle A'CT = \\angle A'CE.\n$$\n\nBecause of the fact that $\\angle CH_A T$, $\\angle CMT$, $\\angle CEA'$, $\\angle CFA'$ are right angles, we have,\n\n$$\n\\begin{aligned}\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} \\\\\n&= \\frac{CE}{CA'} \\cdot \\frac{CA'}{CF} = \\frac{CE}{CF},\n\\end{aligned}\n$$\n\ni.e., $CH_A \\cdot CF = CM \\cdot CE$, so $H_A, F, M$ and $E$ are on the same circle $\\omega_1$.\n\nSimilarly, let $N$ be the projection of $T$ onto $AB$, then $H_A$, $F$, $N$ and $D$ are on the same circle $\\omega_2$. Since $A'FH_A T$ and $A'EMT$ are both right trapezoids, the perpendicular bisector of the segments $H_A F$ and $EM$ meet at the midpoint $K$ of the segment $A'T$, i.e., $K$ is the center of circle $\\omega_1$ and $KF$ is the radius of circle $\\omega_1$. Similarly, $K$ and $KF$ are also the center and the radius of circle $\\omega_2$, respectively. Thus, $\\omega_1$ and $\\omega_2$ are the same, $D, N, F, H_A, E$ and $M$ are on the same circle. So $O_A$ is the midpoint $K$ of $A'T$, $O_A H_A \\parallel AA'$.\n\nSince $\\angle H_C AO + \\angle A H_C H_B = \\frac{\\pi}{2} - \\angle ACB + \\angle ACB = \\frac{\\pi}{2}$, we have $AA' \\perp H_B H_C$, thus $O_A H_A \\perp H_B H_C$, therefore, $O_A H_A$, $O_B H_B$ and $O_C H_C$ all pass through the orthocenter of $\\triangle H_A H_B H_C$. $\\square$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 14117,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all real numbers $x$, $y$, and $z$ which satisfy the simultaneous equations\n\n$$\n\\begin{aligned}\nx^2 - 4y + 7 &= 0 \\\\\ny^2 - 6z + 14 &= 0 \\\\\nz^2 - 2x - 7 &= 0\n\\end{aligned}\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "Sum all three equations to obtain\n\n$$\nx^2 - 4y + 7 + y^2 - 6z + 14 + z^2 - 2x - 7 = 0.\n$$\n\nRearrange to get\n\n$$\n(x - 1)^2 + (y - 2)^2 + (z - 3)^2 = 0.\n$$\n\nSince all squares are nonnegative, this only occurs if $x = 1$, $y = 2$, and $z = 3$.\n\nCheck these values in the original equations:\n\n- $x^2 - 4y + 7 = 1 - 8 + 7 = 0$\n- $y^2 - 6z + 14 = 4 - 18 + 14 = 0$\n- $z^2 - 2x - 7 = 9 - 2 - 7 = 0$\n\nAll are satisfied. Thus, the unique solution is $x = 1$, $y = 2$, $z = 3$.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 14118,
    "subject": "Mathematics (Olympiad)",
    "question": "Given positive integers $n$ and $k \\le n$. Consider an equilateral triangular board with side $n$, which consists of circles: in the first (top) row there is one circle, in the second row there are two circles, ..., in the bottom row there are $n$ circles (see the figure below). Let us place checkers on this board so that any line parallel to a side of the triangle (there are $3n$ such lines) contains no more than $k$ checkers. Denote by $T(k, n)$ the largest possible number of checkers in such a placement.\n\n![](images/BLR_ABooklet_2024_p16_data_6a2fe1b396.png)\n\na) Prove that the following upper bound is true:\n\n$$\nT(k, n) \\le \\left\\lfloor \\frac{k(2n + 1)}{3} \\right\\rfloor.\n$$\n\nb) Find $T(1, n)$ and $T(2, n)$.",
    "options": [],
    "answer": "See solution",
    "solution": "a) Let us prove that in any such placement of checkers on a triangular board with side $n$, the number of checkers $q$ satisfies the inequality\n\n$$\nq \\le \\frac{2n + 1}{3} \\cdot k.\n$$\n\nThe main idea of the proof is double counting. Note that wherever a checker stands, if you count all the cells of the three lines in which it stands, and take into account the cell on which it stands three times, you get $2n + 1$. Let us denote by $S$ this number summed over all checkers. We have\n\n$$\nS = (2n + 1)q.\n$$\n\nNow let's find this amount based on the location of the checkers along the lines. Let there be $x_1, \\dots, x_n$ checkers in the $n$ horizontal lines, $y_1, \\dots, y_n$ checkers in the $n$ lines with angle $60^\\circ$, and in the $n$ lines with angle $120^\\circ$ there are $z_1, \\dots, z_n$ checkers, as shown in Figure 1.\n\n![](images/BLR_ABooklet_2024_p17_data_eefb771b36.png)\n\nThe $i$-th line has exactly $i$ cells, and each of them is in the same line with $x_i, y_i$ or $z_i$ checkers from that line depending on the angle. Therefore,\n\n$$\nS = 1 \\cdot (x_1 + y_1 + z_1) + 2 \\cdot (x_2 + y_2 + z_2) + \\dots + n \\cdot (x_n + y_n + z_n).\n$$\n\nSince there are $q$ checkers in total, but there are no more than $k$ checkers in any row, we have the following conditions:\n\n$$\nx_1 + \\dots + x_n = y_1 + \\dots + y_n = z_1 + \\dots + z_n = q, \\\\\n0 \\le x_i, y_i, z_i \\le \\min(i, k) \\quad \\text{for any } 1 \\le i \\le n.\n$$\n\nLet $q = mk + r$, where $0 \\le r < k$ is the remainder when divided by $k$. It is clear that the expression above is maximized when the sums $x_i + y_i + z_i$ with higher coefficients are the maximum possible (that is, equal to $3k$). Hence, under these conditions we get\n\n$$\n\\begin{aligned}\n(2n + 1)q &= n \\cdot 3k + (n - 1) \\cdot 3k + \\dots + (n - m + 1) \\cdot 3k + (n - m) \\cdot 3r \\\\\n&\\le \\frac{2n - m + 1}{2} m \\cdot 3k + (n - m) \\cdot 3r \\\\\n&\\le \\frac{6nmk - 3m^2k + 3mk}{2} + 3nr - 3mr \\\\\n&\\le \\frac{6n(mk + r) - 3m^2k + 3(mk + r) - 3r - 6mr}{2} \\\\\n&\\le \\frac{(6n + 3)q - 3m^2k - 3r - 6mr}{2}.\n\\end{aligned}\n$$\n\nMoving $(2n + 1)q$ to the right side, multiplying the inequality by $2k$ and substituting $mk = q - r$, we arrive at a quadratic inequality for $q$:\n\n$$\n\\begin{aligned}\n0 &\\le (2n + 1)kq - 3(q - r)^2 - 6(q - r)r - 3rk \\\\\n&= (2n + 1)kq - 3q^2 + 6qr - 3r^2 - 6qr + 6r^2 - 3rk \\\\\n&= -3q^2 + (2n + 1)kq + 3r(r - k),\n\\end{aligned}\n$$\n\nfrom which we obtain\n\n$$\n3q^2 \\le (2n + 1)kq + 3r(r - k) \\le (2n + 1)kq, \\quad \\text{that is} \\quad q \\le \\frac{2n + 1}{3}k,\n$$\n\nbecause $r - k < 0$. Since $q$ is an integer, we get $q \\le \\left\\lfloor \\frac{2n+1}{3}k \\right\\rfloor$, as required.\n\nb) We have $T(1, n) = \\left\\lfloor \\frac{2n+1}{3} \\right\\rfloor$ and $T(2, n) = \\left\\lfloor \\frac{4n+2}{3} \\right\\rfloor$. Placements of checkers, when equality is achieved in the last inequality, for $k=1$ and $k=2$, are shown in figures 2, 3 and 4.\n\n![](images/BLR_ABooklet_2024_p18_data_70b764eb15.png)\n\n![](images/BLR_ABooklet_2024_p18_data_ff85b8c3e2.png)\n\n![](images/BLR_ABooklet_2024_p18_data_66da694828.png)",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 14119,
    "subject": "Mathematics (Olympiad)",
    "question": "Given a real number $a$, consider the sequence $(u_n)$ defined by\n\n$$\nu_1 = a, \\quad u_{n+1} = \\frac{1}{2} + \\sqrt{\\frac{2n+3}{n+1}u_n + \\frac{1}{4}}, \\quad \\forall n \\in \\mathbb{N}^*.$$  \n\n1. If $a = 5$, prove that $(u_n)$ has a finite limit and find that limit.\n\n2. Find all values of $a$ such that the sequence $(u_n)$ is defined and has a finite limit.",
    "options": [],
    "answer": "See solution",
    "solution": "We solve part 2) first, from which part 1) follows. The sequence $(u_n)$ is defined if and only if $u_2$ is defined. Since\n\n$$u_2 = \\frac{1}{2} + \\sqrt{\\frac{5}{2}a + \\frac{1}{4}},$$\n\n$u_2$ is defined if and only if\n\n$$a \\geq -\\frac{1}{10}.$$\n\nWe will prove that $(u_n)$ converges to $3$ for every $a \\geq -\\frac{1}{10}$. It is easy to see that $u_n \\geq \\frac{1}{2}$ for all $n \\geq 2$. Note that $f(x) = \\frac{2x+3}{x+1}$ is strictly decreasing over $\\mathbb{R}^+$, so for every positive integer $n$,\n\n$$\n\\frac{2n+3}{n+1} > \\frac{2(n+1)+3}{(n+1)+1}.\n$$\n\nIf there exists $n_0 \\in \\mathbb{N}$ such that $u_{n_0} \\ge u_{n_0+1}$, then\n\n$$\nu_{n_0+2} = \\frac{1}{2} + \\sqrt{\\frac{2(n_0+1)+3}{(n_0+1)+1}u_{n_0+1} + \\frac{1}{4}} \\le \\frac{1}{2} + \\sqrt{\\frac{2n_0+3}{n_0+1}u_{n_0} + \\frac{1}{4}} = u_{n_0+1}.\n$$\n\nThus, $u_{n_0} \\ge u_{n_0+1} \\ge u_{n_0+2} \\ge \\dots$, so $(u_n)$ is non-increasing from $n_0$ onward and bounded below, so $\\lim_{n \\to \\infty} u_n = L$ exists with $L \\ge \\frac{1}{2}$. Passing to the limit in the recurrence,\n\n$$L = \\frac{1}{2} + \\sqrt{2L + \\frac{1}{4}}.$$\n\nSolving, we find $L = 3$.\n\nIf such $n_0$ does not exist, then $(u_n)$ is strictly increasing. Since $\\frac{2n+3}{n+1} < 3$ for $n \\ge 2$, we have\n\n$$\n\\frac{1}{2} + \\sqrt{3u_n + \\frac{1}{4}} > u_{n+1} > u_n, \\quad \\forall n \\ge 2.\n$$\n\nSolving, $u_n < \\frac{4+\\sqrt{17}}{2}$ for all $n \\ge 2$, so $(u_n)$ is strictly increasing and bounded above, hence convergent. Passing to the limit again, $\\lim_{n \\to \\infty} u_n = 3$.\n\nIn summary, for every $a \\ge -\\frac{1}{10}$, the sequence $(u_n)$ is defined and converges to $3$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 14120,
    "subject": "Mathematics (Olympiad)",
    "question": "Ats and Pets both thought of two positive integers that do not exceed some positive integer $n$. If they both added the numbers they thought of, then both sums gave the same remainder when divided by $n$. But if both of them multiplied the numbers they thought of, then both products also gave equal remainders when divided by $n$. Is it necessarily true that the numbers they thought of were the same, if \na) $n = 99$?\nb) $n = 101$?",
    "options": [],
    "answer": "See solution",
    "solution": "a) No. For example, if Ats thought of numbers 1 and 21, and Pets thought of 10 and 12, both sums are $22$ and both products are $21$ and $120$, respectively. Both $21$ and $120$ give the remainder $21$ when divided by $99$.\n\nb) Yes. Let Ats choose numbers $a$ and $b$, and Pets choose $c$ and $d$. The conditions imply that $(a + b) \\equiv (c + d) \\pmod{101}$ and $ab \\equiv cd \\pmod{101}$. Thus, $(a + b) - (c + d) = 101k$ for some integer $k$, so $a = 101k - b + c + d$. Then:\n\n$$\n\\begin{aligned}\nab - cd &= (101k - b + c + d)b - cd \\\\\n&= 101kb - b^2 + bc + bd - cd \\\\\n&= 101kb - (c - b)(d - b).\n\\end{aligned}\n$$\n\nSo $(c - b)(d - b)$ is divisible by $101$. Since $101$ is prime, it must divide either $c - b$ or $d - b$. W.l.o.g., suppose $c - b$ is divisible by $101$. Since all numbers are between $1$ and $101$, this means $c = b$. Then $(a + b) - (c + d) = a - d$ is divisible by $101$, so $a = d$. Therefore, Ats and Pets must have chosen the same numbers.\n\n*Remark*: In part a), there are many other possibilities to show the answer is no; for example, Ats could have chosen $99$ and $36$ and Pets $33$ and $3$, or Ats $99$ and $20$ and Pets $11$ and $9$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 14121,
    "subject": "Mathematics (Olympiad)",
    "question": "Each player, Andriy and Olesya, has a set of 2019 cards numbered $1, 2, \\ldots, 2019$ (each number appears exactly once for each player). The game proceeds as follows:\n\nAt the start, a card with number $k \\in \\{1, 2, \\ldots, 2019\\}$ is on the table. Players take turns (Andriy goes first) to change one of their cards to the one currently on the table. Additionally, Andriy can replace the card on the table with one from his hand if its number is greater than the current table card, and Olesya can do so if her card has a smaller number than the table card. A player who cannot make a move loses. Who will win if both play optimally?",
    "options": [],
    "answer": "See solution",
    "solution": "Each time Andriy plays a card $n > k$, Olesya responds with the card $n - 1 < n$. Andriy's set becomes:\n\n$$\n\\{1, 2, \\ldots, k, k, k+1, \\ldots, n-1, n+1, n+2, \\ldots, 2018\\}\n$$\n\nHe must then play a card greater than or equal to $n+1$. Olesya plays the card $n$ back. After each move, Andriy must play a higher card, while Olesya always has a card numbered one less than Andriy's play. When Andriy plays the card numbered $2018$, he will no longer have $2018$, and Olesya will have $2017$. She will then change the $2018$ card on the table to $2017$, leaving Andriy unable to move. Thus, Olesya wins if both play optimally.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 14122,
    "subject": "Mathematics (Olympiad)",
    "question": "Given that the smallest and next smallest possible numbers of bulbs to be planted are 52 and 64, and that each participating student is asked to plant at least 1 bulb, what is the possible distribution of students among three grades such that the total number of participating students is 52, and the smallest number of students in any grade is 12?",
    "options": [],
    "answer": "See solution",
    "solution": "Let $A$, $B$, and $C$ be the numbers of participating students in the three grades, with $A \\leq B \\leq C$. Since $A + B + C = 52$ and $2A + B + C = 64$, we find $A = 12$ and $B + C = 40$. Given $A \\leq B \\leq C$, $12 \\leq B \\leq 20$. Checking possible distributions, only $(12, 12, 28)$ yields exactly 6 possible numbers of bulbs under 100 to be planted, matching the conditions. Thus, the only possible distribution is $(12, 12, 28)$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 14123,
    "subject": "Mathematics (Olympiad)",
    "question": "For a natural number $n$, consider the sequence $(a_k)_{k=1}^{\\infty}$ given by $a_1 = \\frac{1}{n}$ and the recurrence relation\n\n$$\na_{k+1} = 3a_k - \\lfloor 2a_k \\rfloor - \\lfloor a_k \\rfloor,\n$$\n\nfor all $k \\ge 1$. Determine all the values of $n$ for which the sequence is eventually constant.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $f$ be the function $f(x) = 3x - \\lfloor 2x \\rfloor - \\lfloor x \\rfloor$, so the defining relation can be written as $a_{k+1} = f(a_k)$.\n\nNote that $f(0) = f(1) = 0$ and $f(\\frac{1}{2}) = \\frac{1}{2}$, so the sequence will be eventually constant whenever $\\frac{1}{2}$ or $1$ occurs in it. Also, $f(x) = 3x$ for $0 < x < \\frac{1}{2}$. By induction, for any non-negative integer $\\alpha$ and $n = 3^{\\alpha}$, we have $a_{\\alpha+1} = 1$; similarly, for $n = 2 \\cdot 3^{\\alpha}$, we have $a_{\\alpha+1} = \\frac{1}{2}$. Thus, integers $n$ of these forms are solutions.\n\nTo show there are no other solutions, suppose the sequence is eventually constant for some $n$. Define $b_k = n a_k$, so $b_1 = 1$ and\n\n$$\nb_{k+1} = 3b_k - n \\left\\lfloor \\frac{2b_k}{n} \\right\\rfloor - n \\left\\lfloor \\frac{b_k}{n} \\right\\rfloor$$\n\nBy induction, $b_k$ is a non-negative integer and $b_k \\equiv 3^{k-1} \\pmod{n}$ for all $k \\ge 1$.\n\nIf the sequence is eventually constant, there exists $\\alpha \\ge 0$ with $b_{\\alpha+2} = b_{\\alpha+1}$, so $3^{\\alpha+1} \\equiv 3^{\\alpha} \\pmod{n}$, which is equivalent to $n \\mid 2 \\cdot 3^{\\alpha}$.\n\n**Conclusion:** The sequence is eventually constant if and only if $n$ is of the form $3^{\\alpha}$ or $2 \\cdot 3^{\\alpha}$ for some non-negative integer $\\alpha$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 14124,
    "subject": "Mathematics (Olympiad)",
    "question": "A quadrilateral $ABCD$ with $\\angle BAD + \\angle ADC > 180^{\\circ}$ is circumscribed around a circle of center $I$. A line through $I$ meets $AB$ and $CD$ at points $X$ and $Y$, respectively. Prove that if $IX = IY$ then $AX \\cdot DY = BX \\cdot CY$.",
    "options": [],
    "answer": "See solution",
    "solution": "Denote by $M$ and $N$ the tangent points of the incircle of $ABCD$ with $AB$ and $CD$, respectively. It follows from $\\angle BAD + \\angle ADC > 180^{\\circ}$ that $AB \\parallel CD$ and $\\angle MIN < 180^{\\circ}$. Also, the equalities $IM = IN$, $\\angle IMX = \\angle INY$, and $IX = IY$ show that $\\triangle IMX \\cong \\triangle INY$, implying $\\angle IYN = \\angle IXM$.\n\nIf $X \\in BM$ and $Y \\in DN$ (or $X \\in AM$ and $Y \\in CN$), then the equality $\\angle IYN = \\angle IXM$ implies $AB \\parallel CD$, a contradiction. Therefore, $X \\in XB$ and $Y \\in NC$. It follows from $AXYD$ that\n\n$$\n\\angle AXI = \\angle DYI = 180^{\\circ} - \\frac{\\angle A}{2} - \\frac{\\angle D}{2},\n$$\n\ngiving $\\angle AIX = \\frac{\\angle D}{2}$ and $\\angle DIY = \\frac{\\angle A}{2}$. Therefore, $\\triangle AIX \\sim \\triangle IDY$, which implies that $AX \\cdot DY = IY \\cdot IX$. Analogously, $\\triangle BIX \\sim \\triangle ICY$, i.e., $BX \\cdot CY = IY \\cdot IX$. Hence $AX \\cdot DY = BX \\cdot CY$.\n\n![](images/broshura_07_english_p24_data_8c20b4ec35.png)",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 14125,
    "subject": "Mathematics (Olympiad)",
    "question": "In triangle $ABC$ with angle $A = 60^\\circ$, the Euler line intersects the circumcircle of the triangle at points $U$ and $V$. $AD$ is the altitude of $\\triangle ABC$, and $\\omega$ is the nine-point circle. The circumcircle $\\Omega$ of triangle $UVD$ intersects $\\omega$ for the second time at point $K$. Let $S$ be a point such that $KS$ is a diameter of $\\omega$. Prove that $AS$ is the symmedian of $\\triangle ABC$.\n\n*An Euler line is a line through the orthocenter, the circumcenter, and the centroid of the triangle.*\n\n*The nine-point circle passes through the feet of the altitudes, the midpoints of the sides, and the midpoints of the segments from the vertices to the orthocenter.*\n\n*The symmedian from a vertex of a triangle is the line symmetric to the median from that vertex about the angle bisector at that vertex.*",
    "options": [],
    "answer": "See solution",
    "solution": "Let $H$ and $O$ be the orthocenter and circumcenter of $\\triangle ABC$, respectively, and let $E$ be the center of $\\omega$. Let $M$ be the midpoint of $BC$. The circle $\\Omega$ intersects $BC$ for the second time at point $X$. Since $\\angle CAB = 60^\\circ$, $\\angle BHC = \\angle BOC = 120^\\circ$, so points $B$, $H$, $O$, and $C$ are concyclic. Let $T$ be the intersection of $UV$ and $BC$. By the power of a point,\n\n$$TD \\cdot TX = TU \\cdot TV = TB \\cdot TC = TH \\cdot TO.$$ \n\nThus, these points lie on the same circle, and since $\\angle HDX = 90^\\circ$, $\\angle HOX = 90^\\circ$, so $XO \\perp HO$. Since $OU = OV$, $XO$ bisects chord $UV$ of $\\Omega$, so $XO$ passes through the center of $\\Omega$ and the point diametrically opposite $X$ on $\\Omega$. But $\\angle XDA = 90^\\circ$, so $Z$ is the point diametrically opposite $X$ on $\\omega$. Thus, $XO$ passes through $Z$.\n\nIn right triangle $BOM$, $\\angle BOM = \\frac{1}{2}\\angle BOC = \\angle BAC = 60^\\circ$, so $\\angle OBM = 30^\\circ$ and $OM = \\frac{1}{2}BO$. Since $AH = 2OM$, $AH = BO = AO$. Therefore, $\\angle BAC = 60^\\circ$ implies $AH = AO$, so $AE \\perp HO$, with $AE$ the bisector of $\\angle HAO$, and thus also the bisector of $\\angle CAB$. The radius of $\\omega$ is $\\frac{1}{2}AO = OM$. Since $\\angle HOZ = 90^\\circ$, $AH = AO$ implies $A$ is the midpoint of $HZ$.\n\nLet $W$ be the point symmetric to $O$ with respect to $M$. Then $W$ is also symmetric to $O$ with respect to $BC$, so $BW = WC$ and $\\angle BWC = 120^\\circ$, so $W$ is the midpoint of the smaller arc $BC$ of the circumcircle of $\\triangle ABC$, and thus $W$ lies on $AE$. Since $OA = OW$ and $OE \\perp AW$, $E$ is the midpoint of $AW$.\n\nIn trapezoid $ZOWH$, $A$ and $M$ are midpoints of bases $ZH$ and $OW$, so the intersection of diagonals $ZW \\cap HO$ lies on $AM$, and $AM \\cap HO = G$ is the centroid of $\\triangle ABC$, so $G$ lies on $ZW$.\n\nTo prove collinearity of $Z$, $K$, and $W$, let $K'$ be the projection of $X$ onto $ZW$. Since $\\angle XK'Z = \\angle XDZ = 90^\\circ$, points $X$, $Z$, $D$, and $K'$ are cyclic, so $K'$ lies on $\\Omega$. Since $K'$ lies on $ZW$, to prove collinearity of $Z$, $K$, and $W$, it suffices to show $K'$ lies on $\\omega$, so $K' = K$, and thus $K$ lies on $ZW$.\n\nWe have $\\angle WK'X = \\angle WMX = 90^\\circ$, so $X$, $M$, $K'$, and $W$ are concyclic, and $\\angle MK'X = \\angle MWX = \\angle MOX$. Since $ZD$ and $OM$ are both perpendicular to $BC$, they are parallel, so $\\angle MOX = \\angle DZX$, i.e., $\\angle MK'X = \\angle DZX$. Since $X$, $Z$, $K'$, and $D$ are cyclic, $\\angle DK'X = 180^\\circ - \\angle DZX$. Therefore,\n\n$$\n\\angle DK'M = \\angle DK'X - \\angle MK'X = 180^\\circ - \\angle DZX - \\angle DZX = 180^\\circ - 2\\angle DZX.\n$$\n\nNote that $AE$ and $ZX$ are both perpendicular to $BC$, so they are parallel, and $\\angle DZX = \\angle DAE$. Thus,\n\n$$\n\\angle DK'M = 180^\\circ - 2\\angle DZX = 180^\\circ - 2\\angle DAE = 180^\\circ - \\angle DAO.\n$$\n\nLet $N$ be the midpoint of $AH$, then $N$ lies on $\\omega$, and $\\angle MDN$ is right, so $MN$ is the diameter of $\\omega$, and $M$, $E$, and $N$ are collinear. Since $AN = OM$ and $AH \\parallel OM$, $AOMN$ is a parallelogram, so $MN \\parallel AO$, and $\\angle DAO = \\angle DNM$. Thus,\n\n$$\n\\angle DK'M = 180^\\circ - \\angle DAO = 180^\\circ - \\angle DNM,\n$$\n\nso $\\angle DK'M + \\angle DNM = 180^\\circ$, so $K'$ lies on the circumcircle of $\\triangle DNM$ ($\\omega$), proving collinearity of $G$, $K$, $Z$, and $W$.\n\nLet $P$ be symmetric to $M$ with respect to $HO$. Since $A$ and $W$ are symmetric with respect to $HO$, $APMW$ is an isosceles trapezoid, so the intersection of diagonals $PW \\cap AM$ lies on $HO$, and $G = HO \\cap AM$, so $T$ lies on $GW$, which passes through $K$. Thus, $S$ is the second intersection of $KE$ and $\\omega$. Then $\\angle KSM = \\angle KTM$, and from the isosceles trapezoid $APMW$, $\\angle KTM = \\angle EAM$, so $\\angle EAM = \\angle ESM$, so $A$, $S$, $E$, and $M$ are concyclic, and $ES = EM$, so $AE$ is the bisector of $\\angle SAM$, i.e., $AS$ is the symmedian of $\\triangle ABC$, as required.\n\n![](images/Ukraine2022-23_p45_data_7ec6846a8b.png)\n\nFig. 23",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 14126,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $p$ and $q$ be positive integers such that\n$$\n\\frac{2011}{2010} = \\frac{p+1}{p} \\cdot \\frac{q+1}{q}.\n$$\nShow that $pq = 2010(p + q + 1)$.",
    "options": [],
    "answer": "See solution",
    "solution": "From the equation $pq = 2010(p + q + 1)$, we solve for $p$:\n\n$$\np = \\frac{2010(q+1)}{q-2010} = 2010 + \\frac{2010 \\cdot 2011}{q-2010}.\n$$\n\nSince $p$ and $q$ are positive integers, $q - 2010$ must be a positive divisor of $2010 \\cdot 2011$. Each such divisor gives a unique pair $(p, q)$. Since $2010 \\cdot 2011 = 2 \\cdot 3 \\cdot 5 \\cdot 67 \\cdot 2011$, it has $2^5 = 32$ divisors. As $(p, q)$ and $(q, p)$ yield the same representation, the number of required representations is $16$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 14127,
    "subject": "Mathematics (Olympiad)",
    "question": "$\\triangle ABC$-ийн гадна талд $A$ оройд хамаарах гадаад багцсан тойргийн төв нь $J$ байг. Энэ тойрог нь $BC$ талыг $M$ цэгт, $AB$ ба $AC$ шулуунуудыг $K$ ба $L$ цэгүүдэд тус тус шүргэнэ. $LM$ ба $BJ$ шулуунууд $F$ цэгт, $KM$ ба $CJ$ шулуунууд $G$ цэгт огтлолцоно. $AF$ ба $BC$ шулуунуудын огтлолын цэгийг $S$, $AG$ ба $BC$ шулуунуудын огтлолын цэгийг $T$ гэе. $M$ цэг нь $ST$ хэрчмийн дундач гэдгийг батал.",
    "options": [],
    "answer": "See solution",
    "solution": "Тэгш өнцөгт гурвалжныг авч үзье. $\\alpha = \\angle CAB$, $\\beta = \\angle ABC$, $\\gamma = \\angle BCA$ гэж авъя. $AJ$ шулуун нь $\\angle CAB$ өнцгийн дундуур татсан тул $\\angle JAK = \\angle JAL = \\frac{\\alpha}{2}$. Мөн $\\angle AKJ = \\angle ALJ = 90^\\circ$ тул $K$ ба $L$ цэгүүд $AJ$ диаметртэй $\\omega$ тойрог дээр оршино.\n\n$KBM$ гурвалжин тэнцүү хажуут гурвалжин бөгөөд $BK$ ба $BM$ нь гадаад багцсан тойргийн шүргэгчид юм. $BJ$ шулуун нь $\\angle KBM$ өнцгийн дундуур татсан тул $\\angle MBJ = 90^\\circ - \\frac{\\alpha}{2}$, $\\angle BMK = \\frac{\\alpha}{2}$. Үүнтэй адил $\\angle MCJ = 90^\\circ - \\frac{\\alpha}{2}$, $\\angle CML = \\frac{\\alpha}{2}$. Мөн $\\angle BMF = \\angle CML$, тэгэхээр\n\n$$\n\\angle LFJ = \\angle MBJ - \\angle BMF = \\left(90^\\circ - \\frac{\\beta}{2}\\right) - \\frac{\\alpha}{2} = \\angle LAJ.\n$$\n\nИймд $F$ цэг $\\omega$ тойрог дээр оршино. (Өнцгийн тооцоогоор $F$ ба $A$ нь $BC$-ийн нэг талд байна.) Үүнтэй адил $G$ ч мөн $\\omega$ дээр оршино. $AJ$ нь $\\omega$-ийн диаметр тул $\\angle AFJ = \\angle AGJ = 90^\\circ$.\n\n![](images/2013-ilovepdf-compressed_p7_data_4d28b6cefa.png)\n\n$AB$ ба $BC$ шулуунууд нь гадаад өнцгийн дундуур татсан $BF$ шулуунаар харьцангуй симметрик. $AF \\perp BF$ ба $KM \\perp BF$ тул $SM$ ба $AK$ нь $BF$-ээр симметрик, тэгэхээр $SM = AK$. Үүнтэй адил $TM = AL$. $AK$ ба $AL$ нь гадаад багцсан тойргийн шүргэгч тул $AK = AL$, тэгэхээр $SM = TM$, батлагдлаа.\n\nТайлбар: $AFKJLG$ тойргийг олсны дараа шийдлийг олон янзаар үргэлжлүүлж болно. Жишээ нь, $JMF_S$ ба $JMGT$ циклик дөрвөлжнүүдээс $\\angle TJS = \\angle STJ = \\frac{\\alpha}{2}$ гэдгийг олж болно. Эсвэл $AS$ ба $GM$ шулуунууд параллель ($BJ$ гадаад өнцгийн дундуур татсан шулуунд перпендикуляр тул), тэгэхээр $\\frac{MS}{MT} = 1$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 14128,
    "subject": "Mathematics (Olympiad)",
    "question": "If the radius of the smaller circle is $r$ cm, and the radius of the larger circle is $12$ cm, what is the value of $r$ if $$\\frac{\\text{area of smaller circle}}{\\text{area of larger circle}} \\approx \\frac{4}{9}?$$",
    "options": [],
    "answer": "See solution",
    "solution": "The ratio of the areas is $$\\frac{\\pi r^2}{\\pi \\cdot 12^2} = \\frac{r^2}{12^2}.$$ Setting this equal to $\\frac{4}{9}$, we get:\n$$\\frac{r^2}{12^2} = \\frac{4}{9}$$\n$$r^2 = 12^2 \\cdot \\frac{4}{9} = 144 \\cdot \\frac{4}{9} = \\frac{576}{9} = 64$$\n$$r = \\sqrt{64} = 8$$\nSo, $r = 8$ cm.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 14129,
    "subject": "Mathematics (Olympiad)",
    "question": "Prove that the equation\n\n$$\n\\frac{1}{\\sqrt{x} + \\sqrt{1006}} + \\frac{1}{\\sqrt{2012 - x} + \\sqrt{1006}} = \\frac{2}{\\sqrt{x} + \\sqrt{2012 - x}}\n$$\n\nhas 2013 integer solutions.",
    "options": [],
    "answer": "See solution",
    "solution": "One can easily check that the given relation holds for any admissible value of $x$. Since $x$ is subject to the conditions $0 \\le x \\le 2012$, the conclusion is easily reached.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 14130,
    "subject": "Mathematics (Olympiad)",
    "question": "In a triangle $\\triangle ABC$, let the incircle be tangent to $BC$, $CA$, and $AB$ at $D$, $E$, and $F$, respectively, and let the excircle opposite to $A$ be tangent to $BC$, $CA$, and $AB$ at $P$, $Q$, and $R$, respectively. Let $EX$ and $FY$ be altitudes in triangle $\\triangle DEF$, and $QZ$ and $RW$ altitudes in triangle $\\triangle PQR$. Prove that the points $X$, $Y$, $Z$, and $W$ are collinear.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $K$ be the orthogonal projection of $B$ onto the angle bisector of $\\angle BAC$. By the Iran lemma and its analogue for the excircle (both easy angle chases), $K$ lies on lines $DE$ and $PQ$. Let $L$ be the orthogonal projection of $K$ onto line $AB$. Our main claim is that $L$ lies on both $XY$ and $WY$, and thus $X$, $Y$, $W$ are collinear. This also suffices since then a symmetric claim with respect to $C$ gives $X$, $Y$, $Z$ collinear. We use directed angles.\n\n**Claim:** $L$ lies on $XY$.\n\n**Proof:** Clearly $KLFY$ is cyclic. Also, letting $I$ be the incenter of $\\triangle ABC$, we see that $BFIK$ is cyclic. Now\n\n$$\n\\angle XYD = \\angle XFE = \\angle CDE\n$$\n\nwhich is easy to chase to be\n\n$$\n\\angle BIK = \\angle BFK = \\angle LFK = \\angle LYK = \\angle LYD.\n$$\n\n**Claim:** $L$ lies on $WY$.\n\n**Proof:** First note that $\\angle QKE = 90^\\circ$ as\n\n$$\n\\angle KEQ + \\angle EQK = \\angle DEC + \\angle CQP = 90^\\circ\n$$\n\nand since $FR$ and $EQ$ are symmetric with respect to $AK$, we also have $\\angle FKR = 90^\\circ$. Now as $KLRW$ is cyclic, we have\n\n$$\n\\angle WLR = \\angle WKR = \\angle QKR\n$$\n\nand by the perpendicularities, we have that this is just\n\n$$\n\\angle EKF = \\angle YKF = \\angle YLF.\n$$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 14131,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $M = \\{1, 2, \\dots, 19\\}$ and $A = \\{a_1, a_2, \\dots, a_k\\} \\subseteq M$. Find the minimum value of $k$ so that there exist $a_i, a_j \\in A$ such that for every $b \\in M$, either $a_i = b$ or $a_i \\pm a_j = b$.",
    "options": [],
    "answer": "See solution",
    "solution": "By the definition of $A$, we have $k(k+1) \\geq 19$, implying $k \\geq 4$.\n\nIf $k=4$, then $k(k+1)=20$. Assume $a_1 < a_2 < a_3 < a_4$. Then $a_4 \\geq 10$.\n\n1. If $a_4 = 10$, then $a_3 = 9$, and $a_2 = 8$ or $7$. If $a_2 = 8$, then $20$, $10-9=1$, $9-8=1$, impossible. If $a_2 = 7$, then $a_1 = 6$ or $5$. Since $20$, $10-9=1$, $7-6=1$ or $20$, $9-7=2$, $7-5=2$, impossible.\n2. If $a_4 = 11$, then $a_3 = 8$, $a_2 = 7$, $a_1 = 6$, impossible.\n3. If $a_4 = 12$, then $a_3 = 7$, $a_2 = 6$, $a_1 = 5$, impossible.\n4. If $a_4 = 13$, then $a_3 = 6$, $a_2 = 5$, $a_1 = 4$, impossible.\n5. If $a_4 = 14$, then $a_3 = 5$, $a_2 = 4$, impossible.\n6. If $a_4 = 15$, then $a_3 = 4$, $a_2 = 3$, $a_1 = 2$, impossible.\n7. If $a_4 = 16$, then $a_3 = 3$, $a_2 = 2$, $a_1 = 1$, impossible.\n8. If $a_4 \\geq 17$, impossible.\n\nSo $k \\geq 5$. Let $A = \\{1, 3, 5, 9, 16\\}$; then $A$ satisfies the conditions of the problem. Therefore, $k_{\\min} = 5$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 14132,
    "subject": "Mathematics (Olympiad)",
    "question": "A sequence $\\langle a_n \\rangle$ of positive integers is given, such that $a_1 = 1$ and $a_{n+1}$ is the smallest positive integer such that\n\n$$\nlcm(a_1, a_2, \\dots, a_n, a_{n+1}) > \\lcm(a_1, a_2, \\dots, a_n).\n$$\n\nWhich numbers are contained in the sequence?",
    "options": [],
    "answer": "See solution",
    "solution": "The first few elements of the sequence are:\n\n$1, 2, 3, 4, 5, 7, 8, 9, 11, \\ldots$\n\nThe first two positive integers not contained in the sequence are $6$ and $10$. These would not have made the lcm larger when it was their turn, and they will not do so later. Thus, a number left out at its turn cannot appear later.\n\nEach prime is included in the sequence: when a prime $p$ is considered, its inclusion always increases the lcm, so all primes and their powers are included. Any integer $N$ with at least two different prime divisors has all its prime power divisors already included, so adding $N$ does not increase the lcm, and such $N$ are not included.\n\nIn summary, the sequence $\\langle a_n \\rangle$ consists of $1$ and all powers of primes in ascending order.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 14133,
    "subject": "Mathematics (Olympiad)",
    "question": "Call a number _interesting_ if it can be represented as the sum of squares of three distinct non-negative integers. For example, the number 5 is interesting, because $5 = 0^2 + 1^2 + 2^2$.\n\nCall a number _special_ if it is not interesting, but can be represented as the product of two distinct interesting numbers.\n\n(a) Find one special number.\n\n(b) Prove that there are infinitely many special numbers.",
    "options": [],
    "answer": "See solution",
    "solution": "The factorization $(2k+1)^2 = 4k^2 + 4k + 1 = 4k(k+1) + 1$, where one of the numbers $k$ and $k+1$ is always even, shows that the square of any odd number gives a remainder of 1 upon division by 8. The square of an even number not divisible by 4 gives a remainder of 4, and the square of an even number divisible by 4 gives a remainder of 0. Thus, the only possible remainders of squares modulo 8 are 0, 1, and 4.\n\nTherefore, the sum of squares of three odd numbers gives a remainder of 3 modulo 8. The sum of squares of one number divisible by 4, one number divisible by 2 but not by 4, and one odd number gives a remainder of 5. Thus, there are infinitely many interesting numbers giving remainders 3 and 5 modulo 8, and therefore also infinitely many products of distinct interesting numbers with a remainder of $3 \\cdot 5 \\equiv 7 \\pmod{8}$. These numbers are indeed special, as three numbers with remainders of 0, 1, or 4 cannot add up to a remainder of 7 modulo 8.\n\nThe smallest example is obtained by choosing the interesting numbers $1^2 + 3^2 + 5^2$ and $0^2 + 1^2 + 2^2$, whose product is $35 \\cdot 5 = 175$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 14134,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $a_1$, $a_2$, $r$, and $s$ be positive integers with $r$ and $s$ odd. The sequence $a_1, a_2, a_3, \\dots$ is defined by\n$$\na_{n+2} = r a_{n+1} + s a_n\n$$\nfor all $n \\ge 1$. Determine the maximum possible number of integers $1 \\le l \\le 2025$ such that $a_l$ divides $a_{l+1}$, over all possible choices of $a_1$, $a_2$, $r$, and $s$.",
    "options": [],
    "answer": "See solution",
    "solution": "1350.\n\nWe first provide the upper bound. We start by dividing out any common factors of $a_1$ and $a_2$ from the whole sequence. Note that since $r$ and $s$ are odd, and $a_1$ and $a_2$ cannot both be divisible by 2, the sequence $a_1, a_2, a_3, \\dots$ (mod 2) must be some cyclic shift of the sequence $1, 1, 0, 1, 1, 0, 1, 1, 0, \\dots$. This means that exactly $\\frac{2025}{3} = 675$ of the values of $l$ satisfy $a_l \\equiv 0 \\pmod{2}$ and $a_{l+1} \\equiv 1 \\pmod{2}$. An even number can never divide an odd number, so we have an upper bound of $2025 \\times \\frac{2}{3} = 1350$.\n\nNow we provide a construction so that $a_l$ divides $a_{l+1}$ for 1350 values of $l$. Let $F_1, F_2, F_3, \\dots$ denote the Fibonacci sequence with $F_1 = F_2 = 1$. Let\n$$\nC = \\frac{F_1 F_2 F_3 F_4 \\cdots F_{2025}}{F_3 F_6 F_9 F_{12} \\cdots F_{2025}} = F_1 F_2 F_4 F_5 \\cdots F_{2024}.\n$$\nNote that $C$ is odd, since it is the product of all odd Fibonacci numbers up to $F_{2025}$. We let $a_n = C^{n-1} F_n$, which satisfies the recurrence with $r = C^2$ and $s = C$. This gives\n$$\n\\frac{a_{l+1}}{a_l} = C \\cdot \\frac{F_{l+1}}{F_l},\n$$\nwhich is an integer whenever $3 \\nmid l$ for $1 \\le l \\le 2025$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 14135,
    "subject": "Mathematics (Olympiad)",
    "question": "In a directed graph $G$, the outgoing degree of each vertex equals $3$ (loops and bidirectional edges are allowed and considered as cycles). Prove that the graph contains $2$ non-intersecting directed cycles.",
    "options": [],
    "answer": "See solution",
    "solution": "We proceed by induction on $n$, the number of vertices.\n\n**Base case:** For $n = 4$, the only possible digraph is the complete digraph, which contains $2$ disjoint cycles of length $2$.\n\n**Inductive step:** Assume any $3$-out digraph with $n-1 \\ge 4$ vertices contains $2$ disjoint cycles. Let $G$ be a digraph on $n$ vertices failing our assumption.\n\nIf $G$ has a bidirectional edge $uw$, then $G \\setminus \\{u, w\\}$ is still $1$-out, so it contains a cycle. Paired with the $2$-cycle made by the bidirectional edge $uw$, this gives the desired disjoint cycles.\n\nThe main idea is to use edge contractions. If there is an edge $uv$ such that $u$ and $v$ have no common parent, we can modify $G$ to a new digraph $G'$ by removing $u$ and $v$ and adding a vertex $w$ whose outgoing edges go to all children of $v$ and whose ingoing edges come from all parents of $u$ and $v$. $G'$ has $n-1$ nodes and is still $3$-out, so by the inductive assumption $G'$ contains $2$ disjoint cycles. If $w$ is not in any of the cycles, they were contained in $G$ and we are done. If $w$ is in one of the cycles, the other cycle is in $G$. Depending on whether the cycle in-edge of $w$ comes from $u$'s in-edge or $v$'s in-edge, replacing $w$ by $uv$ or $v$ gives the other cycle in $G$.\n\nThe only remaining option is when every edge of $G$ has a \"witness\" (a common parent to both its end-vertices). Let $v$ be a vertex with the smallest in-degree $d^-(v)$. Since there are $3n$ edges in total, $d^-(v) \\le 3$.\n\n- **Case 1:** $d^-(v) = 0$. Then the edge starting in $v$ has no witnesses.\n- **Case 2:** $d^-(v) = 1$. Then the edge ending in $v$ has no witnesses.\n- **Case 3:** $d^-(v) = 2$. Let $u, w$ be two parents of $v$. Then $u$ must be the witness to $wv$ and $w$ to $uv$, implying $uw$ is a bidirectional edge, which is a contradiction.\n- **Case 4:** $d^-(v) = 3$. Since the graph is exactly $3$-out and the minimal in-degree is $3$, all vertices have in-degree $3$. Given a vertex $x$ and its parents $u, v, w$, $u, v, w$ must form a $3$-cycle. Each of the witnesses of $ux, vx, wx$ must be among $u, v, w$, so the $u, v, w$ induced subgraph is $1$-in. Since bidirectional edges do not exist, $u, v, w$ make a $3$-cycle. Reversing the edges of $G$ shows that children of $x$ also form a $3$-cycle. As there are no bidirectional edges, the children and parents of $x$ are disjoint and give $2$ disjoint $3$-cycles.",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 14136,
    "subject": "Mathematics (Olympiad)",
    "question": "In a triangle $ABC$, let segment $AP$ bisect $\\angle BAC$, with $P$ on side $BC$, and let segment $BQ$ bisect $\\angle ABC$, with $Q$ on side $CA$. It is known that $\\angle BAC = 60^\\circ$ and that $AB + BP = AQ + QB$. What are the possible angles of triangle $ABC$?",
    "options": [],
    "answer": "See solution",
    "solution": "Let $\\angle ABC = 2x$ and $\\angle BCA = y$. Then $\\angle ABQ = \\angle QBC = x$ and\n$$\n\\angle CAB + \\angle ABC + \\angle BCA = 60^{\\circ} + 2x + y = 180^{\\circ},\n$$\nso\n$$\ny = 120^{\\circ} - 2x. \\qquad (1)\n$$\n\nMany of the following trigonometric or complex number calculations lead to the possible values of the angles of triangle *ABC*. It is not difficult to check that these values satisfy the conditions of the problem. We leave this to the readers.\n\n**First Solution.** (by Reid Barton and Gabriel Carroll) Extend segment $AB$ through $B$ to $R$ so that $BR = BP$, and construct $S$ on ray $AQ$ so that $AS = AR$.\n\n![](images/USA_IMO_2001_p91_data_5278c5a810.png)\n\n**Lemma 1.** Points $B$, $P$, $S$ are collinear, and consequently, $S$ coincides with $C$.\n\n*Proof.* Because $BR = BP$, triangle $BPR$ is isosceles with base angles\n$$\n\\angle BRP = \\angle RPB = \\frac{180^{\\circ} - \\angle PBR}{2} = x = \\angle QBP. \\quad (2)\n$$\nNote that $AS = AR$ and $\\angle RAS = \\angle BAC = 60^{\\circ}$, implying that triangle $ARS$ is equilateral. Since line $AP$ bisects $\\angle RAS$, $R$ and $S$ are symmetric with respect to line $AP$. Thus,\n$$\nPR = PS \\qquad (3)\n$$\nand $\\angle ARP = \\angle PSA$, or $\\angle BRP = \\angle PSQ$. By (2), we have\n$$\n\\angle QBP = \\angle BRP = \\angle PSQ. \\qquad (4)\n$$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 14137,
    "subject": "Mathematics (Olympiad)",
    "question": "Mr. Pisut walks in the first quadrant of the coordinate plane, starting at $ (0, N) $ and ending at $ (M, 0) $, where $ M $ and $ N $ are positive integers. His walk satisfies the following conditions:\n\n* Each of his steps is of 1 unit length in the direction parallel to either the X-axis or the Y-axis.\n* For each point $ (x, y) $ on his path, $ x \\ge 0 $ and $ y \\ge 0 $.\n\nFor each step, he measures the distance from himself to the axis to which his step is parallel. If the step takes him farther away from the origin, he records the distance as a positive value; otherwise it is recorded as negative.\n\nProve that after he finishes his walk, the sum of all distances recorded is zero.",
    "options": [],
    "answer": "See solution",
    "solution": "Suppose that Mr. Pisut walks $k$ steps in total and the $i$-th step is from the point $(x_{i-1}, y_{i-1})$ to the point $(x_i, y_i)$.\n\nNotice that if the $i$-th step is parallel to the X-axis, then $y_i = y_{i-1}$ and he records $y_{i-1}(x_i - x_{i-1})$. Likewise, if the $i$-th step is parallel to the Y-axis, then $x_i = x_{i-1}$ and he records $x_i(y_i - y_{i-1})$.\n\nSo the distance he records for the $i$-th step, regardless of the direction, is $y_{i-1}(x_i - x_{i-1}) + x_i(y_i - y_{i-1})$.\n\nTherefore, the sum of all distances recorded is\n\n$$\n\\sum_{i=1}^{k} y_{i-1}(x_i - x_{i-1}) + x_i(y_i - y_{i-1}) = \\sum_{i=1}^{k} (x_i y_i - x_{i-1} y_{i-1}) \\\\\n= x_k y_k - x_0 y_0 = M \\cdot 0 - 0 \\cdot N = 0.\n$$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 14138,
    "subject": "Mathematics (Olympiad)",
    "question": "In a right square pyramid $P-ABCD$, point $G$ is the centroid of the lateral face $\\triangle PBC$. Let $V_1$ and $V_2$ be the volumes of tetrahedrons $PABG$ and $PADG$, respectively. Find the value of $\\frac{V_1}{V_2}$.",
    "options": [],
    "answer": "See solution",
    "solution": "As shown in the figure below, let $M$ be the midpoint of $BC$, so $G$ lies on the median $PM$ of $\\triangle PBC$.\n\n![](images/Mathematical_Olympiad_In_China_2021_2022___Problems_and_Solutions_2024_p67_data_ffc6683b41.png)\n\nNote that\n\n$$\nV_1 = V_{G-PAB} = \\frac{PG}{PM} \\cdot V_{M-PAB},\n$$\n\n$$\nV_2 = V_{G-PAD} = \\frac{PG}{PM} \\cdot V_{M-PAD}.\n$$\n\nTherefore,\n\n$$\n\\frac{V_1}{V_2} = \\frac{V_{M-PAB}}{V_{M-PAD}} = \\frac{V_{P-MAB}}{V_{P-MAD}} = \\frac{S_{\\triangle MAB}}{S_{\\triangle MAD}} = \\frac{1}{2}.\n$$",
    "topic": "Geometry",
    "subtopic": "Solid Geometry"
  },
  {
    "id": 14139,
    "subject": "Mathematics (Olympiad)",
    "question": "Amy and Bec play the following game. Initially, there are three piles, each containing 2020 stones. The players take turns to make a move, with Amy going first. Each move consists of choosing one of the piles available, removing the unchosen pile(s) from the game, and then dividing the chosen pile into 2 or 3 non-empty piles. A player loses the game if they are unable to make a move.\n\nProve that Bec can always win the game, no matter how Amy plays.",
    "options": [],
    "answer": "See solution",
    "solution": "Call a pile *perilous* if the number of stones in it is one more than a multiple of three, and *safe* otherwise. Bec has a winning strategy by ensuring that she only leaves Amy perilous piles. Bec wins because the number of stones is strictly decreasing, and eventually Amy will be left with two or three piles each with just one stone.\n\nTo see that this is a winning strategy, we prove that Bec can always leave Amy with only perilous piles, and that under such circumstances, Amy must always leave Bec with at least one safe pile.\n\nOn Amy's turn, whenever all piles are perilous it is impossible to choose one such perilous pile and divide it into two or three perilous piles by virtue of the fact that $1+1 \\neq 1 \\pmod{3}$ and $1+1+1 \\neq 1 \\pmod{3}$. Thus Amy must leave Bec with at least one safe pile.\n\nOn Bec's turn, whenever one of the piles is safe, she can divide it into two or three piles, each of which are safe, by virtue of the fact that $2 \\equiv 1+1 \\pmod{3}$ and $0 \\equiv 1+1+1 \\pmod{3}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 14140,
    "subject": "Mathematics (Olympiad)",
    "question": "How many palindromic numbers (numbers that read the same forwards and backwards) are there less than 2012?",
    "options": [],
    "answer": "See solution",
    "solution": "*Classify palindromic numbers by digit count:*\n\n- **4-digit numbers:**\n  - A 4-digit palindromic number has the form $abba$.\n  - For numbers less than 2012, the thousands digit can be 1 or 2.\n  - If the thousands digit is 2, the only possibility is $2002$ (since $2112 > 2012$), so 1 number.\n  - If the thousands digit is 1, the hundreds digit can be $0$ to $9$, giving $10$ numbers ($1001, 1111, \\ldots, 1991$).\n  - Total 4-digit palindromic numbers: $1 + 10 = 11$.\n\n- **3-digit numbers:**\n  - Form: $aba$.\n  - Hundreds digit: $1$ to $9$ ($9$ choices).\n  - Tens digit: $0$ to $9$ ($10$ choices).\n  - Total: $9 \\times 10 = 90$.\n\n- **2-digit numbers:**\n  - Form: $aa$.\n  - Tens digit: $1$ to $9$ ($9$ choices).\n  - Total: $9$.\n\n- **1-digit numbers:**\n  - $1$ to $9$ ($9$ numbers).\n\n*Total palindromic numbers less than 2012:*\n\n$$\n11 + 90 + 9 + 9 = 119\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 14141,
    "subject": "Mathematics (Olympiad)",
    "question": "Given a foundation in the form of a $6 \\times 6$ square, divided into smaller $1 \\times 1$ squares. There is a gap of length $1$ between any two adjacent squares. The foundation is covered by several layers of bricks of size $3 \\times 1$. Every layer consists of $12$ bricks, and each brick fully covers exactly two gaps of length $1$. Such a covering is called *strong* if every gap is covered by a brick in at least one of the layers. Determine the minimum number of layers required for a strong covering.\n\n![](Fig26.png)",
    "options": [],
    "answer": "See solution",
    "solution": "**Answer:** 4 layers.\n\nSuppose, for contradiction, that a strong covering can be achieved with only three layers. Consider a particular square (the grey square) and four marked gaps as illustrated in the figure. Each gap can only be covered by a brick that also covers the grey square. However, any such brick cannot cover two of the marked gaps simultaneously. Therefore, the grey square must be covered by at least four bricks, implying that at least four layers are necessary for a strong covering.",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 14142,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $\\omega$ be the circumcircle of a scalene triangle $ABC$. The tangents to $\\omega$ at $A$ and $C$ meet in $P$, and the line $BP$ intersects $\\omega$ in $D$. Let $BB'$ be a diameter of $\\omega$. The exterior angle bisector of $\\angle ABC$ and the lines $B'A$ and $B'C$ intersect in $A'$ and $C'$, respectively. Prove that $A', B', C', D$ are cyclic.\n\n![](images/MNG_ABooklet_2017_p23_data_fd6547e20a.png)",
    "options": [],
    "answer": "See solution",
    "solution": "Since $PC$ is tangent to $\\omega$, we have $\\triangle PCD \\cong \\triangle PBC$, so $CD \\cdot PB = PC \\cdot CB$. Similarly, $AD \\cdot PB = PA \\cdot AB$. Thus\n$$\n\\frac{CD}{CB} = \\frac{AD}{AB}. \\qquad (*)\n$$\nsince $PA = PC$. We also have that $\\triangle AA'B \\sim \\triangle CC'B$ because $\\angle ABA' = \\angle CBC'$ and $\\angle A'AB = \\angle BCC' = 90^\\circ$. Hence $\\frac{AB}{CB} = \\frac{AA'}{CC'}$. Thus, it follows from $(*)$ that $\\frac{AA'}{CC'} = \\frac{AD}{CD}$. Hence $\\triangle DAA' \\sim \\triangle DCC'$ since $\\angle B'AD = \\angle B'CD$, which follows from $\\angle DAA' = \\angle DCC'$. Therefore, $\\angle AA'D = \\angle B'C'D$, implying that $B', A', C', D$ are cyclic.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 14143,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $x$, $y$, and $z$ be any positive real numbers. Prove the inequality\n\n$$\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$$\n\nFind when equality holds.",
    "options": [],
    "answer": "See solution",
    "solution": "Since the inequality involves the minimum of two positive numbers and since the function $y = x^2$ is increasing on $\\mathbb{R}^+$, our task is to verify\n\n$$\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\n$$\nand\n$$\n(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.\n$$\n\nWe need to find when at least one equality holds. By swapping $x$ and $y$, the second inequality follows from the first, so we can focus on the first inequality. Expanding both sides gives\n\n$$\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$$\n\nLet $a = x/y$, $b = y/z$, $c = z/x$ (all positive). The inequality becomes\n\n$$\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.\n$$\n\nFor any positive $t$,\n$$\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$$\n\nThis is always non-negative, and equality holds if and only if $a = b = c = 1$, i.e., $x = y = z$. This condition is unchanged under swapping $x$ and $y$. Thus, the original inequality is proven, and equality holds if and only if $x = y = z$.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 14144,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $O$ be the circumcenter of triangle $ABC$, and let $O_1, O_2, O_3$ be the circumcenters of triangles $OBC$, $OCA$, and $OAB$, respectively. Let $P_1, P_2, P_3$ be the images of $O$ under inversion with respect to the circumcircle $\\Omega$ of $ABC$. Prove that $O_1, O_2, O_3$ are collinear.\n\n![](images/Saudi_Booklet_2025_p33_data_72cf733caf.png)",
    "options": [],
    "answer": "See solution",
    "solution": "To prove that $O_1, O_2, O_3$ are collinear, it suffices to show that the circle $(P_1P_2P_3)$ passes through $O$, which is equivalent to the perpendicular bisectors of $OP_1, OP_2, OP_3$ being concurrent. Let $OK, OL$ cut $BC$ at $K', L'$, then by the shooting lemma,\n\n$$\nOK \\cdot OK' = OL \\cdot OL' = R^2\n$$\n\nso $K', L'$ are images of $K, L$ through $\\Omega$. Let $X$ be the midpoint of $K'L'$. Notice that $O, O_1, P_1$ are collinear and $OK'L'P_1$ is an isosceles trapezoid, which implies that $X$ belongs to the perpendicular bisector of $OP_1$. The triangles $HBC$ and $OK'L'$ have parallel sides, so $HM \\parallel OX$. Similarly, define $Y, Z$ for the other sides. Thus, it suffices to prove that the lines through $X, Y, Z$ perpendicular to $BC, CA, AB$ concur. By Carnot's theorem, we need to show\n\n$$\n\\sum_{sym} (BX^2 - CX^2) = 0.\n$$\n\nWe have\n\n$$\nBX^2 - CX^2 = \\overline{BC}(\\overline{BX} - \\overline{XC}) = 2\\overline{MX} \\cdot \\overline{BC}.\n$$\n\nSince triangles $OXM$ and $HMD$ are similar,\n\n$$\n\\frac{XM}{MD} = \\frac{OM}{HD} = \\frac{AH}{2HD}.\n$$\n\nTherefore,\n\n$$\n2\\overline{MX} \\cdot \\overline{BC} = \\frac{AH}{HD} \\cdot \\overline{DM} \\cdot \\overline{BC} = \\frac{AH^2 \\cdot 2\\overline{DM} \\cdot \\overline{BC}}{2AH \\cdot HD} = \\frac{AH^2 \\cdot (AC^2 - AB^2)}{|\\mathcal{P}_{H/(O)}|}.\n$$\n\nSince $AC^2 - AB^2 = CD^2 - BD^2$, we need to prove $\\sum_{sym} AH^2(CD^2 - BD^2) = 0$. This holds because\n\n$$\n\\sum_{A,B,C} AH^2(CD^2 - BD^2) = 4 \\sum_{A,B,C} ([HAB]^2 - [HAC]^2) = 0.\n$$\n\nThus, the problem is completely solved. $\\square$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 14145,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $D$, $E$, $F$ be points on the sides $BC$, $CA$, $AB$ respectively of a triangle $ABC$ such that $BD = CE = AF$ and $\\angle BDF = \\angle CED = \\angle AFE$. Prove that $ABC$ is equilateral.",
    "options": [],
    "answer": "See solution",
    "solution": "![](images/Indija_mo_2011_p2_data_e1e1fdcb32.png)\n\nConsider the triangles $BDF$, $CED$, and $AFE$ with $BD$, $CE$, and $AF$ as bases. The sides $DF$, $ED$, and $FE$ make equal angles $\\theta$ with the bases of the respective triangles. If $B \\geq C \\geq A$, then it is easy to see that $FD \\geq DE \\geq EF$. Now, using the triangle $FDE$, we see that $B \\geq C \\geq A$ gives $DE \\geq EF \\geq FD$. Combining, you get $FD = DE = EF$ and hence $A = B = C = 60^\\circ$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 14146,
    "subject": "Mathematics (Olympiad)",
    "question": "ABC гурвалжны $\\angle BAC = 90^\\circ$. А оройгоос BC талд татсан өндрийн суурийг D гэе. ABD болон ACD гурвалжинд багтсан тойргийн төвүүдийг харгалзан $I_1$, $I_2$ гэе. $I_1$ ба $I_2$ цэгүүдээс $AD$ хэрчимд татсан перпендикулярын сууриудыг харгалзан $M$ ба $K$ гэе. Хэрэв $I_1M + I_2K = \\frac{1}{4}BC$ бол $ABC$ гурвалжны өнцгүүдийг ол.",
    "options": [],
    "answer": "See solution",
    "solution": "$m = BD$, $n = CD$, $I_1M = d_1$, $I_2K = d_2$, $AD = h$ гэе.\n\n$$\n\\begin{align*}\nBN &= BP \\text{ ба } AP = AM \\text{ байх нь} \\\\\n\\text{ойломжтой.}\n\\end{align*}\n$$\n\n$$\n\\text{Иймд } AB = AM + BN \\text{ ба}\n$$\n\n$$\nc = h - d_1 + m - d_1. \\text{ Адилаар}\n$$\n\n$$\nb = h - d_2 + n - d_2. \\text{ Эндээс}\n$$\n\n$$\nh = \\frac{bc}{a}; \\quad m + n = a; \\quad d_1 + d_2 = \\frac{a}{4} \\text{ гэдгээс } 4bc + a^2 = 2ab + 2ac \\Rightarrow 2c(2b - a) - a(2b - a) = 0\n$$\n\n$$\n\\rightarrow (2b-a)(2c-a) = 0 \\Rightarrow a = 2b \\text{ эсвэл } a = 2c\n$$\n\n$$\n\\rightarrow \\angle ABC = 30^{\\circ}, \\angle ACB = 60^{\\circ} \\text{ эсвэл } \\angle ABC = 60^{\\circ}, \\angle ACB = 30^{\\circ}\n$$\n\n$$\n\\text{Тэгэхээр: } (30^{\\circ}) \\text{ ба } 60^{\\circ}, \\angle A = 90^{\\circ}.\n$$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 14147,
    "subject": "Mathematics (Olympiad)",
    "question": "We define a *chessboard polygon* as a polygon whose edges lie along lines of the form $x = a$ or $y = b$, where $a$ and $b$ are integers. These lines divide the interior into unit squares, which are shaded alternately grey and white so that adjacent squares have different colors. To tile a chessboard polygon by dominoes is to exactly cover the polygon with non-overlapping $1 \\times 2$ rectangles. A *tasteful tiling* is one that avoids the two configurations of dominoes shown on the left below. Two tilings of a $3 \\times 4$ rectangle are shown; the first is tasteful, while the second is not, due to the vertical dominoes in the upper right corner.\n\n![](images/pamphlet0910_main_p6_data_5c6709e291.png)\n\n(a) Prove that if a chessboard polygon can be tiled by dominoes, then it can be done so tastefully.\n\n(b) Prove that such a tasteful tiling is unique.",
    "options": [],
    "answer": "See solution",
    "solution": "(a) We prove the first part by induction on the number $n$ of dominoes in the tiling. The claim is clearly true for $n=1$. Suppose we have a chessboard polygon that can be tiled by $n > 1$ dominoes. Of all the leftmost squares, select the lowest one and label it $L$; assume $L$ is black. Remove the domino covering $L$, leaving a polygon that can be tiled with $n-1$ dominoes. By induction, this polygon can be tastefully tiled.\n\nReplace the removed domino. If it is horizontal, the tiling remains tasteful, since $L$ is black and there are no squares below it. If it is vertical, a problem arises only if another vertical domino is directly to its right; in this case, rotate the pair to get two horizontal dominoes. Repeat this process as needed to obtain a tasteful tiling.\n\nIf $L$ is white, a similar process applies. Difficulty arises only if the domino covering $L$ is horizontal and another horizontal domino is directly above it. Rotate this pair, remove the now vertical domino covering $L$, tile the remainder tastefully by induction, and restore the vertical domino to finish.\n\n(b) Suppose there are two tasteful tilings of a given chessboard polygon. Overlaying them yields chains of overlapping dominoes, since each square is part of one domino from each tiling. A chain of length one indicates a common domino; a chain of length two cannot occur, since it would require a $2 \\times 2$ block covered by horizontal dominoes in one tiling and vertical dominoes in the other, which is distasteful.\n\nIf the tilings are distinct, a chain of length three or more must occur; let $R$ be the region consisting of such a chain and its interior. The chain must include a horizontal domino along its lowest row. If there are two or more overlapping horizontal dominoes, one will be a WB domino (white square on the left). Otherwise, two adjacent vertical dominoes overlap with a single horizontal domino, and again there must be a WB domino. Focus on the tiling with this WB domino.\n\nThe two squares above the WB domino must be part of $R$. A single horizontal domino cannot cover both, nor can a pair of vertical dominoes, as both cases are distasteful. Thus, a horizontal domino must cover at least one, extending past the WB domino. This implies a horizontal WB domino on the next row up. Repeat this argument until reaching a horizontal WB domino in $R$ for which the two squares above are not both in $R$; this domino is part of the chain defining $R$.\n\nNow, walk along the chain, starting on the white square of the lowest WB domino in $R$, moving toward the black square. Draw an arrow along each domino in the direction of travel. Since squares alternate white and black, arrows always point from white to black. The interior of $R$ is always to the left as the chain follows the boundary.\n\nA contradiction arises: the existence of a horizontal WB domino adjacent to the boundary of $R$, with a square above not in $R$, means this domino must be traversed from right to left, so it must be a BW domino instead. This is impossible, completing the proof.",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 14148,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $n$ be a positive integer. Let $\\sigma(n)$ denote the sum of the positive divisors of $n$, and $\\tau(n)$ denote the number of positive divisors of $n$.\n\nProve that\n$$\n\\frac{\\sigma(n)}{\\tau(n)\\sqrt{n}} \\ge \\frac{3}{2\\sqrt{2}}\n$$\nfor all $n \\ge 2$, with equality if and only if $n = 2$.\n\nAdditionally, show that\n$$\n\\frac{\\sigma(n)}{\\tau(n)} \\le \\frac{n+1}{2}\n$$\nwith equality if and only if $n = 1$ or $n$ is prime.",
    "options": [],
    "answer": "See solution",
    "solution": "We first compare the sums:\n$$\n\\frac{1+p+p^2+\\dots+p^k}{(k+1)p^{k/2}} \\ge \\frac{1+2+\\dots+2^k}{(k+1)2^{k/2}} \\ge \\frac{3}{2\\sqrt{2}}\n$$\nThis leads to\n$$\n\\frac{\\sigma(n)}{\\tau(n)\\sqrt{n}} \\ge \\left( \\frac{3}{2\\sqrt{2}} \\right)^s \\ge \\frac{3}{2\\sqrt{2}}\n$$\nwhere equality holds when $n = 2$.\n\nFor the left inequality, let $1 = d_1 < d_2 < \\dots < d_{\\tau(n)} = n$ be all divisors of $n$. Then $\\sigma(n) = \\sum_{i=1}^{\\tau(n)} d_i$ and also $\\sigma(n) = \\sum_{i=1}^{\\tau(n)} \\frac{n}{d_i}$.\n\nFor each divisor $d_i$:\n$$\nd_i + \\frac{n}{d_i} \\le n + 1 \\iff d_i^2 - d_i(n + 1) + n \\le 0 \\iff (n - d_i)(1 - d_i) \\le 0\n$$\nEquality holds only if $d_i = 1$ or $d_i = n$.\n\nThus,\n$$\n\\begin{aligned}\n\\frac{\\sigma(n)}{\\tau(n)} &= \\frac{\\sigma(n) + \\sigma(n)}{2\\tau(n)} = \\frac{\\sum_{i=1}^{\\tau(n)} d_i + \\sum_{i=1}^{\\tau(n)} \\frac{n}{d_i}}{2\\tau(n)} \\\\\n&= \\frac{\\sum_{i=1}^{\\tau(n)} \\left(d_i + \\frac{n}{d_i}\\right)}{2\\tau(n)} \\le \\frac{\\sum_{i=1}^{\\tau(n)} (n+1)}{2\\tau(n)} \\\\\n&= \\frac{\\tau(n) \\cdot (n+1)}{2\\tau(n)} = \\frac{n+1}{2}.\n\\end{aligned}\n$$\nEquality holds only when $\\tau(n) \\le 2$, i.e., $n = 1$ or $n$ is prime.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 14149,
    "subject": "Mathematics (Olympiad)",
    "question": "Suppose you have two counters, each with two different numbers on them. When you toss both counters, the possible sums you can obtain are 8, 9, 10, and 11. What are the numbers on each counter?",
    "options": [],
    "answer": "See solution",
    "solution": "### Alternative i\n\nThe smallest sum, 8, is the sum of the smaller numbers on the two counters. The largest sum, 11, is the sum of the larger numbers on the two counters. Since $8 = 1+7 = 2+6 = 3+5$, the two smaller numbers on the counters are 1 and 7, 2 and 6, or 3 and 5. Since $11 = 2+9 = 3+8 = 4+7 = 5+6$, the two larger numbers on the counters are 2 and 9, 3 and 8, 4 and 7, or 5 and 6. The four numbers on the two counters are all different. The table shows the only combinations we need to consider:\n\n| Smaller numbers | Larger numbers | Give sums 8, 9, 10, 11? |\n|---|---|---|\n| 1 and 7 | 2 and 9 | yes |\n| 1 and 7 | 3 and 8 | yes |\n| 1 and 7 | 5 and 6 | no |\n| 2 and 6 | 3 and 8 | yes |\n| 2 and 6 | 4 and 7 | yes |\n| 3 and 5 | 2 and 9 | no |\n| 3 and 5 | 4 and 7 | yes |\n\n### Alternative ii\n\nLet the numbers on one counter be $a$ and $a+r$.\nLet the numbers on the other counter be $b$ and $b+s$.\nThen $a+b=8$ and $a+r+b+s=11$. So $r+s=3$.\nHence $(a,b) = (1,7), (2,6), (3,5), (5,3), (6,2)$, or $(7,1)$, and $(r,s) = (1,2)$ or $(2,1)$. From symmetry we may assume $r=1$ and $s=2$.\nSo the two counters are:\n\n- 1/2 and 7/9\n- 2/3 and 6/8\n- 3/4 and 5/7\n- 6/7 and 2/4\n- 7/8 and 1/3\n\n### Alternative iii\n\nThe smallest sum, 8, is the sum of the smaller numbers on the two counters. So these numbers are 1 and 7, 2 and 6, or 3 and 5. The largest sum, 11, is the sum of the larger numbers on the two counters. So we have the following addition tables for the four numbers on the counters:\n\n$$\n\\begin{array}{|c|c|c|} \\hline + & 1 & 2 \\\\ \\hline 7 & 8 & 9 \\\\ \\hline 9 & 10 & 11 \\\\ \\hline \\end{array}\n$$\n\n$$\n\\begin{array}{|c|c|c|} \\hline + & 1 & 3 \\\\ \\hline 7 & 8 & 10 \\\\ \\hline 8 & 9 & 11 \\\\ \\hline \\end{array}\n$$\n\n$$\n\\begin{array}{|c|c|c|} \\hline + & 2 & 3 \\\\ \\hline 6 & 8 & 9 \\\\ \\hline 8 & 10 & 11 \\\\ \\hline \\end{array}\n$$\n\n$$\n\\begin{array}{|c|c|c|} \\hline + & 2 & 4 \\\\ \\hline 6 & 8 & 10 \\\\ \\hline 7 & 9 & 11 \\\\ \\hline \\end{array}\n$$\n\n$$\n\\begin{array}{|c|c|c|} \\hline + & 3 & 4 \\\\ \\hline 5 & 8 & 9 \\\\ \\hline 7 & 10 & 11 \\\\ \\hline \\end{array}\n$$\n\n$$\n\\begin{array}{|c|c|c|} \\hline + & 3 & 5 \\\\ \\hline 5 & 8 & 10 \\\\ \\hline 6 & 9 & 11 \\\\ \\hline \\end{array}\n$$\n",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 14150,
    "subject": "Mathematics (Olympiad)",
    "question": "One writes distinct positive integers into the cells of a $3 \\times 3$ table in such a way that, in each row and in each column, one number equals the sum of the other two numbers. Find the least possible total sum of the numbers written into the table.",
    "options": [],
    "answer": "See solution",
    "solution": "The sum of all numbers written in the cells is at least $1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 = 45$. As one number is the sum of the others in every row, the sum of all numbers in every row is even. Thus, the sum of all numbers in the table must be even, too. Hence, the sum of all numbers in the table cannot be $45$. It is possible that the sum is $46$, as shown below:\n\n![](images/EST_ABooklet_2020_p9_data_e1b99d946c.png)\n\nFig. 6",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 14151,
    "subject": "Mathematics (Olympiad)",
    "question": "Is it possible to construct a sequence of nonzero integers $\\{a_k\\}$ such that for every positive integer $n \\geq 2$ and every $k \\geq 1$,\n\n$$\na_k + 2a_{2k} + \\cdots + n a_{nk} = 0?\n$$\n\nIf so, for which values of $n$ does such a sequence exist?",
    "options": [],
    "answer": "See solution",
    "solution": "We claim that such a sequence exists for all $n$ except $n = 2$.\n\nIf $n = 2$, then\n$$\na_1 = -2a_2 = 4a_4 = -8a_8 = \\dots\n$$\nwhich is impossible since $a_1$ cannot be divisible by arbitrarily large powers of $2$. Thus, no valid sequence exists for $n = 2$.\n\nFor $n \\geq 3$, we proceed as follows.\n\n**Lemma 3.** If there exists a multiplicative function $f(x)$ from $\\mathbb{N}$ to $\\mathbb{Z}$ such that $f(x) \\neq 0$ for all $x$ and\n$$\nf(1) + 2f(2) + \\cdots + n f(n) = 0,\n$$\nthen there exists a sequence $\\{a_i\\}$ satisfying the desired conditions.\n\n*Proof.* Assume $f(x)$ satisfies the hypothesis. Let $a_k = f(k)$ for all $k$. Since $f(x)$ is multiplicative, for all $k$,\n$$\n\\begin{aligned}\na_k + 2a_{2k} + \\dots + n a_{nk} &= f(k) + 2f(2k) + \\dots + n f(nk) \\\\\n&= k f(1) + 2k f(2) + \\dots + n k f(n) \\\\\n&= k (f(1) + 2f(2) + \\dots + n f(n)) \\\\\n&= 0.\n\\end{aligned}\n$$\n$\\square$\n\n**Lemma 4.** If there exist primes $p$ and $q$ with $\\sqrt{n} < q < p \\leq n$ and $p > \\frac{n}{2}$, then there exists a sequence $\\{a_i\\}$ satisfying the desired conditions.\n\n*Proof.* For a prime $r$, let $v_r(k)$ be the largest $v$ such that $r^v \\mid k$. Define $f(x) = a^{v_p(x)} b^{v_q(x)}$ for all $x$, with $a, b \\neq 0$. $f(x)$ is multiplicative. For $\\ell$ with $1 \\leq \\ell \\leq n$,\n$$\nf(\\ell) = \\begin{cases} a & \\text{if } \\ell = p \\\\ b & \\text{if } \\ell = q, 2q, \\dots, m q \\\\ 1 & \\text{otherwise} \\end{cases}\n$$\nwhere $m < q$ since $\\sqrt{n} < q$.\n\nThus,\n$$\nf(1) + 2f(2) + \\dots + n f(n) = a p + b q + 2b q + \\dots + m b q + d = a p + b \\frac{q m(m+1)}{2} + d.\n$$\nSince $\\gcd(p, \\frac{q m(m+1)}{2}) = 1$, by Bézout's identity, we can choose $a, b$ so that $a p + b \\frac{q m(m+1)}{2} = -d$, making the sum zero. We can also ensure $a, b \\neq 0$. Thus, $f(x)$ satisfies Lemma 3, so a valid sequence exists. $\\square$\n\nFor $n \\geq 3$, either a function $f(x)$ as in Lemma 3 or a pair of primes $p, q$ as in Lemma 4 can be found:\n\n- For $n = 3$, let $f(x) = (-1)^{v_3(x)}$. Then $f(1) + 2f(2) + 3f(3) = 0$.\n- For $n = 4$, let $f(x) = (-1)^{v_2(x)}(-1)^{v_3(x)}$.\n- For $n = 5, 6, 7, 8$, let $p = 5$ and $q = 3$.\n\nThus, such a sequence exists for all $n \\geq 3$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 14152,
    "subject": "Mathematics (Olympiad)",
    "question": "令 $Z$ 表示所有整數所成的集合且 $n \\ge 1$ 為一奇數。試求所有的函數 $f: Z \\to Z$ 使得對任意的整數 $x, y$，$(f(x) - f(y))$ 能整除 $(x^n - y^n)$。",
    "options": [],
    "answer": "See solution",
    "solution": "答：$f(x) = \\epsilon x^d + c$，其中 $\\epsilon \\in \\{-1, 1\\}$，正整數 $d$ 為 $n$ 的因數且 $c$ 為一整數。\n\n滿足題設之所有函數為上述之 $f$。\n\n令函數 $f$ 為滿足題設之解。對任意整數 $n$，定義函數 $g(x) = f(x) + n$ 也滿足題設。我們可假設 $f(0) = 0$。\n\n對任意質數 $p$，取 $(x, y) = (p, 0)$ 則 $f(p) \\mid p^n$。因為質數有無窮多個，所以存在整數 $d$ 與 $\\epsilon$ 且 $0 \\le d \\le n$，$\\epsilon \\in \\{-1, 1\\}$ 使得有無窮多個質數 $p$ 滿足 $f(p) = \\epsilon p^d$。記集合 $P$ 如下：\n\n$$\nP = \\{p \\text{ 為質數} \\mid f(p) = \\epsilon p^d\\}\n$$\n\n因函數 $g$ 滿足題設之充要條件為 $(-g)$ 亦滿足題設，不妨假設 $\\epsilon = 1$。\n\n排除 $d=0$ 之情形。假設 $d \\ge 1$ 且 $n = md + r$，其中 $m$ 與 $r$ 是整數使得 $m \\ge 1$ 且 $0 \\le r \\le d-1$。令 $x$ 為一任意整數。對集合 $P$ 中的質數 $p$，$(f(p) - f(x)) \\mid (p^n - x^n)$。利用等式 $f(p) = p^d$，可得\n\n$$\np^n - x^n = p^r (p^d)^m - x^n \\equiv p^r (f(x))^m - x^n \\equiv 0 \\pmod{p^d - f(x)}\n$$\n\n因 $r < d$，對 $P$ 中足夠大的質數 $p$，我們有\n\n$$\n|p^r (f(x))^m - x^n| < p^d - f(x)\n$$\n\n因此 $p^r (f(x))^m - x^n = 0$。由此可推得：$r=0$ 且 $x^n = (x^d)^m = (f(x))^m$。\n\n因 $m$ 是奇數，得 $f(x) = x^d$。",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 14153,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $x_1, x_2, \\dots, x_n$ and $y_1, y_2, \\dots, y_n$ be positive real numbers such that\n$$\nx_1 + x_2 + \\dots + x_n \\geq x_1 y_1 + x_2 y_2 + \\dots + x_n y_n.\n$$\nShow that for any non-negative integer $p$, the following inequality holds:\n$$\n\\frac{x_1}{y_1^p} + \\frac{x_2}{y_2^p} + \\dots + \\frac{x_n}{y_n^p} \\geq x_1 + x_2 + \\dots + x_n.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "Assume by contradiction that\n$$\n\\frac{x_1}{y_1^p} + \\frac{x_2}{y_2^p} + \\dots + \\frac{x_n}{y_n^p} < x_1 + x_2 + \\dots + x_n.\n$$\nOn the other hand, by multiplying the given relation by $p \\in \\mathbb{N}$, we have\n$$\np(x_1 + x_2 + \\dots + x_n) \\geq p(x_1 y_1 + x_2 y_2 + \\dots + x_n y_n).\n$$\nAdding the above inequalities, we get\n$$\n(p+1) \\sum_{i=1}^{n} x_i > \\sum_{i=1}^{n} x_i \\left( p y_i + \\frac{1}{y_i^p} \\right).\n$$\nBut by the AM-GM inequality, we have\n$$\np y_i + \\frac{1}{y_i^p} \\geq p + 1.\n$$\nTherefore,\n$$\n(p+1) \\sum_{i=1}^{n} x_i > (p+1) \\sum_{i=1}^{n} x_i,\n$$\nwhich is a contradiction.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 14154,
    "subject": "Mathematics (Olympiad)",
    "question": "Determine the complex numbers $z$ and $w$ such that\n$$\n|z^{2n} + z^n w^n + w^{2n}| = 2^{2n} + 2^n + 1,\n$$\nfor any positive integer $n$.",
    "options": [],
    "answer": "See solution",
    "solution": "Apply the modulus to both sides of the identity $(z^2 + zw + w^2)(z^2 - zw + w^2) = z^4 + z^2w^2 + w^4$. Using the problem's hypothesis, we get $|z^2 - zw + w^2| = 3$.\n\nNext, apply the modulus to $(z^4 + z^2w^2 + w^4)(z^4 - z^2w^2 + w^4) = z^8 + z^4w^4 + w^8$, and again use the hypothesis to obtain $|z^4 - z^2w^2 + w^4| = 13$.\n\nConsider the identity:\n$$\n\\frac{1}{2}(z^2 + zw + w^2)^2 + \\frac{1}{2}(z^2 - zw + w^2)^2 + (z^4 - z^2w^2 + w^4) = 2(z^4 + z^2w^2 + w^4)\n$$\nUsing the modulus and the previous results, we have:\n$$\n\\frac{1}{2} \\cdot 49 + \\frac{1}{2} \\cdot 9 + 13 \\ge 2 \\cdot 21\n$$\nEquality holds, so there is a real $t$ such that $(z^2 + zw + w^2)^2 = t^2(z^2 - zw + w^2)^2$. Taking moduli, $t = \\pm\\frac{7}{3}$, so $3(z^2 + zw + w^2) = \\pm 7(z^2 - zw + w^2)$.\n\nFor $t = \\frac{7}{3}$: $2z^2 - 5zw + 2w^2 = 0 \\implies (2z - w)(z - 2w) = 0$. Thus, $w = 2z$ or $z = 2w$. Substituting into $|z^2 + zw + w^2| = 7$, in the first case $|z| = 1$, in the second $|w| = 1$.\n\nFor $t = -\\frac{7}{3}$: $5z^2 - 2zw + 5w^2 = 0 \\implies (\\varphi z - w)(z - \\varphi w) = 0$, where $\\varphi = \\frac{1+2\\sqrt{6i}}{5}$, $5\\varphi^2 - 2\\varphi + 5 = 0$, $|\\varphi| = 1$. Thus $w = \\varphi z$ or $z = \\varphi w$. Substituting into $|z^2 + zw + w^2| = 7$, in the first case $|z| = \\sqrt{5}$, in the second $|w| = \\sqrt{5}$. These do not satisfy $|z^4 - z^2w^2 + w^4| = 13$.\n\nTherefore, all pairs of the form $(z, 2z)$ and $(2z, z)$, where $z$ is a complex number with $|z| = 1$, satisfy the property in the statement.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 14155,
    "subject": "Mathematics (Olympiad)",
    "question": "Every two of $n$ towns in a country are connected by either a one-way or a two-way road. It is known that for every $k$ towns, there exists a round trip passing through each of these $k$ towns exactly once. Find the maximal possible number of one-way roads.",
    "options": [],
    "answer": "See solution",
    "solution": "Suppose there exists a town $A$ with $k-1$ one-way roads all pointing in the same direction from $A$. Town $A$ together with all $k-1$ endpoints of these one-way roads form a group of $k$ towns that violates the problem's condition. Thus, for any town, there can be at most $k-2$ one-way roads into the town and at most $k-2$ out of the town. Therefore, the number of one-way roads is at most $n(k-2)$.\n\nNow, when $n \\le 2k-3$, it is possible to have all roads be one-way roads. When $n > 2k-3$, the maximal number of one-way roads equals $n(k-2)$.\n\nLet $n = 2k-3$ and consider the towns as vertices of a regular $n$-gon. For every town $A$, let $k-2$ one-way roads from $A$ point to the next $k-2$ towns clockwise. Since $n = 2k-3$, such an allocation is possible and every two towns are connected by a one-way road. Consider any group of $k$ towns ordered clockwise; between any two neighbors $A$ and $B$, there exists a one-way road from $A$ to $B$, so the desired round trip exists.\n\nWhen $n < 2k-3$, choose any $n$ towns from the above construction.\n\nWhen $n > 2k-3$, proceed similarly: connect every town $A$ with one-way roads to the next $k-2$ towns, and let all remaining roads be two-way roads.\n\nFor any $k$ towns, order them clockwise; between any two neighbors $A$ and $B$, there exists a one-way road from $A$ to $B$, so the desired round trip exists.\n\n**Answer:** When $n \\le 2k-3$, all roads can be one-way; when $n > 2k-3$, there are at most $n(k-2)$ one-way roads.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 14156,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all values of the parameter $a$ such that the parabolas $y = x^2 + 2013x + a$ and $y = -x^2 + ax + 2013$ are tangent to each other.",
    "options": [],
    "answer": "See solution",
    "solution": "**Answer:** $a = 2013$ and $a = 2021$.\n\n**Solution.** Two parabolas with different leading coefficients are tangent if and only if they have exactly one common point, so the equation $x^2 + 2013x + a = -x^2 + ax + 2013$, which is equivalent to $2x^2 + (2013 - a)x + (a - 2013) = 0$, must have exactly one root. Therefore,\n\n$$\nD = (2013 - a)^2 - 8(a - 2013) = (a - 2013)(a - 2021) = 0.\n$$\n\nHence, $a = 2013$ or $a = 2021$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 14157,
    "subject": "Mathematics (Olympiad)",
    "question": "設 $n$ 為正整數。對於滿足 $\\sum_{i=1}^{2n} a_i = \\sum_{j=1}^{2n} b_j = n$ 的 $4n$ 個非負實數 $a_1, \\dots, a_{2n}$ 及 $b_1, \\dots, b_{2n}$，定義兩集合\n\n$$\nA := \\left\\{ \\sum_{j=1}^{2n} \\frac{a_i b_j}{a_i b_j + 1} : i \\in \\{1, \\dots, 2n\\} \\text{ 滿足 } \\sum_{j=1}^{2n} \\frac{a_i b_j}{a_i b_j + 1} \\neq 0 \\right\\},\n$$\n\n$$\nB := \\left\\{ \\sum_{i=1}^{2n} \\frac{a_i b_j}{a_i b_j + 1} : j \\in \\{1, \\dots, 2n\\} \\text{ 滿足 } \\sum_{i=1}^{2n} \\frac{a_i b_j}{a_i b_j + 1} \\neq 0 \\right\\}.\n$$\n\n令 $m$ 為 $A \\cup B$ 的最小值。試求：在所有可能的數組 $a_1, \\dots, a_{2n}, b_1, \\dots, b_{2n}$ 得到的 $m$ 中的最大值。",
    "options": [],
    "answer": "See solution",
    "solution": "最大值為 $n/2$。當且僅當恰有一半的 $a_i$ 和一半的 $b_j$ 為 1，其餘為 0 時可達到。\n\n證明如下：不妨假設 $a_1, \\dots, a_s$ 和 $b_1, \\dots, b_t$ 非零，其餘為零。則 $a_1+\\cdots+a_s = b_1+\\cdots+b_t = n$ 且\n\n$$\n\\min(A \\cup B) \\leq \\frac{1}{\\max(s,t)} \\sum_{i=1}^{s} \\sum_{j=1}^{t} \\frac{a_i b_j}{a_i b_j + 1}. \\quad (*)\n$$\n\n設 $k = st$ 且 $x_{(i-1)t+j} = a_i b_j$，$i = 1, \\dots, s$，$j = 1, \\dots, t$。則 $x_1, \\dots, x_k > 0$ 且 $x_1 + \\cdots + x_k = (a_1 + \\cdots + a_s)(b_1 + \\cdots + b_t) = n^2$。又 $\\max(s,t) \\geq \\sqrt{k}$。\n因此\n\n$$\n\\min(A \\cup B) \\leq \\frac{1}{\\sqrt{k}} \\sum_{i=1}^{k} \\frac{x_i}{x_i + 1}. \\quad (**)\n$$\n\n注意 $f(x) = \\frac{x}{x+1}$ 在 $x > -1$ 時為凹函數，因此\n\n$$\n\\sum_{i=1}^{k} \\frac{x_i}{x_i + 1} \\leq k \\cdot \\frac{\\frac{n^2}{k}}{\\frac{n^2}{k} + 1} = \\frac{k n^2}{n^2 + k}.\n$$\n\n因此，\n\n$$\n\\min(A \\cup B) \\leq \\frac{\\sqrt{k} n^2}{n^2 + k} \\leq \\frac{n}{2} \\quad (***)\n$$\n\n最後一個不等式由 AM-GM 不等式得出。",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 14158,
    "subject": "Mathematics (Olympiad)",
    "question": "On the board are initially written three consecutive positive integers: $n-1$, $n$, and $n+1$. A move consists of choosing two numbers on the board, $a$ and $b$, and replacing them with $2a-b$ and $2b-a$. For which values of $n$ is it possible, after a succession of such moves, to obtain two numbers on the board that are equal to $0$?",
    "options": [],
    "answer": "See solution",
    "solution": "We prove that two zeros can be obtained on the board if and only if $n$ is a power of $3$.\n\nThe sum of the numbers on the board remains constant, so the final configuration must be $(0, 0, 3n)$. If $p \\ne 3$ is a prime divisor of $n$, then in the final configuration all numbers are divisible by $p$. If $p \\mid 2a - b$ and $p \\mid 2b - a$, then $p \\mid (2(2a - b) + (2b - a)) = 3a$, so since $(p, 3) = 1$, it follows $p \\mid a$ and $p \\mid b$. Thus, if all numbers are divisible by a prime $p \\ne 3$ at the end, they were always divisible by $p$. Since $\\gcd(n-1, n, n+1) = 1$, it follows that two zeros cannot be obtained if $n$ has prime divisors other than $3$.\n\nNow consider $n = 3^k$, with $k \\in \\mathbb{N} \\cup \\{0\\}$. For $k = 0$, starting from $(0, 1, 2)$, it is easy to reach $(0, 0, 3)$. For $k \\ge 1$, choose $a = n-1$, $b = n+1$ in the first move to obtain $n-3$, $n$, $n+3$, all multiples of $3$. These can be written as $3(m-1)$, $3m$, $3(m+1)$, and subsequent moves act as if the numbers were $m-1$, $m$, $m+1$. After $j \\le k$ moves, the board shows $3^k - 3^j$, $3^k$, $3^k + 3^j$. After move $k$, we get $0$, $3^k$, $2 \\cdot 3^k$. Choosing $a = 3^k$ and $b = 2 \\cdot 3^k$ yields $0$, $0$, $3^{k+1}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 14159,
    "subject": "Mathematics (Olympiad)",
    "question": "Consider two four-digit numbers, where the second is obtained from the first by reversing its digits. Find the maximum number of digits 5 in the number which is the absolute value of the difference of those two numbers.",
    "options": [],
    "answer": "See solution",
    "solution": "Let the original number be $abcd$ and its reverse be $dcba$. The difference is:\n\n$$\nabcd - dcba = 1000a + 100b + 10c + d - (1000d + 100c + 10b + a) = 999(a - d) + 90(b - c)\n$$\n\nThis difference is divisible by 9. If the absolute value of the difference had four 5's (i.e., 5555 or 555), it would not be divisible by 9, leading to a contradiction. Therefore, the maximum number of digits 5 in the absolute value of the difference is 3.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 14160,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all polynomials $P(x)$ with integer coefficients such that there exists a positive integer $N$ so that for all positive integers $n > N$, we have $P(n) > 0$ and\n\n$$\nn + P(n) \\mid n^{P(n)} + P(n)^n.$$",
    "options": [],
    "answer": "See solution",
    "solution": "The solutions are $P(n) = 1$ and $P(n) = 2^s n^k - n$ for $0 \\le s \\le 3$ and $k \\ge 1$, excluding $P(n) = 0$.\n\nLet $Q(n) = P(n) + n$. Then $Q(n) > n$ for all $n > N$. The divisibility condition can be rewritten as\n\n$$\nn^{Q(n)-n} + (Q(n) - n)^n \\equiv 0 \\pmod{Q(n)}$$\n\nor\n\n$$\nn^{Q(n)-n} + (-n)^n \\equiv 0 \\pmod{Q(n)}.$$ \n\nBy analyzing the structure of $Q(n)$ and using properties of primitive polynomials and divisibility, we deduce that $Q(n)$ must be of the form $2^l n^k$ for some $l \\ge 0$, $k \\ge 1$, or $Q(n) = 1$. This leads to $P(n) = Q(n) - n = 2^s n^k - n$ for $0 \\le s \\le 3$, $k \\ge 1$, or $P(n) = 1$.\n\nThe proof involves several steps:\n- Rewriting the divisibility in terms of $Q(n)$.\n- Using properties of orders modulo primes and the Chinese Remainder Theorem.\n- Applying a lemma about primitive polynomials: if $Q(n) \\mid P(n)$ for infinitely many $n$, then $Q(x)$ divides $P(x)$ as polynomials.\n- Analyzing the possible forms of $Q(n)$ and $P(n)$ using irreducibility and positivity constraints.\n\nThus, the only polynomials $P(x)$ with integer coefficients satisfying the given condition are $P(x) = 1$ and $P(x) = 2^s x^k - x$ for $0 \\le s \\le 3$, $k \\ge 1$ (excluding $P(x) = 0$).",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 14161,
    "subject": "Mathematics (Olympiad)",
    "question": "Determine the least real number $c$ such that for any integer $n \\ge 1$ and any positive real numbers $a_1, a_2, \\dots, a_n$, the following holds:\n\n$$\n\\sum_{k=1}^{n} \\frac{k}{\\frac{1}{a_1} + \\frac{1}{a_2} + \\dots + \\frac{1}{a_k}} < c \\sum_{k=1}^{n} a_k.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "We claim $c_{\\min} = 2$.\n\nTake $a_j = \\frac{1}{j}$ for $j = 1, 2, \\dots, n$. Then\n\n$$\n\\sum_{k=1}^{n} \\frac{k}{\\frac{1}{a_1} + \\frac{1}{a_2} + \\dots + \\frac{1}{a_k}} = \\sum_{k=1}^{n} \\frac{k}{1 + 2 + \\dots + k} = 2 \\sum_{k=1}^{n} \\frac{1}{k+1},\n$$\nwhile\n$$\nc \\sum_{k=1}^{n} a_k = c \\sum_{k=1}^{n} \\frac{1}{k}.\n$$\n\nThus, the inequality holds when $(c-2) \\sum_{k=1}^{n} \\frac{1}{k} > -2 + \\frac{2}{n+1} > -2$, so $c \\ge 2$ (since $\\sum_{k=1}^{n} \\frac{1}{k}$ can be made arbitrarily large as $n$ increases).\n\nNow, we show $c = 2$ works. By the Cauchy-Schwarz inequality,\n\n$$\n\\left(\\sum_{j=1}^{k} j\\right)^2 \\le \\left(\\sum_{j=1}^{k} j^2 a_j\\right) \\left(\\sum_{j=1}^{k} \\frac{1}{a_j}\\right),\n$$\nso\n$$\n\\frac{k}{\\frac{1}{a_1} + \\frac{1}{a_2} + \\dots + \\frac{1}{a_k}} \\le \\frac{4}{k(k+1)^2} \\sum_{j=1}^{k} j^2 a_j.\n$$\n\nTherefore,\n$$\n\\begin{align*}\n\\sum_{k=1}^{n} \\frac{k}{\\frac{1}{a_1} + \\frac{1}{a_2} + \\dots + \\frac{1}{a_k}} &\\le \\sum_{k=1}^{n} \\left( \\frac{4}{k(k+1)^2} \\sum_{j=1}^{k} j^2 a_j \\right) \\\\\n&= \\sum_{j=1}^{n} \\left( j^2 a_j \\sum_{k=j}^{n} \\frac{4}{k(k+1)^2} \\right) \\\\\n&= 2 \\sum_{j=1}^{n} \\left( j^2 a_j \\sum_{k=j}^{n} \\frac{2k}{k^2(k+1)^2} \\right) \\\\\n&< 2 \\sum_{j=1}^{n} \\left( j^2 a_j \\sum_{k=j}^{n} \\frac{2k+1}{k^2(k+1)^2} \\right).\n\\end{align*}\n$$\nBut\n$$\n\\sum_{k=j}^{n} \\frac{2k+1}{k^2(k+1)^2} = \\sum_{k=j}^{n} \\left( \\frac{1}{k^2} - \\frac{1}{(k+1)^2} \\right) = \\frac{1}{j^2} - \\frac{1}{(n+1)^2} < \\frac{1}{j^2},\n$$\nso\n$$\n\\sum_{k=1}^{n} \\frac{k}{\\frac{1}{a_1} + \\frac{1}{a_2} + \\dots + \\frac{1}{a_k}} < 2 \\sum_{k=1}^{n} a_k.\n$$\n\n*Remark.* This inequality is related to Hardy's inequality (for $p = -1$):\n\n$$\n\\sum_{k=1}^{n} \\left( \\frac{1}{k} \\sum_{j=1}^{k} a_j^{1/p} \\right)^p < \\left( \\frac{p}{p-1} \\right)^p \\sum_{k=1}^{n} a_k,\n$$\nwhere $\\left(\\frac{p}{p-1}\\right)^p$ is the best constant.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 14162,
    "subject": "Mathematics (Olympiad)",
    "question": "Prove that for every integer $k \\ge 1$, there exists a positive integer $n$ such that, in the decimal representation of $2^n$, there is a block of exactly $k$ consecutive zeros. That is,\n\n$$\n2^n = \\dots a \\overline{00\\dots0} b \\dots, \\quad \\text{$k$ zeros}\n$$\n\nwhere $a$ and $b$ are nonzero digits.",
    "options": [],
    "answer": "See solution",
    "solution": "First, we show that there are arbitrarily long blocks of zeros in powers of 2. To get at least $k$ zeros in $2^n$, this power must be of the form $y \\cdot 10^{m+k} + z$ with $y, z$ positive integers and $z$ having at most $m$ digits, i.e., $z < 10^m$. Thus, it is sufficient to find $n, m$ with $2^n$ having residue less than $10^m$ modulo $10^{m+k}$.\n\nBy Euler's theorem, for every positive integer $t$ (since $(2, 5^t) = 1$),\n\n$$\n2^{\\varphi(5^t)} \\equiv 1 \\pmod{5^t}.\n$$\n\nMultiplying by $2^t$, we obtain\n\n$$\n2^{t+\\varphi(5^t)} \\equiv 2^t \\pmod{10^t}, \\quad \\text{thus} \\quad 2^{t+\\varphi(5^t)} = y \\cdot 10^t + 2^t\n$$\n\nfor some positive integer $y$. Set $n = t + \\varphi(5^t)$ and $m = t - k$. We want $t$ such that $2^t < 10^{t-k}$. Such a $t$ exists, for example $t = 2k$ (since $2^{2k} = 4^k < 10^k$). Thus,\n\n$$\n2^{2k+\\varphi(5^{2k})} = y \\cdot 10^{2k} + 2^{2k}\n$$\n\ncontains a block of at least $k$ zeros.\n\nNow, for a given $k$, consider a power of 2 (say $2^n$) containing a block of exactly $r$ zeros with $r \\ge k$. We study what happens to this block when considering subsequent powers, i.e., when multiplying the number with the block by 2. For some nonzero digits $a, b$,\n\n$$\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$$\n\nThus,\n\n$$\n2^{n+1} = 2y \\cdot 10^{r+s} + 2z.\n$$\n\nThe number $2z$ has either the same number of digits as $z$, or one more. Hence, on the right side, the block of zeros either does not shrink or shrinks by one zero. On the left side, the block can only extend (when $y$ is divisible by 5). Globally, the length of the block either decreases by 1, does not change, or increases. Repeating this process, the length of the block in each step decreases at most by 1. Thus, the only way to avoid a block of length $k$ is to always have the length greater than $k$, which is impossible. Namely, $y$ has in its prime factorization the prime 5 with some exponent, say $\\alpha$. When we multiply $2^n$ by 2 $\\alpha$ times, under the next multiplication the block will not extend on the left. On the right, at least every fourth multiplication, the block shrinks (since $2^4 > 10$). Hence, after a sufficient number of steps, we obtain a power of 2 with a block of exactly $k$ zeros.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 14163,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all quadratic functions $f(x) = ax^2 + bx + c$ with $a \\neq 0$ such that $f(f(1)) = f(f(0)) = f(f(-1))$.",
    "options": [],
    "answer": "See solution",
    "solution": "First, note that $f(f(x))$ is a quartic. The quartic $g(x) = ax^4 + \\beta x^3 + \\gamma x^2 + \\delta x + \\epsilon$ has the property that $g(1) = g(0) = g(-1)$ if and only if $\\alpha + \\gamma = 0$ and $\\beta + \\delta = 0$.\n\nFor the given function:\n$$\nf(f(x)) = a(ax^2 + bx + c)^2 + b(ax^2 + bx + c) + c\n$$\nSo the property $f(f(1)) = f(f(0)) = f(f(-1))$ gives $a^3 + ab^2 + 2a^2c + ab = 0$ and $2a^2b + 2abc + b^2 = 0$.\n\nSince $a \\neq 0$, the first equation gives $b^2 + b + a^2 + 2ac = 0$. The latter equation gives $b(2a^2 + 2ac + b) = 0$. So $b = 0$ or $b = -2a^2 - 2ac = -2a(a + c)$. We look at these cases separately.\n\n**Case 1:** $b = 0$.\n\nThen $a^2 + 2ac = 0$ and hence $c = -\\frac{a}{2}$ (since $a \\neq 0$). Then $f(x) = a(x^2 - \\frac{1}{2})$. Let us verify that it works. First, $f(0) = -\\frac{a}{2}$ and $f(1) = -\\frac{a}{2}$. But $f$ is even, so $f(-\\frac{1}{2}) = f(\\frac{1}{2})$ and hence $f(f(1)) = f(f(0)) = f(f(-1))$ as required.\n\n**Case 2:** $b = -2a^2 - 2ac = -2a(a + c)$.\n\nThen the first equation gives $4a^2(a + c)^2 - 2a^2 - 2ac + a^2 + 2ac = 0$. Since $a \\neq 0$, this gives $(a + c)^2 = \\frac{1}{4}$. Thus $c = -a \\pm \\frac{1}{2}$ and hence $b = \\mp a$. So we get two solutions:\n\n$$\nf(x) = a(x^2 - x - 1) + \\frac{1}{2}, \\quad f(x) = a(x^2 + x - 1) - \\frac{1}{2}\n$$\nfor any $a \\neq 0$.\n\nWe check that both of these work. For the first function, $f(\\frac{1}{2} - x) = f(\\frac{1}{2} + x)$. Also, $f(0) = \\frac{1}{2}$ and $f(-1) = \\frac{1}{2}$. So $f(f(1)) = f(f(0)) = f(f(-1))$ as required.\n\nFor the second function, $f(-\\frac{1}{2} - x) = f(-\\frac{1}{2} + x)$. Also, $f(0) = -\\frac{1}{2}$ and $f(-1) = -\\frac{1}{2}$. So $f(f(1)) = f(f(0)) = f(f(-1))$ as required.\n\n**In summary, the solutions are:**\n$$\na(x^2 - \\frac{1}{2}), \\quad a(x^2 - x - 1) + \\frac{1}{2}, \\quad a(x^2 + x - 1) - \\frac{1}{2},\n$$\nfor any $a \\neq 0$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 14164,
    "subject": "Mathematics (Olympiad)",
    "question": "Una rolls 6 standard 6-sided dice simultaneously and calculates the product of the 6 numbers obtained. What is the probability that the product is divisible by 4?\n\n(A) $\\frac{3}{4}$  \n(B) $\\frac{57}{64}$  \n(C) $\\frac{59}{64}$  \n(D) $\\frac{187}{192}$  \n(E) $\\frac{63}{64}$",
    "options": [],
    "answer": "See solution",
    "solution": "The product will not be divisible by 4 precisely when all 6 rolls are odd, or exactly one of them is equal to either 2 or 6 and the rest are odd. The probability of this complementary event is\n\n$$\n\\left(\\frac{1}{2}\\right)^6 + 6 \\cdot \\left(\\frac{1}{3}\\right) \\cdot \\left(\\frac{1}{2}\\right)^5 = \\frac{5}{64}.\n$$\n\nThe requested probability is therefore $1 - \\frac{5}{64} = \\frac{59}{64}$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 14165,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $m$ and $n$ be positive integers. For each integer $k$ with $1 \\leq k \\leq n$, consider the number $m + k$. Prove that for each $k$, there exists a prime $p_k$ dividing $m + k$ such that all the $p_k$ are distinct for different $k$.",
    "options": [],
    "answer": "See solution",
    "solution": "We call the number $m + k$ convenient if it has at least $n$ distinct prime divisors; otherwise, we call it inconvenient. We may consider only inconvenient numbers. Take one such number, $m + k = q_1^{a_1} q_2^{a_2} \\dots q_l^{a_l}$, where all $q_i$ are primes and $l \\leq n - 1$. Since\n$$\nq_1^{a_1} q_2^{a_2} \\dots q_l^{a_l} = m + k > n^{n-1} \\geq n^l,\n$$\nthere exists an $i$ with $q_i^{a_i} > n$, so we can choose $p_k = q_i$. Similarly, we can choose prime divisors for other inconvenient numbers. It remains to note that for different inconvenient numbers, the chosen prime divisors are different.\n\nSuppose the same $p$ is chosen for $m + k_1$ and $m + k_2$; let $m + k_1 : p^{a_1}$, $m + k_2 : p^{a_2}$, and without loss of generality $a_1 \\geq a_2$. Then\n$$\nn > |k_1 - k_2| = |(m + k_1) - (m + k_2)| : p^{a_2} > n,\n$$\na contradiction.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 14166,
    "subject": "Mathematics (Olympiad)",
    "question": "Consider the quadratic polynomial $p(x) = x^2 + a x + b$, where $a, b$ are in the interval $[-2, 2]$. Find the range of the real roots of $p(x) = 0$ as $a$ and $b$ vary over $[-2, 2]$.",
    "options": [],
    "answer": "See solution",
    "solution": "Suppose $a, b \\in [-2, 2]$ and $\\alpha$ is a real root of $x^2 + a x + b = 0$. Then\n\n$$\n\\alpha = \\frac{-a \\pm \\sqrt{a^2 - 4b}}{2},\n$$\n\nwhich shows that\n\n$$\n\\alpha \\leq \\frac{2 + \\sqrt{4 + 8}}{2} = 1 + \\sqrt{3}.\n$$\n\n(Take $a = -2$, $b = -2$.) Similarly, $\\alpha \\ge -1 - \\sqrt{3}$ by taking $a = 2$ and $b = -2$.\n\nTake any real number $\\lambda$ such that $|\\lambda| \\le 1$. We have\n\n$$\n(\\lambda \\alpha)^2 + (\\lambda a)(\\lambda \\alpha) + \\lambda^2 b = \\lambda^2 (\\alpha^2 + a \\alpha + b) = 0.\n$$\n\nThus $\\lambda \\alpha$ is also a root of the equation $x^2 + c x + d$, where\n\n$c = \\lambda a$, $d = \\lambda^2 b$.\n\nNote that $|c| \\le |a|$ and $|d| \\le |b|$. Thus, for each real root $\\alpha$ of $x^2 + a x + b = 0$ and each real $\\lambda$ with $|\\lambda| \\le 1$, $\\lambda \\alpha$ is also a root of $x^2 + c x + d$, where $c, d \\in [-2, 2]$.\n\nSince $1 + \\sqrt{3}$ is a root of $x^2 - 2x - 2 = 0$, it follows that $\\lambda (1 + \\sqrt{3})$ is in the range for any $\\lambda$ with $|\\lambda| \\le 1$. Thus, the range is $[-1 - \\sqrt{3}, 1 + \\sqrt{3}]$.",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 14167,
    "subject": "Mathematics (Olympiad)",
    "question": "In triangle $ABC$, points $P$, $Q$, and $R$ lie on sides $BC$, $CA$, and $AB$, respectively. Let $\\omega_A$, $\\omega_B$, and $\\omega_C$ denote the circumcircles of triangles $AQR$, $BRP$, and $CPQ$, respectively. Given that segment $AP$ intersects $\\omega_A$, $\\omega_B$, and $\\omega_C$ again at $X$, $Y$, and $Z$, respectively, prove that\n$$\n\\frac{YX}{XZ} = \\frac{BP}{PC}.\n$$\n\n![](images/USA_IMO_2013-2014_p4_data_4659404503.png)",
    "options": [],
    "answer": "See solution",
    "solution": "Assume that $\\omega_B$ and $\\omega_C$ intersect again at a second point $S$ other than $P$. If not, the degenerate case where $\\omega_B$ and $\\omega_C$ are tangent at $P$ can be dealt with similarly. Because $BPSR$ and $CPSQ$ are cyclic, we have $\\angle RSP = 180^\\circ - \\angle PBR$ and $\\angle PSQ = 180^\\circ - \\angle QCP$. Hence,\n\n$$\n\\angle QSR = 360^\\circ - \\angle RSP - \\angle PSQ = \\angle PBR + \\angle QCP = \\angle CBA + \\angle ACB = 180^\\circ - \\angle BAC,\n$$\n\nfrom which it follows that $ARSQ$ is cyclic. This means that $\\omega_A$, $\\omega_B$, and $\\omega_C$ meet at $S$ (Miquel's theorem).\n\nBecause $BPSY$ is inscribed in $\\omega_B$, $\\angle XYS = \\angle PYS = \\angle PBS$. Because $ARXS$ is inscribed in $\\omega_A$, $\\angle SXY = \\angle SXA = \\angle SRA$. Because $BPSR$ is inscribed in $\\omega_B$, $\\angle SRA = \\angle SPB$. Thus, $\\angle SXY = \\angle SRA = \\angle SPB$. In triangles $SYX$ and $SBP$, $\\angle XYS = \\angle PBS$ and $\\angle SXY = \\angle SPB$. Therefore, triangles $SYX$ and $SBP$ are similar, which implies\n$$\n\\frac{YX}{BP} = \\frac{SX}{SP}.\n$$\n\nSimilarly, triangles $SXZ$ and $SPC$ are similar, so\n$$\n\\frac{SX}{SP} = \\frac{XZ}{PC}.\n$$\n\nCombining these yields the desired result:\n$$\n\\frac{YX}{XZ} = \\frac{BP}{PC}.\n$$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 14168,
    "subject": "Mathematics (Olympiad)",
    "question": "Encontrar cuántas soluciones enteras tiene la ecuación\n\n$$\n|5 - x_1 - x_2| + |5 + x_1 - x_2| + |5 + x_2 + x_3| + |5 + x_2 - x_3| = 20.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "Podemos reescribir la ecuación en la forma\n\n$$\n|y_1| + |y_2 - y_1| + |y_3 - y_2| + |20 - y_3| = 20,\n$$\n\ndonde $y_1 = 5 - x_1 - x_2$, $y_2 = 10 - 2x_2$ y $y_3 = 15 - x_2 + x_3$. Por tanto, toda solución entera de la ecuación original da una solución entera de (1) con $y_2$ un número par. Recíprocamente, toda solución de (1) con $y_2$ par da una solución de la ecuación del enunciado.\n\nObservamos que (1) se puede escribir como\n\n$$\nd(0, y_1) + d(y_1, y_2) + d(y_2, y_3) + d(y_3, 20) = d(0, 20),\n$$\n\ndonde $d(x, y) = |x - y|$ es la distancia entre los números reales $x$ e $y$. Sólo puede darse esta situación si $0 \\leq y_1 \\leq y_2 \\leq y_3 \\leq 20$ (podemos imaginar una regla de carpintero con cuatro segmentos de longitudes que suman 20 unidades y que tienen que cubrir desde 0 a 20, que es una distancia de 20 unidades. La única posibilidad es que la regla esté completamente estirada).\n\nSe trata entonces de contar las ternas de números enteros $0 \\leq y_1 \\leq y_2 \\leq y_3 \\leq 20$ con $y_2$ par. Si escribimos $y_2 = 2k$ con $0 \\leq k \\leq 10$, hay $2k + 1$ posibilidades para $y_1$ y $21 - 2k$ para $y_3$, luego el número de soluciones buscado es\n\n$$\n\\sum_{k=0}^{10} (2k+1)(21-2k) = \\sum_{k=0}^{10} (21+40k-4k^2) = 21 \\times 11 + 40 \\times 55 - 4 \\times 385 = 891.\n$$\n\nPor tanto, el número de soluciones enteras de la ecuación dada es $891$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 14169,
    "subject": "Mathematics (Olympiad)",
    "question": "Given a polynomial $P(x)$ and a function $F(x)$, consider the decimal representation of $P(F(10^e))$ for some integer $e$. The decimal representation is constructed as follows:\n\n* The decimal representation of $c_N - 1$,\n* The decimal representation of $10^e - c_{N-1}$,\n* The decimal representation of $c_{N-2}$, with several leading zeros,\n* The decimal representation of $c_{N-3}$, with several leading zeros,\n* ...\n* The decimal representation of $c_0$, with several leading zeros.\n\nFor example, if $P(F(x)) = 15x^3 - 7x^2 + 4x + 19$, then $P(F(1000)) = 14,993,004,019$.\n\nFind the sum of the digits of $P(F(10^e))$ in terms of $e$ and constants depending only on $P$ and $F$.",
    "options": [],
    "answer": "See solution",
    "solution": "Thus, the sum of the digits of this expression is equal to\n\n$$\nS(P(F(10^e))) = 9e + k\n$$\n\nfor some constant $k$ depending only on $P$ and $F$, independent of $e$. But this will eventually hit a Fibonacci number by the second claim, contradiction.\n\n**Remark.** It is important to control the number of negative coefficients in the created polynomial. If one tries to use this approach on a polynomial $P$ with $m > 0$ negative coefficients, then one would require that the Fibonacci sequence is surjective modulo $9m$ for any $m > 1$, which is not true: for example, the Fibonacci sequence avoids all numbers congruent to $4 \\pmod{11}$ (and thus $4 \\pmod{99}$).",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 14170,
    "subject": "Mathematics (Olympiad)",
    "question": "A regular hexagonal prism $ABCDEF A'B'C'D'E'F'$ has edge $AB = 12$ and height $AA' = 12\\sqrt{3}$. Let $N$ be the midpoint of the edge $CC'$.  \n(a) Prove that the lines $BF'$ and $ND$ are perpendicular.  \n(b) Find the distance between the lines $BF'$ and $ND$.",
    "options": [],
    "answer": "See solution",
    "solution": "(a) Let $M$ be the midpoint of $BB'$. Since $MN \\parallel AD$, the points $A$, $D$, $M$, and $N$ are coplanar. Let $Q$ be the midpoint of $BF$. The intersection of the planes $(BFF')$ and $(ADN)$ is the line $MQ$. Notice that $BFF'B'$ is a square, so $BF' \\perp MQ$ and $AD \\perp (BFF')$, implying $BF' \\perp AD$. Therefore, $BF' \\perp (ADN)$ and thus $BF' \\perp ND$.\n\n(b) Let $S$ be the intersection point of $MQ$ and $BF'$. Let $P$ be the projection of $S$ onto $ND$. Since $BF' \\perp (ADN)$, we have $BF' \\perp SP$, so $SP$ is the common perpendicular of $BF'$ and $ND$.\n\nTo find the length of $SP$, evaluate the area of triangle $SND$ in two ways: $S[SND] = S[MNDQ] - S[SDQ] - S[SMN] = \\frac{1}{2}SP \\cdot ND$. Given $MN = 12$, $QD = 18$, $ND = 6\\sqrt{7}$, $MQ = 6\\sqrt{6}$, we get $SP = \\frac{15\\sqrt{42}}{7}$.",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 14171,
    "subject": "Mathematics (Olympiad)",
    "question": "A grasshopper is jumping along the set $\\mathbb{Z}$ of integers. He starts at the origin, and for each jump, he may decide whether to jump to the left or to the right. For each $n \\in \\mathbb{N}_0$, the $n$-th jump has length $n^2$.\n\nProve or disprove that for each $k \\in \\mathbb{Z}$ the grasshopper can arrive at $k$ starting from the origin.",
    "options": [],
    "answer": "See solution",
    "solution": "Clearly, the grasshopper can arrive at the integers $+1$ and $-1$ in one jump. He can also arrive at the integer $14$ using three jumps to the right ($1 + 4 + 9 = 14$).\n\nNote the following: if the grasshopper can arrive at a number $a \\in \\mathbb{Z}$ using $n$ jumps, then he can also arrive at the numbers $a-4$ and $a+4$ using $n+4$ jumps, by jumping left, right, right, left (for $a-4$), and right, left, left, right (for $a+4$), because:\n\n$$\na - (n + 1)^2 + (n + 2)^2 + (n + 3)^2 - (n + 4)^2 = a - 4,\n$$\n\n$$\na + (n + 1)^2 - (n + 2)^2 - (n + 3)^2 + (n + 4)^2 = a + 4.\n$$\n\nSince the numbers $0$, $+1$, $-1$, and $14$ all have distinct remainders modulo four, the grasshopper can arrive at each integer in a finite number of jumps.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 14172,
    "subject": "Mathematics (Olympiad)",
    "question": "Triangle $ABC$ is said to be perpendicular to triangle $DEF$ if the perpendiculars from $A$ to $EF$, from $B$ to $FD$, and from $C$ to $DE$ are concurrent. Prove that if $ABC$ is perpendicular to $DEF$, then $DEF$ is perpendicular to $ABC$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $U, V, W$ be the feet of the perpendiculars from $A, B, C$ to $EF, FD, DE$ respectively, and let $X, Y, Z$ be the feet of the perpendiculars from $D, E, F$ to $BC, CA, AB$ respectively. Since $U$ and $Z$ both subtend a right angle from $AF$, $AFUZ$ is concyclic and so (with an appropriate sign convention) $\\angle UAB = \\angle UAZ = \\angle UFZ = \\angle EFZ$. Combining this with five similar equalities of angles, we see that\n\n$$\n\\frac{\\sin \\angle UAB \\sin \\angle VBC \\sin \\angle WCA}{\\sin \\angle CAU \\sin \\angle ABV \\sin \\angle BCW} = \\frac{\\sin \\angle EFZ \\sin \\angle FDX \\sin \\angle DEY}{\\sin \\angle ZFD \\sin \\angle XDE \\sin \\angle YEF}\n$$\n\n![](images/BMO_Short_list_2014_p23_data_35d035986d.png)\n\nYet by the angle form of Ceva's theorem, the left hand expression is 1 if and only if the Cevians $AU, BV, CW$ concur, i.e. iff $ABC$ is perpendicular to $DEF$. Similarly, the right hand expression is 1 if and only if $DEF$ is perpendicular to $ABC$. Thus $ABC$ is perpendicular to $DEF$ if and only if $DEF$ is perpendicular to $ABC$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 14173,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $f: \\pi \\to \\mathbb{R}$ be a function from the Euclidean plane to the real numbers such that\n$$\nf(A) + f(B) + f(C) = f(O) + f(G) + f(H)\n$$\nfor any acute triangle $ABC$ with circumcenter $O$, centroid $G$, and orthocenter $H$. Prove that $f$ is constant.",
    "options": [],
    "answer": "See solution",
    "solution": "![](images/BMO2024Shortlist_p58_data_a225d29ae9.png)\n\nLet $G_1 \\neq G_2$ be arbitrary points and let $d$ be the perpendicular bisector of $G_1G_2$. We shall construct two congruent triangles, symmetric with respect to $d$ and with centroids at $G_1$ and $G_2$. Choose an acute triangle $A_1BC$ with centroid $G_1$, and rotate it around $G_1$ if necessary to ensure that $BC \\parallel G_1G_2$. By applying a homothety centered at $G_1$ if necessary, we may ensure that the midpoint $M$ of $BC$ lies on $d$. Let $H_1, O$ be the orthocenter and circumcenter of $\\triangle ABC$. By construction, we have $O \\in d$ and $A_1H_1 \\parallel OM \\perp BC$. If $A_2, H_2$ are the reflections of points $A_1, H_1$ across $d$, then one sees that $\\triangle A_1BC \\equiv \\triangle A_2CB$, and $H_2, G_2, O$ are the orthocenter, centroid and circumcenter of $\\triangle A_2BC$.\n\nTherefore, applying the property of $f$ to $\\triangle A_1BC$ and $\\triangle A_2BC$, we have\n$$\n\\begin{aligned}\nf(A_1) + f(B) + f(C) &= f(O) + f(G_1) + f(H_1) \\\\\nf(A_2) + f(B) + f(C) &= f(O) + f(G_2) + f(H_2)\n\\end{aligned}\n$$\nSubtracting, we obtain\n$$\nf(A_1) - f(A_2) = f(G_1) - f(G_2) + f(H_1) - f(H_2) \\quad (1)\n$$\n\nNow let's reflect the picture across line $l$, the perpendicular bisector of $A_1H_1$ ($l$ is a symmetry axis of rectangle $A_1A_2H_2H_1$), denoting by $X'$ the image of point $X$. It follows that $A'_1 = H_1$, $A'_2 = H_2$ and triangles $H_1B'C'$, $H_2B'C'$ are symmetric about $d$. Using this symmetry, relation (1) becomes\n$$\nf(H_1) - f(H_2) = f(G'_1) - f(G'_2) + f(A_1) - f(A_2) \\quad (2)\n$$\nFrom (2) and (1) we obtain\n$$\nf(G_1) - f(G_2) + f(G'_1) - f(G'_2) = 0 \\quad (3)\n$$\n\nThe argument up to now holds if we *scale* the picture. It follows that for any rectangle $XYZT$ which is similar to the rectangle $G_1G'_1G'_2G_2$, we have\n$$\nf(X) - f(T) + f(Y) - f(Z) = 0 \\quad (4)\n$$\n\n![](images/BMO2024Shortlist_p59_data_057f061049.png)\n\nLet us double up the rectangle $G_1G'_1G'_2G_2$ to a rectangle $G_1KLM$ (see the picture), with $D$ and $E$ the midpoints of $KL$ and $LM$. Applying (4) to rectangle $G_2G'_2EM$, we have $f(G_2) - f(M) + f(G'_2) - f(E) = 0$ which added to (3) yields\n$$\nf(G_1) - f(M) + f(G'_1) - f(E) = 0\n$$\nSimilarly, one obtains\n$$\nf(G'_1) - f(E) + f(K) - f(L) = 0\n$$\nSubtracting the last relations we get\n$$\nf(G_1) - f(M) - f(K) + f(L) = 0\n$$\nOn the other hand, applying (4) to the rectangle $G_1KLM$ we have\n$$\nf(G_1) - f(M) + f(K) - f(L) = 0\n$$\nThe last two relations imply $f(G_1) - f(M) = 0$, and scaling back we finally obtain $f(G_1) - f(G_2) = 0$. The choice of $G_1 \\neq G_2$ being arbitrary, it follows that the function $f$ must be constant. $\\square$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 14174,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $n$ be a positive integer, and let $x_1, x_2, \\dots, x_{2n}$ be nonnegative real numbers such that\n$$\nx_1 + x_2 + \\dots + x_{2n} = 4.\n$$\nProve that there exist non-negative integers $p$ and $q$ with $q \\le n-1$ such that\n$$\n\\sum_{i=1}^{q} x_{p+2i-1} \\le 1, \\quad \\sum_{i=q+1}^{n-1} x_{p+2i} \\le 1.\n$$\n\n*Note 1:* The subscripts are taken modulo $2n$, that is, $k \\equiv l \\pmod{2n}$ implies $x_k = x_l$.\n\n*Note 2:* If $q=0$, the first sum is $0$; if $q = n-1$, the second sum is $0$.",
    "options": [],
    "answer": "See solution",
    "solution": "Divide $x_1, x_2, \\dots, x_{2n}$ into two groups by the parity of the subscripts:\nlet $A = x_1 + x_3 + \\dots + x_{2n-1}$ and $B = x_2 + x_4 + \\dots + x_{2n}$.\nDefine the partial sums $A(0) = B(0) = 0$,\n$$\n\\begin{cases}\nA(2k + 1) = A(2k) + x_{2k+1}, \\\\\nB(2k + 1) = B(2k),\n\\end{cases}\n\\quad\n\\begin{cases}\nA(2k + 2) = A(2k + 1), \\\\\nB(2k + 2) = B(2k + 1) + x_{2k+2},\n\\end{cases}\n\\quad k = 0, 1, 2, \\dots\n$$\nThe subscripts are modulo $2n$; the partial sums increase periodically: $A(k + 2n) = A(k) + A$, $B(k + 2n) = B(k) + B$.\n\nWe turn the partial sums into (piecewise linear) continuous functions: for $t \\ge 0$,\n$$\n\\begin{aligned}\nA(t) &= A(\\lfloor t \\rfloor) + (t - \\lfloor t \\rfloor)[A(\\lfloor t \\rfloor + 1) - A(\\lfloor t \\rfloor)], \\\\\nB(t) &= B(\\lfloor t \\rfloor) + (t - \\lfloor t \\rfloor)[B(\\lfloor t \\rfloor + 1) - B(\\lfloor t \\rfloor)].\n\\end{aligned}\n$$\nThen $A(\\cdot)$ and $B(\\cdot)$ are non-decreasing, continuous, periodic functions on $\\mathbb{R}_{\\ge 0}$.\n\nSince $A + B = 4$, there exists a positive integer $L$ such that $\\lfloor \\frac{L}{A} \\rfloor + \\lfloor \\frac{L}{B} \\rfloor \\ge L$. For $l = 0, 1, \\dots, L$, by continuity of $A(\\cdot)$, we can choose $t_l \\ge 0$ such that $A(t_l) = l$ (let $t_0 = 0$). Since $L \\ge A \\lfloor \\frac{L}{A} \\rfloor = A (2n \\cdot \\lfloor \\frac{L}{A} \\rfloor)$, we may further require $t_L \\ge 2n \\cdot \\lfloor \\frac{L}{A} \\rfloor$.\n\nNow,\n$$\n\\begin{aligned}\nB(t_L) - B(t_0) &= B(t_L) \\ge B(2n \\cdot \\lfloor \\frac{L}{A} \\rfloor) \\ge B \\lfloor \\frac{L}{A} \\rfloor \\\\\n&\\ge B (L - \\lfloor \\frac{L}{B} \\rfloor) \\ge (B-1) L.\n\\end{aligned}\n$$\nSo, there exists some $l = 0, 1, \\dots, L-1$ such that $B(t_{l+1}) - B(t_l) \\ge (B-1)$, or\n$$\nB(t_l + 2n) - B(t_{l+1}) \\le 1.\n$$\nTake nonnegative integers $c$ and $d$ with $2c \\le t_l \\le 2c+2$, $2d-1 \\le t_{l+1} \\le 2d+1$. Then\n$$\n\\begin{align*}\nA(t_l) &\\le A(2c+1) = A(2c+2), \\\\\nA(t_{l+1}) &\\ge A(2d) = A(2d-1); \\\\\nB(t_l) &\\ge B(2c+1) = B(2c), \\\\\nB(t_{l+1}) &\\le B(2d) = B(2d+1).\n\\end{align*}\n$$\nThus,\n$$\n\\begin{align*}\nx_{2c+3} + x_{2c+5} + \\cdots + x_{2d-1} &= A(2d) - A(2c+1) \\le A(t_{l+1}) - A(t_l) = 1, \\\\\nx_{2d+2} + x_{2d+4} + \\cdots + x_{2c+2n} &= B(2c+1+2n) - B(2d) \\\\\n&\\le B(t_l + 2n) - B(t_{l+1}) \\le 1,\n\\end{align*}\n$$\nso $p = 2c + 2$, $q = d - c - 1$ satisfy the problem conditions (if $q < 0$, let $q = 0$; if $q \\ge n$, let $q = n - 1$).",
    "topic": "Number Theory",
    "subtopic": "Modular Arithmetic"
  },
  {
    "id": 14175,
    "subject": "Mathematics (Olympiad)",
    "question": "For each $a \\ge 0$, find the number of solutions of the following equation:\n\n$$\n\\sqrt{a x} - x = \\sqrt{a} - 1.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "For $a = 0$, the only solution is $x = 1$.\n\nSuppose $a > 0$. Then $x \\ge 0$. Note that $x = 1$ is a solution for each $a > 0$.\n\nSuppose $x \\ne 1$. We have\n\n$$\n\\sqrt{a x} - x = \\sqrt{a} - 1 \\Leftrightarrow \\sqrt{a x} - \\sqrt{a} = x - 1 \\Leftrightarrow \\sqrt{a}(\\sqrt{x} - 1) = x - 1 \\Leftrightarrow \\sqrt{a} = \\sqrt{x} + 1.\n$$",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 14176,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $ABCDE$ be a pentagon with $\\hat{A} = \\hat{B} = \\hat{C} = \\hat{D} = 120^\\circ$. Prove that\n\n$$\n4AC \\cdot BD \\ge 3AE \\cdot ED.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "![](images/Saudi_Arabia_booklet_2012_p29_data_c9d1f02c26.png)\n\nIt is clear that $AB \\parallel ED$ and $AE \\parallel CD$. Let $F$ be the intersection point of lines $AB$ and $CD$. The inequality is equivalent to\n\n$$\n\\frac{AC}{AF} \\cdot \\frac{BD}{FD} \\ge \\frac{3}{4}. \\quad (1)\n$$\n\nUsing the Law of Sines in triangles $ACF$ and $BDF$, we get that (1) is equivalent to\n\n$$\n\\frac{\\sqrt{3}}{2} \\cdot \\frac{\\sqrt{3}}{\\sin \\widehat{ACF}} \\cdot \\frac{\\sqrt{3}}{\\sin \\widehat{DBF}} \\ge \\frac{3}{4},\n$$\n\nso $\\sin \\widehat{ACF} \\cdot \\sin \\widehat{DBF} \\le 1$, which is clearly true.\n\nWe have equality if and only if $\\widehat{ACF} = \\widehat{DBF} = 90^\\circ$, hence $BC = AB = CD = BF = CF$. This means $AEDF$ is a rhombus, and $B$ and $C$ are the midpoints of $AF$ and $DF$ respectively.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 14177,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $p$ be a prime number, let $n_1, n_2, \\dots, n_p$ be positive integers, and let $d$ be the greatest common divisor of the numbers $n_1, n_2, \\dots, n_p$. Prove that the polynomial\n\n$$\n\\frac{X^{n_1} + X^{n_2} + \\dots + X^{n_p} - p}{X^d - 1}\n$$\n\nis irreducible in $\\mathbb{Q}[X]$.",
    "options": [],
    "answer": "See solution",
    "solution": "Clearly, $f = \\frac{X^{n_1} + X^{n_2} + \\dots + X^{n_p} - p}{X^d - 1}$ is a polynomial with integer coefficients. If all $n_i$ are equal, then $f = p$, a constant polynomial; therefore, assume at least two of the $n_i$ are distinct. By Gauss' Lemma, it is sufficient to prove $f$ is irreducible in $\\mathbb{Z}[X]$.\n\nWe claim that the roots of $f$ all lie outside the closed unit disc in the complex plane. Assuming the claim, suppose, if possible, that $f = gh$ is a non-trivial factorization of $f$ in $\\mathbb{Z}[X]$. Since $f(0) = p$, a prime, one of the numbers $|g(0)|$, $|h(0)|$ is $1$. To reach a contradiction, notice that $g(0)$ and $h(0)$ are both products of roots of $f$, all of which lie outside the closed unit disc, so $|g(0)|$ and $|h(0)|$ are both greater than $1$.\n\nBack to the claim, write\n\n$$\nX^{n_1} + X^{n_2} + \\dots + X^{n_p} - p = (X^d - 1)f. \\quad (*).\n$$\n\nSuppose, if possible, that $f$ has a root $z$ in the closed unit disc. Then\n\n$$\np = |z^{n_1} + z^{n_2} + \\dots + z^{n_p}| \\leq |z|^{n_1} + |z|^{n_2} + \\dots + |z|^{n_p} \\leq p,\n$$\n\nwhich forces $|z| = 1$ and $z^{n_k - n_\\ell} > 0$ for any $k$ and $\\ell$. Consequently, the $z^{n_k}$ are all equal to some complex number $w$, so\n\n$$\np(w - 1) = z^{n_1} + z^{n_2} + \\dots + z^{n_p} - p = 0;\n$$\n\nthat is, $w = 1$. Now write $d = n_1 t_1 + n_2 t_2 + \\dots + n_p t_p$, for some integers $t_1, t_2, \\dots, t_p$, to get\n\n$$\nz^d = z^{n_1 t_1 + n_2 t_2 + \\dots + n_p t_p} = (z^{n_1})^{t_1} (z^{n_2})^{t_2} \\dots (z^{n_p})^{t_p} = 1.\n$$\n\nFinally, to reach a contradiction, evaluate the formal derivatives of both sides of $(*)$ at $z$. The left-hand side is $(n_1 + n_2 + \\dots + n_p)/z$, while the right-hand side vanishes.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 14178,
    "subject": "Mathematics (Olympiad)",
    "question": "An integer is assigned to each vertex of a regular pentagon so that the sum of the five integers is $2011$. A turn of a solitaire game consists of subtracting an integer $m$ from each of the integers at two neighboring vertices and adding $2m$ to the opposite vertex, which is not adjacent to either of the first two vertices. (The amount $m$ and the vertices chosen can vary from turn to turn.)\n\nThe game is won at a certain vertex if, after some number of turns, that vertex has the number $2011$ and the other four vertices have the number $0$. Prove that for any choice of the initial integers, there is exactly one vertex at which the game can be won.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $a_1, a_2, a_3, a_4$, and $a_5$ represent the integers at vertices $v_1$ to $v_5$ (in order around the pentagon) at the start of the game. We will first show that the game can be won at only one of the vertices.\n\nObserve that the quantity $a_1 + 2a_2 + 3a_3 + 4a_4 \\pmod{5}$ is an invariant of the game. For instance, one move involves replacing $a_1, a_3$ and $a_5$ by $a_1 - m, a_3 + 2m$ and $a_5 - m$. Thus the quantity $a_1 + 2a_2 + 3a_3 + 4a_4$ becomes\n\n$$\n(a_1 - m) + 2a_2 + 3(a_3 + 2m) + 4a_4 = a_1 + 2a_2 + 3a_3 + 4a_4 + 5m,\n$$\n\nwhich is unchanged modulo $5$. The other moves may be checked similarly.\n\nNow suppose that the game may be won at vertex $v_j$. The value of the invariant at the winning position is $2011j$. If the initial value of the invariant is $n$, then we must have $2011j \\equiv n \\pmod{5}$, or $j \\equiv n \\pmod{5}$. Hence the game may only be won at vertex $v_j$, where $j$ is the least positive residue of $n \\pmod{5}$.\n\nBy renumbering the vertices, we may assume without loss of generality that the potentially winning vertex is $v_5$. We will show that the game can be won in four moves by adding a suitable amount $2m_j$ at vertex $v_j$ (and subtracting $m_j$ from the opposite vertices) on the $j$th turn for $j = 1, 2, 3, 4$. The net change at vertex $v_1$ after these four moves is $2m_1 - m_3 - m_4$, which must equal $-a_1$ if we are to finish with $0$ at $v_1$. In this fashion we obtain the system of equations\n\n$$\n\\begin{aligned}\n2m_1 - m_3 - m_4 &= -a_1 \\\\\n2m_2 - m_4 &= -a_2 \\\\\n2m_3 - m_1 &= -a_3 \\\\\n2m_4 - m_1 - m_2 &= -a_4 \\\\\n-m_2 - m_3 &= -a_5 + 2011\n\\end{aligned}\n$$\n\nwhich has an integral solution if and only if the game may be won. The sum of the first four equations is the negative of the fifth equation, so the fifth equation is redundant. Multiplying the first four equations by $-1, 3, -3, 1$ and adding them yields $5m_2 - 5m_3 = a_1 - 3a_2 + 3a_3 - a_4$. But we are assuming $v_5$ is the potentially winning vertex, so we see\n\n$$\na_1 - 3a_2 + 3a_3 - a_4 \\equiv a_1 + 2a_2 + 3a_3 + 4a_4 \\equiv n \\equiv 5 \\equiv 0 \\pmod{5}.\n$$\n\nTherefore we may divide by $5$ to obtain $m_2 - m_3 = \\frac{1}{5}(a_1 - 3a_2 + 3a_3 - a_4)$. We also know that $m_2 + m_3 = a_1 + a_2 + a_3 + a_4$, and one easily confirms that the right-hand sides of these equations are integers with the same parity. Hence the system admits a solution with $m_2$ and $m_3$ integral. The second and third equations then quickly give integer values for $m_1$ and $m_4$ as well, so the system has an integral solution, meaning it is indeed possible to win the game at vertex $v_5$.",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 14179,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $N_0 = 0$. For $1 \\leq i \\leq k$, let $N_i$ be the number of babies born on or before Day $i$, and let $n_i = N_i - N_{i-1}$ be the number of babies born on Day $i$.\n\nLet $S = \\{N_1, \\dots, N_k\\}$ and $T = \\{N_1 + 100, \\dots, N_k + 100\\}$. Because at least one baby is born each day, both sets contain $k$ distinct integers between $1$ and $2114$, inclusive.\n\nThe sets $S$ and $T$ might intersect: in fact, they intersect if and only if there are a pair of indices $i$ and $j$ with $N_j = N_i + 100$ for some $i < j$, which is equivalent to the number of babies born in the period between days $i+1$ and $j$ inclusive being $100$.\n\n(a) Prove that if $k \\geq 1015$, then $S$ and $T$ must intersect.\n\n(b) Show that it is possible for $S$ and $T$ to be disjoint if $k = 1014$.",
    "options": [],
    "answer": "See solution",
    "solution": "Suppose there were at least $12$ different integers in $S$ having the same remainder mod $100$. Then there would also be at least $12$ different integers in $T$ with this remainder. But between $1$ and $2114$, no remainder mod $100$ occurs more than $22$ times, so $S \\cap T$ cannot be empty in this case.\n\nAssume now that no remainder mod $100$ occurs more than $11$ times in $S$. If $k \\geq 1015$, then there are at least $15$ remainders mod $100$ that occur at least $11$ times in $S$, and consequently also at least $11$ times in $T$. Thus $S \\cup T$ has at least $15$ remainders mod $100$ that each occur at least $22$ times (including repetitions, in the case of any remainders that occur in both $S$ and $T$). Since there are only $14$ remainders that occur $22$ times between $1$ and $2114$, and none that occur more frequently than that, some of these occurrences must overlap. Thus, $S$ and $T$ must intersect, proving (a).\n\nFor (b), let $n_i = 1$ if $i$ is not a multiple of $100$, and $n_i = 101$ if $i$ is a multiple of $100$. This pattern ensures that $N_j - N_i \\neq 100$ for all $0 \\leq i < j$. Also, $N_{100s+j} = 200s + j$ for all integers $0 \\leq j < 100$, $s \\geq 0$. In particular, $N_{1014} = 2014$, as required for $k = 1014$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 14180,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $\\triangle ABC$ be a triangle with $\\angle BAC = 2\\angle CBA$ and $AB < AC$. A point $P$ lies on segment $AC$ such that $PC = AB + AP$. Let $O$ be the circumcentre of triangle $ABP$, and let the line through $O$ parallel to $AB$ intersect $BP$ at $Q$. Show that $AQ$ passes through the midpoint of segment $BC$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $M$ be the midpoint of $BC$ and let $D$ be the reflection of $C$ in $P$. From the length condition in the statement, we have $DA = DP - AP = DP - (PC - AB) = AB$. This, together with $PM$ being the midline in $\\triangle CDB$, implies $\\angle MPC = \\angle BDA = \\frac{\\angle BAC}{2} = \\angle MBA$. So $ABMP$ is cyclic. Denote its circumcircle by $\\omega$.\n\nNext, let $R$ denote the intersection of $AB$ and $PM$. We have $\\angle RMB = \\angle PMB = 180^\\circ - \\angle BAP = 180^\\circ - 2\\angle MBR$, so $MR = MB$. Since $MB = MC$ as well, $M$ is the circumcentre of $\\triangle BRC$ and $\\angle BRC = 90^\\circ$.\n\n![](images/2025-SL-b_p2_data_b9ad37e210.png)\n\nLet $\\Gamma$ denote the circumcircle of $\\triangle CBD$, and let $W$ be the centre of $\\Gamma$. Let $N$ be the midpoint of arc $\\overarc{CBD}$ of $\\Gamma$. We showed that $AB$ bisects $\\angle CBD$, so $AB \\perp BN$. Also, since $P$ is the midpoint of $CD$, $\\angle NPA = 90^\\circ$ which shows that $N$ lies on $\\omega$. Thus, $NB$ is the radical axis of $\\Gamma$ and $\\omega$, which yields $NB \\perp OW$. We conclude $AB \\parallel OW$, and since $OQ \\parallel AB$, we have that $Q$ lies on line $OW$.\n\nLet $Q'$ be the intersection of $AM$ and $OW$. We will show that $B$, $Q'$, and $P$ are collinear which, by the previous result, is sufficient to show $Q' = Q$ and finish the problem.\n\nNote that $\\angle WCB = 90^\\circ - \\angle BDA = 90^\\circ - \\angle CBR = \\angle RCB$, so $W$ lies on line $CR$. Let $A'$ be the intersection of $MW$ (which is the perpendicular bisector of $BC$) and $AB$.\n\nApplying Menelaus' theorem to $\\triangle AMC$ and points $Q'$, $B$, and $P$ we have\n\n$$\nQ', B, P \\text{ collinear} \\iff -1 = \\frac{AQ'}{Q'M} \\cdot \\frac{MB}{BC} \\cdot \\frac{CP}{PA} = \\frac{AQ'}{Q'M} \\cdot \\frac{-1}{2} \\cdot \\frac{CP}{PA}\n$$\n\nFrom $Q'W \\parallel AA'$, we get $\\frac{AQ'}{Q'M} = \\frac{A'W}{WM}$. Thus, defining $F$ as the midpoint of $A'W$, we can rewrite the condition as\n\n$$\nQ', B, P \\text{ collinear} \\iff \\frac{AP}{PC} = \\frac{A'W/2}{WM} = \\frac{FW}{WM}. \\qquad (\\dagger)\n$$\n\nSince $F$ is the circumcenter of the right-angled $\\triangle A'RW$, we have\n\n$$\n\\angle FRA' + \\angle BRM = \\angle RA'F + \\angle MBR = \\angle BA'M + \\angle MBA' = 90^\\circ \\implies \\angle MRF = 90^\\circ.\n$$\n\nThis also shows\n\n$$\n\\angle RFM = \\angle RFW = 2\\angle RA'F = 2(90^{\\circ} - \\angle MBA) = \\angle CAR.\n$$\n\nCombining this with $\\angle ARC = 90^{\\circ}$ yields $\\triangle ARC \\sim \\triangle FRM$ and, since $\\angle MRW = \\angle PRC$, $W$ and $P$ are corresponding points in these triangles from which $(\\dagger)$ follows.\n\n**Comment.** A slicker finish from above is defining $\\tilde{Q}$ as the intersection of $AM$ and $BP$ then applying Brokard's theorem to $APMB$ to get $O\\tilde{Q} \\perp CR$. Since we have shown $CR \\perp AB$, this is enough to show $O\\tilde{Q} \\parallel AB$ so in fact, $\\tilde{Q} \\equiv Q$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 14181,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $a, b, c > 0$ and $abc = 1$. Prove that\n\n$$\n\\frac{a^{2014}}{1+2bc} + \\frac{b^{2014}}{1+2ca} + \\frac{c^{2014}}{1+2ab} \\ge \\frac{3}{ab+bc+ca}.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "Rewrite it as $\\left(\\sum \\frac{a^{2014}}{1+2bc}\\right) \\left(\\sum \\frac{1}{a}\\right) \\ge 3$. Observe that $\\frac{a^{2014}}{1+2bc} = \\frac{a^{2015}}{a+2}$ and $\\sum \\frac{1}{a} \\ge \\frac{1}{3} \\sum \\frac{a+2}{a}$. Thus\n\n$$\n\\left(\\sum \\frac{a^{2014}}{1+2bc}\\right) \\left(\\sum \\frac{1}{a}\\right) \\ge \\frac{1}{3} \\left(\\sum \\frac{a^{2015}}{a+2}\\right) \\left(\\sum \\frac{a+2}{a}\\right) \\ge \\frac{1}{3} \\left(\\sum a^{1007}\\right)^2 \\ge \\frac{1}{3} \\cdot 9 = 3. \\quad \\blacksquare\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 14182,
    "subject": "Mathematics (Olympiad)",
    "question": "Suppose there are initially $k$ coloured tiles and $n^2 - k$ uncoloured tiles arranged in a grid. The following process is performed:\n\n1. For each of the $k$ coloured tiles, give $1 to each of its uncoloured neighbours.\n2. If an uncoloured tile accumulates $3, colour it and give $1 to each of its uncoloured neighbours.\n\nIf all tiles are eventually coloured, what is the minimum possible value of $k$ in terms of $n$?",
    "options": [],
    "answer": "See solution",
    "solution": "Let there be initially $k$ coloured tiles and $n^2 - k$ uncoloured tiles. We start giving money to uncoloured tiles as follows.\n\n1. For each of the $k$ coloured tiles, give $1 to each of its uncoloured neighbours.\n2. If an uncoloured tile amasses $3, colour it and give $1 to each of its uncoloured neighbours.\n\nIf all the tiles are eventually coloured, then all of the $n^2 - k$ tiles, which were originally uncoloured, now each have at least $3 in them. Thus $D \\ge 3(n^2 - k)$ where $D$ is the total amount of dollars at the end.\n\nAll dollars in the array come from (1) and (2). The amount of dollars coming from (1) is at most $4k$. The amount of dollars coming from (2) is at most $n^2 - k$. Thus $D \\le 4k + n^2 - k$.\n\nCombining the two inequalities for $D$ we deduce\n\n$$\n\\begin{aligned}\n4k + n^2 - k &\\ge 3(n^2 - k) \\\\\n\\Rightarrow \\quad &k \\ge \\frac{n^2}{3}.\n\\end{aligned}\n$$\n\nHowever, since a corner tile has only two neighbours, at least one of the inequalities for $D$ is strict. Thus the final inequality is strict. $\\square$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 14183,
    "subject": "Mathematics (Olympiad)",
    "question": "Points $P$ and $Q$ lie on side $BC$ of acute-angled $\\triangle ABC$ so that $\\angle PAB = \\angle BCA$ and $\\angle CAQ = \\angle ABC$. Points $M$ and $N$ lie on lines $AP$ and $AQ$, respectively, such that $P$ is the midpoint of $AM$, and $Q$ is the midpoint of $AN$. Prove that lines $BM$ and $CN$ intersect on the circumcircle of $\\triangle ABC$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $S$ be the intersection point of lines $BM$ and $CN$. Denote $\\beta = \\angle QAC = \\angle CBA$, $\\gamma = \\angle PAB = \\angle ACB$. Then $\\triangle ABP \\sim \\triangle CAQ$, thus\n\n$$\n\\frac{BP}{PM} = \\frac{BP}{PA} = \\frac{AQ}{QC} = \\frac{NQ}{QC}.\n$$\n\nSince $\\angle BPM = \\beta + \\gamma = \\angle CQN$, $\\triangle BPM \\sim \\triangle NQC$, thus $\\angle BMP = \\angle NCQ$. Consequently, $\\triangle BPM \\sim \\triangle BSC$, thus $\\angle CSB = \\angle BPM = \\beta + \\gamma = 180^\\circ - \\angle BAC$. $\\Box$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 14184,
    "subject": "Mathematics (Olympiad)",
    "question": "Given the set $P = \\{1, 2, 3, 4, 5\\}$, define\n$$\nf(m, k) = \\sum_{i=1}^{5} \\lfloor m \\sqrt{\\frac{k+1}{i+1}} \\rfloor\n$$\nfor any $k \\in P$ and positive integer $m$, where $\\lfloor a \\rfloor$ denotes the greatest integer less than or equal to $a$.\n\nProve that for any positive integer $n$, there exist $k \\in P$ and a positive integer $m$ such that $f(m, k) = n$.",
    "options": [],
    "answer": "See solution",
    "solution": "Define the set $A = \\{ m \\sqrt{k+1} \\mid m \\in \\mathbb{N}^*,\\ k \\in P \\}$, where $\\mathbb{N}^*$ denotes the set of all positive integers. It is easy to check that for any $k_1, k_2 \\in P$, $k_1 \\neq k_2$, the ratio $\\frac{\\sqrt{k_1+1}}{\\sqrt{k_2+1}}$ is irrational. Therefore, for any $k_1, k_2 \\in P$ and positive integers $m_1, m_2$, the equality $m_1 \\sqrt{k_1+1} = m_2 \\sqrt{k_2+1}$ implies $m_1 = m_2$ and $k_1 = k_2$.\n\nNote that $A$ is an infinite set. Arrange the elements of $A$ in ascending order to form an infinite sequence. For any positive integer $n$, suppose the $n$th term of the sequence is $m \\sqrt{k+1}$. Any term before the $n$th can be written as $m_i \\sqrt{i+1}$, and\n$$\nm_i \\sqrt{i+1} \\leq m \\sqrt{k+1}.\n$$\nEquivalently, $m_i \\leq m \\frac{\\sqrt{k+1}}{\\sqrt{i+1}}$. There are $\\lfloor m \\frac{\\sqrt{k+1}}{\\sqrt{i+1}} \\rfloor$ such $m_i$ for each $i = 1, 2, 3, 4, 5$. Therefore,\n$$\nn = \\sum_{i=1}^{5} \\lfloor m \\frac{\\sqrt{k+1}}{\\sqrt{i+1}} \\rfloor = f(m, k).\n$$\nThe proof is complete.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 14185,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $\\alpha = \\frac{1+\\sqrt{5}}{2}$. Find all continuous functions $f: \\mathbb{R} \\to \\mathbb{R}$ such that, for all $x, y, z \\in \\mathbb{R}$,\n\n$$\n\\begin{aligned}\nf(\\alpha x + y) + f(\\alpha y + z) + f(\\alpha z + x)\n= \\alpha f(x + y + z) + 2f(x) + 2f(y) + 2f(z).\n\\end{aligned}\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "Let $(*)$ be the given functional equation.\n\nSetting $x = y = z = 0$ in $(*)$, we have $f(0) = 0$.\n\nSetting $y = z = 0$ in $(*)$ and simplifying, we have\n\n$$\nf(\\alpha x) = \\alpha^2 f(x) \\text{ for all } x \\in \\mathbb{R}.\n$$\n\nLetting $z = 0$ in the functional equation, we get\n\n$$\nf(\\alpha x + y) + f(\\alpha y) + f(x) = \\alpha f(x + y) + 2f(x) + 2f(y).\n$$\n\nUsing $f(\\alpha y) = \\alpha^2 f(y) = (\\alpha + 1)f(y)$, it follows that\n\n$$\nf(\\alpha x + y) = \\alpha f(x + y) + f(x) + (1 - \\alpha)f(y).\n$$\n\nThen $(*)$ simplifies to\n\n$$\nf(x + y) + f(y + z) + f(z + x) = f(x + y + z) + f(x) + f(y) + f(z).\n$$\n\nPutting $z = -y$ in the above equation, we have\n\n$$\nf(x + y) + f(x - y) = 2f(x) + f(y) + f(-y).\n$$\n\nLet $f_c(x) = \\frac{f(x) + f(-x)}{2}$ and $f_o(x) = \\frac{f(x) - f(-x)}{2}$. Then\n\n$$\n\\begin{aligned}\nf_c(x + y) + f_c(x - y) &= 2f_c(x) + 2f_c(y) \\\\\nf_o(x + y) + f_o(x - y) &= 2f_o(x)\n\\end{aligned}\n$$\n\nwhich respectively are the quadratic and the Jensen functional equations with the continuous solutions $f_c(x) = a x^2$ and $f_o(x) = b x$ (note that $f_o(0) = 0$). Thus, $f(x) = a x^2 + b x$.\n\nSubstituting $f(x) = a x^2 + b x$ into $(*)$, we can see that $b = 0$.\n\nThus $f(x) = a x^2$ is the only continuous solution.\n\n![](images/Tajland_2008_p9_data_60a76012ce.png)",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 14186,
    "subject": "Mathematics (Olympiad)",
    "question": "Determine all positive integers $a$, $b$, $c$ for which there exist positive integers $x$, $y$, $z$ such that:\n\n$$\nab + 1 = x!,\n$$\n$$\nbc + 1 = y!,\n$$\n$$\nca + 1 = z!,\n$$\n\nwhere $n!$ denotes the product $1 \\cdot 2 \\cdot 3 \\ldots \\cdot n$.",
    "options": [],
    "answer": "See solution",
    "solution": "We can easily see that $x, y, z \\geq 2$.\n\nIf $x, y, z \\geq 3$, then observe that $3$ does not divide $a$, $b$, or $c$. Indeed, if for example $3 \\mid a$, then $3$ does not divide $ab + 1$, but $3$ divides $x!$, a contradiction.\n\nThis means that two of the numbers are congruent modulo $3$. Due to the symmetry, suppose that $a \\equiv b \\pmod{3}$. Then, we have\n\n$$\nx! = ab + 1 \\equiv a^2 + 1 \\pmod{3}.\n$$\n\nHowever, $3 \\mid x!$, so $3 \\mid a^2 + 1$, which is absurd, since for all integers $x$, we have $x^2 \\equiv 0, 1 \\pmod{3}$.\n\nIt follows that one of $x, y, z$ is smaller than $3$; let it be $x$. Then $x = 2$ and from $ab + 1 = 2! = 2$, we get $ab = 1$, so $a = b = 1$. Then, from the second relation, $y = z$, so all solutions have the form $(1, 1, y! - 1)$ with its cyclic permutations.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 14187,
    "subject": "Mathematics (Olympiad)",
    "question": "a) Does there exist a set $S$ of natural numbers such that, for every $n \\in \\mathbb{N}$, the number of elements of $S$ that are coprime to $n$ is exactly $n$?\n\nb) If such a set $S$ exists, construct one.",
    "options": [],
    "answer": "See solution",
    "solution": "a) We claim that such a set $S$ cannot exist. Assume to the contrary that there exists a set $S$ satisfying the problem's condition and let $n \\in \\mathbb{N}$, $n \\notin S$. Thus, exactly $n$ elements of $S$ are coprime to $n$. Therefore, there exist infinitely many prime numbers which do not belong to $S$. Let $p$ and $q$ be two of them. Since $p \\notin S$, exactly $p$ elements of $S$ are not divisible by $p$ (i.e., are coprime to $p$). For each $\\alpha > 1$, $p^{\\alpha} \\in S$. This is true because if $p^{\\alpha} \\notin S$, there must be $p^{\\alpha}$ elements of $S$ coprime to $p^{\\alpha}$ (and therefore coprime to $p$), but we have shown that $S$ has exactly $p$ elements of this kind. Finally, the greatest common divisor of $p^{\\alpha}$ and $q$ is $1$, so infinitely many elements of $S$ are coprime to $q \\notin S$, a contradiction!\n\nb) We construct the set $S$ in the following way:\n\nSuppose $a_1, a_2, a_3, \\dots$ are elements of $S$ and at first all $a_i$'s, $i \\in \\mathbb{N}$, are $1$. In each step, we multiply $a_i$ by some primes.\n\nFor $n \\in \\mathbb{N}$ in step $n$:\n\n1. We multiply $a_n$ by some new primes $p_{2n-1}$ and $p_{2n}$ so that $a_n$ becomes greater than $a_{n-1}$ (where $p_n$ is the $n$th prime number).\n2. Some elements among $\\{a_1, a_2, \\dots, a_{n-1}\\}$ are coprime to $a_n$. Let $b_n$ be the number of such elements, so $b_n < n < a_n$. Let $t_n = a_n - b_n$.\n3. Because of the type of construction, infinitely many $a_i$'s, $i > n$, are coprime to $a_n$. We leave the first $t_n$ and assume that the $t_n$th number is $a_m$.\n4. In the sequence $a_{m+1}, a_{m+2}, \\dots$, multiply the first $2^{n-1}$ terms by $p_{2n-1}$, the next $2^{n-1}$ terms by $p_{2n}$, and so on.\n\nTherefore, at the end of step $n$:\n\n- Exactly $a_n$ numbers are coprime to $a_n$.\n- For each sequence $(q_1, q_2, \\dots, q_n)$ of prime numbers such that $q_i \\in \\{p_{2i-1}, p_{2i}\\}$, there exist infinitely many numbers $k$ such that $a_k$ has exactly these prime factors. So for each $a_i$ there exist infinitely many numbers $k$ such that $a_k$ has prime factors different from those of $a_i$ and consequently is coprime to $a_i$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 14188,
    "subject": "Mathematics (Olympiad)",
    "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": "See solution",
    "solution": "By the law of cosines, we get\n\n$$\n\\cos A = \\frac{CA^2 + AB^2 - BC^2}{2 \\cdot CA \\cdot AB} = \\frac{5^2 + 6^2 - 4^2}{2 \\times 5 \\times 6} = \\frac{3}{4}.\n$$\n\nTherefore,\n\n$$\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 \\\\\n&= \\frac{1}{4} + \\frac{3}{4} \\left(\\frac{9}{16}\\right) \\\\\n&= \\frac{1}{4} + \\frac{27}{64} \\\\\n&= \\frac{16}{64} + \\frac{27}{64} \\\\\n&= \\frac{43}{64}.\n\\end{align*}\n$$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 14189,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all functions $f$ from the positive real numbers to the positive real numbers for which $f(x) \\leq f(y)$ whenever $x \\leq y$ and\n$$\nf(x^4) + f(x^2) + f(x) + f(1) = x^4 + x^2 + x + 1\n$$\nfor all $x > 0$.",
    "options": [],
    "answer": "See solution",
    "solution": "Substituting $x = 1$ gives $f(1) = 1$. Replacing $x$ by $x^2$ gives $f(x^8) + f(x^4) + f(x^2) + 1 = x^8 + x^4 + x^2 + 1$. Now subtract the original equation to get $f(x^8) - f(x) = x^8 - x$. Define $c_x = f(x) - x$. Observe that for all $x$, $c_x = c_{x^8}$, and by induction, this implies $c_x = c_{x^{8k}}$ for all integers $k$.\n\nConsider $x < 1$. If $c_x > 0$ then by choosing a large negative $k$, we can make $c_x + x^{8k} > 1$, but then $f(x^{8k}) = c_x + x^{8k} > 1 = f(1)$, which is a contradiction as $x^{8k} < 1$. If $c_x < 0$, we can choose a large positive $k$ so that $x^{8k} < -c_x$, but then $f(x^{8k}) = c_x + x^{8k} < 0$. So we must have $c_x = 0$, so $f(x) = x$ for $x < 1$.\n\nNow consider $x > 1$. If $c_x < 0$ we can choose a large negative $k$ such that $c_x + x^{8k} < 1$, so $f(x^{8k}) < 1$ which is a contradiction. So $c_x \\geq 0$ for all $x > 1$. If $c_x > 0$, then our original equation, after substituting $f(x) = x + c_x$ and so on, becomes\n$$\nx^4 + c_{x^4} + x^2 + c_{x^2} + x + c_x + 1 = x^4 + x^2 + x + 1, \\\\\n\\text{i.e. } c_{x^4} + c_{x^2} + c_x = 0, \\\\\n\\text{and hence } c_{x^4} + c_{x^2} < 0,\n$$\nbut this is a contradiction as both $c_{x^4} \\geq 0$ and $c_{x^2} \\geq 0$. So $c_x = 0$ for all $x > 1$. Therefore $f(x) = x$ for all $x$, and it is clear that this is a valid solution.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 14190,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all pairs $ (x, y) $ of positive integers such that\n\n$$\nx^3 + y^3 = x^2 + 42xy + y^2.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "Let $d = \\gcd(x, y)$ be the greatest common divisor of positive integers $x$ and $y$.\n\nSet $x = ad$, $y = bd$, where $d \\in \\mathbb{N}$, $(a, b) = 1$, $a, b \\in \\mathbb{N}$. Then:\n\n$$\n\\begin{aligned}\nx^3 + y^3 &= x^2 + 42xy + y^2 \\\\\nd^3(a^3 + b^3) &= d^2(a^2 + 42ab + b^2) \\\\\nd(a+b)(a^2 - ab + b^2) &= a^2 + 42ab + b^2 \\\\\n(da + db - 1)(a^2 - ab + b^2) &= 43ab\n\\end{aligned}\n$$\n\nLet $c = da + db - 1 \\in \\mathbb{N}$. The equation $a^2c - abc + b^2c = 43ab$ implies:\n\n$$\n\\begin{aligned}\nb \\mid ca^2 &\\implies b \\mid c \\\\\na \\mid cb^2 &\\implies a \\mid c \\\\\n\\implies ab \\mid c \\\\\nc = mab,\\ m \\in \\mathbb{N}^+ \\\\\nm(a^2 - ab + b^2) = 43 \\\\\na^2 - ab + b^2 \\mid 43 \\\\\na^2 - ab + b^2 = 1 \\text{ or } 43\n\\end{aligned}\n$$\n\nIf $a^2 - ab + b^2 = 1$, then $(a-b)^2 = 1 - ab \\ge 0 \\implies a = b = 1$, $2d = 44$, so $(x, y) = (22, 22)$.\n\nIf $a^2 - ab + b^2 = 43$, by symmetry, suppose $a \\ge b$. Then $43 = a^2 - ab + b^2 \\ge ab \\ge b^2 \\implies b \\in \\{1, 2, 3, 4, 5, 6\\}$.\n\n- If $b=1$, then $a=7$, $d=1$, so $(x, y) = (7, 1)$ or $(1, 7)$.\n- If $b=6$, then $a=7$, $d = \\frac{43}{13} \\notin \\mathbb{N}$.\n- For $b \\in \\{2, 3, 4, 5\\}$, there are no positive integer solutions for $a$.\n\nThus, the solutions are $(x, y) \\in \\{(1, 7), (7, 1), (22, 22)\\}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 14191,
    "subject": "Mathematics (Olympiad)",
    "question": "Solve the equation\n\n$$\nx^4 y^3 (y - x) = x^3 y^4 - 216\n$$\n\nin integers.",
    "options": [],
    "answer": "See solution",
    "solution": "The given equation is equivalent to\n\n$$\nx^3 y^4 + x^4 y^3 (x - y) = 216 \\iff (ry)^3 (x^2 - xy + y) = 6^3.\n$$\n\nBoth $x$ and $y$ must therefore be divisors of $6$, and therefore equal to $\\pm 1$, $\\pm 2$, $\\pm 3$ or $\\pm 6$. Also, $xy \\mid 6$ must hold. This means that $|x|$ and $|y|$ can only both equal $1$ or the set $\\{|x|, |y|\\}$ equal one of the sets $\\{1, 2\\}$, $\\{1, 3\\}$, $\\{1, 6\\}$ or $\\{2, 3\\}$. Altogether, this yields $36$ possible combinations for $(x, y)$ and a straightforward check yields the three solutions $(-3, -2)$, $(2, 3)$ and $(1, 6)$. $\\Box$",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 14192,
    "subject": "Mathematics (Olympiad)",
    "question": "For a positive integer $n$, let $\\tau(n)$ be the number of positive divisors of $n$.\n\n(a) Find all positive integers $n$ such that $\\tau(n) + 2023 = n$.\n\n(b) Prove that there exist infinitely many positive integers $k$ such that there are exactly two positive integers $n$ satisfying $\\tau(kn) + 2023 = n$.",
    "options": [],
    "answer": "See solution",
    "solution": "a) First, we prove the following lemma.\n\n*Lemma.* For any positive integer $n$, $\\tau(n) \\leq 2\\sqrt{n}$.\n\n*Proof.* Let $d_1, d_2, \\dots, d_s$ be all positive divisors of $n$ not exceeding $\\sqrt{n}$. Clearly, $s \\leq \\sqrt{n}$. For any divisor $x \\geq \\sqrt{n}$, $n/x$ is a divisor not exceeding $\\sqrt{n}$. Thus, $\\tau(n) \\leq 2s \\leq 2\\sqrt{n}$. ■\n\nNow, suppose there exists a positive integer $n$ such that $\\tau(n) + 2023 = n$. By the lemma, $n \\leq 2\\sqrt{n} + 2023$. Solving this inequality gives\n\n$$\nn \\leq (\\sqrt{2024} + 1)^2 < 2115.\n$$\n\nAlso, $n = \\tau(n) + 2023 \\geq 2025$. Therefore, $2025 \\leq n \\leq 2114$.\n\nLet $n = p_1^{e_1} p_2^{e_2} \\cdots p_k^{e_k}$, where $p_1, \\dots, p_k$ are distinct primes and $e_1, \\dots, e_k$ are positive integers. Then $\\tau(n) = (e_1 + 1)(e_2 + 1)\\cdots(e_k + 1)$, so\n\n$$\n(e_1 + 1)(e_2 + 1)\\cdots (e_k + 1) + 2023 = p_1^{e_1} p_2^{e_2} \\cdots p_k^{e_k} \\quad (1)\n$$\n\nIf all $e_i$ are even, $n$ is a perfect square in $[2025, 2114]$, so $n = 2025$, but this does not satisfy the equation. Thus, at least one $e_i$ is odd, so all $p_i$ are odd primes. Assume $3 \\leq p_1 < p_2 < \\dots < p_k$.\n\nIf $k \\geq 5$, $n \\geq 3 \\cdot 5 \\cdot 7 \\cdot 11 \\cdot 13 > 2114$, contradiction. So $k \\leq 4$.\n\n*Case 1: $k = 1$.* $n = p_1^{e_1} \\leq 2114$, so $e_1 \\leq 6$. Checking all cases, no $p_1$ satisfies (1).\n\n*Case 2: $k = 2$.* $n = p_1^{e_1} p_2^{e_2} \\leq 2114$. $e_1 + e_2 \\leq 6$. $\\tau(n) \\leq 9$. No $n$ satisfies (1).\n\n*Case 3: $k = 3$.* $n = p_1^{e_1} p_2^{e_2} p_3^{e_3} \\leq 2114$. $e_1 + e_2 + e_3 \\leq 5$. Possible $\\tau(n) \\in \\{8, 12, 16, 18\\}$. No $n$ satisfies (1).\n\n*Case 4: $k = 4$.* $n = p_1 p_2 p_3 p_4$, $\\tau(n) = 16$. No $n$ satisfies (1).\n\nTherefore, there is no positive integer solution to $\\tau(n) + 2023 = n$.\n\nb) All prime numbers $k > 6996$ satisfy the requirement. For any such prime $k$, note that\n\n$$\n\\tau(kn) \\leq 2\\tau(n).\n$$\n\nLet $n = k^e p_1^{e_1} \\cdots p_s^{e_s}$, with $p_i \\neq k$. Then\n\n$$\n\\tau(kn) = (e+2)(e_1+1)\\cdots(e_s+1) \\leq 2(e+1)(e_1+1)\\cdots(e_s+1) = 2\\tau(n).\n$$\n\nIf $n$ is not divisible by $k$ ($e=0$), $\\tau(kn) = 2\\tau(n)$.\n\nNow, for any solution $n$ to $\\tau(kn) + 2023 = n$,\n\n$$\nn = \\tau(kn) + 2023 \\leq 2\\tau(n) + 2023 \\leq 4\\sqrt{n} + 2023.\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 14193,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $ABC$ be an acute-angled triangle with $AC < AB$ and circumradius $R$. Furthermore, let $D$ be the foot of the altitude from $A$ on $BC$ and let $T$ denote the point on the line $AD$ such that $AT = 2R$ holds with $D$ lying between $A$ and $T$. Finally, let $S$ denote the midpoint of the arc $BC$ on the circumcircle that does not include $A$.\n\nProve: $\\angle AST = 90^\\circ$.",
    "options": [],
    "answer": "See solution",
    "solution": "As usual, we denote the angles $\\angle BAC$, $\\angle ABC$, and $\\angle BCA$ by $\\alpha$, $\\beta$, and $\\gamma$, respectively. The center of the circumcircle is denoted by $O$.\n\n![](images/Austria_2015_booklet_p7_data_a073f3b8ec.png)\n\nBy assumption, we have $\\beta < \\gamma$. Let $E$ be the point on the circumcircle of $ABC$ diametrically opposite to $A$.\n\nBy the inscribed angle theorem, we have $\\angle AOB = 2\\gamma$. By definition, $S$ is the intersection of the angular bisector of $\\angle CAB$ and the circumcircle. We note that\n\n$$\n\\angle EAS = \\angle BAS - \\angle BAO = \\frac{\\alpha}{2} - \\frac{1}{2}(180^\\circ - \\angle AOB) = \\frac{\\alpha}{2} - \\frac{1}{2}(180^\\circ - 2\\gamma) = \\frac{\\alpha}{2} + \\gamma - 90^\\circ\n$$\n\nSince we also have\n\n$$\n\\angle SAT = \\angle SAC - \\angle DAC = \\frac{\\alpha}{2} - (90^\\circ - \\angle ACD) = \\frac{\\alpha}{2} + \\gamma - 90^\\circ\n$$\n\nit therefore follows that $\\angle EAS = \\angle TAS$ holds. Since we also have $AE = AT = 2R$, triangles $ASE$ and $AST$ are congruent, and therefore $\\angle AST = \\angle ASE$ follows. Since $AE$ is a diameter of the circumcircle, we have $\\angle ASE = 90^\\circ$, and the claim is proven.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 14194,
    "subject": "Mathematics (Olympiad)",
    "question": "The number inserted into the box located on the $i$-th row from the top and $j$-th column from the left is given by $13(i - 1) + j$ for the first grid, and by $20(13 - j) + i$ for the second grid. If the same number goes into the boxes located at the same position in the two grids, what numbers can appear in both grids at the same position?",
    "options": [],
    "answer": "See solution",
    "solution": "Set $13(i - 1) + j = 20(13 - j) + i$. Simplifying, $12i + 21j = 273$. Since $21j$ and $273$ are multiples of $7$, $12i$ must also be a multiple of $7$, so $i$ is a multiple of $7$. With $1 \\leq i \\leq 20$, possible $i$ are $7$ and $14$. For $i = 7$, $12 \\times 7 + 21j = 273 \\implies 84 + 21j = 273 \\implies 21j = 189 \\implies j = 9$. For $i = 14$, $12 \\times 14 + 21j = 273 \\implies 168 + 21j = 273 \\implies 21j = 105 \\implies j = 5$. The corresponding numbers are $13(7-1) + 9 = 13 \\times 6 + 9 = 78 + 9 = 87$ and $13(14-1) + 5 = 13 \\times 13 + 5 = 169 + 5 = 174$. Thus, the numbers are $87$ and $174$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 14195,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $n$ and $m$ be natural numbers. We want to color each cell of an $n \\times m$ table either white or black. For what $n$ and $m$ can this be done so that each cell has an odd number of neighbouring cells that are of the same color? Two cells are called neighbouring if they have a common side.",
    "options": [],
    "answer": "See solution",
    "solution": "We shall prove that this can be done if and only if at least one of the numbers $n$ and $m$ is even.\n\nFirst, observe the case when one of the numbers is even. Without loss of generality, suppose $m$ is even. We divide a table of size $n \\times m$ ($n$ rows, $m$ columns) into quadrilaterals of size $1 \\times 2$ and color them white and black in turn, in the form of a chessboard. Since $m$ is even, each row can be divided into exactly $\\frac{m}{2}$ such quadrilaterals, which makes the coloring possible. It also holds that each cell has exactly one neighbouring cell of the same color. The table-coloring problem is solved.",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 14196,
    "subject": "Mathematics (Olympiad)",
    "question": "In the country of Squareland, there are stamps with values $1$ cent, $4$ cent, $9$ cent, etc., one type for each square number. Stamps can be combined in all possible ways in Squareland without additional rules.\n\nProve for every positive integer $n$: In Oddland and Squareland there are equally many ways to correctly place stamps of a total value of $n$ cent on an envelope. Rearranging the stamps on an envelope makes no difference.",
    "options": [],
    "answer": "See solution",
    "solution": "We construct a bijection between possible combinations in Oddland and possible combinations in Squareland. Suppose we have a combination of Squareland stamps that sum to $n$ cent, consisting of $a_1$ stamps of value 1 cent, $a_2$ stamps of value 4 cent, ..., $a_M$ stamps of value $M^2$ cent, so that\n\n$$\nn = \\sum_{k=1}^{M} k^2 a_k.\n$$\n\nNow we express $k^2$ as $\\sum_{j=1}^{k} (2j - 1)$ and interchange the order of summation, which yields\n\n$$\nn = \\sum_{k=1}^{M} \\sum_{j=1}^{k} (2j - 1) a_k = \\sum_{j=1}^{M} (2j - 1) \\sum_{k=j}^{M} a_k.\n$$\n\nThis gives us a possible combination of Oddland stamps: By setting $b_j = \\sum_{k=j}^{M} a_k$, we have\n\n$$\nn = \\sum_{j=1}^{M} (2j - 1)b_j.\n$$\n\nThis can be interpreted as a collection of $b_1$ stamps of value 1 cent, $b_2$ stamps of value 3 cent, ..., $b_M$ stamps of value $(2M - 1)$ cent. We have $b_1 \\ge b_2 \\ge \\dots \\ge b_M$ by definition, so this is a legal combination in Oddland.\n\nConversely, if a combination in Oddland is given by the values $b_1, b_2, \\dots, b_M$, we can use the identities $a_1 = b_1 - b_2$, $a_2 = b_2 - b_3$, ..., $a_{M-1} = b_{M-1} - b_M$, $a_M = b_M$ to recover the corresponding combination in Squareland. (Note that these values are nonnegative whenever $b_1 \\ge b_2 \\ge \\dots \\ge b_M$.)\n\nSince these two operations obviously are inverse to one another, we have found a bijection, which proves the statement.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 14197,
    "subject": "Mathematics (Olympiad)",
    "question": "The cells of a $ (n^2 - n + 1) \\times (n^2 - n + 1) $ matrix are coloured using $ n $ colours. A colour is called *dominant* on a row (or a column) if there are at least $ n $ cells of this colour on that row (or column). A cell is called *extremal* if its colour is dominant both on its row and its column. Find all $ n \\ge 2 $ for which there is a colouring with no extremal cells.",
    "options": [],
    "answer": "See solution",
    "solution": "Such colourings exist for $ n = 2 $ (almost trivial), and $ n = 3 $ (by detailed case analysis). For $ n = 4 $, a colouring with no extremal cells can be constructed, though it is laborious. For $ n \\ge 5 $, colourings with no extremal cells can be built inductively, using a symmetric pattern based on the model for $ n - 1 $. For a detailed proof, see Mathematical Reflections, no. 4/2006, pages 22-25: [http://reflections.awesomemath.org/2006_4/2006_4_solutions.pdf](http://reflections.awesomemath.org/2006_4/2006_4_solutions.pdf).",
    "topic": "Algebra",
    "subtopic": "Linear Algebra"
  },
  {
    "id": 14198,
    "subject": "Mathematics (Olympiad)",
    "question": "Consider all sequences of real numbers $x_0, x_1, x_2, \\dots, x_{100}$ satisfying the following conditions:\n\n1. $x_0 = 0$;\n2. For any integer $i$, $1 \\le i \\le 100$, $1 \\le x_i - x_{i-1} \\le 2$ holds.\n\nFind the largest positive integer $k \\le 100$ such that\n\n$$\nx_k + x_{k+1} + \\dots + x_{100} \\ge x_0 + x_1 + \\dots + x_{k-1}\n$$\n\nholds for every such sequence $x_0, x_1, x_2, \\dots, x_{100}$.",
    "options": [],
    "answer": "See solution",
    "solution": "The answer is $67$.\n\nOn one hand, consider the sequence where $x_i = 2i$ for $1 \\le i \\le 34$, and $x_{34+j} = x_{34} + j = 68 + j$ for $1 \\le j \\le 66$. This sequence satisfies the conditions. We have\n\n$$\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$$\n\nThis example shows that when $k \\ge 68$, the requirement is not satisfied.\n\nOn the other hand, for any sequence $x_1, x_2, \\dots, x_{100}$ satisfying the conditions, we have $x_i \\le 2i$ for $1 \\le i \\le 100$. For $0 \\le s < t \\le 100$, $x_t - x_s \\ge t - s$. Therefore,\n\n$$\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$$\n\nIn conclusion, the largest $k$ is $67$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 14199,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $ABC$ be a scalene triangle and let $I$ be its incenter. The projections of $I$ on $BC$, $CA$, and $AB$ are $D$, $E$, and $F$ respectively. Let $K$ be the reflection of $D$ over the line $AI$, and let $L$ be the second point of intersection of the circumcircles of the triangles $BFK$ and $CEK$. If $\\frac{1}{3}BC = AC - AB$, prove that $DE = 2KL$.",
    "options": [],
    "answer": "See solution",
    "solution": "Writing $AE = AF = x$, $BF = BD = y$ and $CE = CD = z$, the condition $\\frac{1}{3}BC = AC - AB$ translates to $y+z = 3(z-y)$ giving $z = 2y$, i.e. $CD = 2BD$.\n\nLetting $B'$ be the reflection of $B$ on $AI$ we have that $B'$ belongs on $AC$ with $B'E = BF = BD = \\frac{1}{2}CD = \\frac{1}{2}CE$ therefore $B'$ is the midpoint of $CE$.\n\n![](images/Bmo_Shortlist_2021_p45_data_eb75ca3c46.png)\n\nUnder reflection on $AI$, the circumcircle $\\omega$ of triangle $DEF$ remains fixed. Its tangent $BD$ maps to $B'K$. So $B'K$ is tangent to $\\omega$. Since $B'E$ is tangent to $\\omega$, then $B'E = B'K = B'C$. Thus $CKE$ is a right-angled triangle with diameter $CE$. If $Q$ is the midpoint of $DE$ then, since $CD = CE$, we have that $\\angle CQE = 90^\\circ$ and therefore the points $C, K, Q, L, E$ are concyclic.\n\nObserve that\n\n$$\n\\begin{align*}\n\\angle BLC &= \\angle BLK + \\angle CLK = \\angle BFK + \\angle CEK = (180^\\circ - \\angle AFK) + (180^\\circ - \\angle AEK) \\\\\n&= \\angle BAC + \\angle FKE = \\angle BAC + \\angle FDE = \\angle BAC + \\left(90^\\circ - \\frac{1}{2}\\angle BAC\\right) \\\\\n&= 90^\\circ + \\frac{1}{2}\\angle BAC = \\angle BIC.\n\\end{align*}\n$$\n\nSo *L* belongs on the circumcircle of triangle *BIC*, i.e. on the *A*-excircle $\\omega_A$ of triangle $ABC$.\n\nLet *J* be the *A*-excenter of triangle $ABC$ and recall that it is the antipodal point of $I$ on $\\omega_A$.\nThen\n\n$$\n\\angle CLJ = \\angle CBJ = 90^\\circ - \\frac{1}{2}\\angle ABC = \\angle BFD = \\angle CEK = \\angle CLK.\n$$\n\nSo $K, L, J$ are collinear and therefore $\\angle ILK = 90^\\circ$.\n\nLet $T$ be the reflection of $L$ on $AI$. Since $L$ belongs on the circle with centre $B'$ containing $E$ and $K$, then $L$ belongs on the circle $\\omega_2$ with centre $B$ containing $F$ and $D$. Let $S$ be the intersection of $IT$ and $BC$. Since $KL \\perp IL$, then $DT \\perp IT$. It follows that $\\angle IDT = 90^\\circ - \\angle DIS = \\angle ISD$. Since $ID$ is tangent on $\\omega_2$, then $S$ belongs on $\\omega_2$. Then $SD = 2BD = DC$ and so the triangles $IDC$ and $IDS$ are equal. Their height $DT$ and $DQ$ must be equal. Therefore $DE = 2DQ = 2DT = 2KL$ as required.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 14200,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $a$, $b$, and $c$ be positive real numbers such that $a + b + c = 1$. Prove that the following inequality holds:\n\n$$\n2(a^2 + b^2 + c^2) \\geq \\frac{1}{9} + 15abc.\n$$\n\nWhen does equality occur?",
    "options": [],
    "answer": "See solution",
    "solution": "Let $t = a^2 + b^2 + c^2$. Since $a + b + c = 1$, we have\n\n$$\n3t = 3(a^2 + b^2 + c^2) \\geq (a + b + c)^2 = 1,\n$$\n\ni.e., $t \\geq \\frac{1}{3}$. Applying the Cauchy-Schwarz inequality to the vectors $(\\sqrt{a}, \\sqrt{b}, \\sqrt{c})$ and $(\\sqrt{a^3}, \\sqrt{b^3}, \\sqrt{c^3})$, we obtain\n\n$$\nt^2 = (a^2 + b^2 + c^2)^2 \\leq (a+b+c)(a^3+b^3+c^3) = a^3+b^3+c^3. \\quad (1)\n$$\n\nOn the other hand,\n\n$$\n\\frac{ab+bc+ca}{abc} = \\frac{1}{a} + \\frac{1}{b} + \\frac{1}{c} = (a+b+c)\\left(\\frac{1}{a} + \\frac{1}{b} + \\frac{1}{c}\\right) \\geq 9,\n$$\n\ngiving that\n\n$$\nab + bc + ca \\geq 9abc. \\quad (2)\n$$\n\nFurther, (1) and (2) imply\n\n$$\nt^2 \\leq a^3+b^3+c^3 = a^2+b^2+c^2-ab-ac-bc+3abc = (a+b+c)^2-3(ab+bc+ca)+3abc \\leq 1-24abc,\n$$\n\ni.e., $\\frac{1-t^2}{24} \\geq abc$. To complete the proof, we need to show that\n\n$$\n2t \\geq \\frac{1}{9} + \\frac{15(1-t^2)}{24}.\n$$\n\nThe latter inequality is equivalent to $45t^2 + 144t - 53 \\geq 0$. Since\n\n$$\n45t^2 + 144t - 53 = (3t - 1)(15t + 53),\n$$\n\nand $t \\geq \\frac{1}{3}$, the proof follows. It is easy to check that equality occurs if and only if $a = b = c = \\frac{1}{3}$.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 14201,
    "subject": "Mathematics (Olympiad)",
    "question": "在 $\\triangle ABC$ 中，設點 $D$ 在 $BC$ 邊上且 $AD$ 平分 $\\angle BAC$，並設 $AD$ 的中點為 $M$。設以 $AC$ 為直徑的圓 $\\omega_1$ 與 $BM$ 交於點 $E$，以 $AB$ 為直徑的圓 $\\omega_2$ 與 $CM$ 交於點 $F$。證明 $B, E, F, C$ 四點共圓。\n\nLet $M$ be the midpoint of the internal bisector $AD$ of $\\triangle ABC$. Circle $\\omega_1$ with diameter $AC$ intersects $BM$ at $E$ and circle $\\omega_2$ with diameter $AB$ intersects $CM$ at $F$. Show that $B, E, F, C$ belong to the same circle.",
    "options": [],
    "answer": "See solution",
    "solution": "如果 $AB = AC$，則命題顯然成立。無損一般性，假設 $AB < AC$。設 $AH$ 為所給兩圓的公共弦，如圖所示。作過 $A$ 且垂直於 $AD$ 的直線，分別與 $\\omega_1$、$\\omega_2$ 交於 $K$、$L$。\n\n證明 $BL$ 通過 $M$。設 $X$ 為 $BL$ 與 $AD$ 的交點。由於 $KB \\parallel AD \\parallel LC$，有：\n\n$$\n\\frac{AX}{KB} = \\frac{LA}{LK}, \\quad \\frac{DX}{CL} = \\frac{BD}{BC}, \\quad \\frac{BD}{BC} = \\frac{KA}{LK}.\n$$\n\n因此，$AX = \\frac{KB \\cdot LA}{LK}$，$DX = \\frac{CL \\cdot KA}{LK}$。又因 $\\angle KAB = \\angle LAC$，且 $\\triangle AKB \\sim \\triangle ALC$，所以 $\\frac{KA}{LA} = \\frac{KB}{LC}$，即 $KA \\cdot LC = KB \\cdot LA$。因此 $AX = DX$，$X$ 與 $M$ 重合。類似地可證 $CK$ 也通過 $M$。\n\n有：\n\n$$\n\\angle DME = \\angle LMA = \\angle CLE = 180^\\circ - \\angle DHE.\n$$\n\n這說明 $E, M, D, H$ 共圓。$KBFH$ 鑲嵌於圓 $\\omega_1$，因此\n\n$$\n\\angle DMF = \\angle KMA = \\angle MKB = 180^{\\circ} - \\angle BHF = \\angle DHF,\n$$\n\n說明 $M, H, D, F$ 共圓。\n\n已證 $M, H, D, F, E$ 共圓。在直角三角形 $HAD$ 中，$HM$ 為中線，故 $MD = MH$。因此\n\n$$\n\\angle MDH = \\angle MHD = \\angle MED = \\angle MFH.\n$$\n\n考慮 $\\triangle MDE$ 和 $\\triangle MBD$，它們有公共角 $M$，且\n\n$$\n180^{\\circ} - \\angle CFE = \\angle MFE = \\angle MDE = \\angle MBD.\n$$\n\n因此 $B, E, F, C$ 共圓，命題得證。",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 14202,
    "subject": "Mathematics (Olympiad)",
    "question": "Given that\n\n$$\n\\frac{1 + 3 + 5 + \\cdots + (2n - 1)}{2 + 4 + 6 + \\cdots + (2n)} = \\frac{2011}{2012},\n$$\n\ndetermine $n$.",
    "options": [],
    "answer": "See solution",
    "solution": "Using the sum formula for arithmetic progressions, we obtain\n\n$$\n\\frac{1 + 3 + 5 + \\cdots + (2n - 1)}{2 + 4 + 6 + \\cdots + (2n)} = \\frac{n \\cdot \\frac{1 + 2n - 1}{2}}{n \\cdot \\frac{2 + 2n}{2}} = \\frac{n}{n + 1} = \\frac{2011}{2012},\n$$\n\nfrom which it follows that $n = 2011$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 14203,
    "subject": "Mathematics (Olympiad)",
    "question": "Тус бүрдээ нэгээс их $a, b$ натурал тоонуудын хувьд $a^b$ хэлбэртэй бичигдэх тоо 1-1000 хүртэлх тоонууд дунд хичнээн байх вэ?",
    "options": [],
    "answer": "See solution",
    "solution": "$a > 1,\\ b > 1$ гэдгээс $a = 2$ гэе.\n\n$$\n1 < 2^b < 1000 \\implies b = 2,\\ 3,\\ 4,\\ 5,\\ 6,\\ 7,\\ 8,\\ 9 \\quad (8\\ \\text{боломж})\n$$\n\n$a = 3$ гэе.\n\n$$\n1 < 3^b < 1000 \\implies b = 2,\\ 3,\\ 4,\\ 5,\\ 6 \\quad (5\\ \\text{боломж})\n$$\n\n$a = 4$ гэе.\n\n$$\n1 < 4^b < 1000 \\implies b = 2,\\ 3,\\ 4 \\quad (3\\ \\text{боломж})\n$$\n\n$a = 5$ гэе.\n\n$$\n1 < 5^b < 1000 \\implies b = 2,\\ 3 \\quad (2\\ \\text{боломж})\n$$\n\n$a = 6$ гэе.\n\n$$\n1 < 6^b < 1000 \\implies b = 2,\\ 3 \\quad (2\\ \\text{боломж})\n$$\n\nҮүнтэй адил $a = 7,\\ 8,\\ 9$ үед $b = 2,\\ 3$ гэсэн 2 боломжтой. Харин $a = 10,\\ 11,\\ \\dots,\\ 31$ үед $b = 2$ гэсэн ганц боломжтой. Эдгээрийг нэмбэл:\n\n$$\n8 + 5 + 3 + 5 \\times 2 + 22 \\times 1 = 50\n$$\n\nТиймээс 1-1000 хүртэлх тоонууд дунд $a^b$ ($a > 1,\\ b > 1$) хэлбэртэй бичигдэх 50 тоо байна. $\\blacktriangle$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 14204,
    "subject": "Mathematics (Olympiad)",
    "question": "The sequence is given by the formula $a_n = 2 \\cdot n! + 1$ for $n \\ge 0$. (We use the usual definition $0! = 1$, which satisfies $1! = 1 \\cdot 0!$, in the same way we have $n! = n \\cdot (n-1)!$ for other positive integers $n$.)\n\nFind all positive integers $m \\ge 2$ such that $\\gcd(m, a_n) = 1$ for all $n \\ge 0$.",
    "options": [],
    "answer": "See solution",
    "solution": "We will prove the formula for $a_n$ by induction. We have $a_0 = 3$, which equals $2 \\cdot 0! + 1$. Now suppose for certain $k \\ge 0$ that $a_k = 2 \\cdot k! + 1$, then\n\n$$\na_{k+1} = a_k + k(a_k - 1) = 2 \\cdot k! + 1 + k \\cdot 2 \\cdot k! = 2 \\cdot k! \\cdot (1+k) + 1 = 2 \\cdot (k+1)! + 1.\n$$\n\nThis finishes the induction.\n\nWe see that $a_n$ is always odd, hence $\\gcd(2, a_n) = 1$ for all $n$. It follows also that $\\gcd(2^i, a_n) = 1$ for all $i \\ge 1$. Hence, $m = 2^i$ with $i \\ge 1$ satisfies the condition.\n\nNow consider an $m \\ge 2$ which is not a power of two. Then $m$ has an odd prime divisor, say $p$. We will show that $p$ is a divisor of $a_{p-3}$. By Wilson's theorem, we have $(p-1)! \\equiv -1 \\pmod{p}$. Hence,\n\n$$\n\\begin{aligned}\n2 \\cdot (p-3)! &\\equiv 2 \\cdot (p-1)! \\cdot ((p-2)(p-1))^{-1} \\\\\n&\\equiv 2 \\cdot -1 \\cdot (-2 \\cdot -1)^{-1} \\equiv 2 \\cdot -1 \\cdot 2^{-1} \\equiv -1 \\pmod{p}.\n\\end{aligned}\n$$\n\nSo indeed we have $a_{p-3} = 2 \\cdot (p-3)! + 1 \\equiv 0 \\pmod{p}$. We conclude that $m$ does not satisfy the condition. Hence, the only values of $m$ satisfying the condition are powers of two. $\\square$",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 14205,
    "subject": "Mathematics (Olympiad)",
    "question": "Изпъкнал 2009-ъгълник е разбит на триъгълници чрез непресичащи се диагонали. Един от тези диагонали е оцветен в зелено. Разрешена е следната операция: за два триъгълника *ABC* и *BCD* от разбиването с обща страна $BC$ можем да заменим диагонала $BC$ с диагонала $AD$, като, ако замененият диагонал е бил зелен, той губи цвета си и заменилият го диагонал става зелен. Да се докаже, че всеки предварително избран диагонал на 2009-ъгълника може да бъде оцветен в зелено чрез прилагане на разрешената операция краен брой пъти.",
    "options": [],
    "answer": "See solution",
    "solution": "Първо ще докажем, че за даден връх на изпъкналия 2009-ъгълник и всяка триангулация, с прилагане на разрешената операция можем да получим триангулацията, получена от прекарването на всички диагонали през този връх. За произволен връх $A$, движейки се обратно на часовниковата стрелка, да означим с $B_1, B_2, \\dots, B_k$ последователните върхове, за които $AB_i$ е страна на дадения многоъгълник или диагонал в дадената триангулация. Ако отсечката $B_iB_{i+1}$ не е страна, тя е диагонал и след извършване на разрешената операция, ще получим нова триангулация, от която излизащите от $A$ диагонали са с един повече. Продължавайки по този начин, ще получим триангулация с диагонали само от върха $A$.\n\nЩе докажем по индукция по $n \\ge 4$, че твърдението е вярно за произволен изпъкнал $n$-ъгълник. При $n = 4, 5$ твърдението се проверява директно. Да допуснем, че твърдението е вярно за някое $k \\ge 5$ и да разгледаме триангулация на изпъкнал $(k+1)$-ъгълник. Без ограничение приемаме, че избраният диагонал е $A_1A_i$. Съгласно доказаното, от дадената триангулация можем да получим триангулацията, получена с прекарването на всички диагонали през $A_1$. Ако при това $A_1A_i$ е станал зелен, задачата е решена. Нека зелен е станал диагоналът $A_1A_j$, като без ограничение считаме, че $j < i$. От индукционното допускане следва, че в многоъгълника $A_1A_2\\ldots A_i$ можем да получим триангулация, в която диагоналът $A_1A_{i-1}$ е зелен. Тъй като $k \\ge 5$ и всяка триангулация на $(k+1)$-ъгълник съдържа $k-2$ диагонала, то в триангулацията освен диагоналите $A_1A_{i-1}$ и $A_1A_i$ има поне още един диагонал. Този диагонал разделя $(k+1)$-ъгълника на два изпъкнали многоъгълника, всеки с по-малко от $k+1$ върха, като диагоналите $A_1A_{i-1}$ и $A_1A_i$ са в един от двата многоъгълника. Остава да приложим индукционното допускане за този многоъгълник.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 14206,
    "subject": "Mathematics (Olympiad)",
    "question": "For positive numbers $x$, $y$, $z$, prove the inequality:\n\n$$\n\\frac{1}{3}(x^3 + y^3 + z^3) \\geq xyz + \\frac{2}{9}(x + y + z)(x - z)^2.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "From the known identity:\n\n$$\nx^3 + y^3 + z^3 - 3xyz = (x + y + z) \\cdot \\frac{(x-y)^2 + (y-z)^2 + (z-x)^2}{2}.\n$$\n\nFrom the inequality between the arithmetic mean and the root mean square, we have:\n\n$$\n\\begin{aligned}\n(x - y)^2 + (y - z)^2 + (z - x)^2 &\\geq \\frac{1}{3}(|x - y| + |y - z| + |z - x|)^2 \\\\ &\\geq \\frac{1}{3}(|x - y| + |y - z| + |x - z|)^2 = \\frac{4}{3}(x - z)^2.\n\\end{aligned}\n$$\n\nFrom this, the required inequality follows.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 14207,
    "subject": "Mathematics (Olympiad)",
    "question": "Prove that among any 20 consecutive positive integers, there exists an integer $d$ such that for each positive integer $n$ we have the inequality\n\n$$\nn\\sqrt{d}\\{n\\sqrt{d}\\} > \\frac{5}{2}$$\n\nwhere $\\{x\\}$ denotes the fractional part of the real number $x$. The fractional part of a real number $x$ is $x$ minus the greatest integer less than or equal to $x$.",
    "options": [],
    "answer": "See solution",
    "solution": "Among the given numbers, there is a number of the form $20k + 15 = 5(4k + 3)$. We shall prove that $d = 5(4k + 3)$ satisfies the statement's condition. Since $d \\equiv -1 \\pmod{4}$, it follows that $d$ is not a perfect square, and thus for any $n \\in \\mathbb{N}$ there exists $a \\in \\mathbb{N}$ such that $a + 1 > n\\sqrt{d} > a$, that is, $(a + 1)^2 > n^2 d > a^2$.\n\nActually, we are going to prove that $n^2 d \\geq a^2 + 5$. Indeed:\n\nIt is known that each positive integer of the form $4s + 3$ has a prime divisor of the same form. Let $p \\mid 4k + 3$ and $p \\equiv -1 \\pmod{4}$. Because of the form of $p$, the numbers $a^2 + 1^2$ and $a^2 + 2^2$ are not divisible by $p$, and since $p \\mid n^2 d$, it follows that $n^2 d \\neq a^2 + 1, a^2 + 4$. On the other hand, $5 \\mid n^2 d$, and since $5 \\nmid a^2 + 2, a^2 + 3$, we conclude $n^2 d \\neq a^2 + 2, a^2 + 3$. Since $n^2 d > a^2$, we must have $n^2 d \\geq a^2 + 5$ as claimed. Therefore,\n\n$$\nn\\sqrt{d}\\{n\\sqrt{d}\\} = n\\sqrt{d}(n\\sqrt{d} - a) \\geq a^2 + 5 - a\\sqrt{a^2 + 5} > a^2 + 5 - \\frac{a^2 + (a^2 + 5)}{2} = \\frac{5}{2},$$\n\nwhich was to be proved. $\\square$",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 14208,
    "subject": "Mathematics (Olympiad)",
    "question": "The obtuse-angled triangle $ABC$ has sides of length $a$, $b$, and $c$ opposite the angles $A$, $B$, and $C$ respectively. Prove that\n\n$$\na^3 \\cos A + b^3 \\cos B + c^3 \\cos C < abc.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "From the cosine rule, $a^2 + b^2 - c^2 = 2ab \\cos C$ is negative if and only if $\\angle C$ is obtuse. Exactly one of $\\angle A$, $\\angle B$, and $\\angle C$ is obtuse, so exactly one of $a^2 + b^2 - c^2$, $b^2 + c^2 - a^2$, and $c^2 + a^2 - b^2$ is negative.\n\nIt follows that\n\n$$\n(a^2 + b^2 - c^2)(b^2 + c^2 - a^2)(c^2 + a^2 - b^2) < 0\n$$\n\nwhich rearranges to\n\n$$\n-a^6 - b^6 - c^6 + a^4 b^2 + a^2 b^4 + b^4 c^2 + b^2 c^4 + c^4 a^2 + c^2 a^4 < 2a^2 b^2 c^2.\n$$\n\nWe may divide by $2abc$, which is positive, and group terms to obtain\n\n$$\n\\frac{a^3 (b^2 + c^2 - a^2)}{2bc} + \\frac{b^3 (c^2 + a^2 - b^2)}{2ca} + \\frac{c^3 (a^2 + b^2 - c^2)}{2ab} < abc,\n$$\n\nand so, applying the cosine rule, we have proved the inequality in the question:\n\n$$\na^3 \\cos A + b^3 \\cos B + c^3 \\cos C < abc.\n$$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 14209,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $a_1 > a_2 > \\dots > a_n > 1$ be positive integers. Let $M$ denote the least common multiple of $a_1, a_2, \\dots, a_n$. For a finite set of integers $X$, define\n\n$$\nf(X) = \\min_{1 \\le i \\le n} \\sum_{x \\in X} \\left\\{ \\frac{x}{a_i} \\right\\}\n$$\n\nwhere $\\{u\\} = u - \\lfloor u \\rfloor$ is the fractional part of the real number $u$. Put $f(\\emptyset) = 0$. We say a set $X$ is *minimal* if for any proper subset $Y \\subsetneq X$, we have $f(Y) < f(X)$.\n\nProve that if $X$ is a minimal finite set of integers and if $f(X) \\ge \\frac{2}{a_n}$, then\n\n$$\n|X| \\le f(X) \\cdot M\n$$\n\nwhere $|X|$ denotes the number of elements in $X$.",
    "options": [],
    "answer": "See solution",
    "solution": "Assume $f(X) = \\lambda \\ge \\frac{2}{a_n}$. Suppose, for contradiction, that $|X| > \\lambda M$. Since $\\lambda$ is of the form $\\frac{k}{a_i}$ for some $k \\in \\mathbb{Z}_{>0}$, $\\lambda M$ is an integer. We will show there exists $x \\in X$ such that $f(X \\setminus \\{x\\}) = f(X)$, contradicting the minimality of $X$.\n\nFor $1 \\le i \\le n$, let $X_i = \\{x \\in X \\mid a_i \\nmid x\\}$.\n\nConsider all indices $i$ with $|X_i| \\le \\lceil \\lambda a_i \\rceil$, and denote these indices by $i_1 < i_2 < \\dots < i_m$. If\n\n$$\nX_{i_1} \\cup X_{i_2} \\cup \\dots \\cup X_{i_m} \\neq X,\n$$\n\ntake $x \\in X \\setminus (X_{i_1} \\cup X_{i_2} \\cup \\dots \\cup X_{i_m})$, and let $Y = X \\setminus \\{x\\}$. Then $f(Y) = f(X)$.\n\nFor $Y_i = \\{y \\in Y \\mid a_i \\nmid y\\}$:\n- If $i \\in \\{i_1, \\dots, i_m\\}$, then $X_i = Y_i$, so\n  $$\n  \\sum_{y \\in Y} \\left\\{ \\frac{y}{a_i} \\right\\} = \\sum_{x \\in X} \\left\\{ \\frac{x}{a_i} \\right\\} \\ge \\lambda.\n  $$\n- If $i \\notin \\{i_1, \\dots, i_m\\}$, then $|Y_i| \\ge |X_i| - 1 \\ge \\lceil \\lambda a_i \\rceil$, so\n  $$\n  \\sum_{y \\in Y} \\left\\{ \\frac{y}{a_i} \\right\\} \\ge |Y_i| \\cdot \\frac{1}{a_i} \\ge \\lambda.\n  $$\n\nThus, $f(Y) \\ge \\lambda$. Since $f(Y) \\le f(X) = \\lambda$, we have $f(Y) = f(X)$.\n\nNow, we prove $X_{i_1} \\cup \\dots \\cup X_{i_m} \\neq X$. For $1 \\le j \\le m$, let $T_j = X_{i_1} \\cup \\dots \\cup X_{i_j}$ and $M_j = \\mathrm{lcm}(a_{i_1}, \\dots, a_{i_j})$. Clearly, $|T_1| \\le \\lceil \\lambda M_1 \\rceil$.\n\nFor $2 \\le j \\le m$, $|T_j \\setminus T_{j-1}| \\le \\lceil \\lambda M_j \\rceil - \\lceil \\lambda M_{j-1} \\rceil$. If $M_j = M_{j-1}$, then $T_j = T_{j-1}$, so $|T_j \\setminus T_{j-1}| = 0$. If $M_j > M_{j-1}$, let $d = \\gcd(a_{i_j}, M_{j-1})$, $M_{j-1} = du$, $a_{i_j} = dv$, $u > v > 1$, $M_j = duv$.\n\n$$\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.\n$$\n\nThe last inequality holds since $\\lambda a_{i_j} \\le \\lambda M_j - \\lambda M_{j-1} - 1$; with $\\lambda \\ge \\frac{2}{a_n} \\ge \\frac{2}{a_{i_j}}$, this is satisfied for $u > v > 1$.\n\nTherefore,\n$$\n|T_m| \\le \\lceil \\lambda M_m \\rceil \\le \\lceil \\lambda M \\rceil = \\lambda M.\n$$\n\nThis proves the claim by contradiction, so $|X| \\le f(X) \\cdot M$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 14210,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $\\triangle ABC$ be an acute triangle. Point $D$ lies on side $BC$. Let $O_B$ and $O_C$ be the circumcenters of triangles $ABD$ and $ACD$, respectively. Suppose that points $B$, $C$, $O_C$, $O_B$ lie on a circle centered at point $X$. Let $H$ be the orthocenter of triangle $ABC$. Prove that $\\angle DAX = \\angle DAH$.\n\n![](images/pamphlet0910_main_p21_data_4a076acd19.png)\n![](images/pamphlet0910_main_p21_data_3df9772cb5.png)",
    "options": [],
    "answer": "See solution",
    "solution": "Assume on the contrary that $AB \\le AC$. Then $\\angle ACB \\le \\angle CBA$, implying that $\\angle DAC = \\angle ADB - \\angle ACB \\ge \\angle CDA - \\angle CBA = \\angle BAD$. We obtain the right-hand side figure shown above. By the proof of (b), we have\n\n$$\n\\angle BO_B O_C = \\angle BO_B A - \\angle O_C O_B A = \\angle CO_C A - \\angle CBA = \\angle O_C CB + \\angle BAO_C = \\angle O_C CD + \\angle BAO_C.\n$$\n\nBecause $O_C$ is the circumcenter of $ACD$, $\\angle O_C CD = 90^\\circ - \\angle DAC$. Consequently, we have\n\n$$\n\\begin{aligned}\n\\angle O_C CB + \\angle BO_B O_C &= 2\\angle O_C CD + \\angle BAO_C = 180^\\circ - 2\\angle DAC + \\angle BAO_C \\\\\n&= 180^\\circ - \\angle DAC - \\angle O_C AC + \\angle BAD \\le 180^\\circ - \\angle O_C AC < 180^\\circ,\n\\end{aligned}\n$$\n\ncontradicting the fact that $BO_B O_C C$ is cyclic. Hence our assumption was wrong and we only need to consider the left configuration shown above.\n\n**Solution 1.** First, we claim that, if $I$ is the incenter of triangle $ABC$, then $AD$ trisects $\\angle HAI$ with $\\angle IAH = 3\\angle IAD$. Extend rays $AI$ and $AH$ to meet side $BC$ at $I_A$ and $H_A$, respectively. We want to show that $D$ lies on segment $I_A H_A$ with $\\angle I_A A H_A = 3\\angle I_A A D$. Since $\\angle A H_A C = 90^\\circ < \\angle ADB$, $D$ lies on segment $B H_A$. Since $B O_B O_C C$ is cyclic, we have\n\n$$\n\\begin{aligned}\n180^\\circ &= \\angle CBO_B + \\angle O_B O_C C = (\\angle ABO_B + \\angle CBA) + (360^\\circ - \\angle AOCO_B - \\angle CO_C A) \\\\\n&= \\angle ACO_C + \\angle CBA + 360^\\circ - \\angle ACB - 2\\angle CDA \\\\\n&= (90^\\circ - \\angle CDA) + \\angle CBA + 360^\\circ - \\angle ACB - 2\\angle CDA,\n\\end{aligned}\n$$\n\nor $\\angle ACB - \\angle CBA = 3(90^\\circ - \\angle CDA)$. Let $H_1$ be the point on side $BC$ such that $\\angle BAH_1 = \\angle CAH_A$. It follows that\n\n$$\n\\begin{aligned}\n\\angle H_1 A H_A &= \\angle BAH_A - \\angle H_A A C = (90^\\circ - \\angle CBA) - (90^\\circ - \\angle ACB) \\\\\n&= \\angle BCA - \\angle ABC = 3(90^\\circ - \\angle ADC) = 3\\angle DAH_A.\n\\end{aligned}\n$$\n\nIt is also clear that $AI_A$ bisects $\\angle H_1 A H_A$. It is not difficult to see that we have the left configuration shown below and that $\\angle D A H_A = 2\\angle I_A A D$, as claimed. To complete the proof of our main result, it suffices to show that $AI_A$ bisects $\\angle X A D$; that is,\n\n$$\n\\angle BAX = \\angle DAC. \\qquad (1)\n$$\n\n![](images/pamphlet0910_main_p22_data_620b74f28c.png)\n\nWe claim that $ABXO_C$ is cyclic. Indeed, by noting that triangles $BXC$ and $O_CXC$ are both isosceles with $BX = O_CX = CX$, we have\n\n$$\n\\begin{align*}\n\\angle XBA + \\angle AOCX &= (\\angle CBA + \\angle CBX) + (\\angle AOCO_B + \\angle O_BOCX) \\\\\n&= \\angle CBA + \\angle BCX + \\angle ACB + \\angle O_BOCX \\\\\n&= \\angle CBA + \\angle BCX + (\\angle O_CCB + \\angle ACO_C) + \\angle O_BOCX \\\\\n&= (\\angle CBA + \\angle ACO_C) + \\angle O_BOCX + (\\angle BCX + \\angle O_CCB) \\\\\n&= (\\angle CBA + \\angle ABO_B) + \\angle O_BOCX + \\angle O_CCX \\\\\n&= \\angle CBO_B + \\angle O_BOCX + \\angle XOC_C = \\angle CBO_B + \\angle O_BOC_C;\n\\end{align*}\n$$\n\nthat is, $ABXO_C$ is cyclic if and only if $BO_BOC_C$ is cyclic; this means $ABXO_C$ is cyclic. (This calculation applies when $X$ and $A$ are on opposite sides of line $BC$; but when they are on the same side of $BC$, the calculation is similar.)\n\nBecause $ABXO_C$ is cyclic and $O_C$ is the circumcenter of $ACD$, it follows that\n\n$$\n\\angle BAX = \\angle BO_CX = 90^\\circ - \\frac{\\angle O_CXB}{2} = 90^\\circ - \\angle O_CCB = 90^\\circ - \\angle O_CCD = \\angle CAD,\n$$\n\nwhich is (1). Our proof is complete.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 14211,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $p$ be a positive integer with $p > 1$. Find the number of $m \\times n$ matrices with entries in the set $\\{1, 2, 3, \\ldots, p\\}$ such that the sum of the elements in each row and each column is not divisible by $p$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $X$ be the set of all $m \\times n$ matrices with entries in $\\{1, 2, \\ldots, p\\}$. Let $A_i$ be the set of matrices where the sum of the elements in row $i$ is divisible by $p$, and $B_j$ the set where the sum in column $j$ is divisible by $p$. We seek the cardinality\n\n$$\nN = \\left| X \\setminus \\left( \\bigcup_{i=1}^{m} A_i \\cup \\bigcup_{j=1}^{n} B_j \\right) \\right|.\n$$\n\nBy the inclusion-exclusion principle,\n\n$$\nN = p^{mn} + \\sum_{i=0}^{m} \\sum_{j=0}^{n} (-1)^{i+j} \\sum_{\\substack{k_1 < \\cdots < k_i \\\\ l_1 < \\cdots < l_j}} |A_{k_1} \\cap \\cdots \\cap A_{k_i} \\cap B_{l_1} \\cap \\cdots \\cap B_{l_j}|,\n$$\n\nwhere the sum is over all $i, j$ with $i + j \\neq 0$.\n\nWe have\n\n$$\n|A_{k_1} \\cap \\cdots \\cap A_{k_i} \\cap B_{l_1} \\cap \\cdots \\cap B_{l_j}| =\n\\begin{cases}\np^{mn-i-j}, & \\text{if } i \\neq m \\text{ or } j \\neq n \\\\\np^{mn-m-n+1}, & \\text{if } i = m \\text{ and } j = n\n\\end{cases}\n$$\n\nTherefore,\n\n$$\n\\begin{align*}\nN &= \\sum_{i=0}^{m} \\sum_{j=0}^{n} (-1)^{i+j} \\binom{m}{i} \\binom{n}{j} p^{mn-i-j} \\\\\n&\\quad + (-1)^{m+n} \\binom{m}{m} \\binom{n}{n} (p^{mn-m-n+1} - p^{mn-m-n}) \\\\\n&= \\sum_{s=0}^{m+n} (-1)^s p^{mn-s} \\sum_{i+j=s} \\binom{m}{i} \\binom{n}{j} + (-1)^{m+n} p^{mn-m-n} (p-1)\n\\end{align*}\n$$\n\nUsing the binomial theorem, this simplifies to\n\n$$\nN = p^{mn-m-n} \\left( (p-1)^{m+n} + (-1)^{m+n} (p-1) \\right).\n$$",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 14212,
    "subject": "Mathematics (Olympiad)",
    "question": "Consider a general problem with cards numbered $1, 2, \\ldots, n$. Let $F_n$ be the number of choices when there are $n$ cards. Given $F_1 = 1$ and $F_2 = 2$. For $k \\geq 3$, the recurrence is:\n\n$$\nF_k = F_{k-1} + F_{k-2} + 1\n$$\n\nCalculate $F_{15}$ using this relation.",
    "options": [],
    "answer": "See solution",
    "solution": "We have the recurrence $F_k = F_{k-1} + F_{k-2} + 1$ with $F_1 = 1$ and $F_2 = 2$.\n\nCalculating step by step:\n\n- $F_3 = F_2 + F_1 + 1 = 2 + 1 + 1 = 4$\n- $F_4 = F_3 + F_2 + 1 = 4 + 2 + 1 = 7$\n- $F_5 = F_4 + F_3 + 1 = 7 + 4 + 1 = 12$\n- $F_6 = F_5 + F_4 + 1 = 12 + 7 + 1 = 20$\n- $F_7 = F_6 + F_5 + 1 = 20 + 12 + 1 = 33$\n- $F_8 = F_7 + F_6 + 1 = 33 + 20 + 1 = 54$\n- $F_9 = F_8 + F_7 + 1 = 54 + 33 + 1 = 88$\n- $F_{10} = F_9 + F_8 + 1 = 88 + 54 + 1 = 143$\n- $F_{11} = F_{10} + F_9 + 1 = 143 + 88 + 1 = 232$\n- $F_{12} = F_{11} + F_{10} + 1 = 232 + 143 + 1 = 376$\n- $F_{13} = F_{12} + F_{11} + 1 = 376 + 232 + 1 = 609$\n- $F_{14} = F_{13} + F_{12} + 1 = 609 + 376 + 1 = 986$\n- $F_{15} = F_{14} + F_{13} + 1 = 986 + 609 + 1 = 1596$\n\nThus, $F_{15} = 1596$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 14213,
    "subject": "Mathematics (Olympiad)",
    "question": "Positive integers $p$, $a$, and $b$ satisfy the equation $p^2 + a^2 = b^2$. Prove that if $p$ is a prime greater than $3$, then $a$ is a multiple of $12$ and $2(p + a + 1)$ is a perfect square.",
    "options": [],
    "answer": "See solution",
    "solution": "Rearranging and factoring, we have that\n\n$$\np^2 = b^2 - a^2 = (b + a)(b - a).\n$$\n\nThe only positive factors of $p^2$ are $1$, $p$, and $p^2$. We cannot have $b + a = b - a = p$ since $a$ is positive, so it must be the case that $b + a = p^2$ and $b - a = 1$. So\n\n$$\nb = \\frac{p^2 + 1}{2} \\quad \\text{and} \\quad a = \\frac{p^2 - 1}{2}\n$$\n\nand therefore\n\n$$\n2a = p^2 - 1 = (p + 1)(p - 1).\n$$\n\nSince $p$ is odd, both $p + 1$ and $p - 1$ are even, and moreover one of them is divisible by $4$. Hence $8$ divides $2a$. Since $p$ is not divisible by $3$, one of $p + 1$ and $p - 1$ is divisible by $3$ and so $3$ divides $2a$. As $8$ and $3$ are coprime, it follows that $24$ divides $2a$ and $12$ divides $a$, as required.\n\nNow\n\n$$\n\\begin{aligned}\n2(p + a + 1) &= 2\\left(p + \\frac{p^2-1}{2} + 1\\right) \\\\\n&= 2p + p^2 - 1 + 2 \\\\\n&= p^2 + 2p + 1 \\\\\n&= (p + 1)^2\n\\end{aligned}\n$$\n\nwhich is to say that $2(p + a + 1)$ is a perfect square, as required.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 14214,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $f : \\mathbb{N} \\to \\mathbb{N}$ be a function such that for all $a, b \\in \\mathbb{N}$:\n\n1. $f(ab) = f(a)f(b)$,\n2. $f(a) + f(b) > f(a + b - 1)$.\n\nFind all such functions $f$.",
    "options": [],
    "answer": "See solution",
    "solution": "If $f(2) = 1$, then let $a = 2$, $b = 1$. The second condition implies that $2f(2) > f(1)$, so $f(1) = 1$. We can now use induction to show that $f(n) = 1$ for all $n$. This is true for the base case. Now, assume that $f(n) = 1$ for some $n \\ge 2$. Using $a = n, b = 2$ in the second condition we get\n\n$$\nf(n+1) < f(n) + f(2) = 2 \\implies f(n+1) = 1.\n$$\n\nThis gives us the solution $f(n) = 1$ for all $n \\in \\mathbb{N}$.\n\nIf $f(2) = 2$, then $f(4) = f(2)^2 = 4$. Moreover, for all $k \\in \\mathbb{N}$, induction on the first condition implies that\n\n$$\nf(2^k) = f(2)f(2^{k-1}) = \\dots = f(2)^k = 2^k.\n$$\n\nWe know from before that $f(4) - f(2) < f(3) < 2f(2)$, which implies that $f(3) = 3$.\n\nWe now use induction to show that $f(n) = n$ for all $n \\ge 2$. The base case is obvious. Now, assume that the induction hypothesis holds for $2, 3, \\ldots, n-1$. Using $a = n-1, b = 2$ we get\n\n$$\nf(n) < f(n-1) + f(2) = n+1 \\implies f(n) \\le n.\n$$\n\nLet $2^r$ be the greatest power of $2$ such that $2^r \\le n$. If $2^r = n$, then the induction step is concluded because $f(2^r) = 2^r$. Otherwise, let $n = 2^r + s$, where $1 \\le s < 2^r$.\n\nWe will use $a = n = 2^r + s$, $b = 2^r - s + 1$ with the second condition. Since $2^r - s + 1 \\ge 2$, the induction hypothesis implies that $f(2^r - s + 1) = 2^r - s + 1$. So,\n\n$$\n\\begin{align*}\nf(n) + f(2^r - s + 1) &> f(2^r + s + 2^r - s + 1 - 1) = f(2^{r+1}) \\\\\n\\implies f(n) &> f(2^{r+1}) - f(2^r - s + 1) = 2^r + s - 1 = n - 1 \\\\\n\\implies f(n) &\\ge n.\n\\end{align*}\n$$\n\nWe have shown that $f(n) = n$, which concludes the induction step. All that remains is to determine $f(1)$. The only condition is that $f(1) < 2f(2) = 4$. So, $f(1)$ can be either $1$, $2$, or $3$.\n\nWe have obtained the solutions $f(n) = 1$ for all $n$ and $f(n) = n$ for $n \\ge 2$ with $f(1) \\in \\{1, 2, 3\\}$. It is easy to check that all of these satisfy the conditions of the problem.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 14215,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $a_1 < a_2 < \\dots$ be the positive divisors of a positive integer $a$ and let $b_1 < b_2 < \\dots$ be the positive divisors of a positive integer $b$. Find all $a, b$ such that\n\n$$\n\\begin{cases} a_{10} + b_{10} = a \\\\ a_{11} + b_{11} = b \\end{cases}\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "Answer: $a = 2^{10}$ and $b = 2^{11}$.\n\nIf $b = b_{11}$ then $a_{11} = 0$, which is impossible. Thus $b \\geq 2b_{11}$.\n\nLet $a = a_{10} \\cdot n$. Since $a > a_{10}$, we have $n \\geq 2$ and\n\n$$\nb_{10} = a - a_{10} = a_{10} \\cdot (n - 1) \\geq a_{10}.\n$$\n\nHence $2b_{10} \\geq b_{10} + a_{10} = a$. It follows that $2b_{11} > 2b_{10} \\geq a \\geq a_{11}$ and thus $3b_{11} > b_{11} + a_{11} = b$. This implies $b = 2b_{11}$ and $a_{11} = b_{11}$.\n\nThen we have $b = 2a_{11} > a > a_{10}$. This means $a = a_{11}$. Thus $a$ has 11 divisors and since 11 is a prime number, $a = p^{10}$ for a prime $p$. Then $b = 2 \\cdot p^{10}$ and it is easy to conclude that $p = 2$. Thus $a = 2^{10}$, $b = 2^{11}$ is the solution.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 14216,
    "subject": "Mathematics (Olympiad)",
    "question": "Let a triangle $ABC$ with $|AC| < |AB|$ be given, together with its circumcircle. Let $D$ be a varying point on the short arc $AC$. Let $E$ be the reflection of $A$ in the internal angle bisector of $\\angle BDC$. Prove that the line $DE$ passes through a fixed point, independent of where $D$ lies.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $M$ be the intersection of the internal angle bisector of $\\angle BDC$ with the circumcircle of $\\triangle ABC$. As $D$ lies on the short arc $AC$, $M$ lies on the arc $BC$ not containing $A$. We have $\\angle BDM = \\angle MDC$ as $DM$ is the internal angle bisector of $\\angle BDC$, so arcs $BM$ and $CM$ have equal lengths. Hence $M$ is independent of $D$.\n\n![](images/NLD_ABooklet_2020_p34_data_aa019ef11c.png)\n\nLet $S$ be the intersection of $DE$ and the circumcircle of $\\triangle ABC$. We show that $S$ is independent of $D$. As $S$ and $M$ lie on the circumcircle of $\\triangle ABC$, we have $\\angle AMD = \\angle ASD = \\angle ASE$. As $E$ is the reflection of $A$ in $DM$, we have $\\angle AME = 2\\angle AMD = 2\\angle ASE$.\n\nConsider the circle with center $M$ passing through $A$. As $E$ is the reflection of $A$ in $DM$, $|MA| = |ME|$, so this circle also passes through $E$. By the inscribed angle theorem, from $\\angle AME = 2\\angle ASE$ it follows that $S$ is also on this circle. Therefore, $S$ is the second intersection point of the circumcircle of $\\triangle ABC$ and the circle with center $M$ passing through $A$. This is a description of $S$ independent of $D$. As $DE$ passes through $S$, the point $S$ is the required fixed point. $\\Box$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 14217,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $ABCDEF$ be a convex hexagon in which triangles $ABC$ and $DEF$ are equilateral, and the diagonals $AD$, $BE$, and $CF$ are concurrent. Prove that $AC \\parallel DF$ or $BE = AD + CF$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let points $X \\in AD$ and $Y \\in CF$ be such that $EX \\parallel AB$ and $EY \\parallel BC$. Denote $\\{T\\} = AD \\cap BE \\cap CF$. Clearly, $\\angle XEY = 60^\\circ$.\n\nIf $X = D$, from $\\angle XEY = 60^\\circ$ we get $Y = F$ and $\\frac{CT}{TF} = \\frac{BT}{TE} = \\frac{AT}{TD}$, thus $AC \\parallel DF$.\n\nIf $X \\ne D$ (and therefore $Y \\ne F$), we have $\\frac{XE}{AB} = \\frac{ET}{TB} = \\frac{YE}{BC}$, from which $XE = YE$.\n\nSince $\\angle XEF \\equiv \\angle YEF$, it follows that $\\triangle XED \\equiv \\triangle YEF$. Hence, $\\angle TFE = \\angle XDE = 180^\\circ - \\angle TDE$, so $T$ lies on the circumscribed circle of triangle $DEF$.\n\nBy symmetry, $T$ also lies on the circumscribed circle of triangle $ABC$.\n\nApplying Ptolemy's theorem, we deduce $TE = TD + TF$ and $TB = TA + TC$.\n\nAdding these two equalities, we get $BE = AD + CF$.\n\n![](images/RMC_2025_p67_data_48e4d8c22c.png)",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 14218,
    "subject": "Mathematics (Olympiad)",
    "question": "Given a natural number $n$. We have $n + 1$ balls numbered $1, 1, 2, 3, \\ldots, 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\n![](path/to/file.png)",
    "options": [],
    "answer": "See solution",
    "solution": "Exactly one of the colors will be used for two of the balls; let their numbers be $a$ and $b$, such that $a \\leq 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\n$$\n\\frac{1}{4} n! (n-1)! n (n-1)(n-2) = \\frac{1}{4} n! (n^2 - 3n + 2)\n$$\n\nIf $a = 1$, then there are $n!$ options for coloring all balls except $a$ in $n$ colors. Now there are $n$ choices for the color of $a$. Thus, in this case there are $n! \\cdot n$ possible colorings.\n\nFinally,\n$$\na_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 \\mid 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$.\n\n$\\boxed{13}$",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 14219,
    "subject": "Mathematics (Olympiad)",
    "question": "Suppose $\\triangle ABC$ has angles $\\angle BAC = 84^\\circ$, $\\angle ABC = 60^\\circ$, and $\\angle ACB = 36^\\circ$. Let $D$, $E$, and $F$ be the midpoints of sides $\\overline{BC}$, $\\overline{AC}$, and $\\overline{AB}$, respectively. The circumcircle of $\\triangle DEF$ intersects $\\overline{BD}$, $\\overline{AE}$, and $\\overline{AF}$ at points $G$, $H$, and $J$, respectively. The points $G$, $D$, $E$, $H$, $J$, and $F$ divide the circumcircle of $\\triangle DEF$ into six minor arcs, as shown. Find $\\overline{DE} + 2 \\cdot \\overline{HJ} + 3 \\cdot \\overline{FG}$, where the arcs are measured in degrees.\n\n![](images/2025AIME_II_Solutions_p3_data_f25ddd33ab.png)",
    "options": [],
    "answer": "See solution",
    "solution": "Because $D$, $E$, and $F$ are the midpoints of the sides, it follows that $\\triangle ABC$ is similar to the four congruent triangles $\\triangle AFE$, $\\triangle FBD$, $\\triangle EDC$, and $\\triangle DEF$. Because $\\angle DFE = \\angle FDG = \\angle FEH$, the Inscribed Angle Theorem implies that $\\overline{FJH} = \\overline{FG} = \\overline{DE} = 2 \\cdot \\angle DFE = 72^\\circ$. Also, $\\overline{DGF} = 2 \\cdot \\angle DEF = 120^\\circ$, so $\\overline{DG} = \\overline{DGF} - \\overline{FG} = 48^\\circ$. Because $\\angle BAC$ includes the arcs $\\overline{EDGF}$ and $\\overline{HJ}$, it follows that $\\angle BAC = \\frac{1}{2} (\\overline{EDGF} - \\overline{HJ})$, so $\\overline{HJ} = (\\overline{DE} + \\overline{DG} + \\overline{FG}) - 2 \\cdot \\angle BAC = 24^\\circ$. The requested value is $72 + 2 \\cdot 24 + 3 \\cdot 72 = 336$.\n\n**Note:** The circumcircle of $\\triangle DEF$ is known as the *Nine-Point Circle*, the *Euler Circle*, or the *Feuerbach Circle*. The points $G$, $H$, and $J$ are the feet of the altitudes of $\\triangle ABC$ from $A$, $B$, and $C$, respectively.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 14220,
    "subject": "Mathematics (Olympiad)",
    "question": "Өгөгдсөн нөхцөлөөс $B$ цэг нь $DA$ ба $DC$ шулуунуудыг шүргэсэн бөгөөд $A$, $C$ цэгүүдийг дайрсан тойрог дээр оршино гэж гарна. $U$ цэг уг тойрог дээр оршино гэдгийг дараах Паскалийн урвуу теорем ашиглан баталъя: Хэрэв гүдгэр зургаан өнцөгтийн 5 нь нэг тойрог дээр орших бөгөөд, эсрэг талуудын огтлолцол болох 3 цэг 1 шулуун дээр оршин байвал 6 дахь цэг уг тойрог дээр оршино. Одоо уг дүнг хэрхэн дөрвөн өнцөгт дээр ашиглахыг авч үзье. Үүнд: 6 өнцөгтийн аль нэгэн эсрэг 2 талыг нь цэг гэж үзээд харгалзах талууд нь тойрог татсан шүргэгч гэж үзнэ. $AD$ ба $CD$ нь $(ABC)$ тойргийн шүргэгч тул дээрх теоремыг хэрэглэвэл $U \\in (ABC)$ буюу $ABCU$ тойрогт багтана. Учир нь $AABCCU$ зургаан өнцөгт болон $F$, $E$, $D$ цэгүүд нэг шулуун дээр оршино гэдгийг ашиглав. Одоо\n\n$$\n\\angle DEU = \\angle UCA = \\angle UAD \\text{ гэж баталъя.}\n$$\n\n$$\n\\begin{array}{l}\nAC // EF \\Rightarrow \\angle DEU = \\angle UCA \\text{ ба} \\\\\nU \\in (ABC) \\hfill (1) \\\\\nAD \\text{ нь (ABC) тойргийн A цэг дээрх шүргэгч} \\hfill (2) \\\\\n\\Rightarrow \\angle UAD = \\angle UCA\n\\end{array}\n$$\n\n$$\n\\Rightarrow (1) \\text{ ба } (2)\\text{-оос } \\angle DEU = \\angle UCA = \\angle UAD.\n$$\n\n$$\n\\Rightarrow A, E, U, D \\text{ нэг тойрог дээр оршино.}\n$$\n\nЯг үүнтэй адилаар $C$, $F$, $D$, $U$ нэг тойрог дээр оршино. $U$ цэгийн $AB$, $BC$, $CD$, $DA$ талууд дээрх проекцуудийг харгалзан $P$, $Q$, $R$, $S$-г ол.",
    "options": [],
    "answer": "See solution",
    "solution": "Мөн $AC$, $EF$ шулуунууд дээрх проекцүүдийг харгалзан $V$, $T$ гэж тэмдэглэе. $U \\in (ABC)$ тул $P$, $V$, $Q$ цэгүүд нэг шулуун дээр оршино. Учир нь уг шулуун нь $(ABC)$-ийн Симпсоны шулуун юм. Иймд $U$ нь $AC$ болон $PQ$ хэрчмүүдийн ерөнхий цэг байна. Яг үүнтэй адилаар $U \\in (AED)$, $U \\in (CDF)$ тул $T \\in EF$, $T \\in PS$, $T \\in QR$ байна. Мөн $A$, $P$, $V$, $U$, $S$ цэгүүд; $C$, $Q$, $R$, $U$, $V$ цэгүүд; $D$, $T$, $S$, $U$, $R$ цэгүүд нэг тойрог дээр орших нь илэрхий.\n\n$$\n\\begin{align*}\n\\angle SPQ &= \\angle SPV = \\angle SAV = \\angle DAC = \\angle DCA \\\\\n&= \\angle RCV = \\angle RQV = \\angle RQP \\\\\n&\\Rightarrow \\angle SPQ = \\angle RQP \\\\\n\\angle TSR &= \\angle TUR = 180^{\\circ} - \\angle RUV = \\angle RQV = \\angle TQP \\\\\n&= \\angle TPQ \\Rightarrow SR // PQ.\n\\end{align*}\n$$\n\n$$\n\\Rightarrow PQRS \\text{ ади пажуу трапец.}\n$$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 14221,
    "subject": "Mathematics (Olympiad)",
    "question": "Is it possible for an integer of the form $44\\ldots41$—consisting of an odd number of fours followed by a $1$—to be a square?",
    "options": [],
    "answer": "See solution",
    "solution": "It is impossible for such an integer to be a square.\n\nTo show this, note that such an integer is of the form $$a_m = 4 \\cdot \\frac{10^{2m}-1}{9} - 3$$ with $m \\ge 1$ an integer. As $10^{2m}-1 = (10^2-1)(10^{2m-2} + 10^{2m-4} + \\dots + 1)$, we have $99 = 10^2 - 1 \\mid 10^{2m} - 1$, therefore $11 \\mid \\frac{10^{2m}-1}{9}$, and therefore $a_m \\equiv -3 \\equiv 8 \\pmod{11}$. However, the residue classes of squares modulo $11$ are $0, 1, 4, 9, 5, 3, 3, 5, 9, 4, 1$, respectively, so $a_m$ cannot be square. $\\square$\n\n![](images/NLD_ABooklet_2025_p30_data_fb335ea6a7.png)",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 14222,
    "subject": "Mathematics (Olympiad)",
    "question": "Given any integer $n \\ge 2$, show that there exists a set of $n$ pairwise coprime composite integers in arithmetic progression.",
    "options": [],
    "answer": "See solution",
    "solution": "Fix a prime $p > n$ and an integer $N \\ge p + (n-1)n!$. Consider the arithmetic progression of length $n$ consisting of the numbers $N! + p + k n!$ for $k = 0, 1, \\dots, n-1$.\n\nSuppose, for contradiction, that a prime $q$ divides two of these numbers. Then $q$ divides their difference, which is of the form $k n!$ for some $k < n$. Thus, $q \\leq n$, so $q$ divides $n!$ and $N!$, and hence also $p$, which is impossible since $p > n$ and is prime. Therefore, the numbers are pairwise coprime. Each number is composite because $N!$ is divisible by all primes up to $N$, so $N! + p + k n!$ is composite for sufficiently large $N$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 14223,
    "subject": "Mathematics (Olympiad)",
    "question": "The positive integers $a_0, a_1, a_2, \\dots, a_{3030}$ satisfy\n\n$$\n2a_{n+2} = a_{n+1} + 4a_n \\quad \\text{for } n = 0, 1, 2, \\dots, 3028.\n$$\n\nProve that at least one of the numbers $a_0, a_1, a_2, \\dots, a_{3030}$ is divisible by $2^{2020}$.",
    "options": [],
    "answer": "See solution",
    "solution": "We prove by induction, for $m \\geq 1$, the proposition that\n\n$a_n \\equiv 0 \\pmod{2^m}$ for $n = \\left\\lfloor \\frac{m}{2} \\right\\rfloor, \\dots, 3030 - m$.\n\n**Base cases:**\n\nFor $m = 1$:\n\nFrom the recurrence,\n$$\n2a_{n+2} = a_{n+1} + 4a_n\n$$\nfor $n = 0, 1, \\dots, 3028$.\n\nModulo $2$, $2a_{n+2} \\equiv 0 \\pmod{2}$, so $a_{n+1} + 4a_n \\equiv 0 \\pmod{2}$, i.e., $a_{n+1} \\equiv 0 \\pmod{2}$ for $n = 0, 1, \\dots, 3028$. Thus,\n\n$a_n \\equiv 0 \\pmod{2}$ for $n = 1, 2, \\dots, 3029$.\n\nFor $m = 2$:\n\nSince $a_{n+2} \\equiv 0 \\pmod{2}$ for $n = 0, 1, \\dots, 3027$, $2a_{n+2} \\equiv 0 \\pmod{4}$, so $a_{n+1} + 4a_n \\equiv 0 \\pmod{4}$, i.e., $a_{n+1} \\equiv 0 \\pmod{4}$ for $n = 0, 1, \\dots, 3027$. Thus,\n\n$a_n \\equiv 0 \\pmod{4}$ for $n = 1, 2, \\dots, 3028$.\n\n**Inductive step:**\n\nSuppose the proposition holds for $m = k-1$ and $m = k$ for some $k \\geq 2$.\n\nThen $a_n \\equiv 0 \\pmod{2^k}$ for $n = \\left\\lceil \\frac{k}{2} \\right\\rceil, \\dots, 3030 - k$ implies $a_{n+2} \\equiv 0 \\pmod{2^k}$ for $n = \\left\\lceil \\frac{k}{2} \\right\\rceil - 2, \\dots, 3028 - k$, so\n\n$2a_{n+2} \\equiv 0 \\pmod{2^{k+1}}$ for $n = \\left\\lceil \\frac{k}{2} \\right\\rceil - 2, \\dots, 3028 - k$.\n\nAlso, $a_n \\equiv 0 \\pmod{2^{k-1}}$ for $n = \\left\\lceil \\frac{k-1}{2} \\right\\rceil, \\dots, 3030 - (k-1)$, i.e., $n = \\left\\lceil \\frac{k+1}{2} \\right\\rceil - 1, \\dots, 3031 - k$, so\n\n$4a_n \\equiv 0 \\pmod{2^{k+1}}$ for $n = \\left\\lceil \\frac{k+1}{2} \\right\\rceil - 1, \\dots, 3031 - k$.\n\nSince $\\left\\lceil \\frac{k+1}{2} \\right\\rceil - 1 > \\left\\lceil \\frac{k}{2} \\right\\rceil - 2$ and $3028 - k < 3031 - k$, from the recurrence we have\n\n$a_{n+1} \\equiv 0 \\pmod{2^{k+1}}$ for $n = \\left\\lceil \\frac{k+1}{2} \\right\\rceil - 1, \\dots, 3028 - k$.\n\nThat is,\n\n$a_n \\equiv 0 \\pmod{2^{k+1}}$ for $n = \\left\\lceil \\frac{k+1}{2} \\right\\rceil, \\dots, 3030 - (k+1)$.\n\nThus, the induction is complete.\n\n**Conclusion:**\n\nLetting $m = 2020$, we have $a_n \\equiv 0 \\pmod{2^{2020}}$ for $n = 1010$. Thus, at least one of $a_0, a_1, \\dots, a_{3030}$ (namely $a_{1010}$) is divisible by $2^{2020}$.\n\n$\\boxed{}$",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 14224,
    "subject": "Mathematics (Olympiad)",
    "question": "In a plane rectangular coordinate system $xOy$, the graph of the parabola $y = ax^2 - 3x + 3$ ($a \\neq 0$) and that of the parabola $y^2 = 2px$ ($p > 0$) are symmetric with respect to the line $y = x + m$. Then the product of the real numbers $a, p, m$ is ______.",
    "options": [],
    "answer": "See solution",
    "solution": "For any point $(x_0, y_0)$ on the parabola $y = ax_0^2 - 3x_0 + 3$ ($a \\neq 0$),\n$$\ny_0 = ax_0^2 - 3x_0 + 3. \\tag{1}\n$$\nSuppose the symmetric point of $(x_0, y_0)$ with respect to the line $y = x + m$ is $(x_1, y_1)$.\n\nBy $\\frac{y_1 + y_0}{2} = \\frac{x_1 + x_0}{2} + m$ and $x_1 + y_1 = x_0 + y_0$, it follows that\n$$\nx_1 = y_0 - m, \\quad y_1 = x_0 + m.\n$$\nSince $(x_1, y_1)$ lies on the parabola $y^2 = 2px$, we have $(x_0 + m)^2 = 2p(y_0 - m)$. This is equivalent to\n$$\ny_0 = \\frac{1}{2p}x_0^2 + \\frac{m}{p}x_0 + \\frac{m^2}{2p} + m. \\tag{2}\n$$\nComparing (1) and (2),\n$$\n\\frac{1}{2p} = a, \\quad \\frac{m}{p} = -3, \\quad \\frac{m^2}{2p} + m = 3.\n$$\nSolving, $\\frac{m}{p} = -3$ and $\\frac{m^2}{2p} + m = 3$. Substitute $p = -\\frac{m}{3}$ into the third equation:\n$$\n\\frac{m^2}{2(-\\frac{m}{3})} + m = 3 \\implies -\\frac{3m}{2} + m = 3 \\implies -\\frac{m}{2} = 3 \\implies m = -6.\n$$\nThen $p = 2$, $a = \\frac{1}{4}$. Thus,\n$$\napm = \\frac{1}{4} \\times 2 \\times (-6) = -3.\n$$",
    "topic": "Discrete Mathematics",
    "subtopic": "Other"
  },
  {
    "id": 14225,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $123_a$ and $146_b$ be numbers written in bases $a$ and $b$, respectively. Find the minimum value of $a + b$ such that $123_a = 146_b$.",
    "options": [],
    "answer": "See solution",
    "solution": "$$\n\\begin{align*}\n123_a = 146_b &\\iff a^2 + 2a + 3 = b^2 + 4b + 6 \\\\\n&\\iff (a+1)^2 + 2 = (b+2)^2 + 2 \\\\\n&\\iff (a+1)^2 = (b+2)^2 \\\\\n&\\iff a+1 = b+2 \\quad (a \\text{ and } b \\text{ are positive}) \\\\\n&\\iff a = b+1\n\\end{align*}\n$$\n\nSince the digits in any number are less than the base, $b \\ge 7$.\n\nIf $b = 7$, then $a = 8$, so $a + b = 8 + 7 = 15$ is the minimum value.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 14226,
    "subject": "Mathematics (Olympiad)",
    "question": "For which positive integers $n$ does the following property hold for every function $f$?  \n\nIf for all $x_1, \\ldots, x_n$,\n$$\n  f\\left(\\frac{x_1 + x_2 + \\dots + x_n}{n}\\right) = \\frac{1}{n}\\left(f(x_1) + f(x_2) + \\dots + f(x_n)\\right),\n$$\nthen $f$ must be linear.",
    "options": [],
    "answer": "See solution",
    "solution": "Suppose that $n$ has the property in the problem. Then for every function $f$, if the property $()$ holds, we have:\n\n$$\n\\begin{aligned}\n  f\\left(\\frac{x_1 + x_2 + \\dots + x_{2n}}{2n}\\right) &= f\\left(\\frac{1}{n}\\left(x_1 + x_2 + \\frac{x_3 + x_4}{2} + \\dots + \\frac{x_{2n-1} + x_{2n}}{2}\\right)\\right) \\\\\n  &= \\frac{1}{n}\\left(f\\left(\\frac{x_1}{2}\\right) + f\\left(\\frac{x_3}{2}\\right) + \\dots + f\\left(\\frac{x_{2n-1}}{2}\\right)\\right) \\\\\n  &= \\frac{1}{n}\\left(\\frac{f(x_1) + f(x_2)}{2} + \\frac{f(x_3) + f(x_4)}{2} + \\dots + \\frac{f(x_{2n-1}) + f(x_{2n})}{2}\\right) \\\\\n  &= \\frac{1}{2n}(f(x_1) + f(x_2) + \\dots + f(x_{2n})),\n\\end{aligned}\n$$\n\nso the property $()$ also holds for $n' = 2n$ and $f$.\n\nLet us substitute $x_n = \\frac{x_1 + x_2 + \\dots + x_{n-1}}{n-1}$ in $()$. Then\n\n$$\n\\begin{aligned}\n  f\\left(\\frac{x_1 + x_2 + \\dots + x_{n-1}}{n-1}\\right) &= f\\left(\\frac{x_1 + x_2 + \\dots + x_n}{n}\\right) \\\\\n  &= \\frac{1}{n}\\left(f(x_1) + f(x_2) + \\dots + f(x_n)\\right) \\\\\n  &= \\frac{1}{n}\\left(f(x_1) + f(x_2) + \\dots + f(x_{n-1}) + f\\left(\\frac{x_1 + x_2 + \\dots + x_{n-1}}{n-1}\\right)\\right).\n\\end{aligned}\n$$\n\nTherefore,\n$$\nf\\left(\\frac{x_1 + x_2 + \\dots + x_{n-1}}{n-1}\\right) = \\frac{f(x_1) + f(x_2) + \\dots + f(x_{n-1})}{n-1},\n$$\nso $f$ and $n' = n-1$ also satisfy $()$. Using induction and backward induction, it is easy to check that the solution is the set of *all* positive integers.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 14227,
    "subject": "Mathematics (Olympiad)",
    "question": "Some numbers are printed on a spool of paper, one below another. Starting from the third number, each number is the sum of the two preceding numbers. The number $2018$ occurs fourth, the number $2020$ occurs eighth. Which are the first and the second numbers?",
    "options": [],
    "answer": "See solution",
    "solution": "Let the first and the second numbers be $x$ and $y$, respectively. Computing the following numbers, we see that the fourth number is $x + 2y$ and the eighth one is $8x + 13y$. The conditions of the problem imply the system of equations\n\n$$\n\\begin{cases}\nx + 2y = 2018 \\\\\n8x + 13y = 2020\n\\end{cases}\n$$\n\nSolving this system gives $x = -7398$ and $y = 4708$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 14228,
    "subject": "Mathematics (Olympiad)",
    "question": "Find the least positive integer $n$ such that $\\sqrt[5]{5n}$, $\\sqrt[6]{6n}$, and $\\sqrt[7]{7n}$ are integers.",
    "options": [],
    "answer": "See solution",
    "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$.\n\nThen:\n- $5n = 2^\\alpha \\cdot 3^\\beta \\cdot 5^{\\gamma+1} \\cdot 7^\\delta \\cdot s$\n- $6n = 2^{\\alpha+1} \\cdot 3^{\\beta+1} \\cdot 5^\\gamma \\cdot 7^\\delta \\cdot s$\n- $7n = 2^\\alpha \\cdot 3^\\beta \\cdot 5^\\gamma \\cdot 7^{\\delta+1} \\cdot s$\n\nFor $\\sqrt[5]{5n}$ to be an integer: $\\alpha$, $\\beta$, $\\gamma + 1$, and $\\delta$ must be divisible by $5$.\n\nFor $\\sqrt[6]{6n}$ to be an integer: $\\alpha + 1$, $\\beta + 1$, $\\gamma$, and $\\delta$ must be divisible by $6$.\n\nFor $\\sqrt[7]{7n}$ to be an integer: $\\alpha$, $\\beta$, $\\gamma$, and $\\delta + 1$ must be divisible by $7$.\n\nThus:\n- $\\alpha$ and $\\beta$ must be divisible by $\\operatorname{lcm}(5,6,7) = 210$ (but see below for minimal values).\n- $\\gamma$ must satisfy $\\gamma + 1 \\equiv 0 \\pmod{5}$, $\\gamma \\equiv 0 \\pmod{6}$, $\\gamma \\equiv 0 \\pmod{7}$.\n- $\\delta$ must satisfy $\\delta \\equiv 0 \\pmod{5}$, $\\delta \\equiv 0 \\pmod{6}$, $\\delta + 1 \\equiv 0 \\pmod{7}$.\n\nBy checking minimal positive solutions:\n- The least $\\alpha$ and $\\beta$ is $35$ (since $35$ is the smallest number divisible by $5$ and $7$, and $35+1=36$ is divisible by $6$).\n- The least $\\gamma$ is $84$ (since $\\gamma \\equiv 0 \\pmod{6}$, $\\gamma \\equiv 0 \\pmod{7}$, $\\gamma + 1 \\equiv 0 \\pmod{5}$; $\\gamma=84$ works).\n- The least $\\delta$ is $90$ (since $\\delta \\equiv 0 \\pmod{5}$, $\\delta \\equiv 0 \\pmod{6}$, $\\delta + 1 \\equiv 0 \\pmod{7}$; $\\delta=90$ works).\n\nTake $s=1$ for minimal $n$.\n\nTherefore, the least such $n$ is:\n$$\nn = 2^{35} \\cdot 3^{35} \\cdot 5^{84} \\cdot 7^{90}\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 14229,
    "subject": "Mathematics (Olympiad)",
    "question": "Two circles $\\Gamma_1$ and $\\Gamma_2$ are given with centres $O_1$ and $O_2$ and common exterior tangents $\\ell_1$ and $\\ell_2$. The line $\\ell_1$ intersects $\\Gamma_1$ in $A$ and $\\Gamma_2$ in $B$. Let $X$ be a point on segment $O_1O_2$, not lying on $\\Gamma_1$ or $\\Gamma_2$. The segment $AX$ intersects $\\Gamma_1$ in $Y \\neq A$ and the segment $BX$ intersects $\\Gamma_2$ in $Z \\neq B$. Prove that the line through $Y$ tangent to $\\Gamma_1$ and the line through $Z$ tangent to $\\Gamma_2$ intersect each other on $\\ell_2$.",
    "options": [],
    "answer": "See solution",
    "solution": "We consider the configuration in which $Y$ lies between $A$ and $X$; the other configurations are treated analogously. Let $C$ be the intersection of $\\ell_2$ and $\\Gamma_1$. Then $C$ is the reflection of $A$ in $O_1O_2$. We get\n\n$$\n\\begin{align*}\n\\angle O_1YX &= 180^\\circ - \\angle O_1YA && \\text{(straight angle)} \\\\\n&= 180^\\circ - \\angle YAO_1 && \\text{($O_1YA$ is isosceles)} \\\\\n&= 180^\\circ - \\angle XAO_1 && \\text{($A$ is the reflection of $C$ in $O_1X$)} \\\\\n&= 180^\\circ - \\angle XCO_1 && \\text{($A$ is the reflection of $C$ in $O_1X$)}\n\\end{align*}\n$$\n\nwhich yields that $O_1CXY$ is cyclic.\n\nNow let $S$ be the intersection of the line through $Y$ tangent to $\\Gamma_1$, and the line $\\ell_2$. Then both $SC$ and $SY$ are tangent to $\\Gamma_1$, hence we have $\\angle SCO_1 = 90^\\circ = \\angle SYO_1$, and $O_1CSY$ is cyclic.\n\nWe see that both $X$ and $S$ lie on the circle through $O_1$, $C$, and $Y$. Therefore, we have $\\angle SXO_1 = \\angle SYO_1 = 90^\\circ$. We conclude that $SX$ is perpendicular to $O_1O_2$. Analogously, for the intersection $S'$ of the line through $Z$ tangent to $\\Gamma_2$, and the line $\\ell_2$, we can deduce that $S'X$ is perpendicular to $O_1O_2$. Because $S$ and $S'$ both lie on $\\ell_2$, we have $S = S'$. Hence the two tangents intersect each other on $\\ell_2$. $\\square$\n\n![](images/NLD_ABooklet_2022_p26_data_9de4fc982a.png)",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 14230,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $c(O, R)$ be a circle and let $A, B$ be two points on the circle such that $R < AB < 2R$. The circle $c_1(A, r)$, $0 < r < R$, meets the circle $c(O, R)$ at points $C$ and $D$ ($C$ belongs to the small arc $\\widehat{AB}$). From point $B$ we draw the tangents $BE$ and $BF$ to the circle $c_1(A, r)$, such that $E$ lies outside of the circle $c(O, R)$. The lines $EC$ and $DF$ intersect at $M$. Prove that the quadrilateral $BCFM$ is cyclic.\n\n![](images/GreekMO2014_booklet_p10_data_60ef07c6e2.png)\n\nFigure 2",
    "options": [],
    "answer": "See solution",
    "solution": "$AEBF$ is cyclic (since $BE$ and $BF$ are tangents and $A\\hat{E}B = A\\hat{F}B = 90^\\circ$). Let $c_2$ be its circumcircle. The segment $CD$ is the common chord of the circles $c$ and $c_1$, and the segment $AB$ is the common chord of the circles $c$ and $c_2$. Finally, $EF$ is the common chord of the circles $c_1$ and $c_2$. Hence, the chords $CD$, $AB$, and $EF$ pass through the radical center, say $L$, of the three circles.\n\nSince $BE$ and $BF$ are tangents to the circle $c_1(A, r)$, $AB$ is the bisector of the angle $E\\hat{B}F$. Moreover, $AB$ is the bisector of the angle $C\\hat{B}D$, because $A\\hat{B}C$ and $A\\hat{B}D$ are inscribed in the circle $c(O, R)$ and correspond to the equal arcs $\\widehat{AC}$ and $\\widehat{AD}$. Hence $E\\hat{B}C = F\\hat{B}D$.\n\nUsing the equality $E\\hat{B}C = F\\hat{B}D$, we will prove that the triangles $BCE$ and $BFD$ are similar. For that, it is enough to prove that:\n\n$$\n\\frac{BC}{BE} = \\frac{BF}{BD} \\Leftrightarrow \\frac{a}{x} = \\frac{y}{a} \\Leftrightarrow xy = a^2,\n$$\n\nwhere $BE = BF = a$, $BC = x$, and $BD = y$. Similarly, from $LCB \\sim LAD$:\n\n$$\n\\frac{LC}{LA} = \\frac{CB}{AD} \\Rightarrow \\frac{LC}{LA} = \\frac{x}{r} \\quad (1),\n$$\n\nand from $LCA \\sim LBD$:\n\n$$\n\\frac{LB}{LC} = \\frac{BD}{CA} \\Rightarrow \\frac{LB}{LC} = \\frac{y}{r} \\quad (2).\n$$\n\nFrom (1) and (2) we have: $\\frac{LB}{LA} = \\frac{xy}{r^2}$ (A), and from the right triangle $ABE$ with $AL$ the altitude to the hypotenuse, we get $\\frac{LB}{LA} = \\frac{EB^2}{EA^2} = \\frac{a^2}{r^2}$ (B).\n\nFrom (A) and (B) we obtain: $xy = a^2$. Therefore, $BCE \\sim BFD$. From the equalities of their corresponding angles, we find: $\\hat{C} = \\hat{F}$. Hence, the quadrilateral $BCFM$ is cyclic.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 14231,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $ABC$ be an isosceles triangle with $AB = BC$. Let $M$ and $N$ be the midpoints of $AC$ and $BM$, respectively. $P$ is the foot of the altitude from $A$ to $AN$ of triangle $AMN$. Prove that triangles $APM$ and $CPB$ are similar.\n\n![](images/MNG2024_p24_data_11716b6c28.png)",
    "options": [],
    "answer": "See solution",
    "solution": "Let us denote $\\angle PNM = \\alpha$. Then $\\angle PMC = \\angle PNB = 180^\\circ - \\alpha$ and $\\angle PAC = \\angle PMB$. Hence $\\triangle APM \\sim \\triangle MPN$. As we have $AM = MC$, $MN = NB$, it follows that $\\triangle APC \\sim \\triangle MPB$. This implies $\\angle APC = \\angle MPB$ and $\\angle MPC = \\angle NPB$. Therefore, $\\angle BPC = 90^\\circ$.\n\nSince $\\angle PBM = \\angle PCM$, quadrilateral $BCMP$ is cyclic. Hence $\\angle MPC = \\angle MBC$. Also, $\\angle PMC = \\angle BNA = \\angle BNC$, which means $\\triangle MPC \\sim \\triangle NBC$.\n\nHence $\\angle BCN = \\angle PCA$ is true. It implies that $\\angle NCA = \\angle PCB$. Also, we have $\\angle APM = \\angle BPC = 90^\\circ$. By the AAA property, we have $\\triangle APM \\sim \\triangle CPB$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 14232,
    "subject": "Mathematics (Olympiad)",
    "question": "The two equilateral triangles $ABC$ and $ADB$ (with $C \\neq D$) share the common side $AB$. The midpoints of $AC$ and $BC$ are denoted by $E$ and $F$, respectively.\n\nShow that $DE$ and $DF$ divide $AB$ into three parts of equal length.\n\n![](images/Austrija_2012_p12_data_cee9fbcc65.png)",
    "options": [],
    "answer": "See solution",
    "solution": "The intersections of $DE$, $DF$, and $DC$ with $AB$ are denoted by $S_1$, $S_2$, and $M$, respectively.\n\nWe consider the triangle $ACD$. In this triangle, $DE$ and $AM$ are medians. Therefore, their intersection $S_1$ is the centroid of this triangle and we have $\\overline{AS_1} = 2 \\cdot \\overline{S_1M}$. An analogous result follows for $S_2$, which proves the assertion. $\\square$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 14233,
    "subject": "Mathematics (Olympiad)",
    "question": "Define the sequence $x_1, x_2, \\dots$ by $x_1 = \\frac{1}{6}$ and\n\n$$\nx_{n+1} = \\frac{n+1}{n+3} \\left( x_n + \\frac{1}{2} \\right),\n$$\n\nfor every $n \\ge 1$. Find $x_{2011}$.",
    "options": [],
    "answer": "See solution",
    "solution": "We have\n\n$$\nx_2 = \\frac{2}{4} \\left( \\frac{1}{6} + \\frac{1}{2} \\right) = \\frac{2}{6}, \\quad x_3 = \\frac{3}{5} \\left( \\frac{2}{6} + \\frac{1}{2} \\right) = \\frac{3}{6}, \\text{ etc.}\n$$\n\nWe will prove by induction that for every $n \\ge 1$, we have $x_n = \\frac{n}{6}$.\n\nIndeed, assuming $x_n = \\frac{n}{6}$, it follows\n\n$$\nx_{n+1} = \\frac{n+1}{n+3} \\left( \\frac{n}{6} + \\frac{1}{2} \\right) = \\frac{n+1}{n+3} \\cdot \\frac{n+3}{6} = \\frac{n+1}{6},\n$$\n\nand we are done.\n\nThe answer is $x_{2011} = \\frac{2011}{6}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 14234,
    "subject": "Mathematics (Olympiad)",
    "question": "Suppose that a positive integer $n$ is divisible by exactly 36 different prime numbers. For $k = 1, 2, \\dots, 5$, let $c_k$ be the number of integers in the interval $\\left[ \\frac{(k-1)n}{5}, \\frac{kn}{5} \\right]$ that are coprime with $n$. It is known that $c_1, c_2, \\dots, c_5$ are not all equal. Prove that\n$$\n\\sum_{1 \\le i < j \\le 5} (c_i - c_j)^2 \\ge 2^{36}.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "**Solution**\n\nSuppose that $n = 2$, where $p_1, \\dots, p_{36}$ are different prime numbers and $\\alpha_1, \\dots, \\alpha_{36}$ are positive integers. Obviously, if $\\frac{kn}{5}$ ($k = 0, 1, 2, 3, 4, 5$) are integers, then they are not coprime with $n$. Since either integers $x$ and $n-x$ are both coprime with $n$ or neither of them are coprime with $n$, it follows that $c_1 = c_5$, $c_2 = c_4$.\n\nFor a positive integer $x$, define the function $\\mu(x)$ as follows: if $x$ is divisible by the square of some prime number, then $\\mu(x) = 0$; if $x$ is the product of $t$ different prime numbers ($t$ can be 0, and at this point $x = 1$), then\n$$\n\\mu(x) = (-1)^t.\n$$\nFor a factor $m$ of $n$ and positive integer $y$, the number of multiples of $m$ in $[1, y]$ is exactly $\\left\\lfloor \\frac{y}{m} \\right\\rfloor$. So by the inclusion-exclusion principle, the number of integers in $[1, y]$ that are coprime with $n$ is\n$$\n\\sum_{m \\mid n} \\mu(m) \\left( \\left\\lfloor \\frac{y}{m} \\right\\rfloor \\right).\n$$\nSubstituting into the problem yields for $k = 0, \\dots, 5$,\n$$\n\\begin{aligned}\nc_k &= \\sum_{m \\mid n} \\mu(m) \\left( \\left\\lfloor \\frac{kn}{5m} \\right\\rfloor - \\left\\lfloor \\frac{(k-1)n}{5m} \\right\\rfloor \\right) \\\\\n&= \\sum_{d \\mid n} \\mu\\left(\\frac{n}{d}\\right) \\left( \\left\\lfloor \\frac{kd}{5} \\right\\rfloor - \\left\\lfloor \\frac{(k-1)d}{5} \\right\\rfloor \\right).\n\\end{aligned}\n$$\nLet $p_1, \\dots, p_{36}$ be all the prime factors of $n$. Note that the above equation only needs to be summed over $d$ satisfying $\\frac{n}{d} \\mid p_1 p_2 \\cdots p_{36}$. For these $d$, denote $r_k(d) := \\left\\lfloor \\frac{kd}{5} \\right\\rfloor - \\left\\lfloor \\frac{(k-1)d}{5} \\right\\rfloor$ and consider the cases that $d$ modulo 5 with different remainders.\n\n![](table.png)\n\n* If $25 \\mid n$, then $d$ that satisfies $\\frac{n}{d}$ divides $p_1 p_2 \\cdots p_{36}$ are all multiples of 5, so all $c_k$ are the same, a contradiction!\n* If there exists some prime factor, say $p_1$, that is congruent to 1 modulo 5, then $d$ satisfying $p_1 \\mid \\frac{n}{d}$ and $d$ satisfying $p_1 \\nmid \\frac{n}{d}$ can be paired by quotient $p_1$. The remainder of $d$ modulo 5 of each pair is the same, and ...",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 14235,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $a, b, c > 0$ and $a + b + c = 3$. Prove that\n$$\n\\frac{1}{a\\sqrt{2(a^2 + bc)}} + \\frac{1}{b\\sqrt{2(b^2 + ca)}} + \\frac{1}{c\\sqrt{2(c^2 + ab)}} \\ge \\frac{1}{a+bc} + \\frac{1}{b+ca} + \\frac{1}{c+ab}\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "Without loss of generality, we may assume that $a \\ge b \\ge c$. Then\n$$\n\\frac{(c-a)(c-b)}{3(c+ab)} \\ge 0\n$$\nand\n$$\n\\frac{(a-b)(a-c)}{3(a+bc)} + \\frac{(b-a)(b-c)}{3(b+ca)} = \\frac{c(a-b)^2}{3} \\left( \\frac{1+a+b-c}{(a+bc)(b+ca)} \\right) \\ge 0\n$$\nTherefore\n$$\n\\sum_{\\text{cyc}} \\frac{(a-b)(a-c)}{3(a+bc)} \\ge 0\n$$\nNow\n$$\n\\begin{align*}\n\\sum_{\\text{cyc}} \\frac{1}{a+bc} &\\le \\frac{9}{2(ab+bc+ca)} \\\\\n\\Leftrightarrow \\sum_{\\text{cyc}} \\frac{1}{a(a+b+c)+3bc} &\\le \\frac{3}{2(ab+bc+ca)} \\\\\n\\Leftrightarrow \\sum_{\\text{cyc}} \\left[ \\frac{1}{2(ab+bc+ca)} - \\frac{1}{a(a+b+c)+3bc} \\right] &\\ge 0 \\\\\n\\Leftrightarrow \\sum_{\\text{cyc}} \\frac{(a-b)(a-c)}{a(a+b+c)+3bc} &= \\sum_{\\text{cyc}} \\frac{(a-b)(a-c)}{3(a+bc)} \\ge 0.\n\\end{align*}\n$$\nBy the AM-GM inequality, we have\n$$\n\\frac{1}{a\\sqrt{2(a^2 + bc)}} = \\frac{\\sqrt{b+c}}{\\sqrt{2a}\\sqrt{(ab+ac)(a^2+bc)}} \\ge \\frac{\\sqrt{2(b+c)}}{\\sqrt{a(a+b)(a+c)}}\n$$\nSo it suffices to prove that\n$$\n\\sum_{\\text{cyc}} \\sqrt{\\frac{b+c}{2a}} \\cdot \\frac{1}{(a+b)(a+c)} \\ge \\frac{9}{4(ab+bc+ca)}\n$$\nSince $\\sqrt{\\frac{b+c}{2a}} \\le \\sqrt{\\frac{c+a}{2b}} \\le \\sqrt{\\frac{a+b}{2c}}$ and\n$$\n\\frac{1}{(a+b)(a+c)} \\le \\frac{1}{(b+c)(b+a)} \\le \\frac{1}{(c+a)(c+b)},\n$$\nby Chebyshev's inequality, we get\n$$\n\\begin{align*}\n\\sum_{\\text{cyc}} \\sqrt{\\frac{b+c}{2a}} \\cdot \\frac{1}{(a+b)(a+c)} &\\ge \\frac{1}{3} \\left( \\sum_{\\text{cyc}} \\sqrt{\\frac{b+c}{2a}} \\right) \\left( \\sum_{\\text{cyc}} \\frac{1}{(a+b)(a+c)} \\right) \\\\\n&= \\frac{2}{(a+b)(b+c)(c+a)} \\left( \\sum_{\\text{cyc}} \\sqrt{\\frac{b+c}{2a}} \\right).\n\\end{align*}\n$$\nIn view of these estimates, we now see that it suffices to show that\n$$\n\\sum_{\\text{cyc}} \\sqrt{\\frac{b+c}{2a}} \\ge \\frac{9(a+b)(b+c)(c+a)}{8(ab+bc+ca)}.\n$$\nLet $t := \\sqrt[6]{\\frac{(a+b)(b+c)(c+a)}{8abc}}$. Clearly $t \\ge 1$. Using the AM-GM inequality, we find that\n$$\n\\sum_{\\text{cyc}} \\sqrt{\\frac{b+c}{2a}} \\ge 3t.\n$$\nIt is easy to verify that $\\frac{9(a+b)(b+c)(c+a)}{8(ab+bc+ca)} = \\frac{27t^6}{8t^6+1}$; so, it is enough to prove that\n$3t \\ge \\frac{27t^6}{8t^6+1}$ or $8t^6 - 9t^5 + 1 \\ge 0$.\nSince $t \\ge 1$, this inequality is true and the proof is completed. $\\square$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 14236,
    "subject": "Mathematics (Olympiad)",
    "question": "There is a figure of a prince on a $6 \\times 6$ square chessboard. The prince can, in one move, jump either horizontally or vertically. The lengths of the jumps alternate between one and two squares, and the jump to the next square is the first one. Decide whether one can choose the initial square for the prince so that, in an appropriate sequence of 35 jumps, the prince visits every square of the chessboard.",
    "options": [],
    "answer": "See solution",
    "solution": "Suppose such a sequence exists. Enumerate the squares of the chessboard as follows:\n\n![]( <table><tr><td>1</td><td>2</td><td>3</td><td>4</td><td>1</td><td>2</td></tr><tr><td>2</td><td>3</td><td>4</td><td>1</td><td>2</td><td>3</td></tr><tr><td>3</td><td>4</td><td>1</td><td>2</td><td>3</td><td>4</td></tr><tr><td>4</td><td>1</td><td>2</td><td>3</td><td>4</td><td>1</td></tr><tr><td>1</td><td>2</td><td>3</td><td>4</td><td>1</td><td>2</td></tr><tr><td>2</td><td>3</td><td>4</td><td>1</td><td>2</td><td>3</td></tr></table> )\n\nThe length-one moves go from odd to even numbers 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}$ as the numbers of visited squares, then among $P_2, P_3, P_4, P_5$ each number (from 1 to 4) appears 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). For the same reasons, any of the four numbers appears among $P_{4k+2}, P_{4k+3}, P_{4k+4}, P_{4k+5}$ for arbitrary $k \\in \\{0, 1, \\dots, 7\\}$. Thus, between $P_2, P_3, \\dots, P_{33}$, each of the numbers 1 to 4 appears exactly eight times.\n\nThe number 4 appears on the chessboard just eight times, so 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 a length-two move apart). Since 4 is not among them, both must be odd. Then $P_{36}$ and $P_1$ must be even, so they must be 2.\n\nThe initial square ($P_1$) must be one of the colored squares on the left chessboard. Repeating the argument for the numbering of the right chessboard (just a rotation of the left one), since no square has number 2 on both chessboards, we reach a contradiction. The initial square cannot be chosen.\n\n![]( <table><tr><td>1</td><td><b>2</b></td><td>3</td><td>4</td><td>1</td><td><b>2</b></td></tr><tr><td><b>2</b></td><td>3</td><td>4</td><td>1</td><td><b>2</b></td><td>3</td></tr><tr><td>3</td><td>4</td><td>1</td><td><b>2</b></td><td>3</td><td>4</td></tr><tr><td>4</td><td>1</td><td><b>2</b></td><td>3</td><td>4</td><td>1</td></tr><tr><td>1</td><td><b>2</b></td><td>3</td><td>4</td><td>1</td><td><b>2</b></td></tr><tr><td><b>2</b></td><td>3</td><td>4</td><td>1</td><td><b>2</b></td><td>3</td></tr></table> )\n\n![]( <table><tr><td><b>2</b></td><td>3</td><td>4</td><td>1</td><td><b>2</b></td><td>3</td></tr><tr><td>1</td><td><b>2</b></td><td>3</td><td>4</td><td>1</td><td><b>2</b></td></tr><tr><td>4</td><td>1</td><td><b>2</b></td><td>3</td><td>4</td><td>1</td></tr><tr><td>3</td><td>4</td><td>1</td><td><b>2</b></td><td>3</td><td>4</td></tr><tr><td><b>2</b></td><td>3</td><td>4</td><td>1</td><td><b>2</b></td><td>3</td></tr><tr><td>1</td><td><b>2</b></td><td>3</td><td>4</td><td>1</td><td><b>2</b></td></tr></table> )",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 14237,
    "subject": "Mathematics (Olympiad)",
    "question": "Numbers $1, 2, 3, \\ldots, 2014$ are written on the board. Andriyko can choose any two numbers $a, b$ on the board and replace them with the number $|a-b|$. After he performs this operation $2013$ times, there will be only one number left. What is the biggest possible value of this number?",
    "options": [],
    "answer": "See solution",
    "solution": "The biggest possible value is $2013$.\n\nIt is easy to see that all the numbers on the board cannot exceed $2014$ at any moment. The parity of $|a-b|$ is the same as the parity of $a+b$, so the parity of the sum of all numbers on the board remains unchanged. Initially, the sum is odd, since there are $1007$ odd numbers ($1, 3, \\ldots, 2013$) and $1007$ even numbers ($2, 4, \\ldots, 2014$). Therefore, the last number must be odd, so it cannot be $2014$ and cannot be greater than $2013$.\n\nWe can achieve $2013$ as follows: split all numbers into pairs $(2, 3), (4, 5), \\ldots, (2012, 2013)$, and $(1, 2014)$. Applying the operation to each pair, we are left with $\\underbrace{1, 1, \\ldots, 1}_{1006}$ and $2013$ on the board. Then, pair the $1$'s into $503$ pairs and replace each with $0$. After any $503$ further operations, the last number will be $2013$.\n\n![](images/Ukrajina_2013_p15_data_3de70e5506.png)\n\nFig. 10",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 14238,
    "subject": "Mathematics (Olympiad)",
    "question": "Los vértices $A$, $B$ y $C$ de un triángulo equilátero de lado $1$ están en la superficie de una esfera de radio $1$ y centro $O$. Sea $D$ la proyección ortogonal de $A$ sobre el plano $\\alpha$ determinado por $B$, $C$ y $O$. Llamamos $N$ a uno de los puntos de intersección de la esfera con la recta perpendicular a $\\alpha$ que pasa por $O$. Halla la medida del ángulo $\\angle DNO$.\n\n(Nota: la proyección ortogonal de $A$ sobre el plano $\\alpha$ es el punto de corte con $\\alpha$ de la recta que pasa por $A$ y es perpendicular a $\\alpha$.)",
    "options": [],
    "answer": "See solution",
    "solution": "Es obvio que $A$, $B$, $C$ y $O$ son vértices de un tetraedro regular de arista igual a $1$, puesto que la distancia entre dos cualesquiera de ellos es $1$. Como $D$ es la proyección ortogonal de $A$ sobre la cara opuesta del tetraedro, $D$ es el centro de la cara $BCO$. Así pues, la distancia de $D$ a $O$ (distancia del centro de un triángulo equilátero de lado $1$ a uno de sus vértices) es\n\n$$\nd(D, O) = \\frac{2\\sqrt{3}}{3} = \\frac{1}{\\sqrt{3}}.\n$$\n\nComo el triángulo $DNO$ es rectángulo en $O$, el cateto $OD$ mide $1/\\sqrt{3}$ y el cateto $ON$ mide $1$, el ángulo buscado es $\\arctan \\frac{1}{\\sqrt{3}} = 30^{\\circ}$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 14239,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $m$ and $n$ be integers with $m, n \\ge 2$. The matrices $A_1, A_2, \\dots, A_m \\in \\mathcal{M}_n(\\mathbb{R})$ are not all nilpotent. Prove that there exists an integer $k > 0$ such that $A_1^k + A_2^k + \\dots + A_m^k \\ne O_n$.",
    "options": [],
    "answer": "See solution",
    "solution": "Denote by $\\lambda_{i1}, \\lambda_{i2}, \\dots, \\lambda_{in}$ the eigenvalues of $A_i$ for $i = 1, 2, \\dots, m$. Suppose $A_1^k + A_2^k + \\dots + A_m^k = O_n$ for all $k \\ge 1$. Then $\\operatorname{tr}(A_1^k) + \\operatorname{tr}(A_2^k) + \\dots + \\operatorname{tr}(A_m^k) = 0$, implying $$\\sum_{j=1}^m \\sum_{i=1}^n \\lambda_{ij}^k = 0.$$ By the Newton relations, the equalities $\\sum_{i,j} \\lambda_{ij}^k = 0$ for all $k \\ge 1$ imply $\\lambda_{ij} = 0$ for any $i = 1, 2, \\dots, m$ and $j = 1, 2, \\dots, n$. Then $A_i^n = O_n$ for $i = 1, 2, \\dots, m$, a contradiction.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 14240,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all functions $g : \\mathbb{N} \\to \\mathbb{N}$ such that for all positive integers $m$ and $n$, the product $(g(m) + n)(m + g(n))$ is always a perfect square.",
    "options": [],
    "answer": "See solution",
    "solution": "The answer is $g(n) = n + C$, where $C$ is a nonnegative integer.\n\nClearly, the function $g(n) = n + C$ satisfies the required property since\n$$\n(g(m) + n)(m + g(n)) = (n + m + C)^2\n$$\nis a perfect square.\n\nWe first prove a lemma.\n\n**Lemma**: If a prime number $p$ divides $g(k) - g(l)$ for some positive integers $k$ and $l$, then $p \\mid k-l$.\n\n*Proof of lemma*: If $p^2 \\mid g(k) - g(l)$, let $g(l) = g(k) + p^2a$, where $a$ is an integer. Choose an integer\n$$\nD > \\max\\{g(k), g(l)\\},\n$$\nand $D$ is not divisible by $p$. Set $n = pD - g(k)$; then $n + g(k) = pD$, and thus\n$$\nn + g(l) = pD + (g(l) - g(k)) = p(D + pa)\n$$\nis divisible by $p$, but not divisible by $p^2$.\n\nBy assumption, $(g(k) + n)(g(n) + k)$ and $(g(l) + n)(g(n) + l)$ are both perfect squares, and therefore they are divisible by $p^2$ since they are divisible by $p$. Hence,\n$$\np \\mid ((g(n) + k) - (g(n) + l)),\n$$\ni.e. $p \\mid k-l$.\n\nIf $p \\mid g(k) - g(l)$ but $p^2$ does not divide $g(k) - g(l)$, choose an integer $D$ as above and set $n = p^3D - g(k)$. Then $g(k) + n = p^3D$ is divisible by $p^3$, but not by $p^4$, and\n$$\ng(l) + n = p^3D + (g(l) - g(k))\n$$\nis divisible by $p$, but not by $p^2$. As with the above argument, we have $p \\mid g(n) + k$ and $p \\mid g(n) + l$, and therefore\n$$\np \\mid ((g(n) + k) - (g(n) + l)),\n$$\ni.e. $p \\mid k-l$. This completes the proof of the lemma.\n\nBack to the original problem: if there exist positive integers $k$ and $l$ such that $g(k) = g(l)$, then the lemma implies that $k-l$ is divisible by any prime number. Hence, $k-l=0$, i.e. $k=l$, and thus $g$ is injective.\n\nNow consider $g(k)$ and $g(k+1)$. Since $(k+1)-k=1$, once again the lemma implies that $g(k+1) - g(k)$ is not divisible by any prime number, and therefore\n$$\n|g(k+1) - g(k)| = 1.\n$$\nLet $g(2) - g(1) = q$, where $|q| = 1$. It follows easily by induction that\n$$\ng(n) = g(1) + (n - 1)q.\n$$\nIf $q = -1$, then $g(n) \\le 0$ for $n \\ge g(1) + 1$, a contradiction. Therefore, we must have $q = 1$ and\n$$\ng(n) = n + (g(1) - 1)\n$$\nfor any $n \\in \\mathbb{N}$, where $g(1) - 1 \\ge 0$. Set $g(1) - 1 = C$ (a constant). Then $g(n) = n + C$, where $C$ is a nonnegative integer.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 14241,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $\\triangle ABC$ be an acute, non-isosceles triangle with centroid $G$ and circumcircle $(O)$. Let $H_a, H_b, H_c$ be the projections of $A, B, C$ onto the opposite sides, and $D, E, F$ be the midpoints of $BC, CA, AB$, respectively. The rays $GH_a, GH_b, GH_c$ meet $(O)$ at $X, Y, Z$.\n\n(a) Prove that the circle $(XCE)$ passes through the midpoint of the segment $BH_a$.\n\n(b) Let $M, N, P$ be the midpoints of $AX, BY, CZ$. Prove that the lines $DM, EN, FP$ are concurrent.",
    "options": [],
    "answer": "See solution",
    "solution": "(a) Let $A_0$ be the intersection of the line through $A$ parallel to $BC$ with the circle $(O)$. Since $AA_0CB$ is an isosceles trapezoid, $A_0C = AB = 2FH_a$. On the other hand,\n\n$$\n\\angle FH_aB = \\angle FBH_a = \\angle A_0CB\n$$\n\nimplies that $FH_a \\parallel A_0C$. Since $GC = 2GF$, $G, A_0, H_a$ are collinear. Let $S$ be the midpoint of $A_0H_a$; then $SE \\parallel AA_0$, so $\\angle AES = \\angle A_0AC = \\angle A_0XC$, thus $SECX$ is cyclic.\n\n![](images/VN_IMO_Booklet_2018_Final_p24_data_d118b5e8cf.png)\n\nLet $T$ be the midpoint of $BH_a$; then $ST \\parallel BA_0$. Hence, $\\angle XST = \\angle XA_0B = \\angle XCB$, so the quadrilateral $XCST$ is also cyclic.\n\nFrom these results, the five points $X, C, E, S, T$ are concyclic, so the circle $(XCE)$ passes through the midpoint of $BH_a$.\n\n(b) We will prove some remarks as follows:\n\n**Remark 1.** $AX, BY, CZ$ are concurrent at a point on $OH$, where $H$ is the orthocenter of triangle $ABC$.\n\n**Proof.** Note that $\\angle AXH_a = \\angle HAO$, so $AO$ is tangent to $(XAH_a)$. Thus $\\mathcal{P}_{O/(XAH_a)} = R^2$, where $R$ is the radius of circle $(O)$. Similarly for the circles $(YBH_b)$, $(ZCH_c)$, so $O$ is the radical center of these circles. On the other hand,\n\n$$\n\\mathcal{P}_{H/(XAH_a)} = \\overline{HA} \\cdot \\overline{HA_a} = \\frac{1}{2} \\mathcal{P}_{H/(O)}\n$$\n\nand similarly, $H$ is the radical center of circles $(XAH_a)$, $(YBH_b)$, $(ZCH_c)$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 14242,
    "subject": "Mathematics (Olympiad)",
    "question": "Find the number of integers $k$ in the set $\\{0, 1, 2, \\dots, 2012\\}$ such that the binomial coefficient $\\binom{2012}{k} = \\frac{2012!}{k!(2012-k)!}$ is a multiple of $2012$.\n\n![](path/to/file.png)",
    "options": [],
    "answer": "See solution",
    "solution": "First, factor $2012 = 4 \\times 503$, where $503$ is prime. For $k$ not a multiple of $503$, $\\binom{2012}{k}$ is divisible by $503$. If $k$ is a multiple of $503$, only $k = 0, 503, 1006, 1509, 2012$ are possible, and in these cases, $\\binom{2012}{k}$ is not divisible by $503$.\n\nNext, consider divisibility by $4$. The power of $2$ in $n!$ is $n - s(n)$, where $s(n)$ is the sum of the binary digits of $n$. $\\binom{2012}{k}$ is odd if the binary addition $k + (2012 - k) = 2012$ has no carrying, which occurs in $2^8 = 256$ cases (since $2012$ has $8$ ones in binary). Thus, $2013 - 256 = 1757$ values of $k$ make $\\binom{2012}{k}$ even.\n\nTo be divisible by $4$, the power of $2$ in $\\binom{2012}{k}$ must be at least $2$. The number of $k$ for which $\\binom{2012}{k}$ is divisible by $4$ is $2013 - 256 - 128 = 1629$, where $128$ is the number of $k$ for which $\\binom{2012}{k}$ is even but not divisible by $4$ (i.e., exactly one carry in binary addition).\n\nSince $2012 = 4 \\times 503$, $\\binom{2012}{k}$ is a multiple of $2012$ if and only if it is divisible by both $4$ and $503$. The $k$ that are multiples of $503$ do not yield binomial coefficients divisible by $503$, so we exclude $k = 0, 503, 1006, 1509, 2012$ from the $1629$ values. Thus, the answer is $1629 - 5 = 1624$.\n\n**Answer:** $1624$ integers $k$ in $\\{0, 1, 2, \\dots, 2012\\}$ make $\\binom{2012}{k}$ a multiple of $2012$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 14243,
    "subject": "Mathematics (Olympiad)",
    "question": "Let\n\n$$\nS = \\{P(a+1), P(a+2), \\dots, P(a+b)\\}.\n$$\n\nFor which values of $b$ does there exist an integer $a$ such that $S$ is *fragrant*? (A set $S$ is called *fragrant* if for every $x \\in S$, there exists $y \\in S$, $y \\ne x$, such that $\\gcd(x, y) > 1$.)\n\nHere, $P(n) = n^2 + n + 1$.",
    "options": [],
    "answer": "See solution",
    "solution": "**Lemma** For any integer $n$ we have the following:\n\n$$\n\\gcd(P(n), P(n+1)) = 1 \\qquad (1)\n$$\n\n$$\n\\gcd(P(n), P(n+2)) > 1 \\Leftrightarrow n \\equiv 2 \\pmod{7} \\qquad (2)\n$$\n\n$$\n\\gcd(P(n), P(n+3)) > 1 \\Leftrightarrow n \\equiv 1 \\pmod{3} \\qquad (3)\n$$\n\n$$\n\\gcd(P(n), P(n+4)) > 1 \\Leftrightarrow n \\equiv 7 \\pmod{19} \\qquad (4)\n$$\n\n*Proof*: The proof of (1), and the proofs of the “⇒” directions of (2) and (3) can be carried out as in solution 1. The proof of the “⇒” direction of (4) is as follows.\n\nSuppose that $P(n) = n^2 + n + 1$ and $P(n + 4) = n^2 + 9n + 21$ are both divisible by $p$ for some odd prime $p$. Then $|P(n + 4) - P(n)| = 8n + 20$. Since $p$ is odd, we have $p \\mid 2n + 5$. It follows that $p \\mid 4(n^2 + n + 1) - (2n + 5)(2n - 3) = 19$. Thus $p = 19$. Furthermore, since $p \\mid 2n + 5$, we have $19 \\mid 2n + 5$. Thus $19 \\mid 2n - 14$ from which we deduce $19 \\mid n - 7$.\n\nThe proofs of the “<” directions of (2), (3), and (4) are straightforward computations. For example, for (2) we verify that $P(n) \\equiv P(n + 2) \\equiv 0 \\pmod{7}$ whenever $n \\equiv 2 \\pmod{7}$. Similar computations apply for (3) and (4). □\n\nThe proof that no fragrant set exists for $b \\le 5$ can be carried out as in solution 1.\n\nTo find a fragrant set for $b = 6$, it is sufficient to observe that if we can partition the set $\\{a + 1, a + 2, a + 3, a + 4, a + 5, a + 6\\}$ into three disjoint pairs so that the respective differences between elements of each pair are 2, 3, and 4, then we can apply (2), (3), and (4) and the Chinese remainder theorem to find a working value of $a$. By inspection there are only two different ways to do this.\n\n**Way 1** $(a + 2, a + 4)$, $(a + 3, a + 6)$, and $(a + 1, a + 5)$.\n\nThese pairs respectively correspond to the following congruences:\n\n$$\na \\equiv 0 \\pmod{7}\n$$\n\n$$\na \\equiv 1 \\pmod{3}\n$$\n\n$$\na \\equiv 6 \\pmod{19}\n$$\n\nSolving yields $a \\equiv 196 \\pmod{399}$.\n\n**Way 2** $(a + 3, a + 5)$, $(a + 1, a + 4)$, and $(a + 2, a + 6)$.\n\nThese pairs respectively correspond to the following congruences:\n\n$$\na \\equiv 6 \\pmod{7}\n$$\n\n$$\na \\equiv 0 \\pmod{3}\n$$\n\n$$\na \\equiv 5 \\pmod{19}\n$$\n\nSolving yields $a \\equiv 195 \\pmod{399}$.\n\nThus $a = 195$ and $a = 196$ are two possible solutions for $b = 6$. □\n\n*Comment*: With just a little more work it can be shown that the full set of solutions for $b = 6$ are $a = 195 + 399k$ and $a = 196 + 399k$, where $k$ ranges over the set of positive integers.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 14244,
    "subject": "Mathematics (Olympiad)",
    "question": "ABCD is a cyclic quadrilateral with $\\angle BAD = 90^\\circ$ and $\\angle ABC > 90^\\circ$. $AB$ is extended to a point $E$ such that $\\angle AEC = 90^\\circ$. If $AB = 7$, $BE = 9$, and $EC = 12$, calculate $AD$.",
    "options": [],
    "answer": "See solution",
    "solution": "Extend $AE$ and $DC$ until they meet at $F$. (These two lines will definitely meet: since $ABCD$ is a cyclic quadrilateral, $\\angle ABC > 90^\\circ$ implies that $\\angle ADC < 90^\\circ$.)\n\n![](images/s3s2023_p0_data_5e59402855.png)\n\nWe have $\\angle FBC = \\angle FDA = \\angle FCE$ (note that $EC \\parallel AD$), so triangles $FAD$, $FCB$, and $FEC$ are all similar. Let $x = AD$ and $y = EF$. From the similarity between triangles $FAD$ and $FEC$, it follows that\n\n$$\n\\frac{x}{y+16} = \\frac{12}{y}. \\qquad (1)\n$$\n\nFurthermore, $BC = \\sqrt{9^2 + 12^2} = 15$, so from the similarity between triangles $FCB$ and $FEC$, it follows that\n\n$$\n\\frac{15}{\\sqrt{12^2 + y^2}} = \\frac{12}{y}. \\qquad (2)\n$$\n\nFrom (2) we get $y = 16$, and then from (1) we find that $x = AD = 24$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 14245,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $\\{a_n\\}_{n \\ge 0}$ be a sequence of positive real numbers such that\n$$\n\\sum_{k=0}^{n} C_{n}^{k} a_{k} a_{n-k} = a_{n}^{2}, \\text{ for any } n \\ge 0.\n$$\nProve that $\\{a_n\\}_{n \\ge 0}$ is a geometric sequence.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $a_0 = a$. It is obvious that $a_1 = 2a$, and, for $n = 2$,\n$$\na_2^2 - 2a a_2 - 8a^2 = 0,\n$$\nimplying $a_2 = 4a$, since $a_2 > 0$.\n\nUse induction on $n$ to prove that $a_n = 2^n a$. Assume that $a_k = 2^k a$ for all $k$, $0 \\le k \\le n$, to prove $a_{n+1} = 2^{n+1} a$. We have\n$$\n2a a_{n+1} + a^2 2^{n+1} \\sum_{k=1}^{n} C_{n+1}^{k} = a_{n+1}^2,\n$$\nor\n$$\n2a a_{n+1} + a^2 2^{n+1} (2^{n+1} - 2) = a_{n+1}^2.\n$$\nSince $a_{n+1}$ is positive, we obtain $a_{n+1} = 2^{n+1} a$, as needed.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 14246,
    "subject": "Mathematics (Olympiad)",
    "question": "A 3-digit number in base 7 is also a 3-digit number when written in base 6, but each digit has increased by 1.\nWhat is the largest value which this number can have when written in base 10?",
    "options": [],
    "answer": "See solution",
    "solution": "$abc_7 = (a+1)(b+1)(c+1)_6$.\n\nThis gives:\n$$49a + 7b + c = 36(a + 1) + 6(b + 1) + (c + 1)$$\nSimplifying:\n$$49a + 7b + c = 36a + 36 + 6b + 6 + c + 1$$\n$$49a + 7b + c = 36a + 6b + c + 43$$\n$$13a + b = 43$$\nSince $a + 1$ and $b + 1$ are less than 6 (so $a, b < 5$), the only solution is $a = 3$, $b = 4$.\n\nThus, the number is $344_7$ or $45(c+1)_6$. Since $c + 1 \\leq 5$, for the largest value, $c = 4$.\n\nTherefore, the number is $344_7 = 179$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 14247,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $ABC$ be a triangle and let points $D, E \\in (BC)$, $F, G \\in (CA)$, $H, I \\in (AB)$ such that $BD = CE$, $CF = AG$ and $AH = BI$. Consider $M, N, P$ the midpoints of the segments $GH, DI, EF$ respectively and let $M'$ be the intersection point of the lines $AM$ and $BC$.\n\n(a) Show that\n$$\n\\frac{BM'}{CM'} = \\frac{AG}{AH} \\cdot \\frac{AB}{AC}.\n$$\n\n(b) Prove that lines $AM$, $BN$ and $CP$ are concurrent.",
    "options": [],
    "answer": "See solution",
    "solution": "(a) Set $m = \\frac{BM'}{CM'}$. We have\n$$\n\\overrightarrow{AM} = \\frac{1}{2}\\overrightarrow{AH} + \\frac{1}{2}\\overrightarrow{AG} = \\frac{1}{2}\\frac{AH}{AB} \\cdot \\overrightarrow{AB} + \\frac{1}{2}\\frac{AG}{AC} \\cdot \\overrightarrow{AC}\n$$\nand\n$$\n\\overrightarrow{AM'} = \\frac{1}{m+1}\\overrightarrow{AB} + \\frac{m}{m+1}\\overrightarrow{AC}.\n$$\nPoints $A$, $M$, $M'$ are collinear, hence\n$$\n\\frac{AH}{AB} \\cdot \\frac{m}{m+1} = \\frac{AG}{AC} \\cdot \\frac{1}{m+1}\n$$\nand the claim follows.\n\n(b) Define similarly the points $N'$, $P'$. Notice that\n$$\n\\frac{BM'}{CM'} \\cdot \\frac{CN'}{AN'} \\cdot \\frac{AP'}{BP'} = 1\n$$\nand apply Ceva's theorem to reach the conclusion.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 14248,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $\\alpha \\neq 0$ be a real number. Find all functions $f: \\mathbb{R}_{>0} \\to \\mathbb{R}_{>0}$ such that\n\n$$\nf(f(x) + y) = \\alpha x + \\frac{1}{f\\left(\\frac{1}{y}\\right)}\n$$\n\nfor all $x, y \\in \\mathbb{R}_{>0}$.",
    "options": [],
    "answer": "See solution",
    "solution": "If $\\alpha = 1$, the only solution is $f(x) = x$. For other values of $\\alpha$, there is no solution.\n\nWe must have $\\alpha > 0$, otherwise, the right-hand side becomes negative for large values of $x$. By using $x$ in the given equation, we can immediately conclude that $f$ is an injective function. Furthermore, since we can choose arbitrary values for $x$ on the right-hand side, we conclude that $f$ is surjective on an interval $(a, \\infty)$. By choosing small values of $x$ and large values of the function on the right-hand side, we conclude that the right-hand side takes all positive values, so the function $f$ is surjective.\n\nNow, we replace $y$ with $f(y)$ and obtain\n\n$$\nf(f(x) + f(y)) = \\alpha x + \\frac{1}{f\\left(\\frac{1}{f(y)}\\right)} \\quad (1)\n$$\n\nThe left-hand side is symmetric in $x$ and $y$, therefore\n\n$$\n\\alpha x + \\frac{1}{f\\left(\\frac{1}{f(y)}\\right)} = \\alpha y + \\frac{1}{f\\left(\\frac{1}{f(x)}\\right)}.\n$$\n\nIf we choose an arbitrary fixed value for $y$, we get\n\n$$\n\\frac{1}{f\\left(\\frac{1}{f(x)}\\right)} = \\alpha x + C.\n$$\n\nWe substitute this identity into Equation (1) and get\n\n$$\nf(f(x) + f(y)) = \\alpha x + \\alpha y + C.\n$$\n\nBecause of injectivity, this implies\n\n$$\nf(x) + f(y) = f(z) + f(w), \\text{ if } x + y = z + w. \\quad (2)\n$$\n\nIn particular, we have\n\n$$\nf(x + 1) + f(y + 1) = f(x + y + 1) + f(1) \\quad \\text{for } x, y \\ge 0.\n$$\n\nWith $g(x) = f(x + 1)$, we get for the function $g: \\mathbb{R}_{\\ge 0} \\to \\mathbb{R}_{\\ge 0}$ that\n\n$$\ng(x) + g(y) = g(x + y) + g(0).\n$$\n\nWe put $h(x) = g(x) - g(0)$ and get $h(x) \\ge -g(0)$ and the Cauchy functional equation\n\n$$\nh(x) + h(y) = h(x + y).\n$$\n\nIf we had $h(t) < 0$ for some $t > 0$, then the values $h(nt) = n h(t)$ would be arbitrarily small for large positive integers $n$. But this is impossible because of the lower bound, so we have $h(x) \\ge 0$ for all $x \\ge 0$ and we get for $0 < u < v$ that $h(v) = h(u) + h(v - u) \\ge h(u)$.\n\nTherefore, the function $h(x)$ is a monotone solution of the Cauchy functional equation which has to be of the form $h(x) = c x$. This implies that for $x > 1$ we also have $f(x) = h(x - 1) + g(0) = c x + d$ for some constants $c$ and $d$. But this is also true for $0 < x \\le 1$ which can be seen by plugging $y = 3, z = 2$ and $w = x + 1$ into Equation (2).\n\nSince $f$ is surjective, the constant term has to be $0$ (otherwise, small positive values could not be reached by $f$ or negative values would be reached). Putting $f(x) = c x$ into the original equation and equating coefficients gives the conditions $c^2 = \\alpha$ and $c^2 = 1$. Since $c$ has to be positive, we obtain the only solution $f(x) = x$ if $\\alpha = 1$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 14249,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $M$, $K$, and $L$ be the midpoints of the sides $PQ$, $PT$, and $TQ$ respectively.\n\n$$\nM\\left(-\\frac{5}{4t},\\ t\\right), \\quad K\\left(-\\frac{1}{2t},\\ \\frac{5}{2}t\\right), \\quad L\\left(\\frac{1}{4t},\\ -\\frac{t}{2}\\right).\n$$\n\nLet $H$ be the centroid of triangle $PQT$. Find the area $S(PQT)$ of triangle $PQT$.",
    "options": [],
    "answer": "See solution",
    "solution": "Comparing the obtained values with (3), we note that $MT$ is parallel to the $Ox$ axis, since $y_M = y_T$, and $QK$ is parallel to $Oy$, since $y_Q = y_K$. Therefore, medians $MT$ and $QK$ in triangle $PQT$ are perpendicular, as required.\n\nLet $H$ be the centroid of triangle $PQT$. Then\n\n$$\n\\begin{aligned}\nS(PQT) = 2S(QKT) &= [TH \\perp QK] = 2 \\cdot \\frac{1}{2} QK \\cdot TH = \\\\\n&= [TH = \\frac{2}{3}MT] = \\frac{2}{3}QK \\cdot TM.\n\\end{aligned}\n$$\n\nSince $TM \\parallel Ox$, we have $TM = \\frac{1}{t} - \\left(-\\frac{5}{4t}\\right) = \\frac{9}{4t}$. Similarly, since $QK \\parallel Oy$, we have $QK = \\frac{5t}{2} - (-2t) = \\frac{9t}{2}$. Therefore,\n\n$$\nS(PQT) = \\frac{2}{3} \\cdot \\frac{9}{4t} \\cdot \\frac{9t}{2} = \\frac{27}{4}.\n$$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 14250,
    "subject": "Mathematics (Olympiad)",
    "question": "Assume $P_1, P_2, \\dots, P_n$ ($n > 10$) are pairwise different polynomials with coefficients $1$, $0$, or $-1$ and such that they do not have integer roots. Additionally, for all $i = 1, \\dots, n$, $|P_i(5)| \\le \\frac{n^2}{2}$. Prove that $P_i + P_j = P_k + P_l$ for some $1 \\le i, j, k, l \\le n$ and $\\{i, j\\} \\neq \\{k, l\\}$. Note that $i$ may be equal to $j$, and $k$ may be equal to $l$.",
    "options": [],
    "answer": "See solution",
    "solution": "Since the polynomials $P_1, \\dots, P_n$ do not have integer roots, $P_i(0) \\neq 0$ for any $i = 1, \\dots, n$, and $P_i(0) = \\pm 1$. Consider all pairs of polynomials $P_k$ and $P_j$ such that $k \\neq j$ and $P_k(0) = P_j(0)$. For each pair, define $Q_{k,j} = P_k - P_j$. If $a$ and $b$ are the numbers of polynomials with $P_i(0) = 1$ and $P_i(0) = -1$ respectively, then the number of such $Q_{k,j}$ is\n\n$$\na(a-1) + b(b-1) = a^2 + b^2 - n \\ge \\frac{1}{2}(a+b)^2 - n = \\frac{1}{2}n(n-2).\n$$\n\nThe coefficients of $Q_{k,j}$ are in $\\{-2, -1, 0, 1, 2\\}$, so the coefficients of $Q_{k,j} - Q_{i,l}$ are at most $4$ in absolute value. Therefore, $Q_{k,j}(5) = Q_{i,l}(5)$ if and only if $Q_{k,j} = Q_{i,l}$, since for all $n \\in \\mathbb{N}$, $5^n > 4(5^{n-1} + \\dots + 5 + 1)$. \n\nAlso, $Q_{k,j}(5)$ is divisible by $5$ and $Q_{k,j}(5) \\neq 0$ (otherwise $Q_{k,j}$ is the zero polynomial, which is impossible), and $|Q_{k,j}(5)| \\le |P_k(5)| + |P_j(5)| \\le n^2$. Thus, $Q_{k,j}(5)$ can take at most $\\frac{2}{5}n^2$ different values. But the total number of $Q_{k,j}$ is greater than $\\frac{2}{5}n^2$, because\n\n$$\n\\frac{1}{2}n(n-2) > \\frac{2}{5}n^2 \\iff 5(n-2) > 4n \\iff n > 10.\n$$\n\nTherefore, by the pigeonhole principle, $Q_{k,j}(5) = Q_{i,l}(5)$ for some distinct pairs, so $Q_{k,j} = Q_{i,l}$, which implies $P_k - P_j = P_i - P_l$, or $P_k + P_l = P_i + P_j$ for some $1 \\le i, j, k, l \\le n$ with $\\{i, j\\} \\neq \\{k, l\\}$. This completes the proof.",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 14251,
    "subject": "Mathematics (Olympiad)",
    "question": "Prove that for any positive real numbers $a, b, c < 1$ satisfying\n\n$$\n2(a+b+c) + 4abc = 3(ab+bc+ca) + 1,\n$$\n\nthe following inequality holds:\n\n$$\na+b+c \\le \\frac{3}{4}.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "We can rewrite the given equality as follows:\n\n$$\n4abc - 4(ab + bc + ca) + 4(a+b+c) - 4 = -(ab + bc + ca) + 2(a+b+c) - 3,\n$$\n\nwhich simplifies to\n\n$$\n4(a-1)(b-1)(c-1) = -(a-1)(b-1) - (b-1)(c-1) - (c-1)(a-1).\n$$\n\nThis leads to\n\n$$\n4 = \\frac{1}{1-a} + \\frac{1}{1-b} + \\frac{1}{1-c}.\n$$\n\nThe function $f(x) = \\frac{1}{1-x}$ is convex on the interval $x < 1$ because $f''(x) = \\frac{2}{(1-x)^3} \\ge 0$. So, by Jensen's inequality,\n\n$$\n4 = f(a) + f(b) + f(c) \\ge 3f\\left(\\frac{a+b+c}{3}\\right) = \\frac{9}{3-(a+b+c)}.\n$$\n\nThe last inequality implies the required one.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 14252,
    "subject": "Mathematics (Olympiad)",
    "question": "求所有正整數 $a$、$b$ 和 $c$，使得 $ab$ 為完全平方數且\n\n$$\na + b + c - 3\\sqrt[3]{abc} = 1.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "答案為 $(n^2, (n+1)^2, n(n+1))$ 或 $((n+1)^2, n^2, n(n+1))$，其中 $n$ 為任意正整數。\n\n我們首先排除 $a = b$ 的可能性。注意到若 $a = b$，則對於所有質數 $p \\mid a$，由於 $p \\mid ab$ 且 $ab$ 為完全平方數，故 $p^2 \\mid ab$，這意味著 $p \\mid a+b-3\\sqrt[3]{abc}$。這表示 $p \\nmid c$，從而 $\\gcd(a, c) = 1$。又基於 $\\sqrt[3]{abc}$ 為整數，必須存在 $m, n \\in \\mathbb{N}$ 使得 $(a, b, c) = (m^3, m^3, n^3)$，從而\n\n$$\na + b + c - 3\\sqrt[3]{abc} = 2m^3 + n^3 - 3m^2n = (m - n)^2(2m + n) = 1,\n$$\n\n此顯然無解，故 $a \\neq b$。\n\n接下來，不失一般性假設 $a > b$。由算幾不等式，我們有\n\n$$\n\\begin{aligned}\n1 &= a + b + c - 3\\sqrt[3]{abc} \\\\\n  &= (\\sqrt{a} - \\sqrt{b})^2 + \\left(\\sqrt{ab} + \\sqrt{ab} + c - 3\\sqrt[3]{\\sqrt{ab}\\sqrt{abc}}\\right) \\\\\n  &\\ge (\\sqrt{a} - \\sqrt{b})^2 > 0,\n\\end{aligned}\n$$\n\n從而 $\\sqrt{a} - \\sqrt{b} = 1$。又由於 $ab$ 為完全平方數，知 $a$ 與 $b$ 皆為完全平方數，故存在 $n \\in \\mathbb{N}$ 使得 $a = (n+1)^2$ 且 $b = n^2$。又由於上述算幾不等式的等號必須成立，故有 $c = \\sqrt{ab} = n(n+1)$。以上解代回驗證成立。證畢。\n\n註：存在滿足該式但 $ab$ 不為完全平方數之解，如 $(a, b, c) = (108, 125, 128)$。\n\n¹例如透過等式 $\\sqrt{a} = (a - b + 1)/2$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 14253,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $ABC$ be an acute-angled triangle with $AB < AC$, and let points $D$ and $E$ be chosen on the sides $AC$ and $BC$ respectively such that $AD = AE = AB$. The circumcircle of $ABE$ intersects the line $AC$ at $A$ and $F$, and the line $DE$ at $E$ and $P$. Prove that $P$ is the circumcentre of $BDF$.",
    "options": [],
    "answer": "See solution",
    "solution": "Since $AD = AE$, triangle $AED$ is isosceles, so $\\angle ADE = \\angle AED$. Combining this with the fact that $ABPE$ and $ABPF$ are cyclic, we obtain\n\n$$\n\\angle ABP = 180^{\\circ} - \\angle AEP = \\angle AED = \\angle ADE = \\angle ADP\n$$\n\nas well as\n\n$$\n\\angle ABP = 180^{\\circ} - \\angle AFP = \\angle DFP,\n$$\n\nso $\\angle FDP = \\angle DFP$, which implies that $PD = PF$. Likewise, $\\angle ABE = \\angle AEB$ since $AB = AE$, which means that\n\n$$\n\\angle APB = \\angle AEB = \\angle ABE = \\angle APE = \\angle APD.\n$$\n\nNow we see that triangles $ABP$ and $ADP$ have the same angles ($\\angle APB = \\angle APD$, $\\angle ABP = \\angle ADP$) and the common side $AP$, so they are congruent. Thus $PB = PD = PF$, which means that $P$ is the circumcentre of $BDF$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 14254,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $I$ be the incenter of 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$.\n\nLemma: 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\n![](images/2022_CGMO_p4_data_95095fb25c.png)\n\n",
    "options": [],
    "answer": "See solution",
    "solution": "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$.\n\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$.\n\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$.\n\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\n![](images/2022_CGMO_p4_data_a8db3efffb.png)\n\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$$\n\\angle BTI = \\angle BTS + \\angle ITS = \\angle ABI + \\angle AIB = 180^{\\circ} - \\angle BAI.\n$$\n\nLet $H$ be the orthocenter of triangle $AIB$. By the properties of the orthocenter, $\\angle BHI = \\angle BAI$. Thus, points $B, T, I, H$ are concyclic. Therefore, $\\angle STH = \\angle BTH - \\angle BTS = \\angle BIH - \\angle ABI = 90^{\\circ}$, and combined with $\\angle AC_1H = 90^{\\circ}$, we have points $A, C_1, T, H$ concyclic, with $AH$ being the diameter of the circle.\n\nThus,\n$$\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$$\n\nMoreover, note that\n$$\n\\angle ILB_1 = \\angle IC_1L + \\angle LIC_1 = \\frac{1}{2}\\angle BAC + \\angle ABA_1 = \\angle BAI + 2\\angle ABI.\n$$\n\nSo $\\angle ITC_1 = \\angle ILB_1$, so $I, T, C_1, L$ are concyclic. From this, we deduce $\\angle ITL = \\angle IC_1L = \\angle BAI$. Therefore, points $B, T, L$ are collinear, and $\\angle LBI = \\angle TBI = \\angle BAS$.\n\nSince $AS$ is the median of triangle $BIJ$, we have $AS \\parallel BJ$, and $\\angle BAS = \\angle ABJ$. Therefore, $\\angle ABJ = \\angle LBI$. The conclusion is proven. $\\square$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 14255,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $a$, $b$, $c$, $d$ be integers with $a, c$ both nonzero, and define $x_n = \\gcd(an + b, cn + d)$ for all positive integers $n$. Show that the sequence $x_1, x_2, x_3, \\dots$ is unbounded if and only if\n\n$$\nad = bc.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "Note that\n\n$$\n\\begin{aligned}\n\\frac{\\gcd(acn + bc, acn + ad)}{ac} &\\leq \\gcd(an + b, cn + d) \\\\\n&\\leq \\gcd(acn + bc, acn + ad).\n\\end{aligned}\n$$\n\nSuppose $ad = bc$. We have\n\n$$\n\\begin{aligned}\n\\gcd(an + b, cn + d) &\\geq \\frac{\\gcd(acn + bc, acn + ad)}{ac} \\\\\n&= \\frac{\\gcd(acn + bc, acn + bc)}{ac} \\\\\n&= n + \\frac{b}{a}.\n\\end{aligned}\n$$\n\nSince $n + \\frac{b}{a}$ is unbounded, $\\gcd(an+b, cn+d)$ is also unbounded as desired.\n\nOn the other hand, suppose $ad \\neq bc$. Then $|ad - bc| \\neq 0$, so\n\n$$\n\\begin{aligned}\n\\gcd(an + b, cn + d) &\\leq \\gcd(acn + bc, acn + ad) \\\\\n&= \\gcd(acn + bc, ad - bc) \\\\\n&\\leq |ad - bc|.\n\\end{aligned}\n$$\n\nTherefore, $\\gcd(an+b, cn+d)$ is bounded above by a fixed number that is independent of $n$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 14256,
    "subject": "Mathematics (Olympiad)",
    "question": "找出所有 $p$ 為質數的正整數數對 $(a, p)$，使得 $p^a + a^4$ 為完全平方數。",
    "options": [],
    "answer": "See solution",
    "solution": "$(a, p) = (1, 3),\n(2, 3),\n(6, 3),\n(9, 3)$ 是所有可能的解。\n\n令 $p^a + a^4 = b^2$，其中 $b$ 為正整數。則有：\n\n$$\np^a = b^2 - a^4 = (b + a^2)(b - a^2)\n$$\n\n因此 $b + a^2$ 和 $b - a^2$ 都是 $p$ 的冪。\n\n設 $b - a^2 = p^x$，則 $b + a^2 = p^{a-x}$ 且 $a - x > x$。因此：\n\n$$\n2a^2 = (b + a^2) - (b - a^2) = p^{a-x} - p^x = p^x(p^{a-2x} - 1)\n$$\n\n分兩種情況討論：\n\n**情況 1 ($p = 2$)：**\n\n$$\na^2 = 2^{x=1}(2^{a-2x} - 1) = 2^{2v_2(a)}(2^{a-2x} - 1)\n$$\n\n其中第二個等式來自 $\\gcd(2, 2^{a-2x} - 1) = 1$。所以 $2^{a-2x} - 1$ 是平方數。\n\n若 $v_2(a) > 0$，則 $2^{a-2x}$ 也是平方數，故 $2^{a-2x} - 1 = 0$，$a = 0$，矛盾。\n\n若 $v_2(a) = 0$，則 $x = 1$，且 $a^2 = 2^{a-2} - 1$。若 $a \\ge 4$，右式模 4 餘 3，不能為平方。檢查 $a = 1, 2, 3$ 也不成立。\n\n因此此情況無解。\n\n**情況 2 ($p \\neq 2$)：**\n\n有 $2v_p(a) = x$。設 $m = v_p(a)$，則 $a^2 = p^{2m} n^2$，$n \\ge 1$。\n\n所以 $2n^2 = p^{a-2x} - 1 = p^{p^m n - 4m} - 1$。\n\n分兩子情況：\n\n*子情況 2-1 ($p \\ge 5$)：*\n\n歸納可證 $p^m \\ge 5^m > 4m$，則\n\n$$\n2n^2 + 1 = p^{p^m n - 4m} > p^{p^m(n-1)} \\ge 5^{5^{m(n-1)}} \\ge 5^{n-1}\n$$\n\n歸納可證 $5^{n-1} > 2n^2 + 1$，$n \\ge 3$。故 $n = 1$ 或 $2$，但此時 $p = 3$，矛盾。無解。\n\n*子情況 2-2 ($p = 3$)：*\n\n有 $2n^2 + 1 = 3^{3m n - 4m}$。若 $m \\ge 2$，歸納可證 $3^m > 4m$，則\n\n$$\n2n^2 + 1 = 3^{3m n - 4m} > 3^{3m(n-1)} \\ge 3^{9(n-1)}\n$$\n\n歸納可證 $3^{9(n-1)} > 2n^2 + 1$，$n \\ge 2$。故 $n = 1$。此時 $3^m - 4m = 1$，唯一解 $m = 2$，$a = 3^2 \\cdot 1 = 9$。\n\n若 $m \\le 1$，分 $m = 0$ 或 $m = 1$：\n\n- $m = 1$：$2n^2 + 1 = 3^{3n-4}$，歸納可證 $3^{3n-4} > 2n^2 + 1$，$n \\ge 3$。檢查 $n = 1, 2$，僅 $n = 2$ 成立，$a = 3^1 \\cdot 2 = 6$。\n- $m = 0$：$2n^2 + 1 = 3^n$，歸納可證 $3^n > 2n^2 + 1$，$n \\ge 3$。檢查 $n = 1, 2$，得 $a = 1, 2$。\n\n因此 $(a, p) = (1, 3), (2, 3), (6, 3), (9, 3)$ 為所有可能解。",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 14257,
    "subject": "Mathematics (Olympiad)",
    "question": "Determine whether there exists a non-empty subset $S$ of a $7 \\times 7$ square grid with the following property: For every tetromino depicted below, there exists a covering of $S$ by copies of that tetromino. Note that we are allowed to flip or turn the tetrominoes around.\n\n![](images/CZE_ABooklet_2024_p2_data_28eb64e3e5.png)",
    "options": [],
    "answer": "See solution",
    "solution": "We shall exhibit an example of such a set $S$.\n\nFirst, note that we can simplify the problem by strengthening the requirement$^1$ and asking that $S$ admits a covering by two kinds of octaminoes, such that each of the octaminoes can be covered by two copies of a certain tetromino.\n\n![](images/CZE_ABooklet_2024_p3_data_6e881b1f13.png)\n\nNow, we can easily see that the following set $S$ admits coverings by both of the desired octaminoes.\n\n![](images/CZE_ABooklet_2024_p3_data_1daee92ccc.png)\n\nTherefore, we have proved the answer to the original question is affirmative.\n\n$^1$Note that this has no a priori reason to be equivalent to the original problem! Moreover, at the time of writing, we are not aware whether it is true that any set $S$ with the property from the problem statement also satisfies this stronger property.",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 14258,
    "subject": "Mathematics (Olympiad)",
    "question": "$$\n(a+\\frac{1}{b})^2 + (b+\\frac{1}{c})^2 + (c+\\frac{1}{a})^2 = a^2 + b^2 + c^2 + \\frac{1}{b^2} + \\frac{1}{c^2} + \\frac{1}{a^2} + 2\\frac{a}{b} + 2\\frac{b}{c} + 2\\frac{c}{a} \n$$\n\nShow that\n$$\n(a+\\frac{1}{b})^2 + (b+\\frac{1}{c})^2 + (c+\\frac{1}{a})^2 \\ge ab + ac + bc + \\frac{1}{bc} + \\frac{1}{ca} + \\frac{1}{ab} + 2\\frac{a}{b} + 2\\frac{b}{c} + 2\\frac{c}{a}.\n$$\n\n*Hint:* The equality holds if and only if $a = b = c = 1$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $a = \\frac{x}{y}$, $b = \\frac{y}{z}$, $c = \\frac{z}{x}$.\n\nThen\n$$\n\\left(\\frac{x}{y} + \\frac{z}{y}\\right)^2 + \\left(\\frac{y}{z} + \\frac{x}{z}\\right)^2 + \\left(\\frac{z}{x} + \\frac{y}{x}\\right)^2 \\geq 3\\left(\\frac{x}{y} + \\frac{y}{z} + \\frac{z}{x} + 1\\right)\n$$\nExpanding and simplifying, we get:\n$$\n(x+z)^2 x^2 z^2 + (y+x)^2 y^2 x^2 + (z+y)^2 z^2 y^2 \\geq 3xyz(x^2z + y^2x + z^2y + xyz)\n$$\nOr,\n$$\nx^4 z^2 + 2x^3 z^3 + x^2 z^4 + x^2 y^4 + 2x^3 y^3 + x^4 y^2 + y^2 z^4 + 2y^3 z^3 + y^4 z^2 \\geq 3x^3 y z^2 + 3x^2 y^3 z + 3x y^2 z^3 + 3x^2 y^2 z^2\n$$\nNow, by the AM-GM inequality:\n\n1) $x^3 y^3 + y^3 z^3 + z^3 x^3 \\geq 3x^2 y^2 z^2$\n\n2) $x^4 z^2 + z^4 x^2 \\geq 2x^3 z^3$\n\n3) $x^4 y^2 + y^4 x^2 + y^3 z^3 \\geq 3x^2 y^2 z$\n\n4) $z^4 y^2 + y^4 z^2 + x^3 z^3 \\geq 3x y^2 z^3$\n\nEquality holds when $x = y = z$, i.e., $a = b = c = 1$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 14259,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $f$ be a primitive polynomial with integer coefficients (their greatest common divisor is $1$) such that $f$ is irreducible in $\\mathbb{Q}[X]$, and $f(X^2)$ is reducible in $\\mathbb{Q}[X]$. Show that $f = \\pm(u^2 - Xv^2)$ for some polynomials $u$ and $v$ with integer coefficients.\n\nFor example, if $a$ and $b$ are coprime integers and $a$ is odd, then $f = a^4 X^2 + 4b^4$ is a primitive polynomial in $\\mathbb{Z}[X]$, irreducible in $\\mathbb{Q}[X]$, $f(X^2) = a^4 X^4 + 4b^4 = (a^2 X^2 - 2abX + 2b^2)(a^2 X^2 + 2abX + 2b^2)$ is reducible in $\\mathbb{Q}[X]$, and $f = (a^2 X + 2b^2)^2 - X (2ab)^2$.",
    "options": [],
    "answer": "See solution",
    "solution": "Unless otherwise stated, we work in $\\mathbb{Q}[X]$. Since the case $\\deg f = 1$ is easily dealt with, let $\\deg f \\geq 2$ and write $f(X^2) = gh$, where $g$ and $h$ both have positive degree, and $g$ is irreducible. Next, write $g = a(X^2) + Xb(X^2)$ and $h = c(X^2) + Xd(X^2)$ to infer (from $f(X^2) = gh$ by an argument on the parity of degrees) that\n\n$$\nad + bc = 0. \\tag{1}$$\n\nSo $f = ac + Xbd$, whence\n\n$$\naf = (a^2 - Xb^2)c. \\tag{2}$$\n\nWe now show that $a$ and $b$ are coprime. Alternatively, $\\delta = \\gcd(a, b)$ is a constant. Write $a = a_1\\delta$ and $b = b_1\\delta$, and refer to the irreducibility of $g$ to deduce that $\\delta(X^2)$ is either associated with $g$, a case to be ruled out, or a constant, in which case we are done.\n\nIn the former case, $a_1(X^2) + Xb_1(X^2)$ is a constant, so $b_1 = 0$, whence $b = 0$ and $g = a(X^2)$, and (1) forces one of $a$ and $d$ to be $0$. Since $g$ is not constant, $a = 0$ is ruled out, so $d = 0$, $f = ac$ and $h = c(X^2)$. Since $f$ is irreducible, one of $a$ and $c$ must be a constant, hence so must one of $g$ and $h$ — a contradiction, since both have positive degree. Thus, $b \\neq 0$.\n\nFurther, $a$ and $X$ are coprime: otherwise, $a(0) = 0$, so $g(0) = 0$, hence $f(0) = 0$, contradicting the irreducibility of $f$ and $\\deg f \\geq 2$.\n\nConsequently, $a$ and $a^2 - Xb^2$ are coprime, so $a$ divides $c$ by (2), and $f = (a^2 - Xb^2)c_1$ for some $c_1$ in $\\mathbb{Q}[X]$. Since $f$ is irreducible, one of $a^2 - Xb^2$ and $c_1$ must be a constant. Since $b \\neq 0$, $a^2 - Xb^2$ cannot be constant, so $c_1$ is a constant.\n\nFrom now on we work in $\\mathbb{Z}[X]$. By the preceding, $nf = m(u^2 - Xv^2)$ for some integers $m$ and $n$, and some $u$ and $v$ in $\\mathbb{Z}[X]$. Fix a prime $p$ and write $m = p^\\mu m_1$, $n = p^\\nu n_1$, $u = p^\\alpha u_1$, $v = p^\\beta v_1$, where $\\alpha, \\beta, \\mu, \\nu$ are non-negative integers, and none of $m_1, n_1, u_1, v_1$ is divisible by $p$. Let $\\alpha \\leq \\beta$; the case $\\alpha > \\beta$ is similar. Since $f$ is primitive, the relation\n\n$$\np^{\\nu} n_{1} f = p^{\\mu+2\\alpha} m_{1} \\left( u_{1}^{2} - X p^{2(\\beta-\\alpha)} v_{1}^{2} \\right)$$\n\nimplies that $\\nu \\geq \\mu + 2\\alpha$. If we show that $\\nu = \\mu + 2\\alpha$, we are done.\n\nSuppose, if possible, that $\\nu > \\mu + 2\\alpha$, so $p$ divides $u_1^2 - Xp^{2(\\beta-\\alpha)}v_1^2$, so it also divides $u_1^2(X^2) - X^2p^{2(\\beta-\\alpha)}v_1^2(X^2) = (u_1(X^2) - Xp^{\\beta-\\alpha}v_1(X^2))(u_1(X^2) + Xp^{\\beta-\\alpha}v_1(X^2))$. Since $p$ is prime, it must divide one of $u_1(X^2) \\pm Xp^{\\beta-\\alpha}v_1(X^2)$, and an argument on the parity of degrees shows that $u_1(X^2)$ must be divisible by $p$, and hence so must $u_1$ — a contradiction. This concludes the proof.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 14260,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $ABC$ be a triangle inscribed in a circle $(O)$, with $AB \\neq AC$. Its incircle touches the side $BC$ at point $D$. Let $I$ be its incenter and let $M$ be the intersection point of the angle bisector $AI$ of $\\angle A$ with the circle $(O)$. The line $DM$ meets again the circle $(O)$ at point $P$. Show that $\\angle API = 90^\\circ$.",
    "options": [],
    "answer": "See solution",
    "solution": "Assume that $AB < AC$.\n\nA well-known result states that $MB = MC = MI$. Indeed, from the given it follows that $\\angle IBM = \\frac{1}{2}\\angle(A + B) = \\angle BIM$.\n\nFurthermore, since $\\angle BPM = \\frac{1}{2}\\angle A = \\angle MBC$, one has $\\triangle BDM \\sim \\triangle BPM$, hence $MB^2 = MD \\cdot MP$. Then $MI^2 = MD \\cdot MP$, so $\\triangle MDI \\sim \\triangle IPM$, and consequently $\\angle MPI = \\angle MID$.\n\nNotice now that $ID \\parallel OM$, implying $\\angle MDI = \\angle AMO$. Let $N$ be the antipode of $M$. Finally, $\\angle API = \\angle APM - \\angle MPI = \\angle APM - \\angle AMN = \\frac{\\widehat{AM} - \\widehat{AN}}{2} = \\frac{\\widehat{MN}}{2} = 90^\\circ$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 14261,
    "subject": "Mathematics (Olympiad)",
    "question": "Prove that there are infinitely many primes $p$ such that each of them divides an integer of the form $3^n - 2$, but does not divide any integer of the form $2^m - 3$, where $m$ and $n$ are positive integers.",
    "options": [],
    "answer": "See solution",
    "solution": "We show the stronger result that the set of primes $p \\equiv \\pm 5 \\pmod{12}$ such that $p \\mid 9^n - 2$ for some $n$ (since $9 = 3^2$, we can work with $9$ instead of $3$, by restricting to even exponents), but $p \\nmid 2^m - 3$ for any $m$, is infinite.\n\nNote that if $p \\mid 9^n - 2$ for some $n$, then $9^{nm} \\equiv 2^m \\equiv 3 \\pmod{p}$ would imply that $3$ is a quadratic residue mod $p$, which contradicts $p \\equiv \\pm 5 \\pmod{12}$. (The latter can be easily checked using the law of quadratic reciprocity.)\n\nThus, it suffices to justify that the set of primes $p \\equiv \\pm 5 \\pmod{12}$ that divide $9^n - 2$ for some $n$ is infinite. Suppose these primes are finite, say $p_1, \\dots, p_k$ (there is at least one such prime, for instance, $p = 7$ for $n = 1$). Then, for $n = \\prod_{i=1}^k (p_i - 1) > 1$, we have $9^n \\equiv 1 \\pmod{p_i}$ by Fermat's little theorem, which implies $p_i \\nmid 9^n - 2$ for each $i$. However, $9^n - 2 \\equiv 7 \\pmod{12}$ must have a prime divisor $q \\equiv \\pm 5 \\pmod{12}$ (if all prime factors were $\\pm 1 \\pmod{12}$, their product with multiplicities yields $9^n - 2 \\equiv \\pm 1 \\pmod{12}$, a contradiction), distinct from each $p_i$. The desired result follows. $\\square$",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 14262,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $\\mathbb{R}$ denote the set of real numbers. Find all functions $f : \\mathbb{R} \\to \\mathbb{R}$ such that\n\n$$\nf(xf(y) + y) + f(-f(x)) = f(yf(x) - y) + y\n$$\n\nfor all $x, y \\in \\mathbb{R}$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $f(0) = c$. We make the following substitutions in the initial equation:\n\n$$\n1) \\quad x = 0, y = 0 \\implies f(0) + f(-c) = f(0) \\implies f(-c) = 0.\n$$\n\n$$\n2) \\quad x = 0, y = -c \\implies f(-c) + f(-c) = f(c - c^2) - c \\implies f(c - c^2) = c.\n$$\n\n$$\n3) \\quad x = -c, y = -c \\implies f(-c) + f(0) = f(c) - c \\implies f(c) = 2c.\n$$\n\n$$\n4) \\quad x = 0, y = c \\implies f(c) + f(-c) = f(c^2 - c) + c \\implies f(c^2 - c) = c.\n$$\n\n$$\n5) \\quad x = -c, y = c^2 - c \\implies f(-c) + f(0) = f(c - c^2) + c^2 - c \\implies c = c^2 \\implies c = 0 \\text{ or } 1.\n$$\n\nSuppose that $c = 0$. Let $f(-1) = d + 1$. We make the following substitutions in the initial equation:\n\n$$\n1) \\quad x = 0 \\implies f(y) + f(0) = f(-y) + y \\implies y - f(y) = -f(-y) \\text{ for any } y \\in \\mathbb{R}.\n$$\n\n$$\n2) \\quad y = 0 \\implies f(0) + f(-f(x)) = f(0) \\implies f(-f(x)) = 0 \\text{ for any } x \\in \\mathbb{R}.\n$$\n\n$$\n3) \\quad x = -1 \\implies f(y - f(y)) + 0 = f(dy) + y \\implies f(dy) = -y + f(-f(-y)) = -y \\text{ for any } y \\in \\mathbb{R}.\n$$\n\nThus, for any $x \\in \\mathbb{R}$ we have $f(x) = f(-f(dx)) = 0$. However, this function does not satisfy the initial equation.\n\nSuppose that $c = 1$. We take $x = 0$ in the initial equation:\n\n$$\nf(y) + f(-c) = f(0) + y \\implies f(y) = y + 1\n$$\n\nfor any $y \\in \\mathbb{R}$. The function satisfies the initial equation.\n\n*Answer:* $f(x) \\equiv x + 1$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 14263,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $d_1, d_2, \\dots, d_n$ be nonnegative real numbers satisfying $1 \\geq d_1 \\geq \\dots \\geq d_n \\geq 0$. Prove\n\n$$\n\\frac{(1 + d_1 + d_2 + \\dots + d_n)^2}{n + 1} \\geq 2 \\cdot \\frac{d_1^2 + 2d_2^2 + \\dots + nd_n^2}{n}.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "Setting $d_0 = 1$, we have $j d_j \\leq d_0 + d_1 + \\dots + d_{j-1}$ for any $1 \\leq j \\leq n$. Hence\n\n$$\n\\sum_{j=1}^{n} j d_j^2 \\leq \\sum_{j=1}^{n} \\sum_{i=0}^{j-1} d_i d_j = \\sum_{0 \\leq i < j \\leq n} d_i d_j \\leq \\frac{n}{2(n+1)} \\left( \\sum_{k=0}^{n} d_k \\right)^2,\n$$\n\nwhere the last part follows from Maclaurin's inequality. Equality holds for $d_1 = d_2 = \\dots = d_n = 1$ only.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 14264,
    "subject": "Mathematics (Olympiad)",
    "question": "Consider any two touching circles of radius $a$ and $b$ that are tangent to a line at points distance $c$ apart.\n\n![](images/2023_Australian_Scene_p56_data_7de53b5ec6.png)\n\nProve by induction that, for all $n \\ge 2$, $x_n = \\frac{n+1}{3n+2}$ and $r_n = \\frac{1}{2(3n+2)^2}$, where $x_n$ and $r_n$ are the $x$-coordinate and radius of the $n$th circle in the sequence, respectively.",
    "options": [],
    "answer": "See solution",
    "solution": "1. It turns out that $x_n = \\frac{n+1}{3n+2}$ is in simplest form for all $n \\ge 2$. To see why, suppose that $n+1$ and $3n+2$ have a common factor $d$, then $d$ divides $3(n+1) - (3n+2) = 1$. Hence $d=1$.\n\n2. The numerators and denominators of consecutive fractions $x_n$ and $x_{n+1}$ satisfy a curious equation. If $a_n = n+1$ and $b_n = 3n+2$, then\n\n$$\na_n b_{n+1} - a_{n+1} b_n = (n+1)(3n+5) - (n+2)(3n+2) = 1.\n$$\n\nFor example, $x_2 = 3/8$, $x_3 = 4/11$, and $3 \\times 11 - 4 \\times 8 = 33 - 32 = 1$.\n\n3. The numerators and denominators of consecutive fractions $x_n$, $x_{n+1}$, $x_{n+2}$, satisfy another curious equation. If $a_n = n+1$ and $b_n = 3n+2$, then\n\n$$\n\\frac{a_n + a_{n+2}}{b_n + b_{n+2}} = \\frac{(n+1) + (n+3)}{(3n+2) + (3n+8)} = \\frac{2n+4}{6n+10} = \\frac{n+2}{3n+5} = \\frac{a_{n+1}}{b_{n+1}}\n$$\n\n4. The circles in this problem are examples of *Ford circles*. A Ford circle is a circle $C[a, b]$ whose centre is $(\\frac{a}{b}, \\frac{1}{2b^2})$, where the greatest common divisor of $a$ and $b$ is 1.\n\nIt turns out that two Ford circles $C[a, b]$ and $C[c, d]$ are tangent if $ad - bc = \\pm 1$ and disjoint otherwise. For example, as we saw above, $C[1, 3]$ is tangent to $C[3, 8]$ and $1 \\times 8 - 3 \\times 3 = -1$, $C[3, 8]$ is tangent to $C[4, 11]$ and $3 \\times 11 - 4 \\times 8 = 1$, but $C_2 = C[3, 8]$ is disjoint from $C_4 = C[5, 14]$ and $3 \\times 14 - 5 \\times 8 = 2 \\neq \\pm 1$.\n\nIt is also true that $C[a, b]$ and $C[c, d]$ are both tangent to $C[a+c, b+d]$.\n\nFor example, $C[1, 3]$ and $C[3, 8]$ are tangent to $C[4, 11]$, and $C[a_n, b_n]$ and $C[a_{n+2}, b_{n+2}]$ are tangent to $C[a_n + a_{n+2}, b_n + b_{n+2}] = C[a_{n+1}, b_{n+1}]$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 14265,
    "subject": "Mathematics (Olympiad)",
    "question": "Suppose there are $2013$ boys and $2013$ girls. Each club consists of some boys and some girls, and each club contains at least one boy and one girl. What is the maximum possible value of $k$, the largest number of members in any club, if the total number of pairs (boy, girl) who are in the same club is at least $2013^2$?",
    "options": [],
    "answer": "See solution",
    "solution": "By the pigeonhole principle, there is a club with $n \\ge \\frac{2013^2}{25}$ pairs. Suppose there are $a$ boys and $b$ girls in this club. Then the number of pairs is at most $ab$. By the AM-GM inequality, we have\n$$\n\\frac{a+b}{2} \\ge \\sqrt{ab} \\ge \\sqrt{n} \\ge \\frac{2013}{5}.\n$$\nThis implies the number of members of this club is $a + b \\ge 806$.\n\nWe now give a construction for which $k \\le 806$. We partition the boys into 5 groups $B_1, B_2, \\dots, B_5$ such that each of $B_1, B_2, B_3$ has 403 boys, while each of $B_4, B_5$ has 402 boys. Similarly, we partition the girls into 5 groups $G_1, G_2, \\dots, G_5$ in a similar way. Suppose each boy in group $B_i$ and each girl in group $G_j$ choose the club $C_{ij}$. Then each club consists of at most $403+403=806$ members. This proves $k \\le 806$.\n\nIt follows that the maximum $k$ is 806.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 14266,
    "subject": "Mathematics (Olympiad)",
    "question": "On the board, 777 pairwise distinct *complex* numbers are written. It turned out that there are exactly 760 ways to choose a pair of numbers $a$ and $b$ written on the board so that\n$$\na^2 + b^2 + 1 = 2ab.\n$$\n(Here pairs are considered unordered, i.e., $(a, b)$ and $(b, a)$ are the same pair.) Prove that one can choose numbers $c$ and $d$ written on the board such that\n$$\nc^2 + d^2 + 2025 = 2cd.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "Note that the condition $a^2 + b^2 + 1 = 2ab$ is equivalent to $(a-b)^2 = -1$, or $a-b = \\pm i$. Consider a graph whose vertices are the numbers written on the board, with an edge connecting two numbers if they differ by $i$. According to the problem's condition, this graph has exactly 760 edges.\n\nEach connected component of this graph forms a path consisting of numbers of the form $z, z+i, z+2i, \\dots, z+(n-1)i$. Suppose the graph has $k$ connected components. Then it contains $777-k$ edges, so $k=17$.\n\nSince $17 \\cdot 45 = 765 < 777$, at least one connected component must have at least 46 vertices. Therefore, there exist two numbers in this component, say $c$ and $d = c + 45i$. Then $(c-d)^2 = 45^2 \\cdot i^2 = -2025$, which implies $c^2 + d^2 + 2025 = 2cd$, as required.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 14267,
    "subject": "Mathematics (Olympiad)",
    "question": "正整數 $x_1, x_2, \\dots, x_n$（$n \\ge 4$）依序排列在圓周上，使得任意 $x_i$ 的左右鄰居之數字和會是 $x_i$ 本身的倍數，也就是分數\n\n$$\n\\frac{x_{i-1} + x_{i+1}}{x_i} = k_i\n$$\n\n是一個整數，其中指定 $x_0 = x_n,\\ x_{n+1} = x_1$。試證：所有倍數和 $k_1 + k_2 + \\dots + k_n$ 滿足不等式\n\n$$\n2n \\leq k_1 + k_2 + \\dots + k_n < 3n.\n$$\n\nSuppose $x_1, x_2, \\dots, x_n$, with $n \\ge 4$, are positive integers arranged in order around a circle so that the sum of the neighbors of each $x_i$ is a multiple of $x_i$ itself, i.e.,\n\n$$\n\\frac{x_{i-1} + x_{i+1}}{x_i} = k_i\n$$\n\nis an integer, where $x_0 = x_n$ and $x_{n+1} = x_1$. Prove that the sum $k_1 + k_2 + \\dots + k_n$ satisfies\n\n$$\n2n \\leq k_1 + k_2 + \\dots + k_n < 3n.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "左邊的不等式可輕易地由 A.M. $\\geq$ G.M. 證出，例如：\n\n$$\n\\begin{aligned}\n\\sum_{j=1}^{n} k_j &= \\sum_{j=1}^{n} \\left( \\frac{x_{j-1}}{x_j} + \\frac{x_{j+1}}{x_j} \\right) \\\\\n&= \\sum_{j=1}^{n} \\left( \\frac{x_{j-1}}{x_j} + \\frac{x_{j+1}}{x_j} \\right) \\\\\n&= \\sum_{j=1}^{n} \\left( \\frac{x_{j-1}}{x_j} + \\frac{x_{j+1}}{x_j} \\right) \\\\\n&\\geq 2 \\sum_{j=1}^{n} \\sqrt{\\frac{x_{j-1}}{x_j} \\cdot \\frac{x_{j+1}}{x_j}} = 2n.\n\\end{aligned}\n$$\n\n以下我們處理右邊的不等式。其實 $n \\geq 4$ 的假設不那麼重要，我們將對所有的正整數 $n$ 證明。我們將中間的連加式記為 $S_n$，要證的就是 $S_n < 3n$。\n\n當 $n=1$ 時，$x_1$ 的左右鄰居就它自己。因此\n\n$$\nS_1 = k_1 = \\frac{x_1 + x_1}{x_1} = 2 < 3 = 3 \\cdot 1.\n$$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 14268,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all functions $f: \\mathbb{N} \\to \\mathbb{N}$ such that for all positive integers $a$ and $b$:\n\n1. $a - b \\mid f(a) - f(b)$,\n2. $f(\\varphi(a)) = \\varphi(f(a))$,\n\nwhere $\\varphi$ denotes Euler's totient function.",
    "options": [],
    "answer": "See solution",
    "solution": "The solutions are $f(x) \\equiv 1$ and $f(x) = x$.\n\nBy setting $a = 1$ in $f(\\varphi(a)) = \\varphi(f(a))$, we get $f(1) = \\varphi(f(1))$, so $f(1) = 1$. Setting $a = 2$ gives $f(1) = \\varphi(f(2))$, so $f(2)$ is either $1$ or $2$.\n\n**Case 1:** $f(2) = 1$.\n\nSince $1 = f(2) = f(\\varphi(3)) = \\varphi(f(3))$, $f(3)$ is either $1$ or $2$. But since $3-1 \\mid f(3) - f(1)$, $f(3) = 1$. By induction, $f(n) = 1$ for all $n$. Assume $f(a) = 1$ for $a = 1, 2, \\dots, n-1$. Since $f(\\varphi(n)) = \\varphi(f(n))$ and $\\varphi(n) \\le n-1$, $\\varphi(f(n)) = 1$, so $f(n)$ is $1$ or $2$. But $n - (n-2) \\mid f(n) - f(n-2)$, so $f(n) = 1$.\n\n**Case 2:** $f(2) = 2$.\n\nBy induction, $f(n) = n$ for all $n$. Assume $f(a) = a$ for $a = 1, 2, \\dots, n-1$. For each $1 \\le a \\le n-1$, $n-a \\mid f(n)-a$, so $n-a \\mid f(n)-n$. Thus, for $1 \\le m \\le n-1$, $m \\mid f(n)-n$. The greatest common divisor $M$ of $1, 2, \\dots, n-1$ divides $f(n)-n$, so $f(n) = Ms + n$ for some integer $s$. Since $\\varphi(n) < n$ and $\\varphi(n) = \\varphi(Ms+n)$, and for each $k$ with $(n, k) = 1$ and $k < n$, $k \\mid M$, so $(k, Ms+n) = 1$. Therefore, $\\varphi(n) \\le \\varphi(Ms+n)$. If $s \\ne 0$, then $(Ms+n-1, Ms+n) = 1$, so $\\varphi(n) < \\varphi(Ms+n)$, a contradiction. Thus, $s = 0$ and $f(n) = n$.",
    "topic": "Number Theory",
    "subtopic": "Number-Theoretic Functions"
  },
  {
    "id": 14269,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $[x_1, x_2]$ and $[y_1, y_2, y_3]$ denote the least common multiple of $x_1, x_2$ and $y_1, y_2, y_3$ respectively. For any positive integers $a, b, c, d$, let $A$ and $B$ be defined as:\n\n$$\nA = [a, b, c] \\cdot [a, b, d] \\cdot [a, c, d] \\cdot [b, c, d]\n$$\n\nand\n\n$$\nB = [a, b] \\cdot [a, c] \\cdot [a, d] \\cdot [b, c] \\cdot [b, d] \\cdot [c, d].\n$$\n\nShow that $A^6 \\geq B^4$.",
    "options": [],
    "answer": "See solution",
    "solution": "Consider a prime $p$ dividing $a$, $b$, $c$, or $d$. Without loss of generality, let the exponents of $p$ in $a$, $b$, $c$, $d$ be $a_1 \\geq b_1 \\geq c_1 \\geq d_1$ respectively. The largest exponent of $p$ dividing $A$ and $B$ are:\n\n$$\np_A = 3a_1 + b_1, \\quad p_B = 3a_1 + 2b_1 + c_1.\n$$\n\nFor $A^6$ and $B^4$, the exponents are:\n\n$$\np'_A = 6p_A = 18a_1 + 6b_1, \\quad p'_B = 4p_B = 12a_1 + 8b_1 + 4c_1.\n$$\n\nSince $a_1 \\geq b_1 \\geq c_1$, it follows that $18a_1 + 6b_1 \\geq 12a_1 + 8b_1 + 4c_1$, so $A^6 \\geq B^4$ for all primes $p$, and thus for all $a, b, c, d$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 14270,
    "subject": "Mathematics (Olympiad)",
    "question": "Each positive integer is coloured red or blue. A function $f$ from the set of positive integers into itself has the following two properties:\n\n(a) If $x \\le y$, then $f(x) \\le f(y)$.\n\n(b) If $x$, $y$, and $z$ are all (not necessarily distinct) positive integers of the same colour and $x + y = z$, then $f(x) + f(y) = f(z)$.\n\nProve that there exists a positive integer $a$ such that $f(x) \\le a x$ for all positive integers $x$.",
    "options": [],
    "answer": "See solution",
    "solution": "For integer $x, y$, by a segment $[x, y]$ we mean the set of all integers $t$ such that $x \\leq t \\leq y$; the length of this segment is $y - x$.\n\nIf for every two positive integers $x, y$ sharing the same colour we have $\\frac{f(x)}{x} = \\frac{f(y)}{y}$, then one can choose $k = \\lceil \\max\\{f(r)/r, f(b)/b\\} \\rceil$, where $r$ and $b$ are arbitrary red and blue numbers, respectively. So we can assume that there are two red numbers $x, y$ such that $\\frac{f(x)}{x} \\neq \\frac{f(y)}{y}$.\n\nSet $m = x y$. Then each segment of length $m$ contains a blue number. Indeed, assume that all the numbers on the segment $[k, k + m]$ are red. Then\n\n$$\nf(k + m) = f(k + x y) = f(k + x(y - 1)) + f(x) = \\dots = f(k) + y f(x),\n$$\n\n$$\nf(k + m) = f(k + x y) = f(k + (x - 1) y) + f(y) = \\dots = f(k) + x f(y),\n$$\n\nso $y f(x) = x f(y)$ — a contradiction. Now we consider two cases.\n\n*Case 1.* Assume that there exists a segment $[k, k + m]$ of length $m$ consisting of blue numbers. Define $D = \\max\\{f(k), \\dots, f(k + m)\\}$. We claim that $f(x) - f(x - 1) \\leq D$ for any $x > k$, and the conclusion follows. Consider the largest blue number $b_1$ not exceeding $x$, so $x - b_1 \\leq m$, and some blue number $b_2$ on the segment $[b_1 + k, b_1 + k + m]$, so $b_2 > x$. Write $f(b_2) = f(b_1) + f(b_2 - b_1) \\leq f(b_1) + D$ to deduce that $f(x + 1) - f(x) \\leq f(b_2) - f(b_1) \\leq D$, as claimed.\n\n*Case 2.* Each segment of length $m$ contains numbers of both colours. Fix any red number $R \\geq 3m$ such that $R + 1$ is blue and set $D = \\max\\{f(R), f(R + 1)\\}$. Now we claim that $f(x + 1) - f(x) \\leq D$ for any $x > 2m$. Consider the largest red number $r$ not exceeding $x$ and the largest blue number $b$ smaller than $r$; then $x - b = (x - r) + (r - b) \\leq 2m$. Let $y = b + R + 1$; then $y > x$. If $y$ is blue, then $f(y) = f(b) + f(R + 1) \\leq f(b) + D$, and $f(x + 1) - f(x) \\leq f(y) - f(b) \\leq D$. Otherwise, $f(y) = f(b + 1) + f(R) \\leq f(b + 1) + D$ hence $f(x + 1) - f(x) \\leq f(y) - f(b + 1) \\leq D$, as claimed.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 14271,
    "subject": "Mathematics (Olympiad)",
    "question": "For every real number $a$, let $[a]$ be the greatest integer that is not greater than $a$. Find all integers $y$ for which there exists a real number $x$ such that\n$$\n\\left[\\frac{x+23}{8}\\right] = [\\sqrt{x}] = y.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "Let $y$ be such a number. Since $[\\sqrt{x}] = y$, we have $y \\leq \\sqrt{x} < y+1$, so $x \\in [y^2, (y+1)^2)$. Also, $\\left[\\frac{x+23}{8}\\right] = y$ implies $y \\leq \\frac{x+23}{8} < y+1$, so $8y - 23 \\leq x < 8y + 8 - 23 = 8y - 15$. Therefore, $x$ must satisfy both $y^2 \\leq x < (y+1)^2$ and $8y - 23 \\leq x < 8y - 15$. The intersection is nonempty if $y^2 < 8y - 15$, or $(y-3)(y-5) < 0$, so $3 < y < 5$. Since $y$ is an integer, $y = 4$. For $y = 4$, $x$ must satisfy $16 \\leq x < 25$ and $9 \\leq x < 17$. The intersection is $16 \\leq x < 17$, so $x = 16$ works. Thus, the only integer $y$ is $4$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 14272,
    "subject": "Mathematics (Olympiad)",
    "question": "Given a rational number $r = \\frac{p}{q} \\in (0, 1)$, where $p$ and $q$ are coprime positive integers and $pq$ divides $3600$, how many such rational numbers $r$ are there?",
    "options": [],
    "answer": "See solution",
    "solution": "Let $\\Omega = \\{ r \\mid r = \\frac{p}{q},\\ p, q \\in \\mathbb{N}_+,\\ (p, q) = 1,\\ pq \\mid 3600 \\}$. The prime factorization of $3600$ is $2^4 \\times 3^2 \\times 5^2$. Write $p = 2^A 3^B 5^C$, $q = 2^a 3^b 5^c$, with $\\min\\{A, a\\} = \\min\\{B, b\\} = \\min\\{C, c\\} = 0$ and $A + a \\le 4$, $B + b \\le 2$, $C + c \\le 2$.\n\nThere are $9$ ways to choose $(A, a)$, $5$ ways for $(B, b)$, and $5$ for $(C, c)$, so $|\\Omega| = 9 \\times 5 \\times 5 = 225$.\n\nSince $r \\in \\Omega$ if and only if $1/r \\in \\Omega$, and $1 \\in \\Omega$, the elements in $\\Omega \\setminus \\{1\\}$ can be paired as $(r, 1/r)$. Each pair contains exactly one number in $(0, 1)$. Thus, the number of such $r$ is\n\n$$\n\\frac{1}{2} (|\\Omega| - 1) = 112.\n$$",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 14273,
    "subject": "Mathematics (Olympiad)",
    "question": "A sequence of natural numbers is *admissible* if its terms are less than or equal to 100 and its sum is greater than 1810. Find the least $d$ such that each admissible sequence has a subsequence sum in the interval $[1810-d,\\ 1810+d]$.",
    "options": [],
    "answer": "See solution",
    "solution": "Consider the sequence $\\alpha$ with 17 terms equal to 98 and 2 terms equal to 96. Its sum is $17 \\cdot 98 + 2 \\cdot 96 = 1858 > 1810$, so $\\alpha$ is admissible. Note that $\\alpha$ has exactly two subsequence sums in the interval $[1810-48,\\ 1810+48] = [1762,\\ 1858]$. They are its extremes: 1858 (the sum of the entire sequence) and 1762 (the sum of all terms except one 96). This example shows that the minimum $d$ in question is at least 48.\n\nWe show that each admissible sequence has a subsequence sum in the interval $[1762,\\ 1858]$, implying that the answer is $d_{\\min} = 48$. Suppose on the contrary that this is false for an admissible sequence $\\beta$. Still more, it is false for any subsequence of $\\beta$. So by possibly removing terms, one may assume that $\\beta$ is minimal, with sum $S > 1810$ but with sum $\\leq 1810$ for each proper subsequence. In fact, the assumption then implies $S \\geq 1859$ and $T \\leq 1761$ for every proper subsequence sum $T$. In particular, if $t$ is any term of $\\beta$, then $S-t \\leq 1761$. Hence the inequalities $S \\geq 1859$ and $S-t \\leq 1761$ imply $t \\geq 1859-1761=98$. Each admissible sequence has at least 19 terms (having sum $> 1810$ and terms $\\leq 100$). Therefore $S \\geq 98 \\cdot 19 = 1862$.\n\nOn the other hand, we proved the inequality $S-t \\leq 1761$ for any term $t$. Since $t \\leq 100$ by hypothesis, it follows that $S \\leq 1761+100=1861$, which yields a contradiction.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 14274,
    "subject": "Mathematics (Olympiad)",
    "question": "Each point in the plane is assigned a real number such that, for any triangle, the number at the center of its inscribed circle is equal to the arithmetic mean of the three numbers at its vertices.\n\nProve that all points in the plane are assigned the same number.",
    "options": [],
    "answer": "See solution",
    "solution": "We prove that for any distinct points *A* and *B*, the numbers that are assigned to those points are equal. Since *A* and *B* are arbitrary, it follows that all points must be assigned the same number. Without loss of generality, we assume that line *AB* is horizontal and that *A* is to the left of *B*.\n\n**Lemma 1.** Let $X, Y, Z, W$ be four collinear points in that order from left to right such that $XY = ZW > YZ$. Then $x - w = 3(y - z)$.\n\n_Proof:_ Let $O$ be the midpoint of segment $YZ$, and $\\ell$ be the line through $O$ perpendicular to line $XW$. Draw two congruent circles, $\\omega_1$ and $\\omega_2$, centered at $Y$ and $Z$ with radius $YZ/2$.\n\n![](images/USA_IMO_2001_p42_data_9af49e212d.png)\n\nNote that $X$ and $W$ are symmetric with respect to line $\\ell$ and that they are outside of circles $\\omega_1$ and $\\omega_2$. Hence, there are points $P$ and $Q$ on line $\\ell$ such that $\\omega_1$ and $\\omega_2$ are incircles of triangles $XPQ$ and $WPQ$, respectively. Then\n\n$$\nx + p + q = 3y \\quad \\text{and} \\quad w + p + q = 3z,\n$$\n\nwhich implies that $x - w = 3(y - z)$.\n\n![](images/USA_IMO_2001_p42_data_f1ead68ffc.png)\n\nLet $M$ be the midpoint of segment $AB$. Let $E, C, D, F$ be collinear points in that order from left to right such that $4AM = 4BM < 2CM = 2DM < EM = FM$. Applying Lemma 1 to the sets of points $\\{C, A, B, D\\}$, $\\{E, A, B, F\\}$, and $\\{E, C, D, F\\}$ yields $c - d = 3(a - b)$, $e - f = 3(a - b)$, and $e - f = 3(c - d)$. Therefore, $3(a - b) = e - f = 3(c - d) = 9(a - b)$, and consequently, $a = b$, as desired.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 14275,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $n$ be a positive integer. Consider triples $(x_1, y_1, z_1)$, $(x_2, y_2, z_2)$, \\dots, $(x_n, y_n, z_n)$ which consist of integers $1, 2, \\dots, 100$ and satisfy the following condition:\n\nFor all infinite sequences $a_1, a_2, \\dots$ which consist of integers $1, 2, \\dots, 100$, there exist a positive integer $i$ and an integer $j$ with $1 \\leq j \\leq n$ such that $(a_i, a_{i+1}, a_{i+2}) = (x_j, y_j, z_j)$.\n\nDetermine the minimum possible value of $n$.",
    "options": [],
    "answer": "See solution",
    "solution": "Set $N = 100$ and $S = \\{(x_1, y_1, z_1), (x_2, y_2, z_2), \\dots, (x_n, y_n, z_n)\\}$.\n\n**Lemma.** Let $x, y, z$ be integers with $1 \\leq x, y, z \\leq N$. Then at least one of $(x, y, z)$, $(y, z, x)$, or $(z, x, y)$ belongs to $S$.\n\n**Proof.** Consider a sequence $a_1 = x, a_2 = y, a_3 = z, a_4 = x, a_5 = y, a_6 = z, \\dots$. Then, for every positive integer $i$, $(a_i, a_{i+1}, a_{i+2})$ is one of $(x, y, z)$, $(y, z, x)$, or $(z, x, y)$. Therefore, by the assumption of the problem, at least one of these triples must belong to $S$. $\u001a0$\n\nConsider the set\n$$\nX = \\{(x_1, y_1, z_1), (x_2, y_2, z_2), \\dots, (x_n, y_n, z_n), (y_1, z_1, x_1), (y_2, z_2, x_2), \\dots, (y_n, z_n, x_n), \\\\\n(z_1, x_1, y_1), (z_2, x_2, y_2), \\dots, (z_n, x_n, y_n)\\}.\n$$\nLet $p, q, r$ be integers with $1 \\leq p, q, r \\leq N$. Applying the lemma to $(x, y, z) = (p, q, r)$, it follows that at least one of $(p, q, r)$, $(q, r, p)$, or $(r, p, q)$ belongs to $S$, say it is $(x_j, y_j, z_j)$ for some $1 \\leq j \\leq n$. Then, $(p, q, r)$ is equal to one of $(x_j, y_j, z_j)$, $(z_j, x_j, y_j)$, or $(y_j, z_j, x_j)$, and therefore, $(p, q, r)$ belongs to $X$. Thus, $X$ is the set of all $N^3$ triples $(p, q, r)$ with $1 \\leq p, q, r \\leq N$.\n\nBy applying the lemma to $(x, y, z) = (1, 1, 1), (2, 2, 2), \\dots, (N, N, N)$, we find that these triples $(1, 1, 1), (2, 2, 2), \\dots, (N, N, N)$ all belong to $S$. If $(x_j, y_j, z_j)$ is equal to one of them, we have $(x_j, y_j, z_j) = (y_j, z_j, x_j) = (z_j, x_j, y_j)$, hence the number of elements of $X$ is less than or equal to $3n - 2N$. Therefore, $3n - 2N \\geq N^3$, thus $n \\geq \\frac{N^3 + 2N}{3}$.\n\nNext, we construct a set $S$ satisfying $n = \\frac{N^3 + 2N}{3}$. Define $S$ to be the set of triples $(x, y, z)$ with $1 \\leq x, y, z \\leq N$ such that either\n\n* $x > y$ and $x \\geq z$, or\n\n[Solution incomplete in input.]\n",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 14276,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all pairs of positive integers $A$ and $B$, having the same number of digits in the decimal representation, such that $2 \\cdot A \\cdot B = \\overline{AB}$.\n\nHere, $\\overline{AB}$ denotes the number obtained by concatenating $A$ and $B$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $n$ be the number of digits of $A$ and $B$. The given relation is equivalent to $2A \\cdot B = \\overline{AB}$, where $\\overline{AB} = 10^n A + B$.\n\nSo:\n$$\n2A B = 10^n A + B\n$$\nRearrange:\n$$\n2A B - B = 10^n A\n$$\n$$\nB(2A - 1) = 10^n A\n$$\nThus, $2A - 1 \\mid 10^n A$. Since $\\gcd(2A - 1, A) = 1$, $2A - 1 \\mid 10^n$.\n\nSince $A$ has $n$ digits, $10^{n-1} \\leq A < 10^n$.\n\nTry $n = 1$:\n$2A - 1 \\mid 10$\nPossible values: $2A - 1 = 1, 2, 5, 10$; so $A = 1, 1.5, 3, 5.5$. Only integer values: $A = 1, 3$.\n\nIf $A = 1$, $B(2 \\cdot 1 - 1) = 10^1 \\cdot 1 \\implies B = 10$. But $B$ must have one digit, contradiction.\nIf $A = 3$, $B(2 \\cdot 3 - 1) = 10^1 \\cdot 3 \\implies B \\cdot 5 = 30 \\implies B = 6$.\nSo $(A, B) = (3, 6)$.\n\nTry $n = 2$:\n$2A - 1 \\mid 100$\nPossible values: $2A - 1 = 1, 2, 4, 5, 10, 20, 25, 50, 100$; $A = 1, 1.5, 2.5, 3, 5.5, 10.5, 13, 25.5, 50.5$.\nOnly integer values: $A = 1, 3, 13$.\nBut $A$ must have two digits, so $A = 13$.\n\n$B(2 \\cdot 13 - 1) = 100 \\cdot 13 \\implies B \\cdot 25 = 1300 \\implies B = 52$.\nSo $(A, B) = (13, 52)$.\n\n**Final answer:**\n- $(A, B) = (3, 6)$\n- $(A, B) = (13, 52)$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 14277,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $P(x) = a x^3 + (b - a) x^2 - (c + b) x + c$ and $Q(x) = x^4 + (b - 1) x^3 + (a - b) x^2 - (c + a) x + c$ be polynomials in the indeterminate $x$, where $a, b, c$ are nonzero real numbers and $b > 0$.\n\nIf the polynomial $P(x)$ has three different real roots $x_0, x_1, x_2$, which are also roots of the polynomial $Q(x)$, then:\n\n(a) Prove that $abc > 28$.\n\n(b) If $a, b, c$ are nonzero integers with $b > 0$, find their possible values.",
    "options": [],
    "answer": "See solution",
    "solution": "(a) The sum of the coefficients of $P(x)$ is $0$, so $1$ is a root. Thus,\n$$\nP(x) = a x^3 + (b - a) x^2 - (c + b) x + c = (x - 1)(a x^2 + b x - c)\n$$\nIf $x_0 = 1$, by Vieta's formulas:\n$$\nx_1 + x_2 = -\\frac{b}{a}, \\quad x_1 x_2 = -\\frac{c}{a} \\neq 0 \\qquad (1)\n$$\nSince $x_0, x_1, x_2$ are also roots of $Q(x)$, consider\n$$\n\\begin{aligned}\nF(x) &= Q(x) - P(x) = x^4 + (b - a - 1)x^3 + 2(a - b)x^2 + (b - a)x \\\\\n&= x(x^3 + (b - a - 1)x^2 + 2(a - b)x + b - a) \\\\\n&= x(x - 1)[x^2 + (b - a)x + (a - b)]\n\\end{aligned}\n$$\nSo $F(x) = 0$ if $x = 0$, $x = 1$, or $x^2 + (b - a)x + (a - b) = 0$. Since $x_0, x_1, x_2 \\neq 0$, we have:\n$$\nx_0 = 1, \\quad x_1 + x_2 = a - b, \\quad x_1 x_2 = a - b \\qquad (2)\n$$\nFrom (1) and (2): $a - b = -\\frac{b}{a} = -\\frac{c}{a} \\implies$\n$$\nb = c \\qquad (3) \\quad \\text{or} \\quad a^2 - a b = -b \\qquad (4)\n$$\nFrom (3) and (4):\n$$\na^2 = b(a - 1) \\implies a > 1 \\ (b > 0), \\quad b = c = \\frac{a^2}{a - 1}\n$$\nThus,\n$$\nabc = a \\left( \\frac{a^2}{a - 1} \\right)^2 = \\frac{a^5}{(a - 1)^2}\n$$\nLet $x = a - 1 \\geq 1$:\n$$\nabc = \\frac{(x + 1)^5}{x^2} = x^3 + 5x^2 + 10x + 10 + \\frac{5x + 1}{x^2}\n$$\nNow,\n- $x^3 > 0$\n- $5x^2 + \\frac{1}{x^2} > 4$ for $x > 0$\n- $10x + \\frac{5}{x} > 14$ for $x > 0$\n\nTherefore, $abc > 28$.\n\n(b) From $b = a + 1 + \\frac{1}{a - 1}$, since $b, a + 1$ are integers, $\\frac{1}{a - 1} \\in \\mathbb{Z}$, so $a - 1 = \\pm 1$.\n\nIf $a - 1 = 1$, $a = 2$; if $a - 1 = -1$, $a = 0$ (rejected since $a \\neq 0$).\n\nFor $a = 2$:\n$$\nb = c = \\frac{a^2}{a - 1} = 4\n$$\nSo the possible values are $a = 2$, $b = 4$, $c = 4$.\n\nFor these values,\n$$\nP(x) = 2x^3 + 2x^2 - 8x + 4 = 2(x - 1)(x^2 + 2x - 2)\n$$\nhas roots $x_0 = 1$, $x_{1,2} = -1 \\pm \\sqrt{3}$, which are also roots of\n$$\nF(x) = Q(x) - P(x) = x(x - 1)(x^2 + 2x - 2)\n$$\nTherefore, they are roots of $Q(x) = F(x) + P(x)$.",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 14278,
    "subject": "Mathematics (Olympiad)",
    "question": "Бүгд хоорондоо тэнцүү биш эерэг $a, b, c, x, y, z$ тоонуудын хувьд\n\n$$\n(a+b+c) \\left( \\frac{1}{a} + \\frac{1}{b} + \\frac{1}{c} \\right) + \\frac{x^2 + y^2 + z^2}{xy + yz + zx} \\cdot \\frac{ab + bc + ca}{a^2 + b^2 + c^2} \\geq 9 + \\frac{x^2 + y^2 + z^2}{xy + yz + zx}\n$$\n\nтэнцэтгэл биш биелдэг бол $x+y \\ge \\frac{\\sqrt{30+4\\sqrt{2}-4\\sqrt{2}}}{2\\sqrt{2}-1} \\cdot z$ тэнцэтгэл.",
    "options": [],
    "answer": "See solution",
    "solution": "Эхлээд\n\n$$\n(a+b+c) \\left( \\frac{1}{a} + \\frac{1}{b} + \\frac{1}{c} \\right) + \\frac{x^2 + y^2 + z^2}{xy + yz + zx} \\cdot \\lambda \\geq 9 + \\lambda \\quad (1)\n$$\n\nтэнцэтгэл биш биелэх $\\lambda$-ийн хамгийн их утгыг ольё.\n\n(1) $\\Leftrightarrow \\sum (a-b)^2 \\left( \\frac{2(a^2+b^2+c^2)}{ab} - \\lambda \\right) \\ge 0$\n\n$b = c$ гэвэл\n\n$(a-b)^2 \\left( \\frac{2a^2+4b^2}{ab} - \\lambda \\right) \\ge 0$\n\n$\\frac{a}{b} = t$ гэвэл\n\n$2t^2 - \\lambda t + 4 \\ge 0 \\Leftrightarrow D \\le 0 \\Leftrightarrow \\lambda^2 - 4 \\cdot 4\\sqrt{2} \\ge 0 \\Rightarrow \\lambda_{\\max} = 4\\sqrt{2}$.\n\nОдоо $\\lambda = 4\\sqrt{2}$ үед биелнэ гэж харуулъя.\n\n(1) $\\Leftrightarrow \\sum \\frac{(a-b)^2}{ab} (a^2 + b^2 + c^2 - 2\\sqrt{2}ab) \\ge 0$\n\n$\\Leftrightarrow \\sum \\frac{(a-b)^2}{ab} [(a+b - (1+2\\sqrt{2}c)^2(2+2\\sqrt{2})(b-c)(c-a))} \\ge 0$\n\n$(\\sum \\frac{(a-b)^2}{ab} (b-c)(c-a) = 0)$\n\n$\\Leftrightarrow \\sum \\frac{(a-b)^2}{ab} (a + b - (1 + 2\\sqrt{2}c)^2) \\ge 0$. Эндээс\n\n$$\n4\\sqrt{2} \\ge \\frac{x^2 + y^2 + z^2}{xy + yz + zx} \\quad (2)\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 14279,
    "subject": "Mathematics (Olympiad)",
    "question": "Sea $a_0 < a_1 < a_2 < \\dots$ una sucesión infinita de números enteros positivos. Demostrar que existe un único entero $n \\ge 1$ tal que\n\n$$\na_n < \\frac{a_0 + a_1 + a_2 + \\dots + a_n}{n} \\le a_{n+1}.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "Definamos\n\n$$\nb_n = n a_n - (a_n + a_{n-1} + \\dots + a_1).\n$$\n\nClaramente, $b_1 = 0$, y $b_{n+1} - b_n = n(a_{n+1} - a_n) > 0$, así que la sucesión $b_1, b_2, \\dots$ es una sucesión infinita y estrictamente creciente de enteros. Nótese además que la condición que se desea imponer a $n$ es equivalente a\n\n$$\nb_n < a_0 \\le b_{n+1}.\n$$\n\nSea ahora $C_k = \\{b_k + 1, b_k + 2, \\dots, b_{k+1}\\}$, donde $k$ recorre todos los enteros positivos. Cada uno de estos conjuntos es no vacío por ser $b_{k+1} > b_k$, luego $b_{k+1} \\ge b_k + 1$. Los conjuntos son disjuntos dos a dos porque si $i > j$, el máximo elemento de $C_j$, que es $b_{j+1}$, es menor que el mínimo elemento de $C_i$, que es $b_i + 1 > b_i \\ge b_{j+1}$ por ser la sucesión de los $b_n$ una sucesión creciente de enteros. Como además el mayor elemento de $C_k$ y el menor elemento de $C_{k+1}$ son consecutivos, y los elementos de cada $C_k$ son consecutivos, cada entero mayor o igual que $b_1 + 1 = 1$ está en alguno de los conjuntos. Luego cada entero positivo pertenece a uno y sólo uno de los $C_k$. En concreto, el entero positivo $a_0$ pertenece a uno y sólo uno de los $C_k$, es decir, existe un único entero positivo $n$ tal que $b_n < a_0 \\le b_{n+1}$, como queríamos demostrar.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 14280,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $a$, $b$, $c$, $d$ be positive numbers such that $abcd = 1$. Prove the inequality\n\n$$\n\\frac{1}{\\sqrt{a + 2b + 3c + 10}} + \\frac{1}{\\sqrt{b + 2c + 3d + 10}} + \\frac{1}{\\sqrt{c + 2d + 3a + 10}} + \\frac{1}{\\sqrt{d + 2a + 3b + 10}} \\le 1.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "Let $x, y, z, t$ be positive numbers such that $a = x^4$, $b = y^4$, $c = z^4$, $d = t^4$.\n\nBy the AM-GM inequality, $x^4 + y^4 + z^4 + 1 \\ge 4xyz$, $y^4 + z^4 + 1 + 1 \\ge 4yz$, and $z^4 + 1 + 1 + 1 \\ge 4z$.\n\nTherefore, we have the following estimation for the first fraction:\n\n$$\n\\frac{1}{\\sqrt{x^4 + 2y^4 + 3z^4 + 10}} \\le \\frac{1}{\\sqrt{4xyz + 4yz + 4z + 4}} = \\frac{1}{2\\sqrt{xyz + yz + z + 1}}.\n$$\n\nAnalogous estimations for the other fractions:\n\n$$\n\\begin{aligned}\n\\frac{1}{\\sqrt{b + 2c + 3d + 10}} &\\le \\frac{1}{2\\sqrt{yzt + zt + t + 1}}, \\\\\n\\frac{1}{\\sqrt{c + 2d + 3a + 10}} &\\le \\frac{1}{2\\sqrt{ztx + tx + x + 1}}, \\\\\n\\frac{1}{\\sqrt{d + 2a + 3b + 10}} &\\le \\frac{1}{2\\sqrt{txy + xy + y + 1}}.\n\\end{aligned}\n$$\n\nThus, the sum does not exceed\n\n$$\n\\frac{1 + \\sqrt{xyz} + \\sqrt{yz} + \\sqrt{z}}{2\\sqrt{xyz + yz + z + 1}}.\n$$\n\nIt remains to apply the inequality $\\sqrt{\\alpha} + \\sqrt{\\beta} + \\sqrt{\\gamma} + \\sqrt{\\delta} \\le 2\\sqrt{\\alpha + \\beta + \\gamma + \\delta}$, which can be proven by squaring both sides or by the inequality between arithmetic and quadratic means.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 14281,
    "subject": "Mathematics (Olympiad)",
    "question": "Let the 9-element set $A = \\{a + bi \\mid a, b \\in \\{1, 2, 3\\}\\}$, where $i$ is the imaginary unit. Let $\\alpha = (z_1, z_2, \\dots, z_9)$ be a permutation of all elements in $A$ such that $|z_1| \\leq |z_2| \\leq \\dots \\leq |z_9|$. How many such permutations $\\alpha$ are there?",
    "options": [],
    "answer": "See solution",
    "solution": "Since\n\n$$\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$$\n\nit follows that\n\n$$\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$$\n\nBy the multiplication principle, the number of permutations $\\alpha$ satisfying the condition is $2^3 = 8$.",
    "topic": "Discrete Mathematics",
    "subtopic": "Combinatorics"
  },
  {
    "id": 14282,
    "subject": "Mathematics (Olympiad)",
    "question": "The real numbers $x, y, z$, with $x \\neq z$, are mutually different and nonzero and they satisfy the following equations:\n\n$$\n(x+y)^2 + (2-xy) = 9, \\\\\n(y+z)^2 - (3+yz) = 4.\n$$\n\nDetermine the value of the expression\n\n$$\nA = \\left( \\frac{x}{y} + \\frac{y^2}{x^2} + \\frac{z^3}{x^2y} \\right) \\left( \\frac{y}{z} + \\frac{z^2}{y^2} + \\frac{x^3}{y^2z} \\right) \\left( \\frac{z}{x} + \\frac{x^2}{z^2} + \\frac{y^3}{z^2x} \\right).\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "The given equalities can be written:\n\n$$\nx^2 + y^2 + xy = 7, \\qquad (1)\n$$\n\n$$\ny^2 + z^2 + yz = 7, \\qquad (2)\n$$\n\nBy subtraction we get:\n\n$$\nx^2 - z^2 + xy - yz = 0 \\Leftrightarrow (x-z)(x+z) + y(x-z) = 0 \\Leftrightarrow (x-z)(x+z+y) = 0.\n$$\n\nSince $x - z \\neq 0$, we get:\n\n$$\nx + y + z = 0. \\qquad (3)\n$$\n\nThen with simple manipulations for the factors of $A$ we find\n\n$$\nA = \\left( \\frac{x^3 + y^3 + z^3}{xyz} \\right)^3. \\qquad (4)\n$$\n\nFrom (3), using Euler's identity, we find\n\n$$\nx^3 + y^3 + z^3 = 3xyz. \\qquad (5)\n$$\n\nHence from (4) and (5) we get $A = 27$.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 14283,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $n > 0$ be an integer. We are given a balance and $n$ weights of weight $2^0, 2^1, \\dots, 2^{n-1}$. We are to place each of the $n$ weights on the balance, one after another, in such a way that the right pan is never heavier than the left pan. At each step, we choose one of the weights that have not yet been placed on the balance and place it on either the left pan or the right pan, until all the weights have been placed.\n\nDetermine the number of ways in which this can be done.",
    "options": [],
    "answer": "See solution",
    "solution": "The number of ways is $(2n-1)!! = 1 \\times 3 \\times 5 \\times \\dots \\times (2n-1)$.\n\nWe prove the problem by induction on $n$.\n\n*Base case*: For $n = 1$, there is only one way: put the single weight on the left pan.\n\n*Inductive step*: Suppose for $n = k$, the number of ways is $(2k-1)!!$.\n\nFor $n = k+1$, consider the $k+1$ weights: $1/2, 1, 2, \\dots, 2^{k-1}$ (after scaling by $1/2$ for convenience). For any $r \\ge 1$,\n\n$$\n2^r > 2^{r-1} + 2^{r-2} + \\dots + 1 + \\frac{1}{2} \\ge \\sum_{i=-1}^{r-1} \\pm 2^i.\n$$\n\nThe heaviest weight must always be on the left pan. Now, consider the position of the weight $1/2$ in the sequence:\n\n(a) If $1/2$ is placed first, it must go on the left pan. The remaining $k$ weights can be arranged in $(2k-1)!!$ ways.\n\n(b) If $1/2$ is placed at step $t = 2, 3, \\dots, k+1$, it can go on either pan, so for each such $t$ there are $(2k-1)!!$ ways.\n\nThus, the total number of ways is $(2k-1)!! + k \\times (2k-1)!! = (1+2k)(2k-1)!! = (2k+1)!!$.\n\nThis completes the induction.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 14284,
    "subject": "Mathematics (Olympiad)",
    "question": "If, for a pair of integers $a, b$ and a positive integer $m$, $a - b$ is divisible by $m$, we write $a - b \\equiv 0 \\pmod{m}$.\n\nFind all positive integers $n$ such that for every integer $x$, if $x^n - 1$ is divisible by a prime $p$, then $x^n - 1$ is divisible by $p^2$ as well.",
    "options": [],
    "answer": "See solution",
    "solution": "We will show that the numbers $n$ we seek are those of the form $kp$, where $k$ is a positive integer.\n\nLet us first show that if $n$ satisfies the condition of the problem, then $n$ must be a multiple of $p$. To see this, let $x = p + 1$. Then from $x^n - 1 \\equiv 1^n - 1 \\equiv 0 \\pmod{p}$, we get that $x^n - 1$ is divisible by $p$, and hence by the condition of the problem, is divisible by $p^2$ as well. By using the binomial expansion, we get $(p+1)^n \\equiv np + 1 \\pmod{p^2}$, and therefore, $x^n - 1 \\equiv np \\pmod{p^2}$. This implies that $np$ is a multiple of $p^2$, and therefore, $n$ must be a multiple of $p$.\n\nConversely, we will show that if $n = kp$ for a positive integer $k$, then it satisfies the condition of the problem. First, we note that $x^n = (x^k)^p \\equiv x^k \\pmod{p}$ by Fermat's Little Theorem. This shows that if $x^n - 1$ is a multiple of $p$, then so is $x^k - 1$. Now we have\n\n$$\n\\frac{x^n - 1}{x^k - 1} = 1 + x^k + x^{2k} + \\dots + x^{(p-1)k} \\equiv \\underbrace{1+1+\\dots+1}_{p} \\equiv 0 \\pmod{p},\n$$\n\nfrom which we conclude that $x^n - 1 = (x^k - 1) \\times \\frac{x^n - 1}{x^k - 1}$ is a multiple of $p^2$. Thus, we see that $n = kp$ satisfies the condition of the problem.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 14285,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $n \\ge 2$. There are $2n$ people standing in a circle, each holding a different painting in an art conference. Each person has their own fixed preference order of the paintings, which can be different from person to person. A trade of paintings between two adjacent people can happen if and only if both people are getting a painting they prefer more. What is the maximum number of trades that can happen?\n",
    "options": [],
    "answer": "See solution",
    "solution": "Answer: $n(2n - 1)$.\n\nConsider the position of the painting for each person in their preference order, and let $S$ denote the sum of these positions. The maximum of $S$ is $2n \\cdot 2n = 4n^2$ and the minimum of $S$ is $2n \\cdot 1 = 2n$. For each trade, $S$ is reduced by at least 2. Hence, the maximum number of trades is $n(2n - 1)$.\n\nNow we show that $n(2n - 1)$ trades can happen. Number the paintings from $1$ to $2n$ in clockwise direction. Let odd-numbered paintings go clockwise and even-numbered paintings go counter-clockwise. After $n$ trades, all odd-numbered paintings are shifted one position clockwise and all even-numbered paintings are shifted one position counter-clockwise. Repeating this $2n$ times returns the paintings to the initial position. Repeating $2n - 1$ times gives $n(2n - 1)$ trades. This determines the preference order for each person.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 14286,
    "subject": "Mathematics (Olympiad)",
    "question": "Pjotr has a bath with a stopper in the bottom that he can fill with two identical taps. The water flows out of both taps at the same constant speed, and the bath empties (if the stopper is not there) at a constant speed.\n\nOn Monday, Pjotr fills the bathtub to the brim by turning one tap fully open, then pulls out the stopper and waits for the bathtub to empty again. Only when the bath is empty does he turn the tap off again.\n\nOn Tuesday, he fills the bath to the brim by turning both taps fully open, after which he pulls out the stopper and waits for the bath to empty again. Only when the bath is empty does he turn the taps off again.\n\nOn both days, Pjotr spent exactly 45 minutes from opening the tap/taps to closing them again.\n\nHow many minutes does it take to completely empty a full bath when both taps are off?",
    "options": [],
    "answer": "See solution",
    "solution": "10",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 14287,
    "subject": "Mathematics (Olympiad)",
    "question": "The excircles of $\\triangle ABC$ touch the sides $AB$, $BC$, and $CA$ at points $M$, $N$, and $P$, respectively. Let $I$ and $O$ be the incenter and the circumcenter of $\\triangle ABC$. Prove that if $AMNP$ is a cyclic quadrilateral, then:\n\na) the points $M$, $P$, and $I$ are collinear;\n\nb) the points $I$, $O$, and $N$ are collinear.",
    "options": [],
    "answer": "See solution",
    "solution": "By Carnot's theorem, the perpendiculars from the points $M$, $N$, and $P$ to the lines $AB$, $BC$, and $CA$, respectively, have a common point $X$. Then the quadrilateral $AMXP$ is cyclic. Now it is easy to see that $X = N$ and hence $AN$ is a diameter of the circumcircle of $AMNP$.\n\na) Denote by $I_B$ and $I_C$ the respective excenters of $\\triangle ABC$. Then Pappus' theorem for the triples of points $(I_C, A, I_B)$ and $(B, N, C)$ implies the desired result.\n\nb) Let the incircle of $\\triangle ABC$ touch the sides $AB$ and $AC$ at points $R$ and $Q$, respectively. Then the bisectors of $AB$ and $AC$ coincide with the bisectors of $RM$ and $QP$, respectively, and pass through the midpoint of $IN$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 14288,
    "subject": "Mathematics (Olympiad)",
    "question": "$n$ natural numbers are written on the board. You can add only natural numbers in the form $\\frac{a+b}{a-b}$ where $a$ and $b$ are numbers already written on the board. It appears that by doing so you can make any natural number appear on the board. Calculate the least value of $n$ and find the numbers initially written (consider all the cases).",
    "options": [],
    "answer": "See solution",
    "solution": "As $(a + b) > (a - b)$, you cannot obtain $1$ by performing the allowed operations. Therefore, $1$ must be written on the board, but one number is not enough. Let's show that two numbers will be enough. Let the other number be $x$. $\\frac{x+1}{x-1}$ is the only number which can be obtained in the first step. Since it is a natural number, $\\frac{x+1}{x-1} \\geq 2 \\Rightarrow (x+1) \\geq 2x-2$ or $x \\leq 3$. Thus, the second number should be $2$ or $3$. We obtain the two possible sets: $\\{1,2\\}$ and $\\{1,3\\}$.\n\nLet's prove that both satisfy the condition. As $\\frac{2+1}{2-1} = 3$ and $\\frac{3+1}{3-1} = 2$, in the first step we obtain the set $\\{1,2,3\\}$ in both cases. Now we have to prove that any natural number greater than $3$ can be obtained from these three numbers.\n\nAssume we've already obtained the set $\\{1,2,3,\\ldots,(2k+1)\\}$. Let's show how to obtain the next two numbers. We obtain $\\frac{(k+2)+(k+1)}{(k+2)-(k+1)} = 2k+3$ from $(k+1)$ and $(k+2)$. Next, we obtain $\\frac{(2k+3)+(2k+1)}{(2k+3)-(2k+1)} = 2k+2$ from $(2k+3)$ and $(2k+1)$. This implies the desired result.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 14289,
    "subject": "Mathematics (Olympiad)",
    "question": "If $x^2 + ax + b = x^2 + bx + a$ and $a \\neq b$, find the value of $a + b$.",
    "options": [],
    "answer": "See solution",
    "solution": "Equalizing the left sides gives:\n\n$$x^2 + ax + b = x^2 + bx + a$$\n\nSubtract $x^2$ from both sides:\n\n$$ax + b = bx + a$$\n\nRearrange:\n\n$$(a - b)x + (b - a) = 0$$\n\n$$(a - b)(x - 1) = 0$$\n\nSince $a \\neq b$, $x - 1 = 0$ so $x = 1$.\n\nSubstitute $x = 1$ into one equation:\n\n$$1^2 + a \\cdot 1 + b = 0$$\n\n$$1 + a + b = 0$$\n\nSo $a + b = -1$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 14290,
    "subject": "Mathematics (Olympiad)",
    "question": "Suppose the numbers $1, 2, \\ldots, 27$ are arranged in a circle. For each pair of consecutive numbers, consider their sum. Let $A$ be the largest of these 27 sums, and $B$ the smallest. What are the possible values of $A - B$?\n\n![](images/Argentina_2019_Booklet_p10_data_59809bf940.png)\n\n![](images/Argentina_2019_Booklet_p10_data_589765301a.png)",
    "options": [],
    "answer": "See solution",
    "solution": "It is easy to see that $A - B \\neq 0$. Indeed, $A - B = 0$ would mean all 27 sums are equal, but for three consecutive numbers $x, y, z$, the sums $x+y$ and $y+z$ differ since $x \\neq z$.\n\nWe now prove $A - B \\neq 1$. If $A - B = 1$, then for three consecutive numbers $x, y, z$, the sums $x+y$ and $y+z$ differ by 1, so $x$ and $z$ differ by 1. Consider where 27 is placed; let the previous numbers be $a, b$ and the following $c, d$. Then both $a$ and $d$ must differ by 1 from 27, so both must be 26, which is impossible.\n\nFinally, we show $A - B = 2$ is possible. Write 1, then alternate odd and even numbers: odd numbers in decreasing order, even numbers in increasing order. The first sum is $1+27=28$, and the following sums alternate between 29 and 27. Thus, $A - B = 29 - 27 = 2$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 14291,
    "subject": "Mathematics (Olympiad)",
    "question": "Jeck and Lisa are playing a game on an $m \\times n$ board, with $m, n > 2$.\n\nLisa starts by putting a knight onto the board. Then, in turn, Jeck and Lisa put a new piece onto the board according to the following rules:\n\n1. Jeck puts a queen on an empty square that is two squares horizontally and one square vertically, or one square horizontally and two squares vertically, away from Lisa's last knight.\n2. Lisa puts a knight on an empty square that is on the same row, column, or diagonal as Jeck's last queen.\n\nThe one who is unable to put a piece on the board loses the game.\n\nFor which pairs $(m, n)$ does Lisa have a winning strategy?",
    "options": [],
    "answer": "See solution",
    "solution": "We will show that Lisa has a winning strategy if and only if $m$ and $n$ are both odd.\n\n**Lisa's winning strategy**\n\nSuppose the game is played on an $m \\times n$ board with $m$ and $n$ both odd. Then Lisa puts her knight in a corner and partitions the remaining squares of the board into \"dominoes\". In each turn, Jeck has to put a queen in one of these dominoes, and Lisa puts a knight on the other square of the domino. As the board is finite, Jeck can't keep finding new dominoes, so Lisa will win.\n\n**Jeck's winning strategy**\n\nSuppose the game is played on an $m \\times n$ board with $m$ or $n$ even. We shall show that Jeck is able to partition the board into pairs of squares that are two squares horizontally and one square vertically, or one square horizontally and two squares vertically, away from each other. In each turn, Lisa has to put a knight in one of these, and Jeck puts a queen on the other square of the pair. As the board is finite, Lisa can't keep finding new pairs, so Jeck will win. Now we prove that Jeck can make the required partition.\n\n**Case 1.** Suppose $4 \\mid m$ or $4 \\mid n$. Any $k \\times 4l$ board ($k \\geq 2$) can be divided into $2 \\times 4$ and $3 \\times 4$ boards (firstly divide the $k \\times 4l$ board into $l$ boards of dimensions $k \\times 4$; after that, every $k \\times 4$ board can be divided into $\\frac{k}{2}$ boards of dimensions $2 \\times 4$, or into $\\frac{k-3}{2}$ boards of dimensions $2 \\times 4$ and one $3 \\times 4$ board, depending on the parity of $k$). The following diagrams show that every $2 \\times 4$ and every $3 \\times 4$ board allows a required partition.\n\n**Case 2.** Suppose $m, n \\equiv 1,2 \\pmod{4}$. Any $(5+4l) \\times (6+4l)$ board can be divided into a $5 \\times 6$ board, a $4k \\times l$ board, a $5 \\times 4l$ board, and a $4k \\times 4l$ board. The following diagram shows that a $5 \\times 6$ board allows a required partition.\n\nAccording to case 1, a $4k \\times 6$ board, a $5 \\times 4l$ board, and a $4k \\times 4l$ board also allow a partition.\n\n**Case 3.** Suppose $m, n \\equiv 2,3 \\pmod{4}$. Any $(3+4k) \\times (6+4l)$ board can be divided into a $3 \\times 6$ board, a $4k \\times 6$ board, a $3 \\times 4l$ board, and a $4k \\times 4l$ board. The following diagram shows that a $3 \\times 6$ board allows a required partition.\n\nAccording to case 1, a $4k \\times 6$ board, a $3 \\times 4l$ board, and a $4k \\times 4l$ board also allow a partition.\n\n**Case 4.** Suppose $m, n \\equiv 2 \\pmod{4}$. Any $(6+4k) \\times (6+4l)$ board can be divided into a $6 \\times 6$ board, a $4k \\times 6$ board, a $6 \\times 4l$ board, and a $4k \\times 4l$ board. The $6 \\times 6$ board can be partitioned into two $3 \\times 6$ boards, which were already solved. According to case 1, a $4k \\times 6$ board, a $6 \\times 4l$ board, and a $4k \\times 4l$ board also allow a partition.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 14292,
    "subject": "Mathematics (Olympiad)",
    "question": "A store sold 235 robots over the course of twelve months. Each month, 16, 20, or 25 robots were sold. Find the number of months in which exactly 20 robots were sold.",
    "options": [],
    "answer": "See solution",
    "solution": "If exactly 16 robots were sold each month, the number of robots sold would have been $12 \\cdot 16 = 192$. The difference of 43 robots comes from the months where 20 robots were sold (4 more each month) and from the months where 25 robots were sold (9 more each month). Denote by $a$ and $b$ the number of months where 20 robots were sold and where 25 robots were sold, respectively. We have $4a + 9b = 43$, therefore $b$ is an odd number which does not exceed 4. We obtain that $b = 3$, therefore exactly 20 robots were sold in $a = 4$ months.\n\n_Alternative solution._ Let $x, y, z$ be the number of months in which 16, 20, and 25 robots were sold, respectively. We have $16x + 20y + 25z = 235$, thus $x = 5(47 - 3x - 4y - 5z)$, therefore $5 \\mid x$, consequently $x \\in \\{5, 10\\}$. If $x = 10$, we obtain $4y + 5z = 15$ and $y + z = 2$, which leads to $z = 7 > y + z = 2$, false. If $x = 5$, we have $4y + 5z = 31$ and $y + z = 7$ and we obtain $z = 3$, therefore exactly 20 robots were sold in $y = 7 - z = 4$ months.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 14293,
    "subject": "Mathematics (Olympiad)",
    "question": "In an exotic country, the National Bank issues coins that can take any value in the interval $[0, 1]$. Find the smallest constant $c > 0$ such that the following holds, no matter the situation in that country:\n\nAny citizen of the exotic country that has a finite number of coins, with a total value of no more than $1000$, can split those coins into $100$ boxes, such that the total value inside each box is at most $c$.",
    "options": [],
    "answer": "See solution",
    "solution": "The answer is $c = \\frac{1000}{91} = 11 - \\frac{11}{1001}$.\n\nClearly, if $c'$ works, so does any $c > c'$. First, we prove that $c = 11 - \\frac{11}{1001}$ is good.\n\nWe start with $100$ empty boxes. First, consider only the coins that individually value more than $\\frac{1000}{1001}$. As their sum cannot exceed $1000$, there are at most $1000$ such coins. Thus, we can put at most $10$ such coins in each of the $100$ boxes. Everything so far is all right: $10 \\cdot \\frac{1000}{1001} < 10 < c = 11 - \\frac{11}{1001}$.\n\nNext, step by step, we take one of the remaining coins and prove there is a box where it can be added. Suppose at some point this algorithm fails. It would mean that at a certain point the total sums in the $100$ boxes are $x_1, x_2, \\dots, x_{100}$ and no matter how we add the coin $x$, where $x \\le \\frac{1000}{1001}$, in any of the boxes, that box would be overflowed, i.e., it would have a total sum of more than $11 - \\frac{11}{1001}$. Therefore,\n\n$$\nx_i + x > 11 - \\frac{11}{1001}\n$$\nfor all $i = 1, 2, \\dots, 100$. Then\n\n$$\nx_1 + x_2 + \\dots + x_{100} + 100x > 100 \\cdot \\left(11 - \\frac{11}{1001}\\right).\n$$\nBut since $1000 \\ge x_1 + x_2 + \\dots + x_{100} + x$ and $\\frac{1000}{1001} \\ge x$, we obtain the contradiction\n\n$$\n1000 + 99 \\cdot \\frac{1000}{1001} > 100 \\cdot \\left(11 - \\frac{11}{1001}\\right) \\iff 1000 \\cdot \\frac{1100}{1001} > 100 \\cdot 11 \\cdot \\frac{1000}{1001}.\n$$\nThus, the algorithm does not fail and since we have finitely many coins, we will eventually reach a happy end.\n\nNow we show that $c = 11 - 11\\alpha$, with $1 > \\alpha > \\frac{1}{1001}$ does not work.\n\nTake $r \\in [\\frac{1}{1001}, \\alpha)$ and let $n = \\lfloor \\frac{1000}{1-r} \\rfloor$. Since $r \\ge \\frac{1}{1001}$, then $\\frac{1000}{1-r} \\ge 1001$, therefore $n \\ge 1001$.\n\nNow take $n$ coins each of value $1-r$. Their sum is $n(1-r) \\le \\frac{1000}{1-r} \\cdot (1-r) = 1000$. No matter how we place them in $100$ boxes, as $n \\ge 1001$, there exist $11$ coins in the same box. But $11(1-r) = 11 - 11r > 11 - 11\\alpha$, so the constant $c = 11 - 11\\alpha$ does not work.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 14294,
    "subject": "Mathematics (Olympiad)",
    "question": "Given a convex pentagon $ABCDE$, where the length of each edge and the diagonals $AC$, $AD$ does not exceed $\\sqrt{3}$, choose 2001 arbitrary distinct points in the interior of the pentagon. Show that there exists a unit disk with center lying on the edges of the pentagon, which contains at least 403 of the chosen points.",
    "options": [],
    "answer": "See solution",
    "solution": "To verify the claim, we will show that it is possible to cover the pentagon $ABCDE$ by 5 unit disks with centers lying on the edges of the pentagon.\n\n**Remark:** It is possible to cover a triangle $XYZ$ with edges of length not exceeding $\\sqrt{3}$ by 3 unit disks with centers at the vertices of the triangle.\n\n**Proof:** Assuming the contrary, there exists a point $M$ belonging to triangle $XYZ$ but not lying in the unit disks with centers at the vertices of the triangle. Then we have $MX > 1$, $MY > 1$, and $MZ > 1$.\n\nClearly, among the angles $\\widehat{XMY}$, $\\widehat{YMZ}$, and $\\widehat{ZMX}$, at least one is larger than $120^\\circ$. Without loss of generality, assume that $\\widehat{XMY} \\geq 120^\\circ$. Using the cosine theorem for triangle $XMY$, we obtain\n\n$$\nXY^2 = MX^2 + MY^2 - 2MX \\cdot MY \\cdot \\cos \\widehat{XMY} > 1 + 1 + 2 \\cdot \\frac{1}{2} = 3 \\quad (\\text{since } \\cos \\widehat{XMY} \\leq -\\frac{1}{2}).\n$$\n\nConsequently, $XY > \\sqrt{3}$, contradicting the assumption. The contradiction yields the claim to be verified.",
    "topic": "Discrete Mathematics",
    "subtopic": "Combinatorics"
  },
  {
    "id": 14295,
    "subject": "Mathematics (Olympiad)",
    "question": "At noon on a certain day, Minneapolis is $N$ degrees warmer than St. Louis. At 4:00 the temperature in Minneapolis has fallen by 5 degrees while the temperature in St. Louis has risen by 3 degrees, at which time the temperatures in the two cities differ by 2 degrees. What is the product of all possible values of $N$?\n\n(A) 10  (B) 30  (C) 60  (D) 100  (E) 120",
    "options": [],
    "answer": "See solution",
    "solution": "At 4:00, the amount by which the temperature in Minneapolis exceeds the temperature in St. Louis has decreased by $5 + 3 = 8$. Therefore, $|N - 8| = 2$. The two solutions to this equation are $N = 6$ and $N = 10$. The requested product is $6 \\cdot 10 = 60$.\n\nAlternatively, let $M$ and $S$ be the temperatures in the two cities at noon. It is given that $M - S = N$. At 4:00, the temperatures are $M - 5$ and $S + 3$, respectively. The difference in temperature at 4:00 is therefore\n\n$$\n|(M - 5) - (S + 3)| = |(M - S) - 8| = |N - 8|\n$$\n\nThe equation $|N - 8| = 2$ has two solutions: $N = 6$ and $N = 10$. The requested product is $6 \\cdot 10 = 60$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 14296,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $a, b$ be two coprime positive integers. Prove that\n\n$$\nN = a^2b + ab^2 - a^2 - b^2 - ab + 1\n$$\n\nis the smallest positive integer such that the equation $a^2x + aby + b^2z = m$ has a natural solution for all $m \\geq N$.",
    "options": [],
    "answer": "See solution",
    "solution": "In case $m \\geq N$, choose $z = m b^{-2}$ (mod $a$) with $0 \\leq z < a$. We need to prove that there exist $x, y \\in \\mathbb{N}$ such that\n\n$$\na x + b y = \\frac{m - b^2 z}{a}\n$$\n\nhas a natural solution. Note that\n\n$$\n\\frac{m - b^2 z}{a} \\geq \\frac{N - b^2(a - 1)}{a} > \\frac{a^2 b - a^2 - a b}{a} = a b - a - b\n$$\n\nso the equation always has a natural solution by Sylvester's theorem.\n\nIf $m < N$, let $m = a^2b + ab^2 - a^2 - b^2 - ab$ and assume there exists a triple $(x, y, z) \\in \\mathbb{N}^3$ such that\n\n$$\na^2x + ab y + b^2 z = m\n$$\n\nthen $a^2x \\equiv -a^2 \\pmod{b}$, $b^2z \\equiv -b^2 \\pmod{a}$, which is equivalent to\n\n$$\nx \\equiv -1 \\pmod{b}, \\quad z \\equiv -1 \\pmod{a}.\n$$\n\nThus $x \\geq b - 1$, $z \\geq a - 1$, which leads to\n\n$$\ny \\leq \\frac{m - a^2(b - 1) - b^2(a - 1)}{ab} = -1.\n$$\n\nThis is a contradiction since $y \\geq 0$.\n\nApplying this result, we conclude that the greatest value of $n$ is $5a^2 + 30a - 36$. $\\square$",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 14297,
    "subject": "Mathematics (Olympiad)",
    "question": "Solve the equation:\n\n$$\n\\cos \\pi x = \\left[ \\frac{x}{2} - \\left[ \\frac{x}{2} \\right] - \\frac{1}{2} \\right].\n$$\n\nHere $[a]$ stands for the greatest integer not exceeding $a$.",
    "options": [],
    "answer": "See solution",
    "solution": "*Answer:* $x = \\frac{3}{2} + 2n$, $n \\in \\mathbb{Z}$.\n\n*Solution.* Since $a - [a] = \\{a\\}$, where $\\{a\\}$ is the fractional part of $a$, we can rewrite the equation as:\n\n$$\n\\cos \\pi x = \\left[ \\frac{\\{x\\}}{2} - \\frac{1}{2} \\right].\n$$\n\nObviously, $0 \\leq \\{ \\frac{x}{2} \\} < 1$ for every real $x$. Consider two cases:\n\n1) If $0 \\leq \\{ \\frac{x}{2} \\} < \\frac{1}{2}$, then $-\\frac{1}{2} \\leq \\{ \\frac{x}{2} \\} - \\frac{1}{2} < 0$, so $[ \\{ \\frac{x}{2} \\} - \\frac{1}{2} ] = -1$. Thus, $\\cos \\pi x = -1$, whose solutions are $x = 1 + 2k$, $k \\in \\mathbb{Z}$. But for such $x$, $\\{ \\frac{x}{2} \\} = \\{ \\frac{1}{2} + k \\} = \\frac{1}{2}$, which contradicts our assumption. So, there are no solutions in this case.\n\n2) If $\\frac{1}{2} \\leq \\{ \\frac{x}{2} \\} < 1$, then $0 \\leq \\{ \\frac{x}{2} \\} - \\frac{1}{2} < \\frac{1}{2}$, so $[ \\{ \\frac{x}{2} \\} - \\frac{1}{2} ] = 0$. Thus, $\\cos \\pi x = 0$, whose solutions are $x = \\frac{1}{2} + k$, $k \\in \\mathbb{Z}$. For such $x$:\n\n$$\n\\frac{x}{2} = \\frac{1}{4} + \\frac{k}{2}\n$$\n\nSo $\\{ \\frac{x}{2} \\} = \\frac{1}{4}$ for even $k$ and $\\frac{3}{4}$ for odd $k$. Only for $k = 2n + 1$, $n \\in \\mathbb{Z}$, do we have $\\frac{1}{2} \\leq \\{ \\frac{x}{2} \\} < 1$. Therefore, $x = \\frac{3}{2} + 2n$, $n \\in \\mathbb{Z}$.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 14298,
    "subject": "Mathematics (Olympiad)",
    "question": "Given an acute triangle $ABC$, let $(c)$ be its circumcircle with center $O$, and let $H$ be the orthocenter of $ABC$. The line $AO$ intersects $(c)$ at the point $D$. Let $D_1$ and $D_2$ be the reflections of $D$ over the lines $AB$ and $AC$, respectively, and let $H_2$ and $H_3$ be the reflections of $H$ over $AB$ and $AC$, respectively. Let $(c_1)$ be the circumcircle of triangle $AD_1D_2$. Suppose the line $AH$ meets $(c_1)$ again at $U$. The line $H_2H_3$ meets segment $D_1D_2$ at $K_1$, and the line $DH_3$ meets segment $UD_2$ at $L_1$. Prove that one of the intersection points of the circumcircles of triangles $D_1K_1H_2$ and $UDL_1$ lies on the line $K_1L_1$.",
    "options": [],
    "answer": "See solution",
    "solution": "It is well known that the reflections $H_1$, $H_2$, $H_3$ of $H$ over the sides $BC$, $AB$, $AC$ of triangle $ABC$ lie on the circle $(c)$.\n\n![](images/2019_bmo_shortlist_p25_data_d1e8294136.png)\n\nLet $L$ be the second intersection point of $(c)$ and $(c_1)$. First, we will prove that the lines $D_1H_2$, $D_2H_3$, and $UD$ pass through $L$.\n\nSuppose the line $AH$ meets $BC$ at $Z$. Since $H_1D \\parallel BC \\parallel D_1D_2$ and $B$, $C$ are the midpoints of $D_1D$ and $D_2D$, respectively, $Z$ is the midpoint of $HH_1$, so $H$ lies on $D_1D_2$. Therefore, $AH \\perp D_1D_2$ and $AU$ is a diameter of $(c_1)$. Thus, $AL \\perp UL$ and $AL \\perp DL$. We have that $U$, $D$, $L$ are collinear. (1)\n\nNow, $\\angle ALD_1 = \\angle AD_2D_1$, $\\angle ALH_2 = \\angle ACH_2$. Since $AHCD_2$ is cyclic, we get\n\n$\\angle ACH_2 = \\angle AD_2D_1$. Therefore, $\\angle ALH_2 = \\angle ALD_1$. So $D_1$, $H_2$, $L$ are collinear. (2)\n\nSimilarly,\n\n$$\n\\angle D_1LD_2 = \\angle D_1AD_2 = 180^\\circ - 2(\\angle AD_1H).\n$$\n\nSince $AD_1BH$ is cyclic, $\\angle AD_1H = \\angle ABH = \\angle ABH_3$. Therefore,\n\n$$\n\\angle D_1LD_2 = 180^\\circ - 2(\\angle ABH_3) = 180^\\circ - 2(\\angle ADH_3) = 180^\\circ - \\angle H_2DH_3.\n$$\n\nThus,\n\n$$\n\\angle D_1LD_2 + \\angle H_2DH_3 = 180^\\circ \\quad \\text{or} \\quad \\angle D_1LD_2 + \\angle H_3LH_2 = 180^\\circ.\n$$\n\nSo $H_3$, $L$, $D_2$ are collinear. (3)\n\nFrom (1), (2), (3), the lines $D_1H_2$, $D_2H_3$, and $UD$ are concurrent at $L$.\n\nAlso,\n\n$$\n\\angle H_3DA = \\angle D_2DA - \\angle CDH_3 = \\angle AD_2D - \\angle CBH_3\n$$\n\nand since $BHD_2C$ is a parallelogram, $\\angle CBH_3 = \\angle HD_2C$. So\n\n$$\n\\angle H_3DA = \\angle AD_2D - \\angle HD_2C = \\angle AD_2D_1 = \\angle AD_1D_2 = \\angle AUD_2.\n$$\n\nTherefore, the circumcircle of triangle $UDL_1$ passes through $A$. Also, $\\angle AD_1K_1 = \\angle D_2D_1A = \\angle D_2UA$. But $AUL_1D$ is cyclic and $\\angle D_2UA = \\angle H_3DA = \\angle H_3BA = \\angle H_3H_2A$. Therefore, $\\angle AD_1K_1 = \\angle H_3H_2A$. Thus, the circumcircle of triangle $D_1K_1H_2$ passes through $A$.\n\nBecause $H_3$, $L$, $D_2$ are collinear by Desargues' theorem, the lines $UD_1$, $L_1K_1$, $DH_2$ are concurrent, say at $M$.\n\nFrom the similarity of triangles $UDL_1$ and $D_1K_1H_2$, we conclude that $M$ is the center of a unique spiral similarity, and since the circumcircles of $D_1K_1H_2$ and $UDL_1$ intersect at $A$, the second intersection point is $M$. Therefore, $M$ lies on the line $K_1L_1$. $\\square$\n\n**Comment.** We can prove the last part in a different way.\n\nLet $M$ be the intersection of the circumcircles of $D_1K_1H_2$ and $UDL_1$. Now,\n\n$$\n\\angle K_1MA = \\angle H_3H_2A = \\angle H_3BA = \\angle ADH_3 = \\angle L_1UA = \\angle L_1MA.\n$$\n\nTherefore, $L_1$, $K_1$, $M$ are collinear. $\\square$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 14299,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $(y_n)_{n \\ge 1}$ be a sequence where $y_n = a_1 \\cdot b_1^n + c_1 \\cdot d_1^n$. Define the set $P$ as the set of all prime divisors of at least one element of the sequence $(y_n)$. Prove that if $P$ is a finite set, then $b_1 = d_1 = 1$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let\n\n$$\n\\begin{align*}\nP_{ad} &= \\{p \\in P : p \\mid a_1 d_1\\}, \\\\\nP_{bc} &= \\{p \\in P : p \\mid b_1 c_1\\}, \\\\\nP_{a+c} &= \\{p \\in P : p \\mid a_1 + c_1\\}, \\\\\nP_0 &= \\{p \\in P : p \\nmid a_1 b_1 c_1 d_1 \\text{ and } p \\nmid a_1 + c_1\\}\n\\end{align*}\n$$\n\nIt can be seen that $P = P_{ad} \\cup P_{bc} \\cup P_{a+c} \\cup P_0$.\n\nLet $p \\in P_{ad}$. If $p \\mid a_1$ then $p \\mid c_1 \\cdot d_1^n$ and hence $p \\mid d_1$. Similarly, if $p \\mid d_1$ then $p \\mid a_1$. For sufficiently large $n$ we get $v_p(y_n) = v_p(a_1)$.\n\nLet $p \\in P_{bc}$. Similarly, for sufficiently large $n$ we have $v_p(y_n) = v_p(c_1)$.\n\nLet $p \\in P_{a+c}$, then $p \\nmid b_1 d_1$. Let $v_p(a_1 + c_1) = \\alpha$. Then by Euler's theorem, for each $n$ such that $\\phi(p^{\\alpha+1}) = p^{\\alpha}(p-1) \\mid n$, we have\n\n$$\ny_n = a_1 \\cdot b_1^n + c_1 \\cdot d_1^n \\equiv a_1 + c_1 \\pmod{p^{\\alpha+1}}\n$$\n\nand hence $v_p(y_n) = \\alpha = v_p(a_1 + c_1)$.\n\nFinally, let $p \\in P_0$. Then by Fermat's theorem, for each $n$ such that $p-1 \\mid n$, we have\n\n$$\ny_n = a_1 \\cdot b_1^n + c_1 \\cdot d_1^n \\equiv a_1 + c_1 \\pmod{p}\n$$\n\nand hence $p \\nmid y_n$.\n\nTherefore, for sufficiently large $k$ and\n\n$$\nn = k \\cdot \\prod_{p \\in P_{a+c}} (p-1) p^{v_p(a_1+c_1)} \\cdot \\prod_{p \\in P_0} (p-1)\n$$\n\nfor each prime divisor $q$ of $y_n$, $v_q(y_n)$ is bounded from above. Since all four sets are finite, $y_n$ is also bounded from above. The only possibility for this is $b_1 = d_1 = 1$. Thus $b_1 = d_1 = 1$, as required.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 14300,
    "subject": "Mathematics (Olympiad)",
    "question": "「虧格」是由三個邊長為 1 的正方形所組成的圖形，或將此圖形經由旋轉及翻轉所得之任意圖形。\n\n(a) 在任何兩個「虧格」彼此不重疊的條件下，用 $3 \\times 671$ 個「虧格」來舖蓋在 $3 \\times 2013$ 的大方格圖形，請問可以完全覆蓋滿嗎？\n\n(b) 在任何兩個「虧格」彼此不重疊的條件下，用 $5 \\times 671$ 個「虧格」來舖蓋在 $5 \\times 2013$ 的大方格圖形，請問可以完全覆蓋滿嗎？",
    "options": [],
    "answer": "See solution",
    "solution": "1. 不可以。\n\n![](images/13-1J-ind_p9_data_fe8339ba60.png)\n\n將 $3 \\times 2013$ 的大方格中，$(1,2k+1)$ 及 $(3,2k+1)$，$k=0,1,2,\\ldots,2006$ 的格子中標上“*”，如上圖所示。每個「虧格」最多可以覆蓋一個“*”，所以至少需要 2014 個「虧格」才能完全覆蓋這些“*”。所以 $3 \\times 671 (= 2013)$ 個「虧格」無法完全覆蓋 $3 \\times 2013$ 的大方格。\n\n2. 可以。\n\n![](images/13-1J-ind_p9_data_869ff56b69.png)\n\n$2 \\times 6, 3 \\times 6$ 的大方格也可以用「虧格」完全覆蓋。由此知 $5 \\times 6$ 的大方格可以用「虧格」完全覆蓋。又 $5 \\times 9$ 的大方格可以被 15 個「虧格」完全覆蓋，上圖是一種可能的方法。所以 $5 \\times (6k+9)$ 的大方格（其中 $k$ 為任意非負整數）都可以被「虧格」完全覆蓋。所以 $5 \\times 2013 = 5 \\times (6 \\cdot 334 + 9)$ 的大方格能夠被「虧格」完全覆蓋。",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 14301,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $\\Gamma$ be the circumcircle of triangle $ABC$, and let $D$ and $E$ be two points (distinct from the vertices) on sides $AB$ and $AC$, respectively. Let $A'$ be the second point where $\\Gamma$ intersects the bisector of angle $\\angle BAC$, and let $P$ and $Q$ be the second points where $\\Gamma$ intersects the lines $A'D$ and $A'E$, respectively. Let $R$ and $S$ be the second points of intersection of the line $AA'$ with the circumcircles of triangles $APD$ and $AQE$, respectively. Show that the lines $DS$, $ER$, and the tangent to $\\Gamma$ at $A$ are concurrent.",
    "options": [],
    "answer": "See solution",
    "solution": "Since $\\angle RPD = \\angle RAD = \\angle A'AC = \\angle A'PC = \\angle DPC$, $P$, $R$, and $C$ are collinear. Then $\\angle PRD = \\angle PAD = \\angle PAB = \\angle PCB$ implies that $DR \\parallel BC$. Similarly, $SE \\parallel BC$, and consequently, $SE \\parallel DR$ and $\\frac{VD}{DA} = \\frac{SR}{RA}$.\n\nLet $\\ell$ be the tangent through $A$ to the circumcircle of $ABC$, and let $T$ and $U$ be the points where $\\ell$ intersects the lines $DS$ and $SE$, respectively. Also, let $V$ be the point of intersection of $AB$ and $SE$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 14302,
    "subject": "Mathematics (Olympiad)",
    "question": "Consider the 12 points on a clock face. Color them with four colors: red, yellow, blue, and green, using each color for exactly three points. Find the largest number $n$ of convex quadrilaterals with vertices among these points such that:\n\n1. No quadrilateral has two vertices of the same color.\n2. Among any three of these quadrilaterals, there is a color such that the vertices of that color are all different.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $A, B, C, D$ represent the four colors, and label the points of each color 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\nConsider color $A$. Suppose in $n$ quadrilaterals, the number of times $a_1, a_2, a_3$ appear as the $A$-colored vertex are $n_1, n_2, n_3$ respectively, so $n_1 + n_2 + n_3 = n$. Assume $n_1 \\geq n_2 \\geq n_3$. If $n \\geq 10$, then $n_1 + n_2 \\geq 7$.\n\nNow, among these seven quadrilaterals (with $A$-vertex $a_1$ or $a_2$), let $m_1, m_2, m_3$ be the counts for $b_1, b_2, b_3$ as the $B$-colored vertex. Then $m_1 + m_2 + m_3 = 7$, and $m_1 \\geq m_2 \\geq m_3$, so $m_3 \\leq 2$, thus $m_1 + m_2 \\geq 5$.\n\nFor these five quadrilaterals (with $A$-vertex $a_1$ or $a_2$ and $B$-vertex $b_1$ or $b_2$), let $k_1, k_2, k_3$ be the counts for $c_1, c_2, c_3$ as the $C$-colored vertex. Then $k_1 + k_2 + k_3 = 5$, $k_1 \\geq k_2 \\geq k_3$, so $k_3 \\leq 1$, thus $k_1 + k_2 \\geq 4$.\n\nConsider these four quadrilaterals, $T_1, T_2, T_3, T_4$ (with $A$-vertex $a_1$ or $a_2$, $B$-vertex $b_1$ or $b_2$, $C$-vertex $c_1$ or $c_2$). Since there are only three $D$-colored points, two quadrilaterals must share the same $D$-vertex. Suppose $T_1$ and $T_2$ both use $d_1$.\n\nThen, among $T_1, T_2, T_3$, for any color, at least two quadrilaterals share the same colored vertex, contradicting condition (2). Thus, $n \\leq 9$.\n\nWe can construct $n = 9$ quadrilaterals as follows:\n\n![](images/Mathematical_Olympiad_in_China_2011-2014_p300_data_919aeda191.png)\n\nDraw three concentric annuli, each with four points on a radius representing the four colors. The nine radii correspond to nine quadrilaterals, satisfying condition (1).\n\nTo check condition (2):\n- If the three quadrilaterals correspond to radii from the same annulus, for each color except $A$, the colored vertices are all different.\n- If from three different annuli, for color $A$, the vertices are all different.\n- If from two annuli, analyze the possible cases (as shown in the tables below):\n\n![](images/Mathematical_Olympiad_in_China_2011-2014_p300_data_76eb60bea6.png)\n\nC\n\n![](images/Mathematical_Olympiad_in_China_2011-2014_p300_data_b4a539d984.png)\n\nD",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 14303,
    "subject": "Mathematics (Olympiad)",
    "question": "從集合 $S = \\{1, 2, 3, \\ldots, 2024\\}$ 中，取出 1000 個數而造出具有 1000 個元素的子集 $T$，而 $T$ 的最小元素是 $k$，求 $k$ 的期望值。",
    "options": [],
    "answer": "See solution",
    "solution": "設 $M$ 是所求的期望值，則有：\n\n$$\n\\begin{aligned}\n\\binom{2024}{1000} M &= 1 \\cdot \\binom{2023}{999} + 2 \\cdot \\binom{2022}{999} + 3 \\cdot \\binom{2021}{999} + \\dots + 1025 \\cdot \\binom{999}{999} \\\\\n&= \\sum_{a+b=1025} \\binom{a}{1} \\binom{b+999}{999} \\\\\n&= \\binom{1025+999+1}{1+999+1} \\\\\n&= \\binom{2025}{1001} \\\\\nM &= \\frac{2025}{1001}\n\\end{aligned}\n$$\n\n*Remark*: 兩類二項係數恆等式來自數列的摺積 (convolution)\n\n$$\n\\sum_{a+b=r} \\binom{m}{a} \\binom{n}{b} = \\binom{m+n}{r}, \\quad \\sum_{a+b=r} \\binom{a+m}{m} \\binom{b+n}{n} = \\binom{m+n+r+1}{m+n+1}\n$$\n\n第一類等式來自\n\n$$\n(1+x)^m = \\sum_{a=0}^{m} \\binom{m}{a} x^a \\quad \\text{與} \\quad (1+x)^n = \\sum_{b=0}^{n} \\binom{n}{b} x^b\n$$\n\n的摺積。第二類等式來自於\n\n$$\n\\frac{1}{(1-x)^{m+1}} = \\sum_{k=0}^{\\infty} \\binom{m+k}{m} x^k, \\quad \\frac{1}{(1-x)^{n+1}} = \\sum_{l=0}^{\\infty} \\binom{n+l}{l} x^l\n$$\n\n的摺積。",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 14304,
    "subject": "Mathematics (Olympiad)",
    "question": "Prove that for all positive integers of the form $n = 2^{2p} - 1$, where $p > 3$ is a prime number, the following hold:\n\n1. $n \\mid 2^n - 8$.\n2. $n$ has at least three distinct prime divisors.",
    "options": [],
    "answer": "See solution",
    "solution": "First, note that $2^n - 8 = 8(2^{n-3} - 1)$. To prove $n \\mid 2^n - 8$, it suffices to show $n \\mid 2^{n-3} - 1$.\n\nSince $n = 2^{2p} - 1$, we have:\n$$\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$$\nfor some integer $k$, so $n \\mid 2^{n-3} - 1$ and thus $n \\mid 2^n - 8$.\n\nBy Fermat's little theorem, $2^p \\equiv 2 \\pmod p$, so:\n$$\nn - 3 = 2^{2p} - 4 = (2^p)^2 - 4 \\equiv 2^2 - 4 = 0 \\pmod p.\n$$\nSince $n-3$ is even and $2$ and $p$ are coprime, $n-3 \\equiv 0 \\pmod{2p}$.\n\nNow, to show $n$ has at least three distinct prime divisors:\n\n$$\nn = 2^{2p} - 1 = (2^p - 1)(2^p + 1)\n$$\n\n$2^p - 1$ and $2^p + 1$ are consecutive odd numbers and thus coprime. $2^p - 1$ has at least one prime divisor. For $2^p + 1$, since $p > 3$ and $p$ is odd, $2^p + 1 \\equiv 0 \\pmod{3}$, but $2^p + 1 \\not\\equiv 0 \\pmod{9}$, so it is not a power of $3$ and must have at least one other prime divisor. Thus, $n$ has at least three distinct prime divisors.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 14305,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $ABC$ be an acute triangle. Let $H$ denote its orthocenter and $D$, $E$, and $F$ the feet of its altitudes from $A$, $B$, and $C$, respectively. Let the common point of $DF$ and the altitude through $B$ be $P$. The line perpendicular to $BC$ through $P$ intersects $AB$ in $Q$. Furthermore, $EQ$ intersects the altitude through $A$ in $N$.\n\nProve that $N$ is the midpoint of $AH$.",
    "options": [],
    "answer": "See solution",
    "solution": "See Figure 3. As usual, let $\\beta = \\angle ABC$ and $\\gamma = \\angle ACB$. Since we know that\n![](images/Austria2017_p7_data_42b07e96e6.png \"Figure 3: Problem 13\")\n$\\angle AFH = \\angle AEH = 90^\\circ$ holds, the quadrilateral $AFHE$ is cyclic, and because $DA$ is parallel to $PQ$ we obtain\n$$\n\\angle FQP = \\angle FAH = \\angle FEH = \\angle FEP.\n$$\nIt follows that $QFPE$ is also cyclic. Since $\\angle AFC = \\angle ADC = 90^\\circ$, $AFDC$ is also cyclic, and we have $\\angle QFP = \\angle AFD = 180^\\circ - \\angle ACD = 180^\\circ - \\gamma$. We therefore have $\\angle QEP = \\gamma$. From this, we obtain $\\angle EAN = 90^\\circ - \\gamma = \\angle AEP - \\angle QEP = \\angle AEN$, which shows us that triangle $ANE$ is isosceles. It therefore follows that $N$ is the circumcenter of the right triangle $AHE$, and we therefore have $NA = NH$, as claimed.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 14306,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $ABC$ be a triangle. Its excircles touch sides $BC$, $CA$, $AB$ at $D$, $E$, $F$, respectively. Prove that the perimeter of triangle $ABC$ is at most twice that of triangle $DEF$.",
    "options": [],
    "answer": "See solution",
    "solution": "We consider the configuration shown in the diagram below. (Our proof uses directed lengths and can be easily modified for different configurations.)\n\n![](images/pamphlet1112_main_p30_data_37e73c25e2.png)\n\nLet $a, b, c$ denote the side lengths of $BC, CA, AB$, and let $A, B, C$ denote $\\angle A, \\angle B, \\angle C$, respectively. Suppose that the incircle touches sides $BC, CA, AB$ at $P, Q, R$, respectively. It is well known that\n\n$$\nFB = CE = QA = AR = \\frac{b+c-a}{2}.\n$$\n\nDenote by $E_a, F_a$ the feet of the perpendiculars from $E, F$ to line $BC$. It is clear that $EF \\geq E_aF_a$. By the above,\n\n$$\nFE \\geq F_aE_a = BC - (BF_a + E_aC) = a - \\frac{b+c-a}{2} (\\cos B + \\cos C).\n$$\n\nSumming the above inequality and its cyclic analogues yields\n\n$$\nEF + FD + DE \\geq a + b + c - \\sum_{\\text{cyc}} \\frac{b+c-a}{2} (\\cos B + \\cos C)\n$$\n\nor\n\n$$\nEF + FD + DE \\geq a + b + c - (a \\cos A + b \\cos B + c \\cos C).\n$$\n\nBy the sum-to-product formulas,\n\n$$\n\\frac{1}{2}(\\sin 2A + \\sin 2B) = \\sin(A+B)\\cos(A-B) = \\sin C \\cos(A-B) \\geq \\sin C.\n$$\n\nSumming this inequality and its cyclic analogues yields\n\n$$\n\\sin A + \\sin B + \\sin C \\geq 2\\sin A \\cos A + 2\\sin B \\cos B + 2\\sin C \\cos C\n$$\n\nMultiplying both sides by $2R$ and applying the extended Law of Sines, we obtain\n\n$$\na + b + c \\geq 2a \\cos A + 2b \\cos B + 2c \\cos C.\n$$\n\nSubstituting this into the previous inequality yields the desired\n\n$$\nEF + FD + DE \\geq \\frac{a+b+c}{2}.\n$$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 14307,
    "subject": "Mathematics (Olympiad)",
    "question": "Find the smallest positive integer that can be inserted between the numbers 20 and 16 so that the resulting number $20\\ldots16$ is a multiple of $2016$.",
    "options": [],
    "answer": "See solution",
    "solution": "Since $2016$ is a multiple of $9$, the digit sum of the resulting number must be divisible by $9$. This means the inserted number should have a digit sum divisible by $9$, i.e., it should be a multiple of $9$. Trying the smallest such numbers: inserting $9$, $18$, and $27$ yields $20916$, $201816$, and $202716$, which are not divisible by $2016$ (or by $16$, considering the last four digits). However, $203616 = 201600 + 2016$ is divisible by $2016$. Thus, the answer is $36$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 14308,
    "subject": "Mathematics (Olympiad)",
    "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": "See solution",
    "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": 14309,
    "subject": "Mathematics (Olympiad)",
    "question": "Suppose two players play a game on a grid with 1403 rows and several columns. The first player fills cells in each column with vertical lines, and the second player fills cells with horizontal lines. The second player aims to form a horizontal line of length 2 in any row. If, after a sequence of moves, all columns except one (column C) become 'almost full' (i.e., each has 1402 filled cells: 701 vertical by the first player and 701 horizontal by the second player), can the second player guarantee a non-losing strategy? Justify your answer.\n\n![](images/IRN_booklet2025_p27_data_f6012b2165.png)",
    "options": [],
    "answer": "See solution",
    "solution": "The second player can always find a move to form a horizontal line of length 2, or else all columns become 'almost full' (1402 filled cells). When two adjacent columns (X and Y) are completely filled (1403 cells each: 701 vertical, 702 horizontal), the Pigeonhole Principle ensures that in at least one row, the second player has placed horizontal lines in both columns X and Y, thus forming a horizontal line of length 2. Therefore, the second player has a non-losing strategy, and the game will end in a draw if both play optimally.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 14310,
    "subject": "Mathematics (Olympiad)",
    "question": "Suppose there are 100 countries with a total of 2002 representatives. Each country has at most 23 representatives, and no two countries with 23 representatives may sit in the same row. Each row can seat at most 45 representatives. What is the minimum number of rows needed to seat all representatives under these conditions?",
    "options": [],
    "answer": "See solution",
    "solution": "At least 86 rows are needed.\n\nSuppose there are 86 (type A) countries with 23 representatives each, 10 countries with 2 representatives each, and 4 countries with 1 representative each. Then there are $86 + 10 + 4 = 100$ countries and $86 \\times 23 + 10 \\times 2 + 4 \\times 1 = 2002$ representatives. No two type A countries may sit in the same row since $23 \\times 2 > 45$. Thus, at least 86 rows are needed.\n\nWe now show that it is always sufficient to use 86 rows. We arrange the countries in decreasing order of the number of representatives. We first put all representatives from country 1 in the first row. Next, if all representatives from country 2 can be seated in the same row as country 1, we put them in that row, or otherwise they are put in the next row. Similarly, if all representatives from country $k+1$ can be seated in the same row as country $k$, we put them in that row, or otherwise they are put in the next row. We do the same thing until all representatives are seated or all 86 rows are used up.\n\nIt suffices to consider the case when all 86 rows are used up but some representatives still do not have a seat. We claim that each row consists of at least 23 representatives. If not, there are at least 23 empty seats in some row $j$, which means the next country to be put has at least 24 representatives. But then there should be at least 24 representatives in row $j$ as well according to the order of the countries. This is a contradiction. It follows that the number of representatives remaining is at most\n\n$$\n2002 - 23 \\times 86 = 24.\n$$\n\nAmong these representatives, suppose $c$ of them come from the same country, where $c$ is maximized. First of all, we have $c \\le 22$ since otherwise at least 99 countries have at least 23 representatives (except possibly the last country), which yields the contradiction that the total number of representatives is at least\n\n$$\n23 \\times 99 = 2277 > 2002.\n$$\n\nNow, we partition the remaining representatives into at most two groups by putting those $c$ representatives in one group. We claim that all groups have at most 22 representatives. If not, since $c \\le 22$, it must be the case that $c = 1$ and there are 23 more representatives. By the choice of $c$, all these 24 representatives come from different countries. But this is impossible since there are at most $100 - 86 = 14$ countries left.\n\nTherefore, it remains to show that there are at least 2 rows with 22 empty seats. Indeed, if this does not hold, then the total number of representatives is at least\n\n$$\n24 \\times 85 + 23 = 2063 > 2002.\n$$\n\nThis is a contradiction, and so we are done.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 14311,
    "subject": "Mathematics (Olympiad)",
    "question": "In a triangle $ABC$ with sides $BC > AC > AB$, consider the angles between the height and median constructed from each vertex. For which vertex is this angle the largest of the three?\n\n![](fig.48.png)",
    "options": [],
    "answer": "See solution",
    "solution": "The angle is largest at vertex $B$.\n\nLet the sides of triangle $ABC$ be $a$, $b$, and $c$, with $a > b > c$. Denote the height and median from vertex $B$ as $h_b = BH$ and $m_b = BM$, respectively. The angle of interest is $\\angle MBH$, and $\\cos \\angle MBH = \\frac{h_b}{m_b}$. Similarly, we can determine the cosines for the other vertices.\n\nTo prove that the angle at $B$ is the largest, it suffices to show that its cosine is the smallest:\n\n$$\n\\frac{h_b}{m_b} < \\min \\left\\{ \\frac{h_a}{m_a}, \\frac{h_c}{m_c} \\right\\}.\n$$\n\nFor this, we prove:\n\n$$\n\\frac{m_b^2}{h_b^2} > \\frac{m_a^2}{h_a^2} \\quad \\text{and} \\quad \\frac{m_b^2}{h_b^2} > \\frac{m_c^2}{h_c^2}.\n$$\n\nRecall the formulas for height and median:\n\n$$\nh_a = \\frac{2S_{ABC}}{a}, \\quad m_a^2 = \\frac{2b^2 + 2c^2 - a^2}{4}.\n$$\n\nThe first inequality is equivalent to:\n\n$$\n\\frac{2a^2 + 2c^2 - b^2}{4} \\cdot \\frac{b^2}{4S^2} > \\frac{2b^2 + 2c^2 - a^2}{4} \\cdot \\frac{a^2}{4S^2}\n$$\n\nwhich simplifies to:\n\n$$\n(2a^2 + 2c^2 - b^2)b^2 > (2b^2 + 2c^2 - a^2)a^2 \\\\\n2b^2c^2 - b^4 > 2a^2c^2 - a^4 \\\\\n(a^2 + b^2)(a^2 - b^2) > 2c^2(a^2 - b^2).\n$$\n\nThis holds since $a > b > c$.\n\nSimilarly, for the second inequality:\n\n$$\n(2a^2 + 2c^2 - b^2)b^2 > (2b^2 + 2a^2 - c^2)c^2 \\\\\n2b^2a^2 - b^4 > 2a^2c^2 - c^4 \\\\\n2a^2(b^2 - c^2) > (b^2 + c^2)(b^2 - c^2).\n$$\n\nAgain, this is true since $a > b > c$.",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 14312,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $a, b$ be such that\n$$\n\\frac{7a^3b^3}{a^6 - 8b^6} = 1.\n$$\nDetermine what the value of $\\frac{a^2 - b^2}{a^2 + b^2}$ can be.",
    "options": [],
    "answer": "See solution",
    "solution": "Rewrite the given equation as\n$$\na^6 - 7a^3b^3 - 8b^6 = 0.\n$$\nThis factors as\n$$\n(a^3 + b^3)(a^3 - 8b^3) = 0.\n$$\nSince $a$ and $b$ cannot both be zero, consider each case:\n\n**Case 1:** $a^3 + b^3 = 0$\n\nThis gives $(a + b)(a^2 - ab + b^2) = 0$. Since $a^2 - ab + b^2 \\neq 0$ for $a, b \\neq 0$, we have $a = -b$.\n\nThen,\n$$\n\\frac{a^2 - b^2}{a^2 + b^2} = \\frac{(-b)^2 - b^2}{(-b)^2 + b^2} = \\frac{b^2 - b^2}{b^2 + b^2} = 0.\n$$\n\n**Case 2:** $a^3 - 8b^3 = 0$\n\nThis gives $(a - 2b)(a^2 + 2ab + 4b^2) = 0$. Since $a^2 + 2ab + 4b^2 \\neq 0$ for $a, b \\neq 0$, we have $a = 2b$.\n\nThen,\n$$\n\\frac{a^2 - b^2}{a^2 + b^2} = \\frac{(2b)^2 - b^2}{(2b)^2 + b^2} = \\frac{4b^2 - b^2}{4b^2 + b^2} = \\frac{3}{5}.\n$$\n\n**Conclusion:**\nThe possible values of $\\frac{a^2 - b^2}{a^2 + b^2}$ are $0$ and $\\frac{3}{5}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 14313,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $ABC$ be an acute-angled triangle with orthocenter $H$. Consider the points $Y$ and $Z$ on the sides $CA$ and $AB$ respectively such that the directed angles $(AC, HY) = -\\pi/3$ and $(AB, HZ) = \\pi/3$. Let $U$ be the circumcenter of $\\triangle HYZ$.\n\nProve that the points $A$, $N$, $U$ are collinear, where $N$ is the nine-point center of $\\triangle ABC$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $O_A$ be the reflection of $O$ into the sideline $BC$. We will prove the result by showing that $A$, $N$, $O_A$ are collinear, and that $A$, $U$, $O_A$ are collinear.\n\n$A$, $N$, $O_A$ are collinear from the fact that $N$ is the midpoint of $HO$ on the Euler line, and that $AH = OO_A$ and both $AH$ and $OO_A$ are perpendicular to $BC$. ($\\triangle AHN \\cong \\triangle O_AON$.)\n\nTo show that $A$, $U$, $O_A$ are collinear, we will first show that $H$, $U$, $A^+$ are collinear, where $A^+$ is a point on the same side as $A$ with respect to $BC$ such that $A^+BC$ is an equilateral triangle.\n\nConsider the case where $\\angle BAC \\neq 60^\\circ$. (The case where $\\angle BAC = 60^\\circ$ can be done similarly using the same idea.) In this case $AZHY$ is not a parallelogram. Let $Y'$, $Z'$ be the intersection points between $AY$ and $HZ$ and between $AZ$ and $HY$, as in the picture.\n\nLet $V$ be the orthocenter of $HYZ$, thus the lines $HU$ and $HV$ are isogonal conjugate with respect to the angle $\\angle ZHY$. Note also that the quadrilateral $YZY'Z'$ is cyclic, since $\\angle Y'ZZ' = \\angle Y'YZ' = 60^\\circ$. Thus, the lines $Y'Z'$ and $YZ$ are antiparallel. Since $HV \\perp YZ$, then $HU \\perp Y'Z'$.\n\nLet $C'$ be the reflection of $C$ in the line $HY'$. Using the directed angle\n\n$$\n\\begin{aligned}\n(HY', HC) &= (HY', CA) + (CA, HC) \\\\\n&= (AB, AC) - 60^\\circ + 90^\\circ - (AB, AC) = 30^\\circ.\n\\end{aligned}\n$$\n\nThis implies that $HCC'$ is an equilateral triangle, and $\\triangle HCC' \\sim \\triangle A^+BC$.\n\nNote also that $\\triangle Y'HC \\sim \\triangle Z'HB$ since $\\angle HY'C = \\angle HZ'B$ and $\\angle HCY' = \\angle HBZ' = 90^\\circ - \\angle BAC$. Since $\\triangle Y'HC'$ is the reflection of $\\triangle Y'HC$, thus $\\triangle Y'HC' \\sim \\triangle Z'HB$ with the common vertex $H$. Therefore, (by spiral transformation or by simple comparison), $\\triangle Z'HY' \\sim \\triangle BHC'$. It follows that\n\n$$\n\\begin{aligned}\n(Y'Z', A^+H) &= (Y'Z', BC') + (BC', A^+H) \\\\\n&= (Z'H, BH) + (BC, A^+C) = 90^\\circ.\n\\end{aligned}\n$$\n\nThis shows that $HA^+ \\perp Y'Z'$, namely, $H$, $U$, $A^+$ are collinear,\n\n![](images/selected_2011_p5_data_05c5d54409.png)\n\nTo show $A$, $U$, $O_A$ are collinear, it suffices to show that $\\triangle AHU \\sim \\triangle A^+O_AU$ since $H$, $U$, $A^+$ are collinear. Since $A^+O \\perp BC$, then $A^+$ lies on $OO_A$ and $AH \\parallel A^+O_A$. Therefore, we only need to show that $\\frac{AH}{HU} = \\frac{A^+O_A}{A^+U}$.\n\nBy the law of sines to triangle $XYZ$,\n\n$$\nUH = \\frac{YZ}{2 \\sin YHZ} = \\frac{YZ}{2 \\sin(120^\\circ - A)}.\n$$\n\nSince $\\triangle HCC'$ and $\\triangle A^+CB$ are equilateral with common vertex $C$, we can easily see that $\\triangle A^+HC \\cong \\triangle BC'C$. From similarity of triangles, $\\triangle Z'HY' \\sim \\triangle BC'H$ and $\\triangle HYZ \\sim HY'Z'$, \n\n$$\n\\begin{align*}\nA^+H = BC' = Y'Z' \\cdot \\frac{BH}{Z'H} &= YZ \\cdot \\frac{Y'H}{ZH} \\frac{BH}{Y'H} = YZ \\cdot \\frac{\\cos 30^\\circ}{|\\cos(A)|} = YZ \\cdot \\frac{\\sqrt{3}}{2|\\cos(A)|} \\\\\n&\\therefore \\quad \\frac{UH}{A^+H} = \\frac{|\\cos(A)|}{\\sqrt{3} \\sin(120^\\circ - A)}\n\\end{align*}\n$$\n\nSince $AH = 2R \\cos(A)$ where $R$ is the circumradius of $\\triangle ABC$,\n\n$$\nAH + A^+O_A = 3R \\cos(A) + BC \\sin 60^\\circ = 2\\sqrt{3}R \\sin(120^\\circ - A).\n$$\n\nTherefore,\n\n$$\n\\frac{AH}{AH + A^+O_A} = \\frac{|\\cos(A)|}{\\sqrt{3}\\sin(120^\\circ - A)} = \\frac{UH}{A^+H} = \\frac{UH}{A^+U + UH}.\n$$\n\nThis implies that $\\frac{AH}{HU} = \\frac{A^+O_A}{A^+U}$.\n\n![](images/selected_2011_p6_data_9eb0bc069e.png)",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 14314,
    "subject": "Mathematics (Olympiad)",
    "question": "In the exterior of the acute-angled triangle $ABC$, we construct the isosceles triangles $DAB$ and $EAC$ with bases $AB$ and $AC$, respectively, such that $\\angle DBC = \\angle ECB = 90^\\circ$. Let $M$ and $N$ be the reflections of $A$ with respect to $D$ and $E$, respectively. Prove that the line $MN$ passes through the orthocenter of the triangle $ABC$.",
    "options": [],
    "answer": "See solution",
    "solution": "Denote by $H$ the orthocenter of triangle $ABC$ and by $I$ the midpoint of segment $AH$. Let $B'$ be the reflection of $B$ with respect to $D$ and $C'$ the reflection of $C$ with respect to $E$. Let lines $AB'$ and $CE$ meet at $P$, lines $BH$ and $CE$ meet at $F$, and lines $AB'$ and $BF$ meet at $Q$.\n\nSince $AD = DB = DB'$, we infer that $\\angle BAB' = 90^\\circ$, therefore $AB' \\perp AB$. Since $CH \\perp AB$, we obtain $AB' \\parallel CH$, so $AP \\parallel CH$.\n\n![](images/RMC_2024_p68_data_df2c567bb9.png)\n\nSince $AH \\perp BC$ and $CP \\perp BC$, we deduce that $AH \\parallel CP$. Therefore, $AHCP$ is a parallelogram, and $CP = AH$.\n\nSimilarly, we deduce that $AHFC'$ is a parallelogram, hence $C'F = AH = CP$. Since $C'E = CE$, we obtain that $EP = EF$.\n\nConsequently, $D$ and $E$ are the midpoints of the bases of the trapezoid $BB'PF$, and $Q$ is the intersection point of the lines $B'P$ and $BF$, therefore $D, E, Q$ are collinear. Since $D$ and $I$ are the midpoints of the bases of the trapezoid $AHBB'$, and $Q$ is the intersection point of the lines $B'A$ and $BH$, it follows that $D, I, Q$ are also collinear, consequently $I$ lies on the line $DE$.\n\n$DI$ and $QI$ are midlines in the triangles $AMH$ and $ANH$, respectively, hence $DI \\parallel MH$ and $IE \\parallel NH$, therefore $H \\in MN$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 14315,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $x$, $y$, and $z$ be real numbers. Prove that\n\n$$\n\\frac{x^2+1}{(x+y)^2+4(z+1)} + \\frac{y^2+1}{(y+z)^2+4(x+1)} + \\frac{z^2+1}{(z+x)^2+4(y+1)} \\ge \\frac{1}{2}.\n$$\n\nDetermine when equality holds.",
    "options": [],
    "answer": "See solution",
    "solution": "Note that $$(x+y)^2 \\le 2(x^2+y^2)$$ and $$4z+4 \\le 2(z^2+3).$$ Therefore,\n\n$$\n\\frac{x^2+1}{(x+y)^2+4(z+1)} \\ge \\frac{x^2+1}{2(x^2+y^2+z^2+3)}.\n$$\n\nSimilarly, analogous inequalities hold for the pairs $(y, z)$ and $(z, x)$. Summing these three inequalities yields\n\n$$\n\\frac{x^2+1}{(x+y)^2+4(z+1)} + \\frac{y^2+1}{(y+z)^2+4(x+1)} + \\frac{z^2+1}{(z+x)^2+4(y+1)} \\ge \\frac{x^2+y^2+z^2+3}{2(x^2+y^2+z^2+3)} = \\frac{1}{2}.\n$$\n\nEquality holds at $x = y = z = 1$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 14316,
    "subject": "Mathematics (Olympiad)",
    "question": "For a positive integer $k$, if it is possible to remove a number from $M_k = \\{1, 2, \\dots, k\\}$ such that the sum of the remaining $k-1$ numbers in $M_k$ is a perfect square, then $k$ is called a \"Taurus number\". For instance, 7 is a Taurus number, as removing 3 from $\\{1, 2, 3, 4, 5, 6, 7\\}$ results in the sum $1 + 2 + 4 + 5 + 6 + 7 = 5^2$.\n\n1. Determine with reasoning whether 2021 is a Taurus number.\n\n2. Find $f(n)$ (in terms of $n$), the number of Taurus numbers among $1, 2, \\dots, n$.",
    "options": [],
    "answer": "See solution",
    "solution": "2021 is a Taurus number, since removing 1190 from $M_{2021} = \\{1, 2, \\dots, 2021\\}$ results in the sum of the remaining 2020 numbers equal to $\\sum_{k=1}^{2021} k - 1190 = 1429^2$.\n\n2. The answer is\n\n$$\nf(n) = \\begin{cases} \\lfloor \\sqrt{\\frac{n(n-1)}{2}} \\rfloor, & \\text{if } \\frac{n(n+1)}{2} \\text{ is a perfect square,} \\\\ \\lfloor \\sqrt{\\frac{n(n+1)}{2}} \\rfloor, & \\text{if } \\frac{n(n+1)}{2} \\text{ is not a perfect square.} \\end{cases}\n$$\n\nAlternatively,\n\n$$\nf(n) = \\left\\lceil \\sqrt{\\frac{n^2 + n - 2}{2}} \\right\\rceil.\n$$\n\nWe give two methods as follows.\n\nMethod 1: To begin, notice that the largest perfect square not exceeding the sum $\\sum_{k=1}^{n} k = \\frac{n(n+1)}{2}$ is $\\left\\lfloor \\sqrt{\\frac{n(n+1)}{2}} \\right\\rfloor^2$. Compare it with $f(n)$ for the first few $n$ values, as illustrated in the table:\n\n![alt](page375.md \"\")\n\ntable: $n$, $n(n + 1)/2$, $[\\sqrt{n(n+1)/2}]^2$, $f(n)$ for $n=1$ to $13$.\n\nWe conjecture the first formula is the answer. Let\n\n$$\n\\varphi(n) = \\frac{n(n+1)}{2} - \\left\\lfloor \\sqrt{\\frac{n(n+1)}{2}} \\right\\rfloor^2.\n$$\n\nIt is clear that $n$ is a Taurus number if and only if $\\varphi(n) \\in M_n$, namely, $1 \\leq \\varphi(n) \\leq n$. For a non-Taurus number $n$, either $\\varphi(n) = 0$ (type A), or $\\varphi(n) \\geq n+1$ (type B).\n\n**Lemma 1**: Define $d = \\sqrt{\\frac{n(n+1)}{2}} - \\sqrt{\\frac{n(n-1)}{2}}$. For $n > 1$, $0 < d < 1$.\n\n*Proof of Lemma 1*: Notice that\n\n$$\n\\begin{align*}\nd &= \\sqrt{\\frac{n(n+1)}{2}} - \\sqrt{\\frac{n(n-1)}{2}} \\\\\n&= \\frac{\\frac{n(n+1)}{2} - \\frac{n(n-1)}{2}}{\\sqrt{\\frac{n(n+1)}{2}} + \\sqrt{\\frac{n(n-1)}{2}}} \\\\\n&= \\frac{n}{\\sqrt{\\frac{n(n+1)}{2}} + \\sqrt{\\frac{n(n-1)}{2}}} = \\frac{\\sqrt{2}}{\\sqrt{1+\\frac{1}{n}} + \\sqrt{1-\\frac{1}{n}}} < 1, \\\\\n&\\qquad \\left( \\text{since } d^2 = \\frac{2}{2+2\\sqrt{1-\\frac{1}{n^2}}} = \\frac{1}{1+\\sqrt{1-\\frac{1}{n^2}}} < 1 \\right)\n\\end{align*}\n$$\n\nand hence $0 < d < 1$.\n\nFrom Lemma 1, the integer parts of $\\sqrt{\\frac{n(n-1)}{2}}$ and $\\sqrt{\\frac{n(n+1)}{2}}$ differ by at most 1. Hence\n\n$$\n\\left[ \\sqrt{\\frac{n(n+1)}{2}} \\right] - \\left[ \\sqrt{\\frac{n(n-1)}{2}} \\right] = 0 \\text{ or } 1.\n$$\n\n**Lemma 2**: For integer $n > 1$, if $\\frac{n(n-1)}{2}$ is not a perfect square, then\n\n$$\n\\frac{n(n-1)}{2} \\geq \\left[ \\sqrt{\\frac{n(n-1)}{2}} \\right]^2 + 1.\n$$\n\n*Proof of Lemma 2*: Given that $\\frac{n(n-1)}{2}$ is not a perfect square, we have\n\n$$\n\\left[ \\sqrt{\\frac{n(n-1)}{2}} \\right] < \\sqrt{\\frac{n(n-1)}{2}}, \\text{ taking squares to yield } \\left[ \\sqrt{\\frac{n(n-1)}{2}} \\right]^2 < \\frac{n(n-1)}{2} \\text{ which is an integer. Therefore, } \\frac{n(n-1)}{2} \\geq \\left[ \\sqrt{\\frac{n(n-1)}{2}} \\right]^2 + 1.\n$$\n\nNow we verify the conjecture by induction on $n$. Obviously it is true when $n=1$. Assume it is true for every positive integer up to $n-1$, and consider the formula for $n$. There are two situations.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 14317,
    "subject": "Mathematics (Olympiad)",
    "question": "Prove that\n\n$$\n3abc + a + b + c \\ge 2(ab + bc + ca)\n$$\n\nholds for all real numbers $a, b, c \\ge 1$.\n\nDetermine all cases for which equality is obtained.",
    "options": [],
    "answer": "See solution",
    "solution": "Let us denote $x = a - 1$, $y = b - 1$, and $z = c - 1$. Then $x, y, z \\ge 0$, and the given inequality transforms into\n\n$$\n3xyz + xy + yz + zx \\ge 0,\n$$\n\nwhich is true since all terms on the left-hand side are non-negative.\n\nEquality holds if and only if $xyz = xy = yz = zx = 0$, which is true if and only if at least two numbers among $x, y, z$ are zero, i.e., if and only if at least two numbers among $a, b, c$ are equal to $1$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 14318,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $G$ be the centroid of triangle $ABC$, and let $A_1$, $B_1$, $C_1$ be the midpoints of sides $BC$, $AC$, and $AB$, respectively. Denote the sides and medians of triangle $ABC$ as follows: $AB = c$, $BC = a$, $CA = b$, $AA_1 = d$, $BB_1 = e$, $CC_1 = f$.\n\nSuppose it is possible to color three of these six segments red and the other three blue so that it is not possible to construct a triangle using segments of the same color. Prove that this situation is not possible.",
    "options": [],
    "answer": "See solution",
    "solution": "Suppose that one can color three of the six segments red and the other three blue so that it is not possible to construct a triangle using segments of the same color. Then, for each color, the length of some segment is not less than the sum of the other two segments of that color. Thus, there exist two of the six segments, say $x$ and $y$, such that $x + y \\geq z + u + v + w$ for the remaining four segments $z, u, v, w$.\n\nWe show that this situation is impossible by considering the following cases:\n\n1. **Both $x$ and $y$ are sides of $\\triangle ABC$**. Without loss of generality, let $a + b \\geq c + d + e + f$. From triangles $BGC$ and $AGC$, we have $\\frac{2f}{3} + \\frac{2e}{3} > a$ and $\\frac{2f}{3} + \\frac{2d}{3} > b$. Summing all three inequalities gives $\\frac{f}{3} > c + \\frac{d}{3} + \\frac{e}{3}$, which contradicts the triangle inequality for medians: $d + e > f$.\n\n2. **Both $x$ and $y$ are medians of $\\triangle ABC$**. Without loss of generality, let $d + e \\geq a + b + c + f$. From triangles $BGC_1$ and $AGC_1$, we have $\\frac{f}{3} + \\frac{c}{2} > \\frac{2e}{3}$ and $\\frac{f}{3} + \\frac{c}{2} > \\frac{2d}{3}$, or equivalently $\\frac{f}{2} + \\frac{3c}{4} > e$ and $\\frac{f}{2} + \\frac{3c}{4} > d$. It follows that $f + \\frac{3c}{2} > d + e \\geq a + b + c + f$, or $\\frac{c}{2} > a + b$, a contradiction.\n\n3a. **$x$ and $y$ are a side and a median of $\\triangle ABC$ sharing a common endpoint**. Without loss of generality, let $a + e \\geq b + c + d + f$. But this contradicts the inequalities $a < b + c$ and $e < d + f$.\n\n3b. **$x$ is a median of $\\triangle ABC$ with its endpoint at the midpoint of side $y$**. Without loss of generality, let $a + d \\geq b + c + e + f$. But this contradicts the inequalities $a < b + c$ and $d < e + f$.\n\nTherefore, such a coloring is not possible.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 14319,
    "subject": "Mathematics (Olympiad)",
    "question": "Prove that any continuous function $f : \\mathbb{R} \\to \\mathbb{R}$ of the form\n$$\nf(x) = \\begin{cases} a_1 x + b_1, & \\text{for } x \\le 1 \\\\ a_2 x + b_2, & \\text{for } x > 1 \\end{cases}\n$$\nwith $a_1, a_2, b_1, b_2 \\in \\mathbb{R}$, can be represented as\n$$\nf(x) = m_1 x + n_1 + \\varepsilon|m_2 x + n_2|, \\text{ for } x \\in \\mathbb{R},\n$$\nwhere $m_1, m_2, n_1, n_2 \\in \\mathbb{R}$ and $\\varepsilon \\in \\{-1, +1\\}$.",
    "options": [],
    "answer": "See solution",
    "solution": "The continuity at $x = 1$ implies $a_1 + b_1 = a_2 + b_2$.\n\nIf $a_1 = a_2$, then $b_1 = b_2$ and we can take $m_1 = a_1 = a_2$, $n_1 = b_1 = b_2$, $m_2 = n_2 = 0$, $\\varepsilon = \\pm 1$.\n\nOtherwise, we search $A$ such that $m_2x + n_2 = A(x - 1)$. The required form can be obtained if $m_1 - \\varepsilon|A| = a_1$, $n_1 + \\varepsilon|A| = b_1$ and $m_1 + \\varepsilon|A| = a_2$, $n_1 - \\varepsilon|A| = b_2$, that is\n$$\nm_1 = \\frac{a_1 + a_2}{2}, \\quad n_1 = \\frac{b_1 + b_2}{2}, \\quad 2\\varepsilon|A| = a_2 - a_1 = b_1 - b_2.\n$$\nChoosing $\\varepsilon = \\operatorname{sign}(a_2 - a_1)$, $m_2 = \\frac{1}{2}(a_1 - a_2)$ and $n_2 = \\frac{1}{2}(a_2 - a_1)$, we get\n$$\nf(x) = \\frac{a_1 + a_2}{2}x + \\frac{b_1 + b_2}{2} + \\operatorname{sign}(a_2 - a_1) \\left| \\frac{a_1 - a_2}{2}x + \\frac{a_2 - a_1}{2} \\right|.\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 14320,
    "subject": "Mathematics (Olympiad)",
    "question": "Does there exist an integer $n$ such that the equation $x^4 - 2011x^2 + n = 0$ has integer solutions for $x$?",
    "options": [],
    "answer": "See solution",
    "solution": "Assume that such $n$ exists. From $x^4 - 2011x^2 + n = 0$, we deduce that\n\n$$\nx^2 = \\frac{2011 \\pm \\sqrt{2011^2 - 4n}}{2}.\n$$\n\nThis must be an integer, so $2011^2 - 4n$ must be a perfect square. Let $2011^2 - 4n = m^2$ for some odd positive integer $m$, so $n = \\frac{2011^2 - m^2}{4}$. Thus, $x^2 = \\frac{2011 \\pm m}{2}$. The numbers $\\frac{2011 + m}{2}$ and $\\frac{2011 - m}{2}$ are perfect squares, and their sum is $2011$.\n\nLet us show that $2011$ cannot be written as a sum of two perfect squares. When divided by $4$, a perfect square can only give the remainder $0$ or $1$. A sum of two perfect squares can therefore only give the remainder $0$, $1$, or $2$. Since $2011$ gives the remainder $3$, it cannot be the sum of two perfect squares. Hence, an integer $n$ with the required properties does not exist.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 14321,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $ABCD$ be a quadrilateral with circumcenter $O$, incenter $I$, and $S$ the midpoint of both diagonals $AC$ and $BD$. Consider the following cases:\n\n- If $S = O$, then $ABCD$ is an inscribed parallelogram. Show that $ABCD$ must be a square.\n- If $S = I$, then the diagonals are angle bisectors. Show that $ABCD$ must be a square.\n- If $I = O$, show that $ABCD$ must be a square.",
    "options": [],
    "answer": "See solution",
    "solution": "If $S = O$, then $AC = BD = 2R$, where $R$ is the circumradius of $ABCD$ and $S$ is the midpoint of both $AC$ and $BD$. This means $ABCD$ is an inscribed parallelogram. The sum of its opposite angles, which are congruent, is $180^\\circ$, so all angles of $ABCD$ are right angles; thus, $ABCD$ is a rectangle. Since there is an inscribed circle of radius $r$ in $ABCD$, all sides are equal to $2r$, so $ABCD$ is a square.\n\nIf $S = I$, then the diagonals $AC$ and $BD$ are bisectors of the angles of $ABCD$, which means every side subtends the same arc of the circumcircle as its neighboring sides. Thus, the four vertices $A, B, C, D$ divide the circle into four equal parts, so $ABCD$ is a square.\n\nIf $I = O$, the quadrilateral can be partitioned into eight congruent right triangles with hypotenuse $R$ and one leg $r$; the angle between $R$ and $r$ in those right triangles always has vertex at $O$, so all central angles $\\angle AOB$, $\\angle BOC$, $\\angle COD$, and $\\angle DOA$ are equal. Again, $A, B, C, D$ divide the circumcircle into four equal arcs, so $ABCD$ is a square.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 14322,
    "subject": "Mathematics (Olympiad)",
    "question": "For every integer $n \\geq 1$, consider the $n \\times n$ table with entry $\\left\\lfloor \\dfrac{i \\cdot j}{n+1} \\right\\rfloor$ at the intersection of row $i$ and column $j$, for every $i = 1, 2, \\dots, n$ and $j = 1, 2, \\dots, n$. Determine all integers $n \\geq 1$ for which the sum of the $n^2$ entries in the table is equal to $\\dfrac{n^2(n-1)}{4}$.",
    "options": [],
    "answer": "See solution",
    "solution": "First, observe that every pair $x, y$ of real numbers for which the sum $x + y$ is an integer satisfies\n\n$$\n\\lfloor x \\rfloor + \\lfloor y \\rfloor \\geq x + y - 1 \\qquad (\\diamond)\n$$\n\nThe inequality is strict if $x$ and $y$ are integers and holds with equality otherwise.\n\nWe estimate the sum $S$ as follows:\n\n$$\n2S = \\sum_{1 \\leq i, j \\leq n} \\left( \\left\\lfloor \\frac{ij}{n+1} \\right\\rfloor + \\left\\lfloor \\frac{(n+1-i)j}{n+1} \\right\\rfloor \\right) \\geq \\sum_{1 \\leq i, j \\leq n} (j - 1) = \\frac{n^2(n-1)}{2}.\n$$\n\nThe inequality in the last line follows from $(\\diamond)$ by setting $x = \\frac{ij}{n+1}$ and $y = \\frac{(n+1-i)j}{n+1}$, then $x + y = j$. Therefore, $S = \\frac{n^2(n-1)}{4}$ if and only if the inequality in the last line holds with equality, which means none of the values $\\frac{ij}{n+1}$ is an integer for $1 \\leq i, j \\leq n$.\n\nIf $n+1$ is composite, for example $n+1 = ab$ where $1 < a, b$, one gets an inequality for $i = a$ and $j = b$. Otherwise, if $n+1$ is prime, then $\\gcd(n+1, ij) = 1$ for every $1 \\leq i, j \\leq n$ and $S = \\frac{n^2(n-1)}{4}$.\n\n$\\boxed{\\text{All } n \\geq 1 \\text{ such that } n+1 \\text{ is prime}}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 14323,
    "subject": "Mathematics (Olympiad)",
    "question": "1. Find all positive integers $n$ of the form $n = r(2r + 3)$, where $r$ is a prime, such that $n = q(q^2 - q - 1)$ for some prime $q$.",
    "options": [],
    "answer": "See solution",
    "solution": "First, $q \\neq r$; otherwise, from\n\n$$\nq(q^2 - q - 1) = r(2r + 3)\n$$\nwe would have $r^2 - r - 1 = 2r + 3$, which gives $r = 4$, not a prime.\n\nThus, $q^2 - q - 1 = kr$ and $2r + 3 = kq$ for some integer $k$. Eliminating $r$ gives:\n\n$$\n2q^2 - (2 + k^2)q + 3k - 2 = 0.\n$$\n\nThe discriminant is $D = k^4 + 4k^2 - 24k + 20$, which must be a perfect square. For $k > 5$, $(k^2)^2 < D < (k^2 + 2)^2$, so $D = (k^2 + 1)^2$ leads to $2k^2 - 24k + 19 = 0$, which has no integer roots.\n\nTrying small odd $k$:\n- $k = 1$: $q = 1$ (not prime)\n- $k = 3$: no integer solutions\n- $k = 5$: $2q^2 - 27q + 13 = 0$ yields $q = 13$, $r = 31$, so $n = r(2r + 3) = 2015$.\n\nThus, the only such $n$ is $\\boxed{2015}$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 14324,
    "subject": "Mathematics (Olympiad)",
    "question": "Determine all integer solutions of the equation\n\n$$\n(x - 1) x (x + 1) + (y - 1) y (y + 1) = 24 - 9xy.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "Since $(x-1) x (x+1) + (y-1) y (y+1) = x^3 + y^3 - x - y$, adding $3xy(x+y)$ to both sides yields the equivalent equation\n\n$$\n(x+y)^3 - (x+y) = 24 + 3xy(x+y-3) \\iff (x+y)^3 - 27 - (x+y-3) = 3xy(x+y-3).\n$$\n\nSince $(x+y)^3 - 27 = (x+y-3)((x+y)^2 + 3(x+y) + 9)$, this is equivalent to\n\n$$\n(x+y-3)((x+y)^2+3(x+y)+9-1-3xy) = 0 \\iff (x+y-3)(x^2-xy+y^2+3x+3y+8) = 0.\n$$\n\nIf $x + y - 3 = 0$, we obtain the set of solutions\n\n$$\n\\{(t, 3-t) : t \\in \\mathbb{Z}\\}.\n$$\n\nIt remains to find all solutions of $x^2 - xy + y^2 + 3x + 3y + 8 = 0$.\n\nConsider $x^2 - (y-3)x + y^2 + 3y + 8 = 0$ as a quadratic in $x$. The discriminant is $(y-3)^2 - 4(y^2 + 3y + 8) = -3y^2 - 18y - 23 = 4 - 3(y+3)^2$.\n\nThis must be a perfect square for integer solutions, which occurs iff $(y+3)^2 = 0$ or $(y+3)^2 = 1$, i.e., $y = -2$, $y = -3$, or $y = -4$.\n\nFor $y = -2$, $x^2 + 5x + 6 = 0$ yields $x = -2$ or $x = -3$; solutions: $(-2, -2)$ and $(-3, -2)$.\n\nFor $y = -3$, $x^2 + 6x + 8 = 0$ yields $x = -2$ or $x = -4$; solutions: $(-2, -3)$ and $(-4, -3)$.\n\nFor $y = -4$, $x^2 + 7x + 12 = 0$ yields $x = -3$ or $x = -4$; solutions: $(-3, -4)$ and $(-4, -4)$.\n\nThus, all integer solutions are:\n\n$$\n\\{(t, 3-t) : t \\in \\mathbb{Z}\\} \\cup \\{(-2, -2), (-3, -2), (-2, -3), (-4, -3), (-3, -4), (-4, -4)\\}.\n$$\n\n$\\Box$",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 14325,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $f$ be a function such that for some constant $a$, there exists $c$ with $f(c) = a$, and for all $x$,\n$$\nf(f(x)) = x f(x) - a x.\n$$\nFind all possible values of $a$ and provide examples of functions $f$ that satisfy these conditions.",
    "options": [],
    "answer": "See solution",
    "solution": "The possible values of $a$ are $a = 0$ or $a = -1$.\n\nBy the given condition, there exists $c$ such that $f(c) = a$. Substituting $c$ for $x$ in the equation,\n$$\nf(f(x)) = x f(x) - a x,\n$$\nwe get $f(a) = f(f(c)) = c f(c) - a c = c a - a c = 0$.\n\nSubstituting $a$ for $x$ gives:\n$$\nf(0) = f(f(a)) = a f(a) - a^2 = -a^2.\n$$\n\nSubstituting $0$ for $x$ gives $f(-a^2) = f(f(0)) = 0 \\cdot f(0) - a \\cdot 0 = 0$.\n\nSubstituting $-a^2$ for $x$ gives:\n$$\nf(0) = f(f(-a^2)) = -a^2 f(-a^2) - a(-a^2) = a^3.\n$$\n\nFrom the above, $-a^2 = a^3$, so $a = 0$ or $a = -1$.\n\nFor $a = -1$, one possible function is:\n$$\nf(x) = \\begin{cases} -1 & \\text{if } x \\ne -1, \\\\ 0 & \\text{if } x = -1 \\end{cases}\n$$\n\nFor $a = 0$, possible functions include:\n$$\nf(x) = \\begin{cases} 0 & \\text{if } x \\ne 1, \\\\ 1 & \\text{if } x = 1 \\end{cases}\n$$\n\nand\n$$\nf(x) = \\begin{cases} x^{\\frac{1+\\sqrt{6}}{2}} & \\text{if } x > 0, \\\\ 0 & \\text{if } x \\le 0 \\end{cases}\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 14326,
    "subject": "Mathematics (Olympiad)",
    "question": "A blue Smartie is randomly placed in one of nine positions arranged in a 3×3 grid. What is the probability that it is placed in the centre position?",
    "options": [],
    "answer": "See solution",
    "solution": "Of the nine positions that the blue Smartie could occupy, only one is in the centre; so the probability is $\\frac{1}{9}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 14327,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $D_1$ be the set of those points $X$ such that\n$$\n\\frac{1}{2} \\cdot \\frac{BX}{BB'} + \\frac{1}{2} \\ge \\alpha \\frac{BX}{BB'}\n$$\nfor some constant $\\alpha$ (for suitable $\\alpha$, $D_1$ is a neighborhood of $B$). Let $D_2$ be the set of those points $X$ in the plane such that\n$$\n\\frac{AX}{BX} < \\frac{\\alpha}{3}\n$$\n(for suitable $\\alpha$, $D_2$ is a neighborhood of $A$). Calculate the areas of $D_1$ and $D_2$ for $\\alpha = \\frac{3}{2}$, and show that\n$$\nT_X \\ge T_{S-D_1-D_2} > 6,\n$$\nwhere $T_K$ represents the area of figure $K$ in the plane.",
    "options": [],
    "answer": "See solution",
    "solution": "To omit $A''X$, we write $\\frac{AA''}{A''X}$ in terms of $AA''$ and $AX$ and treat $B'X$ similarly:\n\n$$\n\\begin{align*}\n\\Rightarrow \\quad \\frac{A''X}{AA''} &= \\frac{1}{2} \\cdot \\frac{XB'}{B'B} \\\\\n\\Rightarrow \\quad \\frac{AX}{AA''} &= 1 - \\frac{1}{2}\\left(1 - \\frac{BX}{BB'}\\right) \\\\\n\\Rightarrow \\quad \\frac{AX}{AA''} &= \\frac{1}{2} \\cdot \\frac{BX}{BB'} + \\frac{1}{2}.\n\\end{align*}\n$$\n\nIf $X \\notin D_1$, then\n$$\n\\frac{AX}{AA''} < \\alpha \\frac{BX}{BB'} \\implies \\frac{AA''}{BB'} > \\frac{1}{\\alpha} \\cdot \\frac{AX}{BX}.\n$$\nIf $X$ is not in either $D_1$ or $D_2$, then\n$$\n\\frac{AA''}{BB'} > \\frac{1}{\\alpha} \\cdot \\frac{\\alpha}{3} = \\frac{1}{3}\n$$\nas desired. So\n$$\nS - D_1 - D_2 \\subseteq S'' \\subseteq S'.\n$$\n\nNow, we calculate the areas of $D_1$ and $D_2$. If $\\alpha < 3$, then $D_2$ is the interior part of the Apollonius circle of $A$ and $B$. Also,\n$$\nX \\in D_1 \\Leftrightarrow \\frac{1}{2} \\cdot \\frac{BX}{BB'} + \\frac{1}{2} \\ge \\alpha \\frac{BX}{BB'} \\Leftrightarrow \\frac{BX}{BB'} < \\frac{1}{2\\alpha - 1}.\n$$\nSo, if $\\alpha > 1$ then $D_1$ is a neighborhood of $B$ similar to $S$. Now, we take $\\alpha = \\frac{3}{2}$. We conclude that the area of $D_1$ is $(\\frac{1}{2})^2$ times the area of $S$ which is $\\frac{10}{4} = 2.5$.\n\nIf $C$ and $D$ are the intersection points of the boundary of $D_2$ with the line $AB$ and $C$ is between $A$ and $B$, we have\n$$\n\\begin{align*}\n\\frac{AD}{DB} &= \\frac{1}{2} \\implies \\frac{AD}{AD+1} = \\frac{1}{2} \\implies AD = 1 \\\\\n\\frac{AC}{CB} &= \\frac{1}{2} \\implies \\frac{AC}{1-AC} = \\frac{1}{2} \\implies AC = \\frac{1}{3}.\n\\end{align*}\n$$\nSo the diameter of $D_2$ is $1 + \\frac{1}{3} = \\frac{4}{3}$ and hence, the area of $D_2$ is $(\\frac{2}{3})^2\\pi < 1.5$.\n\nTherefore,\n$$\n\\begin{align*}\nT_X &\\ge T_{S-D_1-D_2} \\\\\n&\\ge 2T_S - T_{D_1} - T_{D_2} \\\\\n&> 10 - 2.5 - 1.5 = 6.\n\\end{align*}\n$$\nwhere $T_K$ represents the area of figure $K$ in the plane. Hence, the assertion is proved. $\\square$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 14328,
    "subject": "Mathematics (Olympiad)",
    "question": "Given that $\\angle BAP = 60^\\circ$ and $\\angle BAC = 90^\\circ$, find the length of $AC$ if $BC = 4$ and $AB = 2$.",
    "options": [],
    "answer": "See solution",
    "solution": "$\\angle BAP = 60^\\circ$ while $\\angle BAC = 90^\\circ$, so $\\angle PAC = 30^\\circ$. With $\\angle BPA = 60^\\circ$, that means $\\angle C = 30^\\circ$, and so $\\triangle PAC$ is isosceles. Then $BC$ has length $4$, and so by Pythagoras the length of $AC$ is $\\sqrt{4^2 - 2^2}$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 14329,
    "subject": "Mathematics (Olympiad)",
    "question": "For a given positive integer $k$, find, in terms of $k$, the minimum value of $N$ for which there exists a set of $2k+1$ distinct positive integers whose sum is greater than $N$, but every subset of size $k$ has sum at most $N/2$.",
    "options": [],
    "answer": "See solution",
    "solution": "The minimum is $N = 2k^3 + 3k^2 + 3k$.\n\nConsider the set\n$$\n\\{k^2+1,\\ k^2+2,\\ \\dots,\\ k^2+2k+1\\}\n$$\nwhich has sum $2k^3 + 3k^2 + 3k + 1 = N + 1$, exceeding $N$. The sum of the $k$ largest elements is $(2k^3 + 3k^2 + 3k)/2 = N/2$, so this $N$ works.\n\nSuppose $N < 2k^3 + 3k^2 + 3k$ and there are positive integers $a_1 < a_2 < \\dots < a_{2k+1}$ with $a_1 + a_2 + \\dots + a_{2k+1} > N$ and $a_{k+2} + \\dots + a_{2k+1} \\le N/2$. Then\n$$\n(a_{k+1} + 1) + (a_{k+1} + 2) + \\dots + (a_{k+1} + k) \\le a_{k+2} + \\dots + a_{2k+1} \\le \\frac{N}{2} < \\frac{2k^3 + 3k^2 + 3k}{2}.\n$$\nThis gives $2ka_{k+1} \\le N - k^2 - k$ and $a_{k+1} < k^2 + k + 1$, so $a_{k+1} \\le k^2 + k$. Combining these,\n$$\n2(k+1)a_{k+1} \\le N + k^2 + k.\n$$\nAlso,\n$$\n(a_{k+1} - k) + \\dots + (a_{k+1} - 1) + a_{k+1} \\ge a_1 + \\dots + a_{k+1} > \\frac{N}{2}\n$$\nor $2(k+1)a_{k+1} > N + k^2 + k$. This contradicts the previous inequality, so no such set exists for $N < 2k^3 + 3k^2 + 3k$, and the stated value is the minimum.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 14330,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $b$ be a positive integer. Find the value of $234_b$ given that:\n\n$$\n70_{10} = 234_{b+1} - 234_{b-1}\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "Method 1\n\nWe have\n\n$$\n\\begin{aligned}\n70_{10} &= 234_{b+1} - 234_{b-1} \\\\\n&= 2(b+1)^2 + 3(b+1) + 4 - 2(b-1)^2 - 3(b-1) - 4 \\\\\n&= 2(b^2 + 2b + 1) + 3(b+1) + 4 - 2(b^2 - 2b + 1) - 3(b-1) - 4 \\\\\n&= 2b^2 + 4b + 2 + 3b + 3 + 4 - 2b^2 + 4b - 2 - 3b + 3 - 4 \\\\\n&= (2b^2 - 2b^2) + (4b + 3b + 4b - 3b) + (2 + 3 + 4 - 2 + 3 - 4) \\\\\n&= 8b + 6 \\\\\nb &= 8\n\\end{aligned}\n$$\n\nSo $234_b = 234_8 = 2 \\times 64 + 3 \\times 8 + 4 = 128 + 24 + 4 = 156$.\n\nMethod 2\n\nThe largest digit on the left side of the given equation is 4. Hence $b-1$ is at least 5. So $b \\geq 6$.\n\nIf $b=6$, then the left side in base 10 is $234_7 - 234_5 = (2 \\times 49 + 3 \\times 7 + 4) - (2 \\times 25 + 3 \\times 5 + 4) = (98+21+4)-(50+15+4) = 123 - 69 = 54 \\neq 70$.\n\nIf $b=7$, then the left side in base 10 is $234_8 - 234_6 = (2 \\times 64 + 3 \\times 8 + 4) - (2 \\times 36 + 3 \\times 6 + 4) = (128+24+4)-(72+18+4) = 156 - 94 = 62 \\neq 70$.\n\nIf $b=8$, then the left side in base 10 is $234_9 - 234_7 = (2 \\times 81 + 3 \\times 9 + 4) - (2 \\times 49 + 3 \\times 7 + 4) = (162+27+4)-(98+21+4) = 193 - 123 = 70$.\n\nEach time $b$ increases by 1, $4_b$ remains the same, $30_b$ increases by 3, but $200_b$ increases by $2(b+1)^2 - 2b^2 = 4b+2$.\nSo the increase in $234_{b+1}$ is greater than the increase in $234_{b-1}$. Hence $234_{b+1} - 234_{b-1}$ increases with increasing $b$. This means $234_{b+1} - 234_{b-1} > 70_{10}$ for $b > 8$.\n\nSo $234_b = 234_8 = 2 \\times 64 + 3 \\times 8 + 4 = 156$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 14331,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all natural numbers $n \\ge 2$ for which the following statement holds: if the sum of numbers in the sequence of natural numbers $(a_1, a_2, \\ldots, a_n)$ is equal to $2n - 1$, then there exists a block of consecutive members of this sequence containing at least two members, the numbers of which have an arithmetic mean that is an integer.",
    "options": [],
    "answer": "See solution",
    "solution": "**Answer:** $n \\ge 4$.\n\nFor $n = 2, 3$ the counterexamples are the sequences $(1, 2)$ and $(2, 1, 2)$.\n\nLet $(a_1, a_2, \\ldots, a_n)$ be an arbitrary sequence satisfying the conditions. Define the sequence: $s_0 = 0$, $s_k = a_1 + a_2 + \\ldots + a_k - 2k$ for $k = 1, \\ldots, n$. Call a sequence *good* if it does not satisfy the conditions of the problem, i.e., there is no block of at least two consecutive members with integer arithmetic mean. Call a pair $(i, j)$ *divisible* if $(j-i) \\mid (s_j - s_i)$. The sequence $(a_1, \\ldots, a_n)$ is good if and only if there is no divisible pair with $|j-i| \\ge 2$.\n\nLet $n \\ge 4$. By the problem's conditions, $s_n = -1$, and for all $k = 1, \\ldots, n$, $s_{k+1} - s_k = a_k - 2 \\ge -1$. Consider possible cases for $s_2$:\n\n- If $s_2 \\le -2$, since $s_1 \\ge s_0 - 1 = -1$ and $s_2 \\ge s_1 - 1$, we have $s_2 = -1$, $s_1 = -1$, so $n - 1 \\mid s_n - s_1$.\n- If $s_2 = -1$, then $n - 2 \\mid s_n - s_2$.\n- If $s_2 = 0$, then $2 - 0 \\mid s_2 - s_0$.\n- If $s_2 \\ge 1$, since $s_n = -1$ and $s_{k+1} \\ge s_k - 1$, there will be an index $i$ with $s_i = 0$, so $i - 0 \\mid s_i - s_0$.\n\nThus, among the pairs $(1, n)$, $(2, n)$, $(0, 2)$, and $(0, i)$ for $2 < i < n$, at least one pair is divisible. This completes the proof for $n \\ge 4$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 14332,
    "subject": "Mathematics (Olympiad)",
    "question": "Find the minimum value of\n$$\n\\sum_{k=1}^{n-1} kx_k(2n-k)\n$$\ngiven that $x_1, x_2, \\dots, x_{n-1}$ are non-negative integers such that $\\sum_{k=1}^{n-1} kx_k = 2n-2$.",
    "options": [],
    "answer": "See solution",
    "solution": "We have\n$$\n\\begin{align*}\n\\sum_{k=1}^{n-1} kx_k(2n-k) &= 2n(2n-2) - \\sum_{k=1}^{n-1} k^2 x_k \\\\\n&= 2n(2n-2) - \\sum_{k=1}^{n-1} x_k - \\sum_{k=1}^{n-1} (k-1)(k+1)x_k \\\\\n&\\ge 2n(2n-2) - n - \\sum_{k=1}^{n-1} (k-1)nx_k \\\\\n&= 2n(2n-2) - n - n \\sum_{k=1}^{n-1} kx_k + n \\sum_{k=1}^{n-1} x_k \\\\\n&= 2n(2n-2) - n - n(2n-2) + n^2 \\\\\n&= 3n^2 - 3n.\n\\end{align*}\n$$\n\nEquality holds when $x_1 = n-1$, $x_2 = x_3 = \\dots = x_{n-2} = 0$ and $x_{n-1} = 1$. Therefore, the minimum value is $3n^2 - 3n$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 14333,
    "subject": "Mathematics (Olympiad)",
    "question": "Positive real numbers $x, y$ satisfy the following condition: there exist $a \\in [0, x]$, $b \\in [0, y]$ such that\n$$\na^2 + y^2 = 2, \\quad b^2 + x^2 = 1, \\quad ax + by = 1.\n$$\nThen the maximum of $x + y$ is \\_\\_\\_\\_\\_\\_.",
    "options": [],
    "answer": "See solution",
    "solution": "In a plane rectangular coordinate system $xOy$, for positive real number pairs $(x, y)$ that satisfy the condition, take points $L(x, 0)$, $M(x, y)$, $N(0, y)$, and then quadrilateral $OLMN$ is a rectangle. Points $P, Q$ are on sides $LM, MN$, respectively, as shown in the figure below.\n\nSince $a^2 + y^2 = 2$, $b^2 + x^2 = 1$, $ax + by = 1$, we have\n$$\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$$\nThus, $\\triangle OPQ$ is an isosceles right triangle with $P$ as its right-angle vertex.\n\n![](images/Mathematical_Olympiad_In_China_2021_2022___Problems_and_Solutions_2024_p48_data_d45c3ba7c5.png \"Fig. 8.1\")\n\nTherefore, we can set $\\angle LOP = \\theta$, $\\angle QON = \\frac{\\pi}{4} - \\theta$, where $0 \\leq \\theta \\leq \\frac{\\pi}{4}$.\n\nThen\n$$\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) \\\\\n&= 2 \\cos \\theta + \\sin \\theta \\\\\n&= \\sqrt{5} \\sin(\\theta + \\varphi),\n\\end{aligned}\n$$\nwhere $\\varphi = \\arcsin \\frac{2\\sqrt{5}}{5}$.\n\nWhen $\\theta = \\frac{\\pi}{2} - \\varphi$ (correspondingly, $x = \\frac{2\\sqrt{5}}{5}$, $y = \\frac{3\\sqrt{5}}{5}$), $x + y$ takes the maximum $\\sqrt{5}$.\n\n$\\square$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 14334,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $M$ and $N$ be points on the sides $AD$ and $BC$ of the square $ABCD$, such that $AM = BN$. Let $X$ be the foot of the perpendicular from point $D$ onto $AN$. Prove that angle $MXC$ is a right angle.",
    "options": [],
    "answer": "See solution",
    "solution": "Consider the diagonals of $MNCD$. Let $O$ be their intersection point, which is the center of the circle with diameter $DN$. We have the following equalities:\n\n$$\n\\angle NXC = \\angle NDC = \\angle MCD = \\angle MXD.\n$$\n\nHence,\n\n$$\n\\angle MXC = \\angle MXD + \\angle DXC = \\angle CXN + \\angle DXC = \\angle DXN = 90^{\\circ}.\n$$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 14335,
    "subject": "Mathematics (Olympiad)",
    "question": "There are $n$ lamps and $2024$ switches in a room. Each lamp is connected to exactly $1000$ switches. When a switch is pressed, the state of each connected lamp changes from \"ON\" to \"OFF\" or vice versa. It is known that by pressing some of the switches, all lamps can be turned on. Prove that this can be achieved by pressing the switches no more than $1012$ times.",
    "options": [],
    "answer": "See solution",
    "solution": "Obviously, each switch should not be pressed more than once. Let us divide all switches into two groups:\n\n*Group I* — those switches that must be pressed to turn all lamps on, and *Group II* — all other switches. Since each lamp is connected to an even number of switches, if all switches in Group II are pressed, then all lamps will also turn on. Indeed, choose lamp $A$. If the number of switches connected to $A$ is even in Group I, then their number is also even in Group II, and pressing them in either group will not change the state of lamp $A$. Similarly, for an odd number.\n\nIn total, there are $2024$ switches in Groups I and II. Therefore, there are no more than $1012$ switches in at least one of them, which proves the desired answer.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 14336,
    "subject": "Mathematics (Olympiad)",
    "question": "Prove that the equation\n\n$$\na^3 + b^3 + c^3 = a^2 + b^2 + c^2\n$$\nhas infinitely many solutions in integers $a, b, c$ such that the greatest common divisor of $a, b, c$ is $1$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let us choose $b = 1 + x$ and $c = 1 - x$, and substitute into the equation:\n\n$$\na^3 + (1 + x)^3 + (1 - x)^3 = a^2 + (1 + x)^2 + (1 - x)^2.\n$$\n\nExpanding and simplifying, we get:\n\n$$\na^3 + 2(1 + 3x^2) = a^2 + 2(1 + x^2)\n$$\nwhich leads to\n$$\na^3 - a^2 = 4x^2.\n$$\n\nLet $a = 1 - 4p^2$ for integer $p$. Then\n$$\na^3 - a^2 = (1 - 4p^2)^3 - (1 - 4p^2)^2 = 4x^2.\n$$\nLet $x = p(4p^2 - 1)$. Then, the triple\n$$\n(a, b, c) = (1 - 4p^2, 1 + p(4p^2 - 1), 1 - p(4p^2 - 1))\n$$\nsatisfies the equation. Since $p$ can be any integer, there are infinitely many such solutions with $\text{gcd}(a, b, c) = 1$.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 14337,
    "subject": "Mathematics (Olympiad)",
    "question": "Given an ellipse equation $\\frac{x^2}{a^2} + \\frac{y^2}{b^2} = 1$ ($a > b > 0$) in the plane rectangular coordinate system $xOy$, let $A_1, A_2, F_1, F_2$ be its left and right endpoints, and left and right foci, respectively. Let $P$ be any point on the ellipse different from $A_1, A_2$. Suppose there exist points $Q, R$ such that $QA_1 \\perp PA_1$, $QA_2 \\perp PA_2$, $RF_1 \\perp PF_1$, and $RF_2 \\perp PF_2$. Find and prove the relationship between the length of segment $QR$ and $b$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $c = \\sqrt{a^2 - b^2}$. Then $A_1(-a, 0)$, $A_2(a, 0)$, $F_1(-c, 0)$, $F_2(c, 0)$.\n\nDenote $P(x_0, y_0)$, $Q(x_1, y_1)$, $R(x_2, y_2)$, where $\\frac{x_0^2}{a^2} + \\frac{y_0^2}{b^2} = 1$, $y_0 \\neq 0$.\n\nFrom $QA_1 \\perp PA_1$ and $QA_2 \\perp PA_2$, we have:\n\n$$\n\\overrightarrow{A_1Q} \\cdot \\overrightarrow{A_1P} = (x_1 + a)(x_0 + a) + y_1 y_0 = 0\n$$\n\n$$\n\\overrightarrow{A_2Q} \\cdot \\overrightarrow{A_2P} = (x_1 - a)(x_0 - a) + y_1 y_0 = 0\n$$\n\nSubtracting, $2a(x_1 + x_0) = 0$, so $x_1 = -x_0$. Substituting into the first equation:\n\n$$\n(-x_0 + a)(x_0 + a) + y_1 y_0 = 0 \\implies -x_0^2 + a^2 + y_1 y_0 = 0 \\implies y_1 = \\frac{x_0^2 - a^2}{y_0}\n$$\n\nThus, $Q(-x_0, \\frac{x_0^2 - a^2}{y_0})$.\n\nSimilarly, from $RF_1 \\perp PF_1$ and $RF_2 \\perp PF_2$, we obtain $R(-x_0, \\frac{x_0^2 - c^2}{y_0})$.\n\nTherefore,\n\n$$\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$$\n\nSince $|y_0| \\in (0, b]$, then $|QR| \\geq b$, with equality if and only if $|y_0| = b$ (i.e., $P(0, \\pm b)$). $\\square$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 14338,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $f(x) = \\sin^4 x - \\sin x \\cos x + \\cos^4 x$. What is the range of $f(x)$?",
    "options": [],
    "answer": "See solution",
    "solution": "We have\n\n$$\n\\begin{aligned}\nf(x) &= \\sin^4 x - \\sin x \\cos x + \\cos^4 x \\\\\n&= 1 - \\frac{1}{2} \\sin 2x - \\frac{1}{2} \\sin^2 2x.\n\\end{aligned}\n$$\n\nLet $t = \\sin 2x$, then\n\n$$\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$$\n\nFor $-1 \\le t \\le 1$:\n\n$$\n\\min g(t) = g(1) = \\frac{9}{8} - \\frac{1}{2} \\times \\frac{9}{4} = 0,\n$$\n\nand\n\n$$\n\\max g(t) = g\\left(-\\frac{1}{2}\\right) = \\frac{9}{8} - \\frac{1}{2} \\times 0 = \\frac{9}{8}.\n$$\n\nTherefore, the range is $0 \\le f(x) \\le \\frac{9}{8}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 14339,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $n$ be a positive integer. Find all real numbers $x$ such that\n$$\n\\sum_{k=1}^{n} \\frac{(k+1)^2}{x+k} + n x^2 = n x + \\frac{n(n+3)}{2}.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "We start by rewriting the given equation using the identities:\n$$\nnx = x + x + \\cdots + x \\quad \\text{($n$ terms)}, \\quad \\frac{n(n+3)}{2} = (1 + 2 + \\cdots + n) + n.\n$$\nThis leads to:\n$$\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$$\nWe can rearrange as:\n$$\n\\sum_{k=1}^{n} \\left[ \\frac{(k+1)^2}{x+k} - (x+k) \\right] + n x^2 - n = 0.\n$$\nNow, note that:\n$$\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$$\nSo the equation becomes:\n$$\n(1-x) \\left( n + \\sum_{k=1}^{n} \\frac{k+1}{x+k} \\right) + n(x^2-1) = 0.\n$$\nOr:\n$$\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$$\nFor $x = 1$, the equation is satisfied. For $x \\ne 1$, we have:\n$$\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$$\nIf $0 < x < 1$, each term on the left is greater than 1, so the sum is greater than $n$, but $n x < n$.\nIf $x > 1$, each term on the left is less than 1, so the sum is less than $n$, but $n x > n$.\nTherefore, the only solution is $x = 1$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 14340,
    "subject": "Mathematics (Olympiad)",
    "question": "設 $\\triangle ABC$ 為各邊長不相同的銳角三角形，點 $O, H$ 分別為其外心與垂心。\n\n直線 $OA$ 分別與 $\\triangle ABC$ 中過點 $B$ 的高及過點 $C$ 的高交於點 $P, Q$。\n\n試證：三角形 $PQH$ 的外心，落在三角形 $ABC$ 的某一條中線上。\n\n（註：三角形中，頂點與對邊中點的連線，稱為中線。）",
    "options": [],
    "answer": "See solution",
    "solution": "Suppose, without loss of generality, that $AB < AC$. We have\n\n$$\n\\begin{aligned}\n\\angle PQH &= 90^\\circ - \\angle QAB = 90^\\circ - \\angle OAB \\\\\n&= \\frac{1}{2} \\angle AOB = \\angle ACB,\n\\end{aligned}\n$$\n\nand similarly $\\angle QPH = \\angle ABC$. Thus triangles $ABC$ and $HPQ$ are similar. Let $\\Omega$ and $\\omega$ be the circumcircles of $ABC$ and $HPQ$, respectively.\n\nSince\n\n$$\n\\angle AHP = 90^\\circ - \\angle HAC = \\angle ACB = \\angle HPQ,\n$$\n\nline $AH$ is tangent to $\\omega$.\n\n![](images/18-3J_p14_data_9284713e61.png)\n\nLet $T$ be the center of $\\omega$ and let lines $AT$ and $BC$ meet at $M$. We will take advantage of the similarity between $ABC$ and $HPQ$ and the fact that $AH$ is tangent to $\\omega$ at $H$, with $A$ on line $PQ$. Consider the corresponding tangent $AS$ to $\\Omega$, with $S \\in BC$. Then $S$ and $A$ correspond to each other in $\\triangle ABC \\sim \\triangle HPQ$, and therefore\n\n$$\n\\angle OSM = \\angle OAT = \\angle OAM.\n$$\n\nHence quadrilateral $SAOM$ are cyclic, and since the tangent line $AS$ is perpendicular to $AO$, $\\angle OMS = 180^\\circ - \\angle OAS = 90^\\circ$. This means that $M$ is the orthogonal projection of $O$ onto $BC$, which is its midpoint. So $T$ lies on median $AM$ of triangle $ABC$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 14341,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $n$ be an integer with $n \\ge 2$, and let $a_0, a_1, a_2, \\dots, a_n$ be complex numbers with $a_n \\ne 0$. Prove that the following statements are equivalent:\n\n(P) $|a_n z^n + a_{n-1} z^{n-1} + \\dots + a_1 z + a_0| \\le |a_n + a_0|$, for every complex number $z$ of modulus 1.\n\n(Q) $a_1 = a_2 = \\dots = a_{n-1} = 0$ and $\\frac{a_0}{a_n} \\in [0, \\infty)$.",
    "options": [],
    "answer": "See solution",
    "solution": "(Q) $\\Rightarrow$ (P). If $a_1 = a_2 = \\dots = a_{n-1} = 0$ and $\\frac{a_0}{a_n} \\in [0, \\infty)$, then\n\n$$\n\\begin{aligned}\n|a_n z^n + a_{n-1} z^{n-1} + \\dots + a_1 z + a_0| &= |a_n z^n + a_0| \\\\\n&\\le |a_n z^n| + |a_0| = |a_n| + |a_0| = |a_n + a_0|,\n\\end{aligned}\n$$\n\nfor every complex number $z$ of modulus 1.\n\n(P) $\\Rightarrow$ (Q). Let $g(z) = a_{n-1}z^{n-1} + \\cdots + a_1z$, $z \\in \\mathbb{C}$ and $w = a_0 + a_n$. For any $\\varepsilon \\in U_n = \\{z \\in \\mathbb{C} \\mid z^n = 1\\}$ we have $|a_n + g(\\varepsilon) + a_0| \\le |a_n + a_0|$, that is $|w + g(\\varepsilon)| \\le |w|$ or $|g(\\varepsilon)|^2 + \\overline{w}g(\\varepsilon) + w\\overline{g(\\varepsilon)} \\le 0$.\n\nAs $\\sum_{\\varepsilon \\in U_n} \\varepsilon^k = 0$, $k = 1, 2, \\dots, n-1$, we have $\\sum_{\\varepsilon \\in U_n} g(\\varepsilon) = 0$, whence, summing up, $\\sum_{\\varepsilon \\in U_n} |g(\\varepsilon)|^2 + \\overline{w}g(\\varepsilon) + w\\overline{g(\\varepsilon)} = \\sum_{\\varepsilon \\in U_n} |g(\\varepsilon)|^2 \\le 0$, which implies that $g(\\varepsilon) = 0$, for every $\\varepsilon \\in U_n$.\n\nFrom the above, we get $a_k = \\frac{1}{n} \\sum_{\\varepsilon \\in U_n} \\frac{g(\\varepsilon)}{\\varepsilon^k} = 0$, $k = 1, 2, \\dots, n-1$.\n\nThus, we get $|a_n z^n + a_0| \\le |a_n + a_0|$ for every complex number $z$ of modulus 1. If we denote $c = \\frac{a_0}{a_n}$ and $t = z^n$, we have $|t + c| \\le |1 + c|$ for every complex number $z$ of modulus 1. Let $P, M, A$ be the points in the complex plane with complex coordinates $t, -c$, respectively $1$. Then $PM \\le MA$, for every point $P$ on the unit circle, therefore the unit circle is interior to the circle with center $M$ and radius $MA$. As the point $A$ belongs to both circles, we get that they are tangent at the point $A$, and thus the points $M, O$ and $A$ are collinear and $MA \\ge OA = 1$. We get that $-c \\le 0$, and thus $c \\in [0, \\infty)$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 14342,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all integers $n$ that have more than $\\frac{n}{2}$ divisors.",
    "options": [],
    "answer": "See solution",
    "solution": "**Answer:** $n \\in \\{1, 2, 3, 4, 6\\}$.\n\n**Solution.** Clearly, a number cannot have divisors greater than $\\frac{n}{2}$, except for $n$ itself. Therefore, to have more than $\\frac{n}{2}$ divisors, $n$ must be divisible by all numbers from $1$ to $\\frac{n}{2}$ and by $n$. Denote by $m$ the integer equal to either $\\frac{n}{2}$ or $\\frac{n-1}{2}$. Then, if $n \\geq 10$, either $m$ or $m-1$ is coprime with $3$. Hence, $n \\geq 3m$, because it must be divisible by both $3$ and $m$. However, in this case $n \\geq 3 \\cdot \\left(\\frac{n-1}{2} - 1\\right)$, or $2n \\geq 3n - 9$, which contradicts the assumption that $n \\geq 10$. All numbers $n < 10$ can be checked by hand.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 14343,
    "subject": "Mathematics (Olympiad)",
    "question": "For each integer $n \\geq 1$, define $a_n = \\lfloor \\frac{n}{\\sqrt{n}} \\rfloor$, where $\\lfloor x \\rfloor$ denotes the largest integer not exceeding $x$ for any real number $x$. Find the number of all $n$ in the set $\\{1, 2, 3, \\dots, 2010\\}$ for which $a_n > a_{n+1}$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let us examine the first few natural numbers: $1, 2, 3, 4, 5, 6, 7, 8, 9$. Here we see that $a_n = 1, 2, 3, 2, 2, 3, 3, 4, 3$. We observe that $a_n \\leq a_{n+1}$ for all $n$ except when $n+1$ is a square, in which case $a_n > a_{n+1}$. We prove that this observation is valid in general.\n\nConsider the range\n$$\nm^2,\\ m^2+1,\\ m^2+2,\\ \\dots,\\ m^2+m,\\ m^2+m+1,\\ \\dots,\\ m^2+2m.\n$$\nLet $n$ take values in this range so that $n = m^2 + r$, where $0 \\leq r \\leq 2m$. Then $\\lfloor \\sqrt{n} \\rfloor = m$, and hence\n$$\n\\lfloor \\frac{n}{\\sqrt{n}} \\rfloor = \\lfloor \\frac{m^2 + r}{m} \\rfloor = m + \\lfloor \\frac{r}{m} \\rfloor.\n$$\nThus $a_n$ takes the values $\\underbrace{m, m, m, \\dots, m}_{m \\text{ times}},\\ \\underbrace{m+1, m+1, m+1, \\dots, m+1}_{m \\text{ times}},\\ m+2$ in this range.\n\nBut when $n = (m+1)^2$, we see that $a_n = m+1$. This shows that $a_{n-1} > a_n$ whenever $n = (m+1)^2$. When we take $n$ in the set $\\{1, 2, 3, \\dots, 2010\\}$, the only squares are $1^2, 2^2, \\dots, 44^2$ (since $44^2 = 1936$ and $45^2 = 2025$), and $n = (m+1)^2$ is possible for only 43 values of $m$. Thus $a_n > a_{n+1}$ for 43 values of $n$. (These are $2^2-1, 3^2-1, \\dots, 44^2-1$.)",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 14344,
    "subject": "Mathematics (Olympiad)",
    "question": "對於整數數對 $(a, b)$，令 $P(x) = a x^3 + b x$。若對於某個正整數 $m$，以下命題成立：\n\n若整數 $x, y$ 滿足 $m$ 能整除 $P(x) - P(y)$，則 $m$ 能整除 $x - y$。\n\n則稱數對 $(a, b)$ 為「$m$-充分」。若存在無限多個正整數 $k$ 使得 $(a, b)$ 為「$k$-充分」，則稱數對 $(a, b)$ 為「非常充分」。\n\n試問：是否存在數對 $(a, b)$ 使得 $(a, b)$ 是「1110-充分」，但不是「非常充分」？",
    "options": [],
    "answer": "See solution",
    "solution": "答案是否定的！以下證明若 $(a, b)$ 是「1110-充分」，則對於任意 $k = 37^r$，$(a, b)$ 均為 $k$-充分。\n\n1. 若 $(a, b)$ 是 1110-充分，則 $(a, b)$ 是 37-充分。\n\n假設 $(a, b)$ 不是 37-充分，則存在 $x, y$ 使得 $37 \\mid P(x) - P(y)$ 但 $x - y$ 不是 37 的倍數。由中國剩餘定理，存在整數 $x', y'$ 使得 $x'$ 和 $y'$ 都是 30 的倍數，且 $37 \\mid x' - x$，$37 \\mid y' - y$。如此一來有 $30 \\mid P(x') - P(y')$ 且 $37 \\mid P(x') - P(y')$，即 $1110 \\mid P(x') - P(y')$。從而 $1110$ 不整除 $x' - y'$，故得矛盾。\n\n2. 若 $(a, b)$ 是 37-充分，則 $37 \\mid a$。\n\n用反證法，假設 $a$ 不是 37 的倍數。首先若 $b$ 亦不是 37 的倍數，注意 $P(x) - P(y) = (a(x^2 + x y + y^2) + b)(x - y)$，且 $4a(x^2 + x y + y^2) + 4b = a(2x + y)^2 + 3a y^2 + 4b$。令集合\n\n$$\nS = \\{ a s^2 + b \\text{ 除 } 37 \\text{ 的餘數} \\mid s \\text{ 是整數} \\}, \\\\\nT = \\{ -3a t^2 \\text{ 除 } 37 \\text{ 的餘數} \\mid t \\text{ 是整數} \\},\n$$\n\n其中餘數都是在 0 到 36。則 $S$ 和 $T$ 都有 19 個元素，故必然有其中一個重複；即存在整數 $s, t$ 使得 $37 \\mid a s^2 + 3a t^2 + b$。\n\n由於 37 不整除 $b$，$s$ 和 $t$ 不同時為 0，可以取 $t' = t$ 或 $-t$，使得 $3t' - s$ 不是 37 的倍數。取整數 $s'$ 使得 $2s' + t' \\equiv s \\pmod{37}$，則\n\n$$\n4a(s'^2 + s' t' + t'^2) + 4b \\equiv a s + 3 t^2 + b \\equiv 0 \\pmod{37},\n$$\n\n且因為 $3t' - s$ 不是 37 的倍數，有 $3t' - (2s' + t')$ 不是 37 的倍數，即 $s'$ 和 $t'$ 除以 37 的餘數不同。然而我們有 $37 \\mid (a(s'^2 + s' t' + t'^2) + b)(s' - t') = P(s') - P(t')$，矛盾。\n\n若 $b$ 是 37 的倍數，則 $P(10) - P(1) = 999a + 9b$ 是 37 的倍數，亦矛盾，(2) 證畢。\n\n3. 對任意正整數 $r$，$(a, b)$ 是 $37^r$-充分。\n\n由 (2)，$37 \\mid a$。顯然 $b$ 不能是 37 的倍數，否則 $37 \\mid P(1) - P(0)$ 矛盾。對任意整數 $x, y$，若 $37^r \\mid P(x) - P(y) = (a(x^2 + x y + y^2) + b)(x - y)$，則注意到 $a(x^2 + x y + y^2) + b \\equiv b \\pmod{37}$ 不是 37 的倍數，因此 $37^r \\mid x - y$，證畢。",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 14345,
    "subject": "Mathematics (Olympiad)",
    "question": "Given a triangle $ABC$ and two points $D \\in AC$, $E \\in BD$ such that $\\angle DAE = \\angle AED = \\angle ABC$. Show that $BE = 2CD$ if and only if $\\angle ACB = 90^\\circ$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $\\alpha = \\angle A$, $\\beta = \\angle B$, and $\\gamma = \\angle C$. Then\n$$\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$$\nso\n$$\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$ if and only if $\\cos \\gamma = 0$, i.e., $\\gamma = 90^\\circ$. $\\Box$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 14346,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $K$ be the midpoint of the median $AM$ of triangle $ABC$. On side $AB$ there is a point $X$ such that $\\angle KXM = \\angle ACB$ and $AX > BX$, and on side $AC$ there is a point $Y$ such that $\\angle KYM = \\angle ABC$ and $AY > CY$. Prove that the points $B$, $X$, $C$, $Y$ lie on the same circle.\n\n![](images/Ukraine2022-23_p38_data_f2fa4689c2.png)",
    "options": [],
    "answer": "See solution",
    "solution": "Let $N$ be the midpoint of $AB$, and $L$ be the midpoint of $AC$. Then the points $N$, $K$, and $L$ lie on the same line—the midline of $\\triangle ABC$, which is parallel to $BC$. We have $\\angle KXM = \\angle ACB = \\angle KNM$, so the quadrilateral $KNXM$ is cyclic, and similarly $KMYL$ is cyclic. Therefore, $AN \\cdot AX = AK \\cdot AM = AL \\cdot AY$, so the points $Y$, $N$, $L$, and $X$ lie on the same circle. Hence, $\\angle AXY = \\angle ALN = \\angle ACB$, so the quadrilateral $BXYC$ is cyclic, which is what we needed to prove.\n\nRemarks: In this configuration, the quadrilateral $AXMY$ is harmonic.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 14347,
    "subject": "Mathematics (Olympiad)",
    "question": "The vertices of two acute triangles all lie on the same circle. The midpoints of two sides of one triangle both lie on the nine-point circle of the other triangle. Show that the two triangles share the same nine-point circle.",
    "options": [],
    "answer": "See solution",
    "solution": "The proof is based on a well-known fact recalled in the lemma below.\n\n**Lemma.** Let $ABC$ be a triangle and let $O$ and $\\omega$ be its circumcentre and nine-point centre, respectively. Then the reflection $O'$ of $O$ in the line $BC$ lies on the line $\\omega A$, and $\\omega$ is the midpoint of the segment $O'A$.\n\nThis is because $O'$ is the centre of the circle $BCH$, where $H$ is the orthocentre of the triangle $ABC$, and the nine-point circle of the triangle $ABC$ is the image of the circle $BCH$ under the homothety of ratio $1/2$ centred at $A$.\n\n![](images/RMC2013_final_p47_data_7fb8ad207e.png)\n\n![](images/RMC2013_final_p47_data_23f7873a1e.png)\n\nBack to the problem, let $ABC$ and $XYZ$ be two triangles inscribed in a circle $\\Gamma$ such that the midpoints of the sides $XY$ and $XZ$ both lie on the nine-point circle $\\gamma$ of the triangle $ABC$. Alternatively, but equivalently, the points $Y$ and $Z$ both lie in the intersection of $\\Gamma$ and the image $\\Gamma'$ of $\\gamma$ under the homothety of ratio $2$ centred at $X$.\n\nIf $\\Gamma$ and $\\Gamma'$ do not coincide, then they are clearly the reflection of one another in the line $YZ$, and, consequently, so are their centres. Let $O'$ be the centre of $\\Gamma'$. By the lemma, the nine-point centre of the triangle $XYZ$ is the midpoint of the segment $O'X$ which in turn is the centre of $\\gamma$ and the conclusion follows.\n\nIf $\\Gamma$ and $\\Gamma'$ coincide, recall that the homotheties mapping $\\gamma$ to $\\Gamma$ are: one of ratio $2$ centred at the orthocentre $H$ of the triangle $ABC$; and one of ratio $-2$ centred at the centroid of the triangle $ABC$. Consequently, $X = H$, so the triangle $ABC$ is right-angled which case is ruled out by hypothesis.\n\n**Remark.** In the special case where the triangle $ABC$ is right-angled, say at $A$, and $X = A$, the nine-point circle of the triangle $XYZ = AYZ$ is the reflection of $\\gamma$ in the line through the midpoints of the sides $XY = AY$ and $XZ = AZ$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 14348,
    "subject": "Mathematics (Olympiad)",
    "question": "Several points are given in the plane. A child wants to draw $k$ (closed) discs in such a manner that for any two points $A, B$ ($A \\neq B$), there exists a disc that contains only one of these points. What is the minimum $k$ such that for any initial configuration of 2019 points, it is possible to draw the $k$ discs with the above property?",
    "options": [],
    "answer": "See solution",
    "solution": "The minimum is $k = 1010 = \\lfloor n/2 \\rfloor$, where $n$ is the number of points.\n\nWe say that a disc separates two points if it contains only one of them.\n\n**Estimation:** Consider $n$ points placed on a circle. Any disc can separate at most two consecutive pairs, so at least $\\lfloor n/2 \\rfloor$ discs are needed.\n\n**Construction:** Let $S_1$ be the set of $n$ points, and $H_1$ its convex hull. Let $D_1$ be a disc containing exactly $\\lfloor n/2 \\rfloor$ points. There exist points $A_1, B_1 \\in S_1$ on the boundary of $H_1$ such that $A_1 \\in D_1$, $B_1 \\notin D_1$, and the open interval $A_1B_1$ contains no other points of $S_1$. Let $D_2$ be a disc that separates $A_1, B_1$ from the other points of $S_1$.\n\nLet $S_2 = S_1 \\setminus \\{A_1, B_1\\}$ and $H_2$ its convex hull. There exist $A_2, B_2 \\in S_2$ on the boundary of $H_2$ such that $A_2 \\in D_2$, $B_2 \\notin D_2$, and the open interval $A_2B_2$ contains no other points of $S_2$. Let $D_3$ be a disc that separates $A_2, B_2$ from the other points of $S_2$.\n\nContinue this process to obtain $\\lfloor n/2 \\rfloor$ discs. (The last unpaired point is $A_{\\lfloor n/2 \\rfloor}$.) This set of discs has the desired separating property. For any $P, Q \\in S_1$, if $P = A_i$, $Q = B_j$ for some $i, j$, then $D_1$ separates them. If $P = A_i$ and $Q = A_j$ with $i < j$ (or $P = B_i$ and $Q = B_j$ with $i < j$), then $D_{i+1}$ separates them.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 14349,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $\\mathbb{R}_{>0} = \\{x \\in \\mathbb{R} \\mid x > 0\\}$ denote the set of positive real numbers. Find all functions $f: \\mathbb{R}_{>0} \\to \\mathbb{R}_{>0}$ satisfying\n$$\nf(x)f(y + f(x)) = f(1 + xy)\n$$\nfor all $x, y \\in \\mathbb{R}_{>0}$.",
    "options": [],
    "answer": "See solution",
    "solution": "The solutions are $f(x) = 1$ and $f(x) = 1/x$.\n\nIt is easy to check that these are solutions, so we prove there are no other solutions.\n\n**Case 1: $f$ is not injective**\n\nSuppose $a > b > 0$ and $f(a) = f(b)$. Then\n$$\nf(1 + a x) = f(a) f(x + f(a)) = f(b) f(x + f(b)) = f(1 + b x)\n$$\nfor any $x \\in \\mathbb{R}_{>0}$. Let $c = a/b > 1$ and $x = y/b$, then $f(1 + c y) = f(1 + y)$.\n\nTaking $a = 1 + c y$, $b = 1 + y$, we get\n$$\nf(1 + (1 + c y) z) = f(1 + (1 + y) z)\n$$\nfor all $z \\in \\mathbb{R}_{>0}$. Now, for $1 < t < c$, let $y = \\frac{t-1}{c-t} > 0$ and $z = \\frac{c-t}{c-1} > 0$.\n\nThen $(1 + c y) z = t$ and $(1 + y) z = 1$, thus $f(1 + t) = f(2)$. Moreover, by induction, $f(1 + c^n x) = f(1 + x)$ for any $n \\ge 1$, so $f(x) = f(2)$ is constant for all $x > 2$.\n\nFinally, for any $x \\in \\mathbb{R}_{>0}$, let $y = 2 + 1/x$, then $y + f(x) > 2$ and $1 + x y > 2$, thus $f(x) = f(1 + x y)/f(y + f(x)) = 1$.\n\n**Case 2: $f$ is injective**\n\nLet $x > 1$. For $y = \\frac{x-1}{x} > 0$, we have $1 + x y = x$, thus\n$$\nf(1 - 1/x + f(x)) = 1\n$$\nsince $f(x) > 0$. Let $d = f(2) - 1/2$. Then injectivity implies $f(x) = d + 1/x$.\n\nFor $y = x$, we have $x > 1$, $y + f(x) > 1$, $1 + x y > 1$, so\n$$\n\\left(d + \\frac{1}{x}\\right) \\left(d + \\frac{1}{x + d + \\frac{1}{x}}\\right) = d + \\frac{1}{1 + x^2}.\n$$\n\nSimplifying and considering the constant term, we see $d = 0$. Thus $f(x) = 1/x$ for $x > 1$.\n\nNow let $y = f(x)$. Then\n$$\nf(2 f(x)) = \\frac{f(1 + x f(x))}{f(x)}\n$$\nso\n$$\nf\\left(\\frac{2}{x}\\right) = \\frac{x}{1 + x \\cdot \\frac{1}{x}} = \\frac{x}{2}.\n$$\n\nTherefore $f(x) = 1/x$ for $0 < x < 2$ as well.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 14350,
    "subject": "Mathematics (Olympiad)",
    "question": "Five balls are arranged around a circle. Chris chooses two adjacent balls at random and interchanges them. Then Silva does the same, with her choice of adjacent balls to interchange being independent of Chris's. What is the expected number of balls that occupy their original positions after these two successive transpositions?\n\n(A) 1.6  \n(B) 1.8  \n(C) 2.0  \n(D) 2.2  \n(E) 2.4",
    "options": [],
    "answer": "See solution",
    "solution": "Label the balls in order 1 through 5. Assume without loss of generality that the first transposition is $1 \\leftrightarrow 2$, resulting in the order 21345. The following table shows the results of the 5 equally likely second transpositions.\n\n![](images/2021_AMC10B_Solutions_Fall_p7_data_a7faef284e.png)\n\n| 2nd transposition | result | balls in original position | count |\n|-------------------|--------|---------------------------|-------|\n| $2 \\leftrightarrow 1$ | 12345 | 1,2,3,4,5 | 5 |\n| $1 \\leftrightarrow 3$ | 23145 | 4,5 | 2 |\n| $3 \\leftrightarrow 4$ | 21435 | 5 | 1 |\n| $4 \\leftrightarrow 5$ | 21354 | 3 | 1 |\n| $5 \\leftrightarrow 2$ | 51342 | 3,4 | 2 |\n\nThe expected number of balls that occupy their original positions is the average of the numbers in the last column, namely $\\frac{1}{5}(5 + 2 + 1 + 1 + 2) = 2.2$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 14351,
    "subject": "Mathematics (Olympiad)",
    "question": "A group of mathematicians is attending a conference. We say that a mathematician is $k$-content if he is in a room with at least $k$ people he admires or if he is admired by at least $k$ other people in the room. It is known that when all participants are in the same room, they are all at least $3k+1$-content. Prove that you can assign everyone into one of 2 rooms in a way that everyone is at least $k$-content in his room and neither room is empty.\n\n*Admiration is not necessarily mutual and no one admires himself.*",
    "options": [],
    "answer": "See solution",
    "solution": "We will use some basic graph theoretic terms for clarity.\n\nRepresent the situation by a directed graph $G(V, E)$ where each vertex $v \\in V(G)$ represents a mathematician and each edge $e \\in E(G)$ represents an admiration relation. For $v \\in V(G)$, let the out-degree $o(v)$ be the number of mathematicians $v$ admires, and the in-degree $i(v)$ be the number of mathematicians who admire $v$. For $X \\subseteq V$, $G(X)$ denotes the induced subgraph on $X$. A digraph is a $k$-digraph if for every $v \\in V(G)$, $i(v) \\geq k$ or $o(v) \\geq k$.\n\nThe question can be reformulated: Given $G$ is a $3k+1$-digraph, can we split its vertices into 2 disjoint classes so that each induced subgraph is a $k$-digraph?\n\nDefine a subset $X$ of vertices as $k$-tight if for any $Y \\subseteq X$, there is a vertex $v \\in Y$ such that $i_{G(Y)}(v) \\leq k$ and $o_{G(Y)}(v) \\leq k$. A partition $(A_1, A_2)$ is feasible if both $A_1$ and $A_2$ are $k$-tight.\n\nAssume there are no feasible partitions. Consider a minimal subset $A_1 \\subseteq V(G)$ such that $G(A_1)$ is a $k$-digraph, and let $A_2 = V(G) \\setminus A_1$. Any proper subset of $A_1$ is $k$-tight. For $A_1$, removing any $v$ yields a graph where, by minimality, there is a vertex $w$ with $o_G(w) < k$ and $i_G(w) < k$, so $o_{G(A_1)}(w) \\leq k$, $i_{G(A_1)}(w) \\leq k$. Thus, $A_1$ is $k$-tight.\n\n$A_2$ is not $k$-tight, so there exists $A_2' \\subseteq A_2$ such that $A_2'$ is a $(k+1)$-digraph. Now, apply the following proposition:\n\n**Proposition.** If a $2k+1$-digraph $G$ admits a solution pair (disjoint $A, B$ with $G(A)$ and $G(B)$ both $k$-digraphs), then $G$ admits a partition into two $k$-digraphs.\n\n*Proof.* Take a maximal solution pair $(A, B)$. Let $C = V(G) \\setminus (A \\cup B)$. If $C$ is empty, we are done. Otherwise, for $x \\in C$, if $o_{G(B \\cup C)}(x), i_{G(B \\cup C)}(x) < k$, but $i_G(x) \\geq 2k+1$ or $o_G(x) \\geq 2k+1$, so $o_{G(A \\cup \\{x\\})}(x) > k+1$ or $i_{G(A \\cup \\{x\\})}(x) > k+1$, contradicting maximality. Thus, the partition exists.\n\nIf there is a feasible partition $(A, B)$ maximizing $w(A < B) = |E(G(A))| + |E(G(B))|$, then $|A| \\geq k+1$ and $|B| \\geq k+1$. If no $X \\subseteq A$ has $G(X)$ a $k$-digraph, then for any $x \\in B$, $B \\setminus \\{x\\}$ is $k$-tight, and $A \\cup \\{x\\}$ is also $k$-tight. Moving $x$ from $B$ to $A$ increases $w$ by at least $1$, contradicting maximality. Thus, such $X$ exists, and similarly for $Y \\subseteq B$.\n\nApplying the proposition again, we are done.\n\n**Remark.** The same argument, with a modified weight function, can be used for non-symmetric rooms: if the graph is a $(k+l+\\max(k,l)+1)$-digraph, it can be partitioned into $k$-digraph and $l$-digraph parts.",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 14352,
    "subject": "Mathematics (Olympiad)",
    "question": "Iulia and Ștefan shared the 52 playing cards from a deck so each got 26 cards. The cards from 2 to 10 are assigned their own value, the ace is worth 11 points, the jack 12 points, the queen 13 points, and the king 14 points. Ștefan noticed that he had no ace in his stack, no 2, and no four cards of the same value. Iulia noticed that from her cards above 11 points, she doesn't have more than two of the same value. Iulia adds up all her points. What is the lowest value she can get? What is the highest?",
    "options": [],
    "answer": "See solution",
    "solution": "If we denote by $x_i$ the number of cards with the value $i$ that Iulia has in her stack, then $x_i \\geq 1$, $x_{11} = x_4 = 4$, $1 \\leq x_{12} \\leq 2$, $1 \\leq x_{13} \\leq 2$, $1 \\leq x_{14} \\leq 2$.\n\nIf $M$ and $m$ represent the highest and the lowest value that Iulia's stack can have, then:\n\n$$\n\\begin{align*}\nM &= 2 \\cdot 14 + 2 \\cdot 13 + 2 \\cdot 12 + 4 \\cdot 11 + 4 \\cdot 10 + 2 \\cdot 9 + 1 \\cdot 8 + 1 \\cdot 7 + 1 \\cdot 6 + 1 \\cdot 5 + 1 \\cdot 4 + 1 \\cdot 3 + 4 \\cdot 2 = 221, \\\\\nm &= 1 \\cdot 14 + 1 \\cdot 13 + 1 \\cdot 12 + 4 \\cdot 11 + 1 \\cdot 10 + 1 \\cdot 9 + 1 \\cdot 8 + 1 \\cdot 7 + 1 \\cdot 6 + 2 \\cdot 5 + 4 \\cdot 4 + 4 \\cdot 3 + 4 \\cdot 2 = 169.\n\\end{align*}\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 14353,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $f : [0, 1] \\to [0, 1]$ be a function with the property that for every $y \\in [0, 1]$ and every $\\varepsilon > 0$ there exists $x \\in [0, 1]$ so that $|f(x) - y| < \\varepsilon$.\n\na) Prove that if $f$ is continuous on $[0, 1]$, then $f$ is surjective.\n\nb) Give an example of a function $f$ with the given property which is not surjective.",
    "options": [],
    "answer": "See solution",
    "solution": "a) Suppose a continuous function $f : [0, 1] \\to [0, 1]$ has the given property and take $y \\in [0, 1]$. From the hypothesis, there exists a sequence $(x_n)_{n \\ge 1}$ in $[0, 1]$ such that $|f(x_n) - y| < 1/n$ for all $n \\ge 1$. The sequence $(x_n)_{n \\ge 1}$ is bounded, so it has a convergent subsequence $(x_{i_n})_{n \\to \\infty}$. Denote $x := \\lim_{n \\to \\infty} x_{i_n} \\in [0, 1]$. Then $(f(x_{i_n}))_n \\to f(x)$ and $|f(x_{i_n}) - y| < 1/i_n$ for all $n \\ge 1$, hence $|f(x)-y| = 0$, therefore $f(x) = y$. This proves that $f$ is surjective.\n\nb) Consider $f : [0, 1] \\to [0, 1]$ defined by\n$$\n  f(x) = \\begin{cases} x, & x \\in [0, 1] \\cap \\mathbb{Q} \\\\ 0, & x \\in [0, 1] \\setminus \\mathbb{Q} \\end{cases}\n$$\nClearly $f([0, 1]) = [0, 1] \\cap \\mathbb{Q}$, so $f$ is not surjective. Also, $|f(y) - y| = 0$ for all $y \\in [0, 1] \\cap \\mathbb{Q}$. Finally, if $y \\in [0, 1] \\setminus \\mathbb{Q}$ and $\\varepsilon > 0$, then there exists $x \\in [0, 1] \\cap \\mathbb{Q}$ such that $|x - y| < \\varepsilon$, that is, $|f(x) - y| < \\varepsilon$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 14354,
    "subject": "Mathematics (Olympiad)",
    "question": "Given a graph on 28 vertices, where no two vertices of the same degree $i$ are connected by an edge ($1 \\leq i \\leq 27$), determine (with proof) the maximum number of edges in the graph.",
    "options": [],
    "answer": "See solution",
    "solution": "For $0 \\leq i \\leq 27$, let $a_i$ denote the number of vertices of degree $i$. Thus, the total number of edges in the graph is\n\n$$\nN_e = \\frac{0 \\times a_0 + 1 \\times a_1 + \\cdots + 27 \\times a_{27}}{2} = \\frac{(a_{27}) + (a_{27} + a_{26}) + (a_{27} + a_{26} + a_{25}) + \\cdots + (a_{27} + a_{26} + \\cdots + a_1)}{2}.\n$$\n\nNow, consider any vertex $v$ of degree $i$. None of its $i$ neighbouring vertices can have degree $i$, so we have $a_i \\leq 28 - i$. In particular, $a_{27} \\leq 1$, $a_{27} + a_{26} \\leq 1 + 2$, and so on, up until $a_{27} + \\cdots + a_{21} \\leq 1 + 2 + \\cdots + 7$. Also, for each $i \\leq 20$ we have $a_{27} + \\cdots + a_i \\leq 28$, since the sum cannot be larger than the total number of vertices in the graph. This yields the following upper bound on the number of edges:\n\n$$\nN_e \\leq \\frac{(1) + (1+2) + (1+2+3) + \\cdots + (1+2+\\cdots+7) + 20 \\times 28}{2} = 322.\n$$\n\nThis upper bound is attained by the following graph: Partition the 28 vertices into 7 groups of size 1, 2, ..., 7, respectively. Any pair of vertices is connected by an edge if and only if the two vertices lie in different groups. It is easy to see that this graph satisfies the condition of the problem. Because each vertex in the group of size $i$ has degree $28 - i$, the total number of edges in this graph is\n\n$$\nN_e = \\frac{1}{2} \\sum_{i=1}^{7} i(28 - i) = 14 \\sum_{i=1}^{7} i - \\frac{1}{2} \\sum_{i=1}^{7} i^2 = 392 - 70 = 322,\n$$\n\nas required.",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 14355,
    "subject": "Mathematics (Olympiad)",
    "question": "For positive real numbers $a$, $b$, $c$, which of the following statements necessarily implies $a = b = c$?\n\n(I) $a(b^3 + c^3) = b(c^3 + a^3) = c(a^3 + b^3)$\n\n(II) $a(a^3 + b^3) = b(b^3 + c^3) = c(c^3 + a^3)$\n\nJustify your answer.",
    "options": [],
    "answer": "See solution",
    "solution": "Statement (I) does **not** necessarily imply $a = b = c$, while statement (II) **always** implies $a = b = c$.\n\n**For (I):**\n\nLet us analyze $a(b^3 + c^3) = b(c^3 + a^3)$. This gives $c^3(a - b) = ab(a^2 - b^2)$, so either $a = b$ or $ab(a + b) = c^3$. Similarly, $b = c$ or $bc(b + c) = a^3$. If $a \\neq b$ and $b \\neq c$, we obtain:\n\n$$\nab(a + b) = c^3, \\quad bc(b + c) = a^3.$$\n\nFurther manipulation leads to $(a-c)(a^2 + b^2 + c^2 + ab + bc + ca) = 0$. Since $a, b, c > 0$, the only possibility is $a = c$. Thus, there are four possibilities:\n- $a = b = c$\n- $a = b \\neq c$\n- $b = c \\neq a$\n- $c = a \\neq b$\n\nSuppose $a = b \\neq c$. Then $b(c^3 + a^3) = c(a^3 + b^3)$ gives $ac^3 + a^4 = 2ca^3$, which implies $a(a-c)(a^2 - ac - c^2) = 0$. Setting $x = a/c$, we get $x^2 - x - 1 = 0$, so $x = \\frac{1 + \\sqrt{5}}{2}$. Thus,\n\n$$\na = b = \\left( \\frac{1 + \\sqrt{5}}{2} \\right) c, \\quad c > 0.$$\n\nSimilarly, other cases yield:\n$$\nb = c = \\left( \\frac{1 + \\sqrt{5}}{2} \\right) a, \\quad a > 0;$$\n$$\nc = a = \\left( \\frac{1 + \\sqrt{5}}{2} \\right) b, \\quad b > 0.$$\n\nAnd $a = b = c$ is the fourth possibility.\n\n**For (II):**\n\nSuppose $a, b, c$ are mutually distinct. Assume $a = \\max\\{a, b, c\\}$, so $a > b$ and $a > c$. From $a(a^3 + b^3) = b(b^3 + c^3)$, we get $a^3 + b^3 < b^3 + c^3$, so $a^3 < c^3$, which forces $a < c$, a contradiction. Thus, $a, b, c$ cannot all be distinct; at least two must be equal. If $a = b$, then $a(a^3 + b^3) = b(b^3 + c^3)$ gives $a^3 + b^3 = b^3 + c^3$, so $a = c$. Similarly, $b = c$ implies $b = a$, and $c = a$ gives $c = b$. Therefore, $a = b = c$.\n\nAlternatively, adding the three equations:\n\n$$\n\\frac{a^3}{c} + \\frac{b^3}{a} + \\frac{c^3}{b} = a^2 + b^2 + c^2.\n$$\n\nBy Cauchy-Schwarz:\n\n$$\n(a^2 + b^2 + c^2)^2 \\le (a^2 + b^2 + c^2)(ab + bc + ca)\n$$\n\nSo $a^2 + b^2 + c^2 \\le ab + bc + ca$, which implies $(a-b)^2 + (b-c)^2 + (c-a)^2 \\le 0$, hence $a = b = c$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 14356,
    "subject": "Mathematics (Olympiad)",
    "question": "The length of the shortest side of a triangle is $2007$ and the largest angle is twice the smallest. Does such a triangle exist?",
    "options": [],
    "answer": "See solution",
    "solution": "We shall prove that no such triangle satisfies the condition.\n\nIf $\\triangle ABC$ satisfies the condition, let $\\angle A \\leq \\angle B \\leq \\angle C$, then $\\angle C = 2\\angle A$, and $a = 2007$. Draw the bisector of $\\angle ACB$ intersecting $AB$ at point $D$. Then $\\angle BCD = \\angle A$, so $\\triangle CDB \\sim \\triangle ACB$, and it follows that\n\n$$\n\\frac{CB}{AB} = \\frac{BD}{BC} = \\frac{CD}{AC} = \\frac{BD + CD}{BC + AC} = \\frac{AB}{BC + AC}.\n$$\n\nThus,\n\n$$\nc^2 = a(a + b) = 2007(2007 + b),\n$$\n\nwhere $2007 \\leq b \\leq c < 2007 + b$.\n\nSince $a, b, c$ are integers, $2007 \\mid c^2$, so $3 \\cdot 223 \\mid c^2$. Let $c = 669m$. From above, $223m^2 = 2007 + b$. Thus $b = 223m^2 - 2007 \\geq 2007$, so $m \\geq 5$.\n\nBut $c \\geq b$, so $669m \\geq 223m^2 - 2007$, which implies $m < 5$, a contradiction.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 14357,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $f(x) = \\sum_{k=1}^{n} a_k x^k$ and $g(x) = \\sum_{k=1}^{n} \\frac{a_k}{2^k - 1} x^k$ be two polynomials with real coefficients, where $n \\geq 3$. Suppose $1$ and $2^{n+1}$ are roots of $g(x) = 0$. Prove that $f(x) = 0$ has a positive root smaller than $2^n$.",
    "options": [],
    "answer": "See solution",
    "solution": "It is easy to see that $g(2x) - g(x) = f(x)$. Thus\n\n$$\n\\sum_{k=0}^{n} f(2^k) = g(2^{n+1}) - g(1) = 0.\n$$\n\nConsider the relation $f(1) + f(2) + \\dots + f(2^n) = 0$. If $f(2^n) \\neq 0$, then $f(1)$ and $f(2^n)$ have opposite signs. Hence there exists $\\alpha$ in $(1, 2^n)$ such that $f(\\alpha) = 0$. If $f(2^n) = 0$, omit this and the argument applies to $f(1) + f(2) + \\dots + f(2^{n-1}) = 0$.",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 14358,
    "subject": "Mathematics (Olympiad)",
    "question": "A diagonal in a hexagon is considered a \"long\" diagonal if it divides the hexagon into two quadrilaterals. Any two long diagonals divide the hexagon into two triangles and two quadrilaterals.\n\nWe are given a convex hexagon with the property that the division into pieces by any two long diagonals always yields two isosceles triangles with sides of the hexagon as bases.\n\nShow that such a hexagon must have a circumcircle.",
    "options": [],
    "answer": "See solution",
    "solution": "Since any two opposing isosceles triangles (such as $\\textit{ABP}$ and $\\textit{DEP}$) have a common angle at their vertices, they must be similar, and their bases therefore parallel. The angle bisector in their common vertex is therefore also the common altitude.\n\nIf all three diagonals of the hexagon meet at a point $\\textit{M}$, this point is also a common point of all angle bisectors. It must therefore be the same distance from $\\textit{A}$ and $\\textit{B}$, as it lies on the bisector of $\\textit{AB}$, but the same holds for $\\textit{B}$ and $\\textit{C}$, $\\textit{C}$ and $\\textit{D}$, and so on. This point is therefore equidistant from all corners of the hexagon, and is therefore the center of the circumcircle of the hexagon.\n\n![](images/Austria_2010_p12_data_f28b2bff9a.png)\n\nIf the diagonals of the hexagon do not have a common point, they form a triangle. The angle bisectors have a common point, namely the incenter of this triangle, which we again call $\\textit{M}$. The same holds for this point $\\textit{M}$ as in the previous situation, and we once again have established the existence of a circumcircle of the hexagon, as claimed.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 14359,
    "subject": "Mathematics (Olympiad)",
    "question": "Number the rooms according to their capacities. If each of rooms 101, ..., 150 is occupied by 76 guests, and room $i$ is occupied with $i-75$ people for $i \\geq 151$, is it possible to reach the goal? If $n \\leq 8824$, can we choose $1 \\leq i \\leq 50$ such that rooms $100+i$ and $201-i$ are occupied by at most $201-i$ guests in total, and move them all to room $201-i$?",
    "options": [],
    "answer": "See solution",
    "solution": "8824.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 14360,
    "subject": "Mathematics (Olympiad)",
    "question": "Consider an isosceles triangle $ABC$, where $AC$ is a base. Let $P$ be an arbitrary point on $AC$, and let $T$ be the projection of $P$ onto $BC$. Determine the ratio in which a symmedian drawn from vertex $C$ of triangle $\\Delta PBC$ divides $AT$.\n\nA symmedian $CS$, with $S \\in BP$ of $\\Delta PBC$, is a reflection of the median $CF$ over the angle bisector $CL$.\n\n![](images/Ukraine_2020_booklet_p50_data_5ac40a75e5.png)",
    "options": [],
    "answer": "See solution",
    "solution": "**Answer:** $\\frac{AK}{KT} = 2$.\n\n**Solution.** Let $BH$ be an altitude of $\\Delta ABC$, so $H$ is the midpoint of $AC$. $BH$ and $PT$ are altitudes of $\\Delta PBC$ (see Fig. 39). Recall that a symmedian bisects a line segment whose endpoints are the feet of the altitudes. Therefore, a symmedian of $\\Delta PBC$ passes through $M$, the midpoint of $HT$. Let $CM$ intersect $AT$ at $K$. Applying the Menelaus theorem to triangle $\\Delta ATH$, we get:\n\n$$\n\\frac{AK}{KT} \\cdot \\frac{TM}{MH} \\cdot \\frac{HC}{CA} = 1 \\implies \\frac{AK}{KT} = \\frac{AC}{CH} = 2.\n$$",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 14361,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $N \\ge 3$ be an integer. In the country of Sibyl, there are $N^2$ towns arranged as the vertices of an $N \\times N$ grid, with each pair of towns corresponding to an adjacent pair of vertices on the grid connected by a road. Several automated drones are each given the instruction to traverse a rectangular path starting and ending at the same town, following the roads of the country. It turned out that each road was traversed at least once by some drone. Determine the minimum number of drones that must be operating.",
    "options": [],
    "answer": "See solution",
    "solution": "The desired minimum is $N$ if $N$ is odd, and $(N-1)$ if $N$ is even.\n\nWe say a drone covers a vertex if it takes a $90^\\circ$ turn at that vertex.\n\n**Construction:** Let $v_1, \\dots, v_N$ denote the $N$ vertical lines and $h_1, \\dots, h_N$ denote the $N$ horizontal lines of the grid.\n\nFor $N = 2n$, make one drone go along the outer boundary of the grid, and let $(n-1)$ of the drones traverse the rectangles enclosed by $v_{2i}, v_{2i+1}, h_1, h_N$ for all $1 \\le i \\le n-1$, and let $(n-1)$ of the drones traverse the rectangles enclosed by $h_{2i}, h_{2i+1}, v_1, v_N$ for all $1 \\le i \\le n-1$. These $N-1$ drones cover all roads in Sibyl.\n\nFor $N = 2n + 1$, again make one drone go across the boundary, and let $n$ of the drones traverse the rectangles enclosed by $v_{2i}, v_{2i+1}, h_1, h_N$ for all $1 \\le i \\le n$, and let $n$ of the drones traverse the rectangles enclosed by $h_{2i}, h_{2i+1}, v_1, v_N$ for all $1 \\le i \\le n$. One notes that these $N$ drones cover all roads in Sibyl.\n\n**Estimate:** Suppose $k$ drones suffice. We will first show for all $N$ that $k \\ge N-1$. Fix the top-left corner $A$ of the grid and call any drone that passes through this town to be *cornered*. Suppose $C$ is the set of drones that are cornered. Let $\\mathcal{L}_h, \\mathcal{L}_v$ denote the set of drones (not in $C$) that cross a street in $h_1, v_1$ respectively. Note that any drone that covers some vertex in $h_1$ is either in $C$ or $\\mathcal{L}_h$ and any drone in $\\mathcal{L}_h$ covers two vertices in $h_1$ and any drone in $C$ covers one vertex in $h_1$ other than $A$. Since each of the $N$ vertices on the top edge needs to be covered, $1 + |C| + 2|\\mathcal{L}_h| \\ge N$. Similarly, $1 + |C| + 2|\\mathcal{L}_v| \\ge N$. Adding, we get $2 + 2|C| + 2|\\mathcal{L}_h| + 2|\\mathcal{L}_v| \\ge 2N$, hence $k \\ge |C| + |\\mathcal{L}_h| + |\\mathcal{L}_v| \\ge N-1$, proving the claim.\n\nNow suppose $N$ is odd. If possible, suppose $k = N-1$. Following the notation in the last part, we see that equality must hold: $1 + |C| + 2|\\mathcal{L}_h| = N$ and $1 + |C| + 2|\\mathcal{L}_v| = N$. Further, every drone must be in one of these three sets, and no point on the top edge apart from the top-left corner can be covered by two drones (else the bounds would not be tight).\n\nCall a drone a *dominator* if it passes through both $h_1, h_N$ (a vertical dominator) or both $v_1, v_N$ (a horizontal dominator). Since every drone passes through either $h_1$ or $v_1$, and this reasoning applies any other corner, this implies that every drone must be a dominator. Indeed, if some drone is not a dominator, one can pick one of $h_1, h_N$ and one of $v_1, v_N$ so that it doesn't pass through any of the picked lines; and applying the above reasoning to the corner at the intersection of these two leads to a contradiction.\n\nNow if there are two horizontal dominators through $A$, they both cover the top-right corner, contradiction. Similarly, there can be at most one vertical dominator through $A$. But $|C|$ is at least one (some drone needs to cover $A$), and $|C| = 1$ leads to a contradiction modulo $2$ in $1 + |C| + 2|\\mathcal{L}_h| = N$, so $|C|$ is exactly $2$, and there is a horizontal and a vertical dominator through $A$.\n\nBy a similar reasoning, there is a vertical and a horizontal dominator through the bottom-right corner, $D$. But the horizontal dominator through $A$ and the vertical dominator through $D$ both cover the top-right corner point. We noted before that no point on the top edge apart from the top-left corner can be covered by two drones, so this is a contradiction, showing $k \\ge N$. $\\square$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 14362,
    "subject": "Mathematics (Olympiad)",
    "question": "Find the smallest positive integer that cannot be written in the form $\\binom{a}{2} + \\binom{b}{2} + c$ with nonnegative integers $a, b, c$ satisfying $a \\ge b \\ge c$ and $a + b \\le 2019$.",
    "options": [],
    "answer": "See solution",
    "solution": "The number is $m = \\binom{1957}{2} + \\binom{63}{2} + 1 = 1,915,900$.\n\nAssume that $m$ has a representation as above. Then $a \\le 1957$ as $\\binom{1958}{2} > m$. On the other hand, by $\\binom{a}{2} + \\binom{b}{2} + c \\le \\binom{a+1}{2} + \\binom{b-1}{2} + (b-1)$ it follows that the largest number that can be represented as above with $a \\le 1957$ is $\\binom{1957}{2} + \\binom{62}{2} + 62 = m - 1$, a contradiction.\n\nIt remains to show that all natural numbers smaller than $m$ have a representation in the form $\\binom{a}{2} + \\binom{b}{2} + c$ with $a \\ge b \\ge c$ and $a+b \\le 2019$. If some number $k$ has such a representation and $c > 0$ or $c = 0, b \\ge 2$, then $k-1$ can be represented as $\\binom{a}{2} + \\binom{b}{2} + (c-1)$ or $\\binom{a}{2} + \\binom{b-1}{2} + (b-2)$, respectively. Hence, we can represent all integers between $\\binom{a}{2}$ and $k$ in the desired form. Therefore and because we have a representation for $m-1$ already, it suffices to show that the numbers $\\ell_s = \\binom{1957-s}{2} - 1$ with $s = 0, 1, \\dots, 1954$ can be represented. Now the claim follows by $\\ell_0 = \\binom{1956}{2} + \\binom{63}{2} + 2$ and $\\binom{1956-s}{2} < \\ell_s \\le \\binom{1956-s}{2} + \\binom{b}{2}$ with $b = \\min\\{63, 1956-s\\}$ for $s = 1, 2, \\dots, 1954$.\n\n**Remark:**\n\nIf the number 2019 in the problem is replaced by some positive integer $t$, then the answer is $\\binom{p-j}{2} + \\binom{j+1}{2} + 1$, where $j$ is the unique integer with $\\binom{j+2}{2} < t \\le \\binom{j+3}{2}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 14363,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $m$ be a positive integer and $p$ a prime. Suppose $A$ is a unitary ring having exactly $m$ invertible elements and such that\n$$\n\\underbrace{1+1+\\cdots+1}_{p \\text{ times}} = 0.\n$$\n\nProve that $A$ has non-zero nilpotent elements if and only if $p$ divides $m$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $a \\in A$, $a \\neq 0$, be a nilpotent element and $k \\in \\mathbb{N}^*$ such that $a^{p^k} = 0$. From $(a+1)^{p^k} = a^{p^k} + 1 = 1$, we have that $a+1$ is invertible and the order of $a+1$ in the group of units $U(A)$ is a power of $p$. Because the order of $a+1$ divides the order $m$ of $U(A)$, we deduce that $p$ divides $m$.\n\nConversely, by Cauchy's theorem, there is an element $a$ of order $p$ in $U(A)$. Since $(a-1)^p = a^p - 1 = 0$ and $a \\neq 1$, we conclude that $a-1$ is a non-zero nilpotent.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 14364,
    "subject": "Mathematics (Olympiad)",
    "question": "A succession of letters $\\overline{l_1l_2\\dots l_n}$, where $n \\ge 1$, is called a *word*. A word $\\overline{l_1l_2\\dots l_n}$ is called a *palindrome* if $l_k = l_{n-k+1}$ for each $k \\in \\{1, 2, \\dots, n\\}$.\n\nConsider a word $X = \\overline{l_1l_2\\dots l_{2014}}$ consisting of the letters $A$ and/or $B$. Prove that $X$ can be obtained by writing at most 806 palindromes next to each other.",
    "options": [],
    "answer": "See solution",
    "solution": "Let us split $X$ into groups of 5 consecutive letters; this way we obtain 402 groups of 5 letters and one incomplete group of 4 letters.\n\nConsider a 5-letter word $Y$ written only with $A$'s and $B$'s, whose first letter is $A$. So $Y$ is one of the sixteen words of the form $A****$. It is easy to check that all of these words either are palindromes or can be formed by joining two palindromes. Switching $A$ with $B$, a similar observation holds if a 5-letter word $Y$ starts with $B$.\n\nHence, we can concatenate at most $2 \\times 402 = 804$ palindromes to cover the first 2010 letters of $X$. For the last 4 letters, we need at most two palindromes, so 806 is the maximum number of palindromes we need.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 14365,
    "subject": "Mathematics (Olympiad)",
    "question": "In an acute triangle $ABC$ (not isosceles), let $AP$, $BQ$, and $CR$ be the altitudes, and let $H$ be the orthocenter. The line through $A$ parallel to $BC$ meets $RQ$ at $D$. Let $A_1$ be the midpoint of $BC$, and let $K$ be the intersection of $RQ$ and $AA_1$. The line through the midpoint of $AH$ and $K$ meets $DA_1$ at $A_2$. Define $B_2$ and $C_2$ analogously.\n\nSuppose the circumcircle of the non-degenerate triangle $A_2B_2C_2$ is $\\omega$. Prove that there exist three circles $\\odot A'$, $\\odot B'$, and $\\odot C'$ inside $\\omega$ that are tangent to $\\omega$ and satisfy:\n\n1. $\\odot A'$ is tangent to $AB$ and $AC$; $\\odot B'$ is tangent to $BA$ and $BC$; $\\odot C'$ is tangent to $CA$ and $CB$.\n2. The centers $A'$, $B'$, and $C'$ are distinct and collinear.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $A^*$ be the midpoint of $AH$. In the diagram, $\\angle A_1RH = \\angle A_1CR = \\angle RAH$, so $A_1R$ is tangent to the circle $\\Gamma$ with diameter $AH$, and similarly $A_1Q$ is tangent to $\\Gamma$ (with $R$ and $Q$ on $\\Gamma$). Thus,\n\n(a) $$A_1A^* \\perp RQ.$$\n\nConsidering polars with respect to $\\Gamma$, $D$ lies on the polar of $A_1$ with respect to $\\Gamma$, so $A_1$ lies on the polar of $D$. Since $DA$ is tangent to $\\Gamma$, $AA_1$ is the polar of $D$, so\n\n(b) $$DA^* \\perp AA_1.$$\n\nCombining (a) and (b), $K$ is the orthocenter of $\\triangle A^*A_1D$. In particular, $A^*K \\perp A_1D$, so $\\angle A^*A_2A_1 = \\frac{\\pi}{2}$, and $A_2$ lies on the nine-point circle of $\\triangle ABC$. Thus, $\\omega$ is the nine-point circle of $\\triangle ABC$.\n\n![](images/China-TST-2023A_p24_data_aae951c51f.png)\n\nLet $N$ be the center of $\\omega$ and $I$ the incenter of $\\triangle ABC$. Let $BC = a$, $CA = b$, $AB = c$, and $s = \\frac{a+b+c}{2}$. Consider the inversion $f$ centered at $A$ that preserves the nine-point circle. Let $\\odot A'$ be the image of the incircle under $f$; $\\odot A'$ is tangent to $AB$ and $AC$. Similarly define $\\odot B'$ and $\\odot C'$.\n\nLet $M$ be the foot of the perpendicular from $I$ to $AB$. We show $\\odot A'$ and $\\omega$ do not coincide; otherwise, $AR \\cdot AC_1 = AM^2$, which gives $b \\cos A \\cdot \\frac{c}{2} = \\left(\\frac{b+c-a}{2}\\right)^2$, implying $a = b$ or $a = c$, contradicting the non-isosceles condition.\n\nTo verify collinearity of $A'$, $B'$, $C'$, let\n\n$$\n\\overrightarrow{IA'} = p \\cdot \\overrightarrow{IA}, \\quad \\overrightarrow{IB'} = q \\cdot \\overrightarrow{IB}, \\quad \\overrightarrow{IC'} = r \\cdot \\overrightarrow{IC}.\n$$\n\nThen\n\n$$\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$$\n\nTo show $A'$, $B'$, $C'$ are collinear, it suffices to show\n\n$$\n\\frac{a}{p} + \\frac{b}{q} + \\frac{c}{r} = 0.\n$$\n\nLet $L$ be the foot of the perpendicular from $A'$ to $AB$. Then $\\frac{AA'}{AI} = \\frac{AL}{AM}$, $AM = s-a$. By properties of $f$, $f(M) = L$, so\n\n$$\nAM \\cdot AL = AR \\cdot AC_1 = b \\cos A \\cdot \\frac{c}{2} = \\frac{b^2 + c^2 - a^2}{4}.\n$$\n\n$$\n\\frac{AL}{AM} = \\frac{AM \\cdot AL}{AM^2} = \\frac{b^2 + c^2 - a^2}{4(s-a)^2}.\n$$\n\nTherefore,\n\n$$\np = \\frac{IA'}{IA} = \\frac{2(a-b)(a-c)}{(b+c-a)^2}\n$$\n\nSimilarly,\n\n$$\nq = \\frac{2(b-c)(b-a)}{(c+a-b)^2}, \\quad r = \\frac{2(c-a)(c-b)}{(a+b-c)^2}.\n$$\n\nPlugging in, $\\frac{a}{p} + \\frac{b}{q} + \\frac{c}{r} = 0$ (the sum is a quadratic in $a$ vanishing at $a = b$, $a = c$, and $a = b + c$, so it is identically zero). $\\Box$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 14366,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all functions $f: \\mathbb{R} \\to \\mathbb{R}$ such that\n$$\nf(f(y) + 2 + x) + f(f(y) - x) = y f(y)(x + 1)\n$$\nholds for all real numbers $x$ and $y$.",
    "options": [],
    "answer": "See solution",
    "solution": "**Answer:** $f(x) = 0$.\n\n**Solution.**\nLet $x = -2 - t$ for arbitrary real $t$. Then:\n$$\nf(f(y) + 2 + x) + f(f(y) - x) = f(f(y) + 2 - 2 - t) + f(f(y) - (-2 - t)) = f(f(y) - t) + f(f(y) + 2 + t)\n$$\nThe right side becomes:\n$$\ny f(y)(x + 1) = y f(y)((-2 - t) + 1) = y f(y)(-1 - t)\n$$\nSo the equation is:\n$$\nf(f(y) - t) + f(f(y) + 2 + t) = y f(y)(-1 - t)\n$$\nBut from the original equation, swapping $x$ with $t$ gives:\n$$\nf(f(y) + 2 + t) + f(f(y) - t) = y f(y)(t + 1)\n$$\nComparing both, we see:\n$$\ny f(y)(t + 1) = -y f(y)(t + 1)\n$$\nSo $y f(y)(t + 1) = 0$ for all $y, t \\in \\mathbb{R}$. This implies $f(y) = 0$ for all $y \\neq 0$.\n\nNow, substitute $x = -2$, $y = 1$:\n$$\nf(f(1) + 2 - 2) + f(f(1) - (-2)) = 1 \\cdot f(1) \\cdot (-2 + 1) = -f(1)\n$$\nBut since $f(1) = 0$, this gives $f(0) + f(2) = 0$. Since $f(2) = 0$, $f(0) = 0$.\n\nTherefore, the only solution is $f(x) = 0$ for all $x \\in \\mathbb{R}$, which satisfies the original equation.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 14367,
    "subject": "Mathematics (Olympiad)",
    "question": "Consider $m \\ge 3$ positive real numbers $g_1, \\dots, g_m$, each number being less than the sum of the others. For any subset $M \\subseteq \\{1, \\dots, m\\}$, denote\n\n$$\nS_M = \\sum_{k \\in M} g_k.\n$$\n\nFind all $m$ for which it is always possible to partition the indices $1, \\dots, m$ into three sets $A, B, C$, with the property that\n\n$$\nS_A < S_B + S_C, \\quad S_B < S_A + S_C, \\quad \\text{and} \\quad S_C < S_A + S_B.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "The partition is always possible precisely when $m \\ne 4$.\n\nFor $m = 3$ it is trivially possible, and for $m = 4$ the four equal numbers $g, g, g, g$ provide a counter-example. Henceforth, assume $m \\ge 5$.\n\nAmong all possible partitions $A \\sqcup B \\sqcup C = \\{1, \\dots, m\\}$ such that\n\n$$\nS_A \\le S_B \\le S_C,\n$$\n\nselect one for which the difference $S_C - S_A$ is minimal. If there are several such, select one so as to maximise the number of elements in $C$. We will show that $S_C < S_A + S_B$, which is sufficient.\n\nIf $C$ consists of a single element, this number is by assumption less than the sum of the remaining ones, so $S_C < S_A + S_B$ holds.\n\nSuppose now $C$ contains at least two elements, and let $g_c$ be a minimal number indexed by $c \\in C$. We have\n\n$$\nS_C - S_A \\le g_c \\le \\frac{1}{2}S_C.\n$$\n\nThe first is by the minimality of $S_C - S_A$, the second by the minimality of $g_c$. These together yield\n\n$$\nS_A + S_B \\ge 2S_A \\ge 2(S_C - g_c) \\ge S_C.\n$$\n\nIf any of these inequalities is strict, we are done.\n\nSuppose all are equalities, so\n\n$$\nS_A = S_B = \\frac{1}{2}S_C = g_c.\n$$\n\nIt follows that $C = \\{c, d\\}$, where $g_d = g_c$. If $A$ contained more than one element, we could increase the number of elements in $C$ by creating instead a partition\n\n$$\n\\{1, \\dots, m\\} = \\{c\\} \\sqcup B \\sqcup (A \\cup \\{d\\}),\n$$\n\nresulting in the same sums. Similarly for $B$. Thus, $A$ and $B$ must be singleton sets, so\n\n$$\nm = |A| + |B| + |C| = 4.\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 14368,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $x$, $y$, $z$ be real numbers not equal to $1$ satisfying $xyz = 1$.\n\n(a) Prove the inequality\n$$\n\\frac{x^2}{(x-1)^2} + \\frac{y^2}{(y-1)^2} + \\frac{z^2}{(z-1)^2} \\geq 1.\n$$\n\n(b) Show that the equality case holds for infinitely many triples $(x, y, z)$ of rational numbers.",
    "options": [],
    "answer": "See solution",
    "solution": "Set\n$$\na = \\frac{x}{x-1}, \\quad b = \\frac{y}{y-1}, \\quad c = \\frac{z}{z-1};\n$$\nthat is,\n$$\nx = \\frac{a}{a-1}, \\quad y = \\frac{b}{b-1}, \\quad z = \\frac{c}{c-1}.\n$$\nThe desired inequality becomes\n$$\na^2 + b^2 + c^2 \\geq 1. \\qquad (*)\n$$\nThe given condition $xyz = 1$ reads\n$$\nabc = (a-1)(b-1)(c-1) = abc - (ab + bc + ca) + (a+b+c) - 1,\n$$\nso\n$$\n(a+b+c) - 1 = ab + bc + ca.\n$$\nHence,\n$$\n2(a + b + c) - 2 = 2(ab + bc + ca) = (a + b + c)^2 - (a^2 + b^2 + c^2),\n$$\nor\n$$\n(a^2 + b^2 + c^2) - 1 = (a + b + c)^2 - 2(a + b + c) + 1 = (a + b + c - 1)^2 \\geq 0, \\quad (\\dagger)\n$$\nwhich implies the desired inequality $(*)$. Thus, part (a) is established.\n\nFor part (b), since $x$, $y$, $z$ have linear relations with $a$, $b$, $c$, it suffices to show that there are infinitely many rational triples $(a, b, c)$ satisfying equality in $(\\dagger)$; that is, $a + b + c = 1$ and $ab + bc + ca = 0$. Thus,\n$$\n0 = ab + (a+b)c = ab + (a+b)(1-a-b) = ab + (a+b) - a^2 - 2ab - b^2.\n$$\nConsider this as a quadratic in $a$:\n$$\na^2 + (b-1)a + b^2 - b = 0.\n$$\nThe discriminant is\n$$\n\\Delta = (b-1)^2 - 4(b^2 - b) = -3b^2 + 2b + 1 = (3b+1)(1-b).\n$$\nWe seek infinitely many rational $b$ such that $(3b+1)(1-b)$ is a rational square. Let $b = \\frac{p}{q}$ for coprime integers $p$, $q$. We want $(3p+q)(q-p)$ to be a perfect square. Setting $q = p^2 - p + 1$ gives $3p+q = (p+1)^2$ and $q-p = (p-1)^2$, so\n$$\nb = \\frac{p}{p^2 - p + 1}, \\quad \\Delta = \\left( \\frac{p^2 - 1}{p^2 - p + 1} \\right)^2.\n$$\nThus,\n$$\na = \\frac{(1-b) + \\Delta}{2} = \\frac{p^2 - p}{p^2 - p + 1}, \\quad c = 1 - a - b = \\frac{p-1}{p^2 - p + 1}.\n$$\nFor $p > 2$, $x$, $y$, $z$ are not equal to $1$, and distinct $p$ yield distinct rational triples $(x, y, z)$. This completes the proof.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 14369,
    "subject": "Mathematics (Olympiad)",
    "question": "In a regular tetrahedron $ABCD$, consider planes that are parallel to its faces such that each edge is divided into 6 equal segments. These planes determine, on the edges of the tetrahedron, on its faces, and in its interior, a set consisting of 80 points of intersection. Denote this set by $V$.\n\nFind the maximum number of elements of a subset $W$ of the set $V \\cup \\{A, B, C, D\\}$, having the property that any three points of $W$ are not collinear and the plane generated by them is neither parallel to any of the faces of the tetrahedron $ABCD$, nor contains one of its faces.",
    "options": [],
    "answer": "See solution",
    "solution": "Suppose, without loss of generality, that the height of $ABCD$ equals 6. Denote by $k$ the maximal number of elements of the set $W$ and by $M_1, \\dots, M_k$ its elements. For each $i \\in \\{1, 2, \\dots, k\\}$, denote by $a_i, b_i, c_i$, and $d_i$ the distances from the point $M_i$ to the planes $(BCD), (ACD), (ABD)$, and $(ABC)$ respectively. Then $a_i, b_i, c_i, d_i \\in \\{0, 1, 2, \\dots, 6\\}$, and the sum of the distances from an interior point to the faces of a regular tetrahedron equals its height, so $a_i + b_i + c_i + d_i = 6$.\n\nConsequently, if $T = (a_1 + a_2 + \\dots + a_k) + (b_1 + b_2 + \\dots + b_k) + (c_1 + c_2 + \\dots + c_k) + (d_1 + d_2 + \\dots + d_k)$, then $T = 6k$.\n\nLet $s = \\left\\lfloor \\frac{k}{2} \\right\\rfloor$. As no more than two of the numbers $a_1, a_2, \\dots, a_k$ can be equal, we deduce $s \\leq 7$.\n\nFor even $k$, $k = 2s$, we have $a_1 + a_2 + \\dots + a_k \\geq 2 \\cdot 0 + 2 \\cdot 1 + \\dots + 2 \\cdot (s-1) = s^2 - s$. So $T = 4(s^2 - s)$, and $6k = 12s \\geq 4s^2 - 4s$, thus $s \\leq 4$ and $k \\leq 8$.\n\nFor odd $k$, $k = 2s + 1$, we have $a_1 + a_2 + \\dots + a_k \\geq 2 \\cdot 0 + 2 \\cdot 1 + \\dots + 2 \\cdot (s-1) + s = s^2$. Thus $T = 4s^2$, and $6k = 12s + 6 \\geq 4s^2$, implying $s \\leq 3$ and $k \\leq 7$.\n\nTo give an example of an 8-element set, use the notation $(a_i, b_i, c_i, d_i)$ as above and consider as $W$ the set:\n\n$$(0, 1, 2, 3),\\ (1, 0, 2, 3),\\ (0, 1, 3, 2),\\ (1, 0, 3, 2),\\ (2, 3, 0, 1),\\ (2, 3, 1, 0),\\ (3, 2, 1, 0),\\ (3, 2, 0, 1)$$",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 14370,
    "subject": "Mathematics (Olympiad)",
    "question": "Prove that, in the case of $n$ inhabitants and at most $n$ pairs of competitors, there must be $\\lceil \\frac{n}{3} \\rceil$ inhabitants among which no two are competitors.",
    "options": [],
    "answer": "See solution",
    "solution": "The claim holds trivially for $n = 0, 1, 2$.\n\nNow let $n \\ge 3$ and assume the claim is valid for $n-3$ inhabitants. Without loss of generality, assume there are exactly $n$ pairs of competitors. Then there are exactly $2n$ instances of an inhabitant belonging to a pair of competitors.\n\nConsider two cases:\n\n* If every inhabitant has exactly 2 competitors, 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 cancels 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.\n\nThis proves the desired claim.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 14371,
    "subject": "Mathematics (Olympiad)",
    "question": "Benjamin and Anika each sum consecutive integers, but omit some terms. Benjamin sums $1 + 2 + 3 + \\ldots + 2012$, omitting some terms, and Anika sums $1 + 2 + 3 + \\ldots + 2013$, omitting the same terms as Benjamin. If Benjamin's result is divisible by $2011$ and Anika's result is divisible by $2014$, what is the value of $\\frac{N}{A}$, where $N$ is Anika's sum and $A$ is the total sum $1 + 2 + 3 + \\ldots + 2013$?",
    "options": [],
    "answer": "See solution",
    "solution": "Let $x$ be the sum of the terms omitted by Benjamin.\n\nThe total sum for Benjamin is $1 + 2 + 3 + \\ldots + 2012 = \\frac{2012 \\cdot 2013}{2} = 1006 \\cdot 2013$. Benjamin's result is $1006 \\cdot 2013 - x$, which is divisible by $2011$, so $1006 \\cdot 2013 - x = 2011m$ for some integer $m \\geq 0$.\n\nAnika's total sum is $A = 1 + 2 + 3 + \\ldots + 2013 = \\frac{2013 \\cdot 2014}{2} = 2013 \\cdot 1007$. Her result is $N = 2013 \\cdot 1007 - x$, which is divisible by $2014$, so $2013 \\cdot 1007 - x = 2014n$ for some integer $n \\geq 0$.\n\nExpress $x$ from both equations and set them equal:\n\n$$\n1006 \\cdot 2013 - 2011m = 2013 \\cdot 1007 - 2014n\n$$\n\nRearrange:\n\n$$\n2014n - 2011m - 2013 = 0\n$$\n\nSo,\n\n$$\n2011(n - m) = 2013 - 3n\n$$\n\nSince $2014n = 2013 \\cdot 1007 - x \\leq 2013 \\cdot 1007$, $n \\leq \\frac{2013 \\cdot 1007}{2014} < 1007$.\n\nAlso, $2013 - 3n$ must be divisible by both $3$ and $2011$. The only possible value is $2013 - 3n = 0$, so $n = 671$.\n\nTherefore,\n\n$$\n\\frac{N}{A} = \\frac{2014n}{2013 \\cdot 1007} = \\frac{2014 \\cdot 671}{2013 \\cdot 1007} = \\frac{2}{3}\n$$",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 14372,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $ABC$ be an acute triangle with $AB = AC$. Let $D$ be the midpoint of the side $AC$, and let $\\gamma$ be the circumcircle of triangle $ABD$. The tangent to $\\gamma$ at $A$ crosses the line $BC$ at $E$. Let $O$ be the circumcentre of triangle $ABE$. Prove that the midpoint of the segment $AO$ lies on $\\gamma$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $\\Gamma$ be the image of $\\gamma$ under the homothety with centre $A$ and factor $2$. Clearly, $\\Gamma$ is also tangent to $AE$ at $A$, and the conclusion is equivalent to $\\Gamma$ passing through $O$, which is the same as $AE$ being tangent to the circle $ACO$.\n\nAlternatively, but equivalently, this amounts to $\\angle OAE = \\angle OCA$. Write $\\angle OAE = 90^\\circ - \\angle EBA$ and $\\angle OCA = \\angle OCB - \\angle ACB = \\angle OCB - \\angle CBA = \\angle OCB - \\angle EBA$, to infer that the equality of the two angles is equivalent to $C$ being the midpoint of the segment $BE$.\n\n![](images/RMC_2020_p30_data_4a3d5a4740.png)\n\nTo prove the latter, it is sufficient to show that the triangles $ABE$ and $DCB$ are similar, for then $\\dfrac{BE}{BC} = \\dfrac{AB}{CD} = \\dfrac{AC}{CD} = 2$, which implies that $C$ is indeed the midpoint of the segment $BE$.\n\nFinally, to prove the above similarity, write $\\angle EBA = \\angle CBA = \\angle ACB = \\angle BCD$ and $\\angle BDC = \\angle BAD + \\angle DBA = \\angle BAD + \\angle DAE = \\angle BAE$. This completes the proof.\n\n*Remarks.* The condition of $\\triangle ABC$ being acute is not essential, but it avoids considering cases generated by the different relative positions of the points.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 14373,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $x, y$ be integers, $x \\neq -1$, $y \\neq -1$, such that $\\frac{x^4 - 1}{y+1} + \\frac{y^4 - 1}{x+1}$ is also an integer. Prove that $x^4 y^{44} - 1$ is divisible by $x+1$.",
    "options": [],
    "answer": "See solution",
    "solution": "First, we prove that $y^4 - 1$ is divisible by $x+1$.\n\nLet $\\frac{x^4 - 1}{y+1} = \\frac{a}{b}$ and $\\frac{y^4 - 1}{x+1} = \\frac{c}{d}$, with $(a, b) = 1$, $(c, d) = 1$, $b, d > 0$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 14374,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $f : \\mathbb{R} \\to \\mathbb{R}$ be a non-decreasing function and $F : \\mathbb{R} \\to \\mathbb{R}$ a function having right and left finite derivatives at any point in $\\mathbb{R}$ and $F(0) = 0$. Suppose that $\\lim_{x \\uparrow x_0} f(x) \\leq F'_s(x_0)$ and $\\lim_{x \\downarrow x_0} f(x) \\geq F'_d(x_0)$, for any $x_0 \\in \\mathbb{R}$. Prove that $$F(x) = \\int_0^x f(t) \\, dt, \\quad x \\in \\mathbb{R}.$$",
    "options": [],
    "answer": "See solution",
    "solution": "Consider the function $G : \\mathbb{R} \\to \\mathbb{R}$, given by $G(x) = F(x) - \\int_0^x f(t) \\, dt$. Let $x_0 \\in \\mathbb{R}$, $x \\ne x_0$. Then\n\n$$\nF(x) = \\frac{F(x) - F(x_0)}{x - x_0} \\cdot (x - x_0) + F(x_0).\n$$\n\nAs $F$ has right and left finite derivatives in $x_0$, the function $x \\mapsto \\frac{F(x)-F(x_0)}{x-x_0}$, $x \\ne x_0$, is bounded around $x_0$, so $\\lim_{x \\to x_0} F(x) = F(x_0)$. So $G$ is continuous.\n\nLet us show that $G$ has finite right and left derivatives in $x_0 \\in \\mathbb{R}$, $G'_s(x_0) \\geq 0$ and $G'_d(x_0) \\leq 0$. Let $x > x_0$. Because $\\lim_{x \\downarrow x_0} f(x) \\leq f(t) \\leq f(x)$, $x_0 < t \\leq x$, we get\n\n$$\n\\lim_{x \\downarrow x_0} f(x) \\leq \\frac{1}{x-x_0} \\int_{x_0}^{x} f(t) \\, dt \\leq f(x),\n$$\n\nso\n\n$$\n\\lim_{x \\downarrow x_0} \\frac{1}{x-x_0} \\int_{x_0}^{x} f(t) \\, dt = \\lim_{x \\downarrow x_0} f(x).\n$$\n\nConsequently,\n\n$$\n\\begin{aligned}\n\\lim_{x \\downarrow x_0} \\frac{G(x) - G(x_0)}{x - x_0} &= \\lim_{x \\downarrow x_0} \\left( \\frac{F(x) - F(x_0)}{x - x_0} - \\frac{1}{x - x_0} \\int_{x_0}^{x} f(t) \\, dt \\right) \\\\\n&= F'_d(x_0) - \\lim_{x \\downarrow x_0} f(x) \\leq 0.\n\\end{aligned}\n$$\n\nAnalogously, $\\lim_{x \\uparrow x_0} \\frac{G(x) - G(x_0)}{x - x_0} = F'_s(x_0) - \\lim_{x \\uparrow x_0} f(x) \\geq 0$.\n\nIn conclusion, $0 \\leq G'_s(x) < \\infty$ and $-\\infty < G'_d(x) \\leq 0$, for any real $x$, therefore\n\n$$\n\\lim_{\\epsilon \\downarrow 0} \\frac{G(x - \\epsilon) - G(x - \\epsilon/2)}{\\epsilon} = -\\frac{1}{2} G'_s(x) \\leq 0, \\quad x \\in \\mathbb{R},\n$$\n\nand\n\n$$\n\\lim_{\\epsilon \\downarrow 0} \\frac{G(x + \\epsilon) - G(x + \\epsilon/2)}{\\epsilon} = \\frac{1}{2} G'_d(x) \\leq 0, \\quad x \\in \\mathbb{R}.\n$$\n\nLet us show that $G$ is constant. Consider $a < b$, such that $G(a) \\ne G(b)$.\n\nIf $G(a) < G(b)$, consider $\\lambda > (G(a) - G(b))/(b - a)$. The function $H : [a, b] \\to \\mathbb{R}$, $H(x) = G(x) + \\lambda x$, is continuous, $H(a) < H(b)$ and\n\n$$\n\\lim_{\\epsilon \\downarrow 0} \\frac{H(x + \\epsilon) - H(x + \\epsilon/2)}{\\epsilon} = \\frac{1}{2} G'_{d}(x) + \\frac{\\lambda}{2} \\leq \\frac{\\lambda}{2} < 0, \\quad x \\in \\mathbb{R}.\n$$\n\nSo, for any real $x$, there is $\\delta(x) > 0$, such that $H(x + \\epsilon) < H(x + \\epsilon/2)$, $0 < \\epsilon < \\delta(x)$. Let $c \\in [a, b]$, such that $H(c) = \\min\\{H(x) : a \\leq x \\leq b\\}$. Because $H(a) < H(b)$, we have $a \\leq c < b$. Consider a number $\\epsilon < \\min(b - c, \\delta(c))$. Then\n\n$$\nH(c) \\leq H(c + \\epsilon) < H(c + \\epsilon/2) < \\dots < H(c + \\epsilon/2^n) < \\dots \\leq H(c),\n$$\n\n(the last inequality is a consequence of the continuity of $H$), which is a contradiction.\n\nIf $G(a) > G(b)$, proceed analogously: consider $\\lambda < (G(a) - G(b))/(b - a)$. In this case the corresponding continuous function $H(x) = G(x) + \\lambda x$ satisfies $H(a) > H(b)$ and\n\n$$\n\\lim_{\\epsilon \\downarrow 0} \\frac{H(x - \\epsilon) - H(x - \\epsilon/2)}{\\epsilon} = -\\frac{1}{2} G'_{s}(x) - \\frac{\\lambda}{2} \\leq -\\frac{\\lambda}{2} < 0, \\quad x \\in \\mathbb{R}.\n$$\n\nThat is, for any real $x$, there exists $\\delta(x) > 0$, such that $H(x - \\epsilon) < H(x - \\epsilon/2)$, $0 < \\epsilon < \\delta(x)$. Let $c \\in [a, b]$, such that $H(c) = \\min\\{H(x) : a \\leq x \\leq b\\}$. Because $H(a) > H(b)$, we get $a < c \\leq b$. Fix a real positive number $\\epsilon < \\min(c - a, \\delta(c))$. Then\n\n$$\nH(c) \\leq H(c - \\epsilon) < H(c - \\epsilon/2) < \\dots < H(c - \\epsilon/2^n) < \\dots \\leq H(c),\n$$\n\nwhich is a contradiction.\n\nWe thus proved that $G$ is constant. Taking into account $G(0) = 0$, we get $G(x) = 0$, for any real $x$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 14375,
    "subject": "Mathematics (Olympiad)",
    "question": "Show that the number\n\n$$\n\\overbrace{aaa\\dots a}^{2011 \\text{ times}} 0a\n$$\n\nis not a perfect square for any nonzero digit $a$.",
    "options": [],
    "answer": "See solution",
    "solution": "The last digit of a perfect square cannot be $2$, $3$, $7$, or $8$.\n\nIf $a = 5$, the next-to-last digit must be $2$, which is impossible for a perfect square.\n\nIf $a = 6$, the number is of the form $4k + 2$, $k \\in \\mathbb{N}$, which cannot be a perfect square.\n\nThe cases $a = 4$ and $a = 9$ reduce to a number of the form $111\\dots01$ being a perfect square, which is impossible, because such a number is of the form $8k + 5$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 14376,
    "subject": "Mathematics (Olympiad)",
    "question": "Is it possible to cut a regular triangle into:\n\na) three equal quadrilaterals;\n\nb) three equal pentagons?\n\nConvexity of the quadrilaterals and pentagons is not required.",
    "options": [],
    "answer": "See solution",
    "solution": "*Answer:* a), b) Yes, it is possible.\n\n*Solution.*\n\na) Consider a regular triangle $\\triangle ABC$, and denote its midpoints by $M$, $N$, and $K$ respectively. The segments $AN$, $BK$, and $CM$ intersect at $O$. It is easy to see that the quadrilaterals $BMON$, $CKON$, and $AMOK$ satisfy the condition.\n\nb) Now, let us denote the midpoints of $OM$, $ON$, and $OK$ by $P$, $R$, and $Q$ respectively. The pentagons we are looking for are $ABROP$, $ROQCB$, and $APOQC$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 14377,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $X$, $Y$, $Z$ be distinct points on a circle $\\omega$ in that order. At points $A$ and $B$, we construct tangent rays $AC$ and $BD$ to $\\omega$, respectively, such that the rays are both in the clockwise direction. Let $P$ be the intersection of $AB$ and $CD$. We construct tangents $XA$, $YB$, $ZC$ to $\\omega$ such that all have the same length and all are in the clockwise direction. Assume that the line $XY$ meets the segment $AB$ at $M$, and the line $YZ$ meets the segment $BC$ at $N$. Prove that $MN$ is a mid-segment of $ABC$.",
    "options": [],
    "answer": "See solution",
    "solution": "First, we show that $AM = MB$. To prove this, choose the point $T$ on $BY$ beyond $Y$ such that $BY = YT$. Since $XA = YB$ and $TA$, $TY$ are tangents, we see that $TA \\parallel XY$. Since $Y$ is the midpoint of $TB$, the segment $YM$ is the mid-segment of $TAB$. Hence, $AM = MB$. Similarly, $BN = NC$. Therefore, $MN$ is a mid-segment of triangle $ABC$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 14378,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $p$ be a prime number. Find all triples $(a, b, c)$ of integers (not necessarily positive) such that\n$$\na^b b^c c^a = p.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "*Answer:* If $p > 2$ then $(p, 1, 1)$ and $(-p, 1, -1)$ together with cyclic permutations; if $p = 2$ then $(2, 1, 1)$, $(2, 1, -1)$, $(2, 2, -1)$ and $(-2, 2, -1)$ together with cyclic permutations.\n\n*Solution.* Suppose $a, b, c$ satisfy the equation. As $p$ is positive, this implies that $|a|^b |b|^c |c|^a = p$. Clearly none of $a, b, c$ can be zero.\n\nObserve that $\\gcd(a,b,c) = 1$. Indeed, if $d \\mid a, d \\mid b, d \\mid c$, then the exponent of $p$ in the canonical representation of each of the positive rational numbers $|a|^b, |b|^c, |c|^a$ is divisible by $d$. Hence the exponent of $p$ in the canonical representation of the product $|a|^b |b|^c |c|^a$ is divisible by $d$. As this product equals $p$, we get $|d| = 1$.\n\nConsider now an arbitrary prime number $q$ different from $p$. Let $\\alpha, \\beta, \\gamma$ be the exponents of $q$ in the canonical representation of the positive integers $|a|, |b|, |c|$, respectively. Then $\\alpha b + \\beta c + \\gamma a = 0$ whereby not all exponents $\\alpha, \\beta, \\gamma$ are positive because $\\gcd(a,b,c) = 1$. Consequently, if some of $\\alpha, \\beta, \\gamma$ is positive, then there must be exactly two positive exponents among $\\alpha, \\beta, \\gamma$. W.l.o.g., assume $\\alpha > 0, \\beta > 0, \\gamma = 0$. Then $\\alpha b + \\beta c = 0$, implying $\\alpha|b| = \\beta|c|$. Hence $|b|$ divides $\\beta|c|$. As $q^\\beta$ divides $|b|$ while $q^\\beta$ is relatively prime to $|c|$, this implies $q^\\beta \\mid \\beta$ and $q^\\beta \\le \\beta$ which is impossible. This means that actually $\\alpha = \\beta = \\gamma = 0$ and $|a|, |b|, |c|$ are all powers of $p$.\n\nHence the equation rewrites to $p^{\\alpha b} p^{\\beta c} p^{\\gamma a} = p$ where $\\alpha, \\beta, \\gamma$ are now the exponents of $p$ in the canonical representation of $|a|, |b|, |c|$, respectively. This is equivalent to $\\alpha b + \\beta c + \\gamma a = 1$. By $\\gcd(a,b,c) = 1$, one of $\\alpha, \\beta, \\gamma$ must be zero, and clearly, one of the summands $\\alpha b, \\beta c, \\gamma a$ must be positive. W.l.o.g., let $\\alpha b > 0$, i.e., $\\alpha > 0$ and $b > 0$. Now there are three cases.\n\nIf $\\beta = 0$ and $\\gamma = 0$ then $b = 1$ and $|c| = 1$. Furthermore, $\\alpha b + \\beta c + \\gamma a = 1$ reduces to $\\alpha = 1$, whence $|a| = p$. If $p > 2$ then the exponents of $a$ and $c$ in the original equation, $b$ and $a$, are both odd, whence $a$ and $c$ must have the same sign to make the product $a^b b^c c^a$ positive. Both triples $(p, 1, 1)$ and $(-p, 1, -1)$ satisfy the original equation. If $p = 2$ then $c^a$ is positive anyway, hence $a$ must be positive. Both triples $(2, 1, 1)$ and $(2, 1, -1)$ satisfy the original equation.\n\nIf $\\beta = 0$ and $\\gamma > 0$ then $b = 1$. Furthermore, $\\alpha b + \\beta c + \\gamma a = 1$ reduces to $\\alpha + \\gamma a = 1$, whence $a < 0$. We obtain $p^\\alpha \\le \\gamma p^\\alpha = \\gamma|a| = \\alpha - 1 < \\alpha$ which is impossible.\n\nIf $\\beta > 0$ and $\\gamma = 0$ then $|c| = 1$. Furthermore, $\\alpha b + \\beta c + \\gamma a = 1$ reduces to $\\alpha b + \\beta c = 1$, which gives $c = -1$ and $\\alpha p^\\beta = 1 + \\beta$ as the only possibility. If $p > 2$ then this leads to a contradiction similar to the previous case. If $p = 2$ then $\\alpha = \\beta = 1$ is the only solution. This leads to triples $(2, 2, -1)$ and $(-2, 2, -1)$ which both satisfy the original equation.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 14379,
    "subject": "Mathematics (Olympiad)",
    "question": "Пусть $a = x - \\sqrt{2}$, $b = x - 1/x$, $c = x + 1/x$, $d = x^2 + 2\\sqrt{2}$. Найдите все значения $x$, при которых хотя бы два из чисел $a$, $b$, $c$, $d$ являются целыми.",
    "options": [],
    "answer": "See solution",
    "solution": "Заметим, что $b$ и $c$ не могут одновременно быть целыми. Действительно, тогда $b + c = 2x$ также целое, значит, $x$ рационально, поэтому $a$ и $d$ не будут целыми, так как это суммы рационального и иррационального чисел. Итак, одно из чисел $b$ и $c$ нецелое, а тогда $a$ и $d$ должны оба быть целыми.\n\nЗначит, $x = a + \\sqrt{2}$ при целом $a$. Тогда\n$$\nd = x^2 + 2\\sqrt{2} = (a + \\sqrt{2})^2 + 2\\sqrt{2} = a^2 + 2a\\sqrt{2} + 2 + 2\\sqrt{2} = (a^2 + 2) + (2a + 2)\\sqrt{2}\n$$\nЧтобы $d$ было целым, необходимо $2a + 2 = 0$, то есть $a = -1$. Тогда $x = \\sqrt{2} - 1$. Проверим:\n- $a = x - \\sqrt{2} = -1$ (целое)\n- $b = x - 1/x = (\\sqrt{2} - 1) - 1/(\\sqrt{2} - 1) = (\\sqrt{2} - 1) - (\\sqrt{2} + 1) = -2$ (целое)\n- $d = x^2 + 2\\sqrt{2} = (\\sqrt{2} - 1)^2 + 2\\sqrt{2} = (2 - 2\\sqrt{2} + 1) + 2\\sqrt{2} = 3$ (целое)\n\nОтвет: $x = \\sqrt{2} - 1$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 14380,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $p$ be a prime number. Find\n$$\n1! \\cdot 2^2 + 2! \\cdot 3^2 + 3! \\cdot 4^2 + \\dots + (p-3)! \\cdot (p-2)^2 \\pmod{p}.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "It can be shown by induction that\n$$\n1! \\cdot 2^2 + 2! \\cdot 3^2 + 3! \\cdot 4^2 + \\dots + (p-3)! \\cdot (p-2)^2 = (p-1)! - 2\n$$\nwhich is equal to $-3 \\pmod{p}$ by Wilson's theorem.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 14381,
    "subject": "Mathematics (Olympiad)",
    "question": "Prove that for any $p, q \\in \\mathbb{N}$ such that $\\sqrt{11} > \\frac{p}{q}$, the following inequality holds:\n\n$$\n\\sqrt{11} - \\frac{p}{q} > \\frac{1}{2pq}.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "Assume $p$ and $q$ are coprime. Since both sides of the first inequality are positive, we can rewrite it as $11q^2 > p^2$. Similarly, the second inequality becomes:\n\n$$\n11p^2q^2 > p^4 + p^2 + \\frac{1}{4}.\n$$\n\nTo show this holds, we prove a stronger statement:\n\n$$\n11p^2q^2 \\geq p^4 + 2p^2.\n$$\n\nDividing both sides by $p^2$ gives $11q^2 \\geq p^2 + 2$. Since $11q^2 > p^2$, we only need to check that $11q^2 \\neq p^2 + 1$. The possible remainders of squares modulo $11$ are $0, 1, 3, 4, 5, 9$, so $p^2 + 1$ cannot be divisible by $11$. Therefore, $11q^2 \\neq p^2 + 1$.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 14382,
    "subject": "Mathematics (Olympiad)",
    "question": "We are given a balance and 100 weights of weight $1, 2, \\ldots, 100$, respectively. The weights are placed on the two pans of the balance in equilibrium. Prove that two weights can be removed from each pan such that the equilibrium is not broken.",
    "options": [],
    "answer": "See solution",
    "solution": "Denote by $A$ and $B$ the sets of positive integers representing the weights placed on the two pans of the balance. We may assume that $1 \\in A$. Suppose there exist $a, b \\in \\{1, 2, \\ldots, 99\\}$ with $b - a \\ge 2$ as below:\n\n$$\n\\begin{array}{c|cccccc}\nA & 1 & \\ldots & a & ? & b+1 & ? \\\\\n\\hline\nB &     &      & a+1 & ? & b   & ?\n\\end{array}\n$$\n\nSince $a + (b+1) = (a+1) + b$, remove the weights $a$ and $b+1$ from one pan and weights $a+1$ and $b$ from the other to maintain equilibrium.\n\nIf this situation does not occur, the distribution of the weights in $A$ and $B$ must respect one of the following configurations:\n\n![](table.png)\n\nThe first configuration requires $a(a+1) = (1+2+\\cdots+100)/2 = 5050$, but this equation has no integer solutions. The second configuration is impossible as well.\n\nFor the third configuration, we would need $(a+1)(a+2) = (1+2+\\cdots+100)/2 - 2 = 5048$, but this equation has no integer solutions.\n\nFinally, in the fourth case, we have $(a+2)+b = (a+1)+(b+1)$, so we can remove from the two pans the weights $a+2$, $b$ and $a+1$, $b+1$ respectively.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 14383,
    "subject": "Mathematics (Olympiad)",
    "question": "Six points are chosen on the sides of an equilateral triangle $ABC$: $A_1$ and $A_2$ on $BC$, $B_1$ and $B_2$ on $CA$, and $C_1$ and $C_2$ on $AB$. These points are vertices of a convex equilateral hexagon $A_1A_2B_1B_2C_1C_2$. Prove that lines $A_1B_2$, $B_1C_2$, and $C_1A_2$ are concurrent.",
    "options": [],
    "answer": "See solution",
    "solution": "**First Solution:** (Based on work by Hansheng Diao from China)\n\nSet $x = AB$ and $s = A_1A_2$. We construct an equilateral triangle $A_0B_0C_0$ with $A_0B_0 = x - s$. Points $C_4$, $A_4$, and $B_4$ lie on sides $A_0B_0$, $B_0C_0$, and $C_0A_0$, respectively, satisfying $C_4B_0 = C_2B$, $A_4C_0 = A_2C$, and $B_4A_0 = B_2A$. Then it is easy to obtain that $B_0A_4 = BA_1$, $C_0B_4 = CB_1$, and $A_0C_4 = AC_1$. We obtain three pairs of congruent triangles, namely, $AB_2C_1$ and $A_0B_4C_4$, $BC_2A_1$ and $B_0C_4A_4$, and $CA_2B_1$ and $C_0A_4B_4$. (Indeed, we are sliding the three corner triangles together.)\n\n![](images/USA_IMO_2006-2007_p45_data_9d8c9bcb78.png)\n\nIt follows that $A_4B_4 = B_4C_4 = C_4A_4 = s$; that is, triangle $A_4B_4C_4$ is equilateral, implying that $\\angle B_4C_4A_4 = \\angle C_4A_4B_4 = \\angle A_4B_4C_4 = 60^\\circ$. Hence $\\angle A_0B_4C_4 + \\angle A_0C_4B_4 = \\angle B_0C_4A_4 + \\angle A_0C_4B_4 = 120^\\circ$, and so $\\angle A_0B_4C_4 = \\angle B_0C_4A_4$. Hence $\\angle AB_2C_1 = \\angle BC_2A_1$, or $\\angle B_1B_2C_1 = \\angle C_1C_2A_1$. Since the vertex angles of the isosceles triangles $B_1B_2C_1$ and $C_1C_2A_1$ are equal, then two triangles are similar and hence congruent to each other, implying that $C_1B_1 = C_1A_1$. Since $C_1B_1 = C_1A_1$ and $A_2B_1 = A_2A_1$, line $C_1A_2$ is a perpendicular bisector of triangle $A_1B_1C_1$. Likewise, so are lines $A_1B_2$ and $B_1C_2$. Therefore, lines $C_1A_2$, $A_1B_2$, and $B_1C_2$ concur at the circumcenter of triangle $A_1B_1C_1$.\n\nIt is not difficult to see that triangle $A_0B_4C_4$ is congruent to triangle $B_0C_4A_4$ (and to triangle $C_0A_4B_4$).",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 14384,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $a$, $b$, and $c$ be real numbers. Prove that\n\n$$\n\\frac{a^2}{a+b^2} + \\frac{b^2}{b+c^2} + \\frac{c^2}{c+a^2} \\ge \\frac{3}{2}\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "Applying the Cauchy-Schwarz inequality to\n\n$$\n\\frac{a^2}{a+b^2}, \\frac{b^2}{b+c^2}, \\frac{c^2}{c+a^2}, \\quad \\text{and} \\quad a^2(a+b^2), b^2(b+c^2), c^2(c+a^2)\n$$\n\nand dividing by\n\n$$\n a^2 (a+b^2) + b^2 (b+c^2) + c^2 (c+a^2) = a^3 + b^3 + c^3 + a^2 b^2 + b^2 c^2 + c^2 a^2\n$$\n\nwe get\n\n$$\n\\left( \\frac{a^2}{a+b^2} + \\frac{b^2}{b+c^2} + \\frac{c^2}{c+a^2} \\right) \\ge \\frac{(a^2+b^2+c^2)^2}{a^3+b^3+c^3+a^2b^2+b^2c^2+c^2a^2}\n$$\n\nHence it is sufficient to prove:\n\n$$\n\\frac{(a^2 + b^2 + c^2)^2}{a^3 + b^3 + c^3 + a^2 b^2 + b^2 c^2 + c^2 a^2} \\ge \\frac{3}{2}\n$$\n\nThe last inequality is equivalent to\n\n$$\n\\begin{align*}\n& 2 (a^2 + b^2 + c^2)^2 \\ge 3 (a^3 + b^3 + c^3 + a^2 b^2 + b^2 c^2 + c^2 a^2) \\\\\n\\Leftrightarrow & 2 (a^4 + b^4 + c^4) + a^2 b^2 + b^2 c^2 + c^2 a^2 \\ge 3 (a^3 + b^3 + c^3) \\\\\n\\Leftrightarrow & 2 (a^4 + b^4 + c^4) + a^2 b^2 + b^2 c^2 + c^2 a^2 \\ge (a+b+c) (a^3 + b^3 + c^3) \\\\\n\\Leftrightarrow & a^4 + b^4 + c^4 + a^2 b^2 + b^2 c^2 + c^2 a^2 \\ge ab^3 + ac^3 + ba^3 + bc^3 + ca^3 + cb^3. \\tag{4.1}\n\\end{align*}\n$$\n\nUsing the A-G inequality we get:\n\n$$\n\\begin{align*}\n a^4 + a^2 b^2 \\ge 2a^3b, \\quad b^4 + b^2 c^2 \\ge 2b^3c, \\quad c^4 + c^2 a^2 \\ge 2c^3a \\\\\n a^4 + c^2 a^2 \\ge 2a^3c, \\quad b^4 + a^2 b^2 \\ge 2b^3a, \\quad c^4 + b^2 c^2 \\ge 2c^3b.\n\\end{align*}\n$$\n\nSumming up the above inequalities we get\n\n$$\n2 (a^4 + b^4 + c^4 + a^2 b^2 + b^2 c^2 + c^2 a^2) \\ge 2 (ab^3 + ac^3 + ba^3 + bc^3 + ca^3 + cb^3),\n$$\n\nthus we have proved (4.1), and therefore the initial inequality.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 14385,
    "subject": "Mathematics (Olympiad)",
    "question": "Suppose that $p$ is an odd prime number and $M$ is a set of $\\frac{p^2+1}{2}$ integer squares.\n\nInvestigate if one can choose $p$ elements of this set so that the arithmetic mean of these $p$ elements is an integer.",
    "options": [],
    "answer": "See solution",
    "solution": "Yes.\n\nThe idea is to choose from the $\\frac{p^2+1}{2}$ square numbers $p$ numbers that are in the same residue class modulo $p$. Obviously, the sum of these $p$ numbers is then divisible by $p$ and thus the arithmetic mean is an integer.\n\nIt is known that the square numbers do not run through all residue classes modulo $p$, but only through $1 + \\frac{p-1}{2} = \\frac{p+1}{2}$ ones. (On the one hand, this is the residue class $0$ if one squares a number divisible by $p$. Because $a^2 \\equiv (p-a)^2 \\pmod p$, the squares of numbers $a$ that are not divisible by $p$ run through a maximum of half of the $p-1$ nonzero residue classes. On the other hand, $x^2 \\equiv y^2 \\pmod p$ gives the relation $p \\mid (x-y)(x+y)$ and so $x \\equiv y \\pmod p$ or $x \\equiv -y \\pmod p$. Therefore, the squares of numbers $a$, which are not divisible by $p$, run through exactly half of the $p-1$ residue classes different from zero.)\n\nWe now divide the $\\frac{p^2+1}{2}$ square numbers into the $\\frac{p+1}{2}$ residue classes that correspond to square numbers. By the pigeonhole principle, there is therefore a residue class that contains at least\n\n$$\n\\left\\lfloor \\frac{(p^2 + 1)/2}{(p + 1)/2} \\right\\rfloor\n$$\n\nnumbers.\n\nBecause\n\n$$\n\\frac{(p^2 + 1)/2}{(p + 1)/2} = \\frac{p^2 + 1}{p + 1} = \\frac{p^2 + p}{p + 1} - \\frac{p - 1}{p + 1} = p - \\frac{p - 1}{p + 1}\n$$\n\nand $0 < \\frac{p-1}{p+1} < 1$, it follows that\n\n$$\n\\left\\lfloor \\frac{(p^2 + 1)/2}{(p + 1)/2} \\right\\rfloor = p,\n$$\n\nwhich was to be shown.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 14386,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $k$ be a positive integer. Determine the least integer $n \\geq k + 1$ for which the following game can be played indefinitely:\n\nConsider $n$ boxes, labelled $b_1, b_2, \\dots, b_n$. For each index $i$, box $b_i$ initially contains exactly $i$ coins. At each step, the following three substeps are performed in order:\n\n1. Choose $k+1$ boxes.\n2. Of these $k+1$ boxes, choose $k$ and remove at least half of the coins from each, and add to the remaining box, if labelled $b_i$, a number of $i$ coins.\n3. If one of the boxes is left empty, the game ends; otherwise, go to the next step.",
    "options": [],
    "answer": "See solution",
    "solution": "The required minimum is $n = 2^k + k - 1$.\n\nIn this case, the game can be played indefinitely by choosing the last $k+1$ boxes, $b_{2^k-1}, b_{2^k}, \\dots, b_{2^k+k-1}$, at each step. At step $r$, if box $b_{2^k+i-1}$ has exactly $m_i$ coins, then $\\lceil m_i/2 \\rceil$ coins are removed from that box, unless $i \\equiv r-1 \\pmod{k+1}$, in which case $2^k + i - 1$ coins are added. Thus, after step $r$ has been performed, box $b_{2^k+i-1}$ contains exactly $\\lceil m_i/2 \\rceil$ coins, unless $i \\equiv r-1 \\pmod{k+1}$, in which case it contains exactly $m_i + 2^k + i - 1$ coins. This game goes on indefinitely, since each time a box is supplied, at least $2^k - 1$ coins are added, so it will then contain at least $2^k$ coins, good enough to survive the $k$ steps to its next supply.\n\nWe now show that no smaller value of $n$ works. So, let $n \\leq 2^k + k - 2$ and suppose, if possible, that a game can be played indefinitely. Notice that a box currently containing exactly $m$ coins survives at most $w = \\lceil \\log_2 m \\rceil$ withdrawals; this $w$ will be referred to as the *weight* of that box. The sum of the weights of all boxes will be referred to as the *total weight*. The argument hinges on the lemma below, proved at the end of the solution.\n\n**Lemma.** *Performing a step does not increase the total weight. Moreover, supplying one of the first $2^k - 2$ boxes strictly decreases the total weight.*\n\nSince the total weight cannot strictly decrease indefinitely, $n > 2^k - 2$, and from some stage on none of the first $2^k - 2$ boxes is ever supplied. Recall that each step involves a $(k+1)$-box choice. Since $n \\leq 2^k + k - 2$, from that stage on, each step involves a withdrawal from at least one of the first $2^k - 2$ boxes. This cannot go on indefinitely, so the game must eventually come to an end, contradicting the assumption.\n\nConsequently, a game that can be played indefinitely requires $n \\geq 2^k + k - 1$.\n\n**Proof of the Lemma.** Since a withdrawal from a box decreases its weight by at least 1, it is sufficient to show that supplying a box increases its weight by at most $k$; and if the latter is amongst the first $2^k - 2$ boxes, then its weight increases by at most $k-1$. Let the box to be supplied be $b_i$ and let it currently contain exactly $m_i$ coins. Proceed by case analysis:\n\n- If $m_i = 1$, the weight increases by $\\lfloor \\log_2(i+1) \\rfloor \\leq \\lfloor \\log_2(2^k + k - 1) \\rfloor \\leq \\lfloor \\log_2(2^{k+1} - 2) \\rfloor \\leq k$; and if, in addition, $i \\leq 2^k - 2$, then the weight increases by $\\lfloor \\log_2(i+1) \\rfloor \\leq \\lfloor \\log_2(2^k - 1) \\rfloor = k - 1$.\n- If $m_i = 2$, then the weight increases by $\\lfloor \\log_2(i+2) \\rfloor - \\lfloor \\log_2 2 \\rfloor \\leq \\lfloor \\log_2(2^k + k) \\rfloor - 1 \\leq k - 1$.\n- If $m_i \\geq 3$, then the weight increases by\n\n$$\n\\begin{aligned}\n\\lfloor \\log_2(i + m_i) \\rfloor - \\lfloor \\log_2 m_i \\rfloor &\\leq \\lfloor \\log_2(i + m_i) - \\log_2 m_i \\rfloor + 1 \\\\\n&\\leq \\lfloor \\log_2 \\left( 1 + \\frac{2^k + k - 2}{3} \\right) \\rfloor + 1 \\leq k,\n\\end{aligned}\n$$\n\nsince $1 + \\frac{1}{3}(2^k + k - 2) = \\frac{1}{3}(2^k + k + 1) < \\frac{1}{3}(2^k + 2^{k+1}) = 2^k$.\n\nFinally, let $i \\leq 2^k - 2$ to consider the subcases $m_i = 3$ and $m_i \\geq 4$. In the former subcase, the weight increases by\n\n$$\n\\lfloor \\log_2(i + 3) \\rfloor - \\lfloor \\log_2 3 \\rfloor \\leq \\lfloor \\log_2(2^k + 1) \\rfloor - 1 = k - 1,\n$$\n\nand in the latter by\n\n$$\n\\begin{aligned}\n\\lfloor \\log_2(i + m_i) \\rfloor - \\lfloor \\log_2 m_i \\rfloor &\\leq \\lfloor \\log_2(i + m_i) - \\log_2 m_i \\rfloor + 1 \\\\\n&\\leq \\lfloor \\log_2 \\left( 1 + \\frac{2^k - 2}{4} \\right) \\rfloor + 1 \\leq k - 1,\n\\end{aligned}\n$$\n\nsince $1 + \\frac{1}{4}(2^k - 2) = \\frac{1}{4}(2^k + 2) < 2^{k-2} + 1$. This ends the proof and completes the solution.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 14387,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $T$ denote the 15-element set $T = \\{10a + b : 1 \\leq a < b \\leq 6\\}$. Let $S \\subseteq T$ be a subset of $T$ in which all 6 digits appear and in which no 3 members exist which contain together all 6 digits $1, 2, \\ldots, 6$. Determine the largest possible size $n$ of $S$.",
    "options": [],
    "answer": "See solution",
    "solution": "Consider the numbers of $T$ which contain $1$ or $2$. Certainly, no 3 of them can contain all 6 digits and all 6 digits appear. Hence $n \\geq 9$.\n\nConsider the partitions:\n\n$$\n\\begin{aligned}\n&12,\\ 36,\\ 45,\\\\\n&13,\\ 24,\\ 56,\\\\\n&14,\\ 26,\\ 35,\\\\\n&15,\\ 23,\\ 46,\\\\\n&16,\\ 25,\\ 34.\n\\end{aligned}\n$$\n\nSince every row is a partition of $\\{1, 2, \\ldots, 6\\}$, it contains all 6 digits. $S$ can contain at most two numbers of each of the 5 rows, i.e., $n \\leq 10$.\n\nNow we will prove that $n = 9$ is the correct number. Therefore, we assume that $n = 10$ and will exclude this case by contradiction. Certainly, there is a digit, say $1$, which does not appear at least twice (otherwise at most 3 numbers are missing in $S$) and at most 4 times (otherwise this digit does not appear in the members of $S$ at all). Obviously, every row of the above set of partitions contains exactly 2 members of $S$. W.l.o.g., assume that $12, 13 \\notin S$ and $16 \\in S$. Then consider the following partitions, where bold-faced numbers are members of $S$ and numbers in italics are not:\n\n12, 36, 45,\n\n13, 24, 56,\n\n14, 26, 35,\n\n15, 23, 46,\n\n16, 25, 34.\n\nBy $16, 45 \\in S$ it follows $23 \\notin S$ and by $24, 36 \\in S$ it follows $15 \\notin S$. Now $S$ is missing at least 2 members ($15, 23$) of the partition $15, 23, 46$, which is a contradiction.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 14388,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $\\triangle ABC$ be an acute triangle, and let $K$ be a point on the extension of $BC$ beyond $C$ (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$ intersect at another point $T$.\n\n1. Prove that $\\angle APB + \\angle BTC = \\angle CQA$.\n\n2. Prove that $AP \\cdot BT \\cdot CQ = AQ \\cdot BP \\cdot CT$.",
    "options": [],
    "answer": "See solution",
    "solution": "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$.\n\nSimilarly, let $A_2$ be the reflection of $A$ across $C$. We have $\\angle AQC = \\angle A_2KC = \\angle A_2KB$.\n\nHence, the desired conclusion is equivalent to $\\angle BTC = \\angle A_1KA_2$.\n\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.\n\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.\n\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![](images/2023_CMO_p11_data_a9c7fdb101.png)\n\n2. We have\n\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$$\n\nProof completed.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 14389,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $ABC$ be a triangle with $AB < AC$ inscribed in a circle $c$. The tangent to $c$ at $C$ meets the parallel from $B$ to $AC$ at $D$. The tangent to $c$ at $B$ meets the parallel from $C$ to $AB$ at $E$, and the tangent to $c$ at $C$ at $L$. Suppose that the circumcircle $c_1$ of triangle $BDC$ meets $AC$ at $T$, and the circumcircle $c_2$ of triangle $BEC$ meets $AB$ at $S$. Prove that the lines $ST$, $BC$, and $AL$ are concurrent.",
    "options": [],
    "answer": "See solution",
    "solution": "We will first prove that the circle $c_1$ is tangent to $AB$ at $B$. To do this, we need to show that $\\angle BDC = \\angle ABC$. Since $BD \\parallel AC$, $\\angle DBC = \\angle ACB$. Also, $\\angle BCD = \\angle BAC$ (by chord and tangent), so triangles $ABC$ and $BDC$ have two equal angles, and thus the third angles are also equal. Therefore, $\\angle BDC = \\angle ABC$, so $c_1$ is tangent to $AB$ at $B$.\n\nSimilarly, the circle $c_2$ is tangent to $AC$ at $C$.\n\nAs a consequence, $\\angle ABT = \\angle ACB$ (by chord and tangent), and also $\\angle BSC = \\angle ACB$. Thus, $\\angle ABT = \\angle BSC$, so the lines $BT$ and $SC$ are parallel.\n\nLet $ST$ intersect $BC$ at $K$. It suffices to prove that $K$ lies on $AL$. From the trapezoid $BTCS$, we get:\n\n$$\n\\frac{BK}{KC} = \\frac{BT}{SC} \\qquad (1)\n$$\n\nFrom the similar triangles $ABT$ and $ASC$, we have:\n\n$$\n\\frac{BT}{SC} = \\frac{AB}{AS} \\qquad (2)\n$$\n\nBy (1) and (2):\n\n$$\n\\frac{BK}{KC} = \\frac{AB}{AS} \\qquad (3)\n$$\n\nFrom the power of a point theorem:\n\n$$\nAC^2 = AB \\cdot AS \\implies AS = \\frac{AC^2}{AB}\n$$\n\nSubstituting into (3):\n\n$$\n\\frac{BK}{KC} = \\frac{AB^2}{AC^2}\n$$\n\nThus, $K$ lies on the symmedian of triangle $ABC$.\n\nRecall that since $LB$ and $LC$ are tangents, $AL$ is the symmedian of triangle $ABC$, so $K$ lies on $AL$, as required.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 14390,
    "subject": "Mathematics (Olympiad)",
    "question": "Find the average of all 5-digit numbers that satisfy the following:\n\n* The number is of the form $\\overline{ab0cd}$, that is, its third digit is zero.\n* Digits are pairwise distinct.\n* Both numbers $\\overline{ab0cd}$ and $\\overline{dc0ba}$ are divisible by $7$.\n\n![](images/UkraineMO2019_booklet_p6_data_ee872e7cae.png)",
    "options": [],
    "answer": "See solution",
    "solution": "Since $1001$ is divisible by $7$, we can write:\n\n$$\n\\overline{ab0cd} = 1000 \\cdot \\overline{ab} + \\overline{cd} = 1001 \\cdot \\overline{ab} + (\\overline{cd} - \\overline{ab})\n$$\n\nThus, $\\overline{ab0cd}$ is divisible by $7$ if $\\overline{cd} - \\overline{ab}$ is divisible by $7$.\n\nLet $\\overline{ab} = 10a + b$ and $\\overline{cd} = 10c + d$, so:\n\n$$\n\\overline{cd} - \\overline{ab} = 10(c - a) + (d - b)\n$$\n\nSimilarly, for $\\overline{dc0ba}$:\n\n$$\n\\overline{dc0ba} = 1000 \\cdot \\overline{dc} + \\overline{ba} = 1001 \\cdot \\overline{dc} + (\\overline{ba} - \\overline{dc})\n$$\n\nSo $\\overline{ba} - \\overline{dc} = 10(b - d) + (a - c)$ must also be divisible by $7$.\n\nLet $x = c - a$ and $y = d - b$. The conditions become:\n\n$$\n3x + y = 7k \\\\\n3y + x = 7l\n$$\n\nSolving these, we find that $x$ and $y$ must be such that $8x$ and $8y$ are divisible by $7$, which only happens for certain values. The only possible pairs (excluding zero) are $(8, 1)$ and $(9, 2)$. Since zero is already used as the third digit, we exclude pairs containing zero.\n\nThus, the valid numbers are: $12089$, $19082$, $21098$, $28091$, $89012$, $82019$, $98021$, and $91028$.\n\nTheir average is:\n\n$$\n\\frac{12089 + 19082 + 21098 + 28091 + 89012 + 82019 + 98021 + 91028}{8} = 55055\n$$",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 14391,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $a, b, c, d$ be real numbers with $0 < a, b, c, d < 1$ and $a + b + c + d = 2$. Show that\n\n$$\n\\sqrt{(1-a)(1-b)(1-c)(1-d)} \\le \\frac{ac+bd}{2}.\n$$\n\nAre there infinitely many cases of equality?",
    "options": [],
    "answer": "See solution",
    "solution": "Squaring the given inequality and multiplying by $16$, we get\n\n$$\n(2 - 2a)(2 - 2b)(2 - 2c)(2 - 2d) \\le 4(ac + bd)^2.\n$$\n\nWe homogenize by replacing the first $2$ in each parenthesis on the left side by $a + b + c + d$ and get the homogeneous inequality\n\n$$\n(b+d-(a-c))(a+c-(b-d))(b+d+a-c)(a+c+b-d) \\le 4(ac+bd)^2.\n$$\n\nWe evaluate the left-hand side by repeatedly combining two factors and get\n\n$$\n\\begin{align*}\n& (b+d-(a-c))(a+c-(b-d))(b+d+a-c)(a+c+b-d) \\\\\n&= ((a+c)^2 - (b-d)^2)((b+d)^2 - (a-c)^2) \\\\\n&= (2ac + 2bd + a^2 + c^2 - b^2 - d^2)(2ac + 2bd - a^2 - c^2 + b^2 + d^2) \\\\\n&= 4(ac + bd)^2 - (a^2 + c^2 - b^2 - d^2)^2 \\le 4(ac + bd)^2,\n\\end{align*}\n$$\n\nwhich proves the inequality.\n\nEquality holds for $a^2 + c^2 = b^2 + d^2$, in particular for $a = b$ and $c = d = 1-a$ with $0 < a < 1$. Therefore, there are infinitely many equality cases.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 14392,
    "subject": "Mathematics (Olympiad)",
    "question": "In a long line of people arranged left to right, the 1013th person from the left is also the 1010th person from the right. How many people are in the line?\n\n(A) 2021  \n(B) 2022  \n(C) 2023  \n(D) 2024  \n(E) 2025",
    "options": [],
    "answer": "See solution",
    "solution": "There are $1012$ people to the left of the specified person and $1009$ people to the right of that person. There are therefore $$1012 + 1 + 1009 = 2022$$ people in the line.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 14393,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $ABCD$ be a parallelogram, and let $M$ be the midpoint of $AB$. Line $CM$ intersects the circumcircle of triangle $ABC$ at $C$ and $E$. Let $F$ be the point on $BC$ such that $AF \\perp BC$. Prove that $C$, $D$, $E$, and $F$ are concyclic.\n\n![](images/THA_National_2016_p16_data_4f9ac9bc0d.png)",
    "options": [],
    "answer": "See solution",
    "solution": "We use the notation $\\angle ABC$ to mean the directed angle $\\angle (BA, BC)$.\n\nSince $E$, $A$, $C$, $B$ are concyclic and $AD$ is parallel to $BC$, $\\angle EAB = \\angle ECB$ and $\\angle BAD = -\\angle ABC$. Therefore,\n\n$$\n\\begin{align*}\n\\angle EAD &= \\angle EAB + \\angle BAD \\\\\n&= \\angle ECB - \\angle ABC \\\\\n&= \\angle MCF - \\angle ABC. \\tag{13}\n\\end{align*}\n$$\n\nSince $M$ is the midpoint of $AB$ and $AF \\perp BC$, $M$ is the circumcenter of triangle $ABF$. We then have\n\n$$\nMF = MB \\quad \\text{and} \\quad \\angle ABC = -\\angle MFC. \\tag{14}\n$$\n\nFrom (13) and (14),\n\n$$\n\\begin{align*}\n\\angle EAD &= \\angle MCF - \\angle MFC \\\\\n&= \\angle MCF + \\angle CFM \\\\\n&= -\\angle FMC \\\\\n&= \\angle EMF. \\tag{15}\n\\end{align*}\n$$\n\nSince $E$, $A$, $C$, $B$ are concyclic, we get $\\triangle AEM \\sim \\triangle CBM$. Therefore, $\\frac{AE}{EM} = \\frac{CB}{BM}$.\nFrom $CB = AD$ and $BM = MF$, we get $\\frac{AE}{EM} = \\frac{AD}{MF}$. So,\n\n$$\n\\frac{AE}{AD} = \\frac{EM}{MF}. \\tag{16}\n$$\n\nFrom (15) and (16), we obtain $\\triangle AED \\sim \\triangle MEF$. Therefore, $\\angle ADE = \\angle MFE$.\nSo, we get\n\n$$\n\\begin{align*}\n\\angle EDC &= \\angle ADC - \\angle ADE \\\\\n&= \\angle ADC - \\angle MFE. \\tag{17}\n\\end{align*}\n$$\n\nApplying (17) and $\\angle ABC = -\\angle ADC$ from the parallelogram $ABCD$ we obtain,\n\n$$\n\\begin{align*}\n\\angle EDC &= \\angle ADC - \\angle MFE \\\\\n&= \\angle MFC - \\angle MFE \\\\\n&= \\angle EFC.\n\\end{align*}\n$$\n\nTherefore, $C$, $D$, $E$, and $F$ are concyclic.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 14394,
    "subject": "Mathematics (Olympiad)",
    "question": "The country Plato has the shape of a convex polygon, with border towers at each vertex. A trucker must visit every tower, paying one puylyk (the state currency) per kilometre to government swindlers. The route does not need to be closed; the only requirement is to visit each tower, and the trucker may travel in any direction without leaving the state border. The perimeter of the state is 3000 kilometres, and its diameter is 1000 kilometres. What is the guaranteed minimum amount of puylyks the trucker will pay?\n\nThe diameter of a polygon is the largest distance between any pair of vertices.",
    "options": [],
    "answer": "See solution",
    "solution": "If the state is an equilateral triangle, the trucker can travel along two sides, paying only 2000 puylyks.\n\nNow, let's prove that in any case, the trucker must pay at least 2000 puylyks. Suppose there is a convex $n$-gon where the trucker pays less than 2000 puylyks. Consider his shortest route. If we close this route by connecting the start and end points, the total length increases by at least 1000 km (the diameter), so the closed route would be less than the perimeter (3000 km). Let's show this is impossible.\n\n![](images/Ukraine_2016_Booklet_p40_data_381b3d58d7.png)\n\n**Statement 1.** Any part of the route between two points should be a straight segment.\n\nIf not, connecting them directly would shorten the route, contradicting the assumption of minimality.\n\n![](images/Ukraine_2016_Booklet_p40_data_78f117bce5.png)\n\n**Statement 2.** All turns on the route must be at the vertices of the polygon.\n\nIf a turn occurs at a point $A$ not at a vertex, replacing segments $CA$ and $AB$ with $CB$ shortens the route.\n\n**Statement 3.** The route cannot cross itself.\n\nIf segments $AB$ and $CD$ intersect at $O$, replacing $AB$ and $CD$ with $AC$ and $BD$ shortens the route, since\n\n$$\nAB + CD = AO + OB + CO + OD > AC + BD.\n$$\n\nSince the number of possible routes is finite, repeated shortening leads to a route with no self-intersections.\n\nFrom these statements, the route must follow the perimeter. If non-consecutive vertices are connected, it becomes impossible to connect the remaining vertices without intersections. Thus, the shortest route is the perimeter, contradicting the assumption. Therefore, the trucker must pay at least 2000 puylyks.",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 14395,
    "subject": "Mathematics (Olympiad)",
    "question": "We consider the numbers:\n\n$$\nA = \\frac{1}{4} \\cdot \\frac{3}{6} \\cdot \\frac{5}{8} \\cdots \\frac{595}{598} \\cdot \\frac{597}{600} \\text{ and } B = \\frac{2}{5} \\cdot \\frac{4}{7} \\cdot \\frac{6}{9} \\cdots \\frac{596}{599} \\cdot \\frac{598}{601}.\n$$\n\nProve that:\n\n(a) $A < B$\n\n(b) $A < \\frac{1}{5990}$.",
    "options": [],
    "answer": "See solution",
    "solution": "**Solution (a):**\n\nTo each fraction of $A$ of the form $\\frac{2\\nu - 1}{2\\nu + 2}$, $\\nu = 1, 2, \\dots, 299$, corresponds a fraction from $B$ of the form $\\frac{2\\nu}{2\\nu + 3}$, $\\nu = 1, 2, \\dots, 299$. Since\n\n$$\n0 < \\frac{2\\nu - 1}{2\\nu + 2} < \\frac{2\\nu}{2\\nu + 3} \\quad \\text{for every } \\nu \\in \\mathbb{N}^*,\n$$\n\nfor $\\nu = 1, 2, \\dots, 299$, by multiplying the above 299 inequalities we obtain $A < B$.\n\n**(b):**\n\nSince $A > 0$, from $A < B$ we get:\n\n$$\nA^2 < A \\cdot B = \\frac{1 \\cdot 2 \\cdot 3}{599 \\cdot 600 \\cdot 601} < \\frac{1}{100 \\cdot 599^2} = \\frac{1}{5990^2} \\implies A < \\frac{1}{5990}.\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 14396,
    "subject": "Mathematics (Olympiad)",
    "question": "In the quadrilateral $ABCD$, point $E$ is the midpoint of side $AB$, point $F$ is the midpoint of side $BC$, and point $G$ is the midpoint of side $AD$. The segment $GE$ is perpendicular to $AB$, and the segment $GF$ is perpendicular to $BC$. Find the angle $GCD$ given that $\\angle ADC = 70^\\circ$.",
    "options": [],
    "answer": "See solution",
    "solution": "Since the segment $GE$ is both an altitude and a median in $\\triangle ABG$, this triangle is isosceles, so $AG = GB$. Similarly, $\\triangle CBG$ is isosceles, so $BG = GC$. Also, since $G$ is the midpoint of $AD$, we have $DG = AG$. Therefore, $DG = AG = BG = CG$, so $\\triangle CGD$ is isosceles, which means $\\angle GCD = \\angle GDC = 70^\\circ$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 14397,
    "subject": "Mathematics (Olympiad)",
    "question": "On a blackboard, all positive integers from 1 to 30000 are written one after another, forming the sequence of digits:\n\n123456789101112...30000.\n\nFind the number of occurrences of $2023$ in this sequence.",
    "options": [],
    "answer": "See solution",
    "solution": "The sequence $2023$ appears 13 times within the numbers $2023$, $12023$, $22023$, and in $20230$, $20231$, $20232$, ..., $20239$.\n\nAdditionally, $2023$ can appear across two consecutive numbers. The pattern $202|3$ appears once, in $3202|3203$, and the pattern $20|23$ appears 11 times: once in $2320|2321$ and from $23020|23021$ to $23920|23921$.\n\nThe pattern $2|023$ is impossible, since no positive integer starts with $0$.\n\nTherefore, the total number of occurrences of $2023$ is $13 + 11 + 1 = 25$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 14398,
    "subject": "Mathematics (Olympiad)",
    "question": "平面上有不等邊三角形 $ABC$ 與其外一點 $P$。在射線 $AB$, $AC$ 上分別取點 $B'$ 與 $C'$，使得 $AB' = AC$, $AC' = AB$。令點 $Q$ 為 $P$ 對直線 $BC$ 的對稱點。設 $\\triangle BB'P$ 與 $\\triangle CC'P$ 的外接圓再交於點 $P'$，$\\triangle BB'Q$ 與 $\\triangle CC'Q$ 的外接圓再交於點 $Q'$。令點 $O$, $O'$ 分別為 $\\triangle ABC$ 與 $\\triangle AB'C'$ 的外心。證明：\n\n1. $O'$, $P'$, $Q'$ 三點共線。\n2. $O'P' \\cdot O'Q' = OA^2$\n",
    "options": [],
    "answer": "See solution",
    "solution": "![](images/18-1J_p23_data_869c22c7c7.png)\n\n解：\n\n由 1. 及 3.，知 $\\angle QQ'P' = \\angle QPP' = \\angle PAO_1$。又由 2. 及 4.，知 $\\angle QQ'O' = \\angle QO_1A = \\angle PAO_1$。由此證得 $O'$, $P'$, $Q'$ 共線。\n\n因 $\\angle AQ'P' = \\angle QQ'P' = \\angle PAO_1 = \\angle P'AO_1$，得 $O'A$ 切 $\\triangle AP'Q'$ 的外接圓於 $A$，所以 $O'P' \\cdot O'Q' = O'A^2 = OA^2$。證畢。",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 14399,
    "subject": "Mathematics (Olympiad)",
    "question": "Do there exist 2024 nonzero real numbers $a_1, a_2, \\dots, a_{2024}$ such that\n\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$$",
    "options": [],
    "answer": "See solution",
    "solution": "Suppose such $a_i$ exist. The condition rewrites as\n$$\n\\sum_{i=1}^{2024} \\left(a_i + \\frac{1}{a_{i+1}} - 1\\right)^2 = 0,\n$$\nwhich means $a_i + \\frac{1}{a_{i+1}} = 1$ for all $i = 1, 2, \\dots, 2024$. It's easy to see that for any $i$, $a_{i+1} = \\frac{1}{1-a_i}$, $a_{i+2} = 1 - \\frac{1}{a_i}$, so $a_{i+3} = a_i$. Since $3$ does not divide $2024$, all $a_i$ must be equal to some real $k$. However, $k + \\frac{1}{k} = 1$ implies $k^2 - k + 1 = 0$, which has no real solution—a contradiction. $\\square$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 14400,
    "subject": "Mathematics (Olympiad)",
    "question": "As shown in Fig. 1.1, in the quadrilateral $ABCD$, $AB = AD$, $CB = CD$, $\\angle ABC = 90^\\circ$. Let $E, F$ be points on the segments $AB, AD$, respectively, and $P, Q$ be points on the segment $EF$ ($P$ is between $E$ and $Q$) such that $\\frac{AE}{EP} = \\frac{AF}{FQ}$. Drop perpendiculars from $B, D$ to $CP, CQ$, with feet $X, Y$, respectively. Prove that $X, P, Q, Y$ are concyclic.\n\n![](images/Mathematical_Olympiad_In_China_2021_2022___Problems_and_Solutions_2024_p299_data_4c45cd6e50.png)",
    "options": [],
    "answer": "See solution",
    "solution": "To begin, we have $\\triangle ABC$ and $\\triangle ADC$ symmetric about line $AC$, and $\\angle ABC = \\angle ADC = 90^\\circ$.\n\nAs shown in Fig. 1.2, let $AC$ and $EF$ meet at point $K$. As $AK$ bisects $\\angle EAF$, $\\frac{AE}{AF} = \\frac{KE}{KF}$.\n\n![](images/Mathematical_Olympiad_In_China_2021_2022___Problems_and_Solutions_2024_p299_data_62835b4fa4.png)\n\nAlso, $\\frac{AE}{AF} = \\frac{EP}{FQ}$, and thus, $\\frac{KE}{KF} = \\frac{EP}{FQ}$ or $\\frac{KE}{EP} = \\frac{KF}{FQ}$.\n\nExtend $CP$ to meet $AE$ at point $S$, and extend $CQ$ to meet $AF$ at point $T$. Apply Menelaus' theorem to $\\triangle KPC$ and straight line $ESA$, obtaining\n\n$$\n\\frac{KE}{EP} \\cdot \\frac{PS}{CS} \\cdot \\frac{CA}{AK} = 1.\n$$\n\nIn the same manner, we find $\\frac{KF}{FQ} \\cdot \\frac{QT}{CT} \\cdot \\frac{CA}{AK} = 1$. Then,\n\n$$\n\\frac{CS}{PS} = \\frac{KE}{EP} \\cdot \\frac{CA}{AK} = \\frac{KF}{FQ} \\cdot \\frac{CA}{AK} = \\frac{CT}{QT},\n$$\n\nwhich gives $\\frac{CS}{CP} = \\frac{CT}{CQ}$, or $\\frac{CP}{CQ} = \\frac{CS}{CT}$.\n\nNote that $BX \\perp CP$, $DY \\perp CQ$ and $\\angle ABC = \\angle ADC = 90^\\circ$, and hence,\n\n$$\n\\frac{CX \\cdot CP}{CY \\cdot CQ} = \\frac{CX \\cdot CS}{CY \\cdot CT} = \\frac{CB^2}{CD^2} = 1.\n$$\n\nWe conclude that $X, P, Q, Y$ are concyclic. $\\Box$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 14401,
    "subject": "Mathematics (Olympiad)",
    "question": "Determine the largest real number $k$ such that the inequality\n\n$$\n\\left(k + \\frac{a}{b}\\right) \\left(k + \\frac{b}{c}\\right) \\left(k + \\frac{c}{a}\\right) \\leq \\left(\\frac{a}{b} + \\frac{b}{c} + \\frac{c}{a}\\right) \\left(\\frac{b}{a} + \\frac{c}{b} + \\frac{a}{c}\\right)\n$$\n\nholds for all positive real numbers $a$, $b$, and $c$.",
    "options": [],
    "answer": "See solution",
    "solution": "By setting $a = b = c$, it follows that $k \\leq \\sqrt[3]{9} - 1$. We claim that $k = \\sqrt[3]{9} - 1$ is the largest possible number so that the inequality holds. Let\n\n$$\nA = \\frac{a}{b} + \\frac{b}{c} + \\frac{c}{a}, \\quad B = \\frac{b}{a} + \\frac{c}{b} + \\frac{a}{c}.\n$$\n\nBy the AM-GM inequality, we get that\n\n$$\nA \\geq 3\\sqrt[3]{\\frac{a}{b} \\cdot \\frac{b}{c} \\cdot \\frac{c}{a}} = 3,\n$$\n\nand similarly,\n\n$$\nB \\geq 3\\sqrt[3]{\\frac{b}{a} \\cdot \\frac{c}{b} \\cdot \\frac{a}{c}} = 3.\n$$\n\nFor each real number $k \\geq 0$, the following inequalities are true:\n\n$$\n9(k^3 + 1) \\leq (k^3 + 1)AB\n$$\n$$\n9k^2A \\leq 3k^2AB\n$$\n$$\n9kB \\leq 3kAB.\n$$\n\nAdding these inequalities, we get\n\n$$\n9(k^3 + 1 + k^2 A + k B) \\le (k + 1)^3 AB\n$$\n$$\n9 \\left(k + \\frac{a}{b}\\right) \\left(k + \\frac{b}{c}\\right) \\left(k + \\frac{c}{a}\\right) \\le (k + 1)^3 AB.\n$$\n\nSubstituting $k = \\sqrt[3]{9} - 1$ in the last inequality, we discover that\n\n$$\n9 \\left(k + \\frac{a}{b}\\right) \\left(k + \\frac{b}{c}\\right) \\left(k + \\frac{c}{a}\\right) \\le 9AB.\n$$\n\nwhich is precisely the desired inequality. Therefore, the largest possible $k$ is $\\sqrt[3]{9} - 1$ as claimed. $\\square$",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 14402,
    "subject": "Mathematics (Olympiad)",
    "question": "設 $H = \\{\\lfloor m\\sqrt{2} \\rfloor : m \\in \\mathbb{Z}_{>0}\\} = \\{1, 2, 4, 5, 7, \\dots\\}$，且 $N$ 為正整數。證明：若 $A \\subset \\{1, 2, \\dots, N\\}$ 且 $|A| \\ge 1 + \\sqrt{N}$，則存在 $a, b \\in A$ 使得 $a - b \\in H$。\n\n（這裡 $\\mathbb{Z}_{>0}$ 是所有正整數形成的集合，而 $\\lfloor z \\rfloor$ 表示不超過 $z$ 的最大整數。）",
    "options": [],
    "answer": "See solution",
    "solution": "Lemma 1: $n \\in H$ iff $\\left\\{\\frac{n}{\\sqrt{2}}\\right\\} > 1 - \\frac{1}{\\sqrt{2}}$.\n\nProof: $n \\in H$ iff $0 < m\\sqrt{2} - n < 1$ for some $m$ iff $0 < m - n/\\sqrt{2} < 1/\\sqrt{2}$ for some $m$. $\\square$\n\nLemma 2: $\\left\\{\\frac{n}{\\sqrt{2}}\\right\\} > \\frac{1}{2\\sqrt{2n}}$ for any $n \\in \\mathbb{Z}$.\n\nProof: Let $m = \\lfloor \\frac{n}{\\sqrt{2}} \\rfloor$, so $\\left\\{\\frac{n}{\\sqrt{2}}\\right\\} = \\frac{n}{\\sqrt{2}} - m$. Then\n\n$$\n\\left\\{\\frac{n}{\\sqrt{2}}\\right\\} \\cdot \\sqrt{2n} > \\left\\{\\frac{n}{\\sqrt{2}}\\right\\}\\left(\\frac{n}{\\sqrt{2}} + m\\right) = \\frac{n^2 - 2m^2}{2} \\ge \\frac{1}{2}.\n$$\n\n![](images/20-3J_p32_data_5fdca6cda1.png)\n\n假設 $H \\cap \\{a-b : a, b \\in A\\} = \\emptyset$，令 $A = \\{a_1, a_2, \\dots, a_k\\}$ 且 $a_1 < a_2 < \\dots < a_k$。\n\n**解法 1**：注意差集在平移 $A$ 時不變，可假設 $A \\subseteq \\{0, 1, \\dots, N-1\\}$ 且 $a_1 = 0$。由 Lemma 1，對所有 $j > 1$，有 $\\left\\{\\frac{a_j}{\\sqrt{2}}\\right\\} > 1 - \\frac{1}{\\sqrt{2}}$。若 $i < j$ 且 $\\left\\{\\frac{a_j}{\\sqrt{2}}\\right\\} < \\left\\{\\frac{a_i}{\\sqrt{2}}\\right\\}$，則\n\n$$\n-(1 - \\frac{1}{\\sqrt{2}}) < \\left\\{\\frac{a_j}{\\sqrt{2}}\\right\\} - \\left\\{\\frac{a_i}{\\sqrt{2}}\\right\\} < 0.\n$$\n\n所以\n\n$$\n\\left\\{\\frac{a_j - a_i}{\\sqrt{2}}\\right\\} = \\left\\{\\frac{a_j}{\\sqrt{2}}\\right\\} - \\left\\{\\frac{a_i}{\\sqrt{2}}\\right\\} + 1 > \\frac{1}{\\sqrt{2}} > 1 - \\frac{1}{\\sqrt{2}}\n$$\n\n這與 $a_j - a_i \\notin H$ 矛盾。因此 $\\left\\{\\frac{a_j}{\\sqrt{2}}\\right\\}$ 必須嚴格遞增且 $< 1 - \\frac{1}{\\sqrt{2}}$。令 $n_i = a_{i+1} - a_i$，由 Lemma 2 得\n\n$$\n\\left\\{\\frac{a_{i+1}}{\\sqrt{2}}\\right\\} - \\left\\{\\frac{a_i}{\\sqrt{2}}\\right\\} = \\left\\{\\frac{n_i}{\\sqrt{2}}\\right\\} > \\frac{1}{2\\sqrt{2}n_i}.\n$$\n\n因此\n\n$$\n1 - \\frac{1}{\\sqrt{2}} > \\sum_{i} \\left\\{\\frac{a_{i+1}}{\\sqrt{2}}\\right\\} - \\left\\{\\frac{a_i}{\\sqrt{2}}\\right\\} > \\frac{1}{2\\sqrt{2}} \\sum_{i} \\frac{1}{n_i} \\ge \\frac{(k-1)^2}{2\\sqrt{2} \\sum_{i} n_i} \\ge \\frac{(k-1)^2}{2\\sqrt{2}(N-1)}.\n$$\n\n特別地，$\\sqrt{n} > \\sqrt{2\\sqrt{2}-2}\\sqrt{n} > k-1$。\n\n**解法 2**：令 $\\alpha = 2 + \\sqrt{2}$，則 $1/\\alpha + 1/\\sqrt{2} = 1$。因此 $J = \\{\\lfloor m\\alpha \\rfloor : m \\in \\mathbb{Z}_{>0}\\}$ 是 $H$ 的補 Beatty 序列，故 $J = \\mathbb{Z}_{>0} \\setminus H$，且 $A$ 的所有差都在 $J$。設 $a_i - a_1 = \\lfloor b_i \\alpha \\rfloor$。\n\n對任意 $j > i$，有 $a_j - a_i = \\lfloor b_j \\alpha \\rfloor - \\lfloor b_i \\alpha \\rfloor$。由於 $a_j - a_i \\in J$，$\\lfloor b_j \\alpha \\rfloor - \\lfloor b_i \\alpha \\rfloor = \\lfloor t \\alpha \\rfloor$ 對某 $t$。因 $\\lfloor b_j \\alpha \\rfloor - \\lfloor b_i \\alpha \\rfloor$ 為 $\\lfloor (b_j - b_i) \\alpha \\rfloor$ 或 $\\lfloor (b_j - b_i) \\alpha \\rfloor - 1$，必有 $t = b_j - b_i$ 且 $\\lfloor b_j \\alpha \\rfloor - \\lfloor b_i \\alpha \\rfloor = \\lfloor (b_j - b_i) \\alpha \\rfloor$。令 $d_i = b_{i+1} - b_i$，則\n\n$$\n\\lfloor \\alpha \\sum d_i \\rfloor = \\lfloor \\alpha b_k \\rfloor = \\sum_i \\lfloor \\alpha d_i \\rfloor.\n$$\n\n所以 $\\sum_i \\alpha d_i < 1$。又有\n\n$$\n1 + \\lfloor \\alpha \\sum d_i \\rfloor = 1 + a_k - a_1 \\le n\n$$\n\n故 $\\sum_i d_i \\le n/\\alpha$。用解法 1 最後一步類似的論證，可得 $\\sqrt{n} > \\sqrt{2\\sqrt{2}-2}\\sqrt{n} > k-1$。",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 14403,
    "subject": "Mathematics (Olympiad)",
    "question": "Given $2n$ points ($n > 1$) in the plane, and a line $p$ that does not pass through any of these points. Prove that $p$ intersects at most $n^2$ segments whose endpoints are among the given points.",
    "options": [],
    "answer": "See solution",
    "solution": "Let the $2n$ points and the line $p$ lie in the plane. The line $p$ divides the plane into two half-planes. Suppose one half-plane contains $m$ points, and the other contains $2n - m$ points. A segment with both endpoints in the same half-plane does not intersect $p$. Only segments with endpoints in different half-planes are cut by $p$. The number of such segments is $m(2n - m)$. Since $n^2 - m(2n - m) = (n - m)^2 \\geq 0$, we have $n^2 \\geq m(2n - m)$. Therefore, the line $p$ cuts at most $n^2$ segments.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 14404,
    "subject": "Mathematics (Olympiad)",
    "question": "One hundred brownies (girl scouts) are sitting in a big circle around the campfire. Each brownie has one or more chestnuts, and no two brownies have the same number of chestnuts. Each brownie divides her number of chestnuts by the number of chestnuts of her right neighbour and writes down the remainder on a green piece of paper. Each brownie also divides her number by the number of chestnuts of her left neighbour and writes down the remainder on a red piece of paper.\n\nFor example, if Anja has 23 chestnuts and her right neighbour Bregje has 5, then Anja writes $3$ on her green piece of paper and Bregje writes $5$ on her red piece of paper.\n\nIf the number of distinct remainders on the $100$ green pieces of paper equals $2$, what is the smallest possible number of distinct remainders on the $100$ red pieces of paper?",
    "options": [],
    "answer": "See solution",
    "solution": "$100$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 14405,
    "subject": "Mathematics (Olympiad)",
    "question": "There are $2010^{2010}$ children at a mathematics camp. Each child has at most three friends at the camp, and if $A$ is friends with $B$, then $B$ is friends with $A$. The camp leader would like to line the children up so that there are at most $2010$ children between any pair of friends. Is it always possible to do this?",
    "options": [],
    "answer": "See solution",
    "solution": "We will show that it is *not* always possible.\n\nWe consider a person $X$ who has friendships arranged according to this diagram:\n\n![](images/V_Britanija_2010_p24_data_3d56f86b60.png)\n\nThis pattern cannot be maintained forever (we will run out of children!), but we will show that it can go far enough to break the condition that pairs of friends must be separated by at most $2010$ children.\n\nWe will aim for a contradiction: suppose that we do have the children arranged in a line, satisfying this condition. Person $X$ is somewhere along the line. His three friends (on Row 1) must be within $2011$ places of him. This means we have $1 + 3$ children in at most $1 + 2 \\times 2011$ possible places.\n\nNow consider the six people in Row 2. They are all within $2011$ places of their friends in Row 1, and so within $2 \\times 2011$ places of Person $X$. This means we have ten people (person $X$, his three friends in Row 1, and the six people in Row 2) fitting in at most $1 + 2 \\times 2 \\times 2011$ different places.\n\nIf we continue to Row $n$, we have considered a total number of children given by\n\n$$\n\\begin{aligned}\n& 1 + 3 + (3 \\times 2) + (3 \\times 2^2) + \\dots + (3 \\times 2^n) \\\\\n&= 1 + 3(1 + 2 + \\dots + 2^n) \\\\\n&= 1 + 3(2^{n+1} - 1) \\\\\n&= (3 \\times 2^{n+1}) - 2.\n\\end{aligned}\n$$\n\nBut since all the children in rows up to $n$ are connected to Person $X$ by a chain of at most $n$ friendships, they can be at distance at most $2011n$ from Person $X$, and so can be sitting in at most $1 + 2n \\times 2011$ different places.\n\nNow, for sufficiently large $n$, we will have\n\n$$\n(3 \\times 2^{n+1}) - 2 > 2n \\times 2011.\n$$\n\nsince the left-hand side grows exponentially and the right-hand side linearly.\n\nIndeed, take $n = 19$. Then $(3 \\times 2^{20}) - 2 > 3,000,000$ since $2^{20} > 1,000,000$, while $2 \\times 19 \\times 2011 < 80,000$.\n\nSo there are not enough places to seat all these children, breaking the condition. Note that $3 \\times 2^{20}$ is much smaller than $2010^{2010}$, so we have enough children to arrange in this way.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 14406,
    "subject": "Mathematics (Olympiad)",
    "question": "The diagonals of the parallelogram $ABCD$ intersect at $O$, and $M$ is the midpoint of the side $AB$. Let $P$ be a point on the segment $OC$, and $Q$ be the intersection of the lines $MP$ and $BC$. The parallel to $MP$ through $O$ intersects the line $CD$ at the point $N$. Prove that the points $A$, $N$, and $Q$ are collinear if and only if $P$ is the midpoint of the segment $OC$.",
    "options": [],
    "answer": "See solution",
    "solution": "Denote $R$ as the intersection of the lines $QM$ and $CD$.\n\nIf $A$, $N$, $Q$ are collinear, then the fundamental theorem for similarity yields $$\\frac{CR}{MB} = \\frac{QC}{QB} = \\frac{CN}{AB}$$ and from $AB = 2 \\cdot MB$ it follows that $CN = 2 \\cdot CR$. Now, $RP \\parallel ON$ shows that $RP$ is a midline in $\\triangle CON$, therefore $P$ is the midpoint of $OC$.\n\n![](images/RMC_2023_v2_p33_data_2cf3598686.png)\n\nFor the converse, if $P$ is the midpoint of the segment $OC$, then $RP$ is a midline of $\\triangle CON$, therefore $CN = 2 \\cdot CR$. Let the lines $AQ$ and $CD$ meet at $N'$. Then, as above, $CN' = 2 \\cdot CR = CN$, so $N$ coincides with $N'$, that is, $A$, $N$, $Q$ are collinear.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 14407,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $L(z)$ be the smallest prime divisor of $z$.\n\nIf $a, b, c \\in \\mathbb{N}$ such that $a \\neq b$, prove that\n$$\nL\\left(\\frac{a^c - b^c}{a - b}\\right) \\geq L(abc)\n$$\n\nConsider the set\n$$\nA_k = \\{x \\in \\mathbb{N} \\mid L(x) \\geq k\\}\n$$\nIf $a, b, c \\in A_k$ and $d \\mid \\frac{a^c-b^c}{a-b}$, show that $d \\in A_k$ for $k > n$.",
    "options": [],
    "answer": "See solution",
    "solution": "*Proof.* Assume the contrary: there exists an integer $p$ such that $\\gcd(p, a) = \\gcd(p, b) = 1$ and $p \\mid \\frac{a^c-b^c}{a-b}$. Consider two cases:\n\n- If $a \\equiv b \\pmod{p}$, then by the lifting the exponent lemma, $\\nu_p(a^c-b^c) = \\nu_p(a-b)$, which is a contradiction.\n- If $a \\not\\equiv b \\pmod{p}$, then $\\operatorname{ord}_p(ab^{-1}) \\neq 1$ (where $b^{-1}$ is the multiplicative inverse of $b$ modulo $p$). But $\\operatorname{ord}_p(ab^{-1}) \\mid c$ and $p > \\operatorname{ord}_p(ab^{-1}) > L(c)$, which is a contradiction.\n\nThus, the lemma holds.\n\nNow, for the set $A_k = \\{x \\in \\mathbb{N} \\mid L(x) \\geq k\\}$, if $a, b, c \\in A_k$ and $d \\mid \\frac{a^c-b^c}{a-b}$, then by the lemma, $L(d) < L(abc) < n$, so $d \\in A_k$. Therefore, it suffices to take $k > n$. $\\textbf{\\textopenbullet}$",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 14408,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $ABC$ be a triangle. $D$ is the foot of the internal bisector of angle $A$. The perpendicular from $D$ to the tangent $AT$ ($T$ belongs to $BC$) to the circumscribed circle of $ABC$ intersects the altitude $AH_a$ at the point $I$ ($H_a$ belongs to $BC$). If $P$ is the midpoint of $AB$ and $O$ is the circumcenter of $\\triangle ABC$, $TI$ intersects $AB$ at $M$ and $PT$ intersects $AD$ at $F$. Prove that $MF$ is perpendicular to $AO$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $Q$ be the midpoint of $AC$ and $N$ the intersection of $AD$ and $PQ$. Then $N$ is the midpoint of $AD$. As $DE$ is perpendicular to $AT$, with $E$ the intersection point of $DI$ and $AT$, and as $OA$ is perpendicular to $AT$, we get that $DE$ is parallel to $OA$, so the angles $OAN$ and $ADE$ are equal. Indeed, $\\angle OAQ = \\angle BAH$ because $\\angle OAQ = \\frac{180^\\circ - \\angle COA}{2} = 90^\\circ - \\frac{\\angle COA}{2} = 90^\\circ - B = \\angle BAH$. Moreover, $\\angle BAD = \\angle DAC$ and hence $\\angle OAN = \\angle HAD = \\angle ADE$. As a consequence, triangles $ADE$ and $DAH$ are congruent.\n\n![](images/Spanija_b_2016_p36_data_3850f9e218.png)\n\nIn particular, angle $DAT$ equals angle $HAD$, that is, $ATD$ is isosceles and point $I$ is the orthocenter of $\\triangle ABC$. So, $TI$ is perpendicular to $AD$, and the intersection point of $TI$ and $AD$ is the midpoint of $AD$, say $N$. The four points $M, N, I, T$ are collinear. We will apply Ceva's theorem in triangle $APT$ with the cevians $PN$, $AD$, and $TM$. We get\n\n$$\n\\frac{FP}{FT} \\cdot 1 \\cdot \\frac{MA}{PM} \\Leftrightarrow \\frac{PF}{TF} = \\frac{MP}{MA}\n$$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 14409,
    "subject": "Mathematics (Olympiad)",
    "question": "Докажите, что не позже, чем через 10 000 ходов Паша победит в следующей игре: из исходного куска массы 1 кг Паша и Вова по очереди отрезают куски, при этом Паша стремится получить как можно больше кусков массой 0,01 г. После каждого хода Паши Вова может слепить два куска одинаковой массы в один. Победа Паши наступает, когда на столе окажется 100 кусков массой 0,01 г.",
    "options": [],
    "answer": "See solution",
    "solution": "**Второе решение.** Пусть масса исходного куска равна 1 кг. Приведём другой алгоритм действий Паши. Пока это возможно, он будет добиваться выполнения следующего условия: (*) массы всех кусков на столе составляют целое число граммов. Заметим, что Вова своим ходом не может нарушить (*).\n\nЕсли перед ходом Паши на столе есть кусок массой хотя бы 3 г, он может отрезать от него два куска по 1 г, сохраняя (*). Значит, если Паша не может сделать ход, каждый кусок весит либо 1 г, либо 2 г (такой момент обязательно наступит, так как количество кусков перед ходом Паши растёт). Но тогда он уже выиграл: в противном случае общая масса всех кусков (в граммах) меньше $1 \\cdot 100 + 2 \\cdot 100 = 300$, что не так.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 14410,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $ABC$ be an acute triangle. The lines $l_1$ and $l_2$ are perpendicular to $AB$ at the points $A$ and $B$ respectively. The perpendicular lines from the midpoint $M$ of $AB$ to the lines $AC$ and $BC$ intersect $l_1$ and $l_2$ at the points $E$ and $F$, respectively. If $D$ is the intersection point of the lines $EF$ and $MC$, prove that $\\angle ADB = \\angle EMF$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let the circles with diameter $EM$ and $FM$ intersect for a second time at $D'$, and let them intersect the sides $CA$, $CB$ at points $G$ and $K$ respectively. Since\n\n$$\n\\angle ED'M = \\angle FD'M = 90^{\\circ},\n$$\n\nwe have that $E$, $D'$, $F$ are collinear.\n\nSince $EM$ is a diameter and $AG$ is a chord perpendicular to it, we have $MG = MA$, and similarly $MK = MB$. Since $MA = MB$, it follows that $AGKB$ is cyclic.\n\nFrom the above, $CG \\cdot CA = CK \\cdot CB$, so $C$ has equal power to the two circles, meaning it is on the radical axis of them, so $C$, $D'$, $M$ are collinear. Thus, $D' \\equiv D$.\n\nFinally, from the cyclic quadrilaterals $EAMD$ and $DMBF$ we have\n\n$$\n\\angle ADB = 180^{\\circ} - \\angle EDA - \\angle BDF = 180^{\\circ} - \\angle AME - \\angle BMF = \\angle EMF.\n$$\n\n**Solution 2.** Let $H$, $G$ be the points of intersection of $ME$, $MF$ with $AC$, $BC$ respectively. From the similarity of triangles $\\triangle MHA$ and $\\triangle MAE$ we get\n\n$$\n\\frac{MH}{MA} = \\frac{MA}{ME},\n$$\n\nthus,\n\n$$\nMA^2 = MH \\cdot ME. \\tag{1}\n$$\n\nSimilarly, from the similarity of triangles $\\triangle MBG$ and $\\triangle MFB$ we get\n\n$$\n\\frac{MB}{MF} = \\frac{MG}{MB},\n$$\n\nthus,\n\n$$\nMB^2 = MF \\cdot MG. \\tag{2}\n$$\n\nSince $MA = MB$, from (1) and (2) we have that the points $E$, $H$, $G$, $F$ are concyclic.\n\nTherefore, $\\angle FEH = \\angle FEM = \\angle HGM$. Also, the quadrilateral $CHMG$ is cyclic, so $\\angle CMH = \\angle HGC$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 14411,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $x$ be a three-digit number with digits $a$, $b$, and $c$, so that $x = \\overline{abc}$. All three-digit numbers that can be formed from $a$, $b$, and $c$ are $\\overline{abc}$, $\\overline{acb}$, $\\overline{bac}$, $\\overline{bca}$, $\\overline{cab}$, and $\\overline{cba}$. The sum of these numbers is $222(a + b + c)$. Given that the sum of the numbers written on the paper is $3434$ less than this total, what is $x$?",
    "options": [],
    "answer": "See solution",
    "solution": "Let the sum of the numbers be $S = 222(a + b + c)$. We are told that $S - x = 3434$, so:\n\n$$\n3434 = 222(a + b + c) - \\overline{abc} = 122a + 212b + 221c.\n$$\n\nConsider this equation modulo $9$. $3434 \\equiv 5 \\pmod{9}$, and $122a + 212b + 221c \\equiv 5(a + b + c) \\pmod{9}$. Thus, $5(a + b + c) \\equiv 5 \\pmod{9}$, so $a + b + c \\equiv 1 \\pmod{9}$. Since $a + b + c$ is between $6$ and $24$, possible values are $10$ or $19$.\n\nIf $a + b + c = 10$, $3434 = 122a + 212b + 221c < 221 \\times 10 = 2210 < 3434$, which is impossible. So $a + b + c = 19$.\n\nThen $\\overline{abc} = 222 \\times 19 - 3434 = 4218 - 3434 = 784$.\n\nThus, the only possible solution is $x = 784$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 14412,
    "subject": "Mathematics (Olympiad)",
    "question": "Реши ја равенката\n\n$$\n4^{\\log_{10} x} - 32 + x^{\\log_{10} 4} = 0.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "Јасно е дека равенката е определена за $x > 0$. За $x > 0$, $\\log_{10} x$ можеме да го запишеме во облик\n\n$$\n\\log_{10} x = \\frac{\\log_4 x}{\\log_4 10} = (\\log_4 x)(\\log_{10} 4),\n$$\n\nод каде добиваме\n\n$$\n4^{\\log_{10} x} = 4^{(\\log_4 x)(\\log_{10} 4)} = (4^{\\log_4 x})^{\\log_{10} 4} = x^{\\log_{10} 4}.\n$$\n\nСега, равенката можеме да ја запишеме во облик\n\n$$\n4^{\\log_{10} x} - 32 + 4^{\\log_{10} x} = 0,\n$$\n\n$$\n2 \\cdot 4^{\\log_{10} x} - 32 = 0,\n$$\n\n$$\n4^{\\log_{10} x} = 4^2,\n$$\n\n$$\n\\log_{10} x = 2.\n$$\n\nЗначи, решение на равенката е $x = 100$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 14413,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $F$ be the set of all sequences $(a_1, a_2, \\dots, a_{2020})$ with $a_i \\in \\{-1, 1\\}$ for all $i = 1, 2, \\dots, 2020$. Prove that there exists a set $S \\subset F$ with $|S| = 2020$ such that for any $(a_1, a_2, \\dots, a_{2020}) \\in F$, there exists $(b_1, b_2, \\dots, b_{2020}) \\in S$ such that $\\sum_{i=1}^{2020} a_i b_i = 0$.",
    "options": [],
    "answer": "See solution",
    "solution": "For each $i \\in \\{1, 2, \\dots, 2021\\}$, let $e_i = (\\underbrace{1, \\dots, 1}_{i-1}, \\underbrace{-1, \\dots, -1}_{2021-i})$.\n\nFor two sequences $a = (a_1, a_2, \\dots, a_{2020})$, $b = (b_1, b_2, \\dots, b_{2020})$, define\n\n$$\na \\cdot b = a_1 b_1 + a_2 b_2 + \\dots + a_{2020} b_{2020}.\n$$\n\nIt is easy to verify that $a \\cdot b$ is even for each $a, b \\in F$. Let $a = (a_1, a_2, \\dots, a_{2020}) \\in F$ and denote $b_i = a \\cdot e_i$ for each $i \\in \\{1, 2, \\dots, 2020\\}$. Then\n\n$$\nb_{i+1} - b_i = a \\cdot e_{i+1} - a \\cdot e_i = 2a_i \\implies |b_{i+1} - b_i| = 2, \\forall i \\in \\{1, 2, \\dots, 2020\\}.\n$$\n\nLet $S = \\{e_1, e_2, \\dots, e_{2020}\\}$. It is clear that $b_{2021} = -b_1$ and $b_i$ is even for each $i \\in \\{1, 2, \\dots, 2020\\}$. If $b_1 \\neq 0$, there exists an integer $k$ such that $1 < k < 2021$ and $b_k = 0$. Hence, $S$ satisfies the problem's requirements. $\\square$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 14414,
    "subject": "Mathematics (Olympiad)",
    "question": "A convex polygon $\\mathcal{P}$ in the plane is dissected into smaller convex polygons by drawing all of its diagonals. The lengths of all sides and all diagonals of the polygon $\\mathcal{P}$ are rational numbers. Prove that the lengths of all sides of all polygons in the dissection are also rational numbers.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $\\mathcal{P} = A_1A_2\\dots A_n$, where $n \\ge 3$. The problem is trivial for $n=3$ because there are no diagonals and thus no dissections. We assume that $n \\ge 4$. Our proof is based on the following *Lemma*.\n\n*Lemma*: Let $ABCD$ be a convex quadrilateral such that all its sides and diagonals have rational lengths. If segments $AC$ and $BD$ meet at $P$, then segments $AP$, $BP$, $CP$, $DP$ all have rational lengths.\n\n![](images/USA_IMO_2003_p28_data_2fc395e748.png)\n\nIt is clear by the Lemma that the desired result holds when $\\mathcal{P}$ is a convex quadrilateral. Let $A_iA_j$ ($1 \\le i < j \\le n$) be a diagonal of $\\mathcal{P}$. Assume that $C_1, C_2, \\dots, C_m$ are the consecutive division points on diagonal $A_iA_j$ (where point $C_1$ is the closest to vertex $A_i$ and $C_m$ is the closest to $A_j$). Then the segments $C_\\ell C_{\\ell+1}$, $1 \\le \\ell \\le m-1$, are the sides of all polygons in the dissection. Let $C_\\ell$ be the point where diagonal $A_iA_j$ meets diagonal $A_sA_t$. Then quadrilateral $A_iA_sA_jA_t$ satisfies the conditions of the Lemma. Consequently, segments $A_iC_\\ell$ and $C_\\ell A_j$ have rational lengths. Therefore, segments $A_iC_1, A_iC_2, \\dots, A_jC_m$ all have rational lengths. Thus, $C_\\ell C_{\\ell+1} = AC_{\\ell+1} - AC_\\ell$ is rational. Because $i, j, \\ell$ are arbitrarily chosen, we proved that all sides of all polygons in the dissection are also rational numbers.\n\nNow we present two proofs of the Lemma to finish our proof.\n\n*First approach*: We show only that segment $AP$ is rational, the proof for the others being similar. Introduce Cartesian coordinates with $A = (0, 0)$ and $C = (c, 0)$. Put $B = (a, b)$ and $D = (d, e)$. Then by hypothesis, the numbers\n\n$$\n\\begin{aligned}\nAB &= \\sqrt{a^2 + b^2}, & AC &= c, & AD &= \\sqrt{d^2 + e^2}, \\\\\nBC &= \\sqrt{(a-c)^2 + b^2}, & BD &= \\sqrt{(a-d)^2 + (b-e)^2}, \\\\\nCD &= \\sqrt{(d-c)^2 + e^2},\n\\end{aligned}\n$$\n\nare rational. In particular,\n\n$$\nBC^2 - AB^2 - AC^2 = (a-c)^2 + b^2 - (a^2 + b^2) - c^2 = -2ac\n$$\n\nis rational. Because $c \\neq 0$, $a$ is rational. Likewise, $d$ is rational.\n\n![](images/USA_IMO_2003_p29_data_bf64568854.png)\n\nNow we have that $b^2 = AB^2 - a^2$, $e^2 = AD^2 - d^2$, and $(b-e)^2 = BD^2 - (a-d)^2$ are rational, and so $2be = b^2 + e^2 - (b-e)^2$ is rational. Because quadrilateral $ABCD$ is convex, $b$ and $e$ are nonzero and have opposite sign. Hence $b/e = 2be/2b^2$ is rational.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 14415,
    "subject": "Mathematics (Olympiad)",
    "question": "A data set containing 20 numbers, some of which are 6, has mean 45. When all the 6s are removed, the data set has mean 66. How many 6s were in the original data set?\n\n(A) 4  (B) 5  (C) 6  (D) 7  (E) 8",
    "options": [],
    "answer": "See solution",
    "solution": "Let the number of 6s in the original data set be $k$. The sum of the original data set is $20 \\times 45 = 900$. After removing all the 6s, the sum is $900 - 6k$ and the number of remaining numbers is $20 - k$. Thus,\n\n$$\n\\frac{900 - 6k}{20 - k} = 66.\n$$\n\nSolving for $k$:\n\n$$\n900 - 6k = 66(20 - k) \\\\\n900 - 6k = 1320 - 66k \\\\\n60k = 420 \\\\\nk = 7.\n$$\n\nSo, there were 7 sixes in the original data set.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 14416,
    "subject": "Mathematics (Olympiad)",
    "question": "找出所有符合下列條件的整數 $n \\ge 2$：所有總和不被 $n$ 整除、兩兩相異的 $n$ 個整數，都可以被重新排列為 $a_1, a_2, \\dots, a_n$，使得 $n$ 整除 $1 \\cdot a_1 + 2 \\cdot a_2 + \\dots + n \\cdot a_n$。",
    "options": [],
    "answer": "See solution",
    "solution": "所有奇數整數以及所有 2 的冪次皆符合條件。\n\n若 $n = 2^k a$，其中 $a \\ge 3$ 為奇數且 $k$ 為正整數，考慮一組包含 $2^k + 1$ 及 $n-1$ 個同餘於 1（模 $n$）的數。這些數的總和同餘 $2^k$（模 $n$），因此不被 $n$ 整除。對於這些數的任意排列 $(a_1, a_2, \\dots, a_n)$，\n\n$$\n1 \\cdot a_1 + 2 \\cdot a_2 + \\dots + n \\cdot a_n \\equiv 2^{k-1}a(2^k a + 1) \\not\\equiv 0 \\pmod{2^k}\n$$\n\n因此 $1 \\cdot a_1 + 2 \\cdot a_2 + \\dots + n \\cdot a_n$ 也不會被 $n$ 整除。\n\n接下來假設 $n$ 為奇數或 2 的冪次。設 $S$ 為給定的整數集合，$s$ 為 $S$ 的元素和。\n\n**引理 1.** 若存在 $S$ 的一個排列 $(a_i)$ 使 $(n, s)$ 整除 $\\sum_{i=1}^n i a_i$，則存在 $S$ 的一個排列 $(b_i)$ 使 $n$ 整除 $\\sum_{i=1}^n i b_i$。\n\n*證明.* 設 $r = \\sum_{i=1}^n i a_i$。考慮排列 $b_i = a_{i+x}$（$a_{j+n} = a_j$），則\n\n$$\n\\sum_{i=1}^{n} i b_{i} = \\sum_{i=1}^{n} i a_{i+x} \\equiv \\sum_{i=1}^{n} (i-x)a_{i} \\equiv r - s x \\pmod{n}.\n$$\n\n由於 $(n, s)$ 整除 $r$，方程 $r - s x \\equiv 0 \\pmod{n}$ 有解。\n\n**引理 2.** 任意 $km$ 個整數的集合 $T$，$m > 1$，可分成 $m$ 個 $k$ 元子集，使每個子集內元素和不被 $k$ 整除，或所有元素模 $k$ 同餘。\n\n*證明略。*\n\n現在對所有奇數 $n$ 及 $n = 2^k$ 用歸納法證明原命題。\n\n若 $n$ 為質數，由引理 1，$(n, s) = 1$，命題成立。一般情況下，取質數 $p$ 及 $t$ 使 $p^t \\mid n$ 且 $p^t \\nmid s$。由引理 2，可將 $S$ 分成 $p$ 個 $k = n/p$ 元子集，每個子集內元素和不被 $k$ 整除，或元素模 $k$ 同餘。\n\n屬於第一類的子集，歸納假設可得存在排列 $(a_i)$ 使 $k \\mid \\sum_{i=1}^k i a_i$。\n\n若 $n$（及 $k$）為奇數，則對第二類子集的任意排列 $(b_i)$，\n\n$$\n\\sum_{i=1}^{k} i b_i \\equiv b_1 \\frac{k(k+1)}{2} \\equiv 0 \\pmod{k}.\n$$\n\n將所有子集的排列合併，得 $S$ 的一個排列 $(c_i)$ 使 $k \\mid \\sum_{i=1}^n i c_i$。由於 $k \\mid n$ 且 $k \\mid (n, s)$，由引理 1 得證。\n\n若 $n = 2^s$，則 $p = 2, k = 2^{s-1}$。每個子集存在排列 $(a_1, \\dots, a_k)$ 使 $\\sum_{i=1}^k i a_i$ 被 $2^{s-2} = k/2$ 整除：\n\n$$\n\\sum_{i=1}^{k} i a_i \\equiv a_1 \\frac{k(k+1)}{2} \\equiv 0 \\pmod{\\frac{k}{2}}.\n$$\n\n再將每個排列的數分別乘上 $1, 3, \\dots, n-1$ 或 $2, 4, \\dots, n$，使和被 $k$ 整除：\n\n$$\n\\sum_{i=1}^{k} (2i-1)a_i \\equiv \\sum_{i=1}^{k} 2i a_i \\equiv 2 \\sum_{i=1}^{k} i a_i \\equiv 0 \\pmod{k}.\n$$\n\n合併後，得 $S$ 的一個排列 $(c_i)$ 使 $k \\mid \\sum_{i=1}^n i c_i$，由引理 1 得證。\n\n*補充說明*：若去除「總和不被 $n$ 整除」的條件，則命題不成立。例如 $n > 1$ 時，取 $1, -1$ 及 $n-2$ 個 $n$ 的倍數，則不存在所需排列。",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 14417,
    "subject": "Mathematics (Olympiad)",
    "question": "Solve the equation\n\n$$\n\\sqrt{1 + 3\\sin^3 x} = 3 - \\sqrt{1 - \\cos^4 x}.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "The given equation is equivalent to\n\n$$\n\\sqrt{1 + 3\\sin^3 x} + \\sqrt{1 - \\cos^4 x} = 3.\n$$\n\nWe may make these assumptions: $\\sin x \\leq 1$, i.e., $\\sin^3 x \\leq 1$, so $1 + 3\\sin^3 x \\leq 4$; and $\\cos x \\geq 0$, so $-\\cos^4 x \\leq 0$, i.e., $1 - \\cos^4 x \\leq 1$. Then for the left side of the equation we have\n\n$$\n\\sqrt{1 + 3\\sin^3 x} + \\sqrt{1 - \\cos^4 x} \\leq \\sqrt{4} + \\sqrt{1} = 3.\n$$\n\nThe equation is correct only if $1 + 3\\sin^3 x = 4$ and $1 - \\cos^4 x = 1$, i.e., $\\sin x = 1$ and $\\cos x = 0$, from which we get\n\n$$\nx = \\frac{\\pi}{2} + 2k\\pi, \\quad k \\in \\mathbb{Z}.\n$$",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 14418,
    "subject": "Mathematics (Olympiad)",
    "question": "Suppose you have a grid from point $A$ to point $C$, where you must take $2n$ steps: $n$ steps to the right and $n$ steps up. If there are no arcs of circles, how many different routes are there from $A$ to $C$? Now, suppose that before reaching the diagonal $BD$, at each of the first $n$ steps, you have three choices to move from one point to another (including arcs). How many total possible routes are there from $A$ to $C$ with these arcs included?",
    "options": [],
    "answer": "See solution",
    "solution": "Without arcs, each route from $A$ to $C$ consists of $2n$ steps, with $n$ right and $n$ up. The number of such routes is $\\binom{2n}{n}$. \n\nWith arcs, for each of the first $n$ steps before the diagonal, there are $3$ choices, so there are $3^n$ ways to choose these steps. Thus, the total number of possible routes is:\n\n$$\n3^n \\cdot \\binom{2n}{n}\n$$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 14419,
    "subject": "Mathematics (Olympiad)",
    "question": "For a positive integer $n$, let $f(n)$ denote the sum of all positive integers that are not greater than $n$ and not relatively prime to $n$. Show that $f(n+p) \\neq f(n)$ for every positive integer $n$ and every prime $p$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $m, n$ and $k$ be positive integers and $p$ be a prime. Since $(n, k) = 1 \\Leftrightarrow (n, n-k) = 1$, we have\n\n$$\nf(n) = \\frac{n(n+1)}{2} - \\frac{n \\cdot \\phi(n)}{2}\n$$\n\nwhere $\\phi(n)$ is the Euler function.\n\nSuppose $f(m) = f(n)$ for $m > n$ and $(m, n) = 1$. Then\n\n$$\nm(m+1-\\phi(m)) = n(n+1-\\phi(n)) \\implies m \\mid n+1-\\phi(n) \\implies m \\le n,\n$$\n\na contradiction. Therefore, if $f(n+p) = f(n)$, then $n = pk$ for some positive integer $k$ and\n\n$$\n(k+1)(pk+p+1-\\phi(pk+p)) = k(pk+1-\\phi(pk)).\n$$\n\nThis implies\n\n$$\npk+1-\\phi(pk) = a(k+1) \\quad (1)\n$$\n\nand\n\n$$\npk+p+1-\\phi(pk+p) = ak \\quad (2)\n$$\n\nfor some positive integer $a$. In particular,\n\n$$\n(p-a)k = a-1+\\phi(pk) \\implies p > a \\ge 1\n$$\n\nand\n\n$$\na = \\phi(pk+p) - \\phi(pk) - p \\equiv -1 \\pmod{p-1}.\n$$\n\nSo, $a = p - 2$. Substituting this in (1) and (2) we obtain\n\n$$\n\\phi(pk) = 2k+3-p \\quad (3)\n$$\n\nand\n\n$$\n\\phi(pk+p) = 2k+1+p. \\quad (4)\n$$\n\nWe shall prove that $p \\nmid k$ and $p \\nmid k+1$. If $p \\nmid k+1$, the relation (4) implies $p \\nmid k$, a contradiction. On the other hand, if $p \\nmid k$, then (3) implies $p \\nmid 3$. Setting $k = 3^\\alpha k_1$, where $k_1$ is an integer such that $3 \\nmid k_1$, and substituting this in (3) we get $\\phi(k_1) = k_1$. So, $k_1 = 1$ and $k = 3^\\alpha$. Now substituting this in (4) gives a contradiction. Using this observation, we obtain from (3) and (4) the equalities\n\n$$\n\\phi(k) = \\frac{2(k+1)}{p-1} - 1 \\quad (5)\n$$\n\nand\n\n$$\n\\phi(k+1) = \\frac{2(k+1)}{p-1} + 1. \\quad (6)\n$$\n\nIn particular, $p \\ge 5$ and $k+1 \\ge 6$. Moreover $\\phi(k+1) - \\phi(k) = 2$ and either $4 \\nmid \\phi(k)$ or $4 \\nmid \\phi(k+1)$.\n\nIf $4 \\nmid \\phi(k)$, then $k = q^\\beta$ or $k = 2q^\\beta$, where $q$ is an odd prime and $\\beta \\ge 1$ is an integer. From (5) we obtain\n\n$$\n\\frac{2(k+1)}{p-1} - 1 = \\phi(k) \\ge \\frac{1}{2} \\left(1 - \\frac{1}{q}\\right) k \\ge \\frac{1}{2} \\cdot \\frac{2}{3} k = \\frac{k}{3} \\implies p \\le 5 \\implies p=5.\n$$\n\nIf $p=5$, from (5) we obtain $\\phi(k) = \\frac{k-1}{2}$. So, $k$ must be odd, $k+1$ even and $\\phi(k+1) \\le \\frac{k+1}{2}$. This contradicts (6).\n\nIf $4 \\nmid \\phi(k+1)$, then $k+1 = q^\\beta$ or $k+1 = 2q^\\beta$, where $q$ is an odd prime and $\\beta \\ge 1$ is an integer. From (6) we obtain\n\n$$\n\\frac{2(k+1)}{p-1} + 1 = \\phi(k+1) \\ge \\frac{k+1}{3}.\n$$\n\nIf $p > 7$, then\n\n$$\n\\frac{6}{p-7} \\ge \\frac{2(k+1)}{p-1}\n$$\n\nand therefore by (5)\n\n$$\n\\phi(k) \\le \\frac{13-p}{p-7}\n$$\n\nwhich is impossible. On the other hand, if $p \\le 7$, then $p = 7$ and, as $\\frac{p-1}{2}$ divides $k+1$, we also have $q = 3$. Using (6) again, $k+1 = 2 \\cdot 3^0$ leads to a contradiction, and $k+1 = 3^0$ leads to $k+1 = 3$, which was already ruled out. $\\square$",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 14420,
    "subject": "Mathematics (Olympiad)",
    "question": "Show that for any positive integer $n$, the set of all rational numbers of the form $b + \\sum_{k=1}^n \\frac{1}{a_k}$, where $b$ is an integer and $a_1, \\dots, a_n$ are nonzero integers, does not exhaust all rational numbers. In fact, prove that for each $n$, there exist rational numbers $p_n < q_n$ such that no number of this form lies strictly between $p_n$ and $q_n$.",
    "options": [],
    "answer": "See solution",
    "solution": "If the first case holds, then $\\frac{1}{a_l} < p_{n+1} - p_n = \\frac{d}{3}$. If the second case holds, then $\\frac{1}{a_l} < q'_{n+1} - q_n = -\\frac{d}{3}$. In either case, $-\\frac{3}{d} < a_l < \\frac{3}{d}$, so for each $l \\in \\{1, \\dots, n+1\\}$, only finitely many integer values for $a_l$ are possible. Furthermore, for fixed $(a_1, \\dots, a_{n+1})$, only finitely many integers $b$ satisfy $p_{n+1} < b + \\sum_{k=1}^{n+1} \\frac{1}{a_k} < q'_{n+1}$. Thus, there are at most finitely many elements of $A_{n+1}$ between $p_{n+1}$ and $q'_{n+1}$. Define $q_{n+1}$ as the minimum such exceptional number of $A_{n+1}$ if it exists, or $q'_{n+1}$ otherwise. Then $p_{n+1}$ and $q_{n+1}$ satisfy the assertion for $n+1$, completing the induction and the proof.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 14421,
    "subject": "Mathematics (Olympiad)",
    "question": "給定首一實係數多項式 $P_1(x), \\dots, P_n(x)$。對於任意實數 $y$，定義集合\n$$\nS_y = \\{z \\in \\mathbb{R} \\mid \\text{存在某個 } i \\in \\{1, \\dots, n\\} \\text{ 使得 } y = P_i(z)\\}.\n$$\n\n證明：若對於任意兩個相異實數 $y_1, y_2$，集合 $S_{y_1}, S_{y_2}$ 的元素個數相等，則 $P_1, \\dots, P_n$ 有相同的次數。\n\n註：首一多項式即最高次項係數為 1 的多項式。",
    "options": [],
    "answer": "See solution",
    "solution": "不失一般性，假設 $P_1, \\dots, P_n$ 彼此不同。首先證明對任意 $y \\in \\mathbb{R}$，$|S_y| = n$，且每個 $P_i$ 的次數皆為奇數。設 $a$ 為奇次多項式的個數，$b$ 為偶次多項式的個數。令\n$$\nY := \\{P_i(x) \\mid x \\in \\mathbb{R},\\ P_i(x) = P_j(x)\\ \\text{for some } j \\neq i\\}\n$$\n為有限集合。可取 $M > 0$，使 $Y$ 中所有元素絕對值皆小於 $M$。進一步放大 $M$，使對任意 $y > M$ 及 $i = 1, \\dots, n$，有：\n$$\n|\\{x \\in \\mathbb{R} \\mid P_i(x) = y\\}| = \\begin{cases} 1, & 2 \\nmid \\deg P_i(x); \\\\ 2, & 2 \\mid \\deg P_i(x) \\end{cases}\n$$\n對任意 $y < -M$ 及 $i = 1, \\dots, n$，有：\n$$\n|\\{x \\in \\mathbb{R} \\mid P_i(x) = y\\}| = \\begin{cases} 1, & 2 \\nmid \\deg P_i(x); \\\\ 0, & 2 \\mid \\deg P_i(x) \\end{cases}\n$$\n因此 $y > M$ 時 $|S_y| = a + 2b$，$y < -M$ 時 $|S_y| = a$。由假設 $|S_y|$ 恆定，得 $b = 0$，$a = n$，即所有 $P_i$ 皆為奇次，且 $|S_y| = n$。\n\n既然每個 $P_i$ 為奇次，對任意 $y \\in \\mathbb{R}$ 及 $i$，必存在 $x$ 使 $P_i(x) = y$。對 $y \\notin Y$ 及任意 $i$，恰有一個 $x$ 使 $P_i(x) = y$。因此每個 $P_i$ 除有限點外嚴格單調，作為多項式則必處處嚴格單調，故 $Y$ 必為空。若 $\\deg P_i \\ne \\deg P_j$，則 $P_i - P_j$ 仍為奇次，必有實根，導致 $Y$ 非空，矛盾。因此所有 $P_i$ 次數相同，得證。",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 14422,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $n$ be a positive integer. Find the smallest $n$ such that there exists a sequence $a_1, a_2, \\dots, a_n$ of real numbers in the interval $(-1, 1)$ satisfying\n\n- $a_1 + a_2 + \\dots + a_n = 0$\n- $a_1^2 + a_2^2 + \\dots + a_n^2 = 40$.",
    "options": [],
    "answer": "See solution",
    "solution": "Note that $40 = a_1^2 + a_2^2 + \\dots + a_n^2 < 1 + 1 + \\dots + 1 = n$, so $n > 40$ and thus $n \\geq 41$.\n\nSuppose $n = 41$. Then there exists a sequence $a_1, a_2, \\dots, a_{41}$ in $(-1, 1)$ with $a_1 + \\dots + a_{41} = 0$ and $a_1^2 + \\dots + a_{41}^2 = 40$. Assume the sequence is increasing: $a_1 \\leq \\dots \\leq a_{41}$, so $a_1 \\leq 0 \\leq a_{41}$. By minimality, $a_i \\neq 0$ for all $i$, so there is a unique $k$ such that $-1 < a_1 \\leq \\dots \\leq a_k < 0 < a_{k+1} \\leq \\dots \\leq a_{41} < 1$. Also, $-a_1, \\dots, -a_{41}$ is such a sequence, so $k \\leq \\lfloor \\frac{41}{2} \\rfloor = 20$.\n\nFor $k+1 \\leq i \\leq n$, $0 < a_i^2 < a_{i+1}$. Then:\n\n$$\n\\begin{align*}\n40 &= a_1^2 + \\dots + a_{41}^2 \\\\\n&= (a_1^2 + \\dots + a_k^2) + (a_{k+1}^2 + \\dots + a_{41}^2) \\\\\n&< (a_1^2 + \\dots + a_k^2) + (a_{k+1} + \\dots + a_{41}) \\\\\n&= (a_1^2 + \\dots + a_k^2) - (a_1 + \\dots + a_k) \\\\\n&< 2k \\leq 40.\n\\end{align*}\n$$\n\nThis contradiction implies $n \\geq 42$.\n\nTo show $n = 42$ is possible, take $a_i = -\\sqrt{\\frac{20}{21}}$ for $1 \\leq i \\leq 21$, and $a_i = \\sqrt{\\frac{20}{21}}$ for $22 \\leq i \\leq 42$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 14423,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $ABC$ be an acute-angled triangle with $AC > AB$ and let $D$ be the foot of the $A$-angle bisector on $BC$. The reflections of lines $AB$ and $AC$ in line $BC$ meet $AC$ and $AB$ at points $E$ and $F$ respectively. Let $\\ell$ be a line through $D$ meeting $AC$ and $AB$ at $G$ and $H$ respectively such that $G$ lies strictly between $A$ and $C$ while $H$ lies strictly between $B$ and $F$. Prove that the circumcircles of $\\triangle EDG$ and $\\triangle FDH$ are tangent to each other.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $X$ and $Y$ lie on the tangent to the circumcircle of $\\triangle EDG$ on the opposite side to $D$ as shown in the figure below. Regarding diagram dependency, the acute condition with $AC > AB$ ensures $E$ lies on the extension of $CA$ beyond $A$, and $F$ lies on the extension of $AB$ beyond $B$. The condition on $\\ell$ means the points lie in the orders $E, A, G, C$ and $A, B, H, F$.\n\n![](images/BMO2024Shortlist_p45_data_3707b1ea8b.png)\n\nUsing two applications of the alternate segment theorem, the condition that $\\odot EDG$ and $\\odot FDH$ are tangent at $D$ can be rewritten as\n\n$$\n\\triangle XDG \\cong \\triangle YDH \\Leftrightarrow \\triangle DEG \\cong \\triangle DFH.\n$$\n\nSo we can remove $G, H$ from the figure, and we will prove that $\\triangle DEA \\cong \\triangle DFB$.\n\nThe reflection property means that *AD* and *BD* are external angle bisectors in $\\angle EAB$ and hence $D$ is the $E$-excentre of this triangle. Thus $DE$ (internally) bisects $\\angle BEA$, giving\n\n$$\n\\triangle DEA = \\frac{1}{2} \\triangle BEA = \\frac{1}{2} \\triangle BEC.\n$$\n\nSimilarly, in $\\angle FAC$, $AD$ and $CD$ are both internal angle bisectors, and so $D$ is the incentre of this triangle and $FD$ is also the internal angle bisector of $\\angle CFH$, giving\n\n$$\n\\triangle DFB = \\triangle DFA = \\frac{1}{2} \\triangle CFA = \\frac{1}{2} \\triangle CFB.\n$$\n\nNow observe that the pairs of lines ($BE$, $CE$) and ($BF$, $CF$) are reflections in $BC$ thus $E$, $F$ are reflections in $BC$. Hence $\\angle BEC = \\angle CFB$ and so\n\n$$\n\\triangle DEA = \\frac{1}{2} \\triangle BEC = \\frac{1}{2} \\triangle CFB = \\triangle DFB,\n$$\n\nas required.\n\n![](images/BMO2024Shortlist_p46_data_976b296c70.png)",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 14424,
    "subject": "Mathematics (Olympiad)",
    "question": "Denote by $\\mathbb{R}^+$ the set of all positive real numbers. Determine all functions $f: \\mathbb{R}^+ \\to \\mathbb{R}^+$ satisfying the equation:\n\n$$\nf\\left(\\frac{f(y)}{f(x)} + 1\\right) = f\\left(x + \\frac{y}{x} + 1\\right) - f(x) \\quad \\text{for every pair } x, y \\in \\mathbb{R}^+.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "If we substitute $y = x$ in the given equation, we obtain $f(x+2) - f(x) = f(2) > 0$.\n\nLet us first show that $f$ is injective. Let $x_1, x_2$ be positive integers satisfying $x_2 - x_1 > 1$. If we make substitutions $x = x_1$, $y = x_1(x_2 - x_1 - 1)$ in the given equation, then we get a positive number on the left-hand side, and $f(x_2) - f(x_1)$ on the right-hand side. Consequently, if $f(x_1) = f(x_2)$ is satisfied for any pair of positive integers $x_1, x_2$, then $|x_1 - x_2| \\le 1$ must hold.\n\nSuppose for a pair $a, b$ of positive numbers with $a \\ne b$, $f(a) = f(b)$ is satisfied. Then, if we compare the two equations obtained by substituting $x = a$ and $x = b$ in the given equation, we get for any positive number $y$,\n\n$$\nf\\left(a + \\frac{y}{a} + 1\\right) = f\\left(b + \\frac{y}{b} + 1\\right).\n$$\n\nBut if we take $y$ to be sufficiently large, then $|f(a + \\frac{y}{a} + 1)| = |f(b + \\frac{y}{b} + 1)| = |y\\left(\\frac{1}{a} - \\frac{1}{b}\\right) + a - b| > 1$, which contradicts what we have shown above. Therefore, we conclude that $f$ is injective.\n\nBy substituting $x = 2$ in the given identity, we get\n\n$$\nf\\left(\\frac{f(y)}{f(2)} + 1\\right) = f\\left(\\frac{y}{2} + 3\\right) - f(2) = f\\left(\\frac{y}{2} + 1\\right).\n$$\n\nThen, since $f$ is injective, we get $\\frac{f(y)}{f(2)} + 1 = \\frac{y}{2} + 1$, from which it follows that\n\n$$\nf(y) = \\frac{f(2)}{2}y.\n$$\n\nThus, we can write $f(x) = a x$ for a positive constant $a$. Since we can check that for any positive real number $a$, $f(x) = a x$ satisfies the given equation, we conclude that $f(x) = a x$ is the desired answer to the problem.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 14425,
    "subject": "Mathematics (Olympiad)",
    "question": "There are 10 motel rooms in a row, numbered 1 to 10.\n\n![](images/2020_Australian_Scene_W_p51_data_f589117cac.png)\n\nA pair of rooms may have 0 to 8 rooms between them.\n\nFor each possible number of rooms between a pair, the number of such pairs is as follows:\n\n- 0 rooms between: 9 pairs — (1, 2), (2, 3), (3, 4), (4, 5), (5, 6), (6, 7), (7, 8), (8, 9), (9, 10)\n- 1 room between: 8 pairs — (1, 3), (2, 4), (3, 5), (4, 6), (5, 7), (6, 8), (7, 9), (8, 10)\n- 2 rooms between: 7 pairs — (1, 4), (2, 5), (3, 6), (4, 7), (5, 8), (6, 9), (7, 10)\n- 3 rooms between: 6 pairs — (1, 5), (2, 6), (3, 7), (4, 8), (5, 9), (6, 10)\n- 4 rooms between: 5 pairs — (1, 6), (2, 7), (3, 8), (4, 9), (5, 10)\n- 5 rooms between: 4 pairs — (1, 7), (2, 8), (3, 9), (4, 10)\n- 6 rooms between: 3 pairs — (1, 8), (2, 9), (3, 10)\n- 7 rooms between: 2 pairs — (1, 9), (2, 10)\n- 8 rooms between: 1 pair — (1, 10)\n\nIf rooms are allocated at random, all pairs of rooms occupied by Molly and Polly are equally likely. What is the most likely number of rooms between Molly and Polly?",
    "options": [],
    "answer": "See solution",
    "solution": "From the counts above, there are 9 pairs of rooms with 0 rooms between them, 8 pairs with 1 room between, and so on, for a total of $9 + 8 + 7 + \\cdots + 1 = 45$ pairs.\n\nThe probability that there are at most 1 room between Molly and Polly is $\\frac{9 + 8}{45} = \\frac{17}{45} > \\frac{1}{3}$. Thus, the most likely number of rooms between Molly and Polly is zero.\n\n---\n\nFor the extension: If three people (Molly, Polly, and Ollie) are each allocated a room at random, what is the probability that all three are within a block of five consecutive rooms?\n\nThere are $\\binom{10}{3} = 120$ ways to choose 3 rooms, and $6$ ways to assign the people to those rooms, for $720$ total allocations. The number of allocations where all three are within a block of five consecutive rooms is $40$, so the probability is $\\frac{40}{120} = \\frac{1}{3}$.\n\n---\n\nFor the further extension: For what arrangements of Molly and Polly's rooms is there a 50% chance that Ollie is in a room adjacent to one of them?\n\nIf Molly and Polly are in rooms $A$ and $B$ (with $A < B$), and there are at least two rooms between them, and $A \\geq 2$, $B \\leq 9$, then there are $5 + 4 + 3 + 2 + 1 = 15$ such pairs.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 14426,
    "subject": "Mathematics (Olympiad)",
    "question": "Determine all functions $f: \\mathbb{R} \\to \\mathbb{R}$ such that, for all real numbers $x$ and $y$,\n\n$$\nf(x^2 + xy) = f(x)f(y) + yf(x) + xf(x + y).\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "Set $x = 0$:\n\n$$\nf(0) = f(0)f(y) + y f(0).\n$$\n\nIf $f(0) \\neq 0$, then $1 = f(y) + y$, so $f(y) = 1 - y$ for all $y$. Plugging this into the original equation:\n\n$$\n1 - x^2 - x y = (1 - x)(1 - y) + y(1 - x) + x(1 - x - y),\n$$\n\nwhich holds for all $x, y$.\n\nIf $f(0) = 0$, substitute $x + y = z$ to get:\n\n$$\nf(xz) = f(x)f(z - x) + (z - x)f(x) + x f(z). \\quad (1)\n$$\n\nSwitch $x$ and $z$:\n\n$$\nf(zx) = f(z)f(x - z) + (x - z)f(z) + z f(x). \\quad (2)\n$$\n\nSubtracting (1) and (2):\n\n$$\nf(x)f(z - x) + (z - x)f(x) + x f(z) = f(z)f(x - z) + (x - z)f(z) + z f(x),\n$$\n\nwhich simplifies to:\n\n$$\nf(x)(f(z - x) - x) = f(z)(f(x - z) - z). \\quad (3)\n$$\n\nSet $z = 0$:\n\n$$\nf(x)(f(-x) - x) = 0.\n$$\n\nSo for each $x$, either $f(x) = 0$ or $f(-x) = x$.\n\nIf there exists $x \\neq 0$ with $f(x) = 0$, then for $z \\neq 0$, $f(x - z) \\in \\{0, z - x\\}$, so $f(x - z) \\neq z$. Thus, (3) implies $f(z) = 0$ for all $z \\neq 0$. Therefore, $f = 0$ is a solution.\n\nOtherwise, $f(x) \\neq 0$ for all $x \\neq 0$, so $f(-x) = x$ for all $x$, i.e., $f(x) = -x$ for all $x$. This is also a solution.\n\nThus, the solutions are:\n\n- $f(x) = 0$\n- $f(x) = -x$\n- $f(x) = 1 - x$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 14427,
    "subject": "Mathematics (Olympiad)",
    "question": "A sequence $\\{x_n\\}$ is defined as follows:\n\n$$\nx_1 = 2, \\quad x_{n+1} = \\sqrt{x_n + 8} - \\sqrt{x_n + 3},\n$$\nfor all positive integers $n$.\n\n**a)** Prove that $\\{x_n\\}$ has a finite limit and find that limit.\n\n**b)** For every positive integer $n$, prove that\n\n$$\nn \\leq x_1 + x_2 + \\dots + x_n \\leq n + 1.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "**a)** It is easy to see that $x_n > 0$ for all $n \\in \\mathbb{N}^*$. For every positive integer $n$, we have\n\n$$\n\\begin{aligned}\n|x_{n+1} - 1| &= |\\sqrt{x_n + 8} - 3 + 2 - \\sqrt{x_n + 3}| \\\\\n&= |(x_n - 1)\\left(\\frac{1}{\\sqrt{x_n + 8} + 3} - \\frac{1}{\\sqrt{x_n + 3} + 2}\\right)| \\\\\n&\\leq |x_n - 1|\\left(\\frac{1}{\\sqrt{x_n + 8} + 3} + \\frac{1}{\\sqrt{x_n + 3} + 2}\\right) \\\\\n&\\leq |x_n - 1|\\left(\\frac{1}{3} + \\frac{1}{2}\\right) \\\\\n&= \\frac{5}{6}|x_n - 1|.\n\\end{aligned}\n$$\n\nTherefore,\n\n$$\n|x_n - 1| \\leq \\frac{5}{6}|x_{n-1} - 1| \\leq \\dots \\leq \\left(\\frac{5}{6}\\right)^{n-1} |x_1 - 1| = \\left(\\frac{5}{6}\\right)^n, \\quad \\forall n \\in \\mathbb{N}^*.\n$$\n\nNote that $\\lim_{n \\to \\infty} \\left(\\frac{5}{6}\\right)^n = 0$, so $\\lim_{n \\to \\infty} x_n = 1$.\n\n**b)** Consider the function\n\n$$\nf(x) = \\sqrt{x+8} - \\sqrt{x+3} = \\frac{5}{\\sqrt{x+8} + \\sqrt{x+3}},\n$$\n\nwith $x > 0$. We see that $f(x)$ is a continuous and decreasing function on $(0, +\\infty)$. Because $x_1 > 1$, then $x_2 = f(x_1) < f(1) = 1$, and $x_3 = f(x_2) > f(1) = 1$, and so on. In general, we can prove $x_{2k} < 1 < x_{2k-1}$ for all positive integers $k$.\n\nNow, consider the function $g(x) = x + f(x) = x + \\sqrt{x+8} - \\sqrt{x+3}$ with $x > 0$. We get $g(x)$ is a continuous function and\n\n$$\ng'(x) = 1 + \\frac{1}{2\\sqrt{x+8}} - \\frac{1}{2\\sqrt{x+3}} > 1 - \\frac{1}{2\\sqrt{3}} > 0, \\quad \\forall x > 0,\n$$\n\nso $g(x)$ is an increasing function on $(0, \\infty)$. From here, we have the following claims:\n\n- If $x > 1$ then $g(x) > g(1) = 2$.\n- If $0 < x < 1$ then $g(x) < g(1) = 2$.\n\nHence,\n\n$$\nx_{2k-1} + x_{2k} > 2 > x_{2k} + x_{2k+1}, \\quad \\forall k \\in \\mathbb{N}^*.\n$$\n\nNow, we will prove the given inequality. Consider two cases:\n\n- **Case 1:** $n = 2k$ ($k \\in \\mathbb{N}^*$). It is easy to check that $2 < x_1 + x_2 < 3$, so the given inequality is true when $k = 1$. Assume that $k > 1$, we have\n\n$$\nx_1 + x_2 + \\dots + x_n = (x_1 + x_2) + (x_3 + x_4) + \\dots + (x_{2k-1} + x_{2k}) > 2 + 2 + \\dots + 2 = 2k\n$$\nand\n$$\nx_1 + x_2 + \\dots + x_n = x_1 + (x_2 + x_3) + \\dots + (x_{2k-2} + x_{2k-1}) + x_{2k} < 2 + 2 + \\dots + 2 + 1 = 2k + 1.\n$$\n\n- **Case 2:** $n = 2k - 1$ ($k \\in \\mathbb{N}^*$). Clearly, the given inequality is true when $k = 1$. Suppose that $k > 1$, we have\n\n$$\nx_1 + x_2 + \\dots + x_n = (x_1 + x_2) + \\dots + (x_{2k-3} + x_{2k-2}) + x_{2k-1} > 2 + 2 + \\dots + 2 + 1 = 2k - 1\n$$\nand\n$$\nx_1 + x_2 + \\dots + x_n = x_1 + (x_2 + x_3) + \\dots + (x_{2k-2} + x_{2k-1}) < 2 + 2 + \\dots + 2 = 2k.\n$$\n\nTo summarize, we have $n \\leq x_1 + x_2 + \\dots + x_n \\leq n + 1$ for all positive integers $n$. $\\square$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 14428,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $x, y, z$ be positive real numbers. Prove that\n\n$$\n\\frac{x}{\\sqrt{2(x^2 + y^2)}} + \\frac{y}{\\sqrt{2(y^2 + z^2)}} + \\frac{z}{\\sqrt{2(z^2 + x^2)}} < \\frac{4x^2 + y^2}{x^2 + 4y^2} + \\frac{4y^2 + z^2}{y^2 + 4z^2} + \\frac{4z^2 + x^2}{z^2 + 4x^2} < 9.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "To prove the second inequality, we may assume $x = \\max\\{x, y, z\\}$. Then we get\n\n$$\n\\frac{4z^2 + x^2}{z^2 + 4x^2} \\le 1, \\quad \\frac{4x^2 + y^2}{x^2 + 4y^2} < 4, \\quad \\text{and} \\quad \\frac{4y^2 + z^2}{y^2 + 4z^2} < 4.\n$$\n\nSo the desired inequality follows.\n\nNext we prove the first inequality. By the AM-GM inequality, we have $4xy^2 \\le y^3 + 4x^2y$. Then\n\n$$\n\\begin{aligned}\ny^3 + 4x^2y + 3x^3 &> y^3 + 4x^2y \\ge 4xy^2 > 3xy^2 \\\\\n\\iff y^3 + 4x^2y + 4x^3 + xy^2 &> 4xy^2 + x^3 \\\\\n\\iff \\frac{4x^2 + y^2}{x^2 + 4y^2} &> \\frac{x}{x+y}.\n\\end{aligned}\n$$\n\nHence\n\n$$\n\\sum_{cyc} \\frac{x}{x+y} < \\sum_{cyc} \\frac{4x^2 + y^2}{x^2 + 4y^2}.\n$$\n\nNow apply the Cauchy-Schwarz inequality, we have\n\n$$\n\\sum_{cyc} \\frac{x}{\\sqrt{2(x^2 + y^2)}} = \\sum_{cyc} \\frac{x}{\\sqrt{(1+1)(x^2 + y^2)}} \\le \\sum_{cyc} \\frac{x}{x+y} < \\sum_{cyc} \\frac{4x^2 + y^2}{x^2 + 4y^2},\n$$\n\nwhich is the first inequality. $\\square$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 14429,
    "subject": "Mathematics (Olympiad)",
    "question": "We are given the set $M_n = \\{0, 1, 2, \\dots, n\\}$ of all non-negative integers less than or equal to $n$. A subset $S$ of $M_n$ is called *outstanding* if it is not empty and, for every $k \\in S$, there exists a $k$-element subset of $S$. Determine the number of outstanding subsets of $M_n$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $k$ be the largest element of an outstanding subset $S$. Then $S$ must contain $k$ elements. This is possible if $S$ contains all elements not greater than $k$, or all except one. Each outstanding subset of $M_n$ corresponds to an ordered pair $(a, b)$ of integers with $n \\ge a \\ge b \\ge 0$. The case of an $(a+1)$-element subset with maximum element $a$ is denoted by $(a, a)$, and an $a$-element subset with maximum element $a$ and missing the number $b$ is denoted by $(a, b)$. Each such pair corresponds directly to an outstanding subset. Thus, the number of outstanding subsets equals the number of such ordered pairs, which is the number of 2-element subsets of $M_n$ plus the number of elements of $M_n$:\n\n$$\n\\binom{n+1}{2} + (n+1) = \\binom{n+2}{2}\n$$\n\nqed",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 14430,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $k$ and $n$ be arbitrary natural numbers such that $3 \\leq k \\leq n$. Prove that among any $n$ pairwise distinct real numbers, there are either $k$ numbers with a positive sum or $(k-1)$ numbers with a negative sum.",
    "options": [],
    "answer": "See solution",
    "solution": "If there are no positive numbers in the set, then at most one number can be zero, and all others are negative. Thus, there are $(n-1)$ negative numbers, so we can select any subset of $k-1 \\geq 2$ numbers as required. Otherwise, if there is at least one positive number, separate one such number. Among the remaining $(n-1)$ numbers, select any $k-1$ numbers. If their sum is negative, we have found the desired subset. If their sum is non-negative, add the separated positive number to this subset, obtaining $k$ numbers with a positive sum.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 14431,
    "subject": "Mathematics (Olympiad)",
    "question": "Suppose a list $L = (n_i)_{i=1}^N$ is made up only of numbers selected from the (possibly much shorter) list of distinct numbers $(m_i)_{i=1}^k$, and that the products $n_i n_{i+1}$, $1 \\leq i \\leq 2010$, are all distinct. We write down the indices of the numbers $m_i$ occurring in $L$ as another list $M$ also of length $N$, e.g. if $L$ begins $m_3, m_7, m_1, m_9, m_7, \\dots$, then $M$ begins $3, 7, 1, 9, 7, \\dots$.\n\nIf $n_i = m_a$ and $n_{i+1} = m_b$, the product $n_i n_{i+1}$ is determined by the set $\\{a, b\\}$. Hence, any two (not necessarily distinct) numbers $a, b$ from $\\{1, 2, \\dots, k\\}$ can appear as neighbours in the list $M$ at most once. As $M$ contains 2011 elements, there must exist at least 2010 different subsets of $\\{1, 2, \\dots, k\\}$ with one or two elements. Note that, by usual conventions of set theory, $\\{a, a\\} = \\{a\\}$ contains only one element. The number of such subsets is equal to $k(k+1)/2$ and so we need to have $k(k+1)/2 \\geq 2010$, i.e. $k \\geq 63$.\n\n**Question:** In how many ways (up to rotation) can we write the numbers $1, 2, 3, \\dots, k$ at the vertices of a regular $k(k+1)/2$-gon such that there do not exist two distinct edges with the same set of numbers written at the vertices they connect?",
    "options": [],
    "answer": "See solution",
    "solution": "To prove the existence of such a cycle, we first form the set $\\mathcal{T}$ of all subsets of $\\{1, 2, \\dots, k\\}$ with one or two elements. Note that each number appears $k-1$ times in a two-element subset and once in a singleton. We start the sequence with $1, 2, 3, \\dots, k$ and remove from $\\mathcal{T}$ the sets $\\{1, 2\\}, \\{2, 3\\}, \\dots, \\{k-1, k\\}$ which appear as neighbours already. We repeatedly do now the following:\n\n* If $a$ is the last element of our list and there is a set in $\\mathcal{T}$ which contains $a$, we pick such a set, remove it from $\\mathcal{T}$ and append $a$ (if the set was a singleton) or the second element of this set (if it was a two-element set) to the list.\n\n* Because $k-1$ is even, if $\\mathcal{T}$ is not empty but does not contain a set which contains the last element of our list, then the last element of our list must coincide with the first element of our list, i.e. the list is a cycle. In this case, we replace it by an equivalent cycle which has an element as its first and last element which appears in one of the sets in $\\mathcal{T}$. This is possible because we started our list in such a way that it contains all numbers $1, \\dots, k$ at least once. After replacing the cycle by an appropriate equivalent one, we continue as above appending elements to the list.\n\nThis process can be continued until all elements of $\\mathcal{T}$ are used. The result is a list which contains all possible neighbour-subsets exactly once. If $k=63$, the list will have length $63 \\cdot 64/2 + 1 = 2017$. By discarding the last 6 elements we obtain a list $M$ with 2011 elements.\n\nLet $M = (\\mu_i)_{i=1}^{2011}$ be this list and let $m_i$ be the $i$-th prime number. Defining $n_i = m_{\\mu_i}$ we obtain a list $L = (n_i)_{i=1}^{2011}$ which satisfies the conditions of the problem.",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 14432,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $a$, $b$, $c$ be integers such that\n$$\n\\frac{ab}{c} + \\frac{ac}{b} + \\frac{bc}{a}\n$$\nis an integer.\n\nProve that each of the numbers\n$$\n\\frac{ab}{c}, \\frac{ac}{b}, \\text{ and } \\frac{bc}{a}\n$$\nis an integer.",
    "options": [],
    "answer": "See solution",
    "solution": "Set $u := \\frac{ab}{c}$, $v := \\frac{ac}{b}$, and $w := \\frac{bc}{a}$. By assumption, $u + v + w$ is an integer. It is easily seen that $uv + uw + vw = a^2 + b^2 + c^2$ and $uvw = abc$ are integers as well.\n\nAccording to Vieta's formulas, the rational numbers $u$, $v$, $w$ are the roots of a cubic polynomial $x^3 + px^2 + qx + r$ with integer coefficients. As the leading coefficient is $1$, these roots must be integers.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 14433,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $c_0, c_1, \\dots, c_{2n+1}$ be numbers selected from $a_0, a_1, \\dots, a_{2n+1}$ such that:\n\n- $c_{2n+1} = b_{2n+1}$, $c_{2n-1} = b_{2n-1}$, ..., $c_1 = b_{n+1}$\n- $c_2 = b_n$, $c_{2n-2} = b_{n-1}$, ..., $c_0 = b_0$\n\nFor the set $c_0, c_1, \\dots, c_{2n+1}$, we have:\n\n$$\nc_0 \\le c_1 \\ge c_2 \\le c_3 \\ge \\dots \\ge c_{2n} \\le c_{2n+1}.\n$$\n\nConsider the polynomial:\n\n$$\nq(x) = c_{2n+1}x^{2n+1} + c_{2n}x^{2n} + c_{2n-1}x^{2n-1} + \\dots + c_1x + c_0.\n$$\n\nShow that $q(x)$ has at most one real root.",
    "options": [],
    "answer": "See solution",
    "solution": "Since the degree of $q(x)$ is odd and its leading coefficient $c_{2n+1}$ is positive, for large $x$ we have $q(x) > 0$ and $q(-x) < 0$, so $q(x)$ has at least one real root.\n\nIf $q(x)$ is strictly increasing, then it can have at most one real root. It suffices to prove that $q'(x) > 0$ for all $x$.\n\nThe derivative is:\n\n$$\nq'(x) = (2n + 1)c_{2n+1}x^{2n} + 2nc_{2n}x^{2n-1} + \\dots + 2c_2x + c_1.\n$$\n\nFor all $k \\in \\{1, 2, \\dots, n\\}$ and real $x$,\n\n$$\n\\begin{aligned}\n& kc_{2k+1}x^{2k} + 2kc_{2k}x^{2k-1} + kc_{2k-1}x^{2k-2} \\\\\n&\\ge kc_{2k}(|x|^{2k} - 2|x|^{2k-1} + |x|^{2k-2}) = kc_{2k}|x|^{2k-2}(|x|-1)^2 \\ge 0\n\\end{aligned}\n$$\n\nTherefore,\n\n$$\n\\begin{aligned}\nq'(x) &= (n+1)c_{2n+1}x^{2n} + (nc_{2n+1}x^{2n} + 2nc_{2n}x^{2n-1} + nc_{2n-1}x^{2n-2}) \\\\\n&\\quad + ((n-1)c_{2n-1}x^{2n-2} + 2(n-1)c_{2n-2}x^{2n-3} + (n-1)c_{2n-3}x^{2n-4}) + \\dots \\\\\n&\\quad + (c_3x^2 + 2c_2x + c_1) \\\\\n&\\ge (n+1)c_{2n+1}x^{2n} > 0\n\\end{aligned}\n$$\n\nfor $x \\ne 0$. Since $q'(0) = c_1 > 0$, we obtain $q'(x) > 0$ for all real $x$. Thus, $q(x)$ has at most one real root.",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 14434,
    "subject": "Mathematics (Olympiad)",
    "question": "A right pyramid has a regular octagon $ABCDEFGH$ with side length $1$ as its base and apex $V$. Segments $\\overline{AV}$ and $\\overline{DV}$ are perpendicular. What is the square of the height of the pyramid?\n\n(A) $1$  \n(B) $\\frac{1+\\sqrt{2}}{2}$  \n(C) $\\sqrt{2}$  \n(D) $\\frac{3}{2}$  \n(E) $\\frac{2+\\sqrt{2}}{3}$",
    "options": [],
    "answer": "See solution",
    "solution": "**Answer (B):** Let $O$ be the center of the octagon, and let $r = AO$. As can be seen from the figure below, $AD = 1 + \\sqrt{2}$.\n\n![](images/2024_AMC12B_Solutions_p17_data_7eb09b6151.png)\n\nBecause $\\triangle AVD$ is an isosceles right triangle,\n\n$$\nAV = \\frac{\\sqrt{2}}{2} \\cdot AD = \\frac{2+\\sqrt{2}}{2}.\n$$\n\nApplying the Law of Cosines to $\\triangle AOH$ yields\n\n$$\nr^2 + r^2 = 1^2 + 2r^2 \\cos 45^\\circ = 1 + r^2 \\sqrt{2}.\n$$\n\nThus $r^2(2 - \\sqrt{2}) = 1$ and\n\n$$\nr^2 = \\frac{1}{2 - \\sqrt{2}} = \\frac{2 + \\sqrt{2}}{2}.\n$$\n\nThe Pythagorean Theorem applied to $\\triangle VOA$ gives the requested square of the height of the pyramid:\n\n$$\nVO^2 = AV^2 - r^2 = \\left(\\frac{2+\\sqrt{2}}{2}\\right)^2 - \\frac{2+\\sqrt{2}}{2} = \\frac{1+\\sqrt{2}}{2}.\n$$\n\nOR\n\nPlace the figure in a three-dimensional coordinate system with the center $O$ of the base of the pyramid at the origin, the octagon in the $x$-$y$ plane with positive $x$ coordinates for $A$, $B$, $C$, and $D$, the $y$-axis parallel to $\\overline{AD}$, and apex $V(0, 0, h)$ on the positive $z$-axis. See the figure.\n\n![](images/2024_AMC12B_Solutions_p18_data_aa7a96184a.png)\n\nThen consider vectors\n\n$$\n\\overrightarrow{OV} = \\langle 0, 0, h \\rangle, \\quad \\overrightarrow{OA} = \\left\\langle \\frac{1}{2}, -\\frac{1+\\sqrt{2}}{2}, 0 \\right\\rangle, \\quad \\text{and} \\quad \\overrightarrow{OD} = \\left\\langle \\frac{1}{2}, \\frac{1+\\sqrt{2}}{2}, 0 \\right\\rangle.\n$$\n\nIt follows that\n\n$$\n\\overrightarrow{AV} = \\overrightarrow{OV} - \\overrightarrow{OA} = \\left\\langle -\\frac{1}{2}, \\frac{1+\\sqrt{2}}{2}, h \\right\\rangle\n$$\n\nand\n\n$$\n\\overrightarrow{DV} = \\overrightarrow{OV} - \\overrightarrow{OD} = \\left\\langle -\\frac{1}{2}, -\\frac{1+\\sqrt{2}}{2}, h \\right\\rangle.\n$$\n\nBecause $\\overrightarrow{AV}$ and $\\overrightarrow{DV}$ are perpendicular, their dot product is zero. Therefore\n\n$$\n0 = \\left\\langle -\\frac{1}{2}, \\frac{1+\\sqrt{2}}{2}, h \\right\\rangle \\cdot \\left\\langle -\\frac{1}{2}, -\\frac{1+\\sqrt{2}}{2}, h \\right\\rangle = \\frac{1}{4} - \\frac{3+2\\sqrt{2}}{4} + h^2,\n$$\n\nfrom which $h^2 = \\frac{1+\\sqrt{2}}{2}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 14435,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $a, b, c > 0$ such that $ab + bc + ca = 1$. Prove that\n$$\n\\frac{2}{abc} + 9abc \\ge 7(a + b + c).\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "The inequality to prove is equivalent to $2 + 9(abc)^2 \\ge 7abc(a + b + c)$. Let $bc = x$, $ca = y$, $ab = z$. It suffices to show that, for any $x, y, z > 0$ so that $x + y + z = 1$, we have $9xyz + 2 \\ge 7(xy + yz + zx)$.  \n\nSince $x + y + z = 1$, we can rewrite this as $9xyz + 2(x + y + z)^3 \\ge 7(x + y + z)(xy + yz + zx)$. Doing some minor computations, this becomes $2(x^3 + y^3 + z^3) \\ge x^2y + xy^2 + y^2z + yz^2 + z^2x + xz^2$, which can be easily obtained by adding the well-known inequality $x^3 + y^3 \\ge xy(x + y)$ with the similar inequalities obtained through cyclic permutations.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 14436,
    "subject": "Mathematics (Olympiad)",
    "question": "Prove that:\n\na) $\\left(\\frac{1}{2}\\right)^3 + \\left(\\frac{2}{3}\\right)^3 + \\left(\\frac{5}{6}\\right)^3 = 1$;\n\nb) $3^{33} + 4^{33} + 5^{33} < 6^{33}$.",
    "options": [],
    "answer": "See solution",
    "solution": "a) A straightforward computation proves the claim.\n\nb) It suffices to show that\n\n$$\n\\frac{3^{33}}{6^{33}} + \\frac{4^{33}}{6^{33}} + \\frac{5^{33}}{6^{33}} < 1,\n$$\nthat is,\n$$\n\\left(\\frac{1}{2}\\right)^{33} + \\left(\\frac{2}{3}\\right)^{33} + \\left(\\frac{5}{6}\\right)^{33} < 1.\n$$\nBut\n$$\n\\left(\\frac{1}{2}\\right)^{33} + \\left(\\frac{2}{3}\\right)^{33} + \\left(\\frac{5}{6}\\right)^{33} < \\left(\\frac{1}{2}\\right)^{3} + \\left(\\frac{2}{3}\\right)^{3} + \\left(\\frac{5}{6}\\right)^{3} = 1.\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 14437,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $P$ be a point on the nine-point circle of triangle $ABC$. The perpendicular through $P$ to $AP$ meets the line $BC$ at $Q$. The perpendicular through $A$ to $AQ$ meets the line $PQ$ at $X$. Let $H$ be the orthocenter of triangle $ABC$, and let $D$, $M$ be the midpoints of segments $BC$ and $AQ$ respectively. Prove that $XH \\perp DM$.\n\n![](images/China-TST-2025A_p9_data_eb7ee2987d.png)",
    "options": [],
    "answer": "See solution",
    "solution": "Let $N$ be the midpoint of $AH$. By the properties of the nine-point circle, $DN$ is its diameter, so $DP \\perp PN$. Since $AH \\perp BC$ and $AP \\perp PQ$, we have $\\triangle DPQ \\sim \\triangle NPA$. Therefore, $\\frac{DQ}{NA} = \\frac{PQ}{PA}$.\n\nSince $XA \\perp AQ$, we have $\\triangle AQP \\sim \\triangle XAP$, which gives $\\frac{PQ}{PA} = \\frac{AQ}{XA}$. Combining these two results yields $\\frac{DQ}{NA} = \\frac{AQ}{XA}$.\n\nMoreover, we observe that:\n\n$$\n\\angle XAN = 90^\\circ - \\angle NAQ = \\angle AQD\n$$\n\nThis implies that $\\triangle XAN \\sim \\triangle AQD$, and consequently:\n\n$$\nXA \\cdot DQ = AN \\cdot AQ = AH \\cdot QM\n$$\n\nFinally, noting that:\n\n$$\n\\angle XAH = 90^\\circ - \\angle NAQ = \\angle MQD\n$$\n\nwe conclude that $\\triangle XHA \\sim \\triangle MDQ$, which proves that $DM \\perp XH$.\n\n$\\square$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 14438,
    "subject": "Mathematics (Olympiad)",
    "question": "Prove that a convex pentagon with integer side lengths and an odd perimeter can have two right angles, but cannot have more than two right angles.\n\n(A polygon is _convex_ if all of its interior angles are less than $180^\\circ$.)",
    "options": [],
    "answer": "See solution",
    "solution": "First, we establish that such a pentagon with two right angles is possible. A unit equilateral triangle placed on top of a unit square forms a pentagon with perimeter $5$.\n\nFrom the angle sum formula, we see that if a pentagon has four right angles, the fifth angle would be $180^\\circ$, which means that the pentagon actually degenerates into a rectangle (and would have an even perimeter anyway).\n\nSo it remains to show that a pentagon with integer side lengths and three right angles must have an even perimeter. If three out of five angles are right angles, then at least two of them must be adjacent. This gives two cases: all three right angles are adjacent, or only two are adjacent, as shown below:\n\n![](images/2022_Australian_Scene_p86_data_2bd9e67f1d.png)\n\n![](images/2022_Australian_Scene_p86_data_ad52c429eb.png)\n\nFirst, consider Case 1:\n\n![](images/2022_Australian_Scene_p86_data_db4235be42.png)\n\nThe bounding rectangle has integer sides and hence its perimeter has even parity. Furthermore, the dotted section forms a right-angled triangle with integer sides. The perimeter of the pentagon is found by replacing the two dotted sides with the hypotenuse. But since",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 14439,
    "subject": "Mathematics (Olympiad)",
    "question": "Points $E$ and $F$ lie inside a square $ABCD$ such that the two triangles $ABF$ and $BCE$ are equilateral. Show that $DEF$ is an equilateral triangle.",
    "options": [],
    "answer": "See solution",
    "solution": "We have $\\angle FAD = \\angle BAD - \\angle BAF = 90^\\circ - 60^\\circ = 30^\\circ$. Since $AF = AB = AD$, triangle $AFD$ is isosceles, which means that $\\angle ADF = \\angle AFD = \\frac{180^\\circ - \\angle FAD}{2} = 75^\\circ$ and $\\angle CDF = \\angle CDA - \\angle ADF = 90^\\circ - 75^\\circ = 15^\\circ$. By symmetry, we also have $\\angle ADE = 15^\\circ$, thus $\\angle EDF = 90^\\circ - \\angle ADE - \\angle CDF = 60^\\circ$.\n\nAgain by symmetry (with respect to the diagonal $BD$), $DE = DF$, so $DEF$ is an isosceles triangle with an angle of $60^\\circ$. Therefore, $DEF$ is indeed an equilateral triangle.\n\n![](images/SAMF_ANNUAL_REPORT_2015_BOOK_PROOF_p80_data_6b6971b5d4.png)\n\nLet $G$ and $H$ be the midpoints of $AB$ and $CD$ respectively, and let $M$ be the centre of the square. We denote the side length of the square and the two equilateral triangles by $a$. By Pythagoras' Theorem,\n\n$$\nGF^2 = AF^2 - AG^2 = a^2 - \\left(\\frac{a}{2}\\right)^2 = \\frac{3a^2}{4},\n$$\n\nso $GF = \\frac{\\sqrt{3}a}{2}$. Next we find $FH = GH - GF = a - \\frac{\\sqrt{3}a}{2} = \\frac{(2-\\sqrt{3})a}{2}$, $FM = GF - GM = \\frac{\\sqrt{3}a}{2} - \\frac{a}{2} = \\frac{(\\sqrt{3}-1)a}{2}$ and by symmetry $EM = FM = \\frac{(\\sqrt{3}-1)a}{2}$.\n\nApplying Pythagoras' Theorem again, we obtain\n\n$$\nDF^2 = DH^2 + FH^2 = \\left(\\frac{a}{2}\\right)^2 + \\left(\\frac{(2-\\sqrt{3})a}{2}\\right)^2 = a^2 \\left(\\frac{1}{4} + \\frac{4-4\\sqrt{3}+3}{4}\\right) = a^2(2-\\sqrt{3})\n$$\n\nand\n\n$$\nEF^2 = EM^2 + FM^2 = 2 \\left( \\frac{(\\sqrt{3}-1)a}{2} \\right)^2 = 2a^2 \\cdot \\frac{3-2\\sqrt{3}+1}{4} = a^2(2-\\sqrt{3}).\n$$\n\nThus $DF = EF$, and by symmetry $DE = EF$. This means that $DEF$ is an equilateral triangle.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 14440,
    "subject": "Mathematics (Olympiad)",
    "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": "See 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.\n\n![](images/2022_CMO_p12_data_93bce1a489.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$.\n\n![](images/2022_CMO_p12_data_372d5dcae0.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$$\n\\frac{1 + x + (-2)}{2} + \\frac{1 + 3x}{2} = 2,\n$$\n\nalso yields $x = 1$, which is a contradiction.\n\n**Case 2:** None of the segments $A_iA_{i+1}$, $i = 1, 2, 3, 4, 5$ bisect each other. Without loss of generality, assume that the midpoint $B_1$ of $A_1A_2$ lies on $A_3A_4$. Then the midpoint $B_3$ of $A_3A_4$ lies on $A_5A_1$, the midpoint $B_5$ of $A_5A_1$ lies on $A_2A_3$, the midpoint $B_2$ of $A_2A_3$ lies on $A_4A_5$, and the midpoint $B_4$ of $A_4A_5$ lies on $A_1A_2$. This implies that connecting $A_1, A_2, A_3, A_4, A_5$ in sequence yields a five-pointed star.\n\n![](images/2022_CMO_p13_data_7f18de43a7.png)\n\nBy the Law of Sines,\n\n$$\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$$\n\nwhere the subscripts are understood modulo 5. This is a contradiction.\n\nTherefore, the minimum $n$ we seek is 6.\n\n*Remark:* We now present another proof for Case 2 as follows. For $1 \\le i \\le 5$, consider the following equation (which follows from the fact that $B_i$ is the midpoint of $A_iA_{i+1}$) and inequality (which follows from the triangle inequality, where the sum of two sides is greater than the third):\n\n$$\nB_i B_{i+3} + B_{i+3} A_{i+1} = B_i A_{i+1} = A_i B_i, \\quad B_{i+3} A_{i+1} + B_{i+3} B_{i+1} > A_{i+1} B_{i+1}.\n$$\n\nSumming over $i = 1, 2, 3, 4, 5$ for both equations, the left-hand sides are equal, but one is an equation while the other is a strict inequality, which is a contradiction.\n\n*Remark:* The answer is still $n = 6$ if the problem is modified to \"For any $1 \\le i \\le n$, there exists $1 \\le j \\le n$ ($j \\ne i$) such that line $A_jA_{j+1}$ passes through the midpoint of segment $A_iA_{i+1}$.\" The proof for $n = 5$ is essentially the same as in Case 1 (but allowing $x$ to be negative). For Case 2, we know that the vertices of the convex hull of these five points cannot have adjacent indices, otherwise the line passing through them cannot pass through the midpoint of another segment. Therefore, the configuration must be as shown in Case 2.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 14441,
    "subject": "Mathematics (Olympiad)",
    "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.",
    "options": [],
    "answer": "See solution",
    "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$$ where the summation is over pairs of distinct columns $\\{i, j\\}$.\n\nObserve that if a pair of distinct columns $\\{i, j\\}$ is charged $k = \\text{charge}(i, j)$ times, then there are at most $\\frac{k^2}{4}$ pretty sets with squares contained in these columns. This is because for some $a, b$ with $a + b = k$ there are $a$ red-blue and $b$ blue-red mixed pairs within these columns, yielding $ab \\le \\frac{k^2}{4}$ pretty sets. Therefore, the total number of pretty sets is at most\n\n$$\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$$\n\nwhere again the summation is over unordered pairs of distinct columns. This concludes the proof. $\\square$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 14442,
    "subject": "Mathematics (Olympiad)",
    "question": "A sleeping rabbit lies in the interior of a convex 2024-gon. A hunter picks three vertices of the polygon and lays a trap which covers the interior and the boundary of the triangular region determined by them. Determine the minimum number of times he needs to do this to guarantee that the rabbit will be trapped.",
    "options": [],
    "answer": "See solution",
    "solution": "Let the 2024-gon be $A_0A_1\\cdots A_{2023}$. We claim that the answer is $2022$, which is achieved by picking the triangles $\\triangle A_0A_iA_{i+1}$ for $i = 1, 2, \\dots, 2022$.\n\nFor any $0 \\leq i \\leq 2023$, we claim that the entirety of $\\angle A_iA_{i+1}A_{i+2}$ is covered by triangles with one of their vertices at $A_{i+1}$. To prove this, consider one point in the intersection of the triangle $A_iA_{i+1}A_{i+2}$ with each of $A_{i+1}A_jA_{j+1}$ for all $j \\neq i, i-1$. Observe that if the entire angle is not covered, then at least one of these points is not in any triangle.\n\nThus, the sum of all angles covered by the triangles is at least $2022 \\cdot 180^\\circ$, but each triangle covers only $180^\\circ$. Thus, we need at least $2022$ triangles. $\\square$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 14443,
    "subject": "Mathematics (Olympiad)",
    "question": "For which integers $n \\geq 2$ does $n$ divide $\\binom{2n-3}{n-1}$?",
    "options": [],
    "answer": "See solution",
    "solution": "We are going to prove that the answer is: all $n \\geq 2$ that are not a power of 2.\n\nFirst, we have\n$$\n\\binom{2n-3}{n-1} = \\frac{(2n-3)!}{(n-1)!(n-2)!} = \\frac{(2n-3)(2n-4)\\cdots(n+1)n}{(n-2)(n-3)\\cdots2 \\cdot 1}. \\quad (1)\n$$\nLet $p$ be a prime factor of $n$ and suppose that $p > 2$. We know that the following expression is an integer:\n$$\n\\binom{2n-3}{n} = \\frac{(2n-3)!}{n!(n-3)!} = \\frac{(2n-3)(2n-4)\\cdots(n+1)}{(n-3)(n-4)\\cdots2 \\cdot 1}.\n$$\nWrite $e_p(N)$ for the number of factors $p$ in an integer $N$. Now we see that\n$$\ne_p((2n-3)(2n-4)\\cdots(n+1)) \\geq e_p((n-3)(n-4)\\cdots 2 \\cdot 1).\n$$\nFurthermore, since $p \\mid n$ and $p > 2$, we have $p \\nmid n - 2$, so multiplying by $n - 2$ does not add a factor $p$. Therefore we have\n$$\ne_p((2n-3)(2n-4)\\cdots(n+1)n) - e_p((n-2)(n-3)\\cdots 2 \\cdot 1) \\geq e_p(n).\n$$\nSo $\\binom{2n-3}{n-1}$ contains at least $e_p(n)$ factors $p$. Now $n$ is a divisor of $\\binom{2n-3}{n-1}$ if and only if this result also holds for $p=2$.\n\nIn a product $a_1a_2\\cdots a_m$, the total number of factors 2 is equal to the sum of the number of $a_i$'s divisible by 2, the amount $a_i$'s divisible by 4, the amount $a_i$'s divisible by 8, ...\n\nNow, let us first consider $n = 2^k$ with $k \\geq 1$. Then, in (1), the numerator is the product of $2^k, 2^k+1, 2^k+2, \\dots, 2^{k+1}-3$, while the denominator is the product of $1, 2, 3, \\dots, 2^k-2$. For $1 \\leq i \\leq k-1$, the number of integers from $1, 2, 3, \\dots, 2^k-2$ divisible by $2^i$ is exactly equal to the number of numbers from $2^k+1, 2^k+2, 2^k+3, \\dots, 2^{k+1}-2$ divisible by $2^i$. For $i \\geq k$, both numbers are 0. Therefore we have\n$$\ne_2(1 \\cdot 2 \\cdot 3 \\cdots (2^k - 2)) = e_2((2^k + 1)(2^k + 2)(2^k + 3) \\cdots (2^{k+1} - 2)).\n$$\nOn the right hand side, we divide the product by $2^{k+1} - 2$ (with a single factor 2) and multiply by $2^k$ (with $k$ factors 2) to get the product from the numerator. We conclude that\n$$\ne_2(2^k(2^k+1)(2^k+2)(2^k+3)\\cdots(2^{k+1}-3)) - e_2(1\\cdot2\\cdot3\\cdots(2^k-2)) = k-1,\n$$\nand so $\\binom{2n-3}{n-1}$ contains exactly $k-1$ factors 2, which is not enough to be divisible by $n$. So $n = 2^k$ does not satisfy the property in the problem statement.\n\nLet us now consider the case where $n$ is not a power of 2. Let $2^k$ be the greatest power of 2 less than $n$. We know that $\\binom{2n-3}{n-1}$ is an integer, so if $n$ is odd we already have that\n$$\ne_2((2n-3)(2n-4)\\cdots(n+1)n) - e_2((n-2)(n-3)\\cdots 2 \\cdot 1) \\geq 0 = e_2(n).\n$$\nIf $n$ is even, let $\\ell$ be maximal such that $2^\\ell \\mid n$. Since $n$ is not a power of 2, we note that $2^{\\ell+1} < 3 \\cdot 2^\\ell \\leq n < 2^{\\ell+1}$. This implies $\\ell < k$. Furthermore, since $2^k \\leq n-2$ we have $n < 2^{k+1} \\leq 2n-4$ as well, so the numerator of (1) contains $2^{k+1}$. Now, for each $i$, we look at the number of factors in the numerator and denominator divisible by $2^i$:\n\n* $i=1$: there are $n-2$ integers in both products, and $n$ is even, so the number of factors divisible by 2 is the same for numerator and denominator.\n* $2 \\leq i \\leq \\ell$: in the denominator there are $\\lfloor \\frac{n-2}{2^i} \\rfloor$ factors divisible by $2^i$; in the numerator there are $\\lceil \\frac{n-2}{2^i} \\rceil$ factors divisible by $2^i$ since the smallest factor is divisible by $2^i$; we find that the numerator contains precisely one more factor divisible by $2^i$ than the denominator.\n* $\\ell+1 \\leq i \\leq k$: in the denominator, $\\lfloor \\frac{n-2}{2^i} \\rfloor$ factors divisible by $2^i$; in the numerator, there are at least as many.\n* $i=k+1$: in the denominator, no factor is divisible by $2^{k+1}$; in the numerator, there is exactly one such factor.\n* $i > k+1$: both numerator and denominator contain no factors divisible by $2^i$.\n\nAll in all, we conclude that\n$$\ne_2((2n-3)\\cdots(n+1)n) - e_2((n-2)\\cdots 2 \\cdot 1) \\geq \\ell - 1 + 1 = \\ell = e_2(n).\n$$\nSo $n$ divides $\\binom{2n-3}{n-1}$. We conclude that the answer to the problem is: all $n \\geq 2$ that are not a power of 2. $\\square$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 14444,
    "subject": "Mathematics (Olympiad)",
    "question": "在 $\\triangle ABC$ 的邊 $BC$ 的延長線上取一點 $D$ 使得 $CD = AC$。$\\triangle ACD$ 的外接圓與以 $BC$ 為直徑的圓交於 $C, P$ 兩點，直線 $BP$ 與 $AC$ 交於 $E$，直線 $CP$ 與 $AB$ 交於 $F$。求證：$D, E, F$ 三點共線。",
    "options": [],
    "answer": "See solution",
    "solution": "如圖所示。因點 $P$ 在 $\\triangle ACD$ 的外接圓上，且 $AC = CD$，所以\n\n$$\n\\angle APF = \\angle ADC = \\angle CAD = \\angle CPD,\n$$\n\n即 $PC$ 是 $\\angle APD$ 的外角平分線。設直線 $AP$ 交 $BC$ 於 $Q$，則 $PC$ 平分 $\\angle QPD$。又點 $P$ 在以 $BC$ 為直徑的圓上，所以 $BP \\perp PC$，從而 $BP$ 為 $\\angle QPD$ 的外角平分線，因此有\n\n$$\n\\frac{BD}{DC} = -\\frac{BQ}{QC}.\n$$\n\n另一方面，因 $AQ, BE, CF$ 交於點 $P$，由 Ceva 定理得\n\n$$\n\\frac{BQ}{QC} \\cdot \\frac{CE}{EA} \\cdot \\frac{AF}{FB} = 1,\n$$\n\n於是\n\n$$\n\\frac{BD}{DC} \\cdot \\frac{CE}{EA} \\cdot \\frac{AF}{FB} = -\\frac{BQ}{QC} \\cdot \\frac{CE}{EA} \\cdot \\frac{AF}{FB} = -1\n$$\n\n再由 Menelaus 定理知 $D, E, F$ 三點共線。",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 14445,
    "subject": "Mathematics (Olympiad)",
    "question": "In a convex quadrilateral $ABCD$, angles $\\angle ABC$ and $\\angle BCD$ are not less than $120^\\circ$. Prove that\n$$\nAC + BD > AB + BC + CD.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "Let $AB = a$, $BC = b$, $CD = c$ (see the figure below).\n\nThen,\n$$\nAC^2 = a^2 + b^2 - 2ab \\cos \\angle B \\ge a^2 + ab + b^2.\n$$\nBy analogy,\n$$\nBD^2 \\ge b^2 + bc + c^2.\n$$\nTherefore,\n$$\nAC + BD \\ge \\sqrt{a^2 + ab + b^2} + \\sqrt{b^2 + bc + c^2}.\n$$\nSince $\\sqrt{a^2 + ab + b^2} > a + \\frac{1}{2}b$ and $\\sqrt{b^2 + bc + c^2} > c + \\frac{1}{2}b$, it follows that\n$$\nAC + BD > a + b + c = AB + BC + CD.\n$$\n\n![](images/Ukrajina_2010_p27_data_ad2270fc43.png)\n\nFig.18",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 14446,
    "subject": "Mathematics (Olympiad)",
    "question": "A finite set of distinct positive integers is called a $T$-set if each of its members divides the sum of them all. Prove that every finite set of positive integers is a subset of some $T$-set.",
    "options": [],
    "answer": "See solution",
    "solution": "Clearly, any set containing only one element is a $T$-set. Also, since $\\{1, 2, 3\\}$ is a $T$-set, any of its subsets is certainly contained in a $T$-set.\n\nNow let $S$ be a finite set of positive integers with at least two elements, and let $n\\ (> 3)$ be the largest element in $S$. Let $\\sigma(S)$ denote the sum of the elements of $S$.\n\nLet $T_1 = \\{1, 2, \\dots, n\\}$. Note that $\\sigma(T_1) = \\frac{n(n+1)}{2}$. Let $T_2 = T_1 \\cup \\{\\frac{n(n+1)}{2}\\}$. Then $\\sigma(T_2) = n(n+1)$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 14447,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $ABCD$ be a quadrilateral with no two parallel sides, inscribed in the circle $\\Omega$. Let $\\omega_a, \\omega_b, \\omega_c, \\omega_d$ be the incircles of triangles $DAB$, $ABC$, $BCD$, and $CDA$, respectively. Draw the common external tangents $t_1, t_2, t_3, t_4$ to the pairs of circles $\\omega_a$ and $\\omega_b$, $\\omega_b$ and $\\omega_c$, $\\omega_c$ and $\\omega_d$, $\\omega_d$ and $\\omega_a$, respectively, so that none of the lines $t_1, t_2, t_3, t_4$ coincide with the sides of $ABCD$. The quadrilateral whose consecutive sides lie on $t_1, t_2, t_3, t_4$ (in this order) is inscribed in the circle $\\Gamma$. Prove that the three lines connecting the centers of $\\omega_a$ and $\\omega_c$, $\\omega_b$ and $\\omega_d$, and $\\Omega$ and $\\Gamma$ are concurrent.",
    "options": [],
    "answer": "See solution",
    "solution": "Without loss of generality, let rays $AB$ and $DC$ intersect at point $P$, and rays $AD$ and $BC$ intersect at point $Q$. Denote the center of circle $\\omega_a$ as $I_a$, and define points $I_b, I_c, I_d$ similarly. Let the quadrilateral formed by the four tangent lines be $A'B'C'D'$ (where $A'B'$ is the common external tangent to $\\omega_a$ and $\\omega_b$, and similarly for the other three sides).\n\nBy the Three-Pointed Lemma for triangles $ABD$ and $ACD$, points $A$, $D$, $I_a$, $I_d$ lie on the same circle (with center at the midpoint of arc $AD$ of circle $\\Omega$). Let line $I_aI_d$ intersect sides $AB$ and $CD$ at points $U$ and $V$ respectively, and intersect line $AD$ at point $X_{ad}$.\n\nFor quadrilateral $ABCD$, denote $\\angle A = \\alpha$, $\\angle D = \\delta$. By the cyclic properties of quadrilaterals $ABCD$ and $AI_aI_dD$, we have angle relations: $\\angle QCD = \\alpha$, $\\angle UI_aA = \\angle ADI_d = \\delta/2$, $\\angle VI_dD = \\angle DAI_a = \\alpha/2$. Consequently, $\\angle CQD = \\delta - \\alpha$ and $\\angle PUV = \\angle PVU = (\\alpha+\\delta)/2$. In particular, $\\alpha < \\delta$, so point $X_{ad}$ lies on ray $AD$ and $\\angle DX_{ad}I_d = \\angle ADV - \\angle PVU = (\\delta - \\alpha)/2$. This equality implies that line $I_aI_d$ is parallel to the angle bisector of $\\angle CQD$.\n\nSince line $A'D'$ is symmetric to line $AD$ with respect to the line of centers $I_aI_d$, we obtain that $A'D' \\parallel BC$, and line $A'D'$ passes through point $X_{ad}$. Defining points $X_{ab}, X_{bc}, X_{cd}$ similarly, we find these points lie on lines $A'B', B'C', C'D'$ which are parallel to the sides of quadrilateral $ABCD$.\n\nAs established, line $I_aI_d$ is parallel to the bisector of $\\angle CQD$. Similarly, line $I_bI_c$ is parallel to this bisector, while lines $I_aI_b$ and $I_c I_d$ are parallel to the bisector of $\\angle BPC$. Since $\\angle PUV = \\angle PVU$, the bisector of $\\angle BPC$ is perpendicular to line $I_a I_d$. Thus, adjacent sides of quadrilateral $I_a I_b I_c I_d$ are perpendicular, making it a rectangle. Therefore, it is cyclic—denote its circumcircle by $\\omega$, with center at the intersection of its diagonals.\n\nIt remains to prove that the centers of circles $\\omega$, $\\Omega$ and $\\Gamma$ are collinear. We'll show these three circles share a radical axis containing points $X_{ab}, X_{bc}, X_{cd}, X_{da}$ ($\\star$).\n\nLet $\\gamma$ be the circumcircle of quadrilateral $AI_a I_d D$. Then point $X_{ad}$ is the radical center of circles $\\gamma$, $\\omega$ and $\\Omega$, lying on two of their radical axes. Hence, the radical axis of $\\omega$ and $\\Omega$ passes through $X_{ad}$, and similarly through $X_{ab}, X_{bc}, X_{cd}$. Thus these four points are collinear.\n\nLet line $B'C'$ intersect side $AB$ at $S$ and side $CD$ at $T$. Since $B'C' \\parallel AD$, we have $\\angle BST = \\angle BAD = 180^{\\circ} - \\angle BCT$, making quadrilateral $BCTS$ cyclic. As $C'D' \\parallel AB$ and $A'B' \\parallel CD$, by Thales' theorem:\n\n$$\n\\frac{X_{bc}B'}{X_{bc}T} = \\frac{X_{bc}X_{ab}}{X_{bc}X_{cd}} = \\frac{X_{bc}S}{X_{bc}C'}\n$$\n\nFrom these ratios and the cyclicity of $BCTS$, we obtain $X_{bc}B' \\cdot X_{bc}C' = X_{bc}S \\cdot X_{bc}T = X_{bc}B \\cdot X_{bc}C$, showing equal power of $X_{bc}$ with respect to $\\Omega$ and $\\Gamma$. Analogous reasoning applies to $X_{ab}, X_{ad}, X_{cd}$, proving claim ($\\star$) as required.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 14448,
    "subject": "Mathematics (Olympiad)",
    "question": "a) For $m = 1$, show that $k = i^3 - 1$ meets the conditions for each positive integer $i$. For $m > 1$, show that $k = i^3 m^6 + 3i^2 m^3 + 3i$ meets the conditions, i.e.,\nfor $1 \\leq n < m$, prove that $$(im^2 n)^3 < 1 + k n^3 < (im^2 n + 1)^3.$$ \n\nb) Let $m = p^r$ for a prime $p$ and integer $r$. Find all $k$ such that $1 + k p^{3r}$ is a perfect cube.",
    "options": [],
    "answer": "See solution",
    "solution": "a) If $m = 1$, then $k = i^3 - 1$ works for each positive integer $i$. For $m > 1$, let $k = i^3 m^6 + 3i^2 m^3 + 3i$. Then:\n\n$$\n1 + k m^3 = 1 + (i^3 m^6 + 3i^2 m^3 + 3i) m^3 = i^3 m^9 + 3i^2 m^6 + 3i m^3 + 1 = (i m^3 + 1)^3.\n$$\n\nFor $1 \\leq n < m$:\n$$\n1 + k n^3 = 1 + i^3 m^6 n^3 + 3i^2 m^3 n^3 + 3i n^3.\n$$\nTo complete the proof, show that:\n$$(i m^2 n)^3 < 1 + k n^3 < (i m^2 n + 1)^3.$$\nThe first inequality follows directly. Since $n < m$, $(n - m)(i m^3 n + n + m) < 0$, so $i m^3 n^2 + n^2 < i m^4 l + m^2$. Multiplying both sides by $3i n$ and adding $1 + i^3 m^6 n^3$ to both sides gives the desired result.\n\nb) Let $m = p^r$ and $1 + k p^{3r} = t^3$. Then $k p^{3r} = (t - 1)(t^2 + t + 1)$. The greatest common divisor $(t - 1, t^2 + t + 1) = d$ divides $3$. Since $p \\neq 3$, either $p^{3r} \\mid t^2 + t + 1$ or $p^{3r} \\mid t - 1$.\n\nCase 1: $p^{3r} \\mid t^2 + t + 1$. Let $t^2 + t + 1 = p^{3r} j$. Since $p \\neq 2$, $t^2 + t + 1$ is odd. The discriminant $1 - 4(1 - p^{3r} j) = 4 p^{3r} j - 3$ must be a perfect square: $4 p^{3r} j - 3 = z^2$. In modulo $p$, $-3$ must be a quadratic residue for odd primes $p \\equiv 2 \\pmod{3}$, but $-3$ is not.\n\nCase 2: $p^{3r} \\mid t - 1$. Let $t - 1 = p^{3r} j$. Then $1 + k p^{3r} = (p^{3r} j + 1)^3$ yields $k = p^{6r} j^3 + 3 p^{3r} j^2 + 3j$. As in part (a), for $1 \\leq l < p^r$, $(j p^{2r} l)^3 < 1 + k p^{3r} < (j p^{2r} l + 1)^3$. Thus, all required $k$ are $k = p^{6r} j^3 + 3 p^{3r} j^2 + 3j$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 14449,
    "subject": "Mathematics (Olympiad)",
    "question": "Determine all triples $ (x, y, z) $ of real numbers satisfying:\n\n$$\n\\begin{aligned}\nx^2 - yz &= |y - z| + 1, \\\\\ny^2 - zx &= |z - x| + 1, \\\\\nz^2 - xy &= |x - y| + 1.\n\\end{aligned}\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "The system of equations is symmetric: if you swap $x$ and $y$, for example, then the third equation stays the same and the first two equations are swapped. Hence, we can assume without loss of generality that $x \\ge y \\ge z$. Then the system of equations becomes:\n\n$$\n\\begin{aligned}\nx^2 - yz &= y - z + 1, \\\\\ny^2 - zx &= x - z + 1, \\\\\nz^2 - xy &= x - y + 1.\n\\end{aligned}\n$$\n\nSubtracting the second equation from the first, we obtain $x^2 - y^2 + z(x - y) = y - x$, or $(x - y)(x + y + z + 1) = 0$. This yields $x = y$ or $x + y + z = -1$. Subtracting the third equation from the second, we obtain $y^2 - z^2 + x(y - z) = y - z$, or $(y - z)(y + z + x - 1) = 0$. This yields $y = z$ or $x + y + z = 1$.\n\nWe now distinguish two cases: $x = y$ and $x \\neq y$. In the first case, we have $y \\neq z$, as otherwise we would have $x = y = z$ for which the first equation becomes $0 = 1$, a contradiction. Now it follows that $x + y + z = 1$, or $2x + z = 1$. Substituting $y = x$ and $z = 1 - 2x$ in the first equation yields $x^2 - x(1 - 2x) = x - (1 - 2x) + 1$, which can be simplified to $3x^2 - x = 3x$, or $3x^2 = 4x$. We get $x = 0$ or $x = \\frac{4}{3}$. With $x = 0$, we find $y = 0$, $z = 1$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 14450,
    "subject": "Mathematics (Olympiad)",
    "question": "Suppose that integers $a$ and $b$ are greater than $1$ in absolute value. It is known that there are infinitely many positive integers $n$ such that for each of them there exists a positive integer $m$ with the property that $a^m + b$ is divisible by $a^n + 1$. Prove that $|b| = |a|^k$ for some positive integer $k$.\n\n![](path/to/file.png)",
    "options": [],
    "answer": "See solution",
    "solution": "Assume that $|b| \\neq |a|^k$ for every $k \\in \\mathbb{N}$. Consider positive integers $n$ and $m$ such that $a^m + b \\equiv 0 \\pmod{a^n + 1}$. If $m = qn + r$, $0 \\leq r < n$, then\n\n$$\na^n \\equiv -1 \\pmod{a^n + 1}, \\\\\n-b \\equiv a^m = a^{qn + r} \\equiv (-1)^q a^r \\pmod{a^n + 1}, \\text{ and so } a^r + (-1)^q b \\equiv 0 \\pmod{a^n + 1}.\n$$\n\nBy our assumption $|a^r + (-1)^q b| \\neq 0$, so $|a^r + (-1)^q b| \\geq |a^n + 1|$. Then we have:\n\n$$\n|a^{n-1} + |b|| \\geq |a^r| + |b| \\geq |a^r + (-1)^q b| \\geq |a^n| - 1.\n$$\n\nThis implies that $\\frac{1}{|a|} + \\frac{|b| + 1}{|a|^n} \\geq 1$. But for $n$ large enough, the quotient $\\frac{|b| + 1}{|a|^n}$ is less than $1 - \\frac{1}{|a|}$, which means that the last inequality cannot hold for an infinite number of values of $n$. This contradiction completes the proof.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 14451,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all real numbers $x$ such that\n\n$$\nx \\lfloor x \\rfloor + 2022 = \\lfloor x^2 \\rfloor.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "Let $x = n + \\alpha$, where $0 \\le \\alpha < 1$. Inserting it into the main equation\n\n$$\nx \\lfloor x \\rfloor + 2022 = \\lfloor x^2 \\rfloor\n$$\n\ngives\n\n$$\n(n + \\alpha)n + 2022 = \\lfloor (n + \\alpha)^2 \\rfloor \\quad (1)\n$$\n\nThis implies that $(n + \\alpha)n$ is an integer and $n \\ne 0$. Moreover, if $n > 0$ then there exists a nonnegative integer $m < n$ such that $\\alpha = m/n$ for $0 \\le m < n$. Inserting $\\alpha = m/n$ into (1) we get\n\n$$\nn^2 + m + 2022 = \\lfloor \\left(n + \\frac{m}{n}\\right)^2 \\rfloor = \\lfloor n^2 + 2m + \\frac{m^2}{n^2} \\rfloor = n^2 + 2m.\n$$\n\nHence $m = 2022$. Since $m < n$ we get that $n \\ge 2023$. It can be readily seen that for any integer $n \\ge 2023$, $x = n + \\frac{2022}{n}$ satisfies the main equation.\n\n**Answer:** $x = n + \\frac{2022}{n}$, where $n \\ge 2023$ is any integer.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 14452,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $O$ be the circumcenter of triangle $ABC$. Let $H_A$ be the projection of $A$ onto $BC$. The extension of $AO$ intersects the circumcircle of $BOC$ at $A'$. The projections of $A'$ onto $AB$ and $AC$ are $D$ and $E$, respectively, and $O_A$ is the circumcenter of triangle $DH_AE$. Define $H_B, O_B, H_C, O_C$ similarly.\n\nProve that $H_AO_A$, $H_B O_B$, and $H_C O_C$ are concurrent.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $T$ be the reflection of $A$ with respect to $BC$, $F$ be the projection of $A'$ onto $BC$, and $M$ be the projection of $T$ onto $AC$.\n\nSince $AC = CT$, we have $\\angle TCM = 2\\angle TAM$. Since\n\n$$\n\\angle TAM = \\frac{\\pi}{2} - \\angle ACB = \\angle OAB, \\text{ we have}\n$$\n\n$$\n\\angle TCM = 2\\angle OAB = \\angle A'OB = \\angle A'CF, \\text{ and}\n$$\n\n$$\n\\angle TCH_A = \\angle A'CF + \\angle A'CT = \\angle TCM + \\angle A'CT = \\angle A'CE.\n$$\n\nBecause $\\angle CH_A T$, $\\angle CMT$, $\\angle CEA'$, $\\angle CFA'$ are right angles, therefore\n\n$$\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$$\n\ni.e., $CH_A \\cdot CF = CM \\cdot CE$, so $H_A, F, M, E$ are concyclic on the same circle $\\omega_1$.\n\n![](images/Kina_TS_2014_p0_data_55c7fdadca.png)\n\nSimilarly, let $N$ be the projection of $T$ onto $AB$, then $H_A, F, N, D$ are concyclic on the same circle $\\omega_2$. Since $A'FH_A T$ and $A'EMT$ are both right trapezoids, the perpendicular bisectors of the segments $H_AF$ and $EM$ meet at the midpoint $K$ of the segment $A'T$, i.e., $K$ is the center of circle $\\omega_1$, and $KF$ is the radius of $\\omega_1$. Similarly, $K$ and $KF$ are also the center and radius of $\\omega_2$, respectively. Thus, $\\omega_1$ and $\\omega_2$ are the same, and $D, N, F, H_A, E, M$ are concyclic. So $O_A$ is the midpoint $K$ of $A'T$, and $O_A H_A \\parallel AA'$.\n\nSince $\\angle H_CAO + \\angle AH_C H_B = \\frac{\\pi}{2} - \\angle ACB + \\angle ACB = \\frac{\\pi}{2}$, we have $AA' \\perp H_B H_C$, thus $O_A H_A \\perp H_B H_C$, therefore $O_A H_A$, $O_B H_B$, $O_C H_C$ all pass through the orthocenter of $\\triangle H_A H_B H_C$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 14453,
    "subject": "Mathematics (Olympiad)",
    "question": "Consider $A, B \\in M_n(\\mathbb{R})$, and the function $f: M_n(\\mathbb{C}) \\to M_n(\\mathbb{C})$ defined by\n$$\nf(Z) = AZ + B\\bar{Z}, \\quad Z \\in M_n(\\mathbb{C}),\n$$\nwhere $\\bar{Z}$ is the matrix whose entries are the complex conjugates of the entries of $Z$.\n\nProve that the following are equivalent:\n\n1. $f$ is injective;\n2. $f$ is surjective;\n3. The matrices $A+B$ and $A-B$ are non-singular.",
    "options": [],
    "answer": "See solution",
    "solution": "For $Z \\in M_n(\\mathbb{C})$, there exist $X, Y \\in M_n(\\mathbb{R})$ such that $Z = X + iY$ and $\\bar{Z} = X - iY$. Thus,\n$$\nf(Z) = AZ + B\\bar{Z} = A(X + iY) + B(X - iY) = (A+B)X + i(A-B)Y.\n$$\n\n**(1) $\\Rightarrow$ (3):**\nSuppose $A+B$ or $A-B$ is singular. If $A+B$ is singular, then $\\det(A+B) = 0$, so there exists $C \\in M_{n,1}(\\mathbb{R}) \\setminus \\{0\\}$ such that $(A+B)C = 0$. Define $X \\in M_n(\\mathbb{R})$, $X \\neq 0$, with all columns equal to $C$. Then $f(X) = (A+B)X = 0 = f(0)$, contradicting injectivity. The case for $A-B$ is similar.\n\n**(3) $\\Rightarrow$ (1):**\nLet $Z_1 = X_1 + iY_1$ and $Z_2 = X_2 + iY_2$ with $X_1, Y_1, X_2, Y_2 \\in M_n(\\mathbb{R})$, and suppose $f(Z_1) = f(Z_2)$. Then\n$$(A+B)X_1 + i(A-B)Y_1 = (A+B)X_2 + i(A-B)Y_2.$$\nThus, $(A+B)(X_1 - X_2) = 0$ and $(A-B)(Y_1 - Y_2) = 0$. Since $A+B$ and $A-B$ are non-singular, $X_1 = X_2$ and $Y_1 = Y_2$, so $Z_1 = Z_2$.\n\n**(2) $\\Rightarrow$ (3):**\nSuppose $A+B$ or $A-B$ is singular. If $A+B$ is singular, then for any $Z = X + iY \\in M_n(\\mathbb{C})$, $(A+B)X$ cannot be arbitrary, so $f$ cannot be surjective. The same holds if $A-B$ is singular.\n\n**(3) $\\Rightarrow$ (2):**\nLet $Z = X + iY \\in M_n(\\mathbb{C})$. Define $U = (A+B)^{-1}X$ and $V = (A-B)^{-1}Y$. Then\n$$\nf(U + iV) = (A+B)U + i(A-B)V = X + iY = Z,\n$$\nso $f$ is surjective.",
    "topic": "Algebra",
    "subtopic": "Linear Algebra"
  },
  {
    "id": 14454,
    "subject": "Mathematics (Olympiad)",
    "question": "In cyclic quadrilateral $ABCD$, diagonals $AC$ and $BD$ intersect at $P$. Let $E$ and $F$ be the respective feet of the perpendiculars from $P$ to lines $AB$ and $CD$. Segments $BF$ and $CE$ meet at $Q$. Prove that lines $PQ$ and $EF$ are perpendicular to each other.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $H$ lie on $EF$ such that $PH \\perp EF$, and let $R_E$ lie on $PH$ such that $ER_E \\perp BF$.\n\nSince $\\angle HR_EX = \\angle HFX = 90^\\circ$, the quadrilateral $X R_E F H$ is cyclic.\n\n![](images/Saudi_Arabia_booklet_2012_p42_data_4ca11199d0.png)\n\nIt follows that\n\n$$\n\\angle PR_E E = \\angle PR_E X = \\angle HFX = \\angle EFB. \\quad (1)\n$$\n\nAlso note that\n\n$$\n\\angle R_E E P = 90^\\circ - \\angle B E R_E = 90^\\circ - \\angle B E X = \\angle X B E = \\angle F B E. \\quad (2)\n$$\n\nBy (1) and (2), we know that $\\triangle PR_E E \\sim \\triangle EFB$, implying that\n\n$$\n\\frac{PR_E}{EF} = \\frac{EP}{BE}. \\quad (3)\n$$\n\nDefine $R_F$ as the point on $PH$ such that $FR_F \\perp CE$. In exactly the same way, we can show that\n\n$$\n\\frac{PR_F}{EF} = \\frac{FP}{CF}. \\quad (4)\n$$\n\nSince $ABCD$ is cyclic, $\\triangle ABP \\sim \\triangle DCP$, and $E$ and $F$ are corresponding points under this similarity. In particular,\n\n$$\n\\frac{FP}{CF} = \\frac{EP}{BE}. \\quad (5)\n$$\n\nBy (3), (4), and (5), we have\n\n$$\n\\frac{PR_E}{EF} = \\frac{PR_F}{EF},\n$$\n\nso $R_E = R_F = R$. Now we see that $Q$ is the orthocenter of triangle $EFR$. In particular, $RQ \\perp EF$. By definition of $R$, we have $RP \\perp EF$, so $PQ \\perp EF$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 14455,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $ABCD$ be a square. The equilateral triangle $BCS$ is constructed on the exterior of the side $BC$. Let $N$ denote the midpoint of the line segment $AS$ and let $H$ be the midpoint of the side $CD$.\n\nProve: $\\angle NHC = 60^\\circ$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $P$ be the midpoint of $BS$ (see Figure 1). Since triangles $\\triangle SNP$ and $\\triangle SAB$ are similar with factor $2$, the segment $NP$ is parallel to $AB$ and half the length of $AB$. Therefore, $NPCH$ is a parallelogram. As $NP$ and $BC$ are orthogonal and $PC$ and $BS$ are orthogonal, we have $\\angle NPC = \\angle CBP = 60^\\circ$. Thus, we obtain $\\angle NHC = \\angle NPC = 60^\\circ$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 14456,
    "subject": "Mathematics (Olympiad)",
    "question": "A $4 \\times 4$ table is filled with different positive integers so that in each column, the largest number is equal to the sum of the other three numbers in that column. What is the smallest possible value of the largest number on the board?",
    "options": [],
    "answer": "See solution",
    "solution": "For $i = 1, \\dots, 4$, let $a_i$ be the maximum number in column $i$, and let $b_1, b_2, \\dots, b_{12}$ be the remaining 12 numbers written on the board (different from $a_1, a_2, a_3, a_4$). Then, for every $i$, $a_i$ is the sum of the other three numbers in column $i$; therefore,\n\n$$\na_1 + a_2 + a_3 + a_4 = b_1 + b_2 + \\dots + b_{12}. \\qquad (2)\n$$\n\nSince $b_1, b_2, \\dots, b_{12}$ are different positive integers, we have that\n\n$$\nb_1 + b_2 + \\dots + b_{12} \\ge 1 + 2 + \\dots + 12 = 78. \\qquad (3)\n$$\n\nOn the other hand, since $a_1, a_2, a_3, a_4$ are also different positive integers, and $M$ is the largest number on the board, then\n\n$$\na_1 + a_2 + a_3 + a_4 \\le M + (M-1) + (M-2) + (M-3) \\le 4M - 6. \\qquad (4)\n$$\n\nFrom (2), (3) and (4), it follows that\n\n$$\n4M - 6 \\ge a_1 + a_2 + a_3 + a_4 = b_1 + b_2 + \\dots + b_{12} \\ge 78,\n$$\n\nwhich implies that $M \\ge 21$.\n\nThe following is an example with $M = 21$:\n\n![](images/Argentina_2019_Booklet_p11_data_0dbb7a4312.png)\n\n| 1  | 8  | 12 | 21 |\n|----|----|----|----|\n| 7  | 9  | 20 | 4  |\n| 10 | 19 | 3  | 6  |\n| 18 | 2  | 5  | 11 |",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 14457,
    "subject": "Mathematics (Olympiad)",
    "question": "In an acute triangle $\\triangle ABC$ with $|AB| \\neq |AC|$, the perpendicular bisectors of sides $AC$ and $AB$ intersect segment $BC$ at points $D$ and $E$, respectively. The tangents to $\\odot(ABC)$ at the points $B$ and $C$ intersect $\\odot(ABD)$ and $\\odot(ACE)$ at points $Y$ and $Z$, respectively. Suppose that lines $YD$ and $ZE$ intersect at point $X$. Define points $U$ and $V$ to be the intersections of $\\odot(DEX)$ and the lines $AD$ and $AE$, respectively. Prove that the lines $UV$, $BC$ and $XA$ are concurrent.\n\nHere $\\odot(P_1P_2P_3)$ denotes the circumcircle of triangle $\\triangle P_1P_2P_3$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $O$ be the circumcentre of $\\odot(ABC)$. We proceed in several steps.\n\n*Step 1:* Point $O$ lies on $\\odot(ABD)$ and $\\odot(ACE)$.\n\n*Proof.* Note that since $|AD| = |DC|$, we have $\\angle CDA = 180^\\circ - 2\\angle ACB$, therefore $\\angle ADB = 2\\angle ACB = \\angle AOB$, which means that quadrilateral $AODB$ is cyclic. Similarly, we can prove that quadrilateral $AOEC$ is cyclic. $\\square$\n\n*Step 2:* Points $Y, A, Z$ are collinear. Moreover, $YZ$ is tangent to $\\odot(ABC)$ at point $A$.\n\n*Proof.* Note that since $YB$ is tangent to $\\odot(ABC)$, we have that:\n\n$$\n\\angle AOY = \\angle YBA = \\angle ACB.\n$$\n\nSince $\\angle AOB = 2\\angle ACB$, we get that:\n\n$$\n\\angle YAB = \\angle YOB = \\angle AOB - \\angle AOY = \\angle ACB.\n$$\n\nThis means that $YA$ is tangent to $\\odot(ABC)$. Consequently, $|YA| = |YB|$ implies that $Y$ lies on the perpendicular bisector of $AB$. Similarly, we can prove that $AZ$ is tangent to $\\odot(ABC)$ and $Z$ lies on the perpendicular bisector of $AC$. Since the tangent line at a fixed point is unique, we conclude that points $Y, A, Z$ all lie on the tangent to $\\odot(ABC)$ at the point $A$. $\\Box$\n\n*Step 3:* Points $Y, O, E$ and $Z, O, D$ are collinear. Moreover, $O$ is the orthocentre of $\\triangle YZX$.\n\n*Proof.* The first part follows from the previous result that $Y$ lies on the perpendicular bisector of $AB$, which is $OE$. A similar argument applies to $Z, O, D$.\n\nSince the radius of a circle is perpendicular to the corresponding tangent, note that:\n\n$$\n\\angle OBY = \\angle ODY = \\angle YAO = 90^\\circ \\quad \\text{and} \\quad \\angle OCZ = \\angle OEZ = 90^\\circ.\n$$\n\nThis gives us that $ZD \\perp YX$ and $YE \\perp ZX$, implying that $O$ is the orthocentre of $\\triangle YZX$. Also note that this immediately gives us that points $A, O, X$ are collinear, too. $\\Box$\n\nTo finish the problem, note that quadrilateral $YDEZ$ is cyclic from $\\angle YDZ = \\angle YEZ = 90^\\circ$. Therefore:\n\n$$\n\\angle DAO = \\angle DYO = \\angle OZE = \\angle OAE.\n$$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 14458,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $a$ be a non-negative real number and a sequence $(u_n)$ defined as:\n\n$$u_1 = 6, \\quad u_{n+1} = \\frac{2n+a}{n} + \\sqrt{\\frac{n+a}{n} u_n + 4}$$\n\nfor all positive integers $n$.\n\n**a)** For $a = 0$, prove that $(u_n)$ has a finite limit and find its value.\n\n**b)** For $a \\ge 0$, prove that $(u_n)$ has a finite limit.",
    "options": [],
    "answer": "See solution",
    "solution": "**a)** For $a = 0$, the sequence $(u_n)$ is defined by\n\n$$u_1 = 6, \\quad u_{n+1} = 2 + \\sqrt{u_n + 4}, \\quad \\forall n \\in \\mathbb{N}^*.$$ \n\nIt is clear that $u_n \\ge 2$ for all positive integers $n$. Also, $u_2 < u_1$. By induction, $(u_n)$ is decreasing. Hence, $(u_n)$ has a finite limit $l$. Letting $n \\to \\infty$, we get $l = 2 + \\sqrt{l + 4}$, so $l = 5$.\n\n**b)** First, we prove that $(u_n)$ is bounded. Let $n_0$ be a positive integer such that $n_0 > a$. Choose $M > 10$ such that $M > \\max(u_1, u_2, \\dots, u_{n_0})$. Then\n\n$$u_{n_0+1} \\le 3 + \\sqrt{2u_{n_0} + 4} \\le 3 + \\sqrt{2M + 4} \\le M.$$ \n\nBy induction, $u_n \\le M$ for all $n$. Since $u_n > 0$, $(u_n)$ is bounded.\n\nNext, we show that $(u_n)$ is monotone (not necessarily from the first term). If $(u_n)$ is non-decreasing, the statement is proved. Otherwise, there exists $m$ such that $u_m > u_{m+1}$, so\n\n$$\n\\frac{2m+a}{m} + \\sqrt{\\frac{m+a}{m} u_m + 4} > \\frac{2m+2+a}{m+1} + \\sqrt{\\frac{m+1+a}{m+1} u_{m+1} + 4}.\n$$\n\nTherefore, $u_{m+1} > u_{m+2}$. By induction, $(u_n)$ is decreasing from $u_m$. Since $(u_n)$ is bounded, it has a finite limit.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 14459,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $p$ be a prime and $m$ a positive integer. Find the smallest integer $d$ such that there exists a monic polynomial $Q(x)$ of degree $d$ with integer coefficients, and for all integers $n$, $p^m$ divides $Q(n)$.",
    "options": [],
    "answer": "See solution",
    "solution": "We use the following lemma:\n\n**Lemma.** Let $l$ be a fixed non-zero integer and $P(x) = a_nx^n + \\cdots + a_0$ be a polynomial with integer coefficients such that for all integers $k$, $l \\mid P(k)$. Then $l \\mid a_n \\cdot n!$.\n\n*Proof.* We use induction on $n$. For $n=0$ the claim is obvious. Assume the claim is true for $n$.\n\nLet $P(x) = a_{n+1}x^{n+1} + \\cdots + a_0$ be a polynomial of degree $n+1$ such that for all integers $k$, $l \\mid P(k)$. Define\n\n$$\nP(x+1) - P(x) = R(x) = (n+1)a_{n+1}x^n + S(x),\n$$\n\nwhere $R$ is degree $n$ and $S$ has degree less than $n$. By the induction hypothesis, since $l \\mid R(x)$ for all $x$, we have\n\n$$\nl \\mid (n+1)a_{n+1} \\cdot n! = a_{n+1}(n+1)!,\n$$\n\nso the lemma is proved.\n\nApplying the lemma for $l = p^m$ and $Q(x)$ monic of degree $d$, we get $p^m \\mid d!$. Let $s$ be the smallest integer such that $p^m \\mid s!$, so $d \\ge s$.\n\nNow, consider $Q(x) = (x-s)(x-(s-1))\\cdots(x-1)$. For any integer $n$, $Q(n)$ is a product of $s$ consecutive integers, so $s! \\mid Q(n)$, and thus $p^m \\mid Q(n)$ for all $n$.\n\nTherefore, the answer is the smallest $d$ such that $p^m \\mid d!$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 14460,
    "subject": "Mathematics (Olympiad)",
    "question": "Is it possible to write $2017$ as the sum of $128$ positive integers, each of which is either $1$, $2$, or $44$?",
    "options": [],
    "answer": "See solution",
    "solution": "Yes.\n\nNotice that $2017 = 43 \\cdot 46 + 39$. One example is to take $39$ numbers equal to $2$, $46$ numbers equal to $44$, and the remaining numbers equal to $1$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 14461,
    "subject": "Mathematics (Olympiad)",
    "question": "We form a number $n$ by concatenating the integers from $1$ to $100{,}000$ (i.e., $n = 1234567891011\\dots99999100000$). How many times does the sequence $2022$ appear in the decimal representation of $n$?",
    "options": [],
    "answer": "See solution",
    "solution": "To count the number of occurrences of $2022$ in $n$, consider the possible ways $2022$ can appear across the concatenated numbers:\n\n**Case 1: $2022$ appears entirely within a single number (no comma within the digits)**\n\n- As a ten-thousands or thousands digit, each occurs $10$ times, giving $20$ occurrences:\n  - $2022, 20221, 20222, \\dots, 20229$\n  - $2021, 2022, 2023; 12021, 12022, 12023; \\dots; 92021, 92022, 92023$\n\n**Case 2: $202$ at the end of one number, $2$ at the start of the next**\n\n- $202$ may come from a three-digit number ($202$), a four-digit number ($2202$), or a five-digit number starting with $2$ (there are $10$ such numbers), giving $12$ occurrences:\n  - $202, 203$\n  - $2202, 2203$\n  - $20202, 20203; 21202, 21203; \\dots; 29202, 29203$\n\n**Case 3: $20$ at the end of one number, $22$ at the start of the next**\n\n- $20$ may come from a three-digit number ($220$), a four-digit number ($2220$), or a five-digit number starting with $22$ (there are $10$ such numbers), giving $12$ occurrences:\n  - $220, 221$\n  - $2220, 2221$\n  - $22020, 22021, 22120, 22121; \\dots; 22920, 22921$\n\nAdding up all cases: $20 + 12 + 12 = 44$.\n\n**Final answer:** $44$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 14462,
    "subject": "Mathematics (Olympiad)",
    "question": "If $a$, $b$, $c$ are positive real numbers, prove that:\n\n$$\n\\frac{1}{ab(b+1)(c+1)} + \\frac{1}{bc(c+1)(a+1)} + \\frac{1}{ca(a+1)(b+1)} - \\frac{3}{(1+abc)^2}\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "The inequality is equivalent to\n\n$$\n\\frac{c(a+1)+a(b+1)+b(c+1)}{abc(a+1)(b+1)(c+1)} \\geq \\frac{3}{(1+abc)^2},\n$$\n\nor, after simplification,\n\n$$\n(1+abc)^2(ab+bc+ca+a+b+c) \\geq 3abc(ab+bc+ca+a+b+c+abc+1).\n$$\n\nLet $m = a+b+c$, $n = ab+bc+ca$, and $x^3 = abc$. Then,\n\n$$\n(m+n)(1+x^3)^2 \\geq 3x^3(x^3 + m + n + 1),\n$$\nwhich simplifies to\n$$\n(m+n)(x^6 - x^3 + 1) \\geq 3x^3(x^3 + 1).\n$$\n\nBy the arithmetic–geometric mean inequality, $m \\geq 3x$ and $n \\geq 3x^2$, so $m+n \\geq 3x(x+1)$. Thus, it suffices to prove\n\n$$\n3x(x+1)(x^6 - x^3 + 1) \\geq 3x^3(x+1)(x^2 - x + 1),\n$$\nwhich is equivalent to\n$$\n\\begin{aligned}\n& x^6 - x^3 + 1 \\geq x^2(x^2 - x + 1) \\\\\n\\Leftrightarrow & x^6 - x^4 - x^2 + 1 \\geq 0 \\\\\n\\Leftrightarrow & x^4(x^2 - 1) - (x^2 - 1) \\geq 0 \\\\\n\\Leftrightarrow & (x^2 - 1)(x^4 - 1) \\geq 0 \\\\\n\\Leftrightarrow & (x^2 + 1)(x^2 - 1)^2 \\geq 0,\n\\end{aligned}\n$$\nwhich is always true.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 14463,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $\\mu \\times \\nu$ be a rectangle divided into unit squares. What is the minimum number $N$ of unit squares that must initially contain a black token so that, by repeatedly placing a black token in any white square that shares at least two sides with black-token squares, eventually all squares contain black tokens?",
    "options": [],
    "answer": "See solution",
    "solution": "We call the set of all black squares the *black region*. The key observation is that if a white square has at least two black adjacent squares, then after one move, the perimeter of the black region does not increase, as shown below:\n\n![](images/Greece-IMO2019finalbook_p10_data_4a1f3088aa.png)\n\nHere, the perimeter remains 8 after two moves.\n\n![](images/Greece-IMO2019finalbook_p10_data_3e2804b23c.png)\n\nIn the last figure, the perimeter is 12 and then 10, so it reduces by 2. If initially there are $N$ black squares, the total perimeter is at most $4N$ (exactly $4N$ if no black squares are adjacent). If, in the end, all squares are black, the perimeter equals that of the rectangle: $2\\mu + 2\\nu$.\n\nThus,\n$$\n4N \\geq 2\\mu + 2\\nu\n$$\nwhich gives $N \\geq \\frac{\\mu + \\nu}{2}$. Since $N$ is an integer,\n$$\nN \\geq \\left\\lfloor \\frac{\\mu + \\nu + 1}{2} \\right\\rfloor.\n$$\n\nTo show this is achievable, place $\\nu$ black tokens along the main diagonal of the $\\nu \\times \\nu$ square in the upper left. In the remaining rectangle, alternate empty and black-token columns, always placing a black token in the final column. This adds $\\left\\lfloor \\frac{\\mu - \\nu + 1}{2} \\right\\rfloor$ black squares, totaling\n$$\n\\nu + \\left\\lfloor \\frac{\\mu - \\nu + 1}{2} \\right\\rfloor = \\left\\lfloor \\frac{\\mu + \\nu + 1}{2} \\right\\rfloor.\n$$\n\nFor example, in a $5 \\times 10$ rectangle:\n\n![](images/Greece-IMO2019finalbook_p11_data_b1dcfc0cba.png)\n\nWith this placement, after finitely many moves, all squares will become black, as the main diagonal fills the $\\nu \\times \\nu$ square, and the remaining columns are filled with the help of the placed tokens.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 14464,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $G$ and $H$ be the centroid and orthocentre of $\\triangle ABC$, which has an obtuse angle at $\\angle B$. Let $\\omega$ be the circle with diameter $AG$. $\\omega$ intersects the circumcircle $\\odot ABC$ again at $L \\neq A$. The tangent to $\\omega$ at $L$ intersects $\\odot ABC$ at $K \\neq L$.\n\nGiven that $AG = GH$, prove that $\\angle HKG = 90^\\circ$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $L'$ be the midpoint of $AH$. We claim $L'$ lies on $\\odot ABC$.\n\n![](images/2020_BMO_Short_List_p13_data_b53d2c67b2.png)\n\nIndeed, let $D$ be the foot of the $A$-altitude on $BC$. Then:\n\n$$\nAG = GH \\Rightarrow \\angle GL'A = 90^\\circ \\Rightarrow GL' \\parallel BC \\Rightarrow DL' = \\frac{AL'}{2} = \\frac{HL'}{2} \\Rightarrow DL' = HD\n$$\n\nwhere in the last step we have used that if $M$ is the midpoint of $BC$, $AG : GM = 2 : 1$ and that $\\angle B$ is obtuse so $H$ and $A$ lie on opposite sides of line $BC$. This means that $L'$ is the reflection of $H$ in $BC$, which is well-known to lie on $\\odot ABC$. Also $AG = GH \\Rightarrow \\angle GL'A = 90^\\circ$ so $L'$ lies on $\\omega$ and hence in fact $L \\equiv L'$.\n\nLet $O$ be the midpoint of $AG$; then $OL \\perp LK$. Homothety of factor $2$ at $A$ takes $OL$ to $HG$ so $HG \\parallel OL$ and hence $LK \\perp HG$. But the centre of $\\odot ABC$ lies on $HG$ so this means $K$ is the reflection of $L$ across line $HG$ and hence as $\\angle HLG = 90^\\circ$ it follows $\\angle HKG = 90^\\circ$.\n\n**Remark.** The midpoint of $AH$ lies on the circumcircle $\\odot ABC$ iff $\\angle A = 90^\\circ$ or if:\n\n$$\na^4 + a^2(b^2 + c^2) - 2(b^2 - c^2)^2 = 0\n$$\n\nThe latter condition is exactly equivalent to $AG = GH$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 14465,
    "subject": "Mathematics (Olympiad)",
    "question": "In an $n \\times n$ table, columns are numbered from left to right as $1, 2, \\ldots, n$. Each cell is filled with a number from $1$ to $n$ so that every row and every column contains all numbers $1$ to $n$ (i.e., the table is a Latin square). In each column, a cell is painted grey if the number in it is greater than the column's number. The image below shows an example for $n=3$.\n\nIs it possible that the number of grey cells in every row is equal, if\n\n$$\na)\\ n=5;\n$$\n\n$$\nb)\\ n=10.\n$$\n\n![alt](images/ukraine_2015_Booklet_p8_data_c813093e18.png)\n",
    "options": [],
    "answer": "See solution",
    "solution": "**Answer:**\n\na) Yes.\n\nb) No.\n\n**Explanation:**\n\na) For $n=5$, it is possible to construct such an arrangement (see the referenced figure).\n\nb) For $n=10$, the first row contains $9$ grey cells, the second contains $8$, ..., the ninth contains $1$, and the tenth contains none. The total number of grey cells is $9 + 8 + \\ldots + 1 = 45$, which is not divisible by $10$. Therefore, it is impossible for every row to have an equal number of grey cells.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 14466,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all functions $f: \\mathbb{R} \\to \\mathbb{R}$ which satisfy the conditions:\n\n$$\nf(x+y) < f(x) + f(y),\n$$\n$$\nf(f(x)) = [x] + 2.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "Let $f(0) = a$. Then $f(a) = f(f(0)) = 2$, and $f(2) = f(f(a)) = a + 2$. Continuing this procedure, we get $f(2k) = a + 2k$ and $f(a + 2k) = 2k + 2$.\n\nWe have $2k + 2 = f(a + 2k) < f(a) + f(2k) = 2 + a + 2k$, so $a > 0$. If we put $x = y = a$, we get $a + 2a = f(2a) < f(a) + f(a) = 4$, so $3a < 4$, i.e., $a = 1$.\n\nHence, using $f(2k) = a + 2k$ and $f(a + 2k) = 2k + 2$, we get $f(x) = x + 1$ for all natural numbers $x$.\n\nFor $x = y = \\frac{1}{2}$ in the inequality, we get $2 = f(1) = f(\\frac{1}{2} + \\frac{1}{2}) < 2f(\\frac{1}{2})$, so $f(\\frac{1}{2}) > 1$. On the other hand, $1 + f(\\frac{1}{2}) = f(f(\\frac{1}{2})) = 2$, so $f(\\frac{1}{2}) = 1$, which is a contradiction.\n\nIt follows that there exists no function satisfying the required conditions.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 14467,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all pairs $(m, n)$ of positive integers such that for every real polynomial $P(x)$ of degree $m$, there exists a real polynomial $Q(x)$ of degree $n$ such that $Q(P(x))$ is divisible by $Q(x)$ in $\\mathbb{R}[x]$.",
    "options": [],
    "answer": "See solution",
    "solution": "All $(m, n)$ with odd $m$ and arbitrary $n$, or $(m, n)$ with even $m$ and even $n$.\n\n*Proof:*\n\n1. **Case 1: $m$ odd.**\n   - For any $P(x)$ of odd degree, $P(x) - x$ has a real root $a$ (unless $P(x) \\equiv x$). Set $Q(x) = (x - a)^n$. Then $Q(P(x)) = (P(x) - a)^n$, which is divisible by $Q(x)$.\n\n2. **Case 2: $m$ even.**\n   - If $n$ is odd, take $P(x) = x^m + x + 1$. Then $P(x) - x > 0$ for all $x \\in \\mathbb{R}$, so $P(x) - x$ has no real roots. For any $Q(x)$ of odd degree, $Q(x)$ has a real root $c$, but $Q(P(c)) \\ne 0$, so $Q(x)$ does not divide $Q(P(x))$.\n   - If $n$ is even, either $P(x) - x$ has a real root (as above), or all roots are complex. In the latter case, let $z, \\bar{z}$ be a pair of complex conjugate roots of $P(x) - x$, and set $p(x) = (x - z)(x - \\bar{z})$. Then $Q(x) = (p(x))^k$ works, since $Q(P(x))$ is divisible by $Q(x)$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 14468,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $ABC$ be a triangle and let $S_b$ and $S_c$ be the midpoints of the sides $AC$ and $AB$, respectively. Prove that if $AB < AC$ then $\\angle B S_c C < \\angle B S_b C$.",
    "options": [],
    "answer": "See solution",
    "solution": "It suffices to prove that if $AB < AC$ then $S_b$ lies inside the circumcircle $k$ of triangle $B S_c C$.\n\nThe midline $S_b S_c$ is parallel to $BC$. Let the 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$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 14469,
    "subject": "Mathematics (Olympiad)",
    "question": "In the Cartesian plane, consider the set of points with integer coordinates:\n\n$$\nT = \\{(x, y) \\mid x, y \\in \\mathbb{Z};\\ |x|, |y| \\leq 20;\\ (x, y) \\neq (0, 0)\\}\n$$\n\nWe color some points of $T$ such that for each point $(x, y) \\in T$, exactly one of the two points $(x, y)$ and $(-x, -y)$ is colored.\n\nLet $N$ be the number of pairs $((x_1, y_1), (x_2, y_2))$ such that both $(x_1, y_1)$ and $(x_2, y_2)$ are colored and\n$$\nx_1 \\equiv 2x_2 \\pmod{41},\\quad y_1 \\equiv 2y_2 \\pmod{41}.\n$$\n\nFind all possible values of $N$.",
    "options": [],
    "answer": "See solution",
    "solution": "For any $a \\in \\mathbb{Z}$ and $X = (x_1, x_2) \\in T$, denote $aX = (x'_1, x'_2) \\in T$ where $x'_1 \\equiv a x_1$, $x'_2 \\equiv a x_2 \\pmod{41}$.\n\nLet $G$ be a simple, undirected graph with vertices in $T$, connecting each vertex $X$ to $2X$.\n\nSince $2^{20} \\equiv 1 \\pmod{41}$ and $2^{10} \\equiv -1 \\pmod{41}$, the graph $G$ decomposes into edge-disjoint cycles of length 20. Each cycle has the form:\n\n$$\n(X, 2X, \\dots, 2^9 X, -X, -2X, \\dots, -2^9 X)\n$$\n\nFrom the coloring condition, we color 10 vertices in each cycle so that exactly one of $2^i X$ and $-2^i X$ is colored for all $i = 0, 1, \\dots, 9$.\n\nIf we color $X, 2X, \\dots, 2^9 X$, then there are 9 pairs $(U, 2U)$ on the cycle where both are colored. If we color\n\n$$\nX, 2^2 X, 2^4 X, 2^6 X, 2^8 X, 2^9 X, -2X, -2^3 X, -2^5 X, -2^7 X,\n$$\n\nthen only one pair $(U, 2U)$ on the cycle is colored.\n\nChanging the coloring from $U$ to $-U$ changes the number of such pairs by 0, 2, or -2, so all possible values per cycle are odd numbers from 1 to 9. There are $(41^2 - 1)/20 = 84$ cycles. Thus, $N$ can take all even values from $84$ to $84 \\times 9 = 756$.\n\n$\\square$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 14470,
    "subject": "Mathematics (Olympiad)",
    "question": "Define $g(x) = f(3x) - f(x) - 1$. Given that $g(1) = g(3) = g(9) = g(27) = g(81) = 0$, and $g(x)$ is a polynomial of degree at most $5$, find the coefficient of $x$ in $f(x)$.",
    "options": [],
    "answer": "See solution",
    "solution": "Since $g(x)$ has roots at $x = 1, 3, 9, 27, 81$ and degree at most $5$, we can write:\n\n$$\ng(x) = k(x - 1)(x - 3)(x - 9)(x - 27)(x - 81)\n$$\nfor some constant $k$.\n\nGiven $g(0) = -1$, we have:\n$$\n-1 = k(-1)(-3)(-9)(-27)(-81) = k(-1 \\times -3 \\times -9 \\times -27 \\times -81)\n$$\nSo,\n$$\nk = \\frac{1}{1 \\times 3 \\times 9 \\times 27 \\times 81}\n$$\nThus,\n$$\ng(x) = \\frac{(x-1)(x-3)(x-9)(x-27)(x-81)}{1 \\times 3 \\times 9 \\times 27 \\times 81}\n$$\nThe coefficient of $x$ in $g(x)$ is $\\frac{1}{81} + \\frac{1}{27} + \\frac{1}{9} + \\frac{1}{3} + \\frac{1}{1}$.\n\nIf the coefficient of $x$ in $f(x)$ is $c$, then in $f(3x)$ it is $3c$, so in $g(x)$ it is $3c - c = 2c$. Therefore, the coefficient of $x$ in $f(x)$ is half that in $g(x)$:\n$$\n\\frac{1}{2}\\left(\\frac{1}{81} + \\frac{1}{27} + \\frac{1}{9} + \\frac{1}{3} + 1\\right) = \\frac{121}{162}\n$$",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 14471,
    "subject": "Mathematics (Olympiad)",
    "question": "Consider a sequence of integers $a_1, a_2, a_3, \\dots$ such that $a_1 > 1$ and $(2^{a_n} - 1)a_{n+1}$ is a square for all positive integers $n$. Is it possible that two terms of such a sequence be equal?",
    "options": [],
    "answer": "See solution",
    "solution": "The answer is negative. Notice first that if $a_n > 1$, then $2^{a_n} - 1 \\equiv 3 \\pmod{4}$; since $(2^{a_n} - 1)a_{n+1}$ is a perfect square, we should have $a_{n+1} \\equiv 0 \\pmod{4}$ or $a_{n+1} \\equiv 3 \\pmod{4}$, so in particular $a_{n+1} > 1$. As $a_1 > 1$, we conclude that all terms of the sequence are greater than 1.\n\nDenote the largest prime divisor of an integer $k > 1$ by $g(k)$. We will show that $g(a_{n+1}) > g(a_n)$ for all $n$, which yields the desired result. To this end, we use the lemma below.\n\n**Lemma:** For any prime $p$, each prime divisor of $2^p - 1$ is greater than $p$.\n\n*Proof.* Let $q$ be a prime factor of $2^p - 1$; then $q$ is odd. The multiplicative order $d$ of $2$ modulo $q$ divides $p$ and is larger than $1$, so $d = p$. On the other hand, by Fermat's little theorem, $2^{q-1} \\equiv 1 \\pmod{q}$, so $p = d \\mid q - 1$ and the lemma follows.\n\nChoose now any positive integer $n$, and denote, for convenience, $k = a_n$ and $\\ell = a_{n+1}$. Let $p = g(k)$; then $2^p - 1 \\mid 2^k - 1$. Since $2^p - 1 \\equiv 3 \\pmod{4}$, this number is not a square, so there exists a prime $q$ such that $v_q(2^p - 1)$ is odd. By the lemma, $q > p$, so in particular $q \\nmid k$. Therefore, by the Lifting Exponent Lemma,\n\n$$\nv_q(2^k - 1) = v_q(2^p - 1) + v_q(k/p) = v_q(2^p - 1) + 0,\n$$\n\nso $v_q(2^k - 1)$ is odd as well. Since $(2^k - 1)\\ell$ is a perfect square, we should then have $q \\mid \\ell$, so $g(\\ell) \\ge q > p = g(k)$, as desired.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 14472,
    "subject": "Mathematics (Olympiad)",
    "question": "On each side of an equilateral triangle of side $n \\ge 1$, consider $n-1$ points that divide the sides into $n$ equal segments. Through these points, draw lines parallel to the sides of the triangle, obtaining a net of equilateral triangles of side length 1. On each of the vertices of the small triangles, place a coin head up. A *move* consists of flipping over three mutually adjacent coins. Find all values of $n$ for which it is possible to turn all coins tail up after a finite number of moves.",
    "options": [],
    "answer": "See solution",
    "solution": "We shall prove that the admissible values of $n$ (those for which our goal can be achieved) are all the positive integers that are not divisible by 3.\n\nObviously, such turning is possible for $n = 1$. For $n = 2$, flip each of the four 1-sided equilateral triangles once and all the coins will be tail-up.\n\nWe shall now use induction with step 3. Assume that $n$ is an admissible value. Flipping the coins of each unit-sided triangle of an equilateral triangle of side length $n+3$, the coins at the vertices of the big triangle will be flipped once, those along the sides three times, and the interior coins six times each. Consequently, all the exterior coins are turned tail up and all the interior coins are heads up. But the interior coins form the net corresponding to an $n$-sided triangle, so the induction works.\n\nIf $3 \\mid n$, color the coins in red, yellow, and blue so that any three adjacent coins have different colors. Also, any three coins in a row will have different colors. In this case, the corners will all have the same color, say red. Since there are, in total, $\\frac{(n+1)(n+2)}{2} \\equiv 1 \\pmod{3}$ coins, there will be exactly one more red coin than yellow or blue ones. Thus, at the beginning, the parity of the number of red heads is different from the parity of the number of yellow heads. Since each move changes the parity of the number of heads of each color, we cannot end up with the parity of red heads equal to that of yellow or blue heads, which would be the case if all coins showed tails. Thus, the coins cannot all be inverted, so $n$ is not an admissible value.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 14473,
    "subject": "Mathematics (Olympiad)",
    "question": "We consider a $11 \\times 10$ cm rectangular table. The table is divided by parallel lines into 110 squares of side $1$ cm. We have tiles of cross shape, consisting of 6 squares of side $1$ cm, as in the figure. Determine the maximal possible number of non-overlapping tiles we can pack in the table such that each tile is covering exactly 6 small squares of the table.\n\n![](images/Hellenic_Mathematical_Competitions_2011_booklet_p20_data_1a21b06888.png)",
    "options": [],
    "answer": "See solution",
    "solution": "We observe that for the squares having a side on the border of the table, especially the four corner squares, we can put on each corner one tile covering only two squares, while four squares cannot be covered. Therefore, in the four corners, 16 squares will remain uncovered.\n\nFor the rest of the border squares:\n\n* On the sides of length $11$ cm, for each covered square, one also remains uncovered. So on these sides, at least two squares will remain uncovered.\n* On the sides of length $10$ cm, we can cover one square and another will remain uncovered.\n\nThus, in our effort to cover the border squares, at least $16 + 4 + 2 = 22$ squares will remain uncovered. Therefore, it is possible to cover at most $110 - 22 = 88$ squares. Since each tile covers exactly 6 squares, we can pack at most $\\left\\lfloor \\frac{88}{6} \\right\\rfloor = 14$ tiles. Figure 5 shows that such packing is possible. Therefore, the maximal number of tiles we can pack in the table is $14$.\n\n![](images/Hellenic_Mathematical_Competitions_2011_booklet_p21_data_41322539d9.png)\n\nFigure 5",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 14474,
    "subject": "Mathematics (Olympiad)",
    "question": "Prove that\n$$\n\\frac{1}{1 \\cdot 2013} + \\frac{1}{2 \\cdot 2012} + \\frac{1}{3 \\cdot 2011} + \\dots + \\frac{1}{2012 \\cdot 2} + \\frac{1}{2013 \\cdot 1} < 1.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "For arbitrary natural numbers $n \\neq 1 \\neq m$, the inequality $nm \\geq n + m$ holds, since $(n - 1)(m - 1) \\geq 1 \\Rightarrow nm - n - m + 1 \\geq 1 \\Rightarrow nm - n - m \\geq 0 \\Rightarrow nm \\geq n + m$, with equality only when $n = m = 2$.\n\nThen for $n \\geq 2$, we have\n$$\n\\frac{1}{n(2014 - n)} < \\frac{1}{n + 2014 - n} = \\frac{1}{2014}\n$$\nso\n$$\n\\frac{1}{1 \\cdot 2013} + \\frac{1}{2 \\cdot 2012} + \\frac{1}{3 \\cdot 2011} + \\dots + \\frac{1}{2012 \\cdot 2} + \\frac{1}{2013 \\cdot 1} < \\frac{2}{2013} + \\frac{2011}{2014} < \\frac{3}{2014} + \\frac{2011}{2014} = 1.\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 14475,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $n$ be a positive integer. Show that $2\\sqrt{2^n} \\cos\\left(n \\arccos \\frac{\\sqrt{2}}{4}\\right)$ is an odd integer.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $\\alpha = \\arccos \\frac{\\sqrt{2}}{4}$ and $z = \\cos \\alpha + i \\sin \\alpha$. Since $\\cos n\\alpha = \\frac{1}{2}(z^n + \\bar{z}^n)$, we have\n$$\n2\\sqrt{2^n} \\cos n\\alpha = \\sqrt{2^n}(z^n + \\bar{z}^n) = S_n.\n$$\nNotice that $z^n + \\bar{z}^n = (z + \\bar{z})(z^{n-1} + \\bar{z}^{n-1}) - z\\bar{z}(z^{n-2} + \\bar{z}^{n-2})$, so\n$$\nS_n = S_{n-1} - 2S_{n-2}\\quad \\text{for all } n \\ge 3.\n$$\nStart with $S_1 = 1$ and $S_2 = -3$ to get inductively that $S_n$ is an odd integer.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 14476,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $a$, $b$, $c$ be nonzero complex numbers of the same modulus for which the numbers $A = a + b + c$ and $B = abc$ are real. Prove that, for any nonnegative integer $n$, the number $C_n = a^n + b^n + c^n$ is real.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $|a| = |b| = |c| = r > 0$. For a complex number $z$ with modulus $r > 0$, we have $\\bar{z} = \\frac{r^2}{z}$. Since $A$ is real, we obtain:\n\n$$\na + b + c = \\bar{a} + \\bar{b} + \\bar{c} = \\frac{r^2}{a} + \\frac{r^2}{b} + \\frac{r^2}{c} = r^2 \\cdot \\frac{ab + bc + ca}{abc} \\in \\mathbb{R},$$\n\nwhich implies $ab + bc + ca \\in \\mathbb{R}$.\n\nFor $n = 0$ we have $C_0 = 3$, and for $n = 1$ we have $C_1 = A \\in \\mathbb{R}$. Also, for $n = 2$ we have $C_2 = A^2 - 2(ab + bc + ca) \\in \\mathbb{R}$. For $n \\geq 2$, we have:\n\n$$\nC_{n+1} = A \\cdot C_n - (ab + bc + ca) \\cdot C_{n-1} + B \\cdot C_{n-2}.\n$$\n\nBy induction, it now follows that $C_n \\in \\mathbb{R}$ for any $n \\in \\mathbb{N}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 14477,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $m, n$ be integers greater than $1$, and $n$ is not a perfect square. If $n^2 + n + 1$ is divisible by $m$, prove that\n\n$$\n|m - n| > \\sqrt{3n} - 2.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "Let $n = m + k$. Then\n\n$$\nk^2 + k + 1 = n^2 + n + 1 + m(m - 2n - 1) \\equiv n^2 + n + 1 \\equiv 0 \\pmod{m}.\n$$\n\nTherefore, $k^2 + k + 1$ is divisible by $m$. Since $k^2 + k + 1$ is a positive integer, we can write\n\n$$\nk^2 + k + 1 = m t, \\qquad (1)\n$$\n\nwhere $t$ is a positive integer.\n\nIf $t = 1$, then $m = k^2 + k + 1$, so $n = m + k$ is a perfect square, which contradicts the condition. Hence, $t > 1$.\n\nNote that $k^2 + k = k(k + 1)$ is even, so $k^2 + k + 1$ is odd. From (1), $t$ is odd, so $t \\geq 3$.\n\nTherefore, from (1),\n\n$$\nk^2 + k + 1 \\geq 3m = 3(n - k),\n$$\n\nso $3n \\leq k^2 + 4k + 1 < (k + 2)^2$.\n\nThus,\n\n$$\n|k + 2| > \\sqrt{3n},\n$$\n\nand hence $|k| > \\sqrt{3n} - 2$, i.e., $|m - n| > \\sqrt{3n} - 2$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 14478,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $a$, $b$, and $c$ be integers each with absolute value less than or equal to $10$. The cubic polynomial\n$$\nf(x) = x^3 + a x^2 + b x + c\n$$\nsatisfies the property\n$$\n|f(2 + \\, \\sqrt{3})| < 0.0001.\n$$\nDetermine if $2 + \\sqrt{3}$ is a root of $f$.",
    "options": [],
    "answer": "See solution",
    "solution": "Assume $2 + \\sqrt{3}$ is not a root of $f$. Then,\n$$\n\\begin{aligned}\nf(2 + \\sqrt{3}) &= (2 + \\sqrt{3})^3 + a(2 + \\sqrt{3})^2 + b(2 + \\sqrt{3}) + c \\\\\n&= (26 + 7a + 2b + c) + (15 + 4a + b) \\sqrt{3}.\n\\end{aligned}\n$$\nLet $m = 26 + 7a + 2b + c$ and $n = 15 + 4a + b$. Then $|m| < 130$ and $|n| \\leq 65$. Thus,\n$$\n|m + n \\sqrt{3}| \\leq 130 + 65 \\sqrt{3} < 260.\n$$\nSince $f(2 + \\sqrt{3})$ is nonzero, $m + n \\sqrt{3} \\neq 0$. Because $m$ and $n$ are integers and $\\sqrt{3}$ is irrational, $|m^2 - 3n^2| \\geq 1$. Therefore,\n$$\n|f(2 + \\sqrt{3})| = |m + n \\sqrt{3}| = \\left| \\frac{m^2 - 3n^2}{m - n \\sqrt{3}} \\right| \\geq \\frac{1}{260} \\geq 0.001,\n$$\ncontradicting $|f(2 + \\sqrt{3})| < 0.0001$. Thus, $2 + \\sqrt{3}$ must be a root of $f$.",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 14479,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $x$, $y$, and $z$ be positive real numbers such that $xyz = 3(x + y + z)$. Show that\n\n$$\n\\frac{1}{x^2(y+1)} + \\frac{1}{y^2(z+1)} + \\frac{1}{z^2(x+1)} \\geq \\frac{3}{4(x+y+z)}\n$$\n\nand determine the cases of equality.",
    "options": [],
    "answer": "See solution",
    "solution": "By the AM-GM inequality and the given condition, $x + y + z \\geq 9$. Thus,\n\n$$\n\\frac{3}{4(x + y + z)} \\sum (x + 1) = \\frac{3}{4} \\left( 1 + \\frac{3}{x + y + z} \\right) \\leq 1.\n$$\n\nApplying the Cauchy-Schwarz inequality and using the condition,\n\n$$\n\\left(\\sum \\frac{1}{x^2(y+1)}\\right) \\left(\\sum (y+1)\\right) \\geq \\left(\\sum \\frac{1}{x}\\right)^2 \\geq 3 \\sum \\frac{1}{xy} = 1.\n$$\n\nTherefore, the inequality holds. Equality occurs if and only if $x = y = z = 3$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 14480,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $D$ be the midpoint of side $BC$ of triangle $ABC$. Prove that the intersection point of the medians of triangle $ABD$ and that of triangle $ACD$ are equidistant from line $AD$.",
    "options": [],
    "answer": "See solution",
    "solution": "Triangles $ABD$ and $ACD$ have equal area since $|BD| = |CD|$ and the altitudes drawn from $A$ coincide (see the figure below). As these triangles have a common side $AD$, the altitudes drawn from vertices $B$ and $C$, respectively, must also be equal. Thus, $B$ and $C$ are equidistant from line $AD$.\n\n![](images/Estonija_2010_p15_data_d4b4f83d31.png)\n\nSince the point of intersection of medians divides each median in a $2:1$ ratio, the distance from the intersection point of the medians of $ABD$ to line $AD$ is one third the distance from $B$ to line $AD$. An analogous relation holds for triangle $ACD$. Hence, the claim follows.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 14481,
    "subject": "Mathematics (Olympiad)",
    "question": "All the diagonals of the lateral faces of an $n$-sided prism are colored in yellow and blue, so that single-colored diagonals do not have common points. Prove that the sum of the squares of the lengths of all yellow diagonals is equal to the sum of the squares of the lengths of the blue ones.\n\n![](images/Ukraine_booklet_2018_p53_data_8d976a9bbc.png)",
    "options": [],
    "answer": "See solution",
    "solution": "Let the bases of the prism be $A_1, A_2, \\ldots, A_n$ and $B_1, B_2, \\ldots, B_n$, and let the lateral edge vector be $\\vec{h} = \\overrightarrow{B_1A_1} = \\cdots = \\overrightarrow{B_nA_n}$.\n\nAccording to the condition, single-colored diagonals do not have common points. Without loss of generality, suppose the blue diagonals are $B_1A_2, B_2A_3, \\ldots, B_{n-1}A_n, B_nA_1$, and the others are yellow.\n\nSince $\\overrightarrow{B_1A_2} = \\vec{h} + \\overrightarrow{A_1A_2}$, we have:\n\n$$(\\overrightarrow{B_1A_2})^2 = (\\vec{h})^2 + (\\overrightarrow{A_1A_2})^2 + 2\\vec{h} \\cdot \\overrightarrow{A_1A_2}$$\n\nSimilarly, for each blue diagonal:\n\n$$(\\overrightarrow{B_kA_{k+1}})^2 = (\\vec{h})^2 + (\\overrightarrow{A_kA_{k+1}})^2 + 2\\vec{h} \\cdot \\overrightarrow{A_kA_{k+1}}$$\n\nSumming over all blue diagonals:\n\n$$\n\\begin{aligned}\n&\\quad\\; (\\overrightarrow{B_1A_2})^2 + \\cdots + (\\overrightarrow{B_nA_1})^2 \\\\\n&= n(\\vec{h})^2 + (\\overrightarrow{A_1A_2})^2 + \\cdots + (\\overrightarrow{A_nA_1})^2 + 2\\vec{h} \\cdot (\\overrightarrow{A_1A_2} + \\cdots + \\overrightarrow{A_nA_1})\n\\end{aligned}\n$$\n\nBut $\\overrightarrow{A_1A_2} + \\cdots + \\overrightarrow{A_nA_1} = 0$, since the sum of consecutive vectors around a closed polygon is zero. Thus, the cross term vanishes, and the sum simplifies to:\n\n$$n h^2 + A_1A_2^2 + \\cdots + A_nA_1^2$$\n\nA similar calculation for the yellow diagonals yields the same expression. Therefore, the sum of the squares of the lengths of all yellow diagonals equals that of the blue ones, as required.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 14482,
    "subject": "Mathematics (Olympiad)",
    "question": "Two sides of a triangular pyramid are equilateral triangles with length $a\\ \\mathrm{cm}$. The planes of these triangles are normal to each other. Find the area and volume of the pyramid.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $ABCD$ be a pyramid, where $ABC$ and $ABD$ are equilateral triangles with length $a\\ \\mathrm{cm}$, i.e., $\\overline{AC} = \\overline{BC} = \\overline{AB} = \\overline{AD} = \\overline{BD} = a\\ \\mathrm{cm}$. Let $CN$ and $DN$ be the heights in the triangles $ABC$ and $ABD$ respectively. Therefore $CN \\perp AB$ and\n\n![](images/Makedonija_2008_p28_data_ba3fdf2a89.png)\n\nIf the triangle $ABC$ is the pyramid base, then the height of the pyramid is $DN$, so the volume is\n$$\nV = \\frac{1}{3} \\cdot P_{\\Delta ABC} \\cdot \\overline{DN} = \\frac{1}{3} \\cdot \\frac{a^2 \\sqrt{3}}{4} \\cdot \\frac{a\\sqrt{3}}{2} = \\frac{a^3}{8}\\ \\mathrm{cm}^3\n$$\nbecause $CN \\perp DN$. $CND$ is an isosceles right triangle and $\\overline{CN} = \\overline{DN} = \\frac{a\\sqrt{3}}{2}\\ \\mathrm{cm}$.\nTherefore,\n$$\n\\overline{CD} = \\sqrt{\\left(\\frac{a\\sqrt{3}}{2}\\right)^2 + \\left(\\frac{a\\sqrt{3}}{2}\\right)^2} = \\frac{a\\sqrt{3}}{2} \\sqrt{2} = \\frac{a\\sqrt{6}}{2}\\ \\mathrm{cm}\n$$\nThe triangles $CDA$ and $CDB$ are congruent, so\n$$\nP_{\\Delta CDA} = P_{\\Delta CDB} = \\frac{1}{2} \\cdot \\frac{a\\sqrt{6}}{2} \\cdot \\sqrt{a^2 - \\left(\\frac{a\\sqrt{6}}{4}\\right)^2} = \\frac{1}{2} \\cdot \\frac{a\\sqrt{6}}{2} \\cdot \\frac{a\\sqrt{10}}{4} = \\frac{a^2\\sqrt{15}}{8}\\ \\mathrm{cm}^2\n$$\nFinally, the area of the pyramid is\n$$\nP = 2 \\cdot \\frac{a^2\\sqrt{3}}{4} + 2 \\cdot \\frac{a^2\\sqrt{5}}{8} = \\frac{a^2\\sqrt{3}(2+\\sqrt{5})}{4}\\ \\mathrm{cm}^2\n$$",
    "topic": "Geometry",
    "subtopic": "Solid Geometry"
  },
  {
    "id": 14483,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $A \\in \\mathcal{M}_n(\\mathbb{C})$ be a matrix with the property $A^T = -A$, where $A^T$ is the transpose of $A$.\n\na) If $A \\in \\mathcal{M}_n(\\mathbb{R})$ and $A^2 = O_n$, prove that $A = O_n$.\n\nb) If $n$ is an odd natural number and there is a matrix $B \\in \\mathcal{M}_n(\\mathbb{C})$ such that $A$ is the adjoint of $B$, prove that $A^2 = O_n$.",
    "options": [],
    "answer": "See solution",
    "solution": "a) Assume $A = (a_{ij})_{1 \\le i,j \\le n}$ and $A^2 = (m_{ij})_{1 \\le i,j \\le n}$. The property $A^T = -A$ leads to the relations $a_{ji} = -a_{ij}$ for $i, j = 1, \\dots, n$, that is, $A$ is antisymmetric. Then\n\n$$\nm_{ii} = \\sum_{j=1}^{n} a_{ij} a_{ji} = - \\sum_{j=1}^{n} a_{ij}^2, \\quad i = 1, \\dots, n.\n$$\n\nIf $A^2 = O_n$, then $m_{ii} = 0$ for $i = 1, \\dots, n$. Since $A \\in \\mathcal{M}_n(\\mathbb{R})$, we obtain $a_{ij} = 0$ for $i, j = 1, \\dots, n$. So $A = O_n$.\n\nb) From the assumption, $A = B^*$, where $B^*$ is the adjoint of $B$. Since $n$ is an odd integer, we obtain $\\det(A) = \\det(A^T) = \\det(-A) = (-1)^n \\det(A) = -\\det(A)$. Then $\\det(A) = 0$. Therefore $\\det(BB^*) = \\det(B) \\cdot \\det(B^*) = \\det(B) \\cdot \\det(A) = 0$. From the relation $BB^* = \\det(B)I_n$, we get $\\det(B) = 0$. Hence $\\operatorname{rank}(B) \\le n-1$ and $BB^* = O_n$.\n\n**Case 1.** If $\\operatorname{rank}(B) \\le n-2$, then $B^* = O_n$. Thus, $A^2 = (B^*)^2 = O_n$.\n\n**Case 2.** If $\\operatorname{rank}(B) = n-1$, then $B^* \\ne O_n$. From the Sylvester rank inequality, we have $\\operatorname{rank}(B^*) \\le \\operatorname{rank}(BB^*) + n - \\operatorname{rank}(B) = 1$. We conclude $\\operatorname{rank}(B^*) = 1$ and $(B^*)^2 = \\operatorname{tr}(B^*)B^*$. But $\\operatorname{tr}(B^*) = \\operatorname{tr}(A) = 0$ because $a_{ii} = -a_{ii}$ for $i = 1, 2, \\dots, n$. Finally, we get $A^2 = (B^*)^2 = O_n$.",
    "topic": "Algebra",
    "subtopic": "Linear Algebra"
  },
  {
    "id": 14484,
    "subject": "Mathematics (Olympiad)",
    "question": "Given a stripe $1 \\times n$, $n \\geq 4$. In each cell, a positive integer is written (not necessarily the same in each cell). Under each number, write a positive integer equal to the number of times that integer appears in the previous row. Repeat this procedure for each new row.\n\n1. Prove that after a finite number of steps, the rows will stabilize (i.e., will not change).\n2. For $n = 2016$ and for arbitrary $n$, what is the maximal number of steps during which the next row can differ from the previous one?",
    "options": [],
    "answer": "See solution",
    "solution": "Let $k$ be the greatest positive integer such that $2^k \\leq n$. If $n \\neq 2^k + 1$, $n \\neq 2^k + 2$, and $n \\neq 2^k + 4$, then the number of steps before stabilization is $k+1$. If $n=6$, there are 3 steps; if $n=12$, there are 4 steps; otherwise, there are $k$ steps. Since $2016 = 2^{10} + 992$, for $n=2016$ the answer is 11 steps.",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 14485,
    "subject": "Mathematics (Olympiad)",
    "question": "A school has fewer than 400 students in 6th grade. They are divided into several classes. Six of these classes have an equal number of students and together have more than 150 students. In the remaining classes, there are 15\\% more students than in these six classes together. How many 6th grade students are there in the school?",
    "options": [],
    "answer": "See solution",
    "solution": "Let $n$ be the total number of students in the six classes with equal numbers. Then $6 \\mid n$. The remaining classes have $15\\%$ more students than these six classes, so they have $1.15n$ students. The total number of students is $n + 1.15n = 2.15n < 400$, and $n > 150$. Also, $n$ must be divisible by $6$ and $20$ (since $0.15n$ must be integer), so $n = 60k$. The smallest $n > 150$ divisible by $60$ is $180$. For $n = 180$, $2.15 \\times 180 = 387 < 400$. The next possible $n$ is $240$, but $2.15 \\times 240 = 516 > 400$. Thus, the total number of 6th grade students is $387$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 14486,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $d$ be the tangent at $B$ to the circumcircle of the acute scalene triangle $ABC$. Let $K$ be the orthogonal projection of the orthocenter $H$ of triangle $ABC$ onto the line $d$, and let $L$ be the midpoint of the side $AC$. Prove that the triangle $BKL$ is isosceles.",
    "options": [],
    "answer": "See solution",
    "solution": "Without loss of generality, suppose $AB < BC$. Denote by $A'$ and $C'$ the feet of the altitudes from $A$ and $C$, respectively. Points $A'$, $C'$, and $K$ belong to the circle of diameter $BH$. The quadrilateral $ACA'C'$ is cyclic, therefore $\\angle BC'A' = \\angle C = \\angle KBA$, so the trapezoid $BKC'A'$ is cyclic, hence isosceles.\n\n![](images/RMC_2019_var_3_p58_data_be5d76d18d.png)\n\nOn the other hand, $C'L$ and $A'L$ are medians in the right triangles $ACC'$ and $ACA'$, respectively, so $C'L = A'L = \\frac{BC}{2}$ and the triangle $LA'C'$ is isosceles. It is easy to prove the congruence of the triangles $LC'K$ and $LA'B$ (SAS), and the conclusion follows.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 14487,
    "subject": "Mathematics (Olympiad)",
    "question": "What is the least value of $n$ such that $n!$ is a multiple of $2024$?\n\n(A) 11 (B) 21 (C) 22 (D) 23 (E) 253",
    "options": [],
    "answer": "See solution",
    "solution": "The prime factorization of $2024$ is $2^3 \\cdot 11 \\cdot 23$. Therefore, $n!$ is a multiple of $2024$ if and only if $n \\geq 23$. Thus, $23$ is the least value of $n$ such that $n!$ is a multiple of $2024$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 14488,
    "subject": "Mathematics (Olympiad)",
    "question": "$n \\geq 4$ points in the plane are given such that every three of them are not collinear. Prove that there exists a triangle such that all the points are in its interior, and on each of its sides lies exactly one point of the given points.",
    "options": [],
    "answer": "See solution",
    "solution": "The given points are finitely many, so there exists a disk which contains them in its interior.\n\nNamely, we search for the point which is at a maximal distance from the origin of the coordinate system. If we denote that distance by $R$, then the disk centered at the origin with radius $2R$ contains all the given points.\n\nWe draw all possible lines between the $n$ points in the plane. There are finitely many such lines ($\\binom{n}{2}$), so we can choose a point $A$ which does not lie on any of those lines and is outside the disk containing the points. The largest angle under which every segment in the disk is seen from $A$ is acute. We draw an arbitrary line through $A$ which does not intersect the disk. We rotate that line until it passes through a point from the given $n$ points, denoted $A_1$. On that line lies one side of the triangle. We continue rotating the line until it passes through another point from the given points, denoted $A_n$. On that line lies another side of the triangle. Each of the two lines passes only through $A_1$ and $A_n$ respectively, since $A$ does not lie on any of the $\\binom{n}{2}$ lines.\n\nLet $A_i$ denote the point from the given $n$ points which is at the greatest distance from $A$, and let $d$ denote that distance. There may be several such points, but we pick one arbitrarily. We draw a tangent to the circle centered at $A$ with radius $d$. On that line lies the third side of the triangle.\n\n![](images/makedonija2012_p7_data_695dfbff1b.png)",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 14489,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $a$, $b$, and $c$ be real numbers such that $a + b + c = 1$.\n\nProve that\n$$\n\\sqrt{(c+a)^2 + c^2} + \\sqrt{(a+b)^2 + a^2} + \\sqrt{(b+c)^2 + b^2} \\geq \\sqrt{5}.\n$$\n\nEquality holds when $a = b = c = \\frac{1}{3}$.\n\n![](images/1.2_1997-2023_IMO_HK_TST_Solutions_p109_data_fa876ae1da.png)",
    "options": [],
    "answer": "See solution",
    "solution": "Using the condition $a + b + c = 1$, we have:\n\n$$\na^2 - 2bc + 2c = a^2 + 2c(1 - b) = a^2 + 2c(a + c) = (c + a)^2 + c^2.\n$$\n\nSimilarly, the left-hand side of the inequality becomes:\n\n$$\n\\sqrt{(c + a)^2 + c^2} + \\sqrt{(a + b)^2 + a^2} + \\sqrt{(b + c)^2 + b^2}.\n$$\n\nBy the triangle inequality, this is bounded below by:\n\n$$\n\\sqrt{((c + a) + (a + b) + (b + c))^2 + (c + a + b)^2} = \\sqrt{5}.\n$$\n\nEquality holds when $a = b = c = \\frac{1}{3}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 14490,
    "subject": "Mathematics (Olympiad)",
    "question": "Solve the following system of equations:\n\n$$\n\\begin{cases}\n\\sqrt{x^2 - 2x + 6} \\log_3 (6-y) = x \\\\\n\\sqrt{y^2 - 2y + 6} \\log_3 (6-z) = y \\\\\n\\sqrt{z^2 - 2z + 6} \\log_3 (6-x) = z\n\\end{cases}\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "Conditions for $x$, $y$, $z$ for the definition of the expressions in the equation are: $x, y, z < 6$.\n\nThe given system of equations is equivalent to:\n\n$$\n\\begin{aligned}\n\\log_3(6-y) &= \\frac{x}{\\sqrt{x^2-2x+6}} \\quad (1) \\\\\n\\log_3(6-z) &= \\frac{y}{\\sqrt{y^2-2y+6}} \\quad (2) \\\\\n\\log_3(6-x) &= \\frac{z}{\\sqrt{z^2-2z+6}} \\quad (3)\n\\end{aligned}\n$$\n\nThe function $f(x) = \\frac{x}{\\sqrt{x^2-2x+6}}$ is increasing since\n\n$$\nf'(x) = \\frac{6-x}{(x^2 - 2x + 6)\\sqrt{x^2 - 2x + 6}} > 0\n$$\n\nfor $x < 6$.\n\nThe function $g(x) = \\log_3(6-x)$ is decreasing for $x < 6$.\n\nWe now prove that if $(x, y, z)$ is a solution to the given system of equations then $x = y = z = 3$.\n\nWithout loss of generality, suppose that $\\max(x, y, z) = x$. Consider two cases:\n\n1. $x \\geq y \\geq z$\n\nAs $f(x)$ is increasing, (1), (2), (3) show that\n\n$$\n\\log_3(6-y) \\geq \\log_3(6-z) \\geq \\log_3(6-x),\n$$\n\nso $x \\geq z \\geq y$. But $y \\geq z$, so $z = y$. Then (1) and (2) imply $x = y = z$.\n\n2. $x \\geq z \\geq y$\n\nAnalogously, $\\log_3(6-y) \\geq \\log_3(6-x) \\geq \\log_3(6-z)$ and so $z \\geq x \\geq y$. But $x \\geq z$, so $x = z$. Then (1) and (3) imply $x = y = z$.\n\nThe equation $f(x) = g(x)$ has a unique root $x = 3$. Consequently, the given system of equations has a unique solution $x = y = z = 3$.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 14491,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $D$ be a point in the interior of triangle $ABC$ such that $\\angle BAD = 40^\\circ$, $\\angle DAC = 30^\\circ$, $\\angle BCD = 20^\\circ$, and $\\angle DCA = 50^\\circ$. Find $\\angle CBD$.\n\n![](images/ROU_ABooklet_2021_p15_data_17884e1b73.png)",
    "options": [],
    "answer": "See solution",
    "solution": "Observe that $ABC$ is an isosceles triangle, with $BA = BC$ and $\\angle ABC = 40^\\circ$. Suppose that the perpendicular bisector of the triangle's base intersects $AD$ at $T$.\n\nSince $\\angle DAC < \\angle DCA$, we have $CD < DA$, therefore $T \\in (AD)$ and $\\angle TBC = 20^\\circ$.\n\nWe will prove that $D$ is the incenter of triangle $BTC$, hence $BD$ bisects $\\angle TBC$, and it follows that $\\angle CBD = 10^\\circ$.\n\nBecause $TCA$ is an isosceles triangle, we have $\\angle TCA = 30^\\circ$, and hence $\\angle TCD = 20^\\circ = \\angle BCD$. It follows that $CD$ bisects $\\angle BCT$.\n\nA short computation shows that $\\angle CTB = \\angle BTA = 120^\\circ$ and $\\angle DTC = 60^\\circ$, therefore $TD$ is the angle bisector of $\\angle BTC$.\n\nWe conclude that $D$ is the incenter of triangle $BCT$, as desired.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 14492,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $ABC$ be an acute triangle. A circle with diameter $AC$ intersects $[AB]$ and $[BC]$ at $K$ and $L$, respectively. The circumscribed circle of triangle $ABC$ intersects lines $CK$ and $AL$ at $F \\ne C$ and $D \\ne A$, respectively. Let $E$ be a point on the smaller arc $AC$ of the circumscribed circle, and let $N$ be the intersection point of $BE$ and $AC$. Given\n\n$$|AF|^2 + |BD|^2 + |CE|^2 = |AE|^2 + |CD|^2 + |BF|^2$$\n\nprove that $m(\\overline{KNB}) = m(\\overline{BNL})$.\n\n![](path \"title\")",
    "options": [],
    "answer": "See solution",
    "solution": "Since $AC$ is a diameter of the circle passing through $A$, $K$, $L$, $C$, we have $CK \\perp AB$ and $AL \\perp BC$. Since $ABC$ is acute, $K \\in [AB]$ and $L \\in [BC]$. Let $CK \\cap AL = H$, and denote $\\angle BAC = A$, $\\angle BCA = C$. Since all heights are concurrent, $BH \\perp AC$. Therefore, $m(\\angle HBC) = 90^\\circ - m(C)$ or $m(\\angle BHA) = 180^\\circ - m(C)$. Let $F'$ be the point symmetric to $H$ with respect to $AB$. Then $m(\\angle BF'A) = 180^\\circ - m(C)$, so $F'$ lies on the circumscribed circle of $ABC$. Thus, $F = F'$. Similarly, $D$ is the point symmetric to $H$ with respect to $BC$. As a result, $|AF| = |AH|$, $|CD| = |CH|$, and $|BD| = |BH| = |BF|$. Given $|AF|^2 + |BD|^2 + |CE|^2 = |AE|^2 + |CD|^2 + |BF|^2$, we get $|AH|^2 + |CE|^2 = |CH|^2 + |AE|^2$, or $HE \\perp AC$. Since $BH \\perp AC$, $B$, $H$, $E$ are collinear. Therefore, $BN \\perp AC$, and points $A$, $K$, $N$, $H$ lie on the same circle. Thus, $m(KAH) = m(BNL)$, implying $m(KNB) = m(BNL)$. Done.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 14493,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $\\Gamma$ be a connected graph on $r + g + b + 1$ vertices. The edges of $\\Gamma$ are colored red, green, or blue. It is known that $\\Gamma$ has a spanning tree with exactly $r$ red edges, a spanning tree with exactly $g$ green edges, and a spanning tree with exactly $b$ blue edges. Prove that $\\Gamma$ has a spanning tree with exactly $r$ red edges, exactly $g$ green edges, and exactly $b$ blue edges.",
    "options": [],
    "answer": "See solution",
    "solution": "Induct on $n = r + g + b$. The base case, $n = 1$, is clear.\n\nLet now $n > 1$. Let $V$ denote the vertex set of $\\Gamma$, and let $T_r$, $T_g$, and $T_b$ be the trees with exactly $r$ red edges, $g$ green edges, and $b$ blue edges, respectively. Consider two cases.\n\n*Case 1:* There exists a partition $V = A \\cup B$ of the vertex set into two non-empty parts such that the edges joining the parts all bear the same colour, say, blue.\n\nSince $\\Gamma$ is connected, it has a (necessarily blue) edge connecting $A$ and $B$. Let $e$ be one such.\n\nAssume that $T$, one of the three trees, does not contain $e$. Then the graph $T \\cup \\{e\\}$ has a cycle $C$ through $e$. The cycle $C$ should contain another edge $e'$ connecting $A$ and $B$; the edge $e'$ is also blue. Replace $e'$ by $e$ in $T$ to get another tree $T'$ with the same number of edges of each colour as in $T$, but containing $e$.\n\nPerforming such an operation to all three trees, we arrive at the situation where the three trees $T'_r$, $T'_g$, and $T'_b$ all contain $e$. Now shrink $e$ by identifying its endpoints to obtain a graph $\\Gamma^*$, and set $r^* = r$, $g^* = g$, and $b^* = b - 1$. The new graph satisfies the conditions in the statement for those new values — indeed, under the shrinking, each of the trees $T'_r$, $T'_g$, and $T'_b$ loses a blue edge. So $\\Gamma^*$ has a spanning tree with exactly $r$ red, exactly $g$ green, and exactly $b - 1$ blue edges. Finally, pass back to $\\Gamma$ by restoring $e$, to obtain the desired spanning tree in $\\Gamma$.\n\n*Case 2:* There is no such partition.\n\nConsider all possible collections $(R, G, B)$, where $R$, $G$ and $B$ are acyclic sets consisting of $r$ red edges, $g$ green edges, and $b$ blue edges, respectively. By the problem assumptions, there is at least one such collection. Amongst all such collections, consider one such that the graph on $V$ with edge set $R \\cup G \\cup B$ has the smallest number $k$ of components. If $k = 1$, then the collection provides the edges of a desired tree (the number of edges is one less than the number of vertices).\n\nAssume now that $k \\geq 2$; then in the resulting graph some component $K$ contains a cycle $C$. Since $R$, $G$, and $B$ are acyclic, $C$ contains edges of at least two colours, say, red and green. By assumption, the edges joining $V(K)$ to $V \\setminus V(K)$ bear at least two colours; so one of these edges is either red or green. Without loss of generality, consider a red such edge $e$.\n\nLet $e'$ be a red edge in $C$ and set $R' = R \\setminus \\{e'\\} \\cup \\{e\\}$. Then $(R', G, B)$ is a valid collection providing a smaller number of components. This contradicts minimality of the choice above and concludes the proof.",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 14494,
    "subject": "Mathematics (Olympiad)",
    "question": "Prove that for any pair of positive integers $k$ and $n$, there exist $k$ positive integers $m_1, m_2, \\dots, m_k$ (not necessarily different) such that\n\n$$\n1 + \\frac{2^k - 1}{n} = \\left(1 + \\frac{1}{m_1}\\right) \\left(1 + \\frac{1}{m_2}\\right) \\dots \\left(1 + \\frac{1}{m_k}\\right).\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "The proof is by simple induction on $k$.\n\nThe base case, $k=1$, is satisfied by $m_1 = n$.\n\nFor the inductive step, assume that the statement is true for $k = t$.\n\n**Case 1.** If $n$ is even, write $n = 2r$ for some positive integer $r$. The expression for $k = t + 1$ may be written as\n\n$$\n\\begin{aligned}\n1 + \\frac{2^{t+1} - 1}{n} &= \\frac{2r + 2^{t+1} - 1}{2r} \\\\\n&= \\frac{2r + 2^{t+1} - 1}{2r + 2^{t+1} - 2} \\cdot \\frac{2r + 2^{t+1} - 2}{2r} \\\\\n&= \\left(1 + \\frac{1}{2r + 2^{t+1} - 2}\\right) \\left(1 + \\frac{2^t - 1}{r}\\right).\n\\end{aligned}\n$$\n\nIf we set $m_1, m_2, \\dots, m_t$ according to the solution for $k = t$ and $n = r$ and also set $m_{t+1} = 2r + 2^{t+1} - 2$, then we have shown that the statement is true for $k = t + 1$.\n\n**Case 2.** If $n$ is odd, write $n = 2r - 1$ for some positive integer $r$. The expression for $k = t + 1$ may be written as\n\n$$\n\\begin{aligned}\n1 + \\frac{2^{t+1} - 1}{n} &= \\frac{2r + 2^{t+1} - 2}{2r - 1} \\\\\n&= \\frac{2r + 2^{t+1} - 2}{2r} \\cdot \\frac{2r}{2r - 1} \\\\\n&= \\left(1 + \\frac{2^t - 1}{r}\\right) \\left(1 + \\frac{1}{2r - 1}\\right).\n\\end{aligned}\n$$\n\nIf we set $m_1, m_2, \\dots, m_t$ according to the solution for $k = t$ and $n = r$ and also set $m_{t+1} = 2r - 1$, then we have shown that the statement is true for $k = t + 1$.\n\nThus the induction is complete and this concludes the proof.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 14495,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $s \\ge 2$ and $n \\ge k \\ge 2$ be integers, and let $\\mathcal{A}$ be a subset of $\\{1, 2, \\dots, n\\}^k$ of size at least $2sk^2n^{k-2}$ such that any two members of $\\mathcal{A}$ share some entry. Prove that there are an integer $p \\le k$ and $s+2$ members $A_1, A_2, \\dots, A_{s+2}$ of $\\mathcal{A}$ such that $A_i$ and $A_j$ share the $p$-th entry alone, whenever $i \\ne j$.",
    "options": [],
    "answer": "See solution",
    "solution": "Fix a member $A$ of $\\mathcal{A}$. Note that there are at most $\\binom{k}{2} n^{k-2}$ $k$-tuples that share at least two entries with $A$. Indeed, there are $n^{k-2}$ $k$-tuples sharing any two given entries. The bound now follows, since the two entries can be chosen in $\\binom{k}{2}$ different ways.\n\nTherefore, the number of members of $\\mathcal{A}$ that share a single entry with $A$ is at least\n\n$$\n|\\mathcal{A}| - \\binom{k}{2} n^{k-2} \\ge 2sk^2 n^{k-2} - k^2 n^{k-2} > sk^2 n^{k-2}.\n$$\n\nLetting $\\mathcal{B}_p$ be the set of all members of $\\mathcal{A}$ that share the $p$-th entry alone with $A$, the above estimate yields $\\sum_{p=1}^k |\\mathcal{B}_p| > sk^2 n^{k-2}$, and hence $|\\mathcal{B}_p| > sk n^{k-2}$ for some $p$.\n\nLet $B$ be an arbitrary member of this $\\mathcal{B}_p$. Then there are at most $(k-1) n^{k-2}$ $k$-tuples that share the $p$-th entry and some other entry with $B$. Now, let $t$ be maximal with the property that there are $B_1, B_2, \\dots, B_t$ in $\\mathcal{B}_p$ such that any two $B_i$ and $B_j$ share the $p$-th entry alone for $i \\ne j$. Hence any other member $B'$ of $\\mathcal{B}_p$ shares with some $B_i$ at least one entry different from the $p$-th. Consequently, $|\\mathcal{B}_p| \\le t (k-1) n^{k-2}$.\n\nFinally, compare the two bounds for $|\\mathcal{B}_p|$ to get $t > s$, and conclude that $A, B_1, \\dots, B_t$ are $t+1 \\ge s+2$ members of $\\mathcal{A}$ every two of which share the $p$-th entry alone.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 14496,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $n \\ge 2$ be a positive integer. A square with side length $2n - 1$ is divided into $(2n - 1)^2$ unit squares using lines parallel to the sides. Ana and Bogdan play the following game: starting with Ana, the two take turns to colour, Ana with red, Bogdan with blue, in $2n^2$ turns, the $4n^2$ vertices of the unit squares. Then, starting with Ana, each of them connects with a vector a red dot (as the tail) to a blue dot (as the head), obtaining $2n^2$ vectors with distinct tails and heads. If the sum of these vectors is zero, Ana wins. Otherwise, Bogdan wins. Show that Bogdan has a winning strategy.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $O$ be the centre of the square, $B_1, B_2, \\dots, B_{2n^2}$ the blue points, and $R_1, R_2, \\dots, R_{2n^2}$ the red points. Ana will win if\n$$\n\\sum_{i=1}^{2n^2} \\overrightarrow{R_i B_{a_i}} = \\overrightarrow{0},\n$$\nfor some rearrangement $a_1, \\dots, a_{2n^2}$ of $1, \\dots, 2n^2$.\n\nThis can be rewritten as\n$$\n\\sum_{i=1}^{2n^2} \\overrightarrow{OB_{a_i}} - \\overrightarrow{OR_i} = 0 \\iff \\sum_{i=1}^{2n^2} \\overrightarrow{OR_i} = \\sum_{i=1}^{2n^2} \\overrightarrow{OB_i}.\n$$\nSince $O$ is a centre of symmetry for the $4n^2$ coloured points,\n$$\n\\sum_{i=1}^{2n^2} \\overrightarrow{OR_i} + \\sum_{i=1}^{2n^2} \\overrightarrow{OB_i} = 0.\n$$\nThus, Ana wins if and only if\n$$\n\\sum_{i=1}^{2n^2} \\overrightarrow{OR_i} = 0.\n$$\n\nTo prevent this, Bogdan can use the following strategy: after Ana colours her last red point in the penultimate turn, let $\\vec{s} = \\sum_{i=1}^{2n^2-1} \\overrightarrow{OR_i}$. If there is an uncoloured point $X$ such that $\\overrightarrow{OX} = -\\vec{s}$, Bogdan colours $X$ blue. Then,\n$$\n\\sum_{i=1}^{2n^2} \\overrightarrow{OR_i} = \\overrightarrow{OR_{2n^2}} + \\vec{s} = \\overrightarrow{OR_{2n^2}} - \\overrightarrow{OX} = \\overrightarrow{XR_{2n^2}} \\neq 0.\n$$\nOtherwise, Bogdan colours any remaining point, and Ana cannot win. Thus, Bogdan always has a winning strategy.",
    "topic": "Algebra",
    "subtopic": "Linear Algebra"
  },
  {
    "id": 14497,
    "subject": "Mathematics (Olympiad)",
    "question": "令不等邊三角形 $\\triangle ABC$ 的內切圓圓心為 $I$，且該內切圓分別切 $CA, AB$ 於點 $E, F$。設 $\\triangle AEF$ 的外接圓在 $E$ 和 $F$ 的兩條切線交於點 $S$。直線 $EF$ 與 $BC$ 交於點 $T$。證明：以 $ST$ 為直徑的圓垂直於 $\\triangle BIC$ 的九點圓。\n\n註一：三角形的九點圓，即通過三邊中點、三高垂足、三頂點分別與垂心連線的中點等九個點的圓。\n\n註二：兩圓垂直的定義是：以兩圓的任一交點對兩圓所引的切線互相垂直。\n\nIn a scalene triangle $ABC$ with incenter $I$, the incircle is tangent to sides $CA$ and $AB$ at points $E$ and $F$. The tangents to the circumcircle of $\\triangle AEF$ at $E$ and $F$ meet at $S$. Lines $EF$ and $BC$ intersect at $T$. Prove that the circle with diameter $ST$ is orthogonal to the nine-point circle of $\\triangle BIC$.",
    "options": [],
    "answer": "See solution",
    "solution": "令 $D$ 為 $I$ 對 $BC$ 的垂足。令 $X, Y$ 分別為 $B, C$ 對 $CI, BI$ 的垂足。可證 $BIFX, CIEY$ 共圓，因此 $X, Y$ 落在直線 $EF$ 上。設 $M$ 為 $BC$ 的中點，$\\omega$ 為 $DMXY$ 的外接圓。原題等價於證明 $T$ 落在 $S$ 對 $\\omega$ 的極線上。\n\n設 $K$ 為 $AM$ 與 $EF$ 的交點，根據 SL 2005 G6 可得 $K, I, D$ 共線。令 $N$ 為 $EF$ 的中點，$L$ 為 $KS$ 與 $BC$ 的交點。由\n\n$$\n-1 = (A, I; N, S) \\stackrel{K}{=} (T, L; M, D)\n$$\n\n以及\n\n$$\n-1 = (T, D; B, C) \\stackrel{I}{=} (T, K; Y, X)\n$$\n\n得 $T = MD \\cap YX$ 為直線 $KL$ 對 $\\omega$ 的極點，得證。",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 14498,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $ABCD$ be a trapezium with $AB \\parallel CD$. Let $P$ be a point on $AC$ such that $C$ is between $A$ and $P$, and let $X$, $Y$ be the midpoints of $AB$ and $CD$, respectively. Let $PX$ intersect $BC$ at $N$ and $PY$ intersect $AD$ at $M$. Prove that $MN \\parallel AB$.",
    "options": [],
    "answer": "See solution",
    "solution": "Observe that\n\n$$\n\\frac{BN}{NC} = \\frac{[PNB]}{[PNC]}\n$$\n\nHowever, $[PNB] + [XNC] = [PXB] = [PXA] = [PCN] + [ACN] + [AXN]$. Since $[XNB] = [AXN]$, we obtain $[PNB] = [PNC] + [ACN]$. Thus,\n\n$$\n\\frac{BN}{NC} = \\frac{[PNC] + [ACN]}{[PNC]} = 1 + \\frac{[ACN]}{[PNC]} = 1 + \\frac{AC}{PC}.\n$$\n\nSimilarly, we can prove that\n\n$$\n\\frac{AM}{MD} = 1 + \\frac{AC}{CP}.\n$$\n\nComparison shows that $\\frac{AM}{MD} = \\frac{BN}{NC}$. We conclude that $MN \\parallel AB$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 14499,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $P(N)$ be the sum of squares of the product of the subsets of $\\{1, 2, \\ldots, N\\}$. For example:\n\n- For $N = 0$, there is only the empty subset, so $P(0) = 1$.\n- For $N = 1$, the subsets are $\\varnothing$ and $\\{1\\}$, so $P(1) = 2$.\n- For $N = 2$, the subsets are $\\varnothing$, $\\{1\\}$, $\\{2\\}$, and $\\{1,2\\}$, so $P(2) = 6$.\n- For $N = 3$, $P(3) = 24$.\n\nFind a formula for $P(N)$ for all integers $N \\geq 0$.",
    "options": [],
    "answer": "See solution",
    "solution": "We conjecture that $P(N) = (N+1)!$.\n\n*Proof by induction:*\n\nFor $N = 0$, $P(0) = 1 = 1!$.\nFor $N = 1$, $P(1) = 2 = 2!$.\nFor $N = 2$, $P(2) = 6 = 3!$.\nFor $N = 3$, $P(3) = 24 = 4!$.\n\nSuppose $P(n) = (n+1)!$ and $P(n-1) = n!$ for some $n \\geq 1$.\n\nConsider $P(n+1)$. The subsets of $\\{1,2,\\ldots,n+1\\}$ either contain $n+1$ or not.\n- Subsets not containing $n+1$ contribute $P(n)$.\n- Subsets containing $n+1$ correspond to subsets of $\\{1,2,\\ldots,n\\}$, and each product is multiplied by $n+1$; the sum of squares is $(n+1)^2 P(n-1)$.\n\nThus,\n$$\nP(n+1) = P(n) + (n+1)^2 P(n-1)\n$$\nBy the induction hypothesis,\n$$\nP(n+1) = (n+1)! + (n+1)^2 n! = (n+1)! + (n+1)(n+1) n! = (n+1)! + (n+1)(n+1)! = (n+2)!\n$$\nTherefore, $P(N) = (N+1)!$ for all $N \\geq 0$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 14500,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all pairs of real numbers $k, l$ such that the inequality\n\n$$\nka^2 + lb^2 > c^2\n$$\n\nholds for side lengths $a, b, c$ of any triangle.",
    "options": [],
    "answer": "See solution",
    "solution": "Assume that for some $k, l$, the given inequality holds for all side lengths $a, b, c$ of any triangle. Plugging in $a = c = 1$ and arbitrary $b < 2$, we get $k + l b^2 > 1$. If $k < 1$ then we can find sufficiently small $b > 0$ that makes the inequality false. Hence $k \\geq 1$ and likewise $l \\geq 1$.\n\nLet $ABC$ be a triangle in the coordinate plane. Without loss of generality, let $A = [-1, 0]$, $B = [1, 0]$, and denote $C = [x, y]$ ($y \\neq 0$). Using the Pythagorean Theorem, we express the side lengths in terms of $x, y$ and plug them into the given inequality to obtain\n\n$$\nk((x-1)^2 + y^2) + l((x+1)^2 + y^2) > 4\n$$\n\nwhich rewrites as\n\n$$\n(k + l)x^2 + 2(l - k)x + k + l - 4 > -(k + l)y^2. \\quad (1)\n$$\n\nThis inequality has to hold for any $x$ and any $y \\neq 0$. However, varying $y$, the right-hand side attains all negative values (recall $k + l > 0$). Therefore, for any $x$ we have\n\n$$\n(k + l)x^2 + 2(l - k)x + (k + l - 4) \\geq 0, \\quad (2)\n$$\nwhich happens if and only if the discriminant $D = 4(l - k)^2 - 4(k + l - 4)(k + l)$ is not positive. The inequality $D \\leq 0$ rewrites to $kl \\geq k + l$.\n\nWe found that if numbers $k, l$ satisfy the given inequality for any triplet of side lengths of a triangle, then $k \\geq 1$, $l \\geq 1$, and\n\n$$\nkl \\geq k + l. \\tag{3}\n$$\n\nOn the other hand, the conjunction of these three conditions is also sufficient: The third condition implies that (2) is satisfied for any real $x$, and since the right-hand side of (1) is negative for $y \\neq 0$, inequality (1) is satisfied for any $x$ and any $y \\neq 0$. Finally, inequality (1) is equivalent to the given inequality.\n\nThe set of all admissible pairs $k, l$ can also be written as\n\n$$\n\\{(k, l): k > 1 \\land l \\geq \\frac{k}{k - 1}\\}.\n$$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 14501,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $AD$ be the internal bisector of $\\angle A$ in triangle $ABC$. Point $C'$ lies on $AB$. Given that $\\triangle DBC' \\sim \\triangle ABC$, and $\\angle BC'D = \\angle ACB$, together with $\\angle AC'D = \\angle ACB$, it follows that $\\angle BC'D = \\angle AC'D = \\angle ACB = 90^\\circ$.\n\nLet $BC = x$. Then $AC = \\frac{2}{x}$ and $AB = \\sqrt{x^2 + \\frac{4}{x^2}}$. Find the minimum value of the perimeter of $\\triangle ABC$.\n\n![alt](images/Hong_Kong_Booklet_2015_-_2016_p15_data_d86c4c80f4.png)",
    "options": [],
    "answer": "See solution",
    "solution": "Using the AM-GM inequality, the perimeter of $\\triangle ABC$ is\n\n$$\nx + \\frac{2}{x} + \\sqrt{x^2 + \\frac{4}{x^2}} \\ge 2\\sqrt{2} + \\sqrt{2\\sqrt{4}} = 2 + 2\\sqrt{2}.\n$$\n\nThis minimum value is attained when $x = \\frac{2}{x}$ and $x^2 = \\frac{4}{x^2}$, i.e., when $x = \\sqrt{2}$. In this case, $AB = \\sqrt{\\sqrt{2}^2 + \\frac{4}{\\sqrt{2}^2}} = 2$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 14502,
    "subject": "Mathematics (Olympiad)",
    "question": "Suppose $f(x)$ is a polynomial with integer coefficients such that $f(0) = -3$, and $f(x) = x^n + x^{n-1} - x^{n-2}$ for some integer $n \\geq 2$. Is $f(x)$ reducible over the integers?",
    "options": [],
    "answer": "See solution",
    "solution": "Suppose, for the sake of contradiction, that $f(x) = g(x)h(x)$ where $g(x)$ and $h(x)$ both have integer coefficients and degree at least one. Since\n\n$$\ng(0)h(0) = f(0) = -3,\n$$\n\none of $g(0)$, $h(0)$ must be $\\pm 1$, say $g(0) = \\pm 1$. The absolute value of the product of the roots (in the complex numbers) of the polynomial $g(x)$ is $|g(0)| = 1$, so $g(z) = 0$ for some complex number $z$ with $|z| \\le 1$. Now, $f(z) = 0$, so $z^n + z^{n-1} - z^{n-2} = 3$.\n\nHowever,\n$$\n|z^n + z^{n-1} - z^{n-2}| \\le |z^n| + |z^{n-1}| + |z^{n-2}| \\le 3,\n$$\nwith equality implying $|z| = 1$ and that $z^n$, $z^{n-1}$, and $-z^{n-2}$ all have the same argument. This is easily seen to be impossible, so we have reached the desired contradiction.",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 14503,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $S$ be a set of $n = 2020$ points in the plane, no three of which are collinear. Let $M$ be the set of all midpoints of segments joining two distinct points of $S$. What is the minimum possible value of $|M|$?",
    "options": [],
    "answer": "See solution",
    "solution": "There are at least $4037$ points in $M$.\n\nSince there are finitely many points in $S$, we can find a coordinate system such that all points in $S$ have distinct $x$-coordinates. Indeed, we just need to find a line which is not perpendicular to any line joining two of the points in $S$.\n\nNow, let the $n = 2020$ points be $P_j = (x_j, y_j)$ where $j = 1, 2, \\dots, n$. WLOG assume $x_1 < x_2 < \\dots < x_n$. Since\n\n$$\n\\frac{x_1 + x_2}{2} < \\frac{x_1 + x_3}{2} < \\dots < \\frac{x_1 + x_n}{2} < \\frac{x_2 + x_n}{2} < \\dots < \\frac{x_{n-1} + x_n}{2},\n$$\n\nthe midpoints of $P_1P_2, P_1P_3, \\dots, P_1P_n, P_2P_n, \\dots, P_{n-1}P_n$ have pairwise different $x$-coordinates, and so they must be distinct. This shows $|M| \\ge 2n - 3 = 4037$.\n\nThe case $4037$ is attainable. For example, we choose the points $0, 2, 4, 6, \\dots, 2n-2$ on the real number line. Then the midpoints can only be $1, 2, 3, \\dots, 2n-3$. So there are exactly $2n-3 = 4037$ midpoints.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 14504,
    "subject": "Mathematics (Olympiad)",
    "question": "An investor has two rectangular lands, each of size $120 \\times 100$.\n\na) On the first land, she wants to build a house with a rectangular base of size $25 \\times 35$ and nine circular flower pots with diameter $5$ outside the house. Prove that for all positions of the flower pots, the remaining land is still sufficient to build the desired house.\n\nb) On the second land, she wants to construct a polygonal fish pond such that the distance from an arbitrary point on the land, outside the pond, to the nearest pond edge is not over $5$. Prove that the perimeter of the pond is not smaller than $440 - 20\\sqrt{2}$.",
    "options": [],
    "answer": "See solution",
    "solution": "For convenience, instead of writing *a* meters, we write *a*.\n\na) Consider the rectangle $ABCD$ where $AB = CD = 120$ and $AD = BC = 100$. Divide the rectangle into $10$ sub-rectangles of size $30 \\times 40$ as shown below. Consider $9$ centers of the flower pots. By the pigeonhole principle, there exists a sub-rectangle that does not contain any center.\n\nSuppose that rectangle is $XYZT$ where $XY = ZT = 40$, $XT = YZ = 30$. Consider one more rectangle $X'Y'Z'T'$ lying inside $XYZT$ such that the sides of the two rectangles are pairwise parallel and the gaps are equal to $2.5$.\n\n![](images/Vietnamese_mathematical_competitions_p189_data_4de8954c5e.png)\n\nIt is easy to check that $X'Y'Z'T'$ does not share any point with the pots, so we can build a house on this plot.\n\nb) Consider a rectangle $ABCD$ where $AB = CD = 120$ and $AD = BC = 100$. Let $L$ be the perimeter of the lake. According to the problem, there exist points $A', B', C', D'$ in $L$ such that\n\n$$\nAA',\\ BB',\\ CC',\\ DD' \\leq 5.\n$$\n\nSince the lake is a convex polygon, $A'B', B'C', C'D', D'A'$ do not overlap. Hence,\n\n$$\n|L| \\geq A'B' + B'C' + C'D' + D'A'.\n$$\n\nDenote $A_1$ as the projection of $A'$ to $AD$ and $A_2$ as the projection of $A'$ to $AB$. Similarly, define $B_1, B_2, C_1, C_2, D_1, D_2$. We have\n\n$$\nA_1A' + A'B' + B'B_1 \\geq A_1B_1 \\geq AB = 120.\n$$\n\nSimilarly,\n\n$$\nB_2B' + B'C' + C'C_2 \\geq 100,\n$$\n$$\nC_1C' + C'D' + D'D_1 \\geq 120,\n$$\n$$\nD_2D' + D'A' + A'A_2 \\geq 100.\n$$\n\nHence,\n\n$$\nA'B' + B'C' + C'D' + D'A' + (A'A_1 + A'A_2 + B'B_1 + B'B_2 + C'C_1 + C'C_2 + D'D_1 + D'D_2) \\geq 440.\n$$\n\nApplying the Cauchy-Schwarz inequality,\n\n$$\nA'A_1 + A'A_2 \\leq \\sqrt{2(A'A_1^2 + A'A_2^2)} = \\sqrt{2A'A_2^2} \\leq 5\\sqrt{2}.\n$$\n\nSimilarly,\n\n$$\nB'B_1 + B'B_2 \\leq 5\\sqrt{2}, \\quad C'C_1 + C'C_2 \\leq 5\\sqrt{2}, \\quad D'D_1 + D'D_2 \\leq 5\\sqrt{2}.\n$$\n\nFrom these inequalities, it is clear that the length of $L$ does not exceed $440 - 20\\sqrt{2}$. $\\square$",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 14505,
    "subject": "Mathematics (Olympiad)",
    "question": "Let triangle $ABC$ have incentre $I$ and circumcentre $O$. Suppose that $\\angle AIO = 90^\\circ$ and $\\angle CIO = 45^\\circ$. Find the ratio $AB : BC : CA$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $AI$ extended meet $BC$ at $U$. Let $A$, $B$, and $C$ be the angles of $\\triangle ABC$, and let $a$, $b$, and $c$ be its side lengths, labelled in the usual way.\n\nWe know that $\\angle AIC = 180^\\circ - \\frac{A+C}{2} = 90^\\circ + \\frac{B}{2}$. We also know that $\\angle AIC = 90^\\circ \\pm 45^\\circ$, depending on the location of $O$; since $90^\\circ + \\frac{B}{2} > 90^\\circ$, $\\angle AIC = 135^\\circ$ and so\n\n$$\n\\angle ABC = 90^\\circ.\n$$\n\nIt follows that $AC$ is a diameter of the circumcircle, and so $O$ is the midpoint of $AC$.\n\n![](images/V_Britanija_2008_p22_data_04614ec05b.png)\n\nNote that $\\angle CIU = 180^\\circ - 90^\\circ - 45^\\circ = 45^\\circ = \\angle CIO$ and $\\angle OCI = \\frac{C}{2} = \\angle UCI$, so that $\\triangle OCI$ is similar to $\\triangle UCI$. But they have a common length $CI$, and are therefore congruent.\n\nThis tells us that $CU = CO = \\frac{AC}{2}$. By the angle bisector theorem, we also have $\\frac{BU}{CU} = \\frac{AB}{AC}$, so that $BU = \\frac{AB}{2}$. Thus\n\n$$\nBC = \\frac{AC + AB}{2}.\n$$\n\nPutting this together with the Pythagorean relation in $\\triangle ABC$, we get\n\n$$\n2a = b + c \\quad (1)\n$$\n\n$$\nb^2 = a^2 + c^2. \\quad (2)\n$$\n\nRearranging (2) gives $a^2 = b^2 - c^2 = (b+c)(b-c) = 2a(b-c)$, from equation (1), and this simplifies to $a = 2(b-c)$ (as $b-c$ is positive).\n\nSubstituting back into (1),\n\n$$\n4(b-c) = b+c\n$$\n\n$$\n\\therefore 3b = 5c\n$$\n\n$$\n\\therefore \\frac{b}{c} = \\frac{5}{3}\n$$\n\nSo $\\triangle ABC$ is a 3,4,5-triangle, with\n\n$$\nAB : BC : CA = 3 : 4 : 5.\n$$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 14506,
    "subject": "Mathematics (Olympiad)",
    "question": "On an $n \\times n$ chessboard, what are the possible numbers of pawns that can be placed so that every $2 \\times 2$ square contains exactly two pawns?",
    "options": [],
    "answer": "See solution",
    "solution": "Any number from the segment $[2k^2 + k,\\ 2k^2 + 3k + 1]$ for odd $n = 2k+1$; $n^2/2$ for even $n$.\n\nIf $n = 2k$ is even, the board can be partitioned into $k^2 = n^2/4$ of $2 \\times 2$ squares, and there are exactly two pawns in each of them. So the total number of pawns on the board is $2k^2 = n^2/2$.\n\n![](images/Blr2012_p17_data_48c2cae3e4.png)\n\nFig. 1\n\n![](images/Blr2012_p17_data_240d3a1061.png)\n\nFig. 2\n\n![](images/Blr2012_p17_data_fcf19eb243.png)\n\nFig. 3\n\n![](images/Blr2012_p17_data_b53d3e6950.png)\n\nFig. 4",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 14507,
    "subject": "Mathematics (Olympiad)",
    "question": "Are there any positive integers $m, n$ such that $m^{20} + 11^n$ is a square number? Prove your conclusion.",
    "options": [],
    "answer": "See solution",
    "solution": "Assume there are positive integers $m, n$ such that $m^{20} + 11^n = k^2$ for some integer $k$. Then,\n\n$$\n11^n = k^2 - m^{20} = (k - m^{10})(k + m^{10}).\n$$\n\nThus, there exist integers $\\alpha, \\beta \\ge 0$ such that\n\n$$\n\\begin{cases}\nk - m^{10} = 11^{\\alpha}, \\\\\nk + m^{10} = 11^{\\beta}.\n\\end{cases}\n$$\n\nClearly, $\\alpha < \\beta$. Subtracting the first equation from the second gives\n\n$$\n2m^{10} = 11^{\\alpha}(11^{\\beta-\\alpha} - 1).\n$$\n\nLet $m = 11^{\\gamma} m_1$, where $\\gamma \\in \\mathbb{N}$ and $11 \\nmid m_1$. Then\n\n$$\n11^{10\\gamma} \\cdot 2m_1^{10} = 11^{\\alpha}(11^{\\beta-\\alpha} - 1),\n$$\n\nwhich implies $10\\gamma = \\alpha$ and $2m_1^{10} = 11^{\\beta-\\alpha} - 1$.\n\nBy Fermat's Little Theorem, $m_1^{10} \\equiv 1 \\pmod{11}$, so $2m_1^{10} \\equiv 2 \\pmod{11}$. But $11^{\\beta-\\alpha} - 1 \\equiv 10 \\pmod{11}$, a contradiction. Therefore, there are no positive integers $m, n$ such that $m^{20} + 11^n$ is a square number. $\\square$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 14508,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $S(a)$ denote the sum of the digits of the number $a$. Consider the polynomial\n\n$$\nf(x) = \\sum_{a\\text{ is a 55-digit number}} x^{S(a)}.\n$$\n\nIs it possible to partition the set of all 55-digit numbers into 4 parts such that the sum of the digits in each part is always congruent modulo 4?",
    "options": [],
    "answer": "See solution",
    "solution": "It is obvious that\n$$\nf(x) = (x + x^2 + \\ldots + x^9)^{54} = \\frac{x(x^9 - 1)(x^{10} - 1)^{54}}{(x - 1)^{55}}.\n$$\n\nSuppose it is possible to divide the set into 4 parts as described. Then a 4th root of unity $\\epsilon$ would be a root of $f(x)$, and $\\epsilon^3 + \\epsilon^2 + \\epsilon + 1 = 0$. However, $f(x)$ has no such roots, leading to a contradiction. Therefore, it is impossible to partition the set of 55-digit numbers in this way.",
    "topic": "Number Theory",
    "subtopic": "Modular Arithmetic"
  },
  {
    "id": 14509,
    "subject": "Mathematics (Olympiad)",
    "question": "There are 2008 red cards and 2008 white cards, making 4016 cards in total. These cards are shuffled and placed in a pile. Each of the 2008 players is dealt 2 cards from the pile. The players then sit in a circle, facing the center.\n\nAt each turn, every player simultaneously does the following:\n\n- If a player has at least one red card, they pass one red card to the player on their left.\n- If a player has no red cards, they pass one white card to the player on their left.\n\nDetermine the maximum possible number of turns required before, for the first time, every player has exactly one red card and one white card.",
    "options": [],
    "answer": "See solution",
    "solution": "Pick a player and call them $A$. Let $A_1$ be $A$, and for $i = 2, 3, \\dots$, let $A_i$ be the player sitting $i-1$ positions clockwise from $A$. For a non-negative integer $k$ and a positive integer $i$, define $F(k, i)$ as the number obtained by subtracting the total number of red cards possessed by $A_1, A_2, \\dots, A_i$ from the total number of white cards they possess after the $k$-th turn. The 0-th turn is the initial distribution.\n\n**Lemma 1.** If we choose $A$ suitably, we can ensure $F(0, i) \\ge 0$ for all $i \\ge 1$.\n\n*Proof.* Choose any player $B$. Define $B_i$ and $G(k, i)$ analogously. Since the total number of red and white cards among all players is equal, $G(0, i) = G(0, i + 2008)$. Thus, we can pick $m$ ($1 \\le m \\le 2008$) such that $G(0, m)$ is minimal. Let $A = B_{m+1}$. Then $A_i = B_{m+i}$, so $F(0, i) = G(0, m+i) - G(0, m) \\ge 0$.\n\nFrom now on, consider $A$ as above.\n\n**Lemma 2.** $F(k, i) \\ge 0$ for all $k \\ge 0$ and $i \\ge 1$.\n\n*Proof.* Induct on $k$. The base case $k=0$ holds by Lemma 1. Suppose true for $k = l$. If $F(l+1, j) < 0$ for some $j$, since $F(l+1, j) = F(l+1, j+2008)$, assume $j \\ge 2$. $F(l, j)$ is even. The difference $F(l, j) - F(l+1, j)$ is $-2$, $0$, or $2$. If $F(l, j) \\ge 0$ and $F(l+1, j) < 0$, the only possibility is $F(l, j) = 0$ and $F(l+1, j) = -2$, which can only happen if $A_1$ receives a red card and $A_j$ passes a white card. But if $A_j$ passes a white card, they must have had two white cards, so $F(l, j-1) = F(l, j) - 2 = -2 < 0$, contradicting the induction hypothesis. Thus, $F(l+1, i) \\ge 0$ for all $i$.\n\nLetting $i = 2007$, $F(k, 2007) \\ge 0$ for all $k$. Since $F(k, 2008) = 0$ always, $A_{2008}$ always has at least one red card, so $A_1$ receives a red card every turn. By induction, after $n$ turns, each of $A_1, \\dots, A_n$ has at least one red card. Thus, after 2007 turns, every player has at least one red card. Since there are 2008 red cards, each player must have exactly one red card at the end of the 2007th turn, so the desired configuration is reached in at most 2007 turns.\n\nIf initially a player has two white cards, the process can take exactly 2007 turns, so the maximum possible number of turns is $\\boxed{2007}$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 14510,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all pairs $(a, b)$ of positive integers such that $a \\geq b$ and\n\n$$\n\\frac{1}{a} + \\frac{1}{b} = \\frac{1}{2021}.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "*Solution 1:* Let $d = \\gcd(a, b)$, $a = d a'$, $b = d b'$. The given equation reduces to $\\frac{1}{d a'} + \\frac{1}{d b'} = \\frac{1}{2021}$, which is equivalent to\n\n$$\n2021(a' + b') = d a' b'.\n$$\n\nAs $a'$ and $b'$ are relatively prime to each other, they are both relatively prime to $a' + b'$, implying that $a' b'$ and $a' + b'$ are relatively prime. By the above, $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 \\geq b$ holds if and only if $a' \\geq b'$, consider all cases:\n\n- If $a' = 2021$ and $b' = 1$, then $d = 2022$. Consequently, $(a, b) = (2021 \\cdot 2022, 2022)$.\n- If $a' = 47$ and $b' = 43$, then $d = 90$. Consequently, $(a, b) = (47 \\cdot 90, 43 \\cdot 90)$.\n- If $a' = 47$ and $b' = 1$, then $d = 43 \\cdot 48$. Consequently, $(a, b) = (2021 \\cdot 48, 43 \\cdot 48)$.\n- If $a' = 43$ and $b' = 1$, then $d = 47 \\cdot 44$. Consequently, $(a, b) = (2021 \\cdot 44, 47 \\cdot 44)$.\n- If $a' = 1$ and $b' = 1$, then $d = 2021 \\cdot 2$. Thus, $(a, b) = (2021 \\cdot 2, 2021 \\cdot 2)$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 14511,
    "subject": "Mathematics (Olympiad)",
    "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\n(a) Prove that there is a sequence of replacements that will make the final number equal to $2$.\n\n(b) Prove that there is a sequence of replacements that will make the final number equal to $1000$.",
    "options": [],
    "answer": "See solution",
    "solution": "(a) First, 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, replace $2013$ and $2015$ with $2014$ to get $\\{1, 2, \\ldots, 2012, 2014\\}$. Continue this process, replacing the largest two numbers with their average, until you eventually get to $\\{1, 3\\}$. Finish by replacing $1$ and $3$ with $2$.\n\n(b) Using the same construction as in (a), you can find a sequence of replacements that reduces $\\{a, a + 1, \\ldots, b\\}$ to just $\\{a + 1\\}$, or to $\\{b - 1\\}$. In particular, you can reduce $\\{1, 2, \\ldots, 999\\}$ to $\\{998\\}$, and $\\{1001, 1002, \\ldots, 2016\\}$ to $\\{1002\\}$. This leaves $\\{998, 1000, 1002\\}$. 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": 14512,
    "subject": "Mathematics (Olympiad)",
    "question": "There are 100 positive integers written on a board. At each step, Alex composes 50 fractions using each number on the board exactly once, brings these fractions to their irreducible form, and then replaces the 100 numbers on the board with the new numerators and denominators to create 100 new numbers. Find the smallest positive integer $n$ such that, regardless of the initial 100 numbers, after $n$ steps Alex can arrange to have only pairwise coprime numbers on the board.",
    "options": [],
    "answer": "See solution",
    "solution": "Equivalently, consider a graph on 100 vertices with a positive integer on each vertex. At each step, we pick a perfect matching (a set of disjoint edges covering all vertices), and for each edge, we divide the numbers at its endpoints by their greatest common divisor.\n\nSuppose the initial numbers are $p_1, p_2, \\dots, p_{99}$ and $p_1 p_2 \\cdots p_{99}$, where $p_1, \\dots, p_{99}$ are distinct primes. The vertex with $p_1 p_2 \\cdots p_{99}$ must be matched with every other vertex, requiring at least 99 steps.\n\nWe show that 99 steps suffice. It is enough to show that $K_{100}$ (the complete graph on 100 vertices) has a 1-factorization, i.e., its edges can be decomposed into 99 perfect matchings. In general, $K_{2n}$ has a 1-factorization. One way: label the vertices $x, x_1, \\dots, x_{2n-1}$, and for the $i$-th matching ($1 \\leq i \\leq 2n-1$), consider all edges $x_r x_s$ with $1 \\leq r < s \\leq 2n-1$ and $r+s \\equiv i \\pmod{2n-1}$, together with the edge $x x_t$ where $2t \\equiv i \\pmod{2n-1}$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 14513,
    "subject": "Mathematics (Olympiad)",
    "question": "For any non-negative real numbers $x$ and $y$, show the inequality:\n\n$$\nx^2 y^2 + x^2 y + x y^2 \\leq x^4 y + x + y^4.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "If $x = 0$ or $y = 0$, the inequality is trivial. Now assume $x, y > 0$. Divide both sides of the inequality by $x y$:\n\n$$\nxy + x + y \\leq x^3 + \\frac{1}{y} + \\frac{y^3}{x}.\n$$\n\nApply the following inequality, which is a consequence of the Cauchy-Schwarz inequality:\n\n$$\n\\frac{a_1^2}{b_1} + \\frac{a_2^2}{b_2} + \\cdots + \\frac{a_n^2}{b_n} \\geq \\frac{(a_1 + a_2 + \\cdots + a_n)^2}{b_1 + b_2 + \\cdots + b_n}\n$$\n\nNow,\n\n$$\nx^3 + \\frac{1}{y} + \\frac{y^3}{x} \\geq \\frac{(x^2)^2}{x} + \\frac{1^2}{y} + \\frac{(y^2)^2}{x y} \\geq \\frac{(x^2 + 1 + y^2)^2}{x + y + x y} \\geq \\frac{(x + y + x y)^2}{x + y + x y} = x y + x + y,\n$$\n\nwhich completes the proof.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 14514,
    "subject": "Mathematics (Olympiad)",
    "question": "We are given a triangle $ABC$. Let $M$ be the midpoint of side $AB$.\n\nLet $P$ be an interior point of the triangle. Let $Q$ denote the point symmetric to $P$ with respect to $M$.\n\nFurthermore, let $D$ and $E$ be the common points of $AP$ and $BP$ with sides $BC$ and $AC$, respectively.\n\nProve that points $A$, $B$, $D$, and $E$ lie on a common circle if and only if $\\angle ACP = \\angle QCB$ holds.",
    "options": [],
    "answer": "See solution",
    "solution": "Without loss of generality, assume that $P$ lies either on the segment $CM$ or in the interior of triangle $AMC$. If not, exchange the names of vertices $A$ and $B$ and add the angle $\\angle PCQ$ to both given angles.\n\nLet $C'$ be the point symmetric to $C$ with respect to $M$. According to the assumptions, $Q$ and $B$ are the points symmetric to $P$ and $A$ with respect to $M$, respectively. Let $D'$ and $E'$ be the common points of the lines $C'P$ and $CP$ with the sides $AC$ and $AC'$, respectively.\n\n![](images/bwf2015englishSolutions_p1_data_4297f58c85.png)\n\n*Figure 1: equivalence lemma*\n\nWe first prove a lemma.\n\n*Lemma.* Assume that $P$ does not lie on the median $CM$. In this case, the following two facts hold:\n\n1. Angles $\\angle ACP$ and $\\angle BCQ$ are equal if and only if the quadrilateral $C'CD'E'$ is circumscribed.\n2. Angles $\\angle CAP$ and $\\angle C'AQ$ are equal if and only if the quadrilateral $ABDE$ is circumscribed.\n\n*Proof.*\n\n1. Due to symmetry with respect to $M$, we have $\\angle BCQ = \\angle AC'P$. Therefore,\n\n![](images/bwf2015englishSolutions_p1_data_0112e3da1c.png)\n\nbecause of the equal angles on the chord $D'E'$.\n\n2. Since $\\angle C'AQ = \\angle PBC$ also holds because of the symmetry, this follows as above.\n\nWe first prove that if $ABDE$ is circumscribed, then $\\angle ACP = \\angle QCB$.\n\nLet $ABDE$ be circumscribed, as in Figure 2. Since $BQ$ lies symmetric to $AP$ with respect to $M$,\n\n![](images/bwf2015englishSolutions_p2_data_cb7e195a8b.png)\n\n*Figure 2: Problem 2*\n\n$AQBP$ is a parallelogram. Angle $\\angle CED$ is supplementary to $\\angle AED$, which is itself supplementary to $\\angle ABC$ because the quadrilateral is circumscribed, and therefore $\\angle CED = \\angle ABC$. It follows that triangles $CED$ and $CBA$ are similar.\n\nNow reflect along the angle bisector of $\\angle BCA$ and then perform a homothety, such that $D$ is mapped onto $A$. Since the triangles $CED$ and $CBA$ are similar, this must map $E$ onto $B$. Since $AQB$ is congruent to $BPA$, which is itself similar to $DPE$, triangle $DPE$ is mapped onto $AQB$, and therefore $P$ onto $Q$. It therefore follows that $\\angle ACP = \\angle QCB$, as required.\n\nNow, we prove the converse direction under the additional assumption that $P$ does not lie on $CM$. Assume that $\\angle ACP = \\angle BCQ$. The lemma then implies that $C'CD'E'$ is circumscribed. Applying the result on the first direction on triangle $CC'A$ instead of $ABC$ implies $\\angle C'AQ = \\angle CAP$. It therefore follows from the lemma that $ABDE$ lie on a common circle.\n\nIf $P$ lies on $CM$ and $\\angle ACP = \\angle BCQ$ holds, $C$, $P$, $M$, and $Q$ are all points on the angle bisector. It then follows that $ABC$ is isosceles, and since $P$ lies on the axis of symmetry, $ED \\parallel AB$. It follows that $ABDE$ is an isosceles trapezoid, and it is therefore circumscribed.\n\n$\\Box$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 14515,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $F$ be the midpoint of $BE$. Let $DF$ and $AE$ be parallel. Let $M$ be the intersection point of $AE$ and $CD$.\n\n![](images/obm-book_p117_data_d8df9e0962.png)\n\nWhat is the measure of $\\angle BAC$?",
    "options": [],
    "answer": "See solution",
    "solution": "$E$ is the midpoint of $CF$. Since $ME$ is parallel to $DF$, by Thales' theorem $DM = MC$. But triangle $ADM$ is isosceles, so $AM = DM = MC$ and triangle $AMC$ is isosceles as well. Let $\\alpha = \\angle ADC = \\angle BAE$. Then $\\angle AMC = 2\\alpha$ and $\\angle MAC = 90^\\circ - \\alpha$, and consequently, $\\angle BAC = \\angle DAM + \\angle MAC = \\alpha + 90^\\circ - \\alpha = 90^\\circ$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 14516,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $m \\geq 2$ be a positive integer and $p \\geq 5$ be a prime number. Consider the sequence $a_n$ defined by\n$$a_n = (p-2)^n + p^n - m$$\nfor $n \\geq 1$, and let $q$ be the smallest prime that divides a member of the sequence. Determine all pairs $(m, p)$ such that $q \\geq m$.",
    "options": [],
    "answer": "See solution",
    "solution": "The pairs $(m, p)$ such that $q \\geq m$ are $(2, p)$ and $(3, p)$ for any prime $p \\geq 5$.\n\n*Case 1: $m$ is even ($m \\geq 2$)*\n\nConsider $a_1 = (p-2)^1 + p^1 - m = (p-2) + p - m = 2p - 2 - m$. Since $a_1$ is even, $2$ divides $a_1$, so $q = 2$. Thus, $q \\geq m$ only if $m = 2$.\n\n*Case 2: $m = 3$*\n\nFor $m = 3$, all $a_n$ are odd, so the smallest prime divisor $q \\geq 3$. Thus, $q \\geq m$ is satisfied for $m = 3$.\n\n*Case 3: $m \\geq 5$ (odd)*\n\nFor $m \\geq 5$ odd, $a_n$ will have a prime divisor $q < m$ for some $n$. For example, if $m - 2 = p^k$ for some $k \\geq 1$, then for $p = 5$, $k = 1$, $m = 7$, $a_2 = (5-2)^2 + 5^2 - 7 = 9 + 25 - 7 = 27$, which is divisible by $3$, so $q = 3 < 7 = m$.\n\nIn general, for $m \\geq 5$ odd, there will always be a prime divisor $q < m$.\n\n**Conclusion:**\n\nAll pairs $(m, p)$ with $m = 2$ or $m = 3$ and $p \\geq 5$ prime satisfy $q \\geq m$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 14517,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $x, y, z \\in [0, 1]$. What is the maximum value of\n$$\nM = \\sqrt{|x-y|} + \\sqrt{|y-z|} + \\sqrt{|z-x|}?\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "We may assume $0 \\le x \\le y \\le z \\le 1$. Then\n$$\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$$\nThe equality holds if and only if $y-x = z-y$, $x = 0$, $z = 1$ (i.e., $x = 0$, $y = \\frac{1}{2}$, $z = 1$).\n\nTherefore, the answer is $M_{\\max} = \\sqrt{2} + 1$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 14518,
    "subject": "Mathematics (Olympiad)",
    "question": "The quadrilateral $ABCD$ is inscribed. The point $H_1$ is the orthocenter of $\\triangle ABC$ and the points $A_1$ and $B_1$ are symmetric to the points $A$ and $B$ with respect to the lines $BH_1$ and $AH_1$, respectively. The point $O_1$ is the center of the circumscribed circle of $\\triangle A_1B_1H_1$. The point $H_2$ is the orthocenter of $\\triangle ABD$ and the points $A_2$ and $B_2$ are symmetric to the points $A$ and $B$ with respect to the lines $BH_2$ and $AH_2$, respectively. The point $O_2$ is the center of the circumscribed circle of $\\triangle A_2B_2H_2$. Denote the line $O_1O_2$ by $l_{AB}$. The lines $l_{BC}$, $l_{CD}$, and $l_{DA}$ are defined analogously. Let $l_{AB} \\cap l_{BC} = M$, $l_{BC} \\cap l_{CD} = N$, $l_{CD} \\cap l_{DA} = P$, and $l_{DA} \\cap l_{AB} = Q$. Prove that the points $M, N, P$, and $Q$ are concyclic.",
    "options": [],
    "answer": "See solution",
    "solution": "![](images/Broshura-bg_2018_IMO_p2_data_37301d5adf.png)\n\nWe will need the following lemma.\n\n**Lemma.** The point $O_1$ lies on the line $H_1O$, where $O$ is the center of the circumcircle of $\\triangle ABC$. The ratio $H_1O_1 : O_1O$ depends on $\\angle ACB$ only.\n\n**Proof.** Let $K$ and $L$ be the intersection points of $AH_1$ and $BH_1$ with the circumcircle of $\\triangle ABC$. Since $H_1$ and $L$ are symmetric with respect to $AC$, the quadrilateral $H_1ALA_1$ is a rhombus, as $\\angle AH_1A_1 = 2 \\angle AH_1L = 2\\gamma$. Analogously, $H_1BKB_1$ is a rhombus with $\\angle BH_1B_1 = 2\\gamma$. Therefore $H_1ALA_1$ and $H_1BKB_1$ are similar, whence\n\n$$\n\\frac{H_1T}{H_1X} = \\frac{H_1Q}{H_1Y} = \\frac{1}{1 + 2 \\cos 2\\gamma}\n$$\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": 14519,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $v_1, v_2, v_3, v_4$ be four vectors in $\\mathbb{R}^3$.\n\n(a) What is the probability that the cone generated by these four vectors is proper (i.e., contained in a half-space), and what is the probability that the cone is all of $\\mathbb{R}^3$?\n\n(b) What is the probability that one of the four vectors lies in the interior of the cone generated by the other three?",
    "options": [],
    "answer": "See solution",
    "solution": "For part (a):\n\nThe probability that the cone of the four vectors is proper is $\\frac{7}{8}$, so the probability that the cone is all of $\\mathbb{R}^3$ is $\\frac{1}{8}$.\n\nConstruct a vector $u_{12}$ normal to the plane spanned by $v_1$ and $v_2$, oriented so that $v_3 \\cdot u_{12} > 0$. The half-space $\\{w \\mid w \\cdot u_{12} \\ge 0\\}$ contains the cone generated by $\\{v_1, v_2, v_3\\}$. If $v_4 \\cdot u_{12} > 0$, the cone generated by all four vectors is contained in the same half-space. To keep the cone from being proper, we must have $v_4 \\cdot u_{12} < 0$.\n\nSimilarly, define $u_{13}$ orthogonal to $v_1$ and $v_3$ with $v_2 \\cdot u_{13} > 0$, and $u_{23}$ orthogonal to $v_2$ and $v_3$ with $v_1 \\cdot u_{23} > 0$. The cone is proper (contained in a half-space) if and only if at least one of the three values $v_4 \\cdot u_{ij} > 0$. If all three dot products are negative, then the cone covers all of space.\n\nAveraging over all choices of vectors, the probability that all three dot products are negative is $\\frac{1}{8}$.\n\nFor part (b):\n\nGiven $v_1, v_2, v_3$, $v_4$ lies in the interior of the cone generated by those three if and only if $v_4 \\cdot u_{ij} > 0$ for all three such dot products. So there is a $\\frac{1}{8}$ chance that $v_4$ lies in $C(v_1, v_2, v_3)$. Similarly, there is a $\\frac{1}{8}$ chance that $v_2$ lies in $C(v_1, v_3, v_4)$. These events are disjoint: only one vector can be in the interior of the cone of the other three. Thus, the probability is the union of four disjoint events, each of probability $\\frac{1}{8}$, giving a total probability of $\\frac{1}{2}$.",
    "topic": "Geometry",
    "subtopic": "Solid Geometry"
  },
  {
    "id": 14520,
    "subject": "Mathematics (Olympiad)",
    "question": "Halla todas las ternas de enteros positivos $ (x, y, z) $, con $ z > 1 $, que satisfacen simultáneamente que\n\n$$\nx \\text{ divide a } y+1, \\quad y \\text{ divide a } z-1, \\quad z \\text{ divide a } x^2+1.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "Las soluciones son $(1, 1, 2)$, $(2, 1, 5)$ y $(2n+1, 2n, 2n^2+2n+1)$ con $n \\ge 1$.\n\nSi $x=1$, la única solución es $(1, 1, 2)$ ($z$ divide a 2, por lo que $z=2$, y $y$ divide a 1). Si $x=2$, la única solución es $(2, 1, 5)$ ($z$ divide a 5, así que $z=5$, y $y$ divide a 4 y es impar). Supongamos ahora que $x \\ge 3$.\n\nSean $y+1 = rx$, $z-1 = sy$ y $x^2+1 = tz$. Entonces, sustituyendo sucesivamente, obtenemos:\n\n$$\nx^2 + 1 = tz = sty + t = rstx - st + t.\n$$\n\nDe esta igualdad se tiene que $st - t \\equiv -1 \\pmod{x}$. Esto implica que $st - t \\ge x - 1$, y a su vez $st \\ge x$. Si $r \\ge 2$, entonces\n\n$$\nx^2 + 1 = r stx - st + t \\ge (2x - 1)st + 1 \\ge (2x - 1)x + 1,\n$$\n\nlo que, reordenado, deja $x(x-1) \\le 0$, lo que no es cierto para $x \\ge 3$. Por lo tanto, concluimos que $r=1$ en las soluciones restantes. Tenemos entonces\n\n$$\nx^2 + 1 = st(x - 1) + t.\n$$\n\nSi tomamos módulo $x-1$ a ambos lados, tenemos $t \\equiv 2 \\pmod{x-1}$. Si $t=2$, entonces la ecuación es equivalente a $s = \\frac{x+1}{2}$. Entonces $x$ es un número impar (de la forma $2n+1$), $y = rx - 1 = 2n$, y $z = sy + 1 = 2n^2 + 2n + 1$. Si $t \\ne 2$, entonces $t \\ge x+1$. De la ecuación obtenemos $x^2+1 \\ge (x+1)(x-1) + x + 1$, o $x \\le 1$, lo que es una contradicción. No hay más soluciones. Comprobamos que todas las posibles soluciones que hemos obtenido satisfacen las condiciones del enunciado.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 14521,
    "subject": "Mathematics (Olympiad)",
    "question": "The bisector of angle $BAC$ of an acute-angled triangle $ABC$ ($AC \\neq AB$) intersects its circumcircle a second time at point $W$. Let $O$ be the circumcenter of $\\triangle ABC$. The line $AW$ intersects, for a second time, the circumcircles of triangles $OWB$ and $OWC$ at points $N$ and $M$, respectively. Prove that $BN + MC = AW$.\n\n![](fig.18.png)",
    "options": [],
    "answer": "See solution",
    "solution": "**Solution.** Without loss of generality, assume $AC < AB$. From the conditions (see figure), $2\\angle CAW = \\angle COW = \\angle CMW$, hence $\\angle ACM = \\angle CAM$. Thus, $AM = CM$. Similarly, $AN = NB$. Then it is enough to prove that $AM + AN = AW$, thus $AM = NW$. Obviously, $CO = OB$, $CW = WB$. Thus, $\\angle OBW = \\angle OCW$. Hence, $\\angle OBW = \\angle OCW$. Therefore,\n\n$$\n\\angle ONM = \\angle OBW = \\angle OCW = \\angle OMN\n$$\n\nHence, $\\triangle ONM$ is isosceles. It is also obvious that $\\triangle OAW$ is isosceles, so $\\triangle ONW = \\triangle OMA$, thus $AM = NW$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 14522,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $ABC$ be a triangle and let $M$ and $N$ denote the midpoints of $\\overline{AB}$ and $\\overline{AC}$, respectively. Let $X$ be a point such that $\\overline{AX}$ is tangent to the circumcircle of triangle $ABC$. Denote by $\\omega_B$ the circle through $M$ and $B$ tangent to $\\overline{MX}$, and by $\\omega_C$ the circle through $N$ and $C$ tangent to $\\overline{NX}$. Show that $\\omega_B$ and $\\omega_C$ intersect on line $BC$.",
    "options": [],
    "answer": "See solution",
    "solution": "**First solution using symmedians (Merlijn Staps)**\n\nLet $\\overline{XY}$ be the other tangent from $X$ to $(AMN)$.\n\n**Claim.** Line $\\overline{XM}$ is tangent to $(BMY)$; hence $Y$ lies on $\\omega_B$.\n\n![](images/sols-TST-IMO-2019_p1_data_4ad7604b41.png)\n\n*Proof.* Let $Z$ be the midpoint of $\\overline{AY}$. Then $\\overline{MX}$ is the $M$-symmedian in triangle $AMY$. Since $\\overline{MZ} \\parallel \\overline{BY}$, it follows that $\\angle AMX = \\angle ZMY = \\angle BYM$. We conclude that $\\overline{XM}$ is tangent to the circumcircle of triangle $BMY$. $\\square$\n\nSimilarly, $\\omega_C$ is the circumcircle of triangle $CNY$. As $AMYN$ is cyclic too, it follows that $\\omega_B$ and $\\omega_C$ intersect on $\\overline{BC}$, by Miquel's theorem.\n\n**Remark.** The converse of Miquel's theorem is true, which means the problem is equivalent to showing that the second intersection of $\\omega_B$ and $\\omega_C$ moves along $(AMN)$. Thus the construction of $Y$ above is not so unnatural.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 14523,
    "subject": "Mathematics (Olympiad)",
    "question": "Suppose the 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 such sequences is \\_\\_\\_\\_\\_\\_.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $b_i = \\frac{a_{i+1}}{a_i}$ for $1 \\leq i \\leq 8$. For each $\\{a_n\\}$ satisfying the given condition, we have\n\n$$\n\\prod_{i=1}^{8} b_i = \\prod_{i=1}^{8} \\frac{a_{i+1}}{a_i} = \\frac{a_9}{a_1} = 1, \\quad \\text{with } b_i \\in \\{2, 1, -\\frac{1}{2}\\} \\ (1 \\leq i \\leq 8).\n$$\n\nConversely, a sequence of eight terms $\\{b_n\\}$ satisfying this condition uniquely determines a sequence $\\{a_n\\}$ as in the problem.\n\nIn each $\\{b_n\\}$, there must be an even number of $-\\frac{1}{2}$ and the same number of $2$, with the remainder being $1$. In other words, the numbers of $-\\frac{1}{2}$ and $2$ are both $2k$, while the number of $1$ is $8-4k$. It is easy to check that $k$ can only be $0, 1, 2$. Once $k$ is given, there are $\\binom{8}{2k} \\binom{8-2k}{2k}$ ways to construct $\\{b_n\\}$.\n\nTherefore, the total number of $\\{b_n\\}$ satisfying the condition is\n\n$$\nN = 1 + \\binom{8}{2} \\binom{6}{2} + \\binom{8}{4} \\binom{4}{4} = 1 + 28 \\times 15 + 70 \\times 1 = 491.\n$$\n\nThe answer is $491$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 14524,
    "subject": "Mathematics (Olympiad)",
    "question": "![Slovenija_2015_p30_data_c37ab8ba18.png](images/Slovenija_2015_p30_data_c37ab8ba18.png)\n\nLet $D$ and $E$ be the midpoints of the segments $BC$ and $AC$ of triangle $ABC$. The points $P$ and $Q$ are on the circumcircle of triangle $ABC$ such that $DP = EQ$. Prove that $AC = BC$.",
    "options": [],
    "answer": "See solution",
    "solution": "Since $D$ and $E$ are the midpoints of the segments $BC$ and $AC$, the lines $DE$ and $AB$ are parallel. It follows that $\\angle EDA = \\angle BAD$, and by the Angles Subtended by Same Arc Theorem we have $\\angle BAD = \\angle BAP = \\angle BQP$. Therefore $\\angle EQP = \\angle EDA = \\pi - \\angle PDE$, which means that the quadrilateral $EDPQ$ is cyclic, and the triangles $TED$ and $TPQ$ are similar. Let $T$ be the centroid of triangle $ABC$ and $x = |DP| = |EQ|$. From the similarity of triangles $TED$ and $TPQ$ we deduce\n\n$$\n\\frac{|TD| + x}{|TE|} = \\frac{|TP|}{|TE|} = \\frac{|TQ|}{|TD|} = \\frac{|TE| + x}{|TD|}\n$$\n\nWe rearrange this to\n\n$$\n(|TD| - |TE|)(|TD| + |TE| + x) = 0,\n$$\n\nwhich gives $|TE| = |TD|$ since $|TD| + |TE| + x > 0$. The triangle $DET$ is therefore isosceles with apex at vertex $T$. Since the lines $AD$ and $BE$ are parallel, the triangle $ABT$ is also isosceles with apex at vertex $T$, hence $|BE| = |BT| + |TE| = |AT| + |TD| = |AD|$. From this it follows that the triangles $ABE$ and $BAD$ are congruent since they have two pairs of sides of the same length and an angle of the same size between them, $\\angle EBA = \\angle BAD$. Thus $|AE| = |BD|$ and hence $|AC| = |BC|$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 14525,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $ABCD$ be a convex quadrilateral with $AB = 2$, $AD = 7$, and $CD = 3$ such that the bisectors of acute angles $\\angle DAB$ and $\\angle ADC$ intersect at the midpoint of $\\overline{BC}$. Find the square of the area of $ABCD$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $I$ be the midpoint of $\\overline{BC}$, and let $R$ and $S$ denote the reflections of $C$ and $B$ across $\\overline{DI}$ and $\\overline{AI}$, respectively. Note that, because $\\overline{DI}$ and $\\overline{AI}$ are angle bisectors, $R$ and $S$ lie on segment $\\overline{AD}$. Then $AS = 2$ and $DR = 3$, so $RS = 2$. Furthermore, $IR = IC = IB = IS$, so\n\n$$\n\\angle ABI = \\angle ASI = \\angle DRI = \\angle DCI,\n$$\n\nand, if $T$ is the foot of the perpendicular from $I$ to $\\overline{AD}$, then $RT = ST = 1$.\n\n![](images/2022AIME_II_Solutions_p9_data_1513adbcd8.png)\n\nNow let $\\alpha = \\angle BAI = \\angle IAD$, $\\beta = \\angle CDI = \\angle IDA$, and $\\gamma = \\angle ABC = \\angle BCD$. Because the sum of the measures of the interior angles of a quadrilateral is $360^\\circ$, it follows that $\\alpha + \\beta + \\gamma = 180^\\circ$. This implies\n\n$$\n\\angle BIA = 180^\\circ - \\gamma - \\alpha = \\beta,\n$$\n\nand likewise $\\angle CID = \\alpha$. Thus $\\triangle ABI \\sim \\triangle ICD$, so $BI = CI = \\sqrt{AB \\cdot CD} = \\sqrt{6}$.\n\nFinally, applying the Pythagorean Theorem to $\\triangle IRT$ yields $IT = \\sqrt{5}$, so\n\n$$\n\\begin{align*}\n\\text{Area}(ABCD) &= 2 \\cdot \\text{Area}(\\triangle AIS) + \\text{Area}(\\triangle RIS) + 2 \\cdot \\text{Area}(\\triangle DIR) \\\\\n&= 2\\left(\\frac{1}{2} \\cdot 2 \\cdot \\sqrt{5}\\right) + \\frac{1}{2} \\cdot 2 \\cdot \\sqrt{5} + 2\\left(\\frac{1}{2} \\cdot 3 \\cdot \\sqrt{5}\\right) = 6\\sqrt{5}.\n\\end{align*}\n$$\n\nThe square of the area of $ABCD$ is $(6\\sqrt{5})^2 = 180$.\n\n---\n\nAlternatively:\n\nLet $X$ be the intersection of rays $AB$ and $DC$. Then the angle bisectors of $\\angle DAB$ and $\\angle CDA$ meet at the incenter $I$ of $\\triangle XAD$. Because $XI$ is both an angle bisector and a median of $\\triangle BXC$, it is the perpendicular bisector of $BC$. Let $T, E$, and $F$ be the projections of $I$ onto lines $AD, XA$, and $XD$, respectively.\n\n![](images/2022AIME_II_Solutions_p10_data_b240207523.png)\n\nBecause $\\triangle BXI$ is congruent to $\\triangle CXI$, it follows that $CF = BE$. By equal tangents, $DT = DF = 3 + CF = 3 + BE = 1 + AE = 1 + AT$. Because $DT + AT = AD = 7$, $AT = 3$ and $DT = 4$ from which $CF = BE = AE - AB = AT - AB = 3 - 2 = 1$. Let $r$ be the inradius of $\\triangle ADX$. Because $\\triangle BEI \\sim \\triangle IEX$, it follows that $XE = \\frac{IE}{BE} = r^2$.\n\nBecause the inradius of a triangle with sides $a, b$, and $c$ and semiperimeter $s$ is\n\n$$\nr = \\sqrt{\\frac{(s-a)(s-b)(s-c)}{s}},\n$$\n\nit follows that\n\n$$\nr^2(r^2 + 3 + 4) = r^2 \\cdot 3 \\cdot 4,\n$$\n\nimplying that $r = \\sqrt{5}$.\n\nThen\n\n$$\n\\text{Area}(ABCD) = \\text{Area}(\\triangle ABI) + \\text{Area}(\\triangle CDI) + \\text{Area}(\\triangle AID) = \\frac{r \\cdot AB}{2} + \\frac{r \\cdot CD}{2} + \\frac{r \\cdot AD}{2} = 6\\sqrt{5},\n$$\n\nas above.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 14526,
    "subject": "Mathematics (Olympiad)",
    "question": "If $x$, $y$, $z$ are positive real numbers with sum $12$, prove that:\n\n$$\n\\frac{x}{y} + \\frac{y}{z} + \\frac{z}{x} + 3 \\geq \\sqrt{x} + \\sqrt{y} + \\sqrt{z}.\n$$\n\nWhen does equality hold?",
    "options": [],
    "answer": "See solution",
    "solution": "Since $x$, $y$, $z$ are positive real numbers with sum $12$, it is enough to prove that\n\n$$\n\\frac{x}{y} + \\frac{y}{z} + \\frac{z}{x} + \\frac{x + y + z}{4} \\geq \\sqrt{x} + \\sqrt{y} + \\sqrt{z}. \\quad (1)\n$$\n\nFrom the inequality of the arithmetic–geometric mean for the positive real numbers $x$, $y$, $z$, we get\n\n$$\n\\frac{x}{y} + \\frac{y}{4} \\geq 2 \\sqrt{\\frac{x}{y} \\cdot \\frac{y}{4}} = \\sqrt{x}, \\qquad (2)\n$$\n\n$$\n\\frac{y}{z} + \\frac{z}{4} \\geq 2 \\sqrt{\\frac{y}{z} \\cdot \\frac{z}{4}} = \\sqrt{y}, \\qquad (3)\n$$\n\n$$\n\\frac{z}{x} + \\frac{x}{4} \\geq 2 \\sqrt{\\frac{z}{x} \\cdot \\frac{x}{4}} = \\sqrt{z}. \\qquad (4)\n$$\n\nSumming up (2), (3), and (4), we find inequality (1).\n\nEquality holds when all inequalities (2), (3), and (4) hold as equalities, that is, when\n\n$$\n\\begin{align*}\n\\frac{x}{y} = \\frac{y}{4}, \\quad \\frac{y}{z} = \\frac{z}{4}, \\quad \\frac{z}{x} = \\frac{x}{4} &\\Leftrightarrow x = \\frac{y^2}{4}, \\quad y = \\frac{z^2}{4}, \\quad z = \\frac{x^2}{4} \\\\\n&\\Leftrightarrow x = \\frac{y^2}{4}, \\quad y = \\frac{z^2}{4}, \\quad z = \\frac{1}{4} \\left(\\frac{y^2}{4}\\right)^2 = \\frac{y^4}{4^3} \\\\\n&\\Leftrightarrow x = \\frac{y^2}{4}, \\quad y = \\frac{z^2}{4}, \\quad z = \\frac{1}{4^3} \\left(\\frac{z^2}{4}\\right)^4 \\\\\n&\\Leftrightarrow x = \\frac{y^2}{4}, \\quad y = \\frac{z^2}{4}, \\quad z = \\frac{1}{4^7} z^8 \\\\\n&\\Leftrightarrow x = \\frac{y^2}{4}, \\quad y = \\frac{z^2}{4}, \\quad z^7 = 4^7 \\\\\n&\\Leftrightarrow x = y = z = 4.\n\\end{align*}\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 14527,
    "subject": "Mathematics (Olympiad)",
    "question": "On a circle, 2018 points are marked. Each of these points is labeled with an integer. Let each number be larger than the sum of the preceding two numbers in clockwise order.\n\nDetermine the maximal number of positive integers that can occur in such a configuration of 2018 integers.",
    "options": [],
    "answer": "See solution",
    "solution": "Let the points be labeled $a_0, a_1, \\dots, a_{2017}$ clockwise with cyclical notation, i.e., $a_{k+2018} = a_k$ for all integers $k$.\n\n*Lemma.* In a valid configuration, no two neighbouring numbers can be both non-negative.\n\n*Proof.* Assume that there exist neighbouring numbers $a_{k-1}$ and $a_k$ which are both non-negative. We get $a_{k+1} > a_k + a_{k-1} \\ge a_k$, with the first inequality following from the problem statement and the second from $a_{k-1} \\ge 0$. Since now also $a_k$ and $a_{k+1}$ are both non-negative, we analogously get $a_{k+2} > a_{k+1}$, then $a_{k+3} > a_{k+2}$, and so on, until we have $a_{k+2018} > a_{k+2017} > \\dots > a_{k+1} > a_k = a_{k+2018}$, a contradiction. $\\blacksquare$\n\nTherefore at most every second number can be non-negative. Next we will show that these are still too many non-negative numbers.\n\n*Lemma.* In a valid configuration, it is not possible that every second number is non-negative.\n\n*Proof.* Assume that this is the case, so w.l.o.g. $a_{2k} \\ge 0$ and $a_{2k+1} < 0$ for all integers $k$. Then we get $a_3 > a_2 + a_1 \\ge a_1$, where again the first inequality follows from the problem statement and the second from $a_2 \\ge 0$. Analogously we get $a_5 > a_3$, then $a_7 > a_5$, etcetera, until we have $a_1 = a_{2019} > a_{2017} > a_{2015} > \\dots > a_3 > a_1$, a contradiction. $\\blacksquare$\n\nWe can therefore summarize: A configuration with more than 1009 non-negative numbers is not possible because otherwise by the pigeonhole principle we would have two neighbouring non-negative numbers, which is not allowed according to the first lemma. A configuration with exactly 1009 non-negative numbers contradicts either the first or the second lemma.\n\nWith 1008 positive and 1010 negative numbers we find for example the configuration\n\n$$-4035, 1, -4033, 1, -4031, 1, -4029, \\dots, 1, -2021, 1, -2019, -2017,$$\n\nwhich we can easily check for correctness.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 14528,
    "subject": "Mathematics (Olympiad)",
    "question": "The cube $ABCD'A'B'C'D'$ has edges of length $a$. Take the points $E \\in (AB)$ and $F \\in (BC)$ so that $AE + CF = EF$.\n\n(a) Find the measure of the angle between the planes $(D'DE)$ and $(D'DF)$.\n\n(b) Compute the distance from $D'$ to the straight line $EF$.\n\n![](images/RMC2014_p31_data_4e1dad77e8.png)",
    "options": [],
    "answer": "See solution",
    "solution": "a) Extend the segment $BA$ with $AH \\equiv FC$. Then $\\triangle DAH \\equiv \\triangle DCF$ (SAS), whence $\\overline{HDA} \\equiv \\overline{FDC}$, so $m(\\overline{HDF}) = 90^\\circ$. Then $[DH] \\equiv [DF]$, which leads to $\\triangle DHE \\equiv \\triangle DFE$ (SSS), and from here $m(\\overline{FDE}) = m(\\overline{HDE}) = 45^\\circ$. Since $FD \\perp DD'$ and $ED \\perp DD'$, it follows that $m((D'DE), (D'DF)) = m(\\overline{FDE}) = 45^\\circ$.",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 14529,
    "subject": "Mathematics (Olympiad)",
    "question": "*(a)* Find a closed formula for the sequence $a_n$ defined by the recurrence relation with characteristic equation $\\lambda^2 - 5\\lambda + 6 = 0$ and initial conditions $a_0 = 1$, $a_1 = 4$.\n\n*(b)* Find a closed formula for the sequence $b_n$ defined by the recurrence relation with characteristic equation $\\lambda^2 - 4\\lambda + 4 = 0$ and initial conditions $b_0 = 3$, $b_1 = 7$.",
    "options": [],
    "answer": "See solution",
    "solution": "*(a)*\n\nThe characteristic equation is $\\lambda^2 - 5\\lambda + 6 = 0$. The roots are $\\lambda = 3, 2$. Therefore, we have\n\n$$\na_n = A \\cdot 3^n + B \\cdot 2^n\n$$\nfor some constants $A$ and $B$. Using $n = 0$ and $n = 1$:\n\n$$\n\\begin{cases} A + B = 1, \\\\ 3A + 2B = 4. \\end{cases}\n$$\nSolving, $A = 2$, $B = -1$. Thus,\n\n$$\na_n = 2 \\cdot 3^n - 2^n.\n$$\n\n*(b)*\n\nThe characteristic equation is $\\lambda^2 - 4\\lambda + 4 = 0$, with a double root $\\lambda = 2$. Therefore,\n\n$$\nb_n = (Cn + D)2^n\n$$\nfor some constants $C$ and $D$. Using $n = 0$ and $n = 1$:\n\n$$\n\\begin{cases} D = 3, \\\\ 2(C + D) = 7. \\end{cases}\n$$\nSo $C = \\frac{1}{2}$. Thus,\n\n$$\nb_n = (n + 6)2^{n-1}.\n$$",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 14530,
    "subject": "Mathematics (Olympiad)",
    "question": "We have an equilateral triangle $ABC$ with side length $3$, and points $D$, $E$, and $F$ on sides $BC$, $CA$, and $AB$, respectively. Given that $BD = 1$ and $\\angle ADE = \\angle DEF = 60^\\circ$, find the length of segment $AF$.\n\nNote: $XY$ represents the length of segment $XY$.\n\n![](images/Japan2023_p1_data_a6097c1e9c.png)",
    "options": [],
    "answer": "See solution",
    "solution": "$$\n\\frac{7}{9}\n$$\n\nFor the triangles $ABD$, $DCE$, and $EAF$, we have\n\n$$\n\\angle EDC = \\angle ADC - \\angle ADE = (\\angle DAB + \\angle ABD) - 60^\\circ = \\angle DAB\n$$\n\nand similarly,\n\n$$\n\\angle FEA = \\angle DEA - \\angle DEF = (\\angle EDC + \\angle DCE) - 60^\\circ = \\angle EDC\n$$\n\nAlso, since $\\angle ABD = \\angle DCE = \\angle EAF = 60^\\circ$, the triangles $ABD$, $DCE$, and $EAF$ are similar.\nTherefore, we have $\\frac{EA}{AF} = \\frac{DC}{CE} = \\frac{AB}{BD} = 3$, and\n\n$$\nDC = BC - BD = 2, \\quad CE = \\frac{1}{3}DC = \\frac{2}{3}, \\quad EA = CA - CE = \\frac{7}{3}, \\quad AF = \\frac{1}{3}EA = \\frac{7}{9}\n$$\n\nThus, the answer is $\\frac{7}{9}$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 14531,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $ABCD$ be a cyclic quadrilateral. Point $P$ is on line $CB$ such that $CP = CA$ and $B$ lies between $C$ and $P$. Point $Q$ is on line $CD$ such that $CQ = CA$ and $D$ lies between $C$ and $Q$.\n\nProve that the incentre of triangle $ABD$ lies on line $PQ$.\n\n(The incentre of a triangle is the point where its angle bisectors intersect.)",
    "options": [],
    "answer": "See solution",
    "solution": "Let $I$ be the incentre of $\\triangle ABD$. It suffices to show that $APBI$ and $AIDQ$ are cyclic, as that would imply the required collinearity via\n\n$$\n\\angle QIA + \\angle AIP = \\angle QDA + \\angle ABP = 180^{\\circ}.\n$$\n\n![](images/2024_The_Australian_Scene_Final_p85_data_937e4f0f04.png)\n\nSince $\\triangle PCA$ is isosceles and $ABCD$ is cyclic,\n\n$$\n\\angle APC = 90^{\\circ} - \\frac{1}{2}\\angle ACP = 90^{\\circ} - \\frac{1}{2}\\angle ADB.\n$$\n\nUsing the fact that $I$ is the incentre of $\\triangle ABD$,\n\n$$\n\\angle AIB = 180^{\\circ} - \\frac{1}{2}(\\angle DAB + \\angle DBA) = 90^{\\circ} + \\frac{1}{2}\\angle ADB.\n$$\n\nHence $\\angle APC + \\angle AIB = 180^{\\circ}$ and $APBI$ is cyclic. By similar arguments, $AIDQ$ is also cyclic and the proof is complete.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 14532,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $n$ be a positive integer. Elfie the Elf lives in three-dimensional space $\\mathbb{Z}^3$. She starts at the origin $(0, 0, 0)$. On each turn, she can teleport to any point in $\\mathbb{Z}^3$ that is at distance $\\sqrt{n}$ from her current location. However, teleportation is a complicated procedure: Elfie starts off normal, but turns strange with her first teleportation. Each subsequent teleportation alternates her state between normal and strange.\n\nFor which $n$ can Elfie travel to any given point in $\\mathbb{Z}^3$ and be normal when she gets there?",
    "options": [],
    "answer": "See solution",
    "solution": "There are no such $n$.\n\nColor all points in $\\mathbb{Z}^3$ white or black: the point $(x, y, z)$ is white if $x + y + z \\equiv 0 \\pmod{2}$ and black if $x + y + z \\equiv 1 \\pmod{2}$.\n\nAfter the first move, Elfie is at $(a, b, c)$ with $a^2 + b^2 + c^2 = n$. Thus, $a + b + c \\equiv n \\pmod{2}$.\n\nIf $n$ is even, Elfie only visits white points. If $n$ is odd, she alternates between black and white points. Since she is normal after an even number of moves (on a white point), she can never reach a black point while normal. Therefore, there is no $n$ such that Elfie can travel to any given point and be normal when she gets there.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 14533,
    "subject": "Mathematics (Olympiad)",
    "question": "What is the sum of the digits of the greatest prime number that is a divisor of $16,\n383$?\n\n(A) 3  \n(B) 7  \n(C) 10  \n(D) 16  \n(E) 22",
    "options": [],
    "answer": "See solution",
    "solution": "Observe that $16,383 = 2^{14} - 1 = (2^7 + 1)(2^7 - 1) = 129 \\cdot 127 = 3 \\cdot 43 \\cdot 127$. Because all three of these factors are prime, the greatest prime number that is a divisor of $16,383$ is $127$, and the requested sum of digits is $1 + 2 + 7 = 10$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 14534,
    "subject": "Mathematics (Olympiad)",
    "question": "Vertices of a regular $6n+3$-gon, numbered clockwise $1, 2, \\ldots, 6n+3$, form the game field. Vertices numbered $2n+1$, $4n+2$, and $6n+3$ are called *holes*. At the start, there are 3 chips on the field. Two players take turns choosing any one of the 3 chips and moving it clockwise to a neighboring vertex, provided that vertex is not occupied by another chip. The first player wins if, after any turn, at least 2 chips are in holes. Can the first player always win if, at the start, the chips are placed in the form of a regular triangle?",
    "options": [],
    "answer": "See solution",
    "solution": "**Answer:** Yes.\n\nLet's define the *distance to the hole* for each chip as the number of steps needed to reach the nearest hole moving clockwise, assuming no other chips block the way. For example, a chip at vertex $4n$ has distance 2 to the hole if $4n+1$ and $4n+2$ are unoccupied; a chip at $3n$ has distance $n+2$ to the hole $4n+2$ if vertices $3n+1$ through $4n+2$ are free.\n\nWe prove by induction that when two chips have the same distance $k$ to their respective holes, the first player can force a win. Label these two chips as *hot* chips, and the third as *usual*.\n\n**Base case ($k=0$):** Two chips are already in holes, so the first player wins.\n\n**Case $k=1$:** If it's the first player's turn, he moves the usual chip (if possible), maintaining the position. If the second player moves a hot chip into a hole, the first player moves the other hot chip into a hole and wins. Thus, the second player must move the usual chip, and this continues until the usual chip cannot move, resulting in a *zugzwang* position where the second player loses.\n\n**Inductive step:** Assume the statement holds for distance $k$. For distance $k+1$, conceptually shift the holes forward by one position; this reduces the situation to the $k$ case. Thus, the first player can always reach a position where two chips are $k$ away from holes and win.\n\nFinally, the initial configuration (chips forming a regular triangle) satisfies the induction conditions, so the first player can always win.",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 14535,
    "subject": "Mathematics (Olympiad)",
    "question": "In triangle $ABC$, angle $A$ is a right angle. A point $D$ lies on line segment $AB$ in such a way that the angles $ACD$ and $BCD$ are equal. Moreover, $|AD| = 2$ and $|BD| = 3$.\n\n![](images/NLD_ABooklet_2023_p9_data_b6a556a00e.png)\n\nWhat is the length of line segment $CD$?",
    "options": [],
    "answer": "See solution",
    "solution": "$2\\sqrt{6}$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 14536,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $\\triangle ABC$ be an acute-angled triangle with orthocenter $H$ and let $D$ be an arbitrary point on side $BC$. Let $E$ and $Z$ be the points on the segments $AB$ and $AC$ respectively such that the quadrilaterals $ABDZ$ and $ACDE$ are cyclic, and let segments $BZ$ and $CE$ intersect at $P$. Let $L$ be the intersection point of line $HA$ and the tangent to the circumcircle of triangle $\\triangle PBC$ at point $C$. If lines $BH$ and $CP$ intersect at $X$, prove that $D$ lies on the line $LX$.",
    "options": [],
    "answer": "See solution",
    "solution": "We have $\\angle PCD = \\angle ECD = \\angle EAD$ and $\\angle PBD = \\angle ZBD = \\angle ZAD$, therefore\n\n$$\n\\angle BPC = 180^{\\circ} - \\angle EAD - \\angle ZAD = 180^{\\circ} - \\angle BAC = \\angle BHC,\n$$\n\nmeaning that $BHPC$ is cyclic.\n\nWe also have $\\angle PZD = \\angle BZD = \\angle BAD = \\angle PCD$ showing that $DPZC$ is cyclic. Then, using that $BAZD$ is also cyclic, we have\n\n$$\n\\angle DPC = \\angle DZC = \\angle ABC.\n$$\n\nLet $Y$ be the point of intersection of $AH$ with the circumcircle of $BHPC$. Then\n\n$$\n\\angle YPC = \\angle YHC = \\angle ABC = \\angle DPC\n$$\n\nshowing that $D$ belongs on $YP$.\n\n![](images/2025-SL-b_p4_data_7c8ff86a87.png)\n\nFinally, applying Pascal's theorem on the hexagon $BHYPCC$, we get that $\\{X\\} = BH \\cap PC$, $\\{L\\} = HY \\cap CC$ and $\\{D\\} = YP \\cap CB$ are collinear, as required.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 14537,
    "subject": "Mathematics (Olympiad)",
    "question": "Suppose that the line $l$: $y = kx + m$ (where $k, m$ are integers) intercepts the ellipse $$\\frac{x^2}{16} + \\frac{y^2}{12} = 1$$ at two different points $A$ and $B$, and intercepts the hyperbola $$\\frac{x^2}{4} - \\frac{y^2}{12} = 1$$ at two different points $C$ and $D$. Can the line $l$ be such that $\\vec{AC} + \\vec{BD} = 0$? If yes, how many different possibilities are there for the line $l$? If no, explain the reason.",
    "options": [],
    "answer": "See solution",
    "solution": "For the system\n$$\n\\begin{cases}\ny = kx + m \\\\\n\\frac{x^2}{16} + \\frac{y^2}{12} = 1\n\\end{cases}\n$$\neliminating $y$ gives\n$$\n(3 + 4k^2)x^2 + 8kmx + 4m^2 - 48 = 0.\n$$\nLet $A(x_1, y_1)$ and $B(x_2, y_2)$ be the intersection points. Then\n$$\nx_1 + x_2 = -\\frac{8km}{3 + 4k^2}.\n$$\nThe discriminant for real intersection is\n$$\n\\Delta_1 = (8km)^2 - 4(3 + 4k^2)(4m^2 - 48) > 0.\n$$\nFor the system\n$$\n\\begin{cases}\ny = kx + m \\\\\n\\frac{x^2}{4} - \\frac{y^2}{12} = 1\n\\end{cases}\n$$\neliminating $y$ gives\n$$\n(3 - k^2)x^2 - 2kmx - m^2 - 12 = 0.\n$$\nLet $C(x_3, y_3)$ and $D(x_4, y_4)$ be the intersection points. Then\n$$\nx_3 + x_4 = \\frac{2km}{3 - k^2}.\n$$\nThe discriminant for real intersection is\n$$\n\\Delta_2 = (-2km)^2 + 4(3 - k^2)(m^2 + 12) > 0.\n$$\nFrom $\\vec{AC} + \\vec{BD} = 0$, we have $(x_4 - x_2) + (x_3 - x_1) = 0$, which implies $x_1 + x_2 = x_3 + x_4$.\n\nThus,\n$$\n-\\frac{8km}{3 + 4k^2} = \\frac{2km}{3 - k^2}.\n$$\nThis leads to $km = 0$ or $-\\frac{4}{3 + 4k^2} = \\frac{1}{3 - k^2}$ (which is discarded).\n\nSo, possible solutions are $k = 0$ or $m = 0$.\n\n- When $k = 0$, from the discriminants above, $-2\\sqrt{3} < m < 2\\sqrt{3}$. Since $m$ is integer, $m = -3, -2, -1, 0, 1, 2, 3$.\n- When $m = 0$, $-\\sqrt{3} < k < \\sqrt{3}$, so $k = -1, 0, 1$.\n\nCombining these, there are nine lines in total satisfying the given conditions.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 14538,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $A = \\{a_1, a_2, \\ldots, a_{2010}\\}$ and $B = \\{b_1, b_2, \\ldots, b_{2010}\\}$ be two sets of complex numbers such that the equality\n\n$$\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$$\nholds for every $n = 1, 2, \\ldots, 2010$. Prove that $A = B$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $S_k = \\sum_{i=1}^{2010} a_i^k$ and $\\tilde{S}_k = \\sum_{i=1}^{2010} b_i^k$. We first show by induction that $S_k = \\tilde{S}_k$ for $k = 1, 2, \\ldots, 2010$.\n\nSetting $n = 1$ in the given equality, we have $2009 S_1 = 2009 \\tilde{S}_1$, and hence $S_1 = \\tilde{S}_1$. Assume that $S_j = \\tilde{S}_j$ for $j = 1, 2, \\ldots, k-1$, where $2 \\le k \\le 2010$; we are going to show that $S_k = \\tilde{S}_k$.\n\nBy the binomial theorem,\n\n$$\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&= 2009 S_k + \\frac{1}{2} \\sum_{l=1}^{k-1} \\binom{k}{l} S_{k-l} S_l + (2010 - 2^{k-1}) S_k.\n\\end{align*}\n$$\n\nSimilarly,\n\n$$\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.\n$$\n\nSince the sums are equal for all $k$, and by the inductive hypothesis $S_i = \\tilde{S}_i$ for $i = 1, \\ldots, k-1$, it follows that $S_k = \\tilde{S}_k$ (since $2010 - 2^{k-1} \\neq 0$ for $k \\le 2010$).\n\nNow, consider the monic polynomials whose roots are the $a_i$ and $b_i$:\n\n$$\n(x - a_1) \\cdots (x - a_{2010}) = x^{2010} + A_1 x^{2009} + \\cdots + A_{2010},\n$$\n$$\n(x - b_1) \\cdots (x - b_{2010}) = x^{2010} + B_1 x^{2009} + \\cdots + B_{2010}.\n$$\n\nBy Newton's identities,\n\n$$\nS_k + A_1 S_{k-1} + \\cdots + A_{k-1} S_1 + k A_k = 0,\n$$\n$$\n\\tilde{S}_k + B_1 \\tilde{S}_{k-1} + \\cdots + B_{k-1} \\tilde{S}_1 + k B_k = 0,\n$$\nfor $k = 1, 2, \\ldots, 2010$.\n\nSince $S_k = \\tilde{S}_k$ for all $k$, it follows by induction that $A_k = B_k$ for all $k$. Thus, the polynomials are equal, so $A = B$ as multisets.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 14539,
    "subject": "Mathematics (Olympiad)",
    "question": "令 $\\mathbb{N}$ 為所有正整數所成集合。找出所有的函數 $f : \\mathbb{N} \\to \\mathbb{N}$ 使得對於所有正整數 $m, n$，都有\n$$\nmf(m) + (f(f(m)) + n)^2 \\mid 4m^4 + n^2 f(f(n))^2.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "唯一可能的 $f$ 是 $f(n) = n$。\n\n令 $A(m, n) = mf(m) + (f(f(m)) + n)^2$，$B(m, n) = 4m^4 + n^2 f(f(n))^2$，$C(m, n) = m^2 + (m - n^2)$。可以验证，当 $f(n) = n$ 且 $f(m) = m$ 时，$A(m, n)C(m, n) = B(m, n)$，因此 $f(k) = k$ 是一个解。\n\n接下来证明没有其他解。首先，$mf(m)$ 必须是完全平方数。假设 $mf(m)$ 不是完全平方数。\n\n可以找到一个素数 $p$，使得 $p \\equiv 3 \\pmod{4}$ 且 $\\left(\\frac{mf(m)}{p}\\right) = -1$。这是由二次互反律和中国剩余定理保证的，并且根据狄利克雷定理，这样的素数存在。\n\n由于 $\\left(\\frac{-1}{p}\\right) = -1$，存在 $s$ 使得 $s^2 \\equiv -mf(m) \\pmod{p}$。令 $n = s - f(f(m))$，则 $p \\mid mf(m) + s^2 = A(m, n)$，因此 $B(m, n) \\equiv 0 \\pmod{p}$。这与 $p \\equiv 3 \\pmod{4}$ 且 $\\left(\\frac{mf(m)}{p}\\right) = -1$ 矛盾。\n\n接下来证明 $f(p) = p$ 对于形如 $p = 4k + 3$ 的素数成立。\n\n设 $p = 4k + 3$ 是素数。由于 $pf(p)$ 是完全平方数，可设 $f(p) = ps^2$，同理 $f(f(p)) = pt^2$。于是\n$$\nA(p, p) = p(s^2 + (1 + t^2)^2), \\quad B(p, p) = p^4(4 + t^4).\n$$\n由于 $\\left(\\frac{-1}{p}\\right) = -1$，$p$ 与 $t^2 + 1$ 互素，因此与 $q = s^2 + (1 + t^2)^2$ 也互素。因为 $A(p, p) \\mid B(p, p)$，有\n$$\ns^2 + 2t^2 - 3 = q - 4 - t^2 \\equiv -4 - t^2 \\equiv 0 \\pmod{q}.\n$$\n由于 $0 \\leq s^2 + 2t^2 - 3 < q$，只能有 $s = t = 1$，即 $f(p) = p$。\n\n再考虑 $p = 4k + 3$ 为素数时，$A(p, n) = p^2 + (p + n)^2$，$B(p, n) = 4p^4 + n^2 f(f(n))^2$。注意到\n$$\nn^2 f(f(n))^2 - n^4 = B(p, n) - C(p, n)A(p, n) \\equiv 0 \\pmod{A(p, n)}.\n$$\n由于该等式对任意大的 $p$ 成立，必有 $n^2 f(f(n))^2 - n^4 = 0$，即 $f(f(n)) = n$。于是 $A(m, n) = mf(m) + (m + n)^2$，$B(m, n) = 4m^4 + n^4$，且\n$$\n\\frac{B(m, n)}{A(m, n)} - (m^2 - (m - n)^2 + m(m - f(m))) = \\frac{m^2(m - f(m))(m - f(m) - 4n)}{A(m, n)}\n$$\n是整数。固定 $m$，当 $n$ 足够大时，分子为 0，故 $f(m) = m$。",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 14540,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all pairs of integers $ (x, y) $ that satisfy the following equality:\n\n$$\n|x + |x + |x|| | \\cdot ||-y| - y| - y| = 2011.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "Since 2011 is a prime number, the product on the left must be either 1 or 2011 (up to sign).\n\nIf $x \\ge 0$, then $|x + |x + |x|| | = 3x$, which cannot yield a positive product equal to 2011. Similarly, if $y \\le 0$, $||-y| - y| - y| = -3y$, which also cannot yield 2011.\n\nSuppose $x < 0$ and $y > 0$. Then:\n\n$$\n|x + |x + |x|| | = |x + |x - x|| | = |x|,\n$$\n$$\n||-y| - y| - y| = ||y - y| - y| = |-y| = y.\n$$\n\nSo the equation becomes $|x| \\cdot y = 2011$. Since 2011 is prime, the integer solutions are $(x, y) = (-1, 2011)$ and $(-2011, 1)$.\n",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 14541,
    "subject": "Mathematics (Olympiad)",
    "question": "Prove that the following equations do not have integer roots:\n\n$$\n1.\\quad a^3(b-c) + b^3(c-a) + c^3(a-b) = 2023^{2024}.\n$$\n\n$$\n2.\\quad a^3(b-c) + b^3(c-a) + c^3(a-b) = 2024^{2023}.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "First, we have\n\n$$\na^3(b-c) + b^3(c-a) + c^3(a-b) = (a-b)(b-c)(a-c)(a+b+c).\n$$\n\n1. Among numbers $a, b, c$ there must be two numbers with the same parity, so one of the differences $a-b$, $b-c$, or $a-c$ must be even. This implies that the quantity $(a-b)(b-c)(a-c)(a+b+c)$ is always even. Therefore, the given equation has no solution in integers for the first case.\n\n2. Note that the right side is not divisible by $3$. So among $a, b, c$, if there are two numbers congruent modulo $3$, then their difference will be divisible by $3$, which is a contradiction. So they must have different remainders when divided by $3$. This implies that\n\n$$\na + b + c \\equiv 0 + 1 + 2 \\equiv 0 \\pmod{3}.\n$$\n\nThis also leads to a contradiction, so both equations have no integer solutions.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 14542,
    "subject": "Mathematics (Olympiad)",
    "question": "Determine all pairs $\\left(a, b\\right)$ of non-negative integers such that $a^b + b$ divides $a^{2b} + 2b$. (Note that $0^0 = 1$.)",
    "options": [],
    "answer": "See solution",
    "solution": "Let $n = a^b + b$ and $m = a^{2b} + 2b$.\n\n- For $a = b = 0$, $n = m = 1$, so $n \\mid m$. Thus, $(0, 0)$ is a solution.\n- For $a = 0$ and $b > 0$, $n = b$ and $m = 2b$, so $n \\mid m$. Thus, $(0, b)$ is a solution for all $b \\geq 0$.\n- For $a = 1$, $n = b + 1$ and $m = 2b + 1$. $b + 1 \\mid 2b + 1$ only when $b = 0$. Thus, $(1, 0)$ is a solution.\n- For $b = 0$, $n = 1$ and $m = 1$, so $n \\mid m$. Thus, $(a, 0)$ is a solution for all $a \\geq 0$.\n\nNow consider $a > 1$ and $b > 0$:\n\n$$\na^b + b \\mid a^{2b} + 2b = (a^b)^2 - b^2 + (b^2 + 2b) = (a^b + b)(a^b - b) + b(b+2),\n$$\nso $a^b + b \\mid b(b+2)$.\n\n- For $b = 1$, $a + 1 \\mid 3$, so $a = 2$. Thus, $(2, 1)$ is a solution.\n- For $a \\geq 3$, $a^b + b > b(b+2)$ for $b \\geq 1$, so no solutions exist. (This can be shown by induction.)\n- For $a = 2$, $2^b + b \\mid b(b+2)$. For $b > 5$, $2^b > b(b+2)$, so only $b = 1, 2, 3, 4, 5$ need checking. Only $b = 1$ works, as $2^1 + 1 = 3$ and $2^{2} + 2 = 6$; $3 \\mid 6$.\n\nChecking $b = 2, 3, 4$:\n- $b = 2$: $2^2 + 2 = 6$, $2^{4} + 4 = 20$, $6 \\nmid 20$.\n- $b = 3$: $2^3 + 3 = 11$, $2^{6} + 6 = 70$, $11 \\nmid 70$.\n- $b = 4$: $2^4 + 4 = 20$, $2^{8} + 8 = 264$, $20 \\nmid 264$.\n\n**Summary:**\n\nThe solutions are all $(a, 0)$, all $(0, b)$, and $(2, 1)$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 14543,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all functions $f: \\mathbb{N} \\to \\mathbb{R}$ that satisfy the equation\n$$\nf(x + y) = f(x) + f(y)\n$$\nfor all $x, y \\in \\mathbb{N}$ such that $10^6 - 10^{-6} < \\frac{x}{y} < 10^6 + 10^{-6}$.",
    "options": [],
    "answer": "See solution",
    "solution": "All functions of the form $f(x) = cx$ with $c \\in \\mathbb{R}$ are solutions; they are the only ones.\n\nMore generally, let $b > a > 0$, and let the open interval $\\Delta = (a, b)$ contain an integer (in our case $a = 10^6 - 10^{-6}$, $b = 10^6 + 10^{-6}$). Consider any function $f: \\mathbb{N} \\to \\mathbb{R}$ such that $f(x + y) = f(x) + f(y)$ holds whenever $\\frac{x}{y} \\in \\Delta$. We prove that $f(n) = cn$ for all $n \\in \\mathbb{N}$ with a real constant $c$.\n\nTo begin with, let us show that $f(n+1) - f(n) = f(n) - f(n-1)$ for all sufficiently large $n$. The reason is that for each sufficiently large $n \\in \\mathbb{N}$ there is a $z \\in \\mathbb{N}$ such that\n$$\nf(n + 1) - f(n) = f(z + 1) - f(z) = f(n) - f(n - 1).\n$$\nTo ensure the first equality, it is enough to take a $z$ so that $\\frac{z}{n+1} \\in \\Delta$ and $\\frac{z+1}{n} \\in \\Delta$. Then by hypothesis $f(x + y) = f(x) + f(y)$ will hold with $x = z$, $y = n + 1$ and also with $x = z + 1$, $y = n$. Hence $f(z) + f(n+1) = f(n+z+1) = f(z+1) + f(n)$, as desired. Likewise, the second equality will hold provided that $\\frac{z}{n} \\in \\Delta$ and $\\frac{z+1}{n-1} \\in \\Delta$. Since $\\frac{z}{n+1} < \\frac{z}{n} < \\frac{z+1}{n} < \\frac{z+1}{n-1}$, it suffices to find an integer $z$ so that $a < \\frac{z}{n+1}$ and $\\frac{z+1}{n-1} < b$, i.e., $a(n + 1) < z < b(n - 1) - 1$. Such an integer does exist for $n$ large enough. Indeed, $b(n - 1) - 1$ and $a(n + 1)$ differ by $(b - a)n - (a + b + 1)$ which is greater than $1$ for $n > \\frac{a+b+2}{b-a}$.\n\nIn summary, there exists a $k \\in \\mathbb{N}$ such that $f(n+1) - f(n)$ has the same value for all $n \\ge k$. Then by standard induction\n$$\n(*) \\quad f(n) = (n-k)[f(k+1)-f(k)] + f(k) \\quad \\text{for all } n \\ge k.\n$$\nThere are $x, y \\in \\mathbb{N}$ such that $x, y \\ge k$ and $\\frac{x}{y} \\in \\Delta$. For instance, choose a rational $\\frac{r}{s} \\in \\Delta$ ($r, s \\in \\mathbb{N}$) and set $x = rk$, $y = sk$. Take one such pair $x, y$ and compute $f(x), f(y), f(x+y)$ by the formula $(*)$; this can be done because $x, y, x+y \\ge k$. Replace the obtained values in $f(x+y) = f(x)+f(y)$, which holds because $\\frac{x}{y} \\in \\Delta$. Simplification leads to $kf(k+1) = (k+1)f(k)$. Hence $\\frac{f(k)}{k} = \\frac{f(k+1)}{k+1} = c \\in \\mathbb{R}$; equivalently $f(k) = ck$ and $f(k+1) = c(k+1)$. Then $(*)$ takes the form $f(n) = cn$ for all $n \\ge k$. It remains to show that $f(n) = cn$ for all $n \\in \\mathbb{N}$.\n\nOnly finitely many $n \\in \\mathbb{N}$ may possibly disobey $f(n) = cn$. Suppose that such values exist, and let $q$ be the greatest one of them. Choose an integer $w \\in \\Delta$ and set $x = wq$, $y = q$. Then $x/y \\in \\Delta$, hence $f((w+1)q) = f(wq) + f(q)$. If $w > 1$ then $(w+1)q > wq > q$, so by the choice of $q$ we have $f((w+1)q) = c(w+1)q$, $f(wq) = cwq$. However, then $f(q) = c(w+1)q - cwq = cq$, contrary to the assumption that $q$ violates $f(n) = cn$. And if $w = 1$ then $(w+1)q = 2q > q$, so $f(2q) = 2cq$. On the other hand, $f((w+1)q) = f(wq) + f(q)$ takes the form $f(2q) = 2f(q)$ which leads to the impossible $f(q) = cq$ again. This completes the proof.\n\n**Remark.** The main assumption is $f(x + y) = f(x) + f(y)$ whenever $x/y \\in \\Delta$, where $\\Delta$ is an arbitrary open interval with positive endpoints. It ensures $f(n) = cn$ for all sufficiently large values of $n$. However, the additional assumption that $\\Delta$ contains an integer is essential to infer that $f(n) = cn$ for all $n \\in \\mathbb{N}$. Consider, for instance, the function $f: \\mathbb{N} \\to \\mathbb{R}$ defined by $f(n) = n$ for $n \\ge 5$ and $f(n) = 2010$ for $n \\in \\{1, 2, 3, 4\\}$ (in fact $f(1), f(2), f(3), f(4)$ can be arbitrary). The interval $\\Delta = (3/2, 5/3)$ does not contain fractions with denominators $1, 2, 3, 4$. So $x/y \\in \\Delta$ implies $x \\ge y \\ge 5$; the equation $f(x + y) = f(x) + f(y)$ is satisfied for such values. Thus $f$ is a function that satisfies the main assumption and $f(n) = n$ for $n \\ge 5$ but not for all $n$.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 14544,
    "subject": "Mathematics (Olympiad)",
    "question": "Suppose $f: \\mathbb{R} \\to \\mathbb{R}$ is a monotonic function.\n\n(a) Prove that $f$ has one-sided limits at any point $x_0 \\in \\mathbb{R}$.\n\n(b) Define the function $g: \\mathbb{R} \\to \\mathbb{R}$ by $g(x) = \\lim_{t \\to x^-} f(t)$, i.e., $g(x)$ is the left-sided limit at $x$ of the function $f$. Prove that if $g$ is a continuous function, then $f$ is also continuous.",
    "options": [],
    "answer": "See solution",
    "solution": "Suppose, without loss of generality, that $f$ is an increasing function.\n\n**(a)** Let $x_0 \\in \\mathbb{R}$. The set $\\{f(x) \\mid x < x_0\\}$ is upper bounded by $f(x_0)$, since $f$ is increasing. Set $L = \\sup\\{f(x) \\mid x < x_0\\}$. We claim that $L = f(x_0 - 0)$.\n\nLet $\\varepsilon > 0$. There exists $a < x_0$ such that $f(a) > L - \\varepsilon$. Since $f$ is increasing, for any $x \\in (a, x_0)$, $|f(x) - L| = L - f(x) < \\varepsilon$. Thus, $L = f(x_0 - 0)$.\n\nSimilarly, $f(x_0 + 0) = \\inf\\{f(x) \\mid x > x_0\\}$.\n\n**(b)** Let $x_0 \\in \\mathbb{R}$ and $t, s, a, b \\in \\mathbb{R}$ with $t < a < x_0 < s < b$. Then $f(t) \\leq f(a) \\leq f(x_0) \\leq f(s) \\leq f(b)$, and furthermore $g(a) = \\lim_{t \\searrow a} f(t) \\leq f(x_0)$ and $g(b) = \\lim_{s \\nearrow b} f(s) \\geq f(x_0)$, so $g(a) \\leq f(x_0) \\leq g(b)$.\n\nSince $g$ is continuous, $g(x_0) = \\lim_{a \\searrow x_0} g(a) = \\lim_{b \\nearrow x_0} g(b)$, so $g(x_0) \\geq f(x_0) \\geq g(x_0)$, which implies $g(x_0) = f(x_0)$. Consequently, $f = g$ and the claim follows.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 14545,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $ABCDA'B'C'D'$ be a cuboid and $AB' \\cap A'B = \\{O\\}$. Let $N$ be a point on the edge $[BC]$ so that $A'C \\parallel (B'AN)$. It is known that $D'O \\perp (B'AN)$. Prove that $ABCDA'B'C'D'$ is a cube.",
    "options": [],
    "answer": "See solution",
    "solution": "From $D'A' \\perp (ABB')$ and $D'O \\perp AB'$ follows $A'O \\perp AB'$, so $ABB'A'$ is a square. Then $A'C \\parallel (ANB')$ and $(ANB') \\cap (A'BC) = ON$, hence $A'C \\parallel ON$. Since $O$ is the midpoint of the segment $[A'B]$, $N$ is the midpoint of the segment $[BC]$. Rectangle $A'BCD'$ has $D'O \\perp ON$, hence $\\Delta D'A'O \\sim \\Delta OBN$.\n\n![](images/RMC_2015_BT_p14_data_887efe1e44.png)\n\nIt follows that $\\dfrac{D'A'}{A'O} = \\dfrac{OB}{BN}$, whence $2A'D^2 = A'B^2$, that is $A'B = A'D\\sqrt{2}$. This leads to $A'D = A'A$, therefore $AA'D'D$ is a square. This shows that all the edges of $ABCD'A'B'C'D'$ are equal, whence the conclusion.",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 14546,
    "subject": "Mathematics (Olympiad)",
    "question": "Given an integer $n \\ge 2$, prove that\n$$\n\\lfloor \\sqrt{n} \\rfloor + \\lfloor \\sqrt[3]{n} \\rfloor + \\dots + \\lfloor \\sqrt[v]{n} \\rfloor = \\lfloor \\log_2 n \\rfloor + \\lfloor \\log_3 n \\rfloor + \\dots + \\lfloor \\log_n n \\rfloor.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "Let $m = \\lfloor \\log_2 n \\rfloor$. Split the index set $\\{2, \\dots, n\\}$ into $m$ pairwise disjoint subsets:\n$$\nK_j = \\{k : \\lfloor \\log_k n \\rfloor = j\\} = \\{k : n^{1/(j+1)} < k \\le n^{1/j}\\}, \\quad j = 1, \\dots, m,\n$$\nand notice that $|K_j| = \\lfloor n^{1/j} \\rfloor - \\lfloor n^{1/(j+1)} \\rfloor$, so\n$$\n\\begin{aligned}\n\\sum_{k=2}^{n} \\lfloor \\log_k n \\rfloor &= \\sum_{j=1}^{m} j |K_j| = \\sum_{j=1}^{m} j (\\lfloor n^{1/j} \\rfloor - \\lfloor n^{1/(j+1)} \\rfloor) \\\\\n&= n + \\sum_{j=2}^{m} \\lfloor n^{1/j} \\rfloor - m \\lfloor n^{1/(m+1)} \\rfloor.\n\\end{aligned}\n$$\nSince $\\lfloor n^{1/j} \\rfloor = 1$ for $j \\ge m+1$, the conclusion follows.\n\n**Remark.** With the convention $\\log_1 n = n$, we may equally well let $m = \\lfloor n^{1/2} \\rfloor$, partition the index set $\\{2, \\dots, n\\}$ into $m$ subsets:\n$$\nK_j = \\{k : \\lfloor n^{1/k} \\rfloor = j\\} = \\{k : \\log_{j+1} n < k \\le \\log_j n\\}, \\quad j = 1, \\dots, m,\n$$\nand notice that $|K_j| = \\lfloor \\log_j n \\rfloor - \\lfloor \\log_{j+1} n \\rfloor$, so\n$$\n\\begin{aligned}\n\\sum_{k=1}^{n} \\lfloor n^{1/k} \\rfloor &= \\sum_{j=1}^{m} j |K_j| = \\sum_{j=1}^{m} j (\\lfloor \\log_j n \\rfloor - \\lfloor \\log_{j+1} n \\rfloor) \\\\\n&= n + \\sum_{j=2}^{m} \\lfloor \\log_j n \\rfloor - m \\lfloor \\log_{m+1} n \\rfloor.\n\\end{aligned}\n$$\nSince $\\lfloor \\log_j n \\rfloor = 1$ for $j = m+1, \\dots, n$, the conclusion follows.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 14547,
    "subject": "Mathematics (Olympiad)",
    "question": "Given two circles on the plane that do not intersect. Choose diameters $A_1B_1$ and $A_2B_2$ of these circles such that the segments $A_1A_2$ and $B_1B_2$ intersect. Let $A$ and $B$ be the midpoints of segments $A_1A_2$ and $B_1B_2$, and let $C$ be their intersection point. Prove that the orthocenter of triangle $ABC$ belongs to a fixed line that does not depend on the choice of the diameters.\n\n*Comment.* The statement is true if the lines (not necessarily segments) $A_1A_2$ and $B_1B_2$ intersect. We formulate the statement for the segments in order to reduce the number of cases in solutions.",
    "options": [],
    "answer": "See solution",
    "solution": "![](images/bw18shortlist_p35_data_07d7ef66f2.png)\n\nProve that the orthocenter $H$ of $\\triangle ABC$ belongs to their radical axis.\n\nDenote the circles by $s_1$ and $s_2$. Let the line $A_1A_2$ intersect circles $s_1$ and $s_2$ a second time at points $X_1$ and $X_2$ respectively, and the line $B_1B_2$ intersect the circles a second time at points $Y_1$ and $Y_2$.\n\nThe lines $A_1Y_1$ and $A_2Y_2$ are parallel (because both are orthogonal to $B_1B_2$); analogously, $B_1X_1$ and $B_2X_2$ are parallel. Hence, these four lines form a parallelogram $KLMN$ (see figure). It is clear that the perpendiculars from $A$ to $BC$ and from $B$ to $AC$ lie on the midlines of this parallelogram. Therefore, $H$ is the center of parallelogram $KLMN$ and coincides with the midpoint of segment $KM$.\n\nTo prove that $H$ lies on the radical axis of $s_1$ and $s_2$, it is sufficient to show that both points $K$ and $M$ belong to that radical axis.\n\nThe points $X_1$ and $Y_2$ lie on the circle $s_3$ with diameter $B_1A_2$. The line $B_1X_1$ is the radical axis of $s_1$ and $s_3$, and the line $A_2Y_2$ is the radical axis of $s_2$ and $s_3$. Therefore, $K$ is the radical center of these three circles and hence $K$ lies on the radical axis of $s_1$ and $s_2$. Analogously, $M$ lies on the radical axis of $s_1$ and $s_2$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 14548,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $A$ be a set of $n$ real numbers such that for any two distinct elements $a, b \\in A$, $|a - b| \\ge 1$. What is the greatest possible value $M$ of $S = \\sum_{a \\in A} |a|$ in terms of $n$?",
    "options": [],
    "answer": "See solution",
    "solution": "For $n = 4$, consider $A = \\{-1, 0, 1, 2\\}$. Then $S = |-1| + |0| + |1| + |2| = 1 + 0 + 1 + 2 = 4$, so $M = 4$ is achievable. \n\nLet $a < b < c < d$ be the elements in $A$. Since $b - a, c - b, d - c \\ge 1$, we have $d - a \\ge 3$. Thus,\n$$\nS = (|a| + |d|) + (|b| + |c|) \\ge |d - a| + |c - b| \\ge 3 + 1 = 4.\n$$\nSo $M = 4$ for $n = 4$.\n\nFor $n \\ge 5$, consider $A = \\{-1, 0, \\varepsilon, 2\\varepsilon, \\dots, (n-4)\\varepsilon, 1 + (n-4)\\varepsilon, 1 + (n-3)\\varepsilon\\}$ with $\\varepsilon > 0$. As $\\varepsilon \\to 0$, $S$ approaches $3$, so $M \\le 3$.\n\nFor $n = 5$, let $a < b < c < d < e$ be the elements. Since $c-a \\ge 1$ and $e-c \\ge 1$, and at least one of $c-b \\ge 1$ or $d-c \\ge 1$ holds, say $c-b \\ge 1$, then\n$$\nS \\ge (|a| + |e|) + (|b| + |c|) \\ge |e-a| + |c-b| \\ge 2 + 1 = 3.\n$$\nSo $M = 3$ for $n = 5$.\n\nFor $n \\ge 6$, let $a$ be the element with smallest $|a|$. WLOG $a \\ge 0$.\n- If $0 \\le a \\le \\frac{1}{2}$, then for any $x$ with $|x-a| \\ge 1$, $|x| \\ge 1-a$. There are at least three such $x$, so $S \\ge 3(1-a) + (n-3)a \\ge 3$.\n- If $a > \\frac{1}{2}$, then all $|a| > \\frac{1}{2}$, so $S > \\frac{n}{2} \\ge 3$.\n\nThus, $M = 3$ for $n \\ge 5$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 14549,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $ABC$ be an acute-angled triangle and let $D$, $E$, and $F$ be the midpoints of $BC$, $CA$, and $AB$ respectively. Construct a circle, centered at the orthocentre of triangle $ABC$, such that triangle $ABC$ lies in the interior of the circle. Extend $EF$ to intersect the circle at $P$, $FD$ to intersect the circle at $Q$, and $DE$ to intersect the circle at $R$. Show that $AP = BQ = CR$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let the radius of the circle be $r$. Let $X$, $Y$, and $Z$ be the feet of the altitudes from $A$, $B$, and $C$ respectively. Let $PE$ intersect the altitude from $A$ at $U$. We have\n\n$$\nAP^2 = AU^2 + PU^2 = AU^2 + r^2 - UH^2 = r^2 + (AU + UH) \\cdot (AU - UH) = r^2 + AH \\cdot (AU - UH) = r^2 + AH \\cdot (UX - UH) = r^2 + AH \\cdot HX.\n$$\n\nSimilarly, $BQ = r^2 + BH \\cdot HY$, and $CR = r^2 + CH \\cdot HZ$. Since $AH \\cdot HX = BH \\cdot HY = CH \\cdot HZ$, we have $AP = BQ = CR$.\n\n![](images/Singapur_2013_p1_data_8e35375529.png)",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 14550,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all functions $f : \\mathbb{Q}^{+} \\to \\mathbb{Q}^{+}$ such that\n$$\nf(xf(x) + f(y)) = (f(x))^2 + y\n$$\nfor all positive rationals $x, y$.",
    "options": [],
    "answer": "See solution",
    "solution": "The only function satisfying (1) is the identity function. It is not hard to see that the identity function satisfies (1).\n\nLet $f : \\mathbb{Q}^{+} \\to \\mathbb{Q}^{+}$ be a function that satisfies (1) for all $x, y \\in \\mathbb{Q}^{+}$, and let $a = f(1)$. Plugging in $x = 1$ in (1) yields\n$$\nf(a + f(y)) = a^2 + y \\tag{2}\n$$\nfor all $y \\in \\mathbb{Q}^{+}$. Substituting $y$ with $a + f(y)$ in (2) yields\n$$\nf(a + (a^2 + y)) = a^2 + (a + f(y)).\n$$\nLet $b = a^2 + a$. By induction on $n$, we have\n$$\nf(y + nb) = f(y) + nb \\tag{3}\n$$\nfor all $y \\in \\mathbb{Q}^{+}$ and $n \\in \\mathbb{N}$.\n\nSubstitute $x$ with $x + nb$ in (1) to get\n$$\nf((x + nb)f(x + nb) + f(y)) = (f(x + nb))^2 + y.\n$$\nFor an arbitrary $x \\in \\mathbb{Q}^{+}$, choose an integer $n$ such that $nb, nx$, and $nf(x)$ are all integers. From (3), $f(x + nb) = f(x) + nb$, so\n$$\nf(xf(x) + nbf(x) + nbx + n^2b^2 + f(y)) = (f(x))^2 + 2nbf(x) + n^2b^2 + y.\n$$\nSince $nf(x), nx$, and $n^2b$ are integers, it follows from (3) that\n$$\nf(xf(x) + f(y)) + nbf(x) + nbx + n^2b^2 = (f(x))^2 + 2nbf(x) + n^2b^2 + y.\n$$\nThus, $nbx = nbf(x)$, and so $f(x) = x$ for all $x \\in \\mathbb{Q}^{+}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 14551,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $p$ be a prime number. Show that one can choose $p^3$ fields of a $p^2 \\times p^2$ chessboard such that the centers of no four chosen fields are vertices of a rectangle with sides parallel to the sides of the chessboard.",
    "options": [],
    "answer": "See solution",
    "solution": "Label the $p^2$ rows and $p^2$ columns of the chessboard by all pairs $(a, b)$ and $(c, d)$, where $a, b, c, d \\in \\{0, 1, \\dots, p-1\\}$. A field in row $(a, b)$ and column $(c, d)$ is called *good* if and only if\n\n$$\na c \\equiv b + d \\pmod{p}. $$\n\nGiven a pair $(a, b)$, this holds for exactly $p$ pairs $(c, d)$, so there are $p$ good fields in any row and $p^3$ in total.\n\nTo show that no four good fields form the vertices of a rectangle with sides parallel to the chessboard, suppose for contradiction that\n\n$$\na_i c_j \\equiv b_i + d_j \\pmod{p} \\quad \\text{for } i, j \\in \\{1, 2\\}, $$\n\nfor some $(a_1, b_1) \\ne (a_2, b_2)$ and $(c_1, d_1) \\ne (c_2, d_2)$. Subtracting for fixed $i$ gives\n\n$$\na_i(c_2 - c_1) \\equiv d_2 - d_1 \\pmod{p} \\quad \\text{for } i \\in \\{1, 2\\}, $$\n\nwhich leads to\n\n$$\n(a_2 - a_1)(c_2 - c_1) \\equiv 0 \\pmod{p}.\n$$\n\nThus $a_1 = a_2$ or $c_1 = c_2$. By symmetry, assume $c_1 = c_2$. Then $d_1 = d_2$, so $(c_1, d_1) = (c_2, d_2)$, a contradiction.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 14552,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $p, q, r > 0$ and $m > 1$. Prove that\n$$\n\\frac{p^m}{q+r} + \\frac{q^m}{r+p} + \\frac{r^m}{p+q} \\geq \\frac{(p+q+r)^{m-1}}{2 \\cdot 3^{m-2}}.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "By the Cauchy-Schwarz inequality,\n$$\n((q+r) + (r+p) + (p+q)) \\left( \\frac{1}{q+r} + \\frac{1}{r+p} + \\frac{1}{p+q} \\right) \\geq 9.\n$$\nDividing both sides by $2(p+q+r)$ gives\n$$\n\\frac{1}{q+r} + \\frac{1}{r+p} + \\frac{1}{p+q} \\geq \\frac{9}{2(p+q+r)}.\n$$\nAssume $p \\geq q \\geq r$. Then $\\frac{1}{q+r} \\geq \\frac{1}{r+p} \\geq \\frac{1}{p+q}$ and $p^m \\geq q^m \\geq r^m$. By Chebyshev's inequality,\n$$\n\\frac{p^m}{q+r} + \\frac{q^m}{r+p} + \\frac{r^m}{p+q} \\geq \\frac{p^m+q^m+r^m}{3} \\left( \\frac{1}{q+r} + \\frac{1}{r+p} + \\frac{1}{p+q} \\right).\n$$\nBy the power mean inequality (since $m > 1$),\n$$\n\\frac{p^m + q^m + r^m}{3} \\geq \\left( \\frac{p+q+r}{3} \\right)^m.\n$$\nCombining these,\n$$\n\\frac{p^m}{q+r} + \\frac{q^m}{r+p} + \\frac{r^m}{p+q} \\geq \\left(\\frac{p+q+r}{3}\\right)^m \\cdot \\frac{9}{2(p+q+r)} = \\frac{(p+q+r)^{m-1}}{2 \\cdot 3^{m-2}}.\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 14553,
    "subject": "Mathematics (Olympiad)",
    "question": "Circles $A$ and $B$, with six persons in each. Let each acrobat in $B$ stand on the shoulders of two adjacent acrobats of $A$. We call it a *tower* if the label of each acrobat of $B$ is equal to the sum of the labels of the acrobats under his feet. How many different towers can they make?\n\n(Remark: We treat two towers as the same if one can be obtained by rotation or reflection of the other. For example, the following towers are the same, where the labels inside the circle refer to the bottom acrobat, the labels outside the circle refer to the upper acrobat.)\n\n![](images/Mathematical_Olympiad_in_China_2011-2014_p313_data_a0b77afb78.png)",
    "options": [],
    "answer": "See solution",
    "solution": "Denote the sum of labels of $A$ and $B$ by $x$ and $y$, respectively. Then $y = 2x$. Thus, we have\n\n$$\n3x = x + y = 1 + 2 + \\cdots + 12 = 78, \\quad x = 26.\n$$\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 \\geq 3$, $8 \\leq d \\leq 10$ (if $d \\leq 7$, then $a + b + c + d \\leq 4 + 5 + 6 + 7 = 22$, which is a contradiction.)\n\n1. If $d = 8$, then $A = \\{1, 2, a, b, c, 8\\}$, $c \\leq 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![](images/Mathematical_Olympiad_in_China_2011-2014_p313_data_28a107e291.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![](images/Mathematical_Olympiad_in_China_2011-2014_p314_data_af6df9043f.png)\n\n![](images/Mathematical_Olympiad_in_China_2011-2014_p314_data_822e83fc42.png)\n\n2. If $d = 9$, then $A = \\{1, 2, a, b, c, 9\\}$, $c \\leq 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![](images/Mathematical_Olympiad_in_China_2011-2014_p314_data_5c558bfa0b.png)\n\n![](images/Mathematical_Olympiad_in_China_2011-2014_p314_data_dba1f32276.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\n3. If $d = 10$, then $A = \\{1, 2, a, b, c, 10\\}$, where $c \\leq 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![](images/Mathematical_Olympiad_in_China_2011-2014_p314_data_abef29bdff.png)\n\nSumming up, there are six different towers all together. $\\Box$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 14554,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $A_1, A_2, \\dots, A_8$ be fixed points on a circle. Determine the smallest positive integer $n$ such that among any $n$ triangles with these eight points as vertices, two of them will have a common side.",
    "options": [],
    "answer": "See solution",
    "solution": "First, we prove that if there are $r$ triangles such that every two of them have no common side, then the maximum value of $r$ is $8$.\n\nThere are $\\binom{8}{2} = 28$ chords whose endpoints are from the $8$ points.\n\nIf every chord belongs to only one triangle, then the maximum $r \\leq \\left\\lfloor \\frac{28}{3} \\right\\rfloor = 9$. But if there are $9$ triangles such that every two of them have no common side, then there are $27$ vertices among these triangles, so there is a point belonging to $4$ triangles (suppose it is $A_8$). The opposite sides are chords with endpoints among $A_1, A_2, \\dots, A_7$, so there is a point $A_k$ that is the endpoint of two triangles, and the two triangles have a common side $A_8A_k$. Thus, $r \\leq 8$.\n\nOn the other hand, the following example shows that $r$ can be $8$: the triangles $(1,2,8)$, $(1,3,6)$, $(1,4,7)$, $(2,3,4)$, $(2,5,7)$, $(3,5,8)$, $(4,5,6)$, $(6,7,8)$ satisfy the condition. So the maximum value of $r$ is $8$.\n\nTherefore, the minimum $n$ is $8+1=9$.\n\n![](images/Mathematical_Olympiad_in_China_2009-2010_p182_data_52de46134f.png)",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 14555,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all functions $f$ mapping non-negative reals to non-negative reals such that\n\n$$\nf(x_1^2 + \\cdots + x_n^2) = f(x_1)^2 + \\cdots + f(x_n)^2\n$$\n\nfor any choice of numbers $x_1, \\dots, x_n$.",
    "options": [],
    "answer": "See solution",
    "solution": "The solutions are $f(x) = 0$ and $f(x) = x$.\n\nFirst, observe that\n\n$$\nf(1) = f(1^2) = f(1)^2,\n$$\n\nso $f(1)$ is either $0$ or $1$.\n\n**Case 1:** $f(1) = 0$\n\nFor any positive integer $n$,\n\n$$\nf(n) = f(n \\cdot 1^2) = n f(1)^2 = 0.\n$$\n\nGiven any $x \\ge 0$, choose $y$ so that $x^2 + y^2 = n$ for some integer $n$. Then\n\n$$\nf(x)^2 + f(y)^2 = f(x^2 + y^2) = f(n) = 0,\n$$\n\nso $f(x) = 0$ for all $x$.\n\n**Case 2:** $f(1) = 1$\n\nWe will show $f(x) = x$ for all $x \\ge 0$.\n\nFor any positive integer $n$,\n\n$$\nf(n) = f(n \\cdot 1^2) = n f(1)^2 = n.\n$$\n\nFor a non-negative rational $x = \\frac{p}{q}$,\n\n$$\np^2 = f(p^2) = f\\left(q^2 \\left(\\frac{p}{q}\\right)^2\\right) = q^2 f\\left(\\frac{p}{q}\\right)^2,\n$$\n\nso $f(x) = x$ for rational $x$.\n\nFor irrational $x \\ge 0$, pick rational $\\frac{p}{q} > x$ and choose $y$ so that $x^2 + y^2 = \\frac{p^2}{q^2}$. Then\n\n$$\n\\frac{p^2}{q^2} = f\\left(\\frac{p^2}{q^2}\\right) = f(x^2 + y^2) = f(x)^2 + f(y)^2 \\ge f(x)^2,\n$$\n\nso $f(x) \\le \\frac{p}{q}$. Next, pick rational $\\frac{r}{s} < \\sqrt{x}$, so $\\frac{r^2}{s^2} < x$, and choose $z$ so that $\\frac{r^2}{s^2} + z^2 = x$. Then\n\n$$\nf(x) = f\\left(\\frac{r^2}{s^2} + z^2\\right) = f\\left(\\frac{r}{s}\\right)^2 + f(z)^2 = \\frac{r^2}{s^2} + f(z)^2 \\ge \\frac{r^2}{s^2},\n$$\n\nso $f(x) \\ge \\frac{r^2}{s^2}$. Since $f(x)$ is squeezed between all rationals below and above $x$, $f(x) = x$ for all $x \\ge 0$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 14556,
    "subject": "Mathematics (Olympiad)",
    "question": "Encontrar las tres últimas cifras de $7^{2014}$.",
    "options": [],
    "answer": "See solution",
    "solution": "Usaremos el teorema de Euler-Fermat: si $\\gcd(a, m) = 1$, entonces\n\n$$\na^{\\varphi(m)} \\equiv 1 \\pmod{m}.\n$$\n\nEn nuestro caso, queremos calcular $7^{2014} \\pmod{1000}$. Como $1000 = 2^3 \\cdot 5^3$, se tiene que $\\varphi(1000) = 2^2 (2-1) \\cdot 5^2 (5-1) = 400$. Entonces,\n\n$$\n7^{2014} = 7^{5 \\cdot 400 + 14} = (7^{400})^5 \\cdot 7^{14} \\equiv 1^5 \\cdot 7^{14} \\equiv 849 \\pmod{1000}.\n$$\n\nEn consecuencia, las tres últimas cifras de $7^{2014}$ son 849.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 14557,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $\\triangle ABC$ be a triangle such that $\\angle ACB = 90^\\circ$. The point $D$ lies inside triangle $ABC$ and on the circle with centre $B$ that passes through $C$. The point $E$ lies on the side $AB$ such that $\\angle DAE = \\angle BDE$. The circle with centre $A$ that passes through $C$ meets the line through $D$ and $E$ at the point $F$, where $E$ lies between $D$ and $F$.\n\nProve that $\\angle AFE = \\angle EBF$.",
    "options": [],
    "answer": "See solution",
    "solution": "Since $\\angle BAD = \\angle BDE$ by assumption and $\\angle DBA = \\angle EBD$, we know that triangles $BAD$ and $BDE$ are similar.\n\n![](images/Australian_Scene_-2014_p120_data_f301def032.png)\n\nHence, we have the equal ratios\n\n$$\n\\frac{EB}{DB} = \\frac{DB}{AB} \\Rightarrow EB = \\frac{DB^2}{AB}.\n$$\n\nTherefore, we can deduce the following sequence of equalities.\n\n$$\n\\begin{align*}\nAE &= AB - EB \\\\\n   &= AB - \\frac{BD^2}{AB} \\\\\n   &= \\frac{AB^2 - BD^2}{AB} \\\\\n   &= \\frac{AB^2 - BC^2}{AB} && (D \\text{ lies on the circle with centre } B \\text{ through } C) \\\\\n   &= \\frac{AC^2}{AB} && (\\text{Pythagoras' theorem in triangle } ABC) \\\\\n   &= \\frac{AF^2}{AB} && (F \\text{ lies on the circle with centre } A \\text{ through } C)\n\\end{align*}\n$$\n\nThis implies that $\\frac{AF}{AB} = \\frac{AE}{AF}$, which combines with the fact that $\\angle BAF = \\angle FAE$ to show that triangles $AFB$ and $AEF$ are similar.\n\nTherefore, we conclude that\n\n$$\n\\angle AFE = \\angle ABF = \\angle EBF.\n$$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 14558,
    "subject": "Mathematics (Olympiad)",
    "question": "If $n$ is a perfect number, show that $n$ must have at least 3 distinct prime factors.",
    "options": [],
    "answer": "See solution",
    "solution": "If $n = p^k$ for some odd prime $p$ and positive integer $k$, then\n\n$$\n\\frac{\\sigma(n)}{n} = \\frac{1 + p + \\cdots + p^k}{p^k} = 1 + \\frac{1}{p} + \\cdots + \\frac{1}{p^k} < \\frac{1}{1 - \\frac{1}{p}} = \\frac{p}{p-1} < 2.\n$$\n\nIf $n = p^k q^\\ell$ for some odd primes $p, q$ and positive integers $k, \\ell$, then\n\n$$\n\\frac{\\sigma(n)}{n} = \\frac{1 + p + \\cdots + p^k}{p^k} \\cdot \\frac{1 + q + \\cdots + q^\\ell}{q^\\ell} < \\frac{p}{p-1} \\cdot \\frac{q}{q-1} \\leq \\frac{3}{2} \\cdot \\frac{5}{4} < 2.\n$$\n\nThis shows $\\sigma(n) < 2n$ if $n$ has at most 2 distinct prime factors. Therefore, in order that $n$ is a perfect number, it must have at least 3 distinct prime factors.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 14559,
    "subject": "Mathematics (Olympiad)",
    "question": "If $t$ toffees cost $c$ cents, and $r$ rands equals $100r$ cents, how many toffees can be bought for $100r$ cents?",
    "options": [],
    "answer": "See solution",
    "solution": "Each toffee costs $\\frac{c}{t}$ cents. The number of toffees that can be bought for $100r$ cents is:\n$$\n\\frac{100r}{\\frac{c}{t}} = \\frac{100rt}{c}\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 14560,
    "subject": "Mathematics (Olympiad)",
    "question": "Given is the sequence of numbers $a_0, a_1, a_2, \\dots, a_{2020}$ with $a_0 = 0$. Furthermore, the following holds for every $k = 1, 2, \\dots, 2020$:\n\n$$\na_k = \\begin{cases} a_{k-1} \\cdot k & \\text{if } k \\text{ is divisible by } 8, \\\\ a_{k-1} + k & \\text{if } k \\text{ is not divisible by } 8. \\end{cases}\n$$\n\nWhat are the last two digits of $a_{2020}$?",
    "options": [],
    "answer": "See solution",
    "solution": "02",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 14561,
    "subject": "Mathematics (Olympiad)",
    "question": "How many multiplicative functions $f : \\mathbb{Z}/1019\\mathbb{Z} \\to \\mathbb{Z}/1019\\mathbb{Z}$ are there, given that $f(x + 1019) = f(x)$ for all $x$?",
    "options": [],
    "answer": "See solution",
    "solution": "The condition $f(x + 1019) = f(x)$ allows all computations modulo $1019$, a prime. Thus, we seek the number of multiplicative functions on $\\mathbb{Z}/1019\\mathbb{Z}$.\n\nTwo such functions are $f(x) \\equiv 1 \\pmod{1019}$ and $f(x) \\equiv 0 \\pmod{1019}$. Suppose $f$ is different from these.\n\nSubstitute $y = 0$: $f(0) = f(x)f(0) \\implies f(0) = 0$ (unless $f$ is identically $1$). Substitute $y = 1$: $f(x) = f(x)f(1) \\implies f(1) = 1$.\n\nBy Euler's theorem, $x^{\\varphi(1019)} \\equiv 1 \\pmod{1019}$, so $f(x)^{\\varphi(1019)} = f(x^{\\varphi(1019)}) = f(1) = 1$, so $f(x) = \\pm 1$ for $x \\neq 0$.\n\nLet $g$ be a primitive root modulo $1019$. Every $x \\neq 0$ can be written as $g^k$. Then $f(x) = (f(g))^k$.\n\nIf $f(g) = 1$, then $f(x) = 1$ for $x \\neq 0$; if $f(g) = -1$, then $f(x) = 1$ if $k$ is even (i.e., $x$ is a quadratic residue), $-1$ if $k$ is odd (nonresidue).\n\nThus, there are exactly $3$ multiplicative functions: the constant $1$, the constant $0$, and the quadratic character (Legendre symbol).",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 14562,
    "subject": "Mathematics (Olympiad)",
    "question": "$\\{a_n\\}_{n \\ge 0}$ 為公差 $d$ 的無窮整數數列，其中首項滿足 $1 \\leq a_0 \\leq d$。記此數列為 $S_0$，我們以遞迴的方式定義一序列的新數列；數列 $S_{n+1}$ 為 $S_n$ 經由底下兩步操作後得到：\n\n1. 記 $S_n$ 的首項為 $b_n$。將首項移除，並把數列剩下的每一項往前挪。\n2. 從新的首項開始算起，將該數列的前 $b_n$ 項加 $1$。\n\n試證：存在常數 $c$ 使得 $b_n = [ca_n]$ 對 $n \\geq 0$ 恆成立。此處 $[\\cdot]$ 表示高斯符號。",
    "options": [],
    "answer": "See solution",
    "solution": "我們以歸納法證明：\n\n$$\nc = \\frac{1 + \\sqrt{1 + \\frac{4}{d}}}{2},\n$$\n\n它是方程式 $c^2 - c - 1/d = 0$ 的正根。\n\n首先證明 $n = 0$ 時命題成立。依定義 $b_0 = a_0$，而\n\n$$\n\\begin{aligned}\n[ca_0] &= \\left[ \\frac{1 + \\sqrt{1 + 4/d}}{2} a_0 \\right] \\\\\n&= a_0 + \\left[ \\frac{-1 + \\sqrt{1 + 4/d}}{2} a_0 \\right] \\\\\n&= a_0 + \\left[ \\frac{2}{1 + \\sqrt{1 + 4/d}} \\cdot \\frac{a_0}{d} \\right] = a_0,\n\\end{aligned}\n$$\n\n故 $b_0 = [ca_0]$ 成立。\n\n接著以歸納法證明命題對 $n > 0$ 亦成立。依題意，$b_n$ 的值是由 $a_n$ 加若干次 $1$ 得到：進行 $S_k \\rightarrow S_{k+1}$ 的操作時若該項增加 $1$，必有 $b_k + k \\geq n$ 且 $0 \\leq k < n$。對於所有的 $n$，我們把這樣的 $k$ 寫成集合：\n\n$$\nX_n = \\{k \\mid b_k + k \\geq n \\text{ 且 } 0 \\leq k < n\\},\n$$\n\n就有 $b_n = a_n + |X_n|$。關鍵在於集合 $X_n$ 的大小如何計算：根據歸納假設，\n\n$$\n\\begin{aligned}\nX_n &= \\{k \\mid [ca_k] + k \\geq n,\\ 0 \\leq k < n\\} \\\\\n&= \\{k \\mid ca_k + k \\geq n,\\ 0 \\leq k < n\\} \\\\\n&= \\{k \\mid ca_0 + k(1 + cd) \\geq n,\\ 0 \\leq k < n\\} \\\\\n&= \\left\\{k \\mid \\frac{n - ca_0}{1 + cd} \\leq k < n\\right\\}\n\\end{aligned}\n$$\n\n註：即使 $n < ca_0$ 也無妨，因為\n\n$$\n\\frac{n - ca_0}{1 + cd} > -1\n$$\n\n恆成立，不影響 $k$ 的個數。\n\n因此\n\n$$\n\\begin{aligned}\nb_n &= a_n + |X_n| = a_n + \\left[ \\frac{ca_n}{1 + cd} \\right] \\\\\n&= \\left[ \\left(1 + \\frac{c}{1 + cd}\\right) a_n \\right] = [ca_n],\n\\end{aligned}\n$$\n\n原命題得證。",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 14563,
    "subject": "Mathematics (Olympiad)",
    "question": "試求所有正整數對 $ (x, y) $，滿足\n\n$$\n\\sqrt[3]{7x^2 - 13xy + 7y^2} = |x - y| + 1.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "答案為 $x = y = 1$ 與 $\\{x, y\\} = \\{m^3 + m^2 - 2m - 1,\\ m^3 + 2m^2 - m - 1\\}$，其中 $m \\ge 2$。\n\n1. 若 $x = y$，則原式等價於 $x^{2/3} = 1$，故 $x = y = 1$。\n\n2. 若 $x > y$，令 $n = x - y$，則原式可改寫為\n\n$$\n\\sqrt[3]{7(y + n)^2 - 13(y + n)y + 7y^2} = n + 1.\n$$\n\n等號兩邊同時立方並化簡後，我們有\n\n$$\ny^{2} + yn = n^{3} - 4n^{2} + 3n + 1.\n$$\n\n為讓左式配方，我們同乘 4 並同加 $n^2$，得到\n\n$$\n(2y + n)^2 = (n - 2)^2(4n + 1).\n$$\n\n顯然 $n \\le 2$ 是不可能的。當 $n > 2$ 時，基於 $4n + 1$ 必須是完全平方數，必有 $4n + 1 = (2m + 1)^2$，從而\n\n$$\nn = m^2 + m, \\tag{1}\n$$\n\n其中 $m \\ge 2$（因為 $n \\ge 3$）。帶回原式，得\n\n$$\n(2y + m^2 + m)^2 = (2m^3 + 3m^2 - 3m - 2)^2.\n$$\n\n故顯然 $2y + m^2 + m = 2m^3 + 3m^2 - 3m - 2 \\Leftrightarrow y = m^3 + m^2 - 2m - 1$。\n\n帶回即得解。",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 14564,
    "subject": "Mathematics (Olympiad)",
    "question": "Isaac is planning a nine-day holiday. Every day he will go surfing, water skiing, or rest. On any given day, he does just one of these three things. He never does different water-sports on consecutive days. How many schedules are possible for this holiday?",
    "options": [],
    "answer": "See solution",
    "solution": "Let $f(n)$ be the number of possible holidays that conform to Isaac's rules and last $n$ days. We are asked to find $f(9)$.\n\nIf he rests on the $n$th day, then he can surf, ski, or rest on the $(n+1)$th day (three choices). If he water-skis on the $n$th day, then he can rest or water-ski on the $(n+1)$th day (two choices). Similarly, if he surfs, then he can either surf or rest. So, the number of possible holidays of length $n+1$ is equal to three times the number of holidays of length $n$ that end with a rest day, plus two times the number of holidays of length $n$ that end with surfing, plus two times the number of holidays of length $n$ that end with water-skiing.\n\nThus, the number of holidays over $(n+1)$ days is equal to twice the number of holidays over $n$ days, plus the number of holidays of $n$ days that end with a rest day. However, the number of holidays with $n$ days that end with a rest is equal to the number of holidays of length $(n-1)$. That is,\n\n$$\nf(n + 1) = 2f(n) + f(n - 1).\n$$\n\nWe can manually find that $f(1) = 3$ and $f(2) = 7$. Thereafter, we can calculate $f(3) = 17$, $f(4) = 41$, $f(5) = 99$, $f(6) = 239$, $f(7) = 577$, $f(8) = 1393$, and $f(9) = 3363$.\n\nSo the answer is $3363$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 14565,
    "subject": "Mathematics (Olympiad)",
    "question": "There were 25 birds on two branches. After a while, 5 birds from the first branch flew over to the second, and after a while longer, 7 birds from the second flew away. At that moment, there were twice as many birds on the first branch as on the second. How many birds were there on each of the two branches in the beginning?",
    "options": [],
    "answer": "See solution",
    "solution": "When the 7 birds flew away from the second branch, the total number of birds on both branches was $25 - 7 = 18$. Since at that moment there were twice as many birds on the first branch as on the second, there were $18 \\div 3 = 6$ birds on the second branch, and $2 \\cdot 6 = 12$ birds on the first branch.\n\nBefore the 7 birds flew away, the second branch had $6 + 7 = 13$ birds. Before the 5 birds flew from the first to the second branch, the second branch had $13 - 5 = 8$ birds, and the first branch had $12 + 5 = 17$ birds.\n\n**Answer:** There were 17 birds on the first branch and 8 birds on the second branch in the beginning.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 14566,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $n \\ge 3$. Alice and Bob play the following game: Alice chooses $k \\in \\{3, 4, \\dots, n\\}$ and draws a $3 \\times k$ table, then she fills the $k$ cells of the first row with different numbers from $\\{1, 2, \\dots, n\\}$. Then, Bob fills on the second row some of the cells (possibly none) with distinct numbers from $\\{1, 2, \\dots, n\\}$, and the rest of them with $0$. Finally, on each cell of the third row we write the sum of the two cells above. Show that regardless how Alice plays, Bob can guarantee that on the third row he can obtain, in some order, the terms of a non-constant arithmetic progression.\n\nShow that for every integer sequence $a_1 < \\dots < a_k \\le n$ one can choose integers $0 \\le b_1 < \\dots < b_k \\le n$ such that the sums $a_i + b_i$ form a permutation of a non-constant arithmetic progression and all the nonnegative $b_i$'s are distinct.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $1 \\le a_1 < a_2 < \\dots < a_k \\le n$ be the numbers Alice chose. For a sequence $x_1 < x_2 < \\dots < x_k$ of positive integers, we call its *deficit* the set $N \\cap [x_1, x_k] \\setminus \\{x_1, x_2, \\dots, x_k\\}$.\n\nBob has the following strategy: he starts with $a_1 < a_2 < \\dots < a_k$. Let $t$ be the maximum number of its deficit. If we denote $\\delta = t - a_1 < n - 1$, Bob writes under $a_1$ the number $a'_1 = \\delta$. Then $a_1 + a'_1 = t$. If $t < a_2$, then $a_2, a_3, \\dots, a_k$ are consecutive and $t = a_2 - 1$. So Bob writes under all the rest $0$, and he gets on the third row a progression with unit ratio.\n\nOtherwise, we have $t > a_2$ and Bob repeats the process, but for the sequence $t, a_2, \\dots, a_k$. The lowest term is $a_2$ and its deficit does not have $t$, so has lower cardinality. Therefore, each step decreases the cardinality of the deficit. As long as the deficit is not empty, Bob can perform another step, so in the end the deficit will be empty and the numbers on the third row will form an arithmetic progression with unit ratio.\n\nAs the $\\delta$ values are strictly decreasing, Bob fulfills the requirement of using numbers from $1$ to $n$ only once.",
    "topic": "Discrete Mathematics",
    "subtopic": "Combinatorics"
  },
  {
    "id": 14567,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $n \\ge 3$ be a positive integer. Find all non-constant real polynomials $f_1(x), f_2(x), \\dots, f_n(x)$ such that\n\n$$\nf_k(x) f_{k+1}(x) = f_{k+1}(f_{k+2}(x)), \\quad 1 \\le k \\le n\n$$\n\nfor every real $x$ (here $f_{n+1}(x) \\equiv f_1(x)$ and $f_{n+2}(x) \\equiv f_2(x)$).",
    "options": [],
    "answer": "See solution",
    "solution": "Let $\\deg(f_k) = \\alpha_k \\in \\mathbb{N}$ for $1 \\le k \\le n$, with all indices modulo $n$. The condition gives $\\alpha_k + \\alpha_{k+1} = \\alpha_{k+1} \\alpha_{k+2}$, so $\\alpha_{k+1}$ divides $\\alpha_k$ for every $k$. Thus, $\\alpha_1 = \\alpha_2 = \\dots = \\alpha_n = 2$.\n\nLet $f_k(x) = a_k x^2 + b_k x + c_k$ with $a_k \\ne 0$. Comparing $x^4$ coefficients, $a_k = a_{k+2}^2$ for all $k$. If $n$ is even, $a_1 = a_3 = \\dots = a_{2m-1} = 1$ and $a_2 = a_4 = \\dots = a_{2m} = 1$. If $n$ is odd, $a_1 = a_2 = \\dots = a_n = 1$.\n\nComparing $x^3$ coefficients: $b_k + b_{k+1} = 2b_{k+2}$. Let $b = \\min\\{b_1, \\dots, b_n\\} = b_s$. Then $b_{s-2} + b_{s-1} = 2b_s$ implies $b_{s-2} = b_{s-1} = b$, so all $b_k = b$.\n\nComparing $x^2$ coefficients: $c_k + c_{k+1} = 2c_{k+2} + b$. Summing, $nb = 0$ so $b = 0$. Similarly, all $c_k = c$.\n\nThus, $f_k(x) = x^2 + c$. The equation becomes $(x^2 + c)^2 + c = (x^2 + c)(x^2 + c)$, so $c = 0$.\n\nTherefore, the only solution is $f_1(x) = f_2(x) = \\dots = f_n(x) = x^2$.",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 14568,
    "subject": "Mathematics (Olympiad)",
    "question": "It is known that $p, q$ ($q \\neq 0$) are real numbers; the equation $x^2 - px + q = 0$ has two real roots $\\alpha, \\beta$; the sequence $\\{a_n\\}$ satisfies $a_1 = p$, $a_2 = p^2 - q$, $a_n = pa_{n-1} - qa_{n-2}$ ($n = 3, 4, \\dots$).\n\n1. Find the general expression of $\\{a_n\\}$ in terms of $\\alpha, \\beta$.\n\n2. If $p = 1$, $q = -\\frac{1}{4}$, find the sum of the first $n$ terms of $\\{a_n\\}$.",
    "options": [],
    "answer": "See solution",
    "solution": "(1) By Vieta's theorem, $\\alpha \\beta = q \\neq 0$, $\\alpha + \\beta = p$. The recurrence can be rewritten as:\n\n$$\na_n = pa_{n-1} - qa_{n-2} = (\\alpha + \\beta)a_{n-1} - \\alpha\\beta a_{n-2} \\quad (n \\geq 3).\n$$\n\nLet $b_n = a_{n+1} - \\beta a_n$. Then $b_{n+1} = \\alpha b_n$ ($n \\geq 1$), so $\\{b_n\\}$ is geometric with ratio $\\alpha$.\n\nThe first term:\n$$\nb_1 = a_2 - \\beta a_1 = p^2 - q - \\beta p = (\\alpha + \\beta)^2 - \\alpha\\beta - \\beta(\\alpha + \\beta) = \\alpha^2.\n$$\nSo $b_n = \\alpha^{n+1}$, thus $a_{n+1} - \\beta a_n = \\alpha^{n+1}$, or\n$$\na_{n+1} = \\alpha^{n+1} + \\beta a_n.\n$$\n\nIf $\\Delta = p^2 - 4q = 0$ ($\\alpha = \\beta \\neq 0$), then $a_1 = p = 2\\alpha$ and $a_{n+1} = \\alpha^{n+1} + \\alpha a_n$. This leads to $\\frac{a_{n+1}}{\\alpha^{n+1}} - \\frac{a_n}{\\alpha^n} = 1$, so $\\frac{a_n}{\\alpha^n}$ is arithmetic with difference $1$ and first term $2$:\n$$\n\\frac{a_n}{\\alpha^n} = n + 1 \\implies a_n = (n + 1)\\alpha^n.\n$$\n\nIf $\\Delta > 0$ ($\\alpha \\neq \\beta$), then $a_n$ can be expressed as:\n$$\na_n = A \\alpha^n + B \\beta^n,\n$$\nwhere $A$ and $B$ are determined by initial conditions.\n\n(2) For $p = 1$, $q = -\\frac{1}{4}$, solve for $\\alpha, \\beta$:\n$$\nx^2 - x - \\frac{1}{4} = 0 \\implies x = \\frac{1 \\pm \\sqrt{2}}{2}.\n$$\n\nLet $a_n = A \\alpha^n + B \\beta^n$ with $a_1 = 1$, $a_2 = 1^2 - (-\\frac{1}{4}) = \\frac{5}{4}$.\n\nSet up:\n$$\n\\begin{cases}\nA \\alpha + B \\beta = 1 \\\\\nA \\alpha^2 + B \\beta^2 = \\frac{5}{4}\n\\end{cases}\n$$\nSolve for $A, B$ and then sum $S_n = a_1 + a_2 + \\cdots + a_n = A \\sum_{k=1}^n \\alpha^k + B \\sum_{k=1}^n \\beta^k$.\n\nUse geometric series:\n$$\n\\sum_{k=1}^n r^k = r \\frac{1 - r^n}{1 - r}\n$$\nSo,\n$$\nS_n = A \\frac{\\alpha(1 - \\alpha^n)}{1 - \\alpha} + B \\frac{\\beta(1 - \\beta^n)}{1 - \\beta}\n$$\nwhere $A, B, \\alpha, \\beta$ as above.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 14569,
    "subject": "Mathematics (Olympiad)",
    "question": "In triangle $ABC$, $\\angle C = 90^\\circ$. Point $D$ is the foot of the altitude from $C$. Choose a point $X$ in the interior of segment $CD$. Point $K$ lies on segment $AX$ such that $BK = BC$. Point $L$ lies on segment $BX$ such that $AL = AC$. Segments $AL$ and $BK$ meet at $M$. Prove that $MK = ML$.\n\n![](images/USA_IMO_2013-2014_p76_data_0ac39d4bd4.png)",
    "options": [],
    "answer": "See solution",
    "solution": "**Solution 1.** Let $\\omega_A$ be the circle centered at $A$ with radius $AC$, and let $\\omega_B$ be the circle centered at $B$ with radius $BC$ so that $L$ lies on $\\omega_A$ and $K$ lies on $\\omega_B$. Let these circles intersect again at a point $C_1$ other than $C$, so that $D$ and $X$ lie on $CC_1$.\n\nBy symmetry, this statement is true when $AC = BC$. Otherwise, let $P$ be the external center of similitude of $\\omega_A$ and $\\omega_B$. We claim that it is sufficient to prove that $K$, $L$, and $P$ are collinear. If this were true, Menelaus' theorem applied to the line through $K$, $L$, and $P$ and triangle $ABM$ yields\n\n$$\n1 = \\frac{PA}{PB} \\frac{BK}{KM} \\frac{ML}{LA} = \\frac{PA}{PB} \\frac{BC}{CA} \\frac{ML}{KM}.\n$$\n\nBy homothety, we have $\\frac{PA}{PB} = \\frac{CA}{BC}$, allowing us to conclude the desired equality $MK = ML$.\n\nIt remains to show that $K$, $L$, and $P$ are collinear. Because $BC$ and $BE$ are equal tangents to $\\omega_A$, $BCC_1$ is isosceles. The trigonometric form of Ceva's Theorem on point $B$ and triangle $CLC_1$ implies\n\n$$\n1 = \\frac{\\sin \\angle LCB \\sin \\angle CC_1B \\sin \\angle C_1LB}{\\sin \\angle BCC_1 \\sin \\angle BC_1L \\sin \\angle BLC} = \\frac{\\sin \\angle CC_1L \\sin \\angle C_1LX}{\\sin \\angle C_1CL \\sin \\angle XLC},\n$$\n\nfrom which we conclude that\n\n$$\n\\frac{\\sin \\angle XLC}{\\sin \\angle C_1LX} = \\frac{\\sin \\angle CC_1L}{\\sin \\angle C_1CL}.\n$$\n\nBy the Law of Sines, we conclude that\n\n$$\n\\frac{CX}{C_1X} = \\frac{LC \\sin \\angle XLC / \\sin \\angle CXL}{LC_1 \\sin \\angle XLC_1 / \\sin \\angle C_1XL} = \\frac{LC \\sin \\angle XLC}{LC_1 \\sin \\angle C_1XL} = \\frac{LC \\sin \\angle CC_1L}{LC_1 \\sin \\angle C_1CL} = \\left(\\frac{LC}{LC_1}\\right)^2.\n$$\n\nIn fact, $LX$ is the symmedian from $L$ in triangle $CLC_1$, and the equality $\\frac{CX}{C_1X} = \\left(\\frac{LC}{LC_1}\\right)^2$ is true of any symmedian. Similarly, we may show that $\\left(\\frac{KC}{KC_1}\\right)^2 = \\frac{CX}{C_1X}$, so $\\frac{KC}{KC_1} = \\frac{LC}{LC_1}$.\n\nNow, let lines $PC$, $PL$, and $PC_1$ intersect $\\omega_A$ again at $C_2$, $K_1$, and $C_3$, respectively. Because $CC_2K_1L$ and $C_1C_3K_1L$ are cyclic, we see that $PC_2K_1 \\sim PLC$ and $PC_3K_1 \\sim PLC_1$, which implies that\n\n$$\n\\frac{C_2K_1}{CL} = \\frac{PK_1}{PC} = \\frac{PK_1}{PC_1} = \\frac{C_3K_1}{C_1L}\n$$\n\nand hence that\n\n$$\n\\frac{C_2K_1}{C_3K_1} = \\frac{CL}{C_1L} = \\frac{CK}{C_1K}.\n$$\n\nBy equal arcs, we have that $\\angle C_2K_1C_3 = \\angle CKC_1$, which along with our length equality above implies that $C_2K_1C_3 \\sim CKC_1$. Finally, the homothety that takes $\\omega_A$ to $\\omega_B$ takes $C_2$ to $C$ and $C_3$ to $C_1$, so it must take $K_1$ to $K$. This implies that $K$ lies on $PL$, as desired.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 14570,
    "subject": "Mathematics (Olympiad)",
    "question": "In an isosceles triangle $ABC$, $AB = AC$. Let $E$ be a point on $AC$ and extend $AB$ to a point $D$ such that $BD = EC$. Let $F$ be the intersection of $BC$ and $DE$, and let the circumcircle of $ABC$ intersect the circumcircle of $BDF$ at $G$. Prove that $DE$ is perpendicular to $FG$.\n\n![](images/Singapur_2010_p2_data_1c30fcda44.png)",
    "options": [],
    "answer": "See solution",
    "solution": "Note that the points $B$, $D$, $G$, $F$ are concyclic and the points $A$, $B$, $G$, $C$ are also concyclic. Also, since $\\angle EDG = \\angle FDG = \\angle FBG = \\angle CBG = \\angle GAC = \\angle GAE$, the points $A$, $D$, $G$, $E$ are concyclic. Since $\\angle GEC = \\angle ADG = \\angle BDG = \\angle CFG$, the points $G$, $C$, $E$, $F$ are concyclic. Since $BD = EC$, $\\angle CFE = \\angle BFD$, the circle passing through the points $B$, $D$, $G$, $F$ and the circle passing through the points $G$, $C$, $E$, $F$ are equivalent. And since the points $A$, $B$, $G$, $C$ are concyclic, $\\angle DBG = \\angle ECG$ and therefore $DG = GE$. Finally, since the points $B$, $D$, $G$, $F$ are concyclic, $\\angle ABC = \\angle ABF = \\angle FGD$. Together with $\\angle ECF = \\angle ACB = \\angle ABC$, we have $\\angle FGD = \\angle ECF$. Therefore $DF = FE$. Hence $DGF \\equiv EGF$, so $\\angle DFG = \\angle EFG = 90^\\circ$. Thus $DE \\perp FG$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 14571,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $M$ be the greatest integer such that both $M + 1213$ and $M + 3773$ are perfect squares. What is the units digit of $M$?\n\n(A) 1  (B) 2  (C) 3  (D) 6  (E) 8",
    "options": [],
    "answer": "See solution",
    "solution": "Suppose $M + 1213 = j^2$ and $M + 3773 = k^2$ for nonnegative integers $j$ and $k$. Then $(k+j)(k-j) = k^2 - j^2 = 3773 - 1213 = 2560 = 5 \\cdot 2^9$.\n\nBecause $k+j$ and $k-j$ have the same parity and their product is even, they must both be even, and it follows that one of them is $5 \\cdot 2^i$ and the other is $2^{9-i}$ for some $i$ with $1 \\leq i \\leq 8$. Solving for $k$ gives\n\n$$\nk = \\frac{5 \\cdot 2^i + 2^{9-i}}{2}.\n$$\n\nTo maximize $M$ it is sufficient to maximize $k$, and this will occur when $i = 8$ and $k = 5 \\cdot 2^7 + 1 = 641$. Therefore $M = 641^2 - 3773$, and its units digit is 8.\n\nAlternatively,\n\nBecause $(n + 1)^2 - n^2 = 2n + 1$, successive terms in the sequence of squares, $1, 4, 9, 16, \\ldots$, differ by successive odd numbers; and because $(n + 2)^2 - n^2 = 4(n + 1)$, the terms in this sequence that are two apart differ by successive multiples of 4. The two squares required in this problem differ by $3773 - 1213 = 2560$, a multiple of 4. It follows that the greatest such squares are two apart in the sequence of squares, so $n + 1 = \\frac{2560}{4} = 640$. Therefore these squares are $n^2 = 639^2$ and $(n + 2)^2 = 641^2$, and $M + 1213 = 639^2$. Then $M = 639^2 - 1213$, and its units digit is 8.\n\nShown below is a table of $5 \\cdot 2^i$, $2^{9-i}$, $k$, $j$, and $M$ for each $i$ (notation from the first solution). Observe that $M$ is maximized when $k$ is maximized.\n\n![](table)\n\n| i | $5 \\cdot 2^i$ | $2^{9-i}$ | $k$ | $j$ | $M$ |\n|---|----------------|-----------|-----|-----|------|\n| 1 | 10             | 256       | 133 | 123 | 13916 |\n| 2 | 20             | 128       | 74  | 54  | 1703  |\n| 3 | 40             | 64        | 52  | 12  | -1069 |\n| 4 | 80             | 32        | 56  | 24  | -637  |\n| 5 | 160            | 16        | 88  | 72  | 3971  |\n| 6 | 320            | 8         | 164 | 156 | 23123 |\n| 7 | 640            | 4         | 322 | 318 | 99911 |\n| 8 | 1280           | 2         | 641 | 639 | 407108 |",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 14572,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $f(x) = x^2 + a x + b$ be a quadratic polynomial with real coefficients such that $f(x) = 0$ has no nonnegative real solutions. Prove that there exist polynomials $g(x)$ and $h(x)$ with nonnegative real coefficients such that $f(x) = \\frac{g(x)}{h(x)}$ for all $x \\ge 0$.",
    "options": [],
    "answer": "See solution",
    "solution": "First Solution. Since $f(x) = 0$ has no nonnegative real solutions, $b = f(0) > 0$. If $a \\ge 0$, then take $g(x) = f(x)$, $h(x) = 1$. Suppose that $a < 0$. Since $f(x) = 0$ has no nonnegative roots, we must have $a^2 < 4b$. Replacing $f(x)$ by $b^{-1}f(x\\sqrt{b})$, we may assume that $b = 1$.\n\nPut $c_1 = -a$ and $f_1(x) = f(x) = x^2 - c_1x + 1$. Note that $c_1^2 < 4$. Put $f_2(x) = f_1(x)(x^2 + c_1x + 1) = x^4 - c_2x^2 + 1$, where $c_2 = c_1^2 - 2$. If $c_2 \\le 0$, take $g(x) = f_2(x)$, $h(x) = g(x)/f(x)$ to satisfy the conclusion. If $c_2 > 0$, then put\n\n$$\nf_3(x) = f_2(x)(x^4 + c_2x^2 + 1) = x^8 - c_3x^4 + 1,\n$$\n\nso $c_3 = c_2^2 - 2$. Proceeding in this way, suppose that $f_k(x) = x^{2k} - c_k x^{2k-1} + 1$, where $c_k = c_{k-1}^2 - 2$, and that $c_k \\le 0$, then take $g(x) = f_k(x)$, $h(x) = g(x)/f(x)$ to satisfy the conclusion.\n\nIf $c_k > 0$, then put\n\n$$\nf_{k+1}(x) = f_k(x)(x^{2k} + c_k x^{2k-1} + 1) = x^{2k+1} - c_{k+1}x^{2k} + 1.\n$$\n\nSuppose that $c_t > 0$ for all positive integers $t$.\n\nThen, for a positive integer $n$, $c_n^2 > 2$, $c_{n-1}^2 > 2+\\sqrt{2}$, $c_{n-2}^2 > 2+\\sqrt{2+\\sqrt{2}}$, $c_{n-3}^2 > 2+\\sqrt{2+\\sqrt{2+\\sqrt{2}}}$, ..., and in general $c_{n-s}^2 > 2+\\sqrt{2+\\sqrt{2+\\sqrt{2}+\\dots+\\sqrt{2}}}$ (with $s$ terms 2 under the radical signs). By the AGM inequality, $2+\\sqrt{2} > 2\\sqrt{2\\sqrt{2}} = 2^{1+\\frac{1}{2}+\\frac{1}{4}}$, and, in general,\n\n$$\n2 + \\sqrt{2 + \\sqrt{2 + \\sqrt{2 + \\dots + \\sqrt{2}}}} > 2^{1+\\frac{1}{2}+\\frac{1}{4}} + \\dots + \\frac{1}{2^{h+1}},\n$$\n\nwhere there are $h$ terms 2 under the radical sign. Hence $c_1^2 > 2^{1+\\frac{1}{2}+\\frac{1}{4}} + \\dots + \\frac{1}{2^{m+1}}$, for all positive integers $m$.\n\nSince the sum of the geometric progression\n\n$$\n1 + \\frac{1}{2} + \\frac{1}{2^2} + \\dots + \\frac{1}{2^n} + \\dots\n$$\n\nis 2, while $c_1^2 < 4$, this is a contradiction.\n\nHence $c_t \\le 0$ for some positive integer $t$, and the proof is complete.\n\nSecond Solution.\n\nThis is based on the following observation: if\n\n$$\np(x, \\theta) = x^2 - 2 \\cos \\theta x + 1,\n$$\n\nthen\n\n$$\n\\begin{aligned}\np(x, \\theta)p(x, \\theta + \\pi) &= (x^2 - 2 \\cos \\theta x + 1)(x^2 + 2 \\cos \\theta x + 1) \\\\\n&= x^4 - 2(2 \\cos^2 \\theta - 1)x^2 + 1 \\\\\n&= x^4 - 2 \\cos 2\\theta x^2 + 1 \\\\\n&= p(x^2, 2\\theta).\n\\end{aligned}\n$$\n\nTo continue, as in the First Solution, it suffices to consider the case $f(x) = x^2 - cx + 1$, where $0 < c < 2$. Select $\\theta \\in (0, \\pi/2)$, so that $c = 2 \\cos \\theta$. Then $f(x) = p(x, \\theta)$ and so\n\n$$\nf(x)p(x, \\theta + \\pi) = p(x^2, 2\\theta) = x^4 - 2 \\cos 2\\theta x^2 + 1.\n$$\n\nIf $\\cos 2\\theta \\le 0$, we stop and take $g(x) = p(x^2, 2\\theta)$ and $h(x) = p(x, \\theta + \\pi)$. Otherwise, we repeat the process and use the fact that\n\n$$\np(x^2, 2\\theta)p(x^2, 2\\theta + \\pi) = p(x^4, 4\\theta).\n$$\n\nWe stop if $\\cos 4\\theta \\le 0$, and put\n\n$$\ng(x) = p(x^4, 4\\theta), \\quad h(x) = p(x, \\theta + \\pi)p(x^2, 2\\theta + \\pi).\n$$\n\nOtherwise we continue. We stop at the $n$th stage where $\\cos 2^n\\theta \\le 0$, and $\\cos 2^{n-1}\\theta > 0$, at which point we let\n\n$$\ng(x) = p(x^{2^n}, 2^n\\theta), \\quad h(x) = p(x, \\theta + \\pi) \\prod_{k=1}^{2^n-1} p(x^{2^k}, 2^k\\theta + \\pi).\n$$\n\nIt's clear that $g, h$ are polynomials whose coefficients are nonnegative, and that $f = g/h$.",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 14573,
    "subject": "Mathematics (Olympiad)",
    "question": "Prove that if $a$, $b$, and $c$ are nonzero and\n\n$$\na(b + c) + b(c + a) + c(a + b) = ab + bc + ca,\n$$\n\nthen\n\n$$\n\\frac{a^2(b+c) + b^2(a+c) + c^2(a+b)}{abc}\n$$\nis an integer.",
    "options": [],
    "answer": "See solution",
    "solution": "The given equation implies $ab + bc + ca = 0$, so we can write\n\n$$\n\\frac{a^2(b+c) + b^2(a+c) + c^2(a+b)}{abc} = \\frac{a(ab+ac) + b(ba+bc) + c(ca+cb)}{abc}\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 14574,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $ABCD$ be a trapezoid with parallel sides $AB$ and $CD$, with $\\angle BAD = 90^\\circ$ and $AB + CD = BC$. Furthermore, let $M$ be the midpoint of $AD$.\n\n*Prove that $\\angle CMB = 90^\\circ$.*",
    "options": [],
    "answer": "See solution",
    "solution": "We reflect the points $B$ and $C$ in $M$ and obtain the points $E$ and $F$, respectively. We clearly have $EC = BF = AB + AF = AB + CD = BC = EF$, therefore, the quadrilateral $BCEF$ is a rhombus. Since the diagonals in a rhombus are orthogonal, we get $BE \\perp CF$ and we obtain $\\angle BMC = 90^\\circ$ as desired.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 14575,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all pairs of positive integers $(n, m)$ such that\n$$\nn^2 + n + 1 = (m^2 + m - 3)(m^2 - m + 5).\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "We are given:\n$$\nn^2 + n + 1 = (m^2 + m - 3)(m^2 - m + 5) = m^4 + m^2 + 8m - 15.\n$$\n\nConsider the equation:\n$$\nn^2 + n - (m^4 + m^2 + 8m - 16) = 0\n$$\nas a quadratic in $n$. It has positive integer roots only if the discriminant $D = 4m^4 + 4m^2 + 32m - 63$ is a perfect square. But\n$$\nD = 4m^4 + 4m^2 + 32m - 63 = (2m^2 + 2)^2 - 4(m - 4)^2 - 3 < (2m^2 + 2)^2\n$$\nfor any natural number $m$, and\n$$\nD = 4m^4 + 4m^2 + 32m - 63 = (2m^2 + 1)^2 + 32(m - 2) > (2m^2 + 1)^2\n$$\nfor any $m > 2$. Therefore, the equation has natural roots only if $m = 1$ or $m = 2$.\n\nIf $m = 1$, then $n^2 + n + 6 = 0$, so $n = -2$ or $n = -3$.\n\nIf $m = 2$, then $n^2 + n - 20 = 0$, so $n = -5$ or $n = 4$.\n\nThus, $(n, m) = (4, 2)$ is the unique pair of positive integers satisfying the condition.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 14576,
    "subject": "Mathematics (Olympiad)",
    "question": "The numbers $1, 2, 3, 4, \\ldots, 39$ are written on a blackboard. In one step, we are allowed to choose two numbers $a$ and $b$ on the blackboard such that $a$ divides $b$, and replace $a$ and $b$ by the single number $\\frac{b}{a}$. This process is continued until no number on the board divides any other number. Let $S$ be the set of numbers which is left on the board at the end. What is the smallest possible value of $|S|$?",
    "options": [],
    "answer": "See solution",
    "solution": "Let\n\n$$\nS = 39! = 2^{35} \\cdot 3^{18} \\cdot 5^{8} \\cdot 7^{5} \\cdot 11^{3} \\cdot 13^{3} \\cdot 17^{2} \\cdot 19^{2} \\cdot 23 \\cdot 29 \\cdot 31 \\cdot 37.\n$$\n\nSuppose we take numbers $x, y$ from the set such that $x \\mid y$. Let $y = kx$. The new set has the product\n\n$$\n\\frac{S \\cdot k}{x \\cdot y} = \\frac{S \\cdot k}{x \\cdot kx} = \\frac{S}{x^2}.\n$$\n\nThis shows that at each step, the parity of the exponents in the prime factorization of the product is preserved. Hence, at the end, the prime factorization of the last product will be\n\n$$\n2 \\cdot 7 \\cdot 11 \\cdot 13 \\cdot 23 \\cdot 29 \\cdot 31 \\cdot 37.\n$$\n\nHowever, $2 \\cdot 11 = 22$ is a number in the given set. Hence, by judicious choice of $a$ and $b$, we can retain $22$. Thus, the least number of numbers that remains is $7$. We show how to reach the set $\\{7, 22, 13, 23, 29, 31, 37\\}$. Consider the following distribution of the numbers $1, 2, 3, \\ldots, 39$:\n\n(39, 3), (38, 19), 37, (36, 18), (35, 7), (34, 17), (33, 11), (32, 16), 31, (30, 15),\n29, (28, 14), (27, 9), (26, 13), (25, 5), (24, 12), 23, 22, 21, (20, 10), (8, 4), (6, 2), 1.\n\nHere, we have paired numbers such that the second number divides the first and retained all the numbers which could not be paired. Replacing the pair with the respective quotient, we obtain:\n\n13, 2, 37, 2, 5, 2, 3, 2, 31, 2, 29, 2, 3, 2, 5, 2, 23, 22, 21, 2, 2, 3, 1.\n\nAgain, we pair appropriately:\n\n(2, 2), (2, 2), (2, 2), (2, 2), (2, 2), (3, 3), (21, 3), (5, 5), 13, 37, 31, 29, 23, 22.\n\nThis leads to:\n\n1, 1, 1, 1, 1, 7, 1, 13, 37, 31, 29, 23, 22.\n\nNow, we can pair seven $1$'s with the remaining $7$ numbers and get the final set\n\n$\\{7, 13, 22, 23, 29, 31, 37\\}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 14577,
    "subject": "Mathematics (Olympiad)",
    "question": "In a table with $n$ rows and $2n$ columns, where $n$ is a fixed positive integer, we write either zero or one into each cell so that each row has $n$ zeros and $n$ ones. For $1 \\leq k \\leq n$ and $1 \\leq i \\leq n$, define $a_{k,i}$ so that the $i$th zero in the $k$th row is in the $a_{k,i}$th column. Let $\\mathcal{F}$ be the set of such tables with $a_{1,i} \\geq a_{2,i} \\geq \\dots \\geq a_{n,i}$ for every $i$ with $1 \\leq i \\leq n$. We associate another $n \\times 2n$ table $f(C)$ from $C \\in \\mathcal{F}$ as follows: for the $k$th row of $f(C)$, we write $n$ ones in the columns $a_{n,k} - k + 1, a_{n-1,k} - k + 2, \\dots, a_{1,k} - k + n$ (and we write zeros in the other cells in the row).\n\n(a) Show that $f(C) \\in \\mathcal{F}$.\n\n(b) Show that $f(f(f(f(f(C))))) = C$ for any $C \\in \\mathcal{F}$.",
    "options": [],
    "answer": "See solution",
    "solution": "We first give a bijection between tables $C \\in \\mathcal{F}$ and partitions of a fixed regular hexagon of side length $n$ into parallelograms given by two unit equilateral triangles glued together. Call such a partition a *well-partitioned* hexagon. For a well-partitioned hexagon, align one of its edges parallel to the $y$ axis so that the hexagon lies to the right side of this edge and divide it into $n$ unit edges.\n\nConsider the first unit edge on the top, which is an edge of a unit parallelogram. Connect the midpoint $M_0$ of this edge to the midpoint $M_1$ of the opposite edge of this parallelogram. We write 1 in the $(1,1)$ cell of an empty $n \\times 2n$ table $C$ if $M_0 M_1$ has positive slope, and 0 otherwise. Similarly, we take $M_2$ to be the next midpoint and write 1 or 0 in the $(1,2)$ cell if $M_1 M_2$ has positive slope or not, respectively. Iterate this step $2n$ times to fill the first row of $C$ with 0's and 1's. We do the same thing for the second unit edge on the left edge of $H$ to fill the second row of $C$, and so on. The result is an $n \\times 2n$ table whose cells are filled with 0 or 1. An example of this correspondence is below.\n\n$$\nC = \\begin{bmatrix} 1 & 1 & 0 & 0 & 1 & 0 \\\\ 1 & 0 & 1 & 0 & 0 & 1 \\\\ 1 & 0 & 0 & 0 & 1 & 1 \\end{bmatrix} \\longleftrightarrow \\text{Hexagon}\n$$\n\nSince the height of the $k$th unit edge on the left edge of $H$ is the same as the $k$th one on the right edge of $H$, the number of 0's and 1's in the $k$th row is the same, namely $n$. If $a_{k,i} < a_{k+1,i}$ for some $k, i$, let $i_0$ be the minimum of such $i$'s. Then among the $1, 2, \\dots, a_{k,i_0} - 1$st columns, the number of 0's in the $k$th and $k+1$st row are the same. Hence, the $a_{k,i_0} - 1$st edge of the $k$th row is adjacent to the $a_{k,i_0} - 1$st edge of the $k$th and $k+1$th row. But then the next parallelograms of the $k$th and $k+1$th row overlap, a contradiction. Hence, we have that $C \\in \\mathcal{F}$. Similarly, one can check that for any $C \\in \\mathcal{F}$, one can find a corresponding well-partitioned hexagon $H$.\n\nNow, we claim that the well-partitioned hexagon associated to $f(C)$ is obtained by rotating the well-partitioned hexagon of $C$ by 60 degrees clockwise. Starting with the $k$th unit edge in the left bottom edge of $H$, we perform a procedure similar to that described above, but assign 1 if the next edge is in the upper right of the edge and 0 otherwise. We define the 1st strip of the first kind to be the set of parallelograms that the broken line $M_0M_1 \\cdots M_{2n}$ passes through, where $M_0, \\cdots, M_{2n}$ are defined above. Define the $k$th strip of the first kind and the $k$th strip of the second time to be the similar sets starting from the midpoint of the $k$th unit edge on the left edge and the bottom left edge, respectively. Since the $i$th 1 in the $k$th strip of the second kind corresponds to the $k$th 0 in the $n+1-i$th strip of the first kind, the newly obtained table is exactly the same as $f(C)$. On the other hand, our procedure is the same as the procedure of our bijection applied to the rotated hexagon, so our claim is proved.\n\nWe are now ready to address both parts of the problem.\n\n(a) A rotation of a well-partitioned hexagon is still well-partitioned, so $f(C)$ corresponds to a well-partitioned hexagon, hence lies in $\\mathcal{F}$.\n\n(b) The well-partitioned hexagon associated to $f^6(C)$ is simply a rotation of the hexagon associated to $C$ by 360 degrees, hence is the same as the hexagon associated to $C$. Therefore, by our bijection, we find that $C = f^6(C)$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 14578,
    "subject": "Mathematics (Olympiad)",
    "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$$\nfind the maximum value of $xyz$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $u = \\sqrt[6]{xyz}$. By the given condition and the mean inequality,\n$$\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$$\nand thus $u \\le \\frac{\\sqrt{3}}{2}$. Therefore, $xyz \\le \\frac{27}{64}$, and equality holds when $x = y = z = \\frac{3}{4}$. Hence, the maximum is $\\frac{27}{64}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 14579,
    "subject": "Mathematics (Olympiad)",
    "question": "Choose positive integers $b_1, b_2, \\dots$ satisfying\n$$\n1 = \\frac{b_1}{1^2} > \\frac{b_2}{2^2} > \\frac{b_3}{3^2} > \\frac{b_4}{4^2} > \\dots\n$$\nand let $r$ denote the largest real number satisfying $\\frac{b_n}{n^2} \\ge r$ for all positive integers $n$. What are the possible values of $r$ across all possible choices of the sequence $(b_n)$?",
    "options": [],
    "answer": "See solution",
    "solution": "The answer is $0 \\le r \\le \\frac{1}{2}$. Obviously $r \\ge 0$.\n\n**Claim (Greedy bound):** For all integers $n$, we have\n$$\n\\frac{b_n}{n^2} \\le \\frac{1}{2} + \\frac{1}{2n}.\n$$\n\n*Proof.* This is by induction on $n$. For $n=1$ it is given. For the inductive step we have\n$$\n\\begin{align*}\nb_n &< n^2 \\frac{b_{n-1}}{(n-1)^2} \\\\\n&\\le n^2 \\left( \\frac{1}{2} + \\frac{1}{2(n-1)} \\right) = \\frac{n^3}{2(n-1)} \\\\\n&= \\frac{1}{2} \\left[ n^2 + n + 1 + \\frac{1}{n-1} \\right] \\\\\n&= \\frac{n(n+1)}{2} + \\frac{1}{2} \\left[ 1 + \\frac{1}{n-1} \\right] \\\\\n&\\le \\frac{n(n+1)}{2} + 1\n\\end{align*}\n$$\nSo $b_n < \\frac{n(n+1)}{2} + 1$ and since $b_n$ is an integer, $b_n \\le \\frac{n(n+1)}{2}$. This implies the result. $\\square$\n\nWe now give a construction. For $r = \\frac{1}{2}$ we take $b_n = \\frac{1}{2}n(n+1)$; for $r = 0$ we take $b_n = 1$.\n\n**Claim (Explicit construction, given by Nikolai Beluhov):** Fix $0 < r < \\frac{1}{2}$. Let $N$ be large enough that $\\lceil rn^2 + n \\rceil < \\frac{1}{2}n(n+1)$ for all $n \\ge N$. Then the following sequence works:\n$$\nb_n = \\begin{cases} \\lceil rn^2 + n \\rceil & n \\ge N \\\\ \\frac{n^2+n}{2} & n < N. \\end{cases}\n$$\n*Proof.* We certainly have\n$$\n\\frac{b_n}{n^2} = \\frac{rn^2 + n + O(1)}{n^2} \\xrightarrow{n \\to \\infty} r.\n$$\nMainly, we contend $b_n n^{-2}$ is strictly decreasing. We need only check this for $n \\ge N$; in fact\n$$\n\\frac{b_n}{n^2} \\ge \\frac{rn^2 + n}{n^2} > \\frac{[r(n+1)^2 + (n+1)] + 1}{(n+1)^2} > \\frac{b_{n+1}}{(n+1)^2}\n$$\nwhere the middle inequality is true since it rearranges to $\\frac{1}{n} > \\frac{n+2}{(n+1)^2}$. $\\square$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 14580,
    "subject": "Mathematics (Olympiad)",
    "question": "For integers $0 \\leq a \\leq n$, let $f(n, a)$ be the number of coefficients in the expansion of $(x+1)^a (x+2)^{n-a}$ that are divisible by $3$. For example, $(x+1)^3(x+2)^1 = x^4 + 5x^3 + 9x^2 + 7x + 2$, so $f(4, 3)$.\n\nFor a positive integer $n$, define $F(n)$ to be the minimum of $f(n, 0), f(n, 1), \\dots, f(n, n)$.\n\nProve:\n\n1. There exist infinitely many positive integers $n$ such that $F(n) \\geq \\frac{n-1}{3}$.\n2. For any positive integer $n$, there is $f(n) \\leq \\frac{n-1}{3}$.",
    "options": [],
    "answer": "See solution",
    "solution": "**Method 1 (Fu Yunhao)**\n\nWe prove that for $n = 2(3^k - 1)$, $k \\in \\mathbb{N}_+$, there is\n\n$$\nF(n) \\geq \\frac{n-1}{3}.\n$$\n\nWe use mathematical induction. When $k=1$, $n=4$, and by enumeration $F(n) = \\frac{n-1}{3}$. Assume the conclusion holds for $n = 2(3^k - 1)$, and consider $n = 2(3^{k+1} - 1)$. (All congruences are modulo $3$.)\n\nFor $0 \\leq a \\leq 2(3^{k+1} - 1)$:\n\n- If $a \\equiv 0$, let $a = 3b$:\n  $$\n  (x+1)^a (x+2)^{n-a} = (x+2)(x+1)^{3b}(x+2)^{3(2\\cdot3^k-1-b)} \n  \\equiv (x+2)(x^3+1)^b(x^3+2)^{2\\cdot3^k-1-b}\n  $$\n  After expansion, there is no term of the form $x^{3t+2}$. Thus, the coefficients of $x^2, x^5, \\dots, x^{2\\cdot3^{k+1}-4}$ are all multiples of $3$, so\n  $$\n  f(n, a) \\geq \\frac{2 \\cdot 3^{k+1} - 4 - 2}{3} + 1 = 2 \\cdot 3^k - 1 = \\frac{n-1}{3}.\n  $$\n\n- If $a \\equiv 1$, let $a = 3b + 1$:\n  $$\n  (x+1)^a (x+2)^{n-a} = (x+1)(x+1)^{3b}(x+2)^{3(2\\cdot3^k-1-b)} \n  \\equiv (x+1)(x^3+1)^b(x^3+2)^{2\\cdot3^k-1-b}\n  $$\n  Similarly, $f(n, a) \\geq \\frac{n-1}{3}$.\n\n- If $a \\equiv 2$, let $a = 3b + 2$:\n  $$\n  (x+1)^a (x+2)^{n-a} = (x+1)^2 (x+2)^2 (x+1)^{3b} (x+2)^{3(2\\cdot3^k-2-b)} \n  \\equiv (x^4 + x^2 + 1)(x^3 + 1)^b (x^3 + 2)^{2\\cdot3^k-2-b}\n  $$\n  By the induction hypothesis, $f(2 \\cdot 3^k - 2, b) \\geq \\frac{2 \\cdot 3^k - 2 - 1}{3} = 2 \\cdot 3^{k-1} - 1$, so the expansion of $(x^3 + 1)^b (x^3 + 2)^{2 \\cdot 3^k - 2 - b}$ has at most $(2 \\cdot 3^k - 2) + 1 - (2 \\cdot 3^{k-1} - 1) = 4 \\cdot 3^{k-1}$ terms whose coefficients are not multiples of $3$. Hence, the expansion of $(x^4 + x^2 + 1)(x^3 + 1)^b (x^3 + 2)^{2 \\cdot 3^k - 2 - b}$ has at most $4 \\cdot 3^k$ terms whose coefficients are not multiples of $3$, so\n  $$\n  f(n, a) \\geq n + 1 - 4 \\cdot 3^k = \\frac{n-1}{3}.\n  $$\n\nTherefore, $F(n) \\geq \\frac{n-1}{3}$. The proof is complete.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 14581,
    "subject": "Mathematics (Olympiad)",
    "question": "我們稱一個正整數為傑出數，若其等於 $1, 2, \\ldots, n$ 的最小公倍數，其中 $n$ 為正整數。找出所有符合 $x + y = z$ 的傑出數 $x, y, z$。",
    "options": [],
    "answer": "See solution",
    "solution": "所有可能的 $(x, y, z)$ 為 $x = y = [1, 2, \\ldots, 2^k - 1]$ 且 $z = [1, 2, \\ldots, 2^k]$，其中 $k$ 為任意正整數，$[\\ldots]$ 表示最小公倍數。\n\n顯然 $z > x, y$。假設 $x = [1, \\ldots, a]$，$y = [1, \\ldots, b]$ 且 $z = [1, \\ldots, c]$，並不失一般性假設 $a \\leq b$。進一步地，不失一般性讓 $a$ 最大（也就是對於任何 $a' > a$ 有 $x \\neq [1, \\ldots, a']$），$b$ 最大且 $c$ 最小。這必然保證 $c > b$ 且 $y \\mid z - y = x$，故 $x = y$ 且 $z = 2x$。而由最大最小性，這代表 $a = b = c - 1$，同時 $2^{v_2(c)} \\nmid y$ 但 $2^{v_2(c)-1} \\mid y$。這只有可能在 $c$ 是 2 的某個冪次時成立（否則 $2^{v_2(c)} < c$ 會整除 $y$）。故解必型如 $x = y = [1, 2, \\ldots, 2^k - 1]$ 且 $z = [1, 2, \\ldots, 2^k]$，其中 $k$ 為任意正整數。代回檢驗顯然成立。證畢。",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 14582,
    "subject": "Mathematics (Olympiad)",
    "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:\n\n- A claims to have met B, C, F, G in the room;\n- B claims to have met A, C, D, E, F;\n- C claims to have met A, B, E;\n- E claims to have met B, C, F;\n- F claims to have met A, B, D, E;\n- G claims to have met A, D;\n- D claims to have met B, F, G.\n\nInspector Goodenough concluded that one of the servants was lying. Who is he?\n\n![](Lyuben_Lichev.png)",
    "options": [],
    "answer": "See solution",
    "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 loss of generality, 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. $\\square$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 14583,
    "subject": "Mathematics (Olympiad)",
    "question": "Azar, Carl, Jon, and Sergey are the four players left in a singles tennis tournament. They are randomly assigned opponents in the semifinal matches, and the winners of those matches play each other in the final match to determine the winner of the tournament. When Azar plays Carl, Azar will win the match with probability $\\frac{2}{3}$. When either Azar or Carl plays either Jon or Sergey, Azar or Carl will win the match with probability $\\frac{3}{4}$. Assume that outcomes of different matches are independent. The probability that Carl will win the tournament is $\\frac{p}{q}$, where $p$ and $q$ are relatively prime positive integers. Find $p+q$.",
    "options": [],
    "answer": "See solution",
    "solution": "There are two cases, depending on whether Azar and Carl meet in the semifinals. If they do, which occurs with probability $\\frac{1}{3}$, Carl will win the tournament if and only if he beats Azar and goes on to beat the winner of the other semifinal match, which occurs with probability $\\frac{1}{3} \\cdot \\frac{3}{4} = \\frac{1}{4}$. If they do not, which occurs with probability $\\frac{2}{3}$, Carl must beat Jon or Sergey in the semifinal match, which occurs with probability $\\frac{3}{4}$, and go on to win the final match. If Azar wins her semifinal match, which occurs with probability $\\frac{3}{4}$, Carl must beat Azar, which occurs with probability $\\frac{1}{3}$. If Azar loses her semifinal match, which occurs with probability $\\frac{1}{4}$, Carl must beat Azar's opponent, which occurs with probability $\\frac{3}{4}$. Thus Carl will win the tournament with probability\n\n$$\n\\frac{1}{3} \\cdot \\frac{1}{4} + \\frac{2}{3} \\cdot \\frac{3}{4} \\cdot \\left( \\frac{3}{4} \\cdot \\frac{1}{3} + \\frac{1}{4} \\cdot \\frac{3}{4} \\right) = \\frac{29}{96}.\n$$\n\nThe requested sum is $29 + 96 = 125$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 14584,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $ABC$ be an acute, non-isosceles triangle with altitudes $AD$, $BE$, and $CF$. The circle with diameter $AD$ intersects $DE$ and $DF$ at $M$ and $N$. Take $P$ and $Q$ on $AB$ and $AC$ respectively so that $NP$ is perpendicular to $AB$ and $MQ$ is perpendicular to $AC$. Let $(I)$ be the circumcircle of triangle $APQ$.\n\na) Prove that $(I)$ is tangent to $EF$.\n\nb) Let $T$ be the tangency point of the circumcircle of triangle $APQ$ with $EF$, $K$ be the intersection of $DT$ and $MN$, and $L$ be the reflection of $A$ through $MN$. Prove that the circumcircle of triangle $DKL$ passes through the intersection of $MN$ and $EF$.\n\n![](images/Vietnamese_mathematical_competitions_p247_data_680c349261.png)",
    "options": [],
    "answer": "See solution",
    "solution": "a) Let $T$ be the foot of $A$ on $EF$. Note that $FC$ is the internal bisector of $\\angle DFE$, so $FM$ and $FT$ are symmetric with respect to $AB$. On the other hand, $\\angle FNA = \\angle FTA = 90^\\circ$, so $N$ and $T$ are symmetric with respect to $AB$. Therefore, $N$, $P$, $T$ are collinear and $TP \\perp AB$. Similarly, $TQ \\perp AC$ and $T$, $M$, $Q$ are collinear. Hence, $AT$ is the diameter of $(I)$ and $AT \\perp EF$, so $EF$ is tangent to $(I)$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 14585,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $E$ be the foot of the perpendicular from $Y$ to $AX$, and let $P$ be the intersection point of $YE$ and $BD$. Prove that $XP \\perp AY$, or equivalently, $XP \\parallel CD$.\n\n![](images/2016_p19_data_11a8c2472f.png)",
    "options": [],
    "answer": "See solution",
    "solution": "Note that\n$$\nYE \\parallel CB \\Rightarrow \\frac{DY}{YC} = \\frac{DP}{PB}. \\quad (1)\n$$\nOn the other hand, triangles $ADC$ and $ACB$ are similar, so\n$$\n\\frac{DY}{YC} = \\frac{CX}{XB}. \\quad (2)\n$$\nFrom (1) and (2), we have $\\frac{DP}{PB} = \\frac{CX}{XB}$, which implies $XP \\parallel CD$, as desired.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 14586,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $S$ be a set of $n$ different points in the $200 \\times 200$ grid, where each point has integer coordinates $1 \\le x, y \\le 200$. We say that point $A = (x_1, y_1)$ is *stronger* than point $B = (x_2, y_2)$ if $x_1 > x_2$ and $y_1 > y_2$.\n\nWhat is the largest integer $n$ such that every set $S$ of $n$ points contains three distinct points $A, B, C$ such that $A$ is stronger than $B$ and $B$ is stronger than $C$?",
    "options": [],
    "answer": "See solution",
    "solution": "We claim that $n = 200$.\n\n**(a)** For any set $S$ of $200$ different points in the grid, there are three distinct points $A, B, C$ such that $A$ is stronger than $B$ and $B$ is stronger than $C$.\n\nLet $S = \\{(x_1, y_1), (x_2, y_2), \\dots, (x_{200}, y_{200})\\}$, where $1 \\le x_i, y_i \\le 200$ and all points are distinct. By the pigeonhole principle, either there are three points with the same $x$-coordinate, or each $x$ from $1$ to $100$ occurs exactly twice as a first coordinate. Similarly for $y$-coordinates.\n\nIf three points share an $x$-coordinate, say $(x, y_i), (x, y_j), (x, y_k)$ with $y_i < y_j < y_k$, then $(x, y_k)$ is stronger than $(x, y_j)$, which is stronger than $(x, y_i)$.\n\nIf three points share a $y$-coordinate, the argument is symmetric.\n\nIf not, then both (2) and (4) above hold. Then, for example, $(x_\\ell, 100)$ is stronger than $(1, y_j)$, which is stronger than $(1, y_i)$, as constructed in the original argument.\n\n**(b)** There exists a set of $199$ points in the grid with no three distinct points $A, B, C$ such that $A$ is stronger than $B$ and $B$ is stronger than $C$.\n\nLet\n$$\nT_1 = \\{(x, y) : 1 \\le x, y \\le 100 \\text{ and } x + y = 101\\},\n$$\n$$\nT_2 = \\{(x, y) : 2 \\le x, y \\le 100 \\text{ and } x + y = 102\\},\n$$\nand $T = T_1 \\cup T_2$. $T$ has $199$ points, and no three distinct points in $T$ form such a chain.\n\n**(c)** For $n < 199$, a similar set can be constructed by deleting points from $T$. For $n = 1$ or $2$, no set contains three distinct points.\n\nThus, the largest such $n$ is $200$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 14587,
    "subject": "Mathematics (Olympiad)",
    "question": "Prove that the product of every three odd consecutive positive integers can be written as the sum of three consecutive integers.",
    "options": [],
    "answer": "See solution",
    "solution": "Let the three odd consecutive numbers be $2p + 1$, $2p + 3$, and $2p + 5$, where $p$ is a positive integer.\n\nOne of these numbers is divisible by $3$:\n- If $p = 3k$, then $2p + 3 = 6k + 3$ is divisible by $3$.\n- If $p = 3k + 1$, then $2p + 1 = 6k + 3$ is divisible by $3$.\n- If $p = 3k + 2$, then $2p + 5 = 6k + 9$ is divisible by $3$.\n\nIn all cases, the product $P$ is a multiple of $3$, so $P = 3a = (a-1) + a + (a+1)$ for some integer $a$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 14588,
    "subject": "Mathematics (Olympiad)",
    "question": "Троє художників іграшкової фабрики мають розфарбувати 2006 дитячих кубиків. Для розфарбовування однієї грані кубика одному художнику потрібно 5 секунд. Грань має бути повністю розфарбованою тільки одним художником, хоча грані одного кубика можуть розфарбовуватися й різними художниками, але одночасно жодні два з цих художників не можуть розфарбовувати грані одного й того ж самого кубика. За який найменший час художники зможуть впоратися з таким завданням? Відповідь обґрунтуйте.",
    "options": [],
    "answer": "See solution",
    "solution": "Відповідь: 20060 секунд.\n\nЧас, необхідний для розфарбовування, не може бути меншим за\n$$\n\\frac{2006 \\cdot 6 \\cdot 5}{3} = 20060\n$$\nсекунд (зрозуміло, що час розфарбовування буде найменшим, якщо вдасться так розподілити роботу, щоб кожні 5 секунд фарбувалася максимальна кількість граней).\n\nЦього можна досягти, наприклад, так:\n- Спочатку художники розфарбовують повністю по 667 кубиків кожний. На це витрачається\n$$\n667 \\cdot 6 \\cdot 5 = 20010\n$$\nсекунд.\n- Потім протягом 30 секунд двоє розфарбовують ще по одному кубику кожний, а третій у цей же час розфарбовує по дві грані у трьох кубиків, що залишилися.\n- Насамкінець, кожен художник бере собі по одному з кубиків з нерозфарбованими гранями (таких кубиків саме три, і в кожному чотири нерозфарбовані грані), і протягом 20 секунд роботу буде завершено.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 14589,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $f$ be a function defined on the positive rational numbers such that for all positive rational numbers $x$ and $y$, the following equation holds:\n\n$$\nf(f(x)^2 f(y)^2) = f(x)^2 f(y)\n$$\n\nFind all such functions $f$.",
    "options": [],
    "answer": "See solution",
    "solution": "Plugging $(x, y) \\leftarrow (f(x), y)$ into the given equation, it follows that\n\n$$\nf(f(x)^2 f(y)^2) = f(f(x))^2 f(y)\n$$\n\nholds for all positive rational numbers $x$ and $y$, hence $f(f(y))^2 f(x) = f(f(x)^2 f(y)^2) = f(f(x))^2 f(y)$, i.e.\n\n$$\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$$\n\nwhere $f^k$ denotes the $k$th iterate of $f$.\n\nBy mathematical induction, we can show that\n\n$$\n\\frac{f(x)}{f(y)} = \\left( \\frac{f^n(x)}{f^n(y)} \\right)^{2^{n-1}}\n$$\n\nholds, meaning that $f(x)/f(y)$ is the $(2^n)$th power of a rational number for all positive integers $n$. Hence, $f(x) = f(y)$ holds for all positive rational numbers $x$ and $y$, i.e. $f(x)$ is a constant.\n\nFinally, from the given equation we get that $f(x) = 1$ for all positive rational numbers $x$ is the only solution.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 14590,
    "subject": "Mathematics (Olympiad)",
    "question": "The sequence $a_1, a_2, a_3, \\dots$ is defined by $a_1 = 4$, $a_2 = 7$, and\n\n$$\na_{n+1} = 2a_n - a_{n-1} + 2, \\quad \\text{for } n \\ge 2.\n$$\n\nShow that for every $m \\ge 1$, the product $a_m a_{m+1}$ is also a term of the sequence.",
    "options": [],
    "answer": "See solution",
    "solution": "First, we prove the following formula by induction:\n\n$$\na_n = n^2 + 3, \\quad \\text{for } n \\ge 1.\n$$\n\nWe require two base cases to get started. The formula is true for $n = 1$ and $n = 2$ because $a_1 = 4 = 1^2 + 3$ and $a_2 = 7 = 2^2 + 3$.\n\nFor the inductive step, assume that the formula is true for $n = k - 1$ and $n = k$. Then for $n = k + 1$, we have\n\n$$\n\\begin{aligned}\na_{k+1} &= 2a_k - a_{k-1} + 2 && \\text{(given)} \\\\\n&= 2(k^2 + 3) - ((k-1)^2 + 3) + 2 && \\text{(inductive assumption)} \\\\\n&= k^2 + 2k + 4 \\\\\n&= (k+1)^2 + 3.\n\\end{aligned}\n$$\n\nHence the formula is also true for $n = k + 1$. This completes the induction.\n\nUsing the formula, we calculate\n\n$$\n\\begin{aligned}\na_m a_{m+1} &= (m^2 + 3)((m+1)^2 + 3) \\\\\n&= (m^2 + 3)(m^2 + 2m + 4) \\\\\n&= m^4 + 2m^3 + 7m^2 + 6m + 12 \\\\\n&= (m^2 + m + 3)^2 + 3 \\\\\n&= a_{m^2+m+3}.\n\\end{aligned}\n$$\n\nHence $a_m a_{m+1}$ is a term of the sequence. $\\square$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 14591,
    "subject": "Mathematics (Olympiad)",
    "question": "已知圓 $O_1, O_2$ 交於 $M, N$ 兩點。靠近 $M$ 的公切線分別與 $O_1, O_2$ 切於點 $A, B$。點 $C, D$ 分別為 $A, B$ 關於 $M$ 的對稱點，$\\triangle DCM$ 的外接圓與 $O_1, O_2$ 分別交於不同於 $M$ 的點 $E, F$。證明 $\\triangle MEF$ 和 $\\triangle NEF$ 的外接圓半徑相等。",
    "options": [],
    "answer": "See solution",
    "solution": "取 $N'$ 使得四邊形 $NEN'F'$ 為平行四邊形，延長 $AD$ 交圓 $O_1$ 於 $E'$，延長 $BC$ 交圓 $O_2$ 於 $F'$。\n\n由 $M, C, F, E, D$ 五點共圓得 $\\angle MFC = \\angle MDC = \\angle MBA = \\angle MFB$，故 $B, C, F$ 三點共線。即 $F = F'$。同理 $E = E'$。\n\n連結 $NM$ 並延長交 $AB$ 於點 $L$，因為 $LA^2 = LM \\cdot LN$, $LB^2 = LM \\cdot LN$，故 $LA = LB$。\n\n又 $MB = ND$，則 $LM \\parallel AD$，即 $MN \\parallel AE$，同理 $MN \\parallel BF$。於是 $O_1O_2 \\perp AE$, $O_1O_2 \\perp BF$，因此 $A, M, B$ 分別與 $E, N, F$ 關於 $O_1O_2$ 對稱。\n\n從而 $\\angle ENF = \\angle AMB$，故\n\n$$\n\\begin{aligned}\n\\angle EN'F + \\angle EMF &= \\angle ENF + \\angle EMF \\\\\n&= \\angle EMF + \\angle AMB \\\\\n&= \\angle EMN + \\angle FMN + \\angle AMB \\\\\n&= \\angle AEM + \\angle BFM + \\angle AMB \\\\\n&= \\angle BAM + \\angle ABM + \\angle AMB \\\\\n&= \\pi\n\\end{aligned}\n$$\n\n故 $M, E, N', F'$ 四點共圓。\n\n又 $ENFN'$ 是平行四邊形，故 $\\triangle ENF$ 和 $\\triangle FNE'$ 全等。故 $\\triangle ENF$ 和 $\\triangle MEF$ 外接圓半徑相等。",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 14592,
    "subject": "Mathematics (Olympiad)",
    "question": "Points $M$, $N$, and $P$ lie on the sides $BC$, $CA$, and $AB$ of triangle $ABC$. Triangles $CNM$, $APN$, and $BMP$ are acute, and let $H_C$, $H_A$, and $H_B$ be their respective orthocenters. Prove that if the three lines $AH_A$, $BH_B$, and $CH_C$ are concurrent, then $MH_A$, $NH_B$, and $PH_C$ are also concurrent.",
    "options": [],
    "answer": "See solution",
    "solution": "Denote by $A_1$, $B_1$, and $C_1$ the projections of $A$, $B$, and $C$ onto $NP$, $PM$, and $MN$, respectively. It follows from the condition of the problem (Carnot's theorem) that $$(NC_1^2 - MC_1^2) + (MB_1^2 - PB_1^2) + (PA_1^2 - NA_1^2) = 0.$$ Hence, $$0 = (CN^2 - CM^2) + (BM^2 - BP^2) + (AP^2 - AN^2) = (CN^2 - AN^2) + (AP^2 - BP^2) + (BM^2 - CM^2),$$ and the same theorem implies that the perpendiculars from $M$, $N$, and $P$ to $BC$, $AC$, and $AB$ are concurrent at a point $O$.\n\nSince $OMH_CN$ and $OMH_BP$ are parallelograms, we conclude that $PH_BH_CN$ is also a parallelogram. Therefore, the segments $PH_C$ and $NH_B$ bisect each other at some point. The segment $MH_A$ passes through the same point.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 14593,
    "subject": "Mathematics (Olympiad)",
    "question": "某國有 $n$ 座城市，任兩座城市間都有唯一的一條道路，且都被規定只能單向行駛。一條從城市 $X$ 到城市 $Y$ 的路徑為一系列的單向道路，使得一個人可以從 $X$ 經由這些道路移動到 $Y$，且中間不會重複拜訪相同的城市。一組路徑被稱為獨立的，若且唯若這其中的任兩條路徑都沒有使用相同的道路。\n\n對於一個城市 $X$，令 $n_X$ 為由 $X$ 向外的單向道路總數。對於兩個相異城市 $X$ 與 $Y$，令 $N_{XY}$ 為在所有的獨立路徑組中，從 $X$ 到 $Y$ 路徑數的最大可能值。證明：$N_{XY} = N_{YX}$ 若且唯若 $n_X = n_Y$。",
    "options": [],
    "answer": "See solution",
    "solution": "令 $X \\to Y$ 表示 $X$ 有單向道路通往 $Y$，$X \\rightsquigarrow Y$ 表示 $X$ 到 $Y$ 的路徑。定義：\n\n$$\n\\mathcal{F}_X := \\{C : X \\to C\\} \\quad \\mathcal{F}_Y := \\{C : Y \\to C\\}\n$$\n\n為所有單向道路來自（通往）$X$（$Y$）的城市所成集合，且 $n_X = |\\mathcal{F}_X|$。\n\n稱一條路徑是短的，若其道路數 $\\leq 2$，否則稱為長路徑。先證明：\n\n**Lemma:** 令 $\\mathcal{P}$ 為由 $X$ 到 $Y$ 的路徑所成的獨立組，$|\\mathcal{P}| = p$。則存在一組由 $X$ 到 $Y$ 的路徑所成的獨立組 $\\mathcal{P}'$，滿足 $|\\mathcal{P}'| \\geq p$ 且包含所有從 $X$ 到 $Y$ 的短路徑。\n\n*證明略。*\n\n回到原題。任取兩城市 $X$ 與 $Y$，不失一般性假設 $X \\to Y$。任取由 $X$ 到 $Y$ 的路徑組成且大小為 $N_{XY}$ 的獨立路徑組 $\\mathcal{P}$。目標是構造一個由 $Y$ 到 $X$ 的獨立路徑組 $\\mathcal{Q}$，使其大小為 $N_{XY} - (n_X - n_Y)$。\n\n由最大性可得：\n\n$$\nN_{YX} \\geq N_{XY} - (n_X - n_Y) \\quad \\text{且} \\quad N_{XY} \\geq N_{YX} - (n_Y - n_X)\n$$\n\n因此 $N_{XY} - N_{YX} = n_X - n_Y$，證畢。\n\n*註：可用 Ford-Fulkerson 最小割最大流定理證明。*",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 14594,
    "subject": "Mathematics (Olympiad)",
    "question": "A rectangle of size $m \\times n$ consisting of $m \\cdot n$ squares of size $1 \\times 1$ is given. Compute the sum of areas of all subrectangles consisting of some of the $m \\cdot n$ squares.\n\n$$\n\\frac{m(m+1)(m+2)n(n+1)(n+2)}{36}\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "**Solution.** For each square $1 \\times 1$ we will count by how many rectangles it is contained. Take a square from the $i$-th column and $j$-th row. It is contained in $i(m - i + 1)j(n - j + 1)$ rectangles, as we may choose one of $i$ lines on the left side of the square to be the left side of a rectangle, one of $(m - i + 1)$ lines on the right to be the right side, and similarly with top and bottom sides. Then the sum of areas is equal to:\n\n$$\n\\begin{aligned}\n\\sum_{i=1}^{m} \\sum_{j=1}^{n} i(m - i + 1)j(n - j + 1) &= \\left(\\sum_{i=1}^{m} i(m - i + 1)\\right) \\left(\\sum_{j=1}^{n} j(n - j + 1)\\right) \\\\\n&= \\frac{m(m + 1)(m + 2)}{6} \\cdot \\frac{n(n + 1)(n + 2)}{6}.\n\\end{aligned}\n$$\n\nThe equality $\\sum_{j=1}^{n} j(n - j + 1) = \\frac{n(n+1)(n+2)}{6}$ is true as both sides are the number of choices of three numbers from the set $\\{1, 2, \\dots, n + 2\\}$. It may also be proved by induction.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 14595,
    "subject": "Mathematics (Olympiad)",
    "question": "Determine the value of $\\frac{1 \\times 13 \\times 13 \\times 12}{1 + 13 + 13 + 12}$.",
    "options": [],
    "answer": "See solution",
    "solution": "The value is $52$.\n\nThe fraction is:\n$$\n\\frac{13 \\times 13 \\times 12}{39} = \\frac{13 \\times 13 \\times 12}{3 \\times 13} = \\frac{13 \\times 12}{3} = 13 \\times 4 = 52\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 14596,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $n$ be a positive integer. Determine for which $n$ the sum $9^{9^n} + 91^{91^n}$ is divisible by $100$.",
    "options": [],
    "answer": "See solution",
    "solution": "We work modulo $100$ throughout.\n\nFirst, observe the cycles:\n\n$$\n9^1 = 9,\\ 9^2 = 81,\\ 9^3 = 29,\\ 9^4 = 61,\\ 9^5 = 49,\\ 9^6 = 41,\\ 9^7 = 69,\\ 9^8 = 21,\\ 9^9 = 89,\\ 9^{10} = 1\n$$\n\nand\n\n$$\n91^1 = 91,\\ 91^2 = 81,\\ 91^3 = 71,\\ 91^4 = 61,\\ 91^5 = 51,\\ 91^6 = 41,\\ 91^7 = 31,\\ 91^8 = 21,\\ 91^9 = 11,\\ 91^{10} = 1.\n$$\n\nThus, both $9^k$ and $91^k$ are periodic modulo $100$ with period $10$.\n\nSo, $9^{9n}$ and $91^{91n}$ depend on $9n \\bmod 10$ and $91n \\bmod 10$ respectively. Since $9n$ and $91n$ are congruent modulo $10$ (as $9 \\equiv 91 \\pmod{10}$), their exponents cycle together.\n\nChecking the values, we find:\n- If $n$ is even, $9^{9n} \\equiv 9$ and $91^{91n} \\equiv 91$ modulo $100$, so $9^{9n} + 91^{91n} \\equiv 100 \\equiv 0 \\pmod{100}$.\n- If $n$ is odd, $9^{9n} + 91^{91n} \\not\\equiv 0 \\pmod{100}$.\n\n**Conclusion:** $9^{9n} + 91^{91n}$ is divisible by $100$ if and only if $n$ is even.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 14597,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $ABCD$ be a trapezoid with $AD \\parallel BC$ and $\\angle ADC = \\angle BAD$, and let $\\ell$ be a line not intersecting the line segments $AC$ or $BD$. Assume that $\\ell$ intersects the lines $AC$, $AD$, $BC$, $BD$, and $CD$ in the points $P$, $Q$, $R$, $S$, and $T$ respectively. Show that the three circles $\\odot(DQT)$, $\\odot(BRQ)$, and $\\odot(BPS)$ intersect in a common point, where $\\odot(XYZ)$ denotes the circumcircle of $XYZ$.\n\n*Remark.* The condition that $\\ell$ does not intersect the line segments $AC$ or $BD$ is not strictly needed, but it reduces casework.",
    "options": [],
    "answer": "See solution",
    "solution": "![](images/bw18shortlist_p34_data_7388467cc2.png)\n\nNote first that $ABCD$ is concyclic since it is an isosceles trapezoid. We define $X$ to be the intersection of $\\odot(DQT)$ and $\\odot(ABCD)$, and it now suffices to prove that $BRQX$ and $BPSX$ are cyclic quadrilaterals. We have that\n\n$$\n\\begin{aligned}\n\\angle BXQ &= \\angle BXD + \\angle DXQ = (\\pi - \\angle BCD) + \\angle DTQ \\\\\n&= \\angle RCD + \\angle CDR = \\angle RCT + \\angle CTR = \\pi - \\angle CRT\n\\end{aligned}\n$$\n\nand hence $BRQX$ is a cyclic quadrilateral. Now observe that\n\n$$\n\\begin{aligned}\n\\angle AXQ &= \\angle AXD + \\angle DXQ = (\\pi - \\angle ACD) + \\angle DTQ \\\\\n&= \\angle PCD + \\angle CTP = \\angle PCT + \\angle CTP \\\\\n&= \\pi - \\angle CPT = \\pi - \\angle APQ\n\\end{aligned}\n$$\n\nso $APQX$ is concyclic, and it now follows that:\n\n$$\n\\angle SBX = \\angle DBX = \\angle DAX = \\angle QAX = \\angle QPX = \\angle SPX\n$$\n\nHence, $BPSX$ is a cyclic quadrilateral.\n\n*Remark 1.* Note that we only use that $ABCD$ is cyclic, and one can in fact replace $ABCD$ an isosceles trapezoid with the weaker condition $ABCD$ cyclic. The stronger condition only serves the purpose as a red herring.\n\n*Remark 2.* One can entirely avoid angles and instead apply the extended Miquel's theorem three times:\n\nLetting $X = \\odot(DQT) \\cap \\odot(ABCD)$, we get that $BRQX$ is cyclic by application of the extended Miquel's theorem on $\\triangle CRT$ and circles $\\odot(CDB)$, $\\odot(TDQ)$, and $\\odot(RQB)$.\n\nApplying the theorem again on $\\triangle CPT$ and circles $\\odot(CDA)$, $\\odot(TDQ)$, and $\\odot(PQA)$ we get that $APQX$ is cyclic.\n\nHaving established that $APQX$ is cyclic, it also follows that $BPSX$ is cyclic by applying the extended Miquel's theorem on $\\triangle DQS$ and the circles $\\odot(DAB)$, $\\odot(QAP)$, and $\\odot(SBQ)$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 14598,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $ABC$ be an acute-angled triangle and $O$ its circumcenter. Let $K_1$ and $K_2$ be the circumcircles of triangles $ABO$ and $ACO$, respectively. Let $P$ be the point on $K_1$ such that $OP$ is a diameter of $K_1$, and $Q$ the point on $K_2$ such that $OQ$ is a diameter of $K_2$. Let $T$ be the intersection of the line through $P$ tangent to $K_1$ and the line through $Q$ tangent to $K_2$. Let $D$ be the intersection of line $AC$ with $K_1$ other than $A$. Prove that the points $D$, $O$, and $T$ are collinear.",
    "options": [],
    "answer": "See solution",
    "solution": "We use directed angles for generality.\n\nSince $OP$ is a diameter of $K_1$, by Thales' theorem, $\\angle PAO = \\frac{\\pi}{2}$. Similarly, since $OQ$ is a diameter of $K_2$, $\\angle OAQ = \\frac{\\pi}{2}$. Thus, $A$, $P$, and $Q$ are collinear.\n\nSince $|OA| = |OC|$ and $OQ$ is a diameter, in $K_2$ the chord $AC$ is perpendicular to $OQ$. Also, $\\angle OPT = \\angle TQO = \\frac{\\pi}{2}$, so $O$, $P$, $T$, and $Q$ are concyclic.\n\nTherefore, $\\angle TOP = \\angle TQP = \\frac{\\pi}{2} - \\angle AQO = \\angle CAQ = \\angle DAP = \\angle DOP$. Thus, $\\angle TOP = \\angle DOP$, so $D$, $O$, and $T$ are collinear.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 14599,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all functions $f : \\mathbb{R}_+ \\to \\mathbb{R}_+$ such that for all $x, y > 0$,\n\n$$\nf(x + f(y)) = f(x) + y.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "Let $1 + f(1) = a$. Set $x = y = 1$ in\n\n$$\nf(x + f(y)) = f(x) + y, \\quad (*)\n$$\n\nthen $f(a) = a$.\n\nFurther, setting $x = f(z)$ in $(*)$, we have $f(f(z) + f(y)) = f(f(z)) + y$. On the other hand, $f(f(y) + f(z)) = f(f(y)) + z$. So,\n\n$$\nf(f(z)) + y = f(f(y)) + z. \\quad (1)\n$$\n\nNext, if $z = a$, then (1) implies $a + y = f(f(y)) + a$ or\n\n$$\nf(f(y)) = y. \\quad (2)\n$$\n\nNow, setting $y = f(z)$ in $(*)$, we have\n\n$$\nf(x + z) = f(x) + f(z) \\quad \\forall x, z \\in \\mathbb{R}_{+}. \\quad (3)\n$$\n\nNow we claim that $f(x) = x$ for all $x \\in \\mathbb{R}_+$. Assume the converse.\nIf $f(t) > t$ for some $t > 0$, then\n\n$$\nt = f(f(t)) = f(f(t) - t + t) = f(f(t) - t) + f(t) > f(t) > t,\n$$\n\na contradiction.\n\nIf $f(t) < t$ for some $t > 0$, then\n\n$$\nf(t) = f(t - f(t) + f(t)) = f(t - f(t)) + f(f(t)) > f(f(t)) = t,\n$$\n\ncontrary to $f(t) < t$.\n\nHence $f(x) = x$ for all $x \\in \\mathbb{R}_+$. This function obviously satisfies the problem condition.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 14600,
    "subject": "Mathematics (Olympiad)",
    "question": "We call four-digit integers *good* if their digits consist of exactly two types of numbers.\n\nWhat is the smallest good integer greater than $2022$ that is divisible by $3$?",
    "options": [],
    "answer": "See solution",
    "solution": "$2112$",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 14601,
    "subject": "Mathematics (Olympiad)",
    "question": "The sides of a triangle are consecutive terms in an arithmetic progression. Prove that the line connecting the centroid and the center of the incircle is parallel to the side of the triangle with middle length.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $\\triangle ABC$ be a triangle with sides $AB = c$, $BC = c + d$, and $AC = c + 2d$. Let $O$ be the center of the incircle and $O_1$ be the centroid. Let $D$ and $E$ be the feet of the perpendiculars from $O$ to $BC$ and from $O_1$ to $BC$, respectively. Let $OD = r$, and let the area of $\\triangle ABC$ be $P = r s$. Since $s = \\frac{c + d + c + c + 2d}{2} = \\frac{3}{2}(c + d)$, we have $r = \\frac{P}{s} = \\frac{2P}{3(c + d)}$. Now, $P_{O_1BC} = \\frac{1}{3} P_{ABC} = O_1E \\cdot \\frac{1}{2}(c + d)$, i.e., $O_1E = \\frac{2P}{3(c + d)} = r$.\n\nQuadrilateral $DOO_1E$ is a parallelogram, so $OO_1 \\parallel DE$, i.e., $OO_1 \\parallel BC$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 14602,
    "subject": "Mathematics (Olympiad)",
    "question": "We call an affine function $f : \\mathbb{R} \\to \\mathbb{R}$ *useful* if it satisfies:\n\n1. $|f(x)| \\le 2$ for every real $x$ with $|x| \\le 2$;\n2. $|f(x)| \\ge 1$ for every real $x$ with $|x| \\le 1$.\n\nLet $x_0$ be a real number. Prove that there exists a useful function $f$ such that $f(x_0) = 0$ if and only if $|x_0| \\ge 4$.",
    "options": [],
    "answer": "See solution",
    "solution": "($\\Rightarrow$) If $f$ is useful, then $-f$ also satisfies properties (i) and (ii), so it is useful. Thus, we may consider only useful functions $f(x) = ax + b$ with $a > 0$, $b \\in \\mathbb{R}$.\n\nSince $a > 0$, property (i) is equivalent to $f(2) \\le 2$ and $f(-2) \\ge -2$, i.e.,\n$$\n2a + b \\le 2 \\quad \\text{(1)}\n$$\n$$\n-2a + b \\ge -2 \\quad \\text{(2)}\n$$\n\nProperty (ii) leads to two cases:\n\n**Case I:** $f(-1) \\le -1$ and $f(1) \\le -1$.\n\nThis gives:\n$$\n-a + b \\le -1 \\quad \\text{(3)}\n$$\n$$\na + b \\le -1 \\quad \\text{(4)}\n$$\nMultiply (4) by 2 and add to (2):\n$$\n2(a + b) + (-2a + b) \\le 2(-1) + (-2)\n$$\n$$\n2a + 2b - 2a + b \\le -2 - 2\n$$\n$$\n3b \\le -4\n$$\nBut more simply, combining (2) and (4) gives $4a + b \\le 0$, so $x_0 = -\\frac{b}{a} \\ge 4$.\n\n**Case II:** $f(-1) \\ge 1$ and $f(1) \\ge 1$.\n\nThis gives:\n$$\na - b \\le -1 \\quad \\text{(5)}\n$$\n$$\n-a - b \\le -1 \\quad \\text{(6)}\n$$\nMultiply (5) by 2 and add to (1):\n$$\n2(a - b) + (2a + b) \\le 2(-1) + 2\n$$\n$$\n2a - 2b + 2a + b \\le -2 + 2\n$$\n$$\n4a - b \\le 0\n$$\nSo $x_0 = -\\frac{b}{a} \\le -4$.\n\nFrom both cases, $|x_0| \\ge 4$.\n\n($\\Leftarrow$) Let $x_0 \\in \\mathbb{R}$ with $|x_0| \\ge 4$.\n\nIf $x_0 \\ge 4$, take $f(x) = \\frac{1}{x_0-1}x + \\frac{x_0}{1-x_0}$.\n\nIf $x_0 \\le -4$, take $f(x) = \\frac{-1}{1+x_0}x + \\frac{x_0}{x_0+1}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 14603,
    "subject": "Mathematics (Olympiad)",
    "question": "Points D and E are on sides $BC$ and $AC$ of $\\triangle ABC$. Lines $AD$ and $BE$ intersect at point $S$. Point $F$ is on side $AB$, and lines $FE$ and $FD$ intersect line $l$ passing through $C$ and parallel to $AB$ at points $P$ and $Q$. Prove that if $CP = CQ$, then the points $C$, $S$, and $F$ lie on the same line.",
    "options": [],
    "answer": "See solution",
    "solution": "From the similarities $\\triangle PEC \\sim \\triangle FEA$ and $\\triangle CDQ \\sim \\triangle BDF$, we obtain that\n\n$$\n\\frac{CP}{AF} = \\frac{CE}{AE} \\quad \\text{and} \\quad \\frac{CQ}{BF} = \\frac{CD}{BD}.\n$$\n\nTherefore,\n$$\n\\frac{CP}{CQ} = \\frac{CE}{AE} \\cdot \\frac{AF}{BF} \\cdot \\frac{BD}{DC}.\n$$\nBy the condition $CP = CQ$, it follows that\n$$\n\\frac{CE}{AE} \\cdot \\frac{AF}{BF} \\cdot \\frac{BD}{DC} = 1.\n$$\n\nApplying Ceva's theorem for $\\triangle ABC$ and the points $F$, $D$, and $E$, it follows that the lines $AD$, $BE$, and $CF$ intersect in one point, i.e., point $S$ lies on $CF$. $\\square$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 14604,
    "subject": "Mathematics (Olympiad)",
    "question": "![Diagram](images/Ireland_2013_Booklet_p37_data_e36c447ab0.png)\n\nLet $|B'B| = |BA| = |AC| = |CC'|$. Prove that\n$$\n\\frac{1}{|AB|} = \\frac{1}{|B'X|} + \\frac{1}{|BC'|}\n$$\nis equivalent to $|AX| = |BX|$.",
    "options": [],
    "answer": "See solution",
    "solution": "Because $|B'B| = |BA| = |AC| = |CC'|$, we have $|B'X| = |AB| + |BX|$ and $|BC'| = |BC| + |AB|$. For the first equation, we used that $B$ is between $B'$ and $X$. To see this, we have to exclude that $X$ is between $B$ and $B'$ (by assumption, $B'$ is not between $X$ and $B$). If $|AX| = |BX|$, $X$ cannot be between $B$ and $B'$, because the angle $\\angle CBA$ is acute. If\n$$\n\\frac{1}{|AB|} = \\frac{1}{|B'X|} + \\frac{1}{|BC'|},\n$$\nwe have $|B'X| > |AB| = |B'B|$, hence $X$ is not between $B$ and $B'$.\n\nWe see now that the equation\n$$\n\\frac{1}{|AB|} = \\frac{1}{|B'X|} + \\frac{1}{|BC'|}\n$$\nis equivalent to\n$$\n\\frac{1}{|AB|} = \\frac{1}{|AB| + |BX|} + \\frac{1}{|BC| + |AB|}\n$$\nwhich rewrites as\n$$\n(|AB| + |BX|)(|BC| + |AB|) = |AB|(|AB| + |BX|) + |AB|(|BC| + |AB|)\n$$\nwhich simplifies to $|BC| \\cdot |BX| = |AB|^2$, or equivalently $\\frac{|AB|}{|BC|} = \\frac{|BX|}{|AB|}$. Because the triangles $ABX$ and $ABC$ have a common angle at $B$, this last equality is equivalent to these two triangles being similar with sides $AB$ and $BX$ in triangle $ABX$ corresponding to sides $BC$ and $AB$ in triangle $ABC$. Because $\\triangle ABC$ is isosceles, the same is then true for the triangle $ABX$, hence $|AX| = |BX|$. On the other hand, if $|AX| = |BX|$, triangle $ABX$ is isosceles and because it shares an angle with $\\triangle ABC$ at $B$, both triangles are similar. This finishes the proof of the desired equivalence.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 14605,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $ABC$ be an acute-angled triangle inscribed in a circle $c(O, R)$, and let $F$ be a point on the side $AB$ such that $AF < \\frac{AB}{2}$. The circle $c_1(F, FA)$ intersects the line $OA$ at point $A'$, and the circle $c$ at $K$. Prove that the quadrilateral $BKFA'$ is inscribed in a circle passing through $O$.",
    "options": [],
    "answer": "See solution",
    "solution": "The triangle $AFK$ is isosceles, so $\\hat{F}_1 = 2\\hat{A}_1$. The angle $\\hat{A}_1$ is inscribed in the circle $c$, and $\\hat{O}_1 = 2\\hat{A}_1 = \\hat{F}_1$, so the quadrilateral $BKFO$ is cyclic.\n\nNext, we prove that the quadrilateral $OBKA'$ is cyclic. Let $S$ be the point diametrically opposite $A$ in the circle $c_1$. Then triangle $AKS$ is right-angled at $K$, so $\\hat{S}_1 = 90^\\circ - \\hat{A}_1$. The angles $\\hat{S}_1$ and $\\hat{A}'_1$ are equal (both inscribed in $c_1$ and subtending the same arc $KA$). Thus,\n\n$$\n\\hat{A}'_1 = 90^\\circ - \\hat{A}_1 \\tag{1}\n$$\n\nFrom the isosceles triangle $OKB$, we have:\n\n$$\n\\hat{B}_1 = 90^\\circ - \\frac{\\hat{O}_1}{2} = 90^\\circ - \\hat{A}_1 \\tag{2}\n$$\n\nFrom (1) and (2), $\\hat{A}'_1 = \\hat{B}_1$. Hence, the quadrilateral $OBKA'$ is cyclic.\n\n![](images/IMO2017_finalbook_Greece_1_p11_data_8ba2740a59.png \"Figure 6\")",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 14606,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $ABC$ be an acute-angled triangle and let $O$ be its circumcenter. The tangents to the circumcircle of $ABC$ at vertices $B$ and $C$ meet at $P$. The circle of radius $PB$ centered at $P$ meets the internal angle bisector of angle $BAC$ at point $Q$ lying in the interior of triangle $ABC$. The lines $OQ$ and $BC$ meet at $D$. Let $E$ and $F$ be the orthogonal projections of $Q$ onto the lines $AC$ and $AB$, respectively. Prove that the lines $AD$, $BE$, and $CF$ are concurrent.",
    "options": [],
    "answer": "See solution",
    "solution": "The line $AB$ and the circle of radius $PB$ centered at $P$ meet again at some point $R$. Standard angle-chasing shows that the angle $BRC$ is the complement of the angle $BAC$. Hence, the lines $AC$ and $CR$ are perpendicular, so the lines $EQ$ and $CR$ are parallel, and the angles $CQE$ and $QCR$ are equal.\n\nSince the points $B$, $Q$, $C$, $R$ are concyclic, the angles $QBF$ and $QCR$ are equal, so the angles $QBF$ and $CQE$ are equal and the right-angled triangles $BQF$ and $QCE$ are similar: $\\frac{BQ}{QC} = \\frac{BF}{QE} = \\frac{QF}{CE}$. Recall that $AQ$ is the internal angle bisector of angle $BAC$, so $EQ = FQ$, whence $\\frac{BQ^2}{CQ^2} = \\frac{BF}{CE}$.\n\n![](images/RMC2014_p63_data_07804bf4f3.png)\n\nOn the other hand, the line $OQD$ is the $Q$-symmedian of triangle $BCQ$, so $\\frac{BQ^2}{CQ^2} = \\frac{BD}{CD}$. Consequently,\n\n$$\n1 = \\frac{BD}{CD} \\cdot \\frac{CE}{BF} = \\frac{BD}{CD} \\cdot \\frac{CE}{AE} \\cdot \\frac{AF}{BF},\n$$\n\nsince $AQ$ is the internal angle bisector of angle $BAC$, so $AE = AF$. The conclusion follows by Ceva's Theorem.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 14607,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $ABC$ be a triangle with integer side lengths such that $\\angle A = 3\\angle B$. Find the minimum value of its perimeter.",
    "options": [],
    "answer": "See solution",
    "solution": "Let the sides be $a, b, c$. From the sine rule, we have\n\n$$\n\\begin{aligned}\n\\frac{a}{b} &= \\frac{\\sin 3B}{\\sin B} = 4\\cos^2 B - 1 \\\\\n\\frac{c}{b} &= \\frac{\\sin C}{\\sin B} = \\frac{\\sin 4B}{\\sin B} = 8\\cos^3 B - 4\\cos B\n\\end{aligned}\n$$\n\nThus,\n\n$$\n2 \\cos B = \\frac{a^2 + c^2 - b^2}{ac} \\in \\mathbb{Q}.\n$$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 14608,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all positive real numbers $a$, $b$, $c$ which satisfy the equality:\n\n$$\nab\\left(1 - \\frac{c^2}{(a+b)^2}\\right) = bc\\left(1 - \\frac{a^2}{(b+c)^2}\\right) = ca\\left(1 - \\frac{b^2}{(c+a)^2}\\right).\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "All triples of the form $(t, t, t)$ with $t > 0$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 14609,
    "subject": "Mathematics (Olympiad)",
    "question": "Given any positive real number $\\varepsilon$, prove that, for all but finitely many positive integers $v$, any graph on $v$ vertices with at least $(1+\\varepsilon)v$ edges has two distinct simple cycles of equal lengths. (Recall that the notion of a *simple cycle* does not allow repetition of vertices in a cycle.)",
    "options": [],
    "answer": "See solution",
    "solution": "Fix a positive real number $\\varepsilon$, and let $G$ be a graph on $v$ vertices with at least $(1+\\varepsilon)v$ edges, all of whose simple cycles have pairwise distinct lengths.\n\nAssuming $\\varepsilon^2 v \\ge 1$, we exhibit an upper bound linear in $v$ and a lower bound quadratic in $v$ for the total number of simple cycles in $G$, showing thereby that $v$ cannot be arbitrarily large, whence the conclusion.\n\nSince a simple cycle in $G$ has at most $v$ vertices, and each length class contains at most one such, $G$ has at most $v$ pairwise distinct simple cycles. This establishes the desired upper bound.",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 14610,
    "subject": "Mathematics (Olympiad)",
    "question": "A triangle $AB\\Gamma$ is given. Let $c(O, R)$ be its circumcircle, and let $\\Delta$ be a point on the side $B\\Gamma$ different from the midpoint of $B\\Gamma$. The circumcircle of triangle $BO\\Delta$, denoted $c_1$, intersects $c(O, R)$ at $K$ and the line $AB$ at $Z$. The circumcircle of triangle $\\Gamma O\\Delta$, denoted $c_2$, intersects $c(O, R)$ at $M$ and the line $A\\Gamma$ at $E$. Finally, the circumcircle of triangle $AEZ$, denoted $c_3$, intersects $c(O, R)$ at point $N$. Prove that the triangles $AB\\Gamma$ and $KMN$ are congruent.",
    "options": [],
    "answer": "See solution",
    "solution": "We will prove that the circle $c_3$ passes through the center $O$ of $c(O, R)$.\n\nFrom the cyclic quadrilateral $O\\Delta\\Gamma E$ (in the circle $c_3$), we get: $\\hat{O}_1 = \\hat{\\Gamma}$.\n\nFrom the cyclic quadrilateral $O\\Delta BZ$ (in the circle $c_1$), we get: $\\hat{O}_2 = \\hat{B}$.\n\nSumming up the above equations, we find:\n\n$\\hat{O}_1 + \\hat{O}_2 = \\hat{B} + \\hat{\\Gamma} \\Rightarrow EOZ = \\hat{B} + \\hat{\\Gamma} = 180^\\circ - \\hat{A},$\n\nand therefore the quadrilateral $AEOZ$ is cyclic.\n\nNow we are going to prove that the circles $c_1, c_2, c_3$ are congruent.\n\nFrom the cyclic quadrilateral $O\\Delta BZ$, we have: $\\hat{\\Delta}_2 = \\hat{Z}_2$.\n\n![](images/Hellenic_booklet_2013_p7_data_ddd9c28ae1.png)\n\nFrom the cyclic quadrilateral $O\\Delta\\Gamma E$, we have: $\\hat{\\Delta}_2 = \\hat{E}_1$, and hence:\n\n$$\n\\hat{\\Delta}_2 = \\hat{Z}_2 = \\hat{E}_1.\n$$\n\nThese three angles correspond to the equal chords $OB$, $O\\Gamma$, and $OA$ of the circles $c_1, c_2$, and $c_3$, respectively. Therefore, these three circles are congruent.\n\nNow, in the congruent circles $c_1$ and $c_2$, the angles $\\hat{Z}_1$ and $\\hat{\\Delta}_1$ correspond to the equal chords $OK$ and $OM$ ($OK = OM = R$), so $\\hat{Z}_1 = \\hat{\\Delta}_1$.\n\nFrom the last equality, we conclude that the points $K$, $\\Delta$, $M$ are collinear. Similarly, we prove that the points $M$, $E$, $N$ and $N$, $Z$, $K$ are collinear.\n\nFrom the equalities of angles $B\\hat{\\Delta}K = \\Gamma\\hat{\\Delta}M$ and $\\Gamma\\hat{E}M = A\\hat{E}N$, we get the equality of segments $AN = BK = \\Gamma M$ (chords of the circle $c(O, R)$).\n\nThe triangles $AB\\Gamma$ and $KMN$ have common circumcenter $O$, and the triangle $KMN$ is the image of $AB\\Gamma$ under the rotation $R(O, \\omega)$, where\n\n$$\nA\\hat{O}N = B\\hat{O}K = \\Gamma\\hat{O}M = \\hat{\\omega}.\n$$\n\nHence, the triangles are congruent.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 14611,
    "subject": "Mathematics (Olympiad)",
    "question": "Consider an increasing continuous function $f : [0, 1] \\to \\mathbb{R}$ and define the sequence $(a_n)_{n \\ge 1}$ by\n$$\na_n = \\frac{1}{2^n} \\sum_{k=1}^{2^n} f\\left(\\frac{k}{2^n}\\right), \\quad \\text{for all integers } n \\ge 1.\n$$\n\na) Prove that the sequence $(a_n)_{n \\ge 1}$ is increasing.\n\nb) Given that there exists $p \\in \\mathbb{N}^*$ such that $a_p = \\int_0^1 f(x)\\,dx$, prove that $f$ is a constant function.",
    "options": [],
    "answer": "See solution",
    "solution": "a) Notice that\n$$\na_{n+1} = \\frac{1}{2^{n+1}} \\sum_{k=1}^{2^{n+1}} f\\left(\\frac{k}{2^{n+1}}\\right) = \\frac{1}{2^{n+1}} \\left( \\sum_{k=1}^{2^n} f\\left(\\frac{k}{2^n}\\right) + \\sum_{k=1}^{2^n} f\\left(\\frac{2k-1}{2^{n+1}}\\right) \\right)\n$$\nSince $f\\left(\\frac{2k-1}{2^{n+1}}\\right) \\le f\\left(\\frac{k}{2^n}\\right)$ (because $f$ is increasing), we have\n$$\na_{n+1} \\le \\frac{1}{2^n} \\sum_{k=1}^{2^n} f\\left(\\frac{k}{2^n}\\right) = a_n.\n$$\nThus, $(a_n)$ is increasing.\n\nb) Consider $k \\in \\{1, 2, \\dots, 2^p\\}$ and $c \\in \\left(\\frac{k-1}{2^p}, \\frac{k}{2^p}\\right)$. Divide $[0, 1]$ as\n$$\n\\Delta = \\left(0, \\frac{1}{2^p}, \\dots, \\frac{k-1}{2^p}, c, \\frac{k}{2^p}, \\dots, 1\\right).\n$$\nLet $S$ be the upper Darboux sum with respect to $\\Delta$. Then\n$$\n\\begin{aligned}\na_p - S &= \\frac{1}{2^p} f\\left(\\frac{k}{2^p}\\right) - f(c) \\left(c - \\frac{k-1}{2^p}\\right) - f\\left(\\frac{k}{2^p}\\right) \\left(\\frac{k}{2^p} - c\\right) \\\\\n&= \\left(f\\left(\\frac{k}{2^p}\\right) - f(c)\\right) \\left(c - \\frac{k-1}{2^p}\\right) \\ge 0.\n\\end{aligned}\n$$\nSince $\\int_0^1 f(x)\\,dx \\le S \\le a_p = \\int_0^1 f(x)\\,dx$, it follows that $a_p = S$ and thus $f(c) = f\\left(\\frac{k}{2^p}\\right)$. Therefore, $f$ is constant on each interval $\\left(\\frac{k-1}{2^p}, \\frac{k}{2^p}\\right]$, $k = 1, \\dots, 2^p$, whose union is $[0, 1]$. Since $f$ is continuous, $f$ is constant on $[0, 1]$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 14612,
    "subject": "Mathematics (Olympiad)",
    "question": "Triangle *ABC* is such that $\\angle ABC = \\angle ACB = 30^\\circ$. On the side *BC* point *D* is selected. The point *K* is such that *D* is the midpoint of *AK*. It turned out that $\\angle BKA > 60^\\circ$. Prove that $3AD < CB$.\n\n![](images/UkraineMO_2015-2016_booklet_p32_data_04923ae359.png)",
    "options": [],
    "answer": "See solution",
    "solution": "Let us select the points $X$ and $Y$ on the side $BC$ such that $BX = AX$ and $CY = YA$. Then $\\angle AXY = \\angle ABC + \\angle BAX = 2\\angle ABC = 60^\\circ$. Analogously, $\\angle XYA = 60^\\circ$ and then $\\triangle XYA$ is equilateral. Then $BX = AX = XY = AY = YC$, i.e. the points $X$, $Y$ divide $BC$ into three equal parts and $BC = 3AX$. Let $M$ be the midpoint of $BC$. $\\triangle ABC$ is an isosceles triangle, therefore $\\angle AMB = 90^\\circ$. Consider the points $B_1, T, K, N, C_1$, where $B$ is the middle of $AB_1$, $X$ is the middle of $AT$, $M$ is the middle of $AN$, $C$ is the middle of $AC_1$. By Thales' theorem, the points $T$, $K$, $N$ lie on the segment $B_1C_1$. We have that $BX = AX = XT$, and therefore $\\angle ABT = 90^\\circ$. Obviously, $\\angle TNA = 90^\\circ$. Thus, the quadrangle $ABTN$ is inscribed in a circle $w$. Hence, $\\angle ANB = \\angle ATB = 90^\\circ - \\angle BAT = 60^\\circ$. Point $K$ lies on the segment $B_1C_1$, i.e. the points $T$, $K$, $N$ lie in one half-plane with respect to the line $AB$. Then, as the arc $\\cup BTA$ of the circle $w$ has length $60^\\circ$, then from $\\angle AKB > 60^\\circ$ it follows that the point $K$ lies inside the circle $w$. On the other hand, the point $K$ lies on the segment $B_1C_1$, hence $K$ lies inside the segment $TN$. Then $D$ lies inside the segment $XM$. We obtain that $AM \\perp XM$ and the point $D$ is closer to $M$ than to $X$. Then $AD < AX$, and therefore $3AD < 3AX = BC$, Q.E.D.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 14613,
    "subject": "Mathematics (Olympiad)",
    "question": "設凸四邊形 $ABCD$ 有內切圓，其圓心為 $I$。令點 $I_a, I_b, I_c, I_d$ 分別為 $\\triangle DAB, \\triangle ABC, \\triangle BCD$ 與 $\\triangle CDA$ 的內心。設 $\\triangle AI_bI_d$ 與 $\\triangle CI_bI_d$ 的兩外接圓的兩條外公切線交於點 $X$，$\\triangle BI_aI_c$ 與 $\\triangle DI_aI_c$ 的兩外接圓的兩條外公切線交於點 $Y$。\n\n試證：$\\angle XIY$ 為直角。",
    "options": [],
    "answer": "See solution",
    "solution": "設 $\\omega_a, \\omega_b, \\omega_c, \\omega_d$ 分別為 $AI_bI_d, BI_aI_c, CI_bI_d, DI_aI_c$ 的外接圓，其圓心分別為 $O_a, O_b, O_c, O_d$，半徑分別為 $r_a, r_b, r_c, r_d$。\n\n*Claim 1.* $I_bI_d \\perp AC$ 且 $I_aI_c \\perp BD$。\n\n![](images/18-2J_p16_data_99bab27d29.png)\n\n*Proof.* 設 $ABC$ 與 $ACD$ 的內切圓分別與 $AC$ 相切於 $T$ 與 $T'$（見圖 1）。在 $ABC$ 中，$AT = \\frac{1}{2}(AB + AC - BC)$；在 $ACD$ 中，$AT' = \\frac{1}{2}(AD + AC - CD)$；又在四邊形 $ABCD$ 中，$AB - BC = AD - CD$，故\n\n$$\nAT = \\frac{AC + AB - BC}{2} = \\frac{AC + AD - CD}{2} = AT'\n$$\n\n因此 $T = T'$，所以 $I_b I_d \\perp AC$。\n\n第二個結論同理可證。$\\square$\n\n*Claim 2.* 點 $O_a, O_b, O_c, O_d$ 分別在 $AI, BI, CI, DI$ 上。\n\n*Proof.* 只需證明 $O_a$ 的情形（見下圖）。\n\n![](images/18-2J_p17_data_b6aa66a8b0.png)\n\n注意，$ABC$ 與 $ACD$ 的內切圓可由四邊形 $ABCD$ 的內切圓分別以 $B$ 與 $D$ 為中心做相似變換得到，故 $I_b$ 與 $I_d$ 分別在 $BI$ 與 $DI$ 上。\n\n在任意三角形中，從同一頂點出發的高與外接圓直徑關於角平分線對稱。由 Claim 1，$AI_dI_b$ 中，$AT$ 為 $A$ 的高，且 $T$ 在 $I_bI_d$ 上，故外心 $O_a$ 在 $I_bAI_d$ 角域內，且 $\\angle I_bAT = \\angle O_aAI_d$。$I_b, I_d$ 為 $ABC, ACD$ 的內心，故 $AI_b, AI_d$ 分別平分 $\\angle BAC, \\angle CAD$，因此\n\n$$\n\\begin{aligned}\n\\angle O_a AD &= \\angle O_a AI_d + \\angle I_d AD = \\angle I_b AT + \\angle I_d AD \\\\\n&= \\frac{1}{2} \\angle BAC + \\frac{1}{2} \\angle CAD = \\frac{1}{2} \\angle BAD,\n\\end{aligned}\n$$\n\n故 $O_a$ 在 $\\angle BAD$ 的平分線上，即在 $AI$ 上。$\\square$\n\n![](images/18-2J_p18_data_e80d605c3c.png)\n\n點 $X$ 為 $\\omega_a$ 與 $\\omega_c$ 的外部相似中心，設 $U$ 為內部相似中心。$O_a, O_c$ 在 $I_bI_d$ 的垂直平分線上，且 $X, U$ 在同一直線上；由 Claim 2，此線平行於 $AC$。\n\n由 $\\omega_a, \\omega_c$ 的相似性，$O_aI_b = O_aI_d = O_aA = r_a$，$O_cI_b = O_cI_d = O_cC = r_c$，且 $AC \\parallel O_aO_c$ 可知...",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 14614,
    "subject": "Mathematics (Olympiad)",
    "question": "How many 25-tuples $ (x_1, x_2, \\dots, x_{25}) $ of non-negative integers are there which satisfy the following identity?\n\n$$\nx_1^2 + x_2^2 + \\dots + x_{25}^2 = 2 + x_1 x_2 + x_2 x_3 + \\dots + x_{24} x_{25}.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "29900\n\nLet $x_0 = 0$. Then, the given identity can be rewritten as:\n\n$$\n\\begin{aligned}\n& x_0^2 + x_1^2 + \\cdots + x_{25}^2 = 2 + x_0 x_1 + x_1 x_2 + \\cdots + x_{24} x_{25} + x_{25} x_0 \\\\\n& \\iff (x_0 - x_1)^2 + (x_1 - x_2)^2 + \\cdots + (x_{24} - x_{25})^2 + (x_{25} - x_0)^2 = 4.\n\\end{aligned}\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 14615,
    "subject": "Mathematics (Olympiad)",
    "question": "Suppose in a game on the set $\\{0, 1, \\dots, 10\\}$, Breaker and Maker take turns picking elements. A player wins by occupying all elements of an arithmetic progression of length four. Which pairs of elements should Breaker occupy first to guarantee a win, and how does Breaker proceed to ensure victory?",
    "options": [],
    "answer": "See solution",
    "solution": "We observe that there are only two arithmetic progressions of length four that avoid the pair $\\{4, 7\\}$: $(0, 1, 2, 3)$ and $(0, 3, 6, 9)$. If Breaker chooses these as her first two moves, she wins because these progressions only share the points 0 and 3. After Maker's first three moves, at most one progression can be occupied by three of Maker's moves. Breaker then picks the remaining element of that progression as her third move and has time to block the other progression with her fourth move, guaranteeing her win.\n\nBy symmetry, the pair $\\{3, 6\\}$ is equally effective as $\\{4, 7\\}$ for Breaker. Similarly, $\\{4, 9\\}$ is also a critical pair: If Breaker occupies this pair first, the only strictly increasing progressions of length four that avoid $\\{4, 9\\}$ are $(0, 1, 2, 3)$, $(5, 6, 7, 8)$, and $(1, 3, 5, 7)$. The first two are disjoint, and both intersect $(1, 3, 5, 7)$ in two elements. If Maker's first three moves are all from $\\{5, 6, 7, 8\\}$, Breaker blocks $(5, 6, 7, 8)$ and then picks 1 or 3. If one of Maker's first three moves is from $\\{0, 1, 2, 3\\}$, Breaker can block two progressions by picking from $\\{1, 3, 5, 7\\}$ and then block the remaining progression with her fourth move.\n\nBy symmetry, $\\{1, 6\\}$ is also critical. Thus, Maker cannot occupy all critical sets in time: If Maker's first move is $a \\neq 4$, Breaker picks 4 and then occupies either $\\{4, 7\\}$ or $\\{4, 9\\}$ with her second move. If Maker's first pick is 4, Breaker picks 6 and occupies either $\\{3, 6\\}$ or $\\{1, 6\\}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 14616,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $S$ be a set consisting of $n$ points in space, with no four among them lying on the same plane. How many subsets of $S$ can be obtained as the intersection of $S$ with some half-space? Express your answer in terms of $P(n)$.",
    "options": [],
    "answer": "See solution",
    "solution": "A proof of this claim can be obtained similarly to the proof of Lemma 1 above, by replacing **plane** with **space** and **straight lines** with **planes**. Thus, the answer to this problem is given by $P(10) = 260$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 14617,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $x_1, x_2, \\dots, x_9$ be nonnegative real numbers satisfying\n$$\nx_1^2 + x_2^2 + \\dots + x_9^2 \\geq 25.\n$$\n\nProve that there exist three of these numbers with a sum of at least $5$.",
    "options": [],
    "answer": "See solution",
    "solution": "Without loss of generality, assume $x_1 \\ge x_2 \\ge x_3 \\ge x_4 \\ge x_5 \\ge x_6 \\ge x_7 \\ge x_8 \\ge x_9 \\ge 0$. Then $x_1x_2 \\ge x_4^2 \\ge x_5^2$, $x_1x_3 \\ge x_6^2 \\ge x_7^2$, and $x_2x_3 \\ge x_8^2 \\ge x_9^2$. Thus,\n$$\n(x_1 + x_2 + x_3)^2 = x_1^2 + x_2^2 + x_3^2 + 2x_1x_2 + 2x_1x_3 + 2x_2x_3 \\ge x_1^2 + x_2^2 + x_3^2 + x_4^2 + x_5^2 + x_6^2 + x_7^2 + x_8^2 + x_9^2 \\ge 25.\n$$\nTherefore, $x_1 + x_2 + x_3 \\ge 5$, which proves the assertion. $\\Box$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 14618,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all injective functions $f: \\mathbb{R} \\to \\mathbb{R}$ such that for every real number $x$ and every positive integer $n$,\n\n$$\n\\left| \\sum_{i=1}^{n} i(f(x+i+1) - f(x+i)) \\right| < 2016\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "From the condition, we have\n\n$$\n\\left| \\sum_{i=1}^{n-1} i(f(x+i+1) - f(x+i)) \\right| < 2016.\n$$\n\nThen,\n\n$$\n\\begin{aligned}\n&\\left| n(f(x+n+1) - f(x+n)) \\right| \\\\\n&= \\left| \\sum_{i=1}^{n} i(f(x+i+1) - f(x+i)) - \\sum_{i=1}^{n-1} i(f(x+i+1) - f(x+i)) \\right| \\\\\n&< 2 \\cdot 2016 = 4032\n\\end{aligned}\n$$\n\nimplying\n\n$$\n\\left| f(x+n+1) - f(x+n) \\right| < \\frac{4032}{n}\n$$\n\nfor every real number $x$ and every positive integer $n$.\n\nLet $y \\in \\mathbb{R}$ be arbitrary. Then there exists $x$ such that $y = x+n$. We obtain\n\n$$\n\\left| f(y+1) - f(y) \\right| < \\frac{4032}{n}\n$$\n\nfor every real number $y$ and every positive integer $n$. Since this holds for all $n$, we must have $f(y+1) = f(y)$ for all $y$, but since $f$ is injective, this forces $f(y) = y + c$ for some constant $c$. Checking the original condition, only $f(y) = y+1$ satisfies the required bound. Thus, the only solution is $f(y) = y+1$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 14619,
    "subject": "Mathematics (Olympiad)",
    "question": "For a positive integer $n$, let $s(n)$ denote the sum of the proper divisors of $n$ (excluding $n$ itself). Does there exist a positive integer $a$ for which the equation\n\n$$\ns(n) = a + n\n$$\n\nhas infinitely many solutions?",
    "options": [],
    "answer": "See solution",
    "solution": "Yes, for example $a = 12$.\n\nWhenever $p$ is prime,\n\n$$\ns(6p) = 1 + 2 + 3 + 6 + p + 2p + 3p = 12 + 6p.$$\n\nSo $s(6p) = 12 + 6p$ for infinitely many $n = 6p$.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 14620,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all pairs of positive integers $(n, k)$ such that $n! + 1 = (n + 1)^k$.",
    "options": [],
    "answer": "See solution",
    "solution": "The solutions are $n = 1, k = 1$; $n = 2, k = 1$; $n = 4, k = 2$.\n\nIt is easy to check that the solutions above are the only solutions for $n \\leq 4$. So assume $n > 4$. Then $n! + 1 > n + 1$, so $k > 1$. If $n$ is odd, then $n + 1$ is even, but $n! + 1$ is odd, so there are no solutions. So $n$ is even. Hence $n$ is composite, so $n$ divides $(n-1)!$.\n\nUsing the binomial theorem, we have:\n$$\n(n+1)^k - 1 = n^k + \\binom{k}{1} n^{k-1} + \\cdots + \\binom{k}{k-2} n^2 + k n\n$$\nSo,\n$$\n(n-1)! = n(n^{k-2} + n^{k-3} + \\cdots + \\binom{k}{k-2}) + k\n$$\nHence $n$ divides $k$. But that means $k \\geq n$, and $(n+1)^n > n! + 1$. So there are no other solutions.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 14621,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all real numbers $x$, $y$, $z$ such that\n\n$$\n\\begin{cases}\nxy + z = -30 \\\\\nyz + x = 30 \\\\\nzx + y = -18\n\\end{cases}\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "Adding the first two equations yields\n\n$$\n0 = xy + z + yz + x = (x + z)(y + 1).\n$$\n\nSo $y = -1$ or $z = -x$.\n\nIf $y = -1$, then the system reads $z = x - 30$ and $xz = -17$. It follows that $x^2 - 30x + 17 = 0$, and hence $x = 15 \\pm 4\\sqrt{13}$.\n\nIf $z = -x$, then the system becomes $xy - x = -30$, $y = x^2 - 18$, so\n\n$$\nx^3 - 19x + 30 = 0,\n$$\n\nwhose solutions are $x = 2$, $x = 3$, and $x = -5$. Therefore, we have five solutions $(x, y, z)$ in total: $(15 + 4\\sqrt{13}, -1, 15 - 4\\sqrt{13})$, $(15 - 4\\sqrt{13}, -1, 15 + 4\\sqrt{13})$, $(2, -14, -2)$, $(3, -9, -3)$, $( -5, 7, 5)$. $\\square$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 14622,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all integer solutions $(m, n)$ to the equation\n$$\nm^2 = 252 - n^5.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "Since $m^2 = 252 - n^5$ is non-negative, we have $n^5 \\le 252$, so $n < 4$. If $n=1$, we get $m^2 = 251$; if $n=2$, we have $m^2 = 220$; and if $n=3$, we have $m^2 = 9$. The only possible solution is $m = n = 3$.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 14623,
    "subject": "Mathematics (Olympiad)",
    "question": "In isosceles triangle $ABC$ with vertex at $B$, there are altitudes $BH$ and $CL$. Point $D$ is such that $BDCH$ is a rectangle. Find the angle $\\angle DLH$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $O$ be the intersection point of the diagonals of rectangle $HBDC$. Since $\\triangle CBL$ has a right angle, $BO = LO = DO$, therefore $HO = LO = DO$. Hence, $\\triangle DHL$ is a right triangle with hypotenuse $DH$, which yields $\\angle DLH = 90^\\circ$.",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 14624,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $n$ be an integer greater than $2$, and let $x_1, x_2, \\dots, x_n$ be $n$ positive real numbers such that\n$$\n\\sum_{i=1}^{n} \\frac{1}{x_i + 1} = 1,\n$$\nand let $\\alpha$ be a real number greater than $1$. Show that\n$$\n\\sum_{i=1}^{n} \\frac{1}{x_i^{\\alpha} + 1} \\geq \\frac{n}{(n-1)^{\\alpha} + 1}\n$$\nand determine the cases of equality.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $y_i = \\frac{1}{x_i+1}$ for $i = 1, 2, \\dots, n$, so the $y_i$ are positive real numbers that add up to $1$. Upon substitution, the left-hand side of the required inequality becomes\n$$\n\\sum_{i=1}^{n} \\frac{y_i^{\\alpha}}{(1 - y_i)^{\\alpha} + y_i^{\\alpha}} = \\sum_{i=1}^{n} \\frac{y_i^{\\alpha}}{\\left(\\sum_{j \\neq i} y_j\\right)^{\\alpha} + y_i^{\\alpha}},\n$$\nsince $y_1 + y_2 + \\dots + y_n = 1$. Apply Jensen's inequality to the convex function $t \\mapsto t^\\alpha$, $t > 0$, to get\n$$\n\\left( \\sum_{j \\neq i} y_j \\right)^{\\alpha} \\le (n-1)^{\\alpha-1} \\sum_{j \\neq i} y_j^{\\alpha}, \\quad i = 1, 2, \\dots, n,\n$$\nso\n$$\n\\sum_{i=1}^{n} \\frac{y_i^{\\alpha}}{\\left(\\sum_{j \\neq i} y_j\\right)^{\\alpha} + y_i^{\\alpha}} \\ge \\sum_{i=1}^{n} \\frac{y_i^{\\alpha}}{(n-1)^{\\alpha-1} \\sum_{j \\neq i} y_j^{\\alpha} + y_i^{\\alpha}}.\n$$\nNow write $z_i = y_i^\\alpha$ for $i = 1, 2, \\dots, n$, $z = z_1 + z_2 + \\dots + z_n$, and $a = (n-1)^{\\alpha-1}$ to transform the right-hand side of the above inequality to\n$$\n\\sum_{i=1}^{n} \\frac{z_i}{(1-a)z_i + az}.\n$$\nNotice that the function $t \\mapsto \\frac{t}{(1-a)t + az}$, $t < az/(a-1)$, is convex, so by Jensen's inequality:\n$$\n\\sum_{i=1}^{n} \\frac{z_i}{(1-a)z_i + az} \\ge n \\cdot \\frac{\\frac{1}{n} \\sum_{i=1}^{n} z_i}{(1-a)^n \\sum_{i=1}^{n} z_i + az} = \\frac{n}{(n-1)a+1}.\n$$\n\nEquality holds if and only if the $z_i$ are all equal; tracing back, this is the case if and only if the $x_i$ are all equal to $n-1$.\n\n**Remark.** Since the inequality in the statement can be rewritten as\n$$\n\\sum_{i=1}^{n} \\frac{1}{x_i + 1} \\cdot \\frac{x_i + 1}{x_i^{\\alpha} + 1} \\geq \\frac{\\sum_{i=1}^{n} \\frac{x_i}{x_i + 1} + 1}{\\left(\\sum_{i=1}^{n} \\frac{x_i}{x_i + 1}\\right)^{\\alpha} + 1},\n$$\nit is tempting to consider the real-valued function $f(x) = \\frac{x+1}{x^\\alpha + 1}$, $x \\ge 0$, and try to apply Jensen's inequality. Unfortunately, $f$ is concave on the closed unit interval $[0, 1]$ and convex on the ray $x \\ge 1$, as shown by the second derivative:\n$$\nf''(x) = \\frac{\\alpha x^{\\alpha-2}}{(x^{\\alpha} + 1)^3} \\left( (\\alpha - 1)x^{\\alpha+1} + (\\alpha + 1)x^{\\alpha} - (\\alpha + 1)x - (\\alpha - 1) \\right).\n$$\nThe sign of $f''$ is given by\n$$\ng(x) = (\\alpha - 1)x^{\\alpha+1} + (\\alpha + 1)x^{\\alpha} - (\\alpha + 1)x - (\\alpha - 1), \\quad x \\ge 0,\n$$\nwhose first two derivatives are\n$$\ng'(x) = (\\alpha + 1) \\left((\\alpha - 1)x^{\\alpha} + \\alpha x^{\\alpha-1} - 1\\right)\n$$\nand\n$$\ng''(x) = \\alpha(\\alpha^2 - 1)x^{\\alpha-2}(x+1) > 0, \\quad x > 0.\n$$\nHence $g'$ is strictly increasing. Since $g'(0) = -(\\alpha + 1) < 0$, and $g'(1) = 2(\\alpha^2 - 1) > 0$, it follows that $g'$ has a unique zero $x_0 \\in (0, 1)$. Consequently, $g$ is strictly decreasing on the closed interval $[0, x_0]$ and strictly increasing on the ray $x \\ge x_0$. Since $g(0) = 1 - \\alpha < 0$, and $g(1) = 0$, the conclusion follows.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 14625,
    "subject": "Mathematics (Olympiad)",
    "question": "Given a triangle $ABC$, let $D$ be the midpoint of side $AC$ and let $M$ be the point that divides the segment $BD$ in the ratio $1/2$; that is, $MB/MD = 1/2$. The rays $AM$ and $CM$ meet the sides $BC$ and $AB$ at points $E$ and $F$, respectively. Assume the two rays are perpendicular: $AM \\perp CM$. Show that the quadrilateral $AFED$ is cyclic if and only if the line supporting the median from $A$ in triangle $ABC$ meets the line $EF$ at a point situated on the circle $ABC$.",
    "options": [],
    "answer": "See solution",
    "solution": "Denote by $a, b, c$ the side lengths, and by $m_a, m_b, m_c$ the lengths of the medians of triangle $ABC$. Since $MD$ is a median in the right-angled triangle $AMC$, it follows that $2m_b/3 = MD = AD = CD = b/2$, so $m_b = 3b/4$, whence $\\left(\\frac{3b}{4}\\right)^2 = m_b^2 = \\frac{a^2 + c^2}{2} - \\frac{b^2}{4}$; that is, $13b^2 = 8(a^2 + c^2)$.\n\nNext, apply Menelaus' theorem to get $\\frac{EC}{EB} = 4 = \\frac{FA}{FB}$ and deduce thereby that the lines $AC$ and $EF$ are parallel. The quadrilateral $AFED$ is therefore a trapezium; it is cyclic if and only if $AF = DE$.\n\nExpress the two in terms of $a, b,$ and $c$. Recall that $FA/FB = 4$ to obtain $AF = \\frac{4c}{5}$. Next, apply Stewart's theorem in triangle $BCD$ to get $DE^2 = \\frac{b^2}{2} - \\frac{4a^2}{25}$. By the preceding, the quadrilateral $AFED$ is cyclic if and only if $25b^2 - 8a^2 = 32c^2$. Recall that $13b^2 = 8(a^2 + c^2)$ to express $b$ and $c$ in terms of $a$: $b = \\frac{2a\\sqrt{2}}{3}$ and $c = \\frac{2a}{3}$.\n\n![](images/shortlistBMO_2011_p16_data_b4670c1c15.png)\n\nFinally, let $N$ be the midpoint of side $BC$ and let the lines $AN$ and $EF$ meet at $P$. Notice that $EN = \\frac{a}{2} - \\frac{a}{5} = \\frac{3a}{10}$, and the triangles $ANC$ and $PNE$ are similar, so $NP = \\frac{3m_a}{5}$, so\n\n$$\nNA \\cdot NP = \\frac{3m_a^2}{5} = \\frac{3(2(b^2 + c^2) - a^2)}{20} = \\frac{a^2}{4} = NB \\cdot NC.\n$$\n\nThe conclusion follows.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 14626,
    "subject": "Mathematics (Olympiad)",
    "question": "給定一個遞增函數 $f: \\mathbb{N} \\to \\mathbb{R}_{>0}$，其中 $\\mathbb{N}$ 代表全體正整數所成的集合，$\\mathbb{R}_{>0}$ 為所有正實數所成的集合。定義如下：\n\n- 若正整數 $m, n$ 滿足 $f(mn) \\neq f(m)f(n)$，則稱數對 $(m, n)$ 為「不服從的」。\n- 正整數 $m$ 被稱為「極端不服從的」，若且唯若對於任意非負整數 $N$，總存在無限多個正整數 $n$ 使得所有 $(m, n), (m, n + 1), \\dots, (m, n + N)$ 都是「不服從的」。\n\n證明：若存在一個「不服從的」數對，則必存在一個「極端不服從的」正整數。",
    "options": [],
    "answer": "See solution",
    "solution": "**解**\n\n我們證明命題的逆否命題。也就是說，若不存在極端不服從的正整數，則存在非負整數 $\\alpha$ 使得對所有 $n \\in \\mathbb{N}$，有 $f(n) = n^\\alpha$。\n\n首先證明 $f(1) = 1$。由於 1 不是極端不服從的，存在 $n \\in \\mathbb{N}$ 使得 $f(n) = f(n)f(1)$，因此 $f(1) = 1$。\n\n令 $a, b$ 為任意大於 1 的正整數。我們將證明 $\\log f(a)/\\log a = \\log f(b)/\\log b$。由於它們都不是極端不服從的，存在非負整數 $N, M$，使得對任意 $n > M$，$(a, n), \\dots, (a, n + N)$ 和 $(b, n), \\dots, (b, n + N)$ 中各有一對是服從的。\n\n只需證明 $\\log f(a)/\\log a \\geq \\log f(b)/\\log b$。假設相反，則 $\\log f(a)/\\log a < \\log f(b)/\\log b$。因此 $\\log a/\\log b > \\log f(a)/\\log f(b)$，由有理數稠密性可取 $p, q \\in \\mathbb{N}$ 使 $\\log a/\\log b > p/q > \\log f(a)/\\log f(b)$。\n\n設 $x_0 = y_0 = k > M + N$，$k$ 足夠大。對 $i = 1, \\dots, q$，令 $x_i$ 為 $x_{i-1} - N \\leq t \\leq x_{i-1}$ 中某個使 $(a, t)$ 服從的 $t$；對 $j = 1, \\dots, p$，令 $y_j$ 為 $y_{j-1} - N \\leq t \\leq y_{j-1}$ 中某個使 $(b, t)$ 服從的 $t$。則可歸納得 $f(x_q) \\leq f(a)^q f(k)$，$f(y_p) \\geq f(b)^p f(k)$。\n\n又有：\n\n$$\nx_q \\geq a^q k - (a^{q-1} + \\dots + 1)N, \\quad y_p \\leq b^p k + (b^{p-1} + \\dots + 1)N.\n$$\n\n由於 $a^q > b^q$，可取 $k$ 使 $x_q > y_p$。因此 $f(a)^q \\geq f(b)^p$，即 $\\log f(a)/\\log b > p/q$，矛盾。因此 $\\log f(a)/\\log a = \\log f(b)/\\log b$。令 $\\alpha = \\log f(2)/\\log 2$，證畢。",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 14627,
    "subject": "Mathematics (Olympiad)",
    "question": "For every $n = 1, 2, 3, \\ldots$, define $a_n = 3^{3^n-1} + 2$. Prove that there are infinitely many prime numbers $p$ for which there exists a natural number $n$ such that $p$ is a divisor of $a_n$.",
    "options": [],
    "answer": "See solution",
    "solution": "Since $a_n$ is odd and is congruent to $2$ modulo $3$, it always has a prime divisor of the form $3h + 2$ for $h \\in \\mathbb{Z}^+$. Suppose by contradiction that the sequence $(a_n)$ has finitely many prime divisors; then the number of prime divisors of the form $3h + 2$ is also finite, say $p_1 < p_2 < \\dots < p_k$.\n\nSetting $m = p_1p_2\\cdots p_k$, clearly $\\gcd(m, 3) = 1$. By Euler's theorem, $m \\mid 3^{\\varphi(m)} - 1$. We also have\n\n$$\n\\varphi(m) = (p_1 - 1)(p_2 - 1)\\cdots(p_k - 1)\n$$\n\nis a positive integer not divisible by $3$, so setting $n = \\varphi(\\varphi(m))$ and using Euler's theorem again, we get\n\n$$\n\\varphi(m) \\mid 3^{\\varphi(\\varphi(m))} - 1 = 3^n - 1.\n$$\n\nNote that $3^x - 1 \\mid 3^y - 1$ where $x, y$ are positive integers such that $x \\mid y$. Hence,\n\n$$\n3^{\\varphi(m)} - 1 \\mid 3^{3^{n-1}} - 1.\n$$\n\nTherefore $m \\mid 3^{3^{n-1}} - 1$. From here, it follows that\n\n$$\na_n = 3^{3^{n-1}} - 1 + 3 \\equiv 3 \\pmod{m}.\n$$\n\nOn the other hand, $a_n$ will have a prime divisor of the form $3h + 2$, so there exists $i$ with $1 \\le i \\le k$ such that $p_i \\mid a_n$. But $p_i \\mid m$ implies $p_i \\mid 3$, which is a contradiction. Therefore, $(a_n)$ has infinitely many prime divisors.\n\n_Remark_: This problem can be solved directly by Kobayashi's Theorem. $\\Box$",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 14628,
    "subject": "Mathematics (Olympiad)",
    "question": "A $10 \\times 10$ square is divided into $1 \\times 1$ squares. A *point light* at a vertex of a $1 \\times 1$ square lights all the $1 \\times 1$ squares that the vertex belongs to. (A point light can be positioned on a vertex on the edge of the big square.) Find the minimum number of point lights required such that all the squares are lit even if one of the point lights is not functioning.",
    "options": [],
    "answer": "See solution",
    "solution": "The minimum number of lights is $55$.\n\nEach $1 \\times 1$ black square needs at least $2$ lights and each figure that consists of three $1 \\times 1$ squares needs at least $3$ lights (see Picture 1). So we need at least $2 \\times 20 + 3 \\times 5 = 55$ lights.\n\nPicture 2 shows that $55$ lights could be placed as required.\n\n![](images/MNG_ABooklet_2017_p17_data_a24ff744d9.png)\n\nPicture 1\n\n![](images/MNG_ABooklet_2017_p17_data_39f56edf43.png)\n\nPicture 2",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 14629,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $\\mathbb{Z}_{>0}$ be the set of positive integers. Find all functions $f: \\mathbb{Z}_{>0} \\to \\mathbb{Z}_{>0}$ such that, for all positive integers $m$ and $n$, $n \\mid m$ if and only if $f(n) \\mid f(m) - n$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $k$ be a positive integer. Substituting $m = k$ and $n = k$ into the original condition, we have $f(k) \\mid k$.\n\nNow we prove that $f(k) = k$ by induction on $k$.\n\nFor $k = 1$, this claim is trivial. Let $l > 1$ be a positive integer and assume that this claim is true for all $k < l$. Suppose that $f(l) < l$. Since $f(l)$ divides $f(l) - l$ and $f(f(l)) = f(l)$ by the induction hypothesis, $f(l)$ divides $f(f(l)) - l$. On the other hand, $l \\nmid f(l)$, which contradicts the original condition when $m = f(l)$ and $n = l$. Hence, we have $f(l) \\ge l$, and it follows that $f(l) = l$ since $f(l) \\mid l$. This completes the induction.\n\nConversely, the function $f(n) = n$ satisfies the original condition. Thus, this is the only solution.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 14630,
    "subject": "Mathematics (Olympiad)",
    "question": "Inside an equilateral triangle, a circle is drawn that touches all three sides. The radius of the circle is $10$. A second, smaller, circle touches the first circle and two sides of the triangle. A third, even smaller, circle touches the second circle and two sides of the triangle (see the figure). What is the radius of the third circle?\n\n![](images/NLD_ABooklet_2020_p7_data_b588a2be9c.png)",
    "options": [],
    "answer": "See solution",
    "solution": "$\\frac{10}{9}$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 14631,
    "subject": "Mathematics (Olympiad)",
    "question": "The number $2019$ is written on the board. Katia and Mykola are playing the following game: one by one (starting with Katia) they choose any divisor $d$ of the number $N$ written on the board and change the number on the board $N$ to the number $N - (2d - 1)$, if it is a positive integer. Whoever writes number $1$ loses. Who will win in this game and what is the strategy, considering both players want to win?\n\n![](path/to/file.png)",
    "options": [],
    "answer": "See solution",
    "solution": "Firstly, we show that the number on the board decreases with every turn. Clearly, it will be smaller and an integer. It will be positive, since $N = dD \\Rightarrow M = N - (2d - 1) = dD - 2d + 1 = d(D - 2) + 1 \\geq 1$, since if $d < N$ then $D \\geq 2$. Therefore, number $1$ will be written on the board after a finite number of turns.\n\nIt is not hard to notice that every turn changes the parity of the number on the board. Since Katia starts the game, there will be an even number after her turn, and an odd number after Mykola's turn. Therefore, number $1$ will be written on the board after Mykola's turn, so he will lose.",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 14632,
    "subject": "Mathematics (Olympiad)",
    "question": "A circle $\\omega$ is inscribed in a quadrilateral $ABCD$. Let $I$ be the center of $\\omega$. Suppose that\n\n$$\n(AI + DI)^2 + (BI + CI)^2 = (AB + CD)^2.\n$$\n\nProve that $ABCD$ is an isosceles trapezoid.",
    "options": [],
    "answer": "See solution",
    "solution": "and equality holds if and only if $AD \\parallel BC$ and $AB = CD$. Without loss of generality, we may assume that the inradius of $ABCD$ is 1.\n\n![](images/USA_IMO_2004_p39_data_c9d2ebb599.png)\n\n**First Solution:** As shown in the figure above, let $A_1, B_1, C_1$ and $D_1$ be the points of tangency. Because circle $\\omega$ is inscribed in $ABCD$, we can set $\\angle D_1IA = \\angle AIA_1 = x$, $\\angle A_1IB = \\angle BIB_1 = y$, $\\angle B_1IC = \\angle CIC_1 = z$, $\\angle C_1ID = \\angle DID_1 = w$, and $x + y + z + w = 180^\\circ$, or $x + w = 180^\\circ - (y + z)$, with $0^\\circ < x, y, z, w < 90^\\circ$. Then $AI = \\sec x$, $BI = \\sec y$, $CI = \\sec z$, $DI = \\sec w$, $AD = AD_1 + D_1D = \\tan x + \\tan w$, and $BC = BB_1 + B_1C = \\tan y + \\tan z$. The inequality $(*)$ becomes $(\\sec x + \\sec w)^2 + (\\sec y + \\sec z)^2 \\le (\\tan x + \\tan y + \\tan z + \\tan w)^2$.\n\nExpanding both sides of the above inequality and applying the identity $\\sec^2 x = 1 + \\tan^2 x$ gives\n\n$$\n\\begin{aligned}\n& 4 + 2(\\sec x \\sec w + \\sec y \\sec z) \\\\\n\\le & 2 \\tan x \\tan y + 2 \\tan x \\tan z + 2 \\tan x \\tan w \\\\\n& 2 \\tan y \\tan z + 2 \\tan y \\tan w + 2 \\tan z \\tan w,\n\\end{aligned}\n$$\n\nor\n\n$$\n\\begin{aligned}\n& 2 + \\sec x \\sec w + \\sec y \\sec z \\\\\n\\le & \\tan x \\tan w + \\tan y \\tan z + (\\tan x + \\tan w)(\\tan y + \\tan z).\n\\end{aligned}\n$$\n\nNote that by the **Addition-subtraction formulas**,\n\n$$\n1 - \\tan x \\tan w = \\frac{\\cos x \\cos w - \\sin x \\sin w}{\\cos x \\cos w} = \\frac{\\cos(x+w)}{\\cos x \\cos w}.\n$$\n\nHence,\n\n$$\n1 - \\tan x \\tan w + \\sec x \\sec w = \\frac{1 + \\cos(x + w)}{\\cos x \\cos w}.\n$$\n\nSimilarly,\n\n$$\n1 - \\tan y \\tan z + \\sec y \\sec z = \\frac{1 + \\cos(y + z)}{\\cos y \\cos z}.\n$$\n\nAdding the last two equations gives\n\n$$\n\\begin{aligned}\n& 2 + \\sec x \\sec w + \\sec y \\sec z - \\tan x \\tan w - \\tan y \\tan z \\\\\n= & \\frac{1 + \\cos(x + w)}{\\cos x \\cos w} + \\frac{1 + \\cos(y + z)}{\\cos y \\cos z}.\n\\end{aligned}\n$$\n\nIt suffices to show that\n\n$$\n\\frac{1 + \\cos(x + w)}{\\cos x \\cos w} + \\frac{1 + \\cos(y + z)}{\\cos y \\cos z} \\le (\\tan x + \\tan w)(\\tan y + \\tan z),\n$$\n\nor\n\n$$\ns + t \\le (\\tan x + \\tan w)(\\tan y + \\tan z),\n$$\n\nafter setting $s = \\frac{1+\\cos(x+w)}{\\cos x \\cos w}$ and $t = \\frac{1+\\cos(y+z)}{\\cos y \\cos z}$. By the Addition-subtraction formulas, we have\n\n$$\n\\tan x + \\tan w = \\frac{\\sin x \\cos w + \\cos x \\sin w}{\\cos x \\cos w} = \\frac{\\sin(x+w)}{\\cos x \\cos w}.\n$$\n\nSimilarly,\n\n$$\n\\tan y + \\tan z = \\frac{\\sin(y + z)}{\\cos y \\cos z} = \\frac{\\sin(x + w)}{\\cos y \\cos z}.\n$$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 14633,
    "subject": "Mathematics (Olympiad)",
    "question": "Two rhombi $ABCD$ and $AXYZ$ are external to each other with a common vertex $A$. The vertices of the rhombi are labelled clockwise. If $\\angle DAX = \\angle BAZ$, prove that the centres of the rhombi and the midpoint of the segment $BZ$ form an isosceles triangle.",
    "options": [],
    "answer": "See solution",
    "solution": "Let the centres be $P$ and $Q$, and the midpoint of the segment $BZ$ be $R$ as shown in the figure. Since $\\angle DAX = \\angle BAZ$, we have $DZ = BX$. Since $P$ and $R$ are the midpoints of $XZ$ and $BZ$, respectively, $PR = BX/2$. Similarly, $QR = DZ/2$. Therefore, $QR = PR$ and it follows that $\\triangle PQR$ is isosceles.\n\n![](images/combined_25__latex_1_p1_data_083284733a.png)",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 14634,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $n = 2013$. For any real number $r$, denote by $\\lfloor r \\rfloor$ the greatest integer less than or equal to $r$.\n\nConsider $n$ cards numbered $0$ to $n-1$. For each $i$ from $1$ to $n$, an operation is performed: for each card $x$, if $\\lfloor \\frac{i(x+1)}{n} \\rfloor - \\lfloor \\frac{ix}{n} \\rfloor = 1$, the card is turned over during operation $i$; otherwise, it is not.\n\nAfter performing all operations $i = 1, 2, \\ldots, n$, how many cards show their numbered side up?",
    "options": [],
    "answer": "See solution",
    "solution": "We prove two lemmas to analyze the card flipping process.\n\n**Lemma 1.** For $1 \\leq i \\leq n$ and $0 \\leq x \\leq n-1$, $\\lfloor \\frac{i(x+1)}{n} \\rfloor - \\lfloor \\frac{ix}{n} \\rfloor = 1$ if and only if the card numbered $x$ is turned over at operation $i$; otherwise, it is $0$.\n\n*Proof:* For such $(i, x)$, $\\lfloor \\frac{i(x+1)}{n} \\rfloor - \\lfloor \\frac{ix}{n} \\rfloor$ is $0$ or $1$, since $0 \\leq \\frac{i(x+1)}{n} - \\frac{ix}{n} = \\frac{i}{n} < 1$. The card $x$ is turned over at operation $i$ if and only if there exists $j$ such that $\\lfloor \\frac{nj}{i} \\rfloor = x$, which is equivalent to $x \\leq \\frac{nj}{i} < x+1$, or $\\frac{ix}{n} \\leq j < \\frac{i(x+1)}{n}$. This is equivalent to $\\lfloor \\frac{i(x+1)}{n} \\rfloor - \\lfloor \\frac{ix}{n} \\rfloor > 0$.\n\nFor $1 \\leq i \\leq n-1$, consider the pair of operations $\\{i, n-i\\}$.\n\n**Lemma 2.** The card numbered $x$ is turned over once during operations $\\{i, n-i\\}$ if and only if neither $\\frac{i(x+1)}{n}$ nor $\\frac{ix}{n}$ is an integer.\n\n*Proof:* For real $r, s$ with $r+s$ integer, $[r] + [s] = r+s$ if $r$ is integer, $r+s+1$ otherwise. From Lemma 1, the card $x$ is turned over once during $\\{i, n-i\\}$ if\n$$\n\\left( \\left\\lfloor \\frac{i(x+1)}{n} \\right\\rfloor - \\left\\lfloor \\frac{ix}{n} \\right\\rfloor \\right) + \\left( \\left\\lfloor \\frac{(n-i)(x+1)}{n} \\right\\rfloor - \\left\\lfloor \\frac{(n-i)x}{n} \\right\\rfloor \\right) = 1\n$$\nThis reduces to checking whether neither $\\frac{i(x+1)}{n}$ nor $\\frac{ix}{n}$ is integer.\n\nSince $\\frac{ix}{n}$ is integer if and only if $x$ is a multiple of $\\frac{n}{\\gcd(i, n)}$, the card $x$ is turned over once during $\\{i, n-i\\}$ if and only if neither $x$ nor $x+1$ is a multiple of $\\frac{n}{\\gcd(i, n)}$.\n\nNow, $2013 = 3 \\cdot 11 \\cdot 61$. The number of $i$ with $\\gcd(i, 2013)$ equal to $1, 3, 11, 61, 33, 183, 671$ is computed:\n- $\\gcd(i, 2013) = 1$: $1200$ values\n- $= 3$: $600$\n- $= 11$: $120$\n- $= 61$: $20$\n- $= 33$: $60$\n- $= 183$: $10$\n- $= 671$: $2$\n\nFor $i = 1, \\ldots, 1006$, half of each type, so $600, 300, 60, 210, 30, 5, 1$ respectively. Only for $33$ and $671$ are the counts odd.\n\nApplying operations $\\{i, 2013-i\\}$ for even counts leaves the arrangement unchanged. So, only consider:\n(a) $\\{i, 2013-i\\}$ for $\\gcd(i, 2013) = 33$ ($3 \\cdot 11$)\n(b) $\\{i, 2013-i\\}$ for $\\gcd(i, 2013) = 671$ ($11 \\cdot 61$)\n(c) operation $2013$.\n\nBy (a), cards with remainder $1, 2, \\ldots, 9$ mod $11$ change sides. By (b), cards with remainder $1$ mod $3$ change sides. By the Chinese Remainder Theorem, the number of cards with their number facing down after all operations is $(9 \\times 1 + (11-9) \\times (3-1)) \\times 61 = 793$.\n\nThe final operation flips all cards, so the number of cards with their number facing up is $793$.",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 14635,
    "subject": "Mathematics (Olympiad)",
    "question": "a) Give an example of matrices $A$ and $B$ from $M_2(\\mathbb{R})$ such that\n$$\nA^2 + B^2 = \\begin{pmatrix} 2 & 3 \\\\ 3 & 2 \\end{pmatrix}.\n$$\n\nb) Let $A$ and $B$ be matrices from $M_2(\\mathbb{R})$ such that $A^2 + B^2 = \\begin{pmatrix} 2 & 3 \\\\ 3 & 2 \\end{pmatrix}$. Prove that $AB \\neq BA$.",
    "options": [],
    "answer": "See solution",
    "solution": "a) An example is $A = \\frac{\\sqrt{3}}{2} \\begin{pmatrix} 1 & 1 \\\\ 1 & 1 \\end{pmatrix}$ and $B = \\begin{pmatrix} 0 & 1 \\\\ -1 & 0 \\end{pmatrix}$.\n\nb) If the matrices $A$ and $B$ commute, then $A^2 + B^2 = (A + iB)(A - iB)$, hence\n$$\n\\begin{aligned}\n|\\det(A + iB)|^2 &= \\det(A + iB) \\det(A - iB) = \\det(A^2 + B^2) \\\\\n&= \\det\\begin{pmatrix} 2 & 3 \\\\ 3 & 2 \\end{pmatrix} = -5,\n\\end{aligned}\n$$\na contradiction.",
    "topic": "Algebra",
    "subtopic": "Linear Algebra"
  },
  {
    "id": 14636,
    "subject": "Mathematics (Olympiad)",
    "question": "Prove the following inequality for positive real numbers $a, b, c \\ge 1$:\n\n$$\n3abc + a + b + c \\ge 2(ab + bc + ca).\n$$\n\nWhat are the requirements for the inequality to become an equality?",
    "options": [],
    "answer": "See solution",
    "solution": "Let $x = a - 1$, $y = b - 1$, and $z = c - 1$, which are all non-negative. The inequality becomes:\n\n$$\n3xyz + xy + yz + zx \\ge 0,\n$$\n\nwhich is true since each term is non-negative.\n\nEquality holds if and only if at least two of $x, y, z$ are zero; that is, at least two of $a, b, c$ equal $1$.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 14637,
    "subject": "Mathematics (Olympiad)",
    "question": "令 $a$ 和 $b$ 為相異正整數。有一個無限大的棋盤，每格可以被填上一個非負整數。棋盤一開始是空的。我們持續進行以下操作：\n\n1. 如果棋盤上能找到一對相同的數字，則我們挑選相同數字的兩格，把其中一格的數字增加 $a$，另一格增加 $b$。\n2. 如果棋盤上找不到一對相同的數字，則我們在兩個空的格子裡各填上一個 $0$。\n\n試證：不論如何，自某次操作後，都只能再操作第 1 步。",
    "options": [],
    "answer": "See solution",
    "solution": "不失一般性，假設 $\\gcd(a, b) = 1$，因為我們可以將 $a$ 和 $b$ 同除以 $\\gcd(a, b)$。同時，不失一般性假設 $b > a$。\n\n1. 假設經過若干次第 1 步和 $N$ 次第 2 步後，我們被迫要再進行一次第 2 步。對於所有整數 $k$，令 $f_N(k)$ 為到目前為止 $k$ 在棋盤上總計出現的次數。顯然 $f_N(0) = 2N$，且對於所有 $k < 0$ 有 $f_N(k) = 0$。\n\n注意到對於所有 $k$，由於我們本回合被迫進行第 2 步，這表示棋盤上目前至多只有一個 $k - a$；而如果之前回合中有出現其他 $k - a$，表示之前曾經有兩個 $k - a$，且其中一個在之前的某回合被增加為 $k$。同理對 $k - b$ 亦成立。以上觀察告訴我們有以下關係式：\n\n$$\nf_N(k) = \\left\\lfloor \\frac{f_N(k-a)}{2} \\right\\rfloor + \\left\\lfloor \\frac{f_N(k-b)}{2} \\right\\rfloor. \\qquad (1)\n$$\n\n2. 由於 $\\gcd(a, b) = 1$，所有正整數 $x > ab - a - b$ 都可以被表為 $x = as + bt$，其中 $s, t$ 為非負整數。以下證明：\n\n**Lemma 1.** 對於所有 $x = as + bt$，我們有\n\n$$\nf_N(x) > \\frac{f_N(0)}{2^{s+t}} - 2.\n$$\n\n*證明*：我們對 $s+t$ 進行歸納。$s+t=0$ 時顯然。假設 $s+t=v$ 時引理成立，則當 $s+t=v+1$ 時，注意到 $s$ 和 $t$ 至少有一為正，不失一般性假設 $s>0$。則由 (1) 與歸納假設，我們有：\n\n$$\n\\begin{aligned}\nf_N(x) &= f_N(sa + tb) \\ge \\left\\lfloor \\frac{f_N((s-1)a + tb)}{2} \\right\\rfloor \\\\\n&\\ge \\left\\lfloor \\frac{1}{2} \\left( \\frac{f_N(0)}{2^v} - 2 \\right) \\right\\rfloor > \\frac{f_N(0)}{2^{v+1}} - 2,\n\\end{aligned}\n$$\n\n從而引理得證。\n\n3. 現在，假設我們會執行第 2 步無窮多次，亦即 1. 中的 $N$ 可以是任意正整數。取 $n = ab-a-b$，並取 $N = 2^{a+b+1}$。注意到 $n+1, n+2, \\dots, n+b$ 都可以被表為 $sa+tb$，其中 $0 \\leq s \\leq b$ 且 $0 \\leq t \\leq a$。故由 Lemma 1，我們有：\n\n$$\nf_N(n+k) > \\frac{f_N(0)}{2^{s+t}} - 2 = \\frac{2N}{2^{s+t}} - 2 = \\frac{2^{a+b+2}}{2^{s+t}} - 2 \\geq \\frac{2^{a+b+2}}{2^{a+b}} - 2 = 2.\n$$\n\n這表示 $f_N(n+1), \\dots, f_N(n+k)$ 都大於或等於 2，從而由 (1) 與數學歸納法，我們知 $f_N(x) \\geq 2$ 對所有 $x > n$ 都成立。但這意味著我們經過了有限次操作，卻讓棋盤上有無窮多個數字，這是不可能的，故矛盾！",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 14638,
    "subject": "Mathematics (Olympiad)",
    "question": "Publisher Soothsayer published a reference book claiming that for each real number $x$ and positive even number $n$ the equality $$(1+x)^n \\ge 2^n x$$ holds. Is this claim true?",
    "options": [],
    "answer": "See solution",
    "solution": "No.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 14639,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $A$ be a 101-element subset of the set $S = \\{1, 2, \\dots, 1000000\\}$. Prove that there exist numbers $t_1, t_2, \\dots, t_{100}$ in $S$ such that the sets\n\n$$\nA_j = \\{x + t_j \\mid x \\in A\\} \\quad j = 1, 2, \\dots, 100\n$$\n\nare pairwise disjoint.",
    "options": [],
    "answer": "See solution",
    "solution": "**First Solution.** Consider the set $D = \\{x - y \\mid x, y \\in A\\}$. There are at most $101 \\times 100 + 1 = 10101$ elements in $D$ (where the summand $1$ represents the difference $x - y = 0$ for $x = y$). Two sets $A_i$ and $A_j$ have nonempty intersection if and only if $t_i - t_j$ is in $D$. It suffices to choose 100 numbers $t_1, t_2, \\dots, t_{100}$ in such a way that we do not obtain a difference from $D$.\n\nWe select these elements by induction. Choose one element arbitrarily. Assume that $k$ elements, $k \\le 99$, have already been chosen. An element $x$ that is already chosen prevents us from selecting any element from the set $x + D = \\{x + d \\mid d \\in D\\}$. Thus, after $k$ elements are chosen, at most $10101k \\le 999999$ elements are forbidden. Hence we can select one more element. (Note that the numbers chosen are distinct because $0$ is an element in $D$.)",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 14640,
    "subject": "Mathematics (Olympiad)",
    "question": "Prove that\n\n$$\n\\sqrt{a^2 + b^2 - \\sqrt{2} ab} + \\sqrt{b^2 + c^2 - \\sqrt{2} bc} \\ge \\sqrt{a^2 + c^2}\n$$\n\nfor all positive real numbers $a$, $b$, and $c$.",
    "options": [],
    "answer": "See solution",
    "solution": "The inequality results from the triangle inequality, $PQ + PR \\ge QR$, as shown in the figure.\n\n![](images/Tajland_2008_p3_data_fbc62ef35f.png)\n\n![](images/Tajland_2008_p3_data_815aca1c52.png)",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 14641,
    "subject": "Mathematics (Olympiad)",
    "question": "A quadruplet of distinct positive integers $ (a, b, c, d) $ is called $ k $-good if the following conditions hold:\n\n1. Among $ a, b, c, d $, no three form an arithmetic progression.\n2. Among $ a + b, a + c, a + d, b + c, b + d, c + d $, there are $ k $ of them forming an arithmetic progression.\n\nFind:\n\na) A 4-good quadruplet.\n\nb) The maximal $ k $ such that there is a $ k $-good quadruplet.\n\n![](path/to/file.png)",
    "options": [],
    "answer": "See solution",
    "solution": "a) The quadruple $ (7, 6, 4, 3) $ is 4-good because there are no three numbers in it that form an arithmetic progression. Of the six numbers:\n\n$$\n7 + 6 = 13, \\quad 7 + 4 = 11, \\quad 7 + 3 = 10, \\quad 6 + 4 = 10, \\quad 6 + 3 = 9, \\quad 4 + 3 = 7\n$$\n\nthe numbers $ 7, 9, 11, 13 $ form an arithmetic progression.\n\nb) Without loss of generality, let $ a > b > c > d $. Then:\n\n$$\na + b > a + c > \\max(a + d, b + c) > \\min(a + d, b + c) > b + d > c + d.\n$$\n\nNote that if:\n\n1) $ a+b, a+c, a+d $ form an arithmetic progression, then $ 2(a+c) = (a+b) + (a+d) \\iff 2c = b+d $;\n2) $ a+b, a+c, b+c $ form an arithmetic progression, then $ 2(a+c) = (a+b) + (b+c) \\iff 2b = a+c $;\n3) $ a+d, b+d, c+d $ form an arithmetic progression, then $ 2(b+d) = (a+d) + (c+d) \\iff 2b = a+c $;\n4) $ b+c, b+d, c+d $ form an arithmetic progression, then $ 2(b+d) = (b+c) + (c+d) \\iff 2c = b+d $.\n\nIn all four cases, we get a contradiction with the given condition.\n\nFrom the above, it is clear that all 6 numbers cannot form an arithmetic progression.\n\nAssume that 5 of them form an arithmetic progression. Notice that whichever of the numbers $ a+b, a+c, a+d, b+c, b+d, c+d $ we delete, there is always some from progressions 1, 2, 3, or 4, which leads to a contradiction.\n\nIt follows from part (a) that the required $ k $ is $ 4 $. $ \\square $",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 14642,
    "subject": "Mathematics (Olympiad)",
    "question": "Given $a + 2b = 13$ and $5a - 2b = 5$, find the values of $a$ and $b$.",
    "options": [],
    "answer": "See solution",
    "solution": "Add the two equations:\n\n$$\n(a + 2b) + (5a - 2b) = 13 + 5\n$$\n\nThis simplifies to $6a = 18$, so $a = 3$. Substitute $a = 3$ into $a + 2b = 13$:\n\n$$\n3 + 2b = 13 \\implies 2b = 10 \\implies b = 5.\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 14643,
    "subject": "Mathematics (Olympiad)",
    "question": "Let a sequence $(a_n)_{n \\geq 1}$ of positive integers be called *alagoana* if it satisfies the following two conditions:\n\n1. $a_n! = a_n \\cdot a_{(n-1)!}$ for every $n \\geq 1$.\n2. For every $n \\geq 1$, the exponent of any prime $p$ in the prime factorization of $a_n$ is a multiple of $n$.\n\nProve that the only *alagoana* sequence is $a_n = 1$ for every positive integer $n$.",
    "options": [],
    "answer": "See solution",
    "solution": "We will prove that the only *alagoana* sequence is $a_n = 1$ for every positive integer $n$.\n\nSuppose $(a_n)$ is an *alagoana* sequence. Consider a prime $p$. For each $n$, let $\\alpha(n)$ be the exponent of $p$ in $a_n$.\n\nBy the second condition, $\\alpha(n)$ is a multiple of $n$, so $\\alpha(n) = n k_n$ for some $k_n \\geq 0$.\n\nFrom the first condition, $a_n! = a_n \\cdot a_{(n-1)!}$, so $\\alpha(n!) = \\alpha(n) + \\alpha((n-1)!)$, which gives $n k_n = n! k_{n!} - (n-1)! k_{(n-1)!}$.\n\nLet $\\beta(n) = (n-1)!$. Then:\n\n$$\nk_n = \\frac{\\beta(n)}{n} (n k_{n!} - k_{\\beta(n)}). \\qquad (1)\n$$\n\nWe show recursively that for $n \\geq 4$ and any $j \\geq 1$, $\\beta^j(n)$ divides $\\alpha(n)$, where $\\beta^j(n)$ is $\\beta$ applied $j$ times.\n\nBy induction, $k_n = \\frac{\\beta^j(n)}{n} N_j$ for some integer $N_j$.\n\nAs $j$ increases, $\\beta^j(n)$ grows rapidly, so for any fixed $n$, $\\alpha(n)$ is divisible by arbitrarily large numbers, forcing $\\alpha(n) = 0$ for $n \\geq 4$.\n\nFor $n = 1,2,3$, the identity $\\alpha(1) + \\alpha(2) + \\alpha(3) = \\alpha(6) = 0$ implies $\\alpha(1) = \\alpha(2) = \\alpha(3) = 0$.\n\nThus, $a_n$ has no prime factors for any $n$, so $a_n = 1$ for all $n \\geq 1$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 14644,
    "subject": "Mathematics (Olympiad)",
    "question": "Anna has placed real numbers with sum $S$ in the cells of a row. It turned out that she cannot cut the row into two parts so that the sum of the numbers in one part is positive and in the other part is negative. Prove that the modulus of $S$ is not less than any of Anna's numbers.",
    "options": [],
    "answer": "See solution",
    "solution": "Assume $S = 0$. For any division of the row into two parts, their sums add to $0$, so one sum is not less than $0$ and the other is not greater than $0$. If either sum is nonzero, one part is positive and the other negative, contradicting the condition. Thus, all numbers must be $0$.\n\nNow suppose $S \\neq 0$, and without loss of generality, let $S > 0$. Let $a$ be any number in the row. For any division, both sums must be nonnegative, since their total is $S > 0$. Let $x$ and $y$ be the sums of numbers to the left and right of $a$ (if there are no such numbers, set the sum to $0$). Then $x, y \\geq 0$ and $x + a, y + a \\geq 0$. Since $S = x + y + a$, we have $S - a = x + y \\geq 0$ and $S + a = x + y + 2a \\geq 0$. Thus, $S - a \\geq 0$ and $S + a \\geq 0$, so $-S \\leq a \\leq S$, i.e., $|a| \\leq |S|$ for any $a$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 14645,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $M$ be an index set, and let $S$ be a $3 \\times 3$ matrix with entries $x_{ij}$, $i, j \\in \\{1, 2, 3\\}$, and suppose $S$ has property (O). Define the set\n\n$$\nI = \\{k \\in M \\mid x_{1k} > x_{11},\\ x_{3k} > x_{32}\\}\n$$\n\nand suppose $1, 2, 3 \\notin I$. Assume $x_{18}, x_{38} > 1 \\ge x_{11}, x_{32}$, so $8 \\in I$ and $I \\ne \\emptyset$. Let $k^* \\in I$ be such that\n\n$$\nx_{2k^*} = \\max\\{x_{2k} \\mid k \\in I\\}.\n$$\n\nDefine the matrix\n\n$$\nS' = \\begin{bmatrix} x_{11} & x_{12} & x_{1k^*} \\\\ x_{21} & x_{22} & x_{2k^*} \\\\ x_{31} & x_{32} & x_{3k^*} \\end{bmatrix}.\n$$\n\nProve that $S'$ has property (O), and that $S'$ is unique in this sense: if $\\hat{S}$ is another such matrix with property (O), then $\\hat{S} = S'$.",
    "options": [],
    "answer": "See solution",
    "solution": "We first show that $S'$ has property (O).\n\nBy the definition of $I$, we have $x_{1k^*} > x_{11}$ and $x_{3k^*} > x_{32}$. Let $u_1 = x_{11}$ and $u_3 = x_{32}$. By property (O) of $S$, for $k^*$, we have $x_{2k^*} \\le u_2$.\n\nDefine\n$$\nu'_1 = u_1,\\ u'_2 = \\min\\{x_{21}, x_{22}, x_{2k^*}\\} = x_{2k^*},\\ u'_3 = u_3.$$\n\nWe claim that for any $k \\in M$, there exists $i \\in \\{1,2,3\\}$ such that $u'_i \\ge x_{ik}$. Otherwise, $x_{ik} > \\min\\{x_{i1}, x_{i2}\\}$ for $i=1,3$ and $x_{2k} > x_{2k^*}$, contradicting the maximality of $k^*$. Thus, $S'$ has property (O).\n\nFor uniqueness, suppose $\\hat{S}$ is another such matrix with property (O), using $k_0$ in place of $k^*$. By considering the possible cases for the minima in each row and using the properties of $S$ and $S'$, we find that $k_0 = k^*$, so $\\hat{S} = S'$. Thus, $S'$ is unique.",
    "topic": "Algebra",
    "subtopic": "Linear Algebra"
  },
  {
    "id": 14646,
    "subject": "Mathematics (Olympiad)",
    "question": "Compute the sum\n\n$$\n\\sum_{n=1}^{\\infty} \\frac{F_n}{10^{n+1}}\n$$\n\nwhere $F_n$ is the $n$th Fibonacci number given by $F_1 = F_2 = 1$ and $F_{n+1} = F_n + F_{n-1}$ for all $n \\ge 2$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let\n\n$$\nX = \\sum_{n=1}^{\\infty} \\frac{F_n}{10^{n+1}}\n$$\n\nThen\n\n$$\nX = \\frac{1}{10^2} + \\frac{1}{10^3} + \\frac{2}{10^4} + \\frac{3}{10^5} + \\frac{5}{10^6} + \\frac{8}{10^7} + \\frac{13}{10^8} + \\dots\n$$\n\nSo\n\n$$\n10X = \\frac{1}{10} + \\frac{1}{10^2} + \\frac{2}{10^3} + \\frac{3}{10^4} + \\frac{5}{10^5} + \\frac{8}{10^6} + \\frac{13}{10^7} + \\dots\n$$\n\nand\n\n$$\n100X = 1 + \\frac{1}{10} + \\frac{2}{10^2} + \\frac{3}{10^3} + \\frac{5}{10^4} + \\frac{8}{10^5} + \\frac{13}{10^6} + \\dots\n$$\n\nThen $100X - 10X - X = 1$ (using the basic property of the Fibonacci numbers).\n\nSo\n\n$$\nX = \\frac{1}{89}\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 14647,
    "subject": "Mathematics (Olympiad)",
    "question": "Calculate the number of arrangements of 5 girls $G_1, G_2, G_3, G_4$, and $G_5$ and 12 boys in a row, satisfying the following conditions:\n\n1. The order of the girls from left to right is $G_1, G_2, G_3, G_4$, and $G_5$.\n2. There are at least 3 boys between $G_1$ and $G_2$.\n3. There are at least 1 boy and at most 4 boys between $G_4$ and $G_5$.",
    "options": [],
    "answer": "See solution",
    "solution": "Recall that the number of natural solutions to the equation\n\n$$\n\\sum_{i=1}^{n} x_i = m\n$$\n\nis $\\binom{m+n-1}{n-1}$. We use this to count the arrangements.\n\nLet $x_i$ be the number of boys standing between $G_{i-1}$ and $G_i$ for $i \\geq 2$; $x_1$ is the number of boys to the left of $G_1$, and $x_6$ is the number to the right of $G_5$. The constraints are $3 \\leq x_2$, $1 \\leq x_4 \\leq 4$, and\n\n$$\nx_1 + x_2 + x_3 + x_4 + x_5 + x_6 = 12.\n$$\n\nLet $y_i = x_i$ for $i \\neq 2$, $y_2 = x_2 - 3$. Then\n\n$$\ny_1 + y_2 + y_3 + y_4 + y_5 + y_6 = 9,\n$$\n\nwhere $y_i \\geq 0$ and $1 \\leq y_4 \\leq 4$. For each $y_4 = 1, 2, 3, 4$, the number of solutions is\n\n$$\n\\sum_{y_4=1}^{4} \\binom{9-y_4+5-1}{5-1} = \\binom{13}{5} - \\binom{9}{5} = 1161.\n$$\n\nSince the boys can be permuted, the total number of arrangements is $12! \\times 1161$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 14648,
    "subject": "Mathematics (Olympiad)",
    "question": "Consider an $N \\times N$ board, where each unit square is either black or white. Is it possible to change the color of all black unit squares to white, and all white unit squares to black, by repeatedly choosing $2 \\times 2$ squares and toggling the color of each unit square within the chosen $2 \\times 2$ square? For which values of $N$ is this possible?\n\n![](images/Mathematical_competitions_in_Croatia_2017_p15_data_3eca19cdc3.png)\n\n![](images/Mathematical_competitions_in_Croatia_2017_p15_data_3fa87128fe.png)",
    "options": [],
    "answer": "See solution",
    "solution": "We claim that the sought numbers are all multiples of 3.\n\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\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\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.\n\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.\n\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 with 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\n$$2, 2, 0, 1, 1, 0, 2, 2, 0, \\dots$$\n\ntimes, where this sequence is periodical with the period 6.\n\nRegardless of whether $L = 1$ or $L = 2$, it is unambiguously determined how many times we need to choose the $2 \\times 2$ square in the last two columns in order to achieve that the *next to last* unit square in the first row changes its colour from black to white or vice versa. However, that number is different from the number of times we would need to choose that same $2 \\times 2$ square in order to change the colour of the *last* unit square from black to white or vice versa. This completes the proof.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 14649,
    "subject": "Mathematics (Olympiad)",
    "question": "Solve the equation $x[x] = 2016$.\n\nHere $[x]$ is the integer part of the number $x$, i.e., the largest integer not greater than $x$.",
    "options": [],
    "answer": "See solution",
    "solution": "*Answer:* $x = -\\frac{224}{5}$.\n\n*Solution.* Suppose that $x \\ge 0$. Let $t = [x] \\ge 0$, then:\n\n$$\nt \\le x < t+1 \\Rightarrow t^2 \\le x[x] < t^2 + t.\n$$\n\n$44^2 + 44 = 1980 < 2016 < 45^2$, so there are no solutions among positive numbers.\n\nSuppose $x < 0$. Let $t = [x] < 0$, so $-t > 0$. Then:\n\n$$\nt \\le x < t+1 \\Rightarrow -t-1 \\le -x < -t \\Rightarrow t^2 + t \\le x[x] \\le t^2.\n$$\n\n$44^2 = 1936 < 2016$ and $46^2 = 2116 > 2016$, so only $t = -45$ is possible. Indeed, $45^2 - 45 = 1980 < 2016 < 45^2 = 2025$. Let $x = -45 + y$, $0 \\le y < 1$:\n\n$$\nx[x] = -45 \\cdot (-45 + y) = 2025 - 45y = 2016 \\Rightarrow 45y = 9 \\Rightarrow y = \\frac{1}{5}.\n$$\n\nThus, the solution is $x = -45 + \\frac{1}{5} = -\\frac{224}{5}$.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 14650,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $AB$ and $CD$ be two parallel chords of a circle. Prove that if a point $P$ in the plane satisfies $|PA| = |PB|$, then $|PC| = |PD|$.\n\n![](images/IRL_ABooklet_2023_p26_data_333d13e0e2.png)",
    "options": [],
    "answer": "See solution",
    "solution": "A point $X$ is on the perpendicular bisector of $AB$ if and only if $|XA| = |XB|$. Therefore, the center $O$ of the circle is on the perpendicular bisectors of $AB$ and $CD$. Since $AB \\parallel CD$, the two chords share the same perpendicular bisector. If $|PA| = |PB|$, then $P$ must be on this common perpendicular bisector, so $|PC| = |PD|$ as well.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 14651,
    "subject": "Mathematics (Olympiad)",
    "question": "Find the minimum value of $x^2 + y^2 + z^2$ where $x$, $y$, $z$ are real numbers such that $x^3 + y^3 + z^3 - 3xyz = 1$.",
    "options": [],
    "answer": "See solution",
    "solution": "The condition $x^3 + y^3 + z^3 - 3xyz = 1$ can be factorized as\n\n$$\n(x + y + z)(x^2 + y^2 + z^2 - xy - yz - zx) = 1. \\quad (1)\n$$\n\nLet $A = x^2 + y^2 + z^2$ and $B = x + y + z$. Notice that $B^2 - A = 2(xy + yz + zx)$.\n\nNote also that\n\n$$\nx^2 + y^2 + z^2 - xy - yz - zx = \\frac{1}{2} \\left[ (x - y)^2 + (y - z)^2 + (z - x)^2 \\right] \\geq 0,\n$$\n\nand so from (1), $B > 0$.\n\nEquation (1) now becomes\n\n$$\n\\begin{aligned}\nB \\left( A - \\frac{B^2 - A}{2} \\right) &= 1 \\\\\n\\therefore 3A &= B^2 + \\frac{2}{B}.\n\\end{aligned}\n$$\n\nSince $B > 0$, we may apply AM-GM. Therefore,\n\n$$\n\\begin{aligned}\n3A &= B^2 + \\frac{2}{B} \\\\\n&= B^2 + \\frac{1}{B} + \\frac{1}{B} \\geq 3 \\\\\n\\therefore A \\geq 1.\n\\end{aligned}\n$$\n\nThe minimum $A = 1$ is attained by $(x, y, z) = (1, 0, 0)$, for example.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 14652,
    "subject": "Mathematics (Olympiad)",
    "question": "Maria and Bilyana play the following game. Maria has 2024 coins, and Bilyana has 2023 fair coins. Each coin is tossed randomly, with the probability of each individual coin being heads after the toss equal to $\\frac{1}{2}$. Maria wins if there are strictly more heads among her coins than Bilyana's; otherwise, Bilyana wins. What is the probability that Maria wins?",
    "options": [],
    "answer": "See solution",
    "solution": "Let $p$ be the probability that Maria has more heads than Bilyana after tossing the first 2023 of Maria's coins. By symmetry, the probability that Maria has fewer heads than Bilyana is also $p$, so the probability that Maria and Bilyana have an equal number of heads is $1 - 2p$.\n\nIf Maria has fewer heads than Bilyana, her chance of winning is $0$ (regardless of the last coin). If she has more heads, her chance of winning is $1$. If they have an equal number, Maria wins only if her last coin is heads, which has probability $\\frac{1}{2}$.\n\nThus, the probability that Maria wins is:\n$$\np + \\frac{1 - 2p}{2} = \\frac{1}{2}.\n$$\n$\\square$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 14653,
    "subject": "Mathematics (Olympiad)",
    "question": "Written on a blackboard are the 2023 numbers\n\n2023, 2023, ..., 2023.\n\nThe numbers on the blackboard are now modified, in a sequence of moves. In each move, two numbers on the blackboard—call them $x$ and $y$—are chosen, deleted, and replaced by the single number $\\frac{x+y}{4}$. Such moves are carried out until there is only one number left on the blackboard.\n\nProve that this number is always greater than 1.",
    "options": [],
    "answer": "See solution",
    "solution": "The expression $\\frac{x+y}{4}$ reminds us of the arithmetic mean. By the AM-HM inequality, we have\n\n$$\n\\frac{x+y}{2} \\geq \\frac{2}{\\frac{1}{x} + \\frac{1}{y}}\n$$\n\nor\n\n$$\n\\frac{1}{x} + \\frac{1}{y} \\geq \\frac{1}{(x+y)/4}\n$$\n\nThis inequality leads us to consider an argument concerning the reciprocals of the numbers on the board, as the sum of the reciprocals of two of these numbers is at least as large as the reciprocal of the number replacing them. This value remains the same if and only if the two chosen numbers are equal, and is otherwise larger. At the beginning, the sum of all reciprocals is\n\n$$\n\\frac{1}{2023} + \\frac{1}{2023} + \\dots + \\frac{1}{2023} = \\frac{2023}{2023} = 1.\n$$\n\nThis implies the claim, since there is an odd number of 2023s in the beginning that cannot be divided into pairs, so one of them has to be part of a pair with different numbers.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 14654,
    "subject": "Mathematics (Olympiad)",
    "question": "Solve, in the positive integers, the equation\n\n$$\n5^m + n^2 = 3^p.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "Obviously $n$ is even, so let $n = 2a$, $a \\in \\mathbb{N}^*$. Reducing the equation modulo 4 yields $1 = (-1)^p$, so $p$ is even, $p = 2b$, $b \\in \\mathbb{N}^*$.\n\nThen we have $(3^b - 2a)(3^b + 2a) = 5^m$, hence\n\n$$\n\\begin{cases}\n3^b - 2a = 5^x \\\\\n3^b + 2a = 5^y\n\\end{cases}, \\quad \\text{where } x, y \\in \\mathbb{N}, x < y, x+y=m.\n$$\n\nAdding the above equations, we get $2 \\cdot 3^b = 5^x + 5^y = 5^x(1 + 5^{y-x})$, so $x = 0$ and $y = m$. Therefore,\n\n$$\n5^m + 1 = 2 \\cdot 3^b.\n$$\n\nReducing modulo 6, it results that $m$ is an odd number.\n\nSuppose that $b \\ge 2$; working modulo 9, we get that $(-4)^m + 1 = 0$, which implies $4^m \\equiv 1 \\pmod{9}$. Since $4^3 \\equiv 1 \\pmod{9}$, we have $3 \\mid m$. But $m$ is odd, so there exists $t \\in \\mathbb{N}$ so that $m = 6t + 3$. It follows that: $2 \\cdot 3^b = 5^m + 1 = 5^{6t+3} + 1 = 125^{2t+1} + 1 = (7 \\cdot 18 - 1)^{2t+1} + 1 \\equiv (-1)^{2t+1} + 1 \\equiv 0 \\pmod{7}$, absurd. Hence $b=1$, so $m=1$, $n=2$, $p=2$.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 14655,
    "subject": "Mathematics (Olympiad)",
    "question": "At the beginning, there were $x$ cows and $x$ horses on the farm. If the number of cows increases by $50\\%$, how does the ratio of horses to total animals change? What is the new percentage of horses among all animals?",
    "options": [],
    "answer": "See solution",
    "solution": "When the number of cows increases by $50\\%$, there are $x + \\frac{1}{2}x = \\frac{3}{2}x$ cows. The total number of animals is $x$ horses plus $\\frac{3}{2}x$ cows, which is $x + \\frac{3}{2}x = \\frac{5}{2}x$. The ratio of horses to total animals is $\\frac{x}{\\frac{5}{2}x} = \\frac{2}{5}$, which is $40\\%$, not $30\\%$. Therefore, the correct answer is (E).",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 14656,
    "subject": "Mathematics (Olympiad)",
    "question": "給定整數 $M \\ge 2$。試確定最小的實數 $C_M$ 使滿足性質：對於任意 $2023$ 個整數 $a_1, a_2, \\dots, a_{2023}$，總可以決定一個整數 $1 \\le k < M$ 使得\n\n$$\n\\left\\{ \\frac{ka_1}{M} \\right\\} + \\left\\{ \\frac{ka_2}{M} \\right\\} + \\dots + \\left\\{ \\frac{ka_{2023}}{M} \\right\\} \\le C_M.\n$$\n\n這裡 $0 \\le \\{x\\} < 1$ 為 $x$ 的小數部分。",
    "options": [],
    "answer": "See solution",
    "solution": "答案是 $1011 + \\frac{1}{M}$。\n\n我們可以通過選取 $a_{2i+1} = 1$ 和 $a_{2i} = -1$ 來證明 $C_M \\ge 1011 + \\frac{1}{M}$。這樣對於所有 $k = 1, \\dots, M-1$，有：\n\n$$\n\\left\\{ \\frac{ka_1}{M} \\right\\} + \\left\\{ \\frac{ka_2}{M} \\right\\} + \\dots + \\left\\{ \\frac{ka_{2023}}{M} \\right\\} = 1012 \\cdot \\frac{k}{M} + 1011 \\cdot \\frac{M-k}{M} = 1011 + \\frac{k}{M} \\ge 1011 + \\frac{1}{M}.\n$$\n\n為了證明這是最優的，設\n\n$$\nP_i := \\sum_{k=1}^{M-1} \\left\\{ \\frac{ka_i}{M} \\right\\}, \\quad Q_k = \\sum_{i=1}^{2023} \\left\\{ \\frac{ka_i}{M} \\right\\}.\n$$\n\n我們聲稱對於任意整數 $a$，有 $\\sum_{k=1}^{M-1} \\left\\{ \\frac{ka}{M} \\right\\} \\le \\frac{M-1}{2}$，且當 $\\gcd(a, M) = 1$ 時取等號。證明如下：設 $d = \\gcd(a, M)$，則\n\n$$\n\\sum_{k=1}^{M-1} \\left\\{ \\frac{ka}{M} \\right\\} = \\frac{M-1}{2}.\n$$\n\n因此對於每個 $a_i$，有 $P_i \\le \\frac{M-1}{2}$，所以 $\\sum_{k=1}^{M-1} Q_k = \\sum_{i=1}^{2023} P_i \\le \\frac{2023(M-1)}{2}$。\n\n如果存在某個 $Q_k \\le 1011$，則已經滿足條件。否則假設對所有 $k$，$\\lceil Q_k \\rceil \\ge 1012$，則\n\n$$\n\\{-Q_k\\} = \\lceil Q_k \\rceil - Q_k \\ge 1012 - Q_k.\n$$\n\n因此，\n\n$$\n\\sum_{k=1}^{M-1} \\{-Q_k\\} \\ge 1012(M-1) - \\frac{2023}{2}(M-1) = \\frac{1}{2}(M-1).\n$$\n\n注意\n\n$$\n\\{-Q_k\\} = \\left\\{ \\frac{-k(a_1 + \\cdots + a_{2023})}{M} \\right\\}.\n$$\n\n對 $a = a_1 + \\cdots + a_{2023}$ 應用上述結論，得到所有等號成立，故 $\\gcd(a, M) = 1$，且 $\\lceil Q_k \\rceil = 1012$ 對所有 $k$ 成立。特別地，若選 $k$ 為 $a$ 在模 $M$ 下的逆元，則 $Q_k = 1011 + \\frac{1}{M} \\le C_M$，如所需。\n\n**備註：** 若 $2023$ 換成偶數 $2N$，則 $C_M = N$。若換成奇數 $2N+1$，則答案為 $N + \\frac{1}{M}$。若 $M$ 為質數，解法可略簡化。\n\n**補充：** 本題幾乎可由 $Q_k$ 的平均值推得，僅需最後一點微調。",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 14657,
    "subject": "Mathematics (Olympiad)",
    "question": "Given a positive integer $n$, determine the largest real number $\\mu$ such that for every configuration $C$ of $4n$ points in an open unit square $U$, there exists an open rectangle in $U$, with sides parallel to those of $U$, which contains exactly one point of $C$ and has area at least $\\mu$.",
    "options": [],
    "answer": "See solution",
    "solution": "The required maximum is $\\mu = \\frac{1}{2n+2}$.\n\nTo show that the condition fails for $\\mu > \\frac{1}{2n+2}$, let $U = (0,1) \\times (0,1)$, choose a small $\\varepsilon > 0$, and consider the configuration $C$ consisting of $n$ clusters of 4 points each at $(\\frac{i}{n+1} \\pm \\varepsilon, \\frac{1}{2} \\pm \\varepsilon)$ for $i = 1, \\dots, n$, with all sign combinations. Any open rectangle in $U$ containing exactly one point of $C$ has area at most $(\\frac{1}{n+1} + \\varepsilon)(\\frac{1}{2} + \\varepsilon) < \\mu$ for small enough $\\varepsilon$.\n\nNow, for any finite configuration $C$ in $U$, there always exists an open rectangle in $U$ (sides parallel to $U$) containing exactly one point of $C$ and area at least $\\mu_0 = \\frac{2}{|C| + 4}$.\n\nWe use two lemmas:\n\n**Lemma 1.** Let $k$ be a positive integer and $\\lambda < \\frac{1}{|k/2| + 1}$. If $t_1, \\dots, t_k$ are distinct points in $(0,1)$, then some $t_i$ is isolated from the others by an open subinterval of length at least $\\lambda$.\n\n**Lemma 2.** For $k \\ge 2$ and positive integers $m_1, \\dots, m_k$,\n$$\n\\lfloor m_1/2 \\rfloor + \\sum_{i=1}^{k} \\lfloor m_i/2 \\rfloor + \\lfloor m_k/2 \\rfloor \\le \\sum_{i=1}^{k} m_i - k + 2.\n$$\n\nProject $C$ onto the $x$-axis to get $x_1 < \\dots < x_k$ in $(0,1)$, let $\\ell_i$ be the vertical through $x_i$, and $m_i = |C \\cap \\ell_i|$. Set $x_0 = 0$, $x_{k+1} = 1$. If $x_{i+1} - x_{i-1} > (\\lfloor m_i/2 \\rfloor + 1)\\mu_0$ for some $i$, apply Lemma 1 to isolate a point in $C \\cap \\ell_i$ by an open subinterval $x_i \\times J$ of length at least $\\mu_0/(x_{i+1} - x_{i-1})$. Then $(x_{i-1}, x_{i+1}) \\times J$ is an open rectangle in $U$ containing exactly one point of $C$ and area at least $\\mu_0$.\n\nIf $x_{i+1} - x_{i-1} \\le (\\lfloor m_i/2 \\rfloor + 1)\\mu_0$ for all $i$, then $k > 1$ and $x_1 - x_0 < x_2 - x_0 \\le (\\lfloor m_1/2 \\rfloor + 1)\\mu_0$, $x_{k+1} - x_k < x_{k+1} - x_{k-1} \\le (\\lfloor m_k/2 \\rfloor + 1)\\mu_0$. Using Lemma 2,\n$$\n\\begin{align*}\n2 = 2(x_{k+1} - x_0) &= (x_1 - x_0) + \\sum_{i=1}^{k} (x_{i+1} - x_{i-1}) + (x_{k+1} - x_k) \\\\\n&< \\left( \\lfloor m_1/2 \\rfloor + 1 + \\sum_{i=1}^{k} (\\lfloor m_i/2 \\rfloor + 1) + \\lfloor m_k/2 \\rfloor + 1 \\right) \\mu_0 \\\\\n&\\le (|C| + 4)\\mu_0 = 2,\n\\end{align*}\n$$\nwhich is a contradiction.\n\n**Proof of Lemma 1.** Suppose no $t_i$ is isolated by an open subinterval of length at least $\\lambda$. Let $0 = t_0 < t_1 < \\dots < t_k < t_{k+1} = 1$. The interval $(t_{i-1}, t_{i+1})$ isolates $t_i$ and has length $< \\lambda$. If $k$ is odd, $1 = \\sum_{i=0}^{(k-1)/2} (t_{2i+2} - t_{2i}) < \\lambda(1 + (k-1)/2) < 1$; if $k$ is even, $1 < 1 + t_k - t_{k-1} = \\sum_{i=0}^{k/2-1} (t_{2i+2} - t_{2i}) + (t_{k+1} - t_{k-1}) < \\lambda(1 + k/2) < 1$. Contradiction.\n\n**Proof of Lemma 2.** Let $I_0$ be indices $i$ ($2 \\le i \\le k-1$) with $m_i$ even, $I_1$ those with $m_i$ odd. $\\sum_{i=2}^{k-1} m_i \\ge 2|I_0| + |I_1| = 2(k-2) - |I_1|$, so $|I_1| \\ge 2(k-2) - \\sum_{i=2}^{k-1} m_i$. Thus,\n$$\n\\begin{align*}\n\\lfloor m_1/2 \\rfloor + \\sum_{i=1}^{k} \\lfloor m_i/2 \\rfloor + \\lfloor m_k/2 \\rfloor &\\le m_1 + \\left( \\sum_{i=2}^{k-1} m_i/2 - |I_1|/2 \\right) + m_k \\\\\n&\\le m_1 + \\left( \\sum_{i=2}^{k-1} m_i/2 - (k-2) + \\sum_{i=2}^{k-1} m_i/2 \\right) + m_k \\\\\n&= \\sum_{i=1}^{k} m_i - k + 2.\n\\end{align*}\n$$\n\n_Remark_: If $4n$ is replaced by a positive integer $k$ not divisible by 4, the maximal $\\mu$ is unknown.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 14658,
    "subject": "Mathematics (Olympiad)",
    "question": "Suppose $w, x, y, z$ are four consecutive terms in Bowie's list, where $w$ is known to be a multiple of 5. To establish that the pattern continues, we want to show that $z$ is also a multiple of 5.",
    "options": [],
    "answer": "See solution",
    "solution": "**Alternative ii**\n\nConsider the zigzag that can be tiled in $z$ ways. The dashed line is either crossed by a tile or it isn't.\n\n![](images/2024_The_Australian_Scene_Final_p52_data_19db1234ea.png)\n\nIf the dashed line isn't crossed, then there are 5 ways of tiling the first two blocks and $x$ ways of tiling the remainder of the zigzag.\n\nIf the dashed line is crossed by a tile, then that forces the following partial tiling.\n\n![](images/2024_The_Australian_Scene_Final_p52_data_a890c1291f.png)\n\nIn this case, there are 2 ways of tiling the first block and $w$ ways of tiling the remainder of the zigzag.\n\nSo $z = 5x + 2w$ and given that $w$ is a multiple of 5, so is $z$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 14659,
    "subject": "Mathematics (Olympiad)",
    "question": "Andriy and Olesia play the following game:\n\nFirst, Andriy chooses any chess piece (king, queen, rook, bishop, or knight) and places it on the chessboard. Then, they take turns moving the piece according to its standard chess rules. However, it is not allowed to move the piece to a square that was already visited, including the starting square. The player who cannot make a move loses. Who wins if both play optimally?\n\n_Reminder_: A knight moves in an L-shape (two squares in one direction and one in the perpendicular direction). A bishop moves any number of squares diagonally. A rook moves any number of squares horizontally or vertically. A king moves one square in any direction.\n\n![](images/ukraine_2015_Booklet_p7_data_cf39d539ca.png)",
    "options": [],
    "answer": "See solution",
    "solution": "For each chess piece, the chessboard can be partitioned into pairs of squares connected by a legal move of that piece. Olesia's winning strategy is as follows: whenever Andriy moves the piece to a square in a pair, Olesia responds by moving it to the other square in that pair. This ensures that after each of her moves, for every pair, either both squares are used or neither is used, so she always has a move as long as Andriy does.\n\n- For the king, queen, and rook, pair neighboring squares horizontally in columns 1 & 2, 3 & 4, 5 & 6, and 7 & 8 (see figure above).\n- For the knight, pair squares in columns 1 & 3, 2 & 4, 5 & 7, and 6 & 8 (see figure 20).\n- For the bishop, pair neighboring squares along diagonals of the same color, focusing on diagonals with an even number of squares, parallel to the main diagonal.\n\nThus, Olesia can always win if both play optimally.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 14660,
    "subject": "Mathematics (Olympiad)",
    "question": "Si $n$ es un número natural, el $n$\\-ésimo número triangular es $T_n = 1 + 2 + \\cdots + n$. Hallar todos los valores de $n$ para los que el producto de los 16 números triangulares consecutivos $T_n T_{n+1} \\cdots T_{n+15}$ es un cuadrado perfecto.",
    "options": [],
    "answer": "See solution",
    "solution": "Como $T_n = \\dfrac{n(n+1)}{2}$, el producto de los 16 números triangulares es\n$$\nP_n = T_n T_{n+1} \\cdots T_{n+15} = \\frac{n(n+1)}{2} \\cdot \\frac{(n+1)(n+2)}{2} \\cdots \\frac{(n+15)(n+16)}{2}\n$$\nEsto se puede escribir como\n$$\nP_n = \\frac{n(n+1)^2 (n+2)^2 \\cdots (n+15)^2 (n+16)}{2^{16}}\n$$\nObservamos que $(n+1)^2 \\cdots (n+15)^2$ es un cuadrado perfecto, así que $P_n$ será un cuadrado perfecto si y sólo si $n(n+16)$ es un cuadrado perfecto.\n\nComo $n$ y $n+16$ son coprimos salvo posiblemente por el 2, para que $n(n+16)$ sea un cuadrado perfecto, ambos deben ser cuadrados o el doble de un cuadrado. Escribimos $n = 2^a m^2$ y $n+16 = 2^b t^2$ con $a, b = 0$ o $1$.\n\nComo $n$ y $n+16$ tienen la misma paridad, sólo son posibles $a = b = 0$ o $a = b = 1$.\n\n- Si $a = b = 0$, $n = m^2$, $n+16 = t^2$, así que $t^2 - m^2 = 16 \\implies (t-m)(t+m) = 16$. Las soluciones enteras positivas son $t-m = 2$, $t+m = 8$ $\\implies t = 5$, $m = 3$, así que $n = 9$.\n- Si $a = b = 1$, $n = 2m^2$, $n+16 = 2t^2$, así que $2t^2 - 2m^2 = 16 \\implies t^2 - m^2 = 8$. Las soluciones enteras positivas son $t-m = 2$, $t+m = 4$ $\\implies t = 3$, $m = 1$, así que $n = 2$.\n\nPor lo tanto, los valores de $n$ son $n = 2$ y $n = 9$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 14661,
    "subject": "Mathematics (Olympiad)",
    "question": "Show that for any real $x > 0$ and integer $n > 0$ we have\n\n$$\nx^n + \\frac{1}{x^n} - 2 \\ge n^2 \\left(x + \\frac{1}{x} - 2\\right).\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "*Solution.* Without loss of generality, assume that $y = \\sqrt{x} > 1$. The identity\n\n$$\na^2 + \\frac{1}{a^2} - 2 = (a - \\frac{1}{a})^2\n$$\n\nreduces the problem to showing that\n\n$$\ny^n - \\frac{1}{y^n} \\ge n\\left(y - \\frac{1}{y}\\right)\n$$\n\nor $y^{2n} - n(y^{n+1} - y^{n-1}) - 1 \\ge 0$. Upon division by $y - 1 > 0$, this follows from the following computation:\n\n$$\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) \\\\\n&= \\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": 14662,
    "subject": "Mathematics (Olympiad)",
    "question": "a) There are $2n$ rays marked in a plane, with $n$ a natural number. 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 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": "See solution",
    "solution": "b) Yes.\n\nConsider any circle such that all endpoints of the marked rays are inside the circle. Choose a tangent line to the circle that is not parallel to any marked ray. Let this tangent line be $l_0$, and let $l_\\alpha$ be the tangent line obtained 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 $l_\\alpha$. Since $l_0$ and $l_\\pi$ are two parallel lines and all endpoints of the marked rays are between them, we know $f(0) + f(\\pi) = 2n$. Without loss of generality, let $f(0) \\leq n$ and $f(\\pi) \\geq n$. Because no two rays have the same direction, as $\\alpha$ increases continuously, $f(\\alpha)$ can change by at most one at any time. Thus, $f(\\alpha)$ ranges over all integer values between $f(0)$ and $f(\\pi)$, so $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, so it solves both parts of the problem.\n\n**Remark:** Part a) can be solved otherwise. Choose an arbitrary line $l$ that is not parallel to any marked ray or any line connecting the initial points of two marked rays, and from which the initial points of all marked rays lie on one side. Let $l'$ be a line parallel to $l$ such that the initial points of all marked rays lie between $l$ and $l'$. Then every marked ray intersects exactly one of the lines $l$ and $l'$, so the numbers of intersection points always sum to exactly $2n$. Without loss of generality, let the number of intersection points on $l$ be less than or equal to those on $l'$. When shifting $l$ towards $l'$ while keeping its direction unchanged, the number of intersection points on it can change by at most one at any time. Thus, there exists a position where the number of intersection points equals $n$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 14663,
    "subject": "Mathematics (Olympiad)",
    "question": "Vitalii, Michael, and Olexandr were each given $n$ dollars by their mother. Vitalii had to spend all his money on books, Michael on notebooks, and Olexandr on pens. Vitalii bought 1 book, Michael bought 2 notebooks, and Olexandr bought 5 pens. After these purchases, together they had $n$ dollars left. Prove that one of the boys can buy one more item.",
    "options": [],
    "answer": "See solution",
    "solution": "Assume the contrary. Let $q_1$, $q_2$, and $q_3$ be the costs of one book, notebook, and pen, respectively, and $r_1$, $r_2$, $r_3$ the change each boy received. Clearly, $r_1 < q_1$, $r_2 < q_2$, $r_3 < q_3$. From the statement:\n\n$$\n n = q_1 + r_1 \\\\\n n = 2q_2 + r_2 \\\\\n n = 5q_3 + r_3\n$$\n\nWe can write the following inequalities:\n- $n = q_1 + r_1 > 2r_1$ so $r_1 < \\frac{1}{2}n$\n- $n = 2q_2 + r_2 > 3r_2$ so $r_2 < \\frac{1}{3}n$\n- $n = 5q_3 + r_3 > 6r_3$ so $r_3 < \\frac{1}{6}n$\n\nAdding these:\n\n$$\n n = r_1 + r_2 + r_3 < \\frac{n}{2} + \\frac{n}{3} + \\frac{n}{6} = n\n$$\n\nwhich is impossible. This contradiction completes the proof.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 14664,
    "subject": "Mathematics (Olympiad)",
    "question": "定義兩個多項式之間的大小 $f(x) \\geq g(x)$ 為：從最高次項開始向下比較係數大小。\n\n若 $f(x) = \\sum_{i=0}^{n} a_i x^i$, $g(x) = \\sum_{i=0}^{n} b_i x^i$（$a_n, b_n$ 可以為 0），且存在 $r$ 使得 $\\forall i > r, a_i = b_i$ 且 $a_r > b_r$，或是 $f(x) = g(x)$，則稱 $f(x) \\geq g(x)$。\n\n試證：若 $f, g$ 首項係數為正，則 $f(f(x)) + g(g(x)) \\geq f(g(x)) + g(f(x))$。",
    "options": [],
    "answer": "See solution",
    "solution": "容易驗證上述定義的 $\\geq$ 滿足左右同加同減某個多項式。\n\n1. 假設不成立，則以下證明矛盾。\n\n以下使用數學歸納法證明 $a_i = b_i, \\forall i = 1, 2, \\ldots, n$。\n\n先證明 $a_n = b_n$：考慮比較 $x^{n^2}$ 係數，左式為 $a_n^{n+1} + b_n^{n+1}$，右式為 $a_n^n b_n + a_n b_n^n$。左邊 $\\geq$ 右邊且等號成立在 $a_n = b_n$，但由假設可得等號必須成立。\n\n假設 $a_i = b_i, \\forall i = k, k+1, \\ldots, n$ 成立，則原式可以寫為：\n\n$$\n\\begin{aligned}\n& a_n f(x)^n + a_{n-1} f(x)^{n-1} + \\cdots + a_k f(x)^k + \\cdots + a_0 \\\\\n& + b_n g(x)^n + b_{n-1} g(x)^{n-1} + \\cdots + b_k g(x)^k + \\cdots + b_0 \\\\\n\\geq & a_n g(x)^n + a_{n-1} g(x)^{n-1} + \\cdots + a_k g(x)^k + \\cdots + a_0 \\\\\n& + b_n f(x)^n + b_{n-1} f(x)^{n-1} + \\cdots + b_k f(x)^k + \\cdots + b_0\n\\end{aligned}\n$$\n\n等價於\n\n$$\n\\begin{aligned}\n& a_{k-1} f(x)^{k-1} + \\cdots + a_0 + b_{k-1} g(x)^{k-1} + \\cdots + b_0 \\\\\n\\geq & a_{k-1} g(x)^{k-1} + \\cdots + a_0 + b_{k-1} f(x)^{k-1} + \\cdots + b_0\n\\end{aligned}\n$$\n\n不難發現在左右兩式中 $x^q$ 係數會一樣 $\\forall q > n(k-2)+k-1$（因為係數都是在 $a_n$ 到 $a_k$ 中選取），比較 $x^{n(k-2)+k-1}$ 項係數，左式為 $(k-1)a_n^{k-2}(a_{k-1}^2 + b_{k-1}^2)$，右式為 $(k-1)a_n^{k-2} \\times 2a_{k-1}b_{k-1}$，由算幾不等式得左邊 $\\geq$ 右邊且等號成立在 $a_{k-1} = b_{k-1}$，但由假設可得等號必須成立。\n\n由數學歸納法得 $f(x) = g(x)$，矛盾！\n\n2. $f(x) \\geq g(x)$ 等價於存在 $M$ 足夠大使得 $\\forall m > M, f(m) \\geq g(m)$。不失一般性，$f(x) \\geq g(x)$，又原式等價於 $(f-g)(f(x)) \\geq (f-g)(g(x))$，且 $f-g$ 首項係數為正，則存在足夠大的 $M$ 使得 $\\forall m > M$：\n(i) $(f-g)(m)$ 遞增，(ii) $f(m) \\geq g(m)$，則 $(f-g)(f(m)) \\geq (f-g)(g(m))$，故得證。",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 14665,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $ABCD$ be a convex quadrilateral such that $AB^2 + BC^2 = AD^2 + CD^2$. Points $X$ and $Y$ are chosen such that $XD \\perp CD$, $XB \\perp AB$, $YB \\perp BC$ and $YD \\perp AD$. Let lines $AC$ and $XY$ meet at $T$ and $M$ be the midpoint of segment $XY$. Prove that points $T, M, B, D$ lie on a circle.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $P$ be the midpoint of $AC$. Applying the formula for the length of the median on $\\triangle ABC$ and $\\triangle ADC$, and using the fact that $AB^2 + BC^2 = AD^2 + CD^2$, we obtain $BP = DP$.\n\n$$\n\\text{Claim.} \\quad \\angle BXA = \\angle PBD\n$$\n\n*Proof.* We use directed angles.\n\nLet $U$ be the projection of $A$ onto $DX$ and $N$ be the midpoint of $UD$.\n\n![](images/2025-SL-b_p7_data_49aa7f4336.png)\n\nObserve that $AU \\parallel CD$ and $P, N$ are the midpoints of $AC, UD$, so $PN \\parallel CD$. Since $DU \\perp CD$, we get that $PU = PD = PB$, so $P$ is the circumcenter of $\\triangle DUB$. Also observe that $A, U, B, X$ are concyclic ($\\angle AUX = \\angle ABX = 90^\\circ$), so we can infer that\n\n$$\n\\angle PBD = 90^\\circ - \\angle DUB = 90^\\circ - \\angle XUB = 90^\\circ - \\angle XAB = \\angle BXA.\n$$\n\n$\\Box$\n\nBy projecting $C$ onto $YD$, we can similarly prove that $\\angle BYC = \\angle PBD$, so $\\angle BYC = \\angle BXA$. We also have $\\angle XBA = 90^\\circ$ and $\\angle YBC = 90^\\circ$, meaning $\\triangle BXA \\sim \\triangle BYC$. This is a spiral similarity centered at $B$ sending $AX$ to $CY$. It follows that $B$ is also the center of spiral similarity sending $AC$ to $XY$.\n\n![](images/2025-SL-b_p7_data_34e3d6a04b.png)\n\nSince $XB \\perp AB$ and $YB \\perp CB$, the angle of rotation in the spiral similarity $\\triangle BAC \\sim \\triangle BXY$ is $90^{\\circ}$. This means we also have $AC \\perp XY$, thus $T$ is the projection of $P$ onto $XY$, so $PT \\perp TM$. Moreover, the spiral similarity sends $P$ to $M$, so $BP \\perp BM$. We can similarly prove that $DP \\perp DM$, so points $T, M, B, D$ lie on the circle with diameter $PM$ and we are done.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 14666,
    "subject": "Mathematics (Olympiad)",
    "question": "Given an integer $k \\ge 2$, determine the largest number of divisors the binomial coefficient $\\binom{n}{k}$ may have in the range $n-k+1, \\dots, n$, as $n$ runs through the integers greater than or equal to $k$.",
    "options": [],
    "answer": "See solution",
    "solution": "The required maximum is $k-1$ and is achieved, for instance, at $n = k!$. To complete the proof, we now show that at least one of the $k$ numbers $\\frac{1}{n-j}\\binom{n}{k}$, $j = 0, 1, \\dots, k-1$, is not an integer. To this end, we exhibit a $\\mathbb{Z}$-linear combination of these numbers which is not an integer. For instance,\n\n$$\n\\sum_{j=0}^{k-1} (-1)^{k-j-1} \\binom{k-1}{j} \\cdot \\frac{1}{n-j} \\binom{n}{k} = \\frac{1}{k} \\sum_{j=0}^{k-1} \\prod_{i \\neq j} \\frac{n-i}{j-i} = \\frac{1}{k}\n$$\n\nis not an integer, since $k \\ge 2$. The leftmost equality above is easily proved by noticing that the polynomial\n\n$$\n\\sum_{j=0}^{k-1} \\prod_{i \\neq j} \\frac{X-i}{j-i}\n$$\n\nhas degree at most $k-1$, and takes on the value 1 at $k$ distinct points, namely, $0, 1, \\dots, k-1$, so it is identically 1.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 14667,
    "subject": "Mathematics (Olympiad)",
    "question": "Let the circles $k_1$ and $k_2$ intersect at two distinct points $A$ and $B$, and let $t$ be a common tangent of $k_1$ and $k_2$, touching $k_1$ and $k_2$ at $M$ and $N$, respectively. If $t \\perp AM$ and $MN = 2AM$, evaluate $\\angle NMB$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $P$ be the symmetric of $A$ with respect to $M$. Then $AM = MP$ and $t \\perp AP$, so triangle $APN$ is isosceles with $AP$ as its base, thus $\\angle NAP = \\angle NPA$. We have $\\angle BAP = \\angle BAM = \\angle BMN$ and $\\angle BAN = \\angle BNM$.\n\nThus,\n$$\n180^\\circ - \\angle NBM = \\angle BNM + \\angle BMN = \\angle BAN + \\angle BAP = \\angle NAP = \\angle NPA,\n$$\nso quadrilateral $MBNP$ is cyclic (since $B$ and $P$ lie on different sides of $MN$). Hence $\\angle APB = \\angle MPB = \\angle MNB$ and triangles $APB$ and $MNB$ are congruent ($MN = 2AM = AM + MP = AP$). Therefore, $AB = MB$, i.e., triangle $AMB$ is isosceles, and since $t$ is tangent to $k_1$ and perpendicular to $AM$, the center of $k_1$ is on $AM$, so $AMB$ is a right-angled triangle. From this, $\\angle AMB = 45^\\circ$, and so $\\angle NMB = 90^\\circ - \\angle AMB = 45^\\circ$.\n\n![](images/makedonija2012_p9_data_71829650c9.png)\n\nLet $C$ be the intersection point of $MN$ and $AB$. Then $CN^2 = CB \\cdot CA$ and $CM^2 = CB \\cdot CA$, so $CM = CN$. But $MN = 2AM$, so $CM = CN = AM$, thus the right triangle $ACM$ is isosceles, hence $\\angle NMB = \\angle CMB = \\angle BCM = 45^\\circ$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 14668,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $f(x) = a(3a + 2c)x^2 - 2b(2a + c)x + b^2 + (c + a)^2$ ($a, b, c \\in \\mathbb{R}$) be a quadratic function such that $f(x) \\le 1$ for all $x \\in \\mathbb{R}$. Find the maximum of $|ab|$.",
    "options": [],
    "answer": "See solution",
    "solution": "Rewrite $f(x) = a(3a + 2c)x^2 - 2b(2a + c)x + b^2 + (c + a)^2$ as a quadratic polynomial in $c$:\n\n$$\nf(x) = c^2 + 2(ax^2 - bx + a)c + 3(a x)^2 - 4abx + b^2 + a^2 \\le 1.\n$$\n\nComplete the square and simplify, obtaining\n\n$$\n(c + a x^2 - b x + a)^2 + (1 - x^2)(a x - b)^2 \\le 1 \\implies (1 - x^2)(a x - b)^2 \\le 1.\n$$\n\nLet $x = \\pm\\frac{\\sqrt{3}}{3}$, then $\\frac{4\\sqrt{3}}{3}|ab| \\le \\frac{3}{2}$, so $|ab| \\le \\frac{3\\sqrt{3}}{8}$, and the equality holds when $a = \\frac{3\\sqrt{2}}{4}$, $b = -\\frac{\\sqrt{6}}{4}$, $c = -\\frac{5\\sqrt{2}}{4}$. We have\n\n$$\nf(x) = -\\frac{3}{8}x^2 + \\frac{\\sqrt{3}}{4}x + \\frac{7}{8} = -\\frac{3}{8}\\left(x - \\frac{\\sqrt{3}}{3}\\right)^2 + 1 \\le 1.\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 14669,
    "subject": "Mathematics (Olympiad)",
    "question": "Suppose we have an infinite grid of lamps, all initially off. We are allowed to select certain rectangles or squares and toggle (switch on/off) all lamps inside them. \n\n(a) Is it possible, by using only $12 \\times 1$, $20 \\times 1$, and $15 \\times 1$ rectangles, to turn on exactly one lamp (i.e., achieve a configuration where only a single $1 \\times 1$ square has its lamp on)?\n\n(b) For each of the following pairs of allowed moves, is it possible to turn on only the lamps inside a $2 \\times 2$ square (with all other lamps off)? The allowed moves are:\n\n- $3 \\times 3$ and $5 \\times 5$ squares,\n- $3 \\times 3$ and $4 \\times 4$ squares,\n- $4 \\times 4$ and $5 \\times 5$ squares.\n\nJustify your answers.\n\n![](images/ARG_Other_2024_p29_data_b32ac54991.png)",
    "options": [],
    "answer": "See solution",
    "solution": "We can construct a sequence of moves to switch all lamps inside a $20 \\times 1$ rectangle (by overlapping $20 \\times 4$ and $20 \\times 5$ rectangles), or inside a $15 \\times 1$ rectangle (by overlapping $15 \\times 6$ and $15 \\times 5$ rectangles).\n\nSince $\\gcd(12, 20, 15) = 1$, it is possible to get a $1 \\times 1$ square with its lamp *on* by combining those rectangles. For example: we turn 40 lamps *on* using two $20 \\times 1$ rectangles, then we turn the last 15 lamps *off* with a $15 \\times 1$ rectangle, and finally we turn further 24 lamps *off* by using two $12 \\times 1$ rectangles.\n\n**ALGORITHM 2.** With two $4 \\times 4$ squares and two $5 \\times 5$ squares we can achieve that inside a $9 \\times 9$ square all lamps are *on*, except for the one at the center. Now, we can divide the $9 \\times 9$ square into nine $3 \\times 3$ squares and make moves on them to switch all 81 lamps. This leaves only the central lamp *on*, as wanted.\n\n(b) First, consider only $3 \\times 3$ and $5 \\times 5$ squares. Color the columns of the grid with the pattern: two black columns, one white column, two black, one white, etc. Label the rows as ..., $-3$, $-2$, $-1$, $0$, $1$, $2$, $3$, .... For each residue $r$ modulo 5, count how many rows with label $\\equiv r \\pmod 5$ have an odd number of black cells with their lamp *on*. These five numbers are $a, b, c, d, e$. Initially, all are 0. A move with a $3 \\times 3$ square does not alter the parity of these numbers, since each row has either 2 or 0 black cells inside any $3 \\times 3$ square. A move with a $5 \\times 5$ square either leaves all parities unchanged or changes all of them (since there is one row for each residue $r$, and all those rows have the same number of black cells).\n\nThus, $a, b, c, d, e$ are always all even or all odd. Therefore, it is impossible that after a sequence of moves the only lamps that are *on* are those inside a $2 \\times 2$ square if this square has 2 white cells and 2 black cells. By translating the coloring, we can always assume this is the case for our goal $2 \\times 2$ square.\n\nA similar argument works for the remaining two cases:\n\n- If the squares used are $3 \\times 3$ and $4 \\times 4$, color the columns with the pattern: 2 black, 2 white, 2 black, 2 white, ..., and classify rows according to their residue modulo 3.\n- If the squares used are $4 \\times 4$ and $5 \\times 5$, use the same coloring as in the previous case and classify rows modulo 5.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 14670,
    "subject": "Mathematics (Olympiad)",
    "question": "Given four distinct points $A$, $B$, $C$, and $D$ in the plane, suppose that for every triangle formed by any three of these points, the incircle has the same radius. Prove that the quadrilateral $ABCD$ is a rectangle, and that all four triangles formed by any three of the points are congruent.\n\n![](images/Japan_2011_p12_data_8ea66e17fd.png)",
    "options": [],
    "answer": "See solution",
    "solution": "We first prove the following lemma:\n\n*Lemma*: Suppose a triangle $T$ contains a triangle $T'$, and the lengths of the radii of their incircles are the same. Then $T = T'$.\n\n*Proof*: The incircle of a triangle is uniquely determined as the largest circle contained in the triangle. Let $\\Gamma$ and $\\Gamma'$ be the incircles of $T$ and $T'$, respectively. Since $T' \\subset T$, the radius of $\\Gamma'$ is at most that of $\\Gamma$. By assumption, the radii are equal, so $\\Gamma = \\Gamma'$ by uniqueness, and thus $T = T'$.\n\nNow, let $A$, $B$, $C$, $D$ be the four points. Assume $\\triangle ABC$ has the largest area among the four possible triangles. Let $\\ell_A$ be the line through $A$ parallel to $BC$. If $D$ lies on the opposite side of $\\ell_A$ from $B$ and $C$, then $\\triangle DBC$ would have larger area than $\\triangle ABC$, contradicting our assumption. Thus, $D$ must lie on $\\ell_A$ or the same side as $B$ and $C$.\n\nSimilarly, define $\\ell_B$ (through $B$ parallel to $AC$) and $\\ell_C$ (through $C$ parallel to $AB$). By analogous reasoning, $D$ lies inside (or on the boundary of) the triangle formed by $\\ell_A$, $\\ell_B$, and $\\ell_C$. Let $A'$, $B'$, and $C'$ be the intersections of $\\ell_B$ and $\\ell_C$, $\\ell_C$ and $\\ell_A$, and $\\ell_A$ and $\\ell_B$, respectively.\n\nSuppose $D$ lies inside $\\triangle ABC$. Then $\\triangle ABC$ contains $\\triangle ABD$, and by the lemma, since their incircle radii are equal, $D = C$, contradicting the distinctness of the points. Thus, $D$ lies outside $\\triangle ABC$. Without loss of generality, assume $D$ lies inside $\\triangle AB'C$. The quadrilateral $AB'CB$ is a parallelogram, so $\\triangle ABC$ and $\\triangle CB'A$ are congruent, and their incircle radii are equal. Since $\\triangle CDA$ is contained in $\\triangle CB'A$ and their incircle radii are equal, by the lemma, $D = B'$. Thus, $ABCD$ is a parallelogram, and $\\triangle ABC$ and $\\triangle BCD$ have equal area and incircle radii, so their perimeters are equal: $AB + BC + CA = BC + CD + DB$. Since $AB = CD$, it follows that $CA = DB$, so diagonals $AC$ and $BD$ are equal, and the parallelogram is a rectangle. Therefore, all four triangles formed by any three of $A$, $B$, $C$, $D$ are congruent.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 14671,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $p$ be a prime number. Prove that $p^2 + p + 1$ is not a perfect cube.",
    "options": [],
    "answer": "See solution",
    "solution": "Assume that $p^2 + p + 1 = k^3$. Then $p(p+1) = k^3 - 1 = (k-1)(k^2 + k + 1)$. Since $p$ is prime, either $p \\mid k-1$ or $p \\mid k^2 + k + 1$.\n\nIf $p \\mid k-1$, then $p \\leq k-1$ and $k^3 \\geq (p+1)^3 > p^2 + p + 1 = k^3$, which is impossible.\n\nNow consider $p \\mid k^2 + k + 1$. Since $p$ is prime, we have three cases:\n\n1) $p = 3$, which does not give a solution.\n\n2) $p \\equiv 1 \\pmod{3}$. Then $p^2 + p + 1 \\equiv 3 \\pmod{9}$, but no perfect cube is congruent to $3 \\pmod{9}$, so this is impossible.\n\n3) $p \\equiv 2 \\pmod{3}$. For $p = 2$, there is no solution. For odd $p$, $4(k^2 + k + 1)^2 \\equiv 0 \\pmod{p}$, or $(2k+1)^2 \\equiv -3 \\pmod{p}$. This is impossible because $-3$ is not a quadratic residue modulo $p = 3\\ell + 2$. This can be checked using the Legendre symbol:\n\n$$\n\\left(\\frac{-3}{p}\\right) = \\left(\\frac{-1}{p}\\right) \\left(\\frac{3}{p}\\right) = (-1)^{\\frac{p-1}{2}} \\cdot \\left((-1)^{\\frac{3-1}{2}} \\cdot \\frac{p-1}{2} \\left(\\frac{p}{3}\\right)\\right) = \\left(\\frac{2}{3}\\right) = -1.\n$$\n\nThus, $p^2 + p + 1$ is never a perfect cube for any prime $p$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 14672,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $f : \\mathbb{R} \\to \\mathbb{R}$ be a function satisfying the functional equation:\n\n$$\nf(x^2y) + 2f(y^2) = (x^2 + f(y))f(y)\n$$\nfor all $x, y \\in \\mathbb{R}$.\n\nFind all such functions $f$.",
    "options": [],
    "answer": "See solution",
    "solution": "* If $c = 0$ and $d = 0$, then $f(x) = 0$ for all $x$.\n* If $c = 2$ and $d = 2$, then $f(x) = 2x$ for all $x$.\n* If $c = 2$ and $d = -2$, then $f(x) = 2x$ for $x \\ge 0$, and $f(x) = -2x$ for $x < 0$, or, equivalently, $f(x) = 2|x|$ for all $x$.\n\nVerifying each case:\n- For $f(x) = 0$, both sides of the functional equation are $0$.\n- For $f(x) = 2x$, the left side is $2x^2y + 4y^2$, and the right side is $(x^2 + 2y) \\cdot 2y = 2x^2y + 4y^2$.\n- For $f(x) = 2|x|$, the left side is $2|x^2y| + 4|y^2| = 2x^2|y| + 4y^2$, and the right side is $(x^2 + 2|y|) \\cdot 2|y| = 2x^2|y| + 4|y|^2 = 2x^2|y| + 4y^2$.\n\nThus, the three solutions are:\n$$\nf(x) = 0, \\quad f(x) = 2x, \\quad f(x) = 2|x|.\n$$\n$\\square$",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 14673,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $ABC$ be an acute-angled triangle and altitudes $AA_1$ and $BB_1$ intersect at $H$. Consider circles $w_1$ and $w_2$ with centers $H$ and $B$ and with radii $HB_1$ and $BB_1$ respectively. Let $CN$ and $CK$ be the tangent lines from $C$ to circles $w_1$ and $w_2$ respectively ($N \\neq B_1$, $K \\neq B_1$). Prove that $A_1$, $N$, and $K$ are collinear.\n\n![](images/ukraine_2015_Booklet_p20_data_170199cbc2.png)",
    "options": [],
    "answer": "See solution",
    "solution": "Let $CC_1$ be the altitude. Since quadrilateral $AB_1HC_1$ is cyclic, we have $\\angle A = 180^\\circ - \\angle C_1HB_1 = \\angle B_1HC_1$. Since $\\angle B_1HC_1 = \\angle NHC_1$, we obtain $\\angle A = \\angle CHN_1$. Taking into account $\\angle HA_1C = 90^\\circ$, $\\angle HNC = 90^\\circ$, $\\angle AC_1C = 90^\\circ$, we get that quadrilaterals $HA_1NC$ and $AC_1A_1C$ are cyclic. Hence $180^\\circ - \\angle C_1A_1C = \\angle BAC = \\angle CHN = \\angle CA_1N$ and therefore points $C_1$, $A_1$, and $N$ are collinear. Since $BA_1HC_1$ is cyclic, we have $\\angle HC_1A_1 = \\angle HBA_1$. Note that $\\angle A_1BK = \\angle HBA_1$. Since $C$, $C_1$, $B$, $K$ are cyclic, it follows that $\\angle CBK = \\angle KC_1C$. This means that $C_1$, $A_1$, and $K$ are collinear. Notice that points $C_1$ and $A_1$ are different. This proves that $C_1$, $A_1$, $N$, and $K$ are collinear.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 14674,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $T(n)$ be the sum of the cubes of the digits of $n$.\n\nFind an arithmetic progression of length $1402$ such that all the numbers in the progression have the same non-zero digits with the same frequency, and thus $T(a_1) = T(a_2) = \\dots = T(a_{1402})$.",
    "options": [],
    "answer": "See solution",
    "solution": "By adding some number of ones to the left of these numbers, we can keep them as an arithmetic progression and also ensure all of them are 'good' numbers (i.e., with the same non-zero digits and frequencies).\n\nNow, let's construct the required progression. Let $A = \\frac{10^{\\varphi(7^m)} - 1}{7^m}$ for some $m > 10$. We claim that $A, 8A, 15A, \\dots, (1402 \\times 7 + 1)A$ all have the same digits with the same frequency. We use the following lemmas:\n\n**Lemma 1.** $10$ is a primitive root for every power of $7$.\n\n*Proof.* It suffices to check that $10$ is a primitive root modulo $7$ and $7^2$.\n\n**Lemma 2.** For every $1 \\leq a \\leq 7^m$ with $\\gcd(a, 7) = 1$, $aA$ represents the repeating part in the decimal representation of $\\frac{a}{7^m}$. Furthermore, it is a cyclic permutation of the digits in $A$.\n\n*Proof.* Let $\\frac{a}{7^m} = 0.\\overline{a_1\\cdots a_s}$. Then:\n\n$$\n\\overline{a_1 \\cdots a_s} = a \\times \\frac{10^s - 1}{7^m}\n$$\n\nSince the left-hand side is an integer, $7^m \\mid (10^s - 1)a$. Knowing $\\gcd(a, 7) = 1$ and by Lemma 1, we have $s = \\varphi(7^m)$.\n\n$$\n\\overline{a_1 \\cdots a_s} = a \\times \\frac{10^{\\varphi(7^m)} - 1}{7^m} = aA\n$$\n\nAs $10$ is a primitive root modulo $7^m$, there exists $\\alpha$ such that:\n\n$$\na \\equiv 10^{\\alpha} \\pmod{7^m} \\implies \\frac{10^{\\alpha}}{7^m} - \\frac{a}{7^m} \\in \\mathbb{Z}\n$$\n\n$\\frac{a}{7^m}$ is less than $1$, so $\\frac{a}{7^m} = \\left\\{\\frac{10^\\alpha}{7^m}\\right\\}$. The repeating decimal of $\\left\\{\\frac{10^\\alpha}{7^m}\\right\\}$ is a cyclic permutation of the repeating decimal of $\\frac{1}{7^m}$, which is $A$ (the special case $a = 1$).\n",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 14675,
    "subject": "Mathematics (Olympiad)",
    "question": "Does there exist a polynomial $p(x)$ with integer coefficients such that $p(\\sqrt{2}) = \\sqrt{2}$?",
    "options": [],
    "answer": "See solution",
    "solution": "Suppose such a polynomial $p(x)$ with integer coefficients exists. Then $p(\\sqrt{2}) = \\sqrt{2}$ implies $p(-\\sqrt{2}) = -\\sqrt{2}$, so $\\sqrt{2}$ and $-\\sqrt{2}$ are roots of $p(x) - x$. By Bezout's theorem, $p(x) - x$ is divisible by $(x - \\sqrt{2})(x + \\sqrt{2}) = x^2 - 2$ over the rationals. By Gauss's lemma, in $p(x) - x = (x^2 - 2)h(x)$, $h(x)$ has integer coefficients. Substituting $x = 2 + \\sqrt{2}$ and $x = 2 - \\sqrt{2}$ gives:\n\n$$\n2 = 6 \\cdot h(2 + \\sqrt{2}) \\quad \\text{and} \\quad 2 = 6 \\cdot h(2 - \\sqrt{2}).\n$$\n\nMultiplying and dividing by 4 yields:\n\n$$\n1 = 9 \\cdot (h(2 + \\sqrt{2}) h(2 - \\sqrt{2})).\n$$\n\nSince $h(2 + \\sqrt{2}) h(2 - \\sqrt{2})$ is an integer, $1$ is divisible by $9$, a contradiction. Thus, such a polynomial does not exist.",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 14676,
    "subject": "Mathematics (Olympiad)",
    "question": "Joyce has a box with light bulbs, some blue and some red. She counts the bulbs and finds that 10 of the 40 are blue. What percentage of the bulbs are blue?",
    "options": [],
    "answer": "See solution",
    "solution": "The percentage of blue bulbs is:\n\n$$\n\\frac{10}{40} \\times 100\\% = 25\\%\n$$\n\nSo, 25% of the bulbs are blue.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 14677,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $XYZT$ be a parallelogram and $A, B, C, D$ variable points on the sides $XY$, $XT$, $TZ$, $ZY$, respectively, such that $ABCD$ is a cyclic quadrilateral with circumcenter $O$. Suppose $AC \\parallel XT$ and $BD \\parallel XY$. Let $P$ be the intersection of lines $AD$ and $BC$, and $Q$ be the intersection of lines $AB$ and $CD$. Prove that the circle $$(POQ)$$ passes through a fixed point as $A, B, C, D$ vary according to the given restrictions.",
    "options": [],
    "answer": "See solution",
    "solution": "The key idea for this problem lies in the following lemma:\n\n**Lemma:** Let $ABC$ be a triangle and $X$ a point in the interior of angle $\\angle BAC$ such that $\\angle ABX = \\angle ACX$. Define $Y$ such that $BXC$ is a parallelogram. Then $AX$ and $AY$ are isogonal with respect to $\\angle BAC$.\n\n*Proof.* Consider the triangle $ABC$. Since $BXC$ is a parallelogram, we have $BX \\parallel AC$ and $CX \\parallel AB$. By the alternate interior angles theorem, $\\angle XBA = \\angle XCB$. Since $\\angle ABX = \\angle ACX$, we can conclude that $\\angle XAB = \\angle XCA$. Thus, $AX$ and $AY$ are isogonal with respect to $\\angle BAC$. $\\square$\n\nNow, let's proceed with the solution to the main problem. We need to prove that the circle $$(POQ)$$ passes through a fixed point as $A, B, C, D$ vary according to the given restrictions.\n\nLet $O'$ be the intersection of lines $BD$ and $AC$. Since $AC \\parallel XT$ and $BD \\parallel XY$, by the Lemma, $AP$ and $AQ$ are isogonal with respect to $\\angle XO'Y$.\n\nSince $ABCD$ is a cyclic quadrilateral, $\\angle ABC = \\angle ADC$. Thus, $\\angle PBC = \\angle PDC$, which implies $PB$ and $PD$ are isogonal with respect to $\\angle ABC$. Similarly, $QA$ and $QC$ are isogonal with respect to $\\angle ADC$.\n\nTherefore, $\\angle ABP = \\angle DCQ$ and $\\angle BAP = \\angle CQD$. Combining these equalities, $\\angle ABO' = \\angle DCO'$, so $ABO'D$ is a cyclic quadrilateral.\n\nLet $O$ be the circumcenter of $ABO'D$. Since $ABCD$ is a cyclic quadrilateral, $O$ is also the circumcenter of $ABCD$. Therefore, $O$ lies on the circle $$(POQ)$$.\n\nThus, as $A, B, C, D$ vary according to the given restrictions, the circle $$(POQ)$$ passes through the fixed point $O$. This completes the proof.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 14678,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $A(a, \\frac{1}{a})$, $B(b, \\frac{1}{b})$, $C(c, \\frac{1}{c})$ be points on the hyperbola $y = 1/x$.\n\nFind the value of\n\n$$\nT = (a + b + c) \\left( \\frac{1}{a} + \\frac{1}{b} + \\frac{1}{c} \\right)\n$$\n\ngiven that the triangle $ABC$ is equilateral and $a, b, c$ are pairwise distinct.\n\n![](images/BelarusMO_2013_s_p29_data_04daba1a3f.png)",
    "options": [],
    "answer": "See solution",
    "solution": "Since any vertical and any horizontal line meets the hyperbola $y = 1/x$ at most at one point, the numbers $a, b, c$ are pairwise distinct. The triangle $ABC$ is equilateral, so\n\n$$\n\\begin{aligned}\nAB = BC = CA &\\iff (b-a)^2 + (1/b - 1/a)^2 = (c-b)^2 + (1/c - 1/b)^2 = (a-c)^2 + (1/a - 1/c)^2.\n\\end{aligned}\n$$\n\nFrom these equalities, we derive\n\n$$\n\\begin{aligned}\nb^2 - 2ab + a^2 + \\frac{1}{b^2} - \\frac{2}{ab} + \\frac{1}{a^2} &= c^2 - 2bc + b^2 + \\frac{1}{c^2} - \\frac{2}{bc} + \\frac{1}{b^2} \\\\\n&\\implies (a^2 - c^2) - 2b(a-c) + \\frac{c^2 - a^2}{a^2c^2} - \\frac{2(c-a)}{abc} = 0.\n\\end{aligned}\n$$\n\nSince $a \\neq c$, we have\n\n$$\na + c - 2b - \\frac{a + c}{a^2c^2} + \\frac{2}{abc} = 0.\n$$\n\nSimilarly, we obtain two more equalities:\n\n$$\nb + a - 2c - \\frac{b + a}{b^2a^2} + \\frac{2}{abc} = 0, \\quad c + b - 2a - \\frac{c + b}{c^2b^2} + \\frac{2}{abc} = 0.\n$$\n\nSumming all three, we get\n\n$$\n\\frac{6}{abc} - \\frac{a + c}{a^2c^2} - \\frac{b + a}{b^2a^2} - \\frac{c + b}{c^2b^2} = 0,\n$$\nwhich leads to\n\n$$\n\\frac{b}{c} + \\frac{b}{a} + \\frac{c}{a} + \\frac{c}{b} + \\frac{a}{b} + \\frac{a}{c} = 6.\n$$\n\nThus,\n\n$$\nT = (a + b + c) \\left( \\frac{1}{a} + \\frac{1}{b} + \\frac{1}{c} \\right) = 1 + \\frac{a}{b} + \\frac{a}{c} + \\frac{b}{a} + 1 + \\frac{b}{c} + \\frac{c}{a} + \\frac{c}{b} + 1 = 3 + 6 = 9.\n$$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 14679,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all natural numbers $m, n$ such that $\\frac{(m+3n)^2}{m^2+n^2}$ is a perfect square of an integer.",
    "options": [],
    "answer": "See solution",
    "solution": "Suppose that $m, n$ are natural and $\\frac{(m+3n)^2}{m^2+n^2} = k^2$ where $k \\in \\mathbb{N},\\ k > 1$. Then\n\n$$\n(k^2 - 1)m^2 - 6mn + (k^2 - 9)n^2 = 0.\n$$\n\nConsidering this equation as quadratic with respect to $m$, we get that $4n^2(9-(k^2-1)(k^2-9))$ is a perfect square. For $k \\geq 4$ we have $4n^2(9-(k^2-1)(k^2-9)) < 0$. If $k=2$, the number $4n^2(9-(k^2-1)(k^2-9)) = 96n^2$ is not a perfect square. So we are left with the case $k=3$. We have $4m = 3n$, which leads to $m = 3l$, $n = 4l$ and $l \\in \\mathbb{N}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 14680,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all pairs of integers $ (a, b) $ such that there exists an integer $ d \\ge 2 $ for which $ a^n + b^n + 1 $ is divisible by $ d $ for any positive integer $ n $.",
    "options": [],
    "answer": "See solution",
    "solution": "If $a + b$ is odd, then $a^n + b^n + 1$ is even for all $n$, so $d = 2$ works. If $a + b$ is even, then $a^n + b^n + 1$ is always odd, so $d$ must be odd. Since $d \\mid a + b + 1$ and $d \\mid a^2 + b^2 + 1$, we have $d \\mid 2(ab - 1)$, so $d \\mid ab - 1$.\n\nConsider $a^3 + b^3 + 1 \\equiv (a + b)(a^2 + b^2 - ab) + 1 \\pmod d$. Since $a + b + 1 \\equiv 0 \\pmod d$, $a + b \\equiv -1 \\pmod d$, and $a^2 + b^2 + 1 \\equiv 0 \\pmod d$, so $a^2 + b^2 \\equiv -1 \\pmod d$. Thus,\n\n$$(a^3 + b^3 + 1) \\equiv (-1)((-1) - ab) + 1 = (1 + ab) + 1 = ab + 2 \\pmod d.$$\n\nBut since $d \\mid ab - 1$, $ab \\equiv 1 \\pmod d$, so $a^3 + b^3 + 1 \\equiv 1 + 2 = 3 \\pmod d$. Therefore, $d \\mid 3$, so $d = 3$.\n\nNow, $a^2 + b^2 + 1 \\equiv 0 \\pmod 3$. Also, $(a - b)^2 = a^2 + b^2 - 2ab \\equiv -1 - 2 \\equiv 0 \\pmod 3$, so $a \\equiv b \\pmod 3$. Then $a + b + 1 \\equiv 2a + 1 \\equiv 0 \\pmod 3$, so $a \\equiv 1 \\pmod 3$. Thus, $a \\equiv b \\equiv 1 \\pmod 3$.\n\nTherefore, for all $n$, $a^n + b^n + 1 \\equiv 1 + 1 + 1 = 3 \\equiv 0 \\pmod 3$.\n\n**Summary:**\n- If $a + b$ is odd, all pairs $(2k, 2l+1)$ and $(2k+1, 2l)$ for integers $k, l$ work (with $d = 2$).\n- If $a \\equiv b \\equiv 1 \\pmod 3$, all pairs $(3k+1, 3l+1)$ for integers $k, l$ work (with $d = 3$).",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 14681,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $a$ and $b$ be rational numbers such that $a = x + \\sqrt{2}$ and $b = x^4 + \\sqrt{2}$ for some $x$. Does there exist such rational $a$ and $b$?",
    "options": [],
    "answer": "See solution",
    "solution": "Suppose there exist rational $a$ and $b$ such that $a = x + \\sqrt{2}$ and $b = x^4 + \\sqrt{2}$. Then $x = a - \\sqrt{2}$. Substituting into $b$ gives:\n\n$$\nb = a^4 - 4a^3\\sqrt{2} + 12a^2 - 8a\\sqrt{2} + 4 + \\sqrt{2}.\n$$\n\nFor $b$ to be rational, the sum of terms containing $\\sqrt{2}$ must be zero. This leads to $4a^3 + 8a - 1 = 0$. It is easy to check that this equation has no rational solutions. Thus, such $a$ and $b$ do not exist.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 14682,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all integer triples $(n, k, p)$ such that $p^k = |6n^2 - 17n - 39|$, where $p$ is a prime number and $k$ is a positive integer.",
    "options": [],
    "answer": "See solution",
    "solution": "The solutions are $(n, k, p) = (-4, 3, 5), (-2, 1, 19), (-1, 4, 2), (2, 2, 7), (4, 1, 11)$.\n\nWe write\n$$\np^k = |6n^2 - 17n - 39| = |(2n + 3)(3n - 13)|.\n$$\nThis gives\n$$\n2n + 3 = \\pm p^{\\alpha}, \\quad 3n - 13 = \\pm p^{\\beta}\n$$\nfor some non-negative integers $\\alpha, \\beta$. Examining the cases $\\alpha = 0$ or $\\beta = 0$, we find the possibilities $n = -1, n = -2$, and $n = 4$, which give the solutions $(n, k, p) = (-2, 1, 19), (-1, 4, 2), (4, 1, 11)$.\n\nIf $\\alpha, \\beta \\ge 1$, then $p$ divides both $2n+3$ and $3n-13$, so $p$ divides $35$, hence $p \\in \\{5,7\\}$. Note that $p^{\\alpha-\\beta} = \\pm \\frac{2n+3}{3n-13}$. However, if $n \\ge 17$ or $n \\le -5$, then $\\frac{1}{5} < \\frac{2n+3}{3n-13} < 1$, so $\\alpha-\\beta$ is not an integer for $p \\in \\{5,7\\}$, a contradiction.\n\nThus, we examine $-4 \\le n \\le 16$. If $p=5$, then $n \\equiv 1 \\pmod{5}$; if $p=7$, then $n \\equiv 2 \\pmod{7}$. Also, $3n-13$ is odd, so $n$ must be even. Checking $n \\in \\{-4,2,6,16\\}$ gives the solutions $(n, k, p) = (-4, 3, 5), (2, 2, 7)$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 14683,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $a$, $b$, and $c$ be positive real numbers. Prove that\n\n$$\na^3 b^6 + b^3 c^6 + c^3 a^6 + 3a^3 b^3 c^3 \\geq abc(a^3 b^3 + b^3 c^3 + c^3 a^3) + a^2 b^2 c^2 (a^3 + b^3 + c^3).\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "After dividing both sides of the given inequality by $a^3 b^3 c^3$, it becomes\n\n$$\n\\left(\\frac{b}{c}\\right)^3 + \\left(\\frac{c}{a}\\right)^3 + \\left(\\frac{a}{b}\\right)^3 + 3 \\geq \\left(\\frac{a}{c} \\cdot \\frac{b}{c} + \\frac{b}{a} \\cdot \\frac{c}{a} + \\frac{c}{b} \\cdot \\frac{a}{b}\\right) + \\left(\\frac{a}{b} \\cdot \\frac{a}{c} + \\frac{b}{a} \\cdot \\frac{b}{c} + \\frac{c}{a} \\cdot \\frac{c}{b}\\right). \\quad (1)\n$$\n\nTake\n\n$$\n\\frac{a}{b} = x, \\quad \\frac{b}{c} = y, \\quad \\frac{c}{a} = z. \\qquad (2)\n$$\n\nThen\n\n$$\nxyz = 1. \\qquad (3)\n$$\n\nTherefore,\n\n$$\n\\frac{b}{a} = \\frac{1}{x}, \\quad \\frac{c}{b} = \\frac{1}{y}, \\quad \\frac{a}{c} = \\frac{1}{z}. \\qquad (4)\n$$\n\nSubstituting (2) and (4) into (1), we find that\n\n$$\nx^3 + y^3 + z^3 + 3 \\geq \\left(\\frac{y}{z} + \\frac{z}{x} + \\frac{x}{y}\\right) + \\left(\\frac{x}{z} + \\frac{y}{x} + \\frac{z}{y}\\right). \\qquad (5)\n$$\n\nMultiplying the inequality (5) by $xyz$, and using $xyz = 1$ from (3), it is equivalent to\n\n$$\nx^3 + y^3 + z^3 + 3xyz - xy^2 - yz^2 - zx^2 - yx^2 - zy^2 - xz^2 \\geq 0. \\qquad (6)\n$$\n\nFinally, by the special case of Schur's inequality:\n\n$$\nx^r(x - y)(x - z) + y^r(y - x)(y - z) + z^r(z - y)(z - x) \\geq 0, \\quad x, y, z \\geq 0,\\ r > 0,\n$$\n\nwith $r = 1$ there holds\n\n$$\nx(x - y)(x - z) + y(y - x)(y - z) + z(z - y)(z - x) \\geq 0. \\qquad (7)\n$$\n\nAfter expansion, this coincides with (6). $\\square$\n\n**Remark.** The inequality (7) immediately follows by supposing (without loss of generality) that $x \\geq y \\geq z$, and then writing the left-hand side of (7) as\n\n$$\n(x - y)(x(x - z) - y(y - z)) + z(y - z)(x - z),\n$$\n\nwhich is obviously $\\geq 0$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 14684,
    "subject": "Mathematics (Olympiad)",
    "question": "設 $k$ 為一給定實數。試找出所有從實數映至實數的函數 $f(x)$ 滿足對任意實數 $x, y$，均有\n\n$$\nf(x) + (f(y))^2 = k f(x + y^2).\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "若 $k \\neq 1$，將 $y = 0$ 代入原式得到\n\n$$\n(f(0))^2 = (k - 1)f(x).\n$$\n\n因此 $f(x)$ 為常數函數。代入原式後可解出 $f(x) = 0$ 或 $f(x) = \\frac{1}{k-1}$。\n\n若 $k = 1$，將 $x = 0$ 代入原式得到\n\n$$\nf(y)^2 = f(y^2).\n$$\n\n由此可知對任意 $y > 0$ 均有 $f(y) > 0$ 並且 $f(1) = 0$ 或 $1 > f(x)$。將上式代回原題中：\n\n$$\n\\begin{align*}\n& f(x) + (f(y))^2 = f(x + y^2), \\quad \\forall x, y \\in \\mathbb{R} \\\\\n\\Rightarrow \\quad & f(x) + f(y^2) = f(x + y^2), \\quad \\forall x, y \\in \\mathbb{R} \\\\\n\\Rightarrow \\quad & f(x) + f(y) = f(x + y), \\quad \\forall x, y \\in \\mathbb{R},\\ y \\ge 0\n\\end{align*}\n$$\n\n這是一個標準的柯西方程，因此 $f(x) = cx$。由 $f(1) = 0$ 或 $f(x) = x$。",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 14685,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $a$ be a real number. Prove that there exist real numbers $b$ and $c$ such that the inequalities\n$$\n\\min \\{\\sin(x), \\sin(a + x)\\} \\leq b \\sin(x + c) \\leq \\max \\{\\sin(x), \\sin(a + x)\\}\n$$\nhold for all real numbers $x$, and the equalities hold only if $\\sin(x) = \\sin(a + x)$.",
    "options": [],
    "answer": "See solution",
    "solution": "Recall that two real numbers $r$ and $s$ always satisfy the trigonometric identity:\n$$\n\\frac{1}{2}[\\sin(r+s) + \\sin(r-s)] = \\cos(s) \\sin(r)\n$$\nSubstituting $r = x + \\frac{a}{2}$ and $s = \\frac{a}{2}$, we see that $\\cos\\left(\\frac{a}{2}\\right) \\sin\\left(x + \\frac{a}{2}\\right)$ is, for all real numbers $x$, the arithmetic mean of $\\sin(x + a)$ and $\\sin(x)$. Since the arithmetic mean of two numbers lies in the closed interval bounded by the two numbers, $b = \\cos\\left(\\frac{a}{2}\\right)$ and $c = \\frac{a}{2}$ is a suitable choice for $b$ and $c$. Furthermore, the arithmetic mean of two numbers differs from the two numbers if they are not equal. $\\square$",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 14686,
    "subject": "Mathematics (Olympiad)",
    "question": "Three lines in the plane pass through point $A$, dividing the plane into 6 sectors. In the interior of each sector, there are 5 points, with no three of the total 30 points collinear. Prove that there exist at least 1000 triangles, with vertices chosen from these 30 points, such that each triangle contains point $A$ either in its interior or on its sides.",
    "options": [],
    "answer": "See solution",
    "solution": "First, observe that selecting one point from each sector yields a hexagon (convex or non-convex) containing $A$.\n\nFrom these six points, we can form $\\binom{6}{3} = 20$ triangles. We will count how many of these triangles contain $A$.\n\nFor any two points in opposite sectors, the third vertex can be chosen in two ways.\n\n![](images/IMO2017_finalbook_Greece_1_p7_data_a6d71a4350.png)\n\n![](images/IMO2017_finalbook_Greece_1_p7_data_a414dbb696.png)\n\nFor example, for points $B$ and $C$ in the figure, the third vertex can be chosen from the two colored sectors.\n\nThere are three pairs of opposite sectors. For each pair, there are $5 \\times 5$ choices for the base, and $2 \\times 5 = 10$ choices for the third vertex. Thus, there are at least $3 \\times 2 \\times 5^3 = 6 \\times 5^3$ such triangles containing $A$.\n\n![](images/IMO2017_finalbook_Greece_1_p7_data_18f962a668.png)\n\nNow, consider points in non-successive and non-opposite sectors (see the figure). In this case, triangles like $CBD$ or $EFG$ contain $A$.\n\nThere are $5^3$ triangles of each type, so $2 \\times 5^3$ triangles in total in this case.\n\nSumming up, we have at least $6 \\times 5^3 + 2 \\times 5^3 = 8 \\times 5^3 = 1000$ triangles containing $A$ either in their interior or on their sides.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 14687,
    "subject": "Mathematics (Olympiad)",
    "question": "An $n$-type tiles triangle, where $n \\ge 2$, is formed by the cells of a $(2n + 1) \\times (2n + 1)$ array which are situated below its diagonals. For instance, a 3-type tiles triangle is the following:\n\n![](images/RMC_2024_p65_data_d30c7a42fe.png)\n\nDetermine the maximal length of a sequence with pairwise distinct cells in an $n$-type tiles triangle, such that, beginning with the second one, any cell of the sequence has a common side with the previous one.",
    "options": [],
    "answer": "See solution",
    "solution": "We alternately color (as a chessboard) the cells of an $n$-type tiles triangle, as below:\n\n![](images/RMC_2024_p65_data_e9a9541a46.png)\n\nWe have $b_n = 1 + 2 + \\dots + n = \\frac{n(n+1)}{2}$ black cells and $a_n = 1 + 2 + \\dots + (n-1) = \\frac{n(n-1)}{2}$ white cells. In any sequence of cells with the required property, their color alternates, hence its length cannot exceed $2(a_n + 1)$. Therefore, we infer that the maximal length of such a sequence is $L \\le n^2 - n + 1$.\n\nTo complete the solution, we now provide an example of a sequence in an $n$-type tiles triangle which satisfies the required condition and has length $L = n^2 - n + 1$.\n\nWe consider a sequence of cells which begins with the top cell. At each step, we choose the cell below the last cell chosen in the previous step, together with a maximum length sequence of cells situated on the same line, up to one of its ends. For instance, for a 3-type tiles triangle the following succession is a solution:\n\n![](images/RMC_2024_p66_data_840c0c2021.png)\n\nThe length of a sequence as we described before is: $1 + 2 + 4 + \\dots + 2(n - 1) = 1 + n(n - 1) = n^2 - n + 1$.\n\nTherefore, the answer is $L = n^2 - n + 1$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 14688,
    "subject": "Mathematics (Olympiad)",
    "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": "See 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$; for example, if $k \\equiv 0 \\pmod{p-1}$, then $k^2 + k \\equiv 0 \\pmod{p}$.  \n\nNow, let $\\{p_1, p_2, \\dots, p_m\\}$ be the set of all prime numbers not exceeding $2008$. Take $n$ satisfying\n\n$$\nn \\equiv p_j - r(p_j) \\pmod{p_j}, \\quad j = 1, 2, \\dots, m,\n$$\n\nwhich exists by the Chinese Remainder Theorem. This $n$ has the desired property.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 14689,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $T(a)$ be the sum of digits of $a$. For which $R \\in \\mathbb{N}$ does there exist an $n \\in \\mathbb{N}$ such that $\\frac{T(n^2)}{T(n)} = R$?\n",
    "options": [],
    "answer": "See solution",
    "solution": "**Answer:** All positive integers $R$.\n\nLet $R \\in \\mathbb{N}$ and consider the number\n\n$$\nN = \\sum_{k=0}^{R-1} 10^{2^k}.\n$$\n\nWe see that $T(N) = R$. Now\n\n$$\nN^2 = \\left(\\sum_{k=0}^{R-1} 10^{2^k}\\right)^2 = \\sum_{0 \\le a, b < R} 10^{2^a + 2^b},\n$$\n\nand since $2^a + 2^b = 2^c + 2^d$ if and only if $(a, b) = (c, d)$ or $(a, b) = (d, c)$, there is never a carry in the summation $\\sum_{0 \\le a, b < R} 10^{2^a + 2^b}$, and we can write\n\n$$\nT(N^2) = \\sum_{0 \\le a, b < R} T(10^{2^a + 2^b}) = R^2.\n$$\n\nSo $\\frac{T(n^2)}{T(n)} = R$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 14690,
    "subject": "Mathematics (Olympiad)",
    "question": "Consider $2n$ rays (half-lines) in the plane such that no two rays are parallel (the endpoints of the rays may coincide). Prove that there exists a line in the plane that does not pass through any of the endpoints and intersects with exactly $n$ rays.",
    "options": [],
    "answer": "See solution",
    "solution": "Let us choose any circle such that all of the endpoints of the rays are inside the circle. Also, choose a tangent line to the circle that is not parallel to any of the rays. Let this tangent line be $l_0$, and let $l_\\alpha$ be the tangent line obtained by rotating $l_0$ counterclockwise by angle $\\alpha$ about the center of the circle. Let $f(\\alpha)$ be the number of rays that intersect $l_\\alpha$.\n\nSince $l_0$ and $l_\\pi$ are two parallel lines and all endpoints of the 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) \\leq n$ and $f(\\pi) \\geq n$. Because no two rays are parallel, as we continuously increase $\\alpha$ (i.e., rotate the tangent line), $f(\\alpha)$ can change by at most one at any time. Therefore, $f(\\alpha)$ ranges over all integer values between $f(0)$ and $f(\\pi)$, so there exists a value of $\\alpha$ for which $f(\\alpha) = n$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 14691,
    "subject": "Mathematics (Olympiad)",
    "question": "There are 225 children and 105 balls at a football club training. The children are split into several equal groups, and each group receives the same number of balls. \n\n- How many groups can be formed?\n- How many balls does each group get?\n- How many solutions does this problem have?",
    "options": [],
    "answer": "See solution",
    "solution": "The common divisors of $225$ and $105$ are $1$, $3$, $5$, and $15$. Each divisor represents a possible number of groups:\n\n- $1$ group: $225$ children, $105$ balls\n- $3$ groups: $75$ children, $35$ balls each\n- $5$ groups: $45$ children, $21$ balls each\n- $15$ groups: $15$ children, $7$ balls each\n\nThus, there are $4$ possible solutions.",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 14692,
    "subject": "Mathematics (Olympiad)",
    "question": "Do there exist 2024 nonzero real numbers $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$$",
    "options": [],
    "answer": "See solution",
    "solution": "Suppose such $a_i$ exist. The condition rewrites as\n$$\n\\sum_{i=1}^{2024} \\left(a_i + \\frac{1}{a_{i+1}} - 1\\right)^2 = 0,\n$$\nwhich means $a_i + \\frac{1}{a_{i+1}} = 1$ for all $i = 1, 2, \\dots, 2024$. We can deduce recursively that for any $i$, $a_{i+1} = \\frac{1}{1 - a_i}$, $a_{i+2} = 1 - \\frac{1}{a_i}$, so $a_{i+3} = a_i$. Since $3$ does not divide $2024$, all $a_i$ must be equal to some real $k$. However, $k + \\frac{1}{k} = 1$ implies $k^2 - k + 1 = 0$, which has no real solution. Contradiction.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 14693,
    "subject": "Mathematics (Olympiad)",
    "question": "Prove that for all $x, y, z > 0$ with $\\frac{1}{x} + \\frac{1}{y} + \\frac{1}{z} = 1$ and $0 \\leq a, b, c < 1$, the following inequality holds:\n\n$$\n\\frac{x^2 + y^2}{1 - a^2} + \\frac{y^2 + z^2}{1 - b^x} + \\frac{z^2 + x^2}{1 - c^y} \\geq \\frac{6(x + y + z)}{1 - abc}.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "Let $x, y, z, a, b$, and $c$ be as in the statement. We will prove that\n\n$$\n(x^2 + y^2)a^z + (y^2 + z^2)b^x + (z^2 + x^2)c^y \\geq 6(x + y + z)abc.\n$$\n\nIndeed, we have $1 = \\left(\\frac{1}{x} + \\frac{1}{y} + \\frac{1}{z}\\right)^2 \\geq 3\\left(\\frac{1}{xy} + \\frac{1}{yz} + \\frac{1}{zx}\\right) \\Rightarrow xyz \\geq 3(x + y + z)$ and, therefore, $x^2 + y^2 \\geq 2xy = 2xyz \\cdot \\frac{1}{z} \\geq 6(x + y + z) \\cdot \\frac{1}{z}$.\n\nMultiplying by $a^z$ and summing over the analogous inequalities for $b$ and $c$, we obtain\n\n$$\n(x^2 + y^2)a^z + (y^2 + z^2)b^x + (z^2 + x^2)c^y \\geq 6(x + y + z) \\left( \\frac{1}{z}a^z + \\frac{1}{x}b^x + \\frac{1}{y}c^y \\right)\n$$\n\nBy the weighted AM-GM inequality,\n\n$$\n\\left( \\frac{1}{z}a^z + \\frac{1}{x}b^x + \\frac{1}{y}c^y \\right) \\geq abc\n$$\n\nand, therefore,\n\n$$\n(x^2 + y^2)a^z + (y^2 + z^2)b^x + (z^2 + x^2)c^y \\geq 6(x + y + z)abc.\n$$\n\nNow, replace $a, b, c$ by $a^k, b^k, c^k$ and sum the resulting inequalities over all integers $k \\geq 0$. Since, for all $0 \\leq d < 1$, $1 + d + d^2 + \\dots = \\frac{1}{1-d}$, the claim follows.\n\nEquality holds exactly when $x = y = z = 3$ and $a = b = c$.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 14694,
    "subject": "Mathematics (Olympiad)",
    "question": "Given a row of $N > 1$ coins, each showing either a black or white side up, consider the following process: In each move, you may remove a coin (and its square), and flip the coins immediately adjacent to it. Prove that it is possible to remove all coins if and only if the initial number of coins showing black side up is odd.",
    "options": [],
    "answer": "See solution",
    "solution": "Assume the claim holds for all arrangements with fewer than $N$ coins. Let $K$ be the number of coins with black side up in an arrangement of $N$ coins.\n\nIf $K$ is odd, select the first black-up coin $Y$ (not at the ends), with neighbors $X$ (left) and $Z$ (right). Remove $Y$ and flip $X$ and $Z$, splitting the board into two parts. On the left, only $X$ is black-up; by induction, all coins can be removed. On the right, the number of black-up coins remains odd; again, by induction, all coins can be removed. If $Y$ is at an end, only one part remains, and the argument is similar.\n\nIf $K$ is even, suppose all coins can be removed. After the first move, the sum of black-up coins on both parts remains odd, so one part has an even number. By induction, that part cannot be cleared—a contradiction. Thus, all coins can be removed if and only if the initial number of black-up coins is odd.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 14695,
    "subject": "Mathematics (Olympiad)",
    "question": "In the acute triangle $ABC$, there are altitudes $BP$ and $CQ$. Point $T$ is the intersection of the altitudes of $\\triangle PAQ$. It turns out that $\\angle CTB = 90^\\circ$. Find the value of $\\angle BAC$.",
    "options": [],
    "answer": "See solution",
    "solution": "The statement of the problem yields $\\angle BQC = \\angle BTC = \\angle BPC = 90^\\circ$, hence, points $Q$, $T$, $P$ lie on the circle with diameter $BC$, and in this exact order: $B$, $Q$, $T$, $P$, $C$ (since $\\triangle ABC$ is acute).\n\n$$\n\\angle QTP = 180^\\circ - \\angle ABP = 180^\\circ - (90^\\circ - \\angle BAC) = 90^\\circ + \\angle BAC.\n$$\n\nOn the other hand, $QT \\perp AC$, $PT \\perp AB$, so $\\angle QTP = 180^\\circ - \\angle BAC$. Hence,\n\n$$\n90^\\circ + \\angle BAC = 180^\\circ - \\angle BAC \\implies \\angle BAC = 45^\\circ.\n$$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 14696,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $ABC$ be an acute triangle with $AB < AC$ and let $H$ and $O$ be its orthocentre and circumcentre, respectively. Let $\\Gamma$ be the circle $BOC$. The line $AO$ and the circle of radius $AO$ centred at $A$ cross $\\Gamma$ again at $A'$ and $F$, respectively. Prove that $\\Gamma$, the circle on diameter $AA'$, and the circle $AFH$ are concurrent.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $\\Omega$ denote the circle $(ABC)$ (centred at $O$), and let $M$ be the midpoint of the minor arc $BC$ of that circle.\n\nConsider the composition $\\iota$ of an inversion centered at $A$ and the reflection in the bisector $AM$ that swaps $B$ and $C$. Then $\\iota$ swaps the circle $\\Omega$ with the line $BC$, hence it swaps $O$ with the reflection $L$ of $A$ in $BC$. Hence $\\iota(\\Gamma)$ is the circle $\\Gamma^* = (BCL)$,\n\ni.e., the reflection of $\\Omega$ in $BC$ which passes through $H$. Let $S$ and $Q$ be the centers of $\\Gamma$ and $\\Gamma^*$, respectively; then $AM$ is the angle bisector of $\\angle QAS$.\n\nSince $\\iota$ swaps $\\Gamma$ and $\\Gamma^*$, they are seen from $A$ at the same angle, so there exists a rotational homothety $h$ centred at $A$ mapping $\\Gamma$ to $\\Gamma^*$; the angle of $h$ is $\\angle SAQ$. Notice that the rays $AH$ and $AF$ are obtained from $AO$ by reflections in $AM$ and $AS$, respectively, so $\\angle HAF = 2\\angle MAS = \\angle QAS$. This easily yields that $H = h(F)$. Hence the triangles $AHF$ and $AQS$ are similar, and $\\angle AHF = \\angle AQS$.\n\n![](images/RMC_2025_p122_data_ff7892819c.png)\n\nNow, let the circle ($AHF$) meet $\\Gamma$ again at $T$. Using directed angles, we get $\\angle ATF = \\angle AHF = \\angle AQS$; next, since $FO \\perp AS$, we have\n\n$$\n\\angle FTA' = \\angle FOA' = \\pi/2 - \\angle OAS = \\pi/2 - \\angle QAL = \\pi/2 - \\angle AQS,\n$$\n\n(here the equality $\\angle OAS = \\angle QAL$ holds because these angles are symmetric to each other with respect to $AM$), so $\\angle ATA' = \\angle ATF + \\angle FTA' = \\pi/2$, as desired.\n\n**Remark.** Existence of the rotational homothety $h$ may be shown in various ways. E.g., one may notice that $\\Omega$ is an Apollonius circle of the segment $QS$, so the ratio of the radii of $\\Gamma$ and $\\Gamma^*$ is $BS/BQ = AS/AQ$, which also yields that $h$ exists.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 14697,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $\\mathbb{Z}^+$ be the set of all positive integers. Find all functions $f : \\mathbb{Z}^+ \\to \\mathbb{Z}^+$ such that\n\n1. $f(n!) = f(n)!$ for all $n \\in \\mathbb{Z}^+$,\n2. $m - n$ divides $f(m) - f(n)$ for all distinct positive integers $m$ and $n$.",
    "options": [],
    "answer": "See solution",
    "solution": "There are three such functions: the constant functions $f(n) = 1$, $f(n) = 2$, and the identity function $f(n) = n$. These functions clearly satisfy the given conditions.\n\nLet us prove that these are the only ones.\n\nSuppose $f$ has a fixed point $a \\ge 3$, that is, $f(a) = a$. Then $a!$, $(a!)!$, $\\dots$ are all fixed points of $f$, so $f$ has a strictly increasing sequence $a_1 < a_2 < \\dots < a_k < \\dots$ of fixed points. For any positive integer $n$, $a_k - n$ divides $a_k - f(n) = f(a_k) - f(n)$ for every $k \\in \\mathbb{Z}^+$. Also, $a_k - n$ divides $a_k - n$, so it divides $a_k - f(n) - (a_k - n) = n - f(n)$. This is possible only if $f(n) = n$, so in this case $f = \\text{id}_{\\mathbb{Z}^+}$.\n\nNow suppose $f$ has no fixed points greater than $2$. Let $p \\ge 5$ be a prime. By Wilson's Theorem, $(p-2)! \\equiv 1 \\pmod p$, so $p$ divides $(p-2)! - 1$. But $(p-2)! - 1$ divides $f((p-2)!) - f(1)$, so $p$ divides $f((p-2)!) - f(1) = (f(p-2))! - f(1)$. Clearly, $f(1) = 1$ or $f(1) = 2$. Since $p \\ge 5$, the fact that $p$ divides $(f(p-2))! - f(1)$ implies $f(p-2) < p$. Again by Wilson's Theorem, $p$ does not divide $(p-1)! - 1$ or $(p-1)! - 2$, so $f(p-2) \\le p-2$. On the other hand, $p-3 = (p-2)-1$ divides $f(p-2) - f(1) \\le (p-2)-1$. Thus, either $f(p-2) = f(1)$ or $f(p-2) = p-2$. Since $p-2 \\ge 3$, the last case is excluded, as $f$ has no fixed points greater than $2$. It follows $f(p-2) = f(1)$, and this holds for all primes $p \\ge 5$. For any $n$, $p-2-n$ divides $f(p-2) - f(n) = f(1) - f(n)$ for all such $p$, so $f(n) = f(1)$. Thus, $f$ is the constant function $1$ or $2$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 14698,
    "subject": "Mathematics (Olympiad)",
    "question": "考慮所有形如 $f(x) = (x - a_1)(x - a_2)(x - a_3) \\cdots (x - a_{100})$ 的整係數多項式，其中 $a_1, a_2, \\cdots, a_{100}$ 是任意實數。試求 $\\{a_1\\} + \\{a_2\\} + \\cdots + \\{a_{100}\\}$ 可能的最大值。\n\n註：定義 $\\{x\\} = x - [x]$，其中 $[x]$ 為不大於 $x$ 的最大整數。",
    "options": [],
    "answer": "See solution",
    "solution": "因為對任意 $x$ 都有 $\\{x\\} < 1$，因此 $\\{a_1\\} + \\{a_2\\} + \\cdots + \\{a_{100}\\} < 100$。又由根與係數關係知 $-\\sum_{i=1}^{100} a_i$ 為 $f(x)$ 的 $x^{99}$ 項係數，必然是個整數，因此\n\n$$\n\\{a_1\\} + \\{a_2\\} + \\cdots + \\{a_{100}\\} \\le 99.\n$$\n\n以下證明存在 $f(x)$ 使得 $\\{a_1\\} + \\{a_2\\} + \\cdots + \\{a_{100}\\} = 99$。考慮函數\n\n$$\nf(x) = x(x-2)(x-4)(x-6)\\cdots(x-198) + (x-1)(x-3)\\cdots(x-197)\n$$\n\n代入數值檢驗易知 $f(-1)f(0), f(1)f(2), \\ldots, f(197)f(198)$ 皆小於 0，由牛頓勘根可知 $f(x)$ 在 100 個區間 $(-1, 0), (1, 2), (3, 4), \\ldots, (197, 198)$ 中都有根。所以 $f(x)$ 有 100 個實根，滿足題目要求的條件，並且\n\n$$\n[a_1] + [a_2] + \\cdots + [a_{100}] = -1 + 1 + 3 + \\cdots + 197\n$$\n\n但 $f(x)$ 的 $x^{99}$ 項係數為 $-(2+4+6+\\cdots+198)+1$，由根與係數的關係，有\n\n$$\na_1 + a_2 + \\cdots + a_{100} = 2 + 4 + 6 + \\cdots + 198 - 1\n$$\n\n因此\n\n$$\n\\begin{align*}\n\\{a_1\\} + \\{a_2\\} + \\cdots + \\{a_{100}\\} &= (a_1 + a_2 + \\cdots + a_{100}) - ([a_1] + [a_2] + \\cdots + [a_{100}]) \\\\\n&= (2 + 4 + 6 + \\cdots + 198 - 1) \\\\\n&\\quad -(1 + 3 + 5 + \\cdots + 197 - 1) \\\\\n&= 99\n\\end{align*}\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 14699,
    "subject": "Mathematics (Olympiad)",
    "question": "Seien $x_1, x_2, \\dots, x_9$ nichtnegative reelle Zahlen, für die gilt:\n\n$$\nx_1^2 + x_2^2 + \\dots + x_9^2 \\geq 25\n$$\n\nBeweise, dass es drei dieser Zahlen gibt, deren Summe mindestens $5$ ist.",
    "options": [],
    "answer": "See solution",
    "solution": "Sei ohne Beschränkung der Allgemeinheit $x_1 \\ge x_2 \\ge \\dots \\ge x_9 \\ge 0$. Angenommen, $x_1 + x_2 + x_3 < 5$. Dann gilt:\n\n$$\n25 \\le x_1^2 + x_2^2 + \\dots + x_9^2 \\le x_1(x_1 + x_2 + x_3) + x_2(x_4 + x_5 + x_6) + x_3(x_7 + x_8 + x_9) \\le 5x_1 + 5x_2 + 5x_3 < 25.\n$$\n\nDas ist ein Widerspruch, daher gibt es drei Zahlen, deren Summe mindestens 5 ist. $\\Box$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 14700,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $ABC$ be an acute triangle. Circle $\\omega_1$, with diameter $AC$, intersects side $BC$ at $F$ (other than $C$). Circle $\\omega_2$, with diameter $BC$, intersects side $AC$ at $E$ (other than $C$). Ray $AF$ intersects $\\omega_2$ at $K$ and $M$ with $AK < AM$. Ray $BE$ intersects $\\omega_1$ at $L$ and $N$ with $BL < BN$. Prove that lines $AB$, $ML$, and $NK$ are concurrent.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $D$ be the foot of the perpendicular from $C$ to $AB$ and $H$ be the orthocenter of $\\triangle ABC$. Note first that $\\omega_1$ and $\\omega_2$ both intersect $AB$ at $D$. By Power of a Point, $LH \\cdot HN = CH \\cdot HD = KH \\cdot HM$, implying that $KLMN$ is a cyclic quadrilateral. Noting that $AC$ and $BC$ are perpendicular bisectors of the diagonals of $KLMN$, we conclude that the center of its circumcircle is $C$. Observe now that since $\\angle ANC = \\angle ALC = 90^\\circ$, we have that $AN$ and $AL$ are tangent to the circumcircle of $KLMN$. Thus, $H$ is on the polar of $A$. Similarly, $H$ is on the polar of $B$, and so by Brocard's theorem, $KN$ and $LM$ meet on the polar of $H$, which is $AB$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 14701,
    "subject": "Mathematics (Olympiad)",
    "question": "Suppose there are $n$ members in a group, and each member has exactly 3 enemies. No two members can have more than one common enemy. For which values of $n$ is this possible? Give examples for each possible $n$.\n\n![](images/1.2_1997-2023_IMO_HK_TST_Solutions_p156_data_b88993fbd9.png)\n\n![](images/1.2_1997-2023_IMO_HK_TST_Solutions_p156_data_bef10795ff.png)",
    "options": [],
    "answer": "See solution",
    "solution": "We have $n$ members, each with 3 enemies, and no two members share more than one common enemy.\n\nSuppose $A$ and $B$ are enemies. By the given condition, every other member cannot be a friend of both $A$ and $B$. Since each of $A$ and $B$ has 2 more enemies other than themselves, we must have $n \\leq 2 + 2 \\times 2 = 6$. Also, as each member has 3 enemies, we have $n \\geq 4$, and there are $\\frac{3n}{2}$ pairs of enemies. This shows $n$ is even, and hence $n$ can only be 4 or 6.\n\nTo give examples for $n = 4$ and $n = 6$, we use graph theory: let vertices denote members and edges join pairs of enemies. Then $K_4$ and $K_{3,3}$ are examples for $n = 4$ and $n = 6$, respectively.",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 14702,
    "subject": "Mathematics (Olympiad)",
    "question": "На таблата се напишани броевите $1, 2, \\ldots, 2009$. Се бришат неколку од нив и наместо нив на таблата се запишува остатокот на збирот на избришаните броеви при делење со $13$. После одреден број повторувања на оваа постапка, на таблата останале само три броја, од кои двата се $99$ и $999$. Да се определи третиот број кој останал на таблата.",
    "options": [],
    "answer": "See solution",
    "solution": "Нека третиот број е $x$. Јасно е дека после секој чекор, остатокот при делење на збирот на броевите на таблата со $13$ не се менува.\n\nБидејќи $$1 + 2 + 3 + \\ldots + 2009 = \\frac{2009 \\cdot 2010}{2} = 1005 \\cdot 2010,$$ ова дава остаток $2$ при делење со $13$.\n\nПонатаму, $99 + 999 + x$ треба да има остаток $2$ при делење со $13$.\n\nБидејќи $99 + 999 = 1098$, а $1098$ дава остаток $6$ при делење со $13$, следува дека $0 \\leq x < 13$ и мора $x = 9$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 14703,
    "subject": "Mathematics (Olympiad)",
    "question": "Given an acute triangle $ABC$ with circumcenter $O$. Let $P$ be a point on $BC$ such that $BP < \\frac{BC}{2}$, and let $Q$ be a point 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$ and $OE$ meet $BC$ at $K$ and $F$, respectively. Show that $AOKF$ is cyclic.\n\n![](path/to/file.png)",
    "options": [],
    "answer": "See 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$ to be cyclic, or $DP \\cdot DF = DA \\cdot DA' = DB \\cdot DC = DO \\cdot DR$, so we need $OPRF$ to be cyclic, or that $\\angle OFP = \\angle ORP$. By the shooting lemma, $RDEF$ is cyclic, so $\\angle OFP = \\angle ERD$, so we need $AR$ to be 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 theorem 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.\n\n**Remark.** One can show that $AE$ and $NO$ still meet on $(BOC)$ even when $P$ is not the foot of the perpendicular from $A$—just invert about $(ABC)$, and if $NO \\cap (BOC) = T$, then $(AOKF)$ goes to the line $A, E, T$! So, the current problem is the main component of the proof for the generalization of the IGO problem. $\\square$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 14704,
    "subject": "Mathematics (Olympiad)",
    "question": "有一台飛機，上面有一些座位，每個座位都有它的價格，如下圖所示。今天有 $2n-2$ 位乘客要搭這班飛機，但是他們都不想跟其他乘客坐在同一行或同一列。飛機機長發現不管怎麼排這些乘客的座位，收到的錢都一樣多。試證明這件事，並求出他們可以收多少錢。\n\n![](images/15-3J_p8_data_8364436f9c.png)",
    "options": [],
    "answer": "See solution",
    "solution": "將最中間兩列所有數字 $+n-2$，向外兩列 $+n-3$，⋯，最外兩列 $+0$，飛機變成如下圖。如此一來，不論如何安排這些乘客的座位，皆會選到 $1+2+\\dots+(2n-2) = (n-1)(2n-1)$，而不論如何安排座位，都會加上 $(0+1+\\dots+(n-2)) \\times 2 = (n-1)(n-2)$，故不論排法，都會收到 $n^2-1$ 的錢。\n\n![](images/15-3J_p10_data_03a846f462.png)",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 14705,
    "subject": "Mathematics (Olympiad)",
    "question": "Define a domino to be a $1 \\times 2$ rectangular block. A $2023 \\times 2023$ square grid is filled with non-overlapping dominoes, leaving a single $1 \\times 1$ gap. John then repeatedly slides dominoes into the gap; each domino is moved at most once. What is the maximum number of times that John could have moved a domino?\n\n(Example: In the $3 \\times 3$ grid shown below, John could move 2 dominoes: D, followed by A.)\n\n<table><tr><td>A</td><td>B</td><td>C</td></tr><tr><td>D</td><td></td><td></td></tr></table>",
    "options": [],
    "answer": "See solution",
    "solution": "Label the squares in the grid $(0,0)$ to $(2022, 2022)$. Consider the position of the gap after a domino is moved. Note that the parity of the coordinates of the square containing the gap will not change. Also, the same square cannot contain the gap twice, otherwise this implies that a domino was moved into that square, and was moved out of the square again later, which is disallowed.\n\nConsidering coordinates modulo $2$, in the $2023 \\times 2023$ grid, there are $1012^2$ squares labelled $(0,0)$, $1012 \\times 1011$ squares labelled $(0,1)$ or $(1,0)$, and $1011^2$ squares labelled $(1,1)$. Hence, the number of moves is at most $1012^2 - 1$, if all the $(0,0)$ squares are the 'gap square' at some point.\n\nThis is attainable by connecting these squares in a 'snake' pattern, and sliding dominoes accordingly; the remaining $1 \\times 2022$ rectangles can also be tiled with dominoes that do not move throughout.\n\n![](images/Singapore2023-booklet_p2_data_04d272f779.png)",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 14706,
    "subject": "Mathematics (Olympiad)",
    "question": "The numbers\n$$\n\\frac{1}{2016}, \\frac{2}{2016}, \\frac{3}{2016}, \\dots, \\frac{2015}{2016}\n$$\nare written on a blackboard. With each move, one may erase any two numbers $a$ and $b$ and replace them with\n$$\n3ab - 2a - 2b + 2.\n$$\nWhat will be the single remaining number after 2014 moves?",
    "options": [],
    "answer": "See solution",
    "solution": "Note that if $a = \\frac{1344}{2016} = \\frac{2}{3}$, then\n$$\n3ab - 2a - 2b + 2 = \\frac{2}{3},\n$$\nirrespective of the value of $b$. Hence, $\\frac{2}{3}$ will always remain on the blackboard. $\\square$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 14707,
    "subject": "Mathematics (Olympiad)",
    "question": "A and B are points on a given circle. Points C and D move along the circle such that C and D are on the same side of the line $AB$ and the length of the segment $CD$ does not change. $I_1$ and $I_2$ are incenters of the triangles $ABC$ and $ABD$. Prove that there exists a circle such that at every moment the line $I_1I_2$ is tangent to this circle.\n\n![](images/Baltic_Way_SHL_2009-11_13-16_p200_data_a3682f89cc.png)",
    "options": [],
    "answer": "See solution",
    "solution": "Let $M$ be the midpoint of the second arc $AB$. Then it is well-known that $MI_1 = MA = MB$ (lemma of P. Mansion), and similarly $MI_2 = MA = MB$. Observe that $\\angle I_1MI_2 = \\angle CMD = \\text{const}$. Therefore, the points $I_1$ and $I_2$ move along the circle with center $M$ and radius $AM$, and the length $I_1I_2$ is constant. Hence, all possible segments $I_1I_2$ are tangent to some smaller circle with center $M$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 14708,
    "subject": "Mathematics (Olympiad)",
    "question": "In a right triangle $ABC$ with right angle $C$, on the sides $BC$, $AC$, and $AB$, points $D$, $E$, and $F$ are chosen respectively so that $\\angle DAB = \\angle CBE$ and $\\angle BEC = \\angle AEF$. Prove that $DB = DF$.",
    "options": [],
    "answer": "See solution",
    "solution": "Consider the point $K$, symmetric to $B$ with respect to $C$. Then $\\triangle BEC \\cong \\triangle KCE$, and $\\angle KEC = \\angle BEC = \\angle AEF$, so points $K$, $E$, $F$ are collinear. Then it follows that\n\n![](images/Ukraine_2021-2022_p24_data_a4d799e086.png)\n\n$$\n\\angle FAD = \\angle EBC = \\angle EKC = \\angle FKD,\n$$\n\nso points $A$, $K$, $D$, $F$ are concyclic. Therefore,\n\n$$\n\\angle BFD = 180^\\circ - \\angle AFD = \\angle AKD.\n$$\n\nAs $\\triangle ABC \\cong \\triangle AKC$, and $\\angle AKC = \\angle ABC$, the equality above is rewritten as $\\angle BFD = \\angle ABC = \\angle FBD$. Thus, in $\\triangle FBD$, it follows that $DB = DF$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 14709,
    "subject": "Mathematics (Olympiad)",
    "question": "A positive integer $n$ is called *super special* if it can be represented in the form\n$$n = \\frac{x^3 + 2y^3}{u^3 + 2v^3}$$\nfor some positive integers $x, y, u, v$.\n\nProve that:\n\n(a) There are infinitely many super special positive integers.\n\n(b) $2014$ is not super special.",
    "options": [],
    "answer": "See solution",
    "solution": "(a) Every perfect cube $k^3$ of a positive integer is super special because we can write\n$$\nk^3 = k^3 \\frac{x^3 + 2y^3}{x^3 + 2y^3} = \\frac{(kx)^3 + 2(ky)^3}{x^3 + 2y^3}\n$$\nfor some positive integers $x, y$.\n\n(b) Observe that $2014 = 2 \\cdot 19 \\cdot 53$. If $2014$ is super special, then we have\n$$\nx^3 + 2y^3 = 2014(u^3 + 2v^3)\n$$\nfor some positive integers $x, y, u, v$. Assume $x^3 + 2y^3$ is minimal with this property. If $19$ divides $x^3 + 2y^3$, then it divides both $x$ and $y$. Indeed, if $19$ does not divide $x$, then it does not divide $y$ either. The relation $x^3 \\equiv -2y^3 \\pmod{19}$ implies $(x^3)^6 \\equiv (-2y^3)^6 \\pmod{19}$, which is $x^{18} \\equiv 2^6 y^{18} \\pmod{19}$. By Fermat's Little Theorem, $x^{18} \\equiv y^{18} \\equiv 1 \\pmod{19}$, so $1 \\equiv 2^6 \\pmod{19}$, i.e., $19$ divides $63$, which is not possible.\n\nIt follows $x = 19x_1$, $y = 19y_1$ for some positive integers $x_1, y_1$. Substituting into the equation gives\n$$\n19^3 x_1^3 + 2 \\cdot 19^3 y_1^3 = 2014(u^3 + 2v^3)\n$$\nwhich simplifies to\n$$\n19^2(x_1^3 + 2y_1^3) = 2 \\cdot 53(u^3 + 2v^3)\n$$\nso $19$ divides $u^3 + 2v^3$, and thus $u = 19u_1$, $v = 19v_1$. Substituting again,\n$$\nx_1^3 + 2y_1^3 = 2014(u_1^3 + 2v_1^3)\n$$\nwhich contradicts the minimality of $x^3 + 2y^3$. Therefore, $2014$ is not super special.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 14710,
    "subject": "Mathematics (Olympiad)",
    "question": "Points $P$ and $Q$ lie inside acute triangle $ABC$ such that $\\angle PAB = \\angle QAC$ and $\\angle PBA = \\angle QBC$. Point $D$ lies on side $BC$. Prove that\n\n$$\n\\angle DPC + \\angle APB = 180^{\\circ} \\quad \\text{if and only if} \\quad \\angle DQB + \\angle AQC = 180^{\\circ}.\n$$\n\n*Note.* It is well known that from $\\angle PAB = \\angle QAC$ and $\\angle PBA = \\angle QBC$, we may conclude that $\\angle PCB = \\angle QCA$. Indeed, $P$ and $Q$ are isogonal conjugates of each other. This fact can easily be established by trigonometric form of Ceva's theorem. Hence we may set $\\angle BAC = A$, $\\angle ACB = C$, $\\angle CBA = B$, $\\angle PAB = \\angle QAC = x$, $\\angle PBC = \\angle QBA = y$, and $\\angle PCA = \\angle QCB = z$.\n\nWe present three proofs based on the configurations shown below. Our proofs can be slightly modified for other configurations (if $Q$ lies inside triangle $BPC$, for instance). Because the problem statement is symmetric with respect to exchanging the pairs of points $(B, P)$ and $(C, Q)$, it suffices to prove the “only if” part. Hence in the first two proofs, we assume that\n\n$$\n\\angle APB + \\angle CPD = 180^{\\circ} \\quad \\text{or} \\quad \\angle APC + \\angle BPD = 180^{\\circ}. \\qquad (1)\n$$\n\n![](images/pamphlet0910_main_p52_data_0b7ac07461.png)",
    "options": [],
    "answer": "See solution",
    "solution": "**Solution 1.** Extend $BP$ through $P$ to $P_1$ so that $\\angle PAP_1 = \\angle PCD = C - z$. By (1), $\\angle DPC = 180^{\\circ} - \\angle APB = \\angle APP_1$. Hence triangles $PAP_1$ and $PCD$ are similar. In particular, there is a spiral similarity $S_1$ centered at $P$ sending $PAP_1$ to $PCD$. Thus, there is another spiral similarity $S_2$ centered at $P$ sending $PAC$ to $PP_1D$, from which it follows that $\\angle PP_1D = \\angle PAC = A - x = \\angle BAQ$. Combining the last equation with $\\angle ABQ = B - y = \\angle PBQ = \\angle P_1BQ$, we deduce that triangle $QAB$ is similar to triangle $DP_1B$; that is, there is a spiral similarity $S_3$ centered at $B$ sending $BAQ$ to $BP_1D$. Thus, there is another spiral similarity $S_4$ centered at $B$ sending $BAP_1$ to $BQD$. It therefore follows that\n\n$$\n\\angle BQD = \\angle BAP_1 = \\angle BAP + \\angle PAP_1 = \\angle QAC + \\angle PCD = \\angle QAC + \\angle QCA = 180^\\circ - \\angle AQC,\n$$\n\nas desired.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 14711,
    "subject": "Mathematics (Olympiad)",
    "question": "Find the maximum and minimum of the function\n$$\ny = \\sqrt{x+27} + \\sqrt{13-x} + \\sqrt{x}.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "The domain of $y$ is $x \\in [0, 13]$.\n\nWe have\n$$\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$$\nThe equality holds when $4x = 9(13-x) = x + 27$. It is so for $x = 9$. Therefore, the maximum of $y$ is $11$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 14712,
    "subject": "Mathematics (Olympiad)",
    "question": "從集合 $S = \\{1, 2, 3, \\ldots, 2024\\}$ 中，取出 1000 個數而造出具有 1000 個元素的子集 $T$，而 $T$ 的最小元素是 $k$，求 $k$ 的期望值。",
    "options": [],
    "answer": "See solution",
    "solution": "設 $M$ 是所求的期望值，則有：\n\n$$\n\\begin{aligned}\n\\binom{2024}{1000} M &= 1 \\cdot \\binom{2023}{999} + 2 \\cdot \\binom{2022}{999} + 3 \\cdot \\binom{2021}{999} + \\cdots + 1025 \\cdot \\binom{999}{999} \\\\\n&= \\sum_{a+b=1025} \\binom{a}{1} \\binom{b+999}{999} \\\\\n&= \\binom{1025+999+1}{1+999+1} \\\\\n&= \\binom{2025}{1001} \\\\\nM &= \\frac{2025}{1001}\n\\end{aligned}\n$$\n\n*Remark*: 兩類二項係數恆等式來自數列的摺積 (convolution)\n\n$$\n\\sum_{a+b=r} \\binom{m}{a} \\binom{n}{b} = \\binom{m+n}{r}, \\quad \\sum_{a+b=r} \\binom{a+m}{m} \\binom{b+n}{n} = \\binom{m+n+r+1}{m+n+1}\n$$\n\n第一類等式來自\n\n$$\n(1+x)^m = \\sum_{a=0}^{m} \\binom{m}{a} x^a \\quad \\text{與} \\quad (1+x)^n = \\sum_{b=0}^{n} \\binom{n}{b} x^b\n$$\n\n的摺積。第二類等式來自於\n\n$$\n\\frac{1}{(1-x)^{m+1}} = \\sum_{m=0}^{\\infty} \\binom{m+k}{m} x^k, \\quad \\frac{1}{(1-x)^{n+1}} = \\sum_{l=0}^{\\infty} \\binom{n+l}{l} x^l\n$$\n\n的摺積。",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 14713,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all functions $f : \\mathbb{Q} \\to \\mathbb{Q}$ such that\n\n$$\nf(x + f(y + f(z))) = y + f(x + z)\n$$\nfor all $x, y, z \\in \\mathbb{Q}$.",
    "options": [],
    "answer": "See solution",
    "solution": "Set $x = a = f(0)$ in (3), then $f(y+z) = f(z) + f(y) - a$, or $f(y+z) - a = (f(y) - a) + (f(z) - a)$. We see that the function $f(x) - a$ is additive, thus $f(x) - a = \\alpha x$, or $f(x) = \\alpha x + a$. Substituting $f$ in $(*)$ gives $f(x) = x$ and $f(x) = -x + a$, where $a \\in \\mathbb{Q}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 14714,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all functions $f : \\mathbb{R} \\to \\mathbb{R}$ such that for all real numbers $a, b, c, d$, the following inequality holds:\n\n$$\n(f(a) - f(b))(f(c) - f(d)) + (f(a) - f(d))(f(b) - f(c)) \\leq (a - b)(c - d) + (a - d)(b - c).\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "Substituting $a = b = c = d = 0$ gives $2f(0)^2 \\le 0$, so $f(0) = 0$.\n\nNext, let $b = a - x$, $c = a$, $d = a - y$. Then $a - b = x$, $a - c = 0$, $a - d = y$, so:\n\n$$\nf(y)(f(x) + f(-x)) \\le 0 \\qquad (1)\n$$\n\nSuppose $f(y) \\ne 0$ for some $y$. Setting $x = y$ in (1):\n\n$$\n0 < f(y)^2 \\le -f(y)f(-y)\n$$\n\nThus, $f(y)$ and $f(-y)$ have opposite signs. Assume $f(y) > 0$.\n\nNow, for arbitrary $a$ and $y$, set $b = a$, $c = 0$, $d = a - y$:\n\n$$\nf(y)f(a) \\le a f(y) \\implies f(a) \\le a\n$$\n\nSimilarly, with $d = a + y$:\n\n$$\nf(-y)f(a) \\le a f(-y) \\implies f(a) \\ge a\n$$\n\nTherefore, $f(a) = a$ for all $a$.\n\nIf $f(y) = 0$ for all $y$, then $f(x) = 0$ for all $x$.\n\nBoth $f(x) = 0$ and $f(x) = x$ satisfy the original inequality, as can be checked directly:\n\nFor $f(x) = 0$, both sides are $0$.\n\nFor $f(x) = x$:\n\n$$\n(a-b)(c-d) + (a-d)(b-c) = (a-c)(b-d)\n$$\n\nSo equality holds.\n\nThus, the only solutions are $f(x) = 0$ and $f(x) = x$ for all $x$.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 14715,
    "subject": "Mathematics (Olympiad)",
    "question": "求滿足以下條件的最小正整數 $n$，或證明滿足以下條件的正整數 $n$ 不存在：\n\n存在無窮多組由 $n$ 個相異正有理數所構成的數組 $(a_1, a_2, \\cdots, a_n)$，使得\n\n$$\na_1 + a_2 + \\cdots + a_n \\quad \\text{和} \\quad \\frac{1}{a_1} + \\frac{1}{a_2} + \\cdots + \\frac{1}{a_n}\n$$\n\n皆為整數。",
    "options": [],
    "answer": "See solution",
    "solution": "**答案：** $n = 3$\n\n很明顯，當 $n = 1$ 時，唯一的解是 $a_1 = 1$。現在我們來證明：\n\n1. 只有有限組 $(x, y) \\in \\mathbb{Q}_{>0}^2$ 使得 $x + y$ 和 $\\frac{1}{x} + \\frac{1}{y}$ 都是整數。\n\n設 $x = a/b$，$y = c/d$ 為標準形式。則 $x + y \\in \\mathbb{Z}$ 和 $\\frac{1}{x} + \\frac{1}{y} \\in \\mathbb{Z}$ 等價於：\n\n(i) $bd \\mid ad + bc$。\n\n注意這導致 $d \\mid ad + bc \\Rightarrow d \\mid bc \\Rightarrow b \\mid c$（因為 $(c, d) = 1$）。\n\n對稱地，得到 $b \\mid d$，所以 $b = d$。\n\n(ii) $ac \\mid ad + bc$。\n\n同理可得 $a = c$。\n\n因此必須有 $x = y$，問題變成尋找 $x \\in \\mathbb{Q}_{>0}$ 使得 $2x \\in \\mathbb{Z}$ 且 $\\frac{2}{x} \\in \\mathbb{Z}$。顯然唯一解是 $x = 1/2, 1, 2$。\n\n2. 有無窮多組 $(x, y, z) \\in \\mathbb{Q}_{>0}^3$ 使得 $x + y + z$ 和 $\\frac{1}{x} + \\frac{1}{y} + \\frac{1}{z}$ 都是整數。\n\n特別地，設 $x + y + z = 1$，此時可寫成：\n\n$$\n(x, y, z) = \\left( \\frac{a}{a+b+c}, \\frac{b}{a+b+c}, \\frac{c}{a+b+c} \\right)\n$$\n\n其中 $a, b, c \\in \\mathbb{Z}$。要求：\n\n$$\n\\frac{1}{x} + \\frac{1}{y} + \\frac{1}{z} = \\frac{a+b+c}{a} + \\frac{a+b+c}{b} + \\frac{a+b+c}{c} \\in \\mathbb{Z}\n$$\n\n即：\n\n$$\n\\frac{b+c}{a} + \\frac{a+c}{b} + \\frac{a+b}{c} \\in \\mathbb{Z}\n$$\n\n進一步固定 $a = 1$，只需證明存在無窮多整數 $(b, c)$ 使得\n\n$$\n\\frac{1}{b} + \\frac{1}{c} + \\frac{c}{b} + \\frac{b}{c} = 3 \\iff b^2 + c^2 - 3bc + b + c = 0 \\quad (1)\n$$\n\n為證明方程 (1) 有無窮多解，使用 *Vieta 跳躍*（又稱 *根翻轉*）：以 $(b, c) = (2, 3)$ 為起點。以下算法產生無窮多解。令 $c \\ge b$，將 (1) 視為 $b$ 的二次方程：\n\n$$\nb^2 - (3c - 1) b + (c^2 + c) = 0 \\quad (2)\n$$\n\n則存在另一根 $b_0 \\in \\mathbb{Z}$，滿足 $b + b_0 = 3c - 1$ 且 $b \\times b_0 = c^2 + c$。由 $c \\ge b$，\n\n$$\nb_0 = \\frac{c^2 + c}{b} \\ge \\frac{c^2 + c}{c} > c\n$$\n\n因此，從解 $(b, c)$ 可得新解 $(c, b_0)$，且 $b_0 > c$。可重複此跳躍，繼續產生無窮多解。",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 14716,
    "subject": "Mathematics (Olympiad)",
    "question": "Prove that there exist infinitely many integers $x$, $y$, $z$ for which the sum of the digits in the decimal representation of $4x^4 + y^4 - z^2 + 4xyz$ is at most 2.",
    "options": [],
    "answer": "See solution",
    "solution": "We can rewrite\n\n$$\n\\begin{aligned}\n4x^4 + y^4 - z^2 + 4xyz &= (4x^4 + y^4 + 4x^2y^2) - (4x^2y^2 + z^2 - 4xyz) \\\\\n&= (2x^2 + y^2)^2 - (2xy - z)^2 \\\\\n&= (2x^2 + y^2 - 2xy + z)(2x^2 + y^2 + 2xy - z)\n\\end{aligned}\n$$\n\nLet $A = 2x^2 + y^2 - 2xy + z$ and $B = 2x^2 + y^2 + 2xy - z$. Their sum is $A + B = 4x^2 + 2y^2$.\n\nChoose $x = 5^{n+1}$ and $y = 2^n$ for any integer $n \\ge 1$. Set $A = 4x^2 = 4 \\cdot 5^{2n+2}$ and $B = 2y^2 = 2 \\cdot 2^{2n}$. Then $AB = 2 \\cdot 10^{2n+2}$, whose sum of digits is 2.\n\nThe equation $A = 4x^2$ gives $2x^2 + y^2 - 2xy + z = 4x^2$, so\n\n$$\nz = 2x^2 + 2xy - y^2 = 2 \\cdot 5^{2n+2} + 10^{n+1} - 4^n.\n$$\n\nThis $z$ is a positive integer. Thus,\n\n$$\n4x^4 + y^4 - z^2 + 4xyz = 2 \\cdot 10^{2n+2},\n$$\n\nwhose sum of digits is 2. Since $n$ can be any integer $\\ge 1$, there are infinitely many such triples.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 14717,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $a$, $b$, and $c$ be non-negative real numbers, no two of which are equal. Prove that\n\n$$\n\\frac{a^2}{(b-c)^2} + \\frac{b^2}{(c-a)^2} + \\frac{c^2}{(a-b)^2} > 2.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "The left-hand side is symmetric with respect to $a$, $b$, and $c$. Hence, we may assume that $a > b > c \\ge 0$. Note that replacing $(a, b, c)$ with $(a-c, b-c, 0)$ lowers the value of the left-hand side, since the numerators of each of the fractions would decrease and the denominators remain the same. Therefore, to obtain the minimum possible value of the left-hand side, we may assume that $c=0$.\n\nThen the left-hand side becomes\n\n$$\n\\frac{a^2}{b^2} + \\frac{b^2}{a^2},\n$$\n\nwhich yields, by the Arithmetic Mean - Geometric Mean Inequality,\n\n$$\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$$\n\nwith equality if and only if $\\frac{a^2}{b^2} = \\frac{b^2}{a^2}$, or equivalently, $a^4 = b^4$. Since $a, b \\ge 0$, $a = b$. But since no two of $a, b, c$ are equal, $a \\ne b$. Hence, equality cannot hold. This yields\n\n$$\n\\frac{a^2}{b^2} + \\frac{b^2}{a^2} > 2.\n$$\n\nUltimately, this implies the desired inequality. $\\square$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 14718,
    "subject": "Mathematics (Olympiad)",
    "question": "The sequence $a_0, a_1, \\dots$ is defined by the initial conditions $a_0 = 1$, $a_1 = 6$ and the recursion\n$$a_{n+1} = 4a_n - a_{n-1} + 2$$\nfor $n > 1$. Prove that $a_{2^k-1}$ has at least three prime factors for every positive integer $k > 3$.",
    "options": [],
    "answer": "See solution",
    "solution": "Consider the sequence $b_0, b_1, \\dots$, defined by the initial conditions $b_0 = 1$, $b_1 = 2$, and the recursion $b_{n+1} = 4b_n - b_{n-1}$ for $n \\ge 1$. We have $a_n = b_{n+1} - 1$ for all $n \\ge 0$. In particular, $a_{2^k-1} = b_{2^k} - 1$ for $k \\ge 0$.\n\nIt is not hard to see that the general formula for $b_n$ is\n$$\nb_n = \\frac{(2 + \\sqrt{3})^n + (2 - \\sqrt{3})^n}{2}, \\quad n \\ge 2.\n$$\nLet $k \\ge 3$ be fixed. It follows from this formula that\n$$\nb_{2^k} = \\sum_{j=0}^{2^{k-1}} \\binom{2^k}{2j} 2^{2^k-2j} 3^j\n$$\nand thus\n$$\nb_{2^k} \\equiv 1 \\pmod{3} \\quad \\text{and} \\quad b_{2^k} \\equiv 1 \\pmod{4}.\n$$\n\nThe terms of the sequence $c_n = \\frac{(2+\\sqrt{3})^n - (2-\\sqrt{3})^n}{2\\sqrt{3}}$ are positive integers, and satisfy $b_n^2 - 3c_n^2 = 1$ for all $n \\in \\mathbb{N}$. In particular,\n$$\nb_{2k}^2 - 1 = 3c_{2k}^2.\n$$\n\nApplying the identity $(x^{2^s} - y^{2^s})(x^{2^s} + y^{2^s}) = x^{2^{s+1}} - y^{2^{s+1}}$ with $x = 2+\\sqrt{3}$ and $y = 2-\\sqrt{3}$ for $s = 0, 1, \\dots, k-1$, we get\n$$\n2\\sqrt{3} \\prod_{j=0}^{k-1} (2b_{2^j}) = (2 + \\sqrt{3})^{2k} - (2 - \\sqrt{3})^{2k}.\n$$\nTherefore,\n$$\nc_{2^k} = 2^{k+1} b_2 b_{2^2} \\dots b_{2^{k-1}}.\n$$\n\nIt follows that\n$$\n2 \\mid b_{2^k} + 1, \\quad \\gcd(b_{2^j}, 6) = 1 \\text{ for every } j = 1, 2, \\dots, k-1.\n$$\n\nSince $b_{2^k} + 1 \\equiv 2 \\pmod{3}$, we have\n$$\n3 \\mid b_{2^k} - 1 \\quad \\text{and} \\quad 2^{2^{k+1}} \\mid b_{2^k} - 1.\n$$\n\nSuppose there exists $m \\ge 3$ such that $b_{2^m} - 1$ has at most two prime factors. Then, by the above, these must be 2 and/or 3. Furthermore, we must have $b_{2^m} = 2^{2^{m+1}} \\cdot 3 + 1$. Therefore,\n$$\nc_{2^m}^2 = \\frac{b_{2^m}^2 - 1}{3} = 4^{m+1}(3 \\cdot 4^m + 1).\n$$\n\nOn the other hand, $c_{2^m}^2 = 4^{m+1} \\prod_{j=1}^{m-1} b_{2^j}^2 > 4^{m+1}(3c_{2^{m-1}}+1)$, and thus $c_{2^m}^2 > 4^{m+1}(3 \\cdot 4^m + 1)$ as $c_{2^{m-1}} \\ge 2^m$. This contradicts the previous equation, so $a_{2^k-1}$ has at least three prime factors for every positive integer $k \\ge 3$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 14719,
    "subject": "Mathematics (Olympiad)",
    "question": "A primary school has several classes, each consisting of an equal number of boys and girls. The size of each class does not exceed $2n$, where $n$ is a positive integer. The headmaster plans to establish a few activity groups, such that:\n\n1. There are no more than $2n$ groups, and each student belongs to exactly one group.\n2. Any two students in a class belong to different groups.\n3. For each group, the numbers of boys and girls differ by at most $1$.\n\nProve that the plan is always feasible.",
    "options": [],
    "answer": "See solution",
    "solution": "Suppose there are $k$ classes. Apply induction on $n$.\n\n**Base case ($n = 1$):** Each class consists of one boy and one girl. Let $A$ and $B$ be two activity groups: $\\lfloor \\frac{k}{2} \\rfloor$ boys join $A$, and their classmates (girls) join $B$; the remaining $\\lfloor \\frac{k}{2} \\rfloor$ boys join $B$, and their classmates (girls) join $A$. Since $|\\lfloor \\frac{k}{2} \\rfloor - \\lfloor \\frac{k}{2} \\rfloor| \\le 1$, the requirements are met.\n\n**Inductive step:** Suppose when $n = t$ ($t \\in \\mathbb{Z}_+$), the plan is feasible. For $n = t + 1$, each class has no more than $2(t + 1)$ students. Choose one boy and one girl from each class, and let them join groups $A$ and $B$ as in the base case. Now each class has no more than $2t$ students who are to join no more than $2t$ groups. By the induction hypothesis, this is always doable. Clearly, all requirements are met.\n\nThis completes the induction and the proof. $\\square$",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 14720,
    "subject": "Mathematics (Olympiad)",
    "question": "Банкнота од 100 денари треба да се раситни на монети од 2 и 5 денари, при што нивниот број е 32. Ако такво раситнување постои, колку монети од 2 и колку монети од 5 денари се употребени?",
    "options": [],
    "answer": "See solution",
    "solution": "**Прв начин:**\n\nАко избереме сите 32 монети да се од 2 денари, тогаш би имале 64 денари, па банкнотата од 100 денари не е раситнета. Ако една монета од 32-те монети од 2 денари се замени со монета од 5 денари, сумата се зголемува за $5 - 2 = 3$ денари. Значи, сумата од 64 денари треба да ја зголемиме за $100 - 64 = 36$ денари. Според тоа, постапката на замена на монетата од 2 со монета од 5 денари треба да ја повториме $36 \\div 3 = 12$ пати. Така би добиле 12 монети од 5 денари и 20 монети од по 2 денари. Тоа е бараното раситнување.\n\n**Втор начин:**\n\nНека $x$ е бројот на монети од 5 денари. Тогаш $32 - x$ е бројот на монети од 2 денари. Се добива равенката:\n\n$$5x + 2(32 - x) = 100$$\n\nод каде $x = 12$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 14721,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $a$, $b$, and $c$ be positive real numbers such that $a + b + c = 2$. Prove that\n\n$$\n\\frac{(a-1)^2}{b} + \\frac{(b-1)^2}{c} + \\frac{(c-1)^2}{a} \\geq \\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$$",
    "options": [],
    "answer": "See solution",
    "solution": "By the Cauchy-Bunyakovsky-Schwarz inequality, we have\n\n$$\n\\frac{(a-1)^2}{b} + \\frac{(b-1)^2}{c} \\geq \\frac{(2-a-b)^2}{b+c} = \\frac{c^2}{b+c},\n$$\n\nand similar inequalities hold for all pairs. By adding the inequalities, we get\n\n$$\n\\frac{(a-1)^2}{b} + \\frac{(b-1)^2}{c} + \\frac{(c-1)^2}{a} \\geq \\frac{1}{2} \\left( \\frac{b^2}{a+b} + \\frac{c^2}{b+c} + \\frac{a^2}{c+a} \\right).\n$$\n\nNote that the initial claim follows from this, because\n\n$$\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$$\n\nwhich holds since\n\n$$\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$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 14722,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $F'$ be the foot of an altitude from $A$ in the triangle $ABC$.\n\n![](images/BielorrusiaBOOK18-E-A4_p20_data_87d2b3016d.png)\n\nProve that the points $D$, $F'$, and $E$ are colinear. This will imply that $F = F'$, so $AF$ is an altitude of triangle $ABC$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let the lines $EL$ and $CD$ meet at $K$. Since the angles $ADK$ and $AEK$ are right, the points $A$, $K$, $D$, and $E$ lie on the circle with diameter $AK$. Hence $\\angle DKE = \\angle DAE$. Since $AD$ is the bisector, $\\angle DAE = \\angle DAC$, whence $\\angle DKE = \\angle DAC$. The latter is equivalent to $\\angle DKL = \\angle CAL$, whence $A$, $L$, $C$, and $K$ lie on a circle.\n\nConsider the Simson line of $A$ and the triangle $LCK$. The foot of the perpendicular from $A$ to $CK$ is $D$, and to $LK$ is $E$, therefore this line is $DE$. Since $DE$ meets $CL$ at $F$, $F$ is the foot of the perpendicular from $A$ to $CL$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 14723,
    "subject": "Mathematics (Olympiad)",
    "question": "Prove that for every $n \\ge 3$, we have $0.6 < a_n < 0.7$, where\n\n$$\na_n = \\frac{1}{n+1} + \\frac{1}{n+2} + \\dots + \\frac{1}{2n}.\n$$\n\nShow that the first decimal digit of $a_n$ is 6 for all $n \\ge 3$.",
    "options": [],
    "answer": "See solution",
    "solution": "We have\n\n$$\na_{n+1} - a_n = \\frac{1}{2n+1} + \\frac{1}{2n+2} - \\frac{1}{n+1} = \\frac{1}{(2n+1)(2n+2)} > 0,\n$$\n\nso the sequence $(a_n)_{n \\ge 1}$ is strictly increasing.\n\nConsider the sequence $(b_n)_{n \\ge 1}$, defined by\n\n$$\nb_n = 1 + \\frac{1}{2} + \\dots + \\frac{1}{n} - \\ln n.\n$$\n\nIt is clear that\n\n$$\nb_{2n} - b_n = \\frac{1}{n+1} + \\dots + \\frac{1}{2n} - \\ln 2 = a_n - \\ln 2. \\quad (1)\n$$\n\nWe will show that $(b_n)_{n \\ge 1}$ is strictly decreasing. Indeed, we have\n\n$$\nb_{n+1} - b_n = \\frac{1}{n+1} - \\ln \\frac{n+1}{n} = \\frac{1}{n+1} \\left[ 1 - \\ln \\left( 1 + \\frac{1}{n} \\right)^{n+1} \\right] < 0, \\\\\n\\text{since } \\left(1 + \\frac{1}{n}\\right)^{n+1} > e, \\ n = 1, 2, \\dots\n$$\n\nIt follows that $b_{2n} - b_n > 0$, and hence from (1) we obtain $a_n < \\ln 2 < 0.7$. As in the previous solution, for $n \\ge 3$, we have $0.6 \\le a_n < 0.7$, so the first decimal in this case is 6.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 14724,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $x_1, \\dots, x_n$ be non-negative real numbers, not all of which are zero.\n\n1. Prove that\n$$\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$$\n\n2. Show that, for each $n \\ge 1$, both inequalities can hold as equalities.",
    "options": [],
    "answer": "See solution",
    "solution": "Applying the AM-GM inequality gives\n$$\n\\begin{aligned}\n\\left(\\sum_{k=1}^{n} \\frac{x_k}{k}\\right) \\left(\\sum_{k=1}^{n} kx_k\\right) &= \\frac{1}{n} \\left(\\sum_{k=1}^{n} \\frac{nx_k}{k}\\right) \\left(\\sum_{k=1}^{n} kx_k\\right) \\\\ &\\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.\n\\end{aligned}\n$$\n(The last inequality is proved by noting $\\frac{n}{k} + k \\le n + 1$, which is equivalent to $(n-k)(k-1) \\ge 0$.)\n\nThis gives the necessary upper bound; this bound is achieved, for instance, if $x_1 = x_n = 1$ and $x_2 = \\dots = x_{n-1} = 0$.\n\nFor the lower bound, estimate the numerator by the Cauchy-Schwarz inequality:\n$$\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$$\nthe equality holds here if exactly one of the $x_i$ is non-zero.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 14725,
    "subject": "Mathematics (Olympiad)",
    "question": "Consider a circle centered at $O$ with radius $r$ and a line $\\ell$ not passing through $O$. A grasshopper is jumping to and fro between the points of the circle and the line, the length of each jump being $r$. Prove that there are at most 8 points for the grasshopper to reach.\n\n![](images/RMC2013_final_p65_data_71f7461853.png)",
    "options": [],
    "answer": "See solution",
    "solution": "We assume that, when having the choice between only two places to jump to, the grasshopper never jumps back to the point from which he got to that place. Let us denote by $P_1$ the starting point of the grasshopper, with $P_2$ the point on the line on which he has jumped from $P_1$, and so on. As the length of the jumps are all equal to $r$, $OP_1P_2P_3$ is a rhombus (possibly a degenerate one). Similarly, $OP_3P_4P_5$ is also a rhombus. It follows that the triangles $P_1OP_5$ and $P_2P_3P_4$ are congruent (SAS), and from here we obtain that $P_1P_5$ is parallel to $\\ell$. We deduce that $P_5$ is the reflection of $P_1$ across the perpendicular line from $O$ onto $\\ell$. (This fact remains true even in the degenerate cases.) From $P_5$, the grasshopper can get to $P_9$ which, as above, is the reflection of $P_5$ across the perpendicular line from $O$ onto $\\ell$, i.e. $P_1$. In conclusion, the grasshopper can reach only the points $P_k$, $k = \\overline{1,8}$ (which are not necessarily distinct).",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 14726,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all positive real numbers $x, y, z$ such that\n\n$$\n\\frac{1}{x^2+1} + \\frac{1}{y^2+4} + \\frac{1}{z^2+9} = \\frac{7}{12} \\sqrt{\\frac{1}{x^2} + \\frac{1}{y^2} + \\frac{1}{z^2}}.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "We have $x^2+1 \\ge 2x$, $y^2+4 \\ge 4y$, $z^2+9 \\ge 6z$, hence\n\n$$\n\\frac{1}{x^2+1} + \\frac{1}{y^2+4} + \\frac{1}{z^2+9} \\le \\frac{1}{2x} + \\frac{1}{4y} + \\frac{1}{6z}. \\quad (1)\n$$\n\nUsing the Cauchy-Schwarz inequality, it follows that\n\n$$\n\\left(\\frac{1}{2x} + \\frac{1}{4y} + \\frac{1}{6z}\\right)^2 \\le \\left(\\frac{1}{2^2} + \\frac{1}{4^2} + \\frac{1}{6^2}\\right) \\left(\\frac{1}{x^2} + \\frac{1}{y^2} + \\frac{1}{z^2}\\right) = \\frac{1}{4} \\cdot \\frac{49}{36} \\left(\\frac{1}{x^2} + \\frac{1}{y^2} + \\frac{1}{z^2}\\right),\n$$\n\nhence\n\n$$\n\\frac{1}{2x} + \\frac{1}{4y} + \\frac{1}{6z} \\le \\frac{7}{12} \\sqrt{\\frac{1}{x^2} + \\frac{1}{y^2} + \\frac{1}{z^2}}. \\quad (2)\n$$\n\nFrom (1) and (2) we get\n\n$$\n\\frac{1}{x^2+1} + \\frac{1}{y^2+4} + \\frac{1}{z^2+9} \\le \\frac{7}{12} \\sqrt{\\frac{1}{x^2} + \\frac{1}{y^2} + \\frac{1}{z^2}}.\n$$\n\nwith equality if and only if $x = 1$, $y = 2$, $z = 3$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 14727,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $ABC$ be a triangle with $\\angle A = 90^\\circ$. Points $D$ and $E$ lie on sides $AC$ and $AB$, respectively, such that $\\angle ABD = \\angle DBC$ and $\\angle ACE = \\angle ECB$. Segments $BD$ and $CE$ meet at $I$. Determine whether it is possible for segments $AB$, $AC$, $BI$, $ID$, $CI$, $IE$ to all have integer lengths.",
    "options": [],
    "answer": "See solution",
    "solution": "The answer is *no*, it is not possible for segments $AB, AC, BI, ID, CI, IE$ to all have integer lengths.\n\nSuppose on the contrary that these segments do have integer side lengths. Set $\\alpha = \\angle ABD = \\angle DBC$ and $\\beta = \\angle ACE = \\angle ECB$. Note that $I$ is the incenter of triangle $ABC$, and so $\\angle BAI = \\angle CAI = 45^\\circ$. Applying the Law of Sines to triangle $ABI$ yields\n\n$$\n\\frac{AB}{BI} = \\frac{\\sin(45^\\circ + \\alpha)}{\\sin 45^\\circ} = \\sin \\alpha + \\cos \\alpha,\n$$\n\nby the sine addition formula. In particular, we conclude that $s = \\sin \\alpha + \\cos \\alpha$ is rational. It is clear that $\\alpha + \\beta = 45^\\circ$. By the sine and cosine subtraction formulas, we have\n\n$$\ns = \\sin(45^\\circ - \\beta) + \\cos(45^\\circ - \\beta) = \\sqrt{2} \\cos \\beta,\n$$\n\nfrom which it follows that $\\cos \\beta$ is not rational. On the other hand, from right triangle $ACE$, we have $\\cos \\beta = \\frac{AC}{EC}$, which is rational by assumption. Therefore, $\\cos \\beta$ is both rational and irrational, a contradiction. Hence, our assumption was wrong, and not all the segments $AB, AC, BI, ID, CI, IE$ can have integer lengths.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 14728,
    "subject": "Mathematics (Olympiad)",
    "question": "Prove that for any pair of positive integers $k$ and $n$, there exist $k$ positive integers $m_1, m_2, \\dots, m_k$ (not necessarily different) such that\n\n$$\n1 + \\frac{2^k - 1}{n} = \\left(1 + \\frac{1}{m_1}\\right) \\left(1 + \\frac{1}{m_2}\\right) \\cdots \\left(1 + \\frac{1}{m_k}\\right).\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "Method 1. By induction on $k$, the case $k = 1$ is trivial. Suppose the proposition is true for $k = j - 1$; we show it for $k = j$.\n\nIf $n$ is odd, i.e., $n = 2t - 1$ for some positive integer $t$, note that\n\n$$\n1 + \\frac{2^j - 1}{2t - 1} = \\frac{2(t + 2^{j-1} - 1)}{2t} \\cdot \\frac{2t}{2t - 1} = \\left(1 + \\frac{2^{j-1} - 1}{t}\\right) \\left(1 + \\frac{1}{2t - 1}\\right).\n$$\n\nBy the induction hypothesis, we can find $m_1, \\dots, m_{j-1}$ such that\n\n$$\n1 + \\frac{2^{j-1} - 1}{t} = \\left(1 + \\frac{1}{m_1}\\right) \\cdots \\left(1 + \\frac{1}{m_{j-1}}\\right).\n$$\n\nThus, take $m_j = 2t - 1$.\n\nIf $n$ is even, i.e., $n = 2t$ for some positive integer $t$, note that\n\n$$\n1 + \\frac{2^j - 1}{2t} = \\frac{2t + 2^j - 1}{2t + 2^j - 2} \\cdot \\frac{2t + 2^j - 2}{2t} = \\left(1 + \\frac{1}{2t + 2^j - 2}\\right) \\left(1 + \\frac{2^{j-1} - 1}{t}\\right).\n$$\n\nAgain, by induction, we can find $m_1, \\dots, m_{j-1}$ such that\n\n$$\n1 + \\frac{2^{j-1} - 1}{t} = \\left(1 + \\frac{1}{m_1}\\right) \\cdots \\left(1 + \\frac{1}{m_{j-1}}\\right).\n$$\n\nThus, take $m_j = 2t + 2^j - 2$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 14729,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $ABC$ be an isosceles triangle with $AB = AC$ and let $n > 1$ be an integer. Point $M$ lies on the line segment $AB$ such that $n \\cdot AM = AB$. Consider the points $P_1, P_2, \\dots, P_{n-1}$ on the side $BC$ with $BP_1 = P_1P_2 = P_2P_3 = \\dots = P_{n-1}C = \\frac{1}{n} BC$. Prove that\n\n$$\n\\angle MP_1A + \\angle MP_2A + \\dots + \\angle MP_{n-1}A = \\frac{1}{2} \\angle BAC.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "Consider the point $N$ on the side $AC$ such that $n \\cdot AN = AC$. The configuration is symmetric with respect to the perpendicular bisector of the segment $BC$, implying $\\angle MP_iA = \\angle NP_{n-i}A$ for $i = 1, 2, \\dots, n-1$. The claim is equivalent to $\\angle MP_1N + \\angle MP_2N + \\dots + \\angle MP_{n-1}N = \\angle BAC$.\n\nNotice that $BP_1 = P_1P_2 = P_2P_3 = \\dots = P_{n-1}C = MN$. Set $P_0 = B$, and observe that in the parallelograms $P_iMNP_{i+1}$, one has $\\angle MP_{i+1}N = \\angle P_iMP_{i+1}$ for $i = 0, 1, \\dots, n-2$. Therefore, the sum of those angles is equal to $\\angle P_0MP_{n-1}$, which in turn equals $\\angle BAC$, since $MP_{n-1} \\parallel AC$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 14730,
    "subject": "Mathematics (Olympiad)",
    "question": "A configuration of 4027 points in the plane is called *Colombian* if it consists of 2013 red points and 2014 blue points, and no three points of the configuration are collinear. By drawing some lines, the plane is divided into several regions. An arrangement of lines is *good* for a Colombian configuration if the following two conditions are satisfied:\n\n* No line passes through any point of the configuration.\n* No region contains points of both colours.\n\nFind the least value of $k$ such that for any Colombian configuration of 4027 points, there is a good arrangement of $k$ lines.",
    "options": [],
    "answer": "See solution",
    "solution": "The answer is $k = 2013$.\n\nFirstly, we show that $k \\ge 2013$ with an example. Mark 2013 red points and 2013 blue points on a circle alternately. Then there are 4026 arcs on the circle with different endpoint colours. Mark another point in blue elsewhere in the plane. If $k$ lines is good, then each arc must be intersected by some line, and any line intersects the circle at most two points, so there are at least $4026/2 = 2013$ lines.\n\nNext, we show that there exists a good arrangement of 2013 lines.\n\nFor two points $A$ and $B$ of the same colour, we can separate these two points from the others by drawing two sufficiently near parallel lines to $AB$ on each side of $AB$.\n\nLet $P$ be the convex hull of all coloured points. Take two adjacent vertices of $P$, say points $A$ and $B$. The other points are located on one side of line $AB$. If one of them is red, say $A$, then we can draw one line to separate $A$ from the other points. The remaining 2012 red points can be grouped into 1006 pairs; each pair of red points can be separated by two lines. So altogether, 2013 lines meet the requirement. If $A$ and $B$ are both blue, then they can be separated from the other vertices by using a line. The remaining 2012 blue points can be grouped into 1006 pairs; each pair of blue points can be separated by 2 lines. So altogether, 2013 lines meet the requirements.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 14731,
    "subject": "Mathematics (Olympiad)",
    "question": "Andriy, Bogdan, and Olesia were walking by the same road from home to the school. Andriy was walking with velocity equal to $a$ km/h for $(2-b)$ hours, Bogdan was walking with velocity equal to $b$ km/h for $(2-c)$ hours, and Olesia was walking with velocity equal to $c$ km/h for $(2-a)$ hours, where $a, b, c$ are some real numbers. What is the distance between home and school if it is known that it is equal to an integer number?",
    "options": [],
    "answer": "See solution",
    "solution": "Analyzing the problem, we get:\n\n$$\nS = a(2-b), \\quad S = b(2-c), \\quad S = c(2-a),\n$$\n\nwhere $S$ is a positive integer equal to the distance.\n\nWe may assume that $a \\geq b$. If $a > b$, then $2-b < 2-c$ or $b > c$. Analogously, $2-c < 2-a$ or $c > a$. Contradiction. So $a = b = c$. Then we have $S = a(2-a)$. Let us show that $S = a(2-a) \\leq 1$. Indeed, $2a - a^2 \\leq 1 \\Leftrightarrow (a-1)^2 \\geq 0$. Hence, since $S$ is a positive integer which is less than or equal to $1$, $S = 1$.\n\n**Remark.** One can think differently. We have $S^3 = abc(2-a)(2-b)(2-c)$. Since $a(2-a) \\leq 1$ (and analogous inequalities), $S^3 \\leq 1$, so $S = 1$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 14732,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $P$ be a real polynomial of degree $2015$ and $Q$ a real quadratic polynomial. Could it be that the polynomial $P(Q(x))$ has precisely the roots\n\n$-2014, -2013, \\dots, -2, -1, 1, 2, \\dots, 2014, 2015, 2016$?",
    "options": [],
    "answer": "See solution",
    "solution": "The values of $Q$ at the $4030$ points indicated in the problem must be a subset of the zeroes of $P$. But these are at most $2015$ in number, and $Q$, being quadratic, assumes any given value at most twice. Therefore, the $4030$ numbers can be split into $2015$ pairs $(p_i, q_i)$, for which $Q(p_i) = Q(q_i)$ runs through all the $2015$ zeroes of $P$ as $i = 1, \\dots, 2015$.\n\nLet $Q(x) = a x^2 + b x + c$. By Vieta's formula, $p_i + q_i = -\\frac{b}{a}$ for such a pair, so the $2015$ pairs must have equal sums. Since the numbers are integers, this is possible only if their sum is a multiple of $2015$. But it is not; in fact, the sum is\n\n$$\n-1007 \\cdot 2015 + 1008 \\cdot 2017,\n$$\n\nwhich is not even a multiple of $5$. $\\square$",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 14733,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all real numbers $a$ such that there exists a function $f: \\mathbb{R} \\to \\mathbb{R}$ satisfying\n$$\nf(f(x)) = x f(x) + a\n$$\nfor all $x \\in \\mathbb{R}$.",
    "options": [],
    "answer": "See solution",
    "solution": "The only possible value is $a = 0$.\n\nIf $a = 0$, then the function $f \\equiv 0$ satisfies the condition:\n$$\nf(f(x)) = f(0) = 0 = x \\cdot 0 + 0 = x f(x) + a.\n$$\n\nSuppose $a \\ne 0$. Assume $f(\\alpha) = 0$ for some $\\alpha$. Then:\n$$\nf(0) = f(f(\\alpha)) = \\alpha f(\\alpha) + a = a.\n$$\nSo $f(0) = a$. Now,\n$$\nf(a) = f(f(0)) = 0 \\cdot f(0) + a = a.\n$$\nThus $f(a) = a$. Then,\n$$\nf(f(a)) = a f(a) + a = a^2 + a.\n$$\nBut also $f(f(a)) = f(a) = a$, so $a = a^2 + a$, which implies $a = 0$, a contradiction. Therefore, $a = 0$ is the only solution.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 14734,
    "subject": "Mathematics (Olympiad)",
    "question": "If the nonnegative real numbers $x$, $y$, and $z$ have sum $2$, prove that:\n\n$$\nx^2y^2 + y^2z^2 + z^2x^2 + xyz \\le 1.\n$$\n\nFor which values of $x$, $y$, and $z$ does equality hold?",
    "options": [],
    "answer": "See solution",
    "solution": "We use the inequality $2\\alpha\\beta \\le \\alpha^2 + \\beta^2$ for all $\\alpha, \\beta \\in \\mathbb{R}$, with equality when $\\alpha = \\beta$.\n\n$$\n\\begin{align*}\nx^2 y^2 + y^2 z^2 + z^2 x^2 + xyz &= \\frac{1}{2}(2x^2 y^2 + 2y^2 z^2 + 2z^2 x^2 + 2xyz) \\\\\n&= \\frac{1}{2}(2xy \\cdot xy + 2yz \\cdot yz + 2zx \\cdot zx + 2xyz) \\\\\n&\\le \\frac{1}{2}[xy(x^2 + y^2) + yz(y^2 + z^2) + zx(z^2 + x^2) + 2xyz] \\\\\n&= \\frac{1}{2}[(xy + yz + zx)(x^2 + y^2 + z^2) - xyz^2 - yzx^2 - zxy^2 + 2xyz] \\\\\n&= \\frac{1}{2}[(xy + yz + zx)(x^2 + y^2 + z^2) - xyz(x + y + z - 2)] \\\\\n&= \\frac{1}{2}[(xy + yz + zx)(x^2 + y^2 + z^2)], \\quad \\text{since } x + y + z = 2.\n\\end{align*}\n$$\n\nThus,\n\n$$\nx^2 y^2 + y^2 z^2 + z^2 x^2 + xyz \\le \\frac{1}{2} (xy + yz + zx)(x^2 + y^2 + z^2).\n$$\n\nEquality holds when $x = y = z$ or two variables are equal and the third is zero, i.e., $x = y, z = 0$; $y = z, x = 0$; or $z = x, y = 0$.\n\nSince $x + y + z = 2$, equality holds for:\n\n$$\n(x, y, z) = \\left(\\frac{2}{3}, \\frac{2}{3}, \\frac{2}{3}\\right),\\ (1, 1, 0),\\ (1, 0, 1),\\ (0, 1, 1).\n$$\n\nNow, using $\\alpha\\beta \\le \\left(\\frac{\\alpha + \\beta}{2}\\right)^2$ for $\\alpha, \\beta \\in \\mathbb{R}$, with $\\alpha = 2xy + 2yz + 2zx$, $\\beta = x^2 + y^2 + z^2$:\n\n$$\n\\begin{align*}\n\\frac{1}{2} (xy + yz + zx)(x^2 + y^2 + z^2) &= \\frac{1}{4} (2xy + 2yz + 2zx)(x^2 + y^2 + z^2) \\\\\n&\\le \\frac{1}{4} \\left( \\frac{2xy + 2yz + 2zx + x^2 + y^2 + z^2}{2} \\right)^2 \\\\\n&= \\frac{1}{16} (x + y + z)^4 = 1.\n\\end{align*}\n$$\n\nTherefore,\n\n$$\nx^2 y^2 + y^2 z^2 + z^2 x^2 + xyz \\le 1.\n$$\n\nEquality holds when $2xy + 2yz + 2zx = x^2 + y^2 + z^2$, which, together with the sum condition, gives the equality cases:\n\n$$\n(x, y, z) = (1, 1, 0),\\ (1, 0, 1),\\ (0, 1, 1).\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 14735,
    "subject": "Mathematics (Olympiad)",
    "question": "a) Find all functions $f : \\mathbb{Z} \\to \\mathbb{Z}$ such that\n\n$$\nf(x - f(y)) = f(f(x)) - f(y) - 1\n$$\n\nfor all $x, y \\in \\mathbb{Z}$.\n\nb) Find all functions $f : \\mathbb{Z} \\to \\mathbb{Z}$ such that\n\n$$\nf(x - f(y)) = f(f(x)) - f(y) - 2\n$$\n\nfor all $x, y \\in \\mathbb{Z}$.",
    "options": [],
    "answer": "See solution",
    "solution": "a) The solutions are $f(x) = -1$ and $f(x) = x + 1$.\n\n*Proof:* Setting $y = f(x)$ in the given equation,\n\n$$\nf(x - f(f(x))) = -1,\n$$\n\nso there exists $\\lambda$ such that $f(\\lambda) = -1$. Setting $y = \\lambda$ gives\n\n$$\nf(f(x)) = f(x + 1).\n$$\n\nSubstituting $f(x) + f(y)$ for $x$ yields\n\n$$\nf(f(x)) = f(f(x) + f(y)) - f(y) - 1.\n$$\n\nBy symmetry,\n\n$$\nf(f(y)) = f(f(y) + f(x)) - f(x) - 1.\n$$\n\nThus $f(f(x)) - f(x) = f(f(y)) - f(y) = c$, so $f(x+1) - f(x) = c$ for some $c \\in \\mathbb{Z}$.\n\nIf $c = 0$, $f(x)$ is constant, and from the equation $c = -1$, so $f(x) = -1$.\n\nIf $c \\neq 0$, $f(x)$ is injective, and $f(x) = x + 1$.\n\nBoth satisfy the original equation.\n\nb) The solutions are $f(x) = -2$ and $f(x) = x + 2$.\n\n*Proof:* Similarly, for $f : \\mathbb{Z} \\to \\mathbb{Z}$ satisfying\n\n$$\nf(x - f(y)) = f(f(x)) - f(y) - 2,\n$$\n\nwe get\n\n$$\nf(x - f(f(x))) = -2,\\quad f(f(x)) = f(x + 2),\\quad f(x + 2) - f(x) = c.\n$$\n\nIf $c = 0$, $f(x)$ is constant, and $f(x) = -2$.\n\nIf $c \\neq 0$, $f(x)$ is linear. For $c = 2$, $f(x) = x + 2$ is a solution. If $f$ is not injective, then $f(x)$ can be of the form $f(x) = x + 2$ for even $x$, $f(x) = x + d$ for odd $x$ with $d$ odd, and all such functions also solve the equation.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 14736,
    "subject": "Mathematics (Olympiad)",
    "question": "Consider a sequence $a_k$ of integers such that $a_1$ and $a_2$ are arbitrary integers and $a_{k+1}$ for $k \\ge 2$ is determined by the formula\n\n$$\na_{k+1} = \\frac{a_k + a_{k-1}}{2015^i},\n$$\n\nwhere $2015^i$ is the maximal power of $2015$ that divides $a_k + a_{k-1}$. Prove that if this sequence is periodic, then its period is divisible by $3$.",
    "options": [],
    "answer": "See solution",
    "solution": "If all the numbers in the sequence are even, we can divide all the numbers in the sequence by the maximal possible power of $2$. In this way, we obtain a new sequence of integers which is determined by the same recurrence formula and has the same period, but now it necessarily contains an odd number. Consider this new sequence modulo $2$. Since the number $2015$ is odd, it has no influence on the calculations modulo $2$, so we may think that modulo $2$ this sequence is given by the Fibonacci recurrence $a_{k+1} \\equiv a_k + a_{k-1}$. Then it has the following form $\\ldots, 1, 0, 1, 1, 0, 1, 1, 0, 1, \\ldots$, so that the period of such a sequence is divisible by $3$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 14737,
    "subject": "Mathematics (Olympiad)",
    "question": "Assign to each side $b$ of a convex polygon $\\mathcal{P}$ the maximum area of a triangle that has $b$ as a side and is contained in $\\mathcal{P}$. Show that the sum of the areas assigned to the sides of $\\mathcal{P}$ is at least twice the area of $\\mathcal{P}$.",
    "options": [],
    "answer": "See solution",
    "solution": "Define the *weight* of a side $XY$ to be the area assigned to it, and define an *antipoint* of a side of a polygon to be one of the points in the polygon farthest from that side (and consequently forming the triangle with greatest area).\n\n![alt](path \"title\")\n\n**Lemma** For any side $XY$, $Z$ is an antipoint if and only if the line $l$ through $Z$ parallel to $XY$ does not go through the interior of the polygon. (Note that this means we can assume $Z$ is a vertex, as we shall do henceforth).\n\n*Proof:* Clearly, if $Z$ is an antipoint, $l$ must not go through the interior of the polygon. Now if $l$ does not go through the interior of the polygon, assume there is a point $Z'$ farther away from $XY$ than $Z$. Since the polygon is convex, the point $XZ' \\cap l$ is in the interior of the polygon, which is a contradiction. ■\n\nSuppose for the sake of contradiction that the sum of the weights of the sides is less than twice the area of some polygon. Then let $S$ be the non-empty set of all convex polygons for which the sum of the weights is strictly less than twice the area. It is easy to check that no polygon in $S$ can be a triangle, so we may assume all polygons in $S$ have at least 4 sides.\n\nWe first prove by contradiction that there is some polygon in $S$ such that all of its sides are parallel to some other side. Suppose the contrary; then consider one of the polygons in $S$ which has the minimal number of sides not parallel to any other side (this exists by the well-ordering principle). Call this polygon $P = A_1A_2\\cdots A_n$, and WLOG let $A_nA_1$ be a side which is not parallel to any other side of $P$.\n\nThen let $A_i$ be the unique antipoint of $A_nA_1$, and let $A_u$ and $A_v$ be respective antipoints of $A_{i-1}A_i$ and $A_iA_{i+1}$. Define $X$ to be the point such that $A_uX \\parallel A_{i-1}A_i$, $A_vX \\parallel A_iA_{i+1}$.\n\nNow consider the set $T \\subset P$ of points that are strictly on the same side of $A_uA_v$ as $A_nA_1$. First of all, for any side in $T$, $A_i$ must be its antipoint, since the line through $A_i$ parallel to $A_jA_{j+1}$ does not go through the interior of $P$. Similarly, any vertex in $T$ is not the antipoint of any side.\n\nWe now look at the polygon $P' = A_vA_{v+1}\\cdots A_{u-1}A_uX$. First of all, it is clear that $P'$ has fewer sides which are not parallel to any other side than $P$. Using $[\\cdot]$ to denote area, we have\n\n$$\n[P'] - [P] = [A_1A_2\\cdots A_{v-1}A_vXA_uA_{u+1}\\cdots A_n].\n$$\n\nThe weights of the side $A_jA_{j+1}$ is the same in both $P'$ and $P$ for $v \\leq j < u$, but for $P'$, the sum of the weights of the remaining two sides is $[XA_uA_iA_v]$, as $A_i$ is an antipoint of both $A_uX$ and $A_vX$. Meanwhile, the sum of the weights of remaining sides for $P$ is $[A_1A_2\\cdots A_{v-1}A_vA_iA_uA_{u+1}\\cdots A_n]$. Hence the difference in the sums of weights of $P'$ and $P$ is\n\n$$\n[XA_uA_iA_v] - [A_1A_2\\cdots A_{v-1}A_vA_iA_uA_{u+1}\\cdots A_n] = [A_1A_2\\cdots A_{v-1}A_vXA_uA_{u+1}\\cdots A_n],\n$$\n\nthe same as the difference in area (and both differences were positive). Therefore, if the sum of weights of $P$ was less than $2[P]$, then certainly the sum of weights of $P'$ must be less than $2[P']$, so that $P' \\in S$. However, this contradicts the minimality of the number of non-parallel sides in $P$, so there exists a polygon in $S$ with opposite sides parallel.\n\nNow, we will let $R$ be the non-empty set of all polygons in $S$ with all sides parallel to the opposite side. Note that all polygons in $R$ must have an even number of sides. We will show that there is a parallelogram in $R$.\n\nSuppose not, and that $Q = B_1B_2\\cdots B_{2m}$ is one of the polygons in $R$ with the minimal number of sides, and $m \\geq 3$. Let $X = B_1B_2 \\cap B_{2m-1}B_{2m}$ and $Y = B_{m-1}B_m \\cap B_{m+2}B_{m+1}$. Set $Q' = XB_2B_3\\cdots B_{m-1}YB_{m+2}\\cdots B_{2m}$. We propose that the increase in the sum of weights going from $Q$ to $Q'$ is at most twice the increase in area, so that $Q' \\in R$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 14738,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $a$, $b$, and $c$ be positive real numbers such that $abc = 1$. Prove that the following inequality holds:\n\n$$\n\\frac{1}{2}(\\sqrt{a} + \\sqrt{b} + \\sqrt{c}) + \\frac{1}{1+a} + \\frac{1}{1+b} + \\frac{1}{1+c} \\ge 3\n$$\n\nWhen does equality hold?",
    "options": [],
    "answer": "See solution",
    "solution": "Since $(1 - \\sqrt{bc})^2 \\ge 0$, it follows that $1 + bc \\ge 2\\sqrt{bc}$, i.e., $\\frac{1}{2\\sqrt{bc}} \\ge \\frac{1}{1+bc}$. We get that\n\n$$\n\\frac{\\sqrt{a}}{2} + \\frac{1}{1+a} = \\frac{1}{2\\sqrt{bc}} + \\frac{1}{1+a} \\ge \\frac{1}{1+bc} + \\frac{1}{1+a} = \\frac{1}{1+\\frac{1}{a}} + \\frac{1}{1+a} = 1.\n$$\n\nIn the same way, we prove that\n\n$$\n\\frac{\\sqrt{b}}{2} + \\frac{1}{1+b} \\ge 1\n$$\n\nand\n\n$$\n\\frac{\\sqrt{c}}{2} + \\frac{1}{1+c} \\ge 1.\n$$\n\nBy adding these three inequalities, we get the required result. Equality holds if and only if $a = b = c = 1$.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 14739,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $\\triangle ABC$ be an acute triangle with $\\angle ACB > 2\\angle ABC$. Let $I$ be the incenter of $\\triangle ABC$, and let $K$ be the reflection of $I$ about the line $BC$. The lines $BA$ and $KC$ intersect at $D$. The line through $B$ parallel to $CI$ intersects the minor arc $\\widehat{BC}$ on the circumcircle of $\\triangle ABC$ at $E$ ($E \\neq B$). The line through $A$ parallel to $BC$ intersects the line $BE$ at $F$.\n\nProve that if $BF = CE$, then $FK = AD$.\n\n![](images/CHN_TSExams_2022_p9_data_34ab9f7d37.png)",
    "options": [],
    "answer": "See solution",
    "solution": "Extend $CI$ to intersect the circumcircle of $\\triangle ABC$ at point $T$ (where $T$ is the midpoint of arc $\\widehat{AB}$). Then $TI = TA = TB$. Since $BE \\parallel CT$, we have $BT = CE$. So $BF = CE = BT = AT = TI$. This shows that $BETI$ is a parallelogram. So $BI \\parallel FT$ and $BI = ET$. Combining this with $BI = BK$, we deduce that $FT = BK$.\n\nComputing the angles gives\n\n$$\n\\angle FBK = \\angle FBI + \\angle IBK = \\angle ETI + \\angle ATI.\n$$\n\nCombining all above, we get $\\triangle FBK \\cong \\triangle ATF$. So $AF = FK$.\n\nDenote the angles of $\\triangle ABC$ by $A = 2\\alpha$, $B = 2\\beta$, and $C = 2\\gamma$. Since $BI \\parallel FT$ and $BC \\parallel FA$, we deduce that $\\beta = \\angle IBC = \\angle TFA = \\angle BKF$. Moreover, $\\angle EBK = \\angle EBC - \\angle CBK = \\angle BCI - \\beta = \\gamma - \\beta$. So $\\angle BFK = \\angle KBE - \\angle BKF = \\gamma - 2\\beta$.\n\nNote that $\\angle BDK = \\angle BCK - \\angle CBA = \\gamma - 2\\beta = \\angle BFK$. So $B, F, D, K$ are concyclic. We deduce that $\\angle ADF = \\angle BDF = \\angle BKF = \\beta = \\frac{1}{2}\\angle ABC = \\frac{1}{2}\\angle BAF$. Hence $AF = AD$.\n\nSumming these up, $FK = AF = AD$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 14740,
    "subject": "Mathematics (Olympiad)",
    "question": "Let each of the numbers $x_1, x_2, \\ldots, x_n$ be equal to $1$ or $-1$, and also:\n\n$$\nx_1x_2x_3x_4 + x_2x_3x_4x_5 + x_3x_4x_5x_6 + \\ldots + x_{n-2}x_{n-1}x_nx_1 + x_{n-1}x_nx_1x_2 + x_nx_1x_2x_3 = 0\n$$\n\nProve that $n$ is divisible by $4$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $y_k = x_k x_{k+1} x_{k+2} x_{k+3}$ for $k = 1, 2, \\ldots, n$ (with indices modulo $n$). Each $y_k$ is $1$ or $-1$. By the problem's condition, $y_1 + y_2 + \\ldots + y_n = 0$, so $n = 2k$ for some integer $k$, and exactly $k$ of the $y_k$ are $1$, the rest $-1$.\n\nNow, $y_1 \\cdot y_2 \\cdot \\ldots \\cdot y_n = (-1)^k$. But also,\n$$\ny_1 \\cdot y_2 \\cdot \\ldots \\cdot y_n = x_1^4 x_2^4 \\ldots x_n^4 = 1\n$$\nsince $x_i^4 = 1$ for $x_i = \\pm 1$. Thus, $(-1)^k = 1$, so $k$ is even, $k = 2t$, and $n = 4t$. Therefore, $n$ is divisible by $4$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 14741,
    "subject": "Mathematics (Olympiad)",
    "question": "![](images/2013-ilovepdf-compressed_p42_data_4cee8a47db.png)\n\nЗурагт үзүүлсэн дүрс нь хаалгаар багтах ба энэ дүрсийн талбайг ол.\n",
    "options": [],
    "answer": "See solution",
    "solution": "АВ-ээр эхэлж оруулах ба $OD$ хүртэл чигээр нь оруулаад $O \\equiv M$ болгодог. О-г хөдөлгөхгүйгээр D-г $\\overset{\\frown}{EC}$ нумын дагуу $C \\equiv N$ болтол эргүүлээд чигээр нь түлхээд дүрсийг өрөөнд оруулна. Эргүүлхээс өмнө, эргүүлэх явцад мөн эргүүлэхэд AB хэрчим ямар ч саадгүй байх нь $|AB| = 1$ ба $|OA| = |BD| = 1$ гэдгээс гарна.\n\nЗураг дахь дүрсийн талбай нь\n\n$$\nS_1 + S_2 + S_3 = 1 \\cdot 1 + 2 \\cdot 1 + \\frac{(\\pi(1))^2}{4} = 3 + \\frac{\\pi}{4}\n$$\n\n$$\n= \\frac{12+\\pi}{4} > \\frac{15}{4} = 3.75 > \\pi \\approx 3.14\n$$\n\nбайна.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 14742,
    "subject": "Mathematics (Olympiad)",
    "question": "It is known that for natural numbers $a$, $b$, $c$, $d$, and $n$, the following inequalities hold: $a + c < n$ and $\\frac{a}{b} + \\frac{c}{d} < 1$. Prove that\n$$\n\\frac{a}{b} + \\frac{c}{d} < 1 - \\frac{1}{n^3}.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "Since $n > a + c \\geq 2$, then $n \\geq 3$. Also, $a < b$ and $c < d$ because $a < b$, $c < d$.\n\nWe distinguish between the following cases:\n\n*Case 1.* Let $b \\geq n$ and $d \\geq n$. Then\n$$\n\\frac{a}{b} + \\frac{c}{d} \\leq \\frac{a}{n} + \\frac{c}{n} = \\frac{a + c}{n} \\leq \\frac{n - 1}{n} = 1 - \\frac{1}{n} < 1 - \\frac{1}{n^3}.\n$$\n\n*Case 2.* Let $b \\leq n$ and $d \\leq n$. Then the inequality $\\frac{a}{b} + \\frac{c}{d} < 1$ implies $ad + bc < bd$, in other words $ad + bc + 1 \\leq bd$. Now we have\n$$\n\\frac{a}{b} + \\frac{c}{d} \\leq 1 - \\frac{1}{bd} = 1 - \\frac{1}{n^2} < 1 - \\frac{1}{n^3}.\n$$\n\n*Case 3.* Let $b < n < d$. If $d \\leq n^2$, then $bd < n^3$ and thus\n$$\n\\frac{a}{b} + \\frac{c}{d} \\leq 1 - \\frac{1}{bd} < 1 - \\frac{1}{n^3}.\n$$\nIf $d > n^2$, then $\\frac{a}{b} \\leq \\frac{n-2}{n^2} = 1 - \\frac{2}{n^2}$ because $a < n - c \\leq n - 1$, so $a \\leq n - 2$. Suppose that $\\frac{a}{b} + \\frac{c}{d} \\geq 1 - \\frac{1}{n^3}$. Then\n$$\n1 - \\frac{a}{b} \\leq \\frac{c}{d} - \\frac{1}{n^3} \\leq \\frac{1}{n} - \\frac{2}{n^2} + \\frac{1}{n^3} < \\frac{1}{n}.\n$$\nThis implies that $b > n(b - a) \\geq n$, which contradicts $b < n < d$.\n\n*Case 4.* Let $d < n < b$. If $b \\leq n^2$, then $bd < n^3$ and thus\n$$\n\\frac{a}{b} + \\frac{c}{d} \\leq 1 - \\frac{1}{bd} < 1 - \\frac{1}{n^3}.\n$$\nIf $b > n^2$, then $\\frac{c}{d} \\leq \\frac{n-2}{n^2} = 1 - \\frac{2}{n^2}$ because $c < n - a \\leq n - 1$, so $c \\leq n - 2$. Suppose that $\\frac{a}{b} + \\frac{c}{d} \\geq 1 - \\frac{1}{n^3}$. Then\n$$\n1 - \\frac{c}{d} \\leq \\frac{a}{b} - \\frac{1}{n^3} \\leq \\frac{1}{n} - \\frac{2}{n^2} + \\frac{1}{n^3} < \\frac{1}{n}.\n$$\nThis implies that $d > n(d - c) \\geq n$, which contradicts $d < n < b$.\n\nTherefore, the result holds in all cases.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 14743,
    "subject": "Mathematics (Olympiad)",
    "question": "Andriy wrote a 4-digit number. Olesya crossed out the last digit, and it turned out that the difference between the initial number and the obtained number equals $2018$. Which number did Andriy write? Provide all possible answers.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $abcd$ denote the number Andriy wrote. Then:\n\n$$\n\\overline{abcd} - \\overline{abc} = 2018\n$$\n\nThe number $\\overline{abcd} = 1000a + 100b + 10c + d$, and $\\overline{abc} = 100a + 10b + c$. Their difference is:\n\n$$\n(1000a + 100b + 10c + d) - (100a + 10b + c) = 900a + 90b + 9c + d = 2018\n$$\n\nWe seek integer digits $a, b, c, d$ ($a \\neq 0$) such that $900a + 90b + 9c + d = 2018$.\n\nTry $a = 2$:\n\n$$\n900 \\times 2 = 1800\n$$\nSo $90b + 9c + d = 2018 - 1800 = 218$\n\nTry $b = 2$:\n\n$$\n90 \\times 2 = 180\n$$\nSo $9c + d = 218 - 180 = 38$\n\nTry $c = 4$:\n\n$$\n9 \\times 4 = 36\n$$\nSo $d = 38 - 36 = 2$\n\nThus, the number is $2242$.\n\nCheck:\n\n$$\n2242 - 224 = 2018\n$$\n\nNo other digit choices yield a valid 4-digit number. Therefore, the only possible answer is $2242$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 14744,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $m \\ge 5$ be an odd integer, and let $D(m)$ denote the number of quadruples $(a_1, a_2, a_3, a_4)$ of distinct integers with $1 \\le a_i \\le m$ for all $i$ such that $m$ divides $a_1 + a_2 + a_3 + a_4$. There is a polynomial $q(x) = c_3x^3 + c_2x^2 + c_1x + c_0$ such that $D(m) = q(m)$ for all odd integers $m \\ge 5$. What is $c_1$?\n\n(A) $-6$ \n(B) $-1$ \n(C) $4$ \n(D) $6$ \n(E) $11$",
    "options": [],
    "answer": "See solution",
    "solution": "Let $s(m, r)$ denote the number of quadruples $(a_1, a_2, a_3, a_4)$ of distinct residue classes modulo $m$ such that $a_1 + a_2 + a_3 + a_4 \\equiv r \\pmod{m}$. The total number of quadruples of distinct residue classes is $m(m-1)(m-2)(m-3)$, so\n\n$$\n\\sum_{r=1}^{m} s(m, r) = m(m-1)(m-2)(m-3).\n$$\n\nThe pairing $(a_1, a_2, a_3, a_4) \\longleftrightarrow (a_1+1, a_2+1, a_3+1, a_4+1)$ gives a one-to-one correspondence between quadruples of distinct residue classes that sum to $r$ and those that sum to $r+4$. Thus $s(m, r) = s(m, r+4)$.\n\nSince $\\gcd(4, m) = 1$ (as $m$ is odd), the sequence $r, r+4, r+8, \\dots, r+4(m-1)$ forms a complete residue system modulo $m$, so all $s(m, r)$ are equal. Therefore,\n\n$$\ns(m, r) = (m-1)(m-2)(m-3) = m^3 - 6m^2 + 11m - 6\n$$\n\nfor all $r$. Since $D(m) = s(m, 0)$, the required polynomial is $q(x) = x^3 - 6x^2 + 11x - 6$, so $c_1 = 11$.",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 14745,
    "subject": "Mathematics (Olympiad)",
    "question": "Let triangle $ABC$ have orthocenter $H$. Let $B_1, C_1, B_2$ and $C_2$ be collinear points which lie on lines $AB, AC, BH$, and $CH$, respectively. Let $\\omega_B$ and $\\omega_C$ be the circumcircles of triangles $BB_1B_2$ and $CC_1C_2$, respectively. Prove that the radical axis of $\\omega_B$ and $\\omega_C$ intersects the line through their centers on the nine point circle of triangle $ABC$.",
    "options": [],
    "answer": "See solution",
    "solution": "The first important step is to introduce $N$, the circumcenter of $HB_2C_2$.\n\n*Claim* — Lines $NB_2$ and $NC_2$ are tangent to $(BB_1B_2)$ and $(CC_1C_2)$, respectively.\n\n*Proof*. This follows from chasing\n\n$$\n\\angle NB_2B = \\angle NB_2H = 90^\\circ - \\angle HC_2B_2 = \\angle B_2B_1B.\n$$\n\nAn analogous proof works for the other side. $\\square$\n\nSince $NB_2 = NC_2$, it follows that $N$ lies on the radical axis of $(BB_1B_2)$ and $(CC_1C_2)$. Thus we want to show $X$, the foot of $N$ onto $O_BO_C$, lies on the nine point circle.\n\n![](images/TSTST2025Solutions_p13_data_e8a76da432.png)\n\nLet $D, E, F$ be the feet of the altitudes in $\\triangle ABC$, and let $M_B, M_C$ be the midpoints of $BH, CH$. Let $B_3, C_3$ be the second intersections of $(B_2C_2H)$ with $(BB_1B_2), (CC_1C_2)$.\n\n*Claim* — Hexagons $O_BB_2B_3M_BXN$ and $O_CC_2C_3M_CXN$ are cyclic.\n\n*Proof*. Clearly $B_2, B_3$ and $X$ lie on the circle of diameter $O_BN$. To show $M_B$ lies in this circle, we first note that $BB_3HFD$ is cyclic. Indeed, this follows from chasing\n\n$$\n\\angle BB_3H = \\angle B_2B_3H - \\angle B_2B_3B = \\angle B_2C_2H - \\angle B_2B_1B = \\angle B_1FC_2 = \\angle BFH.\n$$\n\nThis circle has center $M_B$, so $O_B M_B$ and $M_B N$ are the perpendicular bisectors of $BB_3$, $B_3 H$. Since $BB_3 \\perp B_3 H$, $O_B M_B \\perp M_B N$, as we wanted to show. $\\square$\n\nNow we are ready to finish. We know $\\angle M_B D M_C = \\angle M_C H M_B = \\angle BAC$. We will now show that $\\angle M_B X M_C$ gives the same value. On one hand\n\n$$\n\\angle M_B X O_B = \\angle M_B B_2 O_B = \\angle B B_2 O_B = 90^\\circ - \\angle B_2 B_1 B = 90^\\circ - \\angle C_1 B_1 A.\n$$\n\nSimilarly $\\angle O_C X M_C = 90^\\circ - \\angle AC_1B_1$, so\n\n$$\n\\angle M_B X M_C = (90^\\circ - \\angle C_1 B_1 A) + (90^\\circ - \\angle AC_1B_1) = \\angle BAC,\n$$\n\nwhich establishes the result.\n\n*Remark*. Since the problem is symmetric under $A \\leftrightarrow H$, a similar solution can be found by considering the circumcenter of $\\triangle AB_1C_1$ (instead of $N$). Moreover, introducing both circumcenters adds more structure to the diagram, which can make it easier to finish. The most important step in this solution is to introduce the point $N$. This can be motivated by noticing that the circles $(BB_1B_2)$ and $(HB_2C_2)$ are orthogonal.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 14746,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all integers $n$ such that $2^n - n^3 = 24$. What is the corresponding value of $2^n + n^3$?",
    "options": [],
    "answer": "See solution",
    "solution": "Because $2^{10} = 1024 = 10^3 + 24$, $n = 10$ satisfies the condition $2^n - n^3 = 24$.\n\nFor $n \\ge 10$, we show that $2^{n+1} - (n+1)^3 > 2^n - n^3 > 0$. First note that\n\n$$\n2^{n+1} - (n+1)^3 = 2 \\cdot 2^n - n^3 - 3n^2 - 3n - 1 \\\\ > 2 \\cdot 2^n - 2 \\cdot n^3 = 2(2^n - n^3)\n$$\n\nbecause $3n^2 + 3n + 1 = n^2\\left(3 + \\frac{3}{n} + \\frac{1}{n^2}\\right) < n^3$ for $n \\ge 6$. Since $2^{10} - 10^3 = 24 > 0$, it follows by induction that $2^n - n^3 > 0$ for $n \\ge 10$, and so $2^{n+1} - (n+1)^3 > 2(2^n - n^3) > 2^n - n^3$ as claimed.\n\nWe have shown that $n = 10$ is the only integer that satisfies $2^n - n^3 = 24$. The only possible value for $2^n + n^3$ therefore is $2^{10} + 10^3 = 2024$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 14747,
    "subject": "Mathematics (Olympiad)",
    "question": "In the convex quadrilateral $ABCD$, the angles at $A$ and $C$ are equal, and the bisector of angle $B$ passes through the midpoint of side $CD$. Given that $CD = 3AD$, find the ratio $\\frac{AB}{BC}$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $M$ be the midpoint of $CD$. Since $BM$ is the bisector of $B$, the reflection $E$ of $C$ in $BM$ lies on the ray $BA$. Because $\\angle MEB = \\angle MCB$ by reflection, and $\\angle MCB = \\angle DAB$ by hypothesis, we have $\\angle MEB = \\angle DAB$. Hence $ME \\parallel DA$; in particular, $E$ is on the side $AB$.\n\nFurthermore, $ME = MC$ by reflection, and $MC = MD$, so $MC = MD = ME$. Therefore, triangle $CDE$ is right with $\\angle CED = 90^\\circ$. Then $DE \\perp CE$ and since $MB \\perp CE$, we obtain $BM \\parallel ED$.\n\nTriangles $BEM$ and $EAD$ have parallel sides, hence they are similar in ratio $\\frac{CD}{AD} = \\frac{3}{1}$. Then $BE = \\frac{3}{2}AE$ and $AB = \\frac{5}{2}AE$. Hence $\\frac{AB}{BC} = \\frac{AB}{BE} = \\frac{5}{3}$.\n\n![](images/3._NATIONAL_XXX_OMA_2013-checkpoint_p5_data_50a5f6c5ac.png)",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 14748,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $f : \\mathbb{R} \\to \\mathbb{R}$ be a function such that:\n\n$$\nf(xy + x + y) + f(xy - x - y) = 2(f(x) + f(y)), \\quad \\forall x, y \\in \\mathbb{R}.\n$$\n\nProve that $f$ fulfills the relation:\n\n$$\nf(x + y) = f(x) + f(y), \\quad \\forall x, y \\in \\mathbb{R}.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "$x = y = 0$ yields $f(0) = 0$ and $y = 0$ yields $f(-x) = -f(x)$, $x \\in \\mathbb{R}$.\n\nThen\n\n$$\ny = 1 \\implies f(2x + 1) = 2f(x) + f(1) \\tag{3}\n$$\n\n$$\ny = -1 \\implies f(2x - 1) = 2f(x) - f(1) \\tag{4}\n$$\n\nDenote $x * y = xy + x + y$ and notice that (ASOC): $x * (y * z) = (x * y) * z$.\n\nFor $x = 1$, $y * 1 = 2y + 1$, hence\n\n$$\nf(y * 1) = 2f(y) + f(1), \\quad \\forall y \\in \\mathbb{R} \\tag{5}\n$$\n\nTherefore,\n\n$$\n\\begin{align*}\nf(x * (y * 1)) &= f(x * (2y + 1)) \\\\\n&= f(x(2y + 1) - x - 2y - 1) + 2f(x) + 2f(2y + 1) \\\\\n&= f(2(xy - y) - 1) + 2f(2y + 1) + 2f(x) \\\\\n&= 2f(xy - y) + f(x) + 2f(y) + f(1) \\tag{6}\n\\end{align*}\n$$\n\nand\n\n$$\n\\begin{align}\nf((x * y) * 1) &= 2f(x * y) + f(1) \\\\\n&= 2(2f(x) + 2f(y)) + 2f(xy - x - y) + f(1) \\\\\n&= 2f(xy - x - y) + 2f(x) + 2f(y) + f(1). \\tag{7}\n\\end{align}\n$$\n\nRelations (6), (7), and (ASOC) yield:\n\n$$\nf(xy - x - y) + f(x) = f(xy - y), \\quad \\forall x, y \\in \\mathbb{R}.\n$$\n\nThe last relation means that $f(u + v) = f(u) + f(v)$ for all $u, v \\in \\mathbb{R}$ which can be written in the form $u = x$, $v = xy - x - y$, that is, there exist $x, y \\in \\mathbb{R}$ so that $x = u$ and $y = \\frac{v+u}{u-1}$; this happens if $u \\neq 1$ or $u = 1 = -v$.\n\nIt remains to check that $f(1 + v) = f(1) + f(v)$, $\\forall v \\in \\mathbb{R}$. Indeed, (4) and $x \\mapsto x + 1$ imply\n\n$$\nf(2x + 1) = 2f(x + 1) - f(1) = 2f(x) + f(1),\n$$\n\nwhence $f(x + 1) = f(x) + f(1)$, $\\forall x \\in \\mathbb{R}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 14749,
    "subject": "Mathematics (Olympiad)",
    "question": "Points $K$ and $M$ are chosen on the sides $AB$ and $BC$ of triangle $ABC$ such that $AK = KM = MC$. Let $N$ be the intersection point of $AM$ and $CK$, $P$ the foot of the perpendicular from $N$ to the line $KM$, and $Q$ a point inside segment $KM$ such that $MQ = KP$. Prove that the incircle of $\\triangle KMB$ touches $KM$ at point $Q$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $I$ be the incenter of $\\triangle KMB$. Then, $KI$ and $MI$ are the angle bisectors of $\\angle BKM$ and $\\angle BMK$, respectively. $\\triangle AKM$ and $\\triangle CMK$ are isosceles, so $\\angle KAM = \\angle KMA = \\angle MKI = \\angle BKI$ and $\\angle MKC = \\angle MCK = \\angle KMI = \\angle BMI$; moreover, these angles are acute.\n\n![](images/Ukrajina_2010_p25_data_9eb3c1cdaa.png)\n\nThis implies that $P$ belongs to the segment $KM$. $\\angle IKM = \\angle KMA$ and $\\angle IMK = \\angle MKC$, thus $KI \\parallel AM$ and $MI \\parallel CK$, so $KIMN$ is a parallelogram. Therefore, $KN = IM$. Since $KP = MQ$ (by the problem's condition), triangles $PKN$ and $QMI$ are congruent (they have two equal corresponding sides and the included angle). This implies that $\\angle IQM = \\angle NPK = 90^\\circ$, so $Q$ is the point of tangency, as required.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 14750,
    "subject": "Mathematics (Olympiad)",
    "question": "What is the maximum number of vertices a convex polygon can have if all its interior angles have integer degree measures?",
    "options": [],
    "answer": "See solution",
    "solution": "The sum of the interior angles of a convex $n$-gon is $180^{\\circ}(n-2)$. The largest possible integer value for an interior angle in a convex polygon is $179^{\\circ}$. Thus, we have:\n\n$$\n180^{\\circ}(n-2) \\leq 179^{\\circ}n\n$$\n\nSolving for $n$:\n\n$$\n180n - 360 \\leq 179n \\\\\n180n - 179n \\leq 360 \\\\\nn \\leq 360\n$$\n\nTherefore, the maximum possible value is $n = 360$. A regular 360-gon provides an example.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 14751,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $x, y, z$ be positive real numbers such that $x + y + z = 1$. Prove that\n$$\n\\frac{(x + y)^3}{z} + \\frac{(y + z)^3}{x} + \\frac{(z + x)^3}{y} + 9xyz \\ge 9(xy + yz + zx).\n$$\nWhen does equality hold?",
    "options": [],
    "answer": "See solution",
    "solution": "We have\n$$\n\\begin{align*}\n\\frac{(x + y)^3}{z} + \\frac{(y + z)^3}{x} + \\frac{(z + x)^3}{y} + 9xyz &\\ge 3\\sqrt[3]{\\frac{(x + y)^3 (y + z)^3 (z + x)^3}{xyz}} + 9xyz \\\\\n&= 3\\frac{(x + y)(y + z)(z + x)}{\\sqrt[3]{xyz}} + 9xyz \\\\\n&\\ge 3\\frac{(x + y)(y + z)(z + x)}{\\frac{x + y + z}{3}} + 9xyz \\\\\n&= 9(x + y)(y + z)(z + x) + 9xyz \\\\\n&= 9(x + y + z)(xy + yz + zx) = 9(xy + yz + zx).\n\\end{align*}\n$$\nEquality holds if and only if $x = y = z = \\frac{1}{3}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 14752,
    "subject": "Mathematics (Olympiad)",
    "question": "We say that a positive integer $n$ is *fantastic* if there exist positive rational numbers $a$ and $b$ such that\n\n$$\nn = a + \\frac{1}{a} + b + \\frac{1}{b}.\n$$\n\n(a) Prove that there exist infinitely many prime numbers $p$ such that no multiple of $p$ is fantastic.\n\n(b) Prove that there exist infinitely many prime numbers $p$ such that some multiple of $p$ is fantastic.\n",
    "options": [],
    "answer": "See solution",
    "solution": "Note that\n\n$$\nr(a, b) := a + \\frac{1}{a} + b + \\frac{1}{b} = \\frac{(a + b)(ab + 1)}{ab}.\n$$\n\nLet $a = \\frac{t}{u}$ and $b = \\frac{v}{w}$, where $t, u, v, w$ are positive integers with $\\gcd(t, u) = 1$ and $\\gcd(v, w) = 1$. Then\n\n$$\nr(a, b) = \\frac{(tv + uw)(tw + uv)}{tuvw},\n$$\n\nso the Diophantine equation\n\n$$\ntu(v^2 + w^2) + vw(t^2 + u^2) = kptuvw\n$$\n\nmust be investigated. Since $\\gcd(tu, t^2 + u^2) = 1$, this implies $tu \\mid vw$. Similarly, $vw \\mid tu$, so\n\n$$\ntu = vw.\n$$\n\nSubstituting, (6) becomes\n\n$$\nt^2 + u^2 + v^2 + w^2 = kptu.\n$$\n\nTherefore, $p$ must divide either $v^2 + t^2$ or $v^2 + u^2$. If $p \\equiv -1 \\pmod{4}$ (i.e., $-1$ is a quadratic non-residue mod $p$), then $p$ divides $v$ (and $t$ or $u$). The same argument applies for $w$ instead of $v$, so $p \\mid v, w$, contradicting $\\gcd(v, w) = 1$. Thus, the infinitely many primes with $p \\equiv -1 \\pmod{4}$ have no fantastic multiple, solving part (a).\n\nFor part (b), choose $v = 1$ and $w = tu$. We seek integers $t, u$ such that\n\n$$\n1 + t^2 + u^2 + t^2u^2 = kptu.\n$$\n\nLet $t = F_{2l+1}$, $u = F_{2l-1}$, where $F_n$ is the $n$th Fibonacci number, and use the identity $1 + F_{2l+1}^2 = F_{2l+3}F_{2l-1}$ to obtain\n\n$$\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$$\n\ni.e., $F_{2l+3} F_{2l-3} = k p$. Therefore, every prime factor of the Fibonacci number $F_{2l+3}$ has a fantastic multiple.\n\nSince $\\gcd(F_a, F_b) = F_{\\gcd(a, b)}$, $F_a$ and $F_b$ are relatively prime if $a$ and $b$ are different primes. Thus, infinitely many primes have a fantastic multiple, solving part (b).\n\n*Footnotes:*\n\n1. It is a well-known problem that $tu \\mid t^2 + u^2 + 1$ with $t > u$ is only possible if $t$ and $u$ are Fibonacci numbers of the form $t = F_{2l+1}$, $u = F_{2l-1}$, in which case $t^2 + u^2 + 1 = 3tu$.\n2. This is a special case of Vajda's identity: $F_{n+i}F_{n+j} - F_nF_{n+i+j} = (-1)^n F_i F_j$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 14753,
    "subject": "Mathematics (Olympiad)",
    "question": "![](images/Macedonia_2017_p9_data_655e751910.png)\n\nFind all pairs $(x, y)$ of positive integers such that\n\n$$\nx^3 + y^3 = x^2 + 42xy + y^2\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "Let $d = \\gcd(x, y)$. Set $x = ad$, $y = bd$, where $d \\in \\mathbb{N}$, $(a, b) = 1$, $a, b \\in \\mathbb{N}$. Then\n\n$$\nx^3 + y^3 = x^2 + 42xy + y^2 \\implies d^3(a^3 + b^3) = d^2(a^2 + 42ab + b^2)\n$$\n\n$$\n\\implies d(a^3 + b^3) = a^2 + 42ab + b^2\n$$\n\n$$\n\\implies d(a + b)(a^2 - ab + b^2) = a^2 + 42ab + b^2\n$$\n\nLet $c = d(a + b) - 1$. Then\n\n$$\na^2c - abc + b^2c = 43ab\n$$\n\nThis implies $b \\mid ca^2 \\implies b \\mid c$ and $a \\mid cb^2 \\implies a \\mid c$, so $ab \\mid c$. Thus, $c = mab$ for some $m \\in \\mathbb{N}^+$. Therefore,\n\n$$\nm(a^2 - ab + b^2) = 43\n$$\n\nSo $a^2 - ab + b^2$ divides $43$, so $a^2 - ab + b^2 = 1$ or $43$.\n\nIf $a^2 - ab + b^2 = 1$, then $(a - b)^2 = 1 - ab \\geq 0 \\implies a = b = 1$. Then $2d = 44 \\implies d = 22$, so $(x, y) = (22, 22)$.\n\nIf $a^2 - ab + b^2 = 43$, by symmetry, assume $a \\geq b$. Then $43 = a^2 - ab + b^2 \\geq ab \\geq b^2 \\implies b \\in \\{1, 2, 3, 4, 5, 6\\}$.\n\n- If $b = 1$, $a^2 - a + 1 = 43 \\implies a = 7$. So $(x, y) = (7, 1)$ or $(1, 7)$.\n- If $b = 6$, $a^2 - 6a + 36 = 43 \\implies a^2 - 6a - 7 = 0$, which has no integer solution.\n- For $b = 2, 3, 4, 5$, there are no positive integer solutions for $a$.\n\nThus, the solutions are $(x, y) \\in \\{(1, 7), (7, 1), (22, 22)\\}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 14754,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all pairs $(a, b)$ of non-negative integers such that\n\n$$\n2017^a = b^6 - 32b + 1.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "The two solutions are $(0, 0)$ and $(0, 2)$.\n\nSince $2017^a$ is always odd, $b$ must be even, so let $b = 2c$ for some integer $c$. Then:\n$$\n2017^a = 64(c^6 - c) + 1\n$$\nThus, $2017^a \\equiv 1 \\pmod{64}$. Now, $2017 \\equiv 33 \\pmod{64}$ and $2017^2 \\equiv (1+32)^2 = 1 + 2 \\cdot 32 + 32^2 \\equiv 1 \\pmod{64}$, so the powers of $2017$ modulo $64$ alternate between $1$ and $33$. Therefore, $a$ is even and $2017^a$ is a perfect square.\n\nLet $r(b) = b^6 - 32b + 1$. For $b > 4$:\n- $r(b) < b^6 = (b^3)^2$\n- $r(b) > (b^3 - 1)^2$ because $b^6 - 32b + 1 > b^6 - 2b^3 + 1$ for $b > 4$\n\nSo $2017^a$ is between two consecutive squares, and there are no solutions for $b > 4$.\n\nSince $b$ is even, check $b = 4, 2, 0$:\n- For $b = 4$, modulo $3$: $1 \\equiv 1 - 2 + 1 = 0$, so no solution.\n- For $b = 2$: $2017^a = 2^6 - 32 \\cdot 2 + 1 = 64 - 64 + 1 = 1$, so $(a, b) = (0, 2)$.\n- For $b = 0$: $2017^a = 0^6 - 32 \\cdot 0 + 1 = 1$, so $(a, b) = (0, 0)$.\n\nTherefore, $(0, 0)$ and $(0, 2)$ are the only solutions.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 14755,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $P$ be an interior point of a triangle $ABC$. Show that\n\n$$\n\\frac{PA}{a} + \\frac{PB}{b} + \\frac{PC}{c} \\geq \\sqrt{3}.\n$$\n\nHere, $a$, $b$, and $c$ are the side lengths opposite vertices $A$, $B$, and $C$, respectively.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $G$ denote the centroid of $\\triangle ABC$. We have\n\n$$\n\\sum \\frac{PA}{a} = \\sum \\frac{PAGA}{aGA} \\geq \\frac{\\sum PAGA}{\\max\\{aGA, bGB, cGC\\}}.\n$$\n\nWe observe that $\\vec{PA} \\cdot \\vec{GA} \\leq PAGA$. Thus,\n\n$$\n\\begin{align*}\n\\sum PAGA &\\geq \\sum \\vec{PA} \\cdot \\vec{GA} \\\\\n&= \\sum (\\vec{PG} + \\vec{GA}) \\cdot \\vec{GA} \\\\\n&= \\vec{PG} \\cdot (\\vec{GA} + \\vec{GB} + \\vec{GC}) + \\sum GA^2 \\\\\n&= GA^2 + GB^2 + GC^2.\n\\end{align*}\n$$\n\nOn the other hand,\n\n$$\n\\begin{align*}\n\\sqrt{3} a m_a &= \\frac{\\sqrt{3}}{2} a (2 m_a) \\\\\n&= \\frac{\\sqrt{3}}{2} a \\sqrt{2(b^2 + c^2) - a^2} \\\\\n&\\leq \\frac{1}{4}(3a^2 + 2b^2 + 2c^2 - a^2) \\\\\n&= \\frac{1}{2}(a^2 + b^2 + c^2).\n\\end{align*}\n$$\n\nThis implies that $3\\sqrt{3} aGA \\leq a^2 + b^2 + c^2$. Hence,\n\n$$\n\\sqrt{3} aGA \\leq \\frac{1}{3}(a^2 + b^2 + c^2) = \\frac{4}{9} \\sum m_a^2 = \\sum GA^2.\n$$\n\nSimilarly,\n\n$$\n\\sqrt{3} bGB \\leq \\sum GA^2, \\quad \\sqrt{3} cGC \\leq \\sum GA^2.\n$$\n\nIt follows that\n\n$$\n\\max\\{aGA, bGB, cGC\\} \\leq \\sum GA^2.\n$$\n\nTherefore,\n\n$$\n\\sum \\frac{PA}{a} \\geq \\frac{\\sum GA^2}{\\sum GA^2 / \\sqrt{3}} = \\sqrt{3}.\n$$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 14756,
    "subject": "Mathematics (Olympiad)",
    "question": "Determine all sequences $\\left(a_1, a_2, \\dots\\right)$ of positive integers satisfying\n\n$$\na_{n+1}^2 = 1 + (n + 2021)a_n\n$$\n\nfor all $n \\ge 1$.",
    "options": [],
    "answer": "See solution",
    "solution": "Clearly, for $C = 1$ we have the solution $\\left(a_n\\right)_{n=1}^{\\infty} = (n + 2019)_{n=1}^{\\infty}$. Let's prove that this is the only value for $C$ that works.\n\nAssume $\\left(a_n\\right)_{n=1}^{\\infty}$ is a solution and let $\\left(b_n\\right)_{n=1}^{\\infty} = (a_n - n)_{n=1}^{\\infty}$. We claim that for $n > |C| + 2021^2$:\n\n1. If $b_n < 2019$, then $b_n < b_{n+1} < 2019$.\n2. If $b_n > 2019$, then $b_n > b_{n+1} > 2019$.\n\nIt is clear that these two claims imply that $b_n = 2019$ for all large $n$ and hence that $C = 1$.\n\nLet us prove the claims:\n\n1. First of all, $b_n \\le 2018$ implies that\n\n$$\n\\begin{aligned}\na_{n+1}^2 &\\le C + (n + 2021)(n + 2018) \\\\\n&= (n + 2020)^2 - n + C + 2018 \\cdot 2021 - 2020^2 \\\\\n&< (n + 2020)^2\n\\end{aligned}\n$$\n\nand hence $a_{n+1} < n + 2020$ so that indeed $b_{n+1} < 2019$.\n\nMoreover, we have\n\n$$\n\\begin{aligned}\na_{n+1}^2 &= C + (n + 2021)(n + b_n) \\\\\n&= (n + 1 + b_n)^2 + (2019 - b_n)n + 2021b_n + C - (b_n + 1)^2 \\\\\n&\\ge (n + 1 + b_n)^2 + n + C - 2019^2 \\\\\n&> (n + 1 + b_n)^2\n\\end{aligned}\n$$\n\nand hence $a_{n+1} > n + 1 + b_n$ so that indeed $b_{n+1} > b_n$.\n\n2. First of all, $b_n \\ge 2020$ implies that\n\n$$\na_{n+1}^2 \\geq C + (n + 2021)(n + 2020) = (n + 2020)^2 + n + C + 2021 > (n + 2020)^2\n$$\n\nand hence $a_{n+1} > n + 2020$ so that indeed $b_{n+1} > 2019$.\n\nMoreover, we have\n\n$$\n\\begin{aligned}\na_{n+1}^2 &= C + (n + 2021)(n + b_n) \\\\\n&= (n + 1 + b_n)^2 + (2019 - b_n)n + 2021b_n + C - (b_n + 1)^2 \\\\\n&\\le (n + 1 + b_n)^2 - n + C \\\\\n&< (n + 1 + b_n)^2\n\\end{aligned}\n$$\n\nand hence $a_{n+1} < n + 1 + b_n$ so that indeed $b_{n+1} < b_n$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 14757,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $(a_n)_{n \\ge 1}$ be a sequence of non-negative integers such that $a_n \\le n$ for all $n \\ge 1$, and\n$$\n\\sum_{k=1}^{n-1} \\cos \\frac{\\pi a_k}{n} = 0\n$$\nfor all $n \\ge 2$. Find a closed formula for the general term of the sequence.",
    "options": [],
    "answer": "See solution",
    "solution": "Notice that $a_1 = 1$ and $\\cos \\frac{\\pi a_1}{3} + \\cos \\frac{\\pi a_2}{3} = 0$ implies $a_2 = 2$.\n\nInduct on $n$ to prove that $a_n = n$ for $n \\ge 1$. Suppose $a_k = k$ for all $k = 1, 2, \\dots, n-1$. The given relation rewrites as\n$$\n\\cos \\frac{\\pi a_n}{n+1} = -\\sum_{k=1}^{n-1} \\cos \\frac{\\pi k}{n+1}.\n$$\nSet $z = \\cos \\frac{\\pi}{n+1} + i \\sin \\frac{\\pi}{n+1}$ and notice that\n$$\nz + z^2 + z^3 + \\dots + z^n = \\frac{z - z^{n+1}}{1 - z} = \\frac{1 + z}{1 - z}.\n$$\nUse $\\bar{z} = \\frac{1}{z}$ to get $\\left(\\frac{1 + z}{1 - z}\\right) = -\\frac{1 + z}{1 - z}$, hence $\\operatorname{Re}\\frac{1 + z}{1 - z} = 0$ and consequently\n$$\n\\sum_{k=1}^n \\cos \\frac{\\pi k}{n+1} = 0.\n$$\nFrom $\\cos \\frac{\\pi a_n}{n+1} = \\cos \\frac{\\pi n}{n+1}$ and $a_n \\le n$ we get $a_n = n$, as claimed.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 14758,
    "subject": "Mathematics (Olympiad)",
    "question": "Evaluate:\n\n$$\n\\sqrt{\\frac{123!}{122! - 121!}}\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "We have\n\n$$\n\\begin{align*}\n\\sqrt{\\frac{123!}{122! - 121!}} &= \\sqrt{\\frac{123 \\cdot 122! - 122!}{122 \\cdot 121! - 121!}} = \\sqrt{\\frac{(123 - 1) \\cdot 122!}{(122 - 1) \\cdot 121!}} \\\\\n&= \\sqrt{\\frac{122 \\cdot 122 \\cdot 121!}{121 \\cdot 121!}} = \\sqrt{\\frac{122^2}{121}} = \\frac{122}{11}.\n\\end{align*}\n$$\n\nSince 122 and 11 are coprime, the answer is $\\frac{122}{11}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 14759,
    "subject": "Mathematics (Olympiad)",
    "question": "正三角形 $XYZ$ 的三個頂點 $X, Y, Z$ 分別落在一銳角三角形 $ABC$ 的三邊 $BC, CA, AB$ 上。\n\n試證：三角形 $ABC$ 的內心落在三角形 $XYZ$ 的內部。",
    "options": [],
    "answer": "See solution",
    "solution": "證明一個更強的結果：三角形 $ABC$ 的內心 $I$ 落在三角形 $XYZ$ 的內切圓內。\n\n記 $d(U, VW)$ 為點 $U$ 到直線 $VW$ 的距離。\n\n令 $O$ 為 $\\triangle XYZ$ 的內心，$r, r'$ 分別為 $\\triangle ABC, \\triangle XYZ$ 的內切圓半徑，$R', R''$ 為 $\\triangle XYZ$ 的外接圓半徑。則 $R' = 2r'$，所求等價於證明不等式 $OI \\leq r'$。假設 $O \\neq I$（若 $O = I$ 則顯然成立）。\n\n令 $\\triangle ABC$ 的內切圓分別與邊 $BC, AC, AB$ 相切於 $A_1, B_1, C_1$。直線 $IA_1, IB_1, IC_1$ 將平面切成六個銳角，每個銳角包含點 $A_1, B_1, C_1$。可假設 $O$ 落在由直線 $IA_1$ 與 $IC_1$ 所定義且包含點 $C_1$ 的角內（如圖 1）。令 $A', C'$ 分別為 $O$ 在直線 $IA_1, IC_1$ 上的投影點。\n\n因 $OX = R'$，$d(O, BC) \\leq R'$。又 $OA' \\parallel BC$，得\n\n$$\nd(A', BC) = A'I + r \\leq R' \\quad \\Rightarrow \\quad A'I \\leq R' - r.\n$$\n\n另一方面，$\\triangle ABC$ 的內切圓落在 $\\triangle ABC$ 內部，故 $d(O, AB) \\geq r'$。同理可得\n\n$$\nd(O, AB) = C'C_1 = r - IC' \\geq r' \\quad \\Rightarrow \\quad IC' \\leq r - r'.\n$$\n\n由於 $\\angle A'$ 與 $\\angle C'$ 皆為直角，四邊形 $IA'OC'$ 為圓內接四邊形（如圖 1）。在此圓上，弧 $A'OC'$ $= 2\\angle A'IC' < 180^\\circ = $ 弧 $OC'I$，故 $180^\\circ \\geq$ 弧 $IC' >$ 弧 $A'O$，其意為 $IC' > A'O$。由此可得\n\n$$\nOI \\leq IA' + A'O < IA' + IC' \\leq (R' - r) + (r - r') = R' - r' = r' \\quad \\text{得證！}\n$$\n\n![](images/11-3J_p12_data_1dc183616b.png)\n\n*圖 1*\n\n![](images/11-3J_p12_data_7aa6f69d09.png)\n\n*圖 2*",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 14760,
    "subject": "Mathematics (Olympiad)",
    "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": "See solution",
    "solution": "a) Let Ann replace on her first move one of the 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 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.\n\n*Remark.* Similarly to the proof presented here, one can show that Ann can win after writing any digit $n \\ge 2$ on her first move.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 14761,
    "subject": "Mathematics (Olympiad)",
    "question": "Amy is playing a game with 15 pool balls numbered 1 to 15. The sum of the numbers on all the balls is $120$.\n\n**a.** What is the total sum of the numbers on all the balls?\n\n**b.** After Amy sinks one ball, can the sum of the remaining balls be a perfect square?\n\n**c.** Give a possible sequence of balls sunk and the corresponding sums left after each shot, indicating whether each subtotal is prime (P) or square (S).\n\n| Ball sunk | 13 | 7 | 11 | 8 | 14 | 3 | 5 |\n|-----------|----|---|----|---|----|---|---|\n| Sum left  | 107|100| 89 | 81| 67 | 64| 59|\n| P or S    | P  | S | P  | S | P  | S | P |\n\n| Ball sunk | 10 | 12 | 6 | 2 | 4 | 9 | 15 | 1 |\n|-----------|----|----|---|---|---|---|----|---|\n| Sum left  | 49 | 37 | 31| 29| 25| 16| 1  | 0 |\n| P or S    | S  | P  | P | P | S | S | S  | S |\n\nThere are many other such sequences.\n\n**d.** What is the lowest prime subtotal Amy can reach, and how can she achieve it? Explain your reasoning.",
    "options": [],
    "answer": "See solution",
    "solution": "The lowest prime subtotal Amy can reach is $53$.\n\nThe first ball that Amy sinks must be odd; otherwise, the subtotal will be an even number greater than $2$, which can't be prime. After that, each ball Amy sinks must be even for the same reason. The seven even balls total $56$.\n\nSo Amy can't reach below $120 - 15 - 56 = 49$. Hence, the smallest prime subtotal that could possibly be reached is $53$ or more. Amy could actually reach $53$ with the following sequence:\n\n$120 \\to 107 \\to 103 \\to 89 \\to 83 \\to 73 \\to 61 \\to 53$\n\nwhere the balls sunk are $13, 4, 14, 6, 10, 12, 8$ in that order.\n\nThere are many other such sequences.\n\n**Alternative reasoning:**\n\nTo produce a prime subtotal, the first ball must be $7$, $11$, or $13$. If all seven even balls are then sunk, the subtotal would be respectively $51$, $53$, and $57$. Only $53$ is prime, so this is the least prime Amy could possibly reach. She could actually reach $53$ with the following sequence:\n\n$120 \\to 109 \\to 107 \\to 103 \\to 89 \\to 83 \\to 73 \\to 61 \\to 53$\n\nwhere the balls sunk are $11, 2, 4, 14, 6, 10, 12, 8$ in that order.\n\nThere are many other such sequences.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 14762,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $f_0$, $f_1$, $f_2$, and $f_3$ be polynomials in $\\mathbb{R}[X]$ such that $f_k(1) = f_{k+1}(0)$ for $k = 0, 1, 2, 3$ (indices are reduced modulo $4$). Show that there exists a polynomial $f$ in $\\mathbb{R}[X, Y]$ such that:\n\n- $f(X, 0) = f_0(X)$,\n- $f(1, Y) = f_1(Y)$,\n- $f(1 - X, 1) = f_2(X)$,\n- $f(0, 1 - Y) = f_3(Y)$.",
    "options": [],
    "answer": "See solution",
    "solution": "The idea is to construct $f(X, Y)$ as a suitable combination of the given polynomials and corrective terms:\n\nConsider $(1 - Y)f_0(X) + Yf_2(1 - X)$ (an $\\mathbb{R}[Y]$-linear combination of $f_0$ and $f_2$), and $Xf_1(Y) + (1 - X)f_3(1 - Y)$ (an $\\mathbb{R}[X]$-linear combination of $f_1$ and $f_3$), plus a degree $2$ corrective term $a_{11}XY + a_{10}X + a_{01}Y + a_{00}$.\n\nSet:\n$$\nf(X, Y) = (1 - Y)f_0(X) + Xf_1(Y) + Yf_2(1 - X) + (1 - X)f_3(1 - Y) + a_{11}XY + a_{10}X + a_{01}Y + a_{00}\n$$\n\nImpose the conditions:\n- $f(X, 0) = f_0(X)$: yields $a_{10} = f_0(0) - f_1(0)$, $a_{00} = -f_0(0)$.\n- $f(0, 1 - Y) = f_3(Y)$: yields $a_{01} = f_0(0) - f_3(0)$.\n- $f(1, Y) = f_1(Y)$: yields $a_{11} = f_1(0) - f_2(0) - f_0(0) + f_3(0)$.\n\nAll other conditions are satisfied by the hypothesis $f_k(1) = f_{k+1}(0)$.\n\nThus, the desired polynomial is:\n$$\nf(X, Y) = (1 - Y)f_0(X) + Xf_1(Y) + Yf_2(1 - X) + (1 - X)f_3(1 - Y) - (f_0(0) - f_1(0) + f_2(0) - f_3(0))XY + (f_0(0) - f_1(0))X + (f_0(0) - f_3(0))Y - f_0(0).\n$$",
    "topic": "Number Theory",
    "subtopic": "Modular Arithmetic"
  },
  {
    "id": 14763,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all pairs of positive integers $(a, b)$ such that\n$$\n(a, b) + 90(a, b) = ab,\n$$\nwhere $(a, b)$ denotes the greatest common divisor of $a$ and $b$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let us solve the equation $ab = 32 + 18(a, b)$ for positive integers $a$ and $b$.\n\nBy the condition, one of the numbers must be divisible by $5$. Moreover, exactly one of the numbers is divisible by $5$; otherwise, both $(a, b)$ and $90(a, b)$ would be divisible by $25$, making $160$ divisible by $25$, which is a contradiction.\n\nAssume without loss of generality that $a$ is divisible by $5$ and $b$ is not. Let $a = 5c$. Then $(a, b) = (5c, b) = (c, b)$, and the equation becomes:\n$$\nbc = 32 + 18(c, b).\n$$\nSince $bc$ and $18(c, b)$ are divisible by $(c, b)$, $(c, b)$ must divide $32$. Thus, $(c, b) = 1, 2, 4, 8, 16, 32$.\n\nIf $(c, b) \\geq 4$, then $bc$ is divisible by $16$, but $bc = 32 + 18(c, b)$, and for these values, $bc$ cannot be a multiple of $16$. Therefore, only $(c, b) = 1$ or $2$ are possible.\n\n**Case 1:** $(c, b) = 1$\n\nThen $bc = 32 + 18 \\cdot 1 = 50$. Since $b$ is not divisible by $5$, the possible pairs are $(b, c) = (1, 50), (2, 25), (5, 10), (10, 5), (25, 2), (50, 1)$. Only those with $b$ not divisible by $5$ are valid, so $(b, c) = (2, 25)$ and $(1, 50)$. Thus, $(a, b) = (125, 2)$ and $(250, 1)$.\n\n**Case 2:** $(c, b) = 2$\n\nThen $bc = 32 + 18 \\cdot 2 = 68$. Write $b = 2b_1$, $c = 2c_1$ with $(b_1, c_1) = 1$, so $b_1c_1 = 17$. The coprime pairs are $(1, 17)$ and $(17, 1)$. Thus, $(b, c) = (2, 34)$ and $(34, 2)$, giving $(a, b) = (10, 34)$ and $(170, 2)$.\n\nSince the equation is symmetric in $a$ and $b$, we also have the pairs $(34, 10)$, $(2, 125)$, $(2, 170)$, and $(1, 250)$.\n\n**Final answer:**\n$$(10, 34),\\ (125, 2),\\ (170, 2),\\ (250, 1),\\ (34, 10),\\ (2, 125),\\ (2, 170),\\ (1, 250).$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 14764,
    "subject": "Mathematics (Olympiad)",
    "question": "Suppose the side of the base and the height of a regular triangular pyramid $PABC$ are $1$ and $\\sqrt{2}$, respectively. Then the radius of the inscribed sphere of the pyramid is \\_\\_\\_\\_\\_.",
    "options": [],
    "answer": "See solution",
    "solution": "As seen in the figure below, suppose the projections of the inscribed sphere's center $O$ on faces $ABC$ and $ABP$ are $H$ and $K$, respectively. Let the midpoint of $AB$ be $M$, and let the radius of the sphere be $r$. Then $P$, $K$, and $M$ are collinear, $\\angle PHM = \\angle PKO = \\frac{\\pi}{2}$, and\n\n![](images/Mathematical_Olympiad_in_China_2011-2014_p60_data_6a1ca0660f.png)\n\n$$\nOH = OK = r, \\quad PO = PH - OH = \\sqrt{2} - r, \\quad MH = \\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$$\n\nThen we have\n\n$$\n\\frac{r}{\\sqrt{2} - r} = \\frac{OK}{PO} = \\sin \\angle KPO = \\frac{MH}{PM} = \\frac{1}{5}.\n$$\n\nTherefore, $r = \\frac{\\sqrt{2}}{6}$.\n\nThe answer is $\\frac{\\sqrt{2}}{6}$.",
    "topic": "Geometry",
    "subtopic": "Solid Geometry"
  },
  {
    "id": 14765,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $ABC$ be an acute triangle. Let $M$ be the midpoint of side $BC$, and let $E$ and $F$ be the feet of the altitudes from $B$ and $C$, respectively. Suppose that the common external tangents to the circumcircles of triangles $BME$ and $CMF$ intersect at a point $K$, and that $K$ lies on the circumcircle of $ABC$. Prove that line $AK$ is perpendicular to line $BC$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $H$ be the orthocenter of $\\triangle ABC$. We use inversion in the circle with diameter $\\overline{BC}$. We identify a few images:\n\n* The circumcircles of $\\triangle BME$ and $\\triangle CMF$ are mapped to lines $BE$ and $CF$.\n* The common external tangents are mapped to the two circles through $M$ which are tangent to lines $BE$ and $CF$.\n* The image of $K$, denoted $K^*$, is the second intersection of these circles.\n* The assertion that $K$ lies on $(ABC)$ is equivalent to $K^*$ lying on $(BHC)$.\n\nHowever, now $K^*$ is simple to identify directly: it's just the reflection of $M$ in the bisector of $\\angle BHC$.\n\n![](images/sols-TST-IMO-2023_p5_data_b89751899c.png)\n\nIn particular, $\\overline{HK}^*$ is a symmedian of $\\triangle BHC$. However, since $K^*$ lies on $(BHC)$, this means $(HK^*; BC) = -1$.\n\nThen, we obtain that $\\overline{BC}$ bisects $\\angle HMK^* \\equiv \\angle HMK$. However, $K$ also lies on $(ABC)$, which forces $K$ to be the reflection of $H$ in $\\overline{BC}$. Thus $\\overline{AK} \\perp \\overline{BC}$, as wanted.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 14766,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $a$ be a real number.\n\n**a)** Find all values of $a$ for which the inequality\n\n$$\nx \\log_{\\frac{1}{2}} a^4 - x^2 > 3 + 2 \\log_2 a^2\n$$\n\nhas a solution.\n\n**b)** Calculate the limit\n\n$$\n\\lim_{a \\to -\\infty} \\left( \\sqrt{a^2 - a + 1} + a \\right).\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "a) Since $\\log_{\\frac{1}{2}}(a^4) = -2 \\cdot \\log_2(a^2)$, let $2\\log_2(a^2) = b$. The inequality becomes $x^2 + b x + 3 + b < 0$. For this quadratic to have a solution, its discriminant must satisfy $D = b^2 - 4b - 12 > 0$, which gives $b < -2$ or $b > 6$. Thus, $\\log_2(a^2) < -1$ or $\\log_2(a^2) > 3$. From logarithm properties, $a^2 < \\frac{1}{2}$ or $a^2 > 8$ and $a \\neq 0$. Therefore,\n\n$$\na \\in (-\\infty, -2\\sqrt{2}) \\cup \\left(-\\frac{\\sqrt{2}}{2}, 0\\right) \\cup \\left(0, \\frac{\\sqrt{2}}{2}\\right) \\cup (2\\sqrt{2}, \\infty).\n$$\n\nb) For $a < 0$,\n\n$$\n\\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$$\n\nTherefore,\n\n$$\n\\lim_{a \\to -\\infty} \\left( \\sqrt{a^2 - a + 1} + a \\right) = \\frac{1}{2}.\n$$",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 14767,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $\\{a, b, c\\}$ be a set of three positive digits with MMR (mean-to-median ratio) equal to $\\frac{4}{5}$, where $a \\leq b \\leq c$. Find all possible sets $\\{a, b, c\\}$ that satisfy this condition.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $\\{a, b, c\\}$ be a set of three positive digits with MMR equal to $\\frac{4}{5}$, where $a \\leq b \\leq c$. The median is $b$ and the mean is $\\frac{a+b+c}{3}$, so\n\n$$\n\\begin{aligned}\n\\frac{a+b+c}{3b} &= \\frac{4}{5} \\\\\n5a+5b+5c &= 12b \\\\\n5a+5c &= 7b.\n\\end{aligned}\n$$\n\nThe left-hand side is a multiple of 5, so $b$ must also be a multiple of 5. Since we only want single-digit numbers, we conclude that $b = 5$.\n\nThen $a + c = 7$. So the only solutions for $\\{a, b, c\\}$ are $\\{1, 5, 6\\}$ and $\\{2, 5, 5\\}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 14768,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $k$ be a circle with midpoint $M$. $T$ is a point on $k$ and $t$ the tangent to $k$ at $T$. $P$ is a point on $t$ with $P \\neq T$, and $g$ is a line containing $P$ with $g \\neq t$. $g$ has the points $U$ and $V$ in common with $k$ ($U \\neq V$), and $S$ is the midpoint of the arc $UV$ not containing $T$. $Q$ is the point symmetric to $P$ with respect to $TS$. Prove that $QTUV$ is a trapezoid.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $R$ be the common point of $t$ and the tangent $s$ to $k$ at $S$. Since $S$ is the midpoint of the arc $UV$, $s$ is parallel to $g$ (and not to $t$). Since $MR$ is perpendicular to $TS$ and bisects $\\angle SRT$, $QP$ bisects $\\angle UPT$. Let $W$ be the common point of $PQ$ and $TS$. Because of the given symmetry, we have $PQ \\perp TS$, and triangle $PWT$ is therefore right-angled.\n\n![](images/Austrija_2011_p2_data_ec1de2b49e.png)\n\nWe therefore have\n$$\n\\begin{aligned}\n\\angle WTP + \\angle WPT &= 90^\\circ \\\\\n\\Leftrightarrow 2 \\cdot \\angle WTP + 2 \\cdot \\angle WPT &= 180^\\circ \\\\\n\\Leftrightarrow \\angle QTP + \\angle UPT &= 180^\\circ,\n\\end{aligned}\n$$\nand $UV$ is therefore parallel to $QT$. $QTUV$ is therefore a trapezoid, as claimed. QED.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 14769,
    "subject": "Mathematics (Olympiad)",
    "question": "There is a number of toy kittens in a box. Each kitten has its head and tail painted with one of $2011$ colors, not necessarily the same. One can take some kittens out of the box to form a collection. A collection is called \"right\" if it consists of exactly $2011$ kittens so that their heads are painted with different colors, and their tails are also painted with different colors. We know that one can choose a \"right\" collection of kittens from the box in more than one way. Prove that some kittens (maybe none) can be removed from the box so that there are exactly two ways to choose a \"right\" collection of kittens from the remaining ones.",
    "options": [],
    "answer": "See solution",
    "solution": "Enumerate the colors with the natural numbers from $1$ to $2011$. Construct a graph with the vertices $A_1, A_2, \\dots, A_{2011}, B_1, B_2, \\dots, B_{2011}$. For each kitten, draw an edge in the graph as follows: if its head is painted with color $i$ and its tail with color $j$, then the edge joins $A_i$ and $B_j$. Thus, each kitten corresponds to an edge in the graph.\n\nA \"right\" collection of kittens corresponds to a subset of $2011$ edges such that no two edges are adjacent to the same vertex. We call this a \"right\" subset of edges. We need to remove several (or none) edges from the graph so that there are exactly two \"right\" subsets of edges in the remaining graph.\n\nBy the problem condition, there exist at least two \"right\" subsets in the graph. Remove all edges that do not belong to any of these two subsets. For every \"right\" subset and every vertex, there is exactly one edge in the subset adjacent to that vertex. So, in the remaining graph, for each vertex, we either have an adjacent edge that belongs to both \"right\" subsets, or two adjacent edges, one from each subset.\n\nIf a vertex has exactly one adjacent edge, then the other vertex adjacent to this edge does not have any other adjacent edges. Thus, the graph is composed of cycles and isolated edges that do not share vertices with other edges. Each such cycle has even length because the edges from the first and second \"right\" subsets must alternate.\n\nEvery \"right\" subset contains all isolated edges and half of the edges from each cycle, chosen so that no two edges share a vertex. There are exactly two ways to choose edges from a cycle for a \"right\" subset. The graph contains at least one cycle, otherwise there would not be two different \"right\" subsets. If there is more than one cycle, remove half of the edges (so that the remaining edges have no common vertices) from all cycles except one. The remaining graph will have exactly one cycle, and hence, exactly two \"right\" subsets.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 14770,
    "subject": "Mathematics (Olympiad)",
    "question": "a. For $(m, n) = (6, 10)$, $(12, 15)$, and $(30, 78)$, compute $p$ where\n\n$$\np = \\frac{n^2 + m^2}{k}\n$$\n\nfor suitable $k$ in each case.\n\nb. Let $k = \\sqrt{n^2 - m^2}$. Show that $(k, m, n)$ forms a Pythagorean triple, and describe all possible prime values of $p$ that can be obtained in this way.",
    "options": [],
    "answer": "See solution",
    "solution": "a. For the given pairs:\n\n$$\n\\frac{10^2 + 6^2}{8} = 17, \\quad \\frac{15^2 + 12^2}{9} = 41, \\quad \\frac{78^2 + 30^2}{72} = 97.\n$$\n\nb. Let $k = \\sqrt{n^2 - m^2}$. Then $(k, m, n)$ is a Pythagorean triple. There exist positive integers $d, x, y$ with $(x, y) = 1$, $n = d(x^2 + y^2)$, and $m = d(x^2 - y^2)$ or $m = 2dxy$.\n\nIf $m = d(x^2 - y^2)$, then $p = \\frac{d(x^4 + y^4)}{xy}$. Since $(x, y) = 1$, $(xy, x^4 + y^4) = 1$, so $xy$ divides $d$. As $x^4 + y^4 \\geq 17$ and $p$ is prime, $d = xy$ and $p = x^4 + y^4$. Since one of $x, y$ is even and the other odd, $p \\equiv 1 \\pmod{8}$.\n\nIf $m = 2dxy$, then $p = d(x^2 - y^2) + \\frac{8dx^2y^2}{x^2 - y^2}$. Since $\\frac{8dx^2y^2}{x^2 - y^2}$ is integer and $(xy, x^2 - y^2) = 1$, $x^2 - y^2$ divides $8d$. As $p$ is prime, $d$ and $\\frac{8d}{x^2 - y^2}$ are coprime, so $x^2 - y^2$ is a multiple of $d$: $x^2 - y^2 = d, 2d, 4d,$ or $8d$.\n\n- If $x^2 - y^2 = d$, then $p = d^2 + 8x^2y^2$ is prime, $d$ is odd, and $p \\equiv 1 \\pmod{8}$.\n- If $x^2 - y^2 = 2d$, then $p = 2(d^2 + 2x^2y^2)$ is not prime.\n- If $x^2 - y^2 = 4d$, then $p = 2(2d^2 + x^2y^2)$ is not prime.\n- If $x^2 - y^2 = 8d$, then $p = 8d^2 + x^2y^2$ is prime only if $x, y$ are odd, so $p \\equiv 1 \\pmod{8}$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 14771,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $ABCD$ be a given convex quadrilateral. Determine the locus of the points $P$ lying inside the quadrilateral $ABCD$ and satisfying\n\n$$\n[PAB] \\cdot [PCD] = [PBC] \\cdot [PDA],\n$$\n\nwhere $[XYZ]$ denotes the area of triangle $XYZ$.",
    "options": [],
    "answer": "See solution",
    "solution": "If $P$ lies on one of the diagonals $AC$ or $BD$, let's say on $AC$, then\n\n$$\n\\frac{[PAB]}{[PBC]} = \\frac{AP}{PC} = \\frac{[PDA]}{[PCD]},\n$$\n\nwhich is the desired equality. We prove that no other point lying inside $ABCD$ satisfies the conditions of the problem.\n\nDenote by $O$ the point of intersection of the diagonals $AC$ and $BD$ and suppose that $P$ lies inside the triangle $ABO$. Let moreover $BP$ and $AC$ meet at $Q$ and $DP$ and $AC$ meet at $R$. Then\n\n$$\n\\frac{[PAB]}{[PBC]} = \\frac{AQ}{QC} \\quad \\text{and} \\quad \\frac{[PDA]}{[PCD]} = \\frac{AR}{RC},\n$$\n\nwhich, since $Q \\neq R$, implies that the given equality cannot hold.\n\nTherefore, the desired locus of the points $P$ consists of the diagonals $AC$ and $BD$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 14772,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all pairs $(p, q)$, where $p, q \\in \\mathbb{N}$, such that\n$$\n(p+1)^{p-1} + (p-1)^{p+1} = q^q.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "First, note that\n$$\n(p+1)^{p-1} + (p-1)^{p+1} \\geq (p+1)^{p-1} \\geq (p-1)^{p-1}\n$$\nand\n$$\n(p+1)^{p-1} + (p-1)^{p+1} < (p+1)^{p+1} + (p+1)^{p-1} = 2(p+1)^{p+1} < (p+2)^{p+2}.\n$$\nFrom these, we get\n$$\n(p-1)^{p-1} \\leq q^q < (p+2)^{p+2}.\n$$\n\n1) Let $q = p-1$:\n$$\n(p+1)^{p-1} + (p-1)^{p+1} = (p-1)^{p-1}.\n$$\nBut $(p-1)^{p+1} = 0$ and $(p+1)^{p-1} = (p-1)^{p-1}$ only if $p=1$, $q=0$. Since $0 \\notin \\mathbb{N}$, there are no solutions in natural numbers for this case.\n\n2) Let $q = p$:\nIf $p=1$,\n$$\n(p+1)^{p-1} + (p-1)^{p+1} = 1, \\quad p^p = 1.\n$$\nSo $(p, q) = (1, 1)$ is a solution.\nIf $p=2$,\n$$\n(p+1)^{p-1} + (p-1)^{p+1} = 4, \\quad p^p = 4.\n$$\nSo $(p, q) = (2, 2)$ is a solution.\nIf $p=3$,\n$$\n(p+1)^{p-1} + (p-1)^{p+1} = 32, \\quad p^p = 27.\n$$\nSo $(p, q) = (3, 3)$ is not a solution.\nFor $p \\geq 4$, $(p-1)^p > p^{p-1}$, so\n$$\n(p+1)^{p-1} + (p-1)^{p+1} > (p+1)^{p-1} + p^{p-1}(p-1) > p^{p-1} + p^{p-1}(p-1) = p^p.\n$$\nThus, for $p \\geq 4$, there are no solutions.\n\n3) Let $q = p+1$:\n$$\n(p+1)^{p-1} + (p-1)^{p+1} = (p+1)^{p+1}\n$$\nThis leads to\n$$\n(p-1)^{p+1} = (p+1)^{p-1}((p+1)^2 - 1) = (p+1)^{p-1} p (p+2).\n$$\nSince $p$ and $p-1$ are coprime, there are no solutions in $\\mathbb{N}$ for this case.\n\n**Conclusion:**\nThe only solutions are $(p, q) = (1, 1)$ and $(p, q) = (2, 2)$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 14773,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $\\mathbb{R}^+$ be the set of positive real numbers. Let $f : \\mathbb{R}^+ \\to \\mathbb{R}^+$ be a function satisfying the following:\n\nFor each positive real number $x$, there exists $y \\in \\mathbb{R}$ such that\n$$\n(x + f(y))(y + f(x)) \\le 4,\n$$\nand the number of such $y$ is finite.\n\nProve that $f(x) > f(y)$ for every pair of positive real numbers $x$ and $y$ with $x < y$.",
    "options": [],
    "answer": "See solution",
    "solution": "For each positive real number $x$, let $A_x$ be the set of positive real numbers $y$ satisfying $(x + f(y))(y + f(x)) \\le 4$. The following are easy consequences by the definition:\n\n1. If $x \\in A_y$, then $y \\in A_x$.\n2. For $x < y$, if $f(x) \\le f(y)$ then $A_y \\subseteq A_x$.\n\nWe prove the following lemma.\n\n**Lemma.** For $x \\in \\mathbb{R}^+$, there are only finitely many positive real numbers $y$ with $y < x$ such that $f(y) < f(x)$.\n\n*Proof.* Since $A_x$ is not empty, there exists $z \\in A_x$. For $y < x$ with $f(y) < f(x)$, we have $A_x \\subseteq A_y$ by (2). So $z$ also belongs to $A_y$, and $y \\in A_z$ by (1). Therefore, if there are infinitely many $y < x$ such that $f(y) < f(x)$, $A_z$ becomes an infinite set, which is a contradiction. $\\square$\n\nAssume that there exist $a, b \\in \\mathbb{R}^+$ such that $a < b$ and $f(a) \\le f(b)$. Let $S = \\{x \\mid a < x < b,\\ f(x) > f(a)\\}$. By the lemma, there are only finitely many $x$ with $a < x < b$ such that $f(x) < f(b)$. So $S$ is an infinite set. For any $s \\in S$, $A_s \\subseteq A_a$ by (2). Since $A_s$ is not empty, there exists $t \\in A_s \\subseteq A_a$. So, $s \\in A_t \\subseteq \\bigcup_{z \\in A_a} A_z$ by (1). That is, $S \\subseteq \\bigcup_{z \\in A_a} A_z$, which yields a contradiction because $S$ is an infinite set but\n$$\n| \\bigcup_{z \\in A_a} A_z | \\le \\sum_{z \\in A_a} |A_z|\n$$\nis finite. $\\square$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 14774,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $F$, $F_A$, $F_B$, and $F_C$ be the Feuerbach points of a non-equilateral triangle $ABC$, and let $I$ be its incenter. Prove that the lines $AF_A$, $BF_B$, $CF_C$, and $IF$ all pass through a single point.\n\nFeuerbach points are the points of tangency of the nine-point circle with the incircle (touches internally) and the three excircles (touch externally) of the triangle: $F$ is the point of tangency with the incircle, $F_A$ is the point of tangency with the excircle touching $BC$, $F_B$ is the point of tangency with the excircle touching $AC$, and $F_C$ is the point of tangency with the excircle touching $AB$.\n\nThe nine-point circle is the circle passing through the midpoint of each side of the triangle.",
    "options": [],
    "answer": "See solution",
    "solution": "Denote by $O_B$ the center of the excircle that touches $AC$, and by $E$ the center of the nine-point circle. Points $E$, $I$, and $F$, as well as $E$, $F_B$, and $O_B$, lie on the same line since $F$ and $F_B$ are the points of tangency of the circles with the corresponding centers. Also, $E \\neq I$, since otherwise the circles would either not have any common points or would coincide. If $E$ lies on the line $IO_B$, i.e., on the angle bisector at $B$, then $F$ and $F_B$ also lie on this line, and the lines $IF$ and $BF_B$ coincide with the bisector.\n\nSuppose $E$ lies outside the line $IO_B$. Point $F_B$ is on the side $EO_B$ of triangle $EIO_B$ since the circles touch externally, while $B$ lies on the extension of side $O_BI$. Thus, the line $BF_B$ intersects the line $EI = IF$ at some point $X_B$, which is an interior point of segment $EI$. By Menelaus's theorem applied to triangle $EIO_B$ and the line through $B$, $X_B$, and $F_B$:\n\n![](images/Ukraine_2021-2022_p38_data_d73d44c0f2.png)\n\n$$\n\\frac{EX_B}{X_B I} \\cdot \\frac{IB}{B O_B} \\cdot \\frac{O_B F_B}{F_B E} = 1.\n$$\n\nLet $r$ be the radius of the incircle of triangle $ABC$, $r_B$ the radius of its excircle touching $AC$, and $R_9$ the radius of the nine-point circle. Dropping perpendiculars $II'$ and $O_B O'_B$ from $I$ and $O_B$ to $AB$ forms similar right triangles $BII'$ and $BO_B O'_B$ with similarity coefficient $\\frac{BO_B}{BI} = \\frac{O_B O'_B}{II'} = \\frac{r_B}{r}$. Therefore,\n\n$$\n\\frac{EX_B}{X_B I} = \\frac{B O_B}{I B} \\cdot \\frac{F_B E}{O_B F_B} = \\frac{r_B}{r} \\cdot \\frac{R_9}{r_B} = \\frac{R_9}{r}.\n$$\n\nThus, line $BF_B$ passes through the unique point $X$ on segment $EI$ such that $\\frac{EX}{XI} = \\frac{R_9}{r}$. The same holds if $E$ lies on $IO_B$, since then all points of $EI$ belong to $BF_B$.\n\nBy similar reasoning, the lines $AF_A$ and $CF_C$ also pass through the same point $X \\in EI$ with $\\frac{EX}{XI} = \\frac{R_9}{r}$. To finish, note that $X$, and the whole segment $EI$, also belongs to the line $IF$.\n\nNote: Traditionally, only one point is called the Feuerbach point—the point of tangency of the nine-point circle with the incircle; the other three points do not have a specific name. All four points exist regardless of the triangle's structure.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 14775,
    "subject": "Mathematics (Olympiad)",
    "question": "For all $k \\ge 2$:\n\nIs it possible to fill a $2k \\times 2k$ table with 0s and 1s (representing even and odd numbers, respectively) so that the sum of each row and each column is even? If so, provide constructions or prove impossibility for various $k$.",
    "options": [],
    "answer": "See solution",
    "solution": "Yet another construction for all $k \\ge 2$ follows. Divide the $2k \\times 2k$ table into $2k$ cyclic diagonals (in fig. 15, one of such cyclic diagonals of the $8 \\times 8$ table is coloured). Each cyclic diagonal contains exactly one cell from each row and each column.\n\nChoose two cyclic diagonals with exactly one diagonal between them. Fill both chosen diagonals with 0 and 1 alternatingly (see fig. 16).\n\n![](images/prob1314_p17_data_edba6f533b.png)\n\nFigure 15\n\n![](images/prob1314_p17_data_68495183dd.png)\n\nFigure 16",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 14776,
    "subject": "Mathematics (Olympiad)",
    "question": "Suppose $f$ and $g$ are functions such that for all $x \\neq y$,\n$$\n(f(x) - f(y))^2 + (g(x) - g(y))^2 \\geq \\frac{1}{2} (|f(x) - f(y)| + |g(x) - g(y)|)^2 > \\frac{1}{2}\n$$\nIs it possible for such functions $f$ and $g$ to exist?",
    "options": [],
    "answer": "See solution",
    "solution": "For every $x$, consider the point $(f(x), g(x))$ in $\\mathbb{R}^2$. The inequality shows that the distance between any two such points is more than $\\frac{1}{\\sqrt{2}}$. For each point $(f(x), g(x))$, draw a circle of radius $\\frac{1}{2\\sqrt{2}}$ centered at that point. No two circles intersect. Since there are uncountably many $x$, there are uncountably many such circles. Each circle contains a point with rational coordinates, so there would be uncountably many points with rational coordinates, which is impossible because the set of points with rational coordinates is countable. This contradiction shows that no two functions with this property exist.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 14777,
    "subject": "Mathematics (Olympiad)",
    "question": "設 $\\langle f_n \\rangle$ 為費氏數列，亦即：$f_0 = 0$、$f_1 = 1$，且對所有非負整數 $n$，$f_{n+2} = f_{n+1} + f_n$ 均成立。\n\n試找出所有的正整數對 $(a, b)$ 滿足 $a < b$，並且對任意的正整數 $n$，$f_n - 2n \\cdot a^n$ 總能被 $b$ 整除。\n\nLet $\\langle f_n \\rangle$ be the Fibonacci sequence, that is, $f_0 = 0$, $f_1 = 1$, and $f_{n+2} = f_{n+1} + f_n$ holds for all nonnegative integers $n$.\n\nFind all pairs $(a, b)$ of positive integers with $a < b$ such that $f_n - 2n \\cdot a^n$ is divisible by $b$ for all positive integers $n$.",
    "options": [],
    "answer": "See solution",
    "solution": "由題設，$b \\mid f_1 - 2a$，即 $b \\mid 1 - 2a$。但因 $b > a$，所以 $b = 2a - 1$。而對任意正整數 $n$，都有\n\n$$\nb \\mid f_n - 2n a^n, \\quad b \\mid f_{n+1} - 2(n+1)a^{n+1}, \\quad b \\mid f_{n+2} - 2(n+2)a^{n+2}.\n$$\n\n由此三式，加上 $f_{n+2} = f_{n+1} + f_n$ 以及 $b = 2a-1$ 等條件，知\n\n$$\nb \\mid (n + 2)a^{n+2} - (n + 1)a^{n+1} - n a^n.\n$$\n\n又因 $b = 2a - 1$ 必與 $a$ 互質，故得\n\n$$\nb \\mid (n + 2)a^2 - (n + 1)a - n.\n$$\n\n在上式中以 $n+1$ 代入，得\n\n$$\nb \\mid (n + 3)a^2 - (n + 2)a - (n + 1).\n$$\n\n兩式相減，可得\n\n$$\nb \\mid a^2 - a - 1,\n$$\n即 $(2a - 1) \\mid a^2 - a - 1$，也就是 $(2a - 1) \\mid 4a^2 - 4a - 4$。\n\n由於 $4a^2 - 4a - 4 = (2a - 1)^2 - 5$，故 $2a - 1$ 整除 $-5$，所以 $2a - 1 = 1$ 或 $5$，只能得到 $a = 3, b = 2a - 1 = 5$ 一組解。（另一組解 $a = 1, b = 1$ 不滿足 $b > a$，捨之。）\n\n最後檢查充分性，也就是 $(a, b) = (3, 5)$ 滿足題設；即對所有的正整數 $n$，$f_n - 2n \\cdot 3^n$ 確能被 $5$ 整除。當 $n = 1, 2$ 時，$f_1 - 2 \\cdot 1 \\cdot 3 = 1 - 6 = -5$，$f_2 - 2 \\cdot 2 \\cdot 3^2 = 1 - 36 = -35$ 都的確被 $5$ 整除。\n\n現在假設當 $n = k, k + 1$ 時結論成立，即 $f_k - 2k \\cdot 3^k$、$f_{k+1} - 2(k + 1) \\cdot 3^{k+1}$ 都能被 $5$ 整除。於是 $5$ 也能整除\n\n$$\n(f_{k+1} - 2(k+1) \\cdot 3^{k+1}) + (f_k - 2k \\cdot 3^k) = f_{k+2} - 2 \\cdot 3^k (4k + 3).\n$$\n\n於是若要 $5$ 能整除 $f_{n+2} - 2(k+2) \\cdot 3^{k+2}$，此條件等價於\n\n$$\n9(k + 2) \\equiv 4k + 3 \\pmod{5}.\n$$\n\n但上式等價於 $5$ 整除 $5k+15$，明顯成立。故本題由數學歸納法得證。",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 14778,
    "subject": "Mathematics (Olympiad)",
    "question": "Given the functions $f(x) = |x - 2| - |x - 4|$ and $g(x) = |x - 8| - 2$, find the area of the figure whose vertices are the intersection points of the graphs of $f(x)$ and $g(x)$, and the intersection points of the graph of $g(x)$ with the $x$-axis.",
    "options": [],
    "answer": "See solution",
    "solution": "The graph of $g(x)$ consists of two rays with a common vertex at $x = 8$. Removing the absolute value, we find its intersection points with the $x$-axis by solving $6 - x = 0$ and $x - 10 = 0$, giving $A(6, 0)$ and $B(10, 0)$.\n\nFor $f(x)$, after removing the absolute values:\n- For $x < 2$: $f(x) = -2$\n- For $2 \\leq x \\leq 4$: $f(x) = 2x - 6$\n- For $x > 4$: $f(x) = 2$\n\nWe solve $f(x) = g(x)$ in each interval and find the intersection points: $D(4, 2)$ and $C(12, 2)$.\n\nThus, the figure $ABCD$ is a trapezoid with bases $8$ and $4$ and height $2$. Its area is:\n\n$$\n\\text{Area} = \\frac{8 + 4}{2} \\times 2 = 12\n$$",
    "topic": "Discrete Mathematics",
    "subtopic": "Other"
  },
  {
    "id": 14779,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $ABCD$ be a parallelogram such that $AC = BC$. A point $P$ is chosen on the extension of the segment $AB$ beyond $B$. The circumcircle of triangle $ACD$ meets the segment $PD$ again at $Q$, and the circumcircle of triangle $APQ$ meets the segment $PC$ again at $R$. Prove that the lines $CD$, $AQ$, and $BR$ are concurrent.\n\n![](images/Saudi_Arabia_booklet_2022_p35_data_630c705128.png)",
    "options": [],
    "answer": "See solution",
    "solution": "Let $E$ be the intersection of $AD$ and $CQ$, $F$ be the intersection of $BD$ and $CP$, and $X$ be the intersection of $CQ$ and $AB$. Apply Desargues's theorem on triangles $ABD$ and $QRC$; one can see that $AQ$, $BR$, and $CD$ concur if and only if $E$, $F$, and $G$ are collinear. Apply Menelaus's theorem on triangle $ABD$; it follows that $E$, $F$, and $G$ are collinear if and only if\n\n$$\n\\frac{AE}{ED} \\cdot \\frac{DF}{FB} \\cdot \\frac{BG}{GA} = 1.\n$$\n\nOn the other hand, by Thales's theorem, we observe that\n\n$$\n\\frac{AE}{ED} = \\frac{AX}{CD}, \\quad \\frac{DF}{FB} = \\frac{DC}{BP}.\n$$\n\nIt suffices to prove that\n\n$$\n\\frac{AX}{BP} = \\frac{AG}{BG}.\n$$\n\nNow, we show that $BXQR$ is a cyclic quadrilateral. Since\n\n$$\n\\begin{aligned}\n\\angle ARC &= \\angle ARQ + \\angle QRC = \\angle APQ + \\angle QAP \\\\\n&= \\angle AQD = \\angle ACD = \\angle ADC = \\angle ABC\n\\end{aligned}\n$$\n\nsince $AC = BC = AD$, thus $ABRC$ is a cyclic quadrilateral. Hence\n\n$$\n\\begin{aligned}\n\\angle BRQ &= \\angle BRC - \\angle QRC = 180^\\circ - \\angle CAB - \\angle QAP \\\\\n&= 180^\\circ - \\angle ADC - \\angle (AQ, CD) = \\angle QAD = 180^\\circ - \\angle QCD = 180^\\circ - \\angle QXR.\n\\end{aligned}\n$$\n\nThen $BXQR$ is a cyclic quadrilateral. Then we have\n\n$$\nGB \\cdot GX = GR \\cdot GQ = GP \\cdot GA \\Leftrightarrow \\frac{GA}{GB} = \\frac{GX}{GP} = \\frac{AX}{BP}\n$$\n\nas desired.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 14780,
    "subject": "Mathematics (Olympiad)",
    "question": "Given an acute triangle $ABC$ with $AB < BC < CA$. Let $D$ be a moving point on side $BC$, and $E$ be a moving point on the minor arc $\\widearc{BC}$ of the circumcircle of $ABC$, such that $\\angle BAD = \\angle BED$.\n\nLet $F$ be the intersection point of the line through $D$ perpendicular to $AB$ with the extension of $AC$. Prove that $\\angle BEF$ is constant.\n\n![](images/2024_CGMO_p7_data_33bf81ba9b.png)",
    "options": [],
    "answer": "See solution",
    "solution": "Let $H$ be the orthocenter of triangle $ABD$. Since $DF \\perp AB$, the points $H$, $D$, $F$ are colinear, as shown:\n\n![](images/2024_CGMO_p7_data_08671e110c.png)\n\n**Case 1:** When $\\angle ADB < 90^\\circ$:\n\n- $H$ lies inside triangle $ABD$.\n- By orthocenter properties, $\\angle BAD$ and $\\angle BHD$ are supplementary.\n- Given $\\angle BED = \\angle BAD$, we have $\\angle BED = 180^\\circ - \\angle BHD$.\n- Thus, $B$, $H$, $D$, $E$ are concyclic.\n\nThis implies:\n\n- $\\angle EHF = \\angle EBD = \\angle EBC = \\angle EAF$.\n- Therefore, $A$, $H$, $E$, $F$ are concyclic.\n- Consequently, $\\angle AEF = \\angle AHF = 180^\\circ - \\angle ABD = 180^\\circ - \\angle ABC$.\n\n**Other Cases:**\n\n- When $\\angle ADB > 90^\\circ$, $D$ lies inside triangle $ABH$.\n- When $\\angle ADB = 90^\\circ$, $D$ coincides with $H$.\n\nIn all cases (using directed angles when necessary), we conclude:\n\n$$\n\\angle BEF = \\angle BEA + \\angle AEF = \\angle BCA + (180^\\circ - \\angle ABC)\n$$\n\nSince $\\angle BCA$ and $\\angle ABC$ are fixed angles of triangle $ABC$, $\\angle BEF$ is constant. $\\square$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 14781,
    "subject": "Mathematics (Olympiad)",
    "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![](images/Estonija_2012_p11_data_9a13dcc768.png)",
    "options": [],
    "answer": "See solution",
    "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 cannot 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 are exactly $k^2$ of them, hence there can be at most $k^2$ snakes.",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 14782,
    "subject": "Mathematics (Olympiad)",
    "question": "Find the least positive integer $n$ for which it is possible to draw an $n$-gon along the lines of a square grid, whose perimeter is $n$ and whose area is $n$. Here, the boundary of the $n$-gon may not visit any point more than once.",
    "options": [],
    "answer": "See solution",
    "solution": "The answer is $n = 32$.\n\nWe start by showing that such a 32-gon exists.\n\nIt remains to prove that this is indeed the smallest possible $n$. We will do that by proving that for all $n < 32$, the area $S$ of an $n$-gon with perimeter $P = n$ is smaller than $n$.\n\nAt first, we note that $P = n$ if and only if all sides of the $n$-gon are of length 1. From this, it follows that $n$ should be an even number, as there should be an even number of horizontal and an even number of vertical sides. Denote $n = 2m$ and the vertices of the $n$-gon as $A_1A_2\\dots A_{2m}$. We call a vertex $A_i$ odd (even) if its index $i$ is odd (even).\n\nNow we can consider the sides of the $n$-gon in pairs. Let vertex $A_i$ be located at the point $(x, y)$. Vertex $A_{i+2}$ is then located at $(x\\pm1, y\\pm1)$. It means that by moving from $A_i$ to $A_{i+2}$ (we count vertices modulo $2m$), we have moved by a unit vector in the “diagonal” coordinate system with unit vectors $(1, 1)$ and $(1, -1)$. We denote these steps $\\rightarrow$ and $\\leftarrow$, respectively.\n\nLet's start at the vertex $A_1$ and travel around the $n$-gon in $m$ steps by stopping at all odd vertices $A_1 \\to A_3 \\to \\dots \\to A_{2m-1} \\to A_1$. Denote the total number of $\\rightarrow$ steps as $2a$ and the total number of $\\leftarrow$ steps as $2b$ (both of them must be even numbers). Therefore $m = 2a + 2b$ and $n = 4a + 4b$, hence $n$ is divisible by 4. It means that we have to prove that $S < n$ for all $P = n \\le 28$.\n\nLet's consider the most “northeastern”, “northwestern”, “southeastern” and “southwestern” vertices of our $n$-gon, denote them $NE, NW, SE, SW$ respectively (see figure). W.l.o.g. assume that $NE$ and at least one more of them are odd. Our $n$-gon is contained in a rectangle whose sides are parallel to vectors $(1, 1)$ and $(1, -1)$ and pass through these four vertices. Denote its width and height as $W$ and $H$. Our $n$-gon's area is no larger than $W \\cdot H - S_g$ where $S_g$ is the area of the “gray triangles”. It can be computed as half of the area of the “square ring” between outer and inner (gray) rectangles:\n\n![](images/BW23_Shortlist_2023-11-01_p51_data_ccf1b27423.png)\n\n$$\nS_g = \\frac{WH - (W - \\sqrt{2})(H - \\sqrt{2})}{2} = \\frac{\\sqrt{2}(W + H)}{2} - 1\n$$\n\nTherefore, the area of the $n$-gon is no larger than\n$$\nS \\le W H - \\frac{\\sqrt{2}(W+H)}{2} + 1.\n$$\nBy using the inequality $WH \\le \\frac{(W+H)^2}{4}$, we can transform it into\n$$\nS \\le \\frac{(W+H)^2}{4} - \\frac{\\sqrt{2}(W+H)}{2} + 1 = \\frac{(W+H-\\sqrt{2})^2}{4} + \\frac{1}{2}. \\quad (*)\n$$\n\nRecall that the $NE$ vertex is odd. Assume at first that $SW$ also is odd. Then, as travelling from $NE$ to $SW$ uses at most $b$ $\\nearrow$ steps, we conclude that $W \\le b\\sqrt{2}$. If the vertex $SW$ is even, then the estimation is slightly worse: we can travel in at most $b$ $\\nearrow$ steps from $NE$ to one of the neighbours of $SW$ and $SW$ itself is further $\\sqrt{2}/2$ away from it in $\\nearrow$ direction, so that $W \\le b\\sqrt{2} + \\sqrt{2}/2$.\n\nIf $SW$ is even, then at least one of vertices $NW$ and $SE$ is odd. In that case, for $H$ we get similar estimation $H \\le a\\sqrt{2} + \\sqrt{2}/2$. But if $SW$ is odd, then it could happen that both $NW$ and $SE$ are even.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 14783,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $ABCD$ be a square in the plane $\\mathcal{P}$. Find the minimum and the maximum values of the function $f: \\mathcal{P} \\to \\mathbb{R}$ defined by\n\n$$\nf(P) = \\frac{PA + PB}{PC + PD}\n$$\n\nwhere $\\mathbb{R}$ is the set of all real numbers.",
    "options": [],
    "answer": "See solution",
    "solution": "We have $f(A) = \\frac{1}{\\sqrt{2} + 1} = \\sqrt{2} - 1$. We will prove that this value is the minimum of the function $f$, or in other words,\n\n$$\nPA + PB \\geq (\\sqrt{2} - 1)(PC + PD)\n$$\n\nfor all $P$.\n\nApplying Ptolemy's inequality for the points $P, A, B, C$, we have $PA + \\sqrt{2} PB \\geq PC$, that is\n\n$$\nPA + \\sqrt{2} PB \\geq PC.\n$$\n\nApplying Ptolemy's inequality for the points $P, A, B, D$, we have $\\sqrt{2} PA + PB \\geq PD$, that is\n\n$$\n\\sqrt{2} PA + PB \\geq PD.\n$$\n\nAdding these inequalities, we get\n\n$$\n(\\sqrt{2} + 1)(PA + PB) \\geq PC + PD,\n$$\n\nhence the desired inequality.\n\nBy Ptolemy's Theorem, it follows that the minimum is attained if and only if the point $P$ belongs to the arc $AB$ of the circumcircle of the square.\n\nIf $P'$ is the symmetric of $P$ with respect to the center of the square, then $f(P) = 1 / f(P')$. It follows that the maximum of the function is $\\frac{1}{\\sqrt{2} - 1} = \\sqrt{2} + 1$, and it occurs exactly at the points on the arc $CD$ of the circumcircle of the square.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 14784,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $m$ and $n$ be two integers and define $a_0 = m$, $a_1 = n$, and $a_{k+1} = 4a_k - 5a_{k-1}$ for $k \\geq 1$. If $p > 5$ is a prime such that $p-1$ is divisible by $4$, show that there exist integers $m$ and $n$ such that $p$ does not divide $a_k$ for any $k \\geq 0$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $t$ be an integer such that $p$ divides $t^2 + 1$. Such an integer exists since $p-1$ is divisible by $4$. Let $m = 1$ and $n = t + 2$. Then\n\n$$\n n^2 = t^2 + 4t + 4 \\equiv 4t + 3 \\equiv 4n - 5m \\pmod{p}.\n$$\n\nTherefore, if $a_0 = 1$ and $a_1 = n$, then $a_2 \\equiv n^2 \\pmod{p}$. By induction, it is easy to see that $a_k \\equiv n_k \\pmod{p}$. Since $p > 5$, it follows that $p$ does not divide $n$. Therefore, $p$ does not divide $a_k$ for any $k \\geq 0$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 14785,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $a$ be a positive real number. Show that there are no real numbers $b$ and $c$, with $b < c$, such that\n$$\n\\left|\\frac{x+y}{x-y}\\right| \\le a\n$$\nfor every $x, y \\in (b, c)$, $x \\ne y$.",
    "options": [],
    "answer": "See solution",
    "solution": "Assume, for the sake of contradiction, that there exist $b, c \\in \\mathbb{R}$, $b < c$, such that\n$$\n\\left|\\frac{x+y}{x-y}\\right| \\le a\n$$\nfor every $x, y \\in (b, c)$, $x \\ne y$. Consider $x \\in (b, c)$, $x \\ne 0$.\n\nIf $x > 0$, there exist infinitely many positive integers $n$ such that $n > \\frac{1}{x-b}$, or equivalently, $x - \\frac{1}{n} > b$. For each such $n$, choose $y_n = x - \\frac{1}{n} \\in (b, c)$. Since $\\left|\\frac{x+y_n}{x-y_n}\\right| \\le a$, it follows that $x \\le \\frac{1+a}{2n}$, so $0 < n \\le \\frac{1+a}{2x}$ for infinitely many positive integers $n$, a contradiction.\n\nIf $x < 0$, there are infinitely many positive integers $n$ such that $n > \\frac{1}{c-x}$, or equivalently, $x + \\frac{1}{n} < c$. For each such $n$, choose $z_n = x + \\frac{1}{n} \\in (b, c)$. Since $\\left|\\frac{x+z_n}{x-z_n}\\right| \\le a$, it follows that $-\\frac{1+a}{2n} \\le x < 0$, so $0 < n \\le -\\frac{1+a}{2x}$ for infinitely many positive integers $n$, again a contradiction. Thus, no such $b$ and $c$ exist.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 14786,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $\\frac{13}{476}$ be written in binary decimal representation as\n\n$$\n\\frac{13}{476} = 0.b_1b_2\\cdots b_s\\overline{a_1a_2\\cdots a_k}\n$$\n\nwhere $a_1a_2\\cdots a_k$ is the repetend and $k$ is its period. Find the period $k$ of the binary expansion of $\\frac{13}{476}$.",
    "options": [],
    "answer": "See solution",
    "solution": "We can write $0.b_1b_2\\cdots b_s = \\frac{B}{2^s}$ and $0.a_1a_2\\cdots a_k = \\frac{A}{2^k}$ for some integers $A$ and $B$. Thus,\n\n$$\n\\frac{13}{476} = \\frac{B}{2^s} + \\frac{A}{2^s(2^k - 1)} = \\frac{B(2^k - 1) + A}{2^s(2^k - 1)}.\n$$\n\nSince $476 = 2^2 \\times 7 \\times 17$, we require $7 \\mid 2^k - 1$ and $17 \\mid 2^k - 1$. The orders of $2$ modulo $7$ and $17$ are $3$ and $8$, so $3 \\mid k$ and $8 \\mid k$. The smallest such $k$ is $24$. Therefore, the period of the binary expansion is $24$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 14787,
    "subject": "Mathematics (Olympiad)",
    "question": "In a triangle $ABC$, the excircle $\\omega_a$ opposite $A$ touches $AB$ at $P$ and $AC$ at $Q$, and the excircle $\\omega_b$ opposite $B$ touches $BA$ at $M$ and $BC$ at $N$. Let $K$ be the projection of $C$ onto $MN$, and let $L$ be the projection of $C$ onto $PQ$.\n\nShow that the quadrilateral $MKLP$ is cyclic.",
    "options": [],
    "answer": "See solution",
    "solution": "Denote by $D$ the intersection point of $MN$ and $PQ$ and let $\\angle CDQ = x$, $\\angle CDN = y$. Notice that $\\angle APQ = 90^\\circ - \\frac{A}{2}$ from the isosceles $\\triangle APQ$ and analogously $\\angle BMN = 90^\\circ - \\frac{B}{2}$.\n\nThen $x + y = \\frac{A + B}{2}$. In the standard notation, we have\n\n$$\n\\frac{CD}{\\cos \\frac{B}{2}} = \\frac{p - a}{\\sin y} \\quad \\text{(sine law for $\\triangle CND$)}\n$$\n\n$$\n\\frac{CD}{\\cos \\frac{A}{2}} = \\frac{p - b}{\\sin x} \\quad \\text{(sine law for $\\triangle CQD$)}\n$$\n\nTherefore,\n$$\n\\frac{\\cos \\frac{A}{2}}{\\cos \\frac{B}{2}} = \\frac{\\sin x \\cdot (p - a)}{\\sin y \\cdot (p - b)}\n$$\nBut $\\frac{p - a}{p - b} = \\frac{\\cotg \\frac{A}{2}}{\\cotg \\frac{B}{2}}$, so\n$$\n\\frac{\\sin x}{\\sin y} = \\frac{\\sin \\frac{A}{2}}{\\sin \\frac{B}{2}}\n$$\n\nThe last equality and $x + y = \\frac{A + B}{2}$ imply that $x = \\frac{A}{2}$ and $y = \\frac{B}{2}$ (set $x + y = t$ and consider $\\frac{\\sin(t - y)}{\\sin y} = \\frac{\\sin\\left(t - \\frac{B}{2}\\right)}{\\sin\\frac{B}{2}}$). Therefore $DC \\perp AB$ and, if $C_1$ is the intersection point of $CD$ and $AB$, we have\n\n$$\n\\overline{DL} \\cdot \\overline{DP} = \\overline{DC} \\cdot \\overline{DC_1} = \\overline{DK} \\cdot \\overline{DM},\n$$\n\nwhich implies that $M, P, K$, and $L$ are concyclic.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 14788,
    "subject": "Mathematics (Olympiad)",
    "question": "甲、乙兩人玩以下的數字遊戲：從甲開始，兩個人輪流自 1 到 9 的數字中不重複地選一個數字出來，並且把選出的數字由左至右依序排成一個七位數（即 $\\overline{A_1B_2A_3B_4A_5B_6A_7}$）。如果排出來的七位數是某個完全七次方數的末七位數字，則甲獲勝；否則的話，乙獲勝。\n\nAlice and Bob play a game. Starting from Alice, two players take turns to choose a digit from 1 to 9 without repetition, and put them from the leftmost to the rightmost to form a 7-digit integer (i.e., $\\overline{A_1B_2A_3B_4A_5B_6A_7}$). If there exists a perfect 7th-power that ends in these 7 digits, then Alice wins; otherwise, Bob wins. Who has the winning strategy?",
    "options": [],
    "answer": "See solution",
    "solution": "甲有必勝策略。由以下引理知，只要甲讓 $A_7 \\in \\{1,3,7,9\\}$，甲必便獲勝；而基於乙最多只能擋其中三個，故甲必勝。\n\n**引理**\n\n若 $0 < a < 10^7$ 且 $\\gcd(a, 10) = 1$，則 $x^7 \\equiv a \\pmod{10^7}$ 有解。\n\n**證明**\n\n我們先證明：\n\n*Claim.* 若 $\\gcd(m, 10) = \\gcd(n, 10) = 1$ 且 $10^7 \\mid m^7 - n^7$，則 $10^7 \\mid m - n$。\n\n*Proof of Claim.* 注意到\n\n$$\nm^7 - n^7 = (m-n)(m^6 + m^5n + m^4n^2 + m^3n^3 + m^2n^4 + mn^5 + n^6).\n$$\n\n因此 $\\gcd(m-n, \\frac{m^7-n^7}{m-n}) = \\gcd(m-n, 7n^6)$。又注意到 $7n^6$ 並非 2 或 5 的倍數，故 $2^7 \\mid m-n$ 或 $2^7 \\mid \\frac{m^7-n^7}{m-n}$，且 $5^7 \\mid m-n$ 或 $5^7 \\mid \\frac{m^7-n^7}{m-n}$。以下證明兩個“或”都必為前者。\n\n1. 由於 $m$ 和 $n$ 都必為奇數，且兩個因子不得具有共同因數，可知 $2^7 \\mid m-n$。\n2. 注意到若 $m \\equiv 1,2,3,4 \\pmod 5$，則 $m^7 \\equiv 1,3,2,4 \\pmod 5$。代入後易知 $5^7 \\mid m^7-n^7$；又 $2^7 \\mid m-n \\Rightarrow 5^7 \\mid m-n$。\n\n綜以上，Claim 得證。\n\n回到原題，由 Claim 知模 $10^7$ 的既約剩餘系，七次方後仍為既約剩餘系，故 $x^7 \\equiv a \\pmod{10^7}$ 恆有解。",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 14789,
    "subject": "Mathematics (Olympiad)",
    "question": "a) What are the possible pairs of seed rectangles that can be combined to form a larger rectangle? What is the smallest such combination?\n\n![](images/Australian_Scene_2010_p31_data_b8c8b8e609.png)\n\nb) The seed rectangles have areas $3, 8, 15, 24,$ and $35\\ \\mathrm{cm}^2$ respectively. Can any two or three of these be combined to form a rectangle with area $42\\ \\mathrm{cm}^2$? If so, how?\n\n![](images/Australian_Scene_2010_p31_data_7ad26336e2.png)\n\nc) The total area of all the rectangles is $88\\ \\mathrm{cm}^2$. What are the possible dimensions of a rectangle with this area, given that the $5 \\times 7$ seed rectangle must be used? How can Kim make such a rectangle?",
    "options": [],
    "answer": "See solution",
    "solution": "a) The only possible pairs of seed rectangles that make a rectangle are $1 \\times 3$ with $3 \\times 5$, $2 \\times 4$ with $4 \\times 6$, and $3 \\times 5$ with $5 \\times 7$. The smallest combination is $1 \\times 3$ with $3 \\times 5$, which make a $3 \\times 6$ rectangle.\n\nb) No two of the areas $3, 8, 15, 24, 35$ give a combined area of $42$, but the following three do: $3, 15, 24$. Join the $3 \\times 6$ rectangle in part (a) to the $4 \\times 6$ rectangle.\n\nc) The total area of all of the rectangles is $3 + 8 + 15 + 24 + 35 + 3 = 88\\ \\mathrm{cm}^2$. Rectangles with an area of $88\\ \\mathrm{cm}^2$ must be $2 \\times 44$, $4 \\times 22$, or $8 \\times 11$. Since the $5 \\times 7$ seed rectangle is used, both the length and width of the large rectangle must be greater than $4$. So the large rectangle must be $8 \\times 11$. Kim can make an $8 \\times 11$ rectangle by first making an $8 \\times 4$ rectangle from the $6 \\times 4$ and $2 \\times 4$ seed rectangles and then an $8 \\times 7$ rectangle from the $5 \\times 7$, $3 \\times 1$, $3 \\times 5$, $3 \\times 1$ rectangles.",
    "topic": "Discrete Mathematics",
    "subtopic": "Combinatorics"
  },
  {
    "id": 14790,
    "subject": "Mathematics (Olympiad)",
    "question": "Count the number of ways to fill the unit squares of a $4 \\times 4$ table with two colours, red and blue, such that no two rows and no two columns are painted the same.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $X$ be the set of colorings of the $4 \\times 4$ table with two colors, red and blue, so that no two rows are painted the same. Since we can color a $1 \\times 4$ row in $2^4 = 16$ ways, we have $|X| = 16 \\cdot 15 \\cdot 14 \\cdot 13$.\n\nNow, let $B_{ij} \\subset X$ be the set of colorings where the $i$-th and $j$-th columns are painted the same. We need to find $|X \\setminus (B_{12} \\cup B_{13} \\cup B_{14} \\cup B_{23} \\cup B_{24} \\cup B_{34})|$.\n\n1. To find $|B_{12}|$ (first and second columns the same, no two rows the same), consider:\n   - (i) Both columns are a single color (red or blue): $2 \\cdot 4!$ ways.\n   - (ii) Three squares one color, one the other: $2 \\cdot \\binom{4}{3} \\cdot 4!$ ways.\n   - (iii) Two squares each color: $6 \\cdot (4 \\cdot 3)^2$ ways.\n\n2. For $B_{ijk}$ (three columns the same):\n   - (i) If three squares are the same color, by the pigeonhole principle, two rows are the same (contradicts the condition).\n   - (ii) Two squares each color: $6 \\cdot 2 \\cdot 2$ ways.\n\n3. For $|B_{12} \\cap B_{34}|$ (first two columns the same, last two the same):\n   - (i) If three squares are the same color, contradiction as above.\n   - (ii) Two squares each color: $6 \\cdot 2 \\cdot 2$ ways.\n\nFor mutually distinct $i, j, k, l$, $|B_{ij}| = |B_{12}|$, $|B_{ijk}| = |B_{123}|$, and $|B_{ij} \\cap B_{kl}| = |B_{12} \\cap B_{34}|$. Thus,\n\n$$\n\\begin{align*}\n|X \\setminus (B_{12} \\cup B_{13} \\cup B_{14} \\cup B_{23} \\cup B_{24} \\cup B_{34})| &= |X| - 6 \\cdot |B_{12}| + 11 \\cdot |B_{123}| + 3 \\cdot |B_{12} \\cap B_{34}| - 4 \\cdot |B_{123}| \\\\\n&= 16 \\cdot 15 \\cdot 14 \\cdot 13 - 6 \\cdot 24 \\cdot 70 + 11 \\cdot 24 \\\\\n&= 33864.\n\\end{align*}\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 14791,
    "subject": "Mathematics (Olympiad)",
    "question": "Find the least natural number $n$ for which $\\cos \\frac{\\pi}{n}$ cannot be expressed in the form $p + \\sqrt{q} + \\sqrt[3]{r}$, where $p$, $q$, and $r$ are rational numbers.",
    "options": [],
    "answer": "See solution",
    "solution": "We show that $n = 7$ is the least such number.\n\nNote that:\n$$\n\\cos \\pi = -1, \\quad \\cos \\frac{\\pi}{2} = 0, \\quad \\cos \\frac{\\pi}{3} = \\frac{1}{2}, \\quad \\cos \\frac{\\pi}{4} = \\frac{\\sqrt{2}}{2}, \\quad \\cos \\frac{\\pi}{6} = \\frac{\\sqrt{3}}{2}.\n$$\nFurther, from $0 = \\cos \\frac{3\\pi}{5} + \\cos \\frac{2\\pi}{5}$, $x_5 = \\cos \\frac{\\pi}{5}$ is a root of the equation\n$$\n0 = 4x^3 - 3x + 2x^2 - 1 = (x+1)(4x^2 - 2x - 1),\n$$\ni.e., $x_5 = \\frac{1+\\sqrt{5}}{4}$.\n\nIt remains to prove that $x_7 = \\cos \\frac{\\pi}{7}$ cannot be expressed in the form $p + \\sqrt{q} + \\sqrt[3]{r}$, where $p$, $q$, and $r$ are rational numbers. Since $0 = \\cos \\frac{4\\pi}{7} + \\cos \\frac{3\\pi}{7}$, we have that $x_7$ is a root of\n$$\n0 = 2(2x^2 - 1)^2 - 1 + 4x^3 - 3x = (x + 1)(8x^3 - 4x^2 - 4x + 1),\n$$\ni.e., $x_7$ is a zero of $P(x) = 8x^3 - 4x^2 - 4x + 1$.\n\nSuppose $x_7 = p + \\sqrt{q} + \\sqrt[3]{r}$, where $q \\ge 0$, $p, r \\in \\mathbb{Q}$. Then $x_7$ is a zero of $Q(x) = (x - p - \\sqrt{q})^3 - r$, with coefficients of the form $a + b\\sqrt{q}$, $a, b \\in \\mathbb{Q}$, i.e., from $\\mathbb{Q}[\\sqrt{q}]$. Since the coefficient of $x$ is nonnegative, $P \\ne 8Q$.\n\nThus, $P = 8Q + R$, where $R$ is a polynomial of degree 1 or 2 with coefficients from $\\mathbb{Q}[\\sqrt{q}]$ and $R(x_7) = 0$. If $\\deg R = 1$, then $x_7 \\in \\mathbb{Q}[\\sqrt{q}]$. If $\\deg R = 2$ and $R$ does not divide $P$, then $x_7$ is a zero of the remainder of $P$ divided by $R$, i.e., again $x_7 \\in \\mathbb{Q}[q]$. If $\\deg R = 2$ and $R$ divides $P$, then the zero of $\\frac{P}{R}$, which is from $\\mathbb{Q}[q]$, is also a zero of $P$. Thus, $P$ has a zero from $\\mathbb{Q}[q]$. It is a zero of a polynomial of degree 1 or 2 with rational coefficients. It follows as above that $P$ has a rational zero.\n\nDirect verification shows that none of the numbers $\\pm 1$, $\\pm \\frac{1}{2}$, $\\pm \\frac{1}{4}$, $\\pm \\frac{1}{8}$ (which are all possible rational zeroes of $P$) is a zero of $P$, a contradiction.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 14792,
    "subject": "Mathematics (Olympiad)",
    "question": "Во триаголна пирамида $SABC$ рамнинските агли при врвот $S$ се прави, а точката $O$ е проекција на врвот $S$ врз рамнината на основата $ABC$. Докажи дека плоштината на триаголникот $ASB$ е геометриска средина на плоштините на триаголниците $ABC$ и $OAB$.\n\n![](images/Makedonija_2009_p37_data_db0b4af111.png)",
    "options": [],
    "answer": "See solution",
    "solution": "Од условот на задачата $CS \\perp AS$, $CS \\perp BS$ и $AS \\perp BS$, а точката $O$ е подножје на нормалата спуштена од врвот $S$ на рамнината на основата $ABC$. Нека $\\alpha$ е аголот меѓу рамнините на триаголниците $SAB$ и $ABC$ (види цртеж). Триаголниците $OAB$, $SAB$ и $CAB$ имаат една заедничка страна $AB$ и имаат различни висини спуштени врз таа страна.\n\nПри тоа за висините $OD$, $SD$ и $CD$ имаме, $\\overline{OD} = \\overline{SD} \\cos \\alpha$, $\\overline{SD} = \\overline{CD} \\cos \\alpha$, од каде што ги добиваме следните равенства:\n\n$$\nP_{OAB} = P_{SAB} \\cos \\alpha, \\quad P_{SAB} = P_{ABC} \\cos \\alpha.\n$$\n\nОд првото равенство добиваме $\\cos \\alpha = \\frac{P_{OAB}}{P_{SAB}}$, и ако замениме во второто равенство,\n\n$$\nP_{SAB} = P_{ABC} \\frac{P_{OAB}}{P_{SAB}},\n$$\n\nодносно $P_{SAB}^2 = P_{ABC} \\cdot P_{OAB}$.\n\nЗначи, $P_{SAB} = \\sqrt{P_{ABC} \\cdot P_{OAB}}$, што и требаше да се докажи.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 14793,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all integers $n \\geq 2$ such that $n$ is divisible by each of the numbers $d_1$, $d_1 + d_2$, $\\dots$, $d_1 + d_2 + \\dots + d_{k-1}$, where $1 = d_1 < d_2 < \\dots < d_{k-1} < d_k = n$ are all the positive divisors of $n$.",
    "options": [],
    "answer": "See solution",
    "solution": "All primes and $n = 6$ satisfy the condition.\n\nFor a prime $n$, its divisors are $1 = d_1 < d_2 = n$, and $n$ is divisible by $d_1$.\n\nSuppose $n$ has $k \\geq 3$ divisors and fulfills the property. If $n$ is odd, all its divisors are odd, but $d_1 + d_2$ is even, so $n$ cannot be odd. Thus, $n$ must be even.\n\nLet $d_1 = 1$, $d_2 = 2$, so $n$ must be divisible by $d_1 + d_2 = 3$, implying $d_3 = 3$. Therefore, $6 \\mid n$ and $\\frac{n}{6}$, $\\frac{n}{3}$, $\\frac{n}{2}$ are divisors of $n$.\n\nSince $n$ is divisible by $d_1 + d_2 + \\dots + d_{k-1}$, we have $n \\geq d_1 + d_2 + \\dots + d_{k-1} \\geq \\frac{n}{6} + \\frac{n}{3} + \\frac{n}{2} = n$. Thus, the only divisors of $n$ less than $n$ are $\\frac{n}{6}$, $\\frac{n}{3}$, and $\\frac{n}{2}$, so $\\frac{n}{6} = d_1 = 1$, $\\frac{n}{3} = d_2 = 2$, $\\frac{n}{2} = d_3 = 3$, yielding $n = 6$, which satisfies the condition.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 14794,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $x$ be the side length of a square tile. If a rectangle is completely tiled with such square tiles, there exist integers $a, b$ such that the side lengths of the rectangle are $ax$ and $bx$. Let $r = 75$ denote the radius of the circle.\n\n![](images/IRL_ABooklet_2021_p42_data_f49825f8b0.png)\n\nThe diagonals of the rectangle are diameters of the circle, so their length is $2r$. Applying the Pythagorean theorem to the right-angled triangle formed by cutting the rectangle along one of its diagonals, we obtain:\n\n$$\n(2r)^2 = x^2(a^2 + b^2).\n$$\n\nWhat is the largest possible integer value of $x$?",
    "options": [],
    "answer": "See solution",
    "solution": "As $x, r, a, b$ are integers, this implies that $x$ is a factor of $2r$, so $c = 2r/x$ is an integer satisfying $c^2 = a^2 + b^2$. To maximize $x = 2r/c$, we seek the smallest $c$ dividing $2r$ that is the hypotenuse of a right triangle with integer sides.\n\nThe positive divisors of $2r = 150 = 2 \\cdot 3 \\cdot 5^2$ are $1, 2, 3, 5, 6, 10, 15, 25, 30, 50, 75$, and $150$. Since $1^2 + 2^2 = 5$, $c = 5$ is the smallest possible value. For $c = 5$, $a = 3$, $b = 4$, so $x = 150/5 = 30$. The rectangle's sides are $ax = 90$ and $bx = 120$. This rectangle fits in a circle of radius $r = 75$ because $150^2 = 90^2 + 120^2$.\n\n![](images/IRL_ABooklet_2021_p43_data_9762bd53a9.png)\n\nTherefore, the largest possible integer side length of a square tile that can be used to completely tile a rectangle inscribed in a circle of radius $75$ is $30$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 14795,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all positive integers $n$ such that $3n-2$ divides $n^{3n-2} - 3n + 1$.",
    "options": [],
    "answer": "See solution",
    "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": 14796,
    "subject": "Mathematics (Olympiad)",
    "question": "Assume that $ABCD$ is a cyclic quadrilateral with circumcircle $\\Omega$. Assume lines $AB$ and $CD$ intersect at point $P$ and lines $AD$ and $BC$ intersect at $Q$. Let $\\Gamma$ be the circumcircle of triangle $APQ$. Then $\\Omega$ and $\\Gamma$ intersect in two points, $A$ is one of them and $R$ is the other. Assume $C \\neq R$. Prove that line $CR$ passes through $M$ where $M$ is the midpoint of line segment $PQ$.",
    "options": [],
    "answer": "See solution",
    "solution": "Take $S$ to be the point such that $CPSQ$ is a parallelogram, as seen in figure 20. For points $X, Y, Z$ let $\\text{rot } XYZ$ denote the morphism on translations induced by the rotation that takes line $XY$ to line $XZ$, modulo half turn. As $ABCD$ is a cyclic quadrilateral it follows that $\\text{rot } BAD = \\text{rot } BCD$. It follows that $\\text{rot } PAD = \\text{rot } BCP$ and hence $\\text{rot } PAQ = \\text{rot } QCP = \\text{rot } PSQ$. It follows that $S$ lies on $\\Gamma$.\n\nUsing the cyclic quadrilateral $AQRS$ it follows that $\\text{rot } RSQ = \\text{rot } RAQ$. Considering the cyclic quadrilateral $ABCD$ it follows that $\\text{rot } RAQ = \\text{rot } RAD = \\text{rot } RCD = \\text{rot } RCP$. Hence, $\\text{rot } RSQ = \\text{rot } RCP$. As lines $CP$ and $SQ$ are parallel it follows that line $SR$ is parallel to line $CR$. Now $R$ is a common point so it follows that line $SR = CR = CS$.\n\nAs lines $CS$ and $PQ$ are diagonals in a parallelogram $CPSQ$ it follows that line $CR = CS$ passes through $M$, the midpoint of line segment $PQ$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 14797,
    "subject": "Mathematics (Olympiad)",
    "question": "已知 $p, q, r$ 是質數，且 $p$ 整除 $qr - 1$，$q$ 整除 $pr - 1$，$r$ 整除 $pq - 1$。試求 $p, q, r$ 之值。",
    "options": [],
    "answer": "See solution",
    "solution": "由 $p \\mid qr - 1$，$q \\mid pr - 1$，$r \\mid pq - 1$，可得：\n\n$$\npqr \\mid p^2q^2r^2 - pqr^2 - pq^2r - p^2qr + pq + pr + qr - 1,\n$$\n\n也就是\n\n$$\npqr \\mid (pq + qr + pr - 1).\n$$\n\n換言之，$k = \\frac{1}{p} + \\frac{1}{q} + \\frac{1}{r} - \\frac{1}{pqr}$ 是一個正整數。但由於 $p, q, r \\ge 2$，顯然有 $k \\le \\frac{3}{2}$，故 $k=1$，即 $pq + qr + rp - 1 = pqr$。\n\n1. 若 $p = q = a$，則有 $a^2 + 2ra - 1 = ra^2$，從而 $1$ 可被 $a \\ge 2$ 整除，不合。由此知 $p, q, r$ 全相異，不失一般性假設 $2 \\le p < q < r$。\n\n2. 若 $q \\ge 5$，則有 $r \\ge 5$，故 $k \\le \\frac{1}{2} + \\frac{1}{5} + \\frac{1}{5} = \\frac{9}{10} < 1$，不合。因此 $q = 3, p = 2$。由此知\n\n$$\n\\frac{1}{2} + \\frac{1}{3} + \\frac{1}{r} - \\frac{1}{6r} = 1.\n$$\n\n故 $(p, q, r) = (2, 3, 5)$ 為唯一解。",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 14798,
    "subject": "Mathematics (Olympiad)",
    "question": "$a, b, c, d$ 是正實數且滿足 $a + b + c + d = 4$。試證明：\n\n$$\n\\frac{a^2}{b} + \\frac{b^2}{c} + \\frac{c^2}{d} + \\frac{d^2}{a} \\geq 4 + (a-d)^2\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "注意到右邊如果沒有 $(a-d)^2$ 這項，那麼\n\n$$\n\\sum_{cyc} \\frac{a^2}{b} \\geq 4\n$$\n\n即\n\n$$\n\\sum_{cyc} \\left(\\frac{a^2}{b} + b\\right) \\geq 8\n$$\n\n這由四個算幾不等式可知成立。回到原題，加上 $(a-d)^2$ 這項後，需要證明\n\n$$\n\\sum_{cyc} \\left(\\frac{a^2}{b} + b\\right) \\geq 8 + (a-d)^2\n$$\n\n即\n\n$$\n\\sum_{cyc} \\left(\\frac{a^2}{b} + b - 2a\\right) \\geq (a-d)^2\n$$\n\n化簡得\n\n$$\n\\sum_{cyc} \\frac{a^2 + b^2 - 2ab}{b} \\geq (a-d)^2\n$$\n\n由柯西不等式知\n\n$$\n\\left(\\sum_{cyc} \\frac{(a-b)^2}{b}\\right) \\left(\\sum_{cyc} b\\right) \\geq \\left(\\sum_{cyc} |a-b|\\right)^2 \\geq \\left((a-b) + (b-c) + (c-d) + (a-d)\\right)^2 = (2a-2d)^2\n$$\n\n所以\n\n$$\n\\sum_{cyc} \\frac{a^2 + b^2 - 2ab}{b} \\geq \\frac{(2a-2d)^2}{a+b+c+d} = \\frac{(2a-2d)^2}{4} = (a-d)^2\n$$\n\n證畢。\n\n註：等號成立條件除了顯然的 $(a, b, c, d) = (1, 1, 1, 1)$ 之外，注意只要滿足柯西不等式的等號成立條件即成立，即\n\n$$\n\\frac{(a-b)^2}{b^2} = \\frac{(b-c)^2}{c^2} = \\frac{(c-d)^2}{d^2} = \\frac{(d-a)^2}{a^2}\n$$\n\n在 $a \\geq b \\geq c \\geq d$ 的情況下可化簡成\n\n$$\n\\frac{a-b}{b} = \\frac{b-c}{c} = \\frac{c-d}{d} = \\frac{a-d}{a}\n$$\n\n所以 $a = b + c + d$ 或 $a - d = 0$。後者是顯然的等號成立條件，而前者可導出 $a = 2$，且\n\n$$\n\\frac{2-b}{b} = \\frac{b-c}{c} = \\frac{c-d}{d} = \\frac{2-d}{2}\n$$\n\n將 $2$ 用 $a+b+c$ 代換後可得\n\n$$\n\\frac{c+d}{b} = \\frac{b-c}{c} = \\frac{c-d}{d} = \\frac{b+c}{b+c+d}\n$$\n\n可以解出一組 $(b, c, d)$ 使上述等號成立。",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 14799,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $a$ be the largest value of the expression $24y - 9y^2$, where $y$ is a rational number, and $b$ be the smallest integer satisfying the inequality\n$$\n(t+3)^3 - (6t-7)^2 - (t-9)^3 < 3.\n$$\n\nFactor into irreducible factors with integer coefficients the expression\n$$\na(x-1)x^3 + bx - 2x - 1.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "We have $24y - 9y^2 = 16 - (3y - 4)^2$, whose largest value $a = 16$ is reached for $y = \\frac{4}{3}$.\n\nThe given inequality is equivalent to\n$$\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$$\ni.e. $t > \\frac{16}{3}$ and $b = 6$.\n\nSubstituting $a = 16$, $b = 6$ in the given expression, we get\n$$\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$$",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 14800,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $ABC$ be a triangle satisfying $\\angle B > \\angle C$ and let $D$ be the point on the side $AC$ such that $\\angle ADC = \\angle C$. Let $I$ be the incenter of $ABC$ and $E$ be the intersection of the circumcircle of $CDI$ and the line $AI$ which is not $I$. Let $P$ be the intersection of the line $BD$ and the line which is parallel to $AB$ and passes through $E$. Let $J$ be the incenter of $ABD$ and $A'$ be the reflection of $A$ with respect to $I$. Suppose that two lines $JP$ and $A'C$ meet at the point $Q$. Show that $QJ = QA'.$",
    "options": [],
    "answer": "See solution",
    "solution": "First, we will show that the line $PJ$ passes through the midpoint of side $AB$.\n\nLet $M$ be the intersection of the lines $PJ$ and $AB$, and $S$ be the intersection of the lines $BP$ and $AI$. By using Menelaus' theorem on triangle $ABS$ and the line $PJ$, we have\n\n$$\n\\frac{AM}{BM} \\cdot \\frac{BP}{PS} \\cdot \\frac{SJ}{JA} = 1. \\qquad (1)\n$$\n\nAlso, by the angle bisector theorem, we get\n\n$$\n\\frac{SJ}{JA} = \\frac{BS}{AB}. \\qquad (2)\n$$\n\nOn the other hand, since $\\angle JBD = \\frac{\\angle C}{2}$ and $\\angle DEI = \\angle DCI = \\frac{\\angle C}{2}$, the four points $J, B, E, D$ are cyclic. Therefore, $\\angle BEJ = \\angle BDJ = \\frac{\\angle B}{2}$. From the fact that $AB \\parallel PE$, we have\n\n$$\n\\angle BEP = \\angle BEJ + \\angle PEA = \\angle BEJ + \\angle EAB = 90^\\circ - \\frac{\\angle C}{2}.\n$$\n\nBy the fact $AB \\parallel PE$ again, we know $\\angle EPB = \\angle PBA = \\angle C$, hence we get\n\n$$\n\\angle PBE = 180^\\circ - \\angle C - \\left(90^\\circ - \\frac{\\angle C}{2}\\right) = 90^\\circ - \\frac{\\angle C}{2},\n$$\n\nso we obtain $\\angle PBE = 90^\\circ - \\frac{\\angle C}{2} = \\angle PEB$ and therefore\n\n$$\nPB = PE. \\qquad (3)\n$$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 14801,
    "subject": "Mathematics (Olympiad)",
    "question": "Anton chooses as starting number an integer $n \\ge 0$ which is not a square. Berta adds to this number its successor $n+1$. If this sum is a perfect square, she has won. Otherwise, Anton adds to this sum the subsequent number $n+2$. If this sum is a perfect square, he has won. Otherwise, it is again Berta's turn and she adds the subsequent number $n+3$, and so on.\n\nProve that Anton wins with infinitely many starting numbers.",
    "options": [],
    "answer": "See solution",
    "solution": "We will prove that Anton wins for infinitely many starting numbers of the form $3x^2 - 1$ with $x \\ge 1$.\n\nSince $3x^2 - 1 \\equiv 2 \\pmod{3}$, it cannot be a perfect square. After Berta adds the subsequent integer $3x^2$, the sum $6x^2 - 1$ is also $\\equiv 2 \\pmod{3}$ and consequently not a perfect square. Now Anton adds the subsequent number $3x^2 + 1$ and obtains the perfect square $9x^2$. Therefore, Anton has won and we have found infinitely many possible starting numbers.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 14802,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $f, g$ be functions from the positive integers to the integers. Vlad the impala is jumping around the integer grid. His initial position is $\\mathbf{x}_0 = (0,0)$, and for every $n \\ge 1$, his jump is\n\n$$\n\\mathbf{x}_n - \\mathbf{x}_{n-1} = (\\pm f(n), \\pm g(n)) \\text{ or } (\\pm g(n), \\pm f(n)),\n$$\n\nwith eight possibilities in total. Is it always possible that Vlad can choose his jumps to return to his initial location $(0,0)$ infinitely many times when\n\n(a) $f, g$ are polynomials with integer coefficients?\n\n(b) $f, g$ are any pair of functions from the positive integers to the integers?\n",
    "options": [],
    "answer": "See solution",
    "solution": "(a) Yes, it is always possible. The key idea is the following: Let $b(n)$ be the number of $1$'s in the binary expansion of $n = 0, 1, 2, \\dots$.\n\n**Lemma:** Given a polynomial $f$ with integer coefficients and degree at most $d$, then\n\n$$\n\\sum_{k=0}^{2^{d+1}-1} (-1)^{b(k)} f(n+k) = 0.\n$$\n\n**Proof of Lemma:** The result is clear for $d=0$. For $d \\ge 1$, we have\n\n$$\n\\sum_{k=0}^{2^{d+1}-1} (-1)^{b(k)} f(n+k) = \\sum_{k=0}^{2^d-1} (-1)^{b(k)} [f(n+k) - f(n+k+2^d)].\n$$\n\nSo set $\\tilde{f}(n) = f(n) - f(n + 2^d)$, which is a polynomial of degree at most $d-1$. Then\n\n$$\n\\sum_{k=0}^{2^{d+1}-1} (-1)^{b(k)} f(n+k) = \\sum_{k=0}^{2^d-1} \\tilde{f}(n+k) = 0,\n$$\n\nby induction, completing the proof of the lemma. $\\square$\n\nIn particular, if we take\n\n$$\n\\mathbf{x}_n - \\mathbf{x}_{n-1} = ((-1)^{b(n)} f(n), (-1)^{b(n)} g(n)),\n$$\n\nthen $\\mathbf{x}_D = \\mathbf{0}$ whenever $D$ is a multiple of $2^{1+\\max(\\deg(f),\\deg(g))}$.\n\n(b) No, it is not always possible. Let $g$ be any suitable function. Then, we construct $f$ inductively. There are at most $8^{n-1}$ possibilities for $\\mathbf{x}_{n-1}$, so choose $f(n)$ to be greater than the magnitude of all of them. Consequently $\\mathbf{x}_n$ cannot be $\\mathbf{0}$.",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 14803,
    "subject": "Mathematics (Olympiad)",
    "question": "A family wears three colours of clothing: red, blue, and green, with a separate laundry bin for each colour. Each week, the family generates a total of $K$ kilogrammes of laundry (the proportion of each colour may vary). The laundry is first sorted by colour and placed in the bins. Next, the heaviest bin is emptied and its contents washed. What is the required storage capacity of the laundry bins so that they never overflow?",
    "options": [],
    "answer": "See solution",
    "solution": "The required bin capacity is $\\frac{5}{2}K$.\n\nEach week, $K$ kg of laundry is added, then the heaviest bin is washed. By the pigeonhole principle, the heaviest bin contains at least $\\frac{1}{3}$ of the total. Let $a_n$ be the total laundry after the $n$th wash. Then:\n\n$$\na_{n+1} = \\frac{2}{3}(a_n + K), \\quad a_0 = 0\n$$\n\nThis sequence is bounded above by $2K$, so post-wash total is less than $2K$, and pre-wash total is less than $3K$.\n\nSuppose pre-wash state $(a, b, c)$ precedes post-wash state $(a, b, 0)$, which precedes pre-wash state $(a', b', c')$. Since $a \\leq c$ and $a' \\leq a + K$:\n\n$$\n3K > a + b + c \\geq 2a \\geq 2(a' - K)\n$$\n\nand similarly for $b'$, so $a', b' < \\frac{5}{2}K$. Also, $c' \\leq K$. Thus, no bin ever contains $\\geq \\frac{5}{2}K$ kg.\n\nFor example, if bins are filled equally, starting from $(0, 0, 0)$, after the first week: $(\\frac{1}{3}K, \\frac{1}{3}K, \\frac{1}{3}K)$ pre-wash, $(\\frac{1}{3}K, \\frac{1}{3}K, 0)$ post-wash; after the second week: $(\\frac{5}{9}K, \\frac{5}{9}K, \\frac{5}{9}K)$ pre-wash, $(\\frac{5}{9}K, \\frac{5}{9}K, 0)$ post-wash, etc. This approaches $(K, K, 0)$ post-wash. If $\\frac{1}{2}K$ is added to each non-empty bin, we get $(\\frac{3}{2}K, \\frac{3}{2}K, 0)$ pre-wash, $(\\frac{3}{2}K, 0, 0)$ post-wash. If the next week's laundry goes into the single non-empty bin, it can reach just under $\\frac{5}{2}K$ kg. $\\square$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 14804,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $n$ be an integer greater than $1$, let $\\mathbf{u}_1, \\dots, \\mathbf{u}_n$ be vectors in some normed vector space over the complex field, and let $\\alpha_1, \\dots, \\alpha_n$ be complex numbers such that $\\alpha = |\\alpha_1 + \\dots + \\alpha_n|$ and $\\delta = \\max_{i,j} |\\alpha_i - \\alpha_j|$ do not vanish simultaneously. Then\n\n$$\n\\max_{\\sigma} \\left| \\sum_{i=1}^{n} \\alpha_i \\mathbf{u}_{\\sigma(i)} \\right| \\geq \\frac{\\alpha \\delta}{2(1 - 1/n)\\alpha + \\delta} \\max_{i} |\\mathbf{u}_i|,\n$$\n\nwhere $\\sigma$ runs through all permutations of $1, 2, \\dots, n$.",
    "options": [],
    "answer": "See solution",
    "solution": "To prove the above inequality, let $M$ denote the maximum over all permutations. Discarding the various trivial cases, we may (and will) assume that\n\n$$\n|\\mathbf{u}_n| = \\max_i |\\mathbf{u}_i| \\neq 0, \\quad |\\mathbf{u}_n - \\mathbf{u}_1| = \\max_i |\\mathbf{u}_n - \\mathbf{u}_i|, \\quad \\alpha \\neq 0 \\quad \\text{and} \\quad |\\alpha_n - \\alpha_1| = \\delta \\neq 0.\n$$\n\nLet\n\n$$\n\\beta = \\frac{|\\mathbf{u}_n - \\mathbf{u}_1|}{|\\mathbf{u}_n|}, \\quad \\mathbf{v} = \\sum_{i=1}^{n} \\alpha_i \\mathbf{u}_i, \\quad \\text{and} \\quad \\mathbf{w} = \\alpha_n \\mathbf{u}_1 + \\sum_{i=2}^{n-1} \\alpha_i \\mathbf{u}_i + \\alpha_1 \\mathbf{u}_n,\n$$\n\nto get\n\n$$\n2M \\geq 2 \\max (|\\mathbf{v}|, |\\mathbf{w}|) \\geq |\\mathbf{v} - \\mathbf{w}| = |\\alpha_n - \\alpha_1| \\cdot |\\mathbf{u}_n - \\mathbf{u}_1| = \\beta \\delta |\\mathbf{u}_n|. \\quad (1)\n$$\n\nNext, with the usual notational convention $\\mathbf{u}_{n+i} = \\mathbf{u}_i$, let\n\n$$\n\\mathbf{v}_j = \\sum_{i=1}^{n} \\alpha_i \\mathbf{u}_{i+j}, \\quad j = 1, 2, \\dots, n,\n$$\n\nto obtain\n\n$$\n\\begin{aligned}\nnM &\\geq \\sum_{j=1}^{n} |\\mathbf{v}_j| \\geq \\left| \\sum_{j=1}^{n} \\mathbf{v}_j \\right| = \\alpha \\left| \\sum_{i=1}^{n} \\mathbf{u}_i \\right| = \\alpha \\left| n\\mathbf{u}_n + \\sum_{i=1}^{n-1} (\\mathbf{u}_i - \\mathbf{u}_n) \\right| \\\\ &\\geq \\alpha \\left( n|\\mathbf{u}_n| - \\left| \\sum_{i=1}^{n-1} (\\mathbf{u}_i - \\mathbf{u}_n) \\right| \\right) \\geq \\alpha \\left( n|\\mathbf{u}_n| - \\sum_{i=1}^{n-1} |\\mathbf{u}_i - \\mathbf{u}_n| \\right) \\\\ &\\geq \\alpha(n|\\mathbf{u}_n| - (n-1)|\\mathbf{u}_n - \\mathbf{u}_1|) = \\alpha(n - (n-1)\\beta)|\\mathbf{u}_n|. \\end{aligned} \\quad (2)\n$$\n\nHence, by (1) and (2),\n\n$$\nM \\geq \\max\\left(\\frac{\\beta\\delta}{2}, \\alpha\\left(1 - (1 - 1/n)\\beta\\right)\\right).\n$$\n\nTo complete the proof, notice that $\\frac{\\alpha\\delta}{2(1 - 1/n)\\alpha + \\delta}$ is the minimum of the function $t \\mapsto \\max\\left(\\frac{\\delta t}{2}, \\alpha\\left(1 - (1 - 1/n)t\\right)\\right)$, $t \\ge 0$.\n\n**Remark.** In the special case in the problem the result can be slightly improved. It can be shown by induction on $n$ or a counting argument that the centroid of some $m$-point subsystem of the $X_i$ lies at least $1/(2m - 1) > 1/(1 + 2m(1 - 1/n))$ units away from the centre of the disc.",
    "topic": "Discrete Mathematics",
    "subtopic": "Combinatorics"
  },
  {
    "id": 14805,
    "subject": "Mathematics (Olympiad)",
    "question": "Пусть $a$ и $b$ — положительные рациональные числа, десятичные записи которых, а также чисел $a-b$ и $a+kb$ (для некоторого целого $k \\ge 1$) — чисто периодические с длиной периода 30, а десятичные записи $a-b$ и $a+kb$ — с длиной периода 15. Найдите наименьшее возможное значение $k$.",
    "options": [],
    "answer": "See solution",
    "solution": "Домножив, если нужно, числа $a$ и $b$ на подходящую степень десятки, мы можем считать, что десятичные записи чисел $a$, $b$, $a-b$ и $a+kb$ — чисто периодические (то есть периоды начинаются сразу после запятой).\n\nВоспользуемся следующим известным фактом: десятичная запись рационального числа $r$ — чисто периодическая с (не обязательно минимальной) длиной периода $T$ тогда и только тогда, когда $r$ имеет вид $\\frac{m}{10^{T}-1}$ для некоторого целого $m$.\n\nПрименительно к условию задачи это значит, что $a = \\frac{m}{10^{30}-1}$ и $b = \\frac{n}{10^{30}-1}$. Нам также известно, что числа $a-b = \\frac{m-n}{10^{30}-1}$ и $a+kb = \\frac{m+kn}{10^{30}-1}$ записываются десятичными дробями с периодом длины 15, то есть могут быть записаны как обыкновенные дроби со знаменателем $10^{15}-1$. Поэтому так может быть записана и их разность $(k+1)b = \\frac{(k+1)n}{10^{30}-1}$. Таким образом, число $(k+1)n$ делится на $10^{15}+1$, а число $n$ — не делится (иначе и $b$ записывалось бы дробью с периодом длины 15). Значит, число $k+1$ делится на некоторый простой делитель числа $10^{15}+1$. Наименьший из таких делителей — это 7. Действительно, число $10^{15}+1$ даёт остаток 1 при делении на 2 и на 5, а также остаток 2 при делении на 3. С другой стороны, оно делится на $10^3+1 = 7 \\cdot 143$. Итак, $k+1 \\ge 7$ и $k \\ge 6$.\n\nВ некотором смысле минимальный пример чисел, удовлетворяющих условию при $k = 6$, получается, если положить $a - b = \\frac{1}{10^{15}-1}$ и $a + 6b = \\frac{2}{10^{15}-1}$. Тогда $a = \\frac{8}{7(10^{15}-1)}$ и $b = \\frac{1}{7(10^{15}-1)}$. Ясно, что длины минимальных периодов чисел $a-b$ и $a+6b$ равны 15. Далее, длины минимальных периодов чисел $a$ и $b$ больше 15 и делятся на 15 (так как $10^T - 1$ должно делиться на $10^{15} - 1$). С другой стороны, так как $10^{30} - 1 = 7(10^{15} - 1)$, числа $a$ и $b$ периодичны с длиной периода 30. Значит, длины их минимальных периодов равны 30.\n\n**Ответ:** $k = 6$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 14806,
    "subject": "Mathematics (Olympiad)",
    "question": "Nonnegative real numbers $a$, $b$, $c$ satisfy $a + b + c = 1$. Prove the inequality\n\n$$\n(1 - a)^2 + (1 - b)^2 + (1 - c)^2 \\geq 6\\sqrt{abc}.\n$$\n\nCan the case of equality occur?",
    "options": [],
    "answer": "See solution",
    "solution": "Expanding the squares:\n\n$$\n(1 - a)^2 + (1 - b)^2 + (1 - c)^2 = (1 - 2a + a^2) + (1 - 2b + b^2) + (1 - 2c + c^2)\n$$\n\nCombine terms:\n\n$$\n= 3 - 2(a + b + c) + (a^2 + b^2 + c^2)\n$$\n\nSince $a + b + c = 1$:\n\n$$\n= 3 - 2 \\cdot 1 + (a^2 + b^2 + c^2) = 1 + a^2 + b^2 + c^2\n$$\n\nWe need to show:\n\n$$\n1 + a^2 + b^2 + c^2 \\geq 6\\sqrt{abc}\n$$\n\nEquality would require $a = b = c$, so $a = b = c = \\frac{1}{3}$. Substitute:\n\n$$\n1 + 3\\left(\\frac{1}{3}\\right)^2 = 1 + 3 \\cdot \\frac{1}{9} = 1 + \\frac{1}{3} = \\frac{4}{3}\n$$\n\nand\n\n$$\n6\\sqrt{\\left(\\frac{1}{3}\\right)^3} = 6 \\cdot \\frac{1}{\\sqrt{27}} = 6 \\cdot \\frac{1}{3\\sqrt{3}} = 2 \\cdot \\frac{1}{\\sqrt{3}} \\approx 1.1547\n$$\n\nSince $\\frac{4}{3} > 1.1547$, the inequality holds, but equality does not occur for any $a, b, c \\geq 0$ with $a + b + c = 1$.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 14807,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $n$ be a natural number and $p$ a prime number such that $\\sqrt{n + \\frac{2}{n}}$ is a natural number. Find all possible values of $n$ and $p$.",
    "options": [],
    "answer": "See solution",
    "solution": "Denote $\\sqrt{n + \\frac{2}{n}} = k$ where $k$ is a natural number. Then $n + \\frac{2}{n} = k^2$, so $n^2 + 2 = k^2 n$. Rearranging, $n^2 - k^2 n + 2 = 0$. This is a quadratic in $n$.\n\nSince $n$ must be a divisor of $p$ and $p$ is prime, $n = 1$ or $n = p$. In both cases, we get $1 + p = k^2$ or $p = k^2 - 1 = (k+1)(k-1)$. Since $p$ is prime, $k+1 = p$ and $k-1 = 1$, so $k = 2$ and $p = 3$.\n\nThus, the possible solutions are $n = 1$ and $p = 3$, or $n = 3$ and $p = 3$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 14808,
    "subject": "Mathematics (Olympiad)",
    "question": "設四邊形 $ABCD$ 內接於一圓 $\\Omega$。過點 $D$ 與 $\\Omega$ 相切的直線分別交射線 $BA, BC$ 於點 $E, F$。於三角形 $ABC$ 內部選取一點 $T$ 使得 $TE \\parallel CD$ 且 $TF \\parallel AD$。設點 $K$ 異於 $D$ 且落在線段 $DF$ 上，滿足 $TD = TK$。\n\n試證：直線 $AC, DT, BK$ 三線共點。",
    "options": [],
    "answer": "See solution",
    "solution": "令 $TE$ 和 $TF$ 分別與 $AC$ 交於 $P$ 和 $Q$。由於 $PE \\parallel CD$ 且 $ED$ 為 $ABCD$ 的外接圓 $\\Omega$ 的切線，有\n\n$$\n\\angle EPA = \\angle DCA = \\angle EDA,\n$$\n\n因此 $A, P, D, E$ 共圓，記為圓 $\\alpha$。同理，$C, Q, D, F$ 共圓，記為圓 $\\gamma$。\n\n接下來要證明 $DT$ 分別為 $\\alpha$ 和 $\\gamma$ 在 $D$ 點的切線。事實上，因為 $\\angle FCD + \\angle EAD = 180^\\circ$，圓 $\\alpha$ 和 $\\gamma$ 在 $D$ 相切。要證明 $T$ 在它們的公切線（即徑軸）上，只需證明 $TP \\cdot TE = TQ \\cdot TF$，即四邊形 $PEFQ$ 共圓。這可由\n\n$$\n\\angle QFE = \\angle ADE = \\angle APE\n$$\n\n得出。\n\n由 $TD = TK$，可知 $\\angle TKD = \\angle TDK$。又因 $TD$ 和 $DE$ 分別為 $\\alpha$ 和 $\\Omega$ 的切線，得\n\n$$\n\\angle TKD = \\angle TDK = \\angle EAD = \\angle BDE,\n$$\n\n因此 $TK \\parallel BD$。\n\n接著證明 $T, P, Q, D, K$ 共圓，記為圓 $\\tau$。因為 $TD$ 為 $\\alpha$ 的切線，有\n\n$$\n\\angle EPD = \\angle TDF = \\angle TKD,\n$$\n\n所以 $P$ 在圓 $(TDK)$ 上。同理，$Q \\in (TDK)$。\n\n最後證明 $PK \\parallel BC$。利用圓 $\\tau$ 和 $\\gamma$，有\n\n$$\n\\angle PKD = \\angle PQD = \\angle DFC,\n$$\n\n因此 $PK \\parallel BC$。\n\n三角形 $TPK$ 和 $DCB$ 對應邊平行，故 $TD, PC, KB$ 三線共點，得證。\n\n![](images/2022-TWNIMO-Problems_p81_data_e801478bc7.png)\n\n**補充說明**：上述解法有多種變形。例如，找到圓 $\\alpha$ 和 $\\gamma$ 後，可注意到存在一個將三角形 $TPQ$ 映射到 $DCA$ 的位似 $h$，其中心為 $Y = AC \\cap TD$。因為\n\n$$\n\\angle DPE = \\angle DAE = \\angle DCB = \\angle DQT,\n$$\n\n四邊形 $TPDQ$ 共圓，記為圓 $\\tau$。有 $h(\\tau) = \\Omega$，所以 $D^* = h(D)$ 在 $\\Omega$ 上。\n\n最後，\n\n$$\n\\angle DCD^* = \\angle TPD = \\angle BAD,\n$$\n\n可知 $B$ 和 $D^*$ 關於通過 $D$ 的 $\\Omega$ 直徑對稱，故 $DB = DD^*$ 且 $BD^* \\parallel EF$，因此 $h(K) = B$，$BK$ 通過 $Y$。",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 14809,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $n \\ge 3$ be an integer. Suppose that $\\alpha, \\beta, \\gamma \\in (0, 1)$ and $a_k, b_k, c_k \\ge 0$ for $k = 1, 2, \\dots, n$ satisfy\n\n$$\n\\sum_{k=1}^n (k+\\alpha)a_k \\le \\alpha,\n$$\n$$\n\\sum_{k=1}^n (k+\\beta)b_k \\le \\beta,\n$$\n$$\n\\sum_{k=1}^n (k+\\gamma)c_k \\le \\gamma.\n$$\n\nFind the minimum value of $\\lambda$ such that\n\n$$\n\\sum_{k=1}^n (k+\\lambda)a_k b_k c_k \\le \\lambda.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "Let $a_1 = \\frac{\\alpha}{1+\\alpha}$, $b_1 = \\frac{\\beta}{1+\\beta}$, $c_1 = \\frac{\\gamma}{1+\\gamma}$, and $a_i, b_i, c_i = 0$ for $i = 2, 3, \\dots, n$. All conditions are satisfied. Thus,\n\n$$\n(1 + \\lambda) \\frac{\\alpha}{1 + \\alpha} \\cdot \\frac{\\beta}{1 + \\beta} \\cdot \\frac{\\gamma}{1 + \\gamma} \\le \\lambda,\n$$\nwhich gives\n$$\n\\lambda \\ge \\frac{\\alpha\\beta\\gamma}{(1+\\alpha)(1+\\beta)(1+\\gamma) - \\alpha\\beta\\gamma}.\n$$\n\nLet $\\lambda_0 = \\frac{\\alpha\\beta\\gamma}{(1+\\alpha)(1+\\beta)(1+\\gamma) - \\alpha\\beta\\gamma}$. We show that for any $a_k, b_k, c_k$ ($k = 1, 2, \\dots, n$),\n$$\n\\sum_{k=1}^{n} (k + \\lambda_0) a_k b_k c_k \\le \\lambda_0.\n$$\n\nBy the problem's conditions,\n$$\n\\sum_{k=1}^{n} \\left( \\frac{k+\\alpha}{\\alpha} a_k \\cdot \\frac{k+\\beta}{\\beta} b_k \\cdot \\frac{k+\\gamma}{\\gamma} c_k \\right)^{\\frac{1}{3}} \\le 1,\n$$\nusing H\"older's Inequality:\n$$\n\\left( \\sum_{i=1}^{n} x_i y_i z_i \\right)^3 \\le \\left( \\sum_{i=1}^{n} x_i^3 \\right) \\left( \\sum_{i=1}^{n} y_i^3 \\right) \\left( \\sum_{i=1}^{n} z_i^3 \\right).\n$$\n\nTo prove the desired inequality, it suffices to show for each $k$:\n$$\n\\frac{k + \\lambda_0}{\\lambda_0} (a_k b_k c_k)^{\\frac{2}{3}} \\le \\left( \\frac{(k + \\alpha)(k + \\beta)(k + \\gamma)}{\\alpha \\beta \\gamma} \\right)^{\\frac{1}{3}}.\n$$\n\nIn fact,\n$$\n\\lambda_0 = \\frac{\\alpha\\beta\\gamma}{1 + (\\alpha + \\beta + \\gamma) + (\\alpha\\beta + \\beta\\gamma + \\gamma\\alpha)} \\ge \\frac{\\alpha\\beta\\gamma}{k^2 + (\\alpha + \\beta + \\gamma)k + (\\alpha\\beta + \\beta\\gamma + \\gamma\\alpha)} = \\frac{k\\alpha\\beta\\gamma}{(k + \\alpha)(k + \\beta)(k + \\gamma) - \\alpha\\beta\\gamma}.\n$$\nThus,\n$$\n\\frac{k + \\lambda_0}{\\lambda_0} \\le \\frac{(k + \\alpha)(k + \\beta)(k + \\gamma)}{\\alpha \\beta \\gamma}.\n$$\n\nSince $(k+\\alpha)a_k \\le \\alpha$, $(k+\\beta)b_k \\le \\beta$, $(k+\\gamma)c_k \\le \\gamma$, we have\n$$\n(a_k b_k c_k)^{\\frac{2}{3}} \\le \\left( \\frac{\\alpha \\beta \\gamma}{(k + \\alpha)(k + \\beta)(k + \\gamma)} \\right)^{\\frac{2}{3}}.\n$$\n\nCombining these, the inequality holds. Therefore,\n$$\n\\lambda_{\\min} = \\lambda_0 = \\frac{\\alpha\\beta\\gamma}{(1+\\alpha)(1+\\beta)(1+\\gamma) - \\alpha\\beta\\gamma}.\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 14810,
    "subject": "Mathematics (Olympiad)",
    "question": "Given natural numbers $a, b, c, d$ for which\n$$\nab^2 + ad^2 + cb^2 = ba^2 + bd^2 + ca^2\n$$\nand the number $a^2 + b^2 + c^2 + d^2$ is prime. Prove that $a = b$.",
    "options": [],
    "answer": "See solution",
    "solution": "Assume the contrary, that $a \\ne b$. The condition $ab^2 + ad^2 + cb^2 = ba^2 + bd^2 + ca^2$ can be rewritten as:\n$$(a-b)(d^2 - ab - ac - bc) = 0.$$\nSince $a \\ne b$, it follows that $d^2 = ab + ac + bc$.\n\nThen:\n$$a^2 + b^2 + c^2 + d^2 = a^2 + b^2 + c^2 + ab + ac + bc = (a+b+c)^2 - d^2 = (a+b+c+d)(a+b+c-d).$$\n\nSince $(a+b+c+d)(a+b+c-d)$ is prime and $a, b, c, d$ are natural numbers, we must have $a+b+c-d=1$. Thus, $ab + ac + bc = d^2 = (a+b+c-1)^2$.\n\nExpanding $(a+b+c-1)^2 = ab + ac + bc$ gives:\n$$a^2 + b^2 + c^2 + 1 + 2ab + 2ac + 2bc - 2a - 2b - 2c = ab + ac + bc$$\nwhich simplifies to:\n$$a(a+b-2) + b(b+c-2) + c(c+a-2) + 1 = 0.$$\n\nBut since $a, b, c$ are natural numbers, the left side is at least $1$. This is a contradiction, so our assumption was wrong and consequently $a = b$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 14811,
    "subject": "Mathematics (Olympiad)",
    "question": "The nonzero real numbers $a$ and $b$ satisfy $$3(a^6 + b^6) = a^2 b^2 (a^2 b^2 + 9).$$ Prove that at least one of the numbers $a$ and $b$ is irrational.",
    "options": [],
    "answer": "See solution",
    "solution": "Rewrite the equation as:\n$$3(a^6 + b^6) = a^2 b^2 (a^2 b^2 + 9)$$\nExpanding and rearranging:\n$$a^4 b^4 - 3a^6 - 3b^6 + 9a^2 b^2 = 0$$\nGroup terms:\n$$a^4(b^4 - 3a^2) - 3b^2(b^4 - 3a^2) = 0$$\nFactor:\n$$(a^4 - 3b^2)(b^4 - 3a^2) = 0$$\nSo either $a^4 = 3b^2$ or $b^4 = 3a^2$. Since $a \\neq 0$ and $b \\neq 0$, this gives $\\sqrt{3} = \\pm \\frac{a^2}{b}$ or $\\sqrt{3} = \\pm \\frac{b^2}{a}$.\n\nIf both $a$ and $b$ are rational, then $\\sqrt{3}$ would be rational, which is false. Therefore, at least one of $a$ or $b$ must be irrational.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 14812,
    "subject": "Mathematics (Olympiad)",
    "question": "(a) Determine the maximum $M$ of $x + y + z$ where $x$, $y$, and $z$ are positive real numbers with\n$$\n16xyz = (x + y)^2 (x + z)^2.\n$$\n\n(b) Prove the existence of infinitely many triples $(x, y, z)$ of positive rational numbers that satisfy $16xyz = (x + y)^2 (x + z)^2$ and $x + y + z = M$.",
    "options": [],
    "answer": "See solution",
    "solution": "(a) The given equation and the AM-GM inequality imply\n$$\n4\\sqrt{xyz} = (x + y)(x + z) = x(x + y + z) + yz \\ge 2\\sqrt{xyz(x + y + z)}.\n$$\nTherefore, $2 \\ge \\sqrt{x + y + z}$, which gives $4 \\ge x + y + z$. Since we will explicitly give infinitely many triples with $x + y + z = 4$ in the second part, $M = 4$ is the maximum.\n\n(b) For $x + y + z = 4$, equality must hold in the AM-GM inequality of the first part, so we have $x(x + y + z) = yz$ and also $x + y + z = 4$. If we choose $y = t$ with rational $t$, we get $4x = t(4 - x - t)$ and therefore $x = \\frac{4t - t^2}{4 + t}$ and $z = 4 - x - y = \\frac{16 - 4t}{4 + t}$. If we take $0 < t < 4$, then all these expressions are positive and rational and are a solution of the given equation.\n\nThe triples $\\left(\\frac{4t - t^2}{4 + t},\\ t,\\ \\frac{16 - 4t}{4 + t}\\right)$ with rational $0 < t < 4$ are infinitely many cases of equality.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 14813,
    "subject": "Mathematics (Olympiad)",
    "question": "An $8 \\times 8$ array is divided into 64 squares. Find the smallest possible number of colours needed to colour the squares so that any four squares which can be covered by the L-shaped tile pictured below\n\n![](images/Slovenija_2009_p15_data_7ebbb0231d.png)\n\n(the tile can be rotated or flipped)\n\nwill have different colours.",
    "options": [],
    "answer": "See solution",
    "solution": "The first and the second figure show interesting $3 \\times 3$ and $5 \\times 5$ arrays. Here, $L$ can be any odd integer and $S$ can be any even integer. It does not matter which odd and even integers we pick (for example, we can choose all odd integers to be equal to $1$ and all even integers to be equal to $2$). The sum of all the numbers in the array is odd in both cases. Any L-shaped subset contains either three odd integers and one even integer or three even integers and one odd integer. Hence, the sum of the numbers inside such a subset is odd as well.\n\nCover the $4 \\times 4$ array using the L-shaped tiles as pictured in the third figure and assume that the sums of the numbers covered by each tile are odd. Then the sum of all numbers in the array is even. Hence, an interesting $4 \\times 4$ array does not exist.\n\n![](images/Slovenija_2009_p16_data_46ea67da18.png)",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 14814,
    "subject": "Mathematics (Olympiad)",
    "question": "Determine all polynomials $P \\in \\mathbb{R}[x, y]$ such that\n$$\nP(a, b^2 - ac) + P(b, c^2 - ab) + P(c, a^2 - bc) = 0\n$$\nfor all $a, b, c \\in \\mathbb{R}$.",
    "options": [],
    "answer": "See solution",
    "solution": "We use the following facts about polynomials:\n\n**Fact 1.** For $P \\in \\mathbb{R}[x, y]$, if a constant $c$ satisfies $P(c, y) = 0$ for infinitely many $y$, then $x - c$ divides $P(x, y)$.\n\n**Fact 2.** For $P \\in \\mathbb{R}[x, y]$, if there exists an infinite set $S \\subset \\mathbb{R}$ such that, for each $c \\in S$, there are infinitely many $y \\in \\mathbb{R}$ where $P(c, y) = 0$, then $P$ is the zero polynomial.\n\n**Fact 3.** For $P \\in \\mathbb{R}[x, y, z]$, if $P(x, y, z) = 0$ for all $x, y, z \\in \\mathbb{R}$, then $P$ is the zero polynomial.\n\nSubstituting $a = b = c$ in (1) gives $P(a, 0) = 0$ for all $a$, so $y$ divides $P(x, y)$. Substituting $a = b = 0$ gives $P(0, c^2) = 0$ for all $c$, so $x$ divides $P(x, y)$. Thus, $P(x, y) = xyQ(x, y)$ for some polynomial $Q$.\n\nSubstituting $a = 0$ in (1):\n$$\nP(b, c^2) = -P(c, -bc)\n$$\nSo $bc^2 Q(b, c^2) = bc^2 Q(c, -bc)$, hence $Q(b, c^2) = Q(c, -bc)$ for all $b, c$.\n\nSubstituting $c = 0$ gives $Q(b, 0) = Q(0, 0)$ for all $b$, so $y$ divides $Q(x, y) - Q(0, 0)$. Let $k_0 = Q(0, 0)$, so $Q(x, y) = yR(x, y) + k_0$.\n\nNow, $Q(b, c^2) = Q(c, -bc)$ becomes:\n$$\nc^2 R(b, c^2) = -bc R(c, -bc)\n$$\nSo $c R(b, c^2) = -b R(c, -bc)$ for all $b, c$.\n\nSubstituting $b = 0$ gives $c R(0, c^2) = 0$ for all $c$, so $R(x, y) = xP_1(x, y)$. Now,\n$$\nbc P_1(b, c^2) = -bc P_1(c, -bc)\n$$\nSo $P_1(b, c^2) = -P_1(c, -bc)$ for all $b, c$.\n\nSetting $b = c = 0$ gives $P_1(0, 0) = 0$. Setting $c = 0$ gives $P_1(b, 0) = 0$. Setting $b = 0$ gives $P_1(0, c^2) = 0$. Thus, $xy$ divides $P_1(x, y)$. Proceeding recursively, we find:\n$$\nP(x, y) = \\sum_{i=0}^{\\infty} k_i (xy)^{2i+1}\n$$\nSubstituting into (1), we find that for $i \\ge 1$, the coefficients must vanish, so $k_i = 0$ for $i \\ge 1$. Thus, $P(x, y) = kxy$ for some $k \\in \\mathbb{R}$, which satisfies the original equation.",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 14815,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $n$ be a positive integer. Let $s : \\mathbb{N} \\to \\{1, \\dots, n\\}$ be a function such that $n$ divides $m - s(m)$ for all positive integers $m$. Let $a_0, a_1, a_2, \\dots$ be a sequence such that $a_0 = 0$ and\n\n$$\na_k = a_{k-1} + s(k) \\text{ for all } k \\ge 1.\n$$\n\nFind all $n$ for which this sequence contains all the residues modulo $(n+1)^2$.",
    "options": [],
    "answer": "See solution",
    "solution": "The answer is $n = 2^k - 1$ for any $k \\in \\mathbb{N}$.\n\nWe begin by noting that the sequence is given by the formula\n\n$$\na_{rn+s} = r \\cdot \\binom{n+1}{2} + \\binom{s+1}{2}$$\n\nfor all $r \\ge 0$ and $s = 1, 2, \\dots, n$. This can be confirmed by induction: $a_0 = 0$ is true, and for fixed $r$, inducting on $s$ gives\n\n$$\na_{rn+s} = a_{rn+s-1} + s = r \\cdot \\binom{n+1}{2} + \\binom{(s-1)+1}{2} + s = r \\cdot \\binom{n+1}{2} + \\binom{s+1}{2},$$\n\nproving the induction step. Also, $a_{(r+1)n} = (r+1)\\binom{n+1}{2} = r \\cdot \\binom{n+1}{2} + \\binom{n+1}{2} = a_{rn+n}$, confirming the induction on $r$ as well.\n\nSuppose $n+1$ has an odd prime factor $p$. Then $p \\mid \\binom{n+1}{2}$, so if all residues mod $(n+1)^2$ are present, then $\\binom{s+1}{2} \\equiv \\frac{s^2+s}{2} \\pmod{p}$ must cover all residues mod $p$. However, the map $x \\mapsto x^2+x \\pmod{p}$ sends $x$ and $-(1+x)$ to the same element mod $p$, and these are distinct unless $x \\equiv \\frac{-1}{2} \\pmod{p}$, so the set of residues represented by $s^2+s \\pmod{p}$ has $\\frac{p+1}{2} < p$ elements, a contradiction. Thus, every $n$ that satisfies the condition must be one less than a power of $2$.\n\nNow, let $n = 2^k - 1$. Then $\\binom{n+1}{2} = 2^{k-1}(2^k - 1) = 2^{2k-1} - 2^{k-1} \\pmod{2^{2k}}$. For $0 \\le r$ and $1 \\le s \\le n$, and $r = 2j + e$ with $e \\in \\{0, 1\\}$,\n\n$$\na_{rn+s} = -j(n+1) + e \\cdot \\binom{n+1}{2} + \\binom{s+1}{2} \\pmod{(n+1)^2}.$$\n\nFor $s = 1, 2, \\dots, n$, the numbers $\\binom{s+1}{2}$ are all distinct mod $n+1$. If $1 \\le s < t \\le n$ and $\\binom{s+1}{2} \\equiv \\binom{t+1}{2} \\pmod{n+1}$, then $(2n+2) \\mid (s-t)(s+t+1)$, but $2n+2 > 2n+1 \\ge s+t+1 > |s-t| > 0$, so this is impossible as $2n+2$ is a power of $2$. Thus, the residues are distinct mod $n+1$ and none are $0$ mod $n+1$. Therefore, $-j(n+1) + \\binom{s+1}{2}$ (for $e=0$) covers all $n^2+n$ residues not of the form $l(n+1) \\pmod{(n+1)^2}$.\n\nFor the remaining residues, take $s = n$ and $e = 1$, so $-j(n+1)+2\\binom{n+1}{2} = -(j+1)(n+1) \\pmod{(n+1)^2}$, which covers the remaining residues as $j$ varies. $\\square$",
    "topic": "Number Theory",
    "subtopic": "Modular Arithmetic"
  },
  {
    "id": 14816,
    "subject": "Mathematics (Olympiad)",
    "question": "The code setting of a cipher lock is established on an $n$-regular polygon with vertices $A_1, A_2, \\dots, A_n$. Each vertex is assigned a number (0 or 1) and a color (red or blue), such that for every pair of adjacent vertices, either the numbers or the colors are the same. How many code-sets can be realized for this lock?\n\n![](images/Mathematical_Olympiad_in_China_2011-2014_p73_data_78b29886eb.png)",
    "options": [],
    "answer": "See solution",
    "solution": "Let us analyze the possible code-sets:\n\nFor each code-set, if two adjacent vertices have different numbers, label the connecting side $a$; if they have different colors, label it $b$; if both numbers and colors are the same, label it $c$. Once the number and color on vertex $A_1$ are set (4 choices), we can set $A_2, \\dots, A_n$ sequentially according to the side labels. To return to the initial state at $A_1$, the number of sides labeled $a$ and $b$ must both be even.\n\nThus, the total number of code-sets is four times the number of labeled-side sequences where the counts of $a$ and $b$ are both even:\n\n$$\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$$\n\nwhere $C_0^0 = 1$.\n\nFor odd $n$, $n - 2i > 0$, so\n$$\n\\sum_{j=0}^{\\left\\lfloor \\frac{n-2i}{2} \\right\\rfloor} C_{n-2i}^{2j} = 2^{n-2i-1}.\n$$\nSubstituting, we get\n$$\n\\begin{aligned}\n4 \\sum_{i=0}^{\\left\\lfloor \\frac{n}{2} \\right\\rfloor} (C_n^{2i} 2^{n-2i-1}) &= 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$$\n\nFor even $n$, if $i < \\frac{n}{2}$, the above holds; if $i = \\frac{n}{2}$, all sides are labeled $a$, which gives one way. Thus,\n$$\n\\begin{aligned}\n4 \\sum_{i=0}^{\\left\\lfloor \\frac{n}{2} \\right\\rfloor} (C_n^{2i} \\sum_{j=0}^{\\left\\lfloor \\frac{n-2i}{2} \\right\\rfloor} C_{n-2i}^{2j}) &= 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$$\n\n**Summary:**\n\nThe number of code-sets for the lock is\n$$\n\\begin{cases}\n3^n + 1 & \\text{if } n \\text{ is odd,} \\\\\n3^n + 3 & \\text{if } n \\text{ is even.}\n\\end{cases}\n$$",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 14817,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $ABC$ be an acute triangle with incenter $I$. The line through $I$ parallel to $AC$ intersects $AB$ at $M$, and the line through $I$ parallel to $AB$ intersects $AC$ at $N$. Let the line $MN$ intersect the circumcircle of $ABC$ at $X$ and $Y$. Let $Z$ be the midpoint of arc $BC$ (not containing $A$). Prove that $I$ is the orthocenter of triangle $XYZ$.",
    "options": [],
    "answer": "See solution",
    "solution": "It is not hard to see that $AMIN$ is a parallelogram. Furthermore, as $\\angle MAI = \\angle NAI$, $AMIN$ must indeed be a rhombus. It follows that $AI$ is perpendicular to $MN$. Since $AI$ bisects $\\angle BAC$, $A$, $I$, $Z$ are collinear. So we have $ZI$ perpendicular to $XY$.\n\n![](images/THA_National_2016_p6_data_af7728ce25.png)\n\nLet $D$ be the intersection of $AI$ and $MN$. Then, $AD = DI$. Since $MN$ is the perpendicular bisector of $AI$, triangle $AYI$ is isosceles.\n\nLet $YI$ intersect $XZ$ at $E$. We have\n\n$$\n\\angle DYE = \\angle DYI = \\angle AYD = \\angle AYX = \\angle AZX = \\angle DZE,\n$$\n\nand so $D$, $Y$, $Z$, $E$ are concyclic. Thus, $\\angle ZEY = \\angle ZDY = 90^\\circ$, and it follows that $I$ is the orthocenter of triangle $XYZ$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 14818,
    "subject": "Mathematics (Olympiad)",
    "question": "Determine all pairs $(x, y)$ of integers that satisfy\n\n$$\n(x + y + 11)^2 = x^2 + y^2 + 11^2.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "Expanding the left-hand side and simplifying, we get:\n\n$$\n(x + y + 11)^2 = x^2 + y^2 + 11^2 \\\\\n(x + y + 11)^2 - x^2 - y^2 - 11^2 = 0\n$$\n\nExpanding and simplifying:\n\n$$\nx^2 + 2xy + 2x \\cdot 11 + y^2 + 2y \\cdot 11 + 11^2 - x^2 - y^2 - 11^2 = 0 \\\\\n2xy + 22x + 22y = 0 \\\\\nxy + 11x + 11y = 0\n$$\n\nAdding $11^2$ to both sides:\n\n$$\nxy + 11x + 11y + 11^2 = 11^2 \\\\\n(x + 11)(y + 11) = 11^2\n$$\n\nThus, $x + 11$ and $y + 11$ are integer pairs whose product is $11^2$. The possible pairs are:\n\n| $x + 11$ | $y + 11$ |\n|:--------:|:--------:|\n| $1$      | $121$    |\n| $11$     | $11$     |\n| $121$    | $1$      |\n| $-1$     | $-121$   |\n| $-11$    | $-11$    |\n| $-121$   | $-1$     |\n\nHence, the required pairs $(x, y)$ are:\n\n$$\n(-10, 110), \\quad (0, 0), \\quad (110, -10), \\quad (-12, -132), \\quad (-22, -22), \\quad (-132, -12).\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 14819,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $ABC$ be an acute-angled triangle with circumcenter $O$. Let $\\Gamma$ be a circle with center on the altitude from $A$ in $ABC$, passing through vertex $A$ and points $P$ and $Q$ on sides $AB$ and $AC$. Assume that $BP \\cdot CQ = AP \\cdot AQ$. Prove that $\\Gamma$ is tangent to the circumcircle of triangle $BOC$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $\\omega$ be the circumcircle of $BOC$. Let $M$ be the point diametrically opposite to $O$ on $\\omega$ and let the line $AM$ intersect $\\omega$ at $M$ and $K$. Since $O$ is the circumcenter of $ABC$, it follows that $OB = OC$ and therefore that $O$ is the midpoint of the arc $\\widehat{BOC}$ of $\\omega$. Since $M$ is diametrically opposite to $O$, it follows that $M$ is the midpoint of the arc $\\widehat{BMC}$ of $\\omega$. This implies, since $K$ is on $\\omega$, that $KM$ is the bisector of $\\angle BKC$. Since $K$ is on $\\omega$, this implies that $\\angle BKM = \\angle CKM$, i.e., $KM$ is the bisector of $\\angle BKC$.\n\nSince $O$ is the circumcenter of $ABC$, it follows that $\\angle BOC = 2\\angle BAC$. Since $B$, $K$, $O$, and $C$ all lie on $\\omega$, it also follows that $\\angle BKC = \\angle BOC = 2\\angle BAC$. Since $KM$ bisects $\\angle BKC$, it follows that $\\angle BKM = \\angle CKM = \\angle BAC$. The fact that $A$, $K$, and $M$ lie on a line therefore implies that $\\angle AKB = \\angle AKC = 180^\\circ - \\angle BAC$. Now it follows that\n\n$$\n\\angle KBA = 180^\\circ - \\angle AKB - \\angle KAB = \\angle BAC - \\angle KAB = \\angle KAC.\n$$\n\nThis implies that triangles $KBA$ and $KAC$ are similar. Rearranging the condition in the problem statement yields that $\\dfrac{BP}{AP} = \\dfrac{AQ}{CQ}$ which, when combined with the fact that $KBA$ and $KAC$ are similar, implies that triangles $KPA$ and $KQC$ are similar. Therefore $\\angle KPA = \\angle KQC = 180^\\circ - \\angle KQA$ which implies that $K$ lies on $\\Gamma$.\n\nNow let $S$ denote the center of $\\Gamma$ and let $T$ denote the center of $\\omega$. Note that $T$ is the midpoint of segment $OM$ and that $TM$ and $AS$, which are both perpendicular to $BC$, are parallel. This implies that $\\angle KMT = \\angle KAS$ since $A$, $K$, and $M$ are collinear. Further, since $KTM$ and $KSA$ are isosceles triangles, it follows that $\\angle TKM = \\angle KMT$ and $\\angle SKA = \\angle KSA$. Therefore $\\angle TKM = \\angle SKA$ which implies that $S$, $T$, and $K$ are collinear. Therefore $\\Gamma$ and $\\omega$ intersect at a point $K$ which lies on the line $ST$ connecting the centers of the two circles. This implies that the circles $\\Gamma$ and $\\omega$ are tangent at $K$.",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 14820,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $\\delta(n)$ denote the number of positive divisors of $n$ and let $\\phi(n)$ denote the number of non-negative integers less than $n$ and relatively prime to $n$.\n\nFind all positive integers $n$ such that $\\delta(n) \\cdot \\phi(n) = n$.",
    "options": [],
    "answer": "See solution",
    "solution": "The solutions are $n = 1$ and $n = 2$.\n\nLet $n = p_1^{\\alpha_1} \\cdots p_k^{\\alpha_k}$ be the canonical representation of $n$, where $\\alpha_1, \\dots, \\alpha_k$ are positive integers. It is known that\n\n$$\n\\delta(n) = (\\alpha_1 + 1) \\cdots (\\alpha_k + 1)\n$$\n\nand\n\n$$\n\\phi(n) = (p_1 - 1) \\cdots (p_k - 1) \\cdot p_1^{\\alpha_1 - 1} \\cdots p_k^{\\alpha_k - 1}.\n$$\n\nHence, the given equation reduces to\n\n$$\n(\\alpha_1 + 1) \\cdots (\\alpha_k + 1) = \\frac{p_1 \\cdots p_k}{(p_1 - 1) \\cdots (p_k - 1)}. \\quad (*)\n$$\n\nNote that $\\alpha_i + 1 \\ge 2 \\ge 1 + \\frac{1}{p_i-1} = \\frac{p_i}{p_i-1}$ for every $i = 1, \\dots, k$, with equality if and only if $\\alpha_i = 1$ and $p_i = 2$. Multiplying the inequalities $\\alpha_i + 1 \\ge \\frac{p_i}{p_i-1}$ for $i = 1, \\dots, k$ leads to the inequality obtained from $(*)$ by replacing equality with inequality. Hence, equality $(*)$ holds if and only if every inequality that was multiplied holds as an equality. That is, if $\\alpha_i = 1$ and $p_i = 2$ for every $i = 1, \\dots, k$. As primes in the canonical representation are distinct, this holds if and only if $k = 0$ or $k = 1$ and $p_1 = 2$, i.e., if $n = 1$ or $n = 2$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 14821,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $n$ be a positive integer and $a_1, \\, \\dots, \\, a_{2n}$ be real numbers in $\\left[-\\frac{1}{2}, \\frac{1}{2}\\right]$. Leaving out any one of the numbers, the sum of the remaining $2n-1$ numbers is always an integer. Prove that $a_1 = \\dots = a_{2n}$.",
    "options": [],
    "answer": "See solution",
    "solution": "Assume that there exist $a_i$ and $a_j$ which are not equal. Let $S = a_1 + \\dots + a_{2n}$. Since $S - a_i$ and $S - a_j$ are integers, their difference $$(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$$ is an integer, $a_k$ must also be either $\\frac{1}{2}$ or $-\\frac{1}{2}$. Hence all numbers $a_i$ are either $\\frac{1}{2}$ or $-\\frac{1}{2}$. It follows that the sum of any two numbers $a_i$ and $a_j$ is an integer. As we have an even number of them, the sum of all the numbers $S$ is also an integer. But then $S - a_i = S \\pm \\frac{1}{2}$ cannot be an integer, a contradiction. Therefore all numbers $a_i$ must be equal.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 14822,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $n > 1$ be an integer. Consider a $2 \\times n$ grid, where the cells are labelled as $(x, y)$ with $1 \\le x \\le 2$ and $1 \\le y \\le n$. A grasshopper lives on the grid and occasionally jumps from one cell to another. The length of a jump from a cell $(x_1, y_1)$ to $(x_2, y_2)$ is defined as $|x_1 - x_2| + |y_1 - y_2|$. Determine the largest possible sum of lengths of jumps the grasshopper can make when it starts from one cell and never jumps to a cell it has visited before.",
    "options": [],
    "answer": "See solution",
    "solution": "The largest possible sum of jump lengths is $n^2 + 2n - 3$.\n\n**Upper Bound:**\n\nConsider the contributions from $x$-wise and $y$-wise movement separately.\n\n- For $x$-wise movement, the grasshopper makes at most $2n - 1$ jumps, so the sum of $|x_1 - x_2|$ over all jumps is at most $2n - 1$.\n- For $y$-wise movement, for $1 \\le i \\le n - 1$, the number of jumps crossing from $y \\le i$ to $y \\ge i + 1$ is at most $2\\min(2i, 2(n-i))$ per cell in the smaller region. For even $n$, this bound is reduced by 1. For $n = 2k$:\n\n$$\n4 \\sum_{i=1}^{n-1} \\min\\{i, n-i\\} - 1 = 4 \\left( 2 \\sum_{i=1}^{k-1} i + k \\right) - 1 = n^2 - 1.\n$$\n\nFor $n = 2k + 1$:\n\n$$\n4 \\sum_{i=1}^{n-1} \\min\\{i, n-i\\} = 8 \\sum_{i=1}^{k} i = n^2 - 1.\n$$\n\nCombining the $x$ and $y$ bounds, the upper bound is $n^2 + 2n - 2$. However, equality cannot be achieved for both bounds simultaneously, so the true maximum is $n^2 + 2n - 3$.\n\n**Lower Bound (Construction):**\n\nFor $n = 2k + 1$, an optimal path is:\n\n$$\n(0, k+1) \\to (1, n) \\to (0, 1) \\to (1, n-1) \\to (0, 2) \\to (1, n-2) \\to \\dots \\\\ \\to (0, k) \\to (1, n-k) = (1, k+1) \\to (0, n) \\to (1, 1) \\to (0, n-1) \\to (1, 2) \\\\\n\\to \\dots \\to (0, k+2) \\to (1, k).\n$$\n\nFor $n = 2k$, an optimal path is:\n\n$$\n(0, k) \\to (1, n) \\to (0, 1) \\to (1, n-1) \\to (0, 2) \\to (1, n-2) \\to \\dots \\\\ \\to (0, k-1) \\to (1, n-k+1) = (1, k+1) \\to (0, k+1) \\to (1, 1) \\to (0, n) \\\\\n\\to (1, 2) \\to \\dots \\to (0, k+2) \\to (1, k).\n$$\n\nIn both cases, the $x$-direction contributes $2n - 1$ (since $x$ changes every step), and the $y$-direction sums to $n^2 - 2$.\n\n*Remark:* For a general $m \\times n$ grid, the maximal path length is:\n\n$$\nm \\cdot \\left\\lfloor \\frac{n^2}{2} \\right\\rfloor + n \\cdot \\left\\lfloor \\frac{m^2}{2} \\right\\rfloor - \\delta\n$$\n\n![](table with cell order for n=7)\n\n![](table with cell order for n=8)",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 14823,
    "subject": "Mathematics (Olympiad)",
    "question": "Обозначим через $a$, $b$, $c$ и $d$ длины сторон четырёхугольника. Пусть $N = 10^{100}$. Известно, что сумма любых трёх сторон делится на четвёртую, то есть для каждой стороны $x$ из $\\{a, b, c, d\\}$ число $a + b + c + d - x$ делится на $x$. Докажите, что четырёхугольник является ромбом.",
    "options": [],
    "answer": "See solution",
    "solution": "Пусть $d$ — наибольшая сторона. По условию, $a + b + c$ делится на $d$, то есть $a + b + c = k d$ для некоторого натурального $k$. Ясно, что $a + b + c > d$ (длина отрезка меньше длины ломаной с теми же концами), поэтому $k > 1$. Кроме того, так как $a \\le d$, $b \\le d$ и $c \\le d$, имеем $a + b + c \\le 3d$, то есть $k \\le 3$.\n\nСлучай $k = 3$ возможен только при $a = b = c = d$. В этом случае четырёхугольник — ромб.\n\nИначе $1 < k < 3$, откуда $k = 2$. Но тогда $N = a + b + c + d = 2d + d = 3d$. Получаем противоречие, поскольку $N$ не делится на $3$.\n\nРассмотрим другой подход. Из условия следует, что каждое из чисел $a, b, c, d$ является делителем числа $N = a + b + c + d$. Значит, $a = N / t_a$, $b = N / t_b$, $c = N / t_c$, $d = N / t_d$ для некоторых натуральных $t_a > 1$, $t_b > 1$, $t_c > 1$, $t_d > 1$.\n\nЗаметим, что $t_a \\ne 2$, иначе длина стороны $a$ равна полупериметру, что невозможно, поскольку $a < b + c + d$.\n\nПоскольку $N$ не делится на $3$, имеем $t_a \\ne 3$, значит, $t_a \\ge 4$ и $a \\le N / 4$. Аналогично, $b \\le N / 4$, $c \\le N / 4$, $d \\le N / 4$. Тогда равенство $N = a + b + c + d$ возможно только в случае $a = b = c = d = N / 4$, то есть четырёхугольник — ромб.\n\n*Замечание.* В полном решении важно использовать:\n\n1. Неравенство многоугольника (то есть тот факт, что $a, b, c, d$ — длины сторон четырёхугольника, а не произвольная четвёрка натуральных чисел с суммой $N$, для которой сумма любых трёх чисел делится на четвёртое). Иначе контрпример: $N / 2$, $N / 4$, $N / 8$, $N / 8$.\n\n2. $N$ не делится на $3$. Для $N$, кратного $6$, контрпример: стороны $N / 3$, $N / 3$, $N / 6$, $N / 6$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 14824,
    "subject": "Mathematics (Olympiad)",
    "question": "a. What is the largest possible total of four different ages if the oldest possible age is 19?\n\nb. Group 2 consists of four students whose ages are consecutive integers and whose total age is 66. What are their ages? (You may use any method.)\n\nc. Before a birthday, the total age of Group 3 was 58, and after the birthday, it was 59. The ages before the birthday were four consecutive integers. If only one person's age increased, what was the oldest age on the birthday?\n\nd. In two years' time, only the twins will be teenagers. The total age of the group will then be 65, and the other two members will be 18 and 19 years old. What are the ages of the twins?",
    "options": [],
    "answer": "See solution",
    "solution": "a. The oldest possible age is 19, so the largest age total with all ages different is $19 + 18 + 17 + 16 = 70$.\n\nb. **Alternative i**\n\nTrying all combinations of 4 consecutive ages, we get the following totals:\n\n$$\n13 + 14 + 15 + 16 = 58\n$$\n\n$$\n14 + 15 + 16 + 17 = 62\n$$\n\n$$\n15 + 16 + 17 + 18 = 66\n$$\n\n$$\n16 + 17 + 18 + 19 = 70\n$$\n\nSo the ages in Group 2 are 15, 16, 17, 18.\n\n**Alternative ii**\n\nThe average age of the 4 students in Group 2 is $66 \\div 4 = 16.5$. Since their ages are consecutive, two are less than 16.5 and two are larger than 16.5. So their ages are 15, 16, 17, 18.\n\n**Alternative iii**\n\nAny set of 4 consecutive ages can be obtained by adding the same number to each of the numbers 0, 1, 2, 3. Since $0 + 1 + 2 + 3 = 6$ and we want the ages to total 66, we must add $(66 - 6) \\div 4 = 15$ to each of 0, 1, 2, 3. So the ages in Group 2 are 15, 16, 17, 18.\n\nc. The age total before the birthday was $59 - 1 = 58$. From the first solution of Part b, the ages in Group 3 before the birthday were 13, 14, 15, 16. Only one of these ages increases on the birthday. If the 16 year old has the birthday then all four ages would remain different. If the 13, 14, or 15 year old has the birthday then there will be two teenagers with the same age, as required. So the oldest age on the birthday is 16.\n\nd. In two years' time only the twins will be teenagers. So the other two teenagers must be either 18 or 19. They cannot be the same age otherwise the total of the four ages would be even, so one must be 18 and the other must be 19. Hence the total of the twins' ages is $65 - 18 - 19 = 28$, so the twins are 14 years old.",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 14825,
    "subject": "Mathematics (Olympiad)",
    "question": "A sequence of real numbers $a_0, a_1, a_2, \\dots, a_{2012}$ satisfies the following conditions:\n\n$$\n|a_0 - a_1| = \\frac{3^1}{2^0} |a_1 - a_2| = \\frac{3^2}{2^1} |a_2 - a_3| = \\dots = \\frac{3^{2011}}{2^{2010}} |a_{2011} - a_{2012}| = \\frac{3^{2012}}{2^{2011}} |a_{2012} - a_0|.\n$$\n\nWhich values can the difference $a_0 - a_{1006}$ take?",
    "options": [],
    "answer": "See solution",
    "solution": "**Answer:** $0$.\n\n**Solution.** Let $|a_0 - a_1| = k$. We can write down the equations:\n\n$$\na_0 - a_1 = \\pm k, \\quad a_1 - a_2 = \\pm \\frac{2^0}{3^1} k, \\quad a_2 - a_3 = \\pm \\frac{2^1}{3^2} k, \\dots, a_{2011} - a_{2012} = \\pm \\frac{2^{2010}}{3^{2011}} k,\n$$\n$$\na_{2012} - a_0 = \\pm \\frac{2^{2011}}{3^{2012}} k.\n$$\n\nIf $k \\neq 0$, add these equations and divide by $k$ to get:\n\n$$\n0 = \\pm 1 \\pm \\frac{2^0}{3^1} \\pm \\frac{2^1}{3^2} \\pm \\dots \\pm \\frac{2^{2010}}{3^{2011}} \\pm \\frac{2^{2011}}{3^{2012}}.\n$$\n\nBut regardless of the choice of signs, the sum cannot be zero, because the sum of the absolute values of all terms except $1$ is less than $1$. This can be shown using the formula for the sum of a geometric progression:\n\n$$\n\\left| \\pm \\frac{2^0}{3^1} \\pm \\frac{2^1}{3^2} \\pm \\dots \\pm \\frac{2^{2010}}{3^{2011}} \\pm \\frac{2^{2011}}{3^{2012}} \\right| \\leq \\frac{2^0}{3^1} + \\frac{2^1}{3^2} + \\dots + \\frac{2^{2010}}{3^{2011}} + \\frac{2^{2011}}{3^{2012}} = \\frac{\\frac{1}{3}\\left(1 - \\left(\\frac{2}{3}\\right)^{2012}\\right)}{1 - \\frac{2}{3}} = 1 - \\left(\\frac{2}{3}\\right)^{2012} < 1\n$$\n\nHence, the case $k \\neq 0$ is impossible. Thus $k = 0$, so\n$$a_0 - a_1 = a_1 - a_2 = \\dots = a_{2011} - a_{2012} = 0,$$\nthat is, $a_0 = a_1 = \\dots = a_{2012}$, and therefore $a_0 - a_{1006} = 0$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 14826,
    "subject": "Mathematics (Olympiad)",
    "question": "Prove that for positive real numbers $x, y, z$,\n\n$$\nx^3(y^2 + z^2)^2 + y^3(z^2 + x^2)^2 + z^3(x^2 + y^2)^2 \\geq xyz[xy(x + y)^2 + yz(y + z)^2 + zx(z + x)^2].\n$$\n\nExpand both sides as follows:\n\n$$\nL = \\sum_{\\text{cyc}} x^3(y^2 + z^2)^2 = \\sum_{\\text{cyc}} x^3(y^4 + z^4) + 2x^2y^2z^2 \\sum_{\\text{cyc}} x\n$$\n\nand\n\n$$\nR = xyz \\sum_{\\text{cyc}} xy(x^2 + y^2) + 2xyz \\sum_{\\text{cyc}} x^2 y^2,\n$$\n\nwhere $\\sum_{\\text{cyc}}$ denotes a cyclic sum in the variables $(x, y, z)$, and $L$ and $R$ denote the left and right sides of the desired inequality.",
    "options": [],
    "answer": "See solution",
    "solution": "**Solution 1.** Multiply both sides of the desired inequality by $\\sum_{\\text{cyc}} x = (x + y + z)$; that is, show that\n\n$$\n(x + y + z) \\cdot L \\geq (x + y + z) \\cdot R\n$$\n\nor\n\n$$\n(x + y + z) \\cdot (L - R) \\geq 0.\n$$\n\nNote that\n\n$$\n\\begin{aligned}\n(x + y + z) L &= (x + y + z) \\left( \\sum_{\\text{cyc}} x^3(y^4 + z^4) + 2x^2y^2z^2 \\sum_{\\text{cyc}} x \\right) \\\\\n&= (x + y + z) \\sum_{\\text{cyc}} x^3(y^4 + z^4) + 2x^2y^2z^2 (x + y + z)^2 \\\\\n&= \\sum_{\\text{cyc}} x^5(y^3 + z^3) + 2 \\sum_{\\text{cyc}} x^4 y^4 + xyz \\sum_{\\text{cyc}} x^3(y^2 + z^2) + 2x^2y^2z^2 \\sum_{\\text{cyc}} x^2 + 4x^2y^2z^2 \\sum_{\\text{cyc}} xy\n\\end{aligned}\n$$\n\nand\n\n$$\n\\begin{aligned}\n(x + y + z) R &= (x + y + z) \\left( xyz \\sum_{\\text{cyc}} xy(x^2 + y^2) + 2xyz \\sum_{\\text{cyc}} x^2 y^2 \\right) \\\\\n&= xyz \\left( \\sum_{\\text{cyc}} x^4(y + z) + \\sum_{\\text{cyc}} x^3(y^2 + z^2) + 2xyz \\sum_{\\text{cyc}} x^2 + 2 \\sum_{\\text{cyc}} x^3(y^2 + z^2) + 4xyz \\sum_{\\text{cyc}} xy \\right).\n\\end{aligned}\n$$\n\nThus,\n\n$$\n(x + y + z)(L - R) = \\sum_{\\text{cyc}} x^5(y^3 + z^3 - yz(y + z)) + 2 \\sum_{\\text{cyc}} x^4 y^4 + 2x^2y^2z^2 \\sum_{\\text{cyc}} xy - 2xyz \\sum_{\\text{cyc}} x^3(y^2 + z^2).\n$$\n\nWe complete the proof by showing this expression is always nonnegative. It is the sum of two non-negative expressions:\n\n$$\nS_1 = \\sum_{\\text{cyc}} x^5(y^3 + z^3 - yz(y+z))\n$$\n\nand\n\n$$\nS_2 = 2 \\sum_{\\text{cyc}} x^4 y^4 + 2x^2 y^2 z^2 \\sum_{\\text{cyc}} xy - 2xyz \\sum_{\\text{cyc}} x^3(y^2 + z^2).\n$$\n\nThe first, $S_1$, is nonnegative because\n\n$$\n3(y^3 + z^3) = (2y^3 + z^3) + (y^3 + 2z^3) \\geq 3y^2z + 3yz^2 = 3yz(y + z),\n$$\n\nby the AM-GM inequality. For $S_2$, use Schur's inequality: set $a = xy$, $b = yz$, $c = zx$, then\n\n$$\nS_2 = 2 \\left( \\sum_{\\text{cyc}} a^4 + abc \\sum_{\\text{cyc}} a - \\sum_{\\text{cyc}} a^3(b+c) \\right) = 2 \\sum_{\\text{cyc}} a^2(a-b)(a-c) \\geq 0,\n$$\n\nby Schur's inequality for $n = 2$.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 14827,
    "subject": "Mathematics (Olympiad)",
    "question": "以三角形 $ABC$ 的三條邊為邊，分別向 $ABC$ 的外面作正三角形 $ABC_1$、$BCA_1$、$CAB_1$。設點 $P$ 為 $ABC_1$ 的外接圓與 $CAB_1$ 的外接圓的另一個交點（$P \\ne A$）。在 $CAB_1$ 的外接圓上找一點 $Q$ 使得 $PQ$ 平行於 $BA_1$。在 $ABC_1$ 的外接圓上找一點 $R$ 使得 $PR$ 平行於 $CA_1$。\n\n證明：三角形 $ABC$ 的重心，與三角形 $PQR$ 的重心連線，會平行於直線 $BC$。",
    "options": [],
    "answer": "See solution",
    "solution": "暴力三角解析法。給直線 $PA$、$PB$、$PC$ 定向，使得由 $PA$ 到 $PB$、由 $PB$ 到 $PC$、由 $PC$ 到 $PA$ 的角度皆為 $\\frac{2\\pi}{3}$。另外也給直線 $PQ$、$PR$ 定向，使得 $\\overrightarrow{BC}$ 與 $\\overrightarrow{PQ}$、$\\overrightarrow{PR}$ 與 $\\overrightarrow{BC}$ 的夾角也是 $\\frac{2\\pi}{3}$。令 $\\angle(\\overrightarrow{BC}, PA) = \\alpha$，$\\angle(\\overrightarrow{BC}, PB) = \\beta$，$\\angle(\\overrightarrow{BC}, PC) = \\gamma$。易知 $\\gamma = \\beta + \\frac{2\\pi}{3} = \\alpha - \\frac{2\\pi}{3}$。題目等價於證明等式：\n\n$$\nPQ \\sin \\frac{2\\pi}{3} - PR \\sin \\frac{2\\pi}{3} = PA \\sin \\alpha + PB \\sin \\beta + PC \\sin \\gamma. \\quad (*)\n$$\n\n(*) 式的左邊可化為 $\\frac{PQ - PR}{2}$。由托勒密定理知\n\n$$\nPQ \\sin \\angle(PC, PA) + PC \\sin \\angle(PA, PQ) + PA \\sin \\angle(PQ, PC) = 0.\n$$\n\n由證明一開始所給出的定向，有 $\\angle(PC, PA) = \\frac{2\\pi}{3}$。上式的第二個角可寫成 $\\angle(PA, PQ) = \\frac{2\\pi}{3} - \\alpha = -\\gamma$，而第三個角可寫成 $\\angle(PQ, PC) = \\angle(\\vec{BC}, PB) = \\beta$。所以\n\n$$\n\\frac{1}{2}PQ = PC \\sin \\gamma - PA \\sin \\beta.\n$$\n\n同理可得\n\n$$\n-\\frac{1}{2}PR = PB \\sin \\beta - PA \\sin \\gamma.\n$$\n\n於是題目等價於證明\n\n$$\n- \\sin \\beta - \\sin \\gamma = \\sin \\alpha,\n$$\n\n即 $\\sin \\alpha + \\sin \\beta + \\sin \\gamma = 0$。此式可由恆等式 $(1+\\omega+\\omega^2)z=0$ 得證，其中 $z$ 為任意複數，$\\omega$ 為 1 的三次原根。$\\square$\n\n**Remark.** In fact, we can fix $\\alpha, \\beta, \\gamma, PB, PC$ and change $PA$ once we verify that the coefficients of $PB$ and $PC$ on both sides match. Then it suffices to pick one $A \\ne F$ and verify that the statement holds there, and we can simply take $A = A_1$. The statement holds there as if we let $F$ be on $A_1BC$ such that $PF$ is parallel to $BC$,\n\nthen the centroid of FQR is exactly the centroid of $A_1BC$, and the line connecting the centroids of FQR and PQR is parallel to PF, which is parallel to BC.\n\n*Alternative proof.* Let $F$ be on $A_1BC$ with $PF$ parallel to $BC$. According to the remark above, it suffices to show that $FQR$ and $ABC$ share the same centroid.\n\n**Lemma.** Let $X, Y, Z$ be moving points on three separate circles with the same angular speed. Then the centroid of $XYZ$ also moves in a circle with the same angular speed, and the center of its locus is the centroid of the centers of the loci of $X, Y, Z$.\n\n*Proof of lemma.* This is obvious by, say, complex coordinates. $\\square$\n\nNow if $X, Y, Z$ are on $A_1BC, AB_1C$, and $ABC_1$，respectively, with $X, Y, Z$ starting at $B, C, A$，then after a while they arrive at $C, A, B$。Therefore the locus of the centroid of $XYZ$ visits the centroid of $ABC$ at least twice in each cycle, showing that in fact the centroid of $XYZ$ is always the same, which has to be the centroid of $ABC$。Now when $X$ travels to $F$, it is easy to chase the angle and show that $Y$ travels to $Q$ and $Z$ travels to $R$。$\\square$\n\n**Remark.** This argument shows that this problem is also easily complex-bashable。",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 14828,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $I$ be the incenter of triangle $ABC$, and let $P$ be any point on the arc $BAC$ of the circumscribed circle. On the tangent to the circumscribed circle $\\omega$ of triangle $API$ at the point $I$, points $K$ and $L$ are selected so that $BK = KI$ and $CL = LI$. Prove that the circumcircle of triangle $PKL$ is tangent to $\\omega$.",
    "options": [],
    "answer": "See solution",
    "solution": "Mark the midpoints of the arcs $AB$, $AC$, and $BC$ that do not contain the other vertices as $W_C$, $W_B$, and $W_A$, respectively.\n\n![](images/Ukraine2022-23_p29_data_1705f7f2fa.png)\n\nWe see that $K$ and $L$ lie on $W_AW_C$ and $W_AW_B$, respectively. Then,\n\n$$\n\\begin{aligned}\n\\angle(KL, PI) &= \\angle(AI, AP) = \\angle(AW_A, AP) = \\angle(W_AW_C, W_CP) \\\\\n&= \\angle(KW_C, W_CP) = \\angle(W_AW_B, W_BP) = \\angle(LW_B, W_BP),\n\\end{aligned}\n$$\n\nso the quadrilaterals $KW_CPI$ and $IPW_BL$ are cyclic. Then,\n\n$$\n\\begin{aligned}\n\\angle(KP, PI) &= \\angle(KW_C, W_CI) = \\angle(W_AW_C, W_CP) = \\angle(W_AA, AC) \\\\\n&= \\angle(BA, AW_A) = \\angle(BW_B, W_BW_A) = \\angle(IW_B, W_BL) = \\angle(IP, PL),\n\\end{aligned}\n$$\n\nTherefore, $PI$ is the angle bisector of $\\angle KPL$, and by the converse of Archimedes' lemma, since $KL$ is tangent to $\\omega$, the circumcircle of triangle $KPL$ is tangent to $\\omega$.\n\n_Archimedes' Lemma._ If a circle is inscribed in a segment of another circle bounded by the chord $BC$ and touches the arc at the point $A_1$ and the chord at the point $A_2$, then the line $A_1A_2$ is the angle bisector of $\\angle BA_1C$.\n\n![](images/Ukraine2022-23_p29_data_b47d157edc.png)",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 14829,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all polynomials $P(x)$ with real coefficients satisfying the relation\n\n$$\n(x^3 + 3x^2 + 3x + 2)P(x - 1) = (x^3 - 3x^2 + 3x - 2)P(x)\n$$\n\nfor every real number $x$.",
    "options": [],
    "answer": "See solution",
    "solution": "We have:\n\n$$\n(x^3 + 3x^2 + 3x + 2)P(x - 1) = (x^3 - 3x^2 + 3x - 2)P(x) \\quad \\forall x \\in \\mathbb{R} \\tag{1}\n$$\n\nThis can be rewritten as:\n\n$$\n(x + 2)(x^2 + x + 1)P(x - 1) = (x - 2)(x^2 - x + 1)P(x) \\quad \\forall x \\in \\mathbb{R} \\tag{2}\n$$\n\nSubstituting $x = -2$ into (2):\n\n$$\n0 = -28 \\cdot P(-2) \\implies P(-2) = 0.\n$$\n\nSubstituting $x = 2$ into (2):\n\n$$\n28 \\cdot P(1) = 0 \\implies P(1) = 0.\n$$\n\nSubstituting $x = -1$ into (1):\n\n$$\n0 = -9 \\cdot P(-1) \\implies P(-1) = 0.\n$$\n\nSubstituting $x = 0$ into (1):\n\n$$\n9 \\cdot P(0) = 0 \\implies P(0) = 0.\n$$\n\nThus,\n\n$$\nP(x) = (x - 1)x(x + 1)(x + 2)Q(x)\n$$\n\nwhere $Q(x)$ is a polynomial with real coefficients.\n\nThen,\n\n$$\nP(x - 1) = (x - 2)(x - 1)x(x + 1)Q(x - 1)\n$$\n\nPlugging into (2):\n\n$$\n(x - 2)(x - 1)x(x + 1)(x + 2)(x^2 + x + 1)Q(x - 1) = (x - 2)(x - 1)x(x + 1)(x + 2)(x^2 - x + 1)Q(x)\n$$\n\nSo,\n\n$$\n(x^2 + x + 1)Q(x - 1) = (x^2 - x + 1)Q(x)\n$$\n\nSince $(x^2 + x + 1)$ and $(x^2 - x + 1)$ are coprime, $Q(x)$ must be divisible by $x^2 + x + 1$:\n\n$$\nQ(x) = (x^2 + x + 1)R(x)\n$$\n\nwhere $R(x)$ is a polynomial. Then,\n\n$$\nQ(x - 1) = (x^2 - x + 1)R(x - 1)\n$$\n\nPlugging back:\n\n$$\n(x^2 + x + 1)(x^2 - x + 1)R(x - 1) = (x^2 - x + 1)(x^2 + x + 1)R(x)\n$$\n\nSo $R(x - 1) = R(x)$ for all $x$, which means $R(x)$ is constant.\n\nTherefore, all solutions are:\n\n$$\nP(x) = c(x - 1)x(x + 1)(x + 2)(x^2 + x + 1)\n$$\n\nwhere $c$ is a real constant.",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 14830,
    "subject": "Mathematics (Olympiad)",
    "question": "Positive integers $a_1, a_2, \\ldots, a_{101}$ are such that $a_i + 1$ is divisible by $a_{i+1}$ for $1 \\leq i \\leq 101$ (with $a_{102} = a_1$). What is the largest value that the maximum of these numbers can attain?",
    "options": [],
    "answer": "See solution",
    "solution": "Let $a_{101}$ be the largest among the numbers. For any $i$ ($1 \\leq i \\leq 100$), $a_i + 1 \\geq a_{i+1}$. Summing these inequalities for $i = 1$ to $100$ gives:\n\n$$a_1 + a_2 + \\cdots + a_{100} + 100 \\geq a_2 + a_3 + \\cdots + a_{101}$$\n\nwhich simplifies to $a_1 \\geq a_{101} - 100$. Since $a_{101} + 1$ is divisible by $a_1$ and $a_{101}$ is the largest, $a_{101} + 1 > a_1$, so $a_{101} + 1 \\geq 2a_1 \\geq 2(a_{101} - 100) = 2a_{101} - 200$. Thus, $a_{101} \\leq 201$.\n\nTo show this bound is attainable, consider $a_1 = 101$, $a_2 = 102$, $\\ldots$, $a_{101} = 201$. Then $a_i + 1 = a_{i+1}$ for $i = 1$ to $100$, and $a_{101} + 1 = 202 = 2 \\times 101 = 2a_1$, so all conditions are satisfied. Therefore, the largest possible value is $201$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 14831,
    "subject": "Mathematics (Olympiad)",
    "question": "Бат, Цэцэг хоёр $8 \\times 8$ хүснэгт дээр дараах тоглоомыг тоглоно ($k \\in \\mathbb{N}$):\n\n- Бат эхлээд $k$ ширхэг нүдийг сонгон авч, эдгээр нүдэнд \"Б\" үсэг тавина.\n- Дараа нь Цэцэг $k+1$ нүдийг сонгон авч, эдгээр нүдэнд \"Ц\" үсэг тавина.\n\nБосоогоор эсвэл хэвтээгээр \"ЦББ\" эсвэл \"ББЦ\" гэсэн үг гарвал Бат хожино, бусад тохиолдолд Цэцэг хожино.\n\nБатад хожих техник олддог байх хамгийн бага $k$ тоог ол.",
    "options": [],
    "answer": "See solution",
    "solution": "Босоо эсвэл хэвтээ $1 \\times 4$ хэмжээст дүрсэд хоёр захын нүдийг муу нүд гэж нэрлэе. Хэрэв Цэцэг муу нүдэнд \"Ц\" үсэг тавибал Бат хожих боломжгүй болно.\n\nМуу нүднүүдийн авч болох хамгийн их утгыг $M_{\\text{max}}$ гэж тэмдэглэе.\n\nТэгвэл $M_{\\text{max}} + k + (k+1) > 64$ байх хамгийн бага $k$-г олъё.\n\n% ![](images/2013-ilovepdf-compressed_p43_data_114212e242.png)\n\nДурын $1 \\times 4$ эсвэл $4 \\times 1$ хэсэгт муу нүдний тоо 2 байна ($M=2$). Нэг \"Б\" үсэгт хамгийн ихдээ 2 муу нүд хамаарна.\n\n% ![](images/2013-ilovepdf-compressed_p43_data_3a90574bc7.png)\n\nИймд $M_{\\text{max}} \\leq 2k$.\n\n$$\n2k + k + k + 1 \\geq M_{\\text{max}} + k + k + 1 > 64 \\implies 4k > 63,\n$$\n\n$$\nk > \\frac{63}{4} \\implies k_{\\text{min}} \\geq 16.\n$$\n\nОдоо $k_{\\text{min}} = 16$ байх нэг хувилбарыг зурж үзүүлье.\n\n% ![](images/2013-ilovepdf-compressed_p43_data_d8e971b80b.png)",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 14832,
    "subject": "Mathematics (Olympiad)",
    "question": "Prove that the equation $2^x + 21^x = y^3$ has no solutions in positive integers, and find all the solutions in nonnegative integers of the equation $2^x + 21^y = z^2$.",
    "options": [],
    "answer": "See solution",
    "solution": "It is obvious that $x \\geq 1$, $y$ is odd, and $21^x \\equiv 0 \\pmod{7}$. Since $2^3 \\equiv 1 \\pmod{7}$, we have $2^{3n} \\equiv 1 \\pmod{7}$, $2^{3n+1} \\equiv 2 \\pmod{7}$, and $2^{3n+2} \\equiv 4 \\pmod{7}$. On the other hand, $y^3 \\equiv 0 \\pmod{7}$ or $y^3 \\equiv \\pm 1 \\pmod{7}$. So, from the preceding, we have $x = 3n$ for all positive integers $n \\geq 1$. Putting $x = 3n$, the first equation becomes\n\n$$\n2^{3n} + 21^{3n} = y^3 \\Leftrightarrow (y - 21^n)(y^2 + y \\cdot 21^n + 21^{2n}) = 2^{3n}\n$$\n\nFrom the preceding, it immediately follows that $(y^2 + y \\cdot 21^n + 21^{2n}) \\mid 2^{3n}$, and this is not possible since $y^2 + y \\cdot 21^n + 21^{2n} > 1$ is an odd positive integer. This completes the proof of the first part of the statement.\n\nTo find the solutions in positive integers of the second equation, we distinguish the following cases:",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 14833,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $a_1, a_2, \\ldots, a_{100}$ be a permutation of $1, 2, \\ldots, 100$. For each triple $\\{a_i, a_{i+1}, a_{i+2}\\}$ of consecutive numbers, $1 \\leq i \\leq 98$, the middle number in the triple is marked. For instance, if $a_i = 7$, $a_{i+1} = 99$, $a_{i+2} = 22$, then $a_{i+2} = 22$ is marked. Let $S$ be the sum of all marked numbers. Find the minimum value of $S$. (Each marked number enters the sum $S$ exactly once, although it may be marked more than once.)",
    "options": [],
    "answer": "See solution",
    "solution": "The desired minimum is $33 \\cdot 34 = 1122$. More generally, for $n = 3k + 1$ instead of $100$, the answer is $S_{\\min} = 2(1 + \\dots + k) = k(k+1)$.\n\nFor clarity, we state separately a fact used later in a proof of the lower bound $S(\\alpha) \\geq 2(1 + \\dots + k)$ for each permutation $\\alpha$ of $1, 2, \\ldots, 3k + 1$.\n\n**Claim.** If $2k$ distinct natural numbers are divided into $k$ pairs $u_j, v_j$ with $u_j < v_j$, $j = 1, \\ldots, k$, then $v_1 + \\dots + v_k \\geq 2(1 + \\dots + k)$.\n\nThe justification is by induction on $k$, with the obvious base case $k=1$.\n\nFor the inductive step $k-1 \\to k$, choose the labeling so that $v_k := \\max_{j=1}^k v_j$. Ignore $u_k$ and $v_k$ for the time being, and apply the inductive hypothesis to the remaining $2k-2$ numbers. This gives $v_1 + \\dots + v_{k-1} \\geq 2(1 + \\dots + (k-1))$. So it is enough to prove $v_k \\geq 2k$ in order to complete the inductive step. We have $v_k > v_j$ for all $j = 1, \\ldots, k-1$ by $v_k = \\max_{j=1}^k v_j$. In addition, observe that $v_k > u_j$ for all $j = 1, \\ldots, k$. This holds for $j=k$ by hypothesis. Suppose that $v_k < u_j$ for some $j=1, \\ldots, k-1$. Then $u_j < v_j$ implies $v_k < v_j$, which contradicts the maximum choice of $v_k$. In summary, there are $2k-1$ distinct natural numbers smaller than $v_k$, namely $u_1, \\ldots, u_k, v_1, \\ldots, v_{k-1}$. Hence $v_k \\geq 2k$, completing the induction.\n\nNow let $\\alpha = (a_1, a_2, \\ldots, a_{3k+1})$ be any permutation of $1, 2, \\ldots, 3k + 1$. Divide $a_1, a_2, \\ldots, a_{3k}$ into $k$ triples $T_j = \\{a_{3j-2}, a_{3j-1}, a_{3j}\\}$, $j = 1, \\ldots, k$. Let $u_j$ and $v_j$ be respectively the smaller number and the middle number in $T_j$. The $2k$ numbers $u_j, v_j$, $j = 1, \\ldots, k$, are distinct. Then the claim above gives $v_1 + \\dots + v_k \\geq 2(1 + \\dots + k)$. Since $v_1, \\ldots, v_k$ are marked numbers (in general there are more of them), the sum $S(\\alpha)$ of all marked numbers in $\\alpha$ also satisfies $S(\\alpha) \\geq 2(1 + \\dots + k)$.\n\nThe equality $S(\\alpha) = 2(1 + \\dots + k)$ is attained for the following permutation of $1, 2, \\ldots, 3k + 1$: $3k + 1, 1, 2, 2k + 1, 4, 3, 2k + 2, 6, 5, 2k + 3, \\ldots, 2k - 2, 2k - 3, 3k - 1, 2k, 2k - 1, 3k$.\n\nThe marked numbers are precisely $2, 4, \\ldots, 2k$, hence $S(\\alpha) = 2(1 + \\dots + k)$. This completes the proof of $S_{\\min} = 2(1 + \\dots + k) = k(k+1)$.",
    "topic": "Discrete Mathematics",
    "subtopic": "Combinatorics"
  },
  {
    "id": 14834,
    "subject": "Mathematics (Olympiad)",
    "question": "Determine all triples $ (x, y, z) $ of real numbers that satisfy\n\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$$",
    "options": [],
    "answer": "See solution",
    "solution": "From the first equation, we get $x + y = z - 1$. Plugging this into the second equation:\n\n$$\n\\begin{aligned}\nx^2 - y^2 + z^2 &= 1 \\\\\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$$\n\nThere are two cases:\n\n**Case 1:** $z = 1$\n\nThen $x + y = 0$, so $x = -y$. Plugging into the third equation:\n\n$$\n-x^3 + y^3 + z^3 = -1 \\\\\n-(-y)^3 + y^3 + 1 = -1 \\\\\n y^3 + y^3 + 1 = -1 \\\\\n2y^3 = -2 \\\\\ny = -1, \\quad x = 1, \\quad z = 1\n$$\n\n**Case 2:** $x - y + z + 1 = 0$\n\nPlugging $z = -x + y - 1$ into the first equation:\n\n$$\nx + y - z = -1 \\\\\nx + y - (-x + y - 1) = -1 \\\\\nx + y + x - y + 1 = -1 \\\\\n2x + 1 = -1 \\\\\nx = -1\n$$\n\nNow $z = -(-1) + y - 1 = 1 + y - 1 = y$, so $z = y$. Plugging into the third equation:\n\n$$\n-x^3 + y^3 + z^3 = -1 \\\\\n-(-1)^3 + y^3 + y^3 = -1 \\\\\n1 + 2y^3 = -1 \\\\\n2y^3 = -2 \\\\\ny = -1, \\quad z = -1, \\quad x = -1\n$$\n\nThus, the solutions are $ (1, -1, 1) $ and $ (-1, -1, -1) $. Direct computation confirms both satisfy the system.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 14835,
    "subject": "Mathematics (Olympiad)",
    "question": "A circle is inscribed in triangle $ABC$, tangent to its sides $BC$, $CA$, and $AB$ at points $A_1$, $B_1$, and $C_1$, respectively. The line passing through the midpoint of segment $A_1B_1$ is perpendicular to $AB$; the line passing through the midpoint of segment $B_1C_1$ is perpendicular to $BC$; and the line passing through the midpoint of segment $C_1A_1$ is perpendicular to $CA$. Prove that these lines intersect at one point.",
    "options": [],
    "answer": "See solution",
    "solution": "Let's prove a more general statement:\n\nLet $O$ be a point inside $\\triangle ABC$. Let $A_1$, $B_1$, and $C_1$ be the feet of the perpendiculars dropped from $O$ to sides $BC$, $CA$, and $AB$ of $\\triangle ABC$, respectively. Let $A_2$, $B_2$, and $C_2$ be the midpoints of segments $B_1C_1$, $C_1A_1$, and $A_1B_1$, respectively. Let $A_3$, $B_3$, and $C_3$ be the feet of the perpendiculars dropped from $A_2$, $B_2$, and $C_2$ to lines $BC$, $CA$, and $AB$, respectively. Prove that lines $A_2A_3$, $B_2B_3$, and $C_2C_3$ intersect at one point.\n\n![](images/Ukrajina_2008_p5_data_b1aece9829.png)\n\nLet's use the well-known Carnot Lemma.\n\n**Carnot Lemma.** Perpendiculars drawn through points $A_3$, $B_3$, and $C_3$ on lines $BC$, $CA$, and $AB$ of $\\triangle ABC$ intersect at one point if and only if\n\n$$\nBA_3^2 - CA_3^2 + CB_3^2 - AB_3^2 + AC_3^2 - BC_3^2 = 0\n$$\n\nAs follows from the median formula for $\\triangle BB_1C_1$ and $\\triangle CB_1C_1$, and the Pythagorean theorem,\n\n$$\nBA_3^2 - CA_3^2 = BA_2^2 - CA_2^2 = \\frac{1}{4}(2BC_1^2 + 2BB_1^2 - B_1C_1^2 - 2CB_1^2 - 2CC_1^2 + B_1C_1^2) = \\frac{1}{2}(BC_1^2 + BB_1^2 - CB_1^2 - CC_1^2).\n$$\n\nSimilarly,\n\n$$\nCB_3^2 - AB_3^2 = \\frac{1}{2}(CA_1^2 + CC_1^2 - AC_1^2 - AA_1^2),\n$$\n\n$$\nAC_3^2 - BC_3^2 = \\frac{1}{2}(AB_1^2 + AA_1^2 - BA_1^2 - BB_1^2).\n$$\n\nSumming all three,\n\n$$\n\\begin{aligned}\n&BA_3^2 - CA_3^2 + CB_3^2 - AB_3^2 + AC_3^2 - BC_3^2 = \\\\\n&\\quad \\frac{1}{2}(BC_1^2 + BB_1^2 - CB_1^2 - CC_1^2 + CA_1^2 + CC_1^2 - AC_1^2 - AA_1^2 + AB_1^2 + AA_1^2 - BA_1^2 - BB_1^2) \\\\\n&= \\frac{1}{2}(BC_1^2 - CB_1^2 + CA_1^2 - AC_1^2 + AB_1^2 - BA_1^2).\n\\end{aligned}\n$$\n\nBut\n\n$$\nAC_1^2 + BA_1^2 + CB_1^2 = AB_1^2 + BC_1^2 + CA_1^2.\n$$\n\nUsing this in the previous result, we find that $BA_3^2 - CA_3^2 + CB_3^2 - AB_3^2 + AC_3^2 - BC_3^2 = 0$. Thus, by Carnot's lemma, the required lines are concurrent. The original problem is the special case where $O$ is the incenter.\n\n![](images/Ukrajina_2008_p6_data_9f31a4ab29.png)",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 14836,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $n \\in \\mathbb{N}$, with $n \\ge 2$, and let $x_1, x_2, \\dots, x_n$ be positive real numbers such that $x_1 + x_2 + \\dots + x_n = 1$. Define\n\n$$\nm = \\min \\left\\{ \\frac{x_1}{1+x_1}, \\frac{x_2}{1+x_1+x_2}, \\dots, \\frac{x_n}{1+x_1+x_2+\\dots+x_n} \\right\\}\n$$\n\nand\n\n$$\nM = \\max \\left\\{ \\frac{x_1}{1+x_1}, \\frac{x_2}{1+x_1+x_2}, \\dots, \\frac{x_n}{1+x_1+x_2+\\dots+x_n} \\right\\}.\n$$\n\nDetermine:\n\na) the greatest possible value of $m$;\n\nb) the smallest possible value of $M$,\n\nfor all possible choices of positive real numbers $x_1, x_2, \\dots, x_n$ with sum $1$.",
    "options": [],
    "answer": "See solution",
    "solution": "a) Let us denote\n\n$$\na_k = \\frac{x_k}{1+x_1+\\cdots+x_k} \\quad \\text{and} \\quad b_k = 1-a_k = \\frac{1+x_1+\\cdots+x_{k-1}}{1+x_1+\\cdots+x_k}, \\quad 1 \\le k \\le n.\n$$\n\nBy the inequality of arithmetic and geometric means, we have\n\n$$\nb_1 + b_2 + \\cdots + b_n \\ge n \\sqrt[n]{b_1 b_2 \\cdots b_n} = \\frac{n}{\\sqrt[n]{1+x_1+\\cdots+x_n}} = \\frac{n}{\\sqrt[n]{2}}.\n$$\n\nTherefore,\n\n$$\nn \\cdot m \\le a_1 + a_2 + \\cdots + a_n = n - (b_1 + b_2 + \\cdots + b_n) \\le n \\left(1 - \\frac{1}{\\sqrt[n]{2}}\\right),\n$$\n\nwhich implies $m \\le 1 - \\frac{1}{\\sqrt[n]{2}}$.\n\nWe will show that there exists a sequence $(x_1, x_2, \\dots, x_n)$ of positive real numbers summing to $1$ such that $a_1 = a_2 = \\dots = a_n = 1 - \\frac{1}{\\sqrt[n]{2}} \\stackrel{\\text{def}}{=} m_0$, which implies that the greatest possible value of $m$ is $m_0$.\n\nThe existence of such a sequence is equivalent to solving the system\n\n$$\nx_1 + \\cdots + x_n = 1, \\quad \\frac{x_1}{1+x_1} = \\frac{x_i}{1+x_1+\\cdots+x_i}, \\quad \\text{for all } i = 2, \\dots, n.\n$$\n\nFor $i = 2$, we get $x_2 = a_1(1 + x_1)$. For $i = 3$, replacing $x_2$, we find $x_3 = x_1(1 + x_1)^2$. By induction on $i$, it follows that $x_i = x_1(1 + x_1)^{i-1}$, for all $i \\le n$. Since their sum is $1$, we get\n\n$$\n1 = x_1 + x_1(1 + x_1) + \\dots + x_1(1 + x_1)^{n-1} = x_1 \\frac{(1+x_1)^n - 1}{(1+x_1) - 1} = (1+x_1)^n - 1.\n$$\n\nHence, $(1+x_1)^n = 2$, so $x_1 = \\sqrt[n]{2}-1$. Substituting back, $x_k = \\sqrt[n]{2^{k-1}}(\\sqrt[n]{2}-1)$, $k=1, 2, \\dots, n$.\n\nb) We will prove that the smallest possible value of $M$ is $M_0 = 1 - \\frac{1}{\\sqrt[n]{2}}$, which is attained if and only if $a_1 = a_2 = \\dots = a_n$.\n\nUsing the same reasoning as before, we have $a_1 = a_2 = \\dots = a_n$ if and only if $x_k = \\sqrt[n]{2^{k-1}}(\\sqrt[n]{2}-1)$, $k = 1, 2, \\dots, n$.\n\nLet $\\alpha_k = \\sqrt[n]{2^{k-1}}(\\sqrt[n]{2}-1)$ for $1 \\le k \\le n$, and consider a positive sequence $(x_1, x_2, \\dots, x_n)$, different from $(\\alpha_1, \\alpha_2, \\dots, \\alpha_n)$, with sum $1$. Then there exists $k \\in \\{1, \\dots, n\\}$ such that $\\alpha_k < x_{k}$.\n\nLet $k_0$ be the smallest such index for which $\\alpha_{k_0} < x_{k_0}$; then $x_i \\le \\alpha_i$, for all $1 \\le i \\le k_0 - 1$. We deduce\n\n$$\n\\begin{aligned}\na_{k_0} &= \\frac{x_{k_0}}{1 + x_1 + \\cdots + x_{k_0}} = \\frac{1}{\\frac{1+x_1+\\cdots+x_{k_0-1}}{x_{k_0}} + 1} > \\frac{1}{\\frac{1+\\alpha_1+\\cdots+\\alpha_{k_0-1}}{\\alpha_{k_0}} + 1} \\\\\n&= \\frac{\\alpha_{k_0}}{1 + \\alpha_1 + \\cdots + \\alpha_{k_0}} = M_0,\n\\end{aligned}\n$$\n\ntherefore, $M = \\max a_k \\ge a_{k_0} > M_0$.\n\nIn conclusion, $M_0 = 1 - \\frac{1}{\\sqrt[n]{2}}$ is the minimal possible value of $M$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 14837,
    "subject": "Mathematics (Olympiad)",
    "question": "Exactly two years ago, the Benson family had 4 members, and their average age was $19$. The Bensons then adopted another child. If the average age of the family today is still $19$, what is the present age of the adopted child?",
    "options": [],
    "answer": "See solution",
    "solution": "Two years ago, the sum of the family's ages was $4 \\times 19 = 76$. In two years, each of the 4 original members aged 2 years, so their total age increased by $2 \\times 4 = 8$. Today, the sum of the ages is $5 \\times 19 = 95$. Thus, the increase in total age is $95 - 76 = 19$. The difference between this and the $8$ years accounted for by the original members is $19 - 8 = 11$. Therefore, the present age of the adopted child is $11$ years.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 14838,
    "subject": "Mathematics (Olympiad)",
    "question": "Given positive integers $n$, $r$, and distinct prime numbers $p_1, p_2, \\dots, p_r$. Initially, there are $(n+1)^r$ numbers on the blackboard: $p_1^{i_1} p_2^{i_2} \\cdots p_r^{i_r}$ where $0 \\le i_1, i_2, \\dots, i_r \\le n$.\n\nAlice and Bob take turns (Alice goes first) to make the following moves, until only one number is left on the blackboard:\n\n* On her turn, Alice erases two numbers (they can be identical) and writes their greatest common divisor on the blackboard.\n* On his turn, Bob erases two numbers (they can be identical) and writes their least common multiple on the blackboard.\n\nFind the least integer $M$ such that Alice can guarantee the remaining number does not exceed $M$.",
    "options": [],
    "answer": "See solution",
    "solution": "The least $M$ is $$(p_1 \\cdots p_r)^{\\lfloor\\frac{n}{2}\\rfloor}.$$ \n\nLet $N = (p_1 \\cdots p_r)^n$ and $M = (p_1 \\cdots p_r)^{\\lfloor\\frac{n}{2}\\rfloor}$. For each divisor $a$ of $N$, consider the pair $(a, \\frac{N}{a})$. Clearly, $\\gcd(a, \\frac{N}{a})$ is a divisor of $M$, while $\\mathrm{lcm}(a, \\frac{N}{a})$ is a multiple of $M$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 14839,
    "subject": "Mathematics (Olympiad)",
    "question": "Given an integer $n \\ge 2$, find the largest number $\\lambda(n)$ with the following property: if a sequence of real numbers $a_0, a_1, a_2, \\dots, a_n$ satisfies\n\n$$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,$$\n\nthen\n\n$$\n\\left(\\sum_{i=1}^{n} i a_{i}\\right)^{2} \\geq \\lambda(n) \\sum_{i=1}^{n} a_{i}^{2}.\n$$\n\n![](path/to/file.png)",
    "options": [],
    "answer": "See solution",
    "solution": "The largest possible value of $\\lambda(n)$ is $\\frac{n(n+1)^2}{4}$.\n\nLet $a_1 = a_2 = \\dots = a_n = 1$. Then $\\lambda(n) \\le \\frac{n(n+1)^2}{4}$.\n\nWe show that for any real numbers $a_0, a_1, \\dots, a_n$ satisfying the given properties, the following holds:\n\n$$\n\\left(\\sum_{i=1}^{n} i a_i\\right)^2 \\geq \\frac{n(n+1)^2}{4} \\left(\\sum_{i=1}^{n} a_i^2\\right).\n$$\n\nFirst, note that\n$$\na_1 \\ge \\frac{a_2}{2} \\ge \\dots \\ge \\frac{a_n}{n}.\n$$\n\nBy assumption, $2i a_i \\ge i(a_{i+1} + a_{i-1})$ for $i = 1, \\dots, n-1$. For $1 \\le l \\le n-1$, summing over $i = 1$ to $l$ gives $(l+1)a_l \\ge l a_{l+1}$, i.e.\n$$\n\\frac{a_l}{l} \\ge \\frac{a_{l+1}}{l+1} \\text{ for } l = 1, \\dots, n-1.\n$$\n\nFor $i, j, k \\in \\{1, \\dots, n\\}$ with $i > j$,\n$$\n\\frac{2ik^2}{i+k} > \\frac{2jk^2}{j+k}.\n$$\nThis is equivalent to $(i-j)k^3 > 0$, which is true.\n\nNow, estimate the lower bound of $a_i a_j$ for $1 \\le i < j \\le n$:\n\nSince $\\frac{a_i}{i} \\ge \\frac{a_j}{j}$, $j a_i - i a_j \\ge 0$. Since $a_i - a_j \\le 0$, $(j a_i - i a_j)(a_j - a_i) \\ge 0$, i.e. $a_i a_j \\ge \\frac{i}{i+j} a_j^2 + \\frac{j}{i+j} a_i^2$.\n\nThus,\n$$\n\\begin{aligned}\n\\left(\\sum_{i=1}^{n} i a_i\\right)^2 &= \\sum_{i=1}^{n} i^2 a_i^2 + 2 \\sum_{1 \\le i < j \\le n} i j a_i a_j \\\\\n&\\ge \\sum_{i=1}^{n} i^2 a_i^2 + 2 \\sum_{1 \\le i < j \\le n} \\left( \\frac{i^2 j}{i+j} a_j^2 + \\frac{i j^2}{i+j} a_i^2 \\right) \\\\\n&= \\sum_{i=1}^{n} \\left( a_i^2 \\sum_{k=1}^{n} \\frac{2 i k^2}{i+k} \\right).\n\\end{aligned}\n$$\n\nLet $b_i = \\sum_{k=1}^{n} \\frac{2 i k^2}{i+k}$. Since $b_1 \\le b_2 \\le \\dots \\le b_n$ and $a_1^2 \\le \\dots \\le a_n^2$, by Chebyshev's inequality,\n$$\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$$\n\nHence,\n$$\n\\left(\\sum_{i=1}^{n} i a_i\\right)^2 \\ge \\frac{1}{n} \\left(\\sum_{i=1}^{n} a_i^2\\right) \\left(\\sum_{i=1}^{n} b_i\\right).\n$$\n\nNow,\n$$\n\\begin{aligned}\n\\sum_{i=1}^{n} b_i &= \\sum_{i=1}^{n} \\sum_{k=1}^{n} \\frac{2 i k^2}{i+k} \\\\\n&= \\sum_{i=1}^{n} i^2 + 2 \\sum_{1 \\le i < j \\le n} i j \\\\\n&= \\left( \\sum_{i=1}^{n} i \\right)^2 = \\frac{n^2 (n+1)^2}{4}.\n\\end{aligned}\n$$\n\nTherefore,\n$$\n\\left(\\sum_{i=1}^{n} i a_i\\right)^2 \\ge \\frac{n(n+1)^2}{4} \\sum_{i=1}^{n} a_i^2.\n$$\n\nThus, the maximum possible value of $\\lambda(n)$ is\n$$\n\\frac{n(n+1)^2}{4}.\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 14840,
    "subject": "Mathematics (Olympiad)",
    "question": "a) Rectangle $ABCD$ is partitioned into squares, each of which has integer perimeter. Is it true that $ABCD$ has integer perimeter?\n\nb) Square $ABCD$ is partitioned into squares, each of which has integer perimeter. Is it true that $ABCD$ has integer perimeter?",
    "options": [],
    "answer": "See solution",
    "solution": "a) We construct a counterexample. Consider two squares $ABMN$ and $NMCD$ with side length $\\frac{1}{4}$. Then, the perimeter of rectangle $ABCD$ is $2 \\cdot \\left(\\frac{1}{2} + \\frac{1}{4}\\right) = \\frac{3}{2}$—not an integer, while both squares have integer perimeter.\n\nb) Consider the side of the external square, and all squares that have one side belonging to the side of the external square (see the figure below). Let $a$ be the side of the external square, and the sides of the small squares be $a_1, a_2, \\ldots, a_n$. Then $a = a_1 + a_2 + \\ldots + a_n$, and each side $a_i$, after multiplying by $4$, is integer. Therefore, the perimeter $P$ of our square $ABCD$ is $P = 4a = 4a_1 + 4a_2 + \\ldots + 4a_n$—an integer.\n\n![](images/Ukrajina_2011_p6_data_c9e4a6cc7b.png)\n\nFig. 1",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 14841,
    "subject": "Mathematics (Olympiad)",
    "question": "Consider $n^2$ numbers chosen from $\\{0, 1, \\dots, n-1\\}$.\n\nFor an arrangement $a_1, a_2, \\dots, a_{n^2}$ of these numbers, define it to be *i*-safe if $\\sum_{j=1}^i a_j$ is not a multiple of $n$, and *i*-not-safe otherwise. An arrangement is *safe-up-to-i* if it is *k*-safe for each $k$, $1 \\leq k \\leq i$.\n\nIs it possible to arrange the $n^2$ numbers so that for every $i$ ($1 \\leq i \\leq n^2 - 1$), the sum $\\sum_{j=1}^i a_j$ is not a multiple of $n$?",
    "options": [],
    "answer": "See solution",
    "solution": "We construct the arrangement step by step:\n\nAt each stage $i$ ($1 \\leq i \\leq n^2 - 1$), the number that would make the arrangement *i*-not-safe is unique among $\\{0, 1, \\dots, n-1\\}$. If, after $(i-1)$ steps, there are at least two distinct numbers (other than 1) left, we can always choose a number to keep the arrangement *i*-safe. Repeating this, we eventually reach a stage where only $n$ 1's and $k$ copies of another number $a$ remain.\n\nWe then place $a$ in the next spot if it keeps the arrangement safe; otherwise, we place a 1. This process ensures that 1's are not placed consecutively while $a$'s remain. If only one $a$ remains, it goes in the last spot, and the arrangement is valid.\n\nIf only $\\ell$ 1's remain, the sum of all $n^2$ numbers is $n \\cdot \\frac{n(n-1)}{2}$, a multiple of $n$. The sum before the last $\\ell$ terms is $An - \\ell$, not a multiple of $n$, so $1 \\leq \\ell \\leq n-1$. Thus, placing the remaining 1's consecutively still keeps the arrangement safe-up-to-$(n^2-1)$.\n\nTherefore, such an arrangement exists.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 14842,
    "subject": "Mathematics (Olympiad)",
    "question": "For a positive integer $n \\ge 3$, plot $n$ equally spaced points around a circle. Label one of them $A$, and place a marker at $A$. One may move the marker forward in a clockwise direction to either the next point or the point after that. Hence there are a total of $2n$ distinct moves available; two from each point. Let $a_n$ count the number of ways to advance around the circle exactly twice, beginning and ending at $A$, without repeating a move. Prove that $a_{n-1} + a_n = 2^n$ for all $n \\ge 4$.\n\n![](path/to/file.png)",
    "options": [],
    "answer": "See solution",
    "solution": "We will show that $a_n = \\frac{1}{3}(2^{n+1} + (-1)^n)$. This would be sufficient, since then we would have\n\n$$\na_{n-1} + a_n = \\frac{1}{3}(2^n + (-1)^{n-1}) + \\frac{1}{3}(2^{n+1} + (-1)^n) = \\frac{1}{3}(2^n + 2 \\cdot 2^n) = 2^n.\n$$\n\n**Lemma 1.** For all positive integers $n$, we have\n\n$$\n\\sum_{k=0}^{\\lfloor n/2 \\rfloor} \\binom{n-k}{k} 2^k = \\frac{1}{3}(2^{n+1} + (-1)^n).\n$$\n\n*Proof.* We argue by strong induction. To begin, the cases $n = 1$ and $n = 2$ are quickly verified. Now suppose that $n \\ge 3$ is odd, say $n = 2m + 1$. We find that\n\n$$\n\\begin{aligned}\n\\sum_{k=0}^{m} \\binom{2m+1-k}{k} 2^k &= 1 + \\sum_{k=1}^{m} \\binom{2m-k}{k} 2^k + \\sum_{k=1}^{m} \\binom{2m-k}{k-1} 2^k \\\\\n&= \\sum_{k=0}^{m} \\binom{2m-k}{k} 2^k + 2 \\sum_{k=0}^{m-1} \\binom{2m-1-k}{k} 2^k \\\\\n&= \\frac{1}{3}(2^{2m+1} + 1) + \\frac{2}{3}(2^{2m} - 1) \\\\\n&= \\frac{1}{3}(2^{2m+2} - 1),\n\\end{aligned}\n$$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 14843,
    "subject": "Mathematics (Olympiad)",
    "question": "Determine all integer solutions $(x, y, z)$ of the equation $x^4 + x^2 = 7^z y^2$.",
    "options": [],
    "answer": "See solution",
    "solution": "Clearly, $(x, y) = (0, 0)$ is a solution for any integer $z$. We claim there are no other solutions.\n\nFirst, suppose $z$ is even. Then $7^z y^2$ is a perfect square, so $x^4 + x^2 = x^2(x^2 + 1)$ must also be a perfect square. The only integer $x$ for which both $x^2$ and $x^2 + 1$ are perfect squares is $x = 0$, so $y = 0$ as well.\n\nNow, suppose $z$ is positive and odd, say $z = 2c + 1$. Since $x^2(x^2 + 1)$ must be divisible by $7$, and $x^2 + 1$ can only be congruent to $1, 2, 3,$ or $5$ modulo $7$, $x$ must be divisible by $7$. Let $x = 7^a u$ and $y = 7^b v$ with $u, v$ not divisible by $7$. The equation becomes\n\n$$\n7^{2a}u^2(7^{2a}u^2 + 1) = 7^{2(b+c)+1}v^2,\n$$\n\nwhich is impossible, since the left side is divisible by an even power of $7$, while the right side is divisible by an odd power.\n\nFinally, suppose $z$ is negative, $z = -w$. Then $7^w x^2(x^2 + 1) = y^2$. If $w$ is even, as before, $x = y = 0$. If $w$ is odd, $w = 2c + 1$, and the same divisibility argument applies, leading to a contradiction.\n\nThus, the only integer solutions are $(x, y, z) = (0, 0, z)$ for any integer $z$.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 14844,
    "subject": "Mathematics (Olympiad)",
    "question": "Let the function $f(x, y)$ be defined recursively by:\n\n$$\nf(0, y) = y + 1,\n$$\n\n$$\nf(x + 1, 0) = f(x, 1),\n$$\n\n$$\nf(x + 1, y + 1) = f(x, f(x + 1, y)).\n$$\n\n(a) Find $f(3, 2005)$.\n\n(b) Let $g(1) = 2$ and $g(n + 1) = 2^{g(n)}$ for $n \\in \\mathbb{Z}^+$. Find $f(4, 2005)$ in terms of $g$.",
    "options": [],
    "answer": "See solution",
    "solution": "For part (a):\n\nBy induction, we find:\n\n$$\nf(1, y) = y + 2,\n$$\n\n$$\nf(2, y) = 2y + 3,\n$$\n\nand\n\n$$\nf(3, y) = 2^{y+3} - 3.\n$$\n\nThus,\n\n$$\nf(3, 2005) = 2^{2008} - 3.\n$$\n\nFor part (b):\n\nWe define $g(1) = 2$ and $g(n + 1) = 2^{g(n)}$ for $n \\in \\mathbb{Z}^+$. By induction,\n\n$$\nf(4, y) = g(y + 3) - 3.\n$$\n\nTherefore,\n\n$$\nf(4, 2005) = g(2008) - 3.\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 14845,
    "subject": "Mathematics (Olympiad)",
    "question": "Solve the equation\n\n$$\n3^n + 4^m = 5^k\n$$\n\nwhere $n$, $m$, and $k$ are nonnegative integers.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $n \\geq 1$, $m \\geq 1$.\n\nFirst, consider the equation modulo $3$:\n$4^m \\equiv 1 \\pmod{3}$. Thus, if $n \\geq 1$, $3^n + 4^m \\equiv 1 \\pmod{3}$. Therefore, $5^k \\equiv 1 \\pmod{3}$, which implies $k = 2k_1$ for some integer $k_1$.\n\nNow, consider the equation modulo $4$:\n$5^k \\equiv 1 \\pmod{4}$, so $5^k - 4^m \\equiv 1 \\pmod{4}$. Therefore, $3^n \\equiv 1 \\pmod{4}$, which implies $n = 2n_1$ for some integer $n_1$.\n\n![fig.2](images/Ukrajina_2008_p3_data_b9f47cef71.png)\n\nThus, the equation becomes:\n$$\n3^{2n_1} + 4^m = 5^{2k_1}\n$$\nwhich can be rewritten as\n$$\n3^{2n_1} = (5^{k_1} - 2^m)(5^{k_1} + 2^m)\n$$\nThis means that\n$$\n\\begin{cases}\n5^{k_1} - 2^m = 3^p \\\\\n5^{k_1} + 2^m = 3^s\n\\end{cases}\n$$\nwhere $0 \\leq p < s$, $p + s = 2n_1$.\n\nAdding the two equations:\n$$\n2 \\cdot 5^{k_1} = 3^p (1 + 3^{s-p})\n$$\nTherefore, $p = 0$. So $s = 2n_1$ and the system becomes\n$$\n\\begin{cases}\n5^{k_1} - 2^m = 1 \\\\\n5^{k_1} + 2^m = 3^{2n_1}\n\\end{cases}\n$$\nSubtracting the two equations:\n$$\n2^{m+1} = 3^{2n_1} - 1 = (3^{n_1} - 1)(3^{n_1} + 1)\n$$\nSo\n$$\n\\begin{cases}\n3^{n_1} - 1 = 2^q \\\\\n3^{n_1} + 1 = 2^t\n\\end{cases}\n$$\nwhere $0 \\leq q < t$, $q + t = 2n_1$. Since $(3^{n_1} - 1, 3^{n_1} + 1) = (2, 1)$, this implies $q = 1$ and $3^{n_1} - 1 = 2 \\Rightarrow n_1 = 1$.\n\nNow the system is\n$$\n\\begin{cases}\n5^{k_1} - 2^m = 1 \\\\\n5^{k_1} + 2^m = 9\n\\end{cases}\n$$\nEnumerating possible values, we find the only possible solution is $k_1 = 1$, $m = 2$. Thus, the solution is $(n, m, k) = (2n_1, m, 2k_1) = (2, 2, 2)$.\n\nNow, consider the cases where $m = 0$, $n = 0$, or $k = 0$:\n\n- If $m = 0$, then $3^n + 1 = 5^k$. But $3^n$ and $5^k$ are both odd for $n, k \\geq 1$, so there is no solution.\n- If $n = 0$, then $1 + 4^m = 5^k$. Consider this equation modulo $3$: $1 + 4^m \\equiv 2 \\pmod{3}$, so $5^k \\equiv 2 \\pmod{3}$, which implies $k = 2k_2 + 1$. Now, modulo $8$: if $m \\geq 2$, $1 + 4^m \\equiv 1 \\pmod{8}$, but $5^{2k_2+1} = 5 \\cdot 25^{k_2} \\equiv 5 \\pmod{8}$, a contradiction. Thus, $m \\leq 1$.\n- For $m = 1$, $1 + 4 = 5^k$, so $k = 1$ and $(n, m, k) = (0, 1, 1)$ is a solution.\n\nTherefore, the solutions in nonnegative integers are:\n\n$$\n(n, m, k) = (2, 2, 2) \\quad \\text{and} \\quad (0, 1, 1)\n$$",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 14846,
    "subject": "Mathematics (Olympiad)",
    "question": "Determine all pairs $(a, p)$ of positive integers with $p$ prime such that $p^a + a^4$ is a perfect square.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $p^a + a^4 = b^2$ for some positive integer $b$. Then:\n\n$$\np^a = b^2 - a^4 = (b + a^2)(b - a^2)\n$$\n\nHence, both $b + a^2$ and $b - a^2$ are powers of $p$. Let $b - a^2 = p^x$ for some integer $x$. Then $b + a^2 = p^{a-x}$ and $a - x > x$. Therefore,\n\n$$\n2a^2 = (b + a^2) - (b - a^2) = p^{a-x} - p^x = p^x (p^{a-2x} - 1). \\quad (\\dagger)\n$$\n\nConsider two cases:\n\n**Case 1:** $p = 2$\n\n$$\na^2 = 2^{x-1} (2^{a-2x} - 1) = 2^{2v_2(a)} (2^{a-2x} - 1)\n$$\n\nwhere the first equality comes from $(\\dagger)$ and the second from $\\gcd(2, 2^{a-2x} - 1) = 1$. So, $2^{a-2x} - 1$ is a square. If $v_2(a) > 0$, then $2^{a-2x}$ is also a square, so $2^{a-2x} - 1 = 0$, and $a = 0$, a contradiction. If $v_2(a) = 0$, then $x = 1$, and $a^2 = 2^{a-2} - 1$. For $a \\ge 4$, the right side is congruent to $3$ modulo $4$, so cannot be a square. For $a = 1, 2, 3$, none satisfy the condition. Thus, no solutions in this case.\n\n**Case 2:** $p \\neq 2$\n\nHere, $2v_p(a) = x$. Let $m = v_p(a)$. Then $a^2 = p^{2m} n^2$ for some integer $n \\ge 1$. So,\n\n$$\n2n^2 = p^{a-2x} - 1 = p^{p^m n - 4m} - 1\n$$\n\nConsider two subcases:\n\n**Subcase 2-1:** $p \\ge 5$\n\nBy induction, $p^m \\ge 5^m > 4m$ for all $m$. Then,\n\n$$\n2n^2 + 1 = p^{p^m n - 4m} > p^{p^m n - p^m} \\ge 5^{5^{m(n-1)}} \\ge 5^{n-1}\n$$\n\nBut $5^{n-1} > 2n^2 + 1$ for all $n \\ge 3$. Thus, $n = 1$ or $2$. If $n = 1$ or $2$, then $p = 3$, a contradiction. So, no solutions in this subcase.\n\n**Subcase 2-2:** $p = 3$\n\nThen $2n^2 + 1 = 3^{3m n - 4m}$. If $m \\ge 2$, $3^m > 4m$. Then,\n\n$$\n2n^2 + 1 = 3^{3m n - 4m} > 3^{3m(n-1)} \\ge 3^{9(n-1)}\n$$\n\nBut $3^{9(n-1)} > 2n^2 + 1$ for all $n \\ge 2$. Thus, $n = 1$. Then $2 \\cdot 1^2 + 1 = 3^{3m-4m}$, so $3 = 3^{3m-4m}$, i.e., $3^m - 4m = 1$. The only solution is $m = 2$, so $a = 3^2 \\cdot 1 = 9$.\n\nIf $m \\le 1$:\n- $m = 1$: $2n^2 + 1 = 3^{3n-4}$. For $n \\ge 3$, $3^{3n-4} > 2n^2 + 1$. For $n = 1, 2$, only $n = 2$ works, so $a = 3^1 \\cdot 2 = 6$.\n- $m = 0$: $a = 2$.\n\nTherefore, the solutions are $(a, p) = (1, 3), (2, 3), (6, 3), (9, 3)$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 14847,
    "subject": "Mathematics (Olympiad)",
    "question": "Consider a triangle $ABC$ with $\\angle BAC = 120^\\circ$ and the isosceles triangles $PAB$ and $NAC$ such that $\\angle APB = \\angle ANC = \\angle BAC$, with line $AB$ separating points $P$ and $C$, and line $AC$ separating points $N$ and $B$. Prove that, if $G$ is the centroid of triangle $ABC$, then $GP = GN = \\frac{AB+AC}{3}$.",
    "options": [],
    "answer": "See solution",
    "solution": "Triangles $PAB$ and $NAC$ are isosceles and $\\angle APB = \\angle ANC = 120^\\circ$. It follows that $\\angle PAB = \\angle NAC = 30^\\circ$ and, since $\\angle BAC = 120^\\circ$, points $P$, $A$, $N$ are collinear.\n\nLet $PF \\parallel AC$, $F \\in AB$ and $NE \\parallel AB$, $E \\in AC$. We have $\\angle APF = \\angle NAC = \\angle FAP$, therefore the triangle $FAP$ is isosceles, with $FA = FP$, (1).\n\n![](images/RMC_2024_p35_data_1962018d51.png)\n\nOn the other hand, from $\\angle APB = 120^\\circ$ and $\\angle APF = 30^\\circ$, it follows that $\\angle FPB = 90^\\circ$, hence the triangle $PBF$ is a $30^\\circ - 60^\\circ - 90^\\circ$ triangle, from which $BF = 2FP$, (2). Using (1) and (2) we conclude $BF = 2FA$, (3).\n\nDenote $BB'$ the median from $B$ of the triangle $ABC$. Since $G$ is the centroid of the triangle $ABC$, we have $BG = 2GB'$ (4).\n\nThe relations (3) and (4) lead, according to the converse of Thales' theorem, to $FG \\parallel AC$, thus points $P$, $F$ and $G$ are collinear. We obtain $\\angle GPN = 30^\\circ$. Similarly, it follows that $\\angle GNP = 30^\\circ$ hence the triangle $GNP$ is isosceles, with $GP = GN$.\n\nSince $AFGE$ is a parallelogram, $GP = GF + FP = AE + FP = \\frac{AB+AC}{3}$, from which the conclusion follows.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 14848,
    "subject": "Mathematics (Olympiad)",
    "question": "Show that there are infinitely many positive integers $n$ such that $n^2 + 1$ has two positive divisors whose difference is $n$.",
    "options": [],
    "answer": "See solution",
    "solution": "Define the sequence $(a_k)_{k \\ge 0}$ by $a_0 = 1$, $a_1 = 2$, and for $k = 0, 1, 2, \\dots$,\n$$\na_{k+2} a_k = a_{k+1}^2 + 1.\n$$\nInductively, all $a_k$ are positive integers, and the sequence $n_k = a_{k+1} - a_k$ is strictly increasing and consists of positive integers. Moreover, both $a_k$ and $a_{k+1}$ divide $n_k^2 + 1$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 14849,
    "subject": "Mathematics (Olympiad)",
    "question": "![Diagram](images/IRL_ABooklet_2022_p30_data_3ac55e7db4.png)\n\nLet $X_0 = O$ and, for $0 \\leq n \\leq 2022$, let $d_n = |OX_n|$. Given $d_0 = 0$ and, for $n \\geq 1$,\n\n$$\nd_{2n-1}^2 + d_{2n}^2 = (2n)^2\n$$\n\nand\n\n$$\nd_{2n-1}^2 + d_{2n-2}^2 = (2n-1)^2.\n$$\n\nFind $|X_{2020}X_{2022}|$.",
    "options": [],
    "answer": "See solution",
    "solution": "Subtracting the two equations:\n\n$$\nd_{2n}^2 - d_{2n-2}^2 = (2n)^2 - (2n-1)^2 = 4n - 1.\n$$\n\nSince $d_0^2 = 0$, we have\n\n$$\nd_{2n}^2 = \\sum_{k=1}^{n} (d_{2k}^2 - d_{2k-2}^2) = \\sum_{k=1}^{n} (4k - 1) = 4 \\sum_{k=1}^{n} k - \\sum_{k=1}^{n} 1 = 2n(n+1) - n = n(2n+1).\n$$\n\nThus,\n\n$$\nd_{2020}^2 = 1010 \\cdot 2021 = 2041210\n$$\n\n$$\nd_{2022}^2 = 1011 \\cdot 2023 = 2045253\n$$\n\nSo,\n\n$$\n|X_{2020}X_{2022}| = \\sqrt{2045253} - \\sqrt{2041210}.\n$$\n\n*Remark.* The answer is very close to $\\sqrt{2}$, the two numbers agreeing to 6 decimal places. To see why, expand using the binomial theorem:\n\n$$\n\\begin{aligned}\nd_{2n} &= \\sqrt{2n^2 + n} = \\sqrt{2n^2} \\left(1 + \\frac{1}{2n}\\right)^{1/2} \\\\\n&= \\sqrt{2} n \\left(1 + \\frac{1}{4n} - \\frac{1}{32n^2} + O(n^{-3})\\right) \\\\\n&= \\sqrt{2} \\left(n + \\frac{1}{4} - \\frac{1}{32n} + O(n^{-2})\\right)\n\\end{aligned}\n$$\n\nSo, for large $n$,\n\n$$\nd_{2n} - d_{2n-2} = \\sqrt{2} \\left(1 + \\frac{1}{32n(n-1)} + O(n^{-2})\\right) = \\sqrt{2} + O(n^{-2}) \\approx \\sqrt{2}.\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 14850,
    "subject": "Mathematics (Olympiad)",
    "question": "Denote by $p(a)$ the first digit of the natural number $a$. Show that each of the sets\n\n$$\nA = \\{n \\in \\mathbb{N} \\mid p(5^n) - p(2^n) > 0\\}, \\quad B = \\{n \\in \\mathbb{N} \\mid p(5^n) - p(2^n) < 0\\}\n$$\nhas infinitely many elements.",
    "options": [],
    "answer": "See solution",
    "solution": "For $k \\in \\mathbb{N}^*$, there exists $n_k \\in \\mathbb{N}^*$ such that $2^{n_k} < 10^k < 2^{n_k+1}$ (where $n_k + 1$ is the smallest $m \\in \\mathbb{N}^*$ with $10^k < 2^m$). Since $2^{n_k} < 10^k < 2^{n_k+1}$, we have $10^k < 2^{n_k+1} < 2 \\cdot 10^k$, so $p(2^{n_k+1}) = 1$.\n\nMultiplying by $5^{n_k+1}$, we get $10^k \\cdot 5^{n_k+1} < 10^{n_k+1} < 2 \\cdot 10^k \\cdot 5^{n_k+1}$. Dividing the first inequality by $10^k$ and the second by $2 \\cdot 10^k$, we obtain $5 \\cdot 10^{n_k-k} < 5^{n_k+1} < 10^{n_k-k+1}$, so $p(5^{n_k+1}) \\ge 5 > p(2^{n_k+1})$.\n\nFrom $2^{n_k} < 10^k < 2^{n_k+1}$, dividing the second inequality by $2$ gives $5 \\cdot 10^{k-1} < 2^{n_k} < 10^k$, so $p(2^{n_k}) \\ge 5$. Multiplying the first inequality by $5^{n_k}$, we get $10^{k-1} \\cdot 5^{n_k+1} < 10^{n_k} < 10^k \\cdot 5^{n_k}$. Dividing the first by $5 \\cdot 10^{k-1}$ and the second by $10^k$, we have $10^{n_k-k} < 5^{n_k} < 2 \\cdot 10^{n_k-k}$, so $p(5^{n_k}) = 1 < p(2^{n_k})$.\n\nThe set $\\{n_k \\mid k \\in \\mathbb{N}^*\\}$ is infinite because $n_k > k$. Since $\\{n_k \\mid k \\in \\mathbb{N}^*\\} \\subset A$ and $\\{n_k + 1 \\mid k \\in \\mathbb{N}^*\\} \\subset B$, the sets $A$ and $B$ are both infinite.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 14851,
    "subject": "Mathematics (Olympiad)",
    "question": "Prove that there exist integers $x_1, x_2, \\dots, x_{10}, y_1, y_2, \\dots, y_{10}$ satisfying:\n\n1. $|x_i| \\le 10^{10}$, $|y_i| \\le 10^{10}$ for $i = 1, 2, \\dots, 10$;\n2. The point set in the plane\n\n$$\nX = \\left\\{ \\left( \\sum_{i=1}^{10} a_i x_i, \\sum_{i=1}^{10} a_i y_i \\right) \\mid a_1, a_2, \\dots, a_{10} \\in \\{0, 1\\} \\right\\}\n$$\n\ncontains exactly 1024 distinct points;\n3. For any two parallel lines in the plane at distance 1, the strip between them (including the lines) contains at most two points from $X$.",
    "options": [],
    "answer": "See solution",
    "solution": "Take $x_i = 3^i$, $y_i = 9^i$ for $i = 1, 2, \\dots, 10$. We show these satisfy the conditions.\n\n(1) and (2) are obvious. For (3), we use:\n\n*Lemma:* If real numbers $z_1, \\dots, z_m$ satisfy $|z_{i+1}| \\ge 2|z_i|$ for $i = 1, 2, \\dots, m-1$, then for any $r_1, \\dots, r_m \\in \\{-1, 0, 1\\}$ not all zero, $\\left|\\sum_{i=1}^m r_i z_i\\right| \\ge |z_1|$.\n\nIf our construction fails (3), some strip $\\{(x, y) \\mid c \\le y - kx \\le c + \\sqrt{k^2+1}\\}$ contains at least 3 points. Let these be:\n\n$$\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$$\n\nThen $\\left|\\sum(u_i - v_i)(9^i - k3^i)\\right| \\le \\sqrt{k^2+1}$, etc.\n\nIf $k \\le 7$, assume $u_1 = v_1$. Since $9^{i+1} - k3^{i+1} > 2(9^i - k3^i)$ for $i \\ge 2$ and $9^2 - k3^2 > \\sqrt{k^2+1}$, contradiction!\n\nIf $k > 7 \\cdot 3^8$, assume $u_{10} = v_{10}$. Since $k3^{i+1} - 9^{i+1} > 2(k3^i - 9^i)$ for $i \\le 8$ and $3k - 9 > \\sqrt{k^2+1}$, contradiction!\n\nFor $7 < k \\le 7 \\cdot 3^8$, let $7 \\cdot 3^d < k \\le 7 \\cdot 3^{d+1}$ ($0 \\le d \\le 7$) and assume $u_{d+2} = v_{d+2}$. We have:\n\n$$\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$$\n\nagain a contradiction. Thus the construction works. $\\Box$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 14852,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $n \\ge 3$ and $a_1, a_2, \\ldots, a_n$ be positive real numbers such that\n$$\n\\frac{1}{1+a_1^4} + \\frac{1}{1+a_2^4} + \\cdots + \\frac{1}{1+a_n^4} = 1.\n$$\nProve that\n$$\na_1 a_2 \\cdots a_n \\ge (n-1)^{n/4}.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "Let $a_i^2 = \\tan^2 x_i$, where $x_i \\in [0, \\frac{\\pi}{2}]$ for $i=1,2,\\ldots,n$. Then $\\sum_{i=1}^{n} \\cos^2 x_i = 1$.\n\nFrom the inequality between the arithmetic and geometric means,\n$$\n\\sin^2 x_i = 1 - \\cos^2 x_i \\ge (n-1) \\left( \\prod_{j=1, j \\ne i}^{n} \\cos x_j \\right)^{2/(n-1)}, \\quad i=1,2,\\ldots,n.\n$$\nMultiplying these $n$ inequalities gives\n$$\n\\prod_{i=1}^{n} \\sin^2 x_i \\ge (n-1)^n \\prod_{i=1}^{n} \\cos^2 x_i.\n$$\nThis is equivalent to\n$$\n\\prod_{i=1}^{n} \\tan x_i \\ge (n-1)^{n/2}.\n$$\nFinally,\n$$\n\\prod_{i=1}^{n} a_i = \\left( \\prod_{i=1}^{n} \\tan x_i \\right)^{1/2} \\ge (n-1)^{n/4},\n$$\nwhich was to be proven.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 14853,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $p_1, p_2, p_3, \\dots$ be all prime numbers in increasing order. Prove that $p_2 + p_4 + \\dots + p_{2n} > 3n^2 - 2n + 1$ for every positive integer $n$.",
    "options": [],
    "answer": "See solution",
    "solution": "For any positive integer $k$, six consecutive integers $6k, 6k+1, 6k+2, 6k+3, 6k+4, 6k+5$ can contain at most two prime numbers. Hence, $p_i \\geq p_{i-2} + 6$ for all $i \\geq 5$, implying $p_{2i} \\geq p_4 + 6(i-2) = 6i - 5$ for all $i \\geq 2$. Thus,\n\n$$\np_2 + p_4 + \\dots + p_{2n} > 2 + (7 + 13 + \\dots + (6n - 5)) = 3n^2 - 2n + 1.\n$$",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 14854,
    "subject": "Mathematics (Olympiad)",
    "question": "在銳角三角形 $ABC$ 中，點 $H$ 為由頂點 $A$ 所引的高的垂足。設 $P$ 為平面上的動點，滿足：$\\angle PBC$ 與 $\\angle PCB$ 的內角平分線，分別記為 $k$ 與 $\\ell$，此兩條角平分線的交點位於 $AH$ 線段上。設 $k$ 與 $AC$ 交於點 $E$；$\\ell$ 與 $AB$ 交於點 $F$；而設 $EF$ 與 $AH$ 交於點 $Q$。證明：不論 $P$ 如何移動（但滿足前述條件），直線 $PQ$ 恆過某一定點。\n\n![](images/2023-TWNIMO-Problems_p65_data_f04164d7ef.png)",
    "options": [],
    "answer": "See solution",
    "solution": "設直線 $BC$ 分別關於 $AB, AC$ 的對稱線交於點 $K$。以下證明 $P, Q, K$ 共線，也就是說，直線 $PQ$ 恆過定點 $K$。\n\n設直線 $BE$ 與 $CF$ 交於點 $I$。對任意點 $O$ 及實數 $d > 0$，將以 $O$ 為圓心、$d$ 為半徑的圓記為 $(O, d)$。定義兩圓 $\\omega_I = (I, IH)$ 及 $\\omega_A = (A, AH)$。再設三角形 $KBC$ 的內切圓為 $\\omega_K$，三角形 $PBC$ 的 $P$-旁切圓為 $\\omega_P$。\n\n由於 $IH \\perp BC$ 及 $AH \\perp BC$，兩圓 $\\omega_I$ 和 $\\omega_A$ 在 $H$ 點相切。所以 $H$ 為 $\\omega_I$ 和 $\\omega_A$ 的外位似中心。由完全四邊形 $BCEF$ 得 $(A, I; Q, H) = -1$，所以 $Q$ 是 $\\omega_I$ 和 $\\omega_A$ 的內位似中心。\n\n因為 $BA$ 和 $CA$ 分別是 $\\angle KBC$ 與 $\\angle KCB$ 的外角平分線，$\\omega_A$ 是三角形 $BKC$ 的 $K$-旁切圓。於是 $K$ 為圓 $\\omega_A$ 和 $\\omega_K$ 的外位似中心。同時明顯有 $P$ 為圓 $\\omega_I$ 和 $\\omega_P$ 的外位似中心。\n\n令點 $T$ 為直線 $BC$ 與圓 $\\omega_P$ 的切點，點 $T'$ 為直線 $BC$ 與圓 $\\omega_K$ 的切點。因為 $\\omega_I$ 與 $\\omega_P$ 分別是三角形 $PBC$ 的內切圓與 $P$-旁切圓，得 $TC = BH$。又因為 $\\omega_K$ 與 $\\omega_A$ 分別是三角形 $KBC$ 的內切圓與 $K$-旁切圓，得 $T'C = BH$。故得 $TC = T'C$，即 $T = T'$。由此推得圓 $\\omega_K$ 與 $\\omega_P$ 在 $T$ 點相切。\n\n令點 $S$ 為 $\\omega_A$ 與 $\\omega_P$ 的內位似中心，而點 $S'$ 為 $\\omega_I$ 與 $\\omega_K$ 的內位似中心。明顯有 $S, S'$ 位於直線 $BC$ 上。令 $r_A, r_I, r_K, r_P$ 分別為圓 $\\omega_A, \\omega_I, \\omega_P, \\omega_K$ 的半徑。熟知若三角形的半周長為 $s = (a + b + c)/2$，$r, r_a$ 分別是內切圓半徑與 $a$-旁切圓半徑，有 $r \\cdot r_a = (s-b)(s-c)$。將此關係套用在三角形 $PBC$ 上，得 $r_I \\cdot r_P = BH \\cdot CH$。又套用到三角形 $KCB$ 上，得 $r_K \\cdot r_A = CT \\cdot BT$。由於 $BH = CT$ 及 $BT = CH$，可知\n\n$$\n\\frac{HS}{ST} = \\frac{r_A}{r_P} = \\frac{r_I}{r_K} = \\frac{HS'}{S'T},\n$$\n\n故得 $S = S'$。\n\n最後，將廣義 Monge 定理套用在圓 $\\omega_A, \\omega_I, \\omega_K$ 上（有兩對內公切線及一對外公切線），可得 $Q, S, K$ 共線。同理可證 $Q, S, P$ 共線，故得證 $P, Q, K$ 共線。$\\square$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 14855,
    "subject": "Mathematics (Olympiad)",
    "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": "See solution",
    "solution": "Denote the card with $i, j$ written on it as $\\{i, j\\}$. It is easy to see that these 10 cards are exactly $\\{i, j\\}$ for $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 can be placed in box 1 are $\\{1, 2\\}$, $\\{1, 3\\}$, $\\{1, 4\\}$, and $\\{1, 5\\}$.\n\n*Case 1:* All 4 of these cards are placed in box 1. In this case, it is not possible for any other box to have 4 cards, so no matter how the remaining 6 cards are placed, the requirement is satisfied. There are $2^6 = 64$ “good” placements in this case.\n\n*Case 2:* Exactly 3 of the 4 cards are in box 1, and the remaining card is in its other possible box. Each box other than box 1 then contains at most 2 cards.\n\nConsider the number $N$ of placements where $\\{1, 2\\}$, $\\{1, 3\\}$, $\\{1, 4\\}$ are in box 1 and $\\{1, 5\\}$ is in box 5.\n\nThere are 8 possible ways to place the cards $\\{2, 3\\}$, $\\{2, 4\\}$, $\\{3, 4\\}$: 6 ways where two cards are placed in one of the boxes 2, 3, 4, and 2 ways where one card is placed in each of boxes 2, 3, 4.\n\nIf two of $\\{2, 3\\}$, $\\{2, 4\\}$, $\\{3, 4\\}$ are in the same box (say, $\\{2, 3\\}$ and $\\{2, 4\\}$ in box 2), then $\\{2, 5\\}$ must be in box 5. Box 5 then has $\\{1, 5\\}$ and $\\{2, 5\\}$, so $\\{3, 5\\}$ and $\\{4, 5\\}$ must go to boxes 3 and 4, respectively. Thus, the placement of $\\{2, 5\\}$, $\\{3, 5\\}$, and $\\{4, 5\\}$ is unique.\n\nIf one of $\\{2, 3\\}$, $\\{2, 4\\}$, $\\{3, 4\\}$ is placed in each of boxes 2, 3, 4, then there are at most 2 cards in each of boxes 2, 3, 4. We must ensure that there are no more than 2 cards in box 5, i.e., there are 0 or 1 of $\\{2, 5\\}$, $\\{3, 5\\}$, $\\{4, 5\\}$ in box 5. The number of such placements is $\\binom{3}{0} + \\binom{3}{1} = 4$.\n\nThus, $N = 6 \\times 1 + 2 \\times 4 = 14$. By symmetry, there are $4N = 56$ “good” placements in *Case 2*.\n\nIn total, there are $64 + 56 = 120$ “good” placements.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 14856,
    "subject": "Mathematics (Olympiad)",
    "question": "За сколько ходов можно с гарантией найти монетку под одним из 100 наперстков, если после каждого хода и перемещения монетки все наперстки (вместе с монеткой) поворачиваются по ходу часовой стрелки на одну позицию? После каждого хода монетка либо остается на месте, либо перемещается на две позиции по часовой стрелке, а наперстки остаются на своих местах.",
    "options": [],
    "answer": "See solution",
    "solution": "Покрасим все наперстки поочередно в белый и черный цвет, и пронумеруем наперстки каждого цвета по порядку против часовой стрелки числами от 0 до 49. Понятно, что цвет наперстка, под которым лежит монетка, не изменяется, а номер либо не изменяется, либо уменьшается на 1 по модулю 50.\n\nПокажем, как найти монетку за 33 хода. Описывая алгоритм, предполагаем, что монетка не обнаружена на всех ходах вплоть до 33-го (в противном случае все уже сделано менее чем за 33 хода).\n\nПервым ходом поднимем черные наперстки с номерами 0, 1, 2, 3. Тогда после перемещения монетка не сможет оказаться под черными наперстками 0, 1, 2. Вторым ходом поднимем черные наперстки 3, 4, 5, 6. Тогда после перемещения монетка не сможет оказаться под черными наперстками 0, 1, \\ldots, 5. Действуем так далее: при $s = 1, 2, \\ldots, 16$ ходом номер $s$ поднимем черные наперстки с номерами $3s - 3, 3s - 2, 3s - 1, 3s$. Тогда после перемещения монетка не сможет оказаться под черными наперстками 0, 1, \\ldots, 3s - 1$.\n\nСемнадцатым ходом поднимем черные наперстки 48 и 49, а также белые наперстки 49 и 0. Теперь мы знаем, что под черными наперстками нет монетки, а также что после перемещения монетка не сможет оказаться под белым наперстком 49. При $s = 1, 2, \\ldots, 15$ ходом номер $17 + s$ поднимем белые наперстки $3s - 3, 3s - 2, 3s - 1, 3s$. Тогда после перемещения монетка не сможет оказаться под белыми наперстками 49, 0, 1, \\ldots, 3s - 1$. Наконец, последним 33-м ходом поднимаем белые наперстки 45, 46, 47, 48; под одним из них обязана быть монетка.\n\nДокажем, что с гарантией обнаружить монету за 32 хода невозможно. Обозначим через $B_k$ множество из четырех наперстков, поднимаемых на $k$-м ходе, а через $A_k$ — множество наперстков, про которые перед выполнением $k$-го хода (после возможного перемещения монетки на $(k-1)$-м ходе) точно известно, что под ними нет монетки. Предполагаем, что пока возможно, под наперстками из $B_k$ нет монетки.\n\nЯсно, что $A_{k+1} \\subset A_k \\cup B_k$ при $k = 1, 2, \\ldots, n-1$, откуда $|A_{k+1}| \\leq |A_k| + 4$. Более того, если множество $A_k \\cup B_k$ не совпадает с множеством всех наперстков, то $|A_{k+1}| \\leq |A_k| + 3$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 14857,
    "subject": "Mathematics (Olympiad)",
    "question": "Each $a_i$ represents the airtime of a ball thrown at time $i$, and $b$ is the total number of balls. The condition $a_i + i \\neq a_j + j$ ensures that no two balls land at the same time. Given $a_i \\leq 2015$ for all $i$, prove that $|\\sum_{i=m+1}^{n} (a_i - b)| \\leq 1007^2$ for any $m < n$.",
    "options": [],
    "answer": "See solution",
    "solution": "**Solution**\n\nLet $S$ be the set of positive integers not of the form $n + a_n$ for any $n$. Let $s_1 < s_2 < \\dots$ be the elements of $S$ in increasing order, and $b = |S|$.\n\n**Lemma:** $|S| \\leq 2015$.\n\n*Proof:* Assume $|S| \\geq 2016$. Choose $n$ so that $a_n + n \\geq s_{2016}$. Since $a_n \\leq 2015$, $s_1, \\dots, s_{2016} \\in \\{1, \\dots, n + 2015\\}$. The $n$ numbers $1 + a_1, \\dots, n + a_n$ are also in $\\{1, \\dots, n + 2015\\}$ and distinct from the $s_i$, so $\\{1, \\dots, n + 2015\\}$ would contain at least $n + 2016$ elements, a contradiction. $\\square$\n\nLet $N$ be larger than all $s_j$. For $n \\geq N$, consider the list $L$ of $n + b$ distinct positive integers:\n\n$$\n1 + a_1, 2 + a_2, \\dots, n + a_n, s_1, \\dots, s_b\n$$\n\nSince these are distinct, their sum is at least $\\sum_{j=1}^{n+b} j$. Thus,\n\n$$\n\\sum_{j=1}^{n} (j + a_j) + \\sum_{j=1}^{b} s_j \\geq \\sum_{j=1}^{n+b} j\n$$\n\nwhich gives\n\n$$\n\\sum_{j=1}^{n} (a_j - b) \\geq \\frac{b^2 + b}{2} - s,\n$$\nwhere $s = \\sum_{j=1}^b s_j$.\n\nFor the upper bound, $s_1, \\dots, s_b$ are in $\\{1, \\dots, n+1\\}$. The remaining $n+1-b$ elements of $\\{1, \\dots, n+1\\}$ are of the form $j + a_j$. The sum of these is $\\sum_{j=1}^{n+1} j - \\sum_{j=1}^b s_j$. The remaining $b-1$ numbers of the form $j + a_j$ are at most $n+2015, n+2014, \\dots, n+2015-b+2$, so their sum is at most $\\sum_{j=1}^{b-1}(n+2016-j)$. Thus,\n\n$$\n\\sum_{j=1}^{n} (a_j - b) \\leq \\frac{4033b - b^2 - 4030}{2} - s.\n$$\n\nTherefore, for $n \\geq N$,\n\n$$\n\\frac{b^2 + b}{2} - s \\leq \\sum_{j=1}^{n} (a_j - b) \\leq \\frac{4033b - b^2 - 4030}{2} - s.\n$$\n\nFor $n > m \\geq N$,\n\n$$\n\\left| \\sum_{j=m+1}^{n} (a_j - b) \\right| \\leq (b-1)(2015-b) \\leq 1007^2.\n$$\n$\\square$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 14858,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $ABC$ be an acute triangle with altitudes $BD$, $CE$. Given that $AE \\cdot AD = BE \\cdot CD$, what is the smallest possible measure of $\\angle BAC$?\n",
    "options": [],
    "answer": "See solution",
    "solution": "Let $\\angle BAC = \\alpha$ and denote the side lengths $b = AC$, $c = AB$.\n\nExpress $AE$, $AD$, $BE$, $CD$ in terms of $\\cos \\alpha$, $b$, and $c$. The given condition becomes:\n\n$$\nb \\cos \\alpha \\cdot c \\cos \\alpha = (c - b \\cos \\alpha)(b - c \\cos \\alpha)\n$$\n\nThis simplifies to:\n\n$$\nbc = (b^2 + c^2) \\cos \\alpha\n$$\n\nSo,\n\n$$\n\\cos \\alpha = \\frac{bc}{b^2 + c^2}\n$$\n\nSince $b, c > 0$, we have $\\cos \\alpha \\le \\frac{1}{2}$, because $(b-c)^2 \\ge 0$ implies $b^2 + c^2 \\ge 2bc$.\n\nTherefore, $\\angle BAC \\ge 60^\\circ$.\n\nFor an equilateral triangle, the condition holds (since $AE = AD = BE = CD$), so the smallest possible measure is $60^\\circ$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 14859,
    "subject": "Mathematics (Olympiad)",
    "question": "For each positive integer $n$, consider a $3n \\times 3n$ chessboard, with its squares colored black and white as follows: starting from the top left corner, every third diagonal is black and all other squares are white, as in the figure.\n\n![](images/Greece_2022_p13_data_0050ec1862.png)\n\nAt each move, we select a $2 \\times 2$ square and change the color of its squares as follows: white squares become orange, oranges (if any) become black, and blacks become white. Our target is, after a finite number of moves, to obtain a chessboard where all initially black squares become white and all initially white squares become black. Prove that:\n\n(a) The target is not feasible for $n = 3$.\n\n(b) The target is feasible for $n = 2$.",
    "options": [],
    "answer": "See solution",
    "solution": "(a) First, observe that if we change the color of a square 3 times, it returns to its original color. Thus, the final configuration does not depend on the order of moves, and applying a move to a square 3 times has no effect. We can assume that each $2 \\times 2$ square is used 0, 1, or 2 times. This means a white square must be changed 2 (mod 3) times, and a black square 1 (mod 3) times.\n\nAssociate each $2 \\times 2$ square with its top left unit square. Consider the first column: its top square is white and included in only one $2 \\times 2$ square, so that square must be used 2 times. The second square in this column is also white and included in its own $2 \\times 2$ square and the one above. Since the first $2 \\times 2$ was used twice, this square has already been turned black, so its own $2 \\times 2$ must be used 0 times.\n\nUsing similar arguments, the next four $2 \\times 2$ squares must be used 1, 1, 1, and 0 times. After that, we have a white square included only in its own $2 \\times 2$ and the one above (used 0 times), so the same as at the start: the next should be used 2 times, and the next (white) 0 times. However, this is the only $2 \\times 2$ containing the last black square, so it cannot change color—a contradiction.\n\n(b) As in (a), we can fill the board with the numbers 0, 1, 2 in each square, indicating how many times to use the corresponding $2 \\times 2$ square.\n\n![](images/Greece_2022_p14_data_2fa6777697.png)",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 14860,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $E$ be an interior point in the convex quadrilateral $ABCD$. Let $F$, $G$, $H$, and $I$ be points opposite the quadrilateral with respect to the lines $AB$, $BC$, $CD$, and $DA$, respectively, such that $\\triangle ABF \\sim \\triangle DCE$, $\\triangle BCG \\sim \\triangle ADE$, $\\triangle CDH \\sim \\triangle BAE$, and $\\triangle DAI \\sim \\triangle CBE$. Let $P$, $Q$, $R$, and $S$ be the projections of $E$ on the lines $AB$, $BC$, $CD$, and $DA$, respectively. Prove that if the quadrilateral $PQRS$ is cyclic, then\n\n$$\nEF \\cdot CD = EG \\cdot DA = EH \\cdot AB = EI \\cdot BC.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "We consider oriented angles modulo $180^\\circ$. From the cyclic quadrilaterals $APES$, $BQEP$, $PQRS$, $CREQ$, $DSER$ and $\\triangle DCE \\sim \\triangle ABF$ we get\n\n$$\n\\begin{align*}\n\\angle AEB &= \\angle EAB + \\angle ABE = \\angle ESP + \\angle PQE \\\\\n            &= \\angle ESR + \\angle RSP + \\angle PQR + \\angle RQE \\\\\n            &= \\angle ESR + \\angle RQE = \\angle EDC + \\angle DCE \\\\\n            &= \\angle DEC = \\angle AFB,\n\\end{align*}\n$$\n\nso the quadrilateral $AEBF$ is cyclic. By Ptolemy we then have\n\n$$\nEF \\cdot AB = AE \\cdot BF + BE \\cdot AF.\n$$\n\nThis transforms by $AB : BF : AF = DC : CE : DE$ into\n\n$$\nEF \\cdot CD = AE \\cdot CE + BE \\cdot DE.\n$$\n\n![](images/Baltic_Way_SHL_2009-11_13-16_p118_data_817295bfc3.png)\n\nSince the expression on the right of this equation is invariant under cyclic permutation of the vertices of the quadrilateral $ABCD$, the asserted equation follows immediately.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 14861,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all positive integers $n$ such that they have at least 4 factors, and $n$ is equal to the sum of the squares of its four smallest factors.",
    "options": [],
    "answer": "See solution",
    "solution": "Let the four smallest factors be $a < b < c < d$. We require:\n\n$$n = a^2 + b^2 + c^2 + d^2$$\n\nThe smallest factor is $a = 1$, and since $n$ is even, $b = 2$. If $4 \\mid n$, then one of $c$ or $d$ is $4$. The other must be odd, but then the sum $a^2 + b^2 + c^2 + d^2$ would be $2 \\pmod{4}$, contradicting $n$ being divisible by $4$. Thus, $4$ is not a factor of $n$.\n\nIf $c$ were even, then $\\frac{c}{2} < c$ would also be a factor, contradicting $c$ being the third smallest. So $c$ is odd, $d$ is even, and $d = 2c$.\n\nThus,\n\n$$n = 1^2 + 2^2 + c^2 + (2c)^2 = 1 + 4 + c^2 + 4c^2 = 5 + 5c^2$$\n\nTrying $c = 3$ gives $d = 6$, but $6$ is not the fourth smallest factor. Try $c = 5$, $d = 10$:\n\n$$n = 5 + 5 \\times 25 = 130$$\n\nThe four smallest factors of $130$ are $1, 2, 5, 10$, and $1^2 + 2^2 + 5^2 + 10^2 = 1 + 4 + 25 + 100 = 130$.\n\nThus, the only such $n$ is $\\boxed{130}$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 14862,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $a$, $b$, and $c$ be positive real numbers. Prove that\n\n$$\n(a^{5} - a^{2} + 3)(b^{5} - b^{2} + 3)(c^{5} - c^{2} + 3) \\geq (a + b + c)^{3}.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "For any positive number $x$, the quantities $x^2 - 1$ and $x^3 - 1$ have the same sign. Thus, we have $0 \\leq (x^3 - 1)(x^2 - 1) = x^5 - x^3 - x^2 + 1$, or\n\n$$\nx^{5} - x^{2} + 3 \\geq x^{3} + 2.\n$$\n\nIt follows that\n\n$$\n(a^{5} - a^{2} + 3)(b^{5} - b^{2} + 3)(c^{5} - c^{2} + 3) \\geq (a^{3} + 2)(b^{3} + 2)(c^{3} + 2).\n$$\n\nIt suffices to show that\n\n$$\n(a^3 + 2)(b^3 + 2)(c^3 + 2) \\geq (a + b + c)^3. \\quad (*)\n$$\n\nWe finish with three approaches.\n\n* **First approach**: Expanding both sides of inequality $(*)$ and cancelling like terms gives\n\n$$\n\\begin{aligned}\n& a^{3}b^{3}c^{3} + 3(a^{3} + b^{3} + c^{3}) + 2(a^{3}b^{3} + b^{3}c^{3} + c^{3}a^{3}) + 8 \\\\\n& \\geq 3(a^{2}b + b^{2}a + b^{2}c + c^{2}b + c^{2}a + ac^{2}) + 6abc.\n\\end{aligned}\n$$\n\nBy the **AM-GM Inequality**, we have $a^3 + a^3 b^3 + 1 \\geq 3a^2b$. Combining similar results, the desired inequality reduces to\n\n$$\na^3 b^3 c^3 + a^3 + b^3 + c^3 + 2 \\geq 6abc,\n$$\n\nwhich is evident by the AM-GM Inequality.\n\n* **Second approach**: We rewrite the left-hand side of inequality $(*)$ as\n\n$$\n(a^3 + 1 + 1)(1 + b^3 + 1)(1 + 1 + c^3).\n$$\n\nBy Hölder's Inequality, we have\n\n$$\n(a^3 + 1 + 1)^{1/3} (1 + b^3 + 1)^{1/3} (1 + 1 + c^3)^{1/3} \\geq (a + b + c),\n$$\n\nfrom which inequality $(*)$ follows.\n\n* **Third approach**: Alternatively, the following double-application of Cauchy-Schwarz Inequality also does the trick:\n\n$$\n\\begin{aligned}\n& \\left[ (a^3 + 1 + 1)(1 + b^3 + 1) \\right] \\left[ (1 + 1 + c^3)(a + b + c) \\right] \\\\\n& \\geq (a^{3/2} + b^{3/2} + 1)^2 (a^{1/2} + b^{1/2} + c^2)^2 \\\\\n& \\geq (a + b + c)^4.\n\\end{aligned}\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 14863,
    "subject": "Mathematics (Olympiad)",
    "question": "For each natural $n$, find the number of sets $(a_1, a_2, \\dots, a_n)$, which are permutations of the numbers $(1, 2, \\dots, n)$ and satisfy the condition:\n\n$$\n(a_1 + a_2 + \\dots + a_k) \\text{ is divisible by } k, \\quad 1 \\leq k \\leq n.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "For $n=1$, there exists one set that satisfies the condition; for $n=3$, there are two sets; for other values of $n$, such sets do not exist.\n\nFor $n=1$, the only permutation is $(1)$, which clearly satisfies the condition.\n\nFor $n=3$, note that $a_1$ is divisible by $1$ and $a_1 + a_2 + a_3 = 6$ is divisible by $3$ for all permutations. The condition $a_1 + a_2 = 6 - a_3$ must be divisible by $2$, which happens if and only if $a_3$ is even, i.e., $a_3 = 2$. Thus, the two permutations that satisfy the condition are $(1, 3, 2)$ and $(3, 1, 2)$.\n\nFor $n=2$ or $n>3$, suppose $(a_1, a_2, \\dots, a_n)$ is a permutation that satisfies the condition. The sum $S_n = a_1 + a_2 + \\dots + a_n = \\frac{1}{2}n(n+1)$ must be divisible by $n$. If $n$ is even, $S_n$ is not divisible by $n$, so no such permutations exist.\n\nIf $n>3$ is odd, $n=2m+1$ with $m \\geq 2$, then $S_n = (m+1)(2m+1)$. For $k=n-1$, the condition $S_{n-1} = S_{2m} = (m+1)(2m+1) - a_{2m+1}$ must be divisible by $2m$. This leads to $a_{2m+1} = m+1$. Similarly, for $k=n-2$, we find $a_{2m} = m+1$. But this is impossible since $m+1$ cannot appear twice in a permutation of $1$ to $n$. Thus, for odd $n>3$, no such permutations exist.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 14864,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $ABC$ be a triangle with $\\angle ABC = 120^\\circ$ and triangle bisectors $AA_1$, $BB_1$, $CC_1$, respectively. $B_1F \\perp A_1C_1$, where $F \\in A_1C_1$. Let $R$, $I$, and $S$ be the centers of circles inscribed in triangles $C_1B_1F$, $C_1B_1A_1$, and $A_1B_1F$, respectively. Let $B_1S \\cap A_1C_1 = \\{Q\\}$. Show that $R$, $I$, $S$, $Q$ are concyclic.",
    "options": [],
    "answer": "See solution",
    "solution": "First, we will show that $\\angle C_1B_1A_1 = 90^\\circ$. Let $K \\in BC$ so that $B \\in (KA_1)$, then $\\angle ABK = 60^\\circ$. Point $C_1$ is on the bisector of $\\angle ACB$ and this implies that $d(C_1, BC) = d(C_1, AC)$ or $C_1F_1 = C_1F_3$, where $F_1$ is the projection of $C_1$ on $BC$ and $F_3$ is the projection of $C_1$ on $AC$. Segment $BA$ is the bisector of $\\angle KBB_1$ implies that $d(C_1, KB) = d(C_1, BB_1)$ or $C_1F_1 = C_1F_2$, where $F_2$ is the projection of $C_1$ on $BB_1$. So, $C_1F_2 = C_1F_3$ and $C_1B_1$ is the bisector of $\\angle BB_1A$. Let us denote $\\angle BB_1C_1 = \\alpha$. Likewise, we prove that $B_1A_1$ is the bisector of $\\angle BB_1C$. Let $\\angle BB_1A_1 = \\angle CB_1A_1 = \\beta$. Then, from $\\angle AB_1C = 180^\\circ$ we have $2\\alpha + 2\\beta = 180^\\circ$ and $\\alpha + \\beta = 90^\\circ$.\n\n![](images/Spanija_b_2013_p27_data_dea81da9ca.png)\n\nLet $r_1$ be the radius of the inscribed circle in $\\triangle A_1B_1C_1$, $r_2$ the radius of the inscribed circle in $\\triangle C_1B_1F$, and $r_3$ the radius of the inscribed circle in $\\triangle A_1B_1F$, respectively. Considering the properties of right triangles, we have\n\n$$\n\\triangle C_1FB_1 \\sim \\triangle C_1B_1A_1 \\Rightarrow \\frac{r_2}{r_1} = \\frac{B_1C_1}{C_1A_1} = \\cos C_1\n$$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 14865,
    "subject": "Mathematics (Olympiad)",
    "question": "Let us call a point in the $xy$-plane a *good point* if each of its coordinates is an integer from $1$ to $2000$. Let us also call polyline $ABCD$ a *Z-shaped polyline* if four points $A(x_1, y_1)$, $B(x_2, y_2)$, $C(x_3, y_3)$, $D(x_4, y_4)$ satisfy all the following conditions:\n\n- $A, B, C, D$ are good points.\n- $x_1 < x_2$, $y_1 = y_2$.\n- $x_2 > x_3$, $y_2 - x_2 = y_3 - x_3$.\n- $x_3 < x_4$, $y_3 = y_4$.\n\nDetermine the smallest possible positive integer $n$ such that there exist Z-shaped polylines $Z_1, Z_2, \\dots, Z_n$ which satisfy the following condition:\n\nAny good point $P$ lies on $Z_i$ for some $1 \\leq i \\leq n$.\n\nNote that polyline $ABCD$ is the union of the line segments (including both endpoints) $AB$, $BC$, and $CD$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let us call a good point on $x = 1$ or $x = 2000$ (excluding $(1, 1)$) a *special point*. Consider a Z-shaped polyline $ABCD$ with $A(x_1, y_1)$, $B(x_2, y_2)$, $C(x_3, y_3)$, $D(x_4, y_4)$. Since $1 \\leq x_1 < x_2 \\leq 2000$, $1 \\leq x_3 < x_2 \\leq 2000$, and $1 \\leq x_3 < x_4 \\leq 2000$, any special point on a Z-shaped polyline $ABCD$ coincides with either $A$, $B$, $C$, or $D$.\n\nAssume that both $B$ and $C$ are special points. Then $x_2 = 2000$, $y_2 \\leq 2000$, $x_3 = 1$, and $y_3 \\geq 2$. Therefore, $y_2 - x_2 \\leq 2000 - 2000 < 2 - 1 \\leq y_3 - x_3$, which contradicts the condition $y_2 - x_2 = y_3 - x_3$. Therefore, at most three special points lie on a Z-shaped polyline, hence we must select at least $\\frac{3999}{3} = 1333$ Z-shaped polylines to meet the condition.\n\nDenote polyline $ABCD$ with $A(x_1, y_1)$, $B(x_2, y_2)$, $C(x_3, y_3)$, $D(x_4, y_4)$ by $(x_1, y_1) - (x_2, y_2) - (x_3, y_3) - (x_4, y_4)$. Define Z-shaped polylines $X_1, X_2, \\dots, X_{666}$, $Y_1, Y_2, \\dots, Y_{666}$, $Z$ as follows:\n\n- For $k = 1, \\dots, 666$, let $X_k$ be $(1, 1334-k) - (1334-2k, 1334-k) - (1, 1+k) - (2000, 1+k)$.\n- For $k = 1, \\dots, 666$, let $Y_k$ be $(1, 2000-k) - (2000, 2000-k) - (667+2k, 667+k) - (2000, 667+k)$.\n- Let $Z$ be $(1, 2000) - (2000, 2000) - (1, 1) - (2000, 1)$.\n\nNote that any good point on $y = 1$ or $y = 2000$ lies on $Z$. For $2 \\leq k \\leq 667$, any good point on $y = k$ lies on $X_{k-1}$. For $1334 \\leq k \\leq 1999$, any good point on $y = k$ lies on $Y_{2000-k}$. Let $668 \\leq k \\leq 1333$ and consider good points on $y = k$:\n\n- When $1 \\leq x < 2k - 1333$, $(x, k)$ lies on $X_{1334-k}$.\n- When $2k - 1333 \\leq x < k$, $(x, k)$ lies on $X_{k-x}$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 14866,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all real numbers $r$ such that there is exactly one real number $a$ satisfying the following system of inequalities:\n\n$$\na + r \\geq 0\n$$\n\n$$\nar \\geq -1\n$$\n\n$$\na \\geq 0\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "We graph the three inequalities. In each case, the shaded region denotes the area defined by each inequality.\n\n![](images/2019_Australian_Scene_W1_p94_data_d57e300fbe.png)\n\n![](images/2019_Australian_Scene_W1_p94_data_9d92d6d030.png)\n\n![](images/2019_Australian_Scene_W1_p94_data_790f39d61f.png)\n\nThe intersection of the shaded regions denotes the region where all three inequalities are true.\n\n![](images/2019_Australian_Scene_W1_p94_data_6dcd4fc2af.png)\n\nFor any vertical line passing through a value of $r$ on the horizontal axis, each point on such a vertical line in the shaded area yields the corresponding values for $a$ satisfying the required inequalities. We seek the vertical lines which have exactly one point in common with the shaded area. The only such place is the indicated intersection point, which satisfies $a + r = 0$, $ar = -1$, and $a \\geq 0$.\n\nSubstituting $r = -a$ into $ar = -1$ yields $a = \\pm 1$. Since $a \\geq 0$, we have $a = 1$. Thus, $r = -1$.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 14867,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all pairs of functions $f, h : \\mathbb{R} \\to \\mathbb{R}$ such that for all $x, y \\in \\mathbb{R}$,\n$$\nf(x^2 + y h(x)) = x h(x) + f(xy).\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "Let $h(0) = a$. Set $x = 0$ in the initial equation:\n\n$$\nf(x^2 + y h(x)) = x h(x) + f(xy) \\quad \\text{for all } x, y \\in \\mathbb{R}. \\quad (*)\n$$\n\nWe obtain $f(a y) = f(0) = c$ for all $y$.\n\nIf $a \\neq 0$, then $a y$ admits all real values, so the function $f(x) = c$ is identically constant. So the equation has the form $c = x h(x) + c$ or $x h(x) = 0$, therefore $h(x) = 0$ for $x \\neq 0$, and $h(0) = a$ admits any value. As it is easy to verify, the obtained pair of functions $(f(x), h(x))$ satisfies $(*)$.\n\nLet $a = 0$, i.e., $h(0) = 0$. Note that if $h(x_0) \\neq x_0$ for some $x_0$ ($x_0 \\neq 0$), then there exists $y_0$ such that $x_0^2 + y_0 h(x_0) = x_0 y_0$ (indeed, it suffices to set $y_0 = \\dfrac{x_0^2}{x_0 - h(x_0)}$). Setting $x = x_0$, $y = y_0$ in the initial equation, we obtain $x_0 h(x_0) = 0$, i.e., $h(x_0) = 0$. Now setting $x = x_0$ in the initial equation, we have $f(x_0^2) = f(x_0 y)$ for all $y$. Since $x_0 \\neq 0$, we see that $x_0 y$ admits any real values, so $f(x) = c$ is identically constant. But this case is already considered above.\n\nIt remains to suppose that $h(x) = x$ for all $x$. Then $(*)$ can be rewritten in the form\n\n$$\nf(x^2 + y x) = x^2 + f(x y) \\quad \\text{for all } x, y \\in \\mathbb{R}. \\quad (**)\n$$\n\nLet $f(0) = b$. Setting $y = 0$ in $(**)$, we obtain $f(x^2) = x^2 + b$ for all $x$, i.e., $f(x) = x + b$ for all nonnegative $x$. If we set $y = -x$, then we obtain $f(-x^2) = -x^2 + b$, i.e., $f(x) = x + b$ for all nonpositive $x$.\n\n**Final answer:**\n- Either $f(x) = c$, $h(x) = \\begin{cases} 0, & x \\neq 0 \\\\ a, & x = 0 \\end{cases}$, where $a$ and $c$ are arbitrary constants,\n- or $f(x) = x + b$, $h(x) = x$, where $b$ is an arbitrary constant.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 14868,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $a_1, a_2, \\dots, a_n$ be a beautiful sequence of length $n$. Take an integer $m$ such that $a_m = 2025$. Then $m \\ge 2$ since $a_1 = 0$.\n\nProve that $n \\le 13$, and for $n = 13$, determine the minimum possible value of $a_{13}$ for such a sequence.",
    "options": [],
    "answer": "See solution",
    "solution": "We first prove $n \\le 13$. Let $t$ be an integer with $1 \\le t \\le n - 2$. By applying the third condition to $(i, j, k) = (t, t+1, n)$, we obtain\n\n$$\n\\frac{a_t + a_n}{2} \\le a_{t+1},\n$$\n\nwhich implies\n\n$$\na_n - a_t \\ge 2(a_n - a_{t+1}).\n$$\n\nIf $m = n$, repeatedly applying this inequality gives\n\n$$\n2025 = a_n - a_1 \\ge 2(a_n - a_2) \\ge \\dots \\ge 2^{n-2}(a_n - a_{n-1}) \\ge 2^{n-2},\n$$\n\nand hence $n \\le 12$. If $m < n$, we have\n\n$$\na_n - a_1 \\ge 2(a_n - a_2) \\ge \\dots \\ge 2^{m-2}(a_n - a_{m-1}) \\ge 2^{m-1}(a_n - a_m),\n$$\n\nand\n\n$$\na_n - a_m \\ge 2(a_n - a_{m+1}) \\ge 2^2(a_n - a_{m+2}) \\ge \\dots \\ge 2^{n-m-1}(a_n - a_{n-1}) \\ge 2^{n-m-1}.\n$$\n\nTherefore\n\n$$\n\\begin{aligned}\n2025 &= a_m - a_1 = (a_n - a_1) - (a_n - a_m) \\ge (2^{m-1} - 1)(a_n - a_m) \\\\\n&\\ge (2^{m-1} - 1)2^{n-m-1} \\ge 2^{m-2} \\cdot 2^{n-m-1} = 2^{n-3}\n\\end{aligned}\n$$\n\nand so we obtain $n \\le 13$.\n\nLet $n = 13$. Then the above argument shows $n > m$. From the previous inequality, we have\n\n$$\n2025 \\ge (2^{m-1} - 1)2^{n-m-1} = 2^{11} - 2^{12-m},\n$$\n\nwhich implies $2^{12-m} \\ge 23$, and hence $2^{12-m} \\ge 32$. Therefore,\n\n$$\na_n = a_m + (a_n - a_m) \\ge a_m + 2^{n-m-1} = 2025 + 2^{12-m} \\ge 2057.\n$$\n\nThus, we have $a_n \\ge 2057$ when $n = 13$.\n\nFinally, we show that the sequence\n\n$$\n(a_1, a_2, \\dots, a_{13}) = (0, 1033, 1545, 1801, 1929, 1993, 2025, 2041, 2049, 2053, 2055, 2056, 2057)\n$$\n\nis a beautiful sequence of length 13. The first and second conditions are clearly satisfied. We now verify the third condition. We have\n\n$$\n(a_{13} - a_1, a_{13} - a_2, \\dots, a_{13} - a_{12}) = (2057, 1024, 512, 256, 128, 64, 32, 16, 8, 4, 2, 1),\n$$\n\nso for all integers $t$ with $1 \\le t \\le 11$, we have $a_{13} - a_t \\ge 2(a_{13} - a_{t+1})$. Therefore, for all integers $1 \\le i < j < k \\le 13$, we have\n\n$$\n\\begin{aligned}\na_j - \\frac{a_i + a_k}{2} &\\ge a_j - \\frac{a_i + a_{13}}{2} = \\frac{(a_{13} - a_i) - 2(a_{13} - a_j)}{2} \\\\\n&\\ge \\frac{(a_{13} - a_i) - (a_{13} - a_{j-1})}{2} = \\frac{a_{j-1} - a_i}{2} \\\\\n&\\ge 0.\n\\end{aligned}\n$$\n\nThis confirms the third condition.\n\nTherefore, the maximum value of the length of a beautiful sequence is $13$ and for beautiful sequences $a_1, a_2, \\dots, a_{13}$ of length $13$, the minimum value of $a_{13}$ is **2057**.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 14869,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $m$ and $n$ be integers such that $m \\geq 2$ and $n \\geq 3$. Prove that there exist $m$ distinct positive integers $a_1, a_2, a_3, \\dots, a_m$, all divisible by $n-1$, such that\n$$\n\\frac{1}{n} = \\frac{1}{a_1} - \\frac{1}{a_2} + \\frac{1}{a_3} - \\dots + (-1)^{m-1} \\frac{1}{a_m}.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "Let $(a_k)_{k \\geq 1}$ be a geometric sequence with ratio $q = n - 1$. Then\n$$\n\\frac{1}{a_1} - \\frac{1}{a_2} + \\dots + (-1)^{p-1} \\frac{1}{a_p} = \\frac{1/a_1 + \\frac{(-1)^{p-1}}{a_p}}{1 + 1/q}, \\quad \\forall p \\in \\mathbb{N}^*.\n$$\nThat is,\n$$\n\\frac{1}{a_1} - \\frac{1}{a_2} + \\dots + (-1)^{p-1} \\frac{1}{a_p} = \\frac{n-1}{n a_1} + \\frac{(-1)^{p-1}}{n a_p}.\n$$\nThe required numbers can be obtained for $a_1 = n-1$, $m = p+1$, and $a_m = n a_p$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 14870,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $n$ be the number of three-digit numbers that satisfy the following properties:\n\n1. No number contains the digit $0$.\n2. The sum of the digits of each number is $9$.\n3. The units digits of any two numbers are different.\n4. The tens digits of any two numbers are different.\n5. The hundreds digits of any two numbers are different.\n\nFind the largest possible value of $n$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $S$ denote the set of three-digit numbers with digit sum equal to $9$ and no digit equal to $0$. First, we find the cardinality of $S$.\n\nWe seek the number of integer solutions to $a + b + c = 9$ where $1 \\leq a, b, c \\leq 9$. Setting $a' = a-1$, $b' = b-1$, $c' = c-1$, we have $a'+b'+c' = 6$ with $0 \\leq a', b', c' \\leq 8$. The number of non-negative integer solutions is:\n\n$$\n\\binom{6+3-1}{3-1} = \\binom{8}{2} = 28.\n$$\n\nSo $S$ contains $28$ numbers.\n\nFrom conditions 3, 4, and 5, if $\\overline{abc}$ is in $S$, then no other number in $S$ can share any digit with $\\overline{abc}$ in the same position. Thus, each digit in each position must be unique among the chosen numbers.\n\nFor each number in $S$, there are $a+b-2$ numbers sharing the same units digit, $a+c-2$ sharing the same tens digit, and $b+c-2$ sharing the same hundreds digit. Summing, we get:\n\n$$\n(a+b-2) + (a+c-2) + (b+c-2) = 2(a+b+c) - 6 = 2 \\times 9 - 6 = 12\n$$\n\ndistinct numbers that cannot be in $S$ if $abc$ is in $S$.\n\nIf $S$ has $n$ numbers, then $12n$ are forbidden, but each number can be forbidden up to $3$ times (once per digit), so:\n\n$$\nn + \\frac{12n}{3} \\leq 28 \\implies n + 4n \\leq 28 \\implies n \\leq \\frac{28}{5}.\n$$\n\nSince $n$ is an integer, $n \\leq 5$.\n\nAn example for $n=5$ is:\n\n$$\nS = \\{144,\\ 252,\\ 315,\\ 423,\\ 531\\}.\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 14871,
    "subject": "Mathematics (Olympiad)",
    "question": "¿De cuántas formas se pueden colorear los vértices de un polígono con $n \\geq 3$ lados usando tres colores, de forma que haya exactamente $m$ lados, $2 \\leq m \\leq n$, cuyos extremos sean de colores diferentes?",
    "options": [],
    "answer": "See solution",
    "solution": "Señalamos los puntos medios de los $m$ lados cuyos extremos deben tener colores diferentes. Esto puede hacerse de $\\binom{n}{m}$ formas. Así, los $n$ vértices quedan divididos en $m$ grupos consecutivos, donde todos los vértices de un grupo tienen el mismo color, pero los grupos adyacentes tienen colores diferentes.\n\nEl número de formas de colorear estos grupos es igual al número de formas de colorear un polígono de $m$ lados usando tres colores, sin que haya ningún lado con extremos del mismo color. Denotemos este número por $C_m$.\n\nPor lo tanto, la solución es $\\binom{n}{m} C_m$.\n\nPara encontrar $C_m$, notamos que $C_2 = C_3 = 6$. Para $m \\geq 4$, se cumple la recurrencia:\n\n$$\nC_m = C_{m-1} + 2C_{m-2}.\n$$\n\nLos primeros valores son: $C_2 = 6$, $C_3 = 6$, $C_4 = 18$, $C_5 = 30$, $C_6 = 66$, $C_7 = 126$, ...\n\nSe observa que $C_m = 2^m + (-1)^m 2$, lo cual se puede demostrar por inducción. Por lo tanto, la respuesta es:\n\n$$\n\\boxed{\\binom{n}{m} \\left(2^m + (-1)^m 2\\right)}\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 14872,
    "subject": "Mathematics (Olympiad)",
    "question": "Given complex numbers $a$, $b$, $c$, let $|a+b| = m$, $|a-b| = n$, and suppose $mn \\neq 0$. Prove that\n$$\n\\max\\{|ac + b|,\\ |a + bc|\\} \\ge \\frac{mn}{\\sqrt{m^2 + n^2}}\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "**Proof**\n$$\n\\begin{aligned}\n\\max\\{ |ac + b|,\\ |a + bc| \\} &\\geq \\frac{ |b| \\cdot |ac + b| + |a| \\cdot |a + bc| }{ |b| + |a| } \\\\\n&\\geq \\frac{ |b(ac + b) - a(a + bc)| }{ |a| + |b| } \\\\\n&= \\frac{ |b^2 - a^2| }{ |a| + |b| } \\\\\n&\\geq \\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| \\} \\geq \\frac{mn}{\\sqrt{m^2 + n^2}}.\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 14873,
    "subject": "Mathematics (Olympiad)",
    "question": "Determine the largest number $m$ such that the inequality\n\n$$\n(a^2 + 4(b^2 + c^2))(b^2 + 4(c^2 + a^2))(c^2 + 4(a^2 + b^2)) \\geq m\n$$\n\nholds for all real numbers $a$, $b$, and $c$ not equal to $0$ and satisfying the condition\n\n$$\n\\left|\\frac{1}{a}\\right| + \\left|\\frac{1}{b}\\right| + \\left|\\frac{1}{c}\\right| \\leq 3.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "The arithmetic-geometric mean inequality gives us\n\n$$\na^2 + 4b^2 + 4c^2 \\geq 9 \\cdot \\sqrt[9]{a^2 b^8 c^8}, \\quad b^2 + 4c^2 + 4a^2 \\geq 9 \\cdot \\sqrt[9]{a^8 b^2 c^8}, \\quad \\text{and} \\quad c^2 + 4a^2 + 4b^2 \\geq 9 \\cdot \\sqrt[9]{a^8 b^8 c^2}.\n$$\n\nFrom this, we obtain\n\n$$\n\\begin{aligned}\n(a^2 + 4(b^2 + c^2))(b^2 + 4(c^2 + a^2))(c^2 + 4(a^2 + b^2)) &\\geq 729 \\cdot \\sqrt[9]{a^{18}b^{18}c^{18}} \\\\\n&= 729 \\cdot (abc)^2 \\\\\n&\\geq 729.\n\\end{aligned}\n$$\n\nSince equality holds for $a = b = c = 1$, we see that the maximum $m$ we are searching for is equal to $729$. Equality holds if the absolute values of all variables are equal to $1$, and we therefore have eight possible triples of variables for which equality holds, namely $(a, b, c) = (\\pm 1, \\pm 1, \\pm 1)$.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 14874,
    "subject": "Mathematics (Olympiad)",
    "question": "給定正整數 $k > 1$ 以及 $n$ 個 $k$ 維座標中的相異點 $a^{(1)} = (a_1^{(1)}, \\,\\cdots, a_k^{(1)}), \\,\\cdots, a^{(n)} = (a_1^{(n)}, \\,\\cdots, a_k^{(n)})$，我們定義 $a^{(i)}$ 的『分數』如下：\n\n$$\n\\prod_{j=1}^{k} \\#\\{i' \\mid 1 \\le i' \\le n \\text{ 使得 } \\pi_j(a^{(i')}) = \\pi_j(a^{(i)})\\}\n$$\n\n其中 $\\#S$ 表示集合 $S$ 的元素個數，而 $\\pi_j : \\mathbb{R}^k \\to \\mathbb{R}^{k-1}$ 是將 $k$ 維座標中的點映到刪除其第 $j$ 個坐標的點的投影映射。找到最大的實數 $t$ 使得所有 $a^{(i)}$ 的『分數』的 $t$ 次幂平均 $\\le n$。\n\n註：正實數 $x_1, \\cdots, x_n$ 的 $t$ 次幂平均定義如下：\n\n$$\n\\left( \\frac{x_1^t + \\cdots + x_n^t}{n} \\right)^{\\frac{1}{t}} \\quad (\\text{若 } t \\neq 0); \\quad \\sqrt[n]{x_1 \\cdots x_n} \\quad (\\text{若 } t = 0)\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "答案是 $\\frac{1}{k-1}$。\n\n為了證明 $t \\le \\frac{1}{k-1}$，考慮所有但一個分量都等於 1，剩下的一個分量為 $1, 2, \\ldots, m$ 的序列。此時 $n = km + 1$，而零序列的分數為 $m^k$。為了使條件成立，必須有\n\n$$\n\\left( \\frac{m^{kt}}{km + 1} \\right)^{\\frac{1}{t}} \\le km + 1,\n$$\n\n即 $m^k \\le (km + 1)^{1+1/t}$。令 $m$ 充分大，得到\n\n$$\nk \\le 1 + \\frac{1}{t} \\implies t \\le \\frac{1}{k-1}.\n$$\n\n接下來用歸納法證明 $t = \\frac{1}{k-1}$ 可行。假設 $k \\ge 3$ 且對 $k-1$ 已成立。對每個實數 $r$，令 $c_r$ 為 $a_1^{(i)} = r$ 的序列數。對每個 $c_r \\ne 0$，收集 $a_1^{(i)} = r$ 的序列並移除第一分量，計算其分數 $s'_1, \\cdots, s'_{c_r}$。對每個截斷序列，計算原序列中產生該截斷序列的個數 $d_1, \\cdots, d_{c_r}$。顯然 $d_1 + \\cdots + d_{c_r} \\le n$，且原序列的分數為 $d_1s'_1, \\cdots, d_{c_r}s'_{c_r}$。根據歸納假設：\n\n$$\ns'_1^{\\frac{1}{k-2}} + \\cdots + s'_{c_r}^{\\frac{1}{k-2}} = c_r^{\\frac{k-1}{k-2}}.\n$$\n\n由 H\"older 不等式：\n\n$$\n\\left((d_1s'_1)^{\\frac{1}{k-1}} + \\cdots + (d_{c_r}s'_{c_r})^{\\frac{1}{k-1}}\\right)^{k-1} \\le (d_1 + \\cdots + d_{c_r}) \\cdot \\left(s'_1^{\\frac{1}{k-2}} + \\cdots + s'_{c_r}^{\\frac{1}{k-2}}\\right)^{k-2} \\le n c_r^{k-1}.\n$$\n\n取 $(k-1)$ 次方根並對所有 $c_r \\ne 0$ 加總，得到 $\\frac{1}{k-1}$ 次幂平均被以下式子界定：\n\n$$\n\\left( \\frac{\\sum_{c_r \\ne 0} n^{\\frac{1}{k-1}} c_r}{n} \\right)^{k-1} = n\n$$\n\n因為\n\n$$\n\\sum_{c_r \\ne 0} c_r = n.\n$$\n\n基礎情形 $k=2$，同理可證，此時\n\n$$\ns'_1 = \\cdots = s'_{c_r} = n\n$$\n\n所以\n\n$$\n(d_1 s'_1 + \\cdots + d_{c_r} s'_{c_r}) = n(d_1 + \\cdots + d_{c_r}) \\le n^2\n$$\n\n其餘證明同理。\n\n**Remark.** 另一個等價的問題敘述：考慮一個 $n$ 點的（不一定完全）簡單圖，每條邊用 $k$ 種顏色之一標記，且每種顏色的子圖是點不交並的團。每個點的分數定義為 $k$ 種顏色團大小的乘積。求使所有分數的 $t$ 次幂平均不超過 $n$ 的最小 $t$。",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 14875,
    "subject": "Mathematics (Olympiad)",
    "question": "The positive integers $a_0, a_1, \\dots, a_9$ and $b_1, b_2, \\dots, b_9$ are such that $a_9 < b_9$, $a_k \\neq b_k$ for $1 \\leq k \\leq 8$. A cash machine is loaded with $n \\geq a_9$ leva. For any $1 \\leq i \\leq 9$, it is allowed to withdraw $a_i$ leva (if the machine has at least $a_i$ leva), and after that the bank puts in the machine $b_i$ leva. It is also allowed to withdraw $a_0$ without any action from the bank. Find all positive integers $n$ for which the cash machine can be emptied by the above described operations.",
    "options": [],
    "answer": "See solution",
    "solution": "Set $d_s = |a_s - b_s|$, $0 \\leq s \\leq 9$, where $b_0 = 0$. Without loss of generality, assume that there exists $k$, $0 \\leq k \\leq 8$ such that $a_s - b_s > 0$ for $0 \\leq s \\leq k$ and $a_s - b_s < 0$ for $k + 1 \\leq s \\leq 9$.\n\nIf $n$ is one of the desired values, then there exist positive integers $x_0, x_1, \\dots, x_9$ for which\n\n$$\nx_0 d_0 + \\dots + x_k d_k - x_{k+1} d_{k+1} - \\dots - x_9 d_9 = n.\n$$\n\nTherefore, $d = \\gcd(d_0, d_1, \\dots, d_9)$ is a divisor of $n$.\n\nWe now prove that if $d$ is a divisor of $n$, then one can empty the cash machine.\n\nIndeed, in this case, it follows from Bézout's theorem that the above equation has a solution $(x_0, x_1, \\dots, x_9)$ in integers. Set $D_1 = d_0 + \\dots + d_k$, $D_2 = d_{k+1} + \\dots + d_9$, $x_s' = x_s + t D_2$ for $0 \\leq s \\leq k$, $x_s' = x_s + t D_1$ for $k+1 \\leq s \\leq 9$. Since $D_1, D_2 > 0$, for large enough $t$, $(x_0', \\dots, x_9')$ is a solution in positive integers and $x_9' > \\max(a_8, \\dots, a_{k+1})$. Consider one such solution and let $x_0'' = x_0' + r D_2$, $x_s'' = x_s'$ for $1 \\leq s \\leq k$, and $x_s'' = x_s' + r d_0$ for $k+1 \\leq s \\leq 9$. For large enough $r$, we obtain a solution $(x_0'', \\dots, x_9'')$ for which\n\n$$\nx_9'' \\geq \\max(a_8, \\dots, a_{k+1})\n$$\n\n$$\nn + x_9'' d_9 + \\dots + x_{k+1}'' d_{k+1} > x_1'' d_1 + \\dots + x_k'' d_k.\n$$\n\nWe draw money as follows: first $x_9''$ times $a_9$ leva, then $x_8''$ times $a_8$ leva, ..., $x_{k+1}''$ times $a_{k+1}$ leva. After that, we take $x_1''$ times $a_1$ leva, ..., $x_k''$ times $a_k$ leva. Since the above conditions hold, all operations are feasible. Now, the equation implies that there are exactly $x_0'' d_0$ leva left in the machine, and we withdraw them by taking $x_0''$ times $a_0$ leva.\n\n**Answer.** All $n \\geq a_9$ divisible by $d$.\n\n**Remark.** The condition that one of the differences $d_s$ is negative is essential. Otherwise, the problem reduces to the so-called Sylvester's problem.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 14876,
    "subject": "Mathematics (Olympiad)",
    "question": "South African Magical Flights (SAMF) operates flights between South African airports. If there is a flight from airport $A$ to airport $B$, there will also be a flight from $B$ to $A$.\n\nThe SAMF headquarters are located in Kimberley. Every airport that is served by SAMF can be reached from Kimberley in precisely one way. This way of reaching Kimberley may involve stopping at other airports on the way. (For example, it may happen that you can get to Kimberley from Durban by flying from Durban to Bloemfontein and then from Bloemfontein to Kimberley. In that case there is no other way to get from Durban to Kimberley. For example, there would be no direct flight from Durban to Kimberley.)\n\nAn airport (other than Kimberley) is called *terminal* if there are flights to (and from) precisely one other airport. Suppose that there are $t$ terminal airports.\n\nDue to budget cuts, SAMF decides to close down $k$ of the airports. It should still be possible to reach each of the remaining airports from Kimberley.\n\nLet $C$ be the number of choices for the $k$ destinations that are discontinued. Prove that\n\n$$\n\\frac{t!}{k!(t-k)!} \\le C \\le \\frac{(t+k-1)!}{k!(t-1)!},\n$$\n\nwhere $n! = 1 \\times 2 \\times \\dots \\times n$ for any positive integer $n$, and $0! = 1$.",
    "options": [],
    "answer": "See solution",
    "solution": "We can depict the situation as a connected graph without cycles, i.e., a tree. The vertices represent the airports and there is an edge between two vertices if and only if there is a direct flight possible between the two airports represented by these two vertices. The leaves of the tree (vertices (other than the vertex representing Kimberley) of degree 1) represent the terminal airports. In Figure 2 an example of such a tree is given where $H$ denotes the Kimberley airport, and there are six terminal airports $T_1, T_2, \\dots, T_6$. All in all, there are 18 airports (including Kimberley) represented by this tree.\n\n![](images/s3s2023_p3_data_9081ed1712.png)\n\nWhen closing an airport, we have to adjust the tree by removing the vertex that represents that airport, as well as all the edges adjoined to that vertex. Note that this implies that, according to the condition that airports can be reached in exactly one way from Kimberley, when a vertex is removed, all vertices further down to the leaf/leaves must also be removed. For example, if vertex $A$ in Figure 2 is removed, then vertices $B$, $C$, $T_2$, $T_3$ must also be removed, since flights to these four airports are only possible *via* $A$.\n\nIf $k > t$ then clearly $0 = \\binom{t}{k} \\le C$. (We use the shorthand notation $\\binom{t}{k}$ for $\\frac{t!}{k!(t-k)!}$.) Otherwise, if $k \\le t$, then one way to remove $k$ vertices (according to the rules), is to remove any $k$ of the $t$ leaves (terminal airports). This can be done in $\\binom{t}{k}$ ways, so that, again, $\\binom{t}{k} \\le C$.\n\nNow for the upper bound $\\binom{t+k-1}{k}$: For each leaf $T_i$ ($1 \\le i \\le t$), let $S_i$ denote either the first vertex from $T_i$ towards $H$ with degree more than 2, or $S_i = H$, whichever comes first. For example, in Figure 2, $S_1 = H$, $S_2 = A$, $S_3 = A$, $S_4 = D$, $S_5 = E$, and $S_6 = E$.\n\nSuppose that the number of vertices between $T_i$ and $S_i$ (including $T_i$ but excluding $S_i$) is at least $k$, for each $i = 1, 2, \\dots, t$. Then we may, for each $i = 1, 2, \\dots, t$, remove the bottom $k_i$ vertices (starting with $T_i$), where $0 \\le k_i \\le k$, and such that $\\sum_{i=1}^t k_i = k$. There are $\\binom{t+k-1}{k}$ solutions to the equation $\\sum_{i=1}^t k_i = k$ in the (non-negative integral) unknowns $k_i$, $1 \\le i \\le t$. This implies that $C = \\binom{t+k-1}{k}$ in this case.\n\nSuppose now that for at least one $i$, $1 \\le i \\le t$, there are fewer than $k$ vertices between $T_i$ and $S_i$ (including $T_i$ but excluding $S_i$). Then, in finding the number of solutions to $\\sum_{i=1}^t k_i = k$, $0 \\le k_i \\le k$, we are now more restricted, since either you may only remove fewer than $k$ vertices between $T_i$ and $S_i$ (to avoid choosing $S_i$), or if you do choose to remove $S_i$, then the complete set of vertices from $S_i$ downwards to another leaf $T_j$, say, will also be removed, without any choice on how to choose $k_j$. Since we are now more restricted in the number of ways we can choose the vertices to be removed, we have $C < \\binom{t+k-1}{k}$ in this case.\n\nHence, for both scenarios, $C \\le \\binom{t+k-1}{k}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 14877,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $I, J$ be intervals and let $\\varphi : J \\to \\mathbb{R}$ be a continuous function which is nonzero on $J$. Let $f, g : I \\to J$ be two differentiable functions such that $f' = \\varphi \\circ f$ and $g' = \\varphi \\circ g$. Prove that if there exists $x_0 \\in I$ such that $f(x_0) = g(x_0)$, then $f$ and $g$ coincide.",
    "options": [],
    "answer": "See solution",
    "solution": "As $\\varphi$ is nonzero and continuous, the function $1/\\varphi$ is well-defined and continuous. Consider an antiderivative $F : J \\to \\mathbb{R}$ of $1/\\varphi$. The given relation for $f$ can then be written as $$(F \\circ f)'(x) = 1$$ for all $x \\in I$. Thus, there exists $a \\in \\mathbb{R}$ such that $F(f(x)) = x + a$ for all $x \\in I$. Similarly, there exists $b \\in \\mathbb{R}$ such that $F(g(x)) = x + b$ for all $x \\in I$. Since $f(x_0) = g(x_0)$, we have $F(f(x_0)) = F(g(x_0))$, so $x_0 + a = x_0 + b$, which implies $a = b$. Therefore, $F(f(x)) = F(g(x))$ for all $x \\in I$.\n\nSince $F'(x) \\neq 0$ for all $x \\in J$, $F$ is strictly monotonic and thus injective. Therefore, $F(f(x)) = F(g(x))$ implies $f(x) = g(x)$ for all $x \\in I$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 14878,
    "subject": "Mathematics (Olympiad)",
    "question": "Natural numbers $a < b$ are written on the board. At each step, two numbers written on the board are wiped out, and their sum and the modulus of their difference are written down instead. At some point, the number $2019$ appeared on the board. What is the smallest possible value of $b$?",
    "options": [],
    "answer": "See solution",
    "solution": "*Answer:* $b = 1010$.\n\nIt is easy to write down all the pairs of numbers that will successively appear on the board:\n\n$$\na, b \\rightarrow b+a, b-a \\rightarrow 2a, 2b \\rightarrow 2(b+a), 2(b-a) \\rightarrow 2a^2, 2b^2 \\dots\n$$\nAs we see, after the appearance of the first four numbers: $a$, $b$, $b+a$, $b-a$, all the other numbers that may appear on the board are even. Since $2019$ is an odd number, the number $2019$ must appear among these first four numbers.\n\n- Case 1: $b = 2019$.\n- Case 2: $a = 2019 < b$.\n- Case 3: $b-a = 2019 < b$.\n- Case 4: $b+a = 2019$, then $2b > b+a = 2019 \\Rightarrow 2b \\ge 2020 \\Rightarrow b \\ge 1010$.\n\nThus, the smallest possible value of $b$ is $b = 1010$. If $b = 1010$ and $a = 1009$, the number $2019$ can appear on the board.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 14879,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all positive integers $n$ such that:\n\n$$\n-2^0 + 2^1 - 2^2 + 2^3 - 2^4 + \\dots - (-2)^n = 4^0 + 4^1 + 4^2 + \\dots + 4^{2010}.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "Using the formula for the sum of a geometric progression:\n\nThe left side is a sum with alternating signs:\n$$\n-2^0 + 2^1 - 2^2 + 2^3 - 2^4 + \\dots - (-2)^n = \\sum_{k=0}^n (-1)^{k+1} 2^k = -\\sum_{k=0}^n (-2)^k\n$$\nThis is a geometric series with first term $-1$ and ratio $-2$:\n$$\n\\sum_{k=0}^n (-2)^k = \\frac{(-2)^{n+1} - 1}{-2 - 1} = \\frac{(-2)^{n+1} - 1}{-3}\n$$\nSo the left side is:\n$$\n-\\sum_{k=0}^n (-2)^k = -\\left(\\frac{(-2)^{n+1} - 1}{-3}\\right) = \\frac{(-2)^{n+1} - 1}{3}\n$$\nThe right side is a geometric series:\n$$\n4^0 + 4^1 + 4^2 + \\dots + 4^{2010} = \\sum_{k=0}^{2010} 4^k = \\frac{4^{2011} - 1}{4 - 1} = \\frac{4^{2011} - 1}{3}\n$$\nEquate both sides:\n$$\n\\frac{(-2)^{n+1} - 1}{3} = \\frac{4^{2011} - 1}{3}\n$$\nSo $(-2)^{n+1} = 4^{2011}$.\n\nSince $(-2)^{n+1}$ is positive only when $n+1$ is even, let $n+1 = 2m$:\n$$\n(-2)^{2m} = (4)^m = 4^{2011}\n$$\nSo $m = 2011$, $n+1 = 4022$, $n = 4021$.\n\n**Answer:** The only positive integer $n$ is $\\boxed{4021}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 14880,
    "subject": "Mathematics (Olympiad)",
    "question": "Find positive integers $a_1, a_2, \\dots, a_{2019}$ which satisfy the equation\n\n$$\na_1 + a_2 + \\dots + a_{2019} = a_1 a_2 \\dots a_{2019} = \\sqrt[2018]{2019^{2019}}.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "**Answer:** $a_1 = a_2 = \\dots = a_{2019} = \\sqrt[2018]{2019}$.\n\n**Solution.** From the Arithmetic Mean - Geometric Mean Inequality,\n\n$$\n\\frac{1}{2019}(a_1 + a_2 + \\dots + a_{2019}) \\geq \\sqrt[2019]{a_1 a_2 \\dots a_{2019}},\n$$\n\nor\n\n$$\n(a_1 + a_2 + \\dots + a_{2019})^{2019} \\geq 2019^{2019} a_1 a_2 \\dots a_{2019}.\n$$\n\nFrom the problem statement,\n\n$$\n(a_1 + a_2 + \\dots + a_{2019})^{2019} = (\\sqrt[2018]{2019^{2019}})^{2019} = 2019^{2019}.\n$$\n\nSo equality holds in the AM-GM inequality, which is possible if and only if\n\n$$\na_1 = a_2 = \\dots = a_{2019} = \\sqrt[2018]{2019}.\n$$",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 14881,
    "subject": "Mathematics (Olympiad)",
    "question": "Prove that if positive numbers $a$, $b$, $c$ satisfy the inequality $5abc > a^3 + b^3 + c^3$, then there is a triangle with sides $a$, $b$, $c$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let positive numbers $a$, $b$, $c$ satisfy the inequality $5abc > a^3 + b^3 + c^3$. Let us show that there exists a triangle with the sides $a$, $b$, $c$.\n\nSuppose, for contradiction, that there is no such triangle. Then at least one of the triangle inequalities fails. Assume, for example, $c \\geq a + b$, i.e., $c = a + b + x$ where $x \\geq 0$.\n\nFrom the initial inequality:\n\n$$\n5ab(a + b + x) > a^3 + b^3 + (a + b + x)^3\n$$\n\nExpanding $(a + b + x)^3$:\n\n$$\n(a + b + x)^3 = (a + b)^3 + 3(a + b)^2 x + 3(a + b)x^2 + x^3\n$$\n\nSo,\n\n$$\n5ab(a + b + x) > a^3 + b^3 + (a + b)^3 + 3(a + b)^2 x + 3(a + b)x^2 + x^3\n$$\n\nRewriting:\n\n$$\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$$\n\nSince the last four terms on the right are nonnegative, we have:\n\n$$\n2a^2b + 2ab^2 > 2a^3 + 2b^3\n$$\n\nWhich is equivalent to:\n\n$$\n\\begin{aligned}\na^3 + b^3 - a^2b - ab^2 &< 0 \\\\\n(a + b)(a - b)^2 &< 0\n\\end{aligned}\n$$\n\nBut $(a + b)(a - b)^2 \\geq 0$ for real $a$, $b$, so this is impossible. Thus, the triangle inequalities must hold, and there exists a triangle with sides $a$, $b$, $c$.\n\n$\\boxed{}$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 14882,
    "subject": "Mathematics (Olympiad)",
    "question": "There are $n > 1$ guests at Georg's birthday party and each guest is friends with at least one other guest. Georg organizes a party game among the guests. Each guest receives a jug of water such that there are no two guests with the same amount of water in their jugs. Every guest now proceeds as follows. The guest takes one cup for each of their friends at the party and distributes all the water from their jug evenly in the cups. They then pass a cup to each of their friends. Each guest having received a cup of water from each of their friends pours the water they have received into their jug. What is the smallest possible number of guests that do not have the same amount of water as they started with?\n\nHere *their, they*, etc. are used as gender-neutral singular pronouns.",
    "options": [],
    "answer": "See solution",
    "solution": "Answer: 2.\n\nIf there are guests $1, 2, \\ldots, n$ and guest $i$ is friends with guest $i-1$ and $i+1$ modulo $n$ (e.g., guest $1$ and guest $n$ are friends). Then if guest $i$ has $i$ amount of water in their jug at the start of the game, only guest $1$ and $n$ end up with a different amount of water than they started with.\n\nTo show that there always will be at least two guests with a different amount of water at the end of the game than they started with, let $x_i$ and $d_i$ be the amount of water and number of friends, respectively, that guest $i$ has. Define $z_v = x_v/d_v$ and assume without loss of generality that the friendship graph of the party is connected. Since every guest has at least one friend, there must exist two guests $a$ and $b$ at the party with the same number of friends by the pigeonhole principle. They must satisfy $z_a \\neq z_b$. Thus, the sets\n\n$$\nS = \\{c \\mid z_c = \\min_d z_d\\} \\text{ and } T = \\{c \\mid z_c = \\max_d z_d\\}\n$$\n\nare non-empty and disjoint. Since we assumed the friendship graph to be connected, there exists a guest $c \\in S$ that has a friend $d$ not in $S$. Let $F$ be the friends of $c$ at the party. Then the amount of water in $c$'s jug at the end of the game is\n\n$$\n\\sum_{f \\in F} z_f \\geq z_d + (d_c - 1)z_c > d_c \\cdot z_c = x_c.\n$$\n\nThus, $c$ ends up with a different amount of water at the end of the game. Similarly, there is a guest in $T$ that ends up with a different amount of water at the end of the game than what they started with.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 14883,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $n$ be a fixed positive integer. Find all triples $(a, b, c)$ of integers satisfying the following system of equations:\n\n$$\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$$",
    "options": [],
    "answer": "See solution",
    "solution": "If $a = 0$, then the first equation implies $b^{n+2}c = 0$. Hence $b = 0$ or $c = 0$; without loss of generality, let $b = 0$. Then the last equation reduces to $c^{n+3} = 0$, which implies $c = 0$. Thus, if $a = 0$, then $a = b = c = 0$. Analogously, if $b = 0$ or $c = 0$, then $a = b = c = 0$. The triple $(0, 0, 0)$ satisfies the equations.\n\nNow consider triples $(a, b, c)$ where all terms are nonzero. Let $p$ be any prime number. If $p \\mid a$, then the first equation implies $p \\mid b^{n+2}c$, so $p \\mid b$ or $p \\mid c$; without loss of generality, $p \\mid b$. Then the last equation gives $p \\mid c^{n+1}$, so $p \\mid c$. Thus, if $p \\mid a$, then $p \\mid a, p \\mid b, p \\mid c$. Similarly, if $p \\mid b$ or $p \\mid c$, then $p \\mid a, p \\mid b, p \\mid c$.\n\nNow, write $a = p a'$, $b = p b'$, $c = p c'$. Dividing both sides of all equations by $p^{n+3}$ gives a similar system for $a', b', c'$. Repeating this process, we eventually reach a solution with no common prime factors, i.e., $a, b, c$ are each $\\pm 1$. We now show such solutions do not exist.\n\nIf $n$ is even, the system reduces to:\n\n$$\n\\begin{cases}\na + c + c + b = 0 \\\\\nb + a + a + c = 0 \\\\\nc + b + b + a = 0\n\\end{cases}\n$$\n\nAdding all equations gives $4(a + b + c) = 0$, so $a + b + c = 0$. But the sum of three odd numbers cannot be zero.\n\nIf $n$ is odd, the system reduces to:\n\n$$\n\\begin{cases}\n1 + bc + 1 + ab = 0 \\\\\n1 + ca + 1 + bc = 0 \\\\\n1 + ab + 1 + ca = 0\n\\end{cases}\n$$\n\nAdding all equations gives $2(ab + bc + ca) + 6 = 0$, so $ab + bc + ca = -3$. Since $|ab| = |bc| = |ca| = 1$, the only possibility is $ab = bc = ca = -1$, which is impossible for integers $a, b, c$.\n\nTherefore, the only solution is $(0, 0, 0)$.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 14884,
    "subject": "Mathematics (Olympiad)",
    "question": "Given an integer $m \\ge 2$, and two real numbers $a, b$ with $a > 0$ and $b \\ne 0$, the sequence $\\{x_n\\}$ is defined by $x_1 = b$ and $x_{n+1} = a x_n^m + b$ for $n = 1, 2, \\dots$.\n\nProve that:\n\n1. When $b < 0$ and $m$ is even, the sequence $\\{x_n\\}$ is bounded if and only if $a b^{m-1} \\ge -2$.\n2. When $b < 0$ and $m$ is odd, or when $b > 0$, the sequence $\\{x_n\\}$ is bounded if and only if $a b^{m-1} \\le \\dfrac{(m-1)^{m-1}}{m^m}$.",
    "options": [],
    "answer": "See solution",
    "solution": "(1) When $b < 0$ and $m$ is even:\n\nIf $a b^{m-1} < -2$, then $a b^m + b > -b > 0$, so $x_2 > 0$. Similarly, $x_3 > x_2 > 0$. Since $a x^m + b$ is increasing on $(0, +\\infty)$, each $x_{n+1} > x_n > 0$ for $n \\ge 2$, and $x_n > -b$. Considering three consecutive terms:\n\n$$\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&> 2m (x_{n+1} - x_n) \\\\\n&> x_{n+1} - x_n.\n\\end{aligned}\n$$\n\nThus, the differences increase and the sequence is unbounded.\n\nIf $a b^{m-1} \\ge -2$, by induction, each $x_n$ stays in $[b, -b]$:\n\n- $x_1 = b \\in [b, -b]$.\n- If $b \\le x_n \\le -b$, then $0 \\le x_n^m \\le b^m$, so\n  $$\nb = a \\cdot 0^m + b \\le x_{n+1} \\le a b^m + b \\le -b.\n  $$\n\nTherefore, the sequence is bounded if and only if $a b^{m-1} \\ge -2$.\n\n(2) When $b > 0$:\n\nEach $x_n > 0$. The sequence is bounded if and only if $a x^m + b = x$ has positive real roots. If not, the minimum of $p(x) = a x^m + b - x$ on $(0, +\\infty)$ is $t > 0$, so $x_{n+1} - x_n \\ge t$ and the sequence is unbounded.\n\nIf $a x^m + b = x$ has a positive root $x_0$, then $x_1 = b < x_0$, and if $x_n < x_0$, then $x_{n+1} = a x_n^m + b < a x_0^m + b = x_0$, so $x_n < x_0$ for all $n$ and the sequence is bounded.\n\nThe equation $a x^m + b = x$ has positive roots if and only if the minimum of $a x^{m-1} + \\frac{b}{x}$ on $(0, +\\infty)$ is $\\le 1$. By the AM-GM inequality:\n\n$$\na x^{m-1} + \\frac{b}{x} \\ge m \\sqrt[m]{\\frac{a b^{m-1}}{(m-1)^{m-1}}}\n$$\n\nSo, the sequence is bounded if and only if\n$$m \\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}.$$\n\nWhen $b < 0$ and $m$ is odd, let $y_n = -x_n$. Then $y_1 = -b > 0$, $y_{n+1} = a y_n^m + (-b)$, so the same reasoning applies and the result holds.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 14885,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all functions $f : \\mathbb{R}^+ \\to \\mathbb{R}^+$ such that\n\n$$\n(x + y)f(2y f(x) + f(y)) = x^3 f(y f(x)), \\quad \\forall x, y \\in \\mathbb{R}^+.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "**Solution.** First, we show that such a function must be injective. Indeed, if $f(a) = f(b)$, then for all $y > 0$,\n\n$$\n\\frac{f(y f(a))}{f(2y f(a) + f(y))} = \\frac{f(y f(b))}{f(2y f(b) + f(y))}\n$$\n\nand by the given relation,\n\n$$\n\\begin{align*}\n\\frac{a + y}{a^3} &= \\frac{b + y}{b^3} \\\\\n\\Leftrightarrow \\quad b^3(a + y) &= a^3(b + y) \\\\\n\\Leftrightarrow \\quad (b - a)(a^2 b + a^2 y + a b^2 + a b y + b^2 y) &= 0 \\\\\n\\Leftrightarrow \\quad a &= b, \\quad \\text{since } a^2 b + a^2 y + a b^2 + a b y + b^2 y > 0.\n\\end{align*}\n$$\n\nNow, notice that for $x > 1$, $x^3 - x > 0$. Setting $y = x^3 - x$ in the relation, we get\n\n$$\nf(2(x^3 - x) f(x) + f(x^3 - x)) = f((x^3 - x) f(x)), \\quad \\forall x > 1.\n$$\n\nInjectivity of $f$ then implies\n\n$$\n\\begin{align*}\n& 2(x^3 - x) f(x) + f(x^3 - x) = (x^3 - x) f(x), \\quad \\forall x \\in \\mathbb{R}^+ \\\\\n\\Leftrightarrow \\quad & f(x^3 - x) = (x - x^3) f(x), \\quad \\forall x \\in \\mathbb{R}^+.\n\\end{align*}\n$$\n\nThe last relation for $x = 2$ gives\n\n$$\nf(6) = f(2^3 - 2) = (2 - 2^3) f(2) = -6 f(2) < 0,\n$$\n\nwhich is absurd. So there exists no such function $f$.\n\n**Remark.** After the injectivity, we can conclude as follows: For $x > 1$, $x^3 - x > 0$, so setting $y = x^3 - x$ in the relation we get $f(2y f(x) + f(y)) = f(y f(x))$, which by injectivity implies $2y f(x) + f(y) = y f(x)$, thus $y f(x) + f(y) = 0$, which cannot hold as $y, f(x), f(y) > 0$.\n\nActually, the same argument shows that no function $f : \\mathbb{R}^+ \\to \\mathbb{R}^-$ exists satisfying the same relation (necessity of injectivity for $f$ is proved again as above).",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 14886,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all integers $n \\ge 3$ such that among any $n$ positive real numbers $a_1, a_2, \\dots, a_n$ with\n\n$$\n\\max(a_1, a_2, \\dots, a_n) \\le n \\cdot \\min(a_1, a_2, \\dots, a_n),\n$$\n\nthere exist three that are the side lengths of an acute triangle.",
    "options": [],
    "answer": "See solution",
    "solution": "The answer is $n \\ge 13$.\n\nFirst, we show that any $n \\ge 13$ satisfies the desired condition. Suppose for the sake of contradiction that $a_1 \\le a_2 \\le \\dots \\le a_n$ are integers such that $\\max(a_1, a_2, \\dots, a_n) \\le n \\cdot \\min(a_1, a_2, \\dots, a_n)$ and no three are the side lengths of an acute triangle. We conclude that\n\n$$\na_{i+2}^2 \\ge a_i^2 + a_{i+1}^2 \\quad (1)\n$$\n\nfor all $i \\le n - 2$. Letting $\\{F_n\\}$ be the Fibonacci numbers, defined by $F_1 = F_2 = 1$ and $F_{n+1} = F_n + F_{n-1}$ for $n \\ge 2$, repeated application of (1) and the ordering of the $\\{a_i\\}$ implies that\n\n$$\na_i^2 \\ge F_i \\cdot a_1^2 \\quad (2)\n$$\n\nfor all $i \\le n$. Noting that $F_{12} = 12^2$, an easy induction shows that $F_n > n^2$ for $n > 12$. Hence, if $n \\ge 13$, (2) implies $a_n^2 > n^2 \\cdot a_1^2$, a contradiction. This shows that any $n \\ge 13$ satisfies the condition of the problem.\n\nOn the other hand, for any $n < 13$, we may take $a_i = \\sqrt{F_i}$ for $1 \\le i \\le n$, so that\n\n$$\n\\max(a_1, a_2, \\dots, a_n) \\le n \\cdot \\min(a_1, a_2, \\dots, a_n)\n$$\n\nholds because $F_n \\le n^2$ for $n \\le 12$. Further, for $i < j$, we have $F_i + F_j \\le F_{j+1}$, which shows that for $i < j < k$, we have $a_k^2 \\ge a_i^2 + a_j^2$. Hence, $\\{a_i, a_j, a_k\\}$ are not the side lengths of an acute triangle. Therefore, all $n < 13$ do not satisfy the conditions of the problem, and the answer is $n \\ge 13$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 14887,
    "subject": "Mathematics (Olympiad)",
    "question": "For any two finite sets $A$ and $B$, define $f(A, B)$ to be the number of elements that are contained either in $A$ or in $B$, but not in both $A$ and $B$.\n\nThree given sets $X$, $Y$ and $Z$ satisfy\n\n$$\nf(X, Y) = f(Y, Z) = f(Z, X).\n$$\n\n(a) Prove that $f(X, Y)$ is even.\n\n(b) Find a set $W$ such that\n\n$$\nf(W, X) = f(W, Y) = f(W, Z) = \\frac{1}{2} f(X, Y).\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "In the Venn diagram, the letters $a$, $b$, $c$, $d$, $e$, $g$ represent the number of elements in the respective regions.\n\n![](images/Brown_Australian_MO_Scene_2013_p45_data_1620fe7cd1.png)\n\n1. We have\n\n$$\nf(X,Y) = a + g + b + e,\n$$\n\n$$\nf(Y,Z) = b + d + c + g,\n$$\n\n$$\nf(Z,X) = c + e + a + d.\n$$\n\nHence,\n\n$$\nf(X,Y) = f(Y,Z) + f(Z,X) - 2(c + d),\n$$\n\n$$\nf(Y,Z) = f(X,Y) + f(Z,X) - 2(a + e),\n$$\n\n$$\nf(Z,X) = f(X,Y) + f(Y,Z) - 2(b + g).\n$$\n\nSince $f(X, Y) = f(Y, Z) = f(Z, X)$, equation above gives $f(X, Y) = 2(c + d)$. Hence $f(X, Y)$ is even.\n\n2. Let $W$ be the set of elements that are in at least two of $X$, $Y$ and $Z$.\n\nThen $f(W, X) = a + e$, $f(W, Y) = b + g$, $f(W, Z) = c + d$. Since $f(X, Y) = f(Y, Z) = f(Z, X)$, the equations above give $c + d = a + e = b + g$.\n\nHence $f(W, X) = f(W, Y) = f(W, Z) = c + d = \\frac{1}{2} f(X, Y)$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 14888,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $\\triangle ABC$ be an acute, non-isosceles triangle with centroid $G$ and circumcircle $(O)$. Let $H_a, H_b, H_c$ be the projections of $A, B, C$ onto the opposite sides, and $D, E, F$ the midpoints of $BC, CA, AB$, respectively. The rays $GH_a, GH_b, GH_c$ meet $(O)$ at $X, Y, Z$.\n\n(a) Prove that the circle $(XCE)$ passes through the midpoint of the segment $BH_a$.\n\n(b) Let $M, N, P$ be the midpoints of $AX, BY, CZ$. Prove that the lines $DM, EN, FP$ are concurrent.",
    "options": [],
    "answer": "See solution",
    "solution": "(a) Let $A_0$ be the intersection of the line through $A$ parallel to $BC$ with the circle $(O)$. Since $AA_0CB$ is an isosceles trapezoid, $A_0C = AB = 2FH_a$. On the other hand,\n\n$$\n\\angle FH_aB = \\angle FBH_a = \\angle A_0CB\n$$\n\nimplies that $FH_a \\parallel A_0C$. Since $GC = 2GF$, $G, A_0, H_a$ are collinear. Let $S$ be the midpoint of $A_0H_a$; then $SE \\parallel AA_0$, so $\\angle AES = \\angle A_0AC = \\angle A_0XC$, thus $SECX$ is cyclic.\n\n![](images/VN_IMO_Booklet_2018_Final_p24_data_d118b5e8cf.png)\n\nLet $T$ be the midpoint of $BH_a$; then $ST \\parallel BA_0$. Hence, $\\angle XST = \\angle XA_0B = \\angle XCB$, so the quadrilateral $XCST$ is also cyclic.\n\nTherefore, the five points $X, C, E, S, T$ are concyclic, so the circle $(XCE)$ passes through the midpoint of $BH_a$.\n\n(b) We prove some remarks:\n\n**Remark 1.** $AX, BY, CZ$ are concurrent at a point on $OH$, where $H$ is the orthocenter of $\\triangle ABC$.\n\n_Proof._ Note that $\\angle AXH_a = \\angle HAO$, so $AO$ is tangent to $(XAH_a)$. Thus $\\mathcal{P}_{O/(XAH_a)} = R^2$, where $R$ is the radius of $(O)$. Similarly for $(YBH_b)$ and $(ZCH_c)$, so $O$ is the radical center of these circles. On the other hand,\n\n$$\n\\mathcal{P}_{H/(XAH_a)} = \\overline{HA} \\cdot \\overline{HA_a} = \\frac{1}{2} \\mathcal{P}_{H/(O)}\n$$\n\nand similarly, $H$ is the radical center of $(XAH_a)$, $(YBH_b)$, $(ZCH_c)$.\n\nHence, $OH$ is the radical axis of these circles. Let $T$ be the intersection of $OH$ and $AX$; then the power of $T$ to the four circles $(XAH_a)$, $(YBH_b)$, $(ZCH_c)$, $(O)$ are equal, which implies that $BY, CZ$ also pass through $T$.\n\nThus, $AX, BY, CZ, OH$ are concurrent at $T$.\n\n![](images/VN_IMO_Booklet_2018_Final_p25_data_81817da0e7.png)\n\n**Remark 2.** Let $X_1$ be the reflection of $X$ with respect to $D$, and $A_1$ the midpoint of $AX_1$. Then $A_1$ belongs to the nine-point circle of $\\triangle ABC$.\n\n_Proof._ $G$ is also the centroid of $\\triangle AXX_1$, so $G \\in MX_1$. Consider $\\Omega$ as the homothety centered at $G$ with ratio $-2$:\n\n$$\n\\Omega : A \\to D,\\ X_1 \\to M,\\ (ABC) \\to (DEF).\n$$\n\nSince $X \\in (ABC)$ and $\\Omega : X \\to A_1$, $A_1 \\in (DEF)$, which is the nine-point circle of $\\triangle ABC$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 14889,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $\\alpha$ be the radian measure of the smallest angle in a $3$-$4$-$5$ right triangle. Let $\\beta$ be the radian measure of the smallest angle in a $7$-$24$-$25$ right triangle. In terms of $\\alpha$, what is $\\beta$?\n\n(A) $\\frac{\\alpha}{3}$  \n(B) $\\alpha - \\frac{\\pi}{8}$  \n(C) $\\frac{\\pi}{2} - 2\\alpha$  \n(D) $\\frac{\\alpha}{2}$  \n(E) $\\pi - 4\\alpha$",
    "options": [],
    "answer": "See solution",
    "solution": "Because $\\alpha$ and $\\beta$ are the smallest angles in these triangles, $\\sin \\alpha = \\frac{3}{5}$, $\\cos \\alpha = \\frac{4}{5}$, $\\sin \\beta = \\frac{7}{25}$, and $\\cos \\beta = \\frac{24}{25}$. By a double angle formula:\n\n$$\n\\sin(2\\alpha) = 2 \\sin \\alpha \\cdot \\cos \\alpha = 2 \\cdot \\frac{3}{5} \\cdot \\frac{4}{5} = \\frac{24}{25} = \\cos \\beta = \\sin\\left(\\frac{\\pi}{2} - \\beta\\right)\n$$\n\nBecause both $2\\alpha$ and $\\beta$ are acute, $2\\alpha = \\frac{\\pi}{2} - \\beta$, so $\\beta = \\frac{\\pi}{2} - 2\\alpha$.\n\nAlternatively, using complex numbers in polar form: $4 + 3i = 5(\\cos \\alpha + i \\sin \\alpha)$. Squaring gives $7 + 24i = 25(\\cos \\alpha + i \\sin \\alpha)^2$. Similarly, $24 + 7i = 25(\\cos \\beta + i \\sin \\beta)$. Multiplying these two equations yields\n\n$$\n(7 + 24i)(24 + 7i) = 25(\\cos \\alpha + i \\sin \\alpha)^2 \\cdot 25(\\cos \\beta + i \\sin \\beta) \\\\\n625i = 625 (\\cos(2\\alpha + \\beta) + i \\sin(2\\alpha + \\beta)) \\\\\n\\cos \\frac{\\pi}{2} + i \\sin \\frac{\\pi}{2} = \\cos(2\\alpha + \\beta) + i \\sin(2\\alpha + \\beta)\n$$\n\nBecause both $2\\alpha$ and $\\beta$ are acute, $2\\alpha + \\beta = \\frac{\\pi}{2}$ and $\\beta = \\frac{\\pi}{2} - 2\\alpha$.\n\n**Note:** The angles are $\\alpha \\approx 36.87^\\circ$ and $\\beta \\approx 16.26^\\circ$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 14890,
    "subject": "Mathematics (Olympiad)",
    "question": "What is the maximum number of integers that can be chosen from $1, 2, \\ldots, 99$ so that the chosen integers can be arranged in a circle with the property that the product of every pair of neighbouring integers is a 3-digit number?",
    "options": [],
    "answer": "See solution",
    "solution": "Since $31 \\times 32 = 992$ and $31 \\times 33 = 1023$, any two numbers larger than $31$ cannot be neighbours. So there must be a number less than $32$ between a pair of such numbers. Also, $1$ cannot be chosen. The two neighbours of $31$ are $32$ or less. So the maximum number of chosen integers is $\\leq 30 \\times 2 - 1 = 59$. This bound can be achieved by the following arrangement, where $11$ follows $31$ to form a cycle:\n\n$11, 83, 12, 76, 13, 71, 14, 66, 15, 62, 16, 58, 17, 55, 18, 52, 19, 49,$\n$20, 47, 3, 99, 2, 98, 4, 97, 5, 96, 6, 95, 7, 94, 8, 93, 9, 92, 10, 46, 21, 45,$\n$22, 43, 23, 41, 24, 39, 25, 38, 26, 37, 27, 35, 28, 34, 29, 33, 30, 32, 31$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 14891,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $ABC$ be a triangle. Let $P$ be the point on the extension of $BC$ beyond $B$ such that $BP = BA$. Let $Q$ be the point on the extension of $BC$ beyond $C$ such that $CQ = CA$. Prove that the circumcenter $O$ of the triangle $APQ$ lies on the angle bisector of the angle $\\angle BAC$.\n\n![](images/AUT_ABooklet_2023_p12_data_57e7658abf.png)",
    "options": [],
    "answer": "See solution",
    "solution": "Since $ACQ$ is an isosceles triangle, the perpendicular bisector of $AQ$ is the angle bisector of $\\angle QCA$. But the perpendicular bisector of $AQ$ also passes through the circumcenter $O$ of triangle $APQ$.\n\nTherefore, $O$ lies on the angle bisector of $\\angle QCA$, which is the exterior angle bisector of $\\angle ACB$ by the definition of $Q$.\n\nAnalogously, the point $O$ lies also on the exterior angle bisector of $\\angle CBA$. Therefore, the point $O$ is the intersection of the two exterior angle bisectors, which makes it the excenter of the excircle of $ABC$ tangent to $BC$. This excenter lies on the angle bisector of $\\angle BAC$ as desired.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 14892,
    "subject": "Mathematics (Olympiad)",
    "question": "Find the smallest and largest integers with decimal representation of the form $ababa$ ($a \\neq 0$) that are divisible by 11.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $N = ababa$, where $a \\neq 0$. The divisibility rule for 11 states that $N$ is divisible by 11 if and only if the alternating sum of its digits is divisible by 11:\n\n$$a - b + a - b + a = 3a - 2b$$\n\nWe seek integers $a$ ($1 \\leq a \\leq 9$) and $b$ ($0 \\leq b \\leq 9$) such that $3a - 2b$ is divisible by 11.\n\n**Smallest $N$:**\n- Try $a = 1$ (smallest possible $a$):\n  - $3 \\cdot 1 - 2b = 3 - 2b$\n  - Set $3 - 2b = -11 \\implies b = 7$ (since $b$ must be an integer between 0 and 9)\n- Thus, $N = 17171$ is the smallest such number.\n\n**Largest $N$:**\n- Try $a = 9$ (largest possible $a$):\n  - $3 \\cdot 9 - 2b = 27 - 2b$\n  - Set $27 - 2b = 11 \\implies b = 8$\n- Thus, $N = 98989$ is the largest such number.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 14893,
    "subject": "Mathematics (Olympiad)",
    "question": "令 $N$ 表示所有正整數所成之集合。求所有函數 $f: N \\to N$，滿足：對於所有 $m, n \\in N$，$f(m) + f(n) - mn \\neq 0$ 且 $f(m) + f(n) - mn$ 能整除 $mf(m) + nf(n)$。",
    "options": [],
    "answer": "See solution",
    "solution": "$f(n) = n^2$。\n\n已知：\n\n$$\n(f(m) + f(n) - mn) \\mid (mf(m) + nf(n)).\n$$\n\n1. 取 $m = n = 1$，得 $2f(1) - 1 \\mid 2f(1)$，所以 $f(1) = 1$。\n\n2. 令 $p$ 為質數，取 $(m, n) = (p, 1)$，則 $f(p) - 1 + p \\mid pf(p) + 1$，即：\n\n$$\nf(p) - p + 1 \\mid pf(p) + 1 - p(f(p) - p + 1) = p^2 - p + 1.\n$$\n\n- 若 $f(p) - p + 1 = p^2 - p + 1$，則 $f(p) = p^2$。\n- 若 $f(p) - p + 1 \\neq p^2 - p + 1$，由 $p^2 - p + 1$ 為奇數，必有 $3(f(p) - p + 1) \\leq p^2 - p + 1$，即：\n\n$$\nf(p) \\leq \\frac{1}{3}(p^2 + 2p - 2)\n$$\n\n3. 取 $m = n = p$，得 $2f(p) - p^2 \\mid 2pf(p)$，即：\n\n$$\n2f(p) - p^2 \\mid 2pf(p) - p(2f(p) - p^2) = p^3.\n$$\n\n由上式及 $f(p) \\geq 1$，有：\n\n$$\n-p^2 < 2f(p) - p^2 \\leq \\frac{2}{3}(p^2 + 2p - 2) - p^2 < -p\n$$\n\n對 $p \\geq 7$，這與 $2f(p) - p^2$ 為 $p^3$ 的因數矛盾。\n\n故 $f(p) = p^2$ 對所有質數 $p \\geq 7$ 成立。\n\n4. 固定 $n$，取質數 $p \\geq 7$，設 $m = p$，則：\n\n$$\np^2 + f(n) - pn \\mid p(p^2 - pn + n^2)\n$$\n\n可取 $p$ 充分大使 $p \\nmid f(n)$，故 $p^2 - pn + f(n) \\nmid p^2 - pn + n^2$。因此 $n^2 - f(n) = 0$，即 $f(n) = n^2$。\n\n5. 驗證 $f(n) = n^2$ 確實滿足條件。",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 14894,
    "subject": "Mathematics (Olympiad)",
    "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 \\leq i < j \\leq n$, are constructed. The points $V_i$ and $V_j$ are *neighbors* if $V_iV_j$ is constructed.\n\nInitially, 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 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.\n\nA set of constructed segments is called *harmonic* if for any initial positions of the chess pieces, each chess piece $C_i$ ($1 \\leq i \\leq n$) is at the point $V_i$ after a finite number of moves.\n\nDetermine the minimum number of segments in a harmonic set.\n\n![](path/to/file.png)",
    "options": [],
    "answer": "See solution",
    "solution": "The answer is $n+1$.\n\nFor a harmonic set, consider a graph $G$ with $V_1, V_2, \\dots, V_n$ as its vertices and the segments in the harmonic set as its edges.\n\nFirst, there must be at least $n$ edges in $G$. $G$ must be connected, and each vertex must have degree at least $2$, because when a chess piece is moved from $V_i$ to $V_j$, another piece is moved from $V_k$ ($k \\neq j$) to $V_i$. Thus, the total degree is at least $2n$, so there are at least $n$ edges.\n\nSecond, there must be at least $n+1$ edges. If there are only $n$ edges, then in this connected graph, each vertex has degree $2$, so it must be a cycle. Suppose the cycle $V_1 \\to V_2 \\to \\dots \\to V_n \\to V_1$ consists of all the edges. 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, because we can only rotate all the pieces along the cycle and cannot change their relative positions.\n\nThird, $n+1$ edges is enough. 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 now has a second cycle $C_2: V_2 \\to V_3 \\to \\dots \\to V_n \\to V_2$. With this extra edge, we can switch the relative positions of the chess pieces along $C_1$. For example, if $C_i$ is at $V_1$ and $C_j$ is at $V_2$ initially, applying rotations on $C_2$ allows us to place $C_j$ at $V_n$, switching the relative positions of $C_i$ and $C_j$ along $C_1$. Because we can switch the positions of any two neighboring pieces in a finite number of moves, we can place $C_1, C_2, \\dots, C_n$ in order on $C_1$, and then move each $C_i$ to $V_i$ by rotating along $C_1$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 14895,
    "subject": "Mathematics (Olympiad)",
    "question": "There are $n$ students standing in a circle, one behind the other. The students have heights $h_1 < h_2 < \\dots < h_n$. If a student with height $h_k$ is standing directly behind a student with height $h_{k-2}$ or less, the two students are permitted to switch places. Prove that it is not possible to make more than $\\binom{n}{3}$ such switches before reaching a position in which no further switches are possible.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $h_i$ also denote the student with height $h_i$. We prove that for $1 \\le i < j \\le n$, $h_j$ can switch with $h_i$ at most $j - i - 1$ times. We proceed by induction on $j - i$, the base case $j - i = 1$ being evident because $h_i$ is not allowed to switch with $h_{i-1}$.\n\nFor the inductive step, note that $h_i, h_{j-1}, h_j$ can be positioned on the circle either in this order or in the order $h_i, h_j, h_{j-1}$. Since $h_{j-1}$ and $h_j$ cannot switch, the only way to change the relative order of these three students is for $h_i$ to switch with either $h_{j-1}$ or $h_j$. Consequently, any two switches of $h_i$ with $h_j$ must be separated by a switch of $h_i$ with $h_{j-1}$. Since there are at most $j - i - 2$ of the latter, there are at most $j - i - 1$ of the former.\n\nThe total number of switches is thus at most\n\n$$\n\\begin{aligned}\n\\sum_{i=1}^{n-1} \\sum_{j=i+1}^{n} (j-i-1) &= \\sum_{i=1}^{n-1} \\sum_{j=0}^{n-i-1} j \\\\\n&= \\sum_{i=1}^{n-1} \\binom{n-i}{2} \\\\\n&= \\sum_{i=1}^{n-1} \\left[ \\binom{n-i+1}{3} - \\binom{n-i}{3} \\right] \\\\\n&= \\binom{n}{3}.\n\\end{aligned}\n$$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 14896,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $ABCD$ be a cyclic quadrilateral, and let $E$, $F$, $G$, and $H$ be the midpoints of $AB$, $BC$, $CD$, and $DA$, respectively. Let $W$, $X$, $Y$, and $Z$ be the orthocenters of triangles $AHE$, $BEF$, $CFG$, and $DGH$, respectively. Prove that quadrilaterals $ABCD$ and $WXYZ$ have the same area.",
    "options": [],
    "answer": "See solution",
    "solution": "**Lemma.** Let $ABCD$ be any quadrilateral with $E$, $F$, $G$, $H$ the midpoints of sides $AB$, $BC$, $CD$, $DA$, respectively. Then $[ABCD] = 2[EFGH]$.\n\n*Proof.* Let $P$ be the intersection of diagonals $AC$ and $BD$. $HE$ is the midline of triangle $ABD$ so $HE$ bisects segment $AP$. It follows that $2[HPE] = [AHPE]$. Similarly, we obtain $2[EPF] = [BEPF]$, $2[FPG] = [CFPG]$, and $2[GPH] = [DGPH]$. Summing these up yields the desired result. $\\square$\n\nLet $O$ denote the circumcenter of $ABCD$, and let $P$, $Q$, $R$, $S$ denote the midpoints of $HE$, $EF$, $FG$, $GH$ respectively. Applying the lemma to $ABCD$ and then $EFGH$, we get $[ABCD] = 2[EFGH] = 4[PQRS]$.\n\nWe have $WH \\parallel OE$ since they are both perpendicular to $AB$, and similarly $WE \\parallel OH$, so $WHOE$ is a parallelogram. It follows that $W$ is the image of the midpoint of $P$ (i.e., the midpoint of $HE$) under dilation from $O$ by a factor of $2$. Similarly, $X$, $Y$, $Z$ are the images of $R$, $S$, $T$ under the same transformation, so $WXYZ$ is the image of $PQRS$ under a dilation of factor $2$. Thus, it follows that $[WXYZ] = 4[PQRS] = [ABCD]$ as desired.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 14897,
    "subject": "Mathematics (Olympiad)",
    "question": "Show that the equation\n\n$$\n\\prod_{i=1}^{504} (x - (4i - 2))(x - (4i - 1)) = \\prod_{i=1}^{504} (x - (4i - 3))(x - 4i)\n$$\n\ndoes not have any real solutions.",
    "options": [],
    "answer": "See solution",
    "solution": "**Case 1:** $x = 1, 2, \\dots, 2016$.\n\nOne side of the equation is zero while the other is not, so there are no solutions in this case.\n\n**Case 2:** $4r - 3 < x < 4r - 2$ or $4r - 1 < x < 4r$ for some $r \\in \\{1, 2, \\dots, 504\\}$.\n\nFor every $i \\in \\{1, 2, \\dots, 504\\}$, $(x - (4i - 2))(x - (4i - 1)) > 0$, so the left side is positive. For $i \\neq r$, $(x - (4i - 3))(x - 4i) > 0$, but for $i = r$, $(x - (4r - 3))(x - 4r) < 0$, so the right side is negative. Thus, no solutions exist in this case.\n\n**Case 3:** $x < 1$, $x > 2016$, or $4r < x < 4r + 1$ for some $r \\in \\{1, 2, \\dots, 503\\}$.\n\nRewrite the equation as\n\n$$\n\\begin{aligned}\n1 &= \\prod_{i=1}^{504} \\frac{(x - (4i - 2))(x - (4i - 1))}{(x - (4i - 3))(x - 4i)} \\\\\n  &= \\prod_{i=1}^{504} \\left( 1 + \\frac{2}{(x - (4i - 3))(x - 4i)} \\right).\n\\end{aligned}\n$$\n\nSince $(x - (4i - 3))(x - 4i) > 0$, the product is greater than 1, so there are no solutions.\n\n**Case 4:** $4r - 2 < x < 4r - 1$ for some $r \\in \\{1, 2, \\dots, 504\\}$.\n\nRewrite the equation as\n\n$$\n\\begin{aligned}\n1 &= \\prod_{i=1}^{504} \\frac{(x - (4i - 3))(x - 4i)}{(x - (4i - 2))(x - (4i - 1))} \\\\\n  &= \\frac{x-1}{x-2} \\cdot \\frac{x-2016}{x-2015} \\cdot \\prod_{i=1}^{503} \\frac{(x-4i)(x - (4i+1))}{(x - (4i - 1))(x - (4i + 2))} \\\\\n  &= \\frac{x-1}{x-2} \\cdot \\frac{x-2016}{x-2015} \\cdot \\prod_{i=1}^{503} \\left( 1 + \\frac{2}{(x - (4i + 2))(x - (4i - 1))} \\right)\n\\end{aligned}\n$$\n\nSince $(x-1)(x-2016) < (x-2)(x-2015)$ for $2 < x < 2015$, $\\frac{x-1}{x-2} \\cdot \\frac{x-2016}{x-2015} > 1$, and $(x - (4i + 2))(x - (4i - 1)) > 0$, so the product is greater than 1. Thus, no solutions exist in this case.\n\nFrom all cases, the equation has no real solutions. $\\square$",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 14898,
    "subject": "Mathematics (Olympiad)",
    "question": "On a circle of radius $r$, the distinct points $A$, $B$, $C$, $D$, and $E$ lie in this order, satisfying $AB = CD = DE > r$. Show that the triangle with vertices at the centroids of the triangles $ABD$, $BCD$, and $ADE$ is obtuse.",
    "options": [],
    "answer": "See solution",
    "solution": "Denote by $P$, $Q$, and $R$ the centroids of the triangles $ABD$, $BCD$, and $ADE$, respectively. Let $K$ and $L$ be the midpoints of the segments $BD$ and $AD$, respectively. Since $P$ and $Q$ are centroids, they divide the medians $AK$ and $CK$ in the same ratio. That is, $AP : PK = CQ : QK = 2 : 1$, and we have $PQ \\parallel AC$. Similarly, $PR \\parallel BE$. Hence, the angle $QPR$ is equal to the angle $CXE$ determined by the lines $AC$ and $BE$ (here, $X$ is the intersection point of $AC$ and $BE$, see Fig. 1).\n\n![](images/SVK_Other_2015_p1_data_08fb6fe126.png)\n\nDenote by $\\varphi$ the measure of the inscribed angle determined by the chord $AB$ of the given circle. Since $CD = DE = AB$, we have $\\angle CAE = 2\\varphi$, and therefore from triangle $AXE$ we conclude\n\n$$\n\\angle CXE = 180^\\circ - \\angle AXE = \\varphi + 2\\varphi = 3\\varphi.\n$$\n\nSince $AB > r$, we have $\\varphi > 30^\\circ$, and so $\\angle QPR = 3\\varphi > 90^\\circ$.\n\n*Remark.* The points $P$, $Q$, $R$ always determine a triangle, i.e., they cannot be collinear. This follows from the fact that the diagonals $AC$ and $BE$ of the cyclic quadrilateral $ABCE$ always determine an angle of measure less than $180^\\circ$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 14899,
    "subject": "Mathematics (Olympiad)",
    "question": "Call _pure_ any positive integer $n$ that does not occur in any integer sequence $c_0, c_1, c_2, \\dots$, where $0 < c_0 < n$ and\n$$\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$$\nfor every $i \\ge 1$. (For instance, 10 is not pure since it occurs in the sequence 5, 14, 7, 20, 10, ...)\n\na) Is every positive multiple of 3 pure?\n\nb) Prove that if an integer $n > 1$ is pure but not divisible by 3, then $n + 1$ is divisible by 6.",
    "options": [],
    "answer": "See solution",
    "solution": "a) Note that $3c_{i-1} - 1$ is never divisible by 3 and if $\\frac{1}{2}c_{i-1}$ is divisible by 3, then also $c_{i-1}$ is divisible by 3. Thus, if some term $c_k = n$ is divisible by 3, then, up to it, only dividing by 2 is used to build the terms (i.e., $c_i = \\frac{1}{2}c_{i-1}$ for every $i$ such that $1 \\le i \\le k$) and, consequently, $c_0 > c_1 > \\dots > c_k = n$. But this contradicts the condition $c_0 < n$. Hence every positive multiple of 3 is pure.\n\nb) If $n$ is not divisible by 3, then $n = 3k + 1$ or $n = 6k + 2$ or $n = 6k + 5$. If $n = 3k + 1$, then taking $c_0 = 2k + 1$ gives $c_1 = 6k + 2$ and $c_2 = 3k + 1 = n$. Thereby $k > 0$ since $n > 1$, therefore $c_0 < n$. Hence none of such numbers $n$ is pure. If $n = 6k + 2$, then taking $c_0 = 2k + 1$ gives $c_1 = 6k + 2 = n$, whereby $c_0 < n$. Hence also none of such numbers $n$ is pure. Hence, among the positive integers $n > 1$ not divisible by 3, only those of the form $n = 6k + 5$ can be pure.\n\n**Remark.** Not every integer of the form $n = 6k + 5$ is pure. For example, $23 = 6 \\cdot 3 + 5$ occurs in the sequence 21, 62, 31, 92, 46, 23, ...",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 14900,
    "subject": "Mathematics (Olympiad)",
    "question": "試求出所有由正整數集映至正整數集的函數對 $(f, g)$ 滿足\n\n$$\nf^{g(n)+1}(n) + g^{f(n)}(n) = f(n+1) - g(n+1) + 1\n$$\n\n對所有正整數 $n$ 皆成立。這裡定義 $f^1(n) = f(n)$，$f^{k+1}(n) = f(f^k(n))$。",
    "options": [],
    "answer": "See solution",
    "solution": "唯一滿足題目敘述的函數對 $(f, g)$ 是 $f(n) = n,\\ g(n) = 1$。\n\n由條件可知對所有正整數 $n$ 都有\n\n$$\nf(f^{g(n)}(n)) < f(n+1).\n$$\n\n將函數 $f$ 能取到的所有值依大小記為 $y_1 < y_2 < \\dots$（這個序列的長度可能是有限或無限），我們接下來要運用數學歸納法證明：\n\n$$\n(i)_n : f(x) = y_n \\text{ 若且唯若 } x = n,\n$$\n\n$$\n(ii)_n : y_n = n.\n$$\n\n證明的順序如下：\n\n$$\n(i)_1 \\to (ii)_1 \\to (i)_2 \\to (ii)_2 \\to \\dots \\to (i)_n \\to (ii)_n \\dots\n$$\n\n我們先證明 $(i)_1$。假設一正整數 $x$ 有 $f(x) = y_1$，如果 $x > 1$ 則有\n\n$$\nf(f^{g(x-1)}(x-1)) < f(x) = y_1,\n$$\n\n這與 $y_1$ 是函數 $f$ 值域中最小的數矛盾。\n\n假設現在對於一正整數 $n$，我們已經證明了 $(i)_n$ 以及 $(i)_{n-1}$ 以前的所有命題。換句話說我們已經知道 $f(n) = y_n$ 若且唯若 $x = n$，並且對於 $1 \\leq a < n$ 有 $f(x) = a = y_a$ 若且唯若 $x = a$。注意到這也表示對於任意 $1 \\leq a < n$ 與正整數 $k$，$f^k(x) = a$ 若且唯若 $x = a$。\n\n由於對於 $y_1, y_2, \\cdots, y_n$ 都恰只有一個正整數帶入 $f$ 後對應到它們，因此 $y_{n+1}$ 存在。取任一使 $f(x) = y_{n+1}$ 的正整數 $x$，則 $x$ 必定大於 $n$（根據 $(i)_1, \\cdots, (i)_n$）。將 $x-1$ 帶入上述不等式得到\n\n$$\nf(f^{g(x-1)}(x-1)) < f(x) = y_{n+1},\n$$\n\n因此若記 $f^{g(x-1)}(x-1) = b$，則有\n\n$$\nb \\in \\{1, \\cdots, n\\}\n$$\n\n如果 $b < n$，那麼我們就會有 $x - 1 = b < n$（因為 $f^k(x - 1) = b < n$ 若且唯若 $x - 1 = b$），這與 $x > n$ 矛盾。因此 $b = n$，從而 $y_n = n$（因為已知 $y_{n-1} = n - 1$，所以 $n$ 是值域中比 $y_{n-1}$ 大的最小正整數），這就證明了 $(ii)_n$。\n\n所以根據 $f^{g(x-1)}(x-1) = n$ 和 $(i)_n$，我們知道 $x - 1 = n$，即 $x$ 唯一可能的值就是 $x = n + 1$，這也證明了 $(i)_{n+1}$。\n\n由數學歸納法我們可以知道 $(i)_n$ 和 $(ii)_n$ 對所有正整數 $n$ 成立。因此 $f(n) = n$。帶回原題目的條件，我們得到 $g^n(n) + g(n+1) = 2$。由於 $g(n)$ 的值域是正整數，我們可以立即推得 $g(n) = 1$。",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 14901,
    "subject": "Mathematics (Olympiad)",
    "question": "The area of a right-angled triangle is $120$ and each of its three sides has integer length.\n\nFind the length of its hypotenuse.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $x, y, z$ be the side lengths of the triangle with $x \\leq y < z$. By the Pythagorean theorem, $x^2 + y^2 = z^2$, and the area condition gives $\\frac{xy}{2} = 120$, so $xy = 240$.\n\nThe possible integer pairs $(x, y)$ with $xy = 240$ are: $(1,240)$, $(2,120)$, $(3,80)$, $(4,60)$, $(5,48)$, $(6,40)$, $(8,30)$, $(10,24)$, $(12,20)$, $(15,16)$.\n\nWe check which of these pairs satisfy $x^2 + y^2 = z^2$ for integer $z$:\n\n| $x$ | $y$  | $x^2 + y^2$ | $z$ | Square? |\n|-----|------|-------------|-----|---------|\n| 1   | 240  | 57601       |     | No      |\n| 2   | 120  | 14404       |     | No      |\n| 3   | 80   | 6409        |     | No      |\n| 4   | 60   | 3616        |     | No      |\n| 5   | 48   | 2309        |     | No      |\n| 6   | 40   | 1636        |     | No      |\n| 8   | 30   | 964         |     | No      |\n| 10  | 24   | 676         | 26  | Yes     |\n| 12  | 20   | 544         |     | No      |\n| 15  | 16   | 481         |     | No      |\n\nThus, the only valid triple is $(10, 24, 26)$, so the length of the hypotenuse is $\\boxed{26}$.\n\nAlternatively, by listing familiar Pythagorean triples, we find $(10, 24, 26)$ is the only one with area $120$ and integer sides.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 14902,
    "subject": "Mathematics (Olympiad)",
    "question": "There are $n$ cities in a country, where $n \\ge 100$ is a positive integer. Some pairs of cities are connected by (two-way) flights. For two cities $A$ and $B$, a path is a sequence of distinct cities $C_0, C_1, C_2, \\dots, C_k, C_{k+1}$, such that there are flights between $C_i$ and $C_{i+1}$ for every $0 \\le i \\le k$, with $C_0 = A$ and $C_{k+1} = B$.\n\nA long path between $A$ and $B$ is defined as a path such that no other path has more vertices. Similarly, a short path is defined as a path with the fewest vertices. In particular, if $A$ and $B$ have a direct flight, that is the shortest path.\n\nAssume that for any pair of cities $A$ and $B$ in the country, there exist a long path and a short path between them that have no cities in common (except $A$ and $B$). For a given $n$, find all possible numbers of flights in the country.",
    "options": [],
    "answer": "See solution",
    "solution": "Use the obvious graph interpretation. We show that any such graph is one of the following: the full graph $K_n$, the circular graph $C_n$, and for $n$ even, the bipartite graph $K_{\\frac{n}{2}, \\frac{n}{2}}$. First, we show that these graphs satisfy the condition.\n\n* For $K_n$, we can choose any long path and the short path is the edge.\n* For $C_n$, we have exactly two paths between any two vertices, and one of them has at most as many vertices as the other.\n* For $K_{\\frac{n}{2}, \\frac{n}{2}}$, if the vertices are on different sides, the short path is the edge. Otherwise, take any long path. We observe that it alternates between the sides and begins and ends on one side. Therefore, there is a vertex on the other side that doesn't appear in the long path. Additionally, there is a short path that passes through this vertex.\n\nNext, we show that only these graphs work for $n$ large enough.\n\nThe graph is clearly connected, as any two vertices belong to a path. Consider a longest path in the graph. Let $p$ be its length and denote the vertices in the path by $V_1, V_2, \\dots, V_p$ in the corresponding order. We can assume that this path is the long path between $V_1$ and $V_p$ that has a corresponding short path through other vertices. We show that the edge $V_1V_p$ belongs to the graph. If the edge doesn't exist, the short path has length at least two, implying that there is a vertex $X$ different from $V_i, i \\in \\{1, \\dots, p\\}$ such that there exists an edge from $V_1$ to $X$. Then the path $XV_1V_2 \\dots V_p$ has length $p+1$, which gives a contradiction.\n\nNext we show that $p = n$, i.e. that the cycle $V_1 \\dots V_p$ contains all the vertices. If there exists another vertex $A$ connected with an edge to a vertex $V_i$, then the path $AV_iV_{i+1} \\dots V_{i-1}$ has length $p+1$, which gives a contradiction. Since the graph is connected, the cycle contains all vertices.\n\nFor two vertices of the graph, we say that they have distance $r$ if there are exactly $r-1$ vertices between them on a side of the cycle. Observe that they also have distance $n-r$. If we relabel the vertices by $A_1, A_2, \\dots, A_n$ in such a way that we know the graph has $n-1$ of the edges $A_iA_{i+1}, i \\in \\{1, \\dots, n\\}$ (where $A_{n+1} = A_1$), then it also has the last one. This is shown same as before.\n\nNext, we show that if we have an edge between $V_i$ and $V_j$, then we also have an edge between $V_{i+1}$ and $V_{j+1}$. Assume $i < j$. Consider the path\n\n$$\nV_{i+1}V_{i+2}\\dots V_jV_iV_{i-1}\\dots V_{j+1}\n$$\n\n![](images/2025-SL-a_p27_data_82d3c5c708.png)\n\nof length $n$. As before, we conclude that there is an edge between $V_{i+1}$ and $V_{j+1}$. Repeating this, we get that if we have an edge between two vertices at distance $r$, then we have edges between any two vertices at distance $r$.\n\nDefine $S$ as the set of numbers $1 \\le r \\le n-1$ such that the graph has the edges of distance $r$. Note that $1, n-1 \\in S$.\n\nFor positive integers $a$ and $b$ with $a+b \\le n-1$, consider the ordering\n\n$$\nV_1, V_{a+b}, V_{a+b-1}, \\dots, V_{a+1}, V_{a+b+1}, V_{a+b+2}, \\dots, V_n, V_a, V_{a-1}, \\dots, V_1.\n$$\n\n![](images/2025-SL-a_p27_data_b094a06522.png)\n\nThe distance between two consecutive vertices in this ordering is $1, a, b$ or $a+b-1$. This implies that if two numbers from the multiset $\\{a, b, a+b-1\\}$ belong to $S$, so does the third one. Now, if $2 \\in S$, we take $b=2$ and easily get that $S$ contains any number from $1$ to $n-1$. This gives us the solution $K_n$.\n\nAssume now $2 \\notin S$. This implies that we do not have two consecutive numbers smaller than $n-2$ in $S$. But as $2 \\notin S$, we also have $n-2 \\notin S$, so $S$ doesn't contain two consecutive integers. If $S = \\{1, n-1\\}$, we get the solution $C_n$. Otherwise, there exists $t \\in S$ such that $3 \\le t \\le n-3$. Consider the path\n\n$$\nV_t V_{t-1} \\dots V_2 V_{t+2} V_{t+1} V_1 V_n \\dots V_{t+3}\n$$\n\n![](images/2025-SL-a_p27_data_948f0eb526.png)\n\nSame as before, we get that there is an edge between $V_t$ and $V_{t+3}$. Therefore, we have $3 \\in S$. Now, taking $b=3$, we get that any odd number smaller than or equal to $n-1$ lies in $S$. Since we assumed $S$ doesn't contain consecutive integers, we get that $n$ is even and $S = \\{1 \\le i \\le n-1 \\mid i \\text{ odd}\\}$. This gives us the solution $K_{\\frac{n}{2}, \\frac{n}{2}}$.\n\nFinally, the number of edges can be $n$, $\\frac{n(n-1)}{2}$, and if $n$ is even it can also be $\\frac{n^2}{4}$.\n\n**Remark:** Even if $n$ is not big enough, we still characterize all such graphs similarly. The condition was added as at some point we choose a number $t$ between $3$ and $n-3$, and this wouldn't make sense for small $n$ and we would need to quickly discuss why those cases also have the same graphs.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 14903,
    "subject": "Mathematics (Olympiad)",
    "question": "There are 13 distinct two-digit multiples of 7. You want to create the longest possible chain from these multiples, where two multiples can only be adjacent if the last digit of the left multiple equals the first digit of the right multiple. Each multiple can be used at most once. For example, $21 \\to 14 \\to 49$ is an admissible chain of length 3.\n\nWhat is the maximum length of an admissible chain?\n\nA) 6  \nB) 7  \nC) 8  \nD) 9  \nE) 10",
    "options": [],
    "answer": "See solution",
    "solution": "B) 7",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 14904,
    "subject": "Mathematics (Olympiad)",
    "question": "Prove that, for a function $f : \\mathbb{R} \\to \\mathbb{R}$, the following statements are equivalent:\n\n1. $f$ is differentiable on $\\mathbb{R}$ with continuous derivative.\n2. For any $a \\in \\mathbb{R}$ and any two sequences $(x_n)_{n \\ge 1}$ and $(y_n)_{n \\ge 1}$, both converging to $a$ and with $x_n \\neq y_n$ for all $n \\in \\mathbb{N}^*$, the sequence $\\left( \\frac{f(x_n) - f(y_n)}{x_n - y_n} \\right)_{n \\ge 1}$ is convergent.",
    "options": [],
    "answer": "See solution",
    "solution": "(1) $\\Rightarrow$ (2): Suppose $f: \\mathbb{R} \\to \\mathbb{R}$ is continuously differentiable on $\\mathbb{R}$.\n\nLet $a \\in \\mathbb{R}$, and two sequences $(x_n)_{n \\ge 1}$ and $(y_n)_{n \\ge 1}$, both converging to $a$ with $x_n \\ne y_n$ for all $n$. By the Mean Value Theorem, there exist points $a_n \\in (\\min\\{x_n, y_n\\}, \\max\\{x_n, y_n\\})$ such that\n$$\n\\frac{f(x_n)-f(y_n)}{x_n-y_n} = f'(a_n).\n$$\nSince $a_n \\to a$ and $f'$ is continuous, $f'(a_n) \\to f'(a)$, so the sequence $\\left(\\frac{f(x_n)-f(y_n)}{x_n-y_n}\\right)$ is convergent.\n\n(2) $\\Rightarrow$ (1): Assume $f$ satisfies property (2). For any $a \\in \\mathbb{R}$, consider $\\ell_a = \\lim_{n \\to \\infty} \\frac{f(a+1/n)-f(a)}{1/n}$. For any sequences $(x_n)$ and $(y_n)$ converging to $a$ with $x_n \\ne y_n$, define $z_{2n-1} = x_n$, $z_{2n} = a+1/n$, $t_{2n-1} = y_n$, $t_{2n} = a$. Both $(z_n)$ and $(t_n)$ converge to $a$ and $z_n \\ne t_n$ for all $n$. By hypothesis, $\\left(\\frac{f(z_n)-f(t_n)}{z_n-t_n}\\right)$ is convergent, so\n$$\n\\lim_{n \\to \\infty} \\frac{f(x_n)-f(y_n)}{x_n-y_n} = \\lim_{n \\to \\infty} \\frac{f(a+1/n)-f(a)}{1/n} = \\ell_a.\n$$\nIn particular, for any sequence $(x_n)$ with $x_n \\to a$ and $x_n \\ne a$, $\\lim_{n \\to \\infty} \\frac{f(x_n)-f(a)}{x_n-a} = \\ell_a$, so $f$ is differentiable at $a$ and $f'(a) = \\ell_a$ for all $a$.\n\nSuppose, by contraposition, that $f'$ is discontinuous at some $a$. Then there exists $\\varepsilon > 0$ and a sequence $(a_n)$ with $a_n \\to a$, $a_n \\ne a$, such that $|f'(a) - f'(a_n)| > \\varepsilon$ for all $n$. Since $f$ is differentiable at $a_n$, choose $x_n \\in (a_n, a_n + 1/n)$ so that $|f'(a_n) - \\frac{f(x_n)-f(a_n)}{x_n-a_n}| < \\varepsilon/2$. Then $x_n \\to a$, $x_n \\ne a_n$, but\n$$\n|f'(a) - \\frac{f(x_n)-f(a_n)}{x_n-a_n}| \\ge |f'(a) - f'(a_n)| - \\frac{\\varepsilon}{2} > \\frac{\\varepsilon}{2},\n$$\ncontradicting the convergence of $\\frac{f(x_n)-f(a_n)}{x_n-a_n}$ to $f'(a)$. Thus, $f'$ is continuous on $\\mathbb{R}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 14905,
    "subject": "Mathematics (Olympiad)",
    "question": "已知實數數列 $a_1, a_2, \\dots$ 滿足下列性質：\n\n1. 存在正整數 $N$ 使得 $a_n = 1$ 對所有 $n \\ge N$ 都成立；\n2. 對所有 $n \\ge 2$ 都有\n\n$$\na_n \\le a_{n-1} + 2^{-n} a_{2n}.\n$$\n\n證明：$a_n > 1 - 2^{-n}$ 對所有正整數 $n$ 均成立。",
    "options": [],
    "answer": "See solution",
    "solution": "對所有 $k$，由於 $a_n$（$n \\ge k$）只有有限多個可能取值，存在 $n_{\\text{max}} \\ge k$ 使得 $a_{n_{\\text{max}}} \\ge a_n$ 對所有 $n \\ge k$ 都成立。\n\n我們可以用歸納法證明：對所有 $k \\le m \\le n_{\\text{max}}$，\n\n$$\na_m \\ge \\left(1 - 2^{-n_{\\text{max}}} - 2^{-n_{\\text{max}}-1} - \\dots - 2^{-(m+1)}\\right)a_{n_{\\text{max}}}\n$$\n\n當 $m = n_{\\text{max}}$ 時顯然成立。假設對 $m+1$ 成立，則\n\n$$\na_{m+1} \\le a_m + 2^{-(m+1)} a_{2(m+1)} \\le a_m + 2^{-(m+1)} a_{n_{\\text{max}}}\n$$\n\n由歸納假設，\n\n$$\na_m \\ge a_{m+1} - 2^{-(m+1)} a_{n_{\\text{max}}} \\ge \\left(1 - 2^{-n_{\\text{max}}} - \\dots - 2^{-(m+1)}\\right)a_{n_{\\text{max}}}\n$$\n\n因此，取 $m = k$ 得\n\n$$\na_k \\ge \\left(1 - 2^{-n_{\\text{max}}} - \\dots - 2^{-(k+1)}\\right)a_{n_{\\text{max}}} \\ge 1 - \\sum_{i=k+1}^{\\infty} 2^{-i} = 1 - 2^{-k}\n$$\n\n其中不等式利用 $a_{n_{\\text{max}}} \\ge a_N \\ge 1$ 以及將有限和擴展為無窮和。$\\square$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 14906,
    "subject": "Mathematics (Olympiad)",
    "question": "Let the internal angle bisector of $\\angle BAC$ of $\\triangle ABC$ meet side $BC$ at $D$. Let $\\Gamma$ be the circle through $A$ tangent to $BC$ at $D$. Suppose $\\Gamma$ meets sides $AB$ and $AC$ at $E$ and $F$ again, respectively. Lines $BF$ and $CE$ meet $\\Gamma$ again at $P$ and $Q$, respectively. Let $AP$ and $AQ$ intersect side $BC$ at $X$ and $Y$, respectively. Prove that $XY = \\frac{1}{2}BC$.",
    "options": [],
    "answer": "See solution",
    "solution": "Since $\\angle AFD = \\angle ADB$ and $\\angle DAF = \\angle BAD$, we have $\\angle ADF = \\angle ABD$. This implies $\\angle AEF = \\angle ADF = \\angle ABD$, so that $EF \\parallel BC$. It follows that $\\angle XBP = \\angle EFP = \\angle BAX$. Thus, $XB$ is a tangent to $(ABP)$. Hence, we have $XB^2 = XP \\cdot XA = XD^2$. This gives $XB = XD$. Similarly, we have $YC = YD$. Therefore, $XY = \\frac{1}{2}BC$. $\\square$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 14907,
    "subject": "Mathematics (Olympiad)",
    "question": "a\n\nPlace 4 cones on a 60 m track so that all pairwise distances between cones are distinct and cover all possible distances.\n\nb\n\nPlace 5 cones on a 90 m track so that all pairwise distances between cones are distinct and cover all possible distances.\n\nc\n\nGiven 4 cones, how many distinct distances can be formed between pairs of cones? What is the maximum possible distance between two cones if the minimum is 10 m?\n\nd\n\nOn a circular track of length 130 m, how should 4 cones be placed so that all pairwise distances (in either direction) between cones are multiples of 10 m up to 60 m? Explain how this covers all possible distances.\n",
    "options": [],
    "answer": "See solution",
    "solution": "a\n\nThe following placement of 4 cones gives the required result for 60 m.\n\n![](images/2019_Australian_Scene_W1_p44_data_8ab027bc5f.png)\n\nA mirror image is also a solution.\n\nb\n\nThe following placement of 5 cones gives the required result for 90 m.\n\n![](images/2019_Australian_Scene_W1_p44_data_29851b3aa0.png)\n\nA mirror image is also a solution.\n\nc\n\nThere are 6 pairs of the 4 cones: $(A, B)$, $(A, C)$, $(A, D)$, $(B, C)$, $(B, D)$, $(C, D)$. Hence the 4 cones can provide at most six different distances.\n\nSo, starting with 10 m, the maximum possible distance between two cones is 60 m.\n\nd\n\nIf the distance on a circular track of length 130 m from one cone to another is $d$ metres, the distance in the other direction is $130 - d$ metres. So we only need to place the four cones for distances 10, 20, 30, 40, 50, and 60 metres. This will also give distances 70, 80, 90, 100, 110, and 120 metres. The solution to Part a gives one way to place the cones. Here is another way:\n\n![](images/2019_Australian_Scene_W1_p44_data_9fd28f743a.png)\n\n(In addition, 130 m can be run as a complete lap starting and finishing at any cone.)",
    "topic": "Geometry",
    "subtopic": "Solid Geometry"
  },
  {
    "id": 14908,
    "subject": "Mathematics (Olympiad)",
    "question": "Prove that the 2015-digit integer $\\overline{11\\ldots1211\\ldots1}$ is composite.",
    "options": [],
    "answer": "See solution",
    "solution": "**Solution.** Rewrite the number as follows:\n\n$$\n\\overline{11\\ldots1211\\ldots1} = \\overline{11\\ldots1100\\ldots0} + \\overline{11\\ldots1} = \\overline{11\\ldots100\\ldots0} + \\overline{11\\ldots1} : \\overline{11\\ldots1}.\n$$\n\nHence, it's not prime.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 14909,
    "subject": "Mathematics (Olympiad)",
    "question": "For any set $A = \\{a_1, a_2, \\dots, a_m\\}$, let $P(A) = a_1 a_2 \\cdots a_m$. In addition, let $n = \\binom{2010}{99}$ and let $A_1, A_2, \\dots, A_n$ be all 99-element subsets of $\\{1, 2, \\dots, 2010\\}$. Prove that $2011 \\mid \\sum_{i=1}^n P(A_i)$.",
    "options": [],
    "answer": "See solution",
    "solution": "One can check that $2011$ is a prime number. Let\n\n$$\nf(x) = (x-1)(x-2)\\cdots(x-2010) - (x^{2010} + 2010!)\n$$\n\nFor $n \\in \\{1, 2, \\dots, 2010\\}$, we have $n^{2010} \\equiv 1 \\pmod{2011}$ (by Fermat's little theorem), and $2010! \\equiv -1 \\pmod{2011}$ (by Wilson's theorem), so\n\n$$\nf(n) = (n-1)(n-2)\\cdots(n-2010) \\equiv 0 \\pmod{2011}.\n$$\n\nThis means that $f(x) \\equiv 0 \\pmod{2011}$ has $2010$ roots, but the degree of $f(x)$ is $2009$. Using Lagrange's theorem, we can see that the coefficients can all be divided by $2011$.\n\n$$\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": 14910,
    "subject": "Mathematics (Olympiad)",
    "question": "You are given $n \\ge 2$ distinct positive integers. A pair of integers is called *elegant* if their sum is $2^k$ for some positive integer $k$. For each $n$, find the largest possible number of elegant pairs.",
    "options": [],
    "answer": "See solution",
    "solution": "We prove that there exist at most $n-1$ *elegant* pairs by induction on $n$.\n\n**Base case:** For $n=2$, the statement is obvious.\n\n**Inductive step:** Assume the statement holds for $n-1$. Consider $n$ numbers, and let $a$ be the largest. Suppose $a$ is in two *elegant* pairs, i.e., $a+b$ and $a+c$ are both powers of two. Let $m$ be such that $2^m \\leq a < 2^{m+1}$. Then $a+b$ and $a+c$ are both greater than $2^m$ but less than $2a < 2^{m+2}$. Since $a+b = a+c = 2^{m+1}$, this contradicts the distinctness of the numbers. Thus, $a$ can be in at most one *elegant* pair. By induction, among the other $n-1$ integers, there are at most $n-2$ *elegant* pairs, so in total at most $n-1$.\n\nTo show this bound is achievable, consider $a_1 = 1$, $a_2 = 2^2-1$, $a_3 = 2^3-1$, ..., $a_n = 2^n-1$. Then the pairs $(a_1, a_i)$ for $i = 2, \\ldots, n$ are all elegant.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 14911,
    "subject": "Mathematics (Olympiad)",
    "question": "Given an acute, scalene triangle $ABC$ with circumcircle $(O)$. Let $G$ be a point on arc $BC$ that does not contain $O$ of the circumcircle $(I)$ of triangle $OBC$. The circumcircle of triangle $ABG$ intersects $AC$ at $E$ and the circumcircle of triangle $ACG$ intersects $AB$ at $F$ ($E \\neq A$, $F \\neq A$).\n\n**a)** Let $K$ be the intersection of lines $BE$ and $CF$. Prove that the lines $AK$, $BC$ and $OG$ are concurrent.\n\n**b)** Let $D$ be a point on arc $BOC$ of $(I)$. $GB$ meets $CD$ at $M$, $GC$ meets $BD$ at $N$. Suppose that $MN$ intersects $(O)$ at $P$, $Q$. Prove that when $G$ moves on the arc $BC$ that does not contain $O$ of $(I)$, the circumcircle of triangle $GPQ$ always passes through two fixed points.",
    "options": [],
    "answer": "See solution",
    "solution": "a) We have $\\angle EGF = \\angle BGE + \\angle CGF - \\angle EGF = 360^\\circ - 2\\angle BAC - (180^\\circ - 2\\angle BAC) = 180^\\circ$, then $E$, $G$ and $F$ are collinear.\n\n![](images/Vietnamese_mathematical_competitions_p168_data_97fb366f6f.png)\n\nSince $\\angle ABK + \\angle ACK = \\angle AGE + \\angle AGF = 180^\\circ$, then $K$ belongs to the circle $(O)$. So $G$ is the Miquel point of the completed quadrilateral $ABKC.EF$, then two triangles $GBA$ and $GKC$ are similar, which implies that $\\angle BGA = \\angle KGC$. We also have $GO$ is the angle bisector of $\\angle BGC$, then $GO$ is the angle bisector of $\\angle AGK$. Combined with $OA = OK$, we can conclude that $AOKG$ is cyclic. By considering the radical axis of circles $(O)$, $(AOKG)$, $(BOC)$, one can check that $AK$, $OG$ and $BC$ are concurrent.\n\nb) In this part, we just need $(O)$, $(I)$ are two fixed circles passing through $B$, $C$, point $D$ is fixed on $(I)$ while $G$ is moving on $(I)$. By applying Pascal's theorem for the tuple $\\begin{pmatrix} B & C & D \\\\ C & B & G \\end{pmatrix}$, one can check that the line $MN$ passes through the intersection of the tangent lines at $B$, $C$, namely $J$ of the fixed circle $(I)$. Suppose that $JD$ meets $(I)$ at the second point $X$, then $X$ is a fixed point and the quadrilateral $BCDX$ is harmonic, then $G(BC, DX) = -1$. Denote $T$ as the intersection of $MN$ and $BC$, then $G(BC, DT) = -1$, which means $GX$ passes through $T$. Thus,\n\n$$\n\\overline{TX} \\cdot \\overline{TG} = \\overline{TB} \\cdot \\overline{TC} = \\overline{TP} \\cdot \\overline{TQ}.\n$$\n\nThis implies that $(GPQ)$ passes through the fixed point $X$. Suppose that $Y$ is the intersection of $(GPQ)$ and $DX$ (which differs from $X$), then $\\overline{JX} \\cdot \\overline{JY} = \\overline{JP} \\cdot \\overline{JQ} = P_{J/(O)}$, which is a constant, then $Y$ is fixed. Therefore, the circles $(GPQ)$ pass through two fixed points $X$, $Y$.\n\n![](images/Vietnamese_mathematical_competitions_p169_data_e7d6d5b563.png)",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 14912,
    "subject": "Mathematics (Olympiad)",
    "question": "Find all polynomials $P(x)$ and $Q(x)$ with real coefficients such that for each real number $a$, $P(a)$ is a solution of the equation\n\n$$\nx^{2023} + Q(a)x^2 + (a^{2024} + a)x + a^3 + 2025a = 0.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "From the hypothesis, for all $x \\in \\mathbb{R}$,\n\n$$\nP(x)^{2023} + Q(x)P(x)^2 + (x^{2024} + x)P(x) + x^3 + 2025x = 0. \\tag{1}\n$$\n\nThis implies that the polynomial $x^3 + 2025x$ is divisible by $P(x)$. Thus, $P(x)$ must be of the form $k$, $kx$, $k(x^2 + 2025)$, or $kx(x^2 + 2025)$ for some real constant $k$. However, if $P(x)$ is divisible by $x$, then from (1), $2025x$ would need to be divisible by $x^2$, which is a contradiction. Therefore, $P(x)$ must be of the form $k$ or $k(x^2 + 2025)$ with $k \\neq 0$.\n\n**Case 1:** $P(x) = k$\n\nSubstituting into (1):\n\n$$\nk^{2023} + k^2 Q(x) + k(x^{2024} + x) + x^3 + 2025x = 0,\n$$\nso\n$$\nQ(x) = \\frac{-k(x^{2024} + x) - x^3 - 2025x - k^{2023}}{k^2}\n$$\nfor all real $x$.\n\n**Case 2:** $P(x) = k(x^2 + 2025)$\n\nSubstituting into (1):\n\n$$\nk^{2023}(x^2 + 2025)^{2023} + k^2(x^2 + 2025)^2 Q(x) + k(x^2 + 2025)(x^{2024} + x) + x^3 + 2025x = 0.\n$$\n\nAfter simplification, we find that $(x^2 + 2025)$ must divide $k(x^{2023} + 1) + 1$. However, since $x^2 + 2025$ and $x$ are coprime, and the degree of $k(x^{2023} + 1) + 1$ is less than $x^2 + 2025$, this is only possible if $k = 0$, which is not allowed. Thus, this case yields no solutions.\n\n**Conclusion:**\n\nThe solutions are all polynomials of the form\n\n$$\nP(x) = k,\n$$\n\nand\n\n$$\nQ(x) = \\frac{-k(x^{2024} + x) - x^3 - 2025x - k^{2023}}{k^2},\n$$\n\nwhere $k$ is any nonzero real number.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 14913,
    "subject": "Mathematics (Olympiad)",
    "question": "Define an _arithmetic permutation_ as a bijective function $f : \\mathbb{Z}^+ \\to \\mathbb{Z}^+$ which satisfies $m \\mid n \\Leftrightarrow f(m) \\mid f(n)$ for all $m, n \\in \\mathbb{Z}^+$. Show that each of the following conditions on bijective functions gives an equivalent definition of an arithmetic permutation:\n\n1. $f(m) \\cdot f(n) = f(mn)$ for all $m, n \\in \\mathbb{Z}^+$.\n2. $\\operatorname{lcm}(f(m), f(n)) = f(\\operatorname{lcm}(m, n))$ for all $m, n \\in \\mathbb{Z}^+$.\n\n% ![](images/Turkey_2018_p38_data_9a43220254.png)",
    "options": [],
    "answer": "See solution",
    "solution": "First, we establish a lemma characterizing arithmetic permutations.\n\n**Lemma.** Let $\\mathbb{P}$ denote the set of prime numbers. An arithmetic permutation $f : \\mathbb{Z}^+ \\to \\mathbb{Z}^+$ maps $\\mathbb{P}$ onto itself bijectively. Furthermore, given a bijective function $\\tilde{f} : \\mathbb{P} \\to \\mathbb{P}$, there is a unique arithmetic permutation $f : \\mathbb{Z}^+ \\to \\mathbb{Z}^+$ which restricts to $\\tilde{f}$ on $\\mathbb{P}$. This unique extension is given by:\n$$\nf(p_1^{\\alpha_1} \\cdots p_k^{\\alpha_k}) = \\tilde{f}(p_1)^{\\alpha_1} \\cdots \\tilde{f}(p_k)^{\\alpha_k}\n$$\n\n*Proof.*\nLet $D(n)$ denote the set of positive divisors of $n$. An arithmetic permutation $f$ maps $D(n)$ onto $D(f(n))$ bijectively for any $n \\in \\mathbb{Z}^+$. Thus, $|D(n)| = |D(f(n))|$. Since primes are characterized by $|D(p)| = 2$, $p$ is prime if and only if $f(p)$ is prime.\n\nWe prove $f(p^n) = f(p)^n$ by induction on $n$. Assume $f(p^{n-1}) = f(p)^{n-1}$. Since $f(p^{n-1}) \\mid f(p^n)$, write $f(p^n) = f(p)^{n-1+a} \\cdot A$ with $a \\ge 0$ and $f(p) \\nmid A$. Then $|D(f(p^n))| = (n+a) \\cdot |D(A)|$. But $|D(f(p^n))| = |D(p^n)| = n+1$, so $(n+a) \\cdot |D(A)| = n+1$. This forces $A=1$, $a=1$, so $f(p^n) = f(p)^n$.\n\nFor $n = p_1^{a_1} \\cdots p_k^{a_k}$, we have $f(p_1)^{a_1} \\cdots f(p_k)^{a_k} \\mid f(p_1^{a_1} \\cdots p_k^{a_k})$. But $|D(f(p_1^{a_1} \\cdots p_k^{a_k}))| = (a_1+1) \\cdots (a_k+1) = |D(f(p_1)^{a_1} \\cdots f(p_k)^{a_k})|$, so equality holds:\n$$\nf(p_1^{a_1} \\cdots p_k^{a_k}) = f(p_1)^{a_1} \\cdots f(p_k)^{a_k}\n$$\nGiven any bijection $\\tilde{f}: \\mathbb{P} \\to \\mathbb{P}$, the function $f$ defined above is an arithmetic permutation.\n\nNow, consider the condition $f(m) \\cdot f(n) = f(mn)$. If $a \\mid b$, then $b = a \\cdot s$ so $f(b) = f(a) \\cdot f(s)$, hence $f(a) \\mid f(b)$. Conversely, if $f(a) \\mid f(b)$, then $f(b) = f(a) \\cdot f(t) = f(at)$, so $b = at$, thus $a \\mid b$. Therefore, a bijection $f$ with $f(m) \\cdot f(n) = f(mn)$ is an arithmetic permutation, and vice versa.\n\nSimilarly, for $\\operatorname{lcm}(f(m), f(n)) = f(\\operatorname{lcm}(m, n))$, if $a \\mid b$, then $b = \\operatorname{lcm}(a, b)$, so $f(b) = f(\\operatorname{lcm}(a, b)) = \\operatorname{lcm}(f(a), f(b))$, hence $f(a) \\mid f(b)$. Conversely, if $f(a) \\mid f(b)$, then $f(b) = \\operatorname{lcm}(f(a), f(b)) = f(\\operatorname{lcm}(a, b))$, so $b = \\operatorname{lcm}(a, b)$, thus $a \\mid b$. Therefore, a bijection $f$ with $\\operatorname{lcm}(f(m), f(n)) = f(\\operatorname{lcm}(m, n))$ is an arithmetic permutation, and vice versa.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 14914,
    "subject": "Mathematics (Olympiad)",
    "question": "Alrededor de una circunferencia están escritos 20 números enteros positivos distintos. Alex divide cada número por el número vecino, recorriendo la circunferencia en el sentido de las agujas del reloj, y anota los restos que obtiene en cada caso. Teo divide cada número por el número vecino, recorriendo la circunferencia en el sentido contrario al de las agujas del reloj, y anota los restos. Si, entre los 20 números que anotó, Alex obtuvo sólo dos restos distintos, determinar la cantidad de restos diferentes que obtendrá Teo.",
    "options": [],
    "answer": "See solution",
    "solution": "Numeramos los 20 números en sentido horario como $a_1, a_2, \\dots, a_{20}$ de modo que $a_{20}$ sea el menor de los 20 números escritos en la circunferencia. Sea $r_{20}$ el resto de dividir $a_{20}$ por $a_1$, es decir, $a_{20} = a_1 \\cdot q + r_{20}$. Como $a_{20}$ es el menor de todos los números escritos, se tiene $a_{20} < a_1$, por lo tanto $q=0$ y $r_{20} = a_{20}$. Por otra parte, el resto de dividir $a_{19}$ por $a_{20}$ se obtiene haciendo $a_{19} = a_{20} \\cdot q' + r_{19}$ con $r_{19} < a_{20}$. Luego $r_{19} < a_{20} = r_{20}$ y estos son los dos únicos restos que obtuvo Alex.\n\nSi $a_i < a_{i+1}$ para algún $i < 20$, entonces al dividir $a_i$ por $a_{i+1}$ se tendrá $a_i = 0 \\cdot a_{i+1} + a_i$, lo que es imposible pues Alex sólo obtiene dos restos distintos y tendríamos $r_{19} < a_{20} < a_i$.\n\nPor lo tanto, debe ocurrir que $a_{20} < a_{19} < \\dots < a_2 < a_1$.\n\nPor lo tanto, Teo obtiene los siguientes restos: $a_i$ para $2 \\le i \\le 20$, cuando divide $a_i$ por $a_{i-1}$, pues $a_i = 0 \\cdot a_{i-1} + a_i$ con $a_i < a_{i-1}$, y el resto de dividir $a_1$ por $a_{20}$, que es menor que $a_{20}$. Por lo tanto, son 20 restos distintos.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 14915,
    "subject": "Mathematics (Olympiad)",
    "question": "Given an acute, non-isosceles triangle $ABC$, let $(O)$ be its circumcircle and $(I)$ its incircle. The incircle $(I)$ touches $BC$, $CA$, and $AB$ at $D$, $E$, and $F$, respectively. Let $K$, $M$, and $N$ be the midpoints of $BC$, $CA$, and $AB$.\n\n**a)** Prove that the lines passing through $D$, $E$, and $F$ and parallel to the lines $IK$, $IM$, and $IN$, respectively, are concurrent.\n\n**b)** Let $T$, $P$, and $Q$ be the midpoints of the larger arcs $BC$, $CA$, and $AB$ of $(O)$. Prove that the lines passing through $D$, $E$, and $F$ and parallel to the lines $IT$, $IP$, and $IQ$, respectively, are concurrent.",
    "options": [],
    "answer": "See solution",
    "solution": "a) Let $DD'$, $EE'$, and $FF'$ be the diameters of $(I)$. Let $AD'$, $BE'$, and $CF'$ meet $BC$, $CA$, and $AB$ at $A'$, $B'$, and $C'$, respectively.\n\nIt is well-known that $CA' = BD$, $CB' = AE$, and $AC' = BF$, and that $AA'$, $BB'$, $CC'$ concur at $N'$, the Nagel point of $ABC$. Note that $I$ and $K$ are the midpoints of $DD'$ and $DA'$, so $IK \\parallel AA'$. Similarly, $IM \\parallel BB'$ and $IN \\parallel CC'$. Hence, the lines passing through $D$, $E$, and $F$ that are parallel to $IK$, $IM$, and $IN$, respectively, concur at $V$, which is the reflection of $N'$ over $I$.\n\n![](images/Vietnamese_mathematical_competitions_p251_data_757b32e1f6.png)\n\nb) Let the line passing through $D$ and parallel to $IT$ meet $IO$ and $OK$ at $X$ and $W$. Note that $IDWT$ is a parallelogram, so $ID \\parallel WT$. Using Thales's theorem, we get\n\n$$\n\\frac{OX}{OI} = \\frac{OW}{OT} = \\frac{OT + TW}{OT} = \\frac{R - r}{R}\n$$\n\n![](images/Vietnamese_mathematical_competitions_p252_data_7bb6a300dd.png)\n\nTherefore, the lines passing through $D$, $E$, and $F$ that are parallel to $IT$, $IP$, and $IQ$, respectively, are concurrent at $X$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 14916,
    "subject": "Mathematics (Olympiad)",
    "question": "The inscribed circle of triangle $ABC$ is tangent to its sides $BC$, $AC$, and $AB$ at points $K$, $L$, and $M$, respectively. Let $P$ be the point of intersection of the bisector of $\\angle BCA$ with the line $MK$. Prove that $AP \\parallel LK$.",
    "options": [],
    "answer": "See solution",
    "solution": "Denote the inscribed circle by $k$, its center by $S$, and its angles by $\\alpha$, $\\beta$, $\\gamma$.\n\nFrom symmetry, $KL \\perp CP$ and $\\angle LPC = \\angle KPC$. Then, with simple calculations, we get $\\angle MKB = 90^\\circ - \\frac{1}{2}\\beta$ and $\\angle LKC = 90^\\circ - \\frac{1}{2}\\gamma$, so $\\angle MKL = 90^\\circ - \\frac{1}{2}\\alpha$. Similarly for other angles.\n\n![](images/Ukraine_2021-2022_p29_data_adb4247192.png)\n\n$$\n\\text{As } \\angle KPC + \\frac{1}{2}\\gamma = \\angle BKP = 90^\\circ - \\frac{1}{2}\\beta,\n$$\n\n$$\n\\angle LPC = \\angle KPC = 90^\\circ - \\frac{1}{2}(\\beta + \\gamma) = \\frac{1}{2}\\alpha.\n$$\n\nThe inscribed circle $k$ of $\\triangle ABC$ is also the circumscribed circle of $\\triangle KLM$, and as we have seen, it's acute. Then $S$ is the interior point of $CP$. As\n\n$$\n\\angle LPC = \\angle LPS = \\angle LAS = \\frac{1}{2}\\alpha,\n$$\n\nthe quadrilateral $LSPA$ is inscribed. As $\\angle ALS = 90^\\circ$, we get $AP \\perp CP \\Rightarrow AP \\parallel KL$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 14917,
    "subject": "Mathematics (Olympiad)",
    "question": "Determine all pairs $(a, b)$ of non-negative integers such that $a^b + b$ divides $a^{2b} + 2b$. (Note that $0^0 = 1$.)",
    "options": [],
    "answer": "See solution",
    "solution": "Let $n = a^b + b$ and $m = a^{2b} + 2b$.\n\n- For $a = b = 0$, $n = m = 1$, so $n \\mid m$. Thus, $(0, 0)$ is a solution.\n- For $a = 0$ and $b > 0$, $n = b$ and $m = 2b$, so $n \\mid m$. Thus, $(0, b)$ is a solution for all $b \\geq 0$.\n- For $a = 1$, $n = b + 1$ and $m = 2b + 1$. The only $b$ such that $b+1 \\mid 2b+1$ is $b = 0$. Thus, $(1, 0)$ is a solution.\n- For $b = 0$, $n = 1$ and $m = 1$, so $n \\mid m$. Thus, $(a, 0)$ is a solution for all $a \\geq 0$.\n\nNow consider $a > 1$ and $b > 0$:\n\n$$\na^b + b \\mid a^{2b} + 2b = (a^b)^2 - b^2 + (b^2 + 2b) = (a^b + b)(a^b - b) + b(b + 2),\n$$\nso $a^b + b \\mid b(b+2)$.\n\n- For $b = 1$, $a + 1 \\mid 3$, so $a = 2$. Thus, $(2, 1)$ is a solution.\n- For $a \\geq 3$, $a^b + b > b(b+2)$ for $b \\geq 1$, so no solutions exist in this case (can be shown by induction).\n- For $a = 2$ and $b > 1$, check $b = 2, 3, 4$:\n    - $b = 2$: $2^2 + 2 = 6$, $2^{4} + 4 = 20$, $6 \\nmid 20$.\n    - $b = 3$: $2^3 + 3 = 11$, $2^{6} + 6 = 70$, $11 \\nmid 70$.\n    - $b = 4$: $2^4 + 4 = 20$, $2^{8} + 8 = 264$, $20 \\nmid 264$.\n  For $b > 5$, $2^b > b(b+2)$, so no solutions.\n\n**Summary:** The solutions are all $(a, 0)$, all $(0, b)$, and $(2, 1)$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 14918,
    "subject": "Mathematics (Olympiad)",
    "question": "What are the powers of 2 that divide 128 and exceed 8?",
    "options": [],
    "answer": "See solution",
    "solution": "The powers of 2 dividing 128 and exceeding 8 are $2^4$, $2^5$, $2^6$, and $2^7$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 14919,
    "subject": "Mathematics (Olympiad)",
    "question": "Нека $x$ е реален број така што броевите $x^3$ и $x^2 + x$ се рационални. Докажи дека бројот $x$ е рационален.",
    "options": [],
    "answer": "See solution",
    "solution": "Нека $a = x^3$, $b = x^2 + x$. Тогаш $a = x^3 = x^2 + x^3 - x^2 - x + x = x b - b + x = x(b + 1) - b$. Јасно е дека $b \\neq -1$, бидејќи во спротивно $x^2 + x = -1$, или $(x + \\frac{1}{2})^2 - \\frac{1}{4} = -1$. Тогаш $(x + \\frac{1}{2})^2 = -\\frac{3}{4}$, што не е возможно. Добиваме дека $x = \\frac{a + b}{b + 1}$, што е рационален број бидејќи броевите $a$ и $b$ се рационални.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 14920,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $ABC$ be an isosceles triangle with $AC = BC$ and $\\angle ACB < 60^\\circ$. We denote the incenter and circumcenter by $I$ and $O$, respectively. The circumcircle of triangle $BIO$ intersects the leg $BC$ also at point $D \\neq B$.\n\n(a) Prove that the lines $AC$ and $DI$ are parallel.\n\n(b) Prove that the lines $OD$ and $IB$ are mutually perpendicular.",
    "options": [],
    "answer": "See solution",
    "solution": "Note that the condition $\\angle ACB < 60^\\circ$ guarantees that $O$ lies between $I$ and $C$.\n\n**(a)** We denote the angles of triangle $ABC$ by $\\alpha = \\angle BAC$, $\\beta = \\angle ABC$, and $\\gamma = \\angle ACB$. Let $K$ and $k$ be the circumcircles of $ABC$ and $BIO$, respectively. The inscribed angle theorem for circle $K$ yields: $\\angle BOC = 2\\alpha$. Therefore, we have $\\angle IOB = 180^\\circ - 2\\alpha$, and because $\\alpha = \\beta$, we obtain $\\angle IOB = \\gamma$. Furthermore, the inscribed angle theorem for circle $k$ gives $\\angle IDB = \\gamma$, whence finally $ID \\parallel AC$.\n\n![](images/gwf2015englishSolutions_p1_data_1873e0e780.png)\n\n**(b)** We denote the point of intersection of lines $OD$ and $IB$ by $F$ and the midpoint of $AB$ by $G$. Since $IODB$ is cyclic, we have $\\angle IOD = 180^\\circ - \\beta/2$, that is, $\\angle DOC = \\beta/2$ or equivalently $\\angle FOI = \\beta/2$. Furthermore, $\\angle GIB = 90^\\circ - \\beta/2$ implies $\\angle OIF = 90^\\circ - \\beta/2$. Therefore, $\\angle IFO = 90^\\circ$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 14921,
    "subject": "Mathematics (Olympiad)",
    "question": "There are $n$ children sitting around a round table. Erika is the oldest among them and she has $n$ candies. No other child has any candy. Erika decides to distribute the candies according to the following rules:\n\n- In every round, all children with at least two candies are eligible to act.\n- Erika chooses one of these children, and that child gives one candy each to both of their immediate neighbors (the children sitting directly to their left and right).\n\n(So, in the first round, only Erika is eligible and she gives one candy to each of her two neighbors.)\n\nFor 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": "See solution",
    "solution": "First, we show that for even $n$, it is impossible to end with every child having one candy.\n\nIn every round, only two candies change position, moving in opposite directions. Consider the sum $S$ of the distances of all candies from Erika, labeling the seats clockwise as $0, 1, \\dots, n-1$ (with Erika at $0$). After each round, $S$ can only change by $\\pm n$ or remain the same, so $S$ is always divisible by $n$.\n\nInitially, $S = 0$. If every child has one candy, then\n$$\nS = 0 + 1 + 2 + \\dots + (n-1) = \\frac{n(n-1)}{2}, \\quad \\text{so} \\quad \\frac{S}{n} = \\frac{n-1}{2}.\n$$\nFor even $n$, $\\frac{n-1}{2}$ is not an integer, so this situation is impossible.\n\nFor odd $n$, let $n = 2k + 1$. We use induction to show it is possible to reach the desired configuration. For each $i = 0, 1, \\dots, k$, we can reach a state where Erika has $n - 2i$ candies, and the first $i$ children to her left and right each have one candy. The base case $i = 0$ is the starting position. At each step, Erika gives one candy to each neighbor, and the process continues as described, eventually reaching the state where every child has one candy.\n\n*Remarks.* For even $n$ not divisible by $4$, a coloring argument shows the impossibility: coloring seats alternately, the parity of the sum of candies on one color does not change, making the final configuration impossible.\n\nFor odd $n$, regardless of choices, the process always ends with every child having one candy. For $n = 2k + 1$, the number of rounds needed is $1^2 + 2^2 + \\dots + k^2$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 14922,
    "subject": "Mathematics (Olympiad)",
    "question": "Each cell of a $100 \\times 100$ array contains a number from $1$ through $100^2$; distinct cells contain distinct numbers. Determine the largest possible integer $c$ satisfying the following condition: every such configuration contains two distinct numbers on the same row or the same column sharing a common divisor at least $c$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $p$ be an odd prime. Each cell of a $(p-1) \\times (p-1)$ array contains a number from $1$ through $(p-1)^2$; distinct cells contain distinct numbers. Determine the largest possible integer $c$ satisfying the following condition: every such configuration contains two distinct numbers on the same row or the same column sharing a common divisor at least $c$.\n\nWe will prove that the required maximum is $c = p - 2$, so, in the special case at hand, it is $c = 99$.\n\nTo prove $c \\ge p - 2$, note that $p - 2$ has $p$ multiples in the range $1$ through $(p-1)^2$, so at least two of these lie on the same row or the same column.\n\nTo prove $c \\le p - 2$, we describe a configuration in which the greatest common divisor of every two distinct numbers on the same row or the same column is at most $p - 2$.\n\nWrite the first $p-1$ numbers on the first column, the next $p-1$ on the second, and so on; explicitly, the numbers on the $j$-th column are $i + (j-1)(p-1)$, $i = 1, 2, \\dots, p-1$.\n\nNote that the difference of any two distinct numbers on the same column is a non-zero integer whose absolute value is strictly less than $p-1$ and hence so is their greatest common divisor; that is, it is at most $p-2$. Clearly, this is preserved under any permutation of the numbers along the column.\n\nNote further that on each column there is exactly one residue modulo $p$ missing; explicitly, no number on the $j$-th column is congruent to $(j-1)(p-1)$ modulo $p$. Also, if $j \\ge 2$, then the $j$-th column contains the number $(j-1)p$.\n\nNow, permute the numbers on each column as follows: on the $j$-th column place $(j-1)p$ in its $(p-j+1)$-st cell and, for $i \\ne p-j+1$, place the number congruent to $i$ modulo $p$ in the $i$-th cell.\n\nConsider two numbers on the same row, say, $a$ and $b$. If one of these numbers is divisible by $p$, then the other is not, so $\\gcd(a, b) \\le a/p \\le (p-1)^2/p < p-1$. Hence $\\gcd(a, b) \\le p-2$.\n\nFinally, if $a$ and $b$ are both coprime to $p$, so is $\\gcd(a, b)$ and $p$ divides $|a-b|$. Hence $|a-b|$ is divisible by $p \\gcd(a, b)$. As $1 \\le |a-b| \\le (p-1)^2 - 1 = p(p-2)$, it follows that $\\gcd(a, b) \\le p-2$.\n\n**Remarks.**\n\n(1) The configuration below exhibits the outcome of the process described in the solution for $p=5$ and achieves the maximum $c=3=5-2$:\n\n![](path/to/file.png)\n\n(2) The configuration described in the solution achieves the maximum $c = p-2$ for any $p$, as the second cells on the last two columns contain the numbers $(p-1)(p-2)$ and $p(p-2)$, respectively; also the second and the fourth cells on the $(p-2)$-nd column contain the numbers $(p-1)(p-2)$ and $(p-2)^2$, respectively.",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 14923,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $a, b, c \\in \\mathbb{R}$ such that $abc = 1$. Prove the inequality\n$$\na^4 + b^4 + c^4 \\geq a + b + c.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "By the trivial inequality $(x-y)^2 + (y-z)^2 + (z-x)^2 \\geq 0$, we have that for every $x, y, z \\in \\mathbb{R}$,\n$$\nx^2 + y^2 + z^2 \\geq xy + yz + zx.\n$$\nApplying this inequality, we get\n$$\n\\begin{align*}\na^4 + b^4 + c^4 &\\geq a^2 b^2 + b^2 c^2 + c^2 a^2 = (ab)^2 + (bc)^2 + (ca)^2 \\\\\n&\\geq (ab)(bc) + (bc)(ca) + (ca)(ab) = abc(a+b+c) = a + b + c\n\\end{align*}\n$$\nwhich finishes the proof. Equality holds if and only if $a = b = c = 1$.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 14924,
    "subject": "Mathematics (Olympiad)",
    "question": "無窮正整數數列 $a_1, a_2, \\dots, a_n, \\dots$ 滿足：對任意正整數 $m, n$，$m+n$ 整除 $a_m + a_n$，且對任意正整數 $n$ 有 $a_n \\leq n^{100}$。試證：存在一個多項式 $f(x)$ 使得對任意正整數 $n$，$f(n) = a_n$。",
    "options": [],
    "answer": "See solution",
    "solution": "首先，對任意 $m < n$，令 $k = n - m$，取正整數 $t$ 使得 $tk > n$。考慮 $tk$ 整除 $a_n + a_{tk-n}$ 及 $(t-1)k$ 整除 $a_m + a_{tk-n}$，可得 $a_n$ 與 $a_m$ 除以 $k$ 的餘數相同，即 $n - m$ 整除 $a_n - a_m$。\n\n由拉格朗日插值法，存在 $100$ 次以內的有理係數多項式 $f(x)$ 使得對於 $k = 1, 2, \\dots, 101$，均有 $f(k) = a_k$。以下我們證明 $f(n) = a_n$。\n\n令 $N$ 為一個使得 $Nf(x)$ 是整係數多項式的正整數。考慮函數 $h$ 在整數集上定義為 $h(n) = N(f(n) - a_n)$，則 $h$ 在 $1$ 到 $101$ 的值為 $0$，且由第一段所述及 $Nf(x)$ 為整係數多項式，可知 $h$ 滿足 $(n-m) \\mid (h(n) - h(m))$。另外，由於 $f(x)$ 是 $100$ 次以內的多項式，且 $a_n \\leq n^{100}$，知存在常數 $M > 0$ 使得 $|h(n)| \\leq M(n^{100} + 1)$。\n\n對任意正整數 $n$，比較 $h(n)$ 和 $h(k)$，這裡 $k = 1, \\dots, 101$。由於 $h(k) = 0$，由 $(n-k) \\mid (h(n) - h(k))$ 得 $h(n)$ 必須是 $(n-101), \\dots, (n-1)$ 的倍數，這些數彼此之間的公因數不大於 $100$，故他們的最小公倍數 $\\geq \\dfrac{(n-101)^{101}}{100^{5050}}$，而 $h(n)$ 必須是他們最小公倍數的倍數。當 $n$ 夠大時，有 $\\dfrac{(n-101)^{101}}{100^{5050}} > M(n^{100}+1)$，此時可知 $h(n) = 0$。\n\n對任意整數 $m$，對任何足夠大的 $n$ 有 $(n-m) \\mid h(m)$，從而 $h(m) = 0$，證畢！",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 14925,
    "subject": "Mathematics (Olympiad)",
    "question": "Touching circles $k_1(S_1, r_1)$ and $k_2(S_2, r_2)$ lie in a right-angled triangle $ABC$ with hypotenuse $AB$ and legs $AC = 4$ and $BC = 3$ in such a way that the sides $AB$, $AC$ are tangent to $k_1$ and the sides $AB$, $BC$ are tangent to $k_2$. Find the radii $r_1$ and $r_2$, if $4r_1 = 9r_2$.\n\n![](images/Czech_and_Slovak_booklet_2013_p8_data_26df49fc01.png)",
    "options": [],
    "answer": "See solution",
    "solution": "The hypotenuse $AB$ has length $AB = 5$ by the Pythagorean theorem. The triangle's angles satisfy $\\cos \\alpha = \\frac{4}{5}$, $\\cos \\beta = \\frac{3}{5}$.\n\n$$\n\\cot \\frac{\\alpha}{2} = \\sqrt{\\frac{1 + \\cos \\alpha}{1 - \\cos \\alpha}} = 3\n$$\n\n$$\n\\cot \\frac{\\beta}{2} = \\sqrt{\\frac{1 + \\cos \\beta}{1 - \\cos \\beta}} = 2\n$$\n\nSince both circles $k_1$ and $k_2$ lie entirely within triangle $ABC$, they are externally tangent. Let $k_1$ and $k_2$ touch $AB$ at points $D$ and $E$, respectively, and let $F$ be the orthogonal projection of $S_2$ onto $S_1D$ (see figure). By the Pythagorean theorem for triangle $FS_2S_1$:\n\n$$\n(r_1 + r_2)^2 = (r_1 - r_2)^2 + DE^2\n$$\n\nwhich gives $DE = 2\\sqrt{r_1 r_2}$.\n\nThe equality $AB = AD + DE + EB$ yields:\n\n$$\n5 = r_1 \\cot \\frac{\\alpha}{2} + 2\\sqrt{r_1 r_2} + r_2 \\cot \\frac{\\beta}{2} = 3r_1 + 2\\sqrt{r_1 r_2} + 2r_2\n$$\n\nGiven $r_1 = \\frac{9}{4} r_2$, substitute to get:\n\n$$\n5 = 3 \\cdot \\frac{9}{4} r_2 + 2 \\sqrt{\\frac{9}{4} r_2 \\cdot r_2} + 2 r_2 = \\frac{27}{4} r_2 + 3 r_2 + 2 r_2\n$$\n\nSo:\n\n$$\n\\frac{27}{4} r_2 + 3 r_2 + 2 r_2 = 5\n$$\n\n$$\n\\frac{47}{4} r_2 = 5 \\implies r_2 = \\frac{20}{47}, \\quad r_1 = \\frac{45}{47}\n$$\n\n**Remark:** Both circles indeed lie within triangle $ABC$ because the incircle has radius $\\rho = \\frac{ab}{a + b + c} = 1$, and both $r_1$ and $r_2$ are less than $1$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 14926,
    "subject": "Mathematics (Olympiad)",
    "question": "$$\n\\min\\{f(x_i), f(x_j)\\} = \\min\\{(x_i + a)(x_i + b), (x_j + a)(x_j + b)\\}\n$$\n\nLet $F = \\sum_{1 \\le i < j \\le n} \\min\\{f(x_i), f(x_j)\\}$, where $x_1, x_2, \\dots, x_n$ are non-negative real numbers such that $x_1 + x_2 + \\dots + x_n = s$ (with $s$ a fixed non-negative real number), and $a, b$ are real constants.\n\nFind the maximum value of $F$ and determine when equality holds.",
    "options": [],
    "answer": "See solution",
    "solution": "Since $F$ is symmetric, we may assume $x_1 \\le x_2 \\le \\cdots \\le x_n$. Note that $f(x)$ is strictly increasing on non-negative real numbers, so\n\n$$\nF = (n-1)f(x_1) + (n-2)f(x_2) + \\cdots + f(x_{n-1}).\n$$\n\nWhen $n=2$, $F = f(x_1) \\le f\\left(\\frac{s}{2}\\right)$, with equality when $x_1 = x_2$. Assume the statement holds for $n$, and consider $n+1$. Applying the inductive hypothesis to $x_2 + x_3 + \\cdots + x_{n+1} = s - x_1$, we have\n\n$$\nF \\le n f(x_1) + \\frac{1}{2} n(n-1) f\\left(\\frac{s-x_1}{n}\\right) = g(x_1),\n$$\n\nwhere $g(x)$ is a quadratic function of $x$. The leading coefficient is $1 + \\frac{n-1}{2n^2}$, and the coefficient of $x$ is $a + b - \\frac{n-1}{2n}(a + b + \\frac{s}{2n})$. Therefore, the axis of symmetry is\n\n$$\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\n(The above inequality is equivalent to $[(n-1)s - 2n(n+1)(a+b)](n+1) \\le 2s(2n^2 + n - 1)$; clearly, the left-hand side $< (n^2-1)s <$ right-hand side.) Therefore, $g\\left(\\frac{s}{n+1}\\right)$ is the maximum of $g(x)$ on $[0, \\frac{s}{n+1}]$. Thus, $F$ attains its maximum when $x_1 = x_2 = \\cdots = x_{n+1} = \\frac{s}{n+1}$, completing the solution. $\\square$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 14927,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $n$ be an odd integer ($n \\geq 3$). Consider a game played on a regular $n$-gon inscribed in a circle, with vertices $A_0, A_1, \\dots, A_{n-1}$ and center $O$. Two players take turns drawing segments between vertices or between a vertex and the center $O$, following certain rules. Prove that the first player can always win if, on their first move, they connect an arbitrary vertex to the center of the circle.",
    "options": [],
    "answer": "See solution",
    "solution": "Suppose at some moment the first player can win if the second player makes a move as before. Let $x$ denote the last segment drawn by the first player, and let $x^*$ denote the segment symmetric to $x$. Let $z$ denote the winning segment for the first player if the second player draws $x^*$. Note that $x$ does not join a vertex with the center $O$, since otherwise, after the move $x^*$, there remains an odd number of components and the move $z$ cannot be winning.\n\nConsider the components before the move $x$. The segment $x$ connects two different components, say $A$ and $B$, and $x^*$ connects the symmetric components $A^*$ and $B^*$. Consider the center $O$ of the circle.\n\nAssume $O$ belongs to one of the components $A$ or $B$, say $A$. Then it also belongs to $A^*$, so these components coincide. This means $A = A^*$ is centrally symmetric and contains $O$. After the move $x^*$, the number of components is odd as before the move $x$, so the move $z$ cannot be winning—a contradiction.\n\nSuppose now that $O$ does not belong to $A$ or $B$. Then there is at least one other component $C$ containing $O$. For a move $z$ to be winning, it must connect $C$ with the component obtained after the moves $x$ and $x^*$.\n\nSince after the symmetric moves $x$ and $x^*$ a common component is obtained that does not contain $O$ ($B = A^*$, $A = B^*$), and the move $x^*$ did not decrease the number of components, the second player can draw $z$ instead of $x^*$ and win.\n\nWe now prove by induction that the first player can always win if they connect an arbitrary vertex to the center $O$ on their first move. The base case $n=3$ is obvious.\n\nSuppose the statement holds for all odd $n$ from $3$ to $2k-1$. Consider $n = 2k+1$. Enumerate the vertices of the $n$-gon in clockwise order: $A_0, A_1, \\dots, A_{2k+1}$. Let the first player connect $A_0$ with $O$ on their first move. Consider the first move of the second player:\n\n- **Connected two vertices distinct from $A_0$:** Suppose the second player connects $A_{i-1}$ with $A_i$ where $2 \\leq i \\leq k+1$. The first player connects $A_i$ with $A_{i+1}$, reducing the game to $n = 2k-1$. Consider $A_{i-1}, A_i, A_{i+1}$ as one vertex, with connections as described, and two spare connections to $O$ that do not affect the game.\n\n- **Connected $A_0$ with adjacent vertex:** Suppose the second player connects $A_1$ with $A_0$. The first player connects $A_0$ with $A_{2k+1}$, reducing the game to $n = 2k-1$, with $A_{2k+1}, A_0, A_1$ considered as one vertex.\n\n- **Connected with $O$ a vertex adjacent to $A_0$:** Suppose the second player connects $A_1$ with $O$. The first player connects $A_{2k+1}$ with $O$, reducing the game to $n = 2k-1$, with $A_{2k+1}, A_0, A_1$ as one vertex. The only difference is that $A_1$ and $A_{2k+1}$ have two spare connections with $A_0$.\n\n- **Connected with $O$ a vertex not adjacent to $A_0$:** Suppose the second player connects $A_{2i}$ with $O$. The first player connects $A_i$ with $O$, reducing the game to $n = 2k - 2i + 1$. Vertices between $A_0$ and $A_{2i}$ are considered as one vertex connected to $O$. If the second player draws a segment within one of the sectors $OA_0A_i$ or $OA_iA_{2i}$, the first player draws the symmetric segment with respect to $OA_i$. There is an even number of spare moves, so the game is correctly reduced.\n\nIn each case, the game is reduced to an equivalent game with a smaller odd number of vertices, proving the induction step. Thus, the first player has a winning strategy for all odd $n \\geq 3$.",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 14928,
    "subject": "Mathematics (Olympiad)",
    "question": "Determine all triples $(a, b, c)$ of integers $a \\ge 0$, $b \\ge 0$, and $c \\ge 0$ that satisfy the equation\n\n$$\na^{b+20}(c-1) = c^{b+21} - 1.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "The solutions are:\n\n$$\n\\{(1, b, 0) : b \\in \\mathbb{Z}_{>0}\\} \\cup \\{(a, b, 1) : a, b \\in \\mathbb{Z}_{>0}\\}\n$$\n\nFirst, factor the right side:\n\n$$\na^{b+20}(c-1) = (c^{b+20} + c^{b+19} + \\dots + c + 1)(c-1).\n$$\n\nConsider $c=1$ separately. For $c \\ne 1$, divide both sides by $c-1$:\n\n$$\na^{b+20} = c^{b+20} + c^{b+19} + \\dots + c + 1.\n$$\n\n**Case 1:** $c=1$\n\n$$\n0 = 0.\n$$\n\nSo any $a, b \\ge 0$ work.\n\n**Case 2:** $c \\ne 1$\n\nWe have\n\n$$\nc^{b+20} + c^{b+19} + \\dots + c + 1 > c^{b+20},\n$$\n\nso $a \\ge c+1$. By the binomial theorem,\n\n$$\n\\begin{aligned}\na^{b+20} &\\ge (c+1)^{b+20} \\\\\n&= c^{b+20} + \\binom{b+20}{1} c^{b+19} + \\dots + \\binom{b+20}{b+19} c + 1 \\\\\n&\\ge c^{b+20} + c^{b+19} + \\dots + c + 1 \\\\\n&= a^{b+20}.\n\\end{aligned}\n$$\n\nEquality holds only if $c=0$. For $c=0$,\n\n$$\na^{b+20} = 1\n$$\n\nso $a=1$ and $b \\ge 0$.\n\nFor $c>0$, $a^{b+20}$ is strictly between $c^{b+20}$ and $(c+1)^{b+20}$, which is impossible for integer $a$.\n\nThus, the only solutions are $(1, b, 0)$ for $b \\ge 0$ and $(a, b, 1)$ for $a, b \\ge 0$.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 14929,
    "subject": "Mathematics (Olympiad)",
    "question": "Two tangents are drawn from a point $M$ to circle $k$, touching it at points $G$ and $H$. If $O$ is the center of $k$ and $K$ is the orthocenter of triangle $MGH$, prove that $\\angle GMH = \\angle OGK$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let us notice that $K$ must lie on $OM$. From $HK \\perp GM$ and $OG \\perp GM$, it follows that $HK \\parallel OG$. Analogously, $OH \\parallel GK$.\n\nFrom $\\overline{OG} = \\overline{OH}$, it follows that $OHKG$ is a rhombus.\n\nNotice that $O$, $H$, $M$, and $G$ lie on the circle with diameter $OM$. Hence $\\angle OGH = \\angle OMH$. Now, the statement of the exercise follows from $\\angle OGK = 2\\angle OGH$ and $\\angle GMH = 2\\angle OMH$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 14930,
    "subject": "Mathematics (Olympiad)",
    "question": "Points $A, V_1, V_2, B, U_2, U_1$ lie fixed on a circle $\\Gamma$, in that order, and such that $BU_2 > AU_1 > BV_2 > AV_1$.\n\nLet $X$ be a variable point on the arc $V_1V_2$ of $\\Gamma$ not containing $A$ or $B$. Line $XA$ meets line $U_1V_1$ at $C$, while line $XB$ meets line $U_2V_2$ at $D$.\n\nProve there exists a fixed point $K$, independent of $X$, such that the power of $K$ to the circumcircle of $\\triangle XCD$ is constant.",
    "options": [],
    "answer": "See solution",
    "solution": "For brevity, let $\\ell_i$ denote line $U_iV_i$ for $i = 1, 2$.\n\nWe first give an explicit description of the fixed point $K$. Let $E$ and $F$ be points on $\\Gamma$ such that $\\overline{AE} \\parallel \\ell_1$ and $\\overline{BF} \\parallel \\ell_2$. The problem conditions imply that $E$ lies between $U_1$ and $A$ while $F$ lies between $U_2$ and $B$. Then we let\n\n$$\nK = \\overline{AF} \\cap \\overline{BE}.\n$$\n\nThis point exists because $AEFB$ are the vertices of a convex quadrilateral.\n\n**Remark (How to identify the fixed point).** If we drop the condition that $X$ lies on the arc, then the choice above is motivated by choosing $X \\in \\{E, F\\}$. Essentially, when one chooses $X \\to E$, the point $C$ approaches an infinity point. So in this degenerate case, the only points whose power is finite to $(XCD)$ are those on line $BE$. The same logic shows that $K$ must lie on line $AF$. Therefore, if the problem is going to work, the fixed point must be exactly $\\overline{AF} \\cap \\overline{BE}$.\n\nWe give two possible approaches for proving the power of $K$ with respect to $(XCD)$ is fixed.\n\n**First approach by Vincent Huang**\n\nWe need the following claim:\n\n**Claim** — Suppose distinct lines $AC$ and $BD$ meet at $X$. Then for any point $K$\n\n$$\n\\mathrm{pow}(K, XAB) + \\mathrm{pow}(K, XCD) = \\mathrm{pow}(K, XAD) + \\mathrm{pow}(K, XBC).\n$$\n\n*Proof.* The difference between the left-hand side and right-hand side is a linear function in $K$, which vanishes at all of $A, B, C, D$. $\\square$\n\nConstruct the points $P = \\ell_1 \\cap \\overline{BE}$ and $Q = \\ell_2 \\cap \\overline{AF}$, which do not depend on $X$.\n\n**Claim** — Quadrilaterals $BPCX$ and $AQDX$ are cyclic.\n\n*Proof.* By Reim's theorem: $\\angle CPB = \\angle AEB = \\angle AXB = \\angle CXB$, etc. $\\square$\n\n![](images/sols-TST-IMO-2021_p3_data_dadec6cd40.png)\n\nNow, for the particular $K$ we choose, we have\n\n$$\n\\begin{aligned}\n\\operatorname{pow}(K, XCD) &= \\operatorname{pow}(K, XAD) + \\operatorname{pow}(K, XBC) - \\operatorname{pow}(K, \\Gamma) \\\\\n&= KA \\cdot KQ + KB \\cdot KP - \\operatorname{pow}(K, \\Gamma).\n\\end{aligned}\n$$\n\nThis is fixed, so the proof is completed.\n\n**Second approach by authors**\n\nLet $Y$ be the second intersection of $(XCD)$ with $\\Gamma$. Let $S = \\overline{EY} \\cap \\ell_1$ and $T = \\overline{FY} \\cap \\ell_2$.\n\n**Claim** — Points $S$ and $T$ lie on $(XCD)$ as well.\n\n*Proof.* By Reim's theorem: $\\angle CSY = \\angle AEY = \\angle AXY = \\angle CXY$, etc. $\\square$\n\n![](images/sols-TST-IMO-2021_p3_data_3c8903a334.png)\n\nNow let $X'$ be any other choice of $X$, and define $C'$ and $D'$ in the obvious way. We are going to show that $K$ lies on the radical axis of $(XCD)$ and $(X'C'D')$.\n\n![](images/sols-TST-IMO-2021_p3_data_8a6f54ebf6.png)\n\nThe main idea is as follows:\n\n**Claim** — The point $L = \\overline{EY} \\cap \\overline{AX}'$ lies on the radical axis. By symmetry, so does the point $M = \\overline{FY} \\cap \\overline{BX}'$ (not pictured).\n\n*Proof.* Again by Reim's theorem, $SC'YX'$ is cyclic. Hence we have\n\n$$\n\\mathrm{pow}(L, X'C'D') = LC' \\cdot LX' = LS \\cdot LY = \\mathrm{pow}(L, XCD). \\quad \\square\n$$\n\nTo conclude, note that by Pascal's theorem on $EYFAX'B$, it follows $K, L, M$ are collinear, as needed.\n\n**Remark.** All the conditions about $U_1, V_1, U_2, V_2$ at the beginning are there to eliminate configuration issues, making the problem less obnoxious to the contestant.\n\nIn particular, without the various assumptions, there exist configurations in which the point $K$ is at infinity. In these cases, the center of $XCD$ moves along a fixed line.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 14931,
    "subject": "Mathematics (Olympiad)",
    "question": "If $a$ is an even positive integer and $A = a^n + a^{n-1} + \\dots + a + 1$, where $n \\in \\mathbb{N}^*$, is a perfect square, prove that $a$ is a multiple of $8$.",
    "options": [],
    "answer": "See solution",
    "solution": "Since $a$ is an even positive integer, $A$ is odd. Therefore, $A$ must be a perfect square of an odd integer, that is,\n\n$$\nA = (2\\kappa + 1)^2 = 4\\kappa^2 + 4\\kappa + 1 = 4\\kappa(\\kappa + 1) + 1,\n$$\n\nwhere $\\kappa$ is a positive integer. Since one of $\\kappa$ or $\\kappa + 1$ is even, we have\n\n$$\nA = 4\\kappa(\\kappa + 1) + 1 = 8\\rho + 1, \\text{ where } \\rho \\text{ is a positive integer}.\n$$\n\nThus,\n\n$$\nA - 1 = a^n + a^{n-1} + \\dots + a = 8\\rho.\n$$\n\nSo,\n\n$$\na(a^{n-1} + \\dots + a + 1) = 8\\rho \\implies 8 \\mid a(a^{n-1} + \\dots + a + 1).\n$$\n\nSince $\\gcd(8, a^{n-1} + \\dots + a + 1) = 1$, it follows that $8 \\mid a$, i.e., $a$ is a multiple of $8$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 14932,
    "subject": "Mathematics (Olympiad)",
    "question": "一場數學會議中只有 $n$ 對夫妻參加（$n \\ge 8$），大會安排所有男士坐在一張有 $n$ 個座位的圓桌，所有女士坐另一張也是有 $n$ 個座位的圓桌。大會主辦單位發現一種人傳人的病毒正在與會者之間傳播，且其傳染途徑如下：設 $P$ 為一位健康的與會者，且其所坐位置兩側鄰居標為 $P_{\\text{left}}$, $P_{\\text{right}}$，其配偶標為 $P_{\\text{mate}}$；只有在 $P_{\\text{left}}, P_{\\text{right}}, P_{\\text{mate}}$ 這三人中至少有兩人感染病毒的情況下，$P$ 才會立刻被傳染而感染到病毒，否則 $P$ 會一直保持健康。設會議一開始的 $2n$ 個人中有某 $s$ 個人已經為病毒感染者，且經由病毒在與會者間不斷傳染，導致最終所有原本健康的與會者皆感染到病毒。試問 $s$ 的最小值為何？",
    "options": [],
    "answer": "See solution",
    "solution": "令 $s^*$ 為 $s$ 的最小可能值。將這場會議中的 $n$ 位男士以 $n$ 個點 $h_1, \\dots, h_n$ 表示，$n$ 位女士以另外 $n$ 個點 $w_1, \\dots, w_n$ 表示，並設其中點 $h_i$ 與點 $w_i$ 為夫妻關係（$i = 1, 2, \\dots, n$）。不失一般性，假設 $h_i$ 的左右兩側鄰居分別為 $h_{i-1}, h_{i+1}$（$i = 1, 2, \\dots, n$，且 $h_0 = h_n, h_{n+1} = h_1$）。令 $S^* = \\{h_1, h_3, h_5, \\dots, h_{2\\lfloor \\frac{n-2}{2} \\rfloor+1}\\} \\cup \\{h_{n-2}, w_{n-1}\\}$。易知，若會議一開始時 $S^*$ 內這 $\\lfloor \\frac{n+2}{2} \\rfloor = \\lceil \\frac{n+1}{2} \\rceil$ 人皆已感染到病毒，則經由題意的傳染方式，必會使所有與會者皆感染到病毒。\n\n令集合 $S$ 收集會議一開始已感染病毒者，接著我們要證明 $|S| > \\lceil \\frac{n+1}{2} \\rceil$。首先，以 $V = \\{h_1, h_2, \\dots, h_n\\} \\cup \\{w_1, w_2, \\dots, w_n\\}$ 為點集建造一個圖 $G$，$G$ 中兩點 $x, y$ 有邊相連若且唯若 $x$ 與 $y$ 為夫妻或 $x$ 與 $y$ 的座位相鄰。易知 $G$ 中共有 $3n$ 條邊，且每一點接 $3$ 條邊。設 $p_1, p_2, \\dots, p_{2n-s}$ 為所有健康與會者感染到病毒的順序。由於 $G$ 中的每一條邊 $e$ 皆满足下列兩性質之一：\n\n性質 1：$e$ 有一端點位於 $S$ 內。\n\n性質 2：$e$ 的兩端點 $p_i, p_j \\in V \\setminus S$ 且 $i < j$。\n\n故 $G$ 中滿足（性質 1）的邊數目小於或等於 $3s$，且滿足（性質 2）的邊數目小於或等於 $2n-s-1$。故 $G$ 中邊的數目滿足 $3n \\le 3s + (2n-s-1)$，亦即 $s \\ge \\frac{n+1}{2}$。\n\n綜合上面討論得知 $s^* = \\lceil \\frac{n+1}{2} \\rceil$。\n\n![](images/13-2J-ind_p7_data_e4a1f9d757.png)",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 14933,
    "subject": "Mathematics (Olympiad)",
    "question": "Find the greatest number of depicted pieces composed of 4 unit squares that can be placed without overlapping on an $n \\times n$ grid (where $n$ is a positive integer) in such a way that it is possible to move from some corner to the opposite corner via uncovered squares (moving between squares requires a common edge). The shapes can be rotated and reflected.\n\n![](images/prob1718_p29_data_7893e4d8d2.png)",
    "options": [],
    "answer": "See solution",
    "solution": "As moving from one corner to the opposite one involves at least $n-1$ horizontal and $n-1$ vertical steps, the path goes through at least $2n-1$ unit squares. Thus the shapes can cover no more than $(n-1)^2$ unit squares, and there can be no more than $\\frac{(n-1)^2}{4}$ for odd $n$ and $\\frac{(n-1)^2-1}{4} = \\frac{n^2-2n}{4}$ for even $n$.\n\nWe show that for $n \\equiv 3 \\pmod 4$ it is not possible to place $\\frac{(n-1)^2}{4}$ shapes. Suppose that this number of shapes has been placed and exactly $2n-1$ unit squares are not covered along the path from one corner to the opposite. Notice that the number of shapes $\\frac{(n-1)^2}{4}$ is odd as $n-1$ is divisible by 2 but not by 4 and consequently 4 is the greatest power of 2 that divides $(n-1)^2$. Let us colour the rows alternatingly black and white and let $m$ and $v$ be the numbers of black and white squares, respectively, covered by shapes. As every shape covers exactly 3 squares of one colour and 1 of the other, $m$ and $v$ must be odd.\n\nIf the path of squares not covered is along the edge (Fig. 44 for $n = 7$), then the area covered by the shapes would form an $(n-1) \\times (n-1)$ grid containing an even number of both black and white squares (each row has an even number of squares of the same colour). Any other configuration of the path can be obtained as a transformation of this by steps of substituting a corner in the path by a square situated diagonally and hence of the opposite colour (Fig. 45). Hence the parities of the numbers of black and white squares along the path change at every step. Covering the unused area on either side of the path requires the number of squares to be divisible by 4 and every step changes the number by 1, thus 4 must divide the number of steps taken. Therefore, the parities of black and white squares under the path equal those at the initial configuration of the path, and so do the parities of black and white squares covered by the shapes equal those for the $(n-1) \\times (n-1)$ grid. Thus $m$ and $v$ are even. The contradiction shows that it is not possible to place $\\frac{(n-1)^2}{4}$ shapes according to the given conditions.\n\nIt is sufficient to place the required number of shapes on an $(n-1) \\times (n-1)$ grid to show that it is possible to place $\\frac{n^2-2n}{4}$ shapes for even $n$, $\\frac{(n-1)^2}{4}$ shapes for $n \\equiv 1 \\pmod 4$ and $\\frac{(n-1)^2}{4} - 1$ shapes for $n \\equiv 3 \\pmod 4$.\n\nIf $n \\equiv 1 \\pmod 4$, then $(n-1) \\times (n-1)$ grid can be completely covered by $2 \\times 4$ rectangles, each composed of 2 shapes (Fig. 46).\n\n![](images/prob1718_p30_data_8c9f3809cb.png)\n\nFor $n \\equiv 3 \\pmod 4$ it is possible to place the shapes on an $(n-1) \\times (n-1)$ grid in such a way that only a $2 \\times 2$ area at the centre of the grid is not covered. This is trivial for $n=3$. For $n > 3$, cover a strip that is 2 squares wide at the edges of the grid, as shown in Fig. 47. This yields a square with side length 4 less than the previous grid that can be covered in the same manner until reaching a square with side length 2.\n\n![](images/prob1718_p30_data_f8928a52ec.png)\n\nFor $n \\equiv 2 \\pmod 4$ or $n \\equiv 0 \\pmod 4$, it is possible to place the shapes on an $(n-1) \\times (n-1)$ grid in such a way that only the middle square is not covered. This is trivial for $n=2$ and shown for $n=4$ on Fig. 48. For $n > 4$, cover a strip that is 2 squares wide at the edges of the grid, as shown in\n\n![](images/prob1718_p30_data_3aa5a8a64e.png)\n\n![](images/prob1718_p30_data_c860fe06b6.png)\n\n![](images/prob1718_p30_data_2b64447265.png)\n\nFigures 49 and 50 for $n \\equiv 2 \\pmod 4$ and $n \\equiv 0 \\pmod 4$, respectively. This yields a square with side length 4 less than the previous grid that can be covered in the same manner until reaching a square with side length 1 or 3.",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 14934,
    "subject": "Mathematics (Olympiad)",
    "question": "In the plane rectangular coordinate system, the graph of the function\n$$\ny = \\frac{x+1}{|x|+1}\n$$\nhas three different points lying on a line $l$, and the sum of the abscissas of these three points is $0$. Find the range of values of the slope of $l$.",
    "options": [],
    "answer": "See solution",
    "solution": "When $x \\ge 0$, $y = 1$; when $x < 0$, $y = \\frac{x+1}{1-x}$, which is strictly increasing in $x$ and less than $1$.\n\nSuppose the line $l$ is $y = kx + b$. The given condition is equivalent to the equation\n$$\nkx + b = \\frac{x+1}{|x|+1}\n$$\nhaving three different real solutions $x_1, x_2, x_3$ (with $x_1 < x_2 < x_3$) such that $x_1 + x_2 + x_3 = 0$.\n\nFirst, $k \\neq 0$, otherwise $l$ would be $y = 1$, but then the sum of the abscissas of any three intersection points would be greater than $0$, which contradicts the condition.\n\nFor $x < 0$, the equation becomes\n$$\nkx^2 - (k - b - 1)x + 1 - b = 0.\n$$\nThis quadratic has at most two negative solutions.\n\nFor $x \\ge 0$, the equation becomes\n$$\nkx + b = 1,\n$$\nwhich has at most one non-negative solution.\n\nThus, the quadratic has two different negative solutions $x_1, x_2$, where\n$$\nx_1 + x_2 = \\frac{k - b - 1}{2}.\n$$\nThe linear equation has a non-negative solution $x_3 = \\frac{1-b}{k}$. Since $x_1 + x_2 + x_3 = 0$, we have $k = 2b$.\n\nTherefore, $x_3 = \\frac{1-b}{2b}$. Since $x_3 \\ge 0$, $0 < b \\le 1$.\n\nThe quadratic becomes\n$$\n2b x^2 + (1-b)x + 1 - b = 0.\n$$\nThe discriminant is\n$$\n(1-b)^2 - 4 \\cdot 2b (1-b) = (1-b)(1-9b) > 0.\n$$\nCombining with $0 < b \\le 1$, we get $0 < b < \\frac{1}{9}$.\n\nTherefore, the range of the slope $k = 2b$ of $l$ is\n$$\n0 < k < \\frac{2}{9}.\n$$",
    "topic": "Discrete Mathematics",
    "subtopic": "Other"
  },
  {
    "id": 14935,
    "subject": "Mathematics (Olympiad)",
    "question": "It is given that there exists a prime $p$ such that $10^{17} \\leq p \\leq 10^{17} + 10$. Find $p$.",
    "options": [],
    "answer": "See solution",
    "solution": "**Answer:** $10^{17} + 3$.\n\nWe need to find the prime $p$ in the interval $[10^{17}, 10^{17} + 10]$.\n\nFirst, note that the last digit of $p$ cannot be even or $5$ (since such numbers are not prime except for $2$ and $5$).\n\n- $10^{17} + 1$ is divisible by $11$.\n- Consider $10^{17} + 9$ modulo $7$:\n  - $10 \\equiv 3 \\pmod{7}$\n  - $10^2 \\equiv 2 \\pmod{7}$\n  - $10^3 \\equiv -1 \\pmod{7}$\n  - $10^{15} = (10^3)^5 \\equiv (-1)^5 = -1 \\pmod{7}$\n  - $10^{17} = 10^{15} \\cdot 10^2 \\equiv (-1) \\cdot 2 = -2 \\pmod{7}$\n  - $10^{17} + 9 \\equiv -2 + 9 = 7 \\equiv 0 \\pmod{7}$\n  - So $10^{17} + 9$ is divisible by $7$.\n- Consider $10^{17} + 7$ modulo $17$:\n  - $10 \\equiv 10 \\pmod{17}$\n  - $10^2 \\equiv -2 \\pmod{17}$\n  - $10^8 \\equiv (10^2)^4 \\equiv (-2)^4 = 16 \\equiv -1 \\pmod{17}$\n  - $10^{16} = (10^8)^2 \\equiv (-1)^2 = 1 \\pmod{17}$\n  - $10^{17} = 10^{16} \\cdot 10 \\equiv 1 \\cdot 10 = 10 \\pmod{17}$\n  - $10^{17} + 7 \\equiv 10 + 7 = 17 \\equiv 0 \\pmod{17}$\n  - So $10^{17} + 7$ is divisible by $17$.\n\nThus, the only candidate for a prime in this interval is $10^{17} + 3$. Since it is given that such a prime exists, $p = 10^{17} + 3$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 14936,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $ABC$ be a triangle with $AB = AC$. The angle bisectors of $\\angle CAB$ and $\\angle ABC$ meet the sides $BC$ and $CA$ at $D$ and $E$, respectively. Let $K$ be the incenter of triangle $ADC$. Suppose that $\\angle BEK = 45^\\circ$. Find all possible values of $\\angle CAB$.\n\n![](images/pamphlet0910_main_p81_data_2dc235a88a.png)\n\n![](images/pamphlet0910_main_p81_data_2c5f1c5207.png)",
    "options": [],
    "answer": "See solution",
    "solution": "**Comment:** The possible values are $\\angle CAB = 60^\\circ$ and $\\angle CAB = 90^\\circ$. We first construct configurations achieving these values.\n\nLet $I$ be the intersection of $AD$ and $BE$. Then $I$ is the incenter of triangle $ABC$. In particular, $CI$ bisects $\\angle BCA$ and $K$ lies on segment $CI$.\n\nIf $\\angle CAB = 60^\\circ$, then $ABC$ is equilateral and it is routine to check that $CDIE$ is a kite with $CD = CE$ and $ID = IE$, implying that $D$ and $E$ are symmetric with respect to line $CI$ and that $\\angle BEK = \\angle IEK = \\angle IDK = 45^\\circ$.\n\nIf $\\angle CAB = 90^\\circ$, then $ABC$ is an isosceles right triangle. It is routine to check that $\\angle AIE = \\angle AEI = 67.5^\\circ$, implying that $AI = AE$. Thus, $AIKE$ is a kite with $AK$ as its symmetry axis and $\\angle BEK = \\angle IEK = \\angle EIK = 45^\\circ$.\n\nWe present two solutions showing that these are the only possible values.\n\n**Solution 1** (By Mea Bombardelli, the leader of the Croatia delegation). Let $E_1$ be the reflection of $E$ across the line $CI$. Because $CI$ is the angle bisector of $\\angle BCA$, $E_1$ lies on ray $CB$. We consider two cases.\n\nIn the first case, we assume that $E_1 = D$. Then $CDIE$ is a kite with $CI$ as the symmetry axis. In particular, $\\angle IEC = \\angle IDC = 90^\\circ$; that is, $BE$ is also an altitude of triangle $ABC$, implying that $AB = BC = CA$ or $\\angle BAC = 60^\\circ$. (See the left-hand side figure shown below.)\n\nIn the second case, we assume that $\\angle D \\neq E_1$. By symmetry, $\\angle IE_1K = \\angle IEK = 45^\\circ = \\angle IDK$, from which it follows that $I, D, E_1, K$ lie on a circle. Hence either $\\angle IKE = \\angle IDE_1 = 90^\\circ$ or $\\angle IKE_1 = 180^\\circ - \\angle IDE = 90^\\circ$; that is, $\\angle IKE_1 = 90^\\circ$. Therefore, we deduce the right-hand side figure shown below. In cyclic quadrilateral, we have $\\angle KIE_1 = \\angle KDE_1 = \\angle KDC = 45^\\circ$. By symmetry, we have $\\angle EIK = \\angle KIE_1 = 45^\\circ$, from which it follows that $\\angle IBC = \\angle ICB = 22.5^\\circ$ and $\\angle BAC = 90^\\circ$.\n\n![](images/pamphlet0910_main_p82_data_5ff2b0ae42.png)\n\n**Solution 2** (By Zuming Feng). Set $\\angle ABC = \\angle BCA = 2x$, with $0^\\circ < x < 45^\\circ$. Then $\\angle EIK = 2x$, $\\angle ECK = \\angle DCK = x$, and $\\angle CEK = 135^\\circ - x$.\n\nIt is clear that $CID$ is a right triangle, and, by the angle-bisector theorem, we have\n\n$$\n\\frac{IK}{KC} = \\frac{ID}{DC} \\quad \\text{or} \\quad \\frac{IK}{KC} = \\frac{\\sin x}{\\cos x}.\n$$\n\nApplying the law of sines in triangles $EIK$ and $CEK$ gives\n\n$$\n\\frac{\\sin 45^\\circ}{IK} = \\frac{\\sin 2x}{EK} \\quad \\text{and} \\quad \\frac{KC}{\\sin(135^\\circ - 3x)} = \\frac{EK}{\\sin x}.\n$$\n\nMultiplying the last three equations yields\n\n$$\n\\frac{\\sin 45^\\circ}{\\sin(135^\\circ - 3x)} = \\frac{\\sin 2x}{\\cos x} = 2 \\sin x \\quad \\text{or} \\quad \\sin 45^\\circ = 2 \\sin x \\sin(135^\\circ - 3x).\n$$\n\nBy the addition-subtraction formulas, we obtain\n\n$$\n\\sin 45^\\circ = 2 \\sin x \\sin(135^\\circ - 3x) = 2 \\sin x(\\sin 135^\\circ \\cos 3x - \\cos 135^\\circ \\sin 3x),\n$$\n\nhence $1 = 2 \\sin x(\\cos 3x + \\sin 3x)$. By the product-to-sum formulas and the double-angle formulas, we have\n\n$$\n\\begin{aligned}\n1 &= 2 \\sin x(\\cos 3x + \\sin 3x) = 2 \\sin x \\cos 3x + 2 \\sin x \\sin 3x \\\\\n&= \\sin 4x - \\sin 2x + \\cos 2x - \\cos 4x \\\\\n&= 2 \\sin 2x \\cos 2x - \\sin 2x + \\cos 2x - 2 \\cos^2 2x + 1 \\\\\n&= 1 + (2 \\cos x - 1)(\\sin 2x - \\cos 2x);\n\\end{aligned}\n$$\n\nthat is, we obtain $(2 \\cos x - 1)(\\sin 2x - \\cos 2x) = 0$. Because $0^\\circ < x < 45^\\circ$, the possible values of $x$ are $x = 30^\\circ$ (corresponding to $2 \\cos x - 1 = 0$) and $x = 22.5^\\circ$ (corresponding to $\\sin 2x - \\cos 2x = 0$), leading to $\\angle CAB = 60^\\circ$ and $\\angle CAB = 90^\\circ$, respectively.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 14937,
    "subject": "Mathematics (Olympiad)",
    "question": "Consider a convex quadrilateral $ABCD$ with\n$$\nAB = CB \\quad \\text{and} \\quad \\angle ABC + 2\\angle CDA = \\pi,\n$$\nand let $E$ be the midpoint of $AC$. Show that $\\angle CDE = \\angle BDA$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let point $X$ lie on line $BE$ such that $\\angle CXE = \\angle CDE$ (where $X$ is the intersection point, other than $C$, of the circumcircle of $\\triangle CDE$ and the line $BE$).\n\nTherefore, the quadrilateral $DECX$ is cyclic, so $\\angle DXE = \\angle DCE = \\pi - \\angle CDA - \\angle CAD$. But $\\angle CDA = \\frac{1}{2}(\\pi - \\angle ABC) = \\angle CAB$, so $\\angle DCE = \\pi - \\angle CAB - \\angle CAD = \\pi - \\angle BAD$, hence the quadrilateral $ABXD$ is cyclic.\n\nIt follows that $\\angle BDA = \\angle BXA = \\angle CXE = \\angle CDE$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 14938,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $ABCD$ be a convex quadrilateral such that $\\angle ABC = \\angle ADC = 135^\\circ$ and\n$$\nAC^2 \\cdot BD^2 = 2AB \\cdot BC \\cdot CD \\cdot DA.\n$$\nProve that the diagonals of quadrilateral $ABCD$ are perpendicular.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $\\alpha$ be the angle formed by diagonals $AC$ and $BD$, where $0^\\circ < \\alpha \\leq 90^\\circ$. We calculate the area of quadrilateral $ABCD$ in two ways. We obtain\n$$\n[ABCD] = \\frac{1}{2}(AC \\cdot BD \\sin \\alpha)\n$$\nand\n$$\n\\begin{align*}\n[ABCD] &= [ABC] + [CDA] \\\\\n&= \\frac{1}{2}(AB \\cdot BC \\sin 135^\\circ) + \\frac{1}{2}(CD \\cdot DA \\sin 135^\\circ) \\\\\n&= \\frac{1}{2\\sqrt{2}}(AB \\cdot BC + CD \\cdot DA).\n\\end{align*}\n$$\nCombining the above two equations yields\n$$\nAC^2 \\cdot BD^2 \\sin^2 \\alpha = \\frac{1}{2}(AB \\cdot BC + CD \\cdot DA)^2.\n$$\nFor real numbers $a$ and $b$, $(a + b)^2 \\geq 4ab$, hence\n$$\nAC^2 \\cdot BD^2 \\sin^2 \\alpha \\geq 2AB \\cdot BC \\cdot CD \\cdot DA.\n$$\nIt follows that $\\sin^2 \\alpha \\geq 1$. Therefore, $\\sin \\alpha = 1$ and $\\alpha = 90^\\circ$, as desired.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 14939,
    "subject": "Mathematics (Olympiad)",
    "question": "As shown in the figure, $BC$ and $AC$ are tangent to the inscribed circle $I$ of $\\triangle ABC$ at $M$ and $N$. $E$ and $F$ are the midpoints of $AB$ and $AC$ respectively. $EF$ intersects $BI$ at $D$. Prove that $M$, $N$, $D$ are collinear.\n\n![](images/Mathematical_Olympiad_in_China_2007-2008-Problems_and_Solutions_p189_data_29d628e098.png)",
    "options": [],
    "answer": "See solution",
    "solution": "Join $AD$, and it is obvious that $\\angle ADB = 90^\\circ$. Then join $AI$ and $DM$. Suppose $DM$ intersects $AC$ at $G$. Since $\\angle ABI = \\angle DBM$, we obtain $\\frac{AB}{BD} = \\frac{BI}{BM}$. Hence, $\\triangle ABI \\sim \\triangle DBM$, and\n\n$$\n\\angle DMB = \\angle AIB = 90^\\circ + \\frac{1}{2} \\angle ACB.\n$$\n\nJoin $IG$, $IC$, $IM$, then\n\n$$\n\\angle IMG = \\angle DMB - 90^\\circ = \\frac{1}{2} \\angle ACB = \\angle GCI.\n$$\n\nTherefore, $I$, $M$, $C$, $G$ are concyclic, and $IG \\perp AC$.\n\nConsequently, since $G$ and $N$ represent the same point, we conclude that $M$, $N$, $D$ are collinear.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 14940,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $ABCD$ be a convex quadrilateral with $BA \\ne BC$. Denote by $\\omega_1$ and $\\omega_2$ the incircles of triangles $ABC$ and $ADC$, respectively. Suppose that there exists a circle $\\omega$ tangent to ray $BA$ beyond $A$ and to ray $BC$ beyond $C$, which is also tangent to lines $AD$ and $CD$. Prove that the common external tangent lines of $\\omega_1$ and $\\omega_2$ intersect on $\\omega$.",
    "options": [],
    "answer": "See solution",
    "solution": "We start our solution with two interesting and well-known geometry facts.\n\n**Lemma 1.** Let $BCA$ be a triangle, and let its incircle $\\omega_1$ touch side $CA$ at $T_B$. Point $B_1$ is diametrically opposite to $T_B$ on $\\omega_1$. Ray $BB_1$ meets side $CA$ at $B_2$. Prove that $CT_B = AB_2$.\n\n![](images/pamphlet0910_main_p72_data_645fd68517.png)\n\n*Proof.* Let $\\ell$ be the line tangent to circle $\\omega_1$ at $B_1$, and let $\\ell$ intersect segments $BC$ and $BA$ at $C_1$ and $A_1$, respectively. Then $\\omega_1$ is the excircle of triangle $BC_1A_1$ opposite $B$. Let $\\mathbf{H}$ denote the homothety (or dilation) with center $B$ and ratio $BB_2/BB_1$. Since $\\ell \\perp T_B B_1$ and $CA \\perp T_B B_1$, $\\ell \\parallel CA$. Hence, $BC/BC_1 = BA/BA_1 = BB_2/BB_1$. Thus, $\\mathbf{H}(B_1) = B_2$, $\\mathbf{H}(C_1) = C$, and $\\mathbf{H}(A_1) = A$. It also follows that the excircle $\\omega_B$ of triangle $BCA$ opposite vertex $B$ is tangent to side $CA$ at $B_2$.\n\nIt is well known that\n\n$$\nCT_B = \\frac{BC + CA - AB}{2}.\n$$\n\nWe compute $AB_2$. Let $A_2$ and $C_2$ denote the points of tangency of circle $\\omega_B$ with rays $BA$ and $BC$, respectively. Then by equal tangents, $BA_2 = BC_2$, $AA_2 = AB_2$, and $CC_2 = CB_2$. Hence,\n\n$$\nBA_2 = \\frac{BA_2 + BC_2}{2} = \\frac{BA + AA_2 + BC + CC_2}{2} = \\frac{BA + AB_2 + B_2C + CB}{2} = \\frac{BA + AC + CB}{2}.\n$$\n\nIt follows that\n\n$$\nAB_2 = AA_2 = BA_2 - BA = \\frac{BC + CA - AB}{2} = CB_2,\n$$\n\nas desired.\n\n$\\Box$\n\n**Lemma 2.** Let $ABCD$ be a convex quadrilateral. Circle $\\omega$ is an *excircle* of quadrilateral $ABCD$ opposite $B$; that is, it is tangent to ray $BA$ (beyond $A$ at $U_A$), ray $BC$ (beyond $C$ at $U_C$), ray $AD$ (beyond $D$ at $V_A$), and ray $CD$ (beyond $D$ at $V_C$). Prove that $BA + AD = BC + CD$.\n\n![](images/pamphlet0910_main_p72_data_43ea9429bf.png)\n\n*Proof.* Extend $AV_A$ through $V_A$ to meet ray $BC$ at $A_3$, and extend segment $CV_C$ through $V_C$ to meet $BA$ at $C_3$. By the convexity of $ABCD$, it is not difficult to reduce our configuration to the figure shown above. By equal tangents, we have\n\n$$\nBU_A = BU_C, \\quad AU_A = AV_A, \\quad C_3U_A = C_3V_C, \\quad CU_C = CV_C, \\quad A_3U_C = A_3V_A, \\quad DV_A = DV_C.\n$$\n\nIt follows that\n\n$$\nBA + AD = BU_A - U_AA + AD = BU_A - V_AA + AD = BU_A - DV_A.\n$$\n\nSimilarly, we can show that $BC + CD = BU_C - DV_C$. Thus, $BA + AD = BC + CD$. $\\square$\n\n![](images/pamphlet0910_main_p73_data_829f96577a.png)\n\n![](images/pamphlet0910_main_p73_data_fea2b82e22.png)\n\nNow we prove the result of the problem. We maintain the notations in Lemmas 1 and 2. Circle $\\omega_2$ touches sides $AD$, $DC$, $CA$ at $S_C$, $S_A$, $S_D$, respectively.\n\nSince $\\omega_2$ is the incircle of triangle $ADC$, we know that\n\n$$\nAS_D = \\frac{AC + AD - CD}{2}.\n$$\n\nBy Lemmas 1 and 2, we conclude that\n\n$$\nAS_D = \\frac{AC + AD - CD}{2} = \\frac{AC + BC - AB}{2} = CT_B = AB_2,\n$$\n\nthat is, $B_2 = S_D$. By Lemma 1, points $B$, $B_1$, $S_D$ are collinear. Ray $BB_1$ meets minor arc $\\widehat{U_A U_C}$ (which is part of $\\omega$) at $H_1$. Construct points $A_4$ and $C_4$ on rays $BA$ and $BC$, respectively, such that $A_4C_4 \\parallel AC$ and $H_1$ lies on $A_4C_4$. Since $S_D = B_2$, the excircle of triangle $BCA$ opposite $B$ is tangent to $CA$ at $B_2$. Consider the homothety $\\mathbf{H}_1$ centered at $B$ sending $AC$ to $A_4C_4$. Since $\\mathbf{H}_1(B_2) = H_1$, circle $\\omega$ is tangent to $A_4C_4$ at $H_1$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 14941,
    "subject": "Mathematics (Olympiad)",
    "question": "Пусть $P(x)$ — многочлен второй степени. Докажите, что если $t$ — общий корень многочленов $P(x)$, $P(P(x))$ и $P(P(P(x)))$, то $P(0) \\cdot P(1) = 0$.",
    "options": [],
    "answer": "See solution",
    "solution": "Пусть $t$ — общий корень данных многочленов. Тогда $0 = P(P(P(t))) = P(P(0))$. Пусть $P(x) = x^2 + a x + b$; тогда $P(0) = b$, $P(1) = a + b + 1$, а значит, $0 = P(P(0)) = P(b) = a b + b^2 + b = b(a + b + 1) = P(0) \\cdot P(1)$, что и требовалось доказать.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 14942,
    "subject": "Mathematics (Olympiad)",
    "question": "求所有正整數有序對 $(a, b, c)$ 使得\n\n$$\na^b + b^c + c^a = a^c + b^a + c^b\n$$\n成立。",
    "options": [],
    "answer": "See solution",
    "solution": "*答案：* $(1, 2, 3)$ 的所有排列，以及 $(x, y, y)$ 的所有排列，其中 $x, y \\in \\mathbb{N}$。\n\n*解答：*\n\n**引理 1.** 對所有 $n \\ge 2$，有 $2 < (1 + \\frac{1}{n})^n < 3$。\n\n*證明*：利用二項式定理，\n\n$$\n2 = 1 + n \\cdot \\frac{1}{n} < 1 + n \\cdot \\frac{1}{n} + \\sum_{i=2}^{n} \\frac{\\binom{n}{i}}{n^i} = (1 + \\frac{1}{n})^n.\n$$\n\n另一方面，\n\n$$\n(1 + \\frac{1}{n})^n = 1 + n \\cdot \\frac{1}{n} + \\sum_{i=2}^{n} \\frac{\\binom{n}{i}}{n^i} < 1 + 1 + \\sum_{i=2}^{n} \\frac{1}{2^{k-1}} < 3.\n$$\n\n**引理 2.** 對所有 $3 \\le x < y \\in \\mathbb{N}$，有 $x^y > y^x$。\n\n*證明*：固定 $x$，用歸納法證明。當 $y = x + 1$ 時，\n\n$$\nx^{x+1} = x \\cdot x^x \\ge 3x^x > \\left(1 + \\frac{1}{x}\\right)^x \\cdot x^x = (x+1)^x.\n$$\n\n假設對 $y = x + 1, \\dots, x + k - 1$ 成立，則\n\n$$\nx^{x+k} = x \\cdot x^{x+k-1} \\ge \\left(1 + \\frac{1}{x+k-1}\\right)^x \\cdot (x+k-1)^x = (x+k)^x.\n$$\n\n因此對所有 $y \\ge x+1$ 成立。\n\n**引理 3.** 若 $a > b > c \\ge 2$，則 $a^b + b^c + c^a < a^c + b^a + c^b$。\n\n*證明*：固定 $b, c$，考慮左右差值。\n\n$$\nb^{a+1} - c^{a+1} + (a+1)^c - (a+1)^b > b^a - c^a + a^c - a^b, \\quad \\text{對所有 } a \\ge b > c \\ge 2.\n$$\n\n化簡得：\n\n$$\n(b-1)b^a - (c-1)c^a + (a+1)^c - a^c > (a+1)^b - b^b\n$$\n\n注意 $(a+1)^c - a^c > 0$ 且 $(a+1)^b < (1+\\frac{1}{a})^b a^b < 3a^b$。若\n\n$$\n(b-1)b^a - (c-1)c^a > 2a^b\n$$\n\n成立，則原不等式成立。由引理 1，$(\\frac{b}{c})^a \\ge (1+\\frac{1}{c})^c > 2$，所以\n\n$$\n(b-1)b^a - (c-1)c^a > \\left(b-1-\\frac{c-1}{2}\\right)b^a \\ge \\left(\\frac{c+1}{2}\\right)b^a.\n$$\n\n若 $c \\ge 3$，$(\\frac{c+1}{2}) b^a \\ge 2b^a > 2a^b$。$c=2, b \\ge 4$ 時同理。\n\n剩下 $b=3, c=2$，即\n\n$$\n2 \\cdot 3^a - 2^a + (a+1)^2 - a^2 > (a+1)^3 - a^3\n$$\n\n顯然成立。$a=b$ 時為等式。\n\n回到原題，假設 $c$ 為 $a, b, c$ 的最小值。若 $c \\ge 2$ 且三數皆不同，分兩種情況：\n\n- $a > b > c$：由引理 3，$LHS < RHS$，無解。\n- $b > a > c$：同理，$RHS < LHS$，無解。\n\n所以只可能是 $(x, y, y)$ 的排列，$x, y \\in \\mathbb{N}$ 且 $\\min\\{x, y\\} = 2$，這確實滿足等式。\n\n若 $c = 1$，則變為\n\n$$\na^b - b^a = a - b\n$$\n\n假設 $a > b$，則 $a^b - b^a = a - b > 0 \\Rightarrow a^b > b^a$。由引理 2，$b < 3$，即 $b = 1$ 或 $2$。\n\n- $b = 1$，$a$ 可為任意正整數。\n- $b = 2$，$2^a = a^2 - a + 2 = (a + 1)(a - 2)$，即 $a + 1, a - 2$ 為 2 的冪。只有 $a = 3$ 滿足。\n\n綜合，所有解為 $(1, 2, 3)$ 的排列，以及 $(x, y, y)$ 的排列，$x, y \\in \\mathbb{N}$。",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 14943,
    "subject": "Mathematics (Olympiad)",
    "question": "Michael is a good chess player. He took part in a competition where all the best chess players of the city were invited. The competition had two stages.\n\nAfter the first stage, Michael found that the number of players ranked higher than him was half the number of players ranked lower than him.\n\nIn the second stage, Michael played better: he managed to surpass four of the players who were ranked higher than him after the first stage, but he was also surpassed by two who were ranked lower than him.\n\nAt the end of the competition, the final results showed that the number of players ranked higher than him was a quarter of the number of players ranked lower than him.\n\nWhat place did Michael get at the end of the competition? Justify your answer.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $x$ be the number of players ranked higher than Michael after the first stage. Then the number of players ranked lower than him is $2x$.\n\nIn the second stage, Michael surpassed 4 players who were ahead of him and was surpassed by 2 players who were behind him. Thus, at the end, the number of players ahead of him is $x - 4 + 2 = x - 2$, and the number behind him is $2x + 4 - 2 = 2x + 2$.\n\nGiven that the number of players ahead of him at the end is a quarter of those behind him:\n\n$$\n4(x - 2) = 2x + 2\n$$\n\nSolving for $x$:\n\n$$\n4x - 8 = 2x + 2 \\\\\n4x - 2x = 2 + 8 \\\\\n2x = 10 \\\\\nx = 5\n$$\n\nSo, after the first stage, 5 players were ahead of Michael. At the end, the number of players ahead of him is $x - 2 = 3$. Therefore, Michael finished in 4th place.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 14944,
    "subject": "Mathematics (Olympiad)",
    "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": "See solution",
    "solution": "Let $Z$ be the midpoint of the longer arc $BC$ of the circumcircle $\\omega$ of triangle $ABC$. The triangles $ZDB$ and $ZEC$ are congruent, because they share the sides $BD = CE$ and $ZB = ZC$, as well as the corresponding angles between them, since both lie over the chord $AZ$ of $\\omega$. It follows that $\\angle ZDA = \\angle ZEA$, which implies that the quadrilateral $ADEZ$ is cyclic. So it remains to be shown that the quadrilateral $AO_1O_2Z$ is cyclic.\n\n![](images/cps_solutions_day1_p1_data_f8eb1e5c1f.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 lengths 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$.\n\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 combination with $AO = ZO$ and $OO_1 = OO_2$, this proves that the quadrilateral $AO_1O_2Z$ is an isosceles trapezium and thus, in particular, cyclic. Thereby the problem is solved.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 14945,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $n$ be an integer, $n \\ge 2$, and $x_1, x_2, \\dots, x_n \\in [0, 1]$. Prove that\n\n$$\n\\sum_{1 \\le k < l \\le n} k x_k x_l \\le \\frac{n-1}{3} \\sum_{k=1}^{n} k x_k.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "As $x_1, x_2, \\dots, x_n \\in [0, 1]$, $x_i x_j \\le x_i$, so we have\n\n$$\n3 \\sum_{1 \\le k < l \\le n} k x_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$$\n\nFor $1 \\le k \\le n$, the coefficient of $x_k$ in the last sum is\n\n$$\n2[1 + 2 + \\dots + (k-1)] + k(n-k) = k(n-1),\n$$\n\nso we have\n\n$$\n\\begin{aligned}\n3 \\sum_{1 \\le k < l \\le n} k x_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&= (n-1) \\sum_{k=1}^{n} k x_k,\n\\end{aligned}\n$$\n\nand hence the desired inequality holds. $\\square$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 14946,
    "subject": "Mathematics (Olympiad)",
    "question": "有一三角形 $ABC$，其外接圓為 $O$。兩圓 $O_1, O_2$ 與射線 $\\overrightarrow{AB}$、$\\overrightarrow{AC}$ 以及圓 $O$ 均相切，其中 $O_1$ 在 $O$ 內部，$O_2$ 在 $O$ 外部。設 $O$、$O_1$ 的公切線與 $O$、$O_2$ 的公切線交於點 $X$，$O$ 上不含 $A$ 點的 $BC$ 弧的中點為 $M$，且線段 $\\overline{AA'}$ 為圓 $O$ 之直徑。證明：$X, M, A'$ 共線。\n\nLet $O$ be the circumcircle of triangle $ABC$. Two circles $O_1, O_2$ are tangent to each of the circle $O$ and the rays $\\overrightarrow{AB}, \\overrightarrow{AC}$, with $O_1$ interior to $O$, $O_2$ exterior to $O$. The common tangent of $O, O_1$ and the common tangent of $O, O_2$ intersect at the point $X$. Let $M$ be the midpoint of the arc $BC$ (not containing the point $A$) on the circle $O$, and the segment $\\overline{AA'}$ be a diameter of $O$. Prove that $X, M, A'$ are collinear.",
    "options": [],
    "answer": "See solution",
    "solution": "(解 1): 令 $O, O_1$ 切點為 $P$，$O, O_2$ 切點為 $Q$；原題等價於 $PMQA'$ 為調和四邊形。考慮變換「對 $A$ 取幂為 $\\overrightarrow{AB} \\times \\overrightarrow{AC}$ 的反演後對 $\\angle BAC$ 角平分線鏡射」，則 $A, B, C$ 不變，$O_1$ 變成 $A$－旁切圓（設圓心為 $I_A$），$O_2$ 變成內切圓（設圓心為 $I$）。令 $D, Y, Z$ 分別為 $A, I, I_A$ 到 $\\overrightarrow{BC}$ 的垂足，$L$ 為 $\\angle BAC$ 角平分線與 $\\overrightarrow{BC}$ 的交點，則 $P$ 變換到 $Z$，$Q$ 變換到 $Y$，$A'$ 變換到 $D$，$M$ 變換到 $L$；原題等價於 $(D, L; Y, Z)$ 為調和點列，又等價於 $(A, L; I, I_A)$ 為調和點列。因 $\\overrightarrow{AI} : \\overrightarrow{IL} = \\overrightarrow{AB} : \\overrightarrow{BL} = \\overrightarrow{AI_A} : \\overrightarrow{LI_A}$，故原命題成立。\n\n(解 2): 只要證明 $MA'$ 為 $O_1, O_2$ 之根軸，就可得 $X$ 為三圓根心，即原命題得證。\n\n令 $MA'$ 交 $AB$ 於 $M_1$，交 $AC$ 於 $M_2$，$O_1$ 切 $AB$ 於 $N_1$，$O_1$ 切 $AC$ 於 $N_2$，$O_2$ 切 $AB$ 於 $W_1$，$O_2$ 切 $AC$ 於 $W_2$。由曼海姆定理可得 $N_1, I, N_2$ 共線，$W_1, I_A, W_2$ 共線且 $N_1N_2 \\perp AI$，$W_1W_2 \\perp AI_A$。由雞爪定理可得 $IM = I_AM$，故 $N_1M_1 = W_1M_1 = N_2M_2 = W_2M_2$，由此知 $MA'$ 確為 $O_1, O_2$ 之根軸。證畢。",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 14947,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $(i, j)$ denote the small square at the intersection of the $i$-th row and $j$-th column of an $8 \\times 8$ square grid. The vertices $A$, $B$, $C$, $D$ of the big square are at $(1, 1)$, $(8, 1)$, $(8, 8)$, and $(1, 8)$, respectively. The square symmetric to $(i, j)$ with respect to diagonal $AC$ is $(j, i)$, and with respect to diagonal $BD$ is $(9-j, 9-i)$. For a pair of integers $\\{i, j\\}$ with $1 \\leq i, j \\leq 8$, define the set $\\{(i, j), (j, i), (9-i, 9-j), (9-j, 9-i)\\}$ as the clique $[i, j]$. If the coloring of small squares is symmetric with respect to both diagonals $AC$ and $BD$, then every square in a clique $[i, j]$ is colored if $(i, j)$ is colored. For each $\\{i, j\\}$, the cliques $[i, j]$, $[j, i]$, $[9-i, 9-j]$, $[9-j, 9-i]$ are identical, and the set of all small squares is partitioned into cliques.\n\n![](images/Japan_2011_p20_data_1a7ceae390.png)\n\nHow many ways are there to color small squares of the $8 \\times 8$ grid, symmetric with respect to both diagonals $AC$ and $BD$, such that in each row and each column at most one square is colored?",
    "options": [],
    "answer": "See solution",
    "solution": "Let us count the number of colorings as described.\n\nFirst, consider coloring off-diagonal cliques (cliques not on either diagonal). Each such clique consists of 4 squares. The number of ways to choose these cliques is as follows:\n\n- No clique chosen: $1$ way.\n- One clique chosen: For each $i$ ($1 \\leq i \\leq 8$), $j$ can be any of $6$ values different from $i$ and $j-1$. Each clique is counted $4$ times, so the number is $\\frac{8 \\times 6}{4} = 12$.\n- Two cliques chosen: For each of the $12$ choices above, there are $2$ ways to choose a second clique (after accounting for overcounting), so $\\frac{12 \\times 2}{2} = 12$ ways.\n\nNext, for each choice of off-diagonal cliques, consider coloring cliques on the diagonals. For $k = 1, 2, 3, 4$:\n\n- If there is already a colored square in the $k$-th row, neither $[k, k]$ nor $[k, 9-k]$ can be chosen: $1$ way.\n- If not, at most one of $[k, k]$ or $[k, 9-k]$ can be chosen: $3$ ways (choose $[k, k]$, $[k, 9-k]$, or neither).\n\nIf $n$ off-diagonal cliques are chosen, $4n$ rows already have a colored square, so $4-2n$ values of $k$ remain for the second case. Thus, there are $3^{4-2n}$ ways to choose diagonal cliques.\n\nSumming over all cases:\n\n$$\n1 \\cdot 3^4 + 12 \\cdot 3^2 + 12 \\cdot 3^0 = 81 + 108 + 12 = 201\n$$\n\n**Answer:** $201$ ways.",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 14948,
    "subject": "Mathematics (Olympiad)",
    "question": "Determine all natural numbers $a$, $b$, $c$ such that $ab + bc + ca$ is a prime number $p$ and $p$ divides $a^2b^2 + b^2c^2 + c^2a^2$.",
    "options": [],
    "answer": "See solution",
    "solution": "From the identity $$a^2b^2 + b^2c^2 + c^2a^2 = (ab + bc + ca)^2 - 2abc(a + b + c)$$ it follows that $p$ divides $abc(a + b + c)$. Because $p$ is a prime number, we get $p \\mid a$, $p \\mid b$, $p \\mid c$, or $p \\mid (a + b + c)$.\n\nSince $a$, $b$, $c < p$, the first three cases are impossible. The fourth situation can be true if and only if $a = b = c = 1$. This case works.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 14949,
    "subject": "Mathematics (Olympiad)",
    "question": "證明存在一個整係數多項式 $f(x)$，使得：\n\n1. $f(x) = 0$ 沒有有理實根。\n2. 對於任何正整數 $n$，均存在整數 $m$，使得 $f(m)$ 是 $n$ 的倍數。",
    "options": [],
    "answer": "See solution",
    "solution": "（注意可能的 $f(x)$ 不唯一。）以下證明：\n\n$$\nf(x) = (x^2 + 1)(x^2 + 2)(x^2 - 2)(x^2 + 7)\n$$\n\n滿足條件。易見 $f(x) = 0$ 沒有有理實根。\n\n1. 首先證明：對於所有奇質數的冪次方 $p^\\alpha$，都存在整數 $m_{p^\\alpha}$ 使得\n$$\np^{\\alpha} \\mid f(m_{p^{\\alpha}}).\n$$\n事實上，我們可以用歸納法證明以下引理：\n\n**引理一.** 對於所有奇質數 $p$，存在 $r_p \\in \\{-1, -2, 2\\}$，使得對於所有 $\\alpha \\ge 1$，必存在 $m_{p^\\alpha} \\in \\mathbb{Z}$ 使得 $p^\\alpha \\mid m_{p^\\alpha}^2 - r_p$。\n\n(a) 首先，對於 $\\alpha = 1$，注意到：\n- 若 $p = 4k + 1$，則 $-1$ 是模 $p$ 的二次剩餘；\n- 若 $p = 8k + 3$，則 $-2$ 是模 $p$ 的二次剩餘；\n- 若 $p = 8k + 7$，則 $2$ 是模 $p$ 的二次剩餘。\n\n綜上，必存在 $m_p \\in \\mathbb{Z}$ 及 $r_p \\in \\{-1, -2, 2\\}$，使得 $p \\mid m_p^2 - r_p$。\n\n(b) 假設命題對 $\\alpha = k$ 成立。當 $\\alpha = k+1$ 時，取 $m_{p^{k+1}} = m_{p^k} + t p^k$，其中 $t$ 待定。注意到：\n$$\n\\begin{aligned}\nm_{p^{k+1}}^2 - r_p &= (m_{p^k} + t p^k)^2 - r_p \\\\\n&\\equiv (m_{p^k}^2 - r_p) + 2 m_{p^k} t p^k \\pmod{p^{k+1}}.\n\\end{aligned}\n$$\n由歸納假設，$p^k \\mid m_{p^k}^2 - r_p$，故僅需再取\n$$\nt \\equiv - \\frac{m_{p^k}^2 - r_p}{p^k} (2 m_{p^k})^{-1} \\pmod{p},\n$$\n即可讓 $p^{k+1} \\mid m_{p^{k+1}}^2 - r_p$（註：顯然 $p$ 不整除 $m_{p^k}$，故 $(2 m_{p^k})^{-1}$ 存在）。\n\n綜上，引理一證畢，從而原命題在 $n$ 為奇質數冪次方時成立。\n\n2. 接著證明：對於所有 $2$ 的冪次方 $2^\\alpha$，都存在整數 $m_{2^\\alpha}$ 使得 $2^\\alpha \\mid f(m_{2^\\alpha})$。\n\n事實上，我們可以用歸納法證明以下引理：\n\n**引理二.** 對於所有 $\\alpha \\ge 1$，必存在 $m_{2^\\alpha} \\in \\mathbb{Z}$ 使得 $2^\\alpha \\mid m_{2^\\alpha}^2 + 7$。\n\n(a) 首先，對於 $\\alpha \\ge 3$，取 $m_{2^\\alpha} = 1$ 即可。\n\n(b) 假設命題對 $\\alpha = k \\ge 3$ 成立。當 $\\alpha = k + 1$ 時，取\n$$\nm_{2^{k+1}} = m_{2^k} + t 2^{k-1},\n$$\n其中 $t$ 待定。注意到：\n$$\n\\begin{aligned}\nm_{2^{k+1}}^2 + 7 &= (m_{2^k} + t 2^{k-1})^2 + 7 \\\\\n&\\equiv (m_{2^k}^2 + 7) + m_{2^k} t 2^k \\pmod{2^{k+1}}.\n\\end{aligned}\n$$\n由歸納假設，$2^k \\mid m_{2^k}^2 + 7$，故僅需再取\n$$\nt \\equiv \\frac{m_{2^k}^2 + 7}{2^k} \\pmod{2},\n$$\n即可讓 $2^{k+1} \\mid m_{2^{k+1}}^2 + 7$。\n\n綜上，引理二證畢，從而原命題在 $n$ 為 $2$ 的冪次方時成立。\n\n3. 最後回到原命題。令 $n = p_1^{\\alpha_1} \\cdots p_s^{\\alpha_s}$。\n\n- 由以上兩個引理，知對於所有 $1 \\le i \\le s$，存在整數 $m_{p_i^{\\alpha_i}}$ 使得 $p_i^{\\alpha_i} \\mid f(m_{p_i^{\\alpha_i}})$。\n- 又對於任何整係數多項式 $f(x)$，任意兩相異整數 $u, v$，都必有 $u - v \\mid f(u) - f(v)$。\n\n故只要讓 $m$ 滿足同餘方程組\n$$\nm \\equiv m_{p_i^{\\alpha_i}} \\pmod{p_i^{\\alpha_i}} \\quad \\forall i = 1, 2, \\dots, s\n$$\n便有\n$$\n\\begin{aligned}\nf(m) &= \\{f(m) - f(m_{p_i^{\\alpha_i}})\\} + f(m_{p_i^{\\alpha_i}}) \\\\\n&\\equiv \\{m - m_{p_i^{\\alpha_i}}\\} + 0 \\equiv 0 \\pmod{p_i^{\\alpha_i}}\n\\end{aligned}\n$$\n對於所有 $i = 1, 2, \\dots, s$ 皆成立；換言之，$n \\mid f(m)$。\n\n綜上，知 $f(x)$ 滿足題意。\n\n*編按一：* 此題的構造部分，應該不能以“構造出完整解答”來給分，而是“構造出符合局部 Case 的解且證明該 Case”的情況來給分。倒是完全沒有證明出任何一種 Case，但有構造出一個滿足全部題意的 $f(x)$ 時要給予幾分可以斟酌。\n\n*編按二：* 此題的關鍵在於是否有往二次剩餘的方向造，光這個概念出來可能都有價值。",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 14950,
    "subject": "Mathematics (Olympiad)",
    "question": "設 $ABCD$ 為凸四邊形，$BC$ 與 $AD$ 兩邊並不平行。假設 $BC$ 邊上有一點 $E$ 使得四邊形 $ABED$ 與四邊形 $AECD$ 都有內切圓。試證：$AD$ 邊上存在一點 $F$ 使得四邊形 $ABCF$ 與四邊形 $BCDF$ 都有內切圓的充要條件是 $AB$ 平行於 $CD$。",
    "options": [],
    "answer": "See solution",
    "solution": "設 $\\omega_1, \\omega_2$ 分別為四邊形 $ABED, AECD$ 的內切圓，點 $O_1, O_2$ 分別為 $\\omega_1, \\omega_2$ 的圓心。存在滿足題設中的一點 $F$ 的充分條件是如果 $\\omega_1, \\omega_2$ 也分別是四邊形 $ABCF, BCDF$ 的內切圓。\n\n自 $B$ 向 $\\omega_2$ 引異於 $BC$ 的切線，並且自 $C$ 向 $\\omega_1$ 引異於 $BC$ 的切線，令此兩條切線分別與 $AD$ 邊交於點 $F_1, F_2$。我們需要證明：$F_1 = F_2$ 的充要條件是 $AB \\parallel CD$。\n\n引理：設兩圓 $\\omega_1, \\omega_2$ 的圓心分別為 $O_1, O_2$，且此兩圓同時內切於一角，並設此角的頂點為 $O$。設點 $P, S$ 在角 $O$ 的同一邊，點 $Q, R$ 在角 $O$ 的另一邊，且設 $\\omega_1$ 是三角形 $PQO$ 的內切圓，$\\omega_2$ 是三角形 $RSO$ 相對於角 $O$ 的旁切圓。令 $p = OO_1 \\cdot OO_2$。則下列的關係中恰有一成立：\n\n$$\nOP \\cdot OR < p < OQ \\cdot OS, \\quad OP \\cdot OR > p > OQ \\cdot OS, \\quad OP \\cdot OR = p = OQ \\cdot OS.\n$$\n\n引理的證明：令 $\\angle OPO_1 = \\alpha, \\angle OQO_1 = \\beta, \\angle OO_2R = \\gamma, \\angle OO_2S = \\delta$, $\\angle POQ = 2\\varphi$。因為線段 $PO_1, QO_1, RO_2, SO_2$ 分別是三角形 $PQO, RSO$ 的內角平分線或外角平分線，所以有\n\n$$\n\\nu + v = x + y(= 90^\\circ - \\varphi). \\qquad (1)\n$$\n\n由正弦定理知\n\n$$\n\\frac{OP}{OO_1} = \\frac{\\sin(u + \\varphi)}{\\sin u} \\quad \\text{and} \\quad \\frac{OO_2}{OR} = \\frac{\\sin(x + \\varphi)}{\\sin x}.\n$$\n\n因為 $x, u, \\varphi$ 都是銳角，所以\n\n$$\n\\begin{aligned}\nOP \\cdot OR \\ge p & \\Leftrightarrow \\frac{OP}{OO_1} \\ge \\frac{OO_2}{OR} & \\Leftrightarrow \\sin x \\sin(u + \\varphi) \\ge \\sin u \\sin(x + \\varphi) \\\\\n& \\Leftrightarrow \\sin(x - u) \\ge 0 & \\Leftrightarrow x \\ge u.\n\\end{aligned}\n$$\n\n由此知 $OP \\cdot OR \\ge p$ 等價於 $x \\ge u$，並且 $OP \\cdot OR = p$ 的充要條件是 $x = u$。\n\n同理可證，$p \\ge OQ \\cdot OS$ 等價於 $v \\ge y$，且 $p = OQ \\cdot OS$ 的充要條件是 $v = y$。另一方面由 (1) 式知 $x \\ge u$ 和 $v \\ge y$ 是等價的，並且 $x = u$ 等價於 $v = y$。所以引理得證。\n\n![](images/13-1J_p11_data_05224467f9.png)\n\n回到問題本身，將引理應用在下列各組的四個點：$\\{B, E, D, F_1\\}$, $\\{A, B, C, D\\}$, $\\{A, E, C, F_2\\}$。先設 $OE \\cdot OF_1 > p$，就可得到\n\n$$\nOE \\cdot OF_1 > p \\Rightarrow OB \\cdot OD < p \\Rightarrow OA \\cdot OC > p \\Rightarrow OE \\cdot OF_2 < p.\n$$\n\n換句話說，$OE \\cdot OF_1 > p$ 可推得\n\n$OB \\cdot OD < p < OA \\cdot OC \\quad \\text{and} \\quad OE \\cdot OF_1 > p > OE \\cdot OF_2$。\n\n同理，若假設 $OE \\cdot OF_1 < p$，則可得到\n\n$OB \\cdot OD > p > OA \\cdot OC \\quad \\text{and} \\quad OE \\cdot OF_1 < p < OE \\cdot OF_2$。\n\n在這些情形下，$F_1 \\neq F_2$，而且 $OB \\cdot OD \\neq OA \\cdot OC$，所以 $AB$ 與 $CD$ 不會平行。\n\n最後剩下 $OE \\cdot OF_1 = p$ 的情形。在此情形下，由引理可推得 $OB \\cdot OD = p = OA \\cdot OC$ 且 $OE \\cdot OF_1 = p = OE \\cdot OF_2$。因此 $F_1 = F_2$ 且 $AB \\parallel CD$。",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 14951,
    "subject": "Mathematics (Olympiad)",
    "question": "A factory increased its production by 25% after modernizing its equipment. Later, after firing some employees, production dropped by 20%. Has the number of products produced changed compared to the original amount?",
    "options": [],
    "answer": "See solution",
    "solution": "Denote the number of products produced by the factory before the changes by $x$. After modernization, they produced $\\frac{125}{100} \\cdot x$ products. After firing employees, the number dropped to $\\frac{80}{100} \\cdot \\frac{125}{100} \\cdot x = x$. Hence, the number of products has not changed.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 14952,
    "subject": "Mathematics (Olympiad)",
    "question": "At Matthijs's table tennis club, one keeps the ping-pong balls on a table with cylindrical ball holders. Here is the side view of a ping-pong ball on top of a ball holder. The underside of the ball exactly touches the table. It is known that the ball holder is 4 centimetres wide and 1 centimetre high.\n\nHow many centimetres is the radius of the ball?\n*Please note that the picture is not to scale.*\n\n![](images/NLD_ABooklet_2025_p5_data_77d6aa365a.png)\n\nA) $2\\frac{1}{3}$  \nB) $2\\frac{1}{2}$  \nC) $2\\frac{2}{3}$  \nD) $2\\frac{5}{6}$  \nE) $3$",
    "options": [],
    "answer": "See solution",
    "solution": "B) $2\\frac{1}{2}$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 14953,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $ABCDE$ be a convex pentagon, where the quadrilateral $ABCD$ is a square. If $\\angle AEC + \\angle BED = 180^\\circ$, show that the pentagon $ABCDE$ is circumscribed by a circle. Here, a convex pentagon means each of the inner angles is less than $180^\\circ$.",
    "options": [],
    "answer": "See solution",
    "solution": "Take a point $E'$ on the opposite side of the line $BC$ from $A$ so that triangle $BCE'$ is congruent to triangle $ADE$. Since $ABCD$ is a square, $AC = BD$ and $\\angle DAC = \\angle CBD$. Also, $AE = BE'$ and $\\angle EAD = \\angle E'BC$. Therefore, $\\angle EAC = \\angle EAD + \\angle DAC = \\angle E'BC + \\angle CBD = \\angle E'BD$, so triangles $ACE$ and $BDE'$ are congruent. Then, $\\angle BED + \\angle BE'D = \\angle BED + \\angle AEC = 180^\\circ$, which implies that $B$, $D$, $E$, $E'$ lie on a circle.\n\nFurthermore,\n\n$$\n\\begin{align*}\n\\angle BEC + \\angle BE'C &= \\angle BEC + \\angle AED \\\\\n&= (\\angle BED - \\angle CED) + (\\angle AEC + \\angle CED) \\\\\n&= \\angle BED + \\angle AEC = 180^\\circ,\n\\end{align*}\n$$\n\nso $B$, $C$, $E$, $E'$ also lie on a circle. Since the circle through $B$, $E$, $E'$ is unique (the circumcircle of triangle $BEE'$), and $C$ and $D$ lie on this circle, it is the circumcircle of square $ABCD$, and $E$ lies on it.\n\n**Alternate Solution:** Let $\\Gamma$ be the circumcircle of square $ABCD$ and $O$ its center. Let $F$ be the intersection of the half-line $OE$ and $\\Gamma$. It suffices to show that if $F$ and $E$ do not coincide, then $\\angle AEC + \\angle BED \\neq 180^\\circ$.\n\nNote $\\angle AFC = \\angle BFD = 90^\\circ$ since $AC$, $BD$ are diameters of $\\Gamma$.\n\nIf $E$ and $F$ are distinct and $O$, $E$, $F$ are collinear in that order, then $\\angle AEC = \\angle AFC - \\angle EAF - \\angle ECF < 90^\\circ$. Similarly, $\\angle BED < 90^\\circ$, so $\\angle AEC + \\angle BED < 180^\\circ$.\n\nIf $O$, $E$, $F$ are collinear in the order $O$, $E$, $F$, then $\\angle AEC = \\angle AFC + \\angle EAF + \\angle ECF > 90^\\circ$, and similarly $\\angle BED > 90^\\circ$, so $\\angle AEC + \\angle BED > 180^\\circ$. Thus, the claim is established.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 14954,
    "subject": "Mathematics (Olympiad)",
    "question": "Solve the following system of equations:\n\n$$\n\\begin{cases}\nx^4 - y^4 = 240 \\\\\nx^3 - 2y^3 = 3(x^2 - 4y^2) - 4(x - 8y)\n\\end{cases}\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "Reformulate the given system as:\n\n$$\n\\begin{cases}\nx^4 + 16 = y^4 + 256 \\\\\nx^3 - 3x^2 + 4x = 2y^3 - 12y^2 + 32y\n\\end{cases}\n$$\n\nMultiplying both sides of equation (2) by $-8$ and adding to (1) gives:\n\n$$\n(x - 2)^4 = (y - 4)^4\n$$\n\nSo,\n\n$$\n(x - 2)^4 = (y - 4)^4 \\implies x - 2 = |y - 4| \\implies \\begin{cases} y = x + 2 \\\\ y = 6 - x \\end{cases}\n$$\n\nPlugging $y = x + 2$ into (1):\n\n$$\nx^4 = (x + 2)^4 + 240\n$$\n\nThis leads to:\n\n$$\nx^3 + 3x^2 + 4x + 32 = 0 \\implies (x + 4)(x^2 - x + 8) = 0 \\implies x = -4\n$$\n\nSo $y = -2$. Thus, $(x, y) = (-4, -2)$.\n\nPlugging $y = 6 - x$ into (1):\n\n$$\nx^4 = (6 - x)^4 + 240\n$$\n\nThis leads to:\n\n$$\nx^3 - 9x^2 + 36x - 64 = 0 \\implies (x - 4)(x^2 - 5x + 16) = 0 \\implies x = 4\n$$\n\nSo $y = 2$. Thus, $(x, y) = (4, 2)$.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 14955,
    "subject": "Mathematics (Olympiad)",
    "question": "A map $f: V \\to \\mathbb{Z}$ on the vertex set $V$ of a graph $G = (V, E)$ is said to be *good* if $f$ satisfies:\n\n1. $$\\sum_{v \\in V} f(v) = |E|;$$\n2. If some vertices are arbitrarily colored red, there exists a red vertex $v$ such that $f(v)$ is not greater than the number of vertices adjacent to $v$ that are not colored red.\n\nLet $m(G)$ be the number of good maps $f$. Show that if each vertex of $V$ is adjacent to at least one other vertex, then $n \\leq m(G) \\leq n!$.",
    "options": [],
    "answer": "See solution",
    "solution": "**Proof**\n\nGiven an ordering $\\tau = (v_1, v_2, \\ldots, v_n)$ on the vertices in $V$, we associate a map $f_\\tau: V \\to \\mathbb{Z}$ as follows: $f_\\tau(v)$ is equal to the number of vertices in $V$ that are ordered preceding $v$. We claim that $f_\\tau$ is good.\n\nEach edge is counted exactly once in $\\sum_{v \\in V} f_\\tau(v)$: for an edge $e \\in E$ with vertices $u, v \\in V$ such that $u$ is ordered before $v$ in $\\tau$, $e$ is counted once in $f_\\tau(v)$. Thus,\n\n$$\n\\sum_{v \\in V} f_{\\tau}(v) = |E|.\n$$\n\nFor any nonempty subset $A \\subseteq V$ of all red vertices, choose $v \\in A$ with the most preceding orderings in $\\tau$. Then by definition, $f_\\tau(v)$ is not greater than the number of vertices adjacent to $v$ that are not colored red. We have verified that $f_\\tau$ is good.\n\nConversely, given any good map $f: V \\to \\mathbb{Z}$, we claim that $f = f_\\tau$ for some ordering $\\tau$ of $V$.\n\nFirst, let the red vertex set $A = V$. By condition (2), there exists $v \\in A$ such that $f(v) \\leq 0$; denote one such vertex by $v_1$. Assuming that we have already chosen $v_1, \\ldots, v_k$ from $V$, if $k < n$, set the red vertex set $A = V \\setminus \\{v_1, \\ldots, v_k\\}$. By condition (2), there exists $v \\in A$ such that $f(v)$ is less than or equal to the number of vertices in $\\{v_1, \\ldots, v_k\\}$ that are adjacent to $v$. Denote one such vertex by $v_{k+1}$. Continuing in this way, we order the vertices by $\\tau = (v_1, v_2, \\ldots, v_n)$. By construction, $f(v) \\leq f_r(v)$ for any $v \\in V$. By condition (1),\n\n$$\n|E| = \\sum_{v \\in V} f(v) \\leq \\sum_{v \\in V} f_r(v) = |E|,\n$$\n\nand therefore $f(v) = f_r(v)$ for any $v \\in V$.\n\nWe have shown that for any ordering $\\tau$, $f_r$ is good, and any good map $f$ is $f_r$ for some $\\tau$. Since the number of orderings on $V$ is $n!$, we see that $m(G) \\leq n!$ (note that two distinct orderings may result in the same map).\n\nNext, we prove that $n \\leq m(G)$. Assume at the moment that $G$ is connected. Pick arbitrarily $v_1 \\in V$. By connectivity, we may choose $v_2 \\in V \\setminus \\{v_1\\}$ such that $v_2$ is adjacent to $v_1$, and again we may choose $v_3 \\in V \\setminus \\{v_1, v_2\\}$ such that $v_3$ is adjacent to at least one of $v_1, v_2$. Continuing in this way, we get an ordering $\\tau = (v_1, v_2, \\ldots, v_n)$ such that $v_k$ is adjacent to at least one of the vertices preceding it under $\\tau$, for any $2 \\leq k \\leq n$. Thus, $f_r(v_1) = 0$ and $f_r(v_k) > 0$ for $2 \\leq k \\leq n$. Since $v_1$ may be arbitrary, we have at least $n$ good maps.\n\nIn general, if $G$ is a union of its connected components $G_1, \\ldots, G_k$, since each vertex is adjacent to at least another vertex, each component has at least two vertices. Denote by $n_1, \\ldots, n_k \\geq 2$ the number of vertices of these components. For each $G_i$, we have at least $n_i$ good maps on its vertices, $i = 1, \\ldots, k$. It is easy to see that patching good maps on $G_i$'s together results in a good map on $G$, and thus\n\n$$\nm(G) \\geq n_1 n_2 \\cdots n_k \\geq n_1 + n_2 + \\cdots + n_k = n.\n$$\n\nWe conclude that $n \\leq m(G) \\leq n!$.",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 14956,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $a$ be an integer different from $1$ such that $a$ divides both $3n-1$ and $2n^2+1$. For which positive integers $n$ with $1 \\leq n \\leq 2015$ is the fraction $\\dfrac{2n^2+1}{3n-1}$ reducible?",
    "options": [],
    "answer": "See solution",
    "solution": "Suppose $a \\neq 1$ divides both $3n-1$ and $2n^2+1$. Then $a$ also divides:\n\n$$\n3(2n^2+1) - 2n(3n-1) = 2n+3\n$$\n\nand thus $a$ divides:\n\n$$\n3(2n+3) - 2(3n-1) = 11.\n$$\n\nSince $11$ is prime, $a=11$. So $11$ divides both $3n-1$ and $2n^2+1$.\n\nLet $3n-1=11k$ for some integer $k$, so $n=\\dfrac{11k+1}{3}$. For $n$ to be integer, $3$ must divide $11k+1$, i.e., $3 \\mid 2k+1$, so $k=3m+1$ for integer $m$.\n\nThus, $n=11m+4$. For $1 \\leq n \\leq 2015$, $0 \\leq m \\leq 182$.\n\nTherefore, there are $183$ positive integers $n$ for which the fraction is reducible.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 14957,
    "subject": "Mathematics (Olympiad)",
    "question": "Let the side of the square be $x$. The square has area $x^2$, and $x$ is also the height of a triangle with base $16$. If the area of the square equals the area of the triangle, what is the area of the square?",
    "options": [],
    "answer": "See solution",
    "solution": "Let $x$ be the side of the square. The area of the square is $x^2$. The triangle has base $16$ and height $x$, so its area is $\\frac{1}{2} \\times 16 \\times x = 8x$. Setting the areas equal:\n\n$$\nx^2 = 8x\n$$\n\nSo $x = 8$. The area of the square is $8^2 = 64$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 14958,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $n > 0$ be an integer. We are given a balance and $n$ weights of weight $2^0, 2^1, \\dots, 2^{n-1}$. We are to place each of the $n$ weights on the balance, one after another, in such a way that the right pan is never heavier than the left pan. At each step, we choose one of the weights that have not yet been placed on the balance and place it on either the left pan or the right pan, until all the weights have been placed.\n\nDetermine the number of ways in which this can be done.",
    "options": [],
    "answer": "See solution",
    "solution": "The number of ways is $(2n-1)!! = 1 \\times 3 \\times 5 \\times \\dots \\times (2n-1)$.\n\nWe prove this by induction on $n$.\n\n*Base case*: For $n = 1$, put the single weight on the left pan. There is only one way.\n\n*Inductive step*: Suppose for $n = k$, the number of ways is $(2k-1)!!$.\n\nFor $n = k+1$, consider the $k+1$ weights: $1/2, 1, 2, \\dots, 2^{k-1}$ (after scaling by $1/2$ for convenience). For any positive integer $r$,\n\n$$\n2^r > 2^{r-1} + 2^{r-2} + \\dots + 1 + \\frac{1}{2} \\geq \\sum_{i=-1}^{r-1} \\pm 2^i.\n$$\n\nThe heaviest weight must be on the left pan. Now, consider the position of the $1/2$ weight:\n\n(a) If $1/2$ is placed first, it must go on the left pan. The remaining $k$ weights have $(2k-1)!!$ ways.\n\n(b) If $1/2$ is placed at step $t = 2, 3, \\dots, k+1$, it can go on either pan, so for each such $t$ there are $(2k-1)!!$ ways.\n\nThus, total ways for $n = k+1$:\n\n$$\n(2k-1)!! + k \\times (2k-1)!! = (1+2k)(2k-1)!! = (2k+1)!!.\n$$\n\nThis completes the induction. $\\square$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 14959,
    "subject": "Mathematics (Olympiad)",
    "question": "Consider an arbitrary triangle $ABC$ with incenter $I$, and let $I_a$, $I_b$, and $I_c$ be the excenters of triangle $ABC$ opposite to vertices $A$, $B$, and $C$, respectively, tangent to sides $BC$, $CA$, and $AB$. Let $E$, $F$, and $G$ be the points of tangency of the incircle with the sides $BC$, $CA$, and $AB$, respectively.\n\nProve that the circumcircles of triangles $IEI_a$, $IFI_b$, and $IGI_c$ intersect in a second common point different from $I$.",
    "options": [],
    "answer": "See solution",
    "solution": "Denote by $J_a$, $J_b$, and $J_c$ the centers of the circumcircles of triangles $IEI_a$, $IFI_b$, and $IGI_c$, respectively. We prove that $J_a$, $J_b$, and $J_c$ are collinear.\n\nLet $L_a$, $L_b$, and $L_c$ be the points diametrically opposite to $I$ on the circumcircles of triangles $IEI_a$, $IFI_b$, and $IGI_c$, respectively. Since $J_a J_b$, $J_b J_c$, and $J_c J_a$ are midlines in triangles $IL_a L_b$, $IL_b L_c$, and $IL_c L_a$, respectively, it suffices to show that $L_a$, $L_b$, and $L_c$ are collinear.\n\nSince $IL_a$ is a diameter in the circumcircle of triangle $IEI_a$, we have $\\angle II_a L_a = 90^\\circ$. Hence $I_a L_a \\perp II_a$. But $II_a \\perp I_b I_c$, so $I_a L_a \\parallel I_b I_c$, implying\n\n$$\n\\angle L_a I_a B = \\angle I_a I_c I_b, \\quad \\text{and} \\quad \\angle L_a I_a C = 180^\\circ - \\angle I_a I_b I_c. \\qquad (1)\n$$\n\nLet $J$ be the midpoint of segment $IE$. Since $JJ_a$ is a midline in triangle $IEL_a$, it follows that $JJ_a \\parallel EL_a$. But $JJ_a$ is the perpendicular bisector of $IE$, so $JJ_a \\perp IE$, and since $BE \\perp IE$, it follows that $JJ_a \\parallel BE$. From $BE \\parallel JJ_a$ and $EL_a \\parallel JJ_a$, we deduce that $L_a \\in BE$, so $L_a \\in BC$. Similarly, we deduce that $L_b \\in CA$ and $L_c \\in AB$.\n\nWe have\n$$\n\\frac{BL_a}{CL_a} = \\frac{S_{BI_a L_a}}{S_{CI_a L_a}} = \\frac{BI_a \\cdot L_a I_a \\cdot \\sin(\\angle L_a I_a B)}{CI_a \\cdot L_a I_a \\cdot \\sin(\\angle L_a I_a C)} \\stackrel{(1)}{=} \\frac{BI_a \\cdot \\sin(\\angle I_a I_c I_b)}{CI_a \\cdot \\sin(\\angle I_a I_b I_c)}.\n$$\n\nSimilarly,\n$$\n\\frac{CL_b}{AL_b} = \\frac{CI_b \\cdot \\sin(\\angle I_b I_a I_c)}{AI_b \\cdot \\sin(\\angle I_b I_c I_a)}, \\quad \\frac{AL_c}{BL_c} = \\frac{AI_c \\cdot \\sin(\\angle I_c I_b I_a)}{BI_c \\cdot \\sin(\\angle I_c I_a I_b)}.\n$$\n\nTherefore,\n$$\n\\frac{BL_a}{CL_a} \\cdot \\frac{CL_b}{AL_b} \\cdot \\frac{AL_c}{BL_c} = 1,\n$$\nand by the converse of Menelaus' theorem, $L_a$, $L_b$, and $L_c$ are collinear, hence $J_a$, $J_b$, and $J_c$ are collinear.\n\nIt follows that the radical axes of any two of the three circles are parallel or coincide. Since $I$ lies on all three circles, we deduce that these circles have the same radical axis. But $I \\notin J_aJ_b$, so the three circles are not tangent, and therefore their radical axis contains a point different from $I$ lying on all three circles.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 14960,
    "subject": "Mathematics (Olympiad)",
    "question": "Петя выбирает $2n$ чисел: $0$, $\\frac{1}{2}$, и $2n-2$ чисел, равных $\\frac{1}{4(n-1)}$. Вася расставляет их по кругу и образует $2n$ пар соседних чисел, вычисляя произведения в каждой паре. Какое наименьшее число обязательно окажется среди этих произведений, независимо от расстановки Васи?",
    "options": [],
    "answer": "See solution",
    "solution": "Ответ: $\\frac{1}{8(n-1)}$.\n\nЕсли Петя выберет числа $0$, $\\frac{1}{2}$, $\\frac{1}{4(n-1)}$, $\\frac{1}{4(n-1)}$, $\\dots$, $\\frac{1}{4(n-1)}$, то, как бы ни расставлял эти числа Вася, число $\\frac{1}{2}$ будет в одной паре с числом $\\frac{1}{4(n-1)}$. Значит, одно из произведений будет равно $\\frac{1}{8(n-1)}$, а остальные будут не больше него. Тогда на доске окажется число $\\frac{1}{8(n-1)}$.\n\nПокажем, как Вася может для любых чисел получить на доске число, не большее $\\frac{1}{8(n-1)}$. Перенумеруем числа в порядке убывания: $x_1 \\ge x_2 \\ge \\dots \\ge x_{2n}$. Поставим в какое-то место на круге число $x_1$, от него по часовой стрелке через пустые места числа $x_2, x_3, \\dots, x_n$. Теперь поставим число $x_{2n}$ между $x_1$ и $x_n$; дальше по часовой стрелке от $x_{2n}$ расставим на пустых местах по очереди числа $x_{2n-1}, x_{2n-2}, \\dots, x_{n+1}$. Тогда произведениями пар соседних чисел будут: $x_n x_{2n}$,\n\n$x_1 x_{2n}, x_2 x_{2n-1}, x_3 x_{2n-2}, \\dots, x_k x_{2n-k+1}, \\dots, x_n x_{n+1}$ и\n\n$x_1 x_{2n-1}, x_2 x_{2n-2}, x_3 x_{2n-3}, \\dots, x_k x_{2n-k}, \\dots, x_{n-1} x_{n+1}$.\n\nПоскольку $x_k x_{2n-k+1} \\le x_k x_{2n-k}$, наибольшее произведение может быть лишь во второй строке.\n\nПокажем, что $a = x_k x_{2n-k} \\le \\frac{1}{8(n-1)}$ при $k \\le n-1$. Действительно, из неравенств $x_k \\le x_{k-1} \\le \\dots \\le x_1$ следует, что $kx_k \\le x_1 + x_2 + \\dots + x_k$, поэтому\n\n$$\nka = kx_k \\cdot x_{2n-k} \\le (x_1 + x_2 + \\dots + x_k) x_{2n-k}.\n$$\n\nАналогично из неравенств\n\n$$\nx_{2n-k} \\le x_{2n-k-1} \\le x_{2n-k-2} \\le \\dots \\le x_{k+1} \\\\\n\\text{следует, что} \\\\\n(2n-2k)x_{2n-k} \\le x_{2n-k} + x_{2n-k-1} + \\dots + x_{k+1} \\le \\\\\n\\le x_{k+1} + x_{k+2} + \\dots + x_{2n} = 1 - x_1 - x_2 - \\dots - x_k.\n$$\n\nПоэтому\n\n$$\n2k(n-k)a \\le (x_1 + x_2 + \\dots + x_k) (1 - x_1 - x_2 - \\dots - x_k) = x(1-x), \\\\\n\\text{где } x = x_1+x_2+\\dots+x_k. \\text{ Поскольку по неравенству о средних для двух чисел } x(1-x) \\le \\left(\\frac{x+(1-x)}{2}\\right)^2 = \\frac{1}{4}, \\text{ получаем неравенство } x_k x_{n-2k} = a \\le \\frac{1}{8k(n-k)}. \\text{ Осталось показать, что } k(n-k) \\ge n-1 \\text{ при } k \\le n-1. \\text{ Но последнее неравенство можно переписать в виде } (k-1)(n-k-1) \\ge 0, \\text{ а обе скобки в последней формуле неотрицательны.}\n$$\n\n*Замечание.* Оптимальная расстановка для Васи не единственна. Однако можно доказать, что при любом $k = 1, 2, \\dots, n-1$ в любой Васиной расстановке среди произведений пар соседних чисел найдётся число, не меньшее $x_k x_{2n-k}$; поэтому оптимальными для Васи окажутся расстановки, в которых наибольшее произведение имеет такой вид.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 14961,
    "subject": "Mathematics (Olympiad)",
    "question": "There are $2n^2$ ($n \\ge 2$) players in a single round-robin chess tournament. It is known that:\n\n1. For any three players $A$, $B$, and $C$, if $A$ beats $B$ and $B$ beats $C$, then $A$ beats $C$.\n2. There are at most $\\frac{n^3}{16}$ draws.\n\nProve that it is possible to choose $n^2$ players and label them $P_{ij}$ ($1 \\le i, j \\le n$), such that for any $i, j, i', j' \\in \\{1, 2, \\dots, n\\}$, if $i < i'$, then $P_{ij}$ beats $P_{i'j'}$.\n\n![](path/to/file.png)",
    "options": [],
    "answer": "See solution",
    "solution": "**Solution**\n\n**Lemma:** Suppose $m$ players participate in a single round-robin tournament with possible draws, and if $A$ beats $B$, $B$ beats $C$, then $A$ beats $C$. Then the $m$ players can be arranged in a row such that for any two players, the one on the left either beats or draws the one on the right.\n\n*Proof of lemma:* Induction on $m$. When $m=1$, the lemma is trivial. Suppose it is true for $m-1$ players, and consider $m$ players. If everyone wins a game, then we can find $x_1$ beats $x_2$, $x_2$ beats $x_3$, and so on, and someone must reappear in the sequence, say $x_i$ beats $x_j$ ($i \\ge j$), which is contradictory. The contradiction indicates that someone has never won a game; put this player in the rightmost position. By the induction hypothesis, the other $m-1$ players can be arranged on the left such that the lemma conditions are satisfied.\n\nFor the original problem, arrange the $2n^2$ players in a row as in the lemma, and then make $n$ groups of players as follows: set the leftmost $n$ players as group $A_1$; for $i=1, \\dots, n-2$, on the right side of $A_i$, set some consecutive $n$ players as $A_{i+1}$; set the rightmost $n$ players as group $A_n$. Moreover, between any two consecutive groups, there are $\\left\\lfloor \\frac{n^2}{n-1} \\right\\rfloor$ or more players.\n\nIf players from different groups never draw, then the proof is done. Otherwise, remove two players who draw and regroup the remaining $2n^2 - 2$ players in the same way as before, except requiring $\\left\\lfloor \\frac{n^2-2}{n-1} \\right\\rfloor$ or more players between consecutive groups. Again, if players from different groups never draw, the proof is done. Otherwise, remove two players and regroup, and so on. In general, when we regroup $2n^2-2i$ ($0 \\le i \\le \\left\\lfloor \\frac{n^2}{2} \\right\\rfloor$) players, it is required that $\\left\\lfloor \\frac{n^2-2i}{n-1} \\right\\rfloor$ or more players are between consecutive groups. During the whole process, we obtain $\\left\\lfloor \\frac{n^2}{2} \\right\\rfloor + 1$ groupings.\n\nFor every $0 \\le i \\le \\lfloor \\frac{n^2}{2} \\rfloor$, assume in the $i$th grouping, $x_i$ and $y_i$ are from different groups and they draw. Then, there are at least $\\lfloor \\frac{n^2 - 2i}{n-1} \\rfloor$ players between them, each of whom draws with either $x_i$ or $y_i$. This implies that the total number of draws that $x_i$ and $y_i$ have is at least $\\lfloor \\frac{n^2 - 2i}{n-1} \\rfloor$, and furthermore the total number of draws in the tournament is at least\n\n$$\nS = \\sum_{0 \\le i \\le \\lfloor \\frac{n^2}{2} \\rfloor} \\left\\lfloor \\frac{n^2 - 2i}{n-1} \\right\\rfloor.\n$$\n\nWhen $n = 2k + 1$, $S = 2k^3 + 3k^2 + 3k + 2 > \\frac{n^3}{4}$; when $n = 2k$, $S > 2k^3 > \\frac{n^3}{4}$. Either way, $S \\le \\frac{n^3}{16}$ is violated and the assumption is untrue. Hence, for $1 \\le i, j \\le n$, let the $j$th player of $A_i$ be $P_{ij}$, and the conditions are satisfied.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 14962,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $n$ be a positive integer. If $\\sigma$ is a permutation of the first $n$ positive integers, let $S(\\sigma)$ be the set of all distinct sums of the form $\\sum_{i=k}^{\\ell} \\sigma(i)$, where $1 \\leq k \\leq \\ell \\leq n$.\n\n1. Exhibit a permutation $\\sigma$ of the first $n$ positive integers such that $|S(\\sigma)| \\geq \\left\\lfloor \\frac{(n+1)^2}{4} \\right\\rfloor$.\n\n2. Show that $|S(\\sigma)| > \\frac{n\\sqrt{n}}{4\\sqrt{2}}$ for all permutations $\\sigma$ of the first $n$ positive integers.",
    "options": [],
    "answer": "See solution",
    "solution": "1. Consider the permutation $\\sigma$ of the first $n$ positive integers defined as follows: for each positive odd $i \\leq n$, $\\sigma(i) = (i+1)/2$; for each positive even $i \\leq n$, $\\sigma(i) = n - i/2 + 1$.\n\nWe claim that the $\\left\\lfloor \\frac{(n+1)^2}{4} \\right\\rfloor$ sums of the form $\\sum_{i=k}^{\\ell} \\sigma(i)$, where $1 \\leq k \\leq \\ell \\leq n$ and $k \\equiv \\ell \\pmod{2}$, are pairwise distinct. Thus, $|S(\\sigma)| \\geq \\left\\lfloor \\frac{(n+1)^2}{4} \\right\\rfloor$.\n\nNotice that $\\sigma(i) + \\sigma(i+1) = n+1$ for every positive odd $i \\leq n$, so\n\n$$\n\\sum_{i=k}^{\\ell} \\sigma(i) = \\begin{cases}\n  (\\ell-k)\\frac{n+1}{2} + \\sigma(\\ell) & \\text{if } k \\text{ is odd,} \\\\\n  \\sigma(k) + (\\ell-k)\\frac{n+1}{2} & \\text{if } k \\text{ is even.}\n\\end{cases}\n$$\n\nSince $|\\sigma(i) - \\sigma(j)| < n$, this formula shows that the assignment $(k, \\ell) \\mapsto \\sum_{i=k}^{\\ell} \\sigma(i)$ is injective on the pairs in question.\n\n2. Let $\\sigma$ be any permutation of the first $n$ positive integers. Split $S(\\sigma)$ into $S_m(\\sigma) = S(\\sigma) \\cap [mn + 1, mn + n]$, where $m$ runs through the integers. $S_m(\\sigma)$ is empty if $m < 0$ or $m > (n-1)/2$, and $|S_0(\\sigma)| \\geq n$.\n\nFor $n \\geq 3$, we show that\n\n$$\n|S_m(\\sigma)| + |S_{m-1}(\\sigma)| > \\sqrt{2n}\n$$\n\nfor every $0 \\leq m \\leq (n+1)/4$. Summing over this range gives\n\n$$\n|S(\\sigma)| \\geq \\frac{1}{2} \\sum_{m=0}^{\\left\\lfloor (n+1)/4 \\right\\rfloor} (|S_m(\\sigma)| + |S_{m-1}(\\sigma)|) > \\frac{1}{2} \\left\\lfloor \\frac{n+5}{4} \\right\\rfloor \\sqrt{2n} > \\frac{n\\sqrt{n}}{4\\sqrt{2}}.\n$$\n\nTo prove the bound, fix $m$ as above. The Minkowski difference $S_m(\\sigma) - (S_m(\\sigma) \\cup S_{m-1}(\\sigma))$ contains every positive integer $\\leq n$; equivalently, it contains every $\\sigma(k)$. Thus, $|S_m(\\sigma)| + |S_{m-1}(\\sigma)| = |S_m(\\sigma) \\cup S_{m-1}(\\sigma)| > \\sqrt{2n}$, since if $X$ is a finite set with $\\{1,2,\\dots,n\\} \\subseteq X-X$, then $|X|(|X|-1)+1 \\geq 2n+1$, so $|X| \\geq (1+\\sqrt{8n+1})/2 > \\sqrt{2n}$.\n\nFinally, for each $k \\leq n$, at least one of $\\sum_{i=1}^k \\sigma(i)$ or $\\sum_{i=k}^n \\sigma(i)$ exceeds $n(n+1)/4 \\geq mn$. Suppose $\\sum_{i=k}^n \\sigma(i) > mn$ (the other case is similar). Let $\\ell \\geq k$ be minimal such that $\\sum_{i=k}^\\ell \\sigma(i) > mn$. Then $\\sum_{i=k}^\\ell \\sigma(i)$ and $\\sum_{i=k+1}^\\ell \\sigma(i)$ belong to $S_m(\\sigma)$ and $S_m(\\sigma) \\cup S_{m-1}(\\sigma)$, respectively, so $\\sigma(k)$ is in $S_m(\\sigma) - (S_m(\\sigma) \\cup S_{m-1}(\\sigma))$.\n\n**Remark.** By probabilistic methods, it can be shown that generically there are at least $cn^2$ such sums, where $c$ is a positive absolute constant.",
    "topic": "Discrete Mathematics",
    "subtopic": "Combinatorics"
  },
  {
    "id": 14963,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $ABCD$ be a cyclic quadrilateral inscribed in a circle with diameter $AC$. Suppose there exist points $E \\in (CD)$ and $F \\in (BC)$ such that the lines $AE$ and $DF$ are perpendicular to the lines $AF$ and $BE$, respectively. Prove that $AB = AD$.",
    "options": [],
    "answer": "See solution",
    "solution": "From the perpendicularity conditions, we have\n\n$$\n\\overrightarrow{AE} \\cdot \\overrightarrow{DF} = 0 \\Leftrightarrow \\overrightarrow{AE} \\cdot (\\overrightarrow{AF} - \\overrightarrow{AD}) = 0,\n$$\n\n$$\n\\overrightarrow{AF} \\cdot \\overrightarrow{BE} = 0 \\Leftrightarrow \\overrightarrow{AF} \\cdot (\\overrightarrow{AE} - \\overrightarrow{AB}) = 0,\n$$\n\nwhence\n\n$$\n\\overrightarrow{AE} \\cdot \\overrightarrow{AF} = \\overrightarrow{AE} \\cdot \\overrightarrow{AD}, \\quad \\overrightarrow{AE} \\cdot \\overrightarrow{AF} = \\overrightarrow{AF} \\cdot \\overrightarrow{AB},\n$$\n\nand thus $\\overrightarrow{AE} \\cdot \\overrightarrow{AD} = \\overrightarrow{AF} \\cdot \\overrightarrow{AB}$. Equivalently, we have\n\n$$\n(\\overrightarrow{AD} + \\overrightarrow{DE}) \\cdot \\overrightarrow{AD} = (\\overrightarrow{AB} + \\overrightarrow{BF}) \\cdot \\overrightarrow{AB} \\Leftrightarrow AD^2 + \\overrightarrow{AD} \\cdot \\overrightarrow{DE} = AB^2 + \\overrightarrow{AB} \\cdot \\overrightarrow{BF}.\n$$\n\nBut $\\angle ADE = \\angle ABF = 90^\\circ$, and hence $\\overrightarrow{AD} \\cdot \\overrightarrow{DE} = \\overrightarrow{AB} \\cdot \\overrightarrow{BF} = 0$, that is $AB = AD$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 14964,
    "subject": "Mathematics (Olympiad)",
    "question": "Determine a polynomial $f(x)$ with integer coefficients which satisfies the following property: There are infinitely many relatively prime positive integers $a, b$ such that $a+b$ divides $f(a) + f(b)$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $f(x)$ be such a polynomial. Define\n$$\ng(x) := \\frac{f(x) - f(-x)}{2}, \\quad h(x) := \\frac{f(x) + f(-x)}{2}.\n$$\nThen $g(x)$ is a polynomial consisting of the odd degree monomials of $f(x)$, and $h(x)$ consists of the even degree monomials (including the constant term), so $f(x) = g(x) + h(x)$.\n\nFor any positive integers $a$ and $b$, since\n$$\na+b \\mid a^{2k+1} + b^{2k+1}\n$$\nfor any nonnegative integer $k$, we have\n$$\na+b \\mid f(a)+f(b) \\iff a+b \\mid h(a)+h(b).\n$$\nHence, we may assume $f(x) = h(x)$ (i.e., $f$ is an even polynomial).\n\nSuppose $f(x) = c x^{2n}$ for some nonnegative integer $n$ and $a+b \\mid f(a)+f(b)$ for some positive integers $a, b$. Note that\n$$\na^{2n} + b^{2n} = a(a^{2n-1} + b^{2n-1}) - b^{2n}(a+b) + 2b^{2n}.\n$$\nHence,\n$$\na+b \\mid f(a)+f(b) \\iff a+b \\mid 2f(b) = 2c b^{2n}.\n$$\nFurthermore, if $\\gcd(a, b) = 1$, then $\\gcd(a+b, b^{2n}) = 1$. Therefore, $a+b \\mid 2c$, which implies there are only finitely many such positive integer pairs $(a, b)$.\n\nNow assume $f(x) = a_{2n} x^{2n} + t(x)$, where $a_{2n} \\neq 0$ and $t(x)$ is a nonzero polynomial consisting of monomials of even degrees greater than $2n$. Let $b$ be any sufficiently large positive integer such that $\\gcd(a_{2n}, b) = 1$, and let $a$ be any positive integer such that\n$$\n|a+b| = a_{2n} + \\frac{t(b)}{b^{2n}}.\n$$\nNote that $\\gcd(a, b) = 1$ and there exist infinitely many such integer pairs $(a, b)$. Since $2f(b) = 2b^{2n}|a+b|$, we have $a+b \\mid f(a)+f(b)$.\n\nTherefore, $f(x)$ satisfies the property if and only if $f(x)$ contains no even degree monomials (i.e., it consists of odd degree monomials only) or contains at least two even degree monomials. $\\square$",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 14965,
    "subject": "Mathematics (Olympiad)",
    "question": "There are five distinct points $A$, $B$, $C$, $D$, $E$ lying in this order on a circle of radius $r$ such that $AC = BD = CE = r$. Consider the triangle whose vertices are the orthocentres of triangles $ACD$, $BCD$, and $BCE$. Prove that this triangle is right-angled.",
    "options": [],
    "answer": "See solution",
    "solution": "In any obtuse triangle $XYZ$ with obtuse angle at $Z$ and orthocentre $W$, the angles $XYZ$ and $XWZ$ are equal, as the angle $YXW$ complements to $90^\\circ$ (see figure 1). Moreover, points $Y$ and $W$ lie in different half-planes determined by $XZ$.\n\n![](images/Cesko-Slovacko-Poljsko_2006_p1_data_5c04f4264c.png)\n\nLet $P$, $Q$, $R$ be the orthocentres of the given triangles in that order. We will show that $\\angle PQR = 90^\\circ$. All three triangles are obtuse at $C$, so $P$, $Q$, $R$ lie on extensions of altitudes through $C$ to the corresponding sides. Because of the positions of these sides, the ray $CQ$ lies between rays $CP$ and $CR$, i.e., in angle $PCR$. Thus,\n\n$$\n\\angle PQR = \\angle RQC + \\angle PQC\n$$\n\n![](images/Cesko-Slovacko-Poljsko_2006_p1_data_5a3a0b7339.png)\n\nBy the fact above, $Q$ and $R$ lie in the same half-plane determined by the line $BC$, and\n\n$$\n\\angle BEC = \\angle BRC \\quad \\text{and} \\quad \\angle BDC = \\angle BQC.\n$$\n\nAngles $BEC$ and $BDC$ are equal since they are inscribed angles subtending the same chord $BC$. Thus, $\\angle BRC = \\angle BQC = \\omega$, and $BCRQ$ is cyclic. So $\\angle RQC = \\angle RBC = \\varphi$. Since $EC = r$, the inscribed angle $\\angle EBC = 30^\\circ$. Let $U$ be the foot on $BE$ in triangle $BEC$. Summing angles in right triangle $BUR$ gives\n\n$$\n\\omega + \\varphi + 30^\\circ + 90^\\circ = 180^\\circ, \\quad \\text{i.e.} \\quad \\angle RQC = \\varphi = 60^\\circ - \\omega = 60^\\circ - \\angle BDC.\n$$\n\nSimilarly, $\\angle PQC = 60^\\circ - \\angle DBC$. Therefore, using the sum of angles in triangle $BCD$ is $180^\\circ$,\n\n$$\n\\angle PQR = \\angle RQC + \\angle PQC = 120^\\circ - (\\angle BDC + \\angle DBC) = \\angle BCD - 60^\\circ. \\quad (1)\n$$\n\nBut also $BD = r$, so $\\angle BCD = 150^\\circ$. Finally, by (1), $\\angle PQR = 90^\\circ$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 14966,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $M$ be a convex 2011-gon. Consider 2011 points lying inside $M$ such that no three of all 4022 points (the vertices of $M$ and the 2011 points inside $M$) are collinear. A coloring of all points in two colors is called *good* if it is possible to connect some of the points by segments such that the following conditions hold:\n\n1. Each segment has its endpoints of one and the same color.\n2. No segments intersect in inner points.\n3. Between any two points of the same color there exists a path through the given segments.\n\nFind the number of all good colorings.",
    "options": [],
    "answer": "See solution",
    "solution": "Let the two colors be blue and red. We first prove the following lemma:\n\n**Lemma.** Consider $\\triangle ABC$ with vertices of both colors. Any coloring of $n$ points inside $\\triangle ABC$ is good.\n\n**Proof.** Without loss of generality, assume that $A$ and $B$ are blue points and $C$ is a red point.\n\nWe proceed by induction on the number $n \\ge 0$ of inner points for the triangle.\n\n- When $n = 0$, we draw a segment with endpoints $A$ and $B$ and we are done.\n- Assume the assertion holds for any triangle with $n = k$ inner points. Consider a triangle with $n = k + 1$ inner points and let these points be colored in an arbitrary manner. If all points are blue, then we connect $A$ with $B$ and with all inner points and the condition holds.\n- If there exists a red point $D$, we apply the induction hypothesis for triangles $ABD$, $BCD$, and $ACD$. It is clear that red points from distinct triangles are connected through $D$, and the blue points from distinct triangles are connected through $A$ or $B$. This completes the proof of the lemma. $\\square$\n\nNote that for any good coloring, all blue points of $M$ are consecutive vertices. Indeed, if this is not the case, we have two red points $A$ and $B$, dividing the contour of $M$ into two parts with two blue points $C$ and $D$ in each of these parts. It is clear now that the red path between $A$ and $B$ intersects the blue path between $C$ and $D$, a contradiction.\n\nLet us first count the number of good colorings of the vertices of $M$. There exist two such colorings with all points having one and the same color. When both colors occur, let $k > 0$ be the number of blue points. For any $k = 1, 2, \\dots, 2010$, there exist 2011 ways of choosing the group of $k$ consecutive blue points. Therefore, we have $2011 \\cdot 2010 + 2$ ways of coloring the vertices of $M$.\n\nWe show now that any coloring of the inner points is good. If all vertices of $M$ have one and the same color (say blue) and all inner points are also blue, we connect one vertex with all remaining points. If there exists a red point $B$, then we consider all triangles with one vertex $B$ and sides the sides of $M$. The assertion of the Lemma completes the proof in this case.\n\nConsider coloring of the vertices of $M$ in which both blue and red points occur. Without loss of generality, assume the blue points are $A_1, A_2, \\dots, A_k$, $k < 2011$. Connect $A_1$ with all red points and $A_{k+1}$ with all blue points. Now $M$ is partitioned into triangles and each triangle has vertices of both colors. Apply the Lemma for each of these triangles. The points having one and the same color from distinct triangles are connected through $A_1$ or $A_{k+1}$ or by the sides of $M$.\n\nTherefore, the number of good colorings equals $2^{2011}(2011 \\cdot 2010 + 2)$.",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 14967,
    "subject": "Mathematics (Olympiad)",
    "question": "Suppose $n$ players are arranged in a circle. For any three players $A$, $B$, $C$, if $A$ and $B$ are adjacent, then at least one of them defeated $C$. Find all possible values of $n$.\n\n![](path/to/file.png)",
    "options": [],
    "answer": "See solution",
    "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 let the players be $A_1, A_2, \\dots, A_{2k+1}$. Arrange the competition results so that $A_i$ ($1 \\leq i \\leq 2k+1$) defeated $A_{i+2}, A_{i+4}, \\dots, A_{i+2k}$ (where $A_{2k+1+j} = A_j$ for $j = 1, 2, \\dots, 2k+1$), and lost to the other players. Arrange the players in a circle: $A_1, A_2, \\dots, A_{2k+1}, A_1$.\n\nGiven any three players $A$, $B$, $C$ with $A$, $B$ adjacent, assume $A = A_i$, $B = A_{i+1}$, $C = A_{i+r}$ ($1 \\leq t \\leq 2k+1$, $2 \\leq r \\leq 2k$). Then either $r$ or $r-1$ is an even number not less than $2k$, so at least one of $A$, $B$ defeated $C$.\n\nNow suppose $n$ is even and at least $4$, and the $n$ players can be arranged in a circle $A_1, A_2, \\dots, A_n, A_1$ meeting the condition. Assume $A_1$ defeated $A_2$. By the requirement, at least one of $A_2, A_3$ defeated $A_1$, so $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 on. Thus, for any $1 \\leq i \\leq n$, $A_i$ defeated $A_{i+1}$ and lost to $A_{i-1}$ (with $A_{n+1} = A_1$, $A_0 = A_n$).\n\nDivide the players after $A_i$ and before $A_{i-1}$ into $\\frac{n-2}{2}$ pairs of adjacent players. In each pair, at least one player defeated $A_i$, so besides $A_{i-1}$, there are at least $\\frac{n-2}{2}$ players who defeated $A_i$. Thus, $A_i$ lost at least $\\frac{n}{2}$ games. So all $n$ players lost at least $\\frac{n^2}{2}$ games in total.\n\nBut the total number of games is $\\binom{n}{2} = \\frac{n(n-1)}{2} < \\frac{n^2}{2}$, a contradiction. Therefore, the possible values of $n$ are all odd numbers not less than $3$.\n\n$\\boxed{}$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 14968,
    "subject": "Mathematics (Olympiad)",
    "question": "Fix a positive integer $x$. For a positive integer $y > x$, define\n\n$$\nb_y = \\frac{a_x a_{x+1} + a_{x+1} a_{x+2} + \\dots + a_{y-1} a_y}{a_x a_y}.\n$$\n\nSuppose that $\\frac{a_{y+1}}{a_y} + \\frac{a_{y+1}}{a_{y+2}} = \\left( \\frac{a_2}{a_1} + \\frac{a_2}{a_3} \\right)$ is a constant with $0 < \\left( \\frac{a_2}{a_1} + \\frac{a_2}{a_3} \\right) < 2$.\n\nFind the smallest positive constant $c$ such that\n\n$$\n\\frac{\\sin n\\theta}{\\sin \\theta} \\leq c\n$$\n\nfor all integers $n \\geq 1$, where $2 \\cos \\theta = \\left( \\frac{a_2}{a_1} + \\frac{a_2}{a_3} \\right)$ and $0 < \\theta < \\frac{\\pi}{2}$.\n\nExpress your answer in terms of $\\frac{a_2}{a_1} + \\frac{a_2}{a_3}$.",
    "options": [],
    "answer": "See solution",
    "solution": "The smallest constant $c$ is\n\n$$\nc = \\frac{1}{\\sin \\theta} = \\frac{2}{\\sqrt{4 - \\left( \\frac{a_2}{a_1} + \\frac{a_2}{a_3} \\right)^2}}.\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 14969,
    "subject": "Mathematics (Olympiad)",
    "question": "Find the smallest positive integer $n \\ge 9$ such that for any group of integers $a_1, a_2, \\dots, a_n$, there always exist $a_{i_1}, a_{i_2}, \\dots, a_{i_9}$ ($1 \\le i_1 < i_2 < \\dots < i_9 \\le n$) and $b_i \\in \\{4, 7\\}$ ($i = 1, 2, \\dots, 9$) such that $b_1 a_{i_1} + b_2 a_{i_2} + \\dots + b_9 a_{i_9}$ is a multiple of 9.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $a_1 = a_2 = 1$, $a_3 = a_4 = 3$, $a_5 = \\dots = a_{12} = 0$.\n\nIt is easy to check that any 9 integers of them will not meet the requirement. So $n \\ge 13$. We will prove that $n = 13$.\n\nWe only need to prove the following statement:\n\nGiven a group of $m$ integers $a_1, a_2, \\dots, a_m$, if there are not three $a_{i_1}, a_{i_2}, a_{i_3}$ in them and $b_1, b_2, b_3 \\in \\{4, 7\\}$ such that $b_1 a_{i_1} + b_2 a_{i_2} + b_3 a_{i_3} \\equiv 0 \\pmod{9}$, then either $m \\le 6$ or $7 \\le m \\le 8$ and there are $a_{i_1}, a_{i_2}, \\dots, a_{i_6}$ in $a_1, a_2, \\dots, a_m$ and $b_1, b_2, \\dots, b_6 \\in \\{4, 7\\}$ such that $9 \\mid b_1 a_{i_1} + b_2 a_{i_2} + \\dots + b_6 a_{i_6}$.\n\nWe define\n\n$$\nA_1 = \\{i \\mid 1 \\le i \\le m, 9 \\mid a_i\\},\n$$\n\n$$\nA_2 = \\{i \\mid 1 \\le i \\le m, a_i \\equiv 3 \\pmod{9}\\},\n$$\n\n$$\nA_3 = \\{i \\mid 1 \\le i \\le m, a_i \\equiv 6 \\pmod{9}\\},\n$$\n\n$$\nA_4 = \\{i \\mid 1 \\le i \\le m, a_i \\equiv 1 \\pmod{3}\\},\n$$\n\n$$\nA_5 = \\{i \\mid 1 \\le i \\le m, a_i \\equiv 2 \\pmod{3}\\}.\n$$\n\nThen $|A_1| + |A_2| + |A_3| + |A_4| + |A_5| = m$ and\n\n1. If $i \\in A_2, j \\in A_3$ then $9 \\mid 4a_i + 4a_j$.\n2. If $i \\in A_4, j \\in A_5$ then one of $4a_i + 4a_j, 4a_i + 7a_j$ and $7a_i + 4a_j$ is a multiple of 9 as all of them are divisible by 3 and they are distinct modulo 9.\n3. If either $i, j, k \\in A_2$ or $i, j, k \\in A_3$ then $9 \\mid 4a_i + 4a_j + 4a_k$.\n4. If either $i, j, k \\in A_4$ or $i, j, k \\in A_5$ then one of $4a_i + 4a_j + 4a_k, 4a_i + 4a_j + 7a_k$ and $4a_i + 7a_j + 7a_k$ is a multiple of 9 as all of them are divisible by 3 and they are distinct modulo 9.\n\nBy the assumption, we have $|A_i| \\le 2$ for $1 \\le i \\le 5$.\n\nIf $|A_1| \\ge 1$, then $|A_2| + |A_3| \\le 2$, $|A_4| + |A_5| \\le 2$. Hence\n\n$$\nm = |A_1| + |A_2| + |A_3| + |A_4| + |A_5| \\le 6.\n$$\n\nNow assume $|A_1| = 0$ and $m \\ge 7$. Then\n\n$$\n7 \\le m = |A_1| + |A_2| + |A_3| + |A_4| + |A_5| \\le 8.\n$$\n\nFurther,\n\n$$\n\\min\\{|A_2|, |A_3|\\} + \\min\\{|A_4|, |A_5|\\} \\ge 3.\n$$\n\nFrom (1) and (2) we know there exist $i_1, i_2, \\dots, i_6 \\in A_2 \\cup A_3 \\cup A_4 \\cup A_5$ ($i_1 < i_2 < \\dots < i_6$) and $b_1, b_2, \\dots, b_6 \\in \\{4, 7\\}$ such that $9 \\mid b_1a_{i_1} + b_2a_{i_2} + \\dots + b_6a_{i_6}$.\n\nThe proof of the statement is complete.\n\nNow, when $n \\ge 13$ it is easy to verify with the statement, for any group of integers $a_1, a_2, \\dots, a_n$, there always exist $a_{i_1}, a_{i_2}, \\dots, a_{i_9}$ ($1 \\le i_1 < i_2 < \\dots < i_9 \\le n$) and $b_i \\in \\{4, 7\\}$ ($i = 1, 2, \\dots, 9$) such that $b_1a_{i_1} + b_2a_{i_2} + \\dots + b_9a_{i_9}$ is a multiple of 9. That completes the proof.",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 14970,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $ABC$ be a right triangle with $\\angle C = 90^\\circ$. Points $D$ and $E$ on the hypotenuse $AB$ are such that $AD = AC$ and $BE = BC$. Points $P$ and $Q$ on $AC$ and $BC$ respectively are such that $AP = AE$ and $BQ = BD$. Let $M$ be the midpoint of segment $PQ$. Find $\\angle AMB$.",
    "options": [],
    "answer": "See solution",
    "solution": "We show that $M$ coincides with the incenter $I$ of the triangle. Since $\\angle A + \\angle B = 90^\\circ$, this implies $\\angle AMB = \\angle AIB = 180^\\circ - \\frac{1}{2}(\\angle A + \\angle B) = 135^\\circ$.\n\n![](images/Argentina2016_booklet_p1_data_4d68c4aa2e.png)\n\nBy hypothesis $AD = AC$, meaning that $D$ is the reflection of $C$ in the bisector $AI$ of $\\angle A$. Likewise, $Q$ is the reflection of $D$ in the bisector $BI$ of $\\angle B$. It follows that $CI = DI = QI$. Analogously, $E$ is the reflection of $C$ in the bisector $BI$ and $P$ is the reflection of $E$ in the bisector $AI$, hence $CI = EI = PI$. We obtain $CI = PI = QI$. Also, $\\angle PCI = \\angle QCI = 45^\\circ$ since $CI$ bisects $\\angle C = 90^\\circ$. Therefore,\n\n$$C\\angle IP = C\\angle IQ = 90^\\circ.$$ \n\nIn conclusion, $P$, $Q$, and $I$ are collinear, and $I$ is the midpoint of $PQ$ as $PI = QI$. Thus $M$ and $I$ coincide, as stated. The solution is complete.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 14971,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $\\mathbb{Z}/n\\mathbb{Z}$ denote the set of integers considered modulo $n$ (so $\\mathbb{Z}/n\\mathbb{Z}$ has $n$ elements). Find all positive integers $n$ for which there exists a bijective function $g: \\mathbb{Z}/n\\mathbb{Z} \\to \\mathbb{Z}/n\\mathbb{Z}$ such that the 101 functions\n\n$$\ng(x),\\quad g(x) + x,\\quad g(x) + 2x,\\quad \\dots,\\quad g(x) + 100x\n$$\n\nare all bijections on $\\mathbb{Z}/n\\mathbb{Z}$.\n\nCall a function $g$ *valiant* if it obeys this condition.",
    "options": [],
    "answer": "See solution",
    "solution": "We claim the answer is all numbers relatively prime to $101!$; the construction is to let $g$ be the identity function.\n\nThe proof is split into two essentially orthogonal claims, stated as lemmas.\n\n**Lemma (Elimination of $g$)**\n\nAssume a valiant $g: \\mathbb{Z}/n\\mathbb{Z} \\to \\mathbb{Z}/n\\mathbb{Z}$ exists. Then\n\n$$\nk! \\sum_{x \\in \\mathbb{Z}/n\\mathbb{Z}} x^k \\equiv 0 \\pmod{n}\n$$\n\nfor $k = 0, 1, \\dots, 100$.\n\n*Proof.* Define $g_x(T) = g(x) + Tx$ for any integer $T$. Viewing $g_x(T)^k$ as a polynomial in $T$ of degree $k$ with leading coefficient $x^k$, the $k$th finite difference gives\n\n$$\nk! x^k = \\binom{k}{0} g_x(k)^k - \\binom{k}{1} g_x(k-1)^k + \\cdots + (-1)^k \\binom{k}{k} g_x(0)^k.\n$$\n\nFor $1 \\leq k \\leq 100$, we have\n\n$$\n\\sum_x g_x(0)^k \\equiv \\sum_x g_x(1)^k \\equiv \\cdots \\equiv \\sum_x g_x(k)^k \\equiv S_k \\stackrel{\\text{def}}{=} 0^k + \\cdots + (n-1)^k \\pmod{n}\n$$\n\nby hypothesis. Thus,\n\n$$\nk! \\sum_x x^k \\equiv \\left[ \\binom{k}{0} - \\binom{k}{1} + \\cdots \\right] S_k \\equiv 0 \\pmod{n}\n$$\n\nfor $1 \\leq k \\leq 100$, and also for $k = 0$. $\\square$\n\n**Lemma (Power sum calculation)**\n\nLet $p$ be a prime, and let $n, M$ be positive integers such that\n\n$$\nM \\text{ divides } 1^k + 2^k + \\cdots + n^k\n$$\n\nfor $k = 0, 1, \\dots, p-1$. If $p \\mid n$ then $\\nu_p(M) < \\nu_p(n)$.\n\n*Proof.* The hypothesis means that any polynomial $f(T) \\in \\mathbb{Z}[T]$ with $\\deg f \\leq p-1$ will have $\\sum_{x=1}^n f(x) \\equiv 0 \\pmod{M}$. In particular,\n\n$$\n\\sum_{x=1}^{n} (x-1)(x-2)\\cdots(x-(p-1)) = (p-1)! \\sum_{x=1}^{n} \\binom{x-1}{p-1} = (p-1)! \\binom{n}{p} \\pmod{M}.\n$$\n\nThus $\\nu_p(M) \\leq \\nu_p\\left(\\binom{n}{p}\\right) = \\nu_p(n) - 1$. $\\square$\n\nNow, assume for contradiction that a valiant $g$ exists and $p \\leq 101$ is the smallest prime dividing $n$. Lemma I implies $k! \\sum_x x^k \\equiv 0 \\pmod{n}$ for $k = 1, \\dots, p-1$, so $\\sum_x x^k \\equiv 0 \\pmod{n}$. Thus $M = n$ in the previous lemma, which is impossible.\n\nTherefore, all $n$ must be relatively prime to $101!$.\n\n*Remark (Motivation for both parts):* For smaller values (e.g., replacing 101 with 2 or 3), similar arguments show $2 \\nmid n$ or $3 \\nmid n$ respectively, by considering sums of $x$ or $x^2$ over $\\mathbb{Z}/n\\mathbb{Z}$.\n\n**A second solution**\n\nBoth lemmas admit variations working modulo $p^e$ rather than $n$.\n\n**Lemma (I')**\n\nAssume valiant $g$ exists. Let $p \\leq 101$ be a prime, $e = \\nu_p(n)$. Then\n\n$$\n\\sum_{x \\in \\mathbb{Z}/n\\mathbb{Z}} x^k \\equiv 0 \\pmod{p^e}\n$$\n\nfor $k = 0, 1, \\dots, p-1$.\n\n*Proof.* Write\n\n$$\n\\sum_x (g(x) + Tx)^k \\equiv \\sum_x x^k \\pmod{p^e}.\n$$\n\nBoth sides are integer polynomials in $T$ vanishing at $T = 0, 1, \\dots, p-1$ by hypothesis. If $f(T) \\in \\mathbb{Z}[T]$ with $\\deg f \\leq p-1$ and $f(0) \\equiv \\cdots \\equiv f(p-1) \\equiv 0 \\pmod{p^e}$, then all coefficients of $f$ are divisible by $p^e$ (by induction on $e$). The leading $T^k$ coefficient is $\\sum_x x^k$ as desired. $\\square$\n\n**Lemma (II')**\n\nIf $e \\geq 1$ and $p$ is a prime,\n\n$$\n\\nu_p (1^{p-1} + 2^{p-1} + \\cdots + (p^e - 1)^{p-1}) = e - 1.\n$$\n\n*Proof.* For $p=2$ or $e=1$, the result is straightforward. For $p > 2, e > 1$, let $g$ be a primitive root modulo $p^e$. Summing terms relatively prime to $p$ gives\n\n$$\nS_0 = \\sum_{\\gcd(x,p)=1} x^{p-1} \\equiv \\sum_{i=1}^{\\varphi(p^e)} g^{(p-1)i} \\equiv \\frac{g^{p^{e-1}(p-1)^2} - 1}{g^{p-1} - 1} \\pmod{p^e}\n$$\n\nwhich implies $\\nu_p(S_0) = e-1$ by lifting the exponent. $\\square$",
    "topic": "Number Theory",
    "subtopic": "Modular Arithmetic"
  },
  {
    "id": 14972,
    "subject": "Mathematics (Olympiad)",
    "question": "a. Each of the rods from 2 to 19 cm can be the side of a triangle, as shown by the following triples: $\\{2, 3, 4\\}$, $\\{5, 6, 7\\}$, $\\{8, 9, 10\\}$, $\\{11, 12, 13\\}$, $\\{14, 15, 16\\}$, $\\{17, 18, 19\\}$, $\\{18, 19, 20\\}$.\n\nWhich rod cannot be the side of a triangle?\n\nb. Using rods of lengths 2 cm to 7 cm, in how many ways can Rachael divide them into two sets of three rods each, so that each set can form a triangle?\n\nc. Find a way to divide the rods from 3 cm to 17 cm into five sets of three rods each, so that each set forms a triangle with even perimeter.\n\nd. Can Rachael divide the rods from 2 cm to 20 cm (excluding the 1 cm rod) into six sets of three rods each, so that each set forms a triangle with equal perimeter? Justify your answer.",
    "options": [],
    "answer": "See solution",
    "solution": "a. The rod that cannot be the side of a triangle is the 1 cm rod. This is because, for any triangle, the shortest side must be longer than the difference of the other two sides. Since the difference in lengths of any two rods is at least 1 cm, the 1 cm rod cannot be used.\n\nb. The 1 cm rod cannot be used, so Rachael must use rods of lengths 2 cm to 7 cm. Dividing these into two sets of three, only one set will contain the 2 cm rod. The only sets of three rods including the 2 cm rod that form a triangle are $\\{2, 3, 4\\}$, $\\{2, 4, 5\\}$, $\\{2, 5, 6\\}$, and $\\{2, 6, 7\\}$. The corresponding sets of the remaining rods are $\\{5, 6, 7\\}$, $\\{3, 6, 7\\}$, $\\{3, 4, 7\\}$, and $\\{3, 4, 5\\}$. Of these, only $\\{3, 4, 7\\}$ cannot form a triangle. So there are just three ways to use the rods to make two triangles.\n\nc. For a triangle to have even perimeter, it must have all three rods of even length or exactly two of odd length. One solution is: $\\{3, 4, 5\\}$, $\\{6, 8, 10\\}$, $\\{7, 15, 16\\}$, $\\{9, 12, 17\\}$, $\\{11, 13, 14\\}$.\n\nd. The 1 cm rod cannot be used. The sum of the lengths of the other 19 rods is $2 + 3 + 4 + \\cdots + 20 = 209$. The sum of the lengths of the 18 rods used must be a multiple of 6, and $209 = 34 \\times 6 + 5$, so the unused rod must have length 5, 11, or 17. The sum of the lengths of the other 18 rods is then 204, 198, or 192, so the perimeter of each triangle is 34, 33, or 32. However, in each case, one triangle must have a side of length 20, and to form a triangle, the sum of the other two sides must be greater than 20, so the perimeter must be greater than 40. Therefore, Rachael cannot form a set of 6 triangles with equal perimeters.\n\nAlternatively, since one of the 18 rods used must be 19 or 20, if a triangle has a side of 19, the sum of the other two sides must be greater than 19, so the perimeter must be greater than 38. Similarly, if a triangle has a side of 20, its perimeter must be greater than 40. The total perimeter of six triangles would then be greater than $6 \\times 38 = 228$, but the total length of 18 rods is at most $20 + 19 + \\cdots + 3 = 207$. Thus, it is not possible.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 14973,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $H$ be a point in triangle $ABC$. The circle passing through $H$ with center at the midpoint of $BC$ intersects the line $BC$ at $A_1$ and $A_2$. Similarly, the circle passing through $H$ with center at the midpoint of $CA$ intersects the line $CA$ at $B_1$ and $B_2$, and the circle passing through $H$ with center at 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.\n\n![](images/Mathematical_Olympiad_in_China_2007-2008-Problems_and_Solutions_p208_data_752bc99abe.png)",
    "options": [],
    "answer": "See solution",
    "solution": "**Proof I**\n\nLet $B_0, C_0$ be the midpoints of $CA$ and $AB$ respectively. Denote $A'$ as the other intersection of the circle centered at $B_0$ passing through $H$ and the circle centered at $C_0$ passing through $H$. We know that $A'H \\perp C_0B_0$. Since $B_0, C_0$ are the midpoints of $CA$ and $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\n$$\nAC_1 \\cdot AC_2 = AA' \\cdot AH = AB_1 \\cdot AB_2,\n$$\n\nso $B_1, B_2, C_1, C_2$ are concyclic.\n\nLet the intersection of the perpendicular bisectors of $B_1B_2$ and $C_1C_2$ be $O$. Then $O$ is the circumcenter of quadrilateral $B_1B_2C_1C_2$, as well as the circumcenter of $\\triangle ABC$. So\n\n$$\nOB_1 = OB_2 = OC_1 = OC_2.\n$$\n\nSimilarly,\n\n$$\nOA_1 = OA_2 = OB_1 = OB_2.\n$$\n\nTherefore, the six points $A_1, A_2, B_1, B_2, C_1, C_2$ all lie on the same circle, whose center is $O$ and radius is $OA_1$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 14974,
    "subject": "Mathematics (Olympiad)",
    "question": "Lucas paints the entire outside of a cube blue. He then saws the cube into 27 equally sized cubes. He neatly stacks these 27 cubes so that he gets a tower of $27 \\times 1 \\times 1$ cubes.\n\nAt most, how many of the 110 side faces of cubes on the outside of his tower are blue?",
    "options": [],
    "answer": "See solution",
    "solution": "48",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 14975,
    "subject": "Mathematics (Olympiad)",
    "question": "How many positive integers, where the only allowed digits are 0 and 1, are less than $1111100100$?",
    "options": [],
    "answer": "See solution",
    "solution": "Solution 1: 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.\n\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$.\n\nSolution 2: Ordering 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 in base 2, then the first number is larger than the second number also in base 10. The number $1111100100$ in base 2 equals $2^{10} + 2^{9} + 2^{8} + 2^{7} + 2^{6} + 2^{2} = 2020$ in 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": 14976,
    "subject": "Mathematics (Olympiad)",
    "question": "a) Is it possible for grand master $N$ to take only the eighth place in a tournament?\n\nb) Is it possible for grand master $N$ not to be among the first eight players, given that he wins more than half of his games?\n",
    "options": [],
    "answer": "See solution",
    "solution": "a) The following table shows an example of a tournament where grand master $N$ takes only the eighth place. So, grand master $A$ can be right.\n\n<table><thead><tr><th rowspan=\"2\"></th><th rowspan=\"2\">N</th><th rowspan=\"2\">1</th><th rowspan=\"2\">2</th><th rowspan=\"2\">3</th><th rowspan=\"2\">4</th><th rowspan=\"2\">5</th><th rowspan=\"2\">6</th><th rowspan=\"2\">7</th><th rowspan=\"2\">8</th><th rowspan=\"2\">9</th><th colspan=\"2\">Points ($\\Sigma$)</th></tr><tr><th>New system</th><th>Old system</th></tr></thead><tbody><tr><td>N</td><td>●</td><td>0</td><td>0</td><td>0</td><td>1</td><td>1</td><td>1</td><td>0</td><td>1</td><td>1</td><td>15</td><td>5</td></tr><tr><td>1</td><td>1</td><td>●</td><td>0.5</td><td>0.5</td><td>0.5</td><td>0</td><td>0.5</td><td>0.5</td><td>1</td><td>1</td><td>14</td><td>5.5</td></tr><tr><td>2</td><td>1</td><td>0.5</td><td>●</td><td>0.5</td><td>0.5</td><td>0.5</td><td>0</td><td>0.5</td><td>1</td><td>1</td><td>14</td><td>5.5</td></tr><tr><td>3</td><td>1</td><td>0.5</td><td>0.5</td><td>●</td><td>0</td><td>0.5</td><td>0.5</td><td>0.5</td><td>1</td><td>1</td><td>14</td><td>5.5</td></tr><tr><td>4</td><td>0</td><td>0.5</td><td>0.5</td><td>1</td><td>●</td><td>0.5</td><td>0.5</td><td>0.5</td><td>1</td><td>1</td><td>14</td><td>5.5</td></tr><tr><td>5</td><td>0</td><td>1</td><td>0.5</td><td>0.5</td><td>0.5</td><td>●</td><td>0.5</td><td>0.5</td><td>1</td><td>1</td><td>14</td><td>5.5</td></tr><tr><td>6</td><td>0</td><td>0.5</td><td>1</td><td>0.5</td><td>0.5</td><td>0.5</td><td>●</td><td>0.5</td><td>1</td><td>1</td><td>14</td><td>5.5</td></tr><tr><td>7</td><td>1</td><td>0.5</td><td>0.5</td><td>0.5</td><td>0.5</td><td>0.5</td><td>0.5</td><td>●</td><td>0.5</td><td>1</td><td>13</td><td>5.5</td></tr><tr><td>8</td><td>0</td><td>0</td><td>0</td><td>0</td><td>0</td><td>0</td><td>0</td><td>0.5</td><td>●</td><td>1</td><td>4</td><td>1.5</td></tr><tr><td>9</td><td>0</td><td>0</td><td>0</td><td>0</td><td>0</td><td>0</td><td>0</td><td>0</td><td>0</td><td>●</td><td>0</td><td>0</td></tr></tbody></table>\n\nb) By condition, grand master $N$ wins more than half of his games, i.e., he wins at least 5 of his games, so he takes at least 5 points in accordance with the old system of marking. If $N$ is not among the first eight players, i.e., the eighth player has more points than $N$, then at least eight players have more points than grand master $N$, i.e., any of them has at least 5.5 points. Thus, the sum of the points of the first eight players is greater than or equal to $8 \\cdot 5.5 = 44$, but the total number of points in the tournament is $\\frac{10 \\cdot 9}{2} = 45$ in accordance with the old marking system, a contradiction. Thus, grand master $B$ cannot be right.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 14977,
    "subject": "Mathematics (Olympiad)",
    "question": "Let the numbers of blue, yellow, and green balls be $b$, $y$, and $g$, respectively, where $b + y + g = 10$. The balls can be arranged in a line in $\\frac{10!}{b!y!g!}$ different ways. If $\\frac{10!}{b!y!g!} = 360$, what is the maximum possible value of $b$?",
    "options": [],
    "answer": "See solution",
    "solution": "We have $\\frac{10!}{b!y!g!} = 360$. This can be rewritten as:\n\n$$\n10 \\cdot 9 \\cdots (b+1) = 360 \\cdot y! \\cdot g!\n$$\n\nSince $360 = 10 \\cdot 9 \\cdot 4$, we get $10 \\cdot 9 \\cdots (b+1) \\geq 360$, so $b+1 \\leq 8$, or $b \\leq 7$.\n\nFor example, if $b = 7$, $y = 2$, $g = 1$:\n\n$$\n\\frac{10!}{7!2!1!} = \\frac{10 \\cdot 9 \\cdot 8}{2} = 360.\n$$\n\nThus, the maximum possible number of blue balls is $7$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 14978,
    "subject": "Mathematics (Olympiad)",
    "question": "The area of a convex polygon in the plane is equally shared by the four standard quadrants, and all non-zero lattice points lie outside the polygon. Show that the area of the polygon is less than $4$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $K$ be a convex polygon in the plane satisfying the conditions above, and let $K_i$ be the intersection of $K$ with the $i$-th standard quadrant (subscripts modulo $4$). It suffices to show that one of the $K_i$ has area less than $1$.\n\nSince $K$ is convex and the $K_i$ have equal areas, the origin $O = (0, 0)$ is interior to $K$; it is a vertex of each $K_i$. Consider the lattice points $A_0 = (1, 0)$, $A_1 = (0, 1)$, $A_2 = (-1, 0)$, and $A_3 = (0, -1)$. Since $K$ is closed, convex, and contains no non-zero lattice points, there exists a line $\\ell_i$ through $A_i$ such that $K$ lies in one of the open half-planes determined by $\\ell_i$, a positive distance away from $\\ell_i$.\n\nIf one of the lines $\\ell_i$ meets the closed segment $A_{i+1}A_{i+3}$, then the triangle determined by the lines $\\ell_i$, $A_iA_{i+2}$, $A_{i+1}A_{i+3}$ contains either $K_i$ or $K_{i+3}$, and its area does not exceed $\\frac{1}{2}$, so either $\\text{area}(K_i) < \\frac{1}{2}$ or $\\text{area}(K_{i+3}) < \\frac{1}{2}$ (the inequality is strict, since $K$ is a positive distance away from $\\ell_i$).\n\nIf no $\\ell_i$ meets the corresponding closed segment $A_{i+1}A_{i+3}$, then the lines in each pair $(\\ell_i, \\ell_{i+1})$ meet at some point $B_i$ situated in the $i$-th quadrant. Notice that $K_i$ is contained in the convex quadrangle $OA_iB_iA_{i+1}$, a positive distance away from $\\ell_i \\cup \\ell_{i+1}$, so\n\n$$\n\\text{area}(K_i) < \\text{area}(OA_iB_iA_{i+1}) = \\text{area}(OA_iA_{i+1}) + \\text{area}(A_iB_iA_{i+1}) = \\frac{1}{2} + \\text{area}(A_iB_iA_{i+1}).\n$$\n\nIt is therefore sufficient to show that the area of one of the four triangles $A_iB_iA_{i+1}$ does not exceed $\\frac{1}{2}$. To this end, notice that the convex quadrangle $B_0B_1B_2B_3$ has at least one non-acute internal angle, say at $B_i$, so $B_i$ does not lie outside the closed disc of diameter $A_iA_{i+1}$, and the area of the triangle $A_iB_iA_{i+1}$ is at most $\\frac{1}{2}$. This completes the proof.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 14979,
    "subject": "Mathematics (Olympiad)",
    "question": "The set $\\{2, 3, 4, \\ldots, 2020\\}$ is partitioned into triples. In each triple $(a, b, c)$, the numbers are arranged in ascending order, i.e., $a < b < c$, and the difference $|b - \\frac{a + c}{2}|$ is called the error of this triple. Find the maximal possible sum of errors of all $673$ triples.",
    "options": [],
    "answer": "See solution",
    "solution": "The maximal possible sum of errors is $$\\frac{3n^2 - 3n + 2 \\lfloor \\frac{n}{2} \\rfloor^2}{4},$$ where $n = 2019$.\n\nNote: Shifting all numbers in the set by the same value does not change the errors, so we can consider partitions of $\\{1, 2, \\ldots, 2019\\}$ and generalize to $3n$ elements.\n\nLet $n \\in \\mathbb{N}$ and consider the set $\\{1, 2, \\ldots, 3n\\}$. We seek the partition maximizing the sum of errors. For a triple $a < b < c$, define it as **left** if $b \\leq (a + c)/2$, otherwise **right**.\n\n**Step 1.** If a right triple $a < b < c$ has $b \\leq c - 2$, swap $b$ and $b+1$ with another triple containing $b+1$. This increases the error, so in the maximal partition, right triples must be of the form $a < b < b+1$ and left triples $a < a+1 < c$.\n\n**Step 2.** For any two triples of different types, swapping elements cannot increase the sum of errors, so the maximal element of each triple is greater than the minimal element of any other triple.\n\n**Step 3.** Further swaps show that the minimal element of each right triple is smaller than the minimal element of each left triple, and the maximal element of each left triple is greater than the maximal element of each right triple.\n\nCombining these properties, the partition maximizing the sum of errors is described as follows: there exist $k, \\ell \\in \\mathbb{N}$ with $k + \\ell = n$ such that the numbers $1, 2, \\ldots, k$ are minimal in right triples, $k+1, k+3, \\ldots, k+2\\ell-1$ are minimal in left triples, $k+2\\ell+2, k+2\\ell+4, \\ldots, 3k+2\\ell$ are maximal in right triples, and $3k+2\\ell+1, \\ldots, 3k+3\\ell$ are maximal in left triples.\n\nThe sum of errors is then\n$$\n\\frac{3n^2 - 3n + 2k\\ell}{4}.\n$$\nThe maximum occurs at $k = \\left\\lfloor \\frac{n}{2} \\right\\rfloor$, so the answer is\n$$\n\\frac{3n^2 - 3n + 2 \\lfloor \\frac{n}{2} \\rfloor^2}{4}.\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 14980,
    "subject": "Mathematics (Olympiad)",
    "question": "Consider the function $f(x) = a(|\\sin x| + |\\cos x|) - 3\\sin 2x - 7$, where $a$ is a real parameter.\n\n1. Prove that $f(x) = f\\left(\\frac{\\pi}{2} - x\\right) = f(\\pi + x) = f\\left(\\frac{3\\pi}{2} - x\\right)$ for every $x$.\n\n2. Find all pairs $(a, n)$, where $n$ is a positive integer, for which the equation $f(x) = 0$ has $2007$ roots in the interval $(0, n\\pi)$.",
    "options": [],
    "answer": "See solution",
    "solution": "**a)**\n\n$$\n\\begin{align*}\nf(x + \\pi) &= a(|\\sin(x + \\pi)| + |\\cos(x + \\pi)|) - 3\\sin(2x + 2\\pi) - 7 \\\\\n&= a(|-\\sin x| + |-\\cos x|) - 3\\sin(2x) - 7 = f(x), \\\\\n\nf\\left(\\frac{\\pi}{2} - x\\right) &= a\\left(|\\sin\\left(\\frac{\\pi}{2} - x\\right)| + |\\cos\\left(\\frac{\\pi}{2} - x\\right)|\\right) - 3\\sin(\\pi - 2x) - 7 \\\\\n&= a\\left(|\\cos x| + |\\sin x|\\right) - 3\\sin(2x) - 7 = f(x), \\\\\n\nf\\left(\\frac{3\\pi}{2} - x\\right) &= a\\left(|\\sin\\left(\\frac{3\\pi}{2} - x\\right)| + |\\cos\\left(\\frac{3\\pi}{2} - x\\right)|\\right) - 3\\sin(3\\pi - 2x) - 7 \\\\\n&= a\\left(|-\\cos x| + |-\\sin x|\\right) - 3\\sin(2x) - 7 = f(x).\n\\end{align*}\n$$\n\n**b)**\n\nFor every integer $k$, $f\\left(\\frac{k\\pi}{2}\\right) = a-7$.\n\nAlso, $f\\left(\\frac{\\pi}{4}\\right) = a\\sqrt{2} - 10$ and $f\\left(\\frac{3\\pi}{4}\\right) = a\\sqrt{2} - 4$.\n\nIf $a \\neq 7$, $a \\neq 5\\sqrt{2}$, and $a \\neq 2\\sqrt{2}$, the equation $f(x) = 0$ has an even number of roots in each interval $(0, \\frac{\\pi}{2})$, $(\\frac{\\pi}{2}, \\pi)$, and thus in $(0, n\\pi)$.\n\n1. If $a = 7$:\n   - $f(x) = 7(|\\sin x| + |\\cos x|) - 3\\sin 2x - 7$ and $f\\left(\\frac{\\pi}{2}\\right) = 0$.\n   - For $x \\in (0, \\frac{\\pi}{2})$, set $y = \\sin x + \\cos x = \\sqrt{2}\\sin\\left(x + \\frac{\\pi}{4}\\right) \\in (1, \\sqrt{2}]$ and $\\sin 2x = y^2 - 1$. The equation $f(x) = 0$ becomes $3y^2 - 7y + 4 = 0$, with roots $y_1 = 1$, $y_2 = \\frac{4}{3}$. Thus, two roots in $(0, \\frac{\\pi}{2})$.\n   - For $x \\in (\\frac{\\pi}{2}, \\pi)$, set $y = \\sin x - \\cos x = \\sqrt{2}\\sin\\left(x + \\frac{\\pi}{4}\\right) \\in (1, \\sqrt{2}]$. The equation $f(x) = 0$ becomes $3y^2 + 7y - 10 = 0$, with roots $y_1 = 1$, $y_2 = -\\frac{10}{3}$, so no solutions in this interval.\n   - Therefore, $f(x) = 0$ has $3$ roots in $(0, \\pi)$. In $(0, n\\pi)$, the total number of roots is $3n + n - 1 = 4n - 1$. Setting $4n - 1 = 2007$ gives $n = 502$.\n\n2. If $a = 5\\sqrt{2}$:\n   - For $x \\in (0, \\frac{\\pi}{2})$, $f(x) = 5\\sqrt{2}(\\sin x + \\cos x) - 3\\sin 2x - 7$. Setting $y = \\sin x + \\cos x$, $f(x) = 0$ is $3y^2 - 5\\sqrt{2}y + 4 = 0$, $y \\in (1, \\sqrt{2}]$. Roots: $y_1 = \\sqrt{2}$, $y_2 = \\frac{2\\sqrt{2}}{3} < 1$, so only $x = \\frac{\\pi}{4}$ is a solution.\n   - For $x \\in (\\frac{\\pi}{2}, \\pi)$, $f(x) = 5\\sqrt{2}(\\sin x - \\cos x) - 3\\sin 2x - 7$. Setting $y = \\sin x - \\cos x$, $f(x) = 0$ is $3y^2 + 5\\sqrt{2}y - 10 = 0$, $y \\in (1, \\sqrt{2}]$. Roots do not belong to $(1, \\sqrt{2}]$. Thus, one root in $(0, \\pi)$ and $n$ roots in $(0, n\\pi)$, so $n = 2007$.\n\n3. If $a = 2\\sqrt{2}$:\n   - Similarly, there is a unique root $x = \\frac{3\\pi}{4}$ in $(0, \\pi)$ and $n$ roots in $(0, n\\pi)$, so $n = 2007$.\n\n**Answer:**\n- $a = 7$, $n = 502$\n- $a = 5\\sqrt{2}$, $n = 2007$\n- $a = 2\\sqrt{2}$, $n = 2007$",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 14981,
    "subject": "Mathematics (Olympiad)",
    "question": "Consider a rectangular $n \\times m$ table where $n \\ge 2$ and $m \\ge 2$ are positive integers. Each cell is colored in one of four colors: white, green, red, or blue. Call such a coloring *interesting* if any $2 \\times 2$ square contains every color exactly once. Find the number of interesting colorings.",
    "options": [],
    "answer": "See solution",
    "solution": "Number the columns of the table by $1, 2, \\ldots, m$ and its rows by $1, 2, \\ldots, n$, and denote the cell in the $i$-th column and $j$-th row by $(i, j)$.\n\nConsider an interesting coloring $S$. First, we show that either every row in $S$ contains only two colors, or every column in $S$ contains only two colors.\n\nSuppose $S$ does not contain a rectangle $1 \\times 3$ (with longer horizontal side) containing three distinct colors. Then, in any row of $S$, two colors must alternate.\n\nNow, suppose there exist three cells $(u, v)$, $(u+1, v)$, and $(u+2, v)$ in a column, colored with three distinct colors $a, b, c$. Then $(u+1, v+1)$ must be colored with the fourth color $d$, $(u, v+1)$ with $c$, $(u+2, v+1)$ with $a$, $(u+1, v+2)$ with $b$, $(u, v+2)$ with $a$, $(u+2, v+1)$ with $c$, and so on. By analogy, $(u+1, v-1)$ is colored $d$, $(u, v-1)$ with $c$, $(u+2, v-1)$ with $a$, etc. We conclude that column $u$ contains only colors $a$ and $c$, column $u+1$ contains only $b$ and $d$, and column $u+2$ contains only $a$ and $c$. Thus, any column with the same parity as $u$ contains only $a$ and $c$, and any column with the parity of $u+1$ contains only $b$ and $d$.\n\nHence, either every row in $S$ contains only two colors, or every column in $S$ contains only two colors.\n\nThe number of interesting colorings such that in every column of odd number two of the colors alternate, and in every column of even number the other two colors alternate, is $\\binom{4}{2} \\times 2^m$. The corresponding colorings when the rows are colored in two colors is $\\binom{4}{2} \\times 2^n$. The colorings in which the color of the cell $(i, j)$ depends only on the parity of $i$ and $j$ are counted in both cases. Therefore, the number of interesting colorings is:\n\n$$\n6 \\times (2^m + 2^n) - 24\n$$",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 14982,
    "subject": "Mathematics (Olympiad)",
    "question": "Determine all positive integers $n$ such that the inequality\n$$\n\\sqrt{x-1} + \\sqrt{x-2} + \\dots + \\sqrt{x-n} < x\n$$\nholds for every real number $x \\ge n$.",
    "options": [],
    "answer": "See solution",
    "solution": "The inequality must hold for $x = n$. Thus,\n$$\n1 + \\sqrt{2} + \\dots + \\sqrt{n-1} < n.\n$$\nBut $1 + \\sqrt{2} + \\sqrt{3} > 4$, and by induction,\n$$\n1 + \\sqrt{2} + \\dots + \\sqrt{n-1} > n \\quad \\text{for all } n \\ge 4.\n$$\nTherefore, it is necessary that $n \\le 3$.\n\nWhen $n = 3$, taking $x = 4$ also gives $\\sqrt{3} + \\sqrt{2} + 1 > 4$, not possible.\n\nWhen $n = 2$, we have\n$$\n\\sqrt{x-1} + \\sqrt{x-2} < 2\\sqrt{x-1} \\le x,\n$$\nsince $x^2 - 4x + 4 = (x-2)^2 \\ge 0$.\n\nWhen $n = 1$, we have $\\sqrt{x-1} < x$, since it is equivalent to\n$$\nx^2 - x + 1 > 0.\n$$\nThe positive integers $n$ satisfying the property are $n = 1$ and $n = 2$.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 14983,
    "subject": "Mathematics (Olympiad)",
    "question": "Prove that for any four numbers selected from the set $\\{1, 2, \\dots, 20\\}$ (with repetition allowed), it is always possible to choose three of them, say $a$, $b$, and $c$, such that the congruence equation $ax \\equiv b \\pmod c$ has integer solutions.",
    "options": [],
    "answer": "See solution",
    "solution": "It is well known that $ax \\equiv b \\pmod c$ has integer solutions if and only if $\\gcd(a, c) \\mid b$.\n\nAssume, for contradiction, that for $1 \\leq a_1, a_2, a_3, a_4 \\leq 20$, the statement is not true. Consider three cases:\n\n1. **One of the numbers is a prime power:**\n   Suppose $a_1 = p^\\alpha$ for some prime $p$ and $\\alpha \\geq 1$. Let $v_p(a_2) = \\beta \\leq v_p(a_3) = \\gamma$ (where $v_p(x)$ is the highest power of $p$ dividing $x$). Take $a = a_1$, $b = a_3$, $c = a_2$. Then $\\gcd(a, c) = p^{\\min(\\alpha, \\beta)} \\mid p^\\gamma$, so $\\gcd(a, c) \\mid b$, contradicting the assumption.\n\n2. **One number divides another:**\n   Suppose $a_1 \\mid a_2$. Take $a = a_1$, $b = a_2$, $c = a_3$. Since $\\gcd(a, c) \\mid a_1$, we have $\\gcd(a, c) \\mid b$, again contradicting the assumption.\n\n3. **No prime powers and no divisibility among the numbers:**\n   The four numbers must have distinct prime factors and be unequal. They must be four of the following seven numbers: $6, 12, 18, 10, 15, 14, 20$. Note that the prime factor $7$ of $14$ does not appear in the other numbers. When checking $\\gcd(a, c) \\mid b$, $14$ is equivalent to $2$ and cannot be selected (otherwise, integer solutions exist). For the remaining six numbers, $6 \\mid 12$, $6 \\mid 18$, $10 \\mid 20$, so only $\\{12, 18, 15, 10\\}$ or $20$ have no divisibility relation. Take $a = 12$, $b = 18$, $c = 15$, then $\\gcd(a, c) = 3 \\mid b$. Again, the assumption is contradicted.\n\nTherefore, the original statement is true. $\\square$\n\n**Remark:** The number $20$ in the problem can be improved to $104$, and $104$ is optimal. If $20$ is replaced by $105$ or larger, then taking the products of any three numbers from $2, 3, 5, 7$ gives $30, 42, 70, 105$, and the greatest common divisor of any two does not divide another, so the statement fails for these numbers. On the other hand, for any four numbers selected from $\\{1, 2, \\dots, 104\\}$, there are two whose greatest common divisor divides another. Suppose not, and $a, b, c, d$ is a counterexample. If $\\gcd(a, b, c, d) = x > 1$, use $\\frac{a}{x}, \\frac{b}{x}, \\frac{c}{x}, \\frac{d}{x}$ instead, and it remains a counterexample. Now, assume $\\gcd(a, b, c, d) = 1$, and one must be odd, say $a$ is odd.\n\nConsider $d_1 = \\gcd(a, b)$, $d_2 = \\gcd(a, c)$, $d_3 = \\gcd(a, d)$. Assume $d_1$, $d_2$, $d_3$ have no divisibility relation (otherwise, say $d_2 \\mid d_1$, then $(a, c) \\mid b$, and $ax \\equiv b \\pmod c$ has integer solutions). This means $a$ has three pairwise non-divisible factors $d_1, d_2, d_3$, so $a$ is not a prime power. If $a$ has three distinct prime factors, then $a \\geq 3 \\times 5 \\times 7 > 104$, impossible. Thus, $a$ has two distinct prime factors, say $a = p^\\alpha q^\\beta$.\n\nLet $d_i = p^{\\alpha_i} q^{\\beta_i}$ for $i = 1, 2, 3$. Since $d_1, d_2, d_3$ have no divisibility relation, $\\alpha_1, \\alpha_2, \\alpha_3$ must be distinct, and $\\beta_1, \\beta_2, \\beta_3$ must be distinct. It follows that $\\alpha \\geq 2$, $\\beta \\geq 2$, but $a \\geq 3^2 \\times 5^2 > 104$, impossible. The assertion is now verified.",
    "topic": "Number Theory",
    "subtopic": "Modular Arithmetic"
  },
  {
    "id": 14984,
    "subject": "Mathematics (Olympiad)",
    "question": "Find the smallest positive integer $n$ such that there exist $n$ real numbers $a_1, a_2, \\ldots, a_n$ satisfying the following three conditions:\n\n1. $a_1 + a_2 + \\ldots + a_n > 0$;\n2. $a_1^3 + a_2^3 + \\ldots + a_n^3 < 0$;\n3. $a_1^5 + a_2^5 + \\ldots + a_n^5 > 0$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $S_k(a) = a_1^k + a_2^k + \\dots + a_n^k$ for a $n$-tuple $a = (a_1, \\ldots, a_n)$.\n\n**Claim 1.** There does not exist a 4-tuple $a = (a_1, a_2, a_3, a_4)$ such that\n$$\nS_1(a) > 0, \\quad S_3(a) < 0, \\quad S_5(a) > 0.\n$$\n*Proof.*\n- If there is exactly one positive number, say $a_1 > 0 \\ge a_2, a_3, a_4$, let $b_i = -a_i$ for $i = 2,3,4$. Then $a_1 < b_2 + b_3 + b_4$, so\n$$\na_1^3 > (b_2 + b_3 + b_4)^3 \\ge b_2^3 + b_3^3 + b_4^3 = -(a_2^3 + a_3^3 + a_4^3),\n$$\nwhich is a contradiction.\n- If there is exactly one negative number, say $a_1 < 0 \\le a_2, a_3, a_4$, let $b_1 = -a_1$. Then\n$$\na_2^3 + a_3^3 + a_4^3 < b_1^3 \\text{ and } a_2^5 + a_3^5 + a_4^5 > b_1^5.\n$$\nThe first inequality gives $b_1 > a_2, a_3, a_4$, so\n$$\nb_1^5 > b_1^2(a_2^3 + a_3^3 + a_4^3) > a_2^5 + a_3^5 + a_4^5,\n$$\nwhich is a contradiction.\n- If there are two positive and two negative numbers, let $a_1 = a$, $a_2 = b$, $-a_3 = c$, $-a_4 = d > 0$. Then\n$$\na + b > c + d, \\quad a^3 + b^3 < c^3 + d^3, \\quad a^5 + b^5 > c^5 + d^5.\n$$\nAssume $a \\ge b$, $c \\ge d$. Comparing means, we get $a - b < c - d$. If $a \\ge c$, then $d < b + c - a \\le b$, so $a^3 + b^3 > c^3 + d^3$, which contradicts the second inequality. Thus $a < c$, so $c > a \\ge b > d$. Set $C = \\frac{c}{d}$, $A = \\frac{a}{d}$, $B = \\frac{b}{d}$, and define\n$$\nf(x) = C^x + 1 - A^x - B^x.\n$$\nThe conditions imply $f(1) < 0$, $f(3) > 0$, $f(5) < 0$. Since $C > A, B$, $f(+\\infty) = +\\infty$. By the mean value theorem, $f$ has at least three roots in $(0, +\\infty)$, so $f'$ has at least two positive roots. But\n$$\nf'(x) = C^x \\ln C - A^x \\ln A - B^x \\ln B = C^x \\left[ \\ln C - \\left(\\frac{A}{C}\\right)^x \\ln A - \\left(\\frac{B}{C}\\right)^x \\ln B \\right],\n$$\nwhich is strictly increasing, so $f'$ has at most one positive root—a contradiction.\n\n**Claim 2.** There exists a 5-tuple that meets all the conditions.\n\n*Proof.* Set $a_1 = a$, $a_2 = b$, $a_3 = c$, $-a_4 = d$, $-a_5 = e > 0$. Let\n$$\na = 2x, \\quad b = 1, \\quad c = 1, \\quad d = e = y + 1, \\text{ where } x > y.\n$$\nWe seek $x, y$ such that\n$$\n(2x)^3 + 2 < 2(y+1)^3, \\quad (2x)^5 + 2 > 2(y+1)^5.\n$$\nChoose $x = 1.5$ and $y' + 1 = \\frac{\\sqrt[3]{29}}{\\sqrt[3]{2}} < x + 1$. Then\n$$\n(2x)^5 + 2 - (2y' + 2)^5 > 0,\n$$\nso there exists $\\varepsilon > 0$ such that $y = y' + \\varepsilon$ and $(x, y) = (1.5, y' + \\varepsilon)$ meet all requirements.\n\nTherefore, the minimum value of $n$ is $5$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 14985,
    "subject": "Mathematics (Olympiad)",
    "question": "Consider a triangle $ABC$ with $\\angle BAC = \\frac{4}{3} \\angle ABC < 90^\\circ$. The ray $AE$ bisects the angle $BAC$, with $E \\in BC$. Point $F$ lies on the ray $AE$ such that $\\angle ABF = \\frac{1}{2} \\angle BAC$ and $AF = AC$. Find the measure of the angle $\\angle BCF$.",
    "options": [],
    "answer": "See solution",
    "solution": "Set $\\angle BAC = 4x$ so $\\angle ABC = 3x$. Then $\\angle BAF = \\angle CAF = \\angle ABF = 2x$ and $\\angle CBF = x$. Consider the point $S$ on the ray $AC$ such that $C \\in AS$ and $\\angle CBS = x$. Notice that $F$ is the incentre of triangle $ABS$, since $BF$ bisects $\\angle ABS$. Then $\\angle BSA = 180^\\circ - 8x$ and $\\angle FSA = 90^\\circ - 4x$.\n\nThe triangle $ACF$ is isosceles and $\\angle FAC = 2x$, so $\\angle ACF = 90^\\circ - x$. Consequently, $\\angle SFC = 3x$.\n\nThe lines $SF$ and $BC$ meet at point $M$. Since $SF$ is the perpendicular bisector of $AB$, triangle $BMA$ is isosceles at $M$, so $\\angle FAM = \\angle FBM = x = \\angle MAC$, hence $AM$ bisects $CF$. Consequently,\n\n$$\n\\angle FBC = \\angle SFC = 3x.\n$$\n\nTo conclude, $\\angle BCS = \\angle ABC + \\angle CAB = 3x + 4x = 7x$ and $180^\\circ = \\angle SCA = \\angle BCS + \\angle BCF + \\angle FCA = 7x + 3x + 90^\\circ - x$ gives $x = 10^\\circ$. Therefore $\\angle BCF = 30^\\circ$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 14986,
    "subject": "Mathematics (Olympiad)",
    "question": "For every pair of positive integers $n, m$ with $n < m$, let $s(n, m)$ be the number of positive integers in the range $[n, m]$ that are coprime with $m$.\n\nFind all positive integers $m \\ge 2$ such that $m$ satisfies these conditions:\n\n1. $\\displaystyle \\frac{s(n,m)}{m-n} \\ge \\frac{s(1,m)}{m}$ for all $n = 1, 2, \\ldots, m-1$,\n2. $2022^m + 1$ is divisible by $m^2$.",
    "options": [],
    "answer": "See solution",
    "solution": "First, we prove that if $m$ satisfies the first condition, then $m$ has only one prime divisor. Assume that $m$ has at least two prime divisors. Let $p$ be the smallest prime divisor of $m$ and $p_1, p_2, \\dots, p_k$ be the remaining prime divisors of $m$. We have\n$$\n\\frac{\\varphi(m)}{m} = \\left(1-\\frac{1}{p}\\right) \\left(1-\\frac{1}{p_1}\\right) \\dots \\left(1-\\frac{1}{p_k}\\right) < 1-\\frac{1}{p} = \\frac{p-1}{p}.\n$$\nHence, by choosing $n = p$ in (1), we get\n$$\n\\frac{s(p, m)}{m-p} = \\frac{\\varphi(m) - (p-1)}{m-p} < \\frac{\\varphi(m) - \\frac{\\varphi(m)}{m} \\cdot p}{m-p} = \\frac{\\varphi(m)}{m} = \\frac{s(1, m)}{m},\n$$\nwhich is a contradiction. Therefore, $m$ must be a power of a prime. Let $m = p^k$. Note that\n$$\n2022^{p^k} + 1 \\equiv 2022 + 1 \\equiv 2023 \\equiv 0 \\pmod{p},\n$$\nthus $p \\mid 2023$ and $p \\in \\{7, 17\\}$.\n\nIf $p=7$, using LTE, we have\n$$\nv_7(2022^{7k} + 1) = v_7(2023) + v_7(7^k) = 1 + k \\ge v_7(7^{2k}) = 2k.\n$$\nFrom this, we conclude that $k=1$ and $m=7$.\n\nSimilarly, for $p=17$, applying LTE, we have\n$$\nv_{17}(2022^{17k} + 1) = v_{17}(2023) + v_{17}(17^k) = 2 + k \\ge v_{17}(17^{2k}) = 2k,\n$$\nso $k \\in \\{1, 2\\}$ and $m \\in \\{17, 289\\}$.\n\nTherefore, $m = 7, 17, 289$ are all desired numbers. $\\square$",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 14987,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $M$ be the set of positive integers which are not divisible by any prime number greater than $3$. For arbitrarily chosen subsets $A_1, A_2, A_3, \\ldots$ of $M$, prove that there exist two distinct positive integers $i$ and $j$ such that:\n\nFor each $x$ in $A_i$, $A_j$ has a divisor of $x$.",
    "options": [],
    "answer": "See solution",
    "solution": "Regard $A_n$ as a set of lattice points by\n\n$$\n2^x 3^y \\mapsto (x, y) \\in \\mathbb{N}_0^2,\n$$\n\nwhere $\\mathbb{N}_0$ is the set of nonnegative integers. For two elements $(a, b), (c, d) \\in \\mathbb{N}_0^2$, we denote $(a, b) \\le (c, d)$ if $a \\le c$ and $b \\le d$. Also, we denote $(a, b) < (c, d)$ if $(a, b) \\le (c, d)$ and $(a, b) \\ne (c, d)$. Then the problem can be interpreted as:\n\nFor any $A_1, A_2, \\ldots \\subset \\mathbb{N}_0^2$, there exist $i, j \\in \\mathbb{N}$ such that $\\forall x \\in A_i$, $\\exists y \\in A_j$ with $y \\le x$.\n\nIn this case, we will say that the collection $A_1, A_2, \\ldots$ is \"nice\".\n\nA reduced set $A$ is a subset of $\\mathbb{N}_0^2$ such that there do not exist $x, y \\in A$ with $x < y$. Since a reduced set may have at most one point in each $x = a$ (resp. $y = b$), it is a finite set. Moreover, if a reduced set $A$ has $(a, b) \\in \\mathbb{N}_0^2$, then $|A| \\le a + b + 1$ holds. Obviously, we may assume that each $A_i$ is a reduced set.\n\n**Lemma 1.** Given a collection $A_1, A_2, \\ldots$, suppose that there are no points $p$ in $\\mathbb{N}_0^2$ which are contained in infinitely many $A_i$'s. Then the collection is nice.\n\nBefore proving the main lemma, we introduce a result for a special case.\n\n**Corollary 2.** Given a collection $A_1, A_2, \\ldots$, if there exists $N \\in \\mathbb{N}$ such that $|A_i| \\le N$ for all $i$, then the collection is nice.\n\n*Proof.* (by mathematical induction) In case $N=1$, it is easy. Assume $N \\ge 2$. If the collection satisfies the condition in Lemma 1, then it is done. If the collection does not satisfy the condition, i.e., $\\exists p$ which is contained in infinitely many $A_i$'s. Set $\\{B_i \\mid i \\in \\mathbb{N}\\}$ to be the collection of $A_i$'s containing $p$. From the induction hypothesis, $\\{B_i - \\{p\\} \\mid i \\in \\mathbb{N}\\}$ is nice, which easily implies that $\\{B_i \\mid i \\in \\mathbb{N}\\}$ is nice. Thus, the collection $A_1, A_2, \\ldots$ is also nice. $\\square$\n\nLet us go back to the original problem. If $A_1, A_2, \\ldots$ satisfies the condition in Lemma 1, then it is done. Otherwise, $\\exists p$ such that $p = (a, b)$ is contained in infinitely many $A_i$'s, then each cardinality of $A_i$ containing $p$ is bounded by $|A_i| \\le a + b + 1$ (recall that $A_i$'s are reduced sets). Thus, from Corollary 2, the collection of $A_i$'s containing $p$ is nice, and so is the original collection $A_1, A_2, \\ldots$, which completes the proof.\n\nNow we only have left the proof of Lemma 1.\n\n*Proof (Lemma 1).* For nonnegative integer $h$, define $V_h := \\bigcup A_i \\cap \\{(h, y) \\mid y \\in \\mathbb{N}_0\\}$. If $V_h$ is a finite set, then by the hypothesis of Lemma 1, only finitely many $A_i$'s contribute to $V_h$. So we may delete these finitely many $A_i$'s.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 14988,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $X = \\{x_0, \\dots, x_{n-1}\\}$ be an $n$-element set of real numbers such that $0 < |x_0| \\leq \\dots \\leq |x_{n-1}|$. Prove that the sums of elements of all subsets of $X$ are $2^n$ consecutive members of an arithmetic sequence in some order if and only if\n\n$$\n|x_0| : \\dots : |x_{n-1}| = 2^0 : \\dots : 2^{n-1}.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "Let $d > 0$ be the difference of the arithmetic sequence of sums of elements of subsets. We prove by induction on $n$ that $|x_i| = d \\cdot 2^i$ for $0 \\leq i < n$. The claim holds trivially for $n = 1$. Assume now that the claim holds for $n-1$ numbers.\n\nLet $s$ be the smallest among the $2^n$ sums and $s'$ be the second smallest. The sum $s$ is obviously obtained by the subset $N$ of all negative elements of $X$. In order to obtain $s'$ as the sum, some (possibly 0) negative elements are excluded and some (possibly 0) positive elements are included. It can be easily verified that the only possibility is either just excluding $x_0$ from $N$ if $x_0 \\in N$ or just including $x_0$ into $N$ if $x_0 \\notin N$ (other changes would increase the sum more). Hence $d = s' - s = |x_0|$.\n\nLeave $x_0$ out from $X$. Exactly half of all sums remain; other sums differ from corresponding sums by $|x_0|$ to the same direction. As $d = |x_0|$, this means that, in the arithmetic sequence of sums, exactly one member out of every two consecutive members is dropped. Thus the result is an arithmetic sequence with difference $2d$. By the induction hypothesis, $|x_i| = 2d \\cdot 2^{i-1} = d \\cdot 2^i$ for $1 \\leq i < n$. Including $|x_0| = d = d \\cdot 2^0$ completes the induction step.\n\nTo prove the reverse direction, let $|x_i| = d \\cdot 2^i$ for $0 \\leq i < n$. Consider two different subsets of $X$; suppose their elements sum up to the same number. Assume without loss of generality that these two subsets are disjoint. Then every $x_k$ that occurs in one or another subset can be expressed as a linear combination of others with coefficients 1 and $-1$. But if $x_k$ is the largest by absolute value term occurring in these two subsets then this is impossible since $|x_k| = d \\cdot 2^k > d \\cdot (2^0 + \\dots + 2^{k-1}) = |x_0| + \\dots + |x_{k-1}|$. Consequently, all subsets of $X$ have different sums of elements.\n\nLet $s$ and $t$ be the sums of all negative and all positive elements of $X$, respectively. Clearly, $s$ is the smallest and $t$ is the largest sum of elements of a subset. Their difference is\n\n$$\nt - s = |x_0| + \\dots + |x_{n-1}| = d \\cdot (2^0 + \\dots + 2^{n-1}) = d \\cdot (2^n - 1).\n$$\n\nAs all sums of elements of subsets are integral multiples of $d$, this implies that exactly $2^n$ of them lie on the interval $[s, t]$. By pairwise distinctness, sums of elements of all subsets of $X$ cover all integral multiples of $d$ between $s$ and $t$, hence forming an arithmetic sequence.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 14989,
    "subject": "Mathematics (Olympiad)",
    "question": "A positive integer $n > 1$ is said to have the property ($p$) if, in its prime factorization\n$$\nn = p_1^{\\alpha_1} \\cdots p_j^{\\alpha_j}\n$$\nat least one of the prime factors $p_1, \\dots, p_j$ has exponent equal to 2.\n\n**a)** Find the largest number $k$ for which there exist $k$ consecutive positive integers that do **not** have the property ($p$).\n\n**b)** Prove that there are infinitely many positive integers $n$ such that $n$, $n + 1$, and $n + 2$ all have the property ($p$).",
    "options": [],
    "answer": "See solution",
    "solution": "a) Among any 8 consecutive integers, there is one of the form $8j + 4$. This number has the property ($p$) because the factor 2 in its prime factorization has exponent 2. Therefore, there can be at most 7 consecutive positive integers that do not have the property ($p$). For example, none of the numbers $29, 30, 31, 32, 33, 34, 35$ has the property ($p$), so the largest $k$ is $k = 7$.\n\nb) The numbers $98 = 2 \\cdot 7^2$, $99 = 3^2 \\cdot 11$, and $100 = 2^2 \\cdot 5^2$ all have the property ($p$). Therefore, the numbers $98 + (7 \\cdot 3 \\cdot 2)^3 \\cdot k$, $99 + (7 \\cdot 3 \\cdot 2)^3 \\cdot k$, and $100 + (7 \\cdot 3 \\cdot 2)^3 \\cdot k$ also have the property ($p$) for any integer $k \\geq 0$.\n\n_Alternative solution:_ By the Chinese remainder theorem, there are infinitely many solutions to the system of congruences:\n$$\nn \\equiv 4 \\pmod{8}, \\quad n \\equiv 8 \\pmod{27}, \\quad n \\equiv 23 \\pmod{125}.\n$$\nThen $n$, $n + 1$, and $n + 2$ all have the property ($p$) because the factors 2, 3, and 5, respectively, have exponent 2 in their prime factorizations.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 14990,
    "subject": "Mathematics (Olympiad)",
    "question": "Sequence $\\{a_n\\}$ is defined by $a_0 = \\frac{1}{2}$, $a_{n+1} = a_n + \\frac{a_n^2}{2012}$ for $n = 0, 1, 2, \\dots$. Find the integer $k$ such that $a_k < 1 < a_{k+1}$.\n\n![](path/to/file.png)",
    "options": [],
    "answer": "See solution",
    "solution": "From the given recurrence, note that $a_0 < a_1 < \\dots < a_{2012}$. Observe:\n\n$$\n\\frac{1}{a_{n+1}} = \\frac{2012}{a_n(a_n + 2012)} = \\frac{1}{a_n} - \\frac{1}{a_n + 2012}\n$$\nso\n$$\n\\frac{1}{a_n} - \\frac{1}{a_{n+1}} = \\frac{1}{a_n + 2012}\n$$\nBy telescoping,\n$$\n\\frac{1}{a_0} - \\frac{1}{a_n} = \\sum_{i=0}^{n-1} \\frac{1}{a_i + 2012}\n$$\nFor $n = 2012$:\n$$\n2 - \\frac{1}{a_{2012}} = \\sum_{i=0}^{2011} \\frac{1}{a_i + 2012} < \\sum_{i=0}^{2011} \\frac{1}{2012} = 1\n$$\nso $a_{2012} < 1$. For $n = 2013$:\n$$\n2 - \\frac{1}{a_{2013}} = \\sum_{i=0}^{2012} \\frac{1}{a_i + 2012} > \\sum_{i=0}^{2012} \\frac{1}{1 + 2012} = 1\n$$\nso $a_{2013} > 1$. Thus, $k = 2012$.\n\n$\\boxed{2012}$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 14991,
    "subject": "Mathematics (Olympiad)",
    "question": "A grid rectangle that is not a square is cut into 8 different (non-congruent) grid polygons along the grid lines. What is its minimal possible area?",
    "options": [],
    "answer": "See solution",
    "solution": "There is one grid polygon of area 1 ($1 \\times 1$ square), one such polygon of area 2 ($1 \\times 2$ rectangle), 2 such polygons of area 3 ($1 \\times 3$ rectangle and an angle of 3 squares). To satisfy the condition, one must then use at least 4 grid polygons of area 4 or greater. Hence, the area of the given rectangle $R$ is at least $1 + 2 + 2 \\cdot 3 + 4 \\cdot 4 = 25$.\n\nObserve that it cannot be exactly 25. Otherwise, $R$ is a $5 \\times 5$ square or a $1 \\times 25$ rectangle. The first case is excluded by hypothesis. In the second, only rectangles $1 \\times k$ can be used in the division, hence $R$ would have area at least $1 + 2 + \\ldots + 8 > 25$.\n\nIn conclusion, $R$ has at least area 26. It can be exactly 26 for a $2 \\times 13$ rectangle.\n\n![]( <table><tr><td>1</td><td>3</td><td>3</td><td>4</td><td>5</td><td>5</td><td>5</td><td>6</td><td>6</td><td>7</td><td>7</td><td>7</td><td>7</td></tr><tr><td>2</td><td>2</td><td>3</td><td>4</td><td>4</td><td>4</td><td>6</td><td>6</td><td>8</td><td>8</td><td>8</td><td>8</td><td>8</td></tr></table> )\n\nThe figure displays a division of such a rectangle into 8 different grid polygons. So the required minimal area is 26.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 14992,
    "subject": "Mathematics (Olympiad)",
    "question": "令 $a_1 < a_2 < a_3 < \\cdots$ 為正整數數列，其中每個 $k \\ge 1$，$a_{k+1}$ 都整除 $2(a_1 + a_2 + \\cdots + a_k)$。假設對於無窮多個質數 $p$，存在某個 $k$ 使得 $p$ 整除 $a_k$。證明：對於每一個正整數 $n$，都存在某個 $k$ 使得 $n$ 整除 $a_k$。",
    "options": [],
    "answer": "See solution",
    "solution": "*解.* 對每個 $k \\ge 2$，定義商 $b_k = \\dfrac{2(a_1 + \\cdots + a_{k-1})}{a_k}$，它必為正整數。先證明序列 $(b_k)$ 的以下性質：\n\n*Claim 1.* $b_{k+1} \\le b_k + 1$。\n\n*Proof.* 由 $b_k a_k = 2(a_1 + \\cdots + a_{k-1})$ 和 $b_{k+1} a_{k+1} = 2(a_1 + \\cdots + a_k)$，可得 $b_{k+1} a_{k+1} = b_k a_k + 2a_k$。由 $a_k < a_{k+1}$，可知 $b_k + 2 > b_{k+1}$。\n\n*Claim 2.* 序列 $(b_k)$ 無上界。\n\n*Proof.* 重寫 $b_{k+1} a_{k+1} = (b_k + 2)a_k$ 為\n\n$$\na_{k+1} = a_k \\cdot \\frac{b_k + 2}{b_{k+1}} \\Rightarrow a_{k+1} \\mid a_k(b_k + 2).\n$$\n\n若 $(b_k)$ 有上界 $B$，則 $(a_k)$ 的質因數只能是小於等於 $B+2$ 的質數或整除 $a_1$、$a_2$ 的質數，這與題設矛盾。\n\n考慮任意正整數 $n$。若 $n > b_2$，否則取大於 $b_2$ 的 $n$ 的倍數。由 Claim 2，存在 $k$ 使 $b_{k+1} \\ge n$。取最小的此 $k$，由 Claim 1 可知 $b_k = n-1$ 且 $b_{k+1} = n$（假設 $n > b_2$ 保證 $k \\ge 2$）。此時\n\n$$\na_{k+1} = a_k \\cdot \\frac{b_k + 2}{b_{k+1}} = a_k \\cdot \\frac{n+1}{n}.\n$$\n\n因 $n$ 與 $n+1$ 互質，故 $a_k$ 必可被 $n$ 整除。",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 14993,
    "subject": "Mathematics (Olympiad)",
    "question": "Determine all real-valued functions $f$ defined on the set of all integers and satisfying the following identity for any pair of integers $m, n$:\n\n$$\nf(m) + f(n) = f(mn) + f(m + n + mn).\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "Let $f(1) = a$. Setting $n = 1$ gives $f(m) + f(1) = f(m) + f(2m + 1)$ for any integer $m$, so $f(d) = a$ for any odd integer $d$.\n\nAny nonzero integer can be written as $2^k d$ where $d$ is odd and $k \\geq 0$. Taking $m = d$, $n = 2^k$, we get $f(d) + f(2^k) = f(2^k d) + f(2^k(d+1) + d)$. Both $d$ and $2^k(d+1) + d$ are odd, so $a + f(2^k) = f(2^k d) + a$, which implies $f(2^k d) = f(2^k)$. Thus, knowing $f(2^k)$ for all $k$ and $f(0)$ determines $f(n)$ for all integers $n$.\n\nFor $k \\geq 2$, substitute $m = 2^k$, $n = 2$ to get $f(2^k) + f(2) = f(2^{k+1}) + f(2^k \\cdot 3 + 2)$. Since $2^k \\cdot 3 + 2$ is an even integer, $f(2^k \\cdot 3 + 2) = f(2)$. Thus, $f(2^k) = f(2^{k+1})$ for $k \\geq 2$. Let $f(2^2) = b$, so $f(2^2) = f(2^3) = \\dots = b$. Setting $m = n = 2$ gives $2f(2) = f(4) + f(8) = 2b$, so $f(2) = b$. Therefore, $f(n) = b$ for all even $n \\neq 0$. Finally, setting $m = n = -2$ gives $2f(-2) = f(4) + f(0)$, so $f(0) = b$.\n\nThus, any function $f$ satisfying the equation must be of the form\n\n$$\nf(n) = \\begin{cases} a & \\text{if } n \\text{ is odd} \\\\ b & \\text{if } n \\text{ is even or } 0. \\end{cases}\n$$\n\nfor some real numbers $a, b$.\n\nConversely, for any $a, b \\in \\mathbb{R}$, the function above satisfies the equation:\n\n- If both $m, n$ are even, $mn$ and $m+n+mn$ are even, so both sides are $2b$.\n- If both $m, n$ are odd, $mn$ and $m+n+mn$ are odd, so both sides are $2a$.\n- If one is even and one is odd, $mn$ is even, $m+n+mn$ is odd, so both sides are $a + b$.\n\nTherefore, all such functions $f$ are solutions.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 14994,
    "subject": "Mathematics (Olympiad)",
    "question": "Suppose $n = \\binom{k}{2} + t \\leq \\binom{k+1}{2}$ for some integer $k$ and $t \\geq 0$. Prove that after some weeks, the number of members in the largest group is $k$.\n\nTo do this, you may add $\\binom{k+1}{2} - \\binom{k}{2} - t$ fake members to form a group of $\\binom{k+1}{2}$ members. Assume that if a group contains at least one real member, fake members cannot become king. Show that the number of real members in any group is at most $k$.\n\nAlso, deduce that\n$$\n\\binom{k}{2} < n \\leq \\binom{k+1}{2} \\implies k \\leq \\sqrt{2n} + \\frac{1}{2} < \\sqrt{2n} + 1.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "Lemma: If $n = \\binom{k}{2}$ for some positive integer $k$, after $\\binom{k}{2}$ weeks, there will be $k-1$ groups such that for each $1 \\leq i \\leq k-1$, there is a group with $i$ members.\n\n*Proof.* The cases $n = 2, 3$ are trivial. Assume the assertion holds for all $1 \\leq k \\leq m$. Now, consider a group with $\\binom{m+1}{2}$ members. By induction, after $\\binom{m}{2}$ weeks, we have $m-1$ groups with $1, 2, \\dots, m-1$ members. The first group then has $\\binom{m+1}{2} - \\binom{m}{2} = m$ members. Thus, the assertion holds.\n\nNow, suppose $n = \\binom{k}{2} + t \\leq \\binom{k+1}{2}$. Add $\\binom{k+1}{2} - \\binom{k}{2} - t$ fake members to reach $\\binom{k+1}{2}$ members. By the lemma, after some weeks, for each $1 \\leq i \\leq k$, there is a group with $i$ members. If a group contains at least one real member, fake members cannot become king, so the number of real members in any group is at most $k$. This proves the assertion.\n\nFinally,\n$$\n\\binom{k}{2} < n \\leq \\binom{k+1}{2} \\implies k^2 - k + \\frac{1}{4} = (k - \\frac{1}{2})^2 < 2n \\implies k \\leq \\sqrt{2n} + \\frac{1}{2} < \\sqrt{2n} + 1.\n$$",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 14995,
    "subject": "Mathematics (Olympiad)",
    "question": "已知 $a, b, c, d$ 為正實數且滿足 $a + b + c + d = 4$。試證：\n\n$$\n\\sum_{\\text{cyc}} \\frac{3a^3}{a^2 + ab + b^2} + \\sum_{\\text{cyc}} \\frac{2ab}{a+b} \\geq 8.\n$$",
    "options": [],
    "answer": "See solution",
    "solution": "解：注意到\n\n$$\n\\sum_{\\text{cyc}} \\frac{a^3 - b^3}{a^2 + ab + b^2} = \\sum_{\\text{cyc}} (a-b) = 0.\n$$\n\n因此\n\n$$\n\\text{LHS} = \\sum_{\\text{cyc}} \\frac{3a^3 + 3b^3}{2(a^2 + ab + b^2)} + \\sum_{\\text{cyc}} \\frac{2ab}{a+b}.\n$$\n\n又\n\n$$\n\\frac{2ab}{a+b} = a+b-\\frac{a^2+b^2}{a+b},\n$$\n\n所以只需證明\n\n$$\n\\sum_{\\text{cyc}} \\frac{3a^3 + 3b^3}{2(a^2 + ab + b^2)} \\geq \\sum_{\\text{cyc}} \\frac{a^2 + b^2}{a+b}.\n$$\n\n這成立，因為\n\n$$\n3(a^3 + b^3)(a+b) \\geq 3(a^2 + b^2)^2 \\geq 2(a^2 + ab + b^2)(a^2 + b^2).\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 14996,
    "subject": "Mathematics (Olympiad)",
    "question": "Given circles $\\omega_1$ and $\\omega_2$ intersecting at points $X$ and $Y$, let $\\ell_1$ be a line through the center of $\\omega_1$ intersecting $\\omega_2$ at points $P$ and $Q$, and let $\\ell_2$ be a line through the center of $\\omega_2$ intersecting $\\omega_1$ at points $R$ and $S$. Prove that if $P, Q, R,$ and $S$ lie on a single circle, then the center of this circle lies on line $XY$.",
    "options": [],
    "answer": "See solution",
    "solution": "Let $\\omega$ denote the circumcircle of $P, Q, R, S$ and let $O$ denote the center of $\\omega$. Line $XY$ is the radical axis of circles $\\omega_1$ and $\\omega_2$. It suffices to show that $O$ has equal power to the two circles; that is, to show that\n\n$$\nOO_1^2 - O_1S^2 = OO_2^2 - O_2Q^2 \\quad \\text{or} \\quad OO_1^2 + O_2Q^2 = OO_2^2 + O_1S^2.\n$$\n\nLet $M$ and $N$ be the intersections of lines $O_2O, \\ell_1$ and $O_1O, \\ell_2$. Because circles $\\omega$ and $\\omega_2$ intersect at points $P$ and $Q$, we have $PQ \\perp OO_2$ (or $\\ell_1 \\perp OO_2$). Hence\n\n$$\nOO_1^2 - OQ^2 = (OM^2 + MO_1^2) - (OM^2 + MQ^2) = (O_2M^2 + MO_1^2) - (O_2M^2 + MQ^2) = O_2O_1^2 - O_2Q^2\n$$\n\nor\n\n$$\nO_2O_1^2 + OQ^2 = OO_1^2 + O_2Q^2.\n$$\n\nLikewise, we have $O_2O_1^2 + OS^2 = OO_2^2 + O_1S^2$. Because $OS = OQ$, we obtain that $OO_1^2 + O_2Q^2 = OO_2^2 + O_1S^2$, which is what was to be proved.\n\n![](images/pamphlet0910_main_p5_data_fd28e32ecc.png)",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 14997,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $K_{2011,2011}$ be a complete bipartite graph in which all vertices of a set $A$ with $|A| = 2011$ are connected to all vertices of $B$ with $|B| = 2011$. Prove that there exists a monochromatic connected subgraph with 212 vertices if the edges of the graph $K_{2011,2011}$ are colored so that all edges incident to any given vertex are colored by at most 19 colors.",
    "options": [],
    "answer": "See solution",
    "solution": "Let the $i$-th color degree of a vertex $v$ be $d_i(v)$. For each connected pair of vertices $(u, v)$, define $f(u, v) = d_i(u) + d_i(v)$ where the edge connecting $u$ and $v$ is colored by the $i$-th color. The set of colors used for coloring all edges incident to $u$ is denoted by $C(u)$. By the Cauchy-Schwarz inequality:\n\n$$\n\\frac{1}{2} \\sum_{u \\in A, v \\in B} f(u, v) = \\sum_{u \\in A, i \\in C(u)} d_i^2(u) \\geq \\frac{\\left( \\sum_{u \\in A, i \\in C(u)} d_i(u) \\right)^2}{2011 \\cdot 19} = 2011^2 \\cdot \\frac{2011}{19}\n$$\n\nTherefore, by the pigeonhole principle, there exist vertices $s, t$ with $f(s, t) \\ge 212$ since $2 \\cdot \\frac{2011}{19} = 211.68$.\n\nFor an example with the greatest monochromatic connected component of size 212 using only 19 colors: partition all vertices of $A$ and $B$ into sets $A_1, \\dots, A_{19}$ and $B_1, \\dots, B_{19}$ of sizes 105 or 106, and color all edges between $A_i$ and $B_j$ with color $c(i, j)$ where $c(i, j) \\equiv i + j \\pmod{19}$ and $1 \\le c(i, j) \\le 19$. The maximal monochromatic connected component is of size 212.",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 14998,
    "subject": "Mathematics (Olympiad)",
    "question": "Three pairwise distinct numbers $a$, $b$, $c$ are such that their product is $80$.\nDetermine the least possible prime sum of these numbers.",
    "options": [],
    "answer": "See solution",
    "solution": "Clearly, their sum is greater than $2$, thus their sum is odd. Since all three numbers cannot be odd (since their product is $80$), then two of the numbers are even and one is odd. There are only two odd divisors of $80 = 2^4 \\cdot 5$: $1$ and $5$. Consider these cases.\n\nIf $c = 1$, the following is possible:\n\n- $a = 2$, $b = 40$, $a + b + c = 43$ is prime.\n- $a = 4$, $b = 20$, $a + b + c = 25$ is not prime.\n- $a = 8$, $b = 10$, $a + b + c = 19$ is prime and less than $43$.\n\nIf $c = 5$, the following is possible:\n\n- $a = 2$, $b = 8$, $a + b + c = 15$ is not prime.\n\nThus, the least possible prime sum is $19$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 14999,
    "subject": "Mathematics (Olympiad)",
    "question": "Given a finite set of boys and girls, a *covering set of boys* is a set of boys such that every girl knows at least one boy in that set; and a *covering set of girls* is a set of girls such that every boy knows at least one girl in that set. Prove that the number of covering sets of boys and the number of covering sets of girls have the same parity. (Acquaintance is assumed to be mutual.)",
    "options": [],
    "answer": "See solution",
    "solution": "Let $B$ denote the set of boys, let $G$ denote the set of girls, and induct on $|B| + |G|$. The assertion is vacuously true if either set is empty.\n\nNext, fix a boy $b$, let $B' = B \\setminus \\{b\\}$, and let $G'$ be the set of all girls who do not know $b$. Notice that:\n\n1. A covering set of boys in $B' \\cup G$ is still one in $B \\cup G$; and\n2. A covering set of boys in $B \\cup G$ which is no longer one in $B' \\cup G$ is precisely the union of a covering set of boys in $B' \\cup G'$ and $\\{b\\}$.\n\nSo the number of covering sets of boys in $B \\cup G$ is the sum of those in $B' \\cup G$ and $B' \\cup G'$.\n\nOn the other hand:\n\n1. A covering set of girls in $B \\cup G$ is still one in $B' \\cup G$; and\n2. A covering set of girls in $B' \\cup G$ which is no longer one in $B \\cup G$ is precisely a covering set of girls in $B' \\cup G'$, \n\nso the number of covering sets of girls in $B \\cup G$ is the difference of those in $B' \\cup G$ and $B' \\cup G'$.\n\nSince the assertion is true for both $B' \\cup G$ and $B' \\cup G'$ by the induction hypothesis, the conclusion follows.",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 15000,
    "subject": "Mathematics (Olympiad)",
    "question": "Let $a$, $b$, and $c$ be nonzero integers. Show that there exists an integer $k$ such that\n\n$$\n\\gcd(a + k b, c) = \\gcd(a, b, c).\n$$\n\n(Note: 'gcd' stands for 'greatest common divisor.')",
    "options": [],
    "answer": "See solution",
    "solution": "We may assume that $a$, $b$, and $c$ are all positive, since if $a$, $b$, $c$ are all positive and $\\gcd(a + k b, c) = \\gcd(a, b, c)$ for some integer $k$, then we immediately have $\\gcd(-a + (-k) b, \\pm c) = \\gcd(-a + k(-b), \\pm c) = \\gcd(a + (-k)(-b), \\pm c) = \\gcd(\\pm a, \\pm b, \\pm c)$. Moreover, if at least one of $a$, $b$, or $c$ is equal to $1$, then the result follows immediately: if $a = 1$ or $c = 1$, choose $k = 0$; otherwise, if $b = 1$, choose $k = 1 - a$. In all these cases, $\\gcd(a + k b, c) = \\gcd(a, b, c) = 1$.\n\nHenceforth, assume that all of $a$, $b$, and $c$ are greater than $1$. This implies that there is a list of prime numbers $p_1, p_2, \\dots, p_n$ and non-negative integers $\\alpha_i, \\beta_i, \\gamma_i$, $i = 1, 2, \\dots, n$, such that\n\n$$\na = p_1^{\\alpha_1} p_2^{\\alpha_2} \\cdots p_n^{\\alpha_n}\n$$\n$$\nb = p_1^{\\beta_1} p_2^{\\beta_2} \\cdots p_n^{\\beta_n}\n$$\n$$\nc = p_1^{\\gamma_1} p_2^{\\gamma_2} \\cdots p_n^{\\gamma_n}.\n$$\n\nLet us first assume that $\\gcd(a, b, c) = 1$. This implies that, for each $i \\in \\{1, 2, \\dots, n\\}$, not all three of $\\alpha_i, \\beta_i$, and $\\gamma_i$ are positive. We may also assume that, for each $i \\in \\{1, 2, \\dots, n\\}$, not all three of $\\alpha_i, \\beta_i$, and $\\gamma_i$ are $0$ (otherwise we could simply discard the primes $p_i$ for which this happens). We now call a prime $p_i$ a *one-prime* if exactly one of $\\alpha_i, \\beta_i$, and $\\gamma_i$ is positive. Likewise, we call a prime $p_i$ a *two-prime* if exactly two of $\\alpha_i, \\beta_i$, and $\\gamma_i$ are positive. Let $E = \\{p_{i_1}, p_{i_2}, \\dots, p_{i_t}\\}$ be the complete list of one-primes among $p_1, p_2, \\dots, p_n$. If there are no one-primes in this list, put $E = \\emptyset$. Let\n\n$$\nk = \\begin{cases} p_{i_1} p_{i_2} \\cdots p_{i_t} & \\text{if } E \\neq \\emptyset \\\\ 1 & \\text{if } E = \\emptyset. \\end{cases}\n$$\n\nWe show that, for this $k$, $\\gcd(a + k b, c) = \\gcd(a, b, c) = 1$.\n\nSince $\\gcd(a + k b, c)$ is a divisor of $c$, the only possible way that $\\gcd(a + k b, c) > 1$ is that it is divisible by at least one of $p_1, p_2, \\dots, p_n$. We show that this is not the case.\n\nFirstly, consider any $p_{i_j} \\in E$ (if $E \\neq \\emptyset$). For the triple $(\\alpha_{i_j}, \\beta_{i_j}, \\gamma_{i_j})$ there are three possibilities:\n\nI. $(\\alpha_{i_j}, \\beta_{i_j}, \\gamma_{i_j}) = (\\alpha_{i_j}, 0, 0)$, with $\\alpha_{i_j} > 0$. Here, $p_{i_j}$ does not divide $c$.\n\nII. $(\\alpha_{i_j}, \\beta_{i_j}, \\gamma_{i_j}) = (0, \\beta_{i_j}, 0)$, with $\\beta_{i_j} > 0$. Here, again, $p_{i_j}$ does not divide $c$.\n\nIII. $(\\alpha_{i_j}, \\beta_{i_j}, \\gamma_{i_j}) = (0, 0, \\gamma_{i_j})$, with $\\gamma_{i_j} > 0$. Here, $p_{i_j}$ does not divide $a + k b$ (where $k = p_{i_1} p_{i_2} \\cdots p_{i_t}$).\n\nSo $\\gcd(a + k b, c)$ is not divisible by any of the one-primes. Secondly, let $p_r$ denote any of the two-primes. For the triple $(\\alpha_r, \\beta_r, \\gamma_r)$ there are three possibilities:\n\nI'. $(\\alpha_r, \\beta_r, \\gamma_r) = (\\alpha_r, \\beta_r, 0)$, with $\\alpha_r, \\beta_r > 0$. Here, $p_r$ does not divide $c$.\n\nII'. $(\\alpha_r, \\beta_r, \\gamma_r) = (\\alpha_r, 0, \\gamma_r)$, with $\\alpha_r, \\gamma_r > 0$. Here, $p_r$ does not divide $a + k b$ (recall that $k$ is not divisible by $p_r$).\n\nIII'. $(\\alpha_r, \\beta_r, \\gamma_r) = (0, \\beta_r, \\gamma_r)$, with $\\beta_r, \\gamma_r > 0$. Here, again, $p_r$ does not divide $a + k b$.\n\nIt follows that $\\gcd(a + k b, c) = 1 = \\gcd(a, b, c)$.\n\nFinally, if $\\gcd(a, b, c) = d > 1$, then $\\gcd\\left(\\frac{a}{d}, \\frac{b}{d}, \\frac{c}{d}\\right) = 1$, and from the above, there exists an integer $k$ such that $\\gcd\\left(\\frac{a}{d} + k \\cdot \\frac{b}{d}, \\frac{c}{d}\\right) = 1 = \\gcd\\left(\\frac{a}{d}, \\frac{b}{d}, \\frac{c}{d}\\right)$. Multiplying both sides by $d$ gives $\\gcd(a + k b, c) = d = \\gcd(a, b, c)$, and we are done.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  }
]